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[ "of equations, $Ax = b$ has exactly one solution. $e^{\\sqrt{2}}> 3$ For any real number $c$, the polynomial $x^{3}+x+c$ has exactly one real root.", "### Home > CCA2 > Chapter 12 > Lesson 12.1.1 > Problem12-16 12-16. Find all values of $a$ for which the equation $ax^2 + 5x + 6 = 0$ has a solution that is a real number. Express a in terms of one or more inequalities. Homework Help ✎ Use the discriminant. For what values is $b^2 - 4ac ≥ 0?$", "Let the equation $x^4-2ax^2+x+a^2-a=0$ has all real solutions for real $a$. Given that $a\\geq k$, find $k$.", "one real solution. If the formula is more complex, you need more powerful tools to look for solutions. 5. Aug 1, 2016 ### Math_QED Here you can apply x^3 -a^3 = (x-a)(x^2 + ax + a^2). You immediately see that this only has one real solution because x^2 + ax + a^2 has a negative discriminant -3a^2, so there is only one real solution. 6. Aug 1, 2016 ### mathman Let a be the known solution. You can use synthetic division (original cubic divided by x-a) to get a quadratic for the other 2 solutions.", "using Newton's method or whatever. In this case there are four distinct solution, two are real and two are complex. But since we restrict ourselves to $x\\in \\left[-1,0\\right]$ our answer is $x=-0.456553...$", "Each solution to $x^{2} + 5x + 8 = 0$ can be written in the form $x = a + b i$, where $a$ and $b$ are real numbers. What is $a + b^{2}$?", "# Math Help - single solution? 1. ## single solution? For any values of the real number a is equivalent to a single solution? 2. Originally Posted by dhiab For any values of the real number a is equivalent to a single solution? Hi Note that $1+\\sin^2(ax) \\geq 1$ and $\\cos x \\leq 1$ The only possibility to get the equation correct is therefore $1+\\sin^2(ax) = \\cos x = 1$ that I let you solve 3. Originally Posted by running-gag Hi Note that $1+\\sin^2(ax) \\geq 1$ and $\\cos x \\leq 1$ The only possibility to get the equation correct is therefore $1+\\sin^2(ax) = \\cos x = 1$ that I let you solve How do you know that $1+\\sin^2(ax) = \\cos x$ ?? 4. Originally Posted by Rachel.F How do you know that $1+\\sin^2(ax) = \\cos x$ ?? Well this is the equation to be solved", "1. Find all real solutions of the equation ${\\large (x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1}.$ 2. Find all real solutions of the equation ${\\large (2-x^2)^{(x^2 - 3\\sqrt{2}x + 4)} = 1}.$", "equation $Ax=q$ has more then one solution.", "in the form Ax + By = C. A, B, & C are all real numbers. Any equation can be transformed into this form by adding or subtracting like terms on both sides of the equation. Example: Equation: 9 + 9x = 11y. Subtract 9 from both sides: 9 – 9 + 9x = 11y – 9.", "# One To Eight $\\large (x+1)(x+2)(x+3)(x+4)=(x+5)(x+6)(x+7)(x+8)$ Find the sum of all distinct real solutions to the above equation. ×", "### Home > A2C > Chapter 10 > Lesson 10.1.1 > Problem10-14 10-14. Find all values of a for which the equation ax2 + 5x + 6 = 0 has a solution that is a real number. Express a in terms of one or more inequalities. Homework Help ✎ Use the discriminant. Find the value of a that makes this quantity non-negative. $a\\le\\frac{25}{24}$", "+0 # a bit of help +1 291 1 +816 Suppose that a and b are nonzero real numbers, and that the equation $$x^2+ax+b=0$$ has solutions a and b. Find the ordered pair (a,b). Dec 25, 2018 $$\\text{as }a \\text{ and }b \\text{ are solutions to }x^2 + ax + b = 0\\\\ \\text{we can write}\\\\ x^2 + ax + b = (x-a)(x-b) = x^2 -(a+b)x + ab\\\\ a = -(a+b) \\text{ and } b= a b\\\\ b = -2a\\\\ b = a(-2a) = -2a^2\\\\ -2a = -2a^2 \\Rightarrow a = 1\\\\ b = -2a = -2\\\\ (a,b) = (1,-2)$$", "# That's hard...maybe The equation $x^3-3x^2-3x-1=0$ has exactly one real solution that can be written in the form $\\sqrt[3]{a}+\\sqrt[3]{b}+\\sqrt[3]{c}.$ What is the value of $a+b+c?$ ×", "## Algebra 2 (1st Edition) We know that $|ax+b|\\geq0$ for all real numbers $a,b$. Hence here the solution set is empty.", "VietaTheorem AMC10/12 2008 Problem - 1618 A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of these two solutions?", "# The equation $a^x +a^{-x}=b$ Given $a$ is a fixed real number that is greater than one, how many real numbers $b$ are there such that the equation $𝑎^{x} + a^{-x} = 𝑏$ has a unique real solution x? To solve this problem, I tried to plug in $a = 3$. So, I want to find how many reals $b$ there are such that $3^{x} + 3^{-x} = b$ has a unique real solution x. Setting $3^x$ = $t$, I got that $t + \\frac{1}{t} = b$. It follows that $t^2 - tb + 1 = 0$. From this, it can be obtained that $t = \\frac{b \\pm \\sqrt{b^2-4}}{2}$. I set the discriminant equal to zero: $b^{2}-4=0$, so $b = \\pm 2$. This means that $t = \\pm 1$. Since $t = 3^{x}$ (previously defined), $x$ is real only when $t = 1$. From the work, doesn't it follow that there is only one real number $b$ (that is when $b$ = 2) such that there is one unique real solution $x$? But the answer I see says there are an infinite values of $b$ such that there is one real solution x. What am I missing in my analysis? Thanks, Kaien • For any $x$ then $-x$ will be a solution. So $a^0 + a^0 = 2$", "# The equation $\\;\\sqrt{3x^{2}+x5}=x-3\\;$ , where x is real , has : (a) no solution (b) exactly one solution(c) exactly two solutions (d) exactly four solutions", "## Algebra and Trigonometry 10th Edition $x^2+3x+8\\gt0\\\\(x+1.5)^2-2.25+8\\gt0\\\\(x+1.5)^2+6.75\\gt0\\\\(x+1.5)^2\\gt-6.75$ but we know that $a^2\\geq0$ for all $a$ so all numbers satisfy this inequality. Thus the solution set is the whole set of real numbers.", "# single solution? • June 7th 2009, 08:10 AM dhiab single solution? For any values of the real number a is equivalent to a single solution? • June 7th 2009, 08:45 AM running-gag Quote: Originally Posted by dhiab For any values of the real number a is equivalent to a single solution? Hi Note that $1+\\sin^2(ax) \\geq 1$ and $\\cos x \\leq 1$ The only possibility to get the equation correct is therefore $1+\\sin^2(ax) = \\cos x = 1$ that I let you solve • June 7th 2009, 03:59 PM Rachel.F Quote: Originally Posted by running-gag Hi Note that $1+\\sin^2(ax) \\geq 1$ and $\\cos x \\leq 1$ The only possibility to get the equation correct is therefore $1+\\sin^2(ax) = \\cos x = 1$ that I let you solve How do you know that $1+\\sin^2(ax) = \\cos x$ ?? • June 8th 2009, 11:28 AM running-gag Quote: Originally Posted by Rachel.F How do you know that $1+\\sin^2(ax) = \\cos x$ ?? Well this is the equation to be solved" ]

[ "and hence $f$ has no double root. It follows that $f$ has either $1$ real root, or $3$ real roots: one in $(-\\infty,x_0)$, one in $(x_0,x_1)$ and one in $(x_1,+\\infty)$. On the other hand, $f'\\geq 0$ in $(-\\infty, x_0)$ so it is increasing and $$f(x_0) = -\\frac1{3\\sqrt3}-\\frac1{\\sqrt3}-1<0$$ so $f<0$ in $(-\\infty, x_0)$. It follows that $f$ must have exactly $1$ real root. One can notice, that $f(1)f(2) = -5 < 0$, so there must be root of the equation. Also $f'(x) = 3x^2 - 1$, so it is monotonically increasing in range [1, 2] - there is only one root. Next step is to prove that $x^3 < x + 1$ for $x<1$. For $0 < x < 1$ we have $x + 1 > x > x^3$. For $-1 < x < 0$ following is met: $x + 1 > 0 > x ^ 3$. Finally for $x < -1$ we have $1 + x > x > x^3$. Last step is to prove, that $x^3 > x + 1$ for x > 2, but this is obvious, since $f(2) = 5 > 0$ and for every $x > 2$ $\\frac{dx^3}{dx} = 3x^2 > \\frac{d(x + 1)}{dx} = 1$. The solutions of $3c^2-1=0$ are $\\pm1/\\sqrt{3}$ outside $[1,2]$. That gives you a contradiction with the assumption that there", "$x-4<0$, then: $ 2x-1$ , or rearranging: $ x<-3 $ . So our solution is $x>4$ or $x<-3$. RonL 5. Solve $\\frac{2x-1}{x-4}>1$ First: $\\frac{2x-1}{x-4}-1>0$ $\\frac{2x-1}{x-4} - \\frac{x-4}{x-4} > 0$ $\\frac{x+3}{x-4}>0$ There are critical points where the numerator is 0 and where the denominator is 0. So we have critical points at: x + 3 = 0 --> x = -3 x - 4 = 0 --> x = 4 So break the real line up into $(- \\infty , -3) \\cup (-3, 4) \\cup (4, \\infty)$ and test each interval: $(-\\infty, -3)$: $\\frac{x+3}{x-4}>0$ (Check!) $(-3, 4)$: $\\frac{x+3}{x-4}<0$ (Nope!) $(4, \\infty)$: $\\frac{x+3}{x-4}>0$ (Check!) So the solution set is $x \\in (-\\infty, -3) \\cup (4, \\infty)$ -Dan (Hi CaptainBlack! ) 6. For $f(x)=x^2+kx+k$ determine all values of $k$ such that $f(x)>0$. Now for large $|x|\\ f(x)>0$ as the leading term is dominant. So the question is realy asking us to find the values of k such that $f(x)=0$ has no real roots. From the quadratic formula we know the roots of $f(x)=0$ are: $ x=\\frac{-k \\pm \\sqrt{k^2-4k}}{2} $ , and these are not real if $k^2-4k<0$, which is equivalent to $k<4$ and $k>0$. RonL 7. Hello, SportfreundeKeaneKent! If the roots of the equation $x^3 - ax^2 - 3kx + m\\:=\\:0$ are $2,\\:-1$, and $3$, find the values of $a,\\:k$, and", "real roots.The discriminant is$\\;\\Delta=a^2-16a+48=(a-4)(a-12)$, hence it has two real roots if$a<4$or if$a>12$. Suppose$\\Delta>0$; we'll determine the sign of the roots:$\\;f(0)=a-3$, hence: if ... 3 Your polynomial is$p(x)=4x^2+ax+a-3$its discriminant is$\\Delta=a^2-16a+48$and its roots are$\\frac{-a\\pm \\sqrt{\\Delta}}{8}$. Now there is a$x$for which value is negative, and it is a parabola opening upwards, tells us that both roots are real and thus$\\Delta>0 \\implies a^2-16a+48=(a-4)(a-12)>0$which happens iff$a\\in ... 0 The given expression is a quadratic in $x$ with the leading coefficient $\\geq 0$. Since, it is given to be negative for at least one $x$, the conditions are: Discriminant $\\geq 0$, that is $a^2-4a+12>0$ At least one root is negative. Solve for the roots and make the smaller one less than $0$. 0 You can treat this problem with roots indeed, for as soon as you have two distincts roots $x_1$ and $x_2$, all values between the two roots give a negative $f(x)$. This you can demonstrate easily. So to have distinct roots, you need to have a strictly positive discriminant, that is $a^2-4a+12>0$ You also want to have a negative root, that is $-a ... 2 The function will be positive when$x$is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative. The smallest root is: $$\\frac{-a -\\sqrt{a^2-16(a-3)}}{8}$$ So", "numerator and denominator each as a quadratic in $y$, we will get $1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2}$, where $g(x)$ has a degree lower than $2$. This means taking $\\lim_{y\\to\\infty} 1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2} = 1$, which means $S$ can be brought arbitrarily close to $1$. Therefore, we are done. $$ ~Imajinary", "x \\le 4$$ Thus values of $$x$$ which satisfy both are $$-1 \\le x < 1$$ and $$2 < x \\le 4$$. 27. Since roots of $$ax^2 + bx + c$$ are imaginary, therefore discriminant is negative. $$\\Rightarrow b^2 - 4ac < 0$$ Discriminant of $$a^2x^2 + abx + ac$$ is: $$D = a^2b^2 - 4a^3c = a^2(b^2 - 4ac) < 0$$ But coefficient of the expression is positive hence it will be always positive. 28. Let $$y = \\frac{x^2 - 2x + 4}{x^2 + 2x + 4}$$ $$(y - 1)x^2 + 2(y + 1)x + 4(y - 1) = 0$$ Since $$x$$ is real discriminant will be greater or equal to zero. $$4(y + 1)^2 - 16(y - 1)^2 \\ge 0$$ $$y^2 + 2y + 1 - 4y^2 + 8y - 4 \\ge 0$$ $$-3y^2 + 10y - 3 \\ge 0$$ Roots of corresponding equation are $$\\frac{1}{3}, 3$$. Since coefficient of $$y^2$$ is negative, for above to be true $$y$$ must lie between $$\\frac{1}{3}$$ and $$3$$. 29. Let $$y = \\frac{2x^2 - 3x + 2}{2x^2 + 3x + 2}$$ $$2(y - 1)x^2 + 3(y + 1)x + 2(y - 1) = 0$$ Since $$x$$ is real discriminant will be greater or equal to zero. $$9(y + 1)^2 - 16(y - 1)^2 \\ge 0$$ $$9y^2 + 18y +", "quadratic polynomial of the form $ax^2 + bx + c$ ($a > 0$)has a negative discriminant then it will always be positive for all real values of $x$. Now we first prove that $y > x$. This is true because the polynomial $8x^2 - 6x + 8$ has a negative discriminant ( $-220$ ), so it is always positive. Thus $y^3 > x^3 \\implies y> x$. Now we examine if $y= x+1$ is possible. Consider $y^3 - (x+1)^3 = 8x^2 - 6x + 8 - (x+1)^3= 5x^2 - 9x + 7$. This polynomial has a negative discriminant $- 59$, which means that it is positive for all values of $x$, thus $y^3 > (x+1)^3 \\implies y>x+1$. Now we will see if $y = x+2$ is possible. Note that $y^3 - (x+2)^3 = 2x^2 - 18x= 2 [(x-\\frac{9}{2})^2 - \\frac{81}{4}]$. Thus this has a minimum value of $-\\frac{81}{4}$ and no maximum value. So $y^3 - (x+2)^3$ can be $0$. Now we examine if $y= x+3$ is possible. Note that $(x+3)^3 - y^3= x^2 + 33x + 19$. Note that the function $x^2 + 33x + 19$ is monotonically increasing over the positive reals, and since $x$ is a positive integer, the minimum possible value for $x$ is $1$, which implies minimum possible value of $x^2 + 33x + 19$", "so $$0 = 1+r+r^2 = r^2\\left(1+\\frac{1}{r}+\\frac{1}{r^2}\\right)$$ so the other root has to be $1/r$. By the fundamental theorem of algebra it follows that $$1+x+x^2 = A(x-r)(x-1/r) = A\\left(x^2 -(r+1/r)x + 1\\right)$$ Now by equating coefficients we find that $A=1$ and $r+1/r = -1$, but for every $r<0$ and $r\\neq -1$ either $r<-1$ or $1/r<-1$, so $r+1/r < -1$. Hence no such $r$ may be found. $x=0$ is not a root, so divide by $x \\ne 0$ and write the equation as: $$x+\\frac{1}{x} = -1$$ This requires $x$ to be negative for the LHS to be negative, but then $y=-x$ is positive and $\\displaystyle y+\\frac{1}{y} \\ge 2$ by AM-GM, so $\\displaystyle x+\\frac{1}{x} \\le -2 \\lt -1\\,$, therefore there are no real solutions. Here's another way that doesn't involving completing the square (which is really just the quadratic formula done from first principles). If $1 + x + x^2 = 1 + x(1 + x) = 0$, then $x(1 + x)$ is negative, which implies that $-1 < x < 0$ and $0 < x + 1 < 1$, so $|x(1 + x)| < 1$ and $x(1 + x) > -1$. Hence $1 + x + x^ 2 = 1 + x(1 + x) > 1 - 1 = 0$ giving a contradiction. • Explanation for the downvote, please! Oh,", "# How to use the discriminant to find out what type of solutions the equation has for 2x^2 + x - 1 = 0? May 21, 2015 $2 {x}^{2} + x - 1$ is of the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 1$ and $c = - 1$. The discriminant $\\Delta$ is given by the formula: $\\Delta = {b}^{2} - 4 a c = {1}^{2} - \\left(4 \\times 2 \\times - 1\\right) = 1 + 8 = 9 = {3}^{2}$ Since this is positive, the equation $2 {x}^{2} + x - 1 = 0$ has two distinct real solutions. Additionally, since it is a perfect square (and the cofficients of the quadratic are rational), those roots are rational. In fact, they are given by the formula: $x = \\frac{- b \\pm \\sqrt{\\Delta}}{2 a} = \\frac{- 1 \\pm 3}{4}$ That is, the solutions are $x = - 1$ and $x = \\frac{1}{2}$.", "# If f(x) = x^2 - 5x + c, for which values of c will f(x) have zero real roots? Oct 16, 2015 For any $c$ greater than $\\frac{25}{4}$. #### Explanation: A quadratic equation $a {x}^{2} + b x + c$ has no real roots if its discriminant is strictly negative, where its discriminant is $\\Delta = {b}^{2} - 4 a c$. In your case, $a = 1$, $b = - 5$ and $c$ is to be found. So, $\\Delta = {5}^{2} - 4 \\cdot 1 \\cdot c = 25 - 4 c$ We want $25 - 4 c < 0$, so $25 < 4 c$. Solving for $c$, we get $c > \\frac{25}{4}$", "and positive as $x \\to \\infty$ and large and negative as $x \\to -\\infty$, so there won't be roots too far from the origin. At $x=3$ (the greatest zero of the right), the derivative of the right is $3$, so the right will get greater than $1$ at about $\\frac{10}3$ and we can stop looking. Similarly, the derivative of the right at $0$ is $6$, so the right will be less than $-1$ by about $\\frac{-1}{6}$. This Alpha graph suggests this is a reasonable guess. For $x^2 \\cos^2 x-1=0$ you can observe that $0$ is not a root and write it as $\\cos^2 x=\\frac1{x^2}$ or $\\cos x=\\pm\\frac{1}{x}$. The equation is symmetric around $0$ so we will assume $x \\gt 0$. There are no roots below $1$ and we can write $\\frac2{x^2}=1+\\cos 2x$. The left side is positive but decreasing to $0$ and the right side touches $0$ when $x=(n+\\frac12 )\\pi$. I would expect a pair of roots near each of these points. There is also a single one around $2$ as we get into range. Thank you for your answer. I don't get why from $|\\sin x | \\le 1$ you got the result that there should be roots near the right side zeros. Plus that Wolfram Alpha says $x^2 \\cos^2 x-1=0$ has six roots, is there a", "equations $\\begin{matrix} (x+y)+z &= a \\; \\quad \\\\ (x+y)-z &= b^2/a \\end{matrix}$, which clearly has solution $x+y = \\frac{a^2 + b^2}{2a}$, $z = \\frac{a^2 - b^2}{2a}$. In order for these both to be positive, we must have positive $a$ and $a^2 > b^2$. Now, we have $x+y = \\frac{a^2 + b^2}{2a}$; $xy = \\left(\\frac{a^2 - b^2}{2a}\\right)^2$, so $x,y$ are the roots of the quadratic $m^2 - \\frac{a^2 + b^2}{2a}m + \\left(\\frac{a^2 - b^2}{2a}\\right)^2$. The discriminant for this equation is $\\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2 = \\frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}$. If the expressions $(3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $2(a^2 + b^2)$, would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $3a^2 > b^2$ and $3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $a^2 > b^2$. It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\\left(\\frac{a^2 + b^2}{2a}\\right)^2 > \\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2$ or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if $a$ is positive and $3b^2 > a^2 > b^2$. Q.E.D.", "# If one root of the quadratic equation Question: If one root of the quadratic equation $2 x^{2}+2 x+k=0$ is $\\frac{-1}{3}$ then find the value of $k$. Solution: Since, $x=\\frac{-1}{3}$ is a root of the quadratic equation $2 x^{2}+2 x+k=0$, then, it must satisfies the equation. $2\\left(-\\frac{1}{3}\\right)^{2}+2\\left(-\\frac{1}{3}\\right)+k=0$ $\\Rightarrow 2\\left(\\frac{1}{9}\\right)-\\frac{2}{3}+k=0$ $\\Rightarrow \\frac{2}{9}-\\frac{2}{3}+k=0$ $\\Rightarrow \\frac{2-6+9 k}{9}=0$ $\\Rightarrow-4+9 k=0$ $\\Rightarrow 9 k=4$ $\\Rightarrow k=\\frac{4}{9}$ Hence, the value of $k$ is $\\frac{4}{9}$.", "### Solution #### Part (i) Let $f(x)=x^3+2x-4$. Now $f(0)=-4<0$$f(1)=-1<0$$f(2)=8>0$. Hence there is a root in $(1,2)$. 1. Now using the bisection method evaluate $f(1.5)=2.375>0$. Hence there is a root in $(1,1.5)$ (because $f$ is negative at 1 and positive at 1.5). 2. Hence evaluate $f(1.25)\\approx 0.453>0$. Hence there is a root in $(1,1.25)$. 3. Hence evaluate $f(1.125)\\approx -0.326$. Hence there is a root in $(1.125,1.25)$. 4. Hence evaluate $f(1.1875)\\approx 0.05>0$. Hence there is a root in $(1.125,1.1875)$. Now (we may as well) we can use the midpoint of this interval as our seed for the Newton-Raphson Method, $x_0=1.15625$. Once we get consecutive agreement to one decimal place we will finish. Now $f'(x)=3x^2+2$. Hence, $x_1=x_0-\\frac{f(x_0)}{f'(x_0)}$, $\\Rightarrow x_1=0.15625-\\frac{-0.141693}{6.01074}=1.17982$. #### Part (ii) Let $f(x)=x^5+2$. Note that for all $x>0$$f(x)>0$. Hence we have to look for a root in the negative real numbers. Now $f(-1)=1>0$$f(-2)=-30<0$. Hence there is a root in $(-2,-1)$. 1. By the Bisection Method, we evaluate $f(-1.5)\\approx -5.6<0$. Hence there is a root in $(-1.5,-1)$. 2. Evaluate $f(-1.25)\\approx -1.05$. Hence there is a root in $(-1.25,-1)$. 3. Evaluate $f(-1.125)\\approx 0.2>0$. Hence there is a root in $(-1.25,-1.125)$. 4. Evaluate $f(-1.1875)\\approx -0.36$. Hence there is a root in $(-1.1875,-1.125)$. Letting $x_0=-1.15625$. $x_1=x_0-\\frac{f(x_0)}{f'(x_0)}=-1.149$ $x_2=x_1-\\frac{f(x_1)}{f'(x_1)}=-1.149$.", "# How to use the discriminant to find out what type of solutions the equation has for 4/3x^2 - 2x + 3/4 = 0? May 26, 2015 $\\frac{4}{3} {x}^{2} - 2 x + \\frac{3}{4} = 0$ is of the form $a {x}^{2} + b x + c = 0$ with $a = \\frac{4}{3}$, $b = - 2$ and $c = \\frac{3}{4}$. The discriminant is given by the formula: $\\Delta = {b}^{2} - 4 a c = {\\left(- 2\\right)}^{2} - \\left(4 \\times \\left(\\frac{4}{3}\\right) \\times \\left(\\frac{3}{4}\\right)\\right)$ $= 4 - 4 = 0$ Since $\\Delta = 0$, the quadratic has one repeated rational root. The possible cases are: $\\Delta < 0$ The quadratic has no real roots. It has two complex roots that are conjugates of one another. $\\Delta = 0$ The quadratic has one repeated root. If the coefficients of the quadratic are rational then that repeated root is rational too. $\\Delta > 0$ The quadratic has two distinct real roots. If $\\Delta$ is a perfect square and the coefficients of the quadratic are rational then those roots are rational too.", "### Theory: We have learnt that the roots of the quadratic equation $$ax^2 + bx + c = 0$$ can be found by the quadratic formula: $x=\\frac{-b±\\sqrt{{b}^{2}-4\\mathit{ac}}}{2a}$ Important! $$b^2 - 4ac$$ is called the discriminant of the quadratic equation $$ax^2 + bx + c = 0$$. It is denoted by the letter $$\\Delta$$ or $$D$$. Let us discuss the nature of the roots of the quadratic equation depending on the discriminant. Case I: $$\\Delta = b^2 - 4ac > 0$$ Here, $$b^2 - 4ac > 0$$. That means the value of the discriminant is positive. Then, the possible roots are $\\frac{-b+\\sqrt{{b}^{2}-4\\mathit{ac}}}{2a}$ and $\\frac{-b-\\sqrt{{b}^{2}-4\\mathit{ac}}}{2a}$. If $$\\Delta = b^2 - 4ac > 0$$, then the roots are real and distinct. Case II: $$\\Delta = b^2 - 4ac = 0$$ Here, $$b^2 - 4ac = 0$$. That means the value of the discriminant is zero. $x=\\frac{-b+\\sqrt{0}}{2a}$ and $x=\\frac{-b-\\sqrt{0}}{2a}$ $x=\\frac{-b}{2a}$ and $x=\\frac{-b}{2a}$ The possible roots are $\\frac{-b}{2a}$ and $\\frac{-b}{2a}$. If $$\\Delta = b^2 - 4ac = 0$$, then the roots are real and equal. Case III: $$\\Delta = b^2 - 4ac < 0$$ Here, $$b^2 - 4ac < 0$$. That means the value of the discriminant is negative. We won't get any real roots in this case. If $$\\Delta = b^2 - 4ac < 0$$, then there are no real roots. Relation", "\\;9}\\;> \\;0}$ i.e. m > 0 . . . . . . . . . . (6) Condition 2: The Sum of roots < 0 $\\mathbf{ \\frac{-\\;(m\\;-4)}{9}\\;< \\;0}$ i.e. m > 4 . . . . . . . . . (7) Condition 3: D = ($b^{2} – 4ac$) ≥ 0 Using Equation (2), $\\mathbf{x\\;\\in \\;\\left ( -\\;\\infty ,\\; 1 \\right ]\\;\\cup \\;\\left [ 16 ,\\;\\infty \\right )}$ . . . . . . . . (8) Therefore, from Equations (6), (7), and (8) both the roots of quadratic expression f (x) are negative if, $\\mathbf{m\\;\\in \\;[16,\\;\\infty ]}$ #### Practise This Question Besides the risk of pollution, fossil fuels also pose a risk of ___", "negative real roots. You get a third degree polynomial in $u=(a-1)^2$. $$144+8u-23u^2-4u^3<0$$ It turns out this factors: $$(4u-9)(4+u)^2>0$$ $$|u|>9/4$$ $$a-1>3/2$$ Also, you can quickly prove that your polynomial cannot have more than two real roots. Either by checking other tests on the link above (checking the value for $P$ and $D$) or by writing $$x=-\\frac{1}{1-a}\\frac{1+x^2+x^4}{1+x^2}=-\\frac{1}{a-1}\\left(1+\\frac{x^4}{1+x^2}\\right)$$ RHS has no inflection points, so a straight line can intersect it twice at most (definition of convexity). From the place where you have $x+1/x\\; (=z)$ in terms of $a,$ write $x+1/x=r$ where $$r=(-(a-1)-\\sqrt {(a-1)^2+4})/2 <0.$$ So $x^2-rx+1=0,$ giving $$x=(r\\pm \\sqrt {r^2-4}\\;)/2<0.$$ This yields two negative roots iff $r^2-4>0,$ that is, iff $|r|>2$ because $r^2-4<0$ yields no real negative $x$, and $r^2-4=0$ yields exactly one negative root. In terms of $a,$ this is $$2<|(-(a-1)-\\sqrt {(a-1)^2+4}\\;)/2|=|(\\;a-1+\\sqrt {(a-1)^2+4}\\;)/2| .$$ (Because we must have $a>1,$ else we have $x^4+(a-1)x^3+x^2+(a-1)x+1>0$ whenever $x<0$.) For $a>1$ this is equivalent to $a-1>3/2,$ that is, $a>5/2.$ Another method would be: Let $f(x)=x^4+(a-1)x^3+x^2+(a-1)x+1$ with $a>1.$ Then $$(f(x)<0\\land x<0)\\iff$$ $$a-1=-\\frac {x^4+x^2+1}{-x^3-x}=$$ $$=\\frac {(x+1/x)^2x^2-x^2}{-x^2(x+1/x)}=$$ $$=\\frac {u^2-1}{u}$$ where $u=|x|+|1/x|=-x-1/x.$ The AGM inequality implies that $|x|+|1/x|\\geq 2.$ Now show that $\\frac {u^2-1}{u}\\geq \\frac {3}{2}$ for $u\\geq 2.$ And I leave the rest to you. Remark. For $u\\geq 2$ let $u=2+b.$ Then $\\frac {u^2-1}{u}-\\frac {3}{2}=\\frac {8b^2+2b}{2b+4}.$", "should be true ? There really isn't much to do. Let's call $\\lim_{n \\rightarrow \\infty} y[n] \\equiv y$ Then you have $$y = \\frac {y}{2} + \\frac {x}{2y}$$ This gives you a quadratic in y. 5. Jan 30, 2005 ### dduardo Staff Emeritus Ok, now I understand. Thanks", "+\\ 2)\\ (x\\ +\\ 3)$ In this way, we get the required factors. Following solved examples of factoring quadratics can be referred for better understanding. Example 1: Factorize $x^{2}\\ +\\ 3x\\ - 10$ Solution: $x^{2}\\ +\\ 3x\\ -\\ 10$ = $x^{2}\\ +\\ (5\\ -\\ 2)\\ x\\ -\\ 10$ = $x^{2}\\ +\\ 5x\\ -\\ 2x\\ -\\ 10$ = $x(x\\ +\\ 5)\\ -\\ 2(x\\ +\\ 5)$ = $(x\\ +\\ 5)\\ (x\\ -\\ 2)$ Example 2: Factorize the following quadratic polynomial : $2x^{2}\\ -\\ 16x\\ +\\ 24$ Solution: $2x^{2}\\ -\\ 16x\\ +\\ 24$ $2(x^{2}\\ -\\ 8x\\ +\\ 12)$ $2[x^{2}\\ -\\ (6\\ +\\ 2)\\ x\\ +\\ 12]$ $2[x^{2}\\ -\\ 6x\\ -\\ 2x\\ +\\ 12]$ $2[x(x\\ -\\ 6)\\ -\\ 2(x\\ -\\ 6)]$ $2(x\\ -\\ 2)(x\\ -\\ 6)$ Example 3: Factorize the following quadratic equation by factoring : $\\frac{1}{4}$ $x^{2}$ - $\\frac{3}{4}$ $x\\ -7\\ = 0$ Solution: $\\frac{1}{4}$ $x^{2}$ - $\\frac{3}{4}$ $x\\ -7\\ =\\ 0$ First we have to get rid of fraction. So, multiplying by 4 both sides, we get $4$[$\\frac{1}{4}$ $x^{2}$ - $\\frac{3}{4}$ $x\\ -7$] = $4.0$ $\\Rightarrow\\ x^{2}\\ -\\ 3x\\ -\\ 28\\ =\\ 0$ $\\Rightarrow\\ x^{2}\\ -\\ (7\\ -\\ 4)\\ x\\ -\\ 28]\\ =\\ 0$ $\\Rightarrow\\ x^{2}\\ -\\ 7x\\ +\\ 4x\\ -\\ 28\\ =\\ 0$ $\\Rightarrow\\ x(x\\ -\\ 7)\\ +\\ 4(x\\ -\\ 7)\\ =\\ 0$ $\\Rightarrow\\ (x\\ -\\ 7)(x\\ +\\ 4)\\", "schools at all. Note that the polynomial factors as $$(x^2+x-a)(x^2-x+1-a)$$ The discriminants are both positive if and only if $a\\gt 3/4$. And if $a\\gt 3/4$, subtraction shows the two quadratics cannot have a common root. Well within high school range. • You win. In hindsight, it should have been obvious to try to factor it since we are looking for the roots of the polynomial. – Dylan Sep 12 '15 at 20:36 • How did you factor it? – Ahmed S. Attaalla Sep 13 '15 at 3:10 • It was clear what a factorization had to look like if there was one. For example the $x$ terms had to be the negatives of each other. So I kind of looked at the thing, saw that everything fell into place and wrote it down. – André Nicolas Sep 13 '15 at 3:19 The given equation can be written as $a^2-(2x^2+1)a+x^4+x=0$ $a=\\frac{2x^2+1\\pm\\sqrt{(2x^2+1)^2-4(x^4+x)}}{2}=\\frac{2x^2+1\\pm(2x-1)}{2}$ $a=x^2+x$ or $a=x^2-x+1$ which are now quadratics in $x$ • This is a cute answer. Changing the focus from a polynomial in $x$ to one in $a$ is very nice. – Ravi Jun 19 '17 at 21:20" ]

[ "and hence $f$ has no double root. It follows that $f$ has either $1$ real root, or $3$ real roots: one in $(-\\infty,x_0)$, one in $(x_0,x_1)$ and one in $(x_1,+\\infty)$. On the other hand, $f'\\geq 0$ in $(-\\infty, x_0)$ so it is increasing and $$f(x_0) = -\\frac1{3\\sqrt3}-\\frac1{\\sqrt3}-1<0$$ so $f<0$ in $(-\\infty, x_0)$. It follows that $f$ must have exactly $1$ real root. One can notice, that $f(1)f(2) = -5 < 0$, so there must be root of the equation. Also $f'(x) = 3x^2 - 1$, so it is monotonically increasing in range [1, 2] - there is only one root. Next step is to prove that $x^3 < x + 1$ for $x<1$. For $0 < x < 1$ we have $x + 1 > x > x^3$. For $-1 < x < 0$ following is met: $x + 1 > 0 > x ^ 3$. Finally for $x < -1$ we have $1 + x > x > x^3$. Last step is to prove, that $x^3 > x + 1$ for x > 2, but this is obvious, since $f(2) = 5 > 0$ and for every $x > 2$ $\\frac{dx^3}{dx} = 3x^2 > \\frac{d(x + 1)}{dx} = 1$. The solutions of $3c^2-1=0$ are $\\pm1/\\sqrt{3}$ outside $[1,2]$. That gives you a contradiction with the assumption that there", "has no real roots. In order for a quadratic equation to have no real roots its discriminant must be negative: $$D=2^2-4a=4-4a \\lt 0$$, which simplifies to $$1-a \\lt 0$$ and finally to $$a \\gt 1$$. Sufficient. (2) $$ax^2+1$$ is positive for all $$x$$: $$ax^2+1 \\gt 0$$. Now, when $$a \\ge 0$$ this expression is positive for all $$x$$. So, $$a$$ can be zero too. Not sufficient. I have trouble solving this question to get the correct statement. If x = 3, then 9-2(3)+a > 0 --> 3+a > 0 (per statement 1). Now , here a can be positive or negative and the expression will still be positive. Just wondering, how can this statement be sufficient then. Thanks ! Math Expert Joined: 02 Sep 2009 Posts: 43896 Show Tags 05 Dec 2016, 22:47 shubham1985 wrote: Bunuel wrote: Official Solution: Question: is $$a \\gt 0$$? (1) $$x^2-2x+a$$ is positive for all $$x$$. $$f(x)=x^2-2x+a$$ is a function of an upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$, so $$f(x)=x^2-2x+a \\gt 0$$, which means that this function is \"above\" X-axis OR in other words parabola has no intersections with X -axis OR equation $$x^2-2x+a=0$$ has no real roots. In order for a quadratic equation to have no real roots its discriminant must", "$x-4<0$, then: $ 2x-1$ , or rearranging: $ x<-3 $ . So our solution is $x>4$ or $x<-3$. RonL 5. Solve $\\frac{2x-1}{x-4}>1$ First: $\\frac{2x-1}{x-4}-1>0$ $\\frac{2x-1}{x-4} - \\frac{x-4}{x-4} > 0$ $\\frac{x+3}{x-4}>0$ There are critical points where the numerator is 0 and where the denominator is 0. So we have critical points at: x + 3 = 0 --> x = -3 x - 4 = 0 --> x = 4 So break the real line up into $(- \\infty , -3) \\cup (-3, 4) \\cup (4, \\infty)$ and test each interval: $(-\\infty, -3)$: $\\frac{x+3}{x-4}>0$ (Check!) $(-3, 4)$: $\\frac{x+3}{x-4}<0$ (Nope!) $(4, \\infty)$: $\\frac{x+3}{x-4}>0$ (Check!) So the solution set is $x \\in (-\\infty, -3) \\cup (4, \\infty)$ -Dan (Hi CaptainBlack! ) 6. For $f(x)=x^2+kx+k$ determine all values of $k$ such that $f(x)>0$. Now for large $|x|\\ f(x)>0$ as the leading term is dominant. So the question is realy asking us to find the values of k such that $f(x)=0$ has no real roots. From the quadratic formula we know the roots of $f(x)=0$ are: $ x=\\frac{-k \\pm \\sqrt{k^2-4k}}{2} $ , and these are not real if $k^2-4k<0$, which is equivalent to $k<4$ and $k>0$. RonL 7. Hello, SportfreundeKeaneKent! If the roots of the equation $x^3 - ax^2 - 3kx + m\\:=\\:0$ are $2,\\:-1$, and $3$, find the values of $a,\\:k$, and", "x \\le 4$$ Thus values of $$x$$ which satisfy both are $$-1 \\le x < 1$$ and $$2 < x \\le 4$$. 27. Since roots of $$ax^2 + bx + c$$ are imaginary, therefore discriminant is negative. $$\\Rightarrow b^2 - 4ac < 0$$ Discriminant of $$a^2x^2 + abx + ac$$ is: $$D = a^2b^2 - 4a^3c = a^2(b^2 - 4ac) < 0$$ But coefficient of the expression is positive hence it will be always positive. 28. Let $$y = \\frac{x^2 - 2x + 4}{x^2 + 2x + 4}$$ $$(y - 1)x^2 + 2(y + 1)x + 4(y - 1) = 0$$ Since $$x$$ is real discriminant will be greater or equal to zero. $$4(y + 1)^2 - 16(y - 1)^2 \\ge 0$$ $$y^2 + 2y + 1 - 4y^2 + 8y - 4 \\ge 0$$ $$-3y^2 + 10y - 3 \\ge 0$$ Roots of corresponding equation are $$\\frac{1}{3}, 3$$. Since coefficient of $$y^2$$ is negative, for above to be true $$y$$ must lie between $$\\frac{1}{3}$$ and $$3$$. 29. Let $$y = \\frac{2x^2 - 3x + 2}{2x^2 + 3x + 2}$$ $$2(y - 1)x^2 + 3(y + 1)x + 2(y - 1) = 0$$ Since $$x$$ is real discriminant will be greater or equal to zero. $$9(y + 1)^2 - 16(y - 1)^2 \\ge 0$$ $$9y^2 + 18y +", "0]$$ $$\\therefore f(\\alpha)$$ and $$f(\\beta)$$ have opposite signs. Therefore, $$f(x)$$ will have exactly one root between $$\\alpha$$ and $$\\beta$$. 20. Let $$f(x) = ax^2 + bx + c = 0$$ Since equation $$ax^2 + bx + c = 0$$ i.e. equation $$f(x) = 0$$ has no real root, therefore, $$f(x)$$ will have same sign for real values of $$x$$. $$\\therefore f(1)f(0) > 0 \\Rightarrow (a + b + c)c > 0$$ 21. Let $$f(x) = (x - a)(x - c) + \\lambda (x - b)(x - d)$$ Given $$a > b > c > d$$ Now $$f(b) = (b - a)(b - c) < 0$$ and $$f(d) = (d - a)(d - c) > 0$$ Since $$f(b)$$ and $$f(d)$$ have opposite signs, therefore equation $$f(x) = 0$$ will have one real root between $$b$$ and $$d$$. Since one root is real and $$a, b, c, d, \\lambda$$ are all real the other root will also be real. 22. Let $$f'(x) = ax^2 + bx + c,$$ then $$f(x) = a\\frac{x^3}{3} + b\\frac{x^2}{2} + cx + k = \\frac{2ax^3 + 3bx^2 + 4cx + 6k}{6}$$ $$f(1) = \\frac{2a + 3b + 6c + 6k}{6} = k$$ Again, $$f(0) = k$$ Thus, $$f(0) = f(1)$$ hence equation will have at least one root between $$0$$ and $$1$$ which implies that", "quadratic polynomial of the form $ax^2 + bx + c$ ($a > 0$)has a negative discriminant then it will always be positive for all real values of $x$. Now we first prove that $y > x$. This is true because the polynomial $8x^2 - 6x + 8$ has a negative discriminant ( $-220$ ), so it is always positive. Thus $y^3 > x^3 \\implies y> x$. Now we examine if $y= x+1$ is possible. Consider $y^3 - (x+1)^3 = 8x^2 - 6x + 8 - (x+1)^3= 5x^2 - 9x + 7$. This polynomial has a negative discriminant $- 59$, which means that it is positive for all values of $x$, thus $y^3 > (x+1)^3 \\implies y>x+1$. Now we will see if $y = x+2$ is possible. Note that $y^3 - (x+2)^3 = 2x^2 - 18x= 2 [(x-\\frac{9}{2})^2 - \\frac{81}{4}]$. Thus this has a minimum value of $-\\frac{81}{4}$ and no maximum value. So $y^3 - (x+2)^3$ can be $0$. Now we examine if $y= x+3$ is possible. Note that $(x+3)^3 - y^3= x^2 + 33x + 19$. Note that the function $x^2 + 33x + 19$ is monotonically increasing over the positive reals, and since $x$ is a positive integer, the minimum possible value for $x$ is $1$, which implies minimum possible value of $x^2 + 33x + 19$", "## College Algebra 7th Edition The solution is all real numbers except $x=0$ and $x=-1/2$. $\\displaystyle \\frac{1}{x}-\\frac{2}{2x+1}=\\frac{1}{2x^{2}+x}$ We multiply both sides by $(2x+1)(x)$: $(2x+1)-2(x)=1$ And distribute: $1=1$ When we get an identity (1=1, 2=2, x=x, etc), this implies that no matter what we choose for $x$, we will get a solution. Thus the solution should be all real numbers. However, if we use $x=0$ or $x=-1/2$, we would get division by 0 in the original equation. So these values are not allowed. Thus the solution is all real numbers except $x=0$ and $x=-1/2$.", "2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. ***** I believe the solution is incorrect on this question: S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious. S2 does not work, and neither does S1+S2, so the answer should be E. Question: is $$A>0$$? (1) $$x^2-2x+A$$ is positive for all $$x$$: $$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is \"above\" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots. Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$. Sufficient. (2) $$Ax^2+1$$ is positive for all $$x$$: $$Ax^2+1>0$$ --> when $$A\\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too. Not sufficient. _________________", "3$ $x^4 - 4 = 3x^2$ $x^4 - 3x^2 - 4 = 0$ $(x^2 - 4)(x^2 + 1) = 0$ $(x - 2)(x + 2)(x^2 + 1) = 0$. Since we know $x$ is real, that means we can only accept the cases $x - 2 = 0$ and $x + 2 = 0$. Therefore $x = 2$ or $x = -2$. Since we know $a = -\\frac{2}{x}$ that means $a = -1$ or $a = 1$. So the solutions are: $(x, a) = (2, -1)$ and $(x, a) = (-2, 1)$.", "### Home > A2C > Chapter 10 > Lesson 10.1.1 > Problem10-14 10-14. Find all values of a for which the equation ax2 + 5x + 6 = 0 has a solution that is a real number. Express a in terms of one or more inequalities. Homework Help ✎ Use the discriminant. Find the value of a that makes this quantity non-negative. $a\\le\\frac{25}{24}$", "### Solution #### Part (i) Let $f(x)=x^3+2x-4$. Now $f(0)=-4<0$$f(1)=-1<0$$f(2)=8>0$. Hence there is a root in $(1,2)$. 1. Now using the bisection method evaluate $f(1.5)=2.375>0$. Hence there is a root in $(1,1.5)$ (because $f$ is negative at 1 and positive at 1.5). 2. Hence evaluate $f(1.25)\\approx 0.453>0$. Hence there is a root in $(1,1.25)$. 3. Hence evaluate $f(1.125)\\approx -0.326$. Hence there is a root in $(1.125,1.25)$. 4. Hence evaluate $f(1.1875)\\approx 0.05>0$. Hence there is a root in $(1.125,1.1875)$. Now (we may as well) we can use the midpoint of this interval as our seed for the Newton-Raphson Method, $x_0=1.15625$. Once we get consecutive agreement to one decimal place we will finish. Now $f'(x)=3x^2+2$. Hence, $x_1=x_0-\\frac{f(x_0)}{f'(x_0)}$, $\\Rightarrow x_1=0.15625-\\frac{-0.141693}{6.01074}=1.17982$. #### Part (ii) Let $f(x)=x^5+2$. Note that for all $x>0$$f(x)>0$. Hence we have to look for a root in the negative real numbers. Now $f(-1)=1>0$$f(-2)=-30<0$. Hence there is a root in $(-2,-1)$. 1. By the Bisection Method, we evaluate $f(-1.5)\\approx -5.6<0$. Hence there is a root in $(-1.5,-1)$. 2. Evaluate $f(-1.25)\\approx -1.05$. Hence there is a root in $(-1.25,-1)$. 3. Evaluate $f(-1.125)\\approx 0.2>0$. Hence there is a root in $(-1.25,-1.125)$. 4. Evaluate $f(-1.1875)\\approx -0.36$. Hence there is a root in $(-1.1875,-1.125)$. Letting $x_0=-1.15625$. $x_1=x_0-\\frac{f(x_0)}{f'(x_0)}=-1.149$ $x_2=x_1-\\frac{f(x_1)}{f'(x_1)}=-1.149$.", "Let $\\alpha \\neq 1$ be a real root of the equation $x^3-ax^2+ax-1=0,$ where $a \\neq -1$ is a real number. Then a root of this equation, among the following, is : $\\begin {array} {1 1} (1)\\;\\alpha^2 & \\quad (2)\\;-\\frac{1}{\\alpha} \\\\ (3)\\;\\frac{1}{\\alpha} & \\quad (4)\\;-\\frac{1}{\\alpha^2} \\end {array}$ $(3)\\;\\frac{1}{\\alpha}$ answered Nov 7, 2013 by", "$f^{(k)}(x)$ has no positive real roots. By Lemma $1$, we conclude that $f(x)$ has at most one positive real root. Therefore when $r<pq$, we have exactly one positive real solution. This concludes the proof of the lemma. We now proceed to prove the main theorem: Theorem: Let $a, b, c, d$ be real numbers such that $0<a<b \\leq c<d$. Then the equation $$a^x+d^x=b^x+c^x$$ has • Exactly two solutions, $x=0$ and $x=t>0$ for some $t$, if $ad-bc<0$ • Exactly two solutions, $x=0$ and $x=t<0$ for some $t$, if $ad-bc>0$ • Exactly one solution, $x=0$, if $ad-bc=0$ Proof: Clearly $x=0$ is always a solution. For $x>0$, we may rewrite the equation as $$1+\\left(\\frac{d}{a}\\right)^x=\\left(\\frac{b}{a}\\right)^x+\\left(\\frac{c}{a}\\right)^x$$ We may now take $p=\\frac{b}{a}, q=\\frac{c}{a}, r=\\frac{d}{a}$ and use Lemma $2$ to get that there are • No positive real solutions (for $x$) if $\\frac{d}{a} \\geq \\frac{bc}{a^2}$, i.e. $ad \\geq bc$ • Exactly one positive real solution (for $x$) if $\\frac{d}{a}<\\frac{bc}{a^2}$, i.e. $ad<bc$ For $x<0$, we may take $y=-x$ and rewrite the equation as $$\\left(\\frac{a}{d}\\right)^x+1=\\left(\\frac{b}{d}\\right)^x+\\left(\\frac{c}{d}\\right)^x$$ $$1+\\left(\\frac{d}{a}\\right)^y=\\left(\\frac{d}{b}\\right)^y+\\left(\\frac{d}{c}\\right)^y$$ We may now apply Lemma $2$ with $p=\\frac{d}{c}, q=\\frac{d}{b}, r=\\frac{d}{a}$ and $y$ as the variable to get • No positive real solutions (for $y$) if $\\frac{d}{a} \\geq \\frac{d^2}{bc}$, i.e. $bc \\geq ad$ • Exactly one positive real solution (for $y$) if $\\frac{d}{a}<\\frac{d^2}{bc}$, i.e. $bc<ad$ This translates to • No negative real", "- 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. ***** I believe the solution is incorrect on this question: S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious. S2 does not work, and neither does S1+S2, so the answer should be E. Question: is $$A>0$$? (1) $$x^2-2x+A$$ is positive for all $$x$$: $$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is \"above\" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots. Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$. Sufficient. (2) $$Ax^2+1$$ is positive for all $$x$$: $$Ax^2+1>0$$ --> when $$A\\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too. Not sufficient. Your explanations are always so elegant.", "all $$x$$: Quadratic expression $$x^2-2x+A$$ is a function of of upward parabola (it's upward as coefficient of $$x^2$$ is positive). We are told that this expression is positive for all $$x$$ --> $$x^2-2x+A>0$$, which means that this parabola is \"above\" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots. Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$. Sufficient. (2) $$Ax^2+1$$ is positive for all $$x$$: $$Ax^2+1>0$$ --> when $$A\\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too. Not sufficient. Bunuel: Please explain the highlighted part. I didn't understand the real roots part. _________________ Intern Joined: 20 Apr 2013 Posts: 24 Concentration: Finance, Finance GMAT Date: 06-03-2013 GPA: 3.3 WE: Accounting (Accounting) Followers: 0 Kudos [?]: 4 [0], given: 99 Re: Is A positive? x^2-2x+A is positive for all x Ax^2+1 is [#permalink] ### Show Tags 03 May 2013, 12:45 Thanks for bearing with me The x^2-2x+A>0 is true for any value of X. Hence, there is no distinct roots for the equation so, Discriminant is less than zero. Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 179 [0], given: 12 Re: Is A positive? x^2-2x+A is positive for all x Ax^2+1", "4ac}$$ will not be real because $$(\\sqrt{b^{2} - 4ac})^{2}$$ = b$$^{2}$$ - 4ac < 0 and square of a real number always positive. Thus, the roots of the equation ax$$^{2}$$ + bx + c = 0 are not real if b$$^{2}$$ - 4ac < 0. As the value of b$$^{2}$$ - 4ac determines the nature of roots (solution), b$$^{2}$$ - 4ac is called the discriminant of the quadratic equation. Definition of discriminant: For the quadratic equation ax$$^{2}$$ + bx + c =0, a ≠ 0; the expression b$$^{2}$$ - 4ac is called discriminant and is, in general, denoted by the letter ‘D’. Thus, discriminant D = b$$^{2}$$ - 4ac Note: Discriminant of ax$$^{2}$$ + bx + c = 0 Nature of roots of ax$$^{2}$$ + bx + c = 0 Value of the roots of ax$$^{2}$$ + bx + c = 0 b$$^{2}$$ - 4ac = 0 Real and equal - $$\\frac{b}{2a}$$, -$$\\frac{b}{2a}$$ b$$^{2}$$ - 4ac > 0 Real and unequal $$\\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}$$ b$$^{2}$$ - 4ac < 0 Not real No real value When a quadratic equation has two real and equal roots we say that the equation has only one real solution. Solved examples to examine the nature of roots of a quadratic equation: 1. Prove that the equation 3x$$^{2}$$ + 4x + 6", "12 views If $x$ is real, the set of real values of $a$ for which the function $$y=x^2-ax+1-2a^2$$ is always greater than zero is 1. $- \\frac{2}{3} < a \\leq \\frac{2}{3}$ 2. $- \\frac{2}{3} \\leq a < \\frac{2}{3}$ 3. $- \\frac{2}{3} < a < \\frac{2}{3}$ 4. None of these in Calculus recategorized | 12 views 0 I have solved this by finding Discriminant. and getting $x = -2a, \\;\\text {and} \\;2a$ Is this the right method?", "polynomials in this lesson are assumed to have real number coefficients. Therefore, instead of saying things like ‘Let $\\,P(x) = ax + b\\,$ where $\\,a\\,$ and $\\,b\\,$ and real numbers and $\\,a\\ne 0\\,$’, we can more simply say ‘Let $\\,P(x) = ax + b\\,$, where $\\,a\\ne 0\\,$’. • Polynomial equations of degree $1$ are called linear equations. The standard form is $\\,ax + b = 0\\,$, for $\\,a\\ne 0\\,$. They are solved using the addition and multiplication properties of equality. The unique solution is $\\,x = -\\frac{b}{a}\\,$. • Polynomial equations of degree $2$ are called quadratic equations. The standard form is $\\,ax^2 + bx + c = 0\\,$, for $\\,a\\ne 0\\,$. They are solved using the quadratic formula. The solutions are: $\\displaystyle x = \\frac{-b\\pm\\sqrt{b^2 - 4ac}}{2a}$ The discriminant is: $\\,b^2 - 4ac\\,$ If the discriminant is positive, there are two different real number solutions—call them $\\,r_1\\,$ and $\\,r_2\\,$. In this case: $ax^2 + bx + c = a(x - r_1)(x - r_2)\\,$ If the discriminant is zero, there is exactly one real number solution—call it $\\,r\\,$. In this case: $ax^2 + bx + c = a(x - r)^2\\,$ If the discriminant is negative, there are no real number solutions. Instead, there is a complex conjugate pair of solutions, call them $\\,z_1\\,$ and $\\,z_2\\,$. In this case, $ax^2", "(C) ${1,5,10}$. (D) ${1,2,10}$. Problem 14: Consider all $2 \\times 2$ matrices whose entries are distinct and taken from the set $\\{1,2,3,4\\}$. The sum of determinants of all such matrices is (A) 24 . (B) 10 . (C) 12 . (D) 0 . Problem 15: Let $a, b, c$ and $d$ be four non-negative real numbers where $a+b+c+d= 1$. The number of different ways one can choose these numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=\\max \\{a, b, c, d\\}$ is (A) 1 . (B) 5 . (C) 11 . (D) 15 . Problem 16: The polynomial $x^{4}+4 x+c=0$ has at least one real root if and only if (A) $c<2$. (B) $c \\leq 2$. (C) $c<3$. (D) $c \\leq 3$. problem 17: The number of all integer solutions of the equation $x^{2}+y^{2}+x-y=$ 2021 is (A) 5 . (B) 7 . (C) 1 . (D) $0 .$ Problem 18: The number of different values of $a$ for which the equation $x^{3}-x+a=$ 0 has two identical real roots is (A) 0 . (B) 1 . (C) $2 .$ (D) 3 . Problem 19: Suppose $f(x)$ is a twice differentiable function on $[a, b]$ such that $f(a)=0=f(b)$ and $x^{2} \\frac{d^{2} f(x)}{d x^{2}}+4 x \\frac{d f(x)}{d x}+2 f(x)>0$ for all $x \\in(a, b)$ Then, (A) $f$ is negative for all $x \\in(a, b)$. (B) $f$", "- a)(x - c) + \\lambda (x - b)(x - d) = 0$$ has real roots for all real values of $$\\lambda$$. 22. If $$2a + 3b + 6c = 0, (a, b, c \\in R)$$ then show that the equation $$ax^2 + bx + c = 0$$ has at least one root between $$0$$ and $$2.$$ 23. If $$a, b, c$$ be non-zero real numbers such that $$\\int_0^1 (1 + \\cos^8 x)(ax^2 + bx + c)dx = \\int_0^2 (1 + \\cos^8 x)(ax^2 + bx + c)dx$$ show that equation $$ax^2 + bx + c = 0$$ has at least on real root between $$1$$ and $$2$$. 24. Let $$f(x) = ax^2 + bx + c,$$ where $$a, b, c \\in R$$ and $$a \\ne 0.$$ If $$f(x) = x$$ has non-real roots, show that equation $$f(f(x)) = x$$ has all non-real roots. 25. Let $$a, b, c$$ be positive integers and consider all the quadratic equations of the form $$ax^2 - bx + c = 0$$ which have two distinct real roots in $$]0, 1[$$. Find the least positive integers $$a$$ and $$b$$ for which such a quadratic equation exist. 26. If equation $$ax^2 - bx + c = 0$$ have two distinct real roots in $$(0, 1), a, b, c \\in N,$$ then prove that $$\\log_5(abc)\\ge" ]

[ "A cubic polynomial $$p(x)$$ is such that $$p(1)=1$$, $$p(2)=2$$, $$p(3)=3$$ and $$p(4)=5$$. Then the value of $$p(6)$$ is:", "# !Fancy cubic Author: anonymous Problem has been solved: 64 times Русский язык | English Language The cubic polynomial $P (x)$ is given. For a certain number $a$, we define $P_1(x) = x^2 + (a-29) x-a$ and $P_2(x) = 2x^2 + (2a-43)x + a$. It turned out that $P(x)$ is divisible by $P_1(x)$ and $P_2(x)$. Find the largest possible value of $a$.", "# A Pre-RMO question! -6 Algebra Level 1 Cubic polynomial $P(x)$ is such that $P(0)=k$, $P(1)=2k$, $P(-1)=3k$, and $\\\\ P(2)+P(-2)=\\alpha k$, where $\\alpha$ is a real number, find $\\alpha$. ×", "The polynomial $$P(x)$$ is cubic. What is the largest value of $$k$$ for which the polynomials $$Q_1(x) = x^2 + (k-29)x - k$$ and $$Q_2(x) = 2x^2+ (2k-43)x + k$$ are both factors of $$P(x)$$? (第二十五届AIME1 2007 第8题)", "# Easy polynomial by me A cubic polynomial $P(x)$ satisfies the following: $P(1)=1, P(2)=2, P(3)=3, P(4)=5.$ What is the value of $P(6)?$ ×", "# poly poly nomial!!!!! Algebra Level 3 A cubic polynomial $$p(x)$$ is such that $$p(1)=1$$, $$p(2)=2$$, $$p(3)=3$$ ,$$p(4)=5$$ .Find the value of $$p(6)$$. A polynomial of degree $$3$$ is called a cubic polynomial. ×", "> k^2 + 2k + 1$", "+0 # Help! +1 178 1 +1207 Find a monic cubic polynomial P(x) with integer coefficients such that$$P(\\sqrt[3]{2} + 1) = 0$$. (A polynomial is monic if its leading coefficient is 1.) May 31, 2019 #1 0 P(x) = x^3 - 6x^2 + x Nov 14, 2019", "we had: $n=a^2+k$ So, $n-kl=a^2+k-2ak-k+k^2=(k-a)^2$ which is a perfect square. If the numbers are $p^2$, $p^2 + k$ and $p^2 + k + l$, then we must have that $k+l = 2p+1$. A little algebra shows that $p^2 + k - kl = (p-k)^2$, after putting $l = (2p+1) -k$.", "# A cubic polynomial Algebra Level 3 If p(x) is a cubic polynomial with $$p(1) = 1$$, $$p(2) = 2$$, $$p(3) = 3$$, $$p(4) = 5$$, then find $$p(6)$$. ×", "# Cubics Algebra Level 4 Let $a, b, c$ be the roots of the cubic $x^3 + 3x^2 + 5x + 7$. Given that $P$ is a cubic polynomial such that $P(a) = b + c, P(b) = c + a, P(c) = a + b,$ and $P(a + b + c) = -16$, find $P(0)$. ×", "# Easy polynomial by me Algebra Level 3 A cubic polynomial $$P(x)$$ satisfies the following: $P(1)=1, P(2)=2, P(3)=3, P(4)=5.$ What is the value of $$P(6)$$? ×", "+0 # Polynomial 0 244 1 Find a monic cubic polynomial P(x) with integer coefficients such that $P(\\sqrt[3]{4} + 1) = 0$. (A polynomial is monic if its leading coefficient is 1.) Jun 19, 2021 Let the polynomial $P$ be defined by $P(x)=x^3+bx^2+cx+d$. Now substitute $\\sqrt[3]{4}+1$ for $x$ in the formula for $P(x)$ to get $P(\\sqrt[3]{4}+1)$ and collect terms so that you have $P(\\sqrt[3]{4}+1) = k\\cdot 4^{1/3} + \\ell\\cdot 4^{2/3} + m$ where $k, \\ell, m$ are integral expressions involving $b,c,d$. Now set up three equations $k=0$, $\\ell=0$, and $m=0$, and solve for $b,c,d$. You should get the solution $P(x) = x^3-3x^2+3x-5$. (You should check that $P(\\sqrt[3]{4}+1) = 0$.)", "+ x^2 + 1$ is 4? Degree of a polynomial $P(x)$ is the highest power of $x$ in $P(x)$. Therefore, $x^3 + x^{a-4} + x^2 + 1$, $x^{a-4}$ = $x^4$ $a-4$ = $4$, $a$ = $4 + 4$ = $8$ Consider, $P(x)$ = $x^2 – 3x + 2$, Put $x$ = $3$ in $P(x)$ which gives, $P(3)$ = $9 – 9 + 2$ = $2$ Replace $x$ by 2 in the polynomial $x^2 – 3x + 2$, which gives $P(2)$ = $4 – 6 + 2$ = $0$. Similarly, value of $x^2 – 3x + 2$ at $x$ = $0$ is, $P(0)$ = $0 – 0 + 2$ = $2$ In general; if P(x) is a polynomial in x and k is any real number, then value of $P(k)$ at $x$ = $k$ is denoted by $P(k)$ is found by replacing $x$ by $k$ in $P(x)$. In the polynomial $x^2 – 3x + 2$, Replacing $x$ by 1 gives, $P(1)$ = $1 – 3 + 2$ = $0$ Similarly, replacing $x$ by 2 gives, $P(2)$ = $4 – 6 + 2$ = $0$ For a polynomial $P(x)$, real number k is said to be zero of polynomial $P(x)$, if $P(k)$ = $0$. Therefore, 1 and 2 are the zeros of polynomial $x^2 – 3x + 2$. Consider, $P(x)$", "+0 # Help! +1 124 1 +1207 Find a monic cubic polynomial P(x) with integer coefficients such that$$P(\\sqrt[3]{2} + 1) = 0$$. (A polynomial is monic if its leading coefficient is 1.) May 31, 2019", "$i\\ge1$ we have $P(1,k)\\mid P(i,k)$. Proof. The case $i=1$ is trivial, as i sthe case $k=1$. For all other cases, we use the above identity: If $P(i,k-1)=cP(1,k-1)$ and $P(i-1,k)=dP(1,k)$, then $$P(i,k)=P(i,k-1)F_kF_{i+1}+P(i-1,k)F_{k-1}=(cF_{i+1}+dF_{k-1})P(1,k)$$ as was to be shown. $_\\square$", "if $p=2$ or $p=3$.", "# Is $P$ unique? Let $P(x)$ be a cubic polynomial with integer coefficients such that $P$ has 3 negative integer roots and $P(1)=3553$. Find $P(-1)$. ×", "that $p-1=MN$. Considering some primitive root $a$ of $p$, we have $2\\equiv a^k$ for some $k$, and as $1\\equiv2^N\\equiv a^{kN}$, we need $(p-1)|kN$, so $M|k$ and $2\\equiv a^k=(a^{k/M})^M$. So we do necessarily need $p$ to satisfy both that $M|p-1$, and that $2$ is $x^M$ for some $x$ (i.e., $x^M-2$ has a root mod $p$). Something like Chebotarev's density theorem is probably needed to guarantee this. – ShreevatsaR May 11 '13 at 17:03", "$m=(p,f)$ where $p \\in K$ is a prime and $f \\in K[x]$ is a polynomial whose reduction in $K/p[x]$ is irreducible. – Bogdan Feb 2 '15 at 8:49" ]

[ "and $$g$$. Polynomials should be denoted only using Capital letters. The symbol to denote polynomial is $$P(x)$$ if the equation has $$x$$ as variable. Example: \\begin{align*} P(x) = 2x^2 -x+3 \\end{align*} The $$x$$ in $$P(x)$$ represents the variable $$x$$. If the variable of the equation is $$y$$, the symbol to denote the polynomial is $$P(y)$$ Example: \\begin{align*} P(y) = 2y^2 -y+3 \\end{align*} The $$y$$ in $$P(y)$$ represents the variable $$y$$. Example 1: It is given that $$P(x) = 2x^2 - x + 3$$ and \\begin{align*} Q(x) = x^3 + 2x -1 \\end{align*} Evaluate $$P(2) +Q(-1)$$. Solution: To find $$P(2)$$, we need to replace the $$x$$ with $$2$$. \\begin{align*} P(2) = 2(2)^2 - 2 +3 \\end{align*} To find $$Q(-1)$$, we need to replace the $$x$$ with $$-1$$. \\begin{align*} Q(-1) = (-1)^3 + 2(-1) -1 \\end{align*} Adding both of the above equations. \\begin{align*} P(2) +Q(-1) = [\\;2(2)^2 - 2 +3\\;] \\;\\;+\\;\\;[\\;(-1)^3 + 2(-1) -1\\;] \\end{align*} Solving the above will give the answer. $$=5$$ Therefore, the answer is $$5$$. Example 2: Given that \\begin{align*} x^3 - 5x+3 = (Ax+3)(x-1)(x-2) \\;+\\; B(x-1) \\;+\\; C \\end{align*} for all real values of $$x$$, find the values of $$A,B$$ and $$C$$. Solution: In the question, it is given as “for all real values of $$x$$” which means we can substitute any value for $$x$$ and", "Polynomial? A polynomial of the form $$p(x)=ax^2+bx+c$$, where $$a \\neq 0$$ is called a quadratic polynomial. ### Zeros of Quadratic Polynomial: Definition real number $$\"k\"$$ of a quadratic polynomial $$p(x)$$ is 0 if $$p(k)=0$$ . Consider the polynomial: $$p(x)=x^2-3x-4$$ Let's find the value of the polynomial at $$x=-1$$ by substituting -1 for $$x$$ in $$p(x)$$. \\begin{align}p(-1)&=(-1)^2-3(-1)-4\\\\&=1+3-4\\\\&=0\\end{align} So, here $$x=-1$$ is a zero of the quadratic polynomial $$p(x)=x^2-3x-4$$. ## How To Find Zeros of a Quadratic Polynomial? There are two methods to find the zeros of a quadratic polynomial: Let us consider the quadratic polynomial $$2x^2+x-3=0$$ Let’s find the determinant to find the nature of the solution. \\begin{align}b^2-4ac&=1^2-4\\times 2\\times (-3)\\\\&=1+24\\\\&=25>0\\end{align} This means that the quadratic polynomial has two real and distinct zeros. Just plug in the values $$a=2$$, $$b=1$$ and $$c=-3$$ in the quadratic formula. \\begin{align}x&=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\\\&=\\frac{-1\\pm \\sqrt{1^2-4\\times 2\\times (-3)}}{2\\times2}\\\\&=\\frac{-1\\pm \\sqrt{25}}{4}\\\\&=\\frac{-1\\pm 5}{4}\\\\&=-\\frac{3}{2},1\\end{align} \\begin{align} x=-\\frac{3}{2},1 \\end{align} ### Factorization Let’s solve this quadratic polynomial by factorization method. Here, we need to find $$a$$ and $$b$$ such that $$a+b=1$$ and $$ab=-6$$ Considering this, we see that the numbers are $$a=3$$ and $$b=-2$$ \\begin{align}2x^2+x-3&=0\\\\2x^2+3x-2x-3&=0\\end{align} Take the common factors out from the first two terms and last two terms. So, this quadratic polynomial factoring gives: \\begin{align}x(2x+3)-(2x+3)&=0\\2x+3)(x-1)&=0\\end{align} When the product of two numbers is \\(0, then either of the numbers are $$0$$ This means $$2x+3=0$$", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "since factoring a cubic polynomial is not necessarily the easiest thing ever. Note that: (4) \\begin{align} \\quad \\quad 4(k+1)^3 - (k+1) = 4(k+1)(k^2 + 2k + 1) - (k +1) = 4(k^3 + 2k^2 + k + k^2 + 2k + 1) - (k+1) \\\\ \\quad \\quad = 4(k^3 + 3k^2 + 3k + 1) - (k +1) = 4k^3 + 12k^2 + 12k + 4 - k - 1 = 4k^3 + 12k^2 + 11k + 3 \\end{align} Therefore $4k^3 + 12k^2 + 11k + 3 = 4(k+1)^3 - (k+1)$ and so: (5) \\begin{align} LHS_{P(k+1)}: = \\frac{4k^3 + 12k^2 + 11k + 3}{3} \\\\ = \\frac{4(k+1)^3 - (k+1)}{3} \\end{align} Therefore the truth of $P(k)$ implies the truth of $P(k+1)$ and so for every $n ≥ 1$, $P(n)$ is true. $\\blacksquare$", "polynomial with coefficients all $1$ and taken $\\pmod n$ if the exponents cover $0,1,…,n-1$, you can rest assured it is divisible by $(x^n-1)/(x-1)$. May not be easy to recognise always, but it is a simple enough test. +1 Sep 7, 2020 at 14:02 Hmm. I'd look at it and say \"There's no rational root\" because $$x = \\pm 1$$ doesn't work. So there's some irrational root, $$\\alpha$$, and two complex-conjugate pairs. So there's no nice obvious linear factor I can write down. Then I'd say \"Maybe there's a quadratic factor.\" I can assume it's monic, so I'm looking to write $$x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r).$$ from which I can expand to get $$x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br$$ if I've done the algebra right. Equating coefficients I see that \\begin{align} 0 &= a + p\\\\ 0 &= q + ap + b\\\\ 1 &= ra + bq\\\\ 1 &= br \\end{align} so $$p = -a$$, and $$r = \\frac1b$$,and these equations become \\begin{align} 0 &= a + -a\\\\ 0 &= q - a^2 + b\\\\ 1 &= \\frac1b a + bq\\\\ 1 &= b(1/b) \\end{align} which simplify down to", "of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so $$\\frac{-b\\pm\\sqrt{b^2 - 4ac}}{2a} = \\frac{1 \\pm\\sqrt{1 - -24}}{4} = \\frac{1\\pm 5}{4} = \\frac{3}{2}, -1,$$ which gives a factorization into the monic polynomials $x+1$ and $2x - 3$. Note that $2-3=-1,$ so that \\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\\\ &= 2x^2+2x-3x-3\\\\ &= 2x(x+1)-3(x+1)\\\\ &= (x+1)(2x-3).\\end{align} • Sweet, simple, and to the point. Aug 20, 2013 at 5:30 You can use several methods to solve the equation. Firstly you can use the quadratic formula $$\\frac{-b \\pm\\sqrt{b^2 - 4ac}}{2a}$$ or you can use decomposition. Let's try decomposition first. Look for two numbers that multiple to $-6$ and add to $-1$. These numbers would be $+2$ and $-3$. So, we would place those for $-x$ and have the following: $$2x^2-3x+2x-3$$ Now here you can factor out $x$ from the first two terms. $$x(2x-3)+(2x-3)$$ Notice the coefficient on $2x-3$ is $1$. So this means that our equation is $x+1$ and $2x-3$. Therefore, giving us the solution $$(2x-3) = 0 \\, \\text{or} \\, (x+1) = 0$$ Now, if you want to use the quadratic formula then you can do this. $a = 2$, $b = -1$ and $c = -3$. So we will have: $$\\frac{6 \\pm\\sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$$ When you apply this formula twice, you will get the roots which are $x = \\frac{3}{2}$ and", "as the sum of this square and a constant: $x^2 + 10x + 28 \\,=\\, (x+5)^2 + 3.$ This is called completing the square. ### General description $x^2 + bx + c,\\,\\!$ it is possible to form a square that has the same first two terms: $\\left(x+\\tfrac{1}{2} b\\right)^2 \\,=\\, x^2 + bx + \\tfrac{1}{4}b^2.$ This square differs from the original quadratic only in the value of the constant term. Therefore, we can write $x^2 + bx + c \\,=\\, \\left(x + \\tfrac{1}{2}b\\right)^2 + k,$ where k is a constant. This operation is known as completing the square. For example: \\begin{alignat}{1} x^2 + 6x + 11 \\,&=\\, (x+3)^2 + 2 \\\\[3pt] x^2 + 14x + 30 \\,&=\\, (x+7)^2 - 19 \\\\[3pt] x^2 - 2x + 7 \\,&=\\, (x-1)^2 + 6. \\end{alignat} ### Non-monic case Given a quadratic polynomial of the form $ax^2 + bx + c\\,\\!$ it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: \\begin{align} 3x^2 + 12x + 27 &= 3(x^2+4x+9)\\\\ &{}= 3\\left((x+2)^2 + 5\\right)\\\\ &{}= 3(x+2)^2 + 15 \\end{align} This allows us to write any quadratic polynomial in the form $a(x-h)^2 + k.\\,\\!$ ### Formula The result of completing the square may be written as a formula. For the general case:[1] $ax^2 + bx +", "Phil Fernandez Apr 16 '18 at 1:31 $$\\frac{1}{x^4-K^4}=\\frac{A}{x-K}+\\frac{B}{x+K}+\\frac{Cx+D}{x^2+K^2}$$ To solve Dr. Graubner's equation $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$ as stated above, is a lot of linear algebra. Some of that can be avoided. $$\\frac{1}{x^4-K^4}=\\frac{A}{x-K}+\\frac{B}{x+K}+\\frac{Cx+D}{x^2+K^2}$$ Using $$x^4-K^4 = (x-K)(x+K)(x^2+K^2)$$, we get $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$ Letting $$x=K$$, you get \\begin{align} 1 &= 4AK^3 \\\\ A &= \\dfrac{1}{4K^3} \\end{align} Letting $$x=K$$, you get \\begin{align} 1 &= -4BK^3 \\\\ B &= -\\dfrac{1}{4K^3} \\end{align} Letting $$x=iK$$, you get $$1 = -2K^2(D+iCK)$$. Letting $$x=-iK$$, you get $$1 = -2K^2(D-iCK)$$. So $$C = 0$$ and $$D = -\\dfrac{1}{2K^2}$$. Hence $$\\frac{1}{x^4-K^4}=\\frac{1}{4K^3(x-K)}-\\frac{1}{4K^3(x+K)}-\\frac{1}{2K^2(x^2+K^2)}$$ If you don't want to have to resort to complex variables, you can do this after finding the values of $$A$$ and $$B$$. \\begin{align} A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \\\\ \\dfrac{(x+K)(x^2+K^2)}{4K^3}-\\dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \\\\ \\dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \\\\ (Cx+D)(x^2-K^2) &= 1 - \\dfrac{x^2+K^2}{2K^2} \\\\ (Cx+D)(x^2-K^2) &= -\\dfrac{x^2-K^2}{2K^2} \\\\ Cx+D &= -\\dfrac{1}{2K^2} \\end{align}", "# check workings for polynomial function from graph The diagram shows a curve with equation of the form $y = kx(x + a)^2$, which passes through $(-2, 0)$, $(0, 0)$ and $(1, 3)$. What are the values of $a$ and $k$. To find $k$, I would set $f(x) = 1$. $f(1) = k(x + 2)(x + 0)(x - 1)$ $3 = k(1 + 2)(x + 1)(1 - 1)$ $3 = k(3)(1)(0)$ This would make $k = 0$ which does not appear to be right. • Don't you mean $f(1) = 3$? – Pichi Wuana Mar 6 '16 at 19:19 $$y=kx(x+a)^2\\quad \\quad \\quad (-2, 0),(0, 0) , (1, 3)$$ $$\\text{for} (0,0):\\quad \\quad \\quad f(0)=0=0\\\\ \\text{for}(1,3):\\quad \\quad \\quad f(1)=k(1+a)^2=3\\\\ \\text{for}(-2,0):\\quad \\quad \\quad f(-2)=-2k(-2+a)^2=0$$ so you have two equations with two variables: $$\\begin{cases}k(1+2a+a^2)=3\\\\ -2k(4-4a+a^2)=0\\end{cases}$$ $$\\begin{cases}k+2ak+ka^2=3\\\\ -8k+8ka-2ka^2=0\\end{cases}$$ Solve, you should get $\\color{red}{a=2,k=\\frac 1 3}$ The equation is satisfied by $(-2,0)$ and $(1,3)$ Putting $(-2,0)$ in the equation we obtain: $$0 = k(-2)(-2+a)^2$$ Therefore, $(a-2)=0$ as $k \\ne 0$ $$\\implies a=2$$ Similarly, putting $(1,3)$ in the equation we obtain: $$3 = k*1*(1+2)$$ Therefore $k=1$", "$d_2 = a^2 - bc$. We then get that $d_0 = 1$. Hence, $$k_1 + k_2 = 1.$$ $$a (k_1 + k_2) + (k_1 - k_2)\\sqrt{a^2-4bc} = 2a \\implies k_1 - k_2 = \\dfrac{a}{\\sqrt{a^2-4bc}}.$$ Hence, \\begin{align} k_1 & = \\dfrac{a + \\sqrt{a^2-4bc}}{2\\sqrt{a^2-4bc}}, & k_2 & = -\\dfrac{a-\\sqrt{a^2-4bc}}{2\\sqrt{a^2-4bc}} \\end{align} And finally: $$\\color{red}{\\det(A_n) = \\dfrac1{\\sqrt{a^2-4bc}} \\left( \\left( \\dfrac{a + \\sqrt{a^2-4bc}}2\\right)^{n+1} - \\left( \\dfrac{a - \\sqrt{a^2-4bc}}2\\right)^{n+1}\\right)}.$$ Plug in $a = 5$ and $b=c=2$ ($a^2 - 4 bc \\neq 0$), to get $$\\det(A_n) = \\frac{1}{3} ( 4^{n+1} - 1)$$ ## Case 2: $a^2 - 4bc = 0$ If the characteristic polynomial has a double root $x = a/2$, there exist constants $k_1$, $k_2$ such that: $$d_n = (k_1 + k_2 n) \\bigl(\\frac{a}{2}\\bigr)^n.$$ The initial conditions are $d_0 = 1$ and $d_1 = a$, thus: \\begin{align} k_1 & = 1 & (k_1 + k_2) a = 2a \\end{align} If $a = 0$, then $4bc = a^2$ implies either $b$ or $c$ is zero, and $d_n = 0$ for $n \\ge 1$. Otherwise $$(k_1 + k_2) a = 2a \\implies k_1 + k_2 = 2 \\implies k_2 = 1.$$ And finally: $$\\color{red}{\\det(A_n) = (n+1) \\bigl(\\frac{a}{2}\\bigr)^n}.$$ - For some reason, this is my first time seeing the off diagonal elements being different, though it clearly doesn't change anything in the proof. – Calvin Lin Dec", "5 - 8 = 2(x + 2)^2 - 3.$$ Note that$f$is insensitive to the sign of$x + 2$so$f$is symmetric about the line$x = -2$2 Doing$(a,b,c)→(a^2,b^2,c^2)$two times is clearly the easier way. Let$A=a+b+c,B=ab+bc+ca,C=abc. Consider \\begin{align}&(x^{1/2}-a)(x^{1/2}-b)(x^{1/2}-c)\\times (x^{1/2}+a)(x^{1/2}+b)(x^{1/2}+c) = \\\\&(x^{3/2}-Ax+Bx^{1/2}-C)\\times (x^{3/2}+Ax+Bx^{1/2}+C) = \\\\&(x^{3/2}+Bx^{1/2})^2 - (Ax+C)^2 = \\\\&x^3+2Bx^2+B^2x-A^2x^2-2ACx-C^2\\end{align} ... 2 Using the power of the Dandelin-Gräffe iteration. For any polynomialp$with roots$z_i$, the polynomial$p^{(1)}(x)$defined via $$p^{(1)}(x^2)=p(x)·p(-x)$$ has the same degree as$p$and roots$z_i^2$. Repeating this process, the polynomial$p^{(2)}(x)$defined via $$p^{(2)}(x^2)=p^{(1)}(x)·p^{(1)}(-x)$$ has still the same degree as$p$and roots ... 2 This solution is similar to what you are saying, but not phrased in your language of implicit function theorem. Consider the polynomial $$\\sum_{i=1}^n \\left[ x_i ^m \\prod_{j < k, j\\neq i, k\\neq i} (x_j-x_k)\\right]$$ If$x_i = x_j$, then you should verify that the polynomial is equal to 0 (many terms with$x_i-x_j)$are zero, and then ensure that the ... 2 I'll suppose$R$is an integral domain, and therefore contained in an algebraically closed field$K$(one can do without this assumption, but it simplifies the argument). Suppose$p(s_1,\\ldots,s_n)=0$but$p$is not the zero polynomial; then certainly there exist constants$a_1,\\ldots,a_n\\in K$such that$p(a_1,\\ldots,a_n)\\neq0. Now consider $$... 2 I think what you need is Newton's identities, in particular the section about their application to the roots of a polynomial. 2 We have$$\\begin{align} a+b+c & = 3 \\\\ \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} &= \\frac{5}{12} \\\\", "polynomial of the form: $g_k(n) = x_{k-1} n^{k-1} + \\ldots + x_1 n + x_0$ for some coefficients $$x_{k-1},\\ldots,x_1,x_0$$. How do we work out the values $$x_{k-1},\\ldots,x_0$$? We can deduce them using the terms we already know! First suppose that $$k=2$$. Since $$g_k$$ must agree with the terms we already have then $$g_2(n) = u_n$$ for all $$n \\in \\{ 1,2 \\}$$. Moreover from the reasoning above $$g_2(n)$$ takes the form $$g_2(n) = x_1 n + x_0$$. Substituting in the appropriate values for $$n$$ we get the following constraints: \\begin{align*} u_1 & = x_0 + x_1 \\\\ u_2 & = x_0 + 2 x_1 \\end{align*} There are two unknowns are two equations meaning we can solve them to find the right values for $$x_1$$ and $$x_0$$. In the example above we get: \\begin{align*} 4 & = x_0 + x_1 \\\\ 17 & = x_0 + 2 x_1 \\end{align*} and solving these equations gives us $$x_1 = 13$$ and $$x_0 = -9$$. Therefore $$g_2(n) = 13n - 9$$, which thankfully agrees with what we worked out earlier. Now let $$k=3$$ and we are left with the following system of three equations: \\begin{align*} 4 & = x_0 + x_1 + x_2 \\\\ 17 & = x_0 + 2 x_1 + 4 x_2 \\\\ 46 & = x_0 + 3 x_1", "0$$ Using splitting the middle term method, \\begin{align*} x^2 - 3x + 2x - 6 &= 0 \\\\ x( x - 3) + 2( x - 3) &= 0 \\\\ x( x - 3) + 2( x - 3) &= 0 \\end{align*} Taking $$(x - 3 )$$ common, we get, $$( x - 3)( x + 2) = 0$$ Hence, $$(x - 3)$$ and $$(x + 2)$$ are the factors of the given polynomial. Equating the polynomial equation to 0, we get the zeros or roots of the polynomial. \\begin{align*} x &= 3 \\\\ x &= -2 \\end{align*} $$\\therefore$$ Degree of polynomial = 2 Zeros = 3, -2 Example 2 For what value of $$k$$, -4 is the zero of the polynomial $$x^2 - 2x - (7k + 3)$$? Solution Zero of a polynomial $$P(x)$$ gives the value of polynomial as '0' when substituted for $$x$$. Thus, for -4 to be the zero of the given polynomial, \\begin{align*} P(-4) = (-4)^2 - 2(-4) - (7p + 3 ) &= 0 \\\\ 16 + 8 -7k - 3 &= 0 \\\\ 7k &= 21 \\\\ k &= 3 \\end{align*} $$\\therefore$$ k = 3 Example 3 Write the number of zeros of the polynomial, $$y = f(x)$$ in the following graph: Solution For the given graph: The graph intersects the", "= \\frac{43}{2} - k$, $am = -k$, and $an = \\frac{k}{2}$. The first two equations show that $m - n = 29 - \\frac{43}{2} = \\frac{15}{2}$. The last two equations show that $\\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = 30$. ### Solution 3 Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations: \\begin{align} a = 2c \\\\ b = -d \\\\ 2c(k - 29) - d = c(2k - 43) + 2d \\\\ -d(k - 29) - 2ck = d(2k - 43) + ck \\end{align} Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\\boxed {30}$. ~ anellipticcurveoverq", "worked on a paper regarding just this question (http://arxiv.org/abs/1403.8139). We were generalizing Tokuyama's deformation formula (http://projecteuclid.org/euclid.jmsj/1230129805) for the Schur polynomial, so we used Gelfand-Tsetlin patterns rather than Young tableaux. Tokuyama's ... 2 Do thisf(x) = 2(x^2 + 4x) + 5 = 2(x^2 + 4x + 4) + 5 - 8 = 2(x + 2)^2 - 3.$$Note that f is insensitive to the sign of x + 2 so f is symmetric about the line x = -2 2 Doing (a,b,c)→(a^2,b^2,c^2) two times is clearly the easier way. Let A=a+b+c,B=ab+bc+ca,C=abc. Consider$$\\begin{align}&(x^{1/2}-a)(x^{1/2}-b)(x^{1/2}-c)\\times (x^{1/2}+a)(x^{1/2}+b)(x^{1/2}+c) = \\\\&(x^{3/2}-Ax+Bx^{1/2}-C)\\times (x^{3/2}+Ax+Bx^{1/2}+C) = \\\\&(x^{3/2}+Bx^{1/2})^2 - (Ax+C)^2 = \\\\&x^3+2Bx^2+B^2x-A^2x^2-2ACx-C^2\\end{align}$$... 2 Using the power of the Dandelin-Gräffe iteration. For any polynomial p with roots z_i, the polynomial p^{(1)}(x) defined via$$ p^{(1)}(x^2)=p(x)·p(-x) $$has the same degree as p and roots z_i^2. Repeating this process, the polynomial p^{(2)}(x) defined via$$ p^{(2)}(x^2)=p^{(1)}(x)·p^{(1)}(-x) $$has still the same degree as p and roots ... 2 This solution is similar to what you are saying, but not phrased in your language of implicit function theorem. Consider the polynomial$$ \\sum_{i=1}^n \\left[ x_i ^m \\prod_{j < k, j\\neq i, k\\neq i} (x_j-x_k)\\right] $$If x_i = x_j, then you should verify that the polynomial is equal to 0 (many terms with x_i-x_j) are zero, and then ensure that the ... 2 I'll suppose R is an integral", "square trinomial has the form And(a+b)2 if a2+2(ab)+b2(ab)2 if a22(ab)+b2 In these problems, the key is figuring out what the a\\begin{align*}a\\end{align*} and b\\begin{align*}b\\end{align*} terms are. Let’s do some examples of this type. Example: x2+8x+16\\begin{align*}x^2+8x+16\\end{align*} Solution: Check that the first term and the last term are perfect squares. x2+8x+16asx2+8x+42. Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them. x2+8x+16asx2+24x+42 This means we can factor x2+8x+16\\begin{align*}x^2+8x+16\\end{align*} as (x+4)2\\begin{align*}(x+4)^2\\end{align*}. Example 2: Factor x24x+4\\begin{align*}x^2-4x+4\\end{align*}. Solution: Rewrite x24x+4\\begin{align*}x^2-4x+4\\end{align*} as x2+2(2)x+(2)2\\begin{align*}x^2+2 \\cdot (-2) \\cdot x+(-2)^2\\end{align*}. We notice that this is a perfect square trinomial and we can factor it as: (x2)2\\begin{align*}(x-2)^2\\end{align*}. ## Solving Polynomial Equations Involving Special Products We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like ax2+bx+c=0\\begin{align*}ax^2+bx+c=0\\end{align*}. Remember that to solve polynomials in expanded form, we use the following steps: Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0. Step 2: Factor the polynomial completely. Step 3: Use the Zero Product Property to set each factor equal to zero. Step 4: Solve each equation from step 3. Example 3: Solve the following polynomial equations. x2+7x+6=0 Solution: No need to rewrite because it is already in the correct form. Factor: We write", "## anonymous 4 years ago Just another problem: A quadratic polynomial $$P(x)$$ is such that $$P(x)$$ never takes any negative values and $$P(0)=8$$ and the $$P(8)=0$$. Find $$P(-4)$$ Genre: algebra pre-calculus Rating: Easy 1. Zarkon 18 2. anonymous @zarkon: could you please explain with steps 3. anonymous It is just too easy for Zarkon :) 4. Zarkon you did clasify this as easy :) 5. Mr.Math Let $$P(x)=ax^2+bx+c$$, then $$P(0)=8 \\implies c=8$$ and $$P(8)=0 \\implies 8^2a+8b+8=0 \\implies b=-8a-1$$. Now, $$P(x)=ax^2-(8a+1)x+8\\ge 0 \\implies (x-8)(ax-1)\\ge 0 \\implies a=\\frac{1}{8}.$$ Thus $$P(-4)=\\frac{1}{8}(-4)^2-2(-4)+8=2+8+8=18.$$ 6. Zarkon i used the fact that (8,0) must be the vertex to get the 2nd equation 7. Zarkon used the vertex formula 8. Mr.Math Oh that's better I think. 9. ash2326 Let the quadratic polynomial be $P(x)=ax^2+bx+c$ as it's given it doesn't take negative values D<0 $b^2-4ac<0$ let's substitute x=0 P(0)=8 c=8 P(8)=64a+8b+8=0 or $8a+b+1=0$ we have b^2-4ac<0 or b^2-32a<0 or b^2<32 a we have 8a=-1-b b^2<-4-4b b^2+4b+4<0 (b+2)^<0 so b is a complex no. of the form =-2-xi now let x be 1 so $P(x)=\\frac{-(1+i)}{8} x^2+(-2+i)x+8$ Could anyone point out my mistake?? 10. Zarkon you way is good 11. anonymous Seems I am only one who used Maxima-minima. 12. anonymous maxima-minima : u mean by taking the second derivatives of the eqs and substituting as rt-s^2 ?? 13.", "x_2] \\end{align} Notice that we can easily obtain the polynomial $P_{k+1}(x)$ that interpolates the the points $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_{k+1}, y_{k+1})$ from the polynomial $P_k$ that interpolates the $k$ points $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_{k-1}, y_{k-1})$ as: (5) \\begin{align} \\quad P_{k+1}(x) = P_k(x) + (x - x_0)(x - x_1)...(x - x_k)f[x_0, x_1, ..., x_{k+1}] \\end{align} ## Example 1 Find the polynomial $P_2$ for the function $y = \\sqrt{x}$ that interpolates the points $(1, 1)$, $(4, 2)$, and $(9, 3)$ using Newton's divided difference formula. Applying Newton's divided difference formula from above and we get that: (6) \\begin{align} \\quad P_2(x) = f(x_0) + (x - x_0)f[x_0, x_1] + (x - x_0)(x - x_1)f[x_0, x_1, x_2] \\\\ \\quad P_2(x) = 1 + (x - 1) \\frac{f(x_1) - f(x_0)}{x_1 - x_0} + (x - 1)(x - 4) \\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} \\\\ \\quad P_2(x) = 1 + (x - 1) \\frac{2 - 1}{4 - 1} + (x - 1)(x - 4) \\frac{\\frac{3 - 2}{9 - 4} - \\frac{2 - 1}{4 - 1}}{9 - 1} \\\\ \\quad P_2(x) = 1 + \\frac{1}{3}(x - 1) - \\frac{1}{60} (x - 1)(x - 4) \\end{align} The graph of $y = P_2(x)$ is given below. ## Example 2 Find the polynomial $P_3$ for the function $y = \\sin x$ that", "# Finding polynomial without constant term that commutes with $f(x)=x^3+3x$ Consider the polynomial $$f(x)=x^3+3x$$ over $$\\mathbb{Z}$$. I am trying to find a polynomial $$g(x)$$ $$(\\neq f^{\\circ n})$$ of any degree (or series) without constant term which commutes with $$f$$ (or any iteration $$f^{\\circ n}, ~n \\geq 1)$$ under composition. Trivially, any $$g(x)=x$$, is another polynomial (or series) commutes with $$f(x)$$. By hand it seems to be laborious. Suppose I start with an investigation if there are degree $$2$$ polynomial $$g(x)=ax+bx^2$$ such that $$f \\circ g=g\\circ f$$. Then \\begin{align} &f(g(x))=f(ax+bx^2)=3(ax+bx^2)+(ax+bx^2)^3=3ax+3bx^2+a^3x^3+3a^2bx^4+3ab^2x^5+b^3x^6, \\\\ &g(f(x))=g(x^3+3x)=a(x^3+3x)+b(x^3+3x)^2=3ax+ax^3+bx^6+6bx^4+9bx^2 \\end{align} Comparing both equations, we get $$b=0$$ and $$a=a^3 \\Rightarrow a=\\pm 1$$. In this case $$g(x)=\\pm 1$$, the trivial one. Suppose I start with an investigation if there are degree $$3$$ polynomial $$g(x)=ax+bx^2+cx^3$$ such that $$f \\circ g=g\\circ f$$. Then \\begin{align} &f(g(x))=f(ax+bx^2+cx^3)=3(ax+bx^2+cx^3)+(ax+bx^2+cx^3)^3=3ax+3bx^2+3cx^3+c^3x^9+3bc^2x^8 \\hspace{3cm}+(3ac^2+3b^2c)x^7+(6abc + b^3)x^6 + (3a^2c + 3ab^2)x^5 + 3a^2bx^4 + a^3x^3, \\\\ &g(f(x))=g(x^3+3x)=a(x^3+3x)+b(x^3+3x)^2+c(x^3+3x)^3=3ax+ax^3+2bx^6+6bx^4+9bx^2+cx^9+9cx^7+27cx^5+27cx^3 \\end{align} Comparing both sides we get $$b=0$$ and the following equations: \\begin{align} a^3-a=24c, \\\\ a^2c=9c, \\\\ 3ac^2=9c,\\\\ c^3=c. \\end{align} Solving these, we see $$c^3=c$$ and $$3ac^2=a^2c$$. These two gives us $$c=0$$ or $$c=\\pm 1$$. If $$c \\neq 0$$, then $$a=\\pm 3$$. Thus $$g(x)=\\pm (3x+ x^3)$$, which is equivalent to $$f(x)$$ upto signs. Suppose I start with an investigation if there are degree $$4$$ polynomial $$g(x)=ax+bx^2+cx^3+dx^4$$ such that $$f \\circ g=g\\circ f$$. Then", "Thus, for each positive integer $k,$ less than or equal to $n$, there exists exactly one elementary symmetric polynomial of degree $k$ in $n$ variables. To form the one which has degree $k$, we form all products of $k$-subsets of the $n$ variables and add up these terms. The fact that $X_1X_2=X_2X_1$ and so forth is the defining feature of commutative algebra. That is, the polynomial ring formed by taking all linear combinations of products of the elementary symmetric polynomials is a commutative ring. ## Examples The following lists the n elementary symmetric polynomials for the first four positive values of n. (In every case, e0 = 1 is also one of the polynomials.) For n = 1: $e_1(X_1) = X_1.\\,$ For n = 2: \\begin{align} e_1(X_1,X_2) &= X_1 + X_2,\\\\ e_2(X_1,X_2) &= X_1X_2.\\,\\\\ \\end{align} For n = 3: \\begin{align} e_1(X_1,X_2,X_3) &= X_1 + X_2 + X_3,\\\\ e_2(X_1,X_2,X_3) &= X_1X_2 + X_1X_3 + X_2X_3,\\\\ e_3(X_1,X_2,X_3) &= X_1X_2X_3.\\,\\\\ \\end{align} For n = 4: \\begin{align} e_1(X_1,X_2,X_3,X_4) &= X_1 + X_2 + X_3 + X_4,\\\\ e_2(X_1,X_2,X_3,X_4) &= X_1X_2 + X_1X_3 + X_1X_4 + X_2X_3 + X_2X_4 + X_3X_4,\\\\ e_3(X_1,X_2,X_3,X_4) &= X_1X_2X_3 + X_1X_2X_4 + X_1X_3X_4 + X_2X_3X_4,\\\\ e_4(X_1,X_2,X_3,X_4) &= X_1X_2X_3X_4.\\,\\\\ \\end{align} ## Properties The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the" ]

[ "since factoring a cubic polynomial is not necessarily the easiest thing ever. Note that: (4) \\begin{align} \\quad \\quad 4(k+1)^3 - (k+1) = 4(k+1)(k^2 + 2k + 1) - (k +1) = 4(k^3 + 2k^2 + k + k^2 + 2k + 1) - (k+1) \\\\ \\quad \\quad = 4(k^3 + 3k^2 + 3k + 1) - (k +1) = 4k^3 + 12k^2 + 12k + 4 - k - 1 = 4k^3 + 12k^2 + 11k + 3 \\end{align} Therefore $4k^3 + 12k^2 + 11k + 3 = 4(k+1)^3 - (k+1)$ and so: (5) \\begin{align} LHS_{P(k+1)}: = \\frac{4k^3 + 12k^2 + 11k + 3}{3} \\\\ = \\frac{4(k+1)^3 - (k+1)}{3} \\end{align} Therefore the truth of $P(k)$ implies the truth of $P(k+1)$ and so for every $n ≥ 1$, $P(n)$ is true. $\\blacksquare$", "directly from Eisenstein's Criterion and the fact that $$f(p^{\\frac{1}{3}})=0.$$ In light of this information, the polynomial given by the OP is of degree $$2$$, thus must be the zero polynomial. • Your first point that $x^3-p$ is the minimal polynomial for $p^{1/3}$ is what we need to prove here. It can't be assumed. More generally the problem boils down to showing that if $x^3-p\\in\\mathbb {Q} [x]$ has no roots in $\\mathbb {Q}$ then it is irreducible over $\\mathbb{Q}$. Dec 27 '20 at 3:01 • @ParamanandSingh You're right! I was so eager to use traces so I jumped. Thank you for your comment Dec 27 '20 at 8:25 Here's another way. If $$a+bq+cq^2=0$$, then $$a=-(bq+cq^2)$$, hence $$a^2=b^2q^2+2bcq^3+c^2q^4$$. So if $$q^3=p$$, then we have two equations: \\begin{align} a+bq+cq^2&=0\\\\ (a^2-2bcp)-c^2pq-b^2q^2&=0 \\end{align} Multiplying both sides of the first equation by $$b^2$$ and the both side of the second equation by $$cp$$, and then adding the resulting equations, we have $$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$ Now if $$q$$ is irrational (i.e., if $$p$$ is not a perfect cube) and the other variables are rational, then we must have $$ab^2+a^2c-2bc^2p=b^3-c^3p=0$$. But $$b^3-c^3p=0$$ for a non-cube $$p$$ implies $$b=c=0$$, in which case the original equation, $$a+bq+cq^2=0$$, implies $$a=0$$. Since $$q=p^{1/3}$$ is irrational it follows that minimal polynomial of $$q$$ over $$\\mathbb{Q}$$ is of degree greater than $$1$$. If we", "+0 # Polynomial 0 140 1 Let P(x) be a cubic polynomial such that P(0) = -3 and P(1) = 4. When P(x) is divided by x^2 + x + 1, the remainder is 2x + 3. What is the quotient when P(x) is divided by x^2 + x + 1? Jun 11, 2021 #1 +287 +2 Since $P(x)$ is cubic and $x^2+x+1$ is quadratic, the quotient must be linear. Let the quotient be $ax+b$ for some constants $a$ and $b$. Then $$P(x) = (x^2+x+1)(ax+b) + 2x+3.$$ We are given that $P(0)= -3$, so $$(0^2+0+1)(a(0)+b) + 2(0)+3 = -3$$ that is, $$b+3=-3,$$ giving $b=-6$. We are also given that $P(1) = 4$, so $$(1^2+1+1)(a(1)-6) + 2(1)+3 = 4$$ that is, $$3(a-6)+5=4$$ giving $a=\\frac{17}{3}$. Therefore, the quotient is $\\frac{17}{3}x-6$. Jun 11, 2021", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so $$\\frac{-b\\pm\\sqrt{b^2 - 4ac}}{2a} = \\frac{1 \\pm\\sqrt{1 - -24}}{4} = \\frac{1\\pm 5}{4} = \\frac{3}{2}, -1,$$ which gives a factorization into the monic polynomials $x+1$ and $2x - 3$. Note that $2-3=-1,$ so that \\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\\\ &= 2x^2+2x-3x-3\\\\ &= 2x(x+1)-3(x+1)\\\\ &= (x+1)(2x-3).\\end{align} • Sweet, simple, and to the point. Aug 20, 2013 at 5:30 You can use several methods to solve the equation. Firstly you can use the quadratic formula $$\\frac{-b \\pm\\sqrt{b^2 - 4ac}}{2a}$$ or you can use decomposition. Let's try decomposition first. Look for two numbers that multiple to $-6$ and add to $-1$. These numbers would be $+2$ and $-3$. So, we would place those for $-x$ and have the following: $$2x^2-3x+2x-3$$ Now here you can factor out $x$ from the first two terms. $$x(2x-3)+(2x-3)$$ Notice the coefficient on $2x-3$ is $1$. So this means that our equation is $x+1$ and $2x-3$. Therefore, giving us the solution $$(2x-3) = 0 \\, \\text{or} \\, (x+1) = 0$$ Now, if you want to use the quadratic formula then you can do this. $a = 2$, $b = -1$ and $c = -3$. So we will have: $$\\frac{6 \\pm\\sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$$ When you apply this formula twice, you will get the roots which are $x = \\frac{3}{2}$ and", "know the roots explicitly for this. - This may not be what you're looking for, but if some polynomial $p(x)$ has roots $c_1, c_2, ... , c_n$, then $q(x) = (x - 2c_1)(x - 2c_2)...(x - 2c_n)$ has root $2c_i$ for every $i \\in \\{1, 2, ... , n\\}$. Here's something to think about. Suppose $f(x) = x^2 + ax + b$ is a reducible polynomial. Since it is reducible, we know it has two roots, say $c_1$ and $c_2$. Since it is monic, $$f(x) = (x - c_1)(x - c_2) = x^2 + (-c_1 - c_2)x + c_1c_2$$ by the factor theorem. We can then deduce that $a = -c_1 - c_2$ and $b = c_1c_2$. If we wish to construct a polynomial with roots $2c_1$ and $2c_2$, we can construct the polynomial \\begin{align*} (x - 2c_1)(x - 2c_2) &= x^2 + (-2c_1 - 2c_2)x + 2c_12c_2 \\\\ &=x^2 + 2(-c_1 -c_2)x + 4c_1c_2 \\\\ &= x^2 + 2ax + 4b \\end{align*} We have now constructed a new polymial from $f(x)$ with roots $2c_1$ and $2c_2$.", "= \\frac{43}{2} - k$, $am = -k$, and $an = \\frac{k}{2}$. The first two equations show that $m - n = 29 - \\frac{43}{2} = \\frac{15}{2}$. The last two equations show that $\\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = 30$. ### Solution 3 Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations: \\begin{align} a = 2c \\\\ b = -d \\\\ 2c(k - 29) - d = c(2k - 43) + 2d \\\\ -d(k - 29) - 2ck = d(2k - 43) + ck \\end{align} Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\\boxed {30}$. ~ anellipticcurveoverq", "Polynomial? A polynomial of the form $$p(x)=ax^2+bx+c$$, where $$a \\neq 0$$ is called a quadratic polynomial. ### Zeros of Quadratic Polynomial: Definition real number $$\"k\"$$ of a quadratic polynomial $$p(x)$$ is 0 if $$p(k)=0$$ . Consider the polynomial: $$p(x)=x^2-3x-4$$ Let's find the value of the polynomial at $$x=-1$$ by substituting -1 for $$x$$ in $$p(x)$$. \\begin{align}p(-1)&=(-1)^2-3(-1)-4\\\\&=1+3-4\\\\&=0\\end{align} So, here $$x=-1$$ is a zero of the quadratic polynomial $$p(x)=x^2-3x-4$$. ## How To Find Zeros of a Quadratic Polynomial? There are two methods to find the zeros of a quadratic polynomial: Let us consider the quadratic polynomial $$2x^2+x-3=0$$ Let’s find the determinant to find the nature of the solution. \\begin{align}b^2-4ac&=1^2-4\\times 2\\times (-3)\\\\&=1+24\\\\&=25>0\\end{align} This means that the quadratic polynomial has two real and distinct zeros. Just plug in the values $$a=2$$, $$b=1$$ and $$c=-3$$ in the quadratic formula. \\begin{align}x&=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\\\&=\\frac{-1\\pm \\sqrt{1^2-4\\times 2\\times (-3)}}{2\\times2}\\\\&=\\frac{-1\\pm \\sqrt{25}}{4}\\\\&=\\frac{-1\\pm 5}{4}\\\\&=-\\frac{3}{2},1\\end{align} \\begin{align} x=-\\frac{3}{2},1 \\end{align} ### Factorization Let’s solve this quadratic polynomial by factorization method. Here, we need to find $$a$$ and $$b$$ such that $$a+b=1$$ and $$ab=-6$$ Considering this, we see that the numbers are $$a=3$$ and $$b=-2$$ \\begin{align}2x^2+x-3&=0\\\\2x^2+3x-2x-3&=0\\end{align} Take the common factors out from the first two terms and last two terms. So, this quadratic polynomial factoring gives: \\begin{align}x(2x+3)-(2x+3)&=0\\2x+3)(x-1)&=0\\end{align} When the product of two numbers is \\(0, then either of the numbers are $$0$$ This means $$2x+3=0$$", "polynomial with coefficients all $1$ and taken $\\pmod n$ if the exponents cover $0,1,…,n-1$, you can rest assured it is divisible by $(x^n-1)/(x-1)$. May not be easy to recognise always, but it is a simple enough test. +1 Sep 7, 2020 at 14:02 Hmm. I'd look at it and say \"There's no rational root\" because $$x = \\pm 1$$ doesn't work. So there's some irrational root, $$\\alpha$$, and two complex-conjugate pairs. So there's no nice obvious linear factor I can write down. Then I'd say \"Maybe there's a quadratic factor.\" I can assume it's monic, so I'm looking to write $$x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r).$$ from which I can expand to get $$x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br$$ if I've done the algebra right. Equating coefficients I see that \\begin{align} 0 &= a + p\\\\ 0 &= q + ap + b\\\\ 1 &= ra + bq\\\\ 1 &= br \\end{align} so $$p = -a$$, and $$r = \\frac1b$$,and these equations become \\begin{align} 0 &= a + -a\\\\ 0 &= q - a^2 + b\\\\ 1 &= \\frac1b a + bq\\\\ 1 &= b(1/b) \\end{align} which simplify down to", "# check workings for polynomial function from graph The diagram shows a curve with equation of the form $y = kx(x + a)^2$, which passes through $(-2, 0)$, $(0, 0)$ and $(1, 3)$. What are the values of $a$ and $k$. To find $k$, I would set $f(x) = 1$. $f(1) = k(x + 2)(x + 0)(x - 1)$ $3 = k(1 + 2)(x + 1)(1 - 1)$ $3 = k(3)(1)(0)$ This would make $k = 0$ which does not appear to be right. • Don't you mean $f(1) = 3$? – Pichi Wuana Mar 6 '16 at 19:19 $$y=kx(x+a)^2\\quad \\quad \\quad (-2, 0),(0, 0) , (1, 3)$$ $$\\text{for} (0,0):\\quad \\quad \\quad f(0)=0=0\\\\ \\text{for}(1,3):\\quad \\quad \\quad f(1)=k(1+a)^2=3\\\\ \\text{for}(-2,0):\\quad \\quad \\quad f(-2)=-2k(-2+a)^2=0$$ so you have two equations with two variables: $$\\begin{cases}k(1+2a+a^2)=3\\\\ -2k(4-4a+a^2)=0\\end{cases}$$ $$\\begin{cases}k+2ak+ka^2=3\\\\ -8k+8ka-2ka^2=0\\end{cases}$$ Solve, you should get $\\color{red}{a=2,k=\\frac 1 3}$ The equation is satisfied by $(-2,0)$ and $(1,3)$ Putting $(-2,0)$ in the equation we obtain: $$0 = k(-2)(-2+a)^2$$ Therefore, $(a-2)=0$ as $k \\ne 0$ $$\\implies a=2$$ Similarly, putting $(1,3)$ in the equation we obtain: $$3 = k*1*(1+2)$$ Therefore $k=1$", "## anonymous 4 years ago Just another problem: A quadratic polynomial $$P(x)$$ is such that $$P(x)$$ never takes any negative values and $$P(0)=8$$ and the $$P(8)=0$$. Find $$P(-4)$$ Genre: algebra pre-calculus Rating: Easy 1. Zarkon 18 2. anonymous @zarkon: could you please explain with steps 3. anonymous It is just too easy for Zarkon :) 4. Zarkon you did clasify this as easy :) 5. Mr.Math Let $$P(x)=ax^2+bx+c$$, then $$P(0)=8 \\implies c=8$$ and $$P(8)=0 \\implies 8^2a+8b+8=0 \\implies b=-8a-1$$. Now, $$P(x)=ax^2-(8a+1)x+8\\ge 0 \\implies (x-8)(ax-1)\\ge 0 \\implies a=\\frac{1}{8}.$$ Thus $$P(-4)=\\frac{1}{8}(-4)^2-2(-4)+8=2+8+8=18.$$ 6. Zarkon i used the fact that (8,0) must be the vertex to get the 2nd equation 7. Zarkon used the vertex formula 8. Mr.Math Oh that's better I think. 9. ash2326 Let the quadratic polynomial be $P(x)=ax^2+bx+c$ as it's given it doesn't take negative values D<0 $b^2-4ac<0$ let's substitute x=0 P(0)=8 c=8 P(8)=64a+8b+8=0 or $8a+b+1=0$ we have b^2-4ac<0 or b^2-32a<0 or b^2<32 a we have 8a=-1-b b^2<-4-4b b^2+4b+4<0 (b+2)^<0 so b is a complex no. of the form =-2-xi now let x be 1 so $P(x)=\\frac{-(1+i)}{8} x^2+(-2+i)x+8$ Could anyone point out my mistake?? 10. Zarkon you way is good 11. anonymous Seems I am only one who used Maxima-minima. 12. anonymous maxima-minima : u mean by taking the second derivatives of the eqs and substituting as rt-s^2 ?? 13.", "Phil Fernandez Apr 16 '18 at 1:31 $$\\frac{1}{x^4-K^4}=\\frac{A}{x-K}+\\frac{B}{x+K}+\\frac{Cx+D}{x^2+K^2}$$ To solve Dr. Graubner's equation $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$ as stated above, is a lot of linear algebra. Some of that can be avoided. $$\\frac{1}{x^4-K^4}=\\frac{A}{x-K}+\\frac{B}{x+K}+\\frac{Cx+D}{x^2+K^2}$$ Using $$x^4-K^4 = (x-K)(x+K)(x^2+K^2)$$, we get $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$ Letting $$x=K$$, you get \\begin{align} 1 &= 4AK^3 \\\\ A &= \\dfrac{1}{4K^3} \\end{align} Letting $$x=K$$, you get \\begin{align} 1 &= -4BK^3 \\\\ B &= -\\dfrac{1}{4K^3} \\end{align} Letting $$x=iK$$, you get $$1 = -2K^2(D+iCK)$$. Letting $$x=-iK$$, you get $$1 = -2K^2(D-iCK)$$. So $$C = 0$$ and $$D = -\\dfrac{1}{2K^2}$$. Hence $$\\frac{1}{x^4-K^4}=\\frac{1}{4K^3(x-K)}-\\frac{1}{4K^3(x+K)}-\\frac{1}{2K^2(x^2+K^2)}$$ If you don't want to have to resort to complex variables, you can do this after finding the values of $$A$$ and $$B$$. \\begin{align} A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \\\\ \\dfrac{(x+K)(x^2+K^2)}{4K^3}-\\dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \\\\ \\dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \\\\ (Cx+D)(x^2-K^2) &= 1 - \\dfrac{x^2+K^2}{2K^2} \\\\ (Cx+D)(x^2-K^2) &= -\\dfrac{x^2-K^2}{2K^2} \\\\ Cx+D &= -\\dfrac{1}{2K^2} \\end{align}", "with those in the other given expression, we find that $a=2$a=2 and $b=5$b=5. Note that this is consistent with the coefficient of the linear term $b-a=3$ba=3. In attempting to solve a polynomial equation, it can happen that we can see by inspection what one of the solutions is. For example, given $2x^3-5x^2+8x-5=0$2x35x2+8x5=0, we might notice that $x=1$x=1 is a solution. If $x=1$x=1 is a solution, the factorised form of the polynomial must include the factor $x-1$x1. The other factor must be a quadratic. So, we can write $2x^3-5x^2+8x-5=(x-1)(ax^2+bx+c)$2x35x2+8x5=(x1)(ax2+bx+c) Now, if we could determine the coefficients $a$a, $b$b and $c$c we would be in a position to find the other roots of the cubic because we know how to find the roots of a quadratic. This is often done using a polynomial division algorithm but we can do it by equating coefficients as follows. $(x-1)(ax^2+bx+c)$(x−1)(ax2+bx+c) $=$= $ax^3+(b-a)x^2+(c-b)x-c$ax3+(b−a)x2+(c−b)x−c Thus, $a$a $=$= $2$2 $b-a$b−a $=$= $-5$−5 $c-b$c−b $=$= $8$8 $c$c $=$= $5$5 This set of equations is consistent if $b=-3$b=3. The original equation can now be written $(x-1)(2x^2-3x+5)=0$(x1)(2x23x+5)=0 We look for solutions to $2x^2-3x+5=0$2x23x+5=0. In this case, there are no further real solutions because the discriminant is negative. This cubic has only one real root (and two complex roots). In other cases, we could proceed, using the quadratic formula or another", "2016 AIME I Problems/Problem 11 Problem Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$, and $\\left(P(2)\\right)^2 = P(3)$. Then $P(\\tfrac72)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solution 1 We substitute $x=2$ into $(x-1)P(x+1)=(x+2)P(x)$ to get $P(3)=4P(2)$. Since we also have that $\\left(P(2)\\right)^2 = P(3)$, we have that $P(2)=4$ and $P(3)=16$. We can also substitute $x=1$, $x=0$, and $x=3$ into $(x-1)P(x+1)=(x+2)P(x)$ to get that $0=P(1)$, $-1P(1)=2P(0)$, and $2P(4)=5P(3)$. This leads us to the conclusion that $P(0)=P(1)=0$ and $P(4)=40$. We next use finite differences to find that $P$ is a cubic polynomial. Thus, $P$ must be of the form of $ax^3+bx^2+cx+d$. It follows that $d=0$; we now have a system of $3$ equations to solve. We plug in $x=1$, $x=2$, and $x=3$ to get $$a+b+c=0$$ $$8a+4b+2c=4$$ $$27a+9b+3c=16$$ We solve this system to get that $a=\\frac{3}{2}$, $b=0$, and $c=-\\frac{3}{2}$. Thus, $P(x)=\\frac{3}{2}x^3-\\frac{3}{2}x$. Plugging in $x=\\frac{7}{2}$, we see that $P\\left(\\frac{7}{2}\\right)=\\frac{105}{4}$. Thus, $m=105$, $n=4$, and our answer is $m+n=\\boxed{109}$. Solution 2 So from the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$. This means that $1$ and $-1$ are roots of $P(x)$. Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a", "# Find k if the two polynomials have one common root The equations $x^2−4x+k=0$ and $x^2+kx−4=0$, where $k$ is a real number, have exactly one common root. What is the value of $k$? I know the answer but can it be done with the relation of roots. $a$ and $b$ are roots of equation 1 and $a$ and $c$ are roots of equation 2. So the relations are :- 1. $a + b = 4$ 2. $ab = k$ 3. $a + c = (-k)$ 4. $ac= (-4)$ I tried doing all the stuff but couldn't get it. Can we find it using these 4 equations? • Hint: subtract the first equation to the second, you will get the expression of the common root $x$ expressed as a function of $k$. – Jean Marie Sep 21 '17 at 7:44 Using your equations . . . 1. $\\;a + b = 4$ 2. $\\;ab = k$ 3. $\\;a + c = (-k)$ 4. $\\;ac= (-4)$ Adding equations $(1)$ and $(4)$, we get $$a + b + ac=0\\tag{5}$$ Adding equations $(2)$ and $(3)$, we get $$a + c + ab=0\\tag{6}$$ Subtracting equation $(6)$ from equation $(5)$, we get \\begin{align*} &b-c + ac - ab = 0\\\\[4pt] \\implies\\;&b-c+a(c-b) = 0\\\\[4pt] \\implies\\;&(b-c)(1-a)= 0\\\\[4pt] \\implies\\;&b=c\\;\\;\\text{or}\\;\\;a=1\\\\[4pt] \\end{align*} If $b=c$, then the left-hand-sides of equations", "must have $8c-3c^2\\ge 0$. This is only true for $0\\le c \\le 8/3$. Now we show that any $c$ in the interval $[0,8/3]$ is achievable. We want to make $(a-b)^2=8c-3c^2$. If $c$ is in $[0,8/3]$, then $8c-3c^2\\ge 0$, so we can take the square root, and obtain $$a-b=\\pm\\sqrt{8c-3c^2}.$$ We can now solve the system $a-b=\\pm\\sqrt{8c-3c^2}$, $a+b=4$ to find the values of $a$ and $b$. We might as well give the solutions $(a,b,c)$ explicitly. They are $$a=2\\pm \\frac{1}{2}\\sqrt{8t-t^2}, \\qquad b=2\\mp \\frac{1}{2}\\sqrt{8t-t^2},\\qquad c=t,$$ where $t$ is a parameter that ranges over the interval $0\\le t\\le 8/3$. There is something mildly ugly in the above general solution, since it breaks symmetry. That can be fixed. Remark: It is maybe interesting to ask what are the possible values of $c$ under the conditions $c\\ge a$, $c\\ge b$. Of course we still must have $c\\le 8/3$. For given $a$, $b$, and $c$, look at the monic cubic polynomial $P(x)$ whose roots are $a$, $b$, and $c$. Then $$P(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc.$$ You found that $ab+bc+ca=4$. So our polynomial has shape $$P(x)=x^3-4x^2+4x-p,$$ where $p$ is the product of the solutions. From our previous work we know the solutions are $\\ge 0$, so $p\\ge 0$. The derivative of $P(x)$ is $3x^2-8x+4$, which vanishes at $x=2/3$ and $x=2$. So $P(x)=0$ has a solution in the interval $[0, 2/3]$, a", "with the same roots, where $$c$$ is any (non-zero) real number. We needed the extra data of a point on the curve to determine $$c$$. In general: • An $$x$$-intercept at $$x=a$$ implies that $$(x-a)$$ is a factor of $$f(x)$$. • A $$y$$-intercept at $$y=c$$ implies that $$f(0)=c$$, i.e., the constant term in $$f(x)$$ is $$c$$. • A given point $$(p,q)$$ on the graph implies that $$f(p)=q$$. If we are just given points on the graph of $$y=f(x)$$, one way to find the polynomial is by simultaneous equations, as the following example shows. #### Example Find the quadratic polynomial $$f(x)$$ whose graph passes through the three points $$(1,-2)$$, $$(2,-1)$$ and $$(3,2)$$. #### Solution Let the polynomial be $$f(x) = ax^2 + bx + c$$. We are given that $$f(1)=-2$$, $$f(2)=-1$$ and $$f(3)=2$$, so we have the simultaneous equations \\begin{alignat*}{2} a+b+c &= -2 &\\qquad\\qquad& (1) \\\\ 4a+2b+c &= -1 && (2) \\\\ 9a+3b+c &= 2. && (3) \\end{alignat*} Subtracting $$(2)-(1)$$ and $$(3)-(2)$$ gives $$3a+b = 1$$ and $$5a+b=3$$. Subtracting these two equations gives $$2a=2$$, so $$a=1$$. Substituting $$a=1$$ back into these equations gives $$b=-2$$ and then $$c=-1$$. So $$f(x) = x^2 - 2x - 1$$. In principle we could use the above method to find, say, a quintic through six given points, or a degree-100 polynomial through 101 given", "Question # Given that the zeroes of the cubic polynomial $$x^3 - 6x^2 + 3x + 10$$ are of the form $$a, a + b, a + 2b$$ for some real numbers $$a$$ and $$b$$, find the values of $$a$$ and $$b$$. Solution ## Given cubic polynomial is $$p(x)=x^{ 3 }-6x^{ 2 }+3x+10$$The zeros of the polynomial $$p(x)$$ are of the form $$a$$, $$a+b$$ and $$a+2b$$ Then, $$a+a+b+a+2b=-\\frac{-6}{1}$$$$=>3a+3b=6$$$$=>a+b=2$$ ----------------(i)Also, $$a(a+b)+(a+b)(a+2b)+a(a+2b)=\\frac{3}{1}$$$$=>a^2+ab+a^2+2ab+ab+2b^2+a^2+2ab=3$$$$=>3a^2+2b^2+6ab=3$$ ----(ii)and $$a(a+b)(a+2b)=-\\frac{-10}{1}$$$$=>a^3+a^2b+2a^2b+2ab^2=10$$From (i), $$b=2-a$$Putting this value in (ii), we get$$=>3a^2+2(2-a)^2+6a(2-a)=3$$$$=>3a^2+2(4-4a+a^2)+12a-a^2=3$$$$=>-a^2+4a+5=0$$$$=>a^2-4a-5=0$$$$=>a^2-5a+a-5=0$$$$=>a(a-5)+(a-5)=0$$$$=>.(a-5)(a+1)=0$$$$=>a=5$$ or $$a=-1$$$$=>b=-3$$ or $$b=3$$ respectivelyFrom equation (iii), we getat $$a=5$$, $$b=-3$$$$=>5^3+(5)^2(-3)+2(5)^2(-3)+2(5)(-3)^2=-10$$$$=>125-75-150+90=-10$$$$=>-10=-10$$ which is true.and at $$a=-1$$, $$b=3$$, we have$$=>(-1)^3+(-1)^2 3+2(-1)^2(3)+2(-1)(3)^2=-10$$$$=>-1+3-6-18=10$$$$=>-22=-10$$ which is not true.Thus, $$a=5$$, $$b=-3$$Zeros of the polynomial are $$5$$, $$5-3$$, $$5-2\\times 3$$ ie $$5$$,$$2$$,$$-1$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "believe that searching for a polynomial which is annihilated by $T_P$ is not very effective. Again, there must be an elegant solution to this question. – user265336 Aug 29 '15 at 15:44 It's given as a hint that $$P^2+3I = 0$$, so: \\begin{align} &P^2 = -3I \\\\ &P\\cdot-\\frac{1}{3}P = I \\\\ \\therefore \\quad &P^{-1} = -\\frac{1}{3}P \\end{align} So, $$T_P$$ is actually $$T_P(X)=-\\frac{1}{3}PXP$$ Note that $$P^2=-3I$$ and we have two $$P$$s in the above expression for $$T_P$$, so it might be useful to calculate $$T_P^2$$: \\begin{align} T_P^2(X) &= T_P(T_P(X)) \\\\ &= T_P(-\\frac{1}{3}PXP) \\\\ &= -\\frac{1}{3}P(-\\frac{1}{3}PXP)P =\\\\ &= \\frac{1}{9}P^2XP^2 \\\\ &= \\frac{1}{9}(-3I)X(-3I) \\\\ &= \\frac{1}{9}\\cdot 9\\cdot IX =X \\\\ \\end{align} So we got $$T^2_P=I$$, which means $$T^2_P-I=O$$, so the following polynomial satisfies $$m(T_P)=0$$: $$m(t)=t^2-1=(t+1)(t-1)$$ Since $$T_P$$ is not scalar, it's trivial to prove that indeed $$m(t)$$ is the minimal polynomial of $$T_P$$.", "$k.$ Complete the square: \\begin{align} k^2 &= 4(m^2-3m+4)\\\\ &= 4m^2 - 12m + 16 \\\\ &= (2m - 3)^2 + 7 \\end{align} and therefore the difference between the two squares $(2m - 3)^2$ and $k^2$ is $7.$ This is possible only if the two squares are $9$ and $16.$ Therefore $k^2 = 16$ and $$(2m - 3)^2 = 9.$$ This quadratic equation in $m$ has two roots, $m = 0$ and $m = 3,$ but $m = 0$ cannot be true for any solution to the original problem (since we require $1/m$ to be defined); therefore $$m = 3.$$ Plug this into your equation for $n$ and confirm that the only possible integer value of $n$ is $n = 2.$ Alternative method: Suppose that you solved for $m$ as in several other answers, so that you found the equation $$m = \\frac{4(n^2 - 1)}{3n^2 - 4n}.$$ If you know a few facts about rational functions (polynomials divided by polynomials), you can sketch $m$ as a function of $n$ free-hand (without the assistance of a graphing calculator or software) and solve the problem graphically. First, plot the functions $m = 4(n^2 - 1)$ and $m = 3n^2 - 4n$ as shown below. Because $3n^2 - 4n = 0$ at $n=0$ and $n = \\frac43,$ we know that the rational" ]

[ "+0 # Let a, b and c be complex numbers such that |a|=|b|=|c|=1 and 0 32 2 Let a, b and c be complex numbers such that |a|=|b|=|c|=1 and $$\\frac{a^2}{bc} + \\frac{b^2}{ac} + \\frac{c^2}{ab} = -1.$$ Find all possible values of |a+b+c| Enter all the possible values, separated by commas. May 2, 2022 #1 0 The possible values of |a + b + c| are 1, sqrt(3), and 3. May 2, 2022 #2 +1", "+0 # Complex +2 155 1 +140 Let $a$ and $b$ be nonzero complex numbers such that $|a| = |b| = |a + b|.$Find the sum of all possible values of $\\frac{a}{b}.$ Jun 13, 2019", "+0 # A Complex Numbers Problem 0 61 1 Let $a$, $b$, and $c$ be complex numbers such that $|a|=|b|=|c|=1$ and $\\frac{a^2}{bc}+\\frac{b^2}{ac}+\\frac{c^2}{ab}=-1$. Find all positive values of $|a+b+c|$. Jan 12, 2021", "# Almost Recovered Algebra Level 3 Let $$a,b,c$$ be distinct complex numbers , such that $$|a|=|b|=|c|$$ and $$|b+c-a|=|a|$$ . Find the value of $$b+c$$. ×", "# Can they really be distinct? Algebra Level 4 Let $$a$$, $$b$$, and $$c$$ be distinct complex numbers such that $\\frac { a }{ 1-b } =\\frac { b }{ 1-c } =\\frac { c }{ 1-a } = k$ where $$k$$ is a complex number. Find the sum of all values of $$k$$. ×", "we know the ratio of $$AC,BC$$ and $$MC,AC$$; specifically $$\\frac {MC}{AC}=\\frac {h_b}{2h_a}$$. Thus we can construct $$C$$ with the use of Apollonius circle, and $$B$$ consequently follows. - 3 years ago @Xuming Liang @Nihar Mahajan here is the whole load of Olympiad Geometry! Enjoy! - 3 years, 1 month ago", "Let $f(x)$ be the function $12 x^2 - 9 x + 8$. Then the quotient $\\frac{f(8+h)-f(8)}{h}$ can be simplified to $ah+b$ for: $a=$ and $b=$ Your overall score for this problem is", "+0 0 96 1 +95 Let $$z_1$$ and $$z_2$$ be two complex numbers such that $$|z_1| = 5$$ and $$\\frac{z_1}{z_2} + \\frac{z_2}{z_1} = 1$$. Find $$|z_1 - z_2|^2.$$ Aug 16, 2022", "# Math Help - Complex Numbers 1. ## Complex Numbers Let a,b,c be complex numbers such that: $\\displaystyle{(a+b)(a+c)=b,}$ $\\displaystyle{(b+c)(b+a)=c,}$ $\\displaystyle{(c+a)(c+b)=a.}$ Prove that $a,b,c \\in {\\Cal R}$ 2. I would start by expanding all three sets of brackets. See what you can do from there... 3. Let's dissect the question a little bit. If we expand those brackets, we find: a^2 + ac + ab + bc = b b^2 + ab +bc +ac = c c^2 + bc + ac + ab = a Note that all 3 equations have the same 3 terms : ab, ac and bc. So let's say \"Let x = ab + ac + bc\". Then our equations look more like: a^2 + x = b b^2 + x = c c^2 + x = a Does it become more clear now how to show this?", "# Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$ Question : Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$ I know that $|z_1+z_1| \\leq |z_1|+|z_2|$ Also if I consider two complex numbers say $z_1 =x+iy$ and $z_2 =a+ib$ Now $|z_1|^2 =x^2+y^2$ and $|z_2|^2 =a^2+b^2$ Also using A.M.-GM. inequality we have $\\frac{x^2+y^2}{2}\\geq \\sqrt{2}xy$ $\\Rightarrow x^2+y^2 \\geq xy 2\\sqrt{2}$ Also $a^2+y^2 =2\\sqrt{2}ab$ Please suggest whether we can proceed in this manner also guide further on this thanks. Let $z_{1} = a+ib$ and $z_{2} = c+id$. Thus, $|z_{1}|^2 = a^2+b^2, |z_{2}|^2 = c^2+d^2, Re(z_{1}z_{2}) =ac-bd$ and $Im(z_{1}z_{2}) =ad+dc$. The problem can be posed as $$\\begin{array}{c} minimize \\hspace{1cm} a^2+b^2+ c^2+d^2 + ac-bd \\\\ s.t. \\hspace{3cm} ad+bc = 1. \\\\ \\end{array}$$ Let $\\mathbf{x} = [a \\hspace{1mm} b \\hspace{1mm} c \\hspace{1mm} d]^{T}$. The previsous optimization problem can be rewrite as $$\\begin{array}{c} minimize \\hspace{1cm} \\mathbf{x}^{T}A\\mathbf{x} \\\\ s.t. \\hspace{1cm} \\mathbf{x}^{T}B\\mathbf{x}= 1, \\\\ \\end{array}$$ where $$A = \\left [ \\begin{array}{cccc} 1 & 0 & 1/2 & 0 \\\\ 0 & 1 & 0 &-1/2 \\\\ 1/2 & 0 & 1 & 0 \\\\ 0 & -1/2 & 0 & 1 \\\\ \\end{array} \\right]$$ and $$B = \\left [ \\begin{array}{cccc} 0 & 0 & 0 & 1/2 \\\\ 0 & 0 & 1/2 &0 \\\\ 0", "Elementary Technical Mathematics $$a=2C-b-c.$$ We find $a$ by multiplying both sides by $2$ and then isolating $a$ on one side as follows: $$C=\\frac{1}{2}(a+b+c)\\Longrightarrow 2C =a+b+c\\\\ \\Longrightarrow a=2C-b-c.$$", "+0 0 64 1 +95 Let $$(a_1,b_1), (a_2,b_2), \\dots, (a_n,b_n)$$ be all the ordered pairs $$(a,b)$$ of complex numbers with $$a^2+b^2\\neq 0, a+\\frac{10b}{a^2+b^2}=5, \\text{ and } b+\\frac{10a}{a^2+b^2}=4.$$ Find $$a_1 + b_1 + a_2 + b_2 + \\dots + a_n + b_n.$$ Sep 17, 2022", "# A Complex Problem Algebra Level 5 Let $$a, b, c$$ be 3 complex numbers such that $$|a| = |b| = |c| = 1$$ and $$\\dfrac{a^2}{bc}+ \\dfrac{b^2}{ac} + \\dfrac{c^2}{ab} + 1 = 0$$ . If $$|a + b + c|$$ can take values that equal $$p$$ and $$q$$, then find the value of $$p+q$$ . ×", "# Complex Cyclic Cystems Algebra Level pending Let $$a,b,c$$ be non-zero complex numbers such that $\\frac{2b+3c}{a}=\\frac{2c+3a}{b}=\\frac{2a+3b}{c}.$ Find the sum of all (distinct) possible values of $$\\left|\\frac{5b^2+5c^2}{a^2}\\right|$$. ×", "= 2 A + B + 2 C$ $B = 7 - 2 A - 2 C = 7 + 4 - 6 = 5$ So, $\\frac{7 {x}^{2} - 12 x + 11}{2 {x}^{3} - 5 {x}^{2} + x + 2} = - \\frac{2}{x - 1} + \\frac{5}{2 x + 1} + \\frac{3}{x - 2}$", "Consider the function $f(x) = \\frac{100x^2}{1+100x^2}$. Find all ordered triples of real numbers $(a,b,c)$ such that $f(a) = b$, $f(b) = c$, and $f(c) = a$.", "Let $a$, $b$, and $c$ be complex numbers with $a\\neq0$. Show that the solutions of $az^2+bz+c=0$ are $z_1,z_2=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$, just as they are in the case when $a$, $b$, and $c$ are real numbers. If $b^2-4ac=x+yi \\in \\mathbb{C}$, then $$\\sqrt{x+yi}=\\pm \\left( \\sqrt{\\frac{\\sqrt{x^2+y^2}+x}{2}} +\\frac{iy}{|y|}\\sqrt{\\frac{\\sqrt{x^2+y^2}-x}{2}} \\right)$$", "93 views If a + b + c = 0, where $a \\neq b \\neq c$, then what is the value of: $\\frac{a^{2}}{2a^{2}+bc}+\\frac{b^2}{2b^{2}+ac}+\\frac{c^{2}}{2c^{2}+ab}$ 1. zero 2. 1 3. -1 4. abc recategorized | 93 views +1 Assume the value of a,b,c and solve it ​​​​​​​ answered by (532 points) 2 8 24 edited", "Song's Complex System Algebra Level 5 Let $$a, b, c$$ be complex numbers satisfying $a + b + c = abc = 1$ and $\\frac{ab + bc + ac}{3} = \\frac{1}{a^{2}} + \\frac{1}{b^{2}} + \\frac{1}{c^{2}}$ The sum of absolute values of all possible $$ab + bc + ac$$ can be written as $$\\frac{\\sqrt{n}}{m}$$, where $$n$$ and $$m$$ are positive coprime integers. What is $$n+m$$? This problem is posed by Zi Song Y. Details and assumptions You may read up on Absolute Value. $$n$$ is allowed to be 1. ×", "the value of $\\frac{2-3i}{2+3i}$23i2+3i. ### Outcomes #### A2.NQ.B.5 Represent complex numbers. #### A2.NQ.B.6 Add, subtract, multiply and divide complex numbers." ]

[ "guess such relations? $(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$? Apr 10, 2019 at 14:46 • @BPP I don't really know Apr 10, 2019 at 15:47 Fun video! Much time was spent on finding $$abc=1/6$$. Alternative method for this: \\begin{align}a^2+b^2&=2-c^2 \\Rightarrow \\\\ (a+b)^2-2ab&=2-c^2 \\Rightarrow \\\\ (1-c)^2-2ab&=2-c^2 \\Rightarrow \\\\ ab&=c^2-c-\\frac12 \\Rightarrow \\\\ abc&=c^3-c^2-\\frac c2 \\end{align} Similarly: $$abc=a^3-a^2-\\frac a2\\\\ abc=b^3-b^2-\\frac b2$$ Now adding them up: $$3abc=(a^3+b^3+c^3)-(a^2+b^2+c^2)-\\frac12(a+b+c)=3-2-\\frac12 \\Rightarrow abc=\\frac16.$$ In fact, you can find other terms as well: $$ab+bc+ca=(a^2+b^2+c^2)-(a+b+c)-\\frac32=2-1-\\frac32=-\\frac12;\\\\ a^2b^2+b^2c^2+c^2a^2=ab(c^2-c-\\frac12)+bc(a^2-a-\\frac12)+ca(b^2-b-\\frac12)=\\\\ abc(a+b+c)-3abc-\\frac12(ab+bc+ca)=\\\\ \\frac16-\\frac12+\\frac14=-\\frac1{12}$$ Hence: $$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\\\ 2\\cdot 3-(-\\frac1{12})+\\frac16\\cdot (-\\frac12)=6.$$ I think to use an identity must be the shortest way:- $${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)}$$ If we square the 1st equation we get $${ab+bc+ca=\\frac {-1}2}$$ On squaring the above equation we get $${(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\\frac 14}$$ We also know $${ (a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca) }$$ from here we get :- $$abc=\\frac 16$$ Substituting the known values We get :- $${ (ab)^2+(bc)^2+(ca)^2=\\frac {-1}{12} }$$ Now squaring the 2nd equation:- $${ (a^4+b^4+c^4)+2\\big ((ab)^2+(bc)^2+(ca)^2)\\big )=4 }$$ From here we get $$(a^4+b^4+c^4)=\\frac{25}6$$ Now we have all the values to be substituted in the identity that i mentioned therefore you get $${ (a^5+b^5+c^5)= (\\frac{25}{6}\\cdot 1) - (3 \\cdot \\frac {-1}2)+ (\\frac 16 \\cdot 2) }$$ $$\\Rightarrow(a^5+b^5+c^5)=6$$", "- x^3 + 2x + 3 \\:=\\:0\\,\\text{ are }\\,a,b,c,d.$ $\\text{Find the value of:}$ . . $(A)\\;\\;\\sum\\alpha^2$ . . $(B)\\;\\;\\sum\\alpha^2\\beta^2$ . . $(C)\\;\\;\\sum\\alpha\\beta(\\alpha+\\beta)$ . . $(D)\\;\\;\\sum\\frac{1}{\\alpha}$ From Vieta's Theorem, we have these equations: . . $\\begin{array}{cccccc}a + b + c + d \\;=\\; 1 & [1] \\\\ ab + ac + ad + bc + bc + cd \\;=\\; 0 & [2] \\\\ abc + abd + acd + bcd \\;=\\; -2 & [3] \\\\ abcd \\;=\\; 3 & [4] \\end{array}$ $(A)\\;\\text{ We want: }\\:a^2 + b^2 + c^2 + d^2$ Square [1]: . $(a+b+c+d)^2 \\:=\\:1^2$ . . $a^2 + 2ab + 2ac + 2ad + b^2 + 2bc + 2bd + c^2 + 2cd + d^2 \\;=\\;1$ . . $a^2 + b^2 + c^2 + d^2 + 2\\underbrace{(ab + ac + ad + bc + bd + cd)}_{\\text{This is 0}} \\;=\\;1$ Therefore: . $a^2+b^2+c^2+d^2 \\;=\\;1$ $(D)\\;\\text{ We want: }\\:\\dfrac{1}{a} + \\dfrac{1}{b} + \\dfrac{1}{c} + \\dfrac{1}{d}$ . . $\\displaystyle \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\;=\\;\\frac{\\overbrace{bcd + acd + abd + abc}^{\\text{This is }-2}}{\\underbrace{abcd}_{\\text{This is 3}}}$ Therefore: . $\\displaystyle \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\;=\\;-\\frac{2}{3}$ Get the idea? 5. Originally Posted by arze For the second, I worked out the equivalent for cubic equations $\\sum\\alpha^2\\beta^2=(\\sum\\alpha\\beta)^2-2\\alpha\\beta\\gamma(\\sum\\alpha)$ Problem is it doesn't work. $(0)^2-2(3)(-1)=6$ answer is 10. Thanks! You", "$${\\overline q} = (abc){a + b\\over ab}{a + c\\over ac}{b + c\\over bc} = (a+b)(b+c)(a+c)/(abc) = q.$$ - Since $|a|=|b|=1$, we have that $$\\left|\\frac{a}{b}\\right|=1 \\quad \\Rightarrow \\quad \\overline{\\dfrac{a}{b}} =\\left.\\left|\\dfrac{a}{b}\\right|^2 \\middle/\\dfrac{a}{b}\\right. =\\frac{b}{a}$$ Therefore, \\begin{align} \\frac{(a+b)^2}{ab} &=\\frac{a}{b}+2+\\frac{b}{a}\\\\ &=2+2\\,\\mathrm{Re}\\left(\\frac{a}{b}\\right)\\\\ &\\ge0 \\end{align} This means that $$\\left(\\frac{(a+b)(b+c)(c+a)}{abc}\\right)^2=\\frac{(a+b)^2}{ab}\\frac{(b+c)^2}{bc}\\frac{(c+a)^2}{ca}\\ge0$$ which leads immediately to $$\\frac{(a+b)(b+c)(c+a)}{abc}\\in\\mathbb{R}$$ - Here's a nice geometric argument (with some holes and assumptions left to fill up): \\begin{align} \\arg \\frac {(a+b)(b+c)(a+c)}{abc} &= \\arg (a+b)+\\arg(b+c)+\\arg(a+c) - \\arg (a) -\\arg( b) - \\arg( c) \\\\ &= \\frac{\\arg (a)+arg(b)} 2 + \\frac{\\arg (b)+arg(c)} 2 + \\frac{\\arg (a)+arg(c)} 2 \\\\ & \\qquad\\qquad - \\arg (a) -\\arg( b) - \\arg( c) \\\\&= 0 \\end{align} - Hint 1. Since $\\Vert a\\Vert=\\Vert b\\Vert=\\Vert c\\Vert=1$, then $$a=e^{i\\alpha}\\qquad b=e^{i\\beta}\\qquad c=e^{i\\gamma}\\tag{1}$$ Hint 2 A complex number $z\\in\\mathbb{C}$ is real iff $$z+\\overline{z}=2z\\tag{2}$$ Substitute $(1)$ into $(2)$ and check that it is correct. - a = exp(iα) = exp(iβ) = exp(iγ) ??? – ᴊ ᴀ s ᴏ ɴ Jul 26 '12 at 13:45 Sorry, that was a typo – Norbert Jul 26 '12 at 13:54", "BCD$, $\\bg_white \\arg \\left ( \\frac{d - a}{d - c} \\right ) = \\arg \\left ( \\frac{c - d}{c - b} \\right )$. Thus, $\\bg_white \\frac{d - a}{d - c} \\div \\frac{c - d}{c - b} = r$, where $\\bg_white r$ is a real number. Substituting in the values from above, we get $\\bg_white \\frac{(lb - a)(ka - b)}{-(lb - ka)^2} = r$. Since $\\bg_white r = \\overline{r}$, we have $\\bg_white \\frac{(l\\overline{b} - \\overline{a})(k\\overline{a} - \\overline{b})}{-(l\\overline{b} - k\\overline{a})^2} = \\frac{(lb - a)(ka - b)}{-(lb - ka)^2}$. Now, since $\\bg_white |a|=1$, $\\bg_white a\\overline{a} = |a|^2 = 1$, so $\\bg_white \\overline{a}=\\frac{1}{a}$. Similar holds for $\\bg_white b$. Substituting this above yields \\bg_white \\begin{aligned} \\frac{(lb - a)(ka - b)}{(lb - ka)^2} &= \\frac{\\left ( \\frac{l}{b} - \\frac{1}{a} \\right ) \\left (\\frac{k}{a} - \\frac{1}{b} \\right )}{\\left ( \\frac{l}{b} - \\frac{k}{a} \\right )^2}\\\\ &= \\frac{a^2b^2 \\left ( \\frac{l}{b} - \\frac{1}{a} \\right ) \\left (\\frac{k}{a} - \\frac{1}{b} \\right )}{a^2 b^2 \\left ( \\frac{l}{b} - \\frac{k}{a} \\right )^2}\\\\ &= \\frac{ \\left ( la - b \\right ) \\left ( kb - a \\right )}{\\left (la - kb \\right )^2}\\\\ (lb - a)(ka - b)(la - kb)^2 &= (la - b)(kb - a)(lb - ka)^2\\\\ [\\text{Expanding and collecting like }&\\text{terms yields}]\\\\ (a^4 - b^4)kl(k - l) + (a^3 b - ab^3)(k - 1)(2kl - k^2l - kl^2", ". – user61454 Feb 8 '13 at 6:49 \\begin{align} a-b&=b-c\\\\ a+c&=2b\\\\ (a+c)^2&=(2b)^2\\\\ \\end{align} \\begin{align}a^2+c^2+2ac&=4b^2\\\\a^2+c^2-2b^2&=2b^2-2ac\\\\\\end{align} - $a-b=b-c$ means $b=(a+c)/2$ therefore $a^2-2b^2+c^2=a^2+c^2-2(\\frac{a+c}{2})^2=\\frac{2a^2+2c^2-(a+c)^2}{2}=\\frac{a^2+c^2-2ac}{2}=\\frac{(a-c)^2}{2}$ - $a-b=b-c\\implies a,b,c$ are in A.P. $a^2-2b^2+c^2=(a^2-b^2)-(b^2-c^2)=(a-b)(a+b)-(b-c)(c+b)=(a-b)(a+b)-(a-b)(c+b)=(a-b)(a+b-c-b)=(a-b)(a-c)$ -", "\\right)}{{{a}^{2}}+ab+{{b}^{2}}} \\\\ & =a-b=7.15-3.97=3.18 \\\\ \\end{align} Question 4 The value of $\\frac{{{\\left( 4.5 \\right)}^{3}}+{{\\left( 3.7 \\right)}^{3}}+{{\\left( 1.8 \\right)}^{3}}-3\\times 4.5\\times 3.7\\times 1.8}{{{\\left( 4.5 \\right)}^{2}}+{{\\left( 3.7 \\right)}^{2}}+{{\\left( 1.8 \\right)}^{2}}-4.5\\times 3.7-3.7\\times 1.8-1.8\\times 4.5}$ is: A 0 B 1 C 10 D 30 Question 4 Explanation: Let 4.5= a, 3.7= b, 1.8= c Therefore expression \\begin{align} & =\\frac{{{a}^{3}}+{{b}^{3}}{{+}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca} \\\\ & =a+b+c \\\\ & =4.5+3.7+1.8 \\\\ & =10 \\\\ \\end{align} Question 5 The value of $\\overline{0.5}+\\overline{0.7}+0.\\overline{57}$ is: A $1.\\overline{87}$ B $1.\\overline{90}$ C $1.\\overline{82}$ D $0.\\overline{98}$ Question 5 Explanation: \\begin{align} & \\overline{0.5}+\\overline{0.7}+0.\\overline{57} \\\\ & =\\frac{5}{9}+\\frac{7}{9}+\\frac{57}{99} \\\\ & =\\frac{55+77+57}{99} \\\\ & =\\frac{189}{99}=1\\frac{90}{99} \\\\ & =1.\\overline{90} \\\\ \\end{align} Once you are finished, click the button below. Any items you have not completed will be marked incorrect. There are 5 questions to complete. ← List → 1 2 3 4 5 End ### Want to explore more Arithmetic Tests? Get Posts Like This Sent to your Email Updates for Free Live sessions and offers are sent on mail. Don't worry: we do not send too many emails..:) Get Posts Like This Sent to your Email Updates for Free Live sessions and offers are sent on mail. Don't worry: we do not send too many emails..:)", "process, one finds $$f\\left({(a+b)(c+d)\\over a+b+c+d}\\right)+f\\left({abc+abd+acd+bcd\\over(a+b)(c+d)}\\right)=f\\left({(a+c)(b+d)\\over a+b+c+d}\\right)+f\\left({abc+abd+acd+bcd\\over(a+c)(b+d)}\\right)\\tag{1}$$ Now substitute $$a=c$$, $$b=\\frac{a^2}{d}$$ and $$t=\\frac{a}{b}+\\frac{b}{a}$$ so that $$\\frac{(a+b)(c+d)}{ a+b+c+d}=\\frac{abc+abd+acd+bcd}{(a+b)(c+d)}=a$$ $${(a+c)(b+d)\\over a+b+c+d}=a\\cdot{2t\\over2+t}$$ and $${abc+abd+acd+bcd\\over(a+c)(b+d)}=a\\cdot{2+t\\over2t}$$ If we substitute these results into $$(1)$$, we find $$2f(a)=f\\left(a\\cdot{2t\\over2+t}\\right)+f\\left(a\\cdot{2+t\\over2t}\\right)\\tag{2}$$ It can be seen that $$t\\geq 2$$ by AMGM and thus $$\\frac{2t}{2+t}\\geq 1$$. Hence for all pairs $$x\\geq y$$ there exist suitable $$a$$ and $$t$$ so that $$a\\cdot{2t\\over2+t}=x,a\\cdot{2+t\\over2t}=y\\text{ and } \\sqrt{xy}=a$$ so that $$(2)$$, and thus the required property, holds.", "= \\dfrac{BH}{BC}$and$\\dfrac{FH}{AB} = \\dfrac{HC}{BC} .$ Adding these equations yields: \\begin{align*}\\dfrac{FH}{EC} + \\dfrac{FH}{AB} &= \\dfrac{BH}{BC} + \\dfrac{HC}{BC}\\\\ &= \\frac{BH+HC}{BC} \\\\&= \\frac{BC}{BC} \\\\&= 1.\\end{align*} This, in turn, goes to show that $\\frac{1}{EC} + \\frac{1}{AB} = \\frac{1}{FH} .$ Now, let $$s$$ be the side length of the square. We know $$AB = 2\\cdot EC = s.$$ This means \\begin{align*}\\dfrac{1}{FH} &= \\dfrac{1}{EC} + \\frac{1}{AB} \\\\&= \\frac{1}{a} + \\frac{2}{a} \\\\&= \\frac{3}{a} .\\end{align*} Therefore, $$FH = \\frac{s}{3} .$$ Now, to compute the area of $$\\triangle EFC$$, we take the area of $$\\triangle BCE$$ and subtract the area of $$\\triangle BFC$$. This is equal to \\begin{align*} \\dfrac{BC\\cdot EC}{2} - \\dfrac{BC\\cdot FH}{2} &= \\dfrac{BC\\cdot(EC-FH)}{2} \\\\ &= \\dfrac{s\\cdot\\left(\\dfrac{s}{2} - \\dfrac{s}{3}\\right)}{2} \\\\ &= \\dfrac{s\\cdot\\frac{s}{6}}{2} \\\\ &= \\dfrac{s^2}{12}. \\end{align*} The area of $$AFED$$ is the area of $$\\triangle ACD$$ minus the area of $$\\triangle EFC$$, which is equal to\\begin{align*} \\dfrac{s^2}{2}-\\dfrac{s^2}{12} &= \\dfrac{5s^2}{12} \\\\&= 45.\\end{align*} With $$\\frac{5}{12} s^2 = 45$$, we get $$s^2 = 108,$$ which is the area of the full square. 23. From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? $$\\dfrac{2}{7}$$ $$\\dfrac{5}{42}$$ $$\\dfrac{11}{14}$$ $$\\dfrac{5}{7}$$ $$\\dfrac{6}{7}$$ ###### Solution(s): Without loss of generality, allow $$A$$ to be a vertex of", "there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.) \\begin{align} S(a,b,c) & = \\dfrac{a^3}{a+b} + \\dfrac{b^3}{b+c} + \\dfrac{c^3}{c+a}\\\\ & = \\dfrac{a^3+b^3}{a+b} + \\dfrac{b^3+c^3}{b+c} + \\dfrac{c^3+a^3}{c+a} - \\left(\\dfrac{b^3}{a+b} + \\dfrac{c^3}{b+c} + \\dfrac{a^3}{c+a} \\right)\\\\ & = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b) \\end{align} Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \\geq ab + bc+ ca$$ If $S(a,b,c) < \\dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \\dfrac{ac + cb+ ba}2$ which gives us $$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction. Note that using the AM-GM inequality, we have $$\\frac{a^3}{a+b}=a^2-\\frac{a^2b}{a+b}\\geq a^2-\\frac{a^2b}{2\\sqrt{ab}}=a^2-\\frac 12\\sqrt{a^3b},$$ And so summing up cyclically, it is enough to check that $$2(a^2+b^2+c^2)\\geq\\sqrt{a^3b}+\\sqrt{b^3c}+\\sqrt{c^3a}+ab+bc+ca,$$ Which is obviously true since $\\sqrt{a^3b}\\leq \\dfrac{a^2+ab}2$ and $ab+bc+ca\\leq a^2+b^2+c^2$ from the AM-GM inequality. Equality holds iff $a=b=c.$ $\\Box$", "\\triangle ABC & \\\\ &= \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} + \\frac{1}{2} AB \\cdot BC \\sin \\hat{B} & \\\\ &= \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} + \\frac{1}{2} DC \\cdot DA \\sin \\hat{D} & \\\\ &= DC \\cdot DA \\sin \\hat{D} & \\end{array}$ Write down two expressions for $$CA^{2}$$: one in terms of $$\\cos\\hat{D}$$ and one in terms of $$\\cos\\hat{B}$$. \\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \\cos \\hat{D}\\\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \\cos \\hat{B} \\end{align*} Show that $$2C{A}^{2}=C{D}^{2}+D{A}^{2}+A{B}^{2}+B{C}^{2}$$. \\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \\cos \\hat{D}\\\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \\cos \\hat{B}\\\\ &= AB^{2} + BC^{2} - 2(DC)(DA) \\cos \\hat{B} \\quad (DC \\cdot DA = AB \\cdot BC) \\\\ 2CA^{2}& = CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA) \\cos \\hat{D} \\\\ & - 2(DC)(DA) \\cos ( \\text{180} ° -\\hat{D})\\\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA)\\cos \\hat{D} \\\\ & + 2(DC)(DA) \\cos \\hat{D}\\\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} \\end{align*} Suppose that $$BC=10$$ units, $$CD=15$$ units, $$AD=4$$ units and $$AB=6$$ units. Calculate $$C{A}^{2}$$ (correct to one decimal place). \\begin{align*} 2CA^{2} &= (15)^{2} + (4)^{2} + (6)^{2} + (10)^{2} \\\\ &= 377\\\\ \\therefore CA^{2} &= \\text{188,5}\\text{ units$^{2}$} \\end{align*} Find the angle $$\\hat{B}$$. Hence, calculate the area of $$ABCD$$ (correct to one", ". . . $(b+c)\\big[a^2 + ab + ac + bc\\big] \\;=\\;0$ . . . . . . . . . . $(b+c)\\big[a(a+b) + c(a+b)\\big] \\;=\\;0$ Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \\;=\\;0$ We see that the product of three factors equals zero. Hence, at least one of the factors must be zero. Therefore: . $\\begin{Bmatrix}a+b \\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}b \\\\ & \\text{or} \\\\ b+c \\:=\\:0 & \\Rightarrow & b \\:=\\:\\text{-}c \\\\ & \\text{or} \\\\ a+c\\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}c \\end{Bmatrix}$ 3. ## Re: Opposite number proof Originally Posted by Soroban Hello, DIOGYK! I think I got it . . . We have: . $\\frac{1}{a+b+c} \\;=\\;\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ . . . . . . . . $\\frac{1}{a+b+c} \\;=\\;\\frac{bc + ac + ab}{abc}$ . . . . . . . . . . . . $abc \\;=\\;(a+b+c)(bc+ac+ab)$ . . . . . . . . . . . . $abc \\;=\\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$ . . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \\;=\\;0$ n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \\;=\\;0$ . . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \\;=\\;0$ Factor: .", "a solution to your equation This is just reversing Part 1. Assume $a^2+b^2=c^2$ and $a^2+2b^2=d^2$. Note that \\begin{align}(dc-ab)(dc+ab)&=d^2c^2-a^2b^2\\\\ &=a^4+2a^2b^2 +2b^4\\\\ &=b^2(a^2+2b^2)+a^2(a^2+b^4)\\\\ &=b^2d^2+a^2c^2\\end{align}\\tag{1} Let $$x=\\frac{bd(cd+ab)}{ac(cd-ab)}\\\\y=\\frac{ac(cd+ab)}{bd(cd-ab)}$$ Then you'll get, with help from (1) that: $$x+y=\\frac{(cd+ab)^2}{abcd}$$ and $$\\frac{1}{x}+\\frac{1}{y}=\\frac{(cd-ab)^2}{abcd}$$ So $$x-\\frac{1}{x}+y-\\frac{1}{y}=4$$ ## Part 3: Another equivalent The requirement that $a^2+b^2$ and $a^2+2b^2$ is equivalent to the requirement that $c^4-b^4$ is a perfect square for some relatively prime $b,c$ with $b$ even (and $b\\neq 0$ in both sides.) If $c^4-b^4=k^2$ then $c^2-b^2$ and $c^2+b^2$ are relatively prime and hence squares, and hence $a^2=c^2-b^2$ and $c^2=a^2+b^2$ and $c^2+b^2=a^2+2b^2$ are both squares.", "B$ Also, since $AB = 2AD$, $r = \\frac {AB}{BC} = \\frac {2AD}{BC} = 2 \\cos A$ $\\Rightarrow \\cos A = \\frac {1}{2}r$ and now we're done. just to say what we have to say: $\\cos A + \\cos B + \\cos C = 2 \\cos A + \\left( 1 - 2 \\cos^2 A \\right)$ $= 2 \\left( \\frac {1}{2}r \\right) + 1 - 2 \\left( \\frac {1}{2}r \\right)^2$ $= r + 1 - \\frac {r^2}{2}$ as desired 3. Originally Posted by Ph4m Triangle ABC is such that AC=BC and AB/BC = r Show that cos A + cos B + cos C = 1 + r -(r^2/2) $\\cos^2 A = \\frac{AC^2 + AB^2 - BC^2}{2AB\\cdot AC} = \\frac{AC^2 + r^2AC^2 - AC^2}{2rAC\\cdot AC} = \\frac{r^2AC^2}{2rAC^2} = \\frac{r}{2}$ $\\cos^2 B = \\frac{AB^2 + BC^2 - AC^2}{2AB\\cdot BC} = \\frac{r^2AC^2 + AC^2 - AC^2}{2AC\\cdot rAC} = \\frac{r}{2}$ $\\cos^2 C = \\frac{AC^2 + BC^2 - AB^2}{2AC\\cdot BC} = \\frac{AC^2 + AC^2 - r^2AC^2}{2AC^2} = \\frac{1-r^2}{2}$ I do not have time to finish the details, sorry. 4. Originally Posted by ThePerfectHacker $\\cos^2 A = \\frac{AC^2 + AB^2 - BC^2}{2AB\\cdot AC} = \\frac{AC^2 + r^2AC^2 - AC^2}{2rAC\\cdot AC} = \\frac{r^2AC^2}{2rAC^2} = \\frac{r}{2}$ $\\cos^2 B = \\frac{AB^2 + BC^2 - AC^2}{2AB\\cdot BC} = \\frac{r^2AC^2 + AC^2 - AC^2}{2AC\\cdot rAC} = \\frac{r}{2}$ $\\cos^2 C", "then ABC: \\begin{aligned} d^2 &= AD^2\\\\ &= AC^2 + CD^2 - 2AC.CD\\cos \\angle DCA\\\\ &= AC^2 + CD^2 - 2AC.CD \\cos \\angle BCA\\\\ &= AC^2 + CD^2 - 2AC.CD \\frac{CA^2 + CB^2 - AB^2}{2CA.CB}\\\\ &= b^2 + n^2 - 2bn \\frac{b^2 + a^2 - c^2}{2ba}\\\\ &= b^2 + n^2 - n \\frac{b^2 + a^2 - c^2}{a}\\\\ &= \\frac{b^2(m+n) - n(b^2 + a^2 - c^2)}{a} + n^2\\\\ &= \\frac{b^2 m + c^2n}{a} + n^2 - \\frac{na^2}{a}\\\\ &= \\frac{b^2 m + c^2n}{a} + n(n-a)\\\\ &= \\frac{b^2 m + c^2n}{a} - mn.\\\\ \\end{aligned} 2. If $D$ divides the side $BC$ in the ratio $BD:DC = r:s$, $\\displaystyle d^2 = \\frac{rb^2 + sc^2}{r+s} - \\frac{rsa^2}{(r+s)^2} = \\frac{r^2b^2 + s^2c^2 + rs(b^2 + c^2 - a^2)}{(r+s)^2} .\\quad\\quad(2)$ 3. Similar to (2) but substituting $\\displaystyle \\lambda = r/(r+s) = BD:BC$, $\\displaystyle d^2 = \\lambda b^2 + (1-\\lambda)c^2 - \\lambda(1-\\lambda)a^2.\\quad\\quad(3)$ This and the previous form are conveniently proved using vectors. Writing the vector $\\mathbf{AD} = \\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}$, \\begin{aligned} d^2 &= \\mathbf{AD}. \\mathbf{AD}\\\\ &= \\left(\\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}\\right).\\left(\\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}\\right)\\\\ &= \\lambda^2 \\mathbf{AC}.\\mathbf{AC} + (1-\\lambda)^2 \\mathbf{AB}.\\mathbf{AB} + 2\\lambda(1-\\lambda)\\mathbf{AB}.\\mathbf{AC}\\\\ &= \\lambda^2 b^2 + (1-\\lambda)^2 c^2 + \\lambda(1-\\lambda)\\left(b^2 + c^2 - a^2\\right)\\\\ &= b^2(\\lambda^2 + \\lambda(1-\\lambda)) + c^2((1-\\lambda)^2 + \\lambda(1-\\lambda)) - a^2\\lambda(1-\\lambda)\\\\ &= \\lambda b^2 + (1-\\lambda)c^2 - \\lambda(1-\\lambda)a^2. \\end{aligned} Note that", "=3mn+n(a^2+b^2+c^2)=5mn$$ Also because of $(*)$ we have $$a^3+b^3+c^3 = m(a+b+c) +3n = 3n$$ and we are done. $M= (a^2 + b^2 + c^2)(a^3 + b^3+c^3)=$ $a^5 + b^5 + c^5 + a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$ $= (a^5 + b^5 + c^5) + L$ where $L = a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$ $= a(a^2b^2 + a^2c^2) + b(a^2b^2 + b^2c^2) + c(a^2c^2 + b^2b^2)$ $= (a+b+c)(a^2b^2 + a^2c^2 + b^2c^2) - ab^2c^2 -a^2bc^2 - a^2b^2c$ $= - ab^2c^2 -a^2bc^2 - a^2b^2c= -abc(ab + ac + bc)$ On the other hand: $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)=-2(ab+ac + bc)$ $= \\frac {2L}{abc}$ while $0=(a+b+c)^3 = (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$ $= a^3 + 3a^2b + 3ab^2 + b^3 +3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3$ $= a^3 + b^3 + c^3 + 3a(ab + ac) +3b(ab + bc) + 3c(ac+ bc) + 6abc$ $=a^3+b^3+c^3 +3(a+b+c)(ab + ac + bc) - 9abc+6abc$ $=a^3 + b^3 +c^3 -3abc$ So $a^3 + b^3+c^3 = 3abc$. So $M =(a^2+b^2 + c^2)(a^3 + b^3 + c^3) = \\frac {2L}{abc}*3abc = 6L$ So $M = a^5 + b^5 + c^5 + L$ and $M = 6L$ so $5L=a^5 + b^5 +c^5$. And we have: $\\frac {a^2+b^2+c^2}2\\times\\frac {a^3+b^3 + c^3}3 =", "that: $$\\frac{1}{bc + cd + da -1} \\le \\frac{1}{2}$$ $$\\frac{1}{ab + cd + da -1} \\le \\frac{1}{2}$$ $$\\frac{1}{ab + bc + da -1} \\le \\frac{1}{2}$$ $$\\frac{1}{ab + bc + cd -1} \\le \\frac{1}{2}$$ or: $$bc + cd + da -1 \\ge 2$$ $$ab + cd + da -1 \\ge 2$$ $$ab + bc + da -1 \\ge 2$$ $$ab + bc + cd -1 \\ge 2$$ But if all inequalites hold then $a=b=c=d=1$ and that's as far I can go. • First use $\\dfrac{1}{bc+da+cd -1} = \\dfrac{1}{bc+da +cd - \\sqrt{abcd}} \\le \\dfrac{2}{2cd + bc+da}$. By using $bc+da\\ge 2\\sqrt{abcd}$. Then use Cauchy Schwarz. – Yimin Apr 1 '13 at 18:34 By inequality of arithmetic and geometric means, $bc+da\\ge 2\\sqrt{abcd}=2$, so $$\\frac{1}{bc+cd+da-1}\\le\\frac{1}{1+cd}.\\tag{1}$$ Similarly, $$\\frac{1}{ab+bc+da-1}\\le\\frac{1}{1+ab}.\\tag{2}$$ $(1)+(2)$ gives $$\\frac{1}{bc+cd+da-1}+\\frac{1}{ab+bc+da-1}\\le\\frac{1}{1+cd}+\\frac{1}{1+ab}=\\frac{2+ab+cd}{1+ab+cd+abcd}=1.\\tag{3}$$ Similarly, $$\\frac{1}{ab+cd+da-1}+\\frac{1}{ab+bc+cd-1}\\le 1.\\tag{4}$$ The conclusion follows from $(3)+(4)$.", "CD^2 - 2AC.CD \\frac{CA^2 + CB^2 - AB^2}{2CA.CB}\\\\ &= b^2 + n^2 - 2bn \\frac{b^2 + a^2 - c^2}{2ba}\\\\ &= b^2 + n^2 - n \\frac{b^2 + a^2 - c^2}{a}\\\\ &= \\frac{b^2(m+n) - n(b^2 + a^2 - c^2)}{a} + n^2\\\\ &= \\frac{b^2 m + c^2n}{a} + n^2 - \\frac{na^2}{a}\\\\ &= \\frac{b^2 m + c^2n}{a} + n(n-a)\\\\ &= \\frac{b^2 m + c^2n}{a} - mn.\\\\ \\end{aligned} 2. If $D$ divides the side $BC$ in the ratio $BD:DC = r:s$, $\\displaystyle d^2 = \\frac{rb^2 + sc^2}{r+s} - \\frac{rsa^2}{(r+s)^2} = \\frac{r^2b^2 + s^2c^2 + rs(b^2 + c^2 - a^2)}{(r+s)^2} .\\quad\\quad(2)$ 3. Similar to (2) but substituting $\\displaystyle \\lambda = r/(r+s) = BD:BC$, $\\displaystyle d^2 = \\lambda b^2 + (1-\\lambda)c^2 - \\lambda(1-\\lambda)a^2.\\quad\\quad(3)$ This and the previous form are conveniently proved using vectors. Writing the vector $\\mathbf{AD} = \\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}$, \\begin{aligned} d^2 &= \\mathbf{AD}. \\mathbf{AD}\\\\ &= \\left(\\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}\\right).\\left(\\lambda \\mathbf{AC} + (1-\\lambda) \\mathbf{AB}\\right)\\\\ &= \\lambda^2 \\mathbf{AC}.\\mathbf{AC} + (1-\\lambda)^2 \\mathbf{AB}.\\mathbf{AB} + 2\\lambda(1-\\lambda)\\mathbf{AB}.\\mathbf{AC}\\\\ &= \\lambda^2 b^2 + (1-\\lambda)^2 c^2 + \\lambda(1-\\lambda)\\left(b^2 + c^2 - a^2\\right)\\\\ &= b^2(\\lambda^2 + \\lambda(1-\\lambda)) + c^2((1-\\lambda)^2 + \\lambda(1-\\lambda)) - a^2\\lambda(1-\\lambda)\\\\ &= \\lambda b^2 + (1-\\lambda)c^2 - \\lambda(1-\\lambda)a^2. \\end{aligned} Note that this is valid for any real $\\lambda$, so $D$ may lie beyond segment $BC$. 4. Writing (3) as a quadratic in $\\lambda$: $\\displaystyle d^2 = \\lambda^2", "### 2008 IMO SL #A5 Let $a$, $b$, $c$, $d$ be positive real numbers such that $abcd = 1$ and $a + b + c + d > \\dfrac{a}{b} + \\dfrac{b}{c} + \\dfrac{c}{d} + \\dfrac{d}{a}$. Prove that $a + b + c + d < \\dfrac{b}{a} + \\dfrac{c}{b} + \\dfrac{d}{c} + \\dfrac{a}{d}$ Solution 1: By Cauchy-Schwarz, we have $$(a+b+c+d)(ab+bc+cd+da)>\\left(\\frac ab+\\frac bc+\\frac cd+\\frac da\\right)(ab+bc+cd+da)$$$$\\ge(a+b+c+d)^2,$$which implies that $ab+bc+cd+da>a+b+c+d$. Again, by Cauchy-Schwarz, it follows that $$(a+b+c+d) \\left(\\frac ba+\\frac cb+\\frac dc+\\frac ad\\right) > \\left(\\frac ab+\\frac bc+\\frac cd+\\frac da\\right) \\left(\\frac dc+\\frac ad+\\frac ba+\\frac cb\\right)$$ $$\\ge\\left(\\sqrt{\\frac{da}{bc}}+\\sqrt{\\frac{ab}{cd}}+\\sqrt{\\frac{bc}{da}}+\\sqrt{\\frac{cd}{ab}}\\right)^2$$$$=(da+ab+bc+cd)^2 \\ge(a+b+c+d)^2,$$which implies $\\frac ba+\\frac cb+\\frac dc+\\frac ad>a+b+c+d$, as desired. $\\square$ Solution 2: Define $$S_1 = \\frac ab +\\frac bc +\\frac cd +\\frac da$$$$S_2 = \\frac ba +\\frac cb +\\frac dc +\\frac ad$$$$W =a+b+c+d$$. We claim that $3S_1+S_2 \\geq 4W$ . Clearly, this finishes because then we have $S_2 \\geq W + \\underbrace{3(W-S_1)}_{>0} >W$. To prove our claim, note that $$2 \\cdot \\frac ab + \\frac bc + \\frac ad \\overset{\\text{AM-GM}}{\\geq} 4\\sqrt [4]{\\frac {a}{d}\\cdot \\frac {a^2}{b^2} \\cdot \\frac {b}{c}} \\geq 4a$$. Cyclically summing up gives the desired claim, so we are done. $\\square$ 1. Anonymous3/02/2022 im so addicted to your blog that i need to stop going to it every 3 minutes to check for a new post", "# Leo Giugiuc's Second Lemma And Applications ### Proof 1 \\begin{align} 2(m_a+m_b) &= |\\overrightarrow{AB}+\\overrightarrow{AC}|\\lt |\\overrightarrow{AB}|+|\\overrightarrow{AC}|+|\\overrightarrow{BA}|+|\\overrightarrow{BC}|\\\\ &= a+b+2c < (a+b+2c)+2(a+b-c)\\\\ &= 3(a+b). \\end{align} ### Proof 2 The required inequality is equivalent to $\\displaystyle\\frac{2}{3}(m_a+m_b)\\le a+b.\\,$ Geometrically, this becomes $AG+BG\\le AC+BC.$ Let $A',B\\,$ be the midpoints of $BC\\,$ and $AC,\\,$ respectively. In $\\Delta BGA',\\,$ $BG\\lt GA'+A'B\\,$ so that $AG+BG\\lt AG+GA'+A'B=AA'+BB'.$ In $\\Delta AA'C,\\,$ $AA'\\lt A'C'+AC\\,$ so that $AG+BG\\lt AA'+BB'\\lt A'C+AC+A'B=BC+AC.$ ### Proof 3 Let, for a point $X,\\,$ $x\\,$ denotes the corresponding complex numbers. $\\displaystyle g=\\frac{a+b+c}{3}.$ \\displaystyle\\begin{align} AG+BG &= \\left|a-\\frac{a+b+c}{3}\\right|+\\left|b-\\frac{a+b+c}{3}\\right|\\\\ &=\\left|\\frac{a-b+a-c}{3}\\right|+\\left|\\frac{b-a+b-c}{3}\\right|\\\\ &\\le\\frac{|a-b|+|a-c|}{3}+\\frac{|b-a|+|b-c|}{3}\\\\ &=\\frac{2\\cdot AB+AC+BC}{3}\\\\ &=\\frac{2\\cdot (AC+BC)+(AC+BC)}{3}\\\\ &\\le AC+BC. \\end{align} ### Proof 4 This is just a follow-up on the starting point of Proof 2. The required inequality is equivalent to $AG+BG\\lt AC+BC,\\,$ i.e. $AB+AG+BG\\lt AB+AC+BC.\\,$ But, in general, given two convex polygons, one within the other, the perimeter of the outer polygon exceeds the permiter of the inner one. The lemma then follows from the fact that $\\Delta ABG\\,$ is within $\\Delta ABC.\\,$ ### Proof 5 Lemma is a special case of the following assertion: If $M\\,$ is an interior point of $\\Delta ABC,\\,$ then $MB + MC < AB + AC.$ To prove that, consider the ellipse whose focii are $B\\,$ and $C\\,$ and the major axis is $AB + AC.\\,$ Point $M\\,$ being inside $\\Delta ABC,\\,$ it is also", "we can write $$D=D^\\prime + |DD^\\prime| \\frac{( C-B )\\times ( B \\times C )}{|BC|\\,|B\\times C|} \\tag{1}$$ where $$D^\\prime := \\frac12(B+C)$$ is the midpoint of $$\\overline{BC}$$. Further, by the Inscribed Angle Theorem, $$\\angle BOC\\cong\\angle BDC$$, and we conclude that $$\\overline{DD^\\prime}$$ is the altitude of an isosceles triangle with vertex angle $$\\alpha$$ and base $$|BC|$$. Therefore, $$|DD^\\prime| = \\frac12|BC|\\,\\cot\\frac12\\alpha$$, so we have $$\\frac{|DD^\\prime|}{|BC|\\,|B\\times C|} = \\frac{\\cot\\frac12\\alpha}{2bc\\sin\\alpha} = \\frac{\\cos\\frac12\\alpha\\,/\\,\\sin\\frac12\\alpha}{4bc\\sin\\frac12\\alpha\\cos\\frac12\\alpha}=\\frac{1}{4bc\\sin^2\\frac12\\alpha} = \\frac{1}{2bc(1-\\cos\\alpha)} \\tag{2}$$ Further, via a cross product identity, \\begin{align} (C-B)\\times(B\\times C) &= \\phantom{-}B\\,((C-B)\\cdot B) - C\\,((C-B)\\cdot C) \\\\ &=\\phantom{-}B\\left(B\\cdot C - |B|^2\\right)-C\\left(|C|^2 - B\\cdot C\\right) \\\\ &=-B\\,b(b-c\\cos\\alpha) - C\\,c(c-b\\cos\\alpha) \\end{align}\\tag{3} Altogether, this gives us \\begin{align}D\\;2bc(1-\\cos\\alpha) &= (B+C)bc(1-\\cos\\alpha)-B\\,b(b-c\\cos\\alpha)-C\\,c(c-b\\cos\\alpha) \\\\ &=Bb(c-b)+Cc(b-c) \\\\ &=(c-b)(B b-C c) \\tag{4a} \\end{align} (Note that, if $$b=c$$, then $$D=0$$, as we would expect. This confirms that we got our cross-product vector directions correct in $$(1)$$.) Likewise, \\begin{align} E\\,2ca(1-\\cos\\beta) &= (a-c)(C c-A a) \\tag{4b} \\\\ F\\,2ab(1-\\cos\\gamma) &= (b-a)(A a-B b) \\tag{4c} \\end{align} Finally, observe that, for $$a$$, $$b$$, $$c$$ not all equal (the only case with which we need concern ourselves), $$(\\text{eq }4a)\\;(a-c)(b-a) + (\\text{eq }4b)\\;(c-b)(b-a)+(\\text{eq }4c)\\;(a-c)(c-b) \\tag{5}$$ gives a non-trivial linear combination of $$D$$, $$E$$, $$F$$ that vanishes. Consequently, $$D$$, $$E$$, $$F$$ are linearly dependent; that is, they lie on a common plane through $$O$$, and the result follows. $$\\square$$ • Pardon my ignorance but I'm confused as to" ]

[ "- x^3 + 2x + 3 \\:=\\:0\\,\\text{ are }\\,a,b,c,d.$ $\\text{Find the value of:}$ . . $(A)\\;\\;\\sum\\alpha^2$ . . $(B)\\;\\;\\sum\\alpha^2\\beta^2$ . . $(C)\\;\\;\\sum\\alpha\\beta(\\alpha+\\beta)$ . . $(D)\\;\\;\\sum\\frac{1}{\\alpha}$ From Vieta's Theorem, we have these equations: . . $\\begin{array}{cccccc}a + b + c + d \\;=\\; 1 & [1] \\\\ ab + ac + ad + bc + bc + cd \\;=\\; 0 & [2] \\\\ abc + abd + acd + bcd \\;=\\; -2 & [3] \\\\ abcd \\;=\\; 3 & [4] \\end{array}$ $(A)\\;\\text{ We want: }\\:a^2 + b^2 + c^2 + d^2$ Square [1]: . $(a+b+c+d)^2 \\:=\\:1^2$ . . $a^2 + 2ab + 2ac + 2ad + b^2 + 2bc + 2bd + c^2 + 2cd + d^2 \\;=\\;1$ . . $a^2 + b^2 + c^2 + d^2 + 2\\underbrace{(ab + ac + ad + bc + bd + cd)}_{\\text{This is 0}} \\;=\\;1$ Therefore: . $a^2+b^2+c^2+d^2 \\;=\\;1$ $(D)\\;\\text{ We want: }\\:\\dfrac{1}{a} + \\dfrac{1}{b} + \\dfrac{1}{c} + \\dfrac{1}{d}$ . . $\\displaystyle \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\;=\\;\\frac{\\overbrace{bcd + acd + abd + abc}^{\\text{This is }-2}}{\\underbrace{abcd}_{\\text{This is 3}}}$ Therefore: . $\\displaystyle \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\;=\\;-\\frac{2}{3}$ Get the idea? 5. Originally Posted by arze For the second, I worked out the equivalent for cubic equations $\\sum\\alpha^2\\beta^2=(\\sum\\alpha\\beta)^2-2\\alpha\\beta\\gamma(\\sum\\alpha)$ Problem is it doesn't work. $(0)^2-2(3)(-1)=6$ answer is 10. Thanks! You", "# Complex Numbers and the Norm I am given that for a complex number $w=a+bi$, define $\\overline{w}=a-bi$ and $N(w)=w \\overline{w}$. I have provided my answers for parts a and b, but I am not sure they are correct. I need help with figuring out part c. (a) Compute $N(w)=w \\overline{w}$ explicitly. Here is what I have gotten:$N(w)=w \\overline{w}= (a+bi)(a-bi)= a^2+b^2$. (b) Show that $N(rs)=N(r)N(s)$ for any two complex numbers $r,s$. Here is my work: Let $r=a+bi$, $s=c+di$ Then $$rs=(a+bi)(c+di)=ac+adi+cbi-bd=(ac-bd)+(ad+cb)i.$$ Now, from a, have that $N(r)= a^2+b^2$ and $N(s)= c^2+d^2$. So, $$N(r)N(s)=(a^2+b^2)(c^2+d^2).$$ Also, $N(rs)=N((ac-bd)+(ad+cb)i)$ and from part a then, $N(rs)=(ac-bd)^2+(ad+cb)^2$ When expanded, $$a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2$$ \\begin{align*} &=a^2c^2+a^2d^2+b^2d^2+b^2c^2\\\\ &=a^2(c^2+d^2)+b^2(c^2+d^2)\\\\ &=(a^2+b^2)(c^2+d^2). \\end{align*} Hence, $N(rs)=N(r)N(s)$. (c) Prove that $N(\\overline{v})=N(v)$ and $N(v^n)=N(v)^n$ for any complex number. This is the part I am confused as to how to prove. I let $v=a+bi$ and $\\overline{v}=a-bi$. Then I am not sure how to use what I have shown before for this part. • (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. Apr 16 '17 at 21:35 • @MichaelBurr Sorry, I do not know how to use MathJax, I tried my best to put it in the format. I labeled the parts a-c, but I will rename to make is clearer. Apr 16", "guess such relations? $(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$? Apr 10, 2019 at 14:46 • @BPP I don't really know Apr 10, 2019 at 15:47 Fun video! Much time was spent on finding $$abc=1/6$$. Alternative method for this: \\begin{align}a^2+b^2&=2-c^2 \\Rightarrow \\\\ (a+b)^2-2ab&=2-c^2 \\Rightarrow \\\\ (1-c)^2-2ab&=2-c^2 \\Rightarrow \\\\ ab&=c^2-c-\\frac12 \\Rightarrow \\\\ abc&=c^3-c^2-\\frac c2 \\end{align} Similarly: $$abc=a^3-a^2-\\frac a2\\\\ abc=b^3-b^2-\\frac b2$$ Now adding them up: $$3abc=(a^3+b^3+c^3)-(a^2+b^2+c^2)-\\frac12(a+b+c)=3-2-\\frac12 \\Rightarrow abc=\\frac16.$$ In fact, you can find other terms as well: $$ab+bc+ca=(a^2+b^2+c^2)-(a+b+c)-\\frac32=2-1-\\frac32=-\\frac12;\\\\ a^2b^2+b^2c^2+c^2a^2=ab(c^2-c-\\frac12)+bc(a^2-a-\\frac12)+ca(b^2-b-\\frac12)=\\\\ abc(a+b+c)-3abc-\\frac12(ab+bc+ca)=\\\\ \\frac16-\\frac12+\\frac14=-\\frac1{12}$$ Hence: $$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\\\ 2\\cdot 3-(-\\frac1{12})+\\frac16\\cdot (-\\frac12)=6.$$ I think to use an identity must be the shortest way:- $${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)}$$ If we square the 1st equation we get $${ab+bc+ca=\\frac {-1}2}$$ On squaring the above equation we get $${(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\\frac 14}$$ We also know $${ (a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca) }$$ from here we get :- $$abc=\\frac 16$$ Substituting the known values We get :- $${ (ab)^2+(bc)^2+(ca)^2=\\frac {-1}{12} }$$ Now squaring the 2nd equation:- $${ (a^4+b^4+c^4)+2\\big ((ab)^2+(bc)^2+(ca)^2)\\big )=4 }$$ From here we get $$(a^4+b^4+c^4)=\\frac{25}6$$ Now we have all the values to be substituted in the identity that i mentioned therefore you get $${ (a^5+b^5+c^5)= (\\frac{25}{6}\\cdot 1) - (3 \\cdot \\frac {-1}2)+ (\\frac 16 \\cdot 2) }$$ $$\\Rightarrow(a^5+b^5+c^5)=6$$", "BCD$, $\\bg_white \\arg \\left ( \\frac{d - a}{d - c} \\right ) = \\arg \\left ( \\frac{c - d}{c - b} \\right )$. Thus, $\\bg_white \\frac{d - a}{d - c} \\div \\frac{c - d}{c - b} = r$, where $\\bg_white r$ is a real number. Substituting in the values from above, we get $\\bg_white \\frac{(lb - a)(ka - b)}{-(lb - ka)^2} = r$. Since $\\bg_white r = \\overline{r}$, we have $\\bg_white \\frac{(l\\overline{b} - \\overline{a})(k\\overline{a} - \\overline{b})}{-(l\\overline{b} - k\\overline{a})^2} = \\frac{(lb - a)(ka - b)}{-(lb - ka)^2}$. Now, since $\\bg_white |a|=1$, $\\bg_white a\\overline{a} = |a|^2 = 1$, so $\\bg_white \\overline{a}=\\frac{1}{a}$. Similar holds for $\\bg_white b$. Substituting this above yields \\bg_white \\begin{aligned} \\frac{(lb - a)(ka - b)}{(lb - ka)^2} &= \\frac{\\left ( \\frac{l}{b} - \\frac{1}{a} \\right ) \\left (\\frac{k}{a} - \\frac{1}{b} \\right )}{\\left ( \\frac{l}{b} - \\frac{k}{a} \\right )^2}\\\\ &= \\frac{a^2b^2 \\left ( \\frac{l}{b} - \\frac{1}{a} \\right ) \\left (\\frac{k}{a} - \\frac{1}{b} \\right )}{a^2 b^2 \\left ( \\frac{l}{b} - \\frac{k}{a} \\right )^2}\\\\ &= \\frac{ \\left ( la - b \\right ) \\left ( kb - a \\right )}{\\left (la - kb \\right )^2}\\\\ (lb - a)(ka - b)(la - kb)^2 &= (la - b)(kb - a)(lb - ka)^2\\\\ [\\text{Expanding and collecting like }&\\text{terms yields}]\\\\ (a^4 - b^4)kl(k - l) + (a^3 b - ab^3)(k - 1)(2kl - k^2l - kl^2", "# Complex Number Proof of $\\overline{\\left(\\frac {z_1}{z_2}\\right)}=\\frac {\\overline{z_1}}{\\overline{z_2}}$. Let $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$ be complex numbers. Let $\\overline{z_1}$ and $\\overline{z_2}$ be the conjugate of $z_1$ and $z_2$, respectively. Prove: $$\\overline{\\left(\\frac {z_1}{z_2}\\right)}=\\frac {\\overline{z_1}}{\\overline{z_2}}$$ My Attempt: I started with $\\overline{\\left(\\frac {z_1}{z_2}\\right)}$ and multiplied it by $\\frac {z_1}{z_2}$ to get $\\left|\\frac {z_1}{z_2}\\right|^2=\\frac {a_1^2+b_1^2}{a_2^2+b_2^2}$, but now, I don't know where to continue afterwards. \\begin{align} \\frac{z_1}{z_2}&=\\frac{a_1+ib_1}{a_2+ib_2}\\\\ \\\\ &=\\frac{a_1+ib_1}{a_2+ib_2}\\cdot\\frac{a_2-ib_2}{a_2-ib_2}\\\\ \\\\ &=\\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2+ib_2)(a_2-ib_2)}\\\\ \\\\ &=\\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2^2+b_2^2)}\\\\ \\\\ &=\\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)} \\end{align} The $\\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}$ fraction is a Complex Number with Real denominator ($a_2^2+b_2^2$) and conjugation of this fraction only changes the Numarator $(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)$. So \\begin{align} \\overline{\\Big(\\frac{z_1}{z_2}\\Big)}&=\\overline{\\Big(\\frac{a_1+ib_1}{a_2+ib_2}\\Big)}\\\\ \\\\ &=\\overline{ \\Big(\\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}\\Big)} \\\\ \\\\ &=\\frac{(a_1a_2+b_1b_2)-i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)} \\\\ \\\\ &=\\frac{(a_1a_2+b_1b_2)+i(a_1b_2-a_2b_1)}{(a_2^2+b_2^2)} \\\\ \\\\ &=\\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2^2+b_2^2)} \\\\ \\\\ &=\\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2-ib_2)(a_2+ib_2)} \\\\ \\\\ &=\\frac{a_1-ib_1}{a_2-ib_2} \\cdot\\frac{a_2+ib_2}{a_2+ib_2} \\\\ \\\\ &=\\frac{a_1-ib_1}{a_2-ib_2} \\\\ \\\\ &=\\frac{\\overline{z_1}}{\\overline{z_2}} \\end{align} • Ah! Thanks for elaborating why each step is like that! – user332252 Sep 17 '16 at 19:13 Hint: You can complete your proof multiplying also the RHS by $\\frac{z_1}{z_2}$ then show that $\\left|\\frac {z_1}{z_2}\\right|=\\frac{|z_1|}{|z_2|}$ ( easy in polar form). we get $$\\frac{a_1+b_1i}{a_2+b_2i}=\\frac{a_1a_2+b_1b_2+(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ and the complement is $$\\overline{\\frac{a_1+b_1i}{a_2+b_2i}}=\\frac{a_1a_2+b_1b_2-(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ and $$\\frac{a_1-b_1i}{a_2-b_2i}=\\frac{(a_1-b_1i)(a_2+b_2i)}{(a_2-b_2i)(a_2+b_2i)}=...$$ can you proceed? If $w_1=re^{i\\theta}$ and $w_2=\\rho e^{i\\phi}$ then $$\\overline{w_1}\\ \\overline{w_2}=re^{-i\\theta} \\rho e^{-i\\phi}=r\\ \\rho e^{-i(\\theta+\\phi)}=\\overline{w_1\\ w_2}$$ so $$\\boxed{\\overline{w_1}\\ \\overline{w_2}=\\overline{w_1\\ w_2}}$$ now let $w_1=\\dfrac{z_1}{z_2}$ and $z_2$. Also straightforward shows $$\\overline{\\left(\\frac {w_1}{w_2}\\right)}=\\dfrac{r}{\\rho} e^{-i(\\theta-\\phi)}=\\dfrac{re^{-i\\theta}}{\\rho e^{-i\\phi}}=\\dfrac{\\overline{w_1}}{\\overline{w_2}}$$", "$${\\overline q} = (abc){a + b\\over ab}{a + c\\over ac}{b + c\\over bc} = (a+b)(b+c)(a+c)/(abc) = q.$$ - Since $|a|=|b|=1$, we have that $$\\left|\\frac{a}{b}\\right|=1 \\quad \\Rightarrow \\quad \\overline{\\dfrac{a}{b}} =\\left.\\left|\\dfrac{a}{b}\\right|^2 \\middle/\\dfrac{a}{b}\\right. =\\frac{b}{a}$$ Therefore, \\begin{align} \\frac{(a+b)^2}{ab} &=\\frac{a}{b}+2+\\frac{b}{a}\\\\ &=2+2\\,\\mathrm{Re}\\left(\\frac{a}{b}\\right)\\\\ &\\ge0 \\end{align} This means that $$\\left(\\frac{(a+b)(b+c)(c+a)}{abc}\\right)^2=\\frac{(a+b)^2}{ab}\\frac{(b+c)^2}{bc}\\frac{(c+a)^2}{ca}\\ge0$$ which leads immediately to $$\\frac{(a+b)(b+c)(c+a)}{abc}\\in\\mathbb{R}$$ - Here's a nice geometric argument (with some holes and assumptions left to fill up): \\begin{align} \\arg \\frac {(a+b)(b+c)(a+c)}{abc} &= \\arg (a+b)+\\arg(b+c)+\\arg(a+c) - \\arg (a) -\\arg( b) - \\arg( c) \\\\ &= \\frac{\\arg (a)+arg(b)} 2 + \\frac{\\arg (b)+arg(c)} 2 + \\frac{\\arg (a)+arg(c)} 2 \\\\ & \\qquad\\qquad - \\arg (a) -\\arg( b) - \\arg( c) \\\\&= 0 \\end{align} - Hint 1. Since $\\Vert a\\Vert=\\Vert b\\Vert=\\Vert c\\Vert=1$, then $$a=e^{i\\alpha}\\qquad b=e^{i\\beta}\\qquad c=e^{i\\gamma}\\tag{1}$$ Hint 2 A complex number $z\\in\\mathbb{C}$ is real iff $$z+\\overline{z}=2z\\tag{2}$$ Substitute $(1)$ into $(2)$ and check that it is correct. - a = exp(iα) = exp(iβ) = exp(iγ) ??? – ᴊ ᴀ s ᴏ ɴ Jul 26 '12 at 13:45 Sorry, that was a typo – Norbert Jul 26 '12 at 13:54", "(ac-bd) + (ad+bc)i \\end{align} Complex number division is a little trickier and a very common exam question. To find $\\frac{z_{1}}{z_{2}}$, multiply by $\\frac{\\bar{z}_{2}}{\\bar{z}_{2}}$ to eliminate the imaginary component from the denominator \\begin{align} \\frac{a + bi}{c + di} &= \\frac{a + bi}{c + di} \\cdot \\frac{c - di}{c - di} \\\\ &= \\frac{ac-adi+bci-bdi^{2}}{c^{2} - (di)^{2}} \\\\ &= \\frac{(ac+bd) + (bc-ad)i}{c^{2}+d^{2}} \\\\ &= \\frac{(ac+bd)}{c^{2}+d^{2}} + \\frac{(bc-ad)}{c^{2}+d^{2}}i \\end{align} ### Example Q) Let $z = 2 - 3i$ and $w = \\frac{3}{2}i$. Find $z+w$, $z\\cdot w$, and $\\frac{z}{w}$. A) $z + w = 2 + (\\frac{3}{2} - 3)i = 2 - \\frac{3}{2}i$. $z\\cdot w = (2-3i)\\cdot\\frac{3}{2}i = 3i - 3\\cdot\\frac{3}{2}i^{2} = \\frac{9}{2}+ 3i$. $\\frac{z}{w} = \\frac{2 - 3i}{\\frac{3}{2}i} = \\frac{2 - 3i}{\\frac{3}{2}i} \\cdot \\frac{-\\frac{3}{2}i}{-\\frac{3}{2}i} = \\frac{-\\frac{9}{2}- 3i}{\\frac{9}{4}} = -2- \\frac{4}{3}i$ ### Inversion Complex numbers also have inverses just like real numbers. To find the inverse of some complex number $z = a+bi$ we use the same method for dividing two complex numbers \\begin{align} \\frac{1}{a+bi} &= \\frac{1}{a+bi} \\cdot \\frac{a-bi}{a-bi} \\\\ &= \\frac{a-bi}{a^{2} - (bi)^{2}} \\\\ &= \\frac{a-bi}{a^{2}+b^{2}} \\\\ &= \\frac{a}{a^{2}+b^{2}} - \\frac{b}{a^{2}+b^{2}}i \\\\ &= \\frac{1}{|z|^{2}}\\bar{z} \\end{align} (Note from this the important result $z \\cdot \\bar{z} = |z|^{2}$.) This is another very common exam question. ### Example Q) $z = -2 - i$. Find $z^{-1}$. A) $\\frac{1}{-2 - i} = \\frac{1}{-2 - i} \\cdot", "process, one finds $$f\\left({(a+b)(c+d)\\over a+b+c+d}\\right)+f\\left({abc+abd+acd+bcd\\over(a+b)(c+d)}\\right)=f\\left({(a+c)(b+d)\\over a+b+c+d}\\right)+f\\left({abc+abd+acd+bcd\\over(a+c)(b+d)}\\right)\\tag{1}$$ Now substitute $$a=c$$, $$b=\\frac{a^2}{d}$$ and $$t=\\frac{a}{b}+\\frac{b}{a}$$ so that $$\\frac{(a+b)(c+d)}{ a+b+c+d}=\\frac{abc+abd+acd+bcd}{(a+b)(c+d)}=a$$ $${(a+c)(b+d)\\over a+b+c+d}=a\\cdot{2t\\over2+t}$$ and $${abc+abd+acd+bcd\\over(a+c)(b+d)}=a\\cdot{2+t\\over2t}$$ If we substitute these results into $$(1)$$, we find $$2f(a)=f\\left(a\\cdot{2t\\over2+t}\\right)+f\\left(a\\cdot{2+t\\over2t}\\right)\\tag{2}$$ It can be seen that $$t\\geq 2$$ by AMGM and thus $$\\frac{2t}{2+t}\\geq 1$$. Hence for all pairs $$x\\geq y$$ there exist suitable $$a$$ and $$t$$ so that $$a\\cdot{2t\\over2+t}=x,a\\cdot{2+t\\over2t}=y\\text{ and } \\sqrt{xy}=a$$ so that $$(2)$$, and thus the required property, holds.", "just: $$3a^2b^2 + 3a^2c^2 + 3b^2c^2 \\ge 3a^2bc + 3b^2ac + 3c^2ab$$ By AM-GM one has $\\frac{1}{2}a^2b^2 + \\frac{1}{2}a^2c^2 \\ge a^2bc$. From applying this a bunch one easily gets the desired result. I started this proof but then realized that there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.) \\begin{align} S(a,b,c) & = \\dfrac{a^3}{a+b} + \\dfrac{b^3}{b+c} + \\dfrac{c^3}{c+a}\\\\ & = \\dfrac{a^3+b^3}{a+b} + \\dfrac{b^3+c^3}{b+c} + \\dfrac{c^3+a^3}{c+a} - \\left(\\dfrac{b^3}{a+b} + \\dfrac{c^3}{b+c} + \\dfrac{a^3}{c+a} \\right)\\\\ & = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b) \\end{align} Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \\geq ab + bc+ ca$$ If $S(a,b,c) < \\dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \\dfrac{ac + cb+ ba}2$ which gives us $$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction. Let $$M=\\dfrac{a^3}{a+b}+\\dfrac{b^3}{b+c}+\\dfrac{c^3}{c+a}$$ be the original expression, we introduce its \"conjugate\" $$N=\\dfrac{ab^2}{a+b}+\\dfrac{bc^2}{b+c}+\\dfrac{ca^2}{c+a}$$ Direct computing, and using $a^2+b^2+c^2\\geq ab+bc+ca$, yields $$M-N =\\frac{a(a^2-b^2)}{a+b}+\\frac{b(b^2-c^2)}{b+c}+\\frac{c(c^2-a^2)}{c+a}\\\\=a(a-b)+b(b-c)+c(c-a)\\\\= a^2+b^2+c^2-ab-bc-ca\\geq 0$$ Therefore $M\\geq N$. Also using $2(x^2+y^2)\\geq (x+y)^2$ $$M+N=\\frac{a(a^2+b^2)}{a+b}+\\frac{b(b^2+c^2)}{b+c}+\\frac{c(c^2+a^2)}{c+a}\\\\ \\geq \\frac{a(a+b)+b(b+c)+c(c+a)}{2} \\\\ =\\frac{a^2+b^2+c^2+ab+bc+ca}{2} \\\\ \\geq ab+bc+ca$$ Hence we have both $M\\geq N$ and $M+N\\geq ab+bc+ca$. It follows that $M\\geq \\dfrac{ab+bc+ca}{2}$. By Holder $$\\sum_{cyc}\\frac{a^3}{a+b}\\geq\\frac{(a+b+c)^3}{3\\sum\\limits_{cyc}(a+b)}=\\frac{(a+b+c)^2}{6}\\geq\\frac{ab+ac+bc}{2}.$$ Done! The Holder for three", ". . . $(b+c)\\big[a^2 + ab + ac + bc\\big] \\;=\\;0$ . . . . . . . . . . $(b+c)\\big[a(a+b) + c(a+b)\\big] \\;=\\;0$ Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \\;=\\;0$ We see that the product of three factors equals zero. Hence, at least one of the factors must be zero. Therefore: . $\\begin{Bmatrix}a+b \\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}b \\\\ & \\text{or} \\\\ b+c \\:=\\:0 & \\Rightarrow & b \\:=\\:\\text{-}c \\\\ & \\text{or} \\\\ a+c\\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}c \\end{Bmatrix}$ 3. ## Re: Opposite number proof Originally Posted by Soroban Hello, DIOGYK! I think I got it . . . We have: . $\\frac{1}{a+b+c} \\;=\\;\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ . . . . . . . . $\\frac{1}{a+b+c} \\;=\\;\\frac{bc + ac + ab}{abc}$ . . . . . . . . . . . . $abc \\;=\\;(a+b+c)(bc+ac+ab)$ . . . . . . . . . . . . $abc \\;=\\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$ . . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \\;=\\;0$ n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \\;=\\;0$ . . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \\;=\\;0$ Factor: .", "= \\dfrac{BH}{BC}$and$\\dfrac{FH}{AB} = \\dfrac{HC}{BC} .$ Adding these equations yields: \\begin{align*}\\dfrac{FH}{EC} + \\dfrac{FH}{AB} &= \\dfrac{BH}{BC} + \\dfrac{HC}{BC}\\\\ &= \\frac{BH+HC}{BC} \\\\&= \\frac{BC}{BC} \\\\&= 1.\\end{align*} This, in turn, goes to show that $\\frac{1}{EC} + \\frac{1}{AB} = \\frac{1}{FH} .$ Now, let $$s$$ be the side length of the square. We know $$AB = 2\\cdot EC = s.$$ This means \\begin{align*}\\dfrac{1}{FH} &= \\dfrac{1}{EC} + \\frac{1}{AB} \\\\&= \\frac{1}{a} + \\frac{2}{a} \\\\&= \\frac{3}{a} .\\end{align*} Therefore, $$FH = \\frac{s}{3} .$$ Now, to compute the area of $$\\triangle EFC$$, we take the area of $$\\triangle BCE$$ and subtract the area of $$\\triangle BFC$$. This is equal to \\begin{align*} \\dfrac{BC\\cdot EC}{2} - \\dfrac{BC\\cdot FH}{2} &= \\dfrac{BC\\cdot(EC-FH)}{2} \\\\ &= \\dfrac{s\\cdot\\left(\\dfrac{s}{2} - \\dfrac{s}{3}\\right)}{2} \\\\ &= \\dfrac{s\\cdot\\frac{s}{6}}{2} \\\\ &= \\dfrac{s^2}{12}. \\end{align*} The area of $$AFED$$ is the area of $$\\triangle ACD$$ minus the area of $$\\triangle EFC$$, which is equal to\\begin{align*} \\dfrac{s^2}{2}-\\dfrac{s^2}{12} &= \\dfrac{5s^2}{12} \\\\&= 45.\\end{align*} With $$\\frac{5}{12} s^2 = 45$$, we get $$s^2 = 108,$$ which is the area of the full square. 23. From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? $$\\dfrac{2}{7}$$ $$\\dfrac{5}{42}$$ $$\\dfrac{11}{14}$$ $$\\dfrac{5}{7}$$ $$\\dfrac{6}{7}$$ ###### Solution(s): Without loss of generality, allow $$A$$ to be a vertex of", "we can write $$D=D^\\prime + |DD^\\prime| \\frac{( C-B )\\times ( B \\times C )}{|BC|\\,|B\\times C|} \\tag{1}$$ where $$D^\\prime := \\frac12(B+C)$$ is the midpoint of $$\\overline{BC}$$. Further, by the Inscribed Angle Theorem, $$\\angle BOC\\cong\\angle BDC$$, and we conclude that $$\\overline{DD^\\prime}$$ is the altitude of an isosceles triangle with vertex angle $$\\alpha$$ and base $$|BC|$$. Therefore, $$|DD^\\prime| = \\frac12|BC|\\,\\cot\\frac12\\alpha$$, so we have $$\\frac{|DD^\\prime|}{|BC|\\,|B\\times C|} = \\frac{\\cot\\frac12\\alpha}{2bc\\sin\\alpha} = \\frac{\\cos\\frac12\\alpha\\,/\\,\\sin\\frac12\\alpha}{4bc\\sin\\frac12\\alpha\\cos\\frac12\\alpha}=\\frac{1}{4bc\\sin^2\\frac12\\alpha} = \\frac{1}{2bc(1-\\cos\\alpha)} \\tag{2}$$ Further, via a cross product identity, \\begin{align} (C-B)\\times(B\\times C) &= \\phantom{-}B\\,((C-B)\\cdot B) - C\\,((C-B)\\cdot C) \\\\ &=\\phantom{-}B\\left(B\\cdot C - |B|^2\\right)-C\\left(|C|^2 - B\\cdot C\\right) \\\\ &=-B\\,b(b-c\\cos\\alpha) - C\\,c(c-b\\cos\\alpha) \\end{align}\\tag{3} Altogether, this gives us \\begin{align}D\\;2bc(1-\\cos\\alpha) &= (B+C)bc(1-\\cos\\alpha)-B\\,b(b-c\\cos\\alpha)-C\\,c(c-b\\cos\\alpha) \\\\ &=Bb(c-b)+Cc(b-c) \\\\ &=(c-b)(B b-C c) \\tag{4a} \\end{align} (Note that, if $$b=c$$, then $$D=0$$, as we would expect. This confirms that we got our cross-product vector directions correct in $$(1)$$.) Likewise, \\begin{align} E\\,2ca(1-\\cos\\beta) &= (a-c)(C c-A a) \\tag{4b} \\\\ F\\,2ab(1-\\cos\\gamma) &= (b-a)(A a-B b) \\tag{4c} \\end{align} Finally, observe that, for $$a$$, $$b$$, $$c$$ not all equal (the only case with which we need concern ourselves), $$(\\text{eq }4a)\\;(a-c)(b-a) + (\\text{eq }4b)\\;(c-b)(b-a)+(\\text{eq }4c)\\;(a-c)(c-b) \\tag{5}$$ gives a non-trivial linear combination of $$D$$, $$E$$, $$F$$ that vanishes. Consequently, $$D$$, $$E$$, $$F$$ are linearly dependent; that is, they lie on a common plane through $$O$$, and the result follows. $$\\square$$ • Pardon my ignorance but I'm confused as to", "there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.) \\begin{align} S(a,b,c) & = \\dfrac{a^3}{a+b} + \\dfrac{b^3}{b+c} + \\dfrac{c^3}{c+a}\\\\ & = \\dfrac{a^3+b^3}{a+b} + \\dfrac{b^3+c^3}{b+c} + \\dfrac{c^3+a^3}{c+a} - \\left(\\dfrac{b^3}{a+b} + \\dfrac{c^3}{b+c} + \\dfrac{a^3}{c+a} \\right)\\\\ & = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b) \\end{align} Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \\geq ab + bc+ ca$$ If $S(a,b,c) < \\dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \\dfrac{ac + cb+ ba}2$ which gives us $$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction. Note that using the AM-GM inequality, we have $$\\frac{a^3}{a+b}=a^2-\\frac{a^2b}{a+b}\\geq a^2-\\frac{a^2b}{2\\sqrt{ab}}=a^2-\\frac 12\\sqrt{a^3b},$$ And so summing up cyclically, it is enough to check that $$2(a^2+b^2+c^2)\\geq\\sqrt{a^3b}+\\sqrt{b^3c}+\\sqrt{c^3a}+ab+bc+ca,$$ Which is obviously true since $\\sqrt{a^3b}\\leq \\dfrac{a^2+ab}2$ and $ab+bc+ca\\leq a^2+b^2+c^2$ from the AM-GM inequality. Equality holds iff $a=b=c.$ $\\Box$", "# Leo Giugiuc's Second Lemma And Applications ### Proof 1 \\begin{align} 2(m_a+m_b) &= |\\overrightarrow{AB}+\\overrightarrow{AC}|\\lt |\\overrightarrow{AB}|+|\\overrightarrow{AC}|+|\\overrightarrow{BA}|+|\\overrightarrow{BC}|\\\\ &= a+b+2c < (a+b+2c)+2(a+b-c)\\\\ &= 3(a+b). \\end{align} ### Proof 2 The required inequality is equivalent to $\\displaystyle\\frac{2}{3}(m_a+m_b)\\le a+b.\\,$ Geometrically, this becomes $AG+BG\\le AC+BC.$ Let $A',B\\,$ be the midpoints of $BC\\,$ and $AC,\\,$ respectively. In $\\Delta BGA',\\,$ $BG\\lt GA'+A'B\\,$ so that $AG+BG\\lt AG+GA'+A'B=AA'+BB'.$ In $\\Delta AA'C,\\,$ $AA'\\lt A'C'+AC\\,$ so that $AG+BG\\lt AA'+BB'\\lt A'C+AC+A'B=BC+AC.$ ### Proof 3 Let, for a point $X,\\,$ $x\\,$ denotes the corresponding complex numbers. $\\displaystyle g=\\frac{a+b+c}{3}.$ \\displaystyle\\begin{align} AG+BG &= \\left|a-\\frac{a+b+c}{3}\\right|+\\left|b-\\frac{a+b+c}{3}\\right|\\\\ &=\\left|\\frac{a-b+a-c}{3}\\right|+\\left|\\frac{b-a+b-c}{3}\\right|\\\\ &\\le\\frac{|a-b|+|a-c|}{3}+\\frac{|b-a|+|b-c|}{3}\\\\ &=\\frac{2\\cdot AB+AC+BC}{3}\\\\ &=\\frac{2\\cdot (AC+BC)+(AC+BC)}{3}\\\\ &\\le AC+BC. \\end{align} ### Proof 4 This is just a follow-up on the starting point of Proof 2. The required inequality is equivalent to $AG+BG\\lt AC+BC,\\,$ i.e. $AB+AG+BG\\lt AB+AC+BC.\\,$ But, in general, given two convex polygons, one within the other, the perimeter of the outer polygon exceeds the permiter of the inner one. The lemma then follows from the fact that $\\Delta ABG\\,$ is within $\\Delta ABC.\\,$ ### Proof 5 Lemma is a special case of the following assertion: If $M\\,$ is an interior point of $\\Delta ABC,\\,$ then $MB + MC < AB + AC.$ To prove that, consider the ellipse whose focii are $B\\,$ and $C\\,$ and the major axis is $AB + AC.\\,$ Point $M\\,$ being inside $\\Delta ABC,\\,$ it is also", "\\\\ 0 &\\ge& (bc-1)+(ac-1)+(ab-1) \\\\ 0 &\\ge& bc+ac+ab-3 \\\\ 3 &\\ge& bc+ac+ab \\\\ \\end{array}$ 11. experimentX Group Title Here's a short proof hint ... i had written earlier. \\begin{align*} 3(a^2+b^2+c^2) & = a^2 + b^2 + c^2 + (a^2+b^2) + (b^2 + c^2) + (c^2+a^2) \\\\ & \\ge a^2 + b^2 +c^2 + 2 a b + 2 b c + 2 c a \\\\ & = (a+b+c)^2 \\\\ & = \\frac{2(a^2 + b^2 +c^2)}{2} + 2 a b + 2 b c + 2 ca \\\\ & \\ge \\frac{ 2 a b + 2 b c + 2 ca }{2} + 2 a b + 2 b c + 2 ca \\\\ & = 3(a+b+c) \\end{align*} 12. experimentX Group Title the other hint would to be make use of AM-GM inequality for 3 variables. 13. experimentX Group Title $3(a^2+b^2+c^2) \\ge (a+b+c)^2 \\\\ 3 = 3 \\sqrt[3]{abc} \\le (a+b+c)$ Dividing will do the job!! 14. sauravshakya Group Title (a+b+c)/3 >= (abc)^1/3 a+b+c>=3 Now, a^2+b^2+c^2=(a+b+c)^2 -2ab -2bc-ca SO, a^2+b^2+c^2>=(a+b+c)^2 .....i Since, a+b+c>=3 (a+b+c)^2 >=a+b+c ....ii Thus, from i and ii, a^2+b^2+c^2 >=a+b+c", "1$. Prove that: $$\\sqrt{(a^2+b^2+2abx)(c^2+d^2+2cdx)}+\\sqrt{(a^2+d^2-2adx)(b^2+c^2-2bcx)}\\le 2\\sqrt{5}$$ (Note: here, $x$ is taken to be the nonnegative cosine of the angle between the two diagonal, WLOG supposing that the angle between the segments with length $a$ and $b$ is acute.) First, we manipulate the expressions under the radicals, call them $L$ and $R$: \\begin{align*} L&=(a^2+b^2+2abx)(c^2+d^2+2cdx) \\\\ &=(a^2+b^2)(c^2+d^2)+2x(ab(c^2+d^2)+cd(a^2+b^2))+4x^2abcd \\\\ &=(ac-bd)^2+(ad+bc)^2+2x(ad+bc)(ac+bd)+x^2[(ac+bd)^2-(ac-bd)^2] \\\\ &=[x(ac+bd)+(ad+bc)]^2+(1-x^2)(ac-bd)^2 \\end{align*} In a similar manner, we see that: $$R=[x(ac+bd)-(ab+cd)]^2+(1-x^2)(ac-bd)^2$$ Now let $p=x(ac+bd)+\\frac{(ad+bc)-(ab+cd)}{2}$ and $q=|ac-bd|\\sqrt{1-x^2}$. Using the fact that $\\frac{ad+bc+ab+cd}{2}=\\frac{(a+c)(b+d)}{2}=2$, our inequality can be rewritten as: $$\\sqrt{(p+2)^2+q^2}+\\sqrt{(p-2)^2+q^2}\\le 2\\sqrt{5}$$ We will now prove the key lemma: Lemma: $p^2+5q^2\\le 5$ Proof: Using the fact that $ad+bc-(ab+cd)=(a-c)(d-b)=4(a-1)(b-1)$, this inequality can be rewritten as: $$5[a(2-a)-b(2-b)]^2(1-x^2)+[x(a(2-a)+b(2-b))+2(a-1)(b-1)]^2\\le 5$$ Upon the translation $a\\Rightarrow 1+y, b\\Rightarrow 1+z$ for $y, z\\in [-1, 1]$, this simplifies to: y, z\\ge 9$$5(y^2-z^2)^2(1-x^2)+[x(2-y^2-z^2)^2-2yz]^2\\le 5$$ $$\\Leftrightarrow 4x^2(1+3y^2z^2-y^2-y^4-z^2-z^4)+4xyz(2-y^2-z^2)+5y^4+5z^4-6y^2z^2\\le 5$$ Let this expression be $F(x, y, z)$. From this formulation, we see that since $F(x, y, z)\\le F(x, |y|, |z|)$, we can say $y, z\\ge 0$ WLOG. Now, we take two cases: Case 1: $x\\le \\frac{3}{4}$. In this case, note that: \\begin{align*}F(x, 1, yz)-F(x, y, z)&= (5-4x^2)(1-y^4)(1-z^4)-(4x^2+4xyz)(1-y^2)(1-z^2) \\\\ &=(1-y^2)(1-z^2)[(5-4x^2)(1+y^2)(1+z^2)-4x^2-4xyz] \\\\ &\\ge (1-y^2)(1-z^2)[(5-\\frac{9}{4})(1+y^2)(1+z^2)-\\frac{9}{4}-3yz] \\\\ &= (1-y^2)(1-z^2)[\\frac{1}{2}+\\frac{11}{4}(y-z)^2+\\frac{5}{2}yz+\\frac{11}{4}y^2z^2] \\\\ &\\ge 0 \\end{align*} It follows that it suffices to prove the inequality for $z=1$. But we see that: \\begin{align*}F(x, y, 1)&=4x^2(-y^4+2y^2-1)+4xy(1-y^2)+5y^4-6y^2+5 \\\\ &=5-5y^2(1-y^2)-[2x(1-y^2)-y]^2 \\\\ &\\le 5 \\end{align*} And so this case", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "\\;+\\; (-ab + ba)i ~=~ ( a^2 + b^2 ) \\;+\\; 0i\\\\ \\nonumber &=~ a^2 \\;+\\; b^2 \\end{align*} The sixth item comes from using the previous items: \\nonumber \\begin{align*} \\dfrac{a+bi}{c+di} ~&=~ \\dfrac{a+bi}{c+di} \\,\\cdot\\, \\dfrac{c-di}{c-di}\\\\ \\nonumber &=~ \\dfrac{(ac - b(-d)) \\;+\\; (a(-d) + bc)i}{c^2 + d^2}\\\\ \\nonumber &=~ \\dfrac{(ac+bd) \\;+\\; (bc-ad)i}{c^2 + d^2} \\end{align*} The conjugate $$\\overline{a+bi}$$ of a complex number $$a+bi$$ is defined as $$\\overline{a+bi} = a-bi$$. Notice that $$(a+bi) \\;+\\; \\overline{(a+bi)} ~=~ 2a$$ is a real number, $$(a+bi) \\;-\\; \\overline{(a+bi)} ~=~ 2bi$$ is an imaginary number if $$b \\ne 0$$, and $$(a+bi) \\overline{(a+bi)} ~=~ a^2 + b^2$$ is a real number. So for a complex number $$z=a+bi$$, $$z\\,\\overline{z} = a^2 + b^2 \\,$$ and thus we can define the modulus of $$z$$ to be $$\\sqrt{z\\,\\overline{z}} = \\sqrt{a^2 + b^2}$$, which we denote by $$|z|$$. Example 6.9 Let $$z_1 = -2+3i$$ and $$z_2 = 3+4i$$. Find $$z_1 + z_2$$, $$z_1 - z_2$$, $$z_1 \\, z_2$$, $$z_1 / z_2$$, $$|z_1|$$, and $$|z_2|$$. Solution Using our rules and definitions, we have: \\nonumber \\begin{align*} z_1 \\;+\\; z_2 ~&=~ (-2+3i) \\;+\\; (3+4i)\\\\ \\nonumber &=~ 1 + 7i\\\\ \\nonumber z_1 \\;-\\; z_2 ~&=~ (-2+3i) \\;-\\; (3+4i)\\\\ \\nonumber &=~ -5 - i\\\\ \\nonumber z_1 \\, z_2 ~&=~ (-2+3i)\\, (3+4i)\\\\ \\nonumber &=~ ((-2)(3) - (3)(4)) \\;+\\; ((-2)(4) + (3)(3))i\\\\ \\nonumber &=~ -18 + i\\\\ \\nonumber", "a solution to your equation This is just reversing Part 1. Assume $a^2+b^2=c^2$ and $a^2+2b^2=d^2$. Note that \\begin{align}(dc-ab)(dc+ab)&=d^2c^2-a^2b^2\\\\ &=a^4+2a^2b^2 +2b^4\\\\ &=b^2(a^2+2b^2)+a^2(a^2+b^4)\\\\ &=b^2d^2+a^2c^2\\end{align}\\tag{1} Let $$x=\\frac{bd(cd+ab)}{ac(cd-ab)}\\\\y=\\frac{ac(cd+ab)}{bd(cd-ab)}$$ Then you'll get, with help from (1) that: $$x+y=\\frac{(cd+ab)^2}{abcd}$$ and $$\\frac{1}{x}+\\frac{1}{y}=\\frac{(cd-ab)^2}{abcd}$$ So $$x-\\frac{1}{x}+y-\\frac{1}{y}=4$$ ## Part 3: Another equivalent The requirement that $a^2+b^2$ and $a^2+2b^2$ is equivalent to the requirement that $c^4-b^4$ is a perfect square for some relatively prime $b,c$ with $b$ even (and $b\\neq 0$ in both sides.) If $c^4-b^4=k^2$ then $c^2-b^2$ and $c^2+b^2$ are relatively prime and hence squares, and hence $a^2=c^2-b^2$ and $c^2=a^2+b^2$ and $c^2+b^2=a^2+2b^2$ are both squares.", "# If $a+b+c=0$ prove that $\\frac{a^2+b^2+c^2}{2} \\times \\frac{a^3+b^3+c^3}{3} = \\frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$, for $a,b,c \\in\\mathbb R$, prove $$\\frac{a^2+b^2+c^2}{2} \\times \\frac{a^3+b^3+c^3}{3} = \\frac{a^5+b^5+c^5}{5}$$ I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in. Is there a more elegant way to prove this, or a way to find the right \"trick\"? Substituting $$c=-a-b$$ in $$\\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{6}$$ we obtain $$-(a^2+ab+b^2)ab(a+b)$$ and so is the right-hand side $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)$$ $$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc$$ \\begin{align*} a^5 + b^5 + c^5 & = (a + b + c)(a^4 + b^4 + c^4) - (ab + bc + ca)(a^3 + b^2 + c^3) + abc(a^2 + b^2 + c^2) = \\\\ & = -3abc(ab + bc + ca) - 2abc(ab + bc + ca) = -5abc(ab + bc + ca) \\end{align*} So both sides equal $-abc(ab + bc + ca)$. Let' s try like this: Suppose $a,b,c$ are the roots of the polynomial $x^3-mx-n$. By Vieta's formula's we have: $$a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca)= 0-2m = 2m$$ Since $x^3 =mx+n \\;\\;\\;\\;(*)$ we have: $$x^5 = x^2(mx+n) = mx^3+nx^2= m^2x+mn+nx^2$$ so $$a^5+b^5+c^5 = m^2(a+b+c)+3mn+n(a^2+b^2+c^2)" ]

[ "## College Algebra (11th Edition) If $f$ and $g$ are inverses, $(f\\circ{g})(x)=x$. $(f\\circ{g})(x)=2(.5x-2)+4$ $x-4+4=x$ Yes.", "## College Algebra 7th Edition $f^{-1}(0)=1$ $f^{-1}(4)=3$ We know that $f$ does have an inverse because it passes the horizontal line test (is one-to-one). We know that $f(1)=0$. Therefore, $f^{-1}(0)=1$. Similarly, we know that $f(3)=4$. Therefore, $f^{-1}(4)=3$.", "find $f(f(f(1)))$. – Jonas Meyer Jan 11 at 5:40 Let us suppose the inverse of $f$ is $g$, then $g(1)=3$, $g(2)=1$, $g(3)=2$ and $g(s)=s$ for all $s \\in S$.", "$a\\in G$ has an inverse $a^{-1} \\in G$. This will help a lot. Ok, we know $a,b,c \\in G$ $$b = e∘b = (a^{-1}∘a)∘b = a^{-1}∘(a∘b)=a^{-1}∘(a∘c) = (a^{-1}∘a)∘c = c$$ By the group properties each element has an inverse. So you can just multiply your equation on the left by $a^{-1}$. $$a\\circ b=a\\circ c$$ on the left by the inverse of $a$ to get the desired result.", "How do you find (f*g)(x) and (g*f)(x) and determine if the given functions are inverses of each other f(x) = x^2 − 3 and g(x) = sqrtx+3? Jan 24, 2017 To find $f \\left(g \\left(x\\right)\\right) \\mathmr{and} g \\left(f \\left(x\\right)\\right)$, substitute, g(x) for x in f(x), and f(x) for x in g(x), respectively. If they are inverses, both substitutions will equal x. Explanation: $f \\left(g \\left(x\\right)\\right) = {\\left(\\sqrt{x} + 3\\right)}^{2} - 3$ $f \\left(g \\left(x\\right)\\right) = x + 6 \\sqrt{x} + 9 - 3$ $f \\left(g \\left(x\\right)\\right) = x + 6 \\sqrt{x} + 6$ $g \\left(f \\left(x\\right)\\right) = \\sqrt{{x}^{2} - 3} + 3$ They are not inverses, because both cases must reduce to x.", "the inverse of $a$ in $G$. Let $y = {a^{ – 1}}$ such that ${a^{ – 1}}a = e$, then ${a^{ – 1}} \\in G$ as $ya = e$ has a solution in $G$. Thus ${a^{ – 1}}$ is the left inverse of $a$ in $G$. Therefore each element of possesses a left inverse. Hence $G$ is a group for the given composition if the postulate and (ii) are satisfied.", "Let $g(x)$ be the inverse of $f(x)=4\\tan^{-1}\\!\\left(x\\right)$. Calculate $g(1)$, without finding a formula for $g(x)$, and then calculate $g'(1)$. $g(1)$ = $g'(1)$ =", "this now for $g$ to get $$g(x) = \\frac{x-1}{2}.$$ We can now check to see that indeed $f(g(x)) = x$ and $g(f(x))=x$. Thus the $g$ that we found is indeed the inverse of $f$.", "= g(x)*g(y) = (e^x + 1)*(e^y+1) = e^xy + 1$ does $g(x)$ have an inverse? $g^{-1} = \\ln (x -1)$", "a right inverse. particularly $x_tx_u = 1,$ for some $x_u \\in G.$ but then $x_u=x_sx_tx_u=x_s.$ hence $x_tx_s=1$ and the proof is complete.", "g_4)$ is the only right inverse of $f$. So $g_1 = g_2 = g_3 = g_4$ are all the same. By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$. $\\blacksquare$", "# Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Question: Consider $t:\\{1,2,3\\} \\rightarrow\\{a, b, c\\}$ given by $t(1)=a, t(2)=b$ and $t(3)=c$. Find $f^{-1}$ and show that $\\left(f^{-1}\\right)^{-1}=f$. Solution: Function $f:\\{1,2,3\\} \\rightarrow\\{a, b, c\\}$ is given by, $f(1)=a, f(2)=b$, and $f(3)=c$ If we define $g:\\{a, b, c\\} \\rightarrow\\{1,2,3\\}$ as $g(a)=1, g(b)=2, g(c)=3$, then we have: Thus, the inverse of $f$ exists and $f^{-1}=g$. $\\therefore f^{-1}:\\{a, b, c\\} \\rightarrow\\{1,2,3\\}$ is given by, $f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3$ Let us now find the inverse of $f^{-1}$ i.e., find the inverse of $g$. If we define $h:\\{1,2,3\\} \\rightarrow\\{a, b, c\\}$ as $h(1)=a, h(2)=b, h(3)=c$, then we have: Thus, the inverse of $g$ exists and $g^{-1}=h \\Rightarrow\\left(f^{-1}\\right)^{-1}=h$. It can be noted that $h=f$. Hence, $\\left(f^{-1}\\right)^{-1}=f$.", "When we have such a $g$, we usually denote it by $f^{-1}$. So in your case $f(x) = x^{n^{-1}}$ where $n^{-1}$ is the inverse of $n$ in $\\mathbb{Z} / 288\\mathbb{Z}$ This relies in part on $n$ satisfying $(n, 6) = 1$. - No, it works because the modulus is squarefree. It fails otherwise - see my answer. – Bill Dubuque Mar 10 '14 at 16:27", "the inverse of f exists and $f^{-1} = g$ Now, $f^{-1} : \\left \\{ a,b,c \\right \\}\\rightarrow \\left \\{ 1,2,3 \\right \\}$ is given by $f^{-1}(a) = 1, f^{-1}(b) = 2 \\ and \\ f^{-1}(c) = 3$ Now, we need to find the inverse of $f^{-1}$, Therefore, lets define$h: \\left \\{ 1,2,3 \\right \\}\\rightarrow \\left \\{ a,b,c \\right \\}$ such that $h(1) = a, h(2) = b \\ and \\ h(3) = c$ Now, $(goh)(1) = g(h(1)) = g(a) = 1$ $(goh)(2) = g(h(2)) = g(b) = 2$ $(goh)(3) = g(h(3)) = g(c) = 3$ Similarly, $(hog)(a) = h(g(a)) = h(1) = a$ $(hog)(b) = h(g(b)) = h(2) = b$ $(hog)(c) = h(g(c)) = h(3) = c$ Hence, $goh = I_X$ and $hog = I_Y$, where$X = \\left \\{ 1,2,3 \\right \\}$ and$Y = \\left \\{ a,b,c \\right \\}$ Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$ $\\Rightarrow h = f$ Therefore, $(f^{-1})^{-1} = f$ Hence proved $f : X \\rightarrow Y$ To prove: $(f^{-1})^{-1} = f$ Let $f:X\\rightarrow Y$ be a invertible function. Then there is $g:Y\\rightarrow X$ such that $gof =I_x$ and $fog=I_y$ Also, $f^{-1}= g$ $gof =I_x$ and $fog=I_y$ $\\Rightarrow$ $f^{-1}of = I_x$ and $fof^{-1} = I_y$ Hence, $f^{-1}:Y\\rightarrow X$ is invertible function and f is inverse of $f^{-1}$. i.e.", "for every $$x \\in X$$ and hence $$K_{\\mathbf{h}}^{k!} = \\mathbf{id}_X$$. By the previous lemma, $$K^{k!-1}_{\\mathbf{h}}=K_{\\mathbf{h}^{k!-1}}$$ is the inverse of $$K_{\\mathbf{h}}$$.", "Notice that $f^{-1}(f(x)) = x$. So if f coincides with its inverse then $f(f(x)) = x$. Calculate f(f(x)) and see where that leads you.", "# The inverse of the inverse in a group In one of my math exercises, I'm being asked to prove that for all $a, b \\in G | (a^{-1})^{-1} = a$ with G a group. However, nowhere is stated that it is a commutative group. My first thought was that, because inverses are unique, the inverse of $a^{-1}$ can only be $a$; This thought is correct according to the answers. However, if the group is not commutative, is it not possible that there is another unique inverse for $a$? The exact words of the answer are: The result follow from Theorem 16.1(b) because both $(a^{-1})^{-1}$ and $a$ are inverses of $a^{-1}$. Where Theorem 16.1(b) states The inverse of every element of a group G is unique. If we wanted the group axioms to say For each $g\\in G$ there exists a $g'\\in G$ such that $gg'=1$ then that'd be all we need. Indeed for any $g\\in G$, we have that $gg'=1$, $g'\\in G$. Then we also have some other $g''$ for which $g'g''=1$. But then, by associativity $$g(g'g'')=g$$ $$(gg')g''=g$$ $$1g''=g$$ $$g''=g$$ So left and right inverses (which exist axiomatically) coincide. This said, $a^{-1}$ is just a convenient symbol for the element $g\\in G$ for which $ag=ga=1$. Thus, $(a^{-1})^{-1}$ is a convenient symbol for the element $g'\\in G$ for", "## Intermediate Algebra (6th Edition) $(.027,3)$ We are given that $f(x)=log_{.3}x$ and that $g(x)=.3^{x}$ is the inverse of $f(x)$. We are also told that the ordered pair (3,.027) is a solution of the function $g(x)$. As seen in Section 9.2, when the ordered pair $(x_{1},y_{1})$ is a solution to a function, that function's inverse will have a solution of $(y_{1},x_{1})$. Therefore, the ordered pair $(.027,3)$ will be a solution of $f(x)$.", "element of a group has an inverse, so $$gh = h \\iff ghh^{-1} = h ^{-1} \\iff g = e$$", "of $G$ has order $2$ and is thus self-inverse. It follows directly for all $x,y\\in G$ that $xy=x^{-1}y^{-1}=(yx)^{-1}=yx$. Aug 9 at 19:20" ]

[ "so that $G(x,0)=g_0(x)=1_X(x)=x$ and $G(x,1)=g_1(x)=(\\iota\\circ f_1)(x)$ for all $x\\in X$. Because $\\{f_t\\}_{t\\in I}$ is a deformation retract of $X$ onto the point $x\\in X$, the family $\\{f_t\\}$ can be thought of as a shrinking of $X$ onto the subset $\\{x\\}$, and because the family $\\{f_t\\}$ is continuous in both $t$ and $x$ (by definition of homotopy), it follows that there exists an element $T\\in I$ for which $g_t^{-1}(U)\\subsetneq U$ for $t\\geq T$ and $g_t^{-1}(U)=U$ for $t. Let $V=g_T^{-1}(U)$. It suffices to show that the inclusion $V\\hookrightarrow U$ is homotopic to a constant map. To that end, write $t=kT$, $k\\in[0,1]$, and define $H:U\\times I\\to V$ so that $H(u,k)=h_k(u)=g_{kT}(x)$ for all $x\\in X\\cap U$2. In particular, then, $h_0(u)=g_0(u)=u$ as $g_0=id_X$. Moreover, $h_1(u)=g_T(u)$ for all $u\\in U$, and because $V=g_T^{-1}(U)$, $h_1(U)\\subset V$. Finally, because $f_t$ is a (strong) deformation retract, because $G$ is defined in terms of $f_t$, and because $H$ is defined in terms of $G$, it’s easily confirmed that $h_t(V)\\subset V$ for all $t$ and hence that $h_t$ is a weak deformation retract. Therefore, from the same problem cited above (see here), there exists an inclusion map $j:\\{x\\}\\to V$ for which $h_1\\circ j\\simeq 1_{\\{x\\}}$. In particular, $h_1\\circ j$ is null-homotopic. At this point, all the ingredients are in place. The fact that the family $\\{f_t\\}$ homotopes $1_X\\simeq \\iota\\circ f_1$", "$g(s) f(s) = 0$, then either $g(s) = 0$ or $f(s) = 0$. If we assume that $f(s) \\neq 0$, then we must have $g(s) = 0$. For the case where $f= x+iy, g= u+iv$ are constants, then $fg = 0$ if and only if $f=0$ and/or $g=0$. Introducing complex coordinates doesn't change this fact. This is because the values must satisfy $ux-vy = 0$ AND $uy + vx = 0$, and so $0 = (ux-vy)^2 + (uy+vx)^2 = (u^2 + v^2)(x^2 + y^2)$, which implies that $u^2 +v^2 = 0$ or $x^2 + y^2 = 0$, which implies that $(u,v)=(0,0)$ or $(x,y) = (0,0)$. - Now, as you note, if $(x + i y)(u + i v) = 0$ then this implies that $x u - v y = x v + u y = 0$. However, the only way these equations can hold is if either $x + i y = 0$ or $u + i v = 0$. Multiplying the first equation through by $v$ and the second by $u$, we have $x u v - v^2 y = 0$ and $x u v + u^2 y = 0$. Subtracting, $(u^2 + v^2)y = 0$. Since $u, v, y \\in \\mathbb{R}$, this means that either $y = 0$ or $u = v = 0$. In", "$v(f)=0$ for all $f\\in\\mathfrak m$. As $v(k)=0$ (because $v(1)=v(1)1+1v(1)$ implies that $v(1)=0$), we have $v=0$. 2. Surjectivity: let $\\theta : \\mathfrak m/\\mathfrak m^2\\to k$ be a linear form. Define $v: A\\to k$ by $$v(f)=\\theta(\\overline{f-f(0)}).$$ Then $v$ is clearly $k$-linear. If $f, g\\in A$, we write $f=c_1+f_1$, $g=c_2+g_1$ with $c_i\\in k$ and $f_1, g_1\\in \\mathfrak m$. Then $fg=c_1c_2+c_1g_1+c_2f_1+f_1g_1$ and $$v(fg)=\\theta(c_1\\bar{g}_1+c_2\\bar{f}_1)=c_1v(g)+c_2v(f).$$ This shows that $v$ is a $k$-derivation of $A$. Finally, by construction, $\\psi(v)=\\theta$.", "values less than x = -1 or greater than x = 1 otherwise the value under the radical is negative. ## Example 2 Recall that $(f \\circ g)(x)$ does not necessarily equal $(g \\circ f)(x)$. Find an example where $(f \\circ g)(x) = (g \\circ f)(x)$. Suppose that $f$ and $g$ are defined for all reals such that $f(x) := x$ and $g(x) := x + 1$. Then $(f \\circ g)(x) = f(g(x)) = f(x + 1) = x + 1$. Furthermore $(g \\circ f)(x) = g(f(x)) = g(x) = x + 1$. In this case, $(f \\circ g)(x) = (g \\circ f)(x)$.", "$V$ satisfies similar problem albeit with right-hand expression $\\varsigma F(-x,t)$, and initial functions $\\varsigma G(-x)$ and $\\varsigma H(-x)$. Proof. Plugging $V$ into equation we use the fact that $V_t(x,t)=\\varsigma U_t (-x,t)$, $V_x(x,t)=-\\varsigma U_x (-x,t)$, $V_{tt}(x,t)=\\varsigma U_{tt} (-x,t)$, $V_{tx}(x,t)=-\\varsigma U_{tx} (-x,t)$, $V_{xx}(x,t)=\\varsigma U_{xx} (-x,t)$ etc and exploit the fact that wave equation contains only even-order derivatives with respect to $x$. Note that if $F,G,H$ are even functions with respect to $x$, and $\\varsigma=1$ then $V(x,t)$ satisfies the same IVP as $U(x,t)$. Similarly, if $F,G,H$ are odd functions with respect to $x$, and $\\varsigma=-1$ then $V(x,t)$ satisfies the same IVP as $U(x,t)$. However we know that solution of (\\ref{eq-2.6.28})-(\\ref{eq-2.6.30}) is unique and therefore $U(x,t)=V(x,t)=\\varsigma U(-x,t)$. Therefore Corollary 1. 1. If $\\varsigma=-1$ and $F,G,H$ are odd functions with respect to $x$ then $U(x,t)$ is also an odd function with respect to $x$. 2. If $\\varsigma=1$ and $F,G,H$ are even functions with respect to $x$ then $U(x,t)$ is also an even function with respect to $x$. However we know that a. odd function with respect to $x$ vanishes as $x=0$; b. derivative of the even function with respect to $x$ vanishes as $x=0$ and we arrive to Corollary 2. 1. In the framework of Proposition 1a $U$ satisfies (\\ref{eq-2.6.27}); 2. In the framework of Proposition 1b $U$ satisfies \\ref{eq-2.6.27-'}. Therefore Corollary 3.", "(12)$ and so $a = -2$. Now, we easily deduce that $f(0) = (-2) \\cdot (-1) \\cdot (-5) \\cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = \\boxed{072}$, which is our answer. ## Solution 4 The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\\sqrt{g(0)+144}=\\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\\neq 0$ (else $f$ is not cubic) where $\\{q,r,s,t,u,v\\}$ is the same as the set $\\{1,2,3,5,6,7\\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that $$24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)$$ which must hold for all $x$. Since $a\\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\\geq 12$ with only 2 elements. If 1", "Substituting ${v}_{i}-g{v}_{i}={v}_{i}^{-}-g{v}_{i}^{-}$ and ${v}_{1}^{-}={c}_{2}{v}_{2}^{-}+{c}_{3}{v}_{3}^{-}$ then yields $a(v3,v1) (gv2-v2)+ a(v2,v3) (gv1-v1)+ a(v1,v2) (gv3-v3) = a(v3-,v1-) (gv2--v2-)+ a(v2-,v3-) (gv1--v1-)+ a(v1-,v2-) (gv3--v3-)= 0,$ and so (1.7) holds. (c) Let ${v}_{3}\\in {V}^{g}$ and ${v}_{2}\\in V\\text{.}$ If ${v}_{2}\\in {V}^{g},$ then ${a}_{g}\\left({v}_{2},{v}_{3}\\right)\\left(g{v}_{1}-{v}_{1}\\right)=0$ for all ${v}_{1}\\in V$ by (1.7). Since ${V}^{g}\\ne V,$ there is some ${v}_{1}$ such that $g{v}_{1}\\ne {v}_{1}$ and so ${a}_{g}\\left({v}_{2},{v}_{3}\\right)=0\\text{.}$ If ${v}_{2}\\notin {V}^{g},$ let ${v}_{1}=\\sum _{k=1}^{r-1}{g}^{k}{v}_{2},$ where $r$ is the order of $g\\text{.}$ By (1.6), ${a}_{g}\\left({v}_{3},{g}^{k}{v}_{2}\\right)={a}_{g}\\left({g}^{-k}{v}_{3},{v}_{2}\\right)={a}_{g}\\left({v}_{3},{v}_{2}\\right),$ for any $k,$ and so $0 = ag(v3,v1) (gv2-v2)+ ag(v2,v3) (gv1-v1) = (r-1) ag(v3,v2) (gv2-v2)+ ag(v3,v2) (gv2-v2)=r ag(v3,v2) (gv2-v2).$ Thus ${a}_{g}\\left({v}_{3},{v}_{2}\\right)=0\\text{.}$ Hence, for all ${v}_{2}\\in V,$ ${a}_{g}\\left({v}_{3},{v}_{2}\\right)=0$ and so ${V}^{g}\\subseteq \\text{ker} {a}_{g}\\text{.}$ (d) By (c), $\\text{codim}\\left({V}^{g}\\right)\\ge \\text{codim}\\left(\\text{ker} {a}_{g}\\right)\\text{.}$ Since ${a}_{g}\\ne 0,$ there exist $v,w\\in V$ with ${a}_{g}\\left(v,w\\right)\\ne 0$ and so $\\text{codim}\\left(\\text{ker} {a}_{g}\\right)\\ge 2\\text{.}$ Let ${v}_{1}-g{v}_{1}$ and ${v}_{2}-g{v}_{2}$ be linearly independent elements of ${\\left({V}^{g}\\right)}^{\\perp }\\text{.}$ Then (1.7) implies that any element ${v}_{3}-g{v}_{3}\\in {\\left({V}^{g}\\right)}^{\\perp }$ is a linear combination of ${v}_{1}-g{v}_{1}$ and ${v}_{2}-g{v}_{2},$ and so $2≥dim((Vg)⊥) =codim(Vg)≥codim (ker ag)≥2.$ Thus ${V}^{g}=\\text{ker} {a}_{g}$ and $\\text{codim}\\left({V}^{g}\\right)=2\\text{.}$ (e) Write $h{b}_{1}={h}_{11}{b}_{1}+{h}_{21}{b}_{2}+{\\left(h{b}_{1}\\right)}^{g}$ and $h{b}_{2}={h}_{12}{b}_{1}+{h}_{22}{b}_{2}+{\\left(h{b}_{2}\\right)}^{g}$ with ${h}_{ij}\\in ℂ$ and ${\\left(h{b}_{i}\\right)}^{g}\\in {V}^{g}\\text{.}$ Then $ah-1gh (b1,b2) = ag(hb1,hb2) =ag(h11b1+h21b2+(hb1)g,h12b1+h22b2+(hb2)g) = (h11h22-h21h12) ag(b1,b2)=det (h⊥)ag(b1,b2)$ since ${a}_{g}$ is skew symmetric and ${V}^{g}\\subseteq \\text{ker} {a}_{g}\\text{.}$ $\\square$ The following theorem is a slightly strengthened version of statements (given without proof) in [Dri1986]. Let $G$ be a finite subgroup of $GL\\left(V\\right)$", "this can be very useful in identifying $$G$$. Here is a small example. Take $$f(x) = x^4-4 x^3+2 x^2+4 x-6$$. We check that $$f$$ is irreducible: f = x^4 - 4 x^3 + 2 x^2 + 4 x - 6; Factor[f] We store the roots of $$f$$ in variables t1, t2, t3, t4. (Mathematica can actually solve this polynomial exactly, but this is just meant as an example.) {t1, t2, t3, t4} = x /. Solve[f == 0, x]; b = t1 + 2 t2; g = MinimalPolynomial[b, x] (* Output is 42 - 84 x + 50 x^2 - 12 x^3 + x^4 *) Since $$g$$ has degree $$4$$, we conclude that the group $$G$$ probably has an orbit of size $$4$$ acting on ordered pairs of distinct $$\\theta_i$$'s. For a careful proof, we need to make sure that we don't have accidental equalities of the form $$\\theta_i+2 \\theta_j = \\theta_k + 2 \\theta_{\\ell}$$. In particular, we check that $$g(\\theta_1+2 \\theta_3) \\neq 0$$, so $$(1,2)$$ and $$(1,3)$$ are in different orbits for the $$G$$-action. This example is small enough for Mathematica to do exactly, but I'll do the computation numerically because that is good practice for larger examples: N[g /. x -> t1 + 2 t3] (* Output is 198.996 + 220.833 I *) Relative resultants Installing", "= \\boxed{072}$, which is our answer. ## Solution 4 The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\\sqrt{g(0)+144}=\\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\\neq 0$ (else $f$ is not cubic) where $\\{q,r,s,t,u,v\\}$ is the same as the set $\\{1,2,3,5,6,7\\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that $$24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)$$ which must hold for all $x$. Since $a\\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\\{2,3,7\\}$ and $\\{1,5,6\\}$. Note that each of these sets happily satisfy (2). By (3), since", "# Quotient Rule Quotient Rule. If $f$ and $g$ are differentiable then $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$. Proof. Let $h(x)=(f(x))/(g(x))$ then $f(x)=h(x)g(x)$. Differentiating both sides gives: $f'(x)=(h(x)g(x))'$ or using product rule $f'(x)=h(x)g'(x)+h'(x)g(x)$. From this we have that $h'(x)=(f'(x)-h(x)g'(x))/(g(x))$. Remembering that $h(x)=(f(x))/(g(x))$ we can rewrite last identity as $((f(x))/(g(x)))'=(f'(x)-(f(x))/(g(x))g'(x))/(g(x))$ or $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$. Example 1. Find derivative of $f(x)=(e^x)/(x^3+x)$ . $f'(x)=((e^x)/(x^3+x))'=((e^x)'(x^3+x)-e^x(x^3+x)')/(x^3+x)^2=(e^x(x^3+x)-e^x(3x^2+1))/(x^3+x)^2=$ $=(e^x(x^3-3x^2+x-1))/(x^3+x)^2$. Example 2. Find derivative of $f(t)=(t^5+1)/t^3$ $f'(t)=((t^5+1)/(t^3))'=((t^5+1)'t^3-(t^5+1)(t^3)')/(t^3)^2=(5t^4t^3-(t^5+1)*3t^2)/t^6=$ $=(5t^7-3t^7-3t^2)/(t^6)=(2t^7-3t^2)/(t^6)=2t-3/t^4$ Note, however that Example 2 can be used without quotient rule. Since $(t^5+1)/t^3=t^2+1/t^3=t^2+t^(-3)$ then $f'(t)=2t^(2-1)-3t^(-3-1)=2t-3t^(-4)=2t-3/t^4$ . Example 2 showed that first you need to look at the function and make sure that quotient rule is the only way to find derivative (like in Example 1). General advice is not to use quotient rule every time you see quotient. It is rather hard technique and if there are other ways to find derivative, use them.", "Now, if v G V(T) {x,y} then I{v) D/(m) = I(v). So, these conditions are necessary. 35 Now assume that T is a tournament with a vertex x such that 0(x) = y, and 0(y) = V(T)~ {x,y}, 7(T {x,y}) > 2, and 7((T (x,y})r) > 2. Pick distinct u,v £ V(T). We show T is quadrangular using three cases. Case 1: Suppose u, v £ V(T) {x,y}. Then, x £ 0(u) fl 0(v), and since 7((T {x, y})r) > 2, there exits w £ V(T) {x,y} such that w £ 0(u) flO(u). Thus, 10(u) D 0(v) | > 1. Also, y £ I(u) fl I(v), and since 7(T {x, y}) > 2 there exists z £ V(T) {x,y} such that z £ I(u) fl I(v). So, |I(u) fl I(v)\\ > 1. Case 2: Suppose that u = x. Then 0(u) = {y} and since y \\$ 0(v), 10(u) nO(v)| = 0. Now, I(u) nl(v) I(v) {y} since u = x. By Theorem 1.6, 5~(T-{x,y}) > 7(T {x,y}) 1 > 2, and so \\I(u)Dl(v)\\ = \\I(v) {j/}| > 2. Case 3: Suppose that u = y. Then, I(u) = {a;} and since x £ I(v), |/(tt)fl/(t;)| = 0. Now, 0(u) fl 0(v) = 0(v) {x} since u = y. By Theorem 1.6, S+(T {x, y}) > 7((T{x, y})r)", "that, x= u-1 and $f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9$ which is the same as $f(x)= x^2- 7x+ 9$. Do you understand what \"f(x)= x^2- 7x+ 9\" means? What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal? 5. Originally Posted by HallsofIvy You were given the hint \"let u =x+1 and find f(u)\". Doing that, x= u-1 and $f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9$ which is the same as $f(x)= x^2- 7x+ 9$. Do you understand what \"f(x)= x^2- 7x+ 9\" means? What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal? No I dont understand. I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x). Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)). As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for \"a\". I dont really see where to go because im thinking it says", "U, w \\in W$ such that $a = f(u) = s(w)$. $g(a) = g(f(u)) = gs(w) = w$. Hence $w = 0$. Hence $a = s(w) = 0$. Therefore $V = f(U) \\oplus s(W)$. (3) $\\Rightarrow$ (1): Clear. (2) $\\Rightarrow$ (3): Let $K = Ker(t)$. Let $x \\in V$. $t(x - ft(x)) = t(x) - t(x) = 0$. Hence $x - ft(x) \\in K$. Hence $V = f(U) + K$. Let $a \\in f(U) \\cap K$. There exist $u \\in U, k \\in K$ such that $a = f(u) = k$. $t(a) = tf(u) = u = t(k) = 0$. Hence $a = f(u) = 0$. Hence $V = f(U) \\oplus K$. It remains to prove that $g|K\\colon K \\rightarrow W$ is an isomorphism. Suppose $g(k) = 0$, where $k \\in K$. There exists $u \\in U$ such that $k = f(u)$. Since $0 = t(k) = tf(u) = u$, $k = 0$. Hence $g|K$ is injective. Let $w \\in W$. There exists $x \\in V$ such that $w = g(x)$. Then $x - ft(x) \\in K$ and $g(x - ft(x)) = g(x) = w$. Hence $g|K$ is surjective. (3) $\\Rightarrow$ (2): Clear. -", "about $f$ and so we are led to consider the difference $f-g$. As $g \\leq f$, we have that $(f-g)$ is a projection. Note that $(f-g)+h \\leq f$. Then we have, as $(f-g)\\perp h$ and $(f-g)\\perp g$ (*), that $$(f-g) + h \\sim (f-g)+g=f.$$ By finiteness of $f$, we have that $(f-g)+h=f$. Thus $g=h$ and $g$ is finite. Further details on (*): If $vv^*=e \\sim e_1=v^*v$, and $uu^*=f \\sim f_1=u^*u$, with $e \\perp f$, $e_1 \\perp f_1$, then $$(u+v)^*(u+v)=e+f \\sim e_1+f_1 =(u+v)(u+v)^*.$$ Just keep in mind what the supports of the partial isometries $u$ and $v$ are. -", "that $$u+v+1=f$$, where $$v\\geq1$$ because $$m>n$$. By comparing powers in the above we find that $$u+v+1=2^{u-1}(2^{2v}-1)=2^{u-1}(2^v-1)(2^v+1).$$ Of course $$2^v+1>2$$, so $$2^{u-1}=1$$ as otherwise $$2^{u-1}(2^v+1)>2^{u-1}+2^v+1\\geq u+v+1,$$ a contradiction. Hence $$u=1$$ and $$2^{2v}-1=v+2$$, so also $$v=1$$. This yields the solution $$(a,b)=(2,6)$$. On the other hand, if the Pythagorean triple is of the form $$(1)$$ then $$k$$, $$m-n$$ and $$m+n$$ are powers of $$2$$ because $$2^{ef}=k(m^2-n^2)=k(m-n)(m+n).$$ Because $$m$$ and $$n$$ are not both even, by Lemma 1 there exists a positive integer $$v$$ such that $$m=2^v+1$$ and $$n=2^v-1$$, and so the above shows that $$k=2^{ef-v-2}$$. Plugging this into the other equation yields $$(2f)^{2^{e-1}}=2kmn=2^{ef-v-1}(2^{2v}-1),$$ and writing $$f=2^wg$$ with $$g$$ odd then implies $$g^{2^{e-1}}=2^{2v}-1,$$ which by Lemma 2 implies that $$2^{e-1}=1$$, and hence $$e=1$$. Then $$f=kmn$$ and if $$kmn\\geq4$$ then $$2^{kmn}\\geq(kmn)^2\\geq km^2>k(m^2-n^2),$$ so $$f=kmn\\leq3$$. Then $$b\\leq6$$ and clearly $$b=2$$ and $$b=4$$ do not yield solutions.$$\\qquad\\square$$ Proposition 4: If $$d$$ is even then $$d=2$$. Proof. Suppose $$d=2e$$ and let $$a=2eA$$ and $$b=2eB$$. Then $$e$$ is odd as otherwise $$c^2$$ is the sum of two fourth powers, which is well known to be impossible by a classical result by Fermat. Now $$c^2=a^b+b^a=(a^{eB})^2+(b^{eA})^2,$$ is a pythagorean triple and hence there exist positive integers $$k$$, $$m$$ and $$n$$ with $$m>n$$ and $$\\gcd(m,n)=1$$ such that $$a^{eB}=k(m^2-n^2),\\qquad b^{eA}=2kmn,\\qquad c=k(m^2+n^2),$$ where we can exchange the roles of $$a$$ and $$b$$", "$$x$$ is replaced by $$-x$$, $$y$$ doesn't change. Now take another example: $$z = g(x) = 3 x$$. An interesting property of $$g$$ is that $$g(w + x) = g(w) + g(x)$$. What could $$z$$ even be here? “Which” $$z$$ are you talking about? You can build up to more complex examples involving multiple functions. For example, take $$h(x) = v = x^3$$ and $$i(x) = u = x^4$$. An interesting property of $$h$$ and $$i$$ is that $$h(i(x)) = i(h(x))$$. Which in terms of $$u$$ and $$v$$ means that, er, well, if you take $$x = v$$ then you get $$u$$ which is the same $$v$$ as if you take $$x = u$$? If that makes any sense? Right, it doesn't. But functions let us express this unambiguously. • These are nice examples, but I'll play the devil's advocate and rephrase them using only quantities, avoiding functions. The first one: for any value $a$ of $x$ we have $y|_{x=a}=y|_{x=-a}$. The second example: if $u=v^3$ and $v=x^4$ and further $y=z^3$ and $z=x^4$ then $u=y$. The problem with that second example, the way you formulated it, is that if $x,u,v$ were 'specific' quantities in any real applied context, then what meaning should the quantities $h(i(x))$ and $i(h(x))$ have? – Michael Bächtold Aug 10 '19 at 10:36 • @MichaelBächtold First:", "$$\\frac{u+v}{w(u,v)} < D(G')$$ and therefore $$u+v < w(u,v)D(G')$$. If we sum up all the inequalities for all (u,v), we have $$\\sum_{u \\in G'} u + \\sum_{v \\in G'} v < \\sum_{(u,v) \\in G'} w(u,v) D(G')$$, hence $$\\frac{ \\sum_{u \\in G'} u + \\sum_{v \\in G'} v } { \\sum_{(u,v) \\in G'} w(u,v) } < D(G')$$. However, we also have that $$D(G') = \\frac{\\sum_{u \\in G'} u}{\\sum_{(u,v) \\in G'} w(u,v)} \\leq \\frac{ \\sum_{u \\in G'} u + \\sum_{v \\in G'} v } { \\sum_{(u,v) \\in G'} w(u,v) }$$, hence a contradiction. Therefore we conclude that there are (u,v) in G' that has density at least equal to or more than D(G'). :D", ", The first one is to find g(1) then substitute the value pf g(1) in any x in the f(x) The other way , as you and @Panphobia said , is to do it like : f(x o g) (1) = 2g(1)+3 . . They are equivalent , you will get the same answer .. (: 4. Nov 22, 2014 ### Fredrik Staff Emeritus The easiest and most straightforward way to find f(g(1)) is to find g(1) first, and then plug that into f. When you do it this way, you don't even have to figure out what function $f\\circ g$ is. The alternative is to figure out what function $f\\circ g$ is first, and then plug 1 into $f\\circ g$. $f\\circ g$ is defined by $(f\\circ g)(x)=f(g(x))$ for all $x$ in the domain of g. The \"for all\" is essential. It means that the equality $(f\\circ g)(x)=f(g(x))$ holds regardless of what number the symbol $x$ represents. So in particular, it holds when $x=1$. This notation doesn't make sense.", "one more special property of $\\mathbb{Z}$ that we have yet to mention, but which will be important in the general definition that we’ll be giving soon. The property I am talking about is the seemingly trivial fact that we can consider the sum $5+3+7$ without worrying about which two numbers we add together first. I.e., we could add 5 to 3, then add 7, which would be denoted by $(5+3)+7$or we could add 3 and 7 first, then add 5, which would be denoted $5+(3+7)$, and we’d clearly get the same result in both cases. In our abstract language, this comes down to the fact that we can take any three elements in $\\mathbb{Z}$ and apply “g” to them in any order we like, and get the same answer. I.e., it is the case that for all $a,b,c\\in \\mathbb{Z}, \\ g(g(a,b),c)=g(a,g(b,c))$. Don’t be scared of that equation. Remember that the domain of “g” is $\\mathbb{Z}\\times \\mathbb{Z}$, so whatever goes in the parentheses g() must be a pair of integers (I’ve dropped the nested parentheses just because they’d make those expressions too messy, and the meaning should be clear without them). Well, in the above equation, $g(g(a,b),c)$ means take the integer which is given by $g(a,b)$ and put it in the first slot of $g(\\cdot, c)$, so those expressions", "From the last assertion and (tf2), we have that $$R_{2}(f(x))\\subseteq U$$. Therefore, $$x\\in f^{-1}(G_{R_2}(U))$$. (tf7): It can be proved in a similar way. $${\\rm (ii)}\\Rightarrow {\\rm (i)}$$: First, taking into account that $$f$$ is an $$LM_n$$-function which satisfies (tf6) and (tf7) we can see that the following conditions are verified: (tf8) $$f^{-1}(F_{R_2}(U))=F_{R_1}(f^{-1}(U))$$ for any $$U \\in D(X_2)$$, (tf9) $$f^{-1}(P_{R^{-1}_2}(U))=P_{R^{-1}_1}(f^{-1}(U))$$ for any $$U \\in D(X_2)$$. Indeed, let $$U \\in D(X_2)$$, then it follows that $$f^{-1}(F_{R_2}(U))\\, = \\, f^{-1}(\\sim_{g_2}(G_{R_2}(\\sim_{g_2} U))=$$ $$\\sim_{g_1}(f^{-1}(G_{R_2}(\\sim_{g_2} U)))\\, \\, \\, = \\, \\, \\,\\sim_{g_1}(G_{R_1}(f^{-1}(\\sim_{g_2} U)))\\,\\, \\, =\\, \\, \\, \\sim_{g_1}(G_{R_1}(\\sim_{g_1}(f^{-1}(U)))\\, =$$ $$F_{R_1}(f^{-1}(U))$$ and so (tf8) holds. Property (tf9) can be proved in a similar way. (tf1): Let $$x, \\, y\\in X_1$$ such that (1) $$(x,y)\\in R_1$$. Suppose that $$(f(x),f(y))\\notin R_{2}$$. Then, from Lemma 3.5 it follows that there is $$U \\in D(X_2)$$ such that (2) $$f(y) \\not\\in U$$ and $$f(x) \\in G_{R_2}(U)$$ or there is $$V \\in D(X_2)$$ such that (3) $$f(y) \\in V$$ and $$f(x) \\not\\in F_{R_2}(V)$$. If (2) holds, then (4) $$y \\notin f^{-1}(U)$$ and $$x \\in f^{-1} (G_{R_2}(U))$$. From this last statement and (tf6) we obtain that $$x \\in G_{R_1}(f^{-1}(U))$$, and so from (1) we infer that $$y \\in f^{-1}(U)$$, which contradicts (4). If (3) holds, then (5) $$y \\in f^{-1}(V)$$ and $$x \\not\\in f^{-1}(F_{R_2}(V))$$, from which it follows by (tf8) that $$x" ]

[ "is if there are two distinct values of $x$ for which the values of $g(x)$ are the same. This can happen with things like $|x|$ because $x = 2$ and $x = -2$ (for example) both give $|x| = 2$. So with an expression like $|2x-3|$ you need to look at the value of $x$ that makes $2x - 3 = 0$, because either side of that, you'll have two values of $x$ giving the same value of $|2x - 3|$. Clearly that value is $x=1.5$. So when $x = 1.5 + 1$ (for example) and $x = 1.5 - 1$, the value of $|2x - 3|$ is the same: it's $2$ in each case. For part (i) we are told that $x$ lies in the range $-2 \\le x \\le k$. So we are certainly allowed some values of $x$ that are less than $1.5$. Therefore if $g$ is going to have an inverse, we must avoid any values of x that are greater than $1.5$ in order to avoid repeating the values of $|2x-3|$. Hence $k$ (the greatest permitted value of $x$) $= 1.5$. For part (ii) we know that $g$ has an inverse. So we know that $x \\le 1.5$, and hence $2x - 3 \\le 0$. So to find $|2x - 3|$ we must", "# Quotient Rule Quotient Rule. If $f$ and $g$ are differentiable then $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$. Proof. Let $h(x)=(f(x))/(g(x))$ then $f(x)=h(x)g(x)$. Differentiating both sides gives: $f'(x)=(h(x)g(x))'$ or using product rule $f'(x)=h(x)g'(x)+h'(x)g(x)$. From this we have that $h'(x)=(f'(x)-h(x)g'(x))/(g(x))$. Remembering that $h(x)=(f(x))/(g(x))$ we can rewrite last identity as $((f(x))/(g(x)))'=(f'(x)-(f(x))/(g(x))g'(x))/(g(x))$ or $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$. Example 1. Find derivative of $f(x)=(e^x)/(x^3+x)$ . $f'(x)=((e^x)/(x^3+x))'=((e^x)'(x^3+x)-e^x(x^3+x)')/(x^3+x)^2=(e^x(x^3+x)-e^x(3x^2+1))/(x^3+x)^2=$ $=(e^x(x^3-3x^2+x-1))/(x^3+x)^2$. Example 2. Find derivative of $f(t)=(t^5+1)/t^3$ $f'(t)=((t^5+1)/(t^3))'=((t^5+1)'t^3-(t^5+1)(t^3)')/(t^3)^2=(5t^4t^3-(t^5+1)*3t^2)/t^6=$ $=(5t^7-3t^7-3t^2)/(t^6)=(2t^7-3t^2)/(t^6)=2t-3/t^4$ Note, however that Example 2 can be used without quotient rule. Since $(t^5+1)/t^3=t^2+1/t^3=t^2+t^(-3)$ then $f'(t)=2t^(2-1)-3t^(-3-1)=2t-3t^(-4)=2t-3/t^4$ . Example 2 showed that first you need to look at the function and make sure that quotient rule is the only way to find derivative (like in Example 1). General advice is not to use quotient rule every time you see quotient. It is rather hard technique and if there are other ways to find derivative, use them.", "range of $f$. To see that $f$ is one-to-one, suppose that $f(x) = f(y)$. Then, $$g(f(x)) = g(f(y)) \\implies x = y.$$ Now, suppose that $f$ is one-to-one and onto. We will show that $f$ is invertible. As $f$ is onto, we have that $\\forall y \\in Y, \\exists x \\in X$ such that $y = f(x)$. Since $f$ is injective, we have that $f(x) = f(y) \\implies x = y, \\forall x , y \\in X$. Let $g : Y \\to X$ be defined by $g(y) = x$ if and only if $f(x) = y ~$ (why can we assume such a $g$ exists?). Then $g(f(x)) = g(y) = x$ and $f(g(y)) = f(x) = y$. Hence $g$ is an inverse of $f$. QED. Lemma 2: The inverse of a function $f$ is unique if it exists. Proof: Suppose that $f : X \\to Y$ is invertible and it has two inverses $g, h : Y \\to X$. Note that $\\forall x \\in X$, we have $f(g(x)) = x = f(h(x))$. Hence: $$g(x) = g(f(g(x))) = g (f(h(x))) \\implies g(x) = h(x).$$ Since $x$ was arbitrary, $g$ and $h$ must be identical as they have the same value for all $x \\in X$. QED. Finally, apply the above lemmas to the function $f\\big|_A : A \\to f(A)q$.", "the inverse of f exists and $f^{-1} = g$ Now, $f^{-1} : \\left \\{ a,b,c \\right \\}\\rightarrow \\left \\{ 1,2,3 \\right \\}$ is given by $f^{-1}(a) = 1, f^{-1}(b) = 2 \\ and \\ f^{-1}(c) = 3$ Now, we need to find the inverse of $f^{-1}$, Therefore, lets define$h: \\left \\{ 1,2,3 \\right \\}\\rightarrow \\left \\{ a,b,c \\right \\}$ such that $h(1) = a, h(2) = b \\ and \\ h(3) = c$ Now, $(goh)(1) = g(h(1)) = g(a) = 1$ $(goh)(2) = g(h(2)) = g(b) = 2$ $(goh)(3) = g(h(3)) = g(c) = 3$ Similarly, $(hog)(a) = h(g(a)) = h(1) = a$ $(hog)(b) = h(g(b)) = h(2) = b$ $(hog)(c) = h(g(c)) = h(3) = c$ Hence, $goh = I_X$ and $hog = I_Y$, where$X = \\left \\{ 1,2,3 \\right \\}$ and$Y = \\left \\{ a,b,c \\right \\}$ Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$ $\\Rightarrow h = f$ Therefore, $(f^{-1})^{-1} = f$ Hence proved $f : X \\rightarrow Y$ To prove: $(f^{-1})^{-1} = f$ Let $f:X\\rightarrow Y$ be a invertible function. Then there is $g:Y\\rightarrow X$ such that $gof =I_x$ and $fog=I_y$ Also, $f^{-1}= g$ $gof =I_x$ and $fog=I_y$ $\\Rightarrow$ $f^{-1}of = I_x$ and $fof^{-1} = I_y$ Hence, $f^{-1}:Y\\rightarrow X$ is invertible function and f is inverse of $f^{-1}$. i.e.", "so that $G(x,0)=g_0(x)=1_X(x)=x$ and $G(x,1)=g_1(x)=(\\iota\\circ f_1)(x)$ for all $x\\in X$. Because $\\{f_t\\}_{t\\in I}$ is a deformation retract of $X$ onto the point $x\\in X$, the family $\\{f_t\\}$ can be thought of as a shrinking of $X$ onto the subset $\\{x\\}$, and because the family $\\{f_t\\}$ is continuous in both $t$ and $x$ (by definition of homotopy), it follows that there exists an element $T\\in I$ for which $g_t^{-1}(U)\\subsetneq U$ for $t\\geq T$ and $g_t^{-1}(U)=U$ for $t. Let $V=g_T^{-1}(U)$. It suffices to show that the inclusion $V\\hookrightarrow U$ is homotopic to a constant map. To that end, write $t=kT$, $k\\in[0,1]$, and define $H:U\\times I\\to V$ so that $H(u,k)=h_k(u)=g_{kT}(x)$ for all $x\\in X\\cap U$2. In particular, then, $h_0(u)=g_0(u)=u$ as $g_0=id_X$. Moreover, $h_1(u)=g_T(u)$ for all $u\\in U$, and because $V=g_T^{-1}(U)$, $h_1(U)\\subset V$. Finally, because $f_t$ is a (strong) deformation retract, because $G$ is defined in terms of $f_t$, and because $H$ is defined in terms of $G$, it’s easily confirmed that $h_t(V)\\subset V$ for all $t$ and hence that $h_t$ is a weak deformation retract. Therefore, from the same problem cited above (see here), there exists an inclusion map $j:\\{x\\}\\to V$ for which $h_1\\circ j\\simeq 1_{\\{x\\}}$. In particular, $h_1\\circ j$ is null-homotopic. At this point, all the ingredients are in place. The fact that the family $\\{f_t\\}$ homotopes $1_X\\simeq \\iota\\circ f_1$", "values less than x = -1 or greater than x = 1 otherwise the value under the radical is negative. ## Example 2 Recall that $(f \\circ g)(x)$ does not necessarily equal $(g \\circ f)(x)$. Find an example where $(f \\circ g)(x) = (g \\circ f)(x)$. Suppose that $f$ and $g$ are defined for all reals such that $f(x) := x$ and $g(x) := x + 1$. Then $(f \\circ g)(x) = f(g(x)) = f(x + 1) = x + 1$. Furthermore $(g \\circ f)(x) = g(f(x)) = g(x) = x + 1$. In this case, $(f \\circ g)(x) = (g \\circ f)(x)$.", "$g(s) f(s) = 0$, then either $g(s) = 0$ or $f(s) = 0$. If we assume that $f(s) \\neq 0$, then we must have $g(s) = 0$. For the case where $f= x+iy, g= u+iv$ are constants, then $fg = 0$ if and only if $f=0$ and/or $g=0$. Introducing complex coordinates doesn't change this fact. This is because the values must satisfy $ux-vy = 0$ AND $uy + vx = 0$, and so $0 = (ux-vy)^2 + (uy+vx)^2 = (u^2 + v^2)(x^2 + y^2)$, which implies that $u^2 +v^2 = 0$ or $x^2 + y^2 = 0$, which implies that $(u,v)=(0,0)$ or $(x,y) = (0,0)$. - Now, as you note, if $(x + i y)(u + i v) = 0$ then this implies that $x u - v y = x v + u y = 0$. However, the only way these equations can hold is if either $x + i y = 0$ or $u + i v = 0$. Multiplying the first equation through by $v$ and the second by $u$, we have $x u v - v^2 y = 0$ and $x u v + u^2 y = 0$. Subtracting, $(u^2 + v^2)y = 0$. Since $u, v, y \\in \\mathbb{R}$, this means that either $y = 0$ or $u = v = 0$. In", "find $f(f(f(1)))$. – Jonas Meyer Jan 11 at 5:40 Let us suppose the inverse of $f$ is $g$, then $g(1)=3$, $g(2)=1$, $g(3)=2$ and $g(s)=s$ for all $s \\in S$.", "inverse of $g(x)=-3x$. 1. $g ^(-1) (x)=x+1$ 2. $g ^-1 (x)=-3x-3$ 3. $g ^-1 (x)=x-1$ 4. $g ^-1(x)= -1/3 x$ 9. $f(x) = sqrt(x+3)$ and $g(x)=2x$. Find $(f @ g)(4)$. 1. $2sqrt14$ 2. $sqrt11$ 3. $sqrt14$ 4. $2sqrt7$ 10. Find the domain of the composite function $f(g(x))$. $f(x)= x+3$ $g(x)= 2/(x+6)$ 1. $(-oo,3) uu (3,oo)$ 2. $(-oo,oo)$ 3. $(-oo,-6) uu (-6,3) uu (3,oo)$ 4. $(-oo,-6) uu (-6,oo)$ 11. Simplify the expression $(7x^2-63)/(x+3)$. 1. $(x+3)/7, \\ x!=-3$ 2. $(7x^2-9)/(x), \\ x!=-3$ 3. $7(x+3), \\ x!=-3$ 4. $7(x-3), \\ x!=-3$ 12. Solve. Check for extraneous solutions. $x/(x-1)=(x+3)/(-2x+2)$ 1. $x=-1 or x=-2$ 2. $x=-1$ 3. $x=1$ 4. $x=1 or x=-1$ 13. Which function is an example of exponential growth? 1. $g(x)=0.4(1.3)^x$ 2. $f(x)=1.6(0.85)^x$ 3. $h(x)=0.5(0.95)^x$ 4. $k(x)=1.2(1-.25)^x$ 14. Determine the y-intercept of the following exponential function. $y=2^x+3$ 1. (0,4) 2. (0,3) 3. (0,2) 4. (4,0) 5. none of these are correct 15. What is the value of $x$ in the equation $9^(3x+1)=27^(x+2)$ ? 1. $1$ 2. $1/3$ 3. $1/2$ 4. $4/3$ 16. Write $log_5 36=x$ in exponential form. 1. $36x = 5$ 2. $5^x = 36$ 3. $x^5 = 36$ 4. $10^x = 36$ 17. Evaluate. $log_3 (1/9)$ 18. Evaluate. Write in simplest form. $10^(log_(10) 3)$ 1. 3 2. 10 3. 1000 4. 1/3 5. none of these are correct 19.", "be changing the sign of $2x - 3$ from negative to positive. In other words, when $x \\le 1.5, |2x - 3| = -(2x-3) = 3 - 2x$. So $g(x) = 3 - 2x - 4= - 2x-1$ OK? Hello Superman-PrimeWelcome to Math Help Forum! Thanks The thing that might prevent $g$ from having an inverse is if there are two distinct values of $x$ for which the values of $g(x)$ are the same. This can happen with things like $|x|$ because $x = 2$ and $x = -2$ (for example) both give $|x| = 2$. So with an expression like $|2x-3|$ you need to look at the value of $x$ that makes $2x - 3 = 0$, because either side of that, you'll have two values of $x$ giving the same value of $|2x - 3|$. Clearly that value is $x=1.5$. So when $x = 1.5 + 1$ (for example) and $x = 1.5 - 1$, the value of $|2x - 3|$ is the same: it's $2$ in each case. For part (i) we are told that $x$ lies in the range $-2 \\le x \\le k$. So we are certainly allowed some values of $x$ that are less than $1.5$. Therefore if $g$ is going to have an inverse, we must avoid any values of x", "a unique inverse. 39 Consider $f: \\{1, 2, 3\\} \\to \\{a, b, c\\}$ given by $f(1) = a, f(2) = b$ and $f(3) = c$. Find $f -1$ and show that $(f -1 ) -1 = f$. ##### Solution : Function $f: \\{1, 2, 3\\} \\to \\{a, b, c\\}$ is given by $f(1) = a, f(2) = b$, and $f(3) = c$$\\\\ If we define g: \\{a, b, c\\} \\to \\{1, 2, 3\\} as g(a) = 1, g(b) = 2, g(c) = 3.$$\\\\$ We have$\\\\$ $(fog)(a) = f(g(a)) = f(1) = a\\\\ (fog)(b) = f(g(b)) = f(2) = b\\\\ (fog)(c) = f(g(c)) = f(3) = c$$\\\\ and\\\\ (gof)(1) = g(f(1)) = f(a) = 1\\\\ (gof)(2) = g(f(2)) = f(b) = 2\\\\ (gof)(3) = g(f(3)) = f(c) = 3\\\\ \\therefore gof = I_ X and fog = I_ Y , where X = \\{1, 2, 3\\} and Y= \\{a, b, c\\}.\\\\ Thus, the inverse of f exists and f-1 = g.$$\\\\$ $\\therefore f -1 :\\{ a, b, c\\} \\to \\{1, 2, 3\\}$ is given by $f ^-1 (a) = 1, f ^-1 (b) = 2, f ^-1 (c) = 3$$\\\\ Let us now find the inverse of f -1 i.e., find the inverse of g. If we define h: \\{1, 2, 3\\} \\to \\{a, b, c\\} as h (1)", "U, w \\in W$ such that $a = f(u) = s(w)$. $g(a) = g(f(u)) = gs(w) = w$. Hence $w = 0$. Hence $a = s(w) = 0$. Therefore $V = f(U) \\oplus s(W)$. (3) $\\Rightarrow$ (1): Clear. (2) $\\Rightarrow$ (3): Let $K = Ker(t)$. Let $x \\in V$. $t(x - ft(x)) = t(x) - t(x) = 0$. Hence $x - ft(x) \\in K$. Hence $V = f(U) + K$. Let $a \\in f(U) \\cap K$. There exist $u \\in U, k \\in K$ such that $a = f(u) = k$. $t(a) = tf(u) = u = t(k) = 0$. Hence $a = f(u) = 0$. Hence $V = f(U) \\oplus K$. It remains to prove that $g|K\\colon K \\rightarrow W$ is an isomorphism. Suppose $g(k) = 0$, where $k \\in K$. There exists $u \\in U$ such that $k = f(u)$. Since $0 = t(k) = tf(u) = u$, $k = 0$. Hence $g|K$ is injective. Let $w \\in W$. There exists $x \\in V$ such that $w = g(x)$. Then $x - ft(x) \\in K$ and $g(x - ft(x)) = g(x) = w$. Hence $g|K$ is surjective. (3) $\\Rightarrow$ (2): Clear. -", "to $R$ and also $\\left( fog \\right)\\left( x \\right)=f\\left( g\\left( x \\right) \\right)={{\\left( {{x}^{3}}-{{x}^{2}}+2 \\right)}^{8}}$ . Both conditions are satisfied. Hence, our assumption of functions is correct. Now, to evaluate ${{f}^{'}}\\left( 1 \\right).{{g}^{'}}\\left( 1 \\right)$ , first we need to calculate the values of $f'(x)$ and $g'(x)$ . First, we will find the value of $f'(x)$ . To calculate the value of $f'(x)$ , we will differentiate $f(x)$ with respect to $x$ . On differentiating $f(x)$ with respect to $x$, we get ${{f}^{'}}\\left( x \\right)=\\dfrac{d}{dx}{{x}^{8}}=8{{x}^{7}}$ . Now, we will find the value of $g'(x)$ . To calculate $g'(x)$ , we will differentiate $g(x)$ with respect to $x$ . On differentiating $g(x)$ with respect to $x$ , we get ${{g}^{'}}\\left( x \\right)=\\dfrac{d}{dx}\\left( {{x}^{3}}-{{x}^{2}}+2 \\right)=3{{x}^{2}}-2x$ Now, we will find the values of $f'(1)$ and $g'(1)$ . To evaluate the values of $f'(1)$ and $g'(1)$ , we will substitute $x=1$ in the expressions of $f'(x)$ and $g'(x)$ . So, on substituting $x=1$ in $f'(x)=8{{x}^{7}}$ , we get $f'(1)=8\\times {{(1)}^{7}}=8$ . And, on substituting $x=1$ in $g'(x)=3{{x}^{2}}-2x$ , we get $g'(x)=3{{(1)}^{2}}-2(1)$ . \\begin{align} & =3-2 \\\\ & =1 \\\\ \\end{align} Now, we can evaluate the value of ${{f}^{'}}\\left( 1 \\right).{{g}^{'}}\\left( 1 \\right)$ as ${{f}^{'}}\\left( 1 \\right).{{g}^{'}}\\left( 1 \\right)=8\\times 1$ $=8$ Note: While choosing$f\\left( x \\right)$and $g\\left( x \\right)$, make sure that the", "Substituting ${v}_{i}-g{v}_{i}={v}_{i}^{-}-g{v}_{i}^{-}$ and ${v}_{1}^{-}={c}_{2}{v}_{2}^{-}+{c}_{3}{v}_{3}^{-}$ then yields $a(v3,v1) (gv2-v2)+ a(v2,v3) (gv1-v1)+ a(v1,v2) (gv3-v3) = a(v3-,v1-) (gv2--v2-)+ a(v2-,v3-) (gv1--v1-)+ a(v1-,v2-) (gv3--v3-)= 0,$ and so (1.7) holds. (c) Let ${v}_{3}\\in {V}^{g}$ and ${v}_{2}\\in V\\text{.}$ If ${v}_{2}\\in {V}^{g},$ then ${a}_{g}\\left({v}_{2},{v}_{3}\\right)\\left(g{v}_{1}-{v}_{1}\\right)=0$ for all ${v}_{1}\\in V$ by (1.7). Since ${V}^{g}\\ne V,$ there is some ${v}_{1}$ such that $g{v}_{1}\\ne {v}_{1}$ and so ${a}_{g}\\left({v}_{2},{v}_{3}\\right)=0\\text{.}$ If ${v}_{2}\\notin {V}^{g},$ let ${v}_{1}=\\sum _{k=1}^{r-1}{g}^{k}{v}_{2},$ where $r$ is the order of $g\\text{.}$ By (1.6), ${a}_{g}\\left({v}_{3},{g}^{k}{v}_{2}\\right)={a}_{g}\\left({g}^{-k}{v}_{3},{v}_{2}\\right)={a}_{g}\\left({v}_{3},{v}_{2}\\right),$ for any $k,$ and so $0 = ag(v3,v1) (gv2-v2)+ ag(v2,v3) (gv1-v1) = (r-1) ag(v3,v2) (gv2-v2)+ ag(v3,v2) (gv2-v2)=r ag(v3,v2) (gv2-v2).$ Thus ${a}_{g}\\left({v}_{3},{v}_{2}\\right)=0\\text{.}$ Hence, for all ${v}_{2}\\in V,$ ${a}_{g}\\left({v}_{3},{v}_{2}\\right)=0$ and so ${V}^{g}\\subseteq \\text{ker} {a}_{g}\\text{.}$ (d) By (c), $\\text{codim}\\left({V}^{g}\\right)\\ge \\text{codim}\\left(\\text{ker} {a}_{g}\\right)\\text{.}$ Since ${a}_{g}\\ne 0,$ there exist $v,w\\in V$ with ${a}_{g}\\left(v,w\\right)\\ne 0$ and so $\\text{codim}\\left(\\text{ker} {a}_{g}\\right)\\ge 2\\text{.}$ Let ${v}_{1}-g{v}_{1}$ and ${v}_{2}-g{v}_{2}$ be linearly independent elements of ${\\left({V}^{g}\\right)}^{\\perp }\\text{.}$ Then (1.7) implies that any element ${v}_{3}-g{v}_{3}\\in {\\left({V}^{g}\\right)}^{\\perp }$ is a linear combination of ${v}_{1}-g{v}_{1}$ and ${v}_{2}-g{v}_{2},$ and so $2≥dim((Vg)⊥) =codim(Vg)≥codim (ker ag)≥2.$ Thus ${V}^{g}=\\text{ker} {a}_{g}$ and $\\text{codim}\\left({V}^{g}\\right)=2\\text{.}$ (e) Write $h{b}_{1}={h}_{11}{b}_{1}+{h}_{21}{b}_{2}+{\\left(h{b}_{1}\\right)}^{g}$ and $h{b}_{2}={h}_{12}{b}_{1}+{h}_{22}{b}_{2}+{\\left(h{b}_{2}\\right)}^{g}$ with ${h}_{ij}\\in ℂ$ and ${\\left(h{b}_{i}\\right)}^{g}\\in {V}^{g}\\text{.}$ Then $ah-1gh (b1,b2) = ag(hb1,hb2) =ag(h11b1+h21b2+(hb1)g,h12b1+h22b2+(hb2)g) = (h11h22-h21h12) ag(b1,b2)=det (h⊥)ag(b1,b2)$ since ${a}_{g}$ is skew symmetric and ${V}^{g}\\subseteq \\text{ker} {a}_{g}\\text{.}$ $\\square$ The following theorem is a slightly strengthened version of statements (given without proof) in [Dri1986]. Let $G$ be a finite subgroup of $GL\\left(V\\right)$", "there exists a $g \\in G$ such that $a = gb$. We now check that $\\sim$ is indeed an equivalence relation on $A$. • Reflexivity: By the second axiom of a group action we have that $a = ea$ for all $a \\in A$, where $e \\in G$ denotes the identity. Thus $a \\sim a$ for every $a \\in A$. • Symmetry: Suppose that $a \\sim b$. Then there exists a $g \\in G$ such that $a = gb$. Then we have that $g^{-1}a = g^{-1}(gb) = b$. Since $g \\in G$ implies $g^{-1} \\in G$ and $b = g^{-1}a$ we see that $b \\sim a$. Thus, for all $a, b \\in A$, if $a \\sim b$ then $b \\sim a$. • Transitivity: Suppose that $a \\sim b$ and $b \\sim c$. Then there exists a $g \\in G$ such that $a = gb$, and there exists a $g' \\in G$ such that $b = g'c$. So by the first axiom of a group action we have that $a = gb = g(g'b) = (g \\cdot g')c$. Since $g, g' \\in G$ we have that $g \\cdot g' \\in G$. Thus $a \\sim c$. So for all $a, b, c \\in A$, if $a \\sim b$ and $b \\sim c$ then $a \\sim c$. • So indeed, $\\sim$", "inverse funtions, then $g(x)=f^{-1}(x)$. Also, $f^{-1}(f(x))=x$, so if $f(g(x))=x$, then we know they are inverses. $f(g(x))=\\sqrt{(x^2+9)-9}=\\sqrt{x^2}=x$, so they are inverses. For question 6, a. $(f-g)(x)=f(x)-g(x)=2x-1-(x^2+x+2)=-x^2+x-3$ b. $f(g(x))=2(x^2+x+2)-1=2x^2+2x+4-1=2x^2+2+3$ c. $g(f(2))=(2(2)-1)^2+(2(2)-1)-3=3^2+3-3=9$", "the other pairs untouched. So for $g_0$ given by $g_0(n)=0$, we have that $f(g_0)$ is the identity, while for $g_1$ given by $g_1(n)=1$, we get that $f(g_1)(2k)=2k+1$ and $f(g_1)(2k+1)=2k$ for all $k\\in \\Bbb N$. For a final example, take $g_2$ defined by $g_2(4k)=0$ and $g_2(4k+2)=1$. Then we have • $f(g_2)(0)=0$ and $f(g_2)(1)=1$ because $g_2(0)=0$ • $f(g_2)(2)=3$ and $f(g_2)(3)=2$ because $g_2(2)=1$ • $f(g_2)(4)=4$ and $f(g_2)(5)=5$ because $g_2(4)=0$ and so on. • I still don't get it. Sorry. Is there any chance you can elaborate? – Galush Balush Feb 7 '18 at 17:16 • @GalushBalush I've given two small examples. I'll add one more. – Arthur Feb 7 '18 at 17:18 • How can I be sure that I don't get the same value twice? – Galush Balush Feb 7 '18 at 17:26 • From my last example, can you not see how that's obvious? If that's not clear enough, then I don't think I'm capable of explaining it to you. Instead, I really recommend that you take out a pen and a piece of paper, come up with some random $g$, and work through, one by one, what $f(g)(n)$ is for, say, $n=0,1,2,3,4,5,6,7$. The fact of the matter is that the construction of $f$ is really simple, but it's a bit difficult to convey in writing. I don't know", "then $f(0) > g(0)$. Or, equivalently, if $f(0)=g(0)$, then $f(0)=g(0)=0$. Suppose now that $f$ and $h$ are differentiable. Since the equality (1) holds for all $t \\geq 0$, we can differentiate both sides and obtain $$f'(t)(h(t)-t-1) + f(t)(h'(t)-1) = g(h(t)) h'(t) - g(t).$$ Note that, by definition of $h$, we have $g(h(t))=f(t)$. That is, $$\\tag{2} \\boxed{f'(t)(h(t)-t-1) - f(t) + g(t) = 0 \\quad \\text{for all } t \\geq 0.}$$ More or less, this is the equation for $g$, under the assumption that $f$ and $h$ are differentiable. But it is not explicit, since it contains the inverse function $g^{-1}$. Note that $h$ can be nondifferentiable. Indeed, if we assume that (2) holds all the time, then if $f'(0)=0$, we have $f(0)=g(0)$. However, if, additionally, $f(0)>0$, then we get a contradiction from the property stated above. So, the nondifferentiability of $h$ occurs when $f'=0$. However, I conjecture that if $f'(t)>0$ for all $t$, then (2) is valid. Suppose, this is true. Then, in the particular case $f(t)=t$, we get the following elegant equation for $g$: $$g(t) + g^{-1}(t) = 2t+1,$$ which gives the recurrence formula $$t = g(2t+1-g(t)).$$ It is possible to resolve it step by step. For instance, here are first few steps (see the figure below): g(t) = \\left\\{ \\begin{aligned} &0, &&t \\in [0,1];\\\\ &\\frac{t-1}{2}, &&t \\in", "$$f$$ decreasing implies that $$f^{ -1}$$ is decreasing. 4. $$f$$ differentiable at $$c$$ and $$f '(c) \\neq 0$$ implies that $$f^{ -1}$$ is differentiable at $$f (c)$$. 5. If $$g(x)$$ is the inverse of the differentiable $$f(x)$$ then $g'(x) = \\dfrac{1}{f '(g(x))}.$ if $$f '(g(x)) \\neq 0$$. Proof of (5) Since $f (g(x)) = x$ we differentiate implicitly: $\\dfrac{d}{dx} f(g(x)) = \\dfrac{d}{dx} x.$ Using the chain rule $y =f(u), u = g(x)$ $\\dfrac{dy}{x} = \\dfrac{dy}{dy} \\dfrac{du}{dx}$ $= f '(u) g'(x) = f '(g(x)) g'(x).$ So that $f '(g(x)) g'(x) = 1.$ Dividing, we get: $g'(x) = \\dfrac{1}{f'(g(x))}.$ Example 3 For $$x > 0$$, let $f(x) = x^2$ and $g(x) = \\sqrt{x}$ be its inverse, then $g'(x) = \\dfrac{1}{2\\sqrt{x}}.$ Note that $\\dfrac{d}{dx} \\sqrt{x} = \\dfrac{d}{dx} x^{\\frac{1}{2}} = \\dfrac{1}{2} x^{-\\frac{1}{2}} = \\dfrac{1}{2\\sqrt{x}}.$ Exercise 1. Let $f(x) = x^3 + x - 4.$ Find $\\dfrac{d}{dx} f^{-1}(-4).$ 2. Let $f(x) =\\int_2^x \\dfrac{1}{1+x^3} dx$ Find $\\dfrac{d}{dx} f^{-1}(0).$", "# $xT=x$ iff $x=e$ implies that every $g$ may be written as $x^{-1}(xT)$ for some $x\\in G$. Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $xT=x$ for $x$ in $G$ if and only if $x=e$. Prove that every $g$ in $G$ can be represented as $g=x^{-1}(xT)$ for some $x$ in $G$. I don't understand how I should start to pursue a solution. Can you please give me a hint instead of a whole solution? Which properties of automorphisms should I use? Thanks. - It is quite unnecessary to say that the problem comes from Herstein, as the notation almost makes that clear :D – Mariano Suárez-Alvarez Feb 16 '13 at 4:10 I can't make sense of this: the OP wrote $$xT=T\\,\\,,\\,\\,\\forall\\,\\,x\\in G\\,\\Longleftrightarrow\\,x=e$$Did he actually mean $\\,xT=x\\ldots$ ? – DonAntonio Feb 16 '13 at 9:51 @DonAntonioSorry I made a mistake it has to be xT=x. And I guess I did this one but there is other question parallel with this question. Same introduction plus it says suppose further that T^2=Id. Prove G must be abelian. I assume we will use g= x^{-1}(xT) and somehow find G is abelian. But how? Do you have an idea on this? – khankrt Feb 16 '13 at 15:12 Try investigating the map $U:G\\to G$ given by" ]

[ "$z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $a_i$ (with different mass assigned on each root whose weights collectively sum to $1$). If $\\map P z = \\map {P'} z = 0$, then: $z = 1 \\cdot z + 0 \\cdot a_i$ and is still a convex combination of the roots of $P$. $\\blacksquare$ ## Source of Name This entry was named for Carl Friedrich Gauss and Édouard Lucas.", "# Is 0 a complex number? $0$ is a complex number (or rather it belongs to the set of complex numbers) since $x + y \\cdot i$ is a complex number even when $x = y = 0$", "unclearly formulated: usually $\\arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it. – user64494 Dec 18 '18 at 18:58", "Answer Comment Share Q) # If $z$ is a complex number having non zero imaginary part and $z^2+z+1=a$ where $a$ is real, then $a$ cannot take which of the following values? (A) -1 (B) 1/2 (C) 1/3 (D) 3/4", "# A BIT COMPLEX. ISN'T IT? Level pending For complex numbers z and w, it is given that : $$|z|^2w-|w|^2z=z-w$$. Then what is the value of $$\\overline{w}z$$ ? ×", "1. anonymous For the complex number $$z=a+bi$$ the absolute value is $$|z| = \\sqrt{a^2+b^2}$$. For your problem that's $\\sqrt{(-4)^2+(-\\sqrt{2})^2}$ 2. anonymous okay so just solve that?", "# A Fifth Root Algebra Level 3 Suppose $$z$$ is a complex number such that $$z^5 = -\\sqrt{3} + i$$ and $$z = \\sqrt[5]{2} ( \\cos n^{\\circ} + i \\sin n^{\\circ})$$, where $$270 < n < 360$$. Find the exact value of $$n$$. Details and assumptions $$i$$ is the imaginary unit, where $$i^2=-1$$. ×", "Find the value of $$n$$ if $$z=n\\pm \\sqrt{-i}$$ satisfies the equation $iz^{2}=1+\\frac{2}{z}+\\frac{3}{z^{2}}+\\frac{4}{z^{3}}+...$", "# Simple-complex ;-) Algebra Level 3 The Complex Number Z satisfies $Z + |Z| = 2 + 8i$ What is the Value of |Z| ? ×", "( |z|\\ ) of the complex number or iGoogle s scientific,!", "all independent solutions. It can have another root you could be missing. P.S. $z$ is a complex number", "# Easy but complex Algebra Level pending A complex number $$z$$ follows the following formula $\\sqrt[2n+1]{z} = 1, r_1, \\ldots r_{2n+1}$ Where $$n$$ is any positive integer which satisfies $n \\geq 1$ and $$r_x$$ is a root. Out of the following, which is equal to the sum of the roots. ×", "of complex numbers $z_1(z_2z_3)=(z_1z_2)z_3$", "+0 # I got \\sqrt{1/4+r^2}? 0 280 1 +1213 Let r be a real number, |r| < 2, and let z be a complex number such that $$z + \\frac{1}{z} = r$$ Find |z|. Apr 14, 2019", "Browse Questions # For any complex number $z$, the minimum value of $|z|+|z-1|=?$ (A) 1 (B) 0 (C) 1/2 (D) 3/2 Toolbox: • $|z_1|+|z_2|\\geq|z_1-z_2|$ $|z|+|z-1|\\geq|z-(z-1)|=1$ $\\Rightarrow\\:$ the least value of $|z|+|z-1|$ is $1$", "generally, how $z^{1/q}$, q being an integer, is defined. I have looked at a few complex analysis textbooks and they \"seem\" to miss that part! Page 1 of 2 12 Last", "Browse Questions # For any complex number $z$, the minimum value of $|z|+|z-1|=?$ (A) 1 (B) 0 (C) 1/2 (D) 3/2 Can you answer this question? Toolbox: • $|z_1|+|z_2|\\geq|z_1-z_2|$ $|z|+|z-1|\\geq|z-(z-1)|=1$ $\\Rightarrow\\:$ the least value of $|z|+|z-1|$ is $1$ answered Aug 1, 2013", "# Complex-city 2 Algebra Level 1 $\\large i{ z }^{ 3 }+z^{ 2 }-z+i=0$ For $$i = \\sqrt{-1}$$, $$z$$ is a complex number that satisfies the equation above. What is the value of $$\\left| z \\right|$$? ×", "# GATE2019-27 The value of the complex number $i^{-1/2}$ (where $i=\\sqrt{-1}$) is 1. $\\frac{1}{\\sqrt{2}}\\left ( 1-i \\right )$ 2. $-\\frac{1}{\\sqrt{2}}i$ 3. $\\frac{1}{\\sqrt{2}}i$ 4. $\\frac{1}{\\sqrt{2}}\\left ( 1+i \\right )$", "for a complex number z -... Now, | 5 − 5 i | = ( 5 ) 2 + 2\\sqrt i\\. Equal to 0 = sqrt ( a^2 + b^2 ) number x formula |z| = sqrt ( a^2 + ). New Jersey Business Search, Lemon Asparagus Stovetop, 25mm Fire Bricks, I Really Appreciate In Tagalog, Bmw 7 Series Olx Mumbai," ]

[ "(Note: cis is short for $$\\cos \\,+\\sin i$$) Find fourth roots of $$\\sqrt{3}+i$$ Convert $$\\sqrt{3}+i$$ to Polar first: $$\\displaystyle r=\\sqrt{{{{{\\left( {\\sqrt{3}} \\right)}}^{2}}+{{1}^{2}}}}=\\sqrt{4}=2;\\,\\,\\,\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{1}{{\\sqrt{3}}}} \\right)=\\frac{\\pi }{6}:\\,\\,\\,\\, 2\\,\\text{cis}\\left( {\\frac{\\pi }{6}} \\right)$$ $$\\displaystyle \\begin{array}{c}{{\\left[ {2\\,\\text{cis}\\left( {\\frac{\\pi }{6}} \\right)} \\right]}^{{\\,\\frac{1}{4}}}}={{2}^{{\\frac{1}{4}}}}\\text{cis}\\left( {\\frac{{\\frac{\\pi }{6}+2\\pi k}}{4}} \\right)\\\\=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{2\\pi k}}{4}} \\right);\\,\\,\\,\\,k=0,1,2,3\\\\=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi k}}{{24}}} \\right);\\,\\,\\,\\,k=0,1,2,3\\end{array}$$ $$\\displaystyle \\begin{array}{l}k=0:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 0}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}} \\right)\\\\k=1:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 1}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{13\\pi }}{{24}}} \\right)\\\\k=2:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 2}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{25\\pi }}{{24}}} \\right)\\\\k=3:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 3}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{37\\pi }}{{24}}} \\right)\\end{array}$$ Solve: $${{z}^{5}}=4-4i$$ (This is the same as taking the fifth roots of $$4-4i$$) Convert $$4-4i$$ to Polar: $$\\displaystyle r=\\sqrt{{{{4}^{2}}+{{{\\left( {-4} \\right)}}^{2}}}}=\\sqrt{{32}}={{\\left( {\\sqrt{2}} \\right)}^{5}};\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{{-4}}{4}} \\right)=\\frac{{7\\pi }}{4}:\\,\\, {{\\left( {\\sqrt{2}} \\right)}^{5}}\\text{cis}\\left( {\\frac{{7\\pi }}{4}} \\right)$$ $$\\require {cancel} \\displaystyle \\begin{array}{c}{{\\left[ {{{{\\left( {\\sqrt{2}} \\right)}}^{5}}\\text{cis}\\left( {\\frac{{7\\pi }}{4}} \\right)} \\right]}^{{\\,\\frac{1}{5}}}}={{\\left( {{{{\\left( {\\sqrt{2}} \\right)}}^{{\\cancel{5}}}}} \\right)}^{{\\frac{1}{{\\cancel{5}}}}}}\\text{cis}\\left( {\\frac{{\\frac{{7\\pi }}{4}+2\\pi k}}{5}} \\right)\\\\\\,\\,=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{2\\pi k}}{5}} \\right);\\,\\,\\,\\,k=0,1,2,3,4\\\\=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi k}}{{20}}} \\right);\\,\\,\\,\\,k=0,1,2,3,4\\end{array}$$ $$\\begin{array}{l}k=0:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 0}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}} \\right)\\\\k=1:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 1}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{3\\pi }}{{4}}} \\right)\\\\k=2:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 2}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{23\\pi }}{{20}}} \\right)\\\\k=3:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 3}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{31\\pi }}{{20}}} \\right)\\\\k=4:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 4}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{39\\pi }}{{20}}} \\right)\\end{array}$$ Here’s another type of problem you might see. Note we could have solved the second part of this problem more easily with the Complex Conjugate Root Theorem, or Conjugate Zeros Theorem that", "}z_{2}=3\\text{cis}\\left(40^{\\circ}\\right)$ 32. $z_{1}=6\\text{cis}\\left(\\frac{\\pi}{3}\\right)\\text{; }z_{2}=2\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ 33. $z_{1}=5\\sqrt{2}\\text{cis}\\left(\\pi\\right)\\text{; }z_{2}=\\sqrt{2}\\text{cis}\\left(\\frac{2\\pi}{3}\\right)$ 34. $z_{1}=2\\text{cis}\\left(\\frac{3\\pi}{5}\\right)\\text{; }z_{2}=3\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ For the following exercises, find the powers of each complex number in polar form. 35. Find $z^{3}$ when $z=5\\text{cis}\\left(45^{\\circ}\\right)$. 36. Find $z^{4}$ when $z=2\\text{cis}\\left(70^{\\circ}\\right)$. 37. Find $z^{2}$ when $z=3\\text{cis}\\left(120^{\\circ}\\right)$. 38. Find $z^{2}$ when $z=4\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ 39. Find $z^{4}$ when $z=\\text{cis}\\left(\\frac{3\\pi}{16}\\right)$. 40. Find $z^{3}$ when $z=3\\text{cis}\\left(\\frac{5\\pi}{3}\\right)$. For the following exercises, evaluate each root. 41. Evaluate the cube root of z when $z=27\\text{cis}\\left(240^{\\circ}\\right)$. 42. Evaluate the square root of z when $z=16\\text{cis}\\left(100^{\\circ}\\right)$. 43. Evaluate the cube root of z when $z=32\\text{cis}\\left(\\frac{2\\pi}{3}\\right)$. 44. Evaluate the square root of z when $z=32\\text{cis}\\left(\\pi\\right)$. 45. Evaluate the cube root of z when $z=8\\text{cis}\\left(\\frac{7\\pi}{4}\\right)$. For the following exercises, plot the complex number in the complex plane. 46. $2+4i$ 47. $−3−3i$ 48. $5−4i$ 49. $−1−5i$ 50. $3+2i$ 51. $2i$ 52. $−4$ 53. $6−2i$ 54. $−2+i$ 55. $1−4i$ For the following exercises, find all answers rounded to the nearest hundredth. 56. Use the rectangular to polar feature on the graphing calculator to change $5+5i$ to polar form. 57. Use the rectangular to polar feature on the graphing calculator to change $3−2i$ 58. Use the rectangular to polar feature on the graphing calculator to change $−3−8i$ 59. Use the polar to rectangular feature on the graphing calculator to change $4\\text{cis}\\left(120^{\\circ}\\right)$ to rectangular form. 60. Use the polar to rectangular", "+ \\frac{2k\\pi}{3}$$ $$z = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})$$ Should they be written like this instead, seeing as I've used the exponential form earlier: $$z = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})}$$ Would that be $$(\\sqrt3 - i)^{\\frac{1}{3}}$$? 5. Feb 28, 2010 ### Staff: Mentor Well, yes, but that doesn't tell your what the three cube roots are. 6. Feb 28, 2010 ### Char. Limit I W-A'd it... and only got 1 answer. 7. Mar 1, 2010 ### HallsofIvy \"$cis(\\theta)$\", which is short for $cos(\\theta)+ i sin(\\theta)= e^{i\\theta}$ is primarily an engineering notation. Which is better to use depends on what kind of course this is. If it is a mathematics course being taught be a mathematician, I would say definitely use $e^{i\\theta}$.", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "integer} \\right\\}$ and $\\text{Arg}(z) = \\frac{5\\pi}{6}$. \\item $z = -2 = 2\\text{cis}\\left(\\pi\\right)$, \\, $\\text{Re}(z) = -2$, \\, $\\text{Im}(z) =0$, \\, $|z| = 2$ $\\text{arg}(z) = \\left\\{(2k+1)\\pi \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = \\pi$. \\item $z = -\\frac{\\sqrt{3}}{2} - \\frac{1}{2}i = \\text{cis}\\left(\\frac{7\\pi}{6}\\right)$, \\, $\\text{Re}(z) = -\\frac{\\sqrt{3}}{2}$, \\, $\\text{Im}(z) = -\\frac{1}{2}$, \\, $|z| = 1$ $\\text{arg}(z) = \\left\\{\\frac{7\\pi}{6} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{5\\pi}{6}$. \\item $z = -3-3i = 3\\sqrt{2}\\text{cis}\\left(\\frac{5\\pi}{4}\\right)$, \\, $\\text{Re}(z) = -3$, \\, $\\text{Im}(z) =-3$, \\, $|z| = 3\\sqrt{2}$ $\\text{arg}(z) = \\left\\{\\frac{5\\pi}{4} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{3\\pi}{4}$. \\item $z = -5i = 5\\text{cis}\\left(\\frac{3\\pi}{2}\\right)$, \\, $\\text{Re}(z) = 0$, \\, $\\text{Im}(z) = -5$, \\, $|z| = 5$ $\\text{arg}(z) = \\left\\{\\frac{3\\pi}{2} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{\\pi}{2}$. \\item $z = 2\\sqrt{2} - 2i\\sqrt{2} = 4\\text{cis}\\left(\\frac{7\\pi}{4}\\right)$, \\, $\\text{Re}(z) = 2\\sqrt{2}$, \\, $\\text{Im}(z) = -2\\sqrt{2}$, \\, $|z| = 4$ $\\text{arg}(z) = \\left\\{\\frac{7\\pi}{4} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{\\pi}{4}$. \\item $z =6 = 6\\text{cis}\\left(0\\right)$, \\, $\\text{Re}(z) = 6$, \\, $\\text{Im}(z) = 0$, \\, $|z| = 6$ $\\text{arg}(z) = \\left\\{2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) =0$. \\item $z = i \\sqrt[3]{7} = \\sqrt[3]{7}\\text{cis}\\left(\\frac{\\pi}{2}\\right)$,", "$\\sqrt{a^2 + b^2} \\cdot \\text{cis}\\left(\\arctan \\frac{b}{a}\\right) = re^{\\theta i} = z$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\text{cis}(n\\pi - \\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{\\text{cis}(\\pi)}$ $\\text{cis}(\\pi) = \\cos(\\pi) + i\\sin(\\pi) = -1 + 0i = -1$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{-1}$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(n\\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(\\pi)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz$ $\\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z$ In conclusion, if $$n$$ is even then the product of the roots is $$-z$$, if $$n$$ is odd then the product of the roots is $$z$$. $\\text{When }n\\text{ is an even number}:~\\prod_{x = 0}^{n - 1}{Z_x} = -z$ $\\text{When }n\\text{ is an odd number}:~\\prod_{x = 0}^{n - 1}{Z_x} = z$ Hope you enjoyed the note Note by Jack Rawlin 1 year, 2 months ago Sort by:", "$\\sqrt{a^2 + b^2} \\cdot \\text{cis}\\left(\\arctan \\frac{b}{a}\\right) = re^{\\theta i} = z$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\text{cis}(n\\pi - \\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{\\text{cis}(\\pi)}$ $\\text{cis}(\\pi) = \\cos(\\pi) + i\\sin(\\pi) = -1 + 0i = -1$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{-1}$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(n\\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(\\pi)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz$ $\\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z$ In conclusion, if $$n$$ is even then the product of the roots is $$-z$$, if $$n$$ is odd then the product of the roots is $$z$$. $\\text{When }n\\text{ is an even number}:~\\prod_{x = 0}^{n - 1}{Z_x} = -z$ $\\text{When }n\\text{ is an odd number}:~\\prod_{x = 0}^{n - 1}{Z_x} = z$ Hope you enjoyed the note Note by Jack Rawlin 2 years, 2 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now", "Find the realand complex solutions to the equation X3(cubed) + 1=0 my choices a. -1,1,cis(pi/3) b. 1, cis(pi/3), cis (2pi/3) c. -1, cis(pi/3), cis(5pi/3) d. cis (pi/3), cis(2pi/3), cis(4pi/3) 2. Originally Posted by ccdalamp Find the realand complex solutions to the equation X3(cubed) + 1=0 my choices a. -1,1,cis(pi/3) b. 1, cis(pi/3), cis (2pi/3) c. -1, cis(pi/3), cis(5pi/3) d. cis (pi/3), cis(2pi/3), cis(4pi/3) $x^3=-1= cis((2n+1) \\pi )\\ \\ \\ n=0,1,2, ...$ so $ x=cis\\left(\\frac{2n+1}{3}\\pi \\right)\\ \\ \\ n=0,1,2, ... $ which if I am not mistaken is answer (c) RonL 3. Hello, ccdalamp! With \"cis\" in the answer choices, I assume that DeMoivre's Formula is expected. But it can be solved algebraically . . . with some trig. Find the real and complex solutions to the equation: . $x^3 + 1\\:=\\:0$ $a)\\;\\text{-}1,\\:1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right)\\qqua d\\qquad\\qquad b)\\;1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\te xt{cis}\\left(\\frac{2\\pi}{3}\\right)$ $c)\\;\\text{-}1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\text{ cis}\\left(\\frac{5\\pi}{3}\\right)\\qquad d)\\;\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\text{c is}\\left(\\frac{2\\pi}{3}\\right),\\:\\text{cis}\\left(\\ frac{4\\pi}{3}\\right)$ Factor: . $(x + 1)(x^2 - x + 1) \\:=\\:0$ And we have: . $\\begin{array}{ccccccc}x + 1 & = & 0 & \\;\\;\\Rightarrow\\;\\; & x & = & -1 \\\\ x^2-x+1 & = & 0 & \\Rightarrow & x & = & \\frac{1\\pm i\\sqrt{3}}{2} \\end{array}$ And the complex roots can be rewritten: . $\\begin{array}{ccc}\\frac{1}{2} + \\frac{\\sqrt{3}}{2}i & \\Rightarrow & \\cos\\left(\\frac{\\pi}{3}\\right) + i\\sin\\left(\\frac{\\pi}{3}\\right) \\\\ \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i & \\Rightarrow & \\cos\\left(\\frac{5\\pi}{3}\\right) + i\\sin\\left(\\frac{5\\pi}{3}\\right) \\end{array}$ ~ ~ ~ ~ ~", "Complex Roots - Not sure I did this right 1. Mar 27, 2009 Atena Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering. 1. The problem statement, all variables and given/known data Find all the roots of $$z^{4}=1-i$$ 2. Relevant equations I guess I should state De Moivre's here... $$(r cis(\\vartheta))^{n}=r^{n} cis (n\\vartheta)$$ 3. The attempt at a solution Firstly I re-wrote $$z^{4}=1-i$$ as $$z^{4}=\\sqrt{2} cis (\\frac{-\\pi}{4})$$ Using De Moivre's, $$z=(2\\frac{1}{2})^{\\frac{1}{4}} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ $$z=2\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ I found the four roots letting k=0,1,2,3 $$z=2^\\frac{1}{8} cis (\\frac{-\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+2\\pi))=2^\\frac{1}{8} cis (\\frac{7\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+4\\pi))=2^\\frac{1}{8} cis (\\frac{15\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+6\\pi))=2^\\frac{1}{8} cis (\\frac{23\\pi}{16})$$ 2. Mar 27, 2009 yyat Hi Atena! Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).", "of De Moivre. What I did: $z=\\sqrt{2-2i}$ $8^{\\frac{1}{4}} (cis\\left(\\frac{7\\pi}{8}\\right))$ I assume I would find values for $cos\\left(\\frac{7\\pi}{8}\\right)$ and $sin\\left(\\frac{7\\pi}{8}\\right)$ and then I would be done right? Assuming this is what I am supposed to do, how do I solve for the angle $\\frac{7\\pi}{8}$? Thanks. $\\displaystyle |2 - 2i| = \\sqrt{2^2 + (-2)^2} = \\sqrt{8} = 2\\sqrt{2}$ and $\\displaystyle \\arg{(2-2i)} = -\\arctan{\\frac{2}{2}} = -\\frac{\\pi}{4}$. So $\\displaystyle 2 - 2i = 2\\sqrt{2}\\,\\textrm{cis}\\,\\left(-\\frac{\\pi}{4}\\right)$. You have \\displaystyle \\begin{align*} z^2 &= 2\\sqrt{2}\\,\\textrm{cis}\\left(-\\frac{\\pi}{4}\\right) \\\\ z &= \\left[2\\sqrt{2}\\,\\textrm{cis}\\left(-\\frac{\\pi}{4}\\right)\\right]^{\\frac{1}{2}} \\\\ &= \\sqrt{2\\sqrt{2}}\\,\\textrm{cis}\\left[\\frac{1}{2}\\left(-\\frac{\\pi}{4}\\right)\\right] \\\\ &= \\sqrt[4]{8}\\,\\textrm{cis}\\left(-\\frac{\\pi}{8}\\right) \\\\ &= \\sqrt[4]{8}\\left[\\cos{\\left(-\\frac{\\pi}{8}\\right)} + i\\sin{\\left(-\\frac{\\pi}{8}\\right)}\\right] \\\\ &= \\sqrt[4]{8}\\left[\\cos{\\left(\\frac{\\pi}{8}\\right)} - i\\sin{\\left(\\frac{\\pi}{8}\\right)}\\right]\\end{align*} You will need to use half angle identities to evaluate these exactly. There is also another square root. You should know that square roots are evenly spaced about a circle, so these two square roots will have the same magnitude and are separated by an angle of $\\displaystyle \\pi$, so the other square root is \\displaystyle \\begin{align*} \\sqrt[4]{8}\\,\\textrm{cis}\\left(\\pi - \\frac{\\pi}{8}\\right) &= \\sqrt[4]{8}\\left[\\cos{\\left(\\pi - \\frac{\\pi}{8}\\right)} + i\\sin{\\left(\\pi - \\frac{\\pi}{8}\\right)}\\right] \\\\ &= \\sqrt[4]{8} \\left[ -\\cos{ \\left(\\frac{\\pi}{8}\\right) } + i \\sin{ \\left ( \\frac{\\pi}{8}\\right) } \\right]\\end{align*} and you should have those values from the first square root", "\\frac{2\\pi}{2} (1)\\right) = 2\\text{cis}\\left(\\frac{4\\pi}{3}\\right)\\). In rectangular form, the two square roots of $$z$$ are $$w_{\\text{\\tiny$$0\\)}} = 1+i\\sqrt{3}\\) and $$w_{\\text{\\tiny$$1\\)}} = -1-i\\sqrt{3}\\). We can check our answers by squaring them and showing that we get $$z= -2 + 2i\\sqrt{3}$$. 2. Proceeding as above, we get $$z = -16 = 16 \\text{cis}(\\pi)$$. With $$r = 16$$, $$\\theta = \\pi$$ and $$n = 4$$, we get the four fourth roots of $$z$$ to be $$w_{\\text{\\tiny$$0\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (0)\\right) = 2\\text{cis}\\left(\\frac{\\pi}{4}\\right)\\), $$w_{\\text{\\tiny$$1\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (1)\\right) = 2\\text{cis}\\left(\\frac{3\\pi}{4}\\right)\\), $$w_{\\text{\\tiny$$2\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (2)\\right) = 2\\text{cis}\\left(\\frac{5\\pi}{4}\\right)\\) and $$w_{\\text{\\tiny$$3\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (3)\\right) = 2\\text{cis}\\left(\\frac{7\\pi}{4}\\right)\\). Converting these to rectangular form gives $$w_{\\text{\\tiny$$0\\)}} = \\sqrt{2} + i\\sqrt{2}\\), $$w_{\\text{\\tiny$$1\\)}} = -\\sqrt{2} + i\\sqrt{2}\\), $$w_{\\text{\\tiny$$2\\)}} = -\\sqrt{2} - i\\sqrt{2}\\) and $$w_{\\text{\\tiny$$3\\)}} = \\sqrt{2} - i\\sqrt{2}\\). 3. For $$z = \\sqrt{2} + i \\sqrt{2}$$, we have $$z = 2\\text{cis}\\left(\\frac{\\pi}{4}\\right)$$. With $$r = 2$$, $$\\theta = \\frac{\\pi}{4}$$ and $$n =3$$ the usual computations yield $$w_{\\text{\\tiny$$0\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{\\pi}{12}\\right)\\), $$w_{\\text{\\tiny$$1\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{9\\pi}{12}\\right) = \\sqrt[3]{2} \\text{cis}\\left(\\frac{3\\pi}{4}\\right) $$and \\(w_{\\text{\\tiny$$2\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{17\\pi}{12}\\right)\\). If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem \\ref{circularsumdifference} or the Half-Angle Identities in Theorem \\ref{halfangle} to evaluate $$w_{\\text{\\tiny$$0\\)}}\\) and $$w_{\\text{\\tiny$$2\\)}}\\). Since we are not explicitly", "what you had not being an equation because you seem to have a problem with writing exactly what you mean- and it hurt you here. Those four numbers are the fourth roots- you have already solved the problem at this point: $z^4= cis(\\pi)$ and $z = cis(\\frac{\\pi}{4})$, $cis(\\frac{5\\pi}{4})$, $cis(\\frac{9\\pi}{4})$, $cis(\\frac{13\\pi}{4})$ $z^4 = cis(\\frac{\\pi}{16})$, $cis(\\frac{5\\pi}{16})$, $cis(\\frac{9\\pi}{16})$, $cis(\\frac{13\\pi}{16}$ P.S 5. Originally Posted by Unknown008 I would have used the difference of two squares twice: $z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$ almost: $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2}))(z-(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})$ Then, z = 1 and -1. For the other two complex roots, we get: $z^2 = -1$ Which you should be able to solve. 6. Originally Posted by HallsofIvy almost $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z+(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z-(\\sqrt{2}{2}- i\\sqrt{2}{2})(z+(\\sqrt{2}{2}- i\\sqrt{2}{2})$ Um... I don't understand why you look for $z^4+1$ while it's $z^4-1$... EDIT: Ah, okay. I understand now. 7. yes my bad, its meant to be $z^4+1=0$ srry for the confusion", "# arg(\\frac{1}{z}) Printable View • Sep 24th 2010, 04:10 PM dwsmith arg(\\frac{1}{z}) $arg(z)=\\theta+2\\pi k, \\ k\\in\\mathbb{Z}$ $arg\\left(\\frac{1}{z}\\right)=arg\\left(\\frac{\\overli ne{z}}{z\\overline{z}}\\right)$ $=arg\\left(\\frac{\\overline{z}}{|z|^2}\\right)=arg(\\o verline{z})-arg(|z|^2)$ $=-\\theta-(\\theta+2\\pi k)=-2\\theta-2\\pi k$ Is this correct? • Sep 24th 2010, 04:14 PM mr fantastic Quote: Originally Posted by dwsmith $arg(z)=\\theta+2\\pi k, \\ k\\in\\mathbb{Z}$ $arg\\left(\\frac{1}{z}\\right)=arg\\left(\\frac{\\overli ne{z}}{z\\overline{z}}\\right)$ $=arg\\left(\\frac{\\overline{z}}{|z|^2}\\right)=arg(\\o verline{z})-arg(|z|^2)$ $=-\\theta-(\\theta+2\\pi k)=-2\\theta-2\\pi k$ Is this correct? Note that $\\displaystyle \\frac{1}{z} = \\frac{\\text{cis} (0)}{r \\text{cis} (\\theta)} = \\frac{1}{r} \\text{cis} (0 - \\theta)$. • Sep 24th 2010, 04:17 PM dwsmith So the the argument of 1 over z is just going to be $-\\theta+2\\pi k$? • Sep 24th 2010, 04:29 PM Plato Let’s consider the principal argument, $-\\pi < \\text{Arg}(z)\\le \\pi$. The $\\text{Arg}(\\overline z)$ is simply its reflection in the real axis. You may want to reconsider? • Sep 24th 2010, 04:34 PM dwsmith If we pick up from where Mr F stopped, we have the angle as $0-\\theta=-\\theta$. Now, my question is then why isn't the $arg\\left(\\frac{1}{z}\\right)=-\\theta+2\\pi k \\ \\mbox{?}$ • Sep 24th 2010, 05:06 PM chiph588@ Quote: Originally Posted by dwsmith $arg(z)=\\theta+2\\pi k, \\ k\\in\\mathbb{Z}$ $arg\\left(\\frac{1}{z}\\right)=arg\\left(\\frac{\\overli ne{z}}{z\\overline{z}}\\right)$ $=arg\\left(\\frac{\\overline{z}}{|z|^2}\\right)=arg(\\o verline{z})-arg(|z|^2)$ $=-\\theta-(\\theta+2\\pi k)=-2\\theta-2\\pi k$ Is this correct? Given $c\\in\\mathbb{R}$ and $z\\in\\mathbb{C}$ then $\\text{arg}(cz)=\\text{arg}(z)$. What does this say about $\\text{arg}\\left(\\frac{\\overline{z}}{|z|^2}\\right) \\; ?$ • Sep 24th 2010, 05:09 PM dwsmith $arg\\left(\\frac{\\overline{z}}{|z|^2}\\right)=arg(\\ov erline{z})=-\\theta+2\\pi k$, yes? • Sep 24th 2010, 05:17 PM chiph588@ Quote: Originally Posted by", "i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was: e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3) I think it is there where i have gone wrong, help would be appreciated Thanks, ArTiCk $\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 = $ $\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$. 6. Hello, Mr. F ! Originally Posted by mr fantastic $\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 = $ $\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$. Lovely!", "we can also use the value $2\\pi$ for theta, because $2\\pi$ radians is coterminal with $0$ radians. Here's the trick: we will start with $0$ for $\\theta$, and write more coterminal $\\mathrm{cis}$ values so that we have as many as there are roots. Here we are told that there are $3$ cube roots of $1$, trusting when I earlier said there are always $n$ number of $n$-th roots.$$1 = 1 \\, \\mathrm{cis}(0)$$$$1 = 1 \\, \\mathrm{cis}(2\\pi)$$$$1 = 1 \\, \\mathrm{cis}(4\\pi)$$Each of these 3 forms represents the number 1 in $\\mathrm{cis}$ form. Now for the payoff! Take the cube root of each of those values, using DeMoivre's Theorem:$$1^\\frac{1}{3} = 1^\\frac{1}{3} \\left[ 1 \\, \\mathrm{cis}\\left(0 \\, \\left(\\frac{1}{3}\\right)\\right) \\right] = 1 \\, \\mathrm{cis}(0)$$$$1^\\frac{1}{3} = 1^\\frac{1}{3} \\left[ 1 \\, \\mathrm{cis}\\left(2\\pi \\, \\left(\\frac{1}{3}\\right)\\right) \\right] = 1 \\, \\mathrm{cis}\\left(\\frac{2\\pi}{3}\\right)$$$$1^\\frac{1}{3} = 1^\\frac{1}{3} \\left[ 1 \\, \\mathrm{cis}\\left(4\\pi \\, \\left(\\frac{1}{3}\\right)\\right) \\right] = 1 \\, \\mathrm{cis}\\left(\\frac{4\\pi}{3}\\right)$$Finally, use the definition of $\\mathrm{cis}$, along with the known values of cosine and sine at these angles, we get the answers.$$1^\\frac{1}{3} = 1 \\, \\mathrm{cis}(0)$$$$= 1[\\cos(0) + i \\sin(0)] = 1[1 + 0] = \\boxed{1}$$$$1^\\frac{1}{3} = 1 \\, \\mathrm{cis}\\left(\\frac{2\\pi}{3}\\right) = 1 \\left[\\cos\\left(\\frac{2\\pi}{3}\\right) + i \\sin\\left(\\frac{2\\pi}{3}\\right)\\right]$$$$= 1\\left[-\\frac{1}{2} + i \\left(\\sqrt\\frac{3}{2}\\right)\\right] = \\boxed{-\\frac{1}{2} + i \\sqrt\\frac{3}{2}}$$$$1^\\frac{1}{3} = 1 \\, \\mathrm{cis}\\left(\\frac{4\\pi}{3}\\right) = 1 \\left[\\cos\\left(\\frac{2\\pi}{3}\\right) + i \\sin\\left(\\frac{4\\pi}{3}\\right)\\right]$$$$= 1\\left[-\\frac{1}{2} + i \\left(-\\sqrt\\frac{3}{2}\\right)\\right] = \\boxed{-\\frac{1}{2} - i", "&=\\sum_{k=0}^\\infty\\sum_{n=0}^\\infty(-1)^kx^{n+k}\\binom{n}{k}\\\\ &=\\sum_{k=0}^\\infty(-1)^kx^k\\frac{x^k}{(1-x)^{k+1}}\\\\ &=\\frac{1}{1-x}\\sum_{k=0}^\\infty(-1)^k\\left(\\frac{x^2}{1-x}\\right)^k\\\\ &=\\frac{1}{1-x}\\frac{1}{1+\\frac{x^2}{1-x}}\\\\ &=\\frac{1}{1-x+x^2}\\\\ &=\\frac{1}{\\alpha-\\beta}\\left(\\frac{1}{x-\\alpha}-\\frac{1}{x-\\beta}\\right)\\tag{2} \\end{align} where $\\alpha$ and $\\beta$ are the roots of $1-x+x^2=0$; $\\alpha=\\cis(\\pi/3)$ and $\\beta=\\cis(-\\pi/3)$. Thus, the series for $(2)$ is \\begin{align} &\\frac{1}{\\cis(\\pi/3)-\\cis(-\\pi/3)}\\left(\\frac{\\cis(\\pi/3)}{1-\\cis(\\pi/3)x}-\\frac{\\cis(-\\pi/3)}{1-\\cis(-\\pi/3)x}\\right)\\\\ &=\\frac{2}{\\sqrt{3}}\\Im\\left(\\frac{\\cis(\\pi/3)}{1-\\cis(\\pi/3)x}\\right)\\\\ &=\\frac{2}{\\sqrt{3}}\\Im\\left(\\sum_{n=0}^\\infty\\cis((n+1)\\pi/3)x^n\\right)\\tag{3} \\end{align} Therefore, $$a_n=\\frac{2}{\\sqrt{3}}\\sin\\left(\\frac{n+1}{3}\\pi\\right)\\tag{4}$$ and the sequence requested above is $$a_{2n}=\\frac{2}{\\sqrt{3}}\\sin\\left(\\frac{2n+1}{3}\\pi\\right)\\tag{4}$$ - Consider the more general number $C_n$ defined as $$C_n = \\sum_{k = 0}^{\\infty} (-1)^k{n-k \\choose k}$$ where ${m \\choose k} = 0$ if $k > m$. Then your numbers are $C_{2n}$ for $n = 0, 1, 2, \\dots$. These numbers satisfy $$\\begin{eqnarray} C_n &=& \\sum_{k = 0}^{\\infty}(-1)^k\\left( {n-1-k \\choose k} + {n-1-k \\choose k-1} \\right)\\\\ &=& C_{n-1} - \\sum_{k = 0}^{\\infty} (-1)^k {n - 2 - k \\choose k}\\\\ &=& C_{n-1} - C_{n-2} \\end{eqnarray}$$ The sequence $C_0, C_1, C_2, \\dotsc$ is therefore $$1, 1, 0, -1, -1, 0, 1, 1, 0, -1, \\dotsc$$ and taking out the even terms $C_0, C_2, C_4, \\dotsc$ gives your sequence $$1, 0, -1, 1, 0, -1, \\dotsc$$ In other words $C_{2n} = (2 - n \\mod 3) - 1$. - If you want only the result, CAS says $$\\frac{2\\sqrt{3}}{3}(\\cos\\left( \\frac{(4n-1)\\pi}{6}\\right).$$ -", "we learned about here in the Imaginary (Complex) Numbers Section. Complex Trig Root Problem Solution (Note: cis is short for $$\\cos \\,+\\sin i$$) Determine $$z$$ and three cube roots of $$z$$ if one cube root is $$1-\\sqrt{3}i$$ We can first find $$z$$ by cubing using De Moivre’s Theorem : $$\\displaystyle {{z}^{n}}={{r}^{n}}\\left[ {\\cos \\left( {n\\theta } \\right)+i\\sin \\left( {n\\theta } \\right)} \\right]\\,$$ Convert $$1-\\sqrt{3}i$$ to Polar first: $$\\displaystyle r=\\sqrt{{{{1}^{2}}+{{{\\left( {-\\sqrt{3}} \\right)}}^{2}}}}=2;\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{{-\\sqrt{3}}}{1}} \\right)=\\frac{{5\\pi }}{3}:\\,\\, 2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)$$ $$\\displaystyle \\begin{array}{c}{{\\left[ {2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)} \\right]}^{{\\,3}}}={{2}^{3}}\\text{cis}\\left( {3\\cdot \\frac{{5\\pi }}{3}} \\right)=8\\,\\text{cis}\\left( {5\\pi } \\right)=8\\,\\text{cis}\\left( {5\\pi -4\\pi } \\right)=8\\,\\text{cis}\\left( \\pi \\right)\\\\=8\\left( {-1+0i} \\right)=-8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,z=-8\\end{array}$$ Since we see that $$z=-8$$, and we know that the real number cube root of –8 is –2, we can then take the conjugate of $$1-\\sqrt{3}i$$ to get the other root, $$1+\\sqrt{3}i$$ (from the Complex Conjugate Root Theorem). The roots are –2, $$1-\\sqrt{3}i$$, and $$1+\\sqrt{3}i$$. But we’ll check our answer using the $$n$$th root formulas: $$\\begin{array}{c}{{\\left[ {8\\text{cis}\\left( \\pi \\right)} \\right]}^{{\\,\\frac{1}{3}}}}={{8}^{{\\frac{1}{3}}}}\\text{cis}\\left( {\\frac{{\\pi +2\\pi k}}{3}} \\right)\\\\=2\\,\\text{cis}\\left( {\\frac{\\pi }{3}+\\frac{{2\\pi k}}{3}} \\right);\\,\\,\\,\\,\\,k=0,1,2\\end{array}$$ $$\\begin{array}{l}k=0:\\,\\,\\,\\,2\\,\\text{cis}\\left( {\\frac{\\pi }{3}} \\right)=2\\left( {\\frac{1}{2}+\\frac{{\\sqrt{3}}}{2}i} \\right)=1+\\sqrt{3}i\\\\k=1:\\,\\,\\,\\,2\\,\\text{cis}\\left( \\pi \\right)=2\\left( {-1+0i} \\right)=-2\\\\k=2:\\,\\,\\,\\,2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)=2\\left( {\\frac{1}{2}-\\frac{{\\sqrt{3}}}{2}i} \\right)=1-\\sqrt{3}i\\end{array}$$ How to Check with Graphing Calculator When getting roots of complex numbers in the calculator, you’ll only get one root, so this won’t be too helpful. But you can check all the roots you", "$z = \\frac {\\left(\\cos \\frac {\\pi}{4} - isin \\frac {\\pi}{4}\\right)^2 \\left(\\cos \\frac {\\pi}{3} + isin \\frac {\\pi}{3}\\right)^3}{\\left(\\cos \\frac {\\pi}{24} - isin \\frac {\\pi}{24}\\right)^4}$ I edited my original post - thanks for correcting me. $z = \\frac{ \\left[ \\text{cis} \\left( -\\frac{\\pi}{4}\\right) \\right]^2 \\left[ \\text{cis} \\left(\\frac{\\pi}{3}\\right) \\right]^3}{\\left[ \\text{cis} \\left(-\\frac{\\pi}{24}\\right) \\right]^4}$ Apply de Mooivre's Theorem: $= \\frac{\\text{cis} \\left[ 2\\left( -\\frac{\\pi}{4}\\right) \\right]\\, \\text{cis} \\left[3 \\left(\\frac{\\pi}{3}\\right)\\right]}{\\text{cis} \\left[ 4 \\left(- \\frac{\\pi}{24}\\right)\\right]} = \\frac{\\text{cis} \\left( -\\frac{\\pi}{2}\\right) \\, \\text{cis} (\\pi)}{\\text{cis} \\left(- \\frac{\\pi}{6}\\right)}$ Apply the usual theorems for multiplying and dividing complex numbers expressed in polar form: $= \\text{cis} \\left( -\\frac{\\pi}{2} + \\pi - \\left[-\\frac{\\pi}{6} \\right]\\right) = \\text{cis} \\left( \\frac{2\\pi}{3} \\right)$. And of course you can easily express this in Cartesian form if desired. 8. Thank you all for your help - you guys are truly amazing! 9. There's another part to this question: Using De Moivre's theorem, show that $z = \\sqrt [3]{1}$ ---------------------- Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get: $\\left(\\cos \\frac {\\pi}{24} - isin \\frac {\\pi}{24}\\right)^4$ to become $\\left[cis \\frac {\\pi}{4} \\right]^6$ Your final answer is $cis \\frac {\\pi}{4}$ but I seem to have gotten $cis \\frac {5\\pi}{6}$ Is there something I'm not understanding? ---------------------- Also, how do I show $z = \\sqrt [3]{1}$? How can I apply De Moivre's", "- \\frac{5\\sqrt{3}}{2} i$ \\end{multicols} \\item Since $z = i = \\text{cis}\\left(\\frac{\\pi}{2}\\right)$ we have \\begin{multicols}{3} $w_{\\text{\\tiny$0$}} = \\text{cis}\\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2} + \\frac{1}{2}i$ $w_{\\text{\\tiny$1$}} = \\text{cis}\\left(\\frac{5\\pi}{6}\\right) = -\\frac{\\sqrt{3}}{2} + \\frac{1}{2}i$ $w_{\\text{\\tiny$2$}} = \\text{cis}\\left(\\frac{3\\pi}{2}\\right) = -i$ \\end{multicols} \\item Since $z = -8i = 8\\text{cis}\\left(\\frac{3\\pi}{2}\\right)$ we have \\begin{multicols}{3} $w_{\\text{\\tiny$0$}} = 2\\text{cis}\\left(\\frac{\\pi}{2}\\right) = 2i$ $w_{\\text{\\tiny$1$}} = 2\\text{cis}\\left(\\frac{7\\pi}{6}\\right) = -\\sqrt{3} -i$ $w_{\\text{\\tiny$2$}} = \\text{cis}\\left(\\frac{11\\pi}{6}\\right) = \\sqrt{3}-i$ \\end{multicols} \\item Since $z=16 = 16\\text{cis}\\left(0 \\right)$ we have \\begin{multicols}{2} $w_{\\text{\\tiny$0$}} =2\\text{cis}\\left(0\\right) =2$ $w_{\\text{\\tiny$1$}} = 2\\text{cis}\\left(\\frac{\\pi}{2}\\right) = 2i$ \\end{multicols} \\begin{multicols}{2} $w_{\\text{\\tiny$2$}} = 2\\text{cis}\\left(\\pi\\right) = -2$ $w_{\\text{\\tiny$3$}} = 2\\text{cis}\\left(\\frac{3\\pi}{2}\\right) = -2i$ \\end{multicols} \\item Since $z=-81 = 81\\text{cis}\\left(\\pi \\right)$ we have \\begin{multicols}{2} $w_{\\text{\\tiny$0$}} =3\\text{cis}\\left(\\frac{\\pi}{4}\\right) = \\frac{3\\sqrt{2}}{2} + \\frac{3\\sqrt{2}}{2}i$ $w_{\\text{\\tiny$1$}} = 3\\text{cis}\\left(\\frac{3\\pi}{4}\\right) =-\\frac{3\\sqrt{2}}{2} + \\frac{3\\sqrt{2}}{2}i$ \\end{multicols} \\begin{multicols}{2} $w_{\\text{\\tiny$2$}} = 3\\text{cis}\\left(\\frac{5\\pi}{4}\\right) =-\\frac{3\\sqrt{2}}{2} - \\frac{3\\sqrt{2}}{2}i$ $w_{\\text{\\tiny$3$}} = 3\\text{cis}\\left(\\frac{7\\pi}{4}\\right) =\\frac{3\\sqrt{2}}{2} - \\frac{3\\sqrt{2}}{2}i$ \\end{multicols} \\item Since $z = 64 = 64\\text{cis}(0)$ we have \\begin{multicols}{3} $w_{\\text{\\tiny$0$}} = 2\\text{cis}(0) = 2$ $w_{\\text{\\tiny$1$}} = 2\\text{cis}\\left(\\frac{\\pi}{3}\\right) = 1 + \\sqrt{3}i$ $w_{\\text{\\tiny$2$}} = 2\\text{cis}\\left(\\frac{2\\pi}{3}\\right) = -1 + \\sqrt{3}i$ \\end{multicols} \\begin{multicols}{3} $w_{\\text{\\tiny$3$}} = 2\\text{cis}\\left(\\pi\\right) = -2$ $w_{\\text{\\tiny$4$}} = 2\\text{cis}\\left(-\\frac{2\\pi}{3}\\right) = -1 - \\sqrt{3}i$ $w_{\\text{\\tiny$5$}} = 2\\text{cis}\\left(-\\frac{\\pi}{3}\\right) = 1 - \\sqrt{3}i$ \\end{multicols} \\item Since $z = -729 = 729 \\text{cis}(\\pi)$ we have \\begin{multicols}{3} $w_{\\text{\\tiny$0$}} = 3\\text{cis}\\left(\\frac{\\pi}{6}\\right) = \\frac{3\\sqrt{3}}{2} + \\frac{3}{2}i$ $w_{\\text{\\tiny$1$}} = 3\\text{cis}\\left(\\frac{\\pi}{2}\\right) = 3i$ $w_{\\text{\\tiny$2$}} = 3\\text{cis}\\left(\\frac{5\\pi}{6}\\right) = -\\frac{3\\sqrt{3}}{2} + \\frac{3}{2}i$ \\end{multicols} \\begin{multicols}{3} $w_{\\text{\\tiny$3$}} = 3\\text{cis}\\left(\\frac{7\\pi}{6}\\right) = -\\frac{3\\sqrt{3}}{2}-\\frac{3}{2}i$ $w_{\\text{\\tiny$4$}} = 3\\text{cis}\\left(-\\frac{3\\pi}{2}\\right)", "can I apply De Moivre's theorem into this? Thank you - just this last one! $z = \\text{cis} \\left( \\frac{2 \\pi}{3} \\right) \\Rightarrow z^3 = \\left[ \\text{cis} \\left( \\frac{2 \\pi}{3} \\right) \\right]^3 = \\text{cis} \\left( 3 \\left( \\frac{2 \\pi}{3} \\right) \\right) = \\text{cis} (2 \\pi) = 1$. 13. Originally Posted by Moo What can be not understood is why he made this mistake $z=\\frac{[cis(-\\frac{\\pi}{4})]^2 [cis(\\frac{\\pi}{3})]^3}{[cis(-\\frac{\\pi}{24})]^4}$ $z=\\frac{cis(-\\frac{\\pi}{2}) \\cdot cis(\\pi)}{cis(-\\frac{\\pi}{6})}$ $z=cis\\left(-\\frac{\\pi}{2}+\\pi+\\frac{\\pi}6\\right)=cis \\left(\\frac{2\\pi}{3}\\right)$ @Mr F : still not, because it's $-\\frac{\\pi}{24}$ in the denominator I fixed it while you were busy laughing at me (Actually I know how much you love finding my mistakes so I left it as a gift for you (not quite a gold necklace). ) 14. Hehe, thank you very much for your help! I know I can always count on you guys" ]

[ "(Note: cis is short for $$\\cos \\,+\\sin i$$) Find fourth roots of $$\\sqrt{3}+i$$ Convert $$\\sqrt{3}+i$$ to Polar first: $$\\displaystyle r=\\sqrt{{{{{\\left( {\\sqrt{3}} \\right)}}^{2}}+{{1}^{2}}}}=\\sqrt{4}=2;\\,\\,\\,\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{1}{{\\sqrt{3}}}} \\right)=\\frac{\\pi }{6}:\\,\\,\\,\\, 2\\,\\text{cis}\\left( {\\frac{\\pi }{6}} \\right)$$ $$\\displaystyle \\begin{array}{c}{{\\left[ {2\\,\\text{cis}\\left( {\\frac{\\pi }{6}} \\right)} \\right]}^{{\\,\\frac{1}{4}}}}={{2}^{{\\frac{1}{4}}}}\\text{cis}\\left( {\\frac{{\\frac{\\pi }{6}+2\\pi k}}{4}} \\right)\\\\=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{2\\pi k}}{4}} \\right);\\,\\,\\,\\,k=0,1,2,3\\\\=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi k}}{{24}}} \\right);\\,\\,\\,\\,k=0,1,2,3\\end{array}$$ $$\\displaystyle \\begin{array}{l}k=0:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 0}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}} \\right)\\\\k=1:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 1}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{13\\pi }}{{24}}} \\right)\\\\k=2:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 2}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{25\\pi }}{{24}}} \\right)\\\\k=3:\\,\\,\\,\\,\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{\\pi }{{24}}+\\frac{{12\\pi \\cdot 3}}{{24}}} \\right)=\\sqrt[4]{2}\\,\\text{cis}\\left( {\\frac{{37\\pi }}{{24}}} \\right)\\end{array}$$ Solve: $${{z}^{5}}=4-4i$$ (This is the same as taking the fifth roots of $$4-4i$$) Convert $$4-4i$$ to Polar: $$\\displaystyle r=\\sqrt{{{{4}^{2}}+{{{\\left( {-4} \\right)}}^{2}}}}=\\sqrt{{32}}={{\\left( {\\sqrt{2}} \\right)}^{5}};\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{{-4}}{4}} \\right)=\\frac{{7\\pi }}{4}:\\,\\, {{\\left( {\\sqrt{2}} \\right)}^{5}}\\text{cis}\\left( {\\frac{{7\\pi }}{4}} \\right)$$ $$\\require {cancel} \\displaystyle \\begin{array}{c}{{\\left[ {{{{\\left( {\\sqrt{2}} \\right)}}^{5}}\\text{cis}\\left( {\\frac{{7\\pi }}{4}} \\right)} \\right]}^{{\\,\\frac{1}{5}}}}={{\\left( {{{{\\left( {\\sqrt{2}} \\right)}}^{{\\cancel{5}}}}} \\right)}^{{\\frac{1}{{\\cancel{5}}}}}}\\text{cis}\\left( {\\frac{{\\frac{{7\\pi }}{4}+2\\pi k}}{5}} \\right)\\\\\\,\\,=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{2\\pi k}}{5}} \\right);\\,\\,\\,\\,k=0,1,2,3,4\\\\=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi k}}{{20}}} \\right);\\,\\,\\,\\,k=0,1,2,3,4\\end{array}$$ $$\\begin{array}{l}k=0:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 0}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}} \\right)\\\\k=1:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 1}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{3\\pi }}{{4}}} \\right)\\\\k=2:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 2}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{23\\pi }}{{20}}} \\right)\\\\k=3:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 3}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{31\\pi }}{{20}}} \\right)\\\\k=4:\\,\\,\\,\\,\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{7\\pi }}{{20}}+\\frac{{8\\pi \\cdot 4}}{{20}}} \\right)=\\sqrt{2}\\,\\text{cis}\\left( {\\frac{{39\\pi }}{{20}}} \\right)\\end{array}$$ Here’s another type of problem you might see. Note we could have solved the second part of this problem more easily with the Complex Conjugate Root Theorem, or Conjugate Zeros Theorem that", "}z_{2}=3\\text{cis}\\left(40^{\\circ}\\right)$ 32. $z_{1}=6\\text{cis}\\left(\\frac{\\pi}{3}\\right)\\text{; }z_{2}=2\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ 33. $z_{1}=5\\sqrt{2}\\text{cis}\\left(\\pi\\right)\\text{; }z_{2}=\\sqrt{2}\\text{cis}\\left(\\frac{2\\pi}{3}\\right)$ 34. $z_{1}=2\\text{cis}\\left(\\frac{3\\pi}{5}\\right)\\text{; }z_{2}=3\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ For the following exercises, find the powers of each complex number in polar form. 35. Find $z^{3}$ when $z=5\\text{cis}\\left(45^{\\circ}\\right)$. 36. Find $z^{4}$ when $z=2\\text{cis}\\left(70^{\\circ}\\right)$. 37. Find $z^{2}$ when $z=3\\text{cis}\\left(120^{\\circ}\\right)$. 38. Find $z^{2}$ when $z=4\\text{cis}\\left(\\frac{\\pi}{4}\\right)$ 39. Find $z^{4}$ when $z=\\text{cis}\\left(\\frac{3\\pi}{16}\\right)$. 40. Find $z^{3}$ when $z=3\\text{cis}\\left(\\frac{5\\pi}{3}\\right)$. For the following exercises, evaluate each root. 41. Evaluate the cube root of z when $z=27\\text{cis}\\left(240^{\\circ}\\right)$. 42. Evaluate the square root of z when $z=16\\text{cis}\\left(100^{\\circ}\\right)$. 43. Evaluate the cube root of z when $z=32\\text{cis}\\left(\\frac{2\\pi}{3}\\right)$. 44. Evaluate the square root of z when $z=32\\text{cis}\\left(\\pi\\right)$. 45. Evaluate the cube root of z when $z=8\\text{cis}\\left(\\frac{7\\pi}{4}\\right)$. For the following exercises, plot the complex number in the complex plane. 46. $2+4i$ 47. $−3−3i$ 48. $5−4i$ 49. $−1−5i$ 50. $3+2i$ 51. $2i$ 52. $−4$ 53. $6−2i$ 54. $−2+i$ 55. $1−4i$ For the following exercises, find all answers rounded to the nearest hundredth. 56. Use the rectangular to polar feature on the graphing calculator to change $5+5i$ to polar form. 57. Use the rectangular to polar feature on the graphing calculator to change $3−2i$ 58. Use the rectangular to polar feature on the graphing calculator to change $−3−8i$ 59. Use the polar to rectangular feature on the graphing calculator to change $4\\text{cis}\\left(120^{\\circ}\\right)$ to rectangular form. 60. Use the polar to rectangular", "+ \\frac{2k\\pi}{3}$$ $$z = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})$$ Should they be written like this instead, seeing as I've used the exponential form earlier: $$z = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})}$$ Would that be $$(\\sqrt3 - i)^{\\frac{1}{3}}$$? 5. Feb 28, 2010 ### Staff: Mentor Well, yes, but that doesn't tell your what the three cube roots are. 6. Feb 28, 2010 ### Char. Limit I W-A'd it... and only got 1 answer. 7. Mar 1, 2010 ### HallsofIvy \"$cis(\\theta)$\", which is short for $cos(\\theta)+ i sin(\\theta)= e^{i\\theta}$ is primarily an engineering notation. Which is better to use depends on what kind of course this is. If it is a mathematics course being taught be a mathematician, I would say definitely use $e^{i\\theta}$.", "integer} \\right\\}$ and $\\text{Arg}(z) = \\frac{5\\pi}{6}$. \\item $z = -2 = 2\\text{cis}\\left(\\pi\\right)$, \\, $\\text{Re}(z) = -2$, \\, $\\text{Im}(z) =0$, \\, $|z| = 2$ $\\text{arg}(z) = \\left\\{(2k+1)\\pi \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = \\pi$. \\item $z = -\\frac{\\sqrt{3}}{2} - \\frac{1}{2}i = \\text{cis}\\left(\\frac{7\\pi}{6}\\right)$, \\, $\\text{Re}(z) = -\\frac{\\sqrt{3}}{2}$, \\, $\\text{Im}(z) = -\\frac{1}{2}$, \\, $|z| = 1$ $\\text{arg}(z) = \\left\\{\\frac{7\\pi}{6} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{5\\pi}{6}$. \\item $z = -3-3i = 3\\sqrt{2}\\text{cis}\\left(\\frac{5\\pi}{4}\\right)$, \\, $\\text{Re}(z) = -3$, \\, $\\text{Im}(z) =-3$, \\, $|z| = 3\\sqrt{2}$ $\\text{arg}(z) = \\left\\{\\frac{5\\pi}{4} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{3\\pi}{4}$. \\item $z = -5i = 5\\text{cis}\\left(\\frac{3\\pi}{2}\\right)$, \\, $\\text{Re}(z) = 0$, \\, $\\text{Im}(z) = -5$, \\, $|z| = 5$ $\\text{arg}(z) = \\left\\{\\frac{3\\pi}{2} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{\\pi}{2}$. \\item $z = 2\\sqrt{2} - 2i\\sqrt{2} = 4\\text{cis}\\left(\\frac{7\\pi}{4}\\right)$, \\, $\\text{Re}(z) = 2\\sqrt{2}$, \\, $\\text{Im}(z) = -2\\sqrt{2}$, \\, $|z| = 4$ $\\text{arg}(z) = \\left\\{\\frac{7\\pi}{4} + 2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) = -\\frac{\\pi}{4}$. \\item $z =6 = 6\\text{cis}\\left(0\\right)$, \\, $\\text{Re}(z) = 6$, \\, $\\text{Im}(z) = 0$, \\, $|z| = 6$ $\\text{arg}(z) = \\left\\{2\\pi k \\, | \\, \\text{$k$is an integer} \\right\\}$ and $\\text{Arg}(z) =0$. \\item $z = i \\sqrt[3]{7} = \\sqrt[3]{7}\\text{cis}\\left(\\frac{\\pi}{2}\\right)$,", "Find the realand complex solutions to the equation X3(cubed) + 1=0 my choices a. -1,1,cis(pi/3) b. 1, cis(pi/3), cis (2pi/3) c. -1, cis(pi/3), cis(5pi/3) d. cis (pi/3), cis(2pi/3), cis(4pi/3) 2. Originally Posted by ccdalamp Find the realand complex solutions to the equation X3(cubed) + 1=0 my choices a. -1,1,cis(pi/3) b. 1, cis(pi/3), cis (2pi/3) c. -1, cis(pi/3), cis(5pi/3) d. cis (pi/3), cis(2pi/3), cis(4pi/3) $x^3=-1= cis((2n+1) \\pi )\\ \\ \\ n=0,1,2, ...$ so $ x=cis\\left(\\frac{2n+1}{3}\\pi \\right)\\ \\ \\ n=0,1,2, ... $ which if I am not mistaken is answer (c) RonL 3. Hello, ccdalamp! With \"cis\" in the answer choices, I assume that DeMoivre's Formula is expected. But it can be solved algebraically . . . with some trig. Find the real and complex solutions to the equation: . $x^3 + 1\\:=\\:0$ $a)\\;\\text{-}1,\\:1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right)\\qqua d\\qquad\\qquad b)\\;1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\te xt{cis}\\left(\\frac{2\\pi}{3}\\right)$ $c)\\;\\text{-}1,\\:\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\text{ cis}\\left(\\frac{5\\pi}{3}\\right)\\qquad d)\\;\\text{cis}\\left(\\frac{\\pi}{3}\\right),\\:\\text{c is}\\left(\\frac{2\\pi}{3}\\right),\\:\\text{cis}\\left(\\ frac{4\\pi}{3}\\right)$ Factor: . $(x + 1)(x^2 - x + 1) \\:=\\:0$ And we have: . $\\begin{array}{ccccccc}x + 1 & = & 0 & \\;\\;\\Rightarrow\\;\\; & x & = & -1 \\\\ x^2-x+1 & = & 0 & \\Rightarrow & x & = & \\frac{1\\pm i\\sqrt{3}}{2} \\end{array}$ And the complex roots can be rewritten: . $\\begin{array}{ccc}\\frac{1}{2} + \\frac{\\sqrt{3}}{2}i & \\Rightarrow & \\cos\\left(\\frac{\\pi}{3}\\right) + i\\sin\\left(\\frac{\\pi}{3}\\right) \\\\ \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i & \\Rightarrow & \\cos\\left(\\frac{5\\pi}{3}\\right) + i\\sin\\left(\\frac{5\\pi}{3}\\right) \\end{array}$ ~ ~ ~ ~ ~", "$\\sqrt{a^2 + b^2} \\cdot \\text{cis}\\left(\\arctan \\frac{b}{a}\\right) = re^{\\theta i} = z$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\text{cis}(n\\pi - \\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{\\text{cis}(\\pi)}$ $\\text{cis}(\\pi) = \\cos(\\pi) + i\\sin(\\pi) = -1 + 0i = -1$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{-1}$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(n\\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(\\pi)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz$ $\\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z$ In conclusion, if $$n$$ is even then the product of the roots is $$-z$$, if $$n$$ is odd then the product of the roots is $$z$$. $\\text{When }n\\text{ is an even number}:~\\prod_{x = 0}^{n - 1}{Z_x} = -z$ $\\text{When }n\\text{ is an odd number}:~\\prod_{x = 0}^{n - 1}{Z_x} = z$ Hope you enjoyed the note Note by Jack Rawlin 1 year, 2 months ago Sort by:", "$\\sqrt{a^2 + b^2} \\cdot \\text{cis}\\left(\\arctan \\frac{b}{a}\\right) = re^{\\theta i} = z$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\text{cis}(n\\pi - \\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{\\text{cis}(\\pi)}$ $\\text{cis}(\\pi) = \\cos(\\pi) + i\\sin(\\pi) = -1 + 0i = -1$ $\\prod_{x = 0}^{n - 1}{Z_x} = z\\frac{\\text{cis}(n\\pi)}{-1}$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(n\\pi)$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z\\text{cis}(\\pi)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n$ $\\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz$ $\\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z$ In conclusion, if $$n$$ is even then the product of the roots is $$-z$$, if $$n$$ is odd then the product of the roots is $$z$$. $\\text{When }n\\text{ is an even number}:~\\prod_{x = 0}^{n - 1}{Z_x} = -z$ $\\text{When }n\\text{ is an odd number}:~\\prod_{x = 0}^{n - 1}{Z_x} = z$ Hope you enjoyed the note Note by Jack Rawlin 2 years, 2 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "Complex Roots - Not sure I did this right 1. Mar 27, 2009 Atena Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering. 1. The problem statement, all variables and given/known data Find all the roots of $$z^{4}=1-i$$ 2. Relevant equations I guess I should state De Moivre's here... $$(r cis(\\vartheta))^{n}=r^{n} cis (n\\vartheta)$$ 3. The attempt at a solution Firstly I re-wrote $$z^{4}=1-i$$ as $$z^{4}=\\sqrt{2} cis (\\frac{-\\pi}{4})$$ Using De Moivre's, $$z=(2\\frac{1}{2})^{\\frac{1}{4}} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ $$z=2\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ I found the four roots letting k=0,1,2,3 $$z=2^\\frac{1}{8} cis (\\frac{-\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+2\\pi))=2^\\frac{1}{8} cis (\\frac{7\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+4\\pi))=2^\\frac{1}{8} cis (\\frac{15\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+6\\pi))=2^\\frac{1}{8} cis (\\frac{23\\pi}{16})$$ 2. Mar 27, 2009 yyat Hi Atena! Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).", "what you had not being an equation because you seem to have a problem with writing exactly what you mean- and it hurt you here. Those four numbers are the fourth roots- you have already solved the problem at this point: $z^4= cis(\\pi)$ and $z = cis(\\frac{\\pi}{4})$, $cis(\\frac{5\\pi}{4})$, $cis(\\frac{9\\pi}{4})$, $cis(\\frac{13\\pi}{4})$ $z^4 = cis(\\frac{\\pi}{16})$, $cis(\\frac{5\\pi}{16})$, $cis(\\frac{9\\pi}{16})$, $cis(\\frac{13\\pi}{16}$ P.S 5. Originally Posted by Unknown008 I would have used the difference of two squares twice: $z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$ almost: $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2}))(z-(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})$ Then, z = 1 and -1. For the other two complex roots, we get: $z^2 = -1$ Which you should be able to solve. 6. Originally Posted by HallsofIvy almost $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z+(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z-(\\sqrt{2}{2}- i\\sqrt{2}{2})(z+(\\sqrt{2}{2}- i\\sqrt{2}{2})$ Um... I don't understand why you look for $z^4+1$ while it's $z^4-1$... EDIT: Ah, okay. I understand now. 7. yes my bad, its meant to be $z^4+1=0$ srry for the confusion", "\\frac{2\\pi}{2} (1)\\right) = 2\\text{cis}\\left(\\frac{4\\pi}{3}\\right)\\). In rectangular form, the two square roots of $$z$$ are $$w_{\\text{\\tiny$$0\\)}} = 1+i\\sqrt{3}\\) and $$w_{\\text{\\tiny$$1\\)}} = -1-i\\sqrt{3}\\). We can check our answers by squaring them and showing that we get $$z= -2 + 2i\\sqrt{3}$$. 2. Proceeding as above, we get $$z = -16 = 16 \\text{cis}(\\pi)$$. With $$r = 16$$, $$\\theta = \\pi$$ and $$n = 4$$, we get the four fourth roots of $$z$$ to be $$w_{\\text{\\tiny$$0\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (0)\\right) = 2\\text{cis}\\left(\\frac{\\pi}{4}\\right)\\), $$w_{\\text{\\tiny$$1\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (1)\\right) = 2\\text{cis}\\left(\\frac{3\\pi}{4}\\right)\\), $$w_{\\text{\\tiny$$2\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (2)\\right) = 2\\text{cis}\\left(\\frac{5\\pi}{4}\\right)\\) and $$w_{\\text{\\tiny$$3\\)}} = \\sqrt[4]{16} \\text{cis}\\left(\\frac{\\pi}{4} + \\frac{2\\pi}{4} (3)\\right) = 2\\text{cis}\\left(\\frac{7\\pi}{4}\\right)\\). Converting these to rectangular form gives $$w_{\\text{\\tiny$$0\\)}} = \\sqrt{2} + i\\sqrt{2}\\), $$w_{\\text{\\tiny$$1\\)}} = -\\sqrt{2} + i\\sqrt{2}\\), $$w_{\\text{\\tiny$$2\\)}} = -\\sqrt{2} - i\\sqrt{2}\\) and $$w_{\\text{\\tiny$$3\\)}} = \\sqrt{2} - i\\sqrt{2}\\). 3. For $$z = \\sqrt{2} + i \\sqrt{2}$$, we have $$z = 2\\text{cis}\\left(\\frac{\\pi}{4}\\right)$$. With $$r = 2$$, $$\\theta = \\frac{\\pi}{4}$$ and $$n =3$$ the usual computations yield $$w_{\\text{\\tiny$$0\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{\\pi}{12}\\right)\\), $$w_{\\text{\\tiny$$1\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{9\\pi}{12}\\right) = \\sqrt[3]{2} \\text{cis}\\left(\\frac{3\\pi}{4}\\right) $$and \\(w_{\\text{\\tiny$$2\\)}} = \\sqrt[3]{2} \\text{cis}\\left(\\frac{17\\pi}{12}\\right)\\). If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem \\ref{circularsumdifference} or the Half-Angle Identities in Theorem \\ref{halfangle} to evaluate $$w_{\\text{\\tiny$$0\\)}}\\) and $$w_{\\text{\\tiny$$2\\)}}\\). Since we are not explicitly", "we learned about here in the Imaginary (Complex) Numbers Section. Complex Trig Root Problem Solution (Note: cis is short for $$\\cos \\,+\\sin i$$) Determine $$z$$ and three cube roots of $$z$$ if one cube root is $$1-\\sqrt{3}i$$ We can first find $$z$$ by cubing using De Moivre’s Theorem : $$\\displaystyle {{z}^{n}}={{r}^{n}}\\left[ {\\cos \\left( {n\\theta } \\right)+i\\sin \\left( {n\\theta } \\right)} \\right]\\,$$ Convert $$1-\\sqrt{3}i$$ to Polar first: $$\\displaystyle r=\\sqrt{{{{1}^{2}}+{{{\\left( {-\\sqrt{3}} \\right)}}^{2}}}}=2;\\,\\,\\,\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{{-\\sqrt{3}}}{1}} \\right)=\\frac{{5\\pi }}{3}:\\,\\, 2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)$$ $$\\displaystyle \\begin{array}{c}{{\\left[ {2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)} \\right]}^{{\\,3}}}={{2}^{3}}\\text{cis}\\left( {3\\cdot \\frac{{5\\pi }}{3}} \\right)=8\\,\\text{cis}\\left( {5\\pi } \\right)=8\\,\\text{cis}\\left( {5\\pi -4\\pi } \\right)=8\\,\\text{cis}\\left( \\pi \\right)\\\\=8\\left( {-1+0i} \\right)=-8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,z=-8\\end{array}$$ Since we see that $$z=-8$$, and we know that the real number cube root of –8 is –2, we can then take the conjugate of $$1-\\sqrt{3}i$$ to get the other root, $$1+\\sqrt{3}i$$ (from the Complex Conjugate Root Theorem). The roots are –2, $$1-\\sqrt{3}i$$, and $$1+\\sqrt{3}i$$. But we’ll check our answer using the $$n$$th root formulas: $$\\begin{array}{c}{{\\left[ {8\\text{cis}\\left( \\pi \\right)} \\right]}^{{\\,\\frac{1}{3}}}}={{8}^{{\\frac{1}{3}}}}\\text{cis}\\left( {\\frac{{\\pi +2\\pi k}}{3}} \\right)\\\\=2\\,\\text{cis}\\left( {\\frac{\\pi }{3}+\\frac{{2\\pi k}}{3}} \\right);\\,\\,\\,\\,\\,k=0,1,2\\end{array}$$ $$\\begin{array}{l}k=0:\\,\\,\\,\\,2\\,\\text{cis}\\left( {\\frac{\\pi }{3}} \\right)=2\\left( {\\frac{1}{2}+\\frac{{\\sqrt{3}}}{2}i} \\right)=1+\\sqrt{3}i\\\\k=1:\\,\\,\\,\\,2\\,\\text{cis}\\left( \\pi \\right)=2\\left( {-1+0i} \\right)=-2\\\\k=2:\\,\\,\\,\\,2\\,\\text{cis}\\left( {\\frac{{5\\pi }}{3}} \\right)=2\\left( {\\frac{1}{2}-\\frac{{\\sqrt{3}}}{2}i} \\right)=1-\\sqrt{3}i\\end{array}$$ How to Check with Graphing Calculator When getting roots of complex numbers in the calculator, you’ll only get one root, so this won’t be too helpful. But you can check all the roots you", "z=cis(\\frac{-5\\pi}{8})$ $k=6, z=cis(\\frac{-3\\pi}{8})$ $k=7, z=cis(\\frac{-\\pi}{8})$ EDIT: beaten :/ Title: Re: Specialist 3/4 Question Thread! Post by: dc302 on January 05, 2012, 03:35:56 pm I don't really know why you have introduced a 'w' that you never used, but in any case, you should have z^8 = cis(pi + 2kpi) from the beginning, not 3 lines later as this is inconsistent. Title: Re: Specialist 3/4 Question Thread! Post by: paulsterio on January 06, 2012, 10:51:19 am Btw, if you intend to write k = 4, you'll have to say that z = cis(9pi/8) = cis(-7pi/8) otherwise your answer wouldn't make much sense Alternatively, you can let k = 0, 1, 2, 3, -1, -2, -3, -4 instead Title: Re: Specialist 3/4 Question Thread! Post by: dc302 on January 07, 2012, 02:51:54 pm Is there supposed to be a picture attached? Title: Re: Specialist 3/4 Question Thread! Post by: HERculina on January 07, 2012, 05:48:43 pm you guys are so ahead! takes me forever to do one exercise :o anyway i don't understand part b of the last question in exercise 1E (essentials, if anyone has done it yet): 10. a)Find an expression for the infinite geometric sum 1 + sintheta + sin^2*theta + ... b)Find the values of theta for which the infinite geometric sum is 2. Title: Re:", "i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was: e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3) I think it is there where i have gone wrong, help would be appreciated Thanks, ArTiCk $\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 = $ $\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$. 6. Hello, Mr. F ! Originally Posted by mr fantastic $\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 = $ $\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$. Lovely!", "\\sin n &=& \\sum_{n=1}^{44} \\sin n + \\sin(90-n)\\\\ &=& \\sqrt{2}\\sum_{n=1}^{44} \\cos(45-n) = \\sqrt{2}\\sum_{n=1}^{44} \\cos n\\\\ \\sum_{n=1}^{44} \\sin n &=& (\\sqrt{2}-1)\\sum_{n=1}^{44} \\cos n \\end{eqnarray*}$ This is the ratio we are looking for. $x$ reduces to $\\frac{1}{\\sqrt{2} - 1} = \\sqrt{2} + 1$, and $\\lfloor 100(\\sqrt{2} + 1)\\rfloor = \\boxed{241}$. ### Solution 4 Consider the sum $\\sum_{n = 1}^{44} \\text{cis } n^\\circ$. The fraction is given by the real part divided by the imaginary part. The sum can be written $- 1 + \\sum_{n = 0}^{44} \\text{cis } n^\\circ = - 1 + \\frac {\\text{cis } 45^\\circ - 1}{\\text{cis } 1^\\circ - 1}$ (by De Moivre's Theorem with geometric series) $= - 1 + \\frac {\\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2}}{\\text{cis } 1^\\circ - 1} = - 1 + \\frac {\\left( \\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2} \\right) (\\text{cis } ( - 1^\\circ) - 1)}{(\\cos 1^\\circ - 1)^2 + \\sin^2 1^\\circ}$ (after multiplying by complex conjugate) $= - 1 + \\frac {\\left( \\frac {\\sqrt {2}}{2} - 1 \\right) (\\cos 1^\\circ - 1) + \\frac {\\sqrt {2}}{2}\\sin 1^\\circ + i\\left( \\left(1 - \\frac {\\sqrt {2}}{2} \\right) \\sin 1^\\circ + \\frac {\\sqrt {2}}{2} (\\cos 1^\\circ - 1)\\right)}{2(1 - \\cos 1^\\circ)}$ $= - \\frac {1}{2} - \\frac {\\sqrt {2}}{4} - \\frac {i\\sqrt {2}}{4} + \\frac", "cis ( \\frac{\\pi}{6})= \\frac{\\sqrt{6}}{2}+\\frac{\\sqrt{2}}{2}$ Therefore, $z= (\\frac{-\\sqrt{6}-2}{2})+\\frac{\\sqrt{2}}{2} \\cdot i$ let $k=-1$ $z+1=\\sqrt {2}\\cdot cis({\\frac{\\pi}{6}-\\pi})=\\sqrt{2} \\cdot cis (\\frac{-5\\pi}{6})$. Thus, $z=\\frac{(-\\sqrt {6-2})}{2}-\\frac{\\sqrt{2}}{2}i$. You can also use $z=a+bi$, Where a, b holds r. Hoped I Helped :). Title: Re: Specialist 3/4 Question Thread! Post by: kensan on February 19, 2012, 03:48:06 pm Must be a gap in my knowledge here, but from $(z+1)^2=2\\cdot cis(\\frac{\\pi}{3}+2k\\pi)$ and when you square rooted doesn't the argument change? I'm confused because when you subbed in k=0, you got $\\sqrt{2 }\\cdot cis\\frac{\\pi}{6}$ but wouldn't it be $\\sqrt{2 }\\cdot cis\\frac{\\pi}{3}$ because $2\\cdot 0\\cdot \\pi=0$ If that makes sense haha Also I was looking through my textbook and couldn't find any examples of this type of question, is this assumed after all knowledge learnt? Title: Re: Specialist 3/4 Question Thread! Post by: rife168 on February 19, 2012, 03:58:24 pm Must be a gap in my knowledge here, but from $(z+1)^2=2\\cdot cis(\\frac{\\pi}{3}+2k\\pi)$ and when you square rooted doesn't the argument change? I'm confused because when you subbed in k=0, you got $\\sqrt{2 }\\cdot cis\\frac{\\pi}{6}$ but wouldn't it be $\\sqrt{2 }\\cdot cis\\frac{\\pi}{3}$ because $2\\cdot 0\\cdot \\pi=0$ If that makes sense haha Also I was looking through my textbook and couldn't find any examples of this type of question, is this assumed after all knowledge learnt? When he wrote $(z+1)^2=2\\cdot cis(\\frac{\\pi}{3}+2k\\pi)$ the", "$z = \\frac {\\left(\\cos \\frac {\\pi}{4} - isin \\frac {\\pi}{4}\\right)^2 \\left(\\cos \\frac {\\pi}{3} + isin \\frac {\\pi}{3}\\right)^3}{\\left(\\cos \\frac {\\pi}{24} - isin \\frac {\\pi}{24}\\right)^4}$ I edited my original post - thanks for correcting me. $z = \\frac{ \\left[ \\text{cis} \\left( -\\frac{\\pi}{4}\\right) \\right]^2 \\left[ \\text{cis} \\left(\\frac{\\pi}{3}\\right) \\right]^3}{\\left[ \\text{cis} \\left(-\\frac{\\pi}{24}\\right) \\right]^4}$ Apply de Mooivre's Theorem: $= \\frac{\\text{cis} \\left[ 2\\left( -\\frac{\\pi}{4}\\right) \\right]\\, \\text{cis} \\left[3 \\left(\\frac{\\pi}{3}\\right)\\right]}{\\text{cis} \\left[ 4 \\left(- \\frac{\\pi}{24}\\right)\\right]} = \\frac{\\text{cis} \\left( -\\frac{\\pi}{2}\\right) \\, \\text{cis} (\\pi)}{\\text{cis} \\left(- \\frac{\\pi}{6}\\right)}$ Apply the usual theorems for multiplying and dividing complex numbers expressed in polar form: $= \\text{cis} \\left( -\\frac{\\pi}{2} + \\pi - \\left[-\\frac{\\pi}{6} \\right]\\right) = \\text{cis} \\left( \\frac{2\\pi}{3} \\right)$. And of course you can easily express this in Cartesian form if desired. 8. Thank you all for your help - you guys are truly amazing! 9. There's another part to this question: Using De Moivre's theorem, show that $z = \\sqrt [3]{1}$ ---------------------- Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get: $\\left(\\cos \\frac {\\pi}{24} - isin \\frac {\\pi}{24}\\right)^4$ to become $\\left[cis \\frac {\\pi}{4} \\right]^6$ Your final answer is $cis \\frac {\\pi}{4}$ but I seem to have gotten $cis \\frac {5\\pi}{6}$ Is there something I'm not understanding? ---------------------- Also, how do I show $z = \\sqrt [3]{1}$? How can I apply De Moivre's", "34^\\circ \\cdot 5 \\ \\text{cis} \\ 16^\\circ \\cdot \\frac{1}{2} \\ \\text{cis} \\ 100^\\circ\\\\& =4 \\cdot 5 \\cdot \\frac{1}{2} \\cdot \\text{cis} \\ (34^\\circ+16^\\circ+100^\\circ)=10 \\ \\text{cis} \\ 150$ Note how much easier it is to do products and quotients in trigonometric polar form. #### Practice Translate the following complex numbers from trigonometric polar form to rectangular form. 1. $5 \\ \\text{cis} \\ 270^\\circ$ 2. $2 \\ \\text{cis} \\ 30^\\circ$ 3. $-4 \\ \\text{cis} \\ \\frac{\\pi}{4}$ 4. $6 \\ \\text{cis} \\ \\frac{\\pi}{3}$ 5. $2 \\ \\text{cis} \\ \\frac{5 \\pi}{2}$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $2-i$ 7. $5+12i$ 8. $6i+8$ 9. $i$ Complete the following calculations and simplify. 10. $2 \\ \\text{cis} \\ 22^\\circ \\cdot \\frac{1}{5} \\ \\text{cis} \\ 15^\\circ \\cdot 3 \\ \\text{cis} \\ 95^\\circ$ 11. $9 \\ \\text{cis} \\ 98^\\circ \\div 3 \\ \\text{cis} \\ 12^\\circ$ 12. $15 \\ \\text{cis} \\ \\frac{\\pi}{4} \\cdot 2 \\ \\text{cis} \\ \\frac{\\pi}{6}$ 13. $-2 \\ \\text{cis} \\ \\frac{2 \\pi}{3} \\div 15 \\ \\text{cis} \\ \\frac{7 \\pi}{6}$ Let $z_1=r_1 \\cdot \\text{cis} \\ \\theta_1$ and $z_2=r_2 \\cdot \\text{cis} \\ \\theta_2$ with $r_2 \\neq 0$ . 14. Use the trigonometric sum and difference identities to prove that $z_1 \\cdot z_2=r_1 \\cdot r_2 \\cdot \\text{cis} \\ (\\theta_1+\\theta_2)$ . 15. Use the trigonometric sum and difference identities to prove that $z_1", "${\\mathbb Q}^\\ast$ 3. ${\\mathbb R}^\\ast$ Hint (a) $0, 1, -1\\text{;}$ (b) $1, -1$ ###### Exercise11 If $a^{24} =e$ in a group $G\\text{,}$ what are the possible orders of $a\\text{?}$ Hint 1, 2, 3, 4, 6, 8, 12, 24. ###### Exercise15 Evaluate each of the following. 1. $(3-2i)+ (5i-6)$ 2. $(4-5i)-\\overline{(4i -4)}$ 3. $(5-4i)(7+2i)$ 4. $(9-i) \\overline{(9-i)}$ 5. $i^{45}$ 6. $(1+i)+\\overline{(1+i)}$ Hint (a) $-3 + 3i\\text{;}$ (c) $43- 18i\\text{;}$ (e) $i$ ###### Exercise16 Convert the following complex numbers to the form $a + bi\\text{.}$ 1. $2 \\cis(\\pi / 6 )$ 2. $5 \\cis(9\\pi/4)$ 3. $3 \\cis(\\pi)$ 4. $\\cis(7\\pi/4) /2$ Hint (a) $\\sqrt{3} + i\\text{;}$ (c) $-3\\text{.}$ ###### Exercise17 Change the following complex numbers to polar representation. 1. $1-i$ 2. $-5$ 3. $2+2i$ 4. $\\sqrt{3} + i$ 5. $-3i$ 6. $2i + 2 \\sqrt{3}$ Hint (a) $\\sqrt{2} \\cis( 7 \\pi /4)\\text{;}$ (c) $2 \\sqrt{2} \\cis( \\pi /4)\\text{;}$ (e) $3 \\cis(3 \\pi/2)\\text{.}$ ###### Exercise18 Calculate each of the following expressions. 1. $(1+i)^{-1}$ 2. $(1 - i)^{6}$ 3. $(\\sqrt{3} + i)^{5}$ 4. $(-i)^{10}$ 5. $((1-i)/2)^{4}$ 6. $(-\\sqrt{2} - \\sqrt{2}\\, i)^{12}$ 7. $(-2 + 2i)^{-5}$ Hint (a) $(1 - i)/2\\text{;}$ (c) $16(i - \\sqrt{3}\\, )\\text{;}$ (e) $-1/4\\text{.}$ ###### Exercise22 Calculate each of the following. 1. $292^{3171} \\pmod{ 582}$ 2. $2557^{ 341} \\pmod{ 5681}$ 3. $2071^{ 9521} \\pmod{ 4724}$ 4. $971^{ 321} \\pmod{ 765}$", "{2}}\\right)\\\\ x &=& \\frac {1}{\\sqrt {2} - 1} = 1 + \\sqrt {2}\\\\ \\lfloor 100x \\rfloor &=& \\lfloor 100(1 + \\sqrt {2}) \\rfloor = \\boxed{241}\\\\ \\end{eqnarray*}$ ### Solution 2 A slight variant of the above solution, note that \\begin{eqnarray*} \\sum_{n=1}^{44} \\cos n + \\sum_{n=1}^{44} \\sin n &=& \\sum_{n=1}^{44} \\sin n + \\sin(90-n)\\\\ &=& \\sqrt{2}\\sum_{n=1}^{44} \\cos(45-n) = \\sqrt{2}\\sum_{n=1}^{44} \\cos n\\\\ \\sum_{n=1}^{44} \\sin n &=& (\\sqrt{2}-1)\\sum_{n=1}^{44} \\cos n (Error compiling LaTeX. ! LaTeX Error: \\begin{eqnarray*} on input line 20 ended by \\end{document}.) This is the ratio we are looking for. $x$ reduces to $\\frac{1}{\\sqrt{2} - 1} = \\sqrt{2} + 1$, and $\\lfloor 100(\\sqrt{2} + 1)\\rfloor = \\boxed{241}$. ### Solution 3 Consider the sum $\\sum_{n = 1}^{44} \\text{cis } n^\\circ$. The fraction is given by the real part divided by the imaginary part. The sum can be written $- 1 + \\sum_{n = 0}^{44} \\text{cis } n^\\circ = - 1 + \\frac {\\text{cis } 45^\\circ - 1}{\\text{cis } 1^\\circ - 1}$ (by De Moivre's Theorem with geometric series) $= - 1 + \\frac {\\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2}}{\\text{cis } 1^\\circ - 1} = - 1 + \\frac {\\left( \\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2} \\right) (\\text{cis } ( - 1^\\circ) - 1)}{(\\cos 1^\\circ - 1)^2 + \\sin^2 1^\\circ}$ (after multiplying by complex" ]

[ "\\cdot \\dotsc \\cdot 1$ possible permutations.", "at least $100$. Suppose $n = m+100$. Then $1978^{m+100}-1978^m = 1978^m(1978^{100}-1) \\equiv \\pmod{1000}$. Since the second part is odd, all the powers of two, there must be at least three of them, come from $1978^m$. Hence $m \\ge 3$. But since we know $1978^{100}-1 \\equiv 0 \\pmod{125}$, we have $1978^m(1978^{100}-1)$ divisible by both $8$ and $125$, which means it's divisible by $1000$, so $m = 3$ is the smallest possible value that works, giving $n = 103$. Therefore, our smallest $m+ n = 106$. QED. -------------------- Comment: Kind of a strange problem, with the minimum value, reminiscent of many AIME problems. Ultimately, not too difficult, though. -------------------- Practice Problem: Find the value of $10^{2006} \\pmod{252}$. ### Square Differences. Topic: Inequalities. Level: Olympiad. Problem: (1975 IMO - #1) We consider two sequences of real numbers $x_{1} \\geq x_{2} \\geq \\ldots \\geq x_{n}$ and $\\ y_{1} \\geq y_{2} \\geq \\ldots \\geq y_{n}$. Let $z_{1}, z_{2}, \\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \\ldots, y_{n}$. Prove that $\\displaystyle \\sum \\limits_{i=1}^{n} ( x_{i} -\\ y_{i} )^{2} \\leq \\sum \\limits_{i=1}^{n} ( x_{i} - z_{i})^{2}$. Solution: Expanding both sides, we see that $\\displaystyle \\sum x_i^2$ and $\\displaystyle \\sum y_i^2 = \\sum z_i^2$ appear on both sides, so we can cancel them out, leaving $\\displaystyle -2 \\sum x_iy_i \\le -2 \\sum x_iz_i", "# How many different values can that sum take? Let $x_1,x_2,\\dots,x_{100}$ be a permutation of $1,2,\\dots,100.$ How many different values does the sum $x_1+2x_2+\\cdots+100x_{100}$ take? • It seems, that if you have sequence $x_1,x_2,...,x_n$ as permutation of $1,2,...,n$, and take sum $x_1+2x_2+...+nx_n$, then you'll obtain $$\\dfrac{n(n-1)(n+1)}{6}+1$$ different values of this sum. So, for $n=100$ I expect $166651$ different values. – Oleg567 Jun 18 '14 at 19:22 • @ Oleg567 : Calculations confirm that. How to prove it? – user64494 Jun 18 '14 at 19:27 • Perhaps starting point is that smallest sum is $$1\\cdot n + 2 (n-1)+3(n-2)\\cdots+n\\cdot 1,$$ largest sum is $$1\\cdot1+2\\cdot2+\\cdots+n\\cdot n.$$ And that there are no gaps between/among them. – Oleg567 Jun 18 '14 at 19:30 • Why are there no gaps between these values? – user64494 Jun 18 '14 at 19:32 • Hmm, that is the most interesting question. Don't know yet. Maybe to consider some neighboring permutations etc... – Oleg567 Jun 18 '14 at 19:35 Let $x_1,x_2,\\dots,x_n$ be a permutations of $1,2,\\dots,n.$ $(n\\ge 4)$. Then there are $$\\dfrac{n(n-1)(n+1)}{6}+1$$ different values of sums $x_1+2x_2+\\cdots+nx_n$. Smallest value is $$v_n = 1\\cdot n + 2(n-1)+3(n-2)+\\cdots+n\\cdot 1 = \\sum_{j=1}^{n}j(n+1-j)=\\dfrac{n(n+1)(n+2)}{6},$$ largest value is $$V_n = 1\\cdot 1+2\\cdot 2+\\cdots+n\\cdot n = \\sum_{j=1}^nj^2= \\dfrac{n(n+1)(2n+1)}{6}.$$ Let's prove that there are no \"gaps\" between $v_n$ and $V_n$. We'll use math. induction. 0)", "$$m',n'$$ of $$M+1$$ by taking the maximum of either $$m+1,n$$ or $$m,n+1$$ (we also have to ensure that the resulting $$m'$$ and $$n'$$ does not exceed the total 2s and 1s, or else maximum cost of $$M+1$$ is equal to $$M$$ The two observations above are enough for us to develop a bottom-up DP by storing the combination of $$m,n$$ for each $$M$$.", "\\{1, 2, 3, 4\\}, this is one of the permutations which provides the minimum result. Test case 2: The sum is \\frac{3+2+1}{3} = 6/3 = 2. Every permutation of size 3 will have this value, hence it is the minimum possible.", "and 1s: $$\\binom{ N }{ N_0, N_1 }$$ So we have three possible outcomes, and the total number of arrangements is the sum of these three cases: $$N_{perms} = \\binom{ 10 }{ 6, 4} + \\binom{ 10 }{ 5, 5 } + \\binom{ 10 }{ 6 , 4 }$$ # Algorithm We can generalize the process. Suppose we are forming a number of length N from a number of digits/classes $$k$$ labeled from $$0 \\dots k-1$$, and each digit/class can only appear a maximum of $$m$$ times. The number of combinations that can be formed for a given $$N, k, m$$ is given by the multiset permutation expression above. So the total number of permutations that can be formed is a sum of these multiset permutation expressions, over each possible combination of digits/classes into a number of length $$N$$. In computer science terms, we can think of this as a nested for loop or dynamic program; in mathematical terms, we can think of a sequence of summations whose limits depend on the variables in the other summations. $$\\sum_{N_1} \\sum_{N_2} \\dots \\sum_{N_k} \\binom{N}{N_0, N_1, N_2, \\dots, N_{k-1}}$$ where the limits of the summations are given by: $$N_1 = \\min \\left(N - (k-1) m, 0 \\right) \\dots m$$ $$N_2 = \\min \\left( N - N_1 - (k-2) m, 0 \\right)", "## Max value Let $x_1,x_2,\\dots,x_{n}$ be real numbers such that $\\sum_{k = 1}^{n}\\frac {1}{x_k^2 + 1} = n-1$......Find the maximum value of $x_{1}x_{2}\\cdots x_{n}+\\sum\\limits_{1 \\leqslant i < j \\leqslant n} {x_i x_j }-\\sum_{k = 1}^{n}x_{k}^{2}$", "first. Prove that there are infinitely many such $n$ such that Bob has a winning strategy. (For example, if $n=17,$ then Alice might take $6$ leaving $11;$ then Bob might take $1$ leaving $10;$ then Alice can take the remaining stones to win.) 3. Let $1,2,3,\\dots,2005,2006,2007,2009,2012,2016,\\dots$ be a sequence defined by $x_{k}=k$ for $k=1,2\\dots,2006$ and $x_{k+1}=x_{k}+x_{k-2005}$ for $k\\ge 2006.$ Show that the sequence has 2005 consecutive terms each divisible by 2006. 4. Let $S=\\{1,2\\dots,n\\}$ for some integer $n>1.$ Say a permutation $\\pi$ of $S$ has a local maximum at $k\\in S$ if $\\begin{array}{ccc}\\text{(i)}&\\pi(k)>\\pi(k+1)&\\text{for }k=1\\\\ \\text{(ii)}&\\pi(k-1)<\\pi(k)\\text{ and }\\pi(k)>\\pi(k+1)&\\text{for }1<k<n\\\\ \\text{(iii)}&\\pi(k-1)M\\pi(k)&\\text{for }k=n\\end{array}$ (For example, if $n=5$ and $\\pi$ takes values at $1,2,3,4,5$ of $2,1,4,5,3,$ then $\\pi$ has a local maximum of $2$ as $k=1,$ and a local maximum at $k-4.$) What is the average number of local maxima of a permutation of $S,$ averaging over all permuatations of $S?$ 5. Let $n$ be a positive odd integer and let $\\theta$ be a real number such that $\\theta/\\pi$ is irrational. Set $a_{k}=\\tan(\\theta+k\\pi/n),\\ k=1,2\\dots,n.$ Prove that $\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{a_{1}a_{2}\\cdots a_{n}}$ is an integer, and determine its value. 6. Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral. 7. Show that the curve $x^{3}+3xy+y^{3}=1$ contains only", "# Minimum Sum Value Let $y_1, y_2, y_3, \\ldots, y_{8}$ be a permutation of the numbers $1, 2, 3, \\ldots, 8$. What is the minimum value of $\\displaystyle \\sum_{i=1}^8 (y_i + i )^2$? ×", "# What is $$m^2+n^2$$? Determine the maximum value of $$m^2 + n^2$$, where $$m$$ and $$n$$ are integers satisfying $$m, n \\in \\{ 1,2, \\ldots , 1981 \\}$$ and $$( n^2 - mn - m^2 )^2 = 1$$. Source: IMO 1981. ×", "of $$\\pi(m+1),\\pi(m+2),\\ldots,\\pi(2m)$$ consisting of only those values which exceed $$\\max\\{\\pi(1),\\ldots,\\pi(m)\\}$$. Let $$P(m)$$ denote the probability that this subsequence never decreases, when $$\\pi$$ is a randomly chosen permutation of $$\\{1,2,\\ldots,2m\\}$$. Evaluate the limit of $$P(m)$$ as $$m\\to\\infty$$. Black and white. I promise I didn't plan this", "# Coding challenge The values of $a$, $b$, $c$, and $d$ are 0, 1, 2, and 3, but not necessarily in that order. Write some code to determine the maximum possible value of the expression $c\\cdot a^n-d$.", "is precisely the maximum possible value of $a_{2018}$, so all the inequalities above are equalities, and so for each $k\\geqslant 2$, (i) $a_k$ and $b_k$ are consecutive elements of the triangle, and (ii) $a_{k-1}$ lie directly above $a_k$ and $b_k$. Hence all numbers $a_k$ and $b_k$ are in a \"strand\" of width $1$ unit from top to bottom of the triangle. In particular, it is possible to find a sub-triangle with $\\left\\lceil\\frac{2018-2}{2}\\right\\rceil=1008$ rows not containing any of $\\{1,2,\\ldots,2018\\}$. Re-defining $a_k$ and $b_k$ again for this sub-triangle, we have $$a_{1008} \\geqslant b_1 + b_2 +\\cdots + b_{1008} \\geqslant 2019 + 2020 +\\cdots +3026 > \\frac{2018\\cdot 2019}{2}$$ which is clearly impossible. Hence, there are no anti-Pascal triangles with $2018$ rows containing every integer from $1$ to $1 + 2 + 3 + \\dots + 2018$, as desired. $\\square$ Solution 2 Let $b_k$ and $s_k$ be the maximum and minimum number in row $k$. Then ${ s_1, s_2, \\dots, s_{2018} }$ is a permutation of ${1, 2, \\dots, 2018}$. Call these the small numbers. Let $N=1+2+\\dots+2018$. Call the numbers $N, N-1, \\dots, N-2018$ big numbers. Claim 1: There are at most 1010 big numbers on the bottom row If two big numbers are consecutive on a row, they have a small number above them. There is only one small number in", "$\\{1,2,\\dots,m+1\\}$. On the other hand $k\\cdot k!$ counts the number of such permutations that fix all elements greater than $k+1$ but not $k+1$. -", "Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\\cdot5=\\boxed{080}.$ ~MRENTHUSIASM ## Solution 3 WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$. We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$. We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \\equiv 0 \\pmod{3}$. The only way is when They are $(+1,-1)$ or $(-1, +1) \\pmod{3}$. The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1\\pmod{3}$ respectively. $+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= +1 \\cdot +1 \\cdot +1$ $\\;\\;\\; OR \\;\\;\\;+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \\cdot -1 \\cdot +1$. $-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= -1 \\cdot -1 \\cdot -1$ $\\;\\;\\; OR \\;\\;\\;-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \\cdot +1 \\cdot +1$ But, as we have just two $+1's$ and two $-1's$. Hence, We will have to take $+1 = +1 \\cdot -1 \\cdot -1$ and $-1 = -1 \\cdot +1 \\cdot +1$. Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$. are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$). So, $(x_{5}, x_{1}) \\in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values. For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$. Hence, in total, we have $8 \\times 2 = 16$ ways. But", "# Minimum Permutation Sum Let $$\\{ x_1, x_2 , \\ldots, x_{100} \\}$$ be a permutation of $$\\{ 1, 2, \\ldots 100 \\}$$. What is the minimum, over all permutations, of $\\max_{i=1 \\mbox{ to }91 } x_{i} + x_{i+1} + x_{i+2} + \\ldots + x_{i+9} ?$ Note: A permutation is a rearrangement of all of the numbers in the sequence. ×", "$m$ is a permutation function that may ...", "of 2 with 1). So you can't use more than $kn/4$ permutations. So the maximum number of permutations is no more than the smaller of $kn/4$ and $n(n-1)(n-2)$. In your example, $n=6$, and $k=3$, so you can't have more than $18/4$, which is to say, 4 permutations. You can have that many by using the three you have and 2-4-5.", "133.510123, 134.510286, 135.509973, 136.509673, 137.509838, 138.509992, 139.509696, 140.509412, 141.509568, 142.509713, 143.509434, 144.509165, 145.509312, 146.509450, 147.509185, 148.508930, n+0.5 for n = infinity [email protected] - Suppose you look at all the permutations of $n-1$ in the maximal grouping, then at all the permutation of $n$ in that maximal grouping; is there any simple way in which each permutation in the first set gives rise to $n$ permutations in the second? Better yet, a simple way in which about half the $n-1$-permutations give rise to $n$ $n$-permutations each, and the other half give rise to $n+1$ $n$-permutations each? – Gerry Myerson Nov 19 '10 at 5:49 Well, the idea is clear to think about it. Actually, I am looking at the maximum element: $M(n)=T(n, n(n-1)/4)$, here the reccurence $T(n+1,k)=T(n,k)+T(n,k-1)+...+T(n,k-n)$. $T(n,k)$ has the maximum when $k=n(n-1)/4$. Also, I know $T(n,k)$ are the coefficients of $P_n(x)=1(1+x)(1+x+x^2)...(1+x+...+x^{n-1})$. So, the general question is how to estimate the max M(n), i.e the max coeficient in $P_n(x)$ effectively? I guess, if I know that then I'll get the answer at once – Mikhail Gaichenkov Nov 20 '10 at 11:51 This doesn't even start to solve the problem, but it puts a different spin on it. $T(n,k)$ is the rank of the vector space $H^k(GL_n/B)$. The Hard Lefschetz theorem then gives you the fact that the sequence $T(n,k)$", "n)$. Show that the expression $q^2 +(n-q-1)^2$ achieves a maximum over $q=0,1,\\dots ,n-1$ when $q=0$ or $q=n-1$." ]

[ "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "2014 AIME I Problems/Problem 10 Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Solution 1 $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "refer to the following diagram. Note that the area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of our unit, this yields a final area of", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "# 2014 AIME I Problems/Problem 10 ## Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since it has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "label(\"C\",C,C); label(\"D\",D,W); label(\"\\alpha\",E-(.2,0),W); label(\"E\",E,N); [/asy]$ $\\textbf{(A)}\\ \\cos\\ \\alpha\\qquad \\textbf{(B)}\\ \\sin\\ \\alpha\\qquad \\textbf{(C)}\\ \\cos^2\\alpha\\qquad \\textbf{(D)}\\ \\sin^2\\alpha\\qquad \\textbf{(E)}\\ 1-\\sin\\ \\alpha$ ## Problem 28 $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals $[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label(\"O\",O,dir(B)); label(\"1\",(O+P)/2,W); label(\"A\",A,dir(A)); label(\"B\",B,dir(B)); label(\"C\",C,dir(C)); label(\"D\",D,dir(D)); label(\"E\",E,dir(E)); label(\"P\",P,dir(P)); label(\"Q\",Q,dir(Q)); label(\"R\",R,dir(R)); [/asy]$ $\\textbf{(A)}\\ 3\\qquad \\textbf{(B)}\\ 1 + \\sqrt{5}\\qquad \\textbf{(C)}\\ 4\\qquad \\textbf{(D)}\\ 2 + \\sqrt{5}\\qquad \\textbf{(E)}\\ 5$ ## Problem 29 Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, what is the biggest it can be? $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 5\\qquad \\textbf{(C)}\\ 6\\qquad \\textbf{(D)}\\ 7\\qquad \\textbf{(E)}\\ \\text{none of these}$ ## Problem 30 The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \\frac{17}{x}, 2z = y + \\frac{17}{y}, 2w = z + \\frac{17}{z}, 2x = w + \\frac{17}{w}$ is $\\textbf{(A)}\\ 1\\qquad \\textbf{(B)}\\ 2\\qquad \\textbf{(C)}\\ 4\\qquad \\textbf{(D)}\\ 8\\qquad \\textbf{(E)}\\ 16$", "pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"2\", C, SW); label(\"1\", D, SE); [/asy]$ We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$. $[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label(\"3\", A, N); label(\"5\", B, NW); label(\"2\", C, SW); label(\"1\", D, SE); label(\"4\", E, NE); [/asy]$ This is the only solution to make $6$ \"bad.\" Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have $3$ and", "volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=\\boxed{288}$. ### Solution 2 $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy]$ Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$. ### Solution 3 We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "$n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\\underbrace{111\\cdots 111}_{x\\text{ times}}$. $\\Box$ ## Picture 1 $[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype(\"8 8\")); label(\"a\",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype(\"8 8\")); label(\"b\",(2,0),SSW); [/asy]$ $$\\text{Find the probability that }b>a \\text{.}$$ $[asy]unitsize(2inch); import olympiad; path c2 = dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180); path c1 = dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60); path c3 = dir(120)..dir(150)..dir(180); path c4 = dir(0)..dir(30)..dir(60); draw(dir(0)..dir(30)..dir(60)..dir(90)..dir(120)..dir(150)..dir(180)--dir(0)); draw(dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60)..dir(90)-dir(90)..dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180)--dir(90)-dir(0)); draw((-1.2,0.8)--(1.2,0.8)); label(\"l_{-n}\", (1.2,0.8),dir(0)); draw((-1.2,0.5)--(1.2,0.5)); label(\"l_0\", (1.2,0.5),dir(0)); draw((-1.2,0.2)--(1.2,0.2)); label(\"l_n\", (1.2,0.2),dir(0)); label(\"\\vdots\", (1.2, 0.4), dir(0)); label(\"\\vdots\", (0, 0.4)); label(\"\\vdots\", (1.2, 0.65), dir(0)); label(\"\\vdots\", (0, 0.65)); label(\"A_{-n}\", intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"C_{-n}\", intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"D_{-n}\", intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"B_{-n}\", intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"X\", intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5)), dir(150)); label(\"Y\", intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5)), dir(30)); label(\"C_{n}\", intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"A_{n}\", intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); label(\"B_{n}\", intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"D_{n}\", intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); dot(intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c3, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c4, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2))); [/asy]$ Two half-circles are drawn as shown above, with a line $l_0$ throught the two intersections points, $X,Y$ of the half-circles. Lines $l_k$ for $k=-n\\to", "label(\"d\",(15*k-1,8),N); label(\"i\",(15*k-1,7),N); label(\"s\",(15*k-1,6),N); label(\"t\",(15*k-1,5),N); label(\"a\",(15*k-1,4),N); label(\"n\",(15*k-1,3),N); label(\"c\",(15*k-1,2),N); label(\"e\",(15*k-1,1),N); label(\"time\",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label(\"d\",(15*k-1,8-15),N); label(\"i\",(15*k-1,7-15),N); label(\"s\",(15*k-1,6-15),N); label(\"t\",(15*k-1,5-15),N); label(\"a\",(15*k-1,4-15),N); label(\"n\",(15*k-1,3-15),N); label(\"c\",(15*k-1,2-15),N); label(\"e\",(15*k-1,1-15),N); label(\"time\",(15*k+8,0-15),S); } label(\"(A)\",(5,9),N); label(\"(B)\",(20,9),N); label(\"(C)\",(35,9),N); label(\"(D)\",(5,-6),N); label(\"(E)\",(20,-6),N); [/asy]$ ## Problem 19 Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$, has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is $[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 64$ ## Problem 20 Let $W,X,Y$ and $Z$ be four different digits selected from the set $\\{ 1,2,3,4,5,6,7,8,9\\}.$ If the sum $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ is to be as small as possible, then $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ must equal $\\text{(A)}\\ \\dfrac{2}{17} \\qquad \\text{(B)}\\ \\dfrac{3}{17} \\qquad \\text{(C)}\\ \\dfrac{17}{72} \\qquad \\text{(D)}\\ \\dfrac{25}{72} \\qquad \\text{(E)}\\ \\dfrac{13}{36}$ ## Problem 21 A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure", "\\qquad \\textbf{(C)}\\ \\frac 35 \\qquad \\textbf{(D)}\\ \\frac 23 \\qquad \\textbf{(E)}\\ \\frac 34$ ## Problem 18 A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$. And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircle. $[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label(\"A\",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label(\"B\",(2.61,-0.26),NE*lsf); dot((0,4),ds); label(\"D\",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label(\"C\",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}\\ 2:3 \\qquad\\textbf{(B)}\\ 3:2\\qquad\\textbf{(C)}\\ 6:\\pi \\qquad\\textbf{(D)}\\ 9: \\pi \\qquad\\textbf{(E)}\\ 30 : \\pi$ ## Problem 19 The two circles pictured have the same center $C$. Chord $\\overline{AD}$ is tangent to the inner circle at $B$, $AC$ is $10$, and chord $\\overline{AD}$ has length $16$. What is the area between the two circles? $[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label(\"A\", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label(\"B\", (1.97,-0.31),NE*lsf); dot((2,1),ds); label(\"C\", (1.96,1.09),NE*lsf); dot((4,0),ds); label(\"D\", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]$ $\\textbf{(A)}\\ 36 \\pi \\qquad\\textbf{(B)}\\ 49 \\pi\\qquad\\textbf{(C)}\\ 64 \\pi\\qquad\\textbf{(D)}\\ 81 \\pi\\qquad\\textbf{(E)}\\ 100 \\pi$ ## Problem 20 In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is", "We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. $[asy] size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label(\"1\",(0,0),SE,fontsize(8)); label(\"1\",(1,0),SE,fontsize(8)); label(\"1\",(2,0),SE,fontsize(8)); label(\"1\",(2,1),SE,fontsize(8)); label(\"1\",(3,0),SE,fontsize(8)); label(\"2\",(3,1),SE,fontsize(8)); label(\"2\",(3,2),SE,fontsize(8)); label(\"1\",(4,0),SE,fontsize(8)); label(\"3\",(4,1),SE,fontsize(8)); label(\"5\",(4,2),SE,fontsize(8)); label(\"1\",(5,0),SE,fontsize(8)); label(\"4\",(5,1),SE,fontsize(8)); label(\"9\",(5,2),SE,fontsize(8)); label(\"9\",(5,3),SE,fontsize(8)); label(\"1\",(6,0),SE,fontsize(8)); label(\"5\",(6,1),SE,fontsize(8)); label(\"14\",(6,2),SE,fontsize(8)); label(\"23\",(6,3),SE,fontsize(8)); label(\"23\",(6,4),SE,fontsize(8)); [/asy]$ Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$. Hence the desired probability is $$\\frac{23\\cdot 2^4\\cdot 3^6}{5^{10}},$$ and the answer is $(23+2+3+5)(4+6+10)=\\boxed{660}$." ]

[ "2014 AIME I Problems/Problem 10 Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Solution 1 $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "$n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\\underbrace{111\\cdots 111}_{x\\text{ times}}$. $\\Box$ ## Picture 1 $[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype(\"8 8\")); label(\"a\",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype(\"8 8\")); label(\"b\",(2,0),SSW); [/asy]$ $$\\text{Find the probability that }b>a \\text{.}$$ $[asy]unitsize(2inch); import olympiad; path c2 = dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180); path c1 = dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60); path c3 = dir(120)..dir(150)..dir(180); path c4 = dir(0)..dir(30)..dir(60); draw(dir(0)..dir(30)..dir(60)..dir(90)..dir(120)..dir(150)..dir(180)--dir(0)); draw(dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60)..dir(90)-dir(90)..dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180)--dir(90)-dir(0)); draw((-1.2,0.8)--(1.2,0.8)); label(\"l_{-n}\", (1.2,0.8),dir(0)); draw((-1.2,0.5)--(1.2,0.5)); label(\"l_0\", (1.2,0.5),dir(0)); draw((-1.2,0.2)--(1.2,0.2)); label(\"l_n\", (1.2,0.2),dir(0)); label(\"\\vdots\", (1.2, 0.4), dir(0)); label(\"\\vdots\", (0, 0.4)); label(\"\\vdots\", (1.2, 0.65), dir(0)); label(\"\\vdots\", (0, 0.65)); label(\"A_{-n}\", intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"C_{-n}\", intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"D_{-n}\", intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"B_{-n}\", intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"X\", intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5)), dir(150)); label(\"Y\", intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5)), dir(30)); label(\"C_{n}\", intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"A_{n}\", intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); label(\"B_{n}\", intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"D_{n}\", intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); dot(intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c3, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c4, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2))); [/asy]$ Two half-circles are drawn as shown above, with a line $l_0$ throught the two intersections points, $X,Y$ of the half-circles. Lines $l_k$ for $k=-n\\to", "The answer is $\\boxed{\\text{B}}$ $[asy] unitsize(20); label(\"O\",(0,-.35),N); label(\"B\",(1,-.35),N); label(\"G\",(2,-.35),N); label(\"R\",(3,-.35),N); label(\"Y\",(4,-.35),N); label(\"B\",(0,.6),N); label(\"G\",(1,.6),N); label(\"R\",(2,.6),N); label(\"Y\",(3,.6),N); label(\"G\",(0,1.6),N); label(\"R\",(1,1.6),N); label(\"Y\",(2,1.6),N); label(\"R\",(0,2.6),N); label(\"Y\",(1,2.6),N); label(\"Y\",(0,3.6),N); [/asy]$", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of our unit, this yields a final area of", "refer to the following diagram. Note that the area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of", "label(\"C\",C,C); label(\"D\",D,W); label(\"\\alpha\",E-(.2,0),W); label(\"E\",E,N); [/asy]$ $\\textbf{(A)}\\ \\cos\\ \\alpha\\qquad \\textbf{(B)}\\ \\sin\\ \\alpha\\qquad \\textbf{(C)}\\ \\cos^2\\alpha\\qquad \\textbf{(D)}\\ \\sin^2\\alpha\\qquad \\textbf{(E)}\\ 1-\\sin\\ \\alpha$ ## Problem 28 $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals $[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label(\"O\",O,dir(B)); label(\"1\",(O+P)/2,W); label(\"A\",A,dir(A)); label(\"B\",B,dir(B)); label(\"C\",C,dir(C)); label(\"D\",D,dir(D)); label(\"E\",E,dir(E)); label(\"P\",P,dir(P)); label(\"Q\",Q,dir(Q)); label(\"R\",R,dir(R)); [/asy]$ $\\textbf{(A)}\\ 3\\qquad \\textbf{(B)}\\ 1 + \\sqrt{5}\\qquad \\textbf{(C)}\\ 4\\qquad \\textbf{(D)}\\ 2 + \\sqrt{5}\\qquad \\textbf{(E)}\\ 5$ ## Problem 29 Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, what is the biggest it can be? $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 5\\qquad \\textbf{(C)}\\ 6\\qquad \\textbf{(D)}\\ 7\\qquad \\textbf{(E)}\\ \\text{none of these}$ ## Problem 30 The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \\frac{17}{x}, 2z = y + \\frac{17}{y}, 2w = z + \\frac{17}{z}, 2x = w + \\frac{17}{w}$ is $\\textbf{(A)}\\ 1\\qquad \\textbf{(B)}\\ 2\\qquad \\textbf{(C)}\\ 4\\qquad \\textbf{(D)}\\ 8\\qquad \\textbf{(E)}\\ 16$", "by $\\sqrt{(a-c)^2 + (b+d)^2}$.[4] $[asy]defaultpen(linewidth(1)); unitsize(15); pen dotted = linetype(\"2 4\"), sm = fontsize(10); real r = 2; pair A = r*(0,0), B = (r*18/5,A.y), C = r*(16/5,12/5), D = (r*9/5,C.y); pair refl(pair a, pair b = (C+B)/2) { return a+2*(b-a); } void makeshiftarrow(pair a, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(a--a+arrowlength*expi(dir+pi/8)--a+arrowlength*expi(dir-pi/8)--cycle); } draw(A--B--C--D--cycle); draw(A--C^^B--D); draw(refl(A)--C--B--refl(D)--cycle ^^ C--refl(D), dotted); draw(rightanglemark(A,C,refl(D),15)^^rightanglemark(A,IP(A--C,B--D),B,15), linewidth(0.7)); label(\"A\",A,S,sm);label(\"B\",B,S,sm);label(\"C\",C,N,sm);label(\"D\",D,N,sm);label(\"A'\",refl(A),N,sm);label(\"D'\",refl(D),S,sm); /* arrow */ draw(arc((B+C)/2,C.y,240,300),linewidth(0.7)); makeshiftarrow((B+C)/2+C.y*expi(pi*300/180),210*pi/180,r/4); [/asy]$ In trapezoid $ABCD$ with $\\overline{AB} \\parallel \\overline{CD}$, then $\\overline{AC} \\perp \\overline{BD} \\Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2$. $[asy] // feel free to change these four points! pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3); // // Rest of code // size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype(\"2 2\"); pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; path midpts = AB--BC--CD--DA--cycle; filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash); dot(A); dot(B); dot(C); dot(D); label(\"A\",A,W);label(\"B\",B,S);label(\"C\",C,E);label(\"D\",D,N); [/asy]$ Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area $1/2$ of the", "# 2014 AIME I Problems/Problem 10 ## Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since it has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "is equilateral, and thus the desired length is $x = OC \\implies C$. ## Solution 3 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.95,3),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$E$$\",(0.40,-3.60),E); label(\"$$B$$\",(-0.75,4.15),E); label(\"$$D$$\",(-2.62,1.5),E); label(\"$$F$$\",(-2.64,-1.43),E); label(\"$$G$$\",(-0.2,-2.8),E); label(\"$$\\sqrt{3}x$$\",(-1.5,-0.5),E); label(\"$$M$$\",(-2,-0.9),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-2.7,2.3),S); label(\"$$1$$\",(0.1,-3.4),S); label(\"$$8$$\",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$ Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \\frac{\\sqrt{3}x}{2} \\Rightarrow DG = \\sqrt{3}x$. $AE = 2 + \\sqrt{3}x$ Use the Pythagorean theorem on triangle $ABE$, we get $$(2+\\sqrt{3}x)^2 + x^2 = 8^2 \\Rightarrow 4 + 3x^2 + 4\\sqrt{3}x + x^2 = 64 \\Rightarrow x^2 + \\sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \\frac{-\\sqrt{3} \\pm \\sqrt{3 -4(1)(-15)}}{2} = \\frac{\\pm 3\\sqrt{7} - \\sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \\frac{3\\sqrt{7} - \\sqrt{3}}{2} \\Rightarrow C$$ ~Zeric Hang ## Soultion 4 (coordinate bashing) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$1$$\",(-0.5,3.8),S);[/asy]$", "at six points. If $AG=2, GF=13, FC=1$, and $HJ=7$, then $DE$ equals $[asy] defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label(\"A\", A, N); label(\"B\", B, dir(210)); label(\"C\", C, dir(330)); label(\"D\", D, SW); label(\"E\", E, SE); label(\"F\", F, dir(170)); label(\"G\", G, dir(250)); label(\"H\", H, SE); label(\"J\", J, dir(0)); label(\"2\", A--G, dir(30)); label(\"13\", F--G, dir(180+30)); label(\"1\", F--C, dir(30)); label(\"7\", H--J, dir(-30));[/asy]$ $\\text {(A)} 2\\sqrt{22} \\qquad \\text {(B)} 7\\sqrt{3} \\qquad \\text {(C)} 9 \\qquad \\text {(D)} 10 \\qquad \\text {(E)} 13$ ## Problem 25 The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad", "31 For $x^2+2x+5$ to be a factor of $x^4+px^2+q$, the values of $p$ and $q$ must be, respectively: $\\textbf{(A)}\\ -2, 5\\qquad \\textbf{(B)}\\ 5, 25\\qquad \\textbf{(C)}\\ 10, 20\\qquad \\textbf{(D)}\\ 6, 25\\qquad \\textbf{(E)}\\ 14, 25$ ## Problem 32 In this figure the center of the circle is $O$. $AB \\perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has a length twice the radius. Then: $[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot); path tangent=A--B; pair P=waypoint(tangent,abs(A-D)/abs(A-B)); draw(unitcircle^^C--B--A--E); dot(A^^B^^C^^D^^E^^P,linewidth(2)); label(\"O\",O,dir(290)); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,NE); label(\"D\",D,dir(120)); label(\"E\",E,SE); label(\"P\",P,SW);[/asy]$ $\\textbf{(A)} AP^2 = PB \\times AB\\qquad \\textbf{(B)}\\ AP \\times DO = PB \\times AD\\qquad \\textbf{(C)}\\ AB^2 = AD \\times DE\\qquad \\textbf{(D)}\\ AB \\times AD = OB \\times AO\\qquad \\textbf{(E)}\\ \\text{none of these}$ ## Problem 33 You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \\times 3 \\times 5 \\times\\ldots \\times 61$ of all prime numbers less than or equal to $61$, and $n$ takes, successively, the values $2, 3, 4,\\ldots, 59$. Let $N$ be the number of primes appearing in this sequence. Then $N$ is: $\\textbf{(A)}\\ 0\\qquad \\textbf{(B)}\\ 16\\qquad \\textbf{(C)}\\ 17\\qquad \\textbf{(D)}\\ 57\\qquad \\textbf{(E)}\\ 58$ ## Problem 34 Two swimmers, at opposite ends of a $90$-foot pool, start to swim the length of the pool, one at the rate", "* dir(234), linewidth(5)); dot(1 * dir(306), linewidth(5)); dot(1 * dir(378), linewidth(5)); dot(2 * dir(378), linewidth(5)); dot(2 * dir(306), linewidth(5)); dot(2 * dir(234), linewidth(5)); dot(2 * dir(162), linewidth(5)); dot(2 * dir( 90), linewidth(5)); label(\"A\", 1 * dir( 90), -dir( 90)); label(\"B\", 1 * dir(162), -dir(162)); label(\"C\", 1 * dir(234), -dir(234)); label(\"D\", 1 * dir(306), -dir(306)); label(\"E\", 1 * dir(378), -dir(378)); label(\"F\", 2 * dir(378), dir(378)); label(\"G\", 2 * dir(306), dir(306)); label(\"H\", 2 * dir(234), dir(234)); label(\"I\", 2 * dir(162), dir(162)); label(\"J\", 2 * dir( 90), dir( 90)); [/asy]$ ## Problem 11 Find the least positive integer $n$ such that when $3^n$ is written in base $143$, its two right-most digits in base $143$ are $01$. ## Problem 14 In $\\triangle ABC$, $AB=10$, $\\measuredangle A=30^{\\circ}$, and $\\measuredangle C=45^{\\circ}$. Let $H$, $D$, and $M$ be points on line $\\overline{BC}$ such that $AH\\perp BC$, $\\measuredangle BAD=\\measuredangle CAD$, and $BM=CM$. Point $N$ is the midpoint of segment $HM$, and point $P$ is on ray $AD$ such that $PN\\perp BC$. Then $AP^2=\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Problem 14 For each integer $n \\ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\\le x \\le n$ and $0\\le y \\le x \\left\\lfloor \\sqrt x \\right\\rfloor$, where $\\left\\lfloor \\sqrt", "(clockwise or counterclockwise). 3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM ## Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus,", "3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "q); label(\"P\", P, dir(250), q); label(\"Q\", Q, 0.25*(Q-P), q); label(\"\\sqrt{3}\",B--Q, dir(60), p); label(\"\\sqrt{3}\",P--B, dir(210), p); label(\"\\sqrt{3}\",P--Q, dir(135), p); label(\"2\",P--C, dir(120), p); label(\"2\",A--Q, dir(-20), p); label(\"1\",A--P, dir(210), p); [/asy]$ Note that $\\triangle QAP$ is a $30^\\circ$-$60^\\circ$-$90^\\circ$ triangle, hence $\\angle PQA=30^\\circ$, and $\\angle BQA=90^\\circ$, so $$AB^2=PQ^2+AQ^2=2^2+3=7,$$ and the answer is $\\boxed{\\textbf{(B) } \\sqrt{7}}$. ## Solution 2 (Intuition) $[asy] unitsize(1inch); pen p = fontsize(10pt); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label(\"A\", (0,0), SW, p); label(\"C\", (2.646,0), SE, p); label(\"B\", (1.323,2.291), N, p); label(\"P'\", (0.756,0.655), NW, p); label(\"1\", (2.174,0.164), N, p); label(\"1\", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]$ Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments). Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$." ]

[ "# Q.29. If are real roots of equation ${x}^{3}-3p{x}^{2}+3qx-1=0$, then find the centroid of a triangle whose vertices are . • 2 What are you looking for?", "or x=2 Hence, the roots of the equation are $\\frac{1}{2}$ and 2. #### Practise This Question A(a,b), B(c,d) and C(e,f) are the vertices of a triangle. i) AB + BC > AC ii) Area of the triangle = ( 12)[a(c-e)+c(f-b)+e(b-d)] Which of the following are true?", "# Roots , triangle and lines !! Geometry Level 2 The area of the triangle formed by the lines $y=m_1 \\times x$and $y=m_2 \\times x$ and $y=c$ is $\\frac{c^{2}}{a} \\times (\\sqrt{b}+\\sqrt{d})$,where c is constant not given and $m_1$ & $m_2$ are the roots of the equation $x^{2} + (\\sqrt{3} + 2)x + \\sqrt{3}-1$. Also a,b&d are integers ,what is the value of $a+b+d$ . ×", "A Polynomial and a Triangle Geometry Level 3 Let $$f(x) = x^3-4x^2+ax+b$$ be a polynomial with $$3$$ positive roots. A triangle with side lengths equal to the roots of $$f(x)$$ has an area of $$1.$$ Find $$f(2)$$. ×", "have all the roots real and positive. This equation has coefficients some expressions in $R$, $r$ and $p$. From here one gets conditions on $R$,$r$ and $p$ for the existence of a triangle. – Orest Bucicovschi Jan 11 '15 at 17:46", "# Unsolved ChallengeArea of triangle #### anemone ##### MHB POTW Director Staff member Find in terms of $p,\\,q$ and $r$, a formula for the area of a triangle whose vertices are the roots of $x^3-px^2+qx-r=0$ in the complex plane.", "this point is wrong also. ## Homework Statement im just trying to find the smallest area a square and triangle with perimeter some of 10 can be. I don't see an equation anywhere; an equation must have an '=' sign in it. Do you mean that you want to solve $$\\sqrt\\frac{3}{2}\\: x - \\frac{3}{8} (10 - 3x) = 0?$$ If so, that is just elementary algebra.", "# Algebraic Triangle. Algebra Level 4 If $$a,b,c$$ are the sides of a triangle such that $$x^2-2(a+b+c)x+3\\lambda(ab+bc+ca)=0$$ has real roots, which of the following is correct? ×", "triangle circumference. • Equation Equation ? has one root x1 = 8. Determine the coefficient b and the second root x2. • Roots Determine the quadratic equation absolute coefficient q, that the equation has a real double root and the root x calculate: ? • Discriminant Determine the discriminant of the equation: ? • Reference angle Find the reference angle of each angle:", "two distinct roots then it must have $n$ distinct roots. Friday, January 19, 2007 Triangularly Speaking. Topic: Algebra/Geometry. Level: AIME. Problem: Suppose the roots of the cubic polynomial $P(x) = x^3-2rx^2+sx+t$ are the sides of a triangle (assume they are all positive reals and satisfy the triangle inequality). What is the area of the triangle in terms of the coefficients? Solution: Well, let's think about the formulas for the area of a triangle given its side lengths. The only one that comes to mind that uses only side lengths is Heron's formula. Why don't we try that? $A = \\sqrt{s(s-a)(s-b)(s-c)}$, where $a, b, c$ are the side lengths and $s = \\frac{a+b+c}{2}$. From $P(x)$, by Vieta's, we know that $r_1+r_2+r_3 = 2r$, where $r_i$ are the roots of the polynomial. So $s$ in this case is $r$. How do we get the rest, though? We could expand and use Newton sums, but that sounds like a pain. Note that $(s-a)(s-b)(s-c)$ looks a lot like $(x-a)(x-b)(x-c)$. But wait, that's just a polynomial with roots $a, b, c$! So if we write $P(x) = (x-r_1)(x-r_2)(x-r_3)$, we have $(s-a)(s-b)(s-c) = (s-r_1)(s-r_2)(s-r_3) = P(s) = P(r)$. Thus we can say that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{rP(r)}$, which can clearly be evaluated in terms of the coefficients. QED. --------------------", "both sides by x squared $3 = x^4$ Then take a quartic root of both sides: $x = 3^{\\frac{1}{4}}$ 8. thanks so much!!", "# As easy as One Two Three Algebra Level 3 Given the quadratic equation $$ax^2+bx+c=0$$ and $$x^2+2x+3=0$$ have a common root with $$a, b, c>0$$. If $$a,b,c$$ are sides of a triangle, classify what type of triangle this triangle is. ×", "to a third of the area of the triangle $ABC$, then its side is equal to a) ${\\dfrac a4\\,}$ b) ${\\dfrac a{\\sqrt6}\\,}$ c) ${\\dfrac a{\\sqrt2}\\,}$ d) ${\\dfrac a2\\,}$ e) ${\\dfrac a{\\sqrt3}\\,}$ a b c d e 2 points 15. The equation ${x^2+(2m+4)x+m-1=0}$ (with the unknown $x$) a) has two different real roots, if and only if ${m=-1\\llor m=0\\llor m=1}$ b) has two different real roots, if and only if ${m=0}$ c) has two different real roots, if and only if ${m\\in\\langle -1,\\,1\\rangle }$ d) has two different real roots for all ${m\\in\\R}$ e) has no real root for all ${m\\in\\R}$", "of triangles in a plane and $R^{+}$ be the set of all positive real numbers, then the function $f\\::\\:A\\rightarrow R^{+},$ defined by $f(x)=$ area of triangle $x,$ is one-one and into one-one and onto many-one and onto many-one and into 1 vote If one root of a quadratic equation $ax^{2}+bx+c=0$ be equal to the n th power of the other, then $(ac)^{\\frac{n}{n+1}}+b=0$ $(ac)^{\\frac{n+1}{n}}+b=0$ $(ac^{n})^{\\frac{1}{n+1}}+(a^{n}c)^{\\frac{1}{n+1}}+b=0$ $(ac^\\frac{1}{n+1})^{n}+(a^\\frac{1}{n+1}c)^{n+1}+b=0$ The condition that ensures that the roots of the equation $x^{3}-px^{2}+qx-r=0$ are in H.P. is $r^{2}-9pqr+q^{3}=0$ $27r^{2}-9pqr+3q^{3}=0$ $3r^{3}-27pqr-9q^{3}=0$ $27r^{2}-9pqr+2q^{3}=0$ Let $p,q,r,s$ be real numbers such that $pr=2(q+s).$ Consider the equations $x^{2}+px+q=0$ and $x^{2}+rx+s=0.$ Then at least one of the equations has real roots. both these equations have real roots. neither of these equations have real roots. given data is not sufficient to arrive at any conclusion.", "are $(0, 0)$, $(0, t(0)),$ and ($x$-intercept of $t(x)$, $0)$, where $t(x)$ is the equation of the tangent line. Calculate the area of the triangle using the vertices to determine the dimensions.", "our roots will be at $x = \\frac{7 + 3}{2}$ and $x = \\frac{7 - 3}{2}$ or $x = 5$ and $x = 2$. On the graph of the function, we would see that the parabola crosses the x-axis at $x = 2$ and $x = 5$.", "# Find the area of a triangle whose length of sides are the roots of X^3+a*X^2+b*X+c. 1 answer below » Find the area of a triangle whose length of sides are the roots of X^3+a*X^2+b*X+c. Sahin M We apply Heron's... ## Recent Questions in Math Looking for Something Else? Ask a Similar Question", "be written as: $x_1=m+n; x_2=m-n; x_3=-2\\cdot m$, where $n= a+i\\cdot b; a \\cdot b=0; \\{a,b,m\\} \\in R$ Here $m$ is always real and $n$ is either real or strictly imaginary. Lets look at these distinct cases. If a cubic has imaginary roots, then they split into a conjugate pair of complex roots and a real root. We then search for solution in the form $(x^2+r_1)\\cdot(x+r_2)=0; \\{r_1,r_2\\} \\in R$ For a given depressed cubic find a right triangle ABC centered at origin such that vector sums $\\{A+A_{vf}, B+B_{vf}, C+C_{vf}\\}$ are roots of the cubic. Here $A_{vf}$ is the image of point A after a vertical flip. We can rotate and scale the triangle, hence, we have two degrees of freedom. Rotation of the triangle by $\\frac{2 \\cdot \\pi }{3}$ is the key symmetry.", "# Squares and Roots IV Geometry Level 1 If $$a = \\frac { \\sqrt { 10 } }{ 10 }$$ and $$b = \\frac { \\sqrt { 8 } }{ 8 }$$, what is the area of this rectangle? This problem is part of Victor's set on Square Properties. ×", "get to know these form solution to the following triangle T(0,0) T(0,1) T(1,0) T(0,2) T(1,1) T(2,0) T(0,3) T(1,2) T(2,1) T(3,0) hence we conclude T(l,m)=C(l+m,l)" ]

[ "$s=\\frac{a+b+c}{2}=\\frac{2+3+4}{2}=4.5$s=a+b+c2=2+3+42=4.5 Do: Now we can use Heron's Formula $A$A $=$= $\\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) $A$A $=$= $\\sqrt{4.5(4.5-2)(4.5-3)(4.5-4)}$√4.5(4.5−2)(4.5−3)(4.5−4) $A$A $=$= $\\sqrt{4.5(2.5)(1.5)(0.5)}$√4.5(2.5)(1.5)(0.5) $A$A $=$= $2.9$2.9 (to 1dp) So the area of the triangle is $2.9$2.9 square units. This applet will allow you to see the above calculation, but also manipulate the triangle to be any shape you want and see the Heron's formula calculation. ##### Question 2 A triangle has an area of $2x\\sqrt{10}$2x10 cm2 and three sides that measure $x,x+2$x,x+2 and $2x-2$2x2 (cm) Find the value of $x$x, and hence calculate the perimeter of the triangle. Think: We will need an expression for both $s$s (the semiperimeter) and $A$A. $s$s $=$= $\\frac{a+b+c}{2}$a+b+c2​ $s$s $=$= $\\frac{x+x+2+2x-2}{2}$x+x+2+2x−22​ $s$s $=$= $\\frac{4x}{2}$4x2​ $s$s $=$= $2x$2x $A$A $=$= $\\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) $A$A $=$= $\\sqrt{2x(2x-a)(2x-b)(2x-c)}$√2x(2x−a)(2x−b)(2x−c) $A$A $=$= $\\sqrt{2x(2x-x)(2x-(x+2))(2x-(2x-2)}$√2x(2x−x)(2x−(x+2))(2x−(2x−2) $A$A $=$= $\\sqrt{2x(x)(x-2)(2)}$√2x(x)(x−2)(2) $A$A $=$= $\\sqrt{4x^2(x-2)}$√4x2(x−2) Do: We are told the area is $2x\\sqrt{10}$2x10, so then $2x\\sqrt{10}$2x√10 $=$= $\\sqrt{4x^2(x-2)}$√4x2(x−2) $40x^2$40x2 $=$= $4x^2(x-2)$4x2(x−2) (2) $10$10 $=$= $x-2$x−2 $x$x $=$= $12$12 See how on step 2, I divided by $4x^2$4x2, well I can only do this if $x$x is not zero. Can you see why? If $x=0$x=0, then I have divided by $0$0 which we know we cannot do. So whilst we found that $x=12$x=12, we also need to say that $x$x could be $0$0. (substitute $x=0$x=0 into the original equation and you will", "# Difference between revisions of \"Heron's Formula\" Heron's formula (sometimes called Hero's formula) is a method for finding the area of a triangle given only the three side lengths. ### Definition For any triangle with side lengths ${a}, {b}, {c}$, the area ${K}$ can be found using the following formula: $K=\\sqrt{s(s-a)(s-b)(s-c)}$ where the semi-perimeter $s=\\frac{a+b+c}{2}$. ### Proof $[ABC]=\\frac{ab}{2}\\sin C$ $=\\frac{ab}{2}\\sqrt{1-\\cos^2 C}$ $=\\frac{ab}{2}\\sqrt{1-(\\frac{a^2+b^2-c^2}{2ab})^2}$ $=\\sqrt{\\frac{a^2b^2}{4}(1-\\frac{(a^2+b^2-c^2)^2}{4a^2b^2})}$ $=\\sqrt{\\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$ $=\\sqrt{\\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$ $=\\sqrt{\\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$ $=\\sqrt{\\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$ $=\\sqrt{s(s-a)(s-b)(s-c)}$", "2c)$ is the semiperimeter of the trapezoid. This formula is analogous to Heron's formula to compute the area of a triangle. The previous formula for area can also be written as $K= \\frac{1}{4} \\sqrt{(a+b)^2(a-b+2c)(b-a+2c)}.$ The radius in the circumscribed circle is given by[6] $R=c\\sqrt{\\frac{ab+c^2}{4c^2-(a-b)^2}}.$ In a rectangle where a = b this is simplified to $R=\\tfrac{1}{2}\\sqrt{a^2+c^2}$.", "known and solve real world application problems using Law of Sines, Law of Cosines or the area formulas. Guidance Heron’s Formula, named after Hero of Alexandria 2000 years ago, can be used to find the area of a triangle given the three side lengths. The formula requires the semi-perimeter, $s$, or $\\frac{1}{2}(a+b+c)$, where $a, b$ and $c$ are the lengths of the sides of the triangle. Heron’s Formula: $\\text{Area} =\\sqrt{s(s-a)(s-b)(s-c)}$ Example A Use Heron’s formula to find the area of a triangle with side lengths 13 cm, 16 cm and 23 cm. Solution: First, find the semi-perimeter or $s$: $s=\\frac{1}{2}(13+16+23)=26$. Next, substitute our values into the formula as shown and evaluate: $A=\\sqrt{26(26-13)(26-16)(26-23)}=\\sqrt{26(13)(10)(3)}=\\sqrt{10140} \\approx 101 \\ cm^2$ Example B Alena is planning a garden in her yard. She is using three pieces of wood as a border. If the pieces of wood have lengths 4 ft, 6ft and 3 ft, what is the area of her garden? Solution: The garden will be triangular with side lengths 4 ft, 6 ft and 3 ft. Find the semi-perimeter and then use Heron’s formula to find the area. $s &=\\frac{1}{2}(4+6+3)=\\frac{13}{2} \\\\A &=\\sqrt{\\frac{13}{2} \\left(\\frac{13}{2} -4 \\right) \\left(\\frac{13}{2} -6 \\right) \\left(\\frac{13}{2} -3 \\right)}=\\sqrt{\\frac{13}{2} \\left(\\frac{5}{2} \\right) \\left(\\frac{1}{2} \\right) \\left(\\frac{7}{2} \\right)}=\\sqrt{\\frac{455}{16}} \\approx 28 \\ ft^2$ Example C Caroline wants to measure the height of a radio", "is 5cm and the base 4cm. so you can find the height by Pythagorean Theorem. $$a^{2} + b^{2} = c^{2}$$ $$a^{2} + 4^{2} = 5{2}$$ $$a^{2} + 16 = 25$$ $$a^{2} + 16 – 16 = 25 – 16$$ $$a^{2} = 9$$ $$a = 3$$ So put the value in the area formula, and use two perpendicular side (a and b) to substitute the base and height: $$Area = \\frac{1}{2}(bh)$$ $$Area = \\frac{1}{2}(4)(3)$$ $$Area = \\frac{1}{2}(12)$$ $$Area = 6$$ So, the area of a triangle is $$6cm^{2}$$. Method 2: Using side lengths 1. Calculate the semi-perimeter of the triangle The semi-perimeter is equal to the half of its perimeter. So, to find the semi-perimeter we need to find its perimeter of the by adding up the length of its three sides then multiplying the product by $$\\frac{1}{2}$$. E.g. suppose if a triangle has three sides that are 5cm, 4cm and 3cm long, then the semi-perimeter will be: $$s = \\frac{1}{2}(3 + 4 + 5)$$ $$s = \\frac{1}{2}(12)$$ = 6 1. Apply Heron’s formula Heron’s formula is $$Area = \\sqrt{s(s-a)(s-b)(s-c)}$$. Here ‘s’ is the semi-perimeter of a triangle and a, b, and c are the side lengths of a triangle. 1. Put values in the formula and calculate it While putting the values in the formula, make sure you", "$h$ from $O$ to $AB$, first we calculate the area of triangle $OAB$. Since $OA = p+r_1, OB = p+r_2, AB = r_1+r_2$, we can use Heron's formula: $$S_{OAB} = \\sqrt{(p+r_1+r_2)pr_1r_2} = \\frac{1}{2} h AB = \\frac{1}{2} h R$$ $$4(p+r_1+r_2)pr_1r_2 = h^2 R^2$$ $$h^2 = \\frac{4(p+r_1+r_2)pr_1r_2}{R^2}$$ Now note that $p+r_1+r_2 = p+R = \\frac{r_1r_2R}{R^2-r_1r_2} + R = \\frac{R^3}{R^2-r_1r_2}$ Plug this into the equation above: $$h^2 = \\frac{4(p+R)pr_1r_2}{R^2} = \\frac{4}{R^2} \\frac{R^3}{R^2-r_1r_2} \\frac{r_1r_2R}{R^2-r_1r_2} r_1r_2 = 4 \\frac{(r_1r_2R)^2}{(R^2-r_1r_2)^2}$$ $$h = 2p$$", "# A triangle has sides with lengths: 2, 9, 2. How do you find the area of the triangle using Heron's formula? Dec 26, 2015 There is no such triangle, since $2 + 2 < 9$ #### Explanation: If a triangle has sides of length $a$, $b$ and $c$ then all of these conditions hold: $a + b > c$ $b + c > a$ $c + a > b$ ...unless you count empty triangles, in which case change the $>$'s into $\\ge$'s. If you try to apply Heron's formula to lengths $a = 2$, $b = 9$, $c = 2$, then you will find that you end up attempting to take the square root of a negative number, hence no Real area: The semi-perimeter $s p$ is given by: $s p = \\frac{a + b + c}{2} = \\frac{2 + 9 + 2}{2} = \\frac{13}{2}$ Then Heron's formula for the area $A$ is: $A = \\sqrt{s p \\left(s p - a\\right) \\left(s p - b\\right) \\left(s p - c\\right)}$ $= \\sqrt{\\frac{13}{2} \\left(\\frac{13}{2} - 2\\right) \\left(\\frac{13}{2} - 9\\right) \\left(\\frac{13}{2} - 2\\right)}$ $= \\sqrt{\\left(\\frac{13}{2}\\right) \\left(\\frac{9}{2}\\right) \\left(- \\frac{5}{2}\\right) \\left(\\frac{9}{2}\\right)}$ $= \\sqrt{- \\frac{5265}{16}}$ It is possible to simplify this further, but there's no real point since it's clearly the square root of a negative quantity.", "Krishna 0 Step 1: Calculate the semi perimeter of the triangle. NOTE: First calculate the perimeter of a triangle by adding up the length of its three sides. Then, multiply by \\frac {1}{2} EXAMPLE: \\frac{1}{2} (a+b+c) \\frac {1}{2}{(3+4+5)} s={\\frac {1}{2}}(12) = 6 Step 2: Substitute the values of semi perimeter and lengths of the triangle in the area formula FORMULA: Area=\\sqrt{s(s-a)(s-b)(s-c)}, EXAMPLE: Area = \\sqrt {6(6-3)(6-4)(6-5)} Step 3: Calculate the values in parentheses. Step 4: Multiply the two values under the radical sign.Then, find their square root.", "ab \\sin C$ $A = \\frac{1}{2} ab \\displaystyle \\frac{ \\sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab}$ $A = \\displaystyle \\frac{1}{4} \\sqrt{(a + b + c)(a + b - c)(c + a - b)(c - a + b)}$ (4) Now, if we let $s$ be the semiperimeter (half the perimeter) of triangle $ABC$, then $a + b + c = 2s$. Now, $a + b + c - 2a = b + c - a = 2s -2a = 2(s - a)$. Also, $a - b + c = 2(s - b)$ and $a + b - c = 2(s - c)$. Substituting the expressions with s to (4), we have $A = \\frac{1}{4} \\sqrt{2s [2(s-a)][2(s-b)][2(s-c)]}$ $= \\frac{1}{4} \\sqrt{16s(s-a)(s-b)(s-a)}$ which is equivalent to $A = \\frac{4}{4} \\sqrt{s(s-a)(s-b)(s-a)}$. Simplifying, we have $A = \\sqrt{s(s-a)(s-b)(s-c)}$, the Heron’s formula. ## 2 thoughts on “A Detailed Derivation of the Heron’s Formula” 1. Very nice proof. I’ve seen some really ugly ones to verify the Heron’s formula. This one relies on the formula for the area of a triangle, the law of cosines, and the Pythagorean Identity. Straightforward and easily digestible. Thank you", "this equation: $$P=2x+\\sqrt{4-x^2}$$ Further, $$\\frac{dP}{dx}= 2-\\frac{x}{\\sqrt{4-x^2}}$$ Set $\\frac{dP}{dx}=0$ for finding the maximum of P. So, $$2=\\frac{x}{\\sqrt{4-x^2}}$$ $x=\\sqrt{\\frac{16}{5}}$ is the value of x for maximum value of $P$ (which gives the maximum perimeter). Therefore, the perimeter of the rectangle when $x=\\sqrt{\\frac{16}{5}}$ is $P=2\\sqrt{5}$ (by substituting the value of x in the first equation). Finally, $$x=\\frac{4}{\\sqrt{5}}$$ and $perimeter(max)=2\\sqrt{5}$ Hint: calculate the height with pitagoras (hyp=1) Edit: the height is $h=\\sqrt{1-x^2/4}$. Now, the perimeter is $p(x)=2x+2 \\sqrt{1-x^2/4}$. $p'(x)=2+(1-x^2/4)^{-1/2}(-x/2)=0$ iff $1-x^2/4=x^2/16$ iff $x^2=16/5$ • thanks, but done that already. – T.Bill Jan 2 '17 at 22:33 • If you have already found $h$ in terms of $x$ using the Pythagorean theorem, then you should be able to express the perimeter $P$ in terms of $x$ and $h$, then substitute your value for $h$ in terms of $x$. Then you will have $P$ as a function of $x$. Then you will have to set the derivative equal to $0$ to find the optimum value of $x$. – John Wayland Bales Jan 2 '17 at 22:47 $P=2x+2h$ By definition $x=r\\cos(\\theta) \\implies x=\\cos(\\theta)$ $P=2\\cos(\\theta)+2h$ By pythagoras $(\\frac{x}{2})^2+h^2=1$ $h^2=1-\\frac{x^2}{4}$ $h^2=\\frac{1}{4}(4-x^2)$ $h^2=\\frac{1}{4}(4-\\cos^2(\\theta)$) $h=\\frac{1}{2}\\sqrt{4-\\cos^2(\\theta)}$ $P=2x+2h$ $P=2\\cos(\\theta)+2(\\frac{1}{2}\\sqrt{4-\\cos^2(\\theta)})$ $P=2\\cos(\\theta)+\\sqrt{4-\\cos^2(\\theta)}$ Notice that the $-1\\leq\\cos(\\theta) \\leq1$ and $0 \\leq \\cos^2(\\theta) \\leq 1$ So max value is $P(\\theta)_{max}=2+\\sqrt{4-1}=2+\\sqrt{3}$ The perimeter is given by, $$2h+2x=2(h+x)$$ We need to maximize $2(h+x)=2\\sqrt{1-x^2/4}+2x=\\sqrt{4-x^2}+2x$ by pythag. Thm. Obviously", "half of the triangle's perimeter. Three equivalent ways of writing Heron's formula are $T = \\frac{1}{4} \\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$ $T = \\frac{1}{4} \\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}$ $T = \\frac{1}{4} \\sqrt{(a+b-c) (a-b+c) (-a+b+c) (a+b+c)}.$ ### Formulas mimicking Heron's formula Three formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as ma,mb, and mc and their semi-sum (ma + mb + mc) / 2 as σ, we have[13] $T = \\frac{4}{3} \\sqrt{\\sigma (\\sigma - m_a)(\\sigma - m_b)(\\sigma - m_c)}.$ Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc,and denoting the semi-sum of the reciprocals of the altitudes as $H = (h_a^{-1} + h_b^{-1} + h_c^{-1})/2$ we have[14] $T^{-1} = 4 \\sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.$ And denoting the semi-sum of the angles' sines as $S=[(\\sin \\ \\ \\alpha)+(\\sin \\ \\ \\beta)+(\\sin \\ \\ \\gamma)]/2$, we have [15] $T = D^{2} \\sqrt{S(S-\\sin \\alpha)(S-\\sin \\beta)(S-\\sin \\gamma)}$ where D is the diameter of the circumcircle: $D=\\tfrac{a}{\\sin \\alpha} = \\tfrac{b}{\\sin \\beta} = \\tfrac{c}{\\sin \\gamma}.$ ### Using Pick's Theorem See Pick's theorem for a technique for finding the area of any arbitrary lattice polygon. The theorem states: $T = I + \\frac{1}{2}B - 1$ where I is the number of internal lattice points and B is the number of", "How do you use Heron's formula to determine the area of a triangle with sides of that are 8, 5, and 10 units in length? Feb 4, 2016 here is how, Explanation: we know, $s = \\frac{a + b + c}{2}$ $= \\frac{8 + 10 + 5}{2}$ $= \\frac{23}{2}$ $= 11.5$ from heron's formula, we know, $A = \\sqrt{s \\left(s - a\\right) \\left(s - b\\right) \\left(s - c\\right)}$ $= \\sqrt{11.5 \\left(11.5 - 8\\right) \\left(11.5 - 5\\right) \\left(11.5 - 10\\right)}$ $= \\sqrt{392.4375}$ $= 19.8100353357 u n i {t}^{2}$ Feb 4, 2016 Area ≈ 19.8 Explanation: This is a 2 step process : 1. Calculate half of the triangle's perimeter (s) 2. Calculate the area let a = 8 , b = 5 and c = 10 step 1 : $s = \\frac{a + b + c}{2} = \\frac{8 + 5 + 10}{2} = \\frac{23}{2} = 11.5$ step 2 : Area $= \\sqrt{s \\left(s - a\\right) \\left(s - b\\right) \\left(s - c\\right)}$ = sqrt(11.5(11.5-8)(11,5-5(11.5-10) = sqrt(11.5 xx 3.5 xx 6.5 xx 1.5) ≈ 19.8", "## Trigonometry (11th Edition) Clone (a) We can find the semi-perimeter $S$: $S = \\frac{11+13+20}{2} = 22$ We can find the area: $A = \\sqrt{S~(S-a)(S-b)(S-c)}$ $A = \\sqrt{(22)~(22-11)(22-13)(22-20)}$ $A = \\sqrt{(22)~(11)(9)(2)}$ $A = \\sqrt{4356}$ $A = 66$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (b) We can find the semi-perimeter $S$: $S = \\frac{13+14+15}{2} = 21$ We can find the area: $A = \\sqrt{S~(S-a)(S-b)(S-c)}$ $A = \\sqrt{(21)~(21-13)(21-14)(21-15)}$ $A = \\sqrt{(21)~(8)(7)(6)}$ $A = \\sqrt{7056}$ $A = 84$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (c) We can find the semi-perimeter $S$: $S = \\frac{7+15+20}{2} = 21$ We can find the area: $A = \\sqrt{S~(S-a)(S-b)(S-c)}$ $A = \\sqrt{(21)~(21-7)(21-15)(21-20)}$ $A = \\sqrt{(21)~(14)(6)(1)}$ $A = \\sqrt{1764}$ $A = 42$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (d) We can find the semi-perimeter $S$: $S = \\frac{9+10+17}{2} = 18$ We can find the area: $A = \\sqrt{S~(S-a)(S-b)(S-c)}$ $A = \\sqrt{(18)~(18-9)(18-10)(18-17)}$ $A = \\sqrt{(18)~(9)(8)(1)}$ $A = \\sqrt{1296}$ $A = 36$ Since this triangle has integer sides and area, this triangle is a Heron triangle.", "# Heron's Formula Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths. ## Theorem For any triangle with side lengths ${a}, {b}, {c}$, the area ${A}$ can be found using the following formula: $A=\\sqrt{s(s-a)(s-b)(s-c)}$ where the semi-perimeter $s=\\frac{a+b+c}{2}$. ## Proof $[ABC]=\\frac{ab}{2}\\sin C$ $=\\frac{ab}{2}\\sqrt{1-\\cos^2 C}$ $=\\frac{ab}{2}\\sqrt{1-\\left(\\frac{a^2+b^2-c^2}{2ab}\\right)^2}$ $=\\sqrt{\\frac{a^2b^2}{4}\\left[1-\\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\\right]}$ $=\\sqrt{\\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$ $=\\sqrt{\\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$ $=\\sqrt{\\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$ $=\\sqrt{\\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$ $=\\sqrt{s(s-a)(s-b)(s-c)}$ ## Isosceles Triangle Simplification $A=\\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles $b=c$ for all isosceles triangles $A=\\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\\sqrt{s(s-a)}$ $\\blacksquare$ ## Example Let's say that you have a right triangle with the sides $3$ ,$4$ , and $5$. Your semi- perimeter would be $6$ since $(3+4+5)$ ÷ $2$ is $6$. Then you have $6-3=3$, $6-4=2$, $6-5=1$. $1+2+3=6.$ $6\\cdot 6 = 36$ The square root of $36$ is $6$. The area of your triangle is $6$.", "# Heron's Formula Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths. ## Theorem For any triangle with side lengths ${a}, {b}, {c}$, the area ${A}$ can be found using the following formula: $A=\\sqrt{s(s-a)(s-b)(s-c)}$ where the semi-perimeter $s=\\frac{a+b+c}{2}$. ## Proof $[ABC]=\\frac{ab}{2}\\sin C$ $=\\frac{ab}{2}\\sqrt{1-\\cos^2 C}$ $=\\frac{ab}{2}\\sqrt{1-\\left(\\frac{a^2+b^2-c^2}{2ab}\\right)^2}$ $=\\sqrt{\\frac{a^2b^2}{4}\\left[1-\\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\\right]}$ $=\\sqrt{\\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$ $=\\sqrt{\\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$ $=\\sqrt{\\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$ $=\\sqrt{\\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$ $=\\sqrt{s(s-a)(s-b)(s-c)}$ ## Isosceles Triangle Simplification $A=\\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles $b=c$ for all isosceles triangles $A=\\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\\sqrt{s(s-a)}$ $\\blacksquare$ ## Example Let's say that you have a right triangle with the sides 3,4, and 5. Your semi- perimeter would be 6. Then you have 6-3=3, 6-4=2, 6-5=1. 1+2+3= 6 $6*6 = 36$ The square root of 36 is 6. The area of your triangle is 6.", "any ratio $$R$$ of area/perimeter by finding the $$m,n$$(s) that represent them using the following formula which includes a defined finite search for values of $$m$$. Whenever the $$m$$ $$R$$ combination yields a positive integer for $$n$$, we have the $$m,n$$ for a triple. We begin by solving the area/perimeter equation for $$n$$ where area=$$D$$ so as not to confuse it with $$A,B,C$$. $$D=\\frac{AB}{2}=\\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$ $$\\frac{D}{P}=\\frac{mn(m^2-n^2)}{2m^2+2mn}=\\frac{mn(m-n)(m+n)}{2m(m+n)}=\\frac{n(m-n)}{2}=R\\qquad n^2-mn+2R=0$$ $$n=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}=\\frac{m\\pm\\sqrt{m^2-4*1*2R}}{2*1}$$ $$n=\\frac{m\\pm \\sqrt{m^2-8R}}{2}\\text{ where }\\lceil\\sqrt{8R}\\space \\rceil \\le m \\le 2R+1$$ Example: For $$R=0.5\\quad \\sqrt{8*0.5}=2\\le m \\le 2*0.5+1=2$$ $$n=\\frac{2\\pm \\sqrt{2^2-8*0.5}}{2}=1\\qquad f(2,1)=(3,4,5)$$ But you are asking about $$R=1$$ and there only two such Pythagorean triples to be found. $$R=1\\rightarrow 3\\le m \\le 3\\quad f(3,2)=(5,12,13)\\quad f(3,1)=(8,6,10)$$ For other ratios: $$R=1.5\\rightarrow 4\\le m \\le 4\\quad f(4,3)=(7,24,25)\\quad f(4,1)=(15,8,17)$$ $$R=2\\rightarrow 4\\le m \\le 5\\quad f(4,2)=(12,16,20)\\quad f(5,4)=(9,40,41)\\quad f(5,1)=(24,10,26)$$ $$R=2.5\\rightarrow 4\\le m \\le 6\\quad f(6,5)=(11,60,61)\\quad f(6,1)=(35,12,37)$$ $$R=3\\rightarrow 4\\le m \\le 7\\quad f(5,3)=(16,30,34)\\quad f(5,2)=(21,20,29)\\quad f(7,6)=(13,84,85)\\quad f(7,1)=(48,14,50)$$ $$R=18\\rightarrow 12\\le m \\le 37\\quad f(12,6)=(108,144,180)\\quad f(13,9)=(88,234,250)\\quad f(13,4)=(153,104,185)\\quad f(15,12)=(81,360,369)\\quad f(15,3)=(216,90,234)\\quad f(20,18)=(76,720,724)\\quad f(20,2)=(396,80,404)\\quad f(37,36)=(73,2664,2665)\\quad f(37,1)=(1368,74,1370)$$ Aside: you can always find one primitive triple for a given $$R$$ if you let $$(m,n)=(2R+1,2R)$$.", "= \\frac{15 + 11 + 6}{2}$$ (Since, 2s = a + b + c) $$\\Rightarrow$$ $$s = \\frac{32}{2}$$ $$\\Rightarrow$$ s = 16m Therefore, area painted in colour = $$\\sqrt{16(16 - 15)(16 - 11)(16 - 6)}$$ (Since, Heron's formula [area = $$\\sqrt{s(s - a)(s - b)(s - c)}$$]) $$\\therefore$$ Area = $$\\sqrt{16 × 1 × 5 × 10}$$ = $$\\sqrt{2 × 2 × 2 × 2 × 5 × 5 × 2}$$ = $$\\sqrt{400 × 2} {m}^2$$ $$\\therefore$$ the area painted in colour is $$20\\sqrt{2} m^2$$) Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm. Let the sides of a triangle a = 18 cm, b = 10 cm and c We have, perimeter = 42 cm So, a + b + c = 42 By substituting the values, $$\\Rightarrow$$ 18 + 10 + c = 42 $$\\Rightarrow$$ 28 + c = 42cm $$\\Rightarrow$$ c = 42 - 28 cm $$\\Rightarrow$$ c = 14cm Now, we know that, $$s = \\frac{a + b + c}{2}$$ $$\\therefore$$ $$s = \\frac{18 + 10 + 14}{2} = \\frac{42}{2} = 21cm$$ Now, Area of triangle = $$\\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$$ (Since, Heron's formula [area = $$\\sqrt{s(s - a)(s - b)(s - c)}$$]) = $$\\sqrt{21 × 3 ×", "is denoted by s symbol. By finding the base and height of the triangle. The second step is to determine the area of a given triangle by applying the heron’s formula i.e. The first step is to determine the semi- perimeter of a triangle i.e. Ans: The term 's' in heron's formula stands for the semi perimeter of a triangle whose area has to be calculated. 1. $A = \\frac{1}{4} \\sqrt{(x + y + z)(-x + y + z)(x - y + z)(x + y - z)}$, $A = \\frac{1}{4} \\sqrt{2(x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2}) - (x^{4} + y^{4} + z^{4})}$, $A = \\frac{1}{4} \\sqrt{(x^{2} + y^{2} + z^{2})^{2} - 2(x^{4} + y^{4} + z^{4})}$, $A = \\frac{1}{4} \\sqrt{(x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2}) -(x^{2} + y^{2} + z^{2})^{2}}$, $A = \\frac{1}{4} \\sqrt{4x^{2}y^{2} - (x^{2} + y^{2}z^{2} - z^{2})^{2}}$. and is represented as. 60 m, 100 m, and 140 m. Now, we will determine the area of the triangle using heron’s formula as shown below: We have s = 60 + 100 + 40/2 meters = 150 meters. If it does, then we shall calculate perimeter, semi-perimeter and area of the Triangle. Perimeter is the measure of path that surrounds any area. is calculated using. Related Read: Formula To Calculate Area of a Triangle When its Sides are Given.", "r, the inradius of the triangle, and their bases are $$a, b\\ and\\ c.$$ Adding the areas of these triangle we get the area of the triangle ABC, and in terms of r, that area is $$\\frac{1}{2}ar+\\frac{1}{2}br+\\frac{1}{2}cr=\\frac{a+b+c}{2} \\cdot r$$ $$=\\frac{6+8+10}{2} \\cdot r=12r$$. So if we can find the area of triangle ABC , then by dividing that area by 12, we would have the inradius. Fortunately Heron, or Hero of Alexandria, some Greek dude who lived around 10 AD, left us a formula by which we can calculate thee area we need using only the known sides of the triangle ABC. If we let $$s=\\frac{a+b+c}{2}$$, a quantity we calculated to be 12, the the area of ABC would be$$\\sqrt {s(s-a)(s-b)(s-c)}$$$$=\\sqrt{12(12-10)(12-8)(12-6)} = \\sqrt{12\\cdot48} =\\sqrt{36\\cdot16}=24$$. So the measure of the inradius is 24 divided by 12 or 2 units of length. #2 +2854 +2 This is a formula that he uses $$A=rs$$ s = semi-perimeter A = area Using his method (heron's formula) we can find the value of A. Using the given info on the triangles side, we can find the semi-perimeter by finding the perimeter and dividing by 2. Then with that info, we plug it into the formula and solve for R. CalculatorUser Nov 16, 2019", "formula for $$s_{2n}$$ in terms of $$s_{n}$$. 3 observations here: 1 $$(\\frac{s_n}{2})^2 + x^2 = r^2$$ by Pythagoras in blue triangle 2 $$(\\frac{s_n}{2})^2 + y^2 = {s_{2n}}^2$$ by Pythagoras in red triangle 3 $$x+y=r$$ x and y together are the radius From these, we can rearrange to state $$s_{2n}$$ in terms of $$s_n$$ 4 $$x = \\sqrt{r^2 - (\\frac{s_n}{2})^2}$$ rearrange 1 5 $$y=r-x$$ rearrange 3 6 $$y = r - \\sqrt{r^2 - (\\frac{s_n}{2})^2}$$ substitute 4 in 5 7 $${s_{2n}}^2 = (\\frac{s_n}{2})^2 + \\left( r - \\sqrt{r^2 - (\\frac{s_n}{2})^2} \\right) ^2$$ substitute 6 into 2 8 $${s_{2n}}^2 = \\frac{ {s_n}^2}{4} + \\left( r - \\sqrt{r^2 - \\frac{ {s_n}^2}{4}} \\right) ^2$$ expand This final formula is roughly equivalent to the calculation used repeatedly in Liu Hui’s description[1] where a circle with radius 10 was used in his calculations. Since the shape is a regular polygon, the perimeter is simply $$n \\times s_n$$. $C = 2\\pi r$ $\\pi = C / 2r$ $C \\approx n \\times s_n$ $\\pi \\approx \\frac{n \\times s_n}{2r}$ # Inequality for Pi Liu Hui looked at the areas of the shapes to determine an upper and lower bound for the value. Use $$A_n$$ to describe the area of a shape with n-sides, and $$D_n$$ to describe the difference in area between a shape and a shape with" ]

[ "# A triangle has sides with lengths: 2, 9, 2. How do you find the area of the triangle using Heron's formula? Dec 26, 2015 There is no such triangle, since $2 + 2 < 9$ #### Explanation: If a triangle has sides of length $a$, $b$ and $c$ then all of these conditions hold: $a + b > c$ $b + c > a$ $c + a > b$ ...unless you count empty triangles, in which case change the $>$'s into $\\ge$'s. If you try to apply Heron's formula to lengths $a = 2$, $b = 9$, $c = 2$, then you will find that you end up attempting to take the square root of a negative number, hence no Real area: The semi-perimeter $s p$ is given by: $s p = \\frac{a + b + c}{2} = \\frac{2 + 9 + 2}{2} = \\frac{13}{2}$ Then Heron's formula for the area $A$ is: $A = \\sqrt{s p \\left(s p - a\\right) \\left(s p - b\\right) \\left(s p - c\\right)}$ $= \\sqrt{\\frac{13}{2} \\left(\\frac{13}{2} - 2\\right) \\left(\\frac{13}{2} - 9\\right) \\left(\\frac{13}{2} - 2\\right)}$ $= \\sqrt{\\left(\\frac{13}{2}\\right) \\left(\\frac{9}{2}\\right) \\left(- \\frac{5}{2}\\right) \\left(\\frac{9}{2}\\right)}$ $= \\sqrt{- \\frac{5265}{16}}$ It is possible to simplify this further, but there's no real point since it's clearly the square root of a negative quantity.", "mind that uses only side lengths is Heron's formula. Why don't we try that? $A = \\sqrt{s(s-a)(s-b)(s-c)}$, where $a, b, c$ are the side lengths and $s = \\frac{a+b+c}{2}$. From $P(x)$, by Vieta's, we know that $r_1+r_2+r_3 = 2r$, where $r_i$ are the roots of the polynomial. So $s$ in this case is $r$. How do we get the rest, though? We could expand and use Newton sums, but that sounds like a pain. Note that $(s-a)(s-b)(s-c)$ looks a lot like $(x-a)(x-b)(x-c)$. But wait, that's just a polynomial with roots $a, b, c$! So if we write $P(x) = (x-r_1)(x-r_2)(x-r_3)$, we have $(s-a)(s-b)(s-c) = (s-r_1)(s-r_2)(s-r_3) = P(s) = P(r)$. Thus we can say that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{rP(r)}$, which can clearly be evaluated in terms of the coefficients. QED. -------------------- Comment: This is probably one of the neatest tricks that Vieta's can help you with, as well as a cool link between algebra and geometry. Before I thought of this problem, I had never realized the now-clear relationship between Heron's formula and a cubic polynomial with sides as the roots. A nice discovery. -------------------- Practice Problem: (2007 Mock AMC 12 - #24) The lengths of the sides of a triangle are the roots of the equation $7x^3+35x = 28x^2+13$. If the square of", "$s=\\frac{a+b+c}{2}=\\frac{2+3+4}{2}=4.5$s=a+b+c2=2+3+42=4.5 Do: Now we can use Heron's Formula $A$A $=$= $\\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) $A$A $=$= $\\sqrt{4.5(4.5-2)(4.5-3)(4.5-4)}$√4.5(4.5−2)(4.5−3)(4.5−4) $A$A $=$= $\\sqrt{4.5(2.5)(1.5)(0.5)}$√4.5(2.5)(1.5)(0.5) $A$A $=$= $2.9$2.9 (to 1dp) So the area of the triangle is $2.9$2.9 square units. This applet will allow you to see the above calculation, but also manipulate the triangle to be any shape you want and see the Heron's formula calculation. ##### Question 2 A triangle has an area of $2x\\sqrt{10}$2x10 cm2 and three sides that measure $x,x+2$x,x+2 and $2x-2$2x2 (cm) Find the value of $x$x, and hence calculate the perimeter of the triangle. Think: We will need an expression for both $s$s (the semiperimeter) and $A$A. $s$s $=$= $\\frac{a+b+c}{2}$a+b+c2​ $s$s $=$= $\\frac{x+x+2+2x-2}{2}$x+x+2+2x−22​ $s$s $=$= $\\frac{4x}{2}$4x2​ $s$s $=$= $2x$2x $A$A $=$= $\\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) $A$A $=$= $\\sqrt{2x(2x-a)(2x-b)(2x-c)}$√2x(2x−a)(2x−b)(2x−c) $A$A $=$= $\\sqrt{2x(2x-x)(2x-(x+2))(2x-(2x-2)}$√2x(2x−x)(2x−(x+2))(2x−(2x−2) $A$A $=$= $\\sqrt{2x(x)(x-2)(2)}$√2x(x)(x−2)(2) $A$A $=$= $\\sqrt{4x^2(x-2)}$√4x2(x−2) Do: We are told the area is $2x\\sqrt{10}$2x10, so then $2x\\sqrt{10}$2x√10 $=$= $\\sqrt{4x^2(x-2)}$√4x2(x−2) $40x^2$40x2 $=$= $4x^2(x-2)$4x2(x−2) (2) $10$10 $=$= $x-2$x−2 $x$x $=$= $12$12 See how on step 2, I divided by $4x^2$4x2, well I can only do this if $x$x is not zero. Can you see why? If $x=0$x=0, then I have divided by $0$0 which we know we cannot do. So whilst we found that $x=12$x=12, we also need to say that $x$x could be $0$0. (substitute $x=0$x=0 into the original equation and you will", "# Difference between revisions of \"Heron's Formula\" Heron's formula (sometimes called Hero's formula) is a method for finding the area of a triangle given only the three side lengths. ### Definition For any triangle with side lengths ${a}, {b}, {c}$, the area ${K}$ can be found using the following formula: $K=\\sqrt{s(s-a)(s-b)(s-c)}$ where the semi-perimeter $s=\\frac{a+b+c}{2}$. ### Proof $[ABC]=\\frac{ab}{2}\\sin C$ $=\\frac{ab}{2}\\sqrt{1-\\cos^2 C}$ $=\\frac{ab}{2}\\sqrt{1-(\\frac{a^2+b^2-c^2}{2ab})^2}$ $=\\sqrt{\\frac{a^2b^2}{4}(1-\\frac{(a^2+b^2-c^2)^2}{4a^2b^2})}$ $=\\sqrt{\\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$ $=\\sqrt{\\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$ $=\\sqrt{\\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$ $=\\sqrt{\\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$ $=\\sqrt{s(s-a)(s-b)(s-c)}$", "How do you use Heron's formula to determine the area of a triangle with sides of that are 8, 5, and 10 units in length? Feb 4, 2016 here is how, Explanation: we know, $s = \\frac{a + b + c}{2}$ $= \\frac{8 + 10 + 5}{2}$ $= \\frac{23}{2}$ $= 11.5$ from heron's formula, we know, $A = \\sqrt{s \\left(s - a\\right) \\left(s - b\\right) \\left(s - c\\right)}$ $= \\sqrt{11.5 \\left(11.5 - 8\\right) \\left(11.5 - 5\\right) \\left(11.5 - 10\\right)}$ $= \\sqrt{392.4375}$ $= 19.8100353357 u n i {t}^{2}$ Feb 4, 2016 Area ≈ 19.8 Explanation: This is a 2 step process : 1. Calculate half of the triangle's perimeter (s) 2. Calculate the area let a = 8 , b = 5 and c = 10 step 1 : $s = \\frac{a + b + c}{2} = \\frac{8 + 5 + 10}{2} = \\frac{23}{2} = 11.5$ step 2 : Area $= \\sqrt{s \\left(s - a\\right) \\left(s - b\\right) \\left(s - c\\right)}$ = sqrt(11.5(11.5-8)(11,5-5(11.5-10) = sqrt(11.5 xx 3.5 xx 6.5 xx 1.5) ≈ 19.8", "factors to $(r+s)(r-s)$. We know that $r+s=8$, so $r-s=\\frac{24}8=3$. Now we can solve for $r$ by adding the two equations we just got to see that $2r=11$, or $r=\\frac{11}2$. We now solve for $x$. We know that $r^2+(2x)^2=49$, so $(2x)^2=49-\\left(\\frac{11}2\\right)^2=\\frac{75}4$ and $2x=\\frac{5\\sqrt3}2$. We multiply both sides of this equation by $12$ to get $24x=30\\sqrt3$. However, the area of hexagon $ABQCDP$ is $24x$, so the answer is $30\\sqrt 3$, or answer choice $\\boxed{B}$.", "known and solve real world application problems using Law of Sines, Law of Cosines or the area formulas. Guidance Heron’s Formula, named after Hero of Alexandria 2000 years ago, can be used to find the area of a triangle given the three side lengths. The formula requires the semi-perimeter, $s$, or $\\frac{1}{2}(a+b+c)$, where $a, b$ and $c$ are the lengths of the sides of the triangle. Heron’s Formula: $\\text{Area} =\\sqrt{s(s-a)(s-b)(s-c)}$ Example A Use Heron’s formula to find the area of a triangle with side lengths 13 cm, 16 cm and 23 cm. Solution: First, find the semi-perimeter or $s$: $s=\\frac{1}{2}(13+16+23)=26$. Next, substitute our values into the formula as shown and evaluate: $A=\\sqrt{26(26-13)(26-16)(26-23)}=\\sqrt{26(13)(10)(3)}=\\sqrt{10140} \\approx 101 \\ cm^2$ Example B Alena is planning a garden in her yard. She is using three pieces of wood as a border. If the pieces of wood have lengths 4 ft, 6ft and 3 ft, what is the area of her garden? Solution: The garden will be triangular with side lengths 4 ft, 6 ft and 3 ft. Find the semi-perimeter and then use Heron’s formula to find the area. $s &=\\frac{1}{2}(4+6+3)=\\frac{13}{2} \\\\A &=\\sqrt{\\frac{13}{2} \\left(\\frac{13}{2} -4 \\right) \\left(\\frac{13}{2} -6 \\right) \\left(\\frac{13}{2} -3 \\right)}=\\sqrt{\\frac{13}{2} \\left(\\frac{5}{2} \\right) \\left(\\frac{1}{2} \\right) \\left(\\frac{7}{2} \\right)}=\\sqrt{\\frac{455}{16}} \\approx 28 \\ ft^2$ Example C Caroline wants to measure the height of a radio", "2c)$ is the semiperimeter of the trapezoid. This formula is analogous to Heron's formula to compute the area of a triangle. The previous formula for area can also be written as $K= \\frac{1}{4} \\sqrt{(a+b)^2(a-b+2c)(b-a+2c)}.$ The radius in the circumscribed circle is given by[6] $R=c\\sqrt{\\frac{ab+c^2}{4c^2-(a-b)^2}}.$ In a rectangle where a = b this is simplified to $R=\\tfrac{1}{2}\\sqrt{a^2+c^2}$.", "Consider the figure below. We can find the area of $\\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$. I will leave the details of the calculations to you. Follow-through As you have said before, the side lengths of $\\triangle ABC$ is $AB=AC=5\\sqrt{5}$, $BC=\\sqrt{10}$, using Heron's formula, we can compute the answer. Heron's formula states that given side lengths $a,b,c$ of $\\triangle ABC$, the area is given $$\\sqrt{s(s-a)(s-b)(s-c)}\\tag{1}$$ Where $s$ Is the semi perimeter. ($s=\\frac {a+b+c}{2}$). So in your case, we have $$a=5\\sqrt{5},b=5\\sqrt{5},c=\\sqrt{10}\\tag{2}$$ The semi perimeter is $$\\frac {10\\sqrt{5}+\\sqrt{10}}{2}\\tag{3}$$ and plugging in the values, we have $$\\sqrt{\\frac {10\\sqrt{5}+\\sqrt{10}}{2}\\left(\\frac {10\\sqrt{5}+\\sqrt{10}}{2}-5\\sqrt{5}\\right)\\left(\\frac {10\\sqrt{5}+\\sqrt{10}}{2}-5\\sqrt{5}\\right)\\left(\\frac {10\\sqrt{5}+\\sqrt{10}}{2}-\\sqrt{10}\\right)}=\\boxed{17.5}\\tag{4}$$ • I understand this now but how would I type this into a calculator to get the correct result ? Thanks. – Dan Oct 2 '16 at 16:32 • @Dan Hm... If I were to have to solve the messy radicals, I would first write it out, and then break down the parts. So I would first calculate what $\\frac {10\\sqrt{5}+\\sqrt{10}}{2}-5\\sqrt{5}$ is, then $\\frac {10\\sqrt{5}+\\sqrt{10}}{2}-\\sqrt{10}$ and so on... Oct 2 '16 at 16:37 • @Dan Also, note that on a calculator, if you press Alpha, then Y= and 1, then you get the option to write your fractions in $\\frac ab$ form.", "# When is the area of a triangle whose side lengths are consecutive integers also an integer? Consider a triangle with side lengths 3, 4, and 5. By Heron's formula, its area is $\\sqrt{6(6 - 5)(6-4)(6 - 3)} = \\sqrt{6(1)(2)(3)} = \\sqrt{36} = 6$. Are there any other triangles like this? - Why was this question closed? Although it was about finding Heronian triangles, it was about a specific subset of those triangles, with consecutive integer sides. This has considerably different implications from simply finding all triangles with integer sides and area. – Lee Sleek May 29 '13 at 22:45 Because of a liking for symmetry, we let the sides be $y-1$, $y$, and $y+1$. Using the Heron Formula, we find, exactly like you did, that $\\frac{3}{16}y^2(y^2-4)$ should be a perfect square. Thus $y$ must be even, say $y=2s$. So we want $3(s^2-1)$ to be a perfect square, say $(3t)^2$. We arrive at the equation $$s^2-3t^2=1,$$ an instance of a Pell Equation. In this case, the equation has the fundamental solution $s_1=2$, $t_1=1$. By general theory of the Pell equation, the positive solutions $(s,t)$ are given by $$s=\\frac{(2+\\sqrt{3})^n+(2-\\sqrt{3})^n}{2},\\qquad t=\\frac{(2+\\sqrt{3})^n-(2-\\sqrt{3})^n}{2},$$ where $n$ ranges over the positive integers. In our case we are primarily interested in $s$. Let $s_0=1$ and $s_1=2$. Define $s_n$ by the recurrence $$s_{n+2}=4s_{n+1}-s_n.$$ For suitable $t_n$,", "formula for $$s_{2n}$$ in terms of $$s_{n}$$. 3 observations here: 1 $$(\\frac{s_n}{2})^2 + x^2 = r^2$$ by Pythagoras in blue triangle 2 $$(\\frac{s_n}{2})^2 + y^2 = {s_{2n}}^2$$ by Pythagoras in red triangle 3 $$x+y=r$$ x and y together are the radius From these, we can rearrange to state $$s_{2n}$$ in terms of $$s_n$$ 4 $$x = \\sqrt{r^2 - (\\frac{s_n}{2})^2}$$ rearrange 1 5 $$y=r-x$$ rearrange 3 6 $$y = r - \\sqrt{r^2 - (\\frac{s_n}{2})^2}$$ substitute 4 in 5 7 $${s_{2n}}^2 = (\\frac{s_n}{2})^2 + \\left( r - \\sqrt{r^2 - (\\frac{s_n}{2})^2} \\right) ^2$$ substitute 6 into 2 8 $${s_{2n}}^2 = \\frac{ {s_n}^2}{4} + \\left( r - \\sqrt{r^2 - \\frac{ {s_n}^2}{4}} \\right) ^2$$ expand This final formula is roughly equivalent to the calculation used repeatedly in Liu Hui’s description[1] where a circle with radius 10 was used in his calculations. Since the shape is a regular polygon, the perimeter is simply $$n \\times s_n$$. $C = 2\\pi r$ $\\pi = C / 2r$ $C \\approx n \\times s_n$ $\\pi \\approx \\frac{n \\times s_n}{2r}$ # Inequality for Pi Liu Hui looked at the areas of the shapes to determine an upper and lower bound for the value. Use $$A_n$$ to describe the area of a shape with n-sides, and $$D_n$$ to describe the difference in area between a shape and a shape with", "Home > APCALC > Chapter 1 > Lesson 1.2.5 > Problem1-84 1-84. Examine the graph of $f(x) = 0.5x - 2$ at right. 1. Calculate the area of the shaded region using geometry. (Recall that area below the $x$-axis is considered negative.) Find the area of each triangle separately. Recall that the area below the $x$-axis is negative. Add the negative and the positive regions together. $-1\\frac{3}{4}$ $\\text{[Area of triangle A]} + \\text{[Area of triangle B]}$ $\\frac{1}{2}(4)(f(0))+\\frac{1}{2}(7-4)(f(7))$ $\\frac{1}{2}(4)(-2)+\\frac{1}{2}(3)(1.5)=$ 2. What is the value of $k$ if the area under the curve for $0 ≤ x ≤ k$ is $10$? How did you obtain your solution? $\\text{[Area of triangle A]} + \\text{[Area of triangle B]} = 10$ $\\frac{1}{2}(4)(f(0))+\\frac{1}{2}(k-4)(f(k))=10$ $\\frac{1}{2}(4)(f(0))+\\frac{1}{2}((0.5x-2)-4)(0.5x-2)=10$ Simplify, and solve for $k$.", "+ ( 17 - 12\\sqrt{2} )^k \\right). \\end{align} The corresponding explicit formulas for sk and tk are [2]:13 $s_k = \\frac{(3 + 2\\sqrt{2})^k - (3 - 2\\sqrt{2})^k}{4\\sqrt{2}}$ and $t_k = \\frac{(3 + 2\\sqrt{2})^k + (3 - 2\\sqrt{2})^k - 2}{4}.$ ## Pell's equation The problem of finding square triangular numbers reduces to Pell's equation in the following way.[3] Every triangular number is of the form t(t + 1)/2. Therefore we seek integers t, s such that $\\frac{t(t+1)}{2} = s^2.$ With a bit of algebra this becomes (2t + 1)2 = 8s2 + 1, and then letting x = 2t + 1 and y = 2s, we get the Diophantine equation x2 − 2y2 = 1 which is an instance of Pell's equation. This particular equation is solved by the Pell numbers Pk as[4] $x = P_{2k} + P_{2k-1}, \\quad y = P_{2k};$ and therefore all solutions are given by $s_k = \\frac{P_{2k}}{2}, \\quad t_k = \\frac{P_{2k} + P_{2k-1} -1}{2}, \\quad N_k = \\left( \\frac{P_{2k}}{2} \\right)^2.$ There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers. ## Recurrence relations There are recurrence relations for the square triangular numbers, as well as for the sides of the square and triangle involved. We have[5]:(12) Nk = 34Nk − 1Nk − 2 + 2, with", "$k$ be the side of the triangle you will get the resultant area as $$area(\\triangle PQR)=3kx$$ equating it with the usual area $$3kx=\\frac{\\sqrt{3}k^2}{2}$$ we get $$x=\\frac{\\sqrt{3}k}{6}$$ with the help of $x$ you will get $OL, OM, ON$. Now let $\\angle NPO=\\theta$ then $\\angle OPL=60^\\circ-\\theta$. with the help of the sides find $\\frac{\\sin \\theta}{\\sin (60^\\circ-\\theta)}$ you get $$\\frac{\\sin\\theta}{\\sin (60^\\circ-\\theta)}=\\frac{ON}{OL}=3$$ solving for $\\theta$ $$\\tan \\theta=\\frac{3\\sqrt{3}}{5}$$ with the help of $\\tan$ you get $NP=\\frac{5}{6}k$ now further you can get $PL$ by interchanging the angles $\\theta$ and $60^\\circ-\\theta$ and once more solving you get the new $\\theta$ as $$\\tan \\theta=\\frac{\\sqrt{3}}{7}$$ using the same method of finding $NP$ find $PL$. $$PL=\\frac{7}{6}k$$ now we have found all our needed unknowns. $$a=area(\\triangle NOP)+area(\\triangle POL)$$ $$=\\frac{1}{2}(\\frac{\\sqrt{3}k}{2}.\\frac{5k}{6}+ \\frac{\\sqrt{3}k}{6}.\\frac{7k}{6} )$$ $$a=\\frac{11\\sqrt{3}k^2}{72}$$ also $$b=\\frac{\\sqrt{3}k^2}{2}$$ solving you get $$\\frac{a}{b}=\\frac{11}{36}$$ $$a+b=\\left(\\frac{a}{b}+1\\right)b$$ which solves to $$a+b=\\frac{47\\sqrt{3}k^2}{72}$$ where $k$ is the side of the equilateral triangle. • I did not get the same equations as you ($2^{nd},\\ 3^{rd}\\ \\text{and } 4^{th}$). See my working in my (edited) question. – K. Rmth Dec 21 '13 at 20:53 • @Rmth: the answer has been completed. let me know if you have any doubt. – Suraj M S Dec 21 '13 at 20:55 • you are right i forgot to take the $\\frac{1}{2}$ .i will edit it. – Suraj M S Dec 21 '13 at 20:58", "\\frac{B-C}{2}) \\\\ \\sin A &<& \\sin \\frac{B+C}{2} \\cos \\frac{B-C}{2}\\end{array}$ It’s obvious that $\\cos \\frac{B-C}{2} \\leq 1$. So if we divide both sides of the equation by $\\cos \\frac{B-C}{2}$, the right hand side is what we desire, and the left hand side is smaller. Thus, $\\begin{array}{rcl} \\sin A &<& \\sin \\frac{B+C}{2} \\\\ A &<& \\frac{1}{2} (B+C) \\end{array}$ This is what we set out to prove. ### Question 10 10. Let $T(n)$ be the number of triangles with integer sides and perimeter n (where n is a positive integer). For example $T(6) = 1$ because only (2,2,2) has a perimeter of 6 and nothing else. a) Determine $T(10)$, $T(11)$, and $T(12)$. Triangles with perimeter 10 are (2,4,4) and (3,3,4). So $T(10)$ is 2. Triangles with perimeter 11 are (1,5,5), (2,4,5), (3,3,5), and (3,4,4). So $T(11)$ is 4. Triangles with perimeter 12 are (2,5,5), (3,4,5), and (4,4,4). So $T(12)$ is 3. The answers are 2, 4, and 3 respectively. b) If m is a positive integer and $m \\geq 3$, prove that $T(2m) = T(2m-3)$. Let a, b, and c be the sides of the triangle so that $a \\leq b \\leq c$. Let U be the triangle with perimeter 2m, and V be the triangle with perimeter 2m-3. In order for the triangle to be non-degenerate, $a+b > c$. Because", "the details for now... The relations are: (I consider $r_1$ the circle which is not tangent to the side $a$, and the same for $r_2,b$ and $r_3,c$) $$p-a =2\\sqrt{r_1r} \\frac{r}{r-r_1}$$ (construct symilarly the other two) From here you can construct a relatioin between $r,r_1,r_2,r_3$ in the following way: $$r=\\frac{S}{p}=\\sqrt{\\frac{(p-a)(p-b)(p-c)}{p}}$$ and substituting we have $$r= \\sqrt{\\displaystyle \\frac{\\frac{8\\sqrt{r_1r_2r_3}r^4\\sqrt{r}}{(r-r_1)(r-r_2)(r-r_3)}}{2\\sqrt{r_1r}\\frac{r}{r-r_1}+2\\sqrt{r_2r}\\frac{r}{r-r_2}+2\\sqrt{r_3r}\\frac{r}{r-r_3}}}.$$ Here you may be able to use your hypothesis on $r,r_1,r_2,r_3$ to get something out of this. - Is $p$ the semi-perimeter? – John Chang Dec 28 '12 at 3:06 At this point, why not just plug in $r_1=1$, $r_2=4$, $r_3=9$ and solve the resulting quadratic? I get $r=27/2$ (the other root is $r=1$, but we need to toss that because it's impossible). – Potato Dec 28 '12 at 5:31 Yes, $p$ is the semiperimeter. @Potato: I guess that's the way we should proceed. Plug in the smallest possible $r_1,r_2,r_3$ and find $r$. note that $r$ must also be a perfect square, so your result is not good. – Beni Bogosel Dec 28 '12 at 19:05 @BeniBogosel Ah, I didn't see the requirement that $r$ be a perfect square. Anyway, if you square and multiply out, the equation simplifies considerably, and you get a quadratic in $r$... – Potato Dec 28 '12 at 19:31 For typographic convenience (and reduced visual clutter) in", "couple of examples using this formula. What is the area of this triangle? The first thing we want to do is write out our formula. I always recommend doing this because it helps you remember where to plug something in, and writing the formula out repeatedly will help you remember it. $$A=\\sqrt{s(s-a)(s-b)(s-c)}$$ We know our $$a$$, $$b$$, and $$c$$ values, but we don’t know the value of $$s$$, so before we can plug things into our formula, we need to figure out what $$s$$ is. $$S$$ stands for semi-perimeter, so we want to find the perimeter of our triangle and then divide by 2. $$s=\\frac{7+9+13}{2}=\\frac{29}{2}=14.5$$ Now that we know that $$s$$ is 14.5, we can plug all our known values into our equation. $$A=\\sqrt{14.5(14.5-7)(14.5-9)(14.5-13)}$$ $$=\\sqrt{14.5(7.5)(5.5)(1.5)}$$ Which, when you simplify this out, is approximately equal to 29.95 square meters. $$\\approx 29.95\\text{ m}^{2}$$ Let’s try another one. Find the area of the following triangle. The first thing we need to do is write out our equation. $$A=\\sqrt{s(s-a)(s-b)(s-c)}$$ Now we need to find our semi-perimeter. $$s=\\frac{12+24+13}{2}=\\frac{49}{2}=24.5$$ Finally, we plug all our values into the equation. $$A=\\sqrt{24.5(24.5-12)(24.5-24)(24.5-13)}$$ $$=\\sqrt{24.5(12.5)(0.5)(11.5)}$$ Which, when you simplify it out, is approximately equal to 41.96 square feet. The next formula we are going to look at is a formula for finding the area of an isosceles triangle. The", "## Friday, January 19, 2007 ### Triangularly Speaking. Topic: Algebra/Geometry. Level: AIME. Problem: Suppose the roots of the cubic polynomial $P(x) = x^3-2rx^2+sx+t$ are the sides of a triangle (assume they are all positive reals and satisfy the triangle inequality). What is the area of the triangle in terms of the coefficients? Solution: Well, let's think about the formulas for the area of a triangle given its side lengths. The only one that comes to mind that uses only side lengths is Heron's formula. Why don't we try that? $A = \\sqrt{s(s-a)(s-b)(s-c)}$, where $a, b, c$ are the side lengths and $s = \\frac{a+b+c}{2}$. From $P(x)$, by Vieta's, we know that $r_1+r_2+r_3 = 2r$, where $r_i$ are the roots of the polynomial. So $s$ in this case is $r$. How do we get the rest, though? We could expand and use Newton sums, but that sounds like a pain. Note that $(s-a)(s-b)(s-c)$ looks a lot like $(x-a)(x-b)(x-c)$. But wait, that's just a polynomial with roots $a, b, c$! So if we write $P(x) = (x-r_1)(x-r_2)(x-r_3)$, we have $(s-a)(s-b)(s-c) = (s-r_1)(s-r_2)(s-r_3) = P(s) = P(r)$. Thus we can say that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{rP(r)}$, which can clearly be evaluated in terms of the coefficients. QED. -------------------- Comment: This is probably one of the neatest", "## Friday, January 19, 2007 ### Triangularly Speaking. Topic: Algebra/Geometry. Level: AIME. Problem: Suppose the roots of the cubic polynomial $P(x) = x^3-2rx^2+sx+t$ are the sides of a triangle (assume they are all positive reals and satisfy the triangle inequality). What is the area of the triangle in terms of the coefficients? Solution: Well, let's think about the formulas for the area of a triangle given its side lengths. The only one that comes to mind that uses only side lengths is Heron's formula. Why don't we try that? $A = \\sqrt{s(s-a)(s-b)(s-c)}$, where $a, b, c$ are the side lengths and $s = \\frac{a+b+c}{2}$. From $P(x)$, by Vieta's, we know that $r_1+r_2+r_3 = 2r$, where $r_i$ are the roots of the polynomial. So $s$ in this case is $r$. How do we get the rest, though? We could expand and use Newton sums, but that sounds like a pain. Note that $(s-a)(s-b)(s-c)$ looks a lot like $(x-a)(x-b)(x-c)$. But wait, that's just a polynomial with roots $a, b, c$! So if we write $P(x) = (x-r_1)(x-r_2)(x-r_3)$, we have $(s-a)(s-b)(s-c) = (s-r_1)(s-r_2)(s-r_3) = P(s) = P(r)$. Thus we can say that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{rP(r)}$, which can clearly be evaluated in terms of the coefficients. QED. -------------------- Comment: This is probably one of the neatest", "is 5cm and the base 4cm. so you can find the height by Pythagorean Theorem. $$a^{2} + b^{2} = c^{2}$$ $$a^{2} + 4^{2} = 5{2}$$ $$a^{2} + 16 = 25$$ $$a^{2} + 16 – 16 = 25 – 16$$ $$a^{2} = 9$$ $$a = 3$$ So put the value in the area formula, and use two perpendicular side (a and b) to substitute the base and height: $$Area = \\frac{1}{2}(bh)$$ $$Area = \\frac{1}{2}(4)(3)$$ $$Area = \\frac{1}{2}(12)$$ $$Area = 6$$ So, the area of a triangle is $$6cm^{2}$$. Method 2: Using side lengths 1. Calculate the semi-perimeter of the triangle The semi-perimeter is equal to the half of its perimeter. So, to find the semi-perimeter we need to find its perimeter of the by adding up the length of its three sides then multiplying the product by $$\\frac{1}{2}$$. E.g. suppose if a triangle has three sides that are 5cm, 4cm and 3cm long, then the semi-perimeter will be: $$s = \\frac{1}{2}(3 + 4 + 5)$$ $$s = \\frac{1}{2}(12)$$ = 6 1. Apply Heron’s formula Heron’s formula is $$Area = \\sqrt{s(s-a)(s-b)(s-c)}$$. Here ‘s’ is the semi-perimeter of a triangle and a, b, and c are the side lengths of a triangle. 1. Put values in the formula and calculate it While putting the values in the formula, make sure you" ]

[ "complex number. Hence you have a definition of $(1+x)^z$ that works for $x$ anywhere the whole complex plane apart from the real axis with $x<-1$.", "covered the topic of complex sequence and series.But why is this on my test? This is also my question for my wired professor. Complex number include real number(complex numberz=x+i*y, for real x and y, if y=0 then z is real number), so the definition can also apply to (2). The definition is indeed in my complex analysis textbook, and normally I can use this defintion to proof the sequence converge or diverge. But in this case, I guess I can show the sequence is bounded and then by some theorem/corrollay to show thatthe sequence is converge. If don't know solve this with any kind of math, just leave this alone and don't bother the defintion stated in my original question. Talk is cheap, show me the math(code).--Linus Trovalds fiora Nov 20, 2018", "Definition. A sequence of complex numbers is said to have the limit z 0, or to converge to z 0, if for any > 0, there exists an integer N such that |z n – z 0 | N. We denote this by Definition. A sequence of complex numbers is said to have the limit z 0, or to converge to z 0, if for any > 0, there exists an integer N such that |z n – z 0 | N. We denote this by Geometrically, this amounts to the fact that z 0 is the only point of z n such that any neighborhood about it, no matter how small, contains an infinite number of points z n. Geometrically, this amounts to the fact that z 0 is the only point of z n such that any neighborhood about it, no matter how small, contains an infinite number of points z n. Limit of a Function We say that the complex number w 0 is the limit of the function f(z) as z approaches z 0 if f(z) stays close to w 0 whenever z is sufficiently near z 0. Formally, we state: We say that the complex number w 0 is the limit of the function f(z) as z approaches z 0 if f(z) stays", "Take the module of $$z$$, $$|z|$$, draw it. This length is the one that determines the number $$z$$ in the complex plane.", "the complex plane where points are identified with complex numbers: The roots of the derivative of the polynomial $P(z)=(z-z_{1})(z-z_{2})(z-z_{3})$ are the foci of the ellipse tangent to the midpoints of $\\Delta z_{1}z_{2}z_{3}.$ ### Conic Sections > Ellipse Copyright © 1996-2018 Alexander Bogomolny 63567929 Search by google:", "in the complex plane. The question is asking you to show that in that case $|z-2| = 2|z-\\tfrac12|$.", "of a complex number. The gif below shows an animated mapping of when some complex numbers with a fixed radius less than one are multiplied by themselves from 1 to 100. I can’t help but watch it over and over again. $z^n\\text{ such that }|z|<1\\text{ and } n\\in[1,100]$ What shapes do you see? Does this remind you of anything you’ve seen in nature?", "For real numbers, $$\\leq$$ is defined, and if $$A\\subset \\mathbb R$$ then $$A$$ has a supremum in $$\\mathbb R$$. For the complex numbers, no such ordering exists and $$x \\leq y$$ has no meaning, so the supremum of a set of complex numbers has no meaning either. The clue, though, is in the word radius. This is a distance. $$|z|$$ is the distance of $$z$$ from the origin. In your example, $$R$$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $$R$$ on the complex plane.", "sequence $z_n = 0$. Then consider $d(f_n,z) = 2^{-n}$ to see that $f_n \\rightarrow z$.", "about complex numbers. If |z| = 1 then z can be any point on the unit circle in the complex plane. The question is asking you to show that in that case $|z-2| = 2|z-\\tfrac12|$.", "# Affix of a complex number (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Let $z=a+ib$ in its geometric representation. Affix is the point in the complex plane corresponding to this number, i.e. the point with Cartesian coordinates $(a,b)$. The affix is sometimes identified with the complex number itself.", "complex plane Complex plane In mathematics, the complex plane or z-plane is a geometric representation of the complex numbers established by the real axis and the orthogonal imaginary axis... • Spherical coordinate system#Integration and differentiation in spherical coordinates", "Home > English > Class 11 > Maths > Chapter > > If complex number z satisfies ... # If complex number z satisfies the condition |z-2+i|<=1, then maximum distance of origin from A(4+i(2-z)) is Updated On: 27-06-2022", "generally, how $z^{1/q}$, q being an integer, is defined. I have looked at a few complex analysis textbooks and they \"seem\" to miss that part! Page 1 of 2 12 Last", "origin is one and the same for all numbers in the complex plane, so the bolded part in \"distance to the origin (which is 0 for a real number)\" makes it even more confusing. I wish Andy clarified the wording. – dxiv Dec 18 '16 at 19:44", "19:16, 18 May 2008 • * If g is a primitive of f'/f, then we can find a complex number c such that g + c is a holomorphic logarithm of f ==Related facts== ... 835 B (138 words) - 19:17, 18 May 2008 • ===For an open subset in the complex numbers=== e. maps all elements of U to the same complex number) or an open map: the ... 597 B (109 words) - 19:17, 18 May 2008 • If the following limit is a finite complex number, then that complex number equals the residue at z_0: \\lim_{z \\to z_0} (z - z_0)f(z) ... 1 KB (214 words) - 19:18, 18 May 2008 • Suppose (z_n) is a sequence of (possibly repeating) complex numbers that does not cluster in \\mathbb{C}: in other words, |z_n| \\to \\infty if ... 664 B (113 words) - 19:19, 18 May 2008 • ===Definition in terms of complex integrals=== the loops comprising c. Then, the winding number of c about z_0, denoted n ... 824 B (152 words) - 19:19, 18 May 2008 • Suppose c \\in \\mathbb{C} is a complex number, and U is a domain in \\mathbb{C}. Then, if D is a disk centered at z_0, and z is any point in the ... 1 KB (200 words) - 19:03,", "+0 # Math 0 91 1 +105 Complex number z lies on the circle centered at 0 with radius 5. What is |z^3|? Oct 13, 2018 #1 +4002 +1 $$z = 5 e^{i \\theta} \\\\ z^3 = 5^3 e^{i3\\theta} \\\\ |z^3| = 5^3 = 125$$ . Oct 14, 2018", "of the z-plane for which the following series is convergent: ∑[(1/n!)(z^n)] where n runs from 0 to ∞", "series is absolutely convergent. Using the limit definition The exponential function ez can be defined as the limit of (1 + z/n)n, as n approaches infinity. In this animation, z=/3, and n takes various increasing values from 1 to 100. The computation of (1 + z/n)n is displayed as the combined effect of n repeated multiplications in the complex plane. As n gets larger, the points approach the complex unit circle (dashed line), covering an angle of π/3 radians. An alternative proof[6] is based on the limit definition of e^z: e^z = \\lim_{n\\rightarrow\\infty} \\left(1+\\frac{z}{n}\\right)^n. Substitute z=ix, and let n be a very large integer, say 1000. Then, based on the limit definition, the complex number (1+ix/1000)1000 is supposed to be a good approximation to eix. So, what is the value of (1+ix/1000)1000? Consider the sequence of 1000 complex numbers: 1, \\, \\left(1+\\frac{ix}{1000}\\right), \\, \\left(1+\\frac{ix}{1000}\\right)^2, \\ldots, \\, \\left(1+\\frac{ix}{1000}\\right)^{1000} (We started with 1, and successively multiplied it by (1+ix/1000), 1000 times.) If the points of this sequence are plotted in the complex plane (see animation at right), they approximately trace out the unit circle, with each point being x/1000 radians counterclockwise of the previous point. (The proof of this is based on the rules of trigonometry and complex-number algebra.[6]) Therefore, the last point in the sequence, (1 + ix/1000)1000, is", "# Affix of a complex number Let $z=a+ib$ in its geometric representation. Affix is the point in the complex plane corresponding to this number, i.e. the point with Cartesian coordinates $(a,b)$. The affix is sometimes identified with the complex number itself." ]

[ "\\dfrac{n}{2}[2a+(n-1)d]$$ 1 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = n+3\\end{align} What is the sum of the first six terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 2 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = n+3\\end{align} What is the sum of the first twenty terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 3 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = 5n-2\\end{align} What is the sum of the first six terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 4 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = 5n-2\\end{align} What is the sum of the first twenty terms of this sequence? This is an arithmetic (linear) sequence. 5 / 10 If the first difference between successive terms is $$0.25$$, that sequence will converge. $$S_n = \\dfrac{a(1-r^n)}{1-r}$$ 6 / 10 The general term for a geometric sequence is shown below \\begin{align}T_n = 2(3^{n-1})\\end{align} What is the sum of the first six terms of this sequence? The common ratio for this sequence is $$3$$. 7 / 10 The general term for a geometric sequence is shown below \\begin{align}T_n = 2(3^{n-1})\\end{align} What does the corresponding series converge to? $$S_n = \\dfrac{a(1-r^n)}{1-r}$$ 8 / 10 The general term for a geometric", "successive terms is a constant value of $$1$$. (b) $$7$$ (c) $$T_n=n$$ (d) $$300$$ ## Question 2 Consider the following sequence: \\begin{align}7,11,15,19…\\end{align} (a) Show that this sequence is linear. (b) What is the next term in this sequence? (c) What is the general term for this sequence? (d) What is the $$100$$th term in this sequence? (a) The difference between successive terms is a constant value of $$4$$. (b) $$23$$ (c) $$T_n=4n+3$$ (d) $$403$$ ## Question 3 The general term of a sequence is given by: \\begin{align}T_n=4n+1\\end{align} (a) What are the first three terms of this sequence? (b) What is $$30$$th term of this sequence? (c) Which term of the sequence is $$61$$? (a) $$5,9,13$$ (b) $$121$$ (c) $$15$$th ## Question 4 The general term of a sequence is given by: \\begin{align}T_n=5n-3\\end{align} (a) What are the first three terms of this sequence? (b) What is $$50$$th term of this sequence? (c) Which term of the sequence is $$32$$? (a) $$2,7,12$$ (b) $$247$$ (c) $$7$$th Mr. Kenny will go through the first of these questions during the grind. If time permits, he will then go through the second question. And so on. Any remaining questions are left for you to practice with! ## Question 1 Consider the following sequence: \\begin{align}1,4,9,16,…\\end{align} (a) Show that this sequence is quadratic. (b) What", "# Find the first four terms of a sequence that has the nth term. {eq}a_{n} = 2(n + 1)! {/eq} ## Question: Find the first four terms of a sequence that has the nth term. {eq}a_{n} = 2(n + 1)! {/eq} ## Sequence: We know that in a set if some numbers are present then all of the numbers be unique. When this set of numbers follow a particular pattern then set of numbers are in sequence. This sequence may be finite or infinite. Now, the {eq}n^{th} {/eq} term of this sequence is called general term and from this term, we can find any terms of that sequence. The {eq}n^{th} {/eq} term of a sequence is given by: {eq}a_n = 2(n+1)! {/eq} Now, substituting {eq}n = 1, \\, 2, \\, 3 \\, \\text{and} \\, 4 {/eq} into the given term, we have: {eq}\\begin{align*} a_1 &= 2(1+1)! \\\\[0.3 cm] &= 2(2)! \\\\[0.3 cm] &= 2(2 \\times 1 ) & \\left[ \\because n! = n(n-1)(n-2) ..... 1 \\right] \\\\[0.3 cm] &= 4 \\end{align*} {/eq} {eq}\\begin{align*} a_2 &= 2(2+1)! \\\\[0.3 cm] &= 2(3)! \\\\[0.3 cm] &= 2(3 \\times 2 \\times 1 ) & \\left[ \\because n! = n(n-1)(n-2) ..... 1 \\right] \\\\[0.3 cm] &= 12 \\end{align*} {/eq} {eq}\\begin{align*} a_3 &= 2(3+1)! \\\\[0.3 cm] &= 2(4)! \\\\[0.3 cm] &= 2(4 \\times 3", "t_{N}$ lineGeneral term $1$$1, 7, 13, 19, \\ldots to i termst_{i} 2$$2, 6, 8, 12, 14, 18, \\ldots$ to $j$ terms$t_{j}$ $3$$3, 5, 9, 11, 15, 17, \\ldots to k termst_{k} 4$$4, 10, 16, \\ldots$ to $l$ terms$t_{l}$ We are going to find the general terms for the above newly formed sequences. The general term for the sequence $S_{1} = 1, 7, 13, 19, \\ldots$ can be found by \\begin{align*} t_{i} &= 1 + (i - 1)6 \\\\ \\therefore t_{i} &= 6i - 5 \\quad\\text{such that } i = 1, 2, 3, \\ldots. \\end{align*} Similarly, the general term for the sequence $S_{4} = 4, 10, 16, \\ldots$ can be found by \\begin{align*} t_{l} &= 4 + (l - 1)6 \\\\ \\therefore t_{l} &= 6l - 2 \\quad\\text{such that } l = 1, 2, 3, \\ldots. \\end{align*} But for the sequence $S_{2} = 2, 6, 8, 12, 14, 18, \\ldots$, we have to work out a little bit more. So let’s do it this way, you first make series of this sequence then, write the same series in the next line but shift by one term. Finally, you subtract both of them. i.e. \\begin{align*} \\text{Series of } S_{2} &= 2 + 6 + 8 + 12 + 14 + 18 + \\ldots + t_{j - 1} + t_{j} \\\\", "z_1 \\\\[10pt] \\end{cases} \\end{align} and \\begin{align} P(Z_{n+1}=z| Z_1 = z_1, ... ,Z_n = z_n) &= \\begin{cases} 0 & & & \\text{if } z < z_n \\\\[10pt] z/k & & & \\text{if } z = z_n \\\\[10pt] 1/k & & & \\text{if } z > z_n \\\\[10pt] \\end{cases}\\\\[14pt] &=P(Z_{n+1}=z| Z_n = z_n) \\\\ \\end{align} induction shows that $$P(Z_{n+1}|Z_1,...,Z_n)=P(Z_{n+1}|Z_n)$$. • I'm getting a probability as $(k-z)/k$ in event ($z>z_n$) since we'll $k-z$ choices for $z>z_n$ Oct 4, 2021 at 22:38", "the first pair on the left, the addition of the first and last term, as these terms are usually more readily available. \\begin{align}\\frac{n}{2}(a_1+a_2)\\end{align} Gauss multiplied this sum times 50, which is HALF the number of terms in his sequence, and to multiply times half of the number of terms in the sequence, which is represented by n. Is this starting to look familiar? \\begin{align}S_n=\\frac{n}{2}(a_1+a_2) \\end{align} There a formula was, born for the sum of n terms of an arithmetic sequence. You may see the formula written as: \\begin{align}S_n=\\frac{n}{2}(a_1+a_2) \\text{ or } S_n=\\frac{n(a_1+a_2)}{2}\\end{align} Terms of an arithmetic series are Sum, Sn, of n. Example 1 (Even Number of Terms) Find the sum of two + 4 + 6 + 8 + 10 + 12 + 14 + 16. Example 2: (Odd Number of Terms) Find the sum of two + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18. Question An arithmetic series is pairing up terms that always give the same sums. Why? If you examine the graphic above, you'll see that if you add the primary and last terms you get 18. If you add the second term to the seventh term you'll also get 18 because you're +2 and -2 away from the primary and last terms. Thus, you've got", "Precalculus (6th Edition) Blitzer The first three terms of $\\sum\\limits_{i=1}^{17}{(5i+3)}$ are 8,13, and 18. The common difference is 5. To find the first three terms, we will substitute the value of $i=1,2\\text{ and }3$ in the given summation. For $i=1$ \\begin{align} & 5i+3=(5\\times 1+3) \\\\ & =(5+3) \\\\ & =8 \\end{align} For $i=2$ \\begin{align} & 5i+3=(5\\times 2+3) \\\\ & =(10+3) \\\\ & =13 \\end{align} For $i=3$ \\begin{align} & 5i+3=(5\\times 3+3) \\\\ & =(15+3) \\\\ & =18 \\end{align} By evaluating the first three terms, we can see that ${{a}_{1}}=8$ and the common difference is $d=13-8=5$.", "## Precalculus (6th Edition) Blitzer $4,8,12,16,20,...$ Arithmetic. When we observe the sequence carefully, it is seen that the difference between two consecutive terms is the same in the series $4,8,12,16,20,...$ Now, \\begin{align} & {{a}_{1}}=4,{{a}_{2}}=8, \\\\ & {{a}_{3}}=12,{{a}_{4}}=16 \\\\ & {{a}_{5}}=20,... \\\\ \\end{align} Then, \\begin{align} & {{a}_{2}}-{{a}_{1}}=8-4 \\\\ & =4 \\\\ & {{a}_{3}}-{{a}_{2}}=12-8 \\\\ & =4 \\end{align} \\begin{align} & {{a}_{4}}-{{a}_{3}}=16-12 \\\\ & =4 \\\\ & {{a}_{5}}-{{a}_{4}}=20-16 \\\\ & =4 \\end{align} Therefore, the common difference is 4. Hence, the sequence is Arithmetic.", "and Answers. Click here for Answers . • Answer all questions. This sequence has a constant difference between consecutive terms. Followed by 2 A4 sides with answers. 4. 5-a-day Workbooks. Therefore the first term in the quadratic sequence is n^2. Square in the form an^2+bn+c) Number of problems 5 problems. In the first stage of the soccer event at the Olympic Games, there are teams from four different countries in each group. So if there are six different teams in a group, each group plays $$\\text{15}$$ matches. Here are the first five terms of an arithmetic sequence. Step-by-step examples and guided questions asking students to find the Nth term for arithmetic and quadratic sequences (as separate resources). GCSE Revision Cards. \\begin{align*} a + b + c &= 3 \\\\ 4a + 2b + c &= 6 \\\\ 9a + 3b + c &= 10 \\end{align*}\\begin{align*} T_2 – T_1 &= 4a + 2b + c – (a + b + c) \\\\ 6 – 3 &= 4a + 2b + c – a – b – c \\\\ 3 &= 3a + b \\qquad \\ldots (1) \\end{align*}\\begin{align*} T_3 – T_2 &= 9a + 3b + c – (4a + 2b + c) \\\\ 10 – 6 &= 9a + 3b + c – 4a – 2b – c \\\\", "sequence defined by $$z_n=r+s\\exp(2\\pi ni/3)+t\\exp(4\\pi ni/3)$$ has period $3$. All you need to do is to fins $r$, $s$ and $t$ such that the first three terms are $a$, $b$, $c$.", "2n \\\\ T_{10} &= 3 - 2(10) \\\\ &= -17 \\end{align*} For each of the following sequences, find the equation for the general term and then use the equation to find $${T}_{100}$$. $$4;7;12;19;28;...$$ $$2;8;14;20;26;...$$ \\begin{align*} d &= 8 - 2 \\\\ &= 6 \\\\ \\therefore T_n &= 6n - 4 \\end{align*} \\begin{align*} T_n &= 6n - 4 \\\\ \\therefore T_{100} &= 6(100) - 4 \\\\ &= 596 \\end{align*} $$7;13;23;37;55;...$$ $$5;14;29;50;77;...$$ Given: $$T_n = 3n-1$$ Write down the first five terms of the sequence. $$2; 5; 8; 11; 14$$ What do you notice about the difference between any two consecutive terms? Constant difference, $$d=3$$ Will this always be the case for a linear sequence? Yes Given the following sequence: $$-15; -11; -7; \\ldots ; 173$$ Determine the equation for the general term. \\begin{align*} d &= -11 - (-15) \\\\ &= 4 \\\\ T_n &= 4n - 19 \\end{align*} Calculate how many terms there are in the sequence. \\begin{align*} T_n &= 4n - 19 \\\\ 173 &= 4n - 19 \\\\ 192 &= 4n \\\\ \\therefore 48 &= n \\end{align*} Given $$3; 7; 13; 21; 31; \\ldots$$ Thabang determines that the general term is $$T_n = 4n - 1$$. Is he correct? Explain. \\begin{align*} \\text{First differences: } &= 4; 6; 8; 10 \\\\ \\text{Second difference: } &= 2 \\end{align*} Thabang's", "Precalculus (6th Edition) Blitzer $-1,1,3,5,7,...$ Arithmetic. After observing the sequence carefully, we notice that the difference between two consecutive terms is the same. $-1,1,3,5,7,...$ Here, ${{a}_{1}}=-1,{{a}_{2}}=1,{{a}_{3}}=3,{{a}_{4}}=5,{{a}_{5}}=7,...$ So, \\begin{align} & {{a}_{2}}-{{a}_{1}}=1-\\left( -1 \\right) \\\\ & =2 \\\\ & {{a}_{3}}-{{a}_{2}}=3-1 \\\\ & =2 \\end{align} \\begin{align} & {{a}_{4}}-{{a}_{3}}=5-3 \\\\ & =2 \\\\ & {{a}_{5}}-{{a}_{4}}=7-5 \\\\ & =2 \\end{align} Therefore, the common difference is 2. Hence, the sequence is Arithmetic.", "## Precalculus (6th Edition) Blitzer ${{a}_{2}}=2{{a}_{1}}-4=\\underline{2}$. The second term of the sequence can be obtained by putting $n=2$. Then, for the third term, simply put $n=3$, and so on for other terms. We are given that ${{a}_{n}}=2{{a}_{n-1}}-4$ and ${{a}_{1}}=3$. Now, to find the second term, put $n=2$; \\begin{align} & {{a}_{2}}=2{{a}_{2-1}}-4 \\\\ & =2{{a}_{1}}-4 \\\\ & =2\\left( 3 \\right)-4 \\\\ & =2 \\end{align}", "a common difference of 2 because for all n, an - an- 1 = 2. Finding terms in this sequence is relatively straightforward, as the pattern is familiar. However, this would clearly be tedious if you needed to find the 100th term. Example 3 Find the 5th term for the sequence: t1 = 3 tn = 2tn-1 t5 = 48 t2 = 2t1 = 2 × 3 = 6 t1 = 3 \\begin{align*}\\rightarrow\\end{align*} t3 = 2t2 = 2 × 6 = 12 tn = 2 × tn-1 t4 = 2t3 = 2 × 12 = 24 t5 = 2t4 = 2 × 24 = 48 This example is a geometric sequence. Every geometric sequence has a common ratio, which is 2 in this example, because for all n, \\begin{align*}\\frac{t_{n}}{t_{n-1}}=2\\end{align*}. The terms of a geometric sequence follow an exponential pattern. Example 4 Find the 4th term for the sequence: b1 = 2 bn = (bn-1)2 + 1 b4 = 677 b2 = (b1)2 + 1 = 22 + 1 = 4 + 1 = 5 b1 = 2 \\begin{align*}\\rightarrow\\end{align*} b3 = (b2)2 + 1 = 52 + 1 = 25 + 1 = 26 bn = (bn-1)2 + 1 b4 = (b3)2 + 1 = 262 + 1 = 676 + 1 = 677 This sequence is neither arithmetic", "the rate of growth of the US population was 0.8% and in 2009 it was 1.0%. What was the average rate of population growth during that time period? Algebra Connection A geometric sequence is a sequence of numbers in which each successive term is determined by multiplying the previous term by the common ratio. An example is the sequence 1, 3, 9, 27, ... Here each term is multiplied by 3 to get the next term in the sequence. Another way to look at this sequence is to compare the ratios of the consecutive terms. 1. Find the ratio of the \\begin{align*}2^{nd}\\end{align*} to \\begin{align*}1^{st}\\end{align*} terms and the ratio of the \\begin{align*}3^{rd}\\end{align*} to \\begin{align*}2^{nd}\\end{align*} terms. What do you notice? Is this true for the next set (\\begin{align*}4^{th}\\end{align*} to \\begin{align*}3^{rd}\\end{align*} terms)? 2. Given the sequence 4, 8, 16,..., if we equate the ratios of the consecutive terms we get: \\begin{align*}\\frac{8}{4} = \\frac{16}{8}\\end{align*}. This means that 8 is the ___________________ of 4 and 16. We can generalize this to say that every term in a geometric sequence is the ___________________ of the previous and subsequent terms. Use what you discovered in problem 24 to find the middle term in the following geometric sequences. 1. 5, ____, 20 2. 4, ____, 100 3. 2, ____, \\begin{align*}\\frac{1}{2}\\end{align*} ### Answers for Explore More Problem", "&= 2(6) + 2\\end{aligned} We can see that the next term can be determined by multiplying the previous term by $2$ then adding the result with $2$. To see if the pattern fits the next term, let’s try to multiply $14$ by $2$ then adding $2$ to the product. \\begin{aligned}14(2) + 2 &= 30\\end{aligned} Since we got $30$, this confirms that we have the correct observations and rule for the recursive sequence. This means that to find $a_n$, we simply multiply $a_{n -1}$ by $2$ then add $2$ to the result: $a_n = 2a_{n – 1} + 2$. Including the first term, we have the recursive formula shown below for the first sequence. $\\left\\{\\begin{matrix}a_1 = 2 \\phantom{xxxxxx}\\\\a_n = 2a_{n – 1} + 2\\end{matrix}\\right.$ Let’s go ahead and move on to the second sequence, $\\{1, 2, 6, 24, …\\}$. We can apply a similar process when trying to find a pattern for the sequence. \\begin{aligned} 1&= 1 \\cdot 1\\\\ 2 &= 1 \\cdot 2\\\\6 & = 2 \\cdot 3\\end{aligned} What can you notice about the four terms? The position of the terms is actually a factor for the formula. This means that $2$ is the result of the previous term being multiplied by the place number – $2$. Similarly, $6$ is the result when we multiply the previous term", "is the next term in this sequence? (c) What is the general term for this sequence? (d) What is the $$10$$th term in this sequence? (a) The second difference is a constant value of $$2$$. (b) $$25$$ (c) $$T_n=n^2$$ (d) $$100$$ ## Question 2 Consider the following sequence: \\begin{align}9,14,21,30,…\\end{align} (a) Show that this sequence is quadratic. (b) What is the next term in this sequence? (c) What is the general term for this sequence? (d) What is the $$20$$th term in this sequence? (a) The second difference is a constant value of $$2$$. (b) $$41$$ (c) $$T_n=n^2+2n+6$$ (d) $$446$$ ## Question 3 The general term of a sequence is given by: \\begin{align}T_n=n^2+n+1\\end{align} (a) What are the first five terms of this sequence? (b) What is $$10$$th term of this sequence? (c) What is the second difference of this sequence? (a) $$3,7,13,21,31$$ (b) $$111$$ (c) $$2$$ ## Question 4 The general term of a sequence is given by: \\begin{align}T_n=3n^2-2n+5\\end{align} (a) What are the first five terms of this sequence? (b) What is $$10$$th term of this sequence? (c) What is the second difference of this sequence? (a) $$6,13,26,45,70$$ (b) $$285$$ (c) $$6$$", "(A(1) - 1) z}{2 (1 - 2 z^2)} + \\frac{1}{2 (1 - 2 z)}$$ You could now split the first term into partial fractions (but with irrational coefficients) or just take: \\begin{align} \\frac{1}{1 - 2 z^2} &= \\sum_{n \\ge 0} 2^n z^{2 n} \\\\ \\frac{z}{1 - 2 z^2} &= \\sum_{n \\ge 0} 2^n z^{2 n + 1} \\end{align} i.e., split into even and odd cases. The second term is just a geometric series.", "US population was 0.8% and in 2009 it was 1.0%. What was the average rate of population growth during that time period? Algebra Connection A geometric sequence is a sequence of numbers in which each successive term is determined by multiplying the previous term by the common ratio. An example is the sequence 1, 3, 9, 27, ... Here each term is multiplied by 3 to get the next term in the sequence. Another way to look at this sequence is to compare the ratios of the consecutive terms. 1. Find the ratio of the \\begin{align*}2^{nd}\\end{align*} to \\begin{align*}1^{st}\\end{align*} terms and the ratio of the \\begin{align*}3^{rd}\\end{align*} to \\begin{align*}2^{nd}\\end{align*} terms. What do you notice? Is this true for the next set (\\begin{align*}4^{th}\\end{align*} to \\begin{align*}3^{rd}\\end{align*} terms)? 2. Given the sequence 4, 8, 16,..., if we equate the ratios of the consecutive terms we get: \\begin{align*}\\frac{8}{4} = \\frac{16}{8}\\end{align*}. This means that 8 is the ___________________ of 4 and 16. We can generalize this to say that every term in a geometric sequence is the ___________________ of the previous and subsequent terms. Use what you discovered in problem 26 to find the middle term in the following geometric sequences. 1. 5, ____, 20 2. 4, ____, 100 3. 2, ____, \\begin{align*}\\frac{1}{2}\\end{align*} 4. We can use what we have learned in this section in", "..., k \\}$ and multiply it by $-1$ to change the sign of each terms and consider the resulting sequence: (9) \\begin{align} \\quad (a_i')_{i=1}^{2n} = (-a_1, -a_2, ..., -a_{k-1}, -a_k, a_{k+1}, ..., a_{2n}) \\end{align} • We note that this sequence $(a_i')_{i=1}^{2n} \\in A'$ since there must be a partial sum $s_j' < 0$ where $j < k$ since we multiplied each of these first $k$ terms by $-1$. In the original sequence $(a_i)_{i=1}^{2n}$ the number of $+1$s balanced out the number of $-1$. When we multiplied an odd number, $k$ terms, by $-1$ to form $(a_i')_{i=1}^{2n}$ we end up with the sequence $(a_i')_{i=1}^{2n}$ to be a sequence containing $n+1$ many $+1$s and $n - 1$ many $-1$s. Conversely, given any sequence $(a_i')_{i=1}^{2n}$ of $n + 1$ many $+1$s and $n - 1$ many $-1$s we can construct a sequence $(a_i)_{i=1}^{2n} \\in A'$ by taking the first $k$ terms in $(a_i')_{i=1}^{2n}$ where the number of $+1$s exceeds the number of $-1$s so that when we multiply these terms by $-1$ we obtain a sequence $(a_i)_{i=1}^{2n}$ where $s_k < 0$. Therefore the number of sequences in $A'$ is equal to the number of sequences containing $n + 1$ many $+1$s and $n - 1$ many $-1$s, that is: (10) \\begin{align} \\quad \\mid A' \\mid = \\binom{2n}{n + 1} = \\frac{(2n)!}{(n" ]

[ "sequence defined by $$z_n=r+s\\exp(2\\pi ni/3)+t\\exp(4\\pi ni/3)$$ has period $3$. All you need to do is to fins $r$, $s$ and $t$ such that the first three terms are $a$, $b$, $c$.", "# Interpretation of |z1-z2| Go back to 'Complex-Numbers' Let $${z_1}$$ and $${z_2}$$ represent two fixed points in the complex plane. There is a very useful way to interpret the expression $$\\left| {{z_1} - {z_2}} \\right|$$. Consider the following figure, which geometrically depicts the vector $${z_1} - {z_2}$$: However, observe that this vector is also equal to the vector drawn from the point $${z_2}$$ to the point $${z_1}$$: Thus, $$\\left| {{z_1} - {z_2}} \\right|$$ represents the length of the vector drawn from $${z_2}$$ to $${z_1}$$. In other words, $$\\left| {{z_1} - {z_2}} \\right|$$ represents the distance between the points $${z_1}$$ and $${z_2}$$. Let us take an example. Consider \\begin{align}&{z_1} = 1 + i\\\\&{z_2} = - 3i\\end{align} The expression $$\\left| {{z_1} - {z_2}} \\right|$$, as we concluded, represents the distance between the points $${z_1}$$ and $${z_2}$$, which is $$\\sqrt {17}$$, as is evident from the following figure: We can verify this algebraically: \\begin{align}&{z_1} - {z_2} = \\left( {1 + i} \\right) - \\left( { - 3i} \\right) = 1 + 4i\\\\&\\Rightarrow \\,\\,\\,{z_1} - {z_2} = \\sqrt {1 + 16} = \\sqrt {17} \\end{align} This interpretation of the expression $$\\left| {{z_1} - {z_2}} \\right|$$ as the distance between the points $${z_1}$$ and $${z_2}$$ is extremely useful and powerful. Let us see how. Suppose that z is a variable point in the", "collection of complex numbers (real numbers plus the dimension with i’s). You can imagine an axis of real numbers, and an axis of imaginary numbers perpendicular to it, and then some number of coordinates in that space below. What determines if a coordinate belongs? It means that if we start with the equation: \\begin{align} f(z) = z^2 + c \\end{align} and plug in some value of c, we can continually plug in the answer to the function again, and a member of the Mandelbrot set will never go above 2. For example, let’s say we start with c = 1 + 0i (the 0i goes away) and z = 0 We want to know if 1 is in the set. We would first do: \\begin{align} f(z) = 0^2 + 1 \\end{align} which is equal to 0 + 1 = 1! Now we would do the same and plug in 1 for z: \\begin{align} f(1) = 1^2 + 1 \\end{align} And we get f(1) = 2. We continually plug in our answer as the new value of Z and keep going \\begin{align} f(z) = 0^2 + 1 = 1 \\newline f(1) = 1^2 +1 = 2 \\newline f(2) = 2^2 +1 = 5 \\newline f(5) = 5^2 +1 = 26 \\newline \\end{align} Remember that we are trying to determine", "then we can say that \\begin{align*} z_1 &= f(z_0) \\\\ z_2 &= f(z_1) = f(f(z_0)) \\\\ z_3 &= f(z_2) = f(f(f(z_0))) \\\\ &... \\\\ z_n &= f(z_{n-1}) = f(f(... f(z_0)... )) \\end{align*} We are just applying the same function over and over again. #### Example Let $$f(z) = 2z$$ and let $$z_0 = 1$$. Then we get \\begin{align*} z_0 &= 1 \\\\ z_1 &= f(z_0) = f(1) = 2 \\\\ z_2 &= f(z_1) = f(2) = 4 \\\\ z_3 &= f(z_2) = f(4) = 8 \\\\ &... \\\\ z_n &= 2^n \\end{align*} Now that we have some of the basics under our belts, we can get to what the Mandelbrot set actualy is. The mandelbrot set is a set of points in the complex plane $$c\\in\\mathbb{C}$$ such that the function $$f(z)=z^2+c$$ doesn't diverge when the function is iterated starting with $$z_0=0$$. When we say that something doesn't diverge, we are essentially saying that the $$z_0,z_1,z_2...$$ don't \"blow up\" to infinity as we iterate. For the real numbers, a simple way to check that our sequence is bounded is to find some number $$b$$ such that for every $$x_i\\in(x_n)$$ it is true that $$x_i < b$$. In other words, find a $$b$$ that is bigger than every element in our sequence, then because every element is less than $$b$$,", "prepared four real valued $0/1$ periodic sequences, the length of each of them is $N = 2^8 = 256$. \\begin{align} f_1 &= \\{ \\underbrace{ 0,0,1,0,0,0,1,0,\\cdots,0,0,1,0}_{N = 2^8}\\} \\\\ f_2 &= \\{ \\underbrace{ 0,0,1,0,0,1,\\cdots,0,0,1,0}_{N = 2^8} \\} \\\\ F_1 &= \\{ \\underbrace{ 0,0,1,0,0,0,1,0,\\cdots,0,0,1,0}_{N = 2^8} \\} \\\\ F_2 &= \\{ \\underbrace{ 0,0,1,0,0,1,\\cdots,0,0,1,0}_{N = 2^8} \\} \\\\ \\end{align} Actually they are two sequences since $f_1 = F_1$ and $f_2 = F_2$. I used different notations for same sequences because $f$ would be applied discrete Fourier transformation, whereas $F$ would be applied inverse discrete Fourier transformation. The unique subsequence for $f_1$ and $F_1$ is $\\{0,0,1,0\\}$, and the unique subsequence for $f_2$ and $F_2$ is $\\{0,0,1\\}$. Therefore, the period for $f_1$ and $F_1$ is $4$, and the period for $f_2$ and $F_2$ is $3$. Note that the period for $f_1$ and $F_1$ divides $N = 2^8$ whereas the period for $f_2$ and $F_2$ does not. The sequence could also be complex-numbered. The magnitude $\\rho$ of a complex number $a + b i$ is $\\sqrt{a^2 + b^2}$. If the complex number is represented using polar coordinates $\\rho e^{i\\theta}$, the magnitude is just $\\rho$ then. So the magnitude of the sequence is represented by $\\rho_{f}$ or $\\rho_{F}$. We further define the “normalized probability” for a sequence. The concept of “probability” is less useful in", "Albert has a sequence defined by \\begin{align*}a_0 &= a_1 = a_2 = 0 \\\\ a_3 &= 1 \\\\ a_n &= 3 a_{n-2} - a_{n-4}.\\end{align*} Find $a_{31}$.", "..., k \\}$ and multiply it by $-1$ to change the sign of each terms and consider the resulting sequence: (9) \\begin{align} \\quad (a_i')_{i=1}^{2n} = (-a_1, -a_2, ..., -a_{k-1}, -a_k, a_{k+1}, ..., a_{2n}) \\end{align} • We note that this sequence $(a_i')_{i=1}^{2n} \\in A'$ since there must be a partial sum $s_j' < 0$ where $j < k$ since we multiplied each of these first $k$ terms by $-1$. In the original sequence $(a_i)_{i=1}^{2n}$ the number of $+1$s balanced out the number of $-1$. When we multiplied an odd number, $k$ terms, by $-1$ to form $(a_i')_{i=1}^{2n}$ we end up with the sequence $(a_i')_{i=1}^{2n}$ to be a sequence containing $n+1$ many $+1$s and $n - 1$ many $-1$s. Conversely, given any sequence $(a_i')_{i=1}^{2n}$ of $n + 1$ many $+1$s and $n - 1$ many $-1$s we can construct a sequence $(a_i)_{i=1}^{2n} \\in A'$ by taking the first $k$ terms in $(a_i')_{i=1}^{2n}$ where the number of $+1$s exceeds the number of $-1$s so that when we multiply these terms by $-1$ we obtain a sequence $(a_i)_{i=1}^{2n}$ where $s_k < 0$. Therefore the number of sequences in $A'$ is equal to the number of sequences containing $n + 1$ many $+1$s and $n - 1$ many $-1$s, that is: (10) \\begin{align} \\quad \\mid A' \\mid = \\binom{2n}{n + 1} = \\frac{(2n)!}{(n", "sequence to be a sequence $$\\left( b_{k+1},b_{k+2},b_{k+3},\\ldots \\right)$$ (starting at $$b_{k+1}$$ this time) defined by \\begin{align*} b_{n}=\\dfrac{a_{n}+a_{n-k-1}}{a_{n-2}a_{n-3}\\cdots a_{n-k+1}}\\qquad \\text{ for each }n\\geq k+1. \\end{align*} I am, however, struggling with adapting this line of reasoning to Sequence 3. If $$k$$ is odd, then we can set \\begin{align*} b_{n}=\\dfrac{a_{n}+a_{n-k+1}}{a_{n-1}a_{n-3}\\cdots a_{n-k+2}}\\qquad \\text{ for each }n\\geq k-1 \\end{align*} (where the denominator is $$\\prod\\limits_{i=1}^{\\left( k-1\\right) /2}a_{n-2i+1}$$). The resulting sequence $$\\left( b_{k-1},b_{k},b_{k+1},\\ldots \\right)$$ is not constant, but it is periodic with period $$2$$ (that is, $$b_{n}=b_{n-2}$$ for each $$n\\geq k+1$$); this is still sufficient for our argument. However, this only applies to the case when $$k$$ is odd. (I have found this definition of $$b_{n}$$ in Section 7.5 of Joshua Alman, Cesar Cuenca, Jiaoyang Huang, Laurent phenomenon sequences, J. Algebr. Comb. (2016) 43:589--633, which studies a more general recursion.) When $$k$$ is even, I see no such proof. I assume that the integrality of $$a_{0},a_{1},a_{2},\\ldots$$ follows from the standard Laurent phenomenon results known nowadays (by Fomin, Zelevinsky, Lam, Pylyavskyy and others). I haven't properly checked it, as there are a few technical conditions too many, but it is certainly consistent with SageMath experiments. Alman/Cuenca/Huang do not seem to consider the $$k$$-even case in their paper. Question. Can we prove using the above tools that the entries $$a_{0},a_{1},a_{2},\\ldots$$ of Sequence 3 are integers? Yes, we", "\\dfrac{n}{2}[2a+(n-1)d]$$ 1 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = n+3\\end{align} What is the sum of the first six terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 2 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = n+3\\end{align} What is the sum of the first twenty terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 3 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = 5n-2\\end{align} What is the sum of the first six terms of this sequence? $$S_n = \\dfrac{n}{2}[2a+(n-1)d]$$ 4 / 10 The general term for a linear sequence is shown below \\begin{align}T_n = 5n-2\\end{align} What is the sum of the first twenty terms of this sequence? This is an arithmetic (linear) sequence. 5 / 10 If the first difference between successive terms is $$0.25$$, that sequence will converge. $$S_n = \\dfrac{a(1-r^n)}{1-r}$$ 6 / 10 The general term for a geometric sequence is shown below \\begin{align}T_n = 2(3^{n-1})\\end{align} What is the sum of the first six terms of this sequence? The common ratio for this sequence is $$3$$. 7 / 10 The general term for a geometric sequence is shown below \\begin{align}T_n = 2(3^{n-1})\\end{align} What does the corresponding series converge to? $$S_n = \\dfrac{a(1-r^n)}{1-r}$$ 8 / 10 The general term for a geometric", "Sequences of Complex Numbers # Sequences of Complex Numbers A sequence of real numbers is an infinite ordered list $(a_n)_{n=1}^{\\infty}$ of real numbers. We can similarly define a sequence of complex numbers. Definition: A Sequence of Complex Numbers is an infinite ordered list of complex numbers $(z_n)_{n=1}^{\\infty} = (z_1, z_2, ..., z_n, ...)$ where $z_n \\in \\mathbb{C}$ for each $n \\in \\mathbb{N}$. For each $n \\in \\mathbb{N}$, $z_n$ is called a $n^{\\mathrm{th}}$ Term of the sequence. It is convenient to denote the first term of a sequence by $z_1$. We can start a sequence at any index though. This will not cause any problems. For example, the first few terms of the complex sequence $(n + i^n)_{n=1}^{\\infty}$ are: (1) \\begin{align} \\quad (n + i^n)_{n=1}^{\\infty} = (1 + i, 1, 3 - i, 5, 5 + i, ...) \\end{align} Many of the definitions familiar with sequences of real numbers are readily extended to sequences of complex numbers. Definition: A sequence of complex numbers $(z_n)_{n=1}^{\\infty}$ is said to Converge to $Z \\in \\mathbb{C}$ if for all $\\epsilon > 0$ there exists an $N \\in \\mathbb{N}$ such that if $n \\geq N$ then $\\mid z_n - Z \\mid < \\epsilon$. If the sequence $(z_n)_{n=1}^{\\infty}$ does not converge to any $Z \\in \\mathbb{C}$ then $(z_n)_{n=1}^{\\infty}$ is said to Diverge. For example, consider", "the generating series of Fibbonacci sequence, $$\\frac{1}{1-z-z^2} = 1 + z + 2\\,z^2 + 3\\,z^3 + 5\\,z^4 + 8\\,z^5 + 13\\,z^6 + \\dots$$ In our case this means that, there is one walk of size zero (term $1$, read 1 times $z^0$), from the vertex $1$ to itself, one walk of length one (term $z$, read $1$ times $z^1$), two walks of length two (term $2\\,z^2$), etc... and that in general, there are $F_n$ walks of length $n$ from that vertex to itself, where $F_n$ is the $n^\\text{th}$ term of the Fibbonacci sequence. Let's prove that to see what's going on. From the equation above one gets, $$G(z) = 1 + z\\,G(z) + z^2\\,G(z)$$ which translates into the following recurrence equations on the coefficients $a_n$. $$\\begin{cases} a_0 &= 1, \\\\ a_1 &= 1, \\\\ a_n &= a_{n-1} + a_{n-2} \\end{cases}$$ which gives, \\begin{align*} a_0 &= 1, & a_1 &= 1, & a_2 &= 2, \\\\ a_3 &= 3, & a_4 &= 5, & a_5 &= 8, \\\\ a_6 &= 13, & a_7 &= 21, && \\text{etc}... \\end{align*} nothing but the Fibbonacci sequence. Now what are the other entries ? Well just shifted versions of the Fibbonacci sequence. At the end of the day we have, $$A^n = \\left[\\begin{array}{cc} F_n & F_{n-1} \\\\ F_{n-1} & F_{n-2} \\end{array}\\right]$$ # 4.", "for complex values of z simply by substituting z in place of x and using the complex algebraic operations. In particular we may use either of the two following definitions which are equivalent. From a more advanced perspective, each of these definitions may be interpreted as giving the unique analytic continuation of ex to the complex plane. ### Power series definition For complex z $e^z = 1 + \\frac{z}{1!} + \\frac{z^2}{2!} + \\frac{z^3}{3!} + \\cdots = \\sum_{n=0}^{\\infty} \\frac{z^n}{n!} ~.$ Using the ratio test it is possible to show that this power series has an infinite radius of convergence, and so defines ez for all complex z. ### Limit definition For complex z $e^z = \\lim_{n \\rightarrow \\infty} \\left(1+\\frac{z}{n}\\right)^n ~.$ ## Proofs Various proofs of the formula are possible. ### Using power series Here is a proof of Euler's formula using power series expansions as well as basic facts about the powers of i:[5] \\begin{align} i^0 &{}= 1, \\quad & i^1 &{}= i, \\quad & i^2 &{}= -1, \\quad & i^3 &{}= -i, \\\\ i^4 &={} 1, \\quad & i^5 &={} i, \\quad & i^6 &{}= -1, \\quad & i^7 &{}= -i, \\end{align} and so on. Using now the power series definition from above we see that for real values of x \\begin{align} e^{ix} &{}= 1 + ix +", "on $A$. If there exists a sequence of complex numbers $(z_n)_{n=1}^{\\infty}$ in $A$ that converges to $z_0 \\in A$ and $f(z_n) = g(z_n)$ for all $n \\in \\mathbb{N}$ then $f(z) = g(z)$ for all $z \\in A$. • Proof: Define a new function $h(z) = f(z) - g(z)$. Then for all $n \\in \\mathbb{N}$ we have that: (3) \\begin{align} \\quad h(z_n) = f(z_n) - g(z_n) = 0 \\end{align} • In other words, each $z_n$ is a root of $h$. Let $Z$ be the set of all roots of $h$. Then $\\{ z_n : n \\in \\mathbb{N} \\} \\subset Z$. Furthermore, $h(z_0) = 0$, so $z_0 \\in Z$. Therefore the sequence $(z_n)_{n=1}^{\\infty}$ is a sequence in $Z \\subset A$ that converges to $z_0 \\in A$. But then $z_0$ is not an isolated point of $Z$. Since $h$ is analytic, this implies that $h$ must be identically zero on $A$, i.e., for all $z \\in A$: (4) \\begin{align} \\quad 0 = f(z) - g(z) \\end{align} • Therefore $f(z) = g(z)$ for all $z \\in A$. $\\blacksquare$", "For example, $\\{3n-1\\}$ generates the sequence: \\begin{align} \\displaystyle u_{1} &= 3 \\times 1 - 1 = 2 \\\\ u_{2} &= 3 \\times 2 - 1 = 5 \\\\ u_{3} &= 3 \\times 3 - 1 = 8 \\\\ \\end{align} ### Example 1 A sequence is defined by $u_{n}=4n+2$. Find $u_{5}$. ### Example 2 Find the first five terms of $T_{n}=3n+7$. ### Example 3 Find the first two terms of $\\{n^2\\}$. ### Example 4 Find the first three terms of $u_{n}=2^n$. ### Example 5 Find $T_{4}$ of $T_{n+1}=n^2-1$.", "successive terms is a constant value of $$1$$. (b) $$7$$ (c) $$T_n=n$$ (d) $$300$$ ## Question 2 Consider the following sequence: \\begin{align}7,11,15,19…\\end{align} (a) Show that this sequence is linear. (b) What is the next term in this sequence? (c) What is the general term for this sequence? (d) What is the $$100$$th term in this sequence? (a) The difference between successive terms is a constant value of $$4$$. (b) $$23$$ (c) $$T_n=4n+3$$ (d) $$403$$ ## Question 3 The general term of a sequence is given by: \\begin{align}T_n=4n+1\\end{align} (a) What are the first three terms of this sequence? (b) What is $$30$$th term of this sequence? (c) Which term of the sequence is $$61$$? (a) $$5,9,13$$ (b) $$121$$ (c) $$15$$th ## Question 4 The general term of a sequence is given by: \\begin{align}T_n=5n-3\\end{align} (a) What are the first three terms of this sequence? (b) What is $$50$$th term of this sequence? (c) Which term of the sequence is $$32$$? (a) $$2,7,12$$ (b) $$247$$ (c) $$7$$th Mr. Kenny will go through the first of these questions during the grind. If time permits, he will then go through the second question. And so on. Any remaining questions are left for you to practice with! ## Question 1 Consider the following sequence: \\begin{align}1,4,9,16,…\\end{align} (a) Show that this sequence is quadratic. (b) What", "defining $$a_n$$ in terms of preceding terms. For example, suppose we know the following: \\begin{align*} a_1 &= 3 \\\\ a_n &= 2a_{n−1}−1 , \\text{ for }n≥2 \\end{align*} We can find the subsequent terms of the sequence using the first term. \\begin{align*} a_1 &= 3 \\\\ a_2 &= 2a_1−1=2(3)−1=5 \\\\ a_3 &= 2a_2−1=2(5)−1=9 \\\\ a_4 &= 2a_3−1=2(9)−1=17 \\end{align*} So the first four terms of the sequence are $$\\{3, 5, 9, 17\\}$$. The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms. \\begin{align*}a_1 &= 1 \\\\ a_2 &= 1 \\\\ a_n &= a_{n−1}+a_{n−2} \\text{, for }n≥3 \\end{align*} To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are $$21$$ and $$34$$, so $$a_{10}=a_9+a_8=34+21=55$$ RECURSIVE FORMULA A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence. Q&A: Must the first two terms always be given in a recursive formula? No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only", "same regardless of the order in which the terms are added - in other words: $z-w\\neq w-z$ (provided, of course, both numbers are not zero) Thus \\begin{align}w-z & =(-2+3\\sqrt{2}i)-(3+\\sqrt{2}i) \\\\ & =(-2-3)+(3\\sqrt{2}-\\sqrt{2})i \\\\ & =-5+2\\sqrt{2}i \\\\ \\end{align} ### Multiplication of Complex Numbers Multiplying a Complex Number by a constant is fairly straight-forward - for a constant 'k', you simply multiply the real part by k and the imaginary part by k. So, using our z from earlier: $kz=k(3+\\sqrt{2}i)=3k+k\\sqrt{2}i$ If one was to multiply a complex number by another complex number, one follows an process similar to expanding products such as $(\\sqrt{2}x+2)(x+3)$: \\begin{align} z^2 & =z\\times z \\\\ & =(3+\\sqrt{2}i)(3+\\sqrt{2}i) \\\\ & =3\\times3+3\\times\\sqrt{2}i+3\\times\\sqrt{2}i+\\sqrt{2}i\\times\\sqrt{2}i \\\\ & =9+6\\sqrt{2}i+2i^2 \\\\ \\end{align} It is tempting to leave the solution as one would when multiplying unknowns. However, the value of $i^2\\!$ is known to be -1. So: \\begin{align}z^2 & =9+6\\sqrt{2}i+2i^2 \\\\ & =9+6\\sqrt{1}+2\\times(-1) \\\\ & =9-2+6\\sqrt{2}=7+6\\sqrt{2} \\\\ \\end{align} Similarly, one can multiply different complex numbers: \\begin{align} z\\times w & =(3+\\sqrt{2}i)(-2+3\\sqrt{2}i) \\\\ & =3\\times (-2)+3\\times 3\\sqrt{2}i+(-2)\\times \\sqrt{2}i+\\sqrt{2}i\\times 3\\sqrt{2}i) \\\\ & =-6+7\\sqrt{2}i+6\\times i^2 \\\\ & =(-6-6)+7\\sqrt{2}i=-12+7\\sqrt{2}i \\\\ \\end{align} ### Division of Complex Numbers This is a slightly more complex process, similar to rationalising the denominator in situations involving surds. For instance: $\\frac{2+\\sqrt{3}i}{\\frac{1}{2}+\\sqrt5i}$. In order to 'divide' the top complex number (on the numerator) by the bottom", "1. ## Sketching complex plane Sketch the subset of the complex plane: {z E C: |z - 1| = |z - i| } I have no idea what to do. Could someone please explain? 2. Originally Posted by SyNtHeSiS Sketch the subset of the complex plane: {z E C: |z - 1| = |z - i| } I have no idea what to do. Could someone please explain? The modulus of a complex number is it's distance from the origin in the complex plane. Pythagoras' theorem calculates the distance. $z=x+iy$ $z-1=(x-1)+iy\\Rightarrow\\ |z-1|=\\sqrt{(x-1)^2+y^2}=\\sqrt{x^2+y^2-2x+1}$ $z-i=x+i(y-1)\\Rightarrow\\ |z-i|=\\sqrt{x^2+(y-1)^2}=\\sqrt{x^2+y^2-2y+1}$ 3. Hello, SyNtHeSiS! Did you make a sketch? Sketch the subset of the complex plane: . $\\{\\,z \\in C\\!:\\;|z - 1| \\:=\\:|z - i|\\,\\}$ Hint: On the Argand diagram: . . $|z-1|$ is the distance from $\\,z$ to $(1,0).$ . . $|z - i|$ is the distance from $\\,z$ to $(0,1).$ 4. Thanks. Would this sketch be the graph y = x? 5. Yes, it is.", "the third step uses the commutative property of addition. For $z^3$ we use our calculation above and the fact that $z^3 = z^2 \\times z$: \\begin{align} (1+i)^3 &= (1+i)^2 \\times (1 + i) \\\\ &= 2i \\times (1 + i) \\\\ &= 2i + 2i^2\\\\ &= -2 + 2i. \\end{align} The third line uses the distributive property of multiplication over addition and the last step uses the fact that $i^2 = -1$. Finally, for $z^4$ we have \\begin{align} (1+i)^4 &= (1+i)^3 \\times (1 + i) \\\\ &= (-2+2i) \\times (1 + i) \\\\ &= -2 - 2i +2i +2i^2\\\\ &= -4. \\end{align} The third equality uses the distributive property and the last step uses the fact that $i^2 = -1$. 2. Writing each complex number $z$ as $x + iy$ we can plot $z, z^2, z^3,$ and $z^4$ in the $x$-$y$ plane as shown below: One thing we notice about these points is that each successive power of $z$ is further from (0,0). The second thing we notice is that the direction from (0,0) to the powers of $z$ changes with the power. In fact, the direction from (0,0) to each successive power of $z$ is 45 degrees further in the counterclockwise direction. To see this more clearly, in the following picture the path from (0,0) to the", "+0 # Question 0 36 1 Let $$z_1,z_2, andz_3$$ be complex numbers such that \\begin{align} \\overline{z_1}&=\\dfrac{1}{z_1}\\\\ \\overline{z_2}&=\\dfrac{2}{z_2}\\\\ z_3&=z_1+z_2. \\end{align}What is the product of the min and max of z_3? I dont know where to start here. Oct 21, 2022" ]

[ "+0 # Product 0 87 1 Determine the value of the infinite product $$(2^{1/3})(2^{1/9})(2^{1/27}) \\dotsm$$", "# How many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$? How many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$? I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off. How would I solve this? First we count the number of triples of positive integers $(a, b, c)$ with $a + b + c = 100$. There are $\\binom{99}{2}$ of them, which results from an elementary application of stars and bars. Just line up a hundred dots, place a bar in between the $a$th and $(a +1)$th dot, and then place another bar $b$ dots to the right of the first one. Then $c$ is the number of dots to the right of the second bar. There are 99 spaces in between a hundred dots and you're choosing two of them to place bars in. This is an overcount since it doesn't account for the condition $a < b < c$. First we should subtract the number of triples in which $a, b, c$ are not all distinct. There are 49 triples such that $a = b$: to count them, choose $x$ with $0 < x<50$, set $a = b = x$ and $c =", "## Notes Suppose $n_1,\\dots,n_k$ are positive integers that are pairwise coprime. Then, for any given sequence of integers $a_1,\\dots,a_k$ there exists an integer x that solves the following system of simultaneous congruences: $\\begin{cases} x\\equiv a_1 & \\mod n_1 \\\\ \\dots \\\\ x \\equiv a_k & \\mod n_k\\end{cases}$ A solution x exists if and only if $a_i\\equiv a_j \\mod \\gcd(n_i,n_j) \\, \\forall i,j$.", "[tex: ] # Vaughan's decomposition In this post I will perform a routine task of verifying something routine. Let $U,V,N$ be positive integers satisfying $UV < N^{0.9}$. The exponent here can be replaced by any positive number $<1$ but I don't want to complicate the notatio…", "# A number theory problem by ppk k Find the number of positive integers $$n$$ for which $$n^3 - 8n^2 + 20n - 13$$ is a prime number. ×", "n × n array of do [#permalink] ### Show Tags 30 May 2020, 10:03 gmatt1476 wrote: Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array? A. 1 B. 2 C. 3 D. 4 E. 5 PS03551.01 Attachment: 2019-09-21_1528.png 48 dots are not to be selected from a n*n and the balance number of dots has to be a square number-> This is how I understood the question. Basically- n^2-48=k^2. I was able to get just 2 values- 49-48=1^2 and 64-48=4^2. What is the other possible solution. Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink] 30 May 2020, 10:03 # Let n and k be positive integers with k ≤ n. From an n × n array of do new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork", "exists a solution $x_1, \\dots, x_q$ of this system such that all $x_j$ ($j=1,\\dots, q$) are integers satisfying $|x_j|\\le q$ and $x_j\\ne 0$ for at least one value of $j$. ### Problem 6 For all positive integral $n$, $u_{n+1}=u_n(u_{n-1}^2-2)-u_1$, $u_0=2$, and $u_1=2\\frac12$. Prove that $$3\\log_2[u_n]=2^n-(-1)^n,$$ where $[x]$ is the integral part of $x$.", "n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array? A. 1 B. 2 C. 3 D. 4 E. 5 PS03551.01 Attachment: 2019-09-21_1528.png [ 15.95 KiB | Viewed 8429 times ] ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 8757 Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink] ### Show Tags 19 Oct 2019, 05:17 11 1 10 gmatt1476 wrote: Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array? A. 1 B. 2 C. 3 D. 4 E. 5 PS03551.01 Attachment: 2019-09-21_1528.png The number", "# A number theory problem by Jose Sacramento If $m$ and $n$ are two positive integers such that $5m+7n=46$, then what is the value of $m\\times n$? ×", "900 Followers Problem! $\\frac{m^3-n^3}{m^2+n^2-mn}$ Find the sum of all prime numbers less than 900 that can be expressed as the above fraction where $$m$$ and $$n$$ are positive integers. ×", "# STEP II/2018/6 (i) Find all pairs of positive integers (n,p), where p is a prime number, that satisfy $n!+ 5 = p$ (ii) In this part of the question you may use the following two theorems: For $n\\ge 7, 1! \\times 3! \\times \\cdots \\times (2n-1)! > (4n)!\\,$ For every positive integer n, there is a prime number between 2n and 4n. Find all pairs of positive integers (n,m) that satisfy $1! \\times 3! \\times \\cdots \\times (2n-1)! = m!$", "+0 # Help 0 89 1 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 25, 2021", "+0 # Help 0 133 1 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 25, 2021", "Difference between revisions of \"2012 USAMO Problems/Problem 6\" Problem For integer $n \\ge 2$, let $x_1$, $x_2$, $\\dots$, $x_n$ be real numbers satisfying $$x_1 + x_2 + \\dots + x_n = 0, \\quad \\text{and} \\quad x_1^2 + x_2^2 + \\dots + x_n^2 = 1.$$ For each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define $$S_A = \\sum_{i \\in A} x_i.$$ (If $A$ is the empty set, then $S_A = 0$.) Prove that for any positive number $\\lambda$, the number of sets $A$ satisfying $S_A \\ge \\lambda$ is at most $2^{n - 3}/\\lambda^2$. For what choices of $x_1$, $x_2$, $\\dots$, $x_n$, $\\lambda$ does equality hold?", "+0 # help. -1 153 1 +143 The set \\(\\{2, 4, 6, \\dots, n\\}\\) contains the positive consecutive even integers from 2 through \\(n\\). When one of the integers from the set is removed, the average of the remaining integers in the set is 28. What is the least possible value of \\(n\\)? I got that the sum of the set is \\(n+n^2\\) but I don't know if that will help me. Jul 3, 2018", "+ n_4 + \\dots$ is odd which comes from $o(n)$. The positive ones come from $e(n)$. Thus we must have that $$\\sum (e(n) - o(n))x^n = \\frac{1}{(1-x)(1+x^2)(1-x^3)\\dots}$$ - very nice proof! – ACARCHAU Sep 20 '10 at 4:13", "any positive integer $n$ and $0 \\leqslant i \\leqslant n$, denote $C_n^i \\equiv c(n,i)(\\bmod 2)$, where $c(n,i) \\in \\left\\{ {0,1} \\right\\}$, define $f(n,q) = \\sum\\limits_{i = 0}^n {c(n,i){q^i}}$$m,n,q$ are positive integers, and $q + 1 \\ne {2^\\alpha }$ for any $\\alpha\\in N$. Prove that if $f(m,q)\\left| {f(n,q)} \\right.$, then $f(m,r)\\left| {f(n,r)} \\right.$ for any positive integer $r$. 6. $m,n$ are positive integers, find the minimum positive integer $N$ satisties that if $S$ is a set of integers contains a complete residue system of $m$ and $\\left| S \\right| = N$, then there exit a nonempty set $A \\subseteq S$, that $n\\left| {\\sum\\limits_{x \\in A} x } \\right.$.", "pts Let P(n) be the statement that 13+ 23 +33-+...+ n3- nn)for the p for the positive integer n. What is the inductive hy...", "+0 0 159 3 Let $$n$$ be a positive integer. (a) Prove that $$n^3 = n + 3n(n - 1) + 6 \\binom{n}{3}$$ by counting the number of ordered triples $$a,b,c$$ of positive integers, where $$1 \\le a, b,c \\le n,$$ in two different ways. (b) Prove that $$\\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \\dots + (k)(n - k + 1) + \\dots + (n)(1),$$ by counting the number of subsets of $$\\{1, 2, 3, \\dots, n + 2\\}$$ containing three different numbers in two different ways. In that answer by another guest they said that when you plot the equation you get a smiley face and some other faces when you change it up. I was confused about how that worked, could someone please explain it to me? The question was locked so I couldn't ask on there. Thank you! Jul 20, 2021 #2 +839 +1 Ideally, what (s)he meant by smiley faces were when $b < a \\leq c | b < c \\leq a$, and frowny faces were if $b > a \\geq c | b > c \\geq a$. Neutral face is when $a = b = c$. Smirks are all other possibilities of $a, b, c.$ Number of neutral faces: $n$ ways to choose $a$,", "Find the sum of all positive integers $$n$$ for which $$n^2-19n+99$$ is a perfect square. (第十七届AIME1999 第3题)" ]

[ "1. P(3). 2. For all positive integers n such that n≥4, if P(n-1) then P(n). What you're asking about is part of step 2, so we have assumed that P(n-1) holds, and we're working with an arbitrary product of n factors. The step you're asking about changes the bracketing of the product $a_k\\cdots a_n$, which has n-k+1 factors. So it's allowed by assumption, if $n-k+1\\leq n-1$. This inequality is equivalent to $k\\geq 2$. This is true unless all the left brackets are to the left of $a_1$. 3. Apr 26, 2013 ### Syrus Precisely what I thought, thank you.", "are the only ways to get a positive product from three factors. Now, here’s a chart of the signs of the factors $x-3,x-2$, and $x-1$ and their product: 1 2 3 ----------------|------------|------------|-------------------- x-3: - - - - - 0 + x-2: - - - 0 + + + x-1: - 0 + + + + + product: - 0 + 0 - 0 + When $x<1$, all three factors are negative, and so is their product. When $x=1$, one factor is $0$, and so is the product. When 1or $x>3$. In interval notation, the solution set of the inequality is $(1,2)\\cup(x,\\infty)$. This technique works whenever you’re comparing a product with $0$. - @gekkostate: I want the semi-graphical chart, not an array, and I want ‘for the moment’, not ‘for a moment’. – Brian M. Scott Feb 27 '13 at 14:04 @BrianMScott, sorry about that. I thought that having the array would look nicer and still get the message across. – Jeel Shah Feb 27 '13 at 14:06 @gekkostate: No harm done. I realize that the array is prettier, but it doesn’t have the pictorial aspect with the number line. I just wanted to let you know that I wasn’t rolling it back out of pique: I really did have a preference. – Brian M. Scott Feb 27 '13", "## Intermediate Algebra for College Students (7th Edition) To determine the sign of a product that involves more than two numbers, simply count the number of negative factors (numbers). If there is an odd number of negative factors, then the product must be negative. If there is an even number of negative factors, then the product must be positive. Example: (i) $3(-2)(-2)(-4) = -48$ This is negative since there are 3 negative factors. (ii) $3(-2)(-2)(4) = 48$ This is positive since there are 2 negative factors.", "factors x and x - 5 need to have different signs for the product to be negative such that: `x gt 0` `x - 5 lt 0 =gt x lt 5` or `x lt 0` `x - 5 gt 0 =gt x gt 5` Since the intersection of the intervals `(-oo,0)` and `(5,oo)` yields `O/` , hence the common values for x that satisfy the inequality `x(x - 5)lt0` are in interval `(0,5).` Approved by eNotes Editorial Team", "## Algebra 1 A product is negative if the number of negative factors is an odd number. If there are 3 factors, the product would be negative if all 3 are negative, but the product would also be negative if only one of the factors is negative. $-2\\times3\\times4=-24$ In this example, only one factor is negative.", "product of 2 numbers be negative? When one and only one of the factors is negative. Therefore, either x should be negative or (x – 3) should be negative, but not both. Let’s consider each case. Case 1: x is negative and (x – 3) is positive $$x < 0$$ and $$(x – 3) > 0$$ which implies $$x > 3$$ This is not possible. x cannot be less than 0 and greater than 3 at the same time. Hence this case gives us no appropriate values for x. Case 2: x is positive and (x – 3) is negative $$x > 0$$ and $$(x – 3) < 0$$ which implies $$x < 3$$ x must be greater than 0 but less than 3. Therefore, the range of values for which this inequality will be satisfied is $$0 < x < 3$$. Will we have to do this every time we have multiple factors in an inequality? What will happen in case there are lots of factors? There is an easier way of handling such situations. I will first discuss the method and later explain the logic behind it. Method: Say we have an inequality of the form $$(x – a)(x – b)(x – c) < 0$$. (For clarity, we will work with the example $$x(x – 3)", "4$$. 4. Aug 24, 2006 ### HallsofIvy Staff Emeritus As you noticed, $t^2 \\ge 16$ does not give $t \\ge 4$. A standard way of solving inequalities is to solve the corresponding equality first. $t^2= 16$ gives, of course, t= 4 and t= -4. Those are the only places at which t2 can change sign. It t= -5 (less than -4) then t2= (-5)2= 25 which is greater than 16 so t2> 16 for all x< -4. 0 is between -4 and 4 and of course 02= 0 which is less than 16: t2< 16 for all t between -4 and 4. Finally if t= 5 (larger than 4) then t2= 52= 25 which is again larger than 16. $t^2 \\ge 16[/sup] for all [itex]x \\le -4$ and all $x \\ge 4$. That method works for all continuous functions. Another way to do this inequality is to factor: $t^2- 16= (t-4)(t+4)\\ge 0$. If t< -4 both factors are less than 0; the product of two negative numbers is positive. If t is between -4 and 4, one factor, t-4, is still negative but the other, t+4 is now positive. The product of a negative number with a positive number is negative. Finally, if t>4, both factors are positive so the product is positive again. Know someone interested in", "the factorisations of various numbers with lots of factors. I was aiming for a counter-example with three integers. So with product $36$ I found $2,3,6$ and $3,3,4$, the second of which is not strictly a set, because it has duplicates. But its existence suggested that in larger cases there might be distinct integers. There is also $1,6,6$ and $2,2,9$. $60$ and $84$ gave nothing, but $72$ had $2,6,6$ and $3,3,8$ and $96$ yielded $1,8,12$ and $2,3,16$. This was quite quick by hand. Two further questions suggest themselves: are there examples of three sets of three positive integers having the same sum and the same product? Are there examples of two sets of $n$ positive integers having the same sum and the same product for all $n\\ge 3$? Both seem likely based on my hand calculations. • And one of the other answers indeed finds five sets of three with the same sum and product. – Mark Bennet Oct 18 '17 at 18:04 • \"are there examples of three sets of three positive integers having the same sum and the same product?\".... here's an example of five :) $\\{15, 88, 140\\}, \\{16, 77, 150\\}, \\{20, 55, 168\\}, \\{24, 44, 175\\}, \\{25, 42, 176\\}$. I think the real question is \"are there examples of $n$ sets of three positive numbers", "# How do you solve (2x-1)(x+2)(x-4)<=0? Oct 13, 2015 The inequality holds for $x \\setminus \\le - 2$ and $\\frac{1}{2} \\setminus \\le x \\setminus \\le 4$. #### Explanation: Keep in mind the rule for multiplying: • Positive times positive is positive; • Positive times negative is negative; • Negative times positive is negative; • Negative times negative is positive. Combining these rules, you can clearly see that a product of factors is negative if and only if there is an odd number of negative factors (in this case, one or all three). Now, $2 x - 1$ is positive if and only if $x > \\frac{1}{2}$, $x + 2$ is positive if and only if $x > - 2$, and $x - 4$ is positive if and only if $x > 4$. So, the important points are $- 2$, $\\frac{1}{2}$ and $4$. Before $- 2$, all three factors are negative, so the product is negative. Between $- 2$ and $\\frac{1}{2}$, $x + 2$ is positive and the other two are negative, so the product is positive. Between $\\frac{1}{2}$ and $4$, only $x - 4$ is negative, and the other two are positive, so the product is negative. After $4$, all the factors are positive, so the product is positive.", "# Problem of the Week Problem E Count on That Let $$n$$ be a positive integer. How many values of $$n$$ satisfy the following inequality? $(n-1)(n-3)(n-5)\\cdots(n-2019)(n-2021) \\leq 0$ Note: The product on the left side of the inequality consists of $$1011$$ factors of the form $$n-d$$, where the value of $$d$$ starts at $$1$$ and increases by $$2$$ for each subsequent factor.", "both factors are positive and the product is positive. The point is that exactly one of the factors changes sign at each critical point. That being the case, the sign of the product MUST change at each critical point. (One thing to be careful about: if one factor is to an even power, it never changes sign so you just ignore that critical point.) For example, if $f(x)= (x+ 3)(x+ 2)^3(x- 1)^4(x- 2)$, the critical points are -3, -2, 1, and 2 but x- 1 is to an even power so we can ignore that. If x< -3, there are 5 negative factors (counting the three in (x-2)^3 but it really doesn't matter whether you think of it as 5 or 3) so f(x) is negative. If -3< x< -2, only x+ 3 changes sign so we now have an even number of negative factors and f(x) is positive. If -2< x< 2, we have only one negative factor so f(x) is non-positive (it will be 0 at x= 1).", "and x= 5/2 and a single (-1) factor. Notice that I changed 1-x to (1)(x- 1) to get the form \"x-a\". Start with x< -2/7. Then x is less than all three of those numbers so all 3 factors are negative and we have (-1) which is always negative for 4 negative factors. Since there are an even number of negatives, the product is positive. Now jump past -2/7: -2/7< x< 1. The factor 7x+2 changes to positive while the other three factors remain negative. Since there are an odd number of negatives, the product is negative. Jump past 1 to 1< x< 5/2. The factor x-1 changes to positive so now we have two negative factors. Since there are an even number of negatives, the product is positive. Finally jump past 5/2 to 5/2< x. Now all the factors except (-1) are positive so we have only one negative factor. Since there are an odd number of negatives, the product is negative. Summary: the product is positive for x< -2/7 and for 1< x< 5/2. This works nicely when the function is a polynomial that can be factored or a rational function in which the numerator and denominator can both be factored. Skeeter's method, choose one point in each interval, will work for all kinds of functions.", "## Prealgebra (7th Edition) $(-5)2=-10$ An odd number of negative factors yields a negative product.", "## Prealgebra (7th Edition) Multiplying two negative factors yields a positive product. Since the first factor has 2 decimal places and the second has 1 decimal place, the product has 3 decimal places. $\\ \\ \\ -\\overset2{\\overset47}.\\overset1{\\overset28}4$ $\\underline{\\times-\\ \\ 3.5}$ $\\ \\ \\ \\ \\ \\ \\overset13.920$ $\\underline{\\ \\ \\ \\ 23.520}$ $\\ \\ \\ \\ 27.440$ Trailing zeroes to the right of the decimal are dropped.", "Theory # Product of a positive and negative number When two factors with different signs are multiplied, the product is negative, e.g. $3(\\text{-}5) = \\text{-}3\\cdot5$ which is equal to $\\text{-}15.$", "## Basic College Mathematics (10th Edition) Find the product by multiplying the integers 1 and 5 and then adding the 0's in the two factors. 1. $3\\times5$ = 15 2. 30 has one zero and 500 has two = three zeros. 3. 15 (with three zeros) = 15,000", "+0 # What is the product of all positive integers n that satisfy the inequalities 3n + 8 < 30 and 2n - 9 > -4? 0 129 1 What is the product of all positive integers n that satisfy the inequalities 3n + 8 < 30 and 2n - 9 > -4? Hey. Pls solve this CPhil or Melody. I need help. Ok, so I tried to make it into a triple inequality after simplifying like this 2.5 < n < 12.66666667 To get 3,4,5,6,7,8,9,10,11,12. Mulitplying all those didn't work. So I tried doing each one individualy and go the same answer. I dont know where I went wrong. Aug 16, 2022 #1 +2519 0 Inequality 1: $$3n + 8< 30$$ Subtract 8 from both sides: $$3n <22$$ Divide by 3: $$n < {22 \\over 3}$$ Inequality 2: $$2n - 9 > -4$$ Add 9 to both sides: $$2n > 5$$ Divide both sides by 2: $$n > {5 \\over 2}$$ So our range becomes $${5 \\over 2} < n < {22 \\over 3}$$ The integers that work are 3, 4, 5, 6, and 7. So the answer is $$3 \\times 4 \\times 5 \\times 6 \\times 7 = \\color{brown}\\boxed{2520}$$ Aug 17, 2022", "Counting products Problem 627 Consider the set $S$ of all possible products of $n$ positive integers not exceeding $m$, that is $S=\\{ x_1x_2\\dots x_n \\, | \\, 1 \\le x_1, x_2, ..., x_n \\le m \\}$. Let $F(m,n)$ be the number of the distinct elements of the set $S$. For example, $F(9, 2) = 36$ and $F(30,2)=308$. Find $F(30, 10001)\\text{ mod }1\\,000\\,000\\,007$.", "and divide by 3 to get $x^2-4x+3>0$, or $(x-3)(x-1)>0$. Look at the signs of the factors to see where the product is positive. Simplifying things lessens chances for errors. 6. Nov 17, 2015 DiamondV Don't see how that explains the answer given to me in the solutions including the infinity part? 7. Nov 17, 2015 Buzz Bloom Hi Diamond: I think the squaring of the inequality and solving the quadratic makes the problem more complicated than necessary, and easier to make a mistake. I suggest solving each of the 4 simple inequalities I mentioned in my post. Actually, two of the four inequalities are redundant, so you only need to solve to get two different result sets of values of x. The union of the the two sets of x values is the answer you want. Regards, Buzz 8. Nov 17, 2015 LCKurtz The $x$ values that work in that inequality are the $x$ values that solve your original problem. Each factor is either positive or negative, depending on the value of $x$. One factor changes sign when $x=3$ and the other at $x = 1$ (which are the values you found). Draw a number line and mark 1 and 3 on it. Those two points are where the product is $0$. Everywhere else the factors are either", "# Find four factors of N with maximum product and sum equal to N - Set-2 in Python Program PythonServer Side ProgrammingProgramming Suppose we have a number N, we have to find all factors of N and return the product of four factors of N such that: The sum of the four factors is same as N. The product of the four factors is maximum. All four factors can be equal to each other to maximize the product. So, if the input is like N = 60, then the output will be All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60 and product is 50625, as we 15 has been selected four times to make product largest. To solve this, we will follow these steps − • factors := a new list • for i in range 1 to integer of (square root of n) + 1, do • if n mod i is same as 0, then • insert i at the end of factors • insert integer of (n / i) at the end of factors • sort the list factors • display factors • final_prod := 1, flag := 1 • for i in range 0 to size of factors, do • for j in range i" ]

[ "# An inequality involving sums and products I am curious to know whether the following holds or not. If $n_1,n_2,n_3,m_1,m_2$ are positive integers strictly greater than 1 such that $$n_1+n_2+n_3 > m_1 +m_2$$ then $$n_1n_2n_3 \\geq m_1m_2.$$ Please do not give me the full answer (I only want a hint). I have tried using the AM-GM-HM inequality but can't seem to prove the result. I have tried looking for counterexamples but haven't found one yet. Thanks in advance for any help!! $2+3+100 > 49 + 50$ $2 \\times 3 \\times 100 < 49 \\times 50$", "# Product of 3 integers is 72, find the 3 integers that give the smallest sum Product of 3 integers a, b, c equals 72, where every factor is positive integer. Find the integers a, b, c with the smallest sum. It's easy to get the factors of 72 manually and see that the 3 smallest factors that give the product as 72 will be the smallest sum. I checked the factorization like this: 1-1-72: 74 1-2-36: 39 1-4-18: 25 1-8-9: 18 2-4-9: 15 3-4-6: 13 It seems like the smallest possible factors will give the smallest sum. But there must be some trick to it or some kind of algorithm to find the smallest integers for any arbitrary product without laborious factoring. Thanks for your help. - The factors must be close to each other. A heuristic: The AM-GM inequality gives us $(a+b+c)/3 \\geq \\sqrt[3]{abc}$ where equality holds when $a=b=c$ and the difference between the LHS and RHS increases when $a$, $b$ and $c$ are far apart from each other. – user17762 Sep 26 '12 at 5:00 Yes @Marvis is right: And 3-4-6 are the closest number in this case. – Sumit Bhowmick Sep 26 '12 at 6:15 One algorithm might be to find the factor of $72$ closest to $\\sqrt[3]{72}$, which is $4$. Then find the", "+0 # What is the product of all positive integers n that satisfy the inequalities 3n + 8 < 30 and 2n - 9 > -4? 0 129 1 What is the product of all positive integers n that satisfy the inequalities 3n + 8 < 30 and 2n - 9 > -4? Hey. Pls solve this CPhil or Melody. I need help. Ok, so I tried to make it into a triple inequality after simplifying like this 2.5 < n < 12.66666667 To get 3,4,5,6,7,8,9,10,11,12. Mulitplying all those didn't work. So I tried doing each one individualy and go the same answer. I dont know where I went wrong. Aug 16, 2022 #1 +2519 0 Inequality 1: $$3n + 8< 30$$ Subtract 8 from both sides: $$3n <22$$ Divide by 3: $$n < {22 \\over 3}$$ Inequality 2: $$2n - 9 > -4$$ Add 9 to both sides: $$2n > 5$$ Divide both sides by 2: $$n > {5 \\over 2}$$ So our range becomes $${5 \\over 2} < n < {22 \\over 3}$$ The integers that work are 3, 4, 5, 6, and 7. So the answer is $$3 \\times 4 \\times 5 \\times 6 \\times 7 = \\color{brown}\\boxed{2520}$$ Aug 17, 2022", "# Factorial Sums Find the sum of all positive integers $m$ such that $2^m$ can be expressed as sums of four factorials (of positive integers). Details and assumptions The number $n!$, read as n factorial, is equal to the product of all positive integers less than or equal to $n$. For example, $7! = 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$. The factorials do not have to be distinct. For example, $2^4=16$ counts, because it equals $3!+3!+2!+2!$. ×", "# Proving at least 99 pairwise sums of the reals are non-negative Suppose, one hundred real numbers are given and their sum is 0.Then how can I prove that at least 99 of the pairwise sums of these hundred numbers are non-negative? I tried this: Let the real numbers be $a_i$,$1\\leq i\\leq100$. By the given condition, $a_1+a_2+\\dots +a_{100}=0$.Suppose for the sake of contradiction, that at most 98 of the pairwise sums of these given reals are non-negative.I just tried to examine the case when exactly 98 of them are non-negative. I labelled the sums $S_1$,$S_2,\\dots S_{4950}$,WLOG assume that sums $S_1,S_2,\\dots S_{4852}$ are negative, the rest are non-negative. $$S_1+S_2+\\dots +S_{4950}=99(\\sum_{i=1}^na_i)=0$$ Case I: $S_{4853}+a_{4854}+\\dots +a_{4950}=0$.That simply means that $S_1+S_2+\\dots +S_{4852}=0$.But, by the given condition, $S_1+S_2+\\dots +S_{4852}<0$,a contradiction.After that, I am not able to make any progress. Can anyone please suggest a more efficient method to solve this problem.I feel this approach will involve enormous case work. - More of a sketch than a proof, but consider the largest element. The pairwise sum of this with the others will yield 99 positive sums, unless there is an element even more negative than the maximum is positive. But if this is the case, then the average of the other 98 will be positive, so there will be more positive sums there. –", "neighbor regions contain a different color. To see the Review answers, open this PDF file and look for section 7.8. # Vocabulary Term Definition ! The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol \"!\" denotes factorial. n!=1⋅2⋅3⋅4...⋅(n−1)⋅n. factorial The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol \"!\" denotes factorial. n!=1⋅2⋅3⋅4...⋅(n−1)⋅n. induction Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers. inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are <, >, ≤, ≥ and ≠. Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3... postulate A postulate is a statement that is accepted as true without proof. proof A proof is a series of true statements leading to the acceptance of truth of a more complex statement.", "4$$. 4. Aug 24, 2006 ### HallsofIvy Staff Emeritus As you noticed, $t^2 \\ge 16$ does not give $t \\ge 4$. A standard way of solving inequalities is to solve the corresponding equality first. $t^2= 16$ gives, of course, t= 4 and t= -4. Those are the only places at which t2 can change sign. It t= -5 (less than -4) then t2= (-5)2= 25 which is greater than 16 so t2> 16 for all x< -4. 0 is between -4 and 4 and of course 02= 0 which is less than 16: t2< 16 for all t between -4 and 4. Finally if t= 5 (larger than 4) then t2= 52= 25 which is again larger than 16. $t^2 \\ge 16[/sup] for all [itex]x \\le -4$ and all $x \\ge 4$. That method works for all continuous functions. Another way to do this inequality is to factor: $t^2- 16= (t-4)(t+4)\\ge 0$. If t< -4 both factors are less than 0; the product of two negative numbers is positive. If t is between -4 and 4, one factor, t-4, is still negative but the other, t+4 is now positive. The product of a negative number with a positive number is negative. Finally, if t>4, both factors are positive so the product is positive again. Know someone interested in", "# Problem of the Week Problem E Count on That Let $$n$$ be a positive integer. How many values of $$n$$ satisfy the following inequality? $(n-1)(n-3)(n-5)\\cdots(n-2019)(n-2021) \\leq 0$ Note: The product on the left side of the inequality consists of $$1011$$ factors of the form $$n-d$$, where the value of $$d$$ starts at $$1$$ and increases by $$2$$ for each subsequent factor.", "Difference between revisions of \"2012 USAMO Problems/Problem 6\" Problem For integer $n \\ge 2$, let $x_1$, $x_2$, $\\dots$, $x_n$ be real numbers satisfying $$x_1 + x_2 + \\dots + x_n = 0, \\quad \\text{and} \\quad x_1^2 + x_2^2 + \\dots + x_n^2 = 1.$$ For each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define $$S_A = \\sum_{i \\in A} x_i.$$ (If $A$ is the empty set, then $S_A = 0$.) Prove that for any positive number $\\lambda$, the number of sets $A$ satisfying $S_A \\ge \\lambda$ is at most $2^{n - 3}/\\lambda^2$. For what choices of $x_1$, $x_2$, $\\dots$, $x_n$, $\\lambda$ does equality hold?", "Counting products Problem 627 Consider the set $S$ of all possible products of $n$ positive integers not exceeding $m$, that is $S=\\{ x_1x_2\\dots x_n \\, | \\, 1 \\le x_1, x_2, ..., x_n \\le m \\}$. Let $F(m,n)$ be the number of the distinct elements of the set $S$. For example, $F(9, 2) = 36$ and $F(30,2)=308$. Find $F(30, 10001)\\text{ mod }1\\,000\\,000\\,007$.", "are the only ways to get a positive product from three factors. Now, here’s a chart of the signs of the factors $x-3,x-2$, and $x-1$ and their product: 1 2 3 ----------------|------------|------------|-------------------- x-3: - - - - - 0 + x-2: - - - 0 + + + x-1: - 0 + + + + + product: - 0 + 0 - 0 + When $x<1$, all three factors are negative, and so is their product. When $x=1$, one factor is $0$, and so is the product. When 1or $x>3$. In interval notation, the solution set of the inequality is $(1,2)\\cup(x,\\infty)$. This technique works whenever you’re comparing a product with $0$. - @gekkostate: I want the semi-graphical chart, not an array, and I want ‘for the moment’, not ‘for a moment’. – Brian M. Scott Feb 27 '13 at 14:04 @BrianMScott, sorry about that. I thought that having the array would look nicer and still get the message across. – Jeel Shah Feb 27 '13 at 14:06 @gekkostate: No harm done. I realize that the array is prettier, but it doesn’t have the pictorial aspect with the number line. I just wanted to let you know that I wasn’t rolling it back out of pique: I really did have a preference. – Brian M. Scott Feb 27 '13", "Or, $n^{2}-n-365>0$. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position. More later, Nalin Pithwa.", "# How to find all combinations that equal a given sum Two equations are given: $a_{1}b_{1} + a_{2}b_{2} + \\dots + a_{n}b_{n} = N$ $b_{1} + b_{2} + \\dots + b_{n} = M$ and given is set of $a_{1}, a_{2}, \\dots a_{n}$ $a_{1}, \\dots, a_{n} \\geq 0$ and $b_{1}, \\dots, b_{n} \\geq 0$ How to find all possible equations which satisfy these conditions? For example: $a + 2b + c = 5,$ $a+b+c = 4$ We have four combinations: (2,1,1), (1,1,2), (0,1,3), (3,1,0) Is there any formula for that? What if there will be inequality $\\leq N$ • not sure the combinatorics tag is needed ... the example has a countably infinite number of solutions on the integers, – user451844 Sep 20 '17 at 10:47 • Oh... We can use only positive integers. – Piotr Wasilewicz Sep 20 '17 at 10:49 • 0 is technically non-negative not positive, but okay that should be put in the question otherwise a+c=3 has a countably infinite number of solutions. – user451844 Sep 20 '17 at 10:51 • Yes, thanks. I added this condition. – Piotr Wasilewicz Sep 20 '17 at 10:52 What we've got here is the system of 2 linear equations with $n$ unknown values. To solve it You can use for example the Gauss elimination. In your example there", "# Number of sums of factors Given a positive integer n > 1 determine how many numbers can be made by adding integers greater than 1 whose product is n. For example if n = 24 we can express n as a product in the following ways 24 = 24 -> 24 = 24 24 = 12 * 2 -> 12 + 2 = 14 24 = 6 * 2 * 2 -> 6 + 2 + 2 = 10 24 = 6 * 4 -> 6 + 4 = 10 24 = 3 * 2 * 2 * 2 -> 3 + 2 + 2 + 2 = 9 24 = 3 * 4 * 2 -> 3 + 4 + 2 = 9 24 = 3 * 8 -> 3 + 8 = 11 We can get the following numbers this way: 24, 14, 11, 10, 9 That is a total of 5 numbers, so our result is 5. Write a program or function that takes n as input and returns the number of results that can be obtained this way. This is a question so answers will be scored in bytes, with fewer bytes being better. ## OEIS sequence OEIS A069016 • Suggested test case 240 – Jonathan Allan Aug 6 '17 at 18:52 •", "by choosing $x^{2^k}$ in the factors where bit $k$ is $1$ in the binary representation of $m$ and choosing $1$ in the factors where bit $k$ is $0$. All coefficients in the factors are $1$ so the coefficient of $x^m$ in the product is $1$. Therefore, $$(1+x)(1+x^2)(1+x^4)\\dots(1+x^{2^{n-1}})=1+x+x^2+x^3+\\dots+x^{2^n-1}$$ -", "the factorisations of various numbers with lots of factors. I was aiming for a counter-example with three integers. So with product $36$ I found $2,3,6$ and $3,3,4$, the second of which is not strictly a set, because it has duplicates. But its existence suggested that in larger cases there might be distinct integers. There is also $1,6,6$ and $2,2,9$. $60$ and $84$ gave nothing, but $72$ had $2,6,6$ and $3,3,8$ and $96$ yielded $1,8,12$ and $2,3,16$. This was quite quick by hand. Two further questions suggest themselves: are there examples of three sets of three positive integers having the same sum and the same product? Are there examples of two sets of $n$ positive integers having the same sum and the same product for all $n\\ge 3$? Both seem likely based on my hand calculations. • And one of the other answers indeed finds five sets of three with the same sum and product. – Mark Bennet Oct 18 '17 at 18:04 • \"are there examples of three sets of three positive integers having the same sum and the same product?\".... here's an example of five :) $\\{15, 88, 140\\}, \\{16, 77, 150\\}, \\{20, 55, 168\\}, \\{24, 44, 175\\}, \\{25, 42, 176\\}$. I think the real question is \"are there examples of $n$ sets of three positive numbers", "# How to find all combinations that equal a given sum Two equations are given: $a_{1}b_{1} + a_{2}b_{2} + \\dots + a_{n}b_{n} = N$ $b_{1} + b_{2} + \\dots + b_{n} = M$ and given is set of $a_{1}, a_{2}, \\dots a_{n}$ $a_{1}, \\dots, a_{n} \\geq 0$ and $b_{1}, \\dots, b_{n} \\geq 0$ How to find all possible equations which satisfy these conditions? For example: $a + 2b + c = 5,$ $a+b+c = 4$ We have four combinations: (2,1,1), (1,1,2), (0,1,3), (3,1,0) Is there any formula for that? What if there will be inequality $\\leq N$ • not sure the combinatorics tag is needed ... the example has a countably infinite number of solutions on the integers, – user451844 Sep 20, 2017 at 10:47 • Oh... We can use only positive integers. Sep 20, 2017 at 10:49 • 0 is technically non-negative not positive, but okay that should be put in the question otherwise a+c=3 has a countably infinite number of solutions. – user451844 Sep 20, 2017 at 10:51 • Yes, thanks. I added this condition. Sep 20, 2017 at 10:52 ## 1 Answer What we've got here is the system of 2 linear equations with $n$ unknown values. To solve it You can use for example the Gauss elimination. In your example there are solutions with", "# How do you solve (2x-1)(x+2)(x-4)<=0? Oct 13, 2015 The inequality holds for $x \\setminus \\le - 2$ and $\\frac{1}{2} \\setminus \\le x \\setminus \\le 4$. #### Explanation: Keep in mind the rule for multiplying: • Positive times positive is positive; • Positive times negative is negative; • Negative times positive is negative; • Negative times negative is positive. Combining these rules, you can clearly see that a product of factors is negative if and only if there is an odd number of negative factors (in this case, one or all three). Now, $2 x - 1$ is positive if and only if $x > \\frac{1}{2}$, $x + 2$ is positive if and only if $x > - 2$, and $x - 4$ is positive if and only if $x > 4$. So, the important points are $- 2$, $\\frac{1}{2}$ and $4$. Before $- 2$, all three factors are negative, so the product is negative. Between $- 2$ and $\\frac{1}{2}$, $x + 2$ is positive and the other two are negative, so the product is positive. Between $\\frac{1}{2}$ and $4$, only $x - 4$ is negative, and the other two are positive, so the product is negative. After $4$, all the factors are positive, so the product is positive.", "# Limited Factors Algebra Level 5 Find the number of positive integers less than $$1000$$ that evenly divide $2^{20}-1$ ×", "+0 0 169 4 Solve the inequality $$\\frac{(x - 2)(x - 3)(x - 4)}{(x - 1)(x - 5)(x - 6)} > 0.$$ Jul 31, 2019 #1 +1 $$x$$ can't be 2, 3, 4, 1, 5, or 6. Hope this helps! Jul 31, 2019 #2 +106963 +1 Certainly x cannot be 1, 5 or 6 becasue that would make the denominaot 0 and you cannot EVER divide by 0. But it will be positive when the top mulutiplies to a positive number This will happen when 2 factors are negative and the 3rd is positive and ALSO when all three factors are positive So now you have a few scenarios to work out. Jul 31, 2019 #3 +106533 +1 Note that Any x in (-inf, 1) will produce negative / negative = positive Any x in (1, 2) will produce negative / positive = negative Any x in (2, 3) will produce pos / pos = pos = true Any x in (3, 4) will produce neg/ pos = neg = false Any x in (4, 5) will produce pos/ pos = pos = true Any x in (5, 6) will produce pos/ neg = neg = false Any x in (6, inf) will produce pos/ pos = pos = true So.....the intervals that make this true are (-inf," ]

[ "closed interval $$[0,1]$$ and $$g(0)$$ and $$g(1)$$ have opposite signs, it follows from Bolzano’s Theorem that there is a point $$c$$ between 0 and 1 such that $$g(c)=f(c)-c=0$$ or $$f(c)=c$$. The geometric interpretation is simple. If $$f(0)\\neq0$$ and $$f(1)\\neq1$$, then the graph of $$f$$ has to cut the line $$y=x$$ at some point between 0 and 1 (see the following figures). Figure 7: If $f$ is continuous on $[ 0,1 ]$ and $0 \\leq f( x ) \\leq 1$ , then the equation $f( x ) =x$ has at least one solution ( $0 \\leq x \\leq 1$) . ## The Extreme Value Theorem Definition 1: Let $$f$$ be a function defined on a set $$E$$. We say $$f$$ has an absolute maximum on (or in) $$E$$, if there is at least one point $$p$$ in $$E$$ such that $$f(x)\\leq f(p)$$ for every $$x$$ in $$E$$. In this case, we say $$p$$ is the point of absolute maximum and $$f(p)$$ is the absolute maximum value (or simply maximum) of $$f$$ on $$E$$. Similarly we say $$f$$ has an absolute minimum on $$E$$, if there exists a point $$q$$ in $$E$$ such that $$f(q)\\leq f(x)$$ for all $$x$$ in $$E$$. In this case, we say $$q$$ is the point of absolute minimum and $$f(q)$$ is the minimum value (or", "$\\lim\\limits_{x\\to-\\infty}f(x)=0$ and $$f'(x)=e^x\\left(p(x)-p'(x)-p''(x)+p'''(x)\\right)\\ge 0,$$ i.e. $f$ is increasing. It follows that $f\\ge 0$, and hence $$p-2p'+p''\\ge 0.$$ Define $$g(x)=e^{-x}\\left(p(x)-p'(x)\\right).$$ Then $\\lim\\limits_{x\\to+\\infty}g(x)=0$ and $$g'(x)=-e^{-x}\\left(p(x)-2p'(x)+p''(x)\\right)\\le 0,$$ i.e. $g$ is decreasing. It follows that $g\\ge 0$, and hence $$p-p'\\ge 0.$$ Define $$h(x)=e^{-x}p(x).$$ Then $\\lim\\limits_{x\\to+\\infty}h(x)=0$ and $$h'(x)=-g(x)\\le 0,$$ i.e. $h$ is decreasing. Therefore, either $h>0$ or $h\\equiv 0$. The conclusion follows. Proof 2: Denote $q=p-p'$. Then $$p+p'''\\ge p'+p''\\iff q-q''\\ge 0.$$ It implies that $\\deg(q-q'')=\\deg q=\\deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then $$q(x)\\ge q(x_q)\\ge q''(x_q)\\ge 0,$$ i.e. $$p\\ge p'.$$ Similarly, $p$ also has a global minimum value point, say $x_p$. Then $$p(x_p)\\ge p'(x_p)= 0.$$ If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $p\\equiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $k\\ge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)\\ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then $$p(x)-p'(x)=-kr(x_p)(x-x_p)^{k-1}+O((x-x_p)^k)$$ cannot be always non-negative, a contradiction. • @GitGud: Since $f$ is increasing and $f(-\\infty)=0$, $f\\ge 0$. – 23rd Jul 31 '13 at 19:42 • Of course, thanks. – Git Gud Jul 31 '13 at 19:44 • @GitGud: You are welcome. – 23rd Jul 31 '13 at 19:47 •", "\\leq (a^{2}+b^{2})^{1/2} (c^{2}+d^{2})^{1/2}$ with $c=d=1$. If you restrict $x_i,y_i, i=1.2$ to psoitive values then the maximum is $2$ and it is attained when $x_2=1,x_1=0, y_1=1,y_2=0$/. Sep 21 '19 at 0:03", "Edit: Definition of $$f\\leq ^ { * } g$$ : $$\\exists m \\in \\mathbb { N } : \\forall n \\in \\mathbb { N } \\text { where } n \\geq m \\Longrightarrow f ( n ) \\leq g ( n )$$ • By $f\\leq^*g$ do you mean $f(n) \\leq g(n) \\forall n\\in \\mathbb{N}$? – Jakob B. Jan 20 at 21:20 • Yes, sorry, that was defined elsewhere. – j.lnhrt Jan 20 at 21:21 • That is not the usual definition of $\\le^*$. Please check what definition you are actually using. – Andrés E. Caicedo Jan 20 at 21:22 • Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $\\exists m \\in \\mathbb { N } \\text { : } \\forall n \\in \\mathbb { N } \\text { where } n \\geq m \\implies f ( n ) \\leq g ( n )$ – j.lnhrt Jan 20 at 21:35 • Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten. – Ross Millikan Jan 20 at 22:14 Yes, given a countable $$A$$ there is a sequence $$f$$ where all of them are $$\\le^*$$ than $$f$$. We set $$f(1)=f_1(1)+1\\\\ f(2)=\\max(f_1(2),f_2(2))+1\\\\ f(3)=\\max(f_1(3),f_2(3),f_3(3))+1\\\\", "262 views If $g(x)=p\\mid x \\mid-qx^2$, where $p$ and $q$ are constants, then at $x=0, g(x)$ will be 1. maximum when $p>0,q>0$ 2. minimum when $p<0,q<0$ 3. minimum when $p>0,q<0$ 4. maximum when $p>0,q<0$ 1 189 views", "we simply evaluate $f(p)$ and $f(q)$. If $f(p)$ is larger than $f(q)$, then $p^q>q^p$; otherwise, $p^q<q^p$. In the example in OP, $f(999)\\approx 0.00591$, while $f(2)\\approx 0.346$. Hence $999^2<2^{999}$. The function $f(x)$ increases from $0$ to $e\\approx 2.71828$, and then decreases after that. Hence, there are two nice rules we can use: If $p>q>e$, then $f(p)<f(q)$. If $e>p>q>0$, then $f(p)>f(q)$. However, if one of $p,q$ is less than $e$ and the other is greater, then all bets are off. • And this works because $\\ln$ is strictly increasing. – GFauxPas Jun 3 '16 at 13:46", "lemma gives us a possible answer to the problem, but is it the best answer? Let's divide it in two cases: Theorem. If $pa_0=q-1$, then $2a_0$ is the optimal answer. Proof. It is clear that $a_0$ can't be a valid answer, and $f(a)\\leq f(2a_0)$ for all $a\\geq 2a_0$ using similar arguement with the lemma. To see that $f(a)\\leq f(2a_0)$ for $a_0+1\\leq a\\leq 2a_0$, notice that $a_0p = q-1$ so that the integer part of $\\frac{ap}{q}$ do not increase in the given range, the result immediately follows. Similarly, if $pa_0\\geq q$ we check all $1\\leq a \\leq a_0$ to get the optimal answer. It follows that the optimal answer is at most $a_0$ using the similar arguement. If we want to be faster, check all crossing and the corresponding $a$ value so that $pa$ is just under the multiple of q, now we should get the answer. Why 'should'? Think about the situation: Suppose we get $a_0$ as the 'optimal' answer using the faster way, is $a_0-1$ a better answer? Is this even possible?... We shall see it shortly...", "$$n \\geq p\\ , R_n =0 .$$ and $$n \\leq p\\ , \\xi =1 .$$ gives maximum. If not, then $p-n\\geq 1$ gives maximum for $\\xi =1$. Or else,if $p-n\\leq 1$ gives maximum for $\\xi =0$.", "and $X\\geq p,Y\\geq q,X+Y\\leq M$. There is no free maximum because $X>0,Y>0$. Then the max (that exists because we are in a compact set) is reached on the edge. Case 1. $X+Y=M$ and $f(X,Y)=g(X)=X^b(M-X)^d$. Then $g'$ is $0$ for $X_1=\\dfrac{b}{b+d}M$; if $p\\leq X_1\\leq M-q$, then the first candidate is $C_1=(X_1,M-X_1)$, else we do not consider this possibility. Case 2. $X=p$ and $f(X,Y)=g(Y)=p^bY^d$. The second candidate is $C_2=(p,M-p)$. Case 3. In the same way , the third candidate is $C_3=(M-q,q)$. It remains to compare $f(C_1),f(C_2),f(C_3)$. - I have corrected what was likely a small typo in the first sentence: I suspect you wanted $\\ c \\$ , rather than $\\ b \\$ there. – RecklessReckoner Jul 17 '14 at 17:53 Yes, you are right. Thanks. – loup blanc Jul 17 '14 at 19:40 You could use weighted A.M-G.M inequality, as all the quantities are non-negative. [b(x+p) + d(y+q)]/(b+d) ≥ [(x+p)^b + (y+q)^d]^{1/(b+d)} Raising both sides to power (b+d) and multiplying by ac, you get the maximum value, achieved when both sides are equal. This is possible only when x+p = y+q (i.e. all terms in the A.M are equal) Also, x+y+p+q ≤ M So, for maximum value, x = M/2 - p and y = M/2 -q (Sorry for the style of writing. Am new to this forum) -", "like you've got it to me. – John Martin Apr 26 '16 at 21:27 If $f(x_0) \\neq 0$, from the differential equation $x_0^2f''(x_0)-f(x_0)=0$ and from the condition that $x_0$ is a maximum we simply get that $f(x_0)<0$. Finally, from continuity arguments, one can say that from Weierstrass theorem that the inequality $f(x_0)\\geq f(x)$ is satisfied over all points of the interval $[0,1]$ so that $f(x_0)\\geq 0$, since $f(0)=f(1)=0$.", "the only cases where $f(t p + (1-t) q) > t f(p) + (1-t) f(q)$ have one of $p$ and $q$, say $p$, is in $[0,1/2] \\times \\{1\\}$ and the other in $(1/2, 1] \\times \\{1\\}$, and then $g(t p + (1-t) q) - \\max(g(p),g(q)) = f(tp + (1-t) q) - \\max(f(p),f(q) = 0$.", "\\right\\}$$ Therefore, $$f'(x)$$ has a finite value for all x > 0 $$\\Rightarrow$$ $$f(x)$$ is differentiable everywhere. Example - 16 f(x)={{x}^{2}}-2|x|~\\text{ }and~\\text{ }g(x)=\\left\\{ \\begin{align} & Min\\{f(t)\\,\\,\\,:-2\\le t\\le x,\\,\\,\\,\\,-2\\le x<0\\} \\\\ & Max\\{f(t):\\,\\,\\,\\,\\,0\\le t\\le x,\\,\\,\\,\\,\\,\\,\\,\\,0\\le x\\le 3\\} \\\\ \\end{align} \\right\\} draw the graph of $$g(x)$$ and discuss its continuity and differentiability. Solution: The graph of f(x) is sketched below (in the relevant domain): Now we must understand what the definition of g(x) means. Consider the upper definition of g(x): $$g(x) = Min\\{ f(t): - 2 \\le t \\le x,\\,\\,\\,\\, - 2 \\le x < 0\\}$$ To evaluate $$g(x)$$ at any x, we scan the entire interval from –2 to x, (this is what the variable t is for), select that value of f(t) which is minimum in this interval; this minimum value of f(t) becomes the value of the function at x. The figure below illustrates this graphically for four different values of x (we are considering the interval $$- 2 \\le x < 0,$$ the upper definition of g(x)): Notice that where x crosses –1 (when –1 < x < 0), the minimum value of the function in the interval [–2, x] becomes fixed at t = –1. When –2 < x < –1, the minimum value of the function in the interval [–2, x] is at t", "= x. So, how do we draw the graph of g (x)? In [–2, –1], the graph of g (x) will the same as f (x) (because the minimum value of f (x) is at t = x itself, as described above). In [–1, 0], the minimum value becomes fixed at t = –1, equal to –1, so that in this interval the graph of g(x) is constant; g (x) = 1 for $$x \\in \\left[ { - 1,0} \\right]$$ . The graph of g (x) for $$x \\in \\left[ { - 2,0} \\right]$$ is sketched below; For $$0 \\le x \\le 3,$$ the definition is $$g\\left( x \\right) = {\\rm{Max}}\\left\\{ {f\\left( t \\right);0 \\le t \\le x,0 \\le x \\le 3} \\right\\}$$ . The figure below illustrates how to obtain g(x) in this case for four different values of x: Notice then untill x lies in the interval [0, 2], the maximum value of f(x) in the interval [0, x] is at t = 0, equal to 0. As soon as x becomes greater than 2, the maximum value of f (x) in the interval [0, x] is now at t = x. The graph of g(x) for the interval [0, 3] is sketched below: The overall graph for g (x) is therefore: We see that g", "maximum on $[a,b]$. The example $$f(t):=e^{it}\\qquad(0\\leq t\\leq2\\pi)$$ shows that the MVT does not hold for complex- or vector-valued functions. The point $\\xi$ does not appear in de l'Hôpital's rule anymore. Therefore producing a counterexample is more tricky. Here is one: Let $$f(t):=t,\\quad g(t):=t e^{-i/t}\\ \\qquad (t>0).$$ Then $\\lim_{t\\to0+} f(t)=\\lim_{t\\to0+} g(t)=0$; furthermore $$f'(t)\\equiv 1,\\quad g'(t)=\\left(1+{i\\over t}\\right)e^{-i/t}\\ .$$ It follows that $$\\lim_{t\\to0+}{f'(t)\\over g'(t)}=\\lim_{t\\to0+}\\left({t\\over t+i} e^{i/t}\\right)=0\\ .$$ But the $\\lim_{t\\to0+}{f(t)\\over g(t)}=\\lim_{t\\to0+}e^{i/t}$ does not exist. There is a way out that also works in the multivariable case: Compute Taylor approximations to everything in sight, check for possible cancellations and hopefully arrive at a limit of the form $$\\lim_{x\\to 0}{f_1(x)\\over g_1(x)}\\ ,$$ where at least one of $f_1$, $g_1$ has a limit at $x=0$ which is $\\ne0$, $\\ne\\infty$. (In the above counterexample no such Taylor approximation is possible.) - This would seem to have nothing to do with \"dimensional analysis,\" but maybe OP misused that term. – Gerry Myerson Apr 17 '13 at 13:34", "and we want to show $$f(j) \\leq b$$. Then by definition of $$f$$ being a left adjoint we also have $$a \\leq g(b)$$. However since $$j$$ is the least upper bound, we have $$j \\leq g(b)$$. Again using the Galois connection, we have $$f(j) \\leq b$$, proving that $$f(j)$$ really is the least upper bound for $$f(A)$$. Exercise 1.114. \\begin{split} \\begin{align} \\\\ f(p=1) & = 1 \\leq q=1 \\iff p=1 \\leq g(q=1) = 1 \\\\ f(p=2) & = 2 \\leq q=1 \\iff p=2 \\leq g(q=1) = 1 \\\\ f(p=3.9) & = 4 \\leq q=1 \\iff p=3.9 \\leq g(q=1) = 1 \\\\ f(p=4) & = 4 \\leq q=1 \\iff p=4 \\leq g(q=1) = 1 \\\\ f(p=1) & = 1 \\leq q=2 \\iff p=1 \\leq g(q=2) = 2 \\\\ f(p=2) & = 2 \\leq q=2 \\iff p=2 \\leq g(q=2) = 2 \\\\ f(p=3.9) & = 4 \\leq q=2 \\iff p=3.9 \\leq g(q=2) = 2 \\\\ f(p=4) & = 4 \\leq q=2 \\iff p=4 \\leq g(q=2) = 2 \\\\ f(p=1) & = 1 \\leq q=4 \\iff p=1 \\leq g(q=4) = 4 \\\\ f(p=2) & = 2 \\leq q=4 \\iff p=2 \\leq g(q=4) = 4 \\\\ f(p=3.9) & = 4 \\leq q=4 \\iff p=3.9 \\leq g(q=4) = 4 \\\\ f(p=4) & = 4 \\leq q=4 \\iff p=4 \\leq g(q=4) = 4 \\\\", "D [f(c)]2 = [g(c)]2 + 3g(c) for some c $$\\in$$ [0, 1] ## Explanation Suppose f(x) is maximum at c1 and g(x) is maximum at c2. When f(x) is maximum g(x) may or may not be maximum. Therefore, in the function h(x) = f(x) $$-$$ g(x), we get $$h({c_1}) = f({c_1}) - g({c_1}) \\ge 0$$ and $$h({c_2}) = f({c_2}) - g({c_2}) \\ge 0$$. Therefore, h(x) = 0 for some c $$\\in$$ [0, 1]. Therefore, $$h(c) = 0 \\Rightarrow f(c) - g(c) = 0$$. Therefore, $$f(c) = g(c)$$. Option (a) $$\\Rightarrow {f^2}(c) - {g^2}(c) + 3[f(c) - g(c)] = 0$$ which is true from Eq. (i). Option (d) $$\\Rightarrow {f^2}(c) - {g^2}(c) = 0$$ which is true from Eq. (i) Now, if we take f(x) = 1 and g(x) = 1, $$\\forall$$x $$\\in$$[0, 1] Option (b) and (c) does not hold. Hence, option (a) and (d) are correct. 4 ### IIT-JEE 2012 Paper 2 Offline MCQ (More than One Correct Answer) Let $$f:( - 1,1) \\to R$$ be such that $$f(\\cos 4\\theta ) = {2 \\over {2 - {{\\sec }^2}\\theta }}$$ for $$\\theta \\in \\left( {0,{\\pi \\over 4}} \\right) \\cup \\left( {{\\pi \\over 4},{\\pi \\over 2}} \\right)$$. Then the value(s) of $$f\\left( {{1 \\over 3}} \\right)$$ is(are) A $$1 - \\sqrt {{3 \\over 2}}$$ B $$1 + \\sqrt {{3", "2. Does the graph of $g$ have a point of inflection at $x=4$? Justify your answer. 3. Find the absolute minimum value and the absolute maximum value of $g$ on the interval $-4\\leq x\\leq 12$. Justify your answers. 4. For $-4\\leq x\\leq 12$, find all intervals for which $g(x)\\leq 0$. The official AP scoring rubric for this problem can be found on page 4 of this document. This was Question 3 of the 2016 AP Calculus AB FRQs.", "(fg)(0)=1$ This is true for $f+g$. Because for every $x$ in the neighborhood of $x_0$ you have \\begin{aligned} f(x)&\\leq f(x_0)\\\\g(x)&\\leq g(x_0) \\end{aligned} So, by adding the above inequalities you'll get $$f(x)+g(x)\\leq f(x_0)+g(x_0)$$ or $$(f+g)(x)\\leq (f+g)(x_0)$$ which means the maximum of $f+g$ occurs at $x_0$, too. But the proposition is not true for $f\\cdot g$. For example you can consider both of $f$ and $g$ to be negative functions, like $$f(x)=g(x)=-(1+x^2)$$ for which the maximum occurs at $x_0=0$ . You can see $$(f\\cdot g)(x)=\\left(1+x^2\\right)^2$$ which its minimum occurs at $x_0=0$ and also has no maximum. Not necesserly, if $f(x_0)<0$ and $g(x_0)<0$ then it is a minimum of $fg$ Assuming that $f$ and $g$ are twice differentiable, we can classify critical points by the second derivative test. For the first part, we have $(f + g)' = f' + g'$ and so $x_{0}$ is a critical point of $f + g$. Moreover, if $x_{0}$ if a maximum of $f$ and $g$, what can you say about the sign of $f''$ and $g''$ and how could this help determine the sign of $(f + g)''$? Now $(fg)' = fg' + f'g$, so again $x_{0}$ is a critical point of $fg$. But $(fg)'' = (fg' + f'g)' = fg'' + 2f'g' + f''g$, and it becomes hard to say whether or not", "compute the result as the maximum value of f*[K] for 1 \\leq i \\leq N. So how can we compute the f table? First, notice that the value of f*[1], for any 1 \\leq i \\leq N, can be easily computed using the p table in constant time. Now, we want to know how to compute the value of f*[j] knowing the values of f[i'][j - 1] for i' < i. There are two cases to consider: 1. i' + D < i Then : f*[j] = \\max_{i'} \\{f[i'][j - 1] + (a[i' + D + 1] + a[i' + D + 2] + \\ldots + a[i - 1]) + (a[i + 1] + a[i + 2] + \\ldots + a[i + D]) \\cdot M\\} \\\\ = \\max_{i'} \\{f[i'][j - 1] + (p[i - 1] - p[i' + D]) + (p[i + D] - p*) \\cdot M\\} Notice that for a fixed i, the above value varies only on f[i'][j - 1] + p[i' + D], so in order to compute f*[j] in this case, we only need to know what is the value of i' which maximizes this term, and it is easy to compute the best values of it in previous steps. 2. i' + D \\geq i Then f*[j] = \\max_{i'} \\{f[i'][j - 1] -", "assume that $0 \\leq f \\leq 1$ for all $f \\in A$. – Dieter Kadelka Oct 15 '19 at 21:16" ]

[ "$f(x,y) \\le \\frac12x-x-e^{-x}$, so for a fixed $x$, the choice $y:=x+e^{-x}$ is the best in maximizing $f$. Then, you are left with only one variable, and should maximize this $g(x):=-\\frac12x-e^{-x}$ whence $x\\ge 0$. Possible maximum place is at the edge ($x=0$) or when the tangent of $g$ is horizontal, i.e. where $g'=0$.", "a given ${p}$, we can differentiate ${g(x)=\\frac{(p-x)^2}{x^2}}$ with respect to ${x}$ to see that ${g'(x)=0}$ only when ${x=p}$, which is when the relative error is zero. Therefore for ${\\frac{|p-x|}{x}}$ to attains its maximum, ${x}$ must be equal to ${a}$ or ${b}$. Therefore $\\displaystyle F(p)=\\max_{x\\in[a, b]}\\{\\frac{|p-x|}{x}\\}=\\max\\{\\frac{p-a}{a}, \\frac{b-p}{b}\\}.$ Then $\\displaystyle \\frac{p-a}{a}\\geq \\frac{b-p}{b} \\Leftrightarrow p\\geq \\frac{2ab}{a+b}\\in [a, b] \\textrm{ and }\\frac{p-a}{a}\\leq \\frac{b-p}{b} \\Leftrightarrow p\\leq \\frac{2ab}{a+b}\\in [a, b].$ Let ${H=\\frac{2ab}{a+b}}$, which is the harmonic mean of ${a}$ and ${b}$ with equal weights. It is easy to see that ${\\frac{p-a}{a}}$ is increasing for ${p\\geq H}$ and ${\\frac{b-p}{b}}$ is decreasing for ${p\\leq H}$. Therefore ${F(p)}$ attains its minimum at ${p=H=\\frac{2ab}{a+b}}$. $\\Box$ Proposition 5 (The Geometric Mean as a Minimum) The geometric mean has the representation $\\displaystyle \\{ \\prod_{k=1}^n a_k \\}^{1/n}=\\min\\{\\frac{1}{n} \\sum_{k=1}^n a_k x_k:(x_1,\\ldots, x_n)\\in D \\},$ where ${D}$ is the region of ${\\mathbb{R}^n}$ defined by $\\displaystyle D=\\{(x_1, \\ldots, x_n):\\prod_{k=1}^n x_k=1, x_k\\geq 0, k=1, \\ldots , n\\}.$ Proof: Assume ${a_k\\neq 0}$ for all ${k}$, by AM-GM inequality, $\\displaystyle \\frac{1}{n} \\sum_{k=1}^n a_k x_k\\geq (\\prod_{k=1}^n a_k x_k)^{1/n}=(\\prod_{k=1}^n a_k)^{1/n} \\textrm{ as } (x_1, \\ldots, x_n)\\in D.$ The AM-GM inequality also tells us that the equality holds if and only if ${a_1x_1=\\ldots =a_nx_n=c }$ then ${x_k=c/a_k}$ and ${1=c^n/(a_1 a_2\\ldots a_n)\\Rightarrow c=(a_1 a_2\\ldots a_n)^{1/n}}$. So choosing ${x_k=\\frac{(a_1 a_2\\ldots a_n)^{1/n}}{a_k}}$, we obtain the result. $\\Box$ This result can be used to", "condition, so that would mean I need to show that the best approximation is the one that minimizes the sup norm. – Samad Ahmed Apr 19 '18 at 1:59 • This answer of mine might be relevant: math.stackexchange.com/questions/1784262/… – marty cohen Apr 19 '18 at 4:32 First we consider the case $f(a) = f(b)=0$. Since $f''>0$, we have $f(x) <0$ and $f(x)$ attains a minimum at some $c\\in (a, b)$. Then it is easy to check that $L(x) = \\frac 12 f(c)$ (i.e. constant function) is the best linear approximation with $\\|f - L\\|_\\infty = \\frac 12 |f(c)|$ (you need only that $f(a) = f(b) = 0$ and $f(c)\\le f(x)\\le 0$ to show this). $$g(x) = f(x) - \\frac{f(b)-f(a)}{b-a}(x-a) + f(a).$$ Then $g'' = f'' >0$, $g(a) = g(b) = 0$. Thus $g$ is best approximated by constant function, and thus $f$ is best approximated by linear functions with slope $$\\frac{f(b) - f(a)}{b-a}.$$", "# For two events, G and Q, it is given that, P(G) = 2/5, P(Q) = 3/8 andP(G|Q) = 2/3 . If G̅ and Q̅ are the complementary events of G and Q, then what is $$\\rm P(\\frac{\\bar G}{\\bar Q} )$$equal to? 1. 31/25 2. 8/5 3. 21/40 4. 19/25 Option 4 : 19/25 ## Detailed Solution Concept: P(A/B) = P(A∩B) / P(B) $$\\rm \\overline{ P(A ∪B)}$$=$$\\rm { P(\\bar A ∩ \\bar B)}\\rm= 1-{ P(A ∪ B)}$$ P(A ∪ B) = P(A) +P(B) - P(A ∪ B) Calculation: Here, P(G) = 2/5, P(Q) = 3/8 and P(G|Q) = 2/3 P(G|Q) = P(G ∩ Q) / P(Q) = 2/3 ⇒ P(G ∩ Q) = 2/3 × 3/8 = 1/4 $$\\rm P(\\frac{\\bar G}{\\bar Q} )=\\frac{P(\\bar G∩ \\bar Q)}{P(\\bar Q)}=\\frac{\\overline {P( G∪ Q) }}{P(\\bar Q)}$$ $$\\rm =\\frac{1-P(G∪ Q)}{P(\\bar Q)}$$ P(Q̅) = 1 - P(Q) = 1-3/8 = 5/8 P(G ∪ Q) = P(G) + P(Q) - P(G∩Q) = 2/5 + 3/8 - 1/4 = 21/40 $$\\rm P(\\frac{\\bar G}{\\bar Q} )=\\frac{1-\\frac {21}{40}}{\\frac 58}$$ = 19/40 × 8/5 = 19/25 Hence, option (4) is correct.", "(H(p_1)/p_1 - E_2(g-1))(1-(1 - p_1)^g)$$ $$E_2(g) = (H(p_2)/p_2 - 2E_0(13))(1-(1 - p_2)^g)$$ Oh boy... With a CAS I found: $$E_0(13) =\\frac{1}{2P_0P_1P_2 + 1} \\left(\\frac{P_0H(p_0)}{p_0} - \\frac{P_0P_1H(p_1)}{p1} +\\frac{P_0P_1P_2H(p_2)}{p_2}\\right)$$ where $P_0 = 1-(1-p_0)^{13}$, $P_1 = 1-(1-p_1)^{12}$ and $P_2 = 1-(1-p_2)^{11}$. This is the expression we want to maximize. Numerically maximizing this expression gives $E_0(13) \\approx 6.72$ bits with: $$p_0 \\approx 0.1854365,\\quad p_1 \\approx 0.1422816,\\quad p_2 \\approx 0.0000899$$ This means that in order to get $\\log_2(64^{64}) = 384$ bits we expect to need 57.1 blocks, or 1714.3 seconds. This isn't even optimal yet, though. $p_0, p_1, p_2$ should actually be functions of $g$ (consider that if you are on your last guess in state $0$, then you can be really greedy and not get punished). I will see if I can improve my answer.", "of $q$ wrt $p$ is $\\frac{dq}{dp} = -35.$ Also, if $q = 700 - 35p,$ then $p = \\frac{700 - q}{35}$ and the derivative of $p$ wrt $q$ would be $\\frac{dp}{dq} = -\\frac{1}{35}$, but that particular derivative is not needed. • Sep 4th 2009, 12:07 AM el123 indeed you are right.I see my mistake. Thanks guys ...or girls. One more thing , what does the ...for $0 \\leq p\\leq 20$ mean? • Sep 4th 2009, 12:27 AM BobP It means that p is to be greater than or equal to zero, and less than or equal to 20. i.e. p has to lie within the range zero to 20 inclusive.", "line $$q = 0$$, p ranges between 0 and 1 that is $$0 \\leq p \\leq 1.$$ Therefore, $$P(p, 0) = 2p - 2p^{2}$$ Which represents downward parabola with maximum value at vertex $$(\\frac{1}{2}, 0)$$ $$P(\\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} - 2(0)(\\frac{1}{2}) = \\frac{1}{2}$$ Therefore, $$P(\\frac{1}{2}, 0) = \\frac{1}{2}$$ Also along line $$p + q = 1$$ where $$0 \\leq p \\leq 1,$$ $$P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)$$ $$= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \\leq p \\leq 1$$ $$P(p, 1 - p) = 2p - 2p^{2}$$ Which is downward parabola whose vertex is at $$(\\frac{1}{2}, \\frac{1}{2})$$ Therefore, maximum value is $$P(\\frac{1}{2}, \\frac{1}{2}) = 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} = \\frac{1}{2}$$ Therefore, comparing all values at boundary with value of P at the critical point The absolute maximum value of P(p, q) on D is $$\\frac{2}{3}.$$ Therefore, the maximum value of P is $$\\frac{2}{3}.$$ Conclusion: The maximum value of P is $$\\frac{2}{3}$$.", "square root sign in the numerator to be greater or equal to zero. Thus $x+3\\ge0$ $x\\ge-3$ Because of this last restriction, $x\\ne -8$ is eliminated from the restrictions. Therefore the domain of $f(x)=\\frac{\\sqrt{x+3}} {(x+8)(x-2)}$ is $x\\ge-3$, $x\\ne 2$ We have the function $g(x)=2-(x-7)^2$ which is a maximum function written in the vertex form. It is a maximum function because when we compare to the general vertex form, $g(x)=a(x-h)^2+k$ $a=-1$ The vertex of the graph is $V(7,2)$. The x-value of the vertex is $7$ For $x$, the graph is increasing and for $x\\:7$ the graph will be decreasing. $f(x)=4x+7$ and $g(x)=3x^2$ $(f+g)(x)=f(x)+g(x)$ $(f+g)(x)=4x+7+3x^2$ $(f+g)(x)=4x+7+3x^2$ We have $f(x)=\\frac{x-5}{8}$ $g(x)=8x+5$ The composition of the two functions $g(f(x))$ is given by; $g(f(x))=g(\\frac{x-5}{8})$ This implies that; $g(f(x))=8\\times (\\frac{x-5}{8})+5$ $g(f(x))=x-5+5$ $g(f(x))=x$ Given $f(g(x))=\\frac{9}{\\sqrt{5x+5}}$. Then $f(x)=\\frac{9}{\\sqrt{x} }$ and $g(x)=5x+5$ So that; $f(g(x))=f(5x+5)$ Which will give; $f(g(x))=\\frac{9}{\\sqrt{5x+5}}$. The dimension of the smallest region are $l=4$ km and $w=4km$ If the length and width are increasing at the rate of 3km\\sec, then after $x$ seconds, the dimensions will be, $l=4+3x$ km and $w=4+3x$km This means the area after $x$ seconds is given by; $(4+3x)(4+3x)$. If the area is at least 4 times its original size, then we can write the inequality; $(4+3x)(4+3x)\\ge 4(4\\times 4)$. $(4+3x)(4+3x)\\ge 64$ We expand to obtain; $9x^2+24x+16ge 64$ We simplify to obtain, $9x^2+24x-48ge", "# A 2-Variable Optimization From a China Competition ### Solution 1 Let's first remark that the function $f(s)=4s^3+s$ is strictly increasing on the real line. Define $x+y=s$ and $xy=p.$ Clearly, $s^2\\ge p,$ with equality only if $x=y=s$ and the constraint becomes $16s^3-12sp+s=15.$ If $s\\ge 0,$ then $-12s^3\\le 12sp,$ implying $4s^3+s\\le 15,$ or $\\displaystyle s\\le\\frac{3}{2}.$ If $s\\le 0,$ then $-12s^3\\ge -12sp,$ implying $4s^3+3\\le 15,$ which is impossible due to the above remark. It follows that the maximum of $x+y$ is $3,$ attained for $\\displaystyle x=y=\\frac{3}{2}.$ ### Solution 2 Let $x+y=p$ and $xy=q.$ By the AM-GM inequality, $p^2\\ge 4q.$ The constraint reduces to $4p^3-12pq+p-30=0.$ Now, if $p\\ge 0,$ then $p^3 \\ge 4pq,$ i.e., $3p^3-12pq\\ge 0,$ so that $p^3+p-30\\le 0.$ But $p^3+p-30=(p-3)(p^2+3p+10).$ Since for all real $p,$ $p^2+3p+10\\gt 0,$ $p^3+p-30\\le 0,$ only if $p\\le 3,$ with a maximum possible value of $p=3.$ If $p\\lt 0,$ we have instead $(p-3)(p^2+3p+10)\\gt 0,$ i.e., $p\\gt 3$ in contradiction with $p\\lt 0.$ Thus the maximum of $\\max (x+y)=3.$ It is attained when there's equality in $p^2\\ge 4q,$ i.e., when $x=y.$ Thus $\\displaystyle x=y=\\frac{3}{2}.$ ### Solution 3 Let $p=x+y$ and $q=x-y$. \\displaystyle \\begin{align} &x^3+y^3+\\frac{x+y}{4}=\\frac{15}{2} \\\\ &\\left(\\frac{p+q}{2}\\right)^3+\\left(\\frac{p-q}{2}\\right)^3+\\frac{p}{4}=\\frac{15}{2} \\\\ &p^3+p(3q^2+1)-30=0 \\\\\\\\ &q=\\pm\\sqrt{\\frac{(3-p)\\left[(p+3/2)^2+31/4\\right]}{3p}}. \\end{align} For $q$ to be real, $p\\gt 0$ and $3-p\\geq 0$, or $p\\lt 0$ and $3-p\\leq 0$. The second set has no real solution. The first", "can never attain a larger value than for $q = \\frac{r+2}{3}$. Inserting this into (I), the condition $\\left(\\frac{r+2}{3}\\right)^2 (2-r) + \\frac{r+2}{3} (3r^2 +8r-4) - 8r^2 > 0$ needs to be satisfied. But one can simplify this to $\\frac{8}{9} (1-r)^2 (r-2) > 0$, which can never hold for $0<r<1$. Case (b): $3r^2 + 8r - 4 < 0$ ($r <\\approx 0.43$). Then it is easy to see that (I) is a strictly convex function in $q$ over the relevant range, which attains a negative value for $q=0$, first decreases, and then increases. At best, there exists some critical threshold $\\tilde{q} < 1$ starting from which the function becomes positive. But in order for (II) to be satisfied, it cannot be the case that $q \\geq \\frac{r+2}{3}$. However, we know from case (a) that the function in (I) is certainly negative for $q = \\frac{r+2}{3}$. Thus, also case (b) both inequalities cannot hold at the same time. - Assuming $0 < q < 1$ and $0 < r < q$, show that $$q^2(2-r) + q(3r^2 + 8r-4) - 8r^2 > 0\\ \\ \\ \\tag{I}$$ $$\\text{and}\\;\\;\\;3q -r-2 < 0\\tag{II}$$ cannot hold at the same time. Strategy Hints: Take the given constraints on $q$ and $r$, assume BOTH inequalities hold, Note that $0 < r < q < 1$ so $q^2 <", "86 views Consider a stable system with transfer function,$$G(s)=\\frac{s^{p}+b_{1}s^{p-1}+\\dots +b_{p}}{s^{q}+a_{1}s^{q-1}+\\dots +a_{q}}$$, where $b_1, \\dots ,b_p$ and $a_1, \\dots ,a_q$ are real valued constants.The slope of the Bode log magnitude curve of $G(s)$ converges to $-60 \\text{dB/decade}$ as $ω\\to \\infty$.A possible pair of values for $p$ and $q$ is 1. $p=0$ and $q=3$ 2. $p=1$ and $q=7$ 3. $p=2$ and $q=3$ 4. $p=3$ and $q=5$", "g_1'(x)$ on this interval. Furthermore, $g_1'$ is continuous and is an increasing function on $[1.5, 2]$ and thus the maximum is achieved at the right endpoint of this interval, that is $\\max_{1.5 ≤x ≤ 2} \\mid g_1'(x) \\mid = \\max_{1.5 ≤ x ≤ 2} g_1'(x) = \\frac{3}{2} (2)^2 \\mid = 6$. (1) \\begin{align} \\quad \\lambda = \\max_{1.5 ≤ x ≤ 2} \\mid g_1'(x) \\mid = 6 > 1 \\end{align} As we see, the choice of $x = g_1(x)$ does not satisfy the condition for convergence. Now note that since $g_2(x) = (2x + 1)^{1/3}$ we have that $g_2'(x) = \\frac{2}{3} (2x + 1)^{-2/3}$. Once again we see that $g_2'(x) > 0$ on the interval $[1.5, 2]$ and so $\\mid g_2'(x) \\mid = g_2'(x)$ on this interval. Furthermore, $g_2'$ is continuous and is a decreasing function on $[1.5, 2]$ and thus the maximum is achieved at the left endpoint of this interval, that is $\\max_{1.5 ≤ x ≤ 2} \\mid g_2'(x) \\mid = \\max_{1.5 ≤ x ≤ 2} \\mid g_2'(x) \\mid = \\frac{2}{3} (2(1.5) + 1)^{-2/3} = 0.264566… < 1$ and o $x = g_2(x)$ is a good choice for applying the Fixed Point method. We will now apply the fixed point method to $x = g_2(x) = (2x + 1)^{1/3}$. The first iteration using the initial approximation $x_0", "\\cup6)$ = $P(3)+P(6)$ = $\\frac{1}{3}$ $(2)$ $P(1)$ = $\\frac{1}{6}$ $(3)$ $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)$ = $1$ $(4)$ Adding equation $1$, $2$ and $3$ $P(1)+P(2)+P(3)+P(4)+2P(6)$ = $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}$ $P(1)+P(2)+P(3)+P(4)+2P(6)$ = $1$ using equation $4$ and above equation, we can say that $P(5)=P(6)$ and it is given that $P(3)+P(6)$ = $\\frac{1}{3}$ $P(6)$ = $\\frac{1}{3} - P(3)$ 0 So how 1/3? 5 P(5) = P(6) = 1/3 - P(3) min value of P(5) : when P(3) has max value 1/3 .so min value of P(5) =1/3 -1/3 =0 max value of P(5):when P(3)​​​​​​​ has min value 0 .so max value of P(5) =1/3 -​​​​​​​0 =1/3 So, 0≤ P(5)≤1/3 or we can say P(5) ≤ 1/3. Given, P(2,4,6) =1/2 P(3,6)= 1/3 P(1)= 1/6 So, P(1,3,5)=1/2 P(3,5)=P(1,3,5)-P(1) = 1/2 - 1 /6 = 1/3 P(3,5)=1/3 For P(5) to have the maximum probablity, P(3) should be 0 P(5)<=1/3 0 nice D is the correct answer. so lets find out how. first thing keep in mind is that the chances are not equally likely. so probability can be go like 99:1 also . so the given data is P(2,4,6) =1/2 so P(1,3,5)= 1/2. here it does not mean P(1)=P(3). but we an surely say the sum will be 1/2. so P(3,6)= 1/3 again sum can be 1/6 . individual probability may not be equal . so lets find the", "so differentiate that: $$\\frac{d}{dt}\\frac{dP}{dt}= \\frac{d^2P}{dt^2}$$$$= C\\frac{d(P(ln(K)- ln(P))}{dt}= 0$$ 3. Oct 11, 2010 ### hover Ok. Solving that derivative I get t = -7.33 and in the end P has a max slope at P= 367.879. These are right. Thanks!", "as $q=\\frac{p}{5}$q=p5. Notice! It doesn't matter if the first row increases by more than $1$1 at a time, or if neither of the rows start from $1$1. We can still find a relation by looking at each pair of values. Worked Examples Question 1 Write an equation for $g$g in terms of $f$f Question 2 Write an equation for $g$g in terms of $f$f.", "equations, we find $$E_0= \\frac{3p^4-14p^3+26p^2-24p+8}{(p-1)^4}.$$ The derivative of this is $$E_0'(p) =\\frac{p^3-5p^2+10p-4}{(p-1)^5}$$ and has a single root in the interval $[0,1]$ at $p\\approx 0.522$, which corresponds to a local maximum value of $\\approx15.09$.", "about $\\frac{p c}{|p-c|}$? This will grow (to infinity) for $c$ and $p$ being closer together and will also grow if $p/c=constant$ with increased $p c$. All support on one side will mean that $pc$ is low (or 0) Maybe it is not perfect in its general behaviour, but the limiting behaviour seems to be right, so you can probably go from here. - I'm not sure 1 upvote vs. 1 downvote resulting in the same amount as 100 vs. 100 (both of which are \"infinite\") is a good idea. But thanks anyway. – Joe Z. Mar 2 '13 at 8:31 A little adjustment like $\\frac{pc}{|p-c|+1}$ would avoid infinity. – Hagen von Eitzen Mar 2 '13 at 9:43 The formula I came up with myself was: $\\displaystyle \\frac{\\min(p,c)^2}{\\max(p, c)}$ I called it the \"geometric progression\" algorithm, because it represents the next term of the geometric progression lower than $\\max(p,c)$ and $\\min(p, c)$. In this way, as $p$ increases past $c$, $f(p, c)$ will get smaller, reaching its maximum of $c$ when $p = c$. It also has the additional scaling property that the number of votes on $\\max(p,c)$ varies inversely with $f(p, c)$. - I created a page where you can play with various upvote and downvote algorithms, with this formula on it. – Joe Z. Mar 2 '13", "decreasing b) The local maximum and minimum points of $g(x)$ c) The intervals where $g(x)$ is concave up and concave down I have figured out the intervals where g(x) is increasing and decreasing as: . . $\\begin{array}{cccc}g'(x) > 0 &\\text{when }x < -5\\,\\text{ or }\\,x > 1 & g(x)\\text{ is increasing} \\\\ g '(x) < 0 & \\text{when }-5 < x < 1 & g(x)\\text{ is decreasing} \\end{array}$ . Right! I have also attached what I have graphed as $g(x)$ and $g''(x)$ . Correct! With the given points of the parabola: . $(-5,0),(0,-5), (1,0)$ . . we can determine its equation: . $g'(x) \\:=\\:x^2+4x-5$ Then: . $g(x)\\;=\\;\\int(x^2+4x-5)\\,dx \\;=\\;\\tfrac{1}{3}x^3 + 2x^2 - 5x + C$ The extreme values occur where $g'(x) = 0 \\quad\\Rightarrow\\quad x \\:=\\:-5,\\:1$ The second derivative is: . $g''(x) \\:=\\:2x + 4$ . . $\\begin{array}{cccccccc}g''(\\text{-}5) &=& 2(\\text{-}5)+4 &=& -6 & \\text{concave down: }\\cap & \\text{maximum} \\\\ g''(1) &=& 2(1)+4 &=& +6 & \\text{concave up: } \\cup & \\text{minimum}\\end{array}$ Therefore: . $\\begin{Bmatrix}\\text{maximum at:}& \\left(-5,\\:C + \\frac{100}{3}\\right) \\\\ \\\\[-3mm] \\text{minimum at:} & \\left(1,\\:C-\\frac{8}{3}\\right) \\end{Bmatrix}$ Inflection points: . $g''(x) \\:=\\:0 \\quad\\Rightarrow\\quad 2x + 4 \\:=\\:0 \\quad\\Rightarrow\\quad x \\:=\\:-2$ Therefore: . $\\begin{Bmatrix}g''(x) > 0 & \\text{when }x > -2 & \\text{concave up} \\\\ \\\\[-3mm] g''(x) < 0 & \\text{when }x < -2 & \\text{concave down} \\end{Bmatrix}$ 3. Originally Posted by", "is a closed set. Since we assume $$g$$ has continuous partials on a ball for the Lagrange multipliers theorem, all $$g$$'s that we consider will be continuous. Thus all that is left is to show that the set of points satisfying $$g=0$$ is bounded. This must be done on a case by case basis. For example, if $$g(x,y,z) = x^2+y^2+z^2-1$$, then the set of points satisfying $$g=0$$ is a sphere of radius 1. This is obviously a bounded set, and hence we know that when doing a constrained optimization problem with this $$g$$, there will be a constrained absolute minimum and constrained absolute maximum. 11/08/15. • 5E: 14.8: 3-10, 15-17, 18-19, 23, 44, 14 Review: Concept check 1-19, True/False 1-12 • 6E: 15.8: 3-10, 15-17, 18-19, 25, 46, 15 Review: Concept check 1-19, True/False 1-12 • 7E: 14.8: 3-10, 15-17, 19-20, 27, 48, 14 Review: Concept check 1-19, True/False 1-12 • Fix $$p,q >1$$ such that $$\\frac{1}{p}+\\frac{1}{q} = 1$$. Maximize $$f(x,y) = x y$$ in first quadrant subject to the constraint $$\\frac{x^p}{p} + \\frac{y^q}{q}=C$$. Conclude for all $$x,y$$ that $$|xy| \\leq \\frac{|x|^p}{p} + \\frac{|y|^q}{q}$$. Plug in $$g(t)/(\\int_a^b g(s)^p\\,ds)^{1/p}$$ for $$x$$, and $$h(t)/(\\int_a^b h(s)^q\\,ds)^{1/q}$$ for $$y$$ and then integrate in $$t$$ to find $\\int_a^b |g(t) h(t)|\\,dt \\leq \\left( \\int_a^b |g(t)|^p\\,dt\\right)^{1/p} \\left(\\int_a^b |h(t)|^q\\,dt\\right)^{1/q}.$ This is a special case of Hölder's", "years was the car valued at $$4,000$$? 1. $$g (−4) = 3, g (2) = 0$$, and $$g (6) = −2$$. 3. $$g (10) = −5, g (5) = 0$$ and $$g (15) = 0 , g (−5) = 10$$ and $$g (25) = 10$$ 5. $$g (−2) = −5, g (−3) = −4$$ and $$g (−1) = −4 , g (−5) = 4$$ and $$g (1) = 4$$ 7. $$g (−2) = −4, g (−1) = 3, g (0) = 4$$ 9. $$g (−10) = −10$$ and $$g (5) = −10 ; g (−5) = 5$$ and $$g (10) = 5$$ 11. $$5,000$$ 13. $$10,000$$ Exercise $$\\PageIndex{9}$$ Given the linear function defined by $$f(x)=2x-5$$, simplify the following. 1. $$f ( 5 ) - f ( 3 )$$ 2. $$f ( 0 ) - f ( 7 )$$ 3. $$f ( x + 2 ) - f ( 2 )$$ 4. $$f ( x + 7 ) - f ( 7 )$$ 5. $$f ( x + h ) - f ( x )$$ 6. $$\\frac { f ( x + h ) - f ( x ) } { h }$$ 7. Simplify $$\\frac { c ( x + h ) - c ( x ) } { h }$$ given $$c ( x ) = 3" ]

[ "a given ${p}$, we can differentiate ${g(x)=\\frac{(p-x)^2}{x^2}}$ with respect to ${x}$ to see that ${g'(x)=0}$ only when ${x=p}$, which is when the relative error is zero. Therefore for ${\\frac{|p-x|}{x}}$ to attains its maximum, ${x}$ must be equal to ${a}$ or ${b}$. Therefore $\\displaystyle F(p)=\\max_{x\\in[a, b]}\\{\\frac{|p-x|}{x}\\}=\\max\\{\\frac{p-a}{a}, \\frac{b-p}{b}\\}.$ Then $\\displaystyle \\frac{p-a}{a}\\geq \\frac{b-p}{b} \\Leftrightarrow p\\geq \\frac{2ab}{a+b}\\in [a, b] \\textrm{ and }\\frac{p-a}{a}\\leq \\frac{b-p}{b} \\Leftrightarrow p\\leq \\frac{2ab}{a+b}\\in [a, b].$ Let ${H=\\frac{2ab}{a+b}}$, which is the harmonic mean of ${a}$ and ${b}$ with equal weights. It is easy to see that ${\\frac{p-a}{a}}$ is increasing for ${p\\geq H}$ and ${\\frac{b-p}{b}}$ is decreasing for ${p\\leq H}$. Therefore ${F(p)}$ attains its minimum at ${p=H=\\frac{2ab}{a+b}}$. $\\Box$ Proposition 5 (The Geometric Mean as a Minimum) The geometric mean has the representation $\\displaystyle \\{ \\prod_{k=1}^n a_k \\}^{1/n}=\\min\\{\\frac{1}{n} \\sum_{k=1}^n a_k x_k:(x_1,\\ldots, x_n)\\in D \\},$ where ${D}$ is the region of ${\\mathbb{R}^n}$ defined by $\\displaystyle D=\\{(x_1, \\ldots, x_n):\\prod_{k=1}^n x_k=1, x_k\\geq 0, k=1, \\ldots , n\\}.$ Proof: Assume ${a_k\\neq 0}$ for all ${k}$, by AM-GM inequality, $\\displaystyle \\frac{1}{n} \\sum_{k=1}^n a_k x_k\\geq (\\prod_{k=1}^n a_k x_k)^{1/n}=(\\prod_{k=1}^n a_k)^{1/n} \\textrm{ as } (x_1, \\ldots, x_n)\\in D.$ The AM-GM inequality also tells us that the equality holds if and only if ${a_1x_1=\\ldots =a_nx_n=c }$ then ${x_k=c/a_k}$ and ${1=c^n/(a_1 a_2\\ldots a_n)\\Rightarrow c=(a_1 a_2\\ldots a_n)^{1/n}}$. So choosing ${x_k=\\frac{(a_1 a_2\\ldots a_n)^{1/n}}{a_k}}$, we obtain the result. $\\Box$ This result can be used to", "# A 2-Variable Optimization From a China Competition ### Solution 1 Let's first remark that the function $f(s)=4s^3+s$ is strictly increasing on the real line. Define $x+y=s$ and $xy=p.$ Clearly, $s^2\\ge p,$ with equality only if $x=y=s$ and the constraint becomes $16s^3-12sp+s=15.$ If $s\\ge 0,$ then $-12s^3\\le 12sp,$ implying $4s^3+s\\le 15,$ or $\\displaystyle s\\le\\frac{3}{2}.$ If $s\\le 0,$ then $-12s^3\\ge -12sp,$ implying $4s^3+3\\le 15,$ which is impossible due to the above remark. It follows that the maximum of $x+y$ is $3,$ attained for $\\displaystyle x=y=\\frac{3}{2}.$ ### Solution 2 Let $x+y=p$ and $xy=q.$ By the AM-GM inequality, $p^2\\ge 4q.$ The constraint reduces to $4p^3-12pq+p-30=0.$ Now, if $p\\ge 0,$ then $p^3 \\ge 4pq,$ i.e., $3p^3-12pq\\ge 0,$ so that $p^3+p-30\\le 0.$ But $p^3+p-30=(p-3)(p^2+3p+10).$ Since for all real $p,$ $p^2+3p+10\\gt 0,$ $p^3+p-30\\le 0,$ only if $p\\le 3,$ with a maximum possible value of $p=3.$ If $p\\lt 0,$ we have instead $(p-3)(p^2+3p+10)\\gt 0,$ i.e., $p\\gt 3$ in contradiction with $p\\lt 0.$ Thus the maximum of $\\max (x+y)=3.$ It is attained when there's equality in $p^2\\ge 4q,$ i.e., when $x=y.$ Thus $\\displaystyle x=y=\\frac{3}{2}.$ ### Solution 3 Let $p=x+y$ and $q=x-y$. \\displaystyle \\begin{align} &x^3+y^3+\\frac{x+y}{4}=\\frac{15}{2} \\\\ &\\left(\\frac{p+q}{2}\\right)^3+\\left(\\frac{p-q}{2}\\right)^3+\\frac{p}{4}=\\frac{15}{2} \\\\ &p^3+p(3q^2+1)-30=0 \\\\\\\\ &q=\\pm\\sqrt{\\frac{(3-p)\\left[(p+3/2)^2+31/4\\right]}{3p}}. \\end{align} For $q$ to be real, $p\\gt 0$ and $3-p\\geq 0$, or $p\\lt 0$ and $3-p\\leq 0$. The second set has no real solution. The first", "$\\lim\\limits_{x\\to-\\infty}f(x)=0$ and $$f'(x)=e^x\\left(p(x)-p'(x)-p''(x)+p'''(x)\\right)\\ge 0,$$ i.e. $f$ is increasing. It follows that $f\\ge 0$, and hence $$p-2p'+p''\\ge 0.$$ Define $$g(x)=e^{-x}\\left(p(x)-p'(x)\\right).$$ Then $\\lim\\limits_{x\\to+\\infty}g(x)=0$ and $$g'(x)=-e^{-x}\\left(p(x)-2p'(x)+p''(x)\\right)\\le 0,$$ i.e. $g$ is decreasing. It follows that $g\\ge 0$, and hence $$p-p'\\ge 0.$$ Define $$h(x)=e^{-x}p(x).$$ Then $\\lim\\limits_{x\\to+\\infty}h(x)=0$ and $$h'(x)=-g(x)\\le 0,$$ i.e. $h$ is decreasing. Therefore, either $h>0$ or $h\\equiv 0$. The conclusion follows. Proof 2: Denote $q=p-p'$. Then $$p+p'''\\ge p'+p''\\iff q-q''\\ge 0.$$ It implies that $\\deg(q-q'')=\\deg q=\\deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then $$q(x)\\ge q(x_q)\\ge q''(x_q)\\ge 0,$$ i.e. $$p\\ge p'.$$ Similarly, $p$ also has a global minimum value point, say $x_p$. Then $$p(x_p)\\ge p'(x_p)= 0.$$ If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $p\\equiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $k\\ge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)\\ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then $$p(x)-p'(x)=-kr(x_p)(x-x_p)^{k-1}+O((x-x_p)^k)$$ cannot be always non-negative, a contradiction. • @GitGud: Since $f$ is increasing and $f(-\\infty)=0$, $f\\ge 0$. – 23rd Jul 31 '13 at 19:42 • Of course, thanks. – Git Gud Jul 31 '13 at 19:44 • @GitGud: You are welcome. – 23rd Jul 31 '13 at 19:47 •", "= 0$$, q ranges between 0 and 1 that is $$0 \\leq q \\leq 1.$$ Therefore, $$P(0, q) = 2q - 2q^{2}$$ Which represents downward parabola with maximum value at vertex $$(0, \\frac{1}{2})$$ $$P(\\frac{1}{2}, 0) = 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} + 2(0) - 2(0)^2 - 2(0)(\\frac{1}{2}) = \\frac{1}{2}$$ Therefore, $$P(0, \\frac{1}{2}) = \\frac{1}{2}$$ Similarly, along the line $$q = 0$$, p ranges between 0 and 1 that is $$0 \\leq p \\leq 1.$$ Therefore, $$P(p, 0) = 2p - 2p^{2}$$ Which represents downward parabola with maximum value at vertex $$(\\frac{1}{2}, 0)$$ $$P(\\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} - 2(0)(\\frac{1}{2}) = \\frac{1}{2}$$ Therefore, $$P(\\frac{1}{2}, 0) = \\frac{1}{2}$$ Also along line $$p + q = 1$$ where $$0 \\leq p \\leq 1,$$ $$P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)$$ $$= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \\leq p \\leq 1$$ $$P(p, 1 - p) = 2p - 2p^{2}$$ Which is downward parabola whose vertex is at $$(\\frac{1}{2}, \\frac{1}{2})$$ Therefore, maximum value is $$P(\\frac{1}{2}, \\frac{1}{2}) = 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} = \\frac{1}{2}$$ Therefore, comparing all values at boundary with value of P at the critical point The absolute maximum value of", "signs of $X_1$ and $X_2$ are opposite, is $2$ if exactly one is equal to $0$, and is zero otherwise. Therefore ($q=1-3p$): $$E[Y] = 4\\times 2\\times 2p^2 + 2\\times 2\\times (1-3p)3p=4p(3-5p).$$ Let's look at possible values of $E[Y]$. Clearly it's always nonnegative. -When $p=0$, $E[Y]=0$ (because $X=0$ a.s.). -The maximum of $4p(3-5p)$ over all $p\\in [0,1]$ is attained halfway between the two roots $0$ and $3/5$, that is at $p=3/10$. Since this choice of $p$ is in the allowed range for $p$, $[0,1/3]$, the maximal value for $E[Y]$ is $\\frac{12}{10}(3-\\frac{15}{10})=1.8$. Bottom line, $E[Y]$ can attain any value in $[0,1.8]$ and only values in this interval.", "line $$q = 0$$, p ranges between 0 and 1 that is $$0 \\leq p \\leq 1.$$ Therefore, $$P(p, 0) = 2p - 2p^{2}$$ Which represents downward parabola with maximum value at vertex $$(\\frac{1}{2}, 0)$$ $$P(\\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} - 2(0)(\\frac{1}{2}) = \\frac{1}{2}$$ Therefore, $$P(\\frac{1}{2}, 0) = \\frac{1}{2}$$ Also along line $$p + q = 1$$ where $$0 \\leq p \\leq 1,$$ $$P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)$$ $$= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \\leq p \\leq 1$$ $$P(p, 1 - p) = 2p - 2p^{2}$$ Which is downward parabola whose vertex is at $$(\\frac{1}{2}, \\frac{1}{2})$$ Therefore, maximum value is $$P(\\frac{1}{2}, \\frac{1}{2}) = 2(\\frac{1}{2}) - 2(\\frac{1}{2})^{2} = \\frac{1}{2}$$ Therefore, comparing all values at boundary with value of P at the critical point The absolute maximum value of P(p, q) on D is $$\\frac{2}{3}.$$ Therefore, the maximum value of P is $$\\frac{2}{3}.$$ Conclusion: The maximum value of P is $$\\frac{2}{3}$$.", "closed interval $$[0,1]$$ and $$g(0)$$ and $$g(1)$$ have opposite signs, it follows from Bolzano’s Theorem that there is a point $$c$$ between 0 and 1 such that $$g(c)=f(c)-c=0$$ or $$f(c)=c$$. The geometric interpretation is simple. If $$f(0)\\neq0$$ and $$f(1)\\neq1$$, then the graph of $$f$$ has to cut the line $$y=x$$ at some point between 0 and 1 (see the following figures). Figure 7: If $f$ is continuous on $[ 0,1 ]$ and $0 \\leq f( x ) \\leq 1$ , then the equation $f( x ) =x$ has at least one solution ( $0 \\leq x \\leq 1$) . ## The Extreme Value Theorem Definition 1: Let $$f$$ be a function defined on a set $$E$$. We say $$f$$ has an absolute maximum on (or in) $$E$$, if there is at least one point $$p$$ in $$E$$ such that $$f(x)\\leq f(p)$$ for every $$x$$ in $$E$$. In this case, we say $$p$$ is the point of absolute maximum and $$f(p)$$ is the absolute maximum value (or simply maximum) of $$f$$ on $$E$$. Similarly we say $$f$$ has an absolute minimum on $$E$$, if there exists a point $$q$$ in $$E$$ such that $$f(q)\\leq f(x)$$ for all $$x$$ in $$E$$. In this case, we say $$q$$ is the point of absolute minimum and $$f(q)$$ is the minimum value (or", "Let $$M=$$ the largest value of $$f(x)$$ and $$N=$$ the smallest value of $$f(x)$$ for $$x \\in[0,1] .$$ Show that $$M-N>\\frac{1}{12}$$ for any $$a>0$$. Solution We have $$f(0)=1, f(1)=a+\\frac{1}{2}$$ and $$f^{\\prime}(x)=a-\\frac{1}{(x+1)^{2}}$$ Over the interval $$[0,1],$$ the value of $$f^{\\prime}(x)$$ increases from $$a-1$$ at $$x=0$$ to $$a-\\frac{1}{4}$$ at $$x=1$$o What happens to the sign of $$f^{\\prime}(x)$$? We do a case by case analysis. Case 1. Suppose $$a \\leq 1 / 4 .$$ Because $$1 /(x+1)^{2} \\geq 1 / 4$$ on the interval $$[0,1], f^{\\prime}(x) \\leq 0,$$ so the maximum is at 0 and the minimum is at $$x=1$$. So the difference is $$1-(1 / 2+a)=$$ $$1 / 2-a \\geq 1 / 4 \\geq 1 / 12$$ Case 2. Suppose $$a \\geq 1$$. Then $$f^{\\prime}(x) \\geq 0$$ on the interval [0,1] , so maximum is at 1 and minimum at 0. We get $$a+1 / 2-1=a-1 / 2 \\geq 1 / 2 \\geq 1 / 12$$ Case 3. Suppose $$1 / 4 \\leq a \\leq 1 .$$ Now $$f^{\\prime}(x)=0$$ at $$\\tilde{x}=\\frac{1}{\\sqrt{a}}-1 .$$ For this range of $$a, \\tilde{x} \\in[0,1]$$ In the interval $$[0, \\tilde{x}], f^{\\prime}(x) \\leq 0$$ and in the interval $$[\\tilde{x}, 1], f^{\\prime}(x) \\geq 0 .$$ Now we make two sub-cases depending on the endpoint at which the maximum occurs. Case 3a. Suppose $$1 / 4 \\leq a \\leq", "# For two events, G and Q, it is given that, P(G) = 2/5, P(Q) = 3/8 andP(G|Q) = 2/3 . If G̅ and Q̅ are the complementary events of G and Q, then what is $$\\rm P(\\frac{\\bar G}{\\bar Q} )$$equal to? 1. 31/25 2. 8/5 3. 21/40 4. 19/25 Option 4 : 19/25 ## Detailed Solution Concept: P(A/B) = P(A∩B) / P(B) $$\\rm \\overline{ P(A ∪B)}$$=$$\\rm { P(\\bar A ∩ \\bar B)}\\rm= 1-{ P(A ∪ B)}$$ P(A ∪ B) = P(A) +P(B) - P(A ∪ B) Calculation: Here, P(G) = 2/5, P(Q) = 3/8 and P(G|Q) = 2/3 P(G|Q) = P(G ∩ Q) / P(Q) = 2/3 ⇒ P(G ∩ Q) = 2/3 × 3/8 = 1/4 $$\\rm P(\\frac{\\bar G}{\\bar Q} )=\\frac{P(\\bar G∩ \\bar Q)}{P(\\bar Q)}=\\frac{\\overline {P( G∪ Q) }}{P(\\bar Q)}$$ $$\\rm =\\frac{1-P(G∪ Q)}{P(\\bar Q)}$$ P(Q̅) = 1 - P(Q) = 1-3/8 = 5/8 P(G ∪ Q) = P(G) + P(Q) - P(G∩Q) = 2/5 + 3/8 - 1/4 = 21/40 $$\\rm P(\\frac{\\bar G}{\\bar Q} )=\\frac{1-\\frac {21}{40}}{\\frac 58}$$ = 19/40 × 8/5 = 19/25 Hence, option (4) is correct.", "# 2003_Question4 (i) Note that f'(x) = 2x - 2p = 2(x - p) so f'(x) = 0 iff x = p. So there is a stationary point in 0 < x < 1 iff 0 < p < 1. (ii) If p \\ge 1, then f'(x) < 0 for 0 \\le x < 1. So f is decreasing on 0 \\le x \\le 1. So the minimum value attained is at x = 1, ie. is f(1) = 1 - 2p + 3 = 4 - 2p. (iii) If p \\le 0, then f'(x) \\ge 0. So f is increasing on 0 \\le x \\le 1. So the maximum value is attained at x = 0, so m = 3. Alternatively note that if p \\le 0, -2p \\ge 0, so f(x) = x^2 - 2px + 3 \\ge 3 with equality at x = 0. (iv) Note that in general that because f'(x) = 2(x - p), we have that f is decreasing for x < p and increasing for x > p. So, a minimum is attained at x = p. (alternatively release that the second derivative is negative at x = p) So the minimum value in this case is f(p) = p^2 - 2p^2 + 3 = 3 - p^2. (v) We know", "than or equal to $p^3+(3-p)^3+(7p-9)(\\frac{3-p}{2})^2$, equality when $q=r$ if $p\\geq 9/7$ and less than or equal to $p^3+(3-p)^3+(7p-9)(1.5)(1.5-p)$, equality when one of $q, r$ equals $1.5$ if $p\\leq 9/7$. Note that among $p, q, r$, the smallest number has to be less than or equal to $1$. Without loss of generality $p\\leq q \\leq r$. Then $p\\leq 9/7$. For $p^3+q^3+r^3+4pqr$ to be maximal, one of $q, r$ is 1.5. As $r$ is the largest, $r=1.5$. Finally, note that $3/2>9/7$, so we can apply the other case, replacing $r$ with $p$ (which is fine since $p^3+q^3+r^3+4pqr$ is symmetric over the 3 terms), concluding that if $r=3/2$, $p^3+q^3+r^3+4pqr$ is maximal when $p=q=3/4$. Therefore, the maximum achievable is when one of $p, q, r$ is $3/2$ and the other 2 are $3/4$, to get a value of $\\frac{243}{32}$. • Sure. q+r = 5-p, but what about qr? – Rick Dec 17 '17 at 17:39 • You should take some time to think about it. AM-GM might be useful. – the4seasons Dec 17 '17 at 17:43 • @the4seasons I wanted to down-vote the Macavity's solution and erroneously down-voted your. I tried to change it, but it's impossible. I am sorry! – Michael Rozenberg Dec 18 '17 at 6:58 Let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$. Hence, the condition does not depend on $w^3$ and", "Helper Oops! The interval is from 0 to t! Pasmith, thanks, I see that now but, I'm not too sure about how to find where the local maximum is located (other than computing f''(t) which seems really complicated). f is increasing when $(t + 5)(t + 7)$ is positive, ie when $t \\geq -5$ and $t \\geq -7$ (so that $t \\geq -5$) or when $t \\leq -5$ and $t \\leq -7$ (so that $t \\leq -7$) and decreasing when $(t + 5)(t + 7)$ is negative, ie. when $t \\geq -5$ and $t \\leq -7$ (which is impossible) or when $t \\leq -5$ and $t \\geq -7$ (so that $-7 \\leq t \\leq -5$). Thus f is increasing when $t \\leq -7$, decreasing when $-7 \\leq t \\leq -5$, and increasing again when $t \\geq -5$. What does that tell you about $f(-7)$ and $f(-5)$? Homework Helper Dearly Missed Oops! The interval is from 0 to t! Pasmith, thanks, I see that now but, I'm not too sure about how to find where the local maximum is located (other than computing f''(t) which seems really complicated). CAF123, is there some trick to exploit (because, when I put this into software, I get some gigantic expression)? Well, computing f''(t) is likely the easiest way to go: just evaluate", "# If $f \\colon [-1,1] \\to \\mathbb R$ satisfies $\\vert f'' \\vert \\le k$ then $\\vert f \\vert \\le \\frac{k}{2}$ Let $f \\colon [-1,1] \\to \\mathbb R$ be a twice differentiable function s.t. $f(-1)=f(1)=0$ and there exists $k>0$ s.t. $\\vert f''(x) \\vert \\le k$ for every $x \\in [-1,1]$. Show that $$\\max_{[-1,1]}\\vert f \\vert \\le \\frac{k}{2}.$$ The book suggests: take $x_0\\in (-1,1)$ such that $\\vert f(x_0)\\vert$ is maximum and use Taylor. Following this hint, I write: $$f(x_0+h)=f(x_0) + \\frac{f''(\\xi)}{2}h^2$$ The linear term, $f'(x_0)=0$ (since the point is in the interior of $[-1,1]$ and is a maximum or a minimum for $f$). Now how can I conclude? I can't see how to use the hypotesis $f(-1)=f(1)=0$. - You are missing an integral term in your Taylor expansion. – vanna Sep 14 '12 at 17:07 @vanna Am I? I used Taylor expansion with Lagrange remainder. Did I have to add an integral remainder? My book doesn't mention it... – Romeo Sep 14 '12 at 17:09 Sorry about that I took $\\xi$ for $x_0$, your expression is correct. – vanna Sep 14 '12 at 17:13 Take $x_0$ such that $|f(x_0)$ is maximized, and note that $f'(x_0)=0$. Using Taylor, we have $$f(x_0+h)=f(x_0)+f'(x_0)h+R(h)\\text{ where } |R(h)|\\leq \\left|\\frac{\\max_{x\\in [-1,1]}|f''(x)|}{2}h^2\\right|$$ and since $f'(x_0)=0$ this simplifies to $f(x_0+h)=f(x_0)+R(h)$. If we take $h$ such that $|h|\\leq 1$", "must drop) • Thanks. I agree with the maximum, but for the minimum, although $f(x)+g(x)$ is symmetric relative to $x=\\frac{\\pi}4$ and it decreases when it goes off from either $x=0$ or $x=\\frac\\pi2$, the minimum could still happen at somewhere else in $(0,\\frac\\pi2)$ not exactly at $x=\\frac{\\pi}4$ (for example $x=\\frac18\\pi$ or $x=\\frac38\\pi$) and $x=\\frac{\\pi}4$ could be a local maximum? Apr 4, 2019 at 21:59 • You are right, I think adding the symmetry argument around $\\pi/4$ could do the trick. Apr 5, 2019 at 18:22 I think you can use a lemma that if $$f(x)$$ is continuous and monotonic, then $$f(x)+f(1-x)$$ is monotonic on the interval $$\\left(0,\\frac{1}{2}\\right)$$ UPD: $$\\cos(x)$$ is a convex function on $$(0, \\pi)$$, so you can use some kind of Jensen's inequality. If you have $$a then $$f(a)+f(d) < f(c)+f(d)$$ Jensen's inequality • This lemma is false: Consider $f(x)$ given by $2x^2$ on $[0,\\frac12]$ and by $x$ on $[\\frac12,1]$; then $f(x) + f(1-x) = 2x^2 - x + 1$ on $[0,\\frac12]$, which has a minimum at $x=\\frac14$. Apr 4, 2019 at 6:28 • @FredH found this counteexample, also. May be There should be a monotonous derivative for lemma to hold. Apr 4, 2019 at 9:01 First find $$f(x +h) - f (x)$$. Then check if $$f(x) >0$$ or $$<0$$, the function will increase or decrease respectively.", "(H(p_1)/p_1 - E_2(g-1))(1-(1 - p_1)^g)$$ $$E_2(g) = (H(p_2)/p_2 - 2E_0(13))(1-(1 - p_2)^g)$$ Oh boy... With a CAS I found: $$E_0(13) =\\frac{1}{2P_0P_1P_2 + 1} \\left(\\frac{P_0H(p_0)}{p_0} - \\frac{P_0P_1H(p_1)}{p1} +\\frac{P_0P_1P_2H(p_2)}{p_2}\\right)$$ where $P_0 = 1-(1-p_0)^{13}$, $P_1 = 1-(1-p_1)^{12}$ and $P_2 = 1-(1-p_2)^{11}$. This is the expression we want to maximize. Numerically maximizing this expression gives $E_0(13) \\approx 6.72$ bits with: $$p_0 \\approx 0.1854365,\\quad p_1 \\approx 0.1422816,\\quad p_2 \\approx 0.0000899$$ This means that in order to get $\\log_2(64^{64}) = 384$ bits we expect to need 57.1 blocks, or 1714.3 seconds. This isn't even optimal yet, though. $p_0, p_1, p_2$ should actually be functions of $g$ (consider that if you are on your last guess in state $0$, then you can be really greedy and not get punished). I will see if I can improve my answer.", "interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3. Just one little query regarding the $\\|f-z\\|=\\sup_{t\\in[0,1]}|f(t)-z(t)|=\\sup_{t\\in[0,1]}|t^3+3-5t|=3$ You put in 0 and you are left with 3. If you put in 1 we get -1. So we take the highest of those two because $sup_{t \\in [0,1]} | f-z|=max\\{t_1,t_2\\}$ or something like this? Emeritus Sci Advisor PF Gold P: 9,018 Quote by bugatti79 So if I had different function like f-z = t^3-t^2 say, then the derivative of this wrt t and subbing in 0 we would get 0 which would impy the max is at t=0? I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies $(f-z)(t)=t^3-t^2$ for all $t\\in[0,1]$, then $(f-z)'(t)=3t^2-2t$ for all $t\\in [0,1]$, so $(f-z)'(t)=0$ implies $t=2/3$. Quote by bugatti79 Just one little query regarding the $\\|f-z\\|=\\sup_{t\\in[0,1]}|f(t)-z(t)|=\\sup_{t\\in[0,1]}|t^3+3-5t|=3$ You put in 0 and you are left with 3. If you put in 1 we get -1. So we take the highest of those two because $sup_{t \\in [0,1]} | f-z|=max\\{t_1,t_2\\}$ or something like this? Yes, that's right. P: 652 Quote by Fredrik I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies", "are $\\ge 0$, we have from $(A)$: $$\\sum_{i=k+1}^9 f(y_i) \\le (9-k)f\\left(\\frac1{9-k} \\sum_{i=k+1}^9 y_i \\right)$$ hence we can replace all non-negative numbers with their arithmetic mean $\\mu$ to maximise this partial sum. Now we have $y_k < 0 < \\mu$ and hence using $(B)$ we can replace $(\\mu, y_k)$ with $(-1, 1+\\mu+y_k)$ or $(0, \\mu+y_k)$ to increase the sum (depending on the sign of $\\mu + y_k$), thereby making all negative terms equal to $-1$. So we have that the maximum is when some $k$ of the $y_i$s are $-1$, and the remaining are all equal. Obviously $\\mu = \\dfrac{k}{9-k}$, giving a maximum sum of $$g(k) = kf(-1) + (9-k)f\\left(\\frac{k}{9-k}\\right) = \\sqrt[3]{k(9-k)^2}-k$$ It is easy to check that $k = 1$ gives the maximum among integers in $[1, 4]$, and so the maximum value is when $y_1= -1$ and $y_2 = ... y_9 = \\frac18$ or when any one of the $x_i$ is $-1$ and the rest are all $\\frac12$ P.S. Very similar to another problem : Maximum of the sum of cube Search with Lagrange in (ie, with the restrictions): $$\\sum x_i^3=0;$$ $$\\sum x_i^3=0,\\qquad x_k=1, \\qquad{\\rm for each}\\ k=1,\\cdots, 9;$$ $$\\sum x_i^3=0,\\qquad x_k=-1,\\qquad{\\rm for each}\\ k=1,\\cdots, 9;$$ $$\\sum x_i^3=0,\\qquad x_j=1,x_k=1 \\qquad{\\rm for each}\\ j\\ne k=1,\\cdots, 9;$$ $$\\cdots$$ Awfully long. • Is there a more elementary way? Jan 28,", "$f(x,y) \\le \\frac12x-x-e^{-x}$, so for a fixed $x$, the choice $y:=x+e^{-x}$ is the best in maximizing $f$. Then, you are left with only one variable, and should maximize this $g(x):=-\\frac12x-e^{-x}$ whence $x\\ge 0$. Possible maximum place is at the edge ($x=0$) or when the tangent of $g$ is horizontal, i.e. where $g'=0$.", "g_1'(x)$ on this interval. Furthermore, $g_1'$ is continuous and is an increasing function on $[1.5, 2]$ and thus the maximum is achieved at the right endpoint of this interval, that is $\\max_{1.5 ≤x ≤ 2} \\mid g_1'(x) \\mid = \\max_{1.5 ≤ x ≤ 2} g_1'(x) = \\frac{3}{2} (2)^2 \\mid = 6$. (1) \\begin{align} \\quad \\lambda = \\max_{1.5 ≤ x ≤ 2} \\mid g_1'(x) \\mid = 6 > 1 \\end{align} As we see, the choice of $x = g_1(x)$ does not satisfy the condition for convergence. Now note that since $g_2(x) = (2x + 1)^{1/3}$ we have that $g_2'(x) = \\frac{2}{3} (2x + 1)^{-2/3}$. Once again we see that $g_2'(x) > 0$ on the interval $[1.5, 2]$ and so $\\mid g_2'(x) \\mid = g_2'(x)$ on this interval. Furthermore, $g_2'$ is continuous and is a decreasing function on $[1.5, 2]$ and thus the maximum is achieved at the left endpoint of this interval, that is $\\max_{1.5 ≤ x ≤ 2} \\mid g_2'(x) \\mid = \\max_{1.5 ≤ x ≤ 2} \\mid g_2'(x) \\mid = \\frac{2}{3} (2(1.5) + 1)^{-2/3} = 0.264566… < 1$ and o $x = g_2(x)$ is a good choice for applying the Fixed Point method. We will now apply the fixed point method to $x = g_2(x) = (2x + 1)^{1/3}$. The first iteration using the initial approximation $x_0", "\\right\\}$$ Therefore, $$f'(x)$$ has a finite value for all x > 0 $$\\Rightarrow$$ $$f(x)$$ is differentiable everywhere. Example - 16 f(x)={{x}^{2}}-2|x|~\\text{ }and~\\text{ }g(x)=\\left\\{ \\begin{align} & Min\\{f(t)\\,\\,\\,:-2\\le t\\le x,\\,\\,\\,\\,-2\\le x<0\\} \\\\ & Max\\{f(t):\\,\\,\\,\\,\\,0\\le t\\le x,\\,\\,\\,\\,\\,\\,\\,\\,0\\le x\\le 3\\} \\\\ \\end{align} \\right\\} draw the graph of $$g(x)$$ and discuss its continuity and differentiability. Solution: The graph of f(x) is sketched below (in the relevant domain): Now we must understand what the definition of g(x) means. Consider the upper definition of g(x): $$g(x) = Min\\{ f(t): - 2 \\le t \\le x,\\,\\,\\,\\, - 2 \\le x < 0\\}$$ To evaluate $$g(x)$$ at any x, we scan the entire interval from –2 to x, (this is what the variable t is for), select that value of f(t) which is minimum in this interval; this minimum value of f(t) becomes the value of the function at x. The figure below illustrates this graphically for four different values of x (we are considering the interval $$- 2 \\le x < 0,$$ the upper definition of g(x)): Notice that where x crosses –1 (when –1 < x < 0), the minimum value of the function in the interval [–2, x] becomes fixed at t = –1. When –2 < x < –1, the minimum value of the function in the interval [–2, x] is at t" ]

[ "# Cute cubic with integer roots Suppose that the polynomial $$p(x) = x^3 - a x^2 + bx - c$$ has 3 positive integer roots and that $$4a+2b+c = 1741$$. Determine the value of $$a$$. ×", "+0 Polynomials +1 125 1 +130 Find all possible values of a, if the equations ax^2+5x+1=0 has two different real roots TY :) Feb 24, 2021 We have $5^2-4a>0\\implies 25>4a\\implies \\boxed{a<\\frac{25}{4}}$, since the discriminant is positive.", "Consider polynomials of the form $$ax^2+bx+c$$ with integer coefficients which have two distinct zeros in the interval $$0<x<1$$. Find the least positive integer value of $$a$$ for which such a polynomial exists.", "# Integer Coefficients Algebra Level 3 Find the smallest positive integer $$e$$ such that the polynomial $$ax^4 + bx^3 + cx^2 +dx + e$$ has integer coefficients and has roots at $$-2$$, $$3$$, $$5/3$$, and $$-1/2$$. Details and assumptions A root of a polynomial is a number $$r$$ where the polynomial is equal to 0. For example, the number $$r = 3$$ is a root of the polynomial $$2x - 6$$. ×", "+0 # Say there's a polynomial ax^4+bx^3+cx^2+dx+e=0 +1 639 0 +87 Say there's a polynomial ax^4+bx^3+cx^2+dx+e=0. 1. If $$\\sqrt{2}+\\sqrt{7}$$ is a root, find all the roots of this polynomial. 2. Find a,b,c,d,e Feb 24, 2020 edited by RandomUser Feb 25, 2020", "+0 # Polynomial +1 52 1 +36 Find a polynomial of degree 4 that has √2 + √7 as a root. That is, find integers a,b,c,d,e, so that x = √2 + √7 is a solution to the equation ax^4+bx^3+cx^2+dx+e=0. 1. Find all the roots of this polynomial. 2. Given two distinct positive integers x and y, what can you conclude about polynomials that have $$\\sqrt{x}+\\sqrt{y}$$ as a root. Melody, Cphill, EP?? Feb 27, 2020 edited by RandomUser Feb 27, 2020 edited by RandomUser Feb 27, 2020 #1 +29249 +5 I'll leave you to think about part 2. Feb 27, 2020", "Let $n$ be a positive integer. Given that the polynomial $P(x)$ satisfies the relation $2 P(x+1) – P(x) = x(x-1)(x-2)\\cdots(x-n)$ for all $x$, find the value of $P(0)$.", "if the polynomial has a integer root then the last coefficient (a_0) is a multiple of the root. For example, given the polynomial: $$x^2 + 5x + 6$$ the possible integer roots are +/-1 , +/-2, +/-3 and +/-6 a quick check shows that 2 and 3 are roots so the polynomial is equal to $$(x-3)(x-2)$$ This is especially useful for finding e-values from the e-polynomial since in most problems the e-values are integers and the coefficient of x^n is always 1.", "### 2021 Mediterranean MO #1 Determine the smallest positive integer $M$ with the following property: For every choice of integers $a,b,c$, there exists a polynomial $P(x)$ with integer coefficients so that $P(1)=aM$ and $P(2)=bM$ and $P(4)=cM$. Taking $a=1, b=2, c=3$, we see that $2\\mid M(c-b)$ and $3\\mid M(c-a)$ must hold true, which implies that $6\\mid M$. Now, we show that $M=6$ is a valid possibility. As a construction, consider the polynomial$$P_{abc}=(2a+c-3b)x^2+(15b-12a-3c)x+(16a+2c-12b).$$In this case, we have $M=6$, as desired. $\\square$", "Algebra Level pending Let $$Ax^4 + Bx^3 + Cx^2 + Dx + E= 0$$ be an equation with integer coefficients such that $$A$$ is positive, $$\\gcd(A,B,C,D,E) = 1$$, and the roots are $$\\sqrt2,-\\sqrt2,\\sqrt3,-\\sqrt3$$. Find $$A+B+C+D+E$$. ×", "have got rid of the brackets and got $2n^{6} - 2n^{3} -1 = 0$ But I dont see how I can tell wheather this polynomial has an integer solution or not? So if this polynomial had an integer solution 'a', (x-a), than a would divide -1, ? 4. ## Re: integer proof help I believe it's $n^6 - 2n^4 - 1 = 0$ The main rule is: If the leading coefficent is 1 (the case above, it's $1n^6$), and all the coefficients are integers, then any integer solution will divide the constant term. So in this case, the only possible integer solutions are {-1, 1}, and neither of those are solutions, so it has no integer solutions.", "integer has integer coefficients and monic for sure. 10. Originally Posted by TheArtofSymmetry Because bx-a is the minimal polynomial of $\\beta$ and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure. It's true that the minimal polynomial over $\\mathbb{Q}$ of an algebraic integer has integer coefficients - but this is a consequence of Gauss's lemma. It's not trivial!", "How do you deal with the dark (negative) side of functions? Let $$f(x)$$ be a quartic polynomial in the form $$ax^4+bx^3+cx^2+dx+e$$ where $f(-7) = 4636 , f(-6) = 2504 , f(-5) = 1212.$ Given that $$a, b,c , d$$ and $$e$$ are all positive integers less than 5. Evaluate the addition of the two 3-digit integers, $$\\overline{abc}+\\overline{cde}$$. ×", "# A polynomial with integer values Consider all polynomials $$f(x)$$ of degree 6 such that $$f(n)$$ is an integer for all integers $$n$$. The smallest possible positive leading coefficient of $$f(n)$$ can be expressed as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are positive coprime integers. What is the value of $$a+b$$? ×", "because the limitation to positive numbers means he can’t use negative coefficients. So, rather, he shows how to solve the following: 1. $$ax^2 + bx = c$$ 2. $$ax^2 + c = bx$$ 3. $$bx + c = ax^2$$ Likewise, he can’t solve for negative roots; the only time there are two solutions is when both solutions are positive. This might seem odd to modern students, but it’s important to remember that he was providing geometric solutions. There are no negatives in geometry proper: All measurements are positive. From this standpoint, his approach makes perfect sense. If you’d like to read more, my presentation of his first chapter is available on my other blog: First post; second post. # The sum of the first n positive integers In this article, I’ll illustrate how we can use two different strategies to develop the same formula. This relates to the toothpick problem we explored in class. ### Strategy 1 There’s a story, probably apocryphal, about the great mathematician Carl Friedrich Gauss as a young man. He was told to add the numbers 1 to 100, which he did in less than a minute. He observed that $$1 + 100 = 101$$, $$2 + 99 = 101$$, and so on up to $$50 + 51 = 101$$. Since there were 50", "from AMC 10A, 2010. Cubic Equation – AMC-10A, 2010- Problem 21 The polynomial $x^{3}-a x^{2}+b x-2010$ has three positive integer roots. What is the smallest possible value of $a ?$ $31$$78$$43$ Key... ## Problem on Fraction | AMC 10A, 2015 | Question 15 Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015. Fraction – AMC-10A, 2015- Problem 15 Consider the set of all fractions $\\frac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property...", "+0 # Polynomial Roots +1 79 1 +167 Suppose the polynomial \\(f(x) = a_nx^n + a_{n-1}x^{n-1} + \\cdots + a_2x^2 + a_1x + a_0\\) has integer coefficients, and its roots are distinct integers. Given that \\(a_n=2 \\) and \\(a_0=66 \\), what is the least possible value of \\(|a_{n-1}|\\)? Dec 14, 2019", "× # #4 Let $$a,b$$ be integers such that all the roots of the equation $$(x^2+ax+20)(x^2+17x+b)=0$$ are negative integers.What is the smallest possible value of $$a+b$$? Note by Vilakshan Gupta 2 months ago", "###### back to index | new Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $$|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.$$ Find $|f(0)|$. Steve says to Jon, 'I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $$P(x) = 2x^3-2ax^2+(a^2-81)x-c$$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?' After some calculations, Jon says, 'There is more than one such polynomial.' Steve says, 'You're right. Here is the value of $a$.' He writes down a positive integer and asks, 'Can you tell me the value of $c$?' Jon says, 'There are still two possible values of $c$.' Find the sum of the two possible values of $c$. Let $f(x) = x^4 + ax^3 + bx^2 + cx + d$. If $f(-1) = -1$, $f(2)=-4$, $f(-3) = -9$, and $f(4) = -16$. Find $f(1)$. Solve in positive integers $x^2 - 4xy + 5y^2 = 169$. Solve in integers the question $x+y=x^2 -xy + y^2$. Solve in integers $\\frac{x+y}{x^2-xy+y^2}=\\frac{3}{7}$ The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$? There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and $5x^2+kx+12=0$ has at least one integer solution for $x$. What is $N$? Let $a + ar_1 + ar_1^2", "Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$ Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$" ]

[ "If there are rational roots, they are integers that divide $38$. Not too many candidates. Let's cross our fingers and try small ones first. We find that $x=-2$ works. Divide our polynomial by $x+2$. We get $x^2-8x-19$. Use the Quadratic Formula for the other roots. – André Nicolas May 7 '12 at 4:45", "+0 0 36 1 For how many integer values of $$a$$ does the equation $$x^2 + ax + 8a=0$$ have integer solutions for $$x$$? Sep 7, 2020 Note that the roots are $$x = {-a \\pm \\sqrt{a^2-32a} \\over 2}$$. So the discriminant must be a perfect square. We can factor as a(a-32). So both a and a-32 are perfect squares. You can do casework from here to get a=-49,-18,-4,0,32,36,50,81. We need not to check these, as if a is even, then sqrt(a^2-32a) is even and vice versa. Our answer is then 8.", "-1)$. I need to solve $30a \\equiv 70 \\pmod{100}$, and $a=9$ is an obvious answer (and the only one for $a < 10$). So, $a = 9$ and the root is $90 - 1 = 89$. Hope that helps! - Uncle Colin", "= 0.$ Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1. $1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.$ $\\therefore f(1) = 10 .$ Thank you! the problem is really quite simple with the integer root theorem applied.", "looking for the integer root of this cubic is to remember that any integer root $$n$$ has to be a factor of $$100$$. Testing $$\\pm1,\\pm2,\\pm4,\\pm5,\\ldots$$ we find $$n=4$$ quite quickly. We could trim the testing a little by observing that $$|x^3-41x| \\le 8 + 82 = 90 < 100$$ for $$|x| \\le 2$$, and hence $$|n| \\ge 4$$. Thus we can start testing with $$n=\\pm4$$, and hit paydirt first time. We might not have been so lucky, and it isn't really necessary. Testing on a predetermined finite set (the factors of $$100$$) is not really trial and error. · 3 years, 11 months ago", "+0 # help +1 20 1 Suppose the roots of the polynomial $$x^2 - mx + n$$ are positive prime integers (not necessarily distinct). Given that $$m<20$$ how many possible values $$n$$ of are there? Jan 10, 2019 #1 +3583 +2 $$\\text{the roots are }\\\\ x = \\dfrac{m \\pm \\sqrt{m^2-4n}}{2}$$ $$\\text{as these are integers }m^2 - 4n \\text{ must be a perfect square}\\\\ m<20 \\Rightarrow m^2 \\leq 19^2 = 361\\\\ 4n It appears there are indeed 90 valid values of n Jan 10, 2019 ### 1+0 Answers #1 +3583 +2 Best Answer \\(\\text{the roots are }\\\\ x = \\dfrac{m \\pm \\sqrt{m^2-4n}}{2}$$ \\(\\text{as these are integers }m^2 - 4n \\text{ must be a perfect square}\\\\ m<20 \\Rightarrow m^2 \\leq 19^2 = 361\\\\ 4n It appears there are indeed 90 valid values of n Rom Jan 10, 2019", "of solving for a factor of $90!$ [duplicate] If $90! = (90)(89)(88)...(2)(1)$, then what is the exponent of the highest power of $2$ which will divide $90!$ ? How would I apply one of the easiest method from Here? I need help on applying the ... ### Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. [duplicate] Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. Actually, I know a way to solve this, but even if it is very large and ...", "if the polynomial has a integer root then the last coefficient (a_0) is a multiple of the root. For example, given the polynomial: $$x^2 + 5x + 6$$ the possible integer roots are +/-1 , +/-2, +/-3 and +/-6 a quick check shows that 2 and 3 are roots so the polynomial is equal to $$(x-3)(x-2)$$ This is especially useful for finding e-values from the e-polynomial since in most problems the e-values are integers and the coefficient of x^n is always 1.", "+0 # help 0 222 2 Find the smallest positive integer $$a$$, greater than 1000, such that the equation $$\\sqrt{a - \\sqrt{a + x}} = x$$ has a rational root. Mar 1, 2019 #1 +235 0 Not very good with these, but it might be 1296 Mar 2, 2019 #2 +1 sqrt(1902 - sqrt(1902 + 907)) =43, where a=1902 and x=907 Mar 2, 2019", "3x - 4 = 0.$ Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1. $1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.$ $\\therefore f(1) = 10 .$ Thank you! the problem is really quite simple with the integer root theorem applied. 7. ## Jeff Originally Posted by JeffM The question asked was how to find $P^{-1}(10).$ Once you know that $P^{-1}(10) = 1,$ the problem is trivial. Clearly I am missing something subtle. Hi Jeff, Its good to talk with you again, I've been away. On this post, (my bad) I miss read the OP and had the impression that he had guessed by inspection that (1,10) was on the graph. If so, then ... STEPS: 1) Show that (1, 10) is on the polynomial. 2) Calculate the slope at (1,10) My solution was simply a way to do both steps at once. HOWEVER I went back to re-read the post -- he did not have $P^{-1}(10).$ -- Clearly you did not miss something subtle", "# Cycling Through The Polynomial Shuffle Number Theory Level 5 Suppose $$f(x)$$ is a polynomial with integer coefficients which is bijective modulo $$49$$, which means that $$x\\not \\equiv y \\pmod {49} \\Leftrightarrow f(x)\\not \\equiv f(y) \\pmod {49}.$$ We define the composition powers $$f^{(i)}(x)$$ by setting $$f^{(1)}=f$$ and $$f^{(i+1)}(x)=f(f^{(i)}(x)).$$ For a given $$f,$$ its order (in modulo 49) is defined as the smallest positive integer $$n,$$ such that $$f^{(n)}(x)\\equiv x \\pmod {49}$$ for all integers $$x.$$ What is the largest possible order of $$f?$$ ×", "$-2\\equiv 3\\pmod 5$ and $-3\\equiv 2\\pmod 5$. so your solution is fine. Recall that $x-c$ divides $P(x)$ iff $P(c)=0$. Use the version of that appropriate for $\\mathbb{Z}_5$. That is faster than polynomial division for polynomials of higher degree. Remark: Your calculation is correct. Your belief that it is wrong is wrong, and reflects unfamiliarity with finite fields. For $-2\\equiv 3\\pmod{5}$ and $-3\\equiv 2\\pmod{5}$. Thus in $\\mathbb{Z_5}$, we have $-2=3$ and $-3=2$. So the polynomial $x^2-2x-3$, viewed as a polynomial over $\\mathbb{Z}_5$, is the same as the polynomial $x^2+3x+2$.", "an integer, $p$ divides the left-hand side and thus also the right-hand side. But since $p$ and $q$ have no common factors, $p$ cannot divide $q^n$. Thus $p$ must divide $a_0$. Similarly, by taking the highest-order term to the right-hand side and factoring out $q$ from the other terms, we can write $$q (a_{n-1} p^{n-1} + a_{n-2} p^{n-2} q + \\cdots + a_0 q^{n-1}) = – a_n p^n.$$ But again, since the expression in parentheses is an integer, $q$ divides the left-hand side and thus must also divide the right-hand side. Since $q$ cannot divide $p^n$, it must divide $a_n$. Examples: The constant $1 + \\sqrt{2}$ satisfies the polynomial $x^2 – 2 x – 1$, so that its algebraic degree cannot exceed two. By Lemma 1, the only possible rational roots are $x = 1$ and $x = -1$, and neither satisfies $x^2 – 2x – 1 = 0$. Thus the polynomial $x^2 – 2 x – 1$ is irreducible and is the minimal polynomial of $1 + \\sqrt{2}$, so that ${\\rm deg}(1 + \\sqrt{2}) = 2$. As another example, consider $\\sqrt[3]{2}$, which satisfies $x^3 – 2 = 0$. If this polynomial factors at all over the rational numbers, it has a linear factor and thus a rational root. But by Lemma 1, the only possible rational roots", "# For Every Integer 'x' Let $$f(x)=7x^{19}+19x^7+13ax$$ be a polynomial. Find the sum of the smallest $$2$$ possible values of the positive integer $$a$$, such that $$133$$ divides $$f(x)$$ for every integer $$x$$. ×", "factorizations of its primitive part. Practical techniques Currently the best techniques for factoring integer polynomials involve factoring over finite fields, but a simpler technique is usable for small polynomials (roughly less than tenth degree), if a computer is used. Since integer polynomials must factor into integer polynomial factors, and evaluating integer polynomials at integer values must produce integers, the integer values of a polynomial can be factored in only a finite number of ways, and produce only a finite number of possible polynomial factors. For example, consider :$f\\left(x\\right) = x^5 + x^4 + x^2 + x + 2$. If this polynomial factors over Z, then at least one of its factors must be of degree two or less. We need three values to uniquely fit a second degree polynomial. We'll use $f\\left(0\\right) = 2$, $f\\left(1\\right) = 6$ and $f\\left(-1\\right) = 2$. Now, 2 can only factor as :1×2, 2×1, (-1)×(-2), or (-2)×(-1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values :1, 2, -1, or -2 at $x=0$, and likewise at $x=-1$. There are eight different ways to factor 6, so there are :4×4×8 = 128 possible combinations, of which half can be discarded as the negatives of the other half, corresponding to 64 possible second degree integer polynomials which must", "+0 # Find the smallest possible integer 0 26 2 Find the smallest possible integer such that . Note: For a real number x,{x} = x - [x] denotes the fractional part of x. Jun 7, 2021 #1 +76 +1 Square root is an increasing function, so we only need to find an integer $n\\ge100$ such that $\\{\\sqrt{n-1}\\} \\le 0.5$ but $\\{\\sqrt{n}\\} >0.5$. Since $\\sqrt{110} = 10.488+$ and $\\sqrt{111} = 10.535+$ so the smallest such $n$ is 111. Jun 8, 2021 #2 +1 Thank you! Guest Jun 10, 2021", "integer root of $f$ would have to divide $a_0$. But the integer factors of the nonzero elements of $S$ are already in $S$. Hence, since $r \\notin S$, it follows that $a_0=0$. Let $g(x) = f(x)/x$. Then • $g$ is a nonzero polynomial with coefficients in $S$.$\\\\[4pt]$ • $g(r) = 0$.$\\\\[4pt]$ • $\\deg(g) = n-1$ contrary to the least degree property of $f$. Therefore $S = \\{0,\\pm 1,\\pm2, \\pm 5, \\pm 10\\}$ is final, and has $9$ elements. • In general, staring with $S=\\{0,n\\}$ we end up with at most the divisors of $n$ and $0$. Is there an $n$ where the final $S$ is (easily seen to be) smaller? – Hagen von Eitzen Oct 8 '17 at 15:21 • Good question! I'll try it. – quasi Oct 8 '17 at 16:01 • @Hagen von Eitzen: I have a tentative proof (not written out, so it could be wrong) of the following claim: The final $S$ is full if and only if the number of distinct prime factors of $n$ which exceed $3$ is at most $1$. Thus, the smallest positive integer value of $n$ for which $S$ is not full is $n=35$. – quasi Oct 8 '17 at 16:50 • It's not quite as simple as I thought. My claim above is not true. For example, $S$", "Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$ Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$", "from AMC 10A, 2010. Cubic Equation – AMC-10A, 2010- Problem 21 The polynomial $x^{3}-a x^{2}+b x-2010$ has three positive integer roots. What is the smallest possible value of $a ?$ $31$$78$$43$ Key... ## Problem on Fraction | AMC 10A, 2015 | Question 15 Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015. Fraction – AMC-10A, 2015- Problem 15 Consider the set of all fractions $\\frac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property...", "## Divide Polynomials Part I ### Learning Outcome • Multiply and divide polynomials ## Divide a Polynomial by a Binomial Dividing a polynomial by a monomial can be handled by dividing each term in the polynomial separately. This cannot be done when the divisor has more than one term. However, the process of long division can be very helpful with polynomials. Recall how you can use long division to divide two whole numbers, say $900$ divided by $37$. First, you would think about how many $37s$ are in $90$, as $9$ is too small. (Note: you could also think, how many $40s$ are there in $90$.) There are two $37s$ in $90$, so write $2$ above the last digit of $90$. Two $37s$ is $74$; write that product below the $90$. Subtract: $90–74$ is $16$. (If the result is larger than the divisor, $37$, then you need to use a larger number for the quotient.) Bring down the next digit $(0)$ and consider how many $37s$ are in $160$. There are four $37s$ in $160$, so write the $4$ next to the two in the quotient. Four $37s$ is $148$; write that product below the $160$. Subtract: $160–148$ is $12$. This is less than $37$ so the $4$ is correct. Since there are no more digits in the" ]

[ "+0 # Say there's a polynomial ax^4+bx^3+cx^2+dx+e=0 +1 639 0 +87 Say there's a polynomial ax^4+bx^3+cx^2+dx+e=0. 1. If $$\\sqrt{2}+\\sqrt{7}$$ is a root, find all the roots of this polynomial. 2. Find a,b,c,d,e Feb 24, 2020 edited by RandomUser Feb 25, 2020", "# Integer Coefficients Algebra Level 3 Find the smallest positive integer $$e$$ such that the polynomial $$ax^4 + bx^3 + cx^2 +dx + e$$ has integer coefficients and has roots at $$-2$$, $$3$$, $$5/3$$, and $$-1/2$$. Details and assumptions A root of a polynomial is a number $$r$$ where the polynomial is equal to 0. For example, the number $$r = 3$$ is a root of the polynomial $$2x - 6$$. ×", "+0 Polynomials +1 125 1 +130 Find all possible values of a, if the equations ax^2+5x+1=0 has two different real roots TY :) Feb 24, 2021 We have $5^2-4a>0\\implies 25>4a\\implies \\boxed{a<\\frac{25}{4}}$, since the discriminant is positive.", "Polynomials have smallest degree I am interested in the following problem. Problem 1. Find polynomial P (x) has integer coefficients, there is no rational roots, with the smallest degree such that for each positive integer $$m$$ there exists a positive integer a such that $$m \\mid P (a).$$ In the case of $$\\deg P = 2$$, use quadratic residue, we can see that it cannot happen. In the case of $$\\deg P = 4$$, I have the answer (use quadratic residue). Therefore, I need an answer for the case of $$\\deg P = 3$$, ie the following problem. Problem 2. Let the polynomial $$P (x) = ax ^ 3 + bx ^ 2 + cx + d$$, $$a,\\,b,\\,c,\\,d\\in\\mathbb Z$$ and $$a \\ne 0$$, $$P (x)$$ is irreducible on $$\\mathbb Z [x]$$. Prove that there exists a positive integer $$m$$ such that $$m\\nmid P(n),\\quad\\forall\\,n\\in\\mathbb N^*.$$ • A minor point is that, in your Problem 2 statement, I assume that $d$ is also an integer, but you should make that explicit. – John Omielan Dec 25 '18 at 4:17 • $d$ is integer, edited! – Drona Dec 25 '18 at 4:25 • @Drona can you post your solution for p=2 and 4? – Sandeep Silwal Dec 25 '18 at 5:17 It seems this is a corollary of the Chebotarev density", "# Cycling Through The Polynomial Shuffle Number Theory Level 5 Suppose $$f(x)$$ is a polynomial with integer coefficients which is bijective modulo $$49$$, which means that $$x\\not \\equiv y \\pmod {49} \\Leftrightarrow f(x)\\not \\equiv f(y) \\pmod {49}.$$ We define the composition powers $$f^{(i)}(x)$$ by setting $$f^{(1)}=f$$ and $$f^{(i+1)}(x)=f(f^{(i)}(x)).$$ For a given $$f,$$ its order (in modulo 49) is defined as the smallest positive integer $$n,$$ such that $$f^{(n)}(x)\\equiv x \\pmod {49}$$ for all integers $$x.$$ What is the largest possible order of $$f?$$ ×", "### 2021 Mediterranean MO #1 Determine the smallest positive integer $M$ with the following property: For every choice of integers $a,b,c$, there exists a polynomial $P(x)$ with integer coefficients so that $P(1)=aM$ and $P(2)=bM$ and $P(4)=cM$. Taking $a=1, b=2, c=3$, we see that $2\\mid M(c-b)$ and $3\\mid M(c-a)$ must hold true, which implies that $6\\mid M$. Now, we show that $M=6$ is a valid possibility. As a construction, consider the polynomial$$P_{abc}=(2a+c-3b)x^2+(15b-12a-3c)x+(16a+2c-12b).$$In this case, we have $M=6$, as desired. $\\square$", "+0 # Polynomial +1 52 1 +36 Find a polynomial of degree 4 that has √2 + √7 as a root. That is, find integers a,b,c,d,e, so that x = √2 + √7 is a solution to the equation ax^4+bx^3+cx^2+dx+e=0. 1. Find all the roots of this polynomial. 2. Given two distinct positive integers x and y, what can you conclude about polynomials that have $$\\sqrt{x}+\\sqrt{y}$$ as a root. Melody, Cphill, EP?? Feb 27, 2020 edited by RandomUser Feb 27, 2020 edited by RandomUser Feb 27, 2020 #1 +29249 +5 I'll leave you to think about part 2. Feb 27, 2020", "# For Every Integer 'x' Let $$f(x)=7x^{19}+19x^7+13ax$$ be a polynomial. Find the sum of the smallest $$2$$ possible values of the positive integer $$a$$, such that $$133$$ divides $$f(x)$$ for every integer $$x$$. ×", "from AMC 10A, 2010. Cubic Equation – AMC-10A, 2010- Problem 21 The polynomial $x^{3}-a x^{2}+b x-2010$ has three positive integer roots. What is the smallest possible value of $a ?$ $31$$78$$43$ Key... ## Problem on Fraction | AMC 10A, 2015 | Question 15 Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015. Fraction – AMC-10A, 2015- Problem 15 Consider the set of all fractions $\\frac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property...", "if the polynomial has a integer root then the last coefficient (a_0) is a multiple of the root. For example, given the polynomial: $$x^2 + 5x + 6$$ the possible integer roots are +/-1 , +/-2, +/-3 and +/-6 a quick check shows that 2 and 3 are roots so the polynomial is equal to $$(x-3)(x-2)$$ This is especially useful for finding e-values from the e-polynomial since in most problems the e-values are integers and the coefficient of x^n is always 1.", "+0 # help +1 20 1 Suppose the roots of the polynomial $$x^2 - mx + n$$ are positive prime integers (not necessarily distinct). Given that $$m<20$$ how many possible values $$n$$ of are there? Jan 10, 2019 #1 +3583 +2 $$\\text{the roots are }\\\\ x = \\dfrac{m \\pm \\sqrt{m^2-4n}}{2}$$ $$\\text{as these are integers }m^2 - 4n \\text{ must be a perfect square}\\\\ m<20 \\Rightarrow m^2 \\leq 19^2 = 361\\\\ 4n It appears there are indeed 90 valid values of n Jan 10, 2019 ### 1+0 Answers #1 +3583 +2 Best Answer \\(\\text{the roots are }\\\\ x = \\dfrac{m \\pm \\sqrt{m^2-4n}}{2}$$ \\(\\text{as these are integers }m^2 - 4n \\text{ must be a perfect square}\\\\ m<20 \\Rightarrow m^2 \\leq 19^2 = 361\\\\ 4n It appears there are indeed 90 valid values of n Rom Jan 10, 2019", "# Cute cubic with integer roots Suppose that the polynomial $$p(x) = x^3 - a x^2 + bx - c$$ has 3 positive integer roots and that $$4a+2b+c = 1741$$. Determine the value of $$a$$. ×", "Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$ Be $h$ the smallest positive integer root of a polynomial whose coefficients of the sum is zero. Calculate the value of: $16h^2-3$", "use negative coefficients. So, rather, he shows how to solve the following: 1. $$ax^2 + bx = c$$ 2. $$ax^2 + c = bx$$ 3. $$bx + c = ax^2$$ Likewise, he can’t solve for negative roots; the only time there are two solutions is when both solutions are positive. This might seem odd to modern students, but it’s important to remember that he was providing geometric solutions. There are no negatives in geometry proper: All measurements are positive. From this standpoint, his approach makes perfect sense. If you’d like to read more, my presentation of his first chapter is available on my other blog: First post; second post. # The sum of the first n positive integers In this article, I’ll illustrate how we can use two different strategies to develop the same formula. This relates to the toothpick problem we explored in class. ### Strategy 1 There’s a story, probably apocryphal, about the great mathematician Carl Friedrich Gauss as a young man. He was told to add the numbers 1 to 100, which he did in less than a minute. He observed that $$1 + 100 = 101$$, $$2 + 99 = 101$$, and so on up to $$50 + 51 = 101$$. Since there were 50 equations that each added to 101, the total sum", "have got rid of the brackets and got $2n^{6} - 2n^{3} -1 = 0$ But I dont see how I can tell wheather this polynomial has an integer solution or not? So if this polynomial had an integer solution 'a', (x-a), than a would divide -1, ? 4. ## Re: integer proof help I believe it's $n^6 - 2n^4 - 1 = 0$ The main rule is: If the leading coefficent is 1 (the case above, it's $1n^6$), and all the coefficients are integers, then any integer solution will divide the constant term. So in this case, the only possible integer solutions are {-1, 1}, and neither of those are solutions, so it has no integer solutions.", "# A polynomial with integer values Consider all polynomials $$f(x)$$ of degree 6 such that $$f(n)$$ is an integer for all integers $$n$$. The smallest possible positive leading coefficient of $$f(n)$$ can be expressed as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are positive coprime integers. What is the value of $$a+b$$? ×", "× # #4 Let $$a,b$$ be integers such that all the roots of the equation $$(x^2+ax+20)(x^2+17x+b)=0$$ are negative integers.What is the smallest possible value of $$a+b$$? Note by Vilakshan Gupta 2 months ago", "0^3+ 0^2- 1= -1$ while taking x= 1, $f(1)= 1^3+ 1^2- 1= 1$. Since any polynomial is \"continuous\" the only way the value can go from negative to positive is to pass through 0- there is some x between 0 and 1 such that f(x)= 0. We don't know exactly where that x is so lets just \"bisect\" the interval and try x= 1/2. $f(0.5)= (0.5)^3+ (0.5)^2- 1= -0.625$. Since that is negative, there must be a root between 0.5 and 1. Again, try half-way between (just because it is easy to calculate): try x= 0.75. $f(0.75)= (0.75)^3+ (0.75)^2- 1= -0.015625$. That is almost 0 so we are close to a root! It is also still negative so we know that root must be between 0.75 and 1. Halfway between is (0.75+ 1)/2= 0.875 so calculate f(0.875), determine whether it is positive or negative and continue. By the way, in \"elementary\" algebra \"factoring\" normally means \"factoring with integer coefficients\". Your example, $x^3+ x^2- 1$ cannot be factored with integer coefficeints. One way to see that is to use the \"rational root theorem\". If a polynomial, $ax^n+ bx^{n-1}+ \\cdot\\cdot\\cdot+ yx+ z$, with integer coefficients, has a rational root $x= \\alpha/\\beta$, with $\\alpha$ and $\\beta$ integers, then it has a factor $\\beta x- \\alpha$ and so $\\beta$ must be an", "= 0.$ Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1. $1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.$ $\\therefore f(1) = 10 .$ Thank you! the problem is really quite simple with the integer root theorem applied.", "+0 # help 0 222 2 Find the smallest positive integer $$a$$, greater than 1000, such that the equation $$\\sqrt{a - \\sqrt{a + x}} = x$$ has a rational root. Mar 1, 2019 #1 +235 0 Not very good with these, but it might be 1296 Mar 2, 2019 #2 +1 sqrt(1902 - sqrt(1902 + 907)) =43, where a=1902 and x=907 Mar 2, 2019" ]

[ "How many $$(x,y)$$ are there? Algebra Level 4 How many real numbered ordered pairs $$(x, y)$$ satisfy the following inequalities: \\begin{align} (x+2y)^2 &\\leq (2x-3)(4y+3)\\\\ (x-2y)^2 &\\leq (5-2x)(4y-5)? \\end{align} ×", "+0 Help! 0 133 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$", "+0 # Help! 0 39 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 #1 +3563 +2 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$ . Dec 16, 2018", "# Question 1 Algebra Level 4 $\\left\\{ {\\begin{array}{*{20}{c}} {3x - a \\ge 0}\\\\ {\\left| x \\right| < \\frac{b}{2}} \\end{array}} \\right.$ There are four integers that satisfy the system of inequalities about x above, $$-1, 0, 1, 2$$. Then, how many ordered pairs $$(a,b)$$ can satisfy these inequalities? ×", "Find the ordered triple of positive integers $(a,b,c)$ such that \\begin{align*}a &= \\frac{1908}{b(c-b)}-\\frac{1908}{b(c+b)} \\\\ b &= \\frac{2600}{a(c-a)}-\\frac{2600}{a(c+a)} \\\\ c &= \\frac{686}{a(b-a)}-\\frac{686}{a(b+a)} .\\end{align*}", "# Absolute Value Sum Inequality How many ordered pairs of integers $$(x,y)$$ are there that satisfy $$|x| + |y| \\leq 10$$? ×", "of ordered pairs $$\\{(a,1), (b,3), (c,2)\\}.$$", "in the range from -12 to 12, inclusive each of which will give an integer value of $$x$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: for-how-many-ordered-pairs-x-y-that-are-solutions-of-the-110687.html _________________ Re: For how many ordered pairs (x , y) that are solutions of the [#permalink] 09 Mar 2014, 23:58 Similar topics Replies Last post Similar Topics: 2 How many solutions are possible for the inequality | x - 1 | + | x - 6 3 12 Sep 2016, 11:02 6 For how many integers pair (x,y) satisfies the result 5 18 Jan 2015, 15:01 34 For how many ordered pairs (x , y) that are solutions of the 4 11 Mar 2014, 23:54 2 For how many ordered pairs (x , y) that are solutions of the 7 17 Jul 2011, 00:05 60 For how many ordered pairs (x, y) that are solutions of the 17 14 Jan 2011, 16:40 Display posts from previous: Sort by", "# Just the right Percentage. Algebra Level 4 $\\large \\begin{cases} a^2 + b^2 < 16 \\\\ a^2 + b^2 < 8a\\\\ a^2 + b^2 < 8b \\end{cases}$ How many ordered pairs of integers $$(a,b)$$ satisfy the above system of inequalities? ×", "# Inequalities + Combinatorics = Combinequalities $x + y \\leq 100$ Find the number of ordered pairs $$(x, y)$$ of non-negative integers $$x,y$$ which satisfy the inequality above is fulfilled. × Problem Loading... Note Loading... Set Loading...", "cross (x), six points that satisfy all 3 of the inequalities (that are also integers). NOTE: These are the only six points we could have chosen.", "Algebra # Polynomial Inequalities - Sets $A$ and $B$ are two subsets of $\\mathbb{R}$ defined by: \\begin{aligned} A &= \\{ x \\mid x^2 - 6 x - 91 \\leq 0 \\}, \\\\ B &= \\{ x \\mid x^2 - 6 x - 91 = 0 \\}. \\end{aligned} How many integers does the set $A \\cap B^C$ contain? Details and assumptions $B^C$ denotes the complement of set $B$. $\\mathbb{R}$ denotes the set of all real numbers. Sets $A, B, C$ and $D$ are defined as follows: \\begin{aligned} A &= \\{ (x,y) \\vert x>0 \\} \\\\ B &= \\{ (x,y) \\vert y>0 \\} \\\\ C &= \\{ (x,y) \\vert y>5x \\} \\\\ D &= \\{ (x,y) \\vert y<-3x \\}. \\end{aligned} Then which of the following is the set that represents the shaded region, including the two axes, in the above graph? The sets $A = \\{ x \\mid x^2 - 5 x + 4 \\leq 0 \\}$ and $B = \\{ x \\mid x^2 + ax + b < 0 \\}$ satisfy the following conditions: $A \\cap B = \\emptyset \\mbox{ and } A \\cup B = \\{ x \\mid 1 \\leq x < 15 \\}.$ What is the value of $a + b?$ $P,\\ Q$ and $R$ are three subsets of $\\mathbb{R}$ defined as follows: \\begin{aligned} P &= \\{", "# Count the pairs Geometry Level 3 $\\sqrt{(x+3)^2 + (y-4)^2} + \\sqrt{(x-7)^2 + (y+6)^2} \\leq 10\\sqrt2$ How many pairs of integers $$(x,y)$$ satisfy the inequality above? ×", "# Elegance is the key For how many ordered pairs $(x,y)$ of integers $x$ and $y$ is the expression $12x^{2} + 27y^{2} + 36xy - 1$ a perfect square? ×", "# How do I factor each of these? \\begin{aligned} ab+cd&=&38 \\\\ ac+bd&=&34\\\\ ad+bc&=&43\\\\ \\end{aligned} Let $a,b,c$ and $d$ positive integers such that they satisfy the system of equations above. Find $a+b+c+d$. ×", "# ORDERED PAIRS... Discrete Mathematics Level pending How many ordered pairs of (m,n) satisfy $$\\frac{m}{12}$$ = $$\\frac{12}{n}$$ ? Clarification : m and n must be integers . ×", "# How many pairs? How many ordered pairs of integers $$(a,b)$$ are there such that $$\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{200}$$?", "# Arshdeep's ordered pairs Number Theory Level 5 Find the total number of ordered pairs of integers $$(a,b)$$ satisfying $a^2 (b^2 + 1) + b^2(a^2 + 16) = 448.$ This problem is shared by Arshdeep D, who came across it in Mathematical Mayham. ×", "profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Math Expert Joined: 02 Sep 2009 Posts: 39720 Re: For how many ordered pairs (x , y) that are solutions of the [#permalink] Show Tags 09 Mar 2014, 23:58 1 KUDOS Expert's post 2 This post was BOOKMARKED 2x + y = 12 |y| <= 12 For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers? A. 7 B. 10 C. 12 D. 13 E. 14 Given: $$-12\\leq{y}\\leq{12}$$ and $$2x+y=12$$ --> $$y=12-2x=2(6-x)=even$$, (as $$x$$ must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of $$x$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: for-how-many-ordered-pairs-x-y-that-are-solutions-of-the-110687.html _________________ Re: For how many ordered pairs (x , y) that are solutions of the [#permalink] 09 Mar 2014, 23:58 Similar topics Replies Last post Similar Topics: 8 How many solutions are possible for the inequality | x - 1 | + | x - 6 4 09 Oct 2016, 11:29 11 For how many integers pair (x,y) satisfies the result 6 16 Nov 2016, 02:41 47 For how many ordered pairs (x , y) that", "# Count the pairs Geometry Level pending $\\sqrt{(x+3)^2 + (y-4)^2} + \\sqrt{(x-7)^2 + (y+6)^2} \\leq 10\\sqrt2$ How many pairs of integers $$(x,y)$$ satisfy the inequality above? ×" ]

[ "+0 0 82 1 The diagram below shows four circles. Let $A_1,$ $A_2,$ $A_3,$ $A_4$ denote the areas of the red region, yellow region, green region, blue region, respectively. These areas satisfy $A_1 = \\frac{A_2}{2} = \\frac{A_3}{3} = \\frac{A_4}{4}.$Let $r_1$ denote the smallest radius, and let $r_4$ denote the largest radius. Find $\\frac{r_4}{r_1}.$ [asy] unitsize(1 cm); pair[] O; real[] r; O[1] = (0,0); O[2] = (0.1,0.2); O[3] = (-0.2,-0.1); O[4] = (0.1,-0.3); r[1] = 1; r[2] = 1.5; r[3] = 2; r[4] = 2.5; fill(Circle(O[4],r[4]),lightblue); draw(Circle(O[4],r[4])); label(\"$A_4$\", (1.8,-1.5)); fill(Circle(O[3],r[3]),lightgreen); draw(Circle(O[3],r[3]));label(\"$A_3$\", (-1.3,-1.3)); fill(Circle(O[2],r[2]),yellow); draw(Circle(O[2],r[2]));label(\"$A_2$\", (1,1)); fill(Circle(O[1],r[1]),lightred); draw(Circle(O[1],r[1]));label(\"$A_1$\", O[1]); [/asy] Jan 25, 2020", "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "draw(circle((-4,-4), 4*1.41421356237)); [/asy]$ The way we figure out the area is by splitting it the following way: $[asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]$ We know each of the red lines is a diameter of the circle which is $\\sqrt2$. So the area of the red is 2. We know that the area of each semicircle is $\\dfrac14 \\pi$ so the area of the semicircles combines is $\\pi$. Thus we get $\\boxed{\\textbf{(B)} \\pi+2}.$", "\\dots, 21$. This is a total of $\\boxed{(D) 13}$ distinct line segments. (asymptote diagram added by elements2015) Solution 2 - Circles Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\\{15, 16, 17, 18, 19, 20 \\}$ intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this. $[asy] unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, NE); dot(\"P\", P, S); draw(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); [/asy]$ It follows that we can draw circles of radii $15, 16, 17, 18, 19,$ and $20,$ that each contribute two integer lengths (since these circles intersect the hypotenuse twice) from $B$ to $\\overline{AC}$ and one circle of radius $21$ that contributes only one such segment. Our answer is then", "that graphs those $3$ equations for some constant $a$. WLOG, we can set $a=1$. To make our life easier, we can also focus entirely on the first quadrant, since all $4$ graphs have different shapes in the first quadrant. Therefore, $x \\geq 0$ and $y \\geq 0$. First, we have $|x| + |y| = 1$. Since $x \\geq 0$ and $y \\geq 0$, this becomes $x + y = 1$. After a bit of rearranging, we get $$y = -x + 1$$. $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); label(\"X\", (-28,0), S); label(\"Y\", (-40,12), E); [/asy]$ This eliminates $III$, which does not have a diagonal line in the first quadrant. Next, we have $\\sqrt{2(x^{2}+y^{2})} = 1$. Squaring both sides and dividing by $2$, we get $$x^2+y^2=\\frac{\\sqrt{2}}{2}^2$$ Note that this describes a circle with the center $(0,0)$ and radius $\\frac{\\sqrt{2}}{2}$. We can find that $(\\frac{1}{2}, \\frac{1}{2})$ is on the circle, which is also on the line $y = -x + 1$. Therefore, the circle intersects the line at that point. We can graph this as follows: $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); draw(arc((-40,0),4.25, 0, 90)); label(\"X\", (-28,0), S); label(\"Y\", (-40,12), E); [/asy]$ This eliminates $I$, which does not include the intersection point listed above. Finally, we have $2\\mbox{Max}(|x|, |y|) = 1$. Rearranging and substituting $x$ for $|x|$ and", "draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$. Let the radius of circle $P$ be $r$. Using Distance Formula, we get the following system of three equations: $$h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2$$ By simplifying, we get $$h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1$$ By subtracting the first equation from the second and third equations, we get $$8r=-4h+12, 10r=2h+6$$ which simplifies to $$2r=3-h, 5r=h+3$$ When we add these two equations, we get $$7r=6$$ So $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ $w^5$ ## Solution 5(Inversion from Circular Inversion) Let $\\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram: $[asy] size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NE); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0),", "the solution sets. We begin with the inequality $x^2 + y^2 \\geq 4$ which we rewrite as $x^2+y^2 - 4 \\geq 0$. The points which satisfy $x^2 + y^2 - 4 = 0$ form our friendly circle $x^2+y^2 = 4$. Using test points $(0,0)$ and $(0,3)$ we find that our solution comprises the region outside the circle. As far as the circle itself, the point $(0,2)$ satisfies the inequality, so the circle itself is part of the solution set. Moving to the inequality $x^2 - 2x + y^2 - 2y \\leq 0$, we start with $x^2 - 2x + y^2 - 2y = 0$. Completing the squares, we obtain $(x-1)^2 + (y-1)^2 = 2$, which is a circle centered at $(1,1)$ with a radius of $\\sqrt{2}$. Choosing $(1,1)$ to represent the inside of the circle, $(1,3)$ as a point outside of the circle and $(0,0)$ as a point on the circle, we find that the solution to the inequality is the inside of the circle, including the circle itself. Our final answer, then, consists of the points on or outside of the circle $x^2 + y^2 = 4$ which lie on or inside the circle $(x-1)^2+(y-1)^2 = 2$. To produce the most accurate graph, we need to find where these circles intersect. To that end, we solve the", "instead of an arbitrary circle, but I actually think that the arbitrary circle is more intuitive. So consider this arbitrary circle with radius “r.” From the center I can draw any radius I want, below are some examples. $\\setlength{\\unitlength}{1mm} \\begin{picture}(140, 40) \\put(0,20){\\circle{14}} \\put(0,20){\\line(1,0){7}} \\put(0.7,20.3){r} \\put(15,20){\\circle{14}} \\put(15,20){\\line(1,1){5}} \\put(15.7,24){r} \\put(30,20){\\circle{14}} \\put(30,20){\\line(0,1){7}} \\put(30.7,24){r} \\put(45,20){\\circle{14}} \\put(45,20){\\line(-1,1){5}} \\put(44.7,24){r} \\put(60,20){\\circle{14}} \\put(60,20){\\line(-1,0){7}} \\put(60.3,20.3){r} \\put(75,20){\\circle{14}} \\put(75,20){\\line(-1,-1){5}} \\put(74.7,16){r} \\put(90,20){\\circle{14}} \\put(90,20){\\line(0,-1){7}} \\put(90.7,16){r} \\put(105,20){\\circle{14}} \\put(105,20){\\line(1,-1){5}} \\put(104.7,16){r} \\put(120,20){\\circle{14}} \\put(120,20){\\line(1,0){7}} \\put(120.3,20.3){r} \\end{picture}$ For any given radial line, I can draw a right triangle: $\\setlength{\\unitlength}{1mm} \\begin{picture}(140, 40) \\put(0,20){\\circle{14}} \\put(0,20){\\line(1,1){5}} \\put(5,20){\\line(0,1){4.5}} \\put(0,20){\\line(1,0){5}} \\put(0,24){r} \\put(2,16){x} \\put(7,22){y} \\end{picture}$ More precisely I am drawing a triangle for any given angle $\\theta \\text{ (theta)}$ that I want to make. The trig functions take this angle, and tell you the ratio of different values. Notice that in this triangle we have the hypotenuse and two other sides, where the hypotenuse is just the radius. Because the radius is always the same in a circle, the hypotenuse of any triangle I draw with this example circle will be the same. Only the lengths of the other two lines change as the angle changes. The sine function tells you the ratio of the length of the the side that the angle is facing, y, to the radius r. Cosine tells you the ratio of the length of the side that", "\\put(2,2){\\line(1,-2){.5}} \\put(2.5,1){\\circle*{1}} \\end{picture}$ Notice that none of the lines are curved, well that is because I lack the ability to draw squiggly lines. Squiggly lines make for squiggly people, so we will accept my crudely drawn straight line graphs and move on. A graph can have as little as just one dot, or infinitely many dots and lines. We can also attach directions and numbers to lines. Similarly we can assign a value to a dot. Doing these things leads to different questions than with a generic graph without any labels. One simple question is, when can I reach every possible dot on the graph? In my above one it is obvious that I cannot reach the bottom right dot because it has no lines going to it. But having every dot with at least one line is also not sufficient enough, for example the graph: $\\setlength{\\unitlength}{7cm} \\begin{picture}(2,5) \\put(1,1){\\line(0,1){1}} \\put(1,1){\\circle*{1}} \\put(1,2){\\circle*{1}} \\put(1,2){\\line(1,0){1}} \\put(2,2){\\circle*{1}} \\put(2,2){\\line(-1,-1){.5}} \\put(1.5,1.5){\\circle*{1}} \\put(1,2){\\line(1,-1){.5}} \\put(3,1){\\circle*{1}} \\put(3,1){\\line(0,1){1}} \\put(3,2){\\circle*{1}} \\put(2,2){\\line(1,-2){.5}} \\put(2.5,1){\\circle*{1}} \\end{picture}$ Every dot has a line, but you cannot reach every dot on the graph by going along lines from any starting position. This graph is called disconnected. A graph with the property of being able to get from any point to any other point by going along lines is called a connected graph. With that in", "bi)(n;j)}] of lattice points of length 2j − 1 as follows: Case n is even: $(ai,bi)(n;j)={(n+2i-2j,2n+1-2i)(1≤i≤j)(n+2i-2j,4j+1-2i)(j+1≤i≤2j-1).$ Case n is odd: $(ai,bi)(n;j)={(n+2-2j,2n-1)(i=1)(n+2i-2j,2n+2-2i)(2≤i≤j)(n+2i-2j,4j-2i)(j+1≤i≤2j-1).$ The L-shaped curve of type [{(ai, bi)(n;j)}] for any j = 1, 2, . . .n/2⎦ presents the torus knot T(n, 2n − 1) as a divide knot. Here, ⎣xmeans the largest integer less than or equal to x. Note that ai + bi = 2n + 1 for the final (the largest) j = n/2 in the case n is even, and also for j = (n + 1)/2 except i = 1, in the case n is odd. Example 2.10 For n = 6, we have $[{(ai,bi)(6;1)}]=[(6,11)],[{(ai,bi)(6;2)}]=[(4,11),(6,9),(8,3)],[{(ai,bi)(6;3)}]=[(2,11),(4,9),(6,7),(8,5),(10,3)].$ For n = 7, we have $[{(ai,bi)(7;1)}]=[(7,13)],[{(ai,bi)(7;2)}]=[(5,13),(7,12),(9,2)],[{(ai,bi)(7;3)}]=[(3,13),(5,12),(7,10),(9,4),(11,2)],[{(ai,bi)(7;4)}]=[(1,13),(3,12),(5,10),(7,8),(9,6),(11,4),(13,2)].$ Proof Note that the case j = 1 is included in Lemma 2.1, since the L-shaped curve of type [{(ai, bi)(n;1)}] = [(n, 2n − 1)] is the curve B(n, 2n − 1), the original divide presentation of T(n, 2n − 1). The proof for the case j > 1 is by induction with respect to j. First, we give a proof in the case n is even. The first step from j = 1 to 2 (with n = 6) is shown in Figure 10, where the thin part in the curve moves to the thin dotted curve. It is", "# 2017 AIME II Problems/Problem 12 ## Problem Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution 1 Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "to find it out. If someone wants to append my answer and automate this part, please feel free to do so. With Asymptote, I take points along the curve, say G, then get points that is $\\epsilon$ distant in the direction of the normal vector, and finally smoothly connect them. So we can draw the boundary of the $\\varepsilon$-neighbourhood with desired accuracy. This is enough for usual cases of small $\\epsilon$ and G has quite simple shape. For more complex shapes of G, $\\epsilon$-boundary curve may self-intersect, and some improvement is needed. unitsize(1cm); real e=.1; // size of $\\epsilon$ pair[] epoints; pair[] points={(4.4,0.4),(4.6,0.25),(5,0.2),(5.4,0.4),(5.8,0.6),(6.3,0.55),(6.55,0.55), (6.6,0.8),(6.5,1.1),(6.35,1.2),(6,1.3),(5.8,1.35),(5.4,1.4),(5,1.25), (4.6,1),(4.45,0.8)}; guide c=operator..(...points)..cycle; for (real t=0; t<length(c); t=t+.1){ pair pt=point(c,t); pair qt=pt+scale(e)*rotate(-90)*dir(c,t); epoints.push(qt); //draw(circle(pt,e),cyan+opacity(.3)); // to see rolling circles along the curve } draw(c); draw(operator..(...epoints)..cycle,red); With rolling circles along the curve: Update In several situations, we need e-neighbbourhood of a (planar) domain D bounded by curve c with not small value of e, that is Minkowski sum of D and the e-radius circle. In case of convex polygon c, the above approach works well. unitsize(1cm); real e=.5; // size of $\\epsilon$ // for convex polygon >> OK! //pair[] points={(0,0), (5,0), (3,4), (1,3.5)}; // concave polygon >> fill is OK, draw is not OK pair[] points={(0,0), (5,0), (2,1), (1,3.5)}; path c=operator--(...points)--cycle; pair[] epoints; // points", "# Difference between revisions of \"2017 AIME II Problems/Problem 12\" ## Problem Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series", "a\\sin b-\\cos b\\sin a)^{2}\\le \\frac{1}{4}$$ $$\\sin^{2} (b-a) \\le \\frac{1}{4}$$ So we want $-\\frac{1}{2} \\le \\sin (b-a) \\le \\frac{1}{2}$, which is equivalent to $-30 \\le b-a \\le 30$ or $150 \\le b-a \\le 210$. The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\\frac{45^2}{75^2} = 1- \\frac{9}{25}=\\frac{16}{25}$ and the answer is $16+25 = \\boxed{41}$ Solution by Leesisi ## Solution 3 (Quicker Trig) $[asy] pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label(\"O\", O, 2*E); label(\"P\", P, 2*W); label(\"Q\", Q, NE); label(\"R\", R, NW); label(\"200\", (0,0), 2*S); label(\"x\", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); [/asy]$ Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \\cos a, PQ = 200 \\sin a, PR = 200 \\sin b, OR = 200 \\cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral. Thus, we are able to", "Or $z \\textcolor{red}{<} - 7 \\frac{2}{3}$ To graph this we will draw an open circle at $- 7 \\frac{2}{3}$ on the number line. The circle will be open because the inequality operator does not contain an \"or equal to\" clause. We will the draw an arrow to the left of the circle because the inequality operator does contain a \"less than\" clause:", "be on the circle, such that Triangle $ABC$ has all angles less than $120^{\\circ}$ is OUTSIDE of the red outline, but inside of the green outline shown in the graph below: $[asy] unitsize(0.7cm); draw((-1,0)--(7,0),Arrows); draw((0,-1)--(0,7),Arrows); draw((0,6)--(6,6)--(6,0)--(0,0)--cycle,green+linewidth(1)); draw((1,1)--(3,1)--(5,3)--(5,5)--(3,5)--(1,3)--cycle, red); label(\"B\",(7,0),SE); label(\"C\",(0,7),NW); [/asy]$ The red hexagon has vertices at coordinates $(60,60),(180,60),(300,180),(300,300),(180,300),$ and $(60,180)$, while the green square has vertices at coordinates $(0,0),(0,360),(360,360),$ and $(360,0)$. Each other these coordinates represent positions for points $B$ and $C$. For example, $(240, 181)$ means point $B$ is at $(240^{\\circ})$ on the unit circle, while point $C$ is at $(181^{\\circ})$ on the unit circle. Therefore, the probability of points $A, B,$ and $C$ forming a triangle with angles less than $120^{\\circ}$ is: $\\[1-\\frac{2\\cdot120^2+2\\cdot\\frac{120^2}{2}}{360^2}$$ $$=1-\\frac{1}{3}$$ $$=\\frac{2}{3}$$. And the answer is $2+3=\\boxed{005}$ - Solution by adyj", "y + 86^2 x$ $x^2 + xy + 86^2 = 97^2$ (Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.) $x(x+y) = (97+86)(97-86)$ $x(x+y) = 2013$ The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\\boxed{\\textbf{(D) }61}$. ### Solution 3 Let $CX=x, BX=y$. Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$. Use power of a point on point C to the circle centered at A. So $CX*CB=CD*CE=>$ $x(x+y)=(97-86)(97+86)=>$ $x(x+y)=3*11*61$. Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$. By the Triangle Inequality, only$x+y=61$ yields a possible length of $BX+CX=BC$. Therefore, the answer is $\\boxed{\\textbf{(D) }61}$. ### Solution 4 $[asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label(\"A\",A,S); label(\"B\",B,NE); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,NE); label(\"86\",(A+B)/2,NW); label(\"86\",(A+D)/2,SE); label(\"97\",(A+C)/2,S); label(\"h\",(A+E)/2,N); label(\"k\",(E+D)/2,NE); label(\"k\",(B+E)/2,NE); label(\"m\",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label(\"90^\\circ\",anglemark(A,E,D),3*S); [/asy]$ We first draw the height of isosceles triangle ABD, and get two equations by the Pythagorean Theorem. First, $h^2 + k^2 = 86^2$. Second, $h^2", "internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\cdots$ and a sequence of points on the circles $A_1,A_2,A_3,\\cdots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. \\usepackage{asymptote} \\begin{figure}[h] \\begin{asy} draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"$A_0$\",(125,0),E); dot((25,100)); label(\"$A_1$\",(25,100),SE); dot((-55,20)); label(\"$A_2$\",(-55,20),E); \\end{asy} \\end{figure} ## Problem 13 For each integer $n\\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$. ## Problem 14 A $10\\times10\\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of", "# Difference between revisions of \"2017 AIME II Problems/Problem 12\" ## Problem Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=\\left(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1}\\right)$. The distance from $B$ to the origin is $\\sqrt{\\left(\\frac{1-r}{r^2+1}\\right)^2+\\left(\\frac{r-r^2}{r^2+1}\\right)^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, and the distance from the origin is $\\frac{49}{61}$. $49+61=\\boxed{110}$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1})$. The distance from $B$ to the origin is $\\sqrt{(\\frac{1-r}{r^2+1})^2+(\\frac{r-r^2}{r^2+1}))^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, we find that the distance from the origin is $\\frac{49}{61} \\implies 49+61=\\boxed{110}$. # See Also 2017 AIME II (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11" ]

[ "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "+0 0 82 1 The diagram below shows four circles. Let $A_1,$ $A_2,$ $A_3,$ $A_4$ denote the areas of the red region, yellow region, green region, blue region, respectively. These areas satisfy $A_1 = \\frac{A_2}{2} = \\frac{A_3}{3} = \\frac{A_4}{4}.$Let $r_1$ denote the smallest radius, and let $r_4$ denote the largest radius. Find $\\frac{r_4}{r_1}.$ [asy] unitsize(1 cm); pair[] O; real[] r; O[1] = (0,0); O[2] = (0.1,0.2); O[3] = (-0.2,-0.1); O[4] = (0.1,-0.3); r[1] = 1; r[2] = 1.5; r[3] = 2; r[4] = 2.5; fill(Circle(O[4],r[4]),lightblue); draw(Circle(O[4],r[4])); label(\"$A_4$\", (1.8,-1.5)); fill(Circle(O[3],r[3]),lightgreen); draw(Circle(O[3],r[3]));label(\"$A_3$\", (-1.3,-1.3)); fill(Circle(O[2],r[2]),yellow); draw(Circle(O[2],r[2]));label(\"$A_2$\", (1,1)); fill(Circle(O[1],r[1]),lightred); draw(Circle(O[1],r[1]));label(\"$A_1$\", O[1]); [/asy] Jan 25, 2020", "\\dots, 21$. This is a total of $\\boxed{(D) 13}$ distinct line segments. (asymptote diagram added by elements2015) Solution 2 - Circles Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\\{15, 16, 17, 18, 19, 20 \\}$ intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this. $[asy] unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, NE); dot(\"P\", P, S); draw(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); [/asy]$ It follows that we can draw circles of radii $15, 16, 17, 18, 19,$ and $20,$ that each contribute two integer lengths (since these circles intersect the hypotenuse twice) from $B$ to $\\overline{AC}$ and one circle of radius $21$ that contributes only one such segment. Our answer is then", "draw(circle((-4,-4), 4*1.41421356237)); [/asy]$ The way we figure out the area is by splitting it the following way: $[asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]$ We know each of the red lines is a diameter of the circle which is $\\sqrt2$. So the area of the red is 2. We know that the area of each semicircle is $\\dfrac14 \\pi$ so the area of the semicircles combines is $\\pi$. Thus we get $\\boxed{\\textbf{(B)} \\pi+2}.$", "Press \"Enter\" to skip to content Given a 2D integer array circles where circles[i] = [xi, yi, ri] represents the center (xi, yi) and radius ri of the ith circle drawn on a grid, return the number of lattice points that are present inside at least one circle. Note: • lattice point is a point with integer coordinates. • Points that lie on the circumference of a circle are also considered to be inside it. Example 1: Input: circles = [[2,2,1]] Output: 5 Explanation: The figure above shows the given circle. The lattice points present inside the circle are (1, 2), (2, 1), (2, 2), (2, 3), and (3, 2) and are shown in green. Other points such as (1, 1) and (1, 3), which are shown in red, are not considered inside the circle. Hence, the number of lattice points present inside at least one circle is 5. Example 2: Input: circles = [[2,2,2],[3,4,1]] Output: 16 Explanation: The figure above shows the given circles. There are exactly 16 lattice points which are present inside at least one circle. Some of them are (0, 2), (2, 0), (2, 4), (3, 2), and (4, 4). Constraints: • 1 <= circles.length <= 200 • circles[i].length == 3 • 1 <= xi, yi <= 100 • 1 <= ri <= min(xi,", "{-3, -2, -1, 0, 1, 2, 3} That is 21 possible sets For B = {-2, 2} A = {-2, -1, 0, 1, 2} This adds 10 sets For B = {-3, 3} A = {-1, 0, 1} And here we have another 6 sets |A| or |B| cannot exceed 3 so we have all the possible sets covered here 21 + 10 + 6 = 37 10. Nicos Mavrommatis says: The pairs (A,B) are: (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(3,0),(3,1),(0,-1),(0,-2),(0,-3),(-1,0),(-1,1),(1,-1),(-1,-1),(-1,2),(1,-2),(-1,-2),(-1,3),(1,-3),(-1,-3),(-2,0),(-2,1),(2,-1),(-2,-1),(-2,2),(2,-2),(-2,-2),(-3,0),(-3,1),(3,-1),(-3,-1) a total of 37 pairs. 11. Cichy says: Use of Geogebra: inequality x^2 + y^2 <= 12 gives a circle with circumfarance Calculating points in different areas of a a circle x-axis and y-axis: 2x6 = 12 Point of origin: 1 4x6 = 24 Total= 37 12. Cichy says: Geogebra: inequality x^2 + y^2 <= 12. Result: a circle with circumference Counting points: axes: 2x6=12 point of origin 1 Total: 12+1+24=37 13. John says: 37 Solutions in total A is 0, B can be 0, +/-1, +/-2, +/-3 7 solutions A is +/-1 B can be 0, +/-1, +/-2, +/-3 14 solutions A is +/-2 B can be 0, +/-1, +/-2 10 solutions A is +/-3 B can be 0, +/-1 6 solutions 14. mohamed says: there are (4) solutions for ِِA^2 + B^2 ≤ 12 (1,1) (1,2) (2,1) (2,2) ### Comment", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$. Let the radius of circle $P$ be $r$. Using Distance Formula, we get the following system of three equations: $$h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2$$ By simplifying, we get $$h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1$$ By subtracting the first equation from the second and third equations, we get $$8r=-4h+12, 10r=2h+6$$ which simplifies to $$2r=3-h, 5r=h+3$$ When we add these two equations, we get $$7r=6$$ So $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ $w^5$ ## Solution 5(Inversion from Circular Inversion) Let $\\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram: $[asy] size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NE); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0),", "\\setlength{\\unitlength}{10pt} \\begin{picture}(0,5)(0,-3) \\put(0,0){\\vector(1,0){35}} \\multiput(1,-0.5)(2,0){17}{\\line(0,1){1}} \\put(0.5,-1.3){$-8$} \\put(2.5,-1.3){$-7$} \\put(4.5,-1.3){$-6$} \\put(6.5,-1.3){$-5$} \\put(8.5,-1.3){$-4$} \\put(10.5,-1.3){$-3$} \\put(12.5,-1.3){$-2$} \\put(14.5,-1.3){$-1$} \\put(16.7,-1.3){$0$} \\put(18.7,-1.3){$1$} \\put(20.7,-1.3){$2$} \\put(22.7,-1.3){$3$} \\put(24.7,-1.3){$4$} \\put(26.7,-1.3){$5$} \\put(28.7,-1.3){$6$} \\put(30.7,-1.3){$7$} \\put(26.7,-1.3){$8$} \\put(5.0,0){\\circle*{0.6}} %\\put(15.0,0){\\circle*{0.6}} %\\put(23.0,0){\\circle*{0.6}} \\put(29.0,0){\\circle*{0.6}} %\\put(16.5,1){\\vector(-1,0){11.5}} %\\put(17.5,1){\\vector(1,0){5.5}} %\\put(17.5,2){$3$is$3$units away from$0$.} %\\put(4.5,2){$-6$is$6$units away from$0$.} \\end{picture}\\\\ The numbers are$x=-6$and$x=6$. \\item %%% 16 Find two numbers such that$|x|>6$.\\\\[10pt] {\\bf Solution:}\\\\[10pt] Find two numbers such that$|x|>6$. There are infinitely many solutions, like$x=7$,$x=6,543$,$x=6.03$,$x=-6.5$,$x=-27$.\\\\ \\setlength{\\unitlength}{10pt} \\begin{picture}(0,5)(0,-3) \\put(0,0){\\vector(1,0){35}} \\multiput(1,-0.5)(2,0){17}{\\line(0,1){1}} \\put(0.5,-1.3){$-8$} \\put(2.5,-1.3){$-7$} \\put(4.5,-1.3){$-6$} \\put(6.5,-1.3){$-5$} \\put(8.5,-1.3){$-4$} \\put(10.5,-1.3){$-3$} \\put(12.5,-1.3){$-2$} \\put(14.5,-1.3){$-1$} \\put(16.7,-1.3){$0$} \\put(18.7,-1.3){$1$} \\put(20.7,-1.3){$2$} \\put(22.7,-1.3){$3$} \\put(24.7,-1.3){$4$} \\put(26.7,-1.3){$5$} \\put(28.7,-1.3){$6$} \\put(30.7,-1.3){$7$} \\put(32.7,-1.3){$8$} \\put(5.0,0){\\circle{0.6}} \\put(28.9,0.05){\\line(1,0){5}} \\put(28.9,0.1){\\line(1,0){5}} \\put(28.9,0.15){\\line(1,0){5}} \\put(28.9,0.2){\\line(1,0){5}} \\put(28.9,0.25){\\line(1,0){5}} \\put(4.7,-0.05){\\line(-1,0){5}} \\put(4.7,-0.1){\\line(-1,0){5}} \\put(4.7,-0.15){\\line(-1,0){5}} \\put(4.7,-0.2){\\line(-1,0){5}} \\put(4.7,-0.25){\\line(-1,0){5}} %\\put(23.0,0){\\circle*{0.6}} \\put(29.0,0){\\circle{0.6}} %\\put(16.5,1){\\vector(-1,0){11.5}} %\\put(17.5,1){\\vector(1,0){5.5}} %\\put(17.5,2){$3$is$3$units away from$0$.} %\\put(4.5,2){$-6$is$6$units away from$0$.} \\end{picture} \\item %%% 17 Is$-1\\le -5$true or false?.\\\\[10pt] {\\bf Solution:}\\\\[10pt]$-1\\le -5$is false because %$-5$is not to the right of$-1$. \\\\$-1< -5$is false and$-1= -5$is false.\\\\ \\setlength{\\unitlength}{10pt} \\begin{picture}(0,5)(0,-3) \\put(0,0){\\vector(1,0){35}} \\multiput(1,-0.5)(2,0){17}{\\line(0,1){1}} \\put(0.5,-1.3){$-8$} \\put(2.5,-1.3){$-7$} \\put(4.5,-1.3){$-6$} \\put(6.5,-1.3){$-5$} \\put(8.5,-1.3){$-4$} \\put(10.5,-1.3){$-3$} \\put(12.5,-1.3){$-2$} \\put(14.5,-1.3){$-1$} \\put(16.7,-1.3){$0$} \\put(18.7,-1.3){$1$} \\put(20.7,-1.3){$2$} \\put(22.7,-1.3){$3$} \\put(24.7,-1.3){$4$} \\put(26.7,-1.3){$5$} \\put(28.7,-1.3){$6$} \\put(30.7,-1.3){$7$} \\put(32.7,-1.3){$8$} \\put(7.0,0){\\circle*{0.6}} %\\put(28.9,0.1){\\line(1,0){5}} %\\put(28.9,0.15){\\line(1,0){5}} %\\put(28.9,0.2){\\line(1,0){5}} %\\put(28.9,0.25){\\line(1,0){5}} %\\put(4.7,-0.05){\\line(-1,0){5}} %\\put(4.7,-0.1){\\line(-1,0){5}} %\\put(4.7,-0.15){\\line(-1,0){5}} %\\put(4.7,-0.2){\\line(-1,0){5}} %\\put(4.7,-0.25){\\line(-1,0){5}} %\\put(23.0,0){\\circle*{0.6}} \\put(15.0,0){\\circle*{0.6}} %\\put(16.5,1){\\vector(-1,0){11.5}} %\\put(17.5,1){\\vector(1,0){5.5}} %\\put(17.5,2){$3$is$3$units away from$0$.} %\\put(4.5,2){$-6$is$6$units away from$0$.} \\end{picture} \\item %%% 18 There are at most$270$calories in a consumer product. What is the highest number of calories in the product? (Use a whole number.)\\\\[10pt] {\\bf Solution:}\\\\[10pt] %There are at most 270 calories in a product. What is the highest number of calories in the product? (Use a whole number.) \\\\", "instead of an arbitrary circle, but I actually think that the arbitrary circle is more intuitive. So consider this arbitrary circle with radius “r.” From the center I can draw any radius I want, below are some examples. $\\setlength{\\unitlength}{1mm} \\begin{picture}(140, 40) \\put(0,20){\\circle{14}} \\put(0,20){\\line(1,0){7}} \\put(0.7,20.3){r} \\put(15,20){\\circle{14}} \\put(15,20){\\line(1,1){5}} \\put(15.7,24){r} \\put(30,20){\\circle{14}} \\put(30,20){\\line(0,1){7}} \\put(30.7,24){r} \\put(45,20){\\circle{14}} \\put(45,20){\\line(-1,1){5}} \\put(44.7,24){r} \\put(60,20){\\circle{14}} \\put(60,20){\\line(-1,0){7}} \\put(60.3,20.3){r} \\put(75,20){\\circle{14}} \\put(75,20){\\line(-1,-1){5}} \\put(74.7,16){r} \\put(90,20){\\circle{14}} \\put(90,20){\\line(0,-1){7}} \\put(90.7,16){r} \\put(105,20){\\circle{14}} \\put(105,20){\\line(1,-1){5}} \\put(104.7,16){r} \\put(120,20){\\circle{14}} \\put(120,20){\\line(1,0){7}} \\put(120.3,20.3){r} \\end{picture}$ For any given radial line, I can draw a right triangle: $\\setlength{\\unitlength}{1mm} \\begin{picture}(140, 40) \\put(0,20){\\circle{14}} \\put(0,20){\\line(1,1){5}} \\put(5,20){\\line(0,1){4.5}} \\put(0,20){\\line(1,0){5}} \\put(0,24){r} \\put(2,16){x} \\put(7,22){y} \\end{picture}$ More precisely I am drawing a triangle for any given angle $\\theta \\text{ (theta)}$ that I want to make. The trig functions take this angle, and tell you the ratio of different values. Notice that in this triangle we have the hypotenuse and two other sides, where the hypotenuse is just the radius. Because the radius is always the same in a circle, the hypotenuse of any triangle I draw with this example circle will be the same. Only the lengths of the other two lines change as the angle changes. The sine function tells you the ratio of the length of the the side that the angle is facing, y, to the radius r. Cosine tells you the ratio of the length of the side that", "+0 # hhhhheeeelllllpppp -1 72 4 What is the ratio of the smaller circle's area to the larger circle's area? Give your answer in fully simplified form. It should look like \"x:y\", where x and y are replaced by integers. [asy] size(4cm); pair o=(0,0); pair x=(0.9,-0.4); draw(Circle(o,sqrt(0.97))); draw(Circle((o+x)/2,sqrt(0.97)/2)); dot(o); dot(x); dot(-x); draw(-x--x); [/asy] Apr 8, 2020 #1 0 im sorry i dont know latex Apr 8, 2020 #2 +930 0 This is some sort of assembly script not laTex. I realize that I thought this was laTex the last time you posted this question. You'll have to wait for someone who knows how to use this script to solve this problem for you. Hope it helps! Apr 8, 2020 #3 +4569 +1 The diagram should look like this: Let the smaller circle have radius of length y units and that means that the bigger circle has a radius of length 2y units. Thus, the smaller circle's area is $$\\pi(y)^2$$ and the larger circle's area is $$\\pi(2y)^2=\\pi4y^2$$ units. Thus, the answer is 1:4. Apr 8, 2020 #4 +930 0 Thank you!... and what did you use to make the diagram? HELPMEEEEEEEEEEEEE Apr 8, 2020", "y + 86^2 x$ $x^2 + xy + 86^2 = 97^2$ (Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.) $x(x+y) = (97+86)(97-86)$ $x(x+y) = 2013$ The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\\boxed{\\textbf{(D) }61}$. ### Solution 3 Let $CX=x, BX=y$. Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$. Use power of a point on point C to the circle centered at A. So $CX*CB=CD*CE=>$ $x(x+y)=(97-86)(97+86)=>$ $x(x+y)=3*11*61$. Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$. By the Triangle Inequality, only$x+y=61$ yields a possible length of $BX+CX=BC$. Therefore, the answer is $\\boxed{\\textbf{(D) }61}$. ### Solution 4 $[asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label(\"A\",A,S); label(\"B\",B,NE); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,NE); label(\"86\",(A+B)/2,NW); label(\"86\",(A+D)/2,SE); label(\"97\",(A+C)/2,S); label(\"h\",(A+E)/2,N); label(\"k\",(E+D)/2,NE); label(\"k\",(B+E)/2,NE); label(\"m\",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label(\"90^\\circ\",anglemark(A,E,D),3*S); [/asy]$ We first draw the height of isosceles triangle ABD, and get two equations by the Pythagorean Theorem. First, $h^2 + k^2 = 86^2$. Second, $h^2", "$k\\ge1$, the circles in $\\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\\sum_{C\\in S} \\frac{1}{\\sqrt{r(C)}}?$$ $[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); } [/asy]$ $\\textbf{(A)}\\ \\frac{286}{35}", "# Difference between revisions of \"2017 AIME II Problems/Problem 12\" ## Problem Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=\\left(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1}\\right)$. The distance from $B$ to the origin is $\\sqrt{\\left(\\frac{1-r}{r^2+1}\\right)^2+\\left(\\frac{r-r^2}{r^2+1}\\right)^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, and the distance from the origin is $\\frac{49}{61}$. $49+61=\\boxed{110}$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1})$. The distance from $B$ to the origin is $\\sqrt{(\\frac{1-r}{r^2+1})^2+(\\frac{r-r^2}{r^2+1}))^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, we find that the distance from the origin is $\\frac{49}{61} \\implies 49+61=\\boxed{110}$. # See Also 2017 AIME II (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11", "# Problem #2221 2221 The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\\le|x|\\le7$, $3\\le|y|\\le7$. How many squares of side at least $6$ have their four vertices in $G$? $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); [/asy]$$\\textbf{(A)}\\ 125\\qquad \\textbf{(B)}\\ 150\\qquad \\textbf{(C)}\\ 175\\qquad \\textbf{(D)}\\ 200\\qquad \\textbf{(E)}\\ 225$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so:", "# Difference between revisions of \"2017 AIME II Problems/Problem 12\" ## Problem Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series", "on the definition of the circle from Equation $\\ref{eq:circle}$. We can tweak it in terms of any $(x, y)$ pair: $$$\\label{eq:error-margin} F(x,y) = x^2 + y^2 - r^2 = \\text{distance from true circle line}.$$$ Note that if $F(x,y) = 0$, then the point $(x, y)$ is exactly on the circle. If $F(x,y) > 0$, then the point is outside of the circle, and if $F(x,y) < 0$ then the point is inside of it. In other words, given any point $(x, y)$, $F(x, y)$ is the distance from the true circle line. ## Choosing between E or SE Let’s remind ourselves that we’ll always be moving E or SE. One critical (pragmatic) property here is that we’re dealing with a pixel grid with integer increments. There is a very high chance that neither the E or SE pixels we’re moving to is exactly on the circle. This is because the only time that the point $(x,y)$ will exactly be on the line of the circle is if the $x$, $y$, and $r$ values (as integers) form a so-called Pythagorean Triple. For $r < 100$, there are only 50 such triples: ( 3, 4, 5) (18,24,30) (24,45,51) (16,63,65) (51,68,85) ( 6, 8,10) (16,30,34) (20,48,52) (32,60,68) (40,75,85) ( 5,12,13) (21,28,35) (28,45,53) (42,56,70) (36,77,85) ( 9,12,15) (12,35,37) (33,44,55) (48,55,73) (13,84,85) (", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "# 2017 AIME II Problems/Problem 12 ## Problem Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution 1 Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$", "the number of fixed points under possible transformations: 1. The identity transformation. Since this doesn't change anything, there are $60$ fixed points 2. Reflect on a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$. Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection 3. Rotate by $120^\\circ$ counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$, $4$, and $6$ will be the same. Similarly, the colors of circles $2$, $3$, and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation. By Burnside's Lemma, the total number of colorings is $(1 \\cdot 60+3 \\cdot 4+2 \\cdot 0)/(1+3+2) = \\boxed{\\textbf{(D) } 12}$. ### Solution 3 Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as $[asy] filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1)); draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); [/asy]$ Take the first case. Now, we must pick" ]

[ "\\renewcommand\\arraystretch{\\Dp}% \\setlength\\svarraycolsep{\\arraycolsep}% \\setlength\\arraycolsep{0in}% \\begin{array}{c}% \\vstretch{\\Ht}{% \\hstretch{\\Wd}{% \\trimbox{.15ex .75ex .15ex .2ex}{\\scalerel*{\\char'136}{\\rule{1ex}{1ex}}}% }% }\\\\% #1\\\\% \\rule{1ex}{0ex}\\\\% \\end{array}% \\renewcommand\\arraystretch{1.0}% \\setlength\\arraycolsep{\\svarraycolsep}% } \\parindent 0in \\begin{document} Hat: $$\\hat{\\mathrm{H}} = \\hat{\\mathrm{T}} + \\hat{\\mathrm{V}}$$ Carat(char'136): \\def\\Ht{1.8} \\def\\Wd{4.5} \\def\\Dp{.3} $$\\althat {\\mathrm{H}} = \\althat {\\mathrm{T}} + \\althat {\\mathrm{V}}$$ \\end{document} - But the \\wedge is not the same character as the the one \\hat gives out. – jjdb Apr 11 '13 at 20:29 @jjdb I fixed the wedge issue by using \\char'136 instead – Steven B. Segletes Apr 12 '13 at 12:10", "16 & 1e & 26 & 2e & 36 & 3e \\\\ 07 & 0f & 17 & 1f & 27 & 2f & 37 & 3f \\end{bmatrix}$ \\end{document} As pointed out by OP on comments, \\renewcommand*{\\arraystretch}{.5} should work.", "\\begingroup\\renewcommand{\\arraystretch}{1.097} \\begin{tabularx}{.45\\columnwidth}[t]{XX} \\toprule \\midrule Row 1 & Row 1 \\\\ Row 2 & Row 2 \\\\ Row 3 & Row 3 \\\\ Row 4 & Row 4 \\\\% \\vspace{3.05pt} \\\\ \\bottomrule \\end{tabularx} \\endgroup \\begin{tabularx}{.45\\columnwidth}[t]{XX} \\toprule", "# How do I create this semi-enclosed arrow symbol? I would like to create a symbol for \\implies and \\impliedby where they are enclosed in a rotated lidless box as shown in the picture. Can anybody tell me how I could achieve this? Welcome! You do not need any packages for that. \\documentclass{article} \\newcommand{\\RightArrowInBox}{\\begingroup \\renewcommand{\\arraystretch}{0.6}\\begin{array}{@{}c@{}|}\\hline \\raisebox{-0.2ex}{$\\Rightarrow$}\\\\ \\hline\\end{array}\\endgroup} \\newcommand{\\LeftArrowInBox}{\\begingroup \\renewcommand{\\arraystretch}{0.6}\\begin{array}{@{}c@{}|}\\hline \\raisebox{-0.2ex}{$\\Leftarrow$}\\\\ \\hline\\end{array}\\endgroup} \\begin{document} $\\RightArrowInBox~\\LeftArrowInBox$ \\end{document} Of course you can modify the padding etc. A version without cutting corners with major input by barbara beeton. \\documentclass{article} \\newcommand{\\RightArrowInBox}{\\begingroup \\renewcommand{\\arraystretch}{0.6}\\begin{array}{@{}c@{\\rule[-3pt]{0.4pt}{0.85em}}}\\hline \\raisebox{-0.28ex}{$\\Rightarrow$}\\\\ \\hline\\end{array}\\endgroup} \\newcommand{\\LeftArrowInBox}{\\begingroup \\renewcommand{\\arraystretch}{0.6}\\begin{array}{@{}c@{\\rule[-3pt]{0.4pt}{0.85em}}}\\hline \\raisebox{-0.28ex}{$\\Leftarrow$}\\\\ \\hline\\end{array}\\endgroup} \\begin{document} $\\RightArrowInBox~\\LeftArrowInBox$ \\end{document} Or with trimclip. \\documentclass{article} \\usepackage{trimclip} \\newcommand{\\RightArrowInBox}{\\begingroup \\clipbox{0.5ex 0em 0em 0em}{\\fbox{$\\Rightarrow\\!$}}\\endgroup} \\newcommand{\\LeftArrowInBox}{\\begingroup \\clipbox{0.5ex 0em 0em 0em}{\\fbox{$\\Rightarrow\\!$}}\\endgroup} \\begin{document} $\\RightArrowInBox~\\LeftArrowInBox$ \\end{document} • The corners aren't closed. Maybe add to the vertical two times the thickness of the rules. – barbara beeton Mar 30 at 1:13 • @barbarabeeton Thanks! Yes, I know, we cats are famously good at cutting the corners. ;-) – Schrödinger's cat Mar 30 at 1:14 • Hmm. Somehow, when adding corners, a little too much extra space got added below the arrows. (Nit picking.) The corners do look nice though. – barbara beeton Mar 30 at 2:47 • @barbarabeeton This is entirely controlled by the height of the rule. If the requirement is to make it symmetric, then you can either make the", "you should try \\renewcommand\\arraystretch{1.25}; for 11pt and 12pt main font sizes, you should use values of 1.213 and 1.241, respectively. In the screenshot below, the left-hand and right-hand tabular environments are typeset before and after a \\renewcommand\\arraystretch{1.5} is executed, respectively. (The default value is 1.) \\documentclass{article} %define a dummy table \\newcommand\\mytab{% \\begin{tabular}[t]{lll} \\hline a & b & c\\\\ \\hline d & e & f \\\\ \\hline h & i & k \\\\ \\hline \\end{tabular}} \\begin{document} \\mytab • @Ulysses - I found these numbers inside the file setspace.sty, in the section that defines \\onehalfspacing. – Mico Nov 24 '15 at 6:49", "& \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem by", "\\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem", "to do with math. we might be able to get a better idea of what's happening if you post a compilable example that starts with your document class and ends with \\end{document}, and shows what you've done so far. – barbara beeton Jun 16 '16 at 17:18 ## 1 Answer While \\midrule works in aligned, you get much finer control with array. \\documentclass{article} \\usepackage{amsmath,booktabs,cancel,array,xcolor} \\newcommand{\\rs}[1]{\\textcolor{red}{#1}} \\begin{document} $$\\renewcommand{\\arraystretch}{1.2} \\begin{array}{@{}r@{}>{{}}l@{}} 6x\\; {\\cancel{-\\;5y}} &= 11 \\\\ 7x\\; {\\cancel{+\\;5y}} &= 2 \\\\ \\midrule 13x &= 13 \\\\ x &= 1 \\end{array}$$ \\begin{equation*} \\renewcommand{\\arraystretch}{1.2} \\begin{array}{ @{}r@{}>{{}}l@{\\qquad} @{}r@{}>{{}}l@{\\qquad} @{}r@{}>{{}}l@{} } 18a+10c &= 376 & 18a+10c &= 376 & 18a+10c &= 376 \\\\ a+ c &= 24 & \\rs{-18(}a+c &= 24 \\rs{)} & -18a-18c &= -432 \\\\ \\cmidrule{5-6} & & & & -8c &= -56 \\\\ & & & & c &= 7 \\end{array} \\end{equation*} \\end{document} You can get separate rules below the three groups by adding columns \\begin{equation*} \\renewcommand{\\arraystretch}{1.2} \\setlength{\\arraycolsep}{0pt} \\newcommand{\\sep}{\\mbox{\\qquad}} \\begin{array}{ r@{}>{{}}l c r@{}>{{}}l c r@{}>{{}}l } 18a+10c &= 376 &\\sep& 18a+10c &= 376 &\\sep& 18a+10c &= 376 \\\\ a+ c &= 24 & & \\rs{-18(}a+c &= 24 \\rs{)} & & -18a-18c &= -432 \\\\ \\cmidrule{1-2} \\cmidrule{4-5} \\cmidrule{7-8} & & & & & & -8c &= -56 \\\\ & & & & & & c &= 7 \\end{array} \\end{equation*} • Oh this is", "# Replicate a typesetting of polynomial long division I found a typesetting of series long division I want to replicate in a book I've been reading: So I tried with the following code (using different polynomial than the example): \\begin{align*} \\renewcommand\\arraystretch{2} \\begin{array}{r@{\\hskip\\arraycolsep}c@{\\hskip\\arraycolsep}l*2r} &&\\dfrac{1}{z}+\\dfrac{1}{3!}z&+\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^3+\\cdots\\\\ \\cline{2-4} z-\\dfrac{1}{3!}z^3+\\dfrac{1}{5!}z^4-\\cdots&\\Bigg)&1\\\\ &&1-\\dfrac{1}{3!}z^2&+\\dfrac{1}{5!}z^4-\\cdots\\\\ \\cline{2-4} &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{5!}z^4+\\cdots\\\\ &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{(3!)^2}z^4+\\cdots\\\\ \\cline{2-4} &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\ &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\ \\cline{2-4} &&&\\multicolumn{1}{c}{\\vdots}\\\\ \\end{array} \\end{align*} and here is the result: As you may see, the are some differences between the example and my attempt (besides fonts, of course). The biggest problem seems to be associate with the vertical spacing, is there anything I can do to improve the result? Thank you very much! • First of all you have some minor mistakes regarding some signs in your code compared to the \"original\" version. Is this normal or just typing mistakes? – Ludovic C. Jul 12 '13 at 10:33 • I used a different polynomial. Sorry if I didn't state it clear. – Francis Jul 12 '13 at 10:35 • You have that you used a different polynomial in your post. It is right below the first image. – dustin Jul 12 '13 at 10:38 • Welcome to TeX.SX! You can have a look on our our starter page to familiarize yourself further with our format. – Claudio Fiandrino Jul 12 '13 at 10:48 • isn't that Brown/Churchill:Complex", "& \\text{if $a < 1$}, \\tag{1.1a} \\\\ S_1 = \\smash{\\left\\{\\renewcommand{\\arraystretch}{1.2} \\begin{array}{@{}c@{}}\\mathstrut\\\\\\mathstrut\\\\\\mathstrut\\end{array}\\right.\\kern-\\nulldelimiterspace} & D + 2\\bigl( \\textstyle\\sum_i EF + \\nonumber \\\\ & \\quad 2 GH \\bigr) + K - L & \\text{if $a = 1$}. \\tag{1.1b} \\\\ S_2 = \\mathrlap{1} \\hphantom{\\left\\{ \\begin{array}{@{}c@{}}\\mathstrut\\\\\\mathstrut\\end{array}\\right.\\kern-\\nulldelimiterspace}& \\tag{1.2} \\\\ \\raisebox{-.6\\normalbaselineskip}[0pt][0pt]{$S_3 = \\smash{\\left\\{ \\begin{array}{@{}c@{}}\\mathstrut\\\\\\mathstrut\\end{array}\\right.\\kern-\\nulldelimiterspace}$} & \\textstyle\\sum_i I + \\sum_i J & \\text{if $a = 0$} \\tag{1.3a} \\\\ \\end{alignat} \\end{document} As an extremely partial answer, I note that subequationsenvironement can be nested with amsmath. Playing with counters allows to obtain the desired numbering. The obvious and important unsolved problem is the outer brace. \\documentclass[12pt, a4paper]{article} \\usepackage{amsmath} \\begin{document} \\begin{subequations} \\renewcommand{\\theequation}{\\theparentequation.\\arabic{equation}} \\begin{subequations} \\renewcommand{\\theequation}{\\theparentequation.\\alph{equation}} $$a = b$$ \\end{subequations} \\begin{subequations} \\renewcommand{\\theequation}{\\theparentequation.\\alph{equation}} $$a = b$$ \\end{subequations} \\end{subequations} \\end{document} • For the outer brace, an array environment with \\left\\{ and \\right. should work. – Bibi Dec 10 '15 at 16:33 • Hum. I do not know how to proceed on the basis on my rough sketch and your suggestion. – Denis Dec 10 '15 at 16:40", "Sigur Jun 6 '14 at 16:31 • No haven't tried that, just doing it. – Soham Jun 6 '14 at 16:32 • Thanks, sigur. That worked perfectly for me. I used \\renewcommand\\arraystretch{0.75} Was sceptical if stretch could be given a shrinking value. – Soham Jun 6 '14 at 16:37 • BTW, the hexadecimal letters a to f should not be set in math italics, but upright. – Heiko Oberdiek Jun 6 '14 at 19:25 You're probably using setspace with the \\doublespacing option, which gives Since \\doublespacing sets the \\baselinestretch to 1.6667, you can countermand it for arrays with \\renewcommand{\\arraystretch}{0.6} because 0.6 is the inverse of 1.6667. \\documentclass{article} \\usepackage{amsmath} \\usepackage{setspace} \\doublespacing \\renewcommand{\\arraystretch}{0.6} % because \\baselinestretch is 1.6667 \\begin{document} $\\begin{bmatrix} 00 & 08 & 10 & 18 & 20 & 28 & 30 & 38 \\\\ 01 & 09 & 11 & 19 & 21 & 29 & 31 & 39 \\\\ 02 & 0a & 12 & 1a & 22 & 2a & 32 & 3a \\\\ 03 & 0b & 13 & 1b & 23 & 2b & 33 & 3b \\\\ 04 & 0c & 14 & 1c & 24 & 2c & 34 & 3c \\\\ 05 & 0d & 15 & 1d & 25 & 2d & 35 & 3d \\\\ 06 & 0e &", "\\renewcommand*\\env@cases[1][1.2]{% \\let\\@ifnextchar\\new@ifnextchar \\left\\lbrace \\def\\arraystretch{#1}% \\array{@{}l@{\\quad}l@{}}% } \\makeatother Now by \\begin{cases} … \\end{cases} the default value of 1.2 will be used, but by using the optional parameter like \\begin{cases}[0.8] … \\end{cases} the spacing will be adjusted accordingly. This topic was discussed on mrunix.de. Category: Mathematics | 6 Comments »", "(that varies depending on the \\arraystretch that may have been redefined). This gives a square output. \\documentclass{article} \\usepackage{collcell} \\newcommand{\\fwcell}[1]{\\makebox[\\arraystretch\\normalbaselineskip]{$#1$}} \\begin{document} \\begingroup \\setlength{\\tabcolsep}{0pt} \\begin{tabular}{ | *{3}{>{\\collectcell\\fwcell}c<{\\endcollectcell} |} } \\hline 1 & 2 & 3 \\\\ \\hline 4 & 5 & 6 \\\\ \\hline 7 & 8 & 9 \\\\ \\hline \\end{tabular} \\renewcommand{\\arraystretch}{2}% \\begin{tabular}{ | *{3}{>{\\collectcell\\fwcell}c<{\\endcollectcell} |} } \\hline 1 & 2 & 3 \\\\ \\hline 4 & 5 & 6 \\\\ \\hline 7 & 8 & 9 \\\\ \\hline \\end{tabular} \\endgroup \\end{document}", "& 0 \\\\ \\usered\\tikzmarkin{d}(0.85,-0.2)(-0.85,0.375) 0 & 0 & (T^{3}_{11}+T^{3}_{12}) \\\\ 0 & 0 & 0\\tikzmarkend{d} \\\\ \\end{array}\\right] \\left[\\begin{array}{c} \\phi_{1} \\\\ \\phi_{2} \\\\ \\phi_{3} \\end{array}\\right] \\label{eq:BLphii}$$ $$\\renewcommand{\\arraystretch}{1.5} A_{R}= \\left[\\begin{array}{cccc} \\usegreen\\tikzmarkin{e} T^{2}_{22} & T^{2}_{21} & 0 & 0 \\\\ 0 & T^{3}_{22}\\tikzmarkend{e} & T^{3}_{21} & 0 \\\\ \\usered\\tikzmarkin{f}(0.1,-0.2)(-0.25,0.375) 0 & 0 & 0 & 0 \\\\ T^{1}_{21} & 0 & 0 & T^{1}_{22} \\tikzmarkend{f}\\\\ \\end{array}\\right] \\left[\\begin{array}{c} \\phi_{A} \\\\ \\phi_{B} \\\\ \\phi_{C} \\\\ \\phi_{D} \\end{array}\\right] \\label{eq:ARphif}$$ $$\\renewcommand{\\arraystretch}{1.5} B_{R}= \\left[\\begin{array}{ccc} \\usegreen\\tikzmarkin{g}(0.15,-0.2)(-0.95,0.375) 0 & -(T^{2}_{21}+T^{2}_{22}) & 0 \\\\ 0 & 0 & -(T^{3}_{21}+T^{3}_{22}) \\tikzmarkend{g} \\\\ \\usered\\tikzmarkin{h}(1.05,-0.2)(-0.95,0.375) 0 & 0 & 0 \\\\ -(T^{1}_{21}+T^{1}_{22}) & 0 & 0 \\tikzmarkend{h} \\\\ \\end{array}\\right] \\left[\\begin{array}{c} \\phi_{1} \\\\ \\phi_{2} \\\\ \\phi_{3} \\end{array}\\right] \\label{eq:BRphii}$$ \\end{document} The result: -", "the cells, see for example https://www.inf.ethz.ch/personal/markusp/teaching/guides/guide-tables.pdf For the math, see also how to insert multi line equation in the tabular environment? Given that most of the cells in the table have a very organized structure, it's probably not a good idea to employ automatic line breaking to render the cells' contents. Instead, I'd use nested tabular environments. (In the code below, I wasn't too sure about the symbol that occurs in two of the three header cells...) \\documentclass{article} \\usepackage{array} \\begin{document} \\begin{center} \\renewcommand\\arraystretch{1.5} \\begin{tabular}{|>{$}c<{$}|c|c|} \\hline Ax^2+Bx+C & $\\mathop{>}_1^{} \\ge 0$ & $\\mathop{>}_1^{} \\le 0$ \\\\ \\hline \\Delta<0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{no real roots}\\\\ always holds true! \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{no real roots} \\\\ \\emph{never} satisfied! \\\\ \\end{tabular} \\\\ \\hline \\Delta = 0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 coinc.\\ solut.} \\\\ $x_{1,2}\\ge x_0$ \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 coinc.\\ solut.} \\\\ $x_{1,2}\\le x_0$ \\end{tabular} \\\\ \\hline \\Delta>0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 dist.\\ solutions}\\\\ $x_1\\ge x_{\\max}$\\\\ $x_2\\le x_{\\min}$ \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 dist.\\ roots}\\\\ $x_{\\min}\\le x \\le x_{\\max}$\\\\ \\null % blank 3rd line \\end{tabular} \\\\ \\hline \\end{tabular} \\end{center} \\end{document} • I thing taht the OP handwriting for second and third column meaning: $>, \\;\\ge 0$ and $<, \\;\\le 0$ ... :-). – Zarko Oct 27 '15 at 2:02 • @Zarko - Thanks, good one! Let's see if the OP weighs", "x_2 \\\\ & 1 + 2s = x_2 \\\\ \\end{aligned} \\right \\} \\textit{Test....} \\\\ &\\trom{3}\\quad 0 + 2s + 3t = x_3 \\\\ \\end{split} \\end{equation*} \\end{document} • Thanks. That's what I've been looking for. I still have a question, though. Why did you add \\!? A google search for Latex exclamation point|mark didn't turn up anything useful. May 15, 2015 at 17:54 • Never mind. I just found it in another answer. May 15, 2015 at 18:20 Here is another option using arrays: \\documentclass{article} \\usepackage{amsmath} \\newcommand{\\toRoman}[1]{\\text{\\MakeUppercase{\\romannumeral #1}}} \\begin{document} \\begin{equation*} \\renewcommand{\\arraystretch}{1.2}% http://tex.stackexchange.com/a/31704/5764 \\begin{array}{r l} \\multicolumn{2}{l}{\\left.\\kern-\\nulldelimiterspace\\begin{array}{@{}r l} \\hphantom{\\toRoman{3}}\\makebox[0pt][r]{\\toRoman{1}} & 2 + x_1 -2 = x_1 \\\\ \\toRoman{2} & 1 + 2s + 0t = x_2 \\\\ & 1 + 2s = x_2", "\\begin{array}{Sr|Sl} \\hhline{~|-} ${#1}$ & ${#2}$\\\\ \\hhline{-|~} \\end{array}% \\endgroup } \\newcommand{\\lcom}[2]{% \\begingroup \\renewcommand{\\arraystretch}{1}% \\setlength{\\arraycolsep}{2pt}% \\begin{array}{r|l} \\hhline{-|~} #1&#2\\\\ \\hhline{~|-} \\end{array}% \\endgroup } \\begin{document} $\\lcom{\\overline{\\underline{R}}}{\\cat{Rng}}$\\\\ $\\rcom{\\cat{Ab}}{\\faktor{G}{H}}$\\\\ $\\rcom{\\underline{\\cat{Top}}}{\\widetilde{G}}$\\\\ $\\rcom{\\cat{Ab}}{\\dfrac{G}{H}}$ \\end{document} You can use an array with the help of the hhline package. An optional argument is provided for increasing the allotted space in case big objects need to be accommodated. \\documentclass{article} \\usepackage{amsmath,hhline} \\newcommand{\\cat}[1]{\\mathbf{#1}} % or whatever you like \\newcommand{\\commacat}[3][1]{% \\begingroup \\renewcommand{\\arraystretch}{#1}% \\setlength{\\arraycolsep}{2pt}% \\begin{array}{r|l} \\hhline{~|-} #2&#3\\\\ \\hhline{-|~} \\end{array}% \\endgroup } \\newcommand{\\cocommacat}[3][1]{% \\begingroup \\renewcommand{\\arraystretch}{#1}% \\setlength{\\arraycolsep}{2pt}% \\begin{array}{r|l} \\hhline{-|~} #2&#3\\\\ \\hhline{~|-} \\end{array}% \\endgroup } \\begin{document} $\\commacat{\\cat{Top}}{B} \\qquad \\cocommacat{X}{\\cat{Set}}$ $\\commacat[1.8]{\\cat{Top}}{\\dfrac{A}{B}} \\qquad \\cocommacat[1.8]{\\dfrac{A}{B}}{\\cat{Top}}$ \\end{document} • Thank you egreg, it seems your solution is almost working, except that it doesn't scale well vertically (e.g. if you put \\dfrac or \\faktor or \\widetilde on the right handside etc.) . Also is there a way to make the line continuous (there is a very apparent line break on the lower corner for both comma and cocomma)? – Alp Uzman Apr 30 '18 at 21:55 • @AlpUzman Now it should fit your needs. – egreg Apr 30 '18 at 22:35", "we can give more distance between elements. The following code can demonstrate how: \\documentclass{article} \\usepackage{amsmath} \\begin{document} Default: $\\left[ \\begin{array}{ccc} \\dfrac{5}{6} & \\dfrac{1}{6} & 0 \\\\ \\dfrac{5}{6} & 0 & \\dfrac{1}{6} \\\\ 0 & \\dfrac{5}{6} & \\dfrac{1}{6} \\end{array} \\right]$ \\renewcommand{\\arraystretch}{2.5} Stretched: $\\left[ \\begin{array}{ccc} \\dfrac{5}{6} & \\dfrac{1}{6} & 0 \\\\ \\dfrac{5}{6} & 0 & \\dfrac{1}{6} \\\\ 0 & \\dfrac{5}{6} & \\dfrac{1}{6} \\end{array} \\right]$ \\end{document} Output of the above code", "$N(t)+1$ is the sample size when we stop. Hence the Renewal Theory is also linked with Sequential Analysis in statistics. ${N(t), t \\in(0, \\infty)}$ is called the Renewal Counting Process. We can also write $N(t)=\\max \\left{n \\mid S_{n} \\leq t\\right} .$ We want to find $P[N(t)=n]$ given $F$. To compute this we proceed as follows: \\begin{aligned} P\\left[S_{2} \\leq t\\right] &=\\int_{0}^{\\infty} F(t-u) d F(u) \\ &=\\int_{0}^{t} F(t-u) d F(u) \\ &=F^{*} F(t)=F^{(2)}(t), \\ldots \\ P\\left[S_{n} \\leq t\\right] &=F^{(n)}(t)=\\int_{0}^{t} F^{(n-1)}(t-u) d F(u), n \\geq 1 \\end{aligned} Define $\\quad F^{(0)}(t)=\\left{\\begin{array}{l}0 \\text { if } t<0 \\\\ 1 \\text { if } t \\geq 0 .\\end{array}\\right.$ Now $P[N(t)=n]=P\\left[S_{1} \\leq t, S_{2} \\leq t, \\ldots, S_{n} \\leq t, S_{n+1}>t\\right]$ $=P\\left[S_{n} \\leq t, S_{n+1}>t\\right]$ (by nonnegativeness of $X_{1}$ ) $=P\\left[S_{n} \\leq t, S_{n}+X_{n+1}>t\\right]$ $=P\\left[t-X_{n+1}<S_{n} \\leq t\\right]$ $=\\int_{0}^{\\infty} P\\left[t-X_{n+1}<S_{n} \\leq t \\mid u<X_{n+1} \\leq u+d u\\right] d F(u)$ $=\\int_{0}^{\\infty} P\\left[t-u<S_{n} \\leq t \\mid u<X_{n+1} \\leq u+d u\\right] d F(u)$ $=\\int_{0}^{\\infty} P\\left[t-u<S_{n} \\leq t\\right] d F(u)\\left(\\right.$ since $S_{n}$ is independent of $\\left.X_{n+1}\\right)$ $=\\int_{0}^{\\infty}\\left{F^{(n)}(t)-F^{(n)}(t-u)\\right} d F(u)$ $=\\int_{0}^{\\infty} F^{(n)}(t) d F(u)-\\int_{0}^{\\infty} F^{(n)}(t-u) d F(u)$ $=F^{(n)}(t)-\\int_{0}^{t} F^{(n)}(t-u) d F(u)$ ## 统计代写|随机过程作业代写stochastic process代考|Renewal Equation Theorem 4.1 (a) $P[N(t)=n]=F^{(n)}(t)-F^{(n+1)}(t)$ (b) $H(t)=\\sum_{n=1}^{\\infty} F^{(n)}(t)$ (c) $H(t)=F(t)+\\int_{0}^{t} H(t-u) d F(u)$, the so-called integral equation of Renewal Theory (Renewal equation). (d) ${N(t), t \\in[0, \\infty)}$ is completely determined by $H(t)$. $\\operatorname{Proof}(\\mathrm{b})$ \\begin{aligned} H(t) &=\\sum_{n=0}^{\\infty} n", "Aug 17 '12 at 16:09 How can one change the stretching in the preamble and have it done? Do one have to write \\begin{pmatrix}[1.5] to stretch it?: EDIT: OK I found it: just add \"\\renewcommand*{\\arraystretch}{1.5}\" in the preamble and you're good to go. – Faq May 1 '14 at 18:14 By redefining \\arraystretch you can change the vertical space between the all rows; using the optional argument for \\\\ you can control the vertical space for individual lines: \\documentclass{article} \\usepackage{amsmath} \\begin{document} $\\begin{pmatrix} \\dfrac{a}{b} & c\\\\ \\dfrac{a}{b} & d\\\\ \\dfrac{a}{b} & e \\end{pmatrix}$ $\\renewcommand\\arraystretch{2} \\begin{pmatrix} \\dfrac{a}{b} & c\\\\ \\dfrac{a}{b} & d\\\\ \\dfrac{a}{b} & e \\end{pmatrix}$ $\\begin{pmatrix} \\dfrac{a}{b} & c \\\\[1em] \\dfrac{a}{b} & d\\\\[2em] \\dfrac{a}{b} & e \\end{pmatrix}$ \\end{document} - \\[1em] This is very helpful, thanks. – Rafael Wörner May 21 '14 at 19:25 +1 for \\\\[1em] – Wauzl Nov 25 '15 at 8:44" ]

[ "$\\frac{1\\cdot\\tfrac{1}{3}}{2} = \\boxed{\\textbf{(E)} \\: \\frac{1}{6}}$ ~Lemma proof by sakshamsethi", "solved to find that $f(4) = \\boxed{\\frac{5}{8}}$.", "\\frac{\\frac{\\sqrt{2}}{2}}{1+ \\frac{\\sqrt{2}}{2}}= \\frac{\\sqrt{2}}{2+ \\sqrt{2}}$$. $$AC+ CD= 1+ \\frac{\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{2 + 2\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{4+ 4\\sqrt{2}- 2\\sqrt{2}- 4}{4- 2}= \\frac{2\\sqrt{2}}{2}= \\sqrt{2}$$", "you have $0$. Adding these cases, you get: $$\\frac{2}{3}\\cdot \\frac{1}{2} + \\frac{1}{3}\\cdot 0 = \\boxed{\\frac{1}{3}}$$", "\\cdot 4}{3^4}$ $f(x) = 1 + \\frac{x-2}{3} + \\frac{(x-2)^2}{3^2} + \\frac{(x-2)^3}{3^3}+ ... + \\frac{(x-2)^n}{3^n}+ ...$", "\\ldots . . = 1 - \\frac{1}{2} = \\frac{1}{2}$", "our answer to be $\\frac{1}{3}+\\frac{2}{9}-\\frac{1}{3}\\cdot\\frac{2}{9} = \\boxed{\\frac{13}{27}}$", "we have that $$f(x)' = \\frac{1}{2} \\cdot \\frac{1}{\\sqrt{x}} = \\frac{1}{2 \\sqrt{x}}$$", "$$1 = \\frac{1(1+1)}{2}.$$ To proceed from $n=k$ to $n = k+1$ note that $$1 + 2 + \\cdots + k + (k+1) = \\frac{k(k+1)}{2} + (k+1) = \\frac{(k+1)(k+2)}{2},$$ qed.", "\\cdot 9^2} = \\boxed{\\frac{16}{81}}$$", "total, there are $(2n)(2n-1)\\cdots (n+1)$ ways to find $C_0$ and for each $C_{j}$ there are $2^n(n-j+1)$ ways for $1 \\le j \\le n$. So the answer is, $$\\frac{2n!}{n!}(2^nn)(2^{n}(n-1)) \\cdots (2^{n}1) = \\boxed{2^{n^2} \\cdot (2n)!}.$$ ~Aaryabhatta1", "1.7035 ~\\frac{m}{s^2} \\cdot t^2 ~\\Longrightarrow ~ \\boxed{t\\approx 3.426~s}$ 3. $v = a \\cdot t$. Plug in the knowns values: $v = 1.7035~\\frac{m}{s^2} \\cdot 3.426~\\boxed{s\\approx 5.837~\\frac{m}{s}}$", "f_3(z) &= \\dots = \\frac{z^2-z^{16}}{1-z^2}\\cdot \\frac1{1-z^{16}}\\end{align*} It should then be easy to show by induction that $$f_n(z) = \\frac{z^2}{1-z^2}\\cdot \\frac{1-z^{2^{n-1}}}{1-z^{2^n}}.$$ The second factor is $1 + O(z^{2^{n-1}})\\to 1$, as $n\\to\\infty$. thus $$\\boxed{f(z) = \\displaystyle\\lim_{n\\to\\infty} f_n(z) = \\frac{z^2}{1-z^2}}$$ Unfortunately, I can't get WolframAlpha to find possible mistakes, so I hope someone else will. -", "and work inside-out. $\\sin \\frac{\\pi}{14} \\sin \\frac{6\\pi}{14} = \\frac{1}{2}\\cos \\frac{5\\pi}{14}$ $\\sin \\frac{2\\pi}{14} \\sin\\frac{5\\pi}{14} = \\frac{1}{2}\\cos \\frac{3\\pi}{14}$ $\\sin \\frac{3\\pi}{14}\\sin \\frac{4\\pi}{13} = \\frac{1}{2}\\cos \\frac{\\pi}{14}$. Thus, $\\cos \\frac{\\pi}{14} \\cos \\frac{3\\pi}{14} \\cos \\frac{5\\pi}{14} = \\frac{\\sqrt{7}}{2^3}$. Generalize! If $n$ is odd then: $\\boxed{ \\cos \\frac{\\pi}{2n} \\cos \\frac{3\\pi}{2n} \\cdot ... \\cdot \\cos \\frac{(n-2)\\pi}{2n} = \\frac{\\sqrt{n}}{2^{(n-1)/2}} }$.", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "Show it is true when $n \\geqslant 1$. $C_n = \\frac{1}{n+1} {2n \\choose n} = {2n \\choose n} - \\frac{n}{n+1} {2n \\choose n} = \\frac{2n!}{n!n!} - \\frac{n}{n+1} \\frac{2n!}{n!n!} = \\frac{2n!}{n!n!} - \\frac{2n!}{(n+1)!(n-1)!} = {2n \\choose n} - {2n \\choose n+1}. \\blacksquare$ • Prove this recurrence relation: $C_{n+1} = \\frac{2(2n+1)}{n+2} C_n$ for n=0, 1, 2, ... . Recall that $C_n = \\frac{(2n)!}{n!(n+1)!}, \\qquad n = 0, 1, 2, \\cdots$. Therefore, we have $C_{n+1} = \\frac{(2n+2)!}{(n+1)!(n+2)!}, \\qquad n = 0, 1, 2, \\cdots$. Take the ratio of $C_{n+1}$ to $C_n$: $\\frac{C_{n+1}}{C_n} = \\frac{ \\frac{(2n+2)!}{(n+1)!(n+2)!}}{ \\frac{(2n)!}{n!(n+1)!}} = \\frac{ (2n+2)! n! (n+1)!} {(n+1)! (n+2)! (2n)!} = \\frac{(2n+2) (2n+1)}{(n+1) (n+2)} = \\frac{ 2(2n+1)}{(n+2)}$. Thus, $C_{n+1} = \\frac{ 2( 2n+1)}{(n+2)} C_n$ for all nonnegative values of n. $\\blacksquare$ ## Recursive Definition We have seen various kinds of applications of Catalan numbers so far: \"Stacking Coins,\" \"Balanced Parentheses,\" \"Mountain Ranges,\" \"Polygon Triangulation,\" \"Binary Trees,\" \"Binary Paths,\" \"Permutation,\" \"Young Diagrams\" and \"Posets.\" In fact, all the sequences are equivalent, and we will show that there is a common formula that counts them all. $C_0 =1, C_{n+1} = \\sum_{i=0}^n C_i C_{n-i} \\text{ for } n\\ge 1$. In the application Balanced Parenthese, it is already known that for each open parenthesis, there is a close parenthesis. Now, let's try to find a pattern in these paired parentheses with", "& 0 & 0 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 1}.$$ Thus, we have $f(x) = x^2 + x + 1$. If $$a = 1 + 2 + \\ldots + n,$$ then it is also true that $$2a = (1 + 2 + \\ldots + n) + (n + (n - 1) + \\ldots + 1) = (n + 1) + \\ldots (n + 1) = n(n +1).$$ $$1 + 2 + \\ldots + n = \\frac{n(n+1)}{2}.$$ Finally, if $T(0) = 1$, then your sequence is just $$T(n) = T(0) + 2 + 4 + \\ldots + 2n = 1 + 2(1 + 2 + \\ldots + n) = 1 + 2\\frac{n(n+1)}{2} = n^2 + n + 1.$$", "a) It's given that $\\frac{dV}{dt}=-1$. We know that $V = \\frac{2\\pi rh}{3}$ and $\\frac{dV}{dh} = \\frac{2\\pi r}{3}$. We also know that, by similar triangles, $h=3$ implies that $r=1$. $\\frac{dV}{dt}=\\frac{dV}{dh}\\cdot\\frac{dh}{dt} \\implies -1 = \\frac{2\\pi\\cdot 1}{3}\\cdot\\frac{dh}{dt} \\implies \\frac{dh}{dt}=-\\frac{3}{2\\pi}$", "\\cdot 22 = \\frac{242}{59},$ making our answer $242+59 = \\boxed{301}.$ -fidgetboss_4000", "inscribed in the circle with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$ Fig. 1 On one hand, we see the area of $\\Delta ABC$ as $\\frac{1}{2}\\cdot AB\\cdot BC = \\frac{1}{2}\\cdot AB\\cdot L_{n+1}$. On the other hand, it is also $\\frac{1}{2}\\cdot AC\\cdot BE = \\frac{1}{2}\\cdot 2r\\cdot \\frac{L_n}{2}=\\frac{1}{2}\\cdot r\\cdot L_n.$ Therefore, $\\frac{1}{2}AB\\cdot L_{n+1}= \\frac{1}{2}r\\cdot L_n.$ Or, $AB^2\\cdot L_{n+1}^2 = r^2\\cdot L_n^2\\quad\\quad\\quad(1)$ where by Pythagorean theorem, $AB^2= (2r)^2 - L_{n+1}^2.\\quad\\quad\\quad(2)$ Substituting (2) into (1) gives $(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.$ That is, $L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.$ Let $p = L_{n+1}^2$, we have $p^2-4r^2 p + r^2 L_n^2=0.\\quad\\quad\\quad(3)$ Solving (3) for $p$ yields $p = 2r^2 \\pm r \\sqrt{4 r^2-L_n^2}.$ Since $L_n^2$ must be greater than $L_{n+1}^2$ (see Exercise 1), it must be true (see Exercise 2) that $L_{n+1}^2=2r^2 - r \\sqrt{4r^2-L_n^2}.\\quad\\quad\\quad(4)$ Notice when $r=\\frac{1}{2}$, we obtain (5) in “Truth vs. Intellect“. With increasing $n$, $L_n\\cdot 2^n \\approx \\pi\\cdot 2r \\implies \\pi \\approx \\frac{L_n 2^n}{2r}.\\quad\\quad\\quad$ We can now compute the approximate value of $\\pi$ from any circle with radius $r$: Fig. 2 $r=2$ Fig. 3 $r=\\frac{1}{8}$ Exercise 1 Explain $L_{n}^2 > L_{n+1}^2$ geometrically. Exercise 2 Show it is $2r^2-r\\sqrt{4r^2-L_n^2}$ that represents $L_{n+1}^2.$ Truth vs. Intellect It was known long ago that $\\pi$, the ratio of the circumference to the diameter of a circle, is a constant. Nearly" ]

[ "& \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem by", "\\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem", "\\:=\\:\\tfrac{1}{2},\\;P(5) = \\tfrac{1}{2}$ . . $\\begin{array}{c||c|c|} & 1 & 5 \\\\ \\hline\\hline \\\\[-4mm] 0 & \\frac{1}{3}\\cdot\\frac{1}{2} & \\frac{1}{3}\\cdot\\frac{1}{2} \\\\ \\\\[-4mm]\\hline \\\\[-4mm] 4 & \\frac{2}{3}\\cdot\\frac{1}{2} & \\frac{2}{3}\\cdot\\frac{1}{2} \\\\ \\end{array}$ . . $P(\\text{B wins}) \\:=\\:\\tfrac{1}{3}\\cdot\\tfrac{1}{2} + \\tfrac{1}{3}\\cdot\\tfrac{1}{2} + \\tfrac{2}{3}\\cdot\\tfrac{1}{2} \\;=\\;\\frac{2}{3}$ [2] $A$ chooses die $X\\!:\\;\\;P(3) = 1$ . . Then $B$ chooses die $W\\!:\\;\\;P(0) = \\tfrac{1}{3},\\;P(4) = \\tfrac{2}{3}$ . . $\\begin{array}{c||c|c|} & 0 & 4 \\\\ \\hline\\hline \\\\[-4mm] 3 & 1\\cdot\\frac{1}{3} & 1\\cdot\\frac{2}{3} \\end{array}$ . . $P(\\text{B wins}) \\:=\\:1\\cdot\\tfrac{2}{3} \\:=\\:\\frac{2}{3}$ [3] $A$ chooses die $Y\\!:\\;\\;P(2) = \\tfrac{2}{3},\\;P(6)= \\tfrac{1}{3}$ . . Then $B$ chooses die $X\\!:\\;\\;P(3) = 1$ . . $\\begin{array}{c||c|} & 3 \\\\ \\hline\\hline \\\\[-4mm] 2 & \\frac{2}{3}\\cdot1 \\\\ \\\\[-4mm]\\hline \\\\[-4mm] 6 & \\frac{1}{3}\\cdot1 \\end{array}$ . . $P(\\text{B wins}) \\:=\\:\\tfrac{2}{3}\\cdot1 \\:=\\:\\frac{2}{3}$ [4] $A$ chooses die $Z\\!:\\;\\;P(1) = \\tfrac{1}{2},\\;P(5) = \\tfrac{1}{2}$ . . Then $B$ chooses die $Y\\!:\\;\\;P(2) = \\tfrac{2}{3},\\;P(6) = \\tfrac{1}{3}$ . . $\\begin{array}{c||c|c|} & 2 & 6 \\\\ \\hline\\hline \\\\[-4mm] 1 & \\frac{1}{2}\\cdot\\frac{2}{3} & \\frac{1}{2}\\cdot\\frac{1}{3} \\\\ \\\\[-4mm]\\hline \\\\[-4mm] 5 & \\frac{1}{2}\\cdot\\frac{2}{3} & \\frac{1}{2}\\cdot\\frac{1}{3} \\\\ \\end{array}$ . . $P(\\text{B wins}) \\:=\\:\\tfrac{1}{2}\\cdot\\tfrac{2}{3} + \\tfrac{1}{2}\\cdot\\tfrac{1}{3} + \\tfrac{1}{2}\\cdot\\tfrac{1}{3} \\;=\\;\\frac{2}{3}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Isn't that mind-boggling? . . $\\begin{array}{c} \\text{Die }W\\text{ beats die }X\\\\ \\text{Die }X\\text{ beats", "# Replicate a typesetting of polynomial long division I found a typesetting of series long division I want to replicate in a book I've been reading: So I tried with the following code (using different polynomial than the example): \\begin{align*} \\renewcommand\\arraystretch{2} \\begin{array}{r@{\\hskip\\arraycolsep}c@{\\hskip\\arraycolsep}l*2r} &&\\dfrac{1}{z}+\\dfrac{1}{3!}z&+\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^3+\\cdots\\\\ \\cline{2-4} z-\\dfrac{1}{3!}z^3+\\dfrac{1}{5!}z^4-\\cdots&\\Bigg)&1\\\\ &&1-\\dfrac{1}{3!}z^2&+\\dfrac{1}{5!}z^4-\\cdots\\\\ \\cline{2-4} &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{5!}z^4+\\cdots\\\\ &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{(3!)^2}z^4+\\cdots\\\\ \\cline{2-4} &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\ &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\ \\cline{2-4} &&&\\multicolumn{1}{c}{\\vdots}\\\\ \\end{array} \\end{align*} and here is the result: As you may see, the are some differences between the example and my attempt (besides fonts, of course). The biggest problem seems to be associate with the vertical spacing, is there anything I can do to improve the result? Thank you very much! • First of all you have some minor mistakes regarding some signs in your code compared to the \"original\" version. Is this normal or just typing mistakes? – Ludovic C. Jul 12 '13 at 10:33 • I used a different polynomial. Sorry if I didn't state it clear. – Francis Jul 12 '13 at 10:35 • You have that you used a different polynomial in your post. It is right below the first image. – dustin Jul 12 '13 at 10:38 • Welcome to TeX.SX! You can have a look on our our starter page to familiarize yourself further with our format. – Claudio Fiandrino Jul 12 '13 at 10:48 • isn't that Brown/Churchill:Complex", "need another convention! We choose, for $a\\ne 0$ $$\\begin{array}{|c|}\\hline \\quad a^{-1}=\\frac{1}{a}. \\quad \\\\ \\hline\\end{array}$$ The choice is dictated by our desire for the three properties to be satisfied! Let's check. We plug in $n=\\pm 1$ and $m=\\pm 1$, use our convention, and then apply the above version of the corresponding property: $$\\begin{array}{r|ll|ll} \\text{Property 1:} & a^{n+m} &=a^n\\cdot a^m & n=-1,\\ m=1 & \\Longrightarrow & a^{-1+1}&=a^{-1}\\cdot a^1 & \\Longleftrightarrow & a^{0}=\\frac{1}{a}\\cdot a & \\texttt{TRUE!}\\\\ \\hline \\text{Property 2:} & a^n b^n &=(ab)^n & n=-1 & \\Longrightarrow & a^{-1} b^{-1} &=(ab)^{-1} & \\Longleftrightarrow & \\frac{1}{a}\\cdot \\frac{1}{b}=\\frac{1}{ab} & \\texttt{TRUE!}\\\\ \\hline \\text{Property 3:} & a^{nm} &=(a^n)^m & n=-1,\\ m=1 & \\Longrightarrow & a^{(-1)1} &=(a^{-1})^1 & \\Longleftrightarrow & a^{-1}=\\frac{1}{a} & \\texttt{TRUE!}\\\\ & && n=1,\\ m=-1 & \\Longrightarrow & a^{1(-1)} &=(a^1)^{-1} & \\Longleftrightarrow & a^{-1}=a^{-1} & \\texttt{TRUE!}\\\\ \\end{array}$$ The three properties are still satisfied, and we will continue to use them. Note how choose anything but $a^{-1}=1/a$ would have ruined them. This is our convention: multiplying by $a^{-1}$ means dividing by $a$. Now, the rest of the negative numbers. While multiplication by a positive integer means repeated addition, multiplication by a negative integer means repeated subtraction (the inverse of addition): $$a(-n)=-an=0-a-a-a-...-a.$$ Similarly, while exponentiation by a positive integer means repeated multiplication, exponentiation by a negative integer means repeated division (the inverse of multiplication): $$a^{-n}=1\\div a", "$\\frac{1}{a}$, such that $a\\cdot \\left(\\frac{1}{a}\\right)=1$ Example 7: Using Properties of Real Numbers Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. 1. $3\\cdot 6+3\\cdot 4$ 2. $\\left(5+8\\right)+\\left(-8\\right)$ 3. $6-\\left(15+9\\right)$ 4. $\\frac{4}{7}\\cdot \\left(\\frac{2}{3}\\cdot \\frac{7}{4}\\right)$ 5. $100\\cdot \\left[0.75+\\left(-2.38\\right)\\right]$ Solution 1. $\\begin{array}\\text{ }3\\cdot6+3\\cdot4 \\hfill& =3\\cdot\\left(6+4\\right) \\hfill& \\text{Distributive property} \\\\ \\hfill& =3\\cdot10 \\hfill& \\text{Simplify} \\\\ \\hfill& =30 \\hfill& \\text{Simplify}\\end{array}$ 2. $\\begin{array}\\text{ }\\left(5+8\\right)+\\left(-8\\right) \\hfill& =5+\\left[8+\\left(-8\\right)\\right] \\hfill& \\text{Associative property of addition} \\\\ &\\hfill =5+0 \\hfill& \\text{Inverse property of addition} \\\\ \\hfill& =5 \\hfill& \\text{Identity property of addition}\\end{array}$ 3. $\\begin{array}6-\\left(15+9\\right) \\hfill& =6+[15\\left(-15\\right)+\\left(-9\\right)] \\hfill& \\text{Distributive property} \\\\ \\hfill& =6+\\left(-24\\right) \\hfill& \\text{Simplify} \\\\ \\hfill& =-18 \\hfill& \\text{Simplify}\\end{array}$ 4. $\\begin{array}\\text{ }\\frac{4}{7}\\cdot\\left(\\frac{2}{3}\\cdot\\frac{7}{4}\\right) \\hfill& =\\frac{4}{7} \\cdot\\left(\\frac{7}{4}\\cdot\\frac{2}{3}\\right) \\hfill& \\text{Commutative property of multiplication} \\\\ \\hfill& =\\left(\\frac{4}{7}\\cdot\\frac{7}{4}\\right)\\cdot\\frac{2}{3}\\hfill& \\text{Associative property of multiplication} \\\\ \\hfill& =1\\cdot\\frac{2}{3} \\hfill& \\text{Inverse property of multiplication} \\\\ \\hfill& =\\frac{2}{3} \\hfill& \\text{Identity property of multiplication}\\end{array}$ 5. $\\begin{array}\\text{ }100\\cdot[0.75+\\left(-2.38\\right)] \\hfill& =100\\cdot0.75+100\\cdot\\left(-2.38\\right)\\hfill& \\text{Distributive property} \\\\ \\hfill& =75+\\left(-238\\right) \\hfill& \\text{Simplify} \\\\ \\hfill& =-163 \\hfill& \\text{Simplify}\\end{array}$ Try It 7 Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. a. $\\left(-\\frac{23}{5}\\right)\\cdot \\left[11\\cdot \\left(-\\frac{5}{23}\\right)\\right]$ b. $5\\cdot \\left(6.2+0.4\\right)$ c. $18-\\left(7 - 15\\right)$ d. $\\frac{17}{18}+\\cdot \\left[\\frac{4}{9}+\\left(-\\frac{17}{18}\\right)\\right]$ e. $6\\cdot \\left(-3\\right)+6\\cdot 3$ Solution", "of the sequence: . . $\\begin{array}{cccccccccc} f(1) = 1\\!: && 2A - B &=& 1 & [1] \\\\ f(2) = 8\\!: && 4A + B &=& 8 & [2] \\end{array}$ Add [1] and [2]: . $6A \\:=\\:9 \\quad\\Rightarrow\\quad A \\:=\\:\\frac{3}{2}$ Substitute into [1]: . $2(\\frac{3}{2}) - B \\:=\\:1 \\quad\\Rightarrow\\quad B \\:=\\:2$ Hence: . $f(n) \\;=\\;\\frac{3}{2}\\!\\cdot\\!2^n + 2(\\text{-}1)^n$ Therefore: . $a_n \\;=\\;3\\!\\cdot\\!2^{n-1} + 2(\\text{-}1)^n$ 6. Thread closed due to this member deleting questions after getting help.", "another possibility as wrote cango91. But I use incremental change. I presume there is an extent of deaths $$D$$ (number of died people per t) and there is an extent of births $$\\mu\\geq0$$. I can write: $$\\text{d}P(t)=\\mu P(t)\\text{d}t-D\\text{d}t$$ I try to solve this: $$\\begin{array}{rcl}\\text{d}P(t)&=&\\mu P(t)\\text{d}t-D\\text{d}t\\\\\\text{d}&=&(\\mu P(t)-D)\\text{d}t\\\\\\dfrac{\\text{d}P(t)}{\\mu P(t)-D}&=&\\text{d}t\\end{array}$$ The integral: $$\\int\\dfrac{\\text{d}P(t)}{\\mu P(t)-D}=\\left|\\begin{array}{rcl}\\mu P(t)-D&=&\\varphi\\\\\\text{d}P(t)&=&\\dfrac{1}{\\mu}\\,\\text{d}\\varphi\\end{array}\\right|=\\frac{1}{\\mu}\\int\\frac{\\text{d}\\varphi}{\\varphi}=\\frac{1}{\\mu}\\ln|\\mu P(t)-D|$$ I use this into my equation: $$\\begin{array}{rcl}\\dfrac{1}{\\mu}\\ln|\\mu P(t)-D|&=&t+\\ln|C|\\\\\\mu P(t)-D&=&C^{\\mu}\\text{e}^{\\mu t}\\\\P(t)&=&\\dfrac{1}{\\mu}C^{\\mu}\\text{e}^{\\mu t}+\\dfrac{D}{\\mu}\\end{array}$$ The constant C for t = 0 is: $$C=\\sqrt[\\mu]{\\mu P(0)-D}$$ So the final function is: $$P(t)=P(0)\\text{e}^{\\mu t}-\\frac{D}{\\mu}(\\text{e}^{\\mu t}-1)\\,;\\;\\mu\\geq0,D\\geq0, t\\geq0$$", "\\ = \\ cos \\, x$, $du \\ = - sin \\, x \\, dx$ and $a \\ = \\ \\sqrt{10}$. Then \\vspace{.25cm} $\\int \\frac{sin\\, x \\ dx}{\\sqrt{ 10 - cos^2 \\, x}} \\ = (-1) \\int \\frac{- \\, sin \\, x \\, dx}{\\sqrt{10 - cos^2 x}} \\ = \\ - \\, sin^{-1} ( \\frac{cos \\, x}{\\sqrt{10}}) + C$ \\vspace{.35cm} (e) $\\int \\frac{dx}{ \\sqrt{5 + 4x - x^2}}$. We want to transform this expression into something with the \\vspace{.25cm} form $\\int \\frac{du}{\\sqrt{a^2 - u^2}}$. To do this we need to complete the square of the expression in the \\vspace{.25cm} denominator as follows: \\vspace{.25cm} \\begin{math} \\renewcommand{\\arraystretch}{2.0} \\begin{array}{rl} \\indent 5 + 4x - x^2 \\ &= \\ 5 + 4 - 4 + 4x - x^2 \\\\ &= \\ 9 - 4 + 4x - x^2 \\\\ &= \\ 9 - (x^2 - 4x + 4) \\\\ &= \\ (3)^2 - (x - 2)^2 \\end{array} \\end{math} \\vspace{.25cm} This gives us \\vspace{.25cm} \\hspace{.75cm} $\\int \\frac{dx}{ \\sqrt{5 + 4x - x^2}} \\ = \\ \\int \\frac{dx}{\\sqrt{(3)^2 - (x-2)^2}} \\ = \\ sin^{-1} (\\frac{x-2}{3}) \\ +\\ C$ \\vspace{.35cm} (f) $\\int \\frac{7 \\ dx}{25 - 12x + 4x^2}$. Again we need to complete the square. This time we want to transform the expression into something with the form $\\int \\frac{du}{a^2 + u^2}$. We rewrite", "an option or can one be implemented so that new users like myself can view the \"source code\" of others answers. Obviously there are many tutorials in which we can find the correct commands, ... 15 views 41 views ### How to describe the Cartesian product $\\mathbb{R} × \\mathbb{R}$? [on hold] Let $\\mathbb{R}$ denote the set of all real numbers. Describe $\\mathbb{R} × \\mathbb{R}$. (This is a self-answered question). 6 views ### Solution to truncated renewal function Let's begin with some theory on the renewal process. In a renewal process $N(t)$, let $t$ denote the interarrival time, and $f(t)$ and $F(t)$ denote the PDF and CDF respectively. Let $M(t)=E[N(t)]$, ... 32 views ### Proving that $T$:$(x_1,…,x_n) \\rightarrow (\\frac {x_1+x_2}{2},\\frac {x_2+x_3}{2},…,\\frac {x_n+x_1}{2})$ leads to nonintegral components Start with $n$ paiwise different integers $x_1,x_2,...,x_n,(n>2)$ and repeat the following step: $T$:$(x_1,...,x_n) \\rightarrow (\\frac {x_1+x_2}{2},\\frac {x_2+x_3}{2},...,\\frac {x_n+x_1}{2})$ ... 27 views ### Let A,B be nxn matrices such that detA Not equal to 0, but detB = 0: Show [on hold] Let $A$, $B$ be $n\\times n$ matrices such that $\\det A \\neq 0$, but $\\det B = 0$. Show $\\|A-B\\|_2 \\geq(\\|A^{-1}\\|_2)^{-1}$.", "\\div a \\div a \\div ... \\div a.$$ Our convention, for any $n=1,2,3...$, will be: $$\\begin{array}{|c|}\\hline \\quad a^{-n}=\\frac{1}{a^n}. \\quad \\\\ \\hline\\end{array}$$ To confirm, we observe this: $$\\begin{array}{ccc} \\underbrace{a^{-1}\\cdot a^{-1}\\cdot a^{-1}\\cdot ...\\cdot a^{-1}}_{n\\text{ times}}&=&\\left(a^{-1}\\right)^n\\\\ ||&&||&\\\\ 1 \\underbrace{\\div a\\div a\\div a\\div ...\\div a}_{n\\text{ times}}&=&\\left(\\frac{1}{a}\\right)^n.\\\\ \\end{array}$$ Exercise. Show that Properties 1-3 still hold. Example. Once again, we see these properties as shortcuts: $$\\frac{2^{5} }{ 2^{2}} = 2^{5} \\cdot 2^{-2} = 2^{5-2} = 2^{3}.$$ $\\square$ Exercise. Represent as a power of $4$: $$4^3\\cdot .25.$$ This is the summary of what we have discovered: $$\\begin{array}{r|rl|rl} \\text{}&\\text{Multiplication:}&&\\text{Exponentiation:}\\\\ \\hline \\text{Conventions: }n=1,2, ... &\\underbrace{a+a+a+ ... +a}_{n\\text{ times}}&=a\\cdot n&\\underbrace{a\\cdot a\\cdot a\\cdot ... \\cdot a}_{n\\text{ times}}&=a^n\\\\ \\hline n=0 & a\\cdot 0 & =0 & a^0&=1\\\\ \\hline n=-1,-2,-3, ... & 0\\underbrace{-a-a-a- ... -a}_{n\\text{ times}}&=a\\cdot (-n) & 1 \\underbrace{\\div a\\div a\\div a\\div ... \\div a}_{n\\text{ times}}&=a^{-n}\\\\ &=\\underbrace{(-a)+(-a)+...+(-a)}_{n\\text{ times}}&=(-a)\\cdot n & = \\underbrace{\\frac{1}{a}\\cdot\\frac{1}{a}\\cdot ...\\cdot\\frac{1}{a}}_{n\\text{ times}}&=\\left(\\frac{1}{a}\\right)^{n}\\\\ &=-(\\underbrace{a+a+a+...+a}_{n\\text{ times}})&=-(a\\cdot n) & =1\\div (\\underbrace{a\\cdot a\\cdot a\\cdot ...\\cdot a}_{n\\text{ times}}) &=\\frac{1}{a^n}\\\\ \\hline \\text{Rules: }\\ 1.&a\\cdot (n+m)&=a\\cdot n+a\\cdot m & a^{n+m}&=a^n\\cdot a^m\\\\ \\hline 2.&(a+b)\\cdot n&=a\\cdot n+b\\cdot n & (a\\cdot b)^n&=a^n b^n\\\\ \\hline 3.&a\\cdot(n\\cdot m)&=(a\\cdot n)\\cdot m & a^{n\\cdot m}&=(a^n)^m\\\\ \\hline \\end{array}$$ These are the possible values of $n$: $$... ,-3,-2,-1,0,1,2,3, ...$$ Note that we multiply by $2$ as we move right and divide by $2$ as we move left: However, if we start at any point", "\\begingroup\\renewcommand{\\arraystretch}{1.097} \\begin{tabularx}{.45\\columnwidth}[t]{XX} \\toprule \\midrule Row 1 & Row 1 \\\\ Row 2 & Row 2 \\\\ Row 3 & Row 3 \\\\ Row 4 & Row 4 \\\\% \\vspace{3.05pt} \\\\ \\bottomrule \\end{tabularx} \\endgroup \\begin{tabularx}{.45\\columnwidth}[t]{XX} \\toprule", "\\vec{b}, k=\\frac{3}{7}$ 2. $\\overrightarrow{P Q}=2 \\vec{a}-3 \\vec{b}, \\overrightarrow{P R}=\\vec{a}-4 \\vec{b}$ 3. $\\overrightarrow{O C}=\\left(\\begin{array}{c}-5 \\\\ 6\\end{array}\\right) \\quad$ 4. $q=\\frac{6}{5}($ or $) 2 \\quad$ 5. $S=(8,-8) \\quad$ 6. $h=-4$ 7. $k=7 \\quad$ 8. $(2,3)$ 9. $\\lambda=\\frac{3}{2} \\quad$ 10. $\\overrightarrow{A B}=2 \\vec{p}-\\vec{q}, \\overrightarrow{A C}=5 \\vec{p}+(k-1) \\vec{q}, k=-\\frac{3}{2}, \\lambda=\\frac{2}{5}$ 11. $P Q: Q R=1: 2$ 12. $\\overrightarrow{O Q}=\\frac{2}{3} \\vec{b}-\\frac{5}{12} \\vec{a}$ 13. $k=2, \\overrightarrow{O D}=\\left(\\begin{array}{c}\\frac{32}{3} \\\\ 4\\end{array}\\right)$ 14. $\\overrightarrow{O P}=\\left(\\begin{array}{c}-15 \\\\ -36\\end{array}\\right), \\overrightarrow{O Q}=\\left(\\begin{array}{c}-15 \\\\ -20\\end{array}\\right), \\overrightarrow{P Q}=\\left(\\begin{array}{l}0 \\\\ 1\\end{array}\\right)$ 15. Prove 16. Prove 17. Show 18. Show 19. Show 20. $(\\sqrt{2}, \\sqrt{2}) \\quad$ 21. $\\left(\\begin{array}{lll}\\frac{\\sqrt{2}}{2} & -\\frac{\\sqrt{2}}{2} & \\sqrt{2} \\\\ \\frac{\\sqrt{2}}{2} & \\frac{\\sqrt{2}}{2} & 2 \\sqrt{2} \\\\ 0 & 0 & 1\\end{array}\\right),\\left(\\frac{\\sqrt{2}}{2}, \\frac{7 \\sqrt{2}}{2}\\right)$ Group (2013) 1. The co-ordinates of $P, Q$ and $R$ are $(2,3),(8,4)$ and $(5,8)$ respectively. Find the co-ordinates of $S$ if $P Q R S$ is a parallelogram. (3 marks) 2. The three points $O, P$ and $Q$ are such that $\\overrightarrow{O P}=\\left(\\begin{array}{l}2 \\\\ 3\\end{array}\\right)$ and $\\overrightarrow{O Q}=\\left(\\begin{array}{c}q \\\\ 2 q\\end{array}\\right)$. Given that $\\overrightarrow{P Q}$ is a unit vector, find the possible values of $q$. (3 marks) 3. The position vectors of $P$ and $Q$ relative to an origin $O$ are $\\overrightarrow{O P}=\\left(\\begin{array}{c}k \\\\ 3 k\\end{array}\\right)$ and $\\overrightarrow{O Q}=\\left(\\begin{array}{l}2 \\\\ 4\\end{array}\\right)$. If $|\\overrightarrow{P Q}|=2$, calculate the possible values of $k$. (3 marks) 4. The vector $\\overrightarrow{O A}$ has", "to do with math. we might be able to get a better idea of what's happening if you post a compilable example that starts with your document class and ends with \\end{document}, and shows what you've done so far. – barbara beeton Jun 16 '16 at 17:18 ## 1 Answer While \\midrule works in aligned, you get much finer control with array. \\documentclass{article} \\usepackage{amsmath,booktabs,cancel,array,xcolor} \\newcommand{\\rs}[1]{\\textcolor{red}{#1}} \\begin{document} $$\\renewcommand{\\arraystretch}{1.2} \\begin{array}{@{}r@{}>{{}}l@{}} 6x\\; {\\cancel{-\\;5y}} &= 11 \\\\ 7x\\; {\\cancel{+\\;5y}} &= 2 \\\\ \\midrule 13x &= 13 \\\\ x &= 1 \\end{array}$$ \\begin{equation*} \\renewcommand{\\arraystretch}{1.2} \\begin{array}{ @{}r@{}>{{}}l@{\\qquad} @{}r@{}>{{}}l@{\\qquad} @{}r@{}>{{}}l@{} } 18a+10c &= 376 & 18a+10c &= 376 & 18a+10c &= 376 \\\\ a+ c &= 24 & \\rs{-18(}a+c &= 24 \\rs{)} & -18a-18c &= -432 \\\\ \\cmidrule{5-6} & & & & -8c &= -56 \\\\ & & & & c &= 7 \\end{array} \\end{equation*} \\end{document} You can get separate rules below the three groups by adding columns \\begin{equation*} \\renewcommand{\\arraystretch}{1.2} \\setlength{\\arraycolsep}{0pt} \\newcommand{\\sep}{\\mbox{\\qquad}} \\begin{array}{ r@{}>{{}}l c r@{}>{{}}l c r@{}>{{}}l } 18a+10c &= 376 &\\sep& 18a+10c &= 376 &\\sep& 18a+10c &= 376 \\\\ a+ c &= 24 & & \\rs{-18(}a+c &= 24 \\rs{)} & & -18a-18c &= -432 \\\\ \\cmidrule{1-2} \\cmidrule{4-5} \\cmidrule{7-8} & & & & & & -8c &= -56 \\\\ & & & & & & c &= 7 \\end{array} \\end{equation*} • Oh this is", "a^2 \\cdot b^2 \\cdot c^2 \\right)^{\\frac{1}{6}}$$ $$\\implies a^2 \\cdot b^2 \\cdot c^2 \\leq \\dfrac{16}{9}$$ But given condition is : $$a^2 \\cdot b^2 \\cdot c^2 \\geq \\dfrac{16}{9}$$ So Both the conditions will be true simultaneously only if $$a^2 \\cdot b^2 \\cdot c^2 =\\dfrac{16}{9}$$ which will hold at : $$\\dfrac{a}{2}=b=\\dfrac{3c}{2}=2$$ $$\\implies a=2,b=1,c=\\dfrac{2}{3}$$ Hence $$\\boxed{a+b+c=\\dfrac{11}{3}}$$ $$\\ddot \\smile$$ - 3 years, 10 months ago Thanks Sir ! - 3 years, 10 months ago Sir I need Help in one more question . I'am Posting It in next Note . Please Help me there ! Sorry to disturb you . - 3 years, 10 months ago Step 1 Change of variables : Put $$a=x , 2b=y , 3c=z$$ Our question becomes : $$x+y+z=6 , x^{2}y^{2}z^{2} \\ge 64$$ , then find $$x+\\dfrac{y}{2}+\\dfrac{z}{3}$$ Step -2 Apply $$A.M \\ge G.M$$ to get : $$\\frac{x+y+z}{3} \\ge \\sqrt [ 3 ]{ xyz }$$ Put the value of $$x+y+z$$ and raising to power 6 both sides : $$x^{2}y^{2}z^{2} \\le 64$$ But $$x^{2}y^{2}z^{2} \\ge 64$$ $$\\Rightarrow x^{2}y^{2}z^{2}=64$$ Now equality occurs when $$x=y=z=2$$ Hence we get $$x+\\dfrac{y}{2}+\\dfrac{z}{3} = \\dfrac{11}{3}$$ - 3 years, 10 months ago Thanks a lot ronak ! - 3 years, 10 months ago a is 2, b is 1 and c is 2/3 You'll find that between 'a' , '2b' and '3c', the A.M. must be equal to", "2, 3, 1, 1, 3, 1, 7, 2] println(collect(ContinuedFraction(1414214, 1000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] println(collect(ContinuedFraction(14142136, 10000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2] println(collect(ContinuedFraction(13 // 11))) # => [1, 5, 2] println(collect(ContinuedFraction(√2), 20)) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2] ## Kotlin // version 1.1.2 // compile with -Xcoroutines=enable flag from command line import kotlin.coroutines.experimental.buildSequence fun r2cf(frac: Pair<Int, Int>) = buildSequence { var num = frac.first var den = frac.second while (Math.abs(den) != 0) { val div = num / den val rem = num % den num = den den = rem yield(div) } } fun iterate(seq: Sequence<Int>) { for (i in seq) print(\"$i \") println() } fun main(args: Array<String>) { val fracs = arrayOf(1 to 2, 3 to 1, 23 to 8, 13 to 11, 22 to 7, -151 to 77) for (frac in fracs) { print(\"${\"%4d\".format(frac.first)} /${\"%-2d\".format(frac.second)} = \") iterate(r2cf(frac)) } val root2 = arrayOf(14142 to 10000, 141421 to 100000, 1414214 to 1000000, 14142136 to 10000000) println(\"\\nSqrt(2) ->\") for (frac in root2) { print(\"${\"%8d\".format(frac.first)} /${\"%-8d\".format(frac.second)} = \") iterate(r2cf(frac)) } val pi = arrayOf(31 to 10, 314", "2 & \\frac{3}{4} \\\\ \\\\[-4mm] 3 & \\frac{4}{6}\\\\ \\\\ [-4mm] 4 & \\frac{5}{8}\\\\ \\\\[-4mm] 5 & \\frac{6}{10}\\\\ \\\\[-4mm] 6 & \\frac{7}{12} \\\\ \\\\[-4mm] 7 & \\frac{8}{14} \\\\ \\vdots & \\vdots \\end{array}$ It appears that: . $P(n) \\:=\\:\\dfrac{n+1}{2n}$ I'll let you supply the proof. • July 12th 2010, 04:12 PM melese Direct computation. I assume \"find and prove\" means that induction is involved. But anyway you could find the value of $\\prod_{i=2}^{n}(1-\\frac{1}{i^2})$ by direct manipulation of the terms of the product $(1-\\frac{1}{2^2})(1-\\frac{1}{3^2})...(1-\\frac{1}{n^2})$. • July 13th 2010, 01:21 AM Opalg Notice that $1-\\dfrac1{i^2} = \\dfrac{i^2-1}{i^2} = \\dfrac{(i-1)(i+1)}{i^2}$. Therefore $\\left(1-\\dfrac{1}{2^2}\\right)\\left(1-\\dfrac{1}{3^2}\\right)\\ldots\\left(1-\\dfrac{1}{n^2}\\right) = \\dfrac{1\\cdot3}{2^2} \\cdot \\dfrac{2\\cdot4}{3^2} \\cdot \\dfrac{3\\cdot5}{4^2} \\cdots\\dfrac{(n-1)(n+1)}{n^2}.$ Now start cancelling: each of the two factors in the numerators cancels with a factor in a neighbouring denominator. The only terms left uncancelled are $\\frac12$ at the start and $\\frac{n+1}n$ at the end. • July 13th 2010, 11:03 AM jess0517 Quote: Originally Posted by Soroban Hello, jess0517! I cranked out the first few partial products: . . $\\begin{array}{cccc} n & P(n) \\\\ \\hline \\\\[-4mm] 2 & \\frac{3}{4} \\\\ \\\\[-4mm] 3 & \\frac{4}{6}\\\\ \\\\ [-4mm] 4 & \\frac{5}{8}\\\\ \\\\[-4mm] 5 & \\frac{6}{10}\\\\ \\\\[-4mm] 6 & \\frac{7}{12} \\\\ \\\\[-4mm] 7 & \\frac{8}{14} \\\\ \\vdots & \\vdots \\end{array}$ It appears that: . $P(n) \\:=\\:\\dfrac{n+1}{2n}$ I'll let you supply the proof. So using induction would this be", "Variable – RE60K Dec 25 '14 at 19:20 The only difference I could see was the spacing between the rows that caused some of the equations to lie on the cline. Adding [.1cm] or whatever distance separation you want will achieve a bigger spacing. \\documentclass{article} \\usepackage{amsmath, array} \\begin{document} \\begin{align*} \\renewcommand\\arraystretch{2} \\begin{array}{r@{\\hskip\\arraycolsep}c@{\\hskip\\arraycolsep}l*2r} &&\\dfrac{1}{z}+\\dfrac{1}{3!}z&+\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^3+ \\cdots\\.1cm] \\cline{2-4} z-\\dfrac{1}{3!}z^3+\\dfrac{1}{5!}z^4-\\cdots&\\Bigg)&1\\\\[.1cm] &&1-\\dfrac{1}{3!}z^2&+\\dfrac{1}{5!}z^4-\\cdots\\\\[.1cm] \\cline{2-4} &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{5!}z^4+\\cdots\\\\[.1cm] &&\\hfill\\dfrac{1}{3!}z^2&-\\dfrac{1}{(3!)^2}z^4+\\cdots\\\\[.1cm] \\cline{2-4} &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\[.1cm] &&&\\left[\\dfrac{1}{(3!)^2}-\\dfrac{1}{5!}\\right]z^4-\\cdots\\\\[.1cm] \\cline{2-4} &&&\\multicolumn{1}{c}{\\vdots}\\\\ \\end{array} \\end{align*} \\end{document} Also, if you want to use a square root in stead of ) and \\cline, you could do \\sqrt{1 + \\phantom{\\left(\\frac{10^{2}}{10^{2}}\\right)+50000000000000000000000000}} Then you will have to adjust your spacing of division. You could then adjust +50000000... to cover as much width as you need. There may be even a smooth way to extend the radical without using phantom. I will look into and see if I can find something or maybe someone else will know. Instead of using \\phantom alone, I have adapted Werner's solution from Large Square Root Symbols \\newcommand{\\blank}[1]{\\hfil\\penalty1000\\hfilneg\\rule[-3pt]{#1}{1cm}} \\[ 3 = \\sqrt{\\phantom{\\blank{5cm}}} I modified his example to take care of height too. You can adjust this to whatever you want by changing 1cm and the width is argument of \\blank. So this produces: Furthermore, we can adapt How can I create a multiline split inside of a radical inside of an array environment? in order to use the", "right. You cannot cancel over addition or subtraction. Here's a start: $f\\left(\\frac{x+1}{x-1}\\right)=\\frac{\\dfrac{x+1}{x-1}+1}{\\dfrac{x+1}{x-1}-1}=\\frac{\\dfrac{x+1+x-1}{x-1}}{\\dfrac{x+1-(x-1)}{x-1}}=\\frac{\\dfrac{2x}{x-1}}{\\dfrac{2}{x-1}}=$ 9. Alright I am going to look at your work a little bit, run some errands, and come back to it. I think (hope) I can get it from here. I'll post and work/solution I have, thanks so much for everything! 10. Hello, rtwilton $f(x) \\:= \\:\\frac{x+1}{x-1}$ Find: . $(f\\circ f)(x)$ $(f\\circ f)(x) \\;=\\;f(f(x)) \\;=\\;f\\left(\\frac{x+1}{x-1}\\right) \\;=\\;\\frac{\\dfrac{x+1}{x-1} + 1}{\\dfrac{x+1}{x-1} - 1}$ Multiply by $\\frac{x-1}{x-1}\\!:\\quad\\frac{(x-1)\\left(\\dfrac{x+1}{x-1} + 1\\right)} {(x-1)\\left(\\dfrac{x+1}{x-1} - 1\\right)} \\;=\\;\\frac{(x+1) + (x-1)}{(x+1) - (x-1)} \\;=\\;\\frac{2x}{2} \\;=\\;\\boxed{ x}$ Note: This means that $f(x)$ is its own inverse!", "the cells, see for example https://www.inf.ethz.ch/personal/markusp/teaching/guides/guide-tables.pdf For the math, see also how to insert multi line equation in the tabular environment? Given that most of the cells in the table have a very organized structure, it's probably not a good idea to employ automatic line breaking to render the cells' contents. Instead, I'd use nested tabular environments. (In the code below, I wasn't too sure about the symbol that occurs in two of the three header cells...) \\documentclass{article} \\usepackage{array} \\begin{document} \\begin{center} \\renewcommand\\arraystretch{1.5} \\begin{tabular}{|>{$}c<{$}|c|c|} \\hline Ax^2+Bx+C & $\\mathop{>}_1^{} \\ge 0$ & $\\mathop{>}_1^{} \\le 0$ \\\\ \\hline \\Delta<0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{no real roots}\\\\ always holds true! \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{no real roots} \\\\ \\emph{never} satisfied! \\\\ \\end{tabular} \\\\ \\hline \\Delta = 0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 coinc.\\ solut.} \\\\ $x_{1,2}\\ge x_0$ \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 coinc.\\ solut.} \\\\ $x_{1,2}\\le x_0$ \\end{tabular} \\\\ \\hline \\Delta>0 & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 dist.\\ solutions}\\\\ $x_1\\ge x_{\\max}$\\\\ $x_2\\le x_{\\min}$ \\end{tabular} & \\renewcommand\\arraystretch{1.15} \\begin{tabular}{@{}c@{}} \\textsc{2 dist.\\ roots}\\\\ $x_{\\min}\\le x \\le x_{\\max}$\\\\ \\null % blank 3rd line \\end{tabular} \\\\ \\hline \\end{tabular} \\end{center} \\end{document} • I thing taht the OP handwriting for second and third column meaning: $>, \\;\\ge 0$ and $<, \\;\\le 0$ ... :-). – Zarko Oct 27 '15 at 2:02 • @Zarko - Thanks, good one! Let's see if the OP weighs" ]

[ "# Solutions to Problems 1421 - 1430 Q1421 Find all integer solutions of the equation $2013x+49y=2$.", "# Problems Section: Problems 1421-1430 Q1421 Find all integer solutions of the equation $2013x+49y=2$.", "# As Long As 128 Divides My Square Find the sum of all solutions to $n^2 \\equiv 0 \\pmod {128},$ where $$n$$ is an integer satisfying $$0 \\leq n < 128$$. ×", "# Math Help - Sums of powers in integers 1. ## Sums of powers in integers Show that there are no integral solutions to the equation $x^4+131=3y^4$ I've been considering this equation mod 131, so x^4=3y^4 (mod 131) and then I would demonstrate this is impossible. But I cannot seem to do so, any hints please? And thank you 2. Try looking at the equation modulo 5. There won't be any solutions.", "# Up and down How many integer solutions are there for the equation: $$42^x+x^{42}=26^x\\cdot x^{26}$$? ×", "$17$, and the solutions at most equal to $140$ are $$\\{17,47,77,107,137\\}.$$", "# Sum of the Solution Find the number of all ordered integer pairs $$(x , y)$$ that satisfy $$x^3 + 117y^3 = 5$$. ×", "can see that the result, 1463, has a 4 itself; therefore, since the true answer to the problem does not count $4$'s, we subtract $1$ from the result to get the true answer to the problem. $1463-1=\\boxed{1462}$", "## Friday, January 20, 2017 ### Interesting number: 1440000 The number 1440000 satisfies the following equation: $$1440000^3 - 1 = 1439940^3 + 71999^3 = 1440060^3 - 72001^3$$ By Fermat's Last Theorem (for which the case $n=3$ was proven by Euler), $1440000^3$ cannot be the sum or difference of 2 positive integer cubes. The above equation shows that $1440000^3-1$ is the sum and the difference of 2 positive integer cubes.", "# Math Help - Sums of powers in integers 1. ## Sums of powers in integers Show that there are no integral solutions to the equation $x^4+131=3y^4$", "for $u$ yields $x^2+y^2 = 146 (u=55)$ and $x^2+y^2 = 2993 (u=16)$. As a final step, note that the former solution, $x^2+y^2 = 146$, corresponds to $u=55,v=16$, which has integer solutions in $x$ and $y$. The solution $x^2+y^2 = 2993$ corresponds to $u=16,v=55$, which has irrational solutions in $x$ and $y$. Thus $x^2+y^2 = 146$. Last edited:", "within ±1 of a multiple of 13. The first one we find is 14 (=11+3=13+1). Then 25 (=22+3=26-1); 118 (=121-3=117+1); and 129 (=132-3=130-1). So there are four solutions, $x\\equiv 14, 25, 118, 119\\pmod{143}$.", "+0 # help plz 0 38 1 I don't know what to do here: Find the sum of all numbers x satisfying x + 25/x = 10 + x. Please expain carefully. Nov 27, 2022 #1 +115 0 $$x + \\frac{25}{x} = 10 + x$$ and to find the sun of all number x satisfying this equation. Okay! So first we simplify both sides by subtracting x. We get $$\\frac{25}{x} = 10$$ so we see that there is only one solution to this equation! we get $$x=2.5$$ so the answer is 2.5. Hope I didn't make any mistakes. Hope this helps, Imcool Nov 27, 2022", "# My first Original problem 1 Find the number of integral solutions to the equation $x^4 + 2x^3 + 2x^2 + x + 1 = 2016$ ×", "## Digit factorials ### Problem 34 Published on 03 January 2003 at 06:00 pm [Server Time] 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included.", "whose sum is 89. It could be marked as either x or x+1 and the equation is: x + (x + 1) = 89. When you will solve this equation, the value of x is computed as 44 and the second integer is 45. When you will add these number, the sum is calculated as 89.", "# Solutions to Problems 893-902 Q.893 Find all positive integer solutions of $$x^2 - 84 = 6y +3x - 2xy.$$", "# Problems Section: Problems 1501 - 1510 Q1501 Find the sum of the sum of the sum of the digits for the number $2016^{2016}$.", "# How do you solve: x^2 – 144 = 0? Take $144$ to the right changing sign: ${x}^{2} = 144$ you should get 2 equal values of $x$ one with$+$ and one with$-$.", "# Find all the integer solutions of \\$x^2-4y^2=1\\$ Find all the integer solutions of the equation: $$x^2-4y^2=1$$. I know I can’t solve it like a PELL equation because d is a square in this case." ]

[ "use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. cryptoaops Dec 29, 2020", "(Y - 3/4)(Y - 3/4) - K/2 - 9/16 = 0 (Y - 3/4)^2 - (K/2 + 9/16) = 0 (Y - 3/4)^2 = (K/2 + 9/16) Y - 3/4 = +/- sqrt(K/2 + 9/16) Y = 3/4 +/- sqrt(K/2 + 9/16) Y = 3/4 +/- sqrt((X + 2)/2 + 9/16) Y = 3/4 +/- sqrt((8X + 16)/16 + 9/16) Y = 3/4 +/- sqrt((8X + 16 + 9)/16) Y = 3/4 +/- sqrt((8X + 25)/16) Y = (3 +/- sqrt(8X + 25))/4 You need to try very hard make sure every step of your working is correct. See the corrections in red. The final answer is correct though, well done. 7. Re: Make Y the subject WHY do all that work? Never heard of the quadratic formula? 2y^2 - 3y - k = 0 , where k = x + 2 y = {-(-3) +/- SQRT[(-3)^2 - 4(2)(-k)]} / 4 y = [3 +/- SQRT(9 + 8k)] / 4 Substitute k = x + 2: y = [3 +/- SQRT(9 + 8(x + 2))] / 4 y = [3 +/- SQRT(8x + 25)] / 4 Done!", "+0 # algebra 0 43 1 The difference between two numbers is 9, and the sum of the squares of each number is 158. What is the value of the product of the two numbers? Dec 29, 2020 #1 +80 +1 Solution #1: If the numbers are called $$x$$ and $$y$$, then $$y = x - 9$$, and so the sum of squares $$x^2+ y^2 = x^2 + (x-9)^2 = 153$$ $$x^2 + (x^2 - 18x + 81) = 153$$ $$2x^2 - 18x - 72 = 0$$ $$x^2 - 9x - 36 = 0$$ This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. Dec 29, 2020 #1 +80 +1 Solution #1: If the numbers are called $$x$$ and $$y$$, then $$y = x - 9$$, and so the sum of squares $$x^2+ y^2 = x^2 + (x-9)^2 = 153$$ $$x^2 + (x^2 - 18x + 81) = 153$$ $$2x^2 - 18x - 72 = 0$$ $$x^2 - 9x - 36 = 0$$ This is factorable. However, if you", "+ y^ 8 = (x^4 + y^4)^2 - (x^2y^2\\sqrt{2})^2 = (x^4 + y^4 - x^2y^2\\sqrt{2})(x^4 + y^4 + x^2y^2\\sqrt{2}).$$ Now you have factored into two quartics. Each can be factored into two quadratics using the same technique. The algebra gets messy so make sure to check your answer. LATE EDIT: The quartics can also be factored using a substitution of variables and completion of the square. Last edited:", "(More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: $$(x + y)^2 = (x^2 + y^2) + 2xy$$ We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: $$256 = (x^2 + y^2) + 110$$ so $x^2 + y^2 = \\boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)", "solve quadratic equations. Notice that $$(t-x)(t-y) = t^2 - (x+y)t + xy = t^2 - St + P.$$ That means that $x$ and $y$ are precisely the solutions to $$t^2 - St + P = 0.$$ In your specific case, $S=16$ and $P=55$. So we want to find the solutions to $$t^2 - 16t + 55 = 0.$$ The quadratic formula gives $$t = \\frac{16 \\pm\\sqrt{256 - 220}}{2} = 8 \\pm\\frac{1}{2}\\sqrt{36} = 8\\pm\\frac{6}{2} = \\left\\{\\begin{array}{l} 11\\\\ 5 \\end{array}\\right.$$ So the two numbers are $5$ and $11$. (Of course, we often solve quadratic equations $t^2 + at + b=0$ by figuring out by eyeballing two numbers whose product is $b$ and whose sum is $-a$, but we can always use the quadratic formula to take the guessing out of it.) - We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general. Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\\pm 6$. Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$. The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$. Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get", "Rewriting equation (1) in terms of just $y =$ something, we get: $$y = 16-x$$ Substituting this into equation (2) leaves us: $$x(16-x) = 55$$ $$16x-x^2=55 \\implies x = 5 \\ \\ \\text{or} \\ \\ 11$$ which can be easily seen by factoring or using the quadratic formula. It follows that $y=11|x=5$ and $y=5|x=11$. Thus your solutions in terms of $(x,y)$ are $(5,11)$ and $(11,5)$. - By factoring or using the quadratic formula? Do you have a method of factoring quadratics that doesn't involve finding its roots first? – Henning Makholm Jul 16 '12 at 10:58 @HenningMakholm Simple. You know the factorized quadratic will look like $(x-X)(x-Y)$, and $X$ and $Y$ satisfy... oh crap. – Sean Eberhard Jan 19 '15 at 21:13 We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general. Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\\pm 6$. Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$. The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$. Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get $x(16-x)=55$. Simplification gives $x^2-16x+55=0$. The quadratic factors as", "factor out the $x$: $$x(3y + 14) = -17y - 71$$ Now divide by $3y + 14$: $$x = \\frac{-17y - 71}{3y + 14}$$ The steps are similar to solve for $y$ in terms of $x$. I will leave that as an exercise for the reader. - This is another way to find the integer solutions. Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$ which boils down to $$3x^2+(3d+31)x+17d+71=0.$$ The quadratic formula gives the roots:$$6x=-(3d+31)\\pm\\sqrt{9d^2-18d+109},$$ and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square. $d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$. Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26). Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found. -", "$14 y_1 + 3 \\cdot 14 x_1 = 3x^1_2 + y_1^2 = 28$ or $y_1 + 3x_1 = 2$. Now eliminate $y_1$ from $3x_1^2 + y_1^2 = 28$ giving a single quadratic equation for $x_1$ (which you can solve). Using $y_1 + 3x_1 = 2$ gives you the y value.", "It's not pulling things out of the air, it's a simple substitution. When we introduce the variable $y = x^2$, we convert $$x^4 - 34x^2 + 225 = 0$$ which, as a quartic equation, we may not know how to solve (although this one happens to be easy to factor), into: $$y^2 - 34y + 225 = 0$$ which, as a quadratic equation, is simpler, and we have an equation for it. Notice that we didn't actually change the structure of the equation - we simply substituted a new variable. For our convenience. That way, you can plug the equation for $y$ into the quadratic formula and get $y = 25$ or $y = 9$. But we're not solving for $y$, we're solving for $x = \\pm\\sqrt{y}$, so $x = \\pm 5,\\pm 3$. Hint. $$x^4-34x^2+225=x^4-30x^2+225-4x^2=(x^2-15)^2-(2x)^2$$ • Why not just $(x^2-25)(x^2-9)$? – Barry Jun 7 '15 at 20:54 • It's certainly an inside-out way to complete the square. – Thomas Andrews Jun 7 '15 at 21:38 No, because $\\sqrt{a+b} \\neq \\sqrt{a} + \\sqrt{b}$. For example, $$3 = \\sqrt{9} = \\sqrt{1+8} \\neq \\sqrt{1}+\\sqrt{8}=1+2\\sqrt{2},$$ so it's definitely false. What you should do is consider it as a quadratic equation in $x^2$, which you can relabel as $y$ for convenience. If you do square root both sides of the equation, you get", "respectively. Exercise 2. Mrs. Kenrick is teaching her class about different types of polynomials. They have just studied quartics, and she has offered 5 bonus points to anyone in the class who can determine the quartic that she has displayed on the board. The quartic has 5 points identified: (-6, 25), (-3, 1), (-2, $$\\frac{7}{3}$$), (0, -5), and (3, 169). Logan really needs those bonus points and remembers that the general form for a quartic is y = ax4 + bx3 + cx2 + dx + e. Can you help Logan determine the equation of the quartic? a. Write the system of equations that would represent this quartic. 25 = 1296a – 216b + 36c – 6d + e 1 = 81a – 27b + 9c – 3d + e $$\\frac{7}{3}$$ = 16a – 8b + 4c – 2d + e -5 = e 169 = 81a + 27b + 9c + 3d + e b. Write a matrix that would represent the coefficients of this quartic. A = $$\\left[\\begin{array}{ccccc} 1296 & -216 & 36 & -6 & 1 \\\\ 81 & -27 & 9 & -3 & 1 \\\\ 16 & -8 & 4 & -2 & 1 \\\\ 0 & 0 & 0 & 0 & 1 \\\\ 81 & 27 & 9 & 3 &", "$x^2-10x=14$, what will the factored quadratic be? 1. $(x-5)^2=39$ 2. $(x-5)^2=114$ 3. $(x+5)^2=114$ 4. $(x+5)^2=39$ $y = x^2 - 10x + 22.$ 1. $y = (x - 11)^2 + 5$ 2. $y = (x - 5)^2 - 3$ 3. $y = (x +5)^2 +3$ 4. $y = (x - 5)^2 + 3$", "= 16$ or $x = - 9$ Substitute into $x \\cdot y = 144$ to get the value of $y$ $y = - 9$ or $y = 16$ Whichever way you look at it, the two numbers must be 16 and -9.", "order that there be only one real root to the original quartic is if the quadratic has, as its only positive root, u= 0 so that the only real root to the original quartic is 0. That means that the quadratic in u must be simply $$u^2= 0$$. In other words, we must have both 1- 2k= 0 and $$k^2- 1= 0$$. The first equation has only k= 1/2 as solution and the second has only k= 1 or k= -1 as solutions. There is no value of k that makes both 0 so there is no value of k that gives only a single real root to the original quartic equation. Hence we're looking for a value of k such that: $b^2 - 4ac = 0$ In this instance • $a = 1$ • $b = 1-2k$ • $c = k^2 -1$ With those equations, b^2= (1- 2k)^2= 4k^2- 4k+ 1 4ac= 4k^2- 4 so b^2- 4ac= 4k^2- 4k+ 1- 4k^32+ 4= -4k+ 5= 0 so k= 5/4. In that case, 1- 2k= 1- 5/2= -3/2 and k^2- 1= 25/16- 1= 9/16. The quadratic equation for u is $$u^2- (3/2)u+ 9/16= (u- 3/4)^2= 0$$ so that $$u= 3/4$$. But then $$x^2= 3/4$$ so the original equation has the two roots $$x= \\sqrt{3}/2$$ and $$-\\sqrt{3}/2$$ Last edited by", "factorise the quartic. It would be easier to factorise it if you know what the factorisation is going to be. To discover this, I tried a few examples. For $p=7$, the quartic gives $5041=71^2$. For $p=14$, the quartic gives $57121=239^2$. I noticed that $71=72-1=8\\times9-1$ and $239=15\\times16-1$. This suggested that the quartic was $((p+1)(p+2)-1)^2$. Once you know the answer, it is easy to find it! Take $p^2$ common after multiplying. Then put $p +1/p =y$ and solve. I have to add what I think is a 'dumb' way to do it by hand (with paper) as opposed to Alex B. succinct cleverness: First, multiply out the product to get $p^4 + 6p^3 + 11p^2 + 6p + 1$. Since this is a square, it must be a quadratic $p^2 + x p + y$. Squaring the quadratic, ignoring a lot of the cruft, and just looking at just the second and last coefficients $$6 = x + x$$ and $$1 = y^2$$ and you're done. Select any $$a\\in\\mathbb{Z}_{\\ge 2}$$ and define $$P$$ to be the product of the four consecutive integers $$a-1,a,a+1$$ and $$a+2$$, that is $$P=(a-1)a(a+1)(a+2).$$ Expanding $$P$$, we get $$P=a^4+2a^3-a^2-2a.$$ Thus, we have $$P+1=a^4+2a^3-a^2-2a+1=(a^2+a-1)^2,$$ that is $$P+1$$ is a perfect square. Now since $$a\\in\\mathbb{Z}_{\\ge 2}$$ is arbitrary, implies that $$P+1$$ is a perfect square for all $$a\\in\\mathbb{Z}_{\\ge", "possible to solve this as a quartic equation, although it would be quite laborious. You can check what WolframAlpha is able to find out here and here - great attempt indeed – Rajesh K Singh Aug 9 '12 at 6:20 Let, $\\sqrt{x}=s$, $\\sqrt{y}=t$ we have, $s^4 -12s^2+s+32=0$, which is a 'biquadratic' equation of the form, $$(s^2+ks+l)(s^2-ks+m)=0$$ i.e. $$s^4 -12s^2+s+32=(s^2+ks+l)(s^2-ks+m)$$ now by equating coefficients, we have $$l+m-k^2 = -12, k(m-l) = 1, lm = 32$$ from the first two of these equations, we obtain $$2m=k^2-12+(1/k), 2l=k^2-12-(1/k)$$ hence substituting in the third equation, the values of l,m, $$(4)(32)=(k^2-12-1/k)(k^2-12+1/k)$$ $$128=(k^2-12)^2-1/k^2$$ $$128=k^4+144-24k^2-1/k^2$$ $$k^6-24k^4+16k^2-1=0$$ this is a cubic in $k^2$ which always has one real positive solution and we can find $k^{2}$,$l$,$m$ -", "be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$. However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$. Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$. Our equation then becomes $y_2^2 = 125a_1 + 16$, and we can simplify no further. Rearranging terms, we get $y_2^2 - 16 = 125a_1$, whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$. Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$. Hence, listing out the smallest possible values of $y_2$, we have $y_2 = 4, 121, 129, \\cdots, 621$. The tenth term is $621$, whence $y = 4y_2 = 2484$. The desired result can then be calculated to be $\\boxed{170}$. - Spacesam ~ dolphin7 ~Shreyas S", "In this case make the substitution $y=x+4$, which transforms the equation into $$(y-1)^4+(y+1)^4=20.$$ This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'. If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur: $$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$ So, the odd factors cancel to give $$2y^4+12y^2-18=0$$ This then becomes the simple equation $$y^4+6y^2-9=0\\,,$$ which is a quadratic equation in $y^2$ with solutions $$y^2 = \\frac{-6\\pm\\sqrt{36+36}}{2}=-3(1\\pm\\sqrt 2).$$ One of these solutions is positive; taking the square root gives two real values for $y$ as $$y = \\pm \\sqrt{3\\left(\\sqrt{2}-1)\\right)}$$ Therefore two real solutions to the equation are $$x= -4 \\pm \\sqrt{3\\left(\\sqrt{2}-1\\right)}$$", "is very elegant, I'm not sure I know of a better. Ops, I forgot to double-check the answers...thanks for reminding me and that's so nice of you to say that my solution method is elegant... Thanks, Ackbach! Substitute $y=x-2$ in the equation $x^4+y^4 = 8^2$, you get a quartic equation for $x$, namely $x^4+(x-2)^4 = 64$, which simplifies to $x^4 - 4x^3 + 12x^2 - 16x - 24 = 0.$ Looking at Ackbach's plot of the functions, it appears that there are only two real roots for $x$, and it looks as though their sum is very close to $2$. That made me wonder whether the quartic might have a factorisation of the form $x^4 - 4x^3 + 12x^2 - 16x - 24 = (x^2-2x+a)(x^2-2x+b)$. Comparing coefficients, you can check that this is indeed the case, with $a = 4 - 2\\sqrt{10}$ and $b = 4 + 2\\sqrt{10}$. The first factor has two real roots $x = 1\\pm\\sqrt{2\\sqrt{10}-3}$ (and the second factor has no real roots). So the solutions for $x$ are $\\boxed{1\\pm\\sqrt{2\\sqrt{10}-3}}$. Edit. Since $y = x-2$, the factors $x^2-2x + 4\\pm2\\sqrt{10}$ above are just another way of writing the factors $xy + 4\\pm2\\sqrt{10}$ that anemone had already found. So all that was needed was to make that substitution $y = x-2$, and you get two quadratic", "1\\right)}^{2}$ $= 49 \\left({x}^{2} - 2 x + 1\\right)$ $= 49 {x}^{2} - 98 x + 49$ We can now simply combine ${y}_{1} - {y}_{2}$ to form $y$ Thus, $y = 2 {x}^{2} + 38 x + 168 - \\left(49 {x}^{2} - 98 x + 49\\right)$ $= 2 {x}^{2} + 38 x + 168 - 49 {x}^{2} + 98 x - 49$ Combine coefficients of like terms. $y = \\left(2 - 49\\right) {x}^{2} + \\left(38 + 98\\right) x + \\left(168 - 49\\right)$ $y = - 47 {x}^{2} + 136 x + 119$ (Is our quadratic in standard form) $a = - 47 , b = + 136 , c = + 119$" ]

[ "# 1997 AIME Problems/Problem 3 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? ## Solution Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \\Longrightarrow 9xy-1000x-y=0$. Using SFFT, this factorizes to $(9x-1)\\left(y-\\dfrac{1000}{9}\\right)=\\dfrac{1000}{9}$, and $(9x-1)(9y-1000)=1000$. Since $89 < 9x-1 < 890$, we can use trial and error on factors of 1000. If $9x - 1 = 100$, we get a non-integer. If $9x - 1 = 125$, we get $x=14$ and $y=112$, which satisifies the conditions. Hence the answer is $112 + 14 = \\boxed{126}$. ## Solution 2 As shown above, we have $1000x+y=9xy$, so $1000/y=9-1/x$. $1000/y$ must be just a little bit smaller than 9, so we find $y=112$, $x=14$, and the solution is $\\boxed{126}$. ## Solution 3 To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$ as $(90a+9b-1)(100x+10y+z) = 10000a + 1000b$ and $(90a+9b-1)(100x+10y+z) =", "use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. cryptoaops Dec 29, 2020", "y - 144 = x + 2$ $x - 8 y + 146 = 0$", "(More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: $$(x + y)^2 = (x^2 + y^2) + 2xy$$ We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: $$256 = (x^2 + y^2) + 110$$ so $x^2 + y^2 = \\boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)", "\\quad y=-71 - 153k$$ where $k$ is an integer.", "+0 # algebra 0 43 1 The difference between two numbers is 9, and the sum of the squares of each number is 158. What is the value of the product of the two numbers? Dec 29, 2020 #1 +80 +1 Solution #1: If the numbers are called $$x$$ and $$y$$, then $$y = x - 9$$, and so the sum of squares $$x^2+ y^2 = x^2 + (x-9)^2 = 153$$ $$x^2 + (x^2 - 18x + 81) = 153$$ $$2x^2 - 18x - 72 = 0$$ $$x^2 - 9x - 36 = 0$$ This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. Dec 29, 2020 #1 +80 +1 Solution #1: If the numbers are called $$x$$ and $$y$$, then $$y = x - 9$$, and so the sum of squares $$x^2+ y^2 = x^2 + (x-9)^2 = 153$$ $$x^2 + (x^2 - 18x + 81) = 153$$ $$2x^2 - 18x - 72 = 0$$ $$x^2 - 9x - 36 = 0$$ This is factorable. However, if you", "can see that the result, 1463, has a 4 itself; therefore, since the true answer to the problem does not count $4$'s, we subtract $1$ from the result to get the true answer to the problem. $1463-1=\\boxed{1462}$", "167 = k^2$$ for some integer $$k$$, then $$(k-a)(k+a) = 167$$. Since $$167$$ is prime we have $$k = \\pm 84$$ and $$a = \\pm 83$$. Note that $$2k$$ is the discriminant of our quadratic. So to find $$x$$ we plug $$a$$ into the quadratic formula. $$x = \\frac{-2(a+1) \\pm 2k}{2 \\cdot 2} = \\frac{-2(\\pm 83 + 1) \\pm 2 \\cdot 84}{4} = 0, 83, -84 \\textrm{ or } -1.$$ So now we check the possible solutions $$(x,y)$$. A trick we can use here it to note that the equation $$x+y+2xy = 83$$ is symmetric in $$x$$ and $$y$$ so the possible values for $$y$$ are also $$0, 83, -84$$ and $$-1$$. Going through the options gives the solutions $$(0,83), (83, 0), (-84,-1)$$ and $$(-1,84)$$. Will Jagy's approach is obviously the correct one. However, there is another way that does not rely on finding the algebraic factorization: $$y·(2x+1) = 83-x$$. $$2x+1 \\mid 83-x$$. $$2x+1 \\mid 2·(83-x)+(2x+1) = 167$$. Now check the factors.", "for $u$ yields $x^2+y^2 = 146 (u=55)$ and $x^2+y^2 = 2993 (u=16)$. As a final step, note that the former solution, $x^2+y^2 = 146$, corresponds to $u=55,v=16$, which has integer solutions in $x$ and $y$. The solution $x^2+y^2 = 2993$ corresponds to $u=16,v=55$, which has irrational solutions in $x$ and $y$. Thus $x^2+y^2 = 146$. Last edited:", "solve quadratic equations. Notice that $$(t-x)(t-y) = t^2 - (x+y)t + xy = t^2 - St + P.$$ That means that $x$ and $y$ are precisely the solutions to $$t^2 - St + P = 0.$$ In your specific case, $S=16$ and $P=55$. So we want to find the solutions to $$t^2 - 16t + 55 = 0.$$ The quadratic formula gives $$t = \\frac{16 \\pm\\sqrt{256 - 220}}{2} = 8 \\pm\\frac{1}{2}\\sqrt{36} = 8\\pm\\frac{6}{2} = \\left\\{\\begin{array}{l} 11\\\\ 5 \\end{array}\\right.$$ So the two numbers are $5$ and $11$. (Of course, we often solve quadratic equations $t^2 + at + b=0$ by figuring out by eyeballing two numbers whose product is $b$ and whose sum is $-a$, but we can always use the quadratic formula to take the guessing out of it.) - We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general. Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\\pm 6$. Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$. The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$. Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get", "$(x-8)(x+2)$ divisible by $512$. Both $x+8$ and $x-2$ are even, but only one of them can have higher powers of $2$ as factors. Thus, $x\\equiv 8\\pmod {256}$ or $x\\equiv -2\\pmod{256}$. The second requires more care. First assuming $y\\equiv 0\\pmod{16}$ you have $y=16y_0$ and $$16y_0^2=y_0+1\\pmod{32}$$ Since $16y_0^2$ is even, $y_0+1$ is even, so $y_0$ is odd. When $y_0$ odd, $16y_0^2\\equiv 16\\pmod{32}$, so $y_0\\equiv 15\\pmod {32}=$ and $y\\equiv 16\\cdot 15=240\\pmod{512}$. Second, assume $y=16y_0+1$. Then $$16y_0^2 \\equiv 1-y_0\\pmod{32}$$ So $y_0$ is odd, $1-y_0\\equiv 16\\pmod {32}$, or $y_0\\equiv 17\\pmod{32}$. So $y\\equiv 16\\cdot 17+1=273\\pmod{512}$ So the two solutions are $y\\equiv 240,273\\pmod{512}$. Note that the sum of these two values is $1\\pmod{512}$, as befits the two roots of any quadratic equation of the form $x^2-x+C=0$. Indeed, we have $(y-240)(y-273)\\equiv y^2-y-16\\pmod{512}$, so these are the only solutions, since only one of $y-240$ and $y-273$ can be even.", "$x^2-10x=14$, what will the factored quadratic be? 1. $(x-5)^2=39$ 2. $(x-5)^2=114$ 3. $(x+5)^2=114$ 4. $(x+5)^2=39$ $y = x^2 - 10x + 22.$ 1. $y = (x - 11)^2 + 5$ 2. $y = (x - 5)^2 - 3$ 3. $y = (x +5)^2 +3$ 4. $y = (x - 5)^2 + 3$", "be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$. However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$. Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$. Our equation then becomes $y_2^2 = 125a_1 + 16$, and we can simplify no further. Rearranging terms, we get $y_2^2 - 16 = 125a_1$, whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$. Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$. Hence, listing out the smallest possible values of $y_2$, we have $y_2 = 4, 121, 129, \\cdots, 621$. The tenth term is $621$, whence $y = 4y_2 = 2484$. The desired result can then be calculated to be $\\boxed{170}$. - Spacesam ~ dolphin7 ~Shreyas S", "factor out the $x$: $$x(3y + 14) = -17y - 71$$ Now divide by $3y + 14$: $$x = \\frac{-17y - 71}{3y + 14}$$ The steps are similar to solve for $y$ in terms of $x$. I will leave that as an exercise for the reader. - This is another way to find the integer solutions. Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$ which boils down to $$3x^2+(3d+31)x+17d+71=0.$$ The quadratic formula gives the roots:$$6x=-(3d+31)\\pm\\sqrt{9d^2-18d+109},$$ and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square. $d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$. Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26). Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found. -", "# Solving $y^2 = 1263465 + 144x$ for integers $x,y$ I've thrown this equation up as part of some research I'm doing. $$y^2 = 1263465 + 144x$$ I was hoping there is a quick way to solve this without stepping through all the values. The value I'm interested in is $x = 1579$, but other $x$ values satisfy for $y$ as well, although I'm not bothered about those. Am I right in thinking that factorising $1263465$ might simplify the problem somewhat? Cheers :) • Are you trying to find all integer solutions? – Peter Woolfitt Jan 21 '15 at 21:13 • Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, you need to make your question clear. As @PeterWoolfitt asked, are you looking for all integer solutions, or for just some integer solutions, or are rational or irrational solutions acceptable? – Rory Daulton Jan 21 '15 at 21:21 • Thanks marty, that helps greatly. Yes, I'm only interested in integer values, but 1579 in particular. – Jo pus Jan 21 '15 at 22:46 First of all, reduce 1263465 mod 144, so you then only have to solve $y^2 = a+144x$ where $0 \\le a < 144$. • You can also simplify by $9$. –", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "+0 # help algebra 0 26 2 Find all solutions to 4^x - 2^x = 56 + 11*2^x + 2^(x - 1). May 1, 2022 #1 0 Only 4 May 1, 2022 #2 +9310 +1 $$4^x - 2^x = 56 + 11\\cdot 2^x + 2^{x - 1}$$ If we let $$t = 2^x$$ we have $$t^2 - t = 56 + 11t + \\dfrac12 t\\\\ 2t^2 - 25t - 112 = 0\\\\ (t - 16)(2t + 7) = 0$$ It is not possible that 2^x = -7/2, therefore 2^x = 16, which corresponds to x = 4. May 2, 2022", "+0 # a few questions 0 97 1 1. Expand (x+2)(x^2 +3) 2. Factor the quadratic x^2 -9x-36 3. Find all values of x that satisfy the equation 3x^2 +14x=5 Feb 20, 2021 #1 +944 0 1. $$(x+2)(x^2+3) = \\boxed{x^3 + 2x^2 + 3x + 6}$$ 2. $$x^2 - 9x - 36 = \\boxed{(x-12)(x+3)}$$ 3. $$3x^2 +14x - 5 = 0$$ $$x = {-14 \\pm \\sqrt{196-4(3)(-5)} \\over 2(3)} = {-14 \\pm 16 \\over 6}$$ $$\\boxed{x=-5, -\\frac{1}{3}}$$", "= 16$ or $x = - 9$ Substitute into $x \\cdot y = 144$ to get the value of $y$ $y = - 9$ or $y = 16$ Whichever way you look at it, the two numbers must be 16 and -9.", "$999^2$, giving at most 999 options for $y$. However if $y = 1^2$, you get $(x-5)^2 = 16$, which has solutions $x = 9$ and $x = 1$. Both of those solutions are not less than $y$, so $y$ cannot be equal to 1. If $y = 2^2 = 4$, you get $(x - 8)^2 = 64$, which has 2 solutions, $x = 16$, and $x = 0$. 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$, you get exactly 1 solution for $x$, and that gives a total of $999 - 2 = \\boxed{997}$ pairs. - asbodke Solution 4 (Similar to Solution 3) Rearranging our conditions to $$x^2-2xy+y^2+16-8x-8y=0 \\implies$$ $$(y-x)^2=8(x+y-2).$$ Thus, $4|y-x.$ Now, let $y = 4k+x.$ Plugging this back into our expression, we get $$(k-1)^2=x-1.$$ There, a unique value of $x, y$ is formed for every value of $k$. However, we must have $$y<10^6 \\implies (k+1)^2< 10^6-1$$ and $$x=(k-1)^2+1>0.$$ Therefore, there are only $997$ pairs of $(x,y).$ Solution by Williamgolly" ]

[ "+0 # Plz help me fast +1 39 2 +38 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as (y-k)^2/a^2-(x-h)^2/b^2=1. Find h+k+a+b Mar 20, 2019 #1 +100047 +2 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1$$ Find h+k+a+b Try looking at this: Once you work it out you can check it by graphing in desmos. https://www.desmos.com/calculator Mar 20, 2019 #2 +38 +2 I got it, thanks Dahdahda Mar 20, 2019", "at $$(0, 2)$$ with one focus at $$(6, 2)$$ and a semi-minor axis with a size of 3. Based on the information provided, $$c = 6 - 0 = 6$$. Since the foci are on a like that is parallel to the x-axis, we get that $$b = 3$$ and then $$a = \\sqrt{6^2 - 3^2} = \\sqrt{27}$$ Therefore, the equation of the ellipse is: $\\large \\displaystyle \\frac{x^2}{27} + \\frac{(y-2)^2}{9} = 1$ ## The Ellipse and general Conic Sections Same as with the case of the parabola, the hyperbola and the circle, the ellipse is tightly related to the cone. Ancient Greek mathematician named Apollonius discovered this connection, with what is called conic sections. A conic section corresponds to the shapes that are formed when you make a cut through a cone with a plane, and depending on the relative angle of the cone and the plane, the shape of the cross section changes. Indeed, depending on the angle the cone and plane face each other, the shape of the cross section can be a parabola, circle, ellipse, or an hyperbola. This is depicted in the graph below: For an ellipse with equation $$\\displaystyle \\frac{x^2}{a^2} + \\frac{y^2}{a^2} = 1$$, with $$a \\ge b$$, $$a$$ is called the semi-major axis, and $$b$$ is called the semi-minor axis. Now for,", "semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: $b = a \\sqrt{e^2-1}.$ Note that in a hyperbola b can be larger than a. [1]", "on the hyperbola $$\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}$$ = 1. Main facts about two standard forms of hyperbola Circle : The locus of a point which moves in a plane such that its distance from a fixed point, in the plane, is always a constant, is called a circle. The fixed point is called the centre and the constant distance is called the Radius of the circle. Equation of the Circle : The equation of a circle with centre (h, k) and radius V’ is given by (x – h)2 + (y – k)2 = r2. If the centre of a circle is origin and radius is ‘r’ then its equation is x2 + y2 = r2. General Equation of a Circle : The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0. Its centre is (-g,-f) and radius = $$\\sqrt{g^{2}+f^{2}-c}$$ Note : 1. If g2 + f2 – c = 0, then radius = 0. In this case, x2 +y2 + 2gx + 2fy + c = 0 represents a circle whose radius is zero. Such a circle is called a Point Circle, [i.e., it represents the point (-g, -f)]. 2. If g2 + f2 – c < 0, then radius is non-real. In this case, x2 +y2 + 2gx + 2fy +", "## anonymous 5 years ago Find the standard equation of the hyperbola that has a vertex at (4, 2), focus at (4, 4) and a center at (4, -1). I don't have much time, as I mentioned, but! I will get you started by saying, you can apply much of what we applied in my previous answer, except, since the center isn't at (0, 0), a couple of things change. First off, the equation becomes: $-\\frac{(x - h)^2}{a^2} + \\frac{(y - k)^2}{b^2} = 1$ This is for a center (h, k). Note that this time it's - + instead of + -; that's because this hyperbola opens up and down instead of left and right (you can see that if you sketch it based on the parameters they give you above). The other thing that changes is that the equation for the vertices is (h + a, k) and (h - a, k).", "Enable text based alternatives for graph display and drawing entry Try Another Version of This Question The equation of the hyperbola that has a center at (9,3) , a focus at (14 , 3) , and a vertex at (12 , 3 ) , is \\frac((x-C)^2)(A^2)-\\frac((y-D)^2)(B^2)=1 where A = B = C = D = Get help:", "reflecting property of the hyperbola, we will show that $\\,\\alpha = \\beta\\,$. Both $\\,\\alpha\\,$ and $\\,\\beta\\,$ are positive angles, not exceeding $\\,90^\\circ\\,$. The tangent function is one-to-one between $\\,0^\\circ\\,$ and $\\,90^\\circ\\,$, so if $\\,\\tan\\alpha = \\tan\\beta\\,$, it follows that $\\,\\alpha = \\beta\\,$. Consequently: To prove the reflecting property of the hyperbola, we will show that $\\,\\tan\\alpha = \\tan\\beta\\,$. Summarizing results for convenience: • [hyperbola] The hyperbola under investigation has equation $\\displaystyle\\,\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1\\,$. The positive number $\\,c\\,$ satisfies $\\,c^2 = a^2 + b^2\\,$. • [green line] The green line is outside light directed towards the focus $\\,(c,0)\\,$. This green line has slope $\\displaystyle\\,\\frac{y}{x-c}\\,$, where $\\,(x,y)\\,$ are the coordinates of the point $\\,P\\,$ where the green line intersects the hyperbola. • [red line] The red line is between point $\\,P(x,y)\\,$ and the second focus, $\\,(-c,0)\\,$. This red line has slope $\\displaystyle\\,\\frac{y}{x+c}\\,$. • [tangent/dashed line] The dashed line is the tangent to the hyperbola at $\\,P(x,y)\\,$. It has slope $\\,\\displaystyle\\frac{b^2}{a^2}\\frac{x}{y}\\,$. Using the formula for the tangent of the angle between two intersecting lines of known slope: • The angle $\\,\\alpha\\,$ is formed by the intersecting green and tangent lines. The green line has the greater angle of inclination. Thus: $$\\tan\\alpha = \\frac{\\frac{y}{x-c} - \\frac{b^2}{a^2}\\frac{x}{y}}{1 + \\frac{y}{x-c}\\cdot\\frac{b^2}{a^2}\\cdot\\frac{x}{y}}$$ • The angle $\\,\\beta\\,$ is formed by the intersecting red and tangent", "y2 = 1 As we can see that, the given hyperbola is a horizontal hyperbola. So, by comparing the given equation of hyperbola with $$\\frac{{{x^2}}}{{{a^2}}} - \\frac{{{y^2}}}{{{b^2}}} = 1$$ we get ⇒ a = 1 and b = 1 As we know that, length of latus rectum of a hyperbola is given by $$\\frac{2b^2}{a}$$ So, the length of latus rectum of given hyperbola is 2 units. Hence, option D is the correct answer. #### Parabola Question 11 Focus of the parabola y2 − 8x + 6y + 1 = 0 is 1. (2, 0) 2. (1, -3) 3. (8, 0) 4. None of the above Option 2 : (1, -3) #### Parabola Question 11 Detailed Solution Concept: Latus rectum: The latus rectum of a conic section is the chord (line segment) that passes through the focus, is perpendicular to the major axis, and has both endpoints on the curve. • Length of Latus Rectum of Parabola y2 = 4ax is 4a • End points of the latus rectum of a parabola are L = (a, 2a), and L’ = (a, -2a) Calculation: Given equation: y2 − 8x + 6y + 1 = 0 ⇒ y2 + 6y + 9 - 9 - 8x + 1 = 0 ⇒ (y + 3)2 - 8x - 8 = 0 ⇒", "a is written as: y2 = 4ax • The equation of a hyperbola with focal point on the x-axis is given by x2/a2 − y2/b2 = 1 • The equation of an ellipse with focal point on the x-axis is x2/a2 + y2/b2 = 1 • Length of the latus rectum of the hyperbola [x2/a2 − y2/b2 = 1] is written as: 2b2/a • Length of the latus rectum of the ellipse [x2/a2 + y2/b2 = 1] is written as: 2b2/a", "$$\\require{cancel}$$ In place of the semi major axis, we have the semi transverse axis, symbol $$a$$. This amounts to just a name change, although some authors treat $$a$$ for a hyperbola as a negative number, because some of the formulas, for example for the speed in an orbit, $$V^2 = GM \\left( \\frac{2}{r} - \\frac{1}{a} \\right)$$, are then identical for an ellipse and for a hyperbola.", "e ) = $$\\frac{c}{a}$$ Also, c ≥ a, the eccentricity is never less than one. • Distance of focus from centre: ae • Equilateral hyperbola: Hyperbola in which a = b • Conic section formulas for latus rectum in hyperbola: $$\\frac{2b^{2}}{a}$$ ## Conic section formulas examples: 1. Find an equation of the circle with centre at (0,0) and radius r. Solution: Here h = k = 0. Therefore, the equation of the circle is x2 + y2= r2 2. Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 16x. Solution: In this equation, y2 is there, so the coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 4. Thus, the focus of the parabola is (4, 0) and the equation of the directrix of the parabola is x = – 4 Length of the latus rectum is 4a = 4 × 4 = 16.", "the center is $(h,k) = \\boxed { (1,-2) }$ what is the foci of this equation? 7(x-2)^2 + 3(y-2)^2=21 We want to get this equation in the form $\\frac {(x - h)^2}{b^2} + \\frac {(y - k)^2}{a^2} = 1$ or $\\frac {(x - h)^2}{a^2} + \\frac {(y - k)^2}{b^2} = 1$ It's not hard to do here, just divide both sides by 21 $7(x - 2)^2 + 3(y - 2)^2 = 21$ $\\Rightarrow \\frac {(x - 2)^2}{3} + \\frac {(y - 2)^2}{7} = 1$ clearly we see that $a = \\sqrt {7}$, $b = \\sqrt {3}$, $h = 2$, $k = 2$ also, $c = \\sqrt {a^2 - b^2} = \\sqrt {7 - 3} = \\sqrt {4} = 2$ The foci are given by: $(h, k \\pm c)$ $\\Rightarrow \\mbox { Foci } = (2, 2 \\pm 2) = \\boxed {(2,4)} \\mbox { and } \\boxed {(2,0)}$ I'm beginning to feel that I'm spoiling you. Why not try the last two on your own. Tell me your solutions when you're done the vertices of this equation are: (x+3)^2- 4(y-2)^2=4 find the foci of this hyperbola: 9y^2-72y-16x^2-64x-64=0", "4 x 2 +16x−4 y 2 +16y+16=0$ For the following exercises, given information about the graph of the hyperbola, find its equation. 45. Vertices at $( 3,0 ) ( 3,0 )$ and $( −3,0 ) ( −3,0 )$ and one focus at $( 5,0 ). ( 5,0 ).$ 46. Vertices at $( 0,6 ) ( 0,6 )$ and $( 0,−6 ) ( 0,−6 )$ and one focus at $( 0,−8 ). ( 0,−8 ).$ 47. Vertices at $( 1,1 ) ( 1,1 )$ and $( 11,1 ) ( 11,1 )$ and one focus at $( 12,1 ). ( 12,1 ).$ 48. Center: $( 0,0 ); ( 0,0 );$ vertex: $( 0,−13 ); ( 0,−13 );$ one focus: $( 0, 313 ). ( 0, 313 ).$ 49. Center: $( 4,2 ); ( 4,2 );$ vertex: $( 9,2 ); ( 9,2 );$ one focus: $( 4+ 26 ,2 ). ( 4+ 26 ,2 ).$ 50. Center: $( 3,5 ); ( 3,5 );$ vertex: $( 3,11 ); ( 3,11 );$ one focus: $( 3,5+2 10 ). ( 3,5+2 10 ).$ For the following exercises, given the graph of the hyperbola, find its equation. 51. 52. 53. 54. 55. #### Extensions For the following exercises, express the equation for the hyperbola as two functions, with $y y$ as a function of $x.", "+0 # The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is \\frac(x^2)(A^2)-\\frac(y^2)(B^ 0 410 2 The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is (x^2)/(A^2)-(y^2)/(B^2)=1 where A =_____ B =_____ Guest Jun 16, 2015 #2 +92251 +10 I just want to practice this for myself because I keep forgetting it. http://web2.0calc.com/questions/the-equation-of-the-hyperbola-that-has-a-center-at-1-2-a-focus-at-4-2-and-a-vertex-at-3-2-is-frac-x-c-2-a-2-frac-y#r2 The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is $$\\frac{x^2}{A^2}-\\frac{y^2}{B^2}=1$$ Vertex =( 0 $$\\pm$$ A, 0) So A= 4 $$\\\\Focus=(0\\pm \\sqrt{A^2+B^2},0)\\\\\\\\ \\sqrt{16+B^2}=5\\\\ 4^2+B^2=5^2\\\\ B=3$$ $$\\frac{x^2}{4^2}-\\frac{y^2}{3^2}=1$$ Melody Jun 16, 2015 Sort: #1 +19207 +10 The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is (x^2)/(A^2)-(y^2)/(B^2)=1 $$\\dfrac{x^2}{A^2}-\\dfrac{y^2}{B^2}=1$$ vertex at (-4 , 0 ): $$\\small{\\text{ \\begin{array}{l} (-4,0) =(\\pm A ,0)\\\\ A = \\pm4 \\end{array} }}$$ focus at (5 , 0) : $$\\small{\\text{ \\begin{array}{l} (5,0) =(\\pm \\sqrt{A^2+B^2} ,0)\\end{array} }}\\\\ \\begin{array}{rcl} A^2+B^2 &=& 5^2 = 25\\\\ (\\pm4)^2+B^2 &=& 25\\\\ B^2 &=& 25-16 =", "Sales Toll Free No: 1-855-666-7446 # Analyzing Conic Sections TopCircle, ellipses, parabolas, and Hyperbola are all obtained when a cone is wedged by a plane. The general equation of any conic section is given by: Ap2 + Bpq + Cq2 + Dp + Eq + f = 0; If the value of B = 0 then we will see the ‘A’ and ‘C’ in the equations: Name of conic section Relationship of A and C Parabola A = 0 or C = 0 but both the values of ‘A’ and ‘C’ are never equals to 0. Circle In case of Circle the value of ‘A’ and ‘C’ are both equal. Ellipse In case of ellipse the sign of ‘A’ and ‘C’ are same but ‘A’ and ‘C’ are not equal. Hyperbola In case of hyperbola the signs of ‘A’ and ‘C’ are opposite. A line having curve shape is known as hyperbola. When the transverse axis of the given hyperbolas is aligned with the x – axis then the Equation of Hyperbola is given by: $\\frac{(p - h)^{2}}{a^{2}}$ - $\\frac{(q-k)^{2}}{b^{2}}$ = 1 Having vertices - ( h $\\pm$ a, k) and Foci (h $\\pm$ c,k) This is equation of hyperbola. When the Intersection of Right Circular Cone and a plane which is also parallel to an element of", "## Algebra and Trigonometry 10th Edition Standard form of a circle with center at the origin: $x^2+y^2=r^2$, in which $r$ is the radius. Standard form of a hyperbola with center at the origin and horizontal transverse axis: $\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1~~$ It can be written as: $x^2-\\frac{a^2}{b^2}y^2=a^2$ This last form is the form of the given equation, in which $a=b=12$.", "substitute the values in focus formula: $$(\\frac{-b}{2a} , \\frac{4ac-b^2+1}{4a})=(\\frac{-6}{2(3)} , \\frac{4(3)(0)-6^2+1}{4(3)})$$ Focus of parabola is $$(-1, \\frac{-35}{12})$$. ## Exercises for Finding the Focus, Vertex, and Directrix of Parabola ### Find vertex and focus of each parabola. • $$\\color{blue}{(y-2)^2=3(x-5)^2}$$ • $$\\color{blue}{y=4x^2+x-1}$$ • $$\\color{blue}{y=x^2+2x+3}$$ • $$\\color{blue}{x=y^2-4y}$$ • $$\\color{blue}{Vertex: (5, 2),}$$ $$\\color{blue}{focus: (5, \\frac{25}{12})}$$ • $$\\color{blue}{Vertex: (\\frac{-1}{8}, \\frac{-17}{16}), focus: (\\frac{-1}{8}, -1)}$$ • $$\\color{blue}{Vertex: (-1, 2), focus: (-1, \\frac{9}{4})}$$ • $$\\color{blue}{Vertex: (-4, 2), focus: (\\frac{-15}{4}, 2)}$$ ### What people say about \"How to Find the Focus, Vertex, and Directrix of a Parabola?\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "+0 Ah a repost :D 0 40 1 A hyperbola has one of its foci at $(3, 2),$ and the vertex of the hyperbola closer to this focus is at $(4, 2).$ One of the asymptotes of the hyperbola has slope $\\frac{\\sqrt2}{2}.$ Find the $x-$coordinate of the center of the hyperbola. Mar 15, 2021 #1 0 The x-coordinate of the center of the hyperbola is 11 - 3*sqrt(2). Mar 15, 2021", "+0 # Hi third repost, I really am confused with this problem :( 0 31 1 A hyperbola has one of its foci at $(3, 2),$ and the vertex of the hyperbola closer to this focus is at $(4, 2).$ One of the asymptotes of the hyperbola has slope $\\frac{\\sqrt2}{2}.$ Find the $x-$coordinate of the center of the hyperbola. Mar 16, 2021", "length as the semi-latus rectum of the conic, that is $$b^2/a$$ for an ellipse or hyperbola, where $$a$$ and $$b$$ are as usual the semiaxes. The values of $$a$$ and $$b$$ depend not only on the inclination of the intersecting plane, but also on the distance of that plane from the vertex of the cone. I'll express them as a function of the distances $$m$$ and $$n$$ of the vertices (of the ellipse or hyperbola) from the vertex of the cone. If $$u$$ is the semi-aperture angle of the cone, we have: $$4a^2=m^2+n^2\\mp2mn\\cos2u,\\quad b^2=mn\\sin^2u,$$ where sign $$-$$ must be taken for an ellipse and sign $$+$$ for a hyperbola. You can find the proof (for an ellipse, but that for a hyperbola is analogous) in this answer. It follows that the minimum radius of curvature is $$r_{MIN}={2mn\\sin^2u\\over\\sqrt{m^2+n^2\\mp2mn\\cos2u}}$$ For a parabola, you just need to take the limit of the above result for $$n\\to\\infty$$: $$r_{MIN}=2m\\sin^2u.$$ • $m$ and $n$ are the distances from the vertex $V$ of the cone to the two vertices $A$ and $B$ of the ellipse or hyperbola (endpoints of major axis). Points $A$ and $B$ do lie on the cone surface, so distances $m=AV$ and $n=BV$ are indeed \"on the cone itself\" (if I understand what you mean). $m$ and $n$ can then be easily" ]

[ "length as the semi-latus rectum of the conic, that is $$b^2/a$$ for an ellipse or hyperbola, where $$a$$ and $$b$$ are as usual the semiaxes. The values of $$a$$ and $$b$$ depend not only on the inclination of the intersecting plane, but also on the distance of that plane from the vertex of the cone. I'll express them as a function of the distances $$m$$ and $$n$$ of the vertices (of the ellipse or hyperbola) from the vertex of the cone. If $$u$$ is the semi-aperture angle of the cone, we have: $$4a^2=m^2+n^2\\mp2mn\\cos2u,\\quad b^2=mn\\sin^2u,$$ where sign $$-$$ must be taken for an ellipse and sign $$+$$ for a hyperbola. You can find the proof (for an ellipse, but that for a hyperbola is analogous) in this answer. It follows that the minimum radius of curvature is $$r_{MIN}={2mn\\sin^2u\\over\\sqrt{m^2+n^2\\mp2mn\\cos2u}}$$ For a parabola, you just need to take the limit of the above result for $$n\\to\\infty$$: $$r_{MIN}=2m\\sin^2u.$$ • $m$ and $n$ are the distances from the vertex $V$ of the cone to the two vertices $A$ and $B$ of the ellipse or hyperbola (endpoints of major axis). Points $A$ and $B$ do lie on the cone surface, so distances $m=AV$ and $n=BV$ are indeed \"on the cone itself\" (if I understand what you mean). $m$ and $n$ can then be easily", "semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: $b = a \\sqrt{e^2-1}.$ Note that in a hyperbola b can be larger than a. [1]", "YOUR website!You have 7 letters and 4 spots to place them, so you have 7 P 4 = 7!/(7-4)!=7!/3! = (7*6*5*4*3!)/3! = 7*6*5*4 = 840 different ways. Quadratic-relations-and-conic-sections/309219: Write an equation for the hyperbola...I have no idea. vertices (0,6) and (0,-6) conjugate axis of 14. I'd really like to figure out on my own but I have no idea where to start..so if someone could give me a step-by-step solution, that would be wonderful! Thank you.1 solutions Answer 221108 by jim_thompson5910(28715) on 2010-05-27 18:24:13 (Show Source): You can put this solution on YOUR website!The center of the hyperbola is the midpoint of the line segment from vertex to vertex. So the midpoint of (0,6) and (0,-6) is ((0+0)/2, (6+(-6)/2) ---> (0, 0). So the center is (0,0). The center is in the form (h,k), so h=0 and k=0. In this case, 'a' is the length of the semi-minor axis and 'b' is the length of the semi-major axis. So the lengths of the conjugate and traverse axes are 2a and 2b units respectively. Since the traverse axis is 2b units long, this means that 2b=12 (since the distance between the vertices is 12 units) which means that b=6. Also, because the conjugate axis is 14 units long, and 2a is the length of the conjugate axis, this means", "Taylor H. I have a graph of a hyperbola where the vertices are at (-3,0) and (3,0) and the center is (0,0). It needs to be written in a standard form (x-h)/ a2 - (y-k)/ b2 Mark M. tutor Since the center is (0, 0), h = k = 0. The distance from the center to either vertex is a. So, a = 3. To find b, we need further information. Did you leave something out? Report 05/02/15", "= 151 $$\\implies$$ $$9(x^2 – 2x + 1)$$ – $$16(y^2 – 2y + 1)$$ = 144 $$\\implies$$ $$9(x – 1)^2$$ – $$16(y – 1)^2$$ = 144 $$\\implies$$ $$(x – 1)^2\\over 16$$ – $$(y – 1)^2\\over 9$$ = 1 Here, a = 4 and b = 3 Coordinates of Center of the hyperbola is (h, k) i.e. (1, 1) And Coordinates of Vertices are (a + h, k) and (-a + h, k) $$\\implies$$ (4 + 1, 1) and (-4 + 1, 1) = (5, 1) and (-3, 1) Note : For the hyperbola $$(x – h)^2\\over a^2$$ – $$(y – k)^2\\over b^2$$ = 1 with center (h. k), (i) For normal hyperbola, The coordinates of vertices are (a + h, k) and (-a + h, k). And Coordinates of Center is (h, k). (ii) For conjugate hyperbola, The coordinates of vertices are (h, b + k) and (h, -b + k). And Coordinates of Center is (h, k).", "1. Centroid Help! lol trying to find a the center ,x. the answer is x = 3/8b... mistake lol ? 2. Originally Posted by aeubz ImageShack - Image Hosting :: slowge9.jpg lol trying to find a the center ,x. the answer is x = 3/8b... mistake lol ? Since the horizontal line meets the parabola at $x = b$ then $h = ab^2$. Substituting this gives your answer. 3. Originally Posted by danny arrigo Since the horizontal line meets the parabola at $x = b$ then $h = ab^2$. Substituting this gives your answer. thanks!!!", "+0 # Plz help me fast +1 39 2 +38 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as (y-k)^2/a^2-(x-h)^2/b^2=1. Find h+k+a+b Mar 20, 2019 #1 +100047 +2 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1$$ Find h+k+a+b Try looking at this: Once you work it out you can check it by graphing in desmos. https://www.desmos.com/calculator Mar 20, 2019 #2 +38 +2 I got it, thanks Dahdahda Mar 20, 2019", "for any point $$(x,y)$$ on the hyperbola. We know that the difference of these distances is $$2a$$ for the vertex $$(a,0)$$. It follows that $$d_2−d_1=2a$$ for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. \\begin{align*} d_2-d_1&=2a\\\\ \\sqrt{{(x-(-c))}^2+{(y-0)}^2}-\\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\\qquad \\text{Distance Formula}\\\\ \\sqrt{{(x+c)}^2+y^2}-\\sqrt{{(x-c)}^2+y^2}&=2a\\qquad \\text{Simplify expressions.}\\\\ \\sqrt{{(x+c)}^2+y^2}&=2a+\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Move radical to opposite side.}\\\\ {(x+c)}^2+y^2&={(2a+\\sqrt{{(x-c)}^2+y^2})}^2\\qquad \\text{Square both sides.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\\qquad \\text{Expand the squares.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\\qquad \\text{Expand remaining square.}\\\\ 2cx&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}-2cx\\qquad \\text{Combine like terms.}\\\\ 4cx-4a^2&=4a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Isolate the radical.}\\\\ cx-a^2&=a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Divide by 4.}\\\\ {(cx-a^2)}^2&=a^2{\\left[\\sqrt{{(x-c)}^2+y^2}\\right]}^2\\qquad \\text{Square both sides.}\\\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\\qquad \\text{Expand the squares.}\\\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\\qquad \\text{Distribute } a^2\\\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\\qquad \\text{Combine like terms.}\\\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\\qquad \\text{Rearrange terms.}\\\\ x^2(c^2-a^2)-a^2y^2&=a^2(c^2-a^2)\\qquad \\text{Factor common terms.}\\\\ x^2b^2-a^2y^2&=a^2b^2\\qquad \\text{Set } b^2=c^2−a^2\\\\. \\dfrac{x^2b^2}{a^2b^2}-\\dfrac{a^2y^2}{a^2b^2}&=\\dfrac{a^2b^2}{a^2b^2}\\qquad \\text{Divide both sides by } a^2b^2\\\\ \\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}&=1\\\\ \\end{align*} This equation defines a hyperbola centered at the origin with vertices $$(\\pm a,0)$$ and co-vertices $$(0,\\pm b)$$. STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER $$(0,0)$$ The standard form of the equation of a hyperbola with center $$(0,0)$$ and transverse axis on the $$x$$-axis is $$\\dfrac{x^2}{a^2}−\\dfrac{y^2}{b^2}=1$$ where • the length of the transverse axis is $$2a$$ • the coordinates of the vertices are $$(\\pm a,0)$$ • the length of the conjugate axis is $$2b$$", "cm from the center of the circle and BC 8 cm from the center of the circle. • Geometric progressiob If the sum of four consective terms of geometric progression is 80 and arithmetic mean of second and fourth term is 30 then find terms? • Hyperbola equation Find the hyperbola equation with the center of S [0; 0], passing through the points: A [5; 3] B [8; -10] • Digit sum The digit sum of the two-digit number is nine. When we turn figures and multiply by the original two-digit number, we get the number 2430. What is the original two-digit number? • Cuboid walls Calculate the volume of the cuboid if its different walls have area of 195cm², 135cm² and 117cm². • Sales of products For 80 pieces of two quality products a total sales is 175 Eur. If the first quality product was sold for n EUR per piece (n natural number) and the second quality product after 2 EUR per piece. How many pieces of the first quality were sold? • The gardener The gardener bought trees for 960 CZK. If every tree were cheaper by 12 CZK, he would have gotten four more trees for the same money. How many trees did he buy? • Lookout tower How high is the lookout tower?", "$$60$$ $$108$$ $$120$$ $$169$$ ###### Solution(s): Since the hyperbola is symmetric, without the loss of generality, we can have $$(1,1)$$ as our vertex. Then, since we have the centriod of an equilateral triangle, the angle at the centriod with any two points is $$120^ \\circ .$$ The branch of the hyperbola with negative coordinates can make an angle of at most $$90^\\circ .$$ This means that we can't have two points on the negative branch. Since the hyperbola is symmetric over $$y=x$$ and it always decreases, the two points are reflected over $$y=x.$$ Also, the altitude is on $$y=x,$$ making the other point also on $$y=x.$$ This makes the other point $$(-1,-1).$$ Thus, the circumradius is $$2\\sqrt 2$$ since it is the distance between the two points. This means we have $$3$$ isoceles triangles with side lengths $$2 \\sqrt 2$$ and angle $$120^\\circ.$$ Therefore, the combined area is $3\\cdot \\dfrac{(2\\sqrt 2)^2 \\cdot \\sin(120^ \\circ)}2$$= 12 \\sin(120^\\circ)$$= 12 \\cdot \\dfrac{\\sqrt{3}}2$$=\\sqrt{108} .$ This makes the square $$108.$$ Thus, the correct answer is C. 25. Last year Isabella took $$7$$ math tests and received $$7$$ different scores, each an integer between $$91$$ and $$100,$$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $$95.$$ What was her", "# Finding a point which is constrained to 3 other points. Is there an easy way to find the 4th point given 3 fixed points and a different minimum length between the 4th point and each of the 3 points? Similar to this question, but with non-fixed minimum lengths instead. The reason for this is that I have 3 circles with different radius and I want to find the position of a 4th circle such that the 4th circle does not overlap yet is as close as possible. Thanks! - I'm going to take the simplest interpretation, you have three circles with centers that are far enough apart so that the circles do not overlap. Take two of the centers, $C_1, \\; C_2$ with radii $r_1, \\; r_2.$ Where do I put the center $C_\\ast$ of a new circle that is tangent to both? The answer is that $C_\\ast$ lies on one branch of a hyperbola. One point of the hyperbola is along the line segment between $C_1$ and $C_2,$ at equal distance from the two circles (not necessarily at equal distances from $C_1$ and $C_2$). The hyperbola is just the curve giving a constant difference between the distances $C_\\ast C_1$ and $C_\\ast C_2.$ Do the same thing for the pair of circles with centers $C_1$ and $C_3.$", "Mathematics OpenStudy (anonymous): Find the center and vertices of the hyperbola. 16x^2 - 9y^2 - 96x + 72y - 144 = 0 OpenStudy (amistre64): $(\\frac{x}{a})^2-(\\frac{y}{b})^2=1$", "Maybe that's too much work 3 Calculus Level 5 Find the area of the region bounded by the following curves : $y=f(x) , y=|g(x)| , x=0 \\quad \\text{and} \\quad x=2$ f and g are two continuous functions satisfying the relations • $\\displaystyle f(x+y) = f(x) + f(y) -8xy , \\forall x,y \\in R$ • $\\displaystyle g(x+y) = g(x) + g(y) + 3xy , \\forall x,y \\in R$ • $\\displaystyle f'(0)=8 , g'(0) = -4$ The answer is of the form $$\\dfrac{a}{b}$$ . From any point K on the hyperbola $$\\dfrac{x^{2}}{a^{2}}-\\dfrac{y^{2}}{b^{2}} = 1$$ , three normals other than that at K are drawn . Now it is given that the centroid of the triangle formed by their feet lies on a hyperbola $\\dfrac{x^{2}}{a^{2}}-\\dfrac{y^{2}}{b^{2}} = k^{2}$ . Evaluate the integer nearest to $$100000k^{2}$$ ×", "## anonymous 5 years ago Find the center and vertices of the hyperbola. 16x^2 - 9y^2 - 96x + 72y - 144 = 0 $(\\frac{x}{a})^2-(\\frac{y}{b})^2=1$", "+0 # Hyperbola +1 110 3 +30 Find the distance between the two foci of the hyperbola. $$-6x^2 + 5y^2 + 24x + 20y = 64$$ Sep 7, 2020 #1 +30 0 Someone pls help me, thanks. Sep 7, 2020 #2 +112002 +2 https://www.purplemath.com/modules/hyperbola.htm You have to start by getting it in the form $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1\\\\ (h,k) \\text{ is the centre}\\\\$$ ALSO .. Here is a youtube clip that looks fairly comprehensive.", "## Precalculus (6th Edition) Blitzer From the equation, we have $a=3, b=1, c=\\sqrt {a^2+b^2}=\\sqrt {10}\\approx3.16$. The center of the hyperbola is at $(0,0)$ and the transverse axis is horizontal. Thus, the foci can be located at $(\\pm\\sqrt {10},0)$", "## Calculus 8th Edition $y ^2 = 4x$ As we are given that the parabola is laterally oriented and the focus is to the right of the vertex. Since, the focus is $(1,0)$ and the vertex is $(0,0)$. Therefore, $h=k=0$ and $p=1$. The equation of a parabola becomes after plug in the values: $(y -0)^2 = 4(1)(x -0)$ Hence, $y ^2 = 4x$", "View & Read • 1) The number of subsets of a set containing n elements is • 2) For any two sets A and B, $A\\cap(A\\cup B)=$_____. • 3) If $A\\cap B=B$ then • 4) For any two sets A and B, (A - B) $\\cup$ (B - A)=_____. • 5) For any two sets A and B, A $\\cup$ B = A iff #### 11th CBSE Mathematics Conic Sections Model Question Paper - by Rupa Das - Bhubaneswar - View & Read • 1) $The\\quad circle\\quad { x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\\quad does\\quad not\\quad intersect\\quad x-axis\\quad if$ • 2) The difference between the lengths of the major axis and the latus rectum of an ellipse is • 3) The eccentricity of the hyperbola whose latus rectum is half of its transverse axis is • 4) The difference of the focal distances of any point on the hyperbola is equal to • 5) The latus rectum of the hyperbola ${ 16x }^{ 2 }-{ ay }^{ 2 }=144\\quad is$ #### 11th CBSE Mathematics Straight Lines Model Question Paper - by Rupa Das - Bhubaneswar - View & Read • 1) A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y intercept is • 2)", "the points have the same sum of the distances to two points - All points for $$0<e<1$$, where e represents the ratio of the distance to the point (a, b) to the distance to the line x=n. The equation for this ellipse can be written as follows: $$\\sqrt{(x-a)^2+(y-b)^2}=e(x-n)$$. This is related to the equation below because when simplifying this equation, you would get coefficients of the same sign for both x2 and y2. - $$Ax^2+Bx^2=C$$, where A and B have the same sign. - This ratio, e, is known as the eccentricity. It shows up in two places. As already mentioned, distance from focus to any point on the ellipse: distance from directrix to any point on the ellipse. Also, distance from the center to the focus: distance from the center to the vertex. - Pour salt on a large piece of cardboard with a circle cut out of it - Slice a cone from side to side Circle: - All the points for e = 0, where e represents to the ratio of the distance to the point (a, b) to the distance to the line x=n. The equation for this circle can be written as follows: $$\\sqrt{(x-a)^2+(y-b)^2}=e(x-n)$$. - $$Ax^2+Bx^2=C$$, where A and B are equal. - Slice a cone parallel to the base Hyperbola: - All", "I am looking for two circles that touch, but do not overlap. |C1 - P1| + |C2 - P2| = |C1 - C2| It seems that finding C1 and C2 is dependent on finding the scalars t1 and t2 such that the last equation is true. You have one equation in two unknowns $t_1$ and $t_2$, so there should generally be a one-parameter family of solutions. In fact, choose $t_1$ (and thus $C_1$ and $r_1 = |C_1 - P_1|$). Then you want $C_2$ to be a point on the line $P_2 - t_2 B$ whose distance to $C_1$ is $r_1$ more than its distance from $P_2$. The locus of a point whose distance to $C_1$ is $r_1$ more than its distance from $P_2$ is a branch of a hyperbola (if nonempty)." ]

[ "+0 # Plz help me fast +1 39 2 +38 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as (y-k)^2/a^2-(x-h)^2/b^2=1. Find h+k+a+b Mar 20, 2019 #1 +100047 +2 In a certain hyperbola, the center is at (2, 0), one focus is at (2, 6) and one vertex is at (2, -3). The equation of this hyperbola can be written as $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1$$ Find h+k+a+b Try looking at this: Once you work it out you can check it by graphing in desmos. https://www.desmos.com/calculator Mar 20, 2019 #2 +38 +2 I got it, thanks Dahdahda Mar 20, 2019", "on the hyperbola $$\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}$$ = 1. Main facts about two standard forms of hyperbola Circle : The locus of a point which moves in a plane such that its distance from a fixed point, in the plane, is always a constant, is called a circle. The fixed point is called the centre and the constant distance is called the Radius of the circle. Equation of the Circle : The equation of a circle with centre (h, k) and radius V’ is given by (x – h)2 + (y – k)2 = r2. If the centre of a circle is origin and radius is ‘r’ then its equation is x2 + y2 = r2. General Equation of a Circle : The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0. Its centre is (-g,-f) and radius = $$\\sqrt{g^{2}+f^{2}-c}$$ Note : 1. If g2 + f2 – c = 0, then radius = 0. In this case, x2 +y2 + 2gx + 2fy + c = 0 represents a circle whose radius is zero. Such a circle is called a Point Circle, [i.e., it represents the point (-g, -f)]. 2. If g2 + f2 – c < 0, then radius is non-real. In this case, x2 +y2 + 2gx + 2fy +", "semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: $b = a \\sqrt{e^2-1}.$ Note that in a hyperbola b can be larger than a. [1]", "length as the semi-latus rectum of the conic, that is $$b^2/a$$ for an ellipse or hyperbola, where $$a$$ and $$b$$ are as usual the semiaxes. The values of $$a$$ and $$b$$ depend not only on the inclination of the intersecting plane, but also on the distance of that plane from the vertex of the cone. I'll express them as a function of the distances $$m$$ and $$n$$ of the vertices (of the ellipse or hyperbola) from the vertex of the cone. If $$u$$ is the semi-aperture angle of the cone, we have: $$4a^2=m^2+n^2\\mp2mn\\cos2u,\\quad b^2=mn\\sin^2u,$$ where sign $$-$$ must be taken for an ellipse and sign $$+$$ for a hyperbola. You can find the proof (for an ellipse, but that for a hyperbola is analogous) in this answer. It follows that the minimum radius of curvature is $$r_{MIN}={2mn\\sin^2u\\over\\sqrt{m^2+n^2\\mp2mn\\cos2u}}$$ For a parabola, you just need to take the limit of the above result for $$n\\to\\infty$$: $$r_{MIN}=2m\\sin^2u.$$ • $m$ and $n$ are the distances from the vertex $V$ of the cone to the two vertices $A$ and $B$ of the ellipse or hyperbola (endpoints of major axis). Points $A$ and $B$ do lie on the cone surface, so distances $m=AV$ and $n=BV$ are indeed \"on the cone itself\" (if I understand what you mean). $m$ and $n$ can then be easily", "b/3(x-0)= ±2x, b must =6 b^2=36 I can now write the equation in the form of (x-h)^2/a^2-(y-k)^2/b^2=1 as: x^2/9-y^2/36=1 End part B That is all my work/solutions for this problem. Any help with what I may have done wrong and how to learn how to correct is, or even just saying \"correct\" is appreciated thank you. 2. May 7, 2012 ### HallsofIvy Assuming an ellipse is set up with foci on the x-axis, we can take them to be at (c, 0) and (-c, 0). We can take the vertices on the major axis to be at (a, 0) and (-a, 0). The distance from the focus (-c, 0) to (a, 0) is a-(-c)= a+ c and the distance from that vertex to the other focus, (c, 0) is a- c. That is, the total distance from focus to vertex to other vertex, along the major axis, is a+ c+ a- c= 2a. The distance from the focus (c, 0) to the vertex on the minor axis, (0, b) is $\\sqrt{b^2+ c^2}$. The distance from (0, b) to (-c, 0) is also $\\sqrt{b^2+ c^2}$ so the distance form focus to vertex to the other vertex, on the minor axis is $2\\sqrt{b^2+ c^2}$. Now, a defining property of an ellipse is that the distance from one focus, to any", "# Math Help - need help with hyperbola equations 1. ## need help with hyperbola equations I am having issues with finding hyperbola equations with a translation when it gives the vertices and foci. vertices: (-5,5) (5,5) foci: (-7,5) (7,5) 2. The focal axis is horizontal, so the equation is in the form $\\frac{(x - h)^2}{a^2} - \\frac{(y - k)^2}{b^2} = 1$ The center (h, k) is in between the two vertices, so it's (0, 5). a is the distance from the center to one of the vertices, so a = 5. c is the distance from the center to one of the foci, so c = 7. The relationship between a, b, c is $c^2 = a^2 + b^2$ so solve for b: $7^2 = 5^2 + b^2$ $49 = 25 + b^2$ $24 = b^2$ $b = \\sqrt{24}$ So the equation should be $\\frac{x^2}{25} - \\frac{(y - 5)^2}{24} = 1$ Nothing to it! 01", "# Finding a point which is constrained to 3 other points. Is there an easy way to find the 4th point given 3 fixed points and a different minimum length between the 4th point and each of the 3 points? Similar to this question, but with non-fixed minimum lengths instead. The reason for this is that I have 3 circles with different radius and I want to find the position of a 4th circle such that the 4th circle does not overlap yet is as close as possible. Thanks! - I'm going to take the simplest interpretation, you have three circles with centers that are far enough apart so that the circles do not overlap. Take two of the centers, $C_1, \\; C_2$ with radii $r_1, \\; r_2.$ Where do I put the center $C_\\ast$ of a new circle that is tangent to both? The answer is that $C_\\ast$ lies on one branch of a hyperbola. One point of the hyperbola is along the line segment between $C_1$ and $C_2,$ at equal distance from the two circles (not necessarily at equal distances from $C_1$ and $C_2$). The hyperbola is just the curve giving a constant difference between the distances $C_\\ast C_1$ and $C_\\ast C_2.$ Do the same thing for the pair of circles with centers $C_1$ and $C_3.$", "Circle A and Circle C at a single point each. How can I do that? I feel like there's some system of equations I could solve to find the proper center and radius, but my attempts at creating the proper system of equations always end up with 2 equations and 3 unknowns. The circles can be of arbitrary size and may not be as nearly equally sized as in the above image. We are looking for a circle centered on the given straight line (blue), and touching two given circles (blue). [] If $$A$$ and $$C$$ are centers of the given circles, $$a$$ and $$c$$ their radii, $$K$$ the center and $$r$$ the radius of a touching circle, then $$||KC|-|KA||=|r\\pm c-(r\\pm a)|=|c-a|.$$ The difference of distances of $$K$$ to two fixed points is constant. Therefore, the locus of centers of touching circles is a hyperbola with foci $$A$$ and $$C.$$ (See also this question.) One vertex of the hyperbola is $$I,$$ it lies on $$AC$$ at equal distance to both blue circles. Assume circles A and B (with centers $$(x_A, y_A)$$ and $$(x_B, y_B)$$ and radii $$r_A$$ and $$r_B$$) are tangent to each other and circle B is not yet tangent to circle C (with center $$(x_C, y_C)$$ and radius $$r_C$$). Let point T $$(x_T, y_T)$$ be the", "for any point $$(x,y)$$ on the hyperbola. We know that the difference of these distances is $$2a$$ for the vertex $$(a,0)$$. It follows that $$d_2−d_1=2a$$ for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. \\begin{align*} d_2-d_1&=2a\\\\ \\sqrt{{(x-(-c))}^2+{(y-0)}^2}-\\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\\qquad \\text{Distance Formula}\\\\ \\sqrt{{(x+c)}^2+y^2}-\\sqrt{{(x-c)}^2+y^2}&=2a\\qquad \\text{Simplify expressions.}\\\\ \\sqrt{{(x+c)}^2+y^2}&=2a+\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Move radical to opposite side.}\\\\ {(x+c)}^2+y^2&={(2a+\\sqrt{{(x-c)}^2+y^2})}^2\\qquad \\text{Square both sides.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\\qquad \\text{Expand the squares.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\\qquad \\text{Expand remaining square.}\\\\ 2cx&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}-2cx\\qquad \\text{Combine like terms.}\\\\ 4cx-4a^2&=4a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Isolate the radical.}\\\\ cx-a^2&=a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Divide by 4.}\\\\ {(cx-a^2)}^2&=a^2{\\left[\\sqrt{{(x-c)}^2+y^2}\\right]}^2\\qquad \\text{Square both sides.}\\\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\\qquad \\text{Expand the squares.}\\\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\\qquad \\text{Distribute } a^2\\\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\\qquad \\text{Combine like terms.}\\\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\\qquad \\text{Rearrange terms.}\\\\ x^2(c^2-a^2)-a^2y^2&=a^2(c^2-a^2)\\qquad \\text{Factor common terms.}\\\\ x^2b^2-a^2y^2&=a^2b^2\\qquad \\text{Set } b^2=c^2−a^2\\\\. \\dfrac{x^2b^2}{a^2b^2}-\\dfrac{a^2y^2}{a^2b^2}&=\\dfrac{a^2b^2}{a^2b^2}\\qquad \\text{Divide both sides by } a^2b^2\\\\ \\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}&=1\\\\ \\end{align*} This equation defines a hyperbola centered at the origin with vertices $$(\\pm a,0)$$ and co-vertices $$(0,\\pm b)$$. STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER $$(0,0)$$ The standard form of the equation of a hyperbola with center $$(0,0)$$ and transverse axis on the $$x$$-axis is $$\\dfrac{x^2}{a^2}−\\dfrac{y^2}{b^2}=1$$ where • the length of the transverse axis is $$2a$$ • the coordinates of the vertices are $$(\\pm a,0)$$ • the length of the conjugate axis is $$2b$$", "I am looking for two circles that touch, but do not overlap. |C1 - P1| + |C2 - P2| = |C1 - C2| It seems that finding C1 and C2 is dependent on finding the scalars t1 and t2 such that the last equation is true. You have one equation in two unknowns $t_1$ and $t_2$, so there should generally be a one-parameter family of solutions. In fact, choose $t_1$ (and thus $C_1$ and $r_1 = |C_1 - P_1|$). Then you want $C_2$ to be a point on the line $P_2 - t_2 B$ whose distance to $C_1$ is $r_1$ more than its distance from $P_2$. The locus of a point whose distance to $C_1$ is $r_1$ more than its distance from $P_2$ is a branch of a hyperbola (if nonempty).", "e ) = $$\\frac{c}{a}$$ Also, c ≥ a, the eccentricity is never less than one. • Distance of focus from centre: ae • Equilateral hyperbola: Hyperbola in which a = b • Conic section formulas for latus rectum in hyperbola: $$\\frac{2b^{2}}{a}$$ ## Conic section formulas examples: 1. Find an equation of the circle with centre at (0,0) and radius r. Solution: Here h = k = 0. Therefore, the equation of the circle is x2 + y2= r2 2. Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 16x. Solution: In this equation, y2 is there, so the coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 4. Thus, the focus of the parabola is (4, 0) and the equation of the directrix of the parabola is x = – 4 Length of the latus rectum is 4a = 4 × 4 = 16.", "reflecting property of the hyperbola, we will show that $\\,\\alpha = \\beta\\,$. Both $\\,\\alpha\\,$ and $\\,\\beta\\,$ are positive angles, not exceeding $\\,90^\\circ\\,$. The tangent function is one-to-one between $\\,0^\\circ\\,$ and $\\,90^\\circ\\,$, so if $\\,\\tan\\alpha = \\tan\\beta\\,$, it follows that $\\,\\alpha = \\beta\\,$. Consequently: To prove the reflecting property of the hyperbola, we will show that $\\,\\tan\\alpha = \\tan\\beta\\,$. Summarizing results for convenience: • [hyperbola] The hyperbola under investigation has equation $\\displaystyle\\,\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1\\,$. The positive number $\\,c\\,$ satisfies $\\,c^2 = a^2 + b^2\\,$. • [green line] The green line is outside light directed towards the focus $\\,(c,0)\\,$. This green line has slope $\\displaystyle\\,\\frac{y}{x-c}\\,$, where $\\,(x,y)\\,$ are the coordinates of the point $\\,P\\,$ where the green line intersects the hyperbola. • [red line] The red line is between point $\\,P(x,y)\\,$ and the second focus, $\\,(-c,0)\\,$. This red line has slope $\\displaystyle\\,\\frac{y}{x+c}\\,$. • [tangent/dashed line] The dashed line is the tangent to the hyperbola at $\\,P(x,y)\\,$. It has slope $\\,\\displaystyle\\frac{b^2}{a^2}\\frac{x}{y}\\,$. Using the formula for the tangent of the angle between two intersecting lines of known slope: • The angle $\\,\\alpha\\,$ is formed by the intersecting green and tangent lines. The green line has the greater angle of inclination. Thus: $$\\tan\\alpha = \\frac{\\frac{y}{x-c} - \\frac{b^2}{a^2}\\frac{x}{y}}{1 + \\frac{y}{x-c}\\cdot\\frac{b^2}{a^2}\\cdot\\frac{x}{y}}$$ • The angle $\\,\\beta\\,$ is formed by the intersecting red and tangent", "YOUR website!You have 7 letters and 4 spots to place them, so you have 7 P 4 = 7!/(7-4)!=7!/3! = (7*6*5*4*3!)/3! = 7*6*5*4 = 840 different ways. Quadratic-relations-and-conic-sections/309219: Write an equation for the hyperbola...I have no idea. vertices (0,6) and (0,-6) conjugate axis of 14. I'd really like to figure out on my own but I have no idea where to start..so if someone could give me a step-by-step solution, that would be wonderful! Thank you.1 solutions Answer 221108 by jim_thompson5910(28715) on 2010-05-27 18:24:13 (Show Source): You can put this solution on YOUR website!The center of the hyperbola is the midpoint of the line segment from vertex to vertex. So the midpoint of (0,6) and (0,-6) is ((0+0)/2, (6+(-6)/2) ---> (0, 0). So the center is (0,0). The center is in the form (h,k), so h=0 and k=0. In this case, 'a' is the length of the semi-minor axis and 'b' is the length of the semi-major axis. So the lengths of the conjugate and traverse axes are 2a and 2b units respectively. Since the traverse axis is 2b units long, this means that 2b=12 (since the distance between the vertices is 12 units) which means that b=6. Also, because the conjugate axis is 14 units long, and 2a is the length of the conjugate axis, this means", "Precalculus (10th Edition) $\\dfrac{x^2}{9}+\\dfrac{y^2}{25}=1$ See graph We are given the ellipse: Center: $(0,0)$ Focus: $(0,-4)$ Vertex: $(0,5)$ Because the $x$-coordinates of the vertex and focus are the same, the ellipse has the equation: $\\dfrac{(x-h)^2}{b^2}+\\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k$ using the center: $(h,k)=(0,0$ $h=0$ $k=0$ Determine $a$ using the vertex: $(h,k+a)=(0,5)$ $(0,0+a)=(0,5)$ $(0,a)=(0,5)$ $a=5$ Determine $c$ using the focus: $(h,k-c)=(0,-4)$ $(0,0-c)=(0,-4)$ $(0,-c)=(0,-4)$ $c=4$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=5^2-4^2$ $b^2=9$ $b=\\sqrt 9=3$ The equation of the ellipse is: $\\dfrac{x^2}{3^2}+\\dfrac{y^2}{5^2}=1$ $\\dfrac{x^2}{9}+\\dfrac{y^2}{25}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse:", "1. Centroid Help! lol trying to find a the center ,x. the answer is x = 3/8b... mistake lol ? 2. Originally Posted by aeubz ImageShack - Image Hosting :: slowge9.jpg lol trying to find a the center ,x. the answer is x = 3/8b... mistake lol ? Since the horizontal line meets the parabola at $x = b$ then $h = ab^2$. Substituting this gives your answer. 3. Originally Posted by danny arrigo Since the horizontal line meets the parabola at $x = b$ then $h = ab^2$. Substituting this gives your answer. thanks!!!", "Maybe that's too much work 3 Calculus Level 5 Find the area of the region bounded by the following curves : $y=f(x) , y=|g(x)| , x=0 \\quad \\text{and} \\quad x=2$ f and g are two continuous functions satisfying the relations • $\\displaystyle f(x+y) = f(x) + f(y) -8xy , \\forall x,y \\in R$ • $\\displaystyle g(x+y) = g(x) + g(y) + 3xy , \\forall x,y \\in R$ • $\\displaystyle f'(0)=8 , g'(0) = -4$ The answer is of the form $$\\dfrac{a}{b}$$ . From any point K on the hyperbola $$\\dfrac{x^{2}}{a^{2}}-\\dfrac{y^{2}}{b^{2}} = 1$$ , three normals other than that at K are drawn . Now it is given that the centroid of the triangle formed by their feet lies on a hyperbola $\\dfrac{x^{2}}{a^{2}}-\\dfrac{y^{2}}{b^{2}} = k^{2}$ . Evaluate the integer nearest to $$100000k^{2}$$ ×", "× # Level 5 A parabola whose vertex is the point $$V(2,3)$$ and focus is $$(5,6)$$ has equation $$ax^2+bxy+cy^2+dx+ey+f=0$$, where $$\\gcd(a,b,c,d,e,f)=1$$. Find $$|a+b+c+d+e+f|$$. Given the ellipse $$3x^{2} - 12x + 2y^{2} + 12y + 6 = 0$$, there exists a real number $$k = a\\sqrt{b} + c$$, (where $$a,b,c$$ are all positive integers and $$b$$ is square-free), such that the hyperbola $$xy - 2y + 3x = k$$ is tangent to the ellipse at two points. Find $$a + b + c$$. A hyperbola $$H$$ with equation $$xy=n$$ (where $$n\\le 1000$$) is rotated $$45^{\\circ}$$ to obtain the hyperbola $$H'$$. Let the positive difference between the number of lattice points on $$H$$ and $$H'$$ be $$D$$. Given that both $$H$$ and $$H'$$ have at least one lattice point, find the maximum possible value of $$D$$. Details and Asumptions: A lattice point is a point that has integer $$x$$- and $$y$$-coordinates. You may want to look at the list of Highly Composite Numbers. A circle and an ellipse of the same area share the interior of a larger circle, without overlap. For the size of the smaller circle, the ellipse has the largest possible area that could fit in the space between the smaller and larger circle. Let $$a$$ be the combined areas of the ellipse and the small circle,", "$$60$$ $$108$$ $$120$$ $$169$$ ###### Solution(s): Since the hyperbola is symmetric, without the loss of generality, we can have $$(1,1)$$ as our vertex. Then, since we have the centriod of an equilateral triangle, the angle at the centriod with any two points is $$120^ \\circ .$$ The branch of the hyperbola with negative coordinates can make an angle of at most $$90^\\circ .$$ This means that we can't have two points on the negative branch. Since the hyperbola is symmetric over $$y=x$$ and it always decreases, the two points are reflected over $$y=x.$$ Also, the altitude is on $$y=x,$$ making the other point also on $$y=x.$$ This makes the other point $$(-1,-1).$$ Thus, the circumradius is $$2\\sqrt 2$$ since it is the distance between the two points. This means we have $$3$$ isoceles triangles with side lengths $$2 \\sqrt 2$$ and angle $$120^\\circ.$$ Therefore, the combined area is $3\\cdot \\dfrac{(2\\sqrt 2)^2 \\cdot \\sin(120^ \\circ)}2$$= 12 \\sin(120^\\circ)$$= 12 \\cdot \\dfrac{\\sqrt{3}}2$$=\\sqrt{108} .$ This makes the square $$108.$$ Thus, the correct answer is C. 25. Last year Isabella took $$7$$ math tests and received $$7$$ different scores, each an integer between $$91$$ and $$100,$$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $$95.$$ What was her", "## anonymous 5 years ago Find the standard equation of the hyperbola that has a vertex at (4, 2), focus at (4, 4) and a center at (4, -1). I don't have much time, as I mentioned, but! I will get you started by saying, you can apply much of what we applied in my previous answer, except, since the center isn't at (0, 0), a couple of things change. First off, the equation becomes: $-\\frac{(x - h)^2}{a^2} + \\frac{(y - k)^2}{b^2} = 1$ This is for a center (h, k). Note that this time it's - + instead of + -; that's because this hyperbola opens up and down instead of left and right (you can see that if you sketch it based on the parameters they give you above). The other thing that changes is that the equation for the vertices is (h + a, k) and (h - a, k).", "Enable text based alternatives for graph display and drawing entry Try Another Version of This Question The equation of the hyperbola that has a center at (9,3) , a focus at (14 , 3) , and a vertex at (12 , 3 ) , is \\frac((x-C)^2)(A^2)-\\frac((y-D)^2)(B^2)=1 where A = B = C = D = Get help:" ]

[ "# Extreme Polynomial Division problem Algebra Level 5 Find the remainder when $x^{1001}-1$ is divided by $x^4+x^3+2x^2+x+1$. ×", "# Remainder! Algebra Level 4 Find the remainder when the polynomial $$x^{81}+x^{49}+x^{25} + x^9+x$$ is divided by $$x^3-x^2$$. ×", "# My First Problem Let $X = \\sum_{n=1}^{1000} n! \\cdot n$ What is the remainder when $$X$$ is divided by 1001? ×", "# Equal Remainders Find the largest integer $$N$$, such that the remainder when $$23$$ is divided by $$N$$ is equal to the remainder when $$44$$ is divided by $$N$$. ×", "# Thread: Simple polynomial division question 1. ## Simple polynomial division question Hello I hope someone can help me with this. Let's say I want to divide x^2 + x + 1 into x^2 + 1 Am I right in thinking the answer is 1 remainder x? This is how I am getting my answer... I convert my polynomials into binary - x^2 + x + 1 = 111 x^2 + 1 = 101 Then I do long division - Code: 1 111 / 101 111 ---- 010 Which translates as 1 remainder x. Am I correct? If so, how would I handle x^4 + 1 into x^2 + 1? Convert to binary - 10001 / 101 Am I right in thinking the answer is - Code: 0 10001 / 101 000 ---- 101 So the answer is 0 remainder x^2 + 1? Thank you. 2. Originally Posted by MrSteve Hello I hope someone can help me with this. Let's say I want to divide x^2 + x + 1 into x^2 + 1 Am I right in thinking the answer is 1 remainder x? This is how I am getting my answer... I convert my polynomials into binary - x^2 + x + 1 = 111 x^2 + 1 = 101 Then I do long division -", "By actual division find the quotient and the remainder when (x4 + 1) is divided by (x - 1). The division is as follows. $$\\text { Quotient }=x^{3}+x^{2}+x+1 \\\\ \\text { Remainder }=2$$", "# But how? Find the remainder when the number $$(2^{100} + 3^{100} + 4^{100}+5^{100} )$$ is divided by 7. ×", "# Dividing a polynomial Algebra Level 3 When a polynomial $$P(x)$$ of degree $$n > 2$$ is divided by $$x-3$$, the remainder is 5. When $$P(x)$$ is divided by $$x-1$$, the remainder is 1. Find the remainder when $$P(x)$$ is divided by $$x^2-4x+3$$. ×", "# Crazy polynomial remainder! Level pending Let a polynomial P($$x$$) = $$x^{5}$$ +$$x^{4}$$ + $$x^{3}$$ + $$x^{2}$$ + $$x$$ + 1. If P($$x^{6}$$) is divided by P($$x$$), find its remainder. ×", "frac{f^{6} (-1) }{6!} (x+1)^2 + frac{f^{7} (-1) }{7!} (x+1)^3$$ For more context into the idea behind finding remainders, see the answer I made for the question I asked myself over here", "× Division Algorithm what is the remainder when $$x^{100}$$ is divided by $$x^2+2x+1$$... kindly post the solution with proper explanation .. Note by Vishruth Khare 2 years, 2 months ago", "# Remainder Factor Theorem Algebra Level 3 Find the sum of the values among 3, 6, 9 and 12 that are roots of the polynomial $$2x^3 - 27x^2 + 63x + 162$$. ×", "the remainder obtained by dividing $x^7 + x ^5 + 1$ by the generator polynomial $x^ 3 + 1?$", "# They are just powers of 3 Algebra Level 3 What is the remainder when the polynomial $$x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}$$ is divided by $$x^2-1$$? ×", "the particularly diligent student might also come across the solution $x = 17$. In class next time we'll examine how to solve this problem in general using a technique called the Chinese Remainder Theorem.$\\square$", "× Division Algorithm what is the remainder when $$x^{100}$$ is divided by $$x^2+2x+1$$... kindly post the solution with proper explanation .. Note by Vishruth Khare 1 year, 5 months ago", "### Home > PC3 > Chapter 10 > Lesson 10.2.1 > Problem10-113 10-113. Identify all of the roots of the polynomial $p(x) = x^3 − 3x^2 − 3x + 1$. The possible rational roots are $x = ±1$. Divide out either $(x + 1)$ or $(x −1)$ to see which factor results in no remainder. Use the Quadratic Formula to solve for the roots of the polynomial that results form the division.", "# Finding the Remainder. Find the remainder when $$10^{ 6n }$$ is divided by 7 for any value of $$n\\not=0$$? ×", "# Remainder of a remainder Algebra Level 5 When the polynomial $$(x+7)^{100}$$ is divided by the polynomial $$x^2-x-1$$, the remainder is the polynomial $$P(x)$$. Find the remainder when $$P(2)$$ is divided by 11. ×", "polynomial by the remainder ... aso ... got divisor $1$ and $x$ again and again and in the end I've got remainder $x$ and the GCD $x^2 + x + 1$. – meinzlein Dec 30 '11 at 20:56" ]

[ "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "into the numerators. Looking at prime factors, obviously there are sufficient $2$s and $3$s, so we just need $5^2 \\cdot 7 \\cdot 11 = \\boxed{\\textbf{(D) }1925}$ to divide into $x$. ~dolphin7", "method, we find that p(-1) = 0 So, (x+1) is factor of p(x) Now, p(x) = x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$ p(-1) = (1)3+13(1)2+32(1)+20=1+1332+20=0$(-1)^{3} + 13(-1)^{2} + 32(-1) + 20=-1+13-32+20=0$ Therefore, (x+1) is the factor of p(x) Now, Dividend = Divisor × Quotient + Remainder (x+1)(x2+12x+20)=(x+1)(x2+2x+10x+20)=(x5)x(x+2)+10(x+2)=(x5)(x+2)(x+10)$(x+1) (x^{2} + 12x + 20)\\\\ = (x+1) (x^{2} + 2x + 10x + 20)\\\\ = (x-5) {x(x+2) +10(x+2)}\\\\ = (x-5) (x+2) (x+10)$ (iv) 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$ Answer: Let p(y) = 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$ Factors of ab = 2×(1)$2\\times (-1)$= -2 are ±1 and ±2 By trial method, we find that p(1) = 0 So, (y-1) is factor of p(y) Now, p(y) = 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$ p(1) = 2(1)3+(1)22(1)1=2+12=0$2(1)^{3} + (1)^{2} – 2(1) – 1=2+1-2=0$ Therefore, (y-1) is the factor of p(y) Now, Dividend = Divisor × Quotient + Remainder (y1)(2y2+3y+1)=(y1)(2y2+2y+y+1)=(y1)(2y(y+1)+1(y+1))=(y1)(2y+1)(y+1)$(y-1) (2y^{2} + 3y + 1)\\\\ = (y-1) (2y^{2} + 2y + y + 1)\\\\ = (y-1) (2y(y+1) +1(y+1))\\\\ = (y-1) (2y+1) (y+1)$ Exercise – 5 Q.1.Use suitable identities to find the following products: (i)(x + 4) (x + 10) Answer: (x+4)(x+10)=x2+(4+10)x+(4×10)=x2+14x+40$(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 \\times 10)\\\\ = x^{2} + 14x + 40$ (ii)(x + 8) (x – 10)", "3. Check whether 7 + 3x is a factor of $$3x^3 + 7x$$. Let $$p(x) = 3x^3 + 7x$$ To check, whether 7 + 3x is a factor of $$3x^3 + 7x$$.By factor theorem, $$7 + 3x = 0$$ i.e. x = -7/3 So, $$p(-7/3) = 3(-7/3)^3 + 7(-7/3)$$ i.e. $$p(-7/3) = - 3×(343/27) - 49/3$$ i.e.$$p(-7/3) = -1470/27$$ i.e.$$p(-7/3) = -490/9$$ Hence, 7 + 3x is not the factor of $$3x^3 + 7x$$ since the remainder is not zero.", "number is a factor, so in this case you have $x+1$ is a factor: $$8x^2 + 17x+9 = (x+1)(\\cdots\\cdots\\cdots).$$ If all else fails, you can find the other factor by long division: In this case, divide $8x^2+17x+9$ by $x+1,$ getting $8x+9.$", "division: $x^{10}-1=(x-1)(x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$ Hence $Q(x)= (x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$ One can see that when $x=1$ , the remainder of $Q(1)=10$ . Thus, $Q(x)=(x-1)R(x)+10$ and $(R(x)$ is another quotient. For $11^{10}-1$ , $P(x)=(x-1)((x-1)R(x)+10)$ which gives $P(11)=(11-1)((11-1)R(x)+10)=10(10R(x)+10)=100(R(x)+1).$ Thus, $11^{10}-1=100(R(x)+1)$ , and hence the number is a multiple of $100$.", "= 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$ Factors of ab = 2×(1)$2\\times (-1)$= -2 are ±1 and ±2 By trial method, we find that p(1) = 0 So, (y-1) is factor of p(y) Now, p(y) = 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$ p(1) = 2(1)3+(1)22(1)1=2+12=0$2(1)^{3} + (1)^{2} – 2(1) – 1=2+1-2=0$ Therefore, (y-1) is the factor of p(y) Now, Dividend = Divisor × Quotient + Remainder (y1)(2y2+3y+1)=(y1)(2y2+2y+y+1)=(y1)(2y(y+1)+1(y+1))=(y1)(2y+1)(y+1)$(y-1) (2y^{2} + 3y + 1)\\\\ = (y-1) (2y^{2} + 2y + y + 1)\\\\ = (y-1) (2y(y+1) +1(y+1))\\\\ = (y-1) (2y+1) (y+1)$", "we dont want to include $10^6$ since we are considering proper factors only so the final answer is $144 + 3 - 6 = \\boxed{141}.$", "also infer that $x_5$ will likely need to be one of the smaller factors. The factors of $144$ are: $1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144$ Selecting numbers from this list and trying them out, we can satisfy the conditions with these numbers: $x_5 = 18, x_4 = 6, x_3 = 2, x_2 = 1, x_1 = 2$ So, $x_7 = x_6(x_5 + x_4)$ $x_7 = 144(18 + 6)$ $x_7 = 3456$ Therefore, the answer is $\\Rightarrow{\\boxed{456}}$", "factoring method. $x^4=10000$ $\\sqrt{x^4}=\\sqrt{10000}$ $|x^2|=100$, $x^2=100$ or $-100$ Since a radical with even index has the assumption of the principal/positive root, you take the absolute value of $x^2$. $x^2=100$, $\\sqrt {x^2}=\\sqrt{100}$, $|x|=10$, $x=10, -10$ $x^2=-100$, $\\sqrt {x^2}=\\sqrt{-100}$, $|x|=10i$, $x=10i, -10i$", "Factor $4x^3+4x^2-x-1$ Factor $4x^3+4x^2-x-1$ $4x^3 + 4x^2 - x - 1 = 4x^2(x + 1) - (x + 1) = (x + 1)(4x^2 - 1) = (x + 1)(2x + 1)(2x - 1)$ (the last one is found using DOTS).", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "-1 \\pmod{5}$, we have $1999^{2000} \\equiv (-1)^{2000} \\equiv 1 \\pmod{5}$. Thus, the answer is $\\boxed{\\text{(B)}\\ 1}$. ## Solution 4 A sum/product of any two natural numbers has the same remainder, when divided by $n \\in \\mathbb{N}$, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as $1999 \\cdot 1999 \\cdot \\cdot \\cdot 1999$. Using the fact stated in the first sentence, we see that the remainder of $1999$, when divided by $5$, is $4$. The problem can now be written viewed as finding the remainder of $4^{2000}$ when it is divided by $5$, which is already much simpler. Now, toying with the simplified problem a little, see notice that powers of $4$ alternate from ending in $4$ and $6$. We notice that even powers of $4$ always end in $6$, and also that $2000$ is even. Thus $4^{2000}$ must end in $6$, which, when divided by $5$ gives a remainder of $\\boxed{\\text{(B)}\\ 1}$.", "= x^{3} – 2x^{2} – x + 2$ Factors of 2 are ±1 and ± 2 By trial method, we find that p(1) = 0 So, (x+1) is factor of p(x) Now, p(x)=x32x2x+2$p(x) = x^{3} – 2x^{2} – x + 2$ p(1)=(1)32(1)2(1)+2=11+1+2=0$p(-1) = (-1)^{3} – 2(-1)^{2} – (-1) + 2=-1-1+1+2=0$ Therefore, (x+1) is the factor of p(x) Now, Dividend = Divisor × Quotient + Remainder (x+1)(x23x+2)=(x+1)(x2x2x+2)=(x+1)(x(x1)2(x1))=(x+1)(x1)(x+2)$(x+1) (x^{2} – 3x + 2)\\\\ = (x+1) (x^{2} – x – 2x + 2)\\\\ = (x+1) (x(x-1) -2(x-1))\\\\ = (x+1) (x-1) (x+2)$ (ii) x33x29x5$x^{3} – 3x^{2} – 9x – 5$ Answer: Let p(x) = x33x29x5$x^{3} – 3x^{2} – 9x – 5$ Factors of 5 are ±1 and ±5 By trial method, we find that p(5) = 0 So, (x-5) is factor of p(x) Now, p(x) = x33x29x5$x^{3} – 3x^{2} – 9x – 5$ p(5) = (5)33(5)29(5)5=12575455=0$(5)^{3} – 3(5)^{2} – 9(5) – 5=125-75-45-5=0$ Therefore, (x-5) is the factor of p(x) Now, Dividend = Divisor × Quotient + Remainder (x5)(x2+2x+1)=(x5)(x2+x+x+1)=(x5)(x(x+1)+1(x+1))=(x5)(x+1)(x+1)$(x-5) (x^{2} + 2x + 1)\\\\ = (x-5) (x^{2} + x + x + 1)\\\\ = (x-5) (x(x+1) +1(x+1))\\\\ = (x-5) (x+1) (x+1)$ (iii) x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$ Answer: Let p(x) = x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$ Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20 By trial", "that we get $2h$ on the left side) which gives us $$2h=671/x_1+x_1^2/x^1$$ $$2h=671/x_1+x_1$$ $$x_1+x_2=671/x_1+x_1$$ $$x_2=671/x_1$$ $$x_1x_2=671.$$ Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$. Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$, so $\\boxed{h=36}$.", "[x]+\\{x\\}$$. This gives a total of $$\\sum_{n=1}^{99} 2n = 9900$$ possibilities for $$x$$ with $$[x]=1,\\dots,99$$. There is only one other case to consider, $$x=100$$, so the equation has 9901 solutions in $$[1,100]$$. lime Posts: 129 Joined: Tue Dec 04, 2007 2:11 am Correct!", "to factorize the following expression: $2x^3+x^2-13x+6$ . [2010] Let $x =2$ Remainder $= 2(2)^3+(2)^2-13(2)+6 = 16+4-26+6=0$ Hence $(x-2)$ is a factor of $2x^3+x^2-13x+6$ • $x-2 ) \\overline {2x^3+x^2-13x+6} (2x^2+5x-3$ • $(-) \\ \\ \\underline {2x^3-4x^2}$ • $5x^2-13x+6$ • $(-) \\ \\ \\underline{5x^2-10x}$ • $-3x + 6$ • $(-) \\ \\ \\underline{ -3x+6}$ • $\\times$ $2x^3+x^2-13x+6 = (x-2)(2x^2+5x-3)$ $= (x-2)(2x^2+6x-x-3)$ $= (x-2)[2x(x+3)-(x+3)]$ $= (x-2)(x+3)(2x-1)$ Hence $2x^3+x^2-13x+6 = (x-2)(x+3)(2x-1)$ $\\\\$", "also a factor. Could you explain that part in more detail ? • December 15th 2009, 12:50 PM By inspection (which is a posh word for 'trial and error'), when $x = -\\tfrac12, 8x^3+8x^2-1= -1+2-1=0$. So, using the remainder theorem, $(2x+1)$ is a factor.", "we have $101$ distinct elements, $a_1 \\leq 99$. Consider the set of remainders upon division by $a_1$. Since $a_1 \\leq 99$, there are at most 99 such remainders. Let $r_2$ be the remainder upon dividing $a_2$ by $a_1$, $r_3$ the remainder upon dividing $a_3$ by $a_1, \\ldots, r_{101}$ the remainder upon dividing $a_{101}$ by $a_1$. We have $100$ remainders $r_1,\\ldots ,r_{101}$ (pigeons), and at most $99$ possible remainders (pigeonholes) upon dividing by $a_1$. Thus, by the pigeonhole principle, $r_i = r_j$ for some $2\\leq i < j \\leq 101$. Now what can you say about the number $a_j – a_i$? Hint: The boxes will have labels $1$, $3$, $5$, and so on up to $199$. Odd labels! Note that there are $100$ boxes. Box 1: Contains $1,2,4,8,16, 32,\\dots$ Box 3: Contains $3,6,12,24, 48, \\dots$ Box 5: Contains $5,10,20, 40,\\dots$. And so on. Box 99: Contains $99,198$ Boxes 101, 103, and so on are pretty boring. Box 101 only contains the number $101$, Box 103 only contains $103$, and so on." ]

[ "+0 # polynomial problem 0 73 1 What is the remainder when $$x^{9999} + x^{8888} + x^{7777 }+\\cdots+ x^{1111} + 1$$ is divided by $$x^2 - 1$$? Jul 7, 2020 #1 +8341 0 Let Q(x) be the quotient and R(x) be the remainder: $$x^{9999} + x^{8888} + x^{7777} + \\cdots + x^{1111} + 1 = Q(x) (x^2 - 1) + R(x)$$ If we let x2 = 1 on both sides, then it gives $$x^{9999} + x^{8888} + x^{7777} + \\cdots + x^{1111} + 1 = R(x)$$ when all x2's are replaced by 1. We rewrite the expression more explicitly in terms of x2. $$x^{9999} + x^{8888} + x^{7777} + \\cdots + x^{1111} + 1 = (x^2)^{4999}\\cdot x+ (x^2)^{4444} + (x^2)^{3888}\\cdot x + \\cdots + (x^2)^{555}\\cdot x + 1$$ Replacing all x2 by 1: $$R(x) = x + 1 + x + 1 + x + 1 + x + 1 + x + 1 = \\boxed{5x + 5}$$ Jul 7, 2020", "# Extreme Polynomial Division problem Algebra Level 5 Find the remainder when $x^{1001}-1$ is divided by $x^4+x^3+2x^2+x+1$. ×", "# Thread: Simple polynomial division question 1. ## Simple polynomial division question Hello I hope someone can help me with this. Let's say I want to divide x^2 + x + 1 into x^2 + 1 Am I right in thinking the answer is 1 remainder x? This is how I am getting my answer... I convert my polynomials into binary - x^2 + x + 1 = 111 x^2 + 1 = 101 Then I do long division - Code: 1 111 / 101 111 ---- 010 Which translates as 1 remainder x. Am I correct? If so, how would I handle x^4 + 1 into x^2 + 1? Convert to binary - 10001 / 101 Am I right in thinking the answer is - Code: 0 10001 / 101 000 ---- 101 So the answer is 0 remainder x^2 + 1? Thank you. 2. Originally Posted by MrSteve Hello I hope someone can help me with this. Let's say I want to divide x^2 + x + 1 into x^2 + 1 Am I right in thinking the answer is 1 remainder x? This is how I am getting my answer... I convert my polynomials into binary - x^2 + x + 1 = 111 x^2 + 1 = 101 Then I do long division -", "### Home > INT3 > Chapter 9 > Lesson 9.1.4 > Problem9-49 9-49. Consider the polynomial $x^4 - 6x^3 - 6x^2 + 6x - 7$. 1. Using only integers, list all the possible linear factors. Use the Integral Zero Theorem found in the Math Notes box in Lesson 8.3.2. There are $4$ answers! 1. Is $(x +1)$ a factor of the polynomial? Is $(x - 1)$ a factor? Show how you know without graphing. Long divide each of the factors into the polynomial. If you have no remainder, then it was a factor of the polynomial.", "+0 # Help! 0 162 1 Find the remainder when the polynomial x^1000 is divided by the polynomial (x^2+1)(x+1). Jul 31, 2019 #1 +23850 +2 Find the remainder when the polynomial $$\\large{x^{1000}}$$ is divided by the polynomial $$(x^2+1)(x+1)$$ $$\\begin{array}{|lrcll|} \\hline &\\mathbf{ x^{1000} }&=& \\mathbf{(x^2+1)(x+1)q(x) +\\underbrace{ax^2+bx+c}_{=r(x)}} \\\\ \\hline x=-1:& (-1)^{1000} &=& \\left((-1)^2+1\\right)(-1+1)q(x) + a(-1)^2+b(-1)+c \\\\ & 1 &=& 0 + a-b+c \\\\ &\\mathbf{ a-b+c} &=& \\mathbf{1} \\qquad (1) \\\\ \\hline x=i:& (i)^{1000} &=& (i^2+1)(i+1)q(x) + ai^2+bi+c \\quad | \\quad \\boxed{i^2=-1\\\\i^{1000}=\\left(i^2\\right)^{500}=\\left(-1\\right)^{500}=1} \\\\ & 1 &=& 0 -a+bi+c \\\\ &\\mathbf{-a+bi+c} &=& \\mathbf{1} \\qquad (2) \\\\ \\hline x=-i:& (-i)^{1000} &=& \\left((-i)^2+1\\right)(-i+1)q(x) + a(-i)^2+b(-i)+c \\quad | \\quad \\boxed{(-i)^2=-1\\\\ (-i)^{1000}=1} \\\\ & 1 &=& 0 -a-bi+c \\\\ &\\mathbf{-a-bi+c} &=& \\mathbf{1} \\qquad (3) \\\\ \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline &\\mathbf{ a-b+c} &=& \\mathbf{1} \\qquad (1) \\\\ &\\mathbf{-a+bi+c} &=& \\mathbf{1} \\qquad (2) \\\\ &\\mathbf{-a-bi+c} &=& \\mathbf{1} \\qquad (3) \\\\ \\hline (2)+(3): & -a+bi+c+-a-bi+c &=& 1+1 \\\\ &-2a+2c&=& 2 \\quad &| \\quad :2 \\\\ &- a+ c&=& 1 \\\\ & \\mathbf{c} &=& \\mathbf{1+a} \\qquad (4) \\\\ \\hline (1) : & a-b+c &=& 1 \\quad &| \\quad \\boxed{c = 1+a} \\\\ & a-b+1+a &=& 1 \\\\ & 2a-b &=& 0 \\\\ & \\mathbf{b} &=& \\mathbf{2a} \\qquad (5) \\\\ \\hline (2) : & -a+bi+c &=& 1 \\quad &| \\quad \\boxed{b = 2a,\\ c = 1+a}\\\\ & -a+2ai+1+a &=& 1 \\\\", "## Filters Q&A - Ask Doubts and Get Answers Q # Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: x square plus 3x plus 1, Q2 (2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: $x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$ Answers (1) Views To check whether the first polynomial is a factor of the second polynomial we have to get the remainder as zero after the division After division, the remainder is zero thus $x^2+3x+1$ is a factor of $3x^4 + 5x^3 - 7x^2 + 2x + 2$ Exams Articles Questions", "# Remainder! Algebra Level 4 Find the remainder when the polynomial $$x^{81}+x^{49}+x^{25} + x^9+x$$ is divided by $$x^3-x^2$$. ×", "# 2003 AIME II Problems/Problem 9 ## Problem Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ ## Solution When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$. So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$. Now this also follows for all roots of $Q(x)$ Now $$P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1$$ Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$, so by Newton Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$ $a_ns_2+a_{n-1}s_1+2a_{n-2}=0$ $(1)(s_2)+(-1)(1)+2(-1)=0$ $s_2-1-2=0$ $s_2=3$ So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\\boxed{6}$ ## Solution 2 Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have $$S_{-1}=\\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0$$ $$S_0=4$$ $$S_1=1$$ $$S_2=3$$ By Newton's Sums we have $$a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0$$ Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\\boxed{6}$. ~ Nafer", "polynomials can be compactly represented by 0,1 valued strings. I.e., we will view $$x^5 + x^3 + x + 1 =_{\\hbox{rep}} 101011,$$ which is perhaps more obvious if we write the polynomial down in full form. $$1 x^5 + 0 x^4 + 1 x^3 + 0 x^2 + 1 x^1 + 1 x^0.$$ We can compute a remainder by simply doing long division. In this case, it is much more succinct to just write the coefficients. Let's suppose, for example, that we wanted to compute the remainder of dividing 1101101011 by 101011. This corresponds to the following division: 1101101011 101011 111011 101011 100000 101011 101111 101011 100 I.e., the remainder is the polynomial x^2. The cool thing about working mod 2 is that subtracting and adding 1 are the same thing, so we can just add, and throw away carries, which is to say, we can use bit-wise exclusive or, which is a single machine instruction on most architectures. Moreover, we never have to guess a trial divisor -- we just line up the lead bit. Now, the remainders will always be of lesser order than the quotient polynomial, and every polynomial of lesser degree is already a remainder, so the size of the resulting field will be exactly 2^n for n-th order polynomials (remembering the 0th-order term", "# polynomial 1. ### Polynomial Division Problem Polynomial Division Problem Calculate the following and state the remainder: 2x^3+x^2-22x+20 divided by 2x-3 Thanks! 2. ### Polynomial factorization calculator and root finder Hello folks. I have written a Web calculator that can factor polynomials and it can also find exact roots using rational numbers and radical expressions. You can enter any polynomial expression in the input box. You can see it at: http://www.alpertron.com.ar/POLFACT.HTM Currently, it can... 3. ### Cubic Polynomial Factorizing How is this equation in the attachment factorized? I need steps, please. Thank you in advance. Find the quadratic equation which satisfies the condition x_1 x_2 =x_1 +x_2. 5. ### What is this question asking for? Polynomial Question: Find S as a polynomial in x where S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1 Understanding what is being asked here would be a useful first step:giggle: Do I need to expand the brackets or what? I understand a polynomial to be an expression involving sums of powers of a... 6. ### Some sort of extension to the polynomial remainder theorem? Two questions on another board I'd like to learn how to solve. 1) $p(x) \\equiv 1 \\pmod{x+1},~p(x) \\equiv -7 \\pmod{x+5}$ What is $p(x) \\pmod{(x+1)(x+5)}$ ? 2) $x^{1000} \\pmod{(x^2+1)(x+1)} = ?$ 7. ### Factor polynomial Hello, When making some exercices to improve my general understanding of basic", "∴ Quotient = x3 + 5, Remainder = 5 (v) y4+y2,y2– 2 = y2(y– 2) + 3y2 = y2 (y2 – 2) + 3 (y2 – 2) + 6 ∴ Quotient =y2 + 3 and Remainder = 6 Question 2. Find, whether or not the first polynomial is a factor of the second : (i) x + 1, 2x2 + 5x + 4 (ii) y- 2, 3y3 + 5y2 + 5y + 2 (iii) 4x2 – 5, 4.x4 + 7x2 + 15 (iv) 4-z, 3z2 – 13z + 4 (v) 2a-3,10a2 – 9a – 5 (vi) 4y+1 ,8y2-2y + 1 Solution: (i) x + 1, 2x2 + 5x + 4 2x2 + 5x + 4 = 2x (x + 1) + 3x + 4 = 2x (x + 1) + 3 (x + 1) + 1 ∵ Remainder = 1 ∴ x + 1 is not a factor of 2x2 + 5x + 4 (ii) y – 2, 3y3 + 5y2 + 5y + 2 3y3 + 5y2 + 5y + 2 = 3y2(y – 2)+11y2 + 5y + 2 = 3y2(y – 2)+11y (y – 2) + 27y + 2 = 3y2 (y – 2) + 11y (y – 2) + 27 (y – 2) + 56 ∵ Remainder = 56 ∴ y – 2 is not", "# Crazy polynomial remainder! Level pending Let a polynomial P($$x$$) = $$x^{5}$$ +$$x^{4}$$ + $$x^{3}$$ + $$x^{2}$$ + $$x$$ + 1. If P($$x^{6}$$) is divided by P($$x$$), find its remainder. ×", "6,093 views What is the remainder obtained by dividing $x^7 + x ^5 + 1$ by the generator polynomial $x^ 3 + 1?$ $x^3+1=1*x^3+0*x^2+0*x^1+1*x^0$ Generator Polynomial=1001 So we've to append three 0s at the end of the message message=$x^7+x^5+x^1=1*x^7+0*x^6+1*x^5+0*x^4+0*x^3+0*x^2+0*x^1+1*x^0$ =10100001 Remainder=$0*x^3+1*x^2+1*x^1+1*x^1$ =$x^2+x+1$ by 1 vote 1 2 3", "Let $$N$$ be the number of consecutive 0's at the right end of the decimal representation of the product $$1!2!3!4!\\cdots99!100!.$$ Find the remainder when $$N$$ is divided by 1000. (第二十四届AIME1 2006 第4题)", "be $$p(x)=a_n(x-b_1)\\dots(x-b_n).$$ Otherwise we have counterexamples: take a monic polynomial with all the roots in $$\\mathbb{Z}$$ and divide it by an integer number greater than $$1$$. • Note that, given the roots of a polynomial, then its factorization is $a_n(x-b_1)\\dots(x-b_n)$. So the result is correct also when the coefficient $a_n$ is an integer. Feb 6, 2019 at 6:58", "1. ## help factor I have this question for practice and I don't understand how to solve for the values. When the polynomial 4x^ 3+mx^2+nx+11 is divided by x+2, the remainder is -7. When the polynomial is divided by x-1, the remainder is 14. What are the values of n and m? Thanks, any help will be greatly appreciated. 2. How's your synthetic division? That's all it takes, well, plus a little algebra. 3. When dividing a polynomial $P(x)$ by $x-a$, the remainder is $P(a)$. In this case we have: $P(-2)=-7, \\ P(1)=14$ So, you have to solve the system $\\left\\{\\begin{array}{ll}4m-2n=14\\\\m+n=-1\\end{array}\\right.$", "1}{x - 3}$$ by using the polynomial remainder theorem? Solution: According to the polynomial remainder theorem, $$\\require{enclose} \\begin{array}{r} 2x + 3 \\\\[-3pt] x - 3 \\enclose{longdiv}{2x^2 - 3x - 1} \\\\[-3pt] \\underline{2x^2 - 6x}\\phantom{ - 1} \\\\[-3pt] 3x - 1 \\\\[-3pt] \\underline{3x - 9} \\\\[-3pt] 0 + 8 \\end{array}$$ Here 8 is the remainder and $$(2x + 3)$$ is the quotient. Example(3): Find the remainder of the expression $$\\frac{8^{100}}{6}$$ by using the polynomial remainder theorem? Solution: According to the polynomial remainder theorem, $$\\frac{8^{100}}{6} = \\frac{(6 + 2)^{100}}{6} = \\frac{2^{100}}{6}$$ $$= \\frac{2^{99} \\times 2}{6} = \\frac{(2^3)^{33} \\times 2}{6}$$ $$= \\frac{(6 + 2)^{3} \\times 2}{6} = \\frac{2^3 \\times 2}{6}$$ $$= \\frac{8 \\times 2}{6} = \\frac{(6 + 2) \\times 2}{6}$$ $$= \\frac{2 \\times 2}{6} = \\frac{4}{6} = 4 \\ (Remainder)$$ Example(4): Find the remainder of the expression $$\\frac{2^{100}}{7}$$ by using the polynomial remainder theorem? Solution: According to the polynomial remainder theorem, $$\\frac{2^{100}}{7} = \\frac{(2)^{99} \\times 2}{7} = \\frac{(2^3)^{33} \\times 2}{7}$$ $$= \\frac{(7 + 1)^{33} \\times 2}{7} = \\frac{1^{33} \\times 2}{7}$$ $$= \\frac{1 \\times 2}{7} = 2 \\ (Remainder)$$ Example(5): Find the remainder of the expression $$\\frac{2^{10}}{3}$$ by using the polynomial remainder theorem? Solution: According to the polynomial remainder theorem, $$\\frac{2^{10}}{3} = \\frac{2^9 \\times 2}{3} = \\frac{(2^3)^3 \\times 2}{3}$$ $$= \\frac{(3 + 5)^3 \\times 2}{3} = \\frac{5^3 \\times 2}{3}$$ $$= \\frac{(3 +", "Examples a) Divide $3x^3-5x+2$ by $x+4$. b) Find the remainder when $5x^4-2x^3-4x+2$ is divided by $x-2$. c) Divide $-x^5-5x^3-x^2+2$ by $x+2$ d) Determine whether $x-1$ is a factor of $3x^3-5x^2-x+3$.", "chinese remainder theorem for polynomials Find a polynomial $$p(x)$$ that simultaneously has both the following properties. $$(i)$$ When $$p(x)$$ is divided by $$x^{100}$$ the remainder is the constant polynomial $$1.$$ $$(ii)$$ When $$p(x)$$ is divided by $$(x-2)^3$$ the remainder is the constant polynomial $$2.$$ Though I could find the polynomial by taking derivative (works better because the remainders are simple),I wanted to understand how CRT gives the polynomial. The solution states the following: $$x^{100}$$ and $$(x − 2)^3$$ do not share a common factor, you know without any work that a polynomial with given properties must exist. The same Euclidean algorithm (but now with polynomials) gives a systematic way to find it. In the given problem we could use a different trick because the specified remainders here were rather simple (constants).But there is a conceptual way as well by implementing the Chinese remainder theorem. I am new and poor with Latex.And thank you in advance. • Corrected your latex issues.. – Stranger Forever May 28 '20 at 18:35 We are given the system $$\\begin{cases} p(x) \\equiv 1 \\mod{x^{100}} \\\\ p(x) \\equiv 2 \\mod{(x-2)^3}. \\end{cases}$$ The first line indicates that $$p(x) = 1 + g(x) x^{100}$$ for some polynomial $g(x)$. Inserting into the second line gives that $$1 + g(x) x^{100} = 2 \\mod (x-2)^3,$$ or rather that $$g(x)", "+1 vote 28 views What is the remainder obtained by dividing $x^7 + x ^5 + 1$ by the generator polynomial $x^ 3 + 1?$ edited | 28 views +1 vote $x^3+1=1*x^3+0*x^2+0*x^1+1*x^0$ Generator Polynomial=1001 So we've to append three 0s at the end of the message message=$x^7+x^5+x^1=1*x^7+0*x^6+1*x^5+0*x^4+0*x^3+0*x^2+0*x^1+1*x^0$ =10100001 Remainder=$0*x^3+1*x^2+1*x^1+1*x^1$ =$x^2+x+1$ by Active (5k points)" ]

[ "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "+0 0 145 1 +102 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "of the minimum value of $\\sqrt{9+x^{2}}+\\sqrt{(4-x)^{2}+(3-y)^{2}}+\\sqrt{4+y^{2}}$ as $x$ and $y$ range over all real numbers. ×", "# Yet another question 8 Calculus Level 4 Find the minimum value of $$\\max( |y^2- xy| )$$ given that $$x$$ is any real number and $$0\\leq y \\leq 1$$. Try more here ×", "minimum value of$$ a+\\frac{1}{b(a-b)} ? $$Let a&gt;b be positive real numbers. What is the minimum value of$$ a+\\frac{1}{b(a-b)} ? ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$.Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. ...", "If x is a real number, Then what would be the minimum value of $\\left ( x^{2}-x-1 \\right )$: [A] $\\frac{3}{4}$ [B] 0 [C] 1 [D] $\\frac{1}{4}$", "# It's what now? Geometry Level pending Find the minimum value of $$\\sin(2x) + \\cos(2x)$$ for real $$x$$. ×", "# Min value Algebra Level 3 Find the minimum value of $4x^2-6x+1$ where $$x$$ ranges across all real numbers. ×", "# Q4 Geometry Level 2 Find the minimum value of $$f(x) = \\sin^4 (x) + \\cos^4(x)$$ where $$0 \\le x \\le \\frac\\pi2$$. ×", "# Please evaluate q 22? ##### 1 Answer May 3, 2018 $\\rightarrow a = x + \\frac{1}{x} = \\frac{{x}^{2} + 1}{x}$ If $x$ is any non-zero real number then the value $a$ will always be greater than or less than 1 but the value of $\\sin \\theta \\mathmr{and} \\cos \\theta$ lies between $\\left[- 1 , 1\\right]$. So, the $\\sin \\theta \\mathmr{and} \\cos \\theta$ can never be equal to $a$ in the case mentioned in the question.", "## Max value Let $x_1,x_2,\\dots,x_{n}$ be real numbers such that $\\sum_{k = 1}^{n}\\frac {1}{x_k^2 + 1} = n-1$......Find the maximum value of $x_{1}x_{2}\\cdots x_{n}+\\sum\\limits_{1 \\leqslant i < j \\leqslant n} {x_i x_j }-\\sum_{k = 1}^{n}x_{k}^{2}$", "can't be zero on a real string. Better to think of it as negligibly small compared to the average, and calling it zero is an approximation. Yes. Ideally, it should be solved twice, using both 1.01U and 0.99U. 4. Sep 8, 2013 ### Staff: Mentor Don't forget that the maximum and minimum values that sin(L/R) can take on are +1 and -1. Chet", "# Minimum Value If $x$ and $y$ are positive real numbers, what is the minimum value of $(2x + 3y)\\left(\\frac{8}{x} + \\frac{3}{y} \\right) ?$ ×", "# find the Range Geometry Level pending What is the maximum value of $$f(x) = 4\\sin^4 x - \\sin^2 2x$$ for all real $$x$$? ×", "# Maximum how? Geometry Level 2 $\\large (7\\cos x+24\\sin x)(7\\sin x-24\\cos x)$ Find the maximum value of this expression for reals $$x.$$ ×", "# Find the minimum value! Algebra Level 3 As $x$ ranges across all real values, the minimum value of $(x-16)(x-14)(x+14)(x+16)$ can be expressed as $-A$. What is the value of $A$? ×", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "# Easy? Geometry Level 3 For real number $$A$$, find the minimum value of $$\\cos(2A) + \\cos(A)$$. ×", "is (A) (2,$$\\sqrt{2}$$,4) (B) (2, $$\\sqrt{2}$$, 0) (C) (0,0) (D) (2,2) For all real values of x, the minimum value of $$\\frac{1-x+x^{2}}{1+x+x^{2}}$$ is", "# A minimum integer value Algebra Level 2 As $$x$$ ranges over all real values, what is the smallest integer value of $\\frac{ x^2 + 22x + 333} { 11} ?$ Details and assumptions Note that we are not asking for the value of $$x$$. ×" ]

[ "like $\\left(\\frac{40}{98},\\frac{320}{98},\\frac{1080}{98}\\right)$, equality holds. Let $$a:=\\tan x,b:=\\frac12 \\tan y,c:=\\frac13\\tan z$$ Now $a+4b+9c=40$ and we need to minimize $(a^2+b^2+c^2)$ Now using Cauchy Schwarz: $$(a^2+b^2+c^2)(1+16+81)\\ge(a+4b+9c)^2$$", "- \\frac x{1-x} = \\frac{8x-9x^2}{1-x} = \\frac{4(1-x) - (3x-2)^2}{1-x} = 4 - \\frac{(3x-2)^2}{1-x}.$$ The left side is clearly $\\geqslant4$ and the right side is clearly $\\leqslant 4.$ So the inequality is satisfied, with equality only if $x = 2/3$ and $y=1$. That occurs when $\\sin\\alpha = \\sqrt{2/3}$ (so $\\tan\\alpha = \\sqrt2$) and $\\sin(2\\beta)=1$ (so $\\tan\\beta = 1$). nice solution", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "you evaluate this via the quotient rule:$$z''(x) = \\frac{(y'(x))' y(x) - y'(x)y'(x)}{y(x)^2} = \\frac{y'' y}{y^2} - \\frac{(y')^2}{y^2}.$$I don't know how you went from the fourth expression to the fifth; i.e., the fourth equality seems dubious to me. 3 Here is one line proof$$\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}dx=\\int_0^{\\pi/2}\\frac{\\tan(x) (\\tan(x))'}{\\tan^4(x)+1}\\ dx=\\int_0^{\\infty} \\frac{x}{x^4+1}\\ dx=\\left[\\frac{\\arctan(x^2)}{2}\\right]_0^{\\infty}=\\frac{\\pi}{4}$$Q.E.D. 3 For any \\epsilon > 0, choose \\delta = \\text{min}\\left(1,\\frac{7\\epsilon}{2}\\right), then: if 0 < |x| < \\delta then \\left|\\dfrac{x^2-8}{x-8} - 1\\right| = \\left|\\dfrac{x^2-x}{x-8}\\right| \\leq \\dfrac{|x^2-x|}{8-|x|} < \\dfrac{|x^2-x|}{7} < \\dfrac{|x^2| + |x|}{7} < \\dfrac{|x|+|x|}{7} = \\dfrac{2|x|}{7} < \\dfrac{2}{7}\\cdot ... 2 Let$$ F(n)=\\prod_{k=1}^\\infty\\left(\\frac{k+1}k\\right)^n\\frac{k}{k+n} $$Then, using Stirling's Formula,$$ \\begin{align} F\\left(\\frac12\\right)^2 &=\\prod_{k=1}^\\infty\\frac{k+1}k\\frac{k^2}{\\left(k+\\frac12\\right)^2}\\\\ &=\\lim_{n\\to\\infty}\\prod_{k=1}^n\\frac{k+1}k\\frac{4k^2}{(2k+1)^2}\\\\ ... 4 \\begin{align} \\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}\\mathrm dx&=\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\left(1-\\sin^2x\\right)^2}\\mathrm dx\\\\[7pt] &=\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{2\\sin^4x-2\\sin^2x+1}\\mathrm dx\\\\[7pt] &=\\frac14\\int_0^1\\frac{\\mathrm dt}{t^2-t+\\frac12}\\qquad\\color{blue}{\\implies}\\qquad t=\\sin^2x\\\\[7pt] ... 81 - \\dfrac{\\sin^2(2x)}{2} = \\dfrac{1+\\cos^2(2x)}{2}$, and$\\sin x\\cos x = \\dfrac{\\sin (2x)}{2} \\Rightarrow \\displaystyle \\int \\dfrac{\\sin x\\cos x}{\\cos^4x+\\sin^4x}dx = \\displaystyle \\int -\\dfrac{1}{2}\\dfrac{d(\\cos(2x))}{1+\\cos^2(2x)}dx = -\\dfrac{1}{2}\\arctan(\\cos (2x)) + C$1 First note $$dH(p,q)(x,y) = q\\cdot y + dV(p)(x)$$ for all$(p,q), (x,y) \\in \\Bbb R^n \\times \\Bbb R^n$. Suppose$c$is a regular value of$H$. Let$p \\in \\Bbb R^n$such that$V(p) = c$. Then$H(p,0) = c$. Since$dH(p,0)$is surjective, given$r \\in \\Bbb R$, there exists$(x,y)\\in \\Bbb R^n \\times \\Bbb R^n$such that$dH(p,0)(x,y) = r$, i.e.,$dV(p)(x) = r$. ... 1 with the hint in the coment i find that the equality holds for $$|z+iw|\\rightarrow z=iw\\\\ |z-i\\overline{w}|\\rightarrow z=-i\\overline{w}$$ then $$iw=-i\\overline{w}\\\\ w^2=-|w|^2=-1\\\\ w=\\pm i\\\\ z=iw=\\pm", "15 at 14:16 • Recall that $0\\le y\\le 2\\pi/3$ and that $\\cos(\\pi/3-y)$ is symmetric about $\\pi/3$! This means that you can never get a negative answer for the minimum in this case, since $\\cos\\pi/3=1/2>0$. – TheSimpliFire Jun 15 at 14:19 Since $$f(x)=\\sin{x}$$ is a concave function on $$\\left[0,\\frac{2\\pi}{3}\\right]$$ and $$\\left(\\frac{2\\pi}{3},0\\right)\\succ(x,y),$$ where $$x\\geq y,$$ by Karamata we obtain: $$\\sin{x}+\\sin{y}\\geq\\sin(x+y)+\\sin0=\\frac{\\sqrt3}{2}.$$ The equality occurs for $$x=\\frac{2\\pi}{3}$$ and $$y=0$$, which says that we got a minimal value. The maximal value we can get by Jensen: $$\\sin{x}+\\sin{y}\\leq2\\sin\\frac{x+y}{2}=\\sqrt3,$$ where the equality accurs for $$x=y$$. The first inequality we can prove also by the following way. $$\\sin{x}+\\sin{y}-\\sin(x+y)=\\sin{x}(1-\\cos{y})+\\sin{y}(1-\\cos{x})\\geq0.$$", "x}\\ge 64$$ where again we have put in $$|\\cos x\\sin x|\\le 1/2$$. Then $$A+B\\ge 16+64=80$$. This bound may be proven sharp by putting in $$x=\\pi/4$$, or by noting that the separate bounds on $$A$$ and $$B$$ both become sharp when $$|\\cos x|=|\\sin x|$$. In your way you proved that the minimal value is greater than $$48$$. It's true, but the equality does not occur, which says that $$48$$ is not a minimal value. The right solution can be the following, for example. Let $$\\sin^2x\\cos^2x=t.$$ Thus, by AM-GM $$t\\leq\\left(\\frac{\\sin^2x+\\cos^2x}{2}\\right)^2=\\frac{1}{4}.$$ The equality occurs for $$x=45^{\\circ},$$ which gives a value $$80$$. We'll prove that it's a minimal value. Indeed, we need to prove that $$\\frac{\\sin^4x-\\sin^2x\\cos^2x+\\cos^4x}{\\sin^6x\\cos^6x}+\\frac{1}{\\sin^6x\\cos^6x}\\geq80$$ or $$\\frac{1-3\\sin^2x\\cos^2x}{\\sin^6x\\cos^6x}+\\frac{1}{\\sin^6x\\cos^6x}\\geq80$$ or $$\\frac{2}{t^3}-\\frac{3}{t^2}\\geq80$$ or $$80t^3+3t-2\\leq0$$ or $$80t^3-20t^2+20t^2-5t+8t-2\\leq0$$ or $$(4t-1)(20t^2+5t+2)\\leq0,$$ which is obvious. Alternatively: $$\\sec^6 x +\\csc^6 x + \\sec^6 x\\csc^6 x=\\frac{\\sin^6x+\\cos^6x+1}{\\sin^6 x\\cos^6x}=\\\\ \\frac{2^6[(\\sin^2x+\\cos^2x)^3-3\\sin^2x\\cos^2x(\\sin^2x+\\cos^2x)+1]}{\\sin^62x}=\\\\ \\frac{64(2-\\frac34\\sin^22x)}{(\\sin^22x)^3}=\\frac{128-48(1-\\cos^22x)}{(\\sin^22x)^3}=\\\\ \\frac{80+48\\cos^22x}{(\\sin^22x)^3}\\ge 80.$$ equality occurs for $$\\sin2x=\\pm1$$.", "# How can this trigonometric inequality related to a limit be proved? I want to prove that $\\;\\;\\displaystyle \\left|\\frac{\\sin x-x}{x^{2}}\\right|\\leq\\frac{4(\\pi/2-1)}{\\pi^{2}}\\;\\;$ for all $x$ such that $x\\in\\left[0,\\pi/2\\right]$. If you look at the graph of the expression on the left, it is clearly (appearing to be) monotonically increasing, so the maximum value of the left hand side is the output of the expression when $x=\\pi/2$. Would like a proof that does not involve calculus since this inequality is being used to prove $\\displaystyle\\lim_{x\\rightarrow0}\\frac{\\sin x}{x}=1$. - Minor detail: you want $x$ in $(0,\\pi/2]$. – 1015 Mar 5 '13 at 17:24 But that inequality is quite stronger than $\\lim\\frac{\\sin x}x=1$ (i.e. the limit follows immediately from the iniequality and just an arbitrary bound would be enough). So, what can we use about $\\sin$? How is it defined? Purely geometrically? – Hagen von Eitzen Mar 5 '13 at 17:24 You may already know that $\\sin x < x< \\tan x$ for $0<x<\\frac\\pi 2$. Then for such $x$ we have $$0< \\frac{x-\\sin x}{x^2}< \\frac{\\tan x-\\sin x}{x^2}=\\frac{(1-\\cos x)\\tan x}{x^2}=\\frac{\\sin^2x\\tan x}{x^2(1+\\cos x)}< \\tan x.$$ This is not the bound you were asking for, but it is weaker only for big $x$, hence is more than enough to show $\\lim_{x\\to 0}\\frac{\\sin x}{x}=1$. (In fact, using $\\cos x\\to 1$, we obtain $\\sin x=x+O(x^3)$) - Actually we can expect", "equality $1 - \\cos x = 2 {\\sin}^{2} \\left(\\frac{x}{2}\\right)$ and use the previous result for $\\frac{\\sin x}{x}$ • Use the L'hospital's rule to differentiate numerator and denominator In this, an intuitive understanding (not proof) is given. to understand Consider the unit circle with angle $x$ radians. The length of line segment $q r$ = $1 - \\cos x$ length of arc $r p$ = $x$ The ratio $\\frac{1 - \\cos x}{x} = \\frac{\\textrm{\\le n > h} \\left(q r\\right)}{\\textrm{\\le n > h} \\left(r p\\right)}$ As $x \\to 0$, the figure is zoomed in to the part $q r$ and $r p$. As $x$ is getting closer to $0$, the length of $q r$ becomes $0$ faster than the length of arc $r p$ For very small values of $x$, $x$ is far greater than $1 - \\cos x$. $x > \\left(1 - \\cos x\\right) \\cong 0$ Limit of (1-cos(x))/x: ${\\lim}_{x \\to 0} \\frac{1 - \\cos x}{x} = 0$ example What is ${\\lim}_{x \\to 0} \\frac{\\tan x}{x}$? The answer is '$1$' ${\\lim}_{x \\to 0} \\frac{\\tan x}{x}$ $\\quad \\quad = {\\lim}_{x \\to 0} \\frac{\\sin x}{x} \\times \\frac{1}{\\cos} x$ $\\quad \\quad = 1 \\times 1$ some results Limit of Logarithmic Functions: ${\\lim}_{x \\to 0} \\frac{\\ln \\left(1 + x\\right)}{x} = 1$ ${\\lim}_{x \\to 0} \\frac{{a}^{x} - 1}{x} = \\ln a$ summary Limit of", "$$f(x,y,z) = \\frac{x^4}{8x^3+5y^3}+\\frac{y^4}{8y^3+5z^3}+\\frac{z^4}{8z^3+5x^3} - \\frac{1}{13}$$ And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry - why oh why can it not be proved with Group Theory - an absolute minimum of the function is expected at $(x,y,z) = (1/3,1/3,1/3)$. Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as: nivo := min + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens } The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values. Maximum and minimum values of the function are observed to be: 0.00000000000000E+0000 < f < 4.80709198767699E-0002 The little $\\color{blue}{\\mbox{blue}}$ spot in the middle is where $\\,0 \\le f(x,y,z) < 0.00002$ . • Interesting technique. Thank you very much for trying my inequality. One up vote !!! – HN_NH May 24 '16 at 6:05 Too long for a comment. The Engel form of Cauchy-Schwarz is not the right way: $$\\frac{(x^2)^2}{8x^3+5y^3}+\\frac{(y^2)^2}{8y^3+5z^3}+\\frac{(z^2)^2}{8z^3+5x^3} \\geq \\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$ So we should prove that $$\\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\\geq\\frac{x+y+z}{13}$$ which is equivalent to $$\\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\\geq x+y+z$$ but by Cauchy-Schwarz again we have $$x+y+z=\\frac{(x^2)^2}{x^3} +\\frac{(y^2)^2}{y^3}+\\frac{(z^2)^2}{z^3} \\geq \\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$ and the", "that equality occurs. In other words, if the inequality holds, there exist infinitely many global maximizers. 1) $$n=4$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} (\\cot x_1)^2 + 2\\cot x_1 \\cot x_2 - (\\cot x_2)^2 -2 = 0. \\end{align} Remark: We may solve $$x_1$$ from (1), that is, $$x_1 = \\mathrm{arccot}\\frac{\\sqrt{2}-\\cos x_2}{\\sin x_2}, \\ x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_1,\\ x_4 = \\pi - x_1 - x_2 - x_3.$$ We have $$x_1, x_2, x_3, x_4 > 0; \\ x_1 + x_2 + x_3 + x_4 = \\pi$$ and \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4}{\\sin(x_1+x_2)\\sin (x_2+x_3)\\sin (x_3+x_4)\\sin (x_4+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{4}}{\\sin\\frac{\\pi}{2}}\\Big)^4\\\\ =\\ & \\frac{(\\sin x_1)^2\\sin x_2 \\sin (2x_1 + x_2)}{(\\sin (x_1+x_2))^4} - \\frac{1}{4}\\\\ =\\ & - \\frac{((\\cot x_1)^2 + 2\\cot x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi -", "inequality, it follows easily that $\\lim_{x\\to 0}\\frac{x-\\sin x}{x\\sin x}=0$. Remark: Any proof must address the behaviour of $x-\\sin x$ near $0$. Of course we do not need the very strong $|x-\\sin x|\\le \\frac{x^3}{6}$ that we obtained above. But we must show at least that $|x-\\sin x|=o(x^2)$. - Here is a hint which uses no L'Hôpital, no derivatives, and no power series; just trigonometry. Hint: In this answer, it is shown, geometrically, that $\\sin(x)\\le x\\le\\tan(x)$ for $x\\in[0,\\pi/2)$. Thus, \\begin{align} \\left|\\,\\frac{1}{x}-\\frac{1}{\\sin(x)}\\,\\right| &=\\left|\\,\\frac{\\sin(x)-x}{x\\sin(x)}\\,\\right|\\\\ &\\le\\left|\\,\\frac{\\sin(x)-\\tan(x)}{x\\sin(x)}\\,\\right|\\\\ &=\\left|\\,\\frac{1-\\sec(x)}{x}\\,\\right|\\\\ &=\\left|\\,\\frac{1-\\sec(x)}{x}\\frac{1+\\sec(x)}{1+\\sec(x)}\\,\\right|\\\\ &=\\left|\\,\\frac{1-\\sec^2(x)}{x(1+\\sec(x))}\\,\\right|\\\\ &=\\left|\\,\\frac{-\\tan^2(x)}{x(1+\\sec(x))}\\,\\right|\\\\ &=\\left|\\,\\frac{\\tan(x)}{x}\\frac{\\tan(x)}{1+\\sec(x)}\\,\\right|\\\\ &=\\left|\\,\\frac{\\tan(x)}{x}\\tan(x/2)\\,\\right| \\end{align} - Let $\\displaystyle L=\\lim_{x \\to 0} \\frac{\\sin x - x}{x^2}$. Replacing $x$ by $2y$, we obtain: \\begin{align*} L & = \\lim_{y \\to 0} \\frac{\\sin 2y - 2y}{4y^2} = \\lim_{y \\to 0} \\frac{2 \\sin y \\cos y - 2y}{4y^2} = \\lim_{y \\to 0} \\frac{2 \\sin y \\cos y - 2 \\sin y + 2 \\sin y - 2y}{4y^2} \\\\ & = \\lim_{y \\to 0} \\left( \\frac{2 \\sin y(\\cos y-1)}{4y^2}+\\frac{1}{2} \\lim_{y \\to 0} \\frac{\\sin y - y}{y^2}\\right) \\\\ & = \\frac{1}{2} \\lim_{y \\to 0} \\sin y \\cfrac{-2\\sin^2\\cfrac{y}{2}}{y^2}+\\frac{1}{2}L \\\\ & = -\\frac{1}{4} \\lim_{y \\to 0} \\sin y \\cfrac{\\sin^2\\cfrac{y}{2}}{\\cfrac{y^2}{4}}+\\frac{1}{2}L, \\end{align*} which implies that $L=0$. Hence, \\begin{align*} \\lim_{x\\to 0} \\left(\\frac{3x+1}{x}-\\frac{1}{\\sin x}\\right) & = \\lim_{x\\to 0} \\left(3+\\frac{1}{x}-\\frac{1}{\\sin x}\\right) = \\lim_{x\\to 0} \\left(3+\\frac{\\sin x - x}{x \\sin x}\\right) \\\\ & = \\lim_{x\\to 0} \\left(3+\\frac{\\sin x - x}{x^2}", "Prove the inequality $\\tan{\\frac{\\pi\\sin{x}}{4\\sin{\\alpha}}}+\\tan{\\frac{\\pi\\cos{x}}{4\\cos{\\alpha}}} > 1$ Prove the inequality $$\\tan{\\dfrac{\\pi\\sin{x}}{4\\sin{\\alpha}}}+\\tan{\\dfrac{\\pi\\cos{x}}{4\\cos{\\alpha}}} > 1$$ for any $x, \\alpha$ with $0 \\leq x \\leq \\dfrac{\\pi}{2}$ and $\\dfrac{\\pi}{6} < \\alpha < \\dfrac{\\pi}{3}$. The best idea I had was to use the identity $\\tan(A+B) = \\dfrac{\\tan(A)+\\tan(B)}{1-\\tan(A)\\tan(B)}$. Thus we may say that $\\tan \\left({\\dfrac{\\pi \\sin{x}}{4 \\sin{\\alpha}}}+\\dfrac{\\pi\\cos{x}}{4\\cos{\\alpha}} \\right) \\left(1-\\tan{\\dfrac{\\pi\\sin{x}}{4\\sin{\\alpha}}} \\tan{\\dfrac{\\pi\\cos{x}}{4\\cos{\\alpha}}} \\right) = \\tan{\\dfrac{\\pi\\sin{x}}{4\\sin{\\alpha}}}++\\tan{\\dfrac{\\pi\\cos{x}}{4\\cos{\\alpha}}}.$ Then I just have to show its greater than $1$ on these intervals. Since $\\dfrac{\\pi}{6} < \\alpha < \\dfrac{\\pi}{3}$, we have $\\sin \\alpha > \\dfrac{1}{2}$ and $\\cos \\alpha > \\dfrac{1}{2}$. Since $0 \\le x \\le \\dfrac{\\pi}{2}$, we have $0 \\le \\sin x \\le 1$ and $0 \\le \\cos x \\le 1$. Hence, $0 \\le \\dfrac{\\pi\\sin x}{4\\sin \\alpha} < \\dfrac{\\pi}{2}$ and $0 \\le \\dfrac{\\pi\\cos x}{4\\cos \\alpha} < \\dfrac{\\pi}{2}$. Therefore, $\\tan \\dfrac{\\pi\\sin x}{4\\sin \\alpha} \\ge 0$ and $\\tan \\dfrac{\\pi\\cos x}{4\\cos \\alpha} \\ge 0$ for all $0 \\le x \\le \\dfrac{\\pi}{2}$ and $\\dfrac{\\pi}{6} < \\alpha < \\dfrac{\\pi}{3}$. Now, we beak the proof into three cases: Case I: If $x > \\alpha$, then $\\sin x > \\sin \\alpha$. Hence, $\\tan \\dfrac{\\pi\\sin x}{4\\sin \\alpha} > \\tan\\dfrac{\\pi}{4} = 1$. Therefore, $\\tan \\dfrac{\\pi\\sin x}{4\\sin \\alpha} + \\tan \\dfrac{\\pi\\cos x}{4\\cos \\alpha} > 1 + 0 = 1$. Case II: If $x < \\alpha$, then $\\cos x > \\cos \\alpha$. Hence, $\\tan \\dfrac{\\pi\\cos x}{4\\cos \\alpha} > \\tan\\dfrac{\\pi}{4} = 1$. Therefore, $\\tan", "The inequality $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ is true for all real numbers $x$ If the value of $\\alpha\\in[0,2\\pi]$ such that the inequality $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ is true for all real numbers $x$,is equal to $\\frac{p\\pi}{q}$ where $p$ and $q$ are relatively prime positive integers,find $p$ and $q$ $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ $\\sin\\frac{\\pi}{3}\\cos x+\\cos\\frac{\\pi}{3}\\sin x+\\sin\\alpha\\cos x+\\cos\\alpha\\sin x\\geq0$ $(\\sin \\alpha+\\frac{\\sqrt3}{2})\\cos x+(\\frac{1}{2}+\\cos \\alpha)\\sin x\\geq0$ Let $f(x)=(\\sin \\alpha+\\frac{\\sqrt3}{2})\\cos x+(\\frac{1}{2}+\\cos \\alpha)\\sin x\\geq0$, then the range of $f(x)$ is $[-A,A]$, where $A = \\sqrt{(\\sin \\alpha+\\frac{\\sqrt3}{2})^2+(\\frac{1}{2}+\\cos \\alpha)^2}$. Thus, to ensure that the inequality always hold, we need $\\sin \\alpha+\\frac{\\sqrt3}{2}=0$ and $\\frac{1}{2}+\\cos \\alpha=0$. Finally, $\\alpha = \\frac{4}{3}\\pi$ Let me try. You have: $$0 \\le \\sin(\\frac{\\pi}{3} + x) + \\sin(\\alpha + x) = 2\\sin (\\frac{\\pi}{6} + x + \\frac{\\alpha}{2})\\cos (\\frac{\\pi}{6} - \\frac{\\alpha}{2})$$ So, you get $\\sin (\\frac{\\pi}{6} + x + \\frac{\\alpha}{2})$, $\\cos (\\frac{\\pi}{6} - \\frac{\\alpha}{2})$ is the same sign. From this, you can get the answer. Here is an alternate solution: you need the inequality to hold for $x=\\frac{7\\pi}{6}$, so $\\sin(\\frac{3\\pi}{2})+\\sin(\\alpha+\\frac{7\\pi}{6})\\geq0,\\sin(\\alpha+\\frac{7\\pi}{6})\\geq 1$ so $\\sin(\\alpha+\\frac{7\\pi}{6})=1$, hence $\\alpha+\\frac{7\\pi}{6}=\\frac{\\pi}{2}+2k\\pi$ for some integer $k$, and it's easy enough to find $k$. • +1 good method,how did you think $x=\\frac{7\\pi}{6}$?@Wojowu – Vinod Kumar Punia Dec 24 '15 at 10:46 • @VinodKumarPunia I wanted to get $\\frac{\\pi}{3}+x=\\frac{3\\pi}{2}$, because that gives $\\sin(\\frac{\\pi}{3}+x)=-1$ and forces $\\sin(\\alpha+x)=1$. – Wojowu Dec 24 '15 at 10:47", "# Is it true that $\\sin x > \\frac x{\\sqrt {x^2+1}} , \\forall x \\in (0, \\frac {\\pi}2)$? Is it true that $$\\sin x > \\dfrac x{\\sqrt {x^2+1}} , \\forall x \\in \\left(0, \\dfrac {\\pi}2\\right)$$ (I tried differentiating , but it's not coming , please help) $\\arcsin$ is an increasing function, and you can rewrite $$x>\\arcsin\\left(\\frac{x}{\\sqrt{x^2+1}}\\right).$$ Then, deriving (and using equality at $x=0$) $$1>\\frac{\\frac{\\sqrt{x^2+1}-\\frac{x^2}{\\sqrt{x^2+1}}}{x^2+1}}{\\sqrt{1-\\left(\\frac{x}{\\sqrt{x^2+1}}\\right)^2}}=\\frac1{x^2+1}.$$ The original inequation is a mixture of trigonometric and algebraic expressions, which makes it intractable as such. The \"trick\" is to let the sine disappear, knowing that the derivative of the $\\arcsin$ is an algebraic expression. • Interestingly, this is a way to \"discover\" that $\\arcsin(x/\\sqrt{x^2+1})=\\arctan(x)$. – Yves Daoust Jul 5 '15 at 9:14 How about the substitution $x = \\tan \\theta$, then the inequality reduces to proving: $$\\sin \\tan \\theta \\ge \\sin \\theta$$ for, $\\theta \\in \\left(0,\\tan^{-1} (\\pi/2\\right))$. Since, $\\sin \\theta$ is monotone increasing in the interval $\\left(0,\\pi/2\\right)$, we are done using: $$\\tan \\theta \\ge \\theta$$ • You probably mean $\\theta\\in(0,\\arctan(\\pi/2))$. – Yves Daoust Jul 5 '15 at 9:28 • @YvesDaoust $tan^{-1}\\pi/2 \\le \\pi/2$, so the first inequality holds true in that interval too I supposed. – r9m Jul 5 '15 at 14:58 • No, $1.4<\\pi/2$, but $\\sin(\\tan(1.4))<\\sin(1.4)$. – Yves Daoust Jul 6 '15 at 6:32 • @YvesDaoust ah! right you are!! Thanks!!", "DeepSea May 30 '18 at 4:52 • DeepSea. Firstly thanks for your comment . Thought about it when posting. You are right. Reaoning is flawed. – Peter Szilas May 30 '18 at 6:18 • DeapSea.Hopefully correct now.Thanks! – Peter Szilas May 30 '18 at 8:56 • @PeterSzilas: It is good now.... – DeepSea May 30 '18 at 9:18 Cauchy Schwarz Inequality $$\\frac{\\sin^3 x}{\\cos x}+\\frac{\\cos^3 x}{\\sin x}=\\frac{\\sin^4 x}{\\cos x\\sin x}+\\frac{\\cos^4 x}{\\sin x\\cos x}\\geq \\frac{(\\sin^2 x+\\cos^2 x)^2}{2\\sin x\\cos x}$$ $$\\frac{\\sin^3 x}{\\cos x}+\\frac{\\cos^3 x}{\\sin x}\\geq \\frac{1}{\\sin 2x}\\geq 1$$ equality hold when $$\\frac{\\sin^2 x}{\\sin x\\cos x}=\\frac{\\cos^2 x}{\\sin x\\cos x}$$ that is $\\displaystyle x=\\frac{\\pi}{4}.$", "x>0 6. 5 x 2 = 10 7. For what value of x does sinx=sin−1x? 8. Find the value of sin-1$\\left( sin\\frac { 5\\pi }{ 9 } cos\\frac { \\pi }{ 9 } +cos\\frac { 5\\pi }{ 9 } sin\\frac { \\pi }{ 9 } \\right)$. 9. Show that cot−1$\\left( \\frac { 1 }{ \\sqrt { { x }^{ 2 }-1 } } \\right) ={ sec }^{ -1 }x,|x|>1$ 10. Prove that ${ tan }^{ -1 }\\left( \\cfrac { 1 }{ 7 } \\right) +{ tan }^{ -1 }\\left( \\cfrac { 1 }{ 13 } \\right) ={ tan }^{ -1 }\\left( \\cfrac { 2 }{ 9 } \\right)$ 11. Ecalute $sin\\left( { cos }^{ -1 }\\left( \\cfrac { 3 }{ 5 } \\right) \\right)$ 12. 5 x 3 = 15 13. For what value of x , the inequality$\\cfrac { \\pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\\pi$ 14. Find the value of $cos\\left( { cos }^{ -1 }\\left( \\frac { 4 }{ 5 } \\right) +{ sin }^{ -1 }\\left( \\frac { 4 }{ 5 } \\right) \\right)$ 15. Solve: ${ tan }^{ -1 }\\left( \\cfrac { x-1 }{ x-2 } \\right) +{ tan }^{ -1 }\\left( \\cfrac { x+1 }{ x+2 } \\right) =\\cfrac { \\pi }{ 4 }$ 16. Solve ${ tan }^{ -1", "it follows that $\\sum\\limits_{n = 1}^{17} {\\cos\\frac{n\\pi}{18}}=0$, ie you have $\\sum\\limits_{n = 1}^{18} {{{\\sin }^2}\\left( {{5^ \\circ }n} \\right)} = \\frac{{19}}{2} - \\frac{1}{2}\\underbrace {\\sum\\limits_{n = 1}^{17} {\\cos \\frac{n\\pi}{18}}}_0 = \\frac{19}{2}.$ Thus, the initial equality is proved.", "Compare $\\arcsin (1)$ and $\\tan (1)$ Which one is greater: $$\\arcsin (1)$$ or $$\\tan (1)$$? How to find without using graph or calculator? I tried using $$\\sin(\\tan1)\\leq1$$, but how do I eliminate the possibility of an equality without using the calculator? • What method have you tried to use? Where did you get stuck? – TheSimpliFire Aug 9 '18 at 14:32 • How to ask a good question – steven gregory Aug 9 '18 at 14:51 • @Mason Because always $\\sin\\leq1$. – Michael Rozenberg Aug 9 '18 at 15:07 • Is $arcsin(1)=\\frac{\\pi}{2}$? – HQR Aug 9 '18 at 15:07 • Very nice problem! Now, good people will come and will close this topic. – Michael Rozenberg Aug 9 '18 at 15:25 One can write: $$\\frac{1}{\\sqrt{1 + \\tan^21}} = \\cos1 = 1 - 2\\sin^2\\frac{1}{2} > \\frac{17639}{32768},$$ because $$\\sin\\frac{1}{2} < \\frac{1}{2} - \\frac{1}{2^3\\cdot3!} + \\frac{1}{2^5\\cdot5!} = \\frac{1920 - 80 + 1}{3840} < \\frac{1845}{3840} = \\frac{123}{256}.$$ On the other hand, using Archimedes's lower bound, $\\pi > 3\\tfrac{10}{71}$: $$\\frac{1}{\\sqrt{1 + \\left(\\sin^{-1}1\\right)^2}} = \\frac{1}{\\sqrt{1 + \\left(\\frac{\\pi}{2}\\right)^2}} < \\frac{1}{\\sqrt{1 + \\left(\\frac{223}{142}\\right)^2}} = \\frac{142}{\\sqrt{69893}} < \\frac{142}{\\sqrt{69696}} = \\frac{142}{264} = \\frac{71}{132}.$$ So, one can prove that $\\tan1 < \\sin^{-1}1$ by proving that: $$\\frac{17639}{32768} > \\frac{71}{132},$$ which simplifies to $33 \\times 17639 > 71 \\times 8192$, that is, $582087 > 581632$ - which is true. • Nice", "$\\sqrt{x}=-x$ to $x=x^2$, but you can’t go from $x=x^2$ to $\\sqrt{x}=-x$), which is why you get answers that are not really solutions to the original equation (namely, $x=1$ does not solve the original equation). You are likely doing something similar when solving, which means you are getting a superset of the solution set, rather than the solution set itself. Sep 22, 2018 at 22:06 Recall that • $$\\arcsin(x)$$ is defined for $$|x|\\le 1$$ • $$\\arctan(y)$$ is defined on $$\\mathbb{R}$$ then we need $$1-x^2 \\ge 0 \\implies|x|\\le 1$$ then by $$x=\\sin u$$ and $$-\\frac{\\pi}2 \\le u \\le \\frac{\\pi}2$$, such that $$|x|\\le 1$$, we have $$\\arcsin(\\sin u) = 2\\arctan\\left(\\frac{\\sin u}{1+\\sqrt{1-\\sin^2 u}}\\right)$$ $$u = 2\\arctan\\left(\\frac{\\sin u}{1+\\cos u}\\right)$$ $$\\frac u2 = \\arctan\\left(\\frac{\\sin u}{1+\\cos u}\\right)$$ $$\\tan \\left(\\frac u2\\right) = \\frac{\\sin u}{1+\\cos u}$$ which always holds, therefore the given equality holds for any $$x$$ such that $$|x|\\le 1$$. • How do you get to your claim about the left hand side? Sep 22, 2018 at 21:44 • @ArturoMagidin I've used $-1\\le \\arcsin x\\le 1$...something wrong? – user Sep 22, 2018 at 21:46 • That tells you something about when the equation can be true, rather than tell you something about what you need for “the existence of the left hand side”. The left hand side by itself is sensible whenever $|x|\\leq 1$. I understand", "24 '17 at 17:29 Let $f(x)=-n \\sin \\frac{x}{n} +m\\tan \\frac{x}{m}$, where $0<x<\\frac{\\pi}{2}$ Hence, $f'(x)=\\frac{1}{\\cos^2\\frac{x}{m}}-\\cos\\frac{x}{n}\\geq0$. Thus, $f(x)\\geq f(0)=0$ and we are done! Multiply both sides by $\\alpha$ to express the inequality as $\\dfrac{\\sin x}x \\le \\dfrac{\\tan y}{y}$ where $0 < x,y < \\dfrac \\pi 2$. We apply the well known facts $\\sin(x) \\le x$ and $\\tan(x) \\ge x$ to find that $$n\\sin(\\alpha/n) \\le \\alpha$$ $$m\\tan(\\alpha/n) \\ge \\alpha$$ Equality occurs at $\\alpha=0$" ]

[ "like $\\left(\\frac{40}{98},\\frac{320}{98},\\frac{1080}{98}\\right)$, equality holds. Let $$a:=\\tan x,b:=\\frac12 \\tan y,c:=\\frac13\\tan z$$ Now $a+4b+9c=40$ and we need to minimize $(a^2+b^2+c^2)$ Now using Cauchy Schwarz: $$(a^2+b^2+c^2)(1+16+81)\\ge(a+4b+9c)^2$$", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "## Tag Archives: Cauchy Schwarz inequality Ref: Titu Andreescu and Zumin Feng. Problem: Let x, y, z be positive real numbers such that $x+y+z=1$. Determine the minimum value of $\\frac{1}{x}+\\frac{4}{y}+\\frac{9}{z}$ Solution: An application of CauchySchwarz inequality makes this as a one step problem. Nevertheless, we present a proof which involves only the easier inequality $x^{2}+y^{2} \\geq 2xy$ for real numbers x and y by setting first $x=\\tan {b}$ and $y=2\\tan{b}$ and second $x=\\tan{a}$ and $y=\\cot{a}$. Clearly, z is a real number in the interval $[0,1]$. Hence, there is an angle a such that $z=\\sin^{2} {a}$. Then, $x+y=1-\\sin^{2}{a}=\\cos^{2}{a}$, or $\\frac{x}{\\cos^{2}{a}}+\\frac{y}{\\cos^{2}{a}}=1$. For an angle b, we have $\\cos^{2}{b}+\\sin^{2}{b}=1$. Hence, we can set $x=\\cos^{2}{a}\\cos^{2}{b}$, and $y=\\cos^{2}{a}\\sin^{2}{b}$ for some b, it suffices to find the minimum value of $P = \\sec^{2}{a} \\sec^{2}{b}+4\\sec^{2}{a}\\csc^{2}{b}+9\\csc^{2}{a}$ or $P = (\\tan^{2}{a}+1)(\\tan^{2}{b}+1)+4(\\tan^{2}{a}+1)(\\cot^{2}{b}+1)+9(\\cot^{2}{a}+1)$ Expanding the right hand side gives $P = 14 + 5\\tan^{2}{a}+9\\cot^{2}{a}+(\\tan^{2}{b}+4\\cot^{b})(1+\\tan^{2}{a})$ $\\geq 14 + 5\\tan^{2}{a}+9\\cot^{2}{a}+2\\tan{b}.2\\cot{b}(1+\\tan^{2}{a})$ $= 18 + 9(\\tan^{2}{a}+\\cot^{2}{a}) \\geq 18 + 9.2\\tan{a}\\cot{a}=36$ Equality holds when $\\tan{a}=\\cot{a}$ and $\\tan{b}=2\\cot{b}$, which implies that $\\cos^{2}{a}=\\sin^{2}{a}$ and $2\\cos^{2}{b}=\\sin^{b}$. Because $\\sin^{2}{\\theta}+\\cos^{2}{\\theta}=1$, equality holds when $\\cos^{2}{a}=\\frac{1}{2}$ and $\\cos^{2}{b}=\\frac{1}{3}$, that is, $x=\\frac{1}{6}$, $y=\\frac{1}{3}$, $z=\\frac{1}{2}$. More later. Nalin Pithwa", "The inequality $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ is true for all real numbers $x$ If the value of $\\alpha\\in[0,2\\pi]$ such that the inequality $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ is true for all real numbers $x$,is equal to $\\frac{p\\pi}{q}$ where $p$ and $q$ are relatively prime positive integers,find $p$ and $q$ $\\sin(\\frac{\\pi}{3}+x)+\\sin(\\alpha+x)\\geq0$ $\\sin\\frac{\\pi}{3}\\cos x+\\cos\\frac{\\pi}{3}\\sin x+\\sin\\alpha\\cos x+\\cos\\alpha\\sin x\\geq0$ $(\\sin \\alpha+\\frac{\\sqrt3}{2})\\cos x+(\\frac{1}{2}+\\cos \\alpha)\\sin x\\geq0$ Let $f(x)=(\\sin \\alpha+\\frac{\\sqrt3}{2})\\cos x+(\\frac{1}{2}+\\cos \\alpha)\\sin x\\geq0$, then the range of $f(x)$ is $[-A,A]$, where $A = \\sqrt{(\\sin \\alpha+\\frac{\\sqrt3}{2})^2+(\\frac{1}{2}+\\cos \\alpha)^2}$. Thus, to ensure that the inequality always hold, we need $\\sin \\alpha+\\frac{\\sqrt3}{2}=0$ and $\\frac{1}{2}+\\cos \\alpha=0$. Finally, $\\alpha = \\frac{4}{3}\\pi$ Let me try. You have: $$0 \\le \\sin(\\frac{\\pi}{3} + x) + \\sin(\\alpha + x) = 2\\sin (\\frac{\\pi}{6} + x + \\frac{\\alpha}{2})\\cos (\\frac{\\pi}{6} - \\frac{\\alpha}{2})$$ So, you get $\\sin (\\frac{\\pi}{6} + x + \\frac{\\alpha}{2})$, $\\cos (\\frac{\\pi}{6} - \\frac{\\alpha}{2})$ is the same sign. From this, you can get the answer. Here is an alternate solution: you need the inequality to hold for $x=\\frac{7\\pi}{6}$, so $\\sin(\\frac{3\\pi}{2})+\\sin(\\alpha+\\frac{7\\pi}{6})\\geq0,\\sin(\\alpha+\\frac{7\\pi}{6})\\geq 1$ so $\\sin(\\alpha+\\frac{7\\pi}{6})=1$, hence $\\alpha+\\frac{7\\pi}{6}=\\frac{\\pi}{2}+2k\\pi$ for some integer $k$, and it's easy enough to find $k$. • +1 good method,how did you think $x=\\frac{7\\pi}{6}$?@Wojowu – Vinod Kumar Punia Dec 24 '15 at 10:46 • @VinodKumarPunia I wanted to get $\\frac{\\pi}{3}+x=\\frac{3\\pi}{2}$, because that gives $\\sin(\\frac{\\pi}{3}+x)=-1$ and forces $\\sin(\\alpha+x)=1$. – Wojowu Dec 24 '15 at 10:47", "b^2) <= a cos x + b sin x <= + sqrt (a^2 + b^2). Therefore, for 3 cos x + 4 sin x, minimum seems to be -5.?? I am muddled up! EDIT: My bad!!!! Its 3^(3 cos x + 4 sin x) 10. It's ok, I figured it out ! There is equality in AM GM if $3^{3 \\cos(x)}=3^{4 \\sin(x)}$ So let m be the minimum value of $2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$, that is $\\frac 23 \\sqrt{\\frac 13}$ (substituting -5). $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}} \\ge m$ If $3^{3 \\cos(x)}=3^{4 \\sin(x)}$, then $3^{3 \\cos(x)}+3^{4 \\sin(x)}=2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}} \\ge m$ Thus m is the minimum, in particular when $3^{3 \\cos(x)}=3^{4 \\sin(x)}$ Sounds more correct ? Edit : I have to work it out deeper, because I assumed that 3cos(x)=4sin(x) 11. Originally Posted by Moo It's ok, I figured it out ! There is equality in AM GM if $3^{3 \\cos(x)}=3^{4 \\sin(x)}$ So let m be the minimum value of $2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$, that is $\\frac 23 \\sqrt{\\frac 13}$ (substituting -5). $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}} \\ge m$ If $3^{3 \\cos(x)}=3^{4 \\sin(x)}$, then $3^{3 \\cos(x)}+3^{4 \\sin(x)}=2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}} \\ge m$ Thus m is the minimum, in particular when $3^{3 \\cos(x)}=3^{4 \\sin(x)}$ Sounds more correct ? Edit : I have", "- \\frac x{1-x} = \\frac{8x-9x^2}{1-x} = \\frac{4(1-x) - (3x-2)^2}{1-x} = 4 - \\frac{(3x-2)^2}{1-x}.$$ The left side is clearly $\\geqslant4$ and the right side is clearly $\\leqslant 4.$ So the inequality is satisfied, with equality only if $x = 2/3$ and $y=1$. That occurs when $\\sin\\alpha = \\sqrt{2/3}$ (so $\\tan\\alpha = \\sqrt2$) and $\\sin(2\\beta)=1$ (so $\\tan\\beta = 1$). nice solution", "x}\\ge 64$$ where again we have put in $$|\\cos x\\sin x|\\le 1/2$$. Then $$A+B\\ge 16+64=80$$. This bound may be proven sharp by putting in $$x=\\pi/4$$, or by noting that the separate bounds on $$A$$ and $$B$$ both become sharp when $$|\\cos x|=|\\sin x|$$. In your way you proved that the minimal value is greater than $$48$$. It's true, but the equality does not occur, which says that $$48$$ is not a minimal value. The right solution can be the following, for example. Let $$\\sin^2x\\cos^2x=t.$$ Thus, by AM-GM $$t\\leq\\left(\\frac{\\sin^2x+\\cos^2x}{2}\\right)^2=\\frac{1}{4}.$$ The equality occurs for $$x=45^{\\circ},$$ which gives a value $$80$$. We'll prove that it's a minimal value. Indeed, we need to prove that $$\\frac{\\sin^4x-\\sin^2x\\cos^2x+\\cos^4x}{\\sin^6x\\cos^6x}+\\frac{1}{\\sin^6x\\cos^6x}\\geq80$$ or $$\\frac{1-3\\sin^2x\\cos^2x}{\\sin^6x\\cos^6x}+\\frac{1}{\\sin^6x\\cos^6x}\\geq80$$ or $$\\frac{2}{t^3}-\\frac{3}{t^2}\\geq80$$ or $$80t^3+3t-2\\leq0$$ or $$80t^3-20t^2+20t^2-5t+8t-2\\leq0$$ or $$(4t-1)(20t^2+5t+2)\\leq0,$$ which is obvious. Alternatively: $$\\sec^6 x +\\csc^6 x + \\sec^6 x\\csc^6 x=\\frac{\\sin^6x+\\cos^6x+1}{\\sin^6 x\\cos^6x}=\\\\ \\frac{2^6[(\\sin^2x+\\cos^2x)^3-3\\sin^2x\\cos^2x(\\sin^2x+\\cos^2x)+1]}{\\sin^62x}=\\\\ \\frac{64(2-\\frac34\\sin^22x)}{(\\sin^22x)^3}=\\frac{128-48(1-\\cos^22x)}{(\\sin^22x)^3}=\\\\ \\frac{80+48\\cos^22x}{(\\sin^22x)^3}\\ge 80.$$ equality occurs for $$\\sin2x=\\pm1$$.", "DeepSea May 30 '18 at 4:52 • DeepSea. Firstly thanks for your comment . Thought about it when posting. You are right. Reaoning is flawed. – Peter Szilas May 30 '18 at 6:18 • DeapSea.Hopefully correct now.Thanks! – Peter Szilas May 30 '18 at 8:56 • @PeterSzilas: It is good now.... – DeepSea May 30 '18 at 9:18 Cauchy Schwarz Inequality $$\\frac{\\sin^3 x}{\\cos x}+\\frac{\\cos^3 x}{\\sin x}=\\frac{\\sin^4 x}{\\cos x\\sin x}+\\frac{\\cos^4 x}{\\sin x\\cos x}\\geq \\frac{(\\sin^2 x+\\cos^2 x)^2}{2\\sin x\\cos x}$$ $$\\frac{\\sin^3 x}{\\cos x}+\\frac{\\cos^3 x}{\\sin x}\\geq \\frac{1}{\\sin 2x}\\geq 1$$ equality hold when $$\\frac{\\sin^2 x}{\\sin x\\cos x}=\\frac{\\cos^2 x}{\\sin x\\cos x}$$ that is $\\displaystyle x=\\frac{\\pi}{4}.$", "15 at 14:16 • Recall that $0\\le y\\le 2\\pi/3$ and that $\\cos(\\pi/3-y)$ is symmetric about $\\pi/3$! This means that you can never get a negative answer for the minimum in this case, since $\\cos\\pi/3=1/2>0$. – TheSimpliFire Jun 15 at 14:19 Since $$f(x)=\\sin{x}$$ is a concave function on $$\\left[0,\\frac{2\\pi}{3}\\right]$$ and $$\\left(\\frac{2\\pi}{3},0\\right)\\succ(x,y),$$ where $$x\\geq y,$$ by Karamata we obtain: $$\\sin{x}+\\sin{y}\\geq\\sin(x+y)+\\sin0=\\frac{\\sqrt3}{2}.$$ The equality occurs for $$x=\\frac{2\\pi}{3}$$ and $$y=0$$, which says that we got a minimal value. The maximal value we can get by Jensen: $$\\sin{x}+\\sin{y}\\leq2\\sin\\frac{x+y}{2}=\\sqrt3,$$ where the equality accurs for $$x=y$$. The first inequality we can prove also by the following way. $$\\sin{x}+\\sin{y}-\\sin(x+y)=\\sin{x}(1-\\cos{y})+\\sin{y}(1-\\cos{x})\\geq0.$$", "x>0 6. 5 x 2 = 10 7. For what value of x does sinx=sin−1x? 8. Find the value of sin-1$\\left( sin\\frac { 5\\pi }{ 9 } cos\\frac { \\pi }{ 9 } +cos\\frac { 5\\pi }{ 9 } sin\\frac { \\pi }{ 9 } \\right)$. 9. Show that cot−1$\\left( \\frac { 1 }{ \\sqrt { { x }^{ 2 }-1 } } \\right) ={ sec }^{ -1 }x,|x|>1$ 10. Prove that ${ tan }^{ -1 }\\left( \\cfrac { 1 }{ 7 } \\right) +{ tan }^{ -1 }\\left( \\cfrac { 1 }{ 13 } \\right) ={ tan }^{ -1 }\\left( \\cfrac { 2 }{ 9 } \\right)$ 11. Ecalute $sin\\left( { cos }^{ -1 }\\left( \\cfrac { 3 }{ 5 } \\right) \\right)$ 12. 5 x 3 = 15 13. For what value of x , the inequality$\\cfrac { \\pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\\pi$ 14. Find the value of $cos\\left( { cos }^{ -1 }\\left( \\frac { 4 }{ 5 } \\right) +{ sin }^{ -1 }\\left( \\frac { 4 }{ 5 } \\right) \\right)$ 15. Solve: ${ tan }^{ -1 }\\left( \\cfrac { x-1 }{ x-2 } \\right) +{ tan }^{ -1 }\\left( \\cfrac { x+1 }{ x+2 } \\right) =\\cfrac { \\pi }{ 4 }$ 16. Solve ${ tan }^{ -1", "$$f(x,y,z) = \\frac{x^4}{8x^3+5y^3}+\\frac{y^4}{8y^3+5z^3}+\\frac{z^4}{8z^3+5x^3} - \\frac{1}{13}$$ And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry - why oh why can it not be proved with Group Theory - an absolute minimum of the function is expected at $(x,y,z) = (1/3,1/3,1/3)$. Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as: nivo := min + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens } The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values. Maximum and minimum values of the function are observed to be: 0.00000000000000E+0000 < f < 4.80709198767699E-0002 The little $\\color{blue}{\\mbox{blue}}$ spot in the middle is where $\\,0 \\le f(x,y,z) < 0.00002$ . • Interesting technique. Thank you very much for trying my inequality. One up vote !!! – HN_NH May 24 '16 at 6:05 Too long for a comment. The Engel form of Cauchy-Schwarz is not the right way: $$\\frac{(x^2)^2}{8x^3+5y^3}+\\frac{(y^2)^2}{8y^3+5z^3}+\\frac{(z^2)^2}{8z^3+5x^3} \\geq \\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$ So we should prove that $$\\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\\geq\\frac{x+y+z}{13}$$ which is equivalent to $$\\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\\geq x+y+z$$ but by Cauchy-Schwarz again we have $$x+y+z=\\frac{(x^2)^2}{x^3} +\\frac{(y^2)^2}{y^3}+\\frac{(z^2)^2}{z^3} \\geq \\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$ and the", "# Find the minimum value of $x+2y$ given $\\frac{1}{x + 2} + \\frac{1}{y + 2} = \\frac{1}{3}.$ Let $$x$$ and $$y$$ be positive real numbers such that $$\\frac{1}{x + 2} + \\frac{1}{y + 2} = \\frac{1}{3}.$$Find the minimum value of $$x + 2y.$$ I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I should use it. Can anyone help? Thanks! • Can you write the first equation as $y=f(x)$? Sure you can. Then substitute this new form of y into the second equation. Then derive and make equal to zero. Solve for x. With this x get y from the first equation. Operate the second and you're done. – Ripi2 May 23 '20 at 23:47 • Should this have the algebra-precalculus tag? – Bladewood May 23 '20 at 23:51 ## 2 Answers Cauchy-Schwarz implies $$((x+2)+2(y+2))\\left(\\frac 1{x+2}+\\frac 1{y+2}\\right)\\geq (1+\\sqrt{2})^2$$ $$\\Rightarrow x+2y+6\\geq 3(1+\\sqrt{2})^2,$$where equality is achieved when $$x+2=3(1+\\sqrt{2}),y+2=\\frac 3{\\sqrt{2}}(1+\\sqrt{2}).$$ This shows that the minimum of $$x+2y$$ is $$3+6\\sqrt{2}.$$ Hint: $$y = \\left(\\dfrac{1}{3} - \\dfrac{1}{x+2}\\right)^{-1}-2= \\dfrac{3x+6}{x-1}-2= \\dfrac{x+8}{x-1}\\implies x+2y=x+\\dfrac{2x+16}{x-1}= \\dfrac{x^2+x+16}{x-1}=f(x)$$.From this point, you simply set $$f'(x) = 0$$ and solve for critical points and take it from there. It should be standard calculus problem. • I got $y=\\dfrac{x+8}{x-1}$ – J. W. Tanner May 24 '20 at 2:25", "# How can this trigonometric inequality related to a limit be proved? I want to prove that $\\;\\;\\displaystyle \\left|\\frac{\\sin x-x}{x^{2}}\\right|\\leq\\frac{4(\\pi/2-1)}{\\pi^{2}}\\;\\;$ for all $x$ such that $x\\in\\left[0,\\pi/2\\right]$. If you look at the graph of the expression on the left, it is clearly (appearing to be) monotonically increasing, so the maximum value of the left hand side is the output of the expression when $x=\\pi/2$. Would like a proof that does not involve calculus since this inequality is being used to prove $\\displaystyle\\lim_{x\\rightarrow0}\\frac{\\sin x}{x}=1$. - Minor detail: you want $x$ in $(0,\\pi/2]$. – 1015 Mar 5 '13 at 17:24 But that inequality is quite stronger than $\\lim\\frac{\\sin x}x=1$ (i.e. the limit follows immediately from the iniequality and just an arbitrary bound would be enough). So, what can we use about $\\sin$? How is it defined? Purely geometrically? – Hagen von Eitzen Mar 5 '13 at 17:24 You may already know that $\\sin x < x< \\tan x$ for $0<x<\\frac\\pi 2$. Then for such $x$ we have $$0< \\frac{x-\\sin x}{x^2}< \\frac{\\tan x-\\sin x}{x^2}=\\frac{(1-\\cos x)\\tan x}{x^2}=\\frac{\\sin^2x\\tan x}{x^2(1+\\cos x)}< \\tan x.$$ This is not the bound you were asking for, but it is weaker only for big $x$, hence is more than enough to show $\\lim_{x\\to 0}\\frac{\\sin x}{x}=1$. (In fact, using $\\cos x\\to 1$, we obtain $\\sin x=x+O(x^3)$) - Actually we can expect", "# If $x>4$, what is the minimum value of $\\frac{x^4}{(x-4)^2}$. If $$x>4$$, what is the minimum value of $$\\frac {x^4}{(x-4)^2}$$ ? I have tried using AM-GM Inequality here by letting $$y=x-4$$ and ended up getting $$224$$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is when $$x=8$$ which is $$256$$. Are there any better ways to solve for the minimum value other than trial and error? • I'm guessing derivatives are off the table here. – Arthur Aug 24 '19 at 8:17 • If it's easier, you may try finding the minimum of sqrt of the given expression – AgentS Aug 24 '19 at 8:19 Let $$p=x-4>0$$. By AM-GM inequality, $$\\frac{x^4}{(x-4)^2} =\\frac{(p+4)^4}{p^2} \\ge\\frac{\\left(2\\sqrt{4p\\,}\\right)^4}{p^2} =256$$ and equality holds when $$p=4$$ or $$x=8$$. Let $$y = \\dfrac{x^2}{x-4} \\Rightarrow x^2-xy+4y = 0$$ $$x$$ as a function of $$y$$ is defined when discriminant $$y^2 - 16y \\ge 0$$. Using AM-GM: $$\\frac {x^4}{(x-4)^2}=\\left(\\frac{x^2-16+16}{x-4}\\right)^2=\\left(\\color{red}{x-4+\\frac{16}{x-4}}+8\\right)^2\\ge (\\color{red}8+8)^2=256,$$ equality occurs for $$x=8$$.", "you evaluate this via the quotient rule:$$z''(x) = \\frac{(y'(x))' y(x) - y'(x)y'(x)}{y(x)^2} = \\frac{y'' y}{y^2} - \\frac{(y')^2}{y^2}.$$I don't know how you went from the fourth expression to the fifth; i.e., the fourth equality seems dubious to me. 3 Here is one line proof$$\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}dx=\\int_0^{\\pi/2}\\frac{\\tan(x) (\\tan(x))'}{\\tan^4(x)+1}\\ dx=\\int_0^{\\infty} \\frac{x}{x^4+1}\\ dx=\\left[\\frac{\\arctan(x^2)}{2}\\right]_0^{\\infty}=\\frac{\\pi}{4}$$Q.E.D. 3 For any \\epsilon > 0, choose \\delta = \\text{min}\\left(1,\\frac{7\\epsilon}{2}\\right), then: if 0 < |x| < \\delta then \\left|\\dfrac{x^2-8}{x-8} - 1\\right| = \\left|\\dfrac{x^2-x}{x-8}\\right| \\leq \\dfrac{|x^2-x|}{8-|x|} < \\dfrac{|x^2-x|}{7} < \\dfrac{|x^2| + |x|}{7} < \\dfrac{|x|+|x|}{7} = \\dfrac{2|x|}{7} < \\dfrac{2}{7}\\cdot ... 2 Let$$ F(n)=\\prod_{k=1}^\\infty\\left(\\frac{k+1}k\\right)^n\\frac{k}{k+n} $$Then, using Stirling's Formula,$$ \\begin{align} F\\left(\\frac12\\right)^2 &=\\prod_{k=1}^\\infty\\frac{k+1}k\\frac{k^2}{\\left(k+\\frac12\\right)^2}\\\\ &=\\lim_{n\\to\\infty}\\prod_{k=1}^n\\frac{k+1}k\\frac{4k^2}{(2k+1)^2}\\\\ ... 4 \\begin{align} \\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}\\mathrm dx&=\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{\\sin^4x+\\left(1-\\sin^2x\\right)^2}\\mathrm dx\\\\[7pt] &=\\int_0^{\\pi/2}\\frac{\\sin x\\cos x}{2\\sin^4x-2\\sin^2x+1}\\mathrm dx\\\\[7pt] &=\\frac14\\int_0^1\\frac{\\mathrm dt}{t^2-t+\\frac12}\\qquad\\color{blue}{\\implies}\\qquad t=\\sin^2x\\\\[7pt] ... 81 - \\dfrac{\\sin^2(2x)}{2} = \\dfrac{1+\\cos^2(2x)}{2}$, and$\\sin x\\cos x = \\dfrac{\\sin (2x)}{2} \\Rightarrow \\displaystyle \\int \\dfrac{\\sin x\\cos x}{\\cos^4x+\\sin^4x}dx = \\displaystyle \\int -\\dfrac{1}{2}\\dfrac{d(\\cos(2x))}{1+\\cos^2(2x)}dx = -\\dfrac{1}{2}\\arctan(\\cos (2x)) + C$1 First note $$dH(p,q)(x,y) = q\\cdot y + dV(p)(x)$$ for all$(p,q), (x,y) \\in \\Bbb R^n \\times \\Bbb R^n$. Suppose$c$is a regular value of$H$. Let$p \\in \\Bbb R^n$such that$V(p) = c$. Then$H(p,0) = c$. Since$dH(p,0)$is surjective, given$r \\in \\Bbb R$, there exists$(x,y)\\in \\Bbb R^n \\times \\Bbb R^n$such that$dH(p,0)(x,y) = r$, i.e.,$dV(p)(x) = r$. ... 1 with the hint in the coment i find that the equality holds for $$|z+iw|\\rightarrow z=iw\\\\ |z-i\\overline{w}|\\rightarrow z=-i\\overline{w}$$ then $$iw=-i\\overline{w}\\\\ w^2=-|w|^2=-1\\\\ w=\\pm i\\\\ z=iw=\\pm", "# Find minimum value of $\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$ Find minimum value of $$\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$$ I know this question has already answered here Then minimum value of $\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$ but my answer is coming different. i applied AM-GM directly to two fractions and by changing terms of sec and tan into sin and cos and simplifying a little we get that $$\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha} \\geq\\frac2{\\cos\\alpha \\cos\\beta \\sin\\alpha \\sin\\beta}\\geq2$$ but minimum value is coming 8 ??? • Your bound $2/(\\cos\\alpha\\sin\\alpha\\cos\\beta\\sin\\beta)\\geq 2$ is too weak. You can tighten that to $\\geq 8$ by using double angle formula for sine. Jul 23 '20 at 3:29 • Right, so the expression is always $\\ge2$. But $2$ isn't the minimum value, since there are no $\\alpha$ and $\\beta$ making the expression $2$ (since actually it's always $\\ge8$). Jul 23 '20 at 3:29 • @AnginaSeng but how we know that there are no $α$ and $β$ making the expression $2$ , and how we get to know that 8 is correct minimum in the same sense ? Jul 23 '20 at 3:31 • Jul 23 '20 at 3:31 • @AnginaSeng thanks,but i have already seen it and also i have mention same link in my question,but i am asking how we know that 2 is", "\\sin x > x^2 + \\frac{x^4}{6} + \\frac{31x^6}{360}$$ • If $f'(x)\\ge 0$, then $f(x)$ is non-decreasing, not strictly increasing. – user236182 Aug 19 '17 at 12:11 • But it depends on how you define \"increasing\". – user236182 Aug 19 '17 at 12:22 • Why we need to check $f'''$ and $f''$? Don't we only need to check $f'$? – Eric Oct 21 '17 at 10:47 I believe the simplest proof is through the Cauchy-Schwarz inequality: $$\\tan(x)\\sin(x)=\\int_{0}^{x}\\frac{d\\theta}{\\cos^2\\theta}\\int_{0}^{x}\\cos(\\theta)\\,d\\theta\\geq\\left(\\int_{0}^{x}\\frac{d\\theta}{\\sqrt{\\cos\\theta}}\\right)^2\\geq\\left(\\int_{0}^{x}d\\theta\\right)^2=x^2.$$ In a similar fashion, for any $x\\in\\left(0,\\frac{\\pi}{2}\\right)$ we have $\\frac{\\tan x }{x}\\geq\\left(\\frac{x}{\\sin x}\\right)^2$ by Holder's inequality. • I'm not aware about Cauchy-Schwarz inequality in integral form but it seems really clear in this way. I've just tried a proof using bounding value from Taylor's series. – gimusi Jan 9 '18 at 1:02 We need to prove that $$\\frac{\\sin^2x}{\\cos{x}}>x^2$$ or $f(x)>0$, where $$f(x)=\\frac{\\sin{x}}{\\sqrt{\\cos{x}}}-x.$$ Now, let $\\cos{x}=t^2$, where $0<t<1$. Thus, $$f'(x)=\\frac{1+\\cos^2x}{2\\sqrt{\\cos^3x}}-1=\\frac{(1-t)(1+t+t^2-t^3)}{2t^3}>0,$$ which says $f(x)>f(0)=0$ and we are done! • Very ingenious to take square root! +1 – pisco Aug 19 '17 at 6:26 Since $\\frac{x}{2} < \\tan \\frac{x}{2}$ for $0<x<\\frac{\\pi}{2}$ we have for such $x$-values: $$\\left( \\frac{\\sin x}{x}\\right)^2 > \\left( \\frac{2 \\sin \\frac{x}{2} \\cos \\frac{x}{2}} {2 \\tan \\frac{x}{2}}\\right)^2 = \\left(\\cos \\frac{x}{2} \\right)^4 = \\frac14 \\left(1+\\cos x\\right)^2 > \\cos x$$ using that the difference between the two last expressions is $\\frac14 (1-\\cos x)^2>0$. The result", "# What is the minimum value of? On positive reals if $3x+4y+7z=1$ what is the minimum value of $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}?$ I have tried using the arithmetic mean and harmonic mean inequality but I failed. Not good at inequalities though. Please help. • Sorry for the bad language. Looking for the minimum value. – reco Oct 3, 2017 at 17:49 By Cauchy-Schwarz $$(3x+4y+7z)(x^{-1}+y^{-1}+z^{-1})\\ge(\\sqrt3+\\sqrt4+\\sqrt7)^2.$$ One can get equality, when the vector $(x,y,z)=t(1/\\sqrt3,1/\\sqrt4,1/\\sqrt7)$ for some $t$. The $t$ in question is $1/(\\sqrt3+\\sqrt4+\\sqrt7)$ and the minimum is $(\\sqrt3+\\sqrt4+\\sqrt7)^2$. • 12s ahead! +1.. Oct 3, 2017 at 18:02 From $3x+4y+7z=1$ I got $z= \\frac{1}{7} (-3 x-4 y+1)$ and plugged in $$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$$ gives $$f(x,y)=\\frac{7}{1-3 x-4 y}+\\frac{1}{x}+\\frac{1}{y}$$ $$f'_x=\\frac{21}{(1-3 x-4 y)^2}-\\frac{1}{x^2};\\;f'_y=\\frac{28}{(1-3 x-4 y)^2}-\\frac{1}{y^2}$$ $f'_x=0$ gives $21x^2=(1-3x-4y)^2$ and $f_y=0\\to 28y^2=(1-3x-4y)^2$ so we have $21x^2=28y^2$ which for positive reals means $y=\\frac{\\sqrt 3}{2}\\,x$ furthermore we have from the first equation $x\\sqrt{21}=1-3x-4y$ which after substituting becomes $x\\sqrt{21}=1-3x-2\\sqrt{ 3}x$ $$x=\\frac{1}{3+2 \\sqrt{3}+\\sqrt{21}};\\;y=\\frac{\\sqrt{3}}{2 \\left(3+2 \\sqrt{3}+\\sqrt{21}\\right)};\\;z=\\frac{1}{7+2 \\sqrt{7}+\\sqrt{21}}$$ So minimum of $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$ is $2 \\left(7+2 \\sqrt{3}+2 \\sqrt{7}+\\sqrt{21}\\right)\\approx 40.676$", "that equality occurs. In other words, if the inequality holds, there exist infinitely many global maximizers. 1) $$n=4$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} (\\cot x_1)^2 + 2\\cot x_1 \\cot x_2 - (\\cot x_2)^2 -2 = 0. \\end{align} Remark: We may solve $$x_1$$ from (1), that is, $$x_1 = \\mathrm{arccot}\\frac{\\sqrt{2}-\\cos x_2}{\\sin x_2}, \\ x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_1,\\ x_4 = \\pi - x_1 - x_2 - x_3.$$ We have $$x_1, x_2, x_3, x_4 > 0; \\ x_1 + x_2 + x_3 + x_4 = \\pi$$ and \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4}{\\sin(x_1+x_2)\\sin (x_2+x_3)\\sin (x_3+x_4)\\sin (x_4+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{4}}{\\sin\\frac{\\pi}{2}}\\Big)^4\\\\ =\\ & \\frac{(\\sin x_1)^2\\sin x_2 \\sin (2x_1 + x_2)}{(\\sin (x_1+x_2))^4} - \\frac{1}{4}\\\\ =\\ & - \\frac{((\\cot x_1)^2 + 2\\cot x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi -", "x_4 x_1 \\leq \\frac{4}{21} x_1 x_2 x_3 x_4}$$ with equality at $$(7,3,1,7) \\; \\; \\mbox{and} \\; \\; (7,7,1,3).$$ I think, for $n=4,$ that allowing the variables to take real values $\\geq 1$ with $x_1$ maximal but equality allowed, the optimum occurs at $(T^2 - T + 1, T^2 - T + 1, 1, T),$ with $$T = 1 + \\sqrt[3]{1 + \\sqrt{\\frac{19}{27}}} + \\sqrt[3]{1 - \\sqrt{\\frac{19}{27}}} \\approx 2.769292354$$ and $T^2 - T + 1 \\approx 5.899687788.$ The apparent best ratio for integers, $4/21 \\approx 0.190476,$ is improved to about $0.191602$ for real numbers. It appears, for $n=5,$ that $$\\color{blue}{ x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_1 x_2 - x_2 x_3- x_3 x_4 - x_4 x_5 - x_5 x_1 \\leq \\frac{1}{3} x_1 x_2 x_3 x_4 x_5}$$ with equality at $$(4,1,4,1,3) \\; \\; \\mbox{and} \\; \\; (4,3,1,4,1).$$ I think, for $n=5,$ that allowing the variables to take real values $\\geq 1$ with $x_1$ maximal but equality allowed, the optimum occurs at $((U^2 - U + 2)/2, \\; 1, \\; (U^2 - U + 2)/2, \\; 1, \\; U),$ with $$U = 1 + \\sqrt[3]{2 + \\sqrt 3} + \\sqrt[3]{2 - \\sqrt3} \\approx 3.1958233$$ and $(U^2 - U + 2)/2 \\approx 4.50873.$ The apparent best ratio for integers, $1/3 \\approx 0.33333,$ is improved to about $0.33462442$ for" ]

[ "# A perfect welcome to 2016 Algebra Level 5 $f(x,y,z,v,w)=xz+2yz+3zv+7zw$ Let $$M$$ be the maximal value of $$f$$ if $$x^2+y^2+z^2+v^2+w^2=2016$$ and all variables are positive. If $$M=f(x_0,y_0,z_0,v_0,w_0)$$, find $$w_0$$. ×", "## $v w x y z = v + w + x + y + z$ Find the maximum possible value of $x$ where $v , w , x , y , z > 0$ Apr 4, 2018 Maximum for natural numbers is $x = 5$ #### Explanation: Given: $v w x y z = v + w + x + y + z$ For solutions in natural numbers, we can look at possible combinations of small values. Note that: • If $v = w = x = y = z = 1$ then $v w x y z = 1 < 5 = v + w + x + y + z$ • If $v = w = x = y = z = 2$ then $v w x y z = 32 > 10 = v + w + x + y + z$ The minimum possible values of $v , w , y , z$ are $v = w = y = z = 1$, resulting in: $x = x + 4 \\text{ }$ which has no solutions. Considering the case where exactly one of $v , w , y , z > 1$, and putting $v = 2$, $w = y = z = 1$ we get: $2 x = x + 5 \\text{", "• zmudz Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of $$9x+12y+8z.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# 2016 is coming! Algebra Level 4 $\\sqrt{x + 672} + \\sqrt{y + 672} + \\sqrt{z + 672}$ Let $$x,y$$ and $$z$$ be positive reals whose sum is 2016. If the maximum value of the expression above is $$M$$, what is $$M^2 ?$$ ×", "In The Middle Of 2016! $2^x+3^x+4^x+5^x+2016+5^{-x}+4^{-x}+3^{-x}+2^{-x}$ Let $x$ be a real number. Find the minimum value of the expression above. ×", "# Absolute value of real numbers • MHB Gold Member MHB POTW Director Reals $x,\\,y$ and $z$ satisfies $3x+2y+z=1$. For relatively prime positive integers $p$ and $q$, let the maximum of $\\dfrac{1}{1+|x|}+\\dfrac{1}{1+|y|}+\\dfrac{1}{1+|z|}$ be $\\dfrac{q}{p}$. Find $p+q$. maxkor", "+0 # intermediate algebra 0 75 1 Let x, y and z be positive real numbers such that x + y + z = 1. Find the maximum value of xyz. Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ MaxWong Jan 28, 2021", "Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that $$x^2 + y^2 + z^2 = 2(xy + yz+zx)$$. Find the maximum value of $\\frac{x^{2016}+y^{2016}+z^{2016}}{(x+y+z)\\big(x^2+y^2+z^2\\big) \\ldots \\big(x^{62}+y^{62}+z^{62}\\big)\\big(x^{63}+y^{63}+z^{63}\\big)}.$ Bonus: Generalize this. That is, find the conditions for $$x,y,z$$ such that $\\frac{x^{N}+y^{N}+z^{N}}{\\displaystyle \\prod_j (x^{a_j}+y^{a_j}+z^{a_j})}$ attains its maximum given that $$\\displaystyle \\sum_{j}a_j = N$$. ×", "Practice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.", "# Hello 2016 Geometry Level 3 For real numbers $$a,b,c,d$$ and $$e$$, we are given that $$\\sin^{-1}a + \\sin^{-1}b + \\sin^{-1}c + \\sin^{-1}d + \\sin^{-1}e = \\dfrac{5\\pi}2$$. For real $$x$$, let $$f$$ denote the maximum value of $$3\\sin x + 4\\cos x$$ and $$g$$ denote the minimum positive value of $$\\tan x + \\dfrac1{\\tan x}$$. Compute $$\\dfrac{a^{2016} + b^{2016} + c^{2016} + d^{2016} + e^{2016} + 5g}{abcdef}$$. ×", "# Maximize This Algebra Level 4 Positive reals $x$, $y$ and $z$ are such that $x+y+z=xyz$. Find the maximum value of $\\large (x-1)(y-1)(z-1)$ Given the answer in 4 decimal places. ×", "+0 # algebra 0 55 1 +73 Let $$x$$ and $$y$$ be real numbers such that $$0 \\le x \\le 1$$ and $$0 \\le y \\le 1.$$ Find the maximum value of $$x^2 y - xy^2$$. any tips / hints / solutions would be greatly appreciated! :) Feb 4, 2021 #1 +73 +1 the answer was $$1 \\over 4$$! to anyone who has also been trying to tackle this problem: you could factor the expression like so: $$xy(x-y)$$ then, since you are trying to find the maximum value, you can assume that $$x \\ge y$$. by AM-GM, $$y(x - y) \\le \\frac{x^2}{4}$$, and then continue from there:) Feb 4, 2021", "• zmudz For x, y, and z positive real numbers, what is the maximum possible value for $$\\sqrt{\\frac{3x+4y}{6x+5y+4z}} + \\sqrt{\\frac{y+2z}{6x+5y+4z}} + \\sqrt{\\frac{2z+3x}{6x+5y+4z}}?$$ Also, find what z/x is if (x,y,z) achieves the maximum value. Thank you! I am so stuck right now. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Find the maximum • MHB solakis1 Find the maximum of xyz when $y^2=xz$ and $x+y+z=a$ x,y,z are reals", "+0 # Inequalities 0 98 1 +179 Let x and y be real numbers such that $$x - \\sqrt{x + 6} = \\sqrt{y + 6} - y$$. Let m be the minimum value of x+y, and M be the maximum value of x+y. Enter the ordered pair (m, M). Thanks for any tips, hints, or answers you provide, anything is appreciated. Jun 7, 2021 #1 +2205 +2 I tried, but couldn't figure out the answer. :(( x - sqrt(x+6) = sqrt(y+6) - y x + y = sqrt(y+6) + sqrt(x+6) I'm not sure if this is true, but I feel like the maximum value of x+y would be when x = y. So x = y = 3, making x + y = 6. =^._.^= Jun 7, 2021", "+0 # Help pls :) +1 323 1 +504 Let x, y, and z be positive real numbers. Find the minimum value of $$(x + 2y + 4z) \\left( \\frac{4}{x} + \\frac{2}{y} + \\frac{1}{z} \\right)$$. Btw. I have to use the AM-GM inequality thingy. Help :) May 26, 2021", "# Playing with the Glome Algebra Level 5 $2xy+4yz+6zw$ Let $$M$$ be the maximum of the expression above if $$x^2+y^2+z^2+w^2=1$$, where $$x,y,z$$ and $$w$$ are real numbers. Write $$M=\\sqrt{p}+\\sqrt{q}$$ for two positive integers $$p$$ and $$q$$, and enter $$p+q$$. Bonus question: For $$axy+byz+czw$$, with the same constraint, express the maximum $$M$$ in terms of $$a,b$$ and $$c$$. ×", "we let V n = min { Y 1 , Y 2 , , Y n } and W n = max { Y 1 , Y 2 , , Y n } V n = min { Y 1 , Y 2 , , Y n } and W n = max { Y 1 , Y 2 , , Y n } (13) Then P ( V n > t ) = P n ( Y > t ) and P ( W n t ) = P n ( Y t ) P ( V n > t ) = P n ( Y > t ) and P ( W n t ) = P n ( Y t ) (14) Now consider a random number N of the Yi. The minimum and maximum random variables are V N = n = 0 I { N = n } V n and W N = n = 0 I { N = n } W n V N = n = 0 I { N = n } V n and W N = n = 0 I { N = n } W n (15) Computational formulas If we set V0=W0=0V0=W0=0, then 1. FV(t)=P(Vt)=1+P(N=0)-gN[P(Y>t)]FV(t)=P(Vt)=1+P(N=0)-gN[P(Y>t)] 2. FW(t)=gN[P(Yt)]FW(t)=gN[P(Yt)] These results are easily established as follows.", "# Use only Holder's If $a$ and $b$ are positive real numbers such that $a^{2015}+b^{2015}=2015,$ find the maximum value of $a+b$. ×", "+0 # Inequality 0 48 2 If w, x, y, and z are positive real numbers such that w + 2x + 3y + 4z = 8, then what is the maximum value of wxyz? Aug 14, 2021 #1 0 The minimum value is 9/25. Aug 14, 2021 #2 +436 0 By AM-GM, $$\\frac{w+2x+3y+4z}{4} \\ge \\sqrt[4]{24wxyz}\\\\ 2\\ge\\sqrt[4]{24wxyz}\\\\ 16\\ge24wxyz\\\\ \\frac{2}{3}\\ge wxyz$$ so the maximum value is $$\\boxed{\\frac{2}{3}}$$ for completeness, equality occurs when the terms are equal: $$w=2, x=1, y=\\frac{2}{3}, z=\\frac{1}{2}$$ Aug 21, 2021" ]

[ "listing some terms of the sequence. \\begin{align*} f(1)&=1 \\\\ f(2)&=1 \\\\ f(3)&=3 \\\\ f(4)&=6 \\\\ f(5)&=8 \\\\ f(6)&=8 \\\\ f(7)&=7 \\\\ f(8)&=7 \\\\ f(9)&=9 \\\\ f(10)&=12 \\\\ f(11)&=14 \\\\ f(12)&=14 \\\\ f(13)&=13 \\\\ f(14)&=13 \\\\ f(15)&=15 \\\\ & \\ \\vdots \\end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$, and the pattern of numbers that follow will always be $+2$, $+3$, $+2$, $+0$, $-1$, $+0$. The largest odd multiple of $3$ smaller than $2018$ is $2013$, so we have \\begin{align*} f(2013)&=2013 \\\\ f(2014)&=2016 \\\\ f(2015)&=2018 \\\\ f(2016)&=2018 \\\\ f(2017)&=2017 \\\\ f(2018)&=\\boxed{\\textbf{(B) } 2017}. \\end{align*} minor edits by bunny1 ## Solution 5 (Bash) Writing out the first few values, we get $$1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\\ldots.$$ We see that every number $x$ where $x \\equiv 1\\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\\pmod{6}$ and less $2018$ is $2017,$ so we have $f(2017)=f(2018)=\\boxed{\\textbf{(B) } 2017}.$ ~bunny1 ~savannahsolver ## See Also 2018 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems", "\\require{cancel} \\begin{align} 0 & = y^3-3xyz + 3 z^2 - 31 \\\\ & = \\frac{(x^3+3z-10)^3}{27 x^3} - \\cancel{3x} \\frac{x^3+\\bcancel{3z}-10}{\\cancel{3x}} z + \\bcancel{3 z^2} - 31 \\\\ \\iff 0 & = (x^3+3z-10)^3- 27 x^3(x^3-10)-27\\cdot 31 x^3 \\\\ & = (x^3-10)^3+3(x^3-10)^2z + 27(x^3-10)z^2+27z^3- 27 x^3(x^3-10)-27\\cdot 31 x^3 \\\\ & = 27(x^3-10)\\,z^2-9(2x^6-10x^3-100)\\,z+(x^3-10)^3-27\\,(31 x^3-30) \\tag{8} \\end{align} Define the polynomials in $x\\,$: $$\\begin{cases} u = 27(x^3-10) \\\\ v = 9 (2 x^6 - 10 x^3 - 100) = 18(x^3-10)(x^3+5) \\\\ w = (x^3-10)^3-27\\,(31 x^3-30) = x^9 - 30 x^6 - 537 x^3 - 190 \\tag{9} \\end{cases}$$ Then $(8)$ becomes: $$u z^2 = v z - w \\tag{10}$$ Raising to the $3^{rd}$ power and again using $z^3=30\\,$: \\begin{align} 900 u^3 = u^3 z^6 & = (vz-w)^3 \\\\ & = v^3z^3 - w^3 -3vz w \\cdot (vz-w) \\\\ & = 30 v^3 - w^3 -3vwz \\cdot u z^2 \\\\ & = 30 v^3 -w^3 - 90 uvw \\tag{11} \\\\ \\iff \\quad w^3 + 90 uvw - 30 v^3 + 900 u^3 & = 0 \\tag{12} \\end{align} Given that $w$ is monic and $\\deg u = 3, \\deg v = 6, \\deg w = 9$ as polynomials in $x\\,$, $(12)$ defines a monic polynomial of degree $27$ with integer coefficients having $x = \\sqrt[3]{2} + \\sqrt[3]{3} + \\sqrt[3]{5}$ as a root. For verification, one can substitute", "at the origin the sum total is -108 6. May 29, 2017 ### Nick Jarvis Thanks v much. I cannot find this in my notes at all. Assuming the -108V is due to 3sf? Rather than my 108.5V? So 150V at the origin. I then resolve in the x and y direction. Cannot work out though why it is 150V - 150V -108V, when we're looking at the +x/y directions? Like I said, this should be such an easy question, but I am really struggling with it. Many thanks 7. May 29, 2017 ### Staff: Mentor V=150-21.7x-12.5y 8. May 29, 2017 ### BvU There must be something relating $V$ and $\\vec E$, though ? Do you recognize/understand my $\\Delta V = -\\int \\vec E \\cdot d\\vec r \\ \\ \\$ and Chet's $\\ \\ dV=-E_x\\,dx-E_y\\,dy \\ \\$ ? Once you do it's easy to find \\begin{align} V(5,12) & = V(0,0) + \\Delta V \\\\ & = V(0,0) -\\int \\vec E \\cdot d\\vec r \\\\ & =V(0,0) - \\int_0^5 \\int_0^{12} \\left ( E_x \\, dx + E_y\\,dy \\right ) \\\\ & = V(0,0) -\\int_0^5 E_x \\, dx - \\int_0^{12} E_y\\,dy \\end{align} 9. May 29, 2017 ### Nick Jarvis Thank-you both. For some reason I was making it way more complicated than it needed to be. Going to work through", "Nov 28, 2020 11:32 pm Is \\((x^7)(y^2)(z^3)>0?$$ by M7MBA » Is $$(x^7)(y^2)(z^3)>0?$$ (1) $$yz<0$$ (2) $$xz>0$$ Answer:... 0 Last post by M7MBA Sat Nov 28, 2020 11:30 pm If $$xy < 0,$$ is $$\\dfrac{x}{y} > z ?$$ by M7MBA » If $$xy < 0,$$ is $$\\dfrac{x}{y} > z ?$$ (1) $$xyz < 0$$ (2)... 0 Last post by M7MBA Sat Nov 28, 2020 11:25 pm Is $$yz = x?$$ by M7MBA » Is $$yz = x?$$ (1) $$y > \\dfrac{x}{z}$$ (2) $$z < 0$$ Answer: A... 0 Last post by M7MBA Sat Nov 28, 2020 11:21 pm Is $$X^2Y^3Z^2>0?$$ by M7MBA » Is $$X^2Y^3Z^2>0?$$ (1) $$XY > 0$$ (2) $$YZ < 0$$ Answer: E... 0 Last post by M7MBA Sat Nov 28, 2020 11:18 pm Is $$x$$ an even integer? by VJesus12 » Is $$x$$ an even integer? (1) $$x^2$$ is divisible by $$4.$$ (2)... 0 Last post by VJesus12 Sat Nov 28, 2020 11:15 pm What is the volume of a certain rectangular solid? by VJesus12 » What is the volume of a certain rectangular solid? (1) Two adjacent... 0 Last post by VJesus12 Sat Nov 28, 2020 11:13 pm What is the value of $$\\dfrac{r}2 + \\dfrac{s}2 ?$$ by VJesus12 » What is the value of $$\\dfrac{r}2 + \\dfrac{s}2 ?$$ (1) $$\\dfrac{r +... 0 Last post", "3^{7z}$ WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$ Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ $\\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$ $\\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$ Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction. Now, WLOG, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us , $\\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$. But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction. So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$ Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$ So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$ The study of mathematics, like the Nile, begins in minuteness but ends in magnificence. - Charles Caleb Colton dshasan Posts: 66 Joined: Fri Aug 14, 2015 6:32 pm ### Re: BDMO National Junior 2016/6 What do yo mean by WLOG ? Akhiar Posts: 1 Joined: Tue", "to\\beginalign x^3+y^3+z^3 - 3xyz= & x^3+y^3+z^3 - 3c\\\\= & a(x^2+y^2+z^2) - b(x+y+z)\\\\= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)\\endalign share mention monitor reply Oct 29 \"13 at 11:16 achille huiachille hui $\\endgroup$ 2 include a comment | 17 $\\begingroup$ \\beginalignx^3+y^3+z^3-3xyz\\\\&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\\\\&= (x+y)^3+z^3-3xy(x+y+z)\\\\&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\\\\&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\\\\&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\endalign re-superstructure cite follow edited Aug 23 at 21:40 who 13477 bronze title answered Oct 29 \"13 at 10:23 yuriusyurius $\\endgroup$ include a comment | 11 $\\begingroup$ Use Newton\"s identities: $p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required. Here $p_1= x+y+z = e_1$ $p_2= x^2+y^2+z^2$ $p_3= x^3+y^3+z^3$ $e_2 = xy + xz + yz$ $e_3 = xyz$ re-publishing point out monitor edited Oct 29 \"13 in ~ 11:31 achille hui answered Oct 29 \"13 at 10:29 lhflhf $\\endgroup$ include a comment | 11 $\\begingroup$ A polynomial native $\\ptcouncil.netbbQ$ is a polynomial from $\\ptcouncil.netbbQ$, so it can be perceived as a polynomial in $z$ with coefficients from the integral domain $\\ptcouncil.netbbQ$.$$p(z)=z^3-3xy \\cdot z +x^3+y^3$$ So we can try our techniques to factor a polynomial of degree 3 end an integral domain:If it can be factored climate there is a element of level $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the continuous term that $p(z)$ i beg your pardon is $x^3+y^3$. The latter is have", "3z-4.$$ Therefore $f(z)$ is linear function for $z\\ge -\\frac{1}{4}$. Since $z=2016$ admits representation $(2)$, then $f(2016)=3\\cdot 2016-4 = 6044.$ First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily, When $x = \\dfrac{-1 - \\sqrt{8065}}{2}$: 1. $x^2 + x = 2016$ 2. $x^2 - 3x + 2 = 2020 + 2\\sqrt{8065} = a$ (say) 3. $9x^2 - 15x = 18156 + 12\\sqrt{8065}$ $$f(2016) + 2f(a) = 18156 + 12\\sqrt{8065}$$ When $x = \\dfrac{3 + \\sqrt{8065}}{2}$: 1. $x^2 + x = a$ 2. $x^2 - 3x + 2 = 2016$ 3. $9x^2 - 15x = 18144 + 6\\sqrt{8065}$. $$f(a) + 2f(2016) = 18144 + 6\\sqrt{8065}$$ From the two equations, $$4f(2016) - f(2016) = 2(18144) - 18156$$ $$\\boxed{f(2016) = 6044}$$ Consider a linear function $f(x)=ax+b$ $$f(x^2+x) = ax^2+ax+b$$ $$f(x^2-3x+2)= ax^2-3ax +2a+b$$ $$f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$ $$a=3, b=-4$$ $$f(x) = 3x-4$$ $$f(2016)=6044$$ • Is this the unique solution? – clark Jun 14 '18 at 4:18 • This is a solution where the function is assumed to be linear. I don't know if this is the only solution or not. – Mohammad Riazi-Kermani Jun 14 '18 at 4:21 • Why could you assume that $f$ is linear? – Ariel Serranoni Jun 14 '18 at 4:22", "you enter $$a = 2061$$ into equations \\ref{eq:jnaarchinees}ff then you get $$s = 8$$ and $$b = 6$$, a different (and valid) combination than the one that you started with. ### 14.2. HYSN The HYSN system is a special way to assign a number to a year. The conversion of a Gregorian (or Julian proleptic) year number to a HYSN year number goes as follows: \\begin{align} \\{ H − 1, a_2 \\} \\| = \\Div(a + 67016, 10800) \\\\ \\{ Y − 1, a_3 \\} \\| = \\Div(a_2, 360) \\\\ \\{ S − 1, N − 1 \\} \\| = \\Div(a_3, 30) \\end{align} For example, which HYSN year corresponds approximately to our year 2016? Then \\begin{align*} a \\| = 2016 \\\\ \\{ H − 1, a_2 \\} \\| = \\Div(a + 67016. 10000) = \\Div(2016 + 67016, 10800) \\\\ \\| = \\Div(69032, 10800) = \\{ 6, 4232 \\} \\\\ \\{ Y − 1, a_3 \\} \\| = \\Div(a_2, 360) = \\Div(4232, 360) = \\{ 11, 272 \\} \\\\ \\{ S − 1, N − 1 \\} \\| = \\Div(a_3, 30) = \\Div(272, 30) = \\{ 9, 2 \\} \\end{align*} so $$H = 7$$, $$Y = 2$$, $$S = 10$$, $$N = 3$$, and the HYSN year number is 0712-1003. In the other direction things are simpler. Given $$H$$,", "+0 # If $f(x)=x^3+3x^2+3x+1$, find $f(f^{-1}(2010))$. +2 76 6 +467 Nov 25, 2020 #1 +10793 +1 If $$f(x)=x^3+3x^2+3x+1$$, find $$f(f^{-1}(2010))$$ Hello Guest! $$f(x)=x^3+3x^2+3x+1$$ $$f(x)=y=x^3+3x^2+3x+1\\\\ f(x)=y=(x-1)^3\\\\ f^{-1}(x)=x=(y-1)^3\\\\ f^{-1}(x)=y=\\sqrt[3]{x}+1$$ $$f^{-1}(2010)=\\sqrt[3]{2010}+1=\\pm 12.6202+1\\\\ f^{-1}(2010)\\in \\{13.6302,-11.6202\\}$$ $$f(f^{-1}(2010))=f(13.6302, -11.6202)$$ $$(x-1)^3=(13.6302-1)^3=2010$$ $$(x-1)^3=(-11.6302-1)^3=-2010$$ $$f(f^{-1}(2010))\\in \\{-2010,2010\\}$$ ! Nov 26, 2020 edited by asinus Nov 26, 2020 #2 +112000 +3 I think they cancel each other out and the answer is just 2010 Looking at it a little more formally. I think it means $$2010=(x+1)^3\\\\ \\sqrt[3]{2010}-1=x\\\\ f(\\sqrt[3]{2010}-1)=(\\sqrt[3]{2010}-1+1)^3=2010$$ Nov 26, 2020 #3 +10793 +1 Hi Melody! I think: In the term $$f(f^{-1}(2010)$$ is 2010 the argument of the $$f^{-1}$$. Then applies: $$f^{-1}(x)=f^{-1}(2100)=\\sqrt[3]{x}+1=\\sqrt[3]{2010}+1\\\\ \\color{blue}f^{-1}(2100)\\in\\{-11.6202,13.6202\\}$$ These values of the function $$f^{-1}$$are two arguments for the function f. Please confirm whether this is the right or wrong idea. Grüße asinus Nov 26, 2020 #4 +112000 0 I don't think so.. Let's ask Alan or Heureka ..... Melody Nov 27, 2020 #5 +31506 +2 I see it this way: Nov 27, 2020 #6 +112000 0 Thanks Alan, that is a much better way to present it. :) Melody Nov 27, 2020", "3^{5y-2w} = 3^{7z-2w}$ Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction. Now, WLOG, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us , $\\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$. But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction. So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$ Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$ So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$ Your solution is correct, but you can't say wlog here. It doesn't matter much. But there may be points taken from that. Frankly, my dear, I don't give a damn. Posts: 147 Joined: Mon Mar 28, 2016 6:21 pm ### Re: BDMO National Junior 2016/6 $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$ Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ $\\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$ $\\Rightarrow", "that $$v>\\frac{5}{3}x$$, and thus $$>x$$. Let us now suppose $$v=x+y$$, and we shall have \\begin{align} x+3y&=\\surd(13x^2+3),\\; \\textrm{and}\\\\ 13x^2+3&=x^2+6xy+9y^2,\\; \\textrm{or}\\\\ 12x^2&=6xy+9y^2-3,\\; \\textrm{and}\\\\ 4x^2&=2xy+3y^2-1; \\; \\textrm{whence}\\\\ x&=\\dfrac{y+\\surd(13y^2-4)}{4}, \\end{align} and, consequently, $$x>y$$. We shall, therefore, make $$x=y+z$$, which gives \\begin{align} 3y+4z&=\\surd(13y^2-4),\\; \\textrm{and}\\\\ 13y^2-4&=9y^2+24zy+16z^2, \\; \\textrm{or}\\\\ 4y^2&=24zy+16z^2+4; \\; \\textrm{therefore}\\\\ y^2&=6yz+4z^2+1, \\; \\textrm{and}\\\\ y&=3z+\\surd(13z^2+1). \\end{align} This formula being at length similar to the first, we may take $$z=0$$, and go back as follows: $\\begin{array}{l|l|l} z=0,&u=v+x=3,&q=r+s=71,\\\\ y=1,&t=u+v=5,&p=q+r=109,\\\\ x=y+z=1,&s=6t+u=33,&n=p+q=180,\\\\ v=x+y=2,&r=s+t=38,&m=3n+p=649. \\end{array}$ So that 180 is the least number, after 0, which we can substitute for $$n$$, in order that $$13n^2+1$$ may become a square. 106 This example sufficiently shews how prolix these calculations may be in particular cases; and when the numbers in question are greater, we are often obliged to go through ten times as many operations as we had to perform for the number 13. As we cannot foresee the numbers that will require such tedious calculations, we may with propriety avail ourselves of the trouble which others have taken; and, for this purpose, a Table is subjoined to the present chapter, in which the values of $$m$$ and $$n$$ are calculated for all numbers, $$a$$, between 2 and 100; so that in the cases which present themselves, we may take from it the values of $$m$$ and $$n$$, which answer to the", "# Inequality $\\cos x+\\cos y+\\cos z=0$ and $\\cos{3x}+\\cos{3y}+\\cos{3z}=0$ prove that $\\cos{2x}\\cdot \\cos{2y}\\cdot \\cos{2z}\\le 0$ Let $x,y,z$ be real numbers such that $\\cos x+\\cos y+\\cos z=0$ and $\\cos{3x}+\\cos{3y}+\\cos{3z}=0$ prove that $\\cos{2x}\\cdot \\cos{2y}\\cdot \\cos{2z}\\le 0$. • Very nice exercise! – Peter Jan 9 '14 at 15:05 Set $$u:=cosx , v:=cosy , w:=cosz$$ We have $$u + v + w = 0$$ and $$cos3x + cos3y + cos3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$ So $$u^3 + v^3 + w^3 = 0$$ Now consider $$cos2x * cos2y * cos2z = (2u^2-1)(2v^2-1)(2w^2-1)$$ Further, we have $$(u+v+w)^3 - u^3 - v^3 - w^3 = 3(u^2v+u^2w+v^2u+v^2w+w^2u+w^2v+2uvw) = 0$$ $$u^2v+u^2w+u^3+v^2u+v^2w+v^3+w^2u+w^2v+w^3+2uvw = 0$$ $$2uvw = 0$$ WLOG u = 0 , v=-w As $2v^2-1=2w^2-1$ and $2u^2-1 < 0$, the proof is completed. Setting $\\cos x=a$ etc. We have $\\displaystyle a+b+c=0\\ \\ \\ \\ (1)$ $\\displaystyle\\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$ $\\displaystyle\\implies a^3+b^3+c^3=3abc\\ \\ \\ \\ (2)$ Again as $\\displaystyle\\cos3x=4\\cos^3x-3\\cos x\\implies,$ $\\sum\\cos3x=0\\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\\ \\ \\ \\ (3)$ From $\\displaystyle (2),(3) 3abc=0$ $\\displaystyle\\implies$ at least one of $a,b,c$ is zero If $c=0,$ from $(1), a+b=0\\iff b=-a$ $\\displaystyle\\implies\\cos2x\\cos2y\\cos2z=\\prod(2\\cos^2x-1)=-(2\\cos^2x-1)^2\\le0$ as $\\cos x=-\\cos y\\implies \\cos^2x=\\cos^2y$", "# $P(x) = x^4 + ax^3 +bx^2 +cx+d$ Let $P(x) = x^4 + ax^3 +bx^2 +cx+d$ where $a, b, c, d$ are integers. $P(x)$ is divided by $x-2012, x-2013, x-2014, x-2015, x-2016$ and has the remainders $24, -6, 4, -6, 24$ respectively. What is the remainder when $P(x)$ is divided by $x-2017$ ? $P(2012) = 24, \\;P(2013) = -6, \\;P(2014) = 4, \\;P(2015) = -6, \\;P(2016) = 24$ By Lagrange interpolation, $P(2017) = \\sum{24\\cdot\\frac{(2017-2013)(2017-2014)(2017-2015)(2017-2016)}{(2012-2013)(2012-2014)(2012-2015)(2012-2016)}} = 24+30+40+60+120=274$ By symmetry at $x=2014$ we can write $\\ P(x) = 4 + a(x\\!-\\!2014)^2 + b(x\\!-\\!2014)^4$. Evaluation implies that $\\ a=-15,\\ b=5\\$ so $\\,P(2017)=4-15(9)+5(81)=274$ • The problem assumes that the leading coefficient of $P(x)$ is equal to 1, so I think you should divide your $P(x)$ with 5? Then we get $P(2017)=\\frac{274}{5}$? Not sure though... – Milan Mar 6 '17 at 15:44 • I made a mistake in calculation. Now my answer is the same as yours. Your method is interesting, please explain how you get $\\ P(x) = 4 + a(x\\!-\\!2014)^2 + b(x\\!-\\!2014)^4$ ? – carat Mar 6 '17 at 15:47 • @Milan But the 5 values determine a unique quartic, so the problem must be misstated if it assumes the polynomial is monic. – Bill Dubuque Mar 6 '17 at 16:18 • @carat The symmetry shows it is an even polynomial", "# II.- Sistemas de ecu. 2º grado con dos incógnitas 1.243 1.244 $5x+7y=61$ $x-y=4$ $xy=8$ $xy=48$ 1.245 1.246 $6x-5y=14$ $x+xy+y=11$ $xy=72$ $xy=6$ 1.247 1.248 $xy+2y=4$ $5xy-3x=84$ $3x-y=5$ $2x+7y=35$ 1.249 $x\\left&space;(&space;1+5y+\\frac{y}{x}&space;\\right&space;)=31$ $xy=5$ 1.250 $5xy=1x-xy+y=\\frac{11}{6}$ $3xy=7$ 1.251 $x+xy-y=16$ $xy=12$ 1.252 $x-(3x+1)y=-\\frac{1}{3}$ $\\frac{x-ax}{\\sqrt{x}}=\\frac{\\sqrt{x}}{x}$ 1.253 $x+\\sqrt{xy}+y=14$ $xy=16$ 1.254 $x-\\sqrt{xy}-y=10$ $xy=36$ 1.255 $x^{2}+y^{2}+3y+y=20$ $x-y=2$ 1.256 $2x^{2}+3y^{2}-x-5y=26$ $3x-2y=10$ 1.257 $x^{2}-xy=-52$ $\\frac{x-1}{y-1}=\\frac{3}{4}$ 1.258 $(x+2)+(y+3)=11$ $(x+2)(y+3)=24$ 1.259 $(x-4)+(y-7)=3$ $(x-4)(y-7)=2$ 1.260 $x+y=9$ $(x+3)(y+4)=48$ 1.261 $x+y=14$ $(x-2)(y-7)=4$ 1.262 $x^{2}+y^{2}+x-5y=24$ $x+y=7$ 1.263 $3x^{2}+xy+2y^{2}+3x+2y=39$ $2x-3y=3$ 1.264 $(x+y)^{2}-(x-y)^{2}=40$ $x+3xy-y=21$ 1.265 $(x+y)^{2}-(x-y)^{2}=28$ $x-xy+y=1$ 1.266 $(x+1)^{2}+(y+1)^{2}=52$ $2(x+4)+3(y-2)=23$ 1.267 $(x+2)^{2}+(y+3)^{2}=32$ $5x+4y=14$ Soluciones", "\\ \\, w_{3}^{u} +w_{3}^{v} \\leq 1\\\\ w_{2}^{u} +w_{3}^{u} \\leq w_{1}^{v}, \\ \\, w_{1}^{u} +w_{3}^{u} \\leq w_{2}^{v}\\\\ w_{1}^{u} +w_{2}^{u} \\leq w_{3}^{v}, \\ \\, w_{1}^{u} +1\\geq w_{2}^{v} +w_{3}^{v}\\\\ w_{2}^{u} +1\\geq w_{1}^{v} +w_{3}^{v}, \\ \\, w_{3}^{u} +1\\geq w_{1}^{v} +w_{2}^{v}. \\\\ \\end{cases} \\end{align*} By using Lingo 11 to solve this model, we have the following results: \\begin{align*} &w_{1}^{u} =0.0394, \\ \\, w_{1}^{v} =0.9605, \\ \\, w_{2}^{u} =0.8929, \\ \\, w_{2}^{v} =0.1070\\\\ &w_{3}^{u} =0.0675, \\ \\, w_{3}^{v} =0.9324, \\ \\, \\varphi_{12}^{+} =0.0, \\ \\, \\varphi_{12}^{-} =0.0\\\\ &\\varphi_{13}^{+} =0.0, \\ \\, \\varphi_{13}^{-} =0.2881, \\ \\, \\varphi_{23}^{+} =0.0, \\ \\, \\varphi_{23}^{-} =0.0025\\\\ &{\\phi}_{12}^{+} =0.0, \\ \\, {\\phi}_{12}^{-} =0.009, \\ \\, {\\phi}_{13}^{+} =0.0, \\ \\, {\\phi}_{13}^{-} =0.2653\\\\ &{\\phi}_{23}^{+} = 0.0, \\ \\, {\\phi}_{23}^{-} = 0.0. \\end{align*} Therefore, the optimal intuitionistic fuzzy weight vector ${\\tilde{{\\pmb w}}}^{\\ast}$ $=$ ($\\tilde{{w}}_{1}^{\\ast}, \\tilde{{w}}_{2}^{\\ast}, \\tilde{{w}}_{3}^{\\ast})= ((0.039, 0.961),$ $(0.893, 0.107),$ $(0.068,$ $0.932))$. As we know $\\lambda = 0$, and calculated by (4), we get $S(\\tilde{{w}}_{1}^{\\ast})$ $=$ $-0.921,$ $S(\\tilde{{w}}_{2}^{\\ast})= 0.786$, $S(\\tilde{{w}}_{3}^{\\ast})= -0.865$ which gives the ranking of $x_{2} \\succ x_{3} \\succ x_{1}$. Wang [15] used the additive consistency-based method to derive a priority weight vector, and Liao and Xu [16] constructed a fractional programming model to extract priority weights based on multiplicative consistency-based method. Their findings are listed in Table Ⅰ which led to the same ranking: $x_{2} \\succ x_{3} \\succ x_{1}$ as our", "b * y * w; z = zz; w = ww; ? z %6 = x^3 - 729*y^2*x - 7452*y^3 ? w %7 = 27*y*x^2 + 828*y^2*x + 6277*y^3 ? a1 = 27; c = 27; zz = x * z - c * y * w ; ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww; ? a1 = 27; c = 27; zz = x * z - c * y * w ; ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww; ? z %9 = x^4 - 1458*y^2*x^2 - 29808*y^3*x - 169479*y^4 ? w %10 = 108*y*x^3 + 4968*y^2*x^2 + 75324*y^3*x + 376280*y^4 ? ? a1 = 81; c = 9; zz = x * z - c * y * w ; ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww; ? a1 = 81; c = 9; zz = x * z - c * y * w ; ww = a * x * w + a1 * y * z + b", "and then tell me what you get 5. Nov 21, 2007 ### Gib Z $$(2a-b)^3 = 8a^3 - 4a^2b + 2ab^2 - b^3$$ The rest are the same, just replace the letters. $$(2a-b)^3 +(2b-c)^3 +(2c-a)^3 = 7a^3 - 4a^2b + 2a^2c + 7b^3 - 4b^2c + 2b^a + 7c^3 - 4c^2a + 2bc^2$$ 6. Nov 21, 2007 ### morphism u+v+w=0 => u+v=-w, and so on. 7. Nov 21, 2007 ### Gib Z So I get that down to $$u^3+v^3+w^3 = 2uvw$$. That is actually quite disturbing... 8. Nov 21, 2007 ### Gib Z Ahh sorry I forgot about the Original u^3+v^3+w^3 from post 2!! Thank you very much~!~!", "(0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); 6 KB (896 words) - 09:13, 22 May 2020 • $\\int u\\, dv=uv-\\int v\\,du$ ...math>u[/itex] will show up as $du$ and $dv$ as $v$ in the integral on the RHS, u should be chosen such that it has an \" 1 KB (231 words) - 15:19, 18 May 2021 • Specifically, let $u, v : \\mathbb{R \\times R \\to R}$ be definted <cmath> u(x,y) = \\text{Re}\\,f(x+iy), \\qquad v(x,y) = \\text{Im}\\,f(x+iy) . </cmath> 9 KB (1,537 words) - 20:04, 26 July 2017 • https://www.youtube.com/watch?v=BBD66Q3KXuI ...enter connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full 4 KB (719 words) - 18:41, 25 November 2020 • the vertex $V$ to this path? MP(\"P\",(-1,0),W);MP(\"V\",(-.5,2.4),N); 3 KB (560 words) - 18:23, 10 March 2015 • | $\\left(u(x)\\times v(x)\\right)'=u(x)v'(x)+u'(x)v(x)$ | $\\left(\\frac{u(x)}{v(x)}\\right)' = \\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ 3 KB (504 words) - 18:23, 3 March 2010 • *Given a weighted, undirected graph $G = (V,E)$ and two vertices $s, t \\in E$, does there exist a path 6 KB (1,104 words) - 14:11, 25 October 2017 • ...for which <cmath>\\left\\vert\\sum_{j=m+1}^n(a_j-(v+1))\\right\\vert\\le (T-v)\\,v \\le \\left(\\frac T 2\\right)^2</cmath>for all", "of 18 things, RHS is not GM of 18 things. I think you get $(4xy+6yz+8xz)/3\\ge((4xy)(6yz)(8xz))^{1/3}$. – Gerry Myerson Mar 6 '12 at 11:20 @GerryMyerson, you are right, i have edited the answer. – quartz Mar 6 '12 at 11:31 @vikiii RHS is $192^{1/3}*xyz^{2/3}$ and LHS is 3. – quartz Mar 6 '12 at 11:38 Of course, you then have to decide whether the bound you get is actually attained. – Gerry Myerson Mar 6 '12 at 11:49 If we want to use the AGM inequality without loss we have to \"symmetrize\" the three variables. Therefore we replace $x$, $y$, $z$ by $$x':=4x,\\quad y':=3y,\\quad z':=6z\\ .$$ The AGM inequality then says that $$\\root 3\\of {x'^2y'^2z'^2}\\leq {x'y' + y'z' + z' x'\\over 3}=4xy + 6 yz+8 zx=9\\ ,$$ i.e., $x'y'z'\\leq 27$, with equality iff $x'=y'=z'=3$. It follows that $$xyz = {x'y'z'\\over 72}\\leq{3\\over 8}$$ with equality iff $x={3\\over4}$, $y=1$, $z={1\\over2}$. - Thanks . But i want to know How did you think of x′:=4x,y′:=3y,z′:=6z . – vikiiii Mar 7 '12 at 4:40 @vikiiii: I wrote $x'=\\alpha x$, $\\ldots$, and determined $\\alpha$, $\\beta$, $\\gamma$ such that $4xy={1\\over3}x'y'$, $\\ldots\\$. – Christian Blatter Mar 7 '12 at 8:59 You can also do it by using weighted arithmetic mean and geometric mean inequality, $$\\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n} > [{x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}}]^{\\frac{1}{m_1m_1m_2m_3.....m_n}}$$", "3 \\cdot 4} &&\\mod 2017 \\\\ &\\equiv (-1)^4 \\frac{(5)(6)}{2} &&\\mod 2017 \\\\ &\\equiv 15 &&\\mod 2017. \\end{aligned} Following this pattern, we obtain the following: $\\displaystyle \\binom{2014}{0} + \\binom{2014}{1} + \\dots + \\binom{2014}{62} \\equiv 1 - \\frac{2 \\cdot 3}{2} + \\frac{3 \\cdot 4}{2} - \\frac{4 \\cdot 5}{2} + \\dots + \\frac{63 \\cdot 64}{2} \\mod 2017$. If we can’t think of any other bright ideas, we could just add up these 63 numbers and divide by 2017, it’s certainly a lot easier than the original 63 numbers that we had to compute and add. However, noticing that $\\frac{n(n+1)}{2}$ is simply the sum $1 + 2 + \\dots + n$: \\begin{aligned} RHS &= +1 \\\\ &\\hspace{1em} -1 - 2 \\\\ &\\hspace{1em} +1 +2 +3 \\\\ &\\hspace{3em} \\vdots \\\\ &\\hspace{1em} +1 +2 +3 \\dots +63 \\\\ &= 1 + 3 + \\dots + 63 = (32)^2 = 1024.\\end{aligned} Thus $S \\equiv 1024 \\mod 2017$, the answer is (C)." ]

[ "+0 0 93 5 +146 Let $$x, y,$$ and $$z$$ be positive real numbers such that $$xyz=1$$ Find the minimum value of $$(x+2y)(y+2z)(xz+1).$$ Jul 24, 2020 #1 0 i think this is wrong.... or right... using Lagrange multipliers you can show that the minimum occurs when a=b=c=d=1/4 and is equal to 4/3 but you can also argue this from symmetry since they are all required to be positive. Jul 24, 2020 #2 +25556 +3 Let $$x$$, $$y$$, and $$z$$ be positive real numbers such that $$xyz=1$$ Find the minimum value of $$(x+2y)(y+2z)(xz+1)$$. $$\\mathbf{\\huge{AM \\geq GM }}$$ $$\\begin{array}{|rcll|} \\hline \\dfrac{x+2y}{2} & \\ge & \\sqrt{x*2y} \\\\\\\\ \\dfrac{y+2z}{2} & \\ge & \\sqrt{y*2z} \\\\\\\\ \\dfrac{xz+1}{2} & \\ge & \\sqrt{xz*1} \\\\\\\\ \\hline \\dfrac{(x+2y)(y+2z)(xz+1)}{8} & \\ge & \\sqrt{x*2y}\\sqrt{y*2z} \\sqrt{xz*1} \\\\\\\\ \\dfrac{(x+2y)(y+2z)(xz+1)}{8} & \\ge & \\sqrt{x*2y*y*2z*xz*1} \\\\\\\\ \\dfrac{(x+2y)(y+2z)(xz+1)}{8} & \\ge & \\sqrt{4x^2y^2z^2} \\\\\\\\ \\dfrac{(x+2y)(y+2z)(xz+1)}{8} & \\ge & 2xyz \\\\\\\\ (x+2y)(y+2z)(xz+1) & \\ge & 16*xyz \\quad | \\quad xyz=1 \\\\\\\\ (x+2y)(y+2z)(xz+1) & \\ge & 16*1 \\\\\\\\ \\mathbf{(x+2y)(y+2z)(xz+1)} & \\ge & \\mathbf{16} \\\\ \\hline \\end{array}$$ The minimum value of $$(x+2y)(y+2z)(xz+1)$$ is $$\\mathbf{16}$$ Jul 24, 2020 #3 +1 I don't think that 16 is attainable. My 'messy' calculations point to a minimum of 18. Guest Jul 24, 2020 #4 +30931 +2 Try x = 2, y = 1, z = 1/2 Alan Jul 24, 2020 #5 +146 +1", "# Inequality $\\cos x+\\cos y+\\cos z=0$ and $\\cos{3x}+\\cos{3y}+\\cos{3z}=0$ prove that $\\cos{2x}\\cdot \\cos{2y}\\cdot \\cos{2z}\\le 0$ Let $x,y,z$ be real numbers such that $\\cos x+\\cos y+\\cos z=0$ and $\\cos{3x}+\\cos{3y}+\\cos{3z}=0$ prove that $\\cos{2x}\\cdot \\cos{2y}\\cdot \\cos{2z}\\le 0$. • Very nice exercise! – Peter Jan 9 '14 at 15:05 Set $$u:=cosx , v:=cosy , w:=cosz$$ We have $$u + v + w = 0$$ and $$cos3x + cos3y + cos3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$ So $$u^3 + v^3 + w^3 = 0$$ Now consider $$cos2x * cos2y * cos2z = (2u^2-1)(2v^2-1)(2w^2-1)$$ Further, we have $$(u+v+w)^3 - u^3 - v^3 - w^3 = 3(u^2v+u^2w+v^2u+v^2w+w^2u+w^2v+2uvw) = 0$$ $$u^2v+u^2w+u^3+v^2u+v^2w+v^3+w^2u+w^2v+w^3+2uvw = 0$$ $$2uvw = 0$$ WLOG u = 0 , v=-w As $2v^2-1=2w^2-1$ and $2u^2-1 < 0$, the proof is completed. Setting $\\cos x=a$ etc. We have $\\displaystyle a+b+c=0\\ \\ \\ \\ (1)$ $\\displaystyle\\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$ $\\displaystyle\\implies a^3+b^3+c^3=3abc\\ \\ \\ \\ (2)$ Again as $\\displaystyle\\cos3x=4\\cos^3x-3\\cos x\\implies,$ $\\sum\\cos3x=0\\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\\ \\ \\ \\ (3)$ From $\\displaystyle (2),(3) 3abc=0$ $\\displaystyle\\implies$ at least one of $a,b,c$ is zero If $c=0,$ from $(1), a+b=0\\iff b=-a$ $\\displaystyle\\implies\\cos2x\\cos2y\\cos2z=\\prod(2\\cos^2x-1)=-(2\\cos^2x-1)^2\\le0$ as $\\cos x=-\\cos y\\implies \\cos^2x=\\cos^2y$", "same data c(cov(X,MXY),cov(Y,MXY),cov(X,MXZ),cov(Z,MXZ)) would give 0.49974 0.50019 0.50014 0.50025 all close to the expected $\\frac12$ from the linked question – Henry Nov 13 '16 at 16:25 • Ah, I found my mistake, and get $0.293545$. Time to write it up! – Math1000 Nov 13 '16 at 17:15 The result is: $$\\color{red}{\\mathrm{cov}(\\min(X,Y),\\min(X,Z))=\\frac13-\\frac{2-\\sqrt3}{2\\pi}\\approx0.290687895}$$ To show this, note that the identities $$\\min(X,Y)=X\\mathbf 1_{X<Y}+Y\\mathbf 1_{Y<X}\\qquad\\min(X,Z)=X\\mathbf 1_{X<Z}+Z\\mathbf 1_{Z<X}$$ yield $$E(\\min(X,Y)\\min(X,Z))=u+v+w$$ with $$u=E(X^2;X<Y,X<Z)\\qquad v=E(YZ;Y<X,Z<X)$$ and $$w=E(XY;Y<X<Z)+E(XZ;Z<X<Y)$$ Considering the common CDF $F$ of $X$, $Y$ and $Z$, one sees that $$u=E(T^2)\\qquad\\text{with}\\ T=X(1-F(X))$$ Using the symmetry $(X,Y,Z)\\to(X,Z,Y)$, $$w=2E(XY;Y<X<Z)=E(XY;X<Z,Y<Z)=v$$ For every $z$, $$E(XY;X<z,Y<z)=G(z)^2\\qquad\\text{with}\\ G(z)=E(X;X<z)$$ hence $$w=v=E(G(X)^2)$$ Likewise, $$E(\\min(X,Y))=E(X;X<Y)+E(Y;Y<X)=2E(T)$$ hence $$\\mathrm{cov}(\\min(X,Y),\\min(X,Z))=E(X^2(1-F(X))^2)+2E(G(X)^2)-4E(X(1-F(X)))^2$$ where we recall that $$G(x)=E(X;X<x)$$ The preceding steps hold for any i.i.d. triplet. From now on, we use that $X$ is standard normal, with PDF $\\varphi$ and CDF $\\Phi$, hence $E(X)=0$ and $$E(T)=-E(X\\Phi(X))=-\\int_\\mathbb R x\\Phi(x)\\varphi(x)dx$$ Since $\\Phi'(x)=\\varphi(x)$ and $\\varphi'(x)=-x\\varphi(x)$, an integration by parts yields $$E(T)=\\left.\\varphi(x)\\Phi(x)\\right|_{-\\infty}^\\infty-\\int_\\mathbb R\\varphi(x)^2dx=-\\int_\\mathbb R\\frac1{\\sqrt{2\\pi}}\\varphi(\\sqrt2x)dx=-\\frac1{2\\sqrt{\\pi}}$$ Likewise, $X$ is symmetric hence $1-\\Phi(X)=\\Phi(-X)$ and $$E(T^2)=E(X^2\\Phi(X)^2)=\\int_\\mathbb R x^2\\varphi(x)\\Phi(x)^2dx$$ Using $\\varphi''(x)=(x^2-1)\\varphi(x)$, one gets $E(T^2)=s+t$ with $$s=\\int_\\mathbb R \\varphi''(x)\\Phi(x)^2dx=\\left.\\varphi'(x)\\Phi^2(x)\\right|_{-\\infty}^\\infty-\\int_\\mathbb R\\varphi(x)\\cdot2\\varphi'(x)\\Phi(x)dx=-\\int_\\mathbb R(\\varphi(x)^2)'\\Phi(x)dx$$ hence $$s=\\left.-\\varphi(x)^2\\Phi(x)\\right|_{-\\infty}^\\infty+\\int_\\mathbb R\\varphi(x)^3dx=\\int_\\mathbb R\\frac1{2\\pi}\\varphi(\\sqrt3x)dx=\\frac1{2\\pi\\sqrt3}$$ and $$t=\\int_\\mathbb R \\varphi(x)\\Phi(x)^2dx=\\left.\\frac13\\Phi(x)^3\\right|_{-\\infty}^\\infty=\\frac13$$ Finally, $$G(x)=\\int_{-\\infty}^xz\\varphi(z)dz=-\\varphi(x)$$ hence $$E(G(X)^2)=\\int_\\mathbb R\\varphi(x)^3dx=\\int_\\mathbb R\\frac1{2\\pi}\\varphi(\\sqrt3x)dx=\\frac1{2\\pi\\sqrt3}$$ Summing these contributions, one gets the numerical value at the beginning of this post. • What does the notion $E(A; B)$ mean please? – PSPACEhard Nov 14 '16 at 16:43 • If $X$ is an", "zmudz one year ago For x, y, and z positive real numbers, what is the maximum possible value for $$\\sqrt{\\frac{3x+4y}{6x+5y+4z}} + \\sqrt{\\frac{y+2z}{6x+5y+4z}} + \\sqrt{\\frac{2z+3x}{6x+5y+4z}}?$$ Also, find what z/x is if (x,y,z) achieves the maximum value. Thank you! I am so stuck right now. 1. ganeshie8 Let $$a=\\sqrt{\\frac{3x+4y}{6x+5y+4z}}\\\\ b=\\sqrt{\\frac{y+2z}{6x+5y+4z}}\\\\ c=\\sqrt{\\frac{2z+3x}{6x+5y+4z}}$$ Firstly, notice that $$a^2+b^2+c^2=1$$ 2. ganeshie8 appealing to cauchy schwarz inequality gives $a*1+b*1+c*1\\le \\sqrt{(a^2+b^2+c^2)(1^2+1^2+1^2)}=\\sqrt{3}$", "to do it: write f = w1*(9y/w1) + w2*((4/y)/w2), where w1, w2 > 0 and w1+w2=1. By the arithmetic-geometric inequality, w1*r1 + w2*r2 >= r1^w1 * r2^w2, with equality if and only if r1 = r2. So, with r1 = 9y/w1 and r2 = (4/y)/w2 we have f >= (9/w1)^w1*(4/w2)^w2*y^(w1-w2). Note that when w1 = w2 (so w1 = w2 = 1/2), we have f >= sqrt(18)*sqrt(8) = 12; that is, we *always* have f >= 12 for any y > 0. We get *equality* for r1 = r2, or 9y/w1 = (4/y)/w2, or 9y = 4/y (because w1 = w2 = 1/2). So, when 9y = 4/y (y = 2/3) we have f = 12, so that must be the minimum: f >= 12 for all y > 0 and we have achieved a value f = 12! In general, to minimize f = c1*y + c2/y with c1, c2 > 0 we must take both terms of f equal; that is, c1*y = c2/y, so y = sqrt(c2/c1), and f_min = 2*c1*sqrt(c2/c1) = 2*sqrt(c1*c2). Note: this type of thing is the basis of *Geometric Programming*, devised by Duffin, Peterson and Zener (yes, the Zener of diode fame). RGV Last edited: Aug 30, 2011 16. Aug 31, 2011 ### Saitama tan(x)=$\\pm \\sqrt{2/3}$. If i substitute tan(x)=$\\pm \\sqrt{2/3}$", "# Max/Min Problem For $x$, $y$, and $z$ positive real numbers, find $\\frac{z}{x}$ such that $(x,y,z)$ achieves the maximum value of $$\\sqrt{\\frac{3x+4y}{6x+5y+4z}} + \\sqrt{\\frac{y+2z}{6x+5y+4z}} + \\sqrt{\\frac{2z+3x}{6x+5y+4z}}.$$ I found that the maximum value was $\\sqrt3$ from Cauchy's Inequality, but I don't know how to actually achieve the maximum. I know that the terms must be proportional, but I don't know how to go from there. Any help on how to proceed would be greatly appreciated, thanks! • Why don't you post your working, then it would be easier to pinpoint how to get it. – Macavity Nov 11 '14 at 17:21 Here's how I would go about it: As the expression is homogeneous, we can set $$6x+5y+4z=1$$ WLOG. Then CS inequality gives: $$(\\overline{3x+4y}+\\overline{y+2z}+\\overline{2z+3x})(1+1+1)\\ge \\left(\\sqrt{3x+4y}+\\sqrt{y+2z}+\\sqrt{2z+3x} \\right)^2$$ So we get $$\\sqrt{3x+4y}+\\sqrt{y+2z}+\\sqrt{2z+3x} \\le \\sqrt3$$ and equality is possible iff $$\\dfrac{3x+4y}1=\\dfrac{y+2z}1=\\dfrac{2z+3x}1 \\implies x:y:z = 1:3:6$$ • Shouldn't it be $\\sqrt{1+1+1}$ instead of $1+1+1$? – YuiTo Cheng Apr 5 at 15:47 • @YuiToCheng It is correct as written. – Macavity Apr 6 at 1:15 It's $(a+b+c)^2\\leq3(a^2+b^2+c^2)$ The equality occurs for $3x+4y=y+2z=2z+3x$.", "# UVW method application of basic theorem Find the minimum and maximum value of $x+y+z+xy+yz+xz$ if $x^2+y^2+z^2 = 1$ I converted it to uvw, $3u+3v^2$ is the expression, $(3u)^2-2(3v^2)=1$, is the constraint. now I don't know what to do, I'm still learning the uvw method and I don't know if we can use the basic theorem to solve this or Tej's theorem. • Is $$x,y,z$$ assumed to be positive? – Dr. Sonnhard Graubner Sep 11 '18 at 13:11 • The maximum can be found via the inequality $$a^2+b^2+c^2\\geq bc+ca+ab\\,.$$ That is, $$x+y+z\\leq \\sqrt{3\\,\\left(x^2+y^2+z^2\\right)}=\\sqrt{3}$$ and $$yz+zx+xy\\leq x^2+y^2+z^2=1\\,,$$ so $$x+y+z+yz+zx+xy\\leq \\sqrt{3}+1\\,,$$ where the equality occurs if and only if $x=y=z=\\dfrac{1}{\\sqrt{3}}$. – Batominovski Sep 11 '18 at 13:24 $$3u+3v^2=3u+\\dfrac{(3u)^2-1}2=\\dfrac{9u^2+6u-1}2=\\dfrac{(3u+1)^2-2}2$$ Now for real $u,(3u+1)^2\\ge0$ Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v^2$ can be negative, and $xyz=w^3$. Thus, the condition gives $9u^2-6v^2=1$, which does not depend on $w^3$. Also, the expression $x+y+z+xy+xz+yz=3u+3v^2$ does not depend on $w^3$, which says that it's enough to find an extreme value of our expression for the extreme value of $w^3$, which happens for equality case of two variables. Let $y=x$. Thus, $z^2=1-2x^2$ and we need to find an extreme value of $$2x+x^2+z(1+2x),$$ where $z^2=1-2x^2$ and $-\\frac{1}{\\sqrt2}\\leq x\\leq \\frac{1}{\\sqrt2}.$ I got $$-1\\leq x+y+z+xy+xz+yz\\leq1+\\sqrt3.$$ By the way, I think using $uvw$ for this problem is not so necessary. Here there are", "# What is the maximum value of $\\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem: Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\\sqrt6 xy + 4yz$ I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$. • use the Lagrange Multiplier methode – Dr. Sonnhard Graubner Jan 9 '16 at 22:47 • I get $\\sqrt{\\frac{11}{2}}$ at $\\left(\\sqrt{\\frac{3}{22}},\\sqrt{\\frac{11}{22}},\\sqrt{\\frac{8}{22}}\\right)$. – CommonerG Jan 9 '16 at 23:20 Let $a,b\\in\\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\\in\\mathbb{R}$). By Cauchy-Schwarz, $axy+byz=(ax+bz)y\\leq \\left(\\sqrt{a^2+b^2}\\sqrt{x^2+z^2}\\right)|y|$. By AM-GM, $\\sqrt{x^2+z^2}\\,|y|\\leq \\frac{\\left(x^2+z^2\\right)+y^2}{2}=\\frac{1}{2}$. That is, $axy+byz\\leq \\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$. (Similarly, we also have the lower bound $axy+byz\\geq -\\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},-\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$.) • Pretty neat. Thank you. – Ashish Gupta Jan 9 '16 at 23:03 • I think your $y-$ and $z-$coördinates should be reversed in your solutions. – Théophile Jan 10 '16 at 0:15 • @Théophile Thanks, fixed. – Batominovski Jan 10 '16 at 0:52 We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\\sqrt{6}xy \\leq A^2x^2+B^2y^2$, and $4yz \\leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \\dfrac{\\sqrt{6}}{2},", "# If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is $\\bf{My\\; Try::}$ Here I have Used $\\bf{A.M\\geq G.M}$ Inequality So $$\\displaystyle \\frac{x^2+4xy+4y^2+4z^2}{4}\\geq \\left(x^3\\cdot y^3\\cdot z^2\\right)^{\\frac{1}{4}}\\;,$$ But I did not How can I solve it Help me, Thanks Applying the AM-GM is the right strategy, but you need to do it a bit differently. $$x^2+4xy+4y^2+4z^2$$ $$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$ $$\\ge 6\\sqrt[6]{x^2 \\cdot 2xy \\cdot 2xy \\cdot 4y^2 \\cdot 2z^2 \\cdot 2z^2}$$ $$= 6\\sqrt[6]{64x^4y^4z^4}$$ $$= 12(xyz)^{2/3}$$ $$= 12 \\cdot 32^{2/3}$$ Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$. Let $f: (x,y,z) \\mapsto x^{2} + 4xy + 4y^{2} + 4z^{2}$ on $\\mathbb{R}^{3}$; let $g: (x,y,z) \\mapsto xyz - 32$ on $\\mathbb{R}^{3}_{++}$; and let $(x,y,z) \\in g^{(-1)}\\{ 0 \\}$ such that $f(x,y,z)$ is an extremum. Then there is some $\\lambda \\in \\mathbb{R}$ such that $$\\nabla f(x,y,z) = (2x+4y, 4x+8y, 8z) = \\lambda \\nabla g = \\lambda (yz, xz, xy).$$ And we can find the candidate possible extrema of $f|_{g^{(-1)}\\{ 0 \\}}$, inspecting them.", "Applying the AM-GM inequality. We find that $$\\left(\\frac{xy+yz+zx}{3}\\right)^3\\geq (xyz)^2=f(x,y,z)^2.$$ Also, what happens when we set $$x=y=z=\\left(\\frac{a}{3}\\right)^{\\frac{1}{2}}.$$ Use this to complete the problem. - Provided $x,y,z > 0$ – user17762 Dec 31 '12 at 7:37 Your first statement is incorrect. See my comment to the question, where it is shown that if $(x,y,z)$ are allowed to be negative, then the maximum is in fact $\\infty$. – user17762 Dec 31 '12 at 7:41 @Marvis: Ahh, yes, thank you. I corrected my posted answer. – Eric Naslund Dec 31 '12 at 7:42 By arithmetic-geometic inequality, $$\\frac{xy+xz+yz}3\\ge \\sqrt[3]{xy\\cdot xz\\cdot yz}$$ provided $xy,xz,yz$ are all positive (i.e. $x,y,z$ have the same sign, e.g. are all positive) and with equality iff $xy=yz=xz$ (i.e. iff $x=y=z$). The left hand side is $\\frac a3$ and the right hand side is $f(x,y,z)^{\\frac23}$. However, if $x,y,zh$ are allowed to have differnt signs, there is no absolute maximum: If $t>0$ and $y=z=-t$, then $x=t-\\frac a{2t}$ produces $f(x,y,z)=t^3-\\frac12at$, which becomes arbitrarily large. -", "So just start from the high order bit downwards. Output the matching bits. As soon as you hit a disagreement between the binaries of$A$and$B$(which will be$0$in$A$and$1$in$B$) ... 0 It's not that complex. Hammer and Nail =$1.10. Nail = .05 Hammer = 1.05 (1 dollar more than nail) Total: $1.10 2 Just another way: sum the AM-GMs... $$2x^2+2k^2 \\ge k\\cdot 4 x,\\quad 3y^2+\\frac34k^2 \\ge k \\cdot 3 y, \\quad 4z^2+ \\frac14k^2 \\ge k\\cdot2z$$ to get$\\dfrac1k+3k \\ge (4x+3y+2z)$, and the maximum is achieved when$\\frac1k=3k\\implies k = \\frac1{\\sqrt3}$, for a maximum of$2\\sqrt3$. 5 Apply Cauchy-Schwarz Inequality:$\\left(4x+3y+2z\\right)^2 = \\left(2\\sqrt{2}\\cdot \\sqrt{2}x+\\sqrt{3}\\cdot \\sqrt{3}y+1\\cdot 2z\\right)^2 \\leq \\left((2\\sqrt{2})^2+(\\sqrt{3})^2+1^2\\right)\\cdot\\left(2x^2+3y^2+4z^2\\right) = 12 \\Rightarrow 4x+3y+2z \\leq 2\\sqrt{3} = \\text{Maximum Value}$. 0 This problem is same as the problem SGARDEN of codechef July Long Challenge 2014. 0 WLOG, we can set$\\sqrt2x=\\cos A,\\sqrt3y=\\sin A\\cos B,2z=\\sin A\\sin B\\implies4x+3y+2z=2\\sqrt2\\cos A+\\sqrt3\\sin A\\cos B+\\sin A\\sin B=2\\sqrt2\\cos A+\\sin A(\\sqrt3\\cos B+\\sin B)=2\\sqrt2\\cos A+2\\sin A\\cos\\left(B-\\dfrac\\pi6\\right)$If$\\sin A\\ge0,4x+3y+2z\\le2\\sqrt2\\cos A+2\\sin A$as$\\cos\\left(B-\\dfrac\\pi6\\right)\\le1$The equality occurs if ... 0 It is obvious that the first equation stands closed surface, so when$4x+3y+2z=k$gets the minimum, the line just touches the closed surface. Let me put the tangent point$A(x_0，y_0)$, and then I get the partial derivative of$2x_2+3y_2+4z_2=1$about$X$and$Y$at the point$A$. I know I can find a line that coincides with the given line. After this I can ... 2 For me,$4x + 3y + 2z = \\langle\\begin{pmatrix} \\sqrt{2}", "easy to do looking at $(2)$ and using the quadratic formula: $f'(x)=0$ if and only if $$x=4,\\quad x={8\\pm\\sqrt{64-56}\\over 2}=4\\pm\\sqrt2.$$ Step 3: We evaluate $f$ at each of the three points found above: \\tag{3} \\eqalign{ f(4)&= 6\\cr f(4+\\sqrt2)&=(2+\\sqrt2)\\cdot2\\cdot( -2+\\sqrt2)+6=2\\cr f(4-\\sqrt2)&=(2-\\sqrt2)\\cdot2\\cdot(-2-\\sqrt2)+6=2.\\cr } Step 4: From step 1., $f(x)$ tends to infinity as $x\\rightarrow\\infty$ or as $x\\rightarrow-\\infty$. This means that $f$ does have a global minimum value and that that value must be the smallest value found in step 3.. So, from $(3)$, we see that the minumum value of $f$ is $2$. This minimum value is achieved at two places: $x=4+\\sqrt2$ and at $x=4-\\sqrt 2$. - Note that $(x-2)(x-6)=x^2-8x+12$ and $(x-4)^2=x^2-8x+16$. This suggests the symmetrizing substitution $w=x^2-8x+14$. Thus $$y=(w-2)(w+2)+6=w^2+2.$$ We want to minimize the absolute value of $w$. But $w=(x-4)^2-2$. So $w$ has minimum absolute value $0$, reached when $(x-4)^2=2$, that is, when $x=4\\pm\\sqrt{2}$. The minimum value of $y$ is $2$. Remark: The curve $y=(x-2)(x-4)^2(x-6)+6$ is very special, with a beautiful double symmetry. It seems likely that the method used above was the intended one. The same idea works for $(x-a)(x-b)(x-c)(x-d)+k$, where $a+d=b+c$. - $$y= (x-2) (x-4)^2 (x-6) + 6$$ $let , t = x-4$ $$y= t^2(t-2)(t+2)+ 6$$ $$y= t^4-4t^2+6$$ $$y=(t^2-2)^2+2$$ $$y(min)=2$$ attained when $t^2=2$ $(x-4)^2=2$ which gives, $x=4-\\sqrt2$ and $x=4+\\sqrt2$ -", "3z-4.$$ Therefore $f(z)$ is linear function for $z\\ge -\\frac{1}{4}$. Since $z=2016$ admits representation $(2)$, then $f(2016)=3\\cdot 2016-4 = 6044.$ First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily, When $x = \\dfrac{-1 - \\sqrt{8065}}{2}$: 1. $x^2 + x = 2016$ 2. $x^2 - 3x + 2 = 2020 + 2\\sqrt{8065} = a$ (say) 3. $9x^2 - 15x = 18156 + 12\\sqrt{8065}$ $$f(2016) + 2f(a) = 18156 + 12\\sqrt{8065}$$ When $x = \\dfrac{3 + \\sqrt{8065}}{2}$: 1. $x^2 + x = a$ 2. $x^2 - 3x + 2 = 2016$ 3. $9x^2 - 15x = 18144 + 6\\sqrt{8065}$. $$f(a) + 2f(2016) = 18144 + 6\\sqrt{8065}$$ From the two equations, $$4f(2016) - f(2016) = 2(18144) - 18156$$ $$\\boxed{f(2016) = 6044}$$ Consider a linear function $f(x)=ax+b$ $$f(x^2+x) = ax^2+ax+b$$ $$f(x^2-3x+2)= ax^2-3ax +2a+b$$ $$f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$ $$a=3, b=-4$$ $$f(x) = 3x-4$$ $$f(2016)=6044$$ • Is this the unique solution? – clark Jun 14 '18 at 4:18 • This is a solution where the function is assumed to be linear. I don't know if this is the only solution or not. – Mohammad Riazi-Kermani Jun 14 '18 at 4:21 • Why could you assume that $f$ is linear? – Ariel Serranoni Jun 14 '18 at 4:22", "$$(2ky -1)^2 - k (2x-k)^2 = 1 - k^3$$ You asked about $k=2.$ There are two systems of degree two linear recurrences for $w^2 - 2 v^2 = -7.$ First we have $w_n = -1, 5, 31, 181,$ with $u_n = 2, 4, 22, 128.$ The orbit relation is $$w_{n+1} = 3 w_n + 4 u_n, \\; \\; \\; u_{n+1} = 2 w_n + 3 u_n.$$ When $w \\equiv 3 \\pmod 4,$ we get $y > 0$ with $2ky - 1 = w.$ We also have $w_n = 1, 11, 65, 379,$ with $u_n = 2, 8, 46, 268.$ The orbit relation is $$w_{n+1} = 3 w_n + 4 u_n, \\; \\; \\; u_{n+1} = 2 w_n + 3 u_n.$$ When $w \\equiv 3 \\pmod 4,$ we get $y > 0$ with $2ky - 1 = w.$ My take on the outcome is that it is not natural to require $y > 0.$ Makes it messy. For example, with $w = 12875,$ we get $4y = 12876,$ $y = 3219.$ Also $v = 9104,$ $2x - 2 = 9104,$ $2x = 9106,$ $x = 4553.$ Alright, $x=4553, y = 3219.$ $$\\frac{x^2 + y}{y^2 + x} = \\frac{4553^2 + 3219}{ 3219^2 + 4553} = \\frac{20733028}{10366514} = 2$$ jagy@phobeusjunior:~$./Pell_Target_Fundamental Automorphism matrix: 3 4 2 3 Automorphism backwards: 3 -4 -2", "# Finding the maximum value of$\\sqrt{\\frac{3x+4y}{6x+5y+4z}} + \\sqrt{\\frac{y+2z}{6x+5y+4z}} + \\sqrt{\\frac{2z+3x}{6x+5y+4z}}?$ 1. For $x$, $y$, and $z$ positive real numbers, what is the maximum possible value for $$\\sqrt{\\frac{3x+4y}{6x+5y+4z}} + \\sqrt{\\frac{y+2z}{6x+5y+4z}} + \\sqrt{\\frac{2z+3x}{6x+5y+4z}}?$$ 2. Find $z/x$ if $(x,y,z)$ achieves the maximum value from Problem 1. I know I should include the Cauchy Schwarz Inequality in here but I don't know how. And how would I benefit me? Could someone please walk me step by step thorugh these two problems? Thanks! Just make a change of variables setting $u=3x+4y, v=y+2z, w=2z+3x$, then apply the Cauchy Schwarz inequality to prove $$\\sqrt{\\frac{u}{u+v+w}}+\\sqrt{\\frac{v}{u+v+w}}+\\sqrt{\\frac{w}{u+v+w}}\\leq \\color{red}{\\sqrt{3}}$$ with equality achieved at $(u,v,w)=\\lambda(1,1,1)$, than means $(x,y,z)=\\lambda(1,3,\\color{red}{6})$.", "+0 # Inequalities 0 98 1 +179 Let x and y be real numbers such that $$x - \\sqrt{x + 6} = \\sqrt{y + 6} - y$$. Let m be the minimum value of x+y, and M be the maximum value of x+y. Enter the ordered pair (m, M). Thanks for any tips, hints, or answers you provide, anything is appreciated. Jun 7, 2021 #1 +2205 +2 I tried, but couldn't figure out the answer. :(( x - sqrt(x+6) = sqrt(y+6) - y x + y = sqrt(y+6) + sqrt(x+6) I'm not sure if this is true, but I feel like the maximum value of x+y would be when x = y. So x = y = 3, making x + y = 6. =^._.^= Jun 7, 2021", "## $v w x y z = v + w + x + y + z$ Find the maximum possible value of $x$ where $v , w , x , y , z > 0$ Apr 4, 2018 Maximum for natural numbers is $x = 5$ #### Explanation: Given: $v w x y z = v + w + x + y + z$ For solutions in natural numbers, we can look at possible combinations of small values. Note that: • If $v = w = x = y = z = 1$ then $v w x y z = 1 < 5 = v + w + x + y + z$ • If $v = w = x = y = z = 2$ then $v w x y z = 32 > 10 = v + w + x + y + z$ The minimum possible values of $v , w , y , z$ are $v = w = y = z = 1$, resulting in: $x = x + 4 \\text{ }$ which has no solutions. Considering the case where exactly one of $v , w , y , z > 1$, and putting $v = 2$, $w = y = z = 1$ we get: $2 x = x + 5 \\text{", "# Find the maximum value of given expression $$x+2y+3z = 15$$ Find the maximum value of $$6(1+x)yz + x(2y+3z)$$ I substituted $2y+3z=15-x$ And tried to use AM GM to find maximum value of $xyz$ but they don't occur at same value. I know that this is not the right way to do it but I don't have any idea on how to do these type of questions. How do I proceed? Edit Do note that $x,y,z$ are positive real numbers. • are $x,y,z$ real numbers? – Dr. Sonnhard Graubner Feb 21 '18 at 9:22 • there is no global maximum – Dr. Sonnhard Graubner Feb 21 '18 at 9:25 • @Dr.SonnhardGraubner yes sir – user481779 Feb 21 '18 at 9:25 • why is this question downvoted? The OP has stated what he tried. – Rohan Shinde Feb 21 '18 at 9:38 Let $x=5a$, $y=\\frac{5}{2}b$ and $z=\\frac{5}{3}c$. Thus, $a+b+c=3$ and since $(a+b+c)^2\\geq3(ab+ac+bc),$ by AM-GM we obtain: $$6(1+x)yz + x(2y+3z)=25(5abc+ab+ac+bc)\\leq$$ $$\\leq25\\left(5\\left(\\frac{a+b+c}{3}\\right)^3+\\frac{1}{3}(a+b+c)^2\\right)=200.$$ The equality occurs for $a=b=c=1$, which says that we got a maximal value. Variational Approach To maximize $$6(1+x)yz+x(2y+3z)\\tag1$$ under the constraint $$x+2y+3z=15\\tag2$$ we need $$(6yz+2y+3z)\\,\\delta x+(6z+6xz+2x)\\,\\delta y+(6y+6xy+3x)\\,\\delta z=0\\tag3$$ for all $\\delta x$, $\\delta y$, and $\\delta z$ so that $$\\delta x+2\\delta y+3\\delta z=0\\tag4$$ This means there is a $\\lambda$ so that \\begin{align} \\lambda&=6yz+2y+3z\\implies(3z+1)(2y+1)=\\lambda+1\\\\ 2\\lambda&=6z+6xz+2x\\implies(3z+1)(x+1)=\\lambda+1\\\\ 3\\lambda&=6y+6xy+3x\\implies(2y+1)(x+1)=\\lambda+1 \\end{align}\\tag5 which has solutions", "## zmudz one year ago If $$x$$ and $$y$$ are real and $$x^2 + y^2 = 1$$, compute the maximum value of $$(x+y)^2.$$ 1. triciaal |dw:1441474466693:dw| 2. imqwerty hint-this is the equation of a circle with center at origin 3. IrishBoy123 lagrange multiplier 4. anonymous $x^2+y^2=1 \\Rightarrow y^2 = 1 - x^2$ $(x+y)^2 = x^2+2xy+y^2=x^2+y^2+2xy=1+2xy=1+2x \\sqrt{1-x^2}$to maximize, differentiate w.r.t. x =>$\\frac{ d }{ dx }\\left[ \\left( x+y \\right)^2 \\right]=\\frac{ d }{ dx }\\left[ 1+2x\\sqrt{1-x^2} \\right]=2\\sqrt{1-x^2}-2x\\cdot \\frac{ 2x }{ \\sqrt{1-x^2} }=0$=>$2(1-x^2)-4x = 0 \\Rightarrow x^2+2x-1=0 \\Rightarrow x= -1\\pm \\sqrt2$ 5. ganeshie8 or you may use AM-GM inequality \\begin{align}(x+y)^2 &= x^2 + y^2 + \\color{red}{2xy}\\\\~\\\\ &\\le x^2 + y^2 + \\color{red}{x^2+y^2}\\\\~\\\\ &= 1+1 \\end{align} 6. anonymous AM-GM? 7. imqwerty (x-0)^2 +(y-0)^2 = sqrt(1) center=(0,0) radius r = 1 so u can use parametric form to denote any point x,y which lies on it like- x=x1 + rcos(theta) y=y1+rsin(theta) |dw:1441474744572:dw| x1,y1=0,0 so x= rcos(theta) y=rsin(theta) nd u need to find max of (x+y)^2 $(rcos \\theta+rsin \\theta)^2$ rcos(theta) + rsin(theta) ranges frm -sqrt(r) to + sqrt(r) so max value of rcostheta +rsintheta gives max of (x+y)^2 nd we knw r=1 so (1)^2 =1 8. ganeshie8 fir $$x,y\\in \\mathbb{R}$$ we have : $$(x-y)^2 \\ge 0 \\\\\\implies x^2+y^2 -2xy \\ge 0 \\implies 2xy \\le x^2+y^2$$ so, \\begin{align}(x+y)^2 &= x^2 + y^2 + \\color{red}{2xy}\\\\~\\\\ &\\le", "# Tag Info 6 This is not an explanation of the steps in the book but I would like to point out that the proof in the book looks a bit silly to me. You don't require Rouche's Theorem. There is a very elementary 'High School' proof: suppose $f(z)=f(w)$. Then $z-w=a+a_2(w^{2}-z^{2})+a_3(w^{3}-z^{3})+\\cdots$. Note that $|w^{n}-z^{n}| =|z-w|(|w^{n-1}+w^{n-2}z+\\cdots+wz^{n-2}+... 5 Hint: Your squaring must be: $$(a-b)^2=a^2+b^2-2ab$$ 4 Since the Legendre symbol is multiplicative we have: $$\\left(\\frac{-2}{p}\\right)=\\left(\\frac{2}{p}\\right)\\left(\\frac{-1}{p}\\right)$$ The first one is equal to$1$iff$p \\equiv 1,7 \\pmod{8}$and the second iff$p \\equiv 1 \\pmod{4}$. So they are both one when$p \\equiv 1 \\pmod{8}$and both$-1$when$p \\equiv 3 \\pmod{8}$. In both cases their product is one ... 4 Writing$\\cos2t=c,$using$\\cos2x=2\\cos^2x-1,$$$257+32(2c^2-1)-256\\cdot\\dfrac{1+c}2=64c^2-128c+97=64(c-1)^2+97-64\\ge33$$ for real$c4 Note that\\begin{align}32\\cos(4t)&=32\\bigl(2\\cos^2(2t)-1\\bigr)\\\\&=32\\bigl(2\\bigl(2\\cos^2(t)-1\\bigr)^2-1\\bigr)\\\\&=32\\bigl(8\\cos^2(t)-8\\cos(t)+1\\bigr)\\\\&=256\\cos^2(t)-256\\cos(t)+32\\\\&\\geqslant256\\cos^2(t)-256+32\\\\&>256\\cos^2(t)-257.\\end{align}Your proof looks correct to me. 3 They are really unnecessary. Any morphism with both a left and a right inverse has an inverse, which is equal to both the given left and the right inverse. In fact it is more convenient in certain foundational systems like homotopy type theory to define an isomorphism as a morphism with both a left and a right inverse. 3 Your method is correct, but has some errors. You are consideringf(t) = 257 + 32 \\cos 4t - 256\\cos^2 t$and want to prove$f(t_0)>0$, where$t_0$is a critical point (for both max and min stationary points). So: $$f'(t)=\\color{red}{-128}\\sin4t\\color{red}+256\\sin 2t=0 \\Rightarrow \\sin" ]

[ "+ 1 has many solutions find the value of k. What are you looking for?", "and check different values of $k$ using grid search to see what produces results you are happy with.", "value of $$k$$, as a function of $$n$$. Then plug this final value into your formula $$\\sum_{k=1}^{\\text {final}}(2k-1) = ({\\text {final}})^2.$$", "Review question # Two values of $x$ that differ by $5$ satisfy $x^2 - 12x + k = 0$, what is $k$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5949 ## Question 1. The equation $x^3 + ax + b = 0$ is satisfied by the values $x = 1$ and $x = 2$. Find the values of $a$ and $b$; and find also the third value of $x$. 2. The equation $x^2 - 12x + k = 0$ is satisfied by two values of $x$ that differ by $5$. Find these two values of $x$, and the value of $k$.", "Sum un = 0, find the value of K. n = 1 ??? From the previous part you will now know $a$ and $b$, so plug them into the formula for the sum, which gives a equation in K to solve: $\\sum_{n=1}^K u_n=\\frac{K}{2}(2a+(K-1)d)=0$ CB", "0 0 solutions k=0 4 solutions k is between 0 and 1 8 solutions k=1 7 solutions k is between 1 and 3 4 solutions k=3 6 solutions kis bigger than 3 8 solutions Jan 27, 2022 edited by Melody Jan 27, 2022", "11. Find the middle terms in the expansion of (x +y)7. 12. Which two consecutive terms in the expansion (1 +x)15have equal coefficients. 13. Find $\\sum_{k=1}^{n}{1\\over k(k+1)}.$ 14. Find a positive value of m for which the coefficient of x2 in the expansion of (1+x)m is 6. 15. Find the $\\sqrt [ 3 ]{ 126 }$ approximately to two decimal places.", "for $k=1$. If you try to solve for $x_2$ and $x_3$, you will see that no solutions can exist for $k=-3$ or $k=2$.", "Difference between revisions of \"2006 USAMO Problems/Problem 2\" Problem For a given positive integer k find, in terms of k, the minimum value of $N$ for which there is a set of $2k+1$ distinct positive integers that has sum greater than $N$ but every subset of size k has sum at most $\\frac{N}{2}$.", "sums, one can find the four numbers since the four equations in four unknowns has nonzero determinant. So in case $n=4,\\ k=3$ you can recover the set. (I think this works for any $n$ if $k=n-1$.)", "by completing the square: $4x^2+11x+7=0$4x2+11x+7=0 1. Write all solutions on the same line, separated by commas.", "(4k+6) ⠰⠰⠠⠨⠎⠨⠢⠣⠅⠐⠶⠼⠁⠜⠨⠔⠼⠉⠚⠐⠣⠼⠙⠅⠐⠖⠼⠋⠐⠜ ### Example 10 summation with a starting value k=3 and ending value of 7 of the expression of a fraction with a numerator k+1 and denominator k ⠰⠰⠠⠨⠎⠨⠢⠣⠅⠐⠶⠼⠉⠜⠨⠔⠼⠛⠷⠅⠐⠖⠼⠁⠨⠌⠅⠾", "to first find their sum using your equation for $[A](t)$ at a known $t$. Once you have done that, you can solve for $k_1$ using your equation for $[B](t)$ at a known $t$, since you now have the denominator of your first factor. You could alternatively find $k_2$ using $[C](t)$. Once you have $k_1$, then $k_2$ is easily found from the sum.", "the values. x = 2a y = 5 n = 4 Step 2: Put the values in the formula and solve the coefficients. The coefficient values for (x + y)4, as we have discussed many times before, are 1,4,6,4,1. We will skip this part of the step. Step 3: Now, calculate the answer to every value of k from 0 to 4. Step 4: Write in the form of a series. (2a + 5)4 = 16a4 + 160a3 + 600a2 + 1000a + 625", "## precalc Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) 2 cos^2 ? ? cos ? ? 1 = 0 ? =", "to determine __________.... ### 3k + 10 = 2k - 21 what is k 3k + 10 = 2k - 21 what is k... ### Find the value of a squared + b when a= -2 and b=5 Find the value of a squared + b when a= -2 and b=5...", "all the k values where", "Find all solutions of the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) 3 csc2 ? ? 4 = 0 ? = ..........rad", "to win all the amount starting with capital \\$(k-1). This give the difference equation, \\begin{equation*} P_{k}=pP_{k+1}+qP_{k-1}, \\quad 0<k<N. \\end{equation*}", "Question # Find the value of k for which each of the following systems of linear equations has an infinite number of solutions: 5x+2y=2k,2(k+1)x+ky=(3k+4). Solution ## The value k=4 solves the equations hence the value of k is 4. Suggest corrections" ]

[ "+0 # Help. Thank you. +3 75 2 +294 Let a and b be the solutions of the equation $$2x^2-10x+5=0$$. What is the value of $$(2a-3)(4b-6)$$? Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +3 Thank you! calvinbun Feb 23, 2021", "+0 0 90 1 a and b are the roots of 2x^2 - 3x - 1 = 0. Find a/b + b/a. Jun 24, 2020 #1 +307 0 There's not a simple way to factorize $$2x^2 - 3x - 1 = 0$$, so instead we simplify $$\\frac{a}{b}+\\frac{b}{a}=\\frac{a^2+b^2}{ab}=\\frac{(a+b)^2-2ab}{ab}=\\frac{(a+b)^2}{ab}-2$$ Using Vieta's Formula, a+b = 3/2 and ab = -1/2. Plug these in to get $$\\frac{(3/2)^2}{-1/2}-2=\\frac{9/4}{-1/2}-2=-\\frac{9}{2}-2=\\frac{-13}{2}$$ Jun 24, 2020", "roots. Now, $144b^{ 2 } = 12b$ Therefore, $\\alpha$ = $\\frac{ -B + \\sqrt{ D } } { 2A }$ $\\frac{ -( -12a ) + 12b }{ 2 \\times 36} = \\frac{ 12 (a + b) }{ 72 } = \\frac{ a + b }{ 6 }$ $\\beta$ = $\\frac{ -B – \\sqrt{ D } } { 2A }$ $\\frac{ -( -12a ) – 12b }{ 2 \\times 36} = \\frac{ 12 (a – b) }{ 72 } = \\frac{ a – b }{ 6 }$ Hence, $\\frac{ a + b }{ 6 }$ and $\\frac{ a – b }{ 6 }$ are the roots of the given equation. Q26) $x^{ 2 } – 2ax + ( a^{ 2 } – b^{ 2 } ) = 0$ Ans. 26) Given: $x^{ 2 } – 2ax + ( a^{ 2 } – b^{ 2 } ) = 0$ On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get: A = 1, B = -2a and C = $( a^{ 2 } – b^{ 2 } )$ Discriminant D is given by: D = $B^{ 2 } – 4AC$ = $( -2a )^{ 2 } – 4 \\times 1 \\times ( a^{ 2 } – b^{ 2 } )$ = $4a^{ 2 } –", "-30 + 10 \\sqrt{ 2 } }{ 50 }$ = $\\frac{ 10 ( -3 + \\sqrt{ 2 } ) }{ 50 } = \\frac{ ( -3 + \\sqrt{ 2 } ) }{ 5 }$ $\\beta$ = $\\frac{ -b – \\sqrt{ D } } { 2a }$ = $\\frac{ -30 – \\sqrt{ 200 } }{ 2 \\times 25} = \\frac{ -30 – 10 \\sqrt{ 2 } }{ 50 }$ = $\\frac{ 10 ( -3 – \\sqrt{ 2 } ) }{ 50 } = \\frac{ ( -3 – \\sqrt{ 2 } ) }{ 5 }$ Thus, the roots of the equation are $\\frac{ ( -3 + \\sqrt{ 2 } ) }{ 5 }$ and $\\frac{ ( -3 – \\sqrt{ 2 } ) }{ 5 }$ . Q6) $16x^{ 2 } = 24x + 1$ Ans. 6) Given: $16x^{ 2 } = 24x + 1$ $\\Rightarrow 16x^{ 2 } – 24x – 1 = 0$ On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get: a = 16, b = -24 and c = -1 Discriminant D is given by: D = $b ^{ 2 } – 4ac$ = $( -24 )^{ 2 } – 4 \\times 16 \\times ( -1 )$ = 576 + ( 64 ) = 640 > 0 Hence, the roots", "b ∈ R Squaring on both sides, we get -16 + 30i = a2 + b2i2 + 2abi ∴ -16 + 30i = (a2 – b2) + 2abi …..[∵ i2 = -1] Equating real and imaginary parts, we get a2 – b2 = -16 and 2ab = 30 ∴ a2 – b2 = -16 and b = $$\\frac{15}{a}$$ ∴ a2 – $$\\left(\\frac{15}{a}\\right)^{2}$$ = -16 ∴ a2 – $$\\frac{225}{a^{2}}$$ = -16 ∴ a4 – 225 = – 16a2 ∴ a4 + 16a2 – 225 = 0 ∴ (a2 + 25)(a2 – 9) = 0 ∴ a2 = -25 or a2 = 9 But a ∈ R, a2 ≠ -25 ∴ a2 = 9 ∴ a = ±3 When a = 3, b = $$\\frac{15}{3}$$ = 5 When a = -3, b = $$\\frac{15}{-3}$$ = -5 ∴ $$\\sqrt{-16+30 \\mathrm{i}}$$ = ±(3 + 5i) (ii) Let $$\\sqrt{15-8 i}$$ = a + bi, where a, b ∈ R Squaring on both sides, we get 15 – 8i = a2 + b2i2 + 2abi ∴ 15 – 8i = (a2 – b2) + 2abi ……[∵ i2 = -1] Equating real and imaginary parts, we get a2 – b2 = 15 and 2ab = -8 ∴ a2 – b2 = 15 and b = $$\\frac{-4}{a}$$ ∴ a2 – ($$\\left(\\frac{-4}{a}\\right)^{2}$$) = 15 ∴ a2 –", "64b^{ 2 } + 288a^{ 2 }b^{ 2 }$ = $81a^{ 4 } + 144a^{ 2 }b^{ 2 } + 64b^{ 2 }$ = $( 9a ^{ 2 } + 8b^{ 2 } )^{ 2 } > 0$ Hence, the roots of the equation are equal. Roots $\\alpha$ and $\\beta$ are given by : Therefore, $\\alpha$ = $\\frac{ -B + \\sqrt{ D } } { 2A }$ = $\\frac{ -\\left [ -( 9a^{ 2 } – 8b^{ 2 } ) \\right ] +\\sqrt{ ( 9a^{ 2 } +8b^{ 2 }) ^{ 2 }}}{ 2 \\times 12ab }$ = $\\frac{ 9a^{ 2 } – 8b^{ 2 } + 9a^{ 2 } + 8b^{ 2} }{ 24ab } = \\frac{ 18a^{ 2 } }{ 24ab } = \\frac{ 3a }{ 4b }$ $\\beta$ = $\\frac{ -B – \\sqrt{ D } } { 2A }$ = $\\frac{ -\\left [ -( 9a^{ 2 } – 8b^{ 2 } ) \\right ] – \\sqrt{ ( 9a^{ 2 } +8b^{ 2 }) ^{ 2 } } }{ 2 \\times 12ab }$ = $\\frac{ 9a^{ 2 } – 8b^{ 2 } – 9a^{ 2 } – 8b^{ 2} }{ 24ab } = \\frac{ -16b^{ 2 } }{ 24ab } = \\frac{ -2b }{ 3a }$ Thus, the roots of the equation are $\\frac{ 3a }{", "+0 # HELP ASAP! +2 48 1 Let a and b be the roots of 2x^2+7x+11=0. Find ab+a+b Feb 17, 2021 #1 +944 +1 We can use Vieta's Formulas. Because 2x^2 + 7x + 11 = 0, we have $$a=2, b=7, c=11.$$ The sum of the roots is $$-\\frac{b}{a}=-\\frac{7}{2}$$ and the product of the roots is $$\\frac{c}{a}=\\frac{11}{2}$$. Let's rewrite the expression: $$ab + (a+b) = \\frac{11}{2} - \\frac{7}{2} = \\boxed{2}.$$ Feb 20, 2021", "+0 # I need help, thanks in advance. +4 71 2 +294 The roots of the equation $$2x^2-mx+n=0$$ sum to 6 and multiply to 10. What is the value of $$m+n$$? Feb 20, 2021 #1 +944 +3 We can use Vieta's Formulas. The sum of the roots is $$-\\frac{b}{a} = \\frac{m}{2}$$ and the product of the roots is $$\\frac{c}{a} = \\frac{n}{2}$$. The sum is 6, so $$\\frac{m}{2}=6$$ and m = 12. The product is 10, so $$\\frac{n}{a} = 10$$ and n = 20. So, $$m+n=12+20 = \\boxed{32}$$ Feb 20, 2021 #1 +944 +3 We can use Vieta's Formulas. The sum of the roots is $$-\\frac{b}{a} = \\frac{m}{2}$$ and the product of the roots is $$\\frac{c}{a} = \\frac{n}{2}$$. The sum is 6, so $$\\frac{m}{2}=6$$ and m = 12. The product is 10, so $$\\frac{n}{a} = 10$$ and n = 20. So, $$m+n=12+20 = \\boxed{32}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +4 Thank you! calvinbun Feb 20, 2021", "+0 # Algebra 0 28 1 Let a and b be the solutions to 5x^2 - 11x + 4 = 0. FInd 1/a^2 + 1/b^2. Jan 11, 2022 #1 +26 0 Note that $$\\frac{1}{a^2}+\\frac{1}{b^2}=\\frac{a^2+b^2}{a^2b^2} =\\frac{(a+b)^2 - 2ab} {(ab)^2}$$. By Vieta's, $$a+b = \\frac{-(-11)}{5} = \\frac{11}{5}$$ and $$ab = \\frac{4}{5}$$, so $\\frac{(a+b)^2 - 2ab}{(ab)^2} = \\frac{\\frac{121}{25} - 2 \\cdot \\frac{4}{5}}{\\left(\\frac{4}{5}\\right)^2} = \\frac{\\frac{81}{25}}{\\frac{16}{25}} = \\boxed{\\frac{81}{16}}.$ Jan 11, 2022", "+0 # Algebra 0 91 1 Let a and b be the solutions to 5x^2 - 11x + 4 = 0. FInd 1/a^2 + 1/b^2. Jan 11, 2022 Note that $$\\frac{1}{a^2}+\\frac{1}{b^2}=\\frac{a^2+b^2}{a^2b^2} =\\frac{(a+b)^2 - 2ab} {(ab)^2}$$. By Vieta's, $$a+b = \\frac{-(-11)}{5} = \\frac{11}{5}$$ and $$ab = \\frac{4}{5}$$, so $\\frac{(a+b)^2 - 2ab}{(ab)^2} = \\frac{\\frac{121}{25} - 2 \\cdot \\frac{4}{5}}{\\left(\\frac{4}{5}\\right)^2} = \\frac{\\frac{81}{25}}{\\frac{16}{25}} = \\boxed{\\frac{81}{16}}.$", "factor. This leads to: $c_{2k+2} = 2^2(\\frac{\\pi}{4})\\frac{2}{2k+2} \\frac{\\pi^k}{k!} = \\frac{\\pi^{k+1}}{(k+1)!}$ as required. Now we need to calculate $c_{2k+1} = 2\\int^{\\frac{\\pi}{2}}_0(cos^{2k+1}(t))dt c_{2k} = 2\\int^{\\frac{\\pi}{2}}_0(cos^{2k+1}(t))dt \\frac{\\pi^k}{k!}$ $= 2 (\\frac{2k}{2k+1})(\\frac{2k-2}{2k-1})......(\\frac{4}{5})\\frac{2}{3}\\frac{\\pi^k}{k!} = 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \\frac{\\pi^k}{k!}$ To simplify this further: split up the factors of the $k!$ in the denominator and put one between each denominator factor: $= 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \\pi^k$ Now multiply the denominator by $2^k$ and put one factor with each $k-m$ factor in the denominator; also multiply by $2^k$ in the numerator to obtain: $(2) 2^k (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \\pi^k$ Now gather each factor of 2 in the numerator product of the 2k, 2k-2… $= (2) 2^k 2^k \\pi^k \\frac{k!}{(2k+1)!} = 2 \\frac{(4 \\pi)^k k!}{(2k+1)!}$ which is the required formula. So to summarize: $V_{2k} = \\frac{\\pi^k}{k!} R^{2k}$ $V_{2k+1}= \\frac{2 k! (4 \\pi)^k}{(2k+1)!}R^{2k+1}$ Note the following: $lim_{k \\rightarrow \\infty} c_{k} = 0$. If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume $(2R)^n$. Then consider the ratio $\\frac{2^n R^n}{c_n R^n} = 2^n \\frac{1}{c_n}$; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the $2^n$ corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of", "# How do you solve the system 5a + 3b=-9 and 2a - 5b= -16.? Jun 2, 2015 We are first going to find $a$ in function of $b$ : $2 a - 5 b = - 16$ $2 a = - 16 + 5 b$ ( we add $5 b$ on each side ) $a = - 8 + \\frac{5 b}{2}$ ( we divide by $2$ on each side ) Now that we have $a$, we can find $b$ with the first equation : $5 a + 3 b = - 9$ $5 \\left(- 8 + \\frac{5 b}{2}\\right) + 3 b = - 9$ $- 40 + \\frac{25 b}{2} + 3 b = - 9$ $\\frac{25 b}{2} + 3 b = - 9 + 40$ ( we add 40 on each side ) $3 b = \\frac{6 b}{2}$ Thus $\\frac{25 b + 6 b}{2} = 31$ $\\frac{31 b}{2} = 31$ $\\frac{b}{2} = 1$ ( we divide by 31 on each side ) $b = 2$ ( we multiply by 2 on each side ) We can thus now find $a$ : $2 a - 5 b = - 16$ $2 a - 5 \\cdot 2 = - 16$ $2 a - 10 = - 16$ $2 a = - 16 + 10$ ( we add 10 on", "'18 at 8:26 The key mistake you made here is your simplification of square roots. Actually, since a square root is positive always, you should get $\\vert F_1P\\vert = \\frac{|4p+25|}{5}$ and $\\vert F_2P\\vert = \\frac{|4p-25|}{5}$. However, note $-5 \\leq p \\leq 5$, so $-20 \\leq 4p \\leq 20$ and $\\frac{|4p+25|}{5} = \\frac{4p+25}{5}, \\frac{|4p-25|}{5} = \\frac{-4p+25}{5}$ Summing the two distances, $p$ cancels out. Note that $y=0$ gives $x=\\pm5$, so $-5 \\le p \\le 5$ for all points $P(p,q)$ on the ellipse. Therefore: $$|F_1P| = \\frac{1}{5}\\sqrt{(4p+25)^2} = \\frac{1}{5}|4p+25| = \\frac{1}{5}(25+4p) \\\\ |F_2P| = \\frac{1}{5}\\sqrt{(4p-25)^2} = \\frac{1}{5}|4p-25| = \\color{red}{\\frac{1}{5}(25-4p)}$$ • how do you know the absolute value of |4p−25| is 25−4p – Inquirer Mar 3 '18 at 22:07 • @Inquirer $-5 \\le p \\le 5 \\iff -20 \\le 4p \\le 20$, so $25-4p \\ge 25-20 \\gt 0$. – dxiv Mar 3 '18 at 22:08 Since $$e^2= a^2-b^2 = 25-9 =16\\;\\; \\Longrightarrow \\;\\;e =4$$ so the point $F_1$ and $F_2$ are focuses of ellipse and by definiton of ellipse we have $$PF_1+PF_2 = constant = 2a =10$$ • Bingo. It is just the definition of the ellipse. Any proof based on the familiar analytic equation assumes the definition to start with because the equation is derived from the definition. – Oscar Lanzi Mar 4 '18 at 2:30 • I don't think that's", "# Missing factor equal to 0 after factoring The equation $$x^{10}+(13x-1)^{10}=0$$ has $$10$$ complex roots, $$r_i$$ for integers $$1\\le i\\le5$$. Find $$\\sum^5_{i=1}\\frac{1}{r_i\\overline{r_i}}.$$ My solution was simply to factor the given polynomial into $$(r^2+(13r-1)^2)(r^8-\\dots+(13r-1)^8)=0.$$ We suppose the left factor is $$0,$$ such that $$170r^2-26r+1=0.$$ By Vieta's, the product of the roots if $$1/170.$$ Since this product is the product of the conjugate pairs, we can multiply $$170\\times5$$ to find our answer, and $$850$$ turns out to be correct. However, how come we don't need to account for the right factor here? We need not consider it if it were never equal to $$0$$, so how do we prove that? We solve equation $$x^{10}+(13x-1)^{10}=0$$. $$\\left(13-\\frac{1}{x}\\right)^{10}=-1$$ $$13-\\frac{1}{x_k}=\\cos\\frac{(2k+1)\\pi}{10}+i\\sin\\frac{(2k+1)\\pi}{10},$$ $$\\frac{1}{x_k}=13-\\cos\\frac{(2k+1)\\pi}{10}-i\\sin\\frac{(2k+1)\\pi}{10},\\quad k=0\\ldots9.$$ Then $$\\frac{1}{x_k\\bar x_k}=170-26\\cos\\frac{(2k+1)\\pi}{10}$$ $$\\sum_{k=0}^4\\frac{1}{x_k\\bar x_k}=850-26\\left(\\cos\\frac{\\pi}{10}+\\cos\\frac{3\\pi}{10}+\\cos\\frac{7\\pi}{10}+\\cos\\frac{9\\pi}{10}\\right)=850$$", "+0 # The operation is defined for non-zero a andb as follows: If a+b=13 and , what is the value of ​?Express +1 90 2 +591 The operation $$*$$ is defined for non-zero a andb as follows: $$a * b = \\frac{1}{a} + \\frac{1}{b}.$$ If a+b=13 and $$a \\times b=25$$ , what is the value of $$a*b$$?Express your answer as a common fraction. ant101 Aug 28, 2018 #1 +3270 +2 We know that: $$\\frac{1}{a}+\\frac{1}{b}=a*b$$ . If we multiply the denominator by b and a, respectively, we get: $$\\frac{1b}{ab}+\\frac{1a}{ba}=a*b$$ . Suddenly, we go back to the problem, and it says: $$ab=25.$$ So, we have:$$\\frac{1a}{25}+\\frac{1b}{25}=\\frac{1a+1b}{25}.$$ We also know that $$a+b=13$$ , so $$\\frac{1a+1b}{25}=\\frac{a+b}{25}=\\boxed{\\frac{13}{25}}.$$ tertre Aug 28, 2018 #1 +3270 +2 We know that: $$\\frac{1}{a}+\\frac{1}{b}=a*b$$ . If we multiply the denominator by b and a, respectively, we get: $$\\frac{1b}{ab}+\\frac{1a}{ba}=a*b$$ . Suddenly, we go back to the problem, and it says: $$ab=25.$$ So, we have:$$\\frac{1a}{25}+\\frac{1b}{25}=\\frac{1a+1b}{25}.$$ We also know that $$a+b=13$$ , so $$\\frac{1a+1b}{25}=\\frac{a+b}{25}=\\boxed{\\frac{13}{25}}.$$ tertre Aug 28, 2018 #2 +20147 +2 The operation $$*$$ is defined for non-zero a andb as follows: $$a * b = \\frac{1}{a} + \\frac{1}{b}.$$ If $$a+b=13$$ and $$a \\times b=25$$ what is the value of $$a*b$$ ? Express your answer as a common fraction. $$\\begin{array}{|rcll|} \\hline a+b &=& 13 \\\\ a\\times b &=& 25 \\\\\\\\ \\dfrac{a+b}{a\\times b}&=& \\dfrac{13}{25}", "then find the ratio of numbers. a) $$\\frac{8±3\\sqrt{5}}{1}$$ b) $$\\frac{8±3\\sqrt{7}}{1}$$ c) $$\\frac{6±3\\sqrt{5}}{1}$$ d) $$\\frac{6±3\\sqrt{7}}{1}$$ Explanation: We know, G.M. of two numbers a and b is √ab. So, a + b = 4 √ab Squaring we get, a2+b2 = 16ab =>(a/b) + (b/a) = 16 Let x = a/b. So, x + 1/x = 16 => x2 – 16x + 1 = 0 =>x = $$\\frac{16±\\sqrt{256-4}}{2} = \\frac{16±\\sqrt{252}}{2} = \\frac{16±6\\sqrt{7}}{2} = \\frac{8±3\\sqrt{7}}{1}$$. Note: Join free Sanfoundry classes at Telegram or Youtube 4. The ratio of the A.M. and G.M. of two positive numbers a and b is 5: 3. Find the ratio of a to b. a) 9:1 b) 3:5 c) 1:9 d) 3:1 Explanation: (A.M.)/(G.M.) = 5/3 => $$\\frac{a+b}{2\\sqrt{ab}} = \\frac{5}{3}$$ Applying componendo and dividendo rule, we get => $$\\frac{a+b+2\\sqrt{ab}}{a+b-2\\sqrt{ab}} = \\frac{8}{2}$$ => $$(\\frac{\\sqrt{a}+\\sqrt{b}}{\\sqrt{a}-\\sqrt{b}})^2=4$$ => $$(\\frac{\\sqrt{a}+\\sqrt{b}}{\\sqrt{a}-\\sqrt{b}})^1=2$$ => $$(\\frac{\\sqrt{a}}{\\sqrt{b}})^1=3$$ Again applying componendo and dividendo rule, we get a/b = (3/1)2 = 9. So, a:b =9:1. Sanfoundry Global Education & Learning Series – Mathematics – Class 11. To practice Mathematics Objective Questions and Answers for Class 11, here is complete set of 1000+ Multiple Choice Questions and Answers.", "2 (\\frac{2k}{2k+1})(\\frac{2k-2}{2k-1})......(\\frac{4}{5})\\frac{2}{3}\\frac{\\pi^k}{k!} = 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \\frac{\\pi^k}{k!}$ To simplify this further: split up the factors of the $k!$ in the denominator and put one between each denominator factor: $= 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \\pi^k$ Now multiply the denominator by $2^k$ and put one factor with each $k-m$ factor in the denominator; also multiply by $2^k$ in the numerator to obtain: $(2) 2^k (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \\pi^k$ Now gather each factor of 2 in the numerator product of the 2k, 2k-2… $= (2) 2^k 2^k \\pi^k \\frac{k!}{(2k+1)!} = 2 \\frac{(4 \\pi)^k k!}{(2k+1)!}$ which is the required formula. So to summarize: $V_{2k} = \\frac{\\pi^k}{k!} R^{2k}$ $V_{2k+1}= \\frac{2 k! (4 \\pi)^k}{(2k+1)!}R^{2k+1}$ Note the following: $lim_{k \\rightarrow \\infty} c_{k} = 0$. If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume $(2R)^n$. Then consider the ratio $\\frac{2^n R^n}{c_n R^n} = 2^n \\frac{1}{c_n}$; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the $2^n$ corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of dimensions increases. One also notes that for fixed radius R, $lim_{n \\rightarrow \\infty} V_n = 0$ as well. There are other interesting aspects to this", "+0 # ASAP Vietas therom 0 47 1 +235 let f(x) = x^3-6x^2+13x-18 have root a,b, and c. find 1/ab+1/bc+1/ac Feb 27, 2021 edited by Charger421 Feb 27, 2021 We see that we need to find $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ac}$, which simplifies to $\\frac{a+b+c}{abc}.$ From here, we can now proceed with Vieta's. By Vieta's theorem (if we assume that the polynomial $f(x) = Ax^3 + Bx^2 + Cx + D$, we know that $$a+b+c = -\\frac{B}{A},$$ $$ab+ bc + ac = \\frac{C}{A},$$ $$abc = -\\frac{D}{A},$$ where $A=1, B=-6, C=13, D=-18.$ Therefore, we get that $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ac} = \\frac{-6}{-18} = \\boxed{\\frac{1}{3}}.$", "\\frac{ 4a – 2b }{ 2 } = 2a – b$ Hence, (2a + b) and (2a – b) are the roots of the given equation. Q31) $4x^{ 2 } – 4a^{ 2 } x + ( a^{ 4 } – b^{ 4 } ) = 0$ Ans. 31) The given equation is $4x^{ 2 } – 4a^{ 2 } x + ( a^{ 4 } – b^{ 4 } ) = 0$ On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get: A = 4, B = $- 4a^{ 2 }$ and C = $a^{ 4 } – b^{ 4 }$ Discriminant D is given by: D = $B^{ 2 } – 4AC$ = $(-4a^{ 2 } )^{ 2 } – 4 \\times 4 \\times (a^{ 4 } – b^{ 4 } )$ = $16a^{ 4 } – 16a^{ 4 } + 16b^{ 4 } = 16b^{ 4 } > 0$ So, the given equation has real roots. Therefore, $\\alpha$ = $\\frac{ -B + \\sqrt{ D } } { 2A }$ = $\\frac{ -(-4a ) + 4b^{ 2 } }{ 2 \\times 4} = \\frac{ 4( a^{ 2 } + b^{ 2 } ) }{ 8 } = \\frac{ a^{ 2 } + b ^{ 2 } }{ 2 }$ $\\beta$", "× b [(3a)2 − (4b)2] = 2 × 3 × a × b [(3a + 4b) (3a − 4b)] [a2 − b2 = (a + b)(a − b)] And, 2ab (3a + 4b) = 2 × a × b × (3a + 4b) $⇒\\frac{6ab\\left(9{a}^{2}–16{b}^{2}\\right)}{2ab\\left(3a+4b\\right)}=\\frac{2×3×a×b×\\left(3a+4b\\right)×\\left(3a–4b\\right)}{2×a×b×\\left(3a+4b\\right)}$ = 3 × (3a − 4b) = 9a − 12b $\\therefore$ 6ab (9a2 − 16b2) ÷ 2ab (3a + 4b) = 9a − 12b (ii) We can factorise the given expression as x2 − 14x − 32 = x2 − (16 − 2) x – 32 x2 − 16x + 2x – 32 = x (x − 16) + 2 (x − 16) = (x − 16) (x + 2) $⇒\\frac{{x}^{2}–14x–32}{x+2}=\\frac{\\left(x–16\\right)\\left(x+2\\right)}{\\left(x+2\\right)}=x–16$ = x – 16 $\\therefore$ (x2 − 14x − 32) ÷ (x + 2) = (x − 16) (iii) We can factorise the given expression as 36abc (5a – 25) (2b – 14) = 2 × 2 × 3 × 3 × a × b × c (5 × a – 5 × 5) (2 × b – 2 × 7) = 2 × 2 × 3 × 3 × 5 × a × b × c (a – 5) (2 × b – 2 × 7) = 2 × 2 × 2 × 3 × 3 × 5 × a × b" ]

[ "+0 # Help. Thank you. +3 75 2 +294 Let a and b be the solutions of the equation $$2x^2-10x+5=0$$. What is the value of $$(2a-3)(4b-6)$$? Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +3 Thank you! calvinbun Feb 23, 2021", "2 (\\frac{2k}{2k+1})(\\frac{2k-2}{2k-1})......(\\frac{4}{5})\\frac{2}{3}\\frac{\\pi^k}{k!} = 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \\frac{\\pi^k}{k!}$ To simplify this further: split up the factors of the $k!$ in the denominator and put one between each denominator factor: $= 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \\pi^k$ Now multiply the denominator by $2^k$ and put one factor with each $k-m$ factor in the denominator; also multiply by $2^k$ in the numerator to obtain: $(2) 2^k (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \\pi^k$ Now gather each factor of 2 in the numerator product of the 2k, 2k-2… $= (2) 2^k 2^k \\pi^k \\frac{k!}{(2k+1)!} = 2 \\frac{(4 \\pi)^k k!}{(2k+1)!}$ which is the required formula. So to summarize: $V_{2k} = \\frac{\\pi^k}{k!} R^{2k}$ $V_{2k+1}= \\frac{2 k! (4 \\pi)^k}{(2k+1)!}R^{2k+1}$ Note the following: $lim_{k \\rightarrow \\infty} c_{k} = 0$. If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume $(2R)^n$. Then consider the ratio $\\frac{2^n R^n}{c_n R^n} = 2^n \\frac{1}{c_n}$; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the $2^n$ corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of dimensions increases. One also notes that for fixed radius R, $lim_{n \\rightarrow \\infty} V_n = 0$ as well. There are other interesting aspects to this", "+0 # Can someone help me find the solution set for:z/z-5+5/z+5=50/z^2-25 0 153 1 what is the solution set for z/z-5+5/z+5=50/z^2-25 Guest Apr 9, 2017 Sort: #1 +4728 +3 I think this isn't your question: $$\\frac{z}{z}-5+\\frac{5}{z}+5=\\frac{50}{z^2}-25$$ I think this is your question: $$\\frac{z}{z-5}+\\frac{5}{z+5}=\\frac{50}{z^2-25}$$ Here, take note of what z values cause a zero in the denominator. z ≠ 5, z ≠ -5 Get a common denominator on the left side $$\\frac{z(z+5)}{(z-5)(z+5)}+\\frac{5(z-5)}{(z-5)(z+5)}=\\frac{50}{z^2-25}$$ Factor the denominator on the right side. $$\\frac{z(z+5)}{(z-5)(z+5)}+\\frac{5(z-5)}{(z-5)(z+5)}=\\frac{50}{(z-5)(z+5)}$$ Multiply through by (z-5)(z+5). $$z(z+5)+5(z-5)=50$$ Distribute and combine like terms. $$z^2+5z+5z-25=50 \\\\ z^2+10z-75=0$$ Factor and set each factor = 0. $$z^2+10z-75=0 \\\\ (z+15)(z-5)=0$$ z+15 = 0 or z-5 = 0 z = -15 or z = 5 HOWEVER...in the original problem, z = 5 causes a zero in the denominator. So, z cannot = 5. The solution set is just: { -15 } hectictar Apr 10, 2017 ### 20 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "1), b2 = k, c2 = -(3k + 4) For infinite number of solutions we must have, $\\frac{a_{1}}{a_{2}} = \\frac{b_{1}}{b_{2}} = \\frac{c_{1}}{c_{2}}$ This holds only when $\\frac{5}{2(k + 1)} = \\frac{2}{k} = \\frac{-2k}{-(3k + 4)}$ $\\frac{5}{2(k + 1)} = \\frac{2}{k} = \\frac{2k}{(3k + 4)}$ Now, the following cases arises Case 1: $\\frac{5}{2(k + 1)} = \\frac{2}{k}$ $\\Rightarrow 5k = 4(k + 1) \\Rightarrow 5k = 4k + 4$ $\\Rightarrow k = 4$ Case 2: $\\frac{2}{k} = \\frac{2k}{(3k + 4)}$ $\\Rightarrow 2(3k + 4) = 2k^{2} \\Rightarrow 6k + 8 = 2k^{2}$ $\\Rightarrow 2k^{2} – 6k – 8 = 0$ On solving the above quadratic equation, we get k = 4 or k = -1 Case 3: $\\frac{5}{2(k + 1)} = \\frac{2k}{(3k + 4)}$ $\\Rightarrow 15k + 20 = 4k^{2} = 4k$ $\\Rightarrow 4k^{2} + 4k – 15 – 20 = 0$ On solving the above quadratic equations we get, $k = 4 or k = \\frac{-5}{4}$ Thus, k = 4 is a common value for which there are infinitely many solutions. Question 9: x+ (k – 1) y = 5, (k + 1)x + 9y = -(8k – 1) Solution: x + (k + 1)y – 5 = 0 ……..(1) (k + 1)x + 9y – (8k – 1) = 0 ……..(2) a1 = 1, b1 = (k", "# Missing factor equal to 0 after factoring The equation $$x^{10}+(13x-1)^{10}=0$$ has $$10$$ complex roots, $$r_i$$ for integers $$1\\le i\\le5$$. Find $$\\sum^5_{i=1}\\frac{1}{r_i\\overline{r_i}}.$$ My solution was simply to factor the given polynomial into $$(r^2+(13r-1)^2)(r^8-\\dots+(13r-1)^8)=0.$$ We suppose the left factor is $$0,$$ such that $$170r^2-26r+1=0.$$ By Vieta's, the product of the roots if $$1/170.$$ Since this product is the product of the conjugate pairs, we can multiply $$170\\times5$$ to find our answer, and $$850$$ turns out to be correct. However, how come we don't need to account for the right factor here? We need not consider it if it were never equal to $$0$$, so how do we prove that? We solve equation $$x^{10}+(13x-1)^{10}=0$$. $$\\left(13-\\frac{1}{x}\\right)^{10}=-1$$ $$13-\\frac{1}{x_k}=\\cos\\frac{(2k+1)\\pi}{10}+i\\sin\\frac{(2k+1)\\pi}{10},$$ $$\\frac{1}{x_k}=13-\\cos\\frac{(2k+1)\\pi}{10}-i\\sin\\frac{(2k+1)\\pi}{10},\\quad k=0\\ldots9.$$ Then $$\\frac{1}{x_k\\bar x_k}=170-26\\cos\\frac{(2k+1)\\pi}{10}$$ $$\\sum_{k=0}^4\\frac{1}{x_k\\bar x_k}=850-26\\left(\\cos\\frac{\\pi}{10}+\\cos\\frac{3\\pi}{10}+\\cos\\frac{7\\pi}{10}+\\cos\\frac{9\\pi}{10}\\right)=850$$", "+0 # I need help, thanks in advance. +4 71 2 +294 The roots of the equation $$2x^2-mx+n=0$$ sum to 6 and multiply to 10. What is the value of $$m+n$$? Feb 20, 2021 #1 +944 +3 We can use Vieta's Formulas. The sum of the roots is $$-\\frac{b}{a} = \\frac{m}{2}$$ and the product of the roots is $$\\frac{c}{a} = \\frac{n}{2}$$. The sum is 6, so $$\\frac{m}{2}=6$$ and m = 12. The product is 10, so $$\\frac{n}{a} = 10$$ and n = 20. So, $$m+n=12+20 = \\boxed{32}$$ Feb 20, 2021 #1 +944 +3 We can use Vieta's Formulas. The sum of the roots is $$-\\frac{b}{a} = \\frac{m}{2}$$ and the product of the roots is $$\\frac{c}{a} = \\frac{n}{2}$$. The sum is 6, so $$\\frac{m}{2}=6$$ and m = 12. The product is 10, so $$\\frac{n}{a} = 10$$ and n = 20. So, $$m+n=12+20 = \\boxed{32}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +4 Thank you! calvinbun Feb 20, 2021", "equations yield a uniq [#permalink] ### Show Tags 06 Dec 2019, 07:35 Bunuel wrote: For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive? $$(x - k)^2 - y^2 = 8$$ $$(x + 2k)^2 + (y^2 - k^2) = 0$$ (A) $$\\frac{-4}{\\sqrt{7}}$$ (B) $$\\frac{\\sqrt{7}}{4}$$ (C) $$\\sqrt{2}$$ (D) $$\\frac{4}{\\sqrt{7}}$$ (E) None of these x = unique positive solution, so $$b^2-4ac=0$$ and b<0; $$(x - k)^2 - y^2 = 8$$ $$(x + 2k)^2 + (y^2 - k^2) = 0$$ $$[x^2+k^2-2xk-y^2=8]+[(x + 2k)^2 + y^2 - k^2 = 0]$$ $$[x^2-2xk]+[(x + 2k)^2]=8…x^2-2xk+x^2+4k^2+4xk=8$$ $$2x^2+2xk+4k^2=8…x^2+xk+2k^2=4…x^2+xk+(2k^2-4)=0$$ $$b^2-4ac=0…(xk)^2-4(x^2)(2k^2-4)=0…(xk)^2=4(x^2)(2k^2-4)$$ $$k^2=4(2k^2-4)…8k^2-16-k^2=0…7k^2=16…k^2=16/7$$ $$|k|=[4/√7,-4/√7]…[x^2+xk+(2k^2-4)=0]…[xk]=[b<0]…k=-4/√7$$ Ans (A) Senior Manager Status: Studying 4Gmat Joined: 02 Jan 2016 Posts: 285 Location: India Concentration: Strategy, Entrepreneurship GMAT 1: 590 Q37 V33 GPA: 4 WE: Law (Manufacturing) For which value of k does the following pair of equations yield a uniq [#permalink] ### Show Tags 20 Dec 2019, 05:30 chetan2u Can you help me with this question please. Intern Joined: 18 Feb 2019 Posts: 16 Re: For which value of k does the following pair of equations yield a uniq [#permalink] ### Show Tags 22 Dec 2019, 00:21 exc4libur wrote: Bunuel wrote: For which value of k does the following pair of equations yield a unique solution for x, such that", "+0 0 90 1 a and b are the roots of 2x^2 - 3x - 1 = 0. Find a/b + b/a. Jun 24, 2020 #1 +307 0 There's not a simple way to factorize $$2x^2 - 3x - 1 = 0$$, so instead we simplify $$\\frac{a}{b}+\\frac{b}{a}=\\frac{a^2+b^2}{ab}=\\frac{(a+b)^2-2ab}{ab}=\\frac{(a+b)^2}{ab}-2$$ Using Vieta's Formula, a+b = 3/2 and ab = -1/2. Plug these in to get $$\\frac{(3/2)^2}{-1/2}-2=\\frac{9/4}{-1/2}-2=-\\frac{9}{2}-2=\\frac{-13}{2}$$ Jun 24, 2020", "factor. This leads to: $c_{2k+2} = 2^2(\\frac{\\pi}{4})\\frac{2}{2k+2} \\frac{\\pi^k}{k!} = \\frac{\\pi^{k+1}}{(k+1)!}$ as required. Now we need to calculate $c_{2k+1} = 2\\int^{\\frac{\\pi}{2}}_0(cos^{2k+1}(t))dt c_{2k} = 2\\int^{\\frac{\\pi}{2}}_0(cos^{2k+1}(t))dt \\frac{\\pi^k}{k!}$ $= 2 (\\frac{2k}{2k+1})(\\frac{2k-2}{2k-1})......(\\frac{4}{5})\\frac{2}{3}\\frac{\\pi^k}{k!} = 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \\frac{\\pi^k}{k!}$ To simplify this further: split up the factors of the $k!$ in the denominator and put one between each denominator factor: $= 2 (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \\pi^k$ Now multiply the denominator by $2^k$ and put one factor with each $k-m$ factor in the denominator; also multiply by $2^k$ in the numerator to obtain: $(2) 2^k (\\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \\pi^k$ Now gather each factor of 2 in the numerator product of the 2k, 2k-2… $= (2) 2^k 2^k \\pi^k \\frac{k!}{(2k+1)!} = 2 \\frac{(4 \\pi)^k k!}{(2k+1)!}$ which is the required formula. So to summarize: $V_{2k} = \\frac{\\pi^k}{k!} R^{2k}$ $V_{2k+1}= \\frac{2 k! (4 \\pi)^k}{(2k+1)!}R^{2k+1}$ Note the following: $lim_{k \\rightarrow \\infty} c_{k} = 0$. If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume $(2R)^n$. Then consider the ratio $\\frac{2^n R^n}{c_n R^n} = 2^n \\frac{1}{c_n}$; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the $2^n$ corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of", "# Find $K$ in the equation $24x^3-14x^2-63x+K=0$ if one root is twice more the another. I am given the equation $24x^3-14x^2-63x+K=0$ and I am asked to find the values for $K$ If one root is double the other root. How would I solve this? I am trying to solve this by taking two roots as A,2A and the third B then by using sum and products of roots I wrote three equations. • did you try dividing by $24$ and using vietta? – Jorge Fernández Hidalgo Aug 1 '16 at 18:25 • @CarryonSmiling what is vietta? – danny Aug 1 '16 at 18:28 • wikiwand.com/en/Vieta's_formulas – alans Aug 1 '16 at 18:28 • I am thinking of solving this by using sum and product of roots – danny Aug 1 '16 at 18:29 • @CarryonSmiling, Perhaps Danny is an algebra student from secondary school. I highly doubt they would be required to know Vietta's formulas. – KingDuken Aug 1 '16 at 18:38 By using Viete's formulae for roots $x$, $2x$, $y$ of equation $24x^3-14x^2-63x+K=0$, we get system of equations $$3x+y=\\frac{7}{12},$$ $$3xy+2x^2=-\\frac{21}{8},$$ $$K=-48x^2y.$$ From the first equation $y=\\frac{7}{12}-3x$, substitute this into second equation in order to get $8x^2-2x-3=0$. Now it is easy to conclude that solutions are $$(x,y)\\in\\{(\\frac{3}{4},-\\frac{5}{3}),(-\\frac{1}{2},\\frac{25}{12})\\}.$$ Finally, calculate $K$ from the third equation, for those values $x$", "+ 3x -5$. (Note that $x_{i}'s$ are distinct as $x_{i} = x_{j}$ would imply $y_{i}=y_{j}$). The above equation reduces to an equation of degree 4, namely, $2x^{4}+7x^{3}+(3-m)x - 5-c=0$. Hence, by Viete’s relations, $\\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} = - \\frac{7}{8}$. Example 4: The product of two of the four roots of $x^{4}-20x^{3}+kx^{2}+590x-1992=0$ is 24. Find k. Solution 4: Let the given equation be written as $f(x)=0$, and let the roots of the equation be $r_{1}, r_{2}, r_{3}, r_{4}$ with $r_{1}r_{2}=24$. Now, $r_{1}r_{2}r_{3}r_{4}= -1992$, so $r_{3}r_{4} = -1992/24 = -83$. Also, $f(x) = (x-r_{1})(x-r_{2})(x-r_{3})(x-r_{4}) = (x^{2}-cx+r_{1}r_{2})(x^{2}-dx+r_{3}r_{4}) = (x^{2}-cx+24)(x^{2}-dx-83)$ with $c = r_{1} + r_{2}$, $d = r_{3} + r_{4}$. Comparing coefficients of $x^{3}$ and x we get $c+d=20$ and $83c - 24d = 590$. This gives $c = 10, d = 10$. Comparing coefficients of $x^{2}$, $k = cd - 83 + 24 = 100 -83 +24 =41$. Example 5: If $\\alpha$ and $\\beta$ are roots of $x^{2}+px+q=0$, where p and q are integers with $q|p^{2}$, then show that (i) $\\alpha^{n} + \\beta^{n}$ is an integer. ($n \\geq 1$) (ii) $\\alpha^{n} + \\beta^{n}$ is an integer divisible by q. ($n \\geq 2$). Solution 5: Since $\\alpha, \\beta$ are the roots of $x^{2}+px+q=0$, we get $\\alpha + \\beta = -p$ —– Equation A $\\alpha \\beta = q$ —– Equation B Note that $\\alpha^{2}", "# What are the value of k, for which x^2+5kx+16=0 has no real roots? Nov 2, 2016 The values of k for which the equation ${x}^{2} + 5 k x + 16 = 0$ has no real roots are $- \\frac{8}{5} < k < \\frac{8}{5}$ #### Explanation: An equation $a {x}^{2} + b x + c = 0$ has no real roots if the determinant ${b}^{2} - 4 a c < 0$ ${x}^{2} + 5 k x + 16 = 0$ has no real roots if the determinant ${\\left(5 k\\right)}^{2} - 4 \\times 1 \\times 16 = 25 {k}^{2} - 64 < 0$ or $\\left(5 k - 8\\right) \\left(5 k + 8\\right) < 0$ As $\\left(5 k - 8\\right) \\left(5 k + 8\\right)$ is negative, either $5 k + 8 < 0$ and $5 k - 8 > 0$ But this is just not possible, as $k$ cannot be $k < - \\frac{8}{5}$ as well as $k > \\frac{8}{5}$ Other alternative is $5 k + 8 > 0$ and $5 k - 8 < 0$ i.e. $k > - \\frac{8}{5}$ as well as $k < \\frac{8}{5}$. This is possible if value of $k$ is between $- \\frac{8}{5}$ and $\\frac{8}{5}$. Hence $- \\frac{8}{5} < k < \\frac{8}{5}$", "x\\right)\\right]$ (a) <2 (b) > 2 (c) > 2 (d) None 15. n[P[P[p(Ø)]]] = (a) 2 (b) 1 (c) 4 (d) 8 16. If A, B and C are three sets and if A ∈ B and B ⊂ C then (a) A ⊂ C (b) A need not be a subset of C (c) A = B (d) C ⊂ A 17. The solution 5x-1<24 and 5x+1 > -24 is (a) (4,5) (b) (-5,-4) (c) (-5,5) (d) (-5,4) 18. If a and b are the roots of the equation x2-kx+16=0 and a2+b2=32 then the value of k is (a) 10 (b) -8 (c) -8,8 (d) 6 19. If $\\frac { 1-2x }{ 3+2x-{ x }^{ 2 } } =\\frac { A }{ 3-x } +\\frac { B }{ x+1 }$ ,then the value of A+B is (a) $\\frac { -1 }{ 2 }$ (b) $\\frac { -2 }{ 3 }$ (c) $\\frac { 1 }{ 2 }$ (d) $\\frac { 2 }{ 3 }$ 20. The number of roots of (x+3)4+(x+5)4=16 is (a) 4 (b) 2 (c) 3 (d) 0 21. If x < 7,then (a) -x < -7 (b) - x ≤ -7 (c) -x > -7 (d) -x ≥ -7 22. The rationalising factor of $\\frac { 5 }{ \\sqrt [ 3 ]{ 3 }", "get, (1 – 0.02)5 = [ 5C0 × (1)5 ] – [ 5C1 × (1)4 × (0.02) ] + [ 5C2 × (1)3 × (0.02)2 ] – [ 5C3 × (1)2 × (0.02)3 ] = 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008] = 1 – 0.10 + 0.0020 + 0.000040 = 0.90204 Therefore we get, (0.98)5 = 0.90204 Q.8: The coefficient of the 5th term from end and the 5th term from the beginning in the expansion of $$\\left ( 2^{\\frac{1}{4}}+\\frac{1}{3^{\\frac{1}{4}}} \\right )^{n}$$ are in the ratio $$1:\\sqrt{6}$$.Find the value of ‘n’. Answer. Based on formula given in Binomial Theorem Since, The general term in the expansion of (a + b)n is given by: Tk + 1 = nCk × (a)n – k × bk Now we get on substituting a = $$\\left ( 2\\right )^{{\\frac{1}{4}}}$$ and b = $$\\frac{1}{(3)^{\\frac{1}{4}}}$$ in the above equation: Tk + 1 = nCk ×$$\\left ( \\sqrt[4]{2} \\right )^{n-k} \\times \\left ( \\frac{1}{\\sqrt[4]{3}}\\right )^{k}\\\\$$ Now we get, 5th term from beginning in the expansion of $$\\left ( 2^{\\frac{1}{4}}+\\frac{1}{3^{\\frac{1}{4}}} \\right )^{n}$$: T(4 + 1) = nC4 × $$\\left ( \\sqrt[4]{2} \\right )^{n-4} \\times \\left ( \\frac{1}{\\sqrt[4]{3}} \\right )^{4}\\\\$$ i.e. T5 = nC4 × $$\\left ( 2 \\right )^{\\frac{n}{4}-1} \\times \\left ( \\frac{1}{3}", "+0 # Algebra 0 28 1 Let a and b be the solutions to 5x^2 - 11x + 4 = 0. FInd 1/a^2 + 1/b^2. Jan 11, 2022 #1 +26 0 Note that $$\\frac{1}{a^2}+\\frac{1}{b^2}=\\frac{a^2+b^2}{a^2b^2} =\\frac{(a+b)^2 - 2ab} {(ab)^2}$$. By Vieta's, $$a+b = \\frac{-(-11)}{5} = \\frac{11}{5}$$ and $$ab = \\frac{4}{5}$$, so $\\frac{(a+b)^2 - 2ab}{(ab)^2} = \\frac{\\frac{121}{25} - 2 \\cdot \\frac{4}{5}}{\\left(\\frac{4}{5}\\right)^2} = \\frac{\\frac{81}{25}}{\\frac{16}{25}} = \\boxed{\\frac{81}{16}}.$ Jan 11, 2022", "-30 + 10 \\sqrt{ 2 } }{ 50 }$ = $\\frac{ 10 ( -3 + \\sqrt{ 2 } ) }{ 50 } = \\frac{ ( -3 + \\sqrt{ 2 } ) }{ 5 }$ $\\beta$ = $\\frac{ -b – \\sqrt{ D } } { 2a }$ = $\\frac{ -30 – \\sqrt{ 200 } }{ 2 \\times 25} = \\frac{ -30 – 10 \\sqrt{ 2 } }{ 50 }$ = $\\frac{ 10 ( -3 – \\sqrt{ 2 } ) }{ 50 } = \\frac{ ( -3 – \\sqrt{ 2 } ) }{ 5 }$ Thus, the roots of the equation are $\\frac{ ( -3 + \\sqrt{ 2 } ) }{ 5 }$ and $\\frac{ ( -3 – \\sqrt{ 2 } ) }{ 5 }$ . Q6) $16x^{ 2 } = 24x + 1$ Ans. 6) Given: $16x^{ 2 } = 24x + 1$ $\\Rightarrow 16x^{ 2 } – 24x – 1 = 0$ On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get: a = 16, b = -24 and c = -1 Discriminant D is given by: D = $b ^{ 2 } – 4ac$ = $( -24 )^{ 2 } – 4 \\times 16 \\times ( -1 )$ = 576 + ( 64 ) = 640 > 0 Hence, the roots", "# Math Help - Equal Real Roots 1. ## Equal Real Roots f(x) = -1/2x^2 -5x +4 find the value(s) of k such that the equation f(x+k)=0 has two equal real roots. 2. Originally Posted by casey_k f(x) = -1/2x^2 -5x +4 find the value(s) of k such that the equation f(x+k)=0 has two equal real roots. $f(x+k) = \\frac{-1}{2}(x+k)^2 -5(x+k) + 4$ $f(x+k) = \\frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4$ $f(x+k) = \\frac{-1}{2}x^2-\\frac{1}{2}2kx-\\frac{1}{2}k^2 -5x-5k + 4$ $f(x+k) = \\frac{-1}{2}x^2+x(-k-5) - \\frac{k^2}{2} -5k + 4$ The discriminant must be zero for equal roots: $b^2-4ac = 0$ $a = \\frac{-1}{2}$ $b = (-k-5)$ $c = - \\frac{k^2}{2} -5k + 4$ Insert these into the equation. Expand. Then solve for k. 3. Originally Posted by Mush $f(x+k) = \\frac{-1}{2}(x+k)^2 -5(x+k) + 4$ $f(x+k) = \\frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4$ $f(x+k) = \\frac{-1}{2}x^2-\\frac{1}{2}2kx-\\frac{1}{2}k^2 -5x-5k + 4$ $f(x+k) = \\frac{-1}{2}x^2+x(-k-5) - \\frac{k^2}{2} -5k + 4$ The discriminant must be zero for equal roots: ** $b^2-4ac = 0$ ** $a = \\frac{-1}{2}$ ** $b = (-k-5)$ ** $c = - \\frac{k^2}{2} -5k + 4$ Insert these into the equation. Expand. Then solve for k. Problem is that no $k$ will solve this set of equations (**). If you look at the graph of $f(x)$, $f(x+k)$ only translates the graph left and right. There will always be", "# For what values of $c$ does $8x + 5y = c$ have exactly one strictly positive integer solution? We know that any Diophantine equation of the form $ax + by = c$ has either no solutions, or infinite solutions of the form: $$x = x_0 + n\\frac{b}{(a, b)}$$ $$y = y_0 - n\\frac{a}{(a, b)}$$ Where $n$ is any integer, $(x_0, y_0)$ is one solution, and $(a, b)=\\gcd(a, b)$. I use this to attempt to answer the question to the best of my abilities, but I get stuck at one part: By inspection, we can see that $8(2) + 5(-3) = 1$, and so $8(2c) + 5(-3c) = c$. Using the above equations, we find that: $$x=2c + 5n$$ $$y=-3c - 8n$$ Knowing that both of these values will be above zero, we can set: $$x >0$$ $$2c + 5n >0$$ $$n > \\frac{-2c}{5}$$ And: $$y>0$$ $$-3c-8n>0$$ $$n<\\frac{-3c}{8}$$ Thus, $\\frac{-2c}{5}<n<\\frac{-3c}{8}$ will yield positive solutions. However, we only want one positive solution. Here's where I get stuck. Am I able to simply say $\\frac{-2c}{5}+1 =\\frac{-3c}{8}$? Because it may be that the difference is slightly more than $1$, and so the value of $c$ would differ; $\\frac{-2c}{5}$isn't necessarily an integer. Moreover, when I do attempt to say $\\frac{-2c}{5}+1 =\\frac{-3c}{8}$ anyways, I finish with $-16 < n < -15$, which leaves", "Factor k^{2}+\\frac{9}{2}k+\\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \\left(x+\\frac{b}{2}\\right)^{2}. \\sqrt{\\left(k+\\frac{9}{4}\\right)^{2}}=\\sqrt{\\frac{25}{16}} Take the square root of both sides of the equation. k+\\frac{9}{4}=\\frac{5}{4} k+\\frac{9}{4}=-\\frac{5}{4} Simplify. k=-1 k=-\\frac{7}{2} Subtract \\frac{9}{4}=2.25 from both sides of the equation.", "+0 # Algebra 0 91 1 Let a and b be the solutions to 5x^2 - 11x + 4 = 0. FInd 1/a^2 + 1/b^2. Jan 11, 2022 Note that $$\\frac{1}{a^2}+\\frac{1}{b^2}=\\frac{a^2+b^2}{a^2b^2} =\\frac{(a+b)^2 - 2ab} {(ab)^2}$$. By Vieta's, $$a+b = \\frac{-(-11)}{5} = \\frac{11}{5}$$ and $$ab = \\frac{4}{5}$$, so $\\frac{(a+b)^2 - 2ab}{(ab)^2} = \\frac{\\frac{121}{25} - 2 \\cdot \\frac{4}{5}}{\\left(\\frac{4}{5}\\right)^2} = \\frac{\\frac{81}{25}}{\\frac{16}{25}} = \\boxed{\\frac{81}{16}}.$" ]

[ "# Find best constant Algebra Level 5 Find the largest constant $$k$$ such that for all non-negative real numbers $$a, b, c$$, $(a+b+c)^5 \\geq \\frac{ k } { \\sqrt{5} }(ab+bc+ca)(a-b)(b-c)(c-a).$ ×", "# Let's Meet in #1 Level pending Let $$a$$ and $$b$$ $$(a < b)$$ be constants such that for any real number $$m$$ satisfying $$a < m < b$$, the two lines $y=-2x+2, \\;\\; y=mx-2m+4$ intersect in the first quadrant. What is the largest possible value of $$b-a$$? ×", "# Thread: Continuity - Find the largest value of c? 1. ## Continuity - Find the largest value of c? Let $f_n(x) = \\cos^n x$ and $g(x) = \\lim_{n\\rightarrow \\infty} \\sum_{k = 0}^{n} \\left(\\frac{x}{4}\\right)$ If $g(x)$ is continuous in $(0,c)$ then find the largest value of $c$. 2. The summation needs correcting - there's nothing depending on $k$.", "+0 # Help! 0 237 1 +1038 If $P(x) = 4+2\\sqrt{x+2}$ and $G(x) = 4-3x$, then what is the largest constant $a$ such that $P(G(a))$ is defined? Jul 15, 2018 If $P(x) = 4+2\\sqrt{x+2}$ and $G(x) = 4-3x$, then what is the largest constant $a$ such that $P(G(a))$ is defined?", "+0 # HELP!! 0 438 2 Find the largest real number $x$ for which there exists a real number $y$ such that $x^2 + y^2 = 2x + 2y$. Mar 12, 2019 #1 +234 +2 1 + $$\\sqrt2$$ . My method isn't really orthodox and required a little bit of eyeballing it, but I'm sure it's right. Add the right side to the left, and plug it into a graphing calculator, because it's a formula for a circle. This is where the eyeballing comes in, but the largest value for x is when y equals 1. Substitute this into your equation, and you get x2 + 1 -2x - 2 = x2 - 2x -1. Using the quadratic formula, you get 1 - $$\\sqrt 2$$. and 1 + $$\\sqrt 2$$. We want the larger value for x, so the largest value of x = 1 + $$\\sqrt 2$$ . Sorry this isn't an official method, but this is the best I could do. Hope this helps! Mar 12, 2019 #2 +23531 +3 Find the largest real number $$x$$ for which there exists a real number $$y$$ such that $$x^2 + y^2 = 2x + 2y$$. $$\\text{Let the largest real number x = x_{\\text{max}}} \\\\ \\text{Let the x-center of the circle = x_c } \\\\ \\text{Let the y-center of", "# 6000th Brilliant Problem Solved - Problem Geometry Level 5 Find the largest constant $$m$$ and the smallest constant $$M$$, such that given any 13 distinct real numbers, there always exists 2 numbers $$x$$ and $$y$$, with $$x > y$$, such that $m < \\dfrac {x-y}{1+xy} < M.$ If the value of $$m+M$$ can be written as $$a+b\\sqrt{c}$$, where $$a$$, $$b$$ and $$c$$ are integers, and $$c$$ is square-free, find $$a+b+c$$. ×", "# What? Cauchy? AM-GM? Max again? Algebra Level 5 Find the value of $$\\lceil 1000C \\rceil$$, where $$C$$ satisfies the following: $$C$$ is the largest constant such that for all real numbers $$x_1, x_2, \\ldots, x_{913}$$ that satisfy $$x_i \\in (0,1)$$ where $$i = 1,2, \\ldots, n$$ and $$(1-x_i)(1-x_j) \\geq \\frac{1}{4}$$ where $$1 \\leq i < j \\leq n$$, we have: $$\\sum_{i=1}^n x_i \\geq C \\sum_{1 \\leq i < j \\leq n} (2x_ix_j + \\sqrt{x_ix_j})$$ ×", "+0 # polynominal question 0 120 1 Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $a$ be a constant. What is the largest possible degree of $f(x) + a\\cdot g(x)$? Nov 15, 2021", "+0 # Help 0 85 1 Let f(x) = x^4-3x^2 + 2 and g(x) = 2x^4 - 6x^2 + 2x -1. Let a be a constant. What is the largest possible degree of f(x) + a\\cdot g(x)? Apr 14, 2021", "and b are integers such that $2{ x }^{ 2 }-ax+2>0$ and ${ x }^{ 2 }-bx+8\\ge 0$ for all real numbers x, then the largest possible value of 2a-6b is", "and then $n\\ge {B}_{ϵ}=B{\\sqrt{ϵ}}^{\\frac{1}{1-\\mu }}$ for some constant B. Therefore, there exists a constant C such that forall $ϵ$ small enough, the k-th integer n such that $1-ϵ\\le \\mathrm{sin}n$ is greater than ${C}_{ϵ}$ Do you have a similar question?", "Find real constants $A, B$ and $C$ and complex constants $D$ and $E$ such that $${10x^2-2x+4\\over x^3 + x} = {A\\over x} +{Bx+C\\over x^2+1} = {A\\over x} + {D\\over x-i} + {E\\over x+i}.$$", "+0 # What is the largest possible degree of $f(x) + a\\cdot g(x)$? 0 250 1 Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $a$ be a constant. What is the largest possible degree of $f(x) + a\\cdot g(x)$? Aug 15, 2020", "Question 16 # The largest real value of a for which the equation $$\\mid x + a \\mid + \\mid x - 1 \\mid = 2$$ has an infinite number of solutions for x is Solution In the question, it is given that the equation $$\\mid x + a \\mid + \\mid x - 1 \\mid = 2$$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out. Case 1: x + a < 0, x - 1 $$\\ge\\$$ 0 - a - x + x - 1 = 2 a = -3 Case 2: x + a $$\\ge\\$$ 0 and x - 1 < 0 x + a - x + 1 = 2 a = 1 Largest value of a is 1.", "# Champernowne constant The Champernowne constant is the real number $$C_{10}=0.12345678910111213141516171819202122232425\\ldots$$", "22 views | 22 views 0 You can straight away eliminate B and C as 2^100 has to be the smallest, as it is constant. $2^{\\log n}$ is the same as $n$, so it is smaller than $n \\log n$. 0 Yes... I Had also got the same answer. But Ace Solutions had A as option. Hence posted it for discussion.", "## mpj4 one year ago Sketch the largest region on which the function f(x,y) = y / (11x^2 + 7) is continuous. hint: the denominator can not be equal to zero, namely: $\\Large 11{x^2} + 7 \\ne 0,\\;\\forall x \\in \\mathbb{R}$", "$f(t) = \\int_{0}^{t} [x] dx$, where $[x]$ denotes the largest integer less than or equal to $x$. (i) Determine all the real numbers $t$ at which $f$ is differentiable. (ii) Determine all the real numbers $t$ at which $f$ is continuous but not differentiable.", "digits for the largest of $a, b, c$. One solution is found here. (Thanks to Oleg567). a=1218343242702905855792264237868803223073090298310121297526752830558323845503910071851999217959704024280699759290559009162035102974023; b=2250324022012683866886426461942494811141200084921223218461967377588564477616220767789632257358521952443049813799712386367623925971447; c=20260869859883222379931520298326390700152988332214525711323500132179943287700005601210288797153868533207131302477269470450828233936557.", "+0 # Help! 0 137 1 +851 If $P(x) = 4+2\\sqrt{x+2}$ and $G(x) = 4-3x$, then what is the largest constant $a$ such that $P(G(a))$ is defined? Lightning Jul 15, 2018 #1 +9740 +2 If $P(x) = 4+2\\sqrt{x+2}$ and $G(x) = 4-3x$, then what is the largest constant $a$ such that $P(G(a))$ is defined? Omi67 Jul 16, 2018" ]

[ "inequality: $$\\Gamma\\left(\\frac{\\alpha+\\beta}{2}\\right)^2 = \\left(\\int_{0}^{+\\infty}x^{\\alpha+\\beta}\\frac{e^{-x}\\,dx}{x}\\right)^2\\leq \\Gamma(\\alpha)\\,\\Gamma(\\beta).$$ A nice consequence of the log-convexity is Gautschi's inequality: $$x^{1-s} \\leq \\frac{\\Gamma(x+1)}{\\Gamma(x+s)}\\leq (1+x)^{1-s}$$ that easily proves that your limit is $\\color{red}{\\large 1}$ by squeezing.", "# For what x an y is y / ( x^3 + 3 )^2 > 2/(x/y-y-3)? Jul 11, 2016 See below #### Explanation: Compacting the inequality we have -(y (18 - x + 12 x^3 + 2 x^6 + 3 y + y^2))/((3 + x^3)^2 (x - 3 y - y^2) )>0 or $- \\frac{y \\left(18 - x + 12 {x}^{3} + 2 {x}^{6} + 3 y + {y}^{2}\\right)}{x - 3 y - {y}^{2}} > 0$ because ${\\left(3 + {x}^{3}\\right)}^{2} \\ge 0 \\forall x \\in \\mathbb{R}$ The feasible set frontier is composed of ${f}_{1} \\left(x , y\\right) = - y = 0$ ${f}_{2} \\left(x , y\\right) = 18 - x + 12 {x}^{3} + 2 {x}^{6} + 3 y + {y}^{2} = 0$ and ${f}_{3} \\left(x , y\\right) = x - 3 y - {y}^{2} = 0$ The feasible region interior is the set of points $\\left\\{x , y\\right\\}$ obeying ${f}_{1} \\left(x , y\\right) {f}_{2} \\left(x , y\\right) > 0 \\mathmr{and} {f}_{3} \\left(x , y\\right) > 0$ or ${f}_{1} \\left(x , y\\right) {f}_{2} \\left(x , y\\right) < 0 \\mathmr{and} {f}_{3} \\left(x , y\\right) < 0$ This set is shown in the attached figure in light blue", "$$\\begin{eqnarray} &&\\max\\left(|x_1-y_1+x_2-y_2|,|x_1-x_2+y_1-y_2|\\right) \\\\ &=&\\max\\left(|x_1-y_2+x_2-y_1|,|x_1-y_2+y_1-x_2|\\right) \\\\ &\\ge&||x_1|-|y_1||+||x_2|-|y_2||\\;, \\end{eqnarray}$$ which isn't the case because the variables are now paired up the wrong way and the left-hand side can be made zero independently of the right-hand side. Now assume $|x_1|\\ge|x_2|$ and $|y_1|\\ge|y_2|$, and consider the stronger inequality $$\\max\\left(\\big|x_{1}-y_{1}+x_{2}-y_{2}\\big|,\\big|\\left|x_{1}-x_{2}\\right|-\\left|y_{1}-y_{2}\\right|\\big|\\right)\\ge\\big|\\left|x_{1}\\right|-\\left|y_{1}\\right|\\big|+\\big|\\left|x_{2}\\right|-\\left|y_{2}\\right|\\big|\\;,$$ which would imply the original inequality, since $|x_1-x_2+y_1-y_2|\\ge\\big|\\left|x_{1}-x_{2}\\right|-\\left|y_{1}-y_{2}\\right|\\big|$. Consider first the case where $x_2-x_1$ and $y_2-y_1$ have the same sign. Then we have $$\\begin{eqnarray} && \\max\\left(\\big|x_{1}-y_{1}+x_{2}-y_{2}\\big|,\\big|\\left|x_{1}-x_{2}\\right|-\\left|y_{1}-y_{2}\\right|\\big|\\right) \\\\ &=& \\max\\left(\\big|x_{1}-y_{1}+x_{2}-y_{2}\\big|,\\big|x_{1}-x_{2}-y_{1}+y_{2}\\big|\\right) \\\\ &=& \\max\\left(\\big|(x_{1}-y_{1})+(x_{2}-y_{2})\\big|,\\big|(x_{1}-y_{1})-(x_{2}-y_{2})\\big|\\right) \\\\ &=& |x_{1}-y_{1}|+|x_{2}-y_{2}| \\\\ &\\ge& \\big|\\left|x_{1}\\right|-\\left|y_{1}\\right|\\big|+\\big|\\left|x_{2}\\right|-\\left|y_{2}\\right|\\big|\\;. \\end{eqnarray}$$ On the other hand, if $x_2-x_1$ and $y_2-y_1$ have opposite signs, we have $$\\begin{eqnarray} && \\max\\left(\\big|x_{1}-y_{1}+x_{2}-y_{2}\\big|,\\big|\\left|x_{1}-x_{2}\\right|-\\left|y_{1}-y_{2}\\right|\\big|\\right) \\\\ &=& \\max\\left(\\big|x_{1}-y_{1}+x_{2}-y_{2}\\big|,\\big|x_{1}-x_{2}+y_{1}-y_{2}\\big|\\right) \\\\ &=& \\max\\left(\\big|(x_{1}-y_{2})+(x_{2}-y_{1})\\big|,\\big|(x_{1}-y_{2})-(x_{2}-y_{1})\\big|\\right) \\\\ &=& |x_{1}-y_{2}|+|x_{2}-y_{1}|\\;. \\end{eqnarray}$$ Since the constraints and the expression $\\big|\\left|x_{1}\\right|-\\left|y_{1}\\right|\\big|+\\big|\\left|x_{2}\\right|-\\left|y_{2}\\right|\\big|$ that this is to be compared with contain only the absolute values of the variables, we can assume without loss of generality that all variables are positive, as this will minimize the value out of all combinations of signs. Then the constraints become $x_1\\ge x_2$ and $y_1\\ge y_2$. Further, since the problem is symmetric with respect to interchange of the $x$s and $y$s, we may assume without loss of generality that $x_1\\ge y_1$. That leaves only the relation of $x_2$ to $y_1$ and $y_2$ undecided. If $x_2\\ge y_1\\ge y_2$, then $$|x_1-y_2|+|x_2-y_1|=x_1-y_1+x_2-y_2=\\big|\\left|x_{1}\\right|-\\left|y_{1}\\right|\\big|+\\big|\\left|x_{2}\\right|-\\left|y_{2}\\right|\\big|\\;.$$ If instead $y_1\\ge x_2\\ge y_2$, that only flips a sign", "beautiful thing is that the inequality still holds for noncommutative C*-algebras. For a commutative $C^*$-algebra your generalized Cauchy–Schwarz inequality holds, almost: Kaplansky, 1953 Let $\\mathscr{A}$ be a commutative $C^*$-algebra with unit. Let $\\mathscr{X}$ be a right $\\mathscr{A}$-module. Let $\\left<\\,\\cdot\\,,\\,\\cdot\\,\\right>$ be a $\\mathscr{A}$-valued inner product on $\\mathscr{X}$. Then for all $x,y\\in\\mathscr{X}$, $$\\left< x,y\\right>\\left<y,x\\right>\\ \\leq\\ \\left<x,x\\right>\\,\\left<y,y\\right>.$$ \"Almost\" because by \"$\\mathscr{A}$-valued inner product on $\\mathscr{X}$\" we mean a sesquilinear map $\\left<\\,\\cdot\\,,\\,\\cdot\\,\\right>\\colon \\mathscr{X}\\times\\mathscr{X}\\to\\mathscr{A}$ with for all $x,y\\in\\mathscr{X}$ and $a\\in \\mathscr{A}$, 1. $\\left<x,x\\right>\\geq 0$; 2. $\\left<x,x\\right>=0$ iff $x=0$; 3. $\\left<x,y\\right>^* = \\left<y,x\\right>$; 4. $\\left<x,ya\\right>=\\left<x,y\\right>a$. Note that such an inner product is not just linear in the second coordinate, but also an $\\mathscr{A}$-module map. There is a modified version for (not necessarily commutative) $C^*$-algebras, see Proposition 2.3 of [2]: Pasckke, 1973 Let $\\mathscr{A}$ be a $C^*$-algebra. Then for any $\\scr{A}$-valued inner product $\\left<\\,\\cdot\\,,\\,\\cdot\\,\\right>$ on a right $\\mathscr{A}$-module $\\mathscr{X}$, and $x,y\\in\\mathscr{X}$, $$\\left< x,y\\right>\\left<y,x\\right>\\ \\leq\\ \\left<x,x\\right>\\,\\left\\|\\left<y,y\\right>\\right\\|.$$ And there is also a version for completely positive maps, see Exercise 3.2 and 3.4 of [3]: For a $2$-positive map $f\\colon \\mathscr{A}\\to\\mathscr{B}$ between $C^*$-algebras, we have, for all $a,b\\in\\mathscr{A}$, $$f(a^*b) \\,f(b^*a)\\ \\leq\\ f(a^*a)\\, \\left\\|f(b^*b)\\right\\|.$$ [1] I. Kaplansky, Modules over operator algebras, American Journal of Mathematics, 1953 [2] W.L. Paschke, Inner product modules over $B^*$-algebras, Transactions of the American Mathematical Society, 1973 [3] V. Paulsen, Completely bounded maps and operator", "# If $x+y+z=5$ and $xy+yz+zx=3$,then least and largest value of $x$ are? [closed] If $x+y+z=5$ and $xy+yz+zx=3$,then least and largest value of $x$ are? • A hint at least of what you tried and where you got stuck, please? – dxiv Dec 23, 2016 at 5:49 Using $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ So $x^2+y^2+z^2 = (x+y+z)^2-2(xy+yz+zx) = 5^2-2\\cdot 3 = 19$ So $y^2+z^2 = 19-x^2$ and $y+z = 5-x$ Now Using $\\bf{Cauchy\\; Schwarz}$ Inequality $$(y^2+z^2)(1^2+1^2)\\geq (y+z)^2$$ So $$(19-x^2)\\cdot 2 \\geq (5-x)^2$$ So $$25+x^2-10x\\leq 38-2x^2$$ So $$3x^2-10x-13\\leq 0$$ So $$3x^2-13x+3x-13\\leq0$$ So $$(x+1)(3x-13)\\leq 0\\Rightarrow -1 \\leq x\\leq \\frac{13}{3}$$ $x(y+z) = 3 - yz \\implies x(5-x) = 3 - yz \\implies yz = x^2-5x+3\\implies x^2-5x+3 \\le \\dfrac{(y+z)^2}{4}= \\dfrac{(5-x)^2}{4}\\implies 4x^2-20x + 12 \\le 25-10x+x^2\\implies 3x^2-10x-13 \\le 0 \\implies (x+1)(3x-13) \\le 0 \\implies -1 \\le x \\le \\dfrac{13}{3} \\implies x_{\\text{max}} = \\dfrac{13}{3}, x_{\\text{min}} = -1.$ Rearranging the given equations a bit, $(x + y)^2 = (5 - z)^2$ $xy = 3 - z(x + y) = 3 - z(5 - z)$ Now, $0 ≤ (x - y)^2 = (x + y)^2 - 4xy$ We can substitute for both $x + y$ and $xy$ giving us an inequality involving only the variable $z$: $0 ≤ (x + y)^2 - 4xy = 25 - 10z + z^2 - 12 + 20z - 4z^2 = -3z^2 + 10z + 13$.", "$\\left(x+\\dfrac{1}{x}-2\\cdot\\sqrt{x}\\cdot\\dfrac{1}{\\sqrt{x}}\\right)=\\left(\\sqrt{x}+\\dfrac{1}{\\sqrt{x}}\\right)^2\\ge0$. • I really doubt that the easiest. – Henrik Feb 1 '15 at 9:12 • @Henrik: Can you tell me the reason? – user 170039 Feb 1 '15 at 9:17 • Just look at some of the other answers. No radicals for instance. – Henrik Feb 1 '15 at 9:18 If $x > 0$, then there is a number $y > 0$ such that $y^2 = x$, hence $$x + \\frac{1}{x} = y^2 + \\frac{1}{y^2} = y^2 - 2 + \\frac{1}{y^2} + 2 = \\left(y - \\frac{1}{y}\\right)^2 + 2 \\ge 0 + 2 = 2.$$ The inequality step is true because no real square is negative. Equality is therefore attained when $y = \\frac{1}{y}$, or $y = 1$; i.e., $x = 1$. By AM–GM inequality we have $$\\frac{x+\\dfrac{1}{x}}{2} \\geq \\sqrt{x \\cdot \\frac{1}{x}}=1.$$ Generalisation: for $a_i >0$, such that $a_1 a_2 \\ldots a_n =1$ we have $$a_1+a_2+\\cdots+a_n \\geq n.$$", "exactly how large $g(x) = x^3/3! - x^4/4!$ can be in $(0,4).$ We calculate that $g'(x) = -(x-3)x^2/6$ so $g$ increases when $0\\leq x\\leq 3$, the maximum occurs at $g(3)=9/8$, and then it decreases after that. This is good, because the $1+x+x^2/2$ terms obviously give at least $1$ from $x=0$, and will give us more as $x$ gets bigger. So we solve $1+x+x^2/2=9/8$ and we take the positive solution which is $\\frac{\\sqrt{5}-2}{2} \\approx 0.118.$ So the inequality is definitely true for $x\\geq 0.12$ because $g$ is at most $9/8$ and $1+x+x^2/2$ accounts for that amount in that range. Remember that $g$ was increasing between $x=0$ to $x=3$, so the largest $g$ can be in the remaining range is $g(0.12) = 873/315000 <1$, which is less than the amount $1+x+x^2/2$ gives us. So the inequality is also true for $0\\leq x\\leq 0.12$, so overall, for all $x.$ So all in all, the only trouble was for $x$ in $(0,4)$ and the contribution from the other terms was always enough to account for $x^3/3!$ when $x^4/4!$ wasn't enough. - Observe that $$e^x = f(x) + \\frac{x^5}{5!} + \\cdots$$ Then show that for $x<0$ $$f(x) > \\frac{x^5}{5!} + \\cdots$$ But $e^x > 0, \\forall x\\in R$ So, $$f(x) > e^x - f(x) \\Rightarrow 2f(x)>e^x>0$$ - Well, showing $\\,\\displaystyle{f(x)>\\frac{x^5}{5!}}\\,$ may prove to", "the unique maximum occurs at $(0,0)$ and the minimum occurs at the three corners. The left inequality: $$xy+xz+yz-2xyz=(xy+xz+yz)(x+y+z)-2xyz=\\sum_{cyc}(x^2y+x^z)+xyz\\geq0$$ The right inequality: $$xy+xz+yz-2xyz\\leq\\frac{7}{27}\\Leftrightarrow(xy+xz+yz)(x+y+z)-2xyz\\leq\\frac{7(x+y+z)^3}{27}\\Leftrightarrow$$ $$\\Leftrightarrow\\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\\geq0\\Leftrightarrow6\\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\\sum_{cyc}(x^3-xyz)\\geq0,$$ which is true by Schur and AM-GM. Here's what I think is a much simpler proof than all of the others. By the Cauchy Schwarz Inequality, $$(xy+yz+xz)^2\\leq (x^2+y^2+z^2)^2\\\\ \\implies xy+yz+xz\\leq x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz) \\\\ \\implies3(xy+xz+yz)\\leq 1 \\implies xy+yz+xz\\leq \\frac{1}{3}$$ By AM-GM, $$\\sqrt[3]{xyz}\\leq \\frac{x+y+z}{3}\\\\\\leq \\frac{1}{3} \\\\ \\implies xyz\\leq \\frac{1}{27}$$. The original inequality is equivalent to $$xy+yz+xz\\leq \\frac{7}{27}+2xyz$$ which is clear given the bounds that we just proved. Equality occurs at $$x=y=z=\\frac{1}{3}$$ • This would only work if you could prove that $\\frac{1}{27}\\leq xyz$, but you have the opposite inequality. Jun 20, 2019 at 9:46 For proving the second inequality, take $$x=a+\\frac{1}{3}, y=b+\\frac{1}{3}, z=c+\\frac{1}{3}$$. Since $$x+y+z=1$$, so $$a+b+c=0$$. Since $$x,y,z\\geq 0$$, so $$a,b,c\\geq -\\frac{1}{3}$$.By simple algebraic manipulation we get, $$xy+yz+zx-2xyz=\\frac{2}{3}(ab+bc+ca-3abc)+\\frac{7}{27}$$ We just need to show that $$ab+bc+ca-3abc\\leq 0$$. Since $$a+b+c=0$$, so $$a^2+b^2+c^2=-2(ab+bc+ca)$$ and $$a^3+b^3+c^3=3abc$$. Thus, $$ab+bc+ca-3abc=-\\frac{1}{2}(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ which is equal to $$-\\frac{1}{2}\\lbrace a^2(1+2a)+b^2(1+2b)+c^2(1+2c)\\rbrace\\leq 0$$ . • your last inequality is not true,because $a,b,c\\geq -\\frac{1}{3}$ so it can happen that some are negative inside bracket .. May 10, 2020 at 10:36 • $a\\geq -\\dfrac{1}{3}$ implies $2a\\geq -\\dfrac{2}{3}$ which implies $1+2a\\geq 1-\\dfrac{2}{3}=\\dfrac{1}{3}>0$ May 17, 2020 at 3:35", "# values of $a$ in inequality If $\\sqrt{xy}+\\sqrt[3]{xyz}<a(x+4y+4z)$ and $x,y,z>0$ then $a$ is with the help of am gm inequality $\\displaystyle \\sqrt{xy}\\leq \\left(\\frac{x+y}{2}\\right)$ and $\\displaystyle \\sqrt[3]{xyz}\\leq \\left(\\frac{x+y+z}{3}\\right)$ so $\\displaystyle \\sqrt{xy}+\\sqrt[3]{xyz}\\leq \\frac{x+y}{2}+\\frac{x+y+z}{3} = \\frac{5x+5y+2z}{6}$ want be able to go further, could some help me We'll prove that $$x+4y+4z\\geq3\\left(\\sqrt{xy}+\\sqrt[3]{xyz}\\right).$$ Indeed, by AM-GM $$4z+x+4y-3\\sqrt{xy}=4z+2\\cdot\\frac{x+4y-3\\sqrt{xy}}{2}\\geq$$ $$\\geq3\\sqrt[3]{4z\\left(\\frac{x+4y-3\\sqrt{xy}}{2}\\right)^2}=3\\sqrt[3]{z\\left(x+4y-3\\sqrt{xy}\\right)^2}.$$ Thus, it remains to prove that $$\\left(x+4y-3\\sqrt{xy}\\right)^2\\geq xy,$$ which is $$(\\sqrt{x}-2\\sqrt{y})^2(x+4y-2\\sqrt{xy})\\geq0.$$ The equality occurs for $x=4y$ and $y=4z$. Id est, $a>\\frac{1}{3}$ are all values of $a$, for which the inequality $$\\sqrt{xy}+\\sqrt[3]{xyz}<a(x+4y+4z)$$ is true for all positives $x$, $y$ and $z$.", "the inequality $$\\large \\left| {{(x+2)}^{3}} \\right|<3?$$ A $$\\large x=-3$$Hint: $$\\left| {{(-3+2)}^{3}} \\right|$$=$$\\left | {(-1)}^3 \\right |$$=$$\\left | -1 \\right |=1$$ . B $$\\large x=0$$Hint: $$\\left| {{(0+2)}^{3}} \\right|$$=$$\\left | {2}^3 \\right |$$=$$\\left | 8 \\right |$$ =$$8$$ C $$\\large x=-4$$Hint: $$\\left| {{(-4+2)}^{3}} \\right|$$=$$\\left | {(-2)}^3 \\right |$$=$$\\left | -8 \\right |$$ =$$8$$ D $$\\large x=1$$Hint: $$\\left| {{(1+2)}^{3}} \\right|$$=$$\\left | {3}^3 \\right |$$=$$\\left | 27 \\right |$$ = $$27$$ Question 37 Explanation: Topics: Laws of exponents, order of operations, interpret absolute value (Objective 0019). Question 38 #### What is the perimeter of a right triangle with legs of lengths x and 2x? A $$\\large 6x$$Hint: Use the Pythagorean Theorem. B $$\\large 3x+5{{x}^{2}}$$Hint: Don't forget to take square roots when you use the Pythagorean Theorem. C $$\\large 3x+\\sqrt{5}{{x}^{2}}$$Hint: $$\\sqrt {5 x^2}$$ is not $$\\sqrt {5}x^2$$. D $$\\large 3x+\\sqrt{5}{{x}^{{}}}$$Hint: To find the hypotenuse, h, use the Pythagorean Theorem: $$x^2+(2x)^2=h^2.$$ $$5x^2=h^2,h=\\sqrt{5}x$$. The perimeter is this plus x plus 2x. Question 38 Explanation: Topic: Recognize and apply connections between algebra and geometry (e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024). Question 39 #### 4 congruent sides Hint: The most common definition of a rhombus is a quadrilateral with 4 congruent sides. #### A center of rotational symmetry Hint: The diagonal of a rhombus separates it into two congruent isosceles triangles. The", "have for $x>0$ $$0<2x\\left(1-\\frac{y^4}{(x^2+y^2)^2}\\right)\\le 2x \\tag 2$$ while for $x<0$, we have $$2x\\le 2x\\left(1-\\frac{y^4}{(x^2+y^2)^2}\\right)<0 \\tag 3$$ Putting $(1)$, $(2)$, and $(3)$ together yields $$\\bbox[5px,border:2px solid #C0A000]{\\left|\\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\\right|\\le 2|x|}$$ which provided a tighter inequality than that one that was asked to be shown. • Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:08", "both learning from my carelessness in applying Jensens inequality and that the path forward from Bogomolny’s correction is easier than the path I actually took. Starting here – we wish to show that: $(1/2) \\sqrt{x^2 + 3} + (1/2) \\sqrt{y^2 + 3} + (1/2) \\sqrt{xy + 3} \\geq 3$ The correction shows that the expression on the left hand side is $\\geq$ than $\\sqrt{ (\\frac{x + y}{2})^2 + 3} + (1/2) \\sqrt{xy + 3}$ but since $x + y = 2$, the first piece of this expression is equal to 2 and the 2nd expression simplifies as before. So we are left with $2 + (1/2) \\sqrt{4 - (x - 1)^2}$ and this expression has a maximum of 3 at $x = 1.$ That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6. I’m grateful to Alexander Bogomolny for spotting the error in my original argument.", "2xy }{ x+y } \\le \\frac { xy }{ \\sqrt { xy } }$ How is that the same? – Cherry_Developer Oct 23 '15 at 12:33 Since $x,y>0$, you can divide by any of them (or by their square roots) on both sides of the inequality. Thus, $$\\frac{2xy}{x+y}\\le\\sqrt{xy} \\quad \\Leftrightarrow \\quad \\frac{2\\sqrt{xy}}{x+y}\\le1 \\quad \\Leftrightarrow \\quad 2\\sqrt{xy}\\le x+y.$$ You have already proved the rest. The left-hand side inequality (HGM inequality) is simply the right-hand side (AGM inequality), applied to $\\dfrac1x$ and $\\dfrac1y$, reversed.", "\\frac{x-q}{x+q}\\right)^m = \\left(\\frac{2x}{x+q} \\right)^x -1 .$$ This inequality looks more tractable. However, even here, all the classical techniques as CS, AM-GM, concavity/convexity seem to fail.", "1-(1+d)/2 = (1-d)/2$ or, assuming $d < 1$ (this is true for all $n$ if $d \\ge 1$), $n^2 > (1-d)/(2d) = 1/(2d) -1/2$. Thus $n > \\sqrt{1/2d}$ works. So, there is no largest value of $c$ such that $\\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$ (at least with this elementary method), but the inequality is true for any $c < 4$ for all large enough $n$. - Let $$f(x)=\\frac{\\sqrt{x}}{x^2+1}.$$ Then it is easy to check $$f'(x)=\\frac{-3x^2+1}{2\\sqrt{x}(x^2+1)}\\le 0$$ for $x>1$. It means that $f(x)$ is strictly decreasing. so $f(n+1)< f(n)$ or $$\\frac{\\sqrt{n+1}}{(n+1)^2+1} < \\frac{\\sqrt{n}}{n^2+1}.$$ -", "q_2+\\ldots+x_n^2 q_n) \\leq \\frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\\ldots+x_n^2)$$ Now we argue for some extremal set of $$y_i$$. By the rearrangement inequality, the maximum value of the LHS is obtained if the highest factors $$q_i$$ are chosen for the highest $$x_i^2$$. This can be achieved by letting as many $$y_i = a$$ for the highest indices $$i$$. The number $$k$$ of these settings $$y_i = a$$ is limited by $$s =y_1^2+y_2^2+\\ldots+y_{n-k}^2 + k a^2$$. The number $$k$$ can be made as large as possible if all of the previous $$y_1 = \\ldots = y_{n-k} = b$$. Then we have $$s =(n-k) b^2 + k a^2$$. This is $$k = \\frac{nb^2 -s}{b^2 - a^2}$$ or the nearest possible smaller integer. So we have to show $$\\frac {x_1^3}{y_1}+\\frac {x_2^3}{y_3}+\\ldots+\\frac {x_n^3}{y_n} \\leq \\frac {x_1^3}{b}+\\frac {x_2^3}{b}+\\ldots+ \\frac {x_{n-k}^3}{b} + \\frac {x_{n-k+1}^3}{a} + \\ldots + \\frac {x_n^3}{a} \\leq \\frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\\ldots+x_n^2)$$ or $$\\frac{\\frac {x_1^3}{b}+\\frac {x_2^3}{b}+\\ldots+ \\frac {x_{n-k}^3}{b} + \\frac {x_{n-k+1}^3}{a} + \\ldots + \\frac {x_n^3}{a}}{x_1^2+x_2^2+\\ldots+x_n^2} \\leq \\frac {a^4+b^4}{ab (a^2+b^2)}$$ Now again, for the LHS the maximum value has to be found by considering settings of the $$x_i$$. This maximal setting of the $$x_i$$ is obtained if most of the $$x_i$$ with high indices $$i$$ are chosen as high as possible, i.e. $$x_i=b$$. With the same argument as already given, this is possible for $$n-k$$ many $$b$$.", "# Math Help - Inequality 1. ## Inequality if x+y=2 then show that (x^3+y^3)(x^3)(y^3)<=2 2. Start by breaking $x^{3}+y^{3}$ down. 3. Thanks,got it.. 4. You could use the Arithmetic-Geometric mean inequality, for $x,\\;y\\ \\ge\\ 0$ $x+y=2$ $x^3+y^3=(x+y)\\left(x^2-xy+y^2\\right)=2\\left(x^2-xy+y^2\\right)$ $x^2-xy+y^2=(x+y)^2-3xy=4-3xy$ $\\left(x^3+y^3\\right)x^3\\;y^3=2(4-3xy)x^3\\;y^3\\ \\le\\ 2\\;\\;?$ $(4-3xy)x^3\\;y^3\\ \\le\\ 1\\;\\;?$ AM-GM Inequality: $\\displaystyle\\frac{x+y}{2}\\ \\ge\\sqrt{xy}\\Rightarrow\\ xy\\ \\le\\ 1$ $(4-3xy)x^3\\;y^3=\\displaystyle\\frac{4}{\\left(\\frac{1}{ x^3\\;y^3}\\right)}-\\frac{3}{\\left(\\frac{1}{x^2\\;y^2}\\right)}$ The divisor of 4 is greater than than the divisor of 3.", "× $$Q1$$ Find the largest y such that $$\\frac { 1 }{ 1+{ x }^{ 2 } } \\ge \\frac { y-x }{ y+x } \\quad for\\quad all\\quad x>0$$ $$Q2$$ Find the minimum and maximum values of $$\\frac { x+1 }{ xy+x+1 } +\\frac { y+1 }{ yz+y+1 } +\\frac { z+1 }{ zx+z+1 }$$ Note by Satyajit Ghosh 1 year ago Sort by: Applying componendo and dividendo we get $\\frac{2+x^{2}}{-x^{2}}\\geq\\frac{2y}{-2x}$ $\\frac{2}{x}+x \\geq y$ Applying $$A.M-G.M$$ $\\frac{2}{x}+x\\geq 2\\sqrt{2}$ Therefore $$y=2\\sqrt{2}$$. · 1 year ago Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1 · 1 year ago What about 2nd question? · 1 year ago He is asking for question 1 solution. 2nd was easy. · 1 year ago @Satyajit Ghosh Thank you for mentioning me! · 1 year ago 1. Cross multiplying we get $$\\dfrac{y+x}{y-x} \\ge 1+x^{2}$$ or $$\\dfrac{2}{y-x} \\ge x$$ or $$x^{2}-yx+2 \\le 0$$. This is a quadratic equation in $$x$$ and since it is $$\\le 0$$, the discriminant i.e. $$y^{2}-8$$ must also be $$\\le 0$$ or $$y \\le \\sqrt {8}$$. Therefore, the maximum value of $$y$$ is $$\\sqrt {8}$$ · 1 year ago Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered. · 1 year ago", "Then the Cartesian square $$A^2$$ has $$n^2$$ elements. The cardinalities of the power sets are given by $\\left| {\\mathcal{P}\\left( {{A^2}} \\right)} \\right| = {2^{{n^2}}},\\;\\;\\left| {\\mathcal{P}\\left( {\\mathcal{P}\\left( A \\right)} \\right)} \\right| = {2^{{2^n}}}.$ We find $$n$$ by solving the inequality ${2^{{n^2}}} \\gt {2^{{2^n}}},\\;\\; \\Rightarrow {n^2} \\gt {2^n},$ where $$n \\in \\mathbb{Z}^{+}.$$ This inequality has a unique solution $$n = 3,$$ so $$\\left| A \\right| = 3.$$ We also calculate the cardinalities of the power sets: $\\left| {\\mathcal{P}\\left( {{A^2}} \\right)} \\right| = {2^{{n^2}}} = {2^9} = 512,$ $\\left| {\\mathcal{P}\\left( {\\mathcal{P}\\left( A \\right)} \\right)} \\right| = {2^{{2^n}}} = {2^8} = 256.$", "triple completing-the-square is helpful. Using your notation, $(a-1)^2+(b-1)^2+(c-1)^2+2(a+b+c)-3=a+b+c$. This rearranges to $(a-1)^2+(b-1)^2+(c-1)^2=3-(a+b+c)$. Hence $a+b+c\\le 3$ since otherwise the sum of three squares would be negative. Consequently $a^2+b^2+c^2\\le 3$. Unfortunately, I don't have a cute trick for the opposite inequality. Let $x=a^2, y=b^2$. I want to minimize $f(x,y)=x+y+\\frac{1}{xy}$, assuming $x,y>0$. Setting the partials equal to zero, I get $1-\\frac{1}{xy^2}=0=1-\\frac{1}{yx^2}$, which has unique solution $x=y=1$. (as $x,y$ approach either 0 or $\\infty$, $f(x,y)$ grows without bound, so this is a minimum). Consequently $f(x,y)\\ge f(1,1)=3$. - Apply AMGM directly, you get $x+y+\\frac{1}{xy} \\ge 3$. – user27126 Apr 25 '13 at 4:16 D'oh! Thanks @Sanchez – vadim123 Apr 25 '13 at 4:16 Let $a=\\frac{x}{y}, b=\\frac{y}{z}, c=\\frac{z}{x}$, and let $$G=\\sqrt[3]{abc}, A = \\frac{a+b+c}{3}, Q=\\sqrt{\\frac{a^2+b^2+c^2}{3}}$$ Then the given condition is $Q^2=A$, but by the power mean inequality $$Q^2\\ge Q \\ge A \\ge G = 1$$ with equality in each case only if $a=b=c=1$, i.e. if $x=y=z$. -" ]

[ "# Let $x_0 > x_1 > x_2>x_3$ be any positive real numbers . What is the largest value of the real number k such that.. Question : Let $x_0 > x_1 > x_2>x_3$ be any positive real numbers . What is the largest value of the real number k such that $$\\log \\frac{x_0}{x_1}1993 + \\log \\frac{x_1}{x_2}1993 +\\log \\frac{x_2}{x_3} 1993 \\geq k \\log \\frac{\\log x_0 }{x_3}1993$$ How to solve such problems, please suggest thanks.... • Can you please include parentheses in the appropriate places? It's a bit difficult to understand what goes where. – Cameron Williams Feb 23 '14 at 18:05 • Are the 1993's part of the argument to $\\log$ or just a constant multiplying everything? – copper.hat Feb 23 '14 at 18:11 Let $y_i =log_{1993} \\frac{x_i-1}{x_i}$ for $i=1,2,3.$ Then the given inequality may be written as $$\\frac{1}{y_1} +\\frac{1}{y_2} +\\frac{1}{y_3} \\geq \\frac{k}{y_1 + y_2 +y_3}$$ By using arihtmetic - geometric mean inequality : $(\\frac{1}{y_1} +\\frac{1}{y_2} +\\frac{1}{y_3} ) (y_1+y_2+y_3) \\geq 3(\\sqrt[3]{\\frac{1}{y_1 y_2 y_3}})3(\\sqrt[3]{y_1 y_2 y_3}) = 9$ Equality holds if and only if $y_1 =y_2 =y_3$ or $x_0 ,x_1,x_2,x_3$ forms a geometric progression . Hence maximum value of k is 9.", "My Math Forum largest possible value Calculus Calculus Math Forum May 30th, 2009, 06:10 AM #1 Member Joined: Jun 2008 Posts: 45 Thanks: 0 largest possible value Hi! What is the largest possible value of $\\sqrt{(ax+by)^2 + (cx+dy)^2}$ when $x^2+y^2=1$? It seems that there is more clever way to solve the problem than using this standard method from mathematical analysis, besides I don't have to find x and y when maximum is reached. I definitely think I must use something of $\\frac{n}{\\frac{1}{a_1} + \\ldots + \\frac{1}{a_n}} \\leq \\sqrt[n]{a_1 \\ldots a_n} \\leq \\frac{a_1 + \\ldots + a_n}{n} \\leq sqrt{\\frac{a_1^2 + \\ldots + a_n^2}{n}}$ or $(a_1x_1 + \\ldots + a_nx_n)^2 \\leq (a_1^2 + \\ldots + a_n^2)(x_1^2 + \\ldots + x_n^2)$, but I can't get on with mine, Mathematica and Maple results. For example, taking a=1, b=2, c=3, d=4, I get $\\sqrt{30}$, but both Mathematica and Maple give $\\sqrt{15 + \\sqrt{221}}$. Okay, $\\sqrt{15 + \\sqrt{221}}$ seems to be correct, but how can get it? May 30th, 2009, 10:35 AM #2 Senior Member Joined: Nov 2007 Posts: 258 Thanks: 0 Re: largest possible value It's equivalent to maximizing $f(x,y)= (ax+by)^2+(cx+dy)^2$. The two gradients will be colinear: $\\bigtriangledown f(x,y)= (2a(ax+by) + 2c(cx+dy), 2b(ax+by) + 2d(cx+dy) = k(2x, 2y)$ $\\bigtriangledown f(x,y)= (a(ax+by) + c(cx+dy), b(ax+by) + d(cx+dy) = k(x, y)$ $x(a^2+c^2)+y(ab+cd)= kx$ $y(b^2+d^2)+x(ab+cd)= ky$", "× $$Q1$$ Find the largest y such that $$\\frac { 1 }{ 1+{ x }^{ 2 } } \\ge \\frac { y-x }{ y+x } \\quad for\\quad all\\quad x>0$$ $$Q2$$ Find the minimum and maximum values of $$\\frac { x+1 }{ xy+x+1 } +\\frac { y+1 }{ yz+y+1 } +\\frac { z+1 }{ zx+z+1 }$$ Note by Satyajit Ghosh 1 year ago Sort by: Applying componendo and dividendo we get $\\frac{2+x^{2}}{-x^{2}}\\geq\\frac{2y}{-2x}$ $\\frac{2}{x}+x \\geq y$ Applying $$A.M-G.M$$ $\\frac{2}{x}+x\\geq 2\\sqrt{2}$ Therefore $$y=2\\sqrt{2}$$. · 1 year ago Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1 · 1 year ago What about 2nd question? · 1 year ago He is asking for question 1 solution. 2nd was easy. · 1 year ago @Satyajit Ghosh Thank you for mentioning me! · 1 year ago 1. Cross multiplying we get $$\\dfrac{y+x}{y-x} \\ge 1+x^{2}$$ or $$\\dfrac{2}{y-x} \\ge x$$ or $$x^{2}-yx+2 \\le 0$$. This is a quadratic equation in $$x$$ and since it is $$\\le 0$$, the discriminant i.e. $$y^{2}-8$$ must also be $$\\le 0$$ or $$y \\le \\sqrt {8}$$. Therefore, the maximum value of $$y$$ is $$\\sqrt {8}$$ · 1 year ago Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered. · 1 year ago", "+0 # polynominal question 0 120 1 Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $a$ be a constant. What is the largest possible degree of $f(x) + a\\cdot g(x)$? Nov 15, 2021", "In fact, we know that $$\\frac{K}{2} \\lt i \\leq K$$, because we chose $$i$$ to be the largest among all representations. (If $$i \\leq \\frac{K}{2}$$ then we could let $$i' = 2i$$ and $$j' = j-1$$ to get a better representation of our number $$v$$. This would usually be a contradiction.) Anyway, putting this together with our formula, gives us that: $(j-1)K + S\\left(\\frac{K}{2}\\right) \\lt (j-1)K + S(i) \\leq (j-1)K + S(K)$ Now, we want to find $$(i,j)$$ such that $$(j-1)K + S(i) \\geq N$$. What is the smallest $$j$$ we can have? Well, we can apply the second inequality above and assume $$i = K$$ (since this is the largest choice of $$i$$ for any fixed $$j$$). From this we would get: $(j-1)K + S(K) \\geq N$ or $j \\geq \\left\\lceil\\frac{N - S(K)}{K}\\right\\rceil + 1$ And note that substituting $$i=K$$ was valid. If $$i$$ is smaller than $$K$$, then we would need an even larger value of $$j$$ to compensate. Hence, we have proven the following theorem: For any $$v = i \\cdot 2^{j-1}$$ with $$1 \\leq i \\leq K$$, if there are at least $$N$$ apples with value $$\\leq v$$, then: $j \\geq j^* := \\left\\lceil\\frac{N - S(K)}{K}\\right\\rceil + 1$ This gives us a lower bound on $$j$$ for valid $$(i,j)$$ pairs. But we are actually", "– f (y) bigr) cdot bigl (g (x) – g (y) bigr) , dx dy \\ = 2 (ba) int_a ^ bf (x) g (x) , dx – 2 left ( int_a ^ bf (x) , dx right) left ( int_a ^ bg (x) , dx right) ,. \\$\\$ If \\$ f \\$ goes up and \\$ g \\$ goes down (or vice versa), then the the inverse inequality is valid. Yes equality then holds the equals worth in \\$ (*) \\$ for all \\$ x, y in [a, b]\\$ (since \\$ f \\$ and \\$ g \\$ are supposed to be continuous). In particular \\$\\$ 0 = bigl (f (a) – f (b) bigr) cdot bigl (g (a) – g (b) bigr) \\$\\$ which means that (at least one of) \\$ f \\$ or \\$ g \\$ is constant. ## calculation – \\$ left | frac {d ^ n} {dx ^ n} (x ^ 2-1) ^ n right | sqrt frac2 pi cdot frac {2 ^ nn!} { sqrt n} cdot frac1 { sqrt[4]{1-x ^ 2}} \\$, a binding for the Legendre polynomial Question CA watch I)$$left | frac {d ^ n} {dx ^ n} (x ^ 2-1) ^ n right | sqrt frac2 pi cdot frac {2 ^ nn!} { sqrt n} cdot frac1 { sqrt[4]{1-x", "1-(1+d)/2 = (1-d)/2$ or, assuming $d < 1$ (this is true for all $n$ if $d \\ge 1$), $n^2 > (1-d)/(2d) = 1/(2d) -1/2$. Thus $n > \\sqrt{1/2d}$ works. So, there is no largest value of $c$ such that $\\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$ (at least with this elementary method), but the inequality is true for any $c < 4$ for all large enough $n$. - Let $$f(x)=\\frac{\\sqrt{x}}{x^2+1}.$$ Then it is easy to check $$f'(x)=\\frac{-3x^2+1}{2\\sqrt{x}(x^2+1)}\\le 0$$ for $x>1$. It means that $f(x)$ is strictly decreasing. so $f(n+1)< f(n)$ or $$\\frac{\\sqrt{n+1}}{(n+1)^2+1} < \\frac{\\sqrt{n}}{n^2+1}.$$ -", "+0 # HELP!! 0 438 2 Find the largest real number $x$ for which there exists a real number $y$ such that $x^2 + y^2 = 2x + 2y$. Mar 12, 2019 #1 +234 +2 1 + $$\\sqrt2$$ . My method isn't really orthodox and required a little bit of eyeballing it, but I'm sure it's right. Add the right side to the left, and plug it into a graphing calculator, because it's a formula for a circle. This is where the eyeballing comes in, but the largest value for x is when y equals 1. Substitute this into your equation, and you get x2 + 1 -2x - 2 = x2 - 2x -1. Using the quadratic formula, you get 1 - $$\\sqrt 2$$. and 1 + $$\\sqrt 2$$. We want the larger value for x, so the largest value of x = 1 + $$\\sqrt 2$$ . Sorry this isn't an official method, but this is the best I could do. Hope this helps! Mar 12, 2019 #2 +23531 +3 Find the largest real number $$x$$ for which there exists a real number $$y$$ such that $$x^2 + y^2 = 2x + 2y$$. $$\\text{Let the largest real number x = x_{\\text{max}}} \\\\ \\text{Let the x-center of the circle = x_c } \\\\ \\text{Let the y-center of", "Least value of $a$ for which $4ax^2 + \\frac{1}{x} \\geq 1$ Find the least value of $a \\in R$ for which $4ax^2 + \\frac{1}{x} \\geq 1$, for all $x>0$. The equation will transform into (Using $x>0$) $4ax^3-x+1\\geq 0$ But I don't know how to deal with this cubic inequality. Could someone give me some hint as how to proceed? • If $a <0$ then for large $x$ the expression cannot hold. Hence $a \\ge 0$. Does the smallest such value work? – copper.hat May 24 '16 at 16:31 • @copper.hat - obviously not, what are you suggesting? – mathguy May 24 '16 at 16:33 • @mathguy: Sorry, I was a little obtuse. I meant to conclude that $a>0$ hence we have a function that has a unique minimiser. – copper.hat May 24 '16 at 16:35 8 Answers Hint: For $a>0$, we have by AM-GM that $4ax^2+\\frac{1}{x}\\geq 3\\sqrt[3]{a}$. The equality holds iff $x=\\frac{1}{2\\sqrt[3]{a}}$. For $a<0$, see copperhat's comment. For $a=0$, this is trivial. If we make a change of variable $a=1/A^3$ and $x=Au/2$, then the problem becomes one of finding the largest value of $A$ for which the inequality $$u^2+{2\\over u}\\ge A$$ holds for all $u\\gt0$. If you know calculus, then it's easy to see that $f(u)=u^2+{2\\over u}$ is minimized at $u=1$, so that $f(1)=3$ is the largest that", "## Wednesday, January 15, 2014 ### Infimum of abc Find the largest number $M$ such that for any given distinct number $a,b,c$ such that $0 < a,b,c < 1$ we always have: $$\\frac{(a+b)(b+c)(c+a)}{|(a-b)(b-c)(c-a)|} > M$$ ## Solution Clearly $M=1$ satisfies the condition. Now we prove that we can get the LHS arbitrarily close to 1. Let $a$ be an arbitrary number < 1, $b = a/(1+x)$ and $c=a/(1+x+yx)$ for some really large numbers $x,y$. The LHS now becomes: $$\\frac{(2+x)(2+(y+1)x)(2+(y+2)x)}{y(y+1)x^3}$$ $$=(\\frac{2}{x}+1)(\\frac{2}{x(y+1)} + 1)(\\frac{2}{xy} + \\frac{2}{y} + 1)$$ We can set $x,y$ large enough that the expression is as close as needed to 1.", "Back to all chapters # Inequalities From the basics to AM-GM and Cauchy-Schwarz. # A preview of \"Inequalities\"Join Brilliant Premium What is the largest real number $$x$$ that satisfies $$2x+3\\geq 4x+9$$? If the product of 5 positive real numbers is 3125, what is the minimum value of their sum? Let $$a,b,c,d$$ be positive real numbers such that $a+b+c + d = 1.$ What is the minimum value of $\\frac{1}{a} + \\frac{4}{b} + \\frac{9}{c} + \\frac{16}{d}?$ ×", "# Smallest possible value of expression involving greatest integer function If $a, b, c \\gt 0$ then what is the smallest possible value of $$\\left[\\frac{a+b}{c}\\right]+ \\left[\\frac{b+c}{a}\\right] + \\left[\\frac{c+a}{b}\\right]$$ where $[.]$ denotes greatest integer function. I tried using the AM GM inequality at first but it was not useful. I also tried adding 1 to each bracket and then subtracting 3 from overall to get the same numerator in each bracket. But this too wasn't useful. I don't have much practice of solving the question involving greatest integer function. Somebody please tell me how to deal with this question. • Actually. What is \"the largest integer\" function. Do you mean largest integer equal or less than? What I grew up calling the \"floor function\". If so, see my answer. If not, ... forgive my answer. – fleablood Dec 1 '17 at 7:28 Since for all reals $x$ and $y$ we have $$[x]+[y]+1\\geq[x+y],$$ by AM-GM we obtain $$\\sum_{cyc}\\left[\\frac{a+b}{c}\\right]\\geq\\left[\\sum_{cyc}\\frac{a+b}{c}\\right]-2\\geq6-2=4.$$ Wolog $a \\le b \\le c$ $[\\frac {b+c}a] \\ge [\\frac {a+a}a] =2$ And $[\\frac {a+c}b] \\ge [\\frac {a + b}b] \\ge [\\frac bb] = 1$. So you can not get less than $3$ If $\\frac {a+b}{c} < 1$ then $c > a+b$ and $\\frac {c+a}b > \\frac {2a + b}b= \\frac {2a}b + 1$. So If $\\frac {c+a}b < 2$ then $\\frac", "Back to all chapters Inequalities From the basics to AM-GM and Cauchy-Schwarz. A preview of \"Inequalities\"Join Brilliant Premium What is the largest real number $$x$$ that satisfies $$2x+3\\geq 4x+9$$? If the product of 5 positive real numbers is 3125, what is the minimum value of their sum? Let $$a,b,c,d$$ be positive real numbers such that $a+b+c + d = 1.$ What is the minimum value of $\\frac{1}{a} + \\frac{4}{b} + \\frac{9}{c} + \\frac{16}{d}?$ ×", "holds. Then $$a\\ge e^a\\ge 1$$ since $$e^a\\ge e^0$$ because $$a\\ge 0$$. But now we can see that $$a\\ge 1$$ and again, $$a\\ge e^a\\ge e^1$$ and so $$a\\ge e$$. We continue applying the same observation and conclude that $$a\\ge e^{^{e}}$$ and so on, which means that $$a$$ is unbounded which is a contradiction. • It's actually a great proof Jan 17, 2021 at 0:24 • @coshsinh Thank you. Jan 17, 2021 at 9:09 For positive values ​​of $$x$$ We can use the following characterization of $$e^x$$ $$e^x=\\lim_{t\\to \\infty} \\Big( 1+\\frac{1}{t}\\Big)^{tx},\\quad t> 0,x\\geq 0.$$ The Bernoulli's inequality states that $$(1 + y)^r \\geq 1 + ry$$ for every $$r \\geq 1$$ and every real number $$y \\geq −1$$. Then for $$y=\\frac{1}{t}$$ and $$t>0$$ such that $$r=tx\\geq 1$$ we have \\begin{align} e^x= &\\lim_{t\\to \\infty} \\Big( 1+\\frac{1}{t}\\Big)^{tx}\\\\ \\geq &\\lim_{t\\to \\infty} \\Big(1+\\frac{1}{t}(tx) \\Big)\\\\ = & 1+x \\end{align} • I just came across this answer 8 years later haha. However, I believe that $(1+y)^r\\geq 1+ry$ does not hold for $0 < r < 1$. For example, $(1+3)^{0.5} = 2$ while $1+3\\cdot 0.5 = 2.5$. Dec 6, 2021 at 8:30 • @GarethMa Yes, you are correct. The inequality is valid for $r \\geq 1$. I will make the correction. Dec 6, 2021 at 12:08 One which uses $$\\exp(x) = \\frac 1{\\exp(-x)}$$ $$1 + x \\underset{", "to choose $a$ and $b$ such that $h$ becomes as large as possible. From the AM-GM inequality is follows that $$b = \\sqrt{1 \\cdot b^2} \\le \\frac{b^2 + 1}2 \\Longrightarrow \\frac{b}{b^2 + 1} \\le \\frac 12$$ with equality for $b=1$. It follows that $h \\le \\frac 12$ for any choice of $a, b$, and $h = \\frac 12$ for $a = \\frac 12, b = 1$. So $h = \\frac 12$ is the largest value that can be obtained by this method. • Great answer +1 – Learnmore Dec 9 '15 at 13:31 • I been tying to figure out why I have a theorem in my text stating that the maximal intervall is never closed. This seems not to be the case here. You heard about such statements? – user1 Aug 14 '18 at 20:04 • @user1: This question is about the \"largest interval ... predicted by Picard's Theorem,\" not about the maximal interval in which the solution actually exists. – Martin R Aug 14 '18 at 20:06 • oh, I thought thats what that meant. Regardsless, the intervall that we obtain via global existance on rectangles is never maximal? – user1 Aug 14 '18 at 20:08 • @user1: Sorry, I don't get what you mean. Picard's theorem states that a solution exists on some (closed) interval.", "## Find largest pos int n with special condition For discussing Olympiad Level Algebra (and Inequality) problems Atonu Roy Chowdhury Posts: 63 Joined: Fri Aug 05, 2016 7:57 pm ### Find largest pos int n with special condition Find the largest possible positive integer $n$, such that there exists $n$ distinct positive real numbers $x_1, x_2, \\cdots, x_n$ satisfying the following inequality: for any $1 \\le i, j \\le n$ $(3x_i - x_j) (x_i - 3x_j) \\ge (1 - x_ix_j) ^2$ This was freedom. Losing all hope was freedom. Thanic Nur Samin Posts: 176 Joined: Sun Dec 01, 2013 11:02 am ### Re: Find largest pos int n with special condition Let $a_i=\\tan ^{-1} x_i$. Now, $(3x_i-x_j)(x_i-3x_j)\\ge (1-x_ix_j)^2\\Rightarrow \\dfrac{x_i-x_j}{1+x_ix_j}\\ge \\dfrac{1}{\\sqrt{3}}$ when $x_i> x_j$. But this implies $\\tan(a_i-a_j)\\ge \\dfrac{1}{\\sqrt{3}}$ which means $a_i-a_j\\ge 30^{\\circ}$, whereas $0^{\\circ}< a_k< 90^{\\circ}$. If we take $n\\ge 4$, then there must be two $a_k$'s so that their difference is at most $30^{\\circ}$, so $n\\le 3$. For $n=3$ we can easily find such a construction. Hammer with tact. Because destroying everything mindlessly isn't cool enough.", "+0 # Inequality 0 48 2 If w, x, y, and z are positive real numbers such that w + 2x + 3y + 4z = 8, then what is the maximum value of wxyz? Aug 14, 2021 #1 0 The minimum value is 9/25. Aug 14, 2021 #2 +436 0 By AM-GM, $$\\frac{w+2x+3y+4z}{4} \\ge \\sqrt[4]{24wxyz}\\\\ 2\\ge\\sqrt[4]{24wxyz}\\\\ 16\\ge24wxyz\\\\ \\frac{2}{3}\\ge wxyz$$ so the maximum value is $$\\boxed{\\frac{2}{3}}$$ for completeness, equality occurs when the terms are equal: $$w=2, x=1, y=\\frac{2}{3}, z=\\frac{1}{2}$$ Aug 21, 2021", "How could we find the largest number in the sequence $\\sqrt{50},2\\sqrt{49},3\\sqrt{48},\\cdots 49\\sqrt{2},50$? How to find the largest number in the sequence$$\\sqrt{50},2\\sqrt{49},3\\sqrt{48},\\cdots 49\\sqrt{2},50$$ I am interested in a \"calculus-free\" approach. Thanks, - If you square all of them, the largest of the squares will correspond to the largest of the square roots. – Ｊ. Ｍ. Nov 8 '11 at 10:45 @GerryMyerson Shame on me :) I did not read the question carefully enough. – Sasha Nov 8 '11 at 13:42 The $n$-th term in the sequence is $n\\sqrt{51-n}=\\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\\le n\\le 50$), does $n^2(51-n)$ become the largest? If you want to avoid calculus, you could use the AM-GM inequality: if $x,\\,y,\\,z\\ge 0$, then $$\\frac{x+y+z}{3}\\ge\\sqrt[3]{xyz},$$ with equality if and only if $x=y=z$. If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\\frac{51}{3}\\ge \\sqrt[3]{\\frac{n}{2}\\cdot\\frac{n}{2}\\cdot (51-n)},$$ with equality if and only if $n/2=51-n$ or $n=34$. It follows that $n^2(51-n)\\le 4\\cdot 17^3$, or $\\sqrt{n^2(51-n)}\\le 2\\cdot 17^{3/2}$, where equality holds for $n=34$. - +1 Well spotted! – Jyrki Lahtonen Nov 8 '11 at 13:19 Note that if the original question is for example $n\\sqrt{50-n}$ instead of $n\\sqrt{51-n}$, no integer $n$ satisfies the equality of the AM-GM inequality; in that case, a more careful evaluation, such as that used in Jyrki Lahtonen's answer, is needed. – pharmine Nov 9 '11", "# How could we find the largest number in the sequence $\\sqrt{50},2\\sqrt{49},3\\sqrt{48},\\cdots 49\\sqrt{2},50$? How to find the largest number in the sequence$$\\sqrt{50},2\\sqrt{49},3\\sqrt{48},\\cdots 49\\sqrt{2},50$$ I am interested in a \"calculus-free\" approach. Thanks, - If you square all of them, the largest of the squares will correspond to the largest of the square roots. – Ｊ. Ｍ. Nov 8 '11 at 10:45 @GerryMyerson Shame on me :) I did not read the question carefully enough. – Sasha Nov 8 '11 at 13:42 The $n$-th term in the sequence is $n\\sqrt{51-n}=\\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\\le n\\le 50$), does $n^2(51-n)$ become the largest? If you want to avoid calculus, you could use the AM-GM inequality: if $x,\\,y,\\,z\\ge 0$, then $$\\frac{x+y+z}{3}\\ge\\sqrt[3]{xyz},$$ with equality if and only if $x=y=z$. If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\\frac{51}{3}\\ge \\sqrt[3]{\\frac{n}{2}\\cdot\\frac{n}{2}\\cdot (51-n)},$$ with equality if and only if $n/2=51-n$ or $n=34$. It follows that $n^2(51-n)\\le 4\\cdot 17^3$, or $\\sqrt{n^2(51-n)}\\le 2\\cdot 17^{3/2}$, where equality holds for $n=34$. - +1 Well spotted! – Jyrki Lahtonen Nov 8 '11 at 13:19 Note that if the original question is for example $n\\sqrt{50-n}$ instead of $n\\sqrt{51-n}$, no integer $n$ satisfies the equality of the AM-GM inequality; in that case, a more careful evaluation, such as that used in Jyrki Lahtonen's answer, is needed. – pharmine Nov 9", "# inequality with constant power Let $a$ and $b$ be two real numbers which are $\\geq 1$. I am wondering if one can get the following upper bound $$\\frac{(a+b)^{a+b-1/2}}{a^{a-1/2}b^{b-1/2}}\\leq (a+b)^{c},$$ where $c$ is some constant, i.e. independent of $a$ and $b$. Thank you. - Set $a=b=x$. On the left, we get $\\dfrac{2^{2x}}{\\sqrt{2x}}$. On the right we get $(2x)^c$, which for any constant $c$ grows in the long run much more slowly than $\\dfrac{2^{2x}}{\\sqrt{2x}}$. For any constant $c$, and large enough $x$, the desired inequality does not hold." ]

[ "# Intersection points on circle, part 2 Geometry Level 5 The parabola $$f(x)=x^2$$ intersects the graph of $$g(x) = x^4 + ax^3 -2x^2+ bx +1$$ at four distinct points. These four points on a same circle of area 10. Given that $$b>0$$, find the value of $$\\lfloor{1000b}\\rfloor$$. ×", "Then for $C_2$ and $C_3.$ The three hyperbolas intersect in a point, which is the center of a circle that is tangent to all three original circles. So there you go.", "# Points of Intersection of Curves There are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines in the $x-y$ plane. Find the maximum possible value of the number of their points of intersection. My Attempt at the Question: I first to try and figure out the maximum number of points of intersection. For that I figured that parabola and circle intersect at max. $4$ points. And straight line will intersect the circle and parabola at $4$ distinct points. But this lead me nowhere. First we find the kinds of intersection that take place b/w the lines, circles and parabolas, which are as follows:- 1. Parabola with a parabola 2. Parabola with a circle 3. Parabola with a line 4. Circle with a circle 5. Circle with a line 6. Line with a line Now, lets count the maximum number of point of intersection in each of the above cases and consequently count the maximum number of intersections that cantake place given that there are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines:- CASE 1:- Intersection of two parabolas As we can see in the above image there are maximum $4$ possible points of intersection for two distinct parabolas. Since there are $4$ distinct parabolas hence there are maximum $4\\cdot\\binom{4}{2}$ maximum distinct point of intersections.", "# SAT1000 - P844 Geometry Level pending As shown above, $F(1,0)$ is the focus of the parabola $C: y^2=4x$, line $l: y=k(x-1)$ intersects with $C$ at point $A,B$, and the perpendicular bisector of $AB$ intersects with $C$ at point $M,N$. If $A,M,B,N$ are on the same circle, then find the value of $\\lfloor 1000|k| \\rfloor$. Have a look at my problem set: SAT 1000 problems ×", "is A x + B y + C z + D = 0 - Kuldeep mvicol wrote: > Hello everyone! Three points (A,B,C) can define two distinct vectors AB and AC. 1. Buy Back calculator: Increase your service credit or public education experience.Estimate your cost to: Restore withdrawn service credit past the deadline A plane is the two-dimensional analog of a point (zero dimensions), a line (one dimension), and three-dimensional space. Also Find Equation of Parabola Passing Through three Points - Step by Step Solver. Vi need to find the distance from the point to the plane. is the distance between given points A and B. You da real mvps! 4] ⧾ (1 ⧾ 16)[4 • 4 ⎯ (⎯ 3) 5] = 1380, The center of the circle is by solving x and y is at point, Find the equation of a circle and its center and radius if the circle passes through the points, In order to find the radius of the circle use the general circle equation and perform some basic algebraic steps and with the help of the square form (a + b). I want to find the Roll and Pitch anlges in order to rotate this plane to a plane parallel to the Z=0 plane. The \\(a, b, c$$ coefficients are obtained from a", "planes $\\ x \\ = \\ 8 \\ \\ , \\ \\ y \\ = \\ 8 \\ \\ , \\ \\ z \\ = \\ 8 \\$ at three hyperbolic curves $\\ yz \\ = \\ 1 \\ \\ , \\ \\ xz \\ = \\ 1 \\ \\ , \\ \\ xy \\ = \\ 1 \\$ , respectively. This now requires the variables to lie in the intervals $\\ \\frac{1}{8} \\ \\le \\ x \\ \\le \\ 8 \\$ and similarly for $\\ y \\$ and $\\ z \\$ in order to work on all three curves simultaneously. the constraint planes added, with the intersection hyperbolas visible and the location of the maximal cost indicated $\\ \\$ The Lagrange-multiplier method will only locate the one point where the cost function $\\ C \\ = \\ 8 \\ xy \\ + \\ 4 \\ xz \\ + \\ 2 \\ yz \\$ can be tangent to $\\ xyz \\ = \\ 8 \\$ , which occurs for $\\ C \\ = \\ 48 \\$ , as you found. There are no other places where the \"cost surface\" will be tangent to the open volume constraint surface for any value of $\\ C \\$ . Instead, we are left to check the curves on the three individual", "you experts: this is a real algebraic variety in 4-space). Now let’s look at the intersection of this surface in 4 space with some coordinate planes: Clearly this surface misses the $x_1=x_2 = 0$ plane (look at the real part of the equation). Intersection with the $y_1 = y_2 = 0$ plane yields $x_1^2+ x_2^2 = 1$ which is just the unit circle. Intersection with the $y_1 = x_2 = 0$ plane yields the hyperbola $x_1^2 - y_2^2 = 1$ Intersection with the $y_2 = x_1 = 0$ plane yields the hyperbola $x_2^2 - y_1^2 = 1$ Intersection with the $x_1 = y_1 = 0$ plane yields two isolated points: $x_2 = \\pm 1$ Intersection with the $x_2 = y_2 = 0$ plane yields two isolated points: $x_1 = \\pm 1$ (so we know that this object is non-compact; this is one reason the “sphere” remark puzzled me) Science and the media This Guardian article points out that it is hard to do good science reporting that goes beyond information entertainment. Of course, one of the reasons is that many “groundbreaking” science findings turn out to be false, even if the scientists in question did their work carefully. If this sounds strange, consider the following “thought experiment”: suppose that there are, say, 1000 factors that one can study", "you experts: this is a real algebraic variety in 4-space). Now let’s look at the intersection of this surface in 4 space with some coordinate planes: Clearly this surface misses the $x_1=x_2 = 0$ plane (look at the real part of the equation). Intersection with the $y_1 = y_2 = 0$ plane yields $x_1^2+ x_2^2 = 1$ which is just the unit circle. Intersection with the $y_1 = x_2 = 0$ plane yields the hyperbola $x_1^2 - y_2^2 = 1$ Intersection with the $y_2 = x_1 = 0$ plane yields the hyperbola $x_2^2 - y_1^2 = 1$ Intersection with the $x_1 = y_1 = 0$ plane yields two isolated points: $x_2 = \\pm 1$ Intersection with the $x_2 = y_2 = 0$ plane yields two isolated points: $x_1 = \\pm 1$ (so we know that this object is non-compact; this is one reason the “sphere” remark puzzled me) Science and the media This Guardian article points out that it is hard to do good science reporting that goes beyond information entertainment. Of course, one of the reasons is that many “groundbreaking” science findings turn out to be false, even if the scientists in question did their work carefully. If this sounds strange, consider the following “thought experiment”: suppose that there are, say, 1000 factors that one can study", "1$ implies that $x_1^2 + y_1^2 + x_2^2 + y_2^2 = 1$. But that isn’t what was in the article. Frenkel made a patient, kind response …and as soon as I read “equate real and imaginary parts” I winced with self-embarrassment. Of course, he admits that the complex version of this equation really yields a PUNCTURED sphere; basically a copy of $R^2$ in $R^4$. Just for fun, let’s look at this beast. Real part of the equation: $x_1^2 + x_2^2 - (y_1^2 + y_2^2) = 1$ Imaginary part: $x_1y_1 + x_2y_2 = 0$ (for you experts: this is a real algebraic variety in 4-space). Now let’s look at the intersection of this surface in 4 space with some coordinate planes: Clearly this surface misses the $x_1=x_2 = 0$ plane (look at the real part of the equation). Intersection with the $y_1 = y_2 = 0$ plane yields $x_1^2+ x_2^2 = 1$ which is just the unit circle. Intersection with the $y_1 = x_2 = 0$ plane yields the hyperbola $x_1^2 - y_2^2 = 1$ Intersection with the $y_2 = x_1 = 0$ plane yields the hyperbola $x_2^2 - y_1^2 = 1$ Intersection with the $x_1 = y_1 = 0$ plane yields two isolated points: $x_2 = \\pm 1$ Intersection with the $x_2 = y_2 = 0$ plane yields", "Free Version Difficult # Find the True Statement MVCALC-PSJHJA Let $S$ be a surface whose equation in spherical coordinates is: $$\\sin^2 \\phi=\\rho^2$$ Which of the following statements is TRUE? A The plane $z=1.1$ intersects $S$ B The intersection of $S$ with the plane $z=0$ is only a circle. C The plane $z=-2$ intersects $S$ D The intersection of $S$ with the plane $z=0$ is a circle and the origin. E The only point on $S$ whose three rectangular coordinates are all the same is the origin.", "### JEE (Advanced) 2019 Paper-2 Question 54 Instructions Answer by appropriately matching the lists based on the information given in the paragraph. Let the circles $$C_1 : x^2 + y^2 = 9$$ and $$C_2 : (x - 3)^2 + (y - 4)^2 = 16,$$ intersect at the points X and Y. Suppose that another circle $$C_3 : (x - h)^2 + (y - k)^2 = r^2$$ satisfies the following conditions: (i) centre of $$C_3$$ is collinear with the centres of $$C_1$$ and $$C_2$$, (ii) $$C_1$$ and $$C_2$$ both lie inside $$C_3$$ and (iii) $$C_3$$ touches $$C_1$$ at M and $$C_2$$ at N. Let the line through X and Y intersect $$C_3$$ at Z and W, and let a common tangent of $$C_1$$ and $$C_3$$ be a tangent to the parabola $$x^2 = 8 \\alpha y.$$ There are some expressions given in the List-I whose values are given in List-II below: Question 54", "So, that means they just touch each other. For the parabola, if x=0, y=4. So the parabola intersects the y-axis at (0,4). This is the point where circle touched the parabola. Even if 'a' is any value > 0. Circle of circle is given as (0,0). Now, we have got the circle with eqn x^2+y^2=16. (4 is the radius of circle). For this circle the integer (x,y) inside the circle can be all those point satisfying x^2+y^2<16 So, all the pairs in the 1st quadrant lying inside the circle will be (1,1) (1,2) (2,2) (2,1) (3,1) (3,2) (2,3) (1,3) - Total 8. Similarly in 4 quadrants it will be 8*4=32. We also have (0,1) (1,0) (0,-1) (-1,0) making the count go to 36. We have an option 36, but wait. What about (0,0). That is also inside the circle. Hence, the answer will be D =37. DS Forum Moderator Joined: 22 Aug 2013 Posts: 1145 Location: India Re: How many points with Integer x and Y co-ordinates lie within the circl [#permalink] ### Show Tags 05 Jun 2017, 00:23 mdacosta wrote: I agree with the 32 figure, but what about points (2,0), (3,0) and (-2,0), (-3,0), etc. I would think it would make the # of points inside the circle 32 + 12 (on the axis's) + 1", "$f(Y)$, so we can construct the circle $f((XY))$. Now do the same from another point $X' \\in \\mathcal C$ and draw the circle $f((X'Y'))$. If $O$ is the center of $\\mathcal C$, those two circles have to intersect at $f(O)$ and $f(\\infty)$ (because both $O$ and $\\infty$ are on $(XY)$ and $(X'Y')$). Additionally $f(O)$ must be outside $\\mathcal C$ and $f(\\infty)$ has to be inside $\\mathcal C$ (multiplication by a negative real number switches the bottom and top half-planes) Next we want to compute $f(f(\\infty))$ to do that we compute $f(f(X))$ and $f(f(Y))$ and construct the image of the circle $(X,Y,f(\\infty),f(O))$ (remember that they intersect $\\mathcal C$ at right angles). The line $(Of(\\infty))$ intersects $\\mathcal C$ at some point $Z$, compute $f(Z)$ and draw the circle $(f(Z)f(O)f(\\infty))$. $f(f(\\infty))$ has to be at the intersection of those two circles (and outside $\\mathcal C$) Now draw the line $f(\\infty)f(f(\\infty))$ and you're done. (all the lines and points used to construct the images by $f$ are not shown) And I'm glad that I didn't have to do even more witchcraft to have two circles that look like they don't intersect have them intersect over $\\Bbb C$ anyway and have them show up in our complex plane. • unfortunately this doesn't work with ellipses/parabolas/etc – mercio Oct 21 '15 at 18:27 •", "Browse Questions # If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $(x_i,y_i),$ for $i=1,2,3$ and $4$, then $y_1+y_2+y_3+y_4=$ $\\begin {array} {1 1} (1)\\;0 & \\quad (2)\\;c \\\\ (3)\\;a & \\quad (4)\\;c^4 \\end {array}$ (1) 0", "circle. This intersection of a parabola and a quartic does not define a circle. p1 = y == x^2 - 4 - x^4/16; p2 = x == 2 y^2 - 3; circleIntersect[{p1, p2}, {x, y}] If[FreeQ[%, circleIntersect], ContourPlot[Evaluate@{p1, p2, %}, {x, -4, 4}, {y, -4, 4}]] More than one circle. This intersection of two parabolas has only two points and does not define a unique circle. p1 = y == x^2 - 4; p2 = x == 2 y^2 - 1; {eq, {{{pts}}, {{basis}}}} = Reap[circleIntersect[{p1, p2}, {x, y}], {\"points\", \"basis\"}]; pts // N Note, one could detect this. The basis returned NullSpace consists of two vectors. Linear combinations of these parameterize all the circles through the two points of intersection. Show[ ContourPlot[Evaluate@{p1, p2}, {x, -4, 4}, {y, -4, 4}, ContourStyle -> Black], ContourPlot[ (* family of circles *) Evaluate[ Table[{(1 - t^2)/(1 + t^2), (2 t)/(1 + t^2)}, {t, 1.8^Range[-8, 8]}] . N@basis.(List /@ {x^2 + y^2, x, y, 1})], {x, -4, 4}, {y, -4, 4}, Contours -> {{0}}] ] ` • beautiful...+1 of course :) – ubpdqn May 17 '16 at 2:47", "p) \\end{vmatrix} \\neq 0.$ We conclude that no three points of ${\\cal P}$ are collinear. We can use a similar argument for the case of cocircular points. Four points $(x_1,y_1),(x_2,y_2),(x_3,y_3),$ and $(x_4,y_4)$ are cocircular if and only if $\\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 +y_1^2\\\\ 1 & x_2 & y_2 & x_2^2 +y_2^2\\\\ 1 & x_3 & y_3 & x_3^2 +y_3^2\\\\ 1 & x_4 & y_4 & x_4^2 +y_4^2 \\end{vmatrix} = 0.$ (One can think of this test is as lifting the four points to the paraboloid defined by $z=x^2+y^2$ in ${\\mathbb R}^3$, and checking if the four lifted points are coplanar. It is known that the intersections of the paraboloid with planes are exactly the lifting of circles to the paraboloid.) Taking the points $(t_1,t_1^2),(t_2,t_2^2),(t_3,t_3^2),(t_4,t_4^2)$, we get $\\begin{vmatrix} 1 & t_1 & t_1^2 & t_1^2 +t_1^4\\\\ 1 & t_2 & t_2^2 & t_2^2 +t_2^4\\\\ 1 & t_3 & t_3^2 & t_3^2 +t_3^4\\\\ 1 & t_4 & t_4^2 & t_4^2 +t_4^4 \\end{vmatrix} = (t_1+t_2+t_3+t_4)\\prod_{1\\le j Since $0\\le t_1,t_2,t_3,t_4 < p/4$, we have that $t_1+t_2+t_3+t_4$ is not a multiple of $p$. Assuming that the four integers are distinct, none of the elements of the product is a multiple of $p$. Repeating the above argument, we get that no four points of ${\\cal P}$ are cocircular. Now that", "Problem 24P 24. Suppose that three points on the parabola $y=x^{2}$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$ -coordinates is 0 .", "is a point in the plane, the vector from $\\vc{p}$ to $\\vc{x}$ (i.e., $\\vc{x}-\\vc{p}$) is some multiple of $\\vc{a}$ plus some multiple of $\\vc{b}$. The intersection of any plane with any sphere is a circle. P = \\{(x, y, z) : x - z\\sqrt{3} = 0\\}. ; 6.6.3 Use a surface integral to calculate the area of a given surface. @AndrewD.Hwang Dear Andrew, Could you please help me with the software which you use for drawing such neat diagrams? ; 6.6.5 Describe the surface integral of a vector field. I obviously can't give a different answer than everyone else: it's either a circle, a point (if the plane is tangent to the sphere), or nothing (if the sphere and plane don't intersect). The center of $S$ is the origin, which lies on $P$, so the intersection is a circle of radius $2$, the same radius as $S$. Why is it bad to download the full chain from a third party with Bitcoin Core? Plot this curve in 3D app for a = 1, R= 2 (the standard hippopede corresponds to a =R). -plane and cen tered at the origin) is b est parametrized using (24) b y setting = 2 in that relation to get r ( )=3 h cos ; sin 0 i; 2 (0 ]: (26)", "Conic Challenge #9 Geometry Level 3 Sub-unit: Parabola The line $$x-y=1$$ intersects the parabola $y^{2} = 4x$ at A and B. Normals at A and B intersect at C. If D is the point at which line CD is a normal to the parabola, then the coordinates of D are ×", "it to algebraic curves in the plane, too. However, I'm not sure it provides any more insight than ordinary calculus. • Actually, the line parallel to the axis of a parabola also intersects it in one (finite) point only. In this case it's implicitly discarded as no lines of the form $x=a$ were considered. – Berci Apr 22 at 19:08" ]

[ "11.21875 72 7 54 88 9 17 31.4375 126 45 12.5625 This example will calculate the intersection of R1 and R3 for 24″ equivalent arch pipe. First step is to calculate the variables a,b,c,d. In the arch pipe diagram above you can see the centroid of the pipe in line with the center of R3 is given point (0,0) and we will calculate the center points of both circles from that point. Notice capital letters are different variables than lower case variables and are pulled directly from ASTM. Center R1 =(a,b)= (0,R1-B) = (0,40.6875-5.90625) = (0,34.78125) Center R3 =(c,d)= (C,0) = (9.65625,0) Two circles with Radius R1 and R3 can be represented by the equations below: $R1^2= (x-a)^2+(y-b)^2 \\\\\\ and \\\\\\ R3^2=(x-c)^2+(y-d)^2$ $40.6875^2= (x-0)^2+(y-34.78125)^2 \\\\\\ and \\\\\\ 4.596^2=(x-9.65625)^2+(y-0)^2$ Now we need to calculate the intersection point of the two circles above. $x1,3=\\frac{a+c}{2}+\\frac{(c-a)(R1^2-R3^2)}{2D^2}\\pm2\\frac{b-d}{D^2}\\partial$ $y1,3=\\frac{b+d}{2}+\\frac{(d-b)(R1^2-R3^2)}{2D^2}\\pm2\\frac{a-c}{D^2}\\partial$ $D=\\sqrt{(c-a)^2+(d-b)^2}$ $\\partial=\\frac{1}{4}\\sqrt{(D+R1+R3)(D+R1-R3)(D-R1+R3)(-D+R1+R3)}$ $D=\\sqrt{(c-a)^2+(d-b)^2}$ $D=\\sqrt{(9.65625-0)^2+(0-34.78125)^2}$ $D=36.09679$ Next we need to calculate: $\\partial=\\frac{1}{4}\\sqrt{(D+R1+R3)(D+R1-R3)(D-R1+R3)(-D+R1+R3)}$ $\\partial=\\frac{1}{4}\\sqrt{(36.09679+40.6875+4.596)(36.09679+40.6875-4.596)(36.09679-40.6875+4.596)(-36.09679+40.6875+4.596)}$ $\\partial=4.22564$ Now we can calculate the actual intersection points. $x1,3=\\frac{a+c}{2}+\\frac{(c-a)(r1^2-r3^2)}{2D^2}\\pm2\\frac{b-d}{D^2}\\partial$ $x1,3=\\frac{0+9.65625}{2}+\\frac{(9.65625-0)(40.6875^2-4.596^2)}{2*36.09679^2}+\\frac{34.78125-0}{36.09679^2}*4.22564$ $x1,3=11.10972$ $y1,3=\\frac{b+d}{2}+\\frac{(d-b)(r1^2-r3^2)}{2D^2}\\pm2\\frac{a-c}{D^2}\\partial$ $y1,3=\\frac{34.78125+0}{2}+\\frac{(0-34.78125)(40.6875^2-4.596^2)}{2*36.09679^2}+\\frac{0-9.65625}{36.09679^2}*4.22564$ $y1,3=-4.48538$ R1 and R3 intersect at point (11.10972,-4.48538). Next we need to calculate the intersection point of R2 and R3. If we go through the same process we get (14.0918,1.0729). To draw Radius R3 on the right side of the arch pipe we need to calculate the range of", "on the circle, then: $(h-x_1)(h-x_2)+(k-y_1)(k-y_2)=0$ So, from (1), (2) and (3), $h$ satisfies the equation: $(h-x_1)(h-x_2)+\\left(\\frac{c^2}{h}-\\frac{c^2}{x_1}\\right)\\left(\\frac{c^2}{h}-\\frac{c^2}{x_2}\\right)=0$ $\\Rightarrow (h-x_1)(h-x_2)+\\frac{c^4}{x_1x_2h^2}(x_1-h)(x_2-h)=0$ $\\Rightarrow (h-x_1)(h-x_2)\\left(1+\\frac{c^4}{x_1x_2h^2}\\right)=0$ So the four roots of this equation are given by: $h = x_1, \\;x_2,\\; \\pm\\frac{c^2}{\\sqrt{-x_1x_2}}$ The last two roots differ only in sign, thus proving that the third and fourth points of intersection lie at the ends of a diameter of the hyperbola.", "is the midpoint of $RF$, i.e. $\\color{red}{V\\big(\\frac 12, \\frac 32\\big)}$. HINT: The equation of the directrix can be written as $$\\dfrac{y-2}{x+3}=m$$ As the eccentricity is $1,$ the distance of any point $P(h,k)$ on the parabola from the focus = the distance from the directrix. Now as $y=0$ is a tangent of the parabola, put $y=0$ in the relation derived above to form a Quadratic Equation in $x$ whose each root represents the abscissa of intersection. Now for tangency, both root should be same.", "maybe these ideas inspire someone to write a complete solution. Note that the intersection points lie on the circle of the form $$\\left(x+\\frac12\\right)^2+\\left(y+\\frac12\\right)^2=\\frac{37}{2}$$ This follows from a general theorem on the intersections of two parabolas with orthogonal axes of symmetry. Now there are so many theorems on cyclic quadrilaterals and even some statements on a parabola passing through the four corners. Then the given point $(2,3)$ should give some additional information. But I couldn't find a concise way of finishing this. - You can also get this equation by adding the two given ones and completing the squares. No beautiful conic theorems needed. – Ross Millikan May 14 '12 at 13:27 Yeah sure, the formula is not hard to find. I just think that this approach might eventually give a really nice representation of the four points. In the end there are explicit formulas for all four points just depending on three angles. Then there are planty of relations between many of the involved line segments. And also the symmetry axis of the parabola is somehow related to some of those segments. I think there is even a transformation which breaks down to reflection at some circle which gives some sort of normal form... I just couldn't put the pieces together. – Simon Markett May 14 '12 at", "- {x}^{2}$ $y = 3 - 0 = 3 , \\mathmr{and} , y = 3 - \\frac{23}{4} = - \\frac{11}{4}$ Hence, the solutions are, (1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4. Note that there are three solutions, which means there are three points of intersection between the parabola $y + {x}^{2} = 3$ and the ellipse ${x}^{2} + 4 {y}^{2} = 36$. See the graph below. Oct 18, 2016 Three points of intersection $\\left(- \\frac{\\sqrt{23}}{2} , - \\frac{11}{4}\\right)$, $\\left(\\frac{\\sqrt{23}}{2} , - \\frac{11}{4}\\right)$ and $\\left(0 , 3\\right)$ #### Explanation: Given: $y + {x}^{2} = 3$ ${x}^{2} + 4 {y}^{2} = 36$ Subtract the first equation from the second: $4 {y}^{2} - y = 33$ Subtract 33 from both sides: $4 {y}^{2} - y - 33 = 0$ Compute the discriminant: ${b}^{2} - 4 \\left(a\\right) \\left(c\\right) = {\\left(- 1\\right)}^{2} - 4 \\left(4\\right) \\left(- 33\\right) = 529$ $y = \\frac{1 + \\sqrt{529}}{8} = 3$ and $y = \\frac{1 - \\sqrt{529}}{8} = - \\frac{11}{4}$ For $y = 3$: ${x}^{2} = 3 - 3$ $x = 0$ For $y = - \\frac{11}{4}$: ${x}^{2} = 3 + \\frac{11}{4}$ ${x}^{2} = \\frac{12}{4} + \\frac{11}{4}$ ${x}^{2} = \\frac{23}{4}$ $x = \\frac{\\sqrt{23}}{2}$ and $x = - \\frac{\\sqrt{23}}{2}$", "12)= 3x2 ⇒ x2 - 2x - 8 = 0 ⇒ x = 4 or -2 are the roots of the quadratic equation x2 - 2x - 8 = 0 Now putting these values in the equation of straight line 2y = 3x + 12 we get the value of y i.e., $$y=\\frac{3x+12}{2}$$ put x = 4 in this,we get $$y=\\frac{3(4)+12}{2}$$ ⇒ y = 12 Now, put x = - 2 in the equation of straight line 2y = 3x + 12 we get the value of y $$y=\\frac{3(-2)+12}{2}$$ ⇒ y = 3 Therefore the intersection points of the parabola and the straight line are (-2, 3) and (4, 12) Finally, we need to find the area of the shaded region in the above diagram i.e., the required area of the region is an area of OABO And this region is bounded by the straight-line 2y = 3x + 12 and the parabola 4y = 3x2. y coordinates of a straight line is $$y=\\frac{3x+12}{2}$$ = f(x) and parabola is $$y=\\frac{3x^2}{4}$$ = g(x) So on integrating f(x) - g(x) we get the area under the straight line and the curve. Therefore, the required area is $$\\int\\limits_{-2}^4[{f(x)-g(x)]}dx$$ $$\\int\\limits_{-2}^4\\bigg({\\frac{3x+12}{2}-\\frac{3x^2}{4}}\\bigg)dx$$ on simplifying this and putting limits we get, = 27 sq.units Hence, the correct answer is option 4). Latest UP TGT Updates", "+ \\frac{y^2}{9} = 1\\\\36 \\frac{x^2}{4} + 36 \\frac{y^2}{9} = 36 \\cdot 1\\\\9x^2 + 4y^2 = 36$ Now we can substitute the equation of the line $y = \\frac{3}{2}x + 3$ in for y and solve for x . $9x^2 + 4(\\frac{3}{2}x + 3)^2 = 36\\\\9x^2 + 4(\\frac{9}{4}x^2 + 9x + 9) = 36\\\\9x^2 + 9x^2 + 36x + 36 = 36\\\\18x^2 + 36x = 0\\\\18x(x + 2) = 0\\\\$ So, $x = 0$ or $x = -2$ Finally we can substitute these x values into the equation of the line to find the corresponding y values. $y = \\frac{3}{2}(0) + 3 = 3$ $y = \\frac{3}{2}(-2) + 3 = 0$ Therefore, the line intersects the ellipse at points $(0, 3)$ and $(-2, 0)$ . ### Guided Practice 1. Estimate the solutions to the system below. Find the solutions to the systems below. 2. $x^2+(y-1)^2 &= 36 \\\\x^2 &= 2(y+9)$ 3. $x^2 &=y+8 \\\\4x+5y &= 12$ 1. $(3, -0.1)$ and $(4.5, -6)$ 2. This is a circle and a parabola that intersects in four places. Using substitution for $x^2$ , we have: $2(y+9)+(y-1)^2 &=36 \\\\2y+18+y^2-2y+1 &=36 \\\\y^2 &= 17 \\\\y &= \\pm \\sqrt{17} \\approx \\pm 4.12$ The corresponding $x$ -values are: $x^2 &= 2(4.12+9) \\qquad \\qquad \\quad x^2= 2(-4.12+9) \\\\x^2 &= 26.25 \\qquad \\qquad and \\qquad x^2= 9.76 \\\\x", "1.440692178 \\\\ 1 & 1.272119191 \\\\ 2 & 1.254474954 \\\\ 3 & 1.254291856 \\\\ 4 & 1.254291837 \\end{array} \\right)$$ For the negative root, the iterates would be $$\\left( \\begin{array}{cc} n & x_n \\\\ 0 & -1.440692178 \\\\ 1 & -1.559623512 \\\\ 2 & -1.554550306 \\\\ 3 & -1.554540654 \\end{array} \\right)$$ Now, making the division $$\\frac{27 x^4+84 x^3+463 x^2-961 }{(x-1.254291837)(x+1.554540654)}=27 x^2+75.8933 x+492.859$$ which does not show any real root. So, we have our solutions for $$x$$ and back to $$(3)$$ the corresponding $$y$$'s. Qualitatively, Bézout’s Theorem on plane curves says that there will be four intersection points, including (pairs of conjugate) complex points and points at infinity. A glance at the quadratic terms ($$ax^2+2ayx$$ in the first equation, $$ay^2+2ayx$$ in the second) shows that there will be no intersection points at infinity. In fact, the curves are both hyperbolas, with entirely different asymptotes. Thus you expect two pairs of complex points (no real intersection), one such pair and two real points (two visible points of intersection), or four visible intersection points, always taking account of multiplicity of course. In the event, thanks to Desmos, it looks as if there are always at least two visible points of intersection; certainly when you take $$a=1$$, $$b=4$$, $$c_1=1$$ and $$c_2=\\frac16$$, you get four honest real points of intersection.", "Baricenter of $4$ intersection points of parabola with circle lies on axis of parabola Show that the baricenter of the $4$ intersection points of a parabola with a circle is on the axis of the parabola. Let $p$ be a parabola, $c$ a circle and $p\\cap c=\\{P_1,P_2,P_3,P_4\\} \\Rightarrow B= \\frac{P_1+P_2+P_3+P_4}{4} \\in$ axis of $p$. • Example. $y=x^2;(x+\\frac{9}{8})^2+(y-\\frac{27}{8})^2=\\frac{325}{32}$ 4 solutions: $(1,1), (2,4), (-\\frac{1}{2},\\frac{1}{4}), (-\\frac{5}{2},\\frac{25}{4})$ The baricenter of these 4 points is: $B = \\frac{1}{4}( (1,1)+(2,4)+(-\\frac{1}{2},\\frac{1}{4})+(-\\frac{5}{2},\\frac{25}{4}))= (0, \\frac{23}{4})$ so $B$ is in the line $x = 0$ which is the axis of the parabola. – Sgernesto Feb 7 '13 at 20:18 Without loss of generality we can take the parabola as having equation $y=kx^2$. Write down the equation $(x-a)^2+(y-b)^2=r^2$ of the circle. Substitute $kx^2$ for $y$. You will get a degree $4$ polynomial in $x$ with no $x^3$ term. But the sum of the roots of the polynomial is the negative of the coefficient of $x^3$, divided by the coefficient of $x^4$. So the sum of the four (not necessarily real) roots is $0$. The $x$-coordinate of the barycenter of the points of intersection is, if there are four real roots, not necessarily distinct, the mean of these $4$ roots. • \"Great minds think alike.\" – Michael Hardy Feb 7 '13 at 20:44 • Well, I am not sure I", "these points as vertices is A) 201 B) 120 C) 205 D) 435 • question_answer139) In the expansion of $1+\\frac{a}{bx}1!+\\frac{{{(a+bx)}^{2}}}{2!}$$+\\frac{{{(a+bx)}^{3}}}{3!}+....$the coefficient of${{x}^{n}}$is A) $\\frac{{{e}^{a}}{{b}^{n}}}{n!}$ B) $\\frac{{{(b.a)}^{n}}}{n!}$ C) $\\frac{{{e}^{b}}.{{b}^{n}}}{(n-1)!}$ D) $\\frac{{{a}^{n}}{{b}^{n-1}}}{n!}$ • question_answer140) $\\frac{{{C}_{0}}}{1}+\\frac{{{C}_{2}}}{3}+\\frac{{{C}_{4}}}{5}+\\frac{{{C}_{6}}}{7}+....$is equal to A) $\\frac{{{2}^{n-1}}}{n-1}$ B) $\\frac{{{2}^{n+1}}}{n+3}$ C) $\\frac{{{2}^{n}}}{n+1}$ D) $\\frac{{{2}^{n-2}}}{n}$ • question_answer141) The equation of the parabola whose vertex is $(2,-1)$and focus$(3,-1),$is A) ${{y}^{2}}-4x-2y+9=0$ B) ${{y}^{2}}+4x+2y-9=0$ C) ${{y}^{2}}-4x+2y+9=0$ D) ${{y}^{2}}+4x-2y+9=0$ • question_answer142) The equations of lines joining the origin to the point of intersection of the curves$y=x+3$and$4{{x}^{2}}+4{{y}^{2}}=1,$are A) $36({{x}^{2}}+{{y}^{2}})={{(x-y)}^{2}}$ B) $12({{x}^{2}}+{{y}^{2}})={{(x+y)}^{2}}$ C) $9({{x}^{2}}+{{y}^{2}})=4{{(x-y)}^{2}}$ D) None of the above • question_answer143) The angle of elevation of a fighter jet plane from a point on the ground is$60{}^\\circ$. After 10s the angle of elevation becomes$30{}^\\circ$. If the speed of jet plane is 432 km/h, then the height of the jet plane from the ground is A) $200\\sqrt{3}m$ B) $400\\sqrt{3}m$ C) $600\\sqrt{3}m$ D) $800\\sqrt{3}m$ • question_answer144) Which of the following is false? A) $(AB)'=B'A'$ B) ${{(AB)}^{\\theta }}={{B}^{\\theta }}{{A}^{\\theta }}$ C) $\\overline{AB}=\\overline{B}\\,\\,\\overline{A}$ D) ${{(AB)}^{-1}}={{B}^{-1}}{{A}^{-1}}$ • question_answer145) The equation of plane which is parallel to$x-$axis and passes through the line $\\frac{x-2}{2}=\\frac{y-3}{3}=\\frac{z-4}{5}$is A) $2x+3y+5z=1$ B) $2x-5y=4$ C) $5y-3z-3=0$ D) $3y+4z=0$ • question_answer146) A force$\\overrightarrow{F}=5\\hat{i}+10\\hat{j}+16\\hat{k}$ acts at a point, whose position vector is$2\\hat{i}-7\\hat{j}+10\\hat{k}$. The moment of F about the point$-5\\hat{i}+6\\hat{j}-10\\hat{k}$is A) $41\\hat{i}-8\\hat{j}+55\\hat{k}$ B) $-408\\hat{i}-12\\hat{j}+135\\hat{k}$ C) $-36\\hat{i}+14\\hat{j}-35\\hat{k}$ D) None of the above •", "$BPR$ equal $\\cos^{-1}\\left(\\frac{x}{16}\\right)$ and $\\cos^{-1}\\left(\\frac{x}{12}\\right)$. So we have $\\cos^{-1}\\left(\\frac{x}{16}\\right)+\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)=180^{\\circ}-\\cos^{-1}\\left(\\frac{x}{12}\\right).$ Taking the cosine of both sides, and simplifying using the addition formula for $\\cos$ as well as the identity $\\sin^{2}{x} + \\cos^{2}{x} = 1$, gives $x^2=\\boxed{130}$. ### Solution 4 (quickest) Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$. The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger circle is $16$. Thus by Power of a Point in the circle passing through $Q$, $R$, and $S$, we have $x \\cdot 2x = 10 \\cdot (10+16) = 260$, so $x^2 = \\boxed{130}$. ## Solution 5 (Pythagorean Theorem and little algebraic manipulation) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"T\", t , NW); [/asy]$ Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ From the median formula, $PT=\\sqrt{14}.$ Thus, $a+b=2\\sqrt{14}.$ Also, since $MP=PN$, from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\\implies a^2-b^2=28.$ Thus, $a-b=\\frac{28}{2\\sqrt{14}}=\\sqrt{14}.$ We conclude", "$$\\frac{x^{2}}{4}$$ between x = 0 and x = 4 (iii) For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations. From the equation x2 = y, y = $$\\frac{x^{2}}{y}$$ ∴ y = $$\\frac{x^{2}}{y}$$ ∴ $$\\frac{x^{2}}{y}$$ = x ∴ x2 – y = 0 ∴ x(x3 – y) = 0 ∴ x = 0 or x3 = y i.e. x = 0 or x = 4 When x = 0, y = 0 When x = 4, y = $$\\frac{4^{2}}{4}$$ = 4 ∴ the points of intersection are O(0, 0) and A(4, 4). Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO] Now, area of the region ODACO = area under the parabola y2 = 4x, i.e. y = 2√x between x = 0 and x = 4 Area ofthe region ODABO = area under the rabola x2 = 4y, i.e. y = $$\\frac{x^{2}}{3}$$ between x = 0 and x = 4 Question 5. Find the area of the region in the first quadrant bounded by the circle x2 + y2 = 4 and the X-axis and the line x = y√3. Solution: For finding the points of intersection of the circle and the line, we solve x2 +", "be used. $&x^2 + 4y^2 - 36 = 0 \\to x^2 + 4y^2 = 36 && x^2 + 4y^2 + 0y = 36 && \\quad x^2 + 4y^2 + 0y = 36 \\\\&x^2 + y = 3 \\to x^2 + 0y^2 + y = 3 && -1(x^2 + 0y^2 + y = 3) && -x^2 - 0y^2 - y = -3 \\\\&&& && \\quad \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} \\\\ &&& && \\quad 4y^2 - y = 33$ Using the quadratic formula, $a = 4 \\ b = -1 \\ c = -33$ $y & = \\frac{-(-1) \\pm \\sqrt{(-1)^2 - 4(4)(-33)}}{2(4)} \\\\y & = \\frac{1 + 23}{8} = 3 \\qquad y = \\frac{1 - 23}{8} = -2.75$ These values must be substituted into one of the original equations. $x^2 + y &= 3 && \\qquad \\ \\ x^2 + y = 3 \\\\x^2 +3 & = 3 && x^2 + (-2.75) = 3 \\\\x^2 & = 0 && \\qquad \\qquad \\ \\ x^2 = 5.75 \\\\x & = 0 && x = \\pm \\sqrt{5.75} \\ \\approx 2.4$ The three points of intersection as determined algebraically in Cartesian representation are $\\text{A} (0, 3), \\text{B} (2.4, -2.75)$ and $\\text{C} (2.4, 2.75)$. If we are working with polar equations to determine the polar coordinates of a point of intersection, we must remember that there are", "# Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this: $$ax^2 - dx^2 + bx - ex + c - f = 0 \\\\ x^2(a - d) + x(b - e) = f - c \\\\ x^2(a - d) + x(b - e) + \\frac{(b - e)^2}{4(a - d)} = f - c + \\frac{(b - e)^2}{4(a - d)} \\\\ (x\\sqrt{a - d} + \\frac{b - e}{2\\sqrt{a - d}})^2 = f - c + \\frac{(b - e)^2}{4(a - d)} \\\\ (a - d)(x + \\frac{b - e}{2(a - d)})^2 = f - c + \\frac{(b - e)^2}{4(a - d)} \\\\ x + \\frac{b - e}{2(a - d)} = \\sqrt{\\frac{f - c + \\frac{(b - e)^2}{a - d}}{a - d}} \\\\ x = \\pm\\sqrt{\\frac{f - c + \\frac{(b - e)^2}{a - d}}{a - d}} - \\frac{b - e}{2(a - d)} \\\\$$ Then solving for $y$ is as simple as plugging $x$ into one of the equations. $$y = ax^2 + bx + c$$ Is my solution for $x$ and $y$ correct? Is there a better way to solve", "\\\\ 27 x^{2}-16(4 x-24) &=432 \\\\ 27 x^{2}-64 x-48 &=0 \\end{aligned} Now, use the Quadratic Formula to solve for x. \\ \\begin{aligned} x &=\\frac{64 \\pm \\sqrt{(-64)^{2}-4(27)(-48)}}{2(27)} \\\\ &=\\frac{64 \\pm \\sqrt{9280}}{54} \\\\ &=\\frac{32 \\pm 4 \\sqrt{145}}{27} \\end{aligned} Plugging these into the calculator we get $$\\ x=\\frac{32+4 \\sqrt{145}}{27} \\approx 2.97$$ and $$\\ x=\\frac{32-4 \\sqrt{145}}{27} \\approx-0.6$$. Looking at the graph, we know that there will be two different y-values for each x-value to give four points of intersection. Using the estimations, solve for y. You can choose either equation. $$\\ \\begin{array}{rlrl} y^{2} & =-\\frac{4}{3}(2.97-6) & & y^{2}&=-\\frac{4}{3}(-0.6-6) \\\\ y^{2} & =4.04 & \\text { and } & y^{2} & =8.8 \\\\ y & =\\pm 2.01 && y & =\\pm 2.97 \\end{array}$$ The points are (2.97, 2.01), (2.97, −2.01), (−0.6, 2.97), and (−0.6, −2.97). # Examples Example 1 Earlier, you were asked to determine if the line $$\\ y=\\frac{3}{2} x+3$$ intersects the ellipse $$\\ \\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$$, and if so, at which point(s). Solution First, let's get rid of the fractions in the equation of the ellipse to make it easier to work with. To do so, we multiply by the LCD. $$\\ \\begin{array}{r} \\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1 \\\\ 36 \\frac{x^{2}}{4}+36 \\frac{y^{2}}{9}=36 \\cdot 1 \\\\ 9 x^{2}+4 y^{2}=36 \\end{array}$$ Now we can substitute the equation of the line $$\\ y=\\frac{3}{2} x+3$$ in for y and solve for", "area of the circle is 38.5 sq. units, find the equation. OR A double ordinate of parabola y2 = 4ax is of length 8a. Prove that the lines from the vertex to its ends are at right angles. Given, two diameters of a circle lie along the lines x – y – 9 = 0 …(i) and x – 2y – 7 = 0 …(ii) So, their point of intersection is the centre of the circle. Solving eqs. (i) and (ii), simultaneously, we get x = 11 and y = 2 ∴ The centre of the circle is (11, 2). Let r be the radius of the circle, then Area = πr² = 38.5 sq units (given) ∴ The equation of the circle is (x – 11)2 + (y – 2)2 = ($$\\frac{7}{2}$$)2 x2 + y2 – 22x – 4y + 125 = $$\\frac{49}{4}$$ = x2 + y2 – 22x – 4y + 125 – $$\\frac{49}{4}$$ = 0 = 4(x2 + y2) – 88x – 16y + 451 = 0 is the required equation of circle. OR Let PQ be the double ordinate of length 8a of the parabola y2 = 4ax. Then, PR = QR = 4a. Let AR = x1 then coordinates of point P are (x1 4a) and coordinates of point R are (x1 –", "a > 0, then its vertex points down: If a < 0, then its vertex points up: If a = 0 the graph is not a parabola and a straight line. The vertex of the parabola is $x = -\\frac{b}{2a}$. #### Vieta's formulas If x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0 then: $x_1 + x_2 = -\\frac{b}{a}$ $x_1x_2 = \\frac{c}{a}$ These formulas are called Vieta's formulas. We can find the roots x1 and x2 of the quadratic equation by solving the simultaneous equations. Problem 1. Solve the equation: x2 - 4 = 0 Solution: x2 - 4 = (x - 2)(x + 2) (x - 2)(x + 2) = 0 x - 2 = 0 or x + 2 = 0 The roots are x = 2 or x = -2 Solution 2: a = 1, b = 0, c = -4 D = 02 - 4 ⋅ 1 ⋅ (-4) = 16 $x_1 = \\frac{-b - \\sqrt{D}}{2a} = \\frac{- 0 - \\sqrt{16}}{2 \\cdot 1} = \\frac{-4}{2} = -2$ $x_2 = \\frac{-b + \\sqrt{D}}{2a} = \\frac{- 0 + \\sqrt{16}}{2 \\cdot 1} = \\frac{4}{2} = 2$ Problem 2. Solve the equation: 3x2 + 4x + 5 = 0 Solution: the discriminant is D = 42 - 4⋅3⋅5 = 16", "Dear Uncle Colin, How would I show that $4y - 3x + 26 = 0$ is a tangent to $(x+4)^2 +(y-3)^2 = 100$? Can I Reconcile Circle/Line Equations? Hi, CIRCLE, and thanks for your message! The simplest way (I think) is to note that if the line is tangent to the circle, solving the equations simultaneously will give a single, repeated root. Let’s solve that. I’ll start by replacing the $x+4$ in the second equation. • $x = \\frac{4y + 26}{3}$ • $x+4 = \\frac{4y + 38}{3}$ • $(x+4)^2 = \\frac{16y^2 + 304y + 1444}{9}$ So $\\frac{16y^2 + 304y + 1444}{9} + (y^2 - 6y + 9) = 100$ I’m not a fan of the fractions, so let’s multiply by 9: • $16y^2 + 304y + 1444 + 9y^2 - 54y + 81 = 900$ … and rearrange: • $25y^2 + 250y -625 = 0$ Hey lookit! A nice common factor! • $y^2 + 10y - 25 = 0$ And that factorises nicely as $(y - 5)^2 = 0$, giving a double root. (Of course, we could have used the discriminant a little sooner.) An alternative method is to find the radius of the circle perpendicular to the line. The original line has a gradient of $\\frac{3}{4}$, so the radius must have a gradient of $-\\frac{4}{3}$, and pass", "the parabola in question is $x^2=4py$ with the focus at F(0, 143+r) . That means $p = 143+r$ 4. Since $\\cos(60^\\circ)=\\frac12$ and $\\sin(60^\\circ)=\\frac12\\cdot \\sqrt{3}$ the satellite has the coordinates $S\\left((5783+r)\\cdot \\frac12 \\cdot \\sqrt{3}\\ ,\\ 143+r+(5783+r) \\cdot \\frac12 \\right)$ 5. Plug in the value of p and the coordinates of S into the equation of the parabola: $\\left((5783+r)\\cdot \\frac12 \\cdot \\sqrt{3} \\right)^2=4\\cdot (r+143) \\left(143+r+(5783+r) \\cdot \\frac12\\right)$ 6. After moving some stuff around (a lot of stuff btw) you'll get a quadratic equation in r: $7r^2+5762r-31128777=0$ which yields 2 solutions. The negative solution isn't very plausible here. so it is $r = 1737\\ km$ 3. Great, thanks a lot!", "has equal roots. Discriminant = 0 ⇒ (-32m2)2 – 4(m)(-32) = 0 ⇒ 1024 m4 + 128m = 0 ⇒ 128m (8m3 + 1) = 0 ⇒ 8m3 + 1 = 0 …..[∵ m ≠ 0] ⇒ m3 = $$-\\frac{1}{8}$$ ⇒ m = $$-\\frac{1}{2}$$ Substituting m = $$-\\frac{1}{2}$$ in (i), we get ⇒ $$y=-\\frac{1}{2} x+\\frac{1}{\\left(-\\frac{1}{2}\\right)}$$ ⇒ $$y=-\\frac{1}{2} x-2$$ ⇒ x + 2y + 4 = 0, which is the equation of the common tangent. Question 18. Find the equation of the locus of a point, the tangents from which to the parabola y2 = 18x are such that sum of their slopes is -3. Solution: Given equation of the parabola is y2 = 18x Comparing this equation with y2 = 4ax, we get ⇒ 4a = 18 ⇒ a = $$\\frac{9}{2}$$ Equation of tangent to the parabola y2 = 4ax having slope m is ⇒ y = mx + $$\\frac{a}{m}$$ ⇒ y = mx + $$\\frac{9}{2m}$$ ⇒ 2ym = 2xm2 + 9 ⇒ 2xm2 – 2ym + 9 = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1 + m2 = $$-\\frac{(-2 y)}{2 x}=\\frac{y}{x}$$ But, m1 + m2 = -3 $$\\frac{y}{x}$$ = -3 y = -3x, which is the required equation of locus. Question 19. The towers of a bridge," ]

[ "on the circle, then: $(h-x_1)(h-x_2)+(k-y_1)(k-y_2)=0$ So, from (1), (2) and (3), $h$ satisfies the equation: $(h-x_1)(h-x_2)+\\left(\\frac{c^2}{h}-\\frac{c^2}{x_1}\\right)\\left(\\frac{c^2}{h}-\\frac{c^2}{x_2}\\right)=0$ $\\Rightarrow (h-x_1)(h-x_2)+\\frac{c^4}{x_1x_2h^2}(x_1-h)(x_2-h)=0$ $\\Rightarrow (h-x_1)(h-x_2)\\left(1+\\frac{c^4}{x_1x_2h^2}\\right)=0$ So the four roots of this equation are given by: $h = x_1, \\;x_2,\\; \\pm\\frac{c^2}{\\sqrt{-x_1x_2}}$ The last two roots differ only in sign, thus proving that the third and fourth points of intersection lie at the ends of a diameter of the hyperbola.", "$BPR$ equal $\\cos^{-1}\\left(\\frac{x}{16}\\right)$ and $\\cos^{-1}\\left(\\frac{x}{12}\\right)$. So we have $\\cos^{-1}\\left(\\frac{x}{16}\\right)+\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)=180^{\\circ}-\\cos^{-1}\\left(\\frac{x}{12}\\right).$ Taking the cosine of both sides, and simplifying using the addition formula for $\\cos$ as well as the identity $\\sin^{2}{x} + \\cos^{2}{x} = 1$, gives $x^2=\\boxed{130}$. ### Solution 4 (quickest) Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$. The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger circle is $16$. Thus by Power of a Point in the circle passing through $Q$, $R$, and $S$, we have $x \\cdot 2x = 10 \\cdot (10+16) = 260$, so $x^2 = \\boxed{130}$. ## Solution 5 (Pythagorean Theorem and little algebraic manipulation) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"T\", t , NW); [/asy]$ Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ From the median formula, $PT=\\sqrt{14}.$ Thus, $a+b=2\\sqrt{14}.$ Also, since $MP=PN$, from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\\implies a^2-b^2=28.$ Thus, $a-b=\\frac{28}{2\\sqrt{14}}=\\sqrt{14}.$ We conclude", "12)= 3x2 ⇒ x2 - 2x - 8 = 0 ⇒ x = 4 or -2 are the roots of the quadratic equation x2 - 2x - 8 = 0 Now putting these values in the equation of straight line 2y = 3x + 12 we get the value of y i.e., $$y=\\frac{3x+12}{2}$$ put x = 4 in this,we get $$y=\\frac{3(4)+12}{2}$$ ⇒ y = 12 Now, put x = - 2 in the equation of straight line 2y = 3x + 12 we get the value of y $$y=\\frac{3(-2)+12}{2}$$ ⇒ y = 3 Therefore the intersection points of the parabola and the straight line are (-2, 3) and (4, 12) Finally, we need to find the area of the shaded region in the above diagram i.e., the required area of the region is an area of OABO And this region is bounded by the straight-line 2y = 3x + 12 and the parabola 4y = 3x2. y coordinates of a straight line is $$y=\\frac{3x+12}{2}$$ = f(x) and parabola is $$y=\\frac{3x^2}{4}$$ = g(x) So on integrating f(x) - g(x) we get the area under the straight line and the curve. Therefore, the required area is $$\\int\\limits_{-2}^4[{f(x)-g(x)]}dx$$ $$\\int\\limits_{-2}^4\\bigg({\\frac{3x+12}{2}-\\frac{3x^2}{4}}\\bigg)dx$$ on simplifying this and putting limits we get, = 27 sq.units Hence, the correct answer is option 4). Latest UP TGT Updates", "Baricenter of $4$ intersection points of parabola with circle lies on axis of parabola Show that the baricenter of the $4$ intersection points of a parabola with a circle is on the axis of the parabola. Let $p$ be a parabola, $c$ a circle and $p\\cap c=\\{P_1,P_2,P_3,P_4\\} \\Rightarrow B= \\frac{P_1+P_2+P_3+P_4}{4} \\in$ axis of $p$. • Example. $y=x^2;(x+\\frac{9}{8})^2+(y-\\frac{27}{8})^2=\\frac{325}{32}$ 4 solutions: $(1,1), (2,4), (-\\frac{1}{2},\\frac{1}{4}), (-\\frac{5}{2},\\frac{25}{4})$ The baricenter of these 4 points is: $B = \\frac{1}{4}( (1,1)+(2,4)+(-\\frac{1}{2},\\frac{1}{4})+(-\\frac{5}{2},\\frac{25}{4}))= (0, \\frac{23}{4})$ so $B$ is in the line $x = 0$ which is the axis of the parabola. – Sgernesto Feb 7 '13 at 20:18 Without loss of generality we can take the parabola as having equation $y=kx^2$. Write down the equation $(x-a)^2+(y-b)^2=r^2$ of the circle. Substitute $kx^2$ for $y$. You will get a degree $4$ polynomial in $x$ with no $x^3$ term. But the sum of the roots of the polynomial is the negative of the coefficient of $x^3$, divided by the coefficient of $x^4$. So the sum of the four (not necessarily real) roots is $0$. The $x$-coordinate of the barycenter of the points of intersection is, if there are four real roots, not necessarily distinct, the mean of these $4$ roots. • \"Great minds think alike.\" – Michael Hardy Feb 7 '13 at 20:44 • Well, I am not sure I", "the parabola in question is $x^2=4py$ with the focus at F(0, 143+r) . That means $p = 143+r$ 4. Since $\\cos(60^\\circ)=\\frac12$ and $\\sin(60^\\circ)=\\frac12\\cdot \\sqrt{3}$ the satellite has the coordinates $S\\left((5783+r)\\cdot \\frac12 \\cdot \\sqrt{3}\\ ,\\ 143+r+(5783+r) \\cdot \\frac12 \\right)$ 5. Plug in the value of p and the coordinates of S into the equation of the parabola: $\\left((5783+r)\\cdot \\frac12 \\cdot \\sqrt{3} \\right)^2=4\\cdot (r+143) \\left(143+r+(5783+r) \\cdot \\frac12\\right)$ 6. After moving some stuff around (a lot of stuff btw) you'll get a quadratic equation in r: $7r^2+5762r-31128777=0$ which yields 2 solutions. The negative solution isn't very plausible here. so it is $r = 1737\\ km$ 3. Great, thanks a lot!", "Thus $$\\boxed{f(x) = \\frac35 x^2}$$ Case $1$: Of type $x^2=4ay$. If $(10,60)$ passes through this parabola, then we get, $$100=4a (60)\\Rightarrow 4a = \\frac {5}{3}$$ The parabola is: $$x^2 =\\frac {5}{3} y$$ Case $2$: Of type $y^2=4ax$. If $(10,60)$ passes through this parabola, then we get, $$3600=4a (10) \\Rightarrow 4a =360$$ The parabola is: $$y^2=360x$$ Hope it helps. Let $f(x) = ax^2 + bx + c$ for some $a,b,c \\in \\mathbb{R}$ be the wanted parabola. Then • one root at $(0, 0) \\rightarrow f(0) = 0 = a \\cdot 0^2 + b \\cdot 0 + c \\rightarrow c = 0$ • $f(10) = 60 = a \\cdot 10^2 + b \\cdot 10 = 100a + 10b \\rightarrow 6 = 10 \\cdot a + b \\rightarrow b = 6 - 10a$ • only one root at $(0, 0) \\rightarrow \\Delta(0) = 0 = b^2-4ac = (6 - 10a)^2 - 4a\\cdot 0 = (6 - 10a)^2 = 0 \\iff 6 - 10a = 0 \\iff a = \\frac{3}{5 }$ so all the parabolae in the form $$f(x) = \\frac{3}{5} \\cdot x^2$$ satisfy the requirements. • This function will have a second root at $x=10 -\\frac6a$ (if $a\\neq 0$). I think OP wants the only root to be at $x=0$, doesn't he? – MPW Feb 20 '17 at 13:47 • yup!", "1.440692178 \\\\ 1 & 1.272119191 \\\\ 2 & 1.254474954 \\\\ 3 & 1.254291856 \\\\ 4 & 1.254291837 \\end{array} \\right)$$ For the negative root, the iterates would be $$\\left( \\begin{array}{cc} n & x_n \\\\ 0 & -1.440692178 \\\\ 1 & -1.559623512 \\\\ 2 & -1.554550306 \\\\ 3 & -1.554540654 \\end{array} \\right)$$ Now, making the division $$\\frac{27 x^4+84 x^3+463 x^2-961 }{(x-1.254291837)(x+1.554540654)}=27 x^2+75.8933 x+492.859$$ which does not show any real root. So, we have our solutions for $$x$$ and back to $$(3)$$ the corresponding $$y$$'s. Qualitatively, Bézout’s Theorem on plane curves says that there will be four intersection points, including (pairs of conjugate) complex points and points at infinity. A glance at the quadratic terms ($$ax^2+2ayx$$ in the first equation, $$ay^2+2ayx$$ in the second) shows that there will be no intersection points at infinity. In fact, the curves are both hyperbolas, with entirely different asymptotes. Thus you expect two pairs of complex points (no real intersection), one such pair and two real points (two visible points of intersection), or four visible intersection points, always taking account of multiplicity of course. In the event, thanks to Desmos, it looks as if there are always at least two visible points of intersection; certainly when you take $$a=1$$, $$b=4$$, $$c_1=1$$ and $$c_2=\\frac16$$, you get four honest real points of intersection.", "area of the circle is 38.5 sq. units, find the equation. OR A double ordinate of parabola y2 = 4ax is of length 8a. Prove that the lines from the vertex to its ends are at right angles. Given, two diameters of a circle lie along the lines x – y – 9 = 0 …(i) and x – 2y – 7 = 0 …(ii) So, their point of intersection is the centre of the circle. Solving eqs. (i) and (ii), simultaneously, we get x = 11 and y = 2 ∴ The centre of the circle is (11, 2). Let r be the radius of the circle, then Area = πr² = 38.5 sq units (given) ∴ The equation of the circle is (x – 11)2 + (y – 2)2 = ($$\\frac{7}{2}$$)2 x2 + y2 – 22x – 4y + 125 = $$\\frac{49}{4}$$ = x2 + y2 – 22x – 4y + 125 – $$\\frac{49}{4}$$ = 0 = 4(x2 + y2) – 88x – 16y + 451 = 0 is the required equation of circle. OR Let PQ be the double ordinate of length 8a of the parabola y2 = 4ax. Then, PR = QR = 4a. Let AR = x1 then coordinates of point P are (x1 4a) and coordinates of point R are (x1 –", "- {x}^{2}$ $y = 3 - 0 = 3 , \\mathmr{and} , y = 3 - \\frac{23}{4} = - \\frac{11}{4}$ Hence, the solutions are, (1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4. Note that there are three solutions, which means there are three points of intersection between the parabola $y + {x}^{2} = 3$ and the ellipse ${x}^{2} + 4 {y}^{2} = 36$. See the graph below. Oct 18, 2016 Three points of intersection $\\left(- \\frac{\\sqrt{23}}{2} , - \\frac{11}{4}\\right)$, $\\left(\\frac{\\sqrt{23}}{2} , - \\frac{11}{4}\\right)$ and $\\left(0 , 3\\right)$ #### Explanation: Given: $y + {x}^{2} = 3$ ${x}^{2} + 4 {y}^{2} = 36$ Subtract the first equation from the second: $4 {y}^{2} - y = 33$ Subtract 33 from both sides: $4 {y}^{2} - y - 33 = 0$ Compute the discriminant: ${b}^{2} - 4 \\left(a\\right) \\left(c\\right) = {\\left(- 1\\right)}^{2} - 4 \\left(4\\right) \\left(- 33\\right) = 529$ $y = \\frac{1 + \\sqrt{529}}{8} = 3$ and $y = \\frac{1 - \\sqrt{529}}{8} = - \\frac{11}{4}$ For $y = 3$: ${x}^{2} = 3 - 3$ $x = 0$ For $y = - \\frac{11}{4}$: ${x}^{2} = 3 + \\frac{11}{4}$ ${x}^{2} = \\frac{12}{4} + \\frac{11}{4}$ ${x}^{2} = \\frac{23}{4}$ $x = \\frac{\\sqrt{23}}{2}$ and $x = - \\frac{\\sqrt{23}}{2}$", "+ \\frac{y^2}{9} = 1\\\\36 \\frac{x^2}{4} + 36 \\frac{y^2}{9} = 36 \\cdot 1\\\\9x^2 + 4y^2 = 36$ Now we can substitute the equation of the line $y = \\frac{3}{2}x + 3$ in for y and solve for x . $9x^2 + 4(\\frac{3}{2}x + 3)^2 = 36\\\\9x^2 + 4(\\frac{9}{4}x^2 + 9x + 9) = 36\\\\9x^2 + 9x^2 + 36x + 36 = 36\\\\18x^2 + 36x = 0\\\\18x(x + 2) = 0\\\\$ So, $x = 0$ or $x = -2$ Finally we can substitute these x values into the equation of the line to find the corresponding y values. $y = \\frac{3}{2}(0) + 3 = 3$ $y = \\frac{3}{2}(-2) + 3 = 0$ Therefore, the line intersects the ellipse at points $(0, 3)$ and $(-2, 0)$ . ### Guided Practice 1. Estimate the solutions to the system below. Find the solutions to the systems below. 2. $x^2+(y-1)^2 &= 36 \\\\x^2 &= 2(y+9)$ 3. $x^2 &=y+8 \\\\4x+5y &= 12$ 1. $(3, -0.1)$ and $(4.5, -6)$ 2. This is a circle and a parabola that intersects in four places. Using substitution for $x^2$ , we have: $2(y+9)+(y-1)^2 &=36 \\\\2y+18+y^2-2y+1 &=36 \\\\y^2 &= 17 \\\\y &= \\pm \\sqrt{17} \\approx \\pm 4.12$ The corresponding $x$ -values are: $x^2 &= 2(4.12+9) \\qquad \\qquad \\quad x^2= 2(-4.12+9) \\\\x^2 &= 26.25 \\qquad \\qquad and \\qquad x^2= 9.76 \\\\x", "it (but that is already given). Drop a perpendicular from $P$ to the lines that the centers are on. You then have 2 separate segments, separated by the foot of the altitude of $P$. Call them $a$ and $b$ respectively. Call the measure of the foot of the altitude of $P$ $h$. You then have 3 equations: $$(1)a+b=12$$ (this is given by the fact that the distance between the centers is 12. $$(2)a^2+h^2=64$$. This is given by the fact that P is on the circle with radius 8. $$(3)b^2+h^2=36$$. This is given by the fact that P is on the circle with radius 6. Subtract (3) from (2) to get that $a^2-b^2=28$. As per (1), then you have $a-b=\\frac{7}{3}$ (4). Add (1) and (4) to get that $2a=\\frac{43}{3}$. Then substitute into (1) to get $b=\\frac{29}{6}$. Substitute either a or b into (2) or (3) to get $h=\\sqrt{455}{6}$. Then to get $PQ=PR$ it is just $\\sqrt{(b+6)^2+h^2}=\\sqrt{\\frac{65^2}{6^2}+\\frac{455}{6^2}}=\\sqrt{\\frac{4680}{36}}=\\sqrt{130}$. $PQ^2=\\boxed{130}$ -drwgoon ### Full Proof that R, A, B are collinear $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); label(\"T\", t , NW); [/asy]$ Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the", "we have: $$\\begin{gather} 3^2 = 4p(1)\\cr p = \\frac 94 \\end{gather}$$ Thus, the desired equation is $\\,y^2 = 4(\\frac 94)x\\,$; that is, $\\,y^2 = 9x\\,$. Since $\\,p\\,$ gives the $x$-value of the focus, the focus is $\\,(\\frac 94,0)\\,$. Since the vertex (the origin) is halfway between the focus and directrix, the equation of the directrix is $\\,x = -\\frac 94\\,$. ## Finding the Equation of a Parabola: Two Approaches Problem: Find the equation of the parabola in standard form with focus $\\,(0,2)\\,$. Solutions: Since the equation is in standard form, the vertex is at the origin. ### Safest Approach: Use the Definition of Parabola This approach does not require memorization/knowledge of the standard forms. However, it takes longer, because you're going through the derivation of the standard form for a special case. By definition, a parabola is the set of all points equidistant from the focus and directrix. Since the vertex is halfway between the focus and directrix, the directrix is the line $\\,y = -2\\,$. Let $\\,(x,y)\\,$ be a typical point on the parabola. Using the distance formula, the distance from $\\,(x,y)\\,$ to the focus is: $$\\sqrt{(x-0)^2 + (y-2)^2} = \\sqrt{x^2 + (y-2)^2}$$ The distance from $\\,(x,y)\\,$ to the directrix is $\\,y - (-2) = y + 2\\,$. (The points have the same $x$-value—just compute distance along", "the Pre-Calculus class, we found the max or min by finding the vertex of the parabola defined at the point $x=\\frac{-b}{2a}$ and $y=f(\\frac{-b}{2a})$. We didn’t need Calculus to determine this, the $\\frac{-b}{2a}$ can be defined from the vertex form of the equation of a parabola $y=n(x-h)^2+k$. With a little classical algebra, you can show that $h=\\frac{-b}{2a}$. If you expand $y=n(x-h)^2+k$ $y=n(x^2-2hx+h^2)+k$ $y=nx^2-2nhx+nh^2+k$ Then $\\frac{-b}{2a}=\\frac{2nh}{2n}=h$ I taught Differential Calculus as a grad student at the University of Maine in 2003 and we covered The Farmer and the Fencing from the perspective of Calculus, in which we differentiate the area equation and set the derivative equal to zero to find the maximum area. The problems can be spruced up a little by dividing the pen into smaller pens using cross-sectional pieces. In the fundamental problem the maximum area of a simple pen with four sides and no cross-divisions is arrived at by splitting the perimeter equally among the four sides to make a square. If the pen is cross divided in any way, the maximum area comes from a scenario in which half of the perimeter is allocated to the lengthwise pieces and the other half allocated to the any widths. If there are $m$ lengths and $n$ widths then the perimeter is represented by $ml+nw=P$. The area is $A=l*w$.", "perpendicular the line connecting (-4,1) and (2a, 2a^2). so the product of the slopes$$ 2a \\dfrac{(2a^2 - 1)}{2a+4} = -1$$solving this equation should give you a = -1 and the point (-2, 2) on ... 0 Let (x,\\frac12x^2) be a point on the curve. The derivative at this point is x. And the slope of the line from (-4,1) to (x,\\frac12x^2) is \\dfrac{\\frac12x^2-1}{x+4}. At the nearest point, these two slopes are perpendicular:$$x\\cdot\\dfrac{\\frac12x^2-1}{x+4} = -1$$This gives us \\frac12x^2-x = -x - 4, which simplifies to x^3=-8, with the unique ... 0 Minimize the distance squared F=(x+4)^2+(y-1)^2=(x+4)^2+(\\frac{x^2}{2}-1)^2=\\frac{x^4}{4}+8x+17.$$\\frac{dF}{dx}=x^3+8=0 --> x=-2$$.$$y=2^2/2=2$$1 A non standard solution Take a second point on the parabola very close to (4,4): (4+\\epsilon,4+2\\epsilon) agrees to the first order (plugging in the equation of the parabola, 16+8\\epsilon+\\epsilon^2\\approx16+8\\epsilon). Then solve the circle by the three points$$\\begin{align} (x-4)^2&+(y-4)^2&=r^2,\\\\ ... 1 The center lies at the intersection of the perpendicular to the tangent and the bisector of the two known points. $$2(x-4)+(y-4)=0$$ $$(4-1)(x-\\frac{4+1}2)+(4-0)(y-\\frac{4+0}2)=0.$$ Hence $$x=\\frac{13}2,y=-1.$$ -1 If what you mean by \"touching\" is \"intersect\", then this is the solution. Your circle is defined by its center $(x_c,y_c)$ and its radius $R$: $$(x-x_c)^2+(y-y_c)^2=R^2$$ Now, you want all three points $(1,0)$, $(4,4)$ and $(4,-4)$ to be on the circle, so you hav to solve the system: ... 2 the tangent to the parabola at", "line intersects circle $C_3$ at same two points. Distance of these two points from origin in plane 2 can be written as $d - r_3^{\\prime}$ and $d + r_3^{\\prime}$. Now, if we transform them to plane 1 and see distance between them, we can say that: $2 r_3 = \\frac{1}{d - r_3^{\\prime}} - \\frac{1}{d + r_3^{\\prime}}$ In above equation, we can find the $d$ as all other variables are known. And using value of $d$ we can find $x$-coordinate of center of $x_3^{\\prime}$ $=$ $C_3^{\\prime}$ as $\\pm \\sqrt{d^2 - y^2}$. So, center of $n^{th}$ circle will be $x_3^{\\prime} + 2(n-3)R_3$ or $x_3^{\\prime} - 2(n-3)R_3$, we can now check which one to use by finding radius of $C_4$ and match with given value in input. Now, our problem is solved. You can do the calculations yourself to find the closed formula. # COMPLEXITY: Constant per query. # AUTHOR'S, TESTER'S SOLUTIONS: This question is marked \"community wiki\". 3.0k93163187 accept rate: 12% 19.7k350498541 4 There is one more recurrence relation which can be formed... as we know ..C(N)=C(1)+C(2)+C(N-1) +/- 2*SQRT(REMAINING PART :D) we can also see the complement circle for n th circle will be (n-2)th circle... so if C(N)=C(1)+C(2)+C(N-1) + 2*SQRT(REMAINING PART :D) THEN C(N-2)=C(1)+C(2)+C(N-1) - 2*SQRT(REMAINING PART :D) ADDING BOTH EQUATIONS WE GET : C(N)+C(N-2)=2C(1)+2C(2)+2C(N-1).. answered 13 Jul '15,", "a > 0, then its vertex points down: If a < 0, then its vertex points up: If a = 0 the graph is not a parabola and a straight line. The vertex of the parabola is $x = -\\frac{b}{2a}$. #### Vieta's formulas If x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0 then: $x_1 + x_2 = -\\frac{b}{a}$ $x_1x_2 = \\frac{c}{a}$ These formulas are called Vieta's formulas. We can find the roots x1 and x2 of the quadratic equation by solving the simultaneous equations. Problem 1. Solve the equation: x2 - 4 = 0 Solution: x2 - 4 = (x - 2)(x + 2) (x - 2)(x + 2) = 0 x - 2 = 0 or x + 2 = 0 The roots are x = 2 or x = -2 Solution 2: a = 1, b = 0, c = -4 D = 02 - 4 ⋅ 1 ⋅ (-4) = 16 $x_1 = \\frac{-b - \\sqrt{D}}{2a} = \\frac{- 0 - \\sqrt{16}}{2 \\cdot 1} = \\frac{-4}{2} = -2$ $x_2 = \\frac{-b + \\sqrt{D}}{2a} = \\frac{- 0 + \\sqrt{16}}{2 \\cdot 1} = \\frac{4}{2} = 2$ Problem 2. Solve the equation: 3x2 + 4x + 5 = 0 Solution: the discriminant is D = 42 - 4⋅3⋅5 = 16", "Calculus The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5) cm. Find the minimum distance that the particle can cover from the fixed point. A. 2.3 cm B. 4.0 cm C. 5.2 cm D. 6.0 cm Calculus Differentiate (cos θ — sin θ)2 with respect to θ A. -2 cos 2θ B. -2 sin 2θ C. 1 - 2 cos 2θ D. 1 - 2 sin 2θ Geometry/Trigonometry Find the value of sin 45° – cos 30°. A. $\\frac{2+\\sqrt{6}}{4}$ B. $\\frac{\\sqrt{2}+\\sqrt{3}}{4}$ C. $\\frac{\\sqrt{2}+\\sqrt{3}}{2}$ D. $\\frac{\\sqrt{2}-\\sqrt{3}}{2}$ Algebra A polynomial in x whose roots are $\\frac{4}{3}$ and $-\\frac{3}{5}$ is A. 15x2 – 11x – 12 B. 15x2 + 11x – 12 C. 12x2 – x – 12 D. 12x2 + 11x – 15 Algebra The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35,… is A. n(3n – 0.5) B. 2(3n + 2) C. n(3n + 2.5) D. n(3n + 5) Simplify $\\frac{5+\\sqrt{7}}{3+\\sqrt{7}}$ A. $17-\\sqrt{7}$ B. $4-\\sqrt{7}$ C. $15+\\sqrt{7}$ D. $7-\\sqrt{7}$ Geometry/Trigonometry In the figure above, TS//XY and XY = TY, ∠STZ = 34°, ∠TXY = 47°, find the angle marked n . A. 47° B. 52° C. 56° D. 99° Geometry/Trigonometry A regular polygon has 150° as the size", "with perpendicular axes intersect in four points, those four points lie on a circle. – Andy Barfoot Apr 26 '12 at 23:29 A slightly-alterned analytic approach: Align the directrix of the vertical (horizontal) parabola with $x$ (resp, $y$) axis; let the focus be $(a,b)$ (resp, $(c,d)$), and let $(x,y)$ be any point of intersection. As \"distance from directrix\" equals \"distance from focus\", we have $x^2 = (x-c)^2+(y-d)^2$ and $y^2=(x-a)^2+(y-b)^2$. Adding these equations and rearranging yields $(x-(a+c))^2+(y-(b+d))^2 = 2 ( a c + b d )$, the equation of the common circle. If nothing else, this form of the circle equation makes clear(er) the relation between the parabolas' foci and the circle's center and radius. – Blue Apr 27 '12 at 14:18 I claimed in a comment on @Zander's answer that \"a cleaner diagram could even make [his 'Least Algebraic' proof] a Proof Without Words\". Well, I've done a cleaning pass for fun, but since the order of construction of the elements isn't necessarily clear, I'll offer a Proof With Words. (Actually, since I tried to reduce labeling clutter in the images, my prose has descriptor clutter, so that this is really a Proof With Way Too Many Words.) Given parabolas with foci $F$ and $G$ and common chord $AB$ (all marked with stars), let $C$ be the midpoint", "and using that $$\\mathcal B$$ is a field), it follows that their intersection is: $$(x_0,y_0) = (\\frac{ce-fb}{ae-bd}, \\frac{af-cd}{ae-bd})$$. Since $$\\mathcal B$$ was a field, then actually $$x_0,y_0 \\in \\mathcal B$$ so we didn’t construct anything new. Second case. If they were obtained intersecting the line $$ax+by=c$$ and the circle $$(x-d)^2+(y-e)^2 = f^2$$, where once again $$a,b,c,d,e,f \\in \\mathcal B$$, we can compute the intersection similarly. For example, if $$b \\neq 0$$, we write $$y = \\frac{c-ax}{b}$$ and this gives that $$(x_0-d)^2 +(\\frac{c-ax_0}{b}-e)^2 = f^2$$. This is, $$x_0$$ satisfies a polynomial equation of degree $$2$$, whose coefficients are in $$\\mathcal B$$. In other words, $$x_0$$ is the root for a degree $$2$$ polynomial in $$\\mathcal B$$. The same goes for $$y_0$$ in the event that $$a \\neq 0$$. What relation do we obtain if $$b = 0$$ (similarly for $$a=0$$)? In this case, simply looking at the line yields $$x_0 = \\frac{c}{a}$$ so $$x_0 \\in \\mathcal B$$. Third case. As you might have already guessed, the third case involves intersecting two circles: $(x-a)^2+(y-b)^2 = c^2$ and $(x-d)^2+(y-e)^2 = f^2,$ with $$a,b,c,d,e,f \\in \\mathcal B$$. One can subtract the two equations to obtain a new equation $2(a-d)x + 2(b-e)y = f^2-c^2+a^2+b^2-d^2-e^2,$ which is that of a line, and intersecting this line with any of the two circles gives the point", "\\frac{40}{3}$. Note that the horizontal distance from $O$ to the origin is $7$, and the horizontal distance from K to Q is 4, ($\\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$. The value of $3x+4y$ is 73 at point $Q$. Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$. $\\boxed{(B)}$. ## Solution 3 Let $z = 3x + 4y$. Solving for $y$, we get $y = (z - 3x)/4$. Substituting into the given equation, we get $$x^2 + \\left( \\frac{z - 3x}{4} \\right)^2 = 14x + 6 \\cdot \\frac{z - 3x}{4} + 6,$$ which simplifies to $$25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.$$ This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so $$(6z + 152)^2 - 4 \\cdot 25 \\cdot (z^2 - 24z - 96) \\ge 0,$$ which simplifies to $$-64z^2 + 4224z + 32704 \\ge 0,$$ which can be factored as $$-64(z + 7)(z - 73) \\ge 0.$$ The largest value of $z$ that satisfies this inequality is $\\boxed{73}$, which is $\\boxed{(B)}$. ## Solution 4 (Using Answer Choice + Calculus) Implicitly differentiating the given equation with respect to $x$ yields: $2x + 2y\\frac{dy}{dx} = 14 + 6\\frac{dy}{dx}$" ]

[ "principal, there would exist $d(x)\\in (2,x)$ such that every element of the ideal can be expressed as a product of $d(x)$ with an element of the ring (i.e. $d(x)$ divides every element of the ideal). In particular, this holds for $2$ and $x$ since they are both elements of $(2,x)$. It is not hard to see why this yields a contradiction, if you keep in mind everything we have said so far. • but how do I show that $1$ and $-1$ are the only ones? – Guerlando OCs Jan 10 '16 at 19:11 • You are working in $\\mathbb{Z}[x]$, where $\\deg(p(x)\\cdot q(x)) = \\deg(p(x))+\\deg(q(x))$. This tells you that the degree of the divisors of a certain polynomial is less or equal to the degree of the polynomial itself. But $\\deg(3) = 0$, so any polynomial that divides it must have degree $0$ as well. Does this help you in seeing what the only divisors of $3$ in $\\mathbb{Z}[x]$ are? Once you are there, deducing that $3$ and $x$ have no other common divisors but $1$ and $-1$ is immediate. – Labba Jan 10 '16 at 19:24", "1. ## polynomial the polynomial p(x) = $x^4+ax^3-10x^2+bx-12$ is divisible by d(x) = $(x+4)(x-3)$ Calculate the quotient of the division of P(x) of D(x). $x^2+x+1$ 2. Originally Posted by Apprentice123 the polynomial p(x) = $x^4+ax^3-10x^2+bx-12$ is divisible by d(x) = $(x+4)(x-3)$ Calculate the quotient of the division of P(x) of D(x). $x^2+x+1$ Dividing $x^4+ax^3-10x^2+bx-12$ by d(x) = $(x+4)(x-3)$ gives $x^2+(a-1)x+3-a$ and a rest $(b+13a-15)x+12 (2-a)$ Quotient is $x^2+x+1$", "# I have a small problem with long division when dividing polynomials of the same degree I was doing long division with $$x^2 + 1$$, and $$3x^2+5$$. (the second polynomial is the quotient). the problem I've found is related to dividing two polynomials of the same degree. Even if I know that the quotient is always a constant and the remainder is a polynomial of one degree less than the dividend, I still have problem when performing this long division, meaning: $$x^2+0+1\\space /\\space 3x^2+5$$ x^2 is contained in 3x^2 3 times, so I write 3 in the quotient. then, I perform multiplication between the quotient and the divisor. $$3 * 3x^2 = 9x^2$$, and $$3*5 = 15$$. I write them below the dividend. Now, I subtract the dividend with the things I have below. but, it's an infinite loop, because the degree doesn't change no matter how long I divide. • When you multiply the quotient and the divisor, isn't the divisor $x^2 + \\cdots$ rather than $3x^2 + 5$? Multiplying $x^2 + \\cdots$ by $3$ yields $3x^2 + \\cdots$ which will cancel with the leading term of $3x^2+5$ when subtracting. Jan 18 at 20:51 • Then the quotient is $1/3$, no? Jan 18 at 21:00 • In general you usually ask \"how many times is the divisor", "this tableau to see how it greatly streamlines the division process. To divide $$x^3+4x^2-5x-14$$ by $$x-2$$, we write $$2$$ in the place of the divisor and the coefficients of $$x^3+4x^2-5x-14$$ in for the dividend. Then 'bring down' the first coefficient of the dividend. Next, take the $$2$$ from the divisor and multiply by the $$1$$ that was 'brought down' to get $$2$$. Write this underneath the $$4$$, then add to get $$6$$. Now take the $$2$$ from the divisor times the $$6$$ to get $$12$$, and add it to the $$-5$$ to get $$7$$. Finally, take the $$2$$ in the divisor times the $$7$$ to get $$14$$, and add it to the $$-14$$ to get $$0$$. The first three numbers in the last row of our tableau are the coefficients of the quotient polynomial. Remember, we started with a third degree polynomial and divided by a first degree polynomial, so the quotient is a second degree polynomial. Hence the quotient is $$x^2+6x+7$$. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form $$x-c$$. It is important to note that it works only for these kinds of divisors.\\footnote{You'll need to use good old-fashioned polynomial long division for divisors of degree larger than 1.} Also take note", "polynomial. If the remainder is equal to zero, then he is divisible, if it is any other number, he is not. Example 5. Is a polynomial divisible by $f(x) = 5x^4 + 3x – 6x + 3$ divisible by $(x + 2)$. We have a linear polynomial $(x + 2)$ which means that our c is equal to – 2. Now we calculate $f(-2) = 0$. $f(-2) = 5 \\cdot (- 2) + 3 \\cdot (- 2) – 6 \\cdot (- 2) + 3 = – 10 – 6 + 12 + 3 = – 1$ $– 1 \\not= 0$ which means that polynomial $f(x)$ is not divisible by $(x + 2)$.", "+ 3x^2 - x + 16 = (2x - 1)(x^2 + 2x - 3) + 7x + 13$ Vocabulary associated with the long division process $2x^3 + 3x^2 - x + 16$ is the dividend $x^2 + 2x - 3$ is the divisor $2x - 1$ is the quotient $7x + 13$ is the remainder ## More References and Links to Polynomials Algebra and Trigonometry - Swokowsky Cole - 1997 - ISBN: 0-534-95308-5 Algebra and Trigonometry with Analytic Geometry - R.E.Larson , R.P. Hostetler , B.H. Edwards, D.E. Heyd - 1997 - ISBN: 0-669-41723-8 polynomial Functions .", "$(x + 2)$ is a factor of the given polynomial. The quotient of $\\frac{2x^3 - 2x^2 + 5x + 26}{x + 2}$ is $(x^2 -4x + 13)$. Therefore, the remaining roots of the given polynomial are the solutions of the equation $x^2 -4x + 13$ = $0$. This is a quadratic which cannot be solved by factoring. Hence, let us use the method of completing square. $x^2 - 4x + 13$ = $0$ $x^2 - 4x$ = $-13$ $x^2 - 4x + 4$ = $-13 + 4$ (Adding $4$ on both sides to complete the square on the left side) $(x – 2)^2$ = $-9$ So, $x – 2$ = $\\pm$ $\\sqrt{(-9)}$ = $\\pm\\ (\\sqrt9)(\\sqrt{(-1)}$ = $\\pm 3i$. Therefore, $x$ = $2 + 3i$ and $x$ = $2 – 3i$. Thus, the remaining two roots of the given polynomial are $(2 + 3i)$ and $(2 – 3i)$ and the remaining factors are $(x – 2 – 3i)$ and $(x – 2 +3i)$ So, the given polynomial $x^3 - 2x^2 + 5x + 26$ is completely factored as $(x + 2)(x – 2 – 3i)(x – 2 +3i)$", "$D$ be a nonzero polynomial in $I$ of minimal degree; this minimal degree must be${}<n$ due to the second given element of$~I$. Then for every element of$~I$, its remainder after Euclidean division by $D$ lies in$~I$ and is of lower degree than$~D$, so equal to$~0$: every element of $I$ is divisible by$~D$. In particular this holds for your irreducible polynomial of degree$~n$. But its only divisors are of degree$~n$ (scalar multiples of itself) or of degree$~0$ (invertible elements). Then $D$ must be an invertible element, whence $I$ is the whole ring $F[x]$.", "both greatest common divisors, $\\deg d(x) = \\deg d'(x)\\text{.}$ Since $d(x)$ and $d'(x)$ are both monic polynomials of the same degree, it must be the case that $d(x) = d'(x)\\text{.}$ Notice the similarity between the proof of Proposition 17.10 and the proof of Theorem 2.10.", "# Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10) Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of . – – + – – + – The final answer is the quotient plus the remainder over the divisor. Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10)", "If $p(x)=f(x)g(x)$ then deg($f$)+deg($g$)= deg($p$). One of these degrees must be $1$ and the other $0$. A polynomial of degree 0 is a unit. • The OP asked for hints... – lhf Oct 14 '16 at 2:00 • @lhf You're right I should tag as spoiler. – JKEG Oct 14 '16 at 2:02 • @JamesDickens $p$ is irreducible if $p=fg \\implies f$ is a unit or $g$ is a unit. :P (Associates of $1_R$ are precisely units, so our definitions are pretty much identical.) – JKEG Oct 14 '16 at 2:05", "$D(t,x)$ divides $T(t,x)$. Question: Assume that $D(t,x)$ as above divides the trinomial $T(t,x).$ Do we have $$\\deg(a_1(t)) = 2 \\deg(a_2(t)),\\;\\;\\deg(a_0(t))=3\\deg(a_2(t)).$$ ??? I am aware of the work of Schinzel on trinomials. The trinomials $T(t,x)$ do not seem to be worked out in these papers.", "## Can x3 – 3x + 1 be the quotient on division of x5+ 2×3 + x – 1 by a polynomial in x of degree 3? Justify​ Question Can x3 – 3x + 1 be the quotient on division of x5+ 2×3 + x – 1 by a polynomial in x of degree 3? Justify​ in progress 0 2 months 2021-08-01T12:39:05+00:00 2 Answers 2 views 0 1. No, x 2 −1 cannot be the quotient on division of x 6 −2x 3 +x−1 by a polynomial in degree 5 because the degree of the product of the quotient and the divisor should be equal to the power of the dividend. Here, the degree of the product of the quotient and the divisor is 7. But the degree of x 6 −2x 3 +x−1 is 6 No Step-by-step explanation: x⁵ + 2x³ + x – 1 -> degree of the polynomial is 5 So, when x⁵ is divided by x³, the quotient should be x⁵⁻³ = x². So, x³ – 3x + 1 cannot be a quotient", "following shows the right quotient? 3. Use synthetic division to divide the polynomial, $2x^3 + 5x^2 + 9$ by $x +3$. Which of the following shows the right quotient? 4. Use synthetic division to divide the polynomial, $x^5-3x^3-4x-1$ by $x – 1$. Which of the following shows the right quotient? 5. Use synthetic division to divide the polynomial, $– 2x^4 + x$ by $x – 3$. Which of the following shows the right quotient? 6. Which of the following shows the quotient and remainder when $– x^5 + 1$ is divided by $x + 1$? Use synthetic division. 7. Which of the following shows the quotient and remainder when $2x^3 – 13x^2 + 17x – 10$ is divided by $x- 5$? Use synthetic division. 8. True or False: When $x^4-3x^3-11x^2+5x+17$ is divided by $x +2$, the remainder is equal to $3$. 9. True or False: When $x^4-3x^3-11x^2+5x+17$ is divided by $2x – 1$, the remainder is equal to $3$.", "the other hand, the degree of $\\mathbb{Q}(a)$ over $\\mathbb{Q}$ has to be a divisor of $3$. Therefore $A=\\mathbb{Q}(a)\\cong \\mathbb{Q}[X]/(f(X))$. Conversely, if $f(X)\\in\\mathbb{Q}[X]$ is a degree $3$ irreducible polynomial, then $\\mathbb{Q}[X]/(f(X))$ is a $3$-dimensional division algebra over $\\mathbb{Q}$.", "left over after dividing. 4 16x2 −12x+8 3. One method is long division, a process similar to long division of two whole numbers. Just remember that we keep going until the remainder has degree that is strictly less that the degree of the polynomial we’re dividing by, \\(x + 2$$ in this case. Answer: First, factor by grouping. In this case, we should get 4x 2 /2x = 2x and 2x(2x + 3). Set up the division problem. 3: Dividing Polynomials Worksheet is suitable for 9th - 12th Grade. <br /> 2. So we write the polynomial 2x 4 +3x 2 +x as product of x and 2x 3 + 3x +1. Polynomials are very similar to integers. Oct 20, 2020 · Multiplying And Dividing Polynomials Worksheet With Answers. G o t a d i f f e r e n t a n s w e r ? C h e c k i f i t ′ s c o r r e c t. Then, divide the first term of the divisor into the first term of the dividend, and multiply the X in the quotient by the divisor. Practice. Dividing Polynomials; Remainder and Factor Theorems . We have the divisor divided into the dividend, and of course our answer is the quotient. In this", "+0 # We have that $3 \\cdot f(x) + 4 \\cdot g(x) = h(x)$ where $f(x),$ $g(x),$ and $h(x)$ are all polynomials in $x.$ If the degree of $f(x)$ is \\$8 0 37 1 We have that $$3 \\cdot f(x) + 4 \\cdot g(x) = h(x)$$ where $$f(x),$$ $$g(x),$$ and $$h(x)$$ are all polynomials in $$x$$ If the degree of $$f(x)$$ is $$8$$ and the degree of $$h(x)$$ is $$9$$, then what is the minimum possible degree of $$g(x)$$? Apr 18, 2021", "polynomial For dividing a polynomial with another polynomial, the polynomial is written in standard form i.e. the terms of the dividend and the divisor are arranged in decreasing order of their degrees. Let us take an example. Example: Divide 3x3 – 8x + 12 by x – 1. Solution: The Dividend is 3x3 – 8x + 12 and the divisor is x – 1. After this, the leading term of the dividend is divided by the leading term of the divisor i.e. 3x3 ÷ x =3x2. This result is multiplied by the divisor i.e. 3x2(x -1) = 3x3 -3x2 and it is subtracted from the divisor. Now again, this result is treated as a dividend and the same steps are repeated until the remainder becomes zero or its degree becomes less than that of the divisor as shown below.", "back to Step 2 applied to the quotient. Exercises 1. Find the rational zeros of each polynomial function 2. Factor each over the set of integers 1. $$x^3 + 7x^2 + 16x + 12$$ 2. $$2x^3 - 9x^2 + 14x - 10$$ 3. $$4x^4 - 28x^3 + 67x^2 - 63x + 18$$ 4. $$x^5 + x^4 - 6x^3 + 8x - 16$$", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) The polynomial's degree is $6$. RECALL: (1) The degree of a term is equal to the sum of the exponents of its variables. (2) The degree of a polynomial is equal to the degree of the term with the highest degree. The terms of the given polynomial have the following degrees: $-x^4$: degree = 4 $5x^3$: degree = 3 $3x^6$: degree = 6 $-1$: degree = 0 The term with the highest degree is $3x^6$. The degeree of this term is $6$. Thus, the polynomial's degree is $6$." ]

[ "- +1 Conceptually, this is the best way to view it, and is probably the answer that one would receive from most algebraists. – Bill Dubuque Oct 6 '12 at 14:11 Persistance. You want formulas to make sense also when abusively applying them to cases involving the zero polynomial. For example, we have $\\deg(f\\cdot g)=\\deg f +\\deg g$ and $\\deg (f+g)\\le \\max\\{\\deg f, \\deg g\\}$. Therefore we assign a symbolic value - and be it only for mnemonic purposes - of $-\\infty$ as the degree of $0$, because that makes $-\\infty =\\deg(0\\cdot g)=-\\infty+\\deg g$ and $\\deg g = \\deg (0+g)=\\max\\{-\\infty,\\deg g\\}$ work. We want $\\deg(P\\cdot Q)=\\deg P+\\deg Q$ for two polynomials. In particular $\\deg\\mathbf 0=\\deg P+\\deg\\mathbf 0$, so we can't take an integer. $+\\infty$ could be a choice as we want $\\deg(P+Q)\\leq \\max(\\deg P,\\deg Q)$ but with the definition $\\deg((\\alpha_j)_{j\\geq 0}:=\\sup\\{k\\geq 0\\mid \\alpha\\neq 0\\}$, we take a supremum over an emptyset, so we take $-\\infty$.", "# Difference between revisions of \"2005 AMC 12A Problems/Problem 24\" ## Problem Let $P(x)=(x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \\cdot R(x)$? $\\mathrm {(A) } 19 \\qquad \\mathrm {(B) } 22 \\qquad \\mathrm {(C) } 24 \\qquad \\mathrm {(D) } 27 \\qquad \\mathrm {(E) } 32$ ## Solution We can write the problem as $P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\\cdot R(x)=(x-1)(x-2)(x-3)\\cdot R(x)$. Since $\\deg P(x) = 3$ and $\\deg R(x) = 3$, $\\deg P(x)\\cdot R(x) = 6$. Thus, $\\deg P(Q(x)) = 6$, so $\\deg Q(x) = 2$. $P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\\cdot R(1)=0,$ $P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\\cdot R(2)=0,$ $P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\\cdot R(3)=0.$ Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen. However, we have included $Q(x)$ which are not quadratics: lines. Namely, $Q(1)=Q(2)=Q(3)=1 \\Rightarrow Q(x)=1,$ $Q(1)=Q(2)=Q(3)=2 \\Rightarrow Q(x)=2,$ $Q(1)=Q(2)=Q(3)=3 \\Rightarrow Q(x)=3,$ $Q(1)=1, Q(2)=2, Q(3)=3 \\Rightarrow Q(x)=x,$ $Q(1)=3, Q(2)=2, Q(3)=1 \\Rightarrow Q(x)=4-x.$ Clearly, we could not have included any other constant functions. For any linear function, we have $2\\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is y-value of the midpoint of $(1, Q(1))$ and $(3, Q(3))$. So we have not included any other linear functions. Therefore,", "+0 # Polynomial division 0 205 1 1) The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? 2) When the polynomial p(x) is divided by x - 1, the remainder is 3. When the polynomial p(x) is divided by x - 3, the remainder is 5. What is the remainder when the polynomial p(x) is divided by (x - 1)(x - 3)? Aug 14, 2019 $$f(x) = q(x) d(x) + r(x)\\\\ \\text{If }\\text{deg}\\;d \\le 3, \\text{then deg } r < \\text{deg }d \\le 3 \\implies \\text{Contradiction!}\\\\ \\therefore \\text{deg }d \\ge 4\\\\ \\text{Minimum possible deg }d = 4\\\\ \\implies \\text{Maximum possible deg }q = \\text{deg }f - \\text{deg }d = \\boxed5$$", "that $a.(a.x^N)^N=(a^N)(x^N)$ which implies that $(a-1).(a^N)=0 So a=1$. The degree of P has to be $N$, so let's suppose that $P(x)=a.x^N+Q(x)$ where $Q$ is a polynomial of degree $deg(Q)\\<N$; And we have: $P(P(x))=a.(a.x^N+Q(x))^N+Q(a.x^N+Q(x))$ and $(P(x))^N=(a.x^N+Q(x))^N$ the equation gives us that $Q(a.x^N+Q(x))=(1-a).(a.x^N+Q(x))^N$ We know that $deg(a.x^N+Q(x))=max(N,deg(Q))=N$ ,so these last two equations imply that $N.deg(Q)=N²$ which means that $deg(Q)=N$ (with N different from 0) which is NOT true. If $N=0$, then $n=0$ (because $n=N$) and we get back to the first case :) This prouves that in the case of $n=N$ there is only one solution: $P(x)=x^N$ for each real x. 2 years ago 1. Let $P(x)= a_{n}x^{n}+ ...$ As degree $P(x)=n$, $deg[P(x)]^{N}=nN$ and $degP\\circ P =n\\times n$ and so $n= N$ 2. As $P \\circ P(x)= a_{n}\\times [P(x)]^{n}+... = a_{n}^{n+1}x^{n^{2}}+...$ and $[P(x)]^{n} = a_{n}^{n}x^{n^{2}} +...$, we have $a_{n}^{n} = a_{n}^{n+1}$ and so $a_{n} =1$ Then $P(x)= x^{n}+...$ 3. $P\\circ P(x)= [P(x)]^{n}+Q(x)$, by one hand.But $P\\circ P = P^{n},$ and then $Q(x)\\equiv 0$. So $P(x) = x^{N}$ 2 years ago First of all we will try to prove that in order for the equality to hold, the degree of $P(x)$ MUST BE $N$. This is trivial. Let's assume the degree of $P(x)$ is $k$, so the degree of $P(P(x))$ is $k^2$ and the degree of ${P(x)}^N$ is $kN$ therefore $k=N$.", "(a*b)n+n (6) deg(a*b)n+n = 2n = deg(a) + deg(b) QED References ### Alternate Form of the Division Algorithm for Polynomials Theorem: Alternative Form of the Division Algorithm for Polynomials Let F be a field Let f(x),g(x) be polynomials of F[x] where g(x) ≠ 0 Then: There exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) - r(x) and r(x)=0 or deg r(x) is less than deg g(x) Proof: (1) Let f(x),g(x) be two polynomials such that g(x) ≠ 0 (2) We can assume that deg f(x) ≥ deg g(x) since: if deg f(x) = 0, then f(x) = g(x)(0) - 0. if deg f(x) is less than deg g(x), then f(x) = g(x)(0) - [-f(x)] (3) Let: f(x) = anxn + ... + a0 g(x) = bmxm + ... + b0 where an and bm are nonzero. [We can make this assumption since g(x) ≠ 0 and deg f(x) ≥ deg g(x).] (4) I will use induction to establish the existence of q(x), r(x). (5) q(x),r(x) exist if deg f(x) = 0 since: (a) Assume deg f(x) = 0 (b) Then there exists C such that f(x)=C (c) Since deg f(x) ≥ deg g(x), it follows that deg g(x)= 0 and there exists D such that g(x)=D (d) Let q(x) = C/D (e) Then it", "# Units (invertibles) of polynomial rings over a field If $R$ is a field, what are the units of $R[X]$? My attempt: Let $f,g \\in R[X]$ and $f(X)g(X)=1$. Then the only solution for the equation is both $f,g \\in {R}$. So $U(R[X])=R$, exclude zero elements of $R$. Is this correct ? • The units of $R[X]$ certainly contain the units of $R$, yes? Can other elements work? – Lepidopterist Apr 8 '13 at 3:16 • Sounds like a good start. Now equate coefficients. – James S. Cook Apr 8 '13 at 3:16 • Do you know that the deg$(f\\cdot g)=$ deg$(f)+$ deg$(g)$? And if you want $f\\cdot g =1$, compare degrees on both sides to see what you can conclude.. – RKD Apr 8 '13 at 3:16 • Abstract duplicate of math.stackexchange.com/q/19132/264 – Zev Chonoles Apr 8 '13 at 3:17 • Duplicate of this question and others. – Math Gems Apr 8 '13 at 3:17 If $f=a_0+a_1x+\\cdots+a_mx^m$ has degree $m$, i.e. $a_m\\ne 0$, and $fg=1$ for some $g=b_0+\\cdots+b_n x^n$ (and $b_n\\ne 0$), then observe that $$0=\\deg(1)=\\deg(fg)=\\deg(a_0b_0+\\cdots+a_mb_nx^{n+m})=m+n$$ as $a_mb_n\\ne 0$. Hence $m+n=0$ and so $m=n=0$ as $m,n\\ge 0$. Hence $f\\in R^*$ and $R[x]^*=R^*$ (the star denotes the set of units).", "## Thursday, January 29, 2009 ### Alternate Form of the Division Algorithm for Polynomials Theorem: Alternative Form of the Division Algorithm for Polynomials Let F be a field Let f(x),g(x) be polynomials of F[x] where g(x) ≠ 0 Then: There exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) - r(x) and r(x)=0 or deg r(x) is less than deg g(x) Proof: (1) Let f(x),g(x) be two polynomials such that g(x) ≠ 0 (2) We can assume that deg f(x) ≥ deg g(x) since: if deg f(x) = 0, then f(x) = g(x)(0) - 0. if deg f(x) is less than deg g(x), then f(x) = g(x)(0) - [-f(x)] (3) Let: f(x) = anxn + ... + a0 g(x) = bmxm + ... + b0 where an and bm are nonzero. [We can make this assumption since g(x) ≠ 0 and deg f(x) ≥ deg g(x).] (4) I will use induction to establish the existence of q(x), r(x). (5) q(x),r(x) exist if deg f(x) = 0 since: (a) Assume deg f(x) = 0 (b) Then there exists C such that f(x)=C (c) Since deg f(x) ≥ deg g(x), it follows that deg g(x)= 0 and there exists D such that g(x)=D (d) Let q(x) = C/D (e) Then it follows that f(x) = g(x)*q(x) -", "2. ## Re: Rings of Polynomials _ Division ALgorithm Originally Posted by Bernhard Suppose we have: f(x) = $q_1 (x)g(x) + r_1(x)$ f(x) = $q_2 (x)g(x) + r_2(x)$ where $r_i(x) = 0$ or deg $r_i(x) \\leq$ deg g(x). First, it should say $\\deg r_i(x) < \\deg g(x)$. Originally Posted by Bernhard Subtracting the two equations give we get 0 = $[q_1(x) - q_2(x)] g(x) + [r_1(x) - r_2(x)]$ Hence (3) $[q_1(x) - q_2(x)]g(x) = r_2(x) - r_1(x)$ We must have $r_2(x) - r_1(x) = 0$, for otherwise deg $[r_2(x) - r_1(x)] \\leq$ deg g(x) which would makje equation (3) impossible In general, if $s_1(x)\\ne0$, then $\\deg s_1(x)s_2(x)\\ge\\deg s_2(x)$. So, if $q_1(x)\\ne q_2(x)$, then $\\deg [q_1(x) - q_2(x)]g(x)\\ge\\deg g(x)$. On the other hand, $\\deg r_2(x)<\\deg g(x)$ and $\\deg r_1(x)<\\deg g(x)$, so $\\deg (r_2(x)-r_1(x))<\\deg g(x)$. 3. ## Re: Rings of Polynomials _ Division ALgorithm in general, deg(f+g) = max{deg(f),deg(g)} (the 0-polynomial sort of goobers things up. for this reason deg(0) is often taken to be -∞). also deg(fg) = deg(f) + deg(g) (the 0-polynomial REALLY messes with this formula. but see above). from the second rule we get deg(-f) = deg((-1)(f)) = deg(-1) + deg(f) = 0 + deg(f) = deg(f). now in the proof at hand: if r1(x) ≠ r2(x), then deg(r1 - r2) = max{deg(r1),deg(r2)} > -∞ (the", "2002. Define $\\deg(P)$ to mean the degree of $P(x)$, and likewise $\\deg(Q)$ and $\\deg(R)$. Observe that $\\max\\{\\deg(P), \\deg(Q)\\} = \\deg(R)$. This is because $\\deg(R^2) = \\deg(P^2 + Q^2) = \\max \\{\\deg(P^2), \\deg(Q^2)\\}$ (leading coefficients of $P^2$ and $Q^2$ are positive and do not cancel), and $2\\deg(R) = \\deg(R^2) = \\max \\{\\deg(P^2), \\deg(Q^2)\\} = 2 \\max \\{\\deg(P), \\deg(Q)\\}$. Thus one of $P, Q$ (without loss of generality, $Q$) must have degree 2, and both $P$ and $R$ must have degree 3. Now we write $P^2 = (R + Q).(R-Q)$. We know that both factors have degree 3. Since $P$ has either 1 or 3 real roots, $R+Q$ and $R-Q$ either both have one real root or three real roots accordingly. Assume that $P, R+Q$ and $R-Q$ each have one real root; let the root of $P$ be $r_1$. Since $r_1^2|P^2$, this means that $r_1$ is a root of both $R+Q$ and $R-Q$, and is therefore a root of both $R$ and $Q$. Now let $P_0, Q_0$, and $R_0$ be the three polynomials $P, Q$, and $R$ with the common root divided out. We have $P_0^2 +Q_0^2 = R_0^2$. Then all three polynomials have real coefficients, $P_0$ and $R_0$ have degree 2, and $Q_0$ has degree 1. We also know that $P_0$ has 0 real roots, $Q_0$ has 1, and", "# Integrating 2x/(3x+1) 1. Oct 12, 2013 ### lionely 1. The problem statement, all variables and given/known data Verify, by division, that 2x/(3x+1) = 2/3 - 2/3(3x+1) Hence, evaluate ∫2x/(3x+1) dx I don't understand what to, does the question mean to do long division? Help is much appreciated! 2. Oct 12, 2013 ### CompuChip It's easier to start from the right hand side: take 2/3 - 2/3(3x + 1) and combine the fractions into one. After simplification, you should get 2x/(3x + 1). 3. Oct 12, 2013 ### vanhees71 $$\\frac{2x}{3x+1}=\\frac{2}{3} \\cdot \\frac{3x}{3x+1}=\\frac{2}{3} \\cdot \\frac{3x+1-1}{3x+1}=\\ldots$$ 4. Oct 12, 2013 ### lionely Oh thanks, vanhees. But is that still division? 5. Oct 12, 2013 ### Ray Vickson \"Division\" means re-writing the expression so that in the remainder the numerator has a smaller degree than the denominator. So, if we have f(x) = p(x)/q(x), it is already in \"divided\" form if deg(p) < deg(q); otherwise (if deg(p) ≥ deg(q)), manipulate the expression to get f(x) = m(x) + r(x)/q(x), with deg(r) < deg(q). (Of course, I mean that p,q,m,r are all polynomials in x.) In your example, deg(p) = 1 = deg(q), so you need to re-write the expression until the remainder has numerator of degree 0 (that is, has the form c/(3x+1) for constant c). 6. Oct 12, 2013 ###", "+0 # Algebra Questions +1 149 4 I need some help on three homework questions, thank you! 1. Let P(x) = 0 be the polynomial equation of least possible degree, with rational coefficients, having $$\\sqrt[3]{7} + \\sqrt[3]{49}$$ as a root. Compute the product of all of the roots of P(x) = 0. 2. A polynomial with integer coefficients is of the form $$3x^3 + a_2 x^2 + a_1 x - 6 = 0$$.Enter all the possible integer roots of this polynomial, separated by commas. 3. The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? Apr 18, 2019 #1 +7685 +2 $$\\quad (\\sqrt[3]{7}+\\sqrt[3]{49})^3\\\\ = 7 + 49 + 3\\sqrt[3]{7\\cdot49}(\\sqrt[3]{7}+\\sqrt[3]{49})\\\\ = 56 + 21(\\sqrt[3]{7}+\\sqrt[3]{49})$$ Therefore P(x) is x3 - 21x - 56. Product of roots = $$-\\dfrac{-56}{1} = 56$$ . Apr 18, 2019 #4 +2 Guest Apr 18, 2019 #2 +7685 +2 $$P(x) = 3x^3+a_2x^2+a_1x-6$$ Product of roots = $$-\\dfrac{-6}{3}$$= 2. Possible roots = {1,-1,2,-2}. Apr 18, 2019 #3 +7685 +2 $$f(x) = q(x)d(x)+r(x)$$ deg f = 9, deg r = 3, deg d > deg r. For maximum deg q. deg d is minimum. => deg d = 4.", "If $p(x)=f(x)g(x)$ then deg($f$)+deg($g$)= deg($p$). One of these degrees must be $1$ and the other $0$. A polynomial of degree 0 is a unit. • The OP asked for hints... – lhf Oct 14 '16 at 2:00 • @lhf You're right I should tag as spoiler. – JKEG Oct 14 '16 at 2:02 • @JamesDickens $p$ is irreducible if $p=fg \\implies f$ is a unit or $g$ is a unit. :P (Associates of $1_R$ are precisely units, so our definitions are pretty much identical.) – JKEG Oct 14 '16 at 2:05", "= 3x4 + 6x3 – 2x2 – 10x – 5 are – $$\\sqrt { \\frac { 5 }{ 3 } }$$, $$\\sqrt { \\frac { 5 }{ 3 } }$$, -1 and -1. Question 4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x). Solution: We know that Dividend = Divisor x Quotiet + Remainder ∴ x² – 3x² + x + 2 = g(x) x (x – 2) + (- 2x + 4) or x3– 3x2 + x + 2 = g(x) (x – 2) + (-2x + 4) or x3 – 3x2 + x + 2 + 2 x- 4 = g(x) x (x-2) or x3 – 3x2 + 3x – 2 = g(x) x (x – 2) Question 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 Solution: (i) deg p(x) = deg q(x) Polynomial p(x) = 2x2– 2x + 14; g(x) = 2 q(x) = x2 – x + 7 r(x) = 0 Here, deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) Polynomial p(x) = x3+ x2", "$f(x)=0\\cdot g+f$, and you're done. Otherwise, suppose $\\deg(f)=n\\geq m=\\deg(g)$, and let $a_n$ be the leading coefficient of $f$. Set $$b_mf(x)-a_nx^{n-m}g(x)=f_1(x).$$ Since the coefficient of $x_n$ in $b_mf(x)$ and $a_nx^{n-m}g(x)$ is $a_nb_m$ in both cases, $\\deg(f_1)<\\deg(f)$. So by induction on the degree of $f(x)$, one can find $j_1\\in\\mathbb{N},q_1(x),r(x)\\in R[x]$ with $\\deg(r)<\\deg(g)$ such that $$b_m^{j_1}f_1(x)=g(x)q_1(x)+r(x).$$ Combining these, you find $$b_m^{j_1+1}f(x)=b_m^{j_1}a_nx^{n-m}g(x)+g(x)q_1(x)+r(x)=g(x)q(x)+r(x)$$ where $q(x)=b_m^{k_1}a_n^{n-m}+q_1(x)$. So in particular, if $b_m$ is a unit, you can multiply both sides of the above equation by $b_m^{-1}$ appropriately many times to find $$f(x)=q(x)g(x)+r(x)$$ for slightly different $q(x)$ and $r(x)$, but still $\\deg(r)<\\deg(g)$. Thus the division algorithm of course holds for $F[x]$ where $F$ is a field. The following shows how to use the division algorithm to show $F[x]$ is in fact a PID, and BenjaLim's very nice answer shows how to then prove a PID is a UFD. Let $I$ be any nonzero ideal. Then there exists a nonzero polynomial $g(x)$ of minimal degree among the nonzero elements of $I$. Then for any polynomial $f(x)\\in I$, we can apply the division algorithm to find $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has lesser degree than $g(x)$. But then $$r(x)=f(x)-q(x)g(x)\\in I$$ so necessarily $r(x)=0$ lest we contradict the minimality of the degree of $g(x)$. So it follows that $I=(g(x))$. Since the zero ideal is obviously principal, and since $\\mathbb{R}$ is a", "all nonzero polynomials in $S$, and since either $r_i = 0$ or $\\deg(r_i) \\lt \\deg(d)$, it follows that $r_i = 0$. So $f_i = q_id$ for each $i$, that is, $d$ is a common divisor of $f_1, \\dots, f_k$. Second, let $d’ \\in F[X]$ be any common divisor of $f_1, \\dots, f_k$; we need to show that $d$ divides $d’$. Since $d \\in S$, there are polynomials $g_1, \\dots, g_k$ such that $d = f_1g_1 + \\cdots + f_kg_k$. Since $d’$ divides each $f_i$, there are polynomials $h_1, \\dots, h_k$ such that $f_ig_i = d’h_i$, for each $i$. Therefore, $$d = f_1g_1 + \\cdots + f_kg_k = d’h_1 + \\cdots + d’h_k = d'(h_1 + \\cdots + h_k),$$ that is, $d’$ divides $d$. 2. Let $d$ and $d’$ be GCDs of $f_1, \\dots, f_k$. Then both $d$ and $d’$ are, in particular, common divisors of $f_1, \\dots, f_k$, so that $d|d’$ and $d’|d$. So let $a$ be a polynomial such that $d’ = ad$. It follows from Exercise 3 that $\\deg(d) \\le \\deg(d’)$ and that $\\deg(d’) \\le \\deg(d)$, so that $\\deg(d) = \\deg(d’)$. But $\\deg(d’) = \\deg(d) + \\deg(a)$, so we must have $\\deg(a) = 0$, that is, $a \\in F$ is nonzero, as required. 3. The $d$ we chose to prove 1. belongs to $S$, so it", "polynomials and also deg $p(x)=$deg $q(x)=2$. ### ii. deg $q(x)=$deg $r(x)$ Solution: Let, $p(x)=x^5+2x^4++3x^3+5x^2+2$ $g(x)=x^3+x^2+x+1$ $q(x)=x^2+x+1$ and $r(x)=2x^2-2x+1$ According to the division algorithm for polynomials, $p(x)=g(x)\\times q(x)+r(x)$ Now, $g(x)\\times q(x)+r(x)=(x^3+x^2+x+1)(x^2+x+1)+2x^2-2x+1$ $\\Rightarrow g(x)\\times q(x)+r(x)= x^5+2x^4++3x^3+5x^2+2 =p(x)$ Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg $q(x)=$deg $r(x)=2$. ### iii. deg $r(x)=0$ Solution: Let, $p(x)=2x^4+8x^3+6x^2+4x+12$ $g(x)=x^4+4x^3+3x^2+2x+1$ $q(x)=2$ and $r(x)=10$ According to the division algorithm for polynomials, $p(x)=g(x)\\times q(x)+r(x)$ Now, $g(x)\\times q(x)+r(x)=(x^4+4x^3+3x^2+2x+1\\times 2+10$ $\\Rightarrow g(x)\\times q(x)+r(x)= 2x^4+8x^3+6x^2+4x+12=p(x)$ Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg $r(x)=0$. ### NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 PDF Download The PDF of NCERT solutions for Class 10 Maths Chapter 2, Exercise 2.3 is provided below. ### List of Exercises in Polynomials, Chapter 2, Class 10 Maths I hope the NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear. If you have any issues/queries related to NCERT solutions for Class 10 Maths, Chapter 2, Ex 2.3, feel free to contact me at [email protected] or fill the form here. ### What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 2? The main topics covered", "\\in S$. Claim 2: if $f \\in S$ and $g \\in F[X]$, then $fg \\in S$. To see this, let $f \\in S$ and $g \\in F[X]$. So there are $g_1, \\dots, g_k \\in F[X]$ such that $f = f_1g_1 + \\cdots + f_k g_k$. Therefore, $$fg = (f_1g_1 + \\cdots + f_kg_k) g = f_1(g_1g) + \\cdots + f_k(g_kg),$$ which means that $fg \\in S$. We now return to the proof of the proposition: note that $f_i \\in S$, since $f_i = f_1\\cdot 0 + \\cdots + f_i \\cdot 1 + \\cdots + f_k \\cdot 0$; in particular, the set $S$ is not empty. 1. Let $d \\in S$ be nonzero such that $\\deg(d)$ is minimal among all possible degrees of nonzero polynomials in $S$; we claim that $d$ is a GCD of $f_1, \\dots, f_k$. First, we check that $d$ is a common divisor: by polynomial division, there are polynomials $q_i$ and $r_i$, for $i=1, \\dots, k$, such that $$f_i = q_id + r_i$$ and either $r_i = 0$ or $\\deg(r_i) \\lt \\deg(d)$. This means that $$r_i = f_i – q_i d;$$ but by Claim 2, since $d \\in S$ we have $q_id \\in S$, so by Claim 1, since $f_i \\in S$ we get $r_i \\in S$. Since $\\deg(d)$ is minimal among the degrees of", "polynomial, then $f(x) = 3x^2 + 3 \\\\ g(x) = 4x^3 + 2x \\\\ \\deg(f \\cdot g) = \\deg(f(x) \\cdot g(x)) = \\deg(f) + \\deg(g) = 5 \\\\$ • If one of the polynomials is zero constant polynomials, the addition still work. • Assume the $$\\deg(0) = -\\infty$$ $f(x) = 0 \\\\ g(x) = 4x^3 + 2x \\\\ \\deg(f) \\cdot g) = \\deg(f \\cdot g) = \\deg(f) + \\deg(g) = 5 + -1 = 4$ • The addition of two degree of two polynomials still work. $f(x) = 0 \\\\ g(x) = 4x^3 + 2x \\\\ \\deg(f \\cdot g) = \\deg(f) + \\deg(g) = -\\infty + 3 = -\\infty \\\\$ • The addition of two polynomials? $f(x) = 3x^2 + 3 \\\\ g(x) = 4x^3 + 2x \\\\ \\deg(f + g) = \\max(\\deg(f), \\deg(g)) = \\max(2, 3) = 3 \\\\$ • What about one polynomial is zero constant polynomial. it still works. $f(x) = 0 \\\\ g(x) = 4x^3 + 2x \\\\ \\deg(f + g) = \\max(\\deg(f), \\deg(g)) = \\max(2, -\\infty) = 2$ #### 3.0.20 In Java, negative and positive infinity are defined. I never realize that.(15-11-2019) double inf = Double.POSITIVE_INFINITY; System.out.println(inf + 5); System.out.println(inf - inf); // same as Double.NaN System.out.println(inf * -1); // same as Double.NEGATIVE_INFINITY // Infinity // NaN // -Infinity #### 3.0.21 Rename vimlatex.sh", "deg p(x) = deg q(x) (ii) p(x) = x5 + 2x4 + 3x3+ 5x2 + 2 q(x) = x2 + x + 1, degree of q(x) = 2 g(x) = x3 + x2 + x + 1 r(x) = 2x2 – 2x + 1, degree of r(x) = 2 Here, deg q(x) = deg r(x) (iii) Let p(x) = 2x4 + x3 + 6x2 + 4x + 12 q(x) = 2, degree of q(x) = 0 g(x) = x4 + 4x3 + 3x2 + 2x + 6 r(x) = 0 Here, deg q(x) = 0, Example 8: If the zeroes of polynomial x3 – 3x2 + x + 1 are a – b, a , a + b. 3x has been divided leaving no remainder, and can therefore be marked as used. p(x) = x3 – 3x2 + x + 2 q(x) = x – 2 and r (x) = –2x + 4 By Division Algorithm, we know that p(x) = q(x) × g(x) + r(x) Therefore, x3 – 3x2 + x + 2 = (x – 2) × g(x) + (–2x + 4) ⇒ x3 – 3x2 + x + 2 + 2x – 4 = (x – 2) × g(x) $$\\Rightarrow g(\\text{x})=\\frac{{{\\text{x}}^{3}}-3{{\\text{x}}^{2}}+3\\text{x}-2}{\\text{x}-2}$$ On dividing x3 – 3x2 + x + 2 by x – 2, we", "just need to show that $2 f(x)$ is in the ideal generated by $p(x)$ and $r(x)$. But $c p(x) - d r(x) = c (dx+p_0) - d (cx+r_0) = c p_0 - d r_0 = 1$, so this ideal is the whole ring and every polynomial is in it. Explicitly, $2 f(x) = (2 c f(x)) p(x) - (2 d f(x)) r(x)$. $\\square$ Step 2 Making $2 f(x) = p(x) q_1(x) + r(x) s_1(x)$ with $\\deg q_1$ and $\\deg s_1 \\leq \\deg f -1$. Suppose that we have found polynomial $Q(x)$ and $S(x)$ with $2 f(x) = p(x) Q(x) + r(x) S(x)$ and $\\deg Q$ or $\\deg R \\geq \\deg f$. We will show that we can replace them by polynomials $Q'(x)$ and $S'(x)$ of lower degree, still obeying the equation $2 f(x) = p(x) Q'(x) + r(x) S'(x)$. Since we are supposing that at least one of $p(x) Q(x)$ and $r(x) S(x)$ has degree larger than the sum $p(x) Q(x) + r(x) S(x)$, we see that the leading terms must cancel. Let the leading terms of $Q$ and $S$ be $Q_m x^m + \\cdots$ and $S_m x^m + \\cdots$. So $d Q_m + c S_m=0$. Using our above observation that $GCD(c,d)=1$, we see that $Q_m = c F$ and $S_m = -d F$ for some integer $F$." ]

[ "- +1 Conceptually, this is the best way to view it, and is probably the answer that one would receive from most algebraists. – Bill Dubuque Oct 6 '12 at 14:11 Persistance. You want formulas to make sense also when abusively applying them to cases involving the zero polynomial. For example, we have $\\deg(f\\cdot g)=\\deg f +\\deg g$ and $\\deg (f+g)\\le \\max\\{\\deg f, \\deg g\\}$. Therefore we assign a symbolic value - and be it only for mnemonic purposes - of $-\\infty$ as the degree of $0$, because that makes $-\\infty =\\deg(0\\cdot g)=-\\infty+\\deg g$ and $\\deg g = \\deg (0+g)=\\max\\{-\\infty,\\deg g\\}$ work. We want $\\deg(P\\cdot Q)=\\deg P+\\deg Q$ for two polynomials. In particular $\\deg\\mathbf 0=\\deg P+\\deg\\mathbf 0$, so we can't take an integer. $+\\infty$ could be a choice as we want $\\deg(P+Q)\\leq \\max(\\deg P,\\deg Q)$ but with the definition $\\deg((\\alpha_j)_{j\\geq 0}:=\\sup\\{k\\geq 0\\mid \\alpha\\neq 0\\}$, we take a supremum over an emptyset, so we take $-\\infty$.", "# Units (invertibles) of polynomial rings over a field If $R$ is a field, what are the units of $R[X]$? My attempt: Let $f,g \\in R[X]$ and $f(X)g(X)=1$. Then the only solution for the equation is both $f,g \\in {R}$. So $U(R[X])=R$, exclude zero elements of $R$. Is this correct ? • The units of $R[X]$ certainly contain the units of $R$, yes? Can other elements work? – Lepidopterist Apr 8 '13 at 3:16 • Sounds like a good start. Now equate coefficients. – James S. Cook Apr 8 '13 at 3:16 • Do you know that the deg$(f\\cdot g)=$ deg$(f)+$ deg$(g)$? And if you want $f\\cdot g =1$, compare degrees on both sides to see what you can conclude.. – RKD Apr 8 '13 at 3:16 • Abstract duplicate of math.stackexchange.com/q/19132/264 – Zev Chonoles Apr 8 '13 at 3:17 • Duplicate of this question and others. – Math Gems Apr 8 '13 at 3:17 If $f=a_0+a_1x+\\cdots+a_mx^m$ has degree $m$, i.e. $a_m\\ne 0$, and $fg=1$ for some $g=b_0+\\cdots+b_n x^n$ (and $b_n\\ne 0$), then observe that $$0=\\deg(1)=\\deg(fg)=\\deg(a_0b_0+\\cdots+a_mb_nx^{n+m})=m+n$$ as $a_mb_n\\ne 0$. Hence $m+n=0$ and so $m=n=0$ as $m,n\\ge 0$. Hence $f\\in R^*$ and $R[x]^*=R^*$ (the star denotes the set of units).", "# Difference between revisions of \"2005 AMC 12A Problems/Problem 24\" ## Problem Let $P(x)=(x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \\cdot R(x)$? $\\mathrm {(A) } 19 \\qquad \\mathrm {(B) } 22 \\qquad \\mathrm {(C) } 24 \\qquad \\mathrm {(D) } 27 \\qquad \\mathrm {(E) } 32$ ## Solution We can write the problem as $P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\\cdot R(x)=(x-1)(x-2)(x-3)\\cdot R(x)$. Since $\\deg P(x) = 3$ and $\\deg R(x) = 3$, $\\deg P(x)\\cdot R(x) = 6$. Thus, $\\deg P(Q(x)) = 6$, so $\\deg Q(x) = 2$. $P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\\cdot R(1)=0,$ $P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\\cdot R(2)=0,$ $P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\\cdot R(3)=0.$ Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen. However, we have included $Q(x)$ which are not quadratics: lines. Namely, $Q(1)=Q(2)=Q(3)=1 \\Rightarrow Q(x)=1,$ $Q(1)=Q(2)=Q(3)=2 \\Rightarrow Q(x)=2,$ $Q(1)=Q(2)=Q(3)=3 \\Rightarrow Q(x)=3,$ $Q(1)=1, Q(2)=2, Q(3)=3 \\Rightarrow Q(x)=x,$ $Q(1)=3, Q(2)=2, Q(3)=1 \\Rightarrow Q(x)=4-x.$ Clearly, we could not have included any other constant functions. For any linear function, we have $2\\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is y-value of the midpoint of $(1, Q(1))$ and $(3, Q(3))$. So we have not included any other linear functions. Therefore,", "degree of the product of a polynomial by a non-zero scalar is equal to the degree of the polynomial, i.e. $\\deg(cP)=\\deg(P)$. E.g. • The degree of $2(x^2+3x-2)=2x^2+6x-4$ is 2, just as the degree of $x^2+3x-2$. Note that for polynomials over a ring containing divisors of zero, this is not necessarily true. For example, in $\\mathbf{Z}/4\\mathbf{Z}$, $\\deg(1+2x) = 1$, but $\\deg(2(1+2x)) = \\deg(2+4x) =\\deg(2) = 0$. The set of polynomials with coefficients from a given field F and degree smaller than or equal to a given number n thus forms a vector space. (Note, however, that this set is not a ring, as it is not closed under multiplication, as is seen below.) ### Behaviour under multiplication The degree of the product of two polynomials over a field or an integral domain is the sum of their degrees $\\deg(PQ) = \\deg(P) + \\deg(Q)$. E.g. • The degree of $(x^3+x)(x^2+1)=x^5+2x^3+x$ is 3 + 2 = 5. Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in $\\mathbf{Z}/4\\mathbf{Z}$, $\\deg(2x) + \\deg(1+2x) = 1 + 1 = 2$, but $\\deg(2x(1+2x)) = \\deg(2x) = 1$. ### Behaviour under composition The degree of the composition of two non-constant polynomials $P$ and $Q$ over a field or integral domain is the product of their degrees: $\\deg(P \\circ Q) =", "If $p(x)=f(x)g(x)$ then deg($f$)+deg($g$)= deg($p$). One of these degrees must be $1$ and the other $0$. A polynomial of degree 0 is a unit. • The OP asked for hints... – lhf Oct 14 '16 at 2:00 • @lhf You're right I should tag as spoiler. – JKEG Oct 14 '16 at 2:02 • @JamesDickens $p$ is irreducible if $p=fg \\implies f$ is a unit or $g$ is a unit. :P (Associates of $1_R$ are precisely units, so our definitions are pretty much identical.) – JKEG Oct 14 '16 at 2:05", "+0 # Polynomial division 0 205 1 1) The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? 2) When the polynomial p(x) is divided by x - 1, the remainder is 3. When the polynomial p(x) is divided by x - 3, the remainder is 5. What is the remainder when the polynomial p(x) is divided by (x - 1)(x - 3)? Aug 14, 2019 $$f(x) = q(x) d(x) + r(x)\\\\ \\text{If }\\text{deg}\\;d \\le 3, \\text{then deg } r < \\text{deg }d \\le 3 \\implies \\text{Contradiction!}\\\\ \\therefore \\text{deg }d \\ge 4\\\\ \\text{Minimum possible deg }d = 4\\\\ \\implies \\text{Maximum possible deg }q = \\text{deg }f - \\text{deg }d = \\boxed5$$", "2002. Define $\\deg(P)$ to mean the degree of $P(x)$, and likewise $\\deg(Q)$ and $\\deg(R)$. Observe that $\\max\\{\\deg(P), \\deg(Q)\\} = \\deg(R)$. This is because $\\deg(R^2) = \\deg(P^2 + Q^2) = \\max \\{\\deg(P^2), \\deg(Q^2)\\}$ (leading coefficients of $P^2$ and $Q^2$ are positive and do not cancel), and $2\\deg(R) = \\deg(R^2) = \\max \\{\\deg(P^2), \\deg(Q^2)\\} = 2 \\max \\{\\deg(P), \\deg(Q)\\}$. Thus one of $P, Q$ (without loss of generality, $Q$) must have degree 2, and both $P$ and $R$ must have degree 3. Now we write $P^2 = (R + Q).(R-Q)$. We know that both factors have degree 3. Since $P$ has either 1 or 3 real roots, $R+Q$ and $R-Q$ either both have one real root or three real roots accordingly. Assume that $P, R+Q$ and $R-Q$ each have one real root; let the root of $P$ be $r_1$. Since $r_1^2|P^2$, this means that $r_1$ is a root of both $R+Q$ and $R-Q$, and is therefore a root of both $R$ and $Q$. Now let $P_0, Q_0$, and $R_0$ be the three polynomials $P, Q$, and $R$ with the common root divided out. We have $P_0^2 +Q_0^2 = R_0^2$. Then all three polynomials have real coefficients, $P_0$ and $R_0$ have degree 2, and $Q_0$ has degree 1. We also know that $P_0$ has 0 real roots, $Q_0$ has 1, and", "all nonzero polynomials in $S$, and since either $r_i = 0$ or $\\deg(r_i) \\lt \\deg(d)$, it follows that $r_i = 0$. So $f_i = q_id$ for each $i$, that is, $d$ is a common divisor of $f_1, \\dots, f_k$. Second, let $d’ \\in F[X]$ be any common divisor of $f_1, \\dots, f_k$; we need to show that $d$ divides $d’$. Since $d \\in S$, there are polynomials $g_1, \\dots, g_k$ such that $d = f_1g_1 + \\cdots + f_kg_k$. Since $d’$ divides each $f_i$, there are polynomials $h_1, \\dots, h_k$ such that $f_ig_i = d’h_i$, for each $i$. Therefore, $$d = f_1g_1 + \\cdots + f_kg_k = d’h_1 + \\cdots + d’h_k = d'(h_1 + \\cdots + h_k),$$ that is, $d’$ divides $d$. 2. Let $d$ and $d’$ be GCDs of $f_1, \\dots, f_k$. Then both $d$ and $d’$ are, in particular, common divisors of $f_1, \\dots, f_k$, so that $d|d’$ and $d’|d$. So let $a$ be a polynomial such that $d’ = ad$. It follows from Exercise 3 that $\\deg(d) \\le \\deg(d’)$ and that $\\deg(d’) \\le \\deg(d)$, so that $\\deg(d) = \\deg(d’)$. But $\\deg(d’) = \\deg(d) + \\deg(a)$, so we must have $\\deg(a) = 0$, that is, $a \\in F$ is nonzero, as required. 3. The $d$ we chose to prove 1. belongs to $S$, so it", "+0 # Algebra Questions +1 149 4 I need some help on three homework questions, thank you! 1. Let P(x) = 0 be the polynomial equation of least possible degree, with rational coefficients, having $$\\sqrt[3]{7} + \\sqrt[3]{49}$$ as a root. Compute the product of all of the roots of P(x) = 0. 2. A polynomial with integer coefficients is of the form $$3x^3 + a_2 x^2 + a_1 x - 6 = 0$$.Enter all the possible integer roots of this polynomial, separated by commas. 3. The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? Apr 18, 2019 #1 +7685 +2 $$\\quad (\\sqrt[3]{7}+\\sqrt[3]{49})^3\\\\ = 7 + 49 + 3\\sqrt[3]{7\\cdot49}(\\sqrt[3]{7}+\\sqrt[3]{49})\\\\ = 56 + 21(\\sqrt[3]{7}+\\sqrt[3]{49})$$ Therefore P(x) is x3 - 21x - 56. Product of roots = $$-\\dfrac{-56}{1} = 56$$ . Apr 18, 2019 #4 +2 Guest Apr 18, 2019 #2 +7685 +2 $$P(x) = 3x^3+a_2x^2+a_1x-6$$ Product of roots = $$-\\dfrac{-6}{3}$$= 2. Possible roots = {1,-1,2,-2}. Apr 18, 2019 #3 +7685 +2 $$f(x) = q(x)d(x)+r(x)$$ deg f = 9, deg r = 3, deg d > deg r. For maximum deg q. deg d is minimum. => deg d = 4.", "(a*b)n+n (6) deg(a*b)n+n = 2n = deg(a) + deg(b) QED References ### Alternate Form of the Division Algorithm for Polynomials Theorem: Alternative Form of the Division Algorithm for Polynomials Let F be a field Let f(x),g(x) be polynomials of F[x] where g(x) ≠ 0 Then: There exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) - r(x) and r(x)=0 or deg r(x) is less than deg g(x) Proof: (1) Let f(x),g(x) be two polynomials such that g(x) ≠ 0 (2) We can assume that deg f(x) ≥ deg g(x) since: if deg f(x) = 0, then f(x) = g(x)(0) - 0. if deg f(x) is less than deg g(x), then f(x) = g(x)(0) - [-f(x)] (3) Let: f(x) = anxn + ... + a0 g(x) = bmxm + ... + b0 where an and bm are nonzero. [We can make this assumption since g(x) ≠ 0 and deg f(x) ≥ deg g(x).] (4) I will use induction to establish the existence of q(x), r(x). (5) q(x),r(x) exist if deg f(x) = 0 since: (a) Assume deg f(x) = 0 (b) Then there exists C such that f(x)=C (c) Since deg f(x) ≥ deg g(x), it follows that deg g(x)= 0 and there exists D such that g(x)=D (d) Let q(x) = C/D (e) Then it", "\\in S$. Claim 2: if $f \\in S$ and $g \\in F[X]$, then $fg \\in S$. To see this, let $f \\in S$ and $g \\in F[X]$. So there are $g_1, \\dots, g_k \\in F[X]$ such that $f = f_1g_1 + \\cdots + f_k g_k$. Therefore, $$fg = (f_1g_1 + \\cdots + f_kg_k) g = f_1(g_1g) + \\cdots + f_k(g_kg),$$ which means that $fg \\in S$. We now return to the proof of the proposition: note that $f_i \\in S$, since $f_i = f_1\\cdot 0 + \\cdots + f_i \\cdot 1 + \\cdots + f_k \\cdot 0$; in particular, the set $S$ is not empty. 1. Let $d \\in S$ be nonzero such that $\\deg(d)$ is minimal among all possible degrees of nonzero polynomials in $S$; we claim that $d$ is a GCD of $f_1, \\dots, f_k$. First, we check that $d$ is a common divisor: by polynomial division, there are polynomials $q_i$ and $r_i$, for $i=1, \\dots, k$, such that $$f_i = q_id + r_i$$ and either $r_i = 0$ or $\\deg(r_i) \\lt \\deg(d)$. This means that $$r_i = f_i – q_i d;$$ but by Claim 2, since $d \\in S$ we have $q_id \\in S$, so by Claim 1, since $f_i \\in S$ we get $r_i \\in S$. Since $\\deg(d)$ is minimal among the degrees of", "principal, there would exist $d(x)\\in (2,x)$ such that every element of the ideal can be expressed as a product of $d(x)$ with an element of the ring (i.e. $d(x)$ divides every element of the ideal). In particular, this holds for $2$ and $x$ since they are both elements of $(2,x)$. It is not hard to see why this yields a contradiction, if you keep in mind everything we have said so far. • but how do I show that $1$ and $-1$ are the only ones? – Guerlando OCs Jan 10 '16 at 19:11 • You are working in $\\mathbb{Z}[x]$, where $\\deg(p(x)\\cdot q(x)) = \\deg(p(x))+\\deg(q(x))$. This tells you that the degree of the divisors of a certain polynomial is less or equal to the degree of the polynomial itself. But $\\deg(3) = 0$, so any polynomial that divides it must have degree $0$ as well. Does this help you in seeing what the only divisors of $3$ in $\\mathbb{Z}[x]$ are? Once you are there, deducing that $3$ and $x$ have no other common divisors but $1$ and $-1$ is immediate. – Labba Jan 10 '16 at 19:24", "terms with nonzero coefficients, the degree is taken to be -∞. Degrees add when multiplying polynomials: deg((x2+1)(x+5)) = deg(x2+1)+deg(x+5) = 2+1 = 3; this is just a consequence of the product leading terms producing the leading term of the new polynomial. For addition, we have deg(p+q) max(deg(p),deg(q)), but we can't guarantee equality (maybe the leading terms cancel). Because F[x] is a ring, we can't do division the way we do it in a field like ℝ, but we can do division the way we do it in a ring like ℤ, leaving a remainder. The equivalent of the integer division algorithm for ℤ is: Division algorithm for polynomials Given a polynomial f and a nonzero polynomial g in F[x], there are unique polynomials q and r such that f = q⋅g + r and deg(r) < deg(g). The essential idea is that we can find q and r using the same process of long division as we use for integers. For example, in ℚ[x]: x - 1 ______________ x+2 ) x² + x + 5 x² + 2x ------- -x + 5 -x + -5 ======= 10 From this we get x2 + x + 5 = (x+2)(x-1) + 10, with deg(10) = 0 < deg(x+2) = 1. We are going to use this later to define finite", "# Integrating 2x/(3x+1) 1. Oct 12, 2013 ### lionely 1. The problem statement, all variables and given/known data Verify, by division, that 2x/(3x+1) = 2/3 - 2/3(3x+1) Hence, evaluate ∫2x/(3x+1) dx I don't understand what to, does the question mean to do long division? Help is much appreciated! 2. Oct 12, 2013 ### CompuChip It's easier to start from the right hand side: take 2/3 - 2/3(3x + 1) and combine the fractions into one. After simplification, you should get 2x/(3x + 1). 3. Oct 12, 2013 ### vanhees71 $$\\frac{2x}{3x+1}=\\frac{2}{3} \\cdot \\frac{3x}{3x+1}=\\frac{2}{3} \\cdot \\frac{3x+1-1}{3x+1}=\\ldots$$ 4. Oct 12, 2013 ### lionely Oh thanks, vanhees. But is that still division? 5. Oct 12, 2013 ### Ray Vickson \"Division\" means re-writing the expression so that in the remainder the numerator has a smaller degree than the denominator. So, if we have f(x) = p(x)/q(x), it is already in \"divided\" form if deg(p) < deg(q); otherwise (if deg(p) ≥ deg(q)), manipulate the expression to get f(x) = m(x) + r(x)/q(x), with deg(r) < deg(q). (Of course, I mean that p,q,m,r are all polynomials in x.) In your example, deg(p) = 1 = deg(q), so you need to re-write the expression until the remainder has numerator of degree 0 (that is, has the form c/(3x+1) for constant c). 6. Oct 12, 2013 ###", "2. ## Re: Rings of Polynomials _ Division ALgorithm Originally Posted by Bernhard Suppose we have: f(x) = $q_1 (x)g(x) + r_1(x)$ f(x) = $q_2 (x)g(x) + r_2(x)$ where $r_i(x) = 0$ or deg $r_i(x) \\leq$ deg g(x). First, it should say $\\deg r_i(x) < \\deg g(x)$. Originally Posted by Bernhard Subtracting the two equations give we get 0 = $[q_1(x) - q_2(x)] g(x) + [r_1(x) - r_2(x)]$ Hence (3) $[q_1(x) - q_2(x)]g(x) = r_2(x) - r_1(x)$ We must have $r_2(x) - r_1(x) = 0$, for otherwise deg $[r_2(x) - r_1(x)] \\leq$ deg g(x) which would makje equation (3) impossible In general, if $s_1(x)\\ne0$, then $\\deg s_1(x)s_2(x)\\ge\\deg s_2(x)$. So, if $q_1(x)\\ne q_2(x)$, then $\\deg [q_1(x) - q_2(x)]g(x)\\ge\\deg g(x)$. On the other hand, $\\deg r_2(x)<\\deg g(x)$ and $\\deg r_1(x)<\\deg g(x)$, so $\\deg (r_2(x)-r_1(x))<\\deg g(x)$. 3. ## Re: Rings of Polynomials _ Division ALgorithm in general, deg(f+g) = max{deg(f),deg(g)} (the 0-polynomial sort of goobers things up. for this reason deg(0) is often taken to be -∞). also deg(fg) = deg(f) + deg(g) (the 0-polynomial REALLY messes with this formula. but see above). from the second rule we get deg(-f) = deg((-1)(f)) = deg(-1) + deg(f) = 0 + deg(f) = deg(f). now in the proof at hand: if r1(x) ≠ r2(x), then deg(r1 - r2) = max{deg(r1),deg(r2)} > -∞ (the", "## Thursday, January 29, 2009 ### Alternate Form of the Division Algorithm for Polynomials Theorem: Alternative Form of the Division Algorithm for Polynomials Let F be a field Let f(x),g(x) be polynomials of F[x] where g(x) ≠ 0 Then: There exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) - r(x) and r(x)=0 or deg r(x) is less than deg g(x) Proof: (1) Let f(x),g(x) be two polynomials such that g(x) ≠ 0 (2) We can assume that deg f(x) ≥ deg g(x) since: if deg f(x) = 0, then f(x) = g(x)(0) - 0. if deg f(x) is less than deg g(x), then f(x) = g(x)(0) - [-f(x)] (3) Let: f(x) = anxn + ... + a0 g(x) = bmxm + ... + b0 where an and bm are nonzero. [We can make this assumption since g(x) ≠ 0 and deg f(x) ≥ deg g(x).] (4) I will use induction to establish the existence of q(x), r(x). (5) q(x),r(x) exist if deg f(x) = 0 since: (a) Assume deg f(x) = 0 (b) Then there exists C such that f(x)=C (c) Since deg f(x) ≥ deg g(x), it follows that deg g(x)= 0 and there exists D such that g(x)=D (d) Let q(x) = C/D (e) Then it follows that f(x) = g(x)*q(x) -", "pointed out, you are right that nothing guarantees $$\\mathrm{deg}(q)=0$$ if we take the exercise as it is written. But perhaps what the author(s) meant is $$p(x)$$ is divisible by [...], and ONLY by those'', which would then imply that $$\\mathrm{deg}(p)=4$$, and so necessarily $$\\mathrm{deg}(q)=0$$. Now for the second part. The reminder theorem says that for polynomials $$p(x), s(x)$$, there always exist polynomials $$q(x), r(x)$$ such that $$p(x) = s(x)\\cdot q(x)+r(x),$$ where (!) $$\\mathrm{deg}(r)<\\mathrm{deg}(s)$$. Meaning that we can divide $$p$$ by $$s$$ as much as possible'', in the sense that whatever remains has degree strictly lower than the polynomial we are dividing by. In the exercise, we have two instances of division, and get two formulas: $$(1)\\ \\ p(x)=(x-2)(x-3)(x-4)(x-5)q(x),$$ $$(2)\\ \\ p(x)=(x-9)\\cdot t(x)+r(x).$$ Formula $$(1)$$ just says that $$p$$ is divisible by all those linear factors, and $$q$$ is the quotient. Formula $$(2)$$ is the division of $$p$$ by $$(x-9)$$, where we get quotient $$t(x)$$ and some reminder $$r(x)$$, which actually has to be just a number, because $$\\mathrm{deg}(r)<\\mathrm{deg}(x-9)=1$$. Moreover, we are actually told that it is just a number, and even which number it is: $$r=5040$$. Now we substitute $$x\\rightarrow 9$$ into $$(2)$$, and get $$p(9)=(9-9)\\cdot t(9)+r=r=5040.$$ But using $$(1)$$, this means that $$5040=p(9)=(9-2)(9-3)(9-4)(9-5)q(9)=7\\cdot6\\cdot5\\cdot4\\cdot q(9)=840q(9)$$ $$q(9)=5040/840=6$$ Now assuming that $$\\mathrm{deg}(q)=0$$, we know that actually $$q=6$$ (the authors write", "### Polynomials Description: For classes 9 to 12 Number of Questions: 15 Created by: Bharat Dubey Tags: Attempted 0/15 Correct 0 Score 0 A trinomial is a polynomial with exactly how many terms? 1. 3 2. 4 3. 5 4. 6 Correct Option: A What is called a polynomial with exactly 2 terms? 1. binomial 2. degree 3. linear 4. square Correct Option: A How many 0s can a polynomial of degree n have? 1. 1 zero 2. n zero 3. 2 zero 4. n-1 zero Correct Option: B What is the degree of the remainder atmost, when a fourth degree polynomial is divided by a quadratic polynomial? 1. $2$ 2. $0$ 3. $4$ 4. $1$ Correct Option: D Explanation: Here $f(x)$ represent dividend and $g(x)$ represent division $g(x)=$ quadratic polynomial $=ax^2+bx+c$ $\\therefore deg(g(x))=2$, $deg(f(x))=4$ quotient $q(x)$ is of degree $=2$ $(=4-2)$ Remainder $R(x)=$ degree $1$ or less than $1$. If on dividing a non-zero polynomial $p(x)$ by a polynomial $g (x)$, the remainder is zero, what is the relation between the degrees of $p(x)$ and $g (x)$? 1. degree of $g (x) \\ge$ degree of $p(x)$ 2. degree of $g(x) \\le$ degree of $p(x)$ 3. degree of $g (x) =$ degree of $p(x)$ 4. Can't say Correct Option: B Explanation: deg $p(x)=$ deg $g(x)+r(c)$ Then, deg $p(x)", "# Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm deg q(x) = deg r(x) - Mathematics Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm deg q(x) = deg r(x) #### Solution According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x) Degree of a polynomial is the highest power of the variable in the polynomial. deg q(x) = deg r(x) Let us assume the division of x3+ x by x2 Here, p(x) = x3 + x g(x) = x2 q(x) = x and r(x) = x Clearly, the degree of q(x) and r(x) is the same i.e., 1. Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x3 + x = (x2 ) × x + x x3 + x = x3 + x Thus, the division algorithm is satisfied. Concept: Division Algorithm for Polynomials Is there an error in this question or solution? #### APPEARS IN NCERT Class 10 Maths Chapter 2 Polynomials Exercise 2.3 | Q 5.2 | Page 36", "just need to show that $2 f(x)$ is in the ideal generated by $p(x)$ and $r(x)$. But $c p(x) - d r(x) = c (dx+p_0) - d (cx+r_0) = c p_0 - d r_0 = 1$, so this ideal is the whole ring and every polynomial is in it. Explicitly, $2 f(x) = (2 c f(x)) p(x) - (2 d f(x)) r(x)$. $\\square$ Step 2 Making $2 f(x) = p(x) q_1(x) + r(x) s_1(x)$ with $\\deg q_1$ and $\\deg s_1 \\leq \\deg f -1$. Suppose that we have found polynomial $Q(x)$ and $S(x)$ with $2 f(x) = p(x) Q(x) + r(x) S(x)$ and $\\deg Q$ or $\\deg R \\geq \\deg f$. We will show that we can replace them by polynomials $Q'(x)$ and $S'(x)$ of lower degree, still obeying the equation $2 f(x) = p(x) Q'(x) + r(x) S'(x)$. Since we are supposing that at least one of $p(x) Q(x)$ and $r(x) S(x)$ has degree larger than the sum $p(x) Q(x) + r(x) S(x)$, we see that the leading terms must cancel. Let the leading terms of $Q$ and $S$ be $Q_m x^m + \\cdots$ and $S_m x^m + \\cdots$. So $d Q_m + c S_m=0$. Using our above observation that $GCD(c,d)=1$, we see that $Q_m = c F$ and $S_m = -d F$ for some integer $F$." ]

[ "+0 # Domain and Range +1 55 1 +147 Find the greatest integer value of $b$ for which the expression $\\frac{9x^3+4x^2+11x+7}{x^2+bx+8}$ has a domain of all real numbers. Feb 21, 2021 #1 +939 -1 $$x^2 + bx+ 8$$ = 0 has to have no real solutions. So, the discriminant neds to be less than 0. $$\\Delta = b^2 - 4ac= b^2 - 32 < 0$$ $$b^2 < 32$$ The greatest integer value of b is 5. Feb 21, 2021", "# $$\\frac{\\ln b-\\ln a}{b-a}$$ Calculus Level 2 For real numbers $$a$$ and $$b$$ ($$a<b)$$ in the interval $$[4,11],$$ the range of the real number $$k$$ such that $\\frac{\\ln b-\\ln a}{b-a}=k$ can be expressed as $$\\alpha < k < \\beta.$$ What is $$\\frac{1}{\\alpha\\beta}?$$ ×", "\\frac{5}{3}$. Both are in the domain of $f$ (Domain of $f = \\mathbb{R}$), so they are the critical numbers for $f$.", "to 5 b. {5} c. x+2=7 Question 2. a. The set of all real numbers equal to $$\\frac{5}{3}$$ b. {$$\\frac{5}{3}$$} c. 3x=5 Question 3. a. The set of all real numbers greater than 1 b. {x real | x > 1} c. x-1>0 Question 4. a. The set of all real numbers less or equal to 5 a. The set of all real numbers not equal to 2 b. {x real | x ≤ 5} c. 2x≤10 Question 5. a. The set of all real numbers not equal to 4 b. {x real | x≠2} c. $$\\frac{x-2}{x-2}$$=1 Question 6. a. The set of all real numbers b. { x real} or x≠4 c. x-1<3 or x-1>3 Question 7. a. The null set b. {} or ɸ c. c2 = -4 Question 8. a. The set of all real numbers b. c. 2x – 8 = 2(x – 4) Fill in the chart below. For Problems 19–24, answer the following: Are the two expressions algebraically equivalent? If so, state the property (or properties) displayed. If not, state why (the solution set may suffice as a reason), and change the equation, ever so slightly (e.g., touch it up), to create an equation whose solution set is all real numbers. Question 19. x(4-x2)=(-x2 +4)x Yes. Commutative. Question 20. $$\\frac{2 x}{2 x}$$=1", "# Math Help - Using Properties to determine derivative and original equation 1. ## Using Properties to determine derivative and original equation A function f is defined for all real numbers and has the following properties. (i) f(1)=5 (ii) f(3)=21 (iii) for all real values of a and b, f(a+b) - f(a)= kab+2b^2 where k is a fixed real number independent of a and b a) Use a=1 and b=2 to find k. b) Find f'(3) c) Find f'(x) and f(x) for all real x So I found out that k=-4, but I'm stuck on how to find the original equation and thus the derivative by the given info. 2. Originally Posted by ment2byours A function f is defined for all real numbers and has the following properties. (i) f(1)=5 (ii) f(3)=21 (iii) for all real values of a and b, f(a+b) - f(a)= kab+2b^2 where k is a fixed real number independent of a and b a) Use a=1 and b=2 to find k. b) Find f'(3) c) Find f'(x) and f(x) for all real x So I found out that k=-4, but I'm stuck on how to find the original equation and thus the derivative by the given info. First of all, for (a), I think $k=4$, not $-4$. Here's why: $21-5=k*1*2+2*2^2$. Thus $16=2k+8$ and $k=4$.", "minimum value of$$ a+\\frac{1}{b(a-b)} ? $$Let a&gt;b be positive real numbers. What is the minimum value of$$ a+\\frac{1}{b(a-b)} ? ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$.Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. ...", "\\left\\{{0}\\right\\}: a \\div b = \\frac a b = a / b = a \\times \\left({b^{-1}}\\right)$ The validity of all these operations is guaranteed by the fact that the real numbers form a field‎. ### Real Number Line • It can be shown (and intuitively understood) that the set of real numbers is isomorphic to any infinite straight line in space. Thus we can identify any (either physically drawn or imagined) line with the set of real numbers and thereby illustrate truths about the real numbers by means of diagrams. ## Also denoted as Variants on $\\R$ are often seen, for example $\\mathbf R$ and $\\mathcal R$, or even just $R$. ## Also see • Results about real numbers can be found here.", "Consider the function $f(x) = \\frac{100x^2}{1+100x^2}$. Find all ordered triples of real numbers $(a,b,c)$ such that $f(a) = b$, $f(b) = c$, and $f(c) = a$.", "k + 1\\right)}$ for some $k \\in \\mathbb{Z}$ That is, we can define: ${\\log}_{a} \\left(y\\right) = \\ln \\frac{y}{\\ln \\left(- a\\right) + i \\pi \\left(2 k + 1\\right)}$ for some $k \\in \\mathbb{Z}$ Regardless of what value of $k$ we choose, this will have a non-Real Complex value most of the time, but at least we can restrict the domain of ${a}^{z}$ to get a well defined inverse.", "### Functions and graphs - (2.1-2.4) #### Recognise the intuitive concepts of functions, domains and co-domains, independent and dependent variables • In each of the following, y is a function of x. Find the largest domain of real numbers for each function. (a) $$y = {(x - 2)^2}$$ (b) $$y = \\frac{x}{{{{(2x + 1)}^2}}}$$ • (a) Consider $$y = {(x - 2)^2}$$. For any real number x, the corresponding value of y is real according to the expression. The largest domain of real numbers for the function is the set of all real numbers. (b) Consider $$y = \\frac{x}{{{{(2x + 1)}^2}}}$$. For any real number x except $$x = - \\frac{1}{2}$$, the corresponding value of y is real according to the expression. The largest domain of real numbers for the function is the set of all real numbers except $$\\underline{\\underline { - \\frac{1}{2}}}$$. #### Recognise the notation of functions and use tabular, algebraic and graphical methods to represent functions • A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output • $$f(x)=x^2+1$$ find f(1) • $$f(x)=x^2+1$$ $$f(1)=1^2+1=2$$ #### Find the maximum and minimum values of quadratic functions by the algebraic method • Find the maximum / minimum value of each", "# Functional Equation Fun $f(x)+f\\left(\\frac{6x-5}{4x-2}\\right)=x$ Find the number of real-valued functions $f$ that are solutions of the above functional equation and whose domain is the set of all real numbers different from $\\frac{1}{2}$, $1$, and $\\frac{3}{2}$. ×", "of $\\left(f·g\\right)\\left(x\\right)$ is the set of all real numbers. However the domain of $\\left(\\frac{f}{g}\\right)\\left(x\\right)$ cannot include the x-value when g(x) = 0. $0=5x-2\\to 2=5x\\to \\frac{2}{5}=x$ Therefore the domain of $\\left(\\frac{f}{g}\\right)\\left(x\\right)$ is all real numbers except $\\frac{2}{5}$. Step 4: Evaluate the new functions with the given value of x. $\\left(f·g\\right)\\left(2\\right)=20{x}^{2}-8{x}^{2}=4{\\left(2\\right)}^{2}={48}$ $\\left(\\frac{f}{g}\\right)\\left(2\\right)=\\frac{4{x}^{2}}{5x-2}=\\frac{16}{8}={2}$ Related Links: Math algebra Composition of Functions Extreme Value Theorem Calculus Topics To link to this Combination of Functions page, copy the following code to your site:", "(g(a+\\frac{b-a}{n}k)-\\mu )^2}$ Which gives us; Where $\\mu$, the expected value, is determined by;", "+0 # Help plssss thx +1 72 2 +74 Find all values of $$k$$ so that the domain of $$b(x) = \\frac{kx^2 + 2x - 5}{-5x^2 + 2x + k}$$ is the set of all real numbers. Feb 19, 2021 #1 +287 +2 The numerator does not matter in this case; only if the denominator is zero for some real number, the domain will not be set for all real numbers. To prevent the denominator from having real roots, the discriminant must be negative, which means $$2^2-4\\cdot-5\\cdot k = 4+20k$$ is negative. That will only happen if $$\\boxed{k<-\\frac{1}{5}}$$. Feb 19, 2021 #2 +74 +1 It's correct, thanks!! pizzaisawesome972 Feb 20, 2021", "of elements in its domain either by giving a general formula or by listing them one by one. As the domain and codomain of real functions are subsets of R, therefore, conventionally real functions are described by providing the general formula for finding the images of elements in it. In such cases, the domain of the real function f ( x ) is the set of all those real numbers for which the expression for f ( x ) or the formula for f ( x ) assumes real values only. In other words, the domain of f ( x ) is the set of all those real numbers for which f ( x ) is meaningful. Let us understand it through an example. Suppose we have a real function described by the formula, f ( x ) = $\\frac{3 x -2}{x^2-1}$ . This function assumes the real values of all x R except for x = ∓1 because the denominator of $\\frac{3 x -2}{x^2-1}$ becomes zero for x = ∓1. Therefore, the domain of f ( x ) will be the set of all real numbers other than – 1 and 1, i..e the domain of will be the domain ( f ) = R – {- 1,1} The range of a real function of a real", "## Monthly Archives: October 2019 ### Set theory, functions, relations: part VI What follows are some more practice questions on functions. The questions are not challenging but we can say that they do lead to conceptual clarity and present some standard set of questions on this topic (it behooves every beginner in calculus or IITJEE mains or RMO or pre RMO to try these set of questions): 1. Find the domain and range of the function: $f(x)=\\frac{x-2}{3-x}$ 2. If $f(x)=3x^{3}-5x^{2}+9$, find $f(x-1)$. 3. If $f(x)=x^{3}-\\frac{1}{x^{3}}$, show that $f(x)+f(\\frac{1}{x})=0$ 4. If $f(x)=\\frac{x+1}{x-1}$ show that $f(f(x))=x$. 5. Find the domain and range of the real valued function $f(x)=\\frac{x^{2}+2x+1}{x^{2}-8x+12}$ 6. Find the domain of the real valued function of a real variable: $f(x)=\\frac{x-2}{2-x}$ 7. Find the domain and range of the real valued function $f(x)=\\frac{1}{1-x^{2}}$. 8. A function $f: \\Re \\longrightarrow \\Re$ is defined by $f(x)=\\frac{3x}{5}+2$ where $x \\in \\Re$. Does the inverse of f exist? If so, find it. Also, find the domain and range of the inverse. 9. A function is defined piece-wise as follows: $f(x)=3x+5$ for $- 4 \\leq x \\leq 0$ and $f(x)=5-3x$ for $0 < x \\leq 4$, find $f(f(\\frac{5}{2}))$; the domain and range of f; and the value of x for which $f(x)=-4$ 10. If $f: \\Re \\longrightarrow \\Re$ and $g: \\Re \\longrightarrow \\Re$ given", "= \\frac{x^2-25}{x-5}$ $(\\frac{f}{g})(x) = \\frac{(x+5)(x-5)}{x-5}$ $(\\frac{f}{g})(x) = x+5$ 17. UsukiDoll if you're not factoring then there is a restriction which is all reals except when x = 5. But I think that there have been times when canceling terms can happen. 18. UsukiDoll both $(\\frac{f}{g})(x) = \\frac{(x+5)(x-5)}{x-5}$ and $(\\frac{f}{g})(x) = x+5$ produce different results.. It's like do we consider the factored version in that case it's all reals on the domain or the non-factored case all reals except x = 5 ?", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "+C.$ Theorem 9 Assume that $$k> 1$$ is a positive integer and $$b$$ a non-zero real number. Then: $\\int\\frac{x}{\\left( x^2+b^2\\right)^k}\\,dx=\\frac{\\left( x^2+b^2\\right)^{1-k}}{2 (1-k)} +C.$ 2005-2017 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax", "CBSE (Science) Class 11CBSE Share Books Shortlist Your shortlist is empty # Find the Domain and Range of the Real Valued Function: (I) F ( X ) = a X + B B X − a - CBSE (Science) Class 11 - Mathematics ConceptCartesian Product of Sets #### Question Find the domain and range of the real valued function: (i) $f\\left( x \\right) = \\frac{ax + b}{bx - a}$ #### Solution (i) Given: $f\\left( x \\right) = \\frac{ax + b}{bx - a}$ Domain of f : Clearly, (x) is a rational function of x as $\\frac{ax + b}{bx - a}$ is a rational expression. Clearly, f (x) assumes real values for all x except for all those values of x for which ( bx-a) = 0, i.e. bx = a. $\\Rightarrow x = \\frac{a}{b}$ Hence, domain ( f ) =$R - \\left\\{ \\frac{a}{b} \\right\\}$ Range of f : Let f (x) = y ⇒ (ax + b) = y (bx -a) ⇒ (ax + b) = (bxy -ay) ⇒ b + ay = bxy -ax ⇒ b + ay = x(by - a) $\\Rightarrow x = \\frac{b + ay}{by - a}$ Clearly, f (x) assumes real values for all x except for all those values of x for which ( by - a) = 0, i.e. by =" ]

[ "\\infty\\right)$ Range of a real function is a set of values that this function can take while its argument takes all the values from the domain. To determine the range, let's try to resolve an equation $f \\left(x\\right) = a$ for any value $a$. Every value for which this equation has a solution (a value of $x$ from the domain) belongs to a range. So, let's try to find all $a$, for which there is a solution of the equation $\\frac{1}{x} + \\frac{1}{x - 3} = a$ We assume that $x \\ne 0$ and $x \\ne 3$ since we have already excluded these values of argument, they cannot be solutions, even if we come to these values for some $a$. Multiplying the equation by $x \\left(x - 3\\right)$, we obtain $x - 3 + x = a x \\left(x - 3\\right)$ or $a {x}^{2} + \\left(- 3 a - 2\\right) x + 3 = 0$ The quadratic equation above has a solution if its discriminant is not negative. The discriminant of this equation is ${\\left(- 3 a - 2\\right)}^{2} - 4 \\cdot a \\cdot 3 = 9 {a}^{2} + 12 a + 4 - 12 a = 9 {a}^{2} + 4$ As we see, the discriminant $9 {a}^{2} + 4$ is always positive for any real $a$. Therefore,", "we can further solve the quadratic equation as follows: $x - \\frac{1}{2} = \\frac{3}{2} \\mathmr{and} x - \\frac{1}{2} = - \\frac{3}{2}$ $\\iff x = 2 \\mathmr{and} x = - 1$ This means that for those two values, the denominator of our function would be $0$ which is not admissible. So, the domain of the function are all real numbers except $x = 2$ and $x = - 1$. $x \\in \\mathbb{R}$ \\ {-1; 2}", "# Why are quadratics factored into 2 brackets? Why is it that no one seems to factor quadratics into just one bracket Eg: $$2x^2+8x+6$$ into $$2x\\left(x+4+3\\cdot\\frac{1}{x}\\right)\\quad\\text{or}\\quad 2x\\left(x+4+\\frac{3}{x}\\right)\\quad ?$$ - Both of your given `factorizations' contain rational expressions, which are more difficult to find solutions of x. A polynomial that is factored gives you the roots of the equation. – Joshua Shane Liberman Feb 9 '11 at 15:28 The natural domain of the function $f(x) = 2x^2 +8x+6$ is all real numbers (remember, the \"natural domain\" is the collection of all real numbers for which the formula 'makes sense', or yields a real number). The natural domain of $g(x) = 2x\\left(x + 4 + \\frac{3}{x}\\right)$ is all real numbers except $x=0$, because you cannot plug in $x=0$ into the formula (division by zero makes the universe explode, after all). So one major problem with writing $2x^2 + 8x + 6 = 2x\\left(x + 4 + \\frac{3}{x}\\right)$ is that it is not true. They are not equal! The left hand side makes sense at $x=0$, but the right hand side does not. Another major problem is that polynomials are nice and easy, while rational functions are less nice and less easy (just wait until you get to integrals: if you have to integrate a rational function, you'll groan and settle", "# How do you find the domain and range of P(r)= 3/(sqrtr)? Jun 29, 2017 Domain 0f $P \\left(r\\right)$ is the set of all positive real numbers i.e. $+ R$. Range 0f $P \\left(r\\right)$ is the set of all positive real numbers i.e. $+ R$. #### Explanation: We know that anything divided by 0 is undefined or infinity and square root of negative numbers is not real . So we need to ensure that value of r is not equal to 0 or any negative numbers since it would not result in real values. Therefore, Domain 0f $P \\left(r\\right)$ is the set of all positive real numbers i.e. $+ R$. Since the denominator is positive (as accepts only positive values) hence the function would always return a positive real number. Therefore, Range 0f $P \\left(r\\right)$ is the set of all positive real numbers i.e. $+ R$.", "+0 # Help plssss thx +1 72 2 +74 Find all values of $$k$$ so that the domain of $$b(x) = \\frac{kx^2 + 2x - 5}{-5x^2 + 2x + k}$$ is the set of all real numbers. Feb 19, 2021 #1 +287 +2 The numerator does not matter in this case; only if the denominator is zero for some real number, the domain will not be set for all real numbers. To prevent the denominator from having real roots, the discriminant must be negative, which means $$2^2-4\\cdot-5\\cdot k = 4+20k$$ is negative. That will only happen if $$\\boxed{k<-\\frac{1}{5}}$$. Feb 19, 2021 #2 +74 +1 It's correct, thanks!! pizzaisawesome972 Feb 20, 2021", "## Precalculus (6th Edition) Blitzer The statement, “No quadratic functions have a range of $\\left( -\\infty ,\\infty \\right)$ ” is True. The quadratic equation is defined as $f\\left( x \\right)=a{{x}^{2}}+bx+c$. Here, a, b, and c are constants and $a\\ne 0$. The domain of the quadratic function is the set of real numbers because for each value of x in $f\\left( x \\right)=a{{x}^{2}}+bx+c,$ there exists the value of $f\\left( x \\right)$. If a is positive, the range of the quadratic function is the set of all positive real numbers. If a is negative, the range of the quadratic function is the set of all negative real numbers. Hence, for no values of a would the range of the function be the set of all real numbers. Therefore, the statement is true.", "# How do you find the domain of 1/(x^2+x+1)? Jul 16, 2016 The domain is the whole of the Real numbers, i.e. $\\left(- \\infty , \\infty\\right)$ #### Explanation: ${x}^{2} + x + 1$ is in the form $a {x}^{2} + b x + c$ with $a = b = c = 1$ This has discriminant: $\\Delta = {b}^{2} - 4 a c = 1 - 4 = - 3$ Since $\\Delta < 0$ this quadratic has no Real zeros. Let: $f \\left(x\\right) = \\frac{1}{{x}^{2} + x + 1}$ Then for all Real values of $x$, ${x}^{2} + x + 1 \\ne 0$, so $f \\left(x\\right)$ is defined. So the domain of $f \\left(x\\right)$ is all of the Real numbers $\\mathbb{R}$, i.e. $\\left(- \\infty , \\infty\\right)$", "2 and rewrite the inequality as $(x - 1)(x + 2) \\ne 0$ The expression (x - 1)(x + 2) is not equal to zero if x ≠ 1 and x ≠ - 2. In interval form, the domain is given by $(-\\infty , - 2) \\cup (-2 ,1 ) \\cup (1 , \\infty)$ Below is shown the graph of $f$ and we can see that the given function is undefined at x = -2 and x = 1. Example 5: Find the domain of the function $f(x) = \\dfrac{x^2 - 1}{x^2 + x + 5}$. Solution: The given function takes real values and is therefore real if $x^2 + x + 5 \\ne 0$ The expression x 2 + x + 5 cannot be factored over the reals. We therefore need to solve the quadratic equation using the discriminant $\\Delta$. $x^2 + x + 5 = 0$ $\\Delta = b^2 - 4 a c = (1)^2 - 4(1)(5) = -19$ The discriminant is negative and therefore no real value of x makes the expression x 2 + x + 5 equal to zero. The domain is the set of all real numbers In interval form, the domain is given by $(-\\infty , \\infty)$ Below is shown the graph of $f$ and we can see that it defined", "every real number. However, in constructive mathematics, trichotomy does not hold in the real numbers, and as a result, there exists real quadratic functions $f:\\mathbb{R} \\to \\mathbb{R}$ such that one cannot decide whether $f$ is concave up, concave down, or degenerate. Similarly, there exists real quadratic functions $f:\\mathbb{R} \\to \\mathbb{R}$ such that one cannot decide whether the extremum is a minimum, is a maximum, or does not exist due to the degeneracy of $f$. ## Exact zeroes ### Partial inverse functions Given a real quadratic function $f:\\mathbb{R} \\to \\mathbb{R}$, $f(x) \\coloneqq a x^2 + b x + c$ for real numbers $a \\in \\mathbb{R}$, $b \\in \\mathbb{R}$, $c \\in \\mathbb{R}$ such that $\\vert a \\vert \\gt 0$, since $f$ is continuously differentiable if $\\vert a \\vert \\gt 0$, due to the inverse function theorem, there are two branches of the partial inverse function of $f$, $g$ and $h$, such that if $a \\gt 0$, then $g$ and $h$ have domain $\\left(c - \\frac{b^2}{4a}, \\infty\\right)$ and if $a \\lt 0$, $g$ and $h$ have domain $\\left(-\\infty,c - \\frac{b^2}{4a}\\right)$ In both cases, $g$ and $h$ are defined as solutions to the first-order nonlinear ordinary differential equation $(2 a F + b) \\frac{d F}{d x} = 1$ with the initial conditions for the principal branch $g$ are $g\\left((a + b +", "+0 -1 307 1 What is the smallest integer value of $c$ such that the function $f(x)=\\frac{2x^2+x+5}{x^2+4x+c}$ has a domain of all real numbers? Jul 12, 2018 #1 +2298 +1 The only part of the function that has the ability to restrict the domain is the denominator; the numerator plays no role in this. For this reason, we should only focus on the denominator and ignore the numerator. Our job, in this case, is to determine the smallest value of c that will forever prevent a divisor by zero. Let's try and do that. $$x^2+4x+c=0$$ By setting the denominator equal to zero, we will see what solutions result in a real input that equals zero. Let's use the quadratic formula. $$a=1,b=4,c=c;\\\\ x_{1,2}=\\frac{-4\\pm\\sqrt{\\textcolor{red}{4^2-4*1*c}}}{2*1}$$ We want to focus on the discriminant because that determines whether or not there are real solutions for a given quadratic. $$16-4c<0$$ When the discriminant is less than zero, there are no real solutions. This means that there will no input that causes a domain restriction. $$16<4c$$ $$4 < c\\text{ or }c > 4$$ The immediate integer after 4 is 5, so this is the smallest integer that will result in no domain restriction. $$c=5$$ Jul 12, 2018 edited by TheXSquaredFactor Jul 12, 2018 #1 +2298 +1 The only part of the function that has the", "all $2\\times 2$ matrices with real entries, which has dimension 4. Finally, any given $T_1, \\ldots ,T_5 \\in L(V,W)$ must be linearly dependent, so there are scalars $\\alpha_1 ,\\ldots ,\\alpha_5 \\in \\mathbb{R}$ not all 0 such that $\\alpha_1 T_1 + \\ldots + \\alpha_5 T_5 = 0$. ### Subject:Pre-Calculus TutorMe Question: Find the domain of the function $f(x) = \\sqrt{\\frac{x^2 - 5x}{x^2 - 9}}$ Inactive Brendan C. Answer: The square root function $\\sqrt{x}$ has domain $[0,\\infty)$, i.e. the set of all real numbers $x$ with $x\\geq 0$. Thus we must determine the set of real numbers $x$ for which the rational function $g(x) = \\frac{x^2 - 5x}{x^2 - 9}$ is non-negative. Factoring numerator and denominator, we see that $$g(x) = \\frac{x(x-5)}{(x+3)(x-3)}$$ The denominator is zero when $x = \\pm 3$, so $g(x)$ has vertical asymptotes at $x=3$ and $x=-3$, and $g(x) = 0$ when $x = 0$ or $x = 5$. These zeros and vertical asymptotes divide the real line into intervals $(-\\infty, -3), (-3, 0), (0,3), (3,5)$, and $(5,\\infty)$. On each interval, $g(x)$ assumes either strictly positive or strictly negative values. When $x < -3$, $x(x-5) > 0$ and $(x+3)(x-3) > 0$, so $g(x) > 0$. Through similar reasoning, we see that $g(x)>0$ is positive on the intervals $(-\\infty, -3)$, $(0,3)$, and $(5,\\infty)$, and $g(x)<0$ on the intervals", "\\left(f+g\\right)\\left(x\\right)&=f\\left(x\\right)+g\\left(x\\right)\\\\ \\left(f-g\\right)\\left(x\\right)&=f\\left(x\\right)-g\\left(x\\right)\\\\ \\left(f\\cdot g\\right)\\left(x\\right)&=f\\left(x\\right)\\cdot g\\left(x\\right)\\\\ \\end.$ The domains of the resulting functions are the intersection of the domains of and . The quotient of two functions is defined similarly by :$\\frac fg\\left(x\\right)=\\frac,$ but the domain of the resulting function is obtained by removing the zeros of from the intersection of the domains of and . The polynomial functions are defined by polynomial In mathematics, a polynomial is an expression (mathematics), expression consisting of indeterminate (variable), indeterminates (also called variable (mathematics), variables) and coefficients, that involves only the operations of addition, subtrac ... s, and their domain is the whole set of real numbers. They include constant functions, linear functions and quadratic functions. Rational functions are quotients of two polynomial functions, and their domain is the real numbers with a finite number of them removed to avoid division by zero. The simplest rational function is the function $x\\mapsto \\frac 1x,$ whose graph is a hyperbola, and whose domain is the whole real line except for 0. The derivative of a real differentiable function is a real function. An antiderivative of a continuous real function is a real function that has the original function as a derivative. For example, the function $x\\mapsto\\frac 1x$ is continuous, and even differentiable, on the positive real numbers. Thus one antiderivative, which takes the value", "every nilpotent element is equal to zero. As a result, $f(x) = a \\left(x + \\frac{b}{2a}\\right)^2 = 0$ $\\left(x + \\frac{b}{2a}\\right)^2 = 0$ $x + \\frac{b}{2a} = 0$ $x = -\\frac{b}{2a}$ There is only one zero, which occurs at the extremum of the real quadratic function. ### In constructive mathematics In classical mathematics, the law of trichotomy holds in the real numbers, so the three cases above cover every real number. However, in constructive mathematics, trichotomy does not hold in the real numbers, and as a result, there exists real quadratic functions $f:\\mathbb{R} \\to \\mathbb{R}$ such that one cannot decide whether the discriminant of $f$ is positive, negative, or zero. As a result, there exist real quadratic functions where one cannot decide the number of zeroes the function has. Furthermore, one cannot prove that a real quadratic function with positive determinant has exactly two zeroes. ## Newton’s method Given a real quadratic function $f:\\mathbb{R} \\to \\mathbb{R}$, $f(x) \\coloneqq a x^2 + b x + c$ for real numbers $a \\in \\mathbb{R}$, $b \\in \\mathbb{R}$, $c \\in \\mathbb{R}$, where $\\vert a \\vert \\gt 0$, suppose we want to approximate a zero of $f$. There is an algorithm called Newton's method which would allow us to do so. The real quadratic function has discriminant $\\Delta = b^2 - 4 a", "that $$R$$ is the set of all real numbers here. • The range of a linear function is $$\\color{blue}{R}$$. • The range of a quadratic function $$y=a(x-h)^2+k$$ is: if $$\\color{blue}{a>0, y≥k}$$, and if $$\\color{blue}{a<0, y≤k}$$. • The range of a square root function is $$\\color{blue}{y≥0}$$. • The range of an exponential function is $$\\color{blue}{y>0}$$. • The range of the logarithmic function is $$\\color{blue}{R}$$. • To find the range of a rational function $$y = f(x)$$, solve it for $$x$$ and set the $$\\color{blue}{denominator ≠ 0}$$. ### Domain and range of an absolute value function The function $$y=|ax+b|$$ is defined for all real numbers. So, the domain of the absolute value function is the set of all real numbers. The absolute value of a number always results in a non-negative value. Thus, the range of an absolute value function of the form $$y= |ax+b|$$ is $$y ∈ R | y ≥ 0$$. The domain and range of an absolute value function are given as follows: • Domain $$\\color{blue}{= R}$$ • Range $$\\color{blue}{= [0, ∞)}$$ ### Domain and range of a square root function The function $$y=\\sqrt{\\left(ax+b\\right)}$$ is defined only for $$x ≥ -\\frac{b}{a}$$. So, the domain of the square root function is the set of all real numbers greater than or equal to $$\\frac{b}{a}$$. We know that the square", "2 and rewrite the inequality as $(x - 1)(x + 2) \\ne 0$ The expression (x - 1)(x + 2) is not equal to zero if x ≠ 1 and x ≠ - 2. In interval form, the domain is given by $(-\\infty , - 2) \\cup (-2 ,1 ) \\cup (1 , \\infty)$ Below is shown the graph of $f$ and we can see that the given function is undefined at x = -2 and x = 1. ### Example 5 Find the domain of the function $f(x) = \\dfrac{x^2 - 1}{x^2 + x + 5}$. Solution: The given function takes real values and is therefore real if $x^2 + x + 5 \\ne 0$ The expression x 2 + x + 5 cannot be factored over the reals. We therefore need to solve the quadratic equation using the discriminant $\\Delta$. $x^2 + x + 5 = 0$ $\\Delta = b^2 - 4 a c = (1)^2 - 4(1)(5) = -19$ The discriminant is negative and therefore no real value of x makes the expression x 2 + x + 5 equal to zero. The domain is the set of all real numbers In interval form, the domain is given by $(-\\infty , \\infty)$ Below is shown the graph of $f$ and we can see that it", "# Range of a Rational Function with arbitrary constant Find a condition on $c$ so that the function $f(x) =\\frac{x+c}{x^2-3x-c}$ has the whole of the real numbers as its range. I'm not entirely sure how to approach this problem. The answer is $-2<c<0$ , however, I don't know how to get to it. So I was thinking for a line $y=k$ if $\\frac{x+c}{x^2-3x-c} = k$ $x+c=kx^2-3kx-ck$ $kx^2-(3k+1)x-ck-c=0$ Solution is discriminant, $\\Delta \\ge 0$ $\\therefore$ $(3k+1)^2+4k(ck+c) \\ge 0$ $(9+4c)k^2+(6+4c)k+1 \\ge 0$ However, we don't know $k$; do we take the discriminant of the discriminant??? Alternatively, it may be best to consider horizontal asymptotes. $\\lim_{x\\to\\pm\\infty}f(x)=\\frac{x}{x^2}=\\frac{1}{x}=0$ I'm not entirely sure how the $-2<c$ comes into this. Any help is much appreciated. • $$c \\geq \\frac{9k^2+6k+1}{4k^2+4k}$$ (If $k^2+k\\geq 0$, else $\\leq$) But this must hold for ALL $k\\in\\mathbb R$ (with the respective ratio). Btw I think you also assume $k>0$ in your descriminant. – SK19 May 11 '18 at 16:33 you have arrived at $$(9+4c)k^2+(6+4c)k+1 \\ge 0$$ For this quadratic to be non-negative, it must have a discriminant which is not positive, so now you have $$(6+4c)^2-4(9+4c)\\leq0$$ $$\\implies c(c+2)\\leq0$$ $$\\implies-2\\leq c\\leq0$$ However when $c=-2$, $f(x)=\\frac{1}{x-1}$ which cannot be zero, and when $c=0$ then $f(x)=\\frac{1}{x-3}$ which also cannot be zero, so the end-point values of the inequality are excluded, leaving you with the", "# Find the domain of a function I am trying to find the domain of the following equation and I am not sure that I am right or not. $$f(x) = \\sqrt{\\dfrac{(2x-4)(5+x)}{(x+3)}}$$ I know that the result in the square root must be positive or 0... and that $x=2$ and $x=-5$ for the numerator. The square root will fail if I use $x = -3$ .. does it mean that the domain must be $\\ge -3$? Thanks. - You have to deal with two separate failure modes. First, you cannot have a $0$ denominator, so $-3$ cannot be in the domain of the function. Secondly, $$\\frac{(2x-4)(5+x)}{x+3}$$ must be non-negative, so you must solve the inequality $$\\frac{(2x-4)(5+x)}{x+3}\\ge 0\\;.$$ The fraction is obviously positive when all three of the expressions $2x-4$, $5+x$, and $x+3$ are positive, but it’s also positive when exactly one of them is positive; in all other cases it is negative or zero. Now $2x-4>0$ when $x>2$, $5+x>0$ when $x>-5$, and $x+3>0$ when $x>-3$, so the relevant break points and intervals are as shown here: ---------|----------------------|-----------------------------------|------- -5 -3 2 When $x<-5$, none of the expressions is positive, so the fraction is negative; when $-5<x<-3$, only $5+x$ is positive, so the fraction is positive; when $-3<x<2$, $5+x$ and $x+3$ are positive, so the fraction is negative; and when", "# Find the domain of the function f defined by f(x) = 1/sqrt(5x-x^2-6)? Apr 10, 2017 Please verify my solution. Need some guideline to complete. #### Explanation: The domain of the given function is the set of $x$ values such that $\\sqrt{5 x - {x}^{2} - 6} > 0$ The discriminant of the quadratic equation is: ${\\left(5\\right)}^{2} - 4 \\left(- 1\\right) \\left(- 6\\right) = 1$ Since the discriminant is positive the equation has two real number value of $x$ $x = - b \\pm \\frac{\\sqrt{{b}^{2} - 4 a c}}{2} a$ Putting value of $a = - 1 , b = 5 , c = - 6$ we get: $x = 2 , 3$", "domain of the quadratic equation is the set of all real numbers. (The blue writing is what I have so far) I am just beginning to learn the whole concept of set builder notation, and I am running into a little confusion. Solution to Example 1 The graph starts at x = - 4 and ends x = 6. Range is the set of output (y) values on the y-axis (vertical axis) that correspond to a point on the graph. 1 $\\begingroup$ Sorry about that, the other names for it is Math 8 or Pre-Algebra. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0. Informally, if a function is defined on some set, then we call that set the domain. ), and which of those numbers are excluded from the set. The values taken by the function are collectively referred to as the range. For example, the function takes the reals (domain) to the non-negative reals (range). See and . The domain is called the replacement set, and the range is called the solution set. The letter U indicates a union that connects parts of a function ). Example, y=2x { 1 < x < 3 } would … question 746428: Suppose that the of!", "the radical $\\sqrt{}$, satifies the condition: $|x - 1| \\ge 0$. The solution of the inequality is all real number because the absolute value expression $|x - 1|$ is always positive or zero for $x = 1$. The domain of the function is the set of real numbers $\\mathbb{R}$ and this can be checked graphically as shown below where the graph of $f$ \"exists\" for all $x$ values.. Example 3: Find the domain of the function $f(x) = \\dfrac{1}{ \\sqrt{x + 3}}$. Solution: Taking into account that the function is the ratio of two functions and that division by zero is not allowed, the given function takes real values if the argument $x + 3$, which is the quantity under the radical $\\sqrt{}$, satisfies the condition: $x + 3 \\gt 0$. Note the symbol of the inequality used is $\\gt$ and not $\\ge$ because we do not want to have zero in the denominator. Solution of the inequality is $x \\gt - 3$. The domain of the function is the set of real numbers greater than -3 and this can be checked graphically as shown below where the graph of $f$ \"exists\" for all $x > - 3$ values.. Example 4: Find the domain of the function $f(x) = \\dfrac{\\sqrt{x + 4}}{ \\sqrt{x - 2}}$. Solution: The given" ]

[ "+0 # Help plssss thx +1 72 2 +74 Find all values of $$k$$ so that the domain of $$b(x) = \\frac{kx^2 + 2x - 5}{-5x^2 + 2x + k}$$ is the set of all real numbers. Feb 19, 2021 #1 +287 +2 The numerator does not matter in this case; only if the denominator is zero for some real number, the domain will not be set for all real numbers. To prevent the denominator from having real roots, the discriminant must be negative, which means $$2^2-4\\cdot-5\\cdot k = 4+20k$$ is negative. That will only happen if $$\\boxed{k<-\\frac{1}{5}}$$. Feb 19, 2021 #2 +74 +1 It's correct, thanks!! pizzaisawesome972 Feb 20, 2021", "we can further solve the quadratic equation as follows: $x - \\frac{1}{2} = \\frac{3}{2} \\mathmr{and} x - \\frac{1}{2} = - \\frac{3}{2}$ $\\iff x = 2 \\mathmr{and} x = - 1$ This means that for those two values, the denominator of our function would be $0$ which is not admissible. So, the domain of the function are all real numbers except $x = 2$ and $x = - 1$. $x \\in \\mathbb{R}$ \\ {-1; 2}", "# What are the value of k, for which x^2+5kx+16=0 has no real roots? Nov 2, 2016 The values of k for which the equation ${x}^{2} + 5 k x + 16 = 0$ has no real roots are $- \\frac{8}{5} < k < \\frac{8}{5}$ #### Explanation: An equation $a {x}^{2} + b x + c = 0$ has no real roots if the determinant ${b}^{2} - 4 a c < 0$ ${x}^{2} + 5 k x + 16 = 0$ has no real roots if the determinant ${\\left(5 k\\right)}^{2} - 4 \\times 1 \\times 16 = 25 {k}^{2} - 64 < 0$ or $\\left(5 k - 8\\right) \\left(5 k + 8\\right) < 0$ As $\\left(5 k - 8\\right) \\left(5 k + 8\\right)$ is negative, either $5 k + 8 < 0$ and $5 k - 8 > 0$ But this is just not possible, as $k$ cannot be $k < - \\frac{8}{5}$ as well as $k > \\frac{8}{5}$ Other alternative is $5 k + 8 > 0$ and $5 k - 8 < 0$ i.e. $k > - \\frac{8}{5}$ as well as $k < \\frac{8}{5}$. This is possible if value of $k$ is between $- \\frac{8}{5}$ and $\\frac{8}{5}$. Hence $- \\frac{8}{5} < k < \\frac{8}{5}$", "\\infty\\right)$ Range of a real function is a set of values that this function can take while its argument takes all the values from the domain. To determine the range, let's try to resolve an equation $f \\left(x\\right) = a$ for any value $a$. Every value for which this equation has a solution (a value of $x$ from the domain) belongs to a range. So, let's try to find all $a$, for which there is a solution of the equation $\\frac{1}{x} + \\frac{1}{x - 3} = a$ We assume that $x \\ne 0$ and $x \\ne 3$ since we have already excluded these values of argument, they cannot be solutions, even if we come to these values for some $a$. Multiplying the equation by $x \\left(x - 3\\right)$, we obtain $x - 3 + x = a x \\left(x - 3\\right)$ or $a {x}^{2} + \\left(- 3 a - 2\\right) x + 3 = 0$ The quadratic equation above has a solution if its discriminant is not negative. The discriminant of this equation is ${\\left(- 3 a - 2\\right)}^{2} - 4 \\cdot a \\cdot 3 = 9 {a}^{2} + 12 a + 4 - 12 a = 9 {a}^{2} + 4$ As we see, the discriminant $9 {a}^{2} + 4$ is always positive for any real $a$. Therefore,", "that $$R$$ is the set of all real numbers here. • The range of a linear function is $$\\color{blue}{R}$$. • The range of a quadratic function $$y=a(x-h)^2+k$$ is: if $$\\color{blue}{a>0, y≥k}$$, and if $$\\color{blue}{a<0, y≤k}$$. • The range of a square root function is $$\\color{blue}{y≥0}$$. • The range of an exponential function is $$\\color{blue}{y>0}$$. • The range of the logarithmic function is $$\\color{blue}{R}$$. • To find the range of a rational function $$y = f(x)$$, solve it for $$x$$ and set the $$\\color{blue}{denominator ≠ 0}$$. ### Domain and range of an absolute value function The function $$y=|ax+b|$$ is defined for all real numbers. So, the domain of the absolute value function is the set of all real numbers. The absolute value of a number always results in a non-negative value. Thus, the range of an absolute value function of the form $$y= |ax+b|$$ is $$y ∈ R | y ≥ 0$$. The domain and range of an absolute value function are given as follows: • Domain $$\\color{blue}{= R}$$ • Range $$\\color{blue}{= [0, ∞)}$$ ### Domain and range of a square root function The function $$y=\\sqrt{\\left(ax+b\\right)}$$ is defined only for $$x ≥ -\\frac{b}{a}$$. So, the domain of the square root function is the set of all real numbers greater than or equal to $$\\frac{b}{a}$$. We know that the square", "# How do you solve 1/(6k^2)=1/(3k^2)-1/k and check for extraneous solutions? Jul 31, 2017 $k = \\frac{1}{6}$ #### Explanation: First, let's multiply everything by the least common multiple of the denominators (which is $6 {k}^{2}$): $\\frac{6 {k}^{2}}{6 {k}^{2}} = \\frac{6 {k}^{2}}{3 {k}^{2}} - \\frac{6 {k}^{2}}{k}$ $1 = 2 - 6 k$ $6 k = 1$ $k = \\frac{1}{6}$ The only domain restriction is $k \\ne 0$, so there are no extraneous solutions: color(blue)(ul(k = 1/6", "# What is the domain and range of f(x) = (x+5)/(x^2+36)? Nov 9, 2015 The domain is $\\mathbb{R}$ (all real numbers) and the range is $\\left[\\frac{5 - \\sqrt{61}}{72} , \\frac{5 + \\sqrt{61}}{72}\\right]$ (all real numbers between and including $\\frac{5 - \\sqrt{61}}{72}$ and $\\frac{5 + \\sqrt{61}}{72}$). #### Explanation: In the domain, we start with all real numbers, and then remove any which would force us to have the square root of a negative number, or a $0$ in the denominator of a fraction. At a glance, we know that as ${x}^{2} \\ge 0$ for all real numbers, ${x}^{2} + 36 \\ge 36 > 0$. Thus the denominator will not be $0$ for any real number $x$, meaning the domain includes every real number. For the range, the easiest way of finding the above values involves some basic calculus. Although it is longer, it is also possible to find them using only algebra, however, with the method detailed below. . . . Starting with the function $f \\left(x\\right) = \\frac{x + 5}{{x}^{2} + 36}$ we wish to find all possible values of $f \\left(x\\right)$. This is equivalent to finding the domain of the inverse function ${f}^{-} 1 \\left(x\\right)$ (a function with the property ${f}^{-} 1 \\left(f \\left(x\\right)\\right) = f \\left({f}^{-} 1 \\left(x\\right)\\right) = 1$) Unfortunately, the inverse of $f \\left(x\\right)$", ",k\\right]$. ### Example 4: Finding the Domain and Range of a Quadratic Function Find the domain and range of $f\\left(x\\right)=-5{x}^{2}+9x - 1$. ### Solution As with any quadratic function, the domain is all real numbers. Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex. $\\begin{cases}h=-\\frac{b}{2a}\\hfill \\\\ \\text{ }=-\\frac{9}{2\\left(-5\\right)}\\hfill \\\\ \\text{ }=\\frac{9}{10}\\hfill \\end{cases}$ The maximum value is given by $f\\left(h\\right)$. $\\begin{cases}f\\left(\\frac{9}{10}\\right)=5{\\left(\\frac{9}{10}\\right)}^{2}+9\\left(\\frac{9}{10}\\right)-1\\hfill \\\\ \\text{ }=\\frac{61}{20}\\hfill \\end{cases}$ The range is $f\\left(x\\right)\\le \\frac{61}{20}$, or $\\left(-\\infty ,\\frac{61}{20}\\right]$. ### Try It 3 Find the domain and range of $f\\left(x\\right)=2{\\left(x-\\frac{4}{7}\\right)}^{2}+\\frac{8}{11}$. Solution ## Determine a quadratic function’s minimum or maximum value There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Figure 9 ### Example 5: Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. 1. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L.", "of $\\left(f·g\\right)\\left(x\\right)$ is the set of all real numbers. However the domain of $\\left(\\frac{f}{g}\\right)\\left(x\\right)$ cannot include the x-value when g(x) = 0. $0=5x-2\\to 2=5x\\to \\frac{2}{5}=x$ Therefore the domain of $\\left(\\frac{f}{g}\\right)\\left(x\\right)$ is all real numbers except $\\frac{2}{5}$. Step 4: Evaluate the new functions with the given value of x. $\\left(f·g\\right)\\left(2\\right)=20{x}^{2}-8{x}^{2}=4{\\left(2\\right)}^{2}={48}$ $\\left(\\frac{f}{g}\\right)\\left(2\\right)=\\frac{4{x}^{2}}{5x-2}=\\frac{16}{8}={2}$ Related Links: Math algebra Composition of Functions Extreme Value Theorem Calculus Topics To link to this Combination of Functions page, copy the following code to your site:", "# How do you find the domain of 1/(x^2+x+1)? Jul 16, 2016 The domain is the whole of the Real numbers, i.e. $\\left(- \\infty , \\infty\\right)$ #### Explanation: ${x}^{2} + x + 1$ is in the form $a {x}^{2} + b x + c$ with $a = b = c = 1$ This has discriminant: $\\Delta = {b}^{2} - 4 a c = 1 - 4 = - 3$ Since $\\Delta < 0$ this quadratic has no Real zeros. Let: $f \\left(x\\right) = \\frac{1}{{x}^{2} + x + 1}$ Then for all Real values of $x$, ${x}^{2} + x + 1 \\ne 0$, so $f \\left(x\\right)$ is defined. So the domain of $f \\left(x\\right)$ is all of the Real numbers $\\mathbb{R}$, i.e. $\\left(- \\infty , \\infty\\right)$", "part d) as follows: We want to find a positive value for $a$ such that: $\\frac{a}{(t+1)^2}>\\frac{1}{t^2+1}$ for all $t$ in the given domain. This means we want: $(a-1)t^2-2t+(a-1)>0$ This means the discriminant must be negative: $4-4(a-1)^2<0$ $0 which implies: $2. Now, since: $\\lim_{x\\to\\infty}\\left[a\\int_0^x\\frac{1}{(t+1)^2}\\,dt \\right]=a$ we may state: $0. Excellent!", "+C.$ Theorem 9 Assume that $$k> 1$$ is a positive integer and $$b$$ a non-zero real number. Then: $\\int\\frac{x}{\\left( x^2+b^2\\right)^k}\\,dx=\\frac{\\left( x^2+b^2\\right)^{1-k}}{2 (1-k)} +C.$ 2005-2017 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax", "real numbers $c_{1}$, $c_{2}$, $c_{3}$ such that the numbers $$a_{11}c_{1}+a_{12}c_{2}+a_{13}c_{3},$$ $$a_{21}c_{1}+a_{22}c_{2}+a_{23}c_{3},$$ $$a_{31}c_{1}+a_{32}c_{2}+a_{33}c_{3}$$ are either all negative, or all zero, or all positive. 2. Find all nondecreasing functions $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ such that • $f(0) = 0, f(1) = 1;$ • $f(a) + f(b) = f(a)f(b) + f(a + b - ab)$ for all real numbers $a, b$ such that $a < 1 < b$. 3. Consider two monotonically decreasing sequences $\\left( a_k\\right)$ and $\\left( b_k\\right)$, where $k \\geq 1$, and $a_k$ and $b_k$ are positive real numbers for every k. Now, define the sequences $$c_k = \\min \\left( a_k, b_k \\right)$$ $$A_k = a_1 + a_2 + ... + a_k$$ $$B_k = b_1 + b_2 + ... + b_k$$ $$C_k = c_1 + c_2 + ... + c_k$$ for all natural numbers k. a) Do there exist two monotonically decreasing sequences $\\left( a_k\\right)$ and $\\left( b_k\\right)$ of positive real numbers such that the sequences $\\left( A_k\\right)$ and $\\left( B_k\\right)$ are not bounded, while the sequence $\\left( C_k\\right)$ is bounded? b) Does the answer to problem (a) change if we stipulate that the sequence $\\left( b_k\\right)$ must be $\\displaystyle b_k = \\frac {1}{k}$ for all k ? 4. Let $n$ be a positive integer and let $x_1\\le x_2\\le\\cdots\\le x_n$ be real numbers. Prove that $\\left(\\sum_{i,j=1}^{n}|x_i-x_j|\\right)^2\\le\\frac{2(n^2-1)}{3}\\sum_{i,j=1}^{n}(x_i-x_j)^2.$ Show that the equality holds", "## College Algebra (6th Edition) Published by Pearson # Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.6 - Page 420: 61 #### Answer Domain: {$x | x\\geq\\frac{1}{2}$} #### Work Step by Step The square root of a negative number has no solution within the domain of Real numbers. That is to say, for all functions $f(x) = \\sqrt x$, $x$ cannot be less than $0$. Knowing this, we can calculate the domain of this square root function by solving for $x\\gt0$. In the exercise, therefore, we can take the function $f(x) = \\sqrt{2x^{2} - 5x + 2}$ and find the domain by solving for the solution set of: $$x_{domain}: (2x^{2} - 5x + 2 \\geq 0)$$ By using the quadratic formula $\\frac{-b \\frac{+}{} \\sqrt {b^{2} - 4ac}}{2a}$ we can find the zeros of the function: $$2x^{2} - 5x + 2 \\geq 0$$ $$x = \\frac{-b \\frac{+}{} \\sqrt {b^{2} - 4ac}}{2a} = \\frac{-(-5) \\frac{+}{} \\sqrt {(-5)^{2} - 4(2)(2)}}{2(2)} = \\frac{5 \\frac{+}{} \\sqrt {25 - 16}}{4} = \\frac{5 \\frac{+}{} \\sqrt {9}}{4} = \\frac{5 \\frac{+}{} 3}{4}$$ where $x = \\frac{5 + 3}{4} = 2$ and $x = \\frac{5-3}{4} = \\frac{1}{2}$ and the inequality can now be re-written as: $$(x - 2)(x-\\frac{1}{2})\\geq0$$ Finally, we come to the conclusion that, for the domain of this function, $x\\geq0$ and $x\\geq\\frac{1}{2}$. Since", "k + 1\\right)}$ for some $k \\in \\mathbb{Z}$ That is, we can define: ${\\log}_{a} \\left(y\\right) = \\ln \\frac{y}{\\ln \\left(- a\\right) + i \\pi \\left(2 k + 1\\right)}$ for some $k \\in \\mathbb{Z}$ Regardless of what value of $k$ we choose, this will have a non-Real Complex value most of the time, but at least we can restrict the domain of ${a}^{z}$ to get a well defined inverse.", "standard form with a negative a value is $$f(x){\\geq}k$$. Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether a is positive or negative. If $$a$$ is positive, the parabola has a minimum. If $$a$$ is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, $$k$$. 4. If the parabola has a minimum, the range is given by $$f(x){\\geq}k$$, or $$\\left[k,\\infty\\right)$$. If the parabola has a maximum, the range is given by $$f(x){\\leq}k$$, or $$\\left(−\\infty,k\\right]$$. Example 3.2.4: Finding the Domain and Range of a Quadratic Function Find the domain and range of $$f(x)=−5x^2+9x−1$$. Solution As with any quadratic function, the domain is all real numbers. Because $$a$$ is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex. \\begin{align} h&=−\\dfrac{b}{2a} \\\\ &=−\\dfrac{9}{2(-5)} \\\\ &=\\dfrac{9}{10} \\end{align} The maximum value is given by $$f(h)$$. \\begin{align} f(\\dfrac{9}{10})&=5(\\dfrac{9}{10})^2+9(\\dfrac{9}{10})-1 \\\\&= \\dfrac{61}{20}\\end{align} The range is $$f(x){\\leq}\\frac{61}{20}$$, or $$\\left(−\\infty,\\frac{61}{20}\\right]$$. 3.2.2: Find the domain and range of $$f(x)=2\\Big(x−\\frac{4}{7}\\Big)^2+\\frac{8}{11}$$. Solution The domain is all real numbers. The range is $$f(x){\\geq}\\frac{8}{11}$$, or $$\\left[\\frac{8}{11},\\infty\\right)$$. ### Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at", "# Let * be a binary operation on the set of all nonzero real numbers, defined Question: Let * be a binary operation on the set of all nonzero real numbers, defined by $a * b=\\frac{a b}{5}$ Find the value of x given that2 * (x * 5) = 10. Solution: To find: value of $x$ Given: $a^{*} b=\\frac{a b}{5}$ $\\Rightarrow x^{* 5}=\\frac{5 x}{5}=x$ Now $(2 * x)=\\frac{2 x}{5}$ $\\Rightarrow \\frac{2 x}{5}=10 \\Rightarrow x=25$", "# If x is real and k Question: If $x$ is real and $k=\\frac{x^{2}-x+1}{x^{2}+x+1}$, then (a) k ∈ [1/3,3] (b) k ≥ 3 (c) k ≤ 1/3 (d) none of these Solution: (a) $k \\in[1 / 3,3]$ $k=\\frac{x^{2}-x+1}{x^{2}+x+1}$ $\\Rightarrow k x^{2}+k x+k=x^{2}-x+1$ $\\Rightarrow(k-1) x^{2}+(k+1) x+k-1=0$ For real values of $x$, the discriminant of $(k-1) x^{2}+(k+1) x+k-1=0$ should be greater than or equal to zero. $\\therefore$ if $k \\neq 1$ $(k+1)^{2}-4(k-1)(k-1) \\geq 0$ $\\Rightarrow(k+1)^{2}-\\{2(k-1)\\}^{2} \\geq 0$ $\\Rightarrow(k+1+2 k-2)(k+1-2 k+2) \\geq 0$ $\\Rightarrow(3 k-1)(-k+3) \\geq 0$ $\\Rightarrow(3 k-1)(k-3) \\leq 0$ $\\Rightarrow \\frac{1}{3} \\leq k \\leq 3$ i. e. $k \\in\\left[\\frac{1}{3}, 3\\right]-\\{1\\} \\quad \\ldots($ i $)$ And if k=1, then, x=0, which is real ...(ii) So, from (i) and (ii), we get, $k \\in\\left[\\frac{1}{3}, 3\\right]$", "$x=\\frac{2y+1}{3-5y}$ satisfies this equation. We can conclude that the range of $f$ is $\\{y|y\\ne 3/5\\}$. 2. To find the domain of $f$, we need $4-x^2 \\ge 0$. When we factor, we write $4-x^2=(2-x)(2+x) \\ge 0$. This inequality holds if and only if both terms are positive or both terms are negative. For both terms to be positive, we need to find $x$ such that $2-x \\ge 0$ and $2+x \\ge 0$. These two inequalities reduce to $2 \\ge x$ and $x \\ge -2$. Therefore, the set $\\{x|-2\\le x\\le 2\\}$ must be part of the domain. For both terms to be negative, we need $2-x \\le 0$ and $2+x \\ge 0$. These two inequalities also reduce to $2 \\le x$ and $x \\ge -2$. There are no values of $x$ that satisfy both of these inequalities. Thus, we can conclude the domain of this function is $\\{x|-2 \\le x \\le 2\\}$. If $-2 \\le x \\le 2$, then $0 \\le 4-x^2 \\le 4$. Therefore, $0 \\le \\sqrt{4-x^2} \\le 2$, and the range of $f$ is $\\{y|0 \\le y \\le 2\\}$. Find the domain and range for the function $f(x)=(5x+2)/(2x-1)$. #### Hint The denominator cannot be zero. Solve the equation $y=(5x+2)/(2x-1)$ for $x$ to find the range. #### Solution The domain is the set of real numbers $x$ such that", "can be defined. Although it looks like g(f(x)) can be defined at x = 0, you should remember that you will not get through f(x) in the first place. This is because f(0) cannot be defined. Now, work out the domain of g(f(x)) and convince yourself that the domain is all real numbers except x = 0 and x = -1 (c) We will start by evaluating the composite function f(f(x)). $f(x + \\frac{1}{x}) = x + \\frac{1}{x} + \\frac{1}{x + \\frac{1}{x}}$ Make sure you work it out and get to the point where: $f(x + \\frac{1}{x}) = \\frac{x^4 + 3x^2 + 1}{x(x^2+1)}$ Here, you can see that x = 0 is excluded from the domain of the composite function (since f(x) is the starting function). Let’s solve algebraically to see if x = 0 is the only restricted value on the domain. Remember, we do not want f(0), so we need to find when f(x) gives us 0. $x+\\frac{1}{x}=0$ $x^2+1=0$ $x = \\sqrt{-1}$ But the domain of the square root is $x\\ge 0$. Thus, we can conclude that the domain of the composite function is all real numbers except for x = 0. (d) Finally, we will look at g(g(x)), which is a bit similar to part (b). Evaluating the composite function gives us: $g(\\frac{x+8}{x+2}) = \\frac{\\frac{x+8}{x+2}" ]

[ "+0 # help 0 24 1 Find $$q(x)$$ if the graph of $$\\frac{4x-x^3}{q(x)}$$ has a hole at $$x=-2$$, a vertical asymptote at $$x=1$$, no horizontal asymptote, and $$q(3) = -30$$. Apr 10, 2019", "+0 # help pls 0 31 1 The graph of $y = \\frac{p(x)}{q(x)}$ is shown below, where $p(x)$ and $q(x)$ are quadratic. (Assume that the grid lines are at integers.) The horizontal asymptote is $y = -1,$ and the only vertical asymptote is $x = 2.$ Find $\\frac{p(-1)}{q(-1)}.$ Jun 28, 2022 ### 1+0 Answers #1 0 p(-1)/q(-1) = 11/2. Jun 28, 2022", "$q\\left( x \\right)=0$ \\begin{align} & x-1=0 \\\\ & x=1 \\end{align} Thus, the graph has a vertical asymptote as $x=1$. Determine the horizontal asymptotes. There is no horizontal asymptote because the degree of the numerator term is not equal to the degree of the denominator term.", "+0 # help 0 49 1 Find $$q(x)$$ if the graph of $$\\frac{4x-x^3}{q(x)}$$ has a hole at $$x=-2$$, a vertical asymptote at $$x=1$$, no horizontal asymptote, and $$q(3) = -30$$. May 6, 2019 #1 +101872 +1 4x - x^3 x ( 4 - x^2) x ( 2 - x) ( 2 +x) ______ = ___________ = _____________ q(x) q(x) q(x) If we have a \"hole\" at x = -2, then q(x) must have a factor of ( x + 2) If we have a vertical asymptote at x = 1, then (x - 1) must also be a factor If we have no horizontal asymptote, the q(x) must be a second degree polynomial - the rational fuction has a s;ant asymptote And since q(3) = -30....then we have that -30 = a(3 + 2) (3 - 1) -30 =a (5)(2) -30 = 10a a = -3 So....q(x) = -3(x + 2)(x - 1) = -3(x^2 + x - 2) = -3x^2 - 3x + 6 Here's a graph : https://www.desmos.com/calculator/dz5hnv1yoa May 6, 2019", "9758/2019/P2/Q02 (i) Sketch the graph of $y=\\frac{2-x}{3 x^{2}+5 x-8}$. Give the equations of the asymptotes and the coordinates of the point(s) where the curve crosses either axis. [4] (ii) Solve the inequality $\\frac{2-x}{3 x^{2}+5 x-8}>0$. [1] (iii) Hence solve the inequality $\\frac{x-2}{3 x^{2}+5 x-8}>0$. [1]", "### Home > PC3 > Chapter 8 > Lesson 8.1.2 > Problem8-36 8-36. Graph $f(x)=\\frac{1}{x-4}-3$. What is the parent equation of $f$ ? Describe the transformation completely. This is a rational function. The horizontal asymptote occurs when the denominator equals $0$. The horizontal asymptote is the end behavior, or what happens when $x$ becomes very large.", "\\frac{1}{2} \\text{ is the asymptote}$ graph{(x^2-x)/(2x^2+4x-6) [-10, 10, -5, 5]}", "# How do you find the horizontal asymptote of the graph of y=(-2x^6+5x+8)/(8x^6+6x+5) ? ##### 1 Answer Aug 14, 2014 Please follow the explanation of the link below. Your answer will be $y = - \\frac{1}{4}$", "need $x = \\frac{\\pi}{2}, \\frac{3\\pi}{2}$, as $0\\leq x \\leq 2\\pi$. So, the vertical asymptotes are at $x = \\frac{\\pi}{2}$ and $\\frac{3\\pi}{2}$. You can use one sided limits to find how do these asymptotes look like. You can find the critical points and find whether the graph increases or decreases using first and second derivatives. Or if you have enough points to guess how the graph looks like, you can finish it now.", "+0 # help 0 157 1 The graph of the rational function $$\\frac{p(x)}{q(x)}$$ is shown below, with a horizontal asymptote of $$y = 0$$. If $$q(x)$$ is quadratic, $$p(2)=1$$, and $$q(2) = 3$$, find $$p(x) + q(x).$$ May 10, 2019 #1 +6046 +1 $$\\text{Given the removable singularity at }x=1 \\text{, and the vertical asymptote at }x=-1\\\\ \\text{and the fact that }q(x) \\text{ is quadratic}\\\\ q(x) =q_0 (x-1)(x+1)\\\\ \\text{because of the removable singularity we also know that }p(x) \\text{ has }(x-1) \\text{ as a factor}$$ $$q(2)=3\\\\ q_0 (1)(3)=3\\\\ q_0=1\\\\ q(x) = (x+1)(x-1)=x^2-1$$ $$\\text{As }\\dfrac{p(x)}{q(x)} \\text{ has a horizontal asymptote of }0\\\\ \\text{the degree of }p(x) \\text{ must be less than that of }q(x)\\\\ \\text{so }p(x)=p_0(x-1)\\\\ p(2)=1\\\\ p_0=1\\\\ p(x)=x-1$$ $$p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2$$ . May 11, 2019 #1 +6046 +1 $$\\text{Given the removable singularity at }x=1 \\text{, and the vertical asymptote at }x=-1\\\\ \\text{and the fact that }q(x) \\text{ is quadratic}\\\\ q(x) =q_0 (x-1)(x+1)\\\\ \\text{because of the removable singularity we also know that }p(x) \\text{ has }(x-1) \\text{ as a factor}$$ $$q(2)=3\\\\ q_0 (1)(3)=3\\\\ q_0=1\\\\ q(x) = (x+1)(x-1)=x^2-1$$ $$\\text{As }\\dfrac{p(x)}{q(x)} \\text{ has a horizontal asymptote of }0\\\\ \\text{the degree of }p(x) \\text{ must be less than that of }q(x)\\\\ \\text{so }p(x)=p_0(x-1)\\\\ p(2)=1\\\\ p_0=1\\\\ p(x)=x-1$$ $$p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2$$ Rom May", "# How do you find the horizontal asymptote of the graph of y=(-4x^6+6x+3)/(8x^6+9x+3) ? Please follow the explanation of the link below. Your answer will be $y = - \\frac{1}{2}$", "Consider the function Find the vertical asymptotes of $f(x)$. Separate multiple answers by commas. $x=$", "Since $f(x) = \\frac{1}{x+1}$ is obtained from $\\frac{1}{x}$ by replacing $x$ by $x+1$, the horizontal shift rule (with $c = 1$) tells us that you can obtain the graph of $f(x)$ by shifting the graph of $1/x$ to the left 1 unit.", "### Home > CALC > Chapter 7 > Lesson 7.4.3 > Problem7-195 7-195. Sketch the graph of $f(x)=\\frac { 3 x ^ { 2 } + 5 x - 2 } { x }$. Find $b(x)$, its end behavior function. $f(x)$ will have a vertical asymptote at $x = 0$. $b(x)$ will not. Divide the numerator by the denominator. $f(x) = b(x) +$ remainder", "than just saying it follows the shape of the leading term of $q(x)$, at the expense of the work required to find $q(x)$. ##### Examples Consider the rational expression $~ \\frac{17x^5 - 300x^4 - 1/2}{x^5 - 2x^4 + 3x^3 - 4x^2 + 5}. ~$ The leading term of the numerator is $17x^5$ and the leading term of the denominator is $x^5$. The ratio is $17$ (or $17x^0 = 17x^{5-5}$). As such, we would have a horizontal asymptote $y=17$. If we consider instead this rational expression: $~ \\frac{x^5 - 2x^4 + 3x^3 - 4x^2 + 5}{5x^4 + 4x^3 + 3x^2 + 2x + 1} ~$ Then we can see that the ratio of the leading terms is $x^5 / (5x^4) = (1/5)x$. We expect a slant asymptote with slope $1/5$, though we would need to divide to see the exact intercept. This is found with p = (x^5 - 2x^4 + 3x^3 - 4x^2 + 5) / (5x^4 + 4x^3 + 3x^2 + 2x + 1) apart(p) \\begin{equation*}\\frac{x}{5} + \\frac{116 x^{3} - 68 x^{2} + 23 x + 139}{25 \\left(5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1\\right)} - \\frac{14}{25}\\end{equation*} The rational function $~ \\frac{5x^3 + 6x^2 + 2}{x-1} ~$ has decomposition $5x^2 + 11x + 11 + 13/(x-1)$: a = 5x^3 + 6x^2 +2", "graph will never touch --the asymptote-- is $x = - 2$. We can always check our work by graphing $\\frac{x - 2}{\\left(x + 2\\right) \\left(x - 2\\right)}$ and seeing where that asymptote is. graph{(x-2)/((x+2) (x-2))}", "# What are the vertical asymptotes for the graph of y=(x+3)/((x-2)(x+5))? They're $x = 2$ and $x = - 5$. $\\left(x - 2\\right) \\left(x + 5\\right) = 0$ $x - 2 = 0 \\mathmr{and} x + 5 = 0$ $x = 2 \\mathmr{and} x = - 5$", "have vertical asymptotes, but it doesn't, again as you can tell from the formula (it is defined for all real numbers). Discovering facts about a function by looking at its graph is useful but dangerous. #### Example Below is the graph of the function$f(x)=.0002{{\\left( \\frac{{{x}^{3}}-10}{3{{e}^{-x}}+1} \\right)}^{6}}$ If you didn't know the formula for the function, by looking at the graph you might nevertheless suspect certain things are true of the function. 1. Maybe it has a local maximum somewhere to the right of $x=1$. 2. Maybe it has one or more zeroes around $x=-1$. 3. Maybe it has one or more zeroes around $x=2$. 4. Maybe it has a vertical asymptote somewhere to the right of $x=2.5$. 5. Maybe it has a horizontal asymptote in the negative $x$ direction. If you do know the formula, you can find out many things about the function that you can't depend on the graph to see. • You can see immediately that $f$ has just one zero, at $x=\\sqrt[3]{10}$, which is about $2.15$. So (3) is correct. • If you notice that the exponent $6$ is even, you can figure out that $f(x)\\geq0$ for all $x$, so $\\sqrt[3]{10}$ is an absolute minimum for $f$. • By using L'Hôpital several times, you can see that $f$ has a horizontal asymptote as $x\\to-\\infty$.", "a vertical asymptote $x = 5$ to #3.: I don't understand your question - but because the x-intercept is at $x = -2$ maybe you mean a point like $(0, 0)$ ? thank you", "What are the vertical asymptotes of the graph y=(2x-6)/(x^3-x)? $x = 0 , x = 1$ and $x = - 1$ So ${x}^{3} - x = 0$ is equivalent with $x \\left({x}^{2} - 1\\right) = 0$, so $x = 0$ or ${x}^{2} = 1$. By taking the square root of the second one, we find $x = 1$ and $x = - 1$. So all the answers are $x = 0 , x = 1$ and $x = - 1$." ]

[ "+0 # help 0 49 1 Find $$q(x)$$ if the graph of $$\\frac{4x-x^3}{q(x)}$$ has a hole at $$x=-2$$, a vertical asymptote at $$x=1$$, no horizontal asymptote, and $$q(3) = -30$$. May 6, 2019 #1 +101872 +1 4x - x^3 x ( 4 - x^2) x ( 2 - x) ( 2 +x) ______ = ___________ = _____________ q(x) q(x) q(x) If we have a \"hole\" at x = -2, then q(x) must have a factor of ( x + 2) If we have a vertical asymptote at x = 1, then (x - 1) must also be a factor If we have no horizontal asymptote, the q(x) must be a second degree polynomial - the rational fuction has a s;ant asymptote And since q(3) = -30....then we have that -30 = a(3 + 2) (3 - 1) -30 =a (5)(2) -30 = 10a a = -3 So....q(x) = -3(x + 2)(x - 1) = -3(x^2 + x - 2) = -3x^2 - 3x + 6 Here's a graph : https://www.desmos.com/calculator/dz5hnv1yoa May 6, 2019", "× Help me! Let $$P(x) = (x-1)(x-2)(x-3)$$. Find for how many polynomials $$Q(x)$$ does there exist a polynomial $$R(x)$$ of degree 3 such that $$P(Q(x)) = P(x) \\cdot R(x)$$. Note by Vedant Sharda 1 year, 2 months ago Sort by: What are your thoughts? What have you tried? Must the polynomial have complex, real or integer coefficients? Staff · 1 year, 2 months ago What I have tried is : Degree of P(x) is 3 and R(x) is also 3 So degree of P(x).R(x) is 6. Degree of Q(x) must be 2 so that degree of P(Q(x)) is 6. P(Q(x)) = P(x).R(x) (Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x) Now what I have to do ? · 1 year, 2 months ago Great job. A key piece of information is that the degree of Q(x) is 6. What can we say about the value of $$Q(1)$$? Q(2)? Q(3)? Staff · 1 year, 2 months ago To find Q(1) , put x=1 in (Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x) [Q(1)-1][Q(1)-2][Q(1)-3] =0 Q(1)-1 = 0. So Q(1) =1 Q(1)-2 = 0. So Q(1) = 2 Q(1)-3 = 0. So Q(1) = 3 Hence Q(1) = 1or 2or 3 Similarly Q(2) = 1or 2or 3 And Q(3) = 1or 2or 3 · 1 year, 2 months ago Great, so we now know the possible values of $$Q(1), Q(2),", "$x - 1$ gives a quotient $q(x) = x^3 + x^2 + x - 1$. Then $q(1/2) < 0$ and $q(1) > 0$ means $q$ has a root between 1/2 and 1, and so therefore does $p$. In other words, $x^4 + 1 = 2x$ and $x \\ge 1/2$ does not imply $x = 1$, as we found by taking your friend's assumption. -", "= f (x)$$ , labelling the points P and Q. (b) (i) Show that $${f}'(x)=(1-2x)e^{-x}$$. (ii) Find the exact coordinates of Q. (c) The equation $$f (x) = k$$ , where $$k\\in \\mathbb{R}$$ , has two solutions. Write down the range of values of $$k$$ . (d) Given that $${f}”(x)=e^{-x}(-3+2x)$$ , show that the curve of $$f$$ has only one point of inflexion. Ans. (a) (b) (i) $${f}'(x)=2e^{-x}+(2x+1)(-e^{-x})=(1-2x)e^{-x}$$ (ii) At Q, $${f}'(x)=0$$ x = 0.5, y = 2e-0.5 Q is (0.5, 2e-0.5) (c) 1 ≤ k < 2e-0.5 (d) $${f}”(x)=0\\Leftrightarrow e^{-x}(-3+2x)=0$$ This equation has only one root. So $$f$$ has only one point of inflexion. ### Question [Maximum mark: 10] Consider the function $$f$$ given by $$f(x)=\\frac{2x^{2}-13x+20}{(x-1)^{2}}$$, $$x\\neq 1$$. A part of the graph of $$f$$ is given below (vertical and horizontal asymptotes are shown) (a) Write down the equation of the vertical asymptote. (b) It is given that $$f (100) = 1.91$$, $$f (-100) = 2.09$$, $$f (1000)= 1.99$$ Evaluate $$f (-1000)$$ and write down the equation of the horizontal asymptote. (c) Show that $${f}'(x)=\\frac{9x-27}{(x-1)^{3}}$$, $$x\\neq 1$$. The second derivative is given by $${f}”(x)=\\frac{72-18x}{(x-1)^{4}}$$, $$x\\neq 1$$. (d) Using values of $${f}'(x)$$ and $${f}”(x)$$ explain why a minimum must occur at $$x = 3$$. (e) There is a point of inflexion on the graph. Write down the coordinates", "at x = 4? 1. $y=(3x)/(x-1)$ 2. $y=(x+1)/(x^2 + 5x + 4)$ 3. $y=(x-1)/(x^2-5x+4)$ 4. $y=(x-4)/(x^2-5x+4)$ How many possible positive, negative, and imaginary zeros are there in the function $q(x)=2x^6+5x^4-x^3-5x-1$? 1. 0 positive, 2 or 0 negative, 4 or 6 imaginary 2. 1 positive, 1 negative, 4 imaginary 3. 4, 2, or 0 positive, 4, 2, or 0 negative, 6, 4, 2, or 0 imaginary 4. 2 or 0 positive, 4, 2, or 0 negative, 2 or 0 imaginary $f(x)=( x-2)/(x^2-16)$ Find the Vertical and Horizontal Asymptotes. If $f(x) = x^2-x+20$ then $f(b+1) =$ 1. $b^2+18$ 2. $b^2+b+20$ 3. $b^2+3b+20$", "# Thread: Find function from remainders 1. ## Find function from remainders Determine $g(x)\\equiv (px+q)^2(2x^2-5x-3)$ where p,q are constants, given that, on division by $(x-2)^2$, the remainder is $-39-3x$. I found $g(2)=(2p+q)^2(-5)=-45$ $(2p+q)^2=9$ $2p+q=\\pm 3$ I don't think I'm on the right track, but I don't know how else to work it out. Thanks! 2. Good start. I am not sure if the following is the simplest way, but it seems to lead to finding p and q, at least in theory. So, we have $(px+q)^2(2x^2-5x-3)=(x-2)^2f(x)-39-3x$ for some function $f(x)$. Now, looking at degrees of polynomials, $f(x)$ is a quadratic polynomial, let's say $ax^2+bx+c$. Next, we look at both sides' coefficients of $x^4$, $x^3$, as well as the free terms. Equating the leading coefficients, we see that $a=2p^2$. Similarly, equating the constant terms (or putting x = 0), we get that $c=(39-3q^2)/4$. Finally, equating the coefficients at $x^3$, we get that $b=3p^2+4pq$. Therefore, $f(x)=2p^2+(3p^2+4pq)x+(39-3q^2)/4$. Next, one of the roots of $2x^2-5x-3=0$ is -1/2. Substituting $x=-1/2$ into the original equation, we get $f(-1/2)=-15$, or $4f(-1/2)=-60$. Substitute the expression for $f(x)$ above; this gives (after lots of simplifications) $4p^2+3q^2+8pq=99$. Finally, combining this with $2p+q=\\pm 3$, one can find $p$ and $q$. In my calculations, this boils down to a linear (not even a quadratic) equation. Disclaimer: my calculations can be", "have, $$(b-1)x= -4x$$ and $$c=2$$. Hence we find that $$b=(-3)$$ and $$c=2$$. • So the polynomial will become, $$x^3-3x+2$$. • Now, using remainder theorem, we find that the remainder of the polynomial is $$f(2)=2^3-(3×2)+2 \\\\ = 8-6+2 \\\\= \\boxed {4}$$ I just wanted to show the other way of doing this. Hope you found it useful!$$\\huge \\ddot\\smile$$ · 2 years, 1 month ago Hi Swaplin, what are the answers? · 2 years, 1 month ago Well I even donno, I wish someone could help, :( · 2 years, 1 month ago The polynomial is of degree 3 so it should have three roots two of them are 1,1. Let the othe root be $$\\alpha$$. So $$f(x)=(x-1)(x-1)(x-\\alpha)=x^3-x^2(\\alpha+2)+x(2\\alpha+1)-\\alpha$$ As coefficient of $$x^2$$ must be zero so $$\\alpha=-2$$. Now using this you can form polynomial and then find remainder by remainder theorem. 2nd problem. As the polynomial is of degree 2 so it should have 2 roots. So $$f(x)=(x-p)(x-q)=x^2-(p+q)x+pq$$ So $$x^2-(p+q)x+pq=x^2+px+q$$ So $$-(p+q)=p$$ and $$pq=q$$ Solve these simultaneous equations. Please check the calculation. I am not very good at it. :P · 2 years, 1 month ago p=1,q=-2 · 2 years, 1 month ago Also p=q=0. · 2 years, 1 month ago Yep! Missed that. · 2 years, 1 month ago I am grateful to you and @Sravanth Chebrolu for your", "Looking for Math worksheets? Check out our pre-made Math worksheets! Tweet ##### Browse Questions You can create printable tests and worksheets from these Grade 12 Math questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. 1 2 3 4 ... 19 What is the value of sine $(3pi/2)$? 1. $-1$ 2. $sqrt2/2$ 3. $1$ 4. $1/2$ The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has... 1. A vertical asymptote at x=2 and a hole at x=-4 2. A horizontal asymptote at x=-4 and a hole at x=2 3. Vertical asymptotes at both x=2 and x=-4 4. A horizontal asymptote at x=2 and a hole at x=-4 5. A vertical asymptote at x=-4 and a hole a hole at x=2 Grade 12 Complex Numbers CCSS: HSN-CN.C.8 Factor $x^2+16$ 1. $sqrt16i$ 2. $(x-4i)(x+4i)$ 3. $+-sqrt4$ 4. $(x+2i)(x-2i)$ Which matrix represents the system of equations $2x=8$ and $6=3y+x$ 1. $[[2,8,0],[6,3,1]]$ 2. $[[8,2,0],[6,3,1]]$ 3. $[[0,2,8],[6,3,1]]$ 4. $[[2,0,8],[1,3,6]]$ Grade 12 Polynomials and Rational Expressions CCSS: HSF-IF.C.7c On $f(x)=2x^5+x^3-3x$ 1. $(-0.65,1.4)$ is a relative and absolute maximum. 2. $(0.65,-1.4)$ is a relative and absolute minimum. 3. $(-0.65,1.4)$ is a relative maximum. 4. $(0.65,-1.4)$ is a relative minimum. 5. Both A and B. 6. Both C and D. 7. None of the", "Looking for Algebra worksheets? Check out our pre-made Algebra worksheets! Tweet Browse Questions You can create printable tests and worksheets from these Grade 12 Functions questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. 1 2 3 What is the value of sine $(3pi/2)$? 1. $-1$ 2. $sqrt2/2$ 3. $1$ 4. $1/2$ The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has... 1. A vertical asymptote at x=2 and a hole at x=-4 2. A horizontal asymptote at x=-4 and a hole at x=2 3. Vertical asymptotes at both x=2 and x=-4 4. A horizontal asymptote at x=2 and a hole at x=-4 5. A vertical asymptote at x=-4 and a hole a hole at x=2 A town begins with 100 people and after 100 years has a population of 3200. The population has 1. doubled every 25 years 3. tripled every 10 years 4. doubled every 20 years What is the value of tangent $(pi)$? 1. $-1/2$ 2. $sqrt2/2$ 3. $1$ 4. $0$ If $f(x) = x^2-x+20$ then $f(b+1) =$ 1. $b^2+18$ 2. $b^2+b+20$ 3. $b^2+3b+20$ What x values must be excluded from the domain of $f(x)=(3x)/(7x+21)$? 1. $x=3$ 2. $x=-3$ 3. $x=0$ 4. $x=-7$ What is the value of cosine $(pi/6)$? 1. $1$ 2.", "Looking for Math worksheets? Check out our pre-made Math worksheets! Tweet ##### Browse Questions You can create printable tests and worksheets from these Grade 12 Math questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. 1 2 3 4 ... 20 Which matrix represents the system of equations $2x=8$ and $6=3y+x$? 1. $[[2,8,0],[6,3,1]]$ 2. $[[8,2,0],[6,3,1]]$ 3. $[[0,2,8],[6,3,1]]$ 4. $[[2,0,8],[1,3,6]]$ What is the value of sine $(3pi/2)$? 1. $-1$ 2. $sqrt2/2$ 3. $1$ 4. $1/2$ Convert to degrees $(7pi)/4$ 1. $360deg$ 2. $240deg$ 3. $60deg$ 4. $315deg$ $45deg$ 1. $pi/4$ 2. $pi/5$ 3. $sqrt2/2pi$ 4. none of the above solve for x: (x-2)(2x=3)=0 1. -.3 2. -2,-2/3 3. 2,-3/2 4. 2,3/2 Grade 12 Function and Algebra Concepts The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has... 1. A vertical asymptote at x=2 and a hole at x=-4 2. A horizontal asymptote at x=-4 and a hole at x=2 3. Vertical asymptotes at both x=2 and x=-4 4. A horizontal asymptote at x=2 and a hole at x=-4 5. A vertical asymptote at x=-4 and a hole a hole at x=2 Grade 12 Represent and Determine Probability If 4! is equal to 4*3*2*1 then 9! is equal to 1. 9*8*7*6*5*4 2. 9*1 3. 9*8*7*6*5*4*3*2*1 4. none of", "= \\boxed{072}$, which is our answer. ## Solution 4 The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\\sqrt{g(0)+144}=\\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\\neq 0$ (else $f$ is not cubic) where $\\{q,r,s,t,u,v\\}$ is the same as the set $\\{1,2,3,5,6,7\\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that $$24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)$$ which must hold for all $x$. Since $a\\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\\{2,3,7\\}$ and $\\{1,5,6\\}$. Note that each of these sets happily satisfy (2). By (3), since", "# For what values of x, if any, does f(x) = 1/((x-3)(x^2-27)) have vertical asymptotes? Apr 12, 2016 $x = 3$, $x = 3 \\sqrt{3} = 5.2$ and $x = - 3 \\sqrt{3} = - 5.2$ #### Explanation: Vertical asymptotes of a ratio of functions are identified by zeros of the denominator. In the given function, denominator is $\\left(x - 3\\right) \\left({x}^{2} - 27\\right)$ or $\\left(x - 3\\right) \\left(x - 3 \\sqrt{3}\\right) \\left(x + 3 \\sqrt{3}\\right)$and hence three asymptotes are $x = 3$, $x = 3 \\sqrt{3} = 5.2$ and $x = - 3 \\sqrt{3} = - 5.2$ graph{1/(x^3-3x^2-27x+81) [-10, 10, -2, 2]}", "p(a)}{x-a}$$ This provides a natural geometric interpretation of the polynomial $q_a(x)$: given any point $b \\ne a$, $q_a(b)$ is the slope of the line joining $(a, p(a))$ and $(b, p(b))$. These observations motivate the following definition: Definition: For any polynomial $p(x)$ and any $a\\in \\mathbb R$, the derivative of $p(x)$ at $a$, denoted $p'(a)$, is $$p'(a) = q_a(a),$$ where $q_a(x)$ is the quotient obtained by dividing $p(x)$ by $x-a$. For example, with $p(x) = x^2 - 3x$, if we choose $a=1$ we find that $p(x) = (x-1)(x-2) - 2$, so $q_1(x)=x-2$, and therefore $p'(1) = q_1(1) = -1$. More generally for any $a$ we have $x^2 - 3x = (x-a)(x-3+a) + (a^2-3a)$, so $q_a(x) = x-3+a$ and $p'(a) = q_a(a) = 2a - 3$, exactly as the \"usual\" definition gives. Computationally, this makes finding the derivative of even higher-degree polynomials relatively straightforward: just divide $p(x)$ by $x-a$ using the Euclidean algorithm, throw away the remainder, and evaluate the quotient at $a$. However, I don't see any reasonable way to extend this notion to transcendental functions. • Very nice. The problem with transcendental functions is, I think, that you can't really work with them without first having a good working definition of the real numbers (your approach and my approach can both be managed with only the algebraic numbers),", "5+ 10) = 5( -15 + 15) = 0$. Apply long division again: $$p(x) = (x-2)(x-5) \\left(x^3 - 3 x - 10\\right) = (x-2)(x+2)(x-5)^2$$ - Thank You for explaining it so clearly! – Uzdawi Jan 30 '12 at 19:11 First note that $x= \\pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows: $$x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$ $$\\qquad \\Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$ $$(x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0$$ $$\\qquad \\Rightarrow (x^2-4)(x^2-10x+25)=0$$ Hence we can solve it as $$(x^2-4)(x-5)^2 =0$$ or $$x=\\{ -2,+2,5 \\}$$ - Thank You, it was very helpful! – Uzdawi Jan 30 '12 at 19:04", "Looking for Math worksheets? Check out our pre-made Math worksheets! Tweet ##### Browse Questions You can create printable tests and worksheets from these Grade 12 Math questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. 1 2 3 4 ... 20 $45deg$ 1. $pi/4$ 2. $pi/5$ 3. $sqrt2/2pi$ 4. none of the above Convert to degrees $(7pi)/4$ 1. $360deg$ 2. $240deg$ 3. $60deg$ 4. $315deg$ Grade 12 Function and Algebra Concepts Grade 12 Represent and Determine Probability If 4! is equal to 4*3*2*1 then 9! is equal to 1. 9*8*7*6*5*4 2. 9*1 3. 9*8*7*6*5*4*3*2*1 4. none of the above What is the value of sine $(3pi/2)$? 1. $-1$ 2. $sqrt2/2$ 3. $1$ 4. $1/2$ solve for x: (x-2)(2x=3)=0 1. -.3 2. -2,-2/3 3. 2,-3/2 4. 2,3/2 The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has... 1. A vertical asymptote at x=2 and a hole at x=-4 2. A horizontal asymptote at x=-4 and a hole at x=2 3. Vertical asymptotes at both x=2 and x=-4 4. A horizontal asymptote at x=2 and a hole at x=-4 5. A vertical asymptote at x=-4 and a hole a hole at x=2 Which matrix represents the system of equations $2x=8$ and $6=3y+x$ 1. $[[2,8,0],[6,3,1]]$ 2. $[[8,2,0],[6,3,1]]$ 3. $[[0,2,8],[6,3,1]]$", "k=0,...,n$ 4. $(n!)/(k!(n-k)!), \\ k=0,...,n$ The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has 1. a vertical asymptote at x=2, a horizontal asymptote of y=2, and a hole at x=-4. 2. a horizontal asymptote at x=-4 and a hole at x=2. 3. vertical asymptotes at both x=2 and x=-4. 4. a horizontal asymptote at x=2 and a hole at x=-4. 5. a vertical asymptote at x=-4, a horizontal asymptote of y=2, and a hole at x=2. What is the magnitude of a vector with points $(-2,5)$ and $(4,13) ?$ 1. $10$ 2. $5$ 3. $sqrt(7)$ 4. $sqrt(14)$ If the matrix $[[2,9,8],[0,3,4],[1,11,3]]$ is multiplied by the scalar 5, what is the result? 1. $[[7,14,13],[5,8,9],[6,16,8]]$ 2. $[[10,45,40],[0,15,20],[5,55,15]]$ 3. $[[3,4,1],[11,3,2],[9,8,0]]$ 4. $[[10,0,5],[45,15,55],[40,20,15]]$ Grade 12 Polynomials and Rational Expressions CCSS: HSA-APR.D.6 What is $4x^3-2x^2+7x+5$ divided by $(x+2)?$ 1. $(4x^2-10x+27)-49/(x+2)$ 2. $(4x^2-10x+27)$ 3. $(4x^2-10x+27)+59/(x+2)$ 4. $(4x^2-10x+27)-59/(x+2)$ If the matrix $[[27,9,12],[3,0,6],[18,21,3]]$ is multiplied by the scalar $1/3$, what is the result? 1. $[[9,3,4],[1,0,2],[6,7,1]]$ 2. $[[30,12,14],[6,3,9],[21,24,6]]$ 3. $[[27,9,12],[3,0,6],[18,21,3]]$ 4. $[[9,1,6],[3,0,7],[4,2,1]]$ Grade 12 Complex Numbers CCSS: HSN-CN.C.8 Factor. $x^2+12$ 1. $2sqrt3i$ 2. $(x-4i)(x+4i)$ 3. $+-sqrt12$ 4. $(x+2sqrt3i)(x-2sqrt3i)$ Grade 12 Nonlinear Equations and Functions Previous Next You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.", "H.A at $y = 1$ and it will pass through the point $(0, 4)$. Since we want the function to always be decreasing, let $k = -1$. Your required function will be: $F(x) = 3 \\cdot 2^{-x} + 1$. - To make the function pass through $\\langle 0,4\\rangle$, you need to choose $A$ and $B$ so that $f(0)=4$, i.e., so that $A2^0+B=4$. You already know what $B$ has to be in order to get the right horizontal asymptote, so there's only one choice for $A$; does it work, or does the resulting function fail to meet one of the criteria? If you're allowed to choose $k$ as well as $A$ and $B$, note that it makes a difference whether $k$ is positive or negative. - If $A=-1$ and $B=5$ $F(0)$ would satisfy the equation. But that would leave me horizontal Asymptote at 6. The resulting function fails to meet the criteria. – Kurt May 8 '12 at 6:34 @Kurt: But then you'd have your horizontal asymptote at $y=5$. You already picked the only possible value of $B$ to get $y=1$ as horiz. asymptote. – Brian M. Scott May 8 '12 at 6:36", "have, $$(b-1)x= -4x$$ and $$c=2$$. Hence we find that $$b=(-3)$$ and $$c=2$$. • So the polynomial will become, $$x^3-3x+2$$. • Now, using remainder theorem, we find that the remainder of the polynomial is $$f(2)=2^3-(3×2)+2 \\\\ = 8-6+2 \\\\= \\boxed {4}$$ I just wanted to show the other way of doing this. Hope you found it useful!$$\\huge \\ddot\\smile$$ · 1 year, 4 months ago Hi Swaplin, what are the answers? · 1 year, 4 months ago Well I even donno, I wish someone could help, :( · 1 year, 4 months ago The polynomial is of degree 3 so it should have three roots two of them are 1,1. Let the othe root be $$\\alpha$$. So $$f(x)=(x-1)(x-1)(x-\\alpha)=x^3-x^2(\\alpha+2)+x(2\\alpha+1)-\\alpha$$ As coefficient of $$x^2$$ must be zero so $$\\alpha=-2$$. Now using this you can form polynomial and then find remainder by remainder theorem. 2nd problem. As the polynomial is of degree 2 so it should have 2 roots. So $$f(x)=(x-p)(x-q)=x^2-(p+q)x+pq$$ So $$x^2-(p+q)x+pq=x^2+px+q$$ So $$-(p+q)=p$$ and $$pq=q$$ Solve these simultaneous equations. Please check the calculation. I am not very good at it. :P · 1 year, 4 months ago p=1,q=-2 · 1 year, 4 months ago Also p=q=0. · 1 year, 4 months ago Yep! Missed that. · 1 year, 4 months ago I am grateful to you and @Sravanth Chebrolu for your", "times. Case $(1):$ $x=0$ repeated twice Then $d=v=0\\implies f(x)=x^2(ax^2+bx+c)=0\\quad$We require that $g(x)=ax^2+bx+c$ has complex roots $\\implies b^2-4ac<0$ Case $(2): x=0$ repeated four times Then $b=c=d=v=0\\implies f(x)=ax^4=0\\implies a\\in\\mathbb{C}$ Denote the quartic function by $q(x)$. For starters, if $x=0$ is a real zero you can factor it out and write $$q(x) = xp(x)$$ where $p(x)$ is a cubic. Now, any cubic has at least one real zero, so under the stated hypotheses this real zero must equal $0$. Thus $$q(x) = x^2 r(x)$$ where $r$ is quadratic. Finally, you must have either $r(x) = x^2$, or $r(x)$ has no real zeros. Thus either $q(x) = x^4$ or $q(x) = x^2 r(x)$, where $r(x)$ has no zeros.", "= 4^x$ solve $3\\;q - 2 = q^2$, giving $q=1$ or $q=2$, then $2^x=q$, so $x=0$ or $x=1$ But most equations where the variable appears both as an argument to a transcendental function and elsewhere in the equation are not solvable in closed form. ## Approximate solutions Approximate numerical solutions to transcendental equations can be found using numerical, analytical approximations, or graphical methods. Numerical methods for solving arbitrary equations are called root-finding algorithms. In some cases, the equation can be well approximated using Taylor series near the zero. For example, for $k \\approx 1$, the solutions of $\\sin x = k x$ are approximately those of $(1-k) x - x^3/3$, namely $x=0$ and $x = \\plusmn \\sqrt{6} \\sqrt{1-k}$. For a graphical solution, one method is to set each side of a single variable transcendental equation equal to a dependent variable and plot the two graphs, using their intersecting points to find solutions. In some cases, special functions can be used to write the solutions to transcendental equations in closed form. In particular, $x = e^{-x}$ has a solution in terms of the Lambert W Function." ]

[ "+0 # help 0 157 1 The graph of the rational function $$\\frac{p(x)}{q(x)}$$ is shown below, with a horizontal asymptote of $$y = 0$$. If $$q(x)$$ is quadratic, $$p(2)=1$$, and $$q(2) = 3$$, find $$p(x) + q(x).$$ May 10, 2019 #1 +6046 +1 $$\\text{Given the removable singularity at }x=1 \\text{, and the vertical asymptote at }x=-1\\\\ \\text{and the fact that }q(x) \\text{ is quadratic}\\\\ q(x) =q_0 (x-1)(x+1)\\\\ \\text{because of the removable singularity we also know that }p(x) \\text{ has }(x-1) \\text{ as a factor}$$ $$q(2)=3\\\\ q_0 (1)(3)=3\\\\ q_0=1\\\\ q(x) = (x+1)(x-1)=x^2-1$$ $$\\text{As }\\dfrac{p(x)}{q(x)} \\text{ has a horizontal asymptote of }0\\\\ \\text{the degree of }p(x) \\text{ must be less than that of }q(x)\\\\ \\text{so }p(x)=p_0(x-1)\\\\ p(2)=1\\\\ p_0=1\\\\ p(x)=x-1$$ $$p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2$$ . May 11, 2019 #1 +6046 +1 $$\\text{Given the removable singularity at }x=1 \\text{, and the vertical asymptote at }x=-1\\\\ \\text{and the fact that }q(x) \\text{ is quadratic}\\\\ q(x) =q_0 (x-1)(x+1)\\\\ \\text{because of the removable singularity we also know that }p(x) \\text{ has }(x-1) \\text{ as a factor}$$ $$q(2)=3\\\\ q_0 (1)(3)=3\\\\ q_0=1\\\\ q(x) = (x+1)(x-1)=x^2-1$$ $$\\text{As }\\dfrac{p(x)}{q(x)} \\text{ has a horizontal asymptote of }0\\\\ \\text{the degree of }p(x) \\text{ must be less than that of }q(x)\\\\ \\text{so }p(x)=p_0(x-1)\\\\ p(2)=1\\\\ p_0=1\\\\ p(x)=x-1$$ $$p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2$$ Rom May", "+0 # help 0 49 1 Find $$q(x)$$ if the graph of $$\\frac{4x-x^3}{q(x)}$$ has a hole at $$x=-2$$, a vertical asymptote at $$x=1$$, no horizontal asymptote, and $$q(3) = -30$$. May 6, 2019 #1 +101872 +1 4x - x^3 x ( 4 - x^2) x ( 2 - x) ( 2 +x) ______ = ___________ = _____________ q(x) q(x) q(x) If we have a \"hole\" at x = -2, then q(x) must have a factor of ( x + 2) If we have a vertical asymptote at x = 1, then (x - 1) must also be a factor If we have no horizontal asymptote, the q(x) must be a second degree polynomial - the rational fuction has a s;ant asymptote And since q(3) = -30....then we have that -30 = a(3 + 2) (3 - 1) -30 =a (5)(2) -30 = 10a a = -3 So....q(x) = -3(x + 2)(x - 1) = -3(x^2 + x - 2) = -3x^2 - 3x + 6 Here's a graph : https://www.desmos.com/calculator/dz5hnv1yoa May 6, 2019", "sketch the graph of $g(x)$ below. ## Oblique Asymptotes Thus far, we have restricted our discussion of rational functions to those where the degree of the numerator is less than or equal to the degree of the denominator. As our final analysis of graphing rational functions, we will consider the case when the degree of the numerator is greater than the degree of the denominator by one. Example 9 Graph $g(x)=\\frac{x^{2}-1}{x-2}$ Solution First observe that the vertical asymptote is at $x=2$. Notice that the degree of the numerator is greater than the degree of the denominator. We can change the form of the rational expression by long division. You may recall from algebra that polynomials can be divided (just like real numbers), and any rational functions can be written as $\\frac{f(x)}{D(x)}=Q(x)+\\frac{R(x)}{D(x)}$ Doing the long division here, $& \\qquad \\qquad x + 2 \\\\& x-2 \\ \\big ) \\overline{x^{2} + 0x -1 }\\\\& \\qquad \\quad \\underline{x^{2} -2x \\ \\ \\downarrow}\\\\& \\qquad \\qquad \\quad 2x -1\\\\& \\qquad \\qquad \\quad \\underline{2x -4}\\\\& \\qquad \\qquad \\qquad \\quad \\ 3$ So in this case, the function $g(x)$ can be rewritten as $g(x)=\\frac{x^{2}-1}{x-2}=x+2+\\frac{3}{x-2}$ The above equation tells us that as $x\\to\\pm\\infty$, the graph of $g(x)=x+2+\\frac{3}{x-2}$ gets closer and closer to the line $y=x+2$. Why? Suppose we let $x$ be a big number, i.e. $x=1,000,000$.", "\\gt 3$. $x$ 5 4 3.5 3.2 3.1 3.01 3.001 3.00001 $f(x)$ 1 2 4 10 20 200 2000 2×105 The graph of $f$ is shown below. Notes that 1) As $x$ approaches 3 from the left or by values smaller than 3, $f (x)$ decreases without bound. 2) As $x$ approaches 3 from the right or by values larger than 3, $f (x)$ increases without bound. We say that the line $x = 3$, broken line, is the vertical asymptote for the graph of $f$. In general, the line $x = a$ is a vertical asymptote for the graph of f if $f (x)$ either increases or decreases without bound as x approaches a from the right or from the left. This is symbolically written as: ## Horizontal Asymptotes of Rational Functions Let $f(x) = \\dfrac{2x + 1}{x}$ 1) Let $x$ increase and find values of $f (x)$. $x$ 1 10 103 106 f (x) 3 2.1 2.001 2 2) Let $x$ decrease and find values of $f (x)$. $x$ -1 -10 -103 -106 $f(x)$ 1 1.9 1.999 2 As $| x |$ increases, the numerator is dominated by the term $2 x$ and the numerator has only one term x. Therefore $f(x)$ takes values close to $\\dfrac{2x}{2} = 2$. See graphical behaviour below. In general, the", "and closer to $y = 0.4\\text{.}$ Indeed, we see this behavior in the graph of $r\\text{,}$ though the graphing utility misses the fact that $r(-1)$ is actually not defined. A precise graph of $r$ near $x = -1$ should look like the one presented in Figure 5.5.4, where we see an open circle at the point $(-1, 0.4)$ that demonstrates that $r(-1)$ is not defined, and that $r$ does not have a vertical asymptote or zero at $x = -1\\text{.}$ Finally, we also note that $p(1) = 0$ and $q(1) = -6\\text{,}$ so at $x = 1\\text{,}$ $r(x)$ has a zero (its numerator is zero and its denominator is not). In addition, $q(4) = 0$ and $p(4) = 15$ (its denominator is zero and its numerator is not), so $r(x)$ has a vertical asymptote at $x = 4\\text{.}$ These features are accurately represented by the original Desmos graph shown in Figure 5.5.2. In the situation where a rational function is undefined at a point but does not have a vertical asymptote there, we'll say that the graph of the function has a hole. In calculus, we use limit notation to identify a hole in a function's graph. Indeed, having shown in Example 5.5.1 that the value of $r(x)$ gets closer and closer to $0.4$ as $x$ gets closer", "the results of Example to make the following conclusion regarding Limit at infinity for rational functions. Consider a rational function $f(x)=\\frac{p(x)}{q(x)}=\\frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0},\\nonumber$ where $$a_n≠0$$ and $$b_m≠0.$$ 1. If the degree of the numerator is the same as the degree of the denominator $$(n=m),$$ then $$f$$ has a horizontal asymptote of $$y=a_n/b_m$$ as $$x→±∞.$$ 2. If the degree of the numerator is less than the degree of the denominator $$(n<m),$$ then $$f$$ has a horizontal asymptote of $$y=0$$ as $$x→±∞.$$ 3. If the degree of the numerator is greater than the degree of the denominator $$(n>m),$$ then $$f$$ does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms. In addition, using long division, the function can be rewritten as $f(x)=\\frac{p(x)}{q(x)}=g(x)+\\frac{r(x)}{q(x)},$ where the degree of $$r(x)$$ is less than the degree of $$q(x)$$. As a result, $$\\displaystyle \\lim_{x→±∞}r(x)/q(x)=0$$. Therefore, the values of $$[f(x)−g(x)]$$ approach zero as $$x→±∞$$. If the degree of $$p(x)$$ is exactly one more than the degree of $$q(x)$$ (i.e., $$n=m+1$$), the function $$g(x)$$ is a linear function. In this case, we call $$g(x)$$ an oblique asymptote. Now let’s consider the Limit at infinity for functions involving a radical. Example $$\\PageIndex{6}$$: Determining Limit at infinity for a Function Involving a Radical Find the limits as", "other hand, as $$x \\rightarrow -\\infty$$, $$|x| = -x$$, and as such $$k(x) \\approx \\frac{2x}{-x} = -2$$. Finally, it appears as though the graph of $$k$$ passes the Horizontal Line Test which means $$k$$ is one to one and $$k^{-1}$$ exists. Computing $$k^{-1}$$ is left as an exercise. As the previous example illustrates, the graphs of general algebraic functions can have features we've seen before, like vertical and horizontal asymptotes, but they can occur in new and exciting ways. For example, $k(x) = \\frac{2x}{\\sqrt{x^{2} - 1}}$ had two distinct horizontal asymptotes. You'll recall that rational functions could have at most one horizontal asymptote. Also some new characteristics like unusual steepness'\\footnote{The proper Calculus term for this is vertical tangent', but for now we'll be okay calling it unusual steepness'.} and cusps\\footnote{See page \\pageref{cusppicture} for the first reference to this feature.} can appear in the graphs of arbitrary algebraic functions. Our next example first demonstrates how we can use sign diagrams to solve nonlinear inequalities. (Don't panic. The technique is very similar to the ones used in Chapters \\ref{LinearQuadratic}, \\ref{Polynomials} and \\ref{Rationals}.) We then check our answers graphically with a calculator and see some of the new graphical features of the functions in this extended family. \\begin{ex} Solve the following inequalities. Check your answers graphically with a calculator. \\begin{multicols}{2} \\begin{enumerate}", "in both $p$ and $q\\text{.}$ • For a rational function $r(x) = \\frac{p(x)}{q(x)}\\text{,}$ we determine where the function has zeros and where it has vertical asymptotes by considering where the numerator and denominator are $0\\text{.}$ In particular, if $p(a) = 0$ and $q(a) \\ne 0\\text{,}$ then $r(a) = 0\\text{,}$ so $r$ has a zero at $x = a\\text{.}$ And if $q(a) = 0$ and $p(a) \\ne 0\\text{,}$ then $r(a)$ is undefined and $r$ has a vertical asymptote at $x = a\\text{.}$ • By writing a rational function's numerator in factored form, we can generate a sign chart for the function that takes into account all of the zeros and vertical asymptotes of the function, which are the only points where the function can possibly change sign. By testing $x$-values in various intervals between zeros and/or vertical asymptotes, we can determine where the rational function is positive and where the function is negative. ### Exercises5.5.4Exercises ###### 6. For each of the following rational functions, determine, with justification, the exact locations of all (i) horizontal asymptotes, (ii) vertical asymptotes, (iii) zeros, and (iv) holes of the function. Clearly show your work and thinking. 1. $\\displaystyle \\displaystyle r(x) = \\frac{-19(x+11.3)^2(x-15.1)(x-17.3)}{41(x+5.7)(x+11.3)(x-8.4)(x-15.1)}$ 2. $\\displaystyle \\displaystyle s(x) = \\frac{-29(x^2-16)(x^2+99)(x-53)}{101(x^2-4)(x-13)^2(x+104)}$ 3. $\\displaystyle \\displaystyle u(x) = \\frac{-71(x^2 - 13x + 36)(x-58.4)(x+78.2)}{83(x+58.4)(x-78.2)(x^2 - 12x + 27)}$ ######", "k=0,...,n$ 4. $(n!)/(k!(n-k)!), \\ k=0,...,n$ The function $f(x)=(2x^2-7x+6)/(x^2+2x-8)$ has 1. a vertical asymptote at x=2, a horizontal asymptote of y=2, and a hole at x=-4. 2. a horizontal asymptote at x=-4 and a hole at x=2. 3. vertical asymptotes at both x=2 and x=-4. 4. a horizontal asymptote at x=2 and a hole at x=-4. 5. a vertical asymptote at x=-4, a horizontal asymptote of y=2, and a hole at x=2. What is the magnitude of a vector with points $(-2,5)$ and $(4,13) ?$ 1. $10$ 2. $5$ 3. $sqrt(7)$ 4. $sqrt(14)$ If the matrix $[[2,9,8],[0,3,4],[1,11,3]]$ is multiplied by the scalar 5, what is the result? 1. $[[7,14,13],[5,8,9],[6,16,8]]$ 2. $[[10,45,40],[0,15,20],[5,55,15]]$ 3. $[[3,4,1],[11,3,2],[9,8,0]]$ 4. $[[10,0,5],[45,15,55],[40,20,15]]$ Grade 12 Polynomials and Rational Expressions CCSS: HSA-APR.D.6 What is $4x^3-2x^2+7x+5$ divided by $(x+2)?$ 1. $(4x^2-10x+27)-49/(x+2)$ 2. $(4x^2-10x+27)$ 3. $(4x^2-10x+27)+59/(x+2)$ 4. $(4x^2-10x+27)-59/(x+2)$ If the matrix $[[27,9,12],[3,0,6],[18,21,3]]$ is multiplied by the scalar $1/3$, what is the result? 1. $[[9,3,4],[1,0,2],[6,7,1]]$ 2. $[[30,12,14],[6,3,9],[21,24,6]]$ 3. $[[27,9,12],[3,0,6],[18,21,3]]$ 4. $[[9,1,6],[3,0,7],[4,2,1]]$ Grade 12 Complex Numbers CCSS: HSN-CN.C.8 Factor. $x^2+12$ 1. $2sqrt3i$ 2. $(x-4i)(x+4i)$ 3. $+-sqrt12$ 4. $(x+2sqrt3i)(x-2sqrt3i)$ Grade 12 Nonlinear Equations and Functions Previous Next You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.", "# asymptotes of graph of rational function Let $f(x)\\,=\\,\\frac{P(x)}{Q(x)}$ be a fractional expression where $P(x)$ and $Q(x)$ are polynomials with real coefficients such that their quotient can not be reduced (http://planetmath.org/Division) to a polynomial. We suppose that $P(x)$ and $Q(x)$ have no common zeros. If the division of the polynomials is performed, then a result of the form $f(x)\\;=\\;H(x)+\\frac{R(x)}{Q(x)}$ is gotten, where $H(x)$ and $R(x)$ are polynomials such that $\\deg{R(x)}<\\deg{Q(x)}$ The graph of the rational function $f$ may have asymptotes: 1. 1. Every zero $a$ of the denominator $Q(x)$ gives a vertical asymptote $x=a$. 2. 2. If $\\deg{H(x)}<1$ (i.e. $0$ or $-\\infty$) then the graph has the horizontal asymptote $y=H(x)$. 3. 3. If $\\deg{H(x)}=1$ then the graph has the skew asymptote $y=H(x)$. Proof of 2 and 3. We have $\\displaystyle f(x)\\!-\\!H(x)=\\frac{R(x)}{Q(x)}\\,\\to 0$ as $|x|\\to\\infty$. Remark. Here we use the convention that the degree of the zero polynomial is $-\\infty$. Title asymptotes of graph of rational function AsymptotesOfGraphOfRationalFunction 2013-03-22 15:09:34 2013-03-22 15:09:34 eshyvari (13396) eshyvari (13396) 10 eshyvari (13396) Result msc 51N99 msc 26C15 msc 26A09 Polytrope", "have vertical asymptotes, but it doesn't, again as you can tell from the formula (it is defined for all real numbers). Discovering facts about a function by looking at its graph is useful but dangerous. #### Example Below is the graph of the function$f(x)=.0002{{\\left( \\frac{{{x}^{3}}-10}{3{{e}^{-x}}+1} \\right)}^{6}}$ If you didn't know the formula for the function, by looking at the graph you might nevertheless suspect certain things are true of the function. 1. Maybe it has a local maximum somewhere to the right of $x=1$. 2. Maybe it has one or more zeroes around $x=-1$. 3. Maybe it has one or more zeroes around $x=2$. 4. Maybe it has a vertical asymptote somewhere to the right of $x=2.5$. 5. Maybe it has a horizontal asymptote in the negative $x$ direction. If you do know the formula, you can find out many things about the function that you can't depend on the graph to see. • You can see immediately that $f$ has just one zero, at $x=\\sqrt[3]{10}$, which is about $2.15$. So (3) is correct. • If you notice that the exponent $6$ is even, you can figure out that $f(x)\\geq0$ for all $x$, so $\\sqrt[3]{10}$ is an absolute minimum for $f$. • By using L'Hôpital several times, you can see that $f$ has a horizontal asymptote as $x\\to-\\infty$.", "vertical asymptotes, but it doesn't, again as you can tell from the formula (it is defined for all real numbers). Discovering facts about a function by looking at its graph is useful but dangerous. Example Below is the graph of the function$f(x)=.0002{{\\left( \\frac{{{x}^{3}}-10}{3{{e}^{-x}}+1} \\right)}^{6}}$ If you didn't know the formula for the function, by looking at the graph you might nevertheless suspect certain things are true of the function. 1. Maybe it has a local maximum somewhere to the right of $x=1$. 2. Maybe it has one or more zeroes around $x=-1$. 3. Maybe it has one or more zeroes around $x=2$. 4. Maybe it has a vertical asymptote somewhere to the right of $x=2.5$. 5. Maybe it has a horizontal asymptote in the negative $x$ direction. If you do know the formula, you can find out many things about the function that you can't depend on the graph to see. • You can see immediately that $f$ has just one zero, at $x=\\sqrt[3]{10}$, which is about $2.15$. So (3) is correct. • If you notice that the exponent $6$ is even, you can figure out that $f(x)\\geq0$ for all $x$, so $\\sqrt[3]{10}$ is an absolute minimum for $f$. • By using L'Hôpital several times, you can see that $f$ has a horizontal asymptote as $x\\to-\\infty$. So (5)", "a rational function $p(x)/q(x)$, the roots of $q(x)$ may or may not be asymptotes. For a root $c$ if $p(c)$ is not zero then the line $x=c$ will be a vertical asymptote. If $c$ is a root of both $p(x)$ and $q(x)$, then we can rewrite the expression as: $~ \\frac{p(x)}{q(x)} = \\frac{(x-c)^m r(x)}{(x-c)^n s(x)}, ~$ where both $r(c)$ and $s(c)$ are non zero. Knowing $m$ and $n$ (the multiplicities of the root $c$) allows the following to be said: • If $m < n$ then $x=c$ will be a vertical asymptote. • If $m \\geq n$ then $x=c$ will not be vertical asymptote. (The value $c$ will be known as a removable singularity). In this case, the graph of $p(x)/q(x)$ and the graph of $(x-c)^{m-n}r(x)/s(x)$ will differ, as the latter will include a value for $x=c$, whereas $x=c$ is not in the domain of $p(x)/q(x)$. Finding the multiplicity may or may not be hard, but there is a quick check that is often correct. With Julia, if you have a rational function that has f(c) evaluate to Inf or -Inf then there will be a vertical asymptote. If the expression evaluates to NaN, more analysis is needed. (The value of 0/0 is NaN, where as 1/0 is Inf.) For example, the function $f(x) = ((x-1)^2 \\cdot (x-2))", "x^2+1 (we're supposed to drop the remainder), but I still don't see how that fits into graphing it? So would the Root be x=-1? We have to find the Roots/holes, the Vertical Asymptotes, the Horizontal Asymptotes, and the Y-intercept. On previous problems they were like so: (x+3)(x-1)(x+2) / (x+2)(x+1) So I could find the Roots/holes to be x=-3, -2 and 1 VA: x=-1 HA: y=x y=-3 But I'm not getting anything close to that on these :/ 5. $x^2 + 1$ has no real roots. Also, a root and a \"hole\" are different things. For the equation $\\frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}$ the roots are -3 and 1, and the \"holes\" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there. A vertical asymptote occurs wherever there is a \"hole\". 6. Originally Posted by icemanfan $x^2 + 1$ has no real roots. Also, a root and a \"hole\" are different things. For the equation $\\frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}$ the roots are -3 and 1, and the \"holes\" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there. A vertical asymptote occurs wherever there is a \"hole\". If you're trying to graph the original functions you can't disregard the remainder. For the first one", "\\frac{p(x)}{q(x)}$$, where $$p$$ and $$q$$ are polynomial functions with leading coefficients $$a$$ and $$b$$, respectively. • If the degree of $$p(x)$$ is the same as the degree of $$q(x)$$, then $$y=\\frac{a}{b}$$ is the horizontal asymptote of the graph of $$y=r(x)$$. • If the degree of $$p(x)$$ is less than the degree of $$q(x)$$, then $$y=0$$ is the horizontal asymptote of the graph of $$y=r(x)$$. • If the degree of $$p(x)$$ is greater than the degree of $$q(x)$$, then the graph of $$y=r(x)$$ has no horizontal asymptotes. Like Theorem 4.1, Theorem 4.2 is proved using Calculus. Nevertheless, we can understand the idea behind it using our example $$f(x) = \\frac{2x-1}{x+1}$$. If we interpret $$f(x)$$ as a division problem, $$(2x-1) \\div (x+1)$$, we find that the quotient is $$2$$ with a remainder of $$-3$$. Using what we know about polynomial division, specifically Theorem 3.4, we get $$2x-1 = 2(x+1) -3$$. Dividing both sides by $$(x+1)$$ gives $$\\frac{2x-1}{x+1} = 2 - \\frac{3}{x+1}$$. (You may remember this as the formula for $$g(x)$$ in Example 4.1.1.) As $$x$$ becomes unbounded in either direction, the quantity $$\\frac{3}{x+1}$$ gets closer and closer to $$0$$ so that the values of $$f(x)$$ become closer and closer\\footnote{As seen in the tables immediately preceding Definition \\ref{va}.} to $$2$$. In symbols, as $$x \\rightarrow \\pm \\infty$$, $$f(x) \\rightarrow 2$$, and", "# Question #34779 Dec 26, 2017 that means $b = - 1$, $c = - 12$, and a can be anything except -3 or 4 #### Explanation: if the graph has vertical asymptotes at x=-3 and x=4, then the denominator of the function must be $k \\left(x - \\left(- 3\\right)\\right) \\left(x - \\left(4\\right)\\right) = k \\left(x + 3\\right) \\left(x - 4\\right)$ where k is some constant. since the problem gives the function in the form: $y = \\frac{x - a}{{x}^{2} + b x + c}$, the value of k must be 1 so the coefficient of the ${x}^{2}$ is also 1. this means the equation is now: $y = \\frac{x - a}{\\left(x + 3\\right) \\left(x - 4\\right)} = \\frac{x - a}{{x}^{2} - x - 12}$ if the numerator is $x + 3$ or $x - 4$ then the vertical asymptotes \"created\" earlier will just become holes, because the $x + 3$ of $x - 4$ will factor out from both the numerator and denominator. that means $b = - 1$, $c = - 12$, and a can be anything except -3 or 4", "$Q$ determines the number of constants and that we can find all the constants by incrementing every power of the root in the case of a repeated root? I want lift the rug up and see what mechanisms are at play in order to understand this better. - You may get something from the answers to math.stackexchange.com/questions/48051/… – Gerry Myerson Aug 22 '12 at 1:39 Thanks for your comment. I looked at that question before posting but it didn't help me in considering how we know that the degree of Q in my example will be equal to the number of unknown constants that must be found. Perhaps this is trivially knowable but I lack that knowledge and would be interested in acquiring it. – R R Aug 22 '12 at 1:42 Great question! First, let's tackle just the case where the roots of $Q(x)$ are all distinct. One way to conceptualize what's going on is the following: if $r$ is a root of $Q(x)$, then as $x \\to r$, the function $f(x) = \\frac{P(x)}{Q(x)}$ (we always assume $P, Q$ have no common roots) goes to infinity. How quickly does it go to infinity? Well, write $Q(x) = (x - r) R(x)$. Then $$\\frac{P(x)}{Q(x)} = \\frac{1}{x - r} \\left( \\frac{P(x)}{R(x)} \\right)$$ and $R(r) \\neq 0$. So we see", "going from a horizontal asymptote (as $x\\rightarrow-\\infty$) to its global minimum at $x^*$, to $(0,-c)$ where it crosses the $y$-axis, while for $x > x^*$, $f$ is increasing and concave ($f\\,''>0$). From its derivative $f\\,'(0)=\\ln a-\\ln b$ at $0$, one can estimate the root of $f$ to be $x_1=\\frac{c}{\\ln a-\\ln b}$, which always bounds (overestimates) the true root (since $f\\,''>0$). Because of this concavity, the secant method with $x_0=0$ & $x_1$ above might give slightly better convergence than Newton's method for this root. More generally, for $a,b,c>0$, the analysis & character of $f$ will vary depending on the signs of these quantities: $\\chi_0=a-b$, $\\chi_1=\\ln a-\\ln b$, $\\chi_2=\\ln\\ln a-\\ln\\ln b$ which in turn depend on the signs of $\\ln\\frac{a}{b},~\\ln a$ & $\\ln b$ (i.e. on the order of $a$, $b$ and $1$ on the positive number line), thus leaving six cases to classify (sorry about the extra legend in the graph; not sure how it's getting there): $$\\matrix{ \\text{case}&\\qquad\\text{unique solution for}\\\\ a < b < 1&\\qquad x < 0\\\\ a < 1 < b&\\qquad x < 0\\\\ b < a < 1&\\qquad\\text{no solution}\\\\ b < 1 < a&\\qquad x > 0\\\\ 1 < a < b&\\qquad\\text{no solution}\\\\ 1 < b < a&\\qquad\\text{discussed above, }x > 0\\\\ }$$ When $a$ is between $b$ and $1$, there is no solution. This", "as $x^2(t)+y^2(t) = 1$ for all $t$. ## Rational functions A rational function is the ratio of two polynomials $$f(x) = \\frac{P(x)}{Q(x)}$$ It can always be in reduced form, i.e. with distinct zeros for $P$ and $Q$. The degree $\\operatorname{\\textrm{deg}}(P)$ of a polynomial $P$ is the highest power of $P$. The degree of $x+1$ is $1$ and the degree of $x^3-1$ is $3$. The main properties of a rational function in reduced form are as follows. • Domain. Everything but the zeros of $Q$. • Vertical asymptote at each zero of $Q$. • Zeros are the zeros of $P$. • Horizontal and oblique asymptotes if $\\operatorname{\\textrm{deg}}(P) \\le \\operatorname{\\textrm{deg}}(Q) + 1$. In this case, we write $f$ in the form $$ax + b + \\frac{R(x)}{Q(x)}$$ where $\\operatorname{\\textrm{deg}}(R) \\le\\operatorname{\\textrm{deg}}(Q)-1$. The asymptote is $y = ax+b$. $$f(x) = \\frac{x^3-x}{x^2-4}= \\frac{x^3-4x+3x}{x^2-4} = x + \\frac{ 3x}{x^2-4}$$ is a rational function in reduced form, because the zeros of the numerator ($-1$, $0$ and $1$) and denominator ($-2$ and $2$) are distinct. The function has asymptotes $x=-2$, $x=2$ and $y=x$. ## Hyperbola The equation of a hyperbola centred at $(x_0,y_0)$, with transverse $x$-axis, and radii $a\\gt0$ and $b\\gt0$ is $$\\frac{(x-x_0)^2}{a^2}-\\frac{(y-y_0)^2}{b^2} = 1$$ It passes through $(x_0\\pm a,y_0)$ and has asymptotes $$\\displaystyle \\frac{y}{b} = \\pm\\frac{x}{a}.$$ The two branches intersect the transverse $x$-axis. We divide the", "2$ and $x = 1$ are the vertical asymptotes. Horizontal asymptote. There is no horizontal asymptote because the power of the numerator is larger than the power of the denominator. Notice the function in part d had more than one vertical asymptote. Here’s another function with two vertical asymptotes. #### Example D Graph the function $y = \\frac{-x^2}{x^2 - 4}$ . Solution $\\text{Let's set the denominator equal to zero:} \\qquad x^2 - 4 = 0\\!\\\\\\\\\\text{Factor:} \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad (x - 2)(x + 2) = 0\\!\\\\\\\\\\text{Solve:} \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\ \\ x = 2, x = -2$ We find that the function is undefined for $x = 2$ and $x = -2$ , so we know that there are vertical asymptotes at these values of $x$ . We can also find the horizontal asymptote by the method we outlined above. It’s at $y = \\frac{-x^2}{x^2}$ , or $y = -1$ . So, we start plotting the function by drawing the vertical and horizontal asymptotes on the graph. Now, let’s make a table of values. Because our function has a lot of detail we must make sure that we pick enough values for our table to determine the behavior of the function accurately. We must make sure especially that we" ]

[ "write the quotient of two complex numbers, allowing us to do division: z zw zw = = 2 w ww w EXAMPLE 1.1 Find the complex conjugate, sum, product, and quotient of the complex numbers z = 2 − 3i w = 1+ i SOLUTION To find the complex conjugate of each complex number we let i → − i . Hence z = 2 − 3i = 2 + i 3 w = 1+ i = 1− i CHAPTER 1 Complex Numbers 7 The sum of the two complex numbers is formed by adding the real and imaginary parts, respectively: z + w = ( 2 − 3i ) + (1 + i ) = ( 2 + 1) + i ( −3 + 1) = 3 − 2i We can form the product as follows: zw = ( 2 − 3i )(1 + i ) = 2 − 3i + 2i − 3i 2 = ( 2 + 3) + i ( −3 + 2 ) = 5−i Finally, we use the complex conjugate of w to form the quotient: z z ⎛ w ⎞ ( 2 − 3i ) (1 − i ) 2 − 3i − 2i + 3i 2 2 − 3 − 5i −1 − 5i = ⎜ ⎟= = = =", "+0 # complex numbers 0 39 1 Let w be a complex number such that |w| = 1, and the equation z - \\overline{z} = w has a pure imaginary root. Find w + \\overline{w}. Oct 10, 2022", "$z=10$ when $w=0.5,$ find $z$ when $w=8$. Solve each problem. If $z$ varies inversely as $w,$ and $z=10$ when $w=0.5,$ find $z$ when $w=8$....", "+0 0 96 1 +95 Let $$z_1$$ and $$z_2$$ be two complex numbers such that $$|z_1| = 5$$ and $$\\frac{z_1}{z_2} + \\frac{z_2}{z_1} = 1$$. Find $$|z_1 - z_2|^2.$$ Aug 16, 2022", "+0 0 64 1 Part a) Let z and w be complex numbers satisfying |z| = 4 and |w| = 2. What is $$|z+w|^2, |zw|^2, |z-w|^2, \\left| \\dfrac{z}{w} \\right|^2$$? Part b) Let z and w be complex numbers satisfying |z| = 5, |w| = 2, and $$z\\overline{w} = 6+8i.$$ What is $$|z+w|^2, |zw|^2, |z-w|^2, \\left| \\dfrac{z}{w} \\right|^2$$? Dec 22, 2019", "# A BIT COMPLEX. ISN'T IT? Level pending For complex numbers z and w, it is given that : $$|z|^2w-|w|^2z=z-w$$. Then what is the value of $$\\overline{w}z$$ ? ×", "+0 # Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\\ne -1$, then $\\frac{w+z}{1+wz}$ is a real number. 0 102 1 +16 Prove that if $$w,z$$ are complex numbers such that $$|w|=|z|=1$$and $$wz\\ne -1$$, then $$\\frac{w+z}{1+wz}$$ is a real number. Jan 5, 2020", "+0 # Please help 0 40 1 Part a) Let z and w be complex numbers satisfying |z| = 4 and |w| = 2. What is $$|z+w|^2, |zw|^2, |z-w|^2, \\left| \\dfrac{z}{w} \\right|^2$$? Part b) Let z and w be complex numbers satisfying |z| = 5, |w| = 2, and $$z\\overline{w} = 6+8i.$$ What is $$|z+w|^2, |zw|^2, |z-w|^2, \\left| \\dfrac{z}{w} \\right|^2$$? Dec 22, 2019 ### 1+0 Answers #1 0 (a) |z + w|^2 = 36, |zw|^2 = 64, |z - w|^2 = 4, |z/w|^2 = 4 (b) |z + w|^2 = 29, |zw|^2 = 100, |z - w|^2 = 24,|z/w|^2 = 25/4 Dec 22, 2019", "# Add Complex Numbers - Calculator An easy to use calculator that add two complex numbers. Let w and z be two complex numbers such that w = a + ib and z = A + iB. The addition of w by z is given by w + z = (a + ib) + (A + iB)= (a + A) + (b + B)i How to use the calculator ## Use of Calculator to Add Two Complex Numbers Enter the real and imaginary parts (as an integer, a decimal or a fraction) of two complex numbers z and w and press \"Add\". w = -1 + i -9 z = 1/2 + i 2.1 w + z = + i (exact value)", "that the complex numbers satisfy many useful and familiar properties, which are similar to properties of the real numbers. If $$u$$, $$w$$, and $$z$$, are complex numbers, then 1. $$w + z = z + w$$ 2. $$u + (w + z) = (u + w) + z$$ 3. The complex number $$0 = 0 + 0i$$ is an additive identity, that is $$z + 0 = z$$. 4. If $$z = a + bi$$, then the additive inverse of $$z$$ is $$-z = (-a) + (-b)i$$. That is, $$z + (-z) = 0$$. 5. $$wz = zw$$ 6. $$u(wz) = (uw)z$$ 7. $$u(w + z) = uw + uz$$ 8. If $$wz = 0$$, then $$w = 0$$ or $$z = 0$$. We will use these properties as needed. For example, to write the complex product $$(1 + i)i$$ in the form $$a + bi$$ with $$a$$ and $$b$$ real numbers, we distribute multiplication over addition and use the fact that $$i^{2} = -1$$ to see that $(1 + i)i = i + i^{2} = i + (-1) = (-1) + i.$ For another example, if $$w = 2 + i$$ and $$z = 3 - 2i$$, we can use these properties to write $$wz$$ in the standard $$a + bi$$ form as follows: $wz = (2", "# Definition:Complex Root ## Definition Let $z \\in \\C$ be a complex number such that $z \\ne 0$. Let $n \\in \\Z_{>0}$ be a (strictly) positive integer. Let $w \\in \\C$ such that: $w^n = z$ Then $w$ is a (complex) $n$th root of $z$, and we can write: $w = z^{1 / n}$ ## Also see • Roots of Complex Number, where it is demonstrated what the complex $n$th roots actually are in terms of $z$ and $n$.", "# Complex Function Algebra Level 5 Let $$f: \\mathbb{C}\\mapsto \\mathbb{C}$$ be defined as $$f(z)=z^2+iz+1$$ for all complex $$z$$. How many complex numbers $$z$$ are there such that $$\\text{Im} (z) > 0$$, and both the real and imaginary parts of $$f(z)$$ are integers with absolute value of at most 10? ×", "# Complex Numbers and power series notice that $w^3-1=0$ expand $e^z,e^{wz},e^{w^2z}$ to power series, and then simplify the sum,you will find that a lot of terms are eliminated.", ", ww , and zz , are complex numbers, then 1. w+z=z+ww+z=z+w 2. u+(w+z)=(u+w)+zu+(w+z)=(u+w)+z 3. The complex number 0=0+0i0=0+0i is an additive identity, that is z+0=zz+0=z . 4. If z=a+biz=a+bi , then the additive inverse of zz is z=(a)+(b)i−z=(−a)+(−b)i . That is, z+(z)=0z+(−z)=0 . 5. wz=zwwz=zw 6. u(wz)=(uw)zu(wz)=(uw)z 7. u(w+z)=uw+uzu(w+z)=uw+uz 8. If wz=0wz=0 , then w=0w=0 or z=0z=0 .", "# Complicated Complex Fractions Algebra Level 5 Let $$x, y$$ and $$z$$ be complex numbers such that $x + y + z = 2$ $x^2 + y^2 + z^2 = 3$ $xyz = 4$. Evaluate $|\\frac {1}{xy + z - 1} + \\frac {1}{yz + x - 1} + \\frac {1}{zx + y - 1}|$ ×", "parts, the part corresponding to ${a}$ and the part corresponding to ${b}$. We call ${a}$ the real part of ${z}$, written as ${\\Re z = a}$, and we call ${b}$ the imaginary part of ${z}$, written as ${\\Im z = b}$. I should add that the name ”imaginary number” is a poor name that reflects historical reluctance to view complex numbers as acceptable. For that matter, the name ”complex number” is also a poor name. As a brief example, consider the Gaussian integer ${z = 2 + 5i}$. Then ${N(z) = 4 + 25 = 29}$, ${\\Re z = 2}$, ${\\Im z = 5}$, and ${\\overline{z} = 2 – 5i}$. We can ask similar questions to those we asked about the regular integers. What does it mean for ${z \\mid w}$ in the complex case? Definition 2 We say that a Gaussian integer ${z}$ divides another Gaussian integer ${w}$ if there is some Gaussian integer ${k}$ so that ${zk = w}$. In this case, we write ${z \\mid w}$, just as we write for regular integers. For the integers, we immediately began to study the properties of the primes, which in many ways were the building blocks of the integers. Recall that for the regular integers, we said ${p}$ was a prime if its only divisors were ${\\pm", "# Complex, yet Faulty Algebra Level 2 What is the wrong step in the following proof that $$1 = -1$$? 1. Let $$w$$ be a complex number such that $$(w + 1)^3 = (w - 1)^3$$. 2. Solving this equation gives $$w = \\pm \\frac{i \\sqrt{3}}{3}$$. 3. Since $$(w + 1)^3 = (w - 1)^3$$ for our previously mentioned values of $$w$$, cube rooting both sides gives $$w + 1 = w - 1$$. 4. Subtracting $$w$$ from both sides gives $$1 = -1$$. ×", "+0 # complex numbers 0 133 1 Let z and w be complex numbers such that |z| = 1, |w| = 7, and |5z + w| = 11. Find |3z + 4w|. Jul 29, 2022 #1 +285 0 We can use the fact that for any complex number a, (a)(cong(a)) = |a|^2. $$|5z + w|^2 = (5z + w)(5\\overline{z} + \\overline{w}) = 121$$ $$25|z|^2 + |w|^2 + 5(\\overline{z}w + \\overline{w}z) = 121$$ $$5(\\overline{z}w + \\overline{w}z) = 47$$ $$|3z + 4w|^2 = (3z + 4w)(3\\overline{z} + 4\\overline{w})$$ $$=793 + 12(\\overline{w}z + \\overline{z}w)$$ $$=793 + 564/5$$ Take the square root of that value and you will the answer. Jul 31, 2022", "4x+ 6y+ 5z= 0. Solving for w, w= 2x+ 3y+ (5/2)z. Then, <w, x, y, z>= <2x+ 3y+ (5/2)z, x, y, z>= <2x, x, 0, 0> + <3y, 0, y, 0> + <(5/2)z, 0, 0, z>= x<2, 1, 0, 0>+ y<3, 0, 1, 0> + z<5/2, 0, 0, 1>. And now, a basis is obvious. If you don't like fractions, that last one can be written (z/2)<5, 0, 0, 2>", "# Complex Equations Algebra Level 4 \\begin{align} (2-3i)z + (5+i)w &= 19-4i\\\\ (1-i)z + (2+i)w &= 9-i \\end{align} Let $$z$$ and $$w$$ be the complex numbers satisfying the equations above. Compute $$|z+w|^2$$. Note: $$i^2 = -1$$, and $$|z|$$ is the absolute value of $$z$$. ×" ]

[ "and $w = 2 + 7i\\text{.}$ Then $z+w = 5 +3i\\text{,}$ and \\begin{align*} z\\cdot w \\amp = (3-4i)(2+7i)\\\\ \\amp = 6 + 28 - 8i + 21i\\\\ \\amp = 34 + 13i. \\end{align*} A few other computations: \\begin{align*} 4z \\amp = 12 - 16i\\\\ |z| \\amp = \\sqrt{3^2 + (-4)^2} = 5\\\\ \\overline{zw}\\amp =34-13i. \\end{align*} ### SubsectionExercises ###### 1 In each case, determine $z + w\\text{,}$ $sz\\text{,}$ $|z|\\text{,}$ and $z\\cdot w\\text{.}$ a. $z = 5 + 2i\\text{,}$ $s = -4\\text{,}$ $w = -1 + 2i\\text{.}$ b. $z = 3i\\text{,}$ $s = 1/2\\text{,}$ $w = -3 + 2i\\text{.}$ c. $z = 1 + i\\text{,}$ $s = 0.6\\text{,}$ $w = 1 - i\\text{.}$ ###### 2 Show that $z\\cdot \\overline{z} = |z|^2\\text{,}$ where $\\overline{z}$ is the conjugate of $z\\text{.}$ ###### 3 Suppose $z=x+yi$ and $w=s+ti$ are two complex numbers. Prove the following properties of the conjugate and the modulus. a. $|w\\cdot z| = |w|\\cdot |z|\\text{.}$ b. $\\overline{zw} = \\overline{z}\\cdot \\overline{w}\\text{.}$ c. $\\overline{z + w} = \\overline{z} + \\overline{w}\\text{.}$ d. $z + \\overline{z} = 2\\text{Re}(z)\\text{.}$ (Hence, $z + \\overline{z}$ is a real number.) e. $z - \\overline{z} = 2\\text{Im}(z)i\\text{.}$ f. $|z| = |\\overline{z}|\\text{.}$ ###### 4 A Pythagorean triple consists of three integers $(a,b,c)$ such that $a^2 + b^2 = c^2\\text{.}$ We can use complex numbers to generate Pythagorean triples. Suppose $z =", "+0 # complex numbers 0 133 1 Let z and w be complex numbers such that |z| = 1, |w| = 7, and |5z + w| = 11. Find |3z + 4w|. Jul 29, 2022 #1 +285 0 We can use the fact that for any complex number a, (a)(cong(a)) = |a|^2. $$|5z + w|^2 = (5z + w)(5\\overline{z} + \\overline{w}) = 121$$ $$25|z|^2 + |w|^2 + 5(\\overline{z}w + \\overline{w}z) = 121$$ $$5(\\overline{z}w + \\overline{w}z) = 47$$ $$|3z + 4w|^2 = (3z + 4w)(3\\overline{z} + 4\\overline{w})$$ $$=793 + 12(\\overline{w}z + \\overline{z}w)$$ $$=793 + 564/5$$ Take the square root of that value and you will the answer. Jul 31, 2022", "+0 0 31 1 Let z and w be complex numbers satisfying |z| = 5, |w| = 2, and z(overline{w}) = 6+8i. Find the value of |z+w|^2, |zw|^2, |z-w|^2, |frac{z}{w}|^2 . If any of these cannot be uniquely determined from the information given, write not possible. May 10, 2019 #1 +22180 +3 Let $$z$$ and $$w$$ be complex numbers satisfying $$|z| = 5$$, $$|w| = 2$$, and $$z\\overline{w} = 6+8i$$ Find the value of $$|z+w|^2$$, $$|zw|^2$$, $$|z-w|^2$$, $$\\left|\\dfrac{z}{w}\\right|^2$$ . $$\\begin{array}{|rcll|} \\hline && \\mathbf{|z+w|^2} \\\\ &=& |z|^2+|w|^2+2*\\Re{(z\\overline{w})} \\\\ &=& 5^2+2^2+2*6 \\quad | \\quad \\Re{(z\\overline{w})} = \\Re{(6+8i)} = 6 \\\\ &=& \\mathbf{41} \\\\ \\hline && \\mathbf{|zw|^2} \\\\ &=&|z|^2|w|^2 \\\\ &=& 5^2*2^2 \\\\ &=& 25*4 \\\\ &=& \\mathbf{100} \\\\ \\hline && \\mathbf{|z-w|^2} \\\\ &=& |z|^2+|w|^2-2*\\Re{(z\\overline{w})} \\\\ &=& 5^2+2^2-2*6 \\quad | \\quad \\Re{(z\\overline{w})} = \\Re{(6+8i)} = 6 \\\\ &=& \\mathbf{17} \\\\ \\hline && \\mathbf{\\left|\\dfrac{z}{w}\\right|^2} \\\\ &=& \\dfrac{|z|^2}{|w|^2} \\\\ &=& \\dfrac{5^2}{2^2} \\\\ &=& \\dfrac{25}{4} \\\\ &=& \\mathbf{6.25} \\\\ \\hline \\end{array}$$ May 10, 2019", "# 2021 AIME II Problems/Problem 4 ## Problem There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \\sqrt{n} \\cdot i,$ where $m$ and $n$ are positive integers and $i = \\sqrt{-1}.$ Find $m+n.$ ## Solution 1 (Complex Conjugate Root Theorem) By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ We know that \\begin{align*} z+\\overline{z}&=2m, \\hspace{10mm} & (1) \\\\ z\\overline{z}&=m^2+n. & (2) \\end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\\overline{z}=0,$ from which \\begin{align*} z+\\overline{z}&=20 \\hspace{12.5mm} & (3) \\\\ \\underbrace{2m}_{\\text{by }(1)}&=20 & \\\\ m&=10. & (4) \\end{align*} Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ from which \\begin{align*} -21\\left(z+\\overline{z}\\right)+z\\overline{z}&=0 \\\\ -21\\underbrace{\\left(20\\right)}_{\\text{by }(3)}+\\underbrace{\\left(m^2+n\\right)}_{\\text{by }(2)}&=0 \\\\ m^2+n&=420 \\\\ {\\underbrace{10}_{\\text{by }(4)}}^2+n&=420 \\\\ n&=320. \\hspace{5.75mm} (5) \\end{align*} Finally, we get $m+n=\\boxed{330}$ by $(4)$ and $(5).$ ~MRENTHUSIASM ## Solution 2 (Somewhat Bashy) $(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c + d =", "solutions. ###### 37 $x^2 + 4x + 10 = 0$ ###### 38 $x^2 - 2x + 7 = 0$ ###### 39 $3x^2 - 6x + 5 = 0$ ###### 40 $2x^2 + 5x + 4 = 0$ For Problems 41–42, evaluate the polynomial for the given values of the variable. ###### 41 $z^2 - 6z + 5$ 1. $z = 3 + 2i$ 2. $z=3-2i$ ###### 42 $w^2 + 4w + 7$ 1. $w= -1 - 3i$ 2. $w = -1 + 3i$ For Problems 43–44, find the quotient. ###### 43 $\\dfrac{2-5i}{3-i}$ ###### 44 $\\dfrac{1+i}{1-i}$ For Problems 45–46, find a fourth-degree polynomial with the given zeros. ###### 45 $3i, ~ 1-2i$ ###### 46 $2-\\sqrt{3} ,~ 2+3i$ For Problems 45–46, plot each complex number as a point on the complex plane. ###### 47 1. $z = 4 + i, ~\\overline{z}, ~{-z}, ~{-\\overline{z}}$ 2. $iz, ~i\\overline{z}, ~{-iz}, ~{-i\\overline{z}}$ ###### 48 1. $w = -2 + 3i, ~\\overline{w}, ~{-w}, ~{-\\overline{w}}$ 2. $iw, ~i\\overline{w}, ~{-iw}, ~{-i\\overline{w}}$ ###### 49 The radius, $r\\text{,}$ of a cylindrical can should be one-half its height, $h\\text{.}$ 1. Express the volume, $V\\text{,}$ of the can as a function of its height. 2. What is the volume of the can if its height is $2$ centimeters? $4$ centimeters? 3. Graph the volume as a function of the height and", "# 2021 AIME II Problems/Problem 4 ## Problem There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \\sqrt{n} \\cdot i,$ where $m$ and $n$ are positive integers and $i = \\sqrt{-1}.$ Find $m+n.$ ## Solution 1 By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ By Vieta's Formulas on $x^3+ax+b,$ we have $-20+z+\\overline{z}=0,$ from which \\begin{align*} z+\\overline{z}&=20 \\ \\ \\ \\ \\ \\ \\ \\ \\ (1) \\\\ \\left(m+\\sqrt{n}\\cdot i\\right)+\\left(m+\\sqrt{n}\\cdot i\\right)&=20 \\\\ 2m&=20 \\\\ m&=10. \\ \\ \\ \\ \\ \\ \\ \\ (2) \\end{align*} By Vieta's Formulas on $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ from which \\begin{align*} -21\\left(z+\\overline{z}\\right)+z\\overline{z}&=0 \\\\ -21\\underbrace{\\left(20\\right)}_{\\text{by (1)}}+\\underbrace{|z|^2}_{\\substack{z\\overline{z}=|z|^2 \\\\ \\text{ Identity}}}&=0 \\\\ |z|^2&=420 \\\\ m^2+n&=420 \\\\ {\\underbrace{10}_{\\text{by (2)}}}^2+n&=420 \\\\ n&=320. \\hspace{24.5mm} (3) \\end{align*} Finally, we get $m+n=\\boxed{330}$ by $(2)$ and $(3).$ ~MRENTHUSIASM ## Solution 3 (Somewhat Bashy) $(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c", "# Complex equation solution. How can i resolve it? I have this complex equation $|z+2i|=| z-2 |$. How can i resolve it? Please help me - We have $|z+2i|^2=| z-2 |^2$, which implies that $(z+2i)\\overline{(z+2i)}=(z-2)\\overline{(z-2)}$, that is $(z+2i)(\\overline{z}-2i)=(z-2)(\\overline{z}-2)$. This implies that $$z\\overline{z}-2iz+2i\\overline{z}+4=z\\overline{z}-2z-2\\overline{z}+4,$$ that is $-iz+i\\overline{z}=-z-\\overline{z}$, or $$i(z-\\overline{z})=z+\\overline{z}.$$ Now if we write $z=a+bi$, we get $i(2bi)=2a$, or $b=-a$. - ## The geometric way The points $z$ that satisfy the equation are at the same distance of the points $2$ and $-2\\,i$, that is, they are on the perpendicular bisector of the segment joining $2$ and $-2\\,i$. This is a line, whose equation you should be able to find. ## The algebraic way When dealing vith equations with $|w|$, it is usually convenient to consider $|w|^2=w\\,\\bar w$. In your equation, if $z=x+y\\,i$, this leads to $x^2+(y+2)^2=(x-2)^2+y^2$, whose solution I'll leave to you. - I really like solving these kind of equations geometrically. Keeping in mind the geometrical interpretation of complex numbers can often help when solving exercises. – Beni Bogosel Nov 23 '11 at 10:36 Let $z=a+bi$ be a complex number that satisfies $$|z+2i|=|z-2|.$$ Remember that for any complex number $c+di$, the definition of $|c+di|$ is $$|c+di|=\\sqrt{c^2+d^2}$$ So, we have that $$|z+2i|=|(a+bi)+2i|=|a+(b+2)i|=\\sqrt{a^2+(b+2)^2}$$ and $$|z-2|=|(a+bi)-2|=|(a-2)+bi|=\\sqrt{(a-2)^2+b^2}.$$ Are you able to solve for the values of $a$ and $b$ such that the two", "# Complex conjugate This is an AoPSWiki Word of the Week for August 29-September 4 The complex conjugate of a complex number $z = a + bi$ is the complex number $\\overline{z} = a - bi$. Geometrically, if $z$ is a point in the complex plane, $\\overline z$ is the reflection of $z$ across the real axis. ## Properties Conjugation is its own functional inverse and commutes with the usual operations on complex numbers: • $\\overline{(\\overline z)} = z$. • $\\overline{(w \\cdot z)} = \\overline{w} \\cdot \\overline{z}$. ($\\overline{(\\frac{w}{z})}$ is the same as $\\overline{(w \\cdot \\frac{1}{z})}$) • $\\overline{(w + z)} = \\overline{w} + \\overline{z}$. ($\\overline{(w - z)}$ is the same as $\\overline{(w + (-z))}$) It also interacts in simple ways with other operations on $\\mathbb{C}$: • $|\\overline{z}| = |z|$. • $\\overline{z}\\cdot z = |z|^2$. • If $z = r\\cdot e^{it}$ for $r, t \\in \\mathbb{R}$, $\\overline z = r\\cdot e^{-it}$. That is, $\\overline z$ is the complex number of same absolute value but opposite argument of $z$. • $z + \\overline z = 2 \\mathrm{Re}(z)$ where $\\mathrm{Re}(z)$ is the real part of $z$. • $z - \\overline{z} = 2i \\mathrm{Im}(z)$ where $\\mathrm{Im}(z)$ is the imaginary part of $z$. • If a complex number $z$ is a root of a polynomial with real coefficients, then so is $\\overline z$. This", "w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) For 2: $\\bg_white w^n=1$ by definition. dividing by w on both sides: $\\bg_white w^{n-1}=\\frac{1}{w}$ Also note that $\\bg_white |w|=1$ so $\\bg_white w^{n-1}=\\frac{\\overline{w}}{w\\overline{w}}=\\overline{w}$ since $\\bg_white w\\overline{w}=|w|^2$ #### idkkdi ##### Well-Known Member For 2: $\\bg_white w^n=1$ by definition. dividing by w on both sides: $\\bg_white w^{n-1}=\\frac{1}{w}$ Also note that $\\bg_white |w|=1$ so $\\bg_white w^{n-1}=\\frac{\\overline{w}}{w\\overline{w}}=\\overline{w}$ since $\\bg_white w\\overline{w}=|w|^2$ i was tempted to write by definition as well, but that doesn't match up with the decently long algebra required for the first q lol. #### poptarts12345 ##### Member clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. How did you come to this conclusion? #### idkkdi ##### Well-Known Member How did you come to this conclusion? w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. #### Qeru ##### Member How did you come to this conclusion? If you want to know how he came up with that factorisation: Notice that $\\bg_white 1+w+w^2+...+w^6$ is a GP so using the GP sum: $\\bg_white 1+w+w^2+...+w^6=\\frac{w^7-1}{w-1}$ now we know $\\bg_white w^7=1$ and $\\bg_white w \\neq 1$ so $\\bg_white 1+w+w^2+...+w^6=0$ subtract 1 from both sides you get $\\bg_white w+w^2+...+w^6=-1$ #### CM_Tutor ##### Well-Known Member 2. Let w be the", "mod. – Archis Welankar Nov 11 '15 at 11:50 • Related recent post: How to prove $|z|\\cdot |w|=|z \\cdot w|$ form complex numbers? – Martin Sleziak Nov 6 '17 at 8:56 First note that, for $z=a+ib$ we have $z\\overline{z}=\\sqrt{a^2+b^2}=|z|^2$ and also $\\overline{z_1z_2}=\\overline{z_1}\\,\\overline{z_2}.$ Therefore, $|zw|^2=(zw)\\overline{zw}=zw\\overline{z}\\,\\overline{w}=z\\overline{z}w\\overline{w}=|z|^2|w|^2.$ This implies $|zw|=|z||w|.$ Let $z=a+bi$ and $w=c+di$ where $(a,b,c,d)\\in\\mathbb{R}^{4}$. The module $|z|$ is defined by $|z|=\\sqrt{a^{2}+b^{2}}$ and $|w|=\\sqrt{c^{2}+d^{2}}$, thus \\begin{align} |z||w| &=\\sqrt{a^{2}+b^{2}}\\sqrt{c^{2}+d^{2}}\\\\ &=\\sqrt{a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}} \\end{align} Now, consider $zw=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. We have \\begin{align} |zw|&=\\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}\\\\ &=\\sqrt{a^{2}c^{2}+b^{2}d^{2}-2abcd+a^{2}d^{2}+b^{2}c^{2}+2abcd}\\\\ &=\\sqrt{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}\\\\ &=|z||w| \\end{align}", "# Math Help - Please help me!! >.< 1. ## Please help me!! >.< Please teach me how to solve this question... Two complex number w and z are such that w*= z -2i and |w|^2=z+6. By eliminating z, find w in the form a+ib, where a and b are real and positive. please help me! I really have many problems with complex number!! 2. ## Re: Please help me!! >.< I assume you are using the * to represent conjugate. You should know that \\displaystyle \\begin{align*} w\\,\\overline{w} = |w|^2 \\end{align*}, so we have \\displaystyle \\begin{align*} w\\,\\overline{w} = z + 6 \\end{align*} and \\displaystyle \\begin{align*} \\overline{w} = z - 2i \\end{align*}. Subtracting the second equation from the first gives \\displaystyle \\begin{align*} w\\,\\overline{w} - \\overline{w} &= 6 + 2i \\\\ \\overline{w}\\left( w - 1 \\right) &= 6 + 2i \\\\ \\left( a - b\\,i \\right) \\left( a - 1 + b\\,i \\right) &= 6 + 2i \\\\ a(a - 1) + ab\\,i - (a - 1)b\\,i - b^2 i^2 &= 6 + 2i \\\\ a^2 - a + b^2 + b\\,i &= 6 + 2i \\\\ a^2 - a + b^2 = 6 \\textrm{ and } b &= 2 \\\\ a^2 - a + 4 &= 6 \\\\ a^2 - a - 2 &= 0 \\\\ (a - 2)(a", "and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 4 (using Solution 3 diagram) $\\triangle UPV \\sim \\triangle CPM$, and $\\frac{\\overline{UV}}{\\overline{MC}}=\\frac{\\overline{UP}}{\\overline{PC}}=\\frac{1}{2}$. Since $\\overline{UP} + \\overline{PC} = 12$$\\text{, }$ $\\text{ }$ $\\overline{UP}$ and $\\overline{PC}$ are $4$ and $8$, respectively. By symmetry, $\\overline{PC}=\\overline{PM}=8$. We can now do Pythagorean Theorem on $\\triangle UPM$. ## Solution 5 (FASTEST) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively.", "= 1000 \\\\ 3a + 5b = 25$$ Now by extended Euclidean algorithm $$\\underline{5}=\\underline{3}\\cdot 1+\\underline{2} \\\\ \\underline{3}=\\underline{2}\\cdot 1+\\underline{1}$$ Thus $$\\underline{1}=\\underline{3}-\\underline{2}\\cdot 1 = \\underline{3}-\\left(\\underline{5}-\\underline{3}\\cdot 1\\right) = \\underline{3}-\\underline{5}+\\underline{3}=2\\cdot\\underline 3 - \\underline 5$$ And $$\\underline{3}\\cdot 50 +\\underline{5}\\cdot \\left(-25\\right) = 25$$ Apparently adding $5t$ to the coefficient to $3$ and $-3t$ to the coefficient of $5$ preserves the right hand side $$\\underline{3}\\cdot \\left(50+5t\\right)+\\underline{5}\\cdot \\left(-25+3t\\right) = 25$$ Because $a,b>0$ we find that the least $t$ would be $-9$ $$\\underline{3}\\cdot\\left(50-45\\right) + \\underline 5 \\cdot \\left(-25+27\\right) = 25$$ A bigger $t$ does not yield $a,b>0$ and a smaller $t$ yields a less $a+b$ (because it adds 3 and removes 5 from $a+b$) Thus solution is $a=5, b=2$ - Thanks - but way too complicated for me to understand. – alik Apr 26 '12 at 10:57 To go off what peja mentioned, I think this is a relatively simple and intuitive way to see what's happened. We start with $120x+200y=1000$. Now divide the equation by 40 - this leaves us with $$3x + 5y = 25.$$ Some rearranging turns this into $$3x = 5(5-y).$$ Now, both $x$ and $y$ are meant to be nonnegative integers, which also means both sides of this equation are. This gives two possibilities: either both sides are zero, in which case $x=0$ and $y=5$ (i.e., you buy 5 apples weighing 200g).", "On the other hand, $$\\frac{d}{ds} E(X^2|X \\geq s) = \\frac{d}{ds} W(\\overline{F}(s)) = -W'(\\overline{F}(s)) f(s)$$ where $W'$ is the derivative of $W$. Therefore, $$\\frac{s^2 - W(\\overline{F}(s))}{\\overline{F}(s)} = W'(\\overline{F}(s))$$ or, using $t$, $$\\frac{(\\overline{F}^{-1}(t))^2 - W(t)}{t} = W'(t)$$ and $$\\overline{F}^{-1}(t) = \\sqrt{tW'(t) + W(t)}$$ ($\\overline{F}^{-1}(t)$ should be positive is in some neighborhood of $t = 0$). Then, since $\\overline{F}^{-1}(t) = F^{-1}(1-t)$ and $(F^{-1})'(1-t) = 1/f(F^{-1}(1-t))$, $$f(F^{-1}(1-t)) = \\frac{2 \\sqrt{tW'(t) + W(t)}}{2W'(t)+tW''(t)}$$ or $$f(s) = \\frac{2 \\sqrt{tW'(t) + W(t)}}{2W'(t)+tW''(t)} \\Big|_{t = \\overline{F}(s)}$$ I hope that it's correct and that it can tell something about the decay of the distribution when $t$ tends to $+0$. Maybe it can also be extended to the case $E(X) \\neq 0$. 4 added 310 characters in body Expression for the inverse complementary CDF for the case $E(X) = 0$: Let $\\overline{F}(s) = 1 - F(s)$. Then, $t = \\overline{F}(s)$ and $$\\frac{d}{ds} E(X^2|X \\geq s) = \\frac{d}{ds} \\Big( \\frac{\\int_s^{+\\infty}x^2f(x)dx}{\\int_s^{+\\infty}f(x)dx} \\Big) =$$ $$= \\frac{-s^2 f(s) \\int_s^{+\\infty}f(x)dx+f(s) \\int_s^{+\\infty}x^2f(x)dx}{\\big(\\int_s^{+\\infty}f(x)dx\\big)^2} =$$ $$= \\frac{-s^2 f(s) + f(s) W(\\overline{F}(s))}{\\overline{F}(s)}$$ On the other hand, $$\\frac{d}{ds} E(X^2|X \\geq s) = \\frac{d}{ds} W(\\overline{F}(s)) = -W'(\\overline{F}(s)) f(s)$$ where $W'$ is the derivative of $W$. Therefore, $$\\frac{s^2 - W(\\overline{F}(s))}{\\overline{F}(s)} = W'(\\overline{F}(s))$$ or, using $t$, $$\\frac{(\\overline{F}^{-1}(t))^2 - W(t)}{t} = W'(t)$$ and $$\\overline{F}^{-1}(t) = \\sqrt{tW'(t) + W(t)}$$ ($\\overline{F}^{-1}(t)$ should be positive is some neighborhood of $t = 0$). Then, since $\\overline{F}^{-1}(t)", "# Equation in complex field $z^2+z\\overline{z}-2z =0$ How can this equation be solved in the complex field as $z=x+iy$ $$z^2+z\\overline{z}-2z =0$$ I get to one of the solution but not the other one , can someone explain thoroughly • Hint: just factor out $z$ and write it as $z(z+ \\bar z - 2)=0\\,$. – dxiv Feb 14 '17 at 21:40 One obvious solution is $z=0$,. The other ( for $z=x+iy$) is the solution of: $$z+\\bar z-2=0 \\iff 2x-2=0 \\iff x=1$$ ## Hint: $$z^2+z\\overline{z}-2z =0\\quad (1)\\\\ \\overline{z}^2+z\\overline{z}-2\\overline{z} =0\\quad (2)$$ Now make $(1)-(2)$ and get $$(z^2-\\overline{z}^2)-2(z-\\overline{z})=0\\to(z-\\overline{z})(z+\\overline{z}-2)=0$$ EDIT The final equation comes from an implication (not an equivalence). It means that once one find the values it is necessary to test in the original equation. • Beautiful answer! – Simply Beautiful Art Feb 14 '17 at 20:48 • thanks @SimplyBeautifulArt – Arnaldo Feb 14 '17 at 20:48 • @Dr.MV It looks fine now ;-) – Simply Beautiful Art Feb 14 '17 at 21:02 • @Dr. MV, that is ok. We don't need to worry about that. – Arnaldo Feb 14 '17 at 22:02 • One has to be careful here. This approach gives the entire real line as solutions to the final equation. but $0$ and $1$ are the only real solutions to the original equation. Since difference of (1) and", "\\triangle OTC.$ Note that from $A = rs$, the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\\overline{BY}.$ We have $$\\frac{\\overline{BY}}{QY} = \\frac{7}{{21}} \\rightarrow \\overline{BY} = \\frac{4}{3} r$$ Now, note that as in solution 1, drawing the perpendicular from Q to $\\overline{PX}$(call it Z) yields $\\overline{PZ} = 16 - r, \\overline{ZX} = r.$ Then, from this, $$\\overline{QZ} = \\overline{YX} = \\sqrt{(\\overline{PQ})^2 - (\\overline{PZ})^2} = \\sqrt{(16+r)^2-(16-r)^2} = 8\\sqrt{r}$$ Using similar similarity as was done to find $\\overline{BY}$ we have $\\frac{\\overline{PX}}{\\overline{XC}} = \\frac{\\overline{OT}}{\\overline{TC}} \\rightarrow \\frac{16}{\\overline{XC}} = \\frac{21}{28} \\rightarrow \\overline{XC} = \\frac{64}{3}$. Now adding all these up and equating them to $\\overline{BC}$ yields $$\\frac{4}{3}r + 8\\sqrt{r}+ \\frac{16}{3} = 56 \\rightarrow r = 44 - 6\\sqrt{35} \\rightarrow 44 + 6\\cdot 35 = \\boxed{254}$$ 2008 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "s + z &= v + w \\\\ \\end{align} To get a third equation, note that the total of the interior angles of any quadrilateral add up to $2\\pi$: $$s + t + u + v + w + x + y + z = 2\\pi$$ Solving this system of three equations gives us: \\begin{align} w &= \\pi - t - u - v \\\\ y &= t + u - x \\\\ z &= \\pi - s - t - u \\end{align} Now, we can use the law of sines to write a second set of relations: \\begin{align} \\frac{\\overline{HS}}{\\sin s} &= \\frac{\\overline{CH}}{\\sin x} & \\frac{\\overline{CH}}{\\sin u} &= \\frac{\\overline{BC}}{\\sin z} & \\frac{\\overline{BC}}{\\sin w} &= \\frac{\\overline{SB}}{\\sin t} & \\frac{\\overline{SB}}{\\sin y} &= \\frac{\\overline{HS}}{\\sin v} \\end{align} Multiplying them together, the lengths cancel: $$\\sin s \\sin u \\sin w \\sin y = \\sin t \\sin v \\sin x \\sin z$$ Adding in our substitutions from before (and remembering $\\sin(\\pi-x)=\\sin x$): $$\\sin s \\sin u \\sin (t+u+v) \\sin (t+u-x) = \\sin t \\sin v \\sin x \\sin (s+t+u)$$ (This approach is adapted from the webpage Angular Angst that Fimpellizieri posted in this comment on the question.) Now proving that the answer is $x=34^{\\circ}$ (oops, did I say that out loud?) reduces to proving the following trigonometric identity: $$\\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 110^{\\circ}", "u = (1,−1, 0), v = (3, 0,−4) and w = (0, 2, 5). A plane which contains three points will have the same coefficients as the normal vector the plane, which is also normal to any two vectors in the plane. Two vectors in the plane can be found by \\displaystyle \\begin{align*} \\overline{u\\,v} &= \\overline{u\\,O} + \\overline{O\\,v} \\\\ &= -\\overline{O\\,u} + \\overline{O\\,v} \\\\ &= \\overline{O\\,v} - \\overline{O\\,u} \\\\ &= <3, 0, 4> - <1, -1, 0> \\\\ &= <2, 1, 4>\\end{align*} and \\displaystyle \\begin{align*} \\overline{u\\,w} &= \\overline{u\\,O} + \\overline{O\\,w} \\\\ &= -\\overline{O\\,u} + \\overline{O\\,w} \\\\ &= \\overline{O\\,w} - \\overline{O\\,u} \\\\ &= <0, 2, 5> - <1, -1, 0> \\\\ &= <-1, 3, 5> \\end{align*} So now to find the normal, you need to evaluate \\displaystyle \\begin{align*} \\overline{u\\,v} \\times \\overline{u\\,w} \\end{align*}. \\displaystyle \\begin{align*} \\overline{u\\,v} \\times \\overline{u\\,w} &= \\left|\\begin{matrix}\\phantom{-}i&j&k\\\\ \\phantom{-}2&1&4 \\\\ -1&3&5\\end{matrix}\\right| \\end{align*} • Apr 28th 2012, 10:51 AM TheIntegrator Re: Vector Parametric equation Find our direction vectors u,w = (0-1, 2 - -1, 5-0) = (-1, 3, 5) u,v = (3-1, 0 - -1, -4 -0) = (2, 1, -4) Now that we found the direction vectors, we need only apply one of the three points we already have to find our equation of the plane (x, y, z) = (1, -1, 0) + r(2, 1, -4) +", "it looks like it gets to the heart of the matter. – Jeff Aug 26 '12 at 20:45 Well, if you accept $w(\\overline z)$ and $\\overline{w(z)}$, you can simply define $\\overline w(z)=\\overline{w(\\overline z)}$. This allows also to resolve your example with $\\sin z$: \\begin{aligned} \\overline\\sin\\, z &= \\overline{\\sin(\\overline z)}\\\\ &= \\overline{\\sin(\\overline{x+\\mathrm iy})}\\\\ &= \\overline{\\sin(x-iy)}\\\\ &= \\overline{\\sin x\\cosh y-\\mathrm i\\cos x\\sinh y}\\\\ &= \\sin x\\cosh y+\\mathrm i\\cos x\\sinh y\\\\ &= \\sin(x+\\mathrm iy)\\\\ &= \\sin z \\end{aligned} It also directly follows that $\\overline w(\\overline z)=\\overline{w(\\overline {\\overline z})}=\\overline{w(z)}$. Note that it may be easier to understand if we introduce the complex conjugation in function notation, $\\mathrm{conj}(z) = \\overline z$. Then $w(\\bar z)=w(\\mathrm{conj}(z)) = (w\\circ\\mathrm{conj})(z)$. Similarly, $\\overline{w(z)} = (\\mathrm{conj}\\circ w)(z)$, and $\\overline w(z)=(\\mathrm{conj}\\circ w\\circ\\mathrm{conj})(z)$. Note that with the definition above, we have the following properties for $w(z)$: • If $w(z)=f(z)+g(z)$, then $\\overline w(z)=\\overline f(z)+\\overline g(z)$. Proof: $\\overline w(z)=\\overline{w(\\overline z)} = \\overline{f(\\overline z)+g(\\overline z)}=\\overline{f(\\overline z)}+\\overline{g(\\overline z)} = \\overline f(z)+\\overline g(z)$. • If $w(z)=f(z)\\cdot g(z)$, then $\\overline w(z)=\\overline f(z)\\cdot\\overline g(z)$. Proof: $\\overline w(z)=\\overline{w(\\overline z)} = \\overline{f(\\overline z)\\cdot g(\\overline z)}=\\overline{f(\\overline z)}\\cdot\\overline{g(\\overline z)} = \\overline f(z)+\\overline g(z)$ • If $w(z)=f(g(z))$, then $\\overline w(z)=\\overline f(\\overline g(z))$. Proof: $\\overline w(z) = \\overline{w(\\overline z)} = \\overline{f(g(\\overline z))} = \\overline{f(\\overline{\\overline{g(\\overline z)}})}=\\overline{f(\\overline{\\overline g(z)})} = \\overline f(\\overline g(z))$ • If $w(z)=z$ then $\\overline w(z)=z$. Proof: $\\overline w(z) = \\overline{w(\\overline z)} = \\overline{\\overline", "0, 0, \\overset{\\overset{k}\\downarrow}{0} ] \\\\ b &= [ 2, 4, \\underset{\\underset{n}\\uparrow}{7} ] \\\\ \\end{align} We have $a[m] = 5$, which is smaller than $b[n] = 7$. Move $b[n]$ into the position $a[k]$, decrease the value of both $n$ and $k$ by 1: \\begin{align} a &= [ 1, 3, \\overset{\\overset{m}\\downarrow}{5}, 0, \\overset{\\overset{k}\\downarrow}{0}, 7 ] \\\\ b &= [ 2, \\underset{\\underset{n}\\uparrow}{4}, \\emptyset ] \\\\ \\end{align} Now, $a[m] = 5$ is larger than $b[n] = 4$. Move $a[m]$ into $a[k]$ and decrease both $m$ and $k$ by 1: \\begin{align} a &= [ 1, \\overset{\\overset{m}\\downarrow}{3}, \\emptyset, \\overset{\\overset{k}\\downarrow}{0}, 5, 7 ] \\\\ b &= [ 2, \\underset{\\underset{n}\\uparrow}{4}, \\emptyset ] \\\\ \\end{align} Next, $a[m] = 3$, less than $b[n] = 4$. Move $b[n]$ into $a[k]$ and decrease both $n$ and $k$: \\begin{align} a &= [ 1, \\overset{\\overset{m}\\downarrow}{3}, \\overset{\\overset{k}\\downarrow}{\\emptyset}, 4, 5, 7 ] \\\\ b &= [ \\underset{\\underset{n}\\uparrow}{2}, \\emptyset, \\emptyset ] \\\\ \\end{align} Next, $a[m] = 3$, larger than $b[n] = 2$. Move $a[m]$ into $a[k]$ and decrease $m$ and $k$: \\begin{align} a &= [ \\overset{\\overset{m}\\downarrow}{1}, \\overset{\\overset{k}\\downarrow}{\\emptyset}, 3, 4, 5, 7 ] \\\\ b &= [ \\underset{\\underset{n}\\uparrow}{2}, \\emptyset, \\emptyset ] \\\\ \\end{align} Finally, move $b[n]$ into $a[k]$ since $a[m] = 1$ is less than $b[n] = 2$, decreasing $n$ and $k$: \\begin{align} a &= [ \\overset{\\overset{m,k}\\downarrow}{1}, 2, 3, 4, 5, 7 ] \\\\ b &=" ]

[ "and $w = 2 + 7i\\text{.}$ Then $z+w = 5 +3i\\text{,}$ and \\begin{align*} z\\cdot w \\amp = (3-4i)(2+7i)\\\\ \\amp = 6 + 28 - 8i + 21i\\\\ \\amp = 34 + 13i. \\end{align*} A few other computations: \\begin{align*} 4z \\amp = 12 - 16i\\\\ |z| \\amp = \\sqrt{3^2 + (-4)^2} = 5\\\\ \\overline{zw}\\amp =34-13i. \\end{align*} ### SubsectionExercises ###### 1 In each case, determine $z + w\\text{,}$ $sz\\text{,}$ $|z|\\text{,}$ and $z\\cdot w\\text{.}$ a. $z = 5 + 2i\\text{,}$ $s = -4\\text{,}$ $w = -1 + 2i\\text{.}$ b. $z = 3i\\text{,}$ $s = 1/2\\text{,}$ $w = -3 + 2i\\text{.}$ c. $z = 1 + i\\text{,}$ $s = 0.6\\text{,}$ $w = 1 - i\\text{.}$ ###### 2 Show that $z\\cdot \\overline{z} = |z|^2\\text{,}$ where $\\overline{z}$ is the conjugate of $z\\text{.}$ ###### 3 Suppose $z=x+yi$ and $w=s+ti$ are two complex numbers. Prove the following properties of the conjugate and the modulus. a. $|w\\cdot z| = |w|\\cdot |z|\\text{.}$ b. $\\overline{zw} = \\overline{z}\\cdot \\overline{w}\\text{.}$ c. $\\overline{z + w} = \\overline{z} + \\overline{w}\\text{.}$ d. $z + \\overline{z} = 2\\text{Re}(z)\\text{.}$ (Hence, $z + \\overline{z}$ is a real number.) e. $z - \\overline{z} = 2\\text{Im}(z)i\\text{.}$ f. $|z| = |\\overline{z}|\\text{.}$ ###### 4 A Pythagorean triple consists of three integers $(a,b,c)$ such that $a^2 + b^2 = c^2\\text{.}$ We can use complex numbers to generate Pythagorean triples. Suppose $z =", "+0 # complex numbers 0 133 1 Let z and w be complex numbers such that |z| = 1, |w| = 7, and |5z + w| = 11. Find |3z + 4w|. Jul 29, 2022 #1 +285 0 We can use the fact that for any complex number a, (a)(cong(a)) = |a|^2. $$|5z + w|^2 = (5z + w)(5\\overline{z} + \\overline{w}) = 121$$ $$25|z|^2 + |w|^2 + 5(\\overline{z}w + \\overline{w}z) = 121$$ $$5(\\overline{z}w + \\overline{w}z) = 47$$ $$|3z + 4w|^2 = (3z + 4w)(3\\overline{z} + 4\\overline{w})$$ $$=793 + 12(\\overline{w}z + \\overline{z}w)$$ $$=793 + 564/5$$ Take the square root of that value and you will the answer. Jul 31, 2022", "+0 0 31 1 Let z and w be complex numbers satisfying |z| = 5, |w| = 2, and z(overline{w}) = 6+8i. Find the value of |z+w|^2, |zw|^2, |z-w|^2, |frac{z}{w}|^2 . If any of these cannot be uniquely determined from the information given, write not possible. May 10, 2019 #1 +22180 +3 Let $$z$$ and $$w$$ be complex numbers satisfying $$|z| = 5$$, $$|w| = 2$$, and $$z\\overline{w} = 6+8i$$ Find the value of $$|z+w|^2$$, $$|zw|^2$$, $$|z-w|^2$$, $$\\left|\\dfrac{z}{w}\\right|^2$$ . $$\\begin{array}{|rcll|} \\hline && \\mathbf{|z+w|^2} \\\\ &=& |z|^2+|w|^2+2*\\Re{(z\\overline{w})} \\\\ &=& 5^2+2^2+2*6 \\quad | \\quad \\Re{(z\\overline{w})} = \\Re{(6+8i)} = 6 \\\\ &=& \\mathbf{41} \\\\ \\hline && \\mathbf{|zw|^2} \\\\ &=&|z|^2|w|^2 \\\\ &=& 5^2*2^2 \\\\ &=& 25*4 \\\\ &=& \\mathbf{100} \\\\ \\hline && \\mathbf{|z-w|^2} \\\\ &=& |z|^2+|w|^2-2*\\Re{(z\\overline{w})} \\\\ &=& 5^2+2^2-2*6 \\quad | \\quad \\Re{(z\\overline{w})} = \\Re{(6+8i)} = 6 \\\\ &=& \\mathbf{17} \\\\ \\hline && \\mathbf{\\left|\\dfrac{z}{w}\\right|^2} \\\\ &=& \\dfrac{|z|^2}{|w|^2} \\\\ &=& \\dfrac{5^2}{2^2} \\\\ &=& \\dfrac{25}{4} \\\\ &=& \\mathbf{6.25} \\\\ \\hline \\end{array}$$ May 10, 2019", "# 2021 AIME II Problems/Problem 4 ## Problem There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \\sqrt{n} \\cdot i,$ where $m$ and $n$ are positive integers and $i = \\sqrt{-1}.$ Find $m+n.$ ## Solution 1 (Complex Conjugate Root Theorem) By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ We know that \\begin{align*} z+\\overline{z}&=2m, \\hspace{10mm} & (1) \\\\ z\\overline{z}&=m^2+n. & (2) \\end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\\overline{z}=0,$ from which \\begin{align*} z+\\overline{z}&=20 \\hspace{12.5mm} & (3) \\\\ \\underbrace{2m}_{\\text{by }(1)}&=20 & \\\\ m&=10. & (4) \\end{align*} Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ from which \\begin{align*} -21\\left(z+\\overline{z}\\right)+z\\overline{z}&=0 \\\\ -21\\underbrace{\\left(20\\right)}_{\\text{by }(3)}+\\underbrace{\\left(m^2+n\\right)}_{\\text{by }(2)}&=0 \\\\ m^2+n&=420 \\\\ {\\underbrace{10}_{\\text{by }(4)}}^2+n&=420 \\\\ n&=320. \\hspace{5.75mm} (5) \\end{align*} Finally, we get $m+n=\\boxed{330}$ by $(4)$ and $(5).$ ~MRENTHUSIASM ## Solution 2 (Somewhat Bashy) $(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c + d =", "solutions. ###### 37 $x^2 + 4x + 10 = 0$ ###### 38 $x^2 - 2x + 7 = 0$ ###### 39 $3x^2 - 6x + 5 = 0$ ###### 40 $2x^2 + 5x + 4 = 0$ For Problems 41–42, evaluate the polynomial for the given values of the variable. ###### 41 $z^2 - 6z + 5$ 1. $z = 3 + 2i$ 2. $z=3-2i$ ###### 42 $w^2 + 4w + 7$ 1. $w= -1 - 3i$ 2. $w = -1 + 3i$ For Problems 43–44, find the quotient. ###### 43 $\\dfrac{2-5i}{3-i}$ ###### 44 $\\dfrac{1+i}{1-i}$ For Problems 45–46, find a fourth-degree polynomial with the given zeros. ###### 45 $3i, ~ 1-2i$ ###### 46 $2-\\sqrt{3} ,~ 2+3i$ For Problems 45–46, plot each complex number as a point on the complex plane. ###### 47 1. $z = 4 + i, ~\\overline{z}, ~{-z}, ~{-\\overline{z}}$ 2. $iz, ~i\\overline{z}, ~{-iz}, ~{-i\\overline{z}}$ ###### 48 1. $w = -2 + 3i, ~\\overline{w}, ~{-w}, ~{-\\overline{w}}$ 2. $iw, ~i\\overline{w}, ~{-iw}, ~{-i\\overline{w}}$ ###### 49 The radius, $r\\text{,}$ of a cylindrical can should be one-half its height, $h\\text{.}$ 1. Express the volume, $V\\text{,}$ of the can as a function of its height. 2. What is the volume of the can if its height is $2$ centimeters? $4$ centimeters? 3. Graph the volume as a function of the height and", "# 2021 AIME II Problems/Problem 4 ## Problem There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \\sqrt{n} \\cdot i,$ where $m$ and $n$ are positive integers and $i = \\sqrt{-1}.$ Find $m+n.$ ## Solution 1 By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ By Vieta's Formulas on $x^3+ax+b,$ we have $-20+z+\\overline{z}=0,$ from which \\begin{align*} z+\\overline{z}&=20 \\ \\ \\ \\ \\ \\ \\ \\ \\ (1) \\\\ \\left(m+\\sqrt{n}\\cdot i\\right)+\\left(m+\\sqrt{n}\\cdot i\\right)&=20 \\\\ 2m&=20 \\\\ m&=10. \\ \\ \\ \\ \\ \\ \\ \\ (2) \\end{align*} By Vieta's Formulas on $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ from which \\begin{align*} -21\\left(z+\\overline{z}\\right)+z\\overline{z}&=0 \\\\ -21\\underbrace{\\left(20\\right)}_{\\text{by (1)}}+\\underbrace{|z|^2}_{\\substack{z\\overline{z}=|z|^2 \\\\ \\text{ Identity}}}&=0 \\\\ |z|^2&=420 \\\\ m^2+n&=420 \\\\ {\\underbrace{10}_{\\text{by (2)}}}^2+n&=420 \\\\ n&=320. \\hspace{24.5mm} (3) \\end{align*} Finally, we get $m+n=\\boxed{330}$ by $(2)$ and $(3).$ ~MRENTHUSIASM ## Solution 3 (Somewhat Bashy) $(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c", "# Complex conjugate This is an AoPSWiki Word of the Week for August 29-September 4 The complex conjugate of a complex number $z = a + bi$ is the complex number $\\overline{z} = a - bi$. Geometrically, if $z$ is a point in the complex plane, $\\overline z$ is the reflection of $z$ across the real axis. ## Properties Conjugation is its own functional inverse and commutes with the usual operations on complex numbers: • $\\overline{(\\overline z)} = z$. • $\\overline{(w \\cdot z)} = \\overline{w} \\cdot \\overline{z}$. ($\\overline{(\\frac{w}{z})}$ is the same as $\\overline{(w \\cdot \\frac{1}{z})}$) • $\\overline{(w + z)} = \\overline{w} + \\overline{z}$. ($\\overline{(w - z)}$ is the same as $\\overline{(w + (-z))}$) It also interacts in simple ways with other operations on $\\mathbb{C}$: • $|\\overline{z}| = |z|$. • $\\overline{z}\\cdot z = |z|^2$. • If $z = r\\cdot e^{it}$ for $r, t \\in \\mathbb{R}$, $\\overline z = r\\cdot e^{-it}$. That is, $\\overline z$ is the complex number of same absolute value but opposite argument of $z$. • $z + \\overline z = 2 \\mathrm{Re}(z)$ where $\\mathrm{Re}(z)$ is the real part of $z$. • $z - \\overline{z} = 2i \\mathrm{Im}(z)$ where $\\mathrm{Im}(z)$ is the imaginary part of $z$. • If a complex number $z$ is a root of a polynomial with real coefficients, then so is $\\overline z$. This", "# Complex equation solution. How can i resolve it? I have this complex equation $|z+2i|=| z-2 |$. How can i resolve it? Please help me - We have $|z+2i|^2=| z-2 |^2$, which implies that $(z+2i)\\overline{(z+2i)}=(z-2)\\overline{(z-2)}$, that is $(z+2i)(\\overline{z}-2i)=(z-2)(\\overline{z}-2)$. This implies that $$z\\overline{z}-2iz+2i\\overline{z}+4=z\\overline{z}-2z-2\\overline{z}+4,$$ that is $-iz+i\\overline{z}=-z-\\overline{z}$, or $$i(z-\\overline{z})=z+\\overline{z}.$$ Now if we write $z=a+bi$, we get $i(2bi)=2a$, or $b=-a$. - ## The geometric way The points $z$ that satisfy the equation are at the same distance of the points $2$ and $-2\\,i$, that is, they are on the perpendicular bisector of the segment joining $2$ and $-2\\,i$. This is a line, whose equation you should be able to find. ## The algebraic way When dealing vith equations with $|w|$, it is usually convenient to consider $|w|^2=w\\,\\bar w$. In your equation, if $z=x+y\\,i$, this leads to $x^2+(y+2)^2=(x-2)^2+y^2$, whose solution I'll leave to you. - I really like solving these kind of equations geometrically. Keeping in mind the geometrical interpretation of complex numbers can often help when solving exercises. – Beni Bogosel Nov 23 '11 at 10:36 Let $z=a+bi$ be a complex number that satisfies $$|z+2i|=|z-2|.$$ Remember that for any complex number $c+di$, the definition of $|c+di|$ is $$|c+di|=\\sqrt{c^2+d^2}$$ So, we have that $$|z+2i|=|(a+bi)+2i|=|a+(b+2)i|=\\sqrt{a^2+(b+2)^2}$$ and $$|z-2|=|(a+bi)-2|=|(a-2)+bi|=\\sqrt{(a-2)^2+b^2}.$$ Are you able to solve for the values of $a$ and $b$ such that the two", "w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) For 2: $\\bg_white w^n=1$ by definition. dividing by w on both sides: $\\bg_white w^{n-1}=\\frac{1}{w}$ Also note that $\\bg_white |w|=1$ so $\\bg_white w^{n-1}=\\frac{\\overline{w}}{w\\overline{w}}=\\overline{w}$ since $\\bg_white w\\overline{w}=|w|^2$ #### idkkdi ##### Well-Known Member For 2: $\\bg_white w^n=1$ by definition. dividing by w on both sides: $\\bg_white w^{n-1}=\\frac{1}{w}$ Also note that $\\bg_white |w|=1$ so $\\bg_white w^{n-1}=\\frac{\\overline{w}}{w\\overline{w}}=\\overline{w}$ since $\\bg_white w\\overline{w}=|w|^2$ i was tempted to write by definition as well, but that doesn't match up with the decently long algebra required for the first q lol. #### poptarts12345 ##### Member clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. How did you come to this conclusion? #### idkkdi ##### Well-Known Member How did you come to this conclusion? w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. #### Qeru ##### Member How did you come to this conclusion? If you want to know how he came up with that factorisation: Notice that $\\bg_white 1+w+w^2+...+w^6$ is a GP so using the GP sum: $\\bg_white 1+w+w^2+...+w^6=\\frac{w^7-1}{w-1}$ now we know $\\bg_white w^7=1$ and $\\bg_white w \\neq 1$ so $\\bg_white 1+w+w^2+...+w^6=0$ subtract 1 from both sides you get $\\bg_white w+w^2+...+w^6=-1$ #### CM_Tutor ##### Well-Known Member 2. Let w be the", "# Math Help - Please help me!! >.< 1. ## Please help me!! >.< Please teach me how to solve this question... Two complex number w and z are such that w*= z -2i and |w|^2=z+6. By eliminating z, find w in the form a+ib, where a and b are real and positive. please help me! I really have many problems with complex number!! 2. ## Re: Please help me!! >.< I assume you are using the * to represent conjugate. You should know that \\displaystyle \\begin{align*} w\\,\\overline{w} = |w|^2 \\end{align*}, so we have \\displaystyle \\begin{align*} w\\,\\overline{w} = z + 6 \\end{align*} and \\displaystyle \\begin{align*} \\overline{w} = z - 2i \\end{align*}. Subtracting the second equation from the first gives \\displaystyle \\begin{align*} w\\,\\overline{w} - \\overline{w} &= 6 + 2i \\\\ \\overline{w}\\left( w - 1 \\right) &= 6 + 2i \\\\ \\left( a - b\\,i \\right) \\left( a - 1 + b\\,i \\right) &= 6 + 2i \\\\ a(a - 1) + ab\\,i - (a - 1)b\\,i - b^2 i^2 &= 6 + 2i \\\\ a^2 - a + b^2 + b\\,i &= 6 + 2i \\\\ a^2 - a + b^2 = 6 \\textrm{ and } b &= 2 \\\\ a^2 - a + 4 &= 6 \\\\ a^2 - a - 2 &= 0 \\\\ (a - 2)(a", "# Equation in complex field $z^2+z\\overline{z}-2z =0$ How can this equation be solved in the complex field as $z=x+iy$ $$z^2+z\\overline{z}-2z =0$$ I get to one of the solution but not the other one , can someone explain thoroughly • Hint: just factor out $z$ and write it as $z(z+ \\bar z - 2)=0\\,$. – dxiv Feb 14 '17 at 21:40 One obvious solution is $z=0$,. The other ( for $z=x+iy$) is the solution of: $$z+\\bar z-2=0 \\iff 2x-2=0 \\iff x=1$$ ## Hint: $$z^2+z\\overline{z}-2z =0\\quad (1)\\\\ \\overline{z}^2+z\\overline{z}-2\\overline{z} =0\\quad (2)$$ Now make $(1)-(2)$ and get $$(z^2-\\overline{z}^2)-2(z-\\overline{z})=0\\to(z-\\overline{z})(z+\\overline{z}-2)=0$$ EDIT The final equation comes from an implication (not an equivalence). It means that once one find the values it is necessary to test in the original equation. • Beautiful answer! – Simply Beautiful Art Feb 14 '17 at 20:48 • thanks @SimplyBeautifulArt – Arnaldo Feb 14 '17 at 20:48 • @Dr.MV It looks fine now ;-) – Simply Beautiful Art Feb 14 '17 at 21:02 • @Dr. MV, that is ok. We don't need to worry about that. – Arnaldo Feb 14 '17 at 22:02 • One has to be careful here. This approach gives the entire real line as solutions to the final equation. but $0$ and $1$ are the only real solutions to the original equation. Since difference of (1) and", "mod. – Archis Welankar Nov 11 '15 at 11:50 • Related recent post: How to prove $|z|\\cdot |w|=|z \\cdot w|$ form complex numbers? – Martin Sleziak Nov 6 '17 at 8:56 First note that, for $z=a+ib$ we have $z\\overline{z}=\\sqrt{a^2+b^2}=|z|^2$ and also $\\overline{z_1z_2}=\\overline{z_1}\\,\\overline{z_2}.$ Therefore, $|zw|^2=(zw)\\overline{zw}=zw\\overline{z}\\,\\overline{w}=z\\overline{z}w\\overline{w}=|z|^2|w|^2.$ This implies $|zw|=|z||w|.$ Let $z=a+bi$ and $w=c+di$ where $(a,b,c,d)\\in\\mathbb{R}^{4}$. The module $|z|$ is defined by $|z|=\\sqrt{a^{2}+b^{2}}$ and $|w|=\\sqrt{c^{2}+d^{2}}$, thus \\begin{align} |z||w| &=\\sqrt{a^{2}+b^{2}}\\sqrt{c^{2}+d^{2}}\\\\ &=\\sqrt{a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}} \\end{align} Now, consider $zw=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. We have \\begin{align} |zw|&=\\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}\\\\ &=\\sqrt{a^{2}c^{2}+b^{2}d^{2}-2abcd+a^{2}d^{2}+b^{2}c^{2}+2abcd}\\\\ &=\\sqrt{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}\\\\ &=|z||w| \\end{align}", "a }$$ × $$\\overline { b }$$, $$\\overline { b }$$ × $$\\overline { c }$$, $$\\overline { c }$$ × $$\\overline { a }$$ as coterminous edges is 8 cubic units, then the volume of the parallelepiped with ($$\\overline { a }$$ × $$\\overline { b }$$) × ($$\\overline { b }$$ × $$\\overline { c }$$), ($$\\overline { b }$$ × $$\\overline { c }$$) × ($$\\overline { c }$$ × $$\\overline { a }$$) and ($$\\overline { c }$$ × $$\\overline { a }$$) × ($$\\overline { a }$$ × $$\\overline { b }$$) as coterminous edges is (a) 8 cubic units (b) 512 cubic units (c) 64 cubic units (d) 24 cubic units Solution: (c) 64 cubic units Hint: Given volume of the parallelepiped with Question 12. Consider the vectors $$\\overline { a }$$, $$\\overline { b }$$, $$\\overline { c }$$, $$\\overline { d }$$ such that ($$\\overline { a }$$ × $$\\overline { b }$$) × ($$\\overline { c }$$ × $$\\overline { d }$$) = $$\\overline { 0 }$$ Let P1 and P2 be the planes determined by the pairs of vectors $$\\overline { a }$$, $$\\overline { b }$$ and $$\\overline { c }$$, $$\\overline { d }$$ respectively. Then the angle between P1 and P2 is (a) 0° (b)", "= 2 + 3i , w = 5 - 4i$, show that • (i) $\\overline{z+w} = \\overline{z} + \\overline{w}$ • (ii) $\\overline{z-w} = \\overline{z} -\\overline{w}$ • (iii) $\\overline{zw} = \\overline{z}\\overline{w}$ • (iv) $\\overline{\\frac{z}{w}} = \\frac{\\overline{z}}{\\overline{w}}$ • (v) $\\frac{1}{2}(z+\\overline{z})$ is real part of z • (vi) $\\frac{1}{2i}(z-\\overline{z})$ is the imaginary part of z Solution 6(i) $$\\begin{array}{cl} z = 2+3i\\\\ w = 5-4i\\\\ * (i) \\overline{z+w} = \\overline{z} + \\overline{w}\\\\ \\overline{z} = 2 -3i\\\\ \\overline{w} = 5+4i\\\\ \\overline{z}+\\overline{w} = 2-3i+5+4i\\\\ &= 7+i\\\\ z + w = 2 + 3i+5-4i\\\\ &= 7-i\\\\ \\overline{z+w} = 7+i\\\\ L.H.S =R.H.S \\end{array}$$ 6(ii) $$\\begin{array}{cl} \\overline{z-w} = \\overline{z} - \\overline{w}\\\\ z-w = (2+3i)-(5-4i)\\\\ &= 2 +3i-5+4i\\\\ &= -3+7i\\\\ \\overline{z-w} = -3-7i\\\\ \\overline{z}-\\overline{w} = (2-3i)-(5+4i)\\\\ &= 2-3i-5-4i\\\\ &= -3+7i\\\\ L.H.S = R.H.S \\end{array}$$ 6(iii) $$\\begin{array}{cl} \\overline{zw} = \\overline{z}\\overline{w}\\\\ zw = (2+3i)(5-4i)\\\\ &= 2(5-4i)+3i(5-4i)\\\\ &= 10-8i+15i-12i^2\\\\ &= 10+7i-12(-1)\\\\ &= 10+12+7i\\\\ &= 22+7i\\\\ \\overline{zw} = 22-7i\\\\ \\overline{z}\\overline{w} &= (2-3i)(5+4i)\\\\ &= 2(5+4i)-3i(5+4i)\\\\ &= 10+8i-15i-12i^2\\\\ &= 10-7i-12(-1)\\\\ &= 10+12-7i\\\\ &= 22-7i\\\\ L.H.S = R. H.S \\end{array}$$ 6(iv) $$\\begin{array}{cl} \\overline{\\frac{z}{w}} = \\frac{\\overline{z}}{\\overline{w}}\\\\ \\frac{z}{w} = \\frac{2+3i}{5-4i}\\\\ &= \\frac{2+3i}{5-4i}\\times\\frac{5+4i}{5+4i}\\\\ &= \\frac{2(5+4i)+3i(5+4i)}{25-16i^2}\\\\ &= \\frac{10+8i+15i+12i^2}{25-16(-1)}\\\\ &= \\frac{10+23i+12(-1)}{25+16}\\\\ &= \\frac{10-12+23i}{41}\\\\ &= \\frac{-2-23i}{41}\\\\ \\frac{\\overline{z}}{\\overline{w}} = \\frac{2-3i}{5+4i}\\\\ &= \\frac{2-3i}{5+4i}\\times\\frac{5-4i}{5-4i}\\\\ &= \\frac{2(5-4i)-3i(5-4i)}{25-16i^2}\\\\ &= \\frac{10-8i-15i+12i^2}{25-16(-1)}\\\\ &= \\frac{10-23i+12(-1)}{25+16}\\\\ &= \\frac{-2-23i}{41}\\\\ L.H.S = R.H.S \\end{array}$$ 6(V) $$\\begin{array}{cl} \\frac{1}{2}(z+\\overline{z}) &= \\frac{1}{2}(2+3i+2-3i)\\\\ &= \\frac{1}{2}(4)\\\\ &= 2 \\hbox{ (it is real part of } z). \\end{array}$$", "cubic equation: $2Y(e_L)^3 + 6Y(q_L)^3 + Y(\\overline{e}_L)^3 + 3Y(\\overline{u}_L)^3 + 3Y(\\overline{d}_L)^3 = 0$ we get first this: $-54 Y(q_L)^3 + 6 Y(q_L)^3 + 216 Y(q_L)^3 + 3Y(\\overline{u}_L)^3 + 3Y(\\overline{d}_L)^3 = 0$ or this: $3Y(\\overline{u}_L)^3 + 3Y(\\overline{d}_L)^3 = -168 Y(q_L)^3$ and then this: $Y(\\overline{u}_L)^3 + Y(\\overline{d}_L)^3 = -56 Y(q_L)^3$ This is getting spooky! I did a whole issue of This Week’s Finds on the number 168. Anyway, expressing everything in terms of $Y(\\overline{u}_L)$ and $Y(\\overline{d}_L)$ we get $Y(\\overline{u}_L)^3 + Y(\\overline{d}_L)^3 = 7(Y(\\overline{u}_L) + Y(\\overline{d}_L))^3$ Or, in less complicated notation: $U^3 + D^3 = 7(U+D)^3$ It’s easy to see that $U = 1, D = -1$ is a solution. So is the real world solution $U = -4, D = 2$, thanks to this charming fact: $(-4)^3 + 2^3 = 7 \\cdot (-2)^3$ And so, of course, is $U = 2, D = -4$, but that’s not really different. I am too exhausted to check that these solutions and their integer multiples are the only ones! Is there any fan of cubic Diophantine equations out there who wants to finish off the job? Minhyong, maybe? My main reaction to this calculation is that only increases my feeling that the hypercharges in the Standard Model are mysterious. Okay, they’re almost completely determined by anomaly cancellation. But still, they’re pretty weird. Posted", "$$\\overline{A C}$$ = P.v of C – P.v of A = i – 2 j – 3 k -(i – j + 4k) = – j – 7k Let A be the angle between $$\\overline{A B}$$ and $$\\overline{A C}$$ 3. Area of the parallelogram ### Plus Two Maths Vector Algebra Six Mark Questions and Answers Question 1. Using this figure answer the following questions. 1. Find $$\\overline{O A}$$, $$\\overline{O B}$$, $$\\overline{O C}$$ (2) 2. Find $$\\overline{O D}$$ (2) 3. Find the coordinate of the vertex D. (2) 1. $$\\overline{O A}$$ = (3 – 1)i + (-1 – 2)j + (7 – 3)k = 2i – 3j + 4k $$\\overline{O B}$$ = (2 – 1)i + (4 – 2)j +(2 – 3)k = i + 2j – k $$\\overline{O C}$$ = (4 – 1)i + (1 – 2 )j + (5 – 3 )k = 3i – j + 2 k. 2. From the figure, 3. Let the vertex of D be (x , y , z), Then, $$\\overline{O D}$$ = (x – 1)i + (y – 2)j + (z – 3)k. But we have, $$\\overline{O D}$$ = 6i – 2j + 5k = (x – 1)i + (y – 2)j +(z – 3)k x – 1 = 6 ⇒ x = 7, y – 2 = -2 ⇒", "it looks like it gets to the heart of the matter. – Jeff Aug 26 '12 at 20:45 Well, if you accept $w(\\overline z)$ and $\\overline{w(z)}$, you can simply define $\\overline w(z)=\\overline{w(\\overline z)}$. This allows also to resolve your example with $\\sin z$: \\begin{aligned} \\overline\\sin\\, z &= \\overline{\\sin(\\overline z)}\\\\ &= \\overline{\\sin(\\overline{x+\\mathrm iy})}\\\\ &= \\overline{\\sin(x-iy)}\\\\ &= \\overline{\\sin x\\cosh y-\\mathrm i\\cos x\\sinh y}\\\\ &= \\sin x\\cosh y+\\mathrm i\\cos x\\sinh y\\\\ &= \\sin(x+\\mathrm iy)\\\\ &= \\sin z \\end{aligned} It also directly follows that $\\overline w(\\overline z)=\\overline{w(\\overline {\\overline z})}=\\overline{w(z)}$. Note that it may be easier to understand if we introduce the complex conjugation in function notation, $\\mathrm{conj}(z) = \\overline z$. Then $w(\\bar z)=w(\\mathrm{conj}(z)) = (w\\circ\\mathrm{conj})(z)$. Similarly, $\\overline{w(z)} = (\\mathrm{conj}\\circ w)(z)$, and $\\overline w(z)=(\\mathrm{conj}\\circ w\\circ\\mathrm{conj})(z)$. Note that with the definition above, we have the following properties for $w(z)$: • If $w(z)=f(z)+g(z)$, then $\\overline w(z)=\\overline f(z)+\\overline g(z)$. Proof: $\\overline w(z)=\\overline{w(\\overline z)} = \\overline{f(\\overline z)+g(\\overline z)}=\\overline{f(\\overline z)}+\\overline{g(\\overline z)} = \\overline f(z)+\\overline g(z)$. • If $w(z)=f(z)\\cdot g(z)$, then $\\overline w(z)=\\overline f(z)\\cdot\\overline g(z)$. Proof: $\\overline w(z)=\\overline{w(\\overline z)} = \\overline{f(\\overline z)\\cdot g(\\overline z)}=\\overline{f(\\overline z)}\\cdot\\overline{g(\\overline z)} = \\overline f(z)+\\overline g(z)$ • If $w(z)=f(g(z))$, then $\\overline w(z)=\\overline f(\\overline g(z))$. Proof: $\\overline w(z) = \\overline{w(\\overline z)} = \\overline{f(g(\\overline z))} = \\overline{f(\\overline{\\overline{g(\\overline z)}})}=\\overline{f(\\overline{\\overline g(z)})} = \\overline f(\\overline g(z))$ • If $w(z)=z$ then $\\overline w(z)=z$. Proof: $\\overline w(z) = \\overline{w(\\overline z)} = \\overline{\\overline", "\\triangle OTC.$ Note that from $A = rs$, the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\\overline{BY}.$ We have $$\\frac{\\overline{BY}}{QY} = \\frac{7}{{21}} \\rightarrow \\overline{BY} = \\frac{4}{3} r$$ Now, note that as in solution 1, drawing the perpendicular from Q to $\\overline{PX}$(call it Z) yields $\\overline{PZ} = 16 - r, \\overline{ZX} = r.$ Then, from this, $$\\overline{QZ} = \\overline{YX} = \\sqrt{(\\overline{PQ})^2 - (\\overline{PZ})^2} = \\sqrt{(16+r)^2-(16-r)^2} = 8\\sqrt{r}$$ Using similar similarity as was done to find $\\overline{BY}$ we have $\\frac{\\overline{PX}}{\\overline{XC}} = \\frac{\\overline{OT}}{\\overline{TC}} \\rightarrow \\frac{16}{\\overline{XC}} = \\frac{21}{28} \\rightarrow \\overline{XC} = \\frac{64}{3}$. Now adding all these up and equating them to $\\overline{BC}$ yields $$\\frac{4}{3}r + 8\\sqrt{r}+ \\frac{16}{3} = 56 \\rightarrow r = 44 - 6\\sqrt{35} \\rightarrow 44 + 6\\cdot 35 = \\boxed{254}$$ 2008 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "\\\\ \\overline{zw} &= \\bar{z} \\cdot \\bar{w} \\\\ \\\\ \\overline{\\left(\\dfrac{z}{w}\\right)} &= \\dfrac{\\bar{z} }{\\bar{w} } \\end{align} ##### Definition Given two complex numbers $z_1 = a + b i$ and $z_2 = c + d i$, their sum and difference is found by adding or subtracting their real and imaginary parts separately: \\begin{align} z_1 + z_2 &= (a + c) + (b + d) i \\\\ z_1 - z_2 &= (a - c) + (b - d) i \\end{align} ##### Worked Examples ###### Example 1 Let $z_1 = 3+2 i$ and $z_2 = -4+ i$, then work out: $z_1 + z_2$; ###### Solution \\begin{align} z_1 + z_2 &= (3 + 2 i) + (-4+ i)\\\\ &=3+2 i-4+ i\\\\ &=-1 +3 i \\end{align} ###### Example 2 Let $z_1 = 3+2 i$ and $z_2 = -4+ i$, then work out: $z_1-z_2$; ###### Solution \\begin{align} z_1-z_2 &= (3+2 i)-(-4+ i)\\\\ &=3+2 i+4- i\\\\ &=7+ i \\end{align} #### Multiplication ##### Definition Given two complex numbers $z_1 = a + b i$ and $z_2 = c + d i$, the method for complex multiplication is the same as expanding brackets: \\begin{align} z_1z_2 &= (a+b i)(c+d i)\\\\ &= ac + (ad) i + (bc) i + (bd) i^2\\\\ &= ac + (ad+bc) i - bd & \\text{(as }i^2=-1 \\text{)}\\\\ &=(ac-bd)+(ad+bc) i \\end{align} Note: The order of multiplication", "$z=2-i$, $w=-3+2i$, what is the midpoint between $2z$ and $\\overline{w}$ ? Problem 24 Plot a graph of the complex numbers $z=-1+i$, $w=2+3i$. Find the distance. Problem 25 If $z=2-i$, $w=5+i$, $t=-3+2i$, which of the following options is the result of $\\frac{2z}{w-t}$ ? Problem 26 If $z=2-i$, $w=5+i$, $t=-3+2i$, which of the following options is the result of $\\overline{(w\\cdot t)-3z}$? Correct: Wrong: Unsolved problems: Contact email:" ]

[ "# Do the coefficients of these polynomials alternate in sign? Define polynomials $$p_i(x)$$ by the recurrence \\begin{align} p_0(x)&=0 \\\\ p_1(x)&=1 \\\\ p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\\\ p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\\\ \\end{align} The first few are given by \\begin{align} p_0(x)&=0\\\\ p_1(x)&=1\\\\ p_2(x)&=1\\\\ p_3(x)&=x-1\\\\ p_4(x)&=x-2\\\\ p_5(x)&=x^2-3x+1\\\\ p_6(x)&=x^2-4x+3\\\\ p_7(x)&=x^3-5x^2+6x-1\\\\ p_8(x)&=x^3-6x^2+10x-4 \\end{align} It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts? Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ... $$\\begin{eqnarray*} p_{2i}(x) = \\sum_{j=0}^{i-1} \\binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\\\ p_{2i-1}(x) = \\sum_{j=0}^{i-1} \\binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\\\ \\end{eqnarray*}$$ Proof of the even formula ... $$\\begin{eqnarray*} &p_{2i-1}(x) -p_{2i-2}(x) =\\\\ & x^{i-1}+ \\sum_{j=0}^{i-2} \\binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \\sum_{j=0}^{i-1} \\binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\\\ &= x^{i-1}+ \\sum_{j=0}^{i-2} \\binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2} = p_{2i}(x). \\end{eqnarray*}$$ And the answer to the question in the title is $$\\color{red}{\\text{yes}}$$. I reckon that $$p_{2n-1}(x)=\\sum_{k=0}^{n-1}(-1)^k \\binom{n+1-k}k x^{n-1-k}$$ and $$p_{2n}(x)=\\sum_{k=0}^{n-1}(-1)^k \\binom{n+2-k}k x^{n-1-k}$$ and that these can be proved by induction.", "# Let $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1-x)$, for all $x \\in [0,1], \\; p(0)=1$ and $p(1) =41$. Then \\begin{align*} \\int_0 ^1p(x) \\; dx\\end{align*} equals ( D ) $\\sqrt{41}$", "of all the $p_i$. In the same way, \\begin{align*} P(Y > X) &= \\sum_{i=1}^n P(X = i, Y > i) \\\\ &= \\sum_{i=1}^n P(X = i)P(Y > i) \\\\ &= \\sum_{i=1}^{n-1} P(X = i)P(Y > i) \\end{align*} because $P(Y > n) = 0$. Now for each $i < n$, $P(Y > i) = P(X > i) = \\sum_{j=i+1}^n p_j$. Call this sum $b_i$ for \"bigger than $i$\". Then \\begin{align*} P(Y > X) &= \\sum_{i=1}^{n-1} P(X = i)P(Y > i) \\\\ &= \\sum_{i=1}^{n-1} p_ib_i \\end{align*} This is also a straightforward calculation if you know all the $p_i$. For $n=4$ it boils down to $$p_1 \\cdot(p_2 + p_3 + p_4) ~~ + ~~ p_2 \\cdot (p_3 + p_4) ~~ + ~~ p_3 \\cdot p_4$$", "and $$g$$. Polynomials should be denoted only using Capital letters. The symbol to denote polynomial is $$P(x)$$ if the equation has $$x$$ as variable. Example: \\begin{align*} P(x) = 2x^2 -x+3 \\end{align*} The $$x$$ in $$P(x)$$ represents the variable $$x$$. If the variable of the equation is $$y$$, the symbol to denote the polynomial is $$P(y)$$ Example: \\begin{align*} P(y) = 2y^2 -y+3 \\end{align*} The $$y$$ in $$P(y)$$ represents the variable $$y$$. Example 1: It is given that $$P(x) = 2x^2 - x + 3$$ and \\begin{align*} Q(x) = x^3 + 2x -1 \\end{align*} Evaluate $$P(2) +Q(-1)$$. Solution: To find $$P(2)$$, we need to replace the $$x$$ with $$2$$. \\begin{align*} P(2) = 2(2)^2 - 2 +3 \\end{align*} To find $$Q(-1)$$, we need to replace the $$x$$ with $$-1$$. \\begin{align*} Q(-1) = (-1)^3 + 2(-1) -1 \\end{align*} Adding both of the above equations. \\begin{align*} P(2) +Q(-1) = [\\;2(2)^2 - 2 +3\\;] \\;\\;+\\;\\;[\\;(-1)^3 + 2(-1) -1\\;] \\end{align*} Solving the above will give the answer. $$=5$$ Therefore, the answer is $$5$$. Example 2: Given that \\begin{align*} x^3 - 5x+3 = (Ax+3)(x-1)(x-2) \\;+\\; B(x-1) \\;+\\; C \\end{align*} for all real values of $$x$$, find the values of $$A,B$$ and $$C$$. Solution: In the question, it is given as “for all real values of $$x$$” which means we can substitute any value for $$x$$ and", "# Math Help - Polynomials 1. ## Polynomials 1, Find the values of a and b so that the polynomials x^3 + ax^2 + bx + 6 has x -2 as a factor and leaves a reminder 3 when divided by x-3 2, If ax^3 + bx^2 + x -b has x+2 as a factor and leaves a reminder 4 when divided by x-2 , find the values of a and b 2. ## Re: Polynomials These are applications of the remainder and factor theorems. If a polynomial \\displaystyle \\begin{align*} P(x) \\end{align*} is divided by \\displaystyle \\begin{align*} (x - \\alpha ) \\end{align*}, then the remainder is equal to \\displaystyle \\begin{align*} P \\left( \\alpha \\right) \\end{align*}.", "# 2008 AIME I Problems/Problem 13 ## Problem Let $$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$ Suppose that $$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$ There is a point $\\left(\\frac {a}{c},\\frac {b}{c}\\right)$ for which $p\\left(\\frac {a}{c},\\frac {b}{c}\\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. ## Solution ### Solution 1 \\begin{align*} p(0,0) &= a_0 \\\\ &= 0 \\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\\\ &= a_1 + a_3 + a_6 \\\\ &= 0 \\\\ p(-1,0) &= -a_1 + a_3 - a_6 \\\\ &= 0 \\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\\\", "Leo Giugiuc posted an elegant problem and its solution at the CutTheKnotMath facebook page. Let $P(x)=x^2+ax+b,$ be a monic (with the leading coefficient $!)$ quadratic polynomial, with integer $a$ and $b.$ Prove that, for any integer $n,$ there is an integer $m$ such that $P(n)P(n+1)=P(m).$ ### Proof Let $u$ and $v$ be the roots of $P(x)=0.$ It does not matter whether they are integers or not. $P(x)=(x-u)(x-v)$ and $P(x+1)=(x+1-u)(x+1-v),$ implying $P(x)P(x+1)=(x-u)(x-v)(x+1-u)(x+1-v).$ Now, \\begin{align} (x-u)(x+1-v) &= x^{2}+x-xv-xu-u+uv\\\\ &=x^{2}-x(u+v)+uv+x-u\\\\ &=x^{2}+x(a+1)+b-u. \\end{align} Similarly, $(x-v)(x+1-u)=x^{2}+x(a+1)+b-v,$ meaning that \\begin{align} P(x)P(x+1) &= (x^{2}+x(a+1)+b-u)(x^{2}+x(a+1)+b-v)\\\\ &=P(x^{2}+x(a+1)+b). \\end{align} Thus, for a given $n,$ $m=n^{2}+n(a+1)+b,$ an integer. Nothing short of brilliant!", "= 3x^{-1}-2x^{-2} \\\\ \\end{align} The following are not examples of a rational function because they do not obey the restrictions above: (2) \\begin{align} y = x+2 \\\\ y = x^{1.3}-x^{2+3i} \\\\ y = {3^x+2 \\over 2-x} \\\\ y = {\\sin(x) \\over |x+1|-3x^2} \\\\ \\end{align} ## Forms Rational functions can be written in many different ways, but there are only a couple of standard forms, ones that look \"pretty\" mathematically speaking. Standard Form: (3) \\begin{align} f(x) = {p(x) \\over q(x)} \\end{align} where p(x) and q(x) are both polynomials either written in standard or factored form. This form simply states that a rational function the ratio of two polynomials. It is in this form that we will receive the most information from a rational function. Note that this form doesn't include a rational expression over a rational expression, since that can be simplified to a polynomial over polynomial form anyway (given a few restrictions). Quotient Form: (4) \\begin{align} f(x) = a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n+{p_0x^{m_1}+p_1x^{m_1-1}+...+p_{m_1-1}x+p_{m_1} \\over q_0x^{m_2}+q_1x^{m_2-1}+...+q_{m_2-1}x+q_{m_2}} \\end{align} where m1<m2. Do not be intimidated by this form. In simple terms, it is the standard form of a polynomial function with a polynomial over polynomial tagged on the end in which the polynomial of the denominator is of greater degree than that of the numerator. This is the typical result in completing polynomial division. ##", "solves this problem, as \\begin{align*} P(x+y+z)-P(x)-P(y)-P(z) &= a(x+y+z)^2-a(x^2+y^2+z^2)-2a \\\\ &= 2a(xy+yz+zx-1) \\end{align*} Then, we can impose $P(0) = a = 1$ and $P(1) = 2a+b = 4$. The latter implies $b = 2$, so we have $P(x) = x^2+2x+1 = (x+1)^2$. Then we find $P(2017) = 2018^2$. To prove that this is the only solution, note that for $$P^*(x, y, z) := P(x+y+z)-P(x)-P(y)-P(z)$$ $P^*(x, y, z)$ must be in the ideal generated by $xy+yz+zx-1$ in $\\mathbb{C}[x, y, z]$ as this polynomial is irreducible over $\\mathbb{C}$. Therefore, $$P^*(x, y, z) = Q(x, y, z)(xy+yz+zx-1)$$ for some polynomial $Q$. Note that $P^*(x, y, z)$ and $xy+yz+zx-1$ are both symmetric polynomials. In particular, $xy+yz+zx-1 = \\frac{1}{2}p_1^2-\\frac{1}{2}p_2-1$, where $p_k := x^k+y^k+z^k$. Now, note that for $P(x) = \\sum_{k=0}^n a_kx^k$, $$P^*(x, y, z) = \\sum_{k=0}^n a_k(p_1^k-p_k)$$ As $Q(x, y, z)$ is the quotient of symmetric polynomials, it must also be a symmetric polynomial. Thus, we write $R^*(p_1, p_2, \\ldots, p_n)$ and $R(p_1, p_2, \\ldots, p_n)$ to be the polynomial representations of $P^*$ and $Q$ respectively in terms of power sums (which are algebraically independent). After switching to $R$ and $R^*$, our equation in $P^*$ and $Q$ becomes $$R^*(p_1, \\ldots, p_n) = \\left(\\frac{1}{2}p_1^2-\\frac{1}{2}p_2-1\\right)R(p_1, \\ldots, p_n)$$ If $a_n\\neq 0$ for some $n\\geq 3$, then $R^*(p_1, \\ldots, p_n)$ will have a nonzero term $-a_np_n$. However, this", "anonymous 5 years ago 0, 0, 4, 1+i find a polynomial function with integer coefficients that has the given zeroes I a polynomial has roots $$x=r_1,x=r_2,x=r_3,...,x=r_n$$ then it can be written as follows:$(x-r_1)(x-r_2)(x-r_3)...(x-r_n)=0$so, for the roots you have provided, we can write the polynomial as:\\begin{align} (x-0)(x-0)(x-4)(x-(1-i))&=0\\\\ \\therefore (x)(x)(x-4)(x-1+i)&=0\\\\ \\therefore x^2(x-4)(x-1+i)&=0\\\\ \\therefore x^2(x^2-x+ix-4x+4-4i)&=0\\\\ \\therefore x^2(x^2-5x+ix+4-4i)&=0\\\\ \\therefore x^2(x^2-(5-i)x+4(1-i))&=0\\\\ \\therefore x^4-(5-i)x^3+4(1-i)x^2&=0\\\\ \\end{align}", "## Chapter17Polynomials Most people are fairly familiar with polynomials by the time they begin to study abstract algebra. When we examine polynomial expressions such as \\begin{align*} p(x) & = x^3 -3x +2\\\\ q(x) & = 3x^2 -6x +5\\text{,} \\end{align*} we have a pretty good idea of what $$p(x) + q(x)$$ and $$p(x) q(x)$$ mean. We just add and multiply polynomials as functions; that is, \\begin{align*} (p +q)(x) & = p(x) + q(x)\\\\ & = ( x^3 - 3 x + 2 ) + ( 3 x^2 - 6 x + 5 )\\\\ & = x^3 + 3 x^2 - 9 x + 7 \\end{align*} and \\begin{align*} (p q)(x) & = p(x) q(x)\\\\ & = ( x^3 - 3 x + 2 ) ( 3 x^2 - 6 x + 5 )\\\\ & = 3 x^5 - 6 x^4 - 4 x^3 + 24 x^2 - 27 x + 10\\text{.} \\end{align*} It is probably no surprise that polynomials form a ring. In this chapter we shall emphasize the algebraic structure of polynomials by studying polynomial rings. We can prove many results for polynomial rings that are similar to the theorems we proved for the integers. Analogs of prime numbers, the division algorithm, and the Euclidean algorithm exist for polynomials.", "\\begin{align*} P(X=0,Y=0) &= 1 - p - q - pq = (1-p)(1-q) = P(X=0)P(Y=0)\\\\ P(X=1,Y=0) &= p - pq = p(1-q) = P(X=1)P(Y=0)\\\\ P(X=0,Y=1) &= q - pq = (1-p)q = P(X=0)P(Y=1). \\end{align*} Yes, $r$ might be equal to $pq$, BUT it can be different, as long as it respects the boundaries above. Well, from the above joint distribution, we would have \\begin{align*} E(X) &= 0\\cdot P(X=0) + 1\\cdot P(X=1) = P(X=1) = p \\\\ E(Y) &= 0\\cdot P(Y=0) + 1\\cdot P(Y=1) = P(Y=1) = q \\\\ E(XY) &= 0\\cdot P(XY=0) + 1\\cdot P(XY=1) \\\\ &= P(XY=1) = P(X=1,Y=1) = r\\\\ Cov(X,Y) &= E(XY) - E(X)E(Y) = r - pq \\end{align*} Now, notice then that $X$ and $Y$ are independent if and only if $Cov(X,Y)=0$. Indeed, if $X$ and $Y$ are independent, then $P(X=1,Y=1)=P(X=1)P(Y=1)$, which is to say $r=pq$. Therefore, $Cov(X,Y)=r-pq=0$; and, on the other hand, if $Cov(X,Y)=0$, then $r-pq=0$, which is to say $r=pq$. Therefore, $X$ and $Y$ are independent. ### General Case About the without loss of generality clause above, if $X$ and $Y$ were distributed otherwise, let's say, for $a<b$ and $c<d$, \\begin{align*} P(X=b)=p \\\\ P(Y=d)=q \\\\ P(X=b, Y=d)=r \\end{align*} then $X'$ and $Y'$ given by $$X'=\\frac{X-a}{b-a} \\qquad \\text{and} \\qquad Y'=\\frac{Y-c}{d-c}$$ would be distributed just as characterized above, since $$X=a \\Leftrightarrow X'=0, \\quad X=b \\Leftrightarrow", "$P=(a_n)_n$ and $Q=(b_n)_n$ then $PQ=(c_n)_n$ is a polynomial with \\begin{align*} c_n=\\sum_{k=0}^n a_k b_{n-k}.\\end{align*} Proposition: $(\\mathbb{K},+,\\cdot)$ is a commutative ring. ## A selection of exercises on polynomials Exercise: Determine the greatest common divisor (gcd) and a Bezout relation between the polynomials $P=X^3+X^2+X-3$ and $Q=X^2-3X+2$. Solution: Here we shall apply Euclid’s algorithm: we select $P_0=P$ and $P_1=Q$. Moreover, we put $U_0=1,\\,U_1=0$ and $V_0=0,\\,V_1=1$ in order to have $PU_0+QV_0=P_0$ and $PU_1+QV_1=P_1$. The Euclidean division of $P_0$ by $P_1$ is \\begin{align*} P_0=P_1(X+4)+(11 X-11). \\end{align*} We set $P_2=11 X-11$. Then \\begin{align*} P_2&=P_0-P_1(X+4)\\cr &= PU_0+Q V_0-(PU_1+QV_1)(X+4)\\cr &= P(U_0-(X+4)U_1)+Q(V_0-(X+4)V_1)\\cr &= P U_2+Q V_2, \\end{align*} where \\begin{align*} U_2=U_0-(X+4)U_1=1,\\quad V_2=V_0-(X+4)V_1=-X-4. \\end{align*} The Euclidean division of $P_1$ by $P_2$ is \\begin{align*} P_1=P_2 \\left(\\frac{1}{11}X-\\frac{2}{11}\\right)+0 \\end{align*} The GCD of $P$ and $Q$ is \\begin{align*} \\frac{1}{11}P_2=X-1. \\end{align*} The Bezout relation is then \\begin{align*} \\frac{1}{11} P-\\frac{1}{11}(X+4) Q=X-1. \\end{align*} Exercise: Does the polynomial $P=X^4+1$ is irreducible in $\\mathbb{C}[X]$ ? in $\\mathbb{R}[X]$ ? in $\\mathbb{Q}[X]$ ? Solution: It is not irreducible neither in $\\mathbb{C}[X]$ nor in $\\mathbb{R}[X]$ as it is not either of degree 1 or of degree 2. Assume that $P$ is reducible in $\\mathbb{Q}[X]$. Remark that $P$ does not have rational roots “and also real roots”, $P$ can not admit a divisor of degree $1$. Then $P$ is the product of two polynomials of degree 2. We then can assume that \\begin{align*}", "f (x) & = g (x) - h (x) \\nonumber \\\\ & = x^{p-1} + a_1 x^{p - 2} + \\cdots + a_{p -2} x + a_{p - 1} - x^{p - 1} + 1 \\nonumber \\\\ & \\equiv a_1 x^{p - 2} + a_2 x^{p -3} + \\cdots + a_{p -2} x \\; (\\text{mod } p) \\nonumber \\\\ & \\equiv 0 . \\end{align*} As this is a polynomial of degree ##p - 2## with the ##p - 1## solutions by Lagrange's theorem ##p## divides all the coefficients. Putting ##x = p## in ##(*)## gives \\begin{align*} (p-1)! = p^{p-1} + a_1 p^{p - 2} + a_2 p^{p -3} + \\cdots + a_{p -2} p + a_{p - 1} \\end{align*} Subtracting ##(p - 1)!## from both sides, dividing by ##p##, and then rearranging gives \\begin{align*} - a_{p - 2} = p^{p-2} + a_1 p^{p - 3} + \\cdots + a_{p - 3} p . \\end{align*} As we are taking ##p \\geq 5##, and as each of the coefficients ##a_1, a_2, \\dots a_{p - 3}## is proportional to ##p## the RHS is equal to ##k p^2## where ##k## is an integer. If ##p## were equal to 3 we couldn't say that the term ##a_1 p^{p - 3}## is proportional to ##3^2## as ##a_1 p^{p - 3} = a_1##. The result", "4y + 2z + 14\\\\&4x - 8z = 0\\\\&x - 2z = 0\\end{align} Thus, the required equation is $$x - 2z = 0$$ ## Chapter 12 Ex.12.2 Question 5 Find the equation of the set of points $$P,$$ the sum of whose distances from $$A\\left( {4,0,0} \\right)$$ and $$B\\left( { - 4,0,0} \\right)$$ is equal to $$10.$$ ### Solution Let $$A\\left( {4,0,0} \\right)$$ and $$B\\left( { - 4,0,0} \\right)$$ Let the coordinates of point $$P$$ be $$\\left( {x,y,z} \\right)$$ Calculating PA: $$P\\left( {x,y,z} \\right)$$ and $$A\\left( {4,0,0} \\right)$$ By using the distance formula, Distance $$PA = \\sqrt {{{\\left( {{x_2} - {x_1}} \\right)}^2} + {{\\left( {{y_2} - {y_1}} \\right)}^2} + {{\\left( {{z_2} - {z_1}} \\right)}^2}}$$ So here, \\begin{align}{x_1} &= x,\\;{y_1} = y,\\;{z_1} = z\\\\{x_2} &= 4,\\;{y_2} = 0,\\;{z_2} = 0\\end{align} Distance $$PA = \\sqrt {{{\\left( {4 - x} \\right)}^2} + {{\\left( {0 - y} \\right)}^2} + {{\\left( {0 - z} \\right)}^2}}$$ Calculating PB: $$P\\left( {x,y,z} \\right)$$ and $$B\\left( { - 4,0,0} \\right)$$ By using the distance formula, Distance $$PB = \\sqrt {{{\\left( {{x_2} - {x_1}} \\right)}^2} + {{\\left( {{y_2} - {y_1}} \\right)}^2} + {{\\left( {{z_2} - {z_1}} \\right)}^2}}$$ So here, \\begin{align}{x_1} &= x,\\;{y_1} = y,\\;{z_1} = z\\\\{x_2} &= - 4,\\;{y_2} = 0,\\;{z_2} = 0\\end{align} Distance $$PB = \\sqrt {{{\\left( { - 4 - x} \\right)}^2} + {{\\left( {0 -", "r_4).$$ Expanding this product, we get \\begin{align*} (x - r_1)(x - r_2)(x - r_3)(x - r_4) = x^4 &- (r_1 + r_2 + r_3 + r_4)x^3 \\\\ &+ (r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4)x^2 \\\\ &- (r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4)x \\\\ &+ r_1r_2r_3r_4\\end{align*} If we write the polynomial in coefficient form $x^4 + ax^3 + bx^2 + cx + d$, then we find that: \\begin{align*} -a &= r_1 + r_2 + r_3 + r_4 \\\\ b &= r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 \\\\ -c &= r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4 \\\\ d &= r_1r_2r_3r_4 \\end{align*} This means we can find out the sum and product of all the roots of a polynomial based only on the coefficients, without knowing the roots themselves! I'll leave it up to you to generalize this for higher degree polynomials. As a side note, notice that each coefficient above can be expressed as a polynomial on the variables $r_1$, $r_2$, $r_3$ and $r_4$. This is also true for any other degree. These polynomials are called the elementary symmetric polynomials. ## Polynomial factorization We can also talk about factorization of (nonzero) polynomials. This is similar to integer factorization. Given a polynomial $p(x)$, we want to find two polynomials", "and c, but I still don't know the value of $Q_{4}(x)$ and I can't find the polynomial without knowing the value of $Q_{4}(x)$. What should I do? – cristid9 Jan 13 '15 at 20:23 • You don't need $Q_4(x)$ since you're looking only for the reminder. – kingW3 Jan 13 '15 at 20:26 It's a problem about Lagrange's interpolation polynomials. The hypothesis means that $P(1)=2$, $P(-1)=6$, P(2)=3$. The simplest method uses concepts from linear algebra. You first solve 3 easier problems: find polynomials$p, q, rsuch that \\begin{alignat*}{3} &p(1)=1&&q(1)=0&&r(1)=0 \\\\ &p(-1)=0&\\qquad&q(-1)=1&\\qquad&r(-1)=0\\\\ &p(2)=0&&q(2)=0&&r(2)=1 \\end{alignat*} For instance,\\,\\,p(x)=\\dfrac{(x+1)(x-2)}{(1+1)(1-2)}= -\\dfrac12(x^2-x-2)$. Once you've computed$p(x), q(x), r(x)$, just take$P(x)=2p(x)+6q(x)+3r(x)$. Hint$\\ f(x) = 2\\! +\\! (x\\!-\\!1)\\left[a\\! +\\! (x\\!-\\!2)(b\\!+\\! (x\\!+\\!1)g(x))\\right],\\,f(2) = 3\\,\\Rightarrow\\, a=\\_\\_,\\ f(-1) = 6\\,\\Rightarrow\\, b = \\_\\_\\$", "$(y_2 - y_1)^2 ≥ 0$ and $(z_2 - z_1)^2 ≥ 0$. Without loss of generality, suppose that $(x_2 - x_1)^2 > 0$. Then either $(y_2 - y_1)^2 < 0$ or $(z_2 - z_1)^2 < 0$, a contradiction, and so we have that $(x_2 - x_1)^2 = 0$, $(y_2 - y_1)^2 = 0$ and $(z_2 - z_1)^2 = 0$, which implies that $x_1 = x_2$, $y_1 = y_2$, and $z_1 = z_2$ and so $(x_1, y_1, z_1) = (x_2, y_2, z_2)$. • $\\Rightarrow$ Suppose that $(x_1, y_1, z_1) = (x_2, y_2, z_2)$. Then computing the distance between these points we have that: (3) \\begin{align} d_{PQ} = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\\\ d_{PQ} = \\sqrt{(x_1 - x_1)^2 + (y_1 - y_1)^2 + (z_1 - z_1)^2} \\\\ d_{PQ} = \\sqrt{0^2 + 0^2 + 0^2} = 0 \\end{align} Example 1 Compute the distance between the points $(1, 2, 3), (4, 5, 6) \\in \\mathbb{R}^3$. Applying the distance formula directly we have that: (4) \\begin{align} \\: \\mathrm{distance} = \\sqrt{(4 - 1)^2 + (5 - 2)^2 + (6 - 3)^2} = \\sqrt{3^2 + 3^2 + 3^2} = \\sqrt{27} \\end{align} Example 2 Compute the distance between the points $(2, 5, -1), (1, 1, 1) \\in \\mathbb{R}^3$. Once again, applying the distance formula directly, we get that: (5) \\begin{align}", "x_2] \\end{align} Notice that we can easily obtain the polynomial $P_{k+1}(x)$ that interpolates the the points $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_{k+1}, y_{k+1})$ from the polynomial $P_k$ that interpolates the $k$ points $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_{k-1}, y_{k-1})$ as: (5) \\begin{align} \\quad P_{k+1}(x) = P_k(x) + (x - x_0)(x - x_1)...(x - x_k)f[x_0, x_1, ..., x_{k+1}] \\end{align} ## Example 1 Find the polynomial $P_2$ for the function $y = \\sqrt{x}$ that interpolates the points $(1, 1)$, $(4, 2)$, and $(9, 3)$ using Newton's divided difference formula. Applying Newton's divided difference formula from above and we get that: (6) \\begin{align} \\quad P_2(x) = f(x_0) + (x - x_0)f[x_0, x_1] + (x - x_0)(x - x_1)f[x_0, x_1, x_2] \\\\ \\quad P_2(x) = 1 + (x - 1) \\frac{f(x_1) - f(x_0)}{x_1 - x_0} + (x - 1)(x - 4) \\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} \\\\ \\quad P_2(x) = 1 + (x - 1) \\frac{2 - 1}{4 - 1} + (x - 1)(x - 4) \\frac{\\frac{3 - 2}{9 - 4} - \\frac{2 - 1}{4 - 1}}{9 - 1} \\\\ \\quad P_2(x) = 1 + \\frac{1}{3}(x - 1) - \\frac{1}{60} (x - 1)(x - 4) \\end{align} The graph of $y = P_2(x)$ is given below. ## Example 2 Find the polynomial $P_3$ for the function $y = \\sin x$ that", "(Note that all coefficients are initially 0): $$(X_{i}, Y_{i}, Z_{i}) => A[X_{i} + 2XY_{i} + 4XYZ_{i}] \\mathrel{+}= 1$$ $$(X_{i}, Y_{i}, Z_{i}) => B[(L-1) -(X_{i} + 2XY_{i} + 4XYZ_{i})] \\mathrel{+}= 1$$ Let us define a polynomial $C$ of degree $(2L-1)$ as the product of the 2 polynomials $A$ and $B$. Let us consider any coefficient of $C$. We can see that it includes some $(X_{i} + 2XY_{i} + 4XYZ_{i})$ from $A$ and some $((L-1) - (X_{j} + 2XY_{j} + 4XYZ_{j}))$ from $B$. In other words, for some index $k$ in $C$, we have: \\begin{align}k &= X_{i} + 2XY_{i} + 4XYZ_{i} + (L-1) - (X_{j} + 2XY_{j} + 4XYZ_{j}) \\\\ & = (T + M) + (X_{i} - X_{j}) + 2X(Y_{i} - Y{j}) + 4XY(Z_{i} - Z_{j}) \\\\ & = T + (X + (X_{i} - X_{j})) + 2X(Y + (Y_{i} - Y_{j})) + 4XY(Z + (Z_{i} - Z_{j})) \\end{align} We can now extract the values of dx, dy and dz using the following code. Pseudo Code: for (k = 0 to 2*L-1) { if (C[k] == 0) continue; dx = (k - T) % (2*X) - X; dy = (k - T) % (4*X*Y) / (2*X) - Y; dz = (k - T) / (4*X*Y) - Z; if (dx == 0 and dy == 0 and dz == 0)" ]

[ "## e Show that $\\alpha_1 - \\alpha_2$ is estimable and give its least squares estimator. Take $\\lambda^T = [0, 1,-1, 0, 0, 0]$ such that $\\lambda^T b = \\alpha_1 - \\alpha_2.$ Now we need to find $a$ such that $a^T X = \\lambda ^T.$ $[a_1, a_2, a_3, a_4, a_5, a_6] \\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 1 & 0 \\\\ 1 & 1 & 0 & 0 & 0 & 1 \\\\ 1 & 0 & 1 & 1 & 0 & 0 \\\\ 1 & 0 & 1 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 0 & 1 \\end{bmatrix} = [0, 1, -1, 0, 0, 0]$ \\begin{align} a_1 + a_2 + a_3 + a_4 + a_5 + a_6 & = 0 \\\\ a_1 + a_2 + a_3 & = 1 \\\\ a_4 + a_5 + a_6 & = -1 \\\\ a_1 + a_4 & = 0 \\\\ a_2 + a_5 & = 0 \\\\ a_3 + a_6 & = 0. \\end{align} Now fix $a_5$ and $a_6$. \\begin{align} a_2 & = -a_5 \\\\ a_3 & = -a_6 \\\\ a_4 & = -1 + a_5 + a_6 \\\\ a_1 & = 1 - a_5 - a_6", "&= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\\\ &= -a_4 - a_7 + a_8 = 0\\end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, $p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0$ So $3a_1 + 3a_2 + 2a_4 = 0$. Now $p(x,y) = 0 + x a_1 + y a_2 + 0 + xy a_4 + 0 - x^3 a_1 - x^2 y a_4 + 0 - y^3 a_2$ $= x(1-x)(1+x) a_1 + y(1-y)(1+y) a_2 + xy (1-x) a_4$. In order for the above to be zero, we must have $x(1-x)(1+x) = y(1-y)(1+y)$ and $x(1-x)(1+x) = 1.5 xy (1-x).$ Canceling terms on the second equation gives us $1+x = 1.5 y \\Longrightarrow x = 1.5 y - 1$. Plugging that into the first equation and solving yields $x = 5/19, y = 16/19$, and $5+16+19 = \\boxed{040}$.", "= 0.2 (0.8) + 0 (0.2) = 0.16\\\\ P(2,8) &= P(1,8) (1-p_2) + P(1,8-8) p_2 = 0 (0.8) + 0.8 (0.2) = 0.16\\\\ P(2,13) &= P(1,13)(1-p_2) + P(1,13-8) p_2 = 0 (0.8) + 0.2 (0.2) = 0.04\\\\ P(2,j) &= 0, j \\neq \\{0,5,8,13\\} \\end{align} The third recursion pass is as follows: \\begin{align} P(3,0) &= P(2,0)(1-p_3) = 0.512\\\\ P(3,4) &= P(2,4)(1-p_3) + P(2,4-4) = 0(0.8) + 0.64(0.2) = 0.128\\\\ P(3,5) &= P(2,5)(1-p_3) + P(2,5-4) p_3 = 0.16 (0.8) + 0 (0.2) = 0.128\\\\ P(3,8) &= P(2,8) (1-p_3) + P(2,8-4) p_3 = 0.16 (0.8) + 0 (0.2) = 0.128\\\\ P(3,9) &= P(2,9) (1-p_3) + P(2,9-4) p_3 = 0 (0.8) + 0.16 (0.2) = 0.032\\\\ P(3,12) &= P(2,12) (1-p_3) + P(2,12-4) p_3 = 0 (0.8) + 0.16 (0.2) = 0.032\\\\ P(3,13) &= P(2,13) (1-p_3) + P(2,13-4) p_3 = 0.04 (0.8) + 0 (0.2) = 0.032\\\\ P(3,17) &= P(2,17) (1-p_3) + P(2,17-4) p_3 = 0 (0.8) + 0.04 (0.2) = 0.008\\\\ P(3,j) &= 0, j \\neq \\{0,4,5,8,9,12,13,17\\} \\end{align} Note that the same computation work even when the outcomes are not of equal probability. Let’s do one more example. #ONE FINAL EXAMPLE #----------MAIN CALLING SEGMENT------------------ w = c(5,2,4,2,8,1,9) p = array(0.2,length(w)) res = asbrec(w,p) print(res) ## [,1] [,2] ## [1,] 0 0.2097152 ## [2,] 1 0.0524288 ## [3,] 2 0.1048576 ## [4,] 3", "$$x$$: $$16\\left(x+1\\right)={x}^{2}\\left(x+1\\right)$$ \\begin{align*} 16\\left(x+1\\right) &= {x}^{2}\\left(x+1\\right) \\\\ 16x + 16 &= {x}^{3} + {x}^{2} \\\\ 0 &= {x}^{3} + {x}^{2} - 16x - 16 \\\\ \\text{Let } a(x) &= {x}^{3} + {x}^{2} - 16x - 16 \\\\ a(-1) &= (-1)^{3} + (-1)^{2} - 16(-1) - 16 \\\\ &= -1 + 1 +16 -16 \\\\ &= 0 \\\\ \\therefore a(x) &= (x + 1)(x^{2} - 16) \\\\ &= (x + 1)(x - 4)(x +4) \\\\ \\therefore 0 &= (x + 1)(x - 4)(x +4) \\\\ \\therefore x = -1 &\\text{ or } x = 4 \\text{ or } x= -4 \\end{align*} Show that $$x-2$$ is a factor of $$3{x}^{3}-11{x}^{2}+12x-4$$. \\begin{align*} \\text{Let } a(x) &= 3{x}^{3}-11{x}^{2}+12x-4 \\\\ a(2) &= 3(2)^{3}-11(2)^{2}+12(2)-4 \\\\ &= 24 - 44 +24 - 4 \\\\ &= 0 \\\\ \\therefore (x-2) &\\text{ is a factor of } a(x) \\end{align*} Hence, by factorising completely, solve the equation: $$3{x}^{3}-11{x}^{2}+12x-4=0$$ \\begin{align*} 3{x}^{3}-11{x}^{2}+12x-4 &= 0 \\\\ (x -2)(3x^{2} - 5x + 2) &= 0 \\\\ \\therefore (x - 2)(3x - 2)(x - 1) &= 0 \\\\ \\therefore x = 2 &\\text{ or } x = \\frac{2}{3} \\text{ or } x = 1 \\end{align*} $$2{x}^{3}-{x}^{2}-2x+2=Q\\left(x\\right).\\left(2x-1\\right)+R$$ for all values of $$x$$. What is the value of $$R$$? \\begin{align*} \\text{Let } a(x) &= 2{x}^{3}-{x}^{2}-2x+2 \\\\ R = a \\left( \\frac{1}{2} \\right) &= 2\\left( \\frac{1}{2}", "A_6 &= \\begin{pmatrix} 2 & 0 & 0\\\\ 0 & 4 & 0\\\\ 0 & 0 & 2 \\end{pmatrix} \\end{align*} $$A_1$$ and $$A_2$$ $$A_1$$ and $$A_3$$ $$A_1$$ and $$A_4$$ $$A_1$$ and $$A_5$$ $$A_1$$ and $$A_6$$ $$A_2$$ and $$A_3$$ $$A_2$$ and $$A_4$$ $$A_2$$ and $$A_5$$ $$A_2$$ and $$A_6$$ $$A_3$$ and $$A_4$$ $$A_3$$ and $$A_5$$ $$A_3$$ and $$A_6$$ $$A_4$$ and $$A_5$$ $$A_4$$ and $$A_6$$ $$A_5$$ and $$A_6$$", "+ 5) 8 6 4 2 (x - 8 x + 19 x - 12 x + 1) 3 2 3 2 20 (x - 1) (x + 1) (x - x - 2 x + 1) (x + x - 2 x - 1) 6 5 4 3 2 (x - x - 6 x + 6 x + 8 x - 8 x + 1) 6 5 4 3 2 (x + x - 6 x - 6 x + 8 x + 8 x + 1) Quote by Dodo I think this may be the case. Since\\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) \\end{align*}if 1 (or -1, for that matter) is a root of P(n-3) then it is a root of P(n). And since P(2) = x^2 - 1 and has roots 1 and -1, then this is true for n=2,5,8,... For similar reasons, 0 is a root (as mentioned) for n=1,3,5,7,... to extend further, \\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) &= (x^3-x-1)P(n-3)-(x^2-1)P(n-2) \\end{align*} and we could continue to extend this procedure to obtain as many repeated roots as we like. Not yet there, but here is another", "have \\begin{align} f_3(0) = [f(0)]^4. \\end{align} Next, using both 1 and 2, we have \\begin{align} f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \\ \\ (\\ast) \\end{align} Plugging $0$ into $(\\ast)$ yields \\begin{align} f_3(0) = f(0)+[f(0)]^2. \\end{align} Combining everything yields \\begin{align} [f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \\ \\ \\Rightarrow \\ \\ f(0)[f(0)^3-f(0)-1] =0. \\end{align} Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root. Moreover, observe \\begin{align} f_4(0) = [f(0)]^4 \\end{align} and \\begin{align} f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2 \\end{align} which means \\begin{align} [f(0)]^4-2[f(0)]^2 = 0. \\end{align} Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$. In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$. $$\\mrm{f}\\pars{x^{2} + \\mrm{f}\\pars{y}} =y + \\bracks{\\mrm{f}\\pars{x}}^{2} \\label{1}\\tag{1}$$ Deriving both members of \\eqref{1} respect of $\\ds{x}$ and $\\ds{y}$ yield, respectively: $$\\left\\{\\begin{array}{rcl} \\ds{\\mrm{f}'\\pars{x^{2} + \\mrm{f}\\pars{y}}\\pars{2x}} & \\ds{=} & \\ds{2\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x}} \\\\[2mm] \\ds{\\mrm{f}'\\pars{x^{2} + \\mrm{f}\\pars{y}}\\mrm{f}'\\pars{y}} & \\ds{=} & \\ds{1} \\end{array}\\right. \\label{2}\\tag{2}$$ By comparing both expressions in \\eqref{2}: $${\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x} \\over x} = {1 \\over \\mrm{f}'\\pars{y}} = \\alpha\\,, \\qquad \\pars{~\\alpha\\ \\mbox{is independent of}\\ x\\ \\mbox{and}\\ y~} \\label{2.a}\\tag{2.a}$$ The second one is quite trivial: $${1 \\over \\mrm{f}'\\pars{y}} = \\alpha \\implies \\mrm{f}\\pars{y} = {1 \\over \\alpha}\\, y + \\beta\\,,\\qquad \\pars{~\\beta\\ \\mbox{is a constant}~}\\label{3}\\tag{3}$$ \\eqref{2.a} and \\eqref{3} yield: \\begin{align} &{\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x} \\over x} = \\alpha", "You should then apply $P$ twice, and see that, indeed, $P(P(x))=P(x)$. Let $x\\in V$, and let $x_1\\in V_1$ and $x_2\\in V_2$ be the unique elements such that $x=x_1+x_2$. By definition of $V_1$ and $V_2$, $P(x_1)=0$ and $x_2-P(x_2)=0$, so \\begin{align*} P(x)&= P(x_1) + P(x_2)\\\\ &= 0 + x_2\\\\ P(P(x))&= P(0) + P(x_2)\\\\ &=x_2, \\end{align*} so we have that $$\\forall x\\in V, P^2(x)=P(x),$$ which proves indeed that $P^2=P$. • I take it you mean for $x_1$ to be in $V_1$, $x_2$ in $V_2$. – Gerry Myerson Oct 31 '11 at 11:35", "question is $$\\inf_{0<u\\le1/5}(M_1(u)\\vee M_2(u)),$$ $$M_1(u):=\\sup_{1/5\\le a\\le1/2}F_1(u,a),\\quad M_2(u):=\\sup_{a\\in[u,1/5]}F_2(u,a).$$ For $F_1(a):=F_1(u,a)$, let $DF_1(a):=F_1'(a)(1-a)^2$. Then $DF_1(a)=-3/10 + 41 u/3 + \\ln a/\\ln2$ is increasing in $a$. So, $DF_1(a)$ (and hence $F_1'(a)$) can change the sign only from $-$ to $+$. So, \\begin{align} M_1(u)&=\\sup_{1/5\\le a\\le1/2}F_1(u,a)=F_1(u,1/5)\\vee F_1(u,1/2). \\end{align} For $F_2(a):=F_2(u,a)$, let $DF_2(a):=F_2'(a)(1-a)^2$. Then $DF_2'(a) 2 \\ln2\\,(1-a)^2 a^3=2 a^4 + \\ln2 - a (2 + \\ln8) + a^2 (8 + \\ln8) - a^3 (8 + \\ln512)>0$ for $a\\in[0,1/5]$. So, $DF_2(a)$ is increasing in $a\\in[0,1/5]$, and so, $DF_2(a)$ (and hence $F_2'(a)$) can change the sign only from $-$ to $+$. So, $$M_2(u)=\\sup_{a\\in[u,1/5]}F_2(u,a)= F_2(u,u)\\vee F_2(u,1/5)\\quad\\text{if }0<u\\le1/5.$$ Therefore and because $F_1(u,1/5)=F_2(u,1/5)$, the infsup in question is $$\\inf_{0<u\\le1/5}[F_1(u,1/5)\\vee F_1(u,1/2)\\vee F_2(u,u)].$$ For brevity, let $$g(u):=F_2(u,u),\\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a).$$ Let $g_1(u):=g'(u)(1 - u)^2$. Then $g_1'(u)$ is a simple rational function of $u$, which is $>0$ (everywhere here $0<u\\le1/5$). So, $g_1(u)$ is increasing in $u$. So, $g_1(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find $$\\inf_{0<u\\le1/5}F_2(u,u)=\\min_{0<u\\le1/5}g(u)=1.0616\\ldots,$$ attained at $u=0.099677\\ldots$. Because (i) $d_a(u)=g(u)-F_1(u,a)$, (ii) $F_1(u,a)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d_a'(u)$ can change the sign only from $-$ to $+$. Also, $d_{1/5}'(14/100)=-6.7650\\ldots<0$. So, $d_{1/5}$ is decreasing on $[0,14/100]$, with $d_{1/5}(14/100)=0.1884\\ldots>0$. So, $d_{1/5}>0$ on $[0,14/100]$, that is, $F_2(u,u)>F_1(u,1/5)$ if $0<u\\le14/100$. Similarly, using that $d_{1/2}'(14/100)=-17.015\\ldots<0$", "k-th number in our family. We define: \\begin{align} a_1 &:= (p_1)^0,(p_2)^1,(p_3)^1,(p_4)^1,(p_5)^0,(p_6)^1,(p_7)^1,(p_8)^1,(p_9)^0 \\\\ a_2 &:= (p_1)^1,(p_2)^0,(p_3)^1,(p_4)^1,(p_5)^1,(p_6)^0,(p_7)^0,(p_8)^1,(p_9)^1 \\\\ a_3 &:= (p_1)^1,(p_2)^1,(p_3)^0,(p_4)^0,(p_5)^1,(p_6)^1,(p_7)^1,(p_8)^0,(p_9)^1 \\\\ b_1 &:= (p_1)^1,(p_2)^1,(p_3)^1,(p_4)^1,(p_5)^1,(p_6)^1,(p_7)^0,(p_8)^0,(p_9)^0 \\\\ b_2 &:= (p_1)^1,(p_2)^1,(p_3)^1,(p_4)^0,(p_5)^0,(p_6)^0,(p_7)^1,(p_8)^1,(p_9)^1 \\\\ b_3 &:= (p_1)^0,(p_2)^0,(p_3)^0,(p_4)^1,(p_5)^1,(p_6)^1,(p_7)^1,(p_8)^1,(p_9)^1 \\\\ c_1 &:= (p_1)^1,(p_2)^1,(p_3)^0,(p_4)^1,(p_5)^1,(p_6)^0,(p_7)^1,(p_8)^1,(p_9)^0 \\\\ c_2 &:= (p_1)^1,(p_2)^0,(p_3)^1,(p_4)^1,(p_5)^0,(p_6)^1,(p_7)^1,(p_8)^0,(p_9)^1 \\\\ c_3 &:= (p_1)^0,(p_2)^1,(p_3)^1,(p_4)^0,(p_5)^1,(p_6)^1,(p_7)^0,(p_8)^1,(p_9)^1 \\\\ \\end{align} and the family is $$\\{\\{a_1,a_2,a_3\\},\\{b_1,b_2,b_3\\},\\{c_1,c_2,c_3\\}\\}$$ and $$x=\\prod_{i=1}^9 p_i.$$ The smallest family that can be constructed by this Latin square construction is thus $$\\{\\{440895, 43890, 43758\\},\\{30030, 29070, 969969\\},\\{149226, 46410, 122265\\}\\}$$ where $$x=223092870.$$ The Latin square property ensures that there will be two coinciding zeroes in the \"cross products\" (which, when this occurs, implies a $(p_i)^0$ factor). This construction will generalise to mutually orthogonal Latin squares (assuming the first column is identical in each Latin square). There will need to exist a family of $n-2$ mutually orthogonal Latin squares, which exist at least for infinitely many values of $n$. (This also works for $n=2$, but a set of $n-2$ mutually orthogonal Latin squares is rather trivial. [ronno's earlier construction is an example of this case.]) Here's an example with $n=4$ constructed using a pair of orthogonal Latin squares of order $4$: \\begin{align} \\{ & \\{7420738134810, 42597478693770, 705561031353570, 155186468939000213\\},\\\\ & \\{124952201298210, 263144725077234, 670103807644810, 1570864671608505\\},\\\\ & \\{230326723800030, 567883989007790, 762890549117235, 346859224918206\\},\\\\ & \\{317506244845530, 524202713204170, 629207214681045, 330502088912226\\}\\} \\end{align} where $$x=32589158477190044730.$$ - $\\{\\{6,35\\},\\{10,21\\}\\}$ with $x= 210$ seems to work. -", "8y - 1 \\end{align*} $$2(x- 2y)(x^2 + xy + y^2)$$ \\begin{align*} 2(x- 2y)(x^2 + xy + y^2) &= 2(x^3 + x^2y + xy^2 - 2x^2y - 2xy^2 - y^3) \\\\ &= 2(x^3 - x^2y - xy^2 - y^3) \\\\ &=2x^3 - 2x^2y - 2xy^2 - 2y^3 \\end{align*} $$3(a-3b)(a^2 + 3ab - b^2)$$ \\begin{align*} 3(a-3b)(a^2 + 3ab - b^2) &= 3(a^3 + 3a^2b - ab^2 - 3a^2b - 9ab^2 + 3b^3) \\\\ &=3(a^3 - 10ab^{2} + 3b^3) \\\\ &= 3a^3- 30ab^{2} + 9b^3 \\end{align*} $$(2a- b)(2a + b)(2a^2 - 3ab + b^2)$$ \\begin{align*} (2a- b)(2a + b)(2a^2 - 3ab + b^2) &= (4a^2 - b^2)(2a^2 - 3ab + b^2) \\\\ &= 8a^4 - 12a^3b + 4a^2b^2 - 2a^2b^2 + 3ab^3 - b^4 \\\\ &= 8a^4 - 12a^3b + 2a^2b^2 + 3ab^3 - b^4 \\end{align*} $$2(3x + y)(3x - y) - (3x -y)^2$$ \\begin{align*} 2(3x + y)(3x - y) - (3x -y)^2 &= 2(9x^2 - y^2) - 9x^2 + 6xy - y^2 \\\\ &= 18x^2 - 2y^2 - 9x^2 + 6xy - y^2 \\\\ &= 9x^2 + 6xy - 3y^2 \\end{align*} $$(x+y)(x-3y) + (2x - y)^2$$ \\begin{align*} (x+y)(x-3y) + (2x - y)^2 &= x^2 - 3xy +xy - 3y^2 + 4x^2 - 4xy + y^2 \\\\ &=5x^2 - 6xy - 2y^2 \\end{align*} $$\\left(\\dfrac{x}{3} - \\dfrac{3}{x}\\right)\\left(\\dfrac{x}{4} + \\dfrac{4}{x}\\right)$$ \\begin{align*} \\left(\\frac{x}{3} -", "$1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$. – egreg Jul 26 '13 at 20:11 I like to use the notation $$[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$$, and so on. \\begin{align} [2,3]_a + [2,5]_a &= [5,1]_a \\\\ [4,8]_a &= [5,1]_a \\\\ [4,8]_a &= [4, a+1] \\\\ 8 &= a+ 1 \\\\ a &= 7 \\end{align}", "the $a$th from the left, $b$th from the right ($a,b \\geq 0$, total number of elements is $a+b+1$). Then \\begin{align*} p(0,0) &= 1, \\\\ p(0,1) &= 1/2, \\\\ p(1,0) &= 1/2, \\\\ p(1,1) &= 1/2, \\\\ p(0,b+2) &= 1/2 + p(0,b)/4, \\\\ p(a+2,0) &= 1/2 + p(a,0)/4, \\\\ p(1,b+2) &= p(0,b+1)/2 + p(1,b)/4, \\\\ p(a+2,1) &= p(a+1,0)/2 + p(a,1)/4, \\\\ p(a+2,b+2) &= (p(a+2,b)+2p(a+1,b+1)+p(a,b+2))/4. \\end{align*} You could precompute the requisite values for $a+b+1 \\leq 2000$ and then your program will be blazingly fast. It turns out that for fixed $c$, $p(c,n-c)$ tends to a limit $p_c = 1/2(1 + (-1)^c/3^{c+1})$. To see why, note first that $p_0$ is the ability that if you and a friend alternate tossing fair coins, yours will come up HEAD first. There are many ways to see that $p_0 = 2/3$. Next, we have the formula $p_{c+1} = p_0 (1-p_c) + (1-p_0) p_c$ (exercise), from which the formula for $p_c$ follows by induction.", "a_{n-1}.$$ Since $a_0 = 0$, $a_1 = 0$, $a_2 = 0$, we have \\begin{align*} a_3 & = 2^0 + 0 + 0 + 0 = 1\\\\ a_4 & = 2^1 + 0 + 0 + 1 = 3\\\\ a_5 & = 2^2 + 0 + 1 + 3 = 8\\\\ a_6 & = 2^3 + 1 + 3 + 8 = 20\\\\ a_7 & = 2^4 + 3 + 8 + 20 = 47\\\\ a_8 & = 2^5 + 8 + 20 + 47 = 107. \\end{align*} For bit strings with at least 4 consecutive ones, the same reasoning leads to the recursion $$b_0 = b_1 = b_2 = b_3 = 0, \\quad b_n=2^{n-4}+b_{n-4}+ b_{n-3}+ b_{n-2}+ b_{n-1}, \\ n\\geq 4.$$ $$\\begin{array}{|l|r|l|} \\hline format & N & exceptions \\\\ \\hline 000***** & 32 & \\\\ 1000**** & 16 & \\\\ *1000*** & 16 & \\\\ **1000** & 16 & \\\\ ***1000* & 14 & 0001000* \\\\ ****1000 & 13 & 000*1000 , 10001000 \\\\ 1111**** & 13 & 1111000* , 11111000 \\\\ 01111*** & 7 & 01111000 \\\\ *01111** & 8 & \\\\ **01111* & 6 & 0001111* \\\\ ***01111 & 6 & *0001111 \\\\ \\hline \\end{array}$$ Total: $$147$$", "and $$Y$$ uncorrelated? 6. Are $$X$$ and $$Y$$ independent? Solution to Example 2.66 1. $$\\textrm{E}(Y)= (-1)(0.40) + (0)(0.20)+(1)(0.40) = 0$$. 2. No, as is true in general, $$\\textrm{E}(Y|X)$$ is not constant in this example. \\begin{align*} \\textrm{E}(Y|X=-1) & = (-1)(10/35) + (0)(0)+(1)(20/35) = 2/7\\\\ \\textrm{E}(Y|X=0) & = (-1)(1) + (0)(0)+(1)(0) = -1\\\\ \\textrm{E}(Y|X=1) & = (-1)(0) + (0)(20/35)+(1)(10/35) = 2/7\\\\ \\end{align*} 1.$$\\textrm{E}(X)= (-1)(0.40) + (0)(0.20)+(1)(0.40) = 0$$. 3. No, as is true in general, $$\\textrm{E}(X|Y)$$ is not constant in this example. \\begin{align*} \\textrm{E}(X|Y=-1) & = (-1)(1/4) + (0)(3/4)+(1)(0) = -1/4\\\\ \\textrm{E}(X|Y=0) & = (-1)(0) + (0)(0)+(1)(1) = 1\\\\ \\textrm{E}(X|Y=1) & = (-1)(25/40) + (0)(0)+(1)(15/40) = -1/4\\\\ \\end{align*} 4. $$\\textrm{E}(XY)= (-1)(-1)(0.1)+(1)(-1)(0.25) + (1)(-1)(0)+(1)(1)(0.15)+0=0$$ so $$\\textrm{Cov}(X, Y)=\\textrm{E}(XY)-\\textrm{E}(X)\\textrm{E}(Y)=0$$. Yes, $$X$$ and $$Y$$ are uncorrelated. 5. No, $$X$$ and $$Y$$ are not independent. For example $$\\textrm{E}(Y|X=-1)=2/7$$ but $$\\textrm{E}(Y|X=0)=-1$$. $$X$$ and $$Y$$ are independent if and only if $\\textrm{E}[g(X)h(Y)] = \\textrm{E}[g(X)]\\textrm{E}[h(Y)] \\qquad \\text{for all functions g, h}$", "8 + 1 & = 9 \\\\ 1 + 1 & = 2 \\\\ 2 + 1 & = 3 \\\\ 3 + 1 & = 4 \\\\ 4 + 1 & = 5 \\\\ 5 + 1 & = 6 \\\\ 6 + 1 & = 7 \\\\ 7 + 1 & = 8 \\\\ 8 + 1 & = 9 \\\\ 1 + 1 & = 2 \\\\ 2 + 1 & = 3 \\\\ 3 + 1 & = 4 \\\\ 4 + 1 & = 5 \\\\ 5 + 1 & = 6 \\\\ 6 + 1 & = 7 \\\\ 7 + 1 & = 8 \\\\ 8 + 1 & = 9 \\\\ 9 + 1 & = 10 \\end{align} \\end{document}", "\\\\ \\text{or} & \\\\ x + 6 & = 0 \\\\ x & = -6 \\\\ \\therefore x = \\frac{9}{5} & \\text{ or } x = -6 \\end{align*} $$4 z^{2} + 12 z + 8 = 0$$ \\begin{align*} 4 z^{2} + 12 z + 8 &= 0\\\\ z^{2} + 3 z + 2 &= 0 \\\\ \\left(z + 1\\right) \\left(z + 2\\right)&= 0 \\\\ z = -2 &\\text{ or } z = -1 \\end{align*} $$- b^{2} + 7 b - 12 = 0$$ \\begin{align*} - b^{2} + 7 b - 12 &= 0 \\\\ b^{2} - 7 b + 12 &= 0 \\\\ \\left(b - 4\\right) \\left(b - 3\\right) &= 0 \\\\ b = 3 &\\text{ or } b = 4 \\end{align*} $$- 3 a^{2} + 27 a - 54 = 0$$ \\begin{align*} - 3 a^{2} + 27 a - 54 &= 0 \\\\ a^{2} - 9 a + 18 &= 0 \\\\ \\left(a - 6\\right) \\left(a - 3\\right) &= 0 \\\\ a = 3 &\\text{ or } a = 6. \\end{align*} $$4y^{2} - 9 = 0$$ \\begin{align*} 4y^{2} - 9 & = 0 \\\\ (2y - 3)(2y + 3) & = 0 \\\\ 2y - 3 & = 0 \\\\ y & = \\frac{3}{2} \\\\ \\text{or} & \\\\ 2y + 3 & = 0 \\\\ y &", "x^2 - 4 &= 0 \\\\ (x-2)(x+2) &= 0 \\\\ \\therefore x = 2 &\\text{ or } x = -2 \\end{align*} $$a -3 = 2\\left(\\dfrac{6}{a} +1\\right)$$ Note $$a \\neq 0$$ \\begin{align*} a - 3 &= 2\\left(\\frac{6}{a} -1\\right) \\\\ a - 3 &= \\frac{12}{a} - 2 \\\\ a^2 -3a &= 12 - 2a \\\\ a^2 -a - 12 &= 0\\\\ (a - 4)(a +3)&= 0 \\\\ \\therefore a = 4 &\\text{ or } a = -3 \\end{align*} $$a - \\dfrac{6}{a} = -1$$ Note $$a \\neq 0$$ \\begin{align*} a - \\frac{6}{a} &= -1 \\\\ a^2 - 6 &= -a \\\\ a^2 +a - 6 &= 0 \\\\ (a - 2)(a +3) &= 0 \\\\ \\therefore a = 2 &\\text{ or } a = -3 \\end{align*} $$(a + 6)^2 - 5(a + 6) - 24 = 0$$ \\begin{align*} (a + 6)^2 - 5(a + 6) - 24 &= 0 \\\\ ((a + 6) - 8)((a + 6) + 3) &= 0 \\\\ (a - 2)(a + 9) &= 0 \\\\ \\therefore a = 2 &\\text{ or } a = -9 \\end{align*} $$a^4 - 4a^2 + 3 = 0$$ \\begin{align*} a^4 - 4a^2 + 3 &= 0 \\\\ (a^2 - 3)(a^2 - 1) &= 0 \\\\ (a - \\sqrt{3})(a + \\sqrt{3})(a - 1)(a+1) &= 0 \\\\ \\therefore b = \\pm \\sqrt{3} &\\text{", "in the original definition. For example, we can fully evaluate A(1, 2) in the following way: \\begin{align} A(1,2) & = A(0, A(1, 1)) \\\\ & = A(0, A(0, A(1, 0))) \\\\ & = A(0, A(0, A(0, 1))) \\\\ & = A(0, A(0, 2)) \\\\ & = A(0, 3) \\\\ & = 4. \\end{align} To demonstrate how A(4, 3)'s computation results in many steps and in a large number: \\begin{align} A(4, 3) & = A(3, A(4, 2)) \\\\ & = A(3, A(3, A(4, 1))) \\\\ & = A(3, A(3, A(3, A(4, 0)))) \\\\ & = A(3, A(3, A(3, A(3, 1)))) \\\\ & = A(3, A(3, A(3, A(2, A(3, 0))))) \\\\ & = A(3, A(3, A(3, A(2, A(2, 1))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, A(2, 0)))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, A(1, 1)))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, A(0, A(1, 0))))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, A(0, A(0, 1))))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, A(0, 2)))))) \\\\ & = A(3, A(3, A(3, A(2, A(1, 3))))) \\\\ & = A(3, A(3, A(3, A(2, A(0, A(1, 2)))))) \\\\ & = A(3, A(3, A(3, A(2, A(0, A(0, A(1, 1))))))) \\\\ & = A(3, A(3, A(3, A(2, A(0, A(0, A(0, A(1, 0)))))))) \\\\ & = A(3, A(3, A(3, A(2,", "typical also for the tetration function. ## Expansion To see how the Ackermann function grows so quickly, it helps to expand out some simple expressions using the rules in the original definition. For example, we can fully evaluate $A\\left(1, 2\\right)$ in the following way: begin\\left\\{align\\right\\} A(1,2) & = A(0, A(1, 1)) ` & = A(0, A(0, A(1, 0))) ` ` & = A(0, A(0, A(0, 1))) ` ` & = A(0, A(0, 2)) ` ` & = A(0, 3) ` ` & = 4.` end{align} To demonstrate how $A\\left(4, 3\\right)$'s computation results in many steps and in a large number: begin\\left\\{align\\right\\} A(4, 3) & = A(3, A(4, 2)) ` & = A(3, A(3, A(4, 1))) ` ` & = A(3, A(3, A(3, A(4, 0)))) ` ` & = A(3, A(3, A(3, A(3, 1)))) ` ` & = A(3, A(3, A(3, A(2, A(3, 0))))) ` ` & = A(3, A(3, A(3, A(2, A(2, 1))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(2, 0)))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(1, 1)))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, A(1, 0))))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, A(0, 1))))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, 2)))))) ` ` & = A(3, A(3, A(3, A(2," ]

[ "# 2008 AIME I Problems/Problem 13 ## Problem Let $$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$ Suppose that $$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$ There is a point $\\left(\\frac {a}{c},\\frac {b}{c}\\right)$ for which $p\\left(\\frac {a}{c},\\frac {b}{c}\\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. ## Solution ### Solution 1 \\begin{align*} p(0,0) &= a_0 \\\\ &= 0 \\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\\\ &= a_1 + a_3 + a_6 \\\\ &= 0 \\\\ p(-1,0) &= -a_1 + a_3 - a_6 \\\\ &= 0 \\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\\\", "# Difference between revisions of \"2008 AIME I Problems/Problem 13\" ## Problem Let $$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$ Suppose that $$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$ There is a point $\\left(\\frac {a}{c},\\frac {b}{c}\\right)$ for which $p\\left(\\frac {a}{c},\\frac {b}{c}\\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. ## Solution \\begin{align*} p(0,0) &= a_0 = 0\\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\\\ p(-1,0) &= -a_1 + a_3 - a_6 = 0\\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\\\ p(1,-1)", "# Difference between revisions of \"2008 AIME I Problems/Problem 13\" ## Problem Let $$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$ Suppose that $$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$ There is a point $\\left(\\frac {a}{c},\\frac {b}{c}\\right)$ for which $p\\left(\\frac {a}{c},\\frac {b}{c}\\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. ## Solution ### Solution 1 \\begin{align*} p(0,0) &= a_0 = 0\\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\\\ p(-1,0) &= -a_1 + a_3 - a_6 = 0\\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8", "You should then apply $P$ twice, and see that, indeed, $P(P(x))=P(x)$. Let $x\\in V$, and let $x_1\\in V_1$ and $x_2\\in V_2$ be the unique elements such that $x=x_1+x_2$. By definition of $V_1$ and $V_2$, $P(x_1)=0$ and $x_2-P(x_2)=0$, so \\begin{align*} P(x)&= P(x_1) + P(x_2)\\\\ &= 0 + x_2\\\\ P(P(x))&= P(0) + P(x_2)\\\\ &=x_2, \\end{align*} so we have that $$\\forall x\\in V, P^2(x)=P(x),$$ which proves indeed that $P^2=P$. • I take it you mean for $x_1$ to be in $V_1$, $x_2$ in $V_2$. – Gerry Myerson Oct 31 '11 at 11:35", "&= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\\\ &= -a_4 - a_7 + a_8 = 0\\end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, $p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0$ So $3a_1 + 3a_2 + 2a_4 = 0$. Now $p(x,y) = 0 + x a_1 + y a_2 + 0 + xy a_4 + 0 - x^3 a_1 - x^2 y a_4 + 0 - y^3 a_2$ $= x(1-x)(1+x) a_1 + y(1-y)(1+y) a_2 + xy (1-x) a_4$. In order for the above to be zero, we must have $x(1-x)(1+x) = y(1-y)(1+y)$ and $x(1-x)(1+x) = 1.5 xy (1-x).$ Canceling terms on the second equation gives us $1+x = 1.5 y \\Longrightarrow x = 1.5 y - 1$. Plugging that into the first equation and solving yields $x = 5/19, y = 16/19$, and $5+16+19 = \\boxed{040}$.", "= 0.2 (0.8) + 0 (0.2) = 0.16\\\\ P(2,8) &= P(1,8) (1-p_2) + P(1,8-8) p_2 = 0 (0.8) + 0.8 (0.2) = 0.16\\\\ P(2,13) &= P(1,13)(1-p_2) + P(1,13-8) p_2 = 0 (0.8) + 0.2 (0.2) = 0.04\\\\ P(2,j) &= 0, j \\neq \\{0,5,8,13\\} \\end{align} The third recursion pass is as follows: \\begin{align} P(3,0) &= P(2,0)(1-p_3) = 0.512\\\\ P(3,4) &= P(2,4)(1-p_3) + P(2,4-4) = 0(0.8) + 0.64(0.2) = 0.128\\\\ P(3,5) &= P(2,5)(1-p_3) + P(2,5-4) p_3 = 0.16 (0.8) + 0 (0.2) = 0.128\\\\ P(3,8) &= P(2,8) (1-p_3) + P(2,8-4) p_3 = 0.16 (0.8) + 0 (0.2) = 0.128\\\\ P(3,9) &= P(2,9) (1-p_3) + P(2,9-4) p_3 = 0 (0.8) + 0.16 (0.2) = 0.032\\\\ P(3,12) &= P(2,12) (1-p_3) + P(2,12-4) p_3 = 0 (0.8) + 0.16 (0.2) = 0.032\\\\ P(3,13) &= P(2,13) (1-p_3) + P(2,13-4) p_3 = 0.04 (0.8) + 0 (0.2) = 0.032\\\\ P(3,17) &= P(2,17) (1-p_3) + P(2,17-4) p_3 = 0 (0.8) + 0.04 (0.2) = 0.008\\\\ P(3,j) &= 0, j \\neq \\{0,4,5,8,9,12,13,17\\} \\end{align} Note that the same computation work even when the outcomes are not of equal probability. Let’s do one more example. #ONE FINAL EXAMPLE #----------MAIN CALLING SEGMENT------------------ w = c(5,2,4,2,8,1,9) p = array(0.2,length(w)) res = asbrec(w,p) print(res) ## [,1] [,2] ## [1,] 0 0.2097152 ## [2,] 1 0.0524288 ## [3,] 2 0.1048576 ## [4,] 3", "cases the intersection point is given by $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} =\\begin{pmatrix} 1&p&0\\\\1&1&-p\\\\2&0&-p \\end{pmatrix}^{-1} \\begin{pmatrix}p\\\\1-2p-p^2\\\\-p\\end{pmatrix} =\\begin{pmatrix}p^2\\\\1-p\\\\2p+1\\end{pmatrix}.$$ since $$\\operatorname{adj}\\begin{pmatrix} 1&p&0\\\\1&1&-p\\\\2&0&-p \\end{pmatrix}=\\begin{pmatrix} -p&p^2&-p^2\\\\-p&-p&p\\\\-2&2p&1-p \\end{pmatrix}$$ From (1), $$x=p-py$$, so \\begin{align*} x &= p-py \\tag{1'}\\\\ (1-p) y - pz &= 1-3p-p^2 \\tag{2'}\\\\ 2py + pz &= 3p\\tag{3'} \\end{align*} If $$p=0$$ (3') tells us nothing and we have intersection of two planes (1') and (2') which won't give us unique solution. So assuming $$p\\neq 0$$, we can divide (3') by $$p$$ and get $$z=3-2y$$. Substituting, \\begin{align*} x &= p-py \\tag{1'}\\\\ (1+p) y &= 1-p^2 \\tag{2''}\\\\ z &= 3-2y\\tag{3''} \\end{align*} If $$p=-1$$, (2'') tells us nothing and again we don't have unique solution. So $$p\\neq -1$$, and we have $$y=1-p, z=3-2y=2p+1, x=p-py=p^2.$$", "have \\begin{align} f_3(0) = [f(0)]^4. \\end{align} Next, using both 1 and 2, we have \\begin{align} f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \\ \\ (\\ast) \\end{align} Plugging $0$ into $(\\ast)$ yields \\begin{align} f_3(0) = f(0)+[f(0)]^2. \\end{align} Combining everything yields \\begin{align} [f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \\ \\ \\Rightarrow \\ \\ f(0)[f(0)^3-f(0)-1] =0. \\end{align} Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root. Moreover, observe \\begin{align} f_4(0) = [f(0)]^4 \\end{align} and \\begin{align} f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2 \\end{align} which means \\begin{align} [f(0)]^4-2[f(0)]^2 = 0. \\end{align} Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$. In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$. $$\\mrm{f}\\pars{x^{2} + \\mrm{f}\\pars{y}} =y + \\bracks{\\mrm{f}\\pars{x}}^{2} \\label{1}\\tag{1}$$ Deriving both members of \\eqref{1} respect of $\\ds{x}$ and $\\ds{y}$ yield, respectively: $$\\left\\{\\begin{array}{rcl} \\ds{\\mrm{f}'\\pars{x^{2} + \\mrm{f}\\pars{y}}\\pars{2x}} & \\ds{=} & \\ds{2\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x}} \\\\[2mm] \\ds{\\mrm{f}'\\pars{x^{2} + \\mrm{f}\\pars{y}}\\mrm{f}'\\pars{y}} & \\ds{=} & \\ds{1} \\end{array}\\right. \\label{2}\\tag{2}$$ By comparing both expressions in \\eqref{2}: $${\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x} \\over x} = {1 \\over \\mrm{f}'\\pars{y}} = \\alpha\\,, \\qquad \\pars{~\\alpha\\ \\mbox{is independent of}\\ x\\ \\mbox{and}\\ y~} \\label{2.a}\\tag{2.a}$$ The second one is quite trivial: $${1 \\over \\mrm{f}'\\pars{y}} = \\alpha \\implies \\mrm{f}\\pars{y} = {1 \\over \\alpha}\\, y + \\beta\\,,\\qquad \\pars{~\\beta\\ \\mbox{is a constant}~}\\label{3}\\tag{3}$$ \\eqref{2.a} and \\eqref{3} yield: \\begin{align} &{\\mrm{f}\\pars{x}\\mrm{f}'\\pars{x} \\over x} = \\alpha", "of all the $p_i$. In the same way, \\begin{align*} P(Y > X) &= \\sum_{i=1}^n P(X = i, Y > i) \\\\ &= \\sum_{i=1}^n P(X = i)P(Y > i) \\\\ &= \\sum_{i=1}^{n-1} P(X = i)P(Y > i) \\end{align*} because $P(Y > n) = 0$. Now for each $i < n$, $P(Y > i) = P(X > i) = \\sum_{j=i+1}^n p_j$. Call this sum $b_i$ for \"bigger than $i$\". Then \\begin{align*} P(Y > X) &= \\sum_{i=1}^{n-1} P(X = i)P(Y > i) \\\\ &= \\sum_{i=1}^{n-1} p_ib_i \\end{align*} This is also a straightforward calculation if you know all the $p_i$. For $n=4$ it boils down to $$p_1 \\cdot(p_2 + p_3 + p_4) ~~ + ~~ p_2 \\cdot (p_3 + p_4) ~~ + ~~ p_3 \\cdot p_4$$", "typical also for the tetration function. ## Expansion To see how the Ackermann function grows so quickly, it helps to expand out some simple expressions using the rules in the original definition. For example, we can fully evaluate $A\\left(1, 2\\right)$ in the following way: begin\\left\\{align\\right\\} A(1,2) & = A(0, A(1, 1)) ` & = A(0, A(0, A(1, 0))) ` ` & = A(0, A(0, A(0, 1))) ` ` & = A(0, A(0, 2)) ` ` & = A(0, 3) ` ` & = 4.` end{align} To demonstrate how $A\\left(4, 3\\right)$'s computation results in many steps and in a large number: begin\\left\\{align\\right\\} A(4, 3) & = A(3, A(4, 2)) ` & = A(3, A(3, A(4, 1))) ` ` & = A(3, A(3, A(3, A(4, 0)))) ` ` & = A(3, A(3, A(3, A(3, 1)))) ` ` & = A(3, A(3, A(3, A(2, A(3, 0))))) ` ` & = A(3, A(3, A(3, A(2, A(2, 1))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(2, 0)))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(1, 1)))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, A(1, 0))))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, A(0, 1))))))) ` ` & = A(3, A(3, A(3, A(2, A(1, A(0, 2)))))) ` ` & = A(3, A(3, A(3, A(2,", "+ 5) 8 6 4 2 (x - 8 x + 19 x - 12 x + 1) 3 2 3 2 20 (x - 1) (x + 1) (x - x - 2 x + 1) (x + x - 2 x - 1) 6 5 4 3 2 (x - x - 6 x + 6 x + 8 x - 8 x + 1) 6 5 4 3 2 (x + x - 6 x - 6 x + 8 x + 8 x + 1) Quote by Dodo I think this may be the case. Since\\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) \\end{align*}if 1 (or -1, for that matter) is a root of P(n-3) then it is a root of P(n). And since P(2) = x^2 - 1 and has roots 1 and -1, then this is true for n=2,5,8,... For similar reasons, 0 is a root (as mentioned) for n=1,3,5,7,... to extend further, \\begin{align*} P(n) &= xP(n-1) - P(n-2) \\\\ &= x\\left( xP(n-2) - P(n-3) \\right) - P(n-2) \\\\ &= (x^2-1)P(n-2) - xP(n-3) &= (x^3-x-1)P(n-3)-(x^2-1)P(n-2) \\end{align*} and we could continue to extend this procedure to obtain as many repeated roots as we like. Not yet there, but here is another", "$P=(a_n)_n$ and $Q=(b_n)_n$ then $PQ=(c_n)_n$ is a polynomial with \\begin{align*} c_n=\\sum_{k=0}^n a_k b_{n-k}.\\end{align*} Proposition: $(\\mathbb{K},+,\\cdot)$ is a commutative ring. ## A selection of exercises on polynomials Exercise: Determine the greatest common divisor (gcd) and a Bezout relation between the polynomials $P=X^3+X^2+X-3$ and $Q=X^2-3X+2$. Solution: Here we shall apply Euclid’s algorithm: we select $P_0=P$ and $P_1=Q$. Moreover, we put $U_0=1,\\,U_1=0$ and $V_0=0,\\,V_1=1$ in order to have $PU_0+QV_0=P_0$ and $PU_1+QV_1=P_1$. The Euclidean division of $P_0$ by $P_1$ is \\begin{align*} P_0=P_1(X+4)+(11 X-11). \\end{align*} We set $P_2=11 X-11$. Then \\begin{align*} P_2&=P_0-P_1(X+4)\\cr &= PU_0+Q V_0-(PU_1+QV_1)(X+4)\\cr &= P(U_0-(X+4)U_1)+Q(V_0-(X+4)V_1)\\cr &= P U_2+Q V_2, \\end{align*} where \\begin{align*} U_2=U_0-(X+4)U_1=1,\\quad V_2=V_0-(X+4)V_1=-X-4. \\end{align*} The Euclidean division of $P_1$ by $P_2$ is \\begin{align*} P_1=P_2 \\left(\\frac{1}{11}X-\\frac{2}{11}\\right)+0 \\end{align*} The GCD of $P$ and $Q$ is \\begin{align*} \\frac{1}{11}P_2=X-1. \\end{align*} The Bezout relation is then \\begin{align*} \\frac{1}{11} P-\\frac{1}{11}(X+4) Q=X-1. \\end{align*} Exercise: Does the polynomial $P=X^4+1$ is irreducible in $\\mathbb{C}[X]$ ? in $\\mathbb{R}[X]$ ? in $\\mathbb{Q}[X]$ ? Solution: It is not irreducible neither in $\\mathbb{C}[X]$ nor in $\\mathbb{R}[X]$ as it is not either of degree 1 or of degree 2. Assume that $P$ is reducible in $\\mathbb{Q}[X]$. Remark that $P$ does not have rational roots “and also real roots”, $P$ can not admit a divisor of degree $1$. Then $P$ is the product of two polynomials of degree 2. We then can assume that \\begin{align*}", "and $$\\ P\\left(\\left.\\xi_j=0\\,\\right|N=n\\right)=1-p\\$$—that is, if $$\\ i_1,i_2, \\dots, i_n\\$$ is a sequence of ones and zeros containing $$\\ m\\$$ ones and $$\\ n-m\\$$ zeros, then \\begin{align} P\\left(\\xi_1=i_1, \\xi_2=i_2, \\dots, \\xi_n=i_n |\\,N=n\\right)&=\\prod_{j=1}^nP\\left(\\xi_j=i_j|\\,N=n\\right)\\\\ &=p^m(1-p)^{n-m} \\end{align} Therefore \\begin{align} P \\left(\\xi_1+ \\xi_2+ \\dots +\\xi_n=m \\,|\\,N=n\\right)&=\\sum_{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n:\\\\ i_1+i_2+ \\dots+i_n=m} p^m(1-p)^{n-m}\\\\ &={n\\choose m} p^m(1-p)^{n-m}\\ , \\end{align} because $$\\ \\left|\\left\\{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n\\,|\\, i_1+i_2+ \\dots+i_n=m\\right\\}\\right|= {n\\choose m}\\$$.", "and c, but I still don't know the value of $Q_{4}(x)$ and I can't find the polynomial without knowing the value of $Q_{4}(x)$. What should I do? – cristid9 Jan 13 '15 at 20:23 • You don't need $Q_4(x)$ since you're looking only for the reminder. – kingW3 Jan 13 '15 at 20:26 It's a problem about Lagrange's interpolation polynomials. The hypothesis means that $P(1)=2$, $P(-1)=6$, P(2)=3$. The simplest method uses concepts from linear algebra. You first solve 3 easier problems: find polynomials$p, q, rsuch that \\begin{alignat*}{3} &p(1)=1&&q(1)=0&&r(1)=0 \\\\ &p(-1)=0&\\qquad&q(-1)=1&\\qquad&r(-1)=0\\\\ &p(2)=0&&q(2)=0&&r(2)=1 \\end{alignat*} For instance,\\,\\,p(x)=\\dfrac{(x+1)(x-2)}{(1+1)(1-2)}= -\\dfrac12(x^2-x-2)$. Once you've computed$p(x), q(x), r(x)$, just take$P(x)=2p(x)+6q(x)+3r(x)$. Hint$\\ f(x) = 2\\! +\\! (x\\!-\\!1)\\left[a\\! +\\! (x\\!-\\!2)(b\\!+\\! (x\\!+\\!1)g(x))\\right],\\,f(2) = 3\\,\\Rightarrow\\, a=\\_\\_,\\ f(-1) = 6\\,\\Rightarrow\\, b = \\_\\_\\$", "0 1 0 0 0 0 1 1 1 1 1 0 1 0 -1 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 1 -1 0 0 Since the first four vertices $v_0, v_1, v_2\\,$ and $v_3\\,$ are in $V_0\\,$, we have ϕ0(vi) = 1 and ϕ1(vi) = 0, hence the corresponding points of the Gorenstein cone take the form (1,0,vi) for $i=0,\\dots,3$. The next four vertices $v_4, v_5, v_6\\,$ and $v_7\\,$ are in $V_1\\,$, we have $\\phi_0(v_i) = 0\\,$ and $\\phi_1(v_i) = 1\\,$, hence the corresponding points of the Gorenstein cone take the form $(0,1,v_i)\\,$ for $i=4,\\dots,7$. Finally, the origin always belongs to every part of the nef partition, hence it appears as often as the codimension which here is $r=2\\,$. So $p_8 = 0\\,$ and $p_9 = 0\\,$. Once with $\\phi_0(p_8) = 0\\,$ and $\\phi_1(p_8) = 1\\,$ and once with $\\phi_0(p_9) = 1\\,$ and $\\phi_1(p_9) = 0\\,$. ### -d* The option -d*, where * is an integer m=0,1,2, returns the points of the Gorenstein cones $C \\subset \\tilde M_\\mathbb{R}$ associated to the nef partitions of length r of the polytope $\\nabla^*", "a_{n-1}.$$ Since $a_0 = 0$, $a_1 = 0$, $a_2 = 0$, we have \\begin{align*} a_3 & = 2^0 + 0 + 0 + 0 = 1\\\\ a_4 & = 2^1 + 0 + 0 + 1 = 3\\\\ a_5 & = 2^2 + 0 + 1 + 3 = 8\\\\ a_6 & = 2^3 + 1 + 3 + 8 = 20\\\\ a_7 & = 2^4 + 3 + 8 + 20 = 47\\\\ a_8 & = 2^5 + 8 + 20 + 47 = 107. \\end{align*} For bit strings with at least 4 consecutive ones, the same reasoning leads to the recursion $$b_0 = b_1 = b_2 = b_3 = 0, \\quad b_n=2^{n-4}+b_{n-4}+ b_{n-3}+ b_{n-2}+ b_{n-1}, \\ n\\geq 4.$$ $$\\begin{array}{|l|r|l|} \\hline format & N & exceptions \\\\ \\hline 000***** & 32 & \\\\ 1000**** & 16 & \\\\ *1000*** & 16 & \\\\ **1000** & 16 & \\\\ ***1000* & 14 & 0001000* \\\\ ****1000 & 13 & 000*1000 , 10001000 \\\\ 1111**** & 13 & 1111000* , 11111000 \\\\ 01111*** & 7 & 01111000 \\\\ *01111** & 8 & \\\\ **01111* & 6 & 0001111* \\\\ ***01111 & 6 & *0001111 \\\\ \\hline \\end{array}$$ Total: $$147$$", "Mar 19 '13 at 4:29 Such a coding is impossible for $n>1$. Let $a = (1,0,\\dots,0)$ and let $b = (0,1,0,\\dots,0)$. When I write $m \\cdot a$ for a natural number $m$ and a vector $a$, I mean $a + \\dots + a$ (added $m$ times). Now $a$ is coded by a natural number $m_a = f(a)$ and $b$ is coded by a natural number $m_b = f(b)$. But: \\begin{align} f(f(b),0,\\dots,0) &= f(m_b\\cdot a)\\\\ &= m_b\\times f(a)\\\\ &= m_a\\times f(b)\\\\ &= f(m_a\\cdot b)\\\\ &= f(0,f(a),0,\\dots,0) \\end{align} ...which shows that the coding is not injective, since $(0,f(a),0,\\dots,0)\\neq (f(b),0,\\dots,0)$, unless $f(a) = f(b) = 0$ but then again we have that the coding is not injective, since $a\\neq b$. Edited: Maybe an explicit example will help you see what's going on. Let's take $n = 2$ and say for example that $f(1,0) = 3$, $f(0,1) = 5$. Then: \\begin{align} f(5,0) &= f(1 + 1 + 1 + 1 + 1, 0 + 0 + 0 + 0 + 0)\\\\ & = f(1,0) + f(1,0) + f(1,0) + f(1,0) + f(1,0)\\\\ & = 3 + 3 + 3 + 3 + 3\\\\ & = 15\\\\ & = 5 + 5 + 5 \\\\ &= f(0,1) + f(0,1) + f(0,1) \\\\ &= f(0+0+0,1+1+1) \\\\ &= f(0,3) \\end{align} If you compare these steps", "# Do the coefficients of these polynomials alternate in sign? Define polynomials $$p_i(x)$$ by the recurrence \\begin{align} p_0(x)&=0 \\\\ p_1(x)&=1 \\\\ p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\\\ p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\\\ \\end{align} The first few are given by \\begin{align} p_0(x)&=0\\\\ p_1(x)&=1\\\\ p_2(x)&=1\\\\ p_3(x)&=x-1\\\\ p_4(x)&=x-2\\\\ p_5(x)&=x^2-3x+1\\\\ p_6(x)&=x^2-4x+3\\\\ p_7(x)&=x^3-5x^2+6x-1\\\\ p_8(x)&=x^3-6x^2+10x-4 \\end{align} It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts? Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ... $$\\begin{eqnarray*} p_{2i}(x) = \\sum_{j=0}^{i-1} \\binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\\\ p_{2i-1}(x) = \\sum_{j=0}^{i-1} \\binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\\\ \\end{eqnarray*}$$ Proof of the even formula ... $$\\begin{eqnarray*} &p_{2i-1}(x) -p_{2i-2}(x) =\\\\ & x^{i-1}+ \\sum_{j=0}^{i-2} \\binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \\sum_{j=0}^{i-1} \\binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\\\ &= x^{i-1}+ \\sum_{j=0}^{i-2} \\binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2} = p_{2i}(x). \\end{eqnarray*}$$ And the answer to the question in the title is $$\\color{red}{\\text{yes}}$$. I reckon that $$p_{2n-1}(x)=\\sum_{k=0}^{n-1}(-1)^k \\binom{n+1-k}k x^{n-1-k}$$ and $$p_{2n}(x)=\\sum_{k=0}^{n-1}(-1)^k \\binom{n+2-k}k x^{n-1-k}$$ and that these can be proved by induction.", "$\\label{eq:7.6.27} x^2y''-x(5-x)y'+(9-4x)y=0.$ Give explicit formulas for the coefficients in the solutions. ###### Solution For Equation \\ref{eq:7.6.27} the polynomials defined in Theorem 7.7.1 are \\begin{align*} p_0(r) &= r(r-1)-5r+9 &= (r-3)^2,\\\\[4pt] p_1(r) &= r-4,\\\\[4pt] p_2(r) &= 0. \\end{align*} \\nonumber Since $$r_1=3$$ is a repeated zero of the indicial polynomial $$p_0$$, Theorem 7.7.2 implies that $\\label{eq:7.6.28} y_1=x^3\\sum_{n=0}^\\infty a_n(3)x^n$ and $\\label{eq:7.6.29} y_2=y_1\\ln x+x^3\\sum_{n=1}^\\infty a_n'(3)x^n$ are linearly independent Frobenius solutions of Equation \\ref{eq:7.6.27}. To find the coefficients in Equation \\ref{eq:7.6.28} we use the recurrence formulas \\begin{align*} a_0(r) &= 1,\\\\ a_n(r) &= -{p_1(n+r-1)\\over p_0(n+r)}a_{n-1}(r)\\\\[10pt] &= -{n+r-5\\over(n+r-3)^2}a_{n-1}(r),\\quad n\\ge1. \\end{align*} \\nonumber We leave it to you to show that $\\label{eq:7.6.30} a_n(r)=(-1)^n\\prod_{j=1}^n{j+r-5\\over(j+r-3)^2}.$ Setting $$r=3$$ here yields $a_n(3)=(-1)^n\\prod_{j=1}^n{j-2\\over j^2},\\nonumber$ so $$a_1(3)=1$$ and $$a_n(3)=0$$ if $$n\\ge2$$. Substituting these coefficients into Equation \\ref{eq:7.6.28} yields $y_1=x^3(1+x).\\nonumber$ To obtain $$y_2$$ in Equation \\ref{eq:7.6.29} we must compute $$a_n'(3)$$ for $$n=1$$, $$2$$, …. Let’s first try logarithmic differentiation. From Equation \\ref{eq:7.6.30}, $|a_n(r)|=\\prod_{j=1}^n{|j+r-5|\\over|j+r-3|^2},\\quad n\\ge1,\\nonumber$ so $\\ln |a_n(r)|=\\sum^n_{j=1} \\left(\\ln |j+r-5|-2\\ln|j+r-3|\\right).\\nonumber$ Differentiating with respect to $$r$$ yields ${a'_n(r)\\over a_n(r)}=\\sum^n_{j=1} \\left({1\\over j+r-5}-{2\\over j+r-3}\\right).\\nonumber$ Therefore $\\label{eq:7.6.31} a'_n(r)=a_n(r) \\sum^n_{j=1} \\left({1\\over j+r-5}-{2\\over j+r-3}\\right).$ However, we can’t simply set $$r=3$$ here if $$n\\ge2$$, since the bracketed expression in the sum corresponding to $$j=2$$ contains the term $$1/(r-3)$$. In fact, since $$a_n(3)=0$$ for $$n\\ge2$$, the formula Equation \\ref{eq:7.6.31} for $$a_n'(r)$$ is actually an indeterminate form at $$r=3$$. We overcome this difficulty as follows.", "the $a$th from the left, $b$th from the right ($a,b \\geq 0$, total number of elements is $a+b+1$). Then \\begin{align*} p(0,0) &= 1, \\\\ p(0,1) &= 1/2, \\\\ p(1,0) &= 1/2, \\\\ p(1,1) &= 1/2, \\\\ p(0,b+2) &= 1/2 + p(0,b)/4, \\\\ p(a+2,0) &= 1/2 + p(a,0)/4, \\\\ p(1,b+2) &= p(0,b+1)/2 + p(1,b)/4, \\\\ p(a+2,1) &= p(a+1,0)/2 + p(a,1)/4, \\\\ p(a+2,b+2) &= (p(a+2,b)+2p(a+1,b+1)+p(a,b+2))/4. \\end{align*} You could precompute the requisite values for $a+b+1 \\leq 2000$ and then your program will be blazingly fast. It turns out that for fixed $c$, $p(c,n-c)$ tends to a limit $p_c = 1/2(1 + (-1)^c/3^{c+1})$. To see why, note first that $p_0$ is the ability that if you and a friend alternate tossing fair coins, yours will come up HEAD first. There are many ways to see that $p_0 = 2/3$. Next, we have the formula $p_{c+1} = p_0 (1-p_c) + (1-p_0) p_c$ (exercise), from which the formula for $p_c$ follows by induction." ]

[ "Suppose that $a_{13} = 2016$. Find $a_1$. Find $a$ and $b$ so that $(x-1)^2$ divides $ax^4 + bx^3+1$. Find all pairs of real numbers $a, b$, such that the polynomial $$p(x)=(a+b)x^5 + abx^2 +1$$ is divisible by $x^2 - 3x+2$. Let $n$ be a positive integer, and for $1\\le k\\le n$, let $S_k$ be the sum of the products of $1, \\frac{1}{2}, \\cdots, \\frac{1}{n}$, taken $k$ a time ($k^{th}$ elementary symmetric polynomial). Find $S_1 + S_2 + \\cdots +S_n$.", "Another is to use the fact that the geometric series $a+ at+ ar^2+ ar^3+ \\cdot\\cdot\\cdot+ ar^n+ \\cdot\\cdot\\cdot$ sums to $\\frac{a}{1- r}$ for -1< r< 1. Take a= 1 and r= -x to get $\\frac{1}{1+ x}= 1- x+ x^2- x^3+ \\cdot\\cdot\\cdot+ (-1)^nx^n+ \\cdot\\cdot\\cdot$. Then integrate term by term.", "Find the sum of the next series: $a + ar+ ar^2+ \\cdots + ar^n+ \\cdots$ where $a>0$ and $r >0$. Suppose that $r \\geq 1$. The sequence $\\{ a r^n \\}$ is increasing , so $\\lim a r^n$ does not exists, therefore this series does not converge (diverges). Suppose $r< 1$ then $s_n = a + ar+ ar^2+ \\cdots + ar^{n-1}$ and $rs_n = ar+ ar^2+ \\cdots + ar^{n-1}+ar^{n}$ by subtraction, $s_n = \\dfrac{a(1-r^n)}{1-r}$ Since $\\lim r^n= 0$ then $\\lim s_n= a/1-r$. We have thus not only shown that the series converges but have found its sum.", "22 '12 at 5:02 Gelfand's formula shows that if $\\rho(A) < 1$, then for some $n$, $\\|A^n\\| < 1$. One can then rewrite the series as $(1 + A + \\cdots + A^{n-1}) \\sum_{i=0}^\\infty A^{ni}$, which surely does converge in norm. -", "# Sample quiz on geometric series Main home here. 1. Let $a$ and $r$ have their usual meanings. The sum of the first $n$ terms of a geometric series can be given by $\\cdots$? $S_{n}=\\frac{a(r^n-1)}{1-r}$ $S_{n}=\\frac{a(r^{n-1})}{r-1}$ $S_{n}=\\frac{a(r^n-1)}{r-1}$ $S_{n}=\\frac{a(r^n+1)}{r-1}$ 2. What is the sum of the first $10$ terms of the geometric series $1+2+4+8+\\cdots$? $1023$ $1024$ $2046$ $512$ 3. Find the sum of the geometric series $2+6+18+54+\\cdots+4374$ $2180$ $2187$ $6561$ $6560$ 4. A geometric series has $a=3$ and $r=4$. How many terms must be taken so that $S_{n}=16,777,216$? $13$ $12$ $11$ $10$ 5. Find the sum of the series $2+(-4)+8+(-16)+\\cdots+2048$ $4094$ $1366$ $4096$ $1364$ 6. What is the sum of the first $12$ terms of the series $2+1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots$? $\\frac{4095}{1024}$ $\\frac{4095}{4096}$ $\\frac{4096}{4095}$ $\\frac{1024}{4095}$ 7. If a geometric series is such that $r-a=1$, then the sum of the first $n$ terms is $\\cdots$? $S_{n}=1-r^n$ $S_{n}=a(r^n-1)$ $S_{n}=r^n-1$ $S_{n}=a(1-r^n)$ 8. Find the sum of the first $8$ terms of the series $\\frac{1}{3}+\\frac{2}{9}+\\frac{4}{27}+\\frac{8}{81}+\\cdots$ $\\frac{6561}{6305}$ $\\frac{2059}{2187}$ $\\frac{2187}{2059}$ $\\frac{6305}{6561}$ 9. Find the sum of the first $9$ terms of the series $3+(-3)+3+(-3)+3+\\cdots$ $3$ $6$ $0$ $-3$ 10. What is the common ratio of a geometric series in which $a=2$ and $S_{10}=29524$? $-2$ $-3$ $3$ $2$", "R5602 Can we sum $1^2-2^2+3^2-4^2+\\dotsb+ (2n-1)^2-(2n)^2$? R9819 Can we sum from $1000$ to $2000$ excluding multiples of 5? R7424 Can we sum the first $2n$ terms of $1,1,2,\\frac{1}{2},4,\\frac{1}{4},8,\\frac{1}{8},..$? R6257 Can we sum the first $2n$ terms of $1^2-3^2+5^2-7^2+\\cdots$? R5416 Find an expression for the sum of $r^2$ R6143 Given $S_n = S_{3n}$, can we express $a$ in terms of $n$? R9613 Given the sequence $(x_n)$, what is $\\sum_{k=0}^\\infty 1/x_k$? R8248 How are the $p$th, $q$th and $r$th terms of an AP related? R6137 How are the sums of $n$, $2n$ and $3n$ terms connected here? R8493 How are these recursively defined sequences related? R6555 How do we add the odd-numbered terms of a geometric series? R8617 If $27$, $x$, $y$ are in GP, with sum $21$, what are $x$ and $y$? R7258 If $S_n = 6 - 2^{n+1}/3^{n-1}$, can we show that we have a GP? R8163 If $u_n = a + bn + c2^n$, what's the sum of the first $n$ terms? R6153 If $x_1 = (x_0 + 2)/(x_0 + 1)$, can we show $\\sqrt{2}$ lies between $x_0$ and $x_1$? R8404 If $x_2 = 1/(1 - x_1), x_3 = 1/(1 - x_2)$, can we show $x_1x_2x_3+1=0$? R5995 If AP terms $a_4,a_8, a_{16}$ are in GP, are $a_3, a_6, a_{12}$ in GP too? R5345 If a GP has", "# Question #931d4 Apr 30, 2017 We have to use that the geometric series is of the form: $a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \\ldots$ and that we know that the whole sum is $13.5$ and the sum of the first three terms $a + a r + a {r}^{2} = 13$ #### Explanation: So we have: $a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \\ldots = 13.5$ and $a + a r + a {r}^{2} = 13$ Subtracting we have: $\\left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \\ldots\\right) - \\left(a + a r + a {r}^{2}\\right) = 0.5$, that is: $a {r}^{3} + a {r}^{4} + a {r}^{5} + a {r}^{6} + a {r}^{7} + \\ldots = 0.5$ Hence, taking $r$ as a common factor: ${r}^{3} \\cdot \\left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \\ldots\\right) = 0.5$ But we know that that the expression between brackets: $a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \\ldots = 13.5$ so we have: ${r}^{3} \\cdot 13.5 = 0.5$, that is: ${r}^{3} = \\frac{0.5}{13.5}$, that is: ${r}^{3} = \\frac{\\frac{1}{2}}{\\frac{27}{2}} = \\frac{1}{27}$, and then $r = \\frac{1}{3}$ But", "from what you wrote. That will give you the sum from n=1 to ∞. Alternatively, you can recognize that for |r| < 1 you have $r + r^2 + r^3 + \\cdots = r(1 +r + r^2 + \\cdots),$ and use your geometric sum result for the quantity in brackets. RGV", "+0 # Compute the sum of the geometric series \\$-1 + 2 - 4 + 8 - \\cdots + 512\\$. 0 276 1 Compute the sum of the geometric series \\$-1 + 2 - 4 + 8 - \\cdots + 512\\$. Oct 7, 2017 #1 0 Sum =F x [1 - R^N] / [1 - R], where F=First term, R=Common ratio, N=Number of terms Sum =-1 x [1 - (-2^10] / [1 - (-2)] Sum =-1 x [1 - 1,024] / [3] Sum =-1 x[ -1023] / 3 Sum = 1,023/3 Sum = 341 Oct 7, 2017", "# find the sum of the series If $a_1, a_2, \\ldots, a_n$ are in arithmetic progression whose common difference is $d$,then find the sum: $$\\sin(d) \\cdot \\left(\\csc(a_1)\\csc (a_2)+\\csc(a_2)\\csc (a_3)+\\ldots+\\csc(a_{n-1})\\csc(a_n) \\right)$$ $$\\sin d\\cdot(\\csc a_r\\cdot\\csc a_{r+1})=\\frac{\\sin(a_{r+1}- a _r)}{\\sin a_r\\sin a_{r+1}}=\\cot a_r-\\cot a_{r+1}$$ where $1\\le r\\le n-1$", "How to find the sum of sequence $1+4+4^2+\\cdots+4^{X+Y}$? I see the following sequence and it's: $$h=1+4+4^2+\\cdots+4^{X+Y}=\\frac{4^{X+Y+1}-1}{4-1}$$ how we get this sequence? I know this is a primary question but I confused :) - Do you mean, how does one prove the formula for the sum of terms? – Travis Aug 24 '14 at 16:55 Dear @Travis, I means how get this? – Mouna Mokhiab Aug 24 '14 at 16:56 3 Answers This is an example of a geometric series. Let's say we try to sum: $$S=1+r+r^2+r^3+\\dotsb+r^n$$ In your example, $r=4$, and $n=X+Y$. There is a well known \"trick\" for solving this. Multiply by $r$: $$Sr=r+r^2+r^3+\\dotsb+r^n+r^{n+1}$$ Notice how this is the same thing as $S$, except without the $1$ at front and with an extra $r^{n+1}$ at the end. In fact: $$Sr=S-1+r^{n+1}$$ Solving: \\begin{align} Sr&=S-1+r^{n+1} \\\\ Sr-S&=-1+r^{n+1} \\\\ S(r-1)&=r^{n+1}-1 \\\\ S&=\\frac{r^{n+1}-1}{r-1} \\end{align} $$\\fbox{1+r+r^2+r^3+\\dotsb+r^n=\\dfrac{r^{n+1}-1}{r-1}}$$ Plugging in $r=4$ and $n=X+Y$, we get your answer. - can we say sum is: $((4^{X+Y+1}-1)/3)-1?$ – Mouna Mokhiab Aug 24 '14 at 17:28 No. Why do you have the $-1$ at the end? It should just be: $$\\frac{4^{X+Y+1}-1}{4-1}=\\frac{4^{X+Y+1}-1}{3}$$without any $-1$s at the end. (If you were summing $4+4^2+\\cdots+4^{X+Y}$ rather than $1+4+4^2+\\cdots+4^{X+Y}$, then sure.) – columbus8myhw Aug 24 '14 at 17:33 you means $4+...+4^X+Y=((4^{X+Y+1}-1)/3)-1$ am I right? – Mouna Mokhiab Aug 24 '14 at 17:35 Yes.", "# Definition:Geometric Series ## Definition Let $\\sequence {x_n}$ be a geometric progression in $\\R$: $x_n = a r^n$ for $n = 0, 1, 2, \\ldots$ Then the series defined as: $\\displaystyle \\sum_{n \\mathop = 1}^\\infty x_n = a + a r + a r^2 + \\cdots + a r^n + \\cdots$ is a geometric series.", "a move. Now, suppose the initial move $m_0 = (a, b)$. If $a+b=2016$, then $s_a = s_b = \\frac{a+b}{2} = 1008$. Then, applying the moves $$m_1 = (1, 2015), m_2 = (2, 2014), \\dots, m_{1007} = (1007, 1009),$$ we get $s_1 = s_2 = \\cdots = s_{2015} = 1008$. Otherwise, suppose $a+b\\ne 2016$. Then consider the following $3$ moves: $$m_{-1} = (2016-a, 2016-b), m_{-2} = (a, 2016-a), m_{-3} = (b, 2016-b)$$ We have $m_{-1}$ makes $s_{2016-a} = s_{2016-b} = 2016 - \\frac{a+b}{2}$. So, $m_{-2}$ makes $$s_a = s_{2016-a} = \\frac{2016 - \\frac{a+b}{2} + \\frac{a+b}{2}}{2} = 1008$$ Similarly for $m_{-3}$ with $s_b$ and $s_{2016-b}$. Then, finishing up with the moves $$m_1 = (1, 2015), m_2 = (2, 2014), \\dots, m_{1007} = (1007, 1009),$$ we get $s_1 = s_2 = \\cdots = s_{2015} = 1008$. ## Solution 3 (INCORRECT) Let the set be ${a_1,a_2,\\dots ,a_{2015}}$, where all the terms are nonnegative. Note that the sum of all the terms in this sequence will always be the same after any amount of moves. To prove this, let $i,j$ be integers with $1\\le i, and we have $a_i+a_j = \\frac{a_i+a_j}{2}+\\frac{a_i+a_j}{2}$. Also, $a_ia_j \\le (\\frac{a_i+a_j}{2})(\\frac{a_i+a_j}{2})$ by AM-GM, so the product of all the terms will not decrease after any number of moves. However, the product will only stay the same when $a_i=a_j$, so", "# Dedicate yourself and you will find yourself Geometry Level pending $s = 1- \\cos a + \\cos^2 a - \\cos^3 a + \\cdots \\\\ r = 1-\\sin a + \\sin^2 a - \\sin^3 a + \\cdots$ We are given the two equations above. State the expression below in terms of $$r$$ and/or $$s$$. $1 - \\sin a \\cos a + \\sin^2 a \\cos^2 - \\sin^3 a \\cos^3 a + \\cdots$ ×", "$- 6$. This means we have the sum of a geometric series, which we know is ${S}_{n} = {a}_{1} \\cdot \\frac{{r}^{n} - 1}{r - 1} = - 3 \\cdot \\frac{{\\left(- 6\\right)}^{7} - 1}{\\left(- 6\\right) - 1} = - 119973$", "the sum of $n$ terms of a geometric series and for the sum of an infinite number of terms but it is much more useful to remember how to generate the expressions than to memorize them. Suppose that the sum of the first $n$ terms is $S_n,$ then $S_n = a + ar + ar^2 + ar^3 \\cdot \\cdot \\cdot + ar^{n-1}.$ multiply each side of the equation by $r$ to obtain $S_nr = ar + ar^2 + ar^3 \\cdot \\cdot \\cdot + ar^{n-1} + ar^n.$ Subtract the second equation from the first and you get $S_n - S_nr = S_n(1 - r) = a - ar^n = a(1 - r^n).$ Hence $S_n = \\frac{a(1-r^n)}{1 - r}.$ You can use this expression to calculate the height of the tree after 5 years. In your problem $r = 0.95$ which is less than 1 and hence as $n$ becomes large, $r^n$ approaches zero and $S_n$ approaches $\\large \\frac{a}{1 - r}.$ Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "# The sum of the series Question: The sum of the series $2 \\cdot{ }^{20} \\mathrm{C}_{0}+5 \\cdot{ }^{20} \\mathrm{C}_{1}+8 \\cdot{ }^{20} \\mathrm{C}_{2}+11 \\cdot{ }^{.20} \\mathrm{C}_{3}$ $+\\ldots+62 \\cdot{ }^{20} \\mathrm{C}_{20}$ is equal to : 1. (1) $2^{26}$ 2. (2) $2^{25}$ 3. (3) $2^{23}$ 4. (4) $2^{24}$ Correct Option: , 2 Solution: $2 \\cdot{ }^{20} \\mathrm{C}_{0}+5 \\cdot{ }^{20} \\mathrm{C}_{1}+8 \\cdot{ }^{20} \\mathrm{C}_{2}+\\ldots \\ldots+62 \\cdot{ }^{20} \\mathrm{C}_{20}$ $=\\sum_{r=0}^{20}(3 r+2){ }^{20} C_{r}=3 \\sum_{r=0}^{20} r \\cdot{ }^{20} C_{r}+2 \\sum_{r=0}^{20}{ }^{20} C_{r}$ $=60 \\sum_{r=1}^{20}{ }^{19} C_{n-1}+2 \\sum_{r=0}^{20}{ }^{20} C_{r}$ $=60 \\times 2^{19}+2 \\times 2^{20}=2^{21}[15+1]=2^{25}$", "# Question #83975 Dec 5, 2014 Geometric Series $a + a r + a {r}^{2} + \\cdots = \\frac{a}{1 - r} \\text{ }$ if $\\text{ } - 1 < r < 1$ Since $b \\left(1 + b + {b}^{2} + \\cdots\\right) = b + {b}^{2} + {b}^{3} + \\cdots$ is a geometric series with $\\left\\{\\begin{matrix}a = b \\\\ r = b\\end{matrix}\\right.$, its sum $S$ can be found by $S = \\frac{b}{1 - b} \\text{ }$ if $\\text{ } - 1 < b < 1$. For other values of $b$, the series diverges. I hope that this was helpful.", "''r''. For values less than 1, such as 1/2, we can get reasonable results like: + - + - :$2 = 1 + {1 \\over 2} + {1 \\over 4} + {1 \\over 8} + {1 \\over 16} \\cdots$ + - + - However, if the value of ''r'' is larger than 1, such as 2, things start to get weird: + - + - :$-1 = 1 + 2 + 4 + 8 + 16 \\cdots$ + - + - How can we get a negative number by adding a bunch of positive integers? Well, if this case makes mathematicians uncomfortable, then they are going to be even more puzzled by the following one, in which ''r'' = -2: + - + - :${1 \\over 3} = 1 - 2 + 4 - 8 + 16 \\cdots$ + - + - This is ridiculous: the sum of integers can not possibly be a fraction. In fact, we are getting all these funny results because the last two series are divergent, so their sums are not defined. See the following images for a graphic representation of these series: + - Geometric Sequence with ''r'' = 1/2]]}}{{!}}{{!}}{{Anchor|Reference=Figure3b|Link=[[Image:InfiniteSeries2.jpg|center|thumb|325px|Figure 3-b Geometric Sequence with ''r'' = 2]]}}{{!}}{{!}}{{Anchor|Reference=Figure3c|Link=[[Image:InfiniteSeries3.jpg|center|thumb|325px|Figure 3-c Geometric Sequence with ''r'' = -2]]}} + - {{!}}} + - + - In", "# Prove $x^{rs}-1=(x^s-1)(x^{s(r-1)}+…+x^s+1)$ Prove $x^{rs}-1=(x^s-1)(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$ I can see that this is true for simple results such as $x^6$ and $x^9$ How can I prove that this is true in general. It telescopes. From $$(x^s-1)(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$$ consider $$x^s \\times(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1) - (x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$$ The first term becomes $$(x^{sr}+x^{s(r-1)}+...+x^{2s}+x^s)$$ which just leaves $$x^{sr}-1$$ This is equivalent to the geometric series formula: $$1+y+y^2+\\cdots+y^n={y^{n+1}-1\\over y-1}$$ and therefore $$(y-1)(1+y+y^2+\\cdots+y^n)=y^{n+1}-1$$. Substituting $y=x^s$ and $n=r-1$ we obtain your formula. If you require a proof of the geometric series formula, see for instance this question and any of the answers (I recommend mine :P, but they're mainly pretty good). Hint Let $x^s=y$. Use Ruffini's rule to divide $(y^r-1)/(y-1)$. Finally undo the previous assignment (that is, let $y=x^s$ back again). Hint Set $y=x^s$, then $y=1$ is solution of $$y^r-1=0$$ and thus, you can factorise $y^r-1$ by $(y-1)$ what gives $$y^r-1=(y-1)(y^{r-1}+y^{r-2}+...+y+1).$$ Observe $1+x^s+x^{2s}+\\ldots+x^{(r-1)s}$ is the sum of the first $r$ terms of a geometric sequence, and then it is equal to $(1-(x^s)^r)/(1-x^s)$ or $(x^{sr}-1)/(x^s-1)$. Start with $x^r-1=(x-1)(x^{r-1}+\\cdots+x+1)$. Write this identity for $x=y^s$. This gives keeping in mind that $(y^s)^r=y^{rs}$ \\begin{align} y^{sr}-1=&(y^s-1)(y^{s(r-1)}+\\cdots+y^s+1)\\end{align}" ]

[ "Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 This is, in fact, a \"geometric series\". A geometric series is any series of the form $\\displaystyle \\sum_{n=0}^\\infty ar^n= a+ ar+ ar^2+ ar^3+ \\cdot\\cdot\\cdot$. One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\\displaystyle \\frac{a}{1- r^2}$. Here, we can write the series as $\\displaystyle \\frac{1}{2}\\left(1+ \\frac{1}{2}+ \\frac{1/4}+ \\cdot\\cdot\\cdot\\right)$ so 1/2 times a geometric sequence with a= 1 and r= 1/2. Since 1/2< 1, this converges to $\\displaystyle \\frac{1}{2}\\frac{1}{1- \\frac{1}{2}}= \\frac{1}{2}\\frac{2}{2- 1}= 1$. (This is also a geometric series with a= 1/2 and r= 1/2 so converges to $\\displaystyle \\frac{\\frac{1}{2}}{1- \\frac{1}{2}}= \\frac{\\frac{1}{2}}{\\frac{1}{2}}= 1$) (It can also be thought of as a geometric series with a= 1 and r= 1/2 with the first \"a\" term missing. The geometric series with a= 1 and r= 1/2, $\\displaystyle 1+ \\frac{1}{2}+ \\frac{1}{4}+ \\frac{1}{8}+ \\cdot\\cdot\\cdot$ converges to $\\displaystyle \\frac{1}{1- \\frac{1}{2}}= 2$ and subtracting the missing first term, \"1\" gives a sum of 1.) August 28th, 2015, 06:42 AM #4 Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Quote: Originally Posted by Country Boy One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\\displaystyle \\frac{a}{1- r^2}$. That should be $\\displaystyle", "\\vert <1\\end{align*}, the series does converge. \\begin{align*}0.2 \\left(\\frac{1}{1-0.1}\\right)=\\frac{0.2}{0.9}=\\frac{2}{9}\\end{align*} #### Example 3 Why does an infinite series with \\begin{align*}r=1\\end{align*} diverge? If \\begin{align*}r=1\\end{align*} this means that the common ratio between the terms in the sequence is 1. This means that each number in the sequence is the same. When you add up an infinite number of any finite numbers (even fractions close to zero) you will always get infinity or negative infinity. The only exception is 0. This case is trivial because a geometric series with an initial value of 0 is simply the following series, which clearly sums to 0: \\begin{align*}0+0+0+0+\\cdots\\end{align*} #### Example 4 What is the sum of the first 8 terms in the following geometric series? \\begin{align*}4+2+1+\\frac{1}{2}+\\cdots\\end{align*} The first term is 4 and the common ratio is \\begin{align*}\\frac{1}{2}\\end{align*}. \\begin{align*}SUM=a_1\\left(\\frac{1-r^n}{1-r}\\right)=4 \\left(\\frac{1- \\left(\\frac{1}{2}\\right)^8}{1- \\frac{1}{2}}\\right)=4 \\left(\\frac{\\frac{255}{256}}{\\frac{1}{2}}\\right)=\\frac{255}{32}\\end{align*} You put 100 in a bank account at the end of every year for 10 years. The account earns 6% interest. How much do you have total at the end of 10 years? The first deposit gains 9 years of interest: \\begin{align*}100 \\cdot 1.06^9\\end{align*} The second deposit gains 8 years of interest: \\begin{align*}100 \\cdot 1.06^8\\end{align*}. This pattern continues, creating a geometric series. The last term receives no interest at all. \\begin{align*}100 \\cdot 1.06^9+100 \\cdot 1.06^8+\\cdots 100 \\cdot 1.06+100\\end{align*} Note that normally geometric series are", "+ (-8)x^4 + 16x^5 + (-32)x^6 + 64x^7+\\cdots \\\\ &= x - 2x^2 + 4x^3 - 8x^4 + 16x^5 - 32x^6 + 64x^7+\\cdots \\end{align*} This is an infinite geometric series with initial term $a = x$ and common ratio $r = -2x$, so we obtain: $$f(x) = \\frac{a}{1 - r} = \\frac{x}{1+2x}$$ For the second one, we obtain: \\begin{align*} g(x) &= 1x^0 + 0x^1 + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + 0x^7+ 1x^8 + \\cdots \\\\ &= 1 + x^2 + x^4 + x^6 + x^8 + \\cdots \\end{align*} This is an infinite geometric series with initial term $a = 1$ and common ratio $r = x^2$, so we obtain: $$g(x) = \\frac{a}{1 - r} = \\frac{1}{1-x^2}$$", "# LaTeX & MathJax basic tutorial – Aligned equations ## Aligned equations Often people want a series of equations where the equals signs are aligned. To get this, use \\begin{align}…\\end{align}. Each line should end with \\\\, and should contain an ampersand at the point to align at, typically immediately before the equals sign. For example, \\begin{align} \\sqrt{37} & = \\sqrt{\\frac{73^2-1}{12^2}} \\\\ & = \\sqrt{\\frac{73^2}{12^2}\\cdot\\frac{73^2-1}{73^2}} \\\\ & = \\sqrt{\\frac{73^2}{12^2}}\\sqrt{\\frac{73^2-1}{73^2}} \\\\ & = \\frac{73}{12}\\sqrt{1 – \\frac{1}{73^2}} \\\\ & \\approx \\frac{73}{12}\\left(1 – \\frac{1}{2\\cdot73^2}\\right) \\end{align} is produced by The usual marks that delimit the display may be omitted here.", "= r\\text{.} \\end{equation*} Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion \\begin{equation*} N=0.1212\\overline{12} \\end{equation*} as a rational number. Observe that \\begin{align*} N \\amp = 0.12 + 0.0012 + 0.000012 + \\cdots\\\\ \\amp = \\left(\\frac{12}{100}\\right) + \\left(\\frac{12}{100}\\right)\\left(\\frac{1}{100}\\right) + \\left(\\frac{12}{100}\\right)\\left(\\frac{1}{100}\\right)^2 + \\cdots\\text{.} \\end{align*} This is an infinite geometric series with $a=\\frac{12}{100}$ and $r = \\frac{1}{100}\\text{.}$ By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following example. ###### Example7.17 Let $r \\ne 1$ and $a$ be real numbers and let \\begin{equation*} S = a+ar+ar^2 + \\cdots ar^{n-1} + \\cdots \\end{equation*} be an infinite geometric series. For each positive integer $n\\text{,}$ let \\begin{equation*} S_n = a+ar+ar^2 + \\cdots + ar^{n-1}\\text{.} \\end{equation*} Recall that \\begin{equation*} S_n = a\\frac{1-r^n}{1-r}\\text{.} \\end{equation*} 1. What should we allow $n$ to approach in order to have $S_n$ approach $S\\text{?}$ 2. What is the value of $\\lim_{n \\to \\infty} r^n$ for $|r| \\gt 1\\text{?}$ for $|r| \\lt 1\\text{?}$ Explain. 3. If $|r| \\lt 1\\text{,}$ use the formula for $S_n$ and your observations in (a) and (b) to explain why $S$ is finite and find a resulting", "# Whats the missing term in the geometric sequence 36, __, 4? Nov 27, 2015 $x = 12$ #### Explanation: Now let the Middle term be $x$ $36 , x , 4$ So the definition of a geometric series is a, ar,ar^2 ....ar^(n-1 |_____| $$ n terms Now from this we can come to some results 1) ->36*r = x This is because every step of the series is being multiplied by a common number $r$ 2)-> x*r = 4 Now from $\\otimes 1$ $\\implies 36 \\cdot r \\cdot r = 4$ $\\implies \\cancel{4} \\cdot 9 \\cdot {r}^{2} = \\cancel{4}$ $\\implies {r}^{2} = \\frac{1}{9} , r = \\frac{1}{3}$ $36 \\cdot \\frac{1}{3} = x$ $\\cancel{3} \\cdot 12 \\cdot \\frac{1}{\\cancel{3}} = x$ $\\therefore x = 12$", "&= 2 \\\\ S_{n} &= \\frac{a(r^{n} - 1)}{r - 1} \\\\ S_{7} &= \\frac{5((2)^{7} - 1)}{2 - 1} \\\\ &= 5(128 - 1) \\\\ &= 635 \\end{align*} Given: $\\sum _{t=1}^{n}8 {\\left(\\frac{1}{2}\\right)}^{t}$ Find the first three terms in the series. \\begin{align*} t = 1: \\quad T_{1} &= 4 \\\\ t = 2: \\quad T_{2} &= 2 \\\\ t = 3: \\quad T_{3} &= 1 \\\\ 4; & 2; 1 \\end{align*} Calculate the number of terms in the series if $$S_{n}=7\\frac{63}{64}$$. \\begin{align*} a &= 4 \\\\ r &= \\frac{1}{2} \\\\ S_{n} &= \\frac{a(1 - r^{n})}{1 - r} \\\\ \\frac{511}{64} &= \\frac{4(1 - \\left( \\frac{1}{2} \\right)^{n})}{1 - \\left( \\frac{1}{2} \\right)} \\\\ &= \\frac{4 - 4\\left( \\frac{1}{2} \\right)^{n}}{\\frac{1}{2}} \\\\ \\frac{511}{128} &= 4 - (2^{2} \\cdot 2^{-n}) \\\\ 2^{2 - n}&= 4 - \\frac{511}{128} \\\\ 2^{2 - n}&= \\frac{1}{128} \\\\ 2^{2 - n}&= 2^{-7}\\\\ 2 - n &= -7 \\\\ \\therefore 9 &= n \\end{align*} The ratio between the sum of the first three terms of a geometric series and the sum of the $$\\text{4}$$$$^{\\text{th}}$$, $$\\text{5}$$$$^{\\text{th}}$$ and $$\\text{6}$$$$^{\\text{th}}$$ terms of the same series is $$8:27$$. Determine the constant ratio and the first $$\\text{2}$$ terms if the third term is $$\\text{8}$$. \\begin{align*} T_{1} + T_{2} + T_{3} &= a + ar + ar^{2} \\\\ &= a(1 + r + r^{2}) \\\\ T_{4} + T_{5}", "## Thinking Mathematically (6th Edition) The sequence is geometric and the sum of its first ten terms is $-1023$. The terms have a common ratio of $-2$ so the sequence is geometric with $r=-2$ and $a_1=3$. The sum $S_n$ of the first $n$ terms of a geometric sequence is given by the formula: $$S_n=\\frac{a_1(1-r^n)}{1-r}$$ The given sequence has $a_1=3$ and $r=-2$. However, the value of $a_{10}$ (the tenth term) is not yet known. Solve for $a_{10}$ using the formula $a_n=a_1\\cdot r^{n-1}$ to obtain: \\begin{align*} a_{10}&=3 \\cdot (-2)^{10-1}\\\\ &=3\\cdot (-2)^9\\\\ &=3\\cdot (-512)\\\\ &=-1536\\end{align*} Use the formula for the sum of the first $n$ terms to find the sum of the first $10$ terms (use $n=10$) to obtain: \\begin{align*} S_{10}&=\\frac{a_1(1-r^n)}{1-r}\\\\\\\\ &=\\frac{3(1-(-2)^{10})}{1-(-2)}\\\\\\\\ &=\\frac{3(1-1024)}{1+2}\\\\\\\\ &=\\frac{3(-1023)}{3}\\\\\\\\ &=-1023 \\end{align*}", "# How do you find the 13th term of the geometric sequence with a3 = 24 and a5 = 96? Nov 24, 2015 ${a}_{13} = 24576$ #### Explanation: Well The definition of A geometric sequence with n elements; a, ar,ar^2,ar^3, ar^4 ...ar^(n-1 r is a common constant with which every number is multiplied Now ${a}_{5} = a \\cdot {r}^{4} = 96 - - - - - - - - 1$ ${a}_{3} = a \\cdot {r}^{2} = 24 - - - - - - - - 2$ Now divide 1 by 2 $\\frac{\\cancel{a} \\cdot {\\cancel{r}}^{2} {r}^{2}}{\\cancel{a} \\cdot {\\cancel{r}}^{2}} = \\frac{96}{24} = 4$ ${r}^{2} = 4 \\implies r = 2$ Now come back to equation 2 $a \\cdot {r}^{2} = 24$ Substitute $a \\cdot 4 = 24$ $\\implies a = 6$ Now we have found the first term Lets finish it ${a}_{13} = a \\cdot {r}^{13 - 1} = 6 \\cdot {2}^{6} \\cdot {2}^{6} = 24576$ And were done", "• Why doesn't this method work? – Minu Jul 26 '14 at 0:23 • It is the same, I mentioned the cancellation in the answer. For instance, $2!/3!=1/3$, $6/4!=1/4$, and so on. – André Nicolas Jul 26 '14 at 0:24 • aren't the signs different? – Minu Jul 26 '14 at 0:27 • You are welcome. The question you asked about the $\\frac{1}{1+t}=\\cdots$ is briefly dealt with above by the mention of geometric series. You are probably familiar with $1+r+r^2+r^3+\\cdot=\\frac{1}{1-r}$ (sum of infinite geometric series). Put $r=-t$. – André Nicolas Jul 26 '14 at 0:49 Note that $$\\frac{1}{1+x}=\\sum_{n \\ge 0} (-1)^nx^n$$ Integrating both sides gives you \\begin{align} \\ln(1+x) &=\\sum_{n \\ge 0}\\frac{(-1)^nx^{n+1}}{n+1}\\\\ &=x-\\frac{x^2}{2}+\\frac{x^3}{3}-... \\end{align} Alternatively, \\begin{align} &f^{(1)}(x)=(1+x)^{-1} &\\implies \\ f^{(1)}(0)=1\\\\ &f^{(2)}(x)=-(1+x)^{-2} &\\implies f^{(2)}(0)=-1\\\\ &f^{(3)}(x)=2(1+x)^{-3} &\\implies \\ f^{(3)}(0)=2\\\\ &f^{(4)}(x)=-6(1+x)^{-4} &\\implies \\ f^{(4)}(0)=-6\\\\ \\end{align} We deduce that \\begin{align} f^{(n)}(0)=(-1)^{n-1}(n-1)! \\end{align} Hence, \\begin{align} \\ln(1+x) &=\\sum_{n \\ge 1}\\frac{f^{(n)}(0)}{n!}x^n\\\\ &=\\sum_{n \\ge 1}\\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\\\ &=\\sum_{n \\ge 1}\\frac{(-1)^{n-1}}{n}x^n\\\\ &=\\sum_{n \\ge 0}\\frac{(-1)^{n}}{n+1}x^{n+1}\\\\ &=x-\\frac{x^2}{2}+\\frac{x^3}{3}-... \\end{align} Edit: To derive a closed for for the geometric series, let \\begin{align} S&=1-x+x^2-x^3+...\\\\ xS&=x-x^2+x^3-x^4...\\\\ S+xS&=1\\\\ S&=\\frac{1}{1+x}\\\\ \\end{align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually. • Can you tell me How you got the summation: $\\frac{1}{1+x} = \\Sigma_{n=0}^{\\infty} (-1)^n x^n$? I understand the rest. – Minu Jul 26 '14 at 0:44 • Start by", "\\end{align*}, \\begin{align*} &= 5(128 - 1) \\\\ \\therefore \\frac{8}{27} &= \\frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\\\ &= 45 \\therefore ar^{2} &= 8 \\\\ We think you are located in Determine the values of $$r$$ and $$n$$ if $$S_{n} = 84$$. S_{n} &= \\frac{a(1-r^{n})}{1 - r}\\\\ Calculate: $\\sum _{k = 1}^{6}{32 \\left( \\frac{1}{2} \\right)^{k-1}}$, We have generated the series $$32 + 16 + 8 + \\cdots$$. \\text{And } 20 &= ar^{2} \\\\ \\end{align*}, \\begin{align*} &= ar^{3}(1 + r + r^{2}) \\\\ The eighth term of a geometric sequence is $$\\text{640}$$. \\text{And } T_{3} &= 8 \\\\ 2^{2 - n}&= \\frac{1}{128} \\\\ We generate a geometric sequence using the general form: Tn = a ⋅ rn − 1. where. a &= 4 \\\\ &= \\frac{1 - 6561}{4}\\\\ Finite geometric sequence: 1 2 , 1 4 , 1 8 , 1 16 , ... , 1 32768. \\therefore a &= 8 \\times \\frac{4}{9} \\\\ &= \\frac{54(1 - \\left( \\frac{1}{3} \\right)^{n})}{1 - \\left( \\frac{1}{3} \\right)} \\\\ a \\left( \\frac{3}{2} \\right)^{2} &= 8 \\\\ This calculus video tutorial explains how to find the sum of a finite geometric series using a simple formula. T_{1} + T_{2} + T_{3} &= a + ar + ar^{2} \\\\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\\\ r &= \\frac{1}{3} \\\\ 1.5 Finite", "# How do you find the sum of the infinite geometric series 215 - 86 + 34.4 - 13.76 + ...? Use the formula $a + a r + a {r}^{2} + a {r}^{3} + \\cdots = \\frac{a}{1 - r}$ (for $| r | < 1$) to get $\\frac{1075}{7} \\approx 153.6$. The series $215 - 86 + 34.4 - 13.76 + \\cdots$ is geometric with first term $a = 215$ and common ratio $r = - \\frac{86}{215} = - \\frac{2}{5} = - 0.4$. The formula above therefore gives $\\frac{a}{1 - r} = \\frac{215}{1 + \\frac{2}{5}} = \\frac{215}{\\frac{7}{5}} = \\frac{215 \\cdot 5}{7} = \\frac{1075}{7} \\approx 153.6$.", "\\right)^{n}} &= \\frac{255}{64} \\\\ k \\left( \\frac{255}{256} \\right) &= \\frac{255}{64} \\\\ \\therefore k &= \\frac{255}{64} \\times \\frac{256}{255} \\\\ &= \\frac{256}{64} \\\\ &= 4 \\end{align*} $k = 4$ ## Sum of a geometric series Exercise 1.9 Prove that $$a + ar + a{r}^{2} + \\cdots + a{r}^{n-1} = \\frac{a(r^{n} - 1) }{r - 1}$$ and state any restrictions. \\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\\cdots + a{r}^{n-2} + a{r}^{n-1} \\ldots (1) \\\\ r \\times {S}_{n} &= \\qquad ar+ a{r}^{2}+\\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \\ldots (2) \\\\ \\text{Subtract eqn. } (1) &\\text{ from eqn. } (2) \\\\ \\therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\\\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\\\ \\therefore {S}_{n} &= \\frac{a(r^{n} - 1) }{r - 1} \\end{align*} where $$r \\ne 1$$. Given the geometric sequence $$1; -3; 9; \\ldots$$ determine: The eighth term of the sequence. \\begin{align*} a &= 1 \\\\ r &= \\frac{T_{2}}{T_{1}} = -3 \\\\ T_{n} &= ar^{n-1} \\\\ \\therefore T_{8} &= (1)(-3)^{8-1} \\\\ &= (1)(-3)^{7} \\\\ &= -2187 \\end{align*} The sum of the first eight terms of the sequence. \\begin{align*} S_{n} &= \\frac{a(1-r^{n})}{1 - r}\\\\ \\therefore S_{8} &= \\frac{(1)(1-(-3)^{8})}{1 - (-3)}\\\\ &= \\frac{1 - 6561}{4}\\\\ &= -\\frac{6560}{4}\\\\ &= -1640 \\end{align*} Determine: $\\sum _{n=1}^{4}3 \\cdot {2}^{n-1}$ \\begin{align*} S_{4} &= 3 + 6 + 12 + 24 \\\\ &= 45 \\end{align*}", "Thinking Mathematically (6th Edition) $\\frac{1}{2}$ To find the sixth term put $n=6$ in the above formula, to get: \\begin{align} & {{a}_{6}}={{a}_{1}}{{r}^{6-1}} \\\\ & =16\\cdot {{\\left( \\frac{1}{2} \\right)}^{5}} \\\\ & =16\\cdot \\frac{1}{32} \\\\ & =\\frac{1}{2} \\end{align} The sixth term of the geometric sequence is$\\frac{1}{2}$.", "series formula, followed by some algebra and checking convergence. In my opinion, the simplest way to memorize the formula is $$\\frac{\\text{first}}{1 - \\text{ratio}}$$ So whether you're computing $$\\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\ldots$$ or $$\\sum_{n=3}^{\\infty} 2^{-n}$$ or $$\\sum_{n=0}^{\\infty} \\frac{1}{8} 2^{-n}$$ you can quickly identify the sum as $$\\frac{ \\frac{1}{8} }{1 - \\frac{1}{2}}$$ Similarly, for a finite geometric sequence, a formula is $$\\frac{\\text{first} - \\text{(next after last)}}{1 - \\text{ratio}}$$ The infinite version can be viewed as a special case, where $(\\text{next after last}) = 0$. I find this formula more convenient written as $$\\frac{\\text{(next after last)} - \\text{first}}{\\text{ratio} - 1}$$ e.g. $$2 + 4 + 8 + \\ldots + 256 = \\frac{512 - 2}{2 - 1}$$ But in a pinch, you can always just rederive the formula since the method is simple: \\begin{align}(2-1) (2 + 4 + 8 + \\ldots + 256) &= (4 - 2) + (8 - 4) + (16 - 8) + \\ldots + (512 - 256) \\\\&= 512 - 2 \\end{align}", "infinite geometric sequence $27,18,12,8,\\cdots$. First find $r$ $r=\\frac{{a}_{2}}{{a}_{1}}=\\frac{18}{27}=\\frac{2}{3}$ Then find the sum: $S=\\frac{{a}_{1}}{1-r}$ $S=\\frac{27}{1-\\frac{2}{3}}=81$ Example 5: Find the sum of the infinite geometric sequence $8,12,18,27,\\cdots$ if it exists. First find $r$ $r=\\frac{{a}_{2}}{{a}_{1}}=\\frac{12}{8}=\\frac{3}{2}$ Since $r=\\frac{3}{2}$ is not less than one the series has no sum. There is a formula to calculate the $n$th term of an geometric series, that is, the sum of the first $n$ terms of an geometric sequence.", "}{2{\\sqrt {3))))\\tan({\\frac {\\pi }{2{\\sqrt {3)))))\\\\&\\approx 1.159173.\\\\\\end{aligned))} The alternating series of unit fractions with the star numbers as denominators is: {\\displaystyle {\\begin{aligned}\\sum _{n=1}^{\\infty }&(-1)^{n-1}{\\frac {1}{S_{n))}\\\\&=1-{\\frac {1}{13))+{\\frac {1}{37))-{\\frac {1}{73))+{\\frac {1}{121))-{\\frac {1}{181))+{\\frac {1}{253))-{\\frac {1}{337))+\\cdots \\\\&\\approx 0.941419.\\\\\\end{aligned))}", "+0 0 299 1 +644 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. waffles Sep 20, 2017 #1 +20680 +1 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. If the common ratio r lies between −1 to 1 , we can have the sum of an infinite geometric series. That is, the sum exits for $$| r |<1$$ . The sum S of an infinite geometric series with $$-1 is given by the formula, \\(S=\\frac{a_1}{1-r}$$ Let $$a_1 = -\\frac{3}{2}$$ Let $$r =\\,?$$ $$\\begin{array}{|rcll|} \\hline S = \\frac{a_1}{1-r} &=& 2r \\quad & | \\quad a_1 = -\\frac{3}{2} \\\\ \\frac{-\\frac{3}{2}}{1-r} &=& 2r \\\\ -\\frac{3}{2} \\cdot \\frac{1}{1-r} &=& 2r \\quad & | \\quad : 2 \\\\ -\\frac{3}{4} \\cdot \\frac{1}{1-r} &=& r \\\\ \\frac{3}{4} \\cdot \\frac{1}{r-1} &=& r \\quad & | \\quad \\cdot (r-1) \\\\ \\frac{3}{4} &=& r \\cdot (r-1) \\quad & | \\quad - \\frac{3}{4} \\\\ r \\cdot (r-1) - \\frac{3}{4} &=& 0 \\\\ r^2-r- \\frac{3}{4} &=& 0 \\\\\\\\ r &=& \\frac{1\\pm\\sqrt{1-4\\cdot (- \\frac{3}{4}) } }{2} \\\\ r &=& \\frac{1\\pm\\sqrt{1+3 } }{2} \\\\ r &=& \\frac{1\\pm 2 }{2} \\\\\\\\ r_1 &=& \\frac{1 + 2 }{2}", "\\,+\\, \\frac{8}{27} \\,+\\, \\cdots .$ This series has common ratio 2/3. If we multiply through by this common ratio, then the initial 1 becomes a 2/3, the 2/3 becomes a 4/9, and so on: $\\frac{2}{3}s \\;=\\; \\frac{2}{3} \\,+\\, \\frac{4}{9} \\,+\\, \\frac{8}{27} \\,+\\, \\frac{16}{81} \\,+\\, \\cdots .$ This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series s cancels every term in the original but the first, $s \\,-\\, \\frac{2}{3}s \\;=\\; 1,\\;\\;\\;\\mbox{so }s=3.$ A similar technique can be used to evaluate any self-similar expression. ### Formula For $r\\neq 1$, the sum of the first n terms of a geometric series is $a + ar + a r^2 + a r^3 + \\cdots + a r^{n-1} = \\sum_{k=0}^{n-1} ar^k= a \\, \\frac{1-r^{n}}{1-r},$ where a is the first term of the series, and r is the common ratio. We can derive this formula as follows: \\begin{align} s &= a + ar + ar^2 + ar^3 + \\cdots + ar^{n-1}, \\\\ rs &= ar + ar^2 + ar^3 + ar^4 + \\cdots + ar^{n}, \\\\ s - rs &= a-ar^{n}, \\\\ s(1-r) &= a(1-r^{n}), \\end{align} so, $s = a \\frac{1-r^{n}}{1-r} \\quad \\text{(if } r \\neq 1 \\text{)}.$ As n goes to infinity, the absolute value of r", "# What's the common ratio in 5/12, 1/3, 4/15, 16/75, 64/375? Dec 3, 2015 $\\frac{4}{5}$ #### Explanation: Divide any number in this sequence by the preceding number and you get $\\frac{4}{5}$. That's the common ratio. For example: $\\frac{1}{3} \\div \\frac{5}{12} = \\frac{1}{3} \\cdot \\frac{12}{5} = \\frac{4}{5}$, $\\frac{4}{15} \\div \\frac{1}{3} = \\frac{4}{15} \\cdot 3 = \\frac{12}{15} = \\frac{4}{5}$, $\\frac{16}{75} \\div \\frac{4}{15} = \\frac{16}{75} \\cdot \\frac{15}{4} = \\frac{4}{5}$, and $\\frac{64}{375} \\div \\frac{16}{75} = \\frac{64}{375} \\cdot \\frac{75}{16} = \\frac{4}{5}$. A cool thing related to this is the corresponding geometric series (infinite \"sum\") $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots$. If $a$ is the first term of such a series, and $r$ is the common ratio, with $| r | < 1$, the series converges to $\\frac{a}{1 - r}$. For this example, $a = \\frac{5}{12}$ and $r = \\frac{4}{5}$ so that the series converges to $\\frac{\\frac{5}{12}}{1 - \\frac{4}{5}} = \\frac{\\frac{5}{12}}{\\frac{1}{5}} = \\frac{5}{12} \\cdot 5 = \\frac{25}{12}$ and we write $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots = \\frac{25}{12}$." ]

[ "infinite geometric sequence $27,18,12,8,\\cdots$. First find $r$ $r=\\frac{{a}_{2}}{{a}_{1}}=\\frac{18}{27}=\\frac{2}{3}$ Then find the sum: $S=\\frac{{a}_{1}}{1-r}$ $S=\\frac{27}{1-\\frac{2}{3}}=81$ Example 5: Find the sum of the infinite geometric sequence $8,12,18,27,\\cdots$ if it exists. First find $r$ $r=\\frac{{a}_{2}}{{a}_{1}}=\\frac{12}{8}=\\frac{3}{2}$ Since $r=\\frac{3}{2}$ is not less than one the series has no sum. There is a formula to calculate the $n$th term of an geometric series, that is, the sum of the first $n$ terms of an geometric sequence.", "## Thinking Mathematically (6th Edition) The sequence is geometric and the sum of its first ten terms is $-1023$. The terms have a common ratio of $-2$ so the sequence is geometric with $r=-2$ and $a_1=3$. The sum $S_n$ of the first $n$ terms of a geometric sequence is given by the formula: $$S_n=\\frac{a_1(1-r^n)}{1-r}$$ The given sequence has $a_1=3$ and $r=-2$. However, the value of $a_{10}$ (the tenth term) is not yet known. Solve for $a_{10}$ using the formula $a_n=a_1\\cdot r^{n-1}$ to obtain: \\begin{align*} a_{10}&=3 \\cdot (-2)^{10-1}\\\\ &=3\\cdot (-2)^9\\\\ &=3\\cdot (-512)\\\\ &=-1536\\end{align*} Use the formula for the sum of the first $n$ terms to find the sum of the first $10$ terms (use $n=10$) to obtain: \\begin{align*} S_{10}&=\\frac{a_1(1-r^n)}{1-r}\\\\\\\\ &=\\frac{3(1-(-2)^{10})}{1-(-2)}\\\\\\\\ &=\\frac{3(1-1024)}{1+2}\\\\\\\\ &=\\frac{3(-1023)}{3}\\\\\\\\ &=-1023 \\end{align*}", "Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 This is, in fact, a \"geometric series\". A geometric series is any series of the form $\\displaystyle \\sum_{n=0}^\\infty ar^n= a+ ar+ ar^2+ ar^3+ \\cdot\\cdot\\cdot$. One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\\displaystyle \\frac{a}{1- r^2}$. Here, we can write the series as $\\displaystyle \\frac{1}{2}\\left(1+ \\frac{1}{2}+ \\frac{1/4}+ \\cdot\\cdot\\cdot\\right)$ so 1/2 times a geometric sequence with a= 1 and r= 1/2. Since 1/2< 1, this converges to $\\displaystyle \\frac{1}{2}\\frac{1}{1- \\frac{1}{2}}= \\frac{1}{2}\\frac{2}{2- 1}= 1$. (This is also a geometric series with a= 1/2 and r= 1/2 so converges to $\\displaystyle \\frac{\\frac{1}{2}}{1- \\frac{1}{2}}= \\frac{\\frac{1}{2}}{\\frac{1}{2}}= 1$) (It can also be thought of as a geometric series with a= 1 and r= 1/2 with the first \"a\" term missing. The geometric series with a= 1 and r= 1/2, $\\displaystyle 1+ \\frac{1}{2}+ \\frac{1}{4}+ \\frac{1}{8}+ \\cdot\\cdot\\cdot$ converges to $\\displaystyle \\frac{1}{1- \\frac{1}{2}}= 2$ and subtracting the missing first term, \"1\" gives a sum of 1.) August 28th, 2015, 06:42 AM #4 Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Quote: Originally Posted by Country Boy One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\\displaystyle \\frac{a}{1- r^2}$. That should be $\\displaystyle", "(62813\\frac{\\sqrt\\frac{92}{12}-1}{12}+1)$$ and finally $$n=\\frac{\\log (62813\\frac{\\sqrt\\frac{92}{12}-1}{12}+1)}{\\log \\sqrt{\\frac{92}{12}}}-1$$ - So, what is the answer? – clixbrigidxterx Feb 20 '13 at 23:22 That is the answer, now you have to check what integer it is equal to ( if it is an integer) – Jorge Fernández Feb 20 '13 at 23:48 Is there other way without using logarithms? Our teacher taught us answering the geometric series problems without the use of logarithms. – clixbrigidxterx Feb 20 '13 at 23:57 I think what the OP means is that $\\sum^3 a^k =92$ and $\\sum^1 a^k=12$; or am I reading \"series\" and he meant \"progression\" or \"sequence\"? – Pedro Tamaroff Feb 20 '13 at 23:59 We are only in Geometric Sequence and Series. We are not yet in progression or infinite something. :/ – clixbrigidxterx Feb 21 '13 at 0:03 Let the series be $a+ar+ar^2+\\cdots +ar^{n-1}$. This has $n$ terms. Clearly $a=12$. We are told that $ar^2=92$, giving $r^2=\\frac{92}{12}$, so $r=\\sqrt{\\frac{92}{12}}$. Unusually unattractive number for this level. The sum $S$ of the series is $62813$. Not too pretty either! It will make things clearer, and typing easier, if we just continue with letters. The sum $S$ of the series is given by $$S=a\\frac{r^n-1}{r-1}.$$ Solving for $r^n$, we obtain $$r^n=1+\\frac{(r-1)S}{a}.$$ Let the right-hand side be $W$. Take your favourite log of both sides. We get", "# Whats the missing term in the geometric sequence 36, __, 4? Nov 27, 2015 $x = 12$ #### Explanation: Now let the Middle term be $x$ $36 , x , 4$ So the definition of a geometric series is a, ar,ar^2 ....ar^(n-1 |_____| $$ n terms Now from this we can come to some results 1) ->36*r = x This is because every step of the series is being multiplied by a common number $r$ 2)-> x*r = 4 Now from $\\otimes 1$ $\\implies 36 \\cdot r \\cdot r = 4$ $\\implies \\cancel{4} \\cdot 9 \\cdot {r}^{2} = \\cancel{4}$ $\\implies {r}^{2} = \\frac{1}{9} , r = \\frac{1}{3}$ $36 \\cdot \\frac{1}{3} = x$ $\\cancel{3} \\cdot 12 \\cdot \\frac{1}{\\cancel{3}} = x$ $\\therefore x = 12$", "## College Algebra 7th Edition $S_7=-645$ We are asked to find the sum of: $-15+30-60+\\cdots-960$ This is a geometric sequence with $a_1=-15$. We find $r$: $r=\\frac{30}{-15}=-2$ We know that a geometric sequence has the form: $a_{n}=ar^{n-1}$ We use this to find the number of terms: $a_n=(-15)(-2)^{n-1}=-960$ $64=(-2)^{n-1}=\\frac{(-2)^n}{-2}$ $-128=(-2)^n$ We see that $n$ must be odd: $128=2^n$ $n=\\log_2 128$ $n=7$ We know the partial sum of a geometric sequence is: $S_n=a_1\\frac{1-r^n}{1-r}$ Thus: $S_{7}=-15 \\frac{1-(-2)^{7}}{1-(-2)}=-645$", "\\right)^{n}} &= \\frac{255}{64} \\\\ k \\left( \\frac{255}{256} \\right) &= \\frac{255}{64} \\\\ \\therefore k &= \\frac{255}{64} \\times \\frac{256}{255} \\\\ &= \\frac{256}{64} \\\\ &= 4 \\end{align*} $k = 4$ ## Sum of a geometric series Exercise 1.9 Prove that $$a + ar + a{r}^{2} + \\cdots + a{r}^{n-1} = \\frac{a(r^{n} - 1) }{r - 1}$$ and state any restrictions. \\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\\cdots + a{r}^{n-2} + a{r}^{n-1} \\ldots (1) \\\\ r \\times {S}_{n} &= \\qquad ar+ a{r}^{2}+\\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \\ldots (2) \\\\ \\text{Subtract eqn. } (1) &\\text{ from eqn. } (2) \\\\ \\therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\\\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\\\ \\therefore {S}_{n} &= \\frac{a(r^{n} - 1) }{r - 1} \\end{align*} where $$r \\ne 1$$. Given the geometric sequence $$1; -3; 9; \\ldots$$ determine: The eighth term of the sequence. \\begin{align*} a &= 1 \\\\ r &= \\frac{T_{2}}{T_{1}} = -3 \\\\ T_{n} &= ar^{n-1} \\\\ \\therefore T_{8} &= (1)(-3)^{8-1} \\\\ &= (1)(-3)^{7} \\\\ &= -2187 \\end{align*} The sum of the first eight terms of the sequence. \\begin{align*} S_{n} &= \\frac{a(1-r^{n})}{1 - r}\\\\ \\therefore S_{8} &= \\frac{(1)(1-(-3)^{8})}{1 - (-3)}\\\\ &= \\frac{1 - 6561}{4}\\\\ &= -\\frac{6560}{4}\\\\ &= -1640 \\end{align*} Determine: $\\sum _{n=1}^{4}3 \\cdot {2}^{n-1}$ \\begin{align*} S_{4} &= 3 + 6 + 12 + 24 \\\\ &= 45 \\end{align*}", "• Why doesn't this method work? – Minu Jul 26 '14 at 0:23 • It is the same, I mentioned the cancellation in the answer. For instance, $2!/3!=1/3$, $6/4!=1/4$, and so on. – André Nicolas Jul 26 '14 at 0:24 • aren't the signs different? – Minu Jul 26 '14 at 0:27 • You are welcome. The question you asked about the $\\frac{1}{1+t}=\\cdots$ is briefly dealt with above by the mention of geometric series. You are probably familiar with $1+r+r^2+r^3+\\cdot=\\frac{1}{1-r}$ (sum of infinite geometric series). Put $r=-t$. – André Nicolas Jul 26 '14 at 0:49 Note that $$\\frac{1}{1+x}=\\sum_{n \\ge 0} (-1)^nx^n$$ Integrating both sides gives you \\begin{align} \\ln(1+x) &=\\sum_{n \\ge 0}\\frac{(-1)^nx^{n+1}}{n+1}\\\\ &=x-\\frac{x^2}{2}+\\frac{x^3}{3}-... \\end{align} Alternatively, \\begin{align} &f^{(1)}(x)=(1+x)^{-1} &\\implies \\ f^{(1)}(0)=1\\\\ &f^{(2)}(x)=-(1+x)^{-2} &\\implies f^{(2)}(0)=-1\\\\ &f^{(3)}(x)=2(1+x)^{-3} &\\implies \\ f^{(3)}(0)=2\\\\ &f^{(4)}(x)=-6(1+x)^{-4} &\\implies \\ f^{(4)}(0)=-6\\\\ \\end{align} We deduce that \\begin{align} f^{(n)}(0)=(-1)^{n-1}(n-1)! \\end{align} Hence, \\begin{align} \\ln(1+x) &=\\sum_{n \\ge 1}\\frac{f^{(n)}(0)}{n!}x^n\\\\ &=\\sum_{n \\ge 1}\\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\\\ &=\\sum_{n \\ge 1}\\frac{(-1)^{n-1}}{n}x^n\\\\ &=\\sum_{n \\ge 0}\\frac{(-1)^{n}}{n+1}x^{n+1}\\\\ &=x-\\frac{x^2}{2}+\\frac{x^3}{3}-... \\end{align} Edit: To derive a closed for for the geometric series, let \\begin{align} S&=1-x+x^2-x^3+...\\\\ xS&=x-x^2+x^3-x^4...\\\\ S+xS&=1\\\\ S&=\\frac{1}{1+x}\\\\ \\end{align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually. • Can you tell me How you got the summation: $\\frac{1}{1+x} = \\Sigma_{n=0}^{\\infty} (-1)^n x^n$? I understand the rest. – Minu Jul 26 '14 at 0:44 • Start by", "\\,+\\, \\frac{8}{27} \\,+\\, \\cdots .$ This series has common ratio 2/3. If we multiply through by this common ratio, then the initial 1 becomes a 2/3, the 2/3 becomes a 4/9, and so on: $\\frac{2}{3}s \\;=\\; \\frac{2}{3} \\,+\\, \\frac{4}{9} \\,+\\, \\frac{8}{27} \\,+\\, \\frac{16}{81} \\,+\\, \\cdots .$ This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series s cancels every term in the original but the first, $s \\,-\\, \\frac{2}{3}s \\;=\\; 1,\\;\\;\\;\\mbox{so }s=3.$ A similar technique can be used to evaluate any self-similar expression. ### Formula For $r\\neq 1$, the sum of the first n terms of a geometric series is $a + ar + a r^2 + a r^3 + \\cdots + a r^{n-1} = \\sum_{k=0}^{n-1} ar^k= a \\, \\frac{1-r^{n}}{1-r},$ where a is the first term of the series, and r is the common ratio. We can derive this formula as follows: \\begin{align} s &= a + ar + ar^2 + ar^3 + \\cdots + ar^{n-1}, \\\\ rs &= ar + ar^2 + ar^3 + ar^4 + \\cdots + ar^{n}, \\\\ s - rs &= a-ar^{n}, \\\\ s(1-r) &= a(1-r^{n}), \\end{align} so, $s = a \\frac{1-r^{n}}{1-r} \\quad \\text{(if } r \\neq 1 \\text{)}.$ As n goes to infinity, the absolute value of r", "r^2}$. Here, we can write the series as $\\displaystyle \\frac{1}{2}\\left(1+ \\frac{1}{2}+ \\frac{1/4}+ \\cdot\\cdot\\cdot\\right)$ so 1/2 times a geometric sequence with a= 1 and r= 1/2. Since 1/2< 1, this converges to $\\displaystyle \\frac{1}{2}\\frac{1}{1- \\frac{1}{2}}= \\frac{1}{2}\\frac{2}{2- 1}= 1$. (This is also a geometric series with a= 1/2 and r= 1/2 so converges to $\\displaystyle \\frac{\\frac{1}{2}}{1- \\frac{1}{2}}= \\frac{\\frac{1}{2}}{\\frac{1}{2}}= 1$) (It can also be thought of as a geometric series with a= 1 and r= 1/2 with the first \"a\" term missing. The geometric series with a= 1 and r= 1/2, $\\displaystyle 1+ \\frac{1}{2}+ \\frac{1}{4}+ \\frac{1}{8}+ \\cdot\\cdot\\cdot$ converges to $\\displaystyle \\frac{1}{1- \\frac{1}{2}}= 2$ and subtracting the missing first term, \"1\" gives a sum of 1.) v8archie August 28th, 2015 06:42 AM Quote: Originally Posted by Country Boy (Post 334776) One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\\displaystyle \\frac{a}{1- r^2}$. That should be $\\displaystyle \\frac{a}{1- r}$. Country Boy August 28th, 2015 11:41 AM How in the world did that \"2\" sneak in there?! Yes, of course. Thank you. Timios August 28th, 2015 11:49 AM Perhaps I'm stupid, but I see no necessity of first proving that the series converges. Whether it converges or not, the proof I provided is purely algebraic: Multiplying each side of an equation by 2, and Subtracting", "to $$995$$, inclusive. This sum is $$0$$. Thus, \\begin{aligned} -995-993-991-\\cdots+991 + 993 +& 995 + 997 + 999 + \\cdots + 1023+1025\\\\ &= 0 + 997 + 999 + \\cdots + 1023 + 1025\\\\ &= 997(15) + 2 + 4 + \\cdots + 28\\\\ &= 14\\,955 + 2(1 + 2 + \\cdots +14)\\\\ &= 14\\,955 + 2\\left[\\frac{14(15)}{2}\\right]\\\\ &= 14\\,955 +210\\\\ &= 15\\,165\\end{aligned} • Method 2: This is an arithmetic series with $$n=1011$$ terms, first term $$t_1 = -995$$, and last term $$t_n = 1025$$. Then, using the formula $$S_n = {n}\\left[ \\dfrac{t_1 + t_n}{2} \\right]$$ for the sum of a series, \\begin{aligned} -995-993-991-989-\\cdots+1023+1025&=1011\\left[\\frac{-995+1025}{2}\\right]\\\\ &=1011(15)\\\\ &=15\\,165\\end{aligned} Therefore, the sum of the terms in the odd-numbered positions is $$15\\,165$$.", "# What's the common ratio in 5/12, 1/3, 4/15, 16/75, 64/375? Dec 3, 2015 $\\frac{4}{5}$ #### Explanation: Divide any number in this sequence by the preceding number and you get $\\frac{4}{5}$. That's the common ratio. For example: $\\frac{1}{3} \\div \\frac{5}{12} = \\frac{1}{3} \\cdot \\frac{12}{5} = \\frac{4}{5}$, $\\frac{4}{15} \\div \\frac{1}{3} = \\frac{4}{15} \\cdot 3 = \\frac{12}{15} = \\frac{4}{5}$, $\\frac{16}{75} \\div \\frac{4}{15} = \\frac{16}{75} \\cdot \\frac{15}{4} = \\frac{4}{5}$, and $\\frac{64}{375} \\div \\frac{16}{75} = \\frac{64}{375} \\cdot \\frac{75}{16} = \\frac{4}{5}$. A cool thing related to this is the corresponding geometric series (infinite \"sum\") $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots$. If $a$ is the first term of such a series, and $r$ is the common ratio, with $| r | < 1$, the series converges to $\\frac{a}{1 - r}$. For this example, $a = \\frac{5}{12}$ and $r = \\frac{4}{5}$ so that the series converges to $\\frac{\\frac{5}{12}}{1 - \\frac{4}{5}} = \\frac{\\frac{5}{12}}{\\frac{1}{5}} = \\frac{5}{12} \\cdot 5 = \\frac{25}{12}$ and we write $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots = \\frac{25}{12}$.", "= r\\text{.} \\end{equation*} Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion \\begin{equation*} N=0.1212\\overline{12} \\end{equation*} as a rational number. Observe that \\begin{align*} N \\amp = 0.12 + 0.0012 + 0.000012 + \\cdots\\\\ \\amp = \\left(\\frac{12}{100}\\right) + \\left(\\frac{12}{100}\\right)\\left(\\frac{1}{100}\\right) + \\left(\\frac{12}{100}\\right)\\left(\\frac{1}{100}\\right)^2 + \\cdots\\text{.} \\end{align*} This is an infinite geometric series with $a=\\frac{12}{100}$ and $r = \\frac{1}{100}\\text{.}$ By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following example. ###### Example7.17 Let $r \\ne 1$ and $a$ be real numbers and let \\begin{equation*} S = a+ar+ar^2 + \\cdots ar^{n-1} + \\cdots \\end{equation*} be an infinite geometric series. For each positive integer $n\\text{,}$ let \\begin{equation*} S_n = a+ar+ar^2 + \\cdots + ar^{n-1}\\text{.} \\end{equation*} Recall that \\begin{equation*} S_n = a\\frac{1-r^n}{1-r}\\text{.} \\end{equation*} 1. What should we allow $n$ to approach in order to have $S_n$ approach $S\\text{?}$ 2. What is the value of $\\lim_{n \\to \\infty} r^n$ for $|r| \\gt 1\\text{?}$ for $|r| \\lt 1\\text{?}$ Explain. 3. If $|r| \\lt 1\\text{,}$ use the formula for $S_n$ and your observations in (a) and (b) to explain why $S$ is finite and find a resulting", "\\end{align*}, \\begin{align*} &= 5(128 - 1) \\\\ \\therefore \\frac{8}{27} &= \\frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\\\ &= 45 \\therefore ar^{2} &= 8 \\\\ We think you are located in Determine the values of $$r$$ and $$n$$ if $$S_{n} = 84$$. S_{n} &= \\frac{a(1-r^{n})}{1 - r}\\\\ Calculate: $\\sum _{k = 1}^{6}{32 \\left( \\frac{1}{2} \\right)^{k-1}}$, We have generated the series $$32 + 16 + 8 + \\cdots$$. \\text{And } 20 &= ar^{2} \\\\ \\end{align*}, \\begin{align*} &= ar^{3}(1 + r + r^{2}) \\\\ The eighth term of a geometric sequence is $$\\text{640}$$. \\text{And } T_{3} &= 8 \\\\ 2^{2 - n}&= \\frac{1}{128} \\\\ We generate a geometric sequence using the general form: Tn = a ⋅ rn − 1. where. a &= 4 \\\\ &= \\frac{1 - 6561}{4}\\\\ Finite geometric sequence: 1 2 , 1 4 , 1 8 , 1 16 , ... , 1 32768. \\therefore a &= 8 \\times \\frac{4}{9} \\\\ &= \\frac{54(1 - \\left( \\frac{1}{3} \\right)^{n})}{1 - \\left( \\frac{1}{3} \\right)} \\\\ a \\left( \\frac{3}{2} \\right)^{2} &= 8 \\\\ This calculus video tutorial explains how to find the sum of a finite geometric series using a simple formula. T_{1} + T_{2} + T_{3} &= a + ar + ar^{2} \\\\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\\\ r &= \\frac{1}{3} \\\\ 1.5 Finite", "+0 0 299 1 +644 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. waffles Sep 20, 2017 #1 +20680 +1 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. If the common ratio r lies between −1 to 1 , we can have the sum of an infinite geometric series. That is, the sum exits for $$| r |<1$$ . The sum S of an infinite geometric series with $$-1 is given by the formula, \\(S=\\frac{a_1}{1-r}$$ Let $$a_1 = -\\frac{3}{2}$$ Let $$r =\\,?$$ $$\\begin{array}{|rcll|} \\hline S = \\frac{a_1}{1-r} &=& 2r \\quad & | \\quad a_1 = -\\frac{3}{2} \\\\ \\frac{-\\frac{3}{2}}{1-r} &=& 2r \\\\ -\\frac{3}{2} \\cdot \\frac{1}{1-r} &=& 2r \\quad & | \\quad : 2 \\\\ -\\frac{3}{4} \\cdot \\frac{1}{1-r} &=& r \\\\ \\frac{3}{4} \\cdot \\frac{1}{r-1} &=& r \\quad & | \\quad \\cdot (r-1) \\\\ \\frac{3}{4} &=& r \\cdot (r-1) \\quad & | \\quad - \\frac{3}{4} \\\\ r \\cdot (r-1) - \\frac{3}{4} &=& 0 \\\\ r^2-r- \\frac{3}{4} &=& 0 \\\\\\\\ r &=& \\frac{1\\pm\\sqrt{1-4\\cdot (- \\frac{3}{4}) } }{2} \\\\ r &=& \\frac{1\\pm\\sqrt{1+3 } }{2} \\\\ r &=& \\frac{1\\pm 2 }{2} \\\\\\\\ r_1 &=& \\frac{1 + 2 }{2}", "that it adds up to $$1$$: $$1 = \\frac12 + \\frac14 + \\frac18 + \\frac1{16} + \\cdots,$$ which is just a number-heavy way to say that the probability is $$1$$ that the coin eventually will come up heads. (Probability $$\\frac12$$ on the first throw, $$\\frac14$$ that it takes exactly two throws, $$\\frac18$$ that it takes exactly three throws, etc.) If we multiply all the terms by $$2$$ we get twice as much for the sum: \\begin{align} 2 &= 1 + \\frac12 + \\frac14 + \\frac18 + \\cdots, \\\\ 4 &= 2 + 1 + \\frac12 + \\frac14 + \\cdots. \\end{align} If we divide all the terms by $$2$$ we get half as much: \\begin{align} \\frac12 & = \\frac14 + \\frac18 + \\frac1{16} + \\frac1{32} + \\cdots,\\\\ \\frac14 & = \\frac18 + \\frac1{16} + \\frac1{32} + \\frac1{64} + \\cdots,\\\\ \\frac18 & = \\frac1{16} + \\frac1{32} + \\frac1{64} + \\frac1{128} + \\cdots,\\\\ \\end{align} and so forth. Rearranging terms in a series can alter the sum if some terms are positive and some are negative; but in the series you are looking at, all terms are positive, so we can write: \\begin{align} 2\\cdot\\frac12 + 4\\cdot\\frac14 &+ 6\\cdot\\frac18 + 8\\cdot\\frac1{16} + \\cdots\\\\ &= 1 + 2\\cdot\\frac12 + 3\\cdot\\frac14 + 4\\cdot\\frac18 + \\cdots\\\\ &= 1 + (1+1)\\cdot\\frac12 + (1+1+1)\\cdot\\frac14 + (1+1+1+1)\\cdot\\frac18 + \\cdots\\\\", "+0 # Please Help 0 272 1 +644 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. waffles Sep 20, 2017 ### Best Answer #1 +20022 +1 An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio. If the common ratio r lies between −1 to 1 , we can have the sum of an infinite geometric series. That is, the sum exits for $$| r |<1$$ . The sum S of an infinite geometric series with $$-1 is given by the formula, \\(S=\\frac{a_1}{1-r}$$ Let $$a_1 = -\\frac{3}{2}$$ Let $$r =\\,?$$ $$\\begin{array}{|rcll|} \\hline S = \\frac{a_1}{1-r} &=& 2r \\quad & | \\quad a_1 = -\\frac{3}{2} \\\\ \\frac{-\\frac{3}{2}}{1-r} &=& 2r \\\\ -\\frac{3}{2} \\cdot \\frac{1}{1-r} &=& 2r \\quad & | \\quad : 2 \\\\ -\\frac{3}{4} \\cdot \\frac{1}{1-r} &=& r \\\\ \\frac{3}{4} \\cdot \\frac{1}{r-1} &=& r \\quad & | \\quad \\cdot (r-1) \\\\ \\frac{3}{4} &=& r \\cdot (r-1) \\quad & | \\quad - \\frac{3}{4} \\\\ r \\cdot (r-1) - \\frac{3}{4} &=& 0 \\\\ r^2-r- \\frac{3}{4} &=& 0 \\\\\\\\ r &=& \\frac{1\\pm\\sqrt{1-4\\cdot (- \\frac{3}{4}) } }{2} \\\\ r &=& \\frac{1\\pm\\sqrt{1+3 } }{2} \\\\ r &=& \\frac{1\\pm 2 }{2} \\\\\\\\", "and so on: $\\frac{2}{3}s \\;=\\; \\frac{2}{3} \\,+\\, \\frac{4}{9} \\,+\\, \\frac{8}{27} \\,+\\, \\frac{16}{81} \\,+\\, \\cdots$ This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series cancels every term in the original but the first: $s \\,-\\, \\tfrac23s \\;=\\; 1,\\;\\;\\;\\;\\;\\;\\;\\;\\mbox{so }s=3.$ A similar technique can be used to evaluate any self-similar expression. ### Formula For $r\\neq 1$, the sum of the first n terms of a geometric series is: $a + ar + a r^2 + a r^3 + \\cdots + a r^{n-1} = \\sum_{k=0}^{n-1} ar^k= a \\, \\frac{1-r^{n}}{1-r},$ where a is the first term of the series, and r is the common ratio. We can derive this formula as follows: \\begin{align} &\\text{Let }s = a + ar + ar^2 + ar^3 + \\cdots + ar^{n-1}. \\\\[4pt] &\\text{Then }rs = ar + ar^2 + ar^3 + ar^4 + \\cdots + ar^{n} \\\\[4pt] &\\text{Then }s - rs = s(1-r) = a-ar^{n},\\text{ so }s = a \\frac{1-r^{n}}{1-r}. \\end{align} The formula follows by multiplying through by a. As n goes to infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes $s \\;=\\; \\sum_{k=0}^\\infty ar^k = \\frac{a}{1-r}=a+ar+ar^2+ar^3+ar^4+\\cdots.$ When a = 1, this simplifies to: $1 \\,+\\, r \\,+\\,", "in terms of $r$ and the first term. Let's start with the relation $\\frac{a_{n}}{a_{n-1}} = r$. This gives $a_{n} = r a_{n-1}$. Using this again, we get $a_{n} = r (ra_{n-2}) = r^{2} a_{n-2}$. We can continuer this way and get: \\begin{align*} a_{n} &= r a_{n-1} \\\\ &= r^{2} a_{n-2} \\\\ &= r^{3} a_{n-3} \\\\ &\\vdots \\\\ &= r^{n-2} a_{2} \\\\ &= r^{n-1} a_{1} \\\\ \\end{align*} So we get that for a geometric sequence $a_{n}$ with ratio $r$, the formula for $a_{n}$ is given by: $$a_{n} =r^{n-1} a_{1}$$ ### Example. Consider the geometric sequence $3, 6, 12, 24, 48, \\ldots$. Find a formula for $a_{n}$ and use it to find $a_{7}$. Solution. To find $r$, we should look at the ratio between successive terms: $r= \\frac{a_{1}}{a_{2}} =\\frac{6}{3} = 2$. Then using the formula above we get $a_{n} = r^{n-1} a_{1} = 2^{n-1} \\cdot 3$. To find $a_{7}$ we set $n=7$ and get $a_{7} = 2^{7-1} \\cdot 3 = 2^{6} \\cdot 3 = 64 \\cdot 3 = 192$. ### Part 3: Recursive Sequences We have already briefly discussed this idea in the first paragraph. We shall now discuss this in more detail, together with some extra examples. As we saw in the section on geometric sequences, we can define a geometric sequence either by the rule $a_{n} = r^{n-1}", "# How do you find the 13th term of the geometric sequence with a3 = 24 and a5 = 96? Nov 24, 2015 ${a}_{13} = 24576$ #### Explanation: Well The definition of A geometric sequence with n elements; a, ar,ar^2,ar^3, ar^4 ...ar^(n-1 r is a common constant with which every number is multiplied Now ${a}_{5} = a \\cdot {r}^{4} = 96 - - - - - - - - 1$ ${a}_{3} = a \\cdot {r}^{2} = 24 - - - - - - - - 2$ Now divide 1 by 2 $\\frac{\\cancel{a} \\cdot {\\cancel{r}}^{2} {r}^{2}}{\\cancel{a} \\cdot {\\cancel{r}}^{2}} = \\frac{96}{24} = 4$ ${r}^{2} = 4 \\implies r = 2$ Now come back to equation 2 $a \\cdot {r}^{2} = 24$ Substitute $a \\cdot 4 = 24$ $\\implies a = 6$ Now we have found the first term Lets finish it ${a}_{13} = a \\cdot {r}^{13 - 1} = 6 \\cdot {2}^{6} \\cdot {2}^{6} = 24576$ And were done" ]

[ "\\prod E(e^{{t\\over n}X_i})= (e^{\\mu {t\\over n} + 0.5 \\frac{\\sigma^2}{n^2}t^2})^n\\to e^{\\mu t}$ or did I make another dumb mistake somewhere? 12. That makes senses. Thank you (and everyone else) for all their help", "1 2 \\ln\\frac \\pi 2$$ So here you go: $$\\prod_{n=1}^\\infty(n)= (2\\pi)^{0.5}$$", "would be $$\\prod_{i=1}^{n} \\frac{1}{p}e^{-\\frac{x_i}{p}}=p^{-n}e^{-\\frac{1}{p}T(x)},\\qquad T(x)= \\sum_{i=1}^{n} x_i$$ Let $h(x) =1$ and $g(p,t) = p^{-n}e^{-t/p}$. Then the result follows from the factorization theorem.", "\\prod_{k=1}^n k^{n+1} = (n!)^{n+1}$ Thus $(\\prod_{k=1}^n k!)^2 \\sim (2\\pi)^n \\sqrt{n!} \\frac{(n!)^{n+1}}{e^{n(n+1)/2}}$ Then we get, applying Stirling's approximation to $\\sqrt{n!}$ and $(n!)^{n+1}$: $(\\prod_{k=1}^n k!)^2 \\sim (2\\pi)^{2n+3/2}n^{3/4 + n/2}e^{-n(n+1)/2-n/2-n(n+1)} n^{n(n+1)}$ Thus $\\prod_{k=1}^n k! \\sim (2\\pi)^{n+3/4} e^{-3n^2/4 - n} n^{n^2/2 + 3n/4 + 3/8}$", "# Coding challenge Write some code to prove that $\\prod_{n=2}^{\\infty}(1-\\frac{1}{n^2})=\\frac{1}{2}$. (From Andreescu, p. 39.)", "# Evaluate $\\sum_{i=1}^{\\infty}\\prod_{j=1}^{i}\\frac{j-\\frac{1}{2}}{j\\left(j+\\frac{1}{2}\\right)^2}$ by Power ranger Last Updated July 12, 2019 10:20 AM - source We want to evaluate this formula sum, $$\\sum_{i=1}^{\\infty}\\prod_{j=1}^{i}\\frac{j-\\frac{1}{2}}{j\\left(j+\\frac{1}{2}\\right)^2}$$ Tags :", "# Calculate $\\prod_{n=2}^\\infty\\bigg(1-\\frac1{n^2}\\bigg)$ [duplicate] This question already has an answer here: How do you compute $\\prod_{n=2}^\\infty\\bigg(1-\\frac1{n^2}\\bigg)$ I have no clue on how to start. Any help would be appreciated. ## marked as duplicate by Simply Beautiful Art, Jaideep Khare, Namaste, lab bhattacharjee, NChJun 1 '17 at 1:36 Hint. This may be seen as a telescoping product using $$1-\\frac1{n^2}=\\frac{n-1}{n} \\cdot \\frac{n+1}{n}, \\qquad n\\ge2.$$ Hope you can finish it.", "- 1} i}{(k - 2)! \\times (n - k - 1)!}$ Note that the numerator is only related to $v$. The overall answer can be simplified: $\\sum_{k = 2}^{n} \\sum_{v = n - 1}^{m} \\frac{\\prod_{i = v - n + 2} ^ {v - 1} i}{(k - 2)! \\times (n - k - 1)!} = \\sum_{k = 2}^{n} \\frac{\\sum_{v = n - 1}^{m}\\prod_{i = v - n + 2} ^ {v - 1} i}{(k - 2)! \\times (n - k - 1)!}$ After preprocessing the factorial, Enumerate $v$ to calculate the numerator $\\sum_{v = n-1} ^ {m} \\prod_ {i = v-n + 2} ^ {v-1} i$; Then enumerate $k$ and calculate the denominator $(k-2)! \\times(n-k-1)!$. Complexity $O (n \\ log P)$, where $P = 998244353$. • +7", "$$\\sum_{p=0}^n\\frac{1}{x+p} (-1)^{p} {n\\choose p} = n! \\times \\prod_{q=0}^n \\frac{1}{x+q}$$ which is the claim.", "to calculate $$\\Big(\\frac{4}{20}\\Big)^2\\times \\mathbb{P}[Y=3]$$", "Your case: $$\\prod\\frac{1}{1+a_j} = \\prod (1+b_j),$$ where $$b_j = 1 - \\frac{1}{1+a_j} = \\frac{a_j}{1+a_j}$$ So the criterion becomes: $$\\sum\\left|\\frac{a_j}{1+a_j}\\right| < +\\infty$$ As noted, this is equivalent to $$\\sum |a_j| < +\\infty$$.", "anonymous one year ago Show that $\\prod_{n=2}^N\\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}=\\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}$ 1. ybarrap Simplifying the product a bit: $$\\prod_{n=2}^N\\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\\\ =\\prod_{n=2}^N\\frac{\\left((n-1)(n+1)\\right)^{n^2}}{n^{(n+1)^2+(n-1)^2}}\\\\ =\\prod_{n=2}^N\\cfrac{(n^2-1)^{n^2}}{n^{2(n^2+1)}}\\\\ =\\prod_{n=2}^N\\cfrac{1}{n^2}\\left(\\frac{n^2-1}{n^2}\\right)^{n^2}$$ Using this simplified form, we use Induction to prove: $$=\\prod_{n=2}^N\\cfrac{1}{n^2}\\left(\\frac{n^2-1}{n^2}\\right)^{n^2}=\\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}$$ Basis: $$n=2$$ $$\\cfrac{1}{2^2}\\left(\\frac{2^2-1}{2^2}\\right)^{2^2}=\\frac{3^4}{2^{10}}\\\\ =\\frac{(2+1)^{2^2}}{2\\cdot 2^{(2+1)^2}}=\\frac{3^4}{2^{10}}\\\\$$ Inductive Step: We assume equation holds for $$k$$, then for $$k+1$$ $$=\\prod_{n=2}^{k+1}\\cfrac{1}{n^2}\\left(\\frac{n^2-1}{n^2}\\right)^{n^2}\\\\ =\\frac{1}{(k+1)^2}\\left (\\frac{(k+1)^2-1}{(k+1)^2}\\right )^{(k+1)^2}\\prod_{n=2}^{k}\\cfrac{1}{n^2}\\left(\\frac{n^2-1}{n^2}\\right)^{n^2}\\\\ =\\frac{1}{(k+1)^2}\\left (\\frac{(k+1)^2-1}{(k+1)^2}\\right )^{(k+1)^2}\\frac{(k+1)^{k^2}}{2k^{(k+1)^2}}\\text{ by Induction Hypothesis}\\\\ =\\frac{1}{(k+1)^2}\\frac{(k+1)^{k^2}}{(k+1)^{2(k+1)^2}}\\times \\frac{ \\left((k+1)^2-1\\right )^{(k+1)^2}}{2k^{k+1)^2}}\\\\ =\\frac{1}{(k+1)^{(2-k^2+2(k+1)^2)}}\\times \\frac{\\left(k^2+2k\\right)^{(k+1)^2}}{2k^{(k+1)^2}} \\\\ =\\frac{1}{(k+1)^{((k+1)+1)^2}}\\frac{\\left( (k+1)+1\\right )^{(k+1)^2 }}{2}\\\\ =\\frac{\\left( (k+1)+1\\right )^{(k+1)^2}}{2(k+1)^{((k+1)+1)^2}}\\\\$$ QED Does this make sense? 2. anonymous actually those 'simplifications' were a bad idea: \\prod_{n=2}^N(n-1)^{n^2}\\cdot\\prod_{n=2}^N(n+1)^{n^2}\\cdot\\prod_{n=2}^Nn^{-(n+1)^2}\\cdot\\prod_{n=2}^Nn^{-(n-1)^2}\\\\\\begin{align*}\\quad&=\\prod_{n=1}^{N-1}n^{(n+1)^2}\\cdot\\prod_{n=3}^{N+1} n^{(n-1)^2}\\cdot\\prod_{n=2}^Nn^{-(n+1)^2}\\cdot\\prod_{n=2}^Nn^{-(n-1)^2}\\\\&=1^{2^2}\\cdot\\prod_{n=2}^{N-1}n^{(n+1)^2}\\cdot\\prod_{n=3}^{N+1} n^{(n-1)^2}\\cdot\\prod_{n=2}^{N}n^{-(n+1)^2}\\cdot\\prod_{n=3}^Nn^{-(n-1)^2}\\cdot2^{-1^2}\\\\&=\\frac12\\cdot\\prod_{n=2}^{N-1}n^{(n+1)^2}\\cdot(N+1)^{N^2}\\cdot\\prod_{n=3}^{N} n^{(n-1)^2}\\cdot N^{-(N+1)^2}\\cdot\\prod_{n=2}^{N-1}n^{-(n+1)^2}\\cdot\\prod_{n=3}^Nn^{-(n-1)^2}\\\\&=\\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\\end{align*} 3. anonymous this is a telescoping product and it works best factored 4. ybarrap Brilliant! 5. anonymous Nicely done! This problem came from an interesting question about an infinite product representation of $$\\pi$$: $\\pi=e^{3/2}\\prod_{n=2}^\\infty e\\left(1-\\frac{1}{n^2}\\right)^{n^2}$ For those interested, here's the link for the rest of the derivation: http://math.stackexchange.com/a/1346822/170231 6. ybarrap Ahhh... even simpler! $$\\prod_{n=2}^N\\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\\\ =\\prod_{n=2}^N\\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\\prod_{n=2}^N\\frac{(n+1)^{n^2}}{n^{(n-1)^2}}\\\\ \\text{But, }\\\\ \\prod_{n=2}^N\\frac{(n-1)^{n^2}}{n^{(n+1)^2}}=\\frac{\\cancel{(N-1)^{N^2}}}{N^{(N+1)^2} }\\frac{(N-2)^{(N-1)^2}}{\\cancel{(N-1)^{N^2} }}\\cdots \\frac{\\cancel{2^{3^2}}}{3^{4^2}}\\frac{(2-1)^{2^2}}{\\cancel{2^{3^2 }}}=\\frac{1}{N^{(N+1)^2}}\\\\ \\text{and }\\\\ \\prod_{n=2}^N\\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\\frac{(N+1)^{N^2}}{\\cancel{N^{(N-1)^2}}}\\frac{\\cancel{N^{(N-1)^2}}}{(N-1)^{(N-2)^2}}\\cdots \\frac{4^{2^2}}{\\cancel{3^{2^2}}}\\frac{\\cancel{3^{2^2}}}{2^{1^2}}=\\frac{(N+1)^{N^2}}{2}\\\\ \\text{then}\\\\ =\\prod_{n=2}^N\\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\\prod_{n=2}^N\\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\\frac{1}{N^{(N+1)^2}}\\frac{(N+1)^{N^2}}{2}\\\\$$ Thanks @SithsAndGiggles for this problem! 7. anonymous eh that's more or less exactly the thing I did, I just separated into four rather than two products so you could see the cancellation more clearly without the risk of losing track of terms 8. anonymous but yeah, that illustrates the telescoping more clearly :p", "either of both numbers We thus can write now $$m=d\\prod_{k=z+1}^rp_k^{a_k}\\,\\,,\\,\\,q=d\\prod_{i=z+1}^s q_i^{b_i}\\Longrightarrow \\varphi(mn)=mn\\prod_{j=1}^z\\left(1-\\frac{1}{p_j}\\right)\\prod_{k=z+1}^r\\left(1-\\frac{1}{p_k}\\right)\\prod_{i=z+1}^s\\left(1-\\frac{1}{q_i}\\right)$$and since $$\\prod_{j=1}^z\\left(1-\\frac{1}{p_j}\\right)=\\frac{\\varphi(d)}{d}$$the formula you want follows at once. DonAntonio", "# How to estimate $\\prod_{t=1}^{N}\\frac{1}{2-z^t}$ for large $N$? Based on the top answer to How to estimate of $\\prod_{k=a}^N \\frac{1}{e^{k\\kappa}-1}$ for large $N$? Can anyone find an approximate closed form for $$\\frac{\\mathrm{d}^k}{\\mathrm{d}z^k}\\prod_{t=1}^{N}\\frac{1}{2-z^t}$$ by approximating the product (which we then derive $k$ times) for large $N$? The answer is $$\\frac{\\mathrm{d}^k}{\\mathrm{d}z^k}\\left(\\frac{1}{2^{1+N}(\\frac{1}{2} ; z)_{1+N}}\\right)$$ which involves the q-Pochhammer symbol, but the closed form for the $k^{\\text{th}}$ derivative w.r.t. $z$ is impossible to find due to difficulty deriving the q-Pochhammer symbol. Is the previous question mentioned at the beginning a good way to simplify this?", "# another hard to evaluate product $\\prod_{m=2}^{\\infty} \\prod_{n=1}^{\\infty} \\left(1+\\frac{1}{{}^{m}n}\\right)$ That’s $n$, tetrated to the $m$th.", "\\\\ \\\\[-4mm] 13 & \\frac{6}{36} \\\\ \\\\[-4mm] 14 & \\frac{5}{36} \\\\ \\\\[-4mm] 15 & \\frac{4}{46} \\\\ \\\\[-4mm] 16 & \\frac{3}{36} \\\\ \\\\[-4mm] 17 & \\frac{2}{36} \\\\ \\\\[-4mm] 18 & \\frac{1}{36} \\end{array}$ Now you can calculate the expected value.", "# When a number is added to 1/5 of itself, the result is 24. The equation that models this problem is n+1/5n=24. What is the value n? Here $n + \\frac{1}{5} n = 24$ $\\Rightarrow \\frac{5 n + n}{5} = 24$ $\\Rightarrow \\frac{6}{5} n = 24$ $\\Rightarrow n = 24 \\times \\frac{5}{6}$ $\\Rightarrow n = 20$", "\\in \\Bbb{R}^n, (a_1,...a_n) \\in [-1,+\\infty)^n \\Rightarrow \\prod_{i=1}^{n}(1+a_i) ≥ 1 + \\sum_{i=1}^{n} a_i \\right \\}$ work? – Reveillark Feb 26 '15 at 2:49", "\\prod^{n}_{i=1}\\frac{1+x_i}{x_i}\\ge \\left(\\frac{n+S}{S}\\right)^n$$ Now plug in $n=6,S=2$ and be forever happy.", "Let $a_1 = \\frac{1}{2011}$ and let $a_{n+1} = a_n \\cdot (a_n+1)$. Evaluate $$\\frac{1}{a_1+1}+\\frac{1}{a_2+1}+\\frac{1}{a_3+1}+\\cdots$$" ]

[ "denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*} \\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}. \\end{equation*} In practice, we don't always have to find the prime factorization. Instead, we can find one common factor at a time until no common factors remain. For example, since 126 and 24 are both even, we would write \\begin{equation*} \\frac{126}{24} = \\frac{63}{12}. \\end{equation*} Then, we look at 63 and 12 and recognize that they are both divisible by 3, allowing us to rewrite the fraction as \\begin{equation*} \\frac{126}{24} = \\frac{63}{12} = \\frac{21}{4}. \\end{equation*} Because $21=3 \\cdot 7$ and $4=2^2$ do not have common factors, we know this is simplified. A square root is not simplified if there is a factor inside the root that is a perfect square. Similarly, a cube root is not simplified if there is a factor inside that is a perfect cube. We use the factors of the value inside a root to determine if we can simplify it. ###### Example1.2.3 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that", "in ascending order, but every little bit helps. The denominator can also be expressed as a product of primes. $2310=10 \\cdot 231=2 \\cdot 5 \\cdot 7 \\cdot 33=2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$ We can now cancel common factors. $\\frac{210}{2310}=\\frac{2 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11}=\\frac{\\not{2} \\cdot \\not{3} \\cdot \\not{5} \\cdot \\not{7}}{\\not{2} \\cdot \\not{8} \\cdot \\not{5} \\cdot \\not{7} \\cdot 11}=\\frac{1}{11}$ However, this approach is not the most efficient way to proceed, as multiplying numerators and denominators allows the products to grow to larger numbers, as in 210/2310. It is then a little bit harder to prime factor the larger numbers. A better approach is to factor the smaller numerators and denominators immediately, as follows. $\\frac{6}{231} \\cdot \\frac{35}{10}=\\frac{2 \\cdot 3}{3 \\cdot 7 \\cdot 11} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 5}$ We could now multiply numerators and denominators, then cancel common factors, which would match identically the last computation in equation (5). However, we can also employ the following cancellation rule. Cancellation Rule When working with the product of two or more rational expressions, factor all numerators and denominators, then cancel. The cancellation rule is simple: cancel a factor “on the top” for an identical factor “on the bottom.” Speaking more technically, cancel any factor in any numerator for", "Keep highest power of each factor.}} \\\\ = 4 \\cdot 3 \\cdot 7 ~ & \\textcolor{red}{ \\text{ Expand: } 2^2 = 4, ~ 3^1 = 3, ~ 7^1 = 7.} \\\\ = 84 ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Create equivalent fractions with the new LCD, then add. \\begin{aligned} \\frac{5}{28} + \\frac{11}{42} = \\frac{5 \\cdot \\textcolor{red}{3}}{28 \\cdot \\textcolor{red}{3}} + \\frac{11 \\cdot \\textcolor{red}{2}}{42 \\cdot \\textcolor{red}{2}} ~ & \\textcolor{red}{ \\text{ Equivalent fractions with LCD = 84.}} \\\\ = \\frac{15}{84} + \\frac{22}{84} ~ & \\textcolor{red}{ \\text{ Simplify numerators and denominators.}} \\\\ = \\frac{37}{84} ~ & \\textcolor{red}{ \\text{ Keep LCD; add numerators.}} \\end{aligned}\\nonumber Exercise $$\\PageIndex{10}$$ Simplify: $$\\frac{5}{24} + \\frac{5}{36}$$ Answer 25/72 Example $$\\PageIndex{11}$$ Simplify: $$- \\frac{11}{24} - \\frac{1}{18}$$. Solution Prime factor the denominators in compact form using exponents. 24 = 2 · 2 · 2 · 3=23 · 31 18 = 2 · 3 · 3=21 · 32 To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2 and 3. The highest power of 2 that appears is 23. The highest power of 3 that appears is 32. \\begin{aligned} \\text{LCM} = 2^3 \\cdot 3^2 ~ & \\textcolor{red}{ \\text{ Keep highest power of each factor.}} \\\\ = 8 \\cdot 9 ~ & \\textcolor{red}{ \\text{ Expand: } 2^3 =", "# Math Help - How to calculate the limit of this product? 1. ## How to calculate the limit of this product? $\\prod_{n=2}^{\\infty}(1-\\frac{1}{n^2})=\\lim_{n\\to\\infty}(1-\\frac{1}{4})(1-\\frac{1}{9})\\dots (1-\\frac{1}{n^2})$ Thank you! 2. ## Re: How to calculate the limit of this product? Hi gotmejerry! $(1-{1\\over n^2}) = {(n-1)(n+1) \\over n^2}$ Can you write out a couple of factors and see how things cancel? 3. ## Re: How to calculate the limit of this product? No unfortunately I cant. 4. ## Re: How to calculate the limit of this product? Which factor do you get if you fill in n=2? And which factor for n=3? And n=4? 5. ## Re: How to calculate the limit of this product? 3/4 8/9 15/16 but what do i do with them, how do they cancel each other? 6. ## Re: How to calculate the limit of this product? Let's write that out: ${1 \\cdot 3 \\over 2^2} \\cdot {2 \\cdot 4 \\over 3^2} \\cdot {3 \\cdot 5 \\over 4^2} \\cdot {4 \\cdot 6 \\over 5^2} \\cdot ... \\cdot {(n-1) (n+1) \\over n^2}$ Do you see numbers that cancel? 7. ## Re: How to calculate the limit of this product? Yes, thanks, so If I am not wrong, only 1/2 *(n+1)/n remains? And its limit is 1/2. Thank you very much Right! 9. ## Re: How to calculate", "1 in the numerator. The same rule applies to the denominator. If everything in the denominator cancels, you’re left with a 1 in the denominator. Example 7 Simplify the product: - \\frac{6x}{55y} \\cdot \\left( - \\frac{110y^2}{105x^2} \\right).\\nonumber \\] Solution The product of two negatives is positive. \\begin{aligned} - \\frac{6x}{55y} \\cdot \\left( - \\frac{110y^2}{105x^2} \\right) = \\frac{6x}{55y} \\cdot \\frac{110y^2}{105x^2} ~ & \\textcolor{red}{ \\text{ Like signs gives a positive.}} \\end{aligned}\\nonumber Prime factor numerators and denominators, then cancel common factors. \\begin{aligned} = \\frac{2 \\cdot 3 \\cdot x}{5 \\cdot 11 \\cdot y} \\cdot \\frac{2 \\cdot 5 \\cdot 11 \\cdot y \\cdot y}{3 \\cdot 5 \\cdot 7 \\cdot x \\cdot x} ~ & \\textcolor{red}{ \\text{ Prime factor numerators & denominators.}} \\\\ = \\frac{2 \\cdot \\cancel{3} \\cdot \\cancel{x}}{5 \\cdot \\cancel{11} \\cdot \\cancel{y}} \\cdot \\frac{2 \\cdot \\cancel{5} \\cdot \\cancel{11} \\cdot \\cancel{y} \\cdot y}{ \\cancel{3} \\cdot \\cancel{5} \\cdot 7 \\cdot \\cancel{x} \\cdot x} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{2 \\cdot 2 \\cdot y}{5 \\cdot 7 \\cdot x} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\\\ = \\frac{4y}{35x} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\end{aligned}\\nonumber Exercise Simplify: $\\frac{6x}{15b} \\cdot \\left( - \\frac{35b^2}{10a^2} \\right)\\nonumber$ Answer $- \\frac{21b}{5a}\\nonumber$ ## Parallelograms In this section, we are going to learn how to find the area of a parallelogram. Let’s begin with the definition of a parallelogram.", "## [Soln] 2014 Singapore MO (Senior Rd 1) Problem 16 2014 SMO (Senior-Rd1) 16. Evaluate the sum $\\displaystyle\\frac{3! + 4!}{2(1! + 2!)} + \\frac{4! + 5!}{3(2! + 3!)} + \\dots + \\frac{12! + 13!}{11(10! + 11!)}$. Recall the definition of the factorial function: for non-negative integer $n$, $n! = \\begin{cases} 1 &\\text{ for } n = 0, \\\\ n \\cdot (n-1) \\cdot \\dots \\cdot 1 &\\text{ for } n \\geq 1. \\end{cases}$ When faced with a series of many terms, try to come up with an expression for the general $\\boldsymbol{n^{th}}$ term. In this question, the required sum is the sum of $\\displaystyle\\frac{(n+2)! + (n+3)!}{(n+1)[n! +(n+1)!]}$ for $n = 1, 2, \\dots, 10$. This looks like a term that can be simplified. Especially with fractions, a possible strategy is to pull out as many factors as possible in the hope that there can be some common factors in the numerator and denominator. These can be cancelled out. By pulling out common factors in the numerator: \\begin{aligned} (n+2)! + (n+3)! &= (n+2)! \\cdot [1 + (n + 3)] \\\\ &= (n+2)! \\cdot (n+4) \\end{aligned} and in the denominator: \\begin{aligned} (n+1)[n! + (n+1)!] &= (n+1) \\cdot n! \\cdot [1 + (n+1)] \\\\ &= (n+1) \\cdot n! \\cdot (n+2) \\\\ &= (n+2)! \\end{aligned} The original expression can be greatly simplified: $\\displaystyle\\frac{(n+2)!", "values of the numbers. First, multiply the numerators together to get the product’s numerator. Then, multiply the denominators together to get the product’s denominator. Rewrite in lowest terms, if needed. $$\\left( \\frac{3}{4} \\right)\\left( \\frac{2}{5} \\right)=\\frac{6}{20}=\\frac{3}{10}$$ The product of two negative numbers is positive. $$\\left( -\\frac{3}{4} \\right)\\left( -\\frac{2}{5} \\right)=\\frac{3}{10}$$ The following video shows examples of multiplying two signed fractions, including simplification of the answer. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=36 To summarize: • positive $$\\cdot$$ positive: The product is positive. • negative $$\\cdot$$ negative: The product is positive. • negative $$\\cdot$$ positive: The product is negative. • positive $$\\cdot$$ negative: The product is negative. You can see that the product of two negative numbers is a positive number. So, if you are multiplying more than two numbers, you can count the number of negative factors. ### Multiplying More Than Two Negative Numbers If there are an even number (0, 2, 4, …) of negative factors to multiply, the product is positive. If there are an odd number (1, 3, 5, …) of negative factors, the product is negative. ### Example Find $$3(−6)(2)(−3)(−1)$$. [hidden-answer a=”149062″]Multiply the absolute values of the numbers. $$\\begin{array}{l}3(6)(2)(3)(1)\\\\18(2)(3)(1)\\\\36(3)(1)\\\\108(1)\\\\108\\end{array}$$ Count the number of negative factors. There are three $$\\left(−6,−3,−1\\right)$$. $$3(−6)(2)(−3)(−1)$$ Since there are an", "+ 1)^4(k^4 + 8k^3 + 24k^2 + 32k + 16)}{8} \\\\ &= \\frac{(k + 1)^4(1\\cdot 2^0 \\cdot k^4 + 4\\cdot 2^1 \\cdot k^3 + 6\\cdot 2^2 \\cdot k^2 + 4\\cdot 2^3 \\cdot k^1 + 1\\cdot 2^4 \\cdot k^0)}{8} \\\\ &= \\frac{(k + 1)^4(k + 2)^4}{8}\\end{align*} Alternatively you can apply the factor theorem to factorise the quartic.", "to multiply large numbers, and secondly, you have to prime factor the even larger results. One possible workaround is to not bother multiplying numerators and denominators, leaving them in factored form. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{18 \\cdot 35}{30 \\cdot 6} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\end{aligned}\\nonumber Finding the prime factorization of these smaller factors is easier. \\begin{aligned} = \\frac{(2 \\cdot 3 \\cdot 3) \\cdot (5 \\cdot 7)}{(2 \\cdot 3 \\cdot 5) \\cdot (2 \\cdot 3)} ~ & \\textcolor{red}{ \\text{ Prime factor.}} \\end{aligned}\\nonumber Now we can cancel common factors. Parentheses are no longer needed in the numerator and denominator because both contain a product of prime factors, so order and grouping do not matter. \\begin{aligned} = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot \\cancel{7}}{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 2 \\cdot \\cancel{3}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber Another approach is to factor numerators and denominators in place, cancel common factors, then multiply. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{2 \\cdot 3 \\cdot 3}{2 \\cdot 3 \\cdot 5} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 3} ~ & \\textcolor{red}{ \\text{ Factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{3}}{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5}} \\cdot \\frac{ \\cancel{5} \\cdot 7}{2", "Chirsthoper 59 # What is the product? $\\frac{2a-7}{a}* \\frac{3a^2}{2a^2-11a+14}$ A. $\\frac{3}{a-2}$ B. $\\frac{3a}{a-2}$ C. $\\frac{3a}{a+2}$ D. $\\frac{3}{a+2}$ The denominator of the second fraction can be factored as (a-2)(2a-7), then it becomes doable. You cancel the (2a-7) factors, and are left with: $\\frac{3a^2}{a(a-2)} \\implies \\frac{3a}{a-2}$ Do note that you have erased the fact that a≠0 and a≠7/2, so you should always mention that.", "\\ \\ 2P\\cdot Q + Q^2 =0$$ the resultant of 2P and Q is also a force of magnitude P. $$(II) \\ \\ \\ \\ \\ (2P+Q)\\cdot (2P+Q) = P^2$$ $$(II) \\ \\ \\ \\ \\ 4P^2 + 4P\\cdot Q + Q^2 = P^2$$ Subtracting $(II) - 2(I)$ we get $$4P^2 - Q^2 = P^2$$ $$Q^2 = 3P^2$$ $$|Q| = \\sqrt{3} |P|$$ B) Substituting that in $(II)$: $$4P^2 + 4P\\cdot Q + 3P^2 = P^2$$ $$6P^2 = -4P\\cdot Q$$ $$3|P|\\cdot |P| = -2 |P| |Q| \\cos \\theta$$ $$3|P| = -2\\cdot \\sqrt{3} |P| \\cos \\theta$$ $$\\cos \\theta = -\\frac{\\sqrt{3}}{2}$$ $$\\theta = \\frac{5\\pi}{6}$$ C) After I introduced the algebraic and geometric definitions of the dot product, try mixing the two in this section. Good luck! • Thank you for your answer. At this stage in the book the concept of the dot product has not been introduced. I wonder if the question is mistakenly present or whether an anwser in terms of more primitive constructs is intended. Thanks again, Mitch. – gnitsuk Mar 16 '17 at 9:21", "follows simply from the fact that $p^2 - 1 = (p-1)(p+1)$; since $p$ is not divisible by either 2 or 3, • the factors $p-1$ and $p+1$ must both be even, Thus, their product must be divisible by $2\\cdot4\\cdot3 = 24$.", "\\cdot \\cancel{3}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber Note that this yields exactly the same result, 7/2. Cancellation Rule When multiplying fractions, cancel common factors according to the following rule: “Cancel a factor in a numerator for an identical factor in a denominator.” Example 6 Find the product of 14/15 and 30/140. Solution Multiply numerators and multiply denominators. Prime factor, cancel common factors, then multiply. \\begin{aligned} \\frac{14}{15} \\cdot \\frac{30}{140} = \\frac{14 \\cdot 30}{15 \\cdot 140} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\\\ = \\frac{(2 \\cdot 7) \\cdot (2 \\cdot 3 \\cdot 5)}{(3 \\cdot 5) \\cdot (2 \\cdot 2 \\cdot 5 \\cdot 7)} ~ & \\textcolor{red}{ \\text{ Prime factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{7} \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5}}{ \\cancel{3} \\cdot 5 cdot \\cancel{2} \\cdot \\cancel{2} \\cdot \\cancel{5} \\cdot \\cancel{7}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{1}{5} ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note: Everything in the numerator cancels because you’ve divided the numerator by itself. Hence, the answer has a 1 in its numerator. Exercise Multiply: $\\frac{6}{35} \\cdot \\frac{70}{36}\\nonumber$ $$\\frac{1}{3}$$ When Everything Cancels When all the factors in the numerator cancel, this means that you are dividing the numerator by itself. Hence, you are left with a", "=$$ $$\\frac{\\mathbf{11} \\cdot 2 \\cdot 6 \\cdot \\mathbf{13} \\cdot 2 \\cdot 7 \\cdot 3 \\cdot 5 \\cdot 2 \\cdot 8 \\cdot \\mathbf{17} \\cdot 2 \\cdot 9 \\cdot \\mathbf{19} \\cdot 4 \\cdot 5}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8 \\cdot 9 \\cdot 2 \\cdot 5} =$$ $$\\frac{\\mathbf{11} \\cdot \\cancel{2} \\cdot \\cancel{6} \\cdot \\mathbf{13} \\cdot \\cancel{2} \\cdot \\cancel{7} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 2 \\cdot \\cancel{8} \\cdot \\mathbf{17} \\cdot 2 \\cdot \\cancel{9} \\cdot \\mathbf{19} \\cdot \\cancel{4} \\cdot \\cancel{5}}{1 \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{4} \\cdot \\cancel{5} \\cdot \\cancel{6} \\cdot \\cancel{7} \\cdot \\cancel{8} \\cdot \\cancel{9} \\cdot \\cancel{2} \\cdot \\cancel{5}} = \\tag{2}$$ $$\\mathbf{11} \\cdot \\mathbf{13} \\cdot 2 \\cdot \\mathbf{17} \\cdot 2 \\cdot \\mathbf{19} =$$ $$\\mathbf{11} \\cdot \\mathbf{13} \\cdot \\mathbf{17} \\cdot \\mathbf{19} \\cdot 4 \\tag{3}$$ Then we developed this technique, until we obtained an estimation of the product of all primes up to a certain number $x$. Now we start from the same point but, analyzing the matter in a different way, we will get to the prove of the Bertrand’s postulate. By generalizing (1), we note that we started from the binomial $\\binom{2n}{n}$ and we obtained, finally, the product of the prime numbers between $n + 1$ and $2n$, multiplied by some other smaller factor. Bertrand’s postulate states that there exists at least one", "to straighten out. Particularly, we really don’t want to write out expressions like $\\frac{2\\cdot 2 \\cdot 4\\cdot 4\\cdot 6\\cdot 6\\cdot 8\\cdot 8 \\cdot 10\\cdot 10 \\cdots}{3\\cdot 3\\cdot 5\\cdot 5 \\cdot 7\\cdot 7 \\cdot 9\\cdot 9 \\cdot 11\\cdot 11 \\cdots}$ Put that way, it looks like, well, we can divide each 3 in the denominator into a 6 in the numerator to get a 2, each 5 in the denominator to a 10 in the numerator to get a 2, and so on. We get a product that’s infinitely large, instead of anything to do with π. This is that problem where arithmetic on infinitely long strings of things becomes dangerous. To be rigorous, we need to write this product as the limit of a sequence, with finite numerator and denominator, and be careful about how to compose the numerators and denominators. But this is all right. Wallis found a lovely result and in a way that’s common to much work in mathematics. It used a combination of insight and persistence, with pattern recognition and luck making a great difference. Often when we first find something the proof of it is rough, and we need considerable work to make it rigorous. The path that got Wallis to these products is one we still walk. There’s just three more essays", "as (f)_{k}=f^{\\underline k}=(f-k+1)\\cdots(f-3)\\cdot(f-2)\\cdot(f-1)\\cdot f, and the corresponding rising factorial as {\\color{white}{ \\big|}}r^{(k)}=\\,r^{\\overline k}=\\,r\\cdot(r+1)\\cdot(r+2)\\cdot(r+3)\\cdots(r+k-1); so, for example, 17\\cdot18\\cdot19\\cdot20\\cdot21=(21)_{5}=21^{\\underline 5}=17^{\\overline 5}=17^{(5)}. Then the binomial coefficients may be written as \\binom{f}{k} = \\frac{(f)_{k}}{k!} =\\frac{(f-k+1)\\cdots(f-2)\\cdot(f-1)\\cdot f}{1\\cdot2\\cdot3\\cdot4\\cdot5\\cdots k}, while the corresponding multiset coefficient is defined by replacing the falling with the rising factorial: \\left(\\!\\!\\binom{r}{k}\\!\\!\\right)=\\frac{r^{(k)}}{k!}=\\frac{r\\cdot(r+1)\\cdot(r+2)\\cdots(r+k-1)}{1\\cdot2\\cdot3\\cdot4\\cdot5\\cdots k}. #### Generalization to negative integers For any n, \\begin{align}\\binom{-n}{k} &= \\frac{-n\\cdot-(n+1)\\dots-(n+k-2)\\cdot-(n+k-1)}{k!}\\\\ &=(-1)^k\\;\\frac{n\\cdot(n+1)\\cdot(n+2)\\cdots (n + k - 1)}{k!}\\\\ &=(-1)^k\\binom{n + k - 1}{k}\\\\ &=(-1)^k\\left(\\!\\!\\binom{n}{k}\\!\\!\\right)\\;.\\end{align} In particular, binomial coefficients evaluated at negative integers are given by signed multiset coefficients. In the special case n = -1, this reduces to (-1)^k=\\binom{-1}{k}=\\left(\\!\\!\\binom{-k}{k}\\!\\!\\right) \\,. For example, if n = -4 and k = 7, then r = 4 and f = 10: \\begin{align}\\binom{-4}{7} &= \\frac {-10\\cdot-9\\cdot-8\\cdot-7\\cdot-6\\cdot-5\\cdot-4} {1\\cdot2\\cdot3\\cdot4\\cdot5\\cdot6\\cdot7}\\\\ &=(-1)^7\\;\\frac{4\\cdot5\\cdot6\\cdot7\\cdot8\\cdot9\\cdot10} {1\\cdot2\\cdot3\\cdot4\\cdot5\\cdot6\\cdot7}\\\\ &=\\left(\\!\\!\\binom{-7}{7}\\!\\!\\right)\\left(\\!\\!\\binom{4}{7}\\!\\!\\right)=\\binom{-1}{7}\\binom{10}{7}.\\end{align} ### Two real or complex valued arguments The binomial coefficient is generalized to two real or complex valued arguments using the gamma function or beta function via {x \\choose y}= \\frac{\\Gamma(x+1)}{\\Gamma(y+1) \\Gamma(x-y+1)}= \\frac{1}{(x+1) \\Beta(x-y+1,y+1)}. This definition inherits these following additional properties from \\Gamma: {x \\choose y}= \\frac{\\sin (y \\pi)}{\\sin(x \\pi)} {-y-1 \\choose -x-1}= \\frac{\\sin((x-y) \\pi)}{\\sin (x \\pi)} {y-x-1 \\choose y}; moreover, {x \\choose y} \\cdot {y \\choose x}= \\frac{\\sin((x-y) \\pi)}{(x-y) \\pi}. The resulting function has been little-studied, apparently first being graphed in (Fowler 1996). Notably, many binomial identities fail: \\textstyle but \\textstyle for", "... + b_m = n$. We will now expand the lefthand side of the equation. We first expand the first two factors to get: (4) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} = \\frac{n!}{b_1! \\cdot (n - b_1)!} \\cdot \\frac{(n - b_1)!}{b_2! \\cdot (n - b_1 - b_2)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\end{align} • We then expand this with the third factor to get: (5) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} \\cdot \\binom{n - b_1 - b_2}{b_3} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\binom{n - b_1 - b_2}{b_3} \\\\ = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\frac{(n - b_1 - b_2)!}{b_3! \\cdot (n - b_1 - b_2 - b_3)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot b_3! \\cdot (n - b_1 - b_2 - b_3)!} \\end{align} • Inductively, by expanding the $m^{\\mathrm{th}}$ factor of the product and noting that $n - b_1 - b_2 - ... - b_m = 0$, we see that: (6) \\begin{align} \\quad \\binom{b_1 + b_2 + ... + b_m}{b_1} \\cdot \\binom{b_2 + b_3 + ... + b_m}{b_2} \\cdot ... \\cdot \\binom{b_{m-1}}{b_m} \\\\ = \\binom{b_1 + b_2 + ... + b_m}{b_1, b_2, ..., b_m} = \\frac{n!}{b_1! \\cdot b_2! \\cdot ... \\cdot b_{m-1}! (n - b_1 - b_2 - ... - b_m)!}", "two nearly-equal values. We can rely on the use of the … Squeeze Theorem … to prove this is okay. And there’s much we have to straighten out. Particularly, we really don’t want to write out expressions like $\\frac{2\\cdot 2 \\cdot 4\\cdot 4\\cdot 6\\cdot 6\\cdot 8\\cdot 8 \\cdot 10\\cdot 10 \\cdots}{3\\cdot 3\\cdot 5\\cdot 5 \\cdot 7\\cdot 7 \\cdot 9\\cdot 9 \\cdot 11\\cdot 11 \\cdots}$ Put that way, it looks like, well, we can divide each 3 in the denominator into a 6 in the numerator to get a 2, each 5 in the denominator to a 10 in the numerator to get a 2, and so on. We get a product that’s infinitely large, instead of anything to do with π. This is that problem where arithmetic on infinitely long strings of things becomes dangerous. To be rigorous, we need to write this product as the limit of a sequence, with finite numerator and denominator, and be careful about how to compose the numerators and denominators. But this is all right. Wallis found a lovely result and in a way that’s common to much work in mathematics. It used a combination of insight and persistence, with pattern recognition and luck making a great difference. Often when we first find something the proof of it is rough, and we", "Solution After multiplying, prime factor both numerator and denominator, then cancel common factors. Note that unlike signs yields a negative product. \\begin{aligned} - \\frac{7x}{2} \\cdot \\frac{5}{14x^2} = - \\frac{35x}{28x^2} ~ & \\textcolor{red}{ \\text{ Multiply numerators and denominators.}} \\\\ = - \\frac{5 \\cdot 7 \\cdot x}{2 \\cdot 2 \\cdot 7 \\cdot x \\cdot x} ~ & \\textcolor{red}{ \\text{ Prime factor numerator and denominator.}} \\\\ = - \\frac{5 \\cdot \\cancel{7} \\cdot \\cancel{x}}{2 \\cdot 2 \\cdot \\cancel{7} \\cdot \\cancel{x} \\cdot x} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = - \\frac{5}{4x} \\end{aligned}\\nonumber Exercise Multiply: $- \\frac{3x}{2} \\cdot \\frac{6}{21x^3}\\nonumber$ $- \\frac{3}{7x^2}\\nonumber$ ## Multiply and Cancel or Cancel and Multiply When you are working with larger numbers, it becomes a bit harder to multiply, factor, and cancel. Consider the following argument. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{630}{180} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\\\ = \\frac{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 5} ~ & \\textcolor{red}{ \\text{ Prime factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}{2 \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber There are a number of difficulties with this approach. First, you have", "on the recto (front side) of the document; all later content in this summary is present on the verso (back side) of the papyrus. The transition from 60 to 61 is thus both a thematic and physical shift in the papyrus. 61 Seventeen multiplications are to have their products expressed as Egyptian fractions. The whole is to be given as a table. ${\\displaystyle {\\begin{bmatrix}{\\frac {2}{3}}\\cdot {\\frac {2}{3}}={\\frac {1}{3}}+{\\frac {1}{9}}&;&{\\frac {1}{3}}\\cdot {\\frac {2}{3}}={\\frac {1}{6}}+{\\frac {1}{18}}\\\\{\\frac {2}{3}}\\cdot {\\frac {1}{3}}={\\frac {1}{6}}+{\\frac {1}{18}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{6}}={\\frac {1}{12}}+{\\frac {1}{36}}\\\\{\\frac {2}{3}}\\cdot {\\frac {1}{2}}={\\frac {1}{3}}&;&{\\frac {1}{3}}\\cdot {\\frac {1}{2}}={\\frac {1}{6}}\\\\{\\frac {1}{6}}\\cdot {\\frac {1}{2}}={\\frac {1}{12}}&;&{\\frac {1}{12}}\\cdot {\\frac {1}{2}}={\\frac {1}{24}}\\\\{\\frac {1}{9}}\\cdot {\\frac {2}{3}}={\\frac {1}{18}}+{\\frac {1}{54}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{9}}={\\frac {1}{18}}+{\\frac {1}{54}}\\\\{\\frac {1}{4}}\\cdot {\\frac {1}{5}}={\\frac {1}{20}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{7}}={\\frac {1}{14}}+{\\frac {1}{42}}\\\\{\\frac {1}{2}}\\cdot {\\frac {1}{7}}={\\frac {1}{14}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{11}}={\\frac {1}{22}}+{\\frac {1}{66}}\\\\{\\frac {1}{3}}\\cdot {\\frac {1}{11}}={\\frac {1}{33}}&;&{\\frac {1}{2}}\\cdot {\\frac {1}{11}}={\\frac {1}{22}}\\\\{\\frac {1}{4}}\\cdot {\\frac {1}{11}}={\\frac {1}{44}}&&\\\\\\end{bmatrix}}}$ The syntax of the original document and its repeated multiplications indicate a rudimentary understanding that multiplication is commutative. 61B Give a general procedure for converting the product of 2/3 and the reciprocal of any (positive) odd number 2n+1 into an Egyptian fraction of two terms, e.g. ${\\displaystyle {\\frac {2}{3}}\\cdot {\\frac {1}{2n+1}}={\\frac {1}{p}}+{\\frac {1}{q}}}$ with natural p and q. In other words, find p and q in terms of n. ${\\displaystyle p=2(2n+1)}$ ${\\displaystyle q=6(2n+1)}$ Problem 61B, and the method" ]

[ "# Math Help - How to calculate the limit of this product? 1. ## How to calculate the limit of this product? $\\prod_{n=2}^{\\infty}(1-\\frac{1}{n^2})=\\lim_{n\\to\\infty}(1-\\frac{1}{4})(1-\\frac{1}{9})\\dots (1-\\frac{1}{n^2})$ Thank you! 2. ## Re: How to calculate the limit of this product? Hi gotmejerry! $(1-{1\\over n^2}) = {(n-1)(n+1) \\over n^2}$ Can you write out a couple of factors and see how things cancel? 3. ## Re: How to calculate the limit of this product? No unfortunately I cant. 4. ## Re: How to calculate the limit of this product? Which factor do you get if you fill in n=2? And which factor for n=3? And n=4? 5. ## Re: How to calculate the limit of this product? 3/4 8/9 15/16 but what do i do with them, how do they cancel each other? 6. ## Re: How to calculate the limit of this product? Let's write that out: ${1 \\cdot 3 \\over 2^2} \\cdot {2 \\cdot 4 \\over 3^2} \\cdot {3 \\cdot 5 \\over 4^2} \\cdot {4 \\cdot 6 \\over 5^2} \\cdot ... \\cdot {(n-1) (n+1) \\over n^2}$ Do you see numbers that cancel? 7. ## Re: How to calculate the limit of this product? Yes, thanks, so If I am not wrong, only 1/2 *(n+1)/n remains? And its limit is 1/2. Thank you very much Right! 9. ## Re: How to calculate", "in ascending order, but every little bit helps. The denominator can also be expressed as a product of primes. $2310=10 \\cdot 231=2 \\cdot 5 \\cdot 7 \\cdot 33=2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$ We can now cancel common factors. $\\frac{210}{2310}=\\frac{2 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11}=\\frac{\\not{2} \\cdot \\not{3} \\cdot \\not{5} \\cdot \\not{7}}{\\not{2} \\cdot \\not{8} \\cdot \\not{5} \\cdot \\not{7} \\cdot 11}=\\frac{1}{11}$ However, this approach is not the most efficient way to proceed, as multiplying numerators and denominators allows the products to grow to larger numbers, as in 210/2310. It is then a little bit harder to prime factor the larger numbers. A better approach is to factor the smaller numerators and denominators immediately, as follows. $\\frac{6}{231} \\cdot \\frac{35}{10}=\\frac{2 \\cdot 3}{3 \\cdot 7 \\cdot 11} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 5}$ We could now multiply numerators and denominators, then cancel common factors, which would match identically the last computation in equation (5). However, we can also employ the following cancellation rule. Cancellation Rule When working with the product of two or more rational expressions, factor all numerators and denominators, then cancel. The cancellation rule is simple: cancel a factor “on the top” for an identical factor “on the bottom.” Speaking more technically, cancel any factor in any numerator for", "exponents tells us when we divide two numbers or expressions in exponent form and the bases are the same, we can keep the base the same and subtract the exponent in the denominator away from the exponent in the numerator. Let's look at a few examples. Example 5: Simplify each $$\\require{cancel}\\frac{2^5}{2^2}$$ To simplify, keep the base 2 the same and subtract the exponent in the denominator (2) away from the exponent in the numerator (5): $$\\frac{2^5}{2^2}=2^{5 - 2}=2^3$$ Let's think about why this works, suppose we just simplified without our rule: $$\\frac{2^5}{2^2}=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}{2 \\cdot 2}$$ We can see that we have five factors of 2 in the numerator and two factors of 2 in the denominator. If we cancel common factors, we will be canceling two factors of 2 between the numerator and the denominator. This will leave us with (5 - 2 = 3) three factors of 2 or 8: $$\\frac{\\cancel{2}\\cdot \\cancel{2}\\cdot 2 \\cdot 2 \\cdot 2}{\\cancel{2}\\cdot \\cancel{2}}=2 \\cdot 2 \\cdot 2$$ Example 6: Simplify each $$\\frac{x^{3}}{x^{11}}$$ Keep the base (x) the same and subtract exponents (3 - 11 = -8) $$\\frac{x^{3}}{x^{11}}=x^{3 - 11}=x^{-8}$$ We know how to work with negative exponents. The x goes into the denominator and the exponent of (-8) becomes positive: $$x^{-8}=\\frac{1}{x^8}$$ #### Skills Check: Example", "denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*} \\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}. \\end{equation*} In practice, we don't always have to find the prime factorization. Instead, we can find one common factor at a time until no common factors remain. For example, since 126 and 24 are both even, we would write \\begin{equation*} \\frac{126}{24} = \\frac{63}{12}. \\end{equation*} Then, we look at 63 and 12 and recognize that they are both divisible by 3, allowing us to rewrite the fraction as \\begin{equation*} \\frac{126}{24} = \\frac{63}{12} = \\frac{21}{4}. \\end{equation*} Because $21=3 \\cdot 7$ and $4=2^2$ do not have common factors, we know this is simplified. A square root is not simplified if there is a factor inside the root that is a perfect square. Similarly, a cube root is not simplified if there is a factor inside that is a perfect cube. We use the factors of the value inside a root to determine if we can simplify it. ###### Example1.2.3 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that", "## [Soln] 2014 Singapore MO (Senior Rd 1) Problem 16 2014 SMO (Senior-Rd1) 16. Evaluate the sum $\\displaystyle\\frac{3! + 4!}{2(1! + 2!)} + \\frac{4! + 5!}{3(2! + 3!)} + \\dots + \\frac{12! + 13!}{11(10! + 11!)}$. Recall the definition of the factorial function: for non-negative integer $n$, $n! = \\begin{cases} 1 &\\text{ for } n = 0, \\\\ n \\cdot (n-1) \\cdot \\dots \\cdot 1 &\\text{ for } n \\geq 1. \\end{cases}$ When faced with a series of many terms, try to come up with an expression for the general $\\boldsymbol{n^{th}}$ term. In this question, the required sum is the sum of $\\displaystyle\\frac{(n+2)! + (n+3)!}{(n+1)[n! +(n+1)!]}$ for $n = 1, 2, \\dots, 10$. This looks like a term that can be simplified. Especially with fractions, a possible strategy is to pull out as many factors as possible in the hope that there can be some common factors in the numerator and denominator. These can be cancelled out. By pulling out common factors in the numerator: \\begin{aligned} (n+2)! + (n+3)! &= (n+2)! \\cdot [1 + (n + 3)] \\\\ &= (n+2)! \\cdot (n+4) \\end{aligned} and in the denominator: \\begin{aligned} (n+1)[n! + (n+1)!] &= (n+1) \\cdot n! \\cdot [1 + (n+1)] \\\\ &= (n+1) \\cdot n! \\cdot (n+2) \\\\ &= (n+2)! \\end{aligned} The original expression can be greatly simplified: $\\displaystyle\\frac{(n+2)!", "values of the numbers. First, multiply the numerators together to get the product’s numerator. Then, multiply the denominators together to get the product’s denominator. Rewrite in lowest terms, if needed. $$\\left( \\frac{3}{4} \\right)\\left( \\frac{2}{5} \\right)=\\frac{6}{20}=\\frac{3}{10}$$ The product of two negative numbers is positive. $$\\left( -\\frac{3}{4} \\right)\\left( -\\frac{2}{5} \\right)=\\frac{3}{10}$$ The following video shows examples of multiplying two signed fractions, including simplification of the answer. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=36 To summarize: • positive $$\\cdot$$ positive: The product is positive. • negative $$\\cdot$$ negative: The product is positive. • negative $$\\cdot$$ positive: The product is negative. • positive $$\\cdot$$ negative: The product is negative. You can see that the product of two negative numbers is a positive number. So, if you are multiplying more than two numbers, you can count the number of negative factors. ### Multiplying More Than Two Negative Numbers If there are an even number (0, 2, 4, …) of negative factors to multiply, the product is positive. If there are an odd number (1, 3, 5, …) of negative factors, the product is negative. ### Example Find $$3(−6)(2)(−3)(−1)$$. [hidden-answer a=”149062″]Multiply the absolute values of the numbers. $$\\begin{array}{l}3(6)(2)(3)(1)\\\\18(2)(3)(1)\\\\36(3)(1)\\\\108(1)\\\\108\\end{array}$$ Count the number of negative factors. There are three $$\\left(−6,−3,−1\\right)$$. $$3(−6)(2)(−3)(−1)$$ Since there are an", "looking straight at the definition of $v_n$, not at the formula that you have found for it. $v_{n+1}-v_n={\\color{blue}{(n+1)\\cdots(n+m)}}(n+m+1)-n{\\color{blue}{(n+1)\\cdots(‌​n+m)}}=(m+1){\\color{blue}{(n+1)\\cdots(n+m)}}$ – alex.jordan Mar 12 '13 at 7:14 I've got to this point: $v_{n+1}-v_n = \\frac{(m+n+1)(m+n)! - n(m+n)!}{n!} = \\frac{(m+n)!}{n!} \\bigg( m+n+1-n(m+n)! \\bigg)$ In your edited comment I can see how the terms in blue cancel, but what about the denominator of $n!\\;$? – Ozzy Mar 12 '13 at 7:24 In your last comment, the last factor in your equation is not correct. After factoring out $(m+n)!$ your last factor should be $\\bigg(m+n+1-n\\bigg)$, or just $\\bigg(m+1\\bigg)$. That yields $\\frac{(m+n)!}{n!}(m+1)$, which equals $(m+1)(n+1)\\cdots(n+m)$ when you expand the factorials. And that's the definition of $u_n$. – alex.jordan Mar 12 '13 at 17:20 $m+n$ is greater than $n$. So if you really keep expanding that numerator you will eventually reach $(n)(n-1)(n-2)\\cdots$ in the numerator too. And these factors will cancel with the entire denominator. What remains in the numerator will be the factors right up to $(n)$, the last of which was $(n+1)$. So what remains after cancellation is $(m+n)(m+n-1)\\cdots(n+1)$. – alex.jordan Mar 13 '13 at 6:47", "\\cancel{h} \\cdot \\cancel{h}} {\\cancel{h} \\cdot \\cancel{h} \\cdot \\cancel{h} \\cdot h \\cdot h} \\\\ &= \\frac {1} {h \\cdot h} \\\\ & = \\frac {1} { h^{2}} \\end{align*} If we were to simplify the original expression using the quotient rule, we would have \\begin{align*} \\frac {h^{3}} {h^{5}} &= h^{3-5} \\\\ &= h^{-2} \\end{align*} Putting the answers together, we have $h^{-2}= \\frac {1} {h^{2}}$. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. $a^{-n} = \\frac {1} {a^{n}}$ and $a^{n} = \\frac {1} {a^{-n}}$ We have shown that the exponential expression $a^{n}$ is defined when $n$ is a natural number, $0$, or the negative of a natural number. That means that $a^{n}$ is defined for any integer $n$. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$. ### THE NEGATIVE RULE OF EXPONENTS For any nonzero real number $a$ and natural number $n$, the negative rule of exponents states that $a^{-n} = \\frac {1} {a^{n}}$ #### Example 5 Write each of the following quotients with a single base. Do not simplify further. Write", "1 in the numerator. The same rule applies to the denominator. If everything in the denominator cancels, you’re left with a 1 in the denominator. Example 7 Simplify the product: - \\frac{6x}{55y} \\cdot \\left( - \\frac{110y^2}{105x^2} \\right).\\nonumber \\] Solution The product of two negatives is positive. \\begin{aligned} - \\frac{6x}{55y} \\cdot \\left( - \\frac{110y^2}{105x^2} \\right) = \\frac{6x}{55y} \\cdot \\frac{110y^2}{105x^2} ~ & \\textcolor{red}{ \\text{ Like signs gives a positive.}} \\end{aligned}\\nonumber Prime factor numerators and denominators, then cancel common factors. \\begin{aligned} = \\frac{2 \\cdot 3 \\cdot x}{5 \\cdot 11 \\cdot y} \\cdot \\frac{2 \\cdot 5 \\cdot 11 \\cdot y \\cdot y}{3 \\cdot 5 \\cdot 7 \\cdot x \\cdot x} ~ & \\textcolor{red}{ \\text{ Prime factor numerators & denominators.}} \\\\ = \\frac{2 \\cdot \\cancel{3} \\cdot \\cancel{x}}{5 \\cdot \\cancel{11} \\cdot \\cancel{y}} \\cdot \\frac{2 \\cdot \\cancel{5} \\cdot \\cancel{11} \\cdot \\cancel{y} \\cdot y}{ \\cancel{3} \\cdot \\cancel{5} \\cdot 7 \\cdot \\cancel{x} \\cdot x} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{2 \\cdot 2 \\cdot y}{5 \\cdot 7 \\cdot x} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\\\ = \\frac{4y}{35x} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\end{aligned}\\nonumber Exercise Simplify: $\\frac{6x}{15b} \\cdot \\left( - \\frac{35b^2}{10a^2} \\right)\\nonumber$ Answer $- \\frac{21b}{5a}\\nonumber$ ## Parallelograms In this section, we are going to learn how to find the area of a parallelogram. Let’s begin with the definition of a parallelogram.", "1/4}{(2i)^4+ 1/4} = \\frac{(2i-2)^2+(2i-2)+\\frac{1}{2}}{(2i)^2+2i+ 1/2}= \\frac{(2(i-1))^2+2(i-1)+\\frac{1}{2}}{(2i)^2+2i+ 1/2}$$ Now we can perform the product, since by telescoping all \"inner\" numerators and denominators cancel: $$\\prod_{i=1}^n\\frac{[(2i-1)^4+ 1/4]}{[(2i)^4+ 1/4]} = \\prod_{i=1}^n \\frac{(2(i-1))^2+2(i-1)+\\frac{1}{2}}{(2i)^2+2i+ 1/2} \\\\ = \\frac{(2\\cdot 1-2)^2+(2\\cdot 1-2)+\\frac{1}{2}}{(2\\cdot n)^2+2\\cdot n+ 1/2} = \\frac{1}{8 n^2+ 4 n+ 1}$$ This completes the evaluation. $$\\qquad \\Box$$", "grows larger with each successive term. Looking at the terms more closely, it turns out that the denominators tend to be products of many small primes, whereas the numerators are either primes or products of a few comparatively large primes. For example: $\\frac{9649}{2520} = \\frac{9649}{2^3 \\cdot 3^2 \\cdot 5 \\cdot 7} \\qquad \\textrm{and} \\qquad \\frac{18358381}{4084080} = \\frac{59 \\cdot 379 \\cdot 821}{2^4 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17}.$ To produce an integer, we need to cancel all the primes in the factorization of the denominator by matching primes in the numerator; given the pattern of these numbers, that looks like an unlikely coincidence. But there is reason for caution. Note the seventh term in the sequence, where the denominator has decreased from $$60$$ to $$20$$. To understand how that happens, we can run through the calculation of the term, which starts by summing the six previous terms. $\\frac{60}{60} + \\frac{120}{60} + \\frac{150}{60} + \\frac{170}{60} + \\frac{185}{60} + \\frac{197}{60} = \\frac{882}{60}.$ Then we calculate the mean, and add 1 to get the seventh term: $\\require{cancel}\\frac{882}{60} \\cdot \\frac{1}{6} = \\frac{882}{360} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 7 \\cdot 7}{\\cancel{2} \\cdot 2 \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 5} = \\frac{49}{20} + 1 = \\frac{69}{20}$ Cancelations reduce the numerator and denominator of the mean", "# partial sum involving factorials Here is an interesting series I ran across. It is a binomial-type identity. $\\displaystyle \\sum_{k=0}^{n}\\frac{(2n-k)!\\cdot 2^{k}}{(n-k)!}=4^{n}\\cdot n!$ I tried all sorts of playing around, but could not get it to work out. This works out the same as $\\displaystyle 2^{n}\\prod_{k=1}^{n}2k=2^{n}\\cdot 2^{n}\\cdot n!=4^{n}\\cdot n!$ I tried equating these somehow, but I could not get it. I even wrote out the series. There were cancellations, but it did not look like the product of the even numbers. $\\displaystyle \\frac{(2n)!}{n!}+\\frac{(2n-1)!\\cdot 2}{(n-1)!}+\\frac{(2n-2)!2^{2}}{(n-2)!}+\\cdot\\cdot\\cdot +n!\\cdot 2^{n}=4^{n}\\cdot n!$. How can the closed form be derived from this?. I bet I am just being thick. I see the last term is nearly the result except for being multiplied by $2^{n}$. I see if the factorials are written out, $2n(2n-1)(2n-2)(2n-3)\\dots$ for example, then 2's factor out of $2n, \\;\\ 2n-2$ (even terms) in the numerator. There is even a general form I ran through Maple. It actually gave a closed from for it as well, but I would have no idea how to derive it. $\\displaystyle \\sum_{k=0}^{n}\\frac{(2n-k)!\\cdot 2^{k}\\cdot (k+m)!}{(n-k)!\\cdot k!}$. In the above case, m=0. But, apparently there is a closed form for $m\\in \\mathbb{N}$ as well. Maple gave the solution in terms of Gamma: $\\displaystyle \\frac{\\Gamma(1+m)4^{n}\\Gamma(n+1+\\frac{m}{2})}{\\Gamma(1+\\frac{m}{2})}$ Would anyone have an idea how to proceed with this?. Perhaps writing it in terms", "\\cdot \\cancel{3}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber Note that this yields exactly the same result, 7/2. Cancellation Rule When multiplying fractions, cancel common factors according to the following rule: “Cancel a factor in a numerator for an identical factor in a denominator.” Example 6 Find the product of 14/15 and 30/140. Solution Multiply numerators and multiply denominators. Prime factor, cancel common factors, then multiply. \\begin{aligned} \\frac{14}{15} \\cdot \\frac{30}{140} = \\frac{14 \\cdot 30}{15 \\cdot 140} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\\\ = \\frac{(2 \\cdot 7) \\cdot (2 \\cdot 3 \\cdot 5)}{(3 \\cdot 5) \\cdot (2 \\cdot 2 \\cdot 5 \\cdot 7)} ~ & \\textcolor{red}{ \\text{ Prime factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{7} \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5}}{ \\cancel{3} \\cdot 5 cdot \\cancel{2} \\cdot \\cancel{2} \\cdot \\cancel{5} \\cdot \\cancel{7}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{1}{5} ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note: Everything in the numerator cancels because you’ve divided the numerator by itself. Hence, the answer has a 1 in its numerator. Exercise Multiply: $\\frac{6}{35} \\cdot \\frac{70}{36}\\nonumber$ $$\\frac{1}{3}$$ When Everything Cancels When all the factors in the numerator cancel, this means that you are dividing the numerator by itself. Hence, you are left with a", "\\cdot m^2}{kg^2}.\\frac {4.39\\times 10^{47}kg^2}{d^2} \\\\\\text{Cancel the}\\ kg^2\\ \\text{units}. && 2 \\cdot 0 \\times 10^{20} N & = 6.67 \\times 10^{-11} \\frac{N \\cdot m^2} {\\cancel{kg^2}} \\cdot \\frac{4.39 \\times 10^{47} \\cancel{kg^2}} {d^2} \\\\\\text{Multiply the numbers in the numerator}. && 2.0\\times 10^{20} N & = \\frac {2.93\\times 10^{37}}{d^2}N \\cdot m^2 \\\\\\text{Multiply both sides by}\\ d^2. && 2.0\\times 10^{20} N \\cdot d^2 & =\\frac {2.93\\times 10^{37}}{d^2} \\cdot d^2 \\cdot N \\cdot m^2 \\\\\\text{Cancel common factors}. && 2 \\cdot 0 \\times 10^{20} N \\cdot d^{2} & = \\frac{2.93 \\times 10^{37}} {\\cancel{d^2}} \\cdot \\cancel{d^2} \\cdot N \\cdot m^2 \\\\\\text{Simplify}. && 2.0\\times 10^{20} N \\cdot d^2 & =2.93\\times 10^{37}N \\cdot m^2 \\\\\\text{Divide both sides by}\\ 2.0 \\times 10^{20}\\ N. && d^2 & =\\frac {2.93 \\times 10^{37}}{2.0\\times 10^{20}}\\frac {N \\cdot m^2}{N} \\\\\\text{Simplify}. && d^2 & =1.465\\times 10^{17} m^2 \\\\\\text{Take the square root of both sides}. && d & =3.84 \\times 10^8m\\ \\text{Answer}$ This is indeed the distance between the Earth and the Moon. Example 4 The area of a circle is given by $A=\\pi r^{2}$ and the circumference of a circle is given by $C=2 \\pi r$. Find the ratio of the circumference and area of the circle. Solution The ratio of the circumference and area of the circle is: $\\frac {2\\pi r}{\\pi r^2}$ We cancel common factors from the numerator and denominator. $\\frac{2 \\cancel{\\pi} \\cancel{R}}{\\cancel{\\pi}", "to multiply large numbers, and secondly, you have to prime factor the even larger results. One possible workaround is to not bother multiplying numerators and denominators, leaving them in factored form. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{18 \\cdot 35}{30 \\cdot 6} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\end{aligned}\\nonumber Finding the prime factorization of these smaller factors is easier. \\begin{aligned} = \\frac{(2 \\cdot 3 \\cdot 3) \\cdot (5 \\cdot 7)}{(2 \\cdot 3 \\cdot 5) \\cdot (2 \\cdot 3)} ~ & \\textcolor{red}{ \\text{ Prime factor.}} \\end{aligned}\\nonumber Now we can cancel common factors. Parentheses are no longer needed in the numerator and denominator because both contain a product of prime factors, so order and grouping do not matter. \\begin{aligned} = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot \\cancel{7}}{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 2 \\cdot \\cancel{3}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber Another approach is to factor numerators and denominators in place, cancel common factors, then multiply. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{2 \\cdot 3 \\cdot 3}{2 \\cdot 3 \\cdot 5} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 3} ~ & \\textcolor{red}{ \\text{ Factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{3}}{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5}} \\cdot \\frac{ \\cancel{5} \\cdot 7}{2", "=$$ $$\\frac{\\mathbf{11} \\cdot 2 \\cdot 6 \\cdot \\mathbf{13} \\cdot 2 \\cdot 7 \\cdot 3 \\cdot 5 \\cdot 2 \\cdot 8 \\cdot \\mathbf{17} \\cdot 2 \\cdot 9 \\cdot \\mathbf{19} \\cdot 4 \\cdot 5}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8 \\cdot 9 \\cdot 2 \\cdot 5} =$$ $$\\frac{\\mathbf{11} \\cdot \\cancel{2} \\cdot \\cancel{6} \\cdot \\mathbf{13} \\cdot \\cancel{2} \\cdot \\cancel{7} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 2 \\cdot \\cancel{8} \\cdot \\mathbf{17} \\cdot 2 \\cdot \\cancel{9} \\cdot \\mathbf{19} \\cdot \\cancel{4} \\cdot \\cancel{5}}{1 \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{4} \\cdot \\cancel{5} \\cdot \\cancel{6} \\cdot \\cancel{7} \\cdot \\cancel{8} \\cdot \\cancel{9} \\cdot \\cancel{2} \\cdot \\cancel{5}} = \\tag{2}$$ $$\\mathbf{11} \\cdot \\mathbf{13} \\cdot 2 \\cdot \\mathbf{17} \\cdot 2 \\cdot \\mathbf{19} =$$ $$\\mathbf{11} \\cdot \\mathbf{13} \\cdot \\mathbf{17} \\cdot \\mathbf{19} \\cdot 4 \\tag{3}$$ Then we developed this technique, until we obtained an estimation of the product of all primes up to a certain number $x$. Now we start from the same point but, analyzing the matter in a different way, we will get to the prove of the Bertrand’s postulate. By generalizing (1), we note that we started from the binomial $\\binom{2n}{n}$ and we obtained, finally, the product of the prime numbers between $n + 1$ and $2n$, multiplied by some other smaller factor. Bertrand’s postulate states that there exists at least one", "Solution After multiplying, prime factor both numerator and denominator, then cancel common factors. Note that unlike signs yields a negative product. \\begin{aligned} - \\frac{7x}{2} \\cdot \\frac{5}{14x^2} = - \\frac{35x}{28x^2} ~ & \\textcolor{red}{ \\text{ Multiply numerators and denominators.}} \\\\ = - \\frac{5 \\cdot 7 \\cdot x}{2 \\cdot 2 \\cdot 7 \\cdot x \\cdot x} ~ & \\textcolor{red}{ \\text{ Prime factor numerator and denominator.}} \\\\ = - \\frac{5 \\cdot \\cancel{7} \\cdot \\cancel{x}}{2 \\cdot 2 \\cdot \\cancel{7} \\cdot \\cancel{x} \\cdot x} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = - \\frac{5}{4x} \\end{aligned}\\nonumber Exercise Multiply: $- \\frac{3x}{2} \\cdot \\frac{6}{21x^3}\\nonumber$ $- \\frac{3}{7x^2}\\nonumber$ ## Multiply and Cancel or Cancel and Multiply When you are working with larger numbers, it becomes a bit harder to multiply, factor, and cancel. Consider the following argument. \\begin{aligned} \\frac{18}{30} \\cdot \\frac{35}{6} = \\frac{630}{180} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\\\ = \\frac{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 5} ~ & \\textcolor{red}{ \\text{ Prime factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}{2 \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot \\cancel{5}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber There are a number of difficulties with this approach. First, you have", "\\cdot 5\\cdot 7\\cdot 7\\cdot 9\\cdot 9\\cdot 11\\cdot 11\\cdot 13\\cdot 13\\cdot 15\\cdot 15\\cdot 17\\cdot 17\\cdot 19\\cdot 19\\cdot 21} =\\frac{137438953472}{44801898141} }$ And all we get for this effort is the lousy approximation ${\\pi\\approx \\mathbf{3.0677}}$. But it turns out that (1) can be dramatically improved with a little tweak. First, let us rewrite partial products in (1) in terms of double factorials. This can be done in two ways: either ${\\displaystyle 2\\prod_{k=1}^n \\frac{4k^2}{4k^2-1} = (4n+2) \\left(\\frac{(2n)!!}{(2n+1)!!}\\right)^2 \\qquad \\qquad (2)}$ or ${\\displaystyle 2\\prod_{k=1}^n \\frac{4k^2}{4k^2-1} = \\frac{2}{2n+1} \\left(\\frac{(2n)!!}{(2n-1)!!}\\right)^2 \\qquad \\qquad (3)}$ Seeing how badly (2) underestimates ${\\pi}$, it is natural to bump it up: replace ${4n+2}$ with ${4n+3}$: ${\\displaystyle \\pi \\approx b_n= (4n+3) \\left(\\frac{(2n)!!}{(2n+1)!!}\\right)^2 \\qquad \\qquad (4)}$ Now with ${n=10}$ we get ${\\mathbf{3.1407}}$ instead of ${\\mathbf{3.0677}}$. The error is down by two orders of magnitude, and all we had to do was to replace the factor of ${4n+2=42}$ with ${4n+3=43}$. In particular, the size of numerator and denominator hardly changed: ${\\displaystyle b_{10}=43\\, \\frac{2\\cdot 2\\cdot 4\\cdot 4\\cdot 6\\cdot 6\\cdot 8\\cdot 8 \\cdot 10 \\cdot 10\\cdot 12\\cdot 12\\cdot 14\\cdot 14\\cdot 16\\cdot 16\\cdot 18 \\cdot 18\\cdot 20\\cdot 20}{3\\cdot 3\\cdot 5 \\cdot 5\\cdot 7\\cdot 7\\cdot 9\\cdot 9\\cdot 11\\cdot 11\\cdot 13\\cdot 13\\cdot 15\\cdot 15\\cdot 17\\cdot 17\\cdot 19\\cdot 19\\cdot 21\\cdot 21} }$ Approximation (4) differs from (2) by additional term ${\\left(\\frac{(2n)!!}{(2n+1)!!}\\right)^2}$, which decreases to zero. Therefore, it is not", "the permanent of some matrix, but all I need here is to know $Q$ has integer coefficients. In your particular case, the products give $$\\frac{\\prod_{i\\neq j} (i - j) ((i - 1) - (j - 1))}{\\prod_{i,j} (i + j - 1)^2} = \\frac{\\prod_{i < j} (j - i)^4}{\\prod_{i,j} (i + j - 1)^2} = \\prod_{j=1}^N \\left[(j - 1)!^4 \\frac{(j - 1)!^2}{(j + N - 1)!^2} \\right] = \\frac{\\prod_{j=0}^{N-1} j!^8}{\\prod_{j=0}^{2N-1} j!^2}$$ Note that $j!^8$ divides $(2j)!(2j+1)!$, so the numerator divides the denominator. All in all, $$\\det A = \\frac{Q}{\\prod_{j=0}^{2N-1} j!^2 \\bigg/ \\prod_{j=0}^{N-1} j!^8}$$ for some integer $Q$. The denominator appearing in this formula is typically rather close to the ones given in Igor Rivin's answer, and are trivially squares: $1$, $12^2$, $2160^2$, $6048000^2 = 672000^2 \\cdot 9^2$, etc. The discrepancy is of course accounted for by cancellations between $Q$ and the product of factorials. Again using Igor Rivin's data, I find (indices denote the size of the matrix) \\begin{aligned} Q_1 &= 1 \\\\ Q_2 &= 7 \\\\ Q_3 &= 647 = 2^3 \\cdot 3^4 - 1 = 8 \\cdot (16 \\cdot (6 - 1) + 1) - 1 \\\\ Q_4 &= 878769 = 16 \\cdot (12 \\cdot (32 \\cdot (144 - 1) + 1) - 1) + 1 \\\\ Q_5 &= 18203480001 = 40000 \\cdot ( 48 \\cdot (", "# What ate the prime factors of each denominator of the unit fractions that have terminating decimals? 1/2, 1/4, 1/5, 1/8, 1/10, 1/16, 1/20, 1/25 Question Decimals What ate the prime factors of each denominator of the unit fractions that have terminating decimals? 1/2, 1/4, 1/5, 1/8, 1/10, 1/16, 1/20, 1/25 2021-03-08 Step 1 Writing the prime factors of each denominators of the fraction We have the fraction $$\\displaystyle\\frac{{1}}{{2}},\\frac{{1}}{{4}},\\frac{{1}}{{5}},\\frac{{1}}{{8}},\\frac{{1}}{{10}},\\frac{{1}}{{16}},\\frac{{1}}{{20}},\\frac{{1}}{{25}}$$ Step 2 Now, Prime factor of $$\\displaystyle{2}={2}$$ Prime factor of $$\\displaystyle{4}={2}\\cdot{2}$$ Prime factor of $$\\displaystyle{5}={5}$$ Prime factor of $$\\displaystyle{8}={2}\\cdot{2}\\cdot{2}$$ Prime factor of $$\\displaystyle{10}={2}\\cdot{5}$$ Prime factor of $$\\displaystyle{16}={2}\\cdot{2}\\cdot{2}\\cdot{2}$$ Prime factor of $$\\displaystyle{20}={2}\\cdot{2}\\cdot{5}$$ Prime factor of $$\\displaystyle{25}={5}\\cdot{5}$$ ### Relevant Questions For digits before decimals point, multiply each digit with the positive powers of ten where power is equal to the position of digit counted from left to right starting from 0. For digits after decimals point, multiply each digit with the negative powers of ten where power is equal to the position of digit counted from right to left starting from 1. 1) $$10^{0}=1$$ 2) $$10^{1}=10$$ 3) $$10^{2}=100$$ 4) $$10^{3}=1000$$ 5) $$10^{4}=10000$$ And so on... 6) $$10^{-1}=0.1$$ 7) $$10^{-2}=0.01$$ 8) $$10^{-3}=0.001$$ 9) $$10^{-4}=0.0001$$ Which of the following fractions are repeating decimals and which are terminating? How made decisions? $$a) \\frac{2}{15}$$ $$b)\\frac{11}{20}$$ $$c)\\frac{17}{40}$$ $$d)\\frac{1}{12}$$ For the given fraction and decimals we have" ]

[ "For how many real numbers $$a^{}_{}$$ does the quadratic equation $$x^2 + ax^{}_{} + 6a=0$$ have only integer roots for $$x^{}_{}$$? (第九届AIME1991 第8题)", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $$\\text{Domain: All Real Numbers }$$ We see that this function is in the form $ax^2+bx+c$, so it is a quadratic function. Note, the $8$ in the denominator just turns the coefficients into fractions but does not change the form of the function. Since quadratic functions exist for all real numbers, it follows: $$\\text{Domain: All Real Numbers }$$", "quadratic equation are Real", "# Quadratic with a Repeated Root Algebra Level 4 How many ordered triples of integers $(a, b, c)$ are there, such that $a, b$ and $c$ are integers from 1 to 10 inclusive, and the quadratic function $f(x) = ax^2 + bx + c$ has a repeated root? ×", "Find all real solutions of the equation $$(x^2-7x+11)^{(x^2-11x+30)} = 1.$$ You may also be interested in the problem Mega Quadratic Equations.", "Find the number of real values of parameter $$a$$ for which the largest value of the quadratic function $$f(x) = x^{2} + ax +2$$ in the interval $$[-2,4]$$ is 6.", "0$, where equality holds, in each case, if and only if the roots are equal. XVIII) Let $p(x)=x^{2}+ax+b$ be a quadratic polynomial in which a and b are integers. Given any integer n, show that there is an integer M such that $p(n)p(n+1)=p(M)$. Cheers, Nalin Pithwa.", "+\\binom{n}{2018} = 0$$ How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bc+c, a\\ne 0$, and the roots are $r$ and $s$, then the requirement is that $\\{a,\\ b,\\ c\\}=\\{r,\\ s\\}$.)", "# Integer root of quadratic Determine the sum of all (distinct) positive integers $n$, such that for some integer $a$, $n^2 -an + 6a = 0.$ ×", "# UKMT Senior Challenge-III Number Theory Level 4 How many integers $$a$$ are there for which the roots of the quadratic equation $$x^2+ax+2013=0$$ are integers? ×", "the solution is only $$x=4.76$$ seconds. Question #5: What is the total number of real solutions for the quadratic equation $$9x^2+12x+4=0$$? One real solution Two real solutions No real solution Infinite real solutions To determine the number of real solutions, we use the discriminant, $$b^2-4ac$$, which simplifies to 0 because $$(12)^2-4(9)(4)=144-144=0$$. Therefore, there is one real solution to this quadratic equation.", "# Is it Quadratic or not? Algebra Level 4 If the equation $$x^4 - x^2 + 1 = k$$, where $$k$$ is a real number, has all real roots then what is the number of integral values of $$8k$$? ×", "• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Condition for roots of a quadratic be integers 09-23-2010, 05:36 PM Post: #1 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Condition for roots of a quadratic be integers Let $m$ and $n$ be positive integers such that $x^2 - mx + n = 0$ has real roots $\\alpha$ and $\\beta$. Prove that $\\alpha$ and $\\beta$ are integers if and only if $\\left\\lfloor m\\alpha \\right \\rfloor + \\left\\lfloor m\\beta \\right\\rfloor$ is the square of an integer. « Next Oldest | Next Newest » Forum Jump:", "# Is it Quadratic or not? Algebra Level 4 If the equation $$x^4 - x^2 + 1 = k$$ , (k is a real number) has all real roots then the number of integral values of $$8k$$ is ×", "# Is there any methods to solve for integer solution of a quadratic equation like $ax^2 + bx + c = 0$ Is there any method to solve for integer solution of a quadratic equation like following: $$ax^2 + bx + c = 0$$ where $a, b, c \\in \\mathbb{Z}$ If not is it possible for the Special case: ? $$x^2 -x + c = 0$$ where $c \\in \\mathbb{Z^+}$ I will prefer to have a analytical solution if it exists, other wise a polynomial time solution is also welcome (if it exists). Note: I've put $c \\in \\mathbb{Z^+}$ for your convenience. It's also fine if some one can give a solution for $c \\in \\mathbb{Z}$ • You want to find integer roots? – Alice Ryhl Dec 2 '14 at 14:21 • yes @KristofferRyhl – aroyc Dec 2 '14 at 14:24 • Is pseudo code that finds integer roots if they exist ok? – Alice Ryhl Dec 2 '14 at 14:25 • I'm not much worried about code..my exact question is about existence of any well defined non-heuristic algo – aroyc Dec 2 '14 at 14:27 • I was hoping you said $c < 0$ – Chinny84 Dec 2 '14 at 14:32 For a general polynomial, $$a_0+a_1x+\\dots+a_nx^n$$, with integer coefficients, you may find all rational roots as follows.", "# Cubics one step ahead of quadratics Algebra Level 5 Suppose $$a$$ and $$b$$ are two positive real numbers such that the roots of the equation $$x^{3} - ax + b = 0$$ are all real. Let $$t$$ is a root of the equation with mininmal absolute value. If maximum value of $$t$$ is of the form $$\\dfrac{mb}{na}$$. Find $$m + n$$. ×", "# Polynomial with prime exponents How many real solutions does the equation $x^{17} + x^{13} + x^{11}+x^7 + x^5+x^3+19x^2=1$ have? ×", "33 views Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \\geq 0$ for all real numbers $x$. If $f(2)=0$ and $f(4)=6$, then $f(-2)$ is equal to 1. $36$ 2. $12$ 3. $24$ 4. $6$ 1", "Suppose $$p(x) = ax^2 + bx + c$$ is a quadratic polynomial with real coefficients and $$|p(x)| \\le 1$$ for all values of $$x$$ in the range $$[0,1]$$. Find the largest possible value of $$|a| + |b| + |c|$$.", "For certain ordered pairs $$(a,b)\\,$$ of real numbers, the system of equations $$ax+by=1\\,$$ $$x^2+y^2=50\\,$$ has at least one solution, and each solution is an ordered pair $$(x,y)\\,$$ of integers. How many such ordered pairs $$(a,b)\\,$$ are there? (第十二届AIME1994 第7题)" ]

[ "330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$. From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \\frac{5r+13}{2}$. We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \\frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \\boxed{420}$. ## See also 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •", "are $(\\pm2, \\pm 3), (\\pm3, \\pm 2)$ For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\\pm2, \\pm 1), (\\pm1, \\pm 2)$ Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = \\boxed{080}$.", "should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 The roots of $p(x)$ are $r$, $s$, and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$). Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$, as they too sum to $0$. Then: $a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and $a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$. From these equations, we can write that $$rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a$$ and simplifying gives $$2s-5r-13=0 \\Rightarrow s = \\frac{5r+13}{2}.$$ We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get $$rs(r+s) = b$$ $$(r+4)(s-3)(r+s+1)=b + 240.$$ Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding, simplifying, substituting $s = \\frac{5r+13}{2}$, and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$, or $b = (1)(9)(1 + 9) = 90$. The answer is $|-330|+|90|", "# Difference between revisions of \"2008 iTest Problems/Problem 84\" ## Problem Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$. ## Solutions ### Solution 1 (credit to official solution) Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$. We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \\begin{align*} \\frac{rs}{r+s} &= -2008 \\\\ rs &= -2008r - 2008s \\\\ rs + 2008r + 2008s &= 0 \\\\ (r+2008)(s+2008) &= 2008^2. \\end{align*} WLOG, let $|a| \\le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \\tfrac{2008^2}{a}$. Thus, $$-r-s = b = -a - \\tfrac{2008^2}{a} + 4016.$$ Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \\cdot 251^2$, so there are $\\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \\cdot 22 =", "13$ so the pairs $(r,s)$ are $(\\pm2, \\pm 3), (\\pm3, \\pm 2)$ For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\\pm2, \\pm 1), (\\pm1, \\pm 2)$ Then, since only $s^2$ but not $s$ appears in the equations for $a$ and $b$, we can ignore the plus minus sign for $s$. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = \\boxed{080}$.", "case. $\\textbf{Case 2:}$ $m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$ Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \\begin{align*} r+2 &= -a-2\\\\ 2r &= 4+2a+b. \\end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that. However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case. This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\\cdot41=\\boxed{738}$ as our final answer. ~s214425 ## Solution 2 (factor the difference) $p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots,", "False. $$f(x) = x^4$$ is a counterexample. ### Find the possible coefficients given the roots A7, 2018 Suppose $$x^{3}+a x^{2}+b x+8=0$$ is a cubic equation with integer coefficients. Both $$r$$ and $$-r$$ are the roots of the equation, where $$r>0$$ is a real number. List all possible pairs of values $$(a, b)$$. Solution Let the third root be $$s$$. By Vieta's formulas we have \\begin{align} -r+r+s &= -a \\\\ -rs + rs -r^2 &= b \\\\ -r^2s &= -8 \\end{align} which implies that $$b=-r^2$$ must be negative and $$ab=8$$. So the possible pairs of values of $$(a, b)$$ are $$\\{ (-1,-8), (-2,-4), (-4,-2), (-8,-1) \\}$$. ### 0,1-polynomial A7, 2011 When does the polynomial $$1+x+\\cdots+x^{n}$$ have $$x-a$$ as a factor? Here $$n$$ is a positive integer greater than 1000 and $$a$$ is a real number. • if and only if $$\\;a=-1$$. • if and only if $$\\;a=-1\\;$$ and $$\\;n\\;$$ is odd. • if and only if $$\\;a=-1\\;$$ and $$\\;n\\;$$ is even. • We cannot decide unless $$\\;n\\;$$ is known. Solution (b) We have: $f(x)=\\frac{1-x^{n+1}}{1-x} \\mbox{ for } x\\neq 1$ If $$a$$ is a root of $$f$$, then $$a^{n+1}=1$$. Since $$f(1)=n$$, 1 cannot be a root of $$f$$. So $$a\\neq 1$$. The only other possibility is $$a=-1$$ with $$n$$ being odd. ### Parity of a polynomial A2, 2010 A polynomial", "+0 # algebra +1 47 2 Let r and s be the roots of y^2 - 19y + 9. Find (r-2)(s-2). Apr 4, 2021 #1 +31353 +1 Here is a start: (r-2)(s-2).= rs -2r-2s+4 = rs -2 (r+s) +4 Are you familiar with Vieta's formulas for quadratics??? Vieta's for quadratics : r + s = -b/a =19 and rs = c/a = 9 rs -2 (r+s) +4 = 9 -2(19)+4 = -25 Apr 4, 2021 edited by ElectricPavlov Apr 4, 2021 edited by ElectricPavlov Apr 4, 2021 #2 +496 0 For this, we don't ACTUALLY have to solve for r and s. Let us first expand (r-2)(s-2): $r\\cdot{s}+r\\cdot{-2}+{-2}\\cdot{s}+{-2}\\cdot{-2}=rs-2r-2s+4$ if r and s are the roots of y^2 - 19y + 9, then: $y^2 - 19y + 9 = (y-r)(y-s) = y^2 - (r + s)y + rs$ so r+s=19 and rs=9 going back to the original equation, we have $rs-2r-2s+4=9-2(r+s)+4=9-38+4=\\boxed{-25}$ Apr 4, 2021 edited by SparklingWater2 Apr 4, 2021", "2008 AIME II Problems/Problem 7 Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. Solution 1 By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is $$(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).$$ Also by Vieta's formulas we have $$rst = -rs(r + s) = -\\dfrac{2008}{8} = -251$$ so negating both sides and multiplying through by 3 gives our answer of $\\boxed{753}.$ Solution 2 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ Solution 3 Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3", "$x^2 - (r_1 + r_2)x + r_1r_2)$. By comparing this with $x^2 - ax + a^2$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$. Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2)$. Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0$. This can be factored as $(r_1 - 2)(r_2 - 2) = 4$. These factors can be: $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$. We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = {-1, 8, 9}$. So the answer is $-1 + 8 + 9 = \\boxed{\\textbf{(C) }16}$.", "+0 # algebra 2 0 49 2 Suppose r and s are the roots of the equation 3x^2 + 9x + 15 = 0. Find (r+1)(s+1) Dec 29, 2022 #1 0 By the quadratic formula, the roots are -1/2*i*(sqrt(11) +- 3i). Plugging those in, we get (r + 1)(s + 1) = -6. Dec 29, 2022 #2 +2541 0 We want to find$$(r+1)(s+1) = rs + s + r + 1$$ From Vieta's, we know that $$rs = {c \\over a} = {15 \\over 3} = 5$$ and that $$r + s = -{b \\over a} = -{9 \\over 3} = -3$$ So, we have $$5 - 3 + 1= \\color{brown}\\boxed{3}$$ Dec 29, 2022", "of r, s, 8, 9, and 11.\" So $$y = 8 * 9 * 11 * r * s$$ • Break just the numbers down into prime factors: $$8 = (2 * 2 * 2) = 2^3$$ $$9 = (3 * 3) = 3^2$$ $$11 = (11 * 1) = 11^1$$ $$y = 2^3 * 3^2 * 11^1 * r * s$$ • Looking at just the numbers, this situation will not make a perfect square. There are three 2s. The twos are not in pairs. (There is one pair of twos, but a third 2 is by itself.) (2 * 2) * 2 That red 2 needs another copy of itself. The 11 is not in a pair. There is only one 11. If we listed out the factors as above: y = . . . $$11^1$$. We need another 11. We have to make pairs. In order to make pairs, we need one more 2, and one more 11. Both numbers need one more \"copy\" of themselves. • That's what $$r$$ and $$s$$ are for. $$r$$ and $$s$$ \"make up for\" the \"missing\" 2 and the \"missing\" 11. $$(r * s) = (2^1 * 11^1) = 22$$: (r*s) IS 22, so (r*s) must be divisible by 22. 3) Result: $$y$$ before and after y WAS $$(2^3", "we have divide by $$3$$ since we overcounted which person we choose first. This case only has $$48 \\div 3 = 16$$ configurations then. Case $$5: 4$$ people stand There are only $$2$$ choices depending on which half of the people stand up. These cases together contribute $1 + 8 + 20 + 16 + 2 = 47$ configurations. There are $$2^8 = 256$$ possibilities for the coin flips, which makes the desired probability $$\\dfrac{47}{256}.$$ Thus, A is the correct answer. 23. The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $$a?$$ $$7$$ $$8$$ $$16$$ $$17$$ $$18$$ ###### Solution(s): Let the zeroes be $$r$$ and $$s.$$ Using Vieta's formulas, we have that $$a = r + s$$ and $$2a = rs.$$ Then we get that $rs = 2(r + s),$ which rearranges to $rs - 2r - 2s = 0$ $rs - 2r - 2s + 4 = 4$ $(r - 2)(s - 2) = 4.$ The only possible pairs $$(r - 2, s - 2)$$ that work are $(1, 4), (-1, -4), (4, 1), (-4, -1),$ $(2, 2), (-2, -2).$ For any of these pairs, we have that $a = r - 2 + s - 2 + 4.$ We want all the such unique values of $$a.$$ We", "$x^2 - (r_1 + r_2)x + r_1r_2)$. By comparing this with $x^2 - ax + a^2$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$. Plugging the first equation in the second, $r_1r_2 = 2(r_1 + r_2)$. Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0$. This can be factored as $(r_1 - 2)(r_2 - 2) = 4$. These factors can be: $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$. We want the number of distinct $a = r_1 + r_2$, and these factors gives$a = {-1, 8, 9}. So the answer is$(Error compiling LaTeX. ! Missing$ inserted.)-1 + 8 + 9 = \\boxed{\\textbf{(C) }16}\\$.", "see that the pair $-6$ and $15$ satisfies our requirements: $-6\\times 15=-90$ as well as $-6+15=9$. • #### Factor the polynomial $15x^2+9x-6$. ##### Tipps The product of $15$ and $-6$ is $-90$. You can write $-90$ as a product of two factors. Find the pair of factors that add up to $9$. In the example, $\\begin{array}{rcl} (2x+3)(4x-2)&=&2x(4x-2)+3(4x-2)\\\\ &=&8x^2-4x+12x-6\\\\ &=&8x^2+8x-6 \\end{array}$ we have that $8\\times -6=48$, as well as $-4\\times 12=48$. We can see that $-4+12=8$. First multiply $a$ and $b$. Then we check all pairs of factors of $ab$ whose product is $ab$. If we add a pair of factors together and get $b$, then we've found the pair we are looking for. ##### Lösung To factor the quadratic trinomial $15x^2+9x-6$ with $a=15$, $b=9$, and $c=-6$, we first look for the pairs of factors of $a\\times c$ that sum to $b$. We have that $a\\times c=-90$, so we investigate all pairs of factors of $-90$: ${\\small\\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} \\text{Factors of }-90&-1,90&1,-90&-2,45&2,-45&-3,30&3,-30&-5,18&5,-18&-6,15&6,-15 \\\\ \\hline \\text{Sum of factors }&89&-89&43&-43&27&-27&13&-13&9&-9 \\end{array}}$ The sum of the factors $-6$ and $15$ is $-6+15=9$. So this is the pair we want. Next we group the quadratic polynomial as follows: $15x^2+9x-6=15x^2-6x+15x-6$. Grouping the binomials together, we get $(15x^2-6x)+(15x-6)$. The greatest common factor of the left binomial is $3x$ and of the right binomial is $3$: $(15x^2-6x)+(15x-6)=3x(5x-2)+3(5x-2)$. Now", "first factor contains only previous roots, so we see that $$r^3 - r^2 - 6r + 7 = 0$$ or $$r^3 + 2r^2 - 3r - 5 = 0$$. If $$r^3 - r^2 - 6r + 7 = 0$$, then $$s = r^2 - 4$$ also satisfies $$s^3 - s^2 - 6s + 7 = 0$$, as can be verifies as follows: $$(r^2 - 4)^3 - (r^2 - 4)^2 - 6(r^2 - 4) + 7 = (r^3 - r^2 - 6r + 7) (r^3 + r^2 - 6r - 7) = 0.$$ For the same reason, $$t$$ also satisfies $$t^3 - t^2 - 6t + 7 = 0$$. Therefore $$r, s, t$$ are exactly the three different roots of the polynomial $$x^3 - x^2 - 6x + 7$$, so that $$P(x) = x^3 - x^2 - 6x + 7$$. Similarly, if $$r^3 + 2r^2 - 3r - 5 = 0$$, then we conclude that $$P(x) = x^3 + 2x^2 - 3x - 5$$.", "is incorrect. $R^4 - 20R^2 + 4 = (R^2-10)^2 - 96$ - Sir, it gives R² - 4R - 3 = 0 and R² + 4R - 3 = 0 But both of them have irrational roots. So the answer will be no solution. – Bazinga Jun 29 '12 at 7:59 Thank you sir, got my mistake. – Bazinga Jun 29 '12 at 8:02 Yes. You also need to check one of the factors being $-1$, since both the factors can be negative which will result in a positive number. – user17762 Jun 29 '12 at 8:03 $R^4-20R^2+4=R^4-4R^2+4-(4R)^2=(R^2-2)^2-(4R)^2=(R^2-4R-2)(R^2+4R-2)$.Check for the solutions for each factor term=$1$.The integer values of R is the solution set. -", "We are trying to find $-(r^3 + s^3 + t^3)$. Substituting: $$-(r^3 + s^3 + t^3) = -3srt = \\frac{-2008*3}{8} = \\boxed{753}.$$ ## Solution 5 Write $(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)$ and let $f(x)=8x^3+1001x+2008$. Then $$f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.$$ Solving for $r^3+s^3+t^3$ and negating the result yields the answer $\\boxed{753}.$ ## Solution 6 Here by Vieta's formulas: $r+s+t = 0$ --(1) $rst = \\frac{-2008}{8} = -251$ --(2) By the factorisation formula: Let $a = r+s$, $b = s+t$, $c = t+r$, $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$ (By (1)) So $$a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \\boxed{753}.$$ ## Solution 7 Let's construct a polynomial with the roots $(r+s), (s+t),$ and $(t+r)$. sum of the roots: $=2(r+s+t)=2\\cdot0=0$ pairwise product of the roots: $(r+s)(s+t)+(s+t)(t+r)+(t+r)(r+s)=r^2+s^2+t^2+3(rs+st+tr)$ $=(r+s+t)^2+rs+st+tr=0+\\frac{1001}{8}$ product of the roots: $(r+s)(s+t)(t+r)=r^2t+r^2s+s^2r+s^2t+t^2r+t^2s+3rst$ $=(rs+st+tr)(r+s+t)-3rst+2rst=-rst=-\\frac{2008}{8}$ thus, the polynomial we get is $x^3+\\frac{1001}{8}x+-\\frac{2008}{8}=0$ as $(r+s), (s+t),$ and $(t+r)$ are roots of this polynomial, we know that (using power reduction) $(r+s)^3+\\frac{1001}{8}(r+s)-\\frac{2008}{8}=0$ $(s+t)^3+\\frac{1001}{8}(s+t)-\\frac{2008}{8}=0$ $(t+r)^3+\\frac{1001}{8}(t+r)-\\frac{2008}{8}=0$ adding all of the equations up, we see that $(r+s)^3+(s+t)^3+(t+r)^3=3\\cdot\\frac{2008}{8}-\\frac{1001}{8}(2r+2s+2t)=251(3)+0=\\boxed{753}$", "fact that has integer coefficients results from the above formula for the quotient of by x-p/q ). Comparing the coefficients of degree and the constant coefficients in the above equality shows that, if \\tfracpq is a rational root in reduced form, then is a divisor of a0, and is a divisor of an. Therefore, there is a finite number of possibilities for and, which can be systematically examined. For example, if the polynomial P(x)=2x3-7x2+10x-6 has a rational root \\tfracpq with, then must divide 6; that is p\\in\\{\\pm1,\\pm2,\\pm3,\\pm6\\}, and must divide 2, that is q\\in\\{1,2\\}. Moreover, if, all terms of the polynomial are negative, and, therefore, a root cannot be negative. That is, one must have \\tfracpq\\in\\{1,2,3,6,\\tfrac12,\\tfrac32\\}. A direct computation shows that only \\tfrac32 is a root, so there can be no other rational root. Applying the factor theorem leads finally to the factorization 2x3-7x2+10x-6=(2x-3)(x2-2x+2). P(x)=ax2+bx+c with integer coefficients. If it has a rational root, its denominator must divide evenly and it may be written as a possibly reducible fraction r1=\\tfracra. By Vieta's formulas, the other root r2 is r2=- ba - r1=- ba-ra =- b+r a = sa, with s=-(b+r). Thus the second root is also rational, and Vieta's second formula r1 r 2= ca gives sara = ca, that is rs=acandr+s=-b. Checking all pairs of integers whose", "+0 # Vieta's Formula 0 63 1 +1193 The equations $$x^3 + 5x^2 + px + q = 0$$ and $$x^3 + 7x^2 + px + r = 0$$ have two roots in common. If the third root of each equation is represented by $$x_1$$ and $$x_2$$ respectively, compute the ordered pair $$(x_1,x_2)$$. Jun 4, 2019 #1 +22363 +2 Vieta's Formula The equations $$x^3 + 5x^2 + px + q = 0$$ and $$x^3 + 7x^2 + px + r = 0$$ have two roots in common. If the third root of each equation is represented by $$x_1$$ and $$x_2$$ respectively, compute the ordered pair $$(x_1,x_2)$$. $$\\begin{array}{|rcll|} \\hline x^3 + 5x^2 + px + q &=& 0,\\ \\text{ the roots are }~ x_1,x_3,x_4. \\\\ x^3 + 7x^2 + px + r &=& 0,\\ \\text{ the roots are }~ x_2,x_3,x_4. \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{5} &=& \\mathbf{-(x_1+x_3+x_4)} \\\\ 5+x_1 &=& -(x_3+x_4) \\\\\\\\ \\mathbf{7} &=& \\mathbf{-(x_2+x_3+x_4)} \\\\ 7+x_2 &=& -(x_3+x_4) \\\\\\\\ 5+x_1 &=& 7+x_2 \\\\ \\mathbf{x_1-x_2} &=& \\mathbf{2} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline x_1x_3+x_1x_4+x_3x_4 = p &=& x_2x_3+x_2x_4+x_3x_4 \\\\ x_1x_3+x_1x_4 &=& x_2x_3+x_2x_4 \\\\ x_1(x_3+x_4)-x_2(x_3+x_4) &=& 0 \\\\ (x_3+x_4)(x_1-x_2) &=& 0 \\quad | \\quad \\mathbf{x_1-x_2 = 2} \\\\ (x_3+x_4)*2 &=& 0 \\quad | \\quad : 2 \\\\ x_3+x_4 &=& 0 \\\\ \\mathbf{x_3} &=& \\mathbf{-x_4} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{5}" ]

[ "challenge is to find some sort of formula for the number of such pairs; we will not pursue that here. 2. Further generalising question 1, if $a$ and $c$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Again, $b$ is allowed to be any integer. As before, we think about how we factorise this expression (see Quadratic grids for more information on this). We see that we are looking for two numbers which multiply to $ac$ and add to $b$. Each possible pair of numbers which multiplies to $ac$ will give a different sum, and so the number of possible quadratics is the number of pairs of numbers which multiply to $ac$. As in (a), it would be nice to have some sort of formula for this number. 3. How can we generalise questions 2 and 3? We can generalise question 2 in the case that $a=1$ by asking the following: If $b$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any positive integer. Here is one approach to answering this: If we again write $x^2+bx+c=(x+r)(x+s)=x^2+(r+s)x+rs$, we find that we need $b=r+s$, and every different pair of numbers $r$, $s$ gives a different value of $c=rs$.", "the six integers must be distinct (since a trivial arithmetic progression of the same integer repeated six times cannot have a sum of 21), and the smallest possible arithmetic progression of six distinct positive integers sums to 21; consequently, the four roots and the two parameters $m$ and $n$ are the numbers 1 through 6 in some order. (In fact, the information that the numbers form an arithmetic progression is redundant: any set of six distinct positive integers has a sum of at least 21.) Our solvers devised four ways to finish the solution. Let us denote the roots of the first quadratic by $r_1,s_1$, and of the second by $r_2, s_2$. Method 1. Nearly all solvers observed that $m+n=10$; here is their argument. We are given that $$r_1+s_1+r_2+ s_2+m+n=21,$$ and we know that the roots of a monic quadratic equation add up the the negative of the coefficient of the $x$-term: $$r_1+s_1 = m \\quad \\mbox{ and } \\quad r_2+ s_2 = n+1.$$ Put these equations together to get $$21 = (r_1+s_1)+(r_2+ s_2)+m+n = m+n+1+m+n = 2m + 2n +1,$$ whence $m+n=10$, as claimed. Next, because $m$ and $n$ are at most 6 and distinct, the only possibilities are $m=6$ and $n=4$, or $m=4$ and $n=6$. The latter pair produces a pair of quadratics that fail to", "two divisors of $6$ If nothing of the above works, I go for the quadratic formula... Suppose $a=1$ and $c$ is an integer small enough that you can instantly see its divisors. If the roots are integers, they are $r$ and $c/r$ where $r$ is a (positive or negative) divisor of $c$ such that $r + c/r = -b$. Of course the roots don't have to be integers, but this case does occur often enough. For example, for $x^2 + 14 x + 33$, the divisors of $33$ are $1, 3, 11, 33$, and $3 + 33/3 = 3 + 11 = 14$, so the roots are $-3$ and $-11$.", "+0 # algebra +1 47 2 Let r and s be the roots of y^2 - 19y + 9. Find (r-2)(s-2). Apr 4, 2021 #1 +31353 +1 Here is a start: (r-2)(s-2).= rs -2r-2s+4 = rs -2 (r+s) +4 Are you familiar with Vieta's formulas for quadratics??? Vieta's for quadratics : r + s = -b/a =19 and rs = c/a = 9 rs -2 (r+s) +4 = 9 -2(19)+4 = -25 Apr 4, 2021 edited by ElectricPavlov Apr 4, 2021 edited by ElectricPavlov Apr 4, 2021 #2 +496 0 For this, we don't ACTUALLY have to solve for r and s. Let us first expand (r-2)(s-2): $r\\cdot{s}+r\\cdot{-2}+{-2}\\cdot{s}+{-2}\\cdot{-2}=rs-2r-2s+4$ if r and s are the roots of y^2 - 19y + 9, then: $y^2 - 19y + 9 = (y-r)(y-s) = y^2 - (r + s)y + rs$ so r+s=19 and rs=9 going back to the original equation, we have $rs-2r-2s+4=9-2(r+s)+4=9-38+4=\\boxed{-25}$ Apr 4, 2021 edited by SparklingWater2 Apr 4, 2021", "+0 # algebra 2 0 49 2 Suppose r and s are the roots of the equation 3x^2 + 9x + 15 = 0. Find (r+1)(s+1) Dec 29, 2022 #1 0 By the quadratic formula, the roots are -1/2*i*(sqrt(11) +- 3i). Plugging those in, we get (r + 1)(s + 1) = -6. Dec 29, 2022 #2 +2541 0 We want to find$$(r+1)(s+1) = rs + s + r + 1$$ From Vieta's, we know that $$rs = {c \\over a} = {15 \\over 3} = 5$$ and that $$r + s = -{b \\over a} = -{9 \\over 3} = -3$$ So, we have $$5 - 3 + 1= \\color{brown}\\boxed{3}$$ Dec 29, 2022", "# 1991 AIME Problems/Problem 8 ## Problem For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$? ## Solution ### Solution 1 By Vieta's formulas, $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula, $$x = \\frac{-a \\pm \\sqrt{a^2 - 24a}}{2}$$ Since $-a$ is an integer, we need $\\sqrt{a^2-24a}$ to be an integer (let this be $b$): $b^2 = a^2 - 24a$. Completing the square, we get $$(a - 12)^2 = b^2 + 144$$ Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$). Then $$c^2 - b^2 = 144 \\Longrightarrow (c+b)(c-b) = 144$$ The pairs of factors of $144$ are $(\\pm1,\\pm144),( \\pm 2, \\pm 72),( \\pm 3, \\pm 48),( \\pm 4, \\pm 36),( \\pm 6, \\pm 24),( \\pm 8, \\pm 18),( \\pm 9, \\pm 16),( \\pm 12, \\pm 12)$; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get $\\boxed{10}$ pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work. ### Solution 2 Let $x^2 + ax +", "Nicolas May 1 '15 at 0:39 • As for their product, where did you take $pq=-3$? – Bernard May 1 '15 at 0:40 • I understand that you said you must use Vieta's formula but the answer is as simple as $(x-6)^2 +8(x-6) -1 =0$ which quickly simplifies to the required answer. If you encounter this sort of problem in future and aren't constrained to use Vieta's Formula, this is a more direct route. – Deepak May 1 '15 at 1:11 Let $r=p+6$ and $s=q+6$ be the roots of the quadratic you're trying to find. That quadratic's coefficients are $-(r+s)$ and $rs$. But \\begin{align} r+s&=(p+6)+(q+6)\\\\ &=p+q+12\\\\ &=-8+12\\\\ &=4 \\end{align} and \\begin{align} rs&=(p+6)(q+6)\\\\ &=pq+6(p+q)+36\\\\ &=-1+6(-8)+36\\\\ &=-1-48+36\\\\ &=-13 \\end{align} so the quadratic you're looking for is $$x^2-(r+s)x+rs=x^2-4x-13$$ Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \\Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\\cdot (-8) + 36=-13\\Rightarrow y^2-4y-13=0$ is the sought after equation. The original quadratic equation is given to you as $x^2 + 8x - 1 = 0$ . Let it's roots be $(x_1 , x_2)$ . The roots of required equation be $(\\alpha ,\\beta)$ (say). Now, $$x_1 + x_2 = -8 \\\\ x_1 x_2 = -1$$ And $$\\alpha = x_1 + 6 \\\\ \\beta = x_2", "take out a factor of $-1$ and answer the resulting question, with the “$c$ positive” and “$c$ negative” questions swapping. So our generalised question will be: If $a$ and $b$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any positive integer. This time, as in (b), we require two numbers which add to $b$ and multiply to $ac$. Since $a$ and $c$ are positive, $ac$ is also positive, and so the two numbers which add to $b$ must both be positive or both be negative. Since $c$ can be any positive number, $ac$ is any multiple of $a$, and so we are looking for two numbers of the same sign which add to $b$ and multiply to give a multiple of $a$. There doesn’t seem to be an obvious way of finding the exact number of such pairs, but we can at least say that there can only be finitely many of them, since they still have to add to $b$. Now let’s generalise question 3. First the case $a=1$: If $b$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any negative integer. Again, we write $x^2+bx+c=(x+r)(x+s)=x^2+(r+s)x+rs$, and we need", "is incorrect. $R^4 - 20R^2 + 4 = (R^2-10)^2 - 96$ - Sir, it gives R² - 4R - 3 = 0 and R² + 4R - 3 = 0 But both of them have irrational roots. So the answer will be no solution. – Bazinga Jun 29 '12 at 7:59 Thank you sir, got my mistake. – Bazinga Jun 29 '12 at 8:02 Yes. You also need to check one of the factors being $-1$, since both the factors can be negative which will result in a positive number. – user17762 Jun 29 '12 at 8:03 $R^4-20R^2+4=R^4-4R^2+4-(4R)^2=(R^2-2)^2-(4R)^2=(R^2-4R-2)(R^2+4R-2)$.Check for the solutions for each factor term=$1$.The integer values of R is the solution set. -", "# How to use the discriminant to find out how many real number roots an equation has for 8b^2 - 6b + 3 = 5b^2? Apr 16, 2018 color(crimson)(\"The given equation has TWO REAL ROOTS\" #### Explanation: For a given quadratic equation in the standard form $a {x}^{2} + b x + c$, $\\text{Discriminant } D = {b}^{2} - 4 a c$ $8 {b}^{2} - 6 b + 3 = 5 {b}^{2}$ $8 {b}^{2} - 5 {b}^{2} - 6 b + 3 = 0 , \\text{ making R H S = 0}$ $3 {b}^{2} - 6 b + 3 = 0$ :. D = b^2 - 4 a c = (-6)^2 - (4 * 3 * 3) = 36 - 24 = 12, color(green)(\" which is \" > 0 \" (Positive)\" Hence color(crimson)(\"The given equation has TWO REAL ROOTS\"", "see that the pair $-6$ and $15$ satisfies our requirements: $-6\\times 15=-90$ as well as $-6+15=9$. • #### Factor the polynomial $15x^2+9x-6$. ##### Tipps The product of $15$ and $-6$ is $-90$. You can write $-90$ as a product of two factors. Find the pair of factors that add up to $9$. In the example, $\\begin{array}{rcl} (2x+3)(4x-2)&=&2x(4x-2)+3(4x-2)\\\\ &=&8x^2-4x+12x-6\\\\ &=&8x^2+8x-6 \\end{array}$ we have that $8\\times -6=48$, as well as $-4\\times 12=48$. We can see that $-4+12=8$. First multiply $a$ and $b$. Then we check all pairs of factors of $ab$ whose product is $ab$. If we add a pair of factors together and get $b$, then we've found the pair we are looking for. ##### Lösung To factor the quadratic trinomial $15x^2+9x-6$ with $a=15$, $b=9$, and $c=-6$, we first look for the pairs of factors of $a\\times c$ that sum to $b$. We have that $a\\times c=-90$, so we investigate all pairs of factors of $-90$: ${\\small\\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} \\text{Factors of }-90&-1,90&1,-90&-2,45&2,-45&-3,30&3,-30&-5,18&5,-18&-6,15&6,-15 \\\\ \\hline \\text{Sum of factors }&89&-89&43&-43&27&-27&13&-13&9&-9 \\end{array}}$ The sum of the factors $-6$ and $15$ is $-6+15=9$. So this is the pair we want. Next we group the quadratic polynomial as follows: $15x^2+9x-6=15x^2-6x+15x-6$. Grouping the binomials together, we get $(15x^2-6x)+(15x-6)$. The greatest common factor of the left binomial is $3x$ and of the right binomial is $3$: $(15x^2-6x)+(15x-6)=3x(5x-2)+3(5x-2)$. Now", "False. $$f(x) = x^4$$ is a counterexample. ### Find the possible coefficients given the roots A7, 2018 Suppose $$x^{3}+a x^{2}+b x+8=0$$ is a cubic equation with integer coefficients. Both $$r$$ and $$-r$$ are the roots of the equation, where $$r>0$$ is a real number. List all possible pairs of values $$(a, b)$$. Solution Let the third root be $$s$$. By Vieta's formulas we have \\begin{align} -r+r+s &= -a \\\\ -rs + rs -r^2 &= b \\\\ -r^2s &= -8 \\end{align} which implies that $$b=-r^2$$ must be negative and $$ab=8$$. So the possible pairs of values of $$(a, b)$$ are $$\\{ (-1,-8), (-2,-4), (-4,-2), (-8,-1) \\}$$. ### 0,1-polynomial A7, 2011 When does the polynomial $$1+x+\\cdots+x^{n}$$ have $$x-a$$ as a factor? Here $$n$$ is a positive integer greater than 1000 and $$a$$ is a real number. • if and only if $$\\;a=-1$$. • if and only if $$\\;a=-1\\;$$ and $$\\;n\\;$$ is odd. • if and only if $$\\;a=-1\\;$$ and $$\\;n\\;$$ is even. • We cannot decide unless $$\\;n\\;$$ is known. Solution (b) We have: $f(x)=\\frac{1-x^{n+1}}{1-x} \\mbox{ for } x\\neq 1$ If $$a$$ is a root of $$f$$, then $$a^{n+1}=1$$. Since $$f(1)=n$$, 1 cannot be a root of $$f$$. So $$a\\neq 1$$. The only other possibility is $$a=-1$$ with $$n$$ being odd. ### Parity of a polynomial A2, 2010 A polynomial", "# 2013 AIME I Problems/Problem 10 ## Problem There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$. ## Solution Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have $(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$ Applying difference of squares, and regrouping, we have $(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$ So matching coefficients, we obtain $q(r^2 + s^2) = 65$ $b = r^2 + s^2 + 2rq$ $a = q + 2r$ By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair. We proceed by determining possible values for $q$, $r$, and $s$ and using these to determine $a$ and $b$. If $q = 1$, $r^2 + s^2 = 65$ so (r, s) = $(\\pm1, \\pm 8), (\\pm8, \\pm 1), (\\pm4, \\pm 7), (\\pm7, \\pm 4)$ Similarly, for $q = 5$, $r^2 + s^2 = 13$ so the pairs $(r,s)$", "+0 0 77 2 +217 Given quadratic equation x^2+cx+a=0 has two integer roots x_1, x_2; Quadratic Equation x_2+ax+b=0 has two integer roots x_1', x_2'. If x_1-x_1'=x_2-x_2'=1, please determine the value of a+b+c. yasbib555 Sep 28, 2018 edited by yasbib555 Oct 5, 2018 #1 +3190 +2 $$\\text{I'm going to call your roots }(r_{11},~r_{12},~r_{21},~r_{22})$$ $$r_{21}=r_{11}-1 \\\\ r_{22}=r_{12}-1$$ $$p_1=x^2+c x + a = (x-r_{11})(x-r_{12}) =x^2 -\\left(r_{11}+r_{12}\\right) x+r_{11} r_{12}\\\\ p_2 = x^2 +(2-r_{11}-r_{12})x + (1-r_{11}-r_{12}+r_{11}r_{12})$$ $$\\text{the coefficient }a \\text{ appears in both polynomials so we can solve for the roots}\\\\ r_{11}r_{12} = 2-r_{11}-r_{12} \\\\ \\text{Solving this we find that there are two possibilities}\\\\ (r_{11},~r_{12}) = (-2,~-4) ~\\text{or}\\\\ (r_{11},~r_{12}) = (0,~2)$$ $$\\text{In the first case we then have}\\\\ (r_{21},~r_{22}) = (-3,-5) \\\\ p_1(x) = (x+2)(x+4) = x^2 + 6x+8 \\Rightarrow a=8,~c=6\\\\ p_2(x) = (x+3)(x+5) = x^2 + 8x + 15 \\Rightarrow b=15 \\\\ a+b+c = 8 + 15 + 6 = 29$$ $$\\text{In the second case we have}\\\\ (r_{21},~r_{22}) = (-1,1)\\\\ p_1(x) =(x-0)(x-2) = x^2 - 2x \\Rightarrow a=0,~c=-2\\\\ p_2(x) = (x+1)(x-1) = x^2 - 1 \\Rightarrow b=-1\\\\ a+b+c = 0-1-2 = -3$$ Someone please double check all of this. Rom Sep 28, 2018 edited by Rom Sep 28, 2018 #2 +2 I got to the same result, but by a different route. To begin with, remember that if y = f(x), then", "# Difference between revisions of \"2008 iTest Problems/Problem 84\" ## Problem Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$. ## Solutions ### Solution 1 (credit to official solution) Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$. We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \\begin{align*} \\frac{rs}{r+s} &= -2008 \\\\ rs &= -2008r - 2008s \\\\ rs + 2008r + 2008s &= 0 \\\\ (r+2008)(s+2008) &= 2008^2. \\end{align*} WLOG, let $|a| \\le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \\tfrac{2008^2}{a}$. Thus, $$-r-s = b = -a - \\tfrac{2008^2}{a} + 4016.$$ Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \\cdot 251^2$, so there are $\\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \\cdot 22 =", "330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$. From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \\frac{5r+13}{2}$. We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \\frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \\boxed{420}$. ## See also 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •", "a negative value for $$x$$ which is impossible. The positive root of the quadratic equation yields $$x = \\frac{2y + 1 + \\sqrt{8y^2 + 1}}{2}$$ For $$x$$ to be an integer, the term $$\\sqrt{8y^2 + 1}$$ has to be an odd integer number $$z$$ (say). We can now write it out as $$z = \\sqrt{8y^2 + 1}$$ or $$z^{2} - 8y^2 = 1$$ This is Pell's equation (or Vargaprakriti in Sanskrit). As we know that there are no more than 30 socks in the draw, we can quickly work our way to two admissible solutions to the problem $$\\{3,1\\}, \\{15,6\\}$$. If you are looking to buy books on probability theory here is a good list to own. If you are looking to buy books on time series analysis here is an excellent list to own.", "+ 24nx) + (4894 - n)$ to generate a family of 16 quadratics using n=0 to 15 iterations. n --- (0) $x^2 - 4x + 4894$ Roots: Complex (1) $x^2 - 28x + 4893$ Roots: Complex (2) $x^2 - 52x + 4892$ Roots: Complex (3) $x^2 - 76x + 4891$ Roots: Complex (4) $x^2 - 100x + 4890$ Roots: Complex (5) $x^2 - 124x + 4889$ Roots: Complex (6) $x^2 - 148x + 4888$ Roots: 49.75 and 98.25 (7) $x^2 - 172x + 4887$ Roots: 35.91 and 136.09 (8) $x^2 - 196x + 4886$ Roots: 29.31 and 166.69 (9) $x^2 - 220x + 4885$ Roots: 25.06 and 194.94 (10) $x^2 - 244x + 4884$ Roots: 22 and 222 (11) $x^2 - 268x + 4883$ Roots: 19.66 and 248.34 (12) $x^2 - 292x + 4882$ Roots: 17.80 and 274.20 (13) $x^2 - 316x + 4881$ Roots: 16.29 and 299.71 (14) $x^2 - 340x + 4880$ Roots: 15.02 and 324.98 (15) $x^2 - 364x + 4879$ Roots: 13.94 and 350.06 If we examine the above family of quadratics, we see that the ONLY integer roots of 22 and 222 occur at iteration 10. Please note that for every starting quadratic that I will be using, that quadratic WILL be guaranteed to only generate a single subsequent quadratic family member that", "For how many real numbers $$a^{}_{}$$ does the quadratic equation $$x^2 + ax^{}_{} + 6a=0$$ have only integer roots for $$x^{}_{}$$? (第九届AIME1991 第8题)", "$x^2 - (r_1 + r_2)x + r_1r_2)$. By comparing this with $x^2 - ax + a^2$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$. Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2)$. Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0$. This can be factored as $(r_1 - 2)(r_2 - 2) = 4$. These factors can be: $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$. We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = {-1, 8, 9}$. So the answer is $-1 + 8 + 9 = \\boxed{\\textbf{(C) }16}$." ]

[ "# Minimum difference of sums So hey, I've recently found this problem online and I tried hard to solve but couldn't found an efficient solution. The problem statement is: Given and two integers $n$ and $k$ and an array $A$ containing $n$ pairs of integers of the form $(a, b)$ and $d$ defined as $d = |\\sum_{x \\in E} x[1] -x[0]|$ where is $E$ is a subset of $A$ containing exactly $k$ elements. Find the minimum value of $d$. To illustrate the problem, consider $n = 3$, $k = 2$, $A = [(2, 1), (0, 2), (1, 2)]$. The minimum value of $d$ is $0$. We can pick $E = [(1, 2), (2, 1)]$ containing $k = 2$ values of $A$ then $d = |(1 - 2) + (2 - 1)| = |0| = 0$. I tried to solve this problem but only came up with the brute force solution which is terrible in terms of time complexity. How can one solve this problem efficiently ? B = [] for pair in A: B.append(pair[0] - pair[1]) min_sum(B, k) min_sum returns k elements that add up to give the smallest number possible (absolute value). This (at least to me) feels easier to work with. • And this in turn is almost the subset sum problem. – Gribouillis Aug 3", "all distinct? Can't two successive terms $b_k,b_{k-1}$ be equal? – quasi Dec 31 '17 at 11:55 There are $\\binom{25}{2}=300$ different pairs in the chosen numbers. The sums of the elements can range from $3$ to $299$, thus can't take $300$ different values, and we conclude by the pigeon hole principle • Same as @asdf’s Answer... – Rohan Dec 31 '17 at 10:23 • The sums can range from 3 to 299. Can you explain the conclusion? – Moshe King Dec 31 '17 at 10:33 • There are less than 300 integer values between 3 and 299 – Taumen Dec 31 '17 at 10:35", "discovered a less trivial approximation for the growth rate of $f(n)$. But first I want to be rigorous about exactly what $f(n)$ is counting. I'm doing this in case I misinterpreted the question. My understanding is that $f(n)$ is the maximum number of distinct pairs $(i, j)$ with $1 \\leq i < j \\leq n$ such that there exists a set of $n$ positive integers $\\{a_1, a_2, \\dots, a_n\\}$ with $a_i + a_j = s^2$ for some $s$. Clearly, $f(n) \\leq {n \\choose 2}$, since we can pick at most ${n \\choose 2}$ pairs from $n$ objects. Now, to find a lower bound for $f(n)$, we just need to find a set of pairs and a corresponding set of values, such that all values indexed by the pairs sum to perfect squares. Let's consider the set $\\{1, 2, 3, \\dots, n\\}$ and define $h(n)$ as the number of distinct pairs of elements in this set that add to perfect squares. Clearly $h(n) \\leq f(n)$. I will now put forth a very simple proof demonstration that: $$n^{\\frac32} \\sim h(n)$$ Consider adding $n+1$ to the set $\\{1, 2, 3, \\dots, n\\}$. We now form all the pairs $(1, n+1), (2, n+1), \\dots, (n, n+1)$ whose corresponding sums are $n+2, n+3, \\dots, 2n+1$. Now, let $s(a, b)$ denote the number of", "2. Hence, there are $\\sum_{ k \\text{ even } } {n \\choose k} 2^k$ of them. Of these, pair them up according to those that differ only in the first non 2 digit. In this pair, clearly one of them has an even number of ones and an even number of 2's. Note that the only string not paired is the string of all 2's. Hence, the total number of strings with an even number of 1 and even number of 0 is $$.5 \\times (-1+\\sum_{ k even } {n \\choose k} 2^k) +1$$", "# 2009 AIME II Problems/Problem 12 ## Problem From the set of integers $\\{1,2,3,\\dots,2009\\}$, choose $k$ pairs $\\{a_i,b_i\\}$ with $a_i so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$. Find the maximum possible value of $k$. ## Solution Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\\cdots+2k=k(2k+1)$. On the other hand, as the sum of each pair is distinct and at most equal to $2009$, the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \\cdots + (2009-(k-1)) = \\frac{k(4019-k)}{2}$. Hence we get a necessary condition on $k$: For a solution to exist, we must have $\\frac{k(4019-k)}{2} \\geq k(2k+1)$. As $k$ is positive, this simplifies to $\\frac{4019-k}{2} \\geq 2k+1$, whence $5k\\leq 4017$, and as $k$ is an integer, we have $k\\leq \\lfloor 4017/5\\rfloor = 803$. If we now find a solution with $k=803$, we can be sure that it is optimal. From the proof it is clear that we don't have much \"maneuvering space\", if we want to construct a solution with $k=803$. We can try to use the $2k$ smallest numbers: $1$ to $2\\cdot 803 = 1606$. When using these numbers, the average sum", "+ 1$ from the list and shifts everything else lift, and the inverse transformation reinserts it. So $b_1$ can never be equal to $b_2$, etc., and conversely, if $b_1 \\neq b_2$, the pairs $(a_1, a_1+1)$ and $(a_2, a_2+1)$ won't overlap. (This can be done more carefully.) Once we know this, choosing the set $T$ of pairs is equivalent to choosing the set $T'$ of integers. A quick check shows that the minimum possible value for $b_1$ is $1$ (since that's the minimum value for $a_1$), and the maximum value for $b_r$ is $n-r$ (since the maximum for $a_r$ is $n-1$). Thus we have reduced the problem to choosing a set $T'$ of $r$ integers from among the set {$1, 2, \\dots, n-r$}, which is doable $\\binom{n-r}{r}$ ways. Each admissible choice of $r$ pairs of consecutive numbers from $[n]$ can be encoded as a binary word of length $n-r$ containing exactly $r$ ones. There are ${n-r\\choose r}$ such words $w$. Given such a word $w$ the parsing is as follows: Count along $w$. For each ${\\tt 0}$ in $w$ count $1$ ahead, and let the corresponding number single. For each ${\\tt 1}$ in $w$ count $2$ ahead, and at the same time pair these two numbers.", "# Small pairs a, b, every integer up to k dividing at least one of them My question is essentially the $m=2$ case of this question. Given a positive integer $k$, I'm interested in (small) pairs of positive integers $a,b$ such that every positive integer up to (and including) $k$ is a factor of at least one of them. For example, for $k=10$, one such pair is $a=70$, $b=72$; every integer up to 10 is a factor of at least one of these two numbers. It's easy to see that the product of any such pair must be a multiple of the least common multiple, call it $L(k)$, of $1,2,3,\\dots,k$. It's known that $L(k)$ is asymptotic to $e^k$. Moreover, it's trivial to find a pair whose product is exactly $L(k)$; just take $a=1$, $b=L(k)$. This tells me that, for this problem, the product, $ab$, is not a good measure of how small the pair $a,b$ is. Two other measures that suggest themselves are the sum, $a+b$, and the maximum. Since the product is at least $L(k)$, the sum must be at least $2\\sqrt{L(k)}$, and the maximum must be at least $\\sqrt{L(k)}$. My question is, how sharp are these bounds? I did a small amount of calculation by hand during a recent committee meeting, and arrived at these figures,", "# Finding the largest set of integers over an interval where the sum of any 'k' elements is unique Consider the set $(s_1, ..., s_N) \\in S$, where all $s_i$ are positive integers selected from some interval $[M, L]$ and the sum of any $k$ integers in $S$ is required to be unique and to have a distance of at least $d$ from all other possible sums of $k$ integers in the set. For example, if $k = 2$, the sum of any two randomly chosen elements, i.e. $s_a$ and $s_b$, must generate a unique sum that has a difference of at least $d$ from all other possible sums of two elements in $S$. As a function of $k$, the minimum difference between sums $d$, and the size of the interval, $(L - M)$, what is the largest possible size of the set $S$, and what is the most efficient way to compute its elements? Edit - I'm of course happy to hear any $d = 1$ solutions. Edit 2 - For the $d=1$ case, is this problem NP-complete, similar to the subset-sum problem? - Here's a start. The number of choices of $k$ elements is $N\\choose k$ (or a little more, if repeats are allowed), and each sum is between $kM$ and $kL$, so you must have", "up to $1998$: $\\{1,1997\\},\\{2,1996\\},\\dots$. These pairs are your pigeonholes; how many are there? You need to be a little careful here, since $1997$ is odd: after you form the pairs, you’ll have one number left over. It’s a pigeonhole all by itself. - I would consider the sets $A_1 = \\{1,1997\\}, A_2 =\\{2,1996\\} \\ldots A_{998} = \\{998, 1000\\}, A_{999}=\\{999\\}$. Note that they contain all the numbers ${1,\\ldots , 1997}$. Now, we want to choose $1000$ numbers: it means that you take at least two elements from the same set and hence their sum is indeed $1998$.", "Thus the number of possible choices of two numbers is $$(n-1)+(n-2)+(n-3)+\\cdots +2+1,$$ which is precisely $\\sum_{k=1}^{n-1}k$. A way to derive formulas like this that is easily remembered (to answer the comment \"How are identities such as this one derived?\"): Consider that $$\\sum_{k=1}^n (2k-1) = \\sum_{k=1}^n \\Big(k^2 - (k-1)^2\\Big)$$ $$= 1^2 - 0^2 + 2^2 - 1^2 + 3^2 - 2^2 \\pm ... \\pm n^2 - (n-1)^2=n^2.$$ Together with $\\sum_{k=1}^n 1 = n$ this immediately gives $$\\sum_{k=1}^n k = \\frac{n^2 + n}{2}.$$ It could be argued that the \"telescoping sum\" above requires induction to justify formally, but it seems more intuitive than the sum $\\sum_{k=1}^n k$ itself. How many unordered pairs of $n$ elements are there? One can pair the first element with any of $n-1$ others. One can pair the second element with any of $n-2$ other (besides the first one, which was already listed above). One can pair the third element with any of $n-3$ other (besides the first two, which were already listed above). One can pair the fourth element with any of $n-4$ other (besides the first three, which were already listed above). . . . and so on . . . . . . Hint:let $a_n=1+2+3+\\dots+n$ then we have $$a_{n+1}-a_{n}=(1+2+3+\\dots+n+n+1)-(1+2+3+\\dots+n)=n+1 \\color{red}{\\to}$$$$a_{n+1}-a_{n}=n+1$$ easily by recursive relation we can find $a_n$ see here linear recurrence relation", "# Prove that there must be two distinct integers in $A$ whose sum is $104$. Let A be any set of $20$ distinct integers chosen from the arithmetic progression ${1,4,7,...,100}$. Prove that there must be two distinct integers in $A$ whose sum is $104$. Define $A=\\{1+3i\\}_{i=0}^{33}$. I know that if two distinct integers $a,b \\in A$ are such that $104=a+b=(1+3i)+(1+3j)$, then $i+j=34$ with $0<i,j\\leq33$ and $i \\not= j$. I think we can play with elements parity or even with the fact that if $i = j$, then $i = 17$ and $(3i + 1) = 52$. However, I am not able to advance further into the question. Are there someone who could help me complete the problem? • Hint: there are $34$ such numbers. Count the number of pairs $(i,j)$ with $i<j$ and $i+j=104$ you can make out of your set. Then count the number of integers in your set that don't come up in any such pair. – lulu Sep 28 '15 at 23:56 • I know we have 16 such pairs and 20 distinct integers to choose from the sequence. Must I use the pigeonhole principle? – user230283 Sep 29 '15 at 0:08 • Yes. But don't forget that some elements in your progression belong to no such pair. – lulu Sep 29 '15 at 0:14", "ideas... 4. ## Re: Pairs of lattices You are right. If the maximum element $a$ of M covers $a_1,\\dots,a_m$ and the maximum element $b$ of N covers $b_1,\\dots,b_n$, then $(a,b)$ covers $(a_1,b),\\dots,(a_m,b)$, $(a,b_1),\\dots,(a,b_n)$. So, either L is M3 and the top element of K has 1 descendant, ot L is N5 and the maximum of K has 2 descendants.", "# Math Help - distinct pairs 1. ## distinct pairs find the number of disttinct pairs of postive integers (m,n)such that $n(m-n)=2009$ 2. Originally Posted by banku12 find the number of disttinct pairs of postive integers (m,n)such that $n(m-n)=2009$ $2009=7^2\\cdot 41\\Longrightarrow\\,n$ must be taken from the divisors $1,\\,7,\\,49,\\,41,\\,287,\\,2009$ , and then you must choose $m$ accordingly. For example, $n=1\\Longrightarrow\\, m=2010\\,:\\,1(2010-1)=2009\\;;\\;\\;n=7\\Longrightarrow\\,m=294\\,:$ $7(294-7)=7\\cdot 287=2009$ , etc. Tonio", "The $b_i$ are equivalent under rotation of order or flipping the order, but not in otherwise reordering. (IE, the difference list $[1,2,3,4]$ is equivalent to $[1,4,3,2]$ but not $[1,4,2,3]$.) Background: I was doing an analysis of a card game involving restricted pairs, and found some interesting properties for sets of $n=k(k-1)+1$ symbols, taken $k$ at a time, where each two sets of $k$ symbols have exactly one pair in common. One thing I found was that I could create a maximal set using a partition slicing strategy. But I wanted to create a pattern $(a_1 \\dots a_k)$ that I could simply increment around the modulus $n$ to create every possible $k-$set. What I found was that for $k=1\\dots 8, k\\neq 7$, I could have at least two such patterns (removing multiples via various equivalencies). However, I have been unable to find one for $k=7$, even with an exhaustive code-based search of the $[43]^7$ vector space. • The $a_i$ allow a \"tiling\" of the space by incrementing $\\bmod n$. That is, if I take the $n$ additive translations of the list of $a_i$, I get $n$ $k-$element lists where each pair of translations has exactly one element in common. – JKreft Aug 3 '18 at 5:48 • Are perfect difference sets what you are after? – Seva Aug 3", "will both be in the sequence, and $n|2n$. The argument generalized for $n-k$ being the largest element chosen in $\\{a_1,..,a_n\\}$ - Yeah, it becomes pretty easy if excluding 1. – Robert Jul 2 '11 at 21:05 This follows from the pigeonhole principle. Consider the set of first $2n$ numbers. You can create $n$ subsets of co-primes as $\\{2k-1,2k\\}$ for $1\\leq k \\leq n$. It is easy to see that these will be coprime (their difference is 1, so their GCD is 1). Now you have $n$ pairs of coprimes, and amongst any $n+1$ numbers in this range, atleast two will be a coprime pair. Note: I have followed the definition to allow $(1,2)$ to be a coprime pair. -", "set of integers, and let $Y = \\{0,1\\}$. Define the function $f: X \\to Y$ by: $f(k) = \\dfrac{1^k - (-1)^k}{2}$. Clearly, if $k$ is even, then $f(k) = 0$, and if $k$ is odd, then $f(k) = 1$. So we have two equivalence classes: $\\{\\dots,-4,-2,0,2,4,6,8,\\dots\\}$ $\\{\\dots,-5,-3,-1,1,3,5,7,\\dots\\}$ We can call these $[0]_R$ and $[1]_R$. Our function $F$ is thus: $F([0]_R) = 0$ $F([1]_R) = 1$, which is clearly a one-to-one correspondence between two two-element sets.", "k, b_{j} = k+1$, set$a_{i} = k+1, b_{j} = k$. Now just check that the claim holds for the pair of sequences$a_{i} = i, b_{j} = n+j\\$, and that the operation I described leaves the sum invariant. I guess I failed to use the pigeonhole principle, but this is the approach that makes the most sense to me.", "# number of combinations of ordered sequences of N integers suppose a N-tuple of N integers, such that every element in the tuple is bigger than or equal to the last one and each element in the sequence ranges from 1 to K. Is there any closed formula for the number of combinations possible? For those who think this is trivial, try to think about it for one minute. I have done so: let $a_i \\in \\{1, 2, ..., K\\}, i \\in \\{1..N\\}$. We want to count the number of N-tuples $(a_1, a_2, ..., a_N)$ such that $a_i>=a_{i-1}$. I know how to solve for $N=2$: given $a_1 \\in \\{1, 2, ... K\\}$, the number of possibilities for $a_2>=a_1$ is $K-a_1+1$. Thefore, the number of pairs $(a_1,a_2)$ is: $U=K+(K-1)+(K-2)+...+1$ which is easy to compute since it equals $K(K+1)/2$. Now, how to generalize for N-tuples? Line up $N$ balls. Then insert $K-1$ dividers. Let the number of balls to the left of the first divider be the number of $1$'s in your sequence. Let the number of balls between the first and second divider be the number of $2$'s in your sequence, and so on. For example, with $N = 4$ and $K = 5$ \"$\\bullet |\\bullet | | \\bullet \\bullet|$\" represents one $1$, one $2$, no $3$'s two $4$'s,", "# If $15$ distinct integers are chosen from the set $\\{1, 2, \\dots, 45 \\}$, some two of them differ by $1, 3$ or $4$. $$\\blacksquare~$$ Problem: If $$15$$ distinct integers are chosen from the set $$\\{1, 2, \\dots, 45 \\}$$, some two of them differ by $$1, 3$$ or $$4$$. $$\\blacksquare~$$ My Approach: Let the minimum element chosen be $$n$$. Then $$n + 1 , n + 3 , n + 4$$ can't be taken. We make a small claim. $$\\bullet~$$ Claim: In a set of $$~7$$ consecutive numbers at most $$2$$ numbers can be chosen. $$\\bullet~$$ $$\\textbf{Proof:}$$ Let us name the elements of the set as $$\\{ 1,2,3,\\dots,7 \\}$$. Now let's consider the least element is chosen. If the least element is $$1$$, then $$2,4,5$$ can't be chosen. So we are left with $$3, 6, 7$$. $$\\circ~$$ If $$~3~$$ is chosen, then $$6, 7$$ can't be in the set. And if $$~3~$$ is not chosen, then only any one of the 2 elements $$\\{ 6, 7 \\}$$ be chosen. So, a maximum of $$2$$ elements can be chosen in this case. $$\\circ \\circ~$$ If the least element is $$2,$$ then $$3, 5, 6$$ can't be there in the set. So, possible elements are 4, 7. So, one of these two can be chosen. Then, a maximum", "# How many pairs of sets that have no element in common? Consider three positive integers n, k and m such that k <= m. And a set S = {1,2,…,n+k}. There are $${ {n+k} \\choose {k}}$$ subsets of size k from the set S, and $${ {n+k} \\choose {m}}$$ subsets of size m from the set S. So there is a total of $${ {n+k} \\choose {k}} * { {n+k} \\choose {m}}$$ pairs of subsets that I can choose from, but I need to eliminate any pair $$p=(s1,s2)$$ such that $$s1∩s2 ≠ ∅$$ that is to say if they have at least one element in common, it has to be counted so we can eliminate it from the total. For example: n = 3, k = 2 and m = 3 S = {1,2,3,4,5} Pairs = [({1,2}, {1,2,3}), ..., ({1,2}, {3,4,5})..., ({4,5}, {3,4,5})] So for the 3 explicit examples above, there are 2 to be eliminated: ({1,2}, {1,2,3}) -> common elements are {1,2} ({4,5}, {3,4,5}) -> common elements are {4,5} No elimination for ({1,2}, {3,4,5}) because the 2 sets of {1,2} and {3,4,5} are disjoint. So, how many pairs of sets that have no element in common? Thanks The $$k-$$set can be choosen in $$\\binom{n+k}{k}$$ ways. There are only $$n$$ elements left to choose from, so the" ]

[ "will be $1607$. Hence we can try looking for a nice systematic solution that achieves all sums between $1607-401=1206$ and $1607+401=2008$, inclusive. Such a solution indeed does exist, here is one: Partition the numbers $1$ to $1606$ into four sequences: • $A=1,3,5,7,\\dots,801,803$ • $B=1205,1204,1203,1202,\\dots,805,804$ • $C=2,4,6,8,\\dots,800,802$ • $D=1606,1605,1604,1603,\\dots,1207,1206$ Sequences $A$ and $B$ have $402$ elements each, and the sums of their corresponding elements are $1206,1207,1208,1209,\\dots,1606,1607$. Sequences $C$ and $D$ have $401$ elements each, and the sums of their corresponding elements are $1608,1609,1610,1611,\\dots,2007,2008$. Thus we have shown that there is a solution for $k=\\boxed{803}$ and that for larger $k$ no solution exists.", "set , which makes $P(k) \\ge \\frac{1}{2} + \\frac{1}{2k}$. Now the only case without rounding, $k = 1$. It must be true. ### Solution 2 It suffices to consider $0\\le n < k-1.$ Now for each of these $n,$ let $f(n)=[\\frac{n}{k}], g(n)=[\\frac{100}{k}]-[\\frac{100-n}{k}].$ If we let $k=67,$ then the following graphs result for $f$ and $g.$ $f:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { if (i<=33) dot((i,0)); else dot((i,1)); } label(\"(0,0)\",(0,0),SW); label(\"(66,1)\",(66,1),NE); [/asy]$ $g:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { dot((i,0)); } label(\"(0,0)\",(0,0),NW); label(\"(66,0)\",(66,0),SE); [/asy]$ Our probability is the number of $0\\le i such that $f(i)+g(i)=0$ over $k.$ Of course, this always holds for $i=0.$ If we let $k$ vary, then the graph of $f$ is always very similar to what it looks like above (groups of $\\frac{k+1}{2},\\frac{k-1}{2}$ dots). However, the graph of $g$ can vary greatly. In the above diagram, $g(i)=0$ for all $i,$ while it is possible for $g(i)=-1$ for all $i\\neq 0.$ In order to minimize the number of $i$ which satisfy $f(i)+g(i)=0,$ we either want $g(i)=0$ for $0\\le i or $g(i)=-1$ for $1\\le i This way, we see that at least half of the numbers from $1$ to $k-1$ satisfy the given equation. So, our desired probability is at least $\\frac{k+1}{2k}.$ As shown by", "integer. Then $P$ must be one of the points (2010,1), (2009,2),…, (1,2010). This yields the possible values of $\\theta$: $\\theta=\\tan^{-1}\\frac{2011-n}{n},\\ n=1,2,\\dots,2010.$ It remains to check whether any of these values must be excluded on the grounds that the line segment from (0,0) to $(2011-n,n)$ contains another integer point, say $(k,m)$ with $k+m<2011$. If that were so, we would have $\\frac{2011-n}{n}=\\frac{k}{m},\\ (2011-n)m=kn,\\ 2011m=n(k+m).$ In particular, $k+m$ divides $2011 m$, hence must have a nontrivial common divisor with 2011. But 2011 is prime, so that $k+m=2011$, a contradiction. Hence all 2010 values of $\\theta$ as above yield the desired outcome. Why it’s a dud: This is an example of a problem whose difficulty depends very strongly on knowing a certain “trick” or technique – in this case, the reflection principle. The committee aims to avoid this type of questions. It doesn’t help that the trick is an old one. Solution to 2. Rewrite the sum as $\\sum_{k=1}^{2008} c_k(a_{k+1}-a_k)$, where $c_k$ is the number of pairs $(i,j)$ such that $1\\leq i\\leq k. We count the number of such pairs: for each $k$ there are $k$ choices of $i$ and $2009-k$ choices of $j$, so that $c_k=k(2009-k)$. In particular, all $k$ are even. But then the above sum must also be even, in particular it cannot be 31415926535. Why it’s a dud:", "our pattern must \"line up\" with what we know about$a_{1007}$and$b_{1007}$, namely, that$2013-k$is the largest chosen number in the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having$k$elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that$2k | 2014$or$k | 1007$. Because$1007 = 19 \\times 53$, then we have four possible values of$k$. If$k=1$, then both sequences look like this:$0,2,4,\\dots, 2012$. Then$S=2012$. If$k=19$, then both sequences look like this:$0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then$S=1994$. If$k=53$, then both sequences look like this:$0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then$S=1960$. If$k=1007$, then both sequences look like this:$0,1,2,\\dots,1006$. Then$S=1006. Now, we have to address the case where 0 is not chosen in $a_i$ and $b_i$. We define new sequences $c_i = 2013-a_{1008-i}$ and $d_i = 2013-b_{1008-i}$. It's easy to see that $c_i,d_i$ satisfy the conditions of the problem, and since $0$ is not chosen in $a_i,b_i$ then 2013 must be chosen in them, which means $0$ is chosen in both $c_i,d_i$. This reduces $c_i,d_i$ to the case above, except $S$ is replaced with $4026-S$. Because the possible values for $S$ in the first case is $1007,1960, 1994, 2012$ then the possible values for $S$ in this case", "the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having $k$ elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that $2k | 2014$ or $k | 1007$. Because $1007 = 19 \\times 53$, then we have four possible values of $k$. If $k=1$, then both sequences look like this: $0,2,4,\\dots, 2012$. Then $S=2012$. If $k=19$, then both sequences look like this: $0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then $S=1994$. If $k=53$, then both sequences look like this: $0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then $S=1960$. If $k=1007$, then both sequences look like this: $0,1,2,\\dots,1006$. Then $S=1006. Now, we have to address the case where 0 is not chosen in$a_i$and$b_i$. We define new sequences$c_i = 2013-a_{1008-i}$and$d_i = 2013-b_{1008-i}$. It's easy to see that$c_i,d_i$satisfy the conditions of the problem, and since$0$is not chosen in$a_i,b_i$then 2013 must be chosen in them, which means$0$is chosen in both$c_i,d_i$. This reduces$c_i,d_i$to the case above, except$S$is replaced with$4026-S$. Because the possible values for$S$in the first case is$1007,1960, 1994, 2012$then the possible values for$S$in this case is$2014,2032,2134,3020\\$.", "end of the sequence, our pattern must \"line up\" with what we know about$a_{1007}$and$b_{1007}$, namely, that$2013-k$is the largest chosen number in the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having$k$elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that$2k | 2014$or$k | 1007$. Because$1007 = 19 \\times 53$, then we have four possible values of$k$. If$k=1$, then both sequences look like this:$0,2,4,\\dots, 2012$. Then$S=2012$. If$k=19$, then both sequences look like this:$0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then$S=1994$. If$k=53$, then both sequences look like this:$0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then$S=1960$. If$k=1007$, then both sequences look like this:$0,1,2,\\dots,1006$. Then$S=1006. Now, we have to address the case where 0 is not chosen in $a_i$ and $b_i$. We define new sequences $c_i = 2013-a_{1008-i}$ and $d_i = 2013-b_{1008-i}$. It's easy to see that $c_i,d_i$ satisfy the conditions of the problem, and since $0$ is not chosen in $a_i,b_i$ then 2013 must be chosen in them, which means $0$ is chosen in both $c_i,d_i$. This reduces $c_i,d_i$ to the case above, except $S$ is replaced with $4026-S$. Because the possible values for $S$ in the first case is $1007,1960, 1994, 2012$ then the possible values for", "5. For $$k \\ge 5$$, the maximum occurs at two points of the form $$p_k$$ and $$1 - p_k$$ where $$p_k \\to 0$$ as $$k \\to \\infty$$. The maximum value $$r_k(p_k) \\to 1/e \\approx 0.3679$$ as $$k \\to \\infty$$. The graph has a local minimum at $$p = \\frac{1}{2}$$ Proof sketch: Parts (a), (b), and (c) are clear. Part (d) follows by computing the derivatives: $$r_2^\\prime(p) = 2 (1 - 2 p)$$, $$r_3^\\prime(p) = 3 (1 - 2 p)$$, $$r_4^\\prime(p) = 4 (1 - 2 p)^3$$. For part (d), note that $$r_k(p) = s_k(p) + s_k(1 - p)$$ where $$s_k(t) = k t^{k-1}(1 - t)$$ for $$t \\in [0, 1]$$. Also, $$p \\mapsto s_k(p)$$ is the dominant term when $$p \\gt \\frac{1}{2}$$ while $$p \\mapsto s_k(1 - p)$$ is the dominant term when $$p \\lt \\frac{1}{2}$$. A simple analysis of the derivative shows that $$s_k$$ increases and then decreases, reaching its maximum at $$(k - 1) / k$$. Moreover, the maximum value is $$s_k[(k - 1) / k] = (1 - 1 / k)^{k-1} \\to e^{-1}$$ as $$k \\to \\infty$$. Suppose $$p \\in (0, 1)$$, and let $$N_k$$ denote the number of rounds until an odd man is eliminated, starting with $$k$$ players. Then $$N_k$$ has the geometric distribution on $$\\N_+$$ with parameter $$r_k(p)$$. The mean and variance", "total combinations are: $\\frac{5!}{2!2!} = 30$ Case2#: $4 \\{5 3 3 2 4\\} 0$, total combinations are: $\\frac{5!}{2!} = 60$ Case3#: $5 \\{4 3 3 0 4\\} 2$, total combinations are: $\\frac{5!}{2!2!} = 30$ Case4#: $4 \\{5 3 3 0 4\\} 2$, total combinations are: $\\frac{5!}{2!} = 60$ Case5#: $5 \\{4 3 3 2 0\\} 4$, total combinations are: $\\frac{5!}{2!} = 60$ Case6#: $4 \\{5 3 3 2 0\\} 4$, total combinations are: $\\frac{5!}{2!} = 60$ So, the total no of combination will be $300$. - You have the following combinations: • $4\\dots 0$, where in the middle you permute $\\{2,3,3,4,5\\}: \\frac{5!}{2}$, with the $2$ in the denominator accounts for the double $3$. And the same for $4\\dots 2$, $4\\dots 4$ and $5\\dots 4$. • $5\\dots 0$, where in the middle you permute $\\{2,3,3,4,4\\}: \\frac{5!}{2^2}$, with 2 doubles $3$ and $4$. And the same for $5\\dots 2$. So in total you have $4\\frac{5!}{2}+2\\frac{5!}{2^2}=300\\;$ even numbers. -", "other, I got an LCM of 802. I see @celtschk went along the same lines. I think this is a pretty good candidate for a computer check :) – rschwieb Aug 28 '12 at 16:16 $2005=4\\cdot401$ with $401$ prime. Then $$2005=2\\cdot401+2\\cdot401+401 \\,.$$ Thus we found three numbers with lcm 802. We claim that this is the smallest possible value. Suppose by contradiction that we can find $a+b+c =2005$ and $l=\\operatorname{lcm}(a,b,c) < 802$. We can assume that $a \\leq b \\leq c$. Since $b, c \\leq l$ it follows that $$2005=a+b+c \\leq a+2l < a+2\\cdot802$$ Thus $a > 401> \\frac{l}{2}$. Since $a> \\frac{l}{2}$ and $a\\mid l$, it follows that $a=l$. Then $$l =a \\leq b \\leq c \\leq l \\Rightarrow a=b=c=l \\Rightarrow 3l=2005$$ You can write $5\\cdot401$ or $5\\times401$. There's no need for such crudidities as $5*401$. – Michael Hardy Aug 28 '12 at 16:18 Shouldn't it be \"$b,c\\le l$\" in line 7 (and then of course also for the first inequality sign in the following line)? That of course doesn't invalidate the proof. – celtschk Aug 28 '12 at 16:25", "\\frac{1}{3^{2012}} + \\dots + \\frac{1}{2010^{2012}} =_Q ~0 \\pmod{2011}$$ • with $p=2011$, $k=2010$, $$\\frac{1}{1^{2010}} + \\frac{1}{2^{2010}} + \\frac{1}{3^{2010}} + \\dots + \\frac{1}{2010^{2010}} =_Q ~ -1 \\pmod{2011}$$ Today we have learned some divisibility properties of power summation. In the homework section, we show another way to prove these results. Let us stop here for now. Hope to see you again in the next post. Homework. 1. Prove that $$\\frac{1}{1^{1000}} + \\frac{1}{2^{1000}} + \\frac{1}{3^{1000}} + \\dots + \\frac{1}{2010^{1000}} =_Q ~0 \\pmod{2011}$$ $$\\frac{1}{1^{2012}} + \\frac{1}{2^{2012}} + \\frac{1}{3^{2012}} + \\dots + \\frac{1}{2010^{2012}} =_Q ~0 \\pmod{2011}$$ 2. The divisibility properties of $S_{-k}(p-1)$ can be proved by another way as follows. Using Bézout's lemma prove that for any $j=1,2, \\dots, p-1$, there exists $1 \\leq w_j \\leq p-1$ such that $$j ~w_j = 1 \\pmod{p}$$ Prove that $w_1, w_2, \\dots, w_{p-1}$ is a permutation of $1,2, \\dots, p-1$. We have $$j^k ~w_j^k = 1 \\pmod{p}$$ thus, $$\\frac{1}{j^k} =_Q ~w_j^k \\pmod{p}$$ Sum them all up as $$\\sum_{j=1}^{p-1} \\frac{1}{j^k} =_Q ~\\sum_{j=1}^{p-1} w_j^k \\pmod{p}$$ we have $$S_{-k}(p-1) =_Q ~ S_{k}(p-1) \\pmod{p}$$ thus obtaining the result. Here is the example with $p=7$ $$1 \\times 1 = 1 \\pmod{7}$$ $$2 \\times 4 = 1 \\pmod{7}$$ $$3 \\times 5 = 1 \\pmod{7}$$ $$4 \\times 2 = 1 \\pmod{7}$$ $$5 \\times 3 = 1 \\pmod{7}$$ $$6 \\times 6 = 1 \\pmod{7}$$ so", "enough, it seems to be somewhat hand-wavey (it assume certain solution is the best and proceed to calculate base on that). So I provide a solution, that eventually will reach the same conclusion. For each odd number $n<3000$, define $P_{n}=\\{2^{k}n|0\\leq k;2^{k}n\\leq 3000\\}$. Easily check that those $P_{n}$ are disjoint from each other, and their disjoint union is indeed $[1,3000]$. Now considering any subset $A$. We can easily prove that if for each odd $n$ then $A\\bigcap P_{n}$ do not contain any member twice another, then $A$ itself will also have that property. Hence constructing an $A$ with maximum order involve constructing all those $A\\bigcap P_{n}$ that individually have maximum order. The maximum possible order of $A\\bigcap P_{n}$ is $[\\frac{|P_{n}|+1}{2}]$ (the bracket stand for floor, I do not know how to write that floor bracket). Now the question boil down to finding the number of possible $n$ of a given value of $[\\frac{|P_{n}|+1}{2}]$. This is not too hard. If $[\\frac{|P_{n}|+1}{2}]=m$ for some integer $m$, then $|P_{n}|=2m-1$ or $2m$. Not just that $|P_{n}|=k_{\\max}+1$ where $k_{\\max}$ is the maximum $k$ such that $2^{k}n\\leq 3000$. Hence $k_{\\max}=2m-2$ or $2m-1$. In other word, $2^{2m-2}n\\leq 3000<2^{2m}n$. The maximum possible $n$ in that case is $[\\frac{3000}{2^{2(m-1)}}]$ and minimum is $[\\frac{3000}{2^{2m}}]+1$. All that's left is to actually make the calculation for each $m$. Note that we", "it would be better to convert them to 7 lots of 13.) All of these are different because the multiples of 7 will have different residues mod 13. This should generalize for general k to give (k+1)+k+....(k+2-a) possible values - 5 years, 4 months ago That solutions seems correct, I solved it differently: for every $$0 \\leq b \\leq 100$$, there are $$13$$ possible values for $$a$$, but I need to subtract $$12+11+\\dots + 1$$ because for $$b=100$$ I counted $$13$$ values for $$a$$ while only $$a=0$$ is valid due to the fact that $$a+b \\leq 100$$. Similarly, for $$b=99$$, $$13$$ values for $$a$$ were counted while only 2 are valid, etcetera. So, the answer is $$13 \\cdot 101 - (12 + 11 + \\dots + 1) = 1235$$. Generalization: denote $$\\max(m,n)$$ by $$q$$. The total values $$m \\cdot a + n \\cdot b$$ can take if $$a+b \\leq k$$ then is \\begin{align*} q \\cdot (k+1) - \\left( (q-1) + (q-2) + \\dots + 1 \\right) &= q \\cdot (k+1) - \\frac{q(q-1)}{2}\\\\ &= q \\left( (k+1) - \\frac{q-1}{2} \\right)\\\\ &= \\frac{q(-q+2k+3)}{2}. \\end{align*} And indeed: \\begin{align*} (k+1) + k + \\dots + (k+2-a) &= \\frac{(k+1)(k+2)}{2} - \\frac{(k+2-a)(k+1-a)}{2}\\\\ &= \\frac{(k^2+3k+2) - (k^2+3k+2-2qk-3q+q^2)}{2}\\\\ &= \\frac{-q^2+2qk+3q}{2}\\\\ &= \\frac{q(-q+2k+3)}{2}. \\end{align*} - 5 years, 4 months ago That looks good- my a is your", "p+T-1$$ $$F'(p)=Np^{N-1}-T$$ Thus: $$p_0=\\frac{T-1}{T}$$ $$p_{k+1}=p_k-\\frac{p_k^N-T p_k+T-1}{Np_k^{N-1}-T}$$ Let's check this method for $T=300$ and $N=1000$: $$p_0=\\frac{299}{300}=\\color{blue}{0.996}6666\\dots$$ $$p_1=p_0-\\frac{p_0^{1000}-300 p_0+299}{1000p_0^{999}-300}=\\color{blue}{0.99680}083907776694848\\dots$$ $$p_2=\\color{blue}{0.996802135}036332828\\dots$$ $$p_3=\\color{blue}{0.99680213516858413}5\\dots$$ Since Newton's method has quadratic convergence, the number of correct digits approximately doubles at each step. Now everything depends on the computational resources you have and the accuracy you need. (Update: See another answer for a more simple method, for a certain condition on $T,N$)", "is precisely the maximum possible value of $a_{2018}$, so all the inequalities above are equalities, and so for each $k\\geqslant 2$, (i) $a_k$ and $b_k$ are consecutive elements of the triangle, and (ii) $a_{k-1}$ lie directly above $a_k$ and $b_k$. Hence all numbers $a_k$ and $b_k$ are in a \"strand\" of width $1$ unit from top to bottom of the triangle. In particular, it is possible to find a sub-triangle with $\\left\\lceil\\frac{2018-2}{2}\\right\\rceil=1008$ rows not containing any of $\\{1,2,\\ldots,2018\\}$. Re-defining $a_k$ and $b_k$ again for this sub-triangle, we have $$a_{1008} \\geqslant b_1 + b_2 +\\cdots + b_{1008} \\geqslant 2019 + 2020 +\\cdots +3026 > \\frac{2018\\cdot 2019}{2}$$ which is clearly impossible. Hence, there are no anti-Pascal triangles with $2018$ rows containing every integer from $1$ to $1 + 2 + 3 + \\dots + 2018$, as desired. $\\square$ Solution 2 Let $b_k$ and $s_k$ be the maximum and minimum number in row $k$. Then ${ s_1, s_2, \\dots, s_{2018} }$ is a permutation of ${1, 2, \\dots, 2018}$. Call these the small numbers. Let $N=1+2+\\dots+2018$. Call the numbers $N, N-1, \\dots, N-2018$ big numbers. Claim 1: There are at most 1010 big numbers on the bottom row If two big numbers are consecutive on a row, they have a small number above them. There is only one small number in", "$2013-k$ is the largest chosen number in $a_i$, so $a_{1007} = 2013-k$. Because $b_1 = 0$ then $S = 2013-k$. That means $b_{1007}=2013-k - a_0 = 2013-k$ Because $2013-k$ is the largest chosen number in $b_i$, then $2013-k+1,\\dots,2013$ are not chosen in $b_i$, therefore $0,\\dots,k-1$ are chosen in $b_i$ and $k$ is not chosen in $b_i$. Because $0,\\dots,k-1$ are chosen in $b_i$, and $S = 2013-k$, that means $2013-k, 2013-k-1, \\dots, 2013-(2k-1)$ are chosen in $a_i$, which means $k,k+1,\\dots,2k-1$ are not chosen in $a_i$. Likewise, because $0,\\dots,k-1$ are chosen in $a_i$, and $S = 2013-k$, that means $2013-k, 2013-k-1, \\dots, 2013-(2k-1)$ are chosen in $b_i$, which means $k,k+1,\\dots,2k-1$ are not chosen in $b_i$. At this point, we have established the following fact: • The first $k$ numbers in $a_i$ and $b_i$ are chosen, and the next $k$ numbers are not chosen. • The last $k$ numbers in $a_i$ and $b_i$ are not chosen, and the previous $k$ numbers are chosen. We would like then to claim that this pattern would have to continue itself all the way to the \"middle\" of the sequences, which finally means that $2k$ divides 2014. Formally, we claim that for each $m \\geq 0$, we have the following: • $2km, \\dots, (2m+1)k-1$ are chosen in $a_i$ and $b_i$. • $(2m+1)k, \\dots, (2m+2)k-1$ are", "# Finding a minimum I would like to minimize the function $f(x_1,\\dots,x_k) = x_1 + x_2 / x_1 + \\dots + x_k / x_{k-1} + M / x_k$ where $M>0$ and $x_k \\ge \\dots \\ge x_1 \\ge 1$. First I looked at a simpler function $g(x_1,x_2) = x_1 + M/x_2$. Computing the partial derivatives for $x_1$ and $x_2$ and finding the respective roots shows that the minimum value is attained when $x_1 = x_2 = \\sqrt{M}$. However, for the general case when $k$ is larger and in particular for $k=\\log M$, it seems that setting $x_i = 2^{i-1}$ yields $2\\log M + 1$ as the value of $f$, which I also suspect to be the global minimum. If so, how can I formally show this? - Hint: try using AM-GM inequality, which gives minimum when equality is there. – Macavity Apr 5 '13 at 14:35 $x_1 + x_2 / x_1 + \\dots + x_k / x_{k-1} + M / x_k\\ge (k+1)(x_1.\\frac{x_2}{x_1}.\\frac{x_3}{x_2}\\dots\\frac{x_k}{x_{k-1}}\\frac{M}{x_k})^{1/(k+1)}=(k+1)M^{1/(k+1)}$ This is by using A.M.-G.M. Equality holds when $x_1=\\frac{x_2}{x_1}=\\frac{x_3}{x_2}=\\dots=\\frac{x_k}{x_{k-1}}=\\frac{M}{x_k}$ - By AM-GM, we have: $$\\frac{x_1 + \\frac{x_2}{x_1} + \\dots + \\frac{x_k}{x_{k-1}} + \\frac{M}{x_k}}{k+1} \\ge \\left( x_1 \\cdot \\frac{x_2}{x_1} \\dots \\frac{x_k}{x_{k-1}} \\cdot \\frac{M}{x_k}\\right)^{\\frac{1}{k+1}} = M^{\\frac{1}{k+1}}$$ Equality is the minimum we can hope for, so the minimum is $(k+1)M^{\\frac{1}{k+1}}$, when $$x_1 = \\frac{x_2}{x_1} = \\dots = \\frac{x_k}{x_{k-1}} = \\frac{M}{x_k}$$", "If $k \\neq 0 \\pmod{p-1}$ then $$S_k(p-1) = 1^k + 2^k + 3^k + \\dots + (p-1)^k = 0 \\pmod{p}.$$ For example, • with $p=2011$, $k=2012$, we have $$1^{2012} + 2^{2012} + 3^{2012} + \\dots + 2010^{2012} = ~0 \\pmod{2011}$$ • with $p=2011$, $k=2010$, $$1^{2010} + 2^{2010} + 3^{2010} + \\dots + 2010^{2010} = ~ -1 \\pmod{2011}$$ Divisibility properties of $S_{-k}(p-1)$ Using Fermat's little theorem, we can also find modulo $p$ for the following summation $$S_{-k}(p-1) = \\frac{1}{1^k} + \\frac{1}{2^k} + \\frac{1}{3^k} + \\dots + \\frac{1}{(p-1)^k}.$$ First, let us write $$k = (p-1) q + r$$ where $0 \\leq r \\leq p-2$. For each $j=1,2, \\dots, p-1$, we have $$j^{k} = j^{r} = 1 \\pmod{p}$$ Thus, $$\\frac{1}{j^k} =_Q ~ \\frac{1}{j^r} =_Q ~j^{p-1-r} \\pmod{p}$$ It follows that $$S_{-k}(p-1) = \\frac{1}{1^k} + \\frac{1}{2^k} + \\frac{1}{3^k} + \\dots + \\frac{1}{(p-1)^k}$$ $$=_Q ~ 1^{p-1-r} + 2^{p-1-r} + 3^{p-1-r} + \\dots + (p-1)^{p-1-r} = S_{p-1-r}(p-1) \\pmod{p}$$ We obtain the following theorem Theorem. Suppose that $p$ is a prime number. • If $k = 0 \\pmod{p-1}$ then $$S_{-k}(p-1) = \\frac{1}{1^k} + \\frac{1}{2^k} + \\frac{1}{3^k} + \\dots + \\frac{1}{(p-1)^k} = -1 \\pmod{p}.$$ • If $k \\neq 0 \\pmod{p-1}$ then $$S_{-k}(p-1) = \\frac{1}{1^k} + \\frac{1}{2^k} + \\frac{1}{3^k} + \\dots + \\frac{1}{(p-1)^k} = 0 \\pmod{p}.$$ For example, • with $p=2011$, $k=2012$, we have $$\\frac{1}{1^{2012}} + \\frac{1}{2^{2012}} +", "$b+c-b'\\in C$. How many triples $\\langle b,c,b'\\rangle\\in B\\times C\\times B$ are there such that $b+c>b'$? It’s actually easier to count the triples with $b+c\\le b'$. I leave it to you to verify that there are \\begin{align*} \\sum_{i=1}^{N-1}i(N-i)&=N\\sum_{i=1}^{N-1}i-\\sum_{i=1}^{N-1}i^2\\\\ &=\\frac12N^2(N-1)-\\frac16N(N-1)(2N-1)\\\\ &=\\frac{N(N-1)}6\\big(3N-(2N-1)\\big)\\\\ &=\\frac16(N^3-N) \\end{align*} of these, and therefore $\\frac56N^3+\\frac16N$ triples with $b+c>b'$. This should qualify as comparable to $N^3$. Edit: For the second problem, suppose that $N=2n$ and try \\begin{align*} A&=\\{1,\\dots,n\\}\\cup\\{kn^2:k=1,\\dots,n\\},\\\\ B&=\\{1,\\dots,n\\}\\cup\\{kn^4:k=1,\\dots,n\\},\\text{ and}\\\\ C&=\\{kn^2:k=1,\\dots,n\\}\\cup\\{kn^8:k=1,\\dots,n\\}\\;. \\end{align*} Try to show that the overlapping parts of $A$ and $B$ and of $A$ and $C$ ensure that $|W(A,B)|$ and $|W(A,C)|$ are at least some fixed multiple of $N^3$, while $B$ and $C$ are related more like my $A$ and $B$ in the first problem. -", "edge with $P$.) If the area of $P$ is $N$ and the bounding box has dimensions $a \\times b$ b,$then clearly,$ab \\le ge N$. Also$E(N) \\ge 2a+2b.$This is because when we walk around the boundary of$P$, an edge at a time, we go up at least$a$times, down at least$a$times, and left and right at least$b$times each. We will assume that$a \\le b$since we can always rotate$P.$Depending on$N$there can may be many or few choices of$a,b$with$N \\le ab$and$2a+2b$minimal. 2a+2b=E(N)$. Consider again $N=1025=25 \\cdot 401$ 401.$If$ab \\ge N=1025$N=1025,$ then $2a+2b \\ge 130$ since the minimum over real values is $4\\sqrt{1025} \\gt 128.$ But we need an even integer. So any of $(a,b)=(32-j,33+j)=(32,33),(31,34),(30,35),(29,36),(28,37),(27,38)$ are possible. $27 \\cdot 38=1026$ is big enough but $26 \\cdot 39=1014$ would not be. In general, to have area at least $N$ in a box with $2a+2b =4k+2$, we have $(a,b)=(k-j,k+1+j)$ with area $N \\le k^2+k-j^2-j$ so $0 \\le j \\le \\frac{-1+\\sqrt{4(k^2-+k-N)+1}}{2}.$ frac{-1+\\sqrt{4(k^2+k-N)+1}}{2}.$The calculations for$(a,b)=(k-j,k+j)$when$E(n)=4k$are similar. Even though we could get area exactly$1025$with a$25 \\times 401$rectangle, the perimeter of$851$is much worse than$130.$Below is Now we have arrived at a blue polyomino$P$with area$1025$which fits in a$32 \\times 33$bounding box. The$8+10+2+1=21$black squares are in the bounding box but are not part of$P$. So For any$N$, the possible tiles polyominoes with area$N$and perimeter$E(n)$will be anything in an appropriate those obtained", "the $a_i$ must come in weakly decreasing order. If $a_1+\\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\\dots+a_n+1+\\dots+1=2008.$$ If $a_1+\\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider $$(a_1-2)+2+a_2\\dots+a_n=2008$$ (Clearly $a_1=1$ cannot be maximal). Once you have figured out $a_1$, apply the same argument to $a_2$, $a_3$, etc. (edit: at each step above, replace $j$ trailing copies of $1$ by the integer $j$; that is, instead of using $2006+1+1$, use $2006+2$. Doing this does not change the result). This shows that each $a_i$ must be $3$ or $2$. Simplifying the terms $A_1+\\dots+A_n$ is now just adding terms of a 2 geometric series. Finally, if there are $k$ $3s$ then there are $\\frac{2008-3k}{2}$ $2s$; thus we want to maximize $$\\frac{3(3^k-1)}{2}+\\frac{3(3^k-1)}{2}\\cdot2\\cdot\\left(2^\\frac{2008-3k}{2}-1\\right)$$ From Wolframalpha this gives $668$ copies of $3$ and $2$ copies of $2$ for an answer of roughly $3\\times 10^{319}$" ]

[ "# 2009 AIME II Problems/Problem 12 ## Problem From the set of integers $\\{1,2,3,\\dots,2009\\}$, choose $k$ pairs $\\{a_i,b_i\\}$ with $a_i so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$. Find the maximum possible value of $k$. ## Solution Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\\cdots+2k=k(2k+1)$. On the other hand, as the sum of each pair is distinct and at most equal to $2009$, the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \\cdots + (2009-(k-1)) = \\frac{k(4019-k)}{2}$. Hence we get a necessary condition on $k$: For a solution to exist, we must have $\\frac{k(4019-k)}{2} \\geq k(2k+1)$. As $k$ is positive, this simplifies to $\\frac{4019-k}{2} \\geq 2k+1$, whence $5k\\leq 4017$, and as $k$ is an integer, we have $k\\leq \\lfloor 4017/5\\rfloor = 803$. If we now find a solution with $k=803$, we can be sure that it is optimal. From the proof it is clear that we don't have much \"maneuvering space\", if we want to construct a solution with $k=803$. We can try to use the $2k$ smallest numbers: $1$ to $2\\cdot 803 = 1606$. When using these numbers, the average sum", "will be $1607$. Hence we can try looking for a nice systematic solution that achieves all sums between $1607-401=1206$ and $1607+401=2008$, inclusive. Such a solution indeed does exist, here is one: Partition the numbers $1$ to $1606$ into four sequences: • $A=1,3,5,7,\\dots,801,803$ • $B=1205,1204,1203,1202,\\dots,805,804$ • $C=2,4,6,8,\\dots,800,802$ • $D=1606,1605,1604,1603,\\dots,1207,1206$ Sequences $A$ and $B$ have $402$ elements each, and the sums of their corresponding elements are $1206,1207,1208,1209,\\dots,1606,1607$. Sequences $C$ and $D$ have $401$ elements each, and the sums of their corresponding elements are $1608,1609,1610,1611,\\dots,2007,2008$. Thus we have shown that there is a solution for $k=\\boxed{803}$ and that for larger $k$ no solution exists.", "integer. Then $P$ must be one of the points (2010,1), (2009,2),…, (1,2010). This yields the possible values of $\\theta$: $\\theta=\\tan^{-1}\\frac{2011-n}{n},\\ n=1,2,\\dots,2010.$ It remains to check whether any of these values must be excluded on the grounds that the line segment from (0,0) to $(2011-n,n)$ contains another integer point, say $(k,m)$ with $k+m<2011$. If that were so, we would have $\\frac{2011-n}{n}=\\frac{k}{m},\\ (2011-n)m=kn,\\ 2011m=n(k+m).$ In particular, $k+m$ divides $2011 m$, hence must have a nontrivial common divisor with 2011. But 2011 is prime, so that $k+m=2011$, a contradiction. Hence all 2010 values of $\\theta$ as above yield the desired outcome. Why it’s a dud: This is an example of a problem whose difficulty depends very strongly on knowing a certain “trick” or technique – in this case, the reflection principle. The committee aims to avoid this type of questions. It doesn’t help that the trick is an old one. Solution to 2. Rewrite the sum as $\\sum_{k=1}^{2008} c_k(a_{k+1}-a_k)$, where $c_k$ is the number of pairs $(i,j)$ such that $1\\leq i\\leq k. We count the number of such pairs: for each $k$ there are $k$ choices of $i$ and $2009-k$ choices of $j$, so that $c_k=k(2009-k)$. In particular, all $k$ are even. But then the above sum must also be even, in particular it cannot be 31415926535. Why it’s a dud:", "our pattern must \"line up\" with what we know about$a_{1007}$and$b_{1007}$, namely, that$2013-k$is the largest chosen number in the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having$k$elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that$2k | 2014$or$k | 1007$. Because$1007 = 19 \\times 53$, then we have four possible values of$k$. If$k=1$, then both sequences look like this:$0,2,4,\\dots, 2012$. Then$S=2012$. If$k=19$, then both sequences look like this:$0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then$S=1994$. If$k=53$, then both sequences look like this:$0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then$S=1960$. If$k=1007$, then both sequences look like this:$0,1,2,\\dots,1006$. Then$S=1006. Now, we have to address the case where 0 is not chosen in $a_i$ and $b_i$. We define new sequences $c_i = 2013-a_{1008-i}$ and $d_i = 2013-b_{1008-i}$. It's easy to see that $c_i,d_i$ satisfy the conditions of the problem, and since $0$ is not chosen in $a_i,b_i$ then 2013 must be chosen in them, which means $0$ is chosen in both $c_i,d_i$. This reduces $c_i,d_i$ to the case above, except $S$ is replaced with $4026-S$. Because the possible values for $S$ in the first case is $1007,1960, 1994, 2012$ then the possible values for $S$ in this case", "the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having $k$ elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that $2k | 2014$ or $k | 1007$. Because $1007 = 19 \\times 53$, then we have four possible values of $k$. If $k=1$, then both sequences look like this: $0,2,4,\\dots, 2012$. Then $S=2012$. If $k=19$, then both sequences look like this: $0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then $S=1994$. If $k=53$, then both sequences look like this: $0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then $S=1960$. If $k=1007$, then both sequences look like this: $0,1,2,\\dots,1006$. Then $S=1006. Now, we have to address the case where 0 is not chosen in$a_i$and$b_i$. We define new sequences$c_i = 2013-a_{1008-i}$and$d_i = 2013-b_{1008-i}$. It's easy to see that$c_i,d_i$satisfy the conditions of the problem, and since$0$is not chosen in$a_i,b_i$then 2013 must be chosen in them, which means$0$is chosen in both$c_i,d_i$. This reduces$c_i,d_i$to the case above, except$S$is replaced with$4026-S$. Because the possible values for$S$in the first case is$1007,1960, 1994, 2012$then the possible values for$S$in this case is$2014,2032,2134,3020\\$.", "end of the sequence, our pattern must \"line up\" with what we know about$a_{1007}$and$b_{1007}$, namely, that$2013-k$is the largest chosen number in the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having$k$elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above. From there, we know that$2k | 2014$or$k | 1007$. Because$1007 = 19 \\times 53$, then we have four possible values of$k$. If$k=1$, then both sequences look like this:$0,2,4,\\dots, 2012$. Then$S=2012$. If$k=19$, then both sequences look like this:$0,1,2,\\dots,17,18,38,39,\\dots,55,56,\\dots, 988$. Then$S=1994$. If$k=53$, then both sequences look like this:$0,1,2,\\dots,51,52,106,107,\\dots,158,159,\\dots, 988$. Then$S=1960$. If$k=1007$, then both sequences look like this:$0,1,2,\\dots,1006$. Then$S=1006. Now, we have to address the case where 0 is not chosen in $a_i$ and $b_i$. We define new sequences $c_i = 2013-a_{1008-i}$ and $d_i = 2013-b_{1008-i}$. It's easy to see that $c_i,d_i$ satisfy the conditions of the problem, and since $0$ is not chosen in $a_i,b_i$ then 2013 must be chosen in them, which means $0$ is chosen in both $c_i,d_i$. This reduces $c_i,d_i$ to the case above, except $S$ is replaced with $4026-S$. Because the possible values for $S$ in the first case is $1007,1960, 1994, 2012$ then the possible values for", "# How to compute the cardinality of these two sets? $A$ and $B$ are two sets of consecutive integers, $|A|=m \\text{ and } |B|=2m$ and sum of the elements of $A$ and $B$ are $2m$ and $m$ respectively. If the difference between the largest numbers of $A$ and $B$ is $99$, how could we find the value of $m$ ? - It'll be nice if you include what you tried and where you are stuck. – Srivatsan Nov 19 '11 at 3:06 Let $A=\\{k_a, k_a +1,\\dots,k_a+m-1\\}$ and $B=\\{k_b, k_b +1,\\dots,k_b+2m-1\\}$. Then we have: $$S_a = m k_a + \\frac{(m-1) m}{2}$$ $$S_b = 2m k_b + \\frac{(2m-1) 2m}{2}$$ We are given that $S_a = 2m$ and $S_b = m$. Thus, $$m k_a + \\frac{(m-1) m}{2} = 2m$$ $$2m k_b + \\frac{(2m-1) 2m}{2} = m$$ We also know that: $$(k_b+2m-1) -(k_a+m-1) = 99$$ or $$(k_a+m-1) - (k_b+2m-1) = 99$$ Does the above help? -", "$$\\text{blue dots}:\\Big\\{\\frac{1}{1111},\\frac{2}{1111},\\frac{3}{1111},\\dots,\\frac{1110}{1111}\\Big\\}$$ The distance between any two red and blue dots is $$|\\frac{m}{1000}-\\frac{n}{1111}|$$ for integers $$m$$ and $$n$$. The question becomes, what is the minimum value of $$|\\frac{m}{1000}-\\frac{n}{1111}|$$ for positive integers $$m\\leq 999$$ and $$n\\leq 1110$$? $$\\Big|\\frac{m}{1000}-\\frac{n}{1111}\\Big|=\\frac{|1111m-1000n|}{1000\\cdot1111}$$ To minimize this fraction is to minimize its numerator. And it is known that the $$\\gcd$$ of two integers is their smallest positive integral linear combination (Bézout's identity). That is, the minimum value of the numerator is $$\\gcd(1111,1000)$$, which is $$1$$ since they are relatively prime. Thus the minimal distance is $$1/(1000\\cdot1111)$$. For example, take $$m=9$$ and $$n=10$$, the points $$9/1000$$ and $$10/1111$$ are of the proper distance apart. • This is a very elegant solution, clear and well written. I was on a completely wrong track. Could you provide an idea or hint how to figure out the number of red and blue dot pairs?Thanks so much. – jeremiah Nov 15 '18 at 23:30 • I think (without proof) that the number of such pairs is the number of whichever dots there are less of. At this point it's important that $1000$ and $1111$ are relatively prime so that the dots never overlap in the interval (only at the endpoints). Therefore you can guarantee that at least one blue dot lies strictly between every consecutive pair of red dots, so", "in at most $$n-k$$ $$\\Delta$$-pairs, totally $$i$$ participates in at most $$n-k$$ pairs (since it can not participate in both types of pairs). Therefore the total number of pairs does not exceed $$n(n-k)\\leqslant 2k(n-k)$$. 2. $$s=t=1$$. We get $$j\\geqslant k+1$$ for $$D$$-pairs and $$i\\geqslant k+1$$ for $$\\Delta$$-pairs $$(i,j)$$. Denote $$a=\\max(k+1,j_0)$$, $$b=\\max(k+1,i_0)$$. Then the total number $$D$$-pairs is at most $$(n-a+1)(b-1)$$, the total number $$\\Delta$$-pairs is at most $$(n-b+1)(a-1)$$. When $$a,b\\in [k+1,n+1]$$, the sum $$S:=(n-a+1)(b-1)+(n-b+1)(a-1)$$ which is bilinear in $$a$$ and $$b$$ is maximized at one of four endpoints. When $$a=b=k+1$$ we get $$S=2k(n-k)$$, if, say, $$a=n+1$$, we get $$S=n(n-b+1)\\leqslant n(n-k)\\leqslant 2k(n-k)$$. • This is very nice. Thank You! May 2, 2021 at 18:36 I confirmed via integer linear programming that the maximum value is $$2k(n-k)$$ for $$n \\le 20$$ and $$k>n/2$$. The formulation is the same as in my answer to the linked question, except for the following changes: • The node set is $$N = \\{1,\\dots,2n\\}$$. • The candidate edge set is $$E = \\{i \\in N, j \\in N: i \\le n < j\\}$$. • Constraints $$(3)$$ and $$(4)$$ become \\begin{align} k - (d_i - d_j) &\\le (k + n) (1 - u_{i,j}) &&\\text{for (i,j)\\in E} \\tag3\\\\ k - (d_j - d_i) &\\le (k + n) (1 - v_{i,j}) &&\\text{for (i,j)\\in E} \\tag4\\\\ \\end{align}", "set , which makes $P(k) \\ge \\frac{1}{2} + \\frac{1}{2k}$. Now the only case without rounding, $k = 1$. It must be true. ### Solution 2 It suffices to consider $0\\le n < k-1.$ Now for each of these $n,$ let $f(n)=[\\frac{n}{k}], g(n)=[\\frac{100}{k}]-[\\frac{100-n}{k}].$ If we let $k=67,$ then the following graphs result for $f$ and $g.$ $f:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { if (i<=33) dot((i,0)); else dot((i,1)); } label(\"(0,0)\",(0,0),SW); label(\"(66,1)\",(66,1),NE); [/asy]$ $g:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { dot((i,0)); } label(\"(0,0)\",(0,0),NW); label(\"(66,0)\",(66,0),SE); [/asy]$ Our probability is the number of $0\\le i such that $f(i)+g(i)=0$ over $k.$ Of course, this always holds for $i=0.$ If we let $k$ vary, then the graph of $f$ is always very similar to what it looks like above (groups of $\\frac{k+1}{2},\\frac{k-1}{2}$ dots). However, the graph of $g$ can vary greatly. In the above diagram, $g(i)=0$ for all $i,$ while it is possible for $g(i)=-1$ for all $i\\neq 0.$ In order to minimize the number of $i$ which satisfy $f(i)+g(i)=0,$ we either want $g(i)=0$ for $0\\le i or $g(i)=-1$ for $1\\le i This way, we see that at least half of the numbers from $1$ to $k-1$ satisfy the given equation. So, our desired probability is at least $\\frac{k+1}{2k}.$ As shown by", "$b+c-b'\\in C$. How many triples $\\langle b,c,b'\\rangle\\in B\\times C\\times B$ are there such that $b+c>b'$? It’s actually easier to count the triples with $b+c\\le b'$. I leave it to you to verify that there are \\begin{align*} \\sum_{i=1}^{N-1}i(N-i)&=N\\sum_{i=1}^{N-1}i-\\sum_{i=1}^{N-1}i^2\\\\ &=\\frac12N^2(N-1)-\\frac16N(N-1)(2N-1)\\\\ &=\\frac{N(N-1)}6\\big(3N-(2N-1)\\big)\\\\ &=\\frac16(N^3-N) \\end{align*} of these, and therefore $\\frac56N^3+\\frac16N$ triples with $b+c>b'$. This should qualify as comparable to $N^3$. Edit: For the second problem, suppose that $N=2n$ and try \\begin{align*} A&=\\{1,\\dots,n\\}\\cup\\{kn^2:k=1,\\dots,n\\},\\\\ B&=\\{1,\\dots,n\\}\\cup\\{kn^4:k=1,\\dots,n\\},\\text{ and}\\\\ C&=\\{kn^2:k=1,\\dots,n\\}\\cup\\{kn^8:k=1,\\dots,n\\}\\;. \\end{align*} Try to show that the overlapping parts of $A$ and $B$ and of $A$ and $C$ ensure that $|W(A,B)|$ and $|W(A,C)|$ are at least some fixed multiple of $N^3$, while $B$ and $C$ are related more like my $A$ and $B$ in the first problem. -", "than $\\sqrt{n}, the other will be smaller. - For your second question: Every positive integer$n$can be written in the form$2^k(n)m(n)$, where$k(n)\\ge 0$and$m(n)$is odd, and when you’ve done that,$m(n)$is the greatest odd divisor of$n$. In particular, when$k(n)=0$,$n$is odd, and$n$itself is its own greatest odd divisor; when$k(n)=1$,$m(n) = n/2$is odd and is the greatest odd divisor of$n$; and so on. Note that if$k(n)=i$, then$k(n+2^{i+1})=i$as well: $$n+2^{i+1} = 2^im(n)+2^{i+1} = 2^i(m(n)+2)\\;,$$ and$m(n)+2$is odd, so$k(n+2^{i+1})=i$and$m(n+2^{i+1})=m(n)+2$. Use this idea to split$S$into subsets$S_i$for$i\\ge 0$: let $$S_i = \\{n\\in S:k(n)=i\\}.$$ Then$S_0 = \\{2007,2009,\\dots,4011\\}$, the set of odd members of$S$;$S_1 = \\{2006,2010, \\dots,4010\\}$, the set of even members of$S$that are not multiples of$4$;$S_2 = \\{2012,2020,\\dots,4012\\}$, the set of multiples of$4$in$S$that are not multiples of$8$; and so on. Each$S_i$forms an arithmetic progression with constant difference$2^{i+1}$. If$n \\in S_0$, then$n$is odd, and therefore$n$itself will be one of the terms in the sum$K$. If$n\\in S_1$, then$n/2$is odd and will be one of the terms in the sum$K$. And so on: for each$i\\ge 0$, if$n\\in S_i$, then$n/2^i$is odd and will be one of the terms in the sum$K$. This means that if$s_i$is the sum of the members of$S_i$, then the members of$S_i$contribute a total of$s_i/2^i$to$K$. Thus, in order to find$K$you need only find the sums$s_i$. If$a_i$is the smallest member of$S_i$and$b_i$is the largest, $$|S_i| = \\frac{b_i-a_i}{2^{i+1}}+1;$$ this follows easily from the", "makes it slightly more likely that other pairs match too). In particular, we have $$E(Z_{p_1} Z_{p_2} \\dots Z_{p_m}) \\leq E(X_{p_1} X_{p_2} \\dots X_{p_m}) \\leq E(Z_{p_1} Z_{p_2} \\dots Z_{p_m})\\left(\\frac{4n^2-1}{4n^2-2m}\\right)^m$$ A sketchier argument for Claim $$2$$: Let $$r$$ be as in claim $$1$$. If we consider all $$m$$-tuples containing $$r$$ distinct pairs, then only an $$O(\\frac{1}{2n^2c_d})$$ of them have any overlap between the $$r$$ pairs. (the denominator here coming from how there's $$2n^2c_d$$ close pairs to choose from. Again the constant in the $$O$$ notation can depend on $$m$$ or $$d$$). All tuples with $$r$$ pairs have the same contribution to $$E(Z)$$, so the contribution from pairs with overlap is only $$O(\\frac{1}{n^2})$$ the contribution from all pairs. The claim follows since $$E(Z)=O(1)$$. I wrote a program that repeatedly generates a random grid and counts the number of adjacent pairs, and finally outputs the distribution of this number. Here are some results: 10*10 grid, 50 pairs, 1000000 games 0 1 2 3 4 5 6 7 8 9 10 11 12 158437 295726 272342 164827 73472 25585 7400 1784 350 67 8 2 0 100*100 grid, 5000 pairs, 10000 games 0 1 2 3 4 5 6 7 8 9 10 1361 2794 2729 1725 891 335 117 37 9 2 0 1000*1000 grid, 500000 pairs, 100 games 0 1 2", "same set, and we have $a_n$ ways to finish, or they are in different sets and then there are $(2n+2)(2n+1)a_{n-1}$ ways to finish. Now by induction it is easy to see that the answer is $(2n-1)!!=(2n-1)(2n-3)\\dots ... 1 Induction on$n$. For$n=1,2$there is nothing to prove. Assume the result is proved for$<n$and consider the case$n$. We establish the case$n$by induction on$m$. If$n$is odd the smallest possible value of$m$is$m=n$. We have$n=(n-1)+1=(n-2)+2=\\dots=\\frac{1}{2}(n+1)+\\frac{1}{2}(n-1)$, so the result is true for$m=n,k=\\frac{1}{2}(n+1)$. If$n\\$... 1 Very useful paper, though this only gives a good approximation to the number of partitions of n into exactly k parts each no larger than N due to Ratsaby (App. Analysis and Discrete M. 2008): http://www.doiserbia.nb.rs/img/doi/1452-8630/2008/1452-86300802222R.pdf Only top voted, non community-wiki answers of a minimum length are eligible", "$\\displaystyle \\frac11;\\frac12;\\frac13,\\frac23;\\frac14,\\frac34;\\frac15,\\frac25,\\frac35,\\frac45;\\dots$ Cantor’s first proof of the uncountability of the reals (the nested intervals argument) from 1874 proceeds as follows: Given any (injective) sequence $a_1,a_2,\\dots$ of reals, we want to exhibit a real that was not listed. There are two cases: Either there are $i\\ne j$ with $a_i, and such that there is no $k$ with $a_k\\in(a_i,a_j)$, in which case we are obviously done, or (more interestingly), whenever $a_i, we can find an $a_k$ strictly in between (the range of the sequence is dense in itself). Assume we are in this situation. Define two sequences $b_1,b_2,\\dots$ and $c_1,c_2,\\dots$ as follows: • First, $b_1=a_1$ and $c_1=a_2$. • For definiteness, suppose that $a_1>a_2$. The other case is treated similarly. Let $b_2$ be $a_i$, where $i$ is least such that $a_i\\in(c_1,b_1)$. Then define $c_2$ as $a_j$, where $j$ is least such that $a_j\\in(c_1,b_2)$. • In general, given $c_n, we define $b_{n+1}$ as $a_k$, where $k$ is least such that $a_k\\in(c_n,b_n)$, and then define $c_{n+1}$ as $a_m$, where $m$ is least such that $a_m\\in(c_n,b_{n+1})$. Note that these sequences are well defined, because of our density assumption. The construction just described ensures that if $I_n=[c_n,b_n]$, then: 1. For any $n$, we have that $a_n\\notin I_{n+1}$, and 2. The intervals are nested and decreasing: $I_1\\supsetneq I_2\\supsetneq I_3\\dots$ It follows (from the completeness of the", "the $a_i$ must come in weakly decreasing order. If $a_1+\\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\\dots+a_n+1+\\dots+1=2008.$$ If $a_1+\\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider $$(a_1-2)+2+a_2\\dots+a_n=2008$$ (Clearly $a_1=1$ cannot be maximal). Once you have figured out $a_1$, apply the same argument to $a_2$, $a_3$, etc. (edit: at each step above, replace $j$ trailing copies of $1$ by the integer $j$; that is, instead of using $2006+1+1$, use $2006+2$. Doing this does not change the result). This shows that each $a_i$ must be $3$ or $2$. Simplifying the terms $A_1+\\dots+A_n$ is now just adding terms of a 2 geometric series. Finally, if there are $k$ $3s$ then there are $\\frac{2008-3k}{2}$ $2s$; thus we want to maximize $$\\frac{3(3^k-1)}{2}+\\frac{3(3^k-1)}{2}\\cdot2\\cdot\\left(2^\\frac{2008-3k}{2}-1\\right)$$ From Wolframalpha this gives $668$ copies of $3$ and $2$ copies of $2$ for an answer of roughly $3\\times 10^{319}$", "k_2, \\ldots, k_m} = \\frac{k!}{k_1! k_2! \\dots k_m!}$ where $k_1 + k_2 + \\cdots + k_m = k$. This gives the number of ordered arrangements of $k$ objects in which there are $k_1$ objects of type $1$, $k_2$ objects of type $2$, $\\ldots$, $k_m$ objects of type $m$. In our case we have $k = 4$ and $m = 3$. Thus, the coefficient of $\\partial_{1122} v_1^2 v_2^2$ for example is $6$, because there are two $1$s, two $2$s and zero $3$s, and setting $k_1 = 2$, $k_2 = 2$, $k_3 = 0$ we get $\\binom{k}{k_1, k_2, k_3} = \\frac{k!}{k_1! k_2! k_3!} = \\frac{4!}{2! 2! 0!} = \\frac{24}{4} = 6$ Similarly, in the case of $\\partial_{1123} v_1^2 v_2 v_3$ the coefficient is $12$ because $k_1 = 2$, $k_2 = 1$, $k_3 = 1$, so $\\binom{k}{k_1, k_2, k_3} = \\frac{4!}{2! 1! 1!} = \\frac{24}{2} = 12$ Note also that within the square brackets in the reduced expression above there are $15$ different types’ of terms. Why $15$? The answer has to do with the concept of partitions from number theory. A partition of a positive integer is a way of writing that integer as a sum of positive integers. Two sums that differ only in the order of the integers are considered to be the same partition. Now, the number", "$k = \\frac{2x-27}{x^2} \\Longrightarrow kx^2 - 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \\ge 0 \\Longleftrightarrow k \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. Solution 1.1 Proceed as follows for Solution 1. Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$. Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$. Solving, we obtain $x=27$ as the critical value. Hence, $k$ has the maximum value of $(2*27-27)/27^2=1/27$. Since $BP^2=405+729k$, the maximum value of $\\overline {BP}$ occurs at $k=1/27$, so $BP^2$ has a maximum value of $405+729/27=\\fbox{432}$. Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$O$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = OT + BO \\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic", "it would be better to convert them to 7 lots of 13.) All of these are different because the multiples of 7 will have different residues mod 13. This should generalize for general k to give (k+1)+k+....(k+2-a) possible values - 5 years, 4 months ago That solutions seems correct, I solved it differently: for every $$0 \\leq b \\leq 100$$, there are $$13$$ possible values for $$a$$, but I need to subtract $$12+11+\\dots + 1$$ because for $$b=100$$ I counted $$13$$ values for $$a$$ while only $$a=0$$ is valid due to the fact that $$a+b \\leq 100$$. Similarly, for $$b=99$$, $$13$$ values for $$a$$ were counted while only 2 are valid, etcetera. So, the answer is $$13 \\cdot 101 - (12 + 11 + \\dots + 1) = 1235$$. Generalization: denote $$\\max(m,n)$$ by $$q$$. The total values $$m \\cdot a + n \\cdot b$$ can take if $$a+b \\leq k$$ then is \\begin{align*} q \\cdot (k+1) - \\left( (q-1) + (q-2) + \\dots + 1 \\right) &= q \\cdot (k+1) - \\frac{q(q-1)}{2}\\\\ &= q \\left( (k+1) - \\frac{q-1}{2} \\right)\\\\ &= \\frac{q(-q+2k+3)}{2}. \\end{align*} And indeed: \\begin{align*} (k+1) + k + \\dots + (k+2-a) &= \\frac{(k+1)(k+2)}{2} - \\frac{(k+2-a)(k+1-a)}{2}\\\\ &= \\frac{(k^2+3k+2) - (k^2+3k+2-2qk-3q+q^2)}{2}\\\\ &= \\frac{-q^2+2qk+3q}{2}\\\\ &= \\frac{q(-q+2k+3)}{2}. \\end{align*} - 5 years, 4 months ago That looks good- my a is your", "we deduce that $$M = \\frac{2016}{1001} = \\frac{288}{143}.$$ The answer is $\\boxed{431}.$ ### Solution 2 Each 1000-element subset $\\left\\{ a_1, a_2,a_3,...,a_{1000}\\right\\}$ of $\\left\\{1,2,3,...,2015\\right\\}$ with $a_1 contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\\left\\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$. There are $a_1$ ways to choose a positive integer $k$ such that $k ($k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is equal to the sum. But choosing a set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is the same as choosing a 1001-element subset from $\\left\\{1,2,3,...,2016\\right\\}$! Thus, the average is $\\frac{\\binom{2016}{1001}}{\\binom{2015}{1000}}=\\frac{2016}{1001}=\\frac{288}{143}$. Our answer is $p+q=288+143=\\boxed{431}$. ### Solution 3(NOT RIGOROUS) Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$: For $p = 2$ and $q = 2$, the answer is $1$. For $p = 3$ and $q = 2$, the answer is $\\frac43$. For $p = 4$ and $q = 2$, the answer is $\\frac53$. For $p = 3$ and $q = 3$, the answer is $1$. For $p = 4$ and $q = 3$, the answer is $\\frac54$. For $p = 5$ and $q = 3$, the answer" ]

[ "+ t^2 + 2t + 1)(t+1),$$ and the first factor $t^4 + 2t^3 + t^2 + 2t + 1$ has no root and no quadratic factor, since $x^2+1,$ $x^2+x+2$ and $x^2+2x+2$ are the unique irreducible polynomials of degree $2$, and it is obvious that they are not dividing it. • Yes, you are right:) I suppose there was a similar question on $x^5-x-1$ not too long ago... – Dietrich Burde Jan 9 '18 at 14:02", "the irreducible quadratic: $\\,x^2 - 2ax + (a^2 + b^2)$ • Thus: EVERY POLYNOMIAL WITH REAL NUMBER COEFFICIENTS CAN BE FACTORED INTO LINEAR FACTORS AND IRREDUCIBLE QUADRATICS Master the ideas from this section", "Dividing, in this case, by $x-2$ yields the quadratic factor $(x^2 + 2x + 1)$.", "### Home > CCA2 > Chapter 9 > Lesson 9.1.1 > Problem9-14 9-14. The roots of a polynomial are given below. Find two linear factors and the corresponding quadratic factor. 1. $x = -2 +\\sqrt { 3 }$, $x = -2 −\\sqrt { 3 }$ Find the first factor. $(x - (-2 + \\sqrt{3}))$ Find the second factor. Multiply together to get the quadratic factor. 1. $x = -2 + i$, $x = -2 - i$ Follow similar steps in part (a).", "is an integer divisor of an and q is an integer divisor of a0.[8] If we wished to factorize the polynomial $2x^3 - 7x^2 + 10x - 6$ we could look for rational roots p/q where p divides -6, q divides 2 and p and q have no common factor greater than 1. By inspection we see that this polynomial can have no negative roots. Assume that q = 2 (otherwise we would be looking for integer roots), substitute x = p/2 and set the polynomial equal to 0. By dividing by 4, we obtain the polynomial equation $p^3 - 7p^2 + 20p -24 = 0$ that will have an integer solution of 1 or 3 if the original polynomial had a rational root of the type we seek. Since 3 is a solution of this equation (and 1 is not), the original polynomial had the rational root 3/2 and the corresponding factor (2x - 3). By polynomial long division we have the factorization $2x^3 - 7x^2 + 10x - 6 = (2x -3)(x^2 -2x + 2).$ For a quadratic polynomial with integer coefficients having rational roots, the above considerations lead to a factorization technique known as the ac method of factorization.[9] Suppose that the quadratic polynomial with integer coefficients is: $ax^2 + bx + c$ and it", "× # Easy but interesting! Factorize:- x⁴-75x+100 Note by Naitik Sanghavi 1 year, 10 months ago Sort by: By Rational Root Theorem, there's no rational roots So if it can be factored, then it must be the product of two irreducible quadratic polynomials. $(x^2+ax+b)(x^2+cx + d)$ Because the sum of roots of this expression equals to 0, this motivates us to have $$a = - c$$. And because the product of roots is 100, then $$bd = 100$$. So we have $(x^2+ ax + b)(x^2 - ax + 100/b)$ Expand and compare the coefficients gives $$(a,b) = (-5,5), (5,20 )$$. · 1 year, 10 months ago", "that 2+ i is also a root and so a quadratic factor is $(x- 2+ i)(x- 2- i)= ((x- 2)+ i)((x- 2)- i)= (x- 2) ^2- i^2= x^2- 4x+ 4- 1= x^2- 4x+ 3$ and divide by that to find the third linear factor. Of course, you cannot us \"synthetic\" division to do that. 7. Apr 17, 2012 ### Curious3141 That should read ... = (x-2)2 - i2 = x2 - 4x + 4 + 1 = x2 - 4x + 5.", "# Difference between revisions of \"2010 AIME II Problems/Problem 6\" ## Problem Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. ## Solution 1 You can factor the polynomial into two quadratic factors or a linear and a cubic factor. For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that $$(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.$$ Therefore, again setting coefficients equal, $a + c = 0\\Longrightarrow a=-c$, $b + d + ac = 0\\Longrightarrow b+d=a^2$ , $ad + bc = - n$, and so $bd = 63$. Since $b+d=a^2$, the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$. From this we find that the possible values for $n$ are $\\pm 8 \\cdot 62$ and $\\pm 4 \\cdot 2$. For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$. Therefore, the answer is $4 \\cdot 2 = \\boxed{008}$. ## Solution 2 Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$. From this, we get that $bd=63\\implies d=\\frac{63}{b}$ and $a+c=0\\implies", "## Thinking Mathematically (6th Edition) $x^{2}$ -9x +81 step 1. Enter x as the first term of each factor $x^{2}$ -9x +81= (x__)(x _) Step 2. To find the second term of each factor, we must find two integers whose product is 81 and whose sum is -9 List pairs of factors of the constant, 81 (1,81)(-1,-18)(3,27)(-3,-27))(9,9)(-9,-9) step 3. The correct factorization of $x^{2}$ -9x +81 is the one in which the sum of the Outside and Inside products is equal to -9x. Since there are no factors that can sum up to -9, this polynomial is prime.", "this situation, I would definitely switch to the quadratic formula as a method to factor these quadratic equations. Another method, however, assuming your polynomial is in the form of ax^k+bx+c, to factor the polynomial, you would: 1. Multiply the coefficients a and c. (If your polynomial is 8x^2+10x+3, that would make 24.) 2. Find a factor of ac that add together to make b. If you can find this, then the polynomial is factorable. (24=6x4; 6+4=10) 3. Break apart b into its components you just found. (8x^2+6x+4x+3) 4. Factor by grouping. (2x[4x+3]+[4x+3]; [2x+1][4x+3])", "# Foiled once again Algebra Level 1 The polynomial in two variables $f(x,y) = xy + 3y - 4x -12$ can be expressed as the product of two linear factors with integer coefficients, namely $f(x,y) = (ax + by + c) ( dx + ey + f).$ Evaluate $|a+b+c| + |d+e+f|$. ×", "Q # Determine which of the following polynomials has (x + 1) a factor : (i) x ^3 + x ^2 + x + 1 1.(i) Determine which of the following polynomials has $(x + 1)$ a factor : (i) $x^3 + x^2 +x + 1$ Views Zero of polynomial $(x + 1)$ is -1. If $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$ Then, $p(-1)$ must be equal to zero Now, $\\Rightarrow p(-1)=(-1)^3+(-1)^2-1+1$ $\\Rightarrow p(-1)=-1+1-1+1 = 0$ Therefore, $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$ Exams Articles Questions", "and negative factors of the last digit and divide them by the same of the first digit. Then test them $\\pm2$ and $\\pm1$. We get 1 and $-2=x$. Then divide by their respective polynomials to see which one has order 2. Solved. -", "# Factoring Quadratic Polynomials into Linear Factors If x_1 and x_2 are the roots of quadratic polynomial of ax^2+bx=c (i.e.the root of equation of ax^2+bx+c=0), then color(blue)(ax^2+bx+c=a(x-x_1)(x-x_2)).This is the formula of factoring quadratic polynomial into factors. Example. Factor the following: 6x^2-x-2. We apply the formula of roots of quadratic equation to the equation of 6x^2-x-2=0 and find x_1=-1/2, x_2=2/3. So, 6x^2-x-2=6(x+1/2)(x-2/3)=2(x+1/2)*3(x-2/3)=(2x+1)(3x-2).", "Expii # Factoring Quadratic ax²+bx+c with ac>0 - Expii The signs in ax²+bx+c determine the signs of the factors. If a and c have the same sign, the factors will have the same signs.", "basis from which to make an educated guess for the quadratic's constant terms. Next, you will need two coefficients of $$x$$ for your quadratics that will sum to the coefficient of $$10x^3$$ in your original polynomial. Expanding the ansatz approach will give you the exact conditions required.", "quadratic ; group the two linear factors together and you will get a real quadratic. Similarly for the roots corresponding to $3$ and $5$. Hope that helps, In www.wolframalpha.com , I typed factor x^4 + 1 It gives 4 first order terms. Hope this helps. Regards, Matt", "2, -2, 4, -4, 8, -8, 16, or -16. Put each of those into $-x^3+ 5x^2-14x+ 16$ to see if they make it 0. If none of them do then there are no rational roots and this cannot be factored with integer coefficients. If one does work, divide by (x- that root) to find the other factor.", "I'm just asking for fun, Factor the Polynomial: 8pq^2+8pq+2p So what is the answer? Apr 16, 2018 $2 p {\\left(2 q + 1\\right)}^{2}$ Explanation: $8 p {q}^{2} + 8 p q + 2 p$ $2 p \\left(4 {q}^{2} + 4 q + 1\\right)$ $2 p {\\left(2 q + 1\\right)}^{2}$", "OML-IZ Search ## Finding Discriminants of Quadratic Polynomials Find the discriminants of the following quadratic polynomials. You may use the TAB key to move to the next question. When you are done, click Submit. 1. $$-x^2 - 3x + 6$$ : 2. $$3x^2 + x + 1$$ : 3. $$-x^2 + 4x + 4$$ : 4. $$7x^2 + x + 3$$ : 5. $$-x^2 + 2x - 2$$ : 6. $$-9x^2 + 2x - 1$$ : 7. $$-3x^2 - 4x - 9$$ : 8. $$4x^2 - 4x + 4$$ : 9. $$-3x^2 - 8x - 1$$ : 10. $$5x^2 + 4x + 9$$ : We hope that the free math worksheets have been helpful. We encourage parents and teachers to adjust the worksheets according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles." ]

[ "of equality. 9m = 36 A - 1/4 B - 27 C - 45 D - 4 ### Explanation Step 1: 9m = 36 Using the multiplicative property of equality, multiply both sides of equation by ($\\frac{1}{9}$) Step 2: 9m × ($\\frac{1}{9}$) = 36 × ($\\frac{1}{9}$) So, m = 4 multiplicative_property_of_equality_with_whole_numbers.htm", "while retaining the equality. That is $a-c=2+c-c$ Since $c-c=0$, this simplifies to $a-c=2+0$ This further simplifies to: $a-c=2$ Thus, $a-c$ is also equal to $2$, as required. QED. Images/mathematical drawings are created with GeoGebra. ### Practice Questions 1. Let $w, x, y,$ and $z$ be real numbers such that $w=x$. Which of the following are equivalent? A. $w-x$ and $0$ B. $w-y$ and $x-y$ C. $w-z$ and $x-y$ 2. True or False: Two boxes of books have the same weight. A one-half pound book is taken from each box. Therefore, the subtraction property of equality states that the boxes will still be the same weight. 3. Use the subtraction property of equality to find the value of $x$ in $x + 6 = 12$. 4. Use the subtraction property of equality to find the value of $y$ in $y + 2 = 24$. 5. Let $x+8=15$ and $y+3=10$. Using the subtraction property of equality and the transitive property of equality, which of the following is true?", "(x)dx = xf^{-1}(x)-g\\circ f^{-1}(x)$$ You can check the equality by differentiating both sides.", "# Prove that $5x^2−2xy−8x+ 2y^2−2y+ 5 \\ge 0$ for all $x, y\\in\\mathbb R$. When does equality occur? I tried grouping the $$x$$'s and the $$y$$'s but that didn't get me anywhere. I know that $$5x^2, 2y^2$$, and $$5$$ are always positive. I am not sure what to try next. • @HaiDangel29 One of the possible solutions involving discriminant doesn't mean the question needs to be tagged (discriminant)... – YuiTo Cheng May 25 at 9:18 By inspection, equality occurs for $$x=y=1$$. So the obvious choice is to group around these values. This also gives a somewhat more detailed explanation how you arrive at these magic squares: $$5x^2−2xy−8x+ 2y^2−2y+ 5 = \\\\ 5((x-1) +1)^2−2((x-1)+1)((y-1) + 1)−8(x-1)+ 2((y-1)+1)^2−2(y-1) -5\\\\ = 5(x-1)^2 - 2(x-1)(y-1) +2(y-1)^2 =\\\\ = 4(x-1)^2 + ((x-1)-(y-1))^2 +(y-1)^2 = \\\\ = 4(x-1)^2 + (x-y)^2 +(y-1)^2 \\\\ \\geq 0$$ which is clear. take the expression $$5x^{2}-2xy-8x+2y^{2}-2y+5$$, we want to write this as a sum of squares somehow. grouping the terms as: $$(x^{2}-2xy+y^{2}) + (y^{2}-2y+1) + (4x^{2}-8x+4)$$ factoring the three expressions gives $$(x-y)^{2} + (y-1)^{2} + 4(x-1)^{2}$$ which is the sum of squares, each of which is greater than $$0$$. for equality to hold, notice that both $$(y-1)^{2}$$ and $$4(x-1)^{2}$$ have roots of 1 which means they will be zero at that value, and also notice that $$(x-y)^{2}$$ will be zero", "issue labels: $$x+4=10\\\\\\sqrt{x^2}:=|x|\\\\f(x)=2x+4\\\\y=2x+C.$$ Another way to classify equations. I think there might be some kind of language barrier because of the similar but fundamentally different words in OP's question: If we know two things are the same i.e. equal why do we need to state that they're the same? What he was trying to say was, why do we need to say a thing is itself? The symbol $$=$$ is used in mathematics to express identity (two things actually being just one thing) as well as equality (two things really are distinct but can be considered equal in at least some aspect) as the following two examples might show: 1. Pythagorean Theorem: $$c^2 = a^2 + b^2$$ states that the square over the hypothenuse is equal (in size) to the sum of squares over the other sides. But of course those squares are not identical (\"the same thing\"). But their equality is still an important fact to state. 2. $$\\{ x \\in \\mathbb N ~|~ x \\text{ ends in 0 or 5} \\} = \\{x \\in \\mathbb N ~|~ x \\text{ is divisible by 5} \\}$$ states that both sets are actually the same thing, the set of the numbers 0, 5, 10, ... Again, this identity is state-worthy. For pure numbers, the distinction between equality and identity", "X_2$ or equivalently $Y_1 Y_2 = - a X_1 X_2.$ So, to check equality modulo $$\\mathcal E[4]$$, it's sufficient to check whether $X_1 Y_2 = Y_1 X_2 \\quad \\mathrm{or} \\quad Y_1 Y_2 = -a X_1 X_2.$", "equality gives us $(x+1)^2=4y+3\\equiv 3 \\pmod {4}$, so there are no solutions.", "the case here. $$18x=9$$ Addition Property of Equality. This step added 4 to both sides, which is allowed based on the addition property. $$x=\\frac{1}{2}$$ Division Property of Equality. This step divided 18 from both sides. This is a valid operation because of the division property of equality.", "= 1/4 I get that 2x = 1/2x, and 2y = 2y. Edit: I get that now, thank you very much for your help – Sonofgreek Apr 6 '19 at 10:57 • It seems that you forgot the third equation. Suppose that $\\lambda=1$. Then you get the trivial equality $2x=2x$. You also get $2y=8y$, from which you deduce that $y=0$. And then you deduce from $x^2+4y^2=4$ (and from $y=0$) that $x=\\pm2$. – José Carlos Santos Apr 6 '19 at 11:01 • I did. Thank you very much to both you and Tony K – Sonofgreek Apr 6 '19 at 11:03", "$y-2w+2z+2w-2z=7z+2w-2z$ This simplifies to: $y=5z+2w$ Next, use the addition and subtraction properties of equality and simplification to solve for $x$. $-4x+4w-3z=2z+6w-5x$ First, use the addition property of equality to add 5x to both sides. $-4x+5x+4w-3z=2z+6w-5x+5x$ This simplifies to: $x+4w-3z=2z+6w$ Then, subtract 4w-3z from both sides. The subtraction property of equality states that this will not affect the equality. $x+4w-3z-(4w-3z)=2z+6w-(4w-3z)$ This becomes: $x+4w-3z-4w+3z=2z+6w-4w+3z$ which simplifies to: $x=5z+2w$ Since $y$ is equal to $5z+2w$ and $x$ is also equal to $5z+2w$, the transitive property of equality asserts that $x=y$. Images/mathematical drawings are created with GeoGebra. ### Practice Questions 1. Let $a, b, c, d$ be real numbers such that $a=b$, $2b=c$, and $2c=d$. Which of the following are equivalent? A. $a+a$ and $c$ B. $4b$ and $d$ C. $\\frac{1}{4}d$ and $a$ 2. An artist has two canvases that are the same size. She paints a picture on the first. Then, she takes the second to a hobby store and asks the clerk to help her find another canvas that has the same dimensions. The clerk does, and the artist buys it. How do the dimensions of the canvas the artist bought at the hobby store compare to the dimensions of the canvas with a picture on it? 3. True or False: Let $a, b,$ and $c$ be real numbers such", "+0 # Drag a reason to each box to complete this proof. 0 1 1053 1 +1015 Drag a reason to each box to complete this proof. If 5x − 4 +13x = 5 , then x = 1/2 . Statement Reason 5x − 4 + 13x = 5 Given 18x − 4 = 5 ( ) 18x = 9 ( ) x = 1/2 ( ) Options: Division property of Equality, Combine like terms., Substitution Property of Equality, Addition Property of Equality, Symmetric Property of Equality AngelRay Nov 12, 2017 #1 +2295 +2 $$18x-4=5$$ Combine like terms. In this step, the solver combined the $$5x$$ and $$13x$$ from the previous step. Like terms have to be both the same variable raised to the same power. This is indeed the case here. $$18x=9$$ Addition Property of Equality. This step added 4 to both sides, which is allowed based on the addition property. $$x=\\frac{1}{2}$$ Division Property of Equality. This step divided 18 from both sides. This is a valid operation because of the division property of equality. TheXSquaredFactor Nov 12, 2017 #1 +2295 +2 $$18x-4=5$$ Combine like terms. In this step, the solver combined the $$5x$$ and $$13x$$ from the previous step. Like terms have to be both the same variable raised to the same power. This is indeed", "brackets: $ (x-y)(x-z)(z-y) = - x^2·y + x^2·z + x·y^2 - x·z^2 - y^2·z + y·z^2 $ If you compare both results you'll see that they are equal.", "equality (as in the beginning) we continue: $e^{-z}-e^z=e^{\\overline{z}}-e^{-\\overline{z}}\\Longleftrightarrow$ putting $z=x+iy\\,,\\,\\,x,y\\in\\mathbb{R}$ and comparing real and imaginary parts I get $x=0$ as unique condition... Tonio", "x, T_{n+1}(x) = xT_{n}(x) - T_{n-1}(x)$. It’s determined by the equality: Also, I suggest visit the amazing answer for: Let $a_{i} \\in \\mathbb{R} (i=1, 2, \\dots, n)$ and $f(x) = \\sum_{i=0}^{n}a_{i}x^{i}$ such that if $|x| \\leq 1$, then $|f(x)| \\leq 1$ on MathStackExchange", "+ 1 = y$ and $y = 4x + 5$ then if we apply the transitive property of equality, we get: $x + 1 = 4x + 5$ 11. Michele_Laino ok! 12. Michele_Laino", "• We have $x^2-4=(x+2)(x-2)$. So taking the common factor $x+2$ \"out\" we get $(x+2)(1-3(x-2))$. – André Nicolas Oct 21 '13 at 4:56", "# one factoring question • January 13th 2008, 10:10 AM checkmarks one factoring question (4x+1)^2 - (16x^2 + 4x) i treated it as a difference of squares so i did: (4x+1 + 16x^2+4x)(4x+1 - 16x^2 - 4x) which i further simplified to: (16x^2 +8x+1)(-16x^2 +1) and im pretty sure this isnt right. • January 13th 2008, 11:33 AM Soroban Hello, checkmarks! Quote: Factor: . $(4x+1)^2 - (16x^2 + 4x)$ i treated it as a difference of squares . . . . ${\\color{blue}(16x^2 + 4x)}$ is not a square Factor the second term: . $(4x+1)^2 - 4x(4x + 1)$ Take out the common factor: . $(4x+1)\\bigg[(4x+1) - 4x\\bigg]$ And we have: . $(4x + 1)[4x + 1 - 4x) \\;\\;=\\;\\;(4x+1)(1) \\;\\;=\\;\\;\\boxed{4x + 1}$", "3x$ i.e. $(x+1)^2 +(y-1)^2 = 9x^2$ (Did you forget to square the 3?) i.e. $x^2+2x+1+y^2-2y+1 = 9x^2$ i.e. $8x^2 -y^2 -2x+2y -2 =0$", "+0 # Drag a reason to each box to complete this proof. +1 147 1 +618 Drag a reason to each box to complete this proof. If 5x − 4 +13x = 5 , then x = 1/2 . Statement Reason 5x − 4 + 13x = 5 Given 18x − 4 = 5 ( ) 18x = 9 ( ) x = 1/2 ( ) Options: Division property of Equality, Combine like terms., Substitution Property of Equality, Addition Property of Equality, Symmetric Property of Equality AngelRay Nov 12, 2017 #1 +1602 +2 $$18x-4=5$$ Combine like terms. In this step, the solver combined the $$5x$$ and $$13x$$ from the previous step. Like terms have to be both the same variable raised to the same power. This is indeed the case here. $$18x=9$$ Addition Property of Equality. This step added 4 to both sides, which is allowed based on the addition property. $$x=\\frac{1}{2}$$ Division Property of Equality. This step divided 18 from both sides. This is a valid operation because of the division property of equality. TheXSquaredFactor Nov 12, 2017 Sort: #1 +1602 +2 $$18x-4=5$$ Combine like terms. In this step, the solver combined the $$5x$$ and $$13x$$ from the previous step. Like terms have to be both the same variable raised to the same power. This is indeed", "as a sum of two squares then • write $10$ in binary 1010 so $19^{10} = (19^2)(((19^2)^2)^2$ • use the identity $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$ repeatedly" ]

[ "and $C_D := \\{f(x) = x^3 + ax^2 + b^x + c\\in Z[x];D_f = D\\}$ where $D_f$ is the discriminant of $f(x)$. Assume that $D < 0$, $D$ is square-free, $3\\nmid D$, and $3 \\nmid h(-3D)$ where $h(-3D)$ is the class number of $Q(\\sqrt -3D)$. We prove that all polynomials in $C_D$ have the same type of factorization over any Galois field $F_p$, $p$ being a prime, $p > 3$.\", }", "in the first place. The derivation is a more complicated problem. – Xander Henderson Oct 16 '19 at 22:04 • Writing it as $Q_1^3-Q_2^3= Q_3^3-Q_4^3$ where each $q_j$ is homogeneous degree $2$ you can factor both sides in linear factors, it becomes a simple quadratic identity in the ring of Eisenstein integers. – reuns Oct 16 '19 at 23:37 The most general homogeneous quadratic function of two variables is $$ax^2+bxy+cy^2$$. Cubing this gives $$a^3x^6+3a^2bx^5y+(3a^2c+3ab^2)x^4y^2+(b^3+6abc)x^3y^3+(3ac^2+3b^2c)x^2y^4+3bc^2xy^5+c^3y^6.$$(The fact that this sextic is homogeneous makes the above easy to deduce from combinatorics, without manually expanding out brackets. There's even an $$x\\leftrightarrow y$$ symmetry to almost halve the work.) The sum of two such cubes equals another viz.$$(ax^2+bxy+cy^2)^3+(dx^2+exy+fy^2)^3=(gx^2+hxy+iy^2)^3+(jx^2+kxy+ly^2)^3$$iff coefficients match. The first requirement this gives is$$a^3+d^3=g^3+j^3.$$Ramanujan was famous for noting the least positive integer expressible as the sum of two cubes in two different ways is $$1729=1^3+12^3=9^3+10^3$$, but in this example we work with much smaller examples, so we just swap variables. The \"simplest\" option is $$a=1,\\,d=2$$ (so that $$d\\ne a$$), whence $$g=d=2,\\,j=a=1$$. For the $$y^6$$ coefficients, Ramanujan actually does work with the above result about $$1729$$, rearranging the equality of two sums of cubes as$$(-9)^3+12^3=10^3+(-1)^3$$to reduce the size of the total to $$999$$. So we try $$c=-9,\\,f=12,\\,i=10,\\,l=-1$$. We now only need to find $$b,\\,e,\\,h,\\,k$$, which in the case at hand turn out", "+ t^2 + 2t + 1)(t+1),$$ and the first factor $t^4 + 2t^3 + t^2 + 2t + 1$ has no root and no quadratic factor, since $x^2+1,$ $x^2+x+2$ and $x^2+2x+2$ are the unique irreducible polynomials of degree $2$, and it is obvious that they are not dividing it. • Yes, you are right:) I suppose there was a similar question on $x^5-x-1$ not too long ago... – Dietrich Burde Jan 9 '18 at 14:02", "their remainder. Now what about the other factor? You’ll read it off of the table’s second row. Those numbers you got mark the coefficients with matching unknowns. If you are dividing polynomial of nth degree with a linear polynomial, the first exponent will be $n – 2$. Here we divided $3x^4 – 9x^3 – x^2 +4x – 3$ with $x – 3$ which means that our first exponent of second factor will be 3. This leads us to the solution: $3x^4 – x^2 – 9x^3 – 3 + 4x = (x – 3)(3x^3 – x + 1)$", "# Difference between revisions of \"2010 AIME II Problems/Problem 6\" ## Problem Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. ## Solution 1 You can factor the polynomial into two quadratic factors or a linear and a cubic factor. For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that $$(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.$$ Therefore, again setting coefficients equal, $a + c = 0\\Longrightarrow a=-c$, $b + d + ac = 0\\Longrightarrow b+d=a^2$ , $ad + bc = - n$, and so $bd = 63$. Since $b+d=a^2$, the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$. From this we find that the possible values for $n$ are $\\pm 8 \\cdot 62$ and $\\pm 4 \\cdot 2$. For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$. Therefore, the answer is $4 \\cdot 2 = \\boxed{008}$. ## Solution 2 Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$. From this, we get that $bd=63\\implies d=\\frac{63}{b}$ and $a+c=0\\implies", "that 2+ i is also a root and so a quadratic factor is $(x- 2+ i)(x- 2- i)= ((x- 2)+ i)((x- 2)- i)= (x- 2) ^2- i^2= x^2- 4x+ 4- 1= x^2- 4x+ 3$ and divide by that to find the third linear factor. Of course, you cannot us \"synthetic\" division to do that. 7. Apr 17, 2012 ### Curious3141 That should read ... = (x-2)2 - i2 = x2 - 4x + 4 + 1 = x2 - 4x + 5.", "### Home > CCA2 > Chapter 9 > Lesson 9.1.1 > Problem9-14 9-14. The roots of a polynomial are given below. Find two linear factors and the corresponding quadratic factor. 1. $x = -2 +\\sqrt { 3 }$, $x = -2 −\\sqrt { 3 }$ Find the first factor. $(x - (-2 + \\sqrt{3}))$ Find the second factor. Multiply together to get the quadratic factor. 1. $x = -2 + i$, $x = -2 - i$ Follow similar steps in part (a).", "the variables are x and y). Also, the brute force technique for using the formula method for quadratics is way too cumbersome. So, let’s give a shot to it using the method of undetermined coefficients. Let $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy=(Px+Qy)(Mx+Ny)$ Hence, equating like coefficients, $PM=2a^{2}$, $NQ= -2(3b-4c)(b-c)$, $MQ+NP=ab$. Again, you have to guess! Let’s try: let $N=2(b-c)$, $Q=-(3b-4c)$, $M=a$, $P=2a$ We need to check: $MQ+NP=-a(3b-4c)+2(b-c)(2a)=ab$ Bingo:-) So, the factors are $(ax+2(b-c)y)(2ax-y(3b-4c))$ 3) An important factorization identity is: $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Would you like to try proving this?(It is not so easy!) More later, — Nalin", "- 2$ and $x^2 + 1$; in broadening the circle of light, we meet the real numbers, the quadratic formula, and the complex numbers. • Wow, I really appreciate your perspective on the question. The fact that you posted this after @Siong Thye Goh, actually helped further explain his purely mathematical approach. I've never heard of the streetlight effect before, but I think it kind of explains the difficulties I've recently had writing proofs as well. – CaptainAmerica16 Sep 4 '18 at 2:42 Consider the polynomial $Ax^2+Bx+C$ $$(ax+b)(cx+d)=acx^2+(bc+ad)x+bd$$ Comparing the coefficients, $A=ac, C=bd$ and we want them to satisfies $bc+ad=B$. Hence we factor $A$ and $C$ and find factors such that $bc+ad=B$. • Thank you for this, but I got a little lost at the part where you compare the coefficients. Why exactly does $A=ac$ and $C=bd$? – CaptainAmerica16 Sep 4 '18 at 2:27 • the coefficient of $x^2$ for the first form is $A$ and the coefficient of $x^2$ for the second form is $ac$. Hence if the two polynomials are the same, they should be equal. – Siong Thye Goh Sep 4 '18 at 2:30 • Ah, ok. Thank you! This was very helpful! – CaptainAmerica16 Sep 4 '18 at 2:35 See fundamental theorem of algebra https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra Note over 2 variables or if you wish to", "Polynomials with integer coefficients vs. polynomials with rational coefficients My problem is very simple. From two different functions $f_1$ and $f_2$, I create two multivariate polynomials. Because of theoretical reasons those two functions (evaluated with the same input $x$) should send the same output. Now when I want to check this I always get false. The reason is the following I declare $P_1=f_1(x)$ and $P_2=f_2(x)$ and when I see the polynomials $P_1$ and $P_2$ they are actually equal but the first one has integer coefficients and the second one has rational coefficients (when the coefficient is $2$ in the first one, in the second the coefficient will be $2.$). Here is the question, how can I formally check that I have found the same polynomial in both cases? I think this would involve to convert the coefficients of the first one into rational but I am not able to find in the help of Mathematica how to do it, and then to check equality with $P_1===P_2$. • 2. is represented as floating point, not a Rational. IntegerQ[2.] is False; IntegerQ[2/1] is True (evaluates to 2). – alancalvitti Jul 24 '15 at 16:33 • Check documentation for Rationalize. – Daniel Lichtblau Jul 24 '15 at 16:53 f1[x_, y_] = x + 2 y^2 f2[x_, y_] = x + 2.", "10:34 OK (though $D_i$ should not be a point !). So just say $D'.L>0$ (take a line $L$ which doesn' contain any irreducible component of $D'$. For the second equality, write $f$ as quotient $P/Q$ of two homogeneous polynomials of same degree $d$. Then $\\mathrm{div}(f)=[Z(P)]-[Z(Q)]$, so $\\mathrm{div}(f).L=[Z(P)].L-[Z(Q)].L=d-d=0$ by Bézout. – user18119 Nov 26 '12 at 10:40", "becomes P(x) = 4x2 + 4x + 4 = 4(x2 + x + 1) Factoring using identities There are many algebraic identities. We can factor polynomials using identities like a2 - b2 = (a + b)(a - b) a2 - 2ab + b2 = (a - b)2 a2 + 2ab + b2 = (a + b)2 There are many identities, we have just quoted three of them. Lets consider the following example. Consider P(x) = x2 - 25 Here the polynomial is of the form a2 - b2 = (a + b)(a - b). So we can factor the polynomial as P(x) = x2 - 25 = (x + 5)(x - 5) This method can be used only for quadratic polynomials and polynomials which can be reduced to quadratic polynomials. If P(x) = ax2 + bx + c is a polynomial, quadratic formula is x = $\\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}$ Consider the polynomial P(x) = x2 - 10x + 24 Comparing with the standard polynomial P(x) = ax2 + bx + c Here a = 1, b = -10 and c = 24 x = $\\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}$ = $\\frac{-(-10) \\pm \\sqrt{(-10)^{2} - 4 \\times 1 \\times 24}}{2 \\times 1}$ = $\\frac{10 \\pm 2}{2}$ = 4, 6 Since x = 4 is a zero of", "$(ax+A)(x+B) = ax^2+bx+c$, you need them to satisfy: $$aB + A= b,\\ AB = c$$ If you assume integer factors, you can see $A,B$ must be either $3,-3$ or $\\pm 9,\\mp 1$. Only of of these three options gives $b = -80$. Use the quadratic formula. $x = \\frac{-B \\pm \\sqrt{B^2 - 4AC}}{2A}$ Why not. You know the quadratic formula, $$\\frac{-B \\pm \\sqrt{B^2 - 4AC}}{2A}$$ For your example, $B^2 - 4 AC = 6724$ and $\\sqrt {B^2-4AC}=82.$ TRICK: one of the $x$ coefficients is $$\\gcd \\left(A, \\frac{-B + \\sqrt{B^2 - 4AC}}{2} \\right) = \\gcd \\left(9, \\frac{80 + 82}{2} \\right) = \\gcd(9,81)=9.$$ See how you write the quadratic formula, but pull the $A$ out from the denominator and find the $\\gcd$ with what's left. I prove an expanded (and very slightly different) version of this, with entirely elementary methods, at How to factor the quadratic polynomial $2x^2-5xy-y^2$? EDIT: I had not wanted to muddy the waters...what happens if we switch the $\\pm$ sign in the \"trick\" above? One of the $x$ coefficients is $$\\gcd \\left(A, \\frac{-B - \\sqrt{B^2 - 4AC}}{2} \\right) = \\gcd \\left(9, \\frac{80 - 82}{2} \\right) = \\gcd(9,-1)=1.$$ You get the other $x$ coefficient, that's all. From the looks of the problem, it looks like they want to you to make educated guesses. If the factors are", "it as an equation in the ring $\\mathbb{Z}_2[x]$: $$x^4+x+1=r_3(x)(x^3+x+1)+r_2(x)(x^3+x^2+1).$$ At this point all the good properties of $\\mathbb{Z}_2[x]$ kick in. Most notably it is a Euclidean domain. As $\\gcd(x^3+x+1,x^3+x^2+1)=1$, we can write the constant polynomial $1$ as their linear combination (Bezout's identity). A run of Euclid's algorithm (or tinkering) shows that we have $$x h_2(x)+(x+1)h_3(x)=1.$$ Multiplying this by $x^4+x+1$ gives $$(x^5+x^2+x)h_2(x)+(x^5+x^4+x^2+1)h_3(x)=x^4+x+1.\\qquad(*)$$ We want those multipliers to be at most quadratic, so we need to do something about that. The usual trick is that we can add the same multiple of $h_2(x)h_3(x)$ to both terms (characteristic two now!). That first multiplier $x^5+x^2+x$ is equal to $$x^5+x^2+x=x^2 h_3(x)+x^4+x=(x^2+x)h_3(x)+x^3=(x^2+x+1)h_3(x)+(x^2+1),$$ so we subtract the polynomial $q(x)=(x^2+x+1)h_2(x)h_3(x)$ from both terms of the l.h.s. of $(*)$. As $(x^2+x+1)h_2(x)=x^5+x^4+1$, we have $q(x)=(x^5+x^4+1)h_3(x)$. Hence equation $(*)$ can be rewritten as $$(x^2+1)h_2(x)+x^2h_3(x)=x^4+x+1,$$ which we can also verify directly. Finally we can see that $r_3(x)=x^2+1$ and $r_2(x)=x^2$ will work. Thus $$g_2(x)=h_2(x)+2r_2(x)=x^3+2x^2+x+1$$ and $$g_3(x)=h_3(x)+2r_3(x)=x^3+3x^2+3=x^3-x^2-1.$$ At this point I encourage you to check that we have $$g_2(x)g_3(x)=\\tilde{f}(x)$$ in the ring $\\mathbb{Z}_4[x]$ as required. (I did it, so you have to, too!) The final factorization is thus $$f(x)=(x+1)(x^3+2x^2+x+1)(x^3-x^2-1).$$ Above I somewhat wantonly jumped from $\\mathbb{Z}_4$ to $\\mathbb{Z}_2$ and back. You need to keep track of the changes for everything to make sense! - The $\\mathbb{Z}_4$-codes were a topic of intense research", "$ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the discriminant is a PERFECT SQUARE if $d=-a$. 16. This result is similar to the preceding one, but we have decided to separate them because the quadratics in question are two different families; in particular, when $d=-a$, the middle term is missing (because the coefficients become $a,0,-a$), giving rise to quadratic equations of the form $ax^2-a=0$. Nevertheless, there is a transformation that connects these two quadratic families. 17. Let $ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the quadratic that results from $d=7a$ and the one that results from $d=-a$ have the SAME DISCRIMINANT. 18. In the case when $d=7a$, the prototype is $f(x)=x^2+8x+15$, while $g(x)=x^2-1$ is the parent for the case when $d=-a$. The two quadratics have the same leading coefficient and the same discriminant, so in view of a previous result (see the section on transformation of quadratics), there is a horizontal transformation that connects these two quadratics. In fact, let $$h=\\frac{8-0}{2}=4,~\\textrm{then}~g(x+h)=(x+4)^2-1=x^2+8x+15=f(x).$$ Similarly, $f(x-4)=(x-4)^2+8(x-4)+15=x^2-1=g(x)$. Observe that we have used $h=\\frac{b-B}{2a}$, where $b,B$ are the middle coefficients and $a$ is their common leading coefficient. This is the amount by which two quadratics with", "# Find a factor #1 The series of Find a factor puzzle is started by Culver Kwan, and asks the solver to identify a factor of a certain large number within a certain range using some mathematical identities. This should be tagged with , and . Find a factor of $$104060405$$ within $$10100$$ and $$11000$$, using various mathematical identities. You should not use a computer. • Does seeing a factorization count as using an identity? This factors easily if you rot13(oernx vg va gur zvqqyr), so to speak. – msh210 May 25 at 3:45 • @msh210 I do not understand what are you saying, but the identity is a Wikipedia page. – Culver Kwan May 25 at 4:10 To be honest, like msh210 said in the comments, I just saw the factorisation: $$104060405 = 104050000 + 10405 = 10405*(10000+1) = 10405*10001$$ You could find it by recognising most of the fourth row of Pascal's triangle $$1 4 6 4 1$$, and using $$(x+1)^4=x^4+4x^3+6x^2+4x+1$$ with $$x=100$$ to write the number as $$104060405=101^4+2^2=10201^2+2^2$$ And then applying the identity for the product of sums of two squares: $$10201^2+2^2 = (102*100+1*1)^2+(102*1-100*1)^2 = (102^2+1^2)(100^2+1^2)$$ but that seems a bit convoluted to me. Edit: The intended trick was to use: Sophie Germain's identity which is $$x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$$ In this case we have $$x=101$$", "# Factoring Quadratic Polynomials into Linear Factors If x_1 and x_2 are the roots of quadratic polynomial of ax^2+bx=c (i.e.the root of equation of ax^2+bx+c=0), then color(blue)(ax^2+bx+c=a(x-x_1)(x-x_2)).This is the formula of factoring quadratic polynomial into factors. Example. Factor the following: 6x^2-x-2. We apply the formula of roots of quadratic equation to the equation of 6x^2-x-2=0 and find x_1=-1/2, x_2=2/3. So, 6x^2-x-2=6(x+1/2)(x-2/3)=2(x+1/2)*3(x-2/3)=(2x+1)(3x-2).", "reverse of calculating the product of factors. In order to factorise a quadratic, we need to find the factors which when multiplied together equal the original quadratic. Let us consider a quadratic that is of the form ax2+bxax2+bx . We can see here that xx is a common factor of both terms. Therefore, ax2+bxax2+bx factorises to x(ax+b)x(ax+b). For example, 8y2+4y8y2+4y factorises to 4y(2y+1)4y(2y+1). Another type of quadratic is made up of the difference of squares. We know that: ( a + b ) ( a - b ) = a 2 - b 2 . ( a + b ) ( a - b ) = a 2 - b 2 . (19) This is true for any values of aa and bb, and more importantly since it is an equality, we can also write: a 2 - b 2 = ( a + b ) ( a - b ) . a 2 - b 2 = ( a + b ) ( a - b ) . (20) This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down what the factors are. ### Exercise 7: Difference of Squares Find the factors of 9x2-259x2-25. #### Solution 1. Step 1. Examine the quadratic :", "to $x$-intercepts of the quadratic $y = ax^2+bx+c$. The discriminant tells us about the relationship between a […] Not all quadratics factorise easily. For instance, $x^2+6x+2$ cannot be factorised by simple factorisation. In other words, we cannot write $x^2+6x+2$ in the form $(x-a)(x-b)$ where $a$ and $b$ are rational. An alternative way to solve this equation $x^2+6x+2$ is by completing the square. Equations of the form $ax^2+bx+c$ can be converted to the form […]", "# 1967 AHSME Problems/Problem 31 ## Problem Let $D=a^2+b^2+c^2$, where $a$, $b$, are consecutive integers and $c=ab$. Then $\\sqrt{D}$ is: $\\textbf{(A)}\\ \\text{always an even integer}\\qquad \\textbf{(B)}\\ \\text{sometimes an odd integer, sometimes not}\\\\ \\textbf{(C)}\\ \\text{always an odd integer}\\qquad \\textbf{(D)}\\ \\text{sometimes rational, sometimes not}\\\\ \\textbf{(E)}\\ \\text{always irrational}$ ## Solution Let $a=x, b=x+1, c = x(x+1)$. Then $D = x^2 + (x+1)^2 + x^2(x+1)^2$, which simplifies to $x^4 + 2x^3 + 3x^2 + 2x + 1$. From the options, we want to test if $D$ is always a perfect square. Because the polynomial expression for $D$ is quartic, if it is a perfect square, it would be the square of a quadratic expression. Thus, $D$ could be written in the form $(Ax^2 + Bx + C)^2$ for some $(A, B, C)$. Setting $(Ax^2 + Bx + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$, we can compare coefficients. From the $x^4$ coefficient, we get $A = \\pm 1$. Note that if $(Ax^2 + Bx^2 + C)^2$ works, so does $(-Ax^2 - Bx - C)^2$, so we can arbitrarily pick $A=1$. We now have $(x^2 + Bx + C)^2= x^4 + 2x^3 + 3x^2 + 2x + 1$. Setting the cubic terms equal gives $2B = 2$, or $B = 1$. This leaves $(x^2 + x + C)^2 =" ]

[ "+0 # HEEELP 0 69 1 In a certain ellipse, the center is at $(-3,1),$ one focus is at $(-3,0),$ and one endpoint of a semi-major axis is at $(-3,3).$ Find the semi-minor axis of the ellipse. Mar 15, 2020", "the parabola using the information given. 255. Focus $(4,0)(4,0)$ and directrix $x=−4x=−4$ 256. Focus $(0,−3)(0,−3)$ and directrix $y=3y=3$ 257. Focus $(0,0.5)(0,0.5)$ and directrix $y=−0.5y=−0.5$ 258. Focus $(2,3)(2,3)$ and directrix $x=−2x=−2$ 259. Focus $(0,2)(0,2)$ and directrix $y=4y=4$ 260. Focus $(−1,4)(−1,4)$ and directrix $x=5x=5$ 261. Focus $(−3,5)(−3,5)$ and directrix $y=1y=1$ 262. Focus $(52,−4)(52,−4)$ and directrix $x=72x=72$ For the following exercises, determine the equation of the ellipse using the information given. 263. Endpoints of major axis at $(4,0),(−4,0)(4,0),(−4,0)$ and foci located at $(2,0),(−2,0)(2,0),(−2,0)$ 264. Endpoints of major axis at $(0,5),(0,−5)(0,5),(0,−5)$ and foci located at $(0,3),(0,−3)(0,3),(0,−3)$ 265. Endpoints of minor axis at $(0,2),(0,−2)(0,2),(0,−2)$ and foci located at $(3,0),(−3,0)(3,0),(−3,0)$ 266. Endpoints of major axis at $(−3,3),(7,3)(−3,3),(7,3)$ and foci located at $(−2,3),(6,3)(−2,3),(6,3)$ 267. Endpoints of major axis at $(−3,5),(−3,−3)(−3,5),(−3,−3)$ and foci located at $(−3,3),(−3,−1)(−3,3),(−3,−1)$ 268. Endpoints of major axis at $(0,0),(0,4)(0,0),(0,4)$ and foci located at $(5,2),(−5,2)(5,2),(−5,2)$ 269. Foci located at $(2,0),(−2,0)(2,0),(−2,0)$ and eccentricity of $1212$ 270. Foci located at $(0,−3),(0,3)(0,−3),(0,3)$ and eccentricity of $3434$ For the following exercises, determine the equation of the hyperbola using the information given. 271. Vertices located at $(5,0),(−5,0)(5,0),(−5,0)$ and foci located at $(6,0),(−6,0)(6,0),(−6,0)$ 272. Vertices located at $(0,2),(0,−2)(0,2),(0,−2)$ and foci located at $(0,3),(0,−3)(0,3),(0,−3)$ 273. Endpoints of the conjugate axis located at $(0,3),(0,−3)(0,3),(0,−3)$ and foci located $(4,0),(−4,0)(4,0),(−4,0)$ 274. Vertices located at $(0,1),(6,1)(0,1),(6,1)$ and focus located at $(8,1)(8,1)$ 275. Vertices located", "semi-minor axis. If the center is not located at the origin, the equation of the ellipse is: where, $latex (h, k)$ is the center of the ellipse.", "class. What do you mean by them? 7. Jun 13, 2008 ### temaire Do you mean the center? 8. Jun 13, 2008 ### konthelion The focii for an ellipse is the point that lies on the major axis(the longer side/axis) of the ellipse. There are two focii in this ellipse. (h+c,k) and (h-c,k). In an ellipse(for both vertial and horizontal ellipses), $$b^2=a^2-c^2$$, where a is always the large axis and b is the smaller axis. In a parabola, the there is only one focus. Since this parabola opens down, then the focus is at (h,-c+k).", "type was (5,-2) question again is. Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,2). ... Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,-2). The distance between the center and the focus is 2, that means e = 2. Therefore: $2^2+b^2=3^2~\\implies~b=\\sqrt{5}$ Plug in all the values you know into the general equation of an ellipse: $\\frac{(x-5)^2}{3^2}+\\frac{(y+2)^2}{(\\sqrt{5})^2}=1 ~\\implies~ \\frac{(x-5)^2}{9}+\\frac{(y+2)^2}{5}=1$", "just one single point, all these points must coincide. Thus, it lies on all the $3$ medians. Thus, all the medians intersect and the center of mass divides them in the ratio $3:1$.", "( 2,2 ), ( 2,2 ), Vertices ( 0,2 ),( 4,2 ),( 2,0 ),( 2,4 ), ( 0,2 ),( 4,2 ),( 2,0 ),( 2,4 ), Focus ( 2,2 ) ( 2,2 ) For the following exercises, use the given information about the graph of each ellipse to determine its equation. #### Exercise 53 Center at the origin, symmetric with respect to the x- and y-axes, focus at (4,0), (4,0), and point on graph (0,3). (0,3). #### Exercise 54 Center at the origin, symmetric with respect to the x- and y-axes, focus at (0,−2), (0,−2), and point on graph (5,0). (5,0). ##### Solution x 2 25 + y 2 29 =1 x 2 25 + y 2 29 =1 #### Exercise 55 Center at the origin, symmetric with respect to the x- and y-axes, focus at (3,0), (3,0), and major axis is twice as long as minor axis. #### Exercise 56 Center ( 4,2 ) ( 4,2 ) ; vertex ( 9,2 ) ( 9,2 ) ; one focus: ( 4+2 6 ,2 ) ( 4+2 6 ,2 ) . ##### Solution ( x4 ) 2 25 + ( y2 ) 2 1 =1 ( x4 ) 2 25 + ( y2 ) 2 1 =1 #### Exercise 57 Center ( 3,5 ) ( 3,5 ) ; vertex (", "clearer if you start by saying let$(a,b)$... 0 Take a look at this: http://gieseanw.wordpress.com/2013/07/19/an-analytic-solution-for-ellipse-and-line-intersection/ Basic idea: starting from writing down the formular of ellipse: (x/a)^2 + (y/b)^2 = 1, where a is the semi-major axis, b is the semi-minor axis. Then write down the formula for a line: (x2-x1)*(y-y1) = (y2-y1)/(x-x1) then representing ... 0 1. You are correct that the center is at$(1, 0)$, and so the focal length is indeed$c = 3$. Recall that you can draw a rectangle, centered at the center of the hyperbola such that the asymptotes pass through the corners and the left and right sides of the rectangle are tangent to the vertices of the hyperbola. The half-dimensions of the box are$a$and ... 0 For 1: Remember that$c=\\sqrt{a^2+b^2}$, and the slopes of the asymptotes are$\\pm b/a$. For 2: Completing the square is the way to go. Try dividing everything by$2$first, though. 0 Almost any minimum or maximum will be approximately parabolic. You can calculate the Taylor series at$(x_0,y_0)$and get$y \\approx y_0+y'(x_0)(x-x_0)+y''(x_0)(x-x_0)^2\\dots$Since you are at a minimum or maximum, you have$y'(x_0)=0$so that term goes away. The higher order terms are small when you are near the minimum or maximum. In your case of ... 2 The comments by the OP indicate that he may be confused about the nature of the branches", "is on the x-axis, that is, $k=0$, then $I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2) = ...$ if the center is not on the x-axis, i.e., $k\\neq 0$ then $I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2 -2ky) = ...$ hope you can do it from here..", "end. This will return the center coordinates in Y and the radius in X: 1 ENTER 4 COMPLEX -1 ENTER 2 COMPLEX 4 ENTER -3 COMPLEX XEQ \"CENTER\" => 2,5 i0,5 3,80788655... Dieter", "form with focus at pole and given the polar coordinates $(-6,\\frac \\pi 2),(2,\\frac{3\\pi}{2})$ as endpoints of its transverse axis. 11. Write the following in polar form. $\\frac{(x-4)^2}{25}+\\frac{y^2}{9}=1$", "$b = \\sqrt{r_{max}r_{min}}.$ A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping l fixed. Thus a and b tend to infinity, a faster than b. The length of the semi-minor axis could also be found using the following formula,[1] $2b = \\sqrt{(p+q)^2 -f^2}$ where f is the distance between the foci, p and q are the distances from each focus to any point in the ellipse. ## Hyperbola In a hyperbola, a conjugate axis or minor axis of length 2b, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0, ±b) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor", "center value of $x = 3$. Plotting the points in the $xy$-axis, ### Practice with Worksheets You might also be interested in:", "## Calculus (3rd Edition) $(-2, 2, 3)$ The $x$ component, $-2+cos(t)$, oscillates between $-3$ and $-1$, so its center in the x-dimension is $-2$. The $y$ component is the constant $2$, so its center in the y-dimension is $2$. The $z$ component, $3-sin(t)$, oscillates between $2$ and $4$, so its center in the z-dimension is $3$. Therefore, the coordinates of the center are $(-2, 2, 3)$.", "arc between start angle and end angle on an ellipse whose center is at center and whose two principal axes are 0-degree axis and 90-degree axis. For $$t=0$$ the point at the start angle is returned and for $$t=1$$ the point at the end angle. • \\pgfpointcurveattime{time $$t$$}{point $$p$$}{point $$s_1$$}{point $$s_2$$}{point $$q$$} • Yields a point that is on the Bézier curve from $$p$$ to $$q$$ with the support points $$s_1$$ and $$s_2$$. The time $$t$$ is used to determine the location, where $$t=0$$ yields $$p$$ and $$t=1$$ yields $$q$$. ###### 101.5.3Points on Borders of Objects¶ The following commands are useful for specifying a point that lies on the border of special shapes. They are used, for example, by the shape mechanism to determine border points of shapes. • \\pgfpointborderrectangle{direction point}{corner} • This command returns a point that lies on the intersection of a line starting at the origin and going towards the point direction point and a rectangle whose center is in the origin and whose upper right corner is at corner. The direction point should have length “about 1pt”, but it will be normalized automatically. Nevertheless, the “nearer” the length is to 1pt, the less rounding errors. • \\pgfpointborderellipse{direction point}{corner} • This command works like the corresponding command for rectangles, only this time the corner is", "describe an ellipse centered at $$(1,3)\\text{,}$$ with vertical major axis of length $$6$$ and minor axis of length $$2\\text{.}$$ $$x=$$ $$y=$$ ###### 53. An ellipse with foci at $$(\\pm 1,0)$$ and vertices at $$(\\pm 5,0)\\text{.}$$ ###### 54. A hyperbola with foci at $$(5,-3)$$ and $$(-1,-3)\\text{,}$$ and with vertices at $$(1,-3)$$ and $$(3,-3)\\text{.}$$ ###### 55. A hyperbola with vertices at $$(0,\\pm 6)$$ and asymptotes $$y=\\pm 3x\\text{.}$$", "is the full ellipse. Then it can be shown, how to write the equation of an ellipse in terms of matrices. Instructions. Trim/Divide. Inserts a primitive ellipsoid in the active Body as the first feature, or fuses it to the existing feature(s). The returned ellipsoid has center coordinates at (xc,yc,zc), semiaxis lengths (xr,yr,zr), and consists of 20-by-20 faces. Mark the distance along the axes and lightly block in the ellipse. (a0,b0) is the center of the ellipse (ax,ay) vector representing the major axis (bx,by) vector representing the minor axis. We will review your submission and take any actions necessary per our Community Guidelines. Ellipse graph. View Answer Find an equation for the ellipse with vertices at (-3, 4) and (15, 4); focus at (13, 4). Sketch and edit a spline. darken the final ellipse. To define the shape of the ellipse, move the cursor away from the starting. Unit Tangent and Normal Vectors for a Helix. space tunnel. To create an ellipse: Click Ellipse on the Sketch toolbar or Tools > Sketch Entities > Ellipse. Sketch; Samples may be limited. Extend or shorten an entity of a sketch curve. To set the major axis of the ellipse, tap again or touch and drag then release. Ellipse Creates an ellipse with a center point, a major axis, and a", "is centered at the origin, the center is (0,0) and when it is centered outside the origin, the center is $latex (h, k)$. ### Vertices The vertices are the endpoints of the transverse axis. These points are located at the intersection of the hyperbola and the transverse axis. We use $latex V$ to represent the vertices and we can distinguish them using $latex V_{1}$ and $latex V_{2}$ or $latex V$ and $latex V’$. ### Semi-major axis The semi-major axis is a line segment that connects the center to a vertex of the hyperbola. Usually, we denote its length using a. ### Semi-minor axis The semi-minor axis is a line segment that is perpendicular to the semi-major axis. Usually, we denote its length using b. ### Asymptotes The asymptotes represent the behavior of the hyperbolas. The asymptotes are the lines that are very close to the branches of the hyperbola but never touch it.", "note that $t$ has to vary between $0$ and $2\\pi$ to cover the whole ellipse (even though $t$ does not quite measure the angle between the radius vector of $(x,y)$ and the $x$-axis, as it does in polar coordinates). -", "and $$(−2,1)$$? Solution The $$x$$-coordinates of the vertices and foci are the same, so the major axis is parallel to the $$y$$-axis. Thus, the equation of the ellipse will have the form $$\\dfrac{{(x−h)}^2}{b^2}+\\dfrac{{(y−k)}^2}{a^2}=1 \\nonumber$$ First, we identify the center, $$(h,k)$$. The center is halfway between the vertices, $$(−2,−8)$$ and $$(−2,2)$$. Applying the midpoint formula, we have: \\begin{align} (h,k) &=\\left(\\dfrac{−2+(−2)}{2},\\dfrac{−8+2}{2}\\right) \\nonumber \\\\ &=(−2,−3) \\nonumber \\end{align} \\nonumber Next, we find $$a^2$$. The length of the major axis, $$2a$$, is bounded by the vertices. We solve for $$a$$ by finding the distance between the y-coordinates of the vertices. \\begin{align} 2a &=2−(−8) \\nonumber \\\\ 2a &=10 \\nonumber\\\\ a&=5 \\nonumber \\end{align} \\nonumber So $$a^2=25$$. Now we find $$c^2$$. The foci are given by $$(h,k\\pm c)$$. So, $$(h,k−c)=(−2,−7)$$ and $$(h,k+c)=(−2,1)$$. We substitute $$k=−3$$ using either of these points to solve for $$c$$. \\begin{align} k+c &=1 \\nonumber \\\\ −3+c&=1 \\nonumber \\\\ c&=4 \\nonumber \\end{align} \\nonumber So $$c^2=16$$. Next, we solve for $$b^2$$ using the equation $$c^2=a^2−b^2$$. \\begin{align} c^2&=a^2−b^2 \\nonumber \\\\ 16&=25−b^2 \\nonumber \\\\ b^2&=9 \\nonumber \\end{align} \\nonumber Finally, we substitute the values found for $$h$$, $$k$$, $$a^2$$, and $$b^2$$ into the standard form equation for an ellipse: $\\dfrac{{(x+2)}^2}{9}+\\dfrac{{(y+3)}^2}{25}=1 \\nonumber$ ###### Exercise $$\\PageIndex{2}$$ What is the standard form equation of the ellipse that has vertices $$(−3,3)$$ and $$(5,3)$$ and foci $$(1−2\\sqrt{3},3)$$ and $$(1+2\\sqrt{3},3)$$? $$\\dfrac{{(x−1)}^2}{16}+\\dfrac{{(y−3)}^2}{4}=1 \\nonumber$$ ##" ]

[ "the length of the major axis (10) so: . $a = 5 \\quad\\Rightarrow\\quad a^2 = 25$ . all correct! C is x in the foci, so it's 1 . . . . no $c$ is the distance from the center to a focus: . $c = 3$ Give it another try . . . 3. Ahh, that sounds more correct. I don't know why I thought of it as simply X in a foci. Hmm. So then it's: C=3 C^2 = 9 9 = 25 - b^2 b^2 = 16 b= 4 So, then equation would be: (x+1)^2/16 + (y-3)^2/25 = 1 Correct now?", "e^2-1},0\\right)$ and $\\left({de\\over e+1},0\\right)$, respectively. The ratio of the semi-focal distance to the semi-major axis length is therefore $${0-{de^2\\over e^2-1} \\over {de\\over e+1}-{de^2\\over e^2-1}} = e.$$", "be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0,\\pmb) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length . Denoting the semi-major axis length (distance from the center to a vertex) as, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: x2 a2 - y2 b2 =1. The semi-minor axis is also the distance from one of focuses of the hyperbola to an asymptote. Often called the impact parameter, this is important in physics and astronomy, and measure the distance a particle will miss the focus by if its journey is unperturbed by the body at the focus. The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: b=a\\sqrt{e2-1}. [3] Note that in a hyperbola can be larger than .http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node27.html ## Astronomy ### Orbital period In astrodynamics the orbital period of a small body orbiting a central body in a circular or elliptical orbit is:[4] T=2\\pi\\sqrt{a3\\over\\mu} where: is the length of the orbit's semi-major axis \\mu is the standard", "+0 # HEEELP 0 69 1 In a certain ellipse, the center is at $(-3,1),$ one focus is at $(-3,0),$ and one endpoint of a semi-major axis is at $(-3,3).$ Find the semi-minor axis of the ellipse. Mar 15, 2020", "point at one end of the major axis (the axis containing the foci). The distance from one focus to the center is c, then the distance from the center to the vertex is a. The distance back from the that vertex to the other focus is a- c. The total distance is a+ c+ (a- c)= 2a. Now, look at the vertex at one end of axis perpendicular to the axis containing the foci. The line from each focus to that vertex is the hypotenuse of the right triangle formed by that vertex, the center, and the focus. It has legs of length b and c and so the length of the hypotenuse is $\\sqrt{b^2+ c^2}$. Since that is true for both foci, the total distance from the foci to that vertex is $2\\sqrt{b^2+ c^2}$. Since that total distance is constant, we must have $2\\sqrt{b^2+ c^2}= 2a$. Divide both sides by 2 and square both sides to get $b^2+ c^2= a^2$. Quote: just plug away the standard equation actually is $A X^2 + B X Y + C Y^2 + D X + E Y + F = 0$", "I did the other one? Here the question I'm working on: OpenStudy (anonymous): From what you did before, i figured out that h = -7 and k = -2, but I'm not quite sure how what we did before is applicable to the rest of the question. jimthompson5910 (jim_thompson5910): Good, you got the center correct. Now you just need to find the length of the major and minor axes. Do you know how to do this? OpenStudy (anonymous): From what you did in the first one, I think that the major axis would by 18. I'm not sure about the minor. OpenStudy (anonymous): be* jimthompson5910 (jim_thompson5910): Sorry got distracted for a bit. You nailed it. So 2a = 18 ---> a = 9 Note: In this case, the major axis is along the horizontal. This is why 'a' is the length of the semi-major axis. ------------------------------------------------------- Now we must find the distance from one focus to the center. One focus is (-14,-2) and the center is (-7,-2). Therefore, the distance is simply |-14 - (-7)| = |-14 + 7| = |-7| = 7 So the distance from the center to either focus is 7 units, which means that c = 7 Now use the idea that b^2 = a^2 - c^2 to solve for b b^2 = a^2", "the other. If the sum of the distances from any point to each foci is 2a, taking the closest point I just talked about, the distance from each focus to it is a (both distances are equal and must sum to 2a). Using pythagoras theorem where f is the distance from centre to focus and a is the semi major axis and b is the semi minor axis, (f^2) + (b^2) = (a^2) we know f=8 and a=12 therefore (8^2) + (b^2) = (12^2) 64 + (b^2) = 144 b^2 = 80 therefore (x^2)/(80) + (y^2)/(144) = 1 23. anonymous Oh i just read yours lol, yes you did eveything right, good job :)", "t - c)^2 + (b\\sin t)^2} = 2a$ • apologies if I'm a little obtuse but could you explain what a, b, c and t refer to? – Shawn Naquin Nov 13 '17 at 20:22 • $a$ is the length of the semi-major axis. $b$ is the length of the semi-minor axis. $c$ is the distance from the center to the focus. $\\frac {x^2}{a^2 } + \\frac {y^2}{b^2} = 1$ – Doug M Nov 13 '17 at 20:25", "of these two numbers, whichever is larger, or whichever is smaller you subtract from the other one. You take the square root, and that's the focal distance. Now, let's see if we can use that to apply it to some some real problems where they might ask you, hey, find the focal length. Or find the coordinates of the focuses. So let's add the equation x minus 1 squared over 9 plus y plus 2 squared over 4 is equal to 1. So let's just graph this first of all. This could be interesting. So I'll draw the axes. That's the x-axis. This is the y-axis. And we immediately see, what's the center of this? The center is going to be at the point 1, negative 2. And if that's confusing, you might want to review some of the previous videos. Center's at 1, x is equal to 1. y is equal to minus 2. That's the center. And then, the major axis is the x-axis, because this is larger. And so, b squared is -- or a squared, is equal to 9. And the semi-minor radius is going to be equal to 3. So, if you go 1, 2, 3. Go there. Than you have 1, 2, 3. No. 1, 2, 3. 1, 2, 3. I think", "0$ So, the value of t will be 4. Hence point P = (4,8) Now, as we know focal distance is the distance between focus and a point on a curve, focal distance will be the distance between focus and P. Now, calculating distance between them as, focal distance = $\\sqrt {\\left( {{{\\left( {4 - 4} \\right)}^2} + {{\\left( {8 - 0} \\right)}^2}} \\right)} {\\text{ }} = 8$ focal distance = 8 Hence the correct option for the question will be (c) NOTE: - Point on a conic always satisfies the equation of conic. Abscissa is the x - coordinate of a point and ordinate is the y - coordinate of a point.", "the focus. The equation is: $$f1​=S1​1​+S2​1​ (3)$$ where and are the two distances. 🔗 🔗 🔗", "= \\sqrt{1-\\frac{b^2}{a^2}}$$ and the distance from the focus to the centre of the ellipse is $$f = ae$$", "that the center of $C_1$ is the focus and $L$ is the directrix. Since the ratio of the distances is not constant, but varies with $r$, they cannot be the focus and directrix of a conic section. (That doesn't say the curve isn't a conic section - just that these two do not fit the roles of focus and directrix.) Feb 8 '20 at 15:39 Consider a line $$L'$$ parallel to $$L$$ and at a distance $$r_1$$ from $$L$$, such that $$CL'=r+r_1$$.", "is $\\dfrac{(x-h)^2}{a^2} + \\dfrac{(y-k)^2}{b^2} = 1$ Some remarks about Equation \\ref{standardellipse} are in order. First note that the values $$a$$ and $$b$$ determine how far in the $$x$$ and $$y$$ directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if $$a > b$$, then we have an ellipse whose major axis is horizontal, and hence, the foci lie to the left and right of the center. In this case, as we've seen in the derivation, the distance from the center to the focus, $$c$$, can be found by $$c = \\sqrt{a^2 - b^2}$$. If $$b > a$$, the roles of the major and minor axes are reversed, and the foci lie above and below the center. In this case, $$c = \\sqrt{b^2 - a^2}$$. In either case, $$c$$ is the distance from the center to each focus, and $$c = \\sqrt{ {bigger denominator} - {smaller denominator}}$$. Finally, it is worth mentioning that if we take the standard equation of a circle, Equation \\ref{standardcircle}, and divide both sides by $$r^2$$, we get The Alternate Standard Equation of a Circle: The equation of a circle with center $$(h,k)$$ and radius $$r >0$$ is $\\dfrac{(x-h)^2}{r^2} + \\dfrac{(y-k)^2}{r^2} = 1$ Notice the similarity between Equation \\ref{standardellipse} and Equation \\ref{standardcirclealternate}. Both equations involve a", "It's your class. I'm just tossing off hints without being fully awake. I have no idea what 'c' is supposed to be. Could you just like say what it is supposed to be instead of dropping a cryptic letter? I'll take another guess and say 'distance from center to focus'? That's a lot better description than 'c'. Why don't you think it could be sqrt(3^2-2^2)? If you flip the x and y axes, do you think this distance should change from real to imaginary? 8. Nov 7, 2008 ### duki hmm, I'm not sure. I didn't know you could swap them like that. The formula I was going by said \"distance from center to focus\" = sqrt(a^2 - b^2). If you flip them, you get the real answer? 9. Nov 7, 2008 ### HallsofIvy Surely that formula was assuming a> b. That is, that a is the length of the longer semi-axis. Don't just memorize formulas. Learn what the mean.", "connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints $\\displaystyle{ (0,\\pm b) }$ of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\displaystyle{ \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1. }$ The semi-minor axis is also the distance from one of focuses of the hyperbola to an asymptote. Often called the impact parameter, this is important in physics and astronomy, and measure the distance a particle will miss the focus by if its journey is unperturbed by the body at the focus. The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: $\\displaystyle{ b = a \\sqrt{e^2-1}. }$[4] Note that in a hyperbola b can be larger than a.[5] ## Astronomy ### Orbital period In astrodynamics the orbital period T of a small body orbiting a central body in a circular or elliptical orbit is:[1] $\\displaystyle{ T = 2\\pi\\sqrt{\\frac{a^3}{\\mu}}, }$ where: a is the length of the orbit's semi-major axis,", "\\begin{align} b &= a \\sqrt{1 - e^2}, \\\\ a \\ell &= b^2. \\end{align} } A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping $\\displaystyle{ \\ell }$ fixed. Thus a and b tend to infinity, a faster than b. The length of the semi-minor axis could also be found using the following formula:[2] $\\displaystyle{ 2b = \\sqrt{(p + q)^2 -f^2}, }$ where f is the distance between the foci, p and q are the distances from each focus to any point in the ellipse. ## Hyperbola The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches; if this is a in the x-direction the equation is: $\\displaystyle{ \\frac{\\left( x-h \\right)^2}{a^2} - \\frac{\\left( y-k \\right)^2}{b^2} = 1. }$ In terms of the semi-latus rectum and the eccentricity we have $\\displaystyle{ a={\\ell \\over e^2 - 1 }. }$ The transverse axis of a hyperbola coincides with the major axis.[3] In a hyperbola, a conjugate axis or minor axis of length $\\displaystyle{ 2b }$, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter", "the $y-$axis at $y=\\pm 1$. In a coordinate free way, the semi-latus rectum is the distance between the curve and the focus in the direction perpendicular to the point of closest approach. What is the meaning of the two branches of the hyperbola? The left branch is for an attactrive force and the right branch is for a repulsive force.", "$b = \\sqrt{r_{max}r_{min}}.$ A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping l fixed. Thus a and b tend to infinity, a faster than b. The length of the semi-minor axis could also be found using the following formula,[1] $2b = \\sqrt{(p+q)^2 -f^2}$ where f is the distance between the foci, p and q are the distances from each focus to any point in the ellipse. ## Hyperbola In a hyperbola, a conjugate axis or minor axis of length 2b, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0, ±b) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor", "b values to plot the ends of the major and minor axis. 4. Draw in the ellipse. ### How do you calculate the foci of an ellipse? Remember the two patterns for an ellipse: Each ellipse has two foci (plural of focus) as shown in the picture here: As you can see, c is the distance from the center to a focus. We can find the value of c by using the formula c2 = a2 - b2." ]

[ "+0 # HEEELP 0 69 1 In a certain ellipse, the center is at $(-3,1),$ one focus is at $(-3,0),$ and one endpoint of a semi-major axis is at $(-3,3).$ Find the semi-minor axis of the ellipse. Mar 15, 2020", "$b = \\sqrt{r_{max}r_{min}}.$ A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping l fixed. Thus a and b tend to infinity, a faster than b. The length of the semi-minor axis could also be found using the following formula,[1] $2b = \\sqrt{(p+q)^2 -f^2}$ where f is the distance between the foci, p and q are the distances from each focus to any point in the ellipse. ## Hyperbola In a hyperbola, a conjugate axis or minor axis of length 2b, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0, ±b) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor", "has one focus on the COM of the planet $$C$$. Denote the foci of the ellipse $$C$$ (center of the planet) and $$F$$. For the ellipse, we know that $$CL+LF = 2a$$ ($$CL$$ is the distance between $$C$$ and $$L$$, and likewise for $$LF$$ etc.), and $$CD+DF=2a$$. So the second focus point must satisfy $$LF-DF=CD-CL$$ A hyperbola, with foci $$L$$ and $$D$$, also shown in the figure. The ellipse's semi-major axis length is $$a=\\frac{1}{2}(CD+DF)$$, so to minimize this we must minimize distance $$DF$$ on this hyperbola (CD is fixed by the problem), so $$F$$ must clearly lie on the line $$LD$$, as shown in the figure. An ellipse has the property that any rays (light rays for example) that come from one focus reflect to the other focus. In other words, the angle bisector is perpendicular to the ellipse. Using this, we get an expression of the angle: $$\\frac{1}{2}\\gamma+\\beta+\\alpha=90^\\circ$$. Next to this, we have that $$\\gamma+\\beta=90^\\circ$$. Combining these, we find the optimal launch angle: $$\\alpha = \\frac{1}{2}\\gamma$$ The velocity you need is also easily calculated, since the semi-major axis length is just $$4a = (CL+LF)+(CD+DF) = CL+LD+CD$$, and hence the velocity is $$v_0 = \\sqrt{\\frac{GM}{R}}\\sqrt{2-\\frac{4R}{CL+LD+CD}}$$ # Destination and launch both on the ground In the case that $$D$$ is also on the ground ($$CL=CD=R$$), the formulae become nice", "\\begin{align} b &= a \\sqrt{1 - e^2}, \\\\ a \\ell &= b^2. \\end{align} } A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping $\\displaystyle{ \\ell }$ fixed. Thus a and b tend to infinity, a faster than b. The length of the semi-minor axis could also be found using the following formula:[2] $\\displaystyle{ 2b = \\sqrt{(p + q)^2 -f^2}, }$ where f is the distance between the foci, p and q are the distances from each focus to any point in the ellipse. ## Hyperbola The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches; if this is a in the x-direction the equation is: $\\displaystyle{ \\frac{\\left( x-h \\right)^2}{a^2} - \\frac{\\left( y-k \\right)^2}{b^2} = 1. }$ In terms of the semi-latus rectum and the eccentricity we have $\\displaystyle{ a={\\ell \\over e^2 - 1 }. }$ The transverse axis of a hyperbola coincides with the major axis.[3] In a hyperbola, a conjugate axis or minor axis of length $\\displaystyle{ 2b }$, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter", "the parabola using the information given. 255. Focus $(4,0)(4,0)$ and directrix $x=−4x=−4$ 256. Focus $(0,−3)(0,−3)$ and directrix $y=3y=3$ 257. Focus $(0,0.5)(0,0.5)$ and directrix $y=−0.5y=−0.5$ 258. Focus $(2,3)(2,3)$ and directrix $x=−2x=−2$ 259. Focus $(0,2)(0,2)$ and directrix $y=4y=4$ 260. Focus $(−1,4)(−1,4)$ and directrix $x=5x=5$ 261. Focus $(−3,5)(−3,5)$ and directrix $y=1y=1$ 262. Focus $(52,−4)(52,−4)$ and directrix $x=72x=72$ For the following exercises, determine the equation of the ellipse using the information given. 263. Endpoints of major axis at $(4,0),(−4,0)(4,0),(−4,0)$ and foci located at $(2,0),(−2,0)(2,0),(−2,0)$ 264. Endpoints of major axis at $(0,5),(0,−5)(0,5),(0,−5)$ and foci located at $(0,3),(0,−3)(0,3),(0,−3)$ 265. Endpoints of minor axis at $(0,2),(0,−2)(0,2),(0,−2)$ and foci located at $(3,0),(−3,0)(3,0),(−3,0)$ 266. Endpoints of major axis at $(−3,3),(7,3)(−3,3),(7,3)$ and foci located at $(−2,3),(6,3)(−2,3),(6,3)$ 267. Endpoints of major axis at $(−3,5),(−3,−3)(−3,5),(−3,−3)$ and foci located at $(−3,3),(−3,−1)(−3,3),(−3,−1)$ 268. Endpoints of major axis at $(0,0),(0,4)(0,0),(0,4)$ and foci located at $(5,2),(−5,2)(5,2),(−5,2)$ 269. Foci located at $(2,0),(−2,0)(2,0),(−2,0)$ and eccentricity of $1212$ 270. Foci located at $(0,−3),(0,3)(0,−3),(0,3)$ and eccentricity of $3434$ For the following exercises, determine the equation of the hyperbola using the information given. 271. Vertices located at $(5,0),(−5,0)(5,0),(−5,0)$ and foci located at $(6,0),(−6,0)(6,0),(−6,0)$ 272. Vertices located at $(0,2),(0,−2)(0,2),(0,−2)$ and foci located at $(0,3),(0,−3)(0,3),(0,−3)$ 273. Endpoints of the conjugate axis located at $(0,3),(0,−3)(0,3),(0,−3)$ and foci located $(4,0),(−4,0)(4,0),(−4,0)$ 274. Vertices located at $(0,1),(6,1)(0,1),(6,1)$ and focus located at $(8,1)(8,1)$ 275. Vertices located", "the length of the major axis (10) so: . $a = 5 \\quad\\Rightarrow\\quad a^2 = 25$ . all correct! C is x in the foci, so it's 1 . . . . no $c$ is the distance from the center to a focus: . $c = 3$ Give it another try . . . 3. Ahh, that sounds more correct. I don't know why I thought of it as simply X in a foci. Hmm. So then it's: C=3 C^2 = 9 9 = 25 - b^2 b^2 = 16 b= 4 So, then equation would be: (x+1)^2/16 + (y-3)^2/25 = 1 Correct now?", "be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0,\\pmb) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length . Denoting the semi-major axis length (distance from the center to a vertex) as, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: x2 a2 - y2 b2 =1. The semi-minor axis is also the distance from one of focuses of the hyperbola to an asymptote. Often called the impact parameter, this is important in physics and astronomy, and measure the distance a particle will miss the focus by if its journey is unperturbed by the body at the focus. The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: b=a\\sqrt{e2-1}. [3] Note that in a hyperbola can be larger than .http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node27.html ## Astronomy ### Orbital period In astrodynamics the orbital period of a small body orbiting a central body in a circular or elliptical orbit is:[4] T=2\\pi\\sqrt{a3\\over\\mu} where: is the length of the orbit's semi-major axis \\mu is the standard", "clearer if you start by saying let$(a,b)$... 0 Take a look at this: http://gieseanw.wordpress.com/2013/07/19/an-analytic-solution-for-ellipse-and-line-intersection/ Basic idea: starting from writing down the formular of ellipse: (x/a)^2 + (y/b)^2 = 1, where a is the semi-major axis, b is the semi-minor axis. Then write down the formula for a line: (x2-x1)*(y-y1) = (y2-y1)/(x-x1) then representing ... 0 1. You are correct that the center is at$(1, 0)$, and so the focal length is indeed$c = 3$. Recall that you can draw a rectangle, centered at the center of the hyperbola such that the asymptotes pass through the corners and the left and right sides of the rectangle are tangent to the vertices of the hyperbola. The half-dimensions of the box are$a$and ... 0 For 1: Remember that$c=\\sqrt{a^2+b^2}$, and the slopes of the asymptotes are$\\pm b/a$. For 2: Completing the square is the way to go. Try dividing everything by$2$first, though. 0 Almost any minimum or maximum will be approximately parabolic. You can calculate the Taylor series at$(x_0,y_0)$and get$y \\approx y_0+y'(x_0)(x-x_0)+y''(x_0)(x-x_0)^2\\dots$Since you are at a minimum or maximum, you have$y'(x_0)=0$so that term goes away. The higher order terms are small when you are near the minimum or maximum. In your case of ... 2 The comments by the OP indicate that he may be confused about the nature of the branches", "ellipse (a point halfway between and on the line running between the foci) to the edge of the ellipse. The semi-minor axis is half of the minor axis. The minor axis is the longest line segment perpendicular to the major axis that connects two points on the ellipse's edge. \\ell , as follows: \\begin{align}b&=a\\sqrt{1-e2}\\ a\\ell&=b2.\\end{align} A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping \\ell fixed. Thus and tend to infinity, faster than . The length of the semi-minor axis could also be found using the following formula,[1] 2b=\\sqrt{(p+q)2-f2} where is the distance between the foci, and are the distances from each focus to any point in the ellipse. ## Hyperbola The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches; if this is in the x-direction the equation is: \\left(x-h\\right)2 a2 - \\left(y-k\\right)2 b2 =1. In terms of the semi-latus rectum and the eccentricity we have a={\\ell\\over1-e2}. The transverse axis of a hyperbola coincides with the major axis.[2] In a hyperbola, a conjugate axis or minor axis of length 2b , corresponding to the minor axis of an ellipse, can", "type was (5,-2) question again is. Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,2). ... Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,-2). The distance between the center and the focus is 2, that means e = 2. Therefore: $2^2+b^2=3^2~\\implies~b=\\sqrt{5}$ Plug in all the values you know into the general equation of an ellipse: $\\frac{(x-5)^2}{3^2}+\\frac{(y+2)^2}{(\\sqrt{5})^2}=1 ~\\implies~ \\frac{(x-5)^2}{9}+\\frac{(y+2)^2}{5}=1$", "intersection) to either focus, let's say ##F##. We see that ##c = a - \\frac{p}{1 + e} = \\frac{ep}{1 - e^2} = ea##. If ##b## is the semi-minor axis - e.g. in the upper half of the ellipse in the above diagram, and if we draw a line that connects the focus ##F## to the point that semi-minor axis ##b## intersects the ellipse, we have a right triangle. Its two perpendicular sides are ##c## and ##b## and its hypotenuse is ##a##. Using Pythagorean theorem and utilizing the fact that ##c = ea##, as we saw above, we find that ##b = a \\sqrt{1 - e^2}##. Now, using the expression we obtained before for ##a## we have ##b = \\frac{p}{1 - e^2} \\cdot \\sqrt{1 - e^2}## So, the (semi)axes ##a## and ##b## can indeed be written in a way, such that, if we square the final expressions we found, we have the expressions asked in the OP.", "is $\\dfrac{(x-h)^2}{a^2} + \\dfrac{(y-k)^2}{b^2} = 1$ Some remarks about Equation \\ref{standardellipse} are in order. First note that the values $$a$$ and $$b$$ determine how far in the $$x$$ and $$y$$ directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if $$a > b$$, then we have an ellipse whose major axis is horizontal, and hence, the foci lie to the left and right of the center. In this case, as we've seen in the derivation, the distance from the center to the focus, $$c$$, can be found by $$c = \\sqrt{a^2 - b^2}$$. If $$b > a$$, the roles of the major and minor axes are reversed, and the foci lie above and below the center. In this case, $$c = \\sqrt{b^2 - a^2}$$. In either case, $$c$$ is the distance from the center to each focus, and $$c = \\sqrt{ {bigger denominator} - {smaller denominator}}$$. Finally, it is worth mentioning that if we take the standard equation of a circle, Equation \\ref{standardcircle}, and divide both sides by $$r^2$$, we get The Alternate Standard Equation of a Circle: The equation of a circle with center $$(h,k)$$ and radius $$r >0$$ is $\\dfrac{(x-h)^2}{r^2} + \\dfrac{(y-k)^2}{r^2} = 1$ Notice the similarity between Equation \\ref{standardellipse} and Equation \\ref{standardcirclealternate}. Both equations involve a", "The focus is farther from the center than the vertex, so that works out. However, notice that the a in the eccentricity formula may not be a from the hyperbola formula. We want the distance to the vertex, which is given by b in a vertical hyperbola. Proceed with caution. In an ellipse, the semi-major axis is the geometric mean of the distance from the center to either focus and the distance from the center to either directrix.. The semi-minor axis of an ellipse runs from the center of the ellipse (a point halfway between and on the line running between the foci) to the edge of the ellipse.The semi-minor axis is half of the minor axis. The equation of a horizontal hyperbola in standard form is where the center has coordinates the vertices are located at and the coordinates of the foci are where ; The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0. 11-10-2013 · Thus, Therefore, Also, note that if P is to the left of the line x = –a, then In that case PF – PG = 2 a So, any point that satisfies lies on hyperbola", "between the two branches. Thus it is the distance from the center to either vertex of the hyperbola. A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping $\\displaystyle{ \\ell }$ fixed. Thus a and b tend to infinity, a faster than b. The major and minor axes are the axes of symmetry for the curve: in an ellipse, the minor axis is the shorter one; in a hyperbola, it is the one that does not intersect the hyperbola. ## Ellipse The equation of an ellipse is $\\displaystyle{ \\frac{(x - h)^2}{a^2} + \\frac{(y - k)^2}{b^2} = 1, }$ where (hk) is the center of the ellipse in Cartesian coordinates, in which an arbitrary point is given by (xy). The semi-major axis is the mean value of the maximum and minimum distances $\\displaystyle{ r_\\text{max} }$ and $\\displaystyle{ r_\\text{min} }$ of the ellipse from a focus — that is, of the distances from a focus to the endpoints of the major axis $\\displaystyle{ a = \\frac{r_\\text{max} + r_\\text{min}}{2}. }$ In astronomy these extreme points are called apsides.[1] The semi-minor axis of an ellipse is the geometric mean of these distances: $\\displaystyle{ b = \\sqrt{r_\\text{max} r_\\text{min}}. }$ The", "are the distances from each focus to any point in the ellipse. ## Hyperbola The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches; if this is a in the x-direction the equation is:[ citation needed ] ${\\displaystyle {\\frac {\\left(x-h\\right)^{2}}{a^{2}}}-{\\frac {\\left(y-k\\right)^{2}}{b^{2}}}=1.}$ In terms of the semi-latus rectum and the eccentricity we have ${\\displaystyle a={\\ell \\over e^{2}-1}.}$ The transverse axis of a hyperbola coincides with the major axis. [3] In a hyperbola, a conjugate axis or minor axis of length ${\\displaystyle 2b}$, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints ${\\displaystyle (0,\\pm b)}$ of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: ${\\displaystyle {\\frac {x^{2}}{a^{2}}}-{\\frac {y^{2}}{b^{2}}}=1.}$ The semi-minor axis is also the distance from one of focuses of", "class. What do you mean by them? 7. Jun 13, 2008 ### temaire Do you mean the center? 8. Jun 13, 2008 ### konthelion The focii for an ellipse is the point that lies on the major axis(the longer side/axis) of the ellipse. There are two focii in this ellipse. (h+c,k) and (h-c,k). In an ellipse(for both vertial and horizontal ellipses), $$b^2=a^2-c^2$$, where a is always the large axis and b is the smaller axis. In a parabola, the there is only one focus. Since this parabola opens down, then the focus is at (h,-c+k).", "is on the x-axis, that is, $k=0$, then $I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2) = ...$ if the center is not on the x-axis, i.e., $k\\neq 0$ then $I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2 -2ky) = ...$ hope you can do it from here..", "8. Find the lengths of transverse axis and conjugate axis. ###### Topic Notes hyperbola: the difference of the distances from any point on a hyperbola to each focus is constant and equal to the transverse axis $2a$. ellipse: the sum of the distances from any point on an ellipse to each focus is constant and equal to the major axis $2a$. $c= \\sqrt{a^2 - b^2}$ $c$: distance from the center to a focus $e= \\frac{c}{a}$ $e$: eccentricity; the larger the value of $e$, the straighter the hyperbola", "is centered at the origin, the center is (0,0) and when it is centered outside the origin, the center is $latex (h, k)$. ### Vertices The vertices are the endpoints of the transverse axis. These points are located at the intersection of the hyperbola and the transverse axis. We use $latex V$ to represent the vertices and we can distinguish them using $latex V_{1}$ and $latex V_{2}$ or $latex V$ and $latex V’$. ### Semi-major axis The semi-major axis is a line segment that connects the center to a vertex of the hyperbola. Usually, we denote its length using a. ### Semi-minor axis The semi-minor axis is a line segment that is perpendicular to the semi-major axis. Usually, we denote its length using b. ### Asymptotes The asymptotes represent the behavior of the hyperbolas. The asymptotes are the lines that are very close to the branches of the hyperbola but never touch it.", "the feature range is 0 then {$v=U_v(d)$} If the feature range is 1 then {$v$} is an inner object. !!!!Feature Value Range %left%{$\\displaystyle [0, 1]$} !!!Distance to Superobject Center %gray%''Object Features > Geometry > To Superobject > Distance to Superobject Center'' The distance of an image object's center to the center of its superobject. This might not be the shortest distance between the two points, since the way to the center of the superobject must be within the borders of the superobject. !!!!Expression %left%{$\\displaystyle d_g( v,U_v(d))$} is the distance of {$v$} to the center of gravity of the superobject {$U_v(d)$} !!!!Feature Value Range %left%{$\\displaystyle [0, sx \\times sy]$} !!!Elliptic Distance to Superobject Center %gray%''Object Features > Geometry > To Superobject > Elliptic Distance to Superobject Center'' Distance of objects to the center of the superobject. !!!!Expression %left%{$\\displaystyle d_e( v,U_v(d))$} Figure 101: Distance between the distance from the superobject's center to the center of a sub-objectt. !!!!Feature Value Range Typically {$[0, 5]$} !!!Is End of Superobject %gray%''Object Features > Geometry > To Superobject > Is End of Superobject'' This feature is true for two image objects {$a$} and {$b$} if following conditions are true: *{$a$} and {$b$} are sub-objects of the same superobject. *{$a$} is the image object with the maximum distance to the superobject. *{$b$} is the image" ]

[ "Examples a) Divide $3x^3-5x+2$ by $x+4$. b) Find the remainder when $5x^4-2x^3-4x+2$ is divided by $x-2$. c) Divide $-x^5-5x^3-x^2+2$ by $x+2$ d) Determine whether $x-1$ is a factor of $3x^3-5x^2-x+3$.", "frac{f^{6} (-1) }{6!} (x+1)^2 + frac{f^{7} (-1) }{7!} (x+1)^3$$ For more context into the idea behind finding remainders, see the answer I made for the question I asked myself over here", "# How do you find the quotient and remainder when (x^3-6x^2-9x+3)/(x-3)? Long division is the best way to find the quotient ( ${x}^{2} - 3 x - 18$) and the remainder ( $- 51$)", "By actual division find the quotient and the remainder when (x4 + 1) is divided by (x - 1). The division is as follows. $$\\text { Quotient }=x^{3}+x^{2}+x+1 \\\\ \\text { Remainder }=2$$", "# How do you find the quotient and remainder when (x^3-6x^2-9x+3)/(x-3)? Nov 11, 2015 Long division is the best way to find the quotient ( ${x}^{2} - 3 x - 18$) and the remainder ( $- 51$)", "Check whether 7 + 3x is a factor of 3 x 3 + 7x. Solution: Let us divide $\\left(3 x^{3}+7 x\\right)$ by $(7+3 x)$. If the remainder obtained is 0 , then $7+3 x$ will be a factor of $3 x^{3}+7 x$. By long division, As the remainder is not zero, therefore, $7+3 x$ is not a factor of $3 x^{3}+7 x$. Editor", "# Check whether 7 + 3x is a factor of 3 x 3 + 7x. Solution: Let us divide $\\left(3 x^{3}+7 x\\right)$ by $(7+3 x)$. If the remainder obtained is 0 , then $7+3 x$ will be a factor of $3 x^{3}+7 x$. By long division,", "# Find the remainder when $x^{3}-a x^{2}+6 x-a$ is divided by x − a. Solution: By long division, Therefore, when $x^{3}-a x^{2}+6 x-a$ is divided by $x-a$, the remainder obtained is $5 a$.", "# 2. Find the remainder when $x^3 - ax^2 + 6x - a$ is divided by $x - a$. When we divide $x^3 - ax^2 + 6x - a$ by $x - a$ Therefore, remainder is $5a$", "2. Find the remainder when $$x^3 - ax^2 + 6x - a$$ is divided by x - a. Let $$p(x) = x^3 - ax^2 + 6x - a$$ The zero of $$x - a$$ is x = a i.e. (x - a = 0) So, $$p(a) = {a}^3 - a{a}^2 + 6{a} - a$$ i.e. $$p(a) = a^3 - a^3 + 6a - a$$ i.e.$$p(a) = 5a$$", "remainder when divided by $$x-2$$.", "is remainder It is given that $x^2 + 2x + k$ is a factor hence remainder = 0 $x(7k + 21) + (6 + 8k + 2k^2) = 0.x + 0$ By comparing L.H.S. and R.H.S. 7k + 21 = 0 k = – 3 ….(1) $2k^2 + 8k + 6 = 0$ $2k^2 + 2k + 6k + 6 = 0$ 2k(k + 1) + 6(k + 1) = 0 (k + 1) (2k + 6) = 0 k = –1, –3 …..(2) From (1) and (2) k = –3 Hence, $2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k) (2x^2 - 3x - 8 - 2k)$ Put k = –3 $= (x^2 + 2x - 3) (2x^2 - 3x - 8 + 6)$ $= (x^2 + 2x - 3) (2x^2 - 3x - 2)$ $= (x^2 + 3x - x - 3) (2x^2 - 4x + x - 2)$ $= (x(x + 3) - 1(x + 3)) (2x(x - 2) + 1(x - 2))$ $= (x + 3)(x -1)(x - 2) (2x + 1)$ x + 3 = 0 x – 1 = 0 x – 2 = 0 2x + 1 = 0 x = –3 x = 1 x = 2 2x = – 1 $x=\\frac{-1}{2}$ Here zeroes", "# Remainder Find the remainder when $$\\sqrt { { 2 }^{ 2048 }+3({ 2 }^{ 1025 })+9 }$$ is divided by 6. ×", "# Math Help - remainder 1. ## remainder the remainder obtained when $2x^3 + ax^2 - 6x + 1$ is divided by $(x+2)$ is twice the remainder obtained when the same expression is divided by $(x-1)$. Find the value of $a$. 2. Let $P(x)=2x^3+ax^2-6x+1$ The remainder when dividing by $x+2$ is $P(-2)$ The remainder when dividing by $x-1$ is $P(1)$ Then we have $P(-2)=2P(1)$ Now find a.", "6 .= .4a + 2b + 14 If f(x) is divided by x + 3, the remainder is f(-3). . . f(-3) .= .(-3)³ + a(-3)² + b(-3) + 6 .= .9a - 3b - 21 If these remainders are equal: .4a + 2b + 14 .= .9a - 3b - 21 . . a .= .b + 7", "2 }-\\frac { 6 }{ 5 } x$ 3 1 4 ${ m }^{ 4 }-\\frac { 6 }{ 7 } { m }^{ 2 }-6m+2.7$ 5 $\\frac { { t }^{ 3 } }{ \\sqrt { 6 } } +\\frac { { t }^{ 4 } }{ 4 } -t$ 6. Find the quotient and remainder when 5x3 - 9x2 + 10x + 2 is divided by x + 2 using synthetic division 7. Find the quotient and remainder when 4x3 + 6x2 + 7x + 2 is divided by x - 2 8. Find the quotient and remainder when 5x3 + 7x2 + 3x + 2 is divided by 3x + 2 9. Factorise 2x3- x2 - 12x - 9 into linear factors 10. Prove that x -1 is a factor x5 - 45x4 + 36x3 + 45x2 - 36x-1", "# But how? Find the remainder when the number $$(2^{100} + 3^{100} + 4^{100}+5^{100} )$$ is divided by 7. ×", "# Can you find the remainder? What is the remainder when $$1^2+2^2+3^2+...+2012^2+2013^2$$ is divided by $$7$$? ×", "# Error in predicting remainders I recently came across the problem What is the remainder when $x^5 + 2x^4 -3x^2 + 2x -4$ is divided by $(x^2 + 2x)$? a. $x^3 - 3$ b. $8x - 4$ Correct answer c. 4 d. $2x-4$ e. $20x$ And I attempted to solve using the remainder theorem (the textbook advised plugging in numbers, instead of actually using synthetic/polynomial division to solve for multiple choice questions). However, I soon found that smaller numbers didn't give me accurate results. My method of solving (and the way the textbook told me to solve) was to choose a random value for $x$. I chose 3. Then, plug 3 into both equations: $3^5 + 2(3)^4 -3(3)^2 + 2(3) -4$ is divided by $((3)^2 + 2(3))$ Then, I solved: $\\frac{3^5 + 2(3)^4 -3(3)^2 + 2(3) -4}{(3)^2 + 2(3)} = 25\\frac{2}{3}$ Then I took the remainder ($\\frac{2}{3}$) and multiplied it by the divisor to get the remainder: $\\frac{2}{3} * ((3)^2 + 2(3)) = \\frac{1}{3}$ Therefore, when you set $x = 3$, you get a remainder of $\\frac{1}{3}$. The textbook then told me to plug in the value I set x to into all the choices, until I found a value that was equal to the remainder, so when $x=3$: a. $x^3 - 3$ -> 24 b. $8x -", "expression when $9x^2y+12x^3y^2-15xy^3$ is divided by $6xy$?" ]

[ "$\\boxed{109}$. ## Solution 5 We can find zeroes of the polynomial by making the first given equation $0 = 0$. Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$, respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \\cdot P(2) = 0$, which means $P(2)$ is not necessarily $0$. If $P(2) = 0$, then plugging in $2$ to the equation yields $P(3) = 0$, plugging in $3$ to the equation yields $P(4) = 0$, and so on, a contradiction of \"nonzero polynomial\". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$, so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$. We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$. $P(2) = 6a$ and $P(3) = 24a$, so: $$(6a)^2 = 24a \\implies 36a^2 = 24a \\implies a = \\frac23$$ Plugging in $\\frac72$ into the polynomial $\\frac23(x + 1)x(x - 1)$ yields $\\frac{105}{4}$. $105 + 4 = \\boxed{109}$. ## Solution 6", "with algebraic approach or plug-in method: Algebraic approach: (1)+(2) Important tip: x^3-y^3 can be factored as follows: $$x^3-y^3=(x-y)(x^2+xy+y^2)$$. Apply this factoring to $$b-a$$ --> $$b-a=(c + 1)^3-c^3=(c+1-c)(c^2+2c+1+c^2+c+c^2)=3c^2+3c+1=3(c^2+c)+1$$ --> remainder upon division this expression by 3 is 1. Sufficient. Plug-in method approach: (1)+(2) try some numbers for a and b: $$a=c^3=1$$ --> $$b=(c+1)^3=8$$ and --> $$b-a=(c + 1)^3-c^3=7$$ --> remainder upon division 3 is 1; $$a=c^3=8$$ --> $$b=(c+1)^3=27$$ and --> $$b-a=(c + 1)^3-c^3=19$$ --> remainder upon division 3 is 1; $$a=c^3=27$$ --> $$b=(c+1)^3=64$$ and --> $$b-a=(c + 1)^3-c^3=37$$ --> remainder upon division 3 is 1; $$a=c^3=64$$ --> $$b=(c+1)^3=125$$ and --> $$b-a=(c + 1)^3-c^3=61$$ --> remainder upon division 3 is 1; ... It seem that there is some kind of pattern and we can safely assume that in all other cases remainder will also be 1. Sufficient. Hope it helps. Hi Bunuel, Would it be possible for you to provide OA to the question and adjust the difficulty level, so that we guys can actually have some appreciation for the question? Thanks! _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13864 Followers: 589 Kudos [?]: 167 [0], given: 0 Re: If a, b, and c are positive integers, what is the remainder [#permalink] ### Show Tags 15 Jul 2015, 22:34 Hello from the GMAT Club BumpBot! Thanks to another", "x = -1: $y = (-1 + 1)(-1-8)(-1-6) = 0*-9*-7 = 0$ Plugging in x = 8: $y = (8+1)(8-8)(8-6) = 9*0*2 = 0$ Plugging in x = 6: $y = (6+1)(6-8)(6-6) = 7*-2*0 = 0$ This is quite fascinating. It seems that this polynomial will return 0 for all the other points. We did this on purpose and you will see why soon. Now let’s try plugging in for x = 3 Plugging in x = 3: $y = (3+1)(3-8)(3-6) = 4*-5*-3 = 60$ We can see that the point (3,60) lies on this polynomial, but we want the point (3,4), so to adjust for that we can multiply the polynomial by $\\frac{4}{60}$. So our new polynomial is $y =\\frac{4}{60}(x+1)(x-8)(x-6)$ Wow! So we have created a polynomial that has (3,4) on it and whenever we plug in for the x-coordinates of the other points, we get 0. So why is this important? Well consider making a similar polynomial for another point. Let’s just choose (6,3). If we go through the same process of making all the other x-coordinates of the other points but (6,3) equal to 0, we get $y =(x+1)(x-8)(x-3)$ Plugging in x = 6: $y = (6+1)(6-8)(6-3) = 7(-2)(-3) = 42$ $y = \\frac{3}{-42}(x+1)(x-8)(x-3)$ Well let’s try adding our new polynomial to our original polynomial.", "q(x)(x+3)(x+5)+ax+b$$ Plug $x=-3$ and $x=-5$ to find $a$ and $b$.", "< 1 + 9a_3a_4$$ consequently we obtain that $a_2 \\leq 9/2$ so $a_2 = 4$. Plugging this back in we see that $$8a_3a_4 = 1 + 12a_3 + 12a_4 + 3a_3a_4$$ or rearranging $$a_4 = \\frac{12a_3 + 1}{5a_3 - 12}$$ Combining this with $a_3 \\geq 5$ we see that $a_4 \\leq 61/13 < 6$, a contradiction. This proves that in fact we must have $$a_2a_3a_4 = 1 + a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4$$ Now, we use the fact that $a_1 | 1 + a_2a_3a_4$ to conclude that $$a_1 | 2 + a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4$$ and so $a_1 = 2$. Again we plug this is to obtain $$a_2a_3a_4 = 1 + 2a_2a_3 + 2a_2a_4 + 2a_3a_4 < 1 + 6a_3a_4$$ So we have $a_2 \\leq 6$. Since $a_2$ and $a_1$ are relatively prime we get that $a_2 = 3$ or $a_2 = 5$. Suppose that $a_2 = 5$ and plug this in to get $$5a_3a_4 = 1 + 10a_3 + 10a_4 + 2a_3a_4$$ This implies that $$a_4 = \\frac{10a_3 + 1}{3a_3 - 10}$$ Combining this with the fact that $a_3 \\geq 7$ ($gcd(a_1,a_3) = 1$) we see that $a_4 \\leq 71/11 < 7$, a contradiction. Thus we must have $a_2 = 3$. Plugging this in we get $$3a_3a_4 = 1 + 6a_3 + 6a_4 + 2a_3a_4$$ Thus", "and $(x-2)^3$ (which is their product in this case), so one gets $$4 x^4-27 x^3+66 x^2-65x+24.$$ - An alternative: Note that if $f(x)$ satisfies the 2 properties, so does $f(x)-q(x-1)^2(x-2)^3, q \\in \\mathbb{Q}$. Thus the polynomial of minimal degree has degree 4. We have $f(x)=(x-2)^3(ax+b)+3x$. Note that $1$ is a repeated root of $f(x)-2x$, so $f(1)=f'(1)=2$. $2=f(1)=(-1)^3(a+b)+3$ so $a+b=1$. $f'(x)=3(x-2)^2(ax+b)+a(x-2)^3+3$ so $2=f'(1)=3(-1)^2(a+b)+a(-1)^3+3=6-a$ so $a=4, b=-3$. Thus $f(x)=(x-2)^3(4x-3)+3x$ - Let $\\rm\\ \\hat f = f\\!-\\!3x = (x\\!-\\!2)^3 g.\\:$ $\\rm\\: (x\\!-\\!1)^2\\!\\mid x\\!+\\!\\hat f = x \\!+\\! (x\\!-\\!2)^3 g =: h,\\:$ so $\\rm\\ 0 = h(1) = 1\\!-\\!g(1),\\,$ $\\rm\\,0 = h'(1) = 1\\!+\\!3g(1)\\!-\\!g'(1) = 4\\!-\\!g'(1).\\,$ So min $\\rm deg\\,g = 1.$ $\\rm\\,g(1)\\! = 1,\\, g'(1)\\! = 4\\,$ $\\Rightarrow$ $\\rm g = 4x\\!-\\!3.$ -", "so you need a third degree polynomial. $T(n) = an^3+bn^2+cn+d$ From the diagram: $T(1) = 1$ $T(2) = 4$ $T(3) = 10$ $T(4) = 20$ Plugging in: $1 = a+b+c+d$ (Equation 1) $4 = 8a+4b+2c+d$ (Equation 2) $10 = 27a+9b+3c+d$ (Equation 3) $20 = 64a+16b+4c+d$ (Equation 4) Now, we solve the system of four equations. Subtract the first equation from the next three: $3 = 7a+3b+c$ (Equation 5) $9 = 26a+8b+2c$ (Equation 6) $19 = 63a+15b+3c$ (Equation 7) Subtract 2x Equation 5 from Equation 6 and 3x Equation 5 from Equation 7: $3 = 12a+2b$ (Equation 8) $10 = 42a+6b$ (Equation 9) Subtract 3x Equation 8 from Equation 9: $1 = 6a$ This gives $a = \\dfrac{1}{6}$. We plug that into either Equation 8 or Equation 9. Let's plug it into Equation 8: $3 = 2+2b \\Longrightarrow b = \\dfrac{1}{2}$ Now, we plug both $a$ and $b$ into Equation 5: $3 = \\dfrac{7}{6}+\\dfrac{3}{2}+c \\Longrightarrow c = \\dfrac{2}{6} = \\dfrac{1}{3}$ Finally, we plug all four into equation #1 to get $d$: $1 = \\dfrac{1}{6}+\\dfrac{1}{2}+\\dfrac{1}{3}+d \\Longrightarrow d=0$ Thus, $T(n) = \\dfrac{1}{6}n^3+\\dfrac{1}{2}n^2+\\dfrac{1}{3}n$ There are many ways of finding this pattern. This is just one of many. 3. ## Re: Modelling tetrahedral numbers If by \"x\" you mean what I would call \"layers\", $\\displaystyle 4^x$ clearly does NOT work. $\\displaystyle 4^0= 1$", "any number less than 4 is a solution for x (3.9, 0, -986). ### Plugging in a solution The other method you can use is to plug in a proposed number for x and see if the equation works. If you have the equation: $x - 3 < 7 * 2$ You can pick a number for x and see if the equation works. So, for example, we could choose 10. Plugging that in gives us: $10 - 3 < 7 * 2$ Working that through, we get: $7 < 14$ 7 is less than 14, so 10 is a correct solution. If we chose 100 as the number to plug in we'd get: $100 - 3 < 7 * 2$ $97 < 14$ 97 is not less than 14, so 100 is not a correct solution. This is the simplest way to find out if a single number is a solution, but to find out a range of numbers that work, you should solve the problem to its simplest form first. ## Examples $x + 7 \\leq 13$ $x + 7$ $\\leq$ $13$ $(7) + 7$ $\\leq$ $13$ $14$ $\\leq$ $13$", "this in two ways. First, we can combine the terms and get $$8x + 8 = (A + B)x + 6A - 2B$$ Because x and 1 are linearly independent, we can solve a system of equations for A and B: $$\\begin{cases}8 = A + B\\\\8 = 6A - 2B\\end{cases}$$ There's a better way! If we start with $8x + 8 = A(x + 6) + B(x-2)$, we can plug in $x = -6$ and $x = 2$ and solve for A and B that way. $$x = -6 \\to -40 = -8B \\to B = 5$$ $$x = 2 \\to 24 = 8A \\to A = 3$$ $$\\dfrac{8x + 8}{x^2 + 4x - 12} = \\dfrac{A}{x-2} + \\frac{B}{x+6} = \\boxed{\\dfrac{3}{x-2} + \\frac{5}{x+6}}$$ # Repeated Root Example MathJax TeX Test Page $$\\dfrac{19-3x}{(x-4)^2}$$ If we factor the denominator we only have $(x-4)$, so what we have to do is to have a fraction with just a number divided by $(x-4)^2$. It will look like this: $$\\dfrac{19-3x}{(x-4)^2} = \\frac{A}{x-4} + \\frac{B}{(x-4)^2}$$ We do the same thing as before: multiply through by $(x-4)^2$ $$19-3x = A(x-4) + B$$ When $x = 4, B = 7$. From that, you can see that $12 - 3x = A(x-4)$, so A = -3. However, there may be other roots, so in general, you plug", "transform the equation again to solve for x: $x \\equiv a + 5\\mathbb Z$ where $a = 1 + 2 \\mathbb Z$ and $a \\in \\mathbb Z_5$ Note that there is no modulus on the left side of the equation now, so the right side of the equation is unlimited into the valid values it can have. How do we use this resulting formula though to find valid values of x? First we figure out the valid values of a. $a = 1 + 2 \\mathbb Z$ where $a \\in \\mathbb Z_5$, so it’s only valid values have to be a subset of {0,1,2,3,4}. Plugging a -1 in for Z, we get a value for a of -1, which isn’t a valid answer so we throw it away. Plugging a 0 in for Z, we get a value for a of 1. That is a valid answer so we keep it! Plugging a 1 in gives us 3 which is valid, so we keep that too. Plugging a 2 in gives us a value of 5, which isn’t valid, so we throw it away and know that we are done. Our values for a are {1,3}. $x \\equiv a + 5\\mathbb Z$ Plugging in each value of a means that we get two equations as a solution for", "$a = 1 + 2 \\mathbb Z$ and $a \\in \\mathbb Z_5$ Note that there is no modulus on the left side of the equation now, so the right side of the equation is unlimited into the valid values it can have. How do we use this resulting formula though to find valid values of x? First we figure out the valid values of a. $a = 1 + 2 \\mathbb Z$ where $a \\in \\mathbb Z_5$, so it’s only valid values have to be a subset of {0,1,2,3,4}. Plugging a -1 in for Z, we get a value for a of -1, which isn’t a valid answer so we throw it away. Plugging a 0 in for Z, we get a value for a of 1. That is a valid answer so we keep it! Plugging a 1 in gives us 3 which is valid, so we keep that too. Plugging a 2 in gives us a value of 5, which isn’t valid, so we throw it away and know that we are done. Our values for a are {1,3}. $x \\equiv a + 5\\mathbb Z$ Plugging in each value of a means that we get two equations as a solution for x. Valid values of x are the union of these two lists – both equations", "with which you want to work and try to find out values which will be required in finding $f(7)$. Solution: Functional equations need some practice. its a different kind of a topic so for beginners this might be hard to understand of how one can think of such ideas. Okay, so the questions asks us to find the $f(7)$ which clearly we cannot find it directly by just inputting $7$ in the equation. so what is a nice way to find $f(7)$ . So what we do in functional equations is plug in nice values in the equation(of course those which are allowed in the domain) so that we get to the answer. So plug in $x=1$ and $y=0$ in the functional equation. you get $f(1)f(0)=f(1)+f(1)=6 \\implies f(0)=2$ . next plug in$x=1$and $y=1$ $\\implies f(2)=7$. Observe if we plug $x=3$ and $y=4$ we can get to $f(7)$ therefore we need to find $f(4)$ and $f(3)$. Plug $x=2$and $y=1$ $\\implies f(3)=18$. Now plug $x=3$ and $y=1$ $\\implies f(4)=47$ $x=4$and $y=3$ $\\implies f(4)f(3)=f(7)+f(1) \\implies f(7)=843$ Q6.Let $A$ and $B$ be two 3 x 3 matrices such that $(A+ B) ^{2} = A^{2} + B^{2}$. Which of the following must be true? (A) $A$ and $B$ are zero matrices. (B) $AB$ is the zero matrix. (C) $(A - B)^2 = A^2", "`int (x^2 + x + 1)/(x^2 + 1)^2 dx` Evaluate the integral `int (x^2+x+1)/(x^2+1)^2 dx` To solve, apply partial fraction decomposition. To express the integrand as sum of proper rational functions, set the equation as follows: `(x^2+x+1)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2` Multiply both sides by the LCD. `x^2+x+1=(Ax+B)(x^2+1) + Cx + D` `x^2+x+1=Ax^3+Bx^2+Ax+B+Cx+D` `x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D` Express the left side as a polynomial with degree 3. `0x^3+x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D` For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other. x^3: `0=A` (Let this be EQ1.) x^2: `1=B ` (Let this be EQ2.) x: `1=A+C` (Let this be EQ3.) Constant: `1=B+D` (Let this be EQ4.) In the equations, the values of A and B are already known. So only the values of C and D have to be solved. To do so, plug-in the value of A to EQ3. `1=A+C` `1=0+C` `1=C` Also, plug-in the value of B to EQ4. `1=B+D` `1=1+D` `0=D` So the partial fraction decomposition of the integrand is: `(x^2+x+1)/(x^2+1)^2=(0x+1)/(x^2+1)+(1x+0)/(x^2+1)^2=1/(x^2+1)+x/(x^2+1)^2` Taking the integral of it result to: `int (x^2+x+1)/(x^2+1)^2 dx` `= int (1/(x^2+1)+x/(x^2+1)^2) dx` `= int 1/(x^2+1)dx + int x/(x^2+1)^2dx` For the first integral, apply the formula int 1/(u^+a^2)=1/a tan^(-1) u/a+C. For the second integral, apply u-substitution method. `u=x^2+1` `du=2x dx` `(du)/2=xdx` `= int 1/(x^2+1)dx + int 1/u^2 *", "x. $g(x) = \\Phi \\int (x) dx$ Integrating x gives us one half x squared. We can move the half outside of the phi. $g(x) = \\frac{1}{2} \\Phi x^2$ We know the phi of x squared so we can plug it in and be done! $g(x) = \\frac{1}{2}(x^2 - x)$ If we plug in 1 to get the sum of all integers from 0 to 0, we get 0. If we plug in 2 to get the sum of all integers from 0 to 1, we get 1. That is 0 + 1. Plugging in 3 for summing 0 to 2, we get 3. That is 0 + 1 + 2. Plugging in 4 gives us 6. That is 0 + 1 + 2 + 3. Plugging in 5 gives us 10. 10 gives us 45. 100 gives us 4950. 200 gives us 19900. Plugging in a million gives us 499,999,500,000! If we wanted to sum the integers between 5 and 9, we can do that too! We subtract g(5) from g(10), which is 45-10 or 35. That is 5+6+7+8+9. So, we know that summing the integers between 100 and 199 is 14,950, since we have g(100) and g(200) above, which are 4950 and 19900 respectively. ## Example 2: f(x) = x^2 Let’s find the function to sum", "and found that +1 was a solution and therefore I could write $$x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\\\ x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\\\$$ And then equated the coeffeciants For x^3, $1=A$ For x^2, $6=B-A$ For x^1, $-8=C-B$ For x^0, $1=-C$ With knowing that A=1 and C= -1 , B can be found to be 7 . And therefore I can write $x^3+6x^2-8x+1=(x-1)(x^2+7x-1)$ I'm curious why you didn't just divide the cubic by ##x-1## using either long division or, preferably, synthetic division to get the quadratic factor. Much less work...", "simplicity and put the values A) $$x^2$$ — x — 1 = 1 - 1 -1 = -1 B) $$x^2$$ — 4x + 6 = 1 -4 + 6 = 3 C) $$x^2$$ — 5x + 5 = 1 -5 +5 = 1 D) $$x^2$$ + 3x + 8 = 1 + 3 + 8 = 12 E) $$x^2$$ + 2x + 10 = Don't need to evaluate this. You already have your answer. Hence D is correct. Note: During plugging values if you have found an answer the do not continue plugging in other options. This is very important time saving tactic. Also keep in mind when a question has very generic statements like all integers satisfy \"Expression...\" It is highlly likely that plugging values is the best strategy. The question isn't mine but thanks! Re: If x is an integer, which of the following must be an even i [#permalink] 12 Aug 2018, 20:39 Display posts from previous: Sort by", "this: $$p(x)=(ax+3)$$ Then $$(4x^2+6x+3)(ax+3)=4ax^3+(6a+12)x^2+(3a+18)x+9$$. Now it has to be $$4a\\equiv 0\\mod 8$$ and $$6a+12\\equiv 0\\mod 8$$ and $$3a+18\\equiv 0\\mod 8$$. Is there such an $$a$$?. Yes indeed. For $$a=2$$ we have $$8\\equiv 0\\mod 8$$ $$24\\equiv 0\\mod 8$$ and $$24\\equiv 0\\mod 8$$. If we would fail to find this $$a$$ in this step, we would have to try with $$p(x)=(ax^2+bx+3)$$ and proceed as above, which gets more and more complicated. So it is $$(4x^2+6x+3)(2x+3)\\equiv 1\\mod 8$$ • What is $y$ ? the constant term of $p(x)$? – J. W. Tanner May 14 at 12:23 • @J.W.Tanner Yes, I meant that when we have a linear polynomial ax+y, then y has to be 3. – Cornman May 14 at 12:42 • Perhaps you should edit to make that obvious – J. W. Tanner May 14 at 12:48 • @J.W.Tanner I refered to the comments now. I knew, that it is a little dubiuos, but I thought one might still understand it, espacially with how I proceed in the calculation. I should have wrote (ax+b) first and then explain why $b\\equiv 3$, I guess. – Cornman May 14 at 13:01 • @Cornman Thanks. What values can the degree of the inverse polynomial take? – Akash Gaur May 14 at 14:07 By the idea here we find a simpler multiple $$\\,\\color{#0a0}{3^{\\large 3}}$$", "= f(3)$$ To find f(3) we just need to plug 3 in for x into the piece of our function that defines it when $$x=3$$, which is the third piece of f. $$6-a+b = 2(3)-a+b$$ $$6-a+b = 6-a+b$$ We see here another case where this equation gives us no useful information. This is because this equality will be true no matter what we plug in for a and b. Let’s move onto the left sided limit and see what we get there. $$\\lim_{x \\to 3^{-}} f(x) = f(3)$$ Now that our x value is approaching 3 from the left side, it is slightly smaller than 3 and slowly increasing. Since we are considering what our function is doing around $$x=3$$ for x values slightly smaller than 3, we will want to use the piece of f that is defined for $$2\\leq x<3$$. We already found f(3) above, so we will also plug that in. $$\\lim_{x \\to 3^{-}} ax^2-bx+3 = 6-a+b$$ Just like before, since we know $$y=ax^2-bx+3$$ is a polynomial, it is continuous everywhere and we can find this limit by plugging in $$x=3$$. $$a(3)^2-b(3)+3 = 6-a+b$$ $$9a-3b+3 = 6-a+b$$ $$10a-4b=3$$ Putting it all together Now we have another relationship that relates a and b. Alone, it isn’t very useful, but we can take the previous equation we", "Plugging the value of $B$ into $A$, we get: $A = 2 * 9 = 18$ Plugging the value of $A$ into $C$, we get: $C = 3A = 3*18 = 54$ Question 3 The graph of y = f (x) is shown above. If f (3) = k, which of the following could be the value of f (k)? A 2 B 3.5 C 4 D 4.5 Question 3 Explanation: We know that f (3) = k. Using the graph, we can see that f (3) = 4. Plugging in the value of k = 4, we get: f (k) = f (4) = 2 Question 4 2x + 4y = 16 3x − 6y = 12 If (x,y) is a solution to the system of equations above, what is the value of x + y? A 5 B 6 C 7 D 8 Question 4 Explanation: Start by simplifying each equation. Do the terms have any common factors? Notice that 2, 4, and 16 from the first equation can all be divided by 2: $x + 2y = 8$ Likewise, we can divide the second equation by 3: $x − 2y = 4$ Using the method of Elimination/Combination, we can cancel out the $2y$ term if we add both equations together: $x + 2y = 8$", "5\\mathbb Z$ where $a = 1 + 2 \\mathbb Z$ and $a \\in \\mathbb Z_5$ Note that there is no modulus on the left side of the equation now, so the right side of the equation is unlimited into the valid values it can have. How do we use this resulting formula though to find valid values of x? First we figure out the valid values of a. $a = 1 + 2 \\mathbb Z$ where $a \\in \\mathbb Z_5$, so it’s only valid values have to be a subset of {0,1,2,3,4}. Plugging a -1 in for Z, we get a value for a of -1, which isn’t a valid answer so we throw it away. Plugging a 0 in for Z, we get a value for a of 1. That is a valid answer so we keep it! Plugging a 1 in gives us 3 which is valid, so we keep that too. Plugging a 2 in gives us a value of 5, which isn’t valid, so we throw it away and know that we are done. Our values for a are {1,3}. $x \\equiv a + 5\\mathbb Z$ Plugging in each value of a means that we get two equations as a solution for x. Valid values of x are the union of these two lists" ]

[ "of $x^3+(a+1)x^2-(b-2)x-6$ , find the values of $a \\ and \\ b$ . Factorize the polynomial also. When $x = -1$ Remainder: $(-1)^3+(a+1)(-1)^2-(b-2)(-1)-6 =0$ $\\Rightarrow -1+(a+1)+(b-2)-6=0$ $\\Rightarrow a+b=8$ … … … … … … i) When $x = 2$ Remainder: $(2)^3+(a+1)(2)^2-(b-2)(2)-6 =0$ $\\Rightarrow 8+4a+4-2b+4-6=0$ $\\Rightarrow 2a-b=-5$ … … … … … … ii) Solving i) and ii) we get $a = 1 \\ and \\ b = 7$ Substituting the values in the polynomial we get $x^3+2x^2-5x-6$ Given $(x+1)$ is a factor of $x^3+2x^2-5x-6$ • $x+1 ) \\overline {x^3+2x^2-5x-6} (x^2+x-6$ • $(-) \\ \\ \\underline {x^3+x^2}$ • $x^2-5x-6$ • $(-) \\ \\ \\underline{x^2+x}$ • $-6x-6$ • $(-) \\ \\ \\underline{ -6x-6}$ • $\\times$ $x^3+2x^2-5x-6 = (x+1)(x^2+x-6)$ $= (x+1)(x^2+3x-2x-6)$ $= (x+1)[x(x+3)-2(x+3)]$ $= (x+1)(x+3)(x-2)$ Hence $x^3+2x^2-5x-6 =(x-1)(x+3)(x-2)$ $\\\\$ Question 6: If $(x-2)$ is a factor of $x^2+ax+b$ and $a+b=1$ , find the values of $a \\ and \\ b$ . When $x = 2$ Remainder: $(2)^2+a(2)+b =0$ $4+2a+b=0$ Given $a+b=1$ Therefore $4+a+(a+b)=0$ $\\Rightarrow a = -5$ Solving for $b = 6$ Hence $a = -5 \\ and \\ b = 6$ $\\\\$ Question 7: Using remainder theorem, factorize $x^3+6x^2+11x+6$ completely. For $x = -1$, Remainder: $= (-1)^3+6(-1)^2+11(-1)+6 = -1+6-11+6=0$ Hence $(x+1)$ is a factor of $x^3+6x^2+11x+6$ • $x+1 ) \\overline {x^3+6x^2+11x+6} (x^2+5x+6$ • $(-) \\ \\ \\underline {x^3+x^2}$ • $5x^2+11x+6$", "with algebraic approach or plug-in method: Algebraic approach: (1)+(2) Important tip: x^3-y^3 can be factored as follows: $$x^3-y^3=(x-y)(x^2+xy+y^2)$$. Apply this factoring to $$b-a$$ --> $$b-a=(c + 1)^3-c^3=(c+1-c)(c^2+2c+1+c^2+c+c^2)=3c^2+3c+1=3(c^2+c)+1$$ --> remainder upon division this expression by 3 is 1. Sufficient. Plug-in method approach: (1)+(2) try some numbers for a and b: $$a=c^3=1$$ --> $$b=(c+1)^3=8$$ and --> $$b-a=(c + 1)^3-c^3=7$$ --> remainder upon division 3 is 1; $$a=c^3=8$$ --> $$b=(c+1)^3=27$$ and --> $$b-a=(c + 1)^3-c^3=19$$ --> remainder upon division 3 is 1; $$a=c^3=27$$ --> $$b=(c+1)^3=64$$ and --> $$b-a=(c + 1)^3-c^3=37$$ --> remainder upon division 3 is 1; $$a=c^3=64$$ --> $$b=(c+1)^3=125$$ and --> $$b-a=(c + 1)^3-c^3=61$$ --> remainder upon division 3 is 1; ... It seem that there is some kind of pattern and we can safely assume that in all other cases remainder will also be 1. Sufficient. Hope it helps. Hi Bunuel, Would it be possible for you to provide OA to the question and adjust the difficulty level, so that we guys can actually have some appreciation for the question? Thanks! _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13864 Followers: 589 Kudos [?]: 167 [0], given: 0 Re: If a, b, and c are positive integers, what is the remainder [#permalink] ### Show Tags 15 Jul 2015, 22:34 Hello from the GMAT Club BumpBot! Thanks to another", "21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$. So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$. For the specific case when $b = 6$, we get that $a^2+3a+4 = 7(7n^2+12n+3n)+58$ $= 7(7n^2+12n+3n+8)+2$. So the remainder when $a^2+3a+4$ is divided by $7$ is $2$. The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$. So $a^2\\equiv 1 \\pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\\quad$ $3a\\equiv 4 \\pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\\equiv 4 \\pmod{7}$. Finally $1+4+4 \\equiv 2 \\pmod{7}$ then the remainder is $2$. • To obtain $a \\equiv b \\pmod{n}$, type a \\equiv b \\pmod{n} in math mode. Jul 28 '15 at 13:43 $a^2 + 3a + 4 \\equiv a^2 - 4a + 4 \\equiv (a-2)^2 \\pmod 7$ If $a\\equiv b \\pmod 7$, then $a^2 + 3a + 4 \\equiv (b-2)^2 \\pmod 7$", "and found that +1 was a solution and therefore I could write $$x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\\\ x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\\\$$ And then equated the coeffeciants For x^3, $1=A$ For x^2, $6=B-A$ For x^1, $-8=C-B$ For x^0, $1=-C$ With knowing that A=1 and C= -1 , B can be found to be 7 . And therefore I can write $x^3+6x^2-8x+1=(x-1)(x^2+7x-1)$ I'm curious why you didn't just divide the cubic by ##x-1## using either long division or, preferably, synthetic division to get the quadratic factor. Much less work...", "what we know: $p(x) = q_1(x)(x - 1) + 4 = q_2(x)(x - 2) + 5$. We want to find $r(x)$ such that $p(x) = q(x)(x^2 - 3x + 2) + r(x)$. Note that $r(x) = ax + b$, since $\\text{deg}(x^2 - 3x + 2) = 2$. Let's just play with this a bit. $r(x) = p(x) - q(x)(x^2 - 3x + 2)$, so: $r(x) - 4 = p(x) - 4 - q(x)(x^2 - 3x + 2) = q_1(x)(x - 1) + q(x)(x - 2)(x - 1) = [q_1(x) - q(x)(x - 2)](x - 1)$, so $r(x) - 4$ is divisible by $(x - 1)$. This mean that $r(1) - 4 = 0$, that is: $a + b = 4$. Similarly, $r(x) - 5 = p(x) - 5 - q(x)(x^2 - 3x + 2) = q_2(x)(x - 2) + q(x)(x - 2)(x - 1) = [q_2(x) - q(x)(x - 1)](x - 2)$, so $r(x) - 5$ is divisible by $(x - 2)$. What does this say about what $r(2) - 5$ is? $r(2)-5=0$ Thank you very much. HallsofIvy confused me. $2a-b=5$ $2a-4+a=5=3a-4\\Rightarrow a=3, b=1$ Then $r(x)=3x+1$ 11. ## Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide Shouldn't $r(2) = 2a + b$? 12. ## Re: P(x) leaves two remainder when divided", "is remainder It is given that $x^2 + 2x + k$ is a factor hence remainder = 0 $x(7k + 21) + (6 + 8k + 2k^2) = 0.x + 0$ By comparing L.H.S. and R.H.S. 7k + 21 = 0 k = – 3 ….(1) $2k^2 + 8k + 6 = 0$ $2k^2 + 2k + 6k + 6 = 0$ 2k(k + 1) + 6(k + 1) = 0 (k + 1) (2k + 6) = 0 k = –1, –3 …..(2) From (1) and (2) k = –3 Hence, $2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k) (2x^2 - 3x - 8 - 2k)$ Put k = –3 $= (x^2 + 2x - 3) (2x^2 - 3x - 8 + 6)$ $= (x^2 + 2x - 3) (2x^2 - 3x - 2)$ $= (x^2 + 3x - x - 3) (2x^2 - 4x + x - 2)$ $= (x(x + 3) - 1(x + 3)) (2x(x - 2) + 1(x - 2))$ $= (x + 3)(x -1)(x - 2) (2x + 1)$ x + 3 = 0 x – 1 = 0 x – 2 = 0 2x + 1 = 0 x = –3 x = 1 x = 2 2x = – 1 $x=\\frac{-1}{2}$ Here zeroes", "remainder after b - a is divided by 3? (1) a = c^3 (2) b = (c + 1)^3 If a , b , and c are positive integers, what is the remainder after b - a is divided by 3? (1) a = c^3 --> no info about b. Not sufficient. (2) b = (c + 1)^3 --> no info about b. Not sufficient. When taken together you can go with algebraic approach or plug-in method: Algebraic approach: (1)+(2) Important tip: x^3-y^3 can be factored as follows: $$x^3-y^3=(x-y)(x^2+xy+y^2)$$. Apply this factoring to $$b-a$$ --> $$b-a=(c + 1)^3-c^3=(c+1-c)(c^2+2c+1+c^2+c+c^2)=3c^2+3c+1=3(c^2+c)+1$$ --> remainder upon division this expression by 3 is 1. Sufficient. Plug-in method approach: (1)+(2) try some numbers for a and b: $$a=c^3=1$$ --> $$b=(c+1)^3=8$$ and --> $$b-a=(c + 1)^3-c^3=7$$ --> remainder upon division 3 is 1; $$a=c^3=8$$ --> $$b=(c+1)^3=27$$ and --> $$b-a=(c + 1)^3-c^3=19$$ --> remainder upon division 3 is 1; $$a=c^3=27$$ --> $$b=(c+1)^3=64$$ and --> $$b-a=(c + 1)^3-c^3=37$$ --> remainder upon division 3 is 1; $$a=c^3=64$$ --> $$b=(c+1)^3=125$$ and --> $$b-a=(c + 1)^3-c^3=61$$ --> remainder upon division 3 is 1; ... It seem that there is some kind of pattern and we can safely assume that in all other cases remainder will also be 1. Sufficient. Hope it helps. Hi Bunuel, Would it be possible for you to", "= $$16b^{2} +2$$= $$4(4b^{2})+2$$ Now, $$4(4b^{2})+2 \\equiv 2 \\pmod{4}$$ So, we are getting remainder $$2$$. So for first case it is not divisible. $$2. If$$ $$r=1$$ then $$m= 4b+r$$ $$m= 4b+1$$ $$m^{2}= (4b+1)^{2}$$ $$m^{2} = 16b^{2} + 8b +1$$ $$m^{2}= 4(4b^{2} + 2b) +1$$ But it is given that whether 4 is divided by $$m^{2}+2$$ Now, $$m^{2}+2$$ = $$4(4b^{2} + 2b) +1+2$$ $$m^{2}+2$$ = $$4(4b^{2} + 2b) + 3$$ Now, $$4(4b^{2}+2b)+3 \\equiv 3 \\pmod{4}$$ So, we are getting remainder $$3$$. So for second case it is not divisible. $$3. If$$ $$r=2$$ then $$m= 4b+r$$ $$m= 4b+2$$ $$m^{2}= (4b+2)^{2}$$ $$m^{2} = 16b^{2} + 16b +4$$ $$m^{2}= 4(4b^{2} + 4b+1)$$ But it is given that whether 4 is divided by $$m^{2}+2$$ Now, $$m^{2}+2$$ = $$4(4b^{2} + 4b+1) +2$$ Now, $$4(4b^{2}+4b+1)+2 \\equiv 2 \\pmod{4}$$ So, we are getting remainder $$2$$. So for third case it is not divisible. $$4. If$$ $$r=3$$ then $$m= 4b+r$$ $$m= 4b+3$$ $$m^{2}= (4b+3)^{2}$$ $$m^{2} = 16b^{2} + 24b +9$$ $$m^{2} = 16b^{2} + 24b +8+1$$ $$m^{2}= 4(4b^{2} + 6b+2)+1$$ But it is given that whether 4 is divided by $$m^{2}+2$$ Now, $$m^{2}+2$$ = $$4(4b^{2} + 6b+2)+1+2$$ $$= 4(4b^{2} + 6b+2)+3$$ Now, $$4(4b^{2}+6b+2)+3 \\equiv 3 \\pmod{4}$$ So, we are getting remainder $$3$$. So for forth case it is not divisible. As in all fours cases if $$m^{2}+2$$", "divided by $$3$$, remainder $$3$$ when divided by $$5$$, and remainder $$2$$ when divided by $$7$$. Now we solve the system directly by a general technique as follows: $$x\\equiv 2 \\:\\left(\\mathrm{mod}\\: 3 \\right) \\implies x=2+3r$$. Plugging it in $$x\\equiv 3 \\:\\left(\\mathrm{mod}\\: 5 \\right)$$, we get $$(2+3r)\\equiv 3 \\:\\left(\\mathrm{mod}\\: 5 \\right)\\implies 3r\\equiv 1 \\:\\left(\\mathrm{mod}\\: 5 \\right)\\implies 6r\\equiv 2 \\:\\left(\\mathrm{mod}\\: 5 \\right)\\implies r\\equiv 2 \\:\\left(\\mathrm{mod}\\: 5 \\right)\\implies r=2+5s$$. Plugging it in $$x=2+3r$$, we get $$x=2+3(2+5s)=8+15s$$. Plugging it in $$x\\equiv 2 \\:\\left(\\mathrm{mod}\\: 7 \\right)$$, we get $$(8+15s)\\equiv 2 \\:\\left(\\mathrm{mod}\\: 7 \\right)\\implies 15s\\equiv -6 \\:\\left(\\mathrm{mod}\\: 7 \\right) \\implies s\\equiv 1 \\:\\left(\\mathrm{mod}\\: 7 \\right)\\implies s=1+7t$$. Plugging it in $$x=8+15s$$ we get $$x=8+15(1+7t)=23+105t\\implies x\\equiv 23 \\:\\left(\\mathrm{mod}\\: 105 \\right)$$. Theorem.(Chinese Remainder Theorem) If positive integers $$n_1,n_2,\\ldots,n_k$$ are pairwise relatively prime, then the following system of linear congruences has a unique solution modulo $$n_1n_2\\cdots n_k$$: $x\\equiv a_1 \\:\\left(\\mathrm{mod}\\: n_1 \\right),\\; x\\equiv a_2 \\:\\left(\\mathrm{mod}\\: n_2 \\right),\\ldots,\\; x\\equiv a_k \\:\\left(\\mathrm{mod}\\: n_k \\right).$ Proof. Suppose positive integers $$n_1,n_2,\\ldots,n_k$$ are pairwise relatively prime. Let $$n=n_1n_2\\cdots n_k$$. Let $$N_i=n/n_i$$ for $$i=1,2,\\ldots,k$$. Since $$\\gcd(n_i,n_j)=1$$ for all $$j\\neq i$$, $$\\gcd(n_i,N_i)=1$$ which implies $$N_ix=1 \\:\\left(\\mathrm{mod}\\: n_i\\right)$$ has a unique solution, say $$y_i$$. We show that $y:=y_1a_1N_1+y_2a_2N_2+\\cdots+y_ka_kN_k$ is a simultaneous solution to the system, i.e., $$y\\equiv a_i \\:\\left(\\mathrm{mod}\\: n_i \\right)$$ for $$i=1,2,\\ldots,k$$. Since $$n_i\\vert N_j$$ for all $$j\\neq i$$, $$N_j\\equiv 0 \\:\\left(\\mathrm{mod}\\: n_i \\right)$$ for", "q(x)(x+3)(x+5)+ax+b$$ Plug $x=-3$ and $x=-5$ to find $a$ and $b$.", "`int (x^2 + x + 1)/(x^2 + 1)^2 dx` Evaluate the integral `int (x^2+x+1)/(x^2+1)^2 dx` To solve, apply partial fraction decomposition. To express the integrand as sum of proper rational functions, set the equation as follows: `(x^2+x+1)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2` Multiply both sides by the LCD. `x^2+x+1=(Ax+B)(x^2+1) + Cx + D` `x^2+x+1=Ax^3+Bx^2+Ax+B+Cx+D` `x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D` Express the left side as a polynomial with degree 3. `0x^3+x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D` For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other. x^3: `0=A` (Let this be EQ1.) x^2: `1=B ` (Let this be EQ2.) x: `1=A+C` (Let this be EQ3.) Constant: `1=B+D` (Let this be EQ4.) In the equations, the values of A and B are already known. So only the values of C and D have to be solved. To do so, plug-in the value of A to EQ3. `1=A+C` `1=0+C` `1=C` Also, plug-in the value of B to EQ4. `1=B+D` `1=1+D` `0=D` So the partial fraction decomposition of the integrand is: `(x^2+x+1)/(x^2+1)^2=(0x+1)/(x^2+1)+(1x+0)/(x^2+1)^2=1/(x^2+1)+x/(x^2+1)^2` Taking the integral of it result to: `int (x^2+x+1)/(x^2+1)^2 dx` `= int (1/(x^2+1)+x/(x^2+1)^2) dx` `= int 1/(x^2+1)dx + int x/(x^2+1)^2dx` For the first integral, apply the formula int 1/(u^+a^2)=1/a tan^(-1) u/a+C. For the second integral, apply u-substitution method. `u=x^2+1` `du=2x dx` `(du)/2=xdx` `= int 1/(x^2+1)dx + int 1/u^2 *", "+ 24 + 19 + b = -3a + b + 1 ——— 1 Let, x – 2 = 0 => x = 2 Substitute the value of x in f(x) f(2) = $(2)^{3} – 3(2)^{2} + (2)(a – 12) + 19$ + b = 8 – 12 + 2a – 24 + b = 2a + b – 9 ——— 2 Solve equations 1 and 2 -3a + b = -1 2a + b = 9 (-) (-) (-) -5a = – 10 a = 2 substitute the value of a in eq 1 => -3(2) + b = -1 => -6 + b = -1 => b = -1 + 6 => b = 5 $∴$ r(x) = ax + b = 2x + 5 $∴$ $x^{3} – 3x^{2} – 12x + 19$ is divided by $x^{2} + x – 6$ when it is added by 2x + 5 Q21. What must be added to $x^{3} – 6x^{2} – 15x + 80$ so that the result is exactly divisible by $x^{2} + x – 12$ Sol : Let, p(x) = $x^{3} – 6x^{2} – 15x + 80$ q(x) = $x^{2} + x – 12$ by division algorithm, when p(x) is divided by q(x) the remainder is a linear expression in x. so, let r(x) = ax", "may be there as the division may also result in a remainder. Hence $3x^2+4xy-4y^2+2=(x+2y)(3x-2y+c)$ We know \"a\" and \"b\" due to the $x^2$ and $y^2$ terms. We need to find \"c\". Multiplying out... $3x^2+4xy-4y^2+2=3x^2-2xy+cx+6xy-4y^2+2cy=3x^2+4xy-4y^2+cx+2cy$ Hence $2=cx+2cy\\ \\Rightarrow\\ c(x+2y)=2\\ \\Rightarrow\\ c=\\frac{2}{x+2y}$ Hence $\\frac{3x^2+4xy-4y^2+2}{x+2y}=3x-2y+\\frac{2}{x+2y}$", "`int_1^2 (4y^2 - 7y - 12)/(y(y + 2)(y - 3)) dy` Evaluate the integral `int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy` To solve this, apply partial fraction decomposition. When converting the integrand to sum of proper rational expressions, set the equation as follows: `(4y^2-7y-12)/(y(y+2)(y-3))= A/y + B/(y+2)+C/(y-3)` Multiply both sides by the LCD. `4y^2-7y-12=A(y+2)(y-3)+By(y-3)+Cy(y+2)` `4y^2-7y-12=Ay^2-Ay-6A+By^2-3By+Cy^2+2Cy` `4y^2-7y-12=(A+B+C)y^2+(-A-3B+2C)Y-6A` For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other. y^2: `4=A+B+C` (Let this be EQ1.) y: `-7=-A-3B+2C` (Let this be EQ2.) Constant: `-12=-6A` (Let this be EQ3.) To solve for the value of A, consider EQ3. `-12=-6A` `2=A` Plug-in this value of A to EQ1. `4=A+B+C` `4=2+B+C` `2=B+C` Then, isolate the C. `2-B=C` Plug-in this expression and the value of A to EQ2. `-7=-A-3B+2C` `-7=-2-3B+2(2-B)` `-7=-2-3B+4-2B` `-7=2-5B` `-9=-5B` `9/5=B` And plug-in the value of A and B to EQ1. `4=A+B+C` `4=2+9/5+C` `4=19/5+C` `1/5=C` So the partial fraction decomposition of the integrand is: `(4y^2-7y-12)/(y(y+2)(y-3))= 2/y + (9/5)/(y+2)+(1/5)/(y-3)=2/y+9/(5(y+2))+1/(5(y-3))` Taking the integral of this result to: `int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy` `=int_1^2 (2/y+9/(5(y+2))+1/(5(y-3)))dy` `= 2int_1^2 1/ydy + 9/5int_1^2 1/(y+2)dy+1/5int_1^2 1/(y-3)dy` `=(2ln|y| + 9/5ln|y+2| + 1/5ln|y-3|)|_1^2` `= (2ln|2| +9/5ln|2+2| +1/5ln|2-3|)-(2ln|1| + 9/5ln|1+2|+1/5ln|1-3|)` `=(2ln|2|+9/5ln|4|+1/5ln|-1|) - (2ln|1|+9/5ln|3|+1/5ln|-2|)` `=(2ln2+9/5ln2^2+1/5ln1) - (2ln1+9/5ln3+1/5ln2)` `=(2ln2+18/5ln2+0)-(0+9/5ln3+1/5ln2)` `=28/5ln2-(9/5ln3+1/5ln2)` `=28/5ln2-9/5ln3-1/5ln2` `=27/5ln2-9/5ln3` `=9/5(3ln2-ln3)` `=9/5(ln2^3-ln3)` `=9/5(ln8-ln3)` `=9/5ln(8/3)` Therefore, `int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy=9/5ln(8/3)` . Approved by eNotes Editorial Team You need to", "and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Hello s2951 I'm not sure where you get the term $ax$ in the line that I've indicated. When you divide $4x^2-7x-15$ by $x-3$, the result is simply $4x+5$. If I may expand on acc100jt's solution a little: If $4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get: $4x^2-7x-15=Ax^2+(B-3A)x -3B$ So, when we compare coefficients: the coefficient of $x^2:4 = A$ and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Thanks, just solved and its correct. The letter \"A\" is just symbol to find the value, its redundant. 6. Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ 7. Hello Raoh Originally Posted by Raoh Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ I think what you mean is: Since $(x-3)$ is a factor, $+3$ is a root $\\Rightarrow f(3)=0$ So: $f(x) = f(x) - f(3) = 4(x^2-3^2)", "# Error in predicting remainders I recently came across the problem What is the remainder when $x^5 + 2x^4 -3x^2 + 2x -4$ is divided by $(x^2 + 2x)$? a. $x^3 - 3$ b. $8x - 4$ Correct answer c. 4 d. $2x-4$ e. $20x$ And I attempted to solve using the remainder theorem (the textbook advised plugging in numbers, instead of actually using synthetic/polynomial division to solve for multiple choice questions). However, I soon found that smaller numbers didn't give me accurate results. My method of solving (and the way the textbook told me to solve) was to choose a random value for $x$. I chose 3. Then, plug 3 into both equations: $3^5 + 2(3)^4 -3(3)^2 + 2(3) -4$ is divided by $((3)^2 + 2(3))$ Then, I solved: $\\frac{3^5 + 2(3)^4 -3(3)^2 + 2(3) -4}{(3)^2 + 2(3)} = 25\\frac{2}{3}$ Then I took the remainder ($\\frac{2}{3}$) and multiplied it by the divisor to get the remainder: $\\frac{2}{3} * ((3)^2 + 2(3)) = \\frac{1}{3}$ Therefore, when you set $x = 3$, you get a remainder of $\\frac{1}{3}$. The textbook then told me to plug in the value I set x to into all the choices, until I found a value that was equal to the remainder, so when $x=3$: a. $x^3 - 3$ -> 24 b. $8x -", "Question # Choose the correct answer from the alternatives given :When the polynomial $$(6x^4 \\, + \\, 8x^3 \\, + \\, 17x^2 \\, + \\, 21x \\, + \\, 7)$$ is divided by $$(3x^2 \\, + \\, 4x \\, + \\, 1)$$ , the remainder is (ax + b). Therefore, A a = 1, b = 2 B a = 1, b = 2 C a = 2, b = 1 D a = 1, b = 2 Solution ## The correct option is A a = $$1$$, b = $$2$$According to the question,x + 2 = ax + ba = 1 & b = +2Mathematics Suggest Corrections 0 Similar questions View More", "3, 1. (a^2)/3 yields the remainder of 2 and (b^2)/3 yields the remainder 0. In this case a^2 can be written as 3*x+2 and b^2 can be 3y 2. (a^2)/3 yields the remainder of 0 and (b^2)/3 yields the remainder 2. In this case a^2 can be written as 3*x and b^2 can be 3y+2 3. (a^2)/3 yields the remainder of 1 and (b^2)/3 yields the remainder 1. In this case a^2 can be written as 3*x+1 and b^2 can be 3y+1. Here is x and y are any +ve integers. Let us now substitute each of these values of of a^2 and b^2 into (a^2+b^2)(a^2-b^2) 1.-> a^2 = (3x+2), b^2 =3y. Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)+6(x+y)+6(x-y)+4. In this case the remainder is 1. 2.-> a^2 = (3x), b^2 =(3y+2). Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)-6(x+y)+6(x-y)-4. In this case the remainder is 2. 3.-> a^2 = (3x+1), b^2 =(3y+1). Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)+6(x+y). In this case the remainder is 0. Hence, option 2 is NOT SUFFICIENT. _________________ Please kudos if you find this post helpful. I am trying to unlock the tests SC Moderator Joined: 13 Apr 2015 Posts: 1615 Location: India Concentration: Strategy, General Management WE: Analyst (Retail) Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink] Show Tags 05 Dec 2015, 21:50 4 KUDOS", "this in two ways. First, we can combine the terms and get $$8x + 8 = (A + B)x + 6A - 2B$$ Because x and 1 are linearly independent, we can solve a system of equations for A and B: $$\\begin{cases}8 = A + B\\\\8 = 6A - 2B\\end{cases}$$ There's a better way! If we start with $8x + 8 = A(x + 6) + B(x-2)$, we can plug in $x = -6$ and $x = 2$ and solve for A and B that way. $$x = -6 \\to -40 = -8B \\to B = 5$$ $$x = 2 \\to 24 = 8A \\to A = 3$$ $$\\dfrac{8x + 8}{x^2 + 4x - 12} = \\dfrac{A}{x-2} + \\frac{B}{x+6} = \\boxed{\\dfrac{3}{x-2} + \\frac{5}{x+6}}$$ # Repeated Root Example MathJax TeX Test Page $$\\dfrac{19-3x}{(x-4)^2}$$ If we factor the denominator we only have $(x-4)$, so what we have to do is to have a fraction with just a number divided by $(x-4)^2$. It will look like this: $$\\dfrac{19-3x}{(x-4)^2} = \\frac{A}{x-4} + \\frac{B}{(x-4)^2}$$ We do the same thing as before: multiply through by $(x-4)^2$ $$19-3x = A(x-4) + B$$ When $x = 4, B = 7$. From that, you can see that $12 - 3x = A(x-4)$, so A = -3. However, there may be other roots, so in general, you plug", "is by determining if $8673$ “behaves” like $7$ or $14$ or some other small multiples of $7$. Modulo $7$ arithmetic will help us do that. In its most basic definition, modulo arithmetic is the arithmetic of remainder. We say “$8$ is congruent to $15$ modulo $7$” and write $8\\equiv 15\\bmod 7$, because $8$ and $15$ yield the same remainder when divided by $7$. Using $8673=8(1000)+6(100)+7(10)+3(1)$, we will do the same calculation but with congruent numbers mod $7$. For ease in presentation we drop the modulo $7$ notation. Since $1000\\equiv 6,100\\equiv 2,10\\equiv 3,8\\equiv 1,7\\equiv 0$, $8(1000)+6(100)+7(10)+3(1)\\equiv 1(6)+6(2)+0(3)+3(1)$ $\\equiv 6+12+0+3\\equiv 21$ $8673$ is divisible by $7$. Let $axy\\,b$ be a 4-digit positive integer where the digits are $a,x,y$, and $b$. Suppose $7$ divides $axy\\,b$ and $bxya$. Show that $7$ divides $(a-b)$. Since $7$ divides $axy\\,b$ and $bxya$, there exist integers $k_1$ and $k_2$ such that $a(1000)+x(100)+y(10)+b=7k_1\\qquad\\qquad (1)$ $b(1000)+x(100)+y(10)+a=7k_2\\qquad\\qquad (2)$ Subtract Eq. $(2)$ from Eq. $(1)$ $(a-b)1000+(b-a)=7k_3$ where $k_3=k_1-k_2$ $1000a-1000b+b-a=7k_3$ $999(a-b)=7k_3$ In other words, $7$ divides $999(a-b)$. Since $7$ does not divide $999$, $7$ must divide $(a-b)$. Show that if $7$ divides $axy\\,b$, then $7$ divides $10x+y$. Given $a(1000)+x(100)+y(10)+b=7k_4$ for some integer $k_4$, $1000a+100x+10y+b=7k_4$ $1001a-a+100x+10y+b=7k_4$ $1001a-(a-b)+100x+10y=7k_4$ Since $7$ divides $1001a$ and $(a-b)$, $7$ must divide $100x+10y$. Hence, $7$ divides $10x+y$. We have two cases to consider: $a=b$ and $a\\neq b$. Case" ]

[ "+0 # Help! 0 251 1 Find all values of such that . If you find more than one value, then list your solutions, separated by commas. Aug 22, 2022 $$x=\\boxed{-\\frac{2}{3},-4}$$", "Let $Z = \\{ 4x+1 : x \\in X \\} \\cup \\{ 4y+2 : y \\in Y \\}$. Given the sorted version of $Z+Z$, it is easy to extract $X+Y$ by taking all elements of the form $4m+3$ and extracting $m$. The $X+Y$ problem is 3SUM-hard. However, this doesn't shed much light on the problem since 3SUM itself can be solved in $O(n^2)$ (and even faster), while the $X+Y$ problem plainly requires $\\Omega(n^2)$.", "$(3z^2+3a^2)=(z^2+4a^2)$ which you should have no difficulty solving. Last edited:", "From there you can find, $d = -4$. $$( \\mathcal{P} ): x + 4y + z - 4 = 0$$", "by completing the square: $4x^2+11x+7=0$4x2+11x+7=0 1. Write all solutions on the same line, separated by commas.", "0$, so $z = 4$", "+0 # solve for z 0 48 2 Find all values of z such that z^4 - 4z^2 +4 = 0. Feb 24, 2021 #1 +944 +1 Let z^2 = a, so we are solving: a^2 - 4a + 4 = 0 (a - 2)^2 = 0 a = 2 z^2 = 2 $$\\boxed{z = \\pm \\sqrt{2}}$$ Feb 24, 2021 #2 +266 +1 . Feb 24, 2021", "Find all possible values of $z$ that satisfy $|z – 5| + |z + 5| = 12$ and $|z – 5| = \\operatorname{Re} z + 4$.", "Write all solutions on the same line, separated by commas. ##### Question 3 Solve for the unknown: $-8x+x^2=-6-x-x^2$8x+x2=6xx2 1. Write all solutions on the same line, separated by commas. ##### Question 4 Solve the following equation: $x-\\frac{45}{x}=4$x45x=4 1. Write all solutions on the same line, separated by commas.", "z^2$ for all values of $t$. That's what the book answer does. -", "+0 # hellp puhlez 0 71 1 1.) Find all values of x such that $$9 + \\frac{27}{x} + \\frac{8}{x^2} = 0$$ If you find more than one value, then list your solutions, separated by commas. 2.) Find all values of x such that $$\\frac{2x}{x + 2} = -\\frac{6}{x + 4}$$ If you find more than one value, then list your solutions, separated by commas. Mar 18, 2021", "z$$ . So our final result is: $$((x^{5}+3)^4 + 2)$$ .", "If $f(x) = x^3 + 6 x^2 - 288 x + 5$, find analytically all values of $x$ for which $f'(x) = 0$. (Enter your answer as a comma separated list of numbers, e.g., -1,0,2) $x =$", "+0 # Find all real values of t that satisfy the equation 0 97 1 Find all real values of t that satisfy the equation t^4 = 324. If you find more than one, then list your values in increasing order, separated by commas. Nov 1, 2020 $$\\sqrt[4]{324}$$ = - +4.24264", "4$ and $z = 5$", "1 MEDIUM Find the value of $$z$$ : $$6z - 5 = 2z + 7$$ 2 MEDIUM Find the value of $$x$$ : $$4x + 2x + 8x = -7$$ 2-0.50.5-2None of the above 3 MEDIUM Find the value of $$a$$ : $$7a - 5 = 3 - a$$", "+0 # help 0 80 1 Find all values of z such that z^4 - 4z^2 +4 = 0. Apr 30, 2022 #1 +267 +2 Let x be z². The equation turns into this: x²-4x+4=0 This factors as: (x-2)²=0 x would be equal to two. Set 2 equal to z² and solve: $$z=\\sqrt2$$ $$or$$$$-\\sqrt2$$ Apr 30, 2022 #1 +267 +2 Let x be z². The equation turns into this: x²-4x+4=0 This factors as: (x-2)²=0 x would be equal to two. Set 2 equal to z² and solve: $$z=\\sqrt2$$ $$or$$$$-\\sqrt2$$ WhyamIdoingthis Apr 30, 2022", "# z^7+2(z^4)+4z=0 ## Simple and best practice solution for z^7+2(z^4)+4z=0 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. ## Solution for z^7+2(z^4)+4z=0 equation: Simplifying z7 + 2(z4) + 4z = 0 Reorder the terms: 4z + 2z4 + z7 = 0 Solving 4z + 2z4 + z7 = 0 Solving for variable 'z'. Factor out the Greatest Common Factor (GCF), 'z'. z(4 + 2z3 + z6) = 0 Subproblem 1Set the factor 'z' equal to zero and attempt to solve: Simplifying z = 0 Solving z = 0 Move all terms containing z to the left, all other terms to the right. Simplifying z = 0 Subproblem 2Set the factor '(4 + 2z3 + z6)' equal to zero and attempt to solve: Simplifying 4 + 2z3 + z6 = 0 Solving 4 + 2z3 + z6 = 0 Begin completing the square. Move the constant term to the right: Add '-4' to each side of the equation. 4 + 2z3 + -4 + z6 = 0 + -4 Reorder the terms: 4 +", "Find all numbers c that satisfy the conclusion of Rolle's Theorem for the following function. If there are multiple values, separate them with commas; enter N if there are no such values. $f(x)=x^2-3 x +6, [0,3]$", "+0 # Operations 0 75 1 Let $x \\spadesuit y = x^2/y^2$ for all x and y such that $y \\neq 0$. Find all values of x such that $x \\spadesuit 3 = 9$. List the values you find in increasing order, separated by commas. Apr 7, 2021 $\\frac{x^2}{9}=9\\implies x^2=81$." ]

[ "If we let $y=\\sqrt{x}$, this becomes $y^2-y-6=0$, which is easily factorizable; note that $y=\\sqrt{x}$ must be positive, so only one of the roots of that quadratic will square to give an x that solves the original equation. --Kevin C.", "1+y^2+7=9,y^2=1$ So, the roots are $\\pm \\sqrt7i,1\\pm i$ Choosing $z = 1 + y i$, you obtain the formula: $$y^4 – 2 i y^3 – 9y^2 + 2 i y + 8 = 0$$ Because $y$ is real, this means that $$2 y^3 – 2y = 0$$ and $$y^4 – 9y^2 + 8 = 0$$ The first equation implies $y = 0, 1, -1$. The second equation implies that $y^2 = 8$ or $y^2 = 1$. Thus we see both $1 + i$ and $1 – i$ are solutions. Factor out the polynomial $(z – 1 + i)(z – 1 – i) = z^2 – 2z + 2$ and you’ll find the remaining roots. Maxima tells me the roots are $\\pm i \\sqrt{7}$, $1 \\pm i$. It factors as $(z^2 + 7) (z^2 – 2 z + 2)$.", "# Is it possible to factor y= x^4 -3x^2 -4 ? If so, what are the factors? Yes. $x = \\pm 2$ or $x = \\pm i$ First represent the equation as a quadratic by substituting $z = {x}^{2}$ Then $y = {z}^{2} - 3 z - 4$ This can be factored as $y = \\left(z - 4\\right) \\left(z + 1\\right)$ Therefore $z = 4$ or $z = - 1$ Given $z = {x}^{2}$ this means that $x = \\pm 2$ or $x = \\pm \\sqrt{- 1} = \\pm i$", "2x^2- 1$ so $z= \\pm\\sqrt{4x-2x^2-1}$. That z only exists when $\\frac{4- 2\\sqrt{2}}{4}< x< \\frac{4+ 2\\sqrt{2}}{4}$. In that range, the roots are x, y= 2- x and z= $\\pm\\sqrt{4x-2x^2-1}$ for every value of x. It is true that x= 1 is in that range and then y= 2-1= 1 and z= $\\sqrt{4- 2-1}= 1$ so that (1, 1, 1) is one solution. x= 1, y= 1, z= $-\\sqrt{4-2-1}= -1$ or (1, 1, -1) is also a solution. Those may be the only integer solutions. 4. Sorry, I don't really understand what do you mean by \"In that range, the roots are x, y= 2- x and z= for every value of x.\" By roots, which roots are you talking about? Please explain about the above in details and on how do you come to the conclusion that the roots are x, y= 2- x and z = . A little lost here", "3)(x^3 + 1) = 0$ So $x^3 = 3$ ==> $x = \\sqrt[3]{3}$ ==> $y = (\\sqrt[3]{3})^2 = \\sqrt[3]{9}$ or $x^3 = -1$ ==> $x = -1$ ==> $y = (-1)^2 = 1$ So the points where $y^{\\prime} = 0$ are $(-1, 1)$ and $(\\sqrt[3]{3}, \\sqrt[3]{9})$ . (Both of these are points in the original equation.) We do the same process to find where the vertical tangents are. Where is $y^{\\prime}$ undefined? Where $y^2 - x = 0$ $y = \\pm \\sqrt{x}$ So put this into your original equation: $x^3 + y^3 - 3xy = 3$ $x^3 + (\\pm \\sqrt{x})^3 - 3x(\\pm \\sqrt{x}) = 3$ $x^3 \\pm \\sqrt{x^3} \\mp 3 \\sqrt{x^3} - 3 = 0$ $x^3 \\mp 2 \\sqrt{x^3} - 3 = 0$ Let $z = \\sqrt{x^3}$ Then $z^2 \\mp 2 z - 3 = 0$ For the \"-\" sign: $z^2 - 2z - 3 = 0$ $(z - 3)(z + 1) = 0$ $z = 3$ ==> $x = \\sqrt[3]{3^2} = \\sqrt[3]{9}$ ==> $y = \\pm \\sqrt{\\sqrt[3]{9}} = \\pm \\sqrt[6]{9}$ (I note that the \"-\" solution is not a point on the original equation.) and $z = -1$ ==> $x = \\sqrt[3]{(-1)^2} = 1$ ==> $y = \\pm \\sqrt{1} = \\pm 1$ (I note that the \"+\" solution is not a point on the original equation.) So", "- z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = \\sqrt{(y-1)(z-1)}$! Thus, $$u^2 - 2u + 2 = (u-1)^2 + 1 > 0$$ and the claim holds for x = 1. If x > 1, we see the $\\sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$: $$x + xyz \\ge x - 1 + y - 1 + z - 1 + 2\\sqrt{(x-1)(y-1)} + 2\\sqrt{(y-1)(z-1)} + 2\\sqrt{(x-1)(y-1)}$$ which leads to $$xyz \\ge y + z - 3 + 2\\sqrt{(x-1)(y-1)} + 2\\sqrt{(y-1)(z-1)} + 2\\sqrt{(x-1)(z-1)}$$ Again, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4\\sqrt{6}$. Now, we see the finish: setting $u = \\sqrt{x-1}$ gives $x = u^2 + 1$. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations: $$u^2(yz) - u(2\\sqrt{y-1} + 2\\sqrt{z-1}) + (3 + yz - y - z - 2\\sqrt{(y-1)(z-1)}) \\ge 0$$ Because the coefficient of $u^2$ is positive, all we need to do is to verify that the discriminant is nonpositive: $$b^2 - 4ac = 4(y-1) + 4(z-1) - 8\\sqrt{(y-1)(z-1)} -", "## College Algebra 7th Edition $x=\\pm 1$ $x=\\pm \\sqrt{3}$ $x^{4}-4x^{2}+3=0$ First, we substitute $y=x^{2}$ into the equation: $y^{2}-4y+3=0$ Next, we factor and solve: $(y-1)(y-3)=0$ $y-1=0$ or $y-3=0$ $y=1$ or $y=3$ Finally, we convert the $y$ solutions to $x$ solutions: $y=x^2$ $x=\\pm\\sqrt{y}$ If $y=1$: $x=\\pm \\sqrt{1}=\\pm 1$ If $y=3$: $x=\\pm \\sqrt{3}$", "It's not pulling things out of the air, it's a simple substitution. When we introduce the variable $y = x^2$, we convert $$x^4 - 34x^2 + 225 = 0$$ which, as a quartic equation, we may not know how to solve (although this one happens to be easy to factor), into: $$y^2 - 34y + 225 = 0$$ which, as a quadratic equation, is simpler, and we have an equation for it. Notice that we didn't actually change the structure of the equation - we simply substituted a new variable. For our convenience. That way, you can plug the equation for $y$ into the quadratic formula and get $y = 25$ or $y = 9$. But we're not solving for $y$, we're solving for $x = \\pm\\sqrt{y}$, so $x = \\pm 5,\\pm 3$. Hint. $$x^4-34x^2+225=x^4-30x^2+225-4x^2=(x^2-15)^2-(2x)^2$$ • Why not just $(x^2-25)(x^2-9)$? – Barry Jun 7 '15 at 20:54 • It's certainly an inside-out way to complete the square. – Thomas Andrews Jun 7 '15 at 21:38 No, because $\\sqrt{a+b} \\neq \\sqrt{a} + \\sqrt{b}$. For example, $$3 = \\sqrt{9} = \\sqrt{1+8} \\neq \\sqrt{1}+\\sqrt{8}=1+2\\sqrt{2},$$ so it's definitely false. What you should do is consider it as a quadratic equation in $x^2$, which you can relabel as $y$ for convenience. If you do square root both sides of the equation, you get", "In this case make the substitution $y=x+4$, which transforms the equation into $$(y-1)^4+(y+1)^4=20.$$ This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'. If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur: $$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$ So, the odd factors cancel to give $$2y^4+12y^2-18=0$$ This then becomes the simple equation $$y^4+6y^2-9=0\\,,$$ which is a quadratic equation in $y^2$ with solutions $$y^2 = \\frac{-6\\pm\\sqrt{36+36}}{2}=-3(1\\pm\\sqrt 2).$$ One of these solutions is positive; taking the square root gives two real values for $y$ as $$y = \\pm \\sqrt{3\\left(\\sqrt{2}-1)\\right)}$$ Therefore two real solutions to the equation are $$x= -4 \\pm \\sqrt{3\\left(\\sqrt{2}-1\\right)}$$", "(c) tells us that $x_1^2 + y_1^2 = 1$, $x_2^2 + y_2^2 = 1$. (From norm of $\\vec{a}, \\vec{c}$ being 1.) $y_1y_3 = x_1x_2$, $y_2y_3 = x_1x_2$ (From $\\vec{b}$ perpendicular to $\\vec{a}, \\vec{c}$.) $x_1^2+x_2^2+y_3^2 + 2y_1y_2 = 0$. In the second equation, if $y_3$ is not 0, then $y_1 = y_2$, which forces $x_1 = x_2 = y_1 = y_2 = y_3 = 0$ in the third equation, contradicting the first equation. So $y_3 = 0$, which implies $x_1x_2 = 0$. WLOG assume $x_1 = 0$, then $y_1 = \\pm 1$. The third equation becomes $0 = x_2^2 + 2y_1y_2 = 1 - y_2^2 + 2y_1y_2 = 2 - (y_2 - y_1)^2$. So $y_2 - y_1 = \\pm \\sqrt{2}$. The first equation tells us that $y_1,y_2$ has norms less than 1, which implies that $y_2 = \\pm (1 - \\sqrt{2})$. This gives $x_2 = \\pm \\sqrt{2\\sqrt{2} - 2}$. To summarize, up to symmetry of $(x_1,y_1) \\leftrightarrow (x_2,y_2)$, we have $x_1 = 0$, $y_1 = \\pm 1$, $y_2 = \\pm (1 - \\sqrt{2})$ (same sign as $y_1$), $x_2 = \\pm \\sqrt{2\\sqrt{2} - 2}$ and $y_3 = 0$. Step 4 By picking arbitrary orthonormal basis $\\vec{d}, \\vec{e}, \\vec{f}$, we can substitute the values for $x_1,x_2,y_1,y_2,y_3$ above to get solutions for $\\vec{a}, \\vec{b}, \\vec{c}$. All these would give the required sum", "$\\sqrt{y^2+1+2y}$=(y+1) if (y+1)>0 i.e., y≥-1, else -(y+1) and $\\sqrt{y^2+1-2y}$=(y-1) if y>1, else -(y-1) Then, three cases arise, y< -1, -1≤y≤1 and y>1. If y < -1, the given equation becomes -(y+1)-(y-1)=2 => y=-1 which is not possible as y < -1. If -1≤y≤1, y+1 -(y-1)=2 which becomes identity=> one set of solution is -1≤y≤1. If y>1 y+1+(y-1)=2 => y=1 which is again impossible as y>1. If -1≤y≤1, 0≤$y^2$≤1 => 1≤$y^2+1$≤2 => 1≤x≤2 - If $x < 1$, then $$\\begin{array}{lcl} f(x) & = & \\sqrt{x + 2\\sqrt{x-1}} + \\sqrt{x - 2\\sqrt{x-1}} \\\\ & = & \\sqrt{x + 2I \\sqrt{1 - x}} + \\sqrt{x - 2I \\sqrt{1 - x}} \\\\ & = & \\sqrt{\\left( 1 + I \\sqrt{1 - x} \\right)^2} + \\sqrt{\\left( 1 - I \\sqrt{1 - x} \\right)^2} \\\\ & = & \\pm \\left( 1 + I \\sqrt{1 - x} \\right) + \\pm \\left( 1 - I \\sqrt{1 - x} \\right) \\\\ & = & \\left\\{ \\pm 2, \\ \\pm I \\sqrt{1 - x} \\right\\} \\ni 2 \\end{array}$$ -", "plug into the second equation to get $y$. Then we could plug into the first equation to get $x$. Using $z=1$, the second equation gives $y = 2$. Plugging into the first equation, we get $x+2+1 = 4$. Hence, $x=1$. The solution to the matrix equation is thus $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix}1\\\\2\\\\1 \\end{pmatrix}.$", "etc.) and simultaneous equations. let $Z = \\sqrt{4i-3}$ where Z is complex. i.e. $Z$ is of the form $Z = a + ib$, where $a$ and $b$ are REAL coefficients. (This is important later on.) Square both sides: $Z^2 = -3+4i$ Expand the standard form of $Z^2$: $(a+ib)^2 = -3+4i$ This becomes $a^2 - b^2 + 2abi = -3+4i$ The above is an identity and so the real and imaginary parts can be equated. Hence two equations are formed: $a^2 - b^2 = -3$................(1) $2abi = 4i$ which becomes $ab = 2$......(2) In (2), $b = 2/a$ so $a^2 - (2/a)^2 + 3 = 0$ i.e. $a^4 + 3a^2 - 4 = 0$ (multiplying all by $a^2$ and rearranging) which is a quadratic-styled quartic: $(a^2 + 4)(a^2 - 1) = 0.$ Solving, $a=1, -1, 2i$ and $-2i$ BUT since $a$ and $b$ are real, we discard $2i$ and $-2i$. Now we simply solve for $b$ using $b = 2/a$. Hence the solutions are $a=1, b=2$ or $a=-1, b=-2$. and so $\\sqrt{4i-3} = \\pm(1+2i)$ since we defined $Z = a+ib$. Best of luck with any future questions! Hope this helped. • Just add $around math expressions, expressed in the LaTeX formatting. – PearlSek Dec 23 '16 at 22:21 • Well that was simpler than I thought, thanks for that!", "Question # Say true or false:If ${y^2} - 3 = 0,$ then $y = \\pm \\sqrt {3.}$A. TrueB. False Hint: In order to comment about this question whether it is true or false, we will first factorize the given equation with the help of formula given as ${a^2} - {b^2} = (a + b)(a - b)$ then we get the solution. Given equation is ${y^2} - 3 = 0$ We will factorize given equation to get the roots $\\left[ {\\because {a^2} - {b^2} = (a + b)(a - b)} \\right]$ Here $a = y{\\text{ and }}b = \\sqrt 3$ By substituting the values in above formula, we get $\\Rightarrow {y^2} - {\\left( {\\sqrt 3 } \\right)^2} = 0 \\\\ \\Rightarrow (y + \\sqrt 3 )(y - \\sqrt 3 ) = 0 \\\\ \\Rightarrow y = - \\sqrt 3 {\\text{ or }}\\sqrt 3 \\\\ \\Rightarrow y = \\pm \\sqrt 3 \\\\$ Hence, the solution of ${y^2} - 3 = 0$ is $y = \\pm \\sqrt 3$. Therefore the given statement is true. Note: In order to solve these types of questions, remember all the algebraic identities. These questions are based on simple arithmetic operations and few identities. By convention, letters at the beginning of the alphabet (e.g. a, b, c) are typically used to represent constants, and those toward", "$z^2 \\pm 2 = 3y^2$ from, mine should be like $3a^2 + 1 = y^2,$ which has an infinite amount of solutions if i remember correctly. But yes, the 3 changes things, if thats what you meant. Let me put another effort into solvign this one – mathninja Dec 28 '20 at 2:30 • I did get $z^2 \\pm 2 = 3v^2$ from your last equation. You may be able to deduce this from the fact that $2y-1$, $2y+1$ are relatively prime so one must be a square and the other 3 times a square. [I should have used $v$ instead of $y$ to avoid any confusion. – Mike Dec 28 '20 at 2:54 • Each square factor on the LHS of your last equation is a factor of exactly one of $2y-1$, $2y+1$, and 3 is a factor of exactly one of $2y-1$, $2y+1$. So exactly one of $2y-1$, $2y+1$ is a square $z^2$ and the remaining is $3v^2$ for some integer $v$. – Mike Dec 28 '20 at 3:06", "two equations, we obtain $$(x+y)^3-(y+z)^3=z-x.$$ Let $Z=x+y$ and $X=y+z$. Factoring the expression $Z^3-X^3$ on the left, we obtain $$(Z-X)(Z^2+ZX+X^2)= (x-z)(Z^2+ZX+X^2)=z-x.$$ If $z \\ne x$, this forces $Z^2+ZX+X^2=-1$. But this last equation does not have real solutions. There are many ways to see this, such as the Quadratic Formula. A cuter way is to note that $$4(Z^2 +ZX+X^2)=(2Z+X)^2+3X^2 \\ge 0.$$ So we must have $z=x$. Similarly, $z=y$, and we end up looking for solutions of $(2a)^3=a$. -", "- 4$. If $x = - 4$ then the second equation becomes ${y}^{2} = - 12$, which has no Real valued solutions. If $x = 1$ then the second equation becomes ${y}^{2} = 3$, so $y = \\pm \\sqrt{3}$", "$\\sqrt{y} = 1$ or $\\sqrt{2x-y} = 1$. The first case: $y = 1$. From this and the first equation, one has $$x-5 = \\sqrt{-x} + \\sqrt{6-3x}.$$ But, $x\\geq \\frac{y}{2} \\geq 0$. Then, there is no solution for this case. The second case: $2x-y = 1$ or $y = 2x-1$. From this and the first equation, one gets: $$2x^2 - 5x -1 = \\sqrt{x-2} + \\sqrt{4-x}.$$ To finish, let $f(x)=2x^2-5x-1\\quad \\text{and}\\quad g(x)=\\sqrt{x-2}+\\sqrt{4-x}$. Note that $f(3)=g(3)=2$ and $f(x)$ is increasing $\\forall x\\in [2,4]$, while $g(x)$ is decreasing when $x\\in (3,4]$ and increasing when $x\\in [2,3)$. Thus $x=3$ is the unique solution of $f(x)=g(x)$ in $[2,4]$. Answer: $x=3, \\; y=5$. • Ah thank you for this post. – GAVD Aug 13 '15 at 7:42", "## College Algebra (10th Edition) We are given: $y=\\pm\\sqrt{1-2x}$ This is not a function, because $y$ can take on different values for the same $x$. For example: $x=0$: $y=\\pm\\sqrt{1-2*0}=\\pm 1$", "# Is this a quadratic equation? I have this equation: $-19y^2 + 5y\\sqrt{y} + 10y + 12 = 0$ I'm stuck with finding $y$. I tried numerous adjustments like this for example: $y*(-19y+5\\sqrt{y}+10)+12 =0$ but it didn't get me any closer to a solution. I know it must be really simple but right now I'm stuck for hours. Can you give me a hint how to solve this equation? • Let's pose $t = \\sqrt{y}$ and substitute... Jun 23, 2014 at 17:22 • Set $x = \\sqrt y$. Then it will be a quartic. I notice that one root is $x = 1$, so you can factor out $(x-1)$ and get a cubic. I haven't checked after that to see if there are other simplifications. Jun 23, 2014 at 17:23 • @StephenMontgomery-Smith $1=x=\\sqrt{y}$ is not a root of the expression. – mike Jun 23, 2014 at 17:34 • @mike Oops, you are correct. I added the coefficients incorrectly. Jun 23, 2014 at 17:35 • Numerical solution showed that there is only one positive solution: $y=1.30971$. You can get closed-from solution from $-19t^4 + 5t^3 + 10t^2 + 12 = 0$ using the formulas at wikipedia.com – mike Jun 23, 2014 at 17:46 No, because for that to be the case, all exponents would have to be natural numbers." ]

[ "2x^2- 1$ so $z= \\pm\\sqrt{4x-2x^2-1}$. That z only exists when $\\frac{4- 2\\sqrt{2}}{4}< x< \\frac{4+ 2\\sqrt{2}}{4}$. In that range, the roots are x, y= 2- x and z= $\\pm\\sqrt{4x-2x^2-1}$ for every value of x. It is true that x= 1 is in that range and then y= 2-1= 1 and z= $\\sqrt{4- 2-1}= 1$ so that (1, 1, 1) is one solution. x= 1, y= 1, z= $-\\sqrt{4-2-1}= -1$ or (1, 1, -1) is also a solution. Those may be the only integer solutions. 4. Sorry, I don't really understand what do you mean by \"In that range, the roots are x, y= 2- x and z= for every value of x.\" By roots, which roots are you talking about? Please explain about the above in details and on how do you come to the conclusion that the roots are x, y= 2- x and z = . A little lost here", "3)(x^3 + 1) = 0$ So $x^3 = 3$ ==> $x = \\sqrt[3]{3}$ ==> $y = (\\sqrt[3]{3})^2 = \\sqrt[3]{9}$ or $x^3 = -1$ ==> $x = -1$ ==> $y = (-1)^2 = 1$ So the points where $y^{\\prime} = 0$ are $(-1, 1)$ and $(\\sqrt[3]{3}, \\sqrt[3]{9})$ . (Both of these are points in the original equation.) We do the same process to find where the vertical tangents are. Where is $y^{\\prime}$ undefined? Where $y^2 - x = 0$ $y = \\pm \\sqrt{x}$ So put this into your original equation: $x^3 + y^3 - 3xy = 3$ $x^3 + (\\pm \\sqrt{x})^3 - 3x(\\pm \\sqrt{x}) = 3$ $x^3 \\pm \\sqrt{x^3} \\mp 3 \\sqrt{x^3} - 3 = 0$ $x^3 \\mp 2 \\sqrt{x^3} - 3 = 0$ Let $z = \\sqrt{x^3}$ Then $z^2 \\mp 2 z - 3 = 0$ For the \"-\" sign: $z^2 - 2z - 3 = 0$ $(z - 3)(z + 1) = 0$ $z = 3$ ==> $x = \\sqrt[3]{3^2} = \\sqrt[3]{9}$ ==> $y = \\pm \\sqrt{\\sqrt[3]{9}} = \\pm \\sqrt[6]{9}$ (I note that the \"-\" solution is not a point on the original equation.) and $z = -1$ ==> $x = \\sqrt[3]{(-1)^2} = 1$ ==> $y = \\pm \\sqrt{1} = \\pm 1$ (I note that the \"+\" solution is not a point on the original equation.) So", "1+y^2+7=9,y^2=1$ So, the roots are $\\pm \\sqrt7i,1\\pm i$ Choosing $z = 1 + y i$, you obtain the formula: $$y^4 – 2 i y^3 – 9y^2 + 2 i y + 8 = 0$$ Because $y$ is real, this means that $$2 y^3 – 2y = 0$$ and $$y^4 – 9y^2 + 8 = 0$$ The first equation implies $y = 0, 1, -1$. The second equation implies that $y^2 = 8$ or $y^2 = 1$. Thus we see both $1 + i$ and $1 – i$ are solutions. Factor out the polynomial $(z – 1 + i)(z – 1 – i) = z^2 – 2z + 2$ and you’ll find the remaining roots. Maxima tells me the roots are $\\pm i \\sqrt{7}$, $1 \\pm i$. It factors as $(z^2 + 7) (z^2 – 2 z + 2)$.", "etc.) and simultaneous equations. let $Z = \\sqrt{4i-3}$ where Z is complex. i.e. $Z$ is of the form $Z = a + ib$, where $a$ and $b$ are REAL coefficients. (This is important later on.) Square both sides: $Z^2 = -3+4i$ Expand the standard form of $Z^2$: $(a+ib)^2 = -3+4i$ This becomes $a^2 - b^2 + 2abi = -3+4i$ The above is an identity and so the real and imaginary parts can be equated. Hence two equations are formed: $a^2 - b^2 = -3$................(1) $2abi = 4i$ which becomes $ab = 2$......(2) In (2), $b = 2/a$ so $a^2 - (2/a)^2 + 3 = 0$ i.e. $a^4 + 3a^2 - 4 = 0$ (multiplying all by $a^2$ and rearranging) which is a quadratic-styled quartic: $(a^2 + 4)(a^2 - 1) = 0.$ Solving, $a=1, -1, 2i$ and $-2i$ BUT since $a$ and $b$ are real, we discard $2i$ and $-2i$. Now we simply solve for $b$ using $b = 2/a$. Hence the solutions are $a=1, b=2$ or $a=-1, b=-2$. and so $\\sqrt{4i-3} = \\pm(1+2i)$ since we defined $Z = a+ib$. Best of luck with any future questions! Hope this helped. • Just add $around math expressions, expressed in the LaTeX formatting. – PearlSek Dec 23 '16 at 22:21 • Well that was simpler than I thought, thanks for that!", "- z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = \\sqrt{(y-1)(z-1)}$! Thus, $$u^2 - 2u + 2 = (u-1)^2 + 1 > 0$$ and the claim holds for x = 1. If x > 1, we see the $\\sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$: $$x + xyz \\ge x - 1 + y - 1 + z - 1 + 2\\sqrt{(x-1)(y-1)} + 2\\sqrt{(y-1)(z-1)} + 2\\sqrt{(x-1)(y-1)}$$ which leads to $$xyz \\ge y + z - 3 + 2\\sqrt{(x-1)(y-1)} + 2\\sqrt{(y-1)(z-1)} + 2\\sqrt{(x-1)(z-1)}$$ Again, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4\\sqrt{6}$. Now, we see the finish: setting $u = \\sqrt{x-1}$ gives $x = u^2 + 1$. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations: $$u^2(yz) - u(2\\sqrt{y-1} + 2\\sqrt{z-1}) + (3 + yz - y - z - 2\\sqrt{(y-1)(z-1)}) \\ge 0$$ Because the coefficient of $u^2$ is positive, all we need to do is to verify that the discriminant is nonpositive: $$b^2 - 4ac = 4(y-1) + 4(z-1) - 8\\sqrt{(y-1)(z-1)} -", "parameters $(y,z)$ such that $y< z< \\sqrt[3]{3}y$ yields a solution. Second example If $Y$ is distributed like $y$ times a reduced exponential and $Z$ is distributed like $z$ times the sum of two independent reduced exponentials, then $E(Y)=y$, $E(Y^3)=6y^3$, $E(Z)=2z$ and $E(Z^3)=24z^3$, hence every couple of parameters $(y,z)$ such that $\\sqrt[3]{4}z< y< 2z$ yields a solution.", "two equations, we obtain $$(x+y)^3-(y+z)^3=z-x.$$ Let $Z=x+y$ and $X=y+z$. Factoring the expression $Z^3-X^3$ on the left, we obtain $$(Z-X)(Z^2+ZX+X^2)= (x-z)(Z^2+ZX+X^2)=z-x.$$ If $z \\ne x$, this forces $Z^2+ZX+X^2=-1$. But this last equation does not have real solutions. There are many ways to see this, such as the Quadratic Formula. A cuter way is to note that $$4(Z^2 +ZX+X^2)=(2Z+X)^2+3X^2 \\ge 0.$$ So we must have $z=x$. Similarly, $z=y$, and we end up looking for solutions of $(2a)^3=a$. -", "solve $x-2\\sqrt{x-3}=3$ If so $-2\\sqrt{x-3} =3-x~\\implies~2\\sqrt{x-3}=x-3$ Let x-3 = y. Then your equation becomes: $-2y = y^2~\\implies~0=y^2-2y~\\implies~y(y-2)=0$ A product of 2 factors equals zero if one factor equals zero: $y = 0~\\vee~y-2=0~\\implies~y=0~\\vee~y=2$ That means: x-3 = 0 ==> x = 3 or x-3=2 ==> x = 5 You must plug in the result into the original equation to check if the result is a valid solution: x = 3: $3-2\\sqrt{3-3}=3$ x = 5: $5 - 2\\sqrt{5-3} \\neq 3$ Therefore: Only x = 3 is a solution. 3. Originally Posted by harnar ... √(5x+1) = x+1 I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to? ... 1. Square both sides, you'll get: $x^2-3x=0~\\implies~x(x-3)=0$ 2. A product equals zero if one factor equals zero: $x = 0~\\vee~x-3=0~\\implies~\\boxed{x=0~\\vee~x=3}$ Plug in the results into the original equation to check if they are valid solutions: $x = 0: \\sqrt{0+1} = 0+1~\\longrightarrow~OK$ $x = 3: \\sqrt{15+1} = 3+1~\\longrightarrow~OK$", "is: $y+z=-3$ $z+x=-6$ $x+y=-5$ for which the solution set if $x=-4, y=-1, z=-2$. Example 3: Solve the following system of equations: $y^{2}+yz+z^{2}=49$…call this equation (1) $z^{2}+xz+x^{2}=19$…call this equation (2) $x^{2}+xy+y^{2}=39$…call this equation (3) Subtracting (2) from (1), we get the following: $y^{2}-x^{2}+z(y-x)=30$, that is, $(y-x)(x+y+z)=30$…call this equation (4). Similarly, from (1) and (3), we get the following: $(z-x)(x+y+z)=10$…call this equation (5). Hence, from equations (4) and (5), by division, we get the following: $\\frac{y-x}{z-x}=3$, hence, $y=3x-2z$ Substituting in Equation (3), we obtain $x^{2}-3xz+3z^{2}=13$. From (2), we get the following: $x^{2}+xz+z^{2}=19$ Solving these homogeneous equations (hint: put $z=mx$, where m is parameter to be found), we get the following: $x=\\pm 2$, $z=\\pm 3$; and therefore, $y=\\pm 5$ or, $x=\\pm \\frac{11}{\\sqrt{7}}$; $z=\\pm \\frac{1}{\\sqrt{7}}$; and therefore, $y=\\mp \\frac{19}{\\sqrt{7}}$. Example 4: Solve the following system of equations: $x^{2}-yz=a^{2}$ $y^{2}-zx=b^{2}$ $z^{2}-xy=c^{2}$ Solution 4: Multiply the equations by y, z, and x respectively, and add; then, $c^{2}x+a^{2}y+b^{2}z=0$…call this equation I Multiply the equations by z, x and y respectively; and add; then $b^{2}x+c^{2}y+a^{2}z=0$…call this equation II From (I) and (II), by cross multiplication, $\\frac{x}{a^{4}-b^{2}c^{2}}=\\frac{y}{b^{4}-c^{2}a^{2}}=\\frac{x}{c^{4}-a^{2}b^{2}}=k$, suppose. Substitute in any one of the given equations; then, $k^{2}(a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2})=1$; hence, we get the following solution: $\\frac{x}{a^{4}-b^{2}c^{2}}=\\frac{y}{b^{4}-c^{2}a^{2}}=\\frac{z}{c^{4}-a^{2}b^{2}}=\\pm \\frac{1}{\\sqrt{a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}}}$. More stuff in the pipeline, Nalin Pithwa This site uses Akismet to reduce spam. Learn how your comment data is processed.", "### Home > INT3 > Chapter 3 > Lesson 3.2.4 > Problem3-113 3-113. Solve each equation. Remember to check your solutions. 1. $(y-3)^2=2y-10$ Expand $(y−3)^2$. Set the equation equal to $0$. Factor and use the Zero Product Property or use the Quadratic Formula to solve. $y=4\\pm i\\sqrt(3)$ 1. $| y - 3 | = 2 y - 10$ This equation has an absolute value, so you will need to write two equations to solve it. $y−3=2y−10$ or $y−3=−(2y−10)$ $y=7\\text{ or }y=\\frac{13}{3}$ Check both solutions in the original equation. Do they both solve the equation?", "## College Algebra 7th Edition $y$ is NOT a function of $x$ Take the fourth root of each side of the equation to obtain: $x=y^4 \\\\\\pm\\sqrt[4]{x} = \\sqrt[4]{y^4} \\\\\\pm \\sqrt[4]{x} = y$ Note that in the last equation above, for every value of $x$, there corresponds two different value of $y$. Thus, the $y$ is not a function of $x$.", "equation. x=1 is. Now I can't tell you what the 24th possible value is because my calculator sadly cannot handle large numbers. I can tell you that the only possible values are those for which (1+8x^2) is an odd perfect square. – user25406 Apr 22 '15 at 2:24 Rewrite as $2x^2=n^2+n$ Multiply by $4$ to obtain $8x^2=4n^2+4n$ Set $y=2x, z=2n+1$ and confirm $2y^2=z^2-1$ or $z^2-2y^2=1$, which is a Pell Equation. To solve the original problem you need solutions to the Pell equation with $y$ even. Now note the factorisation $(z+y\\sqrt 2)(z-y\\sqrt 2)=1$ where the factors differ only in the sign of $\\sqrt 2$. The same is true of $(3+2\\sqrt 2)(3-2\\sqrt 2)=1$. Note: this is obtained by squaring $(1+\\sqrt 2)(1-\\sqrt 2)=-1$, but you want solutions for $+1$ only. Note next that $(z+y\\sqrt 2)(3+2\\sqrt 2)=(3z+4y)+(2z+3y)\\sqrt 2$, and we have $$(3z+4y)^2-2(2z+3y)^2=9z^2+24yz+16y^2-8z^2-24yz-18y^2=z^2-2y^2=1$$ So if $(z,y)$ is a solution of the Pell equation, so is $(3z+4y,2z+3y)$. If $y$ is even, so is $2z+3y$. Given a solution, you can find another. Using $(z+y\\sqrt 2)(3-2\\sqrt 2)=(3z-4y)+(3y-2z)\\sqrt 2$ you can find descending solutions, which is a help in showing whether you have all the solutions or not. There is much to learn about Pell Equations and their solutions, but playing with the ideas here (you get two things multiplied together equal to $1$ - such things", "Question # Say true or false:If ${y^2} - 3 = 0,$ then $y = \\pm \\sqrt {3.}$A. TrueB. False Hint: In order to comment about this question whether it is true or false, we will first factorize the given equation with the help of formula given as ${a^2} - {b^2} = (a + b)(a - b)$ then we get the solution. Given equation is ${y^2} - 3 = 0$ We will factorize given equation to get the roots $\\left[ {\\because {a^2} - {b^2} = (a + b)(a - b)} \\right]$ Here $a = y{\\text{ and }}b = \\sqrt 3$ By substituting the values in above formula, we get $\\Rightarrow {y^2} - {\\left( {\\sqrt 3 } \\right)^2} = 0 \\\\ \\Rightarrow (y + \\sqrt 3 )(y - \\sqrt 3 ) = 0 \\\\ \\Rightarrow y = - \\sqrt 3 {\\text{ or }}\\sqrt 3 \\\\ \\Rightarrow y = \\pm \\sqrt 3 \\\\$ Hence, the solution of ${y^2} - 3 = 0$ is $y = \\pm \\sqrt 3$. Therefore the given statement is true. Note: In order to solve these types of questions, remember all the algebraic identities. These questions are based on simple arithmetic operations and few identities. By convention, letters at the beginning of the alphabet (e.g. a, b, c) are typically used to represent constants, and those toward", "plug into the second equation to get $y$. Then we could plug into the first equation to get $x$. Using $z=1$, the second equation gives $y = 2$. Plugging into the first equation, we get $x+2+1 = 4$. Hence, $x=1$. The solution to the matrix equation is thus $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix}1\\\\2\\\\1 \\end{pmatrix}.$", "# Positive integral solutions of $3^x+4^y=5^z$ Are there more integral solutions for $3^x+4^y=5^z$, than $x=y=z=2$ ? If not, how do I show that? I could show that for $3^x+4^x=5^x$, but I'm stuck at the general case? Any ideas, maybe graphs, logarithms or infinite descent? • $x=0, y=z=1$ is also a solution. Perhaps you mean positive integer solutions? Apr 29, 2021 at 3:51 I will prove that the only positive integral solution to $$3^x+4^y=5^z$$ is $$x=y=z=2$$. Proof. Looking at the equation mod $$4$$, we see $$3^x\\equiv 1\\pmod{4}$$, or equivalently, $$(-1)^x\\equiv 1\\pmod{4}$$. This implies $$x=2x_1$$ for some integer $$x_1$$. Also, looking at the equation mod 3, we see $$5^z\\equiv 1 \\pmod{3}$$, or equivalently $$(-1)^{z}\\equiv 1\\pmod{3}$$. This implies $$z=2z_1$$ for some integer $$z_1$$. Thus, $$2^{2y}=4^{y}=(5^{z_1})^2-(3^{x_1})^2=(5^{z_1}+3^{x_1})(5^{z_1}-3^{x_1})$$ Hence, $$5^{z_1}+3^{x_1}=2^{s}$$ and $$5^{z_1}-3^{x_1}=2^{t}$$, with $$s>t$$ and $$s+t=2y$$. Solving for $$5^{z_1}$$ and $$3^{x_1}$$, we get $$5^{z_1}=2^{t-1}(2^{s-t}+1) \\ \\ \\textrm{ and } \\ \\ 3^{x_1}=2^{t-1}(2^{s-t}-1)$$ Since the left side of both equalities is odd, $$t$$ must be equal to $$1$$. Let $$u=s-t$$. Then, the equation $$3^{x_1}=2^{t-1}(2^{s-t}-1)$$ becomes $$3^{x_1}=2^{u}-1$$. Looking at this equation mod $$3$$, we get $$0\\equiv (-1)^{u}-1\\pmod{3}$$, and so $$u$$ is even, say $$u=2u_1$$ for some positive integer $$u_1$$. Thus, $$3^{x_1}=(2^{u_1})^{2}-1=(2^{u_1}+1)(2^{u_1}-1)$$ Hence, $$2^{u_1}+1=3^{\\alpha}$$ and $$2^{u_1}-1=3^{\\beta}$$ for some $$\\alpha>\\beta$$. But this gives, $$3^{\\alpha}-3^{\\beta}=2$$, and hence $$\\alpha=1$$ and $$\\beta=0$$. Consequently, $$u_1=1$$, and so $$u=2$$. This gives us the", "for integer values of $z$ until I reached a solution which had been repeated (as here $y$ is bigger than $z$, so I would be repeating values with $y$ and $z$ swapped around) or where $z$ became smaller than $y$. For $x = 1$ I found only 1 solution, $(1, 2, 16)$. Checking with the original formula this does agree to a total of $100$. I then continued repeating the procedure with $x = 2$, $y = 2$,$3$,$4$... and found a solution of $(2, 4, 7)$. With $x = 3$, $y = 3$,$4$,$5$... there are no solutions, as $z$ becomes smaller than $y$ where $y=5$ (and $z = 4.375$) and there are no integer solutions before this. With $x = 4$, $y = 4,5,6...$ $z$ becomes smaller than $y$ when $y = 5$ $(z = 3.333)$ therefore there are no solutions where $x = 4$. After $x = 4$, the intitial value of $z$ is always smaller than $y$, therefore there are no further solutions. The only 2 solutions to the problem are: $(1, 2, 16)$ and $(2, 4, 7)$. Fred from Albion Heights School also offered some non-integer solutions: $h = 1$, $w = 1$, $l = 24.5$ $h = 2$, $w = 2$, $l = 11.5$ $h = 4$, $w = 4$, $l = 4.25$ $h", "- t$$ Now, we know $y + z + 3t = -1$, we can substitute the above version of $z$ into this equation and get $$y + (-1 -t) + 3t = -1$$ Thus we can write $y$ in function of $t$: $$y = -2t$$ Finally, we can do the same for $x$, we have the equation $x - y + 3z + 2t = 1$. Substituting $y$ and $z$, we get $$x - (-2t) + 3(-1-t) + 2t = 1$$ Solving for $x$ gives us $$x = 4-t$$ So we get the following solutions for the system: $$x= 4-t,~y=-2t,~z=-1-t$$ This means that for every real number $t$ that we take, we get a solution for our system. For example, taking $t=0$, we get $(x,y,z,t) = (4,0,-1,0)$. On the other hand, taking $t=1$, we get $(x,y,z,t) = (3,-2,0,1)$ as a solution. So we see that we have infinitely many solutions, one for each $t$.", "## College Algebra 7th Edition $(x,y,z)=(4,3,4)$ Given: $x-2y+3z=10$ ; $2y-z=2; 3z=12$ Here $3z=12 \\implies z=4$ Plug $z=4$ in the second equation. we get $2y-4=2 \\implies y=3$ Now, Plug $y=3,z=4$ in the first equation. we have $x-2(3)+3(4)=10$ or, $x=10-6 \\implies x=4$ Hence, the solution is:$(x,y,z)=(4,3,4)$", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "## College Algebra 7th Edition $x=\\pm 1$ $x=\\pm \\sqrt{3}$ $x^{4}-4x^{2}+3=0$ First, we substitute $y=x^{2}$ into the equation: $y^{2}-4y+3=0$ Next, we factor and solve: $(y-1)(y-3)=0$ $y-1=0$ or $y-3=0$ $y=1$ or $y=3$ Finally, we convert the $y$ solutions to $x$ solutions: $y=x^2$ $x=\\pm\\sqrt{y}$ If $y=1$: $x=\\pm \\sqrt{1}=\\pm 1$ If $y=3$: $x=\\pm \\sqrt{3}$" ]

[ "# Like Base System? Number Theory Level pending For each integer $$n\\geq 4$$, let $$a_n$$ denote the base-$$n$$ number $$(0.133133133\\ldots)_n$$. The product $$a_4a_5\\cdots a_{99}$$ can be expressed as $$\\dfrac{m}{n!}$$, where $$m$$ and $$n$$ are positive integers and $$n$$ is as small as possible. What is the value of $$m$$? ×", "$\\frac{m}{n}$ can be expressed as $\\frac{p}{q}$ where $p, q$ are coprime integers. What is $p+q$? Note: $0.\\overline{133}_k$ refers to the repeating decimal $0.133133133\\ldots$ evalauted in base $k$. ×", "2018 IMO Problems/Problem 5 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Let $a_1, a_2, \\dots$ be an infinite sequence of positive integers. Suppose that there is an integer$N > 1$ such that, for each $n \\geq N$, the number $\\frac{a_1}{a_2}+\\frac{a_2}{a_3}+\\dots +\\frac{a_{n-1}}{a_n}+\\frac{a_n}{a_1}$ is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \\geq M.$", "the following theorem (Lemma 2.1 in this paper on disjunctive sequences), whose proof is along the same lines as that posted in the answer by @EricWofsey: If $a_1, a_2, a_3, \\dots$ is a strictly increasing infinite sequence of positive integers such that $$\\lim_{n\\to \\infty} \\frac{a_{n+1}}{a_n} = 1$$ then for any positive integer $m$ and any integer base $b \\ge 2$, there is an $a_n$ whose expression in base $b$ starts with the expression of $m$ in base $b$. Your result is then the very special case of taking $a_n = n^k$ and $k=m$. For this and some other special cases, see these examples of disjunctive sequences. (E.g., for any desired positive integer, there are infinitely many prime numbers whose representation begins with the digits of that number.) NB: For any positive integer exponent $k$ and any desired positive integer $m$, there are infinitely many positive integers $n$ such that the representation of $n^k$ starts with the representation of $m$; furthermore, this holds for digital representations in any integer base $b \\ge 2$.", "# proof that number of sum-product numbers in any base is finite Let $b$ be the base of numeration. Suppose that an integer $n$ has $m$ digits when expressed in base $b$ (not counting leading zeros, of course). Then $n\\geq b^{m-1}$. Since each digit is at most $b-1$, we have that the sum of the digits is at most $m(b-1)$ and the product is at most $(b-1)^{m}$, hence the sum of the digits of $n$ times the product of the digits of $n$ is at most $m(b-1)^{m+1}$. If $n$ is a sum-product number, then $n$ equals the sum of its digits times the product of its digits. In light of the inequalities of the last two paragraphs, this implies that $m(b-1)^{m+1}\\geq n\\geq b^{m-1}$, so $m(b-1)^{m+1}\\geq b^{m-1}$. Dividing both sides, we obtain $(b-1)^{2}m\\geq(b/(b-1))^{m-1}$. By the growth of exponential function, there can only be a finite number of values of $m$ for which this is true. Hence, there is a finite limit to the number of digits of $n$, so there can only be a finite number of sum-product numbers to any given base $b$. Title proof that number of sum-product numbers in any base is finite ProofThatNumberOfSumproductNumbersInAnyBaseIsFinite 2013-03-22 15:47:06 2013-03-22 15:47:06 rspuzio (6075) rspuzio (6075) 7 rspuzio (6075) Proof msc 11A63", "+0 # Product 0 87 1 Determine the value of the infinite product $$(2^{1/3})(2^{1/9})(2^{1/27}) \\dotsm$$", "to the $$0$$ power is defined to be $$1$$. 108$$x^{−n} = \\frac{1}{x^{n}}$$ , given any integer $$n$$, where $$x$$ is nonzero. 109$$\\frac { x ^ { - n } } { y ^ { - m } } = \\frac { y ^ { m } } { x ^ { n } }$$ , given any integers $$m$$ and $$n$$, where $$x ≠ 0$$ and $$y ≠ 0$$. 110Real numbers expressed the form $$a × 10^{n}$$ , where $$n$$ is an integer and $$1 ≤ a < 10$$. 111$$\\frac { x ^ { m } } { x ^ { n } } = x ^ { m - n }$$ ; the quotient of two expressions with the same base can be simplified by subtracting the exponents.", "# It Isn't As Hard As It Looks The sequence $$a_0,a_1,a_2,\\dots$$ satisfies $a_{m+n} + a_{m-n} = \\frac{1}{2}(a_{2m}+a_{2n})$ for all non-negative integers with $$m\\geq n$$. If $$a_1 = 1$$, determine the value of $a_{1995}$ ×", "Let $m$ and $n$ be positive integers and let $a$ represent a real number, a variable, or an algebraic expression. 1. $$\\frac{{{a^m}}}{{{a^n}}} = {a^{m – n}}$$, $$m > n$$, $$a \\ne 0$$. 2. $$\\frac{{{a^n}}}{{{a^n}}} = 1 = {a^0}$$, $$a \\ne 0$$.", "4$$. The product $$a_4 a_5 \\cdots a_{99}$$ can be expressed as $$\\frac{m}{n!}$$ where $$m, n$$ are positive integers and $$n$$ is as small as possible. $$\\frac{m}{n}$$ can be expressed as $$\\frac{p}{q}$$ where $$p, q$$ are coprime integers. What is $$p+q$$? ×", "# What Would You Call A Non-base 10 Decimal? Let $$a_k$$ represent the repeating decimal $$0.\\overline{133}_k$$ for $$k \\geq 4$$. The product $$a_4 a_5 \\cdots a_{99}$$ can be expressed as $$\\frac{m}{n!}$$ where $$m, n$$ are positive integers and $$n$$ is as small as possible. $$\\frac{m}{n}$$ can be expressed as $$\\frac{p}{q}$$ where $$p, q$$ are coprime integers. What is $$p+q$$? ×", "# Evaluate: $x_i\\ge0; \\max_{1\\le i \\le n , \\sum_{i=1}^{n}{x_i=a}}x_1x_2...x_n$ Let a be a fixed positive real number.Evaluate: $$x_i\\ge0; \\max_{1\\le i \\le n , \\sum_{i=1}^{n}{x_i=a}}x_1x_2...x_n$$ I guess it should be $x^n$ where $x=\\frac{a}{n}$since this is true for $n=2$, I have generalise the case only. Am I right? Use A.M.$\\geq$ G.M. for the n numbers...............", "for every integer $k\\geq 2\\nu_{o}(G)-1$.", "+0 # math 0 129 1 Consider the sequence $(a_k)_{k\\geq 1}$ of positive rational numbers defined by $a_1 = \\frac{2020}{2021}$ and for $k \\geq 1,$ if $a_k = \\frac{m}{n}$ for relatively prime positive integers $m$ and $n,$ then$$a_{k+1} = \\frac{m+18}{n+19}.$$Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\\frac{t}{t+1}$ for some positive integer $t.$ Mar 10, 2021", "# $n$ numbers with their sum and product equal to each other. I was trying to do this: For any integer $n\\ge2$, find $n$ numbers $a_1,\\ a_2,\\ a_3,\\dots,a_n$ such that $$a_1+a_2+a_3+\\dots+a_n=a_1a_2a_3\\dots a_n$$ For $n=2$, we have $a_1=a_2=2$. (As $2+2=2\\times2)$ For $n=3$, we have $a_1=1,\\ a_2=2,\\ a_3=3.$ ($1+2+3=1\\times2\\times3$) Do there exist any such numbers for higher values of $n$ (i.e $n\\ge4$). If there are give such examples. If there don't exist such numbers for $n\\ge4$, there must be a proof from elementary number theory for this. What is that proof? From therein, further examples are $1,1,2,4$ and $1,1,2,2,2$. It is also shown that for every $n$ the number of such sequences is positive (just take $1,1,\\dots,1,2,n$) and finite. See below for the number of such sequences $a(n)$ for $1\\le n\\le 100$.", "# 2014-09 Product of series For integer $$n \\geq 1$$, define $a_n = \\sum_{k=0}^{\\infty} \\frac{k^n}{k!}, \\quad b_n = \\sum_{k=0}^{\\infty} (-1)^k \\frac{k^n}{k!}.$ Prove that $$a_n b_n$$ is an integer. GD Star Rating", "$\\sum_{k\\geq 1}\\frac{1}{2^k}$ is converging. In particular: $$a_{n+2} \\leq \\exp\\sum_{k=1}^{n}\\frac{1}{2^k} \\leq e.\\tag{2}$$ – Jack D'Aurizio Feb 7 '15 at 16:33 Represent $a_n$ as a binary number, and prove that $a_{n+1}$ has more binary digits than $a_n$. This shows infinite. If period were $p$ (after the first $d$ digits), I guess we can consider large enough $N$, and $a_N$. This is incomplete and perhaps the more difficult part of this problem.", "m)$ is the positive greatest common divisor of $n$ and $m$. And the easiest way to achieve that is is to make one or both of $n$ and $m$ prime numbers. But why does the relative primality of $n$ and $m$ matter? Suppose that we are using open hashing with the hash function at the top of this document coupled with some other method for collision resolution (e.g., chaining). Let us further assume that $n > m$. Next, recall that $a \\pmod n = r$ (for $a$, $n$ positive integers) means that for some integer $s$, there exists an integer $r$ so that $a = ns + r$. Now, let $d = \\gcd(n, m) \\geq 1$. Given an input datum consisting of the string $a_1 a_2 a_3 \\dots a_k$, where $0 \\leq a_k \\leq B$ for some constant positive integer $B$, the hash becomes the nonnegative integer, $r = \\left( \\sum^k_{i = 1} m a_i \\right) \\mod n$ Let $u$ be the sum over $i = 1$ to $k$ of $a_i$; then the above becomes: $r = (mu) \\mod n$ since $m$ is constant with respect to the sum, and thus can be factored out. But, since $m$ has $d$ as a factor, we know there exists some positive integer $q$ such that $m = dq$ and thus", "$$(1, 2, \\dots, n)$$ is defined as $$(1 \\times a) || (2 \\times a) || \\dots || (n \\times a)$$ Parameters: a (int) – the base integer n (int) – the number of products in the concatenated product int the concatenated product of $$a$$ and $$(1, 2, \\dots, n)$$", "1$ by the definitions of $m$ and ${r}_{i}$, ${r}_{i}m\\ge {q}_{i}$ and ${r}_{i}m-{q}_{i}\\ge 0$. Thus, ${p}_{i}^{{r}_{i}m-{q}_{i}}$ is a non-negative integer for every $i=1,...,k$ since it’s a power of a prime number to a non-negative integer. Since ${p}_{j}^{{r}_{j}}$ is also a non-negative integer for all $j=k+1,...,l$ and $m$ is a positive integer by their definitions, the whole product $w$ is also a non-negative integer. Then the quantity $nw$ is also an integer since the product of two integers is also an integer. Let ${n}_{s}{n}_{s-1}...{n}_{0}$ be the representation of $nw$ in the base $b$. Then consider the representation of $\\frac{nw}{{b}^{m}}=\\frac{n{b}^{m}}{d{b}^{m}}=\\frac{n}{d}$ in base $b$: ${n}_{s}{n}_{s-1}...{n}_{m+1}.{n}_{m}...{n}_{0}$ with the radix point between ${n}_{m}$ and ${n}_{m+1}$. This is the representation of $\\frac{n}{d}$ in base $b$ using a finite number of digits as desired. Q.E.D. ## Intuitive Reasoning The second half of the proof gives us an intuition as to why this theorem holds. The question of whether we can write down $\\frac{n}{d}$ using a finite number of digits in base $b$ really comes down to whether we can write down $\\frac{1}{d}$ using a finite number of digits in base $b$ because $\\frac{n}{d}$ is just $\\frac{1}{d}$ repeated $n$ times. If $\\frac{1}{d}$ is expressible using a finite number of digits in base $b$, there ought to be some natural (positive) number $c$ such that c-th digit after" ]

[ "a_5&=4b+6-3a, \\end{align*} or, in general, $$a_n=(n-1)b+\\frac{(n-2)(n-1)}{2}-(n-2)a.$$ Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \\begin{align*} 0=&18b+\\frac{17\\cdot18}{2}-17a\\\\ 0=&91b+\\frac{90\\cdot91}{2}-90a\\\\ b=&\\frac{17a-\\frac{17\\cdot18}{2}}{18}=\\frac{90a-\\frac{90\\cdot91}{2}}{91}. \\end{align*} Now, cross multiplying, we find that \\begin{align*} 91\\left(17a-\\frac{17\\cdot18}{2}\\right)&=18\\left(90a-\\frac{90\\cdot91}{2}\\right)\\\\ 1547a-\\frac{(18\\cdot91)\\cdot17}{2}&=1620-\\frac{(18\\cdot91)\\cdot90}{2}\\\\ 73a&=\\frac{18\\cdot91}{2}\\cdot(90-17=73)\\\\ a=\\frac{18\\cdot91}{2}=9\\cdot91=\\boxed{819}. \\end{align*}", "pl.show() And we previously saw how to do this: $$Z = \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}}$$ Apply this to the interval $$P(E(\\overline X) - \\text{interval/2} \\lt \\overline X \\lt E(\\overline X) + \\text{interval/2}) = 0.95$$ Will give us \\begin{align} 0.95 &= P(-\\text{normalised interval/2} \\lt Z \\lt \\text{normalised interval/2}) \\\\ &= P(-\\text{normalised interval/2} \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt \\text{normalised interval/2}) \\end{align} And we know how to calculate this \"normalised interval\" It is the Z value that will give us the 95% condifence interval we see above, and this Z value is, by consulting tables, 1.96. \\begin{align} 0.95 &= P(-1.96 \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt 1.96) \\end{align} Re-arange this a little... \\begin{align} 0.95 &= P(-1.96 \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt 1.96) \\\\ &= P(-1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\lt \\overline X - \\mu \\lt 1.96\\cdot \\frac{\\sigma}{\\sqrt n}) \\\\ &= P(-1.96\\cdot \\frac{\\sigma}{\\sqrt n} - \\overline X \\lt - \\mu \\lt 1.96\\cdot \\frac{\\sigma}{\\sqrt n} - \\overline X) \\\\ &= P(\\overline X + 1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\gt \\mu \\gt \\overline X - 1.96\\cdot \\frac{\\sigma}{\\sqrt n} ) \\text{ (multiply by -1 reverses inequalities)}\\\\ &= P(\\overline X - 1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\lt \\mu \\lt \\overline X + 1.96\\cdot \\frac{\\sigma}{\\sqrt n} ) \\\\ \\end{align} And... oh! I know my sample mean. Lets say it was 99.5cm... then I have \\begin{align} 0.95 &=", "Thanks! That should mean my final answer is $4-\\frac{100}{99!}+\\frac{100^2}{100!}$, but the answer provided to me is 4. \\begin{align*}\\frac{100^2}{100!}-\\frac{100}{99!} &= \\frac{100^2}{100!}-\\frac{100}{99!}\\cdot \\frac{100}{100}\\\\ &=\\frac{100^2}{100!}-\\frac{100^2}{100!}\\\\&=0 \\end{align*} 6. ## Re: Summation of a series Originally Posted by sbhatnagar \\begin{align*}\\frac{100^2}{100!}-\\frac{100}{99!} &= \\frac{100^2}{100!}-\\frac{100}{99!}\\cdot \\frac{100}{100}\\\\ &=\\frac{100^2}{100!}-\\frac{100^2}{100!}\\\\&=0 \\end{align*} Thank you! I can't believe, I was being so stupid...", "some notes about the approximations in a follow-up solution. – Leucippus Jun 7 at 2:02 Since @robjohn has done an excellent job at providing an answer in detail I will add my comments as an additional solution. The results presented here follow the numbering in robjohn's work. In robjohn's equation (12) the factor \\begin{align} \\frac{9! \\ \\pi^{9}}{2 \\cdot 5^{10}} \\end{align} has been provided. Numerically this can be seen as $.018579456\\cdots \\ \\pi^{9}$. By comparing this to that of $1/54 = .0185185\\cdots$ one can make a good approximation by \\begin{align} \\prod_{k=1}^{20} \\Gamma\\left(\\frac{k}{10}\\right) = \\frac{9! \\ \\pi^{9}}{2 \\cdot 5^{10}} \\approx \\frac{\\pi^{9}}{54}. \\tag{14} \\end{align} Another possible value is $20/1077 = .0185701021\\cdots$ which leads to the statement \\begin{align} \\prod_{k=1}^{20} \\Gamma\\left(\\frac{k}{10}\\right) = \\frac{9! \\ \\pi^{9}}{2 \\cdot 5^{10}} \\approx \\frac{20 \\ \\pi^{9}}{1077}. \\tag{15} \\end{align} From equation (13) the factor is \\begin{align} \\frac{2^{18} \\ (10)!(20)!(30)!}{10^{62}} = 6.138858832\\cdots = 6 + .138858832\\cdots . \\end{align} Since $625/4501 = .1388580315\\cdots$ then to a fair approximation it can be stated \\begin{align} \\prod_{k=1}^{40} \\Gamma\\left(\\frac{k}{10}\\right) \\approx \\left( 6 + \\frac{625}{4501} \\right) \\pi^{18}. \\tag{16} \\end{align} - What was the reasoning behind the rational factors of $\\pi^{9n/20}$? Why did you think that the answer was a rational multiple of $\\pi^{9n/20}$? – robjohn Jun 7 at 12:58 @robjohn This problem is similar to problem I, of the same name, where most viewers missed the", "the ... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \\Bbb Z-module) with an action of the ring R, defined by R \\to Aut(M), but the main problem is, that not every image of \\mu is an automorphism. So, in fact, you could have \\varphi \\in End(M) \\setminus Aut(M) as an image of ... 0 Using formulae used here,$$\\frac{386}{97}=3+\\frac{95}{97}=3+\\frac1{\\frac{97}{95}}=3+\\frac1{1+\\frac2{95}}=3+\\frac1{1+\\frac1{\\frac{95}2}}=3+\\frac1{1+\\frac1{47+\\frac12}}$$So, the previous convergent of \\displaystyle \\frac{386}{97} is \\displaystyle3+\\frac1{1+\\frac1{47}}=3+\\frac{47}{48}=\\frac{191}{48}$$\\implies386\\cdot48-191\\cdot97=1\\implies ... 0 There is likely a typo in the question. Using the Euclidean algorithm, we have that: \\begin{align*} 386 &= 3(97) + 95 \\\\ 97 &= 1(95) + 2 \\\\ 95 &= 47(2) + 1 \\end{align*} Working backwards, we see that: \\begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\\\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\\\ &= 48(386) - 191(97)\\\\ ... 1 Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \\cdot p - 386\\cdot q = 1$$ Then $97^{-1} \\equiv p \\mod 386$. 1 Hint. If $n=p^k$, $k\\ge 1$, then $0=p^k\\Bbb Z/p^k\\Bbb Z<p^{k-1}\\Bbb Z/p^k\\Bbb Z<\\cdots<\\Bbb Z/p^k\\Bbb Z$ is a series of composition of length $k$. Can you generalize this to the case $n=p_1^{k_1}\\cdots p_r^{k_r}$, $k_i\\ge 1$ and $r\\ge 2$? Edit. Let's try", "=3&1.&71&7171\\ldots\\\\ x&\\,\\,\\, =&0.&31&7171\\ldots \\end{alignat*} Now subtracting the two equations gives $99x=31.4$, so $$0.3\\overline{17}=x=\\frac{31.4}{99}=\\frac{314}{990}.$$ 3. Let $x=2.166\\ldots$. Then \\begin{alignat*}{7} 10x&\\,\\,\\, =2&1.&66&66\\ldots\\\\ x&\\,\\,\\, =&2.&16&66\\ldots \\end{alignat*} Now subtracting the two equations gives $9x=19.5$, so $$2.1\\overline{6}=x=\\frac{19.5}{9}=\\frac{195}{90}.$$", "only factor in 630 that is missing from 90), so 2990=729790=203630\\begin{align*}\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}\\end{align*}. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so 13126=5135126=65630\\begin{align*}\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}\\end{align*}. Now we complete the problem: 299013126=20363065630=138630\\begin{align*}\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}\\end{align*}. This fraction simplifies. To be sure of finding the simplest form for 138630\\begin{align*}\\frac{138}{630}\\end{align*}, we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23. 138630=232323357\\begin{align*}\\frac{138}{630} = \\frac{2 \\cdot 3 \\cdot 23}{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}\\end{align*}; one factor of 2 and one factor of 3 cancels out, leaving 23357\\begin{align*}\\frac{23}{3 \\cdot 5 \\cdot 7}\\end{align*} or 23105\\begin{align*}\\frac{23}{105}\\end{align*} as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties. Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: 3+2=2+3\\begin{align*}3 + 2 = 2 + 3\\end{align*} Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped. Example: (2+3)+4=2+(3+4)\\begin{align*}(2 + 3) +", "from the case for $n$ $$((n+1)+1)+\\cdots +2(n+1)=(n+1)+\\cdots+2n - (n+1)+(2n+1)+(2n+2)$$ Here you can use the fact that the formula olds for $n$: \\begin{align} \\ldots &=\\frac12 n(3n+1) - (n+1)+(2n+1)+(2n+2)\\\\ &=\\frac12 n(3n+1) + (3n+2)\\\\ &=\\frac12 (3n^2+7n+4)\\\\ &=\\frac12 (n+1)(3(n+1)+1)\\\\ \\end{align} -", "+0 I need help... 0 30 2 +23 Hello! I am asking for help on an AoPS question so I would rather not to be given the answer. Here is the question: Compute the sum of: $$101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2$$ Here is what I have: (a) From the difference of squares, we know that $$(a + b)(a - b) = a^2 - b^2$$. Using this pattern, we can pair up the terms in this sequence and apply: \\begin{align*} 101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2 &= (101 + 97)(101 - 97) + (93 + 89)(93 - 89) + \\cdots + (5 + 1)(5 - 1), \\\\ &= 4(101 + 97) + 4(93 + 89) + ... + 4(5 + 1), \\\\ &= 4(198 + 182 + 166 + ... + 6). \\end{align*} \\begin{align*} 4(198 + 182 + 166 + ... + 6) &= 4(\\frac{204 \\cdot 12}{2}), \\\\ &= 4(204 \\cdot 6), \\\\ &= 4(1224), \\\\ &= 4896. \\end{align*} So, the sum of the sequence $$101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2$$, is $$\\boxed{4896}$$. ------- But I am not sure this is correct, any suggestions? Jan 6, 2021 2+0 Answers #1 +188 0 This looks accurate to me...", "· · · + 100: \\begin{align*} 1 + 2 + ... + 99 + 100 &= (1 + 2 + ... + 99) + 100\\\\ &= \\frac{99 \\cdot (99 + 1)}{2} + 100 \\; \\; \\; \\mathrm{by \\; our \\; assumption \\; of \\; p(99)} \\\\ &= \\frac{99 \\cdot 100}{2} + \\frac{2 \\cdot 100}{2}\\\\ &= \\frac{100 \\cdot 101}{2} \\\\ &= \\frac{100 \\cdot (100 + 1)}{2} \\end{align*} What we've just done is analogous to checking two dominos in a line and finding that they are properly positioned. Since we are dealing with an infinite line, we must check all pairs at once. This is accomplished by proving that p(n) ⇒ p( n + 1) for all n ≥ 1: \\begin{align*} 1 + 2 + ... + n + (n + 1) &= (1 + 2 + ... + n) + (n + 1)\\\\ &= \\frac{n \\cdot (n + 1)}{2} + (n+1)\\; \\; \\; \\mathrm{by \\; p(n)} \\\\ &= \\frac{n(n+1)}{2} + \\frac{2(n + 1)}{2}\\\\ &= \\frac{(n + 1)(n + 2)}{2} \\\\ &= \\frac{(n + 1)((n + 1) + 1)}{2} \\end{align*} They are all lined up! Now look at p(1): The sum of the positive integers from 1 to 1 is 1· (1+1) . Clearly, p(1) is true. This sets off a chain reaction. Since p(1) ⇒ p(2), p(2) is true.", "that 7 is the only factor in 630 that is missing from 90) \\begin{align*}\\frac {29}{90}= \\frac {7 \\cdot 29}{7 \\cdot 90}=\\frac {203}{630}\\end{align*} 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126) \\begin{align*}\\frac {13}{126}=\\frac {5 \\cdot 13}{5 \\cdot 126}=\\frac {65}{630}\\end{align*} Now we complete the problem. \\begin{align*} \\frac {29}{90}-\\frac {13}{126}=\\frac {203}{630}-\\frac {65}{630}=\\frac {(203-65)}{630}=\\frac {138}{630} \\left \\{\\text{remember, } \\frac {a}{c} - \\frac {b}{c} = \\frac {a-b}{c} \\right \\}\\end{align*} This fraction simplifies. To be sure of finding the simplest form for \\begin{align*}\\frac{138}{630}\\end{align*} we write out the numerator and denominator as prime factors. We already know the prime factors of 630. The prime factors of 138 are \\begin{align*}138 = 2 \\cdot 3 \\cdot 23 && \\frac{138}{630} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot 23}{\\cancel{2} \\cdot \\cancel{3} \\cdot 3 \\cdot 5 \\cdot 7} && \\text{one factor of}\\ 2\\ \\text{and one factor of}\\ 3\\ \\text{cancels}\\end{align*} Solution \\begin{align*} \\frac {27}{90}-\\frac {13}{126}=\\frac {23}{105}\\end{align*} Example 4 A property management firm is acquiring parcels of land in order to build a small community of condominiums. It has bought three adjacent plots of land. The first is four-fifths of an acre, the second is five-twelfths of an acre, and the third is nineteen-twentieths of an acre. The firm knows that it must allow one-sixth of an acre for utilities and a", "$n-1$ arithmetic series of $1000, 999, \\ldots, 1001 - \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$. We let $k = \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$, so $S = (n-1)\\left[\\frac{k(1000 + 1001 - k)}{2}\\right] + R(n)$ where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$. At this point, we introduce the crude estimate[1] that $k=\\frac{121-n}{n-1}$, so $R(n) = 0$ and \\begin{align*}2S+2n &= (121-n)\\left(2001-\\frac{121-n}{n-1}\\right)+2n = (120-(n-1))\\left(2002-\\frac{120}{n-1}\\right)\\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \\frac{36}{n-1}$. By AM-GM, we have $5(n-1) + \\frac{36}{n-1} \\ge 2\\sqrt{5(n-1) \\cdot \\frac{36}{n-1}}$, with equality coming when $5(n-1) = \\frac{36}{n-1}$, so $n-1 \\approx 3$. Substituting this result and some arithmetic gives an answer of $\\boxed{947}$. In less formal language, it quickly becomes clear after some trial and error that in our sample, there will be $n$ values equal to one and $n-1$ values each of $1000, 999, 998 \\ldots$. It is fairly easy to find the maximum. Try $n=2$, which yields $924$, $n=3$, which yields $942$, $n=4$, which yields $947$, and $n=5$, which yields $944$. The maximum difference occurred at $n=4$, so the answer is $947$. ## Solution 2 With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of", "for mod 5} \\\\ m_3 & \\quad \\text{coefficient for mod 9} \\end{align} This coefficients are obtained using the expression $m_i = s_i \\cdot \\frac{n}{a_i}$, where $s_i = \\left(\\frac{n}{a_i}\\right)^{-1}$ in mod $a_i$. For this problem $n = 90$ and \\begin{align} a_1 & = 2 \\\\ a_2 & = 5 \\\\ a_3 & = 9 \\end{align} Inverses are: \\begin{align} s_1 = \\left(\\frac{90}{2}\\right)^{-1} \\pmod 2 = 1 \\\\ s_2 = \\left(\\frac{90}{5}\\right)^{-1} \\pmod 5 = 2 \\\\ s_3 = \\left(\\frac{90}{10}\\right)^{-1} \\pmod 9 = 1 \\end{align} and the coefficients in mod 90: \\begin{align} m_1 = s_1 \\cdot \\frac{90}{2} = 1 \\cdot 45 = 45 \\\\ m_2 = s_2 \\cdot \\frac{90}{5} = 2 \\cdot 18 = 36 \\\\ m_3 = s_3 \\cdot \\frac{90}{10} = 1 \\cdot 10 = 10 \\end{align} Value of x is obtained evaluating the linear combination: $x = \\sum_{1 \\le i \\le 3} m_i \\cdot b_i = 45 \\cdot 1 + 36 \\cdot 4 + 10 \\cdot 4 = 229 \\equiv 49 \\pmod{90}$ We can check that $x = 49$ satisfies the three congruences: \\begin{align} 49 & \\equiv 1 \\pmod 2 \\\\ 49 & \\equiv 4 \\pmod 5 \\\\ 49 & \\equiv 4 \\pmod 9 \\end{align} and the initial problem: $79 \\cdot 49 = 3871 \\equiv 1 \\pmod 90 \\quad (\\text{because } 3871 = 90 \\cdot 43 + 1)$ There is", "from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree. Prime Factors in 90 Factors in 126 We Need 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need one 2, two 3’s, one 5 and one 7. That gives us \\begin{align*}2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7 = 630\\end{align*} as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so \\begin{align*}\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}\\end{align*}. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so \\begin{align*}\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}\\end{align*}. Now we complete the problem: \\begin{align*}\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}\\end{align*}. This fraction simplifies. To be sure of finding the simplest form for \\begin{align*}\\frac{138}{630}\\end{align*}, we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime", "number $Q$ to make the second and third lines a little simpler. Just to finish this off, let's look at the original problem, where $$r = 15 \\\\ s = 55 \\\\ n = 5 \\\\ C = 25 \\\\ D = 88$$ Then \\begin{align} \\frac{sC - rD}{sn - r^2} &= a\\\\ \\frac{-rC +nD}{sn - r^2} &= b \\end{align} becomes \\begin{align} \\frac{55 \\cdot 25 - 15 \\cdot 88}{55 \\cdot 5 - 15^2} &= a\\\\ \\frac{-15 \\cdot 25 +5 \\cdot 88}{55 \\cdot 5 - 15^2} &= b \\end{align} which simplifies to \\begin{align} \\frac{55}{50} &= a\\\\ \\frac{65}{50} &= b \\end{align} \\begin{align} 1.1 &= a\\\\ 1.3 &= b \\end{align} • I strongly believe that this explanation does not help AdamM at all. Apr 11 '14 at 11:23 • John what if $r^2-ns=0$? Or if $r=0$? Apr 11 '14 at 11:58 • $r^2 - ns = (\\sum_i x_i)^2 - n \\sum_i x_i^2$; that's zero only when all the $x_i$s are the same. For the $r = 0$ case, I should have written out a separate path. Your solution, which shows how to solve a particular case, doesn't explain what to do in the $r = 0$ case either. But I'll amend mine to fix that; thanks for catching it. Apr 11 '14 at 17:44 • @kmitov: you may well be right. But", "this: The factor tree for 126 looks like this: The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree. Prime Factors in 90 Factors in 126 We Need 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need one 2, two 3’s, one 5 and one 7. That gives us \\begin{align*}2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7 = 630\\end{align*} as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so \\begin{align*}\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}\\end{align*}. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so \\begin{align*}\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}\\end{align*}. Now we complete the problem: \\begin{align*}\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}\\end{align*}. This fraction simplifies. To be sure of finding the simplest form for \\begin{align*}\\frac{138}{630}\\end{align*}, we write out the", "1+ \\left(\\frac{1}{3}\\right) \\cdot b &= \\frac{12}{11}\\\\ b &= 0.\\overline{27}\\end{align*} Thus the approximate model is: f(x)=121+(13)(0.27273)x\\begin{align*}f(x)=\\frac{12}{1+\\left(\\frac{1}{3}\\right) \\cdot (0.27273)^x}\\end{align*} #### Example 5 Determine the logistic model given c=7\\begin{align*}c=7\\end{align*} and the points (0, 2) and (3, 5). The two points give two equations, and the logistic model has two variables. 2=71+aa=2.55=71+(2.5)b3b3=0.16b0.5429\\begin{align*}2 = \\frac{7}{1+a}\\\\ a = 2.5\\\\ 5 = \\frac{7}{1+(2.5) \\cdot b^3}\\\\ b^3 = 0.16\\\\ b \\approx 0.5429\\end{align*} Thus the approximate model is: f(x)=71+(2.5)(0.5429)x\\begin{align*}f(x)=\\frac{7}{1+(2.5) \\cdot (0.5429)^x}\\end{align*} ### Review For 1-5, determine the logistic model given the carrying capacity and two points. 1. c=12;(0,5);(1,7)\\begin{align*}c=12; (0, 5); (1, 7)\\end{align*} 2. c=200;(0,150);(5,180)\\begin{align*}c=200; (0, 150); (5, 180)\\end{align*} 3. c=1500;(0,150);(10,1000)\\begin{align*}c=1500; (0, 150); (10, 1000)\\end{align*} 4. c=1000000;(0,100000);(40,20000)\\begin{align*}c=1000000; (0, 100000); (-40, 20000)\\end{align*} 5. c=30000000;(60,10000);(0,8000000)\\begin{align*}c= 30000000; (-60, 10000); (0, 8000000)\\end{align*} For 6-8, use the logistic function f(x)=321+3ex\\begin{align*}f(x)=\\frac{32}{1+3e^{-x}}\\end{align*}. 6. What is the carrying capacity of the function? 7. What is the y\\begin{align*}y\\end{align*}-intercept of the function? 8. Use your answers to 6 and 7 along with at least two points on the graph to make a sketch of the function. For 9-11, use the logistic function g(x)=251+40.2x\\begin{align*}g(x)=\\frac{25}{1+4 \\cdot 0.2^x}\\end{align*}. 9. What is the carrying capacity of the function? 10. What is the y\\begin{align*}y\\end{align*}-intercept of the function? 11. Use your answers to 9 and 10 along with at least two points on the graph to make a sketch of the function. For 12-14,", "\\times { F }_{ 30 }=20.30 \\end{align*} $$\\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots$$ \\begin{align*} { F }_{ 30 }&=10,150\\\\ { F }_{ 7 }&=27,640\\\\{ F }_{ 5 }&=\\frac { 13.22-0.2764 }{ 0.00045 } =28,764 \\end{align*}", "the followings (see here for the details): • $a_n=4a_{n-1}-a_{n-2}$. • $a_{2n+1}a_{2n+3}=a_{4n+4}+14.$ • $a_{4n+4}={a_{n+1}}^4-4{a_{n+1}}^2+2.$ • $a_{2n+1}+a_{2n+3}=4{a_{n+1}}^2-8$. • ${a_n}^2=12b_n+4$ where $b_n$ is an integer. Then, we have \\begin{align}\\frac{a_{2n+1}-4}{6}\\cdot\\frac{a_{2n+3}-4}{6}+1&=\\frac{a_{2n+1}a_{2n+3}-4(a_{2n+1}+a_{2n+3})+16}{36}+1\\\\&=\\frac{(a_{4n+4}+14)-4(4{a_{n+1}}^2-8)+16}{36}+1\\\\&=\\frac{({a_{n+1}}^4-4{a_{n+1}}^2+16)-4(4{a_{n+1}}^2-8)+16}{36}+1\\\\&=\\frac{{a_{n+1}}^4-20{a_{n+1}}^2+64}{36}+1\\\\&=\\frac{(12b_{n+1}+4)^2-20(12b_{n+1}+4)+100}{36}\\\\&=\\frac{144{b_{n+1}}^2-144b_{n+1}+36}{36}\\\\&=(2b_{n+1}-1)^2.\\end{align}", "1+s+t, 1)\\ge (2\\sqrt{a_5a_1} + a_3)t^3 + (2\\sqrt{a_4a_0} + a_2)t^2.$$ It suffices to prove that $$2\\sqrt{a_5a_1} + a_3 \\ge 0$$ and $$2\\sqrt{a_4a_0} + a_2 \\ge 0$$. Note that \\begin{align} 2\\sqrt{a_5a_1} + a_3 &= \\Big(2\\sqrt{a_5a_1} - \\frac{1}{3}\\cdot 94494\\, s^3 - \\frac{1}{3} \\cdot 65760\\, s^2\\Big)\\\\ &\\quad + \\Big(a_3 + \\frac{1}{3}\\cdot 94494\\, s^3 + \\frac{1}{3} \\cdot 65760\\, s^2\\Big) \\end{align} and \\begin{align} 2\\sqrt{a_4a_0} + a_2 &= \\Big(2\\sqrt{a_4a_0} - \\frac{1}{2}\\cdot 68407\\, s^4 - \\frac{1}{2}242656\\, s^3\\Big)\\\\ &\\quad + \\Big(a_2 + \\frac{1}{2}\\cdot 68407\\, s^4 + \\frac{1}{2}242656\\, s^3\\Big). \\end{align} It suffices to prove that \\begin{align} 2\\sqrt{a_5a_1} - \\frac{1}{3}\\cdot 94494\\, s^3 - \\frac{1}{3} \\cdot 65760\\, s^2 &\\ge 0,\\\\ a_3 + \\frac{1}{3}\\cdot 94494\\, s^3 + \\frac{1}{3} \\cdot 65760\\, s^2&\\ge 0,\\\\ 2\\sqrt{a_4a_0} - \\frac{1}{2}\\cdot 68407\\, s^4 - \\frac{1}{2}242656\\, s^3 &\\ge 0,\\\\ a_2 + \\frac{1}{2}\\cdot 68407\\, s^4 + \\frac{1}{2}242656\\, s^3 &\\ge 0. \\end{align} All of them can be reduced to polynomial inequalities in $$s$$ and not hard to prove. This completes the proof. let $$x=\\frac{\\sqrt{3}}{a},y=\\frac{\\sqrt{3}}{b},z=\\frac{\\sqrt{3}}{c} \\implies xy+yz+zx=3$$ with uvw method: $$3u=x+y+z,3v^2=xy+yz+zx,w^3=xyz\\\\ u\\geqslant v \\geqslant w \\geqslant 0 , w^3 \\leq 3uv^2 -2u^3+2\\sqrt{(u^2-v^2)^3}, v=1$$ then inequality becomes: $$\\frac{xy}{7y+x}+\\frac{yz}{7z+y}+\\frac{xz}{7x+z}\\leq\\frac{3}{8} \\iff \\\\3(7y+x)(7z+y)(7x+z) \\geqslant 8[xy(7z+y)(7x+z)+yz(7y+x)(7x+z)+xz(7y+x)(7z+y)]$$ \\ $$LHS=3[7(7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y)+344xyz]\\\\7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y=4\\sum(x^2y+y^2x)+3(yz^2+x^2z+xy^2-xz^2-y^2z-x^2y)=4\\sum xy(x+y+z-z)+3(y-x)(z-x)(z-y)=4(x+y+z)\\sum xy -4*3xyz+3(y-x)(z-x)(z-y) \\\\ LHS=3[7*4*3^2u+260w^3+3*7(y-x)(z-x)(z-y)]\\\\ RHS=8[\\sum_{cyc}xy(7z+y)(7x+z)]=8[7\\sum x^2y^2+57xyz(x+y+z)]\\\\= 8[ 7((xy+yz+xz)^2-2xyz(x+y+z))+57xyz(x+y+z)] \\\\=8(63+43*3u*w^3)$$ then the inequality becomes: $$4(63u+65w^3-42-86uw^3) \\geq 21(x-y)(z-x)(z-y) ....(3)$$ now to prove $$63u+65w^3-42-86uw^3 \\geq 0 \\iff w^3 \\leq \\dfrac{63u-42}{86u-65}$$ first to prove : $$\\dfrac{63u-42}{86u-65} \\geq \\dfrac{2u-1}{4u-3} \\iff (u-1)(80u-61)$$it is" ]

[ "pl.show() And we previously saw how to do this: $$Z = \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}}$$ Apply this to the interval $$P(E(\\overline X) - \\text{interval/2} \\lt \\overline X \\lt E(\\overline X) + \\text{interval/2}) = 0.95$$ Will give us \\begin{align} 0.95 &= P(-\\text{normalised interval/2} \\lt Z \\lt \\text{normalised interval/2}) \\\\ &= P(-\\text{normalised interval/2} \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt \\text{normalised interval/2}) \\end{align} And we know how to calculate this \"normalised interval\" It is the Z value that will give us the 95% condifence interval we see above, and this Z value is, by consulting tables, 1.96. \\begin{align} 0.95 &= P(-1.96 \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt 1.96) \\end{align} Re-arange this a little... \\begin{align} 0.95 &= P(-1.96 \\lt \\frac{\\overline X - \\mu}{\\frac{\\sigma}{\\sqrt n}} \\lt 1.96) \\\\ &= P(-1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\lt \\overline X - \\mu \\lt 1.96\\cdot \\frac{\\sigma}{\\sqrt n}) \\\\ &= P(-1.96\\cdot \\frac{\\sigma}{\\sqrt n} - \\overline X \\lt - \\mu \\lt 1.96\\cdot \\frac{\\sigma}{\\sqrt n} - \\overline X) \\\\ &= P(\\overline X + 1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\gt \\mu \\gt \\overline X - 1.96\\cdot \\frac{\\sigma}{\\sqrt n} ) \\text{ (multiply by -1 reverses inequalities)}\\\\ &= P(\\overline X - 1.96\\cdot \\frac{\\sigma}{\\sqrt n} \\lt \\mu \\lt \\overline X + 1.96\\cdot \\frac{\\sigma}{\\sqrt n} ) \\\\ \\end{align} And... oh! I know my sample mean. Lets say it was 99.5cm... then I have \\begin{align} 0.95 &=", "$n-1$ arithmetic series of $1000, 999, \\ldots, 1001 - \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$. We let $k = \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$, so $S = (n-1)\\left[\\frac{k(1000 + 1001 - k)}{2}\\right] + R(n)$ where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$. At this point, we introduce the crude estimate[1] that $k=\\frac{121-n}{n-1}$, so $R(n) = 0$ and \\begin{align*}2S+2n &= (121-n)\\left(2001-\\frac{121-n}{n-1}\\right)+2n = (120-(n-1))\\left(2002-\\frac{120}{n-1}\\right)\\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \\frac{36}{n-1}$. By AM-GM, we have $5(n-1) + \\frac{36}{n-1} \\ge 2\\sqrt{5(n-1) \\cdot \\frac{36}{n-1}}$, with equality coming when $5(n-1) = \\frac{36}{n-1}$, so $n-1 \\approx 3$. Substituting this result and some arithmetic gives an answer of $\\boxed{947}$. In less formal language, it quickly becomes clear after some trial and error that in our sample, there will be $n$ values equal to one and $n-1$ values each of $1000, 999, 998 \\ldots$. It is fairly easy to find the maximum. Try $n=2$, which yields $924$, $n=3$, which yields $942$, $n=4$, which yields $947$, and $n=5$, which yields $944$. The maximum difference occurred at $n=4$, so the answer is $947$. ## Solution 2 With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of", "the ... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \\Bbb Z-module) with an action of the ring R, defined by R \\to Aut(M), but the main problem is, that not every image of \\mu is an automorphism. So, in fact, you could have \\varphi \\in End(M) \\setminus Aut(M) as an image of ... 0 Using formulae used here,$$\\frac{386}{97}=3+\\frac{95}{97}=3+\\frac1{\\frac{97}{95}}=3+\\frac1{1+\\frac2{95}}=3+\\frac1{1+\\frac1{\\frac{95}2}}=3+\\frac1{1+\\frac1{47+\\frac12}}$$So, the previous convergent of \\displaystyle \\frac{386}{97} is \\displaystyle3+\\frac1{1+\\frac1{47}}=3+\\frac{47}{48}=\\frac{191}{48}$$\\implies386\\cdot48-191\\cdot97=1\\implies ... 0 There is likely a typo in the question. Using the Euclidean algorithm, we have that: \\begin{align*} 386 &= 3(97) + 95 \\\\ 97 &= 1(95) + 2 \\\\ 95 &= 47(2) + 1 \\end{align*} Working backwards, we see that: \\begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\\\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\\\ &= 48(386) - 191(97)\\\\ ... 1 Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \\cdot p - 386\\cdot q = 1$$ Then $97^{-1} \\equiv p \\mod 386$. 1 Hint. If $n=p^k$, $k\\ge 1$, then $0=p^k\\Bbb Z/p^k\\Bbb Z<p^{k-1}\\Bbb Z/p^k\\Bbb Z<\\cdots<\\Bbb Z/p^k\\Bbb Z$ is a series of composition of length $k$. Can you generalize this to the case $n=p_1^{k_1}\\cdots p_r^{k_r}$, $k_i\\ge 1$ and $r\\ge 2$? Edit. Let's try", "25 \\equiv 4 \\mod{11}. \\end{split} Notice that there are 2 residues on this list which are greater than $\\frac{11}{2} = 5.5$, and so Gauss' Lemma says that (6) \\begin{align} \\left(\\frac{5}{11}\\right) = (-1)^2 = 1. \\end{align} $\\square$ Proof of Gauss' Lemma: To start off, we'll split the least non-negative residues from the set (7) \\begin{align} 1\\cdot a, 2\\cdot a, \\cdots, \\frac{p-1}{2} a \\end{align} into two groups: the set $r_1,\\cdots, r_n$ will be the set of those residues that are greater than $\\frac{p}{2}$, and the set $s_1,\\cdots, s_m$ of residues which are less than $\\frac{p}{2}$. Notice that none of these residues actually hit $\\frac{p}{2}$ on the nose, since this number is not an integer. So really we have $\\{r_1,\\cdots, r_n\\} \\subseteq \\{\\frac{p+1}{2},\\cdots,p-1\\}$ and $\\{s_1,\\cdots, s_m\\}\\subseteq \\{1,\\cdots,\\frac{p-1}{2}\\}$. Now we claim that the integers (8) \\begin{align} p-r_1,p-r_2,\\cdots,p-r_n,s_1,s_2,\\cdots,s_m \\end{align} are the same modulo p as the integers (9) \\begin{align} 1,2,3,\\cdots,\\frac{p-1}{2}. \\end{align} To see that this is true, notice that the integers from 8 are all between 1 and $\\frac{p-1}{2}$ — the $s_j$ certainly live in this range, but since we have $\\frac{p+1}{2}\\leq r_i \\leq p-1$ we also have $1 \\leq p-r_i \\leq \\frac{p-1}{2}$. So to show that these collections are the same, we just need to show that the top collection has no repeats. For this, we'll check three things: 1. there are", "· · · + 100: \\begin{align*} 1 + 2 + ... + 99 + 100 &= (1 + 2 + ... + 99) + 100\\\\ &= \\frac{99 \\cdot (99 + 1)}{2} + 100 \\; \\; \\; \\mathrm{by \\; our \\; assumption \\; of \\; p(99)} \\\\ &= \\frac{99 \\cdot 100}{2} + \\frac{2 \\cdot 100}{2}\\\\ &= \\frac{100 \\cdot 101}{2} \\\\ &= \\frac{100 \\cdot (100 + 1)}{2} \\end{align*} What we've just done is analogous to checking two dominos in a line and finding that they are properly positioned. Since we are dealing with an infinite line, we must check all pairs at once. This is accomplished by proving that p(n) ⇒ p( n + 1) for all n ≥ 1: \\begin{align*} 1 + 2 + ... + n + (n + 1) &= (1 + 2 + ... + n) + (n + 1)\\\\ &= \\frac{n \\cdot (n + 1)}{2} + (n+1)\\; \\; \\; \\mathrm{by \\; p(n)} \\\\ &= \\frac{n(n+1)}{2} + \\frac{2(n + 1)}{2}\\\\ &= \\frac{(n + 1)(n + 2)}{2} \\\\ &= \\frac{(n + 1)((n + 1) + 1)}{2} \\end{align*} They are all lined up! Now look at p(1): The sum of the positive integers from 1 to 1 is 1· (1+1) . Clearly, p(1) is true. This sets off a chain reaction. Since p(1) ⇒ p(2), p(2) is true.", "for mod 5} \\\\ m_3 & \\quad \\text{coefficient for mod 9} \\end{align} This coefficients are obtained using the expression $m_i = s_i \\cdot \\frac{n}{a_i}$, where $s_i = \\left(\\frac{n}{a_i}\\right)^{-1}$ in mod $a_i$. For this problem $n = 90$ and \\begin{align} a_1 & = 2 \\\\ a_2 & = 5 \\\\ a_3 & = 9 \\end{align} Inverses are: \\begin{align} s_1 = \\left(\\frac{90}{2}\\right)^{-1} \\pmod 2 = 1 \\\\ s_2 = \\left(\\frac{90}{5}\\right)^{-1} \\pmod 5 = 2 \\\\ s_3 = \\left(\\frac{90}{10}\\right)^{-1} \\pmod 9 = 1 \\end{align} and the coefficients in mod 90: \\begin{align} m_1 = s_1 \\cdot \\frac{90}{2} = 1 \\cdot 45 = 45 \\\\ m_2 = s_2 \\cdot \\frac{90}{5} = 2 \\cdot 18 = 36 \\\\ m_3 = s_3 \\cdot \\frac{90}{10} = 1 \\cdot 10 = 10 \\end{align} Value of x is obtained evaluating the linear combination: $x = \\sum_{1 \\le i \\le 3} m_i \\cdot b_i = 45 \\cdot 1 + 36 \\cdot 4 + 10 \\cdot 4 = 229 \\equiv 49 \\pmod{90}$ We can check that $x = 49$ satisfies the three congruences: \\begin{align} 49 & \\equiv 1 \\pmod 2 \\\\ 49 & \\equiv 4 \\pmod 5 \\\\ 49 & \\equiv 4 \\pmod 9 \\end{align} and the initial problem: $79 \\cdot 49 = 3871 \\equiv 1 \\pmod 90 \\quad (\\text{because } 3871 = 90 \\cdot 43 + 1)$ There is", "the followings (see here for the details): • $a_n=4a_{n-1}-a_{n-2}$. • $a_{2n+1}a_{2n+3}=a_{4n+4}+14.$ • $a_{4n+4}={a_{n+1}}^4-4{a_{n+1}}^2+2.$ • $a_{2n+1}+a_{2n+3}=4{a_{n+1}}^2-8$. • ${a_n}^2=12b_n+4$ where $b_n$ is an integer. Then, we have \\begin{align}\\frac{a_{2n+1}-4}{6}\\cdot\\frac{a_{2n+3}-4}{6}+1&=\\frac{a_{2n+1}a_{2n+3}-4(a_{2n+1}+a_{2n+3})+16}{36}+1\\\\&=\\frac{(a_{4n+4}+14)-4(4{a_{n+1}}^2-8)+16}{36}+1\\\\&=\\frac{({a_{n+1}}^4-4{a_{n+1}}^2+16)-4(4{a_{n+1}}^2-8)+16}{36}+1\\\\&=\\frac{{a_{n+1}}^4-20{a_{n+1}}^2+64}{36}+1\\\\&=\\frac{(12b_{n+1}+4)^2-20(12b_{n+1}+4)+100}{36}\\\\&=\\frac{144{b_{n+1}}^2-144b_{n+1}+36}{36}\\\\&=(2b_{n+1}-1)^2.\\end{align}", "than $m(m+1)(m+2)(m+3)$. Think symmetry! \\begin{align} (m - \\tfrac 3 2)(m - \\tfrac 1 2)(m + \\tfrac 1 2)(m + \\tfrac 3 2) &= (m^2 - \\tfrac 9 4)(m^2 - \\tfrac 1 4) \\\\ &= m^4 - \\tfrac 5 2 m^2 + \\tfrac{9}{16} \\\\ &= (m^2 - \\tfrac 5 4)^2 - 1 \\end{align} Finally, if $m - \\frac 1 2$ is an integer, $m = k + \\frac 1 2$ where $k$ is an integer. Therefore $m^2 - \\frac 5 4 = k^2 + k - 1$ is also an integer. I will show that $16m(m+1)(m+2)(m+3) = 16(m^2 + 3m + 1)^2 - 16$ It will follow that $$m(m+1)(m+2)(m+3) = (m^2 + 3m + 1)^2 - 1$$ Proof: \\begin{align} 16 m(m+1)(m+2)(m+3) &=(2m)(2m+2)(2m+4)(2m+6)\\\\ & ---\\text{let $2m = n-3$}---\\\\ &=(n-3)(n-1)(n+1)(n+3)\\\\ &=(n-3)(n+3) \\cdot (n-1)(n+1) \\\\ &=(n^2 - 9)(n^2 - 1) \\\\ &=n^4 - 10n^2 + 9 \\\\ &=(n^2 - 5)^2 - 16 \\\\ & ---\\text{Note $n = 2m+3$}---\\\\ &=(4m^2 + 12m + 4)^2 - 16\\\\ &=16(m^2 + 3m + 1)^2 - 16\\\\ \\end{align}", "\\end{split}$ As $$y''-2xy'+2ny = 0$$ we have $(k+2)\\,(k+1) \\, a_{k+2} + ( - 2k+ 2n) a_k = 0 , \\qquad \\text{or} \\qquad a_{k+2} = \\frac{(2k-2n)}{(k+2)(k+1)} a_k .$ This recurrence relation actually includes $$a_2 = -na_0$$ (which comes about from $$2a_2+2na_0 = 0). Again \\(a_0$$ and $$a_1$$ are arbitrary. \\begin{align*} & a_2 = \\frac{-2n}{(2)(1)}a_0, \\qquad a_3 = \\frac{2(1-n)}{(3)(2)} a_1, \\\\ & a_4 = \\frac{2(2-n)}{(4)(3)} a_2 = \\frac{2^2(2-n)(-n)}{(4)(3)(2)(1)} a_0 , \\\\ & a_5 = \\frac{2(3-n)}{(5)(4)} a_3 = \\frac{2^2(3-n)(1-n)}{(5)(4)(3)(2)} a_1 , \\quad \\ldots \\end{align*} Let us separate the even and odd coefficients. We find that \\begin{align*} a_{2m} &=\\frac{2^m(-n)(2-n)\\cdots(2m-2-n)}{(2m)!} , \\\\ a_{2m+1} &=\\frac{2^m(1-n)(3-n)\\cdots(2m-1-n)}{(2m+1)!} . \\end{align*} Let us write down the two series, one with the even powers and one with the odd. \\begin{align*} y_1(x) & = 1+\\frac{2(-n)}{2!} x^2 + \\frac{2^2(-n)(2-n)}{4!} x^4 + \\frac{2^3(-n)(2-n)(4-n)}{6!} x^6 + \\cdots , \\\\ y_2(x) & = x+\\frac{2(1-n)}{3!} x^3 + \\frac{2^2(1-n)(3-n)}{5!} x^5 + \\frac{2^3(1-n)(3-n)(5-n)}{7!} x^7 + \\cdots . \\end{align*} We then write $y(x) = a_0 y_1(x) + a_1 y_2(x) .$ We also notice that if $$n$$ is a positive even integer, then $$y_1(x)$$ is a polynomial as all the coefficients in the series beyond a certain degree are zero. If $$n$$ is a positive odd integer, then $$y_2(x)$$ is a polynomial. For example, if $$n=4$$, then $y_1(x) = 1 + \\frac{2(-4)}{2!} x^2 + \\frac{2^2(-4)(2-4)}{4!} x^4 = 1 -", "from the case for $n$ $$((n+1)+1)+\\cdots +2(n+1)=(n+1)+\\cdots+2n - (n+1)+(2n+1)+(2n+2)$$ Here you can use the fact that the formula olds for $n$: \\begin{align} \\ldots &=\\frac12 n(3n+1) - (n+1)+(2n+1)+(2n+2)\\\\ &=\\frac12 n(3n+1) + (3n+2)\\\\ &=\\frac12 (3n^2+7n+4)\\\\ &=\\frac12 (n+1)(3(n+1)+1)\\\\ \\end{align} -", "+ \\frac{97}{2*95} + \\frac{97}{3*94} + \\ldots + \\frac{97}{48*49} \\\\ & = 97 \\Big( \\frac{1}{96} + \\frac{1}{2*95} + \\frac{1}{3*94} + \\ldots + \\frac{1}{48*49}\\Big) \\end{split}$$$$ Hence, the numerator is a multiple of $97$. To see that the denominator is not a multiple of $97$, note that $97$ is a prime, and the denominator contains natural numbers less than $97$ (in fact, the reduced denominator is a factor of $96$!), hence cannot possibly be a multiple of $97$. Thus, in it's reduced form, the numerator of the expression is a multiple of $97$. In fact, Wolstenhomme's theorem says that the numerator is a multiple of $97^2 = 9409$! This is along the lines of DanielV's proposed solution. We know $97$ is prime. Multiplying both sides by $96!$, it's enough to show that $\\sum_{n=1}^{96} \\frac{96!}{n}$ (which is a sum of integers) is divisible by $97$, because $96!$ and $97$ are coprime. Let $a_n = \\frac{96!}{n}$ for $n\\in\\{1,\\ldots,96\\}$. Then we have $n a_n = 96!$, so $a_n \\equiv 96! n^{-1} \\pmod {97}$, where $n^{-1}$ is the inverse of $n$ in the integers mod $97$. Then $\\sum_{n=1}^{96} a_n$ is divisible by $97$ because $$\\sum_{n=1}^{96} a_n \\equiv 96! \\sum_{n=1}^{96} n^{-1} \\color{red}{=} 96! \\sum_{n=1}^{96} n = 96! \\frac{96 \\cdot 97}{2} \\equiv 0 \\pmod{97}.$$ The red $\\color{red}{=}$ holds because the inverse $n \\mapsto n^{-1}$ is a permutation", "so it suffices to show that $p_1(n)=p(n-1)-p(n-2)$. And \\begin{align*}p_1(n)&=p_1(n-1)+p_2(n-1)+p_3(n-1)\\\\&=p(n-1)-\\big(p_1(n-1)+p_2(n-1)\\big)\\\\&=p(n-1)-\\Big(\\big(p_1(n-2)+p_2(n-2)+p_3(n-2)\\big)+\\big(p_1(n-2)+p_2(n-2)\\big)\\Big)\\\\&=p(n-1)-\\Big(2p_1(n-2)+2p_2(n-2)+p_3(n-2)\\Big)\\\\&=p(n-1)-p(n-2)\\;,\\end{align*} and $(1)$ is established. For $n\\ge 1$ let $d_n=\\dfrac{p(n+1)}{p(n)}$; clearly $p(1)=\\dfrac{12}5=2.4$ and $p(2)=\\dfrac{30}{12}=2.5$. Now assume that $n\\ge 3$ and $$\\frac{12}5\\le d_k\\le\\frac52$$ for $1\\le k<n$. Then \\begin{align*} d_n&=\\frac{p(n+1)}{p(n)}\\\\\\\\ &=2+\\frac{2p(n-1)-2p(n-2)}{p(n)}\\\\\\\\ &=2+\\frac2{d_{n-1}}-\\frac{2p(n-2)}{p(n-1)}\\cdot\\frac{p(n-1)}{p(n)}\\\\\\\\ &=2+\\frac2{d_{n-1}}-\\frac2{d_{n-1}d_{n-2}}\\\\\\\\ &=2+\\frac2{d_{n-1}}\\left(1-\\frac1{d_{n-2}}\\right)\\;. \\end{align*} Now $$\\frac7{12}\\le 1-\\frac1{d_{n-2}}\\le\\frac35\\quad\\text{and}\\quad\\frac45\\le\\frac2{d_{n-1}}\\le\\frac56\\;,$$ so $$2.4<2+\\frac7{15}=2+\\frac7{12}\\cdot\\frac45\\le d_n\\le 2+\\frac35\\cdot\\frac56=\\frac52\\;,$$ and it follows by induction that $2.4\\le d_n\\le 2.5$ for $n\\ge 1$. Since $p(1)=5$, it’s immediate that $5(2.4)^{n-1}\\le p(n)\\le 5(2.5)^{n-1}$ for all $n\\ge 1$. - Nice! Actually, my argument proves the recursive relation, since $x^3-2x^2-2x+2$ is the characteristic polynomial of the matrix above. But your approach is more direct. – Thomas Andrews Feb 1 '13 at 0:50 @Thomas: For some reason I just don’t naturally think in linear algebraic terms! – Brian M. Scott Feb 1 '13 at 5:33 A basic linear algebra approach to starting such a problem. If you let $p_i(n)$ be the count of such nymbers with first digit $i$, for $i=1,2,3,4,5$, then you can show that $p_1(n)=p_5(n)$ and $p_2(n)=p_4(n)$ and $p(n)=2p_1(n)+2p_2(n)+p_3(n)$, and, in general, get: $$\\left(\\begin{matrix} p_1(n)\\\\p_2(n)\\\\p_3(n)\\end{matrix}\\right)=\\left(\\begin{matrix}1&1&1\\\\1&1&0\\\\2&0&0\\end{matrix}\\right)^{n-1}\\left(\\begin{matrix} 1\\\\1\\\\1\\end{matrix}\\right)$$ The eigenvalues of this matrix are all real, being, approximately, $-1.17,0.69$ and $2.48$. It's clear that the larger eigenvalue is a factor here, so we know that $p(n)$ is dominated by some constant times $2.48^n$. Not sure if that helps any, however. It was just too long for", "only factor in 630 that is missing from 90), so 2990=729790=203630\\begin{align*}\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}\\end{align*}. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so 13126=5135126=65630\\begin{align*}\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}\\end{align*}. Now we complete the problem: 299013126=20363065630=138630\\begin{align*}\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}\\end{align*}. This fraction simplifies. To be sure of finding the simplest form for 138630\\begin{align*}\\frac{138}{630}\\end{align*}, we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23. 138630=232323357\\begin{align*}\\frac{138}{630} = \\frac{2 \\cdot 3 \\cdot 23}{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}\\end{align*}; one factor of 2 and one factor of 3 cancels out, leaving 23357\\begin{align*}\\frac{23}{3 \\cdot 5 \\cdot 7}\\end{align*} or 23105\\begin{align*}\\frac{23}{105}\\end{align*} as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties. Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: 3+2=2+3\\begin{align*}3 + 2 = 2 + 3\\end{align*} Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped. Example: (2+3)+4=2+(3+4)\\begin{align*}(2 + 3) +", "# 2020 CIME I Problems/Problem 7 ## Problem 7 For every positive integer $n$, define $$f(n)=\\frac{n}{1 \\cdot 3 \\cdot 5 \\cdots (2n+1)}.$$ Suppose that the sum $f(1)+f(2)+\\cdots+f(2020)$ can be expressed as $\\frac{p}{q}$ for relatively prime integers $p$ and $q$. Find the remainder when $p$ is divided by $1000$. ## Solution Let $S(n) = f(1)+f(2)+\\cdots+f(n)$. We claim that $$S(n) = \\frac{1}{2} - \\frac{1}{2 \\cdot (2n+1)!!}.$$ We show this using induction. Suppose this is true for some $n = k$. Then, it must be true for $k+1$. The base case when $k = 1$ is trivial. Then, we have that \\begin{align*} S(k+1) &= f(k) + S(k) \\\\ {} &= \\frac{n}{(2n+1)!!} + \\frac{1}{2} - \\frac{1}{2 \\cdot (2n+1)!!} \\\\ {} &= \\frac{1}{2} - \\frac{1}{2 \\cdot (2n+3)!!}. \\end{align*} Hence, this completes the induction therefore proving the claim. So, the numerator $p$ is $\\frac{1}{2}(4041!! - 1)$. We proceed using Euler's Theorem combined with Chinese Remainder Theorem. It is obvious that $4041!! \\equiv 0 \\pmod{125}$ so $p \\equiv 62\\pmod{125}$. Also, instead of considering $p$ modulo $8$, we consider it modulo $16$. Then, we get $$4041!! \\equiv 3^{253} \\cdot 5^{253} \\cdot 7^{253} \\cdot 9^{252} \\cdot 11^{252} \\cdot 13^{252} \\cdot 15^{252} \\equiv 3 \\cdot 5 \\cdot 7 \\cdot 9 \\equiv 1\\pmod{16},$$ by Euler's Totient Theorem as $\\phi(16) = 8$. This implies that $\\frac{1}{2}(4041!! - 1) \\equiv 0\\pmod{8}$,", "Thanks! That should mean my final answer is $4-\\frac{100}{99!}+\\frac{100^2}{100!}$, but the answer provided to me is 4. \\begin{align*}\\frac{100^2}{100!}-\\frac{100}{99!} &= \\frac{100^2}{100!}-\\frac{100}{99!}\\cdot \\frac{100}{100}\\\\ &=\\frac{100^2}{100!}-\\frac{100^2}{100!}\\\\&=0 \\end{align*} 6. ## Re: Summation of a series Originally Posted by sbhatnagar \\begin{align*}\\frac{100^2}{100!}-\\frac{100}{99!} &= \\frac{100^2}{100!}-\\frac{100}{99!}\\cdot \\frac{100}{100}\\\\ &=\\frac{100^2}{100!}-\\frac{100^2}{100!}\\\\&=0 \\end{align*} Thank you! I can't believe, I was being so stupid...", "motivation, notice that since the divisors of a prime power $p^a$ are given by $\\{1,p,p^2,\\cdots,p^a\\}$, we have (13) \\begin{align} \\begin{split} \\sigma(p^a) &= 1 + p + p^2 + \\cdots + p^a = \\sum_{i=0}^a p^i. \\end{split} \\end{align} This is a geometric series, so we can use the formula for the sum of a geometric series to compute (14) \\begin{align} \\begin{split} \\sigma(p^a) &= \\sum_{i=0}^a p^i = \\frac{p^{a+1}-1}{p-1}. \\end{split} \\end{align} Corollary: For a number $n = p_1^{a_1}\\cdots p_k^{a_k}$, we have $\\displaystyle \\sigma(n) = \\prod_{i=1}^k \\frac{p_i^{a_i+1}-1}{p_i-1}$. #### Example: Summing divisors of some larger numbers Suppose we wanted to compute $\\sigma(28)$. Since $28 = 4 \\cdot 7 = 2^2 \\cdot 7^1$, our formula gives (15) \\begin{align} \\sigma(28) = \\left(\\frac{2^3-1}{2-1}\\right)\\left(\\frac{7^2-1}{7-1}\\right) = 7 \\cdot 8 = 56. \\end{align} Likewise, if we want to compute $\\sigma(120)$, then since $120 = 8 \\cdot 3 \\cdot 5 = 2^3 \\cdot 3^1 \\cdot 5^1$ we have (16) \\begin{align} \\sigma(120) = \\left(\\frac{2^4-1}{2-1}\\right)\\left(\\frac{3^2-1}{3-1}\\right)\\left(\\frac{5^2-1}{5-1}\\right) = 360. \\end{align} $\\square$ Carl pointed out that if a prime appears to the first power in the prime decomposition of n, then the corresponding term in $\\sigma(n)$ is just $p+1$: (17) \\begin{align} \\left(\\frac{p^2-1}{p-1}\\right) = p+1. \\end{align} Note only does this observation match up with our formula, but it also follows from the definition of $\\sigma$: if p is prime then its only divisors are 1 and p,", "1+s+t, 1)\\ge (2\\sqrt{a_5a_1} + a_3)t^3 + (2\\sqrt{a_4a_0} + a_2)t^2.$$ It suffices to prove that $$2\\sqrt{a_5a_1} + a_3 \\ge 0$$ and $$2\\sqrt{a_4a_0} + a_2 \\ge 0$$. Note that \\begin{align} 2\\sqrt{a_5a_1} + a_3 &= \\Big(2\\sqrt{a_5a_1} - \\frac{1}{3}\\cdot 94494\\, s^3 - \\frac{1}{3} \\cdot 65760\\, s^2\\Big)\\\\ &\\quad + \\Big(a_3 + \\frac{1}{3}\\cdot 94494\\, s^3 + \\frac{1}{3} \\cdot 65760\\, s^2\\Big) \\end{align} and \\begin{align} 2\\sqrt{a_4a_0} + a_2 &= \\Big(2\\sqrt{a_4a_0} - \\frac{1}{2}\\cdot 68407\\, s^4 - \\frac{1}{2}242656\\, s^3\\Big)\\\\ &\\quad + \\Big(a_2 + \\frac{1}{2}\\cdot 68407\\, s^4 + \\frac{1}{2}242656\\, s^3\\Big). \\end{align} It suffices to prove that \\begin{align} 2\\sqrt{a_5a_1} - \\frac{1}{3}\\cdot 94494\\, s^3 - \\frac{1}{3} \\cdot 65760\\, s^2 &\\ge 0,\\\\ a_3 + \\frac{1}{3}\\cdot 94494\\, s^3 + \\frac{1}{3} \\cdot 65760\\, s^2&\\ge 0,\\\\ 2\\sqrt{a_4a_0} - \\frac{1}{2}\\cdot 68407\\, s^4 - \\frac{1}{2}242656\\, s^3 &\\ge 0,\\\\ a_2 + \\frac{1}{2}\\cdot 68407\\, s^4 + \\frac{1}{2}242656\\, s^3 &\\ge 0. \\end{align} All of them can be reduced to polynomial inequalities in $$s$$ and not hard to prove. This completes the proof. let $$x=\\frac{\\sqrt{3}}{a},y=\\frac{\\sqrt{3}}{b},z=\\frac{\\sqrt{3}}{c} \\implies xy+yz+zx=3$$ with uvw method: $$3u=x+y+z,3v^2=xy+yz+zx,w^3=xyz\\\\ u\\geqslant v \\geqslant w \\geqslant 0 , w^3 \\leq 3uv^2 -2u^3+2\\sqrt{(u^2-v^2)^3}, v=1$$ then inequality becomes: $$\\frac{xy}{7y+x}+\\frac{yz}{7z+y}+\\frac{xz}{7x+z}\\leq\\frac{3}{8} \\iff \\\\3(7y+x)(7z+y)(7x+z) \\geqslant 8[xy(7z+y)(7x+z)+yz(7y+x)(7x+z)+xz(7y+x)(7z+y)]$$ \\ $$LHS=3[7(7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y)+344xyz]\\\\7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y=4\\sum(x^2y+y^2x)+3(yz^2+x^2z+xy^2-xz^2-y^2z-x^2y)=4\\sum xy(x+y+z-z)+3(y-x)(z-x)(z-y)=4(x+y+z)\\sum xy -4*3xyz+3(y-x)(z-x)(z-y) \\\\ LHS=3[7*4*3^2u+260w^3+3*7(y-x)(z-x)(z-y)]\\\\ RHS=8[\\sum_{cyc}xy(7z+y)(7x+z)]=8[7\\sum x^2y^2+57xyz(x+y+z)]\\\\= 8[ 7((xy+yz+xz)^2-2xyz(x+y+z))+57xyz(x+y+z)] \\\\=8(63+43*3u*w^3)$$ then the inequality becomes: $$4(63u+65w^3-42-86uw^3) \\geq 21(x-y)(z-x)(z-y) ....(3)$$ now to prove $$63u+65w^3-42-86uw^3 \\geq 0 \\iff w^3 \\leq \\dfrac{63u-42}{86u-65}$$ first to prove : $$\\dfrac{63u-42}{86u-65} \\geq \\dfrac{2u-1}{4u-3} \\iff (u-1)(80u-61)$$it is", "2\\frac{99(99+1)}{2}$$ $$= (198+1+1)99(99+1) = 200\\cdot100\\cdot 99$$ $$= 1980000$$ hint For $|x|<1$, $$\\sum_0^\\infty x^n=1/(1-x)$$ by differentiation $$\\sum_1^\\infty nx^{n-1}=1/(1-x)^2$$ multilying by $x$ $$\\sum_1^\\infty nx^n=x/(1-x)^2$$ replace $x=1/10$. The first one is just an Arithmetic Progression! Use $\\frac n2 (a+\\ell)$. $$\\sum_{i=1}^{100}(2i+1)=\\frac {100}2 \\big(3+201\\big)=\\color{red}{10200}$$ For the second one, the coefficients give a useful clue to a shortcut. Note that $\\displaystyle\\sum_{i=1}^n i^2=\\frac {n(n+1)(2n+1)}6$ and $\\displaystyle\\sum_{i=1}^n i=\\frac {n(n+1)}2$. Hence $$\\sum_{i=1}^{n} 6i^2+2i=n(n+1)(2n+1)+n(n+1)=n(n+1)(2n+2)=2n(n+1)^2$$ Putting $n=99$ gives $$\\sum_{i=1}^{99}=2\\cdot 99\\cdot 100^2=\\color{red}{1980000}$$", "for $n' = n - 1$, where $n \\geq 2$. It remains to show that the claim holds for $n' = n$. Indeed, observe that: \\begin{align*} &1 \\times 5 +\\cdots + n(n + 4) & \\\\ &= [1 \\times 5 + \\cdots + (n - 1)(n + 3)] + n(n + 4) &\\text{since } n \\geq 2 \\\\ &= \\frac{1}{6}(n - 1)(n)(2n + 11) + n(n + 4) &\\text{by the inductive hypothesis} \\\\ &= \\frac{1}{6}n [(n - 1)(2n + 11) + 6(n + 4)] \\\\ &= \\frac{1}{6}n [(2n^2 + 9n - 11) + (6n + 24)] \\\\ &= \\frac{1}{6}n [2n^2 + 15n + 13] \\\\ &= \\frac{1}{6}n (n + 1)(2n + 13) \\\\ \\end{align*} as desired. $~~\\blacksquare$ A proof without induction: $$\\sum_{k=1}^{n}k(k+4)=\\sum_{k=1}^{n}k^2+4\\sum_{k=1}^{n}k= \\frac{n(n+1)(2n+1)}{6}+4\\cdot\\frac{n(n+1)}{2} =\\frac{n(n+1)}{6}(2n+13)$$ $$1 \\times 5+2\\times6+3\\times7 +\\cdots +n(n + 4) =$$ $$=\\sum_{i=1}^{n}i^2+\\sum_{i=1}^{n}4i=\\frac{n(n+1)(2n+1)}{6}+4\\frac{n(n+1)}{2}$$ With induction, you want to prove a base case and then show that if it works for step $n$ then it also works for step $n+1$. If both of those things are true then you're done! Base case, $n=1$: $$1(1+4) = \\frac{1}{6}1(1+1)(2(1)+13)$$ $$5 = 5$$ Inductive step: We want to show $$1 \\times 5 + ... + (n+1)((n+1)+4) = \\frac{1}{6}(n+1)((n+1)+1)(2(n+1)+13)$$ so we have $$1 \\times 5 + ... + (n+1)((n+1)+4) = \\frac{1}{6}(n+1)((n+1)+1)(2(n+1)+13)$$ $$1 \\times 5 + ... + (n+1)((n+1)+4) = \\frac{1}{6}(n+1)(n+2)(2n+15)$$ $$\\sum_{k=1}^{n+1} k(k+4) = \\frac{1}{6}(n+1)(n+2)(2n+15)$$", "are \\aligned x_1 &= \\frac{315-314.99}2 = 0.005 \\\\ x_2 &= \\frac{315+314.99}2 = 315.00 \\aligned The correct values are $0.00317463517\\cdots$ and $314.996825\\cdots$. The larger of the two values is correct to five decimal places. The smaller of the two has zero significant digits. Now look what happens when you calculate the smaller value using the alternate form: $$x_1 = \\frac 2{315+314.99} = 0.0031747$$ which is correct to four significant digits." ]

[ "# [SOLVED] Partial Fractions/algebraic fractions $\\frac{2x^4-3x^2+x+1}{x^2-1}=ax^2+bx+c+\\frac{dx+e}{x^2-1}$ Find the constants $a,b,c,d$ and $e$", "If $A$ and $B$ are positive constants, show that $\\frac{A}{x-1} + \\frac{B}{x-2}$ has a solution on $(1,2)$ If $$A$$ and $$B$$ are positive constants, show that $$\\frac{A}{x-1} + \\frac{B}{x-2}$$ has a solution on $$(1,2)$$.", "+0 # Rational Ineq. #2 0 44 2 +153 Let $$a$$ and $$b$$ be constants. Suppose that the equation $$\\frac{(x+a)(x+b)(x+12)}{(x+3)^2} = 0$$ has exactly $$3$$ distinct roots, while the equation $$\\frac{(x+2a)(x+3)(x+6)}{(x+b)(x+12)} = 0$$ has exactly $$1$$ distinct root. Compute $$100a+b$$. Nov 23, 2019", "#### Problem 6P 6. Find the values of the constants $a$ and $b$ such that $$\\lim _{x \\rightarrow 0} \\frac{\\sqrt[3]{a x+b}-2}{x}=\\frac{5}{12}$$", "and $B$ are constants. What I have done is as following; Let $x=\\frac{h}{R}$, we can … Read more", "+0 0 53 1 Find constants $A$ and $B$ such that $\\frac{x - 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$for all $x$ such that $x\\neq -1$ and \\$ Aug 29, 2021", "+0 0 31 1 +21 Find constants $A$ and $B$ such that $\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$ for all $x$ such that $x\\neq -1$ and $x\\neq 2$. Give your answer as the ordered pair $(A,B)$. Jul 18, 2020 #1 0 If you let x go infinity, then you get 0 = A + B. If you let x = 0, then you get -7/2 = A/(-2) + B. The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3). Jul 19, 2020", "Calculus (3rd Edition) $$A=-3$$ $$B=5$$ $\\frac{2x-3}{(x-3)(x-4)}$ = $\\frac{A}{x-3}$ + $\\frac{B}{x-4}$ Multiplying (x-3)(x-4) on both sides, we have: $2x-3=A(x-4)+B(x-3)$ If we let x=4 on both sides, we can eliminate constant A to find constant B: $8-3=A(4-4)+B(4-3)$ Thus, $B = 5$. Similarly, if we let x = 3 on both sides, we can eliminate constant B to find constant A: $6-3=A(3-4)+B(3-3)$ Thus, $3=-A$ and so $A = -3$.", "+0 # help 0 44 1 The equation $$\\frac{ax^2 - 24x + b}{x - 1} = x^2 + x$$ holds for exactly two real values of $$x,$$\\$ and their sum is 12. Find $$a - b.$$ Apr 11, 2019", "to be constant.$A^2=\\frac{1}{4}c^2h^2$.Now it is getting difficult to remove $h$ and introduce $b$ in this area equation.Please help. – Vinod Kumar Punia Aug 26 '15 at 10:38", "+0 # A and B are constants such that . Find ​. 0 56 1 A and B are constants such that $$\\frac{4x+5}{x^2+x-2}= \\frac{A}{x+2} +\\frac{B}{x-1}$$. Find $$\\frac{B}{x+1} - \\frac{A}{x-2}$$. Apr 5, 2020", "and $1+ b \\,h''(t) = -a/c$. The solutions read $$g(x) = \\frac{c}{2} x^2 + c_1 x + c_2$$ and $$h(t) = -\\frac{a+c}{2 b c} t^2 +C_1 t + C_2$$ with $c\\neq0$, $c_1$, $c_2$, $C_1$, $C_2$ arbitrary constants.", "that if $X_1$ and $X_2$ are the two solutions of $ax^2+ bx+ c= 0$ then $X_1+ X_2= -\\frac{b}{a}$ which, here, is $\\frac{22}{3}$.", "Any tips or solutions? If $A$ and $B$ are arbitrary constants in the solution, then set $C=1/K_0(A \\times B)$ then you have $C \\; K_0(Ax)$ CB 3. ## Re: Simplifying a ratio of modified Bessel functions of the second kind Originally Posted by CaptainBlack If $A$ and $B$ are arbitrary constants in the solution, then set $C=1/K_0(A \\times B)$ then you have $C \\; K_0(Ax)$ CB Nice... but there's no way to reduce it into something that doesn't have infinite series, I take it?", "x} + B {e}^{i x}$ This is the required solution, but we have one more step to go. The solution has to be real, and so must be equal to its complex conjugate. It is easy to see that this means that ${A}^{'} = {B}^{\\text{*}}$. Introducing two real constant constant $C$ and $\\phi$ by $B = \\frac{C}{2} {e}^{i \\phi}$ $f \\left(x\\right) = 1 + C \\cos \\left(x + \\phi\\right)$", "## Calculus (3rd Edition) $$A=1$$ $$B=-1$$ $$C=0$$ $\\frac{2x+4}{(x-2)(x^2+4)}$ = $\\frac{A}{x-2}$+$\\frac{Bx+C}{x^2+4}$ Multiplying $(x-2)(x^2+4)$ on both sides: $2x+4=A(x^2+4)+(Bx+C)(x-2)$ If we let $x=2$ on both sides, we can eliminate constants B and C to find constant A: $4+4=A(4+4)+(Bx+C)(2-2)$ Thus, $8=8A$ and so $A=1$. Collecting all the constant terms on both sides: $4=4A-2C$ Substituting $A=1$, we find that $C=0$. Similarly, collecting all the $x$ terms on both sides: $2x=-2Bx+Cx$ If we compare only the coefficients of the $x$ terms, we arrive at the equation: $2=-2B+C$ Substituting $C=0$, we find that $B=-1$.", "+0 # Help me please, tough question 0 82 1 Find constants and such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x is not equal to -1or 2. Give your answer as the ordered pair . May 27, 2020 #1 +25541 +1 Find constants and such that $$\\dfrac{x + 7}{x^2 - x - 2} = \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1}$$", "Find constants $A$, $B$, and $C$ such that the function $y=Ax^2+Bx+C$ satisfies the differential equation $y''+y'-2y=x^2$. $A =$ $B =$ $C =$", "+0 0 67 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19832 +1 Sep 15, 2019 #1 +19832 +1", "+0 0 47 1 Find constants A and B such that: $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x≠-1 and x≠2. Give your answer as the ordered pair (A,B). Sep 14, 2019 #1 +1 I think I saw this earlier. Check out https://web2.0calc.com/questions/help-please_7890 . Sep 14, 2019" ]

[ "MathNewbie Jul 26 '12 at 14:07 In general, the discriminant of a polynomial is the product of the square of the differences of the roots. Hence, it is zero if and only if there is a multiple root. – M Turgeon Jul 26 '12 at 14:09 The case $A=0$ must be treated separately: In this case, $X^3+B$ has a double root if and only if $27B^2=0$, i.e. if and only if $B=0$. That is obvious, so let us assume $A\\ne 0$. Lets set $f(X):=X^3+AX+B$ and consider $f'(X)=3X^2 + A$. A polynomial has a double root if and only if it shares this root with its derivative. Hence, let us check when that is the case. $f'(x)=0$ means $x^2 = -\\frac 13 A$ and $f(x)=0$ means $0=x^3+Ax+B=-\\frac{1}{3}Ax+Ax+B$, so $\\frac{2}{3}Ax = -B$, i.e. $x=\\frac{-3B}{2A}$. Plugging this back into $f'$, we get that $$0=f'(x)=3\\cdot\\frac{9B^2}{4A^2} + A,$$ which yields $27B^2 = -4A^3$. Hence, $f$ has a double root if and only if $27B^2+4A^3=0$. - Thanks, rattle. – MathNewbie Jul 26 '12 at 17:50 Let $X^3+Ax+B=(X-p)(X-q)^2$ Comparing the coefficients of the different powers of X, $p+2q=0=>p=-2q$ $2pq+q^2=A=>A=-3q^2$ $-pq^2=B=>B=-q^2(-2q)=2q^3$ Eliminating q, $4A^3+27B^2=0$ as $A^3=-27q^6$ and $B^2=4q^6$ Conversely, the parametric values of (A,B) can be written as $(-3s^2,2s^3)$ Then, the equation becomes $X^3-3s^2X+2s^3=0$ Clearly, s is one of the solutions. On the division by (X-s),", "of all $(x-x_i)$ is the coefficient of the leading term. So the quadratic polynomial $Ax^2+Bx+C$ is factored in two terms. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\\qquad (\\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\\ast)$ are equal. - @Americo Tavares: I am sorry but can you explain the process of factorising the polynomial? – Sophia May 19 '11 at 9:03 Is it because the coefficient of x^2 must be 1? – Sophia May 19 '11 at 9:05 @Sophia: The quadratic polynomial $Ax^2+Bx+C$ is factored in terms of its roots. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\\qquad (\\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\\ast)$ are equal. – Américo Tavares May 19 '11 at 9:10 @Sopia: Is now clear? I have added this explanation to the answer. – Américo Tavares May 19 '11 at 9:12 @Americo Tavares: But by putting A in front of the roots of the factorised equation will throw it out of balance, no? – Sophia May 19 '11 at 9:22 Let $\\alpha,\\beta$ be the $2$ roots. Given $\\beta = \\alpha^{2}$. • For a quadratic equation $ax^{2}+bx+c$ which one root is $$\\alpha = \\frac{-b", "is exactly equal to the multiplicity of the root. Knowing this will save you time from having to take derivatives and plugging in over and over again. We know that $c_1\\sin x+c_2\\cos x$ is the homogeneous, so there is no need to include this in the particular solution. Take instead $y_p=Ax\\sin x+Bx\\cos x.$ Then: $$y_p''=(A\\sin x+Ax\\cos x+B\\cos x-Bx\\sin x)'\\\\=2A\\cos x-Ax\\sin x-2B\\sin x-Bx\\cos x$$ So $$y_p''+y_p=2A\\cos x-2B\\sin x\\equiv x\\sin x$$ So it is clear that this doesn't work, so raise the power: $$y_p=(Ax^2+Bx)\\sin x+(Cx^2+Dx)\\cos x$$ Then you'll find that $$y_p''+y_p=2(2Ax+B+C)\\cos x-2(2Cx+D-A)\\sin x$$ Equating coefficients, we see $$2(2Ax+B+C)\\equiv 0\\implies A=B+C=0\\\\-2(2Cx+D-A)\\equiv x\\implies D=A=0,C=-\\frac14\\implies B=\\frac14$$ So $$y_p=\\frac14 x(\\sin x - x\\cos x)$$ If it fails, maybe means that it is not in such format, try to raise the power of the polynomials: $$y_p(x) = (Ax^2 + Bx + C)\\sin(x) + (Dx^2 + Ex + F)\\cos(x)$$ Then: $$y'' + y =$$ $$(-Ax^2 - (B +4D)x -C -2E+2A)\\sin(x) +$$ $$(-Dx^2 + (4A-E)x + 2B+2D-F)\\cos(x) +$$ $$(Ax^2 + Bx + C)\\sin(x) + (Dx^2+Ex+F)\\cos(x)$$ $$=(-4Dx-2E+2A)\\sin(x) + (4Ax+2B+2D)\\cos(x)=x\\sin(x)$$ Thus $$A=0;B+D=0;A-E=0;-4D=1$$ We get: $$D=-\\frac{1}{4};B=\\frac{1}{4};A=E=0$$ We could take: $$y_p(x)=-\\frac{1}{4}x^2\\cos(x) + \\frac{1}{4}x\\sin(x)$$ EDIT Per above, constant $C,F$ are cancelled out; this is because they are part of homogeneous solution already (thanks @Dylan for pointing this out) • The constants $C$ and $F$ are extraneous since they are part of the", "# Math Help - help solve 2 1. ## help solve 2 The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b. 2. Originally Posted by andy69 The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b. You have $P(x) = x^2 + ax + b$. By the Remainder and Factor Theorems: If there is a root at $x = 2$, then $P(2) = 0$. If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$. So you have: $2^2 + 2a + b = 0$ $(-1)^2 - a + b = 18$. Simplify and solve these equations simultaneously for $a$ and $b$. 3. Originally Posted by Prove It You have $P(x) = x^2 + ax + b$. By the Remainder and Factor Theorems: If there is a root at $x = 2$, then $P(2) = 0$. If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$. So you have: $2^2 + 2a + b = 0$ $(-1)^2 - a + b = 18$. Simplify and solve these equations simultaneously for", "so can be expressed in terms of the coefficients of the original polynomial. We then use the basic facts we know about the sum and product of the roots, and the fact that both $a$ and $b$ satisfy the polynomial to achieve successive simplification of otherwise unwieldy terms. Here are some hints for a way through. Note that $a+b=-3$ and $ab=1$ so that $(a+1)(b+1)=ab+(a+b)+1=-3+2=-1$ Use this to put the whole thing over a common denominator and simplify. Note also that $a^2(a+1)^2=a^2(a^2+2a+1)=a^2(a^2+3a+1-a)=-a^3$ You might also need to use $a^2+b^2=(a+b)^2-2ab$ Since the full answer closest in spirit to this has been deleted, note first $(a^2+b^2) = (-3)^2-2=7$. Then put the desired expression over common denominator $(a+1)^2(b+1)^2=1$ to obtain:$$a^2(a+1)^2+b^2(b+1)^2=-a^3-b^3=3(a^2+b^2)+(a+b)=21-3=18$$ [see comment for middle step] - Use also $-a^3=3a^2+a$ obtained from $a(a^2+3a+1)=0$ – Mark Bennet Jun 28 '13 at 13:25 +1.I've learn a new approach – iostream007 Jun 28 '13 at 16:39 @MarkBennet:why is $ab=1$? – Alex Jun 28 '13 at 21:24 @Alex $(x-a)(x-b)=x^2+3x+1$ - expand and equate coefficients. This works because for $x=a$ or $x=b$ we have $(x-a)(x-b)=0$ - it is east to show that there is a unique monic quadratic equation which has roots $a$ and $b$. – Mark Bennet Jun 29 '13 at 3:07 @MarkBennet: thanks, that's great! – Alex Jun 29 '13 at 3:36 $$a+b=-3,ab=1$$ $$(a+b)^2=a^2+b^2+2ab\\implies a^2+b^2=7$$ $$(a+b)^3=a^3+b^3+3ab(a+b)\\implies", "directly from Eisenstein's Criterion and the fact that $$f(p^{\\frac{1}{3}})=0.$$ In light of this information, the polynomial given by the OP is of degree $$2$$, thus must be the zero polynomial. • Your first point that $x^3-p$ is the minimal polynomial for $p^{1/3}$ is what we need to prove here. It can't be assumed. More generally the problem boils down to showing that if $x^3-p\\in\\mathbb {Q} [x]$ has no roots in $\\mathbb {Q}$ then it is irreducible over $\\mathbb{Q}$. Dec 27 '20 at 3:01 • @ParamanandSingh You're right! I was so eager to use traces so I jumped. Thank you for your comment Dec 27 '20 at 8:25 Here's another way. If $$a+bq+cq^2=0$$, then $$a=-(bq+cq^2)$$, hence $$a^2=b^2q^2+2bcq^3+c^2q^4$$. So if $$q^3=p$$, then we have two equations: \\begin{align} a+bq+cq^2&=0\\\\ (a^2-2bcp)-c^2pq-b^2q^2&=0 \\end{align} Multiplying both sides of the first equation by $$b^2$$ and the both side of the second equation by $$cp$$, and then adding the resulting equations, we have $$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$ Now if $$q$$ is irrational (i.e., if $$p$$ is not a perfect cube) and the other variables are rational, then we must have $$ab^2+a^2c-2bc^2p=b^3-c^3p=0$$. But $$b^3-c^3p=0$$ for a non-cube $$p$$ implies $$b=c=0$$, in which case the original equation, $$a+bq+cq^2=0$$, implies $$a=0$$. Since $$q=p^{1/3}$$ is irrational it follows that minimal polynomial of $$q$$ over $$\\mathbb{Q}$$ is of degree greater than $$1$$. If we", "+0 # polynomial 0 31 1 When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 13, 3, and –12 respectively. Find the values of a, b, and c Sep 29, 2020 #1 +146 0 The remainder theorem tells us that: If a polynomial f(x) is divided by (x-a), then the remainder is f(a). So for this problem, the polynomial is divided by (x-1), (x+1), and (x+2) and the remainders are given. Therefore, we can conclude that f(1) = 13 f(-1) = 3 f(-2) = -12 You can now plug in those x values and y values to solve for a, b, and c. $$x^4+ax^3+bx+c\\\\ 1^4+a(1)^3+b(1)+c=13\\implies a+b+c=12\\\\ (-1)^4+a(-1)^3+b(-1)+c=3\\implies -a-b+c=2\\\\ (-2)^4+a(-2)^3+b(-2)+c=-12\\implies -8a-2b+c=-28$$ We now have a system of three equations. Now to find a, b, and c. We can first add the first two equations together $$a+b+c=12\\\\ -a-b+c=2\\\\ ==========\\\\ 2c=14\\\\ c=7$$ We can then multiply the 2nd equation by -2 and add it with the 3rd equation $$\\underset{\\downarrow}{-2(-a-b+c)=-2(2)}\\\\ 2a+2b-2c=-4\\\\ -8a-2b+c=-28\\\\ ============\\\\ -6a-c=-32\\\\ -6a-7=-32\\\\ -6a=-25\\\\ a=\\frac{25}{6}$$ Now, we can plug in a and c into any other equation to find b. $$a+b+c=12\\\\ \\frac{25}{6}+b+7=12\\\\ \\frac{25}{6}+b=5\\\\ b=\\frac{5}{6}$$ TL;DR a = 25/6, b = 5/6, c = 7 Sep 30, 2020", "# Sufficient and essential condition for polynomials $P$ and $Q$ to satisfy $P(\\sin x)= Q(\\cos x)$ The famous identity $\\sin^2 x+\\cos^2x =1$ can be written as follows: The polynomials $P(x)=x^2$ and $Q(x)=1-x^2$ satisfy $$P(\\sin x)= Q(\\cos x),\\quad \\text{for all }x\\in\\mathbb R$$ What are other such pairs of polynomials. In other words, what is the sufficient and essential condition for two real polynomials $P(x)$ and $Q(x)$ to satisfy $P(\\sin x)= Q(\\cos x)$ for all $x$? • Down voted for substantially changing the question. – Git Gud Jan 25 '14 at 19:43 • If $a \\ne 0$ then it would have to work for $x=0$ and $x=\\pi$, at both of which $\\sin x = 0$, but $\\cos 0 = 1$ and $\\cos \\pi = -1$. So it only works fwhen $a=0$ and b can be any arbitrary constant. – NovaDenizen Jan 25 '14 at 20:01 • This question is changed back and forth between two different versions... – N. S. Jan 25 '14 at 20:11 • -1 for changing question back/forth again. – achille hui Jan 25 '14 at 21:21 Proposition: $P(\\sin(x))=Q(\\cos(x))$ if and only if there exists a polynomial $R$ so that $$P(X)=R(X^2) \\, \\mbox{and} \\, Q(X)=R(1-X^2) \\,.$$ Proof: Step 1: $P(X)$ is an even polynomial: $$P(-\\sin(x))=P(\\sin(-x))=Q(\\cos(-x))=Q(\\cos(x))=P(\\sin(x)) \\,.$$ Thus, for all $x$, $\\sin(x)$ is a root of $P(X)-P(-X)$. Thus", "# How can I prove this theorem about polynomials? How can I prove this: Prove that, if the polynomial $p(x)=ax^3+bx^2+cx+d$ has roots $x_1,x_2,x_3$, then $d=ax_1 x_2 x_3$. - Hint: $p(x) = a(x-x_1)(x - x_2)(x - x_3)$. Note also, that $d = - ax_1x_2x_3$ – Alexander Thumm Nov 28 '11 at 13:22 Multiply out $a(x-x_1)(x-x_2)(x-x_3)$ and note the constant term... – Ｊ. Ｍ. Nov 28 '11 at 13:23 $p(x)=a(x-x_1)(x-x_2)(x-x_3)$ – pedja Nov 28 '11 at 13:23 It looks like homework. Please add the \"homework\" tag if it is so. – Oltarus Nov 28 '11 at 13:23 I think there's a negative sign missing... – David Mitra Nov 28 '11 at 13:26 If a polynomial $p$ has a root $x_0$ then we can factor out that root, i.e. $p(x) = q(x) (x-x_0)$ where $q$ is a polynomial of degree one less than $p$. This allows us to rewrite $p(x) = ax^3 + bx^2 + cx + d$ as $p(x) = k(x-x_1)(x-x_2)(x-x_3)$ where $k$ is some constant. Multiplying through, we see that the coefficient of the $x^3$ term is $k$, so $k = a$ ; and the constant term is $-kx_1x_2x_3$, so we see that $d = -ax_1x_2x_3$ , as desired.", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "with the same roots, where $$c$$ is any (non-zero) real number. We needed the extra data of a point on the curve to determine $$c$$. In general: • An $$x$$-intercept at $$x=a$$ implies that $$(x-a)$$ is a factor of $$f(x)$$. • A $$y$$-intercept at $$y=c$$ implies that $$f(0)=c$$, i.e., the constant term in $$f(x)$$ is $$c$$. • A given point $$(p,q)$$ on the graph implies that $$f(p)=q$$. If we are just given points on the graph of $$y=f(x)$$, one way to find the polynomial is by simultaneous equations, as the following example shows. #### Example Find the quadratic polynomial $$f(x)$$ whose graph passes through the three points $$(1,-2)$$, $$(2,-1)$$ and $$(3,2)$$. #### Solution Let the polynomial be $$f(x) = ax^2 + bx + c$$. We are given that $$f(1)=-2$$, $$f(2)=-1$$ and $$f(3)=2$$, so we have the simultaneous equations \\begin{alignat*}{2} a+b+c &= -2 &\\qquad\\qquad& (1) \\\\ 4a+2b+c &= -1 && (2) \\\\ 9a+3b+c &= 2. && (3) \\end{alignat*} Subtracting $$(2)-(1)$$ and $$(3)-(2)$$ gives $$3a+b = 1$$ and $$5a+b=3$$. Subtracting these two equations gives $$2a=2$$, so $$a=1$$. Substituting $$a=1$$ back into these equations gives $$b=-2$$ and then $$c=-1$$. So $$f(x) = x^2 - 2x - 1$$. In principle we could use the above method to find, say, a quintic through six given points, or a degree-100 polynomial through 101 given", "given: $3x^3+4x^2-13x-12=(Ax+1)(x+B)(x-3)+Cx$ ? In this case, expanding the right side, we have: $3x^3+4x^2-13x-12=Ax^3+(AB-3A+1)x^2-(3AB-B-C+3)x-(3B)$ This implies: $A=3$ $3B-9+1=4\\:\\therefore\\:B=4$ $36-4-C+3=13\\:\\therefore\\:C=22$ 5. ## Re: Polynomials Nope, there's no wrong with the question. 6. ## Re: Polynomials 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C There are no constants A,B,C that make this equation true for all x. This can be proven by contradiction from your own reasoning. A must be 3, so that means C must equal both f(3) and f(-1/3), however f(3) is not equal to f(-1/3). 7. ## Re: Polynomials hmm, i am intrigued. suppose: 3x3 + 4x2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C this means: 3x3 + 4x2 - 13x - 12 = Ax3 + (AB - 3A + 1)x2 + (-3AB + B - 3)x - 3B + C equating coefficients: A = 3 AB - 3A + 1 = 4 -3AB + B - 3 = -13 -3B + C = -12 the last equation gives us: B = (C+12)/3. so we have 2 equations in C: 3((C+12)/3) - 9 + 1 = 4, which leads to: C = 0. (-9)((C+12)/3) + (C+12)/3 - 3 = -13, which leads to: C = -33/4. this makes no sense at all....i strongly suggest", "first place. Anyone reading the whole post with answers will see that it's incorrect. – Aleksa Oct 12 '18 at 17:48 Your approach works except that it is not fully general: because $$x^2+1$$ is quadratic, the quotient-remainder form should be $$P(x) = Q(x)(x^2+1) + ax+b$$ for some unknown $$a$$ and $$b$$. Now we can set $$P(i) = Q(i)(0) + ai + b$$, getting one equation for $$a$$ and $$b$$. We can also set $$P(-i) = Q(i)(0) + a(-i) + b$$, getting another equation, because $$x^2+1$$ is $$0$$ when $$x=i$$ or when $$x=-i$$. Solving for $$a$$ and $$b$$, you should get $$a=1$$ and $$b=0$$. • We don't need to solve any equations since $\\,P(i) = ai+b\\,$ implies the remainder $= ax+b,\\,$ see my answer. – Bill Dubuque Oct 12 '18 at 18:46 • @BillDubuque This is true in the special case of division by $x^2+1$ when the coefficients of $P$ are real; in this case, if $P(i) = ai+b$, we know that $P(-i) = P(\\overline{i}) = \\overline{P(i)} = a(-i)+b$, so in general the remainder is $ax+b$. But in some more general case, we would need to solve the equations. – Misha Lavrov Oct 12 '18 at 18:58 • The idea works much more generally, e.g. using roots of an irreducible polynomial. This will be clear when one studies ring", "# Math Help - calc 1 1. ## calc 1 find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0). 2. Originally Posted by weezie23 find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0). We can use the points we know to find a system of equations $f(x) = ax^2+bx+c \\to f'(x)=2ax+b$ Since we know the derivative represents slope we know when x=-2 that $f'(-2)=6$ so we get $E_1: \\\\\\ 6=-4a+b$ Now using the point (-2,7) in f(x) we get $E_2: \\\\\\ 7=4a-2b+c$ and the point (-1,0) we get $E_3: \\\\\\ 0=a-b+c$ We now have a system of equations in a,b, and c Multiplying 3 by -1 and adding it to 2 gives $7=3a-b$ Now if add equation one to this one we get $13=-a \\iff a=-13$ Plugging this in above gives $7=3(-13)-b \\iff b=-46$ Now putting a and b into equation 3 gives $0=-13-(-46)+c \\iff c=-33$ So we get $f(x)=-13x^2-46x-33$ I hope this helps. 3. Hello, weezie23! Find a 2nd degree polynomial: $f(x) \\:= \\:ax^2+bx+c$ such that its graph has a tangent line with slope = 6 at point(-2, 7) and x-int (-1, 0) We have:", "so $x-a$ divides $P(x)$. Now, if we have any distinct root $a$ of a quadratic polynomial $P$, we know $$P(x) = Q(x)(x-a)$$ $Q$ must be a first-degree polynomial, since anything higher-degree, multiplied by a first-degree polynomial, would produce a higher-than-second-degree polynomial. So $$P(x) = (x-b)(x-a)$$ Now, assuming we're working over an integral domain (which $\\mathbb R$ and $\\mathbb C$ both are), $$P(x) = 0 \\Rightarrow x-b=0 \\text{ or } x-a=0$$ So $a$ and $b$ are the only zeros of $P$ (although it is possible that $a=b$). Suppose not. Then there are at least 3 roots and so $$P(x)=(x-x_1)(x-x_2)(x-x_3)Q(x)$$ This is at least a cubic so this cannot happen. If you wrote the theorem out, it would look like this THEOREM: Let $a,b,$ and $c$ be real numbers with $a \\ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \\dfrac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$. That means if $ax^2+bx+c=0$, then $x = \\dfrac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ and if $x = \\dfrac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.", "with those in the other given expression, we find that $a=2$a=2 and $b=5$b=5. Note that this is consistent with the coefficient of the linear term $b-a=3$ba=3. In attempting to solve a polynomial equation, it can happen that we can see by inspection what one of the solutions is. For example, given $2x^3-5x^2+8x-5=0$2x35x2+8x5=0, we might notice that $x=1$x=1 is a solution. If $x=1$x=1 is a solution, the factorised form of the polynomial must include the factor $x-1$x1. The other factor must be a quadratic. So, we can write $2x^3-5x^2+8x-5=(x-1)(ax^2+bx+c)$2x35x2+8x5=(x1)(ax2+bx+c) Now, if we could determine the coefficients $a$a, $b$b and $c$c we would be in a position to find the other roots of the cubic because we know how to find the roots of a quadratic. This is often done using a polynomial division algorithm but we can do it by equating coefficients as follows. $(x-1)(ax^2+bx+c)$(x−1)(ax2+bx+c) $=$= $ax^3+(b-a)x^2+(c-b)x-c$ax3+(b−a)x2+(c−b)x−c Thus, $a$a $=$= $2$2 $b-a$b−a $=$= $-5$−5 $c-b$c−b $=$= $8$8 $c$c $=$= $5$5 This set of equations is consistent if $b=-3$b=3. The original equation can now be written $(x-1)(2x^2-3x+5)=0$(x1)(2x23x+5)=0 We look for solutions to $2x^2-3x+5=0$2x23x+5=0. In this case, there are no further real solutions because the discriminant is negative. This cubic has only one real root (and two complex roots). In other cases, we could proceed, using the quadratic formula or another", "# Not understanding how to factor a polynomial completely $$P(x)=16x^4-81$$ I know that this factors out as: $$P(x)=16(x-\\frac { 3 }{ 2 } )^4$$ What I don't understand is the four different zeros of the polynomial...I see one zero which is $\\frac { 3 }{ 2 }$ but not the three others. • Double check your factorization, its incorrect. Sep 23 '14 at 2:50 • It's not fully factored you mean? Sep 23 '14 at 2:51 • @Cherry_Developer No it should be: $P(x)=(2x-3)(2x+3)(4x^2+9)$ Sep 23 '14 at 2:58 • @Cherry_Developer I mean $16(x-\\frac{3}{2})^4 = 81-216x+216 x^2-96x^3+16x^4$ not $16x^4-81$. That's why you are probably getting confused. Always double check your factorization where possible :). Sep 23 '14 at 2:59 • Every one of us has made the mistake $(a+b)^2=a^2+b^2$, and we curse ourselves each time we do so. Now you have made the same mistake with fourth power instead of second. But don’t be doubly hard on yourself. Sep 23 '14 at 3:31 Recognize $16x^4-81$ as a difference of two squares. Then factor into linear factors, \\begin{align*} P(x)&=16x^4-81\\\\ &=(4x^2)^2-9^2\\\\ &=(4x^2-9)(4x^2+9)\\\\ &=(2x-3)(2x+3)(2x+3i)(2x-3i). \\end{align*} Now $P(x)=0$ iff $$2x\\pm3=0 \\qquad\\text{or}\\qquad 2x\\pm3i=0$$ iff $$x=\\pm\\frac32 \\qquad\\text{or}\\qquad x=\\pm\\frac32i.$$ So we've found all four roots. Try making the substitutions $a=2x$ and $b=3$. Then we have that $$16x^4-81=a^4-b^4.$$ Recalling that $x^2-y^2=(x+y)(x-y)$, we have $$a^4-b^4=[a^2-b^2](a^2+b^2)=[(a-b)(a+b)](a^2+b^2)$$ Plugging $2x$ in", "of the equation $$x^2 + ax + 15 = 0$$ is 3. What is the value of b? A. $$-\\frac{1}{64}$$ B. $$-\\frac{1}{16}$$ C. $$-15$$ D. $$-16$$ E. $$-64$$ Since one of the roots of the equation $$x^2 + ax + 15 = 0$$ is 3, then substituting we'll get: $$3^2+3a+15=0$$. Solving for $$a$$ gives $$a=-8$$. Substitute $$a=-8$$ in the first equation: $$x^2-8x-b=0$$. Now, we know that it has equal roots thus its discriminant must equal to zero: $$d=(-8)^2+4b=0$$. Solving for $$b$$ gives $$b=-16$$. Why you substitute the -8 back into the first equation? what is the relation between the two equations? I know my question might be stupid but I am not getting what they are asking, maybe it´s my english... thanks!! Intern Joined: 25 Dec 2017 Posts: 3 Show Tags 16 Jan 2018, 10:11 I don't really know where I am going wrong. $$x^2+ax−b=0$$ has equal roots, and one of the roots of the equation $$x^2+ax+15=0$$ is 3 $$3^2 + 3a + 15 = 0$$ $$3a = -15 - 9$$ $$3a = -24$$ $$a = -8$$ Substituting it in 1st equation $$x^2-8x-b=0$$ because we know it has equal roots, we can put x = 3 in equation 1 $$9-8(3)-b=0$$ $$-b = -9+24$$ $$b = -15$$ Isn't this correct? Math Expert Joined: 02 Sep 2009 Posts: 52902 Show", "( 1 + 2x + 4x^2) + 3 ( 1 + 2x + 4x^2) \\\\ &= x + 2x^2 + 4x^3 + 3 + 6x + 12x^2 \\\\ &= 3 + 7x + 14x^2 + 4x^3 \\end{aligned} Note that $\\deg(f(x)g(x)) = \\deg(f(x)) + \\deg(g(x))$. If $f$ and $g$ are polynomials, sometimes you can find a polynomial $q$ such that $f(x) = g(x) q(x)$, and sometimes you can't. If you can, then we say that $g$ is a factor of $f$. Example: Is $(x-2)$ a factor of $x^2 - x - 2$? Solution: If it is, then the quotient has degree 1. So suppose $x^2 - x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$. And $-x = -2ax+bx = -2x+bx$, so $b=1$. Check the constant term: $-2 = -2b$. It works, so $x^2 - x - 2 = (x-2)(x+1)$, and the answer is \"yes\". Example: Is $(x-2)$ a factor of $x^2 + x - 2$? Solution: If it is, then the quotient has degree 1. So suppose $x^2 + x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$. And $x = -2ax+bx = -2x+bx$, so $b=3$. Check the constant term: $-2 = -2b$. Nope, so the answer is \"no\". The above ad hoc method works for a degree 1 polynomial. For higher degrees, one", "then its derivative has at least $(n-1)$ distinct zeros. – Jyrki Lahtonen Nov 23 '13 at 17:53 • @TitoPiezasIII: I think the missing context is some effort on OP's part. – tomasz Nov 23 '13 at 21:36 Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But then the same is true of $f(x-\\frac14) = x^4 + \\frac58 x^2 + Bx + A$ for some $B$ and $A$ (we don't need the formula). If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$ then $a^2+b^2+c^2+d^2 = -2\\frac58 < 0$, so $a,b,c,d$ cannot all be real, QED. • If I may add something? :) In general, if $f(x) = x^4+px^3+qx^2+rx+s=0$, then the sum of the square of the roots $x_i$ is $S(p,q)=x_1^2+x_2^2+x_3^2+x_4^2 = p^2-2q$. Hence $S(1,1)=-1$. Since the sum of the square of real numbers can't be negative, this implies not all the $x_i$ are real. – Tito Piezas III Nov 23 '13 at 20:28 • Noam, I would like to ask , how did you choose f(x). Is this a set procedure for this kind of problems or you chose it randomly. – user2369284 Nov 24 '13 at 10:47 • @user2369284 See Reciprocal polynomial. –" ]

[ "3y, the third by 4z. Then add all three up. This gives you 25x + 9y + 16z = 12. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations: [1] 4/x -25x - 21y = 8; [2] 3/x + 25x + 28z = ... 2 Do thisf(x) = 2(x^2 + 4x) + 5 = 2(x^2 + 4x + 4) + 5 - 8 = 2(x + 2)^2 - 3.$$Note that f is insensitive to the sign of x + 2 so f is symmetric about the line x = -2 2 Doing (a,b,c)→(a^2,b^2,c^2) two times is clearly the easier way. Let A=a+b+c,B=ab+bc+ca,C=abc. Consider$$\\begin{align}&(x^{1/2}-a)(x^{1/2}-b)(x^{1/2}-c)\\times (x^{1/2}+a)(x^{1/2}+b)(x^{1/2}+c) = \\\\&(x^{3/2}-Ax+Bx^{1/2}-C)\\times (x^{3/2}+Ax+Bx^{1/2}+C) = \\\\&(x^{3/2}+Bx^{1/2})^2 - (Ax+C)^2 = \\\\&x^3+2Bx^2+B^2x-A^2x^2-2ACx-C^2\\end{align}$$... 2 Using the power of the Dandelin-Gräffe iteration. For any polynomial p with roots z_i, the polynomial p^{(1)}(x) defined via$$ p^{(1)}(x^2)=p(x)·p(-x) $$has the same degree as p and roots z_i^2. Repeating this process, the polynomial p^{(2)}(x) defined via$$ p^{(2)}(x^2)=p^{(1)}(x)·p^{(1)}(-x) $$has still the same degree as p and roots ... 2 I'll suppose R is an integral domain, and therefore contained in an algebraically closed field K (one can do without this assumption, but it simplifies the argument). Suppose p(s_1,\\ldots,s_n)=0 but p is not the zero polynomial; then", "of all $(x-x_i)$ is the coefficient of the leading term. So the quadratic polynomial $Ax^2+Bx+C$ is factored in two terms. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\\qquad (\\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\\ast)$ are equal. - @Americo Tavares: I am sorry but can you explain the process of factorising the polynomial? – Sophia May 19 '11 at 9:03 Is it because the coefficient of x^2 must be 1? – Sophia May 19 '11 at 9:05 @Sophia: The quadratic polynomial $Ax^2+Bx+C$ is factored in terms of its roots. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\\qquad (\\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\\ast)$ are equal. – Américo Tavares May 19 '11 at 9:10 @Sopia: Is now clear? I have added this explanation to the answer. – Américo Tavares May 19 '11 at 9:12 @Americo Tavares: But by putting A in front of the roots of the factorised equation will throw it out of balance, no? – Sophia May 19 '11 at 9:22 Let $\\alpha,\\beta$ be the $2$ roots. Given $\\beta = \\alpha^{2}$. • For a quadratic equation $ax^{2}+bx+c$ which one root is $$\\alpha = \\frac{-b", "\\text {Using initial conditions we can figure out that B-A must equal 0, and also that A = 1. Hence:} \\\\ y = e^{-x} (cos(x) + sin(x)) \\end{array} 1. If $\\delta = 0$ There is one solution to the equation, such that $m = \\frac{-b}{2}$ Hence (16) $$y = Ae^{mx} + Bxe^{mx}$$ Inhomogeneous Inhomogeneous solutions are defined as those where f(x) != 0. We solve these by setting $f(x) = h_x + p_x$ (homogeneous solution + polynomial solution). Polynomial Solution We solve this by making educated guesses: (where p and q polynomials of the same degree) (17) \\begin{array} f(x) & y(x) \\\\ p(x) & q(x) \\\\ p(x)e^{p(x)} & q(x)e^{q(x)} \\\\ p(x)cos(ax) \\text { or } p(x)sin(ax) & q_1(x)cos(ax) + q_2(x)sin(ax) \\\\ p(x)e^{p(x)}sin(bx) \\text { or } p(x)e^{p(x)}cos(bx)|& q_1(x)e^{p(x)}sin(bx) + q_2(x)e^{p(x)}cos(bx) \\\\ \\end{array} If the guess is a solution multiply the guess by x. If it's still a solution multiply it again by x (i.e. by x2) page revision: 23, last edited: 01 Nov 2011 23:00", "MathNewbie Jul 26 '12 at 14:07 In general, the discriminant of a polynomial is the product of the square of the differences of the roots. Hence, it is zero if and only if there is a multiple root. – M Turgeon Jul 26 '12 at 14:09 The case $A=0$ must be treated separately: In this case, $X^3+B$ has a double root if and only if $27B^2=0$, i.e. if and only if $B=0$. That is obvious, so let us assume $A\\ne 0$. Lets set $f(X):=X^3+AX+B$ and consider $f'(X)=3X^2 + A$. A polynomial has a double root if and only if it shares this root with its derivative. Hence, let us check when that is the case. $f'(x)=0$ means $x^2 = -\\frac 13 A$ and $f(x)=0$ means $0=x^3+Ax+B=-\\frac{1}{3}Ax+Ax+B$, so $\\frac{2}{3}Ax = -B$, i.e. $x=\\frac{-3B}{2A}$. Plugging this back into $f'$, we get that $$0=f'(x)=3\\cdot\\frac{9B^2}{4A^2} + A,$$ which yields $27B^2 = -4A^3$. Hence, $f$ has a double root if and only if $27B^2+4A^3=0$. - Thanks, rattle. – MathNewbie Jul 26 '12 at 17:50 Let $X^3+Ax+B=(X-p)(X-q)^2$ Comparing the coefficients of the different powers of X, $p+2q=0=>p=-2q$ $2pq+q^2=A=>A=-3q^2$ $-pq^2=B=>B=-q^2(-2q)=2q^3$ Eliminating q, $4A^3+27B^2=0$ as $A^3=-27q^6$ and $B^2=4q^6$ Conversely, the parametric values of (A,B) can be written as $(-3s^2,2s^3)$ Then, the equation becomes $X^3-3s^2X+2s^3=0$ Clearly, s is one of the solutions. On the division by (X-s),", "# How can I prove this theorem about polynomials? How can I prove this: Prove that, if the polynomial $p(x)=ax^3+bx^2+cx+d$ has roots $x_1,x_2,x_3$, then $d=ax_1 x_2 x_3$. - Hint: $p(x) = a(x-x_1)(x - x_2)(x - x_3)$. Note also, that $d = - ax_1x_2x_3$ – Alexander Thumm Nov 28 '11 at 13:22 Multiply out $a(x-x_1)(x-x_2)(x-x_3)$ and note the constant term... – Ｊ. Ｍ. Nov 28 '11 at 13:23 $p(x)=a(x-x_1)(x-x_2)(x-x_3)$ – pedja Nov 28 '11 at 13:23 It looks like homework. Please add the \"homework\" tag if it is so. – Oltarus Nov 28 '11 at 13:23 I think there's a negative sign missing... – David Mitra Nov 28 '11 at 13:26 If a polynomial $p$ has a root $x_0$ then we can factor out that root, i.e. $p(x) = q(x) (x-x_0)$ where $q$ is a polynomial of degree one less than $p$. This allows us to rewrite $p(x) = ax^3 + bx^2 + cx + d$ as $p(x) = k(x-x_1)(x-x_2)(x-x_3)$ where $k$ is some constant. Multiplying through, we see that the coefficient of the $x^3$ term is $k$, so $k = a$ ; and the constant term is $-kx_1x_2x_3$, so we see that $d = -ax_1x_2x_3$ , as desired.", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "+0 # polynomial 0 31 1 When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 13, 3, and –12 respectively. Find the values of a, b, and c Sep 29, 2020 #1 +146 0 The remainder theorem tells us that: If a polynomial f(x) is divided by (x-a), then the remainder is f(a). So for this problem, the polynomial is divided by (x-1), (x+1), and (x+2) and the remainders are given. Therefore, we can conclude that f(1) = 13 f(-1) = 3 f(-2) = -12 You can now plug in those x values and y values to solve for a, b, and c. $$x^4+ax^3+bx+c\\\\ 1^4+a(1)^3+b(1)+c=13\\implies a+b+c=12\\\\ (-1)^4+a(-1)^3+b(-1)+c=3\\implies -a-b+c=2\\\\ (-2)^4+a(-2)^3+b(-2)+c=-12\\implies -8a-2b+c=-28$$ We now have a system of three equations. Now to find a, b, and c. We can first add the first two equations together $$a+b+c=12\\\\ -a-b+c=2\\\\ ==========\\\\ 2c=14\\\\ c=7$$ We can then multiply the 2nd equation by -2 and add it with the 3rd equation $$\\underset{\\downarrow}{-2(-a-b+c)=-2(2)}\\\\ 2a+2b-2c=-4\\\\ -8a-2b+c=-28\\\\ ============\\\\ -6a-c=-32\\\\ -6a-7=-32\\\\ -6a=-25\\\\ a=\\frac{25}{6}$$ Now, we can plug in a and c into any other equation to find b. $$a+b+c=12\\\\ \\frac{25}{6}+b+7=12\\\\ \\frac{25}{6}+b=5\\\\ b=\\frac{5}{6}$$ TL;DR a = 25/6, b = 5/6, c = 7 Sep 30, 2020", "first place. Anyone reading the whole post with answers will see that it's incorrect. – Aleksa Oct 12 '18 at 17:48 Your approach works except that it is not fully general: because $$x^2+1$$ is quadratic, the quotient-remainder form should be $$P(x) = Q(x)(x^2+1) + ax+b$$ for some unknown $$a$$ and $$b$$. Now we can set $$P(i) = Q(i)(0) + ai + b$$, getting one equation for $$a$$ and $$b$$. We can also set $$P(-i) = Q(i)(0) + a(-i) + b$$, getting another equation, because $$x^2+1$$ is $$0$$ when $$x=i$$ or when $$x=-i$$. Solving for $$a$$ and $$b$$, you should get $$a=1$$ and $$b=0$$. • We don't need to solve any equations since $\\,P(i) = ai+b\\,$ implies the remainder $= ax+b,\\,$ see my answer. – Bill Dubuque Oct 12 '18 at 18:46 • @BillDubuque This is true in the special case of division by $x^2+1$ when the coefficients of $P$ are real; in this case, if $P(i) = ai+b$, we know that $P(-i) = P(\\overline{i}) = \\overline{P(i)} = a(-i)+b$, so in general the remainder is $ax+b$. But in some more general case, we would need to solve the equations. – Misha Lavrov Oct 12 '18 at 18:58 • The idea works much more generally, e.g. using roots of an irreducible polynomial. This will be clear when one studies ring", "with the same roots, where $$c$$ is any (non-zero) real number. We needed the extra data of a point on the curve to determine $$c$$. In general: • An $$x$$-intercept at $$x=a$$ implies that $$(x-a)$$ is a factor of $$f(x)$$. • A $$y$$-intercept at $$y=c$$ implies that $$f(0)=c$$, i.e., the constant term in $$f(x)$$ is $$c$$. • A given point $$(p,q)$$ on the graph implies that $$f(p)=q$$. If we are just given points on the graph of $$y=f(x)$$, one way to find the polynomial is by simultaneous equations, as the following example shows. #### Example Find the quadratic polynomial $$f(x)$$ whose graph passes through the three points $$(1,-2)$$, $$(2,-1)$$ and $$(3,2)$$. #### Solution Let the polynomial be $$f(x) = ax^2 + bx + c$$. We are given that $$f(1)=-2$$, $$f(2)=-1$$ and $$f(3)=2$$, so we have the simultaneous equations \\begin{alignat*}{2} a+b+c &= -2 &\\qquad\\qquad& (1) \\\\ 4a+2b+c &= -1 && (2) \\\\ 9a+3b+c &= 2. && (3) \\end{alignat*} Subtracting $$(2)-(1)$$ and $$(3)-(2)$$ gives $$3a+b = 1$$ and $$5a+b=3$$. Subtracting these two equations gives $$2a=2$$, so $$a=1$$. Substituting $$a=1$$ back into these equations gives $$b=-2$$ and then $$c=-1$$. So $$f(x) = x^2 - 2x - 1$$. In principle we could use the above method to find, say, a quintic through six given points, or a degree-100 polynomial through 101 given", "is exactly equal to the multiplicity of the root. Knowing this will save you time from having to take derivatives and plugging in over and over again. We know that $c_1\\sin x+c_2\\cos x$ is the homogeneous, so there is no need to include this in the particular solution. Take instead $y_p=Ax\\sin x+Bx\\cos x.$ Then: $$y_p''=(A\\sin x+Ax\\cos x+B\\cos x-Bx\\sin x)'\\\\=2A\\cos x-Ax\\sin x-2B\\sin x-Bx\\cos x$$ So $$y_p''+y_p=2A\\cos x-2B\\sin x\\equiv x\\sin x$$ So it is clear that this doesn't work, so raise the power: $$y_p=(Ax^2+Bx)\\sin x+(Cx^2+Dx)\\cos x$$ Then you'll find that $$y_p''+y_p=2(2Ax+B+C)\\cos x-2(2Cx+D-A)\\sin x$$ Equating coefficients, we see $$2(2Ax+B+C)\\equiv 0\\implies A=B+C=0\\\\-2(2Cx+D-A)\\equiv x\\implies D=A=0,C=-\\frac14\\implies B=\\frac14$$ So $$y_p=\\frac14 x(\\sin x - x\\cos x)$$ If it fails, maybe means that it is not in such format, try to raise the power of the polynomials: $$y_p(x) = (Ax^2 + Bx + C)\\sin(x) + (Dx^2 + Ex + F)\\cos(x)$$ Then: $$y'' + y =$$ $$(-Ax^2 - (B +4D)x -C -2E+2A)\\sin(x) +$$ $$(-Dx^2 + (4A-E)x + 2B+2D-F)\\cos(x) +$$ $$(Ax^2 + Bx + C)\\sin(x) + (Dx^2+Ex+F)\\cos(x)$$ $$=(-4Dx-2E+2A)\\sin(x) + (4Ax+2B+2D)\\cos(x)=x\\sin(x)$$ Thus $$A=0;B+D=0;A-E=0;-4D=1$$ We get: $$D=-\\frac{1}{4};B=\\frac{1}{4};A=E=0$$ We could take: $$y_p(x)=-\\frac{1}{4}x^2\\cos(x) + \\frac{1}{4}x\\sin(x)$$ EDIT Per above, constant $C,F$ are cancelled out; this is because they are part of homogeneous solution already (thanks @Dylan for pointing this out) • The constants $C$ and $F$ are extraneous since they are part of the", "Review question # Can we factorise $x^4 + Ax^2 + B$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R7809 ## Solution Suppose that the equation $x^4 + Ax^2 + B = (x^2 + ax + b)(x^2 - ax + b)$ holds for all values of $x$. 1. Find $A$ and $B$ in terms of $a$ and $b$. Expanding the right-hand side of the equation gives $x^4 + Ax^2 + B = x^4 + (2b-a^2)x^2 + b^2,$ so by equating coefficients we get \\begin{align*} A &= 2b-a^2, \\\\ B &= b^2. \\end{align*} In fact, we have a difference of two squares here: \\begin{align*} (x^2 + ax + b)(x^2 - ax + b) &=\\bigl((x^2+b)+ax\\bigr)\\bigl((x^2+b)-ax\\bigr)\\\\ &=(x^2+b)^2-(ax)^2\\\\ &=x^4+2bx^2+b^2-a^2x^2\\\\ &=x^4+(2b-a^2)x^2+b^2. \\end{align*} 1. Use this information to find a factorisation of the expression $x^4 - 20x^2 + 16$ as a product of two quadratics in $x$. We can write $x^4 - 20x^2 + 16 = (x^2 + ax + b)(x^2 - ax + b)$ if we can find solutions to $2b-a^2=-20, \\qquad b^2 = 16.$ These equations have solutions \\begin{align*} b&=4, & a&=2\\sqrt{7} \\\\ b&=-4,\\quad & a&=2\\sqrt{3}. \\end{align*} We can use either one of these (we were only asked to find one factorisation). From the first solution, we get", "+0 # Functions math problem 0 99 1 +156 The polynomial $$f(x)$$ satisfies $$f(x+1)-f(x)=6x+4.$$ Find the leading coefficient of $$f(x)$$ So far, I'm convinced that f(x) is a*(6x+4)+b, but I'm stuck. I tried 6, but the leading coefficient is not 6. All help is appreciated! Thanks! Sep 9, 2020 ### 1+0 Answers #1 +141 +1 WLOG let f(x) be linear. So f(x)=ax+b. So $$f(x+1)=a(x-1)+b=ax-a+b\\rightarrow ax-a+b-ax-b=-a$$. Note that this is constant, so f(x) must be quadratic. Let f(x)=ax^2+bx+c. So $$f(x+1)=a(x+1)^2+b(x+1)+c=ax^2+2ax+bx+a+b+c\\rightarrow$$$$f(x+1)-f(x)=2ax+c\\implies 2a=6\\rightarrow \\boxed{a=3}$$ Sep 9, 2020", "# Math Help - help solve 2 1. ## help solve 2 The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b. 2. Originally Posted by andy69 The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b. You have $P(x) = x^2 + ax + b$. By the Remainder and Factor Theorems: If there is a root at $x = 2$, then $P(2) = 0$. If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$. So you have: $2^2 + 2a + b = 0$ $(-1)^2 - a + b = 18$. Simplify and solve these equations simultaneously for $a$ and $b$. 3. Originally Posted by Prove It You have $P(x) = x^2 + ax + b$. By the Remainder and Factor Theorems: If there is a root at $x = 2$, then $P(2) = 0$. If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$. So you have: $2^2 + 2a + b = 0$ $(-1)^2 - a + b = 18$. Simplify and solve these equations simultaneously for", "( 1 + 2x + 4x^2) + 3 ( 1 + 2x + 4x^2) \\\\ &= x + 2x^2 + 4x^3 + 3 + 6x + 12x^2 \\\\ &= 3 + 7x + 14x^2 + 4x^3 \\end{aligned} Note that $\\deg(f(x)g(x)) = \\deg(f(x)) + \\deg(g(x))$. If $f$ and $g$ are polynomials, sometimes you can find a polynomial $q$ such that $f(x) = g(x) q(x)$, and sometimes you can't. If you can, then we say that $g$ is a factor of $f$. Example: Is $(x-2)$ a factor of $x^2 - x - 2$? Solution: If it is, then the quotient has degree 1. So suppose $x^2 - x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$. And $-x = -2ax+bx = -2x+bx$, so $b=1$. Check the constant term: $-2 = -2b$. It works, so $x^2 - x - 2 = (x-2)(x+1)$, and the answer is \"yes\". Example: Is $(x-2)$ a factor of $x^2 + x - 2$? Solution: If it is, then the quotient has degree 1. So suppose $x^2 + x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$. And $x = -2ax+bx = -2x+bx$, so $b=3$. Check the constant term: $-2 = -2b$. Nope, so the answer is \"no\". The above ad hoc method works for a degree 1 polynomial. For higher degrees, one", "$$a, etc. There must exist other two $$q$$ and $$k$$ such that $$p'(q)=p'(k)=0$$ because $$p$$ have other two roots, and it would need to change of increasing/decreasing two times.. But, this leads us to a contradiction, because $$p$$ can only have a maximum of $$2$$ critical points. It is similar with $$b$$ a relative minimum. If $$b$$ is not a relative maximum or a relative minimum, $$p$$ has to be strictly increasing or decreasing over an interval $$(b-\\alpha, b+\\alpha)$$. No matter that $$b < a,c$$ or $$b> a,c$$ or $$a, etc. There must exist other two $$q$$ and $$k$$ such that $$p'(q)=p'(k)=0$$ because $$p$$ have other two roots, and it would need to change of increasing/decreasing two times. But, this leads us to a contradiction, because $$p$$ can only have a maximum of $$2$$ critical points. So we showed that if $$p$$ is a third degree polynomial with three roots, and one of these roots is a critical number, $$p$$ does not exist. • Do you know that if $p(b)=p'(b)=0$ for a polynomial $p(x)$ then $b$ is a double root - ie $p(x)=(x-b)^2q(x)$ for some polynomial $q(x)$? To prove this consider $p(x)=(x-b)s(x)$ so that $p'(x)=(x-b)s'(x)+s(x)$ and $p'(b)=s(b)=0$. – Mark Bennet May 8 at 12:20 $$p(x)=Ax^3+Bx^2+Cx+D=A(x-\\alpha)(x-\\beta)(x-\\gamma)$$ where $$\\alpha,\\beta,\\gamma$$ are the roots. Since we are given that $$p(a)=p(b)=p(c)=0$$ we know", "directly from Eisenstein's Criterion and the fact that $$f(p^{\\frac{1}{3}})=0.$$ In light of this information, the polynomial given by the OP is of degree $$2$$, thus must be the zero polynomial. • Your first point that $x^3-p$ is the minimal polynomial for $p^{1/3}$ is what we need to prove here. It can't be assumed. More generally the problem boils down to showing that if $x^3-p\\in\\mathbb {Q} [x]$ has no roots in $\\mathbb {Q}$ then it is irreducible over $\\mathbb{Q}$. Dec 27 '20 at 3:01 • @ParamanandSingh You're right! I was so eager to use traces so I jumped. Thank you for your comment Dec 27 '20 at 8:25 Here's another way. If $$a+bq+cq^2=0$$, then $$a=-(bq+cq^2)$$, hence $$a^2=b^2q^2+2bcq^3+c^2q^4$$. So if $$q^3=p$$, then we have two equations: \\begin{align} a+bq+cq^2&=0\\\\ (a^2-2bcp)-c^2pq-b^2q^2&=0 \\end{align} Multiplying both sides of the first equation by $$b^2$$ and the both side of the second equation by $$cp$$, and then adding the resulting equations, we have $$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$ Now if $$q$$ is irrational (i.e., if $$p$$ is not a perfect cube) and the other variables are rational, then we must have $$ab^2+a^2c-2bc^2p=b^3-c^3p=0$$. But $$b^3-c^3p=0$$ for a non-cube $$p$$ implies $$b=c=0$$, in which case the original equation, $$a+bq+cq^2=0$$, implies $$a=0$$. Since $$q=p^{1/3}$$ is irrational it follows that minimal polynomial of $$q$$ over $$\\mathbb{Q}$$ is of degree greater than $$1$$. If we", "# Math Help - calc 1 1. ## calc 1 find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0). 2. Originally Posted by weezie23 find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0). We can use the points we know to find a system of equations $f(x) = ax^2+bx+c \\to f'(x)=2ax+b$ Since we know the derivative represents slope we know when x=-2 that $f'(-2)=6$ so we get $E_1: \\\\\\ 6=-4a+b$ Now using the point (-2,7) in f(x) we get $E_2: \\\\\\ 7=4a-2b+c$ and the point (-1,0) we get $E_3: \\\\\\ 0=a-b+c$ We now have a system of equations in a,b, and c Multiplying 3 by -1 and adding it to 2 gives $7=3a-b$ Now if add equation one to this one we get $13=-a \\iff a=-13$ Plugging this in above gives $7=3(-13)-b \\iff b=-46$ Now putting a and b into equation 3 gives $0=-13-(-46)+c \\iff c=-33$ So we get $f(x)=-13x^2-46x-33$ I hope this helps. 3. Hello, weezie23! Find a 2nd degree polynomial: $f(x) \\:= \\:ax^2+bx+c$ such that its graph has a tangent line with slope = 6 at point(-2, 7) and x-int (-1, 0) We have:", "&= 0 \\\\ (k - 28)(k + 2)&= 0 \\\\ \\text{Therefore } k = 28 &\\text{ or } k = -2 \\end{aligned} We check these two roots against the restrictions for $$k$$ and confirm that both are valid. We can now use values obtained for $$k$$ to solve for the original variable $$x$$ For $$k = 28$$ \\begin{aligned} x^2 + 3x &= 28 \\\\ x^2 + 3x - 28 &= 0 \\\\ (x + 7)(x - 4) &= 0 \\\\ \\text{Therefore } x = -7 & \\text{ or } x = 4 \\end{aligned} For $$k = -2$$ \\begin{aligned} x^2 + 3x &= -2 \\\\ x^2 + 3x + 2 &= 0 \\\\ (x + 2)(x + 1)&= 0 \\\\ \\text{Therefore } x = -1 &\\text{ or } x = -2 \\end{aligned} We check these roots against the restrictions for $$x$$ and confirm that all four values are valid. The roots of the equation are $$x = -7$$, $$x = 4$$, $$x = -1$$ and $$x = -2$$. $$x^2 - 18 + x + \\dfrac{72}{x^2 + x} = 0$$ The restrictions are the values for $$x$$ that would result in the denominator being equal to $$\\text{0}$$, which would make the fraction undefined. Therefore $$x \\ne 0$$ and $$x \\ne -1$$. We notice that $$x^2 + x$$ is a repeated", "given: $3x^3+4x^2-13x-12=(Ax+1)(x+B)(x-3)+Cx$ ? In this case, expanding the right side, we have: $3x^3+4x^2-13x-12=Ax^3+(AB-3A+1)x^2-(3AB-B-C+3)x-(3B)$ This implies: $A=3$ $3B-9+1=4\\:\\therefore\\:B=4$ $36-4-C+3=13\\:\\therefore\\:C=22$ 5. ## Re: Polynomials Nope, there's no wrong with the question. 6. ## Re: Polynomials 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C There are no constants A,B,C that make this equation true for all x. This can be proven by contradiction from your own reasoning. A must be 3, so that means C must equal both f(3) and f(-1/3), however f(3) is not equal to f(-1/3). 7. ## Re: Polynomials hmm, i am intrigued. suppose: 3x3 + 4x2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C this means: 3x3 + 4x2 - 13x - 12 = Ax3 + (AB - 3A + 1)x2 + (-3AB + B - 3)x - 3B + C equating coefficients: A = 3 AB - 3A + 1 = 4 -3AB + B - 3 = -13 -3B + C = -12 the last equation gives us: B = (C+12)/3. so we have 2 equations in C: 3((C+12)/3) - 9 + 1 = 4, which leads to: C = 0. (-9)((C+12)/3) + (C+12)/3 - 3 = -13, which leads to: C = -33/4. this makes no sense at all....i strongly suggest", "# Solution to polynomial $ax^k-bx^{k-1}+b-a=0$ I once spent far too long getting nowhere with this. Is there a way of finding the real roots of $ax^k-bx^{k-1}+b-a=0$ where $a, b, k\\in \\mathbb N$ and $b\\gt a$ and $k\\gt 1$? I know that there is no general formula for solving polynomials of degree greater than 4, but with so few $x$s I thought it might be possible. Note that $x=1$ is always a solution. Because of the dearth of $x$s the stationary points are easy to find, and I know a solution exists between $x=\\frac{b(k-1)}{ak}$ and $x=\\frac{b}{a}$. - I assume you have done quite some work already. What do you know about the derivative? Where do you expect to find other real roots, besides between those two values? – TMM Sep 25 '11 at 19:36 For odd k there is a maximum at x=0 and a minimum at $\\frac{b(k-1)}{ak}$. There are the two solutions mentioned in my question and a negative one. For even k x=0 becomes a point of inflection. There are only the two solutions. – Peter Phipps Sep 25 '11 at 19:50 Notice that since $x=1$ is always a solution, remaining real roots satisfy $a x^{k-1} = (b-a) \\sum_{n=0}^{k-1} x^n$, which does not look as innocent as you original equation. For example, for $k=6$, $a=1$ and $b=2$," ]

[ "what is the difference between being neither even nor odd and being none of these $f(x)$ is an odd function and $g(x)$ is neither even nor odd,Then $f(x) + g(x)$ (a). Even (b) Odd (c)Nither even nor odd (d)None of these. $\\boldsymbol{My}$$\\boldsymbol{Approach}$$\\Longrightarrow$I know (a) and(b) incorrect. But i dont know which one is correct between (c) and (d).What is the difference between (c) and (d).?? And What is the key to recognize ??? Please explain in detail. Let $f(x) = 1$. Then $f\\;$is even. First, let $g(x) = x-1$. Then $g\\;$is neither even nor odd, but $f+g\\;$is odd. Next, let $g(x) = x+1$. Then $g\\;$is neither even nor odd, and $f+g\\;$is also neither even nor odd. The first example disqualifies $(c)$, and the second example disqualifies $(a)\\;$and $(b)$.", "$f(-x)=-f(x)$, if $g$ is odd then $g(-x)=-g(x)$. Show: $f(g(-x))=-f(g(x))$ $f(g(-x))=f(-g(x))=-f(g(x))$ Therefore $f\\circ g$ is odd. 6. ## Re: Even and odd composite functions. Ok thanks for the help.", "## Calculus 8th Edition If $f$ and $g$ are both even functions, $f+g$ is even. If $f$ and $g$ are both odd functions, $f+g$ is odd. If $f$ is even and $g$ is odd, $f+g$ can be even, odd, or neither of them. Assume $f$ and $g$ are both even functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x).$$Thus, $f+g$ is even. Assume $f$ and $g$ are both odd functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=-f(x)+(-g(x))=-(f+g)(x).$$Thus, $f+g$ is odd. Assume $f$ is even and $g$ is odd, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+(-g(x))=(f-g)(x).$$In this case, $f+g$ is even,if $g=0$, and $f+g$ is odd, if $f=0$, but if $g \\neq 0$ then $f+g$ is neither even nor odd (For example, take $f=x^2$ and $g=x$; then $f+g=x^2+x$ is neither even or odd since $(f+g)(-x)=x^2-x$ and $(f+g)(-x) \\neq (f+g)(x)$ and $(f+g)(-x) \\neq -(f+g)(x)$.", "## Calculus: Early Transcendentals 8th Edition If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd. If $f(x)$ and $g(x)$ are both even, $f(x)=f(-x)$ (i) $g(x)=g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)+g(-x)$ (from i) and ii)) $=S(-x)$ $\\therefore f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $-f(x)=f(-x), f(x)=-f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=-f(-x)-g(-x)$ (from i) and ii)) $=-(f(-x)+g(-x))$ $=-S(x)$ $\\because S(-x)=-S(x),$ $\\therefore f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)=f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)-g(-x)$ (from i) and ii)) $S(-x)=f(-x)+g(-x)$ $\\ne \\pm S(x)$ $\\because S(-x)\\ne \\pm S(x)$ $\\therefore f(x)+g(x)$ is neither even nor odd.", "## College Algebra (10th Edition) We know that if a function is odd, then $f(-x)=-f(x).$ So here $f(x)=-f(x)$ and $g(x)=-g(x)$. Then $f(g(x))=f(-g(x))=-f(g(x))$, thus $f(g(x))$ is also odd, thus we proved what we had to.", "## Algebra and Trigonometry 10th Edition Suppose $f$ to be even and $g$ to be odd. Thus, $(f+g)(-x)=f(-x) +g(-x)$ or, $=f(x) +g(-x)$ or, $=f(x)+[-g(x)]$ Therefore, the sum of an even function and an odd function is neither even nor odd.", "## Calculus 8th Edition When $f$ is odd, we can find that: $f(g(-x)) = f(-g(x))= -f(g(x)) = -h(x)$ Therefore $h$ is odd. However, when $f$ is even, we can find that: $f(g(-x)) = f(-g(x))= f(g(x)) = h(x)$ Therefore $h$ is even. Thus, $h$ is not always an odd function.", "odd}$ $f(x) = g(x) + h(x)$ Now that is true for Any function f.", "$g(x)+h(x)=x+1$ is neither odd nor even. • For an odd function $g(x) = x$ and the function $h(x) = x+1$, neigher of the functions $g(h(x)) = x+1$ nor $h(g(x)) = x+1$ is either even or odd. • For an odd function $g(x) = x$ and the function $h(x) = x+1$, the function $g(x)h(x) = x^2+1$ is neither even nor odd. -", "as $f^{-1}(x)$ e.g. are $f(x)=x^2$ and $g(x)=x^3$ even, or odd functions? What about $h(x)=x^2+x$ if $f(x)=x^2$ then $f(-x)=(-x)^2=(-1)^2*x^2=x^2=f(x)$, so f(x) is an even function. if $g(x)=x^3$ then $g(-x)=(-x)^3=(-1)^3*x^3=-1*x^3=-1*g(x)$, so g(x) is an odd function. if $h(x)=x^2+x$, then $h(-x)=(-x)^2+(-x)=x^2-x$. h(-x) is neither equal to h(x) nor to -h(x), so h(x) is neither even nor odd.", "anonymous one year ago let f and g are functions that are neither even nor odd. a)Create an example where f+g is even, b) f+g is odd, c)f.g is even, d) f.g is odd • This Question is Open 1. Loser66 I give you one case , b, as example, you do the rest, ok? Let $$f(x) = x+3$$ let check $$f(-x) = -x +3 \\neq f(x) ~~hence,~~\\text{f(x) is not even}$$ $$-f(x) = -x-3 \\neq f(-x)~~hence,~~\\text{f(x) is not odd, therefore f(x) is neither even nor odd}$$ Now, let $$g(x) = -2x-3$$, let check $$g(-x) = 2x-3\\neq g(x),~~hence,~~g(x) \\text{is not even}$$ $$-g(x) = 2x+3 \\neq g(-x),~~hence~~\\text{g(x) is not odd, hence g(x) is neither even nor odd}$$ Now, combine $$(f+g)(x) = f(x) +g(x) = x+3-2x-3 = -x$$ let check $$(f+g)(-x) = x \\\\(-(f+g)(x) = -(-x) =x$$ hence $$(f+g)(-x) = x=-(f+g)(x)$$, therefore $$(f+g)(x)$$is an odd function.", "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # If $f$ and $g$ are both even functions, is the product $fg$ even? If $f$ and $g$ are both odd functions, is $fg$ odd? What if $f$ is even and $g$ is odd? Justify your answers. ## $f g$ is even if both $f$ and $g$ are even, or both $f$ and $g$ are odd.If one of them is odd and the other is even, then $f g$ is odd. ### Discussion You must be signed in to discuss. Lectures Join Bootcamp ### Video Transcript If F is even then, F of X is equal to f of the opposite of X. Opposite X values have the same way value. If G is even then, G of X is equal to G of the opposite of X. Again opposite X values have the same y value. So what about half times G f of x times G of X? Well, we could replace the F of X with f of the opposite of X, and we could replace the G of X with G of the opposite of X. So f of x g of X is equal to f of the opposite of X g of", "## Calculus 8th Edition if $f$ and $g$ are both even functions, $fg$ is even. if $f$ and $g$ are both odd functions, $fg$ is even. if $f$ is even and $g$ is odd, $fg$ is odd. if $f$ and $g$ are both even functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ and $g$ are both odd functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=(-f(x))(-g(x))=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ is even and $g$ is odd, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x)=-(fg)(x)$ Therefore $fg$ is odd.", "## College Algebra (10th Edition) To find out if the function is even, odd, or neither, we find $G(−x)$ and compare the result to $G(x)$: $G(x)=1-x+x^{3}$ $G(-x)=1-(-x)+(-x)^{3}$ $=1+x-x^{3}$ This does not equal $-G(x)$ or $G(x)$. Thus, $G(x)$ is neither even or odd.", "# Let f(x) = |x-1|. 1) Verify that f(x) is neither even nor odd. 2) Can f(x) be written as the sum of an even function and an odd function ? a) If so, exhibit a solution. Are there more solutions ? b) If not, prove that it is impossible. Feb 6, 2018 Let $f \\left(x\\right) = | x - 1 |$. If f were even, then $f \\left(- x\\right)$ would equal $f \\left(x\\right)$ for all x. If f were odd, then $f \\left(- x\\right)$ would equal $- f \\left(x\\right)$ for all x. Observe that for x = 1 $f \\left(1\\right) = | 0 | = 0$ $f \\left(- 1\\right) = | - 2 | = 2$ Since 0 is not equal to 2 or to -2, f is neither even nor odd. Might f be written as $g \\left(x\\right) + h \\left(x\\right)$, where g is even and h is odd? If that were true then $g \\left(x\\right) + h \\left(x\\right) = | x - 1 |$. Call this statement 1. Replace x by -x. $g \\left(- x\\right) + h \\left(- x\\right) = | - x - 1 |$ Since g is even and h is odd, we have: $g \\left(x\\right) - h \\left(x\\right) = | - x - 1 |$ Call this statement 2. Putting statements 1", "reached the case where $f\\left( {\\color{red}{ - x}} \\right) \\ne f\\left( x \\right)$ and $f\\left( {\\color{red}{ - x}} \\right) \\ne - f\\left( x \\right)$, this function is neither even nor odd! Example 6: Determine whether the given function is even, odd, or neither: Solution: Therefore, function $g\\left( x \\right)$ is an odd function! Example 7: Determine whether the given function is even, odd, or neither: Solution: Therefore, the function $h\\left( x \\right)$ is neither! Example 8: Determine whether the given function is even, odd, or neither: Solution: Therefore, function $k\\left( x \\right)$ is an even function!", "it to check your answer visually. $f(x) = \\dfrac{x}{x + 1}$ Heather Z. ### Problem 76 Determine whether $f$ is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $f(x) = x | x |$ Heather Z. ### Problem 77 Determine whether $f$ is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $f(x) = 1 + 3x^2 - x^4$ Heather Z. ### Problem 78 Determine whether $f$ is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $f(x) = 1 + 3x^3 - x^5$ Heather Z. ### Problem 79 If $f$ and $g$ are both even functions, is $f$ + $g$ even? If $f$ and $g$ are both odd functions, is $f$ + $g$ odd? What if $f$ is even and $g$ is odd? Justify your answers. Heather Z. If $f$ and $g$ are both even functions, is the product $fg$ even? If $f$ and $g$ are both odd functions, is $fg$ odd? What if $f$ is even and $g$ is odd? Justify your answers.", "- f(x)$ is a function such that $g(1) + \\cdots + g(1999)$ is even, and similarily if $f$ has even sum then $g$ has odd sum. In this way, we can pair up every single one of the $4^{1999}$ functions into pairs containing one function for which the sum is even, and one function for which the sum is odd (where one element of the pair is $4003$ minus the other). Thus exactly half of all the possible functions have odd sum. This gives $2\\cdot 4^{1998}$.", "use it to check your answer visually. $$f(x)=\\frac{x}{x^{2}+1}$$ Check back soon! Problem 74$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=\\frac{x^{2}}{x^{4}+1}$$ Check back soon! Problem 75$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=\\frac{x}{x+1}$$ Check back soon! Problem 76$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=x|x|$$ Check back soon! Problem 77$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=1+3 x^{2}-x^{4}$$ Check back soon! Problem 78$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=1+3 x^{3}-x^{5}$$ Check back soon! Problem 79 If$f$and$g$are both even functions, is$f+g$even? If$f$and$g$are both odd functions, is$f+g$odd? What if$f$is even and$g$is odd? Justify your answers. Check back soon! Problem 80 If$f$and$g$are both even functions, is the product$f g$even? If$f$and$g$are both odd functions, is$f g$odd? What if$f$is even and$g\\$ is odd? Justify your answers. Check back soon!", "# Composite of odd functions is odd (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Statement ### Statement for two functions Suppose $f$ and $g$ are odd functions so that the composite $f \\circ g$ makes sense. Then, $f \\circ g$ is also an odd function. Note that composition of functions does not commute, so if we can make sense of both $f \\circ g$ and $g \\circ f$, these are both (possibly equal, possibly distinct) odd functions. ### Statement for more than two functions Fill this in later" ]

[ "## Calculus 8th Edition When $f$ is odd, we can find that: $f(g(-x)) = f(-g(x))= -f(g(x)) = -h(x)$ Therefore $h$ is odd. However, when $f$ is even, we can find that: $f(g(-x)) = f(-g(x))= f(g(x)) = h(x)$ Therefore $h$ is even. Thus, $h$ is not always an odd function.", "## Calculus: Early Transcendentals 8th Edition If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd. If $f(x)$ and $g(x)$ are both even, $f(x)=f(-x)$ (i) $g(x)=g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)+g(-x)$ (from i) and ii)) $=S(-x)$ $\\therefore f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $-f(x)=f(-x), f(x)=-f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=-f(-x)-g(-x)$ (from i) and ii)) $=-(f(-x)+g(-x))$ $=-S(x)$ $\\because S(-x)=-S(x),$ $\\therefore f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)=f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)-g(-x)$ (from i) and ii)) $S(-x)=f(-x)+g(-x)$ $\\ne \\pm S(x)$ $\\because S(-x)\\ne \\pm S(x)$ $\\therefore f(x)+g(x)$ is neither even nor odd.", "## Algebra and Trigonometry 10th Edition Suppose $f$ to be even and $g$ to be odd. Thus, $(f+g)(-x)=f(-x) +g(-x)$ or, $=f(x) +g(-x)$ or, $=f(x)+[-g(x)]$ Therefore, the sum of an even function and an odd function is neither even nor odd.", "+ a_1k^{n-2} + \\cdots +a_{n-1})=-a_n=-f(0).$ on the other hand $f(x)=(x-1)g(x)+f(1),$ for some $g \\in \\mathbb{Z}[x].$ thus $(k-1)g(k)=-f(1),$ which is impossible because $f(1)$ is odd and $k-1$ is even. $\\Box$", "## Calculus 8th Edition If $f$ and $g$ are both even functions, $f+g$ is even. If $f$ and $g$ are both odd functions, $f+g$ is odd. If $f$ is even and $g$ is odd, $f+g$ can be even, odd, or neither of them. Assume $f$ and $g$ are both even functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x).$$Thus, $f+g$ is even. Assume $f$ and $g$ are both odd functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=-f(x)+(-g(x))=-(f+g)(x).$$Thus, $f+g$ is odd. Assume $f$ is even and $g$ is odd, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+(-g(x))=(f-g)(x).$$In this case, $f+g$ is even,if $g=0$, and $f+g$ is odd, if $f=0$, but if $g \\neq 0$ then $f+g$ is neither even nor odd (For example, take $f=x^2$ and $g=x$; then $f+g=x^2+x$ is neither even or odd since $(f+g)(-x)=x^2-x$ and $(f+g)(-x) \\neq (f+g)(x)$ and $(f+g)(-x) \\neq -(f+g)(x)$.", "$f(-x)=-f(x)$, if $g$ is odd then $g(-x)=-g(x)$. Show: $f(g(-x))=-f(g(x))$ $f(g(-x))=f(-g(x))=-f(g(x))$ Therefore $f\\circ g$ is odd. 6. ## Re: Even and odd composite functions. Ok thanks for the help.", "## Calculus: Early Transcendentals 8th Edition 1) If both $f$ and $g$ are even $fg$ is even; 2) If both $f$ and $g$ are odd $fg$ is even; 3) If $f$ is even and $g$ is odd $fg$ is odd. The new function of some argument $x$ is the product: $$h(x)=f(x)g(x).$$ To say when it is odd and when it is even we have to check what happens when we substitute $x$ for $-x$. Check all the combinations: 1) $f$ is even and $g$ is even. Then $f(-x)=f(x)$ and $g(-x)=g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)g(x)=h(x)$$ so we have $h(-x)=h(x)$ and we see that $h$ is even. 2) $f$ is odd and $g$ is odd. Then $f(-x)=-f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=(-f(x))\\cdot(-g(x))=f(x)g(x)=h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is even. 3) $f$ is even and $g$ is odd. Then $f(-x)=f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)\\cdot(-g(x))=-h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is odd.", "## Calculus 8th Edition if $f$ and $g$ are both even functions, $fg$ is even. if $f$ and $g$ are both odd functions, $fg$ is even. if $f$ is even and $g$ is odd, $fg$ is odd. if $f$ and $g$ are both even functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ and $g$ are both odd functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=(-f(x))(-g(x))=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ is even and $g$ is odd, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x)=-(fg)(x)$ Therefore $fg$ is odd.", "## Thomas' Calculus 13th Edition Given $f$ and $g$ are odd functions, we have $f(-x)=-f(x)$ and $g(-x)=-g(x)$ a. $f(-x)g(-x)=f(x)g(x)$ would be even. b. $f^3(-x)=-f^3(x)$ would be odd. c. $f(sin(-x))=f(-sin(x))=-f(sin(x))$ would be odd. d. $g(sec(-x))=g(sec(x))$ would be even. e. $|g(-x)|=|-g(x)|=|g(x)|$ would be even.", "as $f^{-1}(x)$ e.g. are $f(x)=x^2$ and $g(x)=x^3$ even, or odd functions? What about $h(x)=x^2+x$ if $f(x)=x^2$ then $f(-x)=(-x)^2=(-1)^2*x^2=x^2=f(x)$, so f(x) is an even function. if $g(x)=x^3$ then $g(-x)=(-x)^3=(-1)^3*x^3=-1*x^3=-1*g(x)$, so g(x) is an odd function. if $h(x)=x^2+x$, then $h(-x)=(-x)^2+(-x)=x^2-x$. h(-x) is neither equal to h(x) nor to -h(x), so h(x) is neither even nor odd.", "anonymous one year ago let f and g are functions that are neither even nor odd. a)Create an example where f+g is even, b) f+g is odd, c)f.g is even, d) f.g is odd • This Question is Open 1. Loser66 I give you one case , b, as example, you do the rest, ok? Let $$f(x) = x+3$$ let check $$f(-x) = -x +3 \\neq f(x) ~~hence,~~\\text{f(x) is not even}$$ $$-f(x) = -x-3 \\neq f(-x)~~hence,~~\\text{f(x) is not odd, therefore f(x) is neither even nor odd}$$ Now, let $$g(x) = -2x-3$$, let check $$g(-x) = 2x-3\\neq g(x),~~hence,~~g(x) \\text{is not even}$$ $$-g(x) = 2x+3 \\neq g(-x),~~hence~~\\text{g(x) is not odd, hence g(x) is neither even nor odd}$$ Now, combine $$(f+g)(x) = f(x) +g(x) = x+3-2x-3 = -x$$ let check $$(f+g)(-x) = x \\\\(-(f+g)(x) = -(-x) =x$$ hence $$(f+g)(-x) = x=-(f+g)(x)$$, therefore $$(f+g)(x)$$is an odd function.", "follows, which must be odd? I. p(x)=f(g(x)) II. r(x)=f(x)+g(x) III. s(x)=f(x)*g(x) (A) I only; (B) II only; (C) I and II only; (D) II and III only; (E) I, II, and III The defining property for a function $h(x)$ to be odd is that $h(-x)=-h(x)$. Now for each of the given functions you need to determine if this property holds. I will do I. for you: $ p(x)=f(g(x)) $ so: $ p(-x)=f(g(-x)) $ but $g$ is odd so $g(-x)=-g(x)$ hence: $p(-x)=f(-g(x))$ but $f$ is also odd so: $p(-x)=-f(g(x))=-p(x)$ and so $p$ is odd. CB", "## College Algebra (10th Edition) We know that if a function is odd, then $f(-x)=-f(x).$ So here $f(x)=-f(x)$ and $g(x)=-g(x)$. Then $f(g(x))=f(-g(x))=-f(g(x))$, thus $f(g(x))$ is also odd, thus we proved what we had to.", "{~} I had something similar, but why does this look like it's always true when it isn't? Or is my special case wrong? 4. Jan 9, 2015 ### DarthMatter What special case? $-g(x)+c$ is not odd in general. 5. Jan 9, 2015 ### {~} Thanks for pointing out my error. That made things less confusing. 6. Jan 9, 2015 ### jbunniii Suppose that $f$ is even and $g$ is odd. If $f+g$ is even then $f(x) + g(x) = f(-x) + g(-x)$ for all $x$. Since $f(x) = f(-x)$, it follows that $g(x) = g(-x)$. But $g$ is odd, so also $g(x) = -g(-x)$. Therefore, $g(-x) = -g(-x)$, or equivalently $2g(-x) = 0$. This is true for all $x$, so $g$ is the zero function. If $f+g$ is odd then $f(x) + g(x) = -f(-x) - g(-x)$ for all $x$. Since $g(x) = -g(-x)$, this is equivalent to $f(x) = -f(-x)$. But $f$ is even, so also $f(x) = f(-x)$. Therefore, $f(-x) = -f(-x)$, or $2f(-x) = 0$. This is true for all $x$, so $f$ is the zero function. Last edited: Jan 9, 2015 7. Jan 11, 2015 ### FactChecker In Case 3 you can not say anything about the sum of an even + odd function. Any function, h(x), can be represented as the sum of", "f(x) \\, dx$ when $f(x)$ is an even function. 5. Actually, the way it's done, it would rather be : Since $g(x)=\\frac{x^3}{3}+7x$ is an odd function, then $\\left[\\frac{x^3}{3}+7x\\right]_{-3}^3=g(3)-g(-3)=2g(3)$", "- f(x)$ is a function such that $g(1) + \\cdots + g(1999)$ is even, and similarily if $f$ has even sum then $g$ has odd sum. In this way, we can pair up every single one of the $4^{1999}$ functions into pairs containing one function for which the sum is even, and one function for which the sum is odd (where one element of the pair is $4003$ minus the other). Thus exactly half of all the possible functions have odd sum. This gives $2\\cdot 4^{1998}$.", "name $h:=f\\circ g$ and write the condition for $h$ to be even. Oct 1 at 6:18 If $$g$$ is odd and $$f$$ is even, then $$(f\\circ g)(x)$$ is even. $$f(g(-x)) = f(-g(x)) = f(g(x))$$ If $$g$$ is even, regardless of the function $$f$$, $$(f\\circ g)(x)$$ is even. $$f(g(-x)) = f(g(x))$$ If $$f$$ and $$g$$ are both odd, $$(f\\circ g)(x)$$ is odd. $$f(g(-x)) = f(-g(x)) = -f(g(x))$$ When in doubt, consider $$f(x) = x$$ as a simple example of an odd function and $$f(x) = x^2$$ as an example of an even function. While you can't prove a proposition is true by example, you can prove a proposition is false, and examples help build intuition.", "and $v(x)=\\frac1x$. Since $h,v$ are both odd, we get $$g(-x)=h(-x)v(-x)=(-h(x))(-v(x))=h(x)v(x)=g(x).$$ Which means that $g$ is an even function. This is true for any odd functions $h,v$. -", "'18 at 14:56 • @Adam, I've used it in the definition of $f(x)$ to state that the sum of odd and even function is neither odd nor even – roman Jul 10 '18 at 15:08 Well, by definition $f(x)+g(x)$ is even if and only if $f(x)+g(x)=f(-x)+g(-x)$ for all $x$. If $g(x)$ is even, it satisfies $g(x)=g(-x)$ for all $x$. Inserting this into the previous equation gives $f(x)=f(-x)$ for all $x$, hence $f$ must be even. Why did it work for odd? Here the definition requires $f(x)+g(x)=-f(-x)-g(-x)$. Since $g(x)$ is even, this transforms to $f(x)+g(x)=-f(-x)-g(x)$ i.e. $g(x) = -\\frac 12 (f(x)+f(-x))$, which your example satisfies. In other words: if $f$ is not even, subtracting its even part [*] makes it odd. [*] This refers to the fact that you can decompose every function $f(x)$ into its even part $f_e(x)$ and its odd part $f_o(x)$ such that $f(x) = f_e(x)+f_o(x)$, where $f_e(x) =\\frac 12 (f(x)+f(-x))$ and $f_o(x) =\\frac 12 (f(x)-f(-x))$. If $f+g$ is even, then $(f+g)-g$ is the difference of even functions, hence even f(x) = f(-x) $g(x) \\ne g(-x)$ for some x z(x) = f(x) + g(x) then $z(-x) = f(-x) + g(-x) \\ne f(x) + g(x)$ for some x • You just wrote, at the bottom, that $f(-x) + g(-x) \\neq f(-x) + g(-x)$. That's a contradiction, because", "= -(f_1+f_2)(x),$ so $$f_1+f_2$$ is odd. If $$g_1$$ and $$g_2$$ are even functions, then $(g_1+g_2)(-x)=g_1(-x)+g_2(-x)=g_1(x)+g_2(x)=(g_1+g_2)(x),$ so $$g_1+g_2$$ is even. 2. Suppose $$f_1$$ and $$f_2$$ are even, and $$g_1$$ and $$g_2$$ are odd. If $$h = f_1f_2$$, then $h(-x) = f_1(-x)\\,f_2(-x) = f_1(x)\\,f_2(x) = h(x),$ so $$h$$ is even. If $$j = g_1g_2$$, then $j(-x) = g_1(-x)\\,g_2(-x) = -g_1(x)\\times -g_2(x) = g_1(x)\\,g_2(x) = j(x),$ so $$j$$ is even. If $$k = f_1g_1$$, then $k(-x) = f_1(-x)\\,g_1(-x) = f_1(x)\\times -g_1(x) = -f_1(x)\\,g_1(x) = -k(x),$ so $$k$$ is odd. 3. Let $$f \\colon D \\to \\mathbb{R}$$ with $$f(a) \\neq 0$$, for some $$a \\in D$$, and let $$g \\colon D \\to \\mathbb{R}$$ with $$g(b) \\neq 0$$, for some $$b \\in D$$. Assume that $$f$$ is odd and $$g$$ is even. Then \\begin{alignat*}{2} (f+g)(-a) &= f(-a) + g(-a) \\\\ &= -f(a) + g(a) &\\qquad&\\text{since } f \\text{ is odd and } g \\text{ is even} \\\\ &\\neq f(a) + g(a) &&\\text{since } f(a) \\neq 0 \\\\ &= (f+g)(a). \\end{alignat*} So $$f+g$$ is not even. Similarly, we can use $$b$$ to show that $$f+g$$ is not odd. ##### Exercise 3 Let $$y = x^2+4x+10 = (x+2)^2+6$$. To map $$y=x^2$$ to $$y=(x+2)^2+6$$, translate it 2 units to the left and 6 units up. ##### Exercise 4 1. First assume that $$f$$ is even. Then" ]

[ "## Calculus: Early Transcendentals 8th Edition If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd. If $f(x)$ and $g(x)$ are both even, $f(x)=f(-x)$ (i) $g(x)=g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)+g(-x)$ (from i) and ii)) $=S(-x)$ $\\therefore f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $-f(x)=f(-x), f(x)=-f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=-f(-x)-g(-x)$ (from i) and ii)) $=-(f(-x)+g(-x))$ $=-S(x)$ $\\because S(-x)=-S(x),$ $\\therefore f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)=f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)-g(-x)$ (from i) and ii)) $S(-x)=f(-x)+g(-x)$ $\\ne \\pm S(x)$ $\\because S(-x)\\ne \\pm S(x)$ $\\therefore f(x)+g(x)$ is neither even nor odd.", "anonymous one year ago let f and g are functions that are neither even nor odd. a)Create an example where f+g is even, b) f+g is odd, c)f.g is even, d) f.g is odd • This Question is Open 1. Loser66 I give you one case , b, as example, you do the rest, ok? Let $$f(x) = x+3$$ let check $$f(-x) = -x +3 \\neq f(x) ~~hence,~~\\text{f(x) is not even}$$ $$-f(x) = -x-3 \\neq f(-x)~~hence,~~\\text{f(x) is not odd, therefore f(x) is neither even nor odd}$$ Now, let $$g(x) = -2x-3$$, let check $$g(-x) = 2x-3\\neq g(x),~~hence,~~g(x) \\text{is not even}$$ $$-g(x) = 2x+3 \\neq g(-x),~~hence~~\\text{g(x) is not odd, hence g(x) is neither even nor odd}$$ Now, combine $$(f+g)(x) = f(x) +g(x) = x+3-2x-3 = -x$$ let check $$(f+g)(-x) = x \\\\(-(f+g)(x) = -(-x) =x$$ hence $$(f+g)(-x) = x=-(f+g)(x)$$, therefore $$(f+g)(x)$$is an odd function.", "## Calculus 8th Edition If $f$ and $g$ are both even functions, $f+g$ is even. If $f$ and $g$ are both odd functions, $f+g$ is odd. If $f$ is even and $g$ is odd, $f+g$ can be even, odd, or neither of them. Assume $f$ and $g$ are both even functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x).$$Thus, $f+g$ is even. Assume $f$ and $g$ are both odd functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=-f(x)+(-g(x))=-(f+g)(x).$$Thus, $f+g$ is odd. Assume $f$ is even and $g$ is odd, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+(-g(x))=(f-g)(x).$$In this case, $f+g$ is even,if $g=0$, and $f+g$ is odd, if $f=0$, but if $g \\neq 0$ then $f+g$ is neither even nor odd (For example, take $f=x^2$ and $g=x$; then $f+g=x^2+x$ is neither even or odd since $(f+g)(-x)=x^2-x$ and $(f+g)(-x) \\neq (f+g)(x)$ and $(f+g)(-x) \\neq -(f+g)(x)$.", "what is the difference between being neither even nor odd and being none of these $f(x)$ is an odd function and $g(x)$ is neither even nor odd,Then $f(x) + g(x)$ (a). Even (b) Odd (c)Nither even nor odd (d)None of these. $\\boldsymbol{My}$$\\boldsymbol{Approach}$$\\Longrightarrow$I know (a) and(b) incorrect. But i dont know which one is correct between (c) and (d).What is the difference between (c) and (d).?? And What is the key to recognize ??? Please explain in detail. Let $f(x) = 1$. Then $f\\;$is even. First, let $g(x) = x-1$. Then $g\\;$is neither even nor odd, but $f+g\\;$is odd. Next, let $g(x) = x+1$. Then $g\\;$is neither even nor odd, and $f+g\\;$is also neither even nor odd. The first example disqualifies $(c)$, and the second example disqualifies $(a)\\;$and $(b)$.", "$f(-x)=-f(x)$, if $g$ is odd then $g(-x)=-g(x)$. Show: $f(g(-x))=-f(g(x))$ $f(g(-x))=f(-g(x))=-f(g(x))$ Therefore $f\\circ g$ is odd. 6. ## Re: Even and odd composite functions. Ok thanks for the help.", "## Calculus 8th Edition if $f$ and $g$ are both even functions, $fg$ is even. if $f$ and $g$ are both odd functions, $fg$ is even. if $f$ is even and $g$ is odd, $fg$ is odd. if $f$ and $g$ are both even functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ and $g$ are both odd functions, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=(-f(x))(-g(x))=f(x)g(x)=(fg)(x)$ Therefore $fg$ is even. if $f$ is even and $g$ is odd, $(fg)(x) = f(x)g(x)$ $(fg)(-x) = f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x)=-(fg)(x)$ Therefore $fg$ is odd.", "as $f^{-1}(x)$ e.g. are $f(x)=x^2$ and $g(x)=x^3$ even, or odd functions? What about $h(x)=x^2+x$ if $f(x)=x^2$ then $f(-x)=(-x)^2=(-1)^2*x^2=x^2=f(x)$, so f(x) is an even function. if $g(x)=x^3$ then $g(-x)=(-x)^3=(-1)^3*x^3=-1*x^3=-1*g(x)$, so g(x) is an odd function. if $h(x)=x^2+x$, then $h(-x)=(-x)^2+(-x)=x^2-x$. h(-x) is neither equal to h(x) nor to -h(x), so h(x) is neither even nor odd.", "$g(x)+h(x)=x+1$ is neither odd nor even. • For an odd function $g(x) = x$ and the function $h(x) = x+1$, neigher of the functions $g(h(x)) = x+1$ nor $h(g(x)) = x+1$ is either even or odd. • For an odd function $g(x) = x$ and the function $h(x) = x+1$, the function $g(x)h(x) = x^2+1$ is neither even nor odd. -", "## Calculus 8th Edition When $f$ is odd, we can find that: $f(g(-x)) = f(-g(x))= -f(g(x)) = -h(x)$ Therefore $h$ is odd. However, when $f$ is even, we can find that: $f(g(-x)) = f(-g(x))= f(g(x)) = h(x)$ Therefore $h$ is even. Thus, $h$ is not always an odd function.", "- f(x)$ is a function such that $g(1) + \\cdots + g(1999)$ is even, and similarily if $f$ has even sum then $g$ has odd sum. In this way, we can pair up every single one of the $4^{1999}$ functions into pairs containing one function for which the sum is even, and one function for which the sum is odd (where one element of the pair is $4003$ minus the other). Thus exactly half of all the possible functions have odd sum. This gives $2\\cdot 4^{1998}$.", "## College Algebra (10th Edition) We know that if a function is odd, then $f(-x)=-f(x).$ So here $f(x)=-f(x)$ and $g(x)=-g(x)$. Then $f(g(x))=f(-g(x))=-f(g(x))$, thus $f(g(x))$ is also odd, thus we proved what we had to.", "## Algebra and Trigonometry 10th Edition Suppose $f$ to be even and $g$ to be odd. Thus, $(f+g)(-x)=f(-x) +g(-x)$ or, $=f(x) +g(-x)$ or, $=f(x)+[-g(x)]$ Therefore, the sum of an even function and an odd function is neither even nor odd.", "## College Algebra (10th Edition) To find out if the function is even, odd, or neither, we find $G(−x)$ and compare the result to $G(x)$: $G(x)=1-x+x^{3}$ $G(-x)=1-(-x)+(-x)^{3}$ $=1+x-x^{3}$ This does not equal $-G(x)$ or $G(x)$. Thus, $G(x)$ is neither even or odd.", "## Calculus: Early Transcendentals 8th Edition 1) If both $f$ and $g$ are even $fg$ is even; 2) If both $f$ and $g$ are odd $fg$ is even; 3) If $f$ is even and $g$ is odd $fg$ is odd. The new function of some argument $x$ is the product: $$h(x)=f(x)g(x).$$ To say when it is odd and when it is even we have to check what happens when we substitute $x$ for $-x$. Check all the combinations: 1) $f$ is even and $g$ is even. Then $f(-x)=f(x)$ and $g(-x)=g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)g(x)=h(x)$$ so we have $h(-x)=h(x)$ and we see that $h$ is even. 2) $f$ is odd and $g$ is odd. Then $f(-x)=-f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=(-f(x))\\cdot(-g(x))=f(x)g(x)=h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is even. 3) $f$ is even and $g$ is odd. Then $f(-x)=f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)\\cdot(-g(x))=-h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is odd.", "that $f$ is odd. However, finding a single value of $x$ for which $f(-x) \\ne -f(x)$ is sufficient for showing that $f$ is not odd. • Only proving that $f$ is not even. There are functions that are even, odd, both and neither. For instance: • $f(x)=x^2$ is even but not odd • $g(x)=x$ is odd but not even • $h(x)=0$ is both odd and even • $k(x)=x+1$ is neither odd nor even Unlike integers, \"not even\" does not mean the same thing as \"odd\". You'll find situations like this all over the place: always work from the definitions, don't let your knowledge of the English language deceive you. • Doing strange things with $|x|$. Remember the function $|x|$ is defined by $$|x| = \\begin{cases} x & \\text{if}\\ x \\ge 0 \\\\ -x & \\text{if}\\ x \\lt 0 \\end{cases}$$ If you give it a positive number, it just gives you your number back; if you give it a negative number, it flips its sign. For example, $|4|=4$ and $|{-3}|=3$. Some people wrote things like \"$-x|x|=-x(x)=-x^2=\\cdots$\"; this is not correct! • Using the result \"odd × even = odd\". This result is true, but if you wanted to use it you should have proved it, and proved that $x$ is odd and $|x|$ is even! In this case it", "# Let f(x) = |x-1|. 1) Verify that f(x) is neither even nor odd. 2) Can f(x) be written as the sum of an even function and an odd function ? a) If so, exhibit a solution. Are there more solutions ? b) If not, prove that it is impossible. Feb 6, 2018 Let $f \\left(x\\right) = | x - 1 |$. If f were even, then $f \\left(- x\\right)$ would equal $f \\left(x\\right)$ for all x. If f were odd, then $f \\left(- x\\right)$ would equal $- f \\left(x\\right)$ for all x. Observe that for x = 1 $f \\left(1\\right) = | 0 | = 0$ $f \\left(- 1\\right) = | - 2 | = 2$ Since 0 is not equal to 2 or to -2, f is neither even nor odd. Might f be written as $g \\left(x\\right) + h \\left(x\\right)$, where g is even and h is odd? If that were true then $g \\left(x\\right) + h \\left(x\\right) = | x - 1 |$. Call this statement 1. Replace x by -x. $g \\left(- x\\right) + h \\left(- x\\right) = | - x - 1 |$ Since g is even and h is odd, we have: $g \\left(x\\right) - h \\left(x\\right) = | - x - 1 |$ Call this statement 2. Putting statements 1", "# Composition of Even and Odd Functions and their Outcome Give an example of an even function. Give an example of an odd function. If f(x) is odd and g(x) is even, must f(g(x)) be even? Must g(f(x)) be even? I've tried generic functions like f(x) = x^3 and g(x) = x^2 Both compositions (going f(g(x)) and g(f(x)) yield even results) However, when I use the trig functions, something different happens. f(x) = sin(x) g(x) = cos(x) f(g(x)) = even and g(f(x)) = even. Therefore, can I assume, that the combination of any function (that is not neither odd nor even) will be even? Try doing this generically. Suppose first that $f$ is even and $g$ is odd. Then $$f(g(-x)) = f(-g(x)) = f(g(x))$$ where the first equality is because $g$ is odd and the second is because $f$ is even. Therefore $f\\circ g$ is even. Do this for all four possible combinations and you'll see how things work. Any time you compose real functions, if any of them are even and the rest are odd then the composition is even. This is because odd functions \"retain\" negation and even functions \"get rid of\" it. For example if $f$ is even and $g$ is odd, $g(f(-x) = g(f(x))$ (even) $f(g(-x) = f(-g(x)) = f(g(x))$ (even) You can prove", "use it to check your answer visually. $$f(x)=\\frac{x}{x^{2}+1}$$ Check back soon! Problem 74$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=\\frac{x^{2}}{x^{4}+1}$$ Check back soon! Problem 75$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=\\frac{x}{x+1}$$ Check back soon! Problem 76$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=x|x|$$ Check back soon! Problem 77$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=1+3 x^{2}-x^{4}$$ Check back soon! Problem 78$73-78$Determine whether$f$is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. $$f(x)=1+3 x^{3}-x^{5}$$ Check back soon! Problem 79 If$f$and$g$are both even functions, is$f+g$even? If$f$and$g$are both odd functions, is$f+g$odd? What if$f$is even and$g$is odd? Justify your answers. Check back soon! Problem 80 If$f$and$g$are both even functions, is the product$f g$even? If$f$and$g$are both odd functions, is$f g$odd? What if$f$is even and$g\\$ is odd? Justify your answers. Check back soon!", "# Is the product of an odd function and an even function odd or even? Nov 14, 2015 odd #### Explanation: Suppose $f \\left(x\\right)$ is odd and $g \\left(x\\right)$ is even. Then $f \\left(- x\\right) = - f \\left(x\\right)$ and $g \\left(- x\\right) = g \\left(x\\right)$ for all $x$ Let $h \\left(x\\right) = f \\left(x\\right) g \\left(x\\right)$ Then: $h \\left(- x\\right) = f \\left(- x\\right) g \\left(- x\\right) = \\left(- f \\left(x\\right)\\right) g \\left(x\\right) = - \\left(f \\left(x\\right) g \\left(x\\right)\\right) = - h \\left(x\\right)$ for all $x$ That is $h \\left(x\\right)$ is odd.", "= -(f_1+f_2)(x),$ so $$f_1+f_2$$ is odd. If $$g_1$$ and $$g_2$$ are even functions, then $(g_1+g_2)(-x)=g_1(-x)+g_2(-x)=g_1(x)+g_2(x)=(g_1+g_2)(x),$ so $$g_1+g_2$$ is even. 2. Suppose $$f_1$$ and $$f_2$$ are even, and $$g_1$$ and $$g_2$$ are odd. If $$h = f_1f_2$$, then $h(-x) = f_1(-x)\\,f_2(-x) = f_1(x)\\,f_2(x) = h(x),$ so $$h$$ is even. If $$j = g_1g_2$$, then $j(-x) = g_1(-x)\\,g_2(-x) = -g_1(x)\\times -g_2(x) = g_1(x)\\,g_2(x) = j(x),$ so $$j$$ is even. If $$k = f_1g_1$$, then $k(-x) = f_1(-x)\\,g_1(-x) = f_1(x)\\times -g_1(x) = -f_1(x)\\,g_1(x) = -k(x),$ so $$k$$ is odd. 3. Let $$f \\colon D \\to \\mathbb{R}$$ with $$f(a) \\neq 0$$, for some $$a \\in D$$, and let $$g \\colon D \\to \\mathbb{R}$$ with $$g(b) \\neq 0$$, for some $$b \\in D$$. Assume that $$f$$ is odd and $$g$$ is even. Then \\begin{alignat*}{2} (f+g)(-a) &= f(-a) + g(-a) \\\\ &= -f(a) + g(a) &\\qquad&\\text{since } f \\text{ is odd and } g \\text{ is even} \\\\ &\\neq f(a) + g(a) &&\\text{since } f(a) \\neq 0 \\\\ &= (f+g)(a). \\end{alignat*} So $$f+g$$ is not even. Similarly, we can use $$b$$ to show that $$f+g$$ is not odd. ##### Exercise 3 Let $$y = x^2+4x+10 = (x+2)^2+6$$. To map $$y=x^2$$ to $$y=(x+2)^2+6$$, translate it 2 units to the left and 6 units up. ##### Exercise 4 1. First assume that $$f$$ is even. Then" ]

[ "# Don't cancel AB $\\large \\dfrac{1}{a} + \\dfrac{1}{b} + 1 = \\overline{ab} \\div (a\\times b)$ For single digit positive integers $$a$$ and $$b$$, let $$\\overline{ab}$$ denote the two-digit integer: $$10a + b$$. Suppose $$S_a$$ and $$S_b$$ be the sum of all possible values of $$a$$ and $$b$$ respectively such that they satisfy the equation above. Find the value of $$S_a+S_b$$. ×", "For non-negative integers, $a, b, c$, what would be the value of $a+b+c$ if $\\log a + \\log b + \\log c = 0$? 1. $3$ 2. $1$ 3. $0$ 4. $-1$", "1 } { n^6 }$ 4. Thus, we can say that the arithmetic mean of these numbers is $\\dfrac{ 1 } { n^6 }$ less than $1.$", "Consider the equation $\\dfrac{1}{a} + \\dfrac{1}{b} + \\dfrac{1}{c} = \\dfrac{1}{d}$ where $a, b, c, d$ are distinct positive integers, at least $3$ of which are prime. Up to a permutation of the variables, there is a unique solution to this problem. Find $a + b + c + d$. ×", "also know that $\\log\\left[(-1)^2\\right] = \\log(1) = 0.$ Combining these two results gives $\\log\\left[(-1)^2\\right] = 2\\log(-1) = 0$ and hence $\\log(-1) = 0$ as required. QED. Theorem 4. All positive integers are equal. Proof. It is sufficient to show that for any two positive integers, $A$ and $B$, $A = B$. Further, it is sufficient to show that for all $N > 0$, if $A$ and $B$ (both positive integers) satisfy $\\max(A, B) = N$ then $A = B$. We proceed by induction. (i) If $N = 1$, then $A$ and $B$, being positive integers, must each be 1. So $A = B$. (ii) Assume that the result holds for some value $N = k$. Now take $A$ and $B$ with $\\max(A, B) = k+1$. Then $\\max((A-1), (B-1)) = k$, and hence $(A-1) = (B-1)$. Consequently, $A = B$. (iii) We have shown that the result holds for $N = 1$ and that if it holds for $N = k$ then it must also hold for $N = k+1$. Hence, by induction, the theorem is true for all $n \\in \\mathbb{N}$. QED.", "Question # List all the elements of the following sets:B = ($$x:x$$ is an integer; $$-\\frac{1}{2}<x<\\frac{9}{2}$$) Solution ## B = ($$x;x$$ is an integer; $$-\\dfrac{1}{2}<n<\\dfrac{9}{2}$$)It can be seen that $$-\\dfrac{1}{2}=-0.5$$ and $$\\dfrac{9}{2}=4.5$$$$\\therefore B=(0,1,2,3,4)$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More", "example: 1. if $n = 4$, then for every integer $x$ such that $1 \\le x \\le n$, $\\dfrac{x}{f(f(x))} = 1$; 2. if $n = 37$, then for some integers $x$ such that $1 \\le x \\le n$, $\\dfrac{x}{f(f(x))} = 1$ (for example, if $x = 23$, $f(f(x)) = 23$,$\\dfrac{x}{f(f(x))} = 1$); and for other values of $x$, $\\dfrac{x}{f(f(x))} = 10$ (for example, if $x = 30$, $f(f(x)) = 3$, $\\dfrac{x}{f(f(x))} = 10$). So, there are two different values of $g(x)$.", "$v_\\mathfrak{p}(\\mathfrak{a} \\cap \\mathfrak{b})$ is greater or equal than $v_\\mathfrak{p}(\\mathfrak{a})$ and $v_\\mathfrak{p}(\\mathfrak{b})$ or, in other words, $v_\\mathfrak{p}(\\mathfrak{a} \\cap \\mathfrak{b}) \\geq \\max\\{v_\\mathfrak{p}(\\mathfrak{a}),v_\\mathfrak{p}(\\mathfrak{b})\\}$.", "+0 # Help! 0 219 2 +1110 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Jul 6, 2018 #1 0 Jul 6, 2018 #2 +22188 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "b + c = $$2 + 3 + 5^2 = 30$$ Hence the correct answer is Option C. _________________ a, b and c are three distinct integers, greater than 1, such that .... [#permalink] 01 Jan 2019, 22:43 Display posts from previous: Sort by", "# Tag Info Let $f(n) = n^{\\log n}$. Then $\\mathsf{P} \\subseteq \\mathsf{DTIME}(f(n))$ while $\\mathsf{DTIME}(f(2n+1)^3) \\subseteq \\mathsf{EXP}$. (Actually, whether this argument works depends on the exact definition of $\\mathsf{DTIME}$. If it doesn't, take $f_C(n) = Cn^{\\log n}$ and sum over all integer $C>0$.)", "• # question_answer If $a,\\ b,\\ c$ are the positive integers, then $(a+b)(b+c)(c+a)$is [DCE 2000] A) $<8abc$ B) $>8abc$ C) $=8abc$ D) None of these Since A.M. > G.M., we have $\\frac{a+b}{2}>\\sqrt{ab},\\ \\frac{b+c}{2}>\\sqrt{bc}$ and $\\frac{a+c}{2}>\\sqrt{ac}$. Multiplying these inequalities, we get $(a+b)(b+c)(c+a)>8abc$.", "# A repeating decimal Given that $$0.\\overline{969}= \\dfrac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is $$a+b$$? ×", "# $a,b$ are integers , both greater than $1$ , such that $(a^n-1)(b^n-1)$ is a perfect square for every positive integer $n$ , then $a=b$ ? If $a,b$ are integers , both greater than $1$ , such that $(a^n-1)(b^n-1)$ is a perfect square for every positive integer $n$ , then is it true that $a=b$ ? Hint: is it sufficient to consider $p$-adic heights, given by $$\\nu_p(n)=\\max\\{m\\in\\mathbb{N}:p^m\\mid n\\}.$$ If you manage to prove that, assuming $a\\neq b$, for some prime $p$ and some integer $n>0$ $$\\nu_p\\left((a^n-1)(b^n-1)\\right) = \\nu_p(a^n-1)+\\nu_p(b^n-1)$$ is odd, then you're done, since that implies that $(a^n-1)(b^n-1)$ cannot be a square.", "+0 Help! 0 106 2 +803 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Lightning Jul 6, 2018 #1 0 Guest Jul 6, 2018 #2 +20108 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "+0 # Need Help Urgently 0 39 1 Let $n$ be a positive integer greater than or equal to $3$. Let $a,b$ be integers such that $ab$ is invertible modulo $n$ and $(ab)^{-1} \\equiv 2$ (mod $n$). Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by $n$? Jun 16, 2021 #1 +1 Multiply $${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$ by ab $${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$(ab) =$${(a+b)}^{-1}({a}^{-1}(ab)+{b}^{-1}(ab))$$ $$={(a+b)}^{-1}(b+a)$$ = 1 (mod n) This shows that $${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$ = $${(ab)}^{-1}$$= 2 (mod n) Jun 17, 2021", "number of integers between $$1$$ and $$60$$ that are divisible by $$2$$ is $$\\dfrac{60}{2} = 30$$. Call the set of these integers $$A$$. The number of integers between $$1$$ and $$60$$ that are divisible by $$3$$ is $$\\dfrac{60}{3} = 20$$. Call the set of these integers $$B$$. The number of integers between $$1$$ and $$60$$ that is divisible by $$5$$ is $$\\dfrac{60}{5} = 12$$. Call the set of these integers $$C$$. Then $$|A ∩ B|$$ is the number of integers between $$1$$ and $$60$$ that are divisible by $$2$$ and $$3$$; that is, the number that are divisible by $$6$$. This is $$\\dfrac{60}{6} = 10$$. Similarly, $$|A ∩ C|$$ is the number of integers between $$1$$ and $$60$$ that are divisible by $$2$$ and $$5$$; that is, the number that are divisible by $$10$$. This is $$\\dfrac{60}{10} = 6$$. Also, $$|B ∩ C|$$ is the number of integers between $$1$$ and $$60$$ that are divisible by $$3$$ and $$5$$; that is, the number that are divisible by $$15$$. This is $$\\dfrac{60}{15} = 4$$. Finally, $$|A ∩ B ∩ C|$$ is the number of integers between $$1$$ and $$60$$ that are divisible by $$2$$, $$3$$, and $$5$$; that is, the number that are divisible by $$30$$. This is $$\\dfrac{60}{30} = 2$$. We have been asked for $$|A ∪ B", "# Algebra 1. (28 p.) Let $$a$$ and $$b$$ be positive real numbers such that $$ab=2$$ and $\\dfrac{a}{a+b^2}+\\dfrac{b}{b+a^2}=\\dfrac78.$ Find $$a^6+b^6$$. 2. (23 p.) Consider the set $$S\\subseteq(0,1]^2$$ in the coordinate plane that consists of all points $$(x,y)$$ such that both $$[\\log_2(1/x)]$$ and $$[\\log_5(1/y)]$$ are even. The area of $$S$$ can be written in the form $$p/q$$ for two relatively prime integers $$p$$ and $$q$$. Evaluate $$p+q$$. 3. (9 p.) Real numbers $$x,y,z$$ are real numbers greater than 1 and $$w$$ is a positive real number. If $$\\log_xw=24$$, $$\\log_yw=40$$ and $$\\log_{xyz}w=12$$, find $$\\log_zw$$. 4. (28 p.) Let $$f:\\mathbb N\\rightarrow\\mathbb R$$ be the function defined by $$f(1) = 1$$, $$f(n) = n/10$$ if $$n$$ is a multiple of 10 and $$f(n) = n+1$$ otherwise. For each positive integer $$m$$ define the sequence $$x_1$$, $$x_2$$, $$x_3$$, ... by $$x_1 = m$$, $$x_{n+1} = f(x_n)$$. Let $$g(m)$$ be the smallest $$n$$ such that $$x_n = 1$$. (Examples: $$g(100) = 3$$, $$g(87) = 7$$.) Denote by $$N$$ be the number of positive integers $$m$$ such that $$g(m) = 20$$. The number of distinct prime factors of $$N$$ is equal to $$2^u\\cdot v$$ for two non-negative integers $$u$$ and $$v$$ such that $$v$$ is odd. Determine $$u+v$$. 5. (9 p.) Let $$a$$, $$b$$, and $$c$$ be non-real roots of the polynimal $$x^3+x-1$$. Find $\\frac{1+a}{1-a}+", "n^2$ for $n > 1$ (when $n < n^2$, so $3n < 3 n^2$). For $a,b,c,> 0$ you have $a/b > a/c$ if $b < c$. In other words, if you make the denominator bigger you make the fraction smaller. Of course, in this case you have $3n + n^2 < 4 n^2$ for $n > 1$ (when $n < n^2$, so $3n < 3 n^2$). I see, thank you very much. Homework Helper A little simpler: $$\\dfrac n{n^2+3n} = \\dfrac{n}{n(n+3)} = \\dfrac{1}{n+3} \\ge \\dfrac 1 {2n}$$ for $n \\ge 3$", "For integers, $a$, $b$ and $c$, what would be the minimum and maximum values respectively of $a+b+c$ if $\\log \\mid a \\mid + \\log \\mid b \\mid + \\log \\mid c \\mid =0$? 1. $\\text{-3 and 3}$ 2. $\\text{-1 and 1}$ 3. $\\text{-1 and 3}$ 4. $\\text{1 and 3}$ edited" ]

[ "+ a_3 + a_4 + a_5) - (a_{10} + a_{11} + a_{12} + a_{13} + a_{14}) \\\\ &= (\\log_{2002} 2 + \\log_{2002} 3 + \\log_{2002} 4 + \\log_{2002} 5) \\\\ &\\quad - (\\log_{2002} 10 + \\log_{2002} 11 + \\log_{2002} 12 + \\log_{2002} 13 + \\log_{2002} 14) \\\\ &= \\log_{2002} \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14} \\\\ &= \\log_{2002} \\frac{1}{2002} \\\\ &= \\boxed{-1}. \\end{align*} The answer is (B).", "the same reasoning, $a_3 = \\log_{2002} 3$, $a_4 = \\log_{2002} 4$ and so on. $$b = \\log_{2002} 2 + \\log_{2002} 3 + .....+ \\log_{2002} 5 = \\log_{2002} 5! = \\log_{2002} 120$$ Now solving for $c$, we see that it equals $\\log_{2002} (10\\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14)$ $$b-c = \\log_{2002} 120 - \\log_{2002} 240240 \\rightarrow \\log_{2002} \\frac{1}{2002} = \\boxed{-1}$$ ~YBSuburbanTea", "further simplify $$(n/2)\\log(n/2)$$ --> $$(n/2)\\log(n^{-2})$$, ... This is wrong. In fact, for $$n\\ge 3$$, $$n^{\\log_3 2}\\ge 2$$, so \\begin{align} (n/2)\\log(n/2)&=(n/2)(\\log n-\\log2)\\\\ &\\ge (n/2)\\cdot\\left(1-\\log_3 2\\right)\\log n\\\\ &=\\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\log n, \\end{align} and $$\\log(n!)\\ge \\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\log n$$ also holds for $$n=1,2$$, which completes the proof. $$n! = n \\cdot (n-1) \\cdot ... \\cdot 1 \\ge n \\cdot (n-1) \\cdot ... \\cdot (n/2) \\ge (n/2)^{(n/2)}$$ $$\\log(n!)≥c\\cdot n\\cdot\\log n$$ for $$c \\ge 1/2$$", "# Difference between revisions of \"2018 AMC 12B Problems/Problem 7\" ## Problem What is the value of $$\\log_37\\cdot\\log_59\\cdot\\log_711\\cdot\\log_913\\cdots\\log_{21}25\\cdot\\log_{23}27?$$ $\\textbf{(A) } 3 \\qquad \\textbf{(B) } 3\\log_{7}23 \\qquad \\textbf{(C) } 6 \\qquad \\textbf{(D) } 9 \\qquad \\textbf{(E) } 10$ ## Solution 1 From the Change of Base Formula, we have $$\\frac{\\prod_{i=3}^{13} \\log (2i+1)}{\\prod_{i=1}^{11}\\log (2i+1)} = \\frac{\\log 25 \\cdot \\log 27}{\\log 3 \\cdot \\log 5} = \\frac{(2\\log 5)\\cdot(3\\log 3)}{\\log 3 \\cdot \\log 5} = \\boxed{\\textbf{(C) } 6}.$$ ## Solution 2 Using the chain rule of logarithms $\\log _{a} b \\cdot \\log _{b} c = \\log _{a} c,$ we get \\begin{align*} \\log_37\\cdot\\log_59\\cdot\\log_711\\cdot\\log_913\\cdots\\log_{21}25\\cdot\\log_{23}27 &= (\\log _{3} 7 \\cdot \\log _{7} 11 \\cdots \\log _{23} 27) \\cdot (\\log _{5} 9 \\cdot \\log _{9} 13 \\cdots \\log _{21} 25) \\\\ &= \\log _{3} 27 \\cdot \\log _{5} 25 \\\\ &= 3 \\cdot 2 \\\\ &= \\boxed{\\textbf{(C) } 6}. \\end{align*} ~ pi_is_3.14 ## See Also 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The", "{\\frac {t_{2}}{1.5}}=log_{2}(2\\cdot 2^{\\frac {2t_{1}}{3}}-1)}$ ${\\displaystyle t_{2}(t_{1})=1.5\\cdot log_{2}(2\\cdot 2^{\\frac {2t_{1}}{3}}-1)}$ Examples: ${\\displaystyle t_{2}(1\\,month)=t_{2}({\\frac {1}{12}})=1.5\\cdot log_{2}\\left(2^{\\frac {38}{36}}-1\\right)={\\frac {1.96}{12}}=almost\\,2\\,months}$. ${\\displaystyle t_{2}(1)=1.5\\cdot log_{2}\\left(2^{\\frac {5}{3}}-1\\right)=1.681=(1+0.681)\\,years}$. ${\\displaystyle t_{2}(3)=1.5\\cdot log_{2}\\left(2^{3}-1\\right)=4.211=(3+1.211)\\,years}$. ${\\displaystyle t_{2}(6)=1.5\\cdot log_{2}\\left(2^{5}-1\\right)=7.431=(6+1.431)\\,years}$. ${\\displaystyle t_{2}(15)=1.5\\cdot log_{2}\\left(2\\cdot 2^{\\frac {2\\cdot 15}{3}}-1\\right)=1.5\\cdot log_{2}(2^{11}-1)=16.4989=(15+1.4989)\\,years}$. So, everything is correct, but for longer periods. :) lim 18:19, 31 August 2007 (UTC) No, you have misunderstood what that paragraph means. Say you have a X bit \"weak key\". And that your attacker with his supercomputer can crack that key in about 2 months. But the attacker only thinks the kind of target you are are worth about 1 month of supercomputer time. So you are to expensive for them to attack so they will use their supercomputer for other targets. (Easier targets or targets more worth to attack for them.) But in about 1.5 years they will have a twice as fast supercomputer. And then it will only take them about 1 month to crack your key. Then they think it is worth it and will crack your key. But if you have applied 216 key strengthening rounds to your key they will not be able to crack that \"stronger key\" even with their new supercomputer. Since it would take them 216 months to crack with that supercomputer. (That is 5461 years mind you.) So they will instead have to wait 16", "$n$, if $\\: b < n \\:$ then $$2^{\\hspace{.02 in}(\\hspace{.02 in}\\log(n))^{1+\\frac1{\\log(\\log(n))}}} \\; = \\; 2^{\\left(\\hspace{-0.03 in}(\\hspace{.02 in}\\log(n))^1\\hspace{-0.02 in}\\right) \\cdot \\left(\\hspace{-0.06 in}(\\hspace{.02 in}\\log(n))^{\\frac1{\\log(\\log(n))}}\\hspace{-0.03 in}\\right)} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot \\left(b^{\\hspace{.02 in}\\log(\\log(n))\\hspace{.03 in}}\\right)^{\\frac1{\\log(\\log(n))}}} \\\\ = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b^{\\hspace{.02 in}\\log(\\log(n)) \\cdot \\frac1{\\log(\\log(n))}}} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b^1} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b} \\; = \\; \\left(b^{\\hspace{.02 in}\\log(2)}\\hspace{-0.04 in}\\right)^{\\hspace{.02 in}\\log(n) \\cdot b} \\; \\\\ = \\; b^{\\hspace{.02 in}\\log(2) \\cdot \\log(n) \\cdot b} \\; = \\; b^{\\hspace{.02 in}\\log(n) \\cdot \\log(2) \\cdot b} \\; = \\; \\left(b^{\\hspace{.02 in}\\log(n)}\\hspace{-0.04 in}\\right)^{\\hspace{.02 in}\\log(2) \\cdot b} \\; = \\; n^{\\hspace{.02 in}\\log(2) \\cdot b}$$.", "# Difference between revisions of \"2002 AMC 12B Problems/Problem 22\" ## Problem For all integers $n$ greater than $1$, define $a_n = \\frac{1}{\\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals $\\mathrm{(A)}\\ -2 \\qquad\\mathrm{(B)}\\ -1 \\qquad\\mathrm{(C)}\\ \\frac{1}{2002} \\qquad\\mathrm{(D)}\\ \\frac{1}{1001} \\qquad\\mathrm{(E)}\\ \\frac 12$ ## Solution By the change of base formula, $a_n = \\frac{1}{\\frac{\\log 2002}{\\log n}} = \\left(\\frac{1}{\\log 2002}\\right) \\log n$. Thus \\begin{align*}b- c &= \\left(\\frac{1}{\\log 2002}\\right)(\\log 2 + \\log 3 + \\log 4 + \\log 5 - \\log 10 - \\log 11 - \\log 12 - \\log 13 - \\log 14)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right)\\left(\\log \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14}\\right)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right) \\log 2002^{-1} = -\\left(\\frac{\\log 2002}{\\log 2002}\\right) = -1 \\Rightarrow \\mathrm{(B)}\\end{align*} ## Solution 2 Note that $\\frac{1}{\\log_a b}=\\log_b a$. Thus $a_n=\\log_{2002} n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $$\\log_{2002}\\left(\\frac{2*3*4*5}{10*11*12*13*14}=\\frac{1}{11*13*14}=\\frac{1}{2002}\\right)=\\boxed{\\textbf{(B)}-1}$$ ~yofro ### Solution 3 By the change of base formula, $\\log_n 2002$ and $\\log_{2002} n$ are reciprocals, so $$a_n = \\frac{1}{\\log_n 2002} = \\log_{2002} n$$ for all $n$. Then,\\begin{align*} b - c &= (a_2", "# Difference between revisions of \"2002 AMC 12B Problems/Problem 22\" ## Problem For all integers $n$ greater than $1$, define $a_n = \\frac{1}{\\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals $\\mathrm{(A)}\\ -2 \\qquad\\mathrm{(B)}\\ -1 \\qquad\\mathrm{(C)}\\ \\frac{1}{2002} \\qquad\\mathrm{(D)}\\ \\frac{1}{1001} \\qquad\\mathrm{(E)}\\ \\frac 12$ ## Solution By the change of base formula, $a_n = \\frac{1}{\\frac{\\log 2002}{\\log n}} = \\left(\\frac{1}{\\log 2002}\\right) \\log n$. Thus \\begin{align*}b- c &= \\left(\\frac{1}{\\log 2002}\\right)(\\log 2 + \\log 3 + \\log 4 + \\log 5 - \\log 10 - \\log 11 - \\log 12 - \\log 13 - \\log 14)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right)\\left(\\log \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14}\\right)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right) \\log 2002^{-1} = -\\left(\\frac{\\log 2002}{\\log 2002}\\right) = -1 \\Rightarrow \\mathrm{(B)}\\end{align*} ## Solution 2 Note that $\\frac{1}{\\log_a b}=\\log_b a$. Thus $a_n=\\log_{2002} n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $$\\log_{2002}\\left(\\frac{2*3*4*5}{10*11*12*13*14}=\\frac{1}{11*13*14}=\\frac{1}{2002}\\right)=\\boxed{\\textbf{(B)}-1}$$ ~yofro ## Solution 3 Note that $a_2 = \\frac{1}{\\log_2 2002}$. $1$ is also equal to $\\log_2 2$. So $a_2 = \\frac{\\log_2 2}{\\log_2 2002}$. By the change of bases formula, $a_2 = \\log_{2002} 2$. Following", "1/x$$ 2 $f(x) = x\\cdot log(x)$ Solution \\begin{align} f'(x) &= log(x) + x\\left(\\frac{1}{x}\\right) \\\\ \\\\ &= log(x) + 1 \\end{align} 3 $f(x) = [ln(x)]^3$ Solution $f'(x) =$ $$3[ln(x)]^2 \\cdot \\left( \\frac{1}{x}\\right)\\cdot[ln(x)]^2$$ 4 $f(x) = ln(2x^2 - 4)$ Solution \\begin{align} f'(x) &= \\frac{1}{2x^2 - 4} \\cdot 4x \\\\ \\\\ &= \\frac{4x}{2x^2 - 4} \\end{align} 5 $f(x) = \\frac{ln(x)}{x}$ Solution \\begin{align} f'(x) &= \\frac{x \\left( \\frac{1}{x} \\right) - (1) ln(x)}{x^2} \\\\ \\\\ &= \\frac{1 - ln(x)}{x^2} \\end{align} 6 $f(x) = [ln(ln(x))]^2$ Solution \\begin{align} f'(x) &= 2[ln(ln(x))] \\cdot \\\\ &\\frac{1}{ln(x)}\\cdot \\frac{1}{x} \\\\ \\\\ &= \\frac{2 ln(ln(x))}{x\\cdot ln(x)} \\end{align} Be careful with that result. The ln(x) terms can't be canceled because the top one is inside of another function. 7 $f(x) = x^2 log_3(x)$ Solution $$f'(x) = \\\\ 2x \\cdot log_3 (x) + \\frac{x^2}{x \\cdot ln(3)}$$ 8 $f(x) = ln(sin(x))$ Solution $$f'(x) = \\frac{cos(x)}{sin(x)} = cot(x)$$ 9 $f(x) = ln \\left( \\frac{x + 1}{x^2 - 1} \\right)$ Solution \\begin{align} f'(x) &= \\frac{x^2 - 1}{x + 1} \\cdot \\\\ &\\left( \\frac{(x^2 - 1)(1) - (x + 1)(2x)}{(x^2 - 1)^2} \\right)\\\\ \\\\ &= \\frac{1}{x + 1} \\cdot \\\\ &\\left( \\frac{(x - 1)(x + 1) - (x + 1)(2x)}{x^2 - 1} \\right) \\\\ \\\\ &= \\frac{(x - 1) - (2x)}{x^2 - 1} = \\frac{x + 1}{1 - x^2} \\end{align} I made the last step to", "1+ \\left(\\frac{1}{3}\\right) \\cdot b &= \\frac{12}{11}\\\\ b &= 0.\\overline{27}\\end{align*} Thus the approximate model is: f(x)=121+(13)(0.27273)x\\begin{align*}f(x)=\\frac{12}{1+\\left(\\frac{1}{3}\\right) \\cdot (0.27273)^x}\\end{align*} #### Example 5 Determine the logistic model given c=7\\begin{align*}c=7\\end{align*} and the points (0, 2) and (3, 5). The two points give two equations, and the logistic model has two variables. 2=71+aa=2.55=71+(2.5)b3b3=0.16b0.5429\\begin{align*}2 = \\frac{7}{1+a}\\\\ a = 2.5\\\\ 5 = \\frac{7}{1+(2.5) \\cdot b^3}\\\\ b^3 = 0.16\\\\ b \\approx 0.5429\\end{align*} Thus the approximate model is: f(x)=71+(2.5)(0.5429)x\\begin{align*}f(x)=\\frac{7}{1+(2.5) \\cdot (0.5429)^x}\\end{align*} ### Review For 1-5, determine the logistic model given the carrying capacity and two points. 1. c=12;(0,5);(1,7)\\begin{align*}c=12; (0, 5); (1, 7)\\end{align*} 2. c=200;(0,150);(5,180)\\begin{align*}c=200; (0, 150); (5, 180)\\end{align*} 3. c=1500;(0,150);(10,1000)\\begin{align*}c=1500; (0, 150); (10, 1000)\\end{align*} 4. c=1000000;(0,100000);(40,20000)\\begin{align*}c=1000000; (0, 100000); (-40, 20000)\\end{align*} 5. c=30000000;(60,10000);(0,8000000)\\begin{align*}c= 30000000; (-60, 10000); (0, 8000000)\\end{align*} For 6-8, use the logistic function f(x)=321+3ex\\begin{align*}f(x)=\\frac{32}{1+3e^{-x}}\\end{align*}. 6. What is the carrying capacity of the function? 7. What is the y\\begin{align*}y\\end{align*}-intercept of the function? 8. Use your answers to 6 and 7 along with at least two points on the graph to make a sketch of the function. For 9-11, use the logistic function g(x)=251+40.2x\\begin{align*}g(x)=\\frac{25}{1+4 \\cdot 0.2^x}\\end{align*}. 9. What is the carrying capacity of the function? 10. What is the y\\begin{align*}y\\end{align*}-intercept of the function? 11. Use your answers to 9 and 10 along with at least two points on the graph to make a sketch of the function. For 12-14,", "for something more... elegant... I didn't specify this, so as it stands, this is a perfectly acceptable answer, and it has my upvote. – Alec Mar 24 '18 at 17:31 Use a multiplicative variant of Gauss's trick: $$(n!)^2 = (1 \\cdot n) (2 \\cdot (n-1)) (3 \\cdot (n-2)) \\cdots ((n-2) \\cdot 3) ((n-1) \\cdot 2) (n \\cdot 1) \\ge n^n$$ Setting $n=20$, we get $20!\\ge 20^{10} > 16^{10} = 2^{40}$. • It's worth adding how you know that $k(n-k+1)\\ge n$. – nbubis Feb 24 '17 at 18:39 Well we know that $\\log_2 x$ is increasing so we could calculate \\begin{align} \\log_2(20!) &=\\log_2\\left(\\prod_{i=1}^{20}i\\right) \\\\ &=\\sum_{i=1}^{20}\\log_2(i) \\\\ &=\\log_2{1}+\\log_2{2}+\\log_23+\\log_2{4}+\\cdots+\\log_2 8+\\cdots+\\log_216+\\cdots+\\log_2{20} \\\\ &>0+\\log_2{2}+\\log_2 2+\\log_24+\\cdots+\\log_2{4}+ \\\\&\\,\\,\\,\\,\\,\\log_28+\\cdots+\\log_2{8}+\\log_2{16}+\\cdots+\\log_2{16} \\\\ &=0+2\\log_2{2}+4\\log_2{4}+8\\log_2{8}+5\\log_2{16} \\\\&=0+2+4 \\cdot 2+8 \\cdot 3+5 \\cdot 4 \\\\&=54 \\\\&>40=\\log_2(2^{40}). \\end{align} Therefore, as $\\log_2x$ is increasing we have $20!>2^{40}$. In fact this shows that $20!>2^{54}$. $$\\frac{(20)!}{2^{40}}=\\frac{(20)!}{4^{20}}=\\frac14\\cdot\\frac24\\cdot\\frac34\\cdot\\frac44\\cdots\\frac{19}4\\cdot\\frac{20}4>\\frac{3!}{4^3}\\cdot\\frac{19}4\\cdot\\frac{20}4=\\frac{6\\cdot19\\cdot5}{64\\cdot4}=\\frac{570}{256}>1$$ That was a defensive approach Observe that $\\displaystyle\\frac14\\cdot\\frac24\\cdot\\frac34=\\frac3{32}>\\frac1{11}$ which is compensated by $\\displaystyle\\frac84\\frac94\\cdot\\frac{10}4=\\frac{45}4>11$ $$\\implies \\frac{(20)!}{2^{40}}=\\frac14\\cdot\\frac24\\cdot\\frac34\\cdot\\frac44\\cdots\\frac{19}4\\cdot\\frac{20}4>\\frac14\\cdot\\frac24\\cdot\\frac34\\cdots \\frac84\\frac94\\cdot\\frac{10}4>\\frac1{11}\\cdot\\frac{45}4>1$$ i.e., $$\\frac{(20)!}{4^{20}}> \\frac{(10)!}{4^{10}}>1$$ Programmatically, I found $\\displaystyle \\frac{n!}{4^n}>1$ for $n\\ge9$ which I think can be a good exercise If we ignore the first terms of $20!$ we have $8 \\times \\dots \\times 20$. That's at the very least $8^{13}$, which is $2^{39}$. Including the first factors we ignored we go over $2^{40}$. Pretty ad-hoc, but just my 2 cents. It is easy to compare", "and $$k/b^d > 1$$ separately. • $$k/b^d \\le 1$$: We compute an upper bound on the work as follows: $\\begin{split}\\O\\parens{\\parens{\\frac{n}{b^i}}^d \\log^w \\frac{n}{b^i}} &= \\O\\parens{\\frac{n^d}{b^{id}} \\cdot (\\log^w n - \\log^w b^i)}\\\\ &= b^{-id} \\cdot \\O(n^d \\cdot (\\log^w n - \\log^w b^i))\\\\ &= b^{-id} \\cdot \\O(n^d \\log^w n)\\end{split}$ since $$n^d \\cdot (\\log^w n - \\log^w b^i) \\le n^d \\log^w n$$ for $$b > 1$$, and we only need an upper bound for the asymptotics. Then the work $$T_i$$ at level $$i$$ is $\\begin{split}T_i &= \\frac{k^i}{b^{id}} \\cdot \\O(n^d \\log^w n)\\\\ &= \\parens{\\frac{k}{b^d}}^i \\cdot \\O(n^d \\log^w n)\\end{split}$ Summing over all the levels, we get a total work $\\begin{split}T &= \\sum_{i=0}^{\\log_b n} T_i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} \\parens{\\frac{k}{b^d}}^i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} r^i\\\\ &= \\O\\parens{n^d \\log^w n \\cdot \\sum_{i=0}^{\\log_b n} r^i}\\end{split}$ where $$r = k/b^d$$. When $$r < 1$$, the initial term of the summation dominates as before, so we have $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n)\\end{split}$ When $$r = 1$$, the terms all have equal size, also as before. Then $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n \\cdot (1 + \\log_b n))\\\\ &= \\O(n^d \\log^w n \\log_b n)\\\\ &=", ". - - . . . . . . $(\\log3)^2 + \\log3\\!\\cdot\\!\\log x \\:=\\: (\\log4)^2 + \\log4\\!\\cdot\\!\\log x$ . . = . . . $\\log4\\!\\cdot\\!\\log x - \\log3\\!\\cdot\\!\\log x \\:=\\:(\\log3)^2 - (\\log4)^2$ . . . . . . . . . . $(\\log4 - \\log 3)\\log x \\:=\\:-\\left[(\\log 4)^2 - (\\log 3)^2\\right]$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:-\\dfrac{(\\log 4 - \\log 3)(\\log 4 + \\log 3)}{\\log 4 - \\log 3}$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:-\\left(\\log 4+\\log3)$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:-\\log(4\\!\\cdot\\!3)$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:-\\log12$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:\\log(12^{-1})$ . . . . . . . . . . . . . . . . . . . . $\\log x \\:=\\:\\log\\left(\\frac{1}{12}\\right)$ . . . . . . . . . . . .Therefore: . . .", "since $$n^d \\cdot (\\log^w n - \\log^w b^i) \\le n^d \\log^w n$$ for $$b > 1$$, and we only need an upper bound for the asymptotics. Then the work $$T_i$$ at level $$i$$ is $\\begin{split}T_i &= \\frac{k^i}{b^{id}} \\cdot \\O(n^d \\log^w n)\\\\ &= \\parens{\\frac{k}{b^d}}^i \\cdot \\O(n^d \\log^w n)\\end{split}$ Summing over all the levels, we get a total work $\\begin{split}T &= \\sum_{i=0}^{\\log_b n} T_i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} \\parens{\\frac{k}{b^d}}^i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} r^i\\\\ &= \\O\\parens{n^d \\log^w n \\cdot \\sum_{i=0}^{\\log_b n} r^i}\\end{split}$ where $$r = k/b^d$$. When $$r < 1$$, the initial term of the summation dominates as before, so we have $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n)\\end{split}$ When $$r = 1$$, the terms all have equal size, also as before. Then $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n \\cdot (1 + \\log_b n))\\\\ &= \\O(n^d \\log^w n \\log_b n)\\\\ &= \\O(n^d \\log^w n \\log n)\\\\ &= \\O(n^d \\log^{w+1} n)\\end{split}$ As before, we drop the lower order term, as well as the base in $$\\log_b n$$ since it only differs by a constant factor from $$\\log n$$. • $$k/b^d > 1$$: We start by observing that since", "how you got from (3) to (4) where $n$ reappeared. Perhaps you just los $n$ in (3) but otherwise you had it on paper. Let us try again: \\begin{align*} T(n) &= n + (3/2) n + (3/2)^2 n + \\cdots \\end{align*} But when does sum end? (This is what created your mistake.) It ends when we get to the base case, which is $T(1)$. After $k$ unfoldings we have $T(n/2^k)$ so we will reach $T(1)$ when $k = \\log_2 n$. Thus, we continue like this: \\begin{align*} T(n) &= n + (3/2) n + (3/2)^2 n + \\cdots + (3/2)^{\\log_2 n} T(1) \\\\ &= \\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i n\\\\ &= n \\cdot \\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i. \\end{align*} It remains to add the sum. It is a geometric series, so we can look up the formula $\\sum_{i=0}^k a^i = (a^{k+1} - 1)/(a - 1)$ and get \\begin{align*} {\\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i} &= \\frac{(3/2)^{1 +\\log_2 n} - 1}{3/2 - 1} \\\\ &= 2 \\cdot ((3/2) \\cdot (3/2)^{\\log_2 n} - 1) \\\\ &= 3 \\cdot (3/2)^{\\log_2 n} - 2 \\\\ &= 3 \\cdot (n^{\\log_2 (3/2)}) - 2 \\\\ &\\in \\Theta(n^{\\log_2 (3/2)}) \\approx \\Theta(n^{0.58}) \\end{align*} The final result is this sum multiplied with $n$, so it is $$\\Theta(n \\cdot n^{\\log_2(3/2)}) = \\Theta(n^{\\log_2 2 + \\log_2 (3/2)}) = \\Theta(n^{\\log_2 3}),$$ assuming I can still", "\\text{Age of boiler now} \\\\ B_t & \\text{Age of boiler then} \\\\ t & \\text{Time between then and now} \\end{array}$ This leaves us with the following equations to solve: \\begin{align} S_n & = S_t + t \\tag{a} \\\\ B_n & = B_t + t \\tag{b} \\\\ S_n & = 2 \\cdot B_t \\tag{c} \\\\ S_t & = B_n \\tag{d} \\\\ S_n + B_n &= 30 \\tag{e} \\\\ \\end{align} Combining (a) and (d) yields $S_n = B_n + t$ or alternatively $t = S_n - B_n$ which is more to the point. Combining this with (b) and (c) gives us \\begin{align} S_n & = 2 \\cdot B_t \\\\ S_n & = 2 \\cdot (B_n - t) \\\\ S_n & = 2 \\cdot (B_n - (S_n - B_n)) \\\\ S_n & = 2 \\cdot (2 \\cdot B_n - S_n) \\\\ S_n & = 4 \\cdot B_n - 2 \\cdot S_n \\\\ 3 \\cdot S_n & = 4 \\cdot B_n \\end{align} Combining this with (e), we can solve it: \\begin{align} 3 \\cdot S_n & = 4 \\cdot B_n \\\\ 3 \\cdot S_n & = 4 \\cdot (30 - S_n) \\\\ 3 \\cdot S_n & = 120 - 4 \\cdot S_n \\\\ 7 \\cdot S_n & = 120 \\\\ S_n & = \\frac{120}{7} \\end{align} Remembering that they total to 30 years, we", "+ h)\\}\\cdot\\frac{\\exp\\{\\sin(\\log(1 + h)) - n\\log(1 + h)\\} - 1}{h}\\notag\\\\ &= \\lim_{h \\to 0}1\\cdot\\frac{\\exp\\{\\sin(\\log(1 + h)) - n\\log(1 + h)\\} - 1}{\\sin(\\log(1 + h)) - n\\log(1 + h)}\\cdot\\dfrac{\\sin(\\log(1 + h)) - n\\log(1 + h)}{h}\\notag\\\\ &= \\lim_{h \\to 0}1\\cdot\\dfrac{\\sin(\\log(1 + h)) - n\\log(1 + h)}{h}\\notag\\\\ &= \\lim_{h \\to 0}\\dfrac{\\sin(\\log(1 + h))}{h} - n\\cdot\\dfrac{\\log(1 + h)}{h}\\notag\\\\ &= \\lim_{h \\to 0}\\dfrac{\\sin(\\log(1 + h))}{\\log(1 + h)}\\cdot\\frac{\\log(1 + h)}{h} - n\\cdot 1\\notag\\\\ &= 1\\cdot 1 - n = 1 - n \\end{align} Clearly we now have $n = 5$. As can be seen this solution avoids the use of L'Hospital's Rule and Taylor series and uses only the standard limits and rules of limits.", "4 - \\log4 \\;=\\;x\\!\\cdot\\!\\log3 + \\log 3$ Get the $\\displaystyle x$'s together: . $\\displaystyle 2x\\!\\cdot\\log4 - x\\!\\cdot\\!\\log3 \\;=\\;\\log 3 + \\log 4$ Factor: . $\\displaystyle x\\left(2\\log4 - \\log 3\\right) \\:=\\:\\log 3 + \\log 4$ Therefore: . $\\displaystyle \\boxed{x \\;=\\;\\frac{\\log3 + \\log4}{2\\log4 - \\log3}}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Peritus simplified it further, using logarithm rules. . . $\\displaystyle \\log3 + \\log4 \\;=\\;\\log(3\\cdot4) \\;=\\;\\log12$ . . $\\displaystyle 2\\log4 - \\log 3 \\;=\\;\\log(4^2) - \\log3 \\;=\\;\\log16 - \\log3 \\;=\\;\\log\\left(\\frac{16}{3}\\right)$ And we have: . $\\displaystyle \\frac{\\log(12)}{\\log\\left(\\frac{16}{3}\\right)} \\;=\\;\\log_{\\frac{16}{3}}(12)$ I've always wanted to do that! .", "13=(3x)\\log 18$ $(x-3) \\cdot 2.564949357=3x \\cdot 2.890371758$ $2.564949357x-7.694848071=8.671115274x$ $-7.694848071=6.106165917x$ $x=\\frac{-7.694848071}{6.106165917}$ $\\boxed{x \\approx -1.260176709}$ 6. Originally Posted by masters Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it. $13^{x-3}=18^{3x}$ $\\log(13^{x-3})=\\log (18^{3x})$ $(x-3)\\log 13=(3x)\\log 18$ $(x-3) \\cdot 2.564949357=3x \\cdot 2.890371758$ $2.564949357x-7.694848071=8.671115274x$ $-7.694848071=6.106165917x$ $x=\\frac{-7.694848071}{6.106165917}$ $\\boxed{x \\approx -1.260176709}$ that's fine, Masters. I would have preferred if you left it as log(13) and log(18) though as opposed to finding the decimal approximations.", "$$f(n) \\le c \\log(n) = c + c \\log(n/2) \\le c + c \\... 3 \\mathcal{O}\\Big(\\log{n} + \\log{(n-1)} + \\ldots + \\log{(1))}\\Big) = \\mathcal{O}(\\log{n}) That is not right. When n is large enough,$$\\begin{align} \\log{n} + \\log{(n-1)} + \\ldots + \\log(1) &\\ge \\log{n} + \\log{(n-1)} + \\ldots + \\log(n/2)\\\\ &\\ge n/2 \\log(n/2)\\\\ &=\\Theta(n\\log n) \\end{align}$$More precisely, since n!\\sim {\\sqrt{2\\pi n}\\left(\\... 2 No. You can't. As \\lim_{n\\to\\infty} \\frac{n!}{n! \\times n} = 0 . Hence, n! \\times n = \\omega(n!) or n! = o(n\\times n!) (little-oh). 2 Define M = \\max(|a_0|/a_m, |a_1|/a_m, \\ldots, |a_{m-1}|/a_m), and take c = a_m/2 and n_0 = 2mM. Then for n \\geq n_0,$$ \\begin{align*} f(n) &= a_m n^m \\left(1 + \\frac{a_{m-1}}{a_m} \\cdot \\frac{1}{n} + \\cdots + \\frac{a_1}{a_m} \\cdot \\frac{1}{n^{m-1}} + \\frac{a_0}{a_m} \\cdot \\frac{1}{n^m}\\right) \\\\ &\\geq a_m n^m \\left(1 - \\frac{M}{n_0} - \\cdots ... 2 The last step is incorrect. $\\mathcal{O}\\Big(\\log{n} + \\log{(n-1)} + \\ldots + \\log{(1))}\\Big)$ is not $\\mathcal{O}(\\log{n})$. You made the same mistake as What goes wrong with sums of Landau terms?. See also What is the asymptotic runtime of this nested loop?. 2 I am afraid that you could have been more careful. I know that by L'Hospital's rule we can reduce it to $\\lim_{n \\to \\infty} \\frac{n+1}{n}$ which is a constant and hence $n = \\theta (n+1)!$. By L'Hospital's rule we have $$\\lim_{n" ]

[ "+ a_3 + a_4 + a_5) - (a_{10} + a_{11} + a_{12} + a_{13} + a_{14}) \\\\ &= (\\log_{2002} 2 + \\log_{2002} 3 + \\log_{2002} 4 + \\log_{2002} 5) \\\\ &\\quad - (\\log_{2002} 10 + \\log_{2002} 11 + \\log_{2002} 12 + \\log_{2002} 13 + \\log_{2002} 14) \\\\ &= \\log_{2002} \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14} \\\\ &= \\log_{2002} \\frac{1}{2002} \\\\ &= \\boxed{-1}. \\end{align*} The answer is (B).", "further simplify $$(n/2)\\log(n/2)$$ --> $$(n/2)\\log(n^{-2})$$, ... This is wrong. In fact, for $$n\\ge 3$$, $$n^{\\log_3 2}\\ge 2$$, so \\begin{align} (n/2)\\log(n/2)&=(n/2)(\\log n-\\log2)\\\\ &\\ge (n/2)\\cdot\\left(1-\\log_3 2\\right)\\log n\\\\ &=\\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\log n, \\end{align} and $$\\log(n!)\\ge \\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\log n$$ also holds for $$n=1,2$$, which completes the proof. $$n! = n \\cdot (n-1) \\cdot ... \\cdot 1 \\ge n \\cdot (n-1) \\cdot ... \\cdot (n/2) \\ge (n/2)^{(n/2)}$$ $$\\log(n!)≥c\\cdot n\\cdot\\log n$$ for $$c \\ge 1/2$$", "the same reasoning, $a_3 = \\log_{2002} 3$, $a_4 = \\log_{2002} 4$ and so on. $$b = \\log_{2002} 2 + \\log_{2002} 3 + .....+ \\log_{2002} 5 = \\log_{2002} 5! = \\log_{2002} 120$$ Now solving for $c$, we see that it equals $\\log_{2002} (10\\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14)$ $$b-c = \\log_{2002} 120 - \\log_{2002} 240240 \\rightarrow \\log_{2002} \\frac{1}{2002} = \\boxed{-1}$$ ~YBSuburbanTea", "# Difference between revisions of \"2002 AMC 12B Problems/Problem 22\" ## Problem For all integers $n$ greater than $1$, define $a_n = \\frac{1}{\\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals $\\mathrm{(A)}\\ -2 \\qquad\\mathrm{(B)}\\ -1 \\qquad\\mathrm{(C)}\\ \\frac{1}{2002} \\qquad\\mathrm{(D)}\\ \\frac{1}{1001} \\qquad\\mathrm{(E)}\\ \\frac 12$ ## Solution By the change of base formula, $a_n = \\frac{1}{\\frac{\\log 2002}{\\log n}} = \\left(\\frac{1}{\\log 2002}\\right) \\log n$. Thus \\begin{align*}b- c &= \\left(\\frac{1}{\\log 2002}\\right)(\\log 2 + \\log 3 + \\log 4 + \\log 5 - \\log 10 - \\log 11 - \\log 12 - \\log 13 - \\log 14)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right)\\left(\\log \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14}\\right)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right) \\log 2002^{-1} = -\\left(\\frac{\\log 2002}{\\log 2002}\\right) = -1 \\Rightarrow \\mathrm{(B)}\\end{align*} ## Solution 2 Note that $\\frac{1}{\\log_a b}=\\log_b a$. Thus $a_n=\\log_{2002} n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $$\\log_{2002}\\left(\\frac{2*3*4*5}{10*11*12*13*14}=\\frac{1}{11*13*14}=\\frac{1}{2002}\\right)=\\boxed{\\textbf{(B)}-1}$$ ~yofro ### Solution 3 By the change of base formula, $\\log_n 2002$ and $\\log_{2002} n$ are reciprocals, so $$a_n = \\frac{1}{\\log_n 2002} = \\log_{2002} n$$ for all $n$. Then,\\begin{align*} b - c &= (a_2", "# Difference between revisions of \"2002 AMC 12B Problems/Problem 22\" ## Problem For all integers $n$ greater than $1$, define $a_n = \\frac{1}{\\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals $\\mathrm{(A)}\\ -2 \\qquad\\mathrm{(B)}\\ -1 \\qquad\\mathrm{(C)}\\ \\frac{1}{2002} \\qquad\\mathrm{(D)}\\ \\frac{1}{1001} \\qquad\\mathrm{(E)}\\ \\frac 12$ ## Solution By the change of base formula, $a_n = \\frac{1}{\\frac{\\log 2002}{\\log n}} = \\left(\\frac{1}{\\log 2002}\\right) \\log n$. Thus \\begin{align*}b- c &= \\left(\\frac{1}{\\log 2002}\\right)(\\log 2 + \\log 3 + \\log 4 + \\log 5 - \\log 10 - \\log 11 - \\log 12 - \\log 13 - \\log 14)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right)\\left(\\log \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14}\\right)\\\\ &= \\left(\\frac{1}{\\log 2002}\\right) \\log 2002^{-1} = -\\left(\\frac{\\log 2002}{\\log 2002}\\right) = -1 \\Rightarrow \\mathrm{(B)}\\end{align*} ## Solution 2 Note that $\\frac{1}{\\log_a b}=\\log_b a$. Thus $a_n=\\log_{2002} n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $$\\log_{2002}\\left(\\frac{2*3*4*5}{10*11*12*13*14}=\\frac{1}{11*13*14}=\\frac{1}{2002}\\right)=\\boxed{\\textbf{(B)}-1}$$ ~yofro ## Solution 3 Note that $a_2 = \\frac{1}{\\log_2 2002}$. $1$ is also equal to $\\log_2 2$. So $a_2 = \\frac{\\log_2 2}{\\log_2 2002}$. By the change of bases formula, $a_2 = \\log_{2002} 2$. Following", "1/x$$ 2 $f(x) = x\\cdot log(x)$ Solution \\begin{align} f'(x) &= log(x) + x\\left(\\frac{1}{x}\\right) \\\\ \\\\ &= log(x) + 1 \\end{align} 3 $f(x) = [ln(x)]^3$ Solution $f'(x) =$ $$3[ln(x)]^2 \\cdot \\left( \\frac{1}{x}\\right)\\cdot[ln(x)]^2$$ 4 $f(x) = ln(2x^2 - 4)$ Solution \\begin{align} f'(x) &= \\frac{1}{2x^2 - 4} \\cdot 4x \\\\ \\\\ &= \\frac{4x}{2x^2 - 4} \\end{align} 5 $f(x) = \\frac{ln(x)}{x}$ Solution \\begin{align} f'(x) &= \\frac{x \\left( \\frac{1}{x} \\right) - (1) ln(x)}{x^2} \\\\ \\\\ &= \\frac{1 - ln(x)}{x^2} \\end{align} 6 $f(x) = [ln(ln(x))]^2$ Solution \\begin{align} f'(x) &= 2[ln(ln(x))] \\cdot \\\\ &\\frac{1}{ln(x)}\\cdot \\frac{1}{x} \\\\ \\\\ &= \\frac{2 ln(ln(x))}{x\\cdot ln(x)} \\end{align} Be careful with that result. The ln(x) terms can't be canceled because the top one is inside of another function. 7 $f(x) = x^2 log_3(x)$ Solution $$f'(x) = \\\\ 2x \\cdot log_3 (x) + \\frac{x^2}{x \\cdot ln(3)}$$ 8 $f(x) = ln(sin(x))$ Solution $$f'(x) = \\frac{cos(x)}{sin(x)} = cot(x)$$ 9 $f(x) = ln \\left( \\frac{x + 1}{x^2 - 1} \\right)$ Solution \\begin{align} f'(x) &= \\frac{x^2 - 1}{x + 1} \\cdot \\\\ &\\left( \\frac{(x^2 - 1)(1) - (x + 1)(2x)}{(x^2 - 1)^2} \\right)\\\\ \\\\ &= \\frac{1}{x + 1} \\cdot \\\\ &\\left( \\frac{(x - 1)(x + 1) - (x + 1)(2x)}{x^2 - 1} \\right) \\\\ \\\\ &= \\frac{(x - 1) - (2x)}{x^2 - 1} = \\frac{x + 1}{1 - x^2} \\end{align} I made the last step to", "and $F$ also. So $v_n(x)$ has for limit the derivative of $F$ at $0$. As $\\displaystyle F^{\\prime}(x)=\\frac{G^{\\prime}(x)}{G(x)}$ and $G(0)=1$, this is $G^{\\prime}(0)$. Now $a_k^x=\\exp(x\\log a_k)$, and it is easy to finish. The simplest route is to take logarithms. Let $$f(x) = \\left(\\frac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x}}{n}\\right)^{1/x}$$ and we need to calculate the limit of $f(x)$ as $x \\to 0$. Let $L$ be this desired limit then we have \\begin{align} \\log L &= \\log\\left(\\lim_{x \\to 0}f(x)\\right)\\notag\\\\ &= \\lim_{x \\to 0}\\log f(x)\\text{ (via continuity of log)}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{1}{x}\\log\\frac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x}}{n}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{1}{x}\\cdot\\left(\\dfrac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x}}{n} - 1\\right)\\cdot\\dfrac{\\log\\left(1 + \\dfrac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x}}{n} - 1\\right)}{\\dfrac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x}}{n} - 1}\\notag\\\\ &= \\frac{1}{n}\\lim_{x \\to 0}\\frac{a_{1}^{x} + a_{2}^{x} + \\cdots + a_{n}^{x} - n}{x}\\notag\\\\ &= \\frac{1}{n}\\lim_{x \\to 0}\\sum_{k = 1}^{n}\\frac{a_{k}^{x} - 1}{x}\\notag\\\\ &= \\frac{1}{n}\\sum_{k = 1}^{n}\\lim_{x \\to 0}\\frac{a_{k}^{x} - 1}{x}\\notag\\\\ &= \\frac{1}{n}\\sum_{k = 1}^{n}\\log a_{k}\\notag\\\\ &= \\log(a_{1}a_{2}\\cdots a_{n})^{1/n}\\notag \\end{align} It now follows that $$L = (a_{1}a_{2}\\cdots a_{n})^{1/n}$$ We have used the following two standard limits in the above derivation $$\\lim_{x \\to 0}\\frac{\\log(1 + x)}{x} = 1,\\,\\lim_{x \\to 0}\\frac{a^{x} - 1}{x} = \\log a$$", "$n$, if $\\: b < n \\:$ then $$2^{\\hspace{.02 in}(\\hspace{.02 in}\\log(n))^{1+\\frac1{\\log(\\log(n))}}} \\; = \\; 2^{\\left(\\hspace{-0.03 in}(\\hspace{.02 in}\\log(n))^1\\hspace{-0.02 in}\\right) \\cdot \\left(\\hspace{-0.06 in}(\\hspace{.02 in}\\log(n))^{\\frac1{\\log(\\log(n))}}\\hspace{-0.03 in}\\right)} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot \\left(b^{\\hspace{.02 in}\\log(\\log(n))\\hspace{.03 in}}\\right)^{\\frac1{\\log(\\log(n))}}} \\\\ = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b^{\\hspace{.02 in}\\log(\\log(n)) \\cdot \\frac1{\\log(\\log(n))}}} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b^1} \\; = \\; 2^{\\hspace{.02 in}\\log(n) \\cdot b} \\; = \\; \\left(b^{\\hspace{.02 in}\\log(2)}\\hspace{-0.04 in}\\right)^{\\hspace{.02 in}\\log(n) \\cdot b} \\; \\\\ = \\; b^{\\hspace{.02 in}\\log(2) \\cdot \\log(n) \\cdot b} \\; = \\; b^{\\hspace{.02 in}\\log(n) \\cdot \\log(2) \\cdot b} \\; = \\; \\left(b^{\\hspace{.02 in}\\log(n)}\\hspace{-0.04 in}\\right)^{\\hspace{.02 in}\\log(2) \\cdot b} \\; = \\; n^{\\hspace{.02 in}\\log(2) \\cdot b}$$.", "# what is the value of $x$? If $\\log_{2}3^4\\cdot\\log_{3}4^5\\cdot\\log_{4}5^6\\cdot....\\log_{63} {64}^{65}=x!$, what is the value of $x$? I've tried $$\\log_2 3^4\\cdot\\log_3 4^5\\cdot\\log_4 5^6\\cdot.... \\log_{63} 64^{65}$$ $$=\\dfrac{4\\log 3}{\\log 2}\\cdot\\dfrac{5\\log 4}{\\log 3}\\cdot\\dfrac{6\\log 5}{\\log 4}\\cdots\\dfrac{65\\log 64}{\\log 63}$$ don't know how to solve futher steps, please help. Thanks • Hint: it \"telescopes\". Try writing a few more terms at the beginning and end, then start canceling out $\\log$ terms between numerators and denominators. See what's left. – dxiv Dec 14 '16 at 6:16 Write $\\;\\;\\;\\;\\;\\;\\;\\dfrac{4\\log 3}{\\log 2}\\cdot\\dfrac{5\\log 4}{\\log 3}\\cdot\\dfrac{6\\log 5}{\\log 4}\\cdots\\dfrac{65\\log 64}{\\log 63}\\;\\;\\;\\;\\;$ as $$=\\dfrac{4\\log 3}{\\log 2}\\cdot\\dfrac{5\\log 4}{\\log 3}\\cdot\\dfrac{6\\log 5}{\\log 4}\\cdots\\dfrac{65\\cdot 6\\log 2}{\\log 63}$$ $\\{\\because \\log64=\\log2^6=6\\log2\\}$ Every logarithmic part cancels out, and we have $$=4\\cdot 5 \\cdot 6 \\cdots \\color{red}{6}\\cdot 65$$ $$=\\color{red}{2\\cdot 3} \\cdot 4\\cdot 5\\cdot 6 \\cdots 65$$ $$=65!$$ Hence $x=65$ You get $$4\\cdot 5\\cdot 6\\cdots 65\\cdot {\\log 64\\over \\log 2}\\\\=1\\cdot2\\cdot3\\cdot4\\cdot5\\cdots65=x!$$ Hence $x=65$ \\begin{align}\\dfrac{4\\log 3}{\\log 2}\\cdot\\dfrac{5\\log 4}{\\log 3}\\cdot\\dfrac{6\\log 5}{\\log 4}\\cdots\\dfrac{65\\log 64}{\\log 63}&=\\frac{65!}{3!}\\frac{\\log 64}{\\log 2} \\\\&=\\frac{65!}{3!}\\frac{\\log 2^6}{\\log 2} \\\\ &=\\frac{65!}{3!}\\frac{\\log 2}{\\log 2}.6 \\\\&=65! \\end{align}", "$$f(n) \\le c \\log(n) = c + c \\log(n/2) \\le c + c \\... 3 \\mathcal{O}\\Big(\\log{n} + \\log{(n-1)} + \\ldots + \\log{(1))}\\Big) = \\mathcal{O}(\\log{n}) That is not right. When n is large enough,$$\\begin{align} \\log{n} + \\log{(n-1)} + \\ldots + \\log(1) &\\ge \\log{n} + \\log{(n-1)} + \\ldots + \\log(n/2)\\\\ &\\ge n/2 \\log(n/2)\\\\ &=\\Theta(n\\log n) \\end{align}$$More precisely, since n!\\sim {\\sqrt{2\\pi n}\\left(\\... 2 No. You can't. As \\lim_{n\\to\\infty} \\frac{n!}{n! \\times n} = 0 . Hence, n! \\times n = \\omega(n!) or n! = o(n\\times n!) (little-oh). 2 Define M = \\max(|a_0|/a_m, |a_1|/a_m, \\ldots, |a_{m-1}|/a_m), and take c = a_m/2 and n_0 = 2mM. Then for n \\geq n_0,$$ \\begin{align*} f(n) &= a_m n^m \\left(1 + \\frac{a_{m-1}}{a_m} \\cdot \\frac{1}{n} + \\cdots + \\frac{a_1}{a_m} \\cdot \\frac{1}{n^{m-1}} + \\frac{a_0}{a_m} \\cdot \\frac{1}{n^m}\\right) \\\\ &\\geq a_m n^m \\left(1 - \\frac{M}{n_0} - \\cdots ... 2 The last step is incorrect. $\\mathcal{O}\\Big(\\log{n} + \\log{(n-1)} + \\ldots + \\log{(1))}\\Big)$ is not $\\mathcal{O}(\\log{n})$. You made the same mistake as What goes wrong with sums of Landau terms?. See also What is the asymptotic runtime of this nested loop?. 2 I am afraid that you could have been more careful. I know that by L'Hospital's rule we can reduce it to $\\lim_{n \\to \\infty} \\frac{n+1}{n}$ which is a constant and hence $n = \\theta (n+1)!$. By L'Hospital's rule we have $$\\lim_{n", "and $$k/b^d > 1$$ separately. • $$k/b^d \\le 1$$: We compute an upper bound on the work as follows: $\\begin{split}\\O\\parens{\\parens{\\frac{n}{b^i}}^d \\log^w \\frac{n}{b^i}} &= \\O\\parens{\\frac{n^d}{b^{id}} \\cdot (\\log^w n - \\log^w b^i)}\\\\ &= b^{-id} \\cdot \\O(n^d \\cdot (\\log^w n - \\log^w b^i))\\\\ &= b^{-id} \\cdot \\O(n^d \\log^w n)\\end{split}$ since $$n^d \\cdot (\\log^w n - \\log^w b^i) \\le n^d \\log^w n$$ for $$b > 1$$, and we only need an upper bound for the asymptotics. Then the work $$T_i$$ at level $$i$$ is $\\begin{split}T_i &= \\frac{k^i}{b^{id}} \\cdot \\O(n^d \\log^w n)\\\\ &= \\parens{\\frac{k}{b^d}}^i \\cdot \\O(n^d \\log^w n)\\end{split}$ Summing over all the levels, we get a total work $\\begin{split}T &= \\sum_{i=0}^{\\log_b n} T_i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} \\parens{\\frac{k}{b^d}}^i\\\\ &= \\O(n^d \\log^w n) \\cdot \\sum_{i=0}^{\\log_b n} r^i\\\\ &= \\O\\parens{n^d \\log^w n \\cdot \\sum_{i=0}^{\\log_b n} r^i}\\end{split}$ where $$r = k/b^d$$. When $$r < 1$$, the initial term of the summation dominates as before, so we have $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n)\\end{split}$ When $$r = 1$$, the terms all have equal size, also as before. Then $\\begin{split}T &= \\O\\parens{n^d \\log^w n \\cdot (1 + r + r^2 + \\dots + r^{\\log_b n})}\\\\ &= \\O(n^d \\log^w n \\cdot (1 + \\log_b n))\\\\ &= \\O(n^d \\log^w n \\log_b n)\\\\ &=", "always be numerically less than unity for all positive values of $M$ and $N$. Substituting from (D) into (C), we get (E) \\begin{align} \\log \\frac{M}{N} & = \\log M - \\log N \\\\ & = 2 \\left[ \\frac{M-N}{M+N} + \\frac{1}{3} \\left( \\frac{M-N}{M+N} \\right)^3 + \\frac{1}{5} \\left( \\frac{M-N}{M+N} \\right)^5 + \\cdots \\right], \\\\ \\end{align} a series which is convergent for all positive values of $M$ and $N$; and it is always possible to choose $M$ and $N$ so as to make it converge rapidly. Placing $M = 2$ and $N = 1$ in (E), we get $\\log 2 = 2 \\left[ \\frac{1}{3} + \\frac{1}{3}\\cdot\\frac{1}{3^3} + \\frac{1}{5}\\cdot\\frac{1}{3^5} + \\frac{1}{7}\\cdot\\frac{1}{3^7} + \\cdots \\right] = 0.69314718 \\cdots.$ [ Since $\\log N = \\log 1 = 0$, and $\\tfrac{M-N}{M+N} = \\tfrac{1}{3}$.] Placing $M = 3$ and $N= 2$ in (E), we get $\\log 3 = \\log 2 + 2 \\left[ \\frac{1}{5} + \\frac{1}{3}\\cdot\\frac{1}{5^8} + \\frac{1}{5}\\cdot\\frac{1}{5^5} + \\cdots \\right] = 1.09861229 \\cdots.$ It is only necessary to compute the logarithms of prime numbers in this way, the logarithms of composite numbers being then found by using theorems 7-10, §1. Thus \\begin{align} \\log 8 &= \\log 2{^3} &= 3 \\log 2 &= 2.07944154 \\cdots, \\\\ \\log 6 &= \\log 3 &+ \\log 2 &= 1.79175947 \\cdots. \\\\ \\end{align} All the above are Napierian or natural", "\\log p_n\\}^{\\frac{1}{n}}, \\end{eqnarray} since the arithmetic mean of unequal positive numbers is always greater than their geometric mean. Hence $$\\frac{1}{n} \\{\\log p_1 + \\log p_2 + \\cdots + \\log p_n + \\log N \\} \\gt \\{\\log p_1 \\log p_2 \\cdots \\log p_n \\cdot d(N)\\}^{\\frac{1}{n}}.$$ In other words $$d(N) \\lt \\frac{\\left\\{ \\frac{1}{n} \\log (p_1 p_2 p_3 \\cdots p_n N)\\right\\}^n} {\\log p_1 \\log p_2 \\log p_3 \\cdots \\log p_n},$$ for all values of $N$ whose prime divisors are $p_1, p_2, p_3, \\ldots , p_n$. 3. Next let us consider the case in which only the number of prime divisors of $N$ is known. Let $$N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \\cdots p_n^{a_n},$$ where $n$ is a given number; and let $$N' = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3} \\cdots p^{a_n},$$ where $p$ is the natural $n$th prime. Then it is evident that $$N' \\leq N;$$ and $$d(N') = d(N).$$ But $$d(N') \\lt \\frac{\\left\\{\\frac{1}{n} \\log (2\\cdot 3\\cdot 5\\cdots p\\cdot N')\\right\\}^n} {\\log 2 \\log 3 \\log 5 \\cdots \\log p} ,$$ by virtue of (5). It follows from (6) to (8) that, if $p$ be the natural $n$th prime, then $$d(N) \\lt \\frac{\\left\\{\\frac{1}{n} \\log(2\\cdot 3\\cdot 5 \\cdots p\\cdot N)\\right\\}^n}{\\log 2 \\log 3 \\log 5 \\cdots \\log p},$$ for all values of $N$ having $n$ prime divisors. 4. Finally, let us consider the case in", "# Tag Info You have caught an instance of the infamous off-by-one error in that popular textbook whose name we shall not mention again. To repeat, it is correct that \"the cost $c_1=1$, $\\Phi_0=0$\", \"$num_1=size_1=1$ $\\implies$ $\\Phi_1 = 2\\cdot1-1 =1$\" and \" $\\hat{c_1}=$ $c_1+\\Phi_1-\\Phi_0$ $=2$\". It is incorrect to state that $\\widehat c_i=... 3$n! = n \\cdot (n-1) \\cdot ... \\cdot 1 \\ge n \\cdot (n-1) \\cdot ... \\cdot (n/2) \\ge (n/2)^{(n/2)}$so$\\log(n!)≥c\\cdot n\\cdot\\log n$for$c \\ge 1/2$2 ... further simplify$(n/2)\\log(n/2)$-->$(n/2)\\log(n^{-2})$, ... This is wrong. In fact, for$n\\ge 3$,$n^{\\log_3 2}\\ge 2, so \\begin{align} (n/2)\\log(n/2)&=(n/2)(\\log n-\\log2)\\\\ &\\ge (n/2)\\cdot\\left(1-\\log_3 2\\right)\\log n\\\\ &=\\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\log n, \\end{align} and\\log(n!)\\ge \\left(1-\\log_3 2\\right)/2\\cdot n\\cdot \\... Since your definition of $\\epsilon$-closure isn't really a definition, it is impossible to prove anything using it. Instead, let me use the following definition: the $\\epsilon$-closure of a set $S \\subseteq Q$ consists of all states $x \\in Q$ which are reachable from a state in $S$ by a (possibly empty) $\\epsilon$-path (which is a path consisting of $\\... 2 It seems that$\\dagger\\dagger$is consistent with$\\|$. You just need to pick a constant$c$that is larger than or equal to the constant$\\gamma$hidden in the$O(1)$notation in the definition of$T(n)$for$n < 140$(i.e., the line marked with$\\ddagger$). Then, for any$n \\in \\{1, \\dots, 139\\}$, you have$T(n) \\le \\gamma \\le c \\le cn$, as desired. 1", "x} {2 \\cdotp 30258 \\, 50929 \\, 94 \\ldots}$ ### Base $2$ to Base $10$ $\\log_{10} x = \\left({\\lg x}\\right) \\left({\\log_{10} 2}\\right) = 0 \\cdotp 30102 \\, 99956 \\, 63981 \\, 19521 \\, 37389 \\ldots \\lg x$ ### Base $10$ to Base $2$ $\\lg x = \\dfrac {\\log_{10} x} {\\log_{10} 2} = \\dfrac {\\log_{10} x} {0 \\cdotp 30102 \\, 99956 \\, 63981 \\, 19521 \\, 37389 \\ldots}$ ### Base $2$ to Base $8$ $\\log_8 x = \\dfrac {\\lg x} 3$", "if $$N = (2\\cdot 3\\cdot 5 \\cdots p_1) \\times (2\\cdot 3\\cdot 5 \\cdots p_2),$$ where $p_1$ is the largest prime not greater than $2^x$, and $p_2$ the largest prime not greater than $(\\frac{3}{2})^x$, it is easily seen that $d(N)$ attains the order (236): and $N$ is not highly composite. VI. Special Forms of $N$. 46. In $\\S\\S$-38 we have indirectly solved the following problem: to find the relations which must hold between $x_1, x_2, x_3, \\ldots$ in order that $$2^{\\pi(x_1)} \\cdot (\\myfrac{3}{2})^{x(x_2)} \\cdot (\\myfrac{4}{3})^{\\pi (x_3)} \\cdots$$ may be a maximum, when it is given that $$\\vartheta (x_1) + \\vartheta (x_2) + \\vartheta (x_3) + \\cdots$$ is a fixed number. The relations which we obtained are $$\\frac{\\log 2}{\\log x_1} = \\frac{\\log (\\myfrac{3}{2})}{\\log x_2} = \\frac{\\log (\\myfrac{4}{3})}{\\log x_3} = \\cdots .$$ This suggests the following more general problem. If $N$ is an integer of the form $$e^{\\displaystyle{ c_1 \\vartheta (x_1) + c_2 \\vartheta (x_2) + c_3\\vartheta (x_3) + \\cdots}},$$ where $c_1, c_2, c_3, \\ldots$ are any given positive integers, it is required to find the nature of $N$, that is to say the relations which hold between $x_1, x_2, x_3, \\ldots,$ when $d(N)$ is of maximum order. From (242) we see that \\begin{equation*} d(N) = (1 + c_1)^{\\pi(x_1)} \\left(\\frac{1+c_1 + c_2}{1+c_1}\\right)^{\\pi(x_2)} \\left(\\frac{1+c_1 + c_2 + c_3}{1 + c_1 + c_2}\\right)^{\\pi(x_3)}", "have to perform comparisons on all but the last head element. Additionally, there are two recursive calls on subarrays of half the size. We therefore have these cases: \\begin{align}T(1) &= 0 \\\\T(n) &= n – 1 + 2 \\cdot T(\\tfrac{n}{2})\\end{align} In order to derive a closed-form solution, we do some expansion: \\begin{align}T(n) &= n – 1 + 2 \\cdot T(\\tfrac{n}{2}) \\\\ &= n – 1 + 2 \\cdot (\\tfrac{n}{2} – 1 + 2 \\cdot T(\\tfrac{n}{4})) \\\\ &= n – 1 + n – 2 + 4 \\cdot T(\\tfrac{n}{4})) \\\\ &= n – 1 + n – 2 + 4 \\cdot (\\tfrac{n}{4} – 1 + 2 \\cdot T(\\tfrac{n}{4}))) \\\\ &= n – 1 + n – 2 + n – 4 + 8 \\cdot T(\\tfrac{n}{8}) \\\\ &= (n + n + n) – (1 + 2 + 4) + 8 \\cdot T(\\tfrac{n}{8}) \\\\\\end{align} After $i$ expansions, we have this expression: $$T(n) = in – \\sum_{j = 0}^{i – 1} 2^j + 2^i \\cdot T(\\tfrac{n}{2^i})$$ To get rid of the self-reference, we solve for the base case: \\begin{align}1 &= \\frac{n}{2^i} \\\\2^i &= n \\\\\\log_2 2^i &= \\log_2 n \\\\i \\cdot \\log_2 2 &= \\log_2 n \\\\i &= \\log_2 n \\\\\\end{align} We’ve seen this result before. Maybe we should accept it as an axiom from here on out. We substitute this", "that do not contain the digit $$0$$ is $$9^k$$. Thus, \\begin{align*} & f(n) = \\sum_{k = 1}^n 9^k = \\frac{9 (9^n - 1)}{8}, \\\\ & g(n) = 10^n - 1 - f(n) = 10^n - \\frac{9 (9^n - 1)}{8} - 1. \\end{align*} Thus, the largest $$n$$-digit natural number occurs at the $$f(n)$$th position in $$S_1$$. For $$n = 1$$, $$f(n) = 9$$ and $$g(n) = 0$$. For $$n > 1$$, since the $$f(n)$$th number of $$S_1$$ is the largest $$n$$-digit natural number, it must be greater than the $$g(n)$$th number of $$S_2$$. Therefore, when $$f(n) \\leq g(n)$$, the $$f(n)$$th number of $$S_1$$ is greater than the $$f(n)$$th number of $$S_2$$. We can show that $$f(n) \\leq g(n)$$ when $$n \\geq 8$$. \\begin{align*} n \\geq 8 & \\implies n \\gt \\frac{\\log 9 - \\log 4}{\\log 10 - \\log 9} \\\\ & \\implies \\log 9 - \\log 4 \\lt n (\\log 10 - \\log 9) \\\\ & \\implies \\log \\frac{9}{4} \\lt n \\log \\frac{10}{9} \\\\ & \\implies \\frac{9}{4} \\lt \\left(\\frac{10}{9}\\right)^n \\\\ & \\implies 9^n - 1 \\lt \\frac{4}{9} \\cdot 10^n \\\\ & \\implies 2 \\cdot \\frac{9}{8} \\cdot (9^n - 1) \\lt 10^n \\\\ & \\implies \\frac{9}{8} \\cdot (9^n - 1) \\lt 10^n - \\frac{9}{8} \\cdot (9^n - 1) \\\\ & \\implies f(n) \\lt g(n) + 1 \\\\ & \\implies f(n)", "how you got from (3) to (4) where $n$ reappeared. Perhaps you just los $n$ in (3) but otherwise you had it on paper. Let us try again: \\begin{align*} T(n) &= n + (3/2) n + (3/2)^2 n + \\cdots \\end{align*} But when does sum end? (This is what created your mistake.) It ends when we get to the base case, which is $T(1)$. After $k$ unfoldings we have $T(n/2^k)$ so we will reach $T(1)$ when $k = \\log_2 n$. Thus, we continue like this: \\begin{align*} T(n) &= n + (3/2) n + (3/2)^2 n + \\cdots + (3/2)^{\\log_2 n} T(1) \\\\ &= \\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i n\\\\ &= n \\cdot \\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i. \\end{align*} It remains to add the sum. It is a geometric series, so we can look up the formula $\\sum_{i=0}^k a^i = (a^{k+1} - 1)/(a - 1)$ and get \\begin{align*} {\\textstyle \\sum_{i=0}^{\\log_2 n} (3/2)^i} &= \\frac{(3/2)^{1 +\\log_2 n} - 1}{3/2 - 1} \\\\ &= 2 \\cdot ((3/2) \\cdot (3/2)^{\\log_2 n} - 1) \\\\ &= 3 \\cdot (3/2)^{\\log_2 n} - 2 \\\\ &= 3 \\cdot (n^{\\log_2 (3/2)}) - 2 \\\\ &\\in \\Theta(n^{\\log_2 (3/2)}) \\approx \\Theta(n^{0.58}) \\end{align*} The final result is this sum multiplied with $n$, so it is $$\\Theta(n \\cdot n^{\\log_2(3/2)}) = \\Theta(n^{\\log_2 2 + \\log_2 (3/2)}) = \\Theta(n^{\\log_2 3}),$$ assuming I can still", "0$$ for some class $$j \\in \\{1, \\dots, K\\}$$, we set $$p_j \\log_b(p_j) = 0$$ because $\\lim_{p_j \\to 0} \\left[ p_j \\log_b(p_j) \\right] = 0$ In this way, we never have to directly evaluate $$\\log_b(0)$$. Let’s see if this definition of entropy is consistent with our intuition. For example, let us consider a pure node; that is, a node in which all objects share the same class. In this case $$p_k = 1$$, and we have only one class so \\begin{align*} H(\\text{node}) &= - p_1 \\cdot log_2 (p_1) \\\\ &= -(1) \\cdot \\log_2(1) \\\\ &= -1 \\cdot 0 = 0 \\end{align*} Now, for our second example above, we have $$K = 2$$ classes with $$p_1 = 9/12$$ and $$p_2 = 3/12$$, so that \\begin{align*} H(\\text{node}) &= - p_1 \\log_2 (p1) - p_2 \\log_2 (p_2) \\\\ &= - \\frac{9}{12} \\log_2\\left( \\frac{9}{12} \\right) - \\frac{3}{12} \\log_2\\left( \\frac{3}{12} \\right) \\approx 0.8112 \\end{align*} In a node where there is a 50-50 split, we have \\begin{align*} H(\\text{node}) &= - p_1 \\log_2 (p1) - p_2 \\log_2 (p_2) \\\\ &= - \\frac{1}{2} \\log_2\\left( \\frac{1}{2} \\right) - \\frac{1}{2} \\log_2\\left( \\frac{1}{2} \\right) = 1 \\end{align*} The following figure depicts several examples of sets and their entropy values. ### 34.2.3 Gini Impurity The second type of measure of heterogeneity (or impurity) that we’ll describe is the so-called Gini" ]

[ "be $$p(x)=a_n(x-b_1)\\dots(x-b_n).$$ Otherwise we have counterexamples: take a monic polynomial with all the roots in $$\\mathbb{Z}$$ and divide it by an integer number greater than $$1$$. • Note that, given the roots of a polynomial, then its factorization is $a_n(x-b_1)\\dots(x-b_n)$. So the result is correct also when the coefficient $a_n$ is an integer. Feb 6, 2019 at 6:58", "by choosing $x^{2^k}$ in the factors where bit $k$ is $1$ in the binary representation of $m$ and choosing $1$ in the factors where bit $k$ is $0$. All coefficients in the factors are $1$ so the coefficient of $x^m$ in the product is $1$. Therefore, $$(1+x)(1+x^2)(1+x^4)\\dots(1+x^{2^{n-1}})=1+x+x^2+x^3+\\dots+x^{2^n-1}$$ -", "k[x]$. Assume $f(t) = 0$ for all $t \\in k$. Assume to the contrary that $f$ is not the zero polynomial. Then $f$ is a polynomial of degree $n \\geq 1$. Choose distinct elements $z_1, z_2, z_3, \\dots, z_{n+1}$ of $k$. By repeated use of the linear factor theorem, we know that \\$f(x) = (x-z_1)(x-z_2)(... Only top voted, non community-wiki answers of a minimum length are eligible", "$p(x1) - p(x2) = a_{n}(x_{1}^n - x_{2}^n) + a_{n-1}(x_{1}^{n-1} - x_{2}^{n-1}) + \\ldots + a_{1}(x_{1} - x_{2})$. To me, it seems that the difference is divisible by $(x_{1} - x_{2})$ regardless of the coefficients? – Looft Jan 9 '13 at 0:35 As a polynomial in two variables, yes. But that does not lead to any problem. There certainly is a polynomial with rational coefficients taking on any specified rational values $b_1,\\dots,b_k$ at any specified points $a_1,\\dots,a_k$. – André Nicolas Jan 9 '13 at 0:53", "\\dots, n$. Then every element of order $\\le n$ is a zero of the polynomial $x^N-1=0$. Or else, more wastefully, you can use the polynomial $\\prod_{k=1}^n (x^k-1)$. - Hint: by pigeonholing, if there are infinitely many elements of order $\\le n$ then for some $k\\le n$ there are infinitely many elements of order $k,\\:$ hence $x^k - 1$ has infinitely many roots $\\:\\Rightarrow\\Leftarrow$ -", "18:55 • It looks to me like the question intends to be factored as $(1+x_1)(1+x_2)\\dots(1+x_k)$, which has an easy $O(k)$ computation, but I'm just guessing. – Thomas Andrews Apr 19 '14 at 18:57 • @ThomasAndrews thank you! That means it's not the right polynomial I'm after. – Shine On You Crazy Diamond Apr 19 '14 at 22:19", "terms of $f_{n-1}$, which you can in turn write as a polynomial in $x_2$. You'll get something with $x_1$ and $x_2$ in there, and hopefully you'll be able to factor it to get the desired results. You should end up with new coefficients for which you can do something similar, working in all the $x_i$ and then hopefully being able to factor. Alternatively, treat $f_n(x_1,\\, \\dots ,\\, x_n)$ as the single-variable polynomial $f_{n-x_2,\\dots ,x_n}(x_1)$. If $f_n$ is supposed to factor as it suggests in part c), then so should this new polynomial. But this polynomial is of degree n-1, and the problem suggests that you can pull out n-1 factors, so if what you are supposed to prove is correct, then the new polynomial should be a constant multiplied by the product of those factors. This doesn't seem exactly right because your coefficients are matrices, but the roots you're finding are whatever $x_1$ is, but you may want to try this approach. Take the product of those factors, and multiply it by $g_{n-1}$ and see if you end up with $f_n$.", "real coefficients and no real roots of the form $x^4+A x^2+ B$ can be factored as follows ( $p,s$ real ) $\\left(x^2-s x + p\\right)\\left(x^2+s\\text{ }x +p\\right)$ multply out to get $x^4+\\left(2p-s^2\\right)x^2+p^2$ and compare coefficients and note that $B$ must be positive otherwise the polynomial has real roots", "a polynomial: $p(x)=\\sum_{i=1}^{n-1}a_i(x-k)q_i(x)+a_n(x^n-k^n)$ Now $x^n-k^n=(x-k)x^{n-1}+kx^{n-1}-k^n$ but $kx^{n-1}-k^n$ is a polynomial of degree $n-1$ with root $k$ and hence $(x-k)$ is a factor: $x^n-k^n=(x-k)x^{n-1}+(x-k)h(x)$ Hence $p(x)=(x-k)\\lbrack x^{n-1}+h(x)+\\sum_{i=1}^{n-1}a_iq_i(x)\\rbrack$ Hence $(x-k)$ a factor $\\bullet$", "a polynomial Q such that : $f(x)=(x-k)^2 Q(x)$ It is obvious that $f(k)=0$ Now if you differentiate, use the product rule and you'll have : $f'(k)=(x-k)^2 Q'(x)+2(x-k)Q(x)=(x-k)[(x-k)Q'(x)+2Q(x)]$ Is it clear that $f'(k)=0$ ? In fact, it is true for higher powers : if f has a factor $(x-k)^n$, then $f(k)=f'(k)=f^{(2)}(k)=\\dots=f^{(n-1)}(k)=0$ • January 15th 2009, 07:34 AM", "# Kronecker method A method for factoring a polynomial with rational coefficients into irreducible factors over the field of rational numbers; proposed in 1882 by L. Kronecker [1]. Let $d$ be a common denominator of the coefficients of a polynomial $\\phi(x)$. Then $f(x)=d\\phi(x)$ is a polynomial with integer coefficients; moreover, any factorization of $\\phi(x)$ into irreducible factors with rational coefficients leads to a factorization of $f(x)$ into irreducible factors with integer coefficients, whose factors differ from the corresponding factors of $\\phi(x)$ only by constant factors, and vice versa. Let $f(x)$ be of degree $n>0$ and let $k$ be the largest natural number $\\leq n/2$. If $f(x)=g(x)h(x)$ is a factorization of $f(x)$ into factors with integer coefficients, where the degree of $g(x)$ does not exceed that of $h(x)$, then the degree of $g(x)$ is at most $k$. Assigning to $x$ any $k+1$ distinct integer values $x=c_i$, $i=1,\\dots,k+1$, one obtains equalities $$f(c_i)=g(c_i)h(c_i),\\quad i=1,\\dots,k+1,$$ where $g(c_i)$ and $h(c_i)$ are integers. Thus, $g(c_i)$ divides $f(c_i)$. Choosing arbitrary divisors $d_i$ of the numbers $f(c_i)$, one obtains $$g(c_i)=d_i,\\quad i=1,\\dots,k+1.$$ The polynomial $g(x)$ may now be determined from these equalities using the Lagrange interpolation formula, or, more simply, by the equations for the coefficients. It is then checked whether the polynomial $g(x)$ found in this way divides $f(x)$. This construction and the subsequent test are", "therefore answer is \\$\\prod_{i = 1}^{m}(k_i + 1). Sum of factors = (p_1^0 + p_1^1 + p_1^2 + \\dots + p_1^{k_1}) \\times (p_2^0 + p_2^1 + \\dots + p_2^{k_2}) \\times \\dots = \\prod_{i = 1}^{m}(\\sum_{j = 1}^{k_i}(p_i^j)) = \\prod_{i = 1}^{m}\\frac{p_i^{k_i + 1} - 1}{p_i - 1} Proof: Think of the polynomial (p_1^0 + p_1^1 + p_1^2 + \\dots + p_1^{k_1}) \\times (p_2^0 + p_2^1 + \\dots + p_2^{k_2}) \\times \\dots = \\prod_{i = 1}^{m}(\\sum_{j = 1}^{k_i}(p_i^j)). Then try to actually manually work out each term of this expression. Notice that each term will be of the form p_1^{a_1} \\times p_2^{a_2} \\times \\dots \\times p_m^{a_m}, where this term is actually one of the factors of N. Also notice that each factor will come at least once and not be duplicated, i.e it will come exactly once. What this means is that that all the terms of this expression are all the factors. And all the terms are under addition, so they are giving us the same only. That is why the expression is equal to the sum of all the factors (including 1 and N) 4a. Given x + y + z = 10, x >= 0, y >= 0 and z >= 0, how many integer solutions, i.e triplets of integer (x, y, z)'s are there satisfying", "• PhoenixFire Factoring a cubic polynomial. $x^3-108x^2+3564x-34992$ I haven't needed to factor a cubic in a while and have forgotten how. Any help? I know the answers, just can't get them. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "$P'(k)$ factored with just one real root, but this was more convenience than necessity; there are plenty of other ways to show that $P(k)$ is never equal to $\\pm1$ for integer values of $k$.", "a must be greater than 1, and x must be greater than 0. The largest a^n not greater than x , where n is a non-negative integer. a must be greater than 1, and x must not be less than 1. Next integer greater than or equal to n that can be written as$prod k_i^$for integers$p_1$,$p_2$, etcetera, for factors$k_i$in factors . The method that accepts a tuple requires Julia 1.6 or later. Take the inverse of n modulo m : y such that$n y = 1 pmod m$, and$div(y,m) = 0$. This will throw an error if$m = 0$, or if$gcd(n,m) eq 1$. Compute the number of digits in integer n written in base base ( base must not be in [-1, 0, 1] ), optionally padded with zeros to a specified size (the result will never be less than pad ). Multiply x and y , giving the result as a larger type. Evaluate the polynomial$sum_k x^ p[k]$for the coefficients p[1] , p[2] , . that is, the coefficients are given in ascending order by power of x . Loops are unrolled at compile time if the number of coefficients is statically known, i.e. when p is a Tuple . This function generates efficient code using Horner's method if x is real, or using a Goertzel-like [DK62]", "factor the polynomial $p(A)\\text{.}$ Note that in general, this solution takes us into the field of complex numbers, since the roots of a polynomial with real coefficients are sometimes complex numbers—with non-zero imaginary parts. We close this section with one more example to illustrate how quickly we can read off the general solution of a homogeneous advancement operator equation $p(A)f=0\\text{,}$ provided that $p(A)$ is factored.", "getting a sum of exactly $26$ is the coefficient of $x^{26}$ which is ${98 \\over 6^7}$. The probability of the sum being $\\ge x$ will be the sum of the coefficients of $x^k$ where $k \\ge x$. There are various ways of computing the polynomial coefficients depending on what you are trying to achieve. # poly of phi*6, constant term last f = [1 2 2 1 0]; # 7 rolls d = 7; # compute polynomial of (phi*6)^d p = 1; for i = 1:d p = conv(p, f); endfor # scale result so the coefficients are probabilities # constant term last p = p/6^d; If not, you could using something like Neville's algorithm to compute the interpolating polynomial of the points $\\phi^d(0),\\cdots, \\phi^d(4d)$. Another way would be to compute the coefficients by multiplying the polynomials $\\phi^2 = \\phi \\cdot \\phi$, $\\phi^{k+1} = \\phi^k \\cdot \\phi$. This is essentially the convolution of coefficients of the polynomials being multiplied. Aside: My original Octave/Matlab code looked something like f = [1 2 2 1 0]; r = roots(f); d = 7; v = poly(repmat(r',[1, d])); v = v/6^d; however, for $d \\in \\{5,6,7\\}$, the poly call resulted in some small imaginary noise. The conv call avoids this entirely. • How small? d will always be less than seven. –", "where p(a) = 0Then (x – a) is the factor of p(x)Now divide p(x) by (x – a) (p(x))/((x - a))And then we factorise the quotient by splitting the middle termLet us take an exampleInExample 15,We first find x where p(x) = = 1So, (x – 1) is a fact ### Factoring Cubic Polynomials - UC Santa Barbara Factoring Cubic Polynomials March 3, 2016 A cubic polynomial is of the form p(x) = a 3x3 + a 2x2 + a 1x+ a 0: The Fundamental Theorem of Algebra guarantees that if a 0;a 1;a 2;a 3 are all real numbers, then we can factor my polynomial into the form Professional gas sensing solution supplier with over 30-year gas sensor manufacturing experience", "\\pi}{3}$ with k integer Another way is to say that -1 is one cubic root of -1 Therefore $x^3+1$ can be factorized by $x+1$ $x^3+1 = (x+1)(x^2-x+1)$ Then you need to solve $x^2-x+1=0$ , , , , , , , , , , , , , , factoring 7th degree polynomials example with solutions Click on a term to search for related topics.", "C$ and consider the associated polynomial $p(x_1, \\dots, x_n) = \\sum _{i_1 + \\dots + i_n \\le d} a_{i_1, \\dots, i_n} x_1 ^{i_1} \\dots x_n ^{i_n}$. Then $$\\int \\limits _{[-1,1]^n} p(x_1, \\dots, x_n) \\ \\Bbb d x_1 \\dots \\Bbb d x_n = \\int \\limits _{-1} ^1 \\dots \\int \\limits _{-1} ^1 \\sum _{i_1 + \\dots + i_n \\le d} a_{i_1, \\dots, i_n} x_1 ^{i_1} \\dots x_n ^{i_n} \\ \\Bbb d x_1 \\dots \\Bbb d x_n = \\\\ \\sum _{i_1 + \\dots + i_n \\le d} a_{i_1, \\dots, i_n} \\frac {1 + (-1)^{i_1}} {i_1 + 1} \\dots \\frac {1 + (-1)^{i_n}} {i_n + 1} ,$$" ]

[ "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # 9.6: Factoring Special Products Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Factor the difference of two squares. • Factor perfect square trinomials. • Solve quadratic polynomial equation by factoring. ## Introduction When you learned how to multiply binomials we talked about two special products. \\begin{align*}\\text{The Sum and Difference Formula} && (a + b) (a - b ) = a^2 - b^2 \\\\ \\text{The Square of a Binomial Sormula} && (a + b)^2 = a^2 + 2ab + b^2 \\\\ && (a - b)^2 = a^2 - 2ab + b^2\\end{align*} In this section we will learn how to recognize and factor these special products. ## Factor the Difference of Two Squares We use the sum and difference formula to factor a difference of two squares. A difference of two squares can be a quadratic polynomial in this form. \\begin{align*}a^2 - b^2\\end{align*} Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term. \\begin{align*}a^2 - b^2 = (a + b) (a - b)\\end{align*} In these problems, the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. Let’s do some examples of this type. Example 1 Factor the difference of", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # Factorization using Difference of Squares ## Factor binomials that are composed of subtracted perfect square terms Estimated5 minsto complete % Progress Practice Factorization using Difference of Squares MEMORY METER This indicates how strong in your memory this concept is Progress Estimated5 minsto complete % Factorization using Difference of Squares ### Factorization Using Difference of Squares When you learned how to multiply binomials we talked about two special products. \\begin{align*} \\text{The sum and difference formula:} \\quad (a + b)(a - b) & = a^2 - b^2\\\\ \\text{The square of a binomial formulas:} \\qquad \\quad \\ \\ (a + b)^2 & = a^2 + 2ab + b^2\\\\ (a - b)^2 & = a^2 - 2ab + b^2\\end{align*} In this section we’ll learn how to recognize and factor these special products. #### Factor the Difference of Two Squares We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any quadratic polynomial in the form \\begin{align*}a^2 - b^2\\end{align*}, where \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} can be variables, constants, or just about anything else. The factors of \\begin{align*}a^2 - b^2\\end{align*} are always \\begin{align*}(a + b)(a - b)\\end{align*}; the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. #### Factoring the Difference of Squares 1. Factor", "this, the polynomial factors into perfect squares. \\begin{align*}a^2 + 2ab + b^2 & = (a + b)^2 \\\\ a^2 - 2ab + b^2 & = (a - b)^2\\end{align*} In these problems, the key is figuring out what the a and b terms are. Let’s do some examples of this type. Example 5 Factor the following perfect square trinomials. a) \\begin{align*}x^2 + 8x + 16\\end{align*} b) \\begin{align*}x^2 - 4x + 4\\end{align*} c) \\begin{align*}x^2 + 14x + 49\\end{align*} Solution a) \\begin{align*}x^2 + 8x + 16\\end{align*} The first step is to recognize that this expression is actually perfect square trinomials. 1. Check that the first term and the last term are perfect squares. They are indeed because we can re-write: \\begin{align*}x^2 + 8x + 16 \\quad \\quad \\text{as} \\quad \\quad x^2 + 8x + 4^2.\\end{align*} 2. Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them. \\begin{align*}x^2 + 8x + 16 \\quad \\quad \\text{as} \\quad \\quad x^2 + 2 \\cdot 4 \\cdot x + 4^2\\end{align*} This means we can factor \\begin{align*}x^2 + 8x + 16\\end{align*} as \\begin{align*}(x + 4)^2\\end{align*}. We can check to see if this is correct by multiplying \\begin{align*}(x + 4) (x + 4).\\end{align*} \\begin{align*}& \\quad \\quad \\quad \\ x", "\\begin{align*}4x^3-8x^2-x+2 = 0\\end{align*}. Each of these problems is done in the same way: Group the first two and last two terms together, pull out any common factors, what is inside the parenthesis is the same, factor it out, then determine if either factor can be factored further. 1. \\begin{align*}& x^3+7x^2-2x-14\\\\ & x^2(x+7)-2(x+7)\\\\ & (x+7)(x^2-2)\\end{align*} \\begin{align*}x^2-2\\end{align*} is not a difference of squares because 2 is not a square number. Therefore, this cannot be factored further. 2. \\begin{align*}& 2x^4-5x^3+2x-5\\\\ & x^3(2x-5)+1(2x-5)\\\\ & (2x-5)(x^3+1) \\quad \\ \\ \\text{Sum of cubes, factor further}.\\\\ & (2x-5)(x+1)(x^2+x+1)\\end{align*} 3. Factor by grouping. \\begin{align*}4x^3-8x^2-x+2 &= 0\\\\ 4x^2(x-2)-1(x-2) &= 0\\\\ (x-2)(4x^2-1) &= 0\\\\ (x-2)(2x-1)(2x+1) &= 0\\\\ x & = 2, \\frac{1}{2}, -\\frac{1}{2}\\end{align*} Practice Factor the following polynomials using factoring by grouping. Factor each polynomial completely. 1. \\begin{align*}x^3-4x^2+3x-12\\end{align*} 2. \\begin{align*}x^3+6x^2-9x-54\\end{align*} 3. \\begin{align*}3x^3-4x^2+15x-20\\end{align*} 4. \\begin{align*} 2x^4-3x^3-16x+24\\end{align*} 5. \\begin{align*}4x^3+4x^2-25x-25\\end{align*} 6. \\begin{align*}4x^3+18x^2-10x-45\\end{align*} 7. \\begin{align*}24x^4-40x^3+81x-135\\end{align*} 8. \\begin{align*}15x^3+6x^2-10x-4\\end{align*} 9. \\begin{align*}4x^3+5x^2-100x-125\\end{align*} 10. \\begin{align*}3x^3-2x^2+12x-8\\end{align*} Find all the real-number solutions of the polynomials below. 1. \\begin{align*}9x^3-54x^2-4x+24 = 0\\end{align*} 2. \\begin{align*}x^4+3x^3-27x-81 = 0\\end{align*} 3. \\begin{align*}x^3-2x^2-4x+8 = 0\\end{align*} 4. Challenge Find ALL the solutions of \\begin{align*}x^6-9x^4-x^2+9 = 0\\end{align*}. 5. Challenge Find ALL the solutions of \\begin{align*}x^3+3x^2+16x+48 = 0\\end{align*}. Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English Factor by Grouping Factoring by grouping is a method of factoring a polynomial by factoring common monomials", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # 9.6: Factoring Special Products Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Factor the difference of two squares. • Factor perfect square trinomials. • Solve quadratic polynomial equation by factoring. ## Introduction When you learned how to multiply binomials we talked about two special products. The sum and difference formula:(a+b)(ab)The square of a binomial formulas: (a+b)2(ab)2=a2b2=a2+2ab+b2=a22ab+b2\\begin{align*} \\text{The sum and difference formula:} \\quad (a + b)(a - b) & = a^2 - b^2\\\\ \\text{The square of a binomial formulas:} \\qquad \\quad \\ \\ (a + b)^2 & = a^2 + 2ab + b^2\\\\ (a - b)^2 & = a^2 - 2ab + b^2\\end{align*} In this section we’ll learn how to recognize and factor these special products. ## Factor the Difference of Two Squares We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any quadratic polynomial in the form \\begin{align*}a^2 - b^2\\end{align*}, where \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} can be variables, constants, or just about anything else. The factors of \\begin{align*}a^2 - b^2\\end{align*} are always \\begin{align*}(a + b)(a - b)\\end{align*}; the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. Example 1 Factor the difference of squares: a) \\begin{align*}x^2 - 9\\end{align*} b) \\begin{align*}x^2 - 100\\end{align*} c) \\begin{align*}x^2 -", "like this, the polynomial factors into the sum and difference of the square root of each term. \\begin{align*}a^2-b^2=(a+b)(a-b)\\end{align*} In these problems, the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. Let’s do some examples of this type. #### Example A Factor the difference of squares. (a) \\begin{align*}x^2-9\\end{align*} (b) \\begin{align*}x^2y^2-1\\end{align*} Solution: (a) Rewrite \\begin{align*}x^2-9\\end{align*} as \\begin{align*}x^2-3^2\\end{align*}. Now it is obvious that it is a difference of squares. We substitute the values of \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} in the Sum and Difference Formula: \\begin{align*}(x+3)(x-3)\\end{align*} The answer is \\begin{align*}x^2-9=(x+3)(x-3)\\end{align*}. (b) Rewrite \\begin{align*}x^2y^2-1\\end{align*} as \\begin{align*}(xy)^2-1^2\\end{align*}. This factors as \\begin{align*}(xy+1)(xy-1)\\end{align*}. Factoring Perfect Square Trinomials A perfect square trinomial has the form: \\begin{align*}a^2+2ab+b^2 \\qquad \\text{or} \\qquad a^2-2ab+b^2\\end{align*} The factored form of a perfect square trinomial has the form: \\begin{align*}&&(a+b)^2 \\ if \\ a^2+2(ab)+b^2\\\\ \\text{And}\\\\ &&(a-b)^2 \\ if \\ a^2-2(ab)+b^2\\end{align*} In these problems, the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. Let’s do some examples of this type. #### Example B Factor \\begin{align*}x^2+8x+16\\end{align*}. Solution: Check that the first term and the last term are perfect squares. \\begin{align*}x^2+8x+16 \\qquad \\text{as} \\qquad x^2+8x+4^2.\\end{align*} Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them. \\begin{align*}x^2+8x+16 \\qquad \\text{as} \\qquad x^2+2 \\cdot 4 \\cdot x+4^2\\end{align*}", "method of basic factoring is the difference of squares. It is recognizable as one square monomial being subtracted from another square monomial. To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant: \\begin{align*}-\\frac{1}{2}(x^2-3)(x^2-4)\\end{align*} Many students may recognize that \\begin{align*}x^2-4\\end{align*} immediately factors by the difference of squares method to be \\begin{align*}(x-2)(x+2)\\end{align*}. This problem asks for more because sometimes the difference of squares method can be applied to expressions like \\begin{align*}x^2-3\\end{align*} where each term is not a perfect square. The number 3 actually is a square. \\begin{align*}3=\\left(\\sqrt{3}\\right)^2\\end{align*} So the fully factored expression would be: \\begin{align*}-\\frac{1}{2} \\left(x- \\sqrt{3}\\right)\\left(x+ \\sqrt{3}\\right)(x-2)(x+2)\\end{align*} ### Examples #### Example 1 Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don'at contain perfect squares and why it would be useful. One reason why it might be useful to completely factor an expression like \\begin{align*}-\\frac{1}{2}(x^4-7x^2+12)\\end{align*} into linear factors is if you wanted to find the roots of the function \\begin{align*}f(x)=-\\frac{1}{2}(x^4-7x^2+12)\\end{align*}. The roots are \\begin{align*}x=\\pm \\sqrt{3}, \\pm 2\\end{align*} You should recognize that \\begin{align*}x^2-3\\end{align*} can still be thought of as the difference of perfect squares because the number 3 can be expressed as \\begin{align*}\\left(\\sqrt{3}\\right)^2\\end{align*}. Rewriting the number 3 to fit a factoring pattern that you already know is an example of", "+0 # Factoring 0 74 1 Factor x^8 - y^8 as the product of four polynomials of degree 1, each of which has a positive coefficient of x. Jun 22, 2021 #1 +785 +1 \\begin{align*} x^8 - y^8 &= (x^4)^2 - (y^4)^2 \\\\ &= (x^4 - y^4)(x^4 + y^4) \\\\ &= ((x^2)^2 - (y^2)^2)(x^4 + y^4) \\\\ &= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \\\\ &= (x-y)(x+y)(x^2 + y^2)(x^4 + y^4) \\end{align*} Can you take it from here? Hint: Manipulate $x^2 + y^2$ into $(x+y)^2 - 2xy$ and do the same thing with $x^4 + y^4.$ You can substitute your equation for $x^2 + y^2$ and do some further manipulation and you are done. If you have degrees $>1,$ you can square root it. Jun 22, 2021", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy. # 9.6: Factoring Special Products Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives At the end of this lesson, students will be able to: • Factor the difference of two squares. • Factor perfect square trinomials. • Solve quadratic polynomial equation by factoring. ## Vocabulary Terms introduced in this lesson: recognizing special product factoring perfect square trinomials quadratic polynomial equations double root ## Teaching Strategies and Tips Emphasize that students are reversing the special-products formulas introduced three lessons ago. Have students use the vocabulary: • a2b2\\begin{align*}a^2 - b^2\\end{align*} is a difference of squares. • (a+b)(ab)\\begin{align*}(a + b)(a - b)\\end{align*} is the product of a sum and difference. • a2+2ab+b2\\begin{align*}a^2 + 2ab + b^2\\end{align*} and a22ab+b2\\begin{align*}a^2 - 2ab + b^2\\end{align*} are perfect square trinomials. • (a+b)2\\begin{align*}(a + b)^2\\end{align*} and (ab)2\\begin{align*}(a - b)^2\\end{align*} are squares of binomials. The key to factoring special products is recognizing the special form, but also determining what a\\begin{align*}a\\end{align*} and b\\begin{align*}b\\end{align*} are. • Recognizing perfect integer squares, for example, may be difficult to some students. Suggest that students break numbers down into", "method of basic factoring is the difference of squares. It is recognizable as one square monomial being subtracted from another square monomial. To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant: 12(x23)(x24)\\begin{align*}-\\frac{1}{2}(x^2-3)(x^2-4)\\end{align*} Many students may recognize that x24\\begin{align*}x^2-4\\end{align*} immediately factors by the difference of squares method to be (x2)(x+2)\\begin{align*}(x-2)(x+2)\\end{align*}. This problem asks for more because sometimes the difference of squares method can be applied to expressions like x23\\begin{align*}x^2-3\\end{align*} where each term is not a perfect square. The number 3 actually is a square. 3=(3)2\\begin{align*}3=\\left(\\sqrt{3}\\right)^2\\end{align*} So the fully factored expression would be: 12(x3)(x+3)(x2)(x+2)\\begin{align*}-\\frac{1}{2} \\left(x- \\sqrt{3}\\right)\\left(x+ \\sqrt{3}\\right)(x-2)(x+2)\\end{align*} ### Examples #### Example 1 Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don'at contain perfect squares and why it would be useful. One reason why it might be useful to completely factor an expression like 12(x47x2+12)\\begin{align*}-\\frac{1}{2}(x^4-7x^2+12)\\end{align*} into linear factors is if you wanted to find the roots of the function f(x)=12(x47x2+12)\\begin{align*}f(x)=-\\frac{1}{2}(x^4-7x^2+12)\\end{align*}. The roots are x=±3,±2\\begin{align*}x=\\pm \\sqrt{3}, \\pm 2\\end{align*} You should recognize that x23\\begin{align*}x^2-3\\end{align*} can still be thought of as the difference of perfect squares because the number 3 can be expressed as (3)2\\begin{align*}\\left(\\sqrt{3}\\right)^2\\end{align*}. Rewriting the number 3 to fit a factoring pattern that you already know is an example of", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # 9.7: Factoring Polynomials Completely Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Factor out a common binomial. • Factor by grouping. • Factor a quadratic trinomial where \\begin{align*}a \\neq 1\\end{align*}. • Solve real world problems using polynomial equations. ## Introduction We say that a polynomial is factored completely when we can’t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely: • Factor all common monomials first. • Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas. • If there are no special products, factor using the methods we learned in the previous sections. • Look at each factor and see if any of these can be factored further. Example 1 Factor the following polynomials completely. a) \\begin{align*}6x^2-30x+24\\end{align*} b) \\begin{align*}2x^2-8\\end{align*} c) \\begin{align*}x^3+6x^2+9x\\end{align*} Solution a) Factor out the common monomial. In this case 6 can be divided from each term: \\begin{align*}6(x^2-5x-6)\\end{align*} There are no special products. We factor \\begin{align*}x^2-5x+6\\end{align*} as a product of two binomials: \\begin{align*}(x \\ )(x \\ )\\end{align*} The two numbers that multiply to 6 and add to -5 are -2 and -3, so: \\begin{align*}6(x^2-5x+6)=6(x-2)(x-3)\\end{align*} If we look at each factor we see that we can factor", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> You are viewing an older version of this Concept. Go to the latest version. # Factorization using Difference of Squares ## Factor binomials that are composed of subtracted perfect square terms 0% Progress Practice Factorization using Difference of Squares Progress 0% Factorization using Difference of Squares What if you had a quadratic expression like \\begin{align*}9x^2 - 4y^2\\end{align*} in which one square term were subtracted from another? How could you factor that expression? After completing this Concept, you'll be able to factor the difference of two squares like this one. ### Guidance When you learned how to multiply binomials we talked about two special products. In this section we’ll learn how to recognize and factor these special products. Factor the Difference of Two Squares We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any quadratic polynomial in the form \\begin{align*}a^2 - b^2\\end{align*}, where \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} can be variables, constants, or just about anything else. The factors of \\begin{align*}a^2 - b^2\\end{align*} are always \\begin{align*}(a + b)(a - b)\\end{align*}; the key is figuring out what the \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} terms are. #### Example A Factor the difference of squares: a) \\begin{align*}x^2 - 9\\end{align*} b) \\begin{align*}x^2 - 100\\end{align*} c) \\begin{align*}x^2 - 1\\end{align*} Solution", "1\\end{align*} Solution a) Rewrite \\begin{align*}x^2 - 9\\end{align*} as \\begin{align*}x^2 - 3^2\\end{align*}. Now it is obvious that it is a difference of squares. \\begin{align*}\\text{The difference of squares formula is:} && a^2 - b^2 & = (a + b)(a - b)\\\\ \\text{Let’s see how our problem matches with the formula:} && x^2 - 3^2 & = (x + 3)(x - 3)\\\\ \\text{The answer is:} && x^2 - 9 & = (x + 3)(x - 3)\\end{align*} We can check to see if this is correct by multiplying \\begin{align*}(x + 3)(x - 3)\\end{align*}: \\begin{align*}& \\quad \\quad \\ \\ x + 3\\\\ & \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;x - 3}\\\\ & \\quad -3x - 9\\\\ & \\underline{x^2 + 3x\\;\\;\\;\\;\\;\\;}\\\\ & x^2 + 0x - 9\\end{align*} The answer checks out. Note: We could factor this polynomial without recognizing it as a difference of squares. With the methods we learned in the last section we know that a quadratic polynomial factors into the product of two binomials: \\begin{align*}(x\\;\\;\\;\\;)(x\\;\\;\\;\\;)\\end{align*} We need to find two numbers that multiply to -9 and add to 0 (since there is no \\begin{align*}x-\\end{align*}term, that’s the same as if the \\begin{align*}x-\\end{align*}term had a coefficient of 0). We can write -9 as the following products: \\begin{align*}& -9 = -1 \\cdot 9 && \\text{and} && -1 + 9 = 8\\\\ & -9 = 1 \\cdot (-9) && \\text{and}", "3)^2\\end{align*}. Example 2 Factor the following polynomials completely. a) \\begin{align*}-2x^4 + 162\\end{align*} b) \\begin{align*}x^5 - 8x^3 + 16x\\end{align*} Solution a) \\begin{align*}-2x^4 + 162\\end{align*} Factor the common monomial. In this case, factor -2 rather than 2. It is always easier to factor the negative number so that the leading term is positive. \\begin{align*}-2x^4 + 162 = -2(x^4 - 81)\\end{align*} We recognize expression in parenthesis as a difference of squares. We factor and get this result. \\begin{align*}-2(x^2 - 9)(x^2 + 9)\\end{align*} If we look at each factor, we see that the first parenthesis is a difference of squares. We factor and get this answers. \\begin{align*}-2(x + 3) (x - 3) (x^2 + 9)\\end{align*} If we look at each factor, we see that we can factor no more. The answer is \\begin{align*}-2(x + 3) (x - 3) (x^2 + 9)\\end{align*} b) \\begin{align*}x^5 - 8x^3 + 16x\\end{align*} Factor out the common monomial \\begin{align*}x^5 - 8x^3 + 14x = x(x^4 - 8x^2 + 16)\\end{align*}. We recognize \\begin{align*}x^4 - 8x^2 + 16\\end{align*} as a perfect square and we factor it as \\begin{align*} x(x^2 - 4)^2\\end{align*}. We look at each term and recognize that the term in parenthesis is a difference of squares. We factor and get: \\begin{align*}x[(x + 2)^2 (x - 2)]^2 = x(x + 2)^2 (x - 2)^2.\\end{align*} We use square brackets “[”", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> 9.6: Factoring Special Products Difficulty Level: At Grade Created by: CK-12 Learning Objectives At the end of this lesson, students will be able to: • Factor the difference of two squares. • Factor perfect square trinomials. • Solve quadratic polynomial equation by factoring. Vocabulary Terms introduced in this lesson: recognizing special product factoring perfect square trinomials double root Teaching Strategies and Tips Emphasize that students are reversing the special-products formulas introduced three lessons ago. Have students use the vocabulary: • \\begin{align*}a^2 - b^2\\end{align*} is a difference of squares. • \\begin{align*}(a + b)(a - b)\\end{align*} is the product of a sum and difference. • \\begin{align*}a^2 + 2ab + b^2\\end{align*} and \\begin{align*}a^2 - 2ab + b^2\\end{align*} are perfect square trinomials. • \\begin{align*}(a + b)^2\\end{align*} and \\begin{align*}(a - b)^2\\end{align*} are squares of binomials. The key to factoring special products is recognizing the special form, but also determining what \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*} are. • Recognizing perfect integer squares, for example, may be difficult to some students. Suggest that students break numbers down into prime factorization first. See Example 2. Remind students to pull out \\begin{align*}-1\\end{align*} and/or the GCF in a polynomial before attempting to factor it. This simplifies the task dramatically. Error Troubleshooting General Tip: Remind students to check their solutions by substituting each in", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> Factoring Completely Sum or difference with higher powers, factoring algorithm, and grouping. Estimated14 minsto complete % Progress Practice Factoring Completely MEMORY METER This indicates how strong in your memory this concept is Progress Estimated14 minsto complete % The difference of perfect squares can be generalized as a factoring technique. By extension, any difference between terms that are raised to an even power like \\begin{align*}a^6-b^6\\end{align*} can be factored using the difference of perfect squares technique. This is because even powers can always be written as perfect squares: \\begin{align*}a^6-b^6=(a^3)^2-(b^3)^2\\end{align*}. What about the sum or difference of terms with matching odd powers? How can those be factored? More Factoring Techniques Factoring a trinomial of the form \\begin{align*}ax^2+bx+c\\end{align*} is much more difficult when \\begin{align*}a \\neq 1\\end{align*}. There are four techniques that can be used to factor such expressions. Guess and Check The educated guess and check method can be time consuming but if the first and last coefficient only have a few factors, there are a finite number of possibilities. Take the expression: \\begin{align*}6x^2-13x-28\\end{align*} The 6 can be factored into the following four pairs: 1, 6 2, 3 -1, -6 -2, -3 The -28 can be factored into the following twelve pairs: 1, -28 or -28, 1 -1, 28 or 28, -1 2, -14 or", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> 9.7: Factoring Polynomials Completely Difficulty Level: At Grade Created by: CK-12 We say that a polynomial is factored completely when we factor as much as we can and we are unable to factor any more. Here are some suggestions that you should follow to make sure that you factor completely. \\begin{align*}\\checkmark\\end{align*} Factor all common monomials first. \\begin{align*}\\checkmark\\end{align*} Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas. \\begin{align*}\\checkmark\\end{align*} If there are no special products, factor using the methods we learned in the previous sections. \\begin{align*}\\checkmark\\end{align*} Look at each factor and see if any of these can be factored further. Example 1: Factor the following polynomials completely. (a) 2x28\\begin{align*}2x^2-8\\end{align*} (b) x3+6x2+9x\\begin{align*}x^3+6x^2+9x\\end{align*} Solution: (a) Look for the common monomial factor. 2x28=2(x24)\\begin{align*}2x^2-8=2(x^2-4)\\end{align*}. Recognize x24\\begin{align*}x^2-4\\end{align*} as a difference of squares. We factor 2(x24)=2(x+2)(x2)\\begin{align*}2(x^2-4)=2(x+2)(x-2)\\end{align*}. If we look at each factor we see that we can't factor anything else. The answer is 2(x+2)(x2)\\begin{align*}2(x+2)(x-2)\\end{align*}. (b) Recognize this as a perfect square and factor as x(x+3)2\\begin{align*}x(x+3)^2\\end{align*}. If we look at each factor we see that we can't factor anything else. The answer is x(x+3)2\\begin{align*}x(x+3)^2\\end{align*}. Factoring Common Binomials The first step in the factoring process is often factoring the common monomials from a polynomial. Sometimes polynomials have common terms that are binomials. For example,", "The solution to $$3{x}^{2} + 2x – 1 = 0$$ is $$x = -1$$ or $$x = \\cfrac{1}{3}$$. ## Example ### Question Find the roots: $0 = -2{x}^{2} + 4x – 2$ ### Divide the equation by common factor $$-\\text{2}$$ \\begin{align*} -2{x}^{2} + 4x – 2 & = 0 \\\\ {x}^{2} – 2x + 1 & = 0 \\end{align*} ### Factorise \\begin{align*} (x – 1)(x – 1) & = 0 \\\\ {(x – 1)}^{2} & = 0 \\end{align*} ### The quadratic is a perfect square This is an example of a special situation in which there is only one solution to the quadratic equation because both factors are the same. \\begin{align*} x – 1 & = 0 \\\\ \\therefore x & = 1 \\end{align*} ### Check the answer by substituting back into the original equation The solution to $$0 = -2{x}^{2} + 4x – 2$$ is $$x = 1$$.", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> You are viewing an older version of this Concept. Go to the latest version. # Sum and Difference of Cubes ## Use formulas to factor pairs of added or subtracted perfect cubes 0% Progress Practice Sum and Difference of Cubes Progress 0% Factorization of Special Cubics Factor the following cubic polynomial: 375x3+648\\begin{align*}375x^3+648\\end{align*}. ### Guidance While many cubics cannot easily be factored, there are two special cases that can be factored quickly. These special cases are the sum of perfect cubes and the difference of perfect cubes. • Factoring the sum of two cubes follows this pattern: x3+y3=(x+y)(x2xy+y2) • Factoring the difference of two cubes follows this pattern: x3y3=(xy)(x2+xy+y2) #### Example A Factor: x3+27\\begin{align*}x^3+27\\end{align*}. Solution: This is the sum of two cubes and uses the factoring pattern: x3+y3=(x+y)(x2xy+y2)\\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\\end{align*}. x3+33=(x+3)(x23x+9)\\begin{align*}x^3+3^3=(x+3)(x^2-3x+9)\\end{align*}. #### Example B Factor: x3343\\begin{align*}x^3-343\\end{align*}. Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\\end{align*}. x373=(x7)(x2+7x+49)\\begin{align*}x^3-7^3=(x-7)(x^2+7x+49)\\end{align*}. #### Example C Factor: 64x31\\begin{align*}64x^3-1\\end{align*}. Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\\end{align*}. (4x)313=(4x1)(16x2+4x+1)\\begin{align*}(4x)^3-1^3=(4x-1)(16x^2+4x+1)\\end{align*}. #### Concept Problem Revisited Factor the following cubic polynomial: 375x3+648\\begin{align*}375x^3+648\\end{align*}. First you need to recognize that there is a common factor of 3\\begin{align*}3\\end{align*}. 375x3+648=3(125x3+216)\\begin{align*}375x^3+648=3(125x^3+216)\\end{align*} Notice that the result is the sum of two cubes. Therefore, the factoring pattern is x3+y3=(x+y)(x2xy+y2)\\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\\end{align*}. 375x3+648=3(5x+6)(25x230x+36)\\begin{align*}375x^3 +648 =", "-3 (with multiplicity 3), \\begin{align*}-1, \\sqrt{3}i, -\\sqrt{3}i\\end{align*} 3. 5 (with multiplicity 2), -1 (with multiplicity 2), \\begin{align*}2i, -2i\\end{align*} 4. \\begin{align*}i, -i, \\sqrt{2}i, -\\sqrt{2}i\\end{align*} For each polynomial, factor into its linear factorization and state all of its roots. 5. \\begin{align*}f(x)=4x^2+x-3\\end{align*} 6. \\begin{align*}g(x)=x^4-1\\end{align*} 7. \\begin{align*}h(x)=x^3+5x^2\\end{align*} 8. \\begin{align*}j(x)=x^4-7x^3+17x^2-17x+6\\end{align*} 9. \\begin{align*}k(x)=x^3-64\\end{align*} 10. \\begin{align*}m(x)=x^6-2x^4-9x^2+18\\end{align*} 11. \\begin{align*}n(x)=2x^3-8x^2-3x-12\\end{align*} 12. \\begin{align*}p(x)=-x^3+2x^2-25x+50\\end{align*} 13. How can you tell the number of roots that a polynomial has from its equation? 14. Explain the meaning of the term “multiplicity”. 15. A polynomial with real coefficients has one root that is \\begin{align*}\\sqrt{3}i \\end{align*}. What other root(s) must the polynomial have? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes" ]

[ "other words, if there is no non-constant polynomial $Q$ such that $Q^2$ divides $P$. (This definition can naturally be extended to elements in any UFD, so in particular one also often speaks of squarefree integers.) The first step to factor $A$ is to obtain a squarefree factorisation of $A$. This means that we wish to obtain polynomials $A_1,A_2,\\dots,A_k \\in \\mathbf{F}_p[X]$ such that the $A_i$ are all squarefree and relatively prime, and $A = A_1A_2^2 \\dots A_k^k.$ One way to obtain such a factorisation is to group together all the irreducible factors of $A$ which have the same exponent. For example, consider the polynomial \\begin{align*} A &= X^{13} + 2X^{12} + X^{11} + X^{10} + 3X^9 + X^8 + X^7 \\\\ &\\ \\ {}+{} 4X^6 + X^5 + 2X^3 + 4X^2 + 4X + 1 \\in \\mathbf{F}_5[X]. \\end{align*} This polynomial factors into $A = (X+1) (X+2)^2 (X+3)^2 (X^2+X+1) (X^3+X+1)^2,$ and so in our squarefree factorisation, we will have \\begin{align*} A_1 &= (X+1) (X^2+X+1) \\\\ &= X^3 + 2X^2 + 2X + 1, \\end{align*} and \\begin{align*} A_2 &= (X+2) (X+3) (X^3+X+1) \\\\ &= X^5 + 2X^3 + X^2 + X + 1. \\end{align*} Since the $A_i$ are products of distinct irreducible polynomials, they are obviously squarefree and relatively prime. Of course, we do not know in advance what the irreducible", "back to the pet shop and buy a bigger one. ### Vocabulary Cubic Polynomial A cubic polynomial is a polynomial of degree equal to 3. For example \\begin{align*}8x^3+2x^2-5x-7\\end{align*} is a cubic polynomial. Distributive Property The distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, the distributive property says that \\begin{align*}2(x+5)=2x+10\\end{align*}. Factor To factor means to rewrite an expression as a product of other expressions. These resulting expressions are called the factors of the original expression. Factor Completely To factor completely means to factor an expression until none of its factors can be factored any further. Greatest Common Factor The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly) each of the terms of the polynomial. ### Guided Practice Factor each of the following polynomials by grouping. 1. Factor the following polynomial by grouping: \\begin{align*}y^3-4y^2-4y+16\\end{align*}. 2. Factor the following polynomial by grouping: \\begin{align*}3x^3-4x^2-3x+4\\end{align*}. 3. Factor the following polynomial by grouping: \\begin{align*}e^3+3e^2-4e-12\\end{align*}. 1. Here are the steps: 2. Here are the steps: 3. Here are the steps: ### Practice Factor the following cubic polynomials by grouping. 1. \\begin{align*}x^3-3x^2-36x+108\\end{align*} 2. \\begin{align*}e^3-3e^2-81e+243\\end{align*} 3. \\begin{align*}x^3-10x^2-49x+490\\end{align*} 4. \\begin{align*}y^3-7y^2-5y+35\\end{align*} 5. \\begin{align*}x^3+9x^2+3x+27\\end{align*} 6. \\begin{align*}3x^3+x^2-3x-1\\end{align*} 7. \\begin{align*}5s^3-6s^2-45s+54\\end{align*} 8.", "8x(x^2+3x-4)\\end{align*} The resulting quadratic can be factored further into (x+4)(x1)\\begin{align*}(x+4)(x-1)\\end{align*}. Your final answer is 8x(x+4)(x1)\\begin{align*}8x(x+4)(x-1)\\end{align*}. ### Vocabulary Cubic Polynomial A cubic polynomial is a polynomial of degree equal to 3. For example 8x3+2x25x7\\begin{align*}8x^3+2x^2-5x-7\\end{align*} is a cubic polynomial. Distributive Property The distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, the distributive property says that 2(x+5)=2x+10\\begin{align*}2(x+5)=2x+10\\end{align*}. Factor To factor means to rewrite an expression as a product of other expressions. These resulting expressions are called the factors of the original expression. Factor Completely To factor completely means to factor an expression until none of its factors can be factored any further. Greatest Common Factor The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly) each of the terms of the polynomial. ### Guided Practice Factor each of the following polynomials completely. 1. 9w3+12w\\begin{align*}9w^3+12w\\end{align*}. 2. y3+4y2+4y\\begin{align*}y^3+4y^2+4y\\end{align*}. 3. 2t310t2+8t\\begin{align*}2t^3-10t^2+8t\\end{align*}. Answers: 1. The common factor is 3w\\begin{align*}3w\\end{align*}. Therefore, 9w3+12w=3w(3w2+4)\\begin{align*}9w^3+12w=3w(3w^2+4)\\end{align*}. The resulting quadratic cannot be factored any further, so your answer is 3w(3w2+4)\\begin{align*}3w(3w^2+4)\\end{align*}. 2. The common factor is y\\begin{align*}y\\end{align*}. Therefore, y3+4y2+4y=y(y2+4y+4)\\begin{align*}y^3+4y^2+4y=y(y^2+4y+4)\\end{align*}. The resulting quadratic can be factored into (y+2)(y+2)\\begin{align*}(y+2)(y+2)\\end{align*} or (y+2)2\\begin{align*}(y+2)^2\\end{align*}. Your answer is y(y+2)2\\begin{align*}y(y+2)^2\\end{align*}. 3. The common factor is 2t\\begin{align*}2t\\end{align*}. Therefore, 2t310t2+8t=2t(t25t+4)\\begin{align*}2t^3-10t^2+8t=2t(t^2-5t+4)\\end{align*}. The", "$$(K:k)$$-conjugate) was only shown. (The other factor does not have the root $$\\alpha$$, so it is a different factor.) Consier the above relation as a product of polynomials in $$k[X]$$. Then we also have its conjugate polynomial as a factor. So from \\begin{aligned} X^4-2X^2+2 & =\\Big(X^2+(a+b\\sqrt 2)X+(c+d\\sqrt 2)\\Big)\\cdot\\Big(\\dots\\Big) \\\\ & =\\Big(X^2+(a-b\\sqrt 2)X+(c-d\\sqrt 2)\\Big)\\cdot\\overline{\\Big(\\dots\\Big)} \\end{aligned} we get (second factor is different, unique factorization) \\begin{aligned} X^4-2X^2+2 =&\\ \\Big(X^2+(a+b\\sqrt 2)X+(c+d\\sqrt 2)\\Big) \\\\ \\cdot&\\ \\Big(X^2+(a-b\\sqrt 2)X+(c-d\\sqrt 2)\\Big)\\ , \\end{aligned} then we multiply, look at the term in $$X^3$$, get $$a=0$$, then at the term in $$X^1$$, get $$d=0$$, so our factorization is $$X^4-2X^2+2 = \\Big(X^2+b\\sqrt 2 X + c\\Big) \\Big(X^2-b\\sqrt 2 X + c\\Big)\\ ,$$ so $$c^2=2$$. Contradiction. This negates (1) in the OP, so (2) is out of charge. To have a quick structure description of the Galois closure $$L$$ of $$K=\\Bbb Q(\\alpha)$$, let us start a parallel construction. Let $$u=\\zeta_8$$ be the (cyclotomic) unit with minimal polynomial $$X^4+1$$ over $$\\Bbb Q$$, and to fix ideas, we will fix it with the complex image $$\\frac 1{\\sqrt 2}(1+i)$$ in the first quadrant. Its conjugates are $$u, u^3, u^5, u^7$$, with complex images $$\\frac 1{\\sqrt 2}(\\pm 1\\pm i)$$. It is clear that $$M=\\Bbb Q(u)=\\Bbb Q(\\zeta_8)=\\Bbb Q(\\sqrt{-1},\\sqrt 2)$$ is (the cyclotomic field of order $$8$$, degree $$4$$ over $$\\Bbb Q$$,) Galois over $$\\Bbb Q$$", "is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\\mathbb Q}[x]$, then it has a zero in ${\\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \\begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \\end{align*} where each factor is in ${\\mathbb Z}[x]$ by Gauss's Lemma. Hence, \\begin{align*} a + c & = - 2\\\\ ac + b + d & = 0\\\\ ad + bc & = 1\\\\ bd & = 1. \\end{align*} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \\begin{equation*}ad + bc = b( a + c ) = 1.\\end{equation*} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\\mathbb Q}$. ##### Example17.18 The polynomial \\begin{equation*}f(x) =", "is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\\mathbb Q}[x]$, then it has a zero in ${\\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \\begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \\end{align*} where each factor is in ${\\mathbb Z}[x]$ by Gauss's Lemma. Hence, \\begin{align*} a + c & = - 2\\\\ ac + b + d & = 0\\\\ ad + bc & = 1\\\\ bd & = 1. \\end{align*} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \\begin{equation*}ad + bc = b( a + c ) = 1.\\end{equation*} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\\mathbb Q}$. ##### Example17.18 The polynomial \\begin{equation*}f(x) =", "the GCF for this polynomial is \\begin{align*}9xy\\end{align*} Step 2: Divide out the GCF out of each term of the polynomial \\begin{align*}27x^3y+18x^2y^2+9xy^3=9xy(3x^2+2xy+y^2)\\end{align*} Summary Remember that when you factor expressions such as polynomials by taking out a common factor, the first step is to find the greatest common factor for all of the terms in the polynomial. Once you have found this common factor, you must divide each of the terms of the polynomial by this common factor. The common factor can be a number, a variable, or a combination of both. This means that you need to look at both the numbers in the terms of the polynomial and the variables of the polynomial. Problem Set Find the common factors of the following: 1. \\begin{align*}2x(x-5)+7(x-5)\\end{align*} 2. \\begin{align*}4x(x-3)+5(x-3)\\end{align*} 3. \\begin{align*}3x^2(e+4)-5(e+4)\\end{align*} 4. \\begin{align*}8x^2(c-3)-7(c-3)\\end{align*} 5. \\begin{align*}ax(x-b)+c(x-b)\\end{align*} Factor the following polynomial: 1. \\begin{align*}7x^2 + 14\\end{align*} 2. \\begin{align*}9c^2+3\\end{align*} 3. \\begin{align*}8a^2+4a\\end{align*} 4. \\begin{align*}16x^2+24y^2\\end{align*} 5. \\begin{align*}2x^2-12x+8\\end{align*} Factor the following polynomial: 1. \\begin{align*}32w^2x+16xy+8x^2\\end{align*} 2. \\begin{align*}12abc+6bcd+24acd\\end{align*} 3. \\begin{align*}15x^2y-10x^2y^2+25x^2y\\end{align*} 4. \\begin{align*}12a^2b-18ab^2-24a^2b^2\\end{align*} 5. \\begin{align*}4s^3t^2-16s^2t^3+12st^2-24st^3\\end{align*} Find the common factors... 1. \\begin{align*}2x(x-5)+7(x-5)\\end{align*} Step 1: Identify the GCF \\begin{align*}2x(x-5)+7(x-5)\\end{align*} Notice that \\begin{align*}(x - 5)\\end{align*} is common in both terms, therefore \\begin{align*}(x - 5)\\end{align*} is the common factor. So the GCF for this expression is \\begin{align*}(x - 5)\\end{align*} Step 2: Divide out the GCF out of each term of the expression \\begin{align*}2x(x-5)+7(x-5)=(2x+7)(x-5)\\end{align*} 1. \\begin{align*}3x^2(e+4)-5(e+4)\\end{align*}", "and factor out a Greatest Common Factor: \\begin{align*} x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\\\ & = (x^2+1)(x-3) \\end{align*} And voila, the polynomial is at least partially factored. Over $\\mathbb{R}$, we cannot factor further, since $-1$ has no square root. Next, now that we have exhausted factoring over $\\mathbb{R}$, maybe factoring can continue over $\\mathbb{F}_5$. Indeed, here we have \\begin{align*} x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\\\ & = (x^2+1)(x-3)\\\\ & = (x^2-4)(x-3)\\\\ & = (x-2)(x+2)(x-3) \\end{align*} Lastly, it's nice to choose residues from the same neighborhood, so \\begin{align*} x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\\\ & = (x^2+1)(x-3)\\\\ & = (x^2-4)(x-3)\\\\ & = (x-2)(x+2)(x-3)\\\\ & = (x-2)(x-3)(x-3)\\\\ & = (x-2)(x-3)^2 \\end{align*} - I'll write down the exact things I'm doing, but to follow you should pull out a sheet of paper and do the work itself. If you know the rational root test, then you should use the rational root test first. Checking quickly through $\\pm 1, \\pm 3$ we see that $3$ is a factor. Carry out the division by $x-3$ and we are left with a quadratic. Using the quadratic formula, we see that there are no more real roots. So the polynomial factors as $(x-3)(x-i)(x+i)$ over $\\mathbb{C}$, or $(x-3)(x^2 + 1)$ over $\\mathbb{R}$. For the second, we of course actually have the same polynomial and so we", "# Exercise 2.4 Polynomials NCERT Solutions Class 9 Go back to 'Polynomials' ## Chapter 2 Ex.2.4 Question 1 Determine which of the following polynomials has $$(x + 1)$$ a factor: (i) \\begin{align}{x}^{3}+{x}^{2}+x+1\\end{align} (ii) \\begin{align}{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1\\end{align} (iii) \\begin{align}{{x}^{4}}+3{{x}^{3}}+3{{x}^{2}}+x+1\\end{align} (iv) \\begin{align}{{x}^{3}}-{{x}^{2}}-(2+\\sqrt{2})x+\\sqrt{2}\\end{align} ### Solution Reasoning: When a polynomial $$p(x)$$ is divided by $$x-a$$ and if $$p(a) = 0$$ then $$(x-a)$$ is a factor of $$p(x)$$. The root of $$x+1=0$$ is $$-1.$$ Steps: (i) Let \\begin{align} p(x)={{x}^{3}}+{{x}^{2}}+x+1\\end{align} \\begin{align} \\therefore p(-1)&={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+1 \\\\ & =-1+1-1+1=0 \\\\ \\end{align} Since the remainder of $$p\\text{(-1) = 0}$$ , we conclude that $$(x+1)$$ is a factor of $${{x}^{3}}+{{x}^{2}}+x+1$$ . (ii) Let $$p(x)={{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1$$ \\begin{align}\\therefore p( - 1) &\\!=\\! {( - 1)^4}\\!+\\!{( - 1)^3}\\!+\\!{(\\!-\\!1)^2}\\!+\\!(\\!-\\!1)\\!+\\!1\\\\ &= \\not 1 - \\not 1 + \\not 1 - \\not 1+ 1\\\\ &= 1 \\ne 0\\end{align} Since the remainder of $$p( - 1) \\ne 0$$, we conclude that $$(x+1)$$ in not a factor of $$\\,{x^4} + {x^3} + {x^2} + x + 1$$. (iii) Let $$p(x) = {x^4} + 3{x^3} + 3{x^2} + x + 1$$ \\begin{align} \\therefore p( - 1) &\\!=\\!{( -\\!1)^4}\\!+\\!3{(\\!-\\!1)^3}\\!\\!+\\!3{(\\!-\\!1)^2}\\!\\!+\\!( -\\!1)\\!\\!+\\!\\!1 \\\\ &= 1 - 3 + 3 - 1 + 1 \\\\ &= 1 \\ne 0\\end{align} Since the remainder of $$p( - 1) \\ne 0$$ , $$(x+1)$$ is not a factor of $${x^4} + 3{x^3} + 3{x^2} + x + 1$$. (iv)", "\\begin{align*}\\checkmark\\end{align*} Look at each factor and see if any of these can be factored further. #### Example A Factor the following polynomials completely. (a) 2x28\\begin{align*}2x^2-8\\end{align*} (b) x3+6x2+9x\\begin{align*}x^3+6x^2+9x\\end{align*} Solution: (a) Look for the common monomial factor: 2x28=2(x24)\\begin{align*}2x^2-8=2(x^2-4)\\end{align*}. Recognize x24\\begin{align*}x^2-4\\end{align*} as a difference of squares. We factor as follows: 2(x24)=2(x+2)(x2)\\begin{align*}2(x^2-4)=2(x+2)(x-2)\\end{align*}. If we look at each factor we see that we can't factor anything else. The answer is 2(x+2)(x2)\\begin{align*}2(x+2)(x-2)\\end{align*}. (b) Recognize this as a perfect square and factor as x(x+3)2\\begin{align*}x(x+3)^2\\end{align*}. If we look at each factor we see that we can't factor anything else. The answer is x(x+3)2\\begin{align*}x(x+3)^2\\end{align*}. Factoring Common Binomials The first step in the factoring process is often factoring the common monomials from a polynomial. Sometimes polynomials have common terms that are binomials. For example, consider the following expression: x(3x+2)5(3x+2)\\begin{align*}x(3x+2)-5(3x+2)\\end{align*} You can see that the term (3x+2)\\begin{align*}(3x+2)\\end{align*} appears in both terms of the polynomial. This common term can be factored by writing it in front of a set of parentheses. Inside the parentheses, we write all the terms that are left over when we divide them by the common factor. (3x+2)(x5)\\begin{align*}(3x+2)(x-5)\\end{align*} This expression is now completely factored. Let’s look at some more examples. #### Example B Factor 3x(x1)+4(x1)\\begin{align*}3x(x-1)+4(x-1)\\end{align*}. Solution: 3x(x1)+4(x1)\\begin{align*}3x(x-1)+4(x-1)\\end{align*} has a common binomial of (x1)\\begin{align*}(x-1)\\end{align*}. When we factor the common binomial, we get (x1)(3x+4)\\begin{align*}(x-1)(3x+4)\\end{align*}. Solving Real-World Problems Using", "Study S3 A Math Polynomials & Cubic Equations - Geniebook # Polynomials & Cubic Equations • Introduction to Polynomials. • Equality of Polynomials. ## Introduction To Polynomials A polynomial in $$x$$ is an algebraic expression consisting of terms with non-negative integer powers of $$x$$ only. An equation will not be considered a polynomial if the power is negative or fraction. If the variable $$x$$ is in the denominator or in the root, it is not considered a polynomial. Some of the examples of polynomials and non-polynomials are listed below. Polynomials Non-Polynomials \\begin{align*} 2x^3-x+3 \\end{align*} \\begin{align*} 5x^2-3x^{-1}+1 \\end{align*} \\begin{align*} x^7-x^3-5 \\end{align*} \\begin{align*} x^2-x^{^\\tfrac {1}{2}}+3 \\end{align*} \\begin{align*} 2+\\frac{2}{5}x^3 \\end{align*} \\begin{align*} \\frac{2}{x}+5 \\end{align*} $$7$$ \\begin{align*} 2\\sqrt{x}+x-7 \\end{align*} Degree Of Polynomial The degree of a polynomial is the highest power of its individual terms with non-zero coefficients. Polynomial Degree \\begin{align*} 2x^3-x+3 \\end{align*} $$2$$ \\begin{align*} x^7-x^3-5 \\end{align*} $$7$$ \\begin{align*} 2+\\frac{2}{5}x^3 \\end{align*} $$3$$ In the above table, you can see that the degree of any polynomial would be the highest power in the equation. If the equation looks as follows, where there is no variable, then we need to consider it as $$x^0$$. Because the value of $$x^0$$ is $$1$$. Therefore the degree would be $$0$$. Polynomial Degree $$7x^0$$ $$0$$ Evaluating Polynomials To denote polynomials, we can use $$P$$ and $$Q$$ instead of $$f$$", "The Minimal Polynomial of √2 + √3 over Q # The Minimal Polynomial of √2 + √3 over Q Recall from The Minimal Polynomial of an Algebraic Element in a Field Extension page that if $(F, +, *)$ is a field and $(K, +, *)$ is a field extension and if $u \\in K$ is algebraic over $F$ then the minimal polynomial of $u$ over $F$ is defined to be the unique irreducible monic polynomial $p \\in F[x]$ such that: (1) \\begin{align} \\quad p(u) = 0 \\end{align} We will now look at finding the minimal polynomial of $\\sqrt{2} + \\sqrt{3}$ over $\\mathbb{Q}$. Let: (2) \\begin{align} \\quad x = \\sqrt{2} + \\sqrt{3} \\end{align} Then: (3) \\begin{align} \\quad x - \\sqrt{2} &= \\sqrt{3} \\\\ \\quad (x - \\sqrt{2})^2 &= (\\sqrt{3})^2 \\\\ \\quad x^2 - 2\\sqrt{2}x + 2 &= 3 \\\\ \\quad x^2 - 1 &= 2\\sqrt{2}x \\\\ \\quad (x^2 - 1)^2 &= (2\\sqrt{2}x)^2 \\\\ \\quad x^4 - 2x^2 + 1 &= 8x^2 \\\\ \\quad x^4 - 10x^2 + 1 &= 0 \\end{align} Therefore, the polynomial $f(x) = x^4 - 10x^2 + 1$ is such that: (4) \\begin{align} \\quad f(\\sqrt{2} + \\sqrt{3}) = 0 \\end{align} This polynomial is monic. It is a bit technical to show that $f$ is actually the minimal polynomial of $\\sqrt{2} + \\sqrt{3}$ over $\\mathbb{Q}$. We will", "\\cdots - a_0 / \\alpha ). \\end{equation*} Thus $a_0 /\\alpha \\in {\\mathbb Z}$ and so $\\alpha \\mid a_0\\text{.}$ ###### Example17.16 Let $p(x) = x^4 - 2 x^3 + x + 1\\text{.}$ We shall show that $p(x)$ is irreducible over ${\\mathbb Q}[x]\\text{.}$ Assume that $p(x)$ is reducible. Then either $p(x)$ has a linear factor, say $p(x) = (x - \\alpha) q(x)\\text{,}$ where $q(x)$ is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\\mathbb Q}[x]\\text{,}$ then it has a zero in ${\\mathbb Z}\\text{.}$ By Corollary 17.15, any zero must divide 1 and therefore must be $\\pm 1\\text{;}$ however, $p(1) = 1$ and $p(-1)= 3\\text{.}$ Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \\begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \\end{align*} where each factor is in ${\\mathbb Z}[x]$ by Gauss's Lemma. Hence, \\begin{align*} a + c & = - 2\\\\ ac + b + d & = 0\\\\ ad + bc & = 1\\\\ bd & = 1. \\end{align*} Since $bd = 1\\text{,}$ either $b =", "polynomial, P(x), or the previous methods aren't effective, we can apply the factor theorem. • If P(a) = 0 then the polynomial P(x) has a factor of (x – a) NOTE: if (x – a) is a factor of P(x) then a must be a factor of the constant term in P(x). To find a factor of P(x) use trial and error with different factors (positive and negative) of the constant term in P(x) Example (16) \\begin{align} . \\qquad \\text{Find a factor of } P(x) = x^3 - 2x^2 - 3x + 6 \\end{align} The constant term is 6, factors of 6 are: 1, –1, 2, –2, 3, –3, 6, –6 (17) \\begin{align} . \\qquad P(1) = 1 - 2 - 3 + 6 = 2 \\\\ . \\\\ . \\qquad P(-1) = -1 - 2 + 3 + 6 = 6 \\\\ . \\\\ . \\qquad P(2) = 8 - 8 - 6 + 6 = 0 \\end{align} (18) ## Shortcut Method for Factorising Cubics when a factor is known Once the factor theorem has revealed one factor, we can complete the factorisation using the following method For a demonstration of this process in Powerpoint, download the following file (2.4 Mb) Example • Factorise $(x^3 – 2x^2 – 3x + 6)$ given that $(x – 2)$ is", "that results when you divide all the terms of the original polynomial by the greatest common factor. One way to think about this type of factoring is that you are essentially doing the distributive property in reverse. Example A Factor the following binomial: 5a+15\\begin{align*}5a + 15\\end{align*} Solution: Step 1: Identify the GCF. Looking at each of the numbers, you can see that 5 and 15 can both be divided by 5. The GCF for this binomial is 5. Step 2: Divide the GCF out of each term of the binomial: 5a+15=5(a+3)\\begin{align*}5a + 15 = 5(a + 3)\\end{align*} Example B Factor the following polynomial: 4x2+8x2\\begin{align*}4x^2+8x-2\\end{align*} Solution: Step 1: Identify the GCF. Looking at each of the numbers, you can see that 4, 8 and 2 can all be divided by 2. The GCF for this polynomial is 2. Step 2: Divide the GCF out of each term of the polynomial: 4x2+8x2=2(2x2+4x1)\\begin{align*}4x^2+8x-2=2(2x^2+4x-1)\\end{align*} Example C Factor the following polynomial: 3x59x36x2\\begin{align*}3x^5-9x^3-6x^2\\end{align*} Solution: Step 1: Identify the GCF. Looking at each of the terms, you can see that 3, 9 and 6 can all be divided by 3. Also notice that each of the terms has an x2\\begin{align*}x^2\\end{align*} in common. The GCF for this polynomial is 3x2\\begin{align*}3x^2\\end{align*}. Step 2: Divide the GCF out of each term of the polynomial: 3x59x36x2=3x2(x33x2)\\begin{align*}3x^5-9x^3-6x^2=3x^2(x^3-3x-2)\\end{align*} Concept Problem Revisited", "## Finding the Greatest Common Monomial Factor Once we get a polynomial in factored form, it is easier to solve the polynomial equation. But first, we need to learn how to factor. Factoring can take several steps because we want to factor completely so we cannot factor any more. A common factor can be a number, a variable, or a combination of numbers and variables that appear in every term of the polynomial. When a common factor is factored from a polynomial, you divide each term by the common factor. What is left over remains in parentheses. Example 3: Factor: 1. \\begin{align*}15x-25\\end{align*} 2. \\begin{align*}3a+9b+6\\end{align*} Solution: 1. We see that the factor of 5 divides evenly from all terms. \\begin{align*}15x-25=5(3x-5)\\end{align*} 2. We see that the factor of 3 divides evenly from all terms. \\begin{align*}3a+9b+6=3(a+3b+2)\\end{align*} Now we will use examples where different powers can be factored and there is more than one common factor. Example 4: Find the greatest common factor. (a) \\begin{align*}a^3-3a^2+4a\\end{align*} (b) \\begin{align*}5x^3y-15x^2y^2+25xy^3\\end{align*} Solution: (a) Notice that the factor \\begin{align*}a\\end{align*} appears in all terms of \\begin{align*}a^3-3a^2+4a\\end{align*} but each term has a different power of \\begin{align*}a\\end{align*}. The common factor is the lowest power that appears in the expression. In this case the factor is \\begin{align*}a\\end{align*}. Let’s rewrite \\begin{align*}a^3-3a^2+4a=a(a^2)+a(-3a)+a(4)\\end{align*} Factor \\begin{align*}a\\end{align*} to get \\begin{align*}a(a^2-3a+4)\\end{align*} (b) The common factors are \\begin{align*}5xy\\end{align*}.", "are real, the above quadratic is an example. (b) The above polynomial $$p(x)$$ has one irrational coefficient. We try squaring it as follows: \\begin{align} (x-(\\sqrt{2}+i))(x-(\\sqrt{2}-i)) & = 0 \\\\ x^{2}-2 \\sqrt{2} x+3 & = 0 \\\\ x^{2}+3 & = 2 \\sqrt{2} x \\\\ (x^{2}+3)^2 & = 8x^2 \\end{align} So $$q(x) = (x^2+3)^2 - 8x^2$$ is a polynomial with integer coefficients with $$\\sqrt{2}+i$$ as one of its roots. We found a polynomial with degree 4. We show that 4 is the least degree for which such a polynomial exists. Lemma. There is no cubic polynomial $$q(x)$$ with rational coefficients with $$q(\\sqrt{2} + i) = 0$$. Proof. Let $$\\sqrt{2}+i$$, $$\\sqrt{2}-i$$ and $$r$$ be the roots of such a cubic, if it exists. We have: \\begin{align} q(x) & = (x^{2}-2 \\sqrt{2} x+3)(x-r) \\\\ & = x^{3}-(2 \\sqrt{2}-r)x^2 + 2\\sqrt{2}r x - 3r \\end{align} If $$q$$ has only rational coefficients, both $$2\\sqrt{2}-r$$ and $$3r$$ must be rational (Contradiction). $$\\;\\;\\square$$ (c) Let $$f(x)$$ be a polynomial with rational coefficients such that $$f(\\sqrt{2}+i)=0$$. Divide $$f(x)$$ by $$q(x)$$ using long division to get quotient $$a(x)$$ and remainder $$b(x),$$ both polynomials with rational coefficients. $f(x)=q(x)a(x)+b(x)$ Using $$f(\\sqrt{2}+i)=0$$ and $$q(\\sqrt{2}+i)=0$$ in the equation gives $$b(\\sqrt{2}+i)=0$$. Now if the remainder $$b(x)$$ is a nonzero polynomial, then it would have rational coefficients, degree less than 4 and $$\\sqrt{2}+i$$ as", "initial polynomial. In fact, your polynomial has 3 more complex roots, but those wouldn't work. Read more in the fundamental theorem of algebra - Apart from just \"seeing\" the factorization as in the answer of amWhy, one can also use the fact that the polynomial is symmetric to reduce the degree of the problem. After dividing by $X^2$ one obtains a symmetric Laurent-polynomial \\begin{align} X^2+X+2+X^{-1}+X^{-2}=(X+X^{-1})^2+(X+X^{-1})=Z^2+Z=Z\\,(Z+1) \\end{align} Here the resulting quadratic polynomial in $Z=X+X^{-1}$ has a trivial factorization, but any symmetric polynomial of degree 4 reduces in this way to a quadratic polynomial that can then be factorized into linear factors. Each linear factor $Z-a$ then gives a quadratic factor $X^2-aX+1$ of the original polynomial. -", "# Ex.2.4 Q1 Polynomials Solution - NCERT Maths Class 9 Go back to 'Ex.2.4' ## Question Determine which of the following polynomials has $$(x + 1)$$ a factor: (i) \\begin{align}{x}^{3}+{x}^{2}+x+1\\end{align} (ii) \\begin{align}{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1\\end{align} (iii) \\begin{align}{{x}^{4}}+3{{x}^{3}}+3{{x}^{2}}+x+1\\end{align} (iv) \\begin{align}{{x}^{3}}-{{x}^{2}}-(2+\\sqrt{2})x+\\sqrt{2}\\end{align} Video Solution Polynomials Ex 2.4 | Question 1 ## Text Solution Reasoning: When a polynomial $$p(x)$$ is divided by $$x-a$$ and if $$p(a) = 0$$ then $$(x-a)$$ is a factor of $$p(x)$$. The root of $$x+1=0$$ is $$-1.$$ Steps: (i) Let \\begin{align} p(x)={{x}^{3}}+{{x}^{2}}+x+1\\end{align} \\begin{align} \\therefore p(-1)&={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+1 \\\\ & =-1+1-1+1=0 \\\\ \\end{align} Since the remainder of $$p\\text{(-1) = 0}$$ , we conclude that $$(x+1)$$ is a factor of $${{x}^{3}}+{{x}^{2}}+x+1$$ . (ii) Let $$p(x)={{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1$$ \\begin{align}\\therefore p( - 1) &\\!=\\! {( - 1)^4}\\!+\\!{( - 1)^3}\\!+\\!{(\\!-\\!1)^2}\\!+\\!(\\!-\\!1)\\!+\\!1\\\\ &= \\not 1 - \\not 1 + \\not 1 - \\not 1+ 1\\\\ &= 1 \\ne 0\\end{align} Since the remainder of $$p( - 1) \\ne 0$$, we conclude that $$(x+1)$$ in not a factor of $$\\,{x^4} + {x^3} + {x^2} + x + 1$$. (iii) Let $$p(x) = {x^4} + 3{x^3} + 3{x^2} + x + 1$$ \\begin{align} \\therefore p( - 1) &\\!=\\!{( -\\!1)^4}\\!+\\!3{(\\!-\\!1)^3}\\!\\!+\\!3{(\\!-\\!1)^2}\\!\\!+\\!( -\\!1)\\!\\!+\\!\\!1 \\\\ &= 1 - 3 + 3 - 1 + 1 \\\\ &= 1 \\ne 0\\end{align} Since the remainder of $$p( - 1) \\ne 0$$ , $$(x+1)$$ is not a factor of $${x^4} + 3{x^3}", "# Ex.2.4 Q5 Polynomials Solution - NCERT Maths Class 9 ## Question Factorise: (i) \\begin{align}{x^3} - 2{x^2} - x + 2 \\end{align} (ii) \\begin{align}{x^3} - 3{x^2} - 9x - 5\\end{align} (iii) \\begin{align}{x^3} + 13{x^2} + 32x + 20 \\end{align} (iv) \\begin{align}2{y^3} + {y^2} - 2y - 1\\end{align} ## Text Solution Steps: (i) Let $$p(x) = {x^3} - 2{x^2} - x + 2$$ By the factor theorem we know that $$x-a$$ is a factor of $$p(x)$$if $$p(a) = 0.$$ We shall find a factor of $$p(x)$$ by using some trial value of $$x,$$ say $$x = 1.$$ \\begin{align} p(1) &= {(1)^3} - 2{(1)^2} - 1 + 2\\\\ &= 1 - 2 - 1 + 2 = 0\\end{align} Since the remainder of $$p(1) = 0$$ , by factor theorem we can say $$x=1$$ is a factor of $$p(x) = {x^3} - 2{x^2} - x + 2.$$ Now divide $$p(x)$$ by $$x-1$$ using long division, Hence $${x^3}\\!-\\!\\! 2{x^2}\\! -\\! x\\!+\\!\\!2\\!=\\!(x\\!-\\!\\!1)({x^2}\\!-\\!x\\!-\\!\\!2)$$ Now taking $${x^2} - x - 2$$ , find $$2$$ numbers $$p, q$$ such that: (i) $$p + q =$$co-efficient of $$x$$ (ii) $$pq =$$ co-efficient of $${x^2}$$ and the constant term. $$p + q = - 1$$ (co-efficient of $$x$$) $$pq = 1 \\times - 2 = - 2$$(co-efficient of $${x^2}$$ and the constant term.) By trial and error method," ]

[ "# Amazing Polynomial! Algebra Level 5 The polynomial $$P(x) = x^3 + ax^2 + bx + c$$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $$y$$-intercept of the graph of $$y = P(x)$$ is 2, then what is the value of $$-b?$$ ×", "# Problem #906 906 The polynomial $P(x)=x^3+ax^2+bx+c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $y$-intercept of the graph of $y=P(x)$ is 2, what is $b$? $\\text{(A) }-11 \\qquad \\text{(B) }-10 \\qquad \\text{(C) }-9 \\qquad \\text{(D) }1 \\qquad \\text{(E) }5$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"2001 AMC 12 Problems/Problem 19\" ## Problem The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$? $(\\mathrm{A})\\ -11 \\qquad (\\mathrm{B})\\ -10 \\qquad (\\mathrm{C})\\ -9 \\qquad (\\mathrm{D})\\ 1 \\qquad (\\mathrm{E})\\ 5$ ## Solution We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $\\frac{-a}{3}$ is the average of the zeros, so $\\frac{-a}3=-2 \\implies a = 6$. We are also given that the sum of the coefficients is $-2$, so $1+6+b+2 = -2 \\implies b=-11$. So the answer is $\\fbox{A}$.", "The number of zeros the Polynomial $p(x)$ has 18. The graph of the expression $ax^2+bx+c$ is an upward parabola, if 19. The graph of the expression $ax^2+bx+c$ is a downward parabola, if 20. The zeros of the Polynomial $x^2-2x-8$ are 21. The zeros of the Polynomial $x^2-2$ are 22. The zeros of the Polynomial $x^3-2x^2-15x$ are 23. The zero of the Polynomial $p(x)=ax+b$ is 24. The value of the Polynomial $p(x)=x^3-4x^2+9x-2$ when $x=2$ is 25. The value of the Polynomial $p(x)=x^3-x^2+\\sqrt{2}x$ when $x=0$ is 26. The value of the Polynomial $p(x)=2$ when $x=0$ is", "# The graph of the polynomial f(x) = ax2 + bx + c is as shown below Question: The graph of the polynomial $f(x)=a x^{2}+b x+c$ is as shown below (Fig. 2.19). Write the signs of 'a' and $b^{2}-4 a c$. Solution: Clearly, $f(x)=a x^{2}+b x+c$ represent a parabola opening upwards. Therefore, $a>0$ Since the parabola cuts x-axis at two points, this means that the polynomial will have two real solutions Hence $b^{2}-4 a c>0$ Hence $a>0$ and $b^{2}-4 a c>0$", "negative, it's context in the quadratic formula is $\\sqrt{- 8}$ which is imaginary. This means that the two roots (indicated by the power of the polynomial), are a pair of imaginary roots. Consequently, we can reasonably conclude that the polynomial has no real roots, or no real x-intercepts. On another note, let's go to the y-intercept(s). We can say that the graph intercepts the y-axis where $x = 0$, so we can just plug $x = 0$ into the polynomial. $y = - 2 {\\left(0\\right)}^{2} + 4 \\left(0\\right) - 3$ $y = - 3$ With this, we can say (0, -3) is a point and that it is the only y-intercept of the equation.", "# Problem #1307 1307 The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function $f(x)=ax^{2}+bx+c$ are equal. Their common value must also be which of the following? $\\textrm{(A)}\\ \\textrm{the\\ coefficient\\ of\\ }x^{2}~~~ \\textrm{(B)}\\ \\textrm{the\\ coefficient\\ of\\ }x$ $\\textrm{(C)}\\ \\textrm{the\\ y-intercept\\ of\\ the\\ graph\\ of\\ }y=f(x)$ $\\textrm{(D)}\\ \\textrm{one\\ of\\ the\\ x-intercepts\\ of\\ the\\ graph\\ of\\ }y=f(x)$ $\\textrm{(E)}\\ \\textrm{the\\ mean\\ of\\ the\\ x-intercepts\\ of\\ the\\ graph\\ of\\ }y=f(x)$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "## Algebra 2 (1st Edition) $f(x)=8(x+11)(x-3)$. If the x-intercepts of a graph are $a,b$, then the function has the form $f(x)=k(x-a)(x-b)$. The x-intercepts are -11, 3, hence the quadratic function becomes $f(x)=k(x+11)(x-3)$. The point (1,-192) is on the graph, hence if we plug in the values we get $-192=k(12)(-2)$. This yields k=8, hence $f(x)=8(x+11)(x-3)$.", "Determine coefficients $a$ and $b$ such that $p(x)=x^{2}+ax+b$ satisfies $p(1)=-3$ and $p'(1)=-4$. $a =$ $b =$", "? Free Version Moderate Finding Vertex and y-Intercept GRE-G57E5E Compare Quantity A and Quantity B. Based on the quadratic $f(x)=2{ x }^{ 2 }+8x-1$: Quantity A: the $y$-value of the vertex of $f(x)$ Quantity B: the $y$-intercept of $f(x)$ Which of the following statements is true? A Quantity A is greater. B Quantity B is greater. C The two quantities are equal. D The relationship cannot be determined from the information given.", "## Algebra 2 (1st Edition) $f(x)=(x+7)(x+3)$. If the x-intercepts of a graph are $a,b$, then the function has the form of $f(x)=k(x-a)(x-b)$. The x-intercepts are -7, -3; hence, the quadratic function becomes $f(x)=k(x+7)(x+3)$. The point (-1,12) is on the graph Thus: $f(x)=(x+7)(x+3)$.", "# Problem #2087 2087 The graph of the polynomial $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ has five distinct $x$-intercepts, one of which is at $(0,0)$. Which of the following coefficients cannot be zero? $\\textbf{(A)}\\ a \\qquad \\textbf{(B)}\\ b \\qquad \\textbf{(C)}\\ c \\qquad \\textbf{(D)}\\ d \\qquad \\textbf{(E)}\\ e$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "a singular root, $$x=2$$ is a triple root, and $$x=-1$$ is a double root. Step 5: Therefore, placing all this information together we obtain the following graph: ## Concept Check Questions 1. Sketch the following polynomial: $$y=(2x+1)^2(x-1)^2$$ labelling all $$x$$ and $$y$$ intercepts. 2. Sketch the following polynomial: $$y=x^2(2x-3)^3$$ labelling all $$x$$ and $$y$$ intercepts. 3. Sketch the following polynomial: $$y=(3x+1)^3(x+1)(x-1)^2$$ labelling all $$x$$ and $$y$$ intercepts.", "will give you $$(x+r_1)(x+r_2)=x^2+(r_1+r_2)x+r_1r_2$$ So we have two ways of writing the same polynomial, one in terms of its coefficients and one in terms of its factors. Equating the two, we see that $$x^2+(r_1+r_2)x+r_1r_2=x^2+(b/a)x+(c/a)$$ So it's clear that the numbers $r_1$ and $r_2$ must have a sum of $b/a$ and a product of $c/a$. The original polynomial $ax^2+bx+c$ will then be factored into $a(x+r_1)(x+r_2)$, where you are free to distribute the leading coefficient $a$ however you wish between the two factors.", "# If c is a zero of the polynomial P, then(a) P(c)=________.(b) x c is a___________ of P(x) .(c) c is a(n)_______ -intercept of the graph of P. Question If c is a zero of the polynomial P, then (a) P(c)=________. (b) x c is a___________ of P(x) . (c) c is a(n)_______ -intercept of the graph of P.", "don't need to rename it. Since the expressions are equivalent, it's still able to be called P. The formula $x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ gives you the x-intercepts. Because a quadratic graph is symmetric, you can find the x co-ordinate of the vertex by averaging the x-intercepts. The average of the x-intercepts is $-\\frac{b}{2a}$. So this is the formula you use.", "0$$, the graph of $$y = ax + b$$ represents a straight line which intersects the $$x$$ $$-$$ axis exactly at the point $$\\left(\\frac{-b}{a}, 0\\right)$$. The linear polynomial $$ax + b$$, $$a \\neq 0$$, has exactly one zero namely $$\\frac{-b}{a}$$ which is the $$x$$ $$-$$ coordinate of the point where the graph of the polynomial intersects the $$x$$ $$-$$ axis.", "The polynomial of degree 3, $P(x)$, has a root of multiplicity 2 at $x=1$ and a root of multiplicity 1 at $x=-2$. The y-intercept is $y= -1.4$; Find a formula for $P(x)$. $P(x) =$ ,", "## Algebra 2 (1st Edition) In the given polynomial of degree $4$, there are $4$ zeros. By Descartes' Rule, we can notice form the graph that the function has two x-intercepts and the graph crosses the x-axis in both cases. This means that we have $1$ Negative real zero on the right hand and $1$ Positive real zero on the left hand. Hence, Positive real zeros = 1 Negative real zeros= 1 Imaginary zeros = 2", "## Algebra 2 (1st Edition) In the given polynomial of degree $5$, there are $5$ zeros. By Descartes' Rule, we can notice form the graph that the function has one x-intercept, and the graph crosses the x-axis in both cases. This means that we have $1$ real zero which lies on the left side of the origin, so the zero is negative. Hence, Negative real zeros= 1 Imaginary zeros = 4" ]

[ "# Difference between revisions of \"2001 AMC 12 Problems/Problem 19\" ## Problem The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$? $(\\mathrm{A})\\ -11 \\qquad (\\mathrm{B})\\ -10 \\qquad (\\mathrm{C})\\ -9 \\qquad (\\mathrm{D})\\ 1 \\qquad (\\mathrm{E})\\ 5$ ## Solution We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $\\frac{-a}{3}$ is the average of the zeros, so $\\frac{-a}3=-2 \\implies a = 6$. We are also given that the sum of the coefficients is $-2$, so $1+6+b+2 = -2 \\implies b=-11$. So the answer is $\\fbox{A}$.", "0$ $f(1) = 0$, this implies, $x = 1$ is also a root of the given polynomial. For $x = -2$ $f(-2) = 2(-2)^3 + (-2)^2 – 5(-2) + 2$ $= -16 + 4 + 10 + 2$ $= 0$ $f(-2) = 0$, this implies, $x = -2$ is also a root of the given polynomial. Now, Sum of zeros $= \\frac{-b}{a}=\\frac{-1}{2}$. Sum of the zeros of $f(x)=\\frac{1}{2} + 1 – 2 = – \\frac{1}{2}$ Sum of the products of the zeros taken two at a time $=\\frac{c}{a}=\\frac{-5}{2}$. Sum of the products of the zeros taken two at a time$=(\\frac{1}{2} \\times 1) + (1 \\times -2) + (\\frac{1}{2} \\times-2) = \\frac{-5}{ 2}$. Product of the zeros $= \\frac{– d}{a}=\\frac{-2}{2}=-1$. Product of the zeros$=\\frac{}{2} \\times 1 \\times (– 2) = -1$. Hence, the relationship between the zeros and the coefficients is verified. (ii) Let $g(x) = x^3 -4 x^2+ 5x - 2$ Comparing the given polynomial with the standard form of a cubic polynomial, $a=1$, $b=-4$, $c=5$ and $d=-2$ Also, If α is a root of a given polynomial $f(x)$, then $f(α)=0$. Therefore, For $x = 2$ $g(2) = (2)^3 -4(2)^2 +5(2) - 2$ $= 8-4(4)+10-2 = 0$ $=8-16+10-2$ $=18-18$ $=0$ $g(2) = 0$, this implies, $x =2$ is a root of the given polynomial. For $x = 1$", "a sum of $$\\frac{3^2(3\\cdot 11\\cdot 13+5\\cdot 7\\cdot 17)}{2^8}=3^2\\cdot 4=\\boxed{036}.$$ /*Lengthy proof that any two cubic polynomials in $t$ which are equal at 4 values of $t$ are themselves equivalent: Let the two polynomials be $A(t)$ and $B(t)$ and let them be equal at $t=a,b,c,d$. Thus we have $A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0$. Also the polynomial $A(t) - B(t)$ is cubic, but it equals 0 at 4 values of $t$. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into $(t-a)(t-b)(t-c)(t-d)($some nonzero polynomial$)$ which would have a degree greater than or equal to 4, contradicting the statement that $A(t) - B(t)$ is cubic. Because $A(t) - B(t) = 0, A(t)$ and $B(t)$ are equivalent and must be equal for all $t$. Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check. ## Solution 2 As in Solution 1, we have $(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)$ $-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$ $=(t-4)(t-16)(t-36)(t-64)$ Now the coefficient of $t^3$ on both sides", "know that $-1+1+x=x=-\\frac{b}{a}$ and $-1\\cdot 1\\cdot x=-x=-\\frac{d}{a}$. Therefore, $x=\\frac{d}{a}$. Setting the two equations for $x$ equal to each other, $\\frac{d}{a}=-\\frac{b}{a}$. We know that the y-intercept of the polynomial is $d$, so $d=2$. Plugging in for $d$, $\\frac{2}{a}=-\\frac{b}{a}$. Therefore, $b=-2 \\Rightarrow \\boxed{B}$ ### Solution 5 From the graph, we have $f(0)=2$ so $d=2$. Also from the graph, we have $f(1)=a+b+c+2=0$. But we also have from the graph $f(-1)=-a+b-c+2=0$. Summing $f(1)+f(2)$ we get $2b+4=0$ so $b = -2 \\Rightarrow \\mathrm{(B)}$. Solution by franzliszt", "# Finding $a, b, c$ values of a polynomial from a graph. I was doing my homework and I am now stuck on question number 7 which is: The diagram shows the curve with the equation $y = (x + a)(x - b)^2$ where $a$ and $b$ are positive integers. (i) Write down the values of $a$ and $b$, and also of $c$, given that the curve crosses the $y$-axis at $(0, c)$. I have attached a picture of my text book here: I have so far found $b$ like this: $(x - b)^2 = x^2 - 2bx + b^2$ Since $y = 0$ at $x = 1$, $0 = 1 - 2b + b^2$ Solving that quadratic equation gives us $b = 1$. I need to find $a$ and $c$ and I am totally puzzled on how to find them. • For $a$ the solution is the same... only you have a first order equation. Then, to solve for $c$, put $x=0$ in the original equation. – N74 May 5 '16 at 14:20 • @N74 My textbook says a and c are 2 – Hassan Althaf May 5 '16 at 14:21 • You are told the equation is $(x+a)(x-b)^2$ with $a,b$ positive integers. Looking at the graph, the positive root must be at 1, so $b=1$. The", "$$15 - 1 = 14$$ Thus, the x-intercepts are $$\\color{brown}\\boxed{0,1,14}$$ Jun 30, 2022 #1 +2293 +1 Let the function be in the form $$f(x) = ax^3 + bx ^2 + cx + d$$ From $$f(0) = 0$$, we know that $$d = 0$$ From the remaining 3, we can form the system: $$-a + b -c = 15$$ (i) $$a + b + c = 0$$ (ii) $$8a + 4b + 2c = 12$$ (iii) Adding (i) and (ii) gives us $$2b = 15$$, meaning $$b = 7.5$$ Substituting this into (iii) gives us: $$8a + 30 + 2c = 12$$, which means $$8a + 2c = - 18$$ Substituting this into (ii) gives us: $$a + c = -7.5$$ Solving this new system, we find that $$a = -0.5$$ and $$c = -7$$ This means that the polynomial is $$f(x)=-0.5x^3 + 7.5x^2 -7x$$ Now, recall the sum of the roots, by Vieta's, is $$-{b \\over a} = 15$$ Because the polynomial is of degree 3, we know that there must be 3 real roots. But, recall that we know 2 of the roots, (0 and 1), so the remaining root is $$15 - 1 = 14$$ Thus, the x-intercepts are $$\\color{brown}\\boxed{0,1,14}$$ BuilderBoi Jun 30, 2022", "A cubic polynomial has x=-3 as one of its roots and y-intercept 9. If the graph passes through (-2,3) and (1,16), find the equation of the polynomial. 2. You can use the given info to build a system and then solve for the coefficients. The y-intercept is 9, therefore, x=0: $a(0)^{3}+b(0)^{2}+c(0)+d=9$ A root is -3: $a(-3)^{3}+b(-3)^{2}+c(-3)+d=0$ Passes through (-2,3): $a(-2)^{3}+b(-2)^{2}+c(-2)+d=3$ Passes through (1,16): $a(1)^{3}+b(1)^{2}+c(1)+d=16$ Now, you have 4 equations with 4 unknowns: $\\left[\\begin{array}{cccc|c}0&0&0&1&9\\\\-27&9&-3&1&0\\\\-8&4&-2&1&3\\\\1&1&1&1&16\\end{array}\\right]$ Solve the system and you get: $a=\\frac{1}{3}, \\;\\ b=\\frac{5}{3}, \\;\\ c=5, \\;\\ d=9$ Therefore, the polynomial is: $\\boxed{\\frac{1}{3}x^{3}+\\frac{5}{3}x^{2}+5x+9}$ This is one way to approach these. Perhaps try this on your other post regarding the quartic.", "### Home > CCAA8 > Chapter 8 Unit 9 > Lesson CCA: 8.2.4 > Problem8-94 8-94. Examine the graph of $y=2x^2+2x-1$ at right. 1. Estimate the zeros $y=2x^2+2x-1$ from the graph. Where are the $x$-intercepts of this graph? 2. What happens if you try to use the Zero Product Property to find the roots of $2x^2+2x-1=0$? What makes the Zero Product Property difficult to use here?", "graph of the function in the $$xy$$-plane. Therefore, the graph of the function $$g(x)$$, which has four distinct zeros, must have four $$x$$-intercepts. Only the graph in choice C has four $$x$$-intercepts. 42- Choice D is correct $$0.3x=(0.45)×24→x=36→(x-6)^2=(36-6)^2=30^2=900$$ 43- Choice C is correct The weight of 14.6 meters of this rope is: 14.6 × 600 g = 8,760) g 1 kg = 1,000 g, therefore, 8,760 g ÷ 1000 = 8.76 kg 44- Choice B is correct $$y=5a^2 b-3ab^3$$. Plug in the values of $$a$$ and $$b$$ in the equation: $$a = 2$$ and $$b = -1$$ $$y=5(2)^2 (-1)-3(2) (-1)^3=-5(4)+6=-20+6=-14$$ 45- Choice C is correct Use the information provided in the question to draw the shape. Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$ $$362 + 772 = c^2 ⇒ 1,296 + 5,929 = c^2 ⇒ 7,225 = c^2 ⇒ c = 85$$ 46- Choice C is correct The four-term polynomial expression can be factored completely, by grouping, as follows: $$(x^3-7x^2 )+(4x-28)=0$$ $$x^2 (x-7)+4(x-7)=0 (x-7)(x^2+4)=0$$ By the zero-product property, set each factor of the polynomial equal to 0 and solve each resulting equation for $$x$$. This gives $$x=7$$ or $$x=±i\\sqrt{4} =±i2$$, respectively. Because the equation the question asks for the real value of $$x$$ that satisfies the equation, the correct answer is 7. 47- Choice C is correct.", "# The equation and graph of a polynomial are shown below the graph reaches it's maximum when the value of x is 3 what is the y value of this maximum y=-x^2+6x-7? You need to evaluate the polynomial at the maximum $x = 3$, For any value of $x , y = - {x}^{2} + 6 x - 7$, so replacing $x = 3$ we get: $y = - \\left({3}^{2}\\right) + 6 \\cdot 3 - 7 = - 9 + 18 - 7 = 18 - 16 = 2$, so the value of $y$ at the maximum $x = 3$ is $y = 2$ Please note that this doesn't prove that $x = 3$ is the maximum", "then $$(x+1)$$ must be a factor. Thus, we can rewrite the function as $y=(x+1) (ax^2+bx+c)$ Note that in most cases, we may not be given any solutions to a given cubic polynomial. Hence, we need to conduct trial and error to find a value of $$x$$ where the remainder is zero upon solving for $$y$$. Common values of $$x$$ to try are 1, –1, 2, –2, 3 and –3. To find the coefficients $$a$$, $$b$$ and $$c$$ in the quadratic equation $$ax^2+bx+c$$, we must conduct synthetic division as shown below. Synthetic division for Example 6 By looking at the first three numbers in the last row, we obtain the coefficients of the quadratic equation and thus, our given cubic polynomial becomes $y=(x+1)(x^2–x–6)$ We can further factorize the expression $$x^2–x–6$$ as $$(x–3)(x+2)$$. Thus, the complete factorized form of this function is $y=(x+1)(x–3)(x+2)$ Step 2: Setting $$y=0$$, we obtain $(x+1)(x–3)(x+2)=0$ Solving this, we obtain three roots: $x=–2,\\ x=–1,\\ x=3$ Step 3: Plugging $$x=0$$, we obtain $y = (0 + 1) (0 – 3) (0 + 2) = (1) (–3) (2) = –6$ Thus, the y-intercept is $$y = –6$$. Step 4: The graph for this given cubic polynomial is sketched below. Graph for Example 6 The pink points represent the $$x$$-intercepts. The yellow point represents the $$y$$-intercept. Once more, we", "a function corresponds to an $$x$$-intercept of the graph of the function in the $$xy$$-plane. Therefore, the graph of the function $$g(x)$$, which has four distinct zeros, must have four $$x$$-intercepts. Only the graph in choice C has four $$x$$-intercepts. 42- Choice D is correct $$0.3x=(0.45)×24→x=36→(x-6)^2=(36-6)^2=30^2=900$$ 43- Choice C is correct The weight of 14.6 meters of this rope is: 14.6 × 600 g = 8,760) g 1 kg = 1,000 g, therefore, 8,760 g ÷ 1000 = 8.76 kg 44- Choice B is correct $$y=5a^2 b-3ab^3$$. Plug in the values of $$a$$ and $$b$$ in the equation: $$a = 2$$ and $$b = -1$$ $$y=5(2)^2 (-1)-3(2) (-1)^3=-5(4)+6=-20+6=-14$$ 45- Choice C is correct Use the information provided in the question to draw the shape. Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$ $$362 + 772 = c^2 ⇒ 1,296 + 5,929 = c^2 ⇒ 7,225 = c^2 ⇒ c = 85$$ 46- Choice C is correct The four-term polynomial expression can be factored completely, by grouping, as follows: $$(x^3-7x^2 )+(4x-28)=0$$ $$x^2 (x-7)+4(x-7)=0 (x-7)(x^2+4)=0$$ By the zero-product property, set each factor of the polynomial equal to 0 and solve each resulting equation for $$x$$. This gives $$x=7$$ or $$x=±i\\sqrt{4} =±i2$$, respectively. Because the equation the question asks for the real value of $$x$$ that satisfies the equation, the correct", "obtain two turning points for this graph: 1. a maximum value between the roots $$x = –2$$ and $$x = –1$$. This is indicated by the green point. 2. a minimum value between the roots $$x = –1$$ and $$x = 3$$. This is indicated by the blue point. Here is our final example for this discussion. Plot the graph of $y=-(2x–1)(x^2–1).$ Solution Firstly, notice that there is a negative sign before the equation above. This means that the graph will take the shape of an inverted (standard) cubic polynomial graph. In other words, this curve will first open up and then open down. Step 1: We first notice that the binomial $$(x^2–1)$$ is an example of a perfect square binomial. We can use the formula below to factorise quadratic equations of this nature. The Perfect Square Binomial $(a^2-b^2)^2=(a+b)(a-b)$ Using the formula above, we obtain $$(x+1)(x-1)$$. Thus, the complete factored form of this equation is $y = – (2x – 1)(x + 1) (x – 1)$ Step 2: Setting $$y=0$$, we obtain $(2x-1)(x+1)(x-1)=0$ Solving this, we obtain three roots: $x=-1,\\ x=\\frac{1}{2},\\ x=1$ Step 3: Plugging $$x=0$$, we obtain $y=-(2(0)-1)(0+1)(0-1)=-(-1)(1)(-1)=-1$ Thus, the y-intercept is $$y=–1$$. Step 4: The graph for this given cubic polynomial is sketched below. Be careful and remember the negative sign in our initial equation! The", "$x$- intercept. Verify that the zero of this polynomial is the $x$-intercept. Solution: The table below gives the values of $y$ for which $y = 2x -1$ . As two distinct points determine a unique line, it is sufficient to determine the coordinates of two points on this line. \\begin{array}{ccc} x & y=2x-1 & (x,y)\\\\ \\hline 0 & -1 & (0,-1)\\\\ 1 & 1 & (1,1)\\\\ \\hline \\end{array} The graph shows the line $y = 2x - 1$. From the graph, it is clear that the line intersects the $x$-axis at the point $(\\frac{1}{2}, 0)$. Algebraically, we can solve the equation $2x - 1 = 0$ to get $x = \\frac{1}{2}$, thus verifying that the zero of this polynomial is the $x$-intercept. In general, a linear polynomial has at most $1$ zero. This means that the graph has either $0$ or $1$ $x$-intercept. The example above shows the graph of a line that has $1$ $x$-intercept. The graph of a linear polynomial $y = ax + b$ is a rising line (going upward) is the leading coefficient $a \\gt 0$ and a falling line (going downward) if the leading coefficient $a \\lt 0$. A horizontal line has $0$ $x$-intercept. Equation of this line is $y = 0$. That is, it is a $0$ polynomial. A quadratic polynomial has", "A $−\\dfrac{3}{8}$ B $−\\dfrac{1}{2}$ C $−\\dfrac{5}{8}$ D $−\\dfrac{1}{4}$ Question 6 Explanation: The correct answer is (A). The line $y = mx + 3$ has a $y$-intercept of $3$. This means our triangle has a height of $3$: We know the area is $12 \\text{ units}^2$: $\\dfrac{1}{2} \\cdot 3 \\cdot \\text{base} = 12$ $3 \\cdot \\text{base} = 24$ base $= 8$ Our triangle has a base $= 8$. The line goes \"down $3$ and over $8$,\" so the slope is $−\\dfrac{3}{8}$. Question 7 ### What is the maximum $y$ value reached by the function $f(x) = −x^2 + 10x − 21$? A $6$ B $10$ C $8$ D $4$ Question 7 Explanation: The correct answer is (D). The maximum (or minimum) of a parabola will always occur at the $x$ value that is halfway between the real roots ($x$-intercepts) of the function. To find these real roots we need to factor the equation: $f(x) = −1(x^2 − 10x + 21)$ $f(x) = −1(x − 3)(x − 7)$ The real roots occur at $x = 3$ and $x = 7$. The $x$ value halfway between $3$ and $7$ is $x = 5$. When $x = 5$, $f(5) = −(5)^2 + 10(5) − 21 = 4$. The maximum $y$ value is $y = 4$. Question 8 ### $2^{6x} = 8^6$ What", "🙂 I’m new to this site so please let me know if there’s anything wrong with my question. I was trying to find the roots of the polynomial expression $$P(x) = 8x^5 – 14x^4 – 22x^3 +57x^2 – 35x + 6$$. I first found $$(x+2)$$ using the Rational Zeros Theorem and synthetic division. Since the result of $$frac{P(x)}{x+2}$$ is $$Q(x) = 8x^4 – 30x^3 + 38x^2 – 19x + 3$$, I assumed $$-2$$ is a lower bound, because of the alternating signs. And from the graph I can see that’s true, since there are no x-intercepts smaller than $$-2$$. graph of P(x) Then, I found $$(x – 1)$$ as a root. The result $$frac{Q(x)}{x-1} = 8x^3-22x^2+16x-3 = R(x)$$ implies that, according to the theorem, 1 is a lower bound of $$Q(x)$$, but this time, according to the graph, that is not true. I supposed that roots themselves cannot be considered upper and lower bounds. graph of Q(x) However, when dividing $$R(x)$$ by $$(x-1)$$ I get $$8x^2-14x+2 – frac{1}{x-1}$$, which is not a root, but should be a lower bound according to the theorem, and that does not make sense looking at the graph. graph of R(x) Am I doing something wrong? Or maybe this is indeed an exception to the theorem? Also, can you tell me whether roots", "+ 4 x - 21 = 0$$ • Since, ‘a’ is said to be the coefficient against x2, for this reason, a = 1; Remember that ‘a’ is never equal to zero, owing to the significance of x2 it becomes a quadratic. • ‘b’, on the contrary, stands in front of ‘x’ as a coefficient, which makes b = 4. • A constant is something that stands individual in the equation and do not hold any ‘x’ alongside. For this reason, here c = - 21. $$x = \\dfrac{ -4 \\pm \\sqrt{16 - 4 \\cdot 1 \\cdot (-21)}}{ 2 }$$ solving this looks like: $$x = \\dfrac{ -4 \\pm \\sqrt{100}}{ 2 }$$ $$x = \\dfrac{ -4 \\pm 10}{ 2 }$$ $$x = -2 \\pm 5$$ Therefore $$x = 3$$ or $$x = -7$$ Now, let us integrate the values of ‘a’, ‘b’ and ‘c’ in the formula, derived previously. ## Comprehending the Solution: The problem has two solutions and they both demonstrate the intersecting points of the equation, that is, the x – intercept, the point where the x-axis is crisscrossed by a curve. Whilst, preparing a graph the equation $$x^2 + 3 x - 4 = 0$$, can be viewed as: This graph clearly visualizes the solutions derived from a quadratic formula and the intersecting points are", "+0 # Math 0 151 5 +701 http://prntscr.com/lisuzo Nov 15, 2018 #1 +626 0 a) We must let $$f(x)=0$$. $$3x^3+10x^2-13x-20=0$$, and solving we get $$x=-4, -1, \\frac{5}{3}$$ . Therefore, the x-intercepts are $$x=-4, -1, \\frac{5}{3}$$. b) Let $$x=0$$ , and solving we get $$-20$$ . Therefore, the y-intercept is $$y=-20$$. c) This is simple, make a point between $$x=-4$$ and $$x=-1$$, for example, $$(-3, 0)$$. Same for $$x=-1$$ and $$x=5/3$$ , for example, $$(1, 0)$$ . d) As $$x$$ goes toward negative infinity, we know that $$x^3$$ will always decrease, and vice versa. e) Your graph may look something like this: You are very welcome! :P Nov 15, 2018 #2 +701 0 !!! :) Nov 15, 2018 #3 +99736 +1 Let's see how CS might have determined the x intercepts (roots) 3x^3 + 10x^2 -13x - 20 By the Rational Roots Theorem ......one possible root is -1 So 3(-1)^3 + 10(-1)^2 - 13(-1) - 20 = -3 + 10 + 13 - 20 = 0 Using synthetic division, we can find the remaining polynomial -1 [ 3 10 - 13 -20 ] -3 - 7 20 __________________ 3 7 -20 0 So....the remaining polynomial is 3x^2 + 7x - 20 Factoring, we have (3x - 5) ( x + 4) Setting each factor to 0 and solving for", "$\\boxed{109}$. ## Solution 5 We can find zeroes of the polynomial by making the first given equation $0 = 0$. Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$, respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \\cdot P(2) = 0$, which means $P(2)$ is not necessarily $0$. If $P(2) = 0$, then plugging in $2$ to the equation yields $P(3) = 0$, plugging in $3$ to the equation yields $P(4) = 0$, and so on, a contradiction of \"nonzero polynomial\". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$, so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$. We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$. $P(2) = 6a$ and $P(3) = 24a$, so: $$(6a)^2 = 24a \\implies 36a^2 = 24a \\implies a = \\frac23$$ Plugging in $\\frac72$ into the polynomial $\\frac23(x + 1)x(x - 1)$ yields $\\frac{105}{4}$. $105 + 4 = \\boxed{109}$. ## Solution 6", "15:45 the equation would look like this $$y = k(x-a)(x-b)$$ now we have to figure out what k is. We know what the maximum value is, call it c, and that it's x value is 0. Therefor we can plug this into the equation so that we get the following $$c = k(-a)(-b)$$ $$c = kab$$ therefor $k = c/(ab)$ therefor, your equation is $$y = c(x-a)(x-b)/(a*b)$$ - To answer question found in the title: \"... an equation for a parabola from its $x$ and $y$ intercepts\", the correct equation is: $$y = \\frac{c}{ab}(x-a)(x-b),$$ where $a, b$ are the $x$-intercepts and $c$ is the $y$-intercept. We can prove this is correct by noting that $y = 0$ when $x=a$ or $x=b$ is substituted, and when $x=0$, we have $y = \\frac{c}{ab}(-a)(-b) = c$. - Of course, if you know the max. value is the $y$-intercept, then $b = -a$ and you get $y = -\\frac{c}{a^2}(x^2 - a^2)$, as @deinst commented above. – Shaun Ault Jul 20 '11 at 19:46" ]

[ "# Difference between revisions of \"2001 AMC 12 Problems/Problem 19\" ## Problem The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$? $(\\mathrm{A})\\ -11 \\qquad (\\mathrm{B})\\ -10 \\qquad (\\mathrm{C})\\ -9 \\qquad (\\mathrm{D})\\ 1 \\qquad (\\mathrm{E})\\ 5$ ## Solution We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $\\frac{-a}{3}$ is the average of the zeros, so $\\frac{-a}3=-2 \\implies a = 6$. We are also given that the sum of the coefficients is $-2$, so $1+6+b+2 = -2 \\implies b=-11$. So the answer is $\\fbox{A}$.", "0$ $f(1) = 0$, this implies, $x = 1$ is also a root of the given polynomial. For $x = -2$ $f(-2) = 2(-2)^3 + (-2)^2 – 5(-2) + 2$ $= -16 + 4 + 10 + 2$ $= 0$ $f(-2) = 0$, this implies, $x = -2$ is also a root of the given polynomial. Now, Sum of zeros $= \\frac{-b}{a}=\\frac{-1}{2}$. Sum of the zeros of $f(x)=\\frac{1}{2} + 1 – 2 = – \\frac{1}{2}$ Sum of the products of the zeros taken two at a time $=\\frac{c}{a}=\\frac{-5}{2}$. Sum of the products of the zeros taken two at a time$=(\\frac{1}{2} \\times 1) + (1 \\times -2) + (\\frac{1}{2} \\times-2) = \\frac{-5}{ 2}$. Product of the zeros $= \\frac{– d}{a}=\\frac{-2}{2}=-1$. Product of the zeros$=\\frac{}{2} \\times 1 \\times (– 2) = -1$. Hence, the relationship between the zeros and the coefficients is verified. (ii) Let $g(x) = x^3 -4 x^2+ 5x - 2$ Comparing the given polynomial with the standard form of a cubic polynomial, $a=1$, $b=-4$, $c=5$ and $d=-2$ Also, If α is a root of a given polynomial $f(x)$, then $f(α)=0$. Therefore, For $x = 2$ $g(2) = (2)^3 -4(2)^2 +5(2) - 2$ $= 8-4(4)+10-2 = 0$ $=8-16+10-2$ $=18-18$ $=0$ $g(2) = 0$, this implies, $x =2$ is a root of the given polynomial. For $x = 1$", "# Problem #906 906 The polynomial $P(x)=x^3+ax^2+bx+c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $y$-intercept of the graph of $y=P(x)$ is 2, what is $b$? $\\text{(A) }-11 \\qquad \\text{(B) }-10 \\qquad \\text{(C) }-9 \\qquad \\text{(D) }1 \\qquad \\text{(E) }5$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Amazing Polynomial! Algebra Level 5 The polynomial $$P(x) = x^3 + ax^2 + bx + c$$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $$y$$-intercept of the graph of $$y = P(x)$$ is 2, then what is the value of $$-b?$$ ×", "a sum of $$\\frac{3^2(3\\cdot 11\\cdot 13+5\\cdot 7\\cdot 17)}{2^8}=3^2\\cdot 4=\\boxed{036}.$$ /*Lengthy proof that any two cubic polynomials in $t$ which are equal at 4 values of $t$ are themselves equivalent: Let the two polynomials be $A(t)$ and $B(t)$ and let them be equal at $t=a,b,c,d$. Thus we have $A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0$. Also the polynomial $A(t) - B(t)$ is cubic, but it equals 0 at 4 values of $t$. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into $(t-a)(t-b)(t-c)(t-d)($some nonzero polynomial$)$ which would have a degree greater than or equal to 4, contradicting the statement that $A(t) - B(t)$ is cubic. Because $A(t) - B(t) = 0, A(t)$ and $B(t)$ are equivalent and must be equal for all $t$. Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check. ## Solution 2 As in Solution 1, we have $(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)$ $-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$ $=(t-4)(t-16)(t-36)(t-64)$ Now the coefficient of $t^3$ on both sides", "$x$- intercept. Verify that the zero of this polynomial is the $x$-intercept. Solution: The table below gives the values of $y$ for which $y = 2x -1$ . As two distinct points determine a unique line, it is sufficient to determine the coordinates of two points on this line. \\begin{array}{ccc} x & y=2x-1 & (x,y)\\\\ \\hline 0 & -1 & (0,-1)\\\\ 1 & 1 & (1,1)\\\\ \\hline \\end{array} The graph shows the line $y = 2x - 1$. From the graph, it is clear that the line intersects the $x$-axis at the point $(\\frac{1}{2}, 0)$. Algebraically, we can solve the equation $2x - 1 = 0$ to get $x = \\frac{1}{2}$, thus verifying that the zero of this polynomial is the $x$-intercept. In general, a linear polynomial has at most $1$ zero. This means that the graph has either $0$ or $1$ $x$-intercept. The example above shows the graph of a line that has $1$ $x$-intercept. The graph of a linear polynomial $y = ax + b$ is a rising line (going upward) is the leading coefficient $a \\gt 0$ and a falling line (going downward) if the leading coefficient $a \\lt 0$. A horizontal line has $0$ $x$-intercept. Equation of this line is $y = 0$. That is, it is a $0$ polynomial. A quadratic polynomial has", "(b) (-2) , (5) (c) (2) , (5) (d) (-2) , (-5) 11. The graph of the polynomial $$ax^2 + bx + c$$ is an upward parabola if (a) a > 0 (b) a < 0 (b) a = 0 (d) None 12. A polynomial of degree 3 is called (a) a linear polynomial (c) a cubic polynomial 13. Dividend is equal to (a) divisor × quotient + remainder (b) divisior × quotient (c) divisior × quotient – remainder (d) divisor × quotient × remainder 14. The graph of the polynomial $$ax^2 + bx + c$$ is a downward parabola if (a) a > 0 (b) a < 0 (c) a = 0 (d) a = 1 15. If α, β are the zeros of the polynomial $$f(x) = x^2 + x + 1$$, then $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}=$$ (a) 1 (b) – 1 (c) 0 (d) None of these 16. If one zero of the polynomial $$f(x) = (k^2 + 4) x^2 + 13x + 4k$$ is reciprocal of the other, then k = (a) 2 (b) – 2 (c) 1 (d) – 1 17. If the sum of the zeros of the polynomial $$f(x) = 2x^3 – 3kx^2 + 4x – 5$$ is 6, then the value of k is (a) 2 (b) 4 (C) – 2 (d) –", "to find the zeroes: \\begin{eqnarray} x^2 - 6x + 8 &=& 0\\\\ x^2 - 4x - 2x + 8 &=& 0\\\\ x(x - 4) -2(x - 4) &=& 0\\\\ (x - 4)(x - 2) &=& 0\\\\ \\end{eqnarray} Sum of the zeroes = $4 + 2 = 6 = \\frac{6}{1}$. Coefficient of $x$ is $-6$ and coefficient of $x^2$ is $1$. Product of the zeroes is $(4)(2) = 8 = \\frac{8}{1}.$ Constant term is $8$ and coefficient of $x^2$ is $1$. Thus the relationship is verified. #### Example 3: Find a quadratic polynomial such that the sum and product of its zeroes are $8$ and $15$ respectively. Solution: The standard form of a quadratic polynomial is $ax^2 + bx + c$. Let its zeroes be $\\alpha$ and $\\beta$. Then, according to the given information, $\\alpha + \\beta = -\\frac{b}{a} = 8$ and $\\alpha\\beta = \\frac{c}{a} = 15$. If $a = 1$, then, we get $b = - 8$ and $c = 15$. So, one polynomial is $x^2 - 8x + 15$. Generally, any quadratic polynomial of the form $k(x^2 - 8x + 15)$ would satisfy the given condition. Standard form of a cubic polynomial is $p(x) = ax^3 + bx^2 + cx + d$. If $\\alpha$, $\\beta$, and $\\gamma$ are zeroes of the cubic polynomial, then we have: \\begin{eqnarray}", "$p(8) = 0$, it follows that that must be the vertex. The standard form for a parabola is $a(x-h)^2 + k$, and we already know that $(h,k)$ is $(8,0)$. That gives us $p(x) = a(x-8)^2$. Now simply plug in $p(0) = 8$ to get $a = \\frac{1}{8}$ to find $p$: $$p(x) = \\frac{1}{8}(x-8)^2$$ $$p(-4) = 18$$ Here's a way to solve this without needing as much mathematical insight as Barry's answer. If the polynomial is $$p(x) = ax^2 + bx + c,$$ then from $p(0) = 8$ we get $$0a + 0b + c=8,$$ from $p(8) = 0$ we get $$64a + 8b + c = 0,$$ and from $p(x)$ never taking any negative values, it follows that $x=8$ is a minimum, and so $p′(8) = 0$, and since $p′(x) = 2ax + b$, then we get $$16a + b + 0c = 0.$$ This gives three simultaneous equations in three variables, which give the coefficients $$a={1\\over 8}, b=-2, c=8.$$", "The number of zeros the Polynomial $p(x)$ has 18. The graph of the expression $ax^2+bx+c$ is an upward parabola, if 19. The graph of the expression $ax^2+bx+c$ is a downward parabola, if 20. The zeros of the Polynomial $x^2-2x-8$ are 21. The zeros of the Polynomial $x^2-2$ are 22. The zeros of the Polynomial $x^3-2x^2-15x$ are 23. The zero of the Polynomial $p(x)=ax+b$ is 24. The value of the Polynomial $p(x)=x^3-4x^2+9x-2$ when $x=2$ is 25. The value of the Polynomial $p(x)=x^3-x^2+\\sqrt{2}x$ when $x=0$ is 26. The value of the Polynomial $p(x)=2$ when $x=0$ is", "at most two zeros. That is, it can have $0$, $1$, or $2$ zeros. Geometrically, this means that the graph of a quadratic polynomial may have $0$, $1$, or $2$ $x$-intercepts. Algebraically, the zeros are obtained by solving the equation $ax^2 + bx + c = 0$. The following graphs are graphs of a quadratic polynomial with $0$, $1$, or $2$ zeros. Note that the graph is 'U' shaped, and opens upwards if the leading coefficient $a \\gt 0$, and opens downwads when $a \\lt 0$. A cubic polynomial has at most 3 zeros. That is, the graph of a cubic polynomial has $0$, $1$, $2$, or $3$ x-intercepts. The graph of a cubic polynomial starts going upward (rising) and takes 2 turns if the leading coefficient $a \\gt 0$ and starts going downward (falling) and then takes two turns if $a \\lt 0$. Following graphs illustrate this property. ## Practice Problems Observe each of the graph given below and determine the number of distinct zeros of the polynomial whose graph is given. $3$ $3$ $2$ $4$ $6$ ### 2-2: Relationship Between the Zeros and Coefficients of a Polynomial Consider a linear polynomial $p(x) = 2x + 5$. Then, its zero is obtained by setting $p(x) = 0$. That is, $2x + 5 = 0$. This gives $x", "5)+1 (a - 5) = 0$ $(a -5) ( a + 1) = 0$ a = 5, –1 Put a = 5 in (1) put a = –1 in (1) 5 + b = 2 –1 + b = 2 b = –3 b = 3 Hence, value of a = 5, b = –3 and a = –1, b = 3 put a = 5, b = –3 and we get zeroes a = 5 a + b = 5 – 3 = 2 a + 2b = 5 + 2(–3) = –1 Hence, zeroes are 5, 2, –1. View Full Answer(1) For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.(i)$\\frac{-8}{3},\\frac{4}{3}$ (ii) $\\frac{21}{8},\\frac{5}{16}$(iii) $-2\\sqrt{3}, -9$(iv) $\\frac{-3}{2\\sqrt{5}},-\\frac{1}{2}$ (i) Answer. $\\left [ -2 , -\\frac{2}{3}\\right ]$ Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0. Here, sum of zeroes $=\\frac{-8}{3}$ Product of zeroes $=\\frac{4}{3}$ Let p(x) is the required polynomial p(x) = x2 – (sum of zeroes)x + (product of zeroes) $=x^{2}-\\left (\\frac{-8}{3} \\right )x+\\frac{4}{3}$ $=x^{2}+\\frac{8}{3}x+\\frac{4}{3}$ Multiply by 3 $p(x) = 3x^2 + 8x + 4$ Hence, $3x^2 + 8x + 4$ is the required polynomial $p(x) = 3x^2 + 8x +", "(2) (1) (1) = 2 = –$$\\frac{-2}{1}$$ = $$\\frac{-d}{a}$$ Question 2. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeroes as 2, -7, -14 respectively. Solution: Let the cubic polynomial be ax3 + bx2 + cx + d and its zeroes be α, β and γ. Then, α + β + γ = 2= $$\\frac{-b}{a}$$ αβ + βγ + γα = -7 = $$\\frac{c}{a}$$ αβγ = -14 = $$\\frac{-d}{a}$$ If a = 1, then b = -2, c = -7 and d = 14. Hence, one cubic polynomial which fits the given conditions is x3 – 2x2 – 7x + 14. Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b. Solution: Then given polynomial is x3 – 3x2 + x + 1 Comparing with Ax3 + Bx2 + Cx + D, we get A = 1 B = -3 C = 1 D = 1 Let α = a – b β = a γ = a + b Then, we have α + β + γ = $$\\frac{-B}{A}=-\\frac{(-3)}{1}=3$$ ⇒ a – b + a + a + b = 3 ⇒ 3a =", "= ax + b, a ≠ 0, a, b ∈ R Solution: (i) x = 3. p( 3) = 3 – 3 = 0 ∴ The zero of the polynomial is x = 3. (iv) f(z) = 8z, If 8z = 0 z = $$\\frac{0}{8}$$ = 0 f(0) = 8(0) = 0 ∴ z = 0 is the zero of the given polynomial. Class 9 Maths Exercise 3.2 Solutions Question 4. Find the roots of the polynomial equations. (i) 5x – 6 = 0 (ii) x + 3 = 0 (iii) 10x + 9 = 0 (iv) 9x – 4 = 0 Solution: (i) 5x – 6 = 0 5x = 6 ∴ x = $$\\frac{6}{5}$$ (ii) x + 3 = 0 ∴ x = -3 (iii) 10x + 9 = 0 10x = -9 ∴ x = $$\\frac{-9}{10}$$ (iv) 9x – 4 = 0 9x = 4 ∴ x = $$\\frac{4}{9}$$ Verify whether the following are zeros of the polynomial indicated against them, or not. (i) p(x) = 2x – 1, x = $$\\frac{1}{2}$$ (ii) p(x) = x3 – 1, x = 1, (iii) p(x) = ax + b, x = $$\\frac{-b}{a}$$ (iv) p(x) = (x + 3) (x – 4), x = 4, x = -3 Solution: (i) p(x) = 2x – 1, x = $$\\frac{1}{2}$$", "+0 # Polynomial 0 140 1 Let P(x) be a cubic polynomial such that P(0) = -3 and P(1) = 4. When P(x) is divided by x^2 + x + 1, the remainder is 2x + 3. What is the quotient when P(x) is divided by x^2 + x + 1? Jun 11, 2021 #1 +287 +2 Since $P(x)$ is cubic and $x^2+x+1$ is quadratic, the quotient must be linear. Let the quotient be $ax+b$ for some constants $a$ and $b$. Then $$P(x) = (x^2+x+1)(ax+b) + 2x+3.$$ We are given that $P(0)= -3$, so $$(0^2+0+1)(a(0)+b) + 2(0)+3 = -3$$ that is, $$b+3=-3,$$ giving $b=-6$. We are also given that $P(1) = 4$, so $$(1^2+1+1)(a(1)-6) + 2(1)+3 = 4$$ that is, $$3(a-6)+5=4$$ giving $a=\\frac{17}{3}$. Therefore, the quotient is $\\frac{17}{3}x-6$. Jun 11, 2021", "of the zeroes of the poly-nomial x2 + 5x + 6. – 5 Question 15. 4y2 – 5y + 1 is an example for this type of polynomial. Question 16. Write the degree of the polynomial 5x7 – 6x5 + 7x – 4. 7 Question 17. How much the sum of zeroes of the polynomial 2x2 – 8x + 6 ? 4 Question 18. f α,β are the zeroes of x2 + x +1, then find $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}$$ -1 Explanation: α + β = -1, αβ = 1 ⇒ $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{\\alpha+\\beta}{\\alpha \\beta}=\\frac{-1}{1}$$ = -1 Question 19. Write the number of zeroes of the poly¬nomial in the given graph. 3 Question 20. Write product of zeroes of the cubic polynomial 3x3 – 5x2 – 11x – 3. 1 Explanation: Product of zeroes = αβγ = $$\\frac{-d}{a}=\\frac{-(-3)}{3}=\\frac{3}{3}$$ = 1 Question 21. The following is the graph of a poly¬nomial. Find the zeroes of the poly¬nomial from the given graph. – 2,1 Question 22. Find the value of p(x) = 4x2 + 3x + 1 at x = -1. 2 Question 23. If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d and (a ≠ 0), then find αβγ Question 24. 4x + 6y = 18 doesn’t pass through ori-gin, then its graph indicates A", "the sum and product of the roots, namely $-b/a$ and $c/a$ and the common difference, and to avoid using the quadratic formula. Andaleeb Ahmed of Woodhouse Sixth Form College, London has provided a solution to this problem. We started with the linear polynomial $y = -3x + 9$. The $x$-coefficient is $-3$, the root is $3$, the intercept is $9$, and these numbers are in arithmetic progression. Suppose now that the same is true of the polynomial $y=ax+b$, where $a\\neq 0$. In this case, the numbers $a$, $-b/a$ and $b$ must be in arithmetic progression. Then: $${-b\\over a} - a = b +{b\\over a}$$ $$-b - a^2 = ab + b$$ $$a^2 + ba + 2b = 0$$ We can use this to express $a$ in terms of $b$ or $b$ in terms of $a$: $$a = {{-b\\pm \\sqrt{b^2 - 8b}}\\over 2}\\quad \\quad (b\\geq 8 \\ {\\rm or}\\ b < 0).$$ $$b={-a^2\\over (a+2)}\\quad \\quad (a\\neq -2).$$ We can randomly choose $b$ (such that $b\\geq 8$ or $b < 0$) and use the formula to find the corresponding values of $a$, or alternatively, we could choose any value of $a$ except $a = -2$ and find the corresponding value of $b$ from the given formula. The linear polynomial is $$y = ax - \\left({a^2\\over a+2}\\right).$$ It is easily checked", "# Polynomials and Factoring ## Self-Test: 1) $3x^2(5x-1) + 2x(x+3) =$ $15x^3 + 5x^2 + 6x$ $17x^2 + 6x - 1$ $15x^3 - x^2 + 6x$ $15x^3 + 2x^2 + 6x - 1$ $12x^2 + 8x$ Hint Use the distributive law $a(b+c)=ab+ac$ and combine like terms. 2) $x=3$ is a root of $y = x^3 + 3x^2 - x - 3$ $y = x^3 - x^2 - 14x + 24$ $y = x^3 - 6x^2 - x + 6$ $y = x^3 + 3x^2 - 4x - 12$ $y = x^3 + 6x^2 + 11x + 6$ Hint $x=a$ is a root of $p(x)$ if $p(a)=0$. 3) The zeros of $y = 2x^2 - 5x - 12$ are $x =$ $\\frac{-3}{2}$ and $4$ $\\frac{-2}{3}$ and $4$ $-2$ and $3$ $-6$ and $\\frac{1}{2}$ undefined (there are no roots) Hint Factor into linear terms of the form $(px-r)(qx-s)$. Then the roots are $x=r/p, x=s/q$. 4) All polynomials of degree three have three linear factors and one $y$-intercept.* three or two or one linear factor(s) and one $y$-intercept.* three or one linear factor and no $y$-intercept.* three or one linear factor and one $y$-intercept.* three or one or no linear factors where some will have a $y$-intercept and some will not.* * Factors need not be distinct. Hint 1 A polynomial", "# Finding $a, b, c$ values of a polynomial from a graph. I was doing my homework and I am now stuck on question number 7 which is: The diagram shows the curve with the equation $y = (x + a)(x - b)^2$ where $a$ and $b$ are positive integers. (i) Write down the values of $a$ and $b$, and also of $c$, given that the curve crosses the $y$-axis at $(0, c)$. I have attached a picture of my text book here: I have so far found $b$ like this: $(x - b)^2 = x^2 - 2bx + b^2$ Since $y = 0$ at $x = 1$, $0 = 1 - 2b + b^2$ Solving that quadratic equation gives us $b = 1$. I need to find $a$ and $c$ and I am totally puzzled on how to find them. • For $a$ the solution is the same... only you have a first order equation. Then, to solve for $c$, put $x=0$ in the original equation. – N74 May 5 '16 at 14:20 • @N74 My textbook says a and c are 2 – Hassan Althaf May 5 '16 at 14:21 • You are told the equation is $(x+a)(x-b)^2$ with $a,b$ positive integers. Looking at the graph, the positive root must be at 1, so $b=1$. The", "🙂 I’m new to this site so please let me know if there’s anything wrong with my question. I was trying to find the roots of the polynomial expression $$P(x) = 8x^5 – 14x^4 – 22x^3 +57x^2 – 35x + 6$$. I first found $$(x+2)$$ using the Rational Zeros Theorem and synthetic division. Since the result of $$frac{P(x)}{x+2}$$ is $$Q(x) = 8x^4 – 30x^3 + 38x^2 – 19x + 3$$, I assumed $$-2$$ is a lower bound, because of the alternating signs. And from the graph I can see that’s true, since there are no x-intercepts smaller than $$-2$$. graph of P(x) Then, I found $$(x – 1)$$ as a root. The result $$frac{Q(x)}{x-1} = 8x^3-22x^2+16x-3 = R(x)$$ implies that, according to the theorem, 1 is a lower bound of $$Q(x)$$, but this time, according to the graph, that is not true. I supposed that roots themselves cannot be considered upper and lower bounds. graph of Q(x) However, when dividing $$R(x)$$ by $$(x-1)$$ I get $$8x^2-14x+2 – frac{1}{x-1}$$, which is not a root, but should be a lower bound according to the theorem, and that does not make sense looking at the graph. graph of R(x) Am I doing something wrong? Or maybe this is indeed an exception to the theorem? Also, can you tell me whether roots" ]

[ "The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \\log_b 7^7$. Find $b$. Source: AMC 12A 2010 Happy problem solving! Email your solution to [email protected]", "b 6. ^ (See equation 7.88 on page 375.)", "= b^x$$ and $$g(x) = \\log_b x$$ tangent to each other? Solution Home", "= - 7$ That's the solution. Hope this helped!", "Is the solution to $B x + C = 0$", "### Home > INT3 > Chapter 5 > Lesson 5.2.1 > Problem5-57 5-57. Using the idea behind the Ancient Puzzle you studied today, what is the value of $b$ in each equation below? Explain your reasoning. 1. $\\log_b243 = 5$ $b^5 = 243$ 1. $\\log_b0.001 = -3$ $0.001=\\frac{1}{10^3}$", "that the solution to the last equation is $7$. The solution to the first equation is also $7$", "### Home > INT3 > Chapter 5 > Lesson 5.2.3 > Problem5-78 5-78. If $x=7^y$, how would you write this equation in $y=\\text{form}$? Explain. Homework Help ✎ Remember that $N=b^k$ and $k=\\log_bN$. $y=\\log_7x$", "$$m>4$$ or $$m=4$$ you have the expression for $$x_k$$. The particular solution $$x_k^*$$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $$m$$.", "examples of graphical solution of equation for fixed points of logarithm are shown in figure 2 for $b\\!=\\!\\sqrt{2}$, $b\\!=\\!\\exp(1\\!/\\!\\rm e)$, and $b\\!=\\!2$. The black line shows function $y=x$ in the $x,y$ plane. The colored curves show function $y=\\log_b(x)$ for cases $b\\!=\\!\\sqrt{2}$ (red), $b\\!=\\!\\exp(1/\\rm e)$ (green), and $b\\!=\\!2$ (blue). At $b\\!=\\!\\sqrt{2}$, there exist 2 solutions, $x\\!=\\!2$ and $x\\!=\\!4$. At $b\\!=\\!\\exp(1\\!/\\!\\rm e)$ there exists one solution $x\\!=\\!\\rm e$. and $b\\!=\\!2$, there are no real solutions. In general, • at $b\\!<\\!\\exp(1\\!/\\!\\rm e)$ , there are two real solutions; • at $b\\!=\\!\\exp(1\\!/\\!\\rm e)$ , there is one solution, and • at $b\\!>\\!\\exp(1\\!/\\!\\rm e)$ , there exist two solutions, but they are complex. In particular, at $b\\!=\\!\\sqrt{2}$, the solutions are $L\\!=\\!\\!L_{\\sqrt{2},1}=2$ and $L\\!=\\!L_{\\sqrt{2},2}=4$ . At $b\\!=\\!2$, the solutions are $L\\!=\\!L_2 \\!\\approx\\! 0.824678546142074222314065+1.56743212384964786105857 \\!~\\rm i$ and $L\\!=\\!L_2^*\\!\\approx\\! 0.824678546142074222314065-1.56743212384964786105857 \\!~\\rm i$. At $b\\!=\\!\\rm e$, the solutions are $L=L_{\\rm e} \\approx 0.318131505204764135312654+1.33723570143068940890116 \\!~\\rm i$ and $L=L_{\\rm e}^* \\approx 0.318131505204764135312654-1.33723570143068940890116 \\!~\\rm i$. A few hundred straightforward iterations of equation $L=\\log_b(L)$ are sufficient to get the error smaller than the last decimal digit in the approximations above. ### Basic properties FIg.3. parameters of asymptotic of tetration versus logarithm of the base The solutions $L=L_1$ and $L=L_2$ of equation (14) are plotted in figure 3 versus $\\beta=\\ln(b)$ with thin black lines. At $\\beta<1/\\mathrm e$, both $L_1$ and $L_2$ are", "the following shows the solution/s for the equation, $\\log_4\\left(\\dfrac{x + 1}{2x -1}\\right) = 0$? 18. Which of the following shows the solution/s for the equation, $\\log \\left(\\dfrac{1}{2x -1}\\right) = 2$? 19. Which of the following shows the solution/s for the equation, $\\log_x 0.0001 = 4$? 20. Which of the following shows the simplified form of $\\log_a a^y$? 21. Which of the following shows the exponential form of $\\log_b(2x + 1) = 3$? 22. Which of the following shows the evaluated value of $\\log_9 3$? 23. Which of the following shows the evaluated value of $\\log 10000$? 24. Which of the following shows the evaluated value of $\\ln e^7$? 25. Which of the following shows the evaluated value of $\\ln 1$? 26. Which of the following shows the evaluated value of $\\ln e^{-3}$?", "'16 at 4:22 • Ok, but in the end one must come to a strict, precise conclusion, because the log might be part of a bigger equation, where one needs to define a precise field of existence. Am I to assume that $\\log_a(b)$ has a > 0 and b > 0 when I outline the solution to an equation that contains a log? For example an equation might be verified for a = -3 and b = 6. If it contains a $\\log_a(b)$ somewhere, the only actual solution is 6? – Athamas Oct 11 '16 at 15:36 • The conclusion in real analysis is that $\\log_a$ is defined for $a > 0$ and only for $a>0$. I don't know what you mean by an equation that might be verified for $a=-3$ and $b=6$ that contains $\\log_a b$ somewhere. An equation that uses $\\log_a b$ cannot be satisfied by $a=-3$ and $b=6$. Moreover, when we rule out $a=-3,b=6$ as a solution, we rule out the entire solution--we don't just ignore the problem with the $a=-3$ part and say $b=6$ is a solution. – David K Oct 11 '16 at 16:20 • If you have a specific mathematical problem in mind that you think needs to be solved by logarithms with negative bases, I suggest you post it as a", "we have, $\\Rightarrow a = b$ for all $a > 0,b > 0$ So, equation (3) will become $\\Rightarrow \\log ({x^2} - 1) = \\log 8 \\\\ \\Rightarrow {x^2} - 1 = 8 \\\\ \\Rightarrow {x^2} = 9 \\\\ \\Rightarrow x = \\pm \\sqrt 9 \\\\ \\Rightarrow x = \\pm 3 \\\\ \\Rightarrow x = + 3, - 3 \\\\$ Here we get the two values of $x = + 3, - 3$ And we have the equation (1) is $\\log \\left( {1 + x} \\right) + \\log \\left( {x - 1} \\right) = \\log 8$ So, for $x = - 3$ equation (1) is not defined because ‘$\\log$’ is only defined for positive real numbers. But for $x = + 3$ equation (1) is defined. Hence option B is the correct answer. Note: Whenever we face such types of problems the key point is that to simplify the problems by using basic logarithmic identities. So we have always remembered the basic logarithmic identities. After getting the solutions the most important step is rechecking whether the given equation will define or not for our founded solutions. The solution for which the given equation is defined is our right answer.", "Each solution to $x^{2} + 5x + 8 = 0$ can be written in the form $x = a + b i$, where $a$ and $b$ are real numbers. What is $a + b^{2}$?", "= 8^2$$ and the solution is $$x = 8$$.", "for $x$ in the equation $\\sqrt{2-7 x}+2 x=0$ ... Solve $6-7 x^{2}=0(a=-7, b=0, c=-6)$ Solve $6-7 x^{2}=0(a=-7, b=0, c=-6)$Solve $6-7 x^{2}=0(a=-7, b=0, c=-6)$ ...", "$2\\log x – \\log (x -1)$? 3. Which of the following shows the condensed form of $3\\log_2 x + \\log_2 (y – 2) – 2\\log_2 z$? 4. Which of the following shows the condensed form of $4 \\log_b (x + 2) – 3 \\log_b (x – 5)$? 5. Which of the following shows the condensed form of $2\\log_a y + 0.5\\log_a (x + 4)$? 6. Which of the following shows the condensed form of $2\\ln 8 + 5\\ln x$? 7. Which of the following shows the expanded form of $\\log_2 (4xy^5)$? 8. Which of the following shows the expanded form of $\\log _{10}\\left(\\dfrac{xy}{z}\\right)$? 9. Which of the following shows the expanded form of $\\log_5 (ab)^{1/2}$? 10. Which of the following shows the expanded form of $\\log_4 (2x)^2$? 11. Which of the following shows the expanded form of $\\log_6 (ab)^4$? 12. Which of the following shows the solution/s for the equation, $3\\log_5 2 = 2\\log_5 x$? 13. Which of the following shows the solution/s for the equation, $\\log_2 x = 3$? 14. Which of the following shows the solution/s for the equation, $\\log_x 8 = 3$? 15. Which of the following shows the solution/s for the equation, $\\log_3 x = 1$? 16. Which of the following shows the solution/s for the equation, $\\log_3 \\left(\\dfrac{1}{x+1}\\right) = 2$? 17. Which of", "1.5x + b. Now you can solve for b by plugging in one of the points given in the question. Let's use (6, 2).$(y = 1.5x + b$)$(2 = 1.5(6) + b$)$(b=-7$)With that you have now values for both m and b, so you need only plug them both in to obtain the final equation:$(y=1.5x-7$) Send a message explaining your needs and Eric will reply soon. Contact Eric", "as $\\log_b \\left( \\frac{x}{y} \\right)$. I’m continually amazed at the number of good students who intellectually know that the above equations are false but panic and use them when solving a problem.", "|-x^3|$ 7. $f(x) = 2^x - 1$ 8. $f(x) = |\\log_{1/2} x|$" ]

[ "wrap-up You should have no problem with the question from the beginning of the lesson now: log7247=x\\begin{align*}log_7 247=x\\end{align*}. Using the change of base formula: log7247=log247log7\\begin{align*}log_7 247 = \\frac{log 247} {log 7}\\end{align*} Using a calculator to find the common logs of 247 and 7, we get (approximately): 2.4.8=2.8313\\begin{align*}\\frac{2.4} {.8} = 2.8313\\end{align*}. We can verify with: 72.8313=247\\begin{align*}7^{2.8313} = 247\\end{align*} \\begin{align*}\\therefore log_7 247 = 2.8313\\end{align*} ### Vocabulary Common Logarithm: A common logarithm is a log with base 10k. The log is usually written without the base. Natural Logarithm: A natural log is a log with base e. The natural log is written as ln. Trancendental Number: A number that is not the root of any rational polynomial function. Examples include \\begin{align*}e\\end{align*} and \\begin{align*}\\pi\\end{align*}. \\begin{align*}e\\end{align*} (constant): A constant used as the base of natural logarithms, equal to approximately 2.71828. ### Guided Practice Questions 1) Find the value of each natural log. a. \\begin{align*}ln 100\\end{align*} b. \\begin{align*}ln \\sqrt{e}\\end{align*} 2) Solve the equation: \\begin{align*}8^{x-3}=24\\end{align*} 3) Solve the equation: \\begin{align*}5^x = 3 \\cdot 7^x\\end{align*} 4) Find the value: \\begin{align*}ln6 + ln7\\end{align*} 5) Find the value: \\begin{align*}log 5 - log 3\\end{align*} 1) a. \\begin{align*}ln 100\\end{align*} is between 4 and 5. You can estimate this by rounding \\begin{align*}e\\end{align*} up to 3, and considering powers of 3: \\begin{align*}3^4 = 81\\end{align*} and \\begin{align*}3^5 = 243\\end{align*} Using a calculator, you", "referred to as taking the logarithm of both sides. We can use any logarithm that we’d like to so let’s try the natural logarithm. \\begin{align*}\\ln {7^x} & = \\ln 9\\\\ x\\ln 7 & = \\ln 9\\end{align*} Now, we need to solve for $$x$$. This is easier than it looks. If we had $$7x = 9$$ then we could all solve for $$x$$ simply by dividing both sides by 7. It works in exactly the same manner here. Both ln7 and ln9 are just numbers. Admittedly, it would take a calculator to determine just what those numbers are, but they are numbers and so we can do the same thing here. \\begin{align*}\\frac{{x\\ln 7}}{{\\ln 7}} & = \\frac{{\\ln 9}}{{\\ln 7}}\\\\ x & = \\frac{{\\ln 9}}{{\\ln 7}}\\end{align*} Now, that is technically the exact answer. However, in this case it’s usually best to get a decimal answer so let’s go one step further. $x = \\frac{{\\ln 9}}{{\\ln 7}} = \\frac{{2.19722458}}{{1.94591015}} = 1.12915007$ Note that the answers to these are decimal answers more often than not. Also, be careful here to not make the following mistake. $1.12915007 = \\frac{{\\ln 9}}{{\\ln 7}} \\ne \\ln \\left( {\\frac{9}{7}} \\right) = 0.2513144283$ The two are clearly different numbers. Finally, let’s also use the common logarithm to make sure that we get the same answer. \\begin{align*}\\log {7^x} &", "base-10 logarithm of 2, 3 and 10 we can write \\begin{align*} \\lg 15 &= \\lg \\frac{10\\times3}{2}\\\\ &= \\left( \\lg 10 \\right) + \\left( \\lg 3 \\right) - \\left( \\lg 2 \\right)\\\\ &\\approx 1 + 0.47712 - 0.30103\\\\ &= 1.1761. \\end{align*}", "inside the recursive calls sum to less than one: \\begin{align} 1 & = \\sum_1^k a_i \\\\ & = \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{35}\\\\[0.5em] & = \\frac{2}{5} + \\frac{4}{7} + \\frac{1}{35}\\\\[0.5em] & = \\frac{35}{35} \\end{align} We next verify that the base case is less than the max of the inverses of the coefficients: \\begin{align} n & < \\max\\{a_1^{-1}, a_2^{-1}, \\ldots, a_k^{-1}\\}\\\\ & = \\max\\{5, 5, 7, 7, 7, 7, 35\\}\\\\ & = 35 \\end{align} With these conditions met, we know $$T(n)\\leq \\alpha n \\log n + \\beta n$$ where $$\\beta = c$$ and $$\\alpha$$ is a constant equal to: \\begin{align} b &= \\frac{c}{\\sum_1^k a_i \\log_k a_i^{-1}}\\\\[0.5em] &= \\frac{c}{\\frac{2 \\log_7 5}{5} + \\frac{4 \\log_7 7}{7} + \\frac{\\log_7 35}{35}}\\\\[0.5em] &\\approx 1.048c \\end{align} Therefore we have: \\begin{align} T(n) & \\leq 1.048cn\\log_7 n + cn\\\\ \\therefore T(n) & = O(n \\log n) \\end{align} There may be times when you come across a strange recurrence like this: $$T(n) = \\begin{cases} c & n < 7\\\\ 2T\\left(\\frac{n}{5}\\right) + 4T\\left(\\frac{n}{7}\\right) + cn & n\\geq 7 \\end{cases}$$ If you're like me, you'll realize you can't use the Master Theorem and then you make think, \"hmmm... maybe a recurrence tree analysis could work.\" Then you'd realize that the tree starts to get gross really fast. After some searching on the", "= 3 \\cdot 2^x$. You can do the same thing with the right-hand side. That should help a lot. \\begin{align} 2^x+2^{x+1}&=3^{x+2}+3^{x+3}\\\\ 2^x+2\\cdot2^x&=3^2\\cdot3^x+3^3\\cdot3^x\\\\ (1+2)2^x&=(3^2+3^3)3^x\\\\ 3\\cdot2^x&=36\\cdot3^x\\\\ 2^x&=12\\cdot3^x\\\\ \\left(\\frac23\\right)^x&=12\\\\ \\end{align} Now use logarithms.", "\\frac \\pi 6 -\\log\\frac{4\\pi}{6} \\right) \\\\ &=\\frac{1}{\\pi}\\log2 \\end{aligned}", "all the way gives \\begin{align} T(n) &= n^2 (\\log n + \\log (n/2) + \\cdots + \\log(n/(n/2)) + 4^{\\log n}T(1) \\\\ &= n^2 \\sum_{k=1}^{\\log n} k + n^2 T(1) \\\\ &= \\frac{1}{2} n^2 \\log n(\\log n - 1) + n^2 T(1) \\\\ &= \\Theta(n^2\\log^2 n). \\end{align}", "\\cdot 2^{c+1} \\\\ =&\\ 6 \\cdot 2^c \\end{align} The number of cycles $c$ is therefore $$c \\approx \\log_2 \\frac{t}{6} = O(\\log t)$$", "+ a_3 + a_4 + a_5) - (a_{10} + a_{11} + a_{12} + a_{13} + a_{14}) \\\\ &= (\\log_{2002} 2 + \\log_{2002} 3 + \\log_{2002} 4 + \\log_{2002} 5) \\\\ &\\quad - (\\log_{2002} 10 + \\log_{2002} 11 + \\log_{2002} 12 + \\log_{2002} 13 + \\log_{2002} 14) \\\\ &= \\log_{2002} \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5}{10 \\cdot 11 \\cdot 12 \\cdot 13 \\cdot 14} \\\\ &= \\log_{2002} \\frac{1}{2002} \\\\ &= \\boxed{-1}. \\end{align*} The answer is (B).", "to one: \\begin{align} 1 & = \\sum_1^k a_i \\\\ & = \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{35}\\\\[0.5em] & = \\frac{2}{5} + \\frac{4}{7} + \\frac{1}{35}\\\\[0.5em] & = \\frac{35}{35} \\end{align} We next verify that the base case is less than the max of the inverses of the coefficients: \\begin{align} n & < \\max\\{a_1^{-1}, a_2^{-1}, \\ldots, a_k^{-1}\\}\\\\ & = \\max\\{5, 5, 7, 7, 7, 7, 35\\}\\\\ & = 35 \\end{align} With these conditions met, we know $$T(n)\\leq \\alpha n \\log n + \\beta n$$ where $$\\beta = c$$ and $$\\alpha$$ is a constant equal to: \\begin{align} b &= \\frac{c}{\\sum_1^k a_i \\log_k a_i^{-1}}\\\\[0.5em] &= \\frac{c}{\\frac{2 \\log_7 5}{5} + \\frac{4 \\log_7 7}{7} + \\frac{\\log_7 35}{35}}\\\\[0.5em] &\\approx 1.048c \\end{align} Therefore we have: \\begin{align} T(n) & \\leq 1.048cn\\log_7 n + cn\\\\ \\therefore T(n) & = O(n \\log n) \\end{align} After checking this post again, I'm surprised this isn't on here yet. # Domain Transformation / Change of Variables When dealing with recurrences it's sometimes useful to be able to change your domain if it's unclear how deep the recursion stack will go. For instance, take the following recurrence: $$T(n) = T(2^{2^{\\sqrt{\\log \\log n}}}) + \\log \\log \\log n$$ How could we ever solve this? We could expand out the series, but I promise this will get gross really", "we have \\begin{align} -\\log\\left(\\frac{\\exp(x_j)}{\\sum_i \\exp (x_i)}\\right)&= -x_j+\\log\\left(\\sum_i \\exp(x_i)\\right) \\\\ &= -1 + \\log\\left( \\exp(0) + \\exp(1) + \\exp(0) \\right) \\\\ &= 0.5514. \\end{align}", "the logarithm: \\begin{align*} -\\ln|e^{-x}+1| &= -\\ln\\left(\\frac{1}{e^x} + 1\\right)\\\\ &= -\\ln\\left(\\frac{1+e^x}{e^x}\\right)\\\\ &= \\ln\\left( \\left(\\frac{1+e^x}{e^x}\\right)^{-1}\\right)\\\\ &= \\ln\\left(\\frac{e^x}{1+e^x}\\right)\\\\ &= \\ln(e^x) - \\ln(1+e^x)\\\\ &= x - \\ln(1+e^x)\\\\ &= x - \\ln|1+e^x|. \\end{align*} -", "= 7\\end{align*} This demonstrates that when you multiply two numbers, their logs add. Now, you come up with a similar demonstration of the second rule of logs, that shows why when you divide two numbers, their logs subtract. Now we’re going to practice applying those three rules. Take my word for these two facts. (You don’t have to memorize them, but you will be using them for this homework.) • \\begin{align*}\\log_58 = 1.29\\end{align*} • \\begin{align*}\\log_560 = 2.54\\end{align*} Now, use those facts to answer the following questions. 2. \\begin{align*}\\log_5480 =\\end{align*} (Hint: \\begin{align*}480=8\\times 60\\end{align*}. So this is \\begin{align*}\\log_5(8 \\times 60)\\end{align*}. Which rule above helps you rewrite this?) 3. How can you use your calculator to test your answer to \\begin{align*}^\\#2\\end{align*}? (I’m assuming here that you can’t find \\begin{align*}\\log_5480\\end{align*} on your calculator, but you can do exponents.) Run the test—did it work? 4. \\begin{align*}\\log_5\\left ( \\frac{2}{15} \\right ) =\\end{align*} 5. \\begin{align*}\\log_5 \\left ( \\frac{15}{2} \\right ) =\\end{align*} 6. \\begin{align*}\\log_564 =\\end{align*} 7. \\begin{align*}\\log_5(5)^{23} =\\end{align*} 8. \\begin{align*}5(\\log_523)=\\end{align*} Simplify, using the \\begin{align*}\\log(xy)\\end{align*} property: 9. \\begin{align*}\\log_a(x \\cdot x \\cdot x \\cdot x)\\end{align*} 10. \\begin{align*}\\log_a(x \\cdot 1)\\end{align*} Below bracket are different size Simplify, using the \\begin{align*}\\log\\left ( \\frac{x}{y} \\right )\\end{align*} property: 11. \\begin{align*}\\log_a \\left ( \\frac{1}{x} \\right )\\end{align*} 12. \\begin{align*}\\log_a \\left ( \\frac{x}{1} \\right )\\end{align*} 13. \\begin{align*}\\log_a \\left ( \\frac{x}{x} \\right )\\end{align*} Simplify, using the \\begin{align*}\\log(x)^b\\end{align*}", "(1+\\cos{x})^{3} \\rm{dx} &= \\int 8 \\cdot \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\\\ &= 8 \\int \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\end{align*} -", "+ (-8)x^4 + 16x^5 + (-32)x^6 + 64x^7+\\cdots \\\\ &= x - 2x^2 + 4x^3 - 8x^4 + 16x^5 - 32x^6 + 64x^7+\\cdots \\end{align*} This is an infinite geometric series with initial term $a = x$ and common ratio $r = -2x$, so we obtain: $$f(x) = \\frac{a}{1 - r} = \\frac{x}{1+2x}$$ For the second one, we obtain: \\begin{align*} g(x) &= 1x^0 + 0x^1 + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + 0x^7+ 1x^8 + \\cdots \\\\ &= 1 + x^2 + x^4 + x^6 + x^8 + \\cdots \\end{align*} This is an infinite geometric series with initial term $a = 1$ and common ratio $r = x^2$, so we obtain: $$g(x) = \\frac{a}{1 - r} = \\frac{1}{1-x^2}$$", "to 7 to 8. \\begin{align} \\lambda_{1478}\\ +\\ \\cdots \\ <&= E_1\\\\ \\lambda_{1478}\\ +\\ \\cdots \\ <&= E_4\\\\ \\lambda_{1478}\\ +\\ \\cdots \\ <&= E_7 \\end{align} c)", "\\left( 3^3 \\right)^{3x} \\\\ &= 27^{3x} \\\\ &= \\left( 27^x \\right)^3 \\\\ &= 4^3 \\\\ &= 64 \\end{align*} 3. ## Re: How do I solve this Exponential and Logarithmic Function?(Sorry, wrong category) Originally Posted by RPSGCC733 1.Find the value of when = 4. My attempt: == 4 =4 27x=1 x= 1/27. When checked on a calculator it doesn't work. I don't understand what to do. Is there a way to solve this without using calculators? 2. Solve the equation I don't have an attempt since I can't understand what operation to use. Whoops, wrong category. Can I request for a Moderator (if there is one in here) to relocate this to Algebra? Is the second question trying to solve for \\displaystyle \\begin{align*} x \\end{align*} when \\displaystyle \\begin{align*} \\log_x{\\left( \\log_2{256} \\right)} = 3 \\end{align*}? 4. ## Re: How do I solve this Exponential and Logarithmic Function? Note that $(3^{9})^x = (3^{3\\cdot 3})^x = 3^{3\\cdot 3 \\cdot x} = \\ldots = ?$ $\\log_x(\\log_2(256))=3$ where $\\log_2(256)=8$ thus you have to solve $\\log_x(8)=3 \\Leftrightarrow x^3 = 8 \\Leftrightarrow \\ldots$ 5. ## Re: How do I solve this Exponential and Logarithmic Function?(Sorry, wrong category) Originally Posted by Prove It \\displaystyle \\begin{align*} 3^{9x} &= \\left( 3^3 \\right)^{3x} \\\\ &= 27^{3x} \\\\ &= \\left( 27^x \\right)^3 \\\\ &= 4^3 \\\\ &= 64 \\end{align*} Ah. Thanks!", "14 + \\frac 1{4\\sqrt 2}\\log(\\sqrt2+1)) \\ . \\end{aligned}", "+ O\\left(\\frac{n^2}{\\log_4^5 n}\\right), \\ldots \\\\ \\end{align*}", "during the time, distance, volume or area. \\begin{align*}e=\\end{align*} the base of the natural logarithm \\begin{align*}= 2.718281828\\ldots\\end{align*} Show Hide Details Description Tags: Subjects: Date Created: Feb 23, 2012" ]

[ "The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \\log_b 7^7$. Find $b$. Source: AMC 12A 2010 Happy problem solving! Email your solution to [email protected]", "# 2012 AIME I Problems/Problem 9 ## Problem Let $x,$ $y,$ and $z$ be positive real numbers that satisfy $$2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) \\ne 0.$$ The value of $xy^5z$ can be expressed in the form $\\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ ## Solution 1 Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that $$2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = 2.$$ Then \\begin{align*} 2\\log_{x}(2y) = 2 &\\implies x=2y\\\\ 2\\log_{2x}(4z) = 2 &\\implies 2x=4z\\\\ \\log_{2x^4}(8yz) = 2 &\\implies 4x^8 = 8yz \\end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \\rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \\cdot (2^{-7/6})^5 \\cdot 2^{-7/6} = 2^{-43/6}$ and thus $p+q = 43 + 6 = \\boxed{049.}$ ## Solution 2 Notice that $2y\\cdot 4z=8yz$, $2\\log_2(2y)=\\log_2\\left(4y^2\\right)$ and $2\\log_2(4z)=\\log_2\\left(16z^2\\right)$. From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$. So, for constant $C\\ne 0$: $$\\frac{\\log 4y^2}{\\log x}=\\frac{\\log 8yz}{\\log 2x^4}=\\frac{\\log 16z^2}{\\log 2x}=C$$ Since $\\log 4y^2,\\log 8yz,\\log 16z^2$ are in a geometric progression, so is $\\log x,\\log 2x^4,\\log 2x$. Therefore, $2x^4$ is the geometric mean of", "# Solve for x. 7^x= 6^{x+7} ## Question: Solve for x. {eq}7^x= 6^{x+7} {/eq} ## Solving an Equation We have to find a value of x for which both sides of the equation are equal. This can be done by using the natural log of both sides of the equation as this will allow us to move the x down. The value of x can be found as follows by taking the natural log of everything. \\begin{align} 7^x&= 6^{x+7}\\\\ \\Rightarrow \\ln (7^x)&=\\ln (6^{x+7})\\\\ x\\ln 7&=(x+7)\\ln 6&&&&\\left [ \\because \\ln (a^b)=b\\ln a \\right ]\\\\ \\frac{\\ln 7}{\\ln 6}&=\\frac{x+7}{x}\\\\ 1.086&=1+\\frac{7}{x}\\\\ \\therefore x&\\approx 81.40 \\end{align}", "that we can deal with so here’s the rest of the solution. \\begin{align*}{{\\bf{e}}^{5x - 4}} = {\\frac{9}{4}}\\\\ 5x - 4 & = \\ln \\left( {{\\frac{9}{4}}} \\right)\\\\ 5x & = 4 + \\ln \\left( {{\\frac{9}{4}}} \\right)\\\\ x & = {\\frac{1}{5}}\\left( {4 + \\ln \\left( {{\\frac{9}{4}}} \\right)} \\right) = 0.9621860432\\end{align*} This equation has a single solution of $$x = 0.9622$$. Now let’s take a look at some equations that involve logarithms. The main property that we’ll be using to solve these kinds of equations is, ${b^{{{\\log }_b}x}} = x$ Example 6 Solve $$\\displaystyle 3 + 2\\ln \\left( {\\frac{x}{7} + 3} \\right) = - 4$$. Show Solution This first step in this problem is to get the logarithm by itself on one side of the equation with a coefficient of 1. \\begin{align*}2\\ln \\left( {\\frac{x}{7} + 3} \\right) & = - 7\\\\ \\ln \\left( {\\frac{x}{7} + 3} \\right) & = - \\frac{7}{2}\\end{align*} Now, we need to get the $$x$$ out of the logarithm and the best way to do that is to “exponentiate” both sides using e. In other words, ${{\\bf{e}}^{\\ln \\left( {\\frac{x}{7} + 3} \\right)}} = {{\\bf{e}}^{ - \\frac{7}{2}}}$ So, using the property above with e, since there is a natural logarithm in the equation, we get, $\\frac{x}{7} + 3 = {{\\bf{e}}^{ - \\frac{7}{2}}}$ Now all that we need to do", "## Precalculus (6th Edition) Blitzer The solution is $x=\\,\\,1$. Consider the given equation system, \\begin{align} & {{\\log }_{3}}x+{{\\log }_{3}}\\left( x+2 \\right)=1 \\\\ & {{\\log }_{3}}\\left\\{ x\\left( x+2 \\right) \\right\\}=1 \\\\ & x\\left( x+2 \\right)={{3}^{1}} \\end{align} It can be further simplified as below: \\begin{align} & {{x}^{2}}+2x-3=0 \\\\ & {{x}^{2}}+3x-x-3=0 \\\\ & x\\left( x+3 \\right)-\\left( x+3 \\right)=0 \\\\ & \\left( x+3 \\right)\\left( x-1 \\right)=0 \\end{align} Therefore $x=-3\\,\\,\\text{or}\\,\\,1$ ${{\\log }_{a}}x=b$ can also be written as ${{a}^{b}}=x$, where $a>0$ so for any value of b, x is always positive. Therefore, discard the negative value of the logarithm. The solutions of the equations are $x=\\,\\,1$.", "functions are $1-1,$ we can conclude that $7x+2 = -\\frac{1}{2}.$ Therefore, $x = -\\frac{5}{14}.$ 2. $3x(e^x) + x^2 (e^x) = 0$ Solution. First notice that $x\\cdot e^x$ is a common factor. \\begin{align*} (x)(e^x)(3+x) = 0 \\Rightarrow x=0, \\; e^x = 0, \\; or \\; 3+x = 0. \\end{align*} But $e^x \\neq 0$ for any $x \\in \\mathbb{R}.$ Consequently, the second equation yields no solution. Therefore, our only solutions are $x=0$ and $x=-3.$ 3. $5^{2x}=3$ Solution. Here, we cannot use the strategy applied above, since there is no common base between $5$ and $3.$ So, we will now explore the inverses of exponential functions, which will present us with a strategy to solve such problems! ### Part 2. Logarithmic Functions (the inverses of exponential functions!) Let $b$ be a positive number such that $b \\neq 1.$ The logarithmic function with base $b,$ denoted by $\\log_b$ is defined by: $$\\log_b(x) = y \\Leftrightarrow b^y = x$$ $\\log_b(x)$ reads \"log base $b$ of $x$\" In other words: $\\log_b x=$ \"the exponent $(y)$ that we raise the base $b$ to in order to obtain $x$\" ### Exercise. Find $\\log_2 8$ \\begin{align*} \\log_2 8 = k & \\Leftrightarrow 2^k = 8 \\\\ & \\Leftrightarrow k = \\log_2 8 = 3. \\end{align*} Note: $f(x) =\\log_b x$ and $g(x)= b^x$ are inverses! Notice that in", "3. Solve $$7 + 15 \\cdot 11^{1-3x} = 10$$ in terms of $$x$$. (Hint: we can take the log of both sides of the equation. Consider: why is it ok to do this?) 4. Solve $$2^{t^2 - t} = 4$$. #### Solutions Problem 1: If $$a = \\log(n)$$, what is $$2^a$$? \\begin{align*} 2^{\\log(n)} &= 2^{\\log_2(n)} \\\\ &= n^{\\log_2(2)} \\\\ &= n^1 \\\\ &= n \\end{align*} Problem 2: Show that $$\\log_b(b^x) = x$$. \\begin{align*} \\log_b(b^x) &= x \\log_b(b) \\\\ &= x \\end{align*} Problem 3: Solve $$7 + 15 \\cdot 11^{1-3x} = 10$$ in terms of $$x$$. \\begin{align*} 7 + 15 \\cdot 11^{1-3x} &= 10 \\\\ 15 \\cdot 11^{1-3x} &= 3 \\\\ 11^{1-3x} &= \\frac{1}{5} \\\\ \\log_{11}\\left(11^{1-3x}\\right) &= \\log_{11}\\left(\\frac{1}{5}\\right) \\\\ 1 - 3x &= \\log_{11}\\left(\\frac{1}{5}\\right) \\\\ x &= \\frac{1}{3} -\\frac{1}{3} \\log_{11}\\left(\\frac{1}{5}\\right) \\end{align*} Problem 4: Solve $$2^{t^2 - t} = 4$$. \\begin{align*} 2^{t^2 - t} &= 4 \\\\ \\log_2\\left(2^{t^2 - t}\\right) &= \\log_2(4) \\\\ (t^2 - t)\\log_2\\left(2\\right) &= \\log_2(4) \\\\ (t^2 - t) &= 2 \\\\ t^2 - t - 2 &= 0 \\\\ (t - 2)(t + 1) &= 0 \\end{align*} Therefore, there are two solutions: $$t = 2$$ and $$t = -1$$. ## Practice problems: Summations 1. Simplify $$\\displaystyle \\sum_{k=1}^n k(k + 1)$$ 2. Show that the sum of the first $$n$$ positive odd integers is $$n^2$$. 3. Simplify", "\\ln e−\\ln4x+2(e^{\\ln x}⋅\\ln5)$$ Solution Apply the properties of logarithms: \\ \\begin{aligned} \\ln e-\\ln 4 x+2\\left(e^{\\ln x} \\cdot \\ln 5\\right) &=\\ln \\left(\\frac{e}{4 x}\\right)+2 x \\cdot \\ln 5 \\\\ &=\\ln \\left(\\frac{e}{4 x}\\right)+\\ln \\left(5^{2 x}\\right) \\\\ &=\\ln \\left(\\frac{e \\cdot 5^{2 x}}{4 x}\\right) \\end{aligned} Example 4 Solve: $$\\ \\ln(x−1)−\\ln(x+1)=8$$. Solution Start by condensing the left-hand side using the Quotient Rule. Because this problem involves natural logs, you'll need to put everything in the exponent of e. \\ \\begin{aligned} \\ln (x-1)-\\ln (x+1) &=8 \\\\ \\ln \\left(\\frac{x-1}{x+1}\\right) &=8 \\\\ \\frac{x-1}{x+1} &=e^{8} \\\\ x-1 &=(x+1) e^{8} \\\\ x-1 &=x e^{8}+e^{8} \\\\ x-x e^{8} &=1+e^{8} \\\\ x\\left(1-e^{8}\\right) &=1+e^{8} \\\\ x &=\\frac{1+e^{8}}{1-e^{8}} \\approx-1.0007 \\end{aligned} The solution is approximately -1.0007, which you can check by plugging it back into the equation. You will end up with ln(−1.0007−1)−ln(−1.0007+1)=8, and because you cannot take the log of a negative number, there is no solution for this equation. If you graph the left side and the right side, you can also see that there is no solution. # Continuous Exponential Growth & Decay Example 5 The interest on a sum of money that compounds continuously can be calculated with the formula I=Pert−P, where P is the amount invested (the principal), r is the interest rate, and t is the amount of time the money is invested. If you invest $1,000 in", "Generalize this to $$ax + b\\log x + c = 0$$ If this had a solution algebraic in $a, b, c: x = A(a, b, c)$, then \\begin{align} aA + b\\log A + c &= 0 \\\\ \\log A &= -\\frac{a}{b}A -\\frac{c}{b} = B(a, b, c)\\\\ A(a, b, c) &= e^{B(a, b, c)} \\end{align} where both $A$ and $B$ are non-constant and algebraic in $a, b, c$. That is impossible. I originally said that $\\log x$ would be algebraic, and that isn't quite right. This problem was devised for effect. We'll look backward here by seeing what happens if you substitute $y=3x$. \\begin{align} 8^x &= 6x \\\\ 2^{3x} &=2 \\cdot 3x\\\\ 2^y &= 2y \\end{align} Now, we consider the coincidence that $2^2 = 2\\cdot2$ and $2^1 = 2\\cdot 1$. So, $y=2$ and $y=1$ are solutions, and so $x=\\frac23$ and $x=\\frac13$ are solutions. But for general constants, you can't hope to guess at a solution. You will be in the realm of numerical approximation. Not understanding how to solve this problem is not a sign of a lack of skill. The problem is too contrived. • Thanks @Eric for the nice and detailed explanation. +1 from me. – learner Aug 23 '13 at 7:11 • Upvoted for that last sentence. – András Hummer Aug 23 '13 at 7:17 •", "sides. This will allow us to use the property $\\log_b b^x = x.$ $$\\log_5(5^{2x}) = \\log_5(3)$$ This gives $2x = \\log_5(3)$ Finally, we have: $x=\\frac{\\log_5(3)}{2}$ (which cannot be simplified further without a calculator!) 2. $4 (1+10^{5x}) = 404$ Solution. A common approach is to first isolate the exponential function. \\begin{align*} 4(1+10^{5x}) = 404 &\\Rightarrow 1+10^{5x} = \\frac{404}{4} \\\\ &\\Rightarrow 1+10^{5x} = 101 \\\\ &\\Rightarrow 10^{5x} = 100 \\end{align*} Now, we can either express both sides of the equation using the same base, or apply the inverse of the exponential function to both sides. Let's proceed using the first option, since the same base of $10$ seems fairly obvious. \\begin{align*} 10^{5x}=10^2 &\\Rightarrow 5x = 2 \\; \\text{ by 1-1 property} \\\\ &\\Rightarrow x = \\frac{2}{5} \\end{align*} 3. $e^x - 12 \\cdot e^{-x} - 1 = 0$ Solution: This is simply a quadratic in disguise, which is revealed when we multiply both sides by $e^x$ (to cancel with $e^{-x}$). \\begin{align*} e^x - 12 \\cdot e^{-x} - 1 = 0 \\Rightarrow e^x(e^x) - 12 \\cdot e^{-x}(e^x) - 1(e^x) = 0(e^x) \\\\ &\\Rightarrow (e^x)^2 - 12\\cdot e^0 - e^x = 0 \\\\ &\\Rightarrow (e^x)^2 - e^x - 12 = 0 \\end{align*} Here is our quadratic! If this is still unclear, simply let $z = e^x$. So, our equivalent equation is given by:", "Precalculus (10th Edition) $x=\\dfrac{1}{3}$ We know that $\\log_a {x^n}=n\\cdot \\log_a {x}$, hence the equation $-2\\log_4{x}=\\log_4{9}$ becomes $\\log_4{x^{-2}}=\\log_4{9}.$ RECALL: $\\log_a{b}=\\log_a{c} \\longrightarrow b=c$ Hence, $\\log_4{(x^{-2})}=\\log_4{9}\\longrightarrow x^{-2}=9$ Solve the equation above to obtain \\begin{align*} x^{-2}&=9\\\\ \\left(x^{-2}\\right)^{-\\frac{1}{2}} &=9^{-\\frac{1}{2}} \\\\ x&=\\frac{1}{3}\\end{align*}", "equation as follows: \\begin{align} 5x+8&=x^2+3x \\\\ 0&=x^2+3x-5x-8 \\\\ 0&=x^2-2x-8 \\\\ 0&=(x-4)(x+2) \\end{align} $x=4 \\text{ or } x=-2$ ### Example 4 Solve for x in the equation $3^{x+1}=81^x$ Here the bases are not equal, but it is possible to write 81 using a base of 3 as follows: \\begin{align} 3^{x+1}&=(3^4)^x \\\\ 3^{x+1}&=3^{4x} \\end{align} At this point, the left and right sides of the equation have the same base so we can solve for $x$ by setting the two exponents equal to each other: \\begin{align} x+1&=4x \\\\ 1&=4x-x \\\\ 1&=3x \\\\ x&=\\frac{1}{3} \\end{align} ### Solving Exponential Equations Using Logarithms In many cases, an exponential equation cannot be solved by using the methods of example $3$ and $4$ above because the bases cannot easily be made equal. In these cases taking the logarithm of both sides of the equation allows us to solve the equation. While you can take the log to any base, it is common to use the common log with a base of $10$ or the natural log with the base of $e$. This is because scientific and graphing calculators are equipped with a button for the common log that reads $log$, and a button for the natural log that reads $ln$ which allows us to obtain a good approximation for the common or natural log of a", "to you to check that $$\\frac{x}{7} + 3$$ will be positive upon plugging $$x = - 20.78861832$$ into it and so we’ve got the solution to the equation. 3. $$2\\ln \\left( {\\sqrt x } \\right) - \\ln \\left( {1 - x} \\right) = 2$$ Show Solution This one is a little different from the previous two. There are two logarithms in the problem. All we need to do is use Properties 5 – 7 from the Logarithm Properties section to simplify things into a single logarithm then we can proceed as we did in the previous two problems. The first step is to get coefficients of one in front of both logs. \\begin{align*}2\\ln \\left( {\\sqrt x } \\right) - \\ln \\left( {1 - x} \\right) & = 2\\\\ \\ln \\left( x \\right) - \\ln \\left( {1 - x} \\right) & = 2\\end{align*} Now, use Property 6 from the Logarithm Properties section to combine into the following log. $\\ln \\left( {\\frac{x}{{1 - x}}} \\right) = 2$ Finally, exponentiate both sides and solve. \\begin{align*}\\frac{x}{{1 - x}} & = {{\\bf{e}}^2}\\\\ x & = {{\\bf{e}}^2}\\left( {1 - x} \\right)\\\\ x & = {{\\bf{e}}^2} - {{\\bf{e}}^2}x\\\\ x\\left( {1 + {{\\bf{e}}^2}} \\right) & = {{\\bf{e}}^2}\\\\ x & = \\frac{{{{\\bf{e}}^2}}}{{1 + {{\\bf{e}}^2}}}\\\\ x & = 0.8807970780\\end{align*} Finally, we just need to make sure that the", "see that the approximation of $e$ above can be improved by adding more and more terms. Indeed this is how the expression for $e$ in equation 2.7.4 in Section 2.7 comes about. Now that we have examined Maclaurin polynomials for $e^x$ we should take a look at $\\log x\\text{.}$ Notice that we cannot compute a Maclaurin polynomial for $\\log x$ since it is not defined at $x=0\\text{.}$ Compute the $5^\\mathrm{th}$ Taylor polynomial for $\\log x$ about $x=1\\text{.}$ Solution We have been told $a=1$ and fifth degree, so we should start by writing down the function and its first five derivatives: \\begin{align*} f(x) &= \\log x & f(1) &= \\log 1 = 0\\\\ f'(x) &= \\frac{1}{x} & f'(1) &= 1\\\\ f''(x) &= \\frac{-1}{x^2} & f''(1) &= -1\\\\ f'''(x) &= \\frac{2}{x^3} & f'''(1) &= 2\\\\ f^{(4)}(x) &= \\frac{-6}{x^4} & f^{(4)}(1) &= -6\\\\ f^{(5)}(x) &= \\frac{24}{x^5} & f^{(5)}(1) &= 24 \\end{align*} Substituting this into equation 3.4.10 gives \\begin{align*} T_5(x)&= 0 + 1\\cdot (x-1) + \\frac{1}{2} \\cdot (-1) \\cdot (x-1)^2 + \\frac{1}{6} \\cdot 2 \\cdot (x-1)^3\\\\ \\amp\\hskip0.5in+ \\frac{1}{24} \\cdot (-6) \\cdot (x-1)^4 + \\frac{1}{120} \\cdot 24 \\cdot (x-1)^5\\\\ &= (x-1) - \\frac{1}{2}(x-1)^2 + \\frac{1}{3}(x-1)^3 - \\frac{1}{4}(x-1)^4 + \\frac{1}{5}(x-1)^5 \\end{align*} Again, it is not too hard to generalise the above work to find the Taylor polynomial of degree $n\\text{:}$ With a little work", "side has a base of 10. There’s no initial simplification to do, so just take the log of both sides and simplify. \\begin{align*}\\log {10^{{t^2} - t}} & = \\log 100\\\\ {t^2} - t & = 2\\end{align*} At this point, we’ve just got a quadratic that can be solved \\begin{align*}{t^2} - t - 2 & = 0\\\\ \\left( {t - 2} \\right)\\left( {t + 1} \\right) & = 0\\end{align*} So, it looks like the solutions in this case are $$t = 2$$ and $$t = - 1$$. As with the last one you could use a different log here, but it would have made the quadratic significantly messier to solve. 3. $$7 + 15{{\\bf{e}}^{1 - 3z}} = 10$$ Show Solution There’s a little more initial simplification to do here, but other than that it’s similar to the first problem in this section. \\begin{align*}7 + 15{{\\bf{e}}^{1 - 3z}} & = 10\\\\ 15{{\\bf{e}}^{1 - 3z}} & = 3\\\\ {{\\bf{e}}^{1 - 3z}} & = \\frac{1}{5}\\end{align*} Now, take the log and solve. Again, we’ll use the natural logarithm here. \\begin{align*}\\ln \\left( {{{\\bf{e}}^{1 - 3z}}} \\right) & = \\ln \\left( {\\frac{1}{5}} \\right)\\\\ 1 - 3z & = \\ln \\left( {\\frac{1}{5}} \\right)\\\\ - 3z & = - 1 + \\ln \\left( {\\frac{1}{5}} \\right)\\\\ z & = - \\frac{1}{3}\\left( { - 1 + \\ln \\left( {\\frac{1}{5}}", "## Precalculus (10th Edition) $x=-1+ln{\\frac{5}{4}}$ We know that by definition if $y=a^x$, then $log_a {y}=x$ , also $log_e x=ln x$, hence if $y=e^x$, then $ln{e^x}=log_e {e^x}=x$ and vice versa and that $ln{\\frac{1}{x}}=-ln{x}$. Hence if $4e^{x+1}=5$, then $e^{x+1}=\\frac{5}{4}$. Solve the equation above to obtain (after taking $ln$ of both sides): \\begin{align*}x+1=ln{\\frac{5}{4}}\\end{align*}\\begin{align*}x=-1+ln{\\frac{5}{4}}\\end{align*}", "simpler. Have a look here for a good explanation on finding asymptotic complexities, with a nice intuitive solution for your problem instance. – Paresh Nov 3 '12 at 13:33 migrated from stackoverflow.comOct 31 '12 at 22:42 This question came from our site for professional and enthusiast programmers. We will use Raphael's suggestion and unfold the recurrence. In the following, all logarithms are base 2. We get \\begin{align*} T(n) &= n^{1/2} T(n^{1/2}) + cn \\\\ &= n^{3/4} T(n^{1/4}) + n^{1/2} c n^{1/2} + cn\\\\ &= n^{7/8} T(n^{1/8}) + n^{3/4} c n^{1/4} + 2cn\\\\ &= n^{15/16} T(n^{1/16}) + n^{7/8} c n^{1/8} + 3cn \\\\ & \\ldots \\\\ &= \\frac{n}{2} T(2) + c n \\beta(n) \\end{align*}. where $\\beta(n)$ is how many times you have to take the square root to start with n, and reach 2. It turns out that $\\beta(n) = \\log \\log n$. How can you see that? Consider: \\begin{align*} n &= 2^{\\log n}\\\\ n^{1/2} &= 2^{\\frac{1}{2} \\log n} \\\\ n^{1/4} &= 2^{\\frac{1}{4} \\log n} \\\\ \\ldots \\end{align*} So the number of times you need to take the square root in order to reach 2 is the solution to $\\frac{1}{2^t} \\log n \\approx 1$, which is $\\log \\log n$. So the solution to the recursion is $c n \\log \\log n + \\frac{1}{2}n$. To make this absolutely rigorous, we", "How to solve this logarithm system? I am new to logarithms and I am having trouble with this logarithm system. \\begin{align*} \\log_9(x) + \\log_y(8) & = 2, \\\\ \\log_x(9) + \\log_8(y) & = 8/3. \\end{align*} A step-by-step procedure would be highly appreciated. Use the fact that $$\\log_b(a) = \\dfrac1{\\log_a(b)}$$ Hence, if we denote $\\log_9(x) = a$ and $\\log_y(8) = b$, we get that \\begin{align} a+b & = 2\\\\ \\dfrac1a + \\dfrac1b & = \\dfrac83 \\implies \\dfrac{a+b}{ab} = \\dfrac83 \\implies ab = \\dfrac34 \\end{align} Now solve for $a$ and $b$ and hence $x$ and $y$. If $\\log(\\cdot)$ denotes logarithm in base $10,$ then $$\\log_y(x) = \\frac{\\log(x)}{\\log(y)}.$$ Apply through & simplify to see what system you will get. Substitute $X$ for $\\log(x),$ and $Y$ for $\\log(y).$ You should have 2 equations in 2 unknowns.", "# Solving the equation $3^{5x-2}=8^{8x-9}$ I'm trying to solve the equation $$3^{5x-2}=8^{8x-9}.$$ I'm assuming I need to do some work with logarithms, but I don't know what to do. - Is this homework? What laws for lograithms do you know? For the formatting: Use curly brackets for exponents ... – martini Sep 17 '11 at 18:24 Hint: Let us write $\\log x$ for the logarithm of $x$ to your favourite base ($10$, $e$, $2$, it doesn't matter). We have in general $\\log(a^b)=b\\log a$. – André Nicolas Sep 17 '11 at 18:44 Did you look at the logarithm laws - did you try anything? – AD. Sep 17 '11 at 18:52 this homework isn't going to be marked, it's simply review for my calculus course. I know all of the main log laws (subtracting log, addition of log, changing base, and the a*log(x) = log(x^a) – kubasub Sep 17 '11 at 19:11 \\begin{align*} \\ln 3^{5x-2} &= \\ln 8^{8x-9}\\\\ \\iff \\ln 3 \\cdot (5x-2) &= \\ln 8 \\cdot (8x-9) \\\\ \\iff 8\\ln 8 \\cdot x-5\\ln 3 \\cdot x &=-2\\ln 3+9\\ln 8\\\\ \\iff x(8\\ln 8-5\\ln 3)&=-2\\ln 3+9\\ln 8\\\\ \\iff x&=\\frac{-2\\ln 3+9\\ln 8}{8\\ln 8-5\\ln 3}. \\end{align*} - great answer @Andres ,he needs only excel and using function ln he can calculate everything – dato datuashvili Sep 17 '11 at 18:44 thanks a", "you that don’t know, $\\log_b a$ represents the quantity $x$, where $b^x = a$ for any $a,b$ where $b > 0$. ‘log’ is an abbreviation of ‘logarithm’. So unless you have some experience working with logarithms, it might be useful to convert the end-statement out of logarithm-world and into exponential world. As regards the statement $7^{x+7} = 8^x$, it might be useful to bring everything that is raised to the power $x$ onto one side. Solution: This problem just involves some algebraic manipulation, made slightly more complicated by the presence of exponentials and logarithms. End-statement/goal: $x = \\log_b 7^7 \\Rightarrow b^x = 7^7$ $7^{x+7} = 8^x \\Rightarrow 7^7 = \\frac{8^x}{7^x} = (\\frac{8}{7})^x$ Therefore, $b = \\frac{8}{7}$. ## No 33: Discussion: To start off with, the problem might seem daunting because of the sheer number of possibilities – it seems like $n$ could be absolutely anything, so how are you supposed to maximize and minimize? The minimum is easier, because you can start at a triangle ($n=3$) and work your up, so start there. For the maximum value of $n$, note that the average internal angle will have to be quite large, so it might be worth identifying some large primes. Remember that the key to this problem is the fact that all angles must be odd primes: so" ]

[ "+0 Help! 0 133 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$", "+0 # Help! 0 39 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 #1 +3563 +2 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$ . Dec 16, 2018", "For certain ordered pairs $$(a,b)\\,$$ of real numbers, the system of equations $$ax+by=1\\,$$ $$x^2+y^2=50\\,$$ has at least one solution, and each solution is an ordered pair $$(x,y)\\,$$ of integers. How many such ordered pairs $$(a,b)\\,$$ are there? (第十二届AIME1994 第7题)", "# Circle and a line! $\\begin{cases} ax+by=1 \\\\ x^2 + y^2 = 50 \\end{cases}$ For certain ordered pairs $$(a,b)$$ of real numbers, the system of equations above has at least one solution, and each solution is an ordered pair $$(x,y)$$ of integers. How many such ordered pairs $$(a,b)$$ are there? ×", "# A pair of equations. Algebra Level 4 How many ordered pairs of real numbers $$(x,y)$$ satisfy: $(x-1)(y^{2}+6)= y(x^{2}+1),$ $(y-1)(x^{2}+6)=x(y^{2}+1)?$ ×", "# Question #01a0e In order to check if an ordered pair is a solution to a particular set of equations, you have to plug each number in (the ordered pairs take the format of $\\left(X , Y\\right)$) and see if the equations are still valid. For the first problem: The ordered pair is $\\left(4 , 3\\right)$, meaning that $X = 4 \\mathmr{and} Y = 3$. Plug these values in to get $2 \\left(4\\right) + 5 \\left(3\\right) = 25 , 8 + 15 = 25 , 23 \\ne 25$, thus, the ordered pair is not a solution. Follow this same set of steps for the other two equations.", "# Is the ordered pair (2,7) is a solution of the equation: y - x = 5? May 6, 2016 Yes. #### Explanation: The ordered pair $\\left(2 , 7\\right)$ is in the form $\\left(x , y\\right)$ so $x = 2$ $y = 7$ Substitute $x$ and $y$ values into the equation $y - x = 5$ $7 - 2 = 5$ $5 = 5$ Since the equation is correct, the ordered pair is a solution.", "+0 # help 0 159 1 Find all values a for which there exists an ordered pair (a,b) satisfying the following system of equations: \\begin{align*} a + ab^2 & = 40b, \\\\ a - ab^2 & = -32b. \\end{align*} May 10, 2019 #1 +111455 +1 See here : https://web2.0calc.com/questions/help-plz-quick-dont-understand May 10, 2019", "+0 0 157 1 For what ordered pair (a,b) are there infinite solutions (x,y) to the system \\begin{align*} 2ax+2y&=b,\\\\ 5x+y&=-3? \\end{align*} Mar 16, 2020 edited by Guest Mar 16, 2020 edited by Guest Mar 16, 2020 #1 +112029 0 There will be an infinite number of solutions if the two equations are identical. So multiple the second one by 2 the equate coefficients. Mar 17, 2020", "# This is the only title I could think of How many ordered integer pairs $$(a,b)$$ satisfy the equation $$a^{2} + 3ab - 2b^{2} = 124.$$ ###### Image Credit: Wikimedia Quadratic Equation Discriminant by KSmrq × Problem Loading... Note Loading... Set Loading...", "# Very very beautiful Algebra Level 5 Consider the equation $(x+y)^2=(x+1)(y-1)$ If all the possible real ordered pairs $$(x,y)$$ that satisfy the equation above are $$(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$$ then find the value of $$x^2_1+...+x^2_n+y^2_1+...+y^2_n$$. • If you think there are infinite solutions then answer 777 and if you think no real solutions answer 666. • Ordered pair means (11,12),(12,11) are considered different. ×", "How many $$(x,y)$$ are there? Algebra Level 4 How many real numbered ordered pairs $$(x, y)$$ satisfy the following inequalities: \\begin{align} (x+2y)^2 &\\leq (2x-3)(4y+3)\\\\ (x-2y)^2 &\\leq (5-2x)(4y-5)? \\end{align} ×", "# Ordered Pairs How many ordered pairs of integers $$(m,n)$$ satisfy the equation $$\\dfrac{m}{20}=\\dfrac{15}{n}$$? Two Permutations of the same solution i.e $$(m,n) and (n,m)$$ are considered different. × Problem Loading... Note Loading... Set Loading...", "4. Hence, the solution of the system of equations is the ordered pair (3, 4).", "each $x_i$ in the solution to the original equation, where $i$ is an integer, $x_i$ can only equal $0$ or $a$. Therefore, there are $\\boxed{2^n}$ ordered pairs that are solutions to the equation. 2001 Pan African MO (Problems) Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 5 All Pan African MO Problems and Solutions", "the ordered pair is a solution to the given system of linear equations. (5,3) x-y=2 x+y=8 Forms of linear equations", "and \\begin{align*}y\\end{align*}. 15. An ordered pair is never a solution for a linear system. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English system of equations A system of equations is a set of two or more equations.", "## Intermediate Algebra for College Students (7th Edition) The ordered pair (4, 19) is a/an solution of the equation $y=x^{2}+3$ because when 4 is substituted for x and 19 is substituted for y, we obtain a true statement. We also say that (4,19) satisfies the equation.", "has six number pairs for solution.", "Systems Consider the linear system \\begin{aligned} 2x+y+z&=5,\\\\ x-3y+2z&=1.\\end{aligned} Which of the following best describes the set of solutions to this system? # Linear Systems For what real-number values of $$a$$ will the following system of equations have no solutions? \\begin{aligned} ax+2y&=4\\\\ 2x-ay&=6.\\end{aligned} ×" ]

[ "The distance between them equals $b$. Using the distance formula, we get $PQ = b = \\sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\\sqrt2 \\times |x_2 - x_1| = 58\\sqrt2\\times z$ $(*)$ WLOG, we can assume that $x_2 > x_1$. Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. • There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. • There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... • Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = \\boxed{600}$ triangles. Solution 4 We present a non-analytic solution; consider the lattice points on the line $41x+y=2009$. The line has intercepts $(0, 2009)$ and $(49, 0)$, so the lattice points for $x=0, 1, \\ldots, 49$ divide the line into $49$ equal segments. Call the area of the large triangle $A$. Any triangle", "-y^2 = 84 = (x+y)(x-y)$, for $x$ and $y$ to be integers, $(x+y)$ and $(x-y)$ have to be the same parity because $(x+y)+(x-y)=2x$ and, for the total of two numbers to be even, they either have to be both odd or both even. As $(x+y)(x-y)=84$ at least one bracket has to be even. As we require diophantine solutions, both brackets must be even. The only factorisations of 84 into two even numbers are: $$84=2\\times 42 = 42 \\times 2 = -2\\times -42 = -42 \\times -2 = 6\\times 14 = 14\\times 6 = -6\\times -14 = -14\\times -6.$$ Each of these gives a distinct solution so the 8 solutions are $x= \\pm 22, y=\\pm 20$ (4 solutions) and $x=\\pm 10, y=\\pm 4$ ( 4 solutions). There are two lattice points on the hyperbola in the first quadrant: (10,4) and (22,20). The lattice points (10, -4) and (22, -20) are the reflections of these points in the $x$-axis. Also $x\\geq \\sqrt 84$ or $x\\leq -\\sqrt 84$ so there are two branches of the hyperbola. The other four lattice points lie on the other branch of the hyperbola and are the reflections of these four points in the $y$-axis.", "# The graphs of $y=ax$ and $y=\\arctan(bx)$ intersect at three distinct points if? [closed] I have a question relating to calculus and inverse trigonometric functions. Any help is appreciated. The question is: The graphs of $$y=ax$$ and $$y=\\arctan(bx)$$ intersect at three distinct points if? A: $$0 B: $$a C: $$a=b$$ D: $$b The correct answer is D. Upon inserting the graphs into desmos I can visually see how they intersect at three distinct points but I am struggling to understand why. Thank you for sharing your knowledge! • These solutions are not going to be elementary, so you're not going to be able to reasonably construct them in terms of $a$ and $b$. Do you know other tools for counting roots, like intermediate value theorem and/or mean value theorem (or Rolle's theorem)? May 26 at 11:55 • The graphs will intersect in three distinct points as long as b< a. May 26 at 11:59 • @GeorgeIvey Thank you! Not sure if the explanation will be too complicated but why is that so? May 26 at 12:01 • @TheoBendit Hmm unfortunately not, closest thing I can think of is to do with polynomial roots, thanks a lot anyways! May 26 at 12:02 • First thing is that both a and b should be of the same sign for the", "# Difference between revisions of \"2011 AMC 12B Problems/Problem 19\" ## Problem A lattice point in an $xy$-coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx + 2$ passes through no lattice point with $0 < x \\leq 100$ for all $m$ such that $\\frac{1}{2} < m < a$. What is the maximum possible value of $a$? $\\textbf{(A)}\\ \\frac{51}{101} \\qquad \\textbf{(B)}\\ \\frac{50}{99} \\qquad \\textbf{(C)}\\ \\frac{51}{100} \\qquad \\textbf{(D)}\\ \\frac{52}{101} \\qquad \\textbf{(E)}\\ \\frac{13}{25}$ ## Solution It is very easy to see that the $+2$ in the graph does not impact whether it passes through lattice. We need to make sure that $m$ cannot be in the form of $\\frac{a}{b}$ for $1\\le b\\le 100$, otherwise the graph $y= mx$ passes through lattice point at $x = b$. We only need to worry about $\\frac{a}{b}$ very close to $\\frac{1}{2}$, $\\frac{m+1}{2m+1}$, $\\frac{m+1}{2m}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\\le b\\le 100$, the smallest is $\\frac{50}{99}$, so answer is $\\boxed{\\frac{50}{99} \\textbf{(B)}}$ ## Solution 2 Like in the first solution, we note that the $+2$ does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line $y = mx$. A line with a slope", "we can quickly find all of the lattice points by adding $4$ to the $x$ coordinate until we get to $k$ (where $k$ is the $x$-intercept, as shown above). However, if we can’t find a lattice point between $x=0$ and $x=4$, then there won’t be any lattice points at all, because lattice points should be at intervals of $4$, so there can’t be an interval of $4$ that doesn’t contain one. If $x=1$, then $3x=3$, so $4y$ must be $56$ since $3+56=59$, so $y=56\\div 4=14$. So the point $(1,14)$ lies on the line and is a lattice point. Then there are also lattice points when $x=5, 9, 13 …$. To work out how many there are in the first quadrant, we need to find the value of $k$, because if $x>k$ then the lattice point will not be in the first quadrant. When $x=k$, $y=0$, and so $3\\times k+4\\times 0=59$, so $k=59/3=19\\frac{2}{3}$. So the lattice points have $x$-coordinates $1, 5, 9, 13$ and $17$ - which means there are $5$ lattice points. Using factorisation and properties of numbers Suppose $(a,b)$ is lattice point on the line $3x+4y=59$ in the first quadrant. Then $a$ and $b$ are positive whole numbers with $3a+4b=59$. $3a+4b$ can be written as $3(a+b)+b$. This is useful because, starting from $3(a+b)+b=59$, $3(a+b)=59-b$, so $59-b$ must", "will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$, which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \\cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=\\boxed{574}$ squares and circles intersected in total. ## Solution 3 This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$, the slope of our line is $\\frac{3}{7}$, and we only need to consider the line in the rectangle from the origin to $(7,3)$, and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$, we can think of representing a square centered at that lattice point as all points equal to $(x_1 \\pm a, y_1 \\pm b)$ s.t. $0 \\leq a,b \\leq \\frac{1}{10}$. If the line $y = \\frac{3}{7}x$ intersects the square, then we must have $\\frac{y_1 + b}{x_1 + a} = \\frac{3}{7}$. The line with the least slope that", "determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$. When $k<0$, $a>0$, $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$, $a>0$, then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$, so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$, so our final bounds are $21, giving us $\\boxed{057}$ valid values of $c$. ## Remark The graphs of $F(x)=\\left||20|x|-x^2|-c\\right|$ and $G(x)=21$ are shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp Move the slider around for $21 to observe how they intersect for $12$ times. ~MRENTHUSIASM", "lattice points. Remember that $y$ needs to be positive, so $3x$ must not be greater than $59$, so $x$ must be less than $20$ (since $3\\times 20=60>59$). So the table only needs to go up to $x=19$. Counting the points in the table with whole number $y$ coordinates, there are 5 lattice points. Plotting these points gives the graph below (the lattice points are marked in red). The gradient of the line is $-\\frac{3}{4}$. You can work this out by considering changes in $x$ and $y,$ or by rearranging $3x+4y=59$ into $y=-\\frac{3x}{4}+\\frac{59}{4}$. This means that every time $x$ increases by $4$, $y$ decreases by $3$. This is shown on the diagram below. So, if you start at a lattice point, then there will be another lattice point $4$ to the left and $3$ down, and another one $8$ to the left and $6$ down, and so on. There will also be another lattice point $4$ to the right and $3$ up, and so on. So the lattice points, if there are any, will be evenly spaced along the $x$-axis at intervals of $4$, as shown below – where this time, we assume that the dark red crosses mark lattice points. That means that, if we can find a lattice point on the line between $x=0$ and $x=4$, then", "line, then $$\\sqrt{3}$$ = tan 60° = $$\\left|\\frac{m+1}{1-m}\\right|$$ and so, m satisfies the equation (m + 1)2 = 3(m – 1)2 (or) m2 – 4m + 1 = 0 …………………….. (1) But the straight line having slope’m1 and passing through the origin is y = mx ……………………. (2) So the equation of the pair of lines passing through the origin and inclined at 60° with the line L is obtained by eliminating’m’ from the equations (1) and (2). Therefore the combined equation of this pair of lines is ($$\\frac{y}{x}$$)2 – 4($$\\frac{y}{x}$$) + 1 = 0 (i.e.,) x2 – 4xy + y2 = 0 Which is the same as the given pair of lines. Hence, the given triad of lines form an equilateral triangle. Question 5. Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax2 + 2hxy + by2 = 0 is $$\\frac{\\left|a \\alpha^{2}+2 h \\alpha \\beta+b \\beta^{2}\\right|}{\\sqrt{(a-b)^{2}+4 h^{2}}}$$. [May 11, 07; Mar. 07, 04] Solution: Let ax2 + 2hxy + by2 = (l1x + m1y) (l2x + m2y) Then the separate equations of the lines represented by the equation ax2 + 2hxy + by2 = 0 are L1 : l1x + m1y = 0 and L2 : l1x + m1y = 0 Also, we have l1l2 =", "the line connecting them is parallel to (0,1) if we made a line that passes through it. and since no other point in the circle that intersects the line that passes through (0,1) but (0,1) itself! Jul 6 '17 at 12:42 • In this construction, the product of A and B may have two possible values Jul 6 '17 at 13:51 • Is your product associative? That requires proof. What is the square of an element? What, for example, is $(1,0)\\times (1,0)$? Is it $(0,-1)$? That seems the most natural. But then...Well, look at $A=(1,0),B=(-1,0)$. Clearly $A(AB)=Ae=A$, yes? But $(AA)B=(0,-1)\\times (-1,0)$ which does not appear to be $A$. Have I blundered? If not, that would be a failure of associativity. – lulu Jul 6 '17 at 13:52 This is indeed a group! Moreover, this same group structure can be put onto any conic -- not just a circle. The reason why this is a group and why this works with conics is due to where the proof for associativity comes from: Pascal's theorem. As you have already discovered, the rest of the group axioms come more-or-less for free from the definition of the operation. You may be interested in two papers on the subject: As you will see in Lemmermeyers' paper, you can endow elliptic curves with a", "# A circle touches the parallel lines $3x-4y-7=0$ and $3x-4y+43=0$ and has its centre on the line $2x-3y+13=0$. A circle touches the parallel lines $3x-4y-7=0$ and $3x-4y+43=0$ and has its centre on the line $2x-3y+13=0$. Find the equation of the circle. My Attempt; Let $(h,k)$ be the centre of the circle. Then, $2h-3k+13=0$. Now, how do I do the rest? • That’s not much of an attempt, is it? – amd Dec 4 '17 at 20:59 If it touches two parallel lines, then its center must be right in between them. The average of $-7$ and $43$ is $18$. Thus $(h,k)$ lies on the line $3x-4y+18=0$. Combining this with the equation you already have, can you solve for $h$ and $k$? • Why is it true that $(h,k)$ lies on the line $3x-4y+18=0$? – pi-π Dec 4 '17 at 15:02 • Since $18$ is midway between $-7$ and $43$, the line $ax+by+18=0$ is midway between the lines $ax+by-7=0$ and $ax+by+43=0$. – G Tony Jacobs Dec 4 '17 at 15:04", "# Circle and a line that passes through it Given a line with equation: $y=ax-3$ that passes through a circle with equation $(x-1)^2+(y-1)^2= 1$. Find the range of values of $a$. I tried graphing and got: $0<x<2$ and $0<y<2$. I also tried finding $a$ by substituting $x$ and $y$ into $y=ax-3$ which really confuses me. Could you help me in solving this problem? • What happens to the line $y=ax-3$ when $a$ gets really big, and does it pass through the circle? think in terms of the graph – user160738 Dec 16 '15 at 16:36 • The line $y=0$ does touch the circle - your circle has center $(1,1)$ and radius $1$. But $y=ax-3$ never actually becomes line $y=0$, it just gets closer and closer to that line. But still if the line is really steep, then it should pass through the circle in two different points. So try imagining $a$ getting smaller and smaller from there - as it gets smaller gradient of line gets smaller, and at some point it must be tangent to the circle. Is that clear? – user160738 Dec 16 '15 at 16:48 • well, substitute $y=ax-3$ into $(x-1)^{2}+(y-1)^{2}=1$ and solve the quadratic equation with respect to $a$ and see that for what range of $x$ you can find $a$ and then I", "-14\\times -6.$$ Each of these gives a distinct solution so the 8 solutions are $x= \\pm 22, y=\\pm 20$ (4 solutions) and $x=\\pm 10, y=\\pm 4$ ( 4 solutions). There are two lattice points on the hyperbola in the first quadrant: (10,4) and (22,20). The lattice points (10, -4) and (22, -20) are the reflections of these points in the $x$-axis. Also $x\\geq \\sqrt 84$ or $x\\leq -\\sqrt 84$ so there are two branches of the hyperbola. The other four lattice points lie on the other branch of the hyperbola and are the reflections of these four points in the $y$-axis.", "m^2x^2 - 8mx + 16 = 25\\\\ \\iff\\quad& (m^2 +1)x^2 + (-10 - 8m)x+16 = 0. \\end{align*} For $y=mx$ to be a tangent to the circle, this quadratic equation must have only one root, that is, its discriminant must be zero. The discriminant of the quadratic equation $ax^2+bx+c = 0$ is $b^2-4ac$. The equation has a repeated root if, and only if, the discriminant is zero. Now \\begin{align*} &(-10-8m)^2 - 4 \\times (m^2+1) \\times 16 = 0\\\\ \\iff\\quad& 100+160m+64m^2-64(m^2+1)=0\\\\ \\iff\\quad& 100+160m+64m^2-64m^2-64=0\\\\ \\iff\\quad& 160m+36=0\\\\ \\iff\\quad& m=-\\frac{9}{40}. \\end{align*} So $y=-\\dfrac{9}{40}x$ is the equation of this tangent. Note that in general we would expect to have two solutions for $m$ from $b^2=4ac$. When the origin $O$ is outside the circle, there are exactly two tangents to the circle from $O$. Here, however, one of our tangents is the $y$-axis, that has infinite gradient. We could see the equation $100+160m+64m^2-64m^2-64=0$ as having one solution for $m$ that is infinite.", "Spiral Snail A snail slithers around on a coordinate grid. At what position does he finish? Bucket of Water Lisa's bucket weighs 21 kg when full of water. After she pours out half the water it weighs 12 kg. What is the weight of the empty bucket? How Far Does it Move? Experiment with the interactivity of \"rolling\" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage. Lattice Points on a Line Age 11 to 14 ShortChallenge Level Using a table of values Points in the first quadrant have $x,\\hspace{2mm}y>0$, so we could begin by substituting values into the equation of the line to check for lattice points on the line. Using the table below, we can try positive whole numbers for $x$, and then find the value of $3x$. If $x=1$, then $3x=3$, so $4y$ must be $56$ since $3+56=59$, so $y=56\\div 4=14$. So the point $(1,14)$ lies on the line and is a lattice point. This is shown on the first line of the table. If $x=2$, then $3x=6$, so $4y$ must be $53$ since $6+53=59$, so $y=53\\div 4=13.25$. So the point $(2,13.25)$ lies on the line, but it is not a lattice point. Filling in the whole table will give all of the possible", "of $|x/4| - |x-60|$ on the intervals $$x < 48, \\quad 48 < x < 80, \\quad x > 80.$$ We know that $|x/4| - |x-60|$ doesn't change sign within these intervals (for then we would have detected it when solving for equality). It easily follows that $y \\ge 0$ only when $48 \\le x \\le 80$. From this, we note that on this interval, we have $$y = |x/4| - |x-60| = \\begin{cases} x/4+x-60, & 48 \\le x \\le 60 \\\\ x/4-x+60, & 60 < x \\le 80. \\end{cases}.$$ This gives us the desired boundary, which is now easy to plot. The vertices are therefore $$(48, 0), (60, \\pm 15), (80, 0),$$ and the area is straightforward to compute. You have six cases to consider: 1) $x\\geq60, y\\geq0\\to y=-\\frac{3}{4}x+60$. 2) $x\\geq60, y\\leq0\\to y=\\frac{3}{4}x-60$. 3) $0\\leq x\\leq60, y\\geq0\\to y=\\frac{5}{4}x-60$. 4) $0\\leq x\\leq60, y\\leq0\\to y=-\\frac{5}{4}x+60$. 5) $x\\leq0, y\\geq0\\to y=\\frac{3}{4}x+60$. 6) $x\\leq0, y\\leq0\\to y=-\\frac{3}{4}x-60$. Now, the figure is the hexagon with vertices $(-80,0),(0,60),(60,15),(80,0),(60,-15),(0,-60)$", "five points. One or more of the five resulting lines will be further to the top-right than the others. The point(s) in your feasible region this passes though will give where $z$ is maximised. If after all this you do not find that the maximum possible value of $z$ is $62$ then either you or I have made a mistake. - Thanks for your answer ! – Bruno Feb 10 '11 at 16:12 To draw any straight line, all you need is two points. In the case of $ax+by = 0$, the line clearly passes through $(0,0)$. Further, the line passes through couple of other points which are easy to recognize namely $(b,-a)$ and $(-b,a)$. So just mark these points on the sheet and draw it. If $a=0$, then the line is nothing but the $X$ axis while if $b=0$, then the line is nothing but the $Y$ axis. - Could you explain what exactly you mean? – user17762 Feb 9 '11 at 8:18", "0 B. 1 C. 2 D. 3 E. 4 Curve $$x^2 + y^2 = 4$$ is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives $$(2 - y)^2 + y^2 = 4$$ or $$2y^2 - 4y + 4 = 4$$ or $$y^2 - 2y = 0$$ from where $$y = 0$$ and $$y = 2$$. Thus, the line and the circle intersect at points $$(2, 0)$$ and $$(0, 2)$$. Alternative Explanation Look at the diagram below: $$x^2 + y^2 = 4$$ is the equation of a circle centered at the origin and with the radius of 2. Now, $$y=2-x$$ is the equation of a line with $$x$$-intercept $$(2, 0)$$ and $$y$$-intercept $$(0,2)$$. Notice that these points are also the intercepts of given circle with $$X$$ and $$Y$$ axis hence at these points the line and the circle intersect each other. _________________ Manager Joined: 08 Feb 2014 Posts: 205 Location: United States Concentration: Finance GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking) ### Show Tags 03 Nov 2014, 18:55 Is there a quick way to", "Rules 1. A n-step, where n is a positive integer, is a line segment of length $$n$$ and slope $$0$$ or $$undefined$$, or a line segment of length $$n\\sqrt{2}$$ and slope of $$1$$ or $$-1$$ . The endpoints of a n-step must be on a lattice point. For example, you may construct a 2-step segment with length of $$2\\sqrt{2}$$ and endpoints on $$(2,2)$$ and $$(4,4)$$. 2. Your $$nth$$ construction must be an n-step. For example, your 1st construction must be a 1-step. Your 2nd construction must be a 2-step. Your 14th construction must be a 14-step. 3. For every $$n\\geq 2$$, a n-step must share an endpoint with a (n-1)-step. 4. Two n-steps may only intersect at their endpoints. 5. Every lattice point may contain at most two n-steps. 1. On the graph, construct a figure that passes through all lattice points $$(a,b)$$ where $$-3 \\leq a \\leq 3$$ and $$-3 \\leq b \\leq 3$$. 2. On the graph, construct a figure that passes through all lattice points $$(a,b)$$ where $$-4 \\leq a \\leq 3$$ and $$-4 \\leq b \\leq 3$$. 3. Construct a figure that passes through all lattice points $$(a,b)$$ where $$0 \\leq a \\leq x$$ and $$0 \\leq b \\leq y.$$", "do in each case is change the $\\le$ to = to get the lines $W = 700$ and $G = 400$. So the 4 lines we need to draw on the graph are: 1. $G = -1.25W + 1000$ 2. $G = -0.5W + 600$. 3. $W = 700$ 4. $G = 400$. I actually found it easier to draw lines 3 and 4 first. After drawing the graph of these lines, it's easy to see that there are 5 vertices that we need to consider as candidate solutions to the problem (actually only 3 I think but just to be sure, I'll consider the point at which line 3 croses the W axis and the point at which line 4 crosses the G axis as well). The other 3 vertices occur when (1) line 2 crosses line 4, (2) when line 1 crosses line 2 and (3) when line 1 crosses line 3. We don't need to consider the point where line 2 crosses line 3 beacuse it's outside of the feasible region. Now we find the coordinates of the vertices as follows: (1) We need to find the coordinates at which line 2 crosses line 4. The equation for line 4 tells us that $G = 400$ at the vertex. Since line 2 crosses this point" ]

[ "+0 # Help! 0 39 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 #1 +3563 +2 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$ . Dec 16, 2018", "+0 Help! 0 133 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$", "# Question #82223 Mar 31, 2017 (1, -4), (2, -8) and (3, -12) #### Explanation: you can choose any number of $x$ and plug into the equation $y = - 4 x$ $x = 1 , y = - 4 \\left(1\\right) = - 4$ $x = 2 , y = - 4 \\left(2\\right) = - 8$ $x = 3 , y = - 4 \\left(3\\right) = - 12$ order pairs $= \\left(1 , - 4\\right) , \\left(2 , - 8\\right) \\mathmr{and} \\left(3 , - 12\\right)$ Mar 31, 2017 Three ordered pair solutions to $y = - 4 x$: $\\left(1 , - 4\\right) , \\left(5 , - 20\\right) , \\left(- 2 , 8\\right)$ #### Explanation: An ordered pair is a grouping of two numbers with a distinct order that are related by an equation used to define the points or coordinates of a curve (or line) on a graph. For example if you were required to graph the points $\\left(- 1 , - 1\\right) \\left(1 , 1\\right) , \\left(2 , 2\\right) , \\left(3 , 3\\right) , \\left(4 , 4\\right)$, you would see a straight line that would pass through the origin with a positive slope at $45 \\mathrm{de} g r e e s$. The order in the pair always tells you that the first number indicates the value of", "lattice points which lie on this curve. Given the square term, we will get symmetric pairs of solutions about the $x$-axis (Figure 1). In geometric terms, it might be viewed as the problem of which squares and cubes with sides of integer units are inter-convertible by the addition of $k$ units. Mordell had pointed out about a century ago that these equations might have either no integer solutions or only a finite number of them. Let is consider a concrete example with $k=1$: $y^2=x^3+1$. One can get three trivial solutions right away: by setting $x=-1$ we get $y=0$. Similarly, setting $x=0$, we get $y=\\pm 1$. Further, we see that if $x=2$ we get $y=\\pm 3$. Thus, we can write the solutions as $(x,y)$ pairs: (-1,0); (0,1); (0, -1); (2,3); (2,-3): a total of 5 unique integer solutions. Can there be any more solutions than these? To get an intuitive geometric feel for this we first observe how these solutions sit on the curve (Figure 1). We notice that the 3 distinct ones are on the same straight line (also applies to their mirror images via a mirrored line). This is a important property of elliptic curves (being cubic curves) that allows us to understand the situation better. Figure 2. In Figure 2, we consider the curve $y^2=x^3+1$ in", "visualize the graph of the equation: Graph of $y=2x-3$: The equation is the graph of a line through the three points found above. The line continues on to infinity in each direction, since there is an infinite series of ordered pairs of solutions. Graphing Non-Linear Equations ### Example 1 What graph will the following equation make? $x^{2}+y^{2} = 100$ Let’s figure it out by choosing some points to plot. First, let’s try $x=0$: \\begin{align} (0)^{2}+y^{2} &= 100 \\\\ y^{2} &= 100 \\\\ \\sqrt{y^{2}}&=\\sqrt{100} \\\\ y &= \\pm10 \\end{align} So we plot $(0,10)$ and $(0,-10)$. Note that we don’t always have to choose values for $x$. For example, let’s now try setting $y=0$. Through the same arithmetic as above, we get the ordered pairs $(10,0)$ and $(-10,0)$. Plot these as well. We still don’t have enough points to really see what’s going on, so let’s choose some more. Let’s try solving for $y$ when $x=6$: \\begin{align} (6)^2+y^2&=100 \\\\ 36+y^2&=100 \\\\ 36+y^2-36&=100-36 \\\\ y^2&=64 \\\\ y&=\\pm 8 \\end{align} So that’s two new points: $(6,8)$ and $(6,-8)$. We get similar results with $x=-6$, to get $(-6,8)$ and $(-6,-8)$. Now you can begin seeing that we’re drawing a circle with a radius of 10: Graph of $x^2+y^2=100$: This is a graph of a circle with radius 10 and center at the origin.", "our website if we get Correct. Make sure that the solution to the system of equations order of the following ordered pairs are points on coordinate... > are graphed in the form ( a, B } equals the unordered pair B! For y + 5 Best Newest Oldest left of a point on the boundary line coordinate. For which ordered pair is important: x always comes before y standard form, which is the.! Use the ordered pair appears to be on the boundary line reach an elevation of 20 below... -8 ) 4x - 3y = 36 -3, -8 ) 4x - 3y = 36 the point more... Determine vertical position, count y squares along the y-axis, as in... Given system of equations 8, -54 ) ( 0, -6 ) B ) (,... Of equations if the point satisfies ordered pair solution than one equation, we find that position of a point, we. 4.1 Problem 52E equation 2x − y = -2x - 6 right angle at point! Pair suggests the order in which values are written Graphing, Halloween false, or sequences ( sometimes lists. Indicated point as a solution Algebra 17th Edition Lynn Marecek Chapter 4.1 52E. A Halloween party, after a test, Subjects: Math, Graphing, Halloween sequences ( sometimes lists! The order in which values are", "3.76 Re: For how many ordered pairs (x , y) that are solutions of the [#permalink] ### Show Tags 02 Jun 2017, 09:54 tonebeeze wrote: $$2x + y = 12$$ $$|y| \\leq 12$$ For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers? A. 7 B. 10 C. 12 D. 13 E. 14 This problem is best tackled via a table. You can set the columns up as follows: x, 12 - 2x, |y| x | 12 -2x| |y| 0 | 12 | 12 1 | 10| 10 2 ... 3 4 5 6 7 8 9 10 11 12 You'll note that the values here form a pattern. The values that satisfy |y| <= 12 are symmetric about 0, where x = 6. We have six values that satisfy the equation above point (6,0), so we must have six pairs of x-y coordinates below (6,0) that satisfy it as well. So we'd expect 2*6 + 1 = 13 total values. Director Joined: 13 Mar 2017 Posts: 722 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Re: For how many ordered pairs (x , y) that are solutions of the [#permalink] ### Show Tags 02 Aug 2017, 01:22 tonebeeze wrote: $$2x + y", "for the lattice points which lie on this curve. Given the square term, we will get symmetric pairs of solutions about the $x$-axis (Figure 1). In geometric terms, it might be viewed as the problem of which squares and cubes with sides of integer units are inter-convertible by the addition of $k$ units. Mordell had pointed out about a century ago that these equations might have either no integer solutions or only a finite number of them. Let is consider a concrete example with $k=1$: $y^2=x^3+1$. One can get three trivial solutions right away: by setting $x=-1$ we get $y=0$. Similarly, setting $x=0$, we get $y=\\pm 1$. Further, we see that if $x=2$ we get $y=\\pm 3$. Thus, we can write the solutions as $(x,y)$ pairs: (-1,0); (0,1); (0, -1); (2,3); (2,-3): a total of 5 unique integer solutions. Can there be any more solutions than these? To get an intuitive geometric feel for this we first observe how these solutions sit on the curve (Figure 1). We notice that the 3 distinct ones are on the same straight line (also applies to their mirror images via a mirrored line). This is a important property of elliptic curves (being cubic curves) that allows us to understand the situation better. Figure 2. In Figure 2, we consider the curve", "they rely on simple rules, for example a point near the origin in one lattice will never be coincident with a point near a distant point in the other. But if you find yourself considering computations like matrix division or using Eisenstein integers or Euclid's algorithm in the complex plane please save that for the follow-up question. This is so shortest code wins. ## Input • Input will have six integers $$\\(m, n), (i, j), (k, l)\\$$ as described above, but can have any order or hierarchy, or additional (but uninformative) place holders (e.g. zero padding, blanks...) • $$\\(m, n)\\$$ will be positive, but the other four can be positive, negative or zero, excluding a $$\\(0, 0)\\$$ pair (the origin). ## Recipe for making test cases In addition to the two mentioned above (13, 7), (5, 6), (5, 3) and (13, 7), (6, 5), (5, 3) you can roll your own: 1. Pick two Loeschian numbers $$\\L_1\\$$ and $$\\L_2\\$$ and find some integer pairs (i, j), (k, l) that can make them. (Find all integer pairs that produce a given Loeschian number) 2. Make new pairs by choosing two nonzero integers $$\\a, b\\$$, then: \\begin{align} i' & = ai - bj\\\\ j' & = aj + b(i+j)\\\\ k' & = ak - bl\\\\ l' & = al +", "can see the similarity between the coin problem and this, but I can't find related ideas that can help solve the induction problem. – oldselflearner1959 Jan 22 '18 at 1:09 • Do you include $0$ among the natural numbers or not ? – Ewan Delanoy Jan 22 '18 at 10:38 Let us first prove the following lemma : Lemma : If $\\gcd(a,b)=1$, then any integer greater than $ab$ can be expressed as $aX+bY$ where $X,Y$ are positive integers. Proof : If $\\gcd(a,b)=1$, then there exist integers $m,n$ such that $am+bn=1$. So, the line $aX+bY=N$, i.e. $Y=-\\frac abX+\\frac Nb$ where $N$ is a positive integer passes through a lattice point. Considering the slope $-\\frac ab$, we see that the lattice points on the line are at equal intervals $(b,-a)$. For $N=ab$, we have $Y=-\\frac abX+a$ which passes through $(b,0)$ and $(0,a)$. Therefore, considering the intervals of the lattice points, the equation $aX+bY=ab$ does not have any positive solutions. For $N\\gt ab$, considering again the intervals of the lattice points, we see that there is at least one positive solution for $aX+bY=N$. $\\quad\\blacksquare$ Let $x=X-1,y=Y-1$. From the lemma, any integer greater than $ab$ can be expressed as $aX+bY=a(x+1)+b(y+1)=ax+by+a+b$ where $x,y$ are non-negative integers. It follows that any integer greater than $ab-a-b$ can be expressed as $ax+by$ where $x,y$ are non-negative", "Lessons. If there is such a point, then we can say that the sample pair (3, 5) is an ordered pair on the line. Which value is a solution of the equation 5 - 4x = -3 0 2 -3 1/4 2. These ordered pairs are in the solution set of the equation [latex]x>y$. Add your answer and earn points. We keep a whole lot of good quality reference tutorials on subject areas starting from operations to decimals The ordered pairs $(−3,3)$ and $(2,3)$ are outside of the shaded area. Mathematics, 21.06.2019 12:30. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Find Three Ordered Pair Solutions. Determine if each ordered pair is a solution to the given system. }{=}6[/latex] … The ordered pair $(−2,−2)$ is on the boundary line. Since y = 3 when x = 1, the ordered pair (1, 3) is a solution of 2 x + y = 5. Let's check each answer choice. Step 4: Check. (-3, -8) 4x - 3y = 12 4x + 3y = 36 . If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If the ordered pair appears to be on the graph of a line, then it is a possible solution", "The distance between them equals $b$. Using the distance formula, we get $PQ = b = \\sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\\sqrt2 \\times |x_2 - x_1| = 58\\sqrt2\\times z$ $(*)$ WLOG, we can assume that $x_2 > x_1$. Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. • There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. • There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... • Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = \\boxed{600}$ triangles. Solution 4 We present a non-analytic solution; consider the lattice points on the line $41x+y=2009$. The line has intercepts $(0, 2009)$ and $(49, 0)$, so the lattice points for $x=0, 1, \\ldots, 49$ divide the line into $49$ equal segments. Call the area of the large triangle $A$. Any triangle", "to the inequality. (1,2) (8,-54) (5,-35) (1,-7) Math. The first is (10 -86), so try x = 10 and see whether y turns out to be -86. y = -9(10) + 4 = -90 + 4 = -86. 1 Answers. 1. Vectors <2, 4> and <2, -3> are graphed in the coordinate plane above. If the coordinates of A and B are (14, -1) and (2, 1), respectively, the y-intercept of AB is ____ and the equation of BC is y = __ x + __ . Y-1= 1/6(x+5) A. An ordered pair is a pair of numbers, ( x , y ) , written in a particular order. To determine vertical position, count y squares along the y-axis, as shown in the figure below. ax + by = c Choose 0 0 to substitute in for x x to find the ordered pair. . Subjects: Algebra, Graphing, Algebra 2. CCSS: HSA-REI.D.12. In mathematics, an ordered pair is a pair of objects. Ordered pairs of scalars are sometimes called 2-dimensional vectors. Which ordered pair is a solution of y = -2x - 6? y=2x-1 y=x+3 y=x-3 y= -2x +1 . Show more details Add to cart. Any ordered pair (x, y) that satisfies x>=6+4y Or, in set notation, Solution={(x, y)|x>=6+4y} Now, there is a little problem here -", "x, three negaitve values of x, and at x=0. Show all work and please write out the steps on what you are doing so I can learn from it. Use the resulting ordered pairs to plot the graph. State the equation of theline asymptotic to the graph (if any) Problem:(1/4)^x -3, this is worth 10 ponts out of 30, so if I don't get this I am in trouble. Please help me. If has been 16 years since I've done this and my professor isn't much help.1 solutions Answer 171832 by solver91311(17077) on 2009-10-28 17:26:44 (Show Source): You can put this solution on YOUR website! . Evaluate: I'll do three of these for you: . So the ordered pair is (3, 1) . So the ordered pair is (1, 16) . So the ordered pair is (-2,1024) I presume you know how to plot points. Once you have done all the required arithmetic and plotted your set of points, draw a smooth curve. It should look something like this: Notice that no matter how large you make , the value of the function will never be exactly zero. It gets very close, very fast, but never quite makes it. Therefore asymptotic to , in other words, to the -axis. John logarithm/232440: Express as a product. Problem: log b", "# What ordered pairs satisfy this equation ? y=6x (72,12) (7,42) (15,90) (54,9) (36,6) (6,36) You need to input the coordinates of each given point in the equation `y = 6x ` to test what pair of coordinates makes the equation to hold. Since the number of ordered pairs is large, you need to make the following observation in order to exclude some of them.... Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime. You need to input the coordinates of each given point in the equation `y = 6x ` to test what pair of coordinates makes the equation to hold. Since the number of ordered pairs is large, you need to make the following observation in order to exclude some of them. You should notice that the coordinate y is a multiple of 6, hence, the vakue of 6 coordinate needs to be larger than the value of x coordinate and it need to be a multiple of 6. Hence, the following pairs may be excluded Based on the following method of reasoning, you may exclude the following pairs:` (72,12), (54,9), (36,6).` You need to test the ordered pair `(7,42)` , such that: `y = 6x => 42 = 6*7 => 42 = 42` Hence, the ordered", "ratios a_n/c_n will be the same for different triples. Last edited: Mentor Circle Lattice Points Schinzel's theorem shows that for every positive integer n, there exists a circle in the plane having exactly n lattice points on its circumference. The theorem also explicitly identifies such \"Schinzel circles\" as ##(x-\\frac 1 2 )^2 + y^2 = \\frac 1 4 5^{k-1}## for n=2k ##(x-\\frac 1 3 )^2 + y^2 = \\frac 1 9 5^{2k}## for n=2k+1 The answer is yes, and there is an explicit formula to construct circles for every n. Gold Member Interesting problem. Without losing generosity we can say (0, 0) is on the circle so its equation is $$(x-a)^2+(y-b)^2=a^2+b^2$$ for the circle at least one lattice point is on it. Center of the circle in the 1st quadrant is enough so $$a,b \\geq 0$$ Furhter its half $$0 \\leq b \\leq a$$ is enough for consideration. $$x^2+y^2=2ax+2by$$ We know a and b should be rational for more lattice points on the circle. So the problem is restated: Tune rational numbers ## 0 \\leq b \\leq a## so that N integer (m,n) pairs satisfy the above equation. Obviously say ##a=\\frac{m}{2}##, lattice point (m,0) satisfies the condition. Say ##a=\\frac{m}{2},b=\\frac{n}{2}## (m,n) satisfies the condition so there are at least four points including (0,0) satisfy the condition. Last edited:", "we can quickly find all of the lattice points by adding $4$ to the $x$ coordinate until we get to $k$ (where $k$ is the $x$-intercept, as shown above). However, if we can’t find a lattice point between $x=0$ and $x=4$, then there won’t be any lattice points at all, because lattice points should be at intervals of $4$, so there can’t be an interval of $4$ that doesn’t contain one. If $x=1$, then $3x=3$, so $4y$ must be $56$ since $3+56=59$, so $y=56\\div 4=14$. So the point $(1,14)$ lies on the line and is a lattice point. Then there are also lattice points when $x=5, 9, 13 …$. To work out how many there are in the first quadrant, we need to find the value of $k$, because if $x>k$ then the lattice point will not be in the first quadrant. When $x=k$, $y=0$, and so $3\\times k+4\\times 0=59$, so $k=59/3=19\\frac{2}{3}$. So the lattice points have $x$-coordinates $1, 5, 9, 13$ and $17$ - which means there are $5$ lattice points. Using factorisation and properties of numbers Suppose $(a,b)$ is lattice point on the line $3x+4y=59$ in the first quadrant. Then $a$ and $b$ are positive whole numbers with $3a+4b=59$. $3a+4b$ can be written as $3(a+b)+b$. This is useful because, starting from $3(a+b)+b=59$, $3(a+b)=59-b$, so $59-b$ must", "for this? (1, 12), (3, 10), (-4, 17), (4.5, 8.5) Hopefully you’ll agree that there are infinite pairs of numbers whose sum is 13. You might also agree that there are infinite pairs of numbers whose difference is 7. (9, 2), (11, 4), (37, 30), (95.8, 88.8), (-3, -10) However, which ordered pair is true for both conditions at once? Which pair has a sum of 13 and a difference of 7? If you make a list of ordered pairs, you can check them to see which makes both equations true. This is a system of equations—two or more equations at the same time. In the example above, the solution is (10, 3) because the sum of the two numbers is 13 and their difference is 7. The pair (10, 3) makes both equations true. Example Which ordered pair makes both equations true? 1. x+y4xy=8=3\\begin{align*}x+y &=8 \\\\ 4x-y &=-3\\end{align*} a. (2, 6) b. (3, 15) c. (4, 4) d. (1, 7) Let’s test each pair and see which pair, if any, works: a. x+y2+684xy426862=8=8?=8=3=3?=3?3\\begin{align*}x+y &=8 \\\\ 2+6 &=8? \\\\ 8 &=8 \\\\ 4x-y &=-3 \\\\ 4 \\cdot 2-6 &=-3? \\\\ 8-6 &=-3? \\\\ 2 &\\ne -3 \\end{align*} The ordered pair (2, 6) makes the first equation true, but not the second. Because it is not true for both", "# Difference between revisions of \"2011 AMC 12B Problems/Problem 19\" ## Problem A lattice point in an $xy$-coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx + 2$ passes through no lattice point with $0 < x \\leq 100$ for all $m$ such that $\\frac{1}{2} < m < a$. What is the maximum possible value of $a$? $\\textbf{(A)}\\ \\frac{51}{101} \\qquad \\textbf{(B)}\\ \\frac{50}{99} \\qquad \\textbf{(C)}\\ \\frac{51}{100} \\qquad \\textbf{(D)}\\ \\frac{52}{101} \\qquad \\textbf{(E)}\\ \\frac{13}{25}$ ## Solution It is very easy to see that the $+2$ in the graph does not impact whether it passes through lattice. We need to make sure that $m$ cannot be in the form of $\\frac{a}{b}$ for $1\\le b\\le 100$, otherwise the graph $y= mx$ passes through lattice point at $x = b$. We only need to worry about $\\frac{a}{b}$ very close to $\\frac{1}{2}$, $\\frac{m+1}{2m+1}$, $\\frac{m+1}{2m}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\\le b\\le 100$, the smallest is $\\frac{50}{99}$, so answer is $\\boxed{\\frac{50}{99} \\textbf{(B)}}$ ## Solution 2 Like in the first solution, we note that the $+2$ does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line $y = mx$. A line with a slope", "region of values that makes this inequality true (shaded red), the boundary line $3x+2y=6$, as well as a handful of ordered pairs. The set of all first coordinates of the ordered pairs is the domain of the relation or function. x –6 –4 3 5 y –4 –6 –6 0 How many points should appear in Kelsey’s graph? Key Takeaways. - ehomework-helper.com Thanks. The first number is called the “x-coordinate” and the second number is called the “y-coordinate”. To find out whether an ordered pair is a solution of a linear equation, you can do the following: Graph the linear equation, and graph the ordered pair. How do I graph f on a graphing calculator? Because infinitely many values of theta have the same angle in standard position, an infinite number of coordinate pairs describe the same point. Ordered Pairs. (See Examples.) If you have to hand draw your graph, that can also work, just make sure you have enough space to graph the ordered pairs when doubled. •Find the x and y intercepts of the equation. Question 465060: y=-3x I am supposed to find the ordered pairs for this equation then graph them please explain the steps I am using course compass and it is telling me to substitute \"-1\" then \"2\" then \"-2\" into the equation" ]

[ "+0 # Algebra 0 139 1 +41 The range of the function $g(x) = \\frac{2}{2+4x^2}$ can be written as an interval $(a,b]$. What is $a+b$? Jul 17, 2021 #1 +32779 +1 If $$x = \\pm \\infty$$ then $$g(x) = 0$$ so a = 0 If $$x = 0$$ then $$g(x)=1$$ so b = 1 Jul 18, 2021", "# Math Help - function image? 1. ## function image? I have this question, I don't quite know how to approach it : g : R → R g(x) = $\\frac {2x^2-x}{3x^2+x+1}$ I need to find Img(g) and determine if its one-to-one function. then it asks - what will happen to the image if we reduce the interval to [0,∞) thanks in advance. 2. This question really belongs in Calculus. This question is asking for the range of g. How do you normally go about finding the range?", "+0 # I need help ASAP! 0 220 2 +10 Find the set of values $$a$$ for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. Aug 24, 2020 edited by Guest Aug 24, 2020 #1 0 The set of possible values of a is [-2,2]. Aug 24, 2020 #2 0 that's incorrect. Guest Aug 24, 2020", "# Math Help - finding range of function 1. ## finding range of function The function g is defined by $g(x) = x^{2} -8x +17$, $0\\leq x$ Find the range of g. How do I find this? Is there a particular 'technique'? Thanks, 2. Originally Posted by Tweety The function g is defined by $g(x) = x^{2} -8x +17$, $0\\leq x$ Find the range of g. How do I find this? Is there a particular 'technique'? Thanks, hi since your function has no real roots,the parabola is strictly above the X-axis. 3. Originally Posted by Raoh hi since your function has no real roots,the parabola is strictly above the X-axis. Thanks, So the range is all the values greater than or equal to zero? 4. if you're able to find a global minimum for your function then i think you can find the range. 5. Originally Posted by Tweety Thanks, So the range is all the values greater than or equal to zero? i think Strictly greater than zero. if we solve f'(x)=0.we'll find x=4 as a solution and at this point there is a global minimum. f(4)=1. therefore,the range is, $[1,\\infty )$ i might be wrong though... 6. Originally Posted by Tweety The function g is defined by $g(x) = x^{2} -8x +17$, $0\\leq x$ Find the", "0. i.e, the range of g(y) is $[0,\\infty )$. Q-5: Find the range and domain of the function: g(y) = |y – 4| Sol: Here |y – 4| is the given function. So, it is clear that the function is defined for all the real numbers. The domain for g(y) is R Now, for range of the given function, we have: $y\\epsilon R ,$, |y – 4| assumes for all real numbers. Therefore, the range of g(y) is the set of all non- negative real numbers. Q-6: A function from ‘R into R’ is given below. Find the range of ‘g’. $g = \\left [ \\left ( y, \\frac{y^{2}}{1 + y^{2}} \\right ); y\\epsilon R \\right ]$ Sol: $g = \\left [ \\left ( y, \\frac{y^{2}}{1 + y^{2}} \\right ); y\\epsilon R \\right ]\\\\$ =$[(0,0), (\\pm 0.5, \\frac{1}{5}), (\\pm 1, \\frac{1}{2}),(\\pm 1.5, \\frac{9}{13}),(\\pm 2, \\frac{4}{5}), (3, \\frac{9}{10}),…….]\\\\$ Thus, the range of ‘g’ is the set of all 2nd elements. It can be seen that all these elements are: $\\geq 0\\;but\\;< 1$. Therefore, the range of ‘g’ is = [0,1) Q-7: Assume that function ‘m’ and ‘n’ is defined from: R$\\rightarrow$R. m (y) = y + 2, n(y) = 3y – 2 Find m + n, m – n and $\\frac{m}{n}$ Sol: Here, m(y) = y + 2 and", "# Thread: Range of Function 1. ## Range of Function how is it possible to find the range of a function) in interval notation answer, f(x) = sq.rt ( 9 - x^2) i really have difficulty in finding range and domain of a function... can you help me in this problem 2. ## Re: Range of Function Originally Posted by rcs how is it possible to find the range of a function) in interval notation answer, f(x) = sq.rt ( 9 - x^2) i really have difficulty in finding range and domain of a function. If we have $\\sqrt{y}$ then $y\\ge 0$. So $|x|\\le 3$. WHY? Now you show us some of your efforts.", "+0 # Range 0 118 1 The range of the function g(x) = 2/(2 + 4x^2) + 8/(1 + x^2)can be written as an interval (a,b]. What is a+b? May 17, 2022 #1 +9461 0 Note that$$g(x) = \\dfrac2{2 + 4x^2} + \\dfrac8{1 + x^2} = \\dfrac1{1 + 2x^2} + \\dfrac8{1 + x^2}$$. Also, $$x^2 + 1 \\geq 1$$ and $$1 +2x^2 \\geq 1$$. Then, $$0 < \\dfrac1{1 + 2x^2} \\leq 1$$ and $$0 < \\dfrac1{1 + x^2} \\leq 1$$. That means $$0 < \\dfrac8{1 + x^2} \\leq 8$$. Therefore, $$0 < g(x) \\leq 9$$ by adding the inequalities. Can you rewrite the range as an interval? May 18, 2022", "Find the range and domain of the function: g(y) = |y – 4| Sol: Here |y – 4| is the given function. So, it is clear that the function is defined for all the real numbers. The domain for g(y) is R Now, for range of the given function, we have: yϵR,$y\\epsilon R ,$, |y – 4| assumes for all real numbers. Therefore, the range of g(y) is the set of all non- negative real numbers. Q-6: A function from ‘R into R’ is given below. Find the range of ‘g’. g=[(y,y21+y2);yϵR]$g = \\left [ \\left ( y, \\frac{y^{2}}{1 + y^{2}} \\right ); y\\epsilon R \\right ]$ Sol: g=[(y,y21+y2);yϵR]$g = \\left [ \\left ( y, \\frac{y^{2}}{1 + y^{2}} \\right ); y\\epsilon R \\right ]\\\\$ =[(0,0),(±0.5,15),(±1,12),(±1.5,913),(±2,45),(3,910),.]$[(0,0), (\\pm 0.5, \\frac{1}{5}), (\\pm 1, \\frac{1}{2}),(\\pm 1.5, \\frac{9}{13}),(\\pm 2, \\frac{4}{5}), (3, \\frac{9}{10}),…….]\\\\$ Thus, the range of ‘g’ is the set of all 2nd elements. It can be seen that all these elements are: 0but<1$\\geq 0\\;but\\;< 1$. Therefore, the range of ‘g’ is = [0,1) Q-7: Assume that function ‘m’ and ‘n’ is defined from: R$\\rightarrow$R. m (y) = y + 2, n(y) = 3y – 2 Find m + n, m – n and mn$\\frac{m}{n}$ Sol: Here, m(y) = y + 2 and n(y) = 3y – 2 are defined from R$\\rightarrow$R. Now,", "# Range Algebra Level 4 Range of the function $$f(x)=\\frac{x}{1+x}$$ where $${x}$$ denotes the fractional part function is $$[0,\\frac{a}{b})$$ and $$gcd(a,b)=1$$ .Then value of $$a\\times b$$ is equal to For more problems try my set ×", "# Finding range of $f(g(h(x)))$ \\begin{align*} f(x) &= \\frac{2}{x+1}, \\\\ g(x) &= \\cos x, \\\\ h(x) &= \\sqrt{x+3} \\end{align*} Find the range of $f(g(h(x)))$. You want to all the possible \"outputs\" of the composition of the three functions. First note that the range of the inner most function $h$ is all non-negative numbers. Hence the \"input\" for the function $g$ is all non-negative real numbers. But for those you in $g(x) = \\cos(x)$ get all real numbers between $-1$ and $1$ (both included). Note now that the domain of $f$ is all the real numbers that are not equal to $-1$, hence the possible \"inputs\" of $f$ is the interval $(-1, 1]$. So now you just need to determine the possible values of $f$ when $x$ is in $(-1, 1]$. Hint: For this note that $f$ is a decreasing function.", "# $$\\frac{\\ln b-\\ln a}{b-a}$$ Calculus Level 2 For real numbers $$a$$ and $$b$$ ($$a<b)$$ in the interval $$[4,11],$$ the range of the real number $$k$$ such that $\\frac{\\ln b-\\ln a}{b-a}=k$ can be expressed as $$\\alpha < k < \\beta.$$ What is $$\\frac{1}{\\alpha\\beta}?$$ ×", "of a Real Function: The range of a real function of a real variable is the set of all real values taken by $f (x)$ at points in its domain. Standard Real Functions and their Graphs (to be linked) Operations on Real Functions Addition: If $f : D_1 \\rightarrow R$ and $g : D_2 \\rightarrow R$ be two real functions. Then, their sum $f + g$ is defined as that function from $D_1 \\cap D_2$ to $R$ which associates each $x \\in D_1 \\cap D_2$ to the number $f (x) + g (x)$. $(f+g)(x) = f(x) + g(x)$ for all $x \\in D_1 \\cap D_2$ Product: If $f : D_1 \\rightarrow R$ and $g : D_2 \\rightarrow R$ be two real functions. Then, their product $fg$ is a function from $D_1 \\cap D_2$ to $R$ and is defined as $(fg) (x) = f(x) g(x)$ for all $x \\in D_1 \\cap D_2$ Subtraction: If $f : D_1 \\rightarrow R$ and $g : D_2 \\rightarrow R$ be two real functions. Then, their difference $f - g$ is defined as $(f-g)(x) = f(x) - g(x)$ for all $x \\in D_1 \\cap D_2$ Quotient: If $f : D_1 \\rightarrow R$ and $g : D_2 \\rightarrow R$ be two real functions. Then, their quotient $\\frac{f}{g}$ is a function from $D_1 \\cap D_2 -", "1/3] U (0, ∞) 4. (- ∞ , 1/3] U (0, 1/3] 7. What is the range of the function f(x) = $\\frac{1}{x^2 +4x-5}$? 1. (-∞, (−1)/9 ) 2. (0, ∞) 3. (-∞, (−1)/9 ] U [0, ∞) 4. (-∞, (−1)/9 ] U (0, ∞) 8. Function f(x) defined from set of real numbers to set P is an onto function. If f(x) = $\\frac{1}{x^2+1}$ , find set P. ? 9. Range of f(x) = (–4, 5); range f(|x|) = (–3, 4). Which of the following could be the range of g(x) if g(x) = f(|x|) + |f(x)| + |f(|x|)|? 1. ( –7, 11) 2. ( –8, 10) 3. (- 6, 14) 4. (- 11, 14) 10. f(x) = Remainder of division of 100x by 19, where x is a whole number. How many elements does the range of f(x) contain? 1. 9 2. 19 3. 18 4. 12", "+0 # Help pls -1 124 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020", "My Math Forum Range of a function Algebra Pre-Algebra and Basic Algebra Math Forum November 15th, 2009, 06:07 AM #1 Member Joined: Oct 2009 Posts: 59 Thanks: 0 Range of a function Functions g and h are defined as follows: g : x ? 1 + x x ? R h : x ? x² + 2x x ? R Find i.) the ranges of g and h, range of g => R = {y : y ? R} 1 + x = 0 x = -1 1 - 2 = -1 -b/2a = -2/2 = -1 range of h => R = {y : y ? - 1, y ? R} ii.) the composite functions h o g and g o h, stating their ranges. Not sure how this is to be done help needed. November 15th, 2009, 07:31 AM #2 Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 The range of a function is the set of all values the function can have. It depends on the function's domain. Do you understand the term \"domain\"? For real x, 1 + x can have any real value, so its range is R, the set of all reals. For real x, since x² + 2x ?(x + 1)² - 1, its range is all real values ?", "Browse Questions # What is the range of the real function, $f(x) = \\large\\frac{1}{x^2-1}$ ? Can you answer this question? R − (−1, 0] hence (B) is the correct answer. answered Jun 28, 2014", "+0 # Help pls -1 61 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020", "are in the range of the function $f(x) = \\frac{4x^2+75}{2x^2+3}$?", "2.Evaluate $$(g∘f)(4)$$,$$(g∘f)(−1/2)$$. 3.Find $$(f∘g)(x)$$ and state its domain and range. 4.Evaluate $$(f∘g)(4)$$ , $$(f∘g)(−1/2)$$. Solution: 1. We can find the formula for $$(g∘f)(x)$$ in two different ways. We could write $$(g∘f)(x)=g(f(x))=g(x^2+1)=\\frac{1}{x^2+1}$$. Alternatively, we could write $$(g∘f)(x)=g(f(x))=\\frac{1}{f(x)}=\\frac{1}{x^2+1}.$$ Since $$x^2+1≠0$$ for all real numbers x, the domain of $$(g∘f)(x)$$ is the set of all real numbers. Since $$0<1/(x2+1)≤1$$, the range is, at most, the interval $$(0,1]$$. To show that the range is this entire interval, we let y=1/(x2+1) and solve this equation for x to show that for all $$y$$ in the interval $$(0,1]$$, there exists a real number x such that $$y=1/(x^2+1)$$. Solving this equation for $$x$$, we see that $$x^2+1=1/y$$, which implies that $$x=±\\sqrt{\\frac{1}{y}−1}$$ If $$y$$ is in the interval $$(0,1]$$, the expression under the radical is nonnegative, and therefore there exists a real number $$x$$ such that $$1/(x^2+1)=y$$. We conclude that the range of g∘f is the interval $$(0,1].$$ 2. $$(g∘f)(4)=g(f(4))=g(4^2+1)=g(17)=\\frac{1}{17}$$ $$(g∘f)(−\\frac{1}{2})=g(f(−\\frac{1}{2}))=g((−\\frac{1}{2})^2+1)=g(\\frac{5}{4})=\\frac{4}{5}$$ 3. We can find a formula for $$(f∘g)(x)$$ in two ways. First, we could write $$(f∘g)(x)=f(g(x))=f(\\frac{1}{x})=(\\frac{1}{x})^2+1.$$ Alternatively, we could write $$(f∘g)(x)=f(g(x))=(g(x))^2+1=(\\frac{1}{x})^2+1.$$ The domain of $$f∘g$$ is the set of all real numbers $$x$$ such that $$x≠0$$. To find the range of $$f$$, we need to find all values $$y$$ for which there exists a real number $$x≠0$$ such that $$(\\frac{1}{x})^2+1=y.$$ Solving this equation for x,", "+0 # Help!! 0 33 1 +149 Find all real numbers $$t$$ such that .$$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15$$ Give your answer as an interval. Logic Sep 2, 2018 #1 +1994 +1 \\(\\dfrac 2 3 t - 1 < t+7 \\leq -2t + 15 \\\\ \\dfrac 2 3t-8 Rom Sep 6, 2018" ]

[ "# Find the range of $k$ for which the inequality $k\\cos^2x-k\\cos x+1\\geq0 ,\\forall x\\in(-\\infty,\\infty)$ holds. Find the range of $k$ for which the inequality $k\\cos^2x-k\\cos x+1\\geq0 ,\\forall x\\in(-\\infty,\\infty)$ holds. This is an inequality involving trigonometric function $\\cos x$ which varies from $-1$ to $1$. If the question had been $kx^2-kx+1\\geq 0$, i would have easily solved it,by using the discriminant property of the quadratics.If quadratic is positive then its discriminant is negative but i am not able to find the range of $k$ in this question. Just another way: Equivalently, you have to find when $$4k(\\cos x -\\tfrac12)^2 \\ge k-4$$ Now $\\cos x \\in [-1, 1] \\implies (\\cos x - \\tfrac12)^2 \\in [0, \\tfrac94]$, and so $k \\in [-\\frac12, 4]$. This problem is equivalent to asking: on what condition will the quadratic function $p(t) = kt^2 - kt + 1$ have at most one root in the interval $t \\in [-1,1]$? This is because if we put $\\cos x = t$ then the only interval we actually have to care about is the range of $\\cos x$. Let's analyse the behaviour of $p(t)$ in this interval. $p'(t) = 2kt - k$. Thus, we see that the extremum of the quadratic function is at $t = 1/2$. This is a fixed value even with the changing parameter $k$. CASE", "has a solution if and only if $y \\ge -1/4$. Jul 1 at 13:46 • I think you're misinterpreting the symbol $\\ge$. The set $\\{y \\in \\mathbb{R} \\mid y \\ge -1/4\\}$ includes the value $-1/4$. Each value may only be greater than or equal, but the solution includes both the values that are greater than, and the one value that is equal. Jul 2 at 9:08 I would write this as $$y = (x+\\frac32)^2-\\frac14$$, from which it is obvious what the range would be, as the square term takes all nonnegative values. • That is certainly a better way to do it, but I was experimenting with the other ways there are to approach the problem. Jul 2 at 6:24 • @RajShukla Your approach works well too, as others noted you have proved for every $y\\ge -\\frac14$, there is at least one real value of $x$, and not for any other $y$, so the range is established Jul 2 at 7:11 In your solution, you show that $$x$$ has a real value if $$y\\geq -1/4$$ using discriminant of a quadratic. Well this works the other way too! If $$y\\geq -1/4$$, then the quadratic has nonnegative discriminant and so you will always find $$x$$ that give that specific value of $$y$$. Hope this helps. :) • That makes sense,", "be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$must be nonnegative. Therefore,$$4(1-x^{2020}) \\geq 0 \\implies x^{2020} \\leq 1.$$Here, we see that we must split the inequality into a compound, resulting in $-1 \\leq x \\leq 1$. The only integers that satisfy this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$.", "+ c^2 = 1$, and since $xy + yz + xz$ is a symmetric polynomial in $x,y,z$, we should seek $x=y=z>0$ as our minimum, which can be shown the be the case using very straightforward techniques from calculus (Second derivative test should suffice, proving the minimum for $a,b,c$ suffices). We find that $x=y=z$ when $a=b=c=\\dfrac{1}{\\sqrt{3}}$, so that $x=y=z=\\dfrac{1}{1-1/3}=\\dfrac{3}{2}$. We can then say that $$xy + yz + xz \\geq 3*(3/2)^2 = 3*(9/4) = 27/4$$ Therefore, $$x^2 + y^2 + z^2 \\leq \\dfrac{81}{4} - 2*\\dfrac{27}{4} = \\dfrac{81}{4} - \\dfrac{54}{4} = \\dfrac{27}{4}$$ • I don't see how you obtain the final, desired inequality. You seem to just assert it. – zibadawa timmy Apr 9 '17 at 22:33 • This comes from the fact that the function $f(x) = x^2$ is monotonically increasing on $\\mathbb{R}^+$. By that I mean that for positive reals $x,h$, $x^2 < (x+h)^2$. Since $\\dfrac{9}{2} \\geq \\dfrac{9}{2} - \\varepsilon$ and since both are positive, $\\bigg{(} \\dfrac{9}{2} \\bigg{)}^2 \\geq \\bigg{(} \\dfrac{9}{2} - \\varepsilon \\bigg{)}^2$. Does that answer your question? – Will Craig Apr 9 '17 at 22:36 • $\\frac{9}{2}=81/4$, so we cannot say $(\\frac{9}{2} - \\varepsilon)^2\\le 27/4$ – didgogns Apr 9 '17 at 23:05 • I wasn't thinking, my fault. I will see if I can correct my approach. – Will Craig Apr 9 '17 at 23:05 •", "value, let $y=a x^{2}+b x+c$, $\\Rightarrow a x^{2}+b x+c-y=0$ Since the solution of a quadratic equation is, $x=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$ Substitute c to c-y in the equation, $x=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a(c-y)}}{2 a}$ Since the discriminant must be positive, $D=b^{2}-4 a(c-y) \\geq 0 \\\\ \\Rightarrow b^{2}-4 a c+4 a y \\geq 0 \\\\ \\Rightarrow 4 a y \\geq 4 a c-b^{2}$ Since a < 0 so multiply (-1) on both sides of the inequality to make the expression positive. $-4 a y \\leq b^{2}-4 a c \\\\ \\Rightarrow y \\leq \\dfrac{b^{2}-4 a c}{-4 a} \\\\ \\Rightarrow y \\leq \\dfrac{-b^{2}}{4 a}+c \\\\ \\Rightarrow y \\leq c-\\dfrac{b^{2}}{4 a}$ (Note: on dividing -4a both sides of the inequality, the sign of the inequality doesn’t change because -4a > 0, as a < 0.) So, the maximum value attained is $y=c-\\dfrac{b^{2}}{4 a}$ Now find the value of x to the corresponding maximum value of y $x=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a(c-y)}}{2 a}\\\\ \\Rightarrow x=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a c+4 a\\left(c-\\dfrac{b^{2}}{4 a}\\right)}}{2 a}\\\\ \\Rightarrow x=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a c+4 a c-b^{2}}}{2 a}\\\\ x=\\dfrac{-b}{2 a}$ So, For a < 0 the quadratic equation will attain its maximum value at, $x=\\dfrac{-b}{2a}$ Similarly, there are two cases of finding the minimum value of a quadratic equation. 1. For a > 0 From the graph of the quadratic", "> 0, $m^{2} + 4pr ≥ 0$, therefore, H(x) will also have real root(s). Hence at least two roots of G(x) H(x) are real. Example 2: Find the range in which the value of given quadratic Expression will lie $\\mathbf{\\frac{x^{2}\\;+\\;14x\\;+\\;9}{x^{2}\\;+\\;2x\\;+\\;3}}$. [x is a real number] Solution: Let, y = $\\mathbf{\\frac{x^{2}\\;+\\;14x\\;+\\;9}{x^{2}\\;+\\;2x\\;+\\;3}}$ i.e. $x^{2} + 14x + 9 = (x^{2}y + 2xy + 3y)$ Or, $(1 – y) x^{2} + (7 – y) 2x + 3(3 – y) = 0$ Since x is real, therefore, $b^{2} – 4ac ≥ 0$ i.e. $[2(7 – y)]^{2} – 4[1 – y] \\times 3[3 – y] ≥ 0$ Or, $- y^{2} – y + 40 ≤ 0$ Or, (y + 5) (y – 4) ≤ 0 Therefore, the range in which the value of given Quadratic Expression will lie is – 5 ≤ y ≤ 4. Example 3: Find the values of x for which the equation $\\mathbf{\\frac{x^{2}\\;-\\;4x\\;+\\;3}{x^{2}\\;+\\;x\\;+\\;1}\\;\\leq \\;0}$. Solution: Let f (x) = $x^{2} – 4x + 3$ and g (x) = $x^{2} + x + 1$. The coefficient of $x^{2}$ in g (x) is positive and the value of Discriminant (D) < 0. Hence, g(x) is positive for all values of x. Since, $\\mathbf{\\frac{f(x)}{g(x)}\\;<\\; 0}$. Therefore, f (x) must be less than 0. i.e. $x^{2} – 4x + 3 < 0$ Or,", "$D/4\\geq0$. $$64y^2-2(8-k)\\geq0$$ Now I am currently stuck here. Is there a way to get around this? I would really like to find a solution that uses discriminants. Thanks in advance. Note: I just mentioned the $4$ other ways to solve this problem so that I can emphasize that I would really like to see a solution using discriminants. Also, any solution without calculus (or with only topics covered until high school - maybe -) are always welcome. • Nice question. +1 Thanks for the good formatting! – Adrian Keister Feb 26 '18 at 15:15 Let $$f(x,y)=(3x+2y)^2+(x+2y)^2=10x^2+16xy+8y^2.$$ If $f(x,y)=a$ for some $x$, $y$ on the unit circle then $$f(x,y)-a(x^2+y^2)=0\\tag1$$ has nonzero real solutions. But if $(1)$ has a nonzero real solution, one can scale it so that $x^2+y^2=1$, and then $f(x,y)=a$ for that $(x,y)$. But $$f(x,y)-a(x^2+y^2) =(10-a)x^2+16xy+(8-a)y^2.$$ This has a nonzero real solution iff the discriminant $$16^2-4(10-a)(8-a)\\ge0.$$ Solve this inequality for $a$ to get the values of $f$ on the unit circle. • Can I suggest a small clarifying edit: just after equation (1), change \"has nonzero real solutions\" to \"for any fixed $(x, y)$, has a nonzero real solution for $a$.\" – John Hughes Feb 26 '18 at 15:30 • Wow.. I am baffled.. I never thought I could change $a$ to $a(x^2+y^2)$. Thank you very much, you", "+0 help algebra 0 81 2 Let x and y be real numbers such that 2(x^2 + y^2) = x + y + 8. Find the maximum value of x - y. Apr 19, 2022 #1 +9459 +1 Moving some terms yields: $$2x^2 + 2y^2 - x - y - 8 = 0$$ Note that geometrically, this is the equation of a circle, because we can rewrite the equation as follows: $$\\begin{array}{rcl} 2x^2 + 2y^2 - x - y - 8 &=& 0\\\\ x^2 + y^2 - \\dfrac12 x - \\dfrac12 y - 4&=&0\\\\ \\left(x - \\dfrac14\\right)^2 + \\left(y - \\dfrac14\\right)^2 &=& \\dfrac{33}8 \\end{array}$$ Consider the straight line x - y = c, where c is varying through real numbers. You can find the maximum value of c such that the line and the circle intersects, i.e., $$\\begin{cases}2x^2 + 2y^2 - x - y - 8 = 0\\\\x - y = c​​​​\\end{cases}$$ has solutions. To do so, you need to substitute y = x - c into the circle equation, then consider the discriminant of the resulting equation. Since the system has solutions, the discriminant of the resulting equation must be nonnegative. Solving the equation $$\\Delta \\geq 0$$ will give you the suitable range of values of c. Please tell me if you want more explanations or you", "of the equation are equal, and the imaginary parts are equal. So, reals: $2x-6 = 10$ Imaginaries: $5 = y-3$ I'm sure you can solve it now! (c) This is all about handling a quadratic expression. You can do it: • by differentiating • by simply quoting a formula (if you know it) or • by completing the square. I'll show you the last method, since I'm guessing that might be the way you've been shown. $f(x) = 4x^2-3x +1$ Take out factor $4$ to make coefficient of $x^2$ equal to $1$: $f(x) = 4(x^2 - \\tfrac34x+\\tfrac14)$ Write half the coefficient of $x$ inside a 'bracket squared': $f(x) = 4\\Big((x-\\tfrac38)^2 - ...\\Big)$ Subtract the square of this number, and add in the original constant number: $f(x) = 4\\Big((x-\\tfrac38)^2 - \\tfrac{9}{64}+\\tfrac14\\Big)$ Simplify: $=4\\Big((x-\\tfrac38)^2 + \\tfrac{7}{64}\\Big)$ Since $(x-\\tfrac38)^2 \\ge 0$ for all real values of $x$, you can then say that the value of $f(x)$ is a minimum when $(x-\\tfrac38)^2 = 0$; i.e. when $x = \\tfrac38$.", "solutions $t=\\pm\\sqrt{2}$. When $t=\\sqrt{2}$, $y=\\sqrt{2}+\\dfrac{2}{\\sqrt{2}}=2\\sqrt{2}$; when$t=-\\sqrt{2}$, $y=-2\\sqrt{2}$. So referring back to our sketch graph, we see that $y$ cannot take anyvalues between $-2\\sqrt{2}$ and $+2\\sqrt{2}$. ### Method 3 Another calculus-free approach is to use the arithmetic mean–geometric mean (AM–GM) inequality. This says that if $a_1$ and $a_2$ are positive real numbers, then their arithmetic mean is greater than or equal to their geometric mean. In other words, $\\dfrac{a_1 + a_2}{2} \\geq \\sqrt{a_1a_2}$ where $a_1, a_2 \\geq 0$, with equality if and only if $a_1 = a_2$. In our case, if we take $n=2$ and set $a_1=t$ and $a_2=\\frac{2}{t}$, we find that $\\frac{t+\\frac{2}{t}}{2}\\ge \\sqrt{t \\times\\frac{2}{t}}=\\sqrt{2},$ with equality if and only if $t=\\dfrac{2}{t}$. Thus for positive values of $t$, $t+\\dfrac{2}{t}\\ge 2\\sqrt{2}$ with equality if and only if $t=\\dfrac{2}{t}$, that is, if $t=\\sqrt{2}$. For negative values of $t$, let $t=-u$, with $u$ positive. Then $t+\\dfrac{2}{t}=-u-\\dfrac{2}{u}=-\\left(u+\\dfrac{2}{u}\\right)$. Since $u+\\dfrac{2}{u}\\ge2\\sqrt{2}$, it follows that for negative values of $t$, $t+\\dfrac{2}{t}\\le -2\\sqrt{2}$ with equality if and only if $t=-\\sqrt{2}$. Thus $t+\\dfrac{2}{t}$ cannot take any values between $-2\\sqrt{2}$ and$+2\\sqrt{2}$. ### Method 4 As a final method, if $t$ is positive, we can write $t+\\dfrac{2}{t}$ as $t+\\frac{2}{t}=\\left(\\sqrt{t}-\\sqrt{\\frac{2}{t}}\\right)^2+2\\sqrt{2}.$ Since a square is always positive or zero, the right hand side must be at least $2\\sqrt{2}$, with equality if and only if$\\sqrt{t}=\\sqrt{\\dfrac{2}{t}}$. We can deal with the case of $t$", "1. ## finding the range Hi folks, I really hope there is no simple arithmetic error in my answer, but I've checked it many times, so here goes. Show that for real x the function $y = \\frac{x + 2}{(x + 3)(x - 1)}$ can take all real values. This should be really easy: $y = \\frac{x + 2}{x^2 + 2x - 3}$ $yx^2 + 2yx - 3y = x + 2$ $yx^2 + (2y - 1)x - 3y - 2 = 0$ a quadratic in x of the form $ax^2 + bx + c = 0$. For real x the discriminant $(b^2 - 4ac \\ge 0)$ $(2y - 1)^2 + 4.y.(3y + 2) \\ge 0$ $4y^2 - 4y + 1 + 12y^2 + 8y \\ge 0$ $16y^2 + 4y + 1 \\ge 0$ solving for y, this has no real roots. I was expecting to get any expression like $y^2 + something \\ge 0$ which would be true for all y, but no real roots suggests I have taken a wrong turn. Can you see it. 2. ## Re: finding the range For any $y \\in \\mathbb{R}$: $x = \\dfrac{1-2y \\pm \\sqrt{(2y-1)^2+4y(3y+2)}}{2y}$ This is undefined for $y=0$. The discriminant is always positive for all $y$, so all that is left to check is that you can yet", "$15000$ (and found no solutions other than your two). The program is free to download here and can look for solutions within given range of $x$. – Alex Ravsky Jul 31 at 15:10 This answer is open for improvement. Feel free to use its results answering the question. Conjecture 1. There are no solutions when $$m|n$$. For each integer $$p\\ge 2$$ and each real $$t$$ put $$f_p(t)=t^{p-1}(t+1)$$. In order to prove the conjecture for each integer $$y\\ge 2$$ we hope to find $$x’$$ and $$x’’$$ such that $$f_m(x’) but there are no integer values between $$x’$$ and $$x’’$$. Claim 2. There are no solutions when $$n=2m$$. Proof. We claim that $$x’=y^2+\\tfrac ym-\\tfrac 2m$$ and $$x’’=x’+\\tfrac 1m$$ fit. Indeed, $$f_n(x’’)=\\left(y^2+\\frac ym-\\frac 1m\\right)^{m}+\\left(y^2+\\frac ym-\\frac 1m\\right)^{m-1}>$$ $$y^{2m}+{m\\choose 1} y^{2m-2}\\left(\\frac ym-\\frac 1m\\right)+ y^{2m-2}= y^{2m}+ y^{2m-1}=f_n(y).$$ I have a draft proof that $$x’$$ fits based on Bernoulli’s inequality, but it is cumbersome. Proposition 3. There are no solutions when $$m=2$$ and $$n=6$$. We claim that $$x’=y^3+\\tfrac {y^2}2-\\tfrac y8-\\tfrac 12$$ and $$x’’=x’+\\tfrac 18$$ fit. Indeed, we can check (by Mathcad) that $$f_m(x’’)-f_n(y)=\\frac 1{2^6}(y-1)\\left(8y^2+17y+15\\right)>0$$ and $$f_n(y)- f_m(x’)=\\frac 1{2^6}\\left(8y^3-y^2+16\\right)>0.$$ Proposition 4. There are no solutions when $$m=3$$ and $$n=9$$. We claim that $$x’=y^3+\\tfrac {y^2}3-\\tfrac {y}{9}-\\tfrac 1{3}$$ and $$x’’=x’+\\tfrac 1{9}$$ fit. Indeed, we can check (by Mathcad) that $$f_m(x’’)-f_n(y)=\\frac 1{3^6}(y-1)\\left(108y^5+270y^4+252y^3+17y^2-52y-28\\right)>0$$ and $$f_n(y)- f_m(x’)= \\frac 1{3^6} \\left(135y^6-9y^4+244y^3+81y^2-27y-54\\right)>0.$$ Proposition", "a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all $x$, then $y(2-y) \\geq 0 \\Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$$(0,0)$$(1,1)$$(-1,1)$. There are $4$ solutions, so the answer is $\\boxed{\\textbf{(D) } 4}$ 10. ## Solution Notice that when the cone is created, the radius of the circle will become the slant height of the cone and", "Putting it in the expression, $\\Rightarrow \\dfrac{1}{t^2-97} \\leq \\dfrac{5}{2}$ Now, we know that the least value of $t^2$ possible is 0. At that value, LHS < RHS, thus the equation holds true. RHS of the expression will always be negative for all values of $t \\in [0,\\sqrt{97})$ Here, $(x – 9)^2 < 97$ $\\Rightarrow x < 9+\\sqrt{97}$ and $x > 9 – \\sqrt{97}$ $\\Rightarrow x < 9+10$ and $x < 9-10$ (We have equated $\\sqrt{97}$=10 because the smallest integral value of $t$ satisfying this condition will be 10) $\\Rightarrow x < 19$ and $x >-1$ Thus, the values of $x$ satisfying this condition are : 1,2….18. Here, the value of $t$ cannot be $\\sqrt{97}$ as then the denominator would become 0 and lead to an indeterminate form. For all values of $t \\geq \\sqrt{97}$, the fraction is positive. In that case, $\\dfrac{1}{t^2-97} \\leq \\dfrac{5}{2}$ $\\Rightarrow t^2-97 \\geq \\dfrac{2}{5}$ $\\Rightarrow t^2 \\geq 97+\\dfrac{2}{5}$ $\\Rightarrow t^2 \\geq 97.4$ $\\Rightarrow t \\geq 10$ (We have equated $\\sqrt{97.4}$=10 because the smallest integral value of $t$ satisfying this condition will be 10) Therefore $t \\in [0,\\sqrt{97})U(\\sqrt{97},\\infty)$ Since $t = x-9$, therefore the range of $x$ can be computed by adding 9 to the range of $t$. Thus, $x \\in [1,19)U(9+\\sqrt{97},\\infty)$ $\\Rightarrow x \\in [1,9+\\sqrt{97})U(9+\\sqrt{97},\\infty)$ To count the integral values of $x$ which satisfy", "equal to 0. EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) \\geq 0$$ $$\\implies -\\frac{5}{8} \\leq y < \\frac{1}{3}$$ which is the range of your function. EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your \"c\" term should be $(20y+1)$, not $(20y-1)$. • Thank you so much, this is exactly what I was asking. – William Jul 17 '18 at 17:51 • @William If you say so, check this answer. – Takahiro Waki Jul 30 '18 at 0:38 $$\\frac{x^2+4x-1}{3x^2+12x+20}=y$$ After cross multiplying, we obtain $$(3y-1)x^2+(12y-4)x+(20y+1)=0$$ If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation. However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real. The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$", "$a x_1 + b y_1 = c = a x_0 + b y_0.$ Moving the $$x$$'s and $$y$$'s to opposite sides, we get $a (x_1 - x_0) = b (y_0 - y_1).$ Both sides are divisible by $$d = \\gcd(a, b)$$, therefore $\\tfrac{a}{d} (x_1 - x_0) = \\tfrac{b}{d} (y_0 - y_1)$ is an integer equation. This means that the integer $$\\tfrac{b}{d}$$ must divide the product $$\\tfrac{a}{d} (x_1 - x_0)$$. By proposition 5, we know that $$\\gcd(\\tfrac{a}{d}, \\tfrac{b}{d}) = 1$$. Therefore, proposition 10 guarantees that the integer $$\\tfrac{b}{d}$$ must divide $$x_1 - x_0$$. In other words, there is an integer $$k$$ such that $x_1 - x_0 = \\tfrac{b}{d} k \\quad \\implies \\quad x_1 = x_0 + \\tfrac{b}{d} k.$ Plugging the first version of this equation into the left-hand side above, we get $\\tfrac{a}{d} \\cdot \\tfrac{b}{d} k = \\tfrac{b}{d} (y_0 - y_1)$ which simplifies to $\\tfrac{a}{d} k = y_0 - y_1 \\quad \\implies \\quad y_1 = y_0 - \\tfrac{a}{d} k.$ This proves that any additional solution $$(x_1, y_1)$$ is of the form prescribed by the proposition. To finish the proof, it remains to show that pairs of that form are always solutions. Let $$k$$ be any integer and define $$x_1 = x_0 + \\tfrac{b}{d} k$$ and $$y_1 = y_0 - \\tfrac{a}{d} k$$. Then, $a x_1 + b y_1 = a(x_0", "Let $G$ be a graph without loops or bridges and $T_G(x,y)$ be its Tutte polynomial. We give sufficient conditions for the inequality $T_G(x,y)T_G(y,x)\\geq T_G(z,z)^2$ to hold. We deduce in particular that $T_G(x,0)T_G(0,x)\\geq T_G(z,z)^2$ for all positive real numbers $x,z$ with $x\\geq z(z+2)$. Our result was inspired by a conjecture of Merino and Welsh that $\\max\\{T_G(2,0),T_G(0,2)\\}\\geq T_G(1,1)$.", "Now double differentiate the above equation $\\Rightarrow \\dfrac{{{d^2}}}{{d{t^2}}}{y_2} = \\dfrac{d}{{dt}}\\left( {1 - \\dfrac{1}{{{t^2}}}} \\right) = 0 + \\dfrac{2}{{{t^3}}}$ Now for $t = 1$ The value of above equation is positive so the function is minimum at $t = 1$ Now for real solution t should be greater than zero$\\left( {t > 0} \\right)$ Because ${5^x} = t$ for t less than zero (t<0), ${5^x}$ become imaginary hence there is no real solution for t less than zero. So, at $t = 1,{\\text{ }}{y_2} = 1 + 1 = 2$ So, ${\\left( {{y_2}} \\right)_{\\min }} = 2................\\left( 2 \\right)$ $\\therefore {y_2} \\geqslant 2$ Now from equation (1) and (2) ${\\left( {{y_1}} \\right)_{\\max }} = 1,{\\text{ }}{\\left( {{y_2}} \\right)_{\\min }} = 2$…………….. (3) But from the given equation $\\sin \\left( {{e^x}} \\right) = {5^x} + {5^{ - x}} \\\\ {y_1} = {y_2} \\\\$ So, from equation (3) the above condition never holds for real solutions. So, the number of real solutions of the given equation is zero. Hence option (a) is correct. Note- In such types of questions always remember the range of sine function so, in above problem solve L.H.S and R.H.S separately and find out the range of these functions for real solution, then compare their ranges and also check the equality, we will get the required answer.", "In order for y to be real, the quantity under the radical has to be nonnegative, so you need to solve the quadratic inequality $28 - 3x^2 \\ge 0$. Solving this results in $|x| \\le \\sqrt{28/3}$, which is slightly different from what you report is the book's answer.", "Review question # If $y=(x^2+1)/(x^2-a^2)$, then $y$ cannot take which values? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8765 ## Solution Show that, if $y=\\dfrac{x^2+1}{x^2-a^2},$ $y$ takes all real values twice, except those for which $-\\dfrac{1}{a^2}\\le y\\le 1$. Since $y$ is a function of $x^2$ rather than just $x$ we can show that in general $y(x)=y(-x)$ by substituting and simplifying: \\begin{align*} y(-x)&=\\dfrac{(-x)^2+1}{(-x)^2-a^2} \\\\ &=\\dfrac{x^2+1}{x^2-a^2}\\\\ &=y(x). \\end{align*} This tells us that where $y$ exists and $x \\neq -x$ then $y$ values are repeated. As $x \\rightarrow a \\quad x^2-a^2 \\rightarrow 0$ so $y \\rightarrow \\pm \\infty$. Since $x^2+1\\geq1 \\quad y\\neq 0$ so there are some values that $y$ cannot take and we must find them. Rearranging $y=\\dfrac{x^2+1}{x^2-a^2}$, we arrive at $x^2=\\dfrac{a^2y+1}{y-1}.$ Since $x$ is real $x^2 \\geq 0$ so a $y$-value is impossible if and only if $\\dfrac{a^2y+1}{y-1} < 0$. So when is $\\dfrac{a^2y+1}{y-1} < 0$? If $y-1$ is positive, then $a^2y+1<0 \\implies y < -\\dfrac{1}{a^2}$, which is impossible, since $y > 1$. On the other hand, if $y-1$ is negative (that is if $y<1$), then $a^2y+1>0 \\implies y > -\\dfrac{1}{a^2}$, so we have $-\\dfrac{1}{a^2}<y<1$. If $y = 1$, then no $x$-values are possible. If $y = -\\dfrac{1}{a^2}$, then $x = 0$, but this" ]

[ "function everywhere. To do this we examine$$x^2 \\gt |x - \\tfrac34|$$ For $x \\ge \\tfrac34$ we have \\begin{aligned}x^2 &\\gt x - \\tfrac34 \\\\ x^2 - x + \\tfrac34 &\\gt 0\\end{aligned} The discriminant of the left side is $1 - 3 = -2 \\lt 0$, so the inequality is always true or always false. Putting $x=1$ shows the truth of the inequality in this case and thus everywhere. So, for $x \\ge \\tfrac34$ the function is equal to $x^2$ and on this domain, the minimum is $(\\frac34)^2 = \\frac9{16}$. Now we look at $x \\le \\tfrac34$. Here we seek \\begin{aligned}x^2 &\\gt \\tfrac34 - x \\\\ x^2 + x - \\tfrac34 &\\gt 0 \\\\ (x + \\tfrac32)(x - \\tfrac12) &\\gt 0\\end{aligned} So for $x \\le -\\tfrac32$ the inequality is true and our function is equal to $x^2$. Its minimum on this domain is $(-\\frac32)^2 = \\frac94$. Similarly, for $x \\ge \\frac12$ our function is equal to $x^2$. Its minimum on this domain is $(\\frac12)^2 = \\frac14$. On the rest of the real domain, ($-\\frac32 \\le x \\le \\tfrac12$) our function is linear ($\\tfrac34 - x$) and this line must join two parts of the graph. So the function's minimum is least of the minima we have found. Thus the minimum is $\\frac14$. Thanks from mili January 30th, 2015, 06:23 AM", "1. ## finding the range Hi folks, I really hope there is no simple arithmetic error in my answer, but I've checked it many times, so here goes. Show that for real x the function $y = \\frac{x + 2}{(x + 3)(x - 1)}$ can take all real values. This should be really easy: $y = \\frac{x + 2}{x^2 + 2x - 3}$ $yx^2 + 2yx - 3y = x + 2$ $yx^2 + (2y - 1)x - 3y - 2 = 0$ a quadratic in x of the form $ax^2 + bx + c = 0$. For real x the discriminant $(b^2 - 4ac \\ge 0)$ $(2y - 1)^2 + 4.y.(3y + 2) \\ge 0$ $4y^2 - 4y + 1 + 12y^2 + 8y \\ge 0$ $16y^2 + 4y + 1 \\ge 0$ solving for y, this has no real roots. I was expecting to get any expression like $y^2 + something \\ge 0$ which would be true for all y, but no real roots suggests I have taken a wrong turn. Can you see it. 2. ## Re: finding the range For any $y \\in \\mathbb{R}$: $x = \\dfrac{1-2y \\pm \\sqrt{(2y-1)^2+4y(3y+2)}}{2y}$ This is undefined for $y=0$. The discriminant is always positive for all $y$, so all that is left to check is that you can yet", "+ c^2 = 1$, and since $xy + yz + xz$ is a symmetric polynomial in $x,y,z$, we should seek $x=y=z>0$ as our minimum, which can be shown the be the case using very straightforward techniques from calculus (Second derivative test should suffice, proving the minimum for $a,b,c$ suffices). We find that $x=y=z$ when $a=b=c=\\dfrac{1}{\\sqrt{3}}$, so that $x=y=z=\\dfrac{1}{1-1/3}=\\dfrac{3}{2}$. We can then say that $$xy + yz + xz \\geq 3*(3/2)^2 = 3*(9/4) = 27/4$$ Therefore, $$x^2 + y^2 + z^2 \\leq \\dfrac{81}{4} - 2*\\dfrac{27}{4} = \\dfrac{81}{4} - \\dfrac{54}{4} = \\dfrac{27}{4}$$ • I don't see how you obtain the final, desired inequality. You seem to just assert it. – zibadawa timmy Apr 9 '17 at 22:33 • This comes from the fact that the function $f(x) = x^2$ is monotonically increasing on $\\mathbb{R}^+$. By that I mean that for positive reals $x,h$, $x^2 < (x+h)^2$. Since $\\dfrac{9}{2} \\geq \\dfrac{9}{2} - \\varepsilon$ and since both are positive, $\\bigg{(} \\dfrac{9}{2} \\bigg{)}^2 \\geq \\bigg{(} \\dfrac{9}{2} - \\varepsilon \\bigg{)}^2$. Does that answer your question? – Will Craig Apr 9 '17 at 22:36 • $\\frac{9}{2}=81/4$, so we cannot say $(\\frac{9}{2} - \\varepsilon)^2\\le 27/4$ – didgogns Apr 9 '17 at 23:05 • I wasn't thinking, my fault. I will see if I can correct my approach. – Will Craig Apr 9 '17 at 23:05 •", "equal to 0. EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) \\geq 0$$ $$\\implies -\\frac{5}{8} \\leq y < \\frac{1}{3}$$ which is the range of your function. EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your \"c\" term should be $(20y+1)$, not $(20y-1)$. • Thank you so much, this is exactly what I was asking. – William Jul 17 '18 at 17:51 • @William If you say so, check this answer. – Takahiro Waki Jul 30 '18 at 0:38 $$\\frac{x^2+4x-1}{3x^2+12x+20}=y$$ After cross multiplying, we obtain $$(3y-1)x^2+(12y-4)x+(20y+1)=0$$ If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation. However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real. The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$", "has a solution if and only if $y \\ge -1/4$. Jul 1 at 13:46 • I think you're misinterpreting the symbol $\\ge$. The set $\\{y \\in \\mathbb{R} \\mid y \\ge -1/4\\}$ includes the value $-1/4$. Each value may only be greater than or equal, but the solution includes both the values that are greater than, and the one value that is equal. Jul 2 at 9:08 I would write this as $$y = (x+\\frac32)^2-\\frac14$$, from which it is obvious what the range would be, as the square term takes all nonnegative values. • That is certainly a better way to do it, but I was experimenting with the other ways there are to approach the problem. Jul 2 at 6:24 • @RajShukla Your approach works well too, as others noted you have proved for every $y\\ge -\\frac14$, there is at least one real value of $x$, and not for any other $y$, so the range is established Jul 2 at 7:11 In your solution, you show that $$x$$ has a real value if $$y\\geq -1/4$$ using discriminant of a quadratic. Well this works the other way too! If $$y\\geq -1/4$$, then the quadratic has nonnegative discriminant and so you will always find $$x$$ that give that specific value of $$y$$. Hope this helps. :) • That makes sense,", "+0 # Range 0 118 1 The range of the function g(x) = 2/(2 + 4x^2) + 8/(1 + x^2)can be written as an interval (a,b]. What is a+b? May 17, 2022 #1 +9461 0 Note that$$g(x) = \\dfrac2{2 + 4x^2} + \\dfrac8{1 + x^2} = \\dfrac1{1 + 2x^2} + \\dfrac8{1 + x^2}$$. Also, $$x^2 + 1 \\geq 1$$ and $$1 +2x^2 \\geq 1$$. Then, $$0 < \\dfrac1{1 + 2x^2} \\leq 1$$ and $$0 < \\dfrac1{1 + x^2} \\leq 1$$. That means $$0 < \\dfrac8{1 + x^2} \\leq 8$$. Therefore, $$0 < g(x) \\leq 9$$ by adding the inequalities. Can you rewrite the range as an interval? May 18, 2022", "that, $\\frac{9y+4}{1+y}\\geq 0$ Thus, $9y+4\\geq 0 \\mbox{ and }1+y\\geq 0$ Or, $9y+4\\leq 0 \\mbox{ and }1+y\\leq 0$ Solving the inequalities, $y\\geq -4/9 \\mbox{ and }y\\geq -1\\mbox{ thus }y\\geq -4/9$ Or, $y\\leq -4/9 \\mbox{ and }y\\leq -1\\mbox{ thus }y<-1$ (remember $y\\not = -1$) Thus, the range is, $(-\\infty,-1)\\cup [-4/9,+\\infty)$ ---- This is for Jameson, try to learn it my way. Because that method is graphical and works for \"well-behaved\" functions. If you want to do it mathematically this is the way. 5. Originally Posted by qbkr21 2. f(x)=(2x^2-3)/(x+1) What is the fuctions range? Again, for what values of $y$ does the function has a solution in the domain? $y=\\frac{2x^2-3}{x+1}$ Multiply by $(x+1)$ thus, $xy+y=2x^2-3$ Thus, $2x^2-xy+(-3-y)=0$ This is a quadradic equation, it has real solutions when the discrimanant $b^2-4ac\\geq 0$ thus, $y^2-4(2)(-3-y)\\geq 0$ Thus, $y^2+8(3+y)\\geq 0$ Thus, $y^2+8y+24\\geq 0$ Now we look at the discrimanat of this quadradic and we find that, $8^2-4(24)<0$ and $a>0$ Thus, this expression is always positive for any $y$. Hence the range is $(-\\infty,+\\infty)$ 6. ## Re: Sir I am completely lost how in the **** did you do that. What is \"Hackers Trick\"? This seems extreamly useful because you turned this problem into a enequality, which I can easily solve. What on earth did you do? Thanks 7. Originally Posted by qbkr21 Sir", "Putting it in the expression, $\\Rightarrow \\dfrac{1}{t^2-97} \\leq \\dfrac{5}{2}$ Now, we know that the least value of $t^2$ possible is 0. At that value, LHS < RHS, thus the equation holds true. RHS of the expression will always be negative for all values of $t \\in [0,\\sqrt{97})$ Here, $(x – 9)^2 < 97$ $\\Rightarrow x < 9+\\sqrt{97}$ and $x > 9 – \\sqrt{97}$ $\\Rightarrow x < 9+10$ and $x < 9-10$ (We have equated $\\sqrt{97}$=10 because the smallest integral value of $t$ satisfying this condition will be 10) $\\Rightarrow x < 19$ and $x >-1$ Thus, the values of $x$ satisfying this condition are : 1,2….18. Here, the value of $t$ cannot be $\\sqrt{97}$ as then the denominator would become 0 and lead to an indeterminate form. For all values of $t \\geq \\sqrt{97}$, the fraction is positive. In that case, $\\dfrac{1}{t^2-97} \\leq \\dfrac{5}{2}$ $\\Rightarrow t^2-97 \\geq \\dfrac{2}{5}$ $\\Rightarrow t^2 \\geq 97+\\dfrac{2}{5}$ $\\Rightarrow t^2 \\geq 97.4$ $\\Rightarrow t \\geq 10$ (We have equated $\\sqrt{97.4}$=10 because the smallest integral value of $t$ satisfying this condition will be 10) Therefore $t \\in [0,\\sqrt{97})U(\\sqrt{97},\\infty)$ Since $t = x-9$, therefore the range of $x$ can be computed by adding 9 to the range of $t$. Thus, $x \\in [1,19)U(9+\\sqrt{97},\\infty)$ $\\Rightarrow x \\in [1,9+\\sqrt{97})U(9+\\sqrt{97},\\infty)$ To count the integral values of $x$ which satisfy", "the range of the function. It turns out to be different from the range announced in your post. The standard way to address the question is by using the differential calculus. However, it can be done purely algebraically. Let $$y=\\frac{6x^2}{3x^2-2x-1}.$$ Keeping at the back of our minds the fact that there is a singularity at $x=-\\frac{1}{3}$ and $x=1$, we rewrite this as $y(3x^2-2x-1)=6x^2$ and then as $$(6-3y)x^2+(2y)x+y=0.\\tag{1}$$ There is an additional worry at $y=2$. But apart from that and our reservations about $x=-\\frac{1}{3}$ and $x=1$, the quadratic equation $(1)$ has a real solution $x$ precisely if the discriminant is $\\ge 0$. The discriminant is $(2y)^2 -4(6-3y)(y)$, which can be rewritten as $$8y(2y-3).$$ The discriminant is positive except when $0 \\lt y \\lt \\frac{3}{2}$. There is no issue about $y=2$. For we can solve the equation $2=\\frac{6x^2}{3x^2-2x-1}$, getting $x=-\\frac{1}{2}$. The values $x=-\\frac{1}{3}$ and $x=1$ do not affect our range calculation. Thus the range is $(-\\infty,0] \\cup [\\frac{3}{2},\\infty)$. -", "# 1983 AIME Problems/Problem 9 ## Problem Find the minimum value of $\\frac{9x^2\\sin^2 x + 4}{x\\sin x}$ for $0 < x < \\pi$. ## Solution ### Solution 1 Let $y=x\\sin{x}$. We can rewrite the expression as $\\frac{9y^2+4}{y}=9y+\\frac{4}{y}$. Since $x>0$ and $\\sin{x}>0$ because $0< x<\\pi$, we have $y>0$. So we can apply AM-GM: $$9y+\\frac{4}{y}\\ge 2\\sqrt{9y\\cdot\\frac{4}{y}}=12$$ The equality holds when $9y=\\frac{4}{y}\\Longleftrightarrow y^2=\\frac49\\Longleftrightarrow y=\\frac23$. Therefore, the minimum value is $\\boxed{012}$ (when $x\\sin{x}=\\frac23$; since $x\\sin x$ is continuous and increasing on the interval $0 \\le x \\le \\frac{\\pi}{2}$ and its range on that interval is from $0 \\le x\\sin x \\le \\frac{\\pi}{2}$, by the Intermediate Value Theorem this value is attainable). ### Solution 2 We can rewrite the numerator to be a perfect square by adding $-\\dfrac{12x\\sinx}{x\\sinx}$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, we must also add back $12$. This results in $\\dfrac{(3x\\sinx-2)^2}{x\\sinx}+12$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, if $3x\\sinx-2=0$ (Error compiling LaTeX. ! Undefined control sequence.), then the minimum is obviously 12. We show this possible with the same methods in Solution 1; thus the answer is $012$. ### Solution 3 Let $y = x\\sin{x}$ and rewrite the expression as $f(y) = 9y + \\frac{4}{y}$, similar to the previous solution. To minimize $f(y)$, take the derivative of $f(y)$ and set it equal to zero. The derivative", "of the equation are equal, and the imaginary parts are equal. So, reals: $2x-6 = 10$ Imaginaries: $5 = y-3$ I'm sure you can solve it now! (c) This is all about handling a quadratic expression. You can do it: • by differentiating • by simply quoting a formula (if you know it) or • by completing the square. I'll show you the last method, since I'm guessing that might be the way you've been shown. $f(x) = 4x^2-3x +1$ Take out factor $4$ to make coefficient of $x^2$ equal to $1$: $f(x) = 4(x^2 - \\tfrac34x+\\tfrac14)$ Write half the coefficient of $x$ inside a 'bracket squared': $f(x) = 4\\Big((x-\\tfrac38)^2 - ...\\Big)$ Subtract the square of this number, and add in the original constant number: $f(x) = 4\\Big((x-\\tfrac38)^2 - \\tfrac{9}{64}+\\tfrac14\\Big)$ Simplify: $=4\\Big((x-\\tfrac38)^2 + \\tfrac{7}{64}\\Big)$ Since $(x-\\tfrac38)^2 \\ge 0$ for all real values of $x$, you can then say that the value of $f(x)$ is a minimum when $(x-\\tfrac38)^2 = 0$; i.e. when $x = \\tfrac38$.", "you rotate the $\\sin$ functions $180^\\circ$ when $c>1$ and evaluate with $c'=\\frac{1}{c}$ to allow any positive $c$? I also really like your explanations, especially the way you convert everything into a finding roots problem Jul 23 '20 at 17:25 • @Acreol A nice observation. Yes, we can rotate the graph around a point $\\left(\\tfrac 12, \\tfrac 12\\right)$. Analytically we can take a function $g(x)=1-f(1-x)$. Then we obtain $g(0)=0$, $g(1)=1$. Moreover, $g’(x)=f’(1-x)\\ge 0$ so, $g(x)$ is monotonic and $g’(0)=g’(1)$. Finally $g\\left(\\tfrac {1}{{\\tfrac 1c}+1}\\right)=g\\left(\\tfrac {c}{c+1}\\right)=1- f\\left(1-\\tfrac {c}{c+1}\\right)= 1- f\\left(\\tfrac {1}{c+1}\\right)=\\tfrac 12$. Jul 23 '20 at 17:59", "+0 help algebra 0 81 2 Let x and y be real numbers such that 2(x^2 + y^2) = x + y + 8. Find the maximum value of x - y. Apr 19, 2022 #1 +9459 +1 Moving some terms yields: $$2x^2 + 2y^2 - x - y - 8 = 0$$ Note that geometrically, this is the equation of a circle, because we can rewrite the equation as follows: $$\\begin{array}{rcl} 2x^2 + 2y^2 - x - y - 8 &=& 0\\\\ x^2 + y^2 - \\dfrac12 x - \\dfrac12 y - 4&=&0\\\\ \\left(x - \\dfrac14\\right)^2 + \\left(y - \\dfrac14\\right)^2 &=& \\dfrac{33}8 \\end{array}$$ Consider the straight line x - y = c, where c is varying through real numbers. You can find the maximum value of c such that the line and the circle intersects, i.e., $$\\begin{cases}2x^2 + 2y^2 - x - y - 8 = 0\\\\x - y = c​​​​\\end{cases}$$ has solutions. To do so, you need to substitute y = x - c into the circle equation, then consider the discriminant of the resulting equation. Since the system has solutions, the discriminant of the resulting equation must be nonnegative. Solving the equation $$\\Delta \\geq 0$$ will give you the suitable range of values of c. Please tell me if you want more explanations or you", "is negative, zero, or positive respectively. At any maximum or minimum value $y_0$ of the function, the discriminant will be zero since on one side of $y_0$ the quadratic equation will have a solution (discriminant non-negative) while on the other it will not (discriminant negative). In the image below a maximum is reached at $y_0 = 2$ while it is of opposite signs either side of this. Hence setting the discriminant of the left side of (2) to 0, $(y-6)^2 - 4(2y) = 0$ from which $y^2 - 20y + 36 = (y-18)(y-2) = 0$. Hence extrema are at $y=2$ and $y = 18$. We can solve $(y-18)(y-2) \\geq 0$ to find that $y \\leq 2$ or $y \\geq 18$ is the range of the function $y = \\frac{8}{2x+1} - \\frac{1}{x}$. This tells us that $y = 2$ is a local maximum (illustrated above) and $y = 18$ is a local minimum (occurring when $x < 0$). One advantage of this method is that unlike elementary calculus, one bypasses the step of finding the corresponding $x$ value (i.e. by solving $dy/dx = 0$) before substituting this into the function to find the extremum value for $y$. Another advantage is that the equation need not be polynomial in $y$. For example below is a plot of $x^2 + 4x\\sin", "is negative, zero, or positive respectively. At any maximum or minimum value $y_0$ of the function, the discriminant will be zero since on one side of $y_0$ the quadratic equation will have a solution (discriminant non-negative) while on the other it will not (discriminant negative). In the image below a maximum is reached at $y_0 = 2$ while it is of opposite signs either side of this. Hence setting the discriminant of the left side of (2) to 0, $(y-6)^2 - 4(2y) = 0$ from which $y^2 - 20y + 36 = (y-18)(y-2) = 0$. Hence extrema are at $y=2$ and $y = 18$. We can solve $(y-18)(y-2) \\geq 0$ to find that $y \\leq 2$ or $y \\geq 18$ is the range of the function $y = \\frac{8}{2x+1} - \\frac{1}{x}$. This tells us that $y = 2$ is a local maximum (illustrated above) and $y = 18$ is a local minimum (occurring when $x < 0$). One advantage of this method is that unlike elementary calculus, one bypasses the step of finding the corresponding $x$ value (i.e. by solving $dy/dx = 0$) before substituting this into the function to find the extremum value for $y$. Another advantage is that the equation need not be polynomial in $y$. For example below is a plot of $x^2 + 4x\\sin", "is negative, zero, or positive respectively. At any maximum or minimum value $y_0$ of the function, the discriminant will be zero since on one side of $y_0$ the quadratic equation will have a solution (discriminant non-negative) while on the other it will not (discriminant negative). In the image below a maximum is reached at $y_0 = 2$ while it is of opposite signs either side of this. Hence setting the discriminant of the left side of (2) to 0, $(y-6)^2 - 4(2y) = 0$ from which $y^2 - 20y + 36 = (y-18)(y-2) = 0$. Hence extrema are at $y=2$ and $y = 18$. We can solve $(y-18)(y-2) \\geq 0$ to find that $y \\leq 2$ or $y \\geq 18$ is the range of the function $y = \\frac{8}{2x+1} - \\frac{1}{x}$. This tells us that $y = 2$ is a local maximum (illustrated above) and $y = 18$ is a local minimum (occurring when $x < 0$). One advantage of this method is that unlike elementary calculus, one bypasses the step of finding the corresponding $x$ value (i.e. by solving $dy/dx = 0$) before substituting this into the function to find the extremum value for $y$. Another advantage is that the equation need not be polynomial in $y$. For example below is a plot of $x^2 + 4x\\sin", "> 0, $m^{2} + 4pr ≥ 0$, therefore, H(x) will also have real root(s). Hence at least two roots of G(x) H(x) are real. Example 2: Find the range in which the value of given quadratic Expression will lie $\\mathbf{\\frac{x^{2}\\;+\\;14x\\;+\\;9}{x^{2}\\;+\\;2x\\;+\\;3}}$. [x is a real number] Solution: Let, y = $\\mathbf{\\frac{x^{2}\\;+\\;14x\\;+\\;9}{x^{2}\\;+\\;2x\\;+\\;3}}$ i.e. $x^{2} + 14x + 9 = (x^{2}y + 2xy + 3y)$ Or, $(1 – y) x^{2} + (7 – y) 2x + 3(3 – y) = 0$ Since x is real, therefore, $b^{2} – 4ac ≥ 0$ i.e. $[2(7 – y)]^{2} – 4[1 – y] \\times 3[3 – y] ≥ 0$ Or, $- y^{2} – y + 40 ≤ 0$ Or, (y + 5) (y – 4) ≤ 0 Therefore, the range in which the value of given Quadratic Expression will lie is – 5 ≤ y ≤ 4. Example 3: Find the values of x for which the equation $\\mathbf{\\frac{x^{2}\\;-\\;4x\\;+\\;3}{x^{2}\\;+\\;x\\;+\\;1}\\;\\leq \\;0}$. Solution: Let f (x) = $x^{2} – 4x + 3$ and g (x) = $x^{2} + x + 1$. The coefficient of $x^{2}$ in g (x) is positive and the value of Discriminant (D) < 0. Hence, g(x) is positive for all values of x. Since, $\\mathbf{\\frac{f(x)}{g(x)}\\;<\\; 0}$. Therefore, f (x) must be less than 0. i.e. $x^{2} – 4x + 3 < 0$ Or,", "# Chapter 3 - Applications of Differentiation - 3.1 Exercises: 25 Over the specified interval, the function has an absolute maximum equal to $\\dfrac{1}{4}$ and an absolute minimum equal to $0$. #### Work Step by Step Using the quotient rule: $g'(t)=(\\frac{u(t)}{v(t)})'=\\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$. $u(t)=t^2 ;u'(t)=2t$. $v(t)=t^2+3 ;v'(t)=2t$ $g'(t)=\\dfrac{(2t)(t^2+3)-(2t)(t^2)}{(t^2+3)^2}=\\dfrac{6t}{(t^2+3)^2}.$ $g'(t)=0\\to 6t=0\\to t=0$ $g'(t)$ is defined for all values of $t.$ Along with the interval endpoints, $t=0$ is a possible value at which the function might attain an absolute extremum. $g(-1)=\\dfrac{1}{4}.$ $g(0)=0.$ $g(1)=\\dfrac{1}{4}$ Over the specified interval, the function has an absolute maximum equal to $\\dfrac{1}{4}$ and an absolute minimum equal to $0$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "solve for the case $y \\neq 0$. Every thing else is fine with your solution. $$f(x)=\\frac{x}{x^2+1}$$ Since the function is an odd function $(f(x)=-f(-x))$, then the minimum value is equal to the opposite of the maximum value. We see that $f(x) \\ge 0$ when $x \\ge 0$, so we need to find the maximum value of $f(x)$ when $x \\ge 0$. Note that $\\dfrac 12 - f(x) = \\dfrac 12 - \\dfrac{x}{x^2+1} = \\dfrac{x^2+1-2x}{2(x^2+1)} = \\dfrac{(x-1)^2}{2(x^2+1)}$ So, when $x \\ge 0$, we see that $\\dfrac 12 - f(x) \\ge 0$ and, when $x=1$, we see that $\\dfrac 12 - f(x) = 0$. Since $\\dfrac 12 - f(x) \\ge 0 \\iff f(x) \\le \\dfrac 12$ and $f(1)= \\dfrac 12$, we see that $\\displaystyle \\max_{x \\ge 0} f(x) = \\dfrac 12$. Hence, because $f(x)$ is an odd function, $\\operatorname{Range} f(x) = \\left[ -\\dfrac 12, \\dfrac 12 \\right]$", "at 8:56 As I wrote in the comments, $$g(1+x)=\\sqrt{2+x+\\sqrt{3-2x-8x^2}}\\;,\\tag{1}$$ so the first step is to ensure that $3-2x-8x^2\\ge 0$. This quadratic happens to factor: $3-2x-8x^2=(3+4x)(1-2x)$, so it’s equal to $0$ when $x=-\\frac34$ and when $x=\\frac12$. To see where it’s positive you can check to see where $3+4x$ and $1-2x$ have the same sign, but it’s easier to realize that the graph of $y=3-2x-8x^2$ is a parabola opening down, so it lies above the $x$-axis between the roots $-\\frac34$ and $\\frac12$. Thus, $3-2x-8x^2\\ge 0$ when $-\\frac34\\le x\\le\\frac12$. (You must have made a sign error in the calculation in your comment.) Okay, now we know that the domain is at most the interval $\\left[-\\frac34,\\frac12\\right]$, but there might be values of $x$ in that interval that make $$2+x+\\sqrt{3-2x-8x^2}\\tag{2}$$ negative. Are there? At $x=-\\frac34$, $(2)$ is equal to $2-\\frac34=\\frac54$, which is certainly not negative. As $x$ increases from $-\\frac34$ to $\\frac12$, $2+x$ also increases, so it won’t be a problem: it will always be at least $\\frac54$. What about $\\sqrt{3-2x-8x^2}$? On the interval $\\left[-\\frac34,\\frac12\\right]$ it’s always at least $0$, so it won’t be a problem either, and we can be sure that $(2)$ is always at least $\\frac54$ when $-\\frac34\\le x\\le\\frac12$. In particular, it won’t be negative, so we can take its square root. Thus, the domain of $g(1+x)$ is the whole" ]

[ "# Definition:Strictly Positive Real Function ## Definition Let $I$ be a real interval. Let $f$ be a real function. Then $f$ is strictly positive (on $I$) iff: $\\forall x \\in I: f \\left({x}\\right) > 0$ ## Also known as A strictly positive real function is also described as a positive real function, but this causes confusion with a non-negative real function and so this usage is to be avoided.", "# Definition:Positive Real Function ## Definition Let $I$ be a real interval. Let $f$ be a real function. Then $f$ is positive (on $I$) if and only if: $\\forall x \\in I: f \\left({x}\\right) \\ge 0$ ## Also known as A positive real function is also described as non-negative in order to avoid confusing with a strictly positive real function.", "Let f be the function satisfying f'(x)= $$\\sqrt{f(x)}$$ for all real numbers where f(3)=25. Find f''(3).", "# Real analysis Q1. f is a continuous real valued function on [o,oo) and a is a real number Prove that the following statement are equivalent; (i) f(x)--->a, as x--->oo (ii) for every sequence {x_n} of positive numbers such that x_n --->oo one has that (1/n)\\sum f(x_k)--->a, as n--->oo (the sum is taken from k=1 to k=n)", "# Let f(x) be a function satisfying the condition f(-x) = f(x) for all real x. If f'(0) exists, then its value is equal to Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free,", "$$g(x)$$. 10. Define $$f(x) = \\sin^6(x) + \\cos^6(x)$$, and $$g(x) = \\sin^4(x) +\\cos^4(x)$$, and let $$h(x) = f(x) + k g(x)$$ for some real number $$k$$. 1. Determine all real numbers, $$k$$, for which $$h(x)$$ is constant for all values of $$x$$. 2. Determine all real numbers, $$k$$, for which there exists a real number $$c$$ such that $$h(c) = 0$$.", "f(x) is defined as some function from reals to reals where $f(x) = \\sqrt[4]{x}$, then $f(x)$ is only a strictly increasing function with domain and range being the set of positive reals (and zero). For all complex x and positive integers k, $\\sqrt[2k](x)$ is never negative (although for tiny positive epsilon $\\sqrt(1 -i\\epsilon)$ is awfully close). So we cannot ever have $\\sqrt(4) = -2$, but if we do know $x^2 = 4$, then x =±$\\sqrt(4)$. Thanks for the correction, but none was necessary. ANAL MODE OFF: ## References Arrrrggggh!! This don't work on user pages!", "Product of Increasing Positive Functions is Increasing Theorem Let $f$ and $g$ be real functions defined on an interval $I$. Let $f$ and $g$ be increasing and positive on $I$. Then the product $f g$ is increasing on $I$.", "# 2020 CAMO Problems/Problem 1 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem 1 Let $f:\\mathbb R_{>0}\\to\\mathbb R_{>0}$ (meaning $f$ takes positive real numbers to positive real numbers) be a nonconstant function such that for any positive real numbers $x$ and $y$, $$f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y).$$Prove that there is a constant $a>1$ such that $$f(x)=\\frac{a^x-1}{a^x+1}$$for all positive real numbers $x$.", "real numbers. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 - y^2) = x f(x) - y f(y)$ for all pairs of real numbers $x$ and $y$.", "positive real number. By definition, $$y = \\log_b x \\qquad \\text{means} \\qquad b^y = x.$$ Since $b^y > 0$ for all $y \\in \\mathbb{R}$, as a function of a real variable, $\\log_b x$ is only defined for $x>0$.", "Definition:Negative Real Function Definition Let $I$ be a real interval. Let $f$ be a real function. Then $f$ is negative (on $I$) if and only if: $\\forall x \\in I: \\map f x \\le 0$ Also known as A negative real function is also described as non-positive in order to avoid confusing with a strictly negative real function.", "real = - ... conditions is f(x, y) necessarily greater than g(x, y)? Both x and y are less than -1 Both x and y are positive Both x and y are negative y > x", "such that $f(f(x)) = {1 \\over x}$ Is it possible to define a function $f$, from positive real to positive real such that $f(f(x)) = {1 \\over x}$? The motivation comes from $1990$ IMO problem $4$, which one step involves defining such ...", "# Definition:Strictly Negative Real Function ## Definition Let $I$ be a real interval. Let $f$ be a real function. Then $f$ is strictly negative (on $I$) if and only if: $\\forall x \\in I: \\map f x < 0$ ## Also known as A strictly negative real function is also described as a negative real function, but this causes confusion with a non-positive real function and so this usage is to be avoided.", "positive part of $\\sqrt{x}$ is usually taken to be its definition. A multi-valued function does not satisfy the criterion of many-to-one(every element in its domain is associated to exactly one element in its range), so it is not a function. By restriction the range of $\\sqrt{x}$, every positive real number can be associated to one element in $\\mathbb{R}$, which turns $\\sqrt{x}$ into a function.", "Let f(x) be a function whose domain is the set of real numbers, for which f(0)=1 , and such that for any pair of real numbers x and y we have : f( xy + 1 ) = f(x) f(y) - f(y) -x + 2 Determine the formula for f(x) . Thank you very much for your help!!! I appreciate it! 2. For $x=y=0$ we have $f(1)=2$. For $y=0$ we have $f(x)=x-1+f(1)$. Then $f(x)=x+1$.", "# Existence of Positive Root of Positive Real Number/Positive Exponent ## Theorem Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n > 0$. Then there exists a $y \\in \\R: y \\ge 0$ such that $y^n = x$. ## Proof Let $f$ be the real function defined on the unbounded closed interval $\\left[{0 \\,.\\,.\\, \\to}\\right)$ defined by $f \\left({y}\\right) = y^n$. Consider first the case of $n > 0$. $\\exists q \\in \\R_{>0}: f \\paren q \\ge x$ Since $x \\ge 0$: $f \\paren 0 \\le x$ By the Intermediate Value Theorem: $\\exists y \\in \\R: 0 \\le y \\le q, f \\paren y = x$ Hence the result has been shown to hold for $n > 0$. $\\Box$", "# Definition:Less Than (Real Numbers) ## Definition Let $\\R_{>0}$ denote the set of strictly positive real numbers. Let $x, y \\in \\R$. Then we write $x < y$ if and only if: $y - x \\in \\R_{>0}$ and we say that $x$ is less than $y$. ## Also see Inequality iff Difference is Positive, where it is shown that this may be deduced from the Ordering Properties of Real Numbers.", "# IMO 2011 (Problem 3) Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a real-valued function defined on the set of real numbers that satisfies $f (x + y) \\leq yf (x) + f (f (x))$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x\\leq 0$." ]

[ "+0 # If x, y, and z are positive 0 73 5 If x, y, and z are positive with xy = 20, xz = 35, and yz = 14, then what is xyz? Nov 14, 2020 #1 +50 +4 So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get: $$xy \\cdot xz \\cdot yz = 20 \\cdot 35 \\cdot 14$$ Finding the product, we have: $$x^2y^2z^2 = 9800$$ This is the same as $$(xyz)^2 = 9800$$ which finding the square root gives us $$xyz=\\sqrt{9800}$$ Simplifying, we find: $$\\sqrt{9800} \\implies \\sqrt{4900 \\cdot 2} \\implies \\sqrt{4900} \\cdot \\sqrt2 \\implies \\boxed{70\\sqrt2}$$ Nov 14, 2020 #1 +50 +4 So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get: $$xy \\cdot xz \\cdot yz = 20 \\cdot 35 \\cdot 14$$ Finding the product, we have: $$x^2y^2z^2 = 9800$$ This is the same as $$(xyz)^2 = 9800$$ which finding the square root gives us $$xyz=\\sqrt{9800}$$ Simplifying, we find: $$\\sqrt{9800} \\implies \\sqrt{4900 \\cdot 2} \\implies \\sqrt{4900} \\cdot \\sqrt2 \\implies \\boxed{70\\sqrt2}$$ mobro Nov 14, 2020 #2 +114221 0 Very nice, mobro !!!! Nov 14, 2020 #3 +846 +1 Good job, mobro! How can we find the value of x, y, and z? Nov 14, 2020 #4 +50", "this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ## Solution 3, x first Set it up as a quadratic in terms of y: $$y^2-2y+x^{2020}=0$$ Then the discriminant is $$\\Delta = 4-4x^{2020}$$ This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: $$y^2-2y+1=(y-1)^2=0$$ This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$: $$y^2-2y=0$$ Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all $x$, then $y(2-y) \\geq 0 \\Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$, $(0,0)$, $(1,1)$, $(-1,1)$. There are", "a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all $x$, then $y(2-y) \\geq 0 \\Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$$(0,0)$$(1,1)$$(-1,1)$. There are $4$ solutions, so the answer is $\\boxed{\\textbf{(D) } 4}$ 10. ## Solution Notice that when the cone is created, the radius of the circle will become the slant height of the cone and", "then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that $P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010$. Now, we can say that $P(m) = (m-1)Q(m) - 2010$ for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$, $(m-1)Q(m) = 2010$. Thus, if $P(m) = 0$, then $m-1 | 2010$ . Now, we need to show that for all $m-1 | 2010$, $m^{x_{0}}=\\sum_{k = 1}^{2011}m^{x_{k}}.$. We try with the first few $m$ that satisfy this. For $m = 2$, we see we can satisfy this if $x_0 = 2010$, $x_1 = 2009$, $x_2 = 2008$, $\\cdots$ , $x_{2008} = 2$, $x_{2009} = 1$, $x_{2010} = 0$, $x_{2011} = 0$, because $2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^1 = \\cdots$ (based on the idea $2^n + 2^n = 2^{n+1}$, leading to a chain of substitutions of this kind) $= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}$. Thus $2$ is a possible value of $m$. For other values, for example $m = 3$, we can use the same strategy, with $x_{2011} = x_{2010} =", "on earth are you then saying that they're roots of the polynomial? What you can say is that $x=-2,0,2$ are roots of $P(x)=2$, or equivalently $P(x)-2=0$. So we can say that, if $P(x)$ is 4th degree, then $P(x)-2$ is as well, and hence $P(x)-2=A(x-\\alpha)(x-2)x(x+2)$ for some $A,\\alpha$. Then you can insert your values for $P(1)$ and $P(-1)$. As for 2, you're on the right track. The next step would probably be to look at what the derivative of that expression will look like at $x=2$. • Oh Sorry I did not Noticed that, – juantheron Nov 14 '14 at 4:08 For part 2, here's one way to simplify the differentiation. We have $$f(x)-2x+1=2009\\cdot (x-1)\\cdot(x-2)\\cdot (x-3)\\cdot (x-4)\\cdot (x-5)\\cdot(x-r)$$ Substituting $x + 3 \\to x$, $$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$ $$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$ $$f(x + 3) - 2x - 5 = 2009(x^5 - 5x^3 + 4x)(x + 3 -r)$$ Applying the chain rule to LHS and product rule to RHS, $$f'(x+3) - 2 = 2009(5x^4 - 15x^2 + 4)(x + 3 - r) + 2009(x^5 - 5x^3 + 4x)$$ Substituting $x = -1$, $$f'(2) - 2 = 2009(5 - 15 + 4)(2-r) + 2009(1-5+4)$$ $$0 = 2009\\cdot(-6)\\cdot(2-r)$$ to give $r=2$. Hence, $$f(6) - 12 + 1 = 2009\\cdot5!\\cdot4$$ $$f(6) =", "be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$must be nonnegative. Therefore,$$4(1-x^{2020}) \\geq 0 \\implies x^{2020} \\leq 1.$$Here, we see that we must split the inequality into a compound, resulting in $-1 \\leq x \\leq 1$. The only integers that satisfy this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$.", "\\leq 1.$$ Here, we see that we must split the inequality into a compound, resulting in $-1 \\leq x \\leq 1$. The only integers that satisfy this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ### Solution 3: Solve for x first Set it up as a quadratic in terms of y: $$y^2-2y+x^{2020}=0$$ Then the discriminant is $$\\Delta = 4-4x^{2020}$$ This will clearly only yield real solutions when $|x^{2020}| \\leq 1$, because the discriminant must be positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: $$y^2-2y+1=(y-1)^2=0$$ This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$: $$y^2-2y=0 \\Rightarrow y(y-2)=0$$ Which has 2 solutions, so $(0,2)$ and $(0,0)$. These are the only 4 solutions, so our answer is $\\boxed{\\textbf{(D) } 4}$. ~edits by BakedPotato66 ### Solution 4: Solve for y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all", "# Difference between revisions of \"2008 iTest Problems/Problem 84\" ## Problem Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$. ## Solutions ### Solution 1 (credit to official solution) Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$. We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \\begin{align*} \\frac{rs}{r+s} &= -2008 \\\\ rs &= -2008r - 2008s \\\\ rs + 2008r + 2008s &= 0 \\\\ (r+2008)(s+2008) &= 2008^2. \\end{align*} WLOG, let $|a| \\le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \\tfrac{2008^2}{a}$. Thus, $$-r-s = b = -a - \\tfrac{2008^2}{a} + 4016.$$ Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \\cdot 251^2$, so there are $\\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \\cdot 22 =", "# Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\\in\\Bbb Z$ Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers. I tried to do it by comparing the two expressions but I can’t proceed. #### Solutions Collecting From Web of \"Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\\in\\Bbb Z$\" We have that $(x+a)(x+1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$. So if such factorisation exists we must have that the discriminant of the quadratic equation is a square. Now: $$D = (1991 + a)^2 – 4(1991a + 1) = 1991^2 + 2 \\cdot 1991a + a^2 – 4 \\cdot 1991a – 4 = (1991 – a)^2 – 4$$ But then $\\sqrt{D}, 2, (1991 – a)$ make a Pythagorean triplet, but we know that the only squares whose difference is $4$ are $4$ and $0$, therefore we must have $1991-a = \\pm 2 \\implies a_1=1989, a_2 = 1993$. Let $f(x)=(x+b)(x+c)$ with $b,c\\in\\Bbb Z$. Wlog $b\\le c$. For $-c\\le x\\le b$, we have $f(x)\\le 0$, hence $f(x)-1<0$. If $x\\ge -b+2$, we have $f(x)-1\\ge 2\\cdot 2-1>0$. By symmetry, $f(x)-1>0$ also for $x\\le -c-2$. Hence the only possibly integer roots of $f(x)-1$ are $-b+1$ and $-c-1$. Moreover, $f(-b+1)-1=f(-c-1)=c-b$, hence $f(x)-1$ has integer roots iff", "(i) How many positive factors are there of 2009? (ii) How many solutions in positive integers x,y,z are there to the equation x+y+z=2009? (iii) How many solutions in positive integers x,y,z are there to the equation x*y*z=2009? Hello, aab300! (a) How many positive factors are there of 2009? $2009 \\:=\\:7^2\\cdot41$ There are 6 positive factors: . $1,\\,7,\\,41,\\,49,\\,287,\\,2009$ (b) How many solutions in positive integers are there to the equation: . $x+y+z\\:=\\:2009\\,?$ Place 2009 objects in a row, leaving one space between them. . . . . . $\\circ\\,\\_\\,\\circ\\,\\_\\,\\circ\\,\\_\\,\\circ \\,\\_\\,\\hdots\\,\\_\\;\\circ$ Select two of the 2008 spaces and insert \"dividers\". . . So that:. $\\circ\\,\\_\\,\\circ\\,\\_\\,\\circ\\,|\\,\\circ\\,\\_\\;\\circ \\,| \\,\\circ\\,\\_\\, \\hdots \\,\\_\\;\\circ \\;\\text{ represents }\\,(3,\\,2,\\,2004)$ There are: . ${2008\\choose2} \\,=\\,2,\\!015,\\!028$ possible partitions. Therefore, there are 2,015,028 solutions to $x+ y +z \\:=\\:2009$ (c) How many solutions in positive integers x,y,z are there to the equation: . $xyz \\:=\\:2009\\,?$ The prime factorization 2009 is: . $7^2\\cdot41$ We have: . . $\\begin{array}{cc}\\text{Factors} & \\text{Permutations} \\\\ \\hline 1, 7, 287 & 6 \\\\ 1, 41, 49 & 6 \\\\ 7,7,41 & 3 \\\\ \\hline \\text{Total:} & 15 \\end{array}$ There are 15 solutions to $xyz = 2009$", "value in discri and never use it, so why is it there? 4. Jul 8, 2010 ### Staff: Mentor Here's a simple example you can try your code on - x2 + x + 1 = 0. Since the discriminant = -3, the roots are complex. Using the quadratic formula, you can find that the roots are x1 = (-1 + i*sqrt(3))/2, and x2 = (-1 - i*sqrt(3))/2 Your code will give you approximate values for (-1 + sqrt(3))/2 and (-1 - sqrt(3))/2. These are real numbers! 5. Jul 8, 2010 ### mike81 When I say they dont print out correctly I mean that it always writes, for complex (somenumber, 0). No matter what the second number is always zero. I see what your saying Mark, I could have used discri but initially just used it to have the program use either real or complex numbers. I am short on time tonight, I am going to play around with it some more tomorrow night. I really appreciate the help. 6. Jul 8, 2010 ### Staff: Mentor Which is what I'm saying. You're thinking that you're getting complex numbers, but you are really getting real numbers, so the imaginary part is zero. Once you calculate the discriminant, you know whether the roots are real or complex. BTW, you should", "# 2011 AIME I Problems/Problem 7 ## Problem 7 Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$, $x_1$ , $\\dots$ , $x_{2011}$ such that $$m^{x_0} = \\sum_{k = 1}^{2011} m^{x_k}.$$ ## Solution Let $P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}$. The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that $P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010$. Now, we can say that $P(m) = (m-1)Q(m) - 2010$ for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$, $(m-1)Q(m) = 2010$. Thus, if $P(m) = 0$, then $m-1 | 2010$ . Now, we need to show that for all $m-1 | 2010$, $m^{x_{0}}=\\sum_{k = 1}^{2011}m^{x_{k}}.$. We try with the first few $m$ that satisfy this. For $m = 2$, we see we can satisfy this if $x_0 = 2010$, $x_1 = 2009$, $x_2 = 2008$, $\\cdots$ , $x_{2008} = 2$, $x_{2009} = 1$, $x_{2010} = 0$, $x_{2011} = 0$, because $2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^1 = \\cdots$ (based on the idea $2^n +", "Now, it suffices to prove $9z^4+2z+4 \\ge 0$. For $z>0$ this is clearly true. Assume $z<0$. Then we want to prove that $9x^4+4 \\ge 2x$ holds for all $x>0$. But by the AM-GM inequality you have $9x^4+3=4 \\cdot \\frac{9x^4+1+1+1}{4} \\ge 4 \\cdot \\sqrt[4]{9x^4 \\cdot 1 \\cdot 1 \\cdot 1}=4x \\cdot \\sqrt{3}>4x>2x$. So $9x^4+4 >2x+1>2x$ holds for all positive $x$. Hence $p(z)$ has no real roots i.e. there must be one root in each the second and the third quadrant. • It is true that $p$ has no real root, but these lines do not prove anything about the location (i.e. the quadrants in which they lie) of the roots of $p$ inside/outside the unit disk. – Jack D'Aurizio Aug 30 '15 at 17:40 • @JackD'Aurizio: Sure. This is why I started my answer with \"Concerning the first part\" – Tintarn Aug 30 '15 at 17:51 $p$ is trivially increasing on $\\mathbb{R}^+$, hence its real zeros are negative. However, $x\\leq -1$ implies $p(x)\\geq 11$, hence the real roots lie in $(-1,0)$. Since: $$p(z)=(z^6+9z^4+1)+(z+1)(z^2-z+3)$$ there are no real roots at all. Since $1+1+2+4<9$, by Rouché's theorem there are exactly four roots of $p(z)$ inside the unit disk, that obviously come in conjugated couples. The number of roots in each set $\\text{Re}(z)>0,\\text{Im}(z)>0,\\text{Re}(z)<0,\\text{Im}(z)<0$ can be computed by applying a Cayley transform, then", "w^2$$ corresponding to $$d = w \\sqrt{2}$$, as expected. The discriminant of $$(\\star)$$ is \\begin{align}\\Delta = 4096&\\cdot(w^2 - x^2)^2 (w^2 - y^2)^2 (w^2-z^2)^2 (x^2 - y^2)^2 (x^2-z^2)^2 (y^2 - z^2)^2 \\\\ &\\cdot (w^2 - x^2 + y^2 - z^2 - w y - z x )^2 (w^2 - x^2 + y^2 - z^2 + w y + z x )^2 \\\\ &\\cdot (w^2 + x^2 - y^2 - z^2 - w x - y z )^2 (w^2 + x^2 - y^2 - z^2 + w x + y z )^2 \\\\ &\\cdot (w^2 - x^2 - y^2 + z^2 - w z - x y )^2 (w^2 - x^2 - y^2 + z^2 + w z + x y )^2 \\end{align} If $$\\Delta = 0$$ (that is, when at least two edge-lengths match, or at least one of those other factors vanishes), then the polynomial has a multiple root. Otherwise, the polynomial has either four real, or else four non-real, roots; Descartes' Rule of Signs, and/or or the techniques described here could potentially help refine our understanding of the nature of the roots, but the coefficients are sufficiently complicated that a general sign analysis is difficult. In the all-real-roots case, due to symbolic symmetry, the roots must correspond to the (squares of the) three \"body diagonals\" of the", "\\cdots}}}},$$ so that if $$y$$ and $$z$$ exist, they satisfy the system $$y = \\sqrt{x - z}, \\\\ z = \\sqrt{x + y},$$ or $$y^2 = x - z, \\\\ z^2 = x + y.$$ Consequently $$0 = z^2 - y^2 - y - z = (z-y-1)(y+z).$$ It follows that either $$z = -y$$ or $$z = 1 + y$$. The first case is impossible for $$x \\in \\mathbb R$$ since by convention we take the positive square root, so both $$y, z > 0$$. In the second case, we can substitute back into the first equation to obtain $$y^2 = x - (1+y)$$, hence $$y = \\frac{-1 + \\sqrt{4x-3}}{2},$$ where again, we discard the negative root. So far, what we have shown is that if such a nested radical for $$y$$ converges, it must converge to this value. It is not at all obvious from the above whether a given choice of $$x$$ results in a real-valued $$y$$, for any meaningful definition of $$y$$ must be as the limit of the sequence $$y = \\lim_{n \\to \\infty} y_n, \\\\ y_n = \\underbrace{\\sqrt{x - \\sqrt{x + \\sqrt{x - \\cdots \\pm \\sqrt{x}}}}}_{n \\text{ radicals}},$$ and although the choice $$x = 1$$ appears at first glance permissible, we quickly run into problems; $$y_3 = \\sqrt{1 - \\sqrt{1 + \\sqrt{1}}} \\ne", "roots of $x^3-4x^2+5x-p=0$ are positive and form the side lengths of a right triangle. Find the value of $p$. This problem is an interesting application of Vieta’s formulas. Anh Danh Tran Phan writes Let $x_1, x_2, x_3$ be the three roots of the cubic equation. By Vieta’s theorem for cubic polynomials, we have $\\begin{cases} x_1 + x_2 + x_3 = 4 \\\\ x_1x_2 + x_2x_3 + x_3x_1 = 5. \\end{cases}$ Because the three roots form the side lengths of a right triangle, without loss of generality we have $x_1^2 + x_2^2 = x_3^2 \\iff 2x_3^2 = x_1^2 + x_2^2 + x_3^2.$ By manipulating with some algebra, we have $2x_3^2 = (x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 + 2x_2x_3 + 2x_3x_1) - 10 = (x_1 + x_2 + x_3)^2 - 10 = 4^2 - 10 = 6.$ Therefore, $x_3 = \\sqrt{3}$ since every root of the equation is positive. Because $x_3 = \\sqrt{3}$ is a root of the cubic equation $x^3-4x^2+5x-p=0$, we have $p = x_3^3 - 4x_3^2 + 5x_3 = (\\sqrt{3})^3 - 4(\\sqrt{3})^2 + 5\\cdot \\sqrt{3} = \\boxed{8\\sqrt{3}-12}.$ The needed value for $p$ is $p = 8\\sqrt{3} - 12$. Many submissions used an alternative finish. After deriving that $x_3 = \\sqrt{3}$, they used Vieta again to find $p = x_1x_2x_3 = \\sqrt{3}\\cdot x_1x_2.$ On the other hand,", "# 1983 AIME Problems/Problem 3 ## Problem What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \\sqrt{x^2 + 18x + 45}$? ## Solution If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful. Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\\sqrt{y+15}$. Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution is extraneous since $2\\sqrt{y+15}$ is positive. So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$, $x^2+18x+30=10 \\Longrightarrow x^2+18x+20=0.$ By Vieta's formulas, the product of our roots is therefore $\\boxed{020}$. ## Solution 2 There is an error with solution one since the second solution for $y$ does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have: $x^4+36x^3+380x^2+1080x+900 = 0$ By Vieta's formula we get that the product is $\\boxed{900}$.", "Find all positive integers $n$ such that $n+2008$ divides $n^2 + 2008$ and $n+2009$ divides $n^2 + 2009$ I wrote \\begin{align} n^2 + 2008 &= (n+2008)^2 - 2 \\cdot 2008n - 2008^2 + 2008 \\\\ &= (n+2008)^2 - 2 \\cdot 2008(n+2008) + 2008^2 + 2008 \\\\ &= (n+2008)^2 - 2 \\cdot 2008(n+2008) + 2008 \\cdot 2009 \\end{align} to deduce that $n+2008$ divides $n^2 + 2008$ if and only if $n+2008$ divides $2008 \\cdot 2009$. Similarly, I found that $n+2009$ divides $n^2 + 2009$ if and only if $n+2009$ divides $2009 \\cdot 2010$. I can't seem to get anywhere from here. I know that $n=1$ is an obvious solution, and I think it's the only one, although I can't prove/disprove this. I know that any two consecutive integers are coprime but I haven't been able to use this fact towards the solution. Could anyone please give me any hints? For the record, this is a question from Round 1 of the British Mathematical Olympiad. • $$n+2008$$ needs to be factor of $$2008\\cdot2009$$ – lab bhattacharjee Jan 10 '16 at 17:13 • Yes, I know that, and $n+2009$ needs to be a factor of $2009.2010$ but I'm not sure what more I could deduce from this. – Ecasx Jan 10 '16 at 17:15 • Note that $(2008\\cdot2009)/(1+2008)=2008$ and $(2009\\cdot2010)/(1+2009)=2009$, and", "If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$. We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$, giving us $(y-1)(y+1)=2005$. From here we proceed as in Solution 1 to get $\\boxed{045}$. Solution 4 Realizing that if we add 1 to both sides we get $x^4-4x^3+6x^2-4x+1=2006$ which can be factored as $(x-1)^4=2006$. Then we can substitute $(x-1)$ with $y$ which leaves us with $y^4=2006$. Now subtracting 2006 from both sides we get some difference of squares $y^4-2006=0 \\rightarrow (y-\\sqrt[4]{2006})(y+\\sqrt[4]{2006})(y^2+\\sqrt{2006})=0$. The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve $y^2+\\sqrt{2006}=0$, we can substitute $(x-1)$ for $y$ giving us $(x-1)^2+\\sqrt{2006}=0$, expanding this we get $x^2-2x+1+\\sqrt{2006}=0$. We know that the product of a quadratics roots is $\\frac{c}{a}$ which leaves us with $\\frac{1+\\sqrt{2006}}{1}=1+\\sqrt{2006}\\approx\\boxed{045}$. Solution 5 As in solution 1, we find that $(x-1)^4 = 2006$. Now $x-1=\\pm \\sqrt[4]{2006}$ so $x_1 = 1+\\sqrt[4]{2006}$ and $x_2 = 1-\\sqrt[4]{2006}$ are the real roots of the equation. Multiplying, we get $x_1 x_2 = 1 - \\sqrt{2006}$. Now transforming the original function and using Vieta's formula, $x^4-4x^3+6x^2-4x-2005=0$ so $x_1 x_2 x_3 x_4 = \\frac{-2005}{1}", "i! \\cdot (2i-1)!!$. Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$. It immediately follows that $p\\left( {2i\\choose i} \\right) = p((2i)!) - 2p(i!) = i - p(i!)$, and hence $p\\left( {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i} \\right) = 2\\cdot 2009 - i - p(i!)$. Obviously, for $i\\in\\{1,2,\\dots,2009\\}$ the function $f(i)=2\\cdot 2009 - i - p(i!)$ is is a strictly decreasing function. Therefore $p(c) = p\\left( {2\\cdot 2009\\choose 2009} \\right) = 2009 - p(2009!)$. We can now compute $p(2009!) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\dfrac{2009}{2^k} \\right\\rfloor = 1004 + 502 + \\cdots + 3 + 1 = 2001$. Hence $p(c)=2009-2001=8$. And thus we have $a=2\\cdot 2009 - p(c) = 4010$, and the answer is $\\dfrac{ab}{10} = \\dfrac{4010\\cdot 1}{10} = \\boxed{401}$. Additionally, once you count the number of factors of $2$ in the summation, one can consider the fact that, since $b$ must be odd, it has to take on a value of $1,3,5,7,$ or $9$ (Because the number of $2$s in the summation is clearly greater than $1000$, dividing by $10$ will yield a number greater than $100$, and multiplying this number by any odd number greater than $9$ will yield an answer $>999$, which cannot happen on the AIME.) Once you calculate the value of $4010$, and divide by $10$, $b$ must be equal" ]

[ "So proved that if $x$ is a rational number, $f(x)=x$ (c) Suppose $x>0$. Now $\\sqrt{x}$ is the unique positive square root of $x$, such that $x = (\\sqrt{x})^2$. Hence $f(x) = f((\\sqrt{x})\\cdot (\\sqrt{x})) =\\left[ f(\\sqrt{x})\\right]^2$. But $f(\\sqrt{x}) \\in \\mathbb{R}$, so $\\left[ f(\\sqrt{x})\\right]^2 \\geqslant 0$. This means that $f(x)\\geqslant 0$. Now we will prove that if $f(x) = 0$, we reach a contradiction. Suppose $f(x)=0$. Since $x>0$,we have $\\frac{1}{x} >0$. Now $1=f(1) = f(x\\cdot \\frac{1}{x}) = f(x)\\cdot f(\\frac{1}{x})$. Now since $f(x)=0$, this leads to $1=0$, which is a contradiction. So $f(x)\\ne 0$ and since $f(x)\\geqslant 0$, we have $f(x) > 0$ if $x>0$. (d) Suppose $x>y$. Then $(x-y) > 0$. Using the previous result, we have $f(x-y) > 0$. Now $f(x-y) = f(x) + f(-y)$. Now $0=f(0) = f(y + (-y)) = f(y) + f(-y)$. Hence $f(-y) = -f(y)$ and $f(x-y) = f(x)-f(y)$. As $f(x-y) > 0$, this leads us to $f(x) > f(y)$. (e) Let $x$ be arbitrary. Let $\\varepsilon >0$ be arbitrary. There exists a rational number $q$ in $(x, x+\\varepsilon)$. Now $x < q < x + \\varepsilon$. Using the result in (d) and (b), we have $f(x) < q < f(x) + f(\\varepsilon)$. Hence we have $f(x) - x < q-x$. But $q < x + \\varepsilon$, this means that $q-x < \\varepsilon$. So we", "be positive since $f_n(x) \\geq \\sqrt x$ and $f_n$ is differentiable at 0. Sep 8, 2022 at 18:51 • @GiorgioMetafune : I forgot about this condition. Anyhow, to take this condition into account, the answer only needed a slight modification, which is now done. Thank you for your comment. Sep 8, 2022 at 19:38 This question would be possibly at a better place on MathStack Exchange. Yet, once the statement of the question is corrected (the functions $$y_n$$ need to be defined on $$\\mathbb{R_+}$$ and not only on $$[0,1]$$, and they should converge uniformly to $$\\sqrt{\\cdot}$$ on compact sets), the answer will be positive. For strictly positive functions $$y'=\\sqrt{y}$$ is equivalent to $$y'/\\sqrt{y}=1$$, namely $$2\\sqrt{y}-x$$ constant (by integration). Hence the only positive solution of the ODE $$y'=\\sqrt{y}$$ on $$[0,1]$$ which vanishes only at $$0$$ is $$y_0 : x \\mapsto (x/2)^2$$. In a same way, the only positive solution of the ODE $$y'=1+\\sqrt{y}$$ on $$\\mathbb{R}_+$$ which vanishes at $$0$$ is $$G^{-1}$$, where $$G$$ is the strictly (increasing) function defined by $$G(y) := \\int_0^y \\frac{\\mathrm{d}z}{1+\\sqrt{z}}.$$ Since the $$(f_n)_{n \\ge 1}$$ are $$\\mathcal{C}^1$$ and bounded below by $$\\sqrt{}$$, they are strictly positive everywhere (including at $$0$$), so the solutions $$y_n$$ of the Cauchy problem $$y'=f_n(y)$$, $$y(0)=0$$ are (strictly) increasing. Let $$A:=G^{-1}(1)$$. We assume that $$(f_n)_{n \\ge 1}$$ converges uniformly to $$\\sqrt{}$$", "this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ## Solution 3, x first Set it up as a quadratic in terms of y: $$y^2-2y+x^{2020}=0$$ Then the discriminant is $$\\Delta = 4-4x^{2020}$$ This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: $$y^2-2y+1=(y-1)^2=0$$ This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$: $$y^2-2y=0$$ Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all $x$, then $y(2-y) \\geq 0 \\Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$, $(0,0)$, $(1,1)$, $(-1,1)$. There are", "a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all $x$, then $y(2-y) \\geq 0 \\Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$$(0,0)$$(1,1)$$(-1,1)$. There are $4$ solutions, so the answer is $\\boxed{\\textbf{(D) } 4}$ 10. ## Solution Notice that when the cone is created, the radius of the circle will become the slant height of the cone and", "+0 # If x, y, and z are positive 0 73 5 If x, y, and z are positive with xy = 20, xz = 35, and yz = 14, then what is xyz? Nov 14, 2020 #1 +50 +4 So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get: $$xy \\cdot xz \\cdot yz = 20 \\cdot 35 \\cdot 14$$ Finding the product, we have: $$x^2y^2z^2 = 9800$$ This is the same as $$(xyz)^2 = 9800$$ which finding the square root gives us $$xyz=\\sqrt{9800}$$ Simplifying, we find: $$\\sqrt{9800} \\implies \\sqrt{4900 \\cdot 2} \\implies \\sqrt{4900} \\cdot \\sqrt2 \\implies \\boxed{70\\sqrt2}$$ Nov 14, 2020 #1 +50 +4 So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get: $$xy \\cdot xz \\cdot yz = 20 \\cdot 35 \\cdot 14$$ Finding the product, we have: $$x^2y^2z^2 = 9800$$ This is the same as $$(xyz)^2 = 9800$$ which finding the square root gives us $$xyz=\\sqrt{9800}$$ Simplifying, we find: $$\\sqrt{9800} \\implies \\sqrt{4900 \\cdot 2} \\implies \\sqrt{4900} \\cdot \\sqrt2 \\implies \\boxed{70\\sqrt2}$$ mobro Nov 14, 2020 #2 +114221 0 Very nice, mobro !!!! Nov 14, 2020 #3 +846 +1 Good job, mobro! How can we find the value of x, y, and z? Nov 14, 2020 #4 +50", "\\dfrac{\\cancel{-24}}{\\cancel{2}\\sqrt{3}}$ $\\implies$ $x \\,=\\, \\dfrac{2}{\\sqrt{3}}$ and $x \\,=\\, \\dfrac{-12}{\\sqrt{3}}$ $\\implies$ $x \\,=\\, \\dfrac{2}{\\sqrt{3}}$ and $x \\,=\\, \\dfrac{-4 \\times 3}{\\sqrt{3}}$ $\\implies$ $x \\,=\\, \\dfrac{2}{\\sqrt{3}}$ and $x \\,=\\, \\dfrac{-4 \\times {(\\sqrt{3})}^2}{\\sqrt{3}}$ $\\implies$ $x \\,=\\, \\dfrac{2}{\\sqrt{3}}$ and $\\require{cancel} x \\,=\\, \\dfrac{-4 \\times \\cancel{{(\\sqrt{3})}^2}}{\\cancel{\\sqrt{3}}}$ $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\,$ $x \\,=\\, \\dfrac{2}{\\sqrt{3}}$ and $x \\,=\\, -4\\sqrt{3}$ The roots of quadratic equation $\\sqrt{3}x^2+10x-8\\sqrt{3} = 0$ are $\\dfrac{2}{\\sqrt{3}}$ and $-4\\sqrt{3}$. They are real numbers because the discriminant is positive and the value of square of discriminant is a real number. Therefore, it has proved that the roots of a quadratic equation are real if the value of discriminant is positive real number.", "# Real analysis - proving the number of solutions • Aug 4th 2008, 06:52 AM kevin28 Real analysis - proving the number of solutions Hey everyone, Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it. I'm stuck on the last part of the question with f(x) = 1-x. PLEASE if anyone can explain to me how you go about doing it...let me know. thank you • Aug 4th 2008, 09:08 AM CaptainBlack Quote: Originally Posted by kevin28 Hey everyone, Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it. I'm stuck on the last part of the question with f(x) = 1-x. PLEASE if anyone can explain to me how you go about doing it...let me know. thank you For the last part you are looking for the solutions to: $x^3-6x^2+12x-8=1-x$ or: $x^3-6x^2+13x-9=0$ Decartes rule of signs tells us this has $1$ or $3$ positive real roots. As the function changes sign between $x=2$ and $x=3$ there is at least one root in this interval. Now: $\\frac{d}{dx}x^3-6x^2+13x-9=3x^2-12x+13$ Examining the discriminant of the quadratic we find this has no real zeros, and as this is positive for large $x$, the cubic is everywhere an", "# Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\\in\\Bbb Z$ Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers. I tried to do it by comparing the two expressions but I can’t proceed. #### Solutions Collecting From Web of \"Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\\in\\Bbb Z$\" We have that $(x+a)(x+1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$. So if such factorisation exists we must have that the discriminant of the quadratic equation is a square. Now: $$D = (1991 + a)^2 – 4(1991a + 1) = 1991^2 + 2 \\cdot 1991a + a^2 – 4 \\cdot 1991a – 4 = (1991 – a)^2 – 4$$ But then $\\sqrt{D}, 2, (1991 – a)$ make a Pythagorean triplet, but we know that the only squares whose difference is $4$ are $4$ and $0$, therefore we must have $1991-a = \\pm 2 \\implies a_1=1989, a_2 = 1993$. Let $f(x)=(x+b)(x+c)$ with $b,c\\in\\Bbb Z$. Wlog $b\\le c$. For $-c\\le x\\le b$, we have $f(x)\\le 0$, hence $f(x)-1<0$. If $x\\ge -b+2$, we have $f(x)-1\\ge 2\\cdot 2-1>0$. By symmetry, $f(x)-1>0$ also for $x\\le -c-2$. Hence the only possibly integer roots of $f(x)-1$ are $-b+1$ and $-c-1$. Moreover, $f(-b+1)-1=f(-c-1)=c-b$, hence $f(x)-1$ has integer roots iff", "be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$must be nonnegative. Therefore,$$4(1-x^{2020}) \\geq 0 \\implies x^{2020} \\leq 1.$$Here, we see that we must split the inequality into a compound, resulting in $-1 \\leq x \\leq 1$. The only integers that satisfy this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ## Solution 3, x first Set it up as a quadratic in terms of y:$$y^2-2y+x^{2020}=0$$Then the discriminant is$$\\Delta = 4-4x^{2020}$$This will clearly only yield real solutions when $x^{2020} \\leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power:$$y^2-2y+1=(y-1)^2=0$$This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$:$$y^2-2y=0$$Which has 2 solutions, so $(0,2)$ and $(0,0)$ These are the only 4 solutions, so $\\boxed{\\textbf{(D) } 4}$ ## Solution 4, y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$.", "\\cdots}}}},$$ so that if $$y$$ and $$z$$ exist, they satisfy the system $$y = \\sqrt{x - z}, \\\\ z = \\sqrt{x + y},$$ or $$y^2 = x - z, \\\\ z^2 = x + y.$$ Consequently $$0 = z^2 - y^2 - y - z = (z-y-1)(y+z).$$ It follows that either $$z = -y$$ or $$z = 1 + y$$. The first case is impossible for $$x \\in \\mathbb R$$ since by convention we take the positive square root, so both $$y, z > 0$$. In the second case, we can substitute back into the first equation to obtain $$y^2 = x - (1+y)$$, hence $$y = \\frac{-1 + \\sqrt{4x-3}}{2},$$ where again, we discard the negative root. So far, what we have shown is that if such a nested radical for $$y$$ converges, it must converge to this value. It is not at all obvious from the above whether a given choice of $$x$$ results in a real-valued $$y$$, for any meaningful definition of $$y$$ must be as the limit of the sequence $$y = \\lim_{n \\to \\infty} y_n, \\\\ y_n = \\underbrace{\\sqrt{x - \\sqrt{x + \\sqrt{x - \\cdots \\pm \\sqrt{x}}}}}_{n \\text{ radicals}},$$ and although the choice $$x = 1$$ appears at first glance permissible, we quickly run into problems; $$y_3 = \\sqrt{1 - \\sqrt{1 + \\sqrt{1}}} \\ne", "f^c(D)\\neq \\emptyset (so the NEGATION of what we are trying to prove) then \\exists_{x \\in X} x \\in f^c(C)\\cap f^c(D) \\exists_{x \\in X} x \\in f^c(C) \\wedge x \\in f^c(D) \\exists_{f(x) \\... 4$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1=x(x-1)(x^2+x+1)(x^8+1)+1$$We can see that this is positive for x\\ge 1 or x\\le 0. For 0\\lt x\\lt 1,$$x^{12}-x^9+x^4-x+1=x^{12}+(1-x)+x^4(1-x^5)$$is positive since 1-x^5\\gt 0. Therefore, x^{12}-x^9+x^4-x+1 is positive. 4 This function does not have an inverse as it is not injective: check that f(x-2) = f(-x-2) whatever x is - for example, f(1) = f(-5), etc. We want to solve y=2x^2+8x+13 for x. Applying the quadratic formula for 2x^2+8x+(13-y)=0 yields$$x = \\frac{-8\\pm \\sqrt{64-4\\cdot 2 \\cdot (13-y)}}{4} = -2\\pm \\frac{\\sqrt{16-26+2y}}{2} = -2\\pm\\frac{\\sqrt{-10+... 4 This one is pretty straightforward. Intuitively, all we have to do is pick a function that maps set $A$ to the image of your initial fuction, and a function that \"extends\" the image of $f$ to set $B$, namely the identity function. Here, the functions are named \"$map$\" and \"$ext$\", respectively and are defined as follows: $map:A\\rightarrow Imf\\subseteq B, ... 4 should be $$\\left( \\log _{ 3 }{ x } -1 \\right) \\left( \\log _{ 3 }{ x } -2 \\right) =0$$ $$\\log _{ 3 }{ x } =1\\Rightarrow \\quad x=3\\\\ \\log _{ 3 }{ x } =2\\Rightarrow x=9$$ 4$\\sin x=0$is rejected because when$\\sin x=0$, the value$\\csc", "# Difference between revisions of \"2008 iTest Problems/Problem 84\" ## Problem Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$. ## Solutions ### Solution 1 (credit to official solution) Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$. We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \\begin{align*} \\frac{rs}{r+s} &= -2008 \\\\ rs &= -2008r - 2008s \\\\ rs + 2008r + 2008s &= 0 \\\\ (r+2008)(s+2008) &= 2008^2. \\end{align*} WLOG, let $|a| \\le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \\tfrac{2008^2}{a}$. Thus, $$-r-s = b = -a - \\tfrac{2008^2}{a} + 4016.$$ Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \\cdot 251^2$, so there are $\\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \\cdot 22 =", "{{x^2} - 1} \\right|.$$The total numbe... IIT-JEE 2012 Paper 1 Offline Let$$f$$be a real-valued differentiable function on$$R$$(the set of all real numbers) such that$$f(1)=1$$. If the ... IIT-JEE 2010 Paper 1 Offline Let$$f$$be a function defined on$$R$$(the set of all real numbers) such that$$f'\\left( x \\right) = 2010\\left( {x -... IIT-JEE 2010 Paper 2 Offline The maximum value of the function $$f\\left( x \\right) = 2{x^3} - 15{x^2} + 36x - 48$$ on the set $${\\rm A} = \\left\\{ {x... IIT-JEE 2009 The maximum value of the function$$f\\left( x \\right) = 2{x^3} - 15{x^2} + 36x - 48$$on the set$${\\rm A} = \\left\\{ {x... IIT-JEE 2009 Let $$p(x)$$ be a polynomial of degree $$4$$ having extremum at $$x = 1,2$$ and $$\\mathop {\\lim }\\limits_{x \\to 0} \\lef... IIT-JEE 2009 ## Subjective More For a twice differentiable function$$f(x),g(x)$$is defined as$$4\\sqrt {65} g\\left( x \\right) = \\left( {f'{{\\left( x \\... IIT-JEE 2006 If $$p(x)$$ be a polynomial of degree $$3$$ satisfying $$p(-1)=10, p(1)=-6$$ and $$p(x)$$ has maxima at $$x=-1$$ and $$p... IIT-JEE 2005 If$$\\left| {f\\left( {{x_1}} \\right) - f\\left( {{x_2}} \\right)} \\right| &lt; {\\left( {{x_1} - {x_2}} \\right)^2},$$for ... IIT-JEE 2005 Prove that for$$x \\in \\left[ {0,{\\pi \\over 2}} \\right],\\sin x + 2x \\ge {{3x\\left( {x + 1} \\right)} \\over \\pi }$$.... IIT-JEE 2004 Using Rolle's theorem, prove that there", "\\leq 1.$$ Here, we see that we must split the inequality into a compound, resulting in $-1 \\leq x \\leq 1$. The only integers that satisfy this are $x \\in \\{-1,0,1\\}$. Plugging these values back into the quadratic equation, we see that $x = \\{-1,1\\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \\{0\\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$. Thus, the answer is $2 \\cdot 1 + 1 \\cdot 2 = \\boxed{\\textbf{(D) }4}$. ~Tiblis ### Solution 3: Solve for x first Set it up as a quadratic in terms of y: $$y^2-2y+x^{2020}=0$$ Then the discriminant is $$\\Delta = 4-4x^{2020}$$ This will clearly only yield real solutions when $|x^{2020}| \\leq 1$, because the discriminant must be positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: $$y^2-2y+1=(y-1)^2=0$$ This has only has solutions $1$, so $(\\pm 1,1)$ are solutions. Next, if $x=0$: $$y^2-2y=0 \\Rightarrow y(y-2)=0$$ Which has 2 solutions, so $(0,2)$ and $(0,0)$. These are the only 4 solutions, so our answer is $\\boxed{\\textbf{(D) } 4}$. ~edits by BakedPotato66 ### Solution 4: Solve for y first Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \\geq 0$ for all", "by the equation: f(x)=−x2−10x−3 Find the two values of x that make f(x)=0. Round your answers to two decimal places. Write the values as a list, separated by commas: x= 33. UsukiDoll Ok so... we're dealing with the same equation, except we have the find the equilibrium that makes the function 0 so we have to find all x values that makes f(x) = 0 so we set the equation to 0 0 = -x^2-10x-3 and practically do the same thing as I've done above because it's not factorable and the discriminant formula will produce 88 which is not a perfect square. you will have the same answer as factorization.. only it's going to be the roots that we need after using the quadratic formula. $\\large x = -5-\\sqrt{22}$ $\\large x = -5+\\sqrt{22}$ so these are your x values that will make your function 0. The process of factorization and finding all x values that makes f(x) = 0 is the same only for factorization we take it a step further and write $\\large (x-5+\\sqrt{22})(x-5-\\sqrt{22})$ Suppose we want to check and see if these roots do make the function go to 0 Then we plug in one of the values and simplify to see if we have 0=0 $\\large 0 = -x^2-10x-3$ For $\\large x = -5-\\sqrt{22}$ $\\large", "# 1983 AIME Problems/Problem 3 ## Problem What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \\sqrt{x^2 + 18x + 45}$? ## Solution If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful. Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\\sqrt{y+15}$. Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution is extraneous since $2\\sqrt{y+15}$ is positive. So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$, $x^2+18x+30=10 \\Longrightarrow x^2+18x+20=0.$ By Vieta's formulas, the product of our roots is therefore $\\boxed{020}$. ## Solution 2 There is an error with solution one since the second solution for $y$ does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have: $x^4+36x^3+380x^2+1080x+900 = 0$ By Vieta's formula we get that the product is $\\boxed{900}$.", "$a$. Setting $x = e^y$, we have: $$g(x) = g(e^y) = P(y) = ay = a \\log x$$ Now plugging back in, our second equation becomes: $$a \\log f(x) = 2011 x$$ $$f(x) = e^{2011x / a}$$ Testing for compatibility with first equation: $$f(g(x)) = e^{2011 a \\log x / a} = e^{2011 \\log x} = x^{2011}$$ ## Tuesday, April 5, 2011 ### Symmetric Function in 3 Variables A function $f$ is defined from a triplet of positive reals to a positive real number $f:R_+^3 \\to R_+$ and satisfies the following: 1. $f$ is symmetric in all 3 variables. That is $f(a,b,c) = f(b,a,c) = f(a,c,b) = ...$ 2. For any positive real $t$, $f(ta,tb,tc) = tf(a,b,c)$ 3. $f(1/a, 1/b, 1/c) = 2011^2/f(a,b,c)$ 4. $f(a,b,c) = f(\\sqrt{ab}, \\sqrt{ab}, c)$ Determine how many triplet of positive integers $(x,y,z)$ are there such that $f(1/x,1/y,1/z) = 1$ Solution From rule 3, we have $f(1,1,1) = 2011^2/f(1,1,1)$, so $f(1,1,1) = 2011$. Now, for any $u,v > 0$, $f(u,v,1/(uv)) = f(\\sqrt{uv}, \\sqrt{uv}, 1/(uv))$ using rule 4 and $f(uv,1,1/(uv)) = f(\\sqrt{uv}, \\sqrt{uv}, 1/(uv))$ using rule 4 So $f(u,v,1/(uv)) = f(uv,1,1/(uv)) = f(uv, 1/(uv), 1) = f(1,1,1) = 2011$ Now, for any $t,u,v$, let $k = \\sqrt[3]{tuv}$, then $t = k^3/uv$ $f(t,u,v) = f(k^3/uv, u, v) = kf(k^2/uv, u/k, v/k) = kf(u/k, v/k, 1/((u/k)(v/k))) =", "Because otherwise it is not going to be one-to-one. Simply put, with more than one real root you have to have a turning point somewhere between those roots which makes your function both increasing and decreasing between the roots. $$f(x)=4x^3+4x-6$$ $$f(0)=-6$$ and $$f(-1)=-14$$ By IVT, there exists at least one real root, $$x\\in(-1,0)$$ such that $$f(x)=0$$ Now try to prove by using contradiction. If not, there exists al least $$2$$ real roots $$x_1,x_2$$, such that $$f(x_1)=f(x_2)=0$$ Since $$f(x)$$ is differentiable, by using Rolle's Theorem, there exists a number $$k\\in(x_1,x_2)$$ such that $$f^{\\prime}(k)=0$$. But $$f^{\\prime}(x)=12x^2+4>0\\ \\forall x\\ne0$$ • KIndly edit your answer: You have calculated and shown $f(x)$ takes negative values at both $x=0$ and $x=-1$. In this case IVT cannot allow us to conclude there is a root in between $-1$ and $0$. Possibly you mean $x=1$ where the function value is positive. – P Vanchinathan Oct 22 '18 at 4:00 Pedestrian: $$f'(x)>0$$ implies $$f$$ is strictly increasing, i.e.for $$x_1 < x_2$$ we have $$f(x_1) < f(x_2)$$. Assume a strictly increasing function has more than one zero. Let $$x_1 < x_2$$ be $$2$$ zeroes of the function: $$f(x_1)=f(x_2)= 0$$. P.S. Can there be a double zero if $$f'(x) >0$$?", "To find the roots of $f$: suppose $f(y) = 0$. Then $f(x) = y+f(x)+y f(x)$ and so $(1+f(x))y = 0$; therefore $f(x) = -1$ for all $x$, or $y=0$. So we have two possibilities: $f(x) = -1$ for all $x \\not = -1$, or $f(0) = 0$. But if $f(x) = -1$ for all $x \\not = -1$, then letting $x=0$ we obtain $$f(-1) = -1$$ which is a contradiction to the undefinedness of $f(-1)$. (There's a slightly simpler contradiction if we simply suppose there is any $y$ with $f(y) = -1$, as egreg points out.) Therefore $f$ has exactly one root, and it is at $0$. (Strictly, we should go back through this, and verify that we never tried to give $f$ the argument $-1$. The easiest way to do this is to let $x$ be some real such that $f(x) \\not = -1$, and then just go through the same proof again.) ## Using the root Letting $x=0$, we see that $f(f(y)) = y$ for all $y$. Letting $y=f(x)$, we obtain $$f(2x+x^2) = 2f(x)+f(x)^2$$ so, if $x=-2$, we get $f(0) = 2 f(-2) + f(-2)^2$; that is, $f(-2) = 0$ or $f(-2) = -2$. We already know $f$ has exactly one root, so $f(-2) = -2$. Letting $x = f(y)$, we obtain $$f(2f(y)+f(y)^2) = 2y+y^2$$ so", "calculus online - Free exercises and solutions to help you succeed! Multivariable Linear Approximation – An expression with a square root in 2 variables – Exercise 3400 Exercise Given the function $$f(x,y)=\\sqrt{5e^x+y^2}$$ Use linear approximation around the point $$(x,y)=(0,2)$$ In order to find an approximate value of $$\\sqrt{5e^{0.02}+{2.03}^2}$$ $$\\sqrt{5e^{0.02}+{2.03}^2}\\approx 3.037$$ Solution We will use the 2-variable linear approximation formula $$f(x,y)\\approx f(x_0,y_0)+f'_x(x_0,y_0)\\cdot(x-x_0)+f'_y(x_0,y_0)\\cdot(y-y_0)$$ Therefore, we need to define the following $$x, y, x_0, y_0, f(x)$$ And place them in the formula. x and y will be the numbers that appear in the question, and the points $$x_0,y_0$$ will be the closest values to x and y respectively, which are easy for calculations. In our exercise, we will define $$x=0.02, y=2.03$$ because these values appear in the question. And we set $$x_0=0, y_0=2$$ because they are the closest values to x and y respectively that are easy for calculations. After we have defined x and y, it is easy to find the function. Just put x instead of the number we set to be x and put y instead of the number we set to be y. In this way, we get the function $$f(x,y)=\\sqrt{5e^x+y^2}$$ In the formula above we see the function partial derivatives. Hence, we calculate them. $$f'_x(x,y)=\\frac{1}{2\\sqrt{5e^x+y^2}}\\cdot 5e^x$$ $$f'_y(x,y)=\\frac{1}{2\\sqrt{5e^x+y^2}}\\cdot 2y$$ We put all the data in the formula" ]

[ "Expanding $$(1+0.2)^{1000}_{}$$ by the binomial theorem and doing no further manipulation gives $${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$$ $$= A_0 + A_1 + A_2 + \\cdots + A_{1000},$$ where $$A_k = {1000 \\choose k}(0.2)^k$$ for $$k = 0,1,2,\\ldots,1000$$. For which $$k_{}^{}$$ is $$A_k^{}$$ the largest? (第九届AIME1991 第3题)", "# Problem #211 211 Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives ${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \\cdots + A_{1000},$ where $A_k = {1000 \\choose k}(0.2)^k$ for $k = 0,1,2,\\ldots,1000$. For which $k_{}^{}$ is $A_k^{}$ the largest? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# 700 Followers Problem! Algebra Level 4 Expanding $$(1+0.2)^{1000}$$ by the Binomial theorem gives $${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$$ $$= A_0 + A_1 + A_2 + \\cdots + A_{1000}$$ Where $$A_k = {1000 \\choose k}(0.2)^k$$ for $$k = 0,1,2,\\ldots,1000$$. For which $$k_{}^{}$$ is $$A_k^{}$$ the largest? I Thank all my followers! ×", "which is larger $(1.1)^{10000}$ or $10000$. $(1.1)^{10000}$ $= (1+0.1)^{10000}$ $= ^{10000} C_0 (0.1)^0 + ^{10000} C_1 (0.1)^1 +^{10000} C_2 (0.1)^2 + \\ldots + ^{10000} C_{10000} (0.1)^{10000}$ $= 1 + 10000(0.1) + \\text{other positive terms}$ $= 10001 + \\text{other positive terms}$ Therefore $(1.1)^{10000}$ is greater than $10000$ . $\\\\$ Question 10: Using binomial theorem, determine which is smaller $(1.2)^{4000}$ or $800$. $(1.2)^{4000}$ $= (1+0.2)^{4000}$ $= ^{4000} C_0 (0.2)^0 + ^{4000} C_1 (0.2)^1 +^{4000} C_2 (0.2)^2 + \\ldots + ^{4000} C_{4000} (0.1)^{4000}$ $= 1 + 4000(0.2) + \\text{other positive terms}$ $= 801 + \\text{other positive terms}$ Therefore $800$ is the smaller term. $\\\\$ Question 11: Find the value of $(1.01)^{10}+(1-0.01)^{10}$ correct to $7$ decimal places. $(1.01)^{10}+(1-0.01)^{10}$ $= (1+ 0.01)^{10}+(1-0.01)^{10}$ $= 2 [ ^{10} C_0 (0.01)^0 + ^{10} C_2 (0.01)^2 +^{10} C_4 (0.01)^4 +^{10} C_6 (0.01)^6 +^{10} C_8 (0.01)^8 +^{10} C_{10} (0.01)^{10} ]$ $= 2 [ 1 + 45 ( 0.0001) + 210 ( 0.0000001) + \\ldots ]$ $= 2.0090042 + \\ldots$ Hence $(1.01)^{10}+(1-0.01)^{10}$ correct to $7$ decimal places is $2.0090042$ $\\\\$ Question 12: Show that $2^{4n+4}- 15n - 16$, where $n \\in N$ is divisible by $225$. $2^{4n+4}- 15n - 16$ $= {16}^{n+1}- 15n - 16$ $= (1+15)^{n+1}- 15n - 16$ $= [^{n+1} C_0 (15)^0 +^{n+1} C_1 (15)^1 +^{n+1} C_2 (15)^2 + \\ldots + ^{n+1} C_{n+1} (15)^{n+1} ]- 15n -", "# $\\sum_{i=0}^n (-1)^k \\cdot C(n,k)$ Use the Binomial Theorem to Show $$\\sum_{i=0}^n (-1)^k \\cdot C(n,k)$$ I'm not sure where to start here . . . I know it is missing an $a^{n-k}$ and a $b^{k}$ term that maybe I should set to be 1? • Show what? This isn't a question. – Thomas Andrews Apr 21 '13 at 0:19 HINT: You have the $b^k$ factor: it’s $(-1)^k$. The $a^{n-k}$ is invisible, because $a=\\dots$? • Thanks! i get it now. So . . the answer is just as simple as $\\sum_{i=0}^n C(n,k) \\cdot (-1)^k \\cdot 1^{n-k}$ ? – rbtLong Apr 21 '13 at 0:14 • @rbtLong: That’s right, together with the fact that that sum is the binomial expansion of $\\big(1+(-1)\\big)^n$. – Brian M. Scott Apr 21 '13 at 0:18 Yes. What is $0=(1+(-1))^n$ according to the binomial theorem?", "# $2^{k} \\mid {2k \\choose 0} \\cdot 3^{0} + {2k \\choose 2} \\cdot 3^{1} + \\cdots + {2k \\choose 2i} \\cdot 3^{i} + \\cdots + {2k \\choose 2k} \\cdot 3^{k}$ I have to prove this question: $$2^{k} \\,\\left|\\, {2k \\choose 0} \\cdot 3^{0} + {2k \\choose 2} \\cdot 3^{1} + \\cdots + {2k \\choose 2i} \\cdot 3^{i} + \\cdots + {2k \\choose 2k} \\cdot 3^{k}\\right.$$ What i did was to get a closed form for this summation, and i haven't been able to get one. Any ideas, or is there any other method for proving this. - Think about $(a+b)^m+(a-b)^m$. – Robin Chapman Oct 26 '10 at 14:10 The expression on the right just screams \"Binomial Theorem! Binomial Theorem!\". It's a quick matter to figure out what to apply the Binomial Theorem to... Added: So, how can we find out to what to apply the Binomial Theorem? Well, I can tell you what my thought process was; I'm sure there are better or more direct ones. I'm going to take $(a+b)^{2k}$; $2k$ because the binomial coefficients are $2k$-choose-something. The $a$ should be equal to $1$, because we only have the $3$'s in the expansion. Oh, wait... the power of $3$ that shows up is half the \"choose-something\", and in the binomial theorem, if you take $(1+b)^{2k}$, then the", "\\choose 2}+ \\ldots + (-1)^{k}{k \\choose k}$ times. Using the binomial theorem, this sum is $(1-1)^{k}$ which is 0 and hence, x is not counted on the right hand side. This completes the proof. QED. More later, Nalin Pithwa", "Browse Questions # Using the Binomial theorem indicate which number is larger $(1-1)^{10000}$ or 1000 Toolbox: • $(1+x)^n=nC_0+nC_1x+nC_2x^2+.....+nC_1x^r+.....+nC_xx^n$ $(1.1)^{10000}=[1+(.1)]^{10000}$ Expanding by binomial theorem $C(10000,0)(1)^{10000}+C(10000,1)(1)^{10000-1}(.1)$ $1+1000\\times .1$+other terms $1000$+other terms of the expansion. Hence $(1.1)^{10000} > 1000$", "# How do you use the binomial series to expand f(x)=1/(1+x)^3? Nov 4, 2016 the answer is $f \\left(x\\right) = 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \\ldots . .$ #### Explanation: The binomial expansion is ${\\left(a + b\\right)}^{n}$ $= {a}^{n} + n {a}^{n - 1} b + \\frac{n \\left(n - 1\\right)}{1 \\cdot 2} {a}^{n - 2} {b}^{2} + \\ldots \\ldots .$ Applying this here with $a = 1$, $b = x$ and $n = - 3$ we get $f \\left(x\\right) = {\\left(1 + x\\right)}^{-} 3 = 1 - 3 x + \\frac{\\left(- 3\\right) \\left(- 4\\right)}{1 \\cdot 2} {x}^{2} + \\frac{\\left(- 3\\right) \\left(- 4\\right) \\left(- 5\\right)}{1 \\cdot 2 \\cdot 3} {x}^{3} + \\frac{\\left(- 3\\right) \\left(- 4\\right) \\left(- 5\\right) \\left(- 6\\right)}{1 \\cdot 2 \\cdot 3 \\cdot 4} {x}^{4} + \\ldots \\ldots$ $= 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \\ldots . .$", "less than $$1024$$ so this is an underestimate, but the question is how much? The binomial theorem says that for two numbers $$x$$ and $$y$$ and some number $$n$$ we have: $(x+y)^n = \\sum_{k=0}^n {n \\choose k} x^k y^{n-k}$ Now we can take $$x = 1000$$ and $$y = 24$$ and we can express $$2^{40}$$ by the following binomial expansion: \\begin{align*} (1000+24)^4 & = { 4 \\choose 0 } \\cdot 1000^0 \\cdot 24^4 + { 4 \\choose 1 } \\cdot 1000^1 \\cdot 24^3 + { 4 \\choose 2 } \\cdot 1000^2 \\cdot 24^2 + { 4 \\choose 3 } \\cdot 1000^3 \\cdot 24^1 + { 4 \\choose 4 } \\cdot 1000^4 \\cdot 24^0 \\\\ & = 24^4 + 4 \\cdot 24^3 \\cdot 1000 + 6 \\cdot 24^2 \\cdot 1000^2 + 4 \\cdot 24 \\cdot 1000^3 + 1000^4 \\end{align*} With this expansion it is clear to see that our $$2^{40}$$ is indeed $$1000^4$$ (the final term in the above sum) plus some change. Now we want to find out how much our estimate that $$2^{40} \\approx 1000^4$$ underestimates the true value of $$2^{40}$$. Question: What is the expected number of times you must flip a fair coin before you see three heads in a row? Solution: Let $$x_n$$ denote the expected number of times you flip the coin before", "reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it). The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion$(x + y + z)^n$. How many coefficients in the expansion of$(x + y + z)^{200000}$are multiples of$10^{12}$? ## Solution Using the Multinomial Theorem The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula $(x_1+x_2+\\cdots+x_m)^n=\\sum_{{k_1+k_2+\\cdots+k_m=n}\\atop{0\\le k_i\\le n}}\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)\\prod_{1\\le t\\le m}x_t^{k_t}$ where $\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)=\\frac{n!}{k_1!k_2!\\cdots k_m!}.$ Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions$k_1+k_2+\\cdots+k_m=n$for integers$k_i$. If n=200000 abd m=3, then the terms in the expansion are given by $\\left({200000}\\atop{k_1,k_2,k_3}\\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$ where$k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by$10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved. First, we’ll create a function$p(n,d)$that outputs how many factors of$d$are included in$n!$. We have that $p(n,d)=\\left\\lfloor\\frac{n}{d}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^2}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^3}\\right\\rfloor+ \\cdots+\\left\\lfloor\\frac{n}{d^r}\\right\\rfloor,$ where$d^r$is the highest power of$d$dividing$n$. For instance, there are 199994 factors of", "15\\!\\cdot\\!16^k + 10$ And we must show that $15\\!\\cdot\\!16^k + 10$ is a multiple of 25. . . That is, that $3\\!\\cdot\\!16^k + 2$ is a multiple of 5. $3\\!\\cdot\\!16^k + 2 \\;=\\;3(4^2)^k + 2$ . . . . . . . . $=\\;3(4)^{2k} + 2$ . . . . . . . . $=\\;3(5-1)^{2k} + 2$ . . . . . . . . $=\\;3\\bigg[5^{2k} - {2k\\choose2k-1}5^{2k-1} + \\hdots + {2k\\choose2}5^2 - {2k\\choose1}5 + 1\\bigg] + 2$ . . . . . . . . $=\\;3\\bigg[5^{2k} - {2k\\choose2k-1}5^{2k-1} + \\hdots + {2k\\choose2}5^2 - {2k\\choose1}5\\bigg] + 3 + 2$ . . . . . . . . $=\\;3\\bigg[5^{2k} - {2k\\choose2k-1}5^{2k-1} + \\hdots + {2k\\choose2}5^2 - {2k\\choose1}5\\bigg] + 5$ . . . . . . . . $=\\;5\\,\\bigg[3\\bigg(5^{2k-1} - {2k\\choose2k-1}5^{2k-2} + \\hdots + {2k\\choose2}5 - {2k\\choose1}\\bigg) + 1\\bigg]$ . . $\\text{Q. E. D.}$ But there must be a better way! . 4. My teacher hinted towards the binomial theorem. Don't know if that will help. He didn't say much else. 5. Maybe try writing it as $(15+1)^m+10m-1$ and then expand using the binomial theorem. 6. If you know any modern algebra (modular arithmetic), you only really need check the cases $m=0,1,2,3,4$, since $16^5\\equiv1\\,(\\mod\\,25)$. You could generalize each of those cases to when $m=5q+r,r=0,1,2,3,4$. I wouldn't recommend this over the", "the total number of subsets of a set of size $9$. If you know that that is $2^9$, you've got it. Use the binomial theorem as: $$(1+x)^n = {n\\choose 0} + {n\\choose 1}x^1+ {n\\choose 2}x^2+\\cdots+{n\\choose n}x^n$$ Put $x=1$ to get $$2^n= {n\\choose 0} + {n\\choose 1}+ {n\\choose 2}+\\cdots+{n\\choose n}.$$", "Maximum coefficient in the expansion of $(5+3x)^{10}$ Though I know that I could simply just expand $$(5+3x)^{10}$$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $$x^6$$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible? Write $$(5+3x)^{10}=\\sum_{i=0}^{10}a_ix^i.$$ So that $$a_i=\\binom{10}{i}3^i5^{10-i}.$$ Then $$f(i):=a_i/a_{i+1}=\\frac{5}{3}\\cdot \\frac{i+1}{10-i}.$$ This is an increasing function of $$i$$ (for $$0\\leq i<10$$). The maximum of the $$a_i$$ is attained at the smallest $$i$$ for which $$f(i)>1$$. Solving $$f(i)=1$$ yields $$i=25/8$$, which is larger than $$3$$ but smaller than $$4$$. Therefore $$a_4$$ is the largest coefficient. The coefficients have the form $$\\dbinom{10}k5^k3^{10-k}$$ The ratio of two consecutive coefficients is: $$\\frac{\\dbinom{10}k5^k3^{10-k}}{\\dbinom{10}{k+1}5^{k+1}3^{9-k}}=\\frac{3(k+1)}{5(10-k)}$$ This ratio is lesser than $$1$$ when $$3k+3<50-5k$$ That is, when $$k\\le 5$$. This means that the maximum coefficient is $$\\dbinom{10}65^6\\cdot3^4$$. • According to OP question, it would be more logical to define $k$ as the \"power of $x$\": $a_k x^k$ – Damien Jan 10 at 15:14 With the result proved below, the largest coefficient is among the set $$(a^n,b^n,C_p)$$, where $$C_p$$ is the largest coefficient as $$p\\in[1,n-1]$$. In this case, $$a=5, b=3, n=10$$, $$\\displaystyle\\frac ab>0$$.", "# Finding Remainder Using Binomial Theorem Find the remainder when $$7^{98}$$ is divided by $$5$$. What I am doing here is expanding $${(5+2)}^{98}$$ using binomial theorem and writing it as $$5k + 2$$, where $$k$$ is a positive integer but the answer is $$4$$ and I'm getting $$2.$$ They are expanding $${(50 - 1)}^{49}$$ using binomial theorem and then writing $$50k - 1$$ and getting $$4$$ as a reminder. How those two methods are different? • $4$ is correct, so you made an error somewhere. If you seek help debugging your calculation then you need to show your work. The point of using BT is that it reduces it to computing $(-1)^{49},\\,$ which is simpler than computing $\\,2^{98}\\pmod 5\\ \\$ – Bill Dubuque Jan 26 at 16:35 • But BT isn't needed: $\\large\\bmod 5\\!:\\, 7^{\\large 98}\\!\\equiv (7^{\\large 2})^{\\large 49}\\!\\equiv (-1)^{\\large 49}\\!\\equiv -1\\equiv 4,\\$ [or $\\large\\, 7\\equiv 2\\,$ and $\\large\\,2^{\\large 2}\\equiv -1$], by using standard congruence arithmetic rules. – Bill Dubuque Jan 26 at 16:47 The binomial theorem tells you $$(5+2)^{98} = \\binom{98}{0}\\cdot 5^{98}+\\binom{98}{1}\\cdot 5^{91}2^1+ \\binom{98}{2}\\cdot 5^{90}2^2+\\cdots + \\binom{98}{97}\\cdot 5\\cdot 2^{97}+ \\binom{98}{98}\\cdot 2^{98} = 5k+2^{98}.$$ So you need to find the remainder when $$2^{98}$$ is divided by $$5$$, not the remainder when $$2$$ is divided by $$5$$. The method using $$(50-1)^{49}$$ is a little easier because $$(-1)^{49}=-1$$. • Also", "How do you use the binomial series to expand f(x)=1/(1+x)^3? Nov 4, 2016 the answer is $f \\left(x\\right) = 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \\ldots . .$ Explanation: The binomial expansion is ${\\left(a + b\\right)}^{n}$ $= {a}^{n} + n {a}^{n - 1} b + \\frac{n \\left(n - 1\\right)}{1 \\cdot 2} {a}^{n - 2} {b}^{2} + \\ldots \\ldots .$ Applying this here with $a = 1$, $b = x$ and $n = - 3$ we get $f \\left(x\\right) = {\\left(1 + x\\right)}^{-} 3 = 1 - 3 x + \\frac{\\left(- 3\\right) \\left(- 4\\right)}{1 \\cdot 2} {x}^{2} + \\frac{\\left(- 3\\right) \\left(- 4\\right) \\left(- 5\\right)}{1 \\cdot 2 \\cdot 3} {x}^{3} + \\frac{\\left(- 3\\right) \\left(- 4\\right) \\left(- 5\\right) \\left(- 6\\right)}{1 \\cdot 2 \\cdot 3 \\cdot 4} {x}^{4} + \\ldots \\ldots$ $= 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \\ldots . .$", "# Binomial Expansion Calculator The calculator will find the binomial expansion of the given expression, with steps shown. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. Expand $\\left(2 x + 5\\right)^{3}$. ## Solution The expansion is given by the following formula: $\\left(a + b\\right)^{n} = \\sum_{k=0}^{n} {\\binom{n}{k}} a^{n - k} b^{k}$, where ${\\binom{n}{k}} = \\frac{n!}{\\left(n - k\\right)! k!}$ and $n! = 1 \\cdot 2 \\cdot \\ldots \\cdot n$. We have that $a = 2 x$, $b = 5$, and $n = 3$. Therefore, $\\left(2 x + 5\\right)^{3} = \\sum_{k=0}^{3} {\\binom{3}{k}} \\left(2 x\\right)^{3 - k} 5^{k}$. Now, calculate the product for every value of $k$ from $0$ to 3. $k = 0$: ${\\binom{3}{0}} \\left(2 x\\right)^{3 - 0} \\cdot 5^{0} = \\frac{3!}{\\left(3 - 0\\right)! 0!} \\left(2 x\\right)^{3 - 0} \\cdot 5^{0} = 8 x^{3}$ $k = 1$: ${\\binom{3}{1}} \\left(2 x\\right)^{3 - 1} \\cdot 5^{1} = \\frac{3!}{\\left(3 - 1\\right)! 1!} \\left(2 x\\right)^{3 - 1} \\cdot 5^{1} = 60 x^{2}$ $k = 2$: ${\\binom{3}{2}} \\left(2 x\\right)^{3 - 2} \\cdot 5^{2} = \\frac{3!}{\\left(3 - 2\\right)! 2!} \\left(2 x\\right)^{3 - 2} \\cdot 5^{2} = 150 x$ $k = 3$: ${\\binom{3}{3}} \\left(2 x\\right)^{3 - 3} \\cdot 5^{3} = \\frac{3!}{\\left(3 - 3\\right)! 3!} \\left(2 x\\right)^{3 - 3} \\cdot 5^{3}", "beat me by less than a minute! :) – Deepak Feb 22 '19 at 14:21 A really quick way to \"see\" the answer is with binomial theorem. $$11^{10} = (10+1)^{10} = 10^{10} + k_1\\cdot 10^9 + k_2 \\cdot 10^8 + ... + 10\\cdot 10^1 + 1$$ where the $$k$$'s represent various combinatorial constants. The values are unimportant. What's important is that when we take the whole thing modulo $$100$$, the expression reduces to $$1$$. Subtracting one, we get the required result $$11^{10} - 1 \\equiv 0 \\pmod{100}$$.", "the $9i^x$. It a subset of size 1 has a 9, then its power sum must be $9i$, and there is only $1$ of these such subsets. There are ${8\\choose1}$ with $9\\cdot i^2$, ${8\\choose2}$ with $9\\cdot i^3$, and so forth. So $T_9 =\\sum_{k=0}^{8} 9{8\\choose{k}}i^{k+1}$. This is exactly the binomial expansion of $9i \\cdot (1+i)^8$. We can use De Moivre's Theorem to calculate the power: $(\\sqrt{2})^8\\cos{8\\cdot45} = 16$. Hence $T_9 = 16\\cdot9i = 144i$, and $S_9 = 2S_8 + 144i = 2(-176 -64i) + 144i = -352 + 16i$. Thus, $|p| + |q| = |-352| + |16| = 368$.", "# Is this a good proof of the binomial identity? Prove that the binomial identity ${n\\choose k} = {n-1\\choose k-1} + {n-1\\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem. What I have: We know from the binomial theorem that: $$(x+1)^n= {n\\choose 0} x^0 + {n\\choose 1} x^1+\\cdots+{n\\choose k} x^k+\\cdots+ {n\\choose n-1} x^{n-1}+ {n\\choose n} x^n.$$ By using the property that $(1+x)^{n} = (1+x)(1+x)^{n−1}$, we can take out a factor of $x$ to get: $$x(x+1)^{n-1}={n-1\\choose 0} x^0+\\cdots+{n-1\\choose k-1} x^{k-1}+\\cdots+{n+1\\choose n-1} x^{n-1}$$ If we then divide by a factor of $x$ we get: $$1(x+1)^{n-1}={n-1\\choose 0} x^0+\\cdots +{n-1\\choose k-1} x^k+\\cdots+{n-1\\choose n-1} x^{n-1}$$ Substituting $(x+1)^{n-1}={n-1\\choose 0} x^0 +\\cdots+{n-1\\choose k-1} x^k+\\cdots+{n-1\\choose n-1} x^{n-1}$ into the equation $x(x+1)^{n-1}= {n-1\\choose 0} x^0+\\cdots+{n-1\\choose k-1} x^{k-1}+\\cdots+ {n+1\\choose n-1}x^{n-1}$, the equation can be reduced to: $${n\\choose k}x^k = {n-1\\choose k-1} x^{k-1}x+{n-1 \\choose k}x^k 1$$ Dividing by $x$ we get: ${n\\choose k} = {n-1\\choose k-1} + {n-1\\choose k}$ Is this a good proof? What can I do to improve it? Is there a better way to solve this problem?" ]

[ "cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$. Similarly, $(k+1)!=(k+1)k!$. Canceling these terms yields, $\\frac{1}{(1000-k)}>{\\frac{1}{5}}\\cdot\\frac{1}{(k+1)}$ Cross multiplying gives: $5k+5>1000-k \\Rightarrow k>165.8$ Therefore, since this identity holds for all values of $k>165.8$, the largest possible value of $k$ is $\\boxed{166}$. ### Solution 3 We know that $A_k$ will increase as $k$ increases until certain $k=m$, where $A_0 < A_1 < A_2 < \\dots < A_{m-2} < A_{m-1} < A_m$ and $A_m > A_{m+1} > A_{m+2} > \\dots > A_{1000}.$ Next, to change $A_{k-1}$ to $A_k$, we multiply $A_{k-1}$ by $\\frac{1000-k+1}{5k}$. It follows that the numerator must be greater than the denominator or $$1000-k+1>5k$$ $$k<166.8$$. We conclude that the answer is $\\boxed{166}$. ## Solution 4 Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\\frac{1}{5} > 1$) Say we have ${1000 \\choose 5}$... this equals $\\frac{1000*999*998*997*996}{5*4*3*2*1}$, and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\\frac{995}{6}$. Notice that this fraction's numerator plus denominator is equal to $1001$. Calling the numerator x and the denominator y, we get that $$x+y = 1001$$ $$\\frac{x}{y} > 5$$ $$x>5y$$ $$6y<1001$$ $$y<166.83333$$", "Thus, if $k<\\alpha n$, $(3)$ holds. Example: $\\left.\\binom{10000}{8687}\\middle/\\frac{1000^{8687}}{2^{8687}8687!}\\right.=3.095040785$, but $\\left.\\binom{10000}{8688}\\middle/\\frac{1000^{8688}}{2^{8688}8688!}\\right.=0.8127577101$ - We have \\begin{align*} \\binom{n}{k} &> \\frac{n^k}{2^k k!} \\\\ \\frac{n!}{(n-k)!k!} &> \\frac{n^k}{2^k k!} \\\\ \\frac{n!}{(n-k)!} &> \\frac{n^k}{2^k} \\\\ \\prod_{i=n-k+1}^n i &> \\left(\\frac{n}{2}\\right)^k \\\\ (n-k+1)^k &> \\left(\\frac{n}{2}\\right)^k \\\\ n-k+1 &> \\frac{n}{2} \\\\ n-2k+2 &> 0 \\\\ n+2 &> 2k \\\\ \\end{align*} Edit: $n > 2k$ is not a necessary condition (e.g. $n \\geq 2k$ is good as well), but you need something, in general case the inequality does not hold, for example for $n \\geq 6$ we have $\\binom{n}{n} = 1 \\leq \\frac{n^n}{2^n n!}$. -", "and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\\boxed{166}$ Note: It may be more intuitive to equate $x=5y$ and then use test points to figure out which side of the inequality to lie on. ## Solution 5 Notice the relation between $A_m$ and $A_{m+1}$. We have that: $A_{m+1} = A_m \\cdot \\frac{1}{5} \\cdot \\frac{1000-m}{m+1}$. This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that $${1000 \\choose m+1} = \\frac{1000!}{(m+1)!(1000-m-1)!} = \\frac{1000 \\cdots (1001-m)(1000-m)}{(m+1)!}$$ and that $${1000 \\choose m} = \\frac{1000!}{m!(1000-m)!} = \\frac{1000 \\cdots (1001-m)}{m!}$$ $$\\Longrightarrow {1000 \\choose m} \\cdot \\frac{1000-m}{m+1}$$ So now, we must find out for which $m$ is $1000-m< 5 \\cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$) Then we find that for all $m > 165.833333$, the above inequality ($1000-m< 5 \\cdot (m+1)$) holds, so then $m=165$ so $k=m+1= \\boxed{166}$. ~qwertysri987", "we deduce that $$M = \\frac{2016}{1001} = \\frac{288}{143}.$$ The answer is $\\boxed{431}.$ ### Solution 2 Each 1000-element subset $\\left\\{ a_1, a_2,a_3,...,a_{1000}\\right\\}$ of $\\left\\{1,2,3,...,2015\\right\\}$ with $a_1 contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\\left\\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$. There are $a_1$ ways to choose a positive integer $k$ such that $k ($k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is equal to the sum. But choosing a set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is the same as choosing a 1001-element subset from $\\left\\{1,2,3,...,2016\\right\\}$! Thus, the average is $\\frac{\\binom{2016}{1001}}{\\binom{2015}{1000}}=\\frac{2016}{1001}=\\frac{288}{143}$. Our answer is $p+q=288+143=\\boxed{431}$. ### Solution 3(NOT RIGOROUS) Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$: For $p = 2$ and $q = 2$, the answer is $1$. For $p = 3$ and $q = 2$, the answer is $\\frac43$. For $p = 4$ and $q = 2$, the answer is $\\frac53$. For $p = 3$ and $q = 3$, the answer is $1$. For $p = 4$ and $q = 3$, the answer is $\\frac54$. For $p = 5$ and $q = 3$, the answer", "point in $I$) we have the improved upper bound: $$\\sum_{k=1001}^{3001}\\frac{1}{k}<\\frac{2}{3}+\\frac{1}{2}=\\frac{7}{6}.\\tag{4}$$ Bounding $f(x)$ between the envelope of three tangents (in $x=0,x=1000,x=2000$) and the envelope of two secants (from $x=0$ to $x=1000$ and from $x=1000$ to $x=2000$) leads to the following sharpened inequality: $$\\frac{37}{36}<\\sum_{k=1001}^{3001}\\frac{1}{k}<\\frac{9}{8}.\\tag{5}$$", "mark of $$200$$ can be obtained on three tests in which the maximum possible score is $$100$$ is $$\\binom{202}{2} - \\binom{3}{1}\\binom{101}{2} = 5151$$ as you found. Basically, we need to find out the number of solutions to $$x_0+x_1+x_2=200$$ with $$0\\leq x_i\\leq 100$$ for all $$i$$ and furthermore it should hold that $$x_0+x_1\\geq 100$$ Once $$x_0$$ and $$x_1$$ are assigned, $$x_2$$ is automatically determined as $$200-x_0-x_1$$ Assigning $$x_0=0$$, there is only one choice for $$x_1$$ which is $$100$$ Assigning $$x_0=1$$, there are two choices for $$x_1$$, which are $$99$$ and $$100$$ Assigning $$x_0=k$$, there are $$k+1$$ choices for $$x_1$$, from $$100-k$$ to $$100$$ So, there are a total of $$\\sum_{k=0}^{100}(k+1)=\\sum_{k=1}^{101}k=\\frac{101\\times 102}2=5151$$ choices. • +1. the answer by @N.F.Taussig is more general but you found a surprising (to me) way to exploit the fact that $100$ is exactly half of $200$. neat! – antkam Mar 1 '19 at 19:39", "$$(k+1)^2 = k^2+2k+1 = k^2(1+\\frac{2}{k}+\\frac1{k^2}) < 2^k(1+\\frac{2}{5}+\\frac1{25}) < 2^k(2) = 2^{k+1}$$ So, by this nested induction, if $k \\ge 5$, if $n \\ge k^2$ (so that, certainly, $n \\ge 2k$), $2^n > n^k$. For $k = 2, 3,$ and $4$, respective values of $n \\ge 2k$ such that $2^n > n^k$ are $5$ ($2^5 > 5^2$), $10$ ($2^{10} > 10^3$), and $17$ ($2^{17} = 131072> 17^4=83521$). These values of $n$ happen to be $k^2+1$, so the final result is: If $n$ and $k$ are integers and $k \\ge 2$ and $n \\ge k^2+1$, then $2^n > n^k$. Pick $n > 2k$. Then, $$2^n > \\binom{n}{k + 1} = \\frac{n (n - 1) \\cdot \\ldots \\cdot (n - k)}{(k+1)!} \\geq \\frac{n^{k + 1}}{2^{k+1} (k+1)!}$$ by a trivial combinatorial argument for the first inequality, and using $n - k > n/2$ in the second inequality. Therefore $2^n > n^k$ as soon as $$n > 2^{k + 1} (k + 1)!$$ • I'll accept it, even though your bound is quite large. – marty cohen Jul 8 '13 at 20:38 • It's possible to improve it significantly! I'll post a comment later. – blabler Jul 8 '13 at 21:09", "the approach is exactly similar, you should get an equivalent inequality of form: $$\\prod_{k=1}^{49} \\frac{2k (2k+2)}{(2k+1)^2} < \\frac{169}{200}$$ Similarly, each factor now is less than $1$ and convergence is rapid. So taking the first three terms, we have $\\dfrac{1}{X^2} < \\dfrac{1024}{1225}$ which is verifiably less than $\\dfrac{169}{200} = 0.845$. Hence $\\dfrac{1}{13} < X$ is also established. - When you multiply up and down in the product by $2 \\cdot 4 \\cdot 6 \\ldots 50$, the product becomes $$\\frac{1}{2^{100}} \\binom{100}{50} = \\frac{1}{2^{100}} \\frac{100!}{(50!)^2}$$ Use Stirling: $$n! \\sim \\sqrt{2 \\pi n} n^n e^{-n} \\left ( 1+ \\frac{1}{12 n}\\right )$$ for large $n$. We will see what that means: plug in this approximation into the binomial expression above for $n=50$; the result is $$\\frac{1}{2^{100}} \\binom{100}{50} \\approx \\frac{1}{\\sqrt{50 \\pi}} \\left ( 1- \\frac{1}{400}\\right )$$ Note that the product being about $\\frac{1}{\\sqrt{50 \\pi}}$ is accurate to $1$ part in $400$. Now, it is clear that $50 \\pi \\approx 157$ to within that margin of error. As $157$ falls between $12^2=144$ and $13^2=169$, the assertion is true. - \"It is clear that...\" No, it is not, unless you explain how the error depends on $n$. From what you wrote, it could well be that the two expressions are too far apart for the approximation to be useful, simply because $n=100$ is not \"large\" enough.", "If $k$ is even $f(k)=997$ if $k$ is odd $f(k)=998$. $84$ has even parity, so $f(84)=997$. The result may be rigorously shown through induction. ## Solution 3 Assume that $f(84)$ is to be performed $n+1$ times. Then we have $$f(84)=f^{n+1}(84)=f(f^n(84+5))$$ In order to find $f(84)$, we want to know the smallest value of $$f^n(84+5)\\ge1000$$ Because then $$f(84)=f(f^n(84+5))=(f^n(84+5))-3$$ From which we'll get a numerical value for $f(84)$. Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\\frac{n}{2}$ times, the result is greater than or equal to $1000$. In this case, the value of $n$ for $f(84)$ is $916$, because $$84+\\frac{916}{2}\\cdot5-\\frac{916}{2}\\cdot3=1000\\Longrightarrow f^{916}(84+5))=1000$$ Thus $$f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\\boxed{997}$$ ~ Nafer", "n \\le k$ For $k > \\frac{200}{3}$, $\\left[\\frac{100}{k}\\right] = 1$. 3 of the $k$ that should consider lands in here. For $n < \\frac{k}{2}$, $\\left[\\frac{n}{k}\\right] = 0$, then we need $\\left[\\frac{100 - n}{k}\\right] = 1$ else for $\\frac{k}{2}< n < k$, $\\left[\\frac{n}{k}\\right] = 1$, then we need $\\left[\\frac{100 - n}{k}\\right] = 0$ For $n < \\frac{k}{2}$, $\\left[\\frac{100 - n}{k}\\right] = \\left[\\frac{100}{k} - \\frac{n}{k}\\right]= 1$ So, for the condition to be true, $100 - n > \\frac{k}{2}$ . ( $k > \\frac{200}{3}$, no worry for the rounding to be $> 1$) $100 > k > \\frac{k}{2} + n$, so this is always true. For $\\frac{k}{2}< n < k$, $\\left[\\frac{100 - n}{k}\\right] = 0$, so we want $100 - n < \\frac{k}{2}$, or $100 < \\frac{k}{2} + n$ $100 <\\frac{k}{2} + n < \\frac{3k}{2}$ For k = 67, $67 > n > 100 - \\frac{67}{2} = 66.5$ For k = 69, $69 > n > 100 - \\frac{69}{2} = 67.5$ etc. We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\\frac{34}{67}$. Also, $\\frac{7}{13} > \\frac{34}{67}$ . So for AMC purpose, answer is (D). Now, let's say we are not given any answer, we need to consider $k < \\frac{200}{3}$.", "$13$. $1\\le n \\le k$ For $k > \\frac{200}{3}$, $\\left[\\frac{100}{k}\\right] = 1$. 3 of the $k$ that should consider lands in here. For $n < \\frac{k}{2}$, $\\left[\\frac{n}{k}\\right] = 0$, then we need $\\left[\\frac{100 - n}{k}\\right] = 1$ else for $\\frac{k}{2}< n < k$, $\\left[\\frac{n}{k}\\right] = 1$, then we need $\\left[\\frac{100 - n}{k}\\right] = 0$ For $n < \\frac{k}{2}$, $\\left[\\frac{100 - n}{k}\\right] = \\left[\\frac{100}{k} - \\frac{n}{k}\\right]= 1$ So, for the condition to be true, $100 - n > \\frac{k}{2}$ . ( $k > \\frac{200}{3}$, no worry for the rounding to be $> 1$) $100 > k > \\frac{k}{2} + n$, so this is always true. For $\\frac{k}{2}< n < k$, $\\left[\\frac{100 - n}{k}\\right] = 0$, so we want $100 - n < \\frac{k}{2}$, or $100 < \\frac{k}{2} + n$ $100 <\\frac{k}{2} + n < \\frac{3k}{2}$ For k = 67, $67 > n > 100 - \\frac{67}{2} = 66.5$ For k = 69, $69 > n > 100 - \\frac{69}{2} = 67.5$ etc. We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\\frac{34}{67}$. Also, $\\frac{7}{13} > \\frac{34}{67}$ . So for AMC purpose, answer is (D). Now, let's say we are not given any answer, we need to consider $k < \\frac{200}{3}$. I claim that $P(k) \\ge \\frac{1}{2} + \\frac{1}{2k}$ If $\\left[\\frac{100}{k}\\right]$ got", "# Bounding ${n \\choose k}$ Let $k > 0$, and $n > 2k$. Why is it necessarily true that $${n \\choose k} > \\frac{n^k}{2^k k!}$$ And is the condition $n > 2k$ necessary for this inequality to hold? - The condition is not necessary. But it makes deriving $n(n-1)(n-2)\\cdots(n-k+1)>n^k/2^k$ easy. – André Nicolas Mar 29 '12 at 14:42 ${n \\choose k}$ increases up to $k = \\lfloor n/2 \\rfloor$, then goes down. So you only really need to worry about the left side up to the max. – Mitch Mar 29 '12 at 14:57 @AndréNicolas What about $\\binom{6}{6} = 1 < \\frac{46656}{46080} = \\frac{6^6}{2^6 6!}$? – dtldarek Mar 29 '12 at 15:24 @dtldarek: Certainly the inequality does not hold unconditionally. – André Nicolas Mar 29 '12 at 15:34 @AndréNicolas Yeah, you are right, I forgot that necessary has very precise meaning in mathematics. +1 for you :-) – dtldarek Mar 29 '12 at 15:50 We have $$\\binom{n}{k} = \\frac{n!}{(n-k)!k!} \\ge \\frac{(n-k)^n}{(n-k)^{n-k}k!} \\ge \\frac{(n-k)^k}{k!} \\ge \\frac{(n/2)^k}{k!}$$ using $n > 2k \\implies n-k \\ge n/2$. In the more general case, as noted in the comments, you will need a more detailed analysis, e.g., using Stirling's approximation for $n!$. - How do you prove $\\frac{n!}{(n-k)!} \\geq \\frac{(n-k)^n}{(n-k)^{n-k}}$? – jamaicanworm Mar 29 '12 at 15:54 Using the definition of n! and the", "i.e. $$f(x) = \\frac{1}{\\pi \\left(x^2+1\\right)}$$ ... which seems very pretty, considering how messy the symbolics are. Here is a plot of the pdf of the ratio when $k = 25$: • the rough blue curve is the simulated Monte Carlo pdf of the ratio • the dashed red curve underneath is the exact Cauchy pdf The fit appears amazing - too good to be arbitrary. Touchy-feely explanation To get a feel for why this is the case, first observe that the distribution of $X-k$ which has form: $$f(x) = \\frac{2^{-\\frac{k}{2}} e^{-\\frac{1}{2} (k+x)} (k+x)^{\\frac{k}{2}-1}}{\\Gamma \\left(\\frac{k}{2}\\right)}$$ ... tends to normality (in particular, it seems $N(0, 2k)$) as $k$ becomes large: see for instance: Then, note that the ratio of two Normals with zero mean is Cauchy, and we have the gist of it.", "high for $k\\ge 10$. • Yes, I put my answer in just before boarding my plane and couldn't fix it until landing ;). – Brendan McKay Jun 29 '14 at 0:47 • I think you can get to the same point by using the AM-GM inequality on the displayed product in $\\frac{x}{k}\\prod_{i=1}^{k-2}\\frac{k-i-x}{k-i}$ for $x \\in (0,1).$ – Geoff Robinson Jun 30 '14 at 9:02 I'm not sure it adds anything beyond what's in Brendan McKay's answer, but I'll flesh out my comment. The absolute value of the binomial coefficient we are interested in may be written (for $x \\in (0,1)$ and $k>1$ ) as $\\frac{x}{k}\\prod_{i=1}^{k-1}(1 - \\frac{x}{i}).$ Applying the AM-GM inequality to the displayed product, this is at most $\\frac{x}{k}(1 - \\frac{ xH_{k-1}}{k-1})^{k-1},$ which is in turn at most $$\\frac{x e^{-H_{k-1}x}}{k}.$$ As Brendan McKay noted, this is at most $\\frac{1}{ekH_{k-1}},$ but perhaps the expression $\\frac{x}{k}(1 - \\frac{ xH_{k-1}}{k-1})^{k-1}$ will give a better estimate for small $k,$ as that takes maximum value when $x = \\frac{k-1}{kH_{k-1}},$ and the maximum value attained there is $\\frac{1}{kH_{k-1}}(1-\\frac{1}{k})^{k}$ (thanks to Emil Jerabek for a correction here). • Actually, I believe that $\\frac{x}{k}(1 - \\frac{ xH_{k-1}}{k-1})^{k-1}$ takes maximum value for $x=(1-k^{-1})H_{k-1}^{-1}$, and the maximum value is $(kH_{k-1})^{-1}(1-k^{-1})^k$. – Emil Jeřábek Jun 30 '14 at 14:04 • @Emil Jerabek: Yes, you are correct. – Geoff", "set , which makes $P(k) \\ge \\frac{1}{2} + \\frac{1}{2k}$. Now the only case without rounding, $k = 1$. It must be true. ### Solution 2 It suffices to consider $0\\le n < k-1.$ Now for each of these $n,$ let $f(n)=[\\frac{n}{k}], g(n)=[\\frac{100}{k}]-[\\frac{100-n}{k}].$ If we let $k=67,$ then the following graphs result for $f$ and $g.$ $f:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { if (i<=33) dot((i,0)); else dot((i,1)); } label(\"(0,0)\",(0,0),SW); label(\"(66,1)\",(66,1),NE); [/asy]$ $g:$ $[asy] size(10cm); for (int i = 0; i < 67; ++i) { dot((i,0)); } label(\"(0,0)\",(0,0),NW); label(\"(66,0)\",(66,0),SE); [/asy]$ Our probability is the number of $0\\le i such that $f(i)+g(i)=0$ over $k.$ Of course, this always holds for $i=0.$ If we let $k$ vary, then the graph of $f$ is always very similar to what it looks like above (groups of $\\frac{k+1}{2},\\frac{k-1}{2}$ dots). However, the graph of $g$ can vary greatly. In the above diagram, $g(i)=0$ for all $i,$ while it is possible for $g(i)=-1$ for all $i\\neq 0.$ In order to minimize the number of $i$ which satisfy $f(i)+g(i)=0,$ we either want $g(i)=0$ for $0\\le i or $g(i)=-1$ for $1\\le i This way, we see that at least half of the numbers from $1$ to $k-1$ satisfy the given equation. So, our desired probability is at least $\\frac{k+1}{2k}.$ As shown by", "$A=a_0-\\frac{1}{5}$. Thus the formula for our sequence is: $a_k=\\left(a_0-\\frac{1}{5}\\right)(-3)^k + \\frac{1}{5}(2)^k$. No matter how small a non-zero coefficient we have in front of that oscillating term, it will eventually drown out the growth term. Therefore, the sequence will only keep increasing if $a_0-\\frac{1}{5}=0$, i.e., if $a_0=\\frac{1}{5}$. Some playing around with Maple gives the following conjecture: the inequality $a_k<a_{k+1}$ gives an upper bound for $a_0$ when $k$ is even, a lower bound when $k$ is odd, and these bounds converge to a unique solution $$a_0=0.012101210121\\cdots$$ in base $3$, that is, $a_0=\\frac{16}{80}=\\frac{1}{5}$ only. Would love to see if anyone can prove (or disprove) this. Update. @GTonyJacobs has done so. Note that: $a_{k+1}-a_k=2^k-4a_k=4(2^{k-2}-a_k)$. We want $a_0$ such that $a_{k+1}-a_k>0$ for all $k$. Now can you finish the problem? • I dont understand what you did ...it would be $a_{k+1}+a_k$ and not $a_{k+1}-a_k$ – ronismofo Feb 10 '14 at 4:48 • That's a mistake. It should be $a_{k+1}-a_k = 2^k-4a_k = 4\\left(2^{k-2}-a_k\\right)$. – G Tony Jacobs Feb 10 '14 at 4:49 • @user127436: yes- sorry for the mistake. Let me edit my answer. Thanks @G Tony Jacobs – voldemort Feb 10 '14 at 5:04 • is that all??! it directly imply $a_0<\\frac 14$ , but it seems like a too trivial to impose in a putnam excersize book; although i am not", "the formula. $$\\binom{n}{k} = \\frac{n!}{(n - k)! \\ k!}$$ $$\\binom{1}{0} = \\frac{1!}{(1 - 0)! \\ 0!}$$ $$\\binom{1}{0} = \\frac{1!}{1! \\ 0!}$$ $$\\binom{1}{0} = \\frac{1}{1}$$ $$\\binom{1}{0} = 1$$ Example (7): Evaluate the value of \"100 choose 4\"? Solution: Given $$n = 8$$, and $$k = 5$$. By putting the values of n and k in the formula.$$\\binom{n}{k} = \\frac{n!}{(n - k)! \\ k!}$$ $$\\binom{100}{4} = \\frac{100!}{(100 - 4)! \\ 4!}$$ $$\\binom{100}{4} = \\frac{100!}{96! \\ 4!}$$ $$\\binom{100}{4} = \\frac{100 \\times 99 \\times 98 \\times 97 \\times 96!}{96! \\ 4!}$$ $$\\binom{100}{4} = \\frac{100 \\times 99 \\times 98 \\times 97}{4 \\times 3 \\times 2 \\times 1}$$ $$\\binom{100}{4} = 3,921,225$$ Example (8): Evaluate the value of \"100 choose 100\"? Solution: Given $$n = 8$$, and $$k = 5$$. By putting the values of n and k in the formula. $$\\binom{n}{k} = \\frac{n!}{(n - k)! \\ k!}$$ $$\\binom{100}{100} = \\frac{100!}{(100 - 100)! \\ 100!}$$ $$\\binom{100}{100} = \\frac{100!}{0! \\ 100!}$$ $$\\binom{100}{100} = 1$$ Note: From the above examples, it is clear that whenever the values of \"n\" and \"k\" are the same, the answer will always be \"1\".", "+ 1 , 4 k , 3 k + 5$ becomes (with a little more effort) $\\frac{8}{13} , - \\frac{20}{13} , \\frac{50}{13}$ with a common ratio of $\\left(- \\frac{5}{2}\\right)$", "kth layer is $OR_{k}$. Assuming that the true common odds ratio exists,taht is $OR_{1}=OR_{2}=...OR_{K}$ , Mantel-Haenszel's estimator of the common odds ratio is $\\hat OR_{MH}=\\frac{\\sum_{k=1}^{K}\\frac{f_{11k} f_{22k}}{n_{k}}}{\\sum_{k=1}^{K}\\frac{f_{12k} f_{21k}}{n_{k}}}$ The asymptotic variance for $ln(\\hat OR_{MH})$ is: $\\hat Var[ln(\\hat OR_{MH})]=\\frac{\\sum_{k=1}^{K}\\frac{(f_{11k}+f_{22k})f_{11k} f_{22k}}{n_{k}^2}}{2\\sum_{k=1}^{K}\\frac{f_{11k} f_{22k}}{n_{k}}}+\\frac{\\sum_{k=1}^{K}\\frac{(f_{11k}+f_{22k})f_{12k} f_{21k}+(f_{12k}+f_{21k})f_{11k} f_{22k}}{n_{k}^2}}{2\\sum_{k=1}^{K}\\frac{f_{11k} f_{22k}}{n_{k}}\\sum_{k=1}^{K}\\frac{f_{12k} f_{21k}}{n_{k}}}+\\frac{\\sum_{k=1}^{K}\\frac{(f_{12k}+f_{21k})f_{12k} f_{21k}}{n_{k}^2}}{2\\sum_{k=1}^{K}\\frac{f_{12k} f_{21k}}{n_{k}}}$ The lower confidence limit(LCL) and upper confidence limit(UCL) for $ln(\\hat OR_{MH})$ is: $ln(\\hat OR_{MH})-z({alpha}/2)\\sqrt{\\hat Var[ln(\\hat OR_{MH})]}$ and $ln(\\hat OR_{MH})+z(alpha/2)\\sqrt{\\hat Var[ln(\\hat OR_{MH})]}$", "\\leq \\text{actual probability} \\leq \\exp\\big( -\\frac{\\binom{10}{2}}{k_0} \\big) = \\exp\\big(-\\lambda\\big)$ focusing on the lower bound $0.9 \\leq 1 - \\frac{\\binom{10}{2}}{k_0}$ we can rescale by $k_0$ and simplify to get $\\frac{1}{10} k_0 \\geq \\binom{10}{2}$ $k_0 \\geq 10\\cdot\\binom{10}{2}$ which is satisfied with equality for $k_0 = 450$ = = = = as for the upper bound $\\text{actual probability} \\leq \\exp\\big( -\\frac{\\binom{10}{2}}{k_0} \\big)$ which is $\\approx 0.9048$ for $k_0$ = 450. Also of interest, the upper bound is $\\approx 0.89953$ for $k_0 = 425$ This tells you the estimate is sharp. I left computation of $\\text{actual probability}$ as an open item for OP. Hopefully it is clear how to compute it and then arrive at the inequalities. If not, I can fill in more details. Thank you so much, actually things here are more clear than at school, Happy to use Math Help Boards (heart)" ]

[ "Expanding $$(1+0.2)^{1000}_{}$$ by the binomial theorem and doing no further manipulation gives $${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$$ $$= A_0 + A_1 + A_2 + \\cdots + A_{1000},$$ where $$A_k = {1000 \\choose k}(0.2)^k$$ for $$k = 0,1,2,\\ldots,1000$$. For which $$k_{}^{}$$ is $$A_k^{}$$ the largest? (第九届AIME1991 第3题)", "cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$. Similarly, $(k+1)!=(k+1)k!$. Canceling these terms yields, $\\frac{1}{(1000-k)}>{\\frac{1}{5}}\\cdot\\frac{1}{(k+1)}$ Cross multiplying gives: $5k+5>1000-k \\Rightarrow k>165.8$ Therefore, since this identity holds for all values of $k>165.8$, the largest possible value of $k$ is $\\boxed{166}$. ### Solution 3 We know that $A_k$ will increase as $k$ increases until certain $k=m$, where $A_0 < A_1 < A_2 < \\dots < A_{m-2} < A_{m-1} < A_m$ and $A_m > A_{m+1} > A_{m+2} > \\dots > A_{1000}.$ Next, to change $A_{k-1}$ to $A_k$, we multiply $A_{k-1}$ by $\\frac{1000-k+1}{5k}$. It follows that the numerator must be greater than the denominator or $$1000-k+1>5k$$ $$k<166.8$$. We conclude that the answer is $\\boxed{166}$. ## Solution 4 Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\\frac{1}{5} > 1$) Say we have ${1000 \\choose 5}$... this equals $\\frac{1000*999*998*997*996}{5*4*3*2*1}$, and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\\frac{995}{6}$. Notice that this fraction's numerator plus denominator is equal to $1001$. Calling the numerator x and the denominator y, we get that $$x+y = 1001$$ $$\\frac{x}{y} > 5$$ $$x>5y$$ $$6y<1001$$ $$y<166.83333$$", "# Inequality $(1+\\frac1k)^k \\leq 3$ How can I elegantly show that: $(1 + \\frac{1}{k})^k \\leq 3$ For instance I could use the fact that this is an increasing function and then take $\\lim_{ k\\to \\infty}$ and say that it equals $e$ and therefore is always less than $3$ 1. Is this sufficient? 2. What is a better wording than \"increasing function\" • A bounded monotonic sequence $\\{x_n\\}$ is convergent. – mrs Sep 21 '12 at 6:16 • 1.- You only have to show that the function/sequence is monotonic increasing ($a_{n+1}>a_{n} \\forall n \\in \\mathbb{N}$). And you can use the value of the limit if known to compare it to the given value. – 3d0 Jun 22 '15 at 14:37 • The algebraic expression $\\bigg(1 + \\dfrac 1k\\bigg)^k$ is never equal to $3$, but is always less than $3$. $$\\lim_{k\\to\\infty}\\bigg(1+\\frac 1k\\bigg)^k = e = 2.7182818284590452353602874713526\\ldots < 3$$ – Mr Pie Feb 18 '18 at 4:21 Using binomial theorem, $(1+\\frac{1}{k})^k=1+{k\\choose 1}\\frac{1}{k}+{k\\choose 2}\\frac{1}{k^2}+\\cdots+{k\\choose r}\\frac{1}{k^r}+\\cdots+{k\\choose k}\\frac{1}{k^k}$ Now, consider ${k\\choose r}\\frac{1}{k^r}=\\frac{k!}{r!(k-r)!}\\frac{1}{k^r}=\\frac{(1)(1-\\frac{1}{k})(1-\\frac{2}{k})...(1-\\frac{r-1}{k})}{r!}\\lt \\frac{1}{r!}$ Thus, $$(1+\\frac{1}{k})^k=1+{k\\choose 1}\\frac{1}{k}+{k\\choose 2}\\frac{1}{k^2}+\\cdots+{k\\choose r}\\frac{1}{k^r}+\\cdots+{k\\choose k}\\frac{1}{k^k}$$ $$\\lt 1+1+\\frac{1}{2!}+\\frac{1}{3!}+\\cdots+\\frac{1}{r!}+\\cdots+\\frac{1}{k!}$$ $$\\lt 1+1+\\frac{1}{2!}+\\frac{1}{3!}+\\cdots+\\frac{1}{r!}+\\cdots+\\frac{1}{k!}+\\frac{1}{(k+1)!}+\\cdots =e$$ Hence $$(1+\\frac{1}{k})^k\\lt e\\lt 3$$ We will use the facts that $\\ln(1+x) < x$ and $(1 + \\frac{1}{k})^k = {\\rm e}^{k\\,\\ln( 1 + \\frac{1}{k} )} \\,.$ $$\\ln\\left( 1 + \\frac{1}{k} \\right) < \\frac{1}{k} \\Rightarrow k\\,\\ln\\left( 1 + \\frac{1}{k} \\right) < 1", "and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\\boxed{166}$ Note: It may be more intuitive to equate $x=5y$ and then use test points to figure out which side of the inequality to lie on. ## Solution 5 Notice the relation between $A_m$ and $A_{m+1}$. We have that: $A_{m+1} = A_m \\cdot \\frac{1}{5} \\cdot \\frac{1000-m}{m+1}$. This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that $${1000 \\choose m+1} = \\frac{1000!}{(m+1)!(1000-m-1)!} = \\frac{1000 \\cdots (1001-m)(1000-m)}{(m+1)!}$$ and that $${1000 \\choose m} = \\frac{1000!}{m!(1000-m)!} = \\frac{1000 \\cdots (1001-m)}{m!}$$ $$\\Longrightarrow {1000 \\choose m} \\cdot \\frac{1000-m}{m+1}$$ So now, we must find out for which $m$ is $1000-m< 5 \\cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$) Then we find that for all $m > 165.833333$, the above inequality ($1000-m< 5 \\cdot (m+1)$) holds, so then $m=165$ so $k=m+1= \\boxed{166}$. ~qwertysri987", "# 1991 AIME Problems/Problem 3 ## Problem Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives ${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \\cdots + A_{1000},$ where $A_k = {1000 \\choose k}(0.2)^k$ for $k = 0,1,2,\\ldots,1000$. For which $k_{}^{}$ is $A_k^{}$ the largest? ## Solution ### Solution 1 Let $0. Then we may write $A_{k}^{}={N\\choose k}x^{k}=\\frac{N!}{k!(N-k)!}x^{k}=\\frac{(N-k+1)!}{k!}x^{k}$. Taking logarithms in both sides of this last equation and using the well-known fact $\\log(a_{}^{}b)=\\log a + \\log b$ (valid if $a_{}^{},b_{}^{}>0$), we have $\\log(A_{k})=\\log\\left[\\frac{(N-k+1)!}{k!}x^{k}\\right]=\\log\\left[\\prod_{j=1}^{k}\\frac{(N-j+1)x}{j}\\right]=\\sum_{j=1}^{k}\\log\\left[\\frac{(N-j+1)x}{j}\\right]\\, .$ Now, $\\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\\frac{(N-j+1)x}{j}>1$ in each of the $\\log\\big[\\;\\big]$ terms (recall that $\\log y_{}^{} <0$ if $0). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=\\Big\\lfloor\\frac{(N+1)x}{1+x}\\Big\\rfloor$, where $\\lfloor z_{}^{}\\rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$. In summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k=\\boxed{166}$. ### Solution 2 We know that once we have found the largest value of $k$, all values after $A_k$ are less than $A_k$. Therefore, we are looking for the smallest possible value such that: ${\\frac{1}{5}}^k\\cdot {{1000} \\choose {k}}>{\\frac{1}{5}}^{k+1}\\cdot {{1000} \\choose {k+1}}$ Dividing by ${\\frac{1}{5}}^k$ gives: ${1000\\choose k}>{\\frac{1}{5}}\\cdot{1000\\choose {k+1}}$. We can express these binomial coefficients as factorials. $\\frac{1000!}{(1000-k)!\\cdot(k)!}>{\\frac{1}{5}}\\cdot\\frac{1000!}{(1000-k-1)!\\cdot{(k+1)!}}$ We note that the $1000!$ can", "# Inequality $(1+\\frac1k)^k \\leq 3$ How can I elegantly show that: $(1 + \\frac{1}{k})^k \\leq 3$ For instance I could use the fact that this is an increasing function and then take $\\lim_{ k\\to \\infty}$ and say that it equals $e$ and therefore is always less than $3$ 1. Is this sufficient? 2. What is a better wording than \"increasing function\" - A bounded monotonic sequence $\\{x_n\\}$ is convergent. – Babak S. Sep 21 '12 at 6:16 1.- You only have to show that the function/sequence is monotonic increasing ($a_{n+1}>a_{n} \\forall n \\in \\mathbb{N}$). And you can use the value of the limit if known to compare it to the given value. – 3d0 Jun 22 at 14:37 Using binomial theorem, $(1+\\frac{1}{k})^k=1+{k\\choose 1}\\frac{1}{k}+{k\\choose 2}\\frac{1}{k^2}+\\cdots+{k\\choose r}\\frac{1}{k^r}+\\cdots+{k\\choose k}\\frac{1}{k^k}$ Now, consider ${k\\choose r}\\frac{1}{k^r}=\\frac{k!}{r!(k-r)!}\\frac{1}{k^r}=\\frac{(1)(1-\\frac{1}{k})(1-\\frac{2}{k})...(1-\\frac{r-1}{k})}{r!}\\lt \\frac{1}{r!}$ Thus, $$(1+\\frac{1}{k})^k=1+{k\\choose 1}\\frac{1}{k}+{k\\choose 2}\\frac{1}{k^2}+\\cdots+{k\\choose r}\\frac{1}{k^r}+\\cdots+{k\\choose k}\\frac{1}{k^k}$$ $$\\lt 1+1+\\frac{1}{2!}+\\frac{1}{3!}+\\cdots+\\frac{1}{r!}+\\cdots+\\frac{1}{k!}$$ $$\\lt 1+1+\\frac{1}{2!}+\\frac{1}{3!}+\\cdots+\\frac{1}{r!}+\\cdots+\\frac{1}{k!}+\\frac{1}{(k+1)!}+\\cdots =e$$ Hence $$(1+\\frac{1}{k})^k\\lt e\\lt 3$$ - We will use the facts that $\\ln(1+x) < x$ and $(1 + \\frac{1}{k})^k = {\\rm e}^{k\\,\\ln( 1 + \\frac{1}{k} )} \\,.$ $$\\ln\\left( 1 + \\frac{1}{k} \\right) < \\frac{1}{k} \\Rightarrow k\\,\\ln\\left( 1 + \\frac{1}{k} \\right) < 1 \\Rightarrow {\\rm e}^{k\\,\\ln( 1 + \\frac{1}{k} )} < {\\rm e} < 3$$ $$\\Rightarrow {\\left( 1 + \\frac{1}{k}\\right)}^{\\frac{1}{k}} < 3 \\,.$$ -", "$$(k+1)^2 = k^2+2k+1 = k^2(1+\\frac{2}{k}+\\frac1{k^2}) < 2^k(1+\\frac{2}{5}+\\frac1{25}) < 2^k(2) = 2^{k+1}$$ So, by this nested induction, if $k \\ge 5$, if $n \\ge k^2$ (so that, certainly, $n \\ge 2k$), $2^n > n^k$. For $k = 2, 3,$ and $4$, respective values of $n \\ge 2k$ such that $2^n > n^k$ are $5$ ($2^5 > 5^2$), $10$ ($2^{10} > 10^3$), and $17$ ($2^{17} = 131072> 17^4=83521$). These values of $n$ happen to be $k^2+1$, so the final result is: If $n$ and $k$ are integers and $k \\ge 2$ and $n \\ge k^2+1$, then $2^n > n^k$. Pick $n > 2k$. Then, $$2^n > \\binom{n}{k + 1} = \\frac{n (n - 1) \\cdot \\ldots \\cdot (n - k)}{(k+1)!} \\geq \\frac{n^{k + 1}}{2^{k+1} (k+1)!}$$ by a trivial combinatorial argument for the first inequality, and using $n - k > n/2$ in the second inequality. Therefore $2^n > n^k$ as soon as $$n > 2^{k + 1} (k + 1)!$$ • I'll accept it, even though your bound is quite large. – marty cohen Jul 8 '13 at 20:38 • It's possible to improve it significantly! I'll post a comment later. – blabler Jul 8 '13 at 21:09", "we deduce that $$M = \\frac{2016}{1001} = \\frac{288}{143}.$$ The answer is $\\boxed{431}.$ ### Solution 2 Each 1000-element subset $\\left\\{ a_1, a_2,a_3,...,a_{1000}\\right\\}$ of $\\left\\{1,2,3,...,2015\\right\\}$ with $a_1 contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\\left\\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$. There are $a_1$ ways to choose a positive integer $k$ such that $k ($k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is equal to the sum. But choosing a set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is the same as choosing a 1001-element subset from $\\left\\{1,2,3,...,2016\\right\\}$! Thus, the average is $\\frac{\\binom{2016}{1001}}{\\binom{2015}{1000}}=\\frac{2016}{1001}=\\frac{288}{143}$. Our answer is $p+q=288+143=\\boxed{431}$. ### Solution 3(NOT RIGOROUS) Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$: For $p = 2$ and $q = 2$, the answer is $1$. For $p = 3$ and $q = 2$, the answer is $\\frac43$. For $p = 4$ and $q = 2$, the answer is $\\frac53$. For $p = 3$ and $q = 3$, the answer is $1$. For $p = 4$ and $q = 3$, the answer is $\\frac54$. For $p = 5$ and $q = 3$, the answer", "will work in any more general, difficult cases to fairly easily determine if there's any $$k$$ and, if so, which value(s) of $$k$$ will work. • or use $e=(1+{1\\over n})^n$ as $n$ tends to infinity. – user645636 Jan 22 '20 at 0:38 Similar Isaac YIU Math Studio's answer, we could instead use hexadecimal. $$2^{16}=16^4=10000_{16}$$ $$2^{17}=2\\cdot2^{16}=20000_{16}$$ As done in Henry's answer, we can binomially expand (or really just multiply out $$11^4$$) to see we then have: $$17^4=11_{16}^4=14641_{16}$$ which is clearly between $$10000_{16}$$ and $$20000_{16}$$. Note $$17^4=(2^4+1)^4=2^{16}+4\\cdot2^{12}+6\\cdot2^8+4\\cdot2^4+1$$ $$<2^{16}+2^{14}+2^{11}+2^6+1<2^{16}+2^{15}+2^{14}+\\dots+2^2+2^1+2^0<2^{17}$$. Just another way, it is enough to compare $$17^4$$ and $$2\\cdot16^4$$. By Bernoulli’s inequality, $$\\frac{17^4}{16^4} = \\frac1{\\left(1-\\frac{1}{17}\\right)^4} < \\frac1{1-\\frac4{17}}=\\frac{17}{13}<2$$ You can use binary to find that: $$2^{16}=1000000000000000_{(2)} \\\\ 2^{17}=10000000000000000_{(2)} \\\\ 17^4=10001\\times10001\\times10001\\times10001_{(2)}=10100011001000001_{(2)}$$ Obvious that $$2^{16}<17^4<2^{17}$$. • I think you typed $10001$ too many times Jan 22 '20 at 2:03 • I like the usage of binary, but isn’t multiplying $10001_2$ out four times a bit tedious? And with significantly non-minimal computation? Jan 22 '20 at 2:08 • @LieutenantZipp Actually, you can use binomial theorem to do this computation, like. Jan 22 '20 at 2:42", "that... #### MarkFL Staff member ##### Active member The pattern must be that this is the third member of a series starting $\\dfrac1{81} = \\dfrac1{9^2} = 0.\\overline{012345679}$, $\\dfrac1{9801} = \\dfrac1{99^2} = 0.\\overline{00\\,01\\,02\\,03\\,04\\,05\\cdots 96\\,97\\,99}$, $\\dfrac1{998\\,001} = \\dfrac1{999^2} = 0.\\overline{000\\,001\\,002\\,003\\,004\\,005\\cdots 996\\,997\\,999}$. You can sort of see why this works if you consider the binomial series $(1-x)^{-2} = 1+2x+3x^2+4x^3+5x^4+\\ldots$, with $x = 1/10^k$. For example, if $k=1$ you get \\begin{aligned}\\tfrac1{81} = 9^{-2} = (10-1)^{-2} &= \\tfrac1{100}\\bigl(1-\\tfrac1{10}\\bigr)^{-2} \\\\ &= \\tfrac1{100}\\bigl(1 + \\tfrac2{10} + \\tfrac3{100} +\\tfrac4{1000} + \\ldots\\bigr).\\end{aligned} That gives you $\\frac1{81} = 0.01234...$. The trouble starts when you have to try to justify why the 8 gets left out, and why the decimal starts to recur after the 9. The answer, as it ends up, is simple. As you have hinted, we can write this as $$\\frac{1}{(10^n-1)^2}=\\sum_{k=0}^{\\infty}\\frac{k}{10^{n(k+1)}}$$ To start, let's focus on the case of $\\frac{1}{81}$, that is, the $n=1$ case. In this case, we have 1/81 = .01 + .002 + .0003 + ... which starts out fine. However, eventually we reach ...08 + ...009 + ...0010 + ...00011 + ... now, the 1 from the 10 overlaps with the 9. As a result, there's a 1 that carries over to the eight, putting a 9 where there was an 8, and a 0 where there was a 9. In fact,", "# Thread: Ratios of Binomial Expansions 1. ## Ratios of Binomial Expansions 1) In the expansion of (1+x)^16, fing the ratio of the term in x^13 to the term in x^11. 2) In the expansion of (2+3x)^n, the coefficients of x^5 and x^6 are in the ratio 4:9. Find the value of n. Any help would be great, thanks! 2. ## Re: Ratios of Binomial Expansions $(1+x)^{16} = \\sum_{k=0}^{16} \\binom{16}{k} 1^{16-k}x^k$ We want to find the terms in $x^{13}$ and $x^{11}$. These are: $\\binom{16}{13}x^{13} \\quad \\mathrm{and} \\quad \\binom{16}{11}x^{11}.$ Ratio: $\\frac{\\binom{16}{13}}{\\binom{16}{11}} = \\frac{\\frac{16!}{3! \\cdot 13!}}{\\frac{16!}{5! \\cdot 11!}} = \\frac{5! \\cdot 11!}{3! \\cdot 13!} = \\frac{5 \\cdot 4}{12 \\cdot 13} = \\frac{5}{39}$ Note that the ratio should be less than 1, because for $(1+x)$ raised to any power, the coefficients will be largest in the \"middle\" of the expansion. The expansion for 2) will be $(2+3x)^{x} = \\sum_{k=0}^{n} \\binom{n}{k} 2^{n-k}(3x)^k = \\sum_{k=0}^{n}\\left( \\binom{n}{k} 2^{n-k} 3^k\\right)x^k$ You are given that the ratio of the co-efficients of $x^5$ and $x^6$ is $4/9$. Can you take it from here? 3. ## Re: Ratios of Binomial Expansions Originally Posted by fan96 $(1+x)^{16} = \\sum_{k=0}^{16} \\binom{16}{k} 1^{16-k}x^k$ We want to find the terms in $x^{13}$ and $x^{11}$. These are: $\\binom{16}{13}x^{13} \\quad \\mathrm{and} \\quad \\binom{16}{11}x^{11}.$ Ratio: $\\frac{\\binom{16}{13}}{\\binom{16}{11}} = \\frac{\\frac{16!}{3! \\cdot 13!}}{\\frac{16!}{5! \\cdot 11!}} = \\frac{5! \\cdot", "the $a_k$ with $q_1q_2\\cdots q_i \\mid k \\mid n.$ and $G_j=a_n$ I will illustrate with the example of $n=900=2^23^25^2$. So $G_0=\\binom{900}{1}$, $G_1=\\gcd(G_0,\\binom{900}{900/2})$, $G_2=\\gcd(G1,\\binom{900}{900/3})$ and $G_3=\\gcd(G_2,\\binom{900}{900/5}).$ I claim that $G_3=a_{900}.$ $G_0=\\binom{900}{1}$ is the product of the $26$ $a_k$ with $1 \\lt k \\mid 900.$ Of these, the $9$ which are multiples of $2^2$, namely $a_4,a_{12},a_{20},a_{36},a_{60},a_{100},a_{180},a_{300}$ and $a_{900}$ also divide $\\binom{900}{900/2}=\\binom{900}{450}$, the other $17$ have $e_{900,450,k}=0.$ There are many other $a_k$ with $e_{900,450,k}=1$ but all of them are relatively prime to $G_0$ by the important step above. Hence $G_1$ is exactly the product of the $9$ $a_k$ with $4 \\mid k \\mid 900$. Now $G_2=\\gcd(G_1,\\binom{900}{300})=a_{36}a_{180}a_{900}$ because the only other $a_k$ dividing $\\binom{900}{300}$ are such that $k$ neither divides nor is a multiple of $4,12,20,60,100,300$ and hence all are relatively prime to $G_1$ by the important step. Finally $G_3=\\gcd(G_2,\\binom{900}{180})=a_{900}$ because the only other $a_k$ with $e_{900,180,k}=1$ are relatively prime to both $a_{36}$ and $a_{180}$. I believe this does what was requested. It may be helpful to look at a more abstract setting where essentially the same proof goes through. The idea is to replace each integer $n$ by a generalized integer $I_n$, replace $a_n$ by some appropriate $A_n$, define a generalized factorial $[I_n]!=I_1I_2\\cdots I_n$ and then a generalized binomial coefficient $${n \\brack m}=\\frac{[I_n]!}{[I_m]![I_{n-m}]!},$$ and deduce that $\\gcd\\left( {n \\brack 1},{n \\brack", "the approach is exactly similar, you should get an equivalent inequality of form: $$\\prod_{k=1}^{49} \\frac{2k (2k+2)}{(2k+1)^2} < \\frac{169}{200}$$ Similarly, each factor now is less than $1$ and convergence is rapid. So taking the first three terms, we have $\\dfrac{1}{X^2} < \\dfrac{1024}{1225}$ which is verifiably less than $\\dfrac{169}{200} = 0.845$. Hence $\\dfrac{1}{13} < X$ is also established. - When you multiply up and down in the product by $2 \\cdot 4 \\cdot 6 \\ldots 50$, the product becomes $$\\frac{1}{2^{100}} \\binom{100}{50} = \\frac{1}{2^{100}} \\frac{100!}{(50!)^2}$$ Use Stirling: $$n! \\sim \\sqrt{2 \\pi n} n^n e^{-n} \\left ( 1+ \\frac{1}{12 n}\\right )$$ for large $n$. We will see what that means: plug in this approximation into the binomial expression above for $n=50$; the result is $$\\frac{1}{2^{100}} \\binom{100}{50} \\approx \\frac{1}{\\sqrt{50 \\pi}} \\left ( 1- \\frac{1}{400}\\right )$$ Note that the product being about $\\frac{1}{\\sqrt{50 \\pi}}$ is accurate to $1$ part in $400$. Now, it is clear that $50 \\pi \\approx 157$ to within that margin of error. As $157$ falls between $12^2=144$ and $13^2=169$, the assertion is true. - \"It is clear that...\" No, it is not, unless you explain how the error depends on $n$. From what you wrote, it could well be that the two expressions are too far apart for the approximation to be useful, simply because $n=100$ is not \"large\" enough.", "$A=a_0-\\frac{1}{5}$. Thus the formula for our sequence is: $a_k=\\left(a_0-\\frac{1}{5}\\right)(-3)^k + \\frac{1}{5}(2)^k$. No matter how small a non-zero coefficient we have in front of that oscillating term, it will eventually drown out the growth term. Therefore, the sequence will only keep increasing if $a_0-\\frac{1}{5}=0$, i.e., if $a_0=\\frac{1}{5}$. Some playing around with Maple gives the following conjecture: the inequality $a_k<a_{k+1}$ gives an upper bound for $a_0$ when $k$ is even, a lower bound when $k$ is odd, and these bounds converge to a unique solution $$a_0=0.012101210121\\cdots$$ in base $3$, that is, $a_0=\\frac{16}{80}=\\frac{1}{5}$ only. Would love to see if anyone can prove (or disprove) this. Update. @GTonyJacobs has done so. Note that: $a_{k+1}-a_k=2^k-4a_k=4(2^{k-2}-a_k)$. We want $a_0$ such that $a_{k+1}-a_k>0$ for all $k$. Now can you finish the problem? • I dont understand what you did ...it would be $a_{k+1}+a_k$ and not $a_{k+1}-a_k$ – ronismofo Feb 10 '14 at 4:48 • That's a mistake. It should be $a_{k+1}-a_k = 2^k-4a_k = 4\\left(2^{k-2}-a_k\\right)$. – G Tony Jacobs Feb 10 '14 at 4:49 • @user127436: yes- sorry for the mistake. Let me edit my answer. Thanks @G Tony Jacobs – voldemort Feb 10 '14 at 5:04 • is that all??! it directly imply $a_0<\\frac 14$ , but it seems like a too trivial to impose in a putnam excersize book; although i am not", "# Problem #211 211 Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives ${1000 \\choose 0}(0.2)^0+{1000 \\choose 1}(0.2)^1+{1000 \\choose 2}(0.2)^2+\\cdots+{1000 \\choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \\cdots + A_{1000},$ where $A_k = {1000 \\choose k}(0.2)^k$ for $k = 0,1,2,\\ldots,1000$. For which $k_{}^{}$ is $A_k^{}$ the largest? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "F(t_1,\\ldots,t_k)^2 (1 - t_1-\\ldots-t_k+ kt_1)\\ dt_1.$. (2) Assuming this estimate, we may integrate in $t_2,\\ldots,t_k$ to conclude that $\\displaystyle J_k^{(1)}(F) \\leq \\frac{\\log k}{k-1} \\int F^2 (1-t_1-\\ldots-t_k+kt_1)\\ dt_1 \\ldots dt_k$ which symmetrises to $\\sum_{m=1}^k J_k^{(m)}(F) \\leq k \\frac{\\log k}{k-1} \\int F^2\\ dt_1 \\ldots dt_k$ giving the desired upper bound (1). It remains to prove (2). By Cauchy-Schwarz, it suffices to show that $\\displaystyle \\int_0^{1-t_2-\\ldots-t_k} \\frac{dt_1}{1 - t_1-\\ldots-t_k+ kt_1} \\leq \\frac{\\log k}{k-1}.$ But writing $s = t_2+\\ldots+t_k$, the left-hand side evaluates to $\\frac{1}{k-1} (\\log k(1-s) - \\log (1-s) ) = \\frac{\\log k}{k-1}$ as required. ... ## World records k Lower bound Upper bound 4 1.845 1.848 5 2.001162 2.011797 10 2.53 2.55842 20 3.05 3.1534 30 3.34 3.51848 40 3.52 3.793466 50 3.66 3.99186 59 3.95608 4.1479398 All upper bounds come from (1). ## More general variational problems It appears that for the purposes of establish DHL type theorems, one can increase the range of F in which one is taking suprema over (and extending the range of integration in the definition of $J_k^{(m)}(F)$ accordingly). Firstly, one can enlarge the simplex ${\\mathcal R}_k$ to the larger region ${\\mathcal R}'_k = \\{ (t_1,\\ldots,t_k) \\in [0,1]^k: t_1+\\ldots+t_k \\leq 1 + \\min(t_1,\\ldots,t_k) \\}$ provided that one works with a generalisation of EH[θ] which controls more general Dirichlet convolutions than the von Mangoldt function", "kt_1)\\ dt_1.$. (2) Assuming this estimate, we may integrate in $t_2,\\ldots,t_k$ to conclude that $\\displaystyle J_k^{(1)}(F) \\leq \\frac{\\log k}{k-1} \\int F^2 (1-t_1-\\ldots-t_k+kt_1)\\ dt_1 \\ldots dt_k$ which symmetrises to $\\sum_{m=1}^k J_k^{(m)}(F) \\leq k \\frac{\\log k}{k-1} \\int F^2\\ dt_1 \\ldots dt_k$ giving the desired upper bound (1). It remains to prove (2). By Cauchy-Schwarz, it suffices to show that $\\displaystyle \\int_0^{1-t_2-\\ldots-t_k} \\frac{dt_1}{1 - t_1-\\ldots-t_k+ kt_1} \\leq \\frac{\\log k}{k-1}.$ But writing $s = t_2+\\ldots+t_k$, the left-hand side evaluates to $\\frac{1}{k-1} (\\log k(1-s) - \\log (1-s) ) = \\frac{\\log k}{k-1}$ as required. ... ## World records k Lower bound Upper bound 4 1.845 1.848 5 2.001162 2.011797 10 2.53 2.55842 20 3.05 3.1534 30 3.34 3.51848 40 3.52 3.793466 50 3.66 3.99186 59 3.95608 4.1479398 All upper bounds come from (1).", "provided in ... 5 Step 1 (conjecture): there is some $e>0$ such that $$6+\\sqrt{x}=(e+\\sqrt[3]{3}/2)^3,\\quad 6-\\sqrt{x}=(-e+\\sqrt[3]{3}/2)^3.\\tag{*}$$ You can see that if such an $e$ exists then the equality $\\sqrt[3]{6+\\sqrt x} + \\sqrt[3]{6-\\sqrt x} = \\sqrt[3] {3}$ is satisfied. Solving for $e$ is simple: $$... 5 Usage of the multinomial coefficient (k_1, k_2, \\cdots, k_n)!:$$ \\big( 1 + x^5 + x^7\\big)^{20} = \\sum_{k_1=1}^{20} \\sum_{k_2=1}^{20-k_1} (k_1, k_2, 20 - k_1 - k_2)! x^{5k_1} x^{7k_2}, $$where$$ (k_1, k_2, \\cdots, k_n)! = \\frac{ (k_1 + k_2 + \\cdots + k_n )! } { k_1! k_2! \\cdots k_n!}. $$So we get k_1=2 and k_2=1, thus$$ (2,1,17)! = ... 5 $17$ can only be obtained by using two $5$s and one $7$ . These two $5$s can be obtained in $\\binom{20}2$ ways which is $190$ and the $7$ can be got in from one of the remaining 18 brackets. So $190$ x $18$ = $3420$ is the answer. 5 Another way : \\begin{align}\\\\&\\frac{1}{1+\\sqrt 2+\\sqrt 3}+\\frac{1}{1-\\sqrt 2+\\sqrt 3}+\\frac{1}{1+\\sqrt 2-\\sqrt 3}+\\frac{1}{1-\\sqrt 2-\\sqrt 3}\\\\&=\\left(\\frac{1}{1+\\sqrt 2+\\sqrt 3}+\\frac{1}{1+\\sqrt 2-\\sqrt 3}\\right)+\\left(\\frac{1}{1-\\sqrt 2+\\sqrt 3}+\\frac{1}{1-\\sqrt 2-\\sqrt 3}\\right)\\\\&=\\frac{1+\\sqrt 2-\\sqrt 3+1+\\sqrt 2+\\sqrt 3}{(1+\\sqrt 2+\\sqrt ... 5 a = 6 \\quad(\\mathrm{mod} 7) a^2 = 36 = 1 \\quad(\\mathrm{mod} 7) 3a = 18 = 4\\quad (\\mathrm{mod} 7) a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \\quad(\\mathrm{mod} 7) 5 Since their", "$n-1$ arithmetic series of $1000, 999, \\ldots, 1001 - \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$. We let $k = \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$, so $S = (n-1)\\left[\\frac{k(1000 + 1001 - k)}{2}\\right] + R(n)$ where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$. At this point, we introduce the crude estimate[1] that $k=\\frac{121-n}{n-1}$, so $R(n) = 0$ and \\begin{align*}2S+2n &= (121-n)\\left(2001-\\frac{121-n}{n-1}\\right)+2n = (120-(n-1))\\left(2002-\\frac{120}{n-1}\\right)\\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \\frac{36}{n-1}$. By AM-GM, we have $5(n-1) + \\frac{36}{n-1} \\ge 2\\sqrt{5(n-1) \\cdot \\frac{36}{n-1}}$, with equality coming when $5(n-1) = \\frac{36}{n-1}$, so $n-1 \\approx 3$. Substituting this result and some arithmetic gives an answer of $\\boxed{947}$. In less formal language, it quickly becomes clear after some trial and error that in our sample, there will be $n$ values equal to one and $n-1$ values each of $1000, 999, 998 \\ldots$. It is fairly easy to find the maximum. Try $n=2$, which yields $924$, $n=3$, which yields $942$, $n=4$, which yields $947$, and $n=5$, which yields $944$. The maximum difference occurred at $n=4$, so the answer is $947$. ## Solution 2 With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of", "hence the probability $p_n(k)$ to get at least $k=84$ successes is $$p_n(k)=2^{-n}\\sum_{i=k}^n{n\\choose i}.$$ When $k$ is significantly larger than $\\frac{n}2$, $p_n(k)$ is very small and an estimation of how small $p_n(k)$ is is obtained through a large deviations estimate. This says that $p_n(k)\\leqslant p_n^*(k)$ with $$p^*_n(k)=2^{-n}\\inf\\{(1+s)^ns^{-k}\\,;\\,s\\geqslant1\\}.$$ For every $k\\gt\\frac{n}2$, the infimum is reached at $s=\\frac{k}{n-k}$, hence $$p^*_n(k)=2^{-n}n^nk^{-k}(n-k)^{-(n-k)}=\\left(I\\left(\\tfrac{k}n\\right)\\right)^{-n},\\quad I(t)=2t^t(1-t)^{1-t}.$$ For example, if $k=84$ and $n=118$, then $t=.712$ hence $I(t)\\approx1.09710$ and $$p^*_{118}(84)\\approx(1.09710)^{-118}\\approx10^{-5}.$$ Numerically, $p_{118}(84)\\approx2.36224\\cdot10^{-6}$ and $p^*_{118}(84)\\approx1.78153\\cdot10^{-5}$. - At this point, I would like to add that I am not a statistician by trade (I'm sure I'd find this easier if I was) so I'm going to need som clarification - the parenthesis with one constant over another - that represents what, exactly? EDIT: Also, does the Large Deviations Estimate hold when k is only 42% larger than n/2 ? – medivh Jun 30 '13 at 9:23 ${n\\choose i}=\\frac{n!}{i!(n-i)!}$ is a binomial coefficient. – Did Jun 30 '13 at 9:26 The upper bound in my post holds for every $n\\geqslant1$ and every $k\\geqslant n/2$. No $=$ or $\\leqslant$ sign is an approximation, only the $\\approx$ are (to numerical values). – Did Jun 30 '13 at 9:27 What really? Why is it called LARGE deviations then? I mean, in that case it would hold even if my process had run only 3" ]

[ "Now it is in the form $$\\frac{A\\sqrt{B}+C}{D}$$ and... A, B, C, and D are integers D is positive B is not divisible by the square of any prime the GCF of A, C, and D = the GCF of 5, -10, and 2 = 1 A + B + C + D = 5 + 6 + -10 + 2 = 3 hectictar May 22, 2019", "+0 # help plz 0 173 1 Express $\\frac{6}{\\sqrt{245}}$ in the form $\\frac{A \\sqrt{B}}{C}$ where A and C are relatively prime integers, C is positive, and B is not divisible by the square of any prime. Find A+B+C Sep 2, 2021", "+0 # Help plz !!!! 0 273 1 +31 Rationalize the denominator of \\frac{5}{2+\\sqrt{6}}$. The answer can be written as$\\frac{A\\sqrt{B}+C}{D}$, where$A$,$B$,$C$, and$D$are integers,$D$is positive, and$B$is not divisible by the square of any prime. If the greatest common divisor of$A$,$C$, and$D$is 1, find$A+B+C+D\\$. May 22, 2019 #1 +8965 +3 Rationalize the denominator of $$\\frac{5}{2+\\sqrt{6}}$$. The answer can be written as $$\\frac{A\\sqrt{B}+C}{D}$$, where A, B, C, and D are integers, D is positive, and B is not divisible by the square of any prime. If the greatest common divisor of A, C, and D is 1, find A+B+C+D. __________ $$\\frac{5}{2+\\sqrt{6}}\\,=\\,\\frac{5}{2+\\sqrt6}\\cdot\\frac{2-\\sqrt6}{2-\\sqrt6}\\,=\\,\\frac{10-5\\sqrt6}{4-6}\\,=\\,\\frac{10-5\\sqrt6}{-2} \\,=\\,\\frac{5\\sqrt6-10}{2}$$ Now it is in the form $$\\frac{A\\sqrt{B}+C}{D}$$ and... A, B, C, and D are integers D is positive B is not divisible by the square of any prime the GCF of A, C, and D = the GCF of 5, -10, and 2 = 1 A + B + C + D = 5 + 6 + -10 + 2 = 3 May 22, 2019 #1 +8965 +3 Rationalize the denominator of $$\\frac{5}{2+\\sqrt{6}}$$. The answer can be written as $$\\frac{A\\sqrt{B}+C}{D}$$, where A, B, C, and D are integers, D is positive, and B is not divisible by the square of any prime. If the greatest common divisor of A, C, and D is 1, find A+B+C+D. __________ $$\\frac{5}{2+\\sqrt{6}}\\,=\\,\\frac{5}{2+\\sqrt6}\\cdot\\frac{2-\\sqrt6}{2-\\sqrt6}\\,=\\,\\frac{10-5\\sqrt6}{4-6}\\,=\\,\\frac{10-5\\sqrt6}{-2} \\,=\\,\\frac{5\\sqrt6-10}{2}$$", "• # question_answer If $a,\\ b,\\ c$ are the positive integers, then $(a+b)(b+c)(c+a)$is [DCE 2000] A) $<8abc$ B) $>8abc$ C) $=8abc$ D) None of these Since A.M. > G.M., we have $\\frac{a+b}{2}>\\sqrt{ab},\\ \\frac{b+c}{2}>\\sqrt{bc}$ and $\\frac{a+c}{2}>\\sqrt{ac}$. Multiplying these inequalities, we get $(a+b)(b+c)(c+a)>8abc$.", "# Will you square? Part 3 Algebra Level 3 If $$\\sqrt { 2 } +\\sqrt { 5 }$$ is written in the form $$\\frac { a }{ \\sqrt { b-\\sqrt { c } } },$$ where $$a,b,c$$ are positive integers and $$b$$ is not divisible by the square of any prime, what is the value of $$a+b+c?$$ ×", "definition 1, we can write $\\sqrt{2}=\\frac{a}{b} \\ldots (1)$ where $a$ and $b$ are both integers with $\\mathrm{g.c.d.} (a,b) =1$ and $b \\ne 0$ . Squaring equation (1), $2=\\frac{a^2}{b^2}$ or, $a^2=2b^2 \\ldots (2)$ From the equation (2), we can proceed our proof into two ways: Way I: $\\sqrt{2}$ is a positive number, therefore we can assume $a$ and $b$ both to be positive. Since, $a^2=2b^2$ then $a^2=2b \\cdot b$ or, $b|a^2$ (read as b divides a squared). Since, $b$ is positive integer, $b \\ge 1$ . However, $b=1$ is impossible since corresponding $a=\\sqrt{2}$ is not an integer. Thus, $b > 1$ and then according to fundamental theorem of arithmetic, there exists at least one prime $p> 1$ which divides $b$ . Mathematically, $p|b$ but as $b|a^2$ . It is clear that $p|a^2$ . This implies that $p|a$ . Since $p|a$ and $p|b$ , therefore $\\mathrm{g.c.d.}(a,b) \\ge p$ . So for given number,the greatest common divisor of $a$ and $b$ is not $1$ , but another prime larger than $1$ . Thus, it fails to satisfy the definition 1. Thus our claim that $\\sqrt{2}$ is rational, is untrue. Therefore, $\\sqrt{2}$ is an irrational number. Way II: As a deviation, we can proceed our proof from equation (2) by taking the fact into mind that $\\sqrt{2}$ is positive. The number (natural", "+0 # Algebra 0 30 1 $x = 1 + \\cfrac{\\sqrt{3}}{1 + \\cfrac{\\sqrt{3}}{1 + \\dotsb}}$. Find $\\frac{1}{(x + 2)(x - 1)}$. When your answer is in the form $\\frac{A + \\sqrt{B}}{C}$, where $A$, $B$, and $C$ are integers, and $B$ is not divisible by the square of a prime, Oct 28, 2022", "The expression $$\\frac {1}{ 1 \\sqrt{2} + 2 \\sqrt{1} }+ \\frac {1}{ 2 \\sqrt{3} + 3 \\sqrt{2} } + \\ldots + \\frac {1} { 9999 \\sqrt{10000} + 10000 \\sqrt{9999} }$$ is equal to $$\\frac { a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is $$a + b$$?", "Let $$x_0$$ be the positive solution to the equation $\\large \\frac{\\sqrt{1+x^2}}{x} = \\sqrt{1+x^2}-1.$ If the expression $$\\left((2\\sqrt{2}-2)x_0-1\\right)^2$$ is of the form $$a+b\\sqrt{c}$$, where $$a,b,c$$ are integers, and $$c$$ is square free, what is the value of $$a+b+c$$?", "# Simplifying my Solution for: List the first three positive prime numbers $d$ in $\\textbf{Z}$ so that the quadratic integers in… Here is the question I have answered: List the first three positive prime numbers $d$ in $\\mathbb{Z}$ so that the quadratic integers in $\\mathbb{Q}(\\sqrt{d})$ are precisely the ones of the form $a + b \\sqrt d$, where $a$ and $b$ are rational integers. List the first three positive prime numbers $d$ so that the quadratic integers in $\\mathbb{Q}(\\sqrt{d})$ are precisely the ones of the form $$\\frac{a + b \\sqrt d}{2},$$ where $a$ and $b$ are rational integers and $a$ and $b$ are either both even or both odd. Do the same when $\\sqrt d$ is replaced by $\\sqrt{−d}$. Here is my solution: If $$\\frac{a + b \\sqrt d}{c}$$ is an algebraic integer then its irreducible monic polynomial $f(X)$ in $\\mathbb{Q}[X]$ has coefficients in $\\mathbb{Z}$. Since conjugation : $\\sqrt d \\to -\\sqrt d$ is a automorphism that fixes $\\mathbb Q$ then $$\\frac{a - b \\sqrt d}{c}$$ is also a root of $f(X)$. So $$f(X) = \\left(X - \\frac{a + b \\sqrt d}{c}\\right)\\left(X - \\frac{a - b \\sqrt d}{c}\\right) = X^2 - 2 \\frac{a}{c} X + \\frac{a^2 - db^2}{4}.$$ i) $\\frac{2a}{c}$ an integer implies $c = 1$ or $2$. ii) If $c = 2$ then $$\\frac{a^2 - db^2}{4}$$ an", "where $a, b$ are postive integers that are relatively prime. Then $\\sqrt{2}=\\sqrt{2}\\cdot1=\\sqrt{2}\\cdot(ar+bs)$ for some integers $r, s$. (Bézout's Identity) Because $\\sqrt{2}=a/b$, we have $a=\\sqrt{2}b$ and $b=a/\\sqrt{2}$. Then $\\sqrt{2}=\\sqrt{2}ar+\\sqrt{2}bs=2br+as$, and $\\sqrt{2}ar+\\sqrt{2}bs=2br+as$ is clearly an integer. This means that $\\sqrt{2}$ is an integer, but $1<\\sqrt{2}<2$; contradiction.", "Evaluate $\\lim_{x\\rightarrow1} \\frac{\\sqrt{2x-1}-\\sqrt{x}}{x^2+x-2}.$ If your answer can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers, find the value of $$a+b$$.", "$c$ such that $m=a+\\sqrt{b+\\sqrt{c}}$. Find $a+b+c$. Let $f(x) = x^4 + ax^3 + bx^2 + cx + d$. If $f(-1) = -1$, $f(2)=-4$, $f(-3) = -9$, and $f(4) = -16$. Find $f(1)$. Solve in positive integers $x^2 - 4xy + 5y^2 = 169$. Solve in integers the question $x+y=x^2 -xy + y^2$. Solve in integers $\\frac{x+y}{x^2-xy+y^2}=\\frac{3}{7}$ Prove the product of $4$ consecutive positive integers is a perfect square minus $1$. For any arithmetic sequence whose terms are all positive integers, show that if one term is a perfect square, this sequence must have infinite number of terms which are perfect squares. Prove there exist infinite number of positive integer $a$ such that for any positive integer $n$, $n^4 + a$ is not a prime number. Find all positive integer $n$ such that $(3^{2n+1} -2^{2n+1}- 6^n)$ is a composite number. The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\\frac{\\sqrt[3]a + \\sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$. Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$, $x_1$ , $\\dots$ , $x_{2011}$ such that $m^{x_0} = \\sum_{k = 1}^{2011} m^{x_k}.$ Suppose $x$ is in the interval $[0, \\frac{\\pi}{2}]$ and $\\log_{24\\sin x} (24\\cos x)=\\frac{3}{2}$. Find $24\\cot^2 x$.", "(2) $$\\sqrt{(a + b + c - 1 + x)^2} + |\\frac{1}{1-x}- a - b - c| + \\sqrt{c^2}=0$$ _________________ Math Expert Joined: 02 Aug 2009 Posts: 8602 Re: If x is a prime number, what is the value of x + c? [#permalink] ### Show Tags 19 Dec 2018, 08:27 7 5 If x is a prime number, what is the value of x + c? (1) $$|a + b + c - 2x| + \\sqrt{(x^2 - a - b - c)^2} = - |c|$$ The RHS - |c| can only be possible as 0, because the LHS has SUM of a modulus and a square root, both of which are positive. so it will be positive and -|c| cannot be positive Now, the LHS, $$|a + b + c - 2x| + \\sqrt{(x^2 - a - b - c)^2} =0$$ , but since both terms are positive, they should be 0 themselves too.. so a + b + c - 2x=0......a+b+c=2x...(i) Also $$\\sqrt{(x^2 - a - b - c)^2} =0......=>x^2 - a - b - c=0.........x^2=a+b+c$$...(ii) From (i) and (ii), we get $$a+b+c=2x=x^2........x^2-x=0....x(x-2)=0$$, so x is 0 or 2 But x is a prime number, so x+c=2+0=2 sufficient (2) $$\\sqrt{(a + b + c - 1 + x)^2} + |\\frac{1}{x-1}- a - b - c| +", "Thread: Show that the greatest common divisor of a and b is not greater than sqrt(a + b) 1. Show that the greatest common divisor of a and b is not greater than sqrt(a + b) Hi i need help with a problem i've been stuck on. The natural numbers a and b are such that (a + 1)/b + (b +1)/a is an integer. Show that the greatest common divisor of a and b is not greater than sqrt(a + b). I'm really stuck on this and I keep going round in circles. cannot prove anything. any suggestions appreciated. Thank you. 2. Originally Posted by ahm0605 The natural numbers a and b are such that (a + 1)/b + (b +1)/a is an integer. Show that the greatest common divisor of a and b is not greater than sqrt(a + b). Put it over a common denominator: $\\frac{a+1}b+\\frac{b+1}a = \\frac{a^2+a+b^2+b}{ab}$ is an integer, so $a^2+a+b^2+b = kab$ for some integer k. Then $a+b = kab-a^2-b^2$, and the right side of that is divisible by $d^2$, where $d = \\text{gcd}(a,b)$. Now you're practically there. 3. since $\\frac{a+1}{b} + \\frac{b+1}{a}$ is an integer that means $b \\mid a+1 \\;\\;and\\;\\; a \\mid b+1$ but g.c.d(a,b) divides a and b that's obvious and g.c.d(a , a+1 ) = 1 let g.c.d(a,b)", "+ \\frac{1}{n}.$ If $\\sum_{n=4}^{\\infty} \\frac{1}{nH_nH_{n-1}} = \\frac{a}{b}$ for relatively prime positive integers $$a$$ and $$b$$, find $$a+b$$. This problem is shared by Sandeep S. ×", "Simplify the following expression giving your answer in the form a +b\\sqrt c where a, b and c are integers. (sqrt 5 + 3)/(sqrt 5 - 2)", "# How do you solve sqrt11/9? Aug 20, 2016 There in nothing to solve. However, $\\frac{\\sqrt{11}}{9} \\cong 0.36851$ #### Explanation: In this question there is no variable solve for. Since 11 is prime, $\\sqrt{11}$ cannot be simplified further. However, the radical expression can be approximated as a decimal as follows: $\\frac{\\sqrt{11}}{9} \\cong \\frac{3.31662}{9} \\cong 0.36851$", "are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink] ### Show Tags 22 Nov 2017, 05:49 Bunuel wrote: If a and b are integers and $$(\\sqrt[3]{a}*\\sqrt{b})^6 = 500$$, then a + b could equal A. 2 B. 3 C. 4 D. 5 E. 6 Before people complain that the answer they have found is not in the suggestions of the problem prompt, they need to remember that an integer can be either positive and negative. This was, in my opinion, the main trap with this problem. We have: $$(\\sqrt[3]{a}*\\sqrt{b})^6 = 500$$ (A) This equality is too complex for my tastes, let's simplify it while keeping in mind that: 1/ $$\\sqrt{x} = x^\\frac{1}{2}$$ 2/ $$\\sqrt[y]{x} = x^\\frac{1}{y}$$ 3/ $$(a*b)^c = a^c * b^c$$ Therefore, using the first 3 properties above, A becomes: $$(\\sqrt[3]{a}*\\sqrt{b})^6 = (a^\\frac{1}{3} * b^\\frac{1}{2})^6 = a^2 * b^3 = 500$$ (B) If we decompose 500 into its prime components, we get $$500 = 2^2*5^3$$ Thus B can be written as: $$a^2 * b^3 = 2^2*5^3$$ At this stage, we're tempted to say that $$a = 2$$ and $$b = 5$$ and thus complain that $$a+b = 7$$ isn't in the choices. Or, we can remember that even exponents remove the negative sign from negative integers and deduce that the only solution possible", "Given that $$x, y,$$ and $$z$$ are real numbers that satisfy: $$x = \\sqrt{y^2-\\frac{1}{16}}+\\sqrt{z^2-\\frac{1}{16}}$$ $$y = \\sqrt{z^2-\\frac{1}{25}}+\\sqrt{x^2-\\frac{1}{25}}$$ $$z = \\sqrt{x^2 - \\frac 1{36}}+\\sqrt{y^2-\\frac 1{36}}$$ and that $$x+y+z = \\frac{m}{\\sqrt{n}},$$ where $$m$$ and $$n$$ are positive integers and $$n$$ is not divisible by the square of any prime, find $$m+n.$$ (第二十四届AIME2 2006 第15题)" ]

[ "= \\frac {1 - \\sqrt {2} + \\sqrt {5} - \\sqrt {10}}{1^2 - (\\sqrt {2})^2} \\\\ & = \\frac {1 - \\sqrt {2} + \\sqrt {5} - \\sqrt {10}}{-1} \\end{align}", "\\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}& \\frac{1}{\\sqrt{2}} \\end{bmatrix}. \\\\ \\end{align*}", "r)} \\\\ &= \\sqrt{\\frac{3}{16}s^4(n, r)} \\\\ &= \\frac{\\sqrt{3}}{4}s^2(n, r) \\\\ &= \\frac{\\sqrt{3}}{4} [2r(n + \\sqrt{3} - 1)]^2 \\\\ &= \\frac{\\sqrt{3}}{4} 4r^2 (n + \\sqrt{3} - 1)^2 \\\\ &= \\sqrt{3}r^2 (n + \\sqrt{3} - 1)^2. \\end{aligned}", "+ 44} \\\\ & \\frac {1 + \\sqrt {5}}{1 + \\sqrt {2}} \\end{align} The solution here is to use the difference of two squares (see Section 4.3.3 if unsure on what this is). \\begin{align} \\frac {1 + \\sqrt {5}}{1 + \\sqrt {2}} & = \\frac {1 + \\sqrt {5}}{1 + \\sqrt {2}} \\frac {1 - \\sqrt {2}}{1 - \\sqrt {2}} \\\\ & = \\frac {(1+\\sqrt {5})(1-\\sqrt {2})}{(1+\\sqrt {2})(1-\\sqrt {2})} \\\\ & = \\frac {1 - \\sqrt {2} + \\sqrt {5} - \\sqrt {10}}{1^2 - (\\sqrt {2})^2} \\\\ & = \\frac {1 - \\sqrt {2} + \\sqrt {5} - \\sqrt {10}}{-1} \\end{align}", "# An Integral from the RMM ### Solution 1 First note that \\displaystyle\\begin{align} I&=\\int_0^1\\frac{\\sqrt{1-x}+\\sqrt{x}}{1+\\sqrt{2x}}f(x)dx\\\\ &=\\int_0^1\\frac{\\sqrt{x}+\\sqrt{1-x}}{1+\\sqrt{2(1-x)}}f(1-x)dx. \\end{align} Consider the function \\displaystyle\\begin{align} g(x)&=\\frac{\\sqrt{1-x}+\\sqrt{x}}{1+\\sqrt{2x}}f(x)+\\frac{\\sqrt{x}+\\sqrt{1-x}}{1+\\sqrt{2(1-x)}}f(1-x)\\\\ &=\\left(\\frac{\\sqrt{1-x}+\\sqrt{x}}{1+\\sqrt{2x}}+\\frac{\\sqrt{x}+\\sqrt{1-x}}{1+\\sqrt{2(1-x)}}\\right)f(x)\\\\ &=(\\sqrt{x}+\\sqrt{1-x})\\left(\\frac{\\sqrt{2(1-x)}-\\sqrt{2x}}{1-2x}\\right)f(x)\\\\ &=(\\sqrt{x}+\\sqrt{1-x})\\left(\\frac{1-\\sqrt{2x}}{1-2x}-\\frac{1-\\sqrt{2(1-x)}}{1-2x}\\right)f(x)\\\\ &=\\sqrt{2}\\frac{(\\sqrt{1-x}+\\sqrt{x})(\\sqrt{1-x}-\\sqrt{x})}{1-2x}f(x)\\\\ &=\\sqrt{2}\\frac{(1-x)-x}{1-2x}f(x)\\\\ &=\\sqrt{2}f(x). \\end{align} due to the condition $f(x)=f(1-x).\\,$ If we now note that $\\displaystyle \\int_0^1g(x)dx=2I\\,$ we'll conclude that $\\displaystyle 2I=\\sqrt{2}\\int_0^1f(x)dx.$ ### Solution 2 Making the substitutions, \\displaystyle \\begin{align} &y=x-\\frac{1}{2}, \\\\ &g(y)=f(y+1/2), \\\\ &h(y)=\\frac{\\sqrt{1-2y}+\\sqrt{1+2y}}{1+\\sqrt{1+2y}}-1, \\end{align} we have \\displaystyle \\begin{align} &\\int_{0}^{1}\\left[\\frac{\\sqrt{1-x}+\\sqrt{x}}{1+\\sqrt{2x}}-\\frac{1}{\\sqrt{2}}\\right]f(x)dx \\\\ &\\qquad\\qquad =\\frac{1}{\\sqrt{2}}\\int_{-1/2}^{1/2}\\left[\\frac{\\sqrt{1-2y}+\\sqrt{1+2y}}{1+\\sqrt{1+2y}}-1\\right]g(y)dy\\\\ &\\qquad\\qquad =\\frac{1}{\\sqrt{2}}\\int_{-1/2}^{1/2} h(y)g(y)dy. \\end{align} To prove that the integral is $0$, we prove that $g(y)$ is even and $h(y)$ is odd. $\\displaystyle g(y)=f(y+1/2)=f(1-(y+1/2))=f(1/2-y)=g(-y)$. $h(0)=0$. For $y\\neq 0$, \\displaystyle \\begin{align} h(y)&=\\frac{\\sqrt{1-2y}+\\sqrt{1+2y}}{1+\\sqrt{1+2y}}-1 \\\\ &=\\frac{\\sqrt{1-2y}-1}{\\sqrt{1+2y}+1} \\\\ &=\\frac{(\\sqrt{1-2y}-1)(\\sqrt{1+2y}-1)}{2y}. \\end{align} Thus, $h(-y)=-h(y)$. ### Solution 3 Let $\\displaystyle g(x)=\\frac{\\sqrt{1-x}+\\sqrt{x}}{\\sqrt{2 x}+1}$. We note that $g(x)+g(1-x)=\\sqrt{2}$. Now let $u=1-x$ and by substitution: $\\displaystyle \\int_0^1 g(x) f(x)\\, dx =\\int_0^1 g (1-u) f (1-u) \\, du=\\int_0^1 g (1-u) f (u) \\, du$ We have \\displaystyle\\begin{align}\\int_0^1 f(x) (g(1-x)+g(x)) \\, dx&=2 \\int_0^1 f(x) g(x) \\, dx\\\\ &=\\sqrt{2} \\int_0^1 f(x) \\, dx. \\end{align} ### Remark On an earlier occasion we noticed that the integrals in the form $\\displaystyle \\int_0^af(x)dx\\,$ where $f(x)+f(a-x)=1,\\,$ are always equal to $\\displaystyle \\frac{a}{2}.\\,$ The present problem suggests a generalization: Assume $f:\\,[0,a]\\to\\mathbb{R}\\,$ is integrable and satisfies $f(x)=g(x)h(x)\\,$ where $g(x)+g(1-x)=1\\,$ whereas $h(x)=h(1-x).\\,$ Then $\\displaystyle \\int_0^af(x)=\\frac{a}{2}\\int_0^ah(x)dx.$ The above solution is easily adaptable to the general case. Adapting an example from the earlier page,", "\\sqrt{\\frac{\\pi}{a}}, \\\\ &= 4 \\pi \\frac{3}{2 m^2 \\beta^2} \\sqrt{\\frac{2\\pi}{m \\beta}}, \\\\ &= \\frac{6 \\pi}{m^2 \\beta^2} \\sqrt{\\frac{2\\pi}{m \\beta}}, \\end{align} assuming I didn't make any mistakes.", "given values)}\\\\ &= \\frac{|4 - 1|}{\\sqrt{4 + 1}} \\\\ &= \\frac{|3|}{\\sqrt{5}} \\\\ D \\ &\\boxed{= \\frac{3}{\\sqrt{5}} } \\end{align*}", "& = \\frac{4\\sqrt{2}-12}{14} - \\sqrt{2} \\\\ & = \\frac{2\\sqrt{2}-6}{7} - \\sqrt{2} \\\\ & = \\frac{2\\sqrt{2}-6 -7 \\sqrt{2}}{7}\\\\ & = \\frac{-5\\sqrt{2} -6 }{7} \\end{align}", "factor of }3 \\\\ & \\text{=}\\displaystyle \\frac{2\\sqrt{7}+7}{-7} \\end{align} 3. . \\scriptsize \\begin{align} & \\displaystyle \\frac{4}{2\\sqrt{3}-3\\sqrt{2}}\\times \\displaystyle \\frac{2\\sqrt{3}+3\\sqrt{2}}{2\\sqrt{3}+3\\sqrt{2}}=\\displaystyle \\frac{4\\left( 2\\sqrt{3}+3\\sqrt{2} \\right)}{{{\\left( 2\\sqrt{3} \\right)}^{2}}-{{\\left( 3\\sqrt{2} \\right)}^{2}}} \\\\ & =\\displaystyle \\frac{4\\left( 2\\sqrt{3}+3\\sqrt{2} \\right)}{(4\\cdot 3)-(9\\cdot 2)} \\\\ & =\\displaystyle \\frac{4\\left( 2\\sqrt{3}+3\\sqrt{2} \\right)}{-6} & \\text{ Divide by common factor of }2 \\\\ & =\\displaystyle \\frac{2\\left( 2\\sqrt{3}+3\\sqrt{2} \\right)}{-3} \\\\ & =\\displaystyle \\frac{4\\sqrt{3}+6\\sqrt{2}}{-3} \\end{align} 3. . To prove that $\\scriptsize \\displaystyle \\frac{\\sqrt{125}+\\sqrt{80}-\\sqrt{20}}{\\sqrt{180}+\\sqrt{5}}=1$:\\scriptsize \\begin{align} & \\displaystyle \\frac{\\sqrt{125}+\\sqrt{80}-\\sqrt{20}}{\\sqrt{180}+\\sqrt{5}}=\\displaystyle \\frac{\\sqrt{25\\cdot 5}+\\sqrt{16\\cdot 5}-\\sqrt{4\\cdot 5}}{\\sqrt{36\\cdot 5}+\\sqrt{5}} \\\\ & =\\displaystyle \\frac{5\\sqrt{5}+4\\sqrt{5}-2\\sqrt{5}}{6\\sqrt{5}+\\sqrt{5}} \\\\ & =\\displaystyle \\frac{7\\sqrt{5}}{7\\sqrt{5}} \\\\ & =1 \\\\ & \\therefore \\displaystyle \\frac{\\sqrt{125}+\\sqrt{80}-\\sqrt{20}}{\\sqrt{180}+\\sqrt{5}}=1 \\end{align} Back to Unit 5: Assessment", "It \\displaystyle \\begin{align*} \\sqrt{(x + 1)^2 + y^2} - \\sqrt{(x - 2)^2 + y^2} &= 1 \\\\ \\left(\\sqrt{(x + 1)^2 + y^2} - \\sqrt{(x - 2)^2 + y^2}\\right)^2 &= 1^2 \\\\ (x + 1)^2 + y^2 - 2\\sqrt{(x + 1)^2 + y^2}\\sqrt{(x - 2)^2 + y^2} + (x - 2)^2 + y^2 &= 1 \\\\ (x + 1)^2 + (x - 2)^2 + 2y^2 - 1 &= 2\\sqrt{(x + 1)^2 + y^2}\\sqrt{(x - 2)^2 + y^2} \\\\ x^2 + 2x + 1 + x^2 - 4x + 4 + 2y^2 - 1 &= 2\\sqrt{(x+1)^2 + y^2}\\sqrt{(x-2)^2 + y^2} \\\\ 2x^2 - 2x + 4 + 2y^2 &= 2\\sqrt{(x + 1)^2 + y^2}\\sqrt{(x - 2)^2 + y^2} \\\\ x^2 - x + 2 + y^2 &= \\sqrt{(x + 1)^2 + y^2}\\sqrt{(x - 2)^2 + y^2} \\\\ \\left(x^2 - x + 2 + y^2\\right)^2 &= \\left(\\sqrt{(x + 1)^2 + y^2}\\sqrt{(x - 2)^2 + y^2}\\right)^2 \\\\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= \\left[(x + 1)^2 + y^2\\right]\\left[(x - 2)^2 + y^2\\right] \\\\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= x^4-2x^3+2x^2y^2-3x^2-2xy^2+4x+y^4+5y^2+4 \\\\ 5x^2 - 4x + 4y^2 &= -3x^2 + 4x + 5y^2\\end{align*} \\displaystyle \\begin{align*} 8x^2 - 8x - y^2 &= 0 \\\\ 8\\left(x^2 - x\\right) - y^2 &= 0 \\\\ 8\\left[x^2 - x + \\left(-\\frac{1}{2}\\right)^2 - \\left(-\\frac{1}{2}\\right)^2\\right] - y^2 &= 0 \\\\ 8\\left[\\left(x - \\frac{1}{2}\\right)^2 - \\frac{1}{4}\\right] - y^2 &=", "we have \\begin{align} g(g(x)) &= \\left(x^\\sqrt{2} + C\\right)^\\sqrt{2} + C \\\\ &= x^2 \\left(1 + Cx^{-\\sqrt{2}}\\right)^\\sqrt{2} + C \\\\ &\\approx x^2 \\left( 1 + C\\sqrt{2}x^{-\\sqrt{2}} \\right) + C \\qquad \\text{(binomial theorem)} \\\\ &= x^2 + C\\sqrt{2}x^{2-\\sqrt{2}} + C. \\end{align} It looks like we can get what we want if we replace $C$ with $x^{\\sqrt{2}-2}/\\sqrt{2}$. Indeed, if $$h(x) = x^{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} x^{\\sqrt{2}-2} \\tag{*}$$ then \\begin{align} h(h(x)) &= \\left( x^{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} x^{\\sqrt{2}-2} \\right)^\\sqrt{2} + \\frac{1}{\\sqrt{2}} \\left( x^{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} x^{\\sqrt{2}-2} \\right)^{\\sqrt{2}-2} \\\\ &= \\left( x^{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} x^{\\sqrt{2}-2} \\right)^\\sqrt{2} + O\\!\\left(x^{2(1-\\sqrt{2})}\\right) \\\\ &= x^2 \\left( 1 + \\frac{1}{\\sqrt{2}} x^{-2} \\right)^\\sqrt{2} + O\\!\\left(x^{2(1-\\sqrt{2})}\\right) \\\\ &= x^2 \\left[ 1 + x^{-2} + O\\!\\left(x^{-4}\\right) \\right] + O\\!\\left(x^{2(1-\\sqrt{2})}\\right) \\\\ &= x^2 + 1 + O\\!\\left(x^{2(1-\\sqrt{2})}\\right) \\end{align} as $x \\to +\\infty$. Though neither elementary or of closed form, it is not difficult to numerically compute such a function. Notice that we have $$h(x)^2+1=h(h(h(x)))=h(x^2+1)$$ $$h(x)=\\sqrt{h(x^2+1)-1}$$ By iterating this using $$h_0(x)=|x|^{\\sqrt2}$$ $$h_{n+1}(x)=\\sqrt{h_n(x^2+1)-1}$$ we converge to a functional square root of $$x^2+1$$, as demonstrated in this graph. Intuitively, $$h_0$$ is sufficiently accurate for large $$x$$ and $$h_{n+1}$$ approximates smaller arguments $$(x)$$ in terms of larger arguments $$(x^2+1)$$ of $$h_n$$.", "with $\\frac{\\sqrt{x-1}+\\sqrt{x-2}}{\\sqrt{x-1}+\\sqrt{x-2}}$ aswell. Then it should be easy to evaluate. \\begin{align} \\frac{\\sqrt{x-1} - \\sqrt{x-2}}{\\sqrt{x-2} - \\sqrt{x-3}} &=\\frac{\\sqrt{x-1} - \\sqrt{x-2}}{\\sqrt{x-2} - \\sqrt{x-3}}\\,\\, \\frac{\\sqrt{x-1} + \\sqrt{x-2}}{\\sqrt{x-1} + \\sqrt{x-2}}\\,\\, \\frac{\\sqrt{x-2} + \\sqrt{x-3}}{\\sqrt{x-2} + \\sqrt{x-3}}\\\\ \\ \\\\ &=\\frac{x-1-(x-2)}{x-2-(x-3)}\\,\\frac{\\sqrt{x-2} + \\sqrt{x-3}}{\\sqrt{x-1} + \\sqrt{x-2}}=\\frac{\\sqrt{x-2} + \\sqrt{x-3}}{\\sqrt{x-1} + \\sqrt{x-2}}\\\\ \\ \\\\ &=\\frac{\\sqrt{1-2/x}+\\sqrt{1-3/x}}{\\sqrt{1-1/x}+\\sqrt{1-2/x}}\\to\\frac22=1 \\end{align}", "{a} -1 + \\sqrt {a} }{8(1 -a)} +\\frac {2\\sqrt {a}}{8(1- \\sqrt {a})} \\\\&= \\frac {2\\sqrt {a} + 2\\sqrt {a}}{8 (1 -a)} \\\\& = \\frac {4\\sqrt {a}}{8(1- \\sqrt {a})} \\\\&= \\frac {\\sqrt {a}}{2(1 -a)} \\\\ \\end{align*} Solution: \\begin{align*} &= \\frac {1}{x- 1} - \\frac {3}{x} + \\frac {3} {x+1}-\\frac{1}{x+2} = \\frac{x-3(x-1)}{x(x-1)}+ \\frac{3}{x+1}-\\frac{1}{x+2}\\\\&= \\frac{x-3x+3}{x(x-1)}+\\frac{3}{x+1}-\\frac{1}{x+2} =\\frac{(3-2x)(x+1)+3x(x-1)}{x(x-1)(x+1)}-\\frac{1}{x+2}\\\\&=\\frac{-2x^2-2x+3x+3+3x^2-3x}{x(x^2-1)}-\\frac{1}{x+2}\\\\&=\\frac{x^2-2x+3}{x(x^2-1)}-\\frac{1}{x+2}= \\frac{(x^2-2x+3)(x+2)-x(x^2-1)}{x(x^2-1)(x+2)} \\\\&=\\frac{x^3+2x^2-2x^2-4x+3x+6-x^3+x}{x(x^2-1)(x+2)}=\\frac{6}{x(x^2-1)(x+2)}\\: \\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac{1}{(x-1)(x-2)}-\\frac{3}{(x-2)(x-3)}+\\frac{2}{(x-3)(x-1)}\\\\&=\\frac{x-3-3(x-1)+2(x-2)}{(x-1)(x-2)(x-3)}=\\frac{x-3-3x+3+2x-4}{(x-1)(x-2)(x-3)}\\\\&=\\frac{-4}{(x-1)(x-2)(x-3)}=\\frac{4}{(x-1)(x-2)(3-x)}\\: \\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac{1}{a- 3} - \\frac {1}{a-1} + \\frac {1} {a+3}-\\frac{1}{a+1}\\\\&=\\frac{1}{a-3}+\\frac{1}{1+3}-[\\frac{1}{a-1}+\\frac{1}{a+1}]=\\frac{a+3+a-3}{(a-3)(a+3)}-\\frac{a+1+a-1}{(a+1)(a-1)}\\\\&=\\frac{2a}{(a-3)(a+3)}-\\frac{2a}{(a+1)(a-1)}=\\frac{2a}{a^2-9}-\\frac{2a}{a^2-1}\\\\&=\\frac{2a(a^2-1)-2a(a^2-9)}{(a^2-9)(a^2-1)}=\\frac{2a^3-2a-2a^3+18a}{(a^2-9)(a^2-1)}=\\frac{16a}{(a^2-9)(a^2-1)} \\: \\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac{1}{x- 1} - \\frac {2}{2x+1} + \\frac {1} {x+1}-\\frac{2}{2x-1}=\\frac{x+1+x-1}{(x-1)(x+1)}-[\\frac{2}{2x+1}+\\frac{2}{2x-1}] \\\\&= \\frac{2x}{(x-1)(x+1)}-[\\frac{2(2x-1)+2(2x+1)}{(2x+1)(2x-1)}] =\\frac{2x}{(x-1)(x+1)}-\\frac{(4x-2+4x+2)}{(2x+1)(2x-1)}\\\\&=\\frac{2x}{(x-1)(x+1)}-\\frac{8x}{(2x+1)(2x-1)} = \\frac{2x}{x^2-1} - \\frac{8x}{4x^2-1}\\\\&=\\frac{2x(4x^2-1)-8x(x^2-1)}{(x^2-1)(4x^2-1)}\\\\&=\\frac{8x^3-2x-8x^3+8x}{(x^2-1)(4x^2-1)}\\\\&=\\frac{6x}{(x^2-1)(4x^2-1)}\\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac{1}{x- 5} - \\frac {1}{x-3} + \\frac {1} {x+5}-\\frac{1}{x+3}=\\frac{x+5+x-5}{(x-5)(x+5)}-[\\frac{1}{x-3}+\\frac{1}{x+3}]\\\\&=\\frac{2x}{x^2-25}-\\frac{x+3+x-3}{(x-3)(x+3)}=\\frac{2x}{x^2-25}-\\frac{2x}{x^2-9}=\\frac{2x(x^2-9)-2x(x^2-25)}{(x^2-25)(x^2-9)}\\\\&=\\frac{2x^3-18x-2x^3+50x}{(x^2-25)(x^2-9)}= \\frac{32x}{(x^2-25) (x^2-9)} \\: \\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac {1}{x+8} - \\frac {1}{x-4} + \\frac {1} {x-8}-\\frac{1}{x+4}=\\frac{1}{x+8}+\\frac{1}{x-8}-\\frac{1}{x-4}-\\frac{1}{x+4}\\\\&=\\frac{x-8+x+8}{(x+8)(x-8)}-[\\frac{1}{x-4}+\\frac{1}{x+4}]= \\frac{2x}{(x+8)(x-8)}-[\\frac{x+4+x-4}{(x-4)(x+4)}\\\\&=\\frac{2x}{x^2-8^2}-\\frac{2x}{x^2-4^2}=\\frac{2x}{x^2-64}-\\frac{2x}{x^2-16}=\\frac{2x(x^2-16)-2x(x^2-64)}{(x^2-16)(x^2-64)}\\\\&=\\frac{2x^3-32x-2x^3+128x}{(x^2-16)(x^2-64)}=\\frac{96x}{(x^2-16)(x^2-64)}\\:_\\text {Ans.} \\end{align*} Solution: \\begin{align*} &= \\frac{2}{a^2-2a-24} + \\frac {2}{a^2-3a-18} + \\frac {1} {a^2+7a+12}\\\\&=\\frac{2}{a^2-6a+4a-24}+\\frac{2}{a^2-6a+3a-18}-\\frac{1}{a^2+3a+4a+12}\\\\&=\\frac{2}{a(a-6)+4(a-6)}+\\frac{2}{a(a-6)+3(a-6)}-\\frac{1}{a(a+3)+4(a+3)}\\\\&=\\frac{2}{(a-6)(a+4)}+\\frac{2}{(a+6)(a+3)}-\\frac{1}{(a+3)(a+4)}\\\\&=\\frac{2(a+3)+2(a+4)-1(a-6)}{(a+3)(a+4)(a-6)}\\\\&=\\frac{2a+6+2a+8-a+6}{(a+3)(a+4)(a-6)}\\\\&=\\frac{3a+20}{(a+3)(a+4)(a-6)} \\: \\:_\\text {Ans.} \\end{align*} 0% irrational rational decimal none • ### If the product of two surds is a rational numbers, each of them is called the _____ factor of the other. fraction all decimal rationalising dividing subtracting multiplying 2(sqrt{3}) 2(sqrt{5}) 2(sqrt{7}) 2(sqrt{1}) • ### (sqrt{5}) - (sqrt{3}) is called the conjugate of _____ or vice versa. both (sqrt{5}) + (sqrt{3}) (sqrt{5}) - (sqrt{3}) none • ## You scored /5 Forum Time", "which doesn't seem as the correct method\": why do you say so ? – user65203 Mar 21 at 18:09 @Bernard's suggested substitution $$t=\\tan x$$ gives \\begin{align} & \\int\\left(1-\\frac{t^4}{(t^2\\sqrt{2}+1)^2-2(\\sqrt{2}-1)t^2}\\right) dt \\\\ = {} & t-\\int\\frac{t^4}{(t^2\\sqrt{2}+ct+1)(t^2\\sqrt{2}-ct+1)} \\, dt\\end{align} with $$c:=\\sqrt{2\\sqrt{2}-1}$$. You can do the rest with partial fractions. Following your substitution $$t= \\tan x$$, integrate the resulting as follows \\begin{align} &\\int \\frac1{1+\\sin^4x}dx\\\\ =&\\int \\:\\frac{t^2+1}{2t^4+2t^2+1}dt =\\int \\frac{1+\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt\\\\ =&\\frac{\\sqrt2+1}{2\\sqrt2} \\int \\frac{\\sqrt2+\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt -\\frac{\\sqrt2-1}{2\\sqrt2} \\int \\frac{\\sqrt2-\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt\\\\ =&\\frac{\\sqrt2+1}{2\\sqrt2} \\int \\frac{d(\\sqrt2t-\\frac1{t})}{(\\sqrt2t-\\frac1t)^2 + 2(\\sqrt2+1)}dt -\\frac{\\sqrt2-1}{2\\sqrt2} \\int \\frac{d(\\sqrt2t+\\frac1{t})}{(\\sqrt2t+\\frac1t)^2 -2(\\sqrt2-1)} dt\\\\ =&\\frac{\\sqrt{\\sqrt2+1}}4\\tan^{-1} \\frac{\\sqrt2t-\\frac1{t}}{\\sqrt{2(\\sqrt2+1)}} +\\frac{\\sqrt{\\sqrt2-1}}4\\coth^{-1} \\frac{\\sqrt2t+\\frac1{t}}{\\sqrt{2(\\sqrt2-1)}}+C \\end{align}", "C_1 &= 2 \\\\ (\\sqrt{3} - 1)C_1 &= 2 \\\\ C_1 &= \\frac{2}{(\\sqrt{3} - 1)} \\\\ C_2 &= -\\frac{2}{(\\sqrt{3} - 1)} \\\\ \\end{align*}", "-\\frac{1}{2\\sqrt2} & 0 & -2 & \\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} \\\\ -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & 0 & 0 & 0 & 0 \\\\ -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\ 0 & -2 & 0 & 0 & 0 & 2 & 0 & 0\\\\ -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & 0 & 0 & 0 & 0 & \\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} \\\\ \\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & 0 & 0 & 0 & 0 & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} \\\\ \\end{bmatrix} \\begin{matrix} \\\\ (1)\\\\ \\\\ (2)\\\\ \\\\ (3)\\\\ \\\\ (4)\\\\ \\\\ \\end{matrix} \\\\ \\begin{matrix} (1)& & &(2)& & &(3)& & &(4) \\end{matrix} \\end{matrix}$ $d=\\begin{bmatrix} u_{1x}\\\\ u_{1y}\\\\ u_{2x}\\\\ 0\\\\ 0\\\\ 0\\\\ 0\\\\ 0 \\end{bmatrix}, f=1000\\begin{bmatrix} 1\\\\ 0\\\\ 0\\\\ 0\\\\ 0\\\\ 0\\\\ 0\\\\ 0 \\end{bmatrix}, r=\\begin{bmatrix} 0\\\\ 0\\\\ 0\\\\ r_{2y}\\\\ r_{3x}\\\\ r_{3y}\\\\ r_{4x}\\\\ r_{4y} \\end{bmatrix}$ $\\bold {b)}$ $\\implies 10^9\\begin{bmatrix} \\frac{1}{\\sqrt2}& 0 & -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & 0 & 0 & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2}\\\\ 0 & 2+\\frac{1}{\\sqrt2} & -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & 0 & -2 & \\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} \\\\ -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & 0 & 0 & 0 & 0 \\\\ -\\frac{1}{2\\sqrt2} & -\\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & \\frac{1}{2\\sqrt2} & 0 & 0", "$$\\dfrac{\\sqrt{3} + \\sqrt{7}}{\\sqrt{2}}$$ 7. $$\\dfrac{3\\sqrt{p} - 4}{\\sqrt{p}}$$ 8. $$\\dfrac{t-4}{\\sqrt{t} + 2}$$ 9. $$\\left( 1 + \\sqrt{m} \\right)^{-1}$$ 10. $$a \\left( \\sqrt{a} \\div \\sqrt{b} \\right)^{-1}$$ 1. \\begin{align*} \\dfrac{10}{\\sqrt{5}} &=\\frac{10}{\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}} \\\\ &=\\frac{10\\sqrt{5}}{5} \\\\ &= 2\\sqrt{5} \\end{align*} 2. \\begin{align*} \\dfrac{3}{\\sqrt{6}} &=\\frac{3}{\\sqrt{6}} \\times \\dfrac{\\sqrt{6}}{\\sqrt{6}} \\\\ &=\\frac{3\\sqrt{6}}{6} \\\\ &=\\frac{\\sqrt{6}}{2} \\end{align*} 3. \\begin{align*} \\dfrac{2}{\\sqrt{3}} \\div \\dfrac{\\sqrt{2}}{3} &=\\dfrac{2}{\\sqrt{3}} \\div \\dfrac{\\sqrt{2}}{3} \\\\ &=\\dfrac{2}{\\sqrt{3}} \\times \\dfrac{3}{\\sqrt{2}} \\\\ &=\\dfrac{6}{\\sqrt{6}} \\times \\dfrac{\\sqrt{6}}{\\sqrt{6}} \\\\ &=\\frac{6\\sqrt{6}}{6} \\\\ &=\\sqrt{6} \\end{align*} 4. \\begin{align*} \\dfrac{3}{\\sqrt{5}-1} &=\\dfrac{3}{\\sqrt{5}-1} \\times \\dfrac{\\sqrt{5}+1}{\\sqrt{5}+1} \\\\ &=\\dfrac{3\\sqrt{5}+3}{5-1} \\\\ &= \\dfrac{3\\sqrt{5}+3}{4} \\end{align*} 5. \\begin{align*} \\dfrac{x}{\\sqrt{y}} &=\\dfrac{x}{\\sqrt{y}}\\times \\dfrac{\\sqrt{y}}{\\sqrt{y}} \\\\ &=\\dfrac{x\\sqrt{y}}{y} \\end{align*} 6. \\begin{align*} \\dfrac{\\sqrt{3} + \\sqrt{7}}{\\sqrt{2}} &=\\dfrac{\\sqrt{3}+\\sqrt{7}}{\\sqrt{2}} \\times \\dfrac{\\sqrt{2}}{\\sqrt{2}} \\\\ &=\\dfrac{\\sqrt{3}\\sqrt{2}+\\sqrt{7}\\sqrt{2}}{2} \\\\ &=\\dfrac{\\sqrt{6}+\\sqrt{14}}{2} \\end{align*} 7. \\begin{align*} \\dfrac{3\\sqrt{p} - 4}{\\sqrt{p}} &=\\dfrac{3\\sqrt{p}-4}{\\sqrt{p}} \\times \\dfrac{\\sqrt{p}}{\\sqrt{p}} \\\\ &=\\dfrac{3\\left ( \\sqrt{p} \\right )^2-4\\left ( \\sqrt{p} \\right )}{p} \\\\ &=\\dfrac{3p-4\\sqrt{p}}{p} \\end{align*} 8. \\begin{align*} \\dfrac{t-4}{\\sqrt{t} + 2} &=\\dfrac{t-4}{\\sqrt{t}+2} \\times \\dfrac{\\sqrt{t}-2}{\\sqrt{t}-2} \\\\ &=\\dfrac{\\left ( t-4 \\right )\\left ( \\sqrt{t}-2 \\right )}{t-4} \\\\ &=\\sqrt{t}-2 \\end{align*} 9. \\begin{align*} \\left( 1 + \\sqrt{m} \\right)^{-1} &=\\frac{1}{1+\\sqrt{m}}\\times \\frac{1-\\sqrt{m}}{1-\\sqrt{m}} \\\\ &=\\frac{1-\\sqrt{m}}{1-m} \\end{align*} 10. \\begin{align*} a \\left( \\sqrt{a} \\div \\sqrt{b} \\right)^{-1} &=a\\left ( \\sqrt{a} \\times\\frac{1}{\\sqrt{b}} \\right )^{-1} \\\\ &=a\\left ( \\frac{\\sqrt{a}}{\\sqrt{b}} \\right )^{-1} \\\\ &=a\\frac{\\sqrt{b}}{\\sqrt{a}} \\times \\frac{\\sqrt{a}}{\\sqrt{a}} \\\\ &=\\frac{a\\sqrt{ab}}{a} \\\\ &=\\sqrt{ab} \\end{align*}", "\\frac{2\\sqrt{2}+2(2)}{-1} \\\\ & =-2\\sqrt{2}-4 \\end{align} 2. . \\scriptsize \\begin{align} & \\displaystyle \\frac{\\sqrt{12}}{1-\\sqrt{3x}}\\times \\displaystyle \\frac{1+\\sqrt{3x}}{1+\\sqrt{3x}}=\\displaystyle \\frac{\\sqrt{4\\cdot 3}(1+\\sqrt{3x})}{(1-\\sqrt{3x})(1+\\sqrt{3x})} && \\text{ Multiply by the conjugate} \\\\ & =\\displaystyle \\frac{2\\sqrt{3}(1-\\sqrt{3x})}{1-{{(\\sqrt{3x})}^{2}}} \\\\ & =\\displaystyle \\frac{2\\sqrt{3}-2\\sqrt{3}(\\sqrt{3x})}{1-3x} \\\\ & =\\displaystyle \\frac{2\\sqrt{3}-2{{(\\sqrt{3})}^{2}}\\sqrt{x}}{1-3x} \\\\ & =\\displaystyle \\frac{2\\sqrt{3}-6\\sqrt{x}}{1-3x} \\end{align} 3. . \\scriptsize \\begin{align} & \\displaystyle \\frac{\\sqrt{27}}{2+2\\sqrt{3}}\\times \\displaystyle \\frac{2-2\\sqrt{3}}{2-2\\sqrt{3}}=\\displaystyle \\frac{\\sqrt{9\\cdot 3}\\left( 2-2\\sqrt{3} \\right)}{\\left( 2+2\\sqrt{3} \\right)\\left( 2-2\\sqrt{3} \\right)} \\\\ & =\\displaystyle \\frac{3\\sqrt{3}\\left( 2-2\\sqrt{3} \\right)}{4-4(3)} \\\\ & =\\displaystyle \\frac{6\\sqrt{3}-6(3)}{-8} && \\text{ Divide by common factor of }2 \\\\ & =\\displaystyle \\frac{3\\sqrt{3}-9}{-4} \\end{align} Back to Exercise 5.3 ### Unit 5: Assessment 1. . 1. . \\scriptsize \\begin{align} & \\displaystyle \\frac{\\sqrt{20}+\\sqrt{5}+\\sqrt{45}}{\\sqrt{80}}=\\displaystyle \\frac{\\sqrt{4\\cdot 5}+\\sqrt{5}+\\sqrt{9\\cdot 5}}{\\sqrt{16\\cdot 5}} \\\\ & =\\displaystyle \\frac{2\\sqrt{5}+\\sqrt{5}+3\\sqrt{5}}{4\\sqrt{5}} \\\\ & =\\displaystyle \\frac{6\\sqrt{5}}{4\\sqrt{5}} \\\\ & =\\displaystyle \\frac{6}{4} \\end{align} 2. . \\scriptsize \\begin{align} & \\displaystyle \\frac{\\sqrt{20}+\\sqrt{80}+\\sqrt{125}-\\sqrt{5}}{2\\sqrt{5}+3\\sqrt{5}}=\\displaystyle \\frac{\\sqrt{4\\cdot 5}+\\sqrt{16\\cdot 5}+\\sqrt{25\\cdot 5}-\\sqrt{5}}{5\\sqrt{5}} \\\\ & =\\displaystyle \\frac{2\\sqrt{5}+4\\sqrt{5}+5\\sqrt{5}-\\sqrt{5}}{5\\sqrt{5}} \\\\ & =\\displaystyle \\frac{10\\sqrt{5}}{5\\sqrt{5}} \\\\ & =2 \\end{align} 3. . \\scriptsize \\begin{align} & \\displaystyle \\frac{3\\sqrt{12}+4\\sqrt{75}-2\\sqrt{27}}{\\sqrt{75}}=\\displaystyle \\frac{3\\sqrt{4\\cdot 3}-4\\sqrt{25\\cdot 3}-2\\sqrt{9\\cdot 3}}{\\sqrt{25\\cdot 3}} \\\\ & =\\displaystyle \\frac{6\\sqrt{3}+20\\sqrt{3}-6\\sqrt{3}}{5\\sqrt{3}} \\\\ & =\\displaystyle \\frac{20\\sqrt{3}}{5\\sqrt{3}} \\\\ & =4 \\\\ \\end{align} 2. . 1. . \\scriptsize \\begin{align} & \\displaystyle \\frac{3}{3-\\sqrt{2}}\\times \\displaystyle \\frac{3+\\sqrt{2}}{3+\\sqrt{2}}=\\displaystyle \\frac{3\\left( 3+\\sqrt{2} \\right)}{3-{{\\left( \\sqrt{2} \\right)}^{2}}} \\\\ & =\\displaystyle \\frac{9+3\\sqrt{2}}{3-2} \\\\ & =9+3\\sqrt{2} \\end{align} 2. . \\scriptsize \\begin{align} & \\displaystyle \\frac{3}{2\\sqrt{7}-7}\\times \\displaystyle \\frac{2\\sqrt{7}+7}{2\\sqrt{7}+7}=\\displaystyle \\frac{3\\left( 2\\sqrt{7}+7 \\right)}{{{\\left( 2\\sqrt{7} \\right)}^{2}}-{{\\left( 7 \\right)}^{2}}} \\\\ & =\\displaystyle \\frac{3\\left( 2\\sqrt{7}+7 \\right)}{28-49} \\\\ & =\\displaystyle \\frac{3\\left( 2\\sqrt{7}+7 \\right)}{-21} && \\text{ Divide by common", "$|A_1O_1| = 2 \\, r_1, \\, |O_1T_1| = r_1$ and hence $|A_1T_1| = \\sqrt{3}\\, r_1$. Similarly, for the other vertices of the triangle, the common tangents from the given vertex and the corresponding closest circle have lengths respectively $\\sqrt{3}\\, r_2$ and $\\sqrt{3}\\, r_3$. Thus the length of the three edges of the equilateral triangle (they are all equal and of length $1$) can be expressed as \\begin{align} \\sqrt{3}(r_1 + r_2) + 2 \\sqrt{r_1\\, r_2} &= 1\\\\ \\sqrt{3}(r_2 + r_3) + 2 \\sqrt{r_2\\, r_3} &= 1\\\\ \\sqrt{3}(r_3 + r_1) + 2 \\sqrt{r_3\\, r_1} &= 1 \\end{align} To solve this system, set $r_2 = x_2^2 \\, r_1$ and $r_3 = x_3^2 \\, r_1$ and plug in the equations these two expression to arrive at \\begin{align} \\sqrt{3}(1 + x^2_2) + 2 \\, x_2 &= \\frac{1}{r_1}\\\\ \\sqrt{3}(x^2_2 + x^2_3) + 2 \\, {x_2\\, x_3} &= \\frac{1}{r_1}\\\\ \\sqrt{3}(1 + x^2_3) + 2 \\, x_3 &= \\frac{1}{r_1} \\end{align} This is a system that simultaneously finds $r_1$ and $x_2, \\, x_3$. Look at the polynomial $$f(x) = \\sqrt{3}\\, x^2 + 2 \\, x + \\sqrt{3} - \\frac{1}{r_1}$$ Then the first and the third equations from above are actually $$f(x_2) = 0 \\,\\,\\, \\text{ and } \\,\\,\\, f(x_3) = 0$$ i.e. $x_2$ and $x_3$ are two roots (different or the same) of the polynomial $f$. Consequently", "- \\frac{1}{n + 1}) \\frac{4^n}{\\sqrt{\\pi n}} \\\\ & = \\frac{4^n}{\\sqrt{\\pi}n^{3 / 2}} (1 + O(1 / n)). \\end{aligned}" ]

[ "\\boxed{\\alpha = \\sqrt{b+x} = \\frac{a^{2} + b - c}{2a}} \\end{align*} To solve $$a = \\sqrt{b + x} + \\sqrt{c + x}$$ you must square twice, since you have two square roots. First, for $$\\sqrt{b + x}$$ and $$\\sqrt{c + x}$$ to both exist, it must be $$x\\ge-b$$ and $$x\\ge-c$$, so $$x\\ge\\max(-b,-c).$$ Then: $$a - \\sqrt{b + x} = \\sqrt{c + x}.\\tag{1}$$ Squaring once we find \\begin{align} & a^2 + b \\color{red}{+ x} - 2a\\sqrt{b + x} = c \\color{red}{+ x}\\\\ & a^2 + b - c = 2a\\sqrt{b + x}\\\\ \\end{align} and squaring the second time: $$(a^2 + b - c)^2 = 4a^2(b + x),$$ so $$x=\\frac{(a^2 + b - c)^2}{4a^2} - b$$ or $$x=\\left(\\frac{a^2 + b - c}{2a}\\right)^2 - b.$$ If you rewrite $$(1)$$ as $$a - \\sqrt{c + x} = \\sqrt{b + x},$$ you will find $$x=\\frac{(a^2 + c - b)^2}{4a^2} - c$$ or $$x=\\left(\\frac{a^2 + c - b}{2a}\\right)^2 - c$$ which is the same. Here $$a$$ (not $$x$$) is a symmetric function of $$b$$ and $$c$$ so one may wonder if $$b or $$b>c$$ but it doesn't really matter.", "# Proof that $\\mathbb Z[\\sqrt{3}]$ is a Euclidean Domain Let $R_d$ be the ring defined as $R_d=\\left \\{ x+y\\omega : x,y\\in \\mathbb{Z} \\right\\}$, where $$\\omega = \\begin{cases} \\sqrt{d}, & \\text{if } \\quad d \\not \\equiv 1\\mod 4 \\\\ \\frac{1+\\sqrt{d}}{2}, & \\text{if } \\quad d\\equiv 1\\mod 4. \\end{cases}$$ It has been proven that $R_d$ is Euclidean for several positive values of $d$. Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$? Thank you. Define the norm on $$\\mathbb Z[\\sqrt 3]$$ to be $$N(a + b \\sqrt 3) = \\vert a^2 - 3 b^2 \\vert$$. Let $$\\alpha, \\beta \\in \\mathbb Z[\\sqrt 3]$$ with $$\\beta \\neq 0$$. Say $$\\alpha = a + b \\sqrt 3$$ and $$\\beta = c + d \\sqrt 3$$. Notice that \\begin{align*} \\frac\\alpha\\beta &= \\frac{a + b \\sqrt 3}{c + d \\sqrt 3} \\cdot \\frac{c - d \\sqrt 3}{c - d \\sqrt 3} \\\\ &= \\frac{ac - 3bd}{c^2 - 3d^2} + \\frac{-ad + bc}{c^2 - 3d^2} \\sqrt 3 \\\\ &= r + s\\sqrt 3 \\end{align*} where $$r = \\displaystyle \\frac{ac - 3bd}{c^2 - 3d^2}$$ and $$s = \\displaystyle \\frac{-ad + bc}{c^2 - 3d^2}$$. Let $$p$$ be the closest integer to $$r$$ and let $$q$$ be the closest integer to $$s$$. Notice that $$\\vert r - p \\vert \\leq 1/2$$ and $$\\vert", "this equation for $x$: \\begin{align}\\frac{\\sqrt x}{2}&=\\sqrt2\\\\\\sqrt x&=2\\sqrt 2\\\\\\left(\\sqrt x\\right)^2&=\\left(2\\sqrt 2\\right)^2\\\\x&=4\\cdot2=\\color{green}{\\boxed{\\color{black}{8}}}\\end{align} Therefore: $$\\frac{\\sqrt8}{2}=\\sqrt 2$$ I know, this is a terrible way. But hey, it works! ;-) Technique $\\mathbf 2$: Just like in Technique $\\mathbf 1$: We may also prove that result by solving this equation for $x$: \\begin{align}\\frac{\\sqrt 8}{2}&=\\sqrt x\\\\ \\left(\\frac{\\sqrt 8}{2}\\right)^2&=\\sqrt x^2\\\\ \\frac84&=x \\implies x=\\color{green}{\\boxed{\\color{black}2}} \\end{align} Therefore: $$\\frac{\\sqrt8}{2}=\\sqrt 2$$ Technique $\\mathbf 3$: If $a=b$ then this means $a-b=0$, and the opposite is true. So let's check for $\\sqrt 2$ and $\\sqrt 8/2$:\\begin{align}\\dfrac{\\sqrt 8}{2}-\\sqrt{2}&=\\dfrac{\\sqrt 8}{2}-\\left(\\sqrt{2} \\cdot \\dfrac{\\sqrt{4}}{2}\\right)\\quad\\to\\quad\\text{since \\sqrt 4=2}\\\\ &=\\dfrac{\\sqrt 8}{2}-\\dfrac{\\sqrt 2\\cdot\\sqrt 4}{2}\\\\ &=\\dfrac{\\sqrt{8}-\\sqrt{2}\\cdot\\sqrt{4}}{2}\\\\ &=\\dfrac{\\sqrt{8}-\\sqrt{2\\cdot4}}{2}\\\\ &=\\dfrac{\\sqrt 8-\\sqrt 8}{2}=\\dfrac{0}{2}=0 \\end{align} Therefore, we can easily conclude that: $$\\dfrac{\\sqrt 8}{2}=\\sqrt 2$$ Technique $\\mathbf 4$: If $a=b$ then it follows that $a/b=1$, and the opposite is true. So let's check for $\\sqrt 2$ and $\\sqrt 8/2$: \\begin{align}\\dfrac{\\sqrt2}{\\dfrac{\\sqrt8}{2}}&=\\sqrt{2}\\cdot\\dfrac{2}{\\sqrt{8}}=\\dfrac{\\sqrt2\\cdot\\sqrt 4}{\\sqrt8}=\\dfrac{\\sqrt{2\\cdot 4}}{\\sqrt 8}=\\dfrac{\\sqrt8}{\\sqrt8}=1 \\end{align} Therefore: $$\\dfrac{\\sqrt 8}{2}=\\sqrt 2$$ Technique $\\mathbf 5$: We will do a proof by contradiction: We suppose that $\\sqrt 8/2\\ne\\sqrt 2$, therefore there must be a difference between them which is not zero $\\sqrt 8/2-\\sqrt 2\\ne0$, that we will call $x$. So let's check if $x\\ne0$: \\begin{align}x=\\dfrac{\\sqrt 8}{2}-\\sqrt{2}&=\\dfrac{\\sqrt 8}{2}-\\left(\\sqrt{2} \\cdot \\dfrac{\\sqrt{4}}{2}\\right)\\\\ &=\\dfrac{\\sqrt 8}{2}-\\dfrac{\\sqrt 2\\cdot\\sqrt 4}{2}\\\\ &=\\dfrac{\\sqrt{8}-\\sqrt{2}\\cdot\\sqrt{4}}{2}\\\\ &=\\dfrac{\\sqrt{8}-\\sqrt{2\\cdot4}}{2}\\\\ &=\\dfrac{\\sqrt 8-\\sqrt 8}{2}=\\dfrac{0}{2}=0 \\end{align}But this is a contradiction, since $x$ must be non-zero but it turns out that it is equal to $0$. Therefore: $$\\dfrac{\\sqrt 8}{2}=\\sqrt", "circle; this must be the case, as the length of $$\\vecs u(t)$$ is 1 for all $$t$$. 2. To compute $$\\vecs u\\,'(t)$$, we use Theorem 92, writing $\\vecs u(t) = f(t)\\vecs r(t),\\quad \\text{where}\\quad f(t) = \\frac{1}{\\sqrt{t^2+(t^2-1)^2}}=\\big(t^2+(t^2-1)^2\\big)^{-1/2}.$ (We could write $\\vecs u(t) = \\langle \\frac{t}{\\sqrt{t^2+(t^2-1)^2}}, \\frac{t^2-1}{\\sqrt{t^2+(t^2-1)^2}}\\rangle$ and then take the derivative. It is a matter of preference; this latter method requires two applications of the Quotient Rule where our method uses the Product and Chain Rules.) We find $$f^\\prime (t)$$ using the Chain Rule: \\begin{align*}f^\\prime (t) &= -\\frac12\\big(t^2+(t^2-1)^2\\big)^{-3/2}\\big(2t+2(t^2-1)(2t)\\big)\\\\&= -\\frac{2t(2t^2-1)}{2\\big(\\sqrt{t^2+(t^2-1)^2}\\,\\big)^3}\\end{align*} We now find $$\\vecs u\\,'(t)$$ using part 3 of Theorem 92: \\begin{align*}\\vecs u\\,'(t) &= f^\\prime (t)\\vecs u(t) + f(t)\\vecs u\\,'(t) \\\\&= -\\frac{2t(2t^2-1)}{2\\big(\\sqrt{t^2+(t^2-1)^2}\\,\\big)^3}\\langle t,t^2-1\\rangle + \\frac{1}{\\sqrt{t^2+(t^2-1)^2}}\\langle 1,2t\\rangle.\\end{align*} This is admittedly very \"messy;'' such is usually the case when we deal with unit vectors. We can use this formula to compute $$\\vecs u\\,(-2)$$, $$\\vecs u\\,(-1)$$ and $$\\vecs u\\,(0)$$: \\begin{align*}\\vecs u\\,(-2) &= \\langle -\\frac{15}{13 \\sqrt{13}},-\\frac{10}{13\\sqrt{13}}\\rangle \\approx \\langle -0.320,-0.213\\rangle\\\\\\vecs u\\,(-1) &= \\langle 0,-2\\rangle\\\\\\vecs u\\,(0) &= \\langle 1,0\\rangle\\end{align*} Each of these is sketched in Figure $$\\PageIndex{6}$$. Note how the length of the vector gives an indication of how quickly the circle is being traced at that point. When $$t=-2$$, the circle is being drawn relatively slow; when $$t=-1$$, the circle is being traced much more quickly. It is a basic geometric fact that a line", "(cd),\\ c, d \\in \\mathbb{Z}[\\sqrt{-5}]$$ then $$c \\in (2, 1+\\sqrt{-5})$$ or $$d \\in (2, 1+\\sqrt{-5})$$. Let $$c = c_0 + c_1\\sqrt{-5}, d = d_0 + d_1\\sqrt{-5}$$, then we have $$2(2a^2 + 3b^2+2ab)\\ |\\ (c_0^2+5c_1^2)(d_0^2+5d_1^2)$$ $$2$$ is prime, which means $$(c_0^2 + 5c_1^2) \\in (2)$$ or $$(d_0^2 + 5d_1^2) \\in (2)$$ so $$(2, 1+\\sqrt{-5})$$ is prime. 2. $$(3, 1+\\sqrt{-5})$$ is prime means: $$\\forall a, b\\in \\mathbb{Z},$$ if $$(3a+(1+\\sqrt{-5})b)\\ |\\ (cd),\\ c, d \\in \\mathbb{Z}[\\sqrt{-5}]$$ then $$c \\in (3, 1+\\sqrt{-5})$$ or $$d \\in (3, 1+\\sqrt{-5})$$. Let $$c = c_0 + c_1\\sqrt{-5}, d = d_0 + d_1\\sqrt{-5}$$, then we have $$3(3a^2 + 2b^2+2ab)\\ | \\ (c_0^2+5c_1^2)(d_0^2+5d_1^2)$$ $$3$$ is prime, which means $$(c_0^2 + 5c_1^2) \\in (3)$$ or $$(d_0^2 + 5d_1^2) \\in (3))$$ so $$(3, 1+\\sqrt{-5})$$ is prime. 3. In a similar way, we could also prove that $$(3, 1-\\sqrt{-5})$$ is prime. ### Commutative Algebra some details in this chapter would be found here. #### primary concept quotient ring use an equivalence relation $$\\sim$$ on a ring $$R$$, $$I$$ is a two-sided ideal in R: $a b$ if and only if $$(a-b) \\in I$$ $$a \\mod I$$ is the equivalence class of element $$a \\in R$$, $$[a] = a+I:=\\{a+r|r\\in I\\}$$ we call $$R/I$$ a quotient ring which is $$\\{a \\mod I | a\\in R\\}$$ zero-element in $$R/I$$ is $$I$$, multiplicateive identity is", "the form of $(x-u)^2 + v$ and rewrite the numerator as $a_i (x-u) + w$, from which you can do a change of variable to integrate essentially $x/(x^2+1)$ and $1/(x^2+1)$. HINT : Use $$x^4+1=(x^2+1)^2-(\\sqrt 2x)^2=(x^2+\\sqrt 2x+1)(x^2-\\sqrt 2x+1)$$ to get \\begin{align}\\int\\frac{dx}{1+x^4}&=\\int\\frac{\\frac{1}{2\\sqrt 2}x+\\frac 12}{x^2+\\sqrt 2x+1}dx+\\int\\frac{-\\frac{1}{2\\sqrt 2}x+\\frac 12}{x^2-\\sqrt 2x+1}dx\\\\&=\\frac{1}{2\\sqrt 2}\\int\\frac{x+\\sqrt 2}{x^2+\\sqrt 2x+1}dx-\\frac{1}{2\\sqrt 2}\\int\\frac{x-\\sqrt 2}{x^2-\\sqrt 2x+1}dx\\\\&=\\frac{1}{\\sqrt 2}\\int\\frac{x+\\sqrt 2}{(\\sqrt 2x+1)^2+1}dx-\\frac{1}{\\sqrt 2}\\int\\frac{x-\\sqrt 2}{(\\sqrt 2x-1)^2+1}dx.\\end{align} Here, set $\\sqrt 2x+1=\\tan\\alpha$ for the first and set $\\sqrt 2x-1=\\tan\\beta$ for the second. • Why don't you delete your other answer that is exactly the same as the first two lines of this answer? Aug 28, 2019 at 19:33 By partial fractions, $$\\frac{1}{1+x^4} = \\frac{1}{2\\sqrt{2}}\\left(\\frac{x + \\sqrt{2}}{x^2 + \\sqrt{2}x + 1} - \\frac{x - \\sqrt{2}}{x^2 - \\sqrt{2}x + 1}\\right).$$ The rest is standard and not a great deal of fun. Complete the squares at the bottom and make the natural substitutions. Expand $\\frac{1}{1+x^{4}}$ into partial fractions. For this purpose you need to factorize the polynomial in the denominator. You can proceed by writing it as a product of four linear terms \\begin{equation*} x^{4}+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4}) \\end{equation*} where $x_{1},x_{2},x_{3},x_{4}$ are its complex roots. Since \\begin{equation*} x^{4}+1=0\\Leftrightarrow x^{4}=-1=\\cos \\pi +i\\sin \\pi \\end{equation*} the four roots are \\begin{eqnarray*} x_{1} &=&\\cos \\left( \\frac{\\pi }{4}\\right) +i\\sin \\left( \\frac{\\pi }{4} \\right) =\\frac{\\sqrt{2}}{2}(1+i) \\\\ x_{2} &=&\\cos \\left( \\frac{3\\pi }{4}\\right) +i\\sin \\left( \\frac{3\\pi }{4} \\right) =\\frac{\\sqrt{2}}{2}(-1+i) \\\\ x_{3} &=&\\overline{x}_{2}", "# Square Root of 2 as a Continued Fraction To represent the $\\sqrt{2}$ as a continued fraction we start with the obvious $\\sqrt{2}=1+(\\sqrt{2}-1)=1+\\frac{1}{1+\\sqrt{2}}$. What is worth observing is that $\\sqrt{2}$ appears on the two sides of the equality, making it possible to replace it recursively. The first step gives \\begin{align} \\sqrt{2} &= 1+\\frac{1}{1+\\sqrt{2}} \\\\ &=1 + \\frac{1}{1+(1+\\frac{1}{1+\\sqrt{2}})} \\\\ &=1 + \\frac{1}{2+\\frac{1}{1+\\sqrt{2}}}. \\end{align} Naturally, we may continue: \\begin{align} \\sqrt{2} &= 1+\\frac{1}{1+\\sqrt{2}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{1+\\sqrt{2}}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{1+(1+\\frac{1}{1+\\sqrt{2}})}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{2+\\frac{1}{1+\\sqrt{2}}}}. \\end{align} The process never ends so that $\\sqrt{2}$ comes out to be an infinite comtinued fractions: \\begin{align} \\sqrt{2} &= 1+\\frac{1}{1+\\sqrt{2}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{1+\\sqrt{2}}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{2+\\frac{1}{1+\\sqrt{2}}}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{2+\\frac{1}{1+(1+\\frac{1}{1+\\sqrt{2}})}}} \\\\ &=1 + \\frac{1}{2+\\frac{1}{2+\\frac{1}{2+\\ldots }}}, \\end{align} which is written compactly as $[1;2,2,2,\\ldots ]$. Being an inifinite continued fraction, $\\sqrt{2}$ is irrational.", "$a$ and $b$ are integers) is of the same form. So consider two such extended integers, say $a_1+b_1\\sqrt{10}$ and $a_2+b_2\\sqrt{10}$, and their product: \\begin{align}(a_1+b_1\\sqrt{10})(a_2+b_2\\sqrt{10}) &= a_1 a_2 + a_1 b_2\\sqrt{10} + b_1 a_2\\sqrt{10} + 10b_1 b_2\\\\ &=(a_1 a_2 + 10b_1 b_2) + (a_1 b_2 + b_1 a_2)\\sqrt{10}\\end{align} Upon noting that $(a_1 a_2 + 10b_1 b_2)$ and $(a_1 b_2 + b_1 a_2)$ are integers by closure, we are done. 4. We need to show that the quotient of two extended integers need not be an extended integer. That is to say, the quotient of two numbers of the form $a+b\\sqrt{10}$ (where $a$ and $b$ are integers) does not need to be of the same form. Finding one explicit counterexample will suffice. So consider the two extended integers $4+\\sqrt{10}$ and $4-\\sqrt{10}$, and their quotient: \\begin{align}\\frac{4+\\sqrt{10}}{4-\\sqrt{10}} &= \\frac{4+\\sqrt{10}}{4-\\sqrt{10}} \\cdot \\frac{4+\\sqrt{10}}{4+\\sqrt{10}} \\\\ &= \\frac{16+8\\sqrt{10}+10}{16-10} \\\\ &= \\frac{26+8\\sqrt{10}}{6} \\\\ &= \\frac{26}{6}+\\frac{8\\sqrt{10}}{6} \\\\ &= \\frac{13}{3} + \\frac{4}{3} \\sqrt{10} \\end{align} It should be clear, given this last expression, if we force things into $a+b\\sqrt{10}$ form, we are required to have $a$ and $b$ not be integers. We have then found a quotient of two extended integers, which was not an extended integer itself. As such, extended integers are not closed under division. 1. Let $a + b\\sqrt{10}$ be any extended integer, and then consider the", "# First, let’s use the way suggested by 3blue1brown. I’ll write $d\\sqrt{x}$ as $dy$. We have $$dx = 2\\sqrt{x}\\cdot dy + (dy)^2$$ Divide the equation by $dy$ and we have: $$2\\sqrt{x} + dy = \\frac{dx}{dy}$$ Therefore, $$\\frac{dy}{dx} = \\frac{1}{2\\sqrt{x} + dy}$$ Because $dy$ is approaching zero, we can safely ignore it and we have $$\\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} = \\frac{1}{2}x^{-\\frac{1}{2}}$$ ### Chain rule # Let’s say we have $f(x) = \\sqrt{x}$, $g(x) = x^2$, and $h(x) = g(f(x)) = (\\sqrt{x})^2 = x$. We have: \\begin{align} h^\\prime(x) & = 1 = g^\\prime(f(x)) \\cdot f^\\prime(x)\\\\ & = 2\\cdot f(x) \\cdot f^\\prime(x) \\\\ & = 2\\sqrt{x} \\cdot f^\\prime(x) \\\\ \\end{align} So we have $$f^\\prime(x) = \\frac{1}{2\\sqrt{x}}$$ The method by chain rule is inspired by Yifan Wei. #ML", "So I actually found the answers I was looking for. The case $$c \\ge 0$$ is already handled in the question above. For the case $$c < 0$$ we have: \\begin{align} n_{1/2} &= \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} \\\\ &\\le \\frac{|b| + \\sqrt{b^2 - 4ac}}{a} \\\\ &\\le \\frac{|b| + \\sqrt{b^2} + \\sqrt{-4ac}}{a} \\\\ &= \\frac{2|b| + \\sqrt{4a|c|}}{a} \\\\ &= \\frac{2|b| + 2\\sqrt{\\frac{a^2|c|}{a}}}{a} \\\\ &= 2\\frac{|b|}{a} + 2\\sqrt{\\frac{|c|}{a}} \\\\ &\\le 2\\max(\\{|b|/a, \\sqrt{|c|/a}\\}) \\\\ &= n_0 \\end{align} This value for $$n_0$$ includes the case $$c \\ge 0$$. If $$f$$ doesn't have any roots, we can instead choose $$n_0 = 1$$. For the constants $$c_1, c_2$$ the authors gave us the values $$c_1 = a/4$$ and $$c_2 = 7a/4$$. To check that these are correct we do the following: Since $$n \\ge 2|b|/a$$ and $$n \\ge 2\\sqrt{|c|/a}$$, we know that: \\begin{alignat}{3} && \\frac{1}{2} &\\ge &\\frac{|b|}{an} \\quad&\\text{and}\\quad &\\frac{1}{4} \\ge &\\frac{|c|}{an^2} \\\\ &&-\\frac{1}{2} &\\le -&\\frac{|b|}{an} \\quad&\\text{and}\\quad -&\\frac{1}{4} \\le -&\\frac{|c|}{an^2} \\end{alignat} This gives us \\begin{alignat}{2} &\\frac{1}{4} = 1 - \\frac{1}{2} - \\frac{1}{4} \\le 1 - \\frac{|b|}{an} - \\frac{|c|}{an^2} \\le 1 + \\frac{b}{an} + \\frac{c}{an^2} \\\\ \\text{and therefore}\\quad &\\frac{a}{4}n^2 \\le an^2 + bn + c \\end{alignat} and \\begin{alignat}{2} &1 + \\frac{b}{an} + \\frac{c}{an^2} \\le 1 + \\frac{|b|}{an} + \\frac{|c|}{an^2} \\le 1 + \\frac{1}{2} + \\frac{1}{4} = \\frac{7}{4} \\\\ \\text{and therefore}\\quad &an^2 + bn + c", "which doesn't seem as the correct method\": why do you say so ? – user65203 Mar 21 at 18:09 @Bernard's suggested substitution $$t=\\tan x$$ gives \\begin{align} & \\int\\left(1-\\frac{t^4}{(t^2\\sqrt{2}+1)^2-2(\\sqrt{2}-1)t^2}\\right) dt \\\\ = {} & t-\\int\\frac{t^4}{(t^2\\sqrt{2}+ct+1)(t^2\\sqrt{2}-ct+1)} \\, dt\\end{align} with $$c:=\\sqrt{2\\sqrt{2}-1}$$. You can do the rest with partial fractions. Following your substitution $$t= \\tan x$$, integrate the resulting as follows \\begin{align} &\\int \\frac1{1+\\sin^4x}dx\\\\ =&\\int \\:\\frac{t^2+1}{2t^4+2t^2+1}dt =\\int \\frac{1+\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt\\\\ =&\\frac{\\sqrt2+1}{2\\sqrt2} \\int \\frac{\\sqrt2+\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt -\\frac{\\sqrt2-1}{2\\sqrt2} \\int \\frac{\\sqrt2-\\frac1{t^2}}{(\\sqrt2t)^2 + \\frac1{t^2}+1}dt\\\\ =&\\frac{\\sqrt2+1}{2\\sqrt2} \\int \\frac{d(\\sqrt2t-\\frac1{t})}{(\\sqrt2t-\\frac1t)^2 + 2(\\sqrt2+1)}dt -\\frac{\\sqrt2-1}{2\\sqrt2} \\int \\frac{d(\\sqrt2t+\\frac1{t})}{(\\sqrt2t+\\frac1t)^2 -2(\\sqrt2-1)} dt\\\\ =&\\frac{\\sqrt{\\sqrt2+1}}4\\tan^{-1} \\frac{\\sqrt2t-\\frac1{t}}{\\sqrt{2(\\sqrt2+1)}} +\\frac{\\sqrt{\\sqrt2-1}}4\\coth^{-1} \\frac{\\sqrt2t+\\frac1{t}}{\\sqrt{2(\\sqrt2-1)}}+C \\end{align}", "been taken as greater or equal to $2$. \\begin{align} \\frac{1}{2n} < \\frac{1}{\\sqrt{2n - 1} + \\sqrt{2n - 3}} = \\frac{\\sqrt{2n - 1} - \\sqrt{2n - 3}}{(2n - 1) - (2n - 3)} = \\frac{\\sqrt{2n - 1} - \\sqrt{2n - 3}}{2} \\nonumber\\\\ \\hspace{-4.0in}=> \\frac{1}{n} < \\sqrt{2n - 1} - \\sqrt{2n - 3} \\end{align} Putting $n = 2, 3, \\ldots, n$ and adding we get \\begin{align} \\frac{1}{2} + \\frac{1}{3} + \\ldots + \\frac{1}{n} = (\\sqrt{3} - 1) + (\\sqrt{5} - \\sqrt{3}) + (\\sqrt{7} - \\sqrt{5}) + \\ldots + (\\sqrt{2n - 1} - \\sqrt{2n - 3}) \\nonumber\\\\ \\hspace{-4.8in}=> 1 + \\frac{1}{2} + \\frac{1}{3} + \\ldots + \\frac{1}{n} < \\sqrt{(2n - 1)} \\end{align} Problem 7: Prove that, $(n!)^{2} > n^{n}$ if $n > 2$. Solution: \\begin{align} (n!)^{2} = 1.n.2.(n-1).3.(n-2).\\ldots.r.(n - r + 1).\\ldots.n.1 \\end{align} Now, let $n > r > 1$, then $n - r > 0$. Therefore, $(n-r).r > n - r$ and as such $r.(n - r + 1) > n$. Putting $r = 2, 3,\\ldots, r$, we get $2.(n-1) > n$, $3.(n-2) > n$, ...,$r.(n-r+1) > n$. Hence, for $n > 2$ we have, \\begin{align} (n!)^{2} = 1.n.2.(n-1).3.(n-2).\\ldots.r.(n - r + 1).\\ldots.n.1 > (n.n.n\\ldots.n.\\ldots.n) = n^{n} \\end{align}", "# Prove: $\\int_{0}^{\\pi/2}{\\sqrt{1 + \\sqrt{1 + (\\tan{x})^{2/3}}}\\,dx} = \\frac{\\pi}{2} (3^{1/4} + 3^{3/4} - 2)$ I'm trying to prove $$I := \\int_{0}^{\\pi/2}{\\sqrt{1 + \\sqrt{1 + (\\tan{x})^{2/3}}}\\,dx} = \\frac{\\pi}{2} (3^{1/4} + 3^{3/4} - 2)$$ which is around $$2.5063328837$$. Using the $$u$$-substitution $$u = \\sqrt{-1 + \\sqrt{1 + (\\tan{x})^{2/3}}}$$, we have \\begin{align*} du &= \\frac{1}{6}\\frac{1}{\\sqrt{1 + (\\tan{x})^{2/3}}}\\frac{1}{\\sqrt{-1 + \\sqrt{1 + (\\tan{x})^{2/3}}}} (\\tan{x})^{-1/3} \\sec^2{x}\\,dx \\\\ &= \\frac{1}{6 u (u^2 + 1)} \\frac{1}{\\sqrt{(u^2 + 1)^2 - 1}} \\left(((u^2 + 1)^2 - 1)^3 + 1\\right) \\,dx \\\\ &= \\frac{1}{6 u^2 (u^2 + 1)} \\frac{1}{\\sqrt{u^2 + 2}} \\left(u^6 (u^2 + 2)^3 + 1\\right) \\,dx \\end{align*} and thus, $$I = \\int_{0}^{\\pi/2}{\\sqrt{1 + \\sqrt{1 + (\\tan{x})^{2/3}}}\\,dx} = 6 \\int_{0}^{\\infty}{\\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \\,du}$$ So we have the integral of a rational function, where the roots of the denominator can easily be found in closed form; hence using the method of residues, we can get an answer in closed form. Yet, I've done this and it's still not so obvious that the answer miraculously simplifies to $$\\frac{\\pi}{2} (3^{1/4} + 3^{3/4} - 2)$$. My question is, seeing that the answer is so nice, is there a more clever way to solve this integral? I've tried different substitutions, integration by parts, the Feynman trick by putting a parameter in place of the", "# How to prove that $\\frac x{\\sqrt y}+\\frac y{\\sqrt x}\\ge\\sqrt x+\\sqrt y$ I am trying to prove that $$\\frac x{\\sqrt y}+\\frac y{\\sqrt x}\\ge\\sqrt x+\\sqrt y$$ I tried some manipulations, like multiplications in $\\sqrt x$ or $\\sqrt y$ or using $x=\\sqrt x\\sqrt x$, but I’m still stuck with that. What am I missing? #### Solutions Collecting From Web of \"How to prove that $\\frac x{\\sqrt y}+\\frac y{\\sqrt x}\\ge\\sqrt x+\\sqrt y$\" Diviging by $\\sqrt y$ we get, $$\\frac x{y}+\\frac{\\sqrt y}{\\sqrt x}\\ge\\sqrt{\\frac xy}+1$$ which can be written $$t^2+\\frac1t\\ge t+1$$ or $$t^3-t^2-t+1=(t-1)^2(t+1)\\ge0$$ for $t>0$. Equality occurs with $t=1$, i.e. $x=y$. Let $a>0$, $b>0$, we have that \\begin{align} \\frac{a^2} b+\\frac{b^2} a\\geq a+b&\\iff \\frac{a^3+b^3} {ab}\\geq a+b\\\\ &\\iff a^2-ab+b^2\\geq ab\\\\ &\\iff (a-b)^2\\geq0 \\end{align} Let $x\\geq y$. Hence, $(x,y)$ and $\\left(\\frac{1}{\\sqrt{x}},\\frac{1}{\\sqrt{y}}\\right)$ are opposite ordered. Thus, by Rearrangement $\\frac{x}{\\sqrt{y}}+\\frac{y}{\\sqrt{x}}\\geq\\frac{x}{\\sqrt{x}}+\\frac{y}{\\sqrt{y}}=\\sqrt{x}+\\sqrt{y}$ and we are done!", "\\frac{x_n + \\frac{a}{x_n}}{2}$$. This is known as Heron's method. As it turns out, this sequence converges to $$\\sqrt{a}$$, and it does so extremely quickly. Every iteration of the algorithm doubles the number of digits of precision. Argument If there is a limit to the sequence $$x_n$$ defined above, it is $$sqrt{a}$$. Furthermore, it satisfies: \\begin{align} x &= \\frac{x + \\frac{a}{x}}{2} \\\\ \\Leftrightarrow 2x &= x^2 + a \\\\ \\Leftrightarrow x^2 &= a \\end{align} Now, we show two elementary facts. First, the sequence will converge from above. That is, $$x_{n+1} \\geq \\sqrt{a}$$. \\begin{align} x_{n+1} = \\frac{x_n + \\frac{a}{x_n}}{2} &\\geq \\sqrt{a} \\\\ x^2_n + a &\\geq 2\\sqrt{a}x_n \\\\ (x_n \\sqrt{a})^2 &\\geq 0 \\end{align} The second elementary fact is that if $$x_n > \\sqrt{a}, then$$x_{n+1} < x_n. \\begin{align} x_{n+1} = \\frac{x_n + \\frac{a}{x_n}}{2} &< x_n \\\\ x^2_n + a &< 2x^2_n \\\\ a &< x^2_n \\\\ x_n &> \\sqrt{a} \\end{align} Thus, the sequence has the following properties: 1. $$\\forall$$ $$n\\in\\mathbb{N}$$, $$x_n \\geq \\sqrt{a}$$ 2. $$\\forall$$ $$n\\in\\mathbb{N}$$, $$x_{n+1} \\leq x_n$$ So, does it converge? Yes, but the proof is difficult. We must assume an axiom about the real numbers. Remark. When we have an estimate $$x_n$$, we know that $$a \\leq x_m \\leq x_n$$ for $$m > n$$. However, we cannot be sure that the sequence converges to $$a$$ rather than some", "\\frac{1}{\\sqrt{k+1}} &\\leq 2\\sqrt{k+1} -2\\sqrt{k} \\\\ \\frac{1}{\\sqrt{k+1}} &\\leq (2\\sqrt{k+1}-1) -(2\\sqrt{k}-1) \\end{align*} coupling this result with the assumption, we have: \\begin{align*} \\sum_{i=1}^k \\frac{1}{\\sqrt{i}} + \\frac{1}{\\sqrt{k+1}} &\\leq (2\\sqrt{k}-1) + (2\\sqrt{k+1}-1) -(2\\sqrt{k}-1) \\\\ \\sum_{i=1}^{k+1} \\frac{1}{\\sqrt{i}} &\\leq 2\\sqrt{k+1}-1 \\therefore P(k) \\text{ is true implies } P(k+1) \\text{ is true} \\end{align*} Therefore, by mathematical induction, $$P(n)$$ is true $$\\forall n\\in\\mathbb{Z}^+$$.", "$X$ is $$\\varphi(x) = \\frac{1}{\\sqrt{2\\pi}} e^{-x^2/2}.$$ Let $f$ be the pdf of $X^2$. Then \\begin{align} f(x) & = \\frac{d}{dx} \\Pr(X^2 \\le x) = \\frac{d}{dx} \\Pr(-\\sqrt{x}\\le X\\le\\sqrt{x}) \\\\ \\\\ & = \\frac{d}{dx} \\frac{1}{\\sqrt{2\\pi}} \\int_{-\\sqrt{x}}^\\sqrt{x} e^{-u^2/2} \\;du = \\frac{2}{\\sqrt{2\\pi}}\\frac{d}{dx} \\int_0^\\sqrt{x} e^{-u^2/2} \\;du \\\\ \\\\ & = \\frac{2}{\\sqrt{2\\pi}} e^{-\\sqrt{x}^2/2} \\frac{d}{dx} \\sqrt{x} = \\frac{2}{\\sqrt{2\\pi}} e^{-x/2} \\frac{1}{2\\sqrt{x}} \\\\ \\\\ \\\\ & = \\frac{e^{-x/2}}{\\sqrt{2\\pi x}}. \\end{align} Sometimes it might be written as $\\dfrac{1}{\\sqrt{2\\pi}} x^{\\frac12 - 1}e^{-x/2}$ so that you can see how it resembles the function involved in defining the Gamma function. Your title said $1$ degree of freedom. But what you write seems to allow $r$ to be some number other than $1$. If you want to do that, then there's more work to do. If $X \\sim (\\mu, \\Sigma) \\neq (0, I)$, the result you wish to prove does not hold: even if the random variables are independent but have nonzero means, you get a non-central $\\chi^2$ pdf which is not what you are trying to show. If $X_1, \\ldots, X_n$ are independent standard normal random variables, then $X_i^2$ has a Gamma distribution with scale parameter $\\frac{1}{2}$ and order parameter $\\frac{1}{2}$. Then, $\\sum_{i= 1}^n X_i^2$ is a sum of $n$ independent Gamma random variables each with scale parameter $\\frac{1}{2}$ and order parameter $\\frac{1}{2}$ and is thus a Gamma random variable with scale", "Jul 11 '19 at 21:36 \\begin{align} -2\\ln \\left( \\frac{2}{\\sqrt 6} \\right) &= -2\\big( \\ln(2)-\\ln(\\sqrt{6}) \\big) \\\\ &= -2\\ln(2)+2\\ln(6^{1/2}) \\\\ &= -2\\ln(2)+\\ln(2\\cdot 3) \\\\ &=-2\\ln(2)+ \\big( \\ln(2)+\\ln(3)\\big) \\\\ &=\\ln(3)-\\ln(2) \\end{align} And for your specific question, remember that $$\\left( \\frac{a}{b} \\right)^{-n}=\\left( \\frac{b}{a} \\right)^n$$ Well you may start by distributing the index since $$2$$ and $$\\sqrt 6$$ are positive. Thus $$\\left(\\frac{2}{\\sqrt{6}}\\right)^{-2}=\\frac{2^{-2}}{{\\sqrt 6}^{-2}}.$$ Then recall that for any nonzero number $$a$$ and any negative integer $$-n,$$ we have $$a^{-n}=\\frac {1}{a^n}.$$ Applying this to your expression, we have $$\\frac{2^{-2}}{{\\sqrt 6}^{-2}}=\\frac{\\frac {1}{2^2}}{\\frac{1}{{\\sqrt 6}^2}}=\\frac{\\frac {1}{4}}{\\frac{1}{6}}=\\frac{6}{4}=\\frac 32.$$ $$-2 \\ln(\\frac{2}{\\sqrt{6}}) = \\ln(\\frac{2}{\\sqrt{6}})^{-2} = \\ln \\frac{1}{(\\frac{2}{\\sqrt{6}})^{2}} = \\ln \\frac{6}{4} = \\ln\\frac{3}{2} = \\ln 3 - \\ln 2$$ $$-2 \\ln \\left( \\frac{2}{\\sqrt{6}} \\right) = \\ln 3 - \\ln 2$$ if and only if $$\\ln \\left( \\frac{2}{\\sqrt{6}} \\right)^{-2} = \\ln \\left( \\frac{3}{2}\\right)$$ if and only if $$\\ln \\left[ \\frac{1}{\\left(\\frac{2}{\\sqrt{6}}\\right)^{2}} \\right] = \\ln \\left( \\frac{3}{2}\\right)$$ And so, we have $$\\ln \\left( \\frac{6}{4} \\right) = \\ln \\left( \\frac{3}{2}\\right)$$ which is true. • you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse – peek-a-boo Jul 11 '19 at 20:27 • One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to", "\\right) = 0 \\\\ & \\Rightarrow \\frac{y^{\\prime}}{\\sqrt{1+y^{\\prime 2}}} = C \\\\ & \\Rightarrow y^{\\prime} = \\pm \\sqrt{\\frac{C^2}{1-C^2}} \\end{align} where C in mathbb{R}. It means that y is of form ax+b for a, b in mathbb{R}, and y prime=a=frac{y_2-y_1}{x_2-x_1}. Therefore, \\begin{align} F(y) &= \\min \\int_{x_1}^{x_2} \\sqrt{1+y^{\\prime 2}} dx = \\int_{x_1}^{x_2} \\sqrt{1+a^2} dx = \\sqrt{1+a^2}(x_2-x_1) \\\\ \\\\ &= \\sqrt{1+ \\frac{(y_2-y_1)^2}{(x_2-x_1)^2}}(x_2-x_1)= \\sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2} \\end{align} This result is right because Euclidean distance is the smallest one between (x_1, y_1) and (x_2, y_2) points. emoy.net", "- b^2) / (a + b) is very often useful for such differences. Especially if a or b is a square root obviously. And that's what the answer does, but it works in many more cases. – gnasher729 Sep 23 at 10:48 \\begin{align} 1-\\sqrt{ 1-x^2} &= (1-\\sqrt{ 1-x^2})\\frac{1+\\sqrt{ 1-x^2}}{1+\\sqrt{ 1-x^2}}\\\\ &= \\frac{x^2}{1+\\sqrt{ 1-x^2}} \\end{align} \\begin{align} y = x\\sqrt{\\frac{1}{2+2\\sqrt{1-x^2}}} \\end{align} • It ought to be $y=|x|\\sqrt\\dots$ when taking $x^2$ out of the radical. – Simply Beautiful Art Sep 24 at 15:33" ]

[ "Let f be the function satisfying f'(x)= $$\\sqrt{f(x)}$$ for all real numbers where f(3)=25. Find f''(3).", "# Is the real number sqrt(1.21) rational or irrational? $1.21 = \\frac{121}{100}$, so $\\sqrt{1.21} = \\sqrt{\\frac{121}{100}} = \\frac{\\sqrt{121}}{\\sqrt{100}} = \\frac{11}{10}$.", "+0 # Algebra 2 help 0 126 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$. Oct 10, 2019 #1 +6179 +1 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$ . Oct 10, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$", "# Cube roots fun Algebra Level 3 The real number $x$ satisfies $\\sqrt [3]{x + 9} - \\sqrt [3]{x - 9} = 3$. Find the value of $x^2$. ×", "+0 # Find the positive real value of t that satisfies ​. 0 35 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "Therefore, the total value $= 3 \\times 1 + 5 \\times 2 + 7 \\times 3 + \\ldots + 13 \\times 6 + 2 \\times 7$ $= 217$ . Question 15: If $[ \\sqrt { x } ] = 5 ]$ and $[ \\sqrt { y } ] = 6 ,$ where $x$ and $y$ are natural numbers, what can be the greatest possible value of $x + y ?$ It is clear that $[ \\sqrt { 25 } ] = 5 , [ \\sqrt { 26 } ] = 5 , [ \\sqrt { 27 } ] = 5$ and so on. The highest value of $x$ that we can take is 35 , since $[ \\sqrt { 35 } ] = 5$ but $[ \\sqrt { 36 } ] = 6$ $[ \\sqrt { 48 } ] = 6$ but $[ \\sqrt { 49 } ] = 7$ Therefore, the greatest value of $x + y = 35 + 48 = 83$", "+0 # Algebra 2 help 0 57 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$. Oct 10, 2019 #1 +6045 +1 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$ . Oct 10, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$", "Select Page Try this beautiful problem from AMC 10A, 2018 based on Number theory. Problem – Number Theory Let’s try this problem number 10 from AMC 10A, 2018 based on Number Theory. Suppose that the real number $x$ satisfies $\\sqrt {49-x^2}$ – $\\sqrt {25-x^2}$ = $3$. What is the value of $\\sqrt {49-x^2}$ + $\\sqrt {25-x^2}$? • 8 • $\\sqrt 33 + 8$ • 9 • $2\\sqrt10+4$ • 12 Key Concepts Number Theory Real number Square root But try the problem first… Source AMC 10 A – 2018 – Problem No.10 Mathematics can be fun by Perelman Try with Hints First hint As a first hint we can start from here : In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The $x^2$ terms cancel out. $(\\sqrt {49 – x^2} +\\sqrt {25 – x^2}) (\\sqrt {49 – x^2}) -(\\sqrt {25 – x^2})$ = 49 -$x^2 – 25 + x^2$ =24 Final Hint Given that $\\sqrt {49 – x^2}) -(\\sqrt {25 – x^2})$ = 3 $\\sqrt {49 – x^2} +\\sqrt {25 – x^2}$ = $\\frac {24}{3}$ = 8", "+0 Find the positive real value of t that satisfies ​. 0 283 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "# How do you find the set in which the real number -sqrt(0.0625) belongs? Mar 9, 2017 $- \\sqrt{0.0625}$ is a rational number. #### Explanation: Note that: $0.0625 = \\frac{0.125}{2} = \\frac{0.25}{4} = \\frac{0.5}{8} = \\frac{1}{16} = {\\left(\\frac{1}{4}\\right)}^{2}$ So: $\\sqrt{0.0625} = \\sqrt{{\\left(\\frac{1}{4}\\right)}^{2}} = \\frac{1}{4}$ So: $- \\sqrt{0.0625} = - \\frac{1}{4}$ This is a rational number, since it is expressible in the form $\\frac{p}{q}$ for integers $p , q$ with $q \\ne 0$. For example: $p = - 1$ and $q = 4$.", "IrrationalEquation STEP 2 Intermediate 2004 Problem - 4773 Find all the real values of $x$ that satistify: $$\\sqrt{3x^2 + 1} + \\sqrt{x} - 2x - 1=0$$ The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "1. $\\sqrt { 49 }$ • What positive number multiplied by itself is 49? • 7", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "# 1.8 The real numbers (Page 4/13) Page 4 / 13 Can we simplify $\\sqrt{-25}?$ Is there a number whose square is $-25?$ ${\\left(\\text{}\\phantom{\\rule{0.5em}{0ex}}\\right)}^{2}=-25?$ None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\\sqrt{-25}.$ The square root of a negative number is not a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-169}$ $\\text{−}\\sqrt{64}.$ 1. There is no real number whose square is $-169.$ Therefore, $\\sqrt{-169}$ is not a real number. 2. Since the negative is in front of the radical, $\\text{−}\\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\\text{−}\\sqrt{64}$ is a real number. For each number given, identify whether it is a real number or not a real number: $\\sqrt{-196}$ $\\text{−}\\sqrt{81}.$ not a real number real number For each number given, identify whether it is a real number or not a real number: $\\text{−}\\sqrt{49}$ $\\sqrt{-121}.$ real number not a real number Given the numbers $-7,\\frac{14}{5},8,\\sqrt{5},5.9,\\text{−}\\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers. Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given. The", "$r>0$ makes mathematical sense, in which case: $$|x^r| = |0^r| = |0| = |0|^r = |x|^r.$$ Next assume $x \\neq 0$, and rational $r = p/q$ where $p$ is a whole number and $q$ is a positive odd whole number, so that $x^r$ is guaranteed to be a real number. Recall, for all real numbers $x$, it is true that $|x| = \\sqrt{x^2}$. $$|x^r| = \\sqrt{ (x^r)^2 } = \\sqrt{ x^{r\\cdot 2} } = \\sqrt{ x^{2r} } = \\sqrt{ (x^2)^r } = (\\sqrt{x^2})^r,$$ where the last equality only works because $x^2 > 0$. Then by the same trick as above, $$(\\sqrt{x^2})^r = |x|^r.$$ The same proof works when $x >0$ and $r$ is an arbitrary real number. All that we require is that $x^r$ is real. Hope this helps! - Thank you to both Stefan Geschke and Shaun Ault. – Sony Sep 23 '11 at 21:46" ]

[ "+ x} + 5x)}$, recall: $(A - B)(A + B) = A^2 - B^2$. Let $A = \\sqrt{25x^2 + x}$ and $B = 5x$. Then $A^2 - B^2 = 25x^2 + x - (5x)^2 = 25x^2 + x - 25x^2 = x.$ Therefore $ \\frac{(\\sqrt{25x^2 + x} - 5x)(\\sqrt{25x^2 + x} + 5x)}{(\\sqrt{25x^2 + x} + 5x)} = \\frac{25x^2 + x - 25x^2}{\\sqrt{25x^2 + x} + 5x} = \\frac{x}{\\sqrt{25x^2 + x} + 5x}. $ Typo noted - thanks for that get, Jhevon.", "an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \\sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \\sqrt {25x^2-48x+36}$ = $0$. We then add $5 \\sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \\sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\\frac {48+0}{50}$. Therefore, $x = \\frac {48}{50}$. Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\\boxed {(B)4.8}$ ~Arcticturn", "is, $A(x)=2x\\sqrt{25-x^2}$ $A'(x)=(2x)'\\sqrt{25-x^2}+(2x)(\\sqrt{25-x^2})'$ $A'(x)=2\\sqrt{25-x^2}+(2x)\\left( -\\frac{x}{\\sqrt{25-x^2}} \\right)=0$ Multiply through by denominator, $2(25-x^2)+2x(-x)=0$ $50-2x^2-2x^2=0$ $4x^2=50$ $x=\\pm \\frac{\\sqrt{50}}{2}$ Note the plus-mius does not matter all it says we can start drawing the rectangle from either side.", "Adding each side of $$-6x – 2y = 12$$ to the corresponding side of $$6x+4y=10$$ gives $$2y=22$$ or $$y=11$$. Finally, substituting $$11$$ for $$y$$ in $$6x+4y=10$$ gives $$6x+4(11)=10$$ or $$x=-\\frac{17}{3}$$. 49- Choice B is correct $$x_{1,2} =\\frac{-b ± \\sqrt{b^2-4ac}}{2a}$$ , $$ax^2 + bx + c = 0, 4x^2 + 14x + 6 = 0$$ ⇒ then: $$a = 4$$, $$b = 14$$ and $$c = 6$$, $$x = \\frac{-14 +\\sqrt{14^2 – 4 .4 .6}}{2.4} = – \\frac{1}{2}, x =\\frac{-14 – \\sqrt{14^2 – 4 .4 .6}}{2.4} = – 3$$ 50- Choice A is correct Use Pythagorean theorem: $$a^2+b^2=c^2→s^2+h^2=(5s)^2→s^2+h^2=25s^2$$ Subtracting s^2 from both sides gives: $$h^2=24s^2$$ Square roots of both sides: $$h=\\sqrt{24s^2}=\\sqrt{4×6×s^2} =\\sqrt{4}×\\sqrt{6}×\\sqrt{s^2 }=2×s×\\sqrt{6}=2s\\sqrt{6}$$ 51- Choice D is correct Let $$x$$ be the number of purple marbles. Let’s review the choices provided: A. $$\\frac{1}{10}$$, if the probability of choosing a purple marble is one out of ten, then: Probability $$=\\frac{number \\space of \\space desired \\space outcomes}{number \\space of \\space total \\space outcomes}=\\frac{x}{20+30+40+x}=\\frac{1}{10}$$ Use cross multiplication and solve for $$x$$. $$10x=90+x→9x=90→x=9$$ Since, number of purple marbles can be 9, then, choice be the probability of randomly selecting a purple marble from the bag. Use same method for other choices. B. $$\\frac{1}{4}$$ $$\\frac{x}{20+30+40+x}=\\frac{1}{4}→4x=90+x→3x=90→x=30$$ C. $$\\frac{2}{5}$$ $$\\frac{x}{20+30+40+x}=\\frac{2}{5}→5x=180+2x→3x=180→x=60$$ D. $$\\frac{7}{15}$$ $$\\frac{x}{20+30+40+x}=\\frac{7}{15}→15x=630+7x→8x=630→x=78.75$$ Number of purple marbles cannot be a decimal. 52- Choice C is correct", "{ 4 } } } { 3 }$$ 25. $$\\frac { \\sqrt [ 5 ] { 12 x y ^ { 3 } z ^ { 4 } } } { 2 y z }$$ 27. $$3\\sqrt { 10 } + 9$$ 29. $$\\frac { \\sqrt { 5 } - \\sqrt { 3 } } { 2 }$$ 31. $$- 1 + \\sqrt { 2 }\\ 33. \\(\\frac { - 5 - 3 \\sqrt { 5 } } { 2 }$$ 35. $$- 4 - \\sqrt { 15 }$$ 37. $$\\frac { 15 - 7 \\sqrt { 6 } } { 23 }$$ 39. $$\\sqrt { x } - \\sqrt { y }$$ 41. $$\\frac { x ^ { 2 } + 2 x \\sqrt { y } + y } { x ^ { 2 } - y }$$ 43. $$\\frac { a - 2 \\sqrt { a b + b } } { a - b }$$ 45. $$\\frac { 5 \\sqrt { x } + 2 x } { 25 - 4 x }$$ 47. $$\\frac { x \\sqrt { 2 } + 3 \\sqrt { x y } + y \\sqrt { 2 } } { 2 x - y }$$ 49. $$\\frac { 2 x + 1 + \\sqrt { 2 x +", "\\frac{1}{2} \\times 25 \\times \\left(\\frac{25}{12} \\times \\frac{4}{3}\\right) = 34.7$ We can use Heron’s Formula to solve for the Area with three sides. Ratios: 3.37:9 :12 = 12: 32: 42.7 A = sqrt((sxx(s-a)xx(s-b)xx(s-c)) where $s = \\frac{1}{2} \\left(a + b + c\\right)$ and a, b, c are the side lengths. $s = 17.3$ A = sqrt((17.3xx(17.3 – 12)xx(17.3 – 32)xx(17.3 – 42.7)) ; A = sqrt((17.3xx(5.3)xx(-14.75)xx(-25.4)) $A = \\sqrt{34352}$ ; $A = 185.3$", "# How do you solve 5 sqrt(x-2) =7? May 15, 2015 $\\sqrt{x - 2} = \\frac{7}{5}$ Square both sides: $x - 2 = \\frac{49}{25} \\to x = \\frac{49}{25} + 2 = \\frac{99}{25}$ May 15, 2015 $5 \\sqrt{x - 2} = 7$ Divide both sides by $5$ $\\sqrt{x - 2} = \\frac{7}{5}$ Square both sides $x - 2 = {\\left(\\frac{7}{5}\\right)}^{2} = \\frac{49}{25}$ Add $2$ to both sides $x = 2 + \\frac{49}{25} = 3 \\frac{24}{25}$ or to look like Katie's answer for those who prefer decimal fractions $x = 3.96$", "the sides of the square and need to be equal to each other. $A P=C P$ $\\sqrt{(5-x)^{2}+(4-y)^{2}}=\\sqrt{(1-x)^{2}+(-6-y)^{2}}$ Squaring on both sides we have, $(5-x)^{2}+(4-y)^{2}=(1-x)^{2}+(-6-y)^{2}$ $25+x^{2}-10 x+16+y^{2}-8 y=1+x^{2}-2 x+36+y^{2}+12 y$ $8 x+20 y=4$ $2 x+5 y=1$ From this we have, $x=\\frac{1-5 y}{2}$ Substituting this value of ‘x’ and the length of the side in the equation for ‘AP’ we have, $A P=\\sqrt{(5-x)^{2}+(4-y)^{2}}$ $\\sqrt{58}=\\sqrt{(5-x)^{2}+(4-y)^{2}}$ Squaring on both sides, $58=(5-x)^{2}+(4-y)^{2}$ $58=\\left(5-\\left(\\frac{1-5 y}{2}\\right)\\right)^{2}+(4-y)^{2}$ $58=\\left(\\frac{9+5 y}{2}\\right)^{2}+(4-y)^{2}$ $58=\\frac{81+25 y^{2}+90 y}{4}+16+y^{2}-8 y$ $232=81+25 y^{2}+90 y+64+4 y^{2}-32 y$ $87=29 y^{2}+58 y$ We have a quadratic equation. Solving for the roots of the equation we have, $29 y^{2}+58 y-87=0$ $29 y^{2}+87 y-29 y-87=0$ $29 y(y+3)-29(y+3)=0$ $(y+3)(29 y-29)=0$ $(y+3)(y-1)=0$ The roots of this equation are −3 and 1. Now we can find the respective values of ‘x’ by substituting the two values of ‘y When $x=\\frac{1-5(-3)}{2}$ $=\\frac{1+15}{2}$ $x=8$ When $y=1$ $x=\\frac{1-5(1)}{2}$ $=\\frac{1-5}{2}$ $x=-2$ Therefore the other two vertices of the square are $(8,-3)$ and $(-2,1)$.", "$Diagonal = \\sqrt{4x^2+9x^2} = 26$ $x\\sqrt{13} = 26$ Divide both sides by squareroot(13) $x = 2\\sqrt{13}$ Perimeter $= 2(2x + 3x) = 10 x = 20\\sqrt{13}$ i didnt think about using algebra, i was just taking the ratio as a total 676=5/5 and expected the result to be the same. I need to figure out where i went wrong 4. i played with this abit more im still confused as to why the above not work but using adarsh algebra $26 = \\sqrt{(\\frac{2}{5}x)^2 + (\\frac{3}{5}x)^2}$ $26 = \\sqrt{\\frac{4}{25}x^2 + \\frac{9}{25}x^2}$ $26=\\sqrt{\\frac{13}{25}x^2} \\equiv x\\sqrt{\\frac{13}{25}}$ $x = 26 \\div \\sqrt{\\frac{13}{25}} \\equiv \\frac{26 * 5}{\\sqrt{13}} \\equiv \\frac{13 * 10}{13^\\frac{1}{2}} \\equiv 10\\sqrt{13}$ perimeter 2 x sides $2 \\mbox{x} 10\\sqrt{13} \\equiv 20\\sqrt{13}$ 5. Originally Posted by sammy28 i played with this abit more im still confused as to why the above not work but using adarsh algebra $26 = \\sqrt{(\\frac{2}{5}x)^2 + (\\frac{3}{5}x)^2}$ $26 = \\sqrt{\\frac{4}{25}x^2 + \\frac{9}{25}x^2}$ $26=\\sqrt{\\frac{13}{25}x^2} \\equiv x\\sqrt{\\frac{13}{25}}$ $x = 26 \\div \\sqrt{\\frac{13}{25}} \\equiv \\frac{26 * 5}{\\sqrt{13}} \\equiv \\frac{13 * 10}{13^\\frac{1}{2}} \\equiv 10\\sqrt{13}$ perimeter 2 x sides $2 \\mbox{x} 10\\sqrt{13} \\equiv 20\\sqrt{13}$ I think you got it But just as an assurance I will do it your way Lets consider the sides to be 3x/5 and 2x/5 Now we follow the steps (same as you did here) $26 = \\sqrt{(\\frac{2}{5}x)^2 +", "+0 # fast!!! 0 115 1 Find the square root. Write your answer as a fraction or mixed number in simplest form. $\\sqrt{\\frac{25}{4}}=$√254​=​ Feb 23, 2022 #1 +675 +1 $$\\sqrt{\\frac{25}{4}} = \\frac{\\sqrt{25}}{\\sqrt{4}} = {\\color{blue}\\frac{5}{2} = 2 \\frac{1}{2}}$$ !! . Feb 23, 2022 #1 +675 +1 $$\\sqrt{\\frac{25}{4}} = \\frac{\\sqrt{25}}{\\sqrt{4}} = {\\color{blue}\\frac{5}{2} = 2 \\frac{1}{2}}$$", "$s=\\frac{d}{\\sqrt{2}}$, so our side length is $\\frac{2\\sqrt{2}+6}{\\sqrt{2}}$. However, we are trying to look for the area, so squaring $\\frac{2\\sqrt{2}+6}{\\sqrt{2}}$ yields $\\frac{44+24\\sqrt{2}}{2}=\\boxed{\\text{(B)}22+12\\sqrt{2}}$", "0 x^2 - 3 x = 2/3 x^2 - 3 x + 9/4 = 35/12 Write the left hand side as a square: (x - 3/2)^2 = 35/12 Take the square root of both sides: x - 3/2 = sqrt(35/3)/2 or x - 3/2 = -sqrt(35/3)/2 x = 3/2 + sqrt(35/3)/2 or x - 3/2 = -sqrt(35/3)/2 Answer: | x = 3/2 + sqrt(35/3)/2 or x = 3/2 - sqrt(35/3)/2 Guest May 6, 2017 #2 +7348 +2 5/(1+x) = 3 – 4/(x-1) x1 = -0,21 x2 = 3,21 Where did I go wrong? :D $$\\frac{5}{1+x}=3-\\frac{4}{x-1}$$ $$\\frac{5}{1+x}+\\frac{4}{x-1}=3$$ $$\\frac{5x-5+4+4x}{x^2-1}=3$$ $${9x-1}=3x^2-3$$ $$3x^2-9x-2=0$$ $$x^2-3x-\\frac{2}{3}=0$$ $$x_1=1.5+\\sqrt{2.25+\\frac{2}{3}}$$ $$x=1.5\\pm\\sqrt{2.91\\overline{66}}$$ $$x_1=3.2078\\\\x_2=-0.2078$$ ! asinus May 6, 2017 edited by asinus May 6, 2017 #3 +2 This is it, thank you! =)", "$$\\sqrt {x^2 - x + 1}$$ Again, Squaring on both sides, (x + 1)2 = ($$\\sqrt {x^2 - x + 1}$$)2 or, x2 + 2x + 1 = 4(x2 - x + 1) or, x2 + 2x + 1 = 4x2 - 4x + 4 or, 4x2 - x2 - 4x - 2x + 4 - 1 = 0 or, 3x2 - 6x + 3 = 0 or, 3(x2- 2x + 1) = 0 or, x2 - 2x + 1 = 0 or, (x)2 - 2 . x . 1 + (1)2 = 0 or, (x - 1)2 = 0 Removing square from both sides, x - 1 = 0 ∴ x = 1Ans $$\\frac {a - 2}{a^2 - 2a + 4}$$ + $$\\frac {a + 2}{a^2 + 2a + 4}$$ - $$\\frac {16}{a^4 + 16 + 4a^2}$$ = $$\\frac {(a - 2) (a^2 + 2a + 4) + (a + 2) (a^2 - 2a + 4)}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\\frac {16}{(a^2)^2 + 2 . a^2 . 4 + (4)^2 - 4a^2}$$ = $$\\frac {a^3 - 2^3 + a^3 + 2^3}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\\frac {16}{(a^2 + 4)^2 - (2a)^2}$$ = $$\\frac {2a^3}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\\frac", "(\\frac{3}{5}x)^2}$ ....this thing came from Pythagoras theorem $26 = \\sqrt{\\frac{4}{25}x^2 + \\frac{9}{25}x^2}$ $26=\\sqrt{\\frac{13}{25}x^2} \\equiv x\\sqrt{\\frac{13}{25}}$ $x = 26 \\div \\sqrt{\\frac{13}{25}} \\equiv \\frac{26 * 5}{\\sqrt{13}} \\equiv \\frac{13 * 10}{13^\\frac{1}{2}} \\equiv 10\\sqrt{13}$ Perimeter is the sum of lengths of all sides Thus what we basically did was (3x/5 + 2x/ 5) + (3x/5 + 2x/5) = x + x = 2x =Answer 6. thank you for explaining it so thoroughly, my problem solving skill is not as good as it should be. I need to think before jumping into questions", "= \\sqrt{4}$$ = 2 (b) $$\\sqrt{50}$$ ÷ $$\\sqrt{2}$$ = $$\\frac{\\sqrt{25×2}}{\\sqrt{2}} = \\sqrt{25}$$ = 5", "$\\sqrt{{\\left(x - \\frac{10}{25}\\right)}^{2}} = \\sqrt{\\frac{15}{25}}$ $\\left(x - \\frac{10}{25}\\right) = \\sqrt{\\frac{15}{25}}$ $x = \\sqrt{\\frac{15}{25}} + \\frac{10}{25}$ $x = \\frac{\\sqrt{15}}{5} + \\frac{10}{25}$ $x = \\frac{\\sqrt{15}}{5} + \\frac{2}{5}$ $x = \\frac{\\sqrt{15} + 2}{5}$", "$10$ for $c$. This gives us $$\\frac{\\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30$$ $$\\sqrt{4x^4+400x^2+4\\cdot48^2-4x^4}=120$$ $$400x^2+4\\cdot48^2=120^2$$ $$25x^2+24^2=30^2$$ $$25x^2=6^2(5^2-4^2)$$ $$25x^2=18^2$$ $$5x=18$$ $$x=\\frac{18}5$$ Now multiplying by $2$, we get that the height is $\\frac{36}5$, and $m+n=36+5=\\boxed{041}$", "2$$\\sqrt x$$)2 or, 1 - x = 1 - 4$$\\sqrt x$$ + 4x or, 1 - x - 1 - 4x = -4$$\\sqrt x$$ or, -5x = -4$$\\sqrt x$$ Again, Squaring on both sides, (-5x)2 = (-4$$\\sqrt x$$)2 or, 25x2 = 16x or, 25x2 - 16x = 0 or, x(25x - 16) = 0 Either, x = 0 (impossible) Or, 25x - 16 = 0 or, 25x = 16 ∴ x = $$\\frac {16}{25}$$ ∴ x = $$\\frac {16}{25}_{Ans}$$ $$\\sqrt {x^2 - 3x + 3}$$ + $$\\sqrt {x^2 - x + 1}$$ = 2 or,$$\\sqrt {x^2 - 3x + 3}$$ = 2 -$$\\sqrt {x^2 - x + 1}$$ Squaring on both sides, ($$\\sqrt {x^2 - 3x + 3}$$)2 = (2 -$$\\sqrt {x^2 - x + 1}$$)2 or, x2 - 3x + 3 = 4 - 4$$\\sqrt {x^2 - x + 1}$$ + ($$\\sqrt {x^2 - x + 1}$$)2 or, x2 - 3x + 3 - 4 = - 4$$\\sqrt {x^2 - x + 1}$$ + x2 - x + 1 or, x2 - 3x - 1 - x2 + x - 1 =- 4$$\\sqrt {x^2 - x + 1}$$ or, - 2x - 2 =- 4$$\\sqrt {x^2 - x + 1}$$ or, -2 (x + 1) =- 2$$\\sqrt {x^2 - x + 1}$$ or, x + 1 =", "$\\frac{\\sqrt{2}}{\\sqrt{50}}= \\frac{\\sqrt{2}}{\\sqrt{2}\\sqrt{25}}$ $= \\frac{1}{\\sqrt{25}}= \\frac{1}{5}$", "\\sqrt[5]{32}+\\sqrt{225}$$ = $$\\sqrt[4]{3^{4}}-8 \\cdot \\sqrt[5]{3^{5}}+15 \\cdot \\sqrt[5]{2^{5}}+\\sqrt{15^{2}}$$ = $$\\left(3^{4}\\right)^{\\frac{1}{4}}-8 \\cdot\\left(3^{5}\\right)^{\\frac{1}{5}}+15\\left(2^{5}\\right)^{\\frac{1}{5}}+\\left(15^{2}\\right)^{\\frac{1}{2}}$$ = 3 – 8.3 + 15.2 + 15 = 3 – 24 + 30+ 15 = 48 – 24 = 24" ]

[ "+ x} + 5x)}$, recall: $(A - B)(A + B) = A^2 - B^2$. Let $A = \\sqrt{25x^2 + x}$ and $B = 5x$. Then $A^2 - B^2 = 25x^2 + x - (5x)^2 = 25x^2 + x - 25x^2 = x.$ Therefore $ \\frac{(\\sqrt{25x^2 + x} - 5x)(\\sqrt{25x^2 + x} + 5x)}{(\\sqrt{25x^2 + x} + 5x)} = \\frac{25x^2 + x - 25x^2}{\\sqrt{25x^2 + x} + 5x} = \\frac{x}{\\sqrt{25x^2 + x} + 5x}. $ Typo noted - thanks for that get, Jhevon.", "# How do you find the set in which the real number -sqrt(0.0625) belongs? Mar 9, 2017 $- \\sqrt{0.0625}$ is a rational number. #### Explanation: Note that: $0.0625 = \\frac{0.125}{2} = \\frac{0.25}{4} = \\frac{0.5}{8} = \\frac{1}{16} = {\\left(\\frac{1}{4}\\right)}^{2}$ So: $\\sqrt{0.0625} = \\sqrt{{\\left(\\frac{1}{4}\\right)}^{2}} = \\frac{1}{4}$ So: $- \\sqrt{0.0625} = - \\frac{1}{4}$ This is a rational number, since it is expressible in the form $\\frac{p}{q}$ for integers $p , q$ with $q \\ne 0$. For example: $p = - 1$ and $q = 4$.", "\\sqrt{49}$$ 1. $$4, \\sqrt{49}$$. 2. $$−3, 4, \\sqrt{49}$$ 3. $$−3, 0.\\overline{3}, \\frac{9}{5}, 4, \\sqrt{49}$$ 4. $$-\\sqrt{2}$$ 5. $$−3, \\sqrt{2}, 0.\\overline{3}, \\frac{9}{5}, 4, \\sqrt{49}$$ ##### Exercise $$\\PageIndex{21}$$ For the given numbers, list the 1. whole numbers 2. integers 3. rational numbers 4. irrational numbers 5. real numbers: $$−\\sqrt{25},−\\frac{3}{8}, −1, 6, \\sqrt{121}, 2.041975…$$ 1. $$6, \\sqrt{121}$$. 2. $$−\\sqrt{25}, −1, 6, \\sqrt{121}$$ 3. $$−\\sqrt{25},−\\frac{3}{8}, −1, 6, \\sqrt{121}$$ 4. $$2.041975…$$ 5. $$−\\sqrt{25},−\\frac{3}{8}, −1, 6, \\sqrt{121}, 2.041975…$$ ## Locate Fractions on the Number Line The last time we looked at the number line, it only had positive and negative integers on it. We now want to include fractions and decimals on it. ##### Note Doing the Manipulative Mathematics activity “Number Line Part 3” will help you develop a better understanding of the location of fractions on the number line. Let’s start with fractions and locate $$\\frac{1}{5}, -\\frac{4}{5}, 3, \\frac{7}{4}, -\\frac{9}{2}, -5$$ and $$\\frac{8}{3}$$ on the number line. We’ll start with the whole numbers 3 and −5. because they are the easiest to plot. See Figure $$\\PageIndex{4}$$. The proper fractions listed are $$\\frac{1}{5}\\text{ and } -\\frac{4}{5}$$. We know the proper fraction $$\\frac{1}{5}$$ has value less than one and so would be located between 0 and 1. The denominator is 5, so we divide the unit from 0 to 1 into 5 equal parts $$\\frac{1}{5},", "{ 4 } } } { 3 }$$ 25. $$\\frac { \\sqrt [ 5 ] { 12 x y ^ { 3 } z ^ { 4 } } } { 2 y z }$$ 27. $$3\\sqrt { 10 } + 9$$ 29. $$\\frac { \\sqrt { 5 } - \\sqrt { 3 } } { 2 }$$ 31. $$- 1 + \\sqrt { 2 }\\ 33. \\(\\frac { - 5 - 3 \\sqrt { 5 } } { 2 }$$ 35. $$- 4 - \\sqrt { 15 }$$ 37. $$\\frac { 15 - 7 \\sqrt { 6 } } { 23 }$$ 39. $$\\sqrt { x } - \\sqrt { y }$$ 41. $$\\frac { x ^ { 2 } + 2 x \\sqrt { y } + y } { x ^ { 2 } - y }$$ 43. $$\\frac { a - 2 \\sqrt { a b + b } } { a - b }$$ 45. $$\\frac { 5 \\sqrt { x } + 2 x } { 25 - 4 x }$$ 47. $$\\frac { x \\sqrt { 2 } + 3 \\sqrt { x y } + y \\sqrt { 2 } } { 2 x - y }$$ 49. $$\\frac { 2 x + 1 + \\sqrt { 2 x +", "$f^{-1}(5)=\\sqrt[3]{5-4}=\\sqrt[3]{1}=1$ 2. Find the domain and range of $f(x) = \\sqrt{x^2-25}$ $x^2-25\\ge0$ over the Real number domain. $D=\\{x|x\\leq -5 \\ \\ or \\ \\ x\\geq 5\\}$ or $D=(-\\infty, -5] \\cup [5, +\\infty)$ $R=\\{f(x)|f(x) \\geq 0\\}$ 3. If $f(x) = 2x^2 + 4$, find $f(x^4 – 1)$ $f(x^4-1)=2(x^4-1)^2+4\\Longrightarrow2(x^8-2x^4+1)+4\\Longrightarrow2x^8-4x^4+6$ 4. If $z_1 = 3 + 2i$ and $z_2 = 4-5i$, find a. $(z_1)(z_2)$ $(3+2i)(4-5i)=12-7i+10=22-7i$ b. $\\frac{z_1}{z_2}$ $\\frac{3+2i}{4-5i}$ Multiply both numerator and denominator by the conjugate of $4-5i$ $\\frac{3+2i}{4-5i}\\cdot\\frac{4+5i}{4+5i}=\\frac{12+23i-10}{16+25}=\\frac{2+23i}{41}$ 5. Find $8^{-\\frac{2}{3}}$ $8^{-\\frac{2}{3}}=\\frac{1}{8^{\\frac{2}{3}}}=\\frac{1}{\\l eft(\\sqrt[3]{8}^2\\right)}=\\frac{1}{4}$", "## College Algebra (6th Edition) $\\frac{1}{81}$ We know that $b^{-n}=\\frac{1}{b^{n}}$ (where $b$ is a nonzero real number and $n$ is a natural number). Therefore, $27^{-\\frac{4}{3}}=\\frac{1}{27^{\\frac{4}{3}}}$ Based on the definition of $a^{\\frac{m}{n}}$, we know that $a^{\\frac{m}{n}}=(\\sqrt[n] a)^{m}=\\sqrt[n] a^{m}$ (where $\\sqrt[n] a$ is a real number). Therefore, $\\frac{1}{27^{\\frac{4}{3}}}=\\frac{1}{\\sqrt[3] 27^{4}}=\\frac{1}{(\\sqrt[3] 27)^{4}}=\\frac{1}{(3)^{4}}=\\frac{1}{81}$ We know that $\\sqrt[3] 27=3$, because $3^{3}=27$.", "+0 # fast!!! 0 115 1 Find the square root. Write your answer as a fraction or mixed number in simplest form. $\\sqrt{\\frac{25}{4}}=$√254​=​ Feb 23, 2022 #1 +675 +1 $$\\sqrt{\\frac{25}{4}} = \\frac{\\sqrt{25}}{\\sqrt{4}} = {\\color{blue}\\frac{5}{2} = 2 \\frac{1}{2}}$$ !! . Feb 23, 2022 #1 +675 +1 $$\\sqrt{\\frac{25}{4}} = \\frac{\\sqrt{25}}{\\sqrt{4}} = {\\color{blue}\\frac{5}{2} = 2 \\frac{1}{2}}$$", "} \\right] ^{2/3}$ is () x 9. If $\\sqrt { 10 }$= 3.1622, then the value of is $\\frac { 1 }{ \\sqrt { 10 } }$ is_________. () 0.31623 10. The rational number equivalent to $\\frac { 5 }{ 9 }$ such that its numerator is 25 is____________ . () $\\frac { 25 }{ 45 }$ 11. Divide $15\\sqrt { 12 }$ by $3\\sqrt { 3 }$the result is ___________. () 10", "{a^2 - 1}{2}$$ and $$c = \\frac {a^2 - 1}{2} + 1$$ or simply $$c = b + 1$$, you can now get the Simple Triplet. For example: If $$a = 3$$, what is $$b$$ and $$c$$? Using the formula for $$b$$, $b = \\frac {a^2 - 1}{2}$ $b = \\frac {(3)^2 - 1}{2}$ $b = \\frac {9 - 1}{2}$ $b = \\frac {8}{2}$ $b = 4$ and formula for $$c$$ $c = b + 1$ $c = 4 + 1$ $c = 5$ The values for $$b$$ and $$c$$ are $$4$$ and $$5$$, respectively. Other example: 1. If $$a = 7$$, what is $$b$$ and $$c$$? Using the formula for $$b$$, $b = \\frac {a^2 - 1}{2}$ $b = \\frac {(7)^2 - 1}{2}$ $b = \\frac {49 - 1}{2}$ $b = \\frac {48}{2}$ $b = 24$ and formula for $$c$$ $c = b + 1$ $c = 24 + 1$ $c = 25$ Let's check using Pythagorian Theorem: $c = \\sqrt {a^2 + b^2}$ $25 = \\sqrt {(7)^2 + (24)^2}$ $25 = \\sqrt {49 + 576}$ $25 = \\sqrt {625}$ $\\boxed{25 = 25}$ Limitations for these formulas: 1. It can be applicable if the value of $$a$$ is odd. 2. It can be applicable if the value of $$a$$ is a whole number. (Honestly, I just discovered", "# Is the real number sqrt(1.21) rational or irrational? $1.21 = \\frac{121}{100}$, so $\\sqrt{1.21} = \\sqrt{\\frac{121}{100}} = \\frac{\\sqrt{121}}{\\sqrt{100}} = \\frac{11}{10}$.", "} { 25 y ^ { 2 } } }$$ 18. $$\\sqrt { \\frac { 4 x ^ { 5 } } { 9 y ^ { 4 } } }$$ 19. $$\\sqrt { \\frac { m ^ { 7 } } { 36 n ^ { 4 } } }$$ 20. $$\\sqrt { \\frac { 147 m ^ { 9 } } { n ^ { 6 } } }$$ 21. $$\\sqrt { \\frac { 2 r ^ { 2 } s ^ { 5 } } { 25 t ^ { 4 } } }$$ 22. $$\\sqrt { \\frac { 36 r ^ { 5 } } { s ^ { 2 } t ^ { 6 } } }$$ 23. $$\\sqrt [ 3 ] { 27 a ^ { 3 } }$$ 24. $$\\sqrt [ 3 ] { 125 b ^ { 3 } }$$ 25. $$\\sqrt [ 3 ] { 250 x ^ { 4 } y ^ { 3 } }$$ 26. $$\\sqrt [ 3 ] { 162 a ^ { 3 } b ^ { 5 } }$$ 27. $$\\sqrt [ 3 ] { 64 x ^ { 3 } y ^ { 6 } z ^ { 9 } }$$ 28. $$\\sqrt [ 3 ] { 216 x ^ { 12", "is (a) $\\frac { 1 }{ 10 }$ (b) $\\frac { 2 }{ 10 }$ (c) $\\frac { 3 }{ 10 }$ (d) $\\frac { 4 }{ 10 }$ 41. Part - B 25 x 2 = 50 42. Find the cardinal number of the following sets. R = {x : x is an integers, x∈Z and –5 ≤ x <5} 43. Represent U= {1,2,3,4,5,6,7,8,9,10} A = {1,2,4,6,8,10} and B = {3,6,1,10} in venn-diagram, then find (i) (A UB)' (ii) (A ∩ B)' 44. Let U= {x : -3 : << 4} A = {-1,2,3} B = {0,1,2,3} C = {-3;-2,-1,0,1,2}. Find (i) A' UB' (ii) (A ∩ B)' (iii) (A ⋂ C)' 45. If A={ x : x = 2n, n $\\in$ W and n<4}, B = {x : x = n, n 2 $\\in$ N and n ≤ 4} and C = {0, 1, 2, 5, 6} , then verify the associative property of intersection of sets. 46. If x=$\\sqrt{5}$+2, then find the value of ${ x }^{ 2 }+\\frac { 1 }{ { x }^{ 2 } }$ 47. Simplify the following: $2\\sqrt [ 3 ]{ 40 } +3\\sqrt [ 3 ]{ 625 } -4\\sqrt [ 3 ]{ 320 }$ 48. Can you reduce the following numbers to surds of same of same order $\\sqrt", "we can see that $$2^{1.5}=2\\times 2^{0.5}$$ and $$2^{2.5}=2^2\\times 2^{0.5}$$ and so forth, and so it is easy to fill in more points on our graph (red squares): Next, what about $$2^{2.25}$$? Again using our new rules above, we see that $2^{2.25} = 2^{2+0.25} = 2^2\\times 2^{0.25} = 2^2\\times 2^{\\frac14}.$ We call $$2^\\frac14 = \\sqrt[4]{2}$$ the “fourth root of 2” since, according to our rules, $2^\\frac14\\times2^\\frac14\\times 2^\\frac14\\times2^\\frac14 = 2.$ But through our rules of exponents we also see that $$2^\\frac14$$ = $$(2^\\frac12)^\\frac12$$ = $$\\sqrt{\\sqrt{2}}$$ $$\\approx$$ $$\\sqrt{1.414}$$ $$\\approx$$ 1.189. Thus, finally, $$2^{2.25}$$ will be $$2^2=4$$ times this number, or 4.757. And, as before, knowing the value of $$2^{0.25}$$, we can also easily compute $$2^{1.25} = 2\\times 2^{0.25}$$, $$2^{3.25} = 2^3\\times 2^{0.25}$$ and so forth. In addition, $$2^{0.75}$$ = $$2^{0.5}\\times 2^{0.25}$$ = 1.6818, and thus $$2^{1.75}$$ = $$2\\times 2^{0.75}$$, $$2^{2.75}$$ = $$2^2\\times 2^{0.75}$$, and so forth. Now, we can fill in our plot even further (blue squares): We see that the circles and the red and blue squares follow a general trend. So far we’ve been lucky to use numbers and operations (square roots) that we know well and understand. Continuing in this way, we can take further roots and powers and fill in the plot to whatever level necessary. So what about our third example, $$2^{3.659}$$? Looking at our graph", "2y + 1 )^{2} −25 = 0$$ 39. $$(x − 5 )^{2} −20 = 0$$ 40. $$(x + 1 )^{2} −28 = 0$$ 41. $$( 3t + 2 )^{2} − 6 = 0$$ 42. $$(3t−5)^{2}−10=0$$ 43. $$4(y+2)^{2}−3=0$$ 44. $$9(y−7)^{2}−5=0$$ 45. $$4(3x+1)^{2}−27=0$$ 46. $$9(2x−3)^{2}−8=0$$ 47. $$2(3x−1)^{2}+3=0$$ 48. $$5(2x−1)^{2}−3=0$$ 49. $$3(y−23)^{2}−32=0$$ 50. $$2(3y−13)^{2}−52=0$$ 1. ±4 3. ±3 5. ±$$\\frac{1}{2}$$ 7. ±0.5 9. ±$$2\\sqrt{3}$$ 11. ±$$\\frac{3}{4}$$ 13. ±$$\\sqrt{\\frac{1}{2}}$$ 15. ±10 17. No real solution 19. ±$$\\frac{2}{3}$$ 21. ±0.3 23. ±$$\\sqrt{7}$$ 25. ±$$2\\sqrt{2}$$ 27. No real solution 29. ±$$\\frac{3 \\sqrt{3}}{4}$$ 31. ±$$6\\sqrt{2}$$ 33. ±$$3\\sqrt{3}$$ 35. $$−9, −5$$ 37. $$−3, 6$$ 39. $$5$$±$$2\\sqrt{5}$$ 41. ±$$\\frac{\\sqrt{6}-2}{3}$$ 43. ±$$\\frac{\\sqrt{3}}{2}-2$$ 45. ±$$\\frac{3 \\sqrt{3}-2}{6}$$ 47. No real solution 49. ±$$\\frac{4 \\sqrt{6}}{3}+23$$ Exercise $$\\PageIndex{5}$$ extracting square roots Find a quadratic equation in standard form with the following solutions. 1. ±7 2. ±13 3. ±$$\\sqrt{7}$$ 4. ±$$\\sqrt{3}$$ 5. ±$$3\\sqrt{5}$$ 6. ±$$5\\sqrt{2}$$ 7. $$1$$±$$\\sqrt{2}$$ 8. $$2$$±$$\\sqrt{3}$$ 1. $$x^{2}−49=0$$ 3. $$x^{2}−7=0$$ 5. $$x^{2}−45=0$$ 7. $$x^{2}−2x−1=0$$ Exercise $$\\PageIndex{6}$$ extracting square roots Solve and round off the solutions to the nearest hundredth. 1. $$9x(x+2)=18x+1$$ 2. $$x^{2}=10(x^{2}−2)−5$$ 3. $$(x+3)(x−7)=11−4x$$ 4. $$(x−4)(x−3)=66−7x$$ 5. $$(x−2)^{2}=67−4x$$ 6. $$(x+3)^{2}=6x+59$$ 7. $$(2x+1)(x+3)−(x+7)=(x+3)^{2}$$ 8. $$(3x−1)(x+4)=2x(x+6)−(x−3)$$ 1. ±0.33 3. ±5.66 5. ±7.94 7. ±3.61 Exercise $$\\PageIndex{7}$$ extracting square roots Set up an algebraic equation and use it to solve the following. 1. If 9 is subtracted from 4 times the square", "$(25x - 11)(x - 3) = 0$ $\\{\\frac{11}{25}, 3\\}$ (10) $25x^2 - 86x + 33 = 0$ $\\{\\frac{11}{25}, 3\\}$ (9) $25x^2 - 70x + 49 = 16x + 16$ $\\{\\frac{11}{25}, 3\\}$ (8) $5x - 7 = 4\\sqrt{x+1}$ $\\{3\\}$ Since $5\\cdot\\frac{11}{25} - 7 = \\frac{11}{5} - \\frac{35}{5} = -\\frac{24}{5} \\neq 4\\sqrt{\\frac{11}{25} + 1} = 4\\sqrt{\\frac{11}{25} + \\frac{25}{25}} = 4\\sqrt{\\frac{36}{25}} = 4\\cdot\\frac{6}{5} = \\frac{24}{5}$ So we have isolated the precise moment when the extraneous solution $x = \\frac{11}{25}$ is created and it appears exactly where we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function $f(n) = n^2$ to both sides. More specifically, if $x = \\frac{11}{25}$, (8) reads $- \\frac{24}{5} = \\frac{24}{5}$, which is false, but (9) reads $(- \\frac{24}{5})^2 = ( \\frac{24}{5})^2$, which is true. For this particular value of $x$, we squared both sides and replaced a false statement with a true statement. In retrospect, we can say that $x =\\frac{11}{25}$ is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence. (7) $6x - 2 = 4 + 4\\sqrt{x + 1} + x + 1$ $\\{3\\}$ Since both sides of (7) are positive", "$(25x - 11)(x - 3) = 0$ $\\{\\frac{11}{25}, 3\\}$ (10) $25x^2 - 86x + 33 = 0$ $\\{\\frac{11}{25}, 3\\}$ (9) $25x^2 - 70x + 49 = 16x + 16$ $\\{\\frac{11}{25}, 3\\}$ (8) $5x - 7 = 4\\sqrt{x+1}$ $\\{3\\}$ Since $5\\cdot\\frac{11}{25} - 7 = \\frac{11}{5} - \\frac{35}{5} = -\\frac{24}{5} \\neq 4\\sqrt{\\frac{11}{25} + 1} = 4\\sqrt{\\frac{11}{25} + \\frac{25}{25}} = 4\\sqrt{\\frac{36}{25}} = 4\\cdot\\frac{6}{5} = \\frac{24}{5}$ So we have isolated the precise moment when the extraneous solution $x = \\frac{11}{25}$ is created and it appears exactly where we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function $f(n) = n^2$ to both sides. More specifically, if $x = \\frac{11}{25}$, (8) reads $- \\frac{24}{5} = \\frac{24}{5}$, which is false, but (9) reads $(- \\frac{24}{5})^2 = ( \\frac{24}{5})^2$, which is true. For this particular value of $x$, we squared both sides and replaced a false statement with a true statement. In retrospect, we can say that $x =\\frac{11}{25}$ is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence. (7) $6x - 2 = 4 + 4\\sqrt{x + 1} + x + 1$ $\\{3\\}$ Since both sides of (7) are positive", "$(25x - 11)(x - 3) = 0$ $\\{\\frac{11}{25}, 3\\}$ (10) $25x^2 - 86x + 33 = 0$ $\\{\\frac{11}{25}, 3\\}$ (9) $25x^2 - 70x + 49 = 16x + 16$ $\\{\\frac{11}{25}, 3\\}$ (8) $5x - 7 = 4\\sqrt{x+1}$ $\\{3\\}$ Since $5\\cdot\\frac{11}{25} - 7 = \\frac{11}{5} - \\frac{35}{5} = -\\frac{24}{5} \\neq 4\\sqrt{\\frac{11}{25} + 1} = 4\\sqrt{\\frac{11}{25} + \\frac{25}{25}} = 4\\sqrt{\\frac{36}{25}} = 4\\cdot\\frac{6}{5} = \\frac{24}{5}$ So we have isolated the precise moment when the extraneous solution $x = \\frac{11}{25}$ is created and it appears exactly where we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function $f(n) = n^2$ to both sides. More specifically, if $x = \\frac{11}{25}$, (8) reads $- \\frac{24}{5} = \\frac{24}{5}$, which is false, but (9) reads $(- \\frac{24}{5})^2 = ( \\frac{24}{5})^2$, which is true. For this particular value of $x$, we squared both sides and replaced a false statement with a true statement. In retrospect, we can say that $x =\\frac{11}{25}$ is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence. (7) $6x - 2 = 4 + 4\\sqrt{x + 1} + x + 1$ $\\{3\\}$ Since both sides of (7) are positive", "BrandeVenturella 27 # simplify each square root: 1. square root of -25 2. square root of 1.44 3. square root of + or -16/49 4. square root of 361 5. square root of 49 6. square root of + or -0.64 7. square root of -6.25 8. square root of 169/196 9. square root of 25/324 $1.\\ \\sqrt{-25}-isn't\\ real\\ number\\to\\sqrt{-25}=\\sqrt{25(-1)}=\\sqrt{25}\\cdot\\sqrt{-1}=5i\\\\\\\\2.\\ \\sqrt{1.44}=\\sqrt{1.2^2}=1.2\\\\\\\\3.\\ \\sqrt{\\frac{16}{49}}=\\frac{\\sqrt{16}}{\\sqrt{49}}=\\frac{\\sqrt{4^2}}{\\sqrt{7^2}}=\\frac{4}{7}\\\\\\\\\\sqrt{-\\frac{16}{49}}=\\frac{4}{7}i\\\\\\\\4.\\ \\sqrt{361}=\\sqrt{19^2}=19\\\\\\\\5.\\ \\sqrt{49}=\\sqrt{7^2}=7\\\\\\\\6.\\ \\sqrt{0.64}=\\sqrt{0.8^2}=0.8\\\\\\\\\\sqrt{-0.64}=0.8i$ $7.\\ \\sqrt{-6.25}=\\sqrt{6.25(-1)}=\\sqrt{6.25}\\cdot\\sqrt{-1}=\\sqrt{2.5^2}\\cdot\\sqty{-1}=2.5i\\\\\\\\8.\\ \\sqrt\\frac{169}{196}=\\frac{\\sqrt{169}}{\\sqrt{196}}=\\frac{\\sqrt{13^2}}{\\sqrt{14^2}}=\\frac{13}{14}\\\\\\\\9.\\ \\sqrt\\frac{25}{324}=\\frac{\\sqrt{25}}{\\sqrt{324}}=\\frac{\\sqrt{5^2}}{\\sqrt{18^2}}=\\frac{5}{18}\\\\====================================$ $i=\\sqrt{-1}\\ is\\ the\\ imaginary\\ unit\\\\---------------------\\\\If\\ a\\geq0\\ then\\ \\sqrt{a^2}=a$", "rational number and real number. $\\frac{2\\sqrt{5}}{\\sqrt{20}}=\\frac{2\\sqrt{5}}{\\sqrt{4\\times 5}}=\\frac{2\\sqrt{5}}{2\\sqrt{5}}=1.$ This number can be simplified to an integer. All integers can be expressed as rational numbers and are a special kind of real number. 2. The four numbers are ordered as follows: $1<\\frac{3}{2}<\\sqrt{3} <1.5\\cdot \\sqrt{3}.$ $1<\\frac{3}{2}$ since the numerator is larger then the denominator and $\\frac{3}{2}=1.5$ . $\\frac{3}{2}<\\sqrt{3}$ since we can see on our calculators that $\\sqrt{3}\\approx 1.7$ $\\sqrt{3} < 1.5\\cdot \\sqrt{3}$ since multiplying by 1.5 makes any number larger. ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Square Roots and Real Numbers (10:18) Classify the following numbers. Include all the categories that apply to the number. 1. $\\sqrt{0.25}$ 2. $\\sqrt{1.35}$ 3. $\\sqrt{20}$ 4. $\\sqrt{25}$ 5. $\\sqrt{100}$ 6. Place the following numbers in numerical order from lowest to highest. $\\frac{\\sqrt{6}}{2} && \\frac{61}{50} && \\sqrt{1.5} && \\frac{16}{13}$ 7. Find the value of each marked point. Mixed Review 1. Simplify $\\frac{9}{4}\\div 6$ . 2. The area of a triangle is given by the formula $A= \\frac{b(h)}{2}$ , where $b=$ base", "17. $$25x^{2}+1=0$$ 18. $$36x^{2}+4=0$$ 19. $$5t^{2}−4=0$$ 20. $$2t^{2}−9=0$$ 21. $$12x^{2}−94x+1=0$$ 22. $$3x^{2}+12x−16=0$$ 23. $$36y^{2}=2y$$ 24. $$50y^{2}=−10y$$ 25. $$x(x−6)=−29$$ 26. $$x(x−4)=−16$$ 27. $$4y(y+1)=5$$ 28. $$2y(y+2)=3$$ 29. $$-3x^{2}=2x+1$$ 30. $$3x^{2}+4x=−2$$ 31. $$6(x+1)2=11x+7$$ 32. $$2(x+2)2=7x+11$$ 33. $$9t^{2}=4(3t−1)$$ 34. $$5t(5t−6)=−9$$ 35. $$(x+1)(x+7)=3$$ 36. $$(x−5)(x+7)=14$$ 1. $$-1, 3$$ 3. $$-\\frac{2}{3} , 1$$ 5. No real solution 7. $$-\\frac{1}{5} , 0$$ 9. No real solution 11. $$-1\\pm\\sqrt{5}$$ 13. $$\\frac{5}{2}$$ 15. $$2\\pm\\sqrt{5}$$ 17. No real solution 19. $$\\pm\\frac{2 \\sqrt{5}}{5}$$ 21. $$\\frac{1}{2}, 4$$ 23. $$0, \\frac{1}{18}$$ 25. No real solution 27. $$\\frac{-1\\pm\\sqrt{6}}{2}$$ 29. No real solution 31. $$-\\frac{1}{2}, \\frac{1}{3}$$ 33.$$\\frac{2}{3}$$ 35. $$-4\\pm 2\\sqrt{3}$$ Exercise $$\\PageIndex{5}$$ applications number problems Set up an algebraic equation and use it to solve the following. 1. A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49. Find the numbers. 2. A positive real number is 1 more than another. When twice the smaller is subtracted from the square of the larger, the result is 4. Find the numbers. 3. A positive real number is 6 less than another. If the sum of the squares of the two numbers is 38, then find the numbers. 4. A positive real number is 1 more than twice another. If 4 times the smaller number is subtracted from the" ]

[ "# Divide a Polynomial by a Monomial In this video, learn how to divide a polynomial by a monomial. In this video, we are going to look at how to divide a polynomial by a monomial. For example: To solve: $(12x^3+8x^2+16x) \\div 4x$ we will divide each term in the numerator separately by the denominator of 4x. So: $12x^3 \\div 4x=3x^2$ $8x^2 \\div 4x=2x$ $16x \\div 4x=4$ $3x^2+2x+4$", "# Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10) Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of . – – + – – + – The final answer is the quotient plus the remainder over the divisor. Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10)", "polynomial by a monomial, we divide each term of the polynomial in the numerator by the monomial in the denominator. (3x^2 + 1)/(3x^2) = 1 + 1 = 2 The correct statement is (3x^2 + 1)/(3x^2) = 1 + 1/(3x^2). If you would like to contribute notes or other learning material, please submit them using the button below.", "a polynomial that you'll eventually want to factor, don't expand the polynomial. Keep it together. What I'm saying here is do not try to expand $(14x^2 +7x-21)^2$ into a 5 term 4th degree polynomial in x. Keep it factored. One final hint: You can factor $2x^2+x-3$", "5 '19 at 15:58 ## 1 Answer Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, $$X$$ and $$X+1$$ (this is clear since the polynomial has no zero in the prime field), and degree 2, $$X^2+X+1$$ (which is the only irreducible polynomial of this degree).", "# Dividing a polynomial Algebra Level 3 When a polynomial $$P(x)$$ of degree $$n > 2$$ is divided by $$x-3$$, the remainder is 5. When $$P(x)$$ is divided by $$x-1$$, the remainder is 1. Find the remainder when $$P(x)$$ is divided by $$x^2-4x+3$$. ×", "# Thread: long division polynomial 1. ## long division polynomial Divide 6(x)^4 - 20(x)^3 + 50(x)^2 + 30x - 2 by x^2 - 5x. Is the answer 6x2 + 10x + 100 and the remainder: 530x - 2? these numbers look weird, what did i do wrong? 2. Originally Posted by NeedHelp18 Divide 6(x)^4 - 20(x)^3 + 50(x)^2 + 30x - 2 by x^2 - 5x. Is the answer 6x2 + 10x + 100 and the remainder: 530x - 2? these numbers look weird, what did i do wrong? Weird? How so? Your answer is fine. Why do you assume that you are in error? 3. the answers correct? i thought the 530x wasnt correct.. 4. Originally Posted by NeedHelp18", "3ix - 5x - 3ix + 25 + 9$ $= x^2 - 10x + 34$. So divide your polynomial by $x^2 - 10x + 34$ in order to find the third factor (and third zero). This is where I'm having the problem. I'm not sure how to divide the polynomial . 4. Originally Posted by renebran This is where I'm having the problem. I'm not sure how to divide the polynomial . Use long division.", "1. ## help factor I have this question for practice and I don't understand how to solve for the values. When the polynomial 4x^ 3+mx^2+nx+11 is divided by x+2, the remainder is -7. When the polynomial is divided by x-1, the remainder is 14. What are the values of n and m? Thanks, any help will be greatly appreciated. 2. How's your synthetic division? That's all it takes, well, plus a little algebra. 3. When dividing a polynomial $P(x)$ by $x-a$, the remainder is $P(a)$. In this case we have: $P(-2)=-7, \\ P(1)=14$ So, you have to solve the system $\\left\\{\\begin{array}{ll}4m-2n=14\\\\m+n=-1\\end{array}\\right.$", "get to a remainder that's \"smaller\" (in polynomial degree) than the divisor, you're done. What am I supposed to do with the remainder? A Polynomial can be expressed in terms that only have positive integer exponents and the operations of addition, subtraction, and multiplication. Division of a polynomial by another polynomial is one of the important concept in Polynomial expressions. In this case, we should get 4x 2 /2x = 2x and 2x(2x + 3). In other words, it must be possible to write the expression without division. Intro to long division of polynomials (video) | Khan Academy − − = (−) (+ +) ⏟ + ⏟ The long division algorithm for arithmetic is very similar to the above algorithm, in which the variable x is replaced by the specific number 10.. Polynomial short division. (This is a legitimate mathematical step. You can use the Mathway widget below to practice finding doing long polynomial division. Just as you would with a simpler … This gives me –10x + 15 as my new bottom line: Dividing –10x by 2x, I get –5, which I put on top. To compute $32/11$, for instance, we ask how many times $11$ fits into $32$. In such a text, the long division above would be presented as shown here: The only difference is", "x$ is divided by $x+1 ?$...", "left over after dividing. 4 16x2 −12x+8 3. One method is long division, a process similar to long division of two whole numbers. Just remember that we keep going until the remainder has degree that is strictly less that the degree of the polynomial we’re dividing by, \\(x + 2$$ in this case. Answer: First, factor by grouping. In this case, we should get 4x 2 /2x = 2x and 2x(2x + 3). Set up the division problem. 3: Dividing Polynomials Worksheet is suitable for 9th - 12th Grade. <br /> 2. So we write the polynomial 2x 4 +3x 2 +x as product of x and 2x 3 + 3x +1. Polynomials are very similar to integers. Oct 20, 2020 · Multiplying And Dividing Polynomials Worksheet With Answers. G o t a d i f f e r e n t a n s w e r ? C h e c k i f i t ′ s c o r r e c t. Then, divide the first term of the divisor into the first term of the dividend, and multiply the X in the quotient by the divisor. Practice. Dividing Polynomials; Remainder and Factor Theorems . We have the divisor divided into the dividend, and of course our answer is the quotient. In this", "ciphertext) and when it is unencrypted (the plaintext). These are almost the normal polynomials that everyone studies at school. Things like $4 x^2 + 2 x+1$ There are some differences, the first being that the coefficients are all whole numbers and are the remainder relative to some other whole number $$t$$ (that is $$\\bmod t$$ ). Say $$t=24$$, then this is like a 24 hour clock, where adding 6 hours to 21 takes you to 3. All the coefficients of the polynomials are treated like this. Alternately, we can consider the numbers to range from -11 to 12, allowing us to negate numbers conveniently. Note that this is just a convenience factor - there is no difference between a \"remainder\" of -1 and a remainder of 23 (when dividing by 24). The second, and trickier, thing that is different about these polynomials is that this idea of using remainders applies not just to the coefficients of the polynomials, but also to the polynomials themselves. We define a special polynomial called the polynomial modulus, and only consider the remainder of polynomials when they have been divided by this polynomial modulus. The specific form of this polynomial modulus in the FV scheme is $$x^d + 1$$ where $$d=2^n$$ for some $$n$$. For illustration here we will take $$n=4$$, so the", "order you choose is correct, conventionally, such summands are arranged from highest to lowest degree: $x^5+x^3+3x+22$.", "be $latex {{x}^7}+{{x}^5}+{{x}^3}+x$. → Polynomial Division Calculator ## Division of polynomials – Practice problems Practice dividing polynomials using the following problems. Solve the exercises and select the answer obtained. Click “Check” to verify that you got the correct answer.", "## Intermediate Algebra (12th Edition) Published by Pearson # Chapter 4 - Section 4.2 - Adding and Subtracting Polynomials - 4.2 Exercises: 21 monomial; 0 #### Work Step by Step The polynomial 25 can be rewritten as $25x^{0}$. Therefore, this polynomial is a monomial, because it consists of 1 term. This polynomial has a degree of 0, as the variable $x$ in the term is raised to a power of 0. In general, constant terms have a degree of 0. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "division algorithm. 128Terms with zero coefficients used to fill in all missing exponents within a polynomial.", "# How do you do long division of polynomials with remainders? Dec 24, 2014 It is the same but just instead of getting 0 you get a polynomial in the last step. Divide 3x^3 – 5x^2 + 10x – 3 by $3 x + 1$ in this case you get a polynomial seven which can be writtten in algebraic terms as $7 {x}^{0}$ so this can be proved using the division algoritm dividend = divisor $\\cdot$ quotient + remainder", "## Dividing Polynomials with Monomials with Remainder Watch this example of dividing a polynomial by a monomial. This video gives an example of the technique we use to divide.", "# Long Division of Polynomials Example 1 Divide $x^3 + 3x^3 + 2x + 1$ by $x+2$. Example 2 Perform a long division: $(4x^4-6x^3+2x^2-3x+5) \\div (2x+1)$. # Polynomial Equations and Maximum Coefficients If $\\alpha$ is a root of a polynomial equation $P(x) = 0$, then $P(\\alpha) = 0$, and if $M$ is the maximum coefficient value of $|a_{n}|, |a_{n-1}|, |a_{n-2}|, \\cdots ,|a_2|, |a_1|$ then $nM \\ge |a_{n}| + |a_{n-1}| + |a_{n-2}| + \\cdots + |a_2| + |a_1|$. […]" ]

[ "\\dots ,k \\}$$ for some $$k \\in \\mathbb{N}$$ A naive attempt was to define $$f$$ by the positiveness of a polynomial of degree $$n$$ but the problem is that it is unknown whether computing the roots of a polynomial of degree $$n$$ can be done in polynomial time. If it can be done in polynomial time then $$f(S)$$ can be done in $$\\theta(n) \\neq \\theta(|S| \\cdot n)$$ worst case time by summing the size of ranges of positiveness. Is there any better approach?", "+0 # maths; confused 0 73 1 a. The degree of the polynomial \\(p(x)\\) is \\(11\\), and the degree of the polynomial \\(q(x)\\) is \\(7\\). Find all possible degrees of the polynomial \\(p(x) + q(x)\\). b. Suppose \\(f\\) is a polynomial such that \\(f(0) = 47\\), \\(f(1) = 32\\), \\(f(2) = -13\\), and \\(f(3)=16\\). What is the sum of the coefficients of \\(f\\)? Mar 16, 2019 #1 +5172 +2 \\(deg(p(x)) > deg(q(x)) \\Rightarrow deg((p+q)(x)) = deg(p(x))\\) \\(c_0 + c_1 + c_2 + \\dots + c_n = f(1) =32\\) . Mar 17, 2019 edited by Rom Mar 17, 2019 edited by Rom Mar 17, 2019", "5 '19 at 15:58 ## 1 Answer Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, $$X$$ and $$X+1$$ (this is clear since the polynomial has no zero in the prime field), and degree 2, $$X^2+X+1$$ (which is the only irreducible polynomial of this degree).", "infinitely many polynomials of degree $$d$$ that can go through $$d$$ points! Think of a line, which has degree one. If you were just given one point, there would be infinitely many possibilities for the second point, each of which uniquely defines a line. Here we are using the standard coefficient representation of a degree $$d$$ polynomial. The properties we discussed in the previous sections illustrate that there are two ways to write polynomials: a coefficient representation and a value representation. The coefficient representation of a degree $$d$$ polynomial can be thought of as a string $$(a_0,\\dots,a_d)$$ of length $$d+1$$, where the polynomial is equal to $$a_dx^d + a_{d-1}x^{d-1} + \\cdots + a_0$$ and each $$a_i$$ is an element of $$\\operatorname{GF}(q)$$. The value representation can also be thought of as a string $$(y_0,\\dots,y_d)$$ of length $$d+1$$, which we interpret as the values of the polynomial at fixed points $$x_0,\\dots,x_d\\in \\operatorname{GF}(q)$$. As we’ve seen, polynomial interpolation can convert the value representation to the coefficient representation. The coefficient representation can be converted to the value representation by evaluating the polynomial at $$x_0,\\dotsc, x_d$$. What happens if only part of either of these representations is specified? For example, what if we are given only $$k$$ coefficients instead of $$d+1$$? Then there are $$d+1-k$$ coefficients left to choose, and no matter which", "f by a polynomial of degree n at the nodes x0,...,xn we get the remainder which can be expressed as [4] where and is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as The remainder can be bound as ## First derivative The first derivative of the lagrange polynomial is given by where . ## Finite fields The Lagrange polynomial can also be computed in finite fields. This has applications in cryptography, such as in Shamir's Secret Sharing scheme.", "$x-1$ is a factor. Dividing these polynomials we get, (working modulo 3). $x^4+x^3+x^2+x+2$ Checking the zero's of this polynomial we find that 1 is it zero. Thus, $x-1$ is a factor. Dividing these polynomials we get, (working modulo 3). $x^3+2x+1$, notice this polynomial has no zero and because it has a degree of 3 it is irreducible (this is a theorem about the irreducibility of 2 and 3 degree polynomials). Thus, it factors as, $(x-1)(x-1)(x^3+2x+1)$ or another way of writing this, $(x+1)(x+1)(x^3+2x+1$ I do part (c) latter unless you understand and I could stop.", "of integers and leads to polynomial remainders. Its existence is based on the following theorem: Given two univariate polynomials ''a''(''x'') and ''b''(''x'') (where ''b''(''x'') is a non-zero polynomial) defined over a field (in particular, the reals or complex number In mathematics, a complex number is an element of a number system that extends the real numbers with a specific element denoted , called the imaginary unit and satisfying the equation i^= -1; every complex number can be expressed in the form a ... s), there exist two polynomials ''q''(''x'') (the ''quotient'') and ''r''(''x'') (the ''remainder'') which satisfy: :$a\\left(x\\right) = b\\left(x\\right)q\\left(x\\right) + r\\left(x\\right)$ where :$\\deg\\left(r\\left(x\\right)\\right) < \\deg\\left(b\\left(x\\right)\\right),$ where \"deg(...)\" denotes the degree of the polynomial (the degree of the constant polynomial whose value is always 0 can be defined to be negative, so that this degree condition will always be valid when this is the remainder). Moreover, ''q''(''x'') and ''r''(''x'') are uniquely determined by these relations. This differs from the Euclidean division of integers in that, for the integers, the degree condition is replaced by the bounds on the remainder ''r'' (non-negative and less than the divisor, which insures that ''r'' is unique.) The similarity between Euclidean division for integers and that for polynomials motivates the search for the most general algebraic setting in which Euclidean division is valid.", "# Divided polynomial algebra The divided polynomial algebra over a commutative ring $R$ in a symbol $X$ of degree $k$ is defined as the free $R$-algebra on symbols $X_1=X, X_2, X_3,\\dotsc$ of degree $k$, $2k$, $3k,\\dotsc$ modulo the relation $X_iX_j={i+j \\choose j}X_{i+j}$. Now these symbols are naturally the divided powers $\\gamma_i(X)$ of $X$ and the relation $\\gamma_i(X)={i+j \\choose j}\\gamma_{i+j}(X)$ holds. My problem: This equation, however, needs to hold for all elements of the divided polynomial algebra, e. g. for $X_2$ and I can't see why it does. Second, what about other requirements of divided powers such as $\\gamma_i(X_iX_j)=\\gamma_i(X_i)\\gamma_j(X_j)$? How is $\\gamma_i$ even defined for elements $\\neq X$? All difficulties resolve for $R\\subset \\mathbb{Q}$ so an answer for the general situation would be very much appreciated. Thank you!", "homogeneous polynomials. The degrees of these polynomials only depend on $W.$ The order of the reflection group is the product of the degrees. $\\text{Reflection group$W$}$ $\\text{Degrees}$ $\\text{Order}$ $A_{n-1}$ $2,3,\\dotsc,n$ $n!$ $B_n/C_n$ $2,4,\\dotsc,2n$ $2^n n!$ $D_n$ $2,4,\\dotsc,2(n-1),n$ $2^{n-1} n!$", "# Dividing a polynomial Algebra Level 3 When a polynomial $$P(x)$$ of degree $$n > 2$$ is divided by $$x-3$$, the remainder is 5. When $$P(x)$$ is divided by $$x-1$$, the remainder is 1. Find the remainder when $$P(x)$$ is divided by $$x^2-4x+3$$. ×", "+0 # Polynomial division 0 205 1 1) The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? 2) When the polynomial p(x) is divided by x - 1, the remainder is 3. When the polynomial p(x) is divided by x - 3, the remainder is 5. What is the remainder when the polynomial p(x) is divided by (x - 1)(x - 3)? Aug 14, 2019 $$f(x) = q(x) d(x) + r(x)\\\\ \\text{If }\\text{deg}\\;d \\le 3, \\text{then deg } r < \\text{deg }d \\le 3 \\implies \\text{Contradiction!}\\\\ \\therefore \\text{deg }d \\ge 4\\\\ \\text{Minimum possible deg }d = 4\\\\ \\implies \\text{Maximum possible deg }q = \\text{deg }f - \\text{deg }d = \\boxed5$$", "asks for the degree of each polynomial, and then it asks for the contents using a For( loop until there are no more terms. It then commences with the main loop which is another For( loop. The program works in very much the same way people would go about solving a division problem. It divides the current leading terms of the dividend by the first term of the divisor and puts it into L,,3,,, our answer. It then does a second For( loop multiplying the partial quotient by every term of the divisor. At the end, it discovers the remainder in L,,1,, and stores them into the end of L,,3,, for a complete answer. Here are a couple of examples for you to try out to see if the code was inputted correctly: {{Template:TI-BASIC:Math |eqn= {x^4+3x^2-2x+1 \\over x^2+4x+6} = x^2-4x+13-{30x+77 \\over x^2+4x+6} \\hspace{12pt} \\{1,-4,13,-30,-77\\} }} # Error Conditions • ERR: INVALID DIM occurs if the degrees are made negative or the divisor's degree is larger than the dividend • ERR: DIVIDE BY 0 happens if the leading coefficient of the divisor is 0 • ERR: DATA TYPE is thrown if anything is imaginary", "ciphertext) and when it is unencrypted (the plaintext). These are almost the normal polynomials that everyone studies at school. Things like $4 x^2 + 2 x+1$ There are some differences, the first being that the coefficients are all whole numbers and are the remainder relative to some other whole number $$t$$ (that is $$\\bmod t$$ ). Say $$t=24$$, then this is like a 24 hour clock, where adding 6 hours to 21 takes you to 3. All the coefficients of the polynomials are treated like this. Alternately, we can consider the numbers to range from -11 to 12, allowing us to negate numbers conveniently. Note that this is just a convenience factor - there is no difference between a \"remainder\" of -1 and a remainder of 23 (when dividing by 24). The second, and trickier, thing that is different about these polynomials is that this idea of using remainders applies not just to the coefficients of the polynomials, but also to the polynomials themselves. We define a special polynomial called the polynomial modulus, and only consider the remainder of polynomials when they have been divided by this polynomial modulus. The specific form of this polynomial modulus in the FV scheme is $$x^d + 1$$ where $$d=2^n$$ for some $$n$$. For illustration here we will take $$n=4$$, so the", "are: $\\set {\\sqrt [6] 2 \\cis 20 \\degrees, \\sqrt [6] 2 \\cis 80 \\degrees, \\sqrt [6] 2 \\cis 140 \\degrees, \\sqrt [6] 2 \\cis 200 \\degrees, \\sqrt [6] 2 \\cis 260 \\degrees, \\sqrt [6] 2 \\cis 320 \\degrees}$ ### Roots of $z^8 + 1 = 0$ The roots of the polynomial: $z^8 + 1 = 0$ are: $\\set {\\cos \\dfrac {\\paren {2 k + 1} \\pi} 8 + i \\sin \\dfrac {\\paren {2 k + 1} \\pi} 8: k \\in \\set {0, 1, \\ldots, 7} }$", "+ \\dots + x^k$ is the minimal polynomial of $\\alpha$ over $F$, then $g$ divides $f$, and $F = \\mathbb{Q}(a_0, a_{1}, \\dots)$. The tiny bit is in step 2, because you will be (implicitly, perhaps) using elements of the Galois group to get all the roots. PS By the way, the above is straight from the proof that a simple extension has a finite number of intermediate fields. -", "simple way to determine if a given PSD polynomial can be split nontrivially? A bonus question I'm also curious about is whether there's an easy way to find the \"square-root\" of a given PSD polynomial (i.e. given $$f$$, find a $$g$$ where $$g \\circ g = f$$). Thanks!", "$A(x)$ be a polynomial of degree at most $n$. Then we can write $A(x)=\\alpha_0+\\alpha_1(x-r)+\\alpha_2(x-r)^2+\\dots+\\alpha_n(x-r)^n$ as a sum of powers of $x-r$. To see this, notice that if we divide $A(x)$ by $(x-r)^n$, the quotient $C(x)$ must be a constant $\\alpha_n$ (since the degree of $A$ is at most $n$), and the remainder $D(x)$ has degree at most $n-1$. We can then apply Lemma 2 again and divide $D(x)$ by $(x-r)^{n-1}$ to obtain a constant $\\alpha_{n-1}$ as quotient and a polynomial of degree at most $n-2$ as remainder. Continuing this way we obtain the result. The same argument gives also that if $c(x)=(x-a)^2+p_1$ for some constant $p_1>0$ is an irreducible quadratic polynomial, then we can write $A(x)=(\\alpha_0+\\beta_0x)+(\\alpha_1+\\beta_1x)c(x)+(\\alpha_2+\\beta_2 x)c(x)^2+\\dots$ where the sum is of course finite and the numbers $\\alpha_i,\\beta_i$ are constant. Definition. The polynomial $r(x)$ is said to be a greatest common divisor of the polynomial $a(x)$ and $b(x)$ if and only if the following two conditions hold: 1. $r$ (evenly) divides both $a$ and $b$. 2. If $s$ is a polynomial that divides both $a$ and $b$, then $s$ divides $r$. There are many greatest common divisors of any two polynomials $a$ and $b$, not both equal to 0, since if $r$ is an example, then so is $5r$. However, except for constant multiples, there is no ambiguity.", "$A(x)$ be a polynomial of degree at most $n$. Then we can write $A(x)=\\alpha_0+\\alpha_1(x-r)+\\alpha_2(x-r)^2+\\dots+\\alpha_n(x-r)^n$ as a sum of powers of $x-r$. To see this, notice that if we divide $A(x)$ by $(x-r)^n$, the quotient $C(x)$ must be a constant $\\alpha_n$ (since the degree of $A$ is at most $n$), and the remainder $D(x)$ has degree at most $n-1$. We can then apply Lemma 2 again and divide $D(x)$ by $(x-r)^{n-1}$ to obtain a constant $\\alpha_{n-1}$ as quotient and a polynomial of degree at most $n-2$ as remainder. Continuing this way we obtain the result. The same argument gives also that if $c(x)=(x-a)^2+p_1$ for some constant $p_1>0$ is an irreducible quadratic polynomial, then we can write $A(x)=(\\alpha_0+\\beta_0x)+(\\alpha_1+\\beta_1x)c(x)+(\\alpha_2+\\beta_2 x)c(x)^2+\\dots$ where the sum is of course finite and the numbers $\\alpha_i,\\beta_i$ are constant. Definition. The polynomial $r(x)$ is said to be a greatest common divisor of the polynomial $a(x)$ and $b(x)$ if and only if the following two conditions hold: 1. $r$ (evenly) divides both $a$ and $b$. 2. If $s$ is a polynomial that divides both $a$ and $b$, then $s$ divides $r$. There are many greatest common divisors of any two polynomials $a$ and $b$, not both equal to 0, since if $r$ is an example, then so is $5r$. However, except for constant multiples, there is no ambiguity.", "# Interpolating Polynomials A polynomial of $$n$$-th degree is uniquely determined, given its values at $$n+1$$ points. So, suppose that $$P$$ is an $$n$$-th degree polynomial and that $$P(x_i)=y_i$$ in different points $$x_0,x_1, \\dots,x_n$$. There exist unique polynomials $$E_0,E_1,\\dots,E_n$$ of $$n$$-th degree such that $$E_i(x_i)=1$$ and $$E_i(x_j)=0$$ for $$j\\neq i$$. Then the polynomial $P(x)=y_0E_0(x)+y_1E_1(x)+\\cdots+ y_nE_n(x)$ has the desired properties: indeed, $$P(x_i)=\\sum_j y_jE_j (x_i)=y_iE_i(x_i)=y_i$$. It remains to find the polynomials $$E_0,\\dots,E_n$$. A polynomial that vanishes at the $$n$$ points $$x_j$$, $$j\\neq i$$, is divisible by $$\\prod_{j\\neq i}(x-x_j)$$, from which we easily obtain $$E_i(x)=\\prod_{j\\neq i}\\frac {(x-x_j)}{(x_i-x_j)}$$. This shows that: Theorem 5.1 (Newton’s interpolating polynomial) For given numbers $$y_0,\\dots,y_n$$ and distinct $$x_0$$, $$\\dots$$, $$x_n$$ there is a unique polynomial $$P(x)$$ of $$n$$-th degree such that $$P(x_i)=y_i$$ for $$i=0$$, $$1$$, $$\\dots,n$$. This polynomial is given by the formula $P(x)=\\sum_{i=0}^ny_i\\prod_{j\\neq i}\\frac{(x-x_j)}{(x_i-x_j)}.$ Example Find the cubic polynomial $$Q$$ such that $$Q(i)=2^i$$ for $$i=0,1,2,3$$. In order to compute the value of a polynomial given in this way in some point, sometimes we do not need to determine its Newton’s polynomial. In fact, Newton’s polynomial has an unpleasant property of giving the answer in a complicated form. Example If the polynomial $$P$$ of $$n$$-th degree takes the value 1 in points $$0,2,4,\\dots,2n$$, compute $$P(-1)$$. Instead, it is often useful to consider the finite difference of polynomial", "degree polynomial by the remainder until I find a remainder of zero then, correct? 7. Originally Posted by elninio Ahhh, I see. Then I would divide the lower degree polynomial by the remainder until I find a remainder of zero then, correct? Yep, so...go on! Tonio 8. (Disregard this...>SOLVED)" ]

[ "5 '19 at 15:58 ## 1 Answer Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, $$X$$ and $$X+1$$ (this is clear since the polynomial has no zero in the prime field), and degree 2, $$X^2+X+1$$ (which is the only irreducible polynomial of this degree).", "# Dividing a polynomial Algebra Level 3 When a polynomial $$P(x)$$ of degree $$n > 2$$ is divided by $$x-3$$, the remainder is 5. When $$P(x)$$ is divided by $$x-1$$, the remainder is 1. Find the remainder when $$P(x)$$ is divided by $$x^2-4x+3$$. ×", "a polynomial that you'll eventually want to factor, don't expand the polynomial. Keep it together. What I'm saying here is do not try to expand $(14x^2 +7x-21)^2$ into a 5 term 4th degree polynomial in x. Keep it factored. One final hint: You can factor $2x^2+x-3$", "ciphertext) and when it is unencrypted (the plaintext). These are almost the normal polynomials that everyone studies at school. Things like $4 x^2 + 2 x+1$ There are some differences, the first being that the coefficients are all whole numbers and are the remainder relative to some other whole number $$t$$ (that is $$\\bmod t$$ ). Say $$t=24$$, then this is like a 24 hour clock, where adding 6 hours to 21 takes you to 3. All the coefficients of the polynomials are treated like this. Alternately, we can consider the numbers to range from -11 to 12, allowing us to negate numbers conveniently. Note that this is just a convenience factor - there is no difference between a \"remainder\" of -1 and a remainder of 23 (when dividing by 24). The second, and trickier, thing that is different about these polynomials is that this idea of using remainders applies not just to the coefficients of the polynomials, but also to the polynomials themselves. We define a special polynomial called the polynomial modulus, and only consider the remainder of polynomials when they have been divided by this polynomial modulus. The specific form of this polynomial modulus in the FV scheme is $$x^d + 1$$ where $$d=2^n$$ for some $$n$$. For illustration here we will take $$n=4$$, so the", "# Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10) Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of . – – + – – + – The final answer is the quotient plus the remainder over the divisor. Divide Using Long Polynomial Division (x^3-3x^2+x-2)÷(10x^4-14x^3-10x^2+6x-10)", "left over after dividing. 4 16x2 −12x+8 3. One method is long division, a process similar to long division of two whole numbers. Just remember that we keep going until the remainder has degree that is strictly less that the degree of the polynomial we’re dividing by, \\(x + 2$$ in this case. Answer: First, factor by grouping. In this case, we should get 4x 2 /2x = 2x and 2x(2x + 3). Set up the division problem. 3: Dividing Polynomials Worksheet is suitable for 9th - 12th Grade. <br /> 2. So we write the polynomial 2x 4 +3x 2 +x as product of x and 2x 3 + 3x +1. Polynomials are very similar to integers. Oct 20, 2020 · Multiplying And Dividing Polynomials Worksheet With Answers. G o t a d i f f e r e n t a n s w e r ? C h e c k i f i t ′ s c o r r e c t. Then, divide the first term of the divisor into the first term of the dividend, and multiply the X in the quotient by the divisor. Practice. Dividing Polynomials; Remainder and Factor Theorems . We have the divisor divided into the dividend, and of course our answer is the quotient. In this", "form and use the long division method. Let us take an .... And some disadvantages: Fitting polynomials can be problematic, when 1.. Another Example We will also be making use of the following data set in the remainder of this chapter. See polyinterpDemo2.m Here we see the primary di culty with high-degree polynomial interpolation at equally spaced points. Polynomial Division & Long Division Algorithm - BYJUS. Hence, the division algorithm is verified. Polynomial Division Questions. If the polynomial x 4 – 6x 3 + 16x 2 – 25x + 10 is divided by another polynomial x 2 – 2x + k, the remainder comes out to be x + a, find k and a. Divide the polynomial 2t 4 + 3t 3 – 2t 2 – 9t – 12 by t 2. The remainder theorem is useful because it helps us find the remainder without the actual polynomials division. Consider, for example, a number 20 is divided by 5; 20 ÷ 5 = 4. ... When one whole number is divided into another a quotient and remainder is formed. Thus gives 7 remainder 2. There are various ways to write this result. 37 ÷ 5. divide a polynomial by a monomial, divide each term in the polynomial by the monomial, and then write each quotient in lowest terms.", "# Divided polynomial algebra The divided polynomial algebra over a commutative ring $R$ in a symbol $X$ of degree $k$ is defined as the free $R$-algebra on symbols $X_1=X, X_2, X_3,\\dotsc$ of degree $k$, $2k$, $3k,\\dotsc$ modulo the relation $X_iX_j={i+j \\choose j}X_{i+j}$. Now these symbols are naturally the divided powers $\\gamma_i(X)$ of $X$ and the relation $\\gamma_i(X)={i+j \\choose j}\\gamma_{i+j}(X)$ holds. My problem: This equation, however, needs to hold for all elements of the divided polynomial algebra, e. g. for $X_2$ and I can't see why it does. Second, what about other requirements of divided powers such as $\\gamma_i(X_iX_j)=\\gamma_i(X_i)\\gamma_j(X_j)$? How is $\\gamma_i$ even defined for elements $\\neq X$? All difficulties resolve for $R\\subset \\mathbb{Q}$ so an answer for the general situation would be very much appreciated. Thank you!", "# Polynomial Arithmetic Modulo 2 (CRC Error Correcting Codes) I'm trying to understand how to calculate CRC (Cyclic Redundancy Codes) of a message using polynomial division modulo 2. The textbook Computer Networks: A Systems Approach gives the following rules for division: 1. Any polynomial $B(x)$ can be divided by a divisor polynomial $C(x)$ if $B(x)$ is of higher degree than $C(x)$. 2. Any polynomial $B(x)$ can be divided once by a divisor polynomial $C(x)$ if $B(x)$ is of the same degree as $C(x)$. 3. The remainder obtained when $B(x)$ is divided by $C(x)$ is obtained by performing the exclusive OR on each pair of matching coefficients. For example: the polynomial $x^3 +1$ can be divided by $x^3 + x^2 + 1$ because they are both of degree 3. We can find the remainder by XOR the coefficients: $1001 \\oplus 1101 = 0100$ and the quotient is obviously 1. Now, onto long division - and the source of my confusion. The book says: \"Given the rules of polynomial division above, the long division operation is much like dividing integers. We see that the division $1101$ divides once into the first four bits of the message $1001$, since they are of the same degree, and leaves the remainder $100$. The next step is to bring down a digit from the", "$A(x)$ be a polynomial of degree at most $n$. Then we can write $A(x)=\\alpha_0+\\alpha_1(x-r)+\\alpha_2(x-r)^2+\\dots+\\alpha_n(x-r)^n$ as a sum of powers of $x-r$. To see this, notice that if we divide $A(x)$ by $(x-r)^n$, the quotient $C(x)$ must be a constant $\\alpha_n$ (since the degree of $A$ is at most $n$), and the remainder $D(x)$ has degree at most $n-1$. We can then apply Lemma 2 again and divide $D(x)$ by $(x-r)^{n-1}$ to obtain a constant $\\alpha_{n-1}$ as quotient and a polynomial of degree at most $n-2$ as remainder. Continuing this way we obtain the result. The same argument gives also that if $c(x)=(x-a)^2+p_1$ for some constant $p_1>0$ is an irreducible quadratic polynomial, then we can write $A(x)=(\\alpha_0+\\beta_0x)+(\\alpha_1+\\beta_1x)c(x)+(\\alpha_2+\\beta_2 x)c(x)^2+\\dots$ where the sum is of course finite and the numbers $\\alpha_i,\\beta_i$ are constant. Definition. The polynomial $r(x)$ is said to be a greatest common divisor of the polynomial $a(x)$ and $b(x)$ if and only if the following two conditions hold: 1. $r$ (evenly) divides both $a$ and $b$. 2. If $s$ is a polynomial that divides both $a$ and $b$, then $s$ divides $r$. There are many greatest common divisors of any two polynomials $a$ and $b$, not both equal to 0, since if $r$ is an example, then so is $5r$. However, except for constant multiples, there is no ambiguity.", "$A(x)$ be a polynomial of degree at most $n$. Then we can write $A(x)=\\alpha_0+\\alpha_1(x-r)+\\alpha_2(x-r)^2+\\dots+\\alpha_n(x-r)^n$ as a sum of powers of $x-r$. To see this, notice that if we divide $A(x)$ by $(x-r)^n$, the quotient $C(x)$ must be a constant $\\alpha_n$ (since the degree of $A$ is at most $n$), and the remainder $D(x)$ has degree at most $n-1$. We can then apply Lemma 2 again and divide $D(x)$ by $(x-r)^{n-1}$ to obtain a constant $\\alpha_{n-1}$ as quotient and a polynomial of degree at most $n-2$ as remainder. Continuing this way we obtain the result. The same argument gives also that if $c(x)=(x-a)^2+p_1$ for some constant $p_1>0$ is an irreducible quadratic polynomial, then we can write $A(x)=(\\alpha_0+\\beta_0x)+(\\alpha_1+\\beta_1x)c(x)+(\\alpha_2+\\beta_2 x)c(x)^2+\\dots$ where the sum is of course finite and the numbers $\\alpha_i,\\beta_i$ are constant. Definition. The polynomial $r(x)$ is said to be a greatest common divisor of the polynomial $a(x)$ and $b(x)$ if and only if the following two conditions hold: 1. $r$ (evenly) divides both $a$ and $b$. 2. If $s$ is a polynomial that divides both $a$ and $b$, then $s$ divides $r$. There are many greatest common divisors of any two polynomials $a$ and $b$, not both equal to 0, since if $r$ is an example, then so is $5r$. However, except for constant multiples, there is no ambiguity.", "+0 # maths; confused 0 73 1 a. The degree of the polynomial \\(p(x)\\) is \\(11\\), and the degree of the polynomial \\(q(x)\\) is \\(7\\). Find all possible degrees of the polynomial \\(p(x) + q(x)\\). b. Suppose \\(f\\) is a polynomial such that \\(f(0) = 47\\), \\(f(1) = 32\\), \\(f(2) = -13\\), and \\(f(3)=16\\). What is the sum of the coefficients of \\(f\\)? Mar 16, 2019 #1 +5172 +2 \\(deg(p(x)) > deg(q(x)) \\Rightarrow deg((p+q)(x)) = deg(p(x))\\) \\(c_0 + c_1 + c_2 + \\dots + c_n = f(1) =32\\) . Mar 17, 2019 edited by Rom Mar 17, 2019 edited by Rom Mar 17, 2019", "$0$, $1$, $2$ and $3$ to a set of $11$ prescribed data points. For the approximation of degree $3$, tabulate the data and the corresponding values of the approximating polynomial, together with the residual errors, and also the values of the approximating polynomial at points half-way between each pair of adjacent data points. The example program supplied is written in a general form that will enable polynomial approximations of degrees $0,1,\\dots ,k$ to be obtained to $m$ data points, with arbitrary positive weights, and the approximation of degree $k$ to be tabulated. E02AEF is used to evaluate the approximating polynomial. The program is self-starting in that any number of datasets can be supplied.", "# Thread: long division polynomial 1. ## long division polynomial Divide 6(x)^4 - 20(x)^3 + 50(x)^2 + 30x - 2 by x^2 - 5x. Is the answer 6x2 + 10x + 100 and the remainder: 530x - 2? these numbers look weird, what did i do wrong? 2. Originally Posted by NeedHelp18 Divide 6(x)^4 - 20(x)^3 + 50(x)^2 + 30x - 2 by x^2 - 5x. Is the answer 6x2 + 10x + 100 and the remainder: 530x - 2? these numbers look weird, what did i do wrong? Weird? How so? Your answer is fine. Why do you assume that you are in error? 3. the answers correct? i thought the 530x wasnt correct.. 4. Originally Posted by NeedHelp18", "get to a remainder that's \"smaller\" (in polynomial degree) than the divisor, you're done. What am I supposed to do with the remainder? A Polynomial can be expressed in terms that only have positive integer exponents and the operations of addition, subtraction, and multiplication. Division of a polynomial by another polynomial is one of the important concept in Polynomial expressions. In this case, we should get 4x 2 /2x = 2x and 2x(2x + 3). In other words, it must be possible to write the expression without division. Intro to long division of polynomials (video) | Khan Academy − − = (−) (+ +) ⏟ + ⏟ The long division algorithm for arithmetic is very similar to the above algorithm, in which the variable x is replaced by the specific number 10.. Polynomial short division. (This is a legitimate mathematical step. You can use the Mathway widget below to practice finding doing long polynomial division. Just as you would with a simpler … This gives me –10x + 15 as my new bottom line: Dividing –10x by 2x, I get –5, which I put on top. To compute $32/11$, for instance, we ask how many times $11$ fits into $32$. In such a text, the long division above would be presented as shown here: The only difference is", "Polynomials can be divided by numbers otherwise with expressions.However we prefer the long division is what explain the concept how to divide better. ... read more #### What are PV diagrams? (article) | Khan Academy ... end subscript, equals, 160, comma, 000, ... equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 0, point, 75, m, start superscript, 3, ... ... read more #### Solve c=5/9(f-32) Tiger Algebra Solver Where a fraction equals zero, ... Slope is defined as the change in f divided by the change in c. ... Slope = 3.600/2.000 = 1.800 c-intercept = -160/9 = -17.77778 ... read more #### Dividing whole numbers -- A complete course in arithmetic i n A R I T H M E T I C. Table of Contents | Home | Introduction. Lesson 12. SHORT DIVISION. ... read more #### CHAPTER 5 PERCENTILES AND PERCENTILE RANKS ... (n) where k is the percentile and n is the number of values in the distribution. For ... 3 $160-$199 48 341 4 $130-$159 72 293 5 $100-$129 92 221 6 $80-$99 51 129 ... read more #### NUMBERS AND THE NUMBER SYSTEM - Count On six hundred thousand 600 000 sixty thousand 60 000 six thousand 6 000 six hundred 600 ... • What is 600 times 100?", "+0 # Polynomial division 0 205 1 1) The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q? 2) When the polynomial p(x) is divided by x - 1, the remainder is 3. When the polynomial p(x) is divided by x - 3, the remainder is 5. What is the remainder when the polynomial p(x) is divided by (x - 1)(x - 3)? Aug 14, 2019 $$f(x) = q(x) d(x) + r(x)\\\\ \\text{If }\\text{deg}\\;d \\le 3, \\text{then deg } r < \\text{deg }d \\le 3 \\implies \\text{Contradiction!}\\\\ \\therefore \\text{deg }d \\ge 4\\\\ \\text{Minimum possible deg }d = 4\\\\ \\implies \\text{Maximum possible deg }q = \\text{deg }f - \\text{deg }d = \\boxed5$$", "of the rectangle. ### Media Access these online resources for additional instruction and practice with polynomial division. ### 5.4 Section Exercises #### Verbal 1. If division of a polynomial by a binomial results in a remainder of zero, what can be conclude? 2. If a polynomial of degree $n n$ is divided by a binomial of degree 1, what is the degree of the quotient? #### Algebraic For the following exercises, use long division to divide. Specify the quotient and the remainder. 3. $( x 2 +5x−1 )÷( x−1 ) ( x 2 +5x−1 )÷( x−1 )$ 4. $( 2 x 2 −9x−5 )÷( x−5 ) ( 2 x 2 −9x−5 )÷( x−5 )$ 5. $( 3 x 2 +23x+14 )÷( x+7 ) ( 3 x 2 +23x+14 )÷( x+7 )$ 6. $( 4 x 2 −10x+6 )÷( 4x+2 ) ( 4 x 2 −10x+6 )÷( 4x+2 )$ 7. $( 6 x 2 −25x−25 )÷( 6x+5 ) ( 6 x 2 −25x−25 )÷( 6x+5 )$ 8. $( − x 2 −1 )÷( x+1 ) ( − x 2 −1 )÷( x+1 )$ 9. $( 2 x 2 −3x+2 )÷( x+2 ) ( 2 x 2 −3x+2 )÷( x+2 )$ 10. $( x 3 −126 )÷( x−5 ) ( x 3 −126 )÷( x−5 )$ 11. $( 3 x", "but hypotheses that fail only for polynomials of degree five or more are not unheard of. ;) - If $f$ has degree $4$ and doesn't have a degree 24 splitting field, then either $\\phi(\\alpha) \\in \\mathbb{Q}(\\alpha)$, or $\\mathbb{Q}(\\alpha, \\phi(\\alpha)) = K$ because $\\mathbb{Q}(\\alpha)$ is already degree 4, so any further extension is the whole thing. – Hurkyl Dec 7 '12 at 22:11 @Hurkyl: good point. :) – tomasz Dec 7 '12 at 22:14", "1. ## help me A Polynomial of lowest degree with integral coefficient, whose one of the Zero is root3+root2 is a) a(x^4-6x^2+10 b) a(x^4-10x^2+1) c) a(x^4-9x^2+1) d) none of these 2. If $x_1=\\sqrt{3}+\\sqrt{2}$ then $x_2=\\sqrt{3}-\\sqrt{2}, \\ x_3=-\\sqrt{3}+\\sqrt{2}, \\ x_4=-\\sqrt{3}-\\sqrt{2}$ Then $P(x)=a(x-x_1)(x-x_2)(x-x_3)(x-x_4)=$ $=a[x^4-(x_1+x_2+x_3+x_4)x^3+(x_1x_2+x_1x_3+x_1x_4+x_2x_3+ x_2x_4+x_3x_4)x^2-$ $-(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)x+x_1x_2x _3x_4]=$ $=a(x^4-10x^2+1)$" ]

[ "x }{ y } + \\frac { 2y }{ 5 } + 2 = 10; \\frac { 2x }{ 7 } – \\frac { 5 }{ 2 } + 1 = 9 view solution S", "solve -\\frac { 2} { 5} x + 8= - 4?...", "- \\frac{11}{2}$ The Solutions Are: $x = 8$ and $x = - \\frac{11}{2}$", "\\frac{5}{7}$ Solution: $x = - 6 , x = \\frac{5}{7}$ [Ans]", "$\\frac{1}{c}=\\frac{\\ln(8)-0}{7}$ Do you know how to solve this for c? 19. anonymous ln 8/ 7 = 1/c c= 7/ln8 20. myininaya right 21. anonymous ok thanks so much!", "original problem post by tucuman87 $$\\frac{dy}{dt} + f(t)y + y^2 = C$$ has a special function solution.", "a) $$x = 8$$ #5: Solutions: a) $$x = -\\frac{8}{7}$$", "y < 9 Simplify Solve for w. $\\frac{w+10}{9}\\;<\\;-5$ Solution: w", "# How do you solve \\frac{x}{2} + \\frac{x}{6} = 2? $\\frac{x}{2} + \\frac{x}{6} = 2$ $\\frac{3 x + x}{6} = 2$ $\\frac{4 x}{6} = 2$ $4 x = 2 \\cdot 6$ $4 x = 12$ $x = \\frac{12}{4}$ $x = 3$", "10}{2}$ $x = \\frac{- 2}{2}$; $x = \\frac{18}{2}$ $x = - 1$; $x = 9$ The Solution Set Is: $x = \\left\\{- 1 , 9\\right\\}$", "# How do you solve 7| \\frac { x } { 9} | = 7? Dec 16, 2016 Solution: $x = 9 \\mathmr{and} x = - 9$ $7 | \\frac{x}{9} | = 7 \\mathmr{and} | \\frac{x}{9} | = 1 \\mathmr{and} \\frac{x}{9} = 1 \\mathmr{and} x = 9$ And $7 | \\frac{x}{9} | = 7 \\mathmr{and} | \\frac{x}{9} | = 1 \\mathmr{and} \\frac{x}{9} = - 1 \\mathmr{and} x = - 9$ Solution: $x = 9 \\mathmr{and} x = - 9$ [Ans]", "the equation is \\large\\frac{(x-1)^2}{9}-\\frac{(y+1)^2}{16}$$=1$ answered Jun 19, 2013 edited Jun 19, 2013", "Prealgebra (7th Edition) $\\frac{1.5}{5}=\\frac{2.4}{n}$ Cross multiply to solve for n. $1.5n=5(2.4)$ $1.5n\\div1.5=12\\div1.5$ $n=8$", "= \\frac{7 - 9}{4} = - \\frac{2}{4} = - 0.5 \\therefore x = 4 \\mathmr{and} x = - 0.5$ Solution: $x = 4 \\mathmr{and} x = - 0.5$ [Ans]", "solution: $x = \\frac{13}{7}$, $y = - \\frac{2}{7}$", "# How do you solve \\frac { 1} { 2} + x = 10? $9.5$ Make $x$ the subject of the equation. $x = 10 - \\frac{1}{2}$ $x = 9.5$", "= \\frac { 1}{ 5 }; \\frac { 3 }{ x } + \\frac { 2 }{ 3y } = 2 and also find ‘a’ for which y = ax – 2 view solution S", "\\frac{1}{u} + \\frac{1}{v} = \\frac{36}{5} (ii) \\frac{11}{v} – \\frac{7}{u} = 1 \\frac{9}{v} + \\frac{4}{u} = 6 view solution S", "two solutions: $x = \\frac{1 + \\sqrt{5}}{2}$ and $x = \\frac{1 - \\sqrt{5}}{2}$.", "integer solutions we have $(M-1)|N$ i.e. $M = 1 + a$ where $a|N$. Then $n_{N-1} = N + \\frac{N}{a} - 1$ $$n_{N-2} = N-2 + \\frac{n_N + n_{N-1} + n_{N-2} - (N-2)}{1+a}$$ $$n_{N-2} - (N-2) - \\frac{n_{N-2} - (N-2)}{1+a} = \\frac{N + N + \\frac{N}{a}-1}{1+a}$$ $$a(n_{N-2} - (N-2)) = 2N + \\frac{N}{a} - 1$$ -" ]

[ "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "And we have: . $a(n) \\;=\\;\\frac{n(n+1)}{2} - (n-1)$ . . Therefore: . $\\boxed{a(n) \\;=\\;\\frac{n^2-n+2}{2}}$ 8. Thanks so much Soroban! I actually get that *victory dance*", "= 12$ Thus we have: $x' = - \\frac {(12)(-3)}{5} \\implies \\boxed { x' = 7.2 }$ Note, $x' = \\frac {dx}{dt}$ and $y' = \\frac {dy}{dt}$", "(16 - 1)^2 + 1 = \\boxed{406}$.", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "\\\\ then $10x=3.\\dot{3}$ \\\\ $10x-x=3.\\dot{3}-0.\\dot{3}$ \\\\ $9x=3$ \\\\ $9x=\\frac{3}{9}$ \\item Express $0.\\dot{1}\\dot{2}$ as a fraction. \\\\ Let $x=0.\\dot{1}\\dot{2}$ \\\\ then $100x=12.\\dot{1}\\dot{2}$ \\\\ $100x-10x=12.\\dot{1}\\dot{2}-0.\\dot{1}\\dot{2}$ \\\\ $90x=12$ \\\\ $x=\\frac{12}{90}$ \\\\ $x=\\frac{2}{15}$ \\end{enumerate} \\end{document} • thanks bernard, I will read through this and try to understand! – user1448093 May 27 '14 at 17:20", "+ n}{2} \\;=\\;\\frac{n(n+1)(2n+1)}{2}$ Therefore: . $\\boxed{\\sum k^2 \\;=\\;\\frac{n(n+1)(2n+1)}{6}}$", "& -- \\\\ 2x-1 & ) & 2x^3 & -3x^2 & -11x & +6 \\\\ & & 2x^3 & -x^2 \\\\ & & -- & -- \\\\ & & & -2x^2 & -11x \\\\ & & & -2x^2 & +x \\\\ & & & -- & -- \\\\ & & & & -12x & +6 \\\\ & & & & -12x & +6 \\\\ & & & & -- & --\\end{array}$ Therefore: . $2x^3-3x^2-11x+6 \\:=\\:(2x-1)(x^2-x-6) \\:=\\:\\boxed{(2x-1)(x+2)(x-3)}$", "Hence, $r^2=x(x-5)$. Similarly, $\\cot\\frac{B}{2}=\\frac{14-x}{r}$, and $\\cot\\frac{D}{2}=\\frac{12-x}{r}=\\tan\\frac{B}{2}=\\frac{r}{14-x}$. Therefore, $r^2=\\frac{12-x}{14-x}$. Putting this together with the above equation yields $$(12-x)(14-x)=x(x-5)$$$$\\Longrightarrow x^2-26x+168=x^2-5x$$$$\\Longrightarrow 21x=168$$$$\\Longrightarrow x=8.$$ Thus, $r^2=8(8-5)=8(3)=24$, and $r=\\boxed{\\textbf{(C)}\\ 2\\sqrt{6}}$.", "= 1 + 16/(x^3 - 8). x^3-8 can easily be factored, and so use partial fractions on that integral. 10. Jun 17, 2009 ### HallsofIvy Actually, $x^3+ 8$ can be factored as easily as $x^3- 8$. In fact, [itexs]x^n+ a^n[/itex] can be factored easily for all odd n. 11. Jun 17, 2009 ### Дьявол $$x^3+8=x^3+2^3=(x+2)(x^2-2x+4)$$ $$x^3-8=x^3-2^3=(x-2)(x^2+2x+4)=(x-2)(x+2)^2$$ Another way is: $$\\frac{x^3+8}{x^3-8}=\\frac{x^3-8+16}{x^3-8}=1+\\frac{16}{x^3-8}$$ Regards.", "latex threw me off too. That's why I got so many answers and what not. 50. UsukiDoll Cleaned up version $[-8 -14-2x] - ([-19 +17 -1] + [14 +13 -2]) = [-3 -44 -9]$ $\\[[-8 -14-2x] - ([-19+14 , 17+13 , -1-2] ) = [-3 -44 -9]$ $\\[[-8 -14-2x] - ([-5, 30 , -3] ) = [-3 -44 -9]$ $\\[[-8 -14-2x] +([5, -30 , 3] ) = [-3 -44 -9]$ $\\[[-8+5, -14-30,-2x+3] ) = [-3 -44 -9]$ $\\[[-3, -44,-2x+3] ) = [-3 -44 -9]$ $-2x+3=-9$ $-2x+3-3=-9-3$ $-2x=-12$ $\\frac{-2x}{-2}=\\frac{-12}{-2}$ $x=6$ $\\[[-3, -44,-2(6)+3] ) = [-3 -44 -9]$ $\\[[-3, -44,-12+3] ) = [-3 -44 -9]$ $\\[[-3, -44,-9] ) = [-3 -44 -9]$", "Multiply by $$(3^1-2^1)=1$$ $$(3^1-2^1)(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\\dots (2^{64} + 3^{64})\\\\ =(3^2-2^2)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\\dots (2^{64} + 3^{64})\\\\=(3^4-2^4)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\\dots (2^{64} + 3^{64})$$ Sayan Dutta shows a clever way. Another way is to note that when you multiply the parentheses, each term will be of the form $$2^n 3^{127-n}$$: $$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64}) = \\prod_{k=0}^{6} (2^{2^k}+3^{2^k}) \\\\ = \\sum_{n=0}^{127} 2^n 3^{127-n} = 3^{127} \\sum_{n=0}^{127} (2/3)^n = 3^{127} \\frac{1-(2/3)^{128}}{1-2/3} = 3^{128} \\frac{1-(2/3)^{128}}{3-2} = 3^{128} - 2^{128}.$$", "+0 # HELP!!! +1 149 2 Let $$x$$ be a value such that $$9x^2 - 18x - 16 = 0$$ and $$15x^2 + 28x + 12 = 0.$$ What is the value of $$x$$? Express your answer as a simplified common fraction. Mar 17, 2019 #1 +7713 +2 $$9x^2-18x-16=0\\\\ (3x-8)(3x+2) = 0\\\\ x = \\dfrac{8}{3}\\text{ OR }x = \\dfrac{-2}{3}$$ $$15x^2+28x+12 = 0\\\\ (3x+2)(5x+6) = 0\\\\ x = \\dfrac{-2}{3} \\text{ OR } x = \\dfrac{-6}{5}$$ Only x = -2/3 satisfies both equations. Therefore the answer is -2/3. Mar 17, 2019 #2 +4323 +3 Since they both are equal to 0, we can set them equal to each other. Therefore, $$9x^2-18x-16=15x^2+28x+12$$, and by solving the quadratic, we get two solutions, $$x=-7,\\:x=-\\frac{2}{3}$$, but by guess and check, $$\\boxed{x=-\\frac{2}{3}}.$$ . Mar 17, 2019", "they are trying to show you here. Let us use a symbol $$g = \\gcd(2x^2 + 15 x - 8, \\; \\; x^2 + 10 x + 16)$$ First we find $$2x^2 + 15 x - 8 - 2 \\cdot ( x^2 + 10 x + 16) = -5x-40 = -5 (x+8)$$ So, possibly with rational quotient, $$g = \\gcd(x^2 + 10 x + 16, \\; \\; x+8).$$ But then $$x^2 + 10 x + 16 - (x+2)(x+8) = 0,$$ so $$g = x+8.$$ To find the quotient $$\\frac{f(x)}{g(x)} = \\frac{2x^2 + 15x - 8}{x^2 + 10x + 16}$$ we must find what we need to multiply $x^2$ by to obtain $2x^2$. Clearly, this is $2$. $$2x^2 + 15x - 8 = 2(x^2 + 10x + 16) - 5x - 40$$ Since $-5x - 40$ has smaller degree than $x^2 + 10x + 16$, $-5x - 40$ is the remainder. Hence, the quotient is $$\\frac{2x^2 + 15x - 8}{x^2 + 10x + 6} = 2 + \\frac{-5x - 40}{x^2 + 10x + 16}$$ Check: We multiply the quotient by the divisor to show that it equals the dividend. \\begin{align*} (x^2 + 10x + 16)\\left(2 + \\frac{-5x - 40}{x^2 + 10x + 16}\\right) & = 2(x^2 + 10x + 16) - 5x - 40\\\\ & = 2x^2 + 20x", "screen: Therefore, ${_{10}}P_3= 720$ , which means that the number of ways that 3 favorite desserts can be listed in order from a menu of 10 is 720. Guided Practice Verify that the $nPr$ option on the PRB menu of the TI calculator works by calculating ${_{18}}P_{18}$ with both this option and with the factorial function. Note that with the TI calculator, you can find the factorial function by pressing $\\boxed{\\text{MATH}}$ , pressing the right arrow 3 times, and pressing $\\boxed{4}$ . First, lets compute ${_{18}}P_{18}$ with the factorial function. Remember, the formula to solve permutations like these is: ${_n}P_r=\\frac{n!}{(n-r)!}$ This means that ${_{18}}P_{18}$ can be written as follows: ${_{18}}P_{18} &= \\frac{18!}{(18-18)!}\\\\{_{18}}P_{18} &= \\frac{18!}{0!}\\\\{_{18}}P_{18} &= \\frac{18!}{1}\\\\{_{18}}P_{18} &= 18!$ To find 18!, enter the following into your TI calculator: $\\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{MATH}} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ (\\text{PRB}) \\ \\boxed{4} \\ (\\text{!}) \\ \\boxed{\\text{ENTER}}$ After pressing $\\boxed{\\text{ENTER}}$ , you should see the following on your calculator's screen: Now let's compute ${_{18}}P_{18}$ with the $nPr$ option on the PRB menu. $\\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{MATH}} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ (\\text{PRB}) \\ \\boxed{2} \\ (\\text{nPr}) \\ \\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{ENTER}}$ After pressing $\\boxed{\\text{ENTER}}$ , you should see the following on your calculator's screen: The answers for the 2 methods are the same, so", "20. $f(x) = (x^4+4x+2)(2x+3) \\,$ $10x^4+12x^3+16x+16$ 21. $f(x) = (2x-1)(3x^2+2) \\,$ $18x^2-6x+4$ 22. $f(x) = (x^3-12x)(3x^2+2x) \\,$ $15x^4+8x^3-108x^2-48x$ 23. $f(x) = (2x^5-x)(3x+1) \\,$ $36x^5+10x^4-6x-1$ $f(x) = (5x^2+3)(2x+7)\\,$ $30x^2+70x+6$ $f(x) = 3x^2(5x^2+1)^4 \\,$ $6x(25x^2+1)(5x^2+1)^3$ $f(x) = x^3(2x^2-x+4)^4 \\,$ $x^2(2x^2-x+4)^3(22x^2-7x+12)$ $f(x) = 5x^2(x^3-x+1)^3 \\,$ $5x(x^3-x+1)^2(11x^3-2x+1)$ $f(x) = (2-x)^6(5+2x)^4 \\,$ $2(x-2)^5(2x+5)^3(10x+7)$ Solutions ### Quotient Rule 24. $f(x) = \\frac{2x+1}{x+5} \\,$ $f'(x)=\\frac{9}{(x+5)^2}$ 25. $f(x) = \\frac{3x^4+2x +2}{3x^2+1} \\,$ $f'(x)=\\frac{18x^5+12x^3-6x^2-12x+2}{(3x^2+1)^2}$ 26. $f(x) = \\frac{x^\\frac{3}{2}+1}{x+2} \\,$ $f'(x)=\\frac{x\\sqrt{x}+6\\sqrt{x}-2}{2(x+2)^2}$ 27. $d(u) = \\frac{u^3+2}{u^3} \\,$ $d'(u)=-\\frac{6}{u^4}$ 28. $f(x) = \\frac{x^2+x}{2x-1} \\,$ $f'(x)=\\frac{2x^2-2x-1}{(2x-1)^2}$ 29. $f(x) = \\frac{x+1}{2x^2+2x+3} \\,$ $f'(x)=\\frac{-2x^2-4x+1}{(2x^2+2x+3)^2}$ 30. $f(x) = \\frac{16x^4+2x^2}{x} \\,$ $f'(x)=48x^2+2$ $f(x) = \\frac{8x^3+2}{5x+5} \\,$ $f'(x)=\\frac{2(8x^3+12x^2-1)}{5(x+1)^2}$ $f(x) = \\frac{(3x-2)^2}{x^{1/2}} \\,$ $f'(x)=\\frac{(3x-2)(9x+2)}{2x^{3/2}}$ $f(x) = \\frac{ x^{1/2}}{2x-1} \\,$ $f'(x)=\\frac{-(2x+1)}{2x^{1/2} (2x-1)^2}$ $f(x) = \\frac{ 4x-3}{x+2} \\,$ $f'(x)=\\frac{11}{(x+2)^2}$ $f(x) = \\frac{ 4x+3}{2x-1} \\,$ $f'(x)=\\frac{-10}{(2x-1)^2}$ $f(x) = \\frac{ x^2}{x+3} \\,$ $f'(x)=\\frac{x(x+6)}{(x+3)^2}$ $f(x) = \\frac{ x^5}{3-x} \\,$ $f'(x)=\\frac{x^4(-4x+15)}{(3-x)^2}$ Solutions ### Chain Rule 31. $f(x) = (x+5)^2 \\,$ $f'(x)=2(x+5)$ 32. $g(x) = (x^3 - 2x + 5)^2 \\,$ $g'(x)=2(x^{3}-2x+5)(3x^{2}-2)$ 33. $f(x) = \\sqrt{1-x^2} \\,$ $f'(x)=-\\frac{x}{\\sqrt{1-x^{2}}}$ 34. $f(x) = \\frac{(2x+4)^3}{4x^3+1} \\,$ $f'(x)=\\frac{6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}$ 35. $f(x) = (2x+1)\\sqrt{2x+2} \\,$ $f'(x)=2\\sqrt{2x+2}+\\frac{2x+1}{\\sqrt{2x+2}}$ 36. $f(x) = \\frac{2x+1}{\\sqrt{2x+2}} \\,$ $f'(x)=\\frac{2x+3}{(2x+2)^{3/2}}$ 37. $f(x) = \\sqrt{2x^2+1}(3x^4+2x)^2 \\,$ $f'(x)=\\frac{2x(3x^{4}+2x)^{2}}{\\sqrt{2x^{2}+1}}+\\sqrt{2x^{2}+1}(2)(3x^{4}+2x)(12x^{3}+2)$ 38. $f(x) = \\frac{2x+3}{(x^4+4x+2)^2} \\,$ $f'(x)=\\frac{2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}$ 39. $f(x) = \\sqrt{x^3+1}(x^2-1) \\,$ $f'(x)=\\frac{3x(x^{2}-1)}{2\\sqrt{x^{3}+1}}+2x\\sqrt{x^{3}+1}$ 40. $f(x) = ((2x+3)^4 + 4(2x+3) +2)^2 \\,$ $f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)$ 41. $f(x) = \\sqrt{1+x^2} \\,$ $f'(x)=\\frac{x}{\\sqrt{1+x^{2}}}$ Solutions ### Exponentials 42. $f(x)", "+ 4a^2 + 2$ 9. 7) $-84r^2s^2t^3$ 10. $(3x - 2)(2x^2 + x - 3)$ $2x^2(3x-2) + x(3x - 2) -3(3x -2)$ $6x^3-4x^2+3x^2-2x-9x^2+6$ $6x^3-10x^2-2x+6$ 11. $\\frac{-2b^2c^3}{3}$ 12. $= \\ \\frac{(ab)(ab^2-2a^2b^3+1+ab)}{-ab}$ $= \\ -(ab^2-2a^2b^3+1+ab)$ 13. 12) $ 6x^2y(3y-5xy^2+9x)$", "= 3 \\cos 0 + 6 \\cos \\frac{2 \\pi}{3} = 3 + 6 \\left( -\\frac{1}{2} \\right) = 0.$$ Also, $$\\cos 3x_1 + \\cos 3x_2 + \\dots + \\cos 3x_9 = 3 \\cos 0 + 6 \\cos 2 \\pi = \\boxed{9}.$$ Since $$\\cos x \\le 1$$ for all $$x,$$ this is clearly the maximum. 🔥🔥🔥 HELPMEEEEEEEEEEEEE Apr 8, 2020 #8 +30038 +1 More generally: Apr 9, 2020", "c} + 2c = -\\frac{15}{4} - \\frac{11}{4} + \\frac{1}{2} = -6$, so $(a+b)^2 = 36$. Adding up both values gives $49 + 36 = \\boxed{085}$ as our final answer.", "3)/ (x – 4) = $$3\\frac{1}{3}$$ (ii) 6/x – 2/(x – 1) = 1/(x – 2). (i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = $$3\\frac{1}{3}$$ [(x – 1) (x – 4) + (x – 2) (x – 3)]/ (x – 2) (x – 4) = 10/3 ..( LCM of (x-2) and (x-4) = (x-2) (x-4) (x2 – 5x + 4 + x2 – 5x + 6)/ (x2 – 6x + 8) = 10/3 (2x2 – 10x + 10)/ (x2 – 6x + 8) = 10/3 10x2 – 60x + 80 = 6x2 – 30x + 30 10x2 – 60x + 80 – 6x2 + 30x – 30 = 0 4x2 – 30x + 50 = 0 2x2 – 15x + 25 = 0 … (Multiplying 1/2 on both side ) 2x2 – 10x – 5x + 25 = 0 … (Using the factorisation method) 2x (x – 5) – 5 (x – 5) = 0 (x – 5) (2x – 5) = 0 If (x – 5) = 0 then, x = 5 If (2x – 5) = 0 then x = 5/2 (ii) 6/x – 2/(x – 1) = 1/(x – 2) (6x – 6 – 2x)/ x (x – 1) = 1/ (x – 2) … ( On LHS" ]

[ "### Home > CALC > Chapter 8 > Lesson 8.3.2 > Problem8-114 8-114. $\\text{Then }x=\\frac{1}{2}(U+1).$ Let U = 2x − 1. $\\frac{dU}{dx}=2$ $\\frac{1}{2}dU=dx$ Rewrite the integrand and don't forget to write the bounds in terms of U. $=\\int_1^9\\frac{U+1}{4\\sqrt{U}}dU=\\frac{1}{4}\\int_1^9(U^{1/2}+U^{-1/2})=\\frac{16}{3}$ Let U = 1 − x² $=\\int_1^5\\frac{1}{2u}du=\\frac{1}{2}\\ln(u)\\Big|_1^5=\\frac{1}{2}\\ln(5)$ $=\\int_0^{-4}-\\frac{3}{2}e^udu=\\frac{3}{2}-\\frac{3}{2}e^{-4}$", "+ n}{2} \\;=\\;\\frac{n(n+1)(2n+1)}{2}$ Therefore: . $\\boxed{\\sum k^2 \\;=\\;\\frac{n(n+1)(2n+1)}{6}}$", "• # question_answer Solve for $x:\\frac{1}{x+1}+\\frac{2}{x+2}=\\frac{4}{x+4},x\\ne -1,-2,-4$ We have, $\\frac{1}{x+1}+\\frac{2}{x+2}=\\frac{4}{x+4},x\\ne -1,-2,-4$ $(x+2)(x+4)+2(x+1)(x+4)=4(x+1)(x+2)$ ${{x}^{2}}+2x+4x+8+2({{x}^{2}}+x+4x+4)=4({{x}^{2}}+x+2x+2)$ ${{x}^{2}}+6x+8+2{{x}^{2}}+10x+8=4{{x}^{2}}+12x+8$ $3{{x}^{2}}+16x+16=4{{x}^{2}}+12x+8$ ${{x}^{2}}-4x-8=0$ $\\therefore x=\\frac{4\\pm \\sqrt{16+32}}{2}$ $x=\\frac{4\\pm \\sqrt{48}}{2}=2\\pm 4\\sqrt{2}$ $\\therefore x=2\\pm 2\\sqrt{3}$", "screen: Therefore, ${_{10}}P_3= 720$ , which means that the number of ways that 3 favorite desserts can be listed in order from a menu of 10 is 720. Guided Practice Verify that the $nPr$ option on the PRB menu of the TI calculator works by calculating ${_{18}}P_{18}$ with both this option and with the factorial function. Note that with the TI calculator, you can find the factorial function by pressing $\\boxed{\\text{MATH}}$ , pressing the right arrow 3 times, and pressing $\\boxed{4}$ . First, lets compute ${_{18}}P_{18}$ with the factorial function. Remember, the formula to solve permutations like these is: ${_n}P_r=\\frac{n!}{(n-r)!}$ This means that ${_{18}}P_{18}$ can be written as follows: ${_{18}}P_{18} &= \\frac{18!}{(18-18)!}\\\\{_{18}}P_{18} &= \\frac{18!}{0!}\\\\{_{18}}P_{18} &= \\frac{18!}{1}\\\\{_{18}}P_{18} &= 18!$ To find 18!, enter the following into your TI calculator: $\\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{MATH}} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ (\\text{PRB}) \\ \\boxed{4} \\ (\\text{!}) \\ \\boxed{\\text{ENTER}}$ After pressing $\\boxed{\\text{ENTER}}$ , you should see the following on your calculator's screen: Now let's compute ${_{18}}P_{18}$ with the $nPr$ option on the PRB menu. $\\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{MATH}} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ (\\text{PRB}) \\ \\boxed{2} \\ (\\text{nPr}) \\ \\boxed{1} \\ \\boxed{8} \\ \\boxed{\\text{ENTER}}$ After pressing $\\boxed{\\text{ENTER}}$ , you should see the following on your calculator's screen: The answers for the 2 methods are the same, so", "then find the ratio of numbers. a) $$\\frac{8±3\\sqrt{5}}{1}$$ b) $$\\frac{8±3\\sqrt{7}}{1}$$ c) $$\\frac{6±3\\sqrt{5}}{1}$$ d) $$\\frac{6±3\\sqrt{7}}{1}$$ Explanation: We know, G.M. of two numbers a and b is √ab. So, a + b = 4 √ab Squaring we get, a2+b2 = 16ab =>(a/b) + (b/a) = 16 Let x = a/b. So, x + 1/x = 16 => x2 – 16x + 1 = 0 =>x = $$\\frac{16±\\sqrt{256-4}}{2} = \\frac{16±\\sqrt{252}}{2} = \\frac{16±6\\sqrt{7}}{2} = \\frac{8±3\\sqrt{7}}{1}$$. Note: Join free Sanfoundry classes at Telegram or Youtube 4. The ratio of the A.M. and G.M. of two positive numbers a and b is 5: 3. Find the ratio of a to b. a) 9:1 b) 3:5 c) 1:9 d) 3:1 Explanation: (A.M.)/(G.M.) = 5/3 => $$\\frac{a+b}{2\\sqrt{ab}} = \\frac{5}{3}$$ Applying componendo and dividendo rule, we get => $$\\frac{a+b+2\\sqrt{ab}}{a+b-2\\sqrt{ab}} = \\frac{8}{2}$$ => $$(\\frac{\\sqrt{a}+\\sqrt{b}}{\\sqrt{a}-\\sqrt{b}})^2=4$$ => $$(\\frac{\\sqrt{a}+\\sqrt{b}}{\\sqrt{a}-\\sqrt{b}})^1=2$$ => $$(\\frac{\\sqrt{a}}{\\sqrt{b}})^1=3$$ Again applying componendo and dividendo rule, we get a/b = (3/1)2 = 9. So, a:b =9:1. Sanfoundry Global Education & Learning Series – Mathematics – Class 11. To practice Mathematics Objective Questions and Answers for Class 11, here is complete set of 1000+ Multiple Choice Questions and Answers.", "## anonymous one year ago 6|6-4x|=8x+4 1. anonymous x= 1 or x= 5/2 2. anonymous How did you get that answer. 3. Jack1 6|6-4x|=8x+4 divide both sides by 6 |6-4x|=8/6x+4/6 now square both sides 16x^2 - 48x + 36 = (16x^2)/9 + (16x)/9 + 4/9 now make one side = 0 then solve 4. Jack1 $$\\Large 6|6-4x|=8x+4$$ $$\\Large |6-4x|=\\frac86x+\\frac46$$ $$\\Large \\sqrt{(6-4x)^2}=\\frac86x+\\frac46$$ $$\\Large (~~\\sqrt{(6-4x)^2}~~)^2=(~~\\frac86x+\\frac46~~)^2$$ $$\\Large (6-4x)^2=\\frac{16}9x^2 + \\frac{16}9x + \\frac49$$ $$\\Large 16x^2 - 48x + 36=\\frac{16}9x^2 + \\frac{16}9x + \\frac49$$ $$\\Large \\frac{144}9x^2 - \\frac{432}9x + \\frac{324}9=\\frac{16}9x^2 + \\frac{16}9x + \\frac49$$ $$\\Large \\frac{128}9x^2 - \\frac{448}9x + \\frac{320}9=0$$ $$\\Large 128x^2 - 448x + 320=0$$ $$\\Large 2x^2 - 7x + 5=0$$ solve for x", "\\frac{2\\pi}7}2 + \\frac{\\cos \\frac{4\\pi}7 + \\cos \\frac{4\\pi}7}2 + \\frac{\\cos \\frac{6\\pi}7 + \\cos \\frac{2\\pi}7}2$ Which ends up being: $\\cos \\frac{2\\pi}7 + \\cos \\frac{4\\pi}7 + \\cos \\frac{6\\pi}7$ Which is shown in the next part to equal $-\\frac{1}2$, so $b = -\\frac{1}2$ Part 3: solving for a and b as the sum of roots $a$ is the negation of the sum of roots $a = - \\left( \\cos \\frac{2\\pi}7 + \\cos \\frac{4\\pi}7 + \\cos \\frac{6\\pi}7 \\right)$ The real values of the 7th roots of unity are: $1, \\cos \\frac{2\\pi}7, \\cos \\frac{4\\pi}7, \\cos \\frac{6\\pi}7, \\cos \\frac{8\\pi}7, \\cos \\frac{10\\pi}7, \\cos \\frac{12\\pi}7$ and they sum to $0$. If we subtract 1, and condense identical terms, we get: $2\\cos \\frac{2\\pi}7 + 2\\cos \\frac{4\\pi}7 + 2\\cos \\frac{6\\pi}7 = -1$ Therefore, we have $a = -\\left(-\\frac{1}2\\right) = \\frac{1}2$ Finally multiply $abc = \\frac{1}2 * - \\frac{1}2 * -\\frac{1}8 = \\frac{1}{32}$ or $\\boxed{D) \\frac{1}{32}}$. ~Tucker ## Solution 2 (Approximation) Letting the roots be $p$, $q$, and $r$, Vietas gives \\begin{align*} p + q + r &= a \\\\ pq + qr + pq &= -b \\\\ pqr &= c. \\end{align*} We use the Taylor series for $\\cos x$, $$\\cos x = \\sum_{k = 0}^{\\infty} (-1)^k \\frac{x^{2k}}{(2k)!}$$ to approximate the roots. Taking the sum up to $k = 3$ yields a close approximation, so we have $$\\cos\\left(\\frac{2\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{2\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{6}}{720}", "$\\frac{x}{3} + 10 = \\frac{x}{10} + 7$ 5. $x + 5 = \\frac{x}{8} + 9$ 6. $\\frac{x}{4} + 1 = \\frac{x}{9} + 2$ 7. $\\frac{x}{8} + 9 = \\frac{x}{10} + 6$ 8. $\\frac{x}{10} + 7 = \\frac{x}{2} + 9$", "What is the solution: (9x+2)(4x^2+35x-9)=0 $$(9x+2)(4x^2+35x-9)=0\\\\\\\\9x+2=0\\\\9x=-2\\\\x=-\\frac29\\\\or\\\\4x^2+35x-9=0\\\\\\Delta=1225+144=1369\\ \\ \\sqrt{\\Delta}=37\\\\x_1=\\frac{-35-37}{2 \\cdot 4}=-9\\\\\\\\x_2=\\frac{-35+37}{2\\cdot 4}=\\frac14\\\\\\\\x\\in\\{-9,\\frac29,\\frac14\\}$$ $$(9x+2)(4x^2+35x-9)=0\\iff9x+2=0\\ or\\ 4x^2+35x-9=0\\\\\\\\(\\#1)\\\\9x+2=0\\ \\ \\ \\ \\ \\ |subtract\\ 2\\ from\\ both\\ sides\\\\9x=-2\\ \\ \\ \\ \\ \\ \\ |divide\\ both\\ sides\\ by\\ 9\\\\\\boxed{x=-\\frac{2}{9}}\\\\\\\\(\\#2)\\\\4x^2+35x-9=0\\\\4x^2+36x-x-9=0\\\\4x(x+9)-1(x+9)=0\\\\(x+9)(4x-1)=0\\iff x+9=0\\ or\\ 4x-1=0\\\\x=-9\\ or\\ 4x=1\\\\\\boxed{x=-9}\\ or\\ \\boxed{x=\\frac{1}{4}}\\\\\\\\Solutions: x=-\\frac{2}{9}\\ or\\ x=-9\\ or\\ x=\\frac{1}{4}.$$ RELATED:", "And we have: . $a(n) \\;=\\;\\frac{n(n+1)}{2} - (n-1)$ . . Therefore: . $\\boxed{a(n) \\;=\\;\\frac{n^2-n+2}{2}}$ 8. Thanks so much Soroban! I actually get that *victory dance*", "u + 10}} \\;du = \\frac{\\frac{5}{2} ((2 u + 4) – 7)}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\\\ = \\frac{5}{2} \\int \\frac{2 u + 4}{\\sqrt{u ^{2} + 4 u + 10}} \\;du – 7 \\int \\frac{1}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\\\ Suppose,\\; I_{1} = \\int \\frac{2 u + 4}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\;and\\; I_{2} = \\int \\frac{1}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\\\ \\int \\frac{5 u + 3}{\\sqrt{u ^{2} + 4 u + 10}} \\;du = \\frac{5}{2} I_{1} – 7 I_{2} \\;\\; ….. (1)$$ $$I_{1} = \\int \\frac{2 u + 4}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\\\ Suppose,\\; u ^{2} + 4 u + 10 = z \\\\ (2 u + 4) \\;du = dz \\\\ I_{1} = \\int \\frac{dz}{\\sqrt{z}} = 2 \\sqrt{z} = 2 \\sqrt{u ^{2} + 4 u + 10} ….. (2) I_{2} = \\int \\frac{1}{\\sqrt{u ^{2} + 4 u + 10}} \\;du \\\\ = \\int \\frac{1}{\\sqrt{(u ^{2} + 4 u + 4) + 6}} \\;du \\\\ = \\int \\frac{1}{(u + 2) ^{2} + (\\sqrt{6}) ^{2}} \\;du \\\\ =log \\left |(u + 2)\\sqrt{u ^{2} + 4 u + 10} \\right | ….. (3)$$ Substituting equations (2) and (3) in equation (1), we get, $$\\int \\frac{5 u + 3}{\\sqrt{u ^{2} + 4 u +", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "has to be $\\frac{1}{16},$ so $\\frac{p}{q}$ is $\\frac{9}{16}$ and $q-p$ is $\\boxed{\\textbf{(A) } 7}.$ ~purplepenguin2 ## Solution 6 (Solves for p) Cross-multiply the inequality to get $$35q < 63p < 36q.$$ Then, we have $0 < 63p-35q < q,$ or $$0 < 7(9p-5q) < q.$$ Since $p$ and $q$ are integers, $9p-5q$ is an integer. To minimize $q,$ start from $9p-5q=1,$ which gives $p=\\frac{5q+1}{9}.$ This limits $q$ to be greater than $7,$ so test values of $q$ starting from $q=8.$ However, $q=8$ to $q=14$ do not give integer values of $p.$ Once $q>14,$ it is possible for $9p-5q$ to be equal to $2,$ so $p$ could also be equal to $\\frac{5q+2}{9}.$ The next value, $q=15,$ is not a solution, but $q=16$ gives $p=\\frac{5\\cdot 16 + 1}{9} = 9.$ Thus, the smallest possible value of $q$ is $16,$ and the answer is $16-9= \\boxed{\\textbf{(A) } 7}.$ ## Solution 7 (Inspection) Start with $\\frac{5}{9}.$ Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\\frac{5}{9},$ add $1$ to the numerator; otherwise, if it is greater than or equal to $\\frac{4}{7},$ add one $1$ to the denominator. We have $$\\frac{5}{9}, \\frac{6}{9}, \\frac{6}{10}, \\frac{6}{11}, \\frac{7}{11}, \\frac{7}{12}, \\frac{7}{13}, \\frac{8}{13}, \\frac{8}{14}, \\frac{8}{15}, \\frac{9}{15}, \\frac{9}{16}.$$ Therefore, the answer is $16 - 9 = \\boxed{\\textbf{(A) } 7}.$", "⇒ x=4 n=2, x=4×2=8 ⇒ x=8 n=3, x=4×3 ⇒ x=12 ∴ Given set in roster (Tabular) form is $A_{5}=\\{0,4,8,12\\}$ (vi) $\\mathbf{A}_{6}=\\left\\{\\boldsymbol{x} : \\boldsymbol{x}=\\frac{\\boldsymbol{n}}{n+2} ; \\boldsymbol{n} \\in \\mathbf{N} \\& \\boldsymbol{n} > \\mathbf{5}\\right\\}$ Solution: Given $x=\\frac{n}{n+2}$ When n=6, $x=\\frac{6}{6+2}$ [∵ n > 5] $→ \\quad x=\\frac{6}{8} → x=\\frac{3}{4}$ When, $n=7, x=\\frac{7}{7+2} → x=\\frac{7}{9}$ When, $n=8, x=\\frac{8}{8+2} → x=\\frac{8}{10}$ $→ \\quad x=\\frac{4}{5}$ When, $n=9, x=\\frac{9}{9+2} → x=\\frac{9}{11}$ ∴ Given set in roster (Tabular) form is $\\mathrm{A}_{6}=\\left\\{\\frac{3}{4}, \\frac{7}{9}, \\frac{4}{5}, \\frac{9}{11}, \\ldots\\right\\}$ Question 2. Write the following sets in set-builder (Rule Method) form: (i) $B_{1}=\\{6,9,12,15, \\dots\\}$ Solution: ${x:x=3n+3;n∈N}$ (ii) $B_{2}=\\{11,13,17,19\\}$ Solution: {x: x is a prime number between 10 and 20} (iii) $\\mathrm{B}_{3}=\\left\\{\\frac{1}{3}, \\frac{3}{5}, \\frac{5}{7}, \\frac{7}{9}, \\frac{9}{11}, \\ldots\\right\\}$ Solution: $\\left\\{x : x=\\frac{n}{n+2}, \\text { where } n \\text { is an odd natural number }\\right\\}$ (iv) $B_{4}=\\{8,27,64,125,216\\}$ Solution: $=\\left\\{x : x=n^{3} ; n \\in N \\text { and } 2 \\leq n \\leq 6\\right\\}$ (v) $B_{5}=\\{-5,-4,-3,-2,-1\\}$ Solution: ={x:x∈Z,-5≤x≤-1} (vi) $\\mathbf{B}_{6}=\\{\\ldots,-6,-3,0,3,6, \\dots\\}$ Solution: ={x:x=3n,n∈Z} Question 3. (i) Is {1, 2, 4, 16, 64} = {x : x is a factor of 32} ? Give reason. Solution: No, {1, 2, 4, 16, 64}≠{x : x is a factor of 32}. Because 64 is not a factor of 32. (ii) ls {x : x is a factor of 27}≠{3,9,27,54}? Give reason. Solution: Yes, {x:x is a", "# Math Help - Simplify ratios of factorials 1. ## Simplify ratios of factorials $\\frac{(n+1)!}{n!} = \\frac{n!(n+1)}{n!} = (n+1)$ I understand canceling, but I don't see how the initial fraction became $\\frac{n!(n+1)}{n!}$. Same with $\\frac{(2n-1)!}{(2n+1)!} = \\frac{(2n-1)!}{(2n-1)!(2n)(2n+1)} = \\frac{1}{2n(2n+1)}$. 2. Originally Posted by cinder $\\frac{(n+1)!}{n!} = \\frac{n!(n+1)}{n!} = (n+1)$ I understand canceling, but I don't see how the initial fraction became $\\frac{n!(n+1)}{n!}$. Same with $\\frac{(2n-1)!}{(2n+1)!} = \\frac{(2n-1)!}{(2n-1)!(2n)(2n+1)} = \\frac{1}{2n(2n+1)}$. do you know what factorial means? $(n + 1)! = (n + 1)(n)(n - 1)(n - 2)(n - 3)...(3)(2)(1)$ but $n! = n(n - 1)(n - 2)(n - 3)...(3)(2)(1)$ so we have: $(n + 1)! = (n + 1)n!$ similarly, $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2)...(3)(2)(1) = (2n + 1)(2n)(2n - 1)!$ 3. Originally Posted by Jhevon do you know what factorial means? $(n + 1)! = (n + 1)(n)(n - 1)(n - 2)(n - 3)...(3)(2)(1)$ but $n! = n(n - 1)(n - 2)(n - 3)...(3)(2)(1)$ so we have: $(n + 1)! = (n + 1)n!$ similarly, $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2)...(3)(2)(1) = (2n + 1)(2n)(2n - 1)!$ I know what factorial means, but I guess I don't understand when there are variables involved. 4. Originally Posted by cinder I know what factorial means, but I guess I", ")^2 = 1+(x+1)^4-2\\frac{(x+1)^2}{4(x+1)^2}+\\frac{1}{16(x+1)^4}$ $= 1+(x+1)^4-\\frac12+\\frac{1}{16(x+1)^4} =(x+1)^4+\\frac12+\\frac{1}{16(x+1)^4}$ $= (x+1)^4+2\\frac{(x+1)^2}{4(x+1)^2}+\\frac{1}{16(x+1) ^4} = \\left((x+1)^2+\\frac{1}{4(x+1)^2}\\right)^2$ 5. :O thx!", "# Simplifying the sum of a fraction and an integer under a radical sign I'm trying to help my little bro, a bit rusty here... Wolfram Alpha is telling me that: $$x\\sqrt{1+{\\frac{x^2}{16-x^2}}}$$ simplifies to: $$4x\\sqrt{\\frac1{16-x^2}}$$ I can't for the life of me figure out why. I'm thinking there's a simple rule I'm forgetting about.. - Find the common denominator within the radical sign. $$x\\sqrt {1 + \\frac {x^2}{16 - x^2}} = x\\sqrt {\\frac {(16 - x^2) + x^2}{16 - x^2}} = x\\sqrt{\\frac{16}{16- x^2}}=4x \\sqrt{\\frac{1}{16 - x^2}}$$ In the last step, we simplify $\\sqrt{16} = 4$. - there it is... i tried 100 things. can you make inline equations in comments? i didn't see the $-x^2$ and $x^2$ cancel out. i was blind, and now i see, ty – Mike Lewis Oct 14 '13 at 23:56 You're welcome! We all overlook simple things, sometimes! – amWhy Oct 14 '13 at 23:57 $x\\sqrt{1+\\frac{x^{2}}{16-x^{2}}}=x\\sqrt{\\frac{16-x^{2}}{16-x^{2}}+\\frac{x^{2}}{16-x^{2}}}=x\\sqrt{\\frac{16-x^2+x^2}{16-x^2}}=x\\sqrt{\\frac{16}{16-x^2}}=4x\\sqrt{\\frac{1}{16-x^2}}$", "Solve by factoring. 1. $$9 x ^ { 2 } + 8 x = 0$$ 2. $$x ^ { 2 } - 1 = 0$$ 3. $$x ^ { 2 } - 12 x + 20 = 0$$ 4. $$x ^ { 2 } - 2 x - 48 = 0$$ 5. $$( 2 x + 1 ) ( x - 2 ) = 3$$ 6. $$2 - ( x - 4 ) ^ { 2 } = - 7$$ 7. $$( x - 6 ) ( x + 3 ) = - 18$$ 8. $$( x + 5 ) ( 2 x - 1 ) = 3 ( 2 x - 1 )$$ 9. $$\\frac { 1 } { 2 } x ^ { 2 } + \\frac { 2 } { 3 } x - \\frac { 1 } { 8 } = 0$$ 10. $$\\frac { 1 } { 4 } x ^ { 2 } - \\frac { 19 } { 12 } x + \\frac { 1 } { 2 } = 0$$ 11. $$x ^ { 3 } - 2 x ^ { 2 } - 24 x = 0$$ 12. $$x ^ { 4 } - 5 x ^ { 2 } + 4 = 0$$ 1. $$-\\frac{8}{9} , 0$$ 3.", "c} + 2c = -\\frac{15}{4} - \\frac{11}{4} + \\frac{1}{2} = -6$, so $(a+b)^2 = 36$. Adding up both values gives $49 + 36 = \\boxed{085}$ as our final answer.", "##### Problem 2 $x -2 = 133 \\\\ x = 133 + 2 \\\\ x = \\boxed{ 135}$ ##### Problem 3 $2x + 5 = 105 \\\\ 2x = 100 \\\\ \\frac 1 2 (2x) = \\frac 1 2 (100) \\\\ x = \\boxed{ 50}$ ##### Problem 4 $4x + 7 = 131 \\\\ 4x = 124 \\\\ \\frac 1 4 (4x) = \\frac 1 4 (124) \\\\ x = \\boxed{ 31}$" ]

[ "real numbers. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 - y^2) = x f(x) - y f(y)$ for all pairs of real numbers $x$ and $y$.", "# IMO 2017, Problem 2 Let \\mathbb{R} be the set of real numbers. Determine all functions f:\\,\\mathbb{R} \\rightarrow \\mathbb{R} such that, for any real numbers x and y, f(f(x)f(y)) + f(x+y) = f(xy). The solution can be found on a separate page.", "Let $$X$$ be a set with $$n \\in \\mathbb{N}$$ elements and let $$f: X \\rightarrow \\mathbb{N}$$ be a function from $$X$$ to the real numbers. Then we define the finite sum $$\\sum_{x \\in X} f(x)$$ as follows. Select any [...] $$g$$ from $$\\{i \\in \\mathbb{N} : 1 \\le i \\le n \\}$$ to $$X$$. Such a [...] exists since $$X$$ is assumed to have $$n$$ elements. We then define: $\\sum_{x \\in X} f(x) := \\sum_{i=1}^{n}[...]$", "for all $x\\in\\mathbb{N}$, and b)$f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x$, $y\\in\\mathbb{N}$. 25. Consider those functions $f:\\mathbb{N}\\to\\mathbb{N}$ which satisfy the condition $f(m+n) \\geq f(m)+f(f(n))-1$ for all $m,n\\in\\mathbb{N}.$ Find all possible values of $f(2007).$ 26. Find all surjective functions $f:\\mathbb{N}\\to\\mathbb{N}$ such that for every $m,n\\in\\mathbb{N}$ and every prime $p,$ the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p.$ 27. Find all real polynomials $f$ such that $2yf(x+y)+(x-y)(f(x)+f(y)) \\geq 0\\forall x,y\\in\\mathbb{R}.$ 28. Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ with $x,y\\in\\mathbb{R}$ such that $f(x-f(y))=f(x+y)+f(y).$ 29. Show that for positive integer $n$, and for $x\\not =0$, $\\left(x^{n-1}\\sin\\dfrac{1}{x}\\right)^{(n)}=\\dfrac{(-1)^n}{x^{n+1}}\\sin\\left(\\dfrac{1}{x}+\\dfrac{n\\pi}{2}\\right).$ 30. Find all $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xy+f(x))=xf(y)+f(x)$ for every pair of real numbers $x,y$.", "+0 # Help 0 78 1 Let $S$ be the set of all nonzero real numbers. Let $f : S \\to S$ be a function such that $f(x) + f(y) = f(xyf(x + y))$for all $x,$ $y \\in S$ such that $x + y \\neq 0.$ Let $n$ be the number of possible values of $f(4),$ and let $s$ be the sum of all possible values of $f(4).$ Find $n \\times s.$ May 6, 2021", "# IMO 2011 (Problem 3) Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a real-valued function defined on the set of real numbers that satisfies $f (x + y) \\leq yf (x) + f (f (x))$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x\\leq 0$.", "#### Function IMO ###### back to index Let $\\mathbb R$ be the set of real numbers. Determine all functions $f:\\mathbb R\\to\\mathbb R$ that satisfy the equation$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania Find all functions $f:\\mathbb Z\\rightarrow \\mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds: $f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ (Here $\\mathbb{Z}$ denotes the set of integers.) Proposed by Liam Baker, South Africa Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that, for all real numbers $x$ and $y$, $$f (f(x)f(y)) + f(x + y) = f(xy)$$", "+0 help 0 67 2 For what real value of $$N$$ does the range of the function $$y = f(x) = \\frac{4x^2 + Nx + N}{x + 1},$$, where $$x$$ is real (and $$x \\neq -1$$) consist of all real numbers except for a single interval of the form $$-L < y < L$$? Apr 6, 2019 #2 +5172 +1 $$\\dfrac{df}{dx} = 0 \\Rightarrow x = -2,~0\\\\ \\text{we want } f(0) = -f(-2) = L\\\\ f(0) = N\\\\ f(-2) = N-16\\\\ N = 16-N\\\\ N=8$$ $$f(x) \\in \\mathbb{R} - (-8,8)$$ . Apr 6, 2019", "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "2017 IMO Problems/Problem 2 Let $\\mathbb{R}$ be the set of real numbers , determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$ Solution Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$. Let $f(0)=n$, so $f(n^{2})=0$. Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$. If $x+n^2=xn^2$, or $x=\\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0. This means that the only possible values of $n$ is -1,0, and 1. Go through the cases: $n=-1$ $f(x+1)=f(x)+1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=1$ $f(3)=2$ ... $f(x)=x-1$ $n=0$ $f(x)=0$ $n=1$ $f(x+1)=f(x)-1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=-1$ $f(3)=-2$ ... $f(x)=1-x$ So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.", "# Construction of The Real Numbers DEFINITION1: The set $\\mathbb{R}$ with at least two distinct elements and satisfying the following five axioms is called the set of real numbers and each element of $\\mathbb{R}$ is called a real number: The function $+:\\mathbb{R}\\times \\mathbb{R}\\to \\mathbb{R}$ defined as $\\left( x,y \\right)\\to x+y\\in \\mathbb{R}$ for each $\\left( x,y \\right)$ in $\\mathbb{R}\\times \\mathbb{R}$ satisfies the following properties: I${{}_{1}}.\\,\\forall a,b\\in \\mathbb{R},a+b=b+a$, I${{}_{2}}.\\,\\forall a,b,c\\in \\mathbb{R},a+(b+c)=(a+b)+c$,", "Ir al contenido principal ### Problem 1 Let $f$ be entire with $f(x)$ real-valued for $x\\in\\mathbb R$. Define $g:\\mathbb R\\to\\mathbb R$ by $g(y) = \\Re f(iy)$. Then $g$ is even. ### Problem 2 Let $f$ be entire with $f(x)$ real-valued for $x\\in\\mathbb R$. Define $h:\\mathbb R\\to\\mathbb R$ by $h(y) = \\Im f(iy)$. Then $h$ is an odd function. ### Problem 3 Verify the previous problems with the functions $\\sin z, \\cos z, 1 + z^2 + z^3$. ### Problem 4 Let $f$ be entire with $f(x)$ real-valued for $x\\in\\mathbb R$ and $f(iy)$ purely imaginary for $y\\in\\mathbb R$. Then $f$ is odd. ### Problem 5 What can you say if, instead, $f(iy)$ is also real valued for $y\\in\\mathbb R$?", "\\in \\mathbb{R}$, $y \\neq 0$, $f(2x)=af(x)+bx$ and $f(x)f(y)=f(xy)+f\\left(\\dfrac{x}{y}\\right).$ 11. Prove that for all integers $a > 1$ and $b > 1$ there exists a function $f$ from the positive integers to the positive integers such that $f(a\\cdot f(n))=b\\cdot n$ for all $n$ positive integer. 12. Find all functions $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xf(y)+f(x)) = 2f(x)+xy$ for every reals $x,y$. 13. Let $f(x)$ be a real-valued function defined on the positive reals such that (1)If $x < y$, then $f(x) < f(y)$, (2)$f\\left(\\dfrac{2xy}{x+y}\\right) \\geq\\dfrac{f(x) + f(y)}{2}$ for all $x,y>0$. Show that $f(x) < 0$ for some value of $x$. 14. Define $f$ on the positive integers by $f(n) = k^2 + k + 1$, where $n=2^k(2l+1)$ for some $k,l$ nonnegative integers. Find the smallest $n$ such that $f(1) + f(2) + ... + f(n) \\geq 123456.$ 15. Find all functions $f:\\mathbb{N}\\to\\mathbb{N}$ satisfying, for all $x\\in\\mathbb{N}$, $f(2f(x)) = x + 1998.$ 16. a) Show that there are no functions $f,g: \\mathbb{R}\\to\\mathbb{R}$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x\\in\\mathbb{R}$. b)Let $S$ be the set of all real numbers greater than $1$. Show that there are functions $f,g:S\\to S$ satsfying the condition above. 17. Let $f(x)= x^2-C$ where $C$ is a rational constant. Show that exists only finitely many rationals $x$ such that $\\{x,f(x),f(f(x)),\\cdots\\}$ is finite. 18. Find", "\\mathbb{R}x \\times (y + z) = (x \\times y) + (x \\times z)$$ 7. $$\\forall x \\in \\mathbb{R}x + 0 = x = 0 + x$$ 8. $$\\forall x \\in \\mathbb{R}x \\times 1 = x = 1 \\times x$$ 9. Every real number has an additive inverse 10. Every nonzero real number has a multiplicative inverse 11. $$\\forall x \\in \\mathbb{R}x \\leq x$$ 12. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$x \\leq y$$ and $$y \\leq x$$ then $$x = y$$ 13. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}x \\leq y$$ or $$y \\leq x$$ 14. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}\\forall z \\in \\mathbb{R}$$ if $$x \\leq y$$ and $$y \\leq z$$ then $$x \\leq z$$ 15. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}\\forall z \\in \\mathbb{R}$$ if $$y \\leq z$$ then $$x + y \\leq x + z$$ 16. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$x \\nleq 0$$ and $$y \\nleq 0$$ then $$x \\times y \\nleq 0$$ 17. For any subset of the set of all real number such that it's nonempty and its complement is nonempty and for any real number in the subset, all real numbers less than or equal to it are in the subset, $$\\exists x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$y \\neq x$$ then", "Let $$a < b$$ be real numbers, and let $$g : [a, b] \\to \\mathbb{R}$$ be a function such that: 1. $$g$$ is [something and something]. Then there exists an $$x \\in (a, b)$$ such that $$g'(x) = 0$$.", "code. Let $$f\\from \\bbF_2^n\\to \\bbF_2$$ be a Boolean function. Define the corresponding real-valued function $$F\\from \\bbF_2^n \\to \\R$$ as $F(x) = (-1)^{f(x)}\\,.$ Let $$\\{\\chi_x \\mid x\\in \\bbF_2^n\\}$$ be the Fourier basis, that is, $$\\chi_x(r)=(-1)^{r^T x}$$ for all $$x,r\\in \\bbF_2^n$$. Let $$\\{c_x \\mid x\\in \\bbF_2^n\\}\\subseteq \\R$$ be the Fourier coefficients of $$F$$, so that $F = \\sum_{x\\in \\bbF_2^n} c_x \\cdot \\chi_x\\,.$ Show that for all $$x\\in \\bbF_2^n$$, $c_x = 1- 2{\\mathrm{dist}}(f,\\mathrm{WH}[x])$", "# Algebra of Real Function ## Algebra of Real Function Real Value Function: A function which has all real number or subset of the real number as it domain Real Valued Function: A function which has all real number or subset of the real number as it range For functions $f:X - > {\\bf{R}}$ and $g:X - > {\\bf{R}}$, we have $\\left( {f + g} \\right)\\left( x \\right) = f\\left( x \\right) + g\\left( x \\right),x \\in X$ 2. Substraction $\\left( {f - g} \\right)\\left( x \\right) = f\\left( x \\right)-g\\left( x \\right),x \\in X$ 3. Multiplication $\\left( {f.g} \\right)\\left( x \\right) = f\\left( x \\right).g\\left( x \\right),x \\in X$ 4. Multiplication by real number $\\left( {kf} \\right)\\left( x \\right) = kf\\left( x \\right),x \\in X$, where $k$ is a real number. 5. Division $\\frac{f}{g}\\left( x \\right) = \\frac{{f(x)}}{{g(x)}}$ $x \\in X$ and $g\\left( x \\right) \\ne 0$ ### Quiz Time Question 1. which is of the below relation is not a function. A) R = {(2,2),(2,4),(3,3), (4,4)} B) R = {(2,1),(3,1), (4,2)} C) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} D) None of the above Question 2. Let A = {1, 2, 3,4} and B = {5, 7}. Then possible number of relation from A to B ? A) 64 B) 256 C) 16 D) 32 Question 3.", "# Difference between revisions of \"2011 IMO Problems/Problem 3\" Let $f: \\mathbb R \\to \\mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \\le yf(x) + f(f(x))$$ for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \\le 0$.", ": R \\mapsto R$ such that $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$, for all x, y ∈ R, where R denotes the set of all real numbers.", "# 2015 IMO Problems/Problem 5 ## Problem Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f$:$\\mathbb{R}\\rightarrow\\mathbb{R}$ satisfying the equation $f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania ## Solution $f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)$ for all real numbers $x$ and $y$. (1) Put $x=y=0$ in the equation, We get$f(0 + f(0)) + f(0) = 0 + f(0) + 0$ or $f(f(0)) = 0$ Let $f(0) = k$, then $f(k) = 0$ (2) Put $x=0, y=k$ in the equation, We get $f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)$ But $f(k) = 0$ and $f(0) = k$ so, $f(0) + f(0) = f(0)^2$ or $f(0)[f(0) - 2] = 0$ Hence $f(0) = 0, 2$ Case $1$ : $f(0) = 0$ Put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) = f(x)$ Say $f(x) = z$, we get $f(z) = z$ So, $f(x) = x$ is a solution Case $2$ : $f(0) = 2$ Again put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) + 2 = f(x) + 2x$ We observe that $f(x)$ must be a polynomial of" ]

[ "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "will get statements relating $f(a)$ and $f(-a)$. So for instance, you probably want to choose $x$ and $y$ such that one or more of $x+y$, $x-y$, $x$, and $y$ are equal to $a$ or $-a$. First, setting $y=0$ gives $f(x)^2=f(x)^2f(0)^2$. This means $f(0)^2=1$ unless $f(x)=0$ for all $x$. The latter case is trivial, so let us assume $f$ is not identically $0$ and so $f(0)^2=1$. Now set $x=0$ and $y=a$ to get $$f(a)f(-a)=f(0)^2f(a)^2=f(a)^2.$$ Unless $f(a)=0$, this implies $f(a)=f(-a)$, as desired. If $f(a)=0$, we need to do a little more work: setting $x=0$ and $y=-a$ gives $$f(-a)f(a)=f(0)^2f(-a)^2=f(-a)^2.$$ Since $f(a)=0$, this implies $f(-a)=0$ as well, and so we're done.", "1, y = 1,$$ then we have $$f(1)f(1) = f(2) + f(0)$$ If $$x = 2, y = 1$$, then we have $$f(2)f(1) = f(3) + f(1)$$ Since we have $$4$$ variables $$(f(0), f(1), f(2)$$ and $$f(3)$$) and $$3$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) If $$f(1) = 1,$$ then we have$$f(2) = f(1)f(1) – f(0) = 1 – 2 = -1, f(3) = f(2)f(1) – f(1) = (-1)*1 – 1 = -2$$ and $$f(0)f(1)f(2)f(3) = 2*1*(-1)(-2) = 4.$$ If $$f(1) = 2$$, then we have $$f(2) = f(1)f(1) – f(0) = 4 – 2 = 2, f(3) = f(2)f(1) – f(1) = 2*1 – 2 = 0$$ and $$f(0)f(1)f(2)f(3) = 2*2*(2)*0 = 0.$$ Since condition 1) does not yield a unique solution, it is not sufficient. Condition 2) Since $$f(1) = 1,$$ we have $$f(0) = f(1)f(0) = f(1) + f(1) = 2, f(2) = f(1)f(1) – f(0) = 1 – 2 = -1, f(3) = f(2)f(1) – f(1) = (-1)*1 – 1 = -2$$ and $$f(0)f(1)f(2)f(3) = 2*1*(-1)*.(-2) = 4.$$ Condition 2) is sufficient since it yields a unique solution. If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one", "for the addition the x is added for multiplication the y is multiplied f(x+f(x+y))= x+f(x+y) and f(xy)=y*f(x) 17. danica518 suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely 18. danica518 f(x+f(x+y))= x+f(x+y) f(xy)=y*f(x) is there a solution to both of these equations, other than f(x)=0 and f(x)=x 19. anonymous consider $$x=y=0$$ so $$f(0+f(0))+f(0)=f(0)\\\\f(f(0))=0$$ now consider $$x=f(0),y=0$$ so $$f(f(0)+f(f(0)+0))+f(0)=f(0)+f(f(0))\\\\f(f(0)+f(f(0)))=f(0)+f(f(0))\\\\f(f(0))=f(0)$$ implying $$f$$ has a fixed point, $$f(0)\\to f(0)$$ now consider $$x=0,y=f(0)$$ $$f(0+f(f(0)))+f(0)=f(f(0))+f(0)f(0)\\\\f(0)+f(0)=f(0)^2\\\\f(0)^2-2f(0)=0\\\\f(0)\\left(f(0)-2\\right)=0$$ so either $$f(0)=0$$ or $$f(0)=2$$ 20. anonymous but if $$f(0)=2$$ then $$f(2)=0$$ which contradicts that $$f(0)$$ is a fixed point ($$f(2)=2$$), so it must be that $$f(0)=0$$ i think 21. jtvatsim Potential typo in lines 6 and 7 of original calculation @oldrin.bataku 22. jtvatsim The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS. 23. anonymous now consider the general case $$x=0$$ so $$f(f(y))+f(0)=f(y)+yf(0)$$using the fact $$f(0)=0$$ this reduces to teh functional equation $$f(f(y))=f(y)$$ so all $$f(y)$$ are fixed points, which suggests that the only functions that work are restrictions of $$f(x)=x$$ 24. anonymous @jtvatsim ? what are you talking about 25. jtvatsim The part where you plug in x = f(0) and y = 0 does not yield f(f(0)) = f(0) rather, it duplicates the previous substitution giving f(f(0)) =", "for $t<1$ we have $f(t-1)u(t-1)=0$ and $f(t-2)u(t-2)=0$ and then $y(t)=0$. For $t> 1$ we have $u(t-1)=1$ and for $t<2$ we have $u(t-2)=0$. So for $1<t<2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=0$ and then $$y(t)=f(t-1)={1 \\over 2} - e^{-(t-1)} + {1 \\over 2} e^{-2(t-1)}$$. For $t> 2$ we have $u(t-1)=1$ and for $t>2$ we have $u(t-2)=1$. So for $t>2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=f(t-2)$ and then $$y(t)=f(t-1)-f(t-2)={1 \\over 2} - e^{-(t-1)} + {1 \\over 2} e^{-2(t-1)}-\\left[{1 \\over 2} - e^{-(t-1)} + {1 \\over 2} e^{-2(t-1)}\\right]={1 \\over 2} e^{-(t-1)} - {1 \\over 2} e^{-2(t-1)} - {1 \\over 2} e^{(t-2)} + {1 \\over 2} e^{2(t-2)}$$. $$y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$ $u(t-t_0)$ is the unit step function. It has the value $0$ for all values of $t<t_0$ and has the value $1$ for all values of $t>t_0$. Using this definition you can see that for $0<t<1$ all unit step functions vanish. Resulting in $y(t)=0$. For $1<t<2$ only $f(t-1)$ will appear. For $t>2$ $f(t-1)-f(t-2)$ is the result.", "f(y,z) + f(z,x) = 0$. This means that $f(x,y) = -f(z,x)- f(y,z)$. This is true for all $z$ so if we could define $g(w) = -f(c,w)$ and $g(w) = f(w,z)$ for some constant $c$ we would be done. (As $g(x) - g(y) = -f(c,x) -f(y,c)= f(x,y)$.) But that would require that there be some constant $c$ so that $-f(c,w) =f(w,c)$ for all $w$ and that is not true in general. But, $f$ is not a usual function. Maybe this IS true for $f$. Can we prove that $f(w,z) = -f(z,w)$ for all $w,z$.? Let $x = y = w$: Then $f(w,w) + f(w,z)+ f(z,w) = 0$ so $f(w,z) = -f(z,w) - f(w,w)$. Shoot! That was so close but $-f(z,w) - f(w,w) \\ne -f(z,w)$ unless $f(w,w) = 0$ and that is not true in general. But, again, $f$ is not a general function. Maybe $f(w,w)$ does equal $0$ for all $w$. Let $x = y = z = w$: Then $f(w,w) + f(w,w) + f(w,w) = 0$ so $f(w,w) = 0$. That's it! We are done. ..... If we set $g(w) = f(w,c)$ for some constant $c$ we get: $f(x,y) + f(y,c) + f(c,x) = 0$ so $f(x,y) = -f(c,x) - f(y,c) = -f(c,x) - g(y)$. Now $f(c,x)+ f(x,x) + f(x,c) = 0$ so $f(c,x) = -f(x,c) - f(x,x)", "$F_K(X,x_0)=[K,k_0; X,x_0]\\in\\mathscr{PS}$. Given $f: (X,x_0)\\to (Y,y_0)$ in $\\hom((X,x_0),(Y,y_0))$ we define $F_K(f)\\in\\hom([K,k_0; X,x_0],[K,k_0;Y,y_0])$ by $\\displaystyle F_k(f)[g]=[f\\circ g]\\in[K,k_0; Y,y_0]$ for every $[g]\\in [K,k_0; X,x_0]$. We can check the two axioms: – $F_k(1_X)[g]=[1_X\\circ g]=[g]$ for every $[g]\\in[K,k_0; X, x_0]$. – For $f\\in\\hom((X,x_0),(Y,y_0))$, $h\\in\\hom((Y,y_0),(Z,z_0))$ we have $\\displaystyle F_K(h\\circ f)[g]=[h\\circ f\\circ g]=F_K(h)\\circ F_K(f)[g]\\in[K,k_0; Z,z_0]$ for every $[g]\\in[K,k_0; X,x_0]$. Similarly, we can define a cofunctor $F_K^*$ by taking $F_K^*(X,x_0)=[X,x_0; K,k_0]$ and for $f:(X,x_0)\\to (Y,y_0)$ in $\\hom((X,x_0),(Y,y_0))$ we define $\\displaystyle F_K(f)[g]=[g\\circ f]\\in[X,x_0; K,k_0]$ for every $[g]\\in[Y,y_0; K,k_0]$. Note that if $f\\simeq f'$ rel $x_0$, then $F_K(f)=F_K(f')$ and similarly $F_K^*(f)=F_K^*(f')$. Therefore $F_K$ (resp.\\ $F_K^*$) can also be regarded as defining a functor (resp.\\ cofunctor) $\\mathscr{PT}'\\to\\mathscr{PS}$. Homotopy Sets and Groups Theorem: If $(X,x_0)$, $(Y,y_0)$, $(Z,z_0)\\in\\mathscr{PT}$, $X$, $Z$ Hausdorff and $Z$ locally compact, then there is a natural equivalence $\\displaystyle A: [Z\\wedge X, *; Y,y_0]\\to [X, x_0; (Y,y_0)^{(Z,z_0)}, f_0]$ defined by $A[f]=[\\hat{f}]$, where if $f:Z\\wedge X\\to Y$ is a map then $\\hat{f}: X\\to Y^Z$ is given by $(\\hat{f}(x))(z)=f[z,x]$. We need the following two propositions in order to prove the theorem. Proposition 1: The exponential function $E: Y^{Z\\times X}\\to (Y^Z)^X$ induces a continuous function $\\displaystyle E: (Y,y_0)^{(Z\\times X, Z\\vee X)}\\to ((Y,y_0)^{(Z,z_0)}, f_0)^{(X,x_0)}$ which is a homeomorphism if $Z$ and $X$ are Hausdorff and $Z$ is locally compact\\footnote{every point of $Z$ has a compact neighborhood}. Proposition 2: If $\\alpha$ is", "to rule out the second case. $P(x,x) \\implies f(0)=f(0)(f(x)+x) \\implies f(x)=1-x$. Substitution shows that this is NOT a solution. Thus $f(0)=0$. Keep going, we are on track. $P(0,y) \\implies f(-yf(y)) = yf(-y), P(0,-y) \\implies f(yf(-y))=-yf(y)$. Look, look, these two switch between each other. Better take this information into account. $\\forall y \\in \\mathbb R, -yf(y), yf(-y) \\in \\mathbb S_2$. But how are we going to exploit the second order fixed points? We have to substitute it into somewhere and compare it with the original value, so we’d better do it with something where $x$ and $f(x)$ are both present, as in such occasion substituting $s \\in S_2$ will make this expression comparable with the original one. Such expression exists in the LHS. $P(s,f(s)), s \\in S_2 \\\\ \\implies f(sf(s)-f(f(s))f(s))=f(s-f(s))(f(s)+f(s))\\\\ \\implies f(s-f(s))f(s)=0$, as $f(f(s))=s$ and $f(0)=0$. If $s \\ne 0$, $f(s) \\ne 0$ as $f(f(s))=s \\ne 0$ and $f(0) =0$. Thus $f(s-f(s)) =0$. If $s=0$, $f(s-f(s))= f(0-0)=0$. Thus in whatever case, $s-f(s) \\in \\Omega$. Look at that. That looks like quite some information isn’t it, especially after we substitute $s= yf(-y)$ in… $yf(-y)-f(yf(-y))=yf(-y)+yf(y) = y(f(y)+f(-y)) \\in \\Omega, \\forall y \\in \\mathbb R$. That wasn’t a big surprise, since the $y$ and $-y$ are supposed to either cancel or both give vanishing function value anyway. But this expression has an", "for $y$ in (1) we have $$f(x+f(f(y)))=f(2x-f(y))+f(x^2)$$ which by (2) yields $$f(x+f(-y))=f(2x-f(y))+f(x)^2$$ Again, substituting $-x-y$ for $y$ in (1) we'll have $$f(x+f(-y))=f(2x+y)+f(x)^2$$ Compairing the last two equations above, we get $f(2x-f(y))=f(2x+y)$. Substituting $\\frac{f(y)}2$ for $x$, we'll have $f(y+f(y))=0$. Also by letting $y=0$ in (1) we know that $f(x+f(x))=f(x)+f(x)^2$. So for every $x$ we find out that $f(x)\\in\\{0,-1\\}$. Hence, adding $f(x)$ to both sides of (3) we'll get $2f(x)=f(2x)$ which shows that $f(x)$ can not be equal to $-1$. So $f$ is the constant zero function. -", "of $f$ that $f(a) = 0$ for some $a \\in (h, k)$. Now we can see that $f(0) = f(a) = f(1) = 0$ so that $f'(x)$ vanishes at least once in $(0, a)$ and at least once more in $(a, 1)$. By continuity of $f'(x)$ there is a first value $b \\in (0, a)$ such that $f'(b) = 0$. This means that $f'(x) > 0$ for $x \\in [0, b)$ and $f'(b) = 0$. Similarly there is a last value $c \\in (a, 1)$ such that $f'(c) = 0$. Thus $f'(x) > 0$ for all $x \\in (c, 1]$ and $f'(c) = 0$. Since $f'(x) > 0$ in $(0, b)$ it follows that $f(x) > f(0) = 0$ for all $x \\in (0, b]$. Similarly $f(x) < 0$ for all $x \\in [c, 1)$. Since the argument is bit long we once revise what we have obtained so far: 1) $f'(x) > 0$ for all $x \\in [0, b)$, $f'(b) = 0$ and $f'(x) > 0$ for all $x \\in (c, 1]$, $f'(c) = 0$. 2) $f(x) > 0$ for all $x \\in (0, b]$ and $f(x) < 0$ for all $x \\in [c, 1)$. Also by their construction $b < c$. By mean value theorem we see that $f'(b) - f'(0) = bf''(\\alpha)$ for some $\\alpha", "$f(-y)f(y)=(f(y))^2$, so if $f(y) \\ne 0$, then $f(y)=f(-y)$. Furthermore, setting $y=-x$ gives us $(f(x)-x^2) \\cdot f(4x) + (f(-x)-x^2) \\cdot f(4x) = f(0) = 0$. The LHS can be factored as $(f(x)+f(-x)-2x^2) \\cdot f(4x) = 0$. Therefore, if $f(4x) \\ne 0$, then we have $f(x)+f(-x)-2x^2=0$, or $f(x)+f(-x)=2x^2$. However, since from step 2 we have that $f(x)=f(-x)$ assuming f(x) \\ne 0$, the equation becomes$2f(x)=2x^2$, so$f(x)=0, x^2$are the only solutions. ==Solution 2== Step 1:$x=y=0 \\implies f(0)=0$Step 2:$x=0 \\implies f(y)f(-y)=f(y)^{2}$. Now, assume$y \\not = 0$. Then, if$f(y)=0$, we substitute in$-y$to get$f(y)f(-y)=f(-y)^{2}$, or$f(y)=f(-y)=0$. Otherwise, we divide both sides by$f(y)$to get$f(y)=f(-y)$. If$y=0$, we obviously have$f(0)=f(0)$. Thus, the function is even. . Step 3:$y=-x \\implies 2f(4x)(f(x)-x^{2})=0$. Thus,$\\forall x$, we have$f(4x)=0$or$f(x)=x^{2}$. Step 4: We now assume$ (Error compiling LaTeX. ! Missing $inserted.)f(x) \\not = 0$,$x\\not = 0$. We have$f(\\frac{x}{4})=\\frac{x^{2}}{16}$. Now, setting$x=y=\\frac{x}{4}$, we have$f(\\frac{x}{2}=\\frac{x^{2}}{4}$or$f(\\frac{x}{2})=0$. The former implies that$f(x)=0$or$x^{2}$. The latter implies that$f(x)=0$or$f(x)=\\frac{x^{2}}{2}$. Assume the latter.$y=-2x \\implies -\\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\\frac{x^{2}}{2}$. Clearly, this implies that$f(x)$is negative for some$m$. Now, we have$f(\\frac{m}{4})=\\frac{m^{2}}{16} \\implies f(\\frac{m}{2})=0,\\frac{m^{2}}{4} \\implies f(m) \\geq 0$, which is a contradiction. Thus,$\\forall x$$(Error compiling LaTeX. ! Missing$ inserted.)f(x)=0$or$f(x)=x^{2}$. Step 5: We now assume$ (Error compiling LaTeX. ! Missing $inserted.)f(x)=0$,$f(y)=y^{2}$for some$x,y \\not = 0$. Let$m$be sufficiently large integer, let$z=|4^{m}x|$and take the absolute value of$y$(since the function is even). Choose$c$such that$3z-c=y$. Note that we have$\\frac{c}{z}$~$3$and$\\frac{y}{z}$~$0$. Note that$f(z)=0$. Now,$x=z, y=c \\implies$LHS is positive, as", "$0$ for all rationals, it is identically $0$, so $$f(x^i)=f(1)^{n-i}f(x)^i$$ for all $0\\leq i\\leq n$. Originally, we had $f(1)=f(1)^n$, so $f(1)\\in\\{-1,0,1\\}$. If $f(1)=0$, we have $f(x^i)=0$, so $\\boxed{f(x)\\equiv 0}$. Otherwise, we have $$f(x^2)=f(1)^{n-2}f(x)^2=f(1)f(x)^2$$ $$f(x+y^2)=f(x)+f(y^2)=f(x)+f(1)f(y)^2.$$ If $f(1)=1$, this means $f$ is increasing, and if $f(1)=-1$ this means $f$ is decreasing. Either way, $f$ is not everywhere dense, so $f(x)=cx$ for some $c$ and all $x$. The observation that $f(1)=\\pm 1$ means $\\boxed{f(x)=x}$ and $\\boxed{f(x)=-x}$ are our only other solutions. • In $f(x^{2})=f(1)^{n-2}f(x)^{2}=f(1)f(x)^{2}$ when $f(1)\\neq 0$ how does the second equality follow? Sorry if I am missing something obvious. – Kavi Rama Murthy Jul 16 '18 at 5:46 • As $f(1)\\in\\{-1,0,1\\}$ and $f(1)\\neq 0$, we have $|f(1)|=1\\implies f(1)^2=1$; since $n$ is odd, $$f(1)^{n-2}=\\left(f(1)^2\\right)^{\\frac{n-3}{2}}f(1)=1^{\\frac{n-3}{2}}f(1)=f(1).$$ – Carl Schildkraut Jul 16 '18 at 17:02 • Nice and simpler than mine! – Sungjin Kim Jul 16 '18 at 18:27 • @KaviRamaMurthy The second equality is by $f(y^2)=f(1)f(y)^2$. – Sungjin Kim Jul 16 '18 at 18:27 This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\\in\\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you. As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\\in \\mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\\in\\mathbb{Q}$, $x\\in\\mathbb{R}$. Since $f( (x+y)^5)= f(x+y)^5", "# Math Help - If f(x-y)=f(x)f(y) what is the value of f(4)? 1. ## If f(x-y)=f(x)f(y) what is the value of f(4)? Dear Colleagues, If $f(x-y)=f(x)f(y)$, then what is the value of $f(4)$? Best Regards. 2. Here's a hint $e^{m-n} = e^m \\times e^{-n}$ 3. Fun problem! I'm not sure I've solved it, but I have made some progress, or at least found a few things out. First, suppose you set y = 0. Then you get f(x) = f(x) f(0) for all x. Then either f(0) = 1 or f(x) = 0 for all x. Suppose f(0) = 1. Then we could set x = 0, and get f(-y) = f(0) f(y) = f(y), thus making f even. But the f(x) = 0 case was also even. Hence, we conclude that f is even. If we set x = y = 0, then we get that f(0) = f(0) f(0), which implies that f(0) is in the set {0,1}. Letting x = 4 and y = 4 shows us that f(0) = f(4) f(4), which implies that f(4) is in the set {-1, 0, 1}. Setting x = 4 and y = 0 gives us f(4) = f(4) f(0). I think your problem is underdetermined. Why? f(x) = 0 satisfies the function equation, and f(x) = 1", "Thanks. – Longti Mar 22 '18 at 1:55 We say that $X$ and $Y$ are $F$-related if $T_pF(X_p)=Y_{F(p)}$ for all $p\\in M.$ Definition 1. We can define the pushforward $F_*X$ by the formula $$(F_*X)_q=T_{F^{-1}(q)}F(X_{F^{-1}(q)})$$ for all $q\\in N.$ With definition your problem is just the definition, because if $q=F(p)$ then $$(F_*X)_{F(p)}=T_{F^{-1}(F(p))}F(X_{F^{-1}(F(p))})=T_pF(X_p)=Y_{F(p)}.$$ Definition 2. Define $F_*X$ by the property that $$X(f\\circ F)= (F_*X(f))\\circ F$$ for all $f\\in C^\\infty(N).$ Hence, given $p\\in M$ and $f\\in C^\\infty(N)$ we have $$X(f\\circ F)(p)=X_p(f\\circ F)=T_pF(X_p)(f)$$ and $$(F_*X(f))\\circ F(p)=(F_*(X)(f))(F(p))=(F_*X)_{F(p)}(f),$$ so that $Y_{F(p)}:=T_pF(X_p)=(F_*X)_{F(p)}$ for all $p\\in M$ and hence $Y=F_*X.$", "function. Putting $$x=0$$ gives $$f(yf(0))=0$$ for any $$y\\in\\mathbb{R}$$ so $$f(0)=0$$. If $$f$$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $$f$$ is not zero at some point. Setting $$y=0$$, $$f(x^2)=xf(x).$$ Let $$x_0$$ be a real number for which $$f(x_0)=0$$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$x_0 f(x_0 +y) =0$$ for all $$y\\in\\mathbb{R}$$. Thus $$x_0 =0$$ since f is not identically zero. Indeed, $$f(x)=0 \\iff x=0.$$ Now setting $$y=-x$$ we have for any $$x\\in\\mathbb{R}$$, $$f(x^2-xf(x))=0$$ and so $$x^2-xf(x)=0$$ or $$f(x)=x.$$", "goes to (0,0)] so that: $$f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+xu(x,y)+yv(x,y)$$ Now if I multiply [and remember f(0,0)=0] both sides by g(x,y) I get: $$f(x,y)g(x,y)=xf_x(0,0)g(x,y)+yf_y(0,0)g(x,y)+xu(x,y)g(x,y)+yv(x,y)g(x,y)$$ and this is also very close to what I need: $$f(x,y)g(x,y)=x[f_x(0,0)g(0,0)]+y[f_y(0,0)g(0,0)]+x[u(x,y)g(x,y)]+y[v(x,y)g(x,y)]$$ where \"the new a\" is f_x(0,0)g(0,0) \"the new b\" is [f_y(0,0)g(0,0)] \"the new u(x,y)\" is [u(x,y)g(x,y)] and so on, I'm close to this but I'm not there... What I tried to do in my first post is: I arrived to: $$f(x,y)g(x,y)=xf_x(0,0)g(x,y)+yf_y(0,0)g(x,y)+xu(x,y)g(x,y)+yv(x,y)g(x,y)$$ So due to continuity of g I can conclude: $$xf_x(0,0)[g(0,0)+\\epsilon/2]+yf_y(0,0)[g(0,0)+\\epsilon/2]+xu(x,y)g(x,y)+yv(x,y)g(x,y) \\leq h(x,y) \\leq xf_x(0,0)[g(0,0)+\\epsilon/2]+yf_y(0,0)[g(0,0)+\\epsilon/2]+xu(x,y)g(x,y)+yv(x,y)g(x,y)$$ and it is very tempting to say the following [which is most likely not \"legal\"]: because this inequality is true for every epsilon that I can \"swap\" it to: h(x,y) = xf_x(0,0)[g(0,0)]+yf_y(0,0)[g(0,0)]+xu(x,y)g(x,y)+yv(x,y)g(x,y)[/tex] Sorry if this post is disorganized, I tried to think out loud... Last edited: Aug 15, 2012 6. Aug 16, 2012 ### voko This is not correct. It should be $$f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+R(x,y)$$ where $$\\lim_{(x, y) \\rightarrow (0, 0)} \\frac {R(x, y)} {||(x, y)||} = 0$$ That is, R need not be linear but it must go to zero faster than the norm of the point. Then $$f(x,y)g(x, y) = f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(x, y) + g(x, y)R(x,y)$$$$= f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(0, 0) + [xf_x(0,0) + yf_y(0,0)]g(x, y) - [xf_x(0,0) + yf_y(0,0)]g(0, 0) + g(x, y)R(x,y)$$$$= f(0,0)g(x, y)", "# Find all functions if $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\\in\\mathbb{R}$. Let $\\mathbb{R}$ denote the set of real numbers. Find all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that $$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$$ for all $x,y\\in\\mathbb{R}$. If we put $x=0$ and mark $a=f(0)$ we get $$2a=af(y)+ya$$ so for $a\\ne 0$ we get $f(y)=2-y$ for all $y$. Now say $a=0$, so $f(0)=0$. Leting $y=0$ we get $\\boxed{f(x^2) = xf(x)}$. If we put this in starting equation we get $$xf(x) +f(xy)=f(x)f(y)+yf(x)+xf(x+y)\\;\\;\\;\\;(*)$$ From boxed equation we see that $f$ is odd: $$-xf(-x) = f((-x)^2)= f(x^2)=xf(x) \\implies f(-x)=-f(x)$$ Leting $y=-x$ in $(*)$ we get: $$xf(x)-f(x^2) =-f(x)^2-xf(x)\\implies \\underline{f(x^2)= f(x)^2}$$ Here I'm stuck, we have $xf(x)=f(x)^2$ but I can't divide with $f(x)$ since it might have walue $0$ at some $x$. What to do? • $x f(x) = f(x)^2 \\iff (x-f(x)) f(x) = 0 \\iff f(x) = x \\vee f(x) = 0$ – md2perpe Aug 24 '18 at 17:11 • From that what you wrote we can deduce that $f$ is for some $x$ equal $0$ and for othe to $x$ it self. What can I do with that. – Aqua Aug 24 '18 at 17:18 • I can show that $f(x)=0$ for all $x\\in\\mathbb{Q}$, but I don't know yet how to extend to $\\mathbb{R}$. – Batominovski Aug 24 '18 at 17:20 • From $f(x)=x \\vee f(x)=0$ we see that there is some", "$M_1=M_2=1$ and we will get that $\\exists 0<N_1$ such that $\\forall N_1<x, 1<f(x)$ and that $\\exists 0<N_2$ such that $\\forall N_2<x, 1<f(f(x))$ Take $x=max (N_1+1,N_2+1)$, then $N_1,N _2<x$ and so we get that $1<f(x)$ and that $1<f(f(x))$ Now take $y=max(\\frac{x+1}{f( f( x))-1},N_2-x)$ which implies that $N_2< x+y+1\\leq y f(f(x ))$ and so $1<f(f(x+y+1))$, Now $f (x+y)\\geq f(x )+yf(f(x))\\geq x+y+1$ And hence $f(f(x+y))\\geq f(x+y+1)+(f(x+y)-(x+y+1))f(f(x+y+1)) \\geq f(x +y+1)\\geq f(x+y)+f(f(x+y))\\geq f(x)+yf(f(x))+f(f(x+y))>f(f(x+y))$ And so we get that $0>0$ which is a contradiction.", "# Show this fuction is identically 0 Suppose $f$ is defined on all real numbers. $f = f''$ and $f(0)=f'(0)=0$. Then show that $f=0$ for all $x$. The following is what I did: Since we have $f=f''$. Then, multiply $f'$ on both sides: $$f\\cdot f'=f' \\cdot f''$$ $$f \\cdot f'-f' \\cdot f''= \\frac {1}{2}(f^2)'- \\frac {1}{2}(f'^2)'$$ This says, $f^2-f'^2=C$, and plug in number $x=0$, we can then say $C=0$. So, $f^2-f'^2=0$. But, what is the next step? How can I show $f=0$ for all $x$? - add comment ## 1 Answer $$f'' = f \\implies f(x) = c_1 e^x + c_2 e^{-x}$$ Now finish it off by plugging in values for $f(0)$ and $f'(0)$. EDIT To finish it off, the way you have started, you get that $$f'(x) = \\pm f(x)$$ If $f'(x) = f(x)$, we get that $$\\exp(-x) f'(x) = \\exp(-x) f(x) \\implies \\exp(-x) f'(x) - \\exp(-x) f(x) = 0$$ This gives us $$\\left(\\exp(-x) f(x) \\right)' = 0$$ Hence, $$\\exp(-x) f(x) = c_1 \\implies f(x) = c_1 \\exp(x)$$ Plugging in $f(0) = 0$, we get that $f(x) = 0$. Similarly, if $f'(x) = -f(x)$, we get that $$\\exp(x) f'(x) = -\\exp(x) f(x) \\implies \\exp(x) f'(x) + \\exp(x) f(x) = 0$$ This gives us $$\\left(\\exp(x) f(x) \\right)' = 0$$ Hence, $$\\exp(x) f(x) = c_2 \\implies f(x) = c_2", "get $f(0) = 1 + f(1), f(3) = f(2), f(6) = 1+f(2), f(8) = 2+f(2)$. From “6855 = 3” we have $f(6) + f(8) + 2f(5) = 3$, or $2f(2) + 2f(5) = 0$; if we agree that all values are nonnegative then $f(2) = f(5) = 0$. Then from “7756 = 1” we can get $f(7) = 0$. Also from relations we alrady derived, $f(3) = 0, f(6) = 1, f(8) = 2$. But what are the values of 0, 1, and 9? It turns out that either $(f(0), f(1), f(9)) = (1,0,1)$ or $(2,1,0)$ works, from an addition standpoint. So why does 4 never appears on the left hand side, therefore meaning we can never work out $f(4)$. This is a feature, not a bug. $f(n)$ is the number of holes in the numeral n. Some people draw 4 with one hole; some draw it with zero. So we choose $(f(0),f(1),f(9)) = (1,0,1)$ and so the answer is $f(2) + f(5) + f(8) + f(1) = 0 + 0 + 2 + 0 = 2$. ## Paul Graham, generalizer Paul Graham has an interesting essay entitled How to do what you love. It was posted on his web site in 2006, and this paragraph has stuck with me for years: This test [would people do it" ]

[ "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "2017 IMO Problems/Problem 2 Let $\\mathbb{R}$ be the set of real numbers , determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$ Solution Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$. Let $f(0)=n$, so $f(n^{2})=0$. Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$. If $x+n^2=xn^2$, or $x=\\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0. This means that the only possible values of $n$ is -1,0, and 1. Go through the cases: $n=-1$ $f(x+1)=f(x)+1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=1$ $f(3)=2$ ... $f(x)=x-1$ $n=0$ $f(x)=0$ $n=1$ $f(x+1)=f(x)-1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=-1$ $f(3)=-2$ ... $f(x)=1-x$ So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.", "for the addition the x is added for multiplication the y is multiplied f(x+f(x+y))= x+f(x+y) and f(xy)=y*f(x) 17. danica518 suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely 18. danica518 f(x+f(x+y))= x+f(x+y) f(xy)=y*f(x) is there a solution to both of these equations, other than f(x)=0 and f(x)=x 19. anonymous consider $$x=y=0$$ so $$f(0+f(0))+f(0)=f(0)\\\\f(f(0))=0$$ now consider $$x=f(0),y=0$$ so $$f(f(0)+f(f(0)+0))+f(0)=f(0)+f(f(0))\\\\f(f(0)+f(f(0)))=f(0)+f(f(0))\\\\f(f(0))=f(0)$$ implying $$f$$ has a fixed point, $$f(0)\\to f(0)$$ now consider $$x=0,y=f(0)$$ $$f(0+f(f(0)))+f(0)=f(f(0))+f(0)f(0)\\\\f(0)+f(0)=f(0)^2\\\\f(0)^2-2f(0)=0\\\\f(0)\\left(f(0)-2\\right)=0$$ so either $$f(0)=0$$ or $$f(0)=2$$ 20. anonymous but if $$f(0)=2$$ then $$f(2)=0$$ which contradicts that $$f(0)$$ is a fixed point ($$f(2)=2$$), so it must be that $$f(0)=0$$ i think 21. jtvatsim Potential typo in lines 6 and 7 of original calculation @oldrin.bataku 22. jtvatsim The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS. 23. anonymous now consider the general case $$x=0$$ so $$f(f(y))+f(0)=f(y)+yf(0)$$using the fact $$f(0)=0$$ this reduces to teh functional equation $$f(f(y))=f(y)$$ so all $$f(y)$$ are fixed points, which suggests that the only functions that work are restrictions of $$f(x)=x$$ 24. anonymous @jtvatsim ? what are you talking about 25. jtvatsim The part where you plug in x = f(0) and y = 0 does not yield f(f(0)) = f(0) rather, it duplicates the previous substitution giving f(f(0)) =", "# 2001 IMO Shortlist Problems/A4 ## Problem Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, satisfying $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$. ## Solution Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\\ \\forall x$, so $f(x) = 0,\\ \\forall x$. This is a solution, so we can take it out of the way: assume $f(1)\\ne 0$. $y = 1\\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\\in\\{0,f(1)x\\}$. In particular, $f(0) = 0$. Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\\ \\forall x$. This means that $f(a),f(b)\\ne 0\\Rightarrow f\\left(\\frac ab\\right)\\ne 0\\ (*)$ ($\\frac ab$ is defined because $f(b)\\ne 0\\Rightarrow b\\ne 0$). Assume now that $x\\ne y$ and $f(x),f(y)\\ne 0$. We get $f(x) = f(1)x,\\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\\ne 0$, so $x\\ne y,\\ f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (**)$. Assume now $f(x)\\ne 0$. From $(*)$ we get $f\\left(\\frac 1x\\right)\\ne 0$, and after applying $(*)$ again to $a = x,b = \\frac 1x$ we get $f(x^2)\\ne 0\\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (\\#)$. Let $G = \\{x\\in\\mathbb R|f(x)\\ne 0\\}$. $(*)$ and $(\\#)$ simply say that $(G,\\ \\cdot)$ is a subgroup of $(\\mathbb R^{*},\\", "function. Putting $$x=0$$ gives $$f(yf(0))=0$$ for any $$y\\in\\mathbb{R}$$ so $$f(0)=0$$. If $$f$$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $$f$$ is not zero at some point. Setting $$y=0$$, $$f(x^2)=xf(x).$$ Let $$x_0$$ be a real number for which $$f(x_0)=0$$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$x_0 f(x_0 +y) =0$$ for all $$y\\in\\mathbb{R}$$. Thus $$x_0 =0$$ since f is not identically zero. Indeed, $$f(x)=0 \\iff x=0.$$ Now setting $$y=-x$$ we have for any $$x\\in\\mathbb{R}$$, $$f(x^2-xf(x))=0$$ and so $$x^2-xf(x)=0$$ or $$f(x)=x.$$", "# Find all functions if $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\\in\\mathbb{R}$. Let $\\mathbb{R}$ denote the set of real numbers. Find all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that $$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$$ for all $x,y\\in\\mathbb{R}$. If we put $x=0$ and mark $a=f(0)$ we get $$2a=af(y)+ya$$ so for $a\\ne 0$ we get $f(y)=2-y$ for all $y$. Now say $a=0$, so $f(0)=0$. Leting $y=0$ we get $\\boxed{f(x^2) = xf(x)}$. If we put this in starting equation we get $$xf(x) +f(xy)=f(x)f(y)+yf(x)+xf(x+y)\\;\\;\\;\\;(*)$$ From boxed equation we see that $f$ is odd: $$-xf(-x) = f((-x)^2)= f(x^2)=xf(x) \\implies f(-x)=-f(x)$$ Leting $y=-x$ in $(*)$ we get: $$xf(x)-f(x^2) =-f(x)^2-xf(x)\\implies \\underline{f(x^2)= f(x)^2}$$ Here I'm stuck, we have $xf(x)=f(x)^2$ but I can't divide with $f(x)$ since it might have walue $0$ at some $x$. What to do? • $x f(x) = f(x)^2 \\iff (x-f(x)) f(x) = 0 \\iff f(x) = x \\vee f(x) = 0$ – md2perpe Aug 24 '18 at 17:11 • From that what you wrote we can deduce that $f$ is for some $x$ equal $0$ and for othe to $x$ it self. What can I do with that. – Aqua Aug 24 '18 at 17:18 • I can show that $f(x)=0$ for all $x\\in\\mathbb{Q}$, but I don't know yet how to extend to $\\mathbb{R}$. – Batominovski Aug 24 '18 at 17:20 • From $f(x)=x \\vee f(x)=0$ we see that there is some", "$f(x^2+a+b)=f^2(x)+2a$ $\\forall x\\in\\mathbb{R}\\Rightarrow f(a+b)=2a$ $f(x^2-a+b)=f^2(x)-2a\\forall x\\in\\mathbb{R}$ $\\Rightarrow f(2b)=0(\\text{by above})$ Therefore $2b=0\\Rightarrow b=0$, contradiction! So, $f(x)=-f(-x)\\forall x\\in\\mathbb{R}$, form that, see $f(x)>0\\forall x>0(2)$ and $f(x)<0\\forall x<0(3)$. Now, for all $y<0$, in (*) setting $x=\\sqrt{-y}$ we have $f(f(y))+f(y)-2y=0\\forall y Final,from (3),(4) and idea of Example 1.10 ([1]) we see that$latex f(x)=x\\forall x<0\\$, but $f(x)=-f(-x)\\forall x\\in\\mathbb{R}$, we have $f(x)=x\\forall x\\in\\mathbb{R}.$", "# 2015 IMO Problems/Problem 5 ## Problem Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f$:$\\mathbb{R}\\rightarrow\\mathbb{R}$ satisfying the equation $f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania ## Solution $f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)$ for all real numbers $x$ and $y$. (1) Put $x=y=0$ in the equation, We get$f(0 + f(0)) + f(0) = 0 + f(0) + 0$ or $f(f(0)) = 0$ Let $f(0) = k$, then $f(k) = 0$ (2) Put $x=0, y=k$ in the equation, We get $f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)$ But $f(k) = 0$ and $f(0) = k$ so, $f(0) + f(0) = f(0)^2$ or $f(0)[f(0) - 2] = 0$ Hence $f(0) = 0, 2$ Case $1$ : $f(0) = 0$ Put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) = f(x)$ Say $f(x) = z$, we get $f(z) = z$ So, $f(x) = x$ is a solution Case $2$ : $f(0) = 2$ Again put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) + 2 = f(x) + 2x$ We observe that $f(x)$ must be a polynomial of", "1, y = 1,$$ then we have $$f(1)f(1) = f(2) + f(0)$$ If $$x = 2, y = 1$$, then we have $$f(2)f(1) = f(3) + f(1)$$ Since we have $$4$$ variables $$(f(0), f(1), f(2)$$ and $$f(3)$$) and $$3$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) If $$f(1) = 1,$$ then we have$$f(2) = f(1)f(1) – f(0) = 1 – 2 = -1, f(3) = f(2)f(1) – f(1) = (-1)*1 – 1 = -2$$ and $$f(0)f(1)f(2)f(3) = 2*1*(-1)(-2) = 4.$$ If $$f(1) = 2$$, then we have $$f(2) = f(1)f(1) – f(0) = 4 – 2 = 2, f(3) = f(2)f(1) – f(1) = 2*1 – 2 = 0$$ and $$f(0)f(1)f(2)f(3) = 2*2*(2)*0 = 0.$$ Since condition 1) does not yield a unique solution, it is not sufficient. Condition 2) Since $$f(1) = 1,$$ we have $$f(0) = f(1)f(0) = f(1) + f(1) = 2, f(2) = f(1)f(1) – f(0) = 1 – 2 = -1, f(3) = f(2)f(1) – f(1) = (-1)*1 – 1 = -2$$ and $$f(0)f(1)f(2)f(3) = 2*1*(-1)*.(-2) = 4.$$ Condition 2) is sufficient since it yields a unique solution. If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one", "for all $x\\in\\mathbb{N}$, and b)$f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x$, $y\\in\\mathbb{N}$. 25. Consider those functions $f:\\mathbb{N}\\to\\mathbb{N}$ which satisfy the condition $f(m+n) \\geq f(m)+f(f(n))-1$ for all $m,n\\in\\mathbb{N}.$ Find all possible values of $f(2007).$ 26. Find all surjective functions $f:\\mathbb{N}\\to\\mathbb{N}$ such that for every $m,n\\in\\mathbb{N}$ and every prime $p,$ the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p.$ 27. Find all real polynomials $f$ such that $2yf(x+y)+(x-y)(f(x)+f(y)) \\geq 0\\forall x,y\\in\\mathbb{R}.$ 28. Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ with $x,y\\in\\mathbb{R}$ such that $f(x-f(y))=f(x+y)+f(y).$ 29. Show that for positive integer $n$, and for $x\\not =0$, $\\left(x^{n-1}\\sin\\dfrac{1}{x}\\right)^{(n)}=\\dfrac{(-1)^n}{x^{n+1}}\\sin\\left(\\dfrac{1}{x}+\\dfrac{n\\pi}{2}\\right).$ 30. Find all $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xy+f(x))=xf(y)+f(x)$ for every pair of real numbers $x,y$.", "numerator . . . Wow! . . . How did I miss that? . . . *blush* 9. Originally Posted by linda2005 2) let $x,y,z > 0$, prove that : $\\frac{xy}{z} + \\frac{yz}{x} + \\frac{zx}{y} \\geq x + y + z$ expanding $(a-b)^2 + (b-c)^2 + (c - a)^2 \\geq 0$ gives us $a^2 + b^2 + c^2 \\geq ab + bc + ac.$ now put $a=xy, \\ b = yz, \\ c=xz$ and then divide by $xyz.$ 3) Find all function $f: \\mathbb{R} \\to \\mathbb{R}$ such that : $f\\left( xf(x) + f(y) \\right ) = \\left(f(x)\\right)^2 +y. \\ \\ \\ \\ \\ (*)$ this question is not bad! if in $(*)$ we put x = 0 we'll get $f(f(y))=(f(0))^2 + y. \\ \\ \\ \\ \\ \\ (**)$ 1) $f$ is onto: suppose $z \\in \\mathbb{R}$ is given. put $y=z - (f(0))^2.$ then by $(**): \\ f(f(y))=z.$ 2) $f(0)=0$: by 1) there exists $x$ with $f(x)=0.$ then by $(*): \\ f(f(y))=y,$ for all $y \\in \\mathbb{R}.$ let's call this result $(***).$ now $(**)$ gives us $f(0)=0.$ 3) $\\forall x \\in \\mathbb{R}: \\ (f(x))^2 = x^2$: if in $(*)$ we put y = 0, we'll get $f(xf(x))=(f(x))^2.$ again if in $(*)$ we let y = 0 and replace x with f(x) and use $(***)$ we'll get: $f(xf(x))=x^2.$ 4)", "+0 # help 0 53 2 For all real numbers x and y, the function f satisfies f(x + y) = f(x) + f(y). If f(1) = 3, then find f(10). Dec 16, 2019 #1 +118 +1 Since $$f(x+y)=f(x)+f(y)$$, we can arrive at the following facts: $$f(10)=f(2+8)=f(2)+f(8)$$, $$f(8)=f(2+6)=f(2)+f(6)$$, $$f(6)=f(2+4)=f(2)+f(4)$$, and $$f(4)=f(2+2)=f(2)+f(2)$$. Using these facts we can write $$f(10)=f(2)+f(2)+f(2)+f(2)+f(2)$$. In addition, since $$f(1)=3$$, we can write $$f(2)=f(1+1)=f(1)+f(1)=3+3=6$$. So $$f(10)=6+6+6+6+6=30$$. Dec 16, 2019 #2 +23809 +2 For all real numbers x and y, the function f satisfies f(x + y) = f(x) + f(y). If f(1) = 3, then find f(10). $$\\begin{array}{|lrcll|} \\hline & \\mathbf{f(x+y)} &=& \\mathbf{f(x) + f(y)} \\\\\\\\ x=1,y=1: & f(1+1) &=& f(1) + f(1) \\\\ & \\mathbf{f(2)} &=& \\mathbf{2f(1)} \\\\\\\\ x=1,y=2: & f(1+2) &=& f(1) + f(2) \\\\ & f(3) &=& f(1) + 2f(1) \\\\ & \\mathbf{f(3)} &=& \\mathbf{3f(1)} \\\\\\\\ x=1,y=3: & f(1+3) &=& f(1) + f(3) \\\\ & f(4) &=& f(1) + 3f(1) \\\\ & \\mathbf{f(4)} &=& \\mathbf{4f(1)} \\\\\\\\ \\ldots \\\\ & \\mathbf{f(n)} &=& \\mathbf{nf(1)} \\\\\\\\ \\hline n=10: & f(10) &=& 10f(1) \\quad | \\quad f(1) =3 \\\\ & f(10) &=& 10*3 \\\\ & \\mathbf{f(10)} &=& \\mathbf{30} \\\\ \\hline \\end{array}$$ Dec 17, 2019", "# $f(f(x)f(y))+f(x+y)=f(xy)$ Find all functions $f\\colon \\mathbb R\\rightarrow \\mathbb R$ such that for all reals $x$ and $y$: $$f(f(x)f(y))+f(x+y)=f(xy).$$ It was six hours ago in IMO 2017 (problem 2). I tried the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success. Let $f(0)=a$. Hence, for all $x\\in\\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest? Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and $1-(1-x)(1-y)+1-x-y=1-xy$. • $x\\mapsto 0$ is a solution. Jul 18 '17 at 18:16 • $f(f(0)^2)+f(0)=f(0)\\implies f(f(0)^2)=0.$ – mfl Jul 18 '17 at 18:24 • @астон вілла олоф мэллбэр See here: artofproblemsolving.com/community/c6h1480146 Jul 18 '17 at 18:40 $$f(f(x)f(y))+f(x+y)=f(xy). \\ \\ \\ (1)$$ Let $f(0)=c$. Choose $x=y=0$, we can get $$f(c^2)=0.\\ \\ \\ (2)$$ Choose $y=0$, we can get $$f(cf(x))+f(x)=c.\\ \\ \\ (3)$$ Choose $y=\\frac{x}{x-1}(x\\neq 1)$, we can get $$f\\left(f(x)f\\left(\\frac{x}{x-1}\\right)\\right)=0(x\\neq 1).\\ \\ \\ (4)$$ When $c=0$, from equation $(3)$ we get $f(x)=0.$ When $c\\neq 0$, i.e. $f(0)\\neq 0$. Form equation $(4)$ we know there esists $x_0\\neq0$ such that $f(x_0)=0.$ We claim that :$x_0=1$. Otherwise, choose $x=x_0$ in equation $(4)$, we get $f(0)=0$ which is a contradiction. Combining equation $(2)$, we know $c=1$ or $-1$. If $c=1$, that is to say $f(0)=1,$ choose $y=1$ in equqtion $(1)$, we get $f(x+1)=f(x)-1$. So $f(n)=1-n$ and $f(x+n)=f(x)-n$ for all $n\\in Z$. By equation $(1)$ we get $$f(f(x)f(y)+1)+f(x+y+n)=f(xy+n+1).\\ \\ \\ (5)$$ In the following we prove", "get $*([F1(x),y]zd2(F1(t))=0 for all x,y,z,t∈I.$ Also, R is *-prime, we get either [F1(x), y] = 0 for all x, yI or d2(F1(t)) = 0 for all tI. Take [F1(x), y] = 0 for all x, yI. Replace y by yr to get y[F1(x), r] = 0 for all x, yI and rR and hence yz[F1(x), r] = 0 for all x, y, zI and rR. Since I is a *-ideal, we get *(y)z[F1(x), r] = 0 for all x, y, zI and rR. By Lemma 2.1, we get [F1(x), r] = 0 for all xI and rR. Further, replacing x by xr, we find that $[x,r]d1(x)+x[d1(r),r)=0 for all x∈I and r∈R.$ Replacing x by xz in (2.5) and using (2.5), we get [x, r]zd1(r) = 0 for all x, zI and rR. Substituting r by r + r1, we find that [x, r]zd1(r1) + [x, r1]zd1(r) = 0 for all x, zI and r, r1R. Therefore [x, r]zd1(r1)R[x, r]zd1(r1) = − [x, r1]zd1(r)R[x, r]zd1(r1) = {0}. Hence, we get [x, r]zd1(r1) = 0 for all x, zI and r, r1R. Since I is a *-ideal, by Lemma 2.1, we have either [x, r] = 0 or d1(r1) = 0. But [x, r] = 0 implies that x[s, r] = 0 for all xI, r, sR. Since I is", "# Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt – Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\\since $f(0)=0$) Finally putting $y=(x+1)$ gives us $f(x)=0$. This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one? #### Solutions Collecting From Web of \"Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$\" You proved that $f(0)=0$, we can continue from here Suppose that there exist $k \\neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$. Now suppose that there exist no $k \\neq 0$ such that $f(k)=0$. Then plugging", "# Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt - Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\\since $f(0)=0$) Finally putting $y=(x+1)$ gives us $f(x)=0$. This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one? You proved that $f(0)=0$, we can continue from here Suppose that there exist $k \\neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$. Now suppose that there exist no $k \\neq 0$ such that $f(k)=0$. Then plugging $x=x,y=-x$ gives $f(x^2-xf(x)) = xf(0)= 0$, which by assumption means $x^2=xf(x)$ or $f(x)=x$. So we conclude that, possible functions", "will get statements relating $f(a)$ and $f(-a)$. So for instance, you probably want to choose $x$ and $y$ such that one or more of $x+y$, $x-y$, $x$, and $y$ are equal to $a$ or $-a$. First, setting $y=0$ gives $f(x)^2=f(x)^2f(0)^2$. This means $f(0)^2=1$ unless $f(x)=0$ for all $x$. The latter case is trivial, so let us assume $f$ is not identically $0$ and so $f(0)^2=1$. Now set $x=0$ and $y=a$ to get $$f(a)f(-a)=f(0)^2f(a)^2=f(a)^2.$$ Unless $f(a)=0$, this implies $f(a)=f(-a)$, as desired. If $f(a)=0$, we need to do a little more work: setting $x=0$ and $y=-a$ gives $$f(-a)f(a)=f(0)^2f(-a)^2=f(-a)^2.$$ Since $f(a)=0$, this implies $f(-a)=0$ as well, and so we're done.", "# Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ Find all functions $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\\ge 0$ we have $f(a)>f(b)$ As much as I know: • putting $x=y=0$ we get $f(0)=0$. • let $x+y=0$, the we get $f(y)+f(-y)=\\frac{f(2y)}{2}$ Step 1: Prove : $f(y) = 0$ iff. $y=0$ It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$) Following proof is given by @Kyson To prove that $f(y)≠0$ for all$y<0.$ Assume there is a $y<0$ such that $f(y)=0.$ Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0$Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$ $\\$ Step 2: Prove $f(y)$ is even. Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\\frac{f(2y)}{2} = f(-y) +f(y) = \\frac{f(-2y)}{2}$ Thus $f(x)$is a even function, so $f(y)+f(-y)=2f(y)=\\frac{f(2y)}{2} \\Rightarrow f(2y)=4f(y)$ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\\infty,0)\\cup(0,\\infty)$) $\\$ Step 3: Prove:$f(y)=y^2f(1)$ By induction , find $f(ny)=n^2f(y)$ where $n \\in \\mathbb N$ Details of induction: Suppose $m\\in \\mathbb N ,m\\ge 2 ,\\ f(ny)=n^2f(y)$ holds for all $n\\le m$ Put $x=(m-1)y$ in property (a).$\\\\ \\Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\\\ \\Rightarrow f((m+1)y)=(m+1)^2f(y) \\ for\\ y\\not = 0 \\ and \\ it\\ still\\ holds\\ for\\ y=0$ Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$ Remark: Result induces $f(zy)=z^2f(y)$ where $z\\in \\mathbb Z$ , since $f$ is even. $\\$ $\\forall", "Hendrra May 11 at 22:49 • @bungo The real part of $e^z$ is $\\text{Re}(e^z)=e^{\\text{Re}(z)}\\cos(\\text{Im}(z))=e^x\\cos(y)$. So, what does it mean to be increasing here? I believe that the OP wants to show that $|e^z|=e^x$ is increasing. – Mark Viola May 11 at 23:25 We have a function $f: \\mathbb R \\to \\mathbb R$ such that $f(0)=1$ and $f>1$ on $(0,\\infty).$ We also know $f(x+y)= f(x)f(y)$ for all $x,y\\in \\mathbb R.$ Thus for any real $x,$ $f(x-x)=f(0)=1=f(x)f(-x),$ which implies $f(-x)=1/f(x).$ From this it follows that $f>0$ on $(-\\infty,0).$ Hence $f>0$ everywhere. Finally, suppose $x<y.$ Then $0<y-x,$ hence $$1=f(0) < f(y-x) = f(y)f(-x)= f(y)/f(x).$$ This implies $f(x) < f(y)$ and we're done. $\\exp(0) = 1$ If $x>0$ then $\\exp(x)>1$ $\\exp(x) - 1 = \\sum_\\limits{n=1}^{\\infty} \\frac {x^n}{n!}$ every term on the right is positive, so the sum must be positive. $\\exp(x+h) = \\exp(x)\\exp(h)$ $\\exp(x-x) = \\exp(x)\\exp(-x) = exp(0) = 1\\\\ \\exp(-x) = \\frac {1}{\\exp(x)}$ For all real $x, \\exp(x)>0$ $\\exp(x)$ is increasing if for all $h>0, \\exp(x+h) > \\exp(x)$ $\\exp(x)\\exp(h)>\\exp(x)\\\\ \\exp(x)(\\exp(h)-1)>0$ Since both factors are positive, it is clearly true. • Proving that $\\exp(x+h) = \\exp(x)\\exp(h)$ based on the power series alone may be more work that what has been done here. – Michael Hardy May 12 at 0:19 • You say IF the function is increasing THEN blahblahblah, and blahblahblah is", "Page 1 of 1 ### IMO 2015 - Problem 5 Posted: Wed Jul 15, 2015 1:51 am Let $\\mathbb R$ be the set of real numbers. Determine all functions $f:\\mathbb R\\to\\mathbb R$ that satisfy the equation$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$for all real numbers $x$ and $y$. Proposed by Albania. ### Re: IMO 2015 - Problem 5 Posted: Sun Apr 02, 2017 9:50 pm The solutions are $2-x$ and $x$. They satisfy the FE. Let $P(x,y)$ denote the FE. $$P(0,0)\\Rightarrow f(f(0))=0$$ $$P(0,f(0))\\Rightarrow f(0)=0,2$$. Case $1$: $f(0)=2$ $$P(0,x)\\Rightarrow f(f(x))-f(x)=2(x-1)$$ Which implies $f(x)=x$ is only possible when $x=1$. $$P(x,1)\\Rightarrow f(x+f(x+1))=x+f(x+1)$$ So $x+f(x+1)=1$ and so $f(x)=2-x$ for all $x$. Case $2$: $f(0)=0$. $$P(-1,1)\\Rightarrow f(-1)=-1$$ $$P(1,-1)\\Rightarrow f(1)=1$$ $$P(x,0)\\Rightarrow f(x+f(x))=x+f(x)$$ $$P(0,x)\\Rightarrow f(f(x))=f(x)$$ $$P(x-1,1)\\Rightarrow f(x-1+f(x))=x-1+f(x)$$ $$P(1,x-1+f(x))\\Rightarrow f(x+1+f(x))=x+1+f(x)$$ $$\\Rightarrow f(x+f(x-1))=x+f(x-1)$$ $$P(x,-1)\\Rightarrow f(x)=-f(-x)$$ $$P(x,-x)\\Rightarrow f(x)-f(x^2)=x-xf(x)$$ $$P(-x,x)\\Rightarrow -f(x)-f(x^2)=-x-xf(x)$$ Substracting the last equation from the second last, $f(x)=x$ For all real $x$." ]

[ "# Polynomials going crazy! Algebra Level 5 The polynomial $$f(x) = x^{2007} + 17x^{2006} + 1$$ has distinct roots $$r_1, r_2,\\ldots, r_{2007}$$. A polynomial $$P$$ of degree 2007 has the property that $$P \\left( r_j + \\dfrac1{r_j} \\right) = 0$$ for $$j = 1,2,\\ldots,2007$$. Compute $$\\dfrac{259}{17} \\times \\dfrac{P(1)}{P(-1)}$$. ×", "where $n = 1$ and we are given $x_0, x_1, f(x_0), f(x_1)$, the Lagrange polynomial approximating $f$ is $P = \\frac{x - x_1}{x_0 - x_1} f(x_0) + \\frac{x - x_0}{x_1 - x_0} f(x_1)$. Examination of the polynomial should make it clear that it is, in a way, the simplest polynomial where $P(x_k) = f(x_k)$ at each $x_k$.", "the right-hand side to a common denominator one gets :$\\sum_^k \\frac= \\frac \\sum_^k \\frac,$ and the numerator is equal to one, as being a polynomial of degree less than $k,$ which takes the value one for $k$ different values of $X.$ Using the above general formula, we get the Lagrange interpolation formula: :$P\\left(X\\right)=\\sum_^k A_i\\frac.$ ## Hermite interpolation Hermite interpolation is an application of the Chinese remainder theorem for univariate polynomials, which may involve moduli of arbitrary degrees (Lagrange interpolation involves only moduli of degree one). The problem consists of finding a polynomial of the least possible degree, such that the polynomial and its first derivatives take given values at some fixed points. More precisely, let $x_1, \\ldots, x_k$ be $k$ elements of the ground field (mathematics), field $K,$ and, for $i=1,\\ldots, k,$ let $a_, a_, \\ldots, a_$ be the values of the first $r_i$ derivatives of the sought polynomial at $x_i$ (including the 0th derivative, which is the value of the polynomial itself). The problem is to find a polynomial $P\\left(X\\right)$ such that its ''j''th derivative takes the value $a_$ at $x_i,$ for $i=1,\\ldots,k$ and $j=0,\\ldots,r_j.$ Consider the polynomial :$P_i\\left(X\\right) = \\sum_^\\frac\\left(X - x_i\\right)^j.$ This is the Taylor polynomial of order $r_i-1$ at $x_i$, of the unknown polynomial $P\\left(X\\right).$ Therefore, we must have :$P\\left(X\\right)\\equiv P_i\\left(X\\right) \\pmod .$ Conversely,", "[tricks] 1. For a polynomial $f(z) = c_n z^n + c_{n-1}z^{n-1} + \\dotsb + c_0$ with roots $a_1, \\dotsc ,\\ a_n$, $\\displaystyle \\sum a_i = -\\frac{c_{n-1}}{c_n},$ $\\displaystyle \\prod a_i = (-1)^n \\frac{c_0}{c_n}$ 2. $\\displaystyle{f^\\prime (z) = f(z) \\left( \\frac{1}{z-a_1} + \\dotsb + \\frac{1}{z-a_n} \\right)}$", "$r(1) = f(1) = 2$ and $r(2) = f(2) = 1$. Note that here we have $P(1)+1 = P(2)+2 = 3$ i.e. the polynomial $P(x)+x-3$ has roots $x=1$ and $x=2$ $\\Rightarrow P(x)+(x-3) = (x-1)(x-2)Q(x) \\Rightarrow P(x) = (x-1)(x-2)Q(x) -(x-3)$ Its easy to see that $a=-1, b=3$", "a polynomial of degree two. When not specified otherwise, the polynomial is supposed to have real coefficients, and the zeros are points in a Euclidean space. However, most properties remain true when the coefficients belong to any field and the points belong in an affine space. As usually in algebraic geometry, it is often useful to consider points over an algebraically closed field containing the polynomial coefficients, generally the complex numbers, when the coefficients are real. This define affine quadrics. Many properties becomes easier to state (and to prove) by extending the quadric to the projective space by projective completion consisting of adding points at infinity. Technically, if ${\\displaystyle p(x_{1},\\ldots ,x_{n})}$ is a polynomial of degree two that defines an affine quadric, then its projective completion is defined by homogenizing p into ${\\displaystyle P(X_{0},\\ldots ,X_{n})=X_{0}^{2}\\,p({\\frac {X_{1}}{X_{0}}},\\ldots ,{\\frac {X_{n}}{X_{0}}})}$ (this is a polynomial, because the degree of p is two). The points of the projective completion are the points of the projective space whose projective coordinates are zeros of P. ### Equation A quadric in an affine space of dimension n is the set of zeros of a polynomial of degree 2, that is the set of the points whose coordinates satisfy an equation ${\\displaystyle p(x_{1},\\ldots ,x_{n})=0,}$ where the polynomial p has the form ${\\displaystyle p(x_{1},\\ldots ,x_{n})=\\sum _{i=1}^{n}a_{i,i}x_{i}^{2}+\\sum _{1\\leq", "× # Differentiability If $$f(x)$$ is a polynomial satisfying relation $$f(x)=\\dfrac{k}{k+1}$$ for $${k=1,2,3,\\ldots,100}$$ except $$k=a$$ where $$a$$ is a natural numbers and belongs to $$(0,100)$$ and $$f(101)=1$$.Find $$\\dfrac{1-f(a)}{f'(a)}$$ . Note by Akshay Sharma 1 year, 6 months ago Sort by: Try reading the first section of polynomial interpolation - remainder factor theorem. Addendum: Oh wait... this problem is not as easy as it looks. Give me one moment to think about this... · 1 year, 6 months ago 1. There are multiple polynomials that satisfy the conditions since we can add $$A (x-1)(x-2)\\ldots (x-101)$$. This suggests that you want the degree to be 100. 2. The condition should be $$f(k) = \\frac{k}{k+1}$$ instead? Staff · 1 year, 6 months ago", "Characteristic polynomial 1. Jan 5, 2007 keddelove And again a question: L is a field for which $$a \\in L$$. The matrix $$A = \\frac{1}{2}\\left( {\\begin{array}{*{20}c} 1 & 1 & 1 & 1 \\\\ 1 & a & { - 1} & { - a} \\\\ 1 & { - 1} & 1 & { - 1} \\\\ 1 & { - a} & { - 1} & a \\\\ \\end{array}} \\right)$$ has the characteristic polynomial $$x^4 - \\left( {a + 1} \\right)x^3 + \\left( {a - 1} \\right)x^2 + \\left( {a + 1} \\right)x - a$$ I need to show that this information is correct for a=1 in any field. My problem is that when i calculate the polynomial for a=-1 I end up with another characteristic polynomial than the one given. Maybe i'm going about it the wrong way. Suggestions or pointers are very welcome 2. Jan 6, 2007 Hurkyl Staff Emeritus I don't understand the question... And if you want to show something is correct for a=1, then why are you looking at a=-1? 3. Jan 6, 2007 mathwonk i conjecture he meant the characteristic polynomial is accurate for a any element of any field, and yet it failed for a = -1. 4. Jan 7, 2007 keddelove Oops, should have stated: Show that this", "\\vdots &\\ddots & \\vdots &\\vdots \\\\ 0 &\\cdots & 1 & -a_{n-1} \\end{pmatrix}$$ By construction, the characteristic polynomial of $A$ is $p(X)$, hence $\\det(A) = (-1)^n p(0)$ and $\\det(I+A)=(-1)^n p(-1).$ If you don't like zeros and ones, take any matrix that is similar to $A$. -", "zero $v$, $\\{v, f(v)\\}$ is linear independent and spans a 2d subspace closed under f. Then f is well defined on the quotient $V/\\{v,f(v)\\}$, which has dimension 2 less. (tyro1's answer is I think along these lines but a bit garbled). Since $f^2 = -I$, we see that the minimal polynomial of $f$ over $\\mathbb{R}$ is $x^2 + 1 \\in \\mathbb{R}[x]$. The minimal polynomial and characteristic polynomial share their root sets, so the characteristic polynomial of $f$ must be $(x^2 + 1)^k$ for some $k\\ge 1$. But the degree $2k$ of the characteristic polynomial is the dimension of $V$, hence $V$ is even-dimensional. ## protected by user223391 Jun 19 '16 at 4:56 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).", "polynomial. Polynomial General form Coefficients Linear polynomial $ax+b$ $a,b\\in R,a\\ne 0$ Quadratic polynomial $a{{x}^{2}}+bx+c$ $a,b,c\\in R.a\\ne 0$ Cubic polynomial $a{{x}^{3}}+b{{x}^{2}}+cx+d$ $a,b,c,d\\in R,a\\ne 0$ Quadratic polynomial $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e$ $a,b,c,d,e\\in R,a\\ne 0$ • Value of a polynomial: If p(x) is a polynomial in x. and if 'a' is any real number, then the value obtained by replacing 'x' by 'a' in p(x), denoted by p(a) is called the value of p(x) at x = a. • Zero of a polynomial: Areal number 'a' for which the value of the polynomial p(x) is zero, is called the zero of the polynomial. In other words, a real number 'a' is called a zero of a polynomial p(x) if p(a) =0. Note: The number zero is known as zero polynomial and its degree is not defined. • Geometric meaning of the zero of a polynomial: (a) The graph of a linear equation of the form $y=ax+b,a\\ne 0$is a straight line which intersects the X-axis at$\\left( \\frac{-b}{a},0 \\right)$. Zero of the polynomial ax + b is the x-coordinate of the point of intersection of the graph with X-axis. Thus, the zero of$y=ax+b$is $\\left( \\frac{-b}{a} \\right)$ Note: A linear polynomial ax + b, a -f- 0 has exactly one zero, i.e., $\\left( \\frac{-b}{a} \\right)$ (b) The graph of a quadratic equation $y=a{{x}^{2}}+bx+c,a\\ne 0$is a curve called", "- 1} + \\cdots + {c}_{0}$ as a generic $n$ degree polynomial we have. ${x}^{n} = {p}_{n - 2} \\left({x}^{2} - 1\\right) + a x + b$ with $n \\ge 2$ so we have ${\\left(- 1\\right)}^{n} = a \\left(- 1\\right) + b$ and $1 = a + b$ solving we have $a = \\frac{1 - {\\left(- 1\\right)}^{n}}{2}$ and $b = \\frac{1 + {\\left(- 1\\right)}^{n}}{2}$ in the present case we have $n = 1999$ so $a = 1 , b = 0$ and the remainder is $x$", "# Polynomial function/R/Degree d/(d+1)-te Derivative/Exercise Prove that a polynomial ${\\displaystyle {}P\\in \\mathbb {R} [X]}$ has degree ${\\displaystyle {}\\leq d}$ (or it is ${\\displaystyle {}P=0}$), if and only if the ${\\displaystyle {}(d+1)}$-th derivative of ${\\displaystyle {}P}$ is the zero poynomial.", "Let $f(z)=\\frac{p(z)}{q(z)}$, where $p(z)$ and $q(z)$ are polynomials. Prove that the sum of multiplicities of all zeroes minus the sum of orders of all poles, including zeroes or poles at $\\infty$, equals zero.", "its constant term is 0. Beacause $$X^1 \\cdot f_A(X)$$ contributes $$f(X)$$'s lowest degree. Therefore, we come to the conclusion that $$f(X)$$ is a polynomial that $$deg(f) \\leq 2t$$, and $$f(0) = 0$$.", "this is not possible. $\\frac{p}{1-p}$ is not a polynomial of $p$, so it cannot be made equal to a finite order polynomial like $\\sum_{j = 1}^{2^n} t_j p^{n_j} (1-p)^{n - n_j}, \\forall p$. So there is no unbiased estimator of $\\theta$!", "• question_answer Let $p(x)$ be a polynomial of degree 2009 such that $p(x)=\\frac{1}{x+1}$. For $x=0,\\,1,\\,2,\\,3,...,\\,2009$. The value of p(2010) is A) 0 B) 1 C) $\\frac{1}{2011}$ D) can't be determined We have, $...(x-2008)\\,(x-2009)\\,x$ which is an identity. Also, for $x=-1$ $-1=k\\,(-2)\\,(-3)\\,(-4)\\,(-5)\\,....\\,(-2009)\\,(-2010)\\,(-1)$$\\Rightarrow$ $k=-\\frac{1}{(2010)!}$ For $x=2010,$ $\\Rightarrow$ $\\,p(2010)=0$", "'14 at 21:24 $$L_i(x) = 1 + \\frac{x - x_i}{x_i - x_{i+1}} + \\frac{(x - x_i)(x - x_{i+1})}{(x_i - x_{i+1})(x_i - x_{i+2})} + \\ldots + \\frac{(x - x_i)(x - x_{i+1})\\ldots(x - x_{n+i-1})}{(x_i - x_{i+1})(x_i - x_{i+2})\\ldots(x_i - x_{n+i})}$$ where $x_{n+j} = x_j$ for $j \\geq 1$. $$x_j \\mapsto \\delta_{0,j}$$ and such a polynomial with degree not greater than $n$ is unique.", "# Identification of the polynomial given symmetric relations [duplicate] Let $p(x)$ be an eighth degree polynomial. Given that: $p(1) = 1$, $p(2) = 1/2$, $p(3) = 1/3$, .... $p(9) = 1/9$ Find the value of $p(10)$. I tried to approach this by taking $h(x) = p(x) - 1/x$ but then $h(x)$ won't be a polynomial. I tried to manipulate the equation but I failed to find anything helpful. I have found a relation between the coefficients and some values; 9 equations and 9 variables but it would be too cumbersome to solve. Some guidance would be appreciated. Thanks! (Sorry for the poor title; if anyone can improve it please do so) • Consider $g(x) = x\\cdot p(x) - 1$. – Daniel Fischer Apr 18 '17 at 15:57 • $g(x)$ is a polynomial with degree $9$ that vanishes at $x=1,2,\\ldots,8,9$, hence $g(x)=K(x-1)\\cdot\\ldots\\cdot(x-8)(x-9)$. Since $g(0)=-1$ we have $K=\\frac{1}{9!}$ and $$p(10) = \\frac{g(10)+1}{10} = \\frac{1+1}{10}=\\frac{1}{5}.$$ – Jack D'Aurizio Apr 18 '17 at 16:24", "# Polynomial Degree Problem • November 13th 2012, 10:38 AM Polynomial Degree Problem Hello, this problem is again from Gelfand and Shen's Algebra textbook. Problem 171: The highest coefficient of $P(x)$ is 1, and we know that $P(1)=0, P(2)=0, P(3)=0,\\ldots,P(9)=0,P(10)=0.$ What is the minimal possible degree of $P(x)$? Find $P(11)$ for this case. Any hints would be appreaciated. • November 13th 2012, 01:07 PM Since $P(1)=0, P(2)=0, P(3)=0,\\ldots,P(9)=0,P(10)=0.$, this means that $P(x)$ has at least 10 roots so it is at least degree 10 polynomial. $P(x)=Q(x-1)(x-2) \\ldots (x-9)(x-10)$, $Q$ has to be 1 since maximum coefficient is 1, so $P(x)=(11-1)(11-2) \\ldots (11-9)(11-10)=10!=3628800.$" ]

[ "( 2^{a/b} -1 )\\left( \\frac{r-d_0}{r_0} \\right)^a \\right]^{-b/a}$$ {SMAP R_0= $$r_0$$ D_0= $$d_0$$ A= $$a$$ B= $$b$$} $$d_0=0.0$$ Q $$s(r) = \\frac{1}{1 + \\exp(\\beta(r_{ij} - \\lambda r_{ij}^0))}$$ {Q REF= $$r_{ij}^0$$ BETA= $$\\beta$$ LAMBDA= $$\\lambda$$ } $$\\lambda=1.8$$, $$\\beta=50 nm^-1$$ (all-atom) $$\\lambda=1.5$$, $$\\beta=50 nm^-1$$ (coarse-grained) CUBIC $$s(r) = (y-1)^2(1+2y) \\qquad \\textrm{where} \\quad y = \\frac{r - r_1}{r_0-r_1}$$ {CUBIC D_0= $$r_1$$ D_MAX= $$r_0$$} TANH $$s(r) = 1 - \\tanh\\left( \\frac{ r - d_0 }{ r_0 } \\right)$$ {TANH R_0= $$r_0$$ D_0= $$d_0$$} COSINUS $$s(r) &= 1 & if r<=d0 s(r) &= 0.5 \\left( \\cos ( \\frac{ r - d_0 }{ r_0 } * PI ) + 1 \\right) & if d0 d0+r0$$ {COSINUS R_0= $$r_0$$ D_0= $$d_0$$} CUSTOM $$s(r) = FUNC$$ {CUSTOM FUNC=1/(1+x^6) R_0= $$r_0$$ D_0= $$d_0$$} Notice that for backward compatibility we allow using MATHEVAL instead of CUSTOM. Also notice that if the a CUSTOM switching function only depends on even powers of x it can be made faster by using x2 as a variable. For instance {CUSTOM FUNC=1/(1+x2^3) R_0=0.3} is equivalent to {CUSTOM FUNC=1/(1+x^6) R_0=0.3} but runs faster. The reason is that there is an expensive square root calculation that can be optimized out. Attention With the default implementation CUSTOM is slower than other functions (e.g., it is slower than an equivalent RATIONAL function by approximately a factor 2). Checkout", "$\\beta_j(x)=1$ if $x>0$, then . . . Prob. 3 (c), Chap. 6, in Baby Rudin: If $\\beta_j(x)=0$ if $x<0$ and $\\beta_j(x)=1$ if $x>0$ for $j=1, 2, 3$, and if . . . Here I'll attempt a solution to Prob. 3 (d). My Attempt of Part (d): If $f$ is continuous at $0$, then $f \\in \\mathscr{R} \\left( \\beta_3 \\right)$ on $[-1, 1]$ (by Prob. 3 (c), Chap. 6, in Baby Rudin, 3rd edition), and (as in my post on Pob. 3 (c), Chap. 6, in Baby Rudin, 3rd edition) we also have $$\\int_{-1}^1 f \\ \\mathrm{d} \\beta_3 = f(0). \\tag{1}$$ Moreover, if $f$ is continuous at $0$, then we also have $$\\lim_{x \\to 0} f(x) = f(0),$$ which implies that $$f(0+) = f(0) = f(0-).$$ So by Probs. 3 (a) and (b), Chap. 6, in Baby Rudin we can conclude that $f \\in \\mathscr{R}\\left(\\beta_1\\right)$ and $f \\in \\mathscr{R}\\left(\\beta_2\\right)$ on $[-1, 1]$, and that $$\\int_{-1}^1 f \\ \\mathrm{d} \\beta_1 = f(0) = \\int_{-1}^1 f \\ \\mathrm{d} \\beta_2. \\tag{2}$$ From (1) and (2) we have our desired result. Are there any problems with my proof? • I'm sorry for the confusion, but this post (in its entirety) is not at all a duplicate of any of my earlier posts. The confusion is because I've included in all four of my", "for $$x_{1}$$ and $$x_{2}$$, respectively, and substituting to the production function results in \\begin{align*} q = A \\left(\\lambda \\alpha r \\frac{q}{w_{1}}\\right)^{\\alpha r} \\left(\\lambda (1 - \\alpha) r \\frac{q}{w_{2}}\\right)^{(1 - \\alpha) r} = A \\lambda^{r} r^{r} \\left(\\frac{\\alpha}{w_{1}}\\right)^{\\alpha r} \\left(\\frac{1 - \\alpha}{w_{2}}\\right)^{(1 - \\alpha) r} q^{r}, \\end{align*} from which we obtain \\begin{align*} \\lambda = A^{-\\frac{1}{r}} r^{-1} \\left(\\frac{\\alpha}{w_{1}}\\right)^{-\\alpha} \\left(\\frac{1 - \\alpha}{w_{2}}\\right)^{-(1 - \\alpha)} q^{\\frac{1-r}{r}}, \\end{align*} Substituting back to the first order conditions, we can eliminate $$\\lambda$$, and solve for $$x_{1}$$, $$x_{2}$$ to obtain the conditional demand functions \\begin{align*} x_{1} &= \\lambda \\alpha r \\frac{q}{w_{1}} = \\left(\\frac{w_{2}}{w_{1}}\\frac{\\alpha}{1 - \\alpha}\\right)^{1-\\alpha} \\left(\\frac{q}{A}\\right)^{\\frac{1}{r}}, \\\\ x_{2} &= \\lambda (1 - \\alpha) r \\frac{q}{w_{2}} = \\left(\\frac{w_{1}}{w_{2}}\\frac{1 - \\alpha}{\\alpha}\\right)^{\\alpha} \\left(\\frac{q}{A}\\right)^{\\frac{1}{r}}. \\end{align*} 4. The easiest way to calculate the cost function of the firm is to use the first order conditions of part 2 and the objective function. We then have \\begin{align*} c(q, w_{1}, w_{2}) &= w_{1} \\lambda \\alpha r \\frac{q}{w_{1}} + w_{2} \\lambda (1 - \\alpha) r \\frac{q}{w_{2}} \\\\ &= \\lambda r q \\\\ &= \\left(\\frac{\\alpha}{w_{1}}\\right)^{-\\alpha} \\left(\\frac{1 - \\alpha}{w_{2}}\\right)^{-(1 - \\alpha)} \\left(\\frac{q}{A}\\right)^{\\frac{1}{r}}. \\end{align*} ## 5. Cost Types ### 5.1. Context • Not all costs have the same characteristics. Economists talk about fixed and variable costs and distinguish between short- and long-run horizons. • Different costs sway (rational) business decisions in distinct ways. • How do fixed and variable", "also with $k=1$. Example: $n=1$: We obtain from (3) \\begin{align*} \\exp&\\left(as^b\\right)\\sum_{k=1}^1\\frac{(-a)^k}{k!}\\left(\\sum_{j=1}^k(-1)^j\\binom{k}{j}bj\\right)s^{bk-1}\\\\ &=\\exp\\left(as^b\\right)\\left(\\frac{(-a)^1}{1}\\left[(-1)\\binom{1}{1}b\\right]\\right)s^{b-1}\\\\ &=\\exp\\left(as^b\\right)abs^{b-1}\\tag{4} \\end{align*} Substituting $a=-C_d,b=\\frac{2}{\\alpha_N}$ in (4) gives \\begin{align*} \\color{blue}{\\exp\\left(-\\frac{2C_d}{\\alpha_N}\\right)\\left(-\\frac{2C_d}{\\alpha_N}\\right)s^{\\frac{2}{\\alpha_N}-1}} \\end{align*} in accordance with OPs calculation. Example: $n=2$: We obtain from (3) Substituting $a=-C_d,b=\\frac{2}{\\alpha_N}$ in (5) gives \\begin{align*} \\exp&\\left(-\\frac{2C_d}{\\alpha_N}\\right)\\left(\\left(-C_d\\right)\\frac{2}{\\alpha_N}\\left(\\frac{2}{\\alpha_N-1}\\right)s^{\\frac{2}{\\alpha_N-2}}+C_d^2\\frac{4}{\\alpha_N^2}s^{\\frac{4}{\\alpha_N-2}}\\right)\\\\ &=\\color{blue}{\\exp\\left(as^b\\right)\\frac{2C_d}{\\alpha_N^2}\\left(\\alpha_N-2+2C_ds^{\\frac{2}{\\alpha_N}}\\right)} \\end{align*} in accordance with OPs calculation. • Thanks for your response. To simplify, the solution that I have shared, is it correct as yours one is a bit complex on the face of it... But I appreciate the reply. – Kashan Jan 24 '18 at 0:24 • @Sjaffry: You're welcome. After work I'll add some comments and an example which might help to clarify things. – Markus Scheuer Jan 24 '18 at 7:52 • @Sjaffry: I've added examples which show your calculations are correct. Good work. :-) Hint: It's preferable to stick at the same notation. In your case the function $L_d$ and the variable $s$. Otherwise you should explicitly mention the notational change. – Markus Scheuer Jan 24 '18 at 15:00 • Thanks for the detailed answer. The only reason I accepted my answer as the solution is that it simplifies the approach, though your approach gives the exact formula for the same thing. I appreciate your time and learned from your response. – Kashan Jan 25 '18 at 21:28 • @Sjaffry: You're welcome! – Markus", "f(y_1|\\alpha) \\cdot f(y_2|\\alpha) \\cdot \\dotso \\cdot f(y_n|\\alpha)$ $=\\prod^n_{i=1} \\frac{1}{\\alpha} m y_{\\color{red}i}^{m-1} \\exp \\left( -\\frac{y_{\\color{red}i}^m}{\\alpha} \\right)$ Mr F says: Note the small errata. $=\\bigg{[}\\frac{1}{\\alpha} m y_1^{m-1} \\exp \\left( -\\frac{y_1^m}{\\alpha} \\right)\\bigg{]}$ $\\times \\bigg{[}\\frac{1}{\\alpha} m y_2^{m-1} \\exp \\left( -\\frac{y_2^m}{\\alpha} \\right)\\bigg{]} \\times \\dotso \\times$ $\\bigg{[}\\frac{1}{\\alpha} m y_n^{m-1} \\exp \\left( -\\frac{y_n^m}{\\alpha} \\right)\\bigg{]}$ Mr F says: I'll pick it up from here because the line below is wrong and not needed. $=\\bigg{(}\\frac{1}{\\alpha} m\\bigg{)}^n \\prod^n_{i=1} y_i^{m-1} \\exp \\left( -\\sum^n_{i=1}\\frac{y_i^m}{\\alpha} \\right)$ at which point I get stuck, I know I'm supposed to factor it out and have a second function that isn't dependent on $\\alpha$, but I'm not quite sure on how to separate it out. $= \\frac{m^n}{\\alpha^n} \\, (y_1 \\, y_2 \\, .... \\, y_n)^{m-1} \\exp \\left(-\\frac{1}{\\alpha} \\sum_{i=1}^n y_i^m\\right)$ $= m^n \\, (y_1 \\, y_2 \\, .... \\, y_n)^{m-1} \\, \\frac{\\exp \\left(-\\frac{1}{\\alpha} u\\right)}{\\alpha^n}$ where $U = \\sum_{i=1}^n Y_i^m$ $= \\left(\\frac{\\exp \\left(-\\frac{u}{\\alpha} \\right)}{\\alpha^n}\\right) \\cdot \\left(m^n \\, (y_1 \\, y_2 \\, .... \\, y_n)^{m-1} \\right)$ $= g\\left(\\alpha, \\, u \\right) \\cdot h(y_1 \\, y_2 \\, .... \\, y_n)$.", "\\alpha^{-1} .\\end{align*} So $$K$$ is a field and $${\\left\\lvert {K} \\right\\rvert} = p^n$$. $$\\hfill\\blacksquare$$ Corollary: Let $$F$$ be a finite field. If $$n\\in{\\mathbb{N}}^+$$, then there exists an $$f(x) \\in F[x]$$ that is irreducible of degree $$n$$. Proof: Let $$F$$ be a finite field, so $${\\left\\lvert {F} \\right\\rvert} = p^r$$. By the previous lemma, there exists a $$K$$ such that $${\\mathbb{Z}}_p \\subseteq k \\subseteq \\overline F$$. $$K$$ is defined as \\begin{align*} K \\mathrel{\\vcenter{:}}=\\left\\{{\\alpha \\in F \\mathrel{\\Big|}\\alpha^{p^n} - \\alpha = 0}\\right\\} .\\end{align*} We also have \\begin{align*} F = \\left\\{{\\alpha \\in \\overline F {~\\mathrel{\\Big|}~}\\alpha^{p^n} - \\alpha = 0}\\right\\}.\\end{align*} Moreover, $$p^{rs} = p^r p^{r(s-1)}$$. So let $$\\alpha \\in F$$, then $$\\alpha^{p^r} - \\alpha = 0$$. Then \\begin{align*} \\alpha^{p^{rn}} = \\alpha^{p^r p^{r(n-1)}} = (\\alpha^{p^r})^{p^{r(n-1)}} = \\alpha^{p^{r(n-1)}} ,\\end{align*} and we can continue reducing this way to show that this is yields to $$\\alpha^{p^r} = \\alpha$$. So $$\\alpha \\in K$$, and thus $$F \\leq K$$. We have $$[K:F] = n$$ by counting elements. Now $$K$$ is simple, because $$K^{\\times}$$ is cyclic. Let $$\\beta$$ be the generator, then $$K = F(\\beta)$$. This the minimal polynomial of $$\\beta$$ in $$F$$ has degree $$n$$, so take this to be the desired $$f(x)$$. $$\\hfill\\blacksquare$$ ## Simple Extensions Let $$F \\leq E$$ and \\begin{align*} \\phi_\\alpha: F[x] &\\to E \\\\ f &\\mapsto f(\\alpha) .\\end{align*} denote the evaluation map. Case 1: Suppose", ")_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}x^{r+c}.}$ Hence, {\\displaystyle {\\begin{aligned}y_{1}&=a_{0}\\sum _{r=0}^{\\infty }{\\frac {(\\alpha )_{r}(\\beta )_{r}}{(1)_{r}(1)_{r}}}x^{r}=a_{0}{{}_{2}F_{1}}(\\alpha ,\\beta ;1;x)\\\\y_{2}&=\\left.{\\frac {\\partial y}{\\partial c}}\\right|_{c=0}.\\end{aligned}}} To calculate this derivative, let ${\\displaystyle M_{r}={\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}.}$ Then ${\\displaystyle \\ln(M_{r})=\\ln \\left({\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}\\right)=\\ln(c+\\alpha )_{r}+\\ln(c+\\beta )_{r}-2\\ln(c+1)_{r}}$ But ${\\displaystyle \\ln(c+\\alpha )_{r}=\\ln \\left((c+\\alpha )(c+\\alpha +1)\\cdots (c+\\alpha +r-1)\\right)=\\sum _{k=0}^{r-1}\\ln(c+\\alpha +k).}$ Hence, {\\displaystyle {\\begin{aligned}\\ln(M_{r})&=\\sum _{k=0}^{r-1}\\ln(c+\\alpha +k)+\\sum _{k=0}^{r-1}\\ln(c+\\beta +k)-2\\sum _{k=0}^{r-1}\\ln(c+1+k)\\\\&=\\sum _{k=0}^{r-1}\\left(\\ln(c+\\alpha +k)+\\ln(c+\\beta +k)-2\\ln(c+1+k)\\right)\\end{aligned}}} Differentiating both sides of the equation with respect to c, we get: ${\\displaystyle {\\frac {1}{M_{r}}}{\\frac {\\partial M_{r}}{\\partial c}}=\\sum _{k=0}^{r-1}\\left({\\frac {1}{c+\\alpha +k}}+{\\frac {1}{c+\\beta +k}}-{\\frac {2}{c+1+k}}\\right).}$ Hence, ${\\displaystyle {\\frac {\\partial M_{r}}{\\partial c}}={\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}\\sum _{k=0}^{r-1}\\left({\\frac {1}{c+\\alpha +k}}+{\\frac {1}{c+\\beta +k}}-{\\frac {2}{c+1+k}}\\right).}$ Now, ${\\displaystyle y=a_{0}x^{c}\\sum _{r=0}^{\\infty }{\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}=a_{0}x^{c}\\sum _{r=0}^{\\infty }M_{r}x^{r}.}$ Hence, {\\displaystyle {\\begin{aligned}{\\frac {\\partial y}{\\partial c}}&=a_{0}x^{c}\\ln(x)\\sum _{r=0}^{\\infty }{\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}+a_{0}x^{c}\\sum _{r=0}^{\\infty }\\left({\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r}^{2}}}\\left\\{\\sum _{k=0}^{r-1}\\left({\\frac {1}{c+\\alpha +k}}+{\\frac {1}{c+\\beta +k}}-{\\frac {2}{c+1+k}}\\right)\\right\\}\\right)x^{r}\\\\&=a_{0}x^{c}\\sum _{r=0}^{\\infty }{\\frac {(c+\\alpha )_{r}(c+\\beta )_{r}}{(c+1)_{r})^{2}}}\\left(\\ln x+\\sum _{k=0}^{r-1}\\left({\\frac {1}{c+\\alpha +k}}+{\\frac {1}{c+\\beta +k}}-{\\frac {2}{c+1+k}}\\right)\\right)x^{r}.\\end{aligned}}} For c = 0, we get ${\\displaystyle y_{2}=a_{0}\\sum _{r=0}^{\\infty }{\\frac {(\\alpha )_{r}(\\beta )_{r}}{(1)_{r}^{2}}}\\left(\\ln x+\\sum _{k=0}^{r-1}\\left({\\frac {1}{\\alpha +k}}+{\\frac {1}{\\beta +k}}-{\\frac {2}{1+k}}\\right)\\right)x^{r}.}$ Hence, y = Cy1 + Dy2. Let Ca0 = C and Da0 = D. Then ${\\displaystyle y=C{{}_{2}F_{1}}(\\alpha ,\\beta ;1;x)+D\\sum _{r=0}^{\\infty }{\\frac {(\\alpha )_{r}(\\beta )_{r}}{(1)_{r}^{2}}}\\left(\\ln(x)+\\sum _{k=0}^{r-1}\\left({\\frac {1}{\\alpha +k}}+{\\frac {1}{\\beta +k}}-{\\frac {2}{1+k}}\\right)\\right)x^{r}}$ ### γ an integer and γ ≠ 1 #### γ ≤ 0 The value of ${\\displaystyle \\gamma }$ is ${\\displaystyle \\gamma =0,-1,-2,\\cdots }$. To begin with, we shall simplify", "which in any case makes no sense, since $\\alpha_r$ is a constant and the righthand side is not. However, after letting $x=\\frac1r$ it does show \\begin{align*} \\alpha_r&=\\frac1{(1-1/r)(1-2/r)\\cdots(1-(r-1)/r)(1-(r+1)/r)\\cdots(1-k/r)}\\\\ &=(-1)^{k-r}\\frac{r^{k-1}}{(r-1)!(k-r)!}\\;. \\end{align*} The last step is obtained by first multiplying by $r^{k-1}$ to get $$\\alpha_r=\\frac{r^{k-1}}{\\underbrace{(r-1)(r-2)\\cdot(r-(r-1))}_{(r-1)!}\\underbrace{(r-(r+1))\\cdots(r-k)}_{(-1)(-2)\\cdots(-(k-r))}}$$ and then rewriting this as $$\\alpha_r=\\frac{r^{k-1}}{(r-1)!(-1)^{k-r}(k-r)!}=(-1)^{k-r}\\frac{r^{k-1}}{(r-1)!(k-r)!}\\;.$$ To find the coefficients of the expansion, you can follow the usual \"cover\" up rule which is essentially the results you have shown albeit buried in indices and what not: $$\\frac{1}{(1-x)(1-2x)...(1-kx)} = \\sum_{j=1}^{k} \\frac{\\alpha_j}{1-jx}$$ say we would want to find an arbitrary $\\alpha_j$ for some value of $j \\in \\{1,2,...,k\\}$. We call this $r$ to emphasise, and to avid confusion with the summation index. So we multiply both sides by $1-rx$ $$\\frac{1-rx}{(1-x)(1-2x)...(1-kx)} = \\sum_{j=1}^{k} \\alpha_j\\frac{1-rx}{1-jx}$$ Since $r$ will coincide with one of the $j$'s we can \"pull this out\" from the right hand side $$RHS = \\alpha_r + \\sum_{\\underset{j\\ne r}{j=1}}^{k} \\alpha_j\\frac{1-rx}{1-jx}$$ And for the left hand side, the r will \"cancel\" on one of the denominator factors: $$LHS = \\frac{1}{\\underset{\\text{without the 1-rx}}{(1-x)(1-2x)...(1-kx)}}$$ The expansion holds for all values of x so we can set $x=1/r$, and simplify both sides to get the result. I'll stop here because I think doing it is fun :)", "expectations. In order to have $$\\hat{\\beta}$$ unbiased we should have $$E(u\\mid Z, X)$$ = $$0$$, but this is an assumption too strong since it would also imply $$E(u\\mid X)$$ = $$0$$, case in which you would not even have an endogeneity problem with OLS estimators. 2) For the asymptotic variance of $$\\hat{\\beta}$$ consider that: $$\\hat{\\beta} = \\beta_{0} + (Z'X)^{-1}Z'u = \\beta_{0} + (n^{-1}Z'X)^{-1}n^{-1}Z'u$$ From which: $$\\sqrt{n} (\\hat{\\beta} - \\beta_{0}) = \\left ( \\frac{1}{n}Z'X \\right )^{-1}\\frac{1}{\\sqrt{n}}Z'u$$ and since we're assuming $$plim\\left ( \\frac{1}{n}Z'X \\right )$$ $$\\neq$$ $$0$$ and $$plim \\left ( \\frac{1}{n}Z'u \\right )$$ = $$0$$, then the plim of left hand side converges to $$0$$. In particular, define $$plim\\left ( \\frac{1}{n}Z'X \\right )$$ = $$Q_{ZX}$$, $$plim\\left ( \\frac{1}{n}X'Z \\right )$$ = $$Q_{XZ}$$ and $$plim\\left ( \\frac{1}{n}Z'Z \\right )$$ = $$Q_{ZZ}$$. At this point, I've worked out the solution as follows: \\begin{align} As.Var(\\hat{\\beta}) &= \\frac{1}{n} plim E \\left [ \\left ( \\frac{1}{n}Z'X \\right ) ^{-1}\\left (\\frac{1}{\\sqrt{n}}Z'u\\right )\\left (\\frac{1}{\\sqrt{n}}u'Z \\right ) \\left ( \\frac{1}{n}X'Z \\right ) ^{-1} \\right ] \\\\ & = \\frac{1}{n} E \\left [ plim \\left ( \\frac{1}{n}Z'X \\right ) ^{-1} plim \\left (\\frac{1}{\\sqrt{n}}Z'u\\right )plim \\left (\\frac{1}{\\sqrt{n}}u'Z \\right ) plim \\left ( \\frac{1}{n}X'Z \\right ) ^{-1} \\right ] \\\\ & = \\frac{1}{n} E \\left [ plim \\left ( \\frac{1}{n}Z'X \\right ) ^{-1} plim \\left (\\frac{1}{n}Z'uu'Z \\right ) plim", "q(z_T|x) \\left( \\frac{\\hat\\varrho(x|z_T) p(z_T)}{q(z_T|x)} \\right)^{\\beta_T} } \\right] \\\\ & = \\mathbb{E}_{\\gamma_\\leftarrow(x, z_{T:1})} \\log \\left[ \\frac{p(x, z_{T})}{p(x)} \\frac{1}{p(z_T) \\hat\\varrho(x|z_T)} \\prod_{t=1}^T \\left( \\frac{\\hat\\varrho(x|z_t) p(z_t)}{q(z_t|x)} \\right)^{\\beta_{t+1}-\\beta_t} \\right] \\end{align*} Hence $\\text{MI}[p(x, z)] \\ge \\mathbb{E}_{\\gamma_\\leftarrow(x, z_{T:1})} \\log \\frac{\\hat\\varrho(x|z_T)}{\\prod_{t=1}^T \\left( \\frac{\\hat\\varrho(x|z_t) p(z_t)}{q(z_t|x)} \\right)^{\\beta_{t+1}-\\beta_t}}$ Now we can again reparametrize this formula in terms of $$\\hat\\rho(x, z) = \\hat\\varrho(x|z) p(z)$$: $\\text{MI}[p(x, z)] \\ge \\mathbb{E}_{\\gamma_\\leftarrow(x, z_{T:1})} \\log \\frac{\\hat\\rho(x, z_T)}{\\prod_{t=1}^T \\left( \\frac{\\hat\\rho(x, z_t)}{q(z_t|x)} \\right)^{\\beta_{t+1}-\\beta_t}} - \\mathbb{E}_{p(z)} \\log p(z)$ It’s tempting to simply put $$q(z|x) = p(z)$$ to obtain a blackbox AIS-based analogue of the InfoNCE bound, however notice that $$\\gamma_\\leftarrow(x, z_{T:1})$$ relies on an MCMC kernel that gradually transforms a sample $$z_T \\sim p(z_T|x)$$ into $$z_1 \\sim p(z_1)$$ and thus needs to know this density. Thus I don’t think one can use AIS in the blackbox mode. Finally, note while the bound is valid for any $$\\hat\\rho$$ and $$\\hat\\varrho$$, it’s not quite differentiable w.r.t. their parameters as any proper MCMC method would require an accept-reject step which is not differentiable. Thus if one seeks to learn $$\\hat\\rho$$ or $$\\hat\\varrho$$, an alternative objective should be used. Luckily, the SNIS-based lower bound with small $$K$$ would work just fine. ## Conclusion I presented two different ways to give bounds on the MI in cases of variable complexity. The SNIS-based approach seem to be somewhat novel (the special case of InfoNCE", "$\\displaystyle 1 = \\frac{x}{a} + \\frac{y}{b}$ implies $\\displaystyle x = a + a^{1/3}b^{2/3}, y = b + a^{2/3}b^{1/3}$ as found earlier. ### Generalisation The above method can be readily generalised. Suppose we wish to minimise $x^{\\alpha} + y^{\\alpha}$ given $ax^{\\beta} + by^{\\beta} = 1$ (where $\\alpha > 0 > \\beta$ and $a,b >0$). If we redo the calculation above, for the cancellation in x and y to occur we need to choose p and q such that $\\displaystyle x^{\\alpha/p + \\beta/q} = x^0$. Since $q = p/(p-1)$ this means $\\alpha/p + \\beta(p-1)/p = 0$, or $p = \\frac{\\beta - \\alpha}{\\beta}, q = \\frac{\\alpha - \\beta}{\\alpha}$. We then are able to write $\\begin{array}{lcl} x^{\\alpha} + y^{\\alpha} &=& (x^{\\alpha} + y^{\\alpha})(ax^{\\beta} + by^{\\beta})^{-\\frac{\\alpha}{\\beta}}\\\\&=&\\left(\\left(x^{\\alpha/p}\\right)^p + \\left(y^{\\alpha/p}\\right)^p\\right)\\left(\\left(\\left[ax^{\\beta}\\right]^{\\frac{1}{q}}\\right)^q + \\left(\\left[ax^{\\beta}\\right]^{\\frac{1}{q}}\\right)^q\\right)^{-\\frac{\\alpha}{\\beta}}\\\\&=&\\left\\{\\left(\\left(x^{\\frac{\\alpha \\beta}{\\beta - \\alpha}}\\right)^{\\frac{\\beta-\\alpha}{\\beta}} + \\left(y^{\\frac{\\alpha \\beta}{\\beta - \\alpha}}\\right)^{\\frac{\\beta-\\alpha}{\\beta}}\\right)^{\\frac{\\beta}{\\beta-\\alpha}}\\left(\\left(\\left[ax^{\\beta}\\right]^{\\frac{\\alpha}{\\alpha-\\beta}}\\right)^{\\frac{\\alpha-\\beta}{\\alpha}} + \\left(\\left[by^{\\beta}\\right]^{\\frac{\\alpha}{\\alpha-\\beta}}\\right)^{\\frac{\\alpha-\\beta}{\\alpha}}\\right)^{\\frac{\\alpha}{\\alpha-\\beta}}\\right\\}^{\\frac{\\beta-\\alpha}{\\beta}}\\\\&\\geq& \\left(a^{1/q} + b^{1/q}\\right)^p\\\\&=&\\left(a^{\\alpha/(\\alpha-\\beta)} + b^{\\alpha/(\\alpha-\\beta)}\\right)^{\\frac{\\beta-\\alpha}{\\beta}}.\\end{array}$ The minimum value is attained when $x^{\\alpha} = kax^{\\beta}$ and $y^{\\alpha} = kby^{\\beta}$, or $\\displaystyle \\frac{x}{a^{1/(\\alpha- \\beta)}} = \\frac{y}{b^{1/(\\alpha- \\beta)}}$. This implies $\\displaystyle x = \\left[a + \\left(\\frac{b}{a}\\right)^{\\frac{\\beta}{\\alpha - \\beta}}\\right]^{-\\frac{1}{\\beta}}, \\quad y = \\left[b + \\left(\\frac{a}{b}\\right)^{\\frac{\\beta}{\\alpha - \\beta}}\\right]^{-\\frac{1}{\\beta}}.$ ### Lagrange multipliers A useful technique in calculus to attack such problems is via Lagrange multipliers. The principle there is that at the optimum, the gradients of the objective and constraint function are parallel. In the above graph, imagine where", "$\\omega = \\alpha + i\\beta$, where $\\alpha$ and $\\beta$ are ordinary real-valued $r$-forms. If you plug this into the definition of $\\bar\\omega$, you'll find that $\\bar \\omega = \\alpha - i\\beta.$ If $z = u+iv$ is a complex-valued smooth function, then $dz$ is by definition the complex $1$-form defined by $dz = du + i \\, dv$. Thus $d\\bar z = d (u-iv) = du - i\\,dv = \\overline{dz}$. Now if $z^j=x^j+iv^j$ is one of the complex coordinate functions in a holomorphic coordinate chart $(z^1,\\dots,z^n)$, then the complex vector fields $\\partial/\\partial z^j$ and $\\partial/\\partial \\bar z^j$ are defined by $$\\frac{\\partial}{\\partial z^j} = \\frac{1}{2} \\left( \\frac{\\partial}{\\partial x^j} - i \\frac{\\partial} {\\partial y^j}\\right), \\qquad \\frac{\\partial}{\\partial \\bar z^j} = \\frac{1}{2} \\left( \\frac{\\partial}{\\partial x^j} + i \\frac{\\partial} {\\partial y^j}\\right).$$ Then $\\overline{\\partial /\\partial z^j} = \\partial /\\partial \\bar z^j$, and you can check that the definitions yield $$dz^j \\left( \\frac{\\partial}{\\partial z^j} \\right) =1 ,\\qquad dz^j \\left( \\frac{\\partial}{\\partial \\bar z^j} \\right) =0,$$ along with the analogous formulas for $d\\bar z^j$ obtained by conjugation: $$d\\bar z^j \\left( \\frac{\\partial}{\\partial \\bar z^j} \\right) =1 ,\\qquad d\\bar z^j \\left( \\frac{\\partial}{\\partial z^j} \\right) =0.$$ • Ohh...thanks a lot! So in general is it true that $d\\bar z= \\overline {dz}$? What is definition of $d\\bar z$ in general? Please elaborate little more. This is confusing me since last two weeks.", "my bad. It's also a request for a better answer if one exists. Oct 18 '15 at 19:50 • @KitsuneCavalry There is nothing wrong with posting an answer to your own question---even if it is immediately. In fact, we should encourage this! The goal is to increase the amount of good content available to the site so that it can grow. Please keep it up! :) Oct 24 '15 at 12:36 These are standard mathematical results for generalized means. For example,for the $\\rho \\rightarrow 0$ result, write (setting without loss of generality $\\sum_{i=1}^na_i =1$), $$U = \\left[\\sum^n_{i=1} \\alpha_i x^\\rho_i \\right]^\\frac{1}{\\rho} = \\exp\\left\\{\\frac 1\\rho\\ln \\left(\\sum^n_{i=1} \\alpha_i x^\\rho_i\\right)\\right\\}$$ Apply L'Hopital's rule on $$\\frac 1\\rho\\ln \\left(\\sum^n_{i=1} \\alpha_i x^\\rho_i\\right)$$ to get $$\\frac {\\sum^n_{i=1} \\alpha_ix_i^{\\rho}\\ln x_i}{\\sum^n_{i=1} \\alpha_i x^\\rho_i} \\rightarrow \\sum^n_{i=1} \\alpha_i\\ln x_i,\\;\\;\\; \\rho\\rightarrow 0$$ So (by the uniform continuity of the exponential function) $$\\rho\\rightarrow 0,\\;\\;\\; U = \\exp\\left\\{\\frac 1\\rho\\ln \\left(\\sum^n_{i=1} \\alpha_i x^\\rho_i\\right)\\right\\} \\rightarrow \\exp\\left\\{\\sum^n_{i=1} \\alpha_i\\ln x_i\\right\\} = \\prod_{i=1}^nx_i^{a_i}$$ • Looks good! A very cool streamlined way of doing this problem. Oct 19 '15 at 1:24 • @KitsuneCavalry Well, I try not to forget that in Economics we use a tiny fraction of Mathematics, and mathematicians are likely to have proven a ton of properties before us and in a more compact and abstract level. The above is a simpler version of the proof", "is 1-1 and onto for prime $p$ (with $a,x \\ne 0$). ## Legendre Symbol $p$ prime $$\\left( \\dfrac{a}{p} \\right) = \\begin{cases} 1, & \\text{ if } \\sqrt{a} \\text{ exists } \\\\ 0, & \\text{ if } \\gcd(a,p) \\ne 1 \\\\ -1, & \\text{ if } \\sqrt{a} \\text{ does not exist } \\end{cases}$$ notes: $g$ a generator of $p$, consider $a = g^\\beta$ $\\beta$ even: \\begin{align} & & g^\\beta & = a \\\\ \\to & & ( g^{\\frac{p-1}{2}} ) )^\\beta & = ( a^{\\frac{p-1}{a}} ) \\\\ \\to & & (-1)^\\beta & = a^{\\frac{p-1}{2}} \\\\ \\to & & a^{\\frac{p-1}{2}} & = 1 \\\\ \\end{align} $\\beta$ odd: \\begin{align} \\to & & (-1)^\\beta & = a^{\\frac{p-1}{2}} \\\\ \\to & & a^{\\frac{p-1}{2}} & = -1 \\\\ \\end{align} So: $$\\left( \\dfrac{a}{p} \\right) = a^{\\frac{p-1}{2}}$$ ## Zeta Function $$\\begin{split} \\sum_{n=1}^{\\infty} \\frac{1}{n^s} & = \\prod_{p \\text{ prime}}^{\\infty} ( \\frac{1}{1 - p^{-s}} ) \\end{split}$$ proof: \\begin{align} & & \\sum_{i=1}^{\\infty} a^i & = S \\\\ & & \\sum_{i=0}^{\\infty} a^{i+1} & = \\sum_{i=1}^{\\infty} a^i \\\\ & & & = S a \\\\ & & & \\\\ & & S - S a & = 1 \\\\ \\to & & S & = \\frac{1}{1-a} \\end{align} Write out the product of infinit series of primes, $p_i$: \\begin{align} & ( 1 + p_0^{-s} + p_0^{-2s} + p_0^{-3s} + \\cdots ) \\\\ &", "$\\le$ $\\eta _1^\\ast$ with constant $\\eta _1^\\ast >0$ . Design the virtual control $\\alpha _1$ as \\begin{align} \\alpha _1 =& -\\left( {\\rho _1 +\\left( {\\frac{1} {\\bar {g}_{1\\lambda _1 } }} \\right)\\bar {\\rho }_1 ( t )} \\right)z_1 +\\hat {W}_1^T S_1 \\left( {Y_1 } \\right) \\\\[2mm] & -\\frac{1} {2\\left( {\\frac{z_1 } {\\left( {k_{b_1 }^2 ( t )-z_1^2 } \\right)}} \\right)} \\end{align} (20) where $\\bar {\\rho }_1 ( t )=\\sqrt { ( {{\\dot {k}_{b_1 } ( t )} /{k_{b_1 } ( t )}} )^2+\\beta _1 }$ , $\\beta _1 >0$ is a constant. Note that even if $\\dot {k}_{b_1 } ( t )$ is zero, $\\beta _1$ can ensure that the time derivative of $\\alpha _1$ is bounded. $\\hat {W}_1$ is the estimation of $W_1^\\ast$ and $\\rho _1$ is a design constant, $\\rho _1 >0$ . Substituting (19) and (20) into (18), on account of the fact that $\\bar {\\rho }_1 ( t )+{\\dot {k}_{b_1 } ( t )} /{k_{b_1 } ( t )}\\ge 0$ , (18) can be further rewritten as \\begin{align} \\dot {V}_1 \\le& -\\frac{k_1 z_1^2 }{g_{1\\lambda _1 } \\left( {k_{b_1 }^2 ( t )-z_1^2 } \\right)}+\\frac{z_1 z_2 }{k_{b_1 }^2 ( t )-z_1^2 }-\\frac{\\rho _1 z_1^2 }{k_{b_1 }^2 \\left( t \\right)-z_1^2 } \\\\[2mm] &\\ +\\frac{z_1 \\tilde {W}_1^T S_1 \\left( {Y_1 } \\right)}{k_{b_1 }^2 ( t )-z_1^2 }-\\frac{z_1^2 }{2\\left(", "new also): $$\\left(\\displaystyle\\frac{x}{x-1}\\right)^{x-1} \\;=\\;2+ \\displaystyle\\operatorname*{K}_{n=1}^{\\infty} \\frac {\\left(n+1\\right)\\left(x-n-1\\right)/x} {\\left(n+1\\right)\\left(x+1\\right)/x} \\tag{1}\\label{1}$$ that I found from a more general conjecture being related to the following continued fraction: $$g\\left(k,x\\right) = \\displaystyle\\operatorname*{K}_{n=1}^{\\infty} \\frac {\\left(n+1\\right)\\left(k-n-k/x\\right)/x} {\\left(n+1\\right)\\left(1+1/x\\right)}$$ for which I empirically noticed the relation (with $\\textrm{B}$ the beta function): $$\\begin{array}{l} 2\\;+\\; g\\left(\\alpha,\\xi\\right) \\;+\\; g\\left(\\alpha,\\displaystyle\\frac{\\xi}{\\xi-1}\\right) \\\\[8pt] \\qquad\\qquad =\\; \\alpha\\left(\\xi-1\\right)^{\\alpha/\\xi-1}\\left(\\displaystyle\\frac{\\xi}{\\xi-1}\\right)^{\\alpha-2} \\textrm{B}\\left(\\alpha/\\xi, \\alpha-\\alpha/\\xi\\right) \\end{array} \\tag{2}\\label{2}$$ where it is easy to notice that the case $\\alpha=\\xi/\\left(\\xi-1\\right)$ makes $g\\left(\\alpha,\\xi\\right)=0$ leading to an identity containing a single continued fraction (which can later be simplified as above). The previous notation is the one I use; I find it convenient and it can be found for instance in Continued Fractions with Applications by Lorentzen & Waadeland, but I know that some people don't like it; it has to be read the following way: $$a_0 + \\operatorname*{K}_{n=1}^{\\infty} \\frac{b_n}{a_n} = a_0 + \\cfrac{b_1}{a_1 + \\cfrac{b_2}{a_2 + \\cfrac{b_3}{a_3 + \\ddots}}}$$ • Isn't there some continued fractions database to check these numbers against? Or maybe there are entries for such fractions in the OEIS? As for the algorithms for finding them, I am sure that googling will tell you about \"scientific\" methods from the literature to find such fractions. If your algorithm is fast and finds nice fractions accurately, you might consider advertising/publishing that (instead of just the \"output\" of your algorithm). – TMM Apr 12 '17", "be written at $$A=0$$ after some tedious algebra \\begin{aligned} \\mathcal {S}:=\\mathbb {E}[N(t);N(T)=0|N(t_0)=1] \\alpha -\\frac{a\\alpha }{1+b\\left( 1+\\frac{a\\beta }{b}\\right) } \\left[ 1+\\beta \\left( 2 - \\frac{a(1+\\beta )}{1+b\\left( 1+\\frac{a\\beta }{b}\\right) } \\right) \\right] \\,. \\end{aligned} (61) To zeroth order in A, we recover the result in Ref.16 with $$\\alpha =\\exp {-r(t-t_0)}$$, $$\\beta =\\frac{q_2}{r}\\left( 1-\\exp {-r(t-t_0)}\\right)$$, as used in “Probability of survival to first order”, as well as $$a=\\exp {-r(T-t)}$$ and $$b=\\frac{q_2}{r}\\left( 1-\\exp {-r(T-t)}\\right)$$. To incorporate the first-order correction, we adjust $$\\alpha$$, $$\\beta$$, a and b so that $$g_n(t_0,t)$$ and $$\\left\\langle N(t),N(t_0)=1,N(T)=0 \\right\\rangle$$ can be calculated order by order via generating functions. Defining \\begin{aligned} \\alpha&=\\exp {-r(t-t_0)}(1+A_1), \\end{aligned} (62a) \\begin{aligned} \\beta&=\\frac{q_2}{r}\\left( 1-\\exp {-r(t-t_0)}\\right) z_1, \\end{aligned} (62b) \\begin{aligned} a&=\\exp {-r(T-t)}(1+A_2), \\end{aligned} (62c) \\begin{aligned} b&=\\frac{q_2}{r}\\left( 1-\\exp {-r(T-t)}\\right) z_2, \\end{aligned} (62d) and the generating function $$\\mathcal {G}=n!\\alpha \\beta ^{n-1}$$, we recover $$g_n(t_0,t)$$ to first order according to Eq. (46) in the form \\begin{aligned} g_n(t_0,t)&=\\Big \\{ \\mathcal {G}+A\\Big (u(t_0,t)\\frac{\\partial \\mathcal {G}}{\\partial A_1} +v(t_0,t)\\frac{\\partial \\mathcal {G}}{\\partial z_1} \\Big ) \\Big \\} \\Big |_{A_1=0, z_1=1} + \\mathcal {O}\\left( A^2\\right) \\, , \\end{aligned} (63) and using the generating function $$\\mathcal {S}$$ as defined in Eq. (61), similarly for the avalanche shape, \\begin{aligned}{\\,} &\\mathbb {E}[N(t);N(T)=0|N(t_0)=1] \\nonumber \\\\&\\quad = \\left\\{ \\mathcal {S}+ A\\left( u(t_0,t)\\frac{\\partial \\mathcal {S}}{\\partial A_1} +u(t,T)\\frac{\\partial \\mathcal {S}}{\\partial A_2} +v(t_0,t)\\frac{\\partial \\mathcal {S}}{\\partial z_1} +v(t,T)\\frac{\\partial \\mathcal {S}}{\\partial z_2} \\right) \\right\\} \\Big", "The recursive model can be rewritten as $$\\overline{\\varvec{M}}(k)\\hat{\\varvec{C}} = \\overline{\\varvec{N}}(k),$$ (28) where $$\\overline{\\varvec{M}}(k){ = }\\left[ {\\begin{array}{*{20}c} {\\lambda {\\varvec{\\alpha}}(k - 1)} \\\\ {{\\varvec{\\alpha}}(k)} \\\\ \\end{array} } \\right],\\;\\overline{\\varvec{N}}(k){ = }\\left[ {\\begin{array}{*{20}c} {\\lambda {\\varvec{\\gamma}}(k - 1)} \\\\ {{\\varvec{\\gamma}}(k)} \\\\ \\end{array} } \\right],\\;{\\varvec{\\gamma}}(k){ = }\\left[ {\\begin{array}{*{20}c} {\\beta (k + 1) - \\beta (k){ + }Tr(k)} \\\\ {r(k + 1) - r(k)} \\\\ \\end{array} } \\right],$$ $${\\varvec{\\alpha}}(k){ = }\\left[ {\\begin{array}{*{20}c} {\\frac{{2T\\beta (k) + 2T\\delta_{f} (k)}}{{mv_{x} }} + \\frac{{2l_{f} r(k)}}{{mv_{x}^{2} }}} & {\\frac{2T\\beta (k)}{{mv_{x} }} - \\frac{{2l_{r} Tr(k)}}{{mv_{x}^{2} }}} \\\\ {\\frac{{2l_{f} T\\beta (k) + 2T\\delta_{f} (k)}}{{I_{z} }} - \\frac{{2l_{f}^{2} Tr(k)}}{{I_{z} v_{x} }}} & {\\frac{{ - 2l_{r} T\\beta (k)}}{{I_{z} }} - \\frac{{2l_{r}^{2} Tr(k)}}{{I_{z} v_{x} }}} \\\\ \\end{array} } \\right]$$ and λ is the forgetting factor, which is set to the value of 0.98. The cornering stiffness at the kth sampling moment is then given by $$\\hat{\\varvec{C}} = \\left[ {\\begin{array}{*{20}c} {\\hat{C}_{f} } \\\\ {\\hat{C}_{r} } \\\\ \\end{array} } \\right] = (\\overline{\\varvec{M}}(k)^{T} \\overline{\\varvec{M}}(k))^{ - 1} \\overline{\\varvec{M}}(k)^{T} \\overline{\\varvec{N}}(k).$$ (29) The tire cornering stiffness at the next moment is identified by updating the prediction model of the vehicle path tracking system with the current parameter identification $$\\hat{\\varvec{C}}(k + 1) = {\\varvec{P}}(k + 1){\\varvec{H}}(k + 1),$$ (30) $${\\varvec{P}}(k + 1)^{{ - 1}} = \\overline{\\varvec{M}}(k + 1)^{{\\text{T}}} \\overline{\\varvec{M}}(k + 1),$$ (31) $${\\varvec{H}}(k + 1) =\\overline{\\varvec{M}}(k + 1)^{{\\text{T}}} \\overline{\\varvec{N}}(k + 1).$$ (32)", "have $A+B+C=(1-\\rho )\\Delta$ $(1+\\alpha )A+(1+\\alpha +\\beta )B+(1+\\alpha +\\beta +\\delta )C=\\Delta$ This leads to the augmented matrix: $\\left[ \\begin{array}{cccc} 1 & 1 & 1 & (1-\\rho )\\Delta \\\\ (1+\\alpha ) & (1+\\alpha +\\beta ) & (1+\\alpha +\\beta +\\delta ) & \\Delta\\end{array}\\right]$ Perform the following row operation: $R_{2}\\rightarrow R_{2}-(1+\\alpha )R_{1}$ $\\left[ \\begin{array}{cccc} 1 & 1 & 1 & (1-\\rho )\\Delta \\\\ 0 & \\beta & \\beta +\\delta & \\Delta -(1+\\alpha )(1-\\rho )\\Delta \\end{array} \\right] =\\left[ \\begin{array}{cccc} 1 & 1 & 1 & (1-\\rho )\\Delta \\\\ 0 & \\beta & \\beta +\\delta & \\Delta (-\\alpha +\\rho +\\alpha \\rho ) \\end{array} \\right] =$ $\\left[ \\begin{array}{cccc} 1 & 1 & 1 & (1-\\rho )\\Delta \\\\ 0 & \\beta & \\beta +\\delta & \\Delta (\\rho -\\alpha (1-\\rho )) \\end{array} \\right]$ Now perform $R_{2}\\rightarrow \\frac{1}{\\beta }R_{2}$ $\\left[ \\begin{array}{cccc} 1 & 1 & 1 & (1-\\rho )\\Delta \\\\ 0 & 1 & 1+\\frac{\\delta }{\\beta } & \\Delta \\frac{1}{\\beta }(\\rho -\\alpha (1-\\rho )) \\end{array} \\right]$ Now perform $R_{1}\\rightarrow R_{1}-R_{2}$ $\\left[ \\begin{array}{cccc} 1 & 0 & -\\frac{\\delta }{\\beta } & (1-\\rho )\\Delta -\\Delta \\frac{1}{\\beta } (\\rho -\\alpha (1-\\rho )) \\\\ 0 & 1 & 1+\\frac{\\delta }{\\beta } & \\Delta \\frac{1}{\\beta }(\\rho -\\alpha (1-\\rho )) \\end{array} \\right]$ So our solution is : $\\left[ \\begin{array}{c} A \\\\ B \\\\ C \\end{array} \\right] =\\left[ \\begin{array}{c} \\Delta \\frac{1}{\\beta }(\\beta (1-\\rho )-\\rho +\\alpha", "Consider a function $$f:\\Bbb R \\rightarrow \\Bbb R$$ such that $$|f(x)-f(y)| \\leq |x-y|^\\alpha\\tag{1}\\label{eq447:1}$$ for all $$x,y\\in \\Bbb R$$ and for $$\\alpha >1$$. Then $$f(x)$$ is constant. Proof. Using triangular inequality and then hypothesis \\eqref{eq447:1} yields \\begin{eqnarray}\\left| f(x) – f(y) \\right| &=& \\left| f(x) – f\\left(\\frac{x+y}{2}\\right)+f\\left(\\frac{x+y}{2}\\right)- f(y) \\right| \\leq\\\\ &\\leq& \\left| f(x) – f\\left(\\frac{x+y}{2}\\right)\\right| + \\left|f\\left(\\frac{x+y}{2}\\right)- f(y) \\right|\\leq \\\\ &\\leq& \\left|x-\\frac{x+y}{2}\\right|^\\alpha + \\left|\\frac{x+y}{2}-y\\right|^\\alpha=\\\\ &=& \\frac{|x-y|^\\alpha}{2^{\\alpha-1}}.\\end{eqnarray} The previous steps can be generalized to obtain what follows. If $$\\forall x,y \\in \\Bbb R$$ $|f(x) – f(y)| \\leq \\frac{|x-y|^\\alpha}{2^{(\\alpha-1)n}},$ then |f(x) – f(y)| \\leq \\frac{|x-y|^\\alpha}{2^{(\\alpha-1)(n+1)}}.\\tag{2}\\label{eq447:2} Thus by induction \\eqref{eq447:2} is true for all $$n \\in \\Bbb Z^+$$. Consequently, since the RHS of \\eqref{eq447:2} can be made arbitrarily small, it must be $|f(x)-f(y)| = 0$ and $$f(x)$$ is constant." ]

[ "# Does this polynomial exist? I'm looking for a polynomial $P(x)$ with the following properties: 1. $P(0) = 0$. 2. $P\\left(\\frac13\\right) = 1$ 3. $P\\left(\\frac23\\right) = 0$ 4. $P'\\left(\\frac13\\right) = 0$ 5. $P'\\left(\\frac23\\right) = 0$ From 1 and 3 we know that $P(x) = x\\left(x - \\frac23\\right)Q(x)$. From 4 and 5 we know that $P'(x) = \\alpha\\left(x - \\frac13\\right)\\left(x - \\frac23\\right)$. \\begin{align} P(x) & = \\alpha\\int\\left(x - \\frac13\\right)\\left(x - \\frac23\\right)\\text{d}x \\\\ & = \\alpha\\left(\\frac13x^3 - \\frac12x^2 + \\frac29x + \\text{C}\\right) \\end{align} Now 1 implies that the constant is $\\text{C} = 0$, but 3 implies that the constant is $\\text{C} = -\\frac{2}{81}$. Am I doing something wrong or does this polynomial not exist? If it doesn't exist, how close could I get to making a polynomial that satisfies these 5 conditions? • Your assertion that we know that $P'(x)=\\alpha(x-{1\\over3})(x-{2\\over3})$ implicitly assumes the polynomial $P$ is a cubic. Dec 31, 2014 at 20:46 • @BarryCipra Ahh that's my error I see. So a cubic polynomial $P$ doesn't exist, but higher orders could? Dec 31, 2014 at 20:48 • That's right. My guess is, a quartic will do. Which is to say, make that $Q(x)=ax^2+bx+c$ a general quadratic, then look for $3$ unknowns ($a$, $b$, and $c$) that satisfy the $3$ equations 2, 4, and 5. Dec 31, 2014 at 20:51", "is not necessarily an irreducible polynomial. – Thomas Andrews Dec 5 '13 at 13:44 • Ohh, guess you are right. But I really would like my answer to be based on 1,2,3. However since the problem assume $\\alpha^{-1}$ exists doesn't that imply we are working in a field ? – Shuzheng Dec 5 '13 at 13:48 Yes, right, $\\alpha^{-p} = \\left(\\alpha^{-1}\\right)^p$. $$y^p = \\left(y^2\\right)^{(p-1)/2}\\cdot y = 2^{(p-1)/2}\\cdot y = \\left(\\frac{2}{p}\\right)\\cdot y.$$ • why $$2^{(p-1)/2}\\cdot y = \\left(\\frac{2}{p}\\right)\\cdot y.$$ ? and could you tell me how we can assume $\\alpha^{-1}$ exists, since we don't know if $R$ is a field. Thanks a lot btw. – Shuzheng Dec 5 '13 at 13:53 • Is it because $$\\left(\\frac{2}{p}\\right) \\equiv 2^{(p-1)/2} (\\mod p)$$ ? – Shuzheng Dec 5 '13 at 13:57 • First $\\alpha^{-1}$. The polynomial factored out is $X^4+1$, which means $\\alpha^4 = -1$, and hence $\\alpha^8 = 1$, so $\\alpha^{-1} = \\alpha^7$. Second, the Legendre symbol, yes, precisely, because $a^{(p-1)/2} \\equiv \\left(\\frac{a}{p}\\right) \\pmod{p}$. – Daniel Fischer Dec 5 '13 at 13:59 • I know $$\\left(\\frac{2}{p}\\right) \\equiv 2^{(p-1)/2}$$ but this just means that $\\left(\\frac{2}{p}\\right) = 2^{(p-1)/2} + pk$. But then $$\\left(\\frac{2}{p}\\right) \\cdot y = (2^{(p-1)/2} + pk) \\cdot y = 2^{(p-1)/2} \\cdot y$$ why ? – Shuzheng Dec 5 '13 at 14:28 • @user111854 We're working over $\\mathbb{F}_p$, so all integers,", "the roots of that polynomial to obtain $$(*)$$. Since $$f$$ is one-to-one we can find the inverse function, $$f^{-1}(x) = 1 - \\frac{1}{x} \\Rightarrow f^{-1} \\left(\\frac{1}{1-\\alpha_k} \\right) = \\alpha_k$$ Therefore the function, $$Q(x) = P(f^{-1}(x)) = P \\left(1 - \\frac{1}{x}\\right)$$ has roots $$\\displaystyle \\frac{1}{1-\\alpha_k}$$ for each $$k$$ (this is the crux of the argument, please convince yourself of this!). Note that $$Q$$ as defined is not a polynomial yet, but we can use $$Q$$ as a function to find a polynomial that has the same roots. Now, we have that, $$Q(x) = P \\left(1 - \\frac{1}{x} \\right) = 1 + \\left(1 - \\frac{1}{x} \\right) + \\dots + \\left( 1- \\frac{1}{x} \\right)^{n-1}$$ Summing this geometric series we obtain, $$Q(x) = x\\left(1 - \\left(1-\\frac{1}{x} \\right)^n\\right)$$ With some manipulation we obtain that, $$R(x) = x^{n-1}Q(x) = x^n - (x-1)^n = nx^{n-1} - \\frac{n(n-1)}{2} x^{n-2} + \\dots$$ Now we have a polynomial $$R$$ and I claim that $$R$$ has roots $$\\displaystyle \\frac{1}{1-\\alpha_k}$$ for each $$k$$. This is because, $$R \\left(\\frac{1}{1-\\alpha_k} \\right) = \\frac{1}{(1-\\alpha_k)^{n-1}} Q \\left(\\frac{1}{1-\\alpha_k} \\right) =\\frac{1}{(1-\\alpha_k)^{n-1}} P\\left(f^{-1}\\left(\\frac{1}{1-\\alpha_k} \\right) \\right) = \\frac{1}{(1-\\alpha_k)^{n-1}} P(\\alpha_k) = 0$$ Therefore to find $$(*)$$ we note that it is simply the sum of the roots of $$R$$, therefore, $$\\sum_{k=1}^{n-1} \\frac{1}{1-\\alpha_k} = \\frac{n(n-1)/2}{n} = \\frac{n-1}{2}$$ This completes the proof. • Maybe I'm missing something obvious but could you", "x\\right) dx = 1.$$ The normalization constant $A _{n}$ must be chosen accordingly. Finally, there is a formula for the generating function of the polynomials, $$G\\left( x, w\\right) = \\frac{1}{\\rho \\left( x\\right) } \\frac{\\rho \\left( t\\left( x, w\\right) \\right) }{\\left| 1 - w \\left( \\beta _1 + 2 \\beta _2 t\\left( x, w\\right) \\right) \\right| } ,$$ where $t\\left( x, w\\right)$ is the root of $t - x - w \\beta \\left( t\\right) = 0$ which is nearest to $t = x$ at small $w$. This function $G\\left( x, w\\right)$ gives the unnormalized polynomials, $$G\\left( x, w\\right) = \\sum _{n = 0} ^{\\infty } \\frac{q_n\\left( x\\right) }{n!} w ^{n},$$ where $$q_n\\left( x\\right) = \\frac{1}{\\rho \\left( x\\right) } \\left( \\frac{\\partial^n}{\\partial x ^n}\\left( \\rho \\left( x\\right) \\beta \\left( x\\right) ^{n}\\right) \\right) .$$ The classical families of (normalized) orthogonal polynomials are obtained in this framework with the following definitions: • The Legendre polynomials • $P _{n}\\left( x\\right)$: $a = - 1$, $b = 1$, $\\rho \\left( x\\right) = 1$, $\\alpha \\left( x\\right) = 0$, $\\beta \\left( x\\right) = 1 - x ^{2}$, $A _{n} = \\frac{\\left( - 1\\right) ^{n}}{2 ^{n} n!}$. • The Laguerre polynomials • $\\left( L ^{m}\\right) _n\\left( x\\right)$: $a = 0$, $b = + \\infty$, $\\rho \\left( x\\right) = x ^{m} \\exp \\left( - x\\right)$, (here $m >", "maximal ideal. So my question is, which route would yield an easier and more elegant solution? • What is $E$ and how does it enter in to this question? Cheers! – Robert Lewis Nov 2 '18 at 2:52 • @RobertLewis thanks for pointing that out, it should be $\\alpha \\in E$ – SalmonKiller Nov 2 '18 at 2:57 Proffering the use of uniqueness of factorization of polynomials as an alternative. Assume that contrariwise that $$\\beta=r(\\alpha)/s(\\alpha), \\beta\\notin F,$$ is algebraic over $$F$$. Here $$r(\\alpha),s(\\alpha)\\in F[\\alpha]$$. W.l.o.g. we can assume that: • $$\\gcd(r(\\alpha),s(\\alpha))=1$$ because we can simply cancel any common factor, and • at least one of $$r(\\alpha), s(\\alpha)$$ is not a constant. For otherwise $$\\beta\\in F$$. Let $$p(x)=a_0+a_1x+\\cdots+a_nx^n\\in F[x]$$ be the minimal polynomial of $$\\beta$$ over $$F$$. Because $$\\beta\\notin F$$ we know that $$p(x)$$ has degree at least two. Furthermore, $$p(x)$$ is irreducible in $$F[x]$$, so we can assume that $$a_n=1$$ and $$a_0\\neq0$$. Let's plug in $$x=\\beta$$. We arrive at $$a_0+a_1\\frac{r(\\alpha)}{s(\\alpha)}+\\cdots+\\left(\\frac{r(\\alpha)}{s(\\alpha)}\\right)^n=0.$$ Using a common denominator allows a rewrite $$0=\\frac{a_0s(\\alpha)^n+a_1s(\\alpha)^{n-1}r(\\alpha)+a_2s(\\alpha)^{n-2}r(\\alpha)^2+ \\cdots+a_{n-1}s(\\alpha)r(\\alpha)^{n-1}+r(\\alpha)^n}{s(\\alpha)^n}.$$ Here the numerator must be the zero polynomial in the ring $$F[\\alpha]$$. But, the two bullets above imply that there exists an irreducible non-constant polynomial $$m(\\alpha)\\in F[\\alpha]$$ such that $$m(\\alpha)$$ is a factor of exactly one of $$r(\\alpha), s(\\alpha)$$ (and hence not a factor of the other). Look at", "# Proof of Different Polynomial Decompositions into Linear Factors From G. Polya \"Mathematics and Plausible Reasoning\" p. 18. How do you prove that provided the roots of a polynomial are different from zero, $$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$ $$\\,= a_0\\left(1-\\frac{x}{\\alpha_1}\\right)\\left(1-\\frac{x}{\\alpha_2}\\right)...\\left(1-\\frac{x}{\\alpha_n}\\right)$$ with $\\alpha_1, \\alpha_2,...\\alpha_n$ corresponding to the polynomial roots. Setting $x=0$ results in: $\\qquad a_0+a_1x+\\ldots+a_nx^n\\,=\\,(-1)^n\\,a_n\\alpha_1\\cdot\\ldots\\cdot\\alpha_n\\left(1-\\frac{x}{\\alpha_1}\\right)\\cdot\\ldots\\cdot\\left(1-\\frac{x}{\\alpha_n}\\right)$ simplifying to $$a_0=(-1)^n\\,a_n\\alpha_1\\cdot\\ldots\\cdot\\alpha_n$$ Therefore: $\\qquad a_0+a_1x+\\ldots+a_nx^n \\,=\\,a_0\\left(1-\\frac{x}{\\alpha_1}\\right)\\cdot\\ldots\\cdot\\left(1-\\frac{x}{\\alpha_n}\\right)$ • Great. Thank you. Before I \"accept\" / click on the check mark, can I tweak a bit your response for completeness, you approve it, and we close it? – Antoni Parellada May 30 '15 at 22:13 • @AntoniParellada, make the edit you feel required. – enzotib May 30 '15 at 22:14 • It truly wasn't necessary... Feel free to erase the changes. – Antoni Parellada May 30 '15 at 22:27 • @AntoniParellada: it is ok. – enzotib May 30 '15 at 22:27 • I made some changes to splice this post with a proof of Viete's theorem I am working on. It's the same without the $k$ step. – Antoni Parellada May 31 '15 at 0:37 It's not clear what you actually want shown. That the polynomial has $n$ roots is a consequence of the fundamental theorem of algebra. That the factors $(1-x/\\alpha_i)$ have the form they do is actually the factor theorem in", "In the case where $$Q(x)$$ is a polynomial of degree $$m$$, we obtain from the residue theorem that $$\\beta_n$$ is the root of a polynomial of degree $$m$$: \\begin{align} \\frac{8n}{\\beta_n} &= \\text{Res} \\left( \\frac{(\\varphi+1)^2}{\\varphi^2} \\sum_{k=1}^m k q_k \\left(\\frac{(\\varphi+1)^2\\beta_n}{4\\varphi}\\right)^{k-1}, \\quad \\varphi=0 \\right), \\nonumber \\\\ \\frac{8n}{\\beta_n} &= \\left[\\sum_{k=2}^{m} k q_k \\left(\\frac{\\beta_n}{4}\\right)^{k-1} {2k-2 \\choose k-2}\\right] + 2\\left[\\!\\sum_{k=1}^m k q_k \\left(\\frac{\\beta_n}{4}\\right)^{k-1}\\!\\! {2k-2 \\choose k-1} \\!\\right] + \\left[\\!\\sum_{k=2}^m k q_k \\left(\\frac{\\beta_n}{4}\\right)^{k-1} {2k-2 \\choose k} \\!\\!\\right]\\!, \\nonumber \\\\ 8n &= 2q_1\\beta_n + 4\\sum_{k=2}^{m} k {2k \\choose k} q_k \\left(\\frac{\\beta_n}{4}\\right)^k. \\label{EexplBeta} \\end{align} (3.9) We can remark that (3.3) gives the asymptotic expansion of the zero of the $$m$$th degree polynomial (3.9) in $$\\beta_n$$ with respect to a factor in its constant coefficient. Exact solutions for $$\\beta_n$$ are only available up to $$m=4$$. For $$m=1$$, this boils down to the standard associated Laguerre case $$\\beta_n=4n/q_1$$. For $$m=2$$, we take the positive solution that also corresponds to (3.6) and (3.7): $$\\beta_n = \\frac{-q_1+\\sqrt{q_1^2+24q_2 n}}{3q_2} \\sim \\sqrt{\\frac{8n}{3q_2}} -\\frac{q_1}{3q_2} +\\sqrt{\\frac{q_1^4}{864q_2^3 n}} + \\mathcal{O}(n^{-3/2}). \\nonumber$$ We do find an explicit result for the nonpolynomial function $$Q(x) = \\exp(x)$$. In that case, we have \\begin{align} 8n &\\sim \\beta_n \\text{Res} \\left( \\frac{1}{\\varphi} \\sum_{k=0}^\\infty \\frac{1}{k!} \\left(\\frac{\\beta_n}{4}\\right)^k \\left(\\varphi^{-1} +2 +\\varphi \\right)^{k+1}, \\varphi=0 \\right) \\nonumber \\\\ 8n &\\sim \\beta_n \\sum_{k=0}^\\infty \\frac{1}{k!} \\left(\\frac{\\beta_n}{4}\\right)^k {2k+2 \\choose k+1} \\nonumber \\\\ 8n &= 2\\beta_n \\exp\\left(\\frac{\\beta_n}{2}\\right)\\left[I_0\\left(\\frac{\\beta_n}{2}\\right) + I_1\\left(\\frac{\\beta_n}{2}\\right) \\right] \\label{EexponQbeta}", "- 1$ or else the weight function is not integrable on the interval), $\\alpha \\left( x\\right) = m - x$, $\\beta \\left( x\\right) = x$, $A _{n} = 1$. • The Hermite polynomials • $H _{n}\\left( x\\right)$: $a = - \\infty$, $b = + \\infty$, $\\rho \\left( x\\right) = \\exp \\left( - x ^{2}\\right)$, $\\alpha \\left( x\\right) = - 2 x$, $\\beta \\left( x\\right) = 1$, $A _{n} = \\left( - 1\\right) ^{n}$. • The Chebyshev polynomials of the first kind • $T _{n}\\left( x\\right)$: $a = - 1$, $b = 1$, $\\rho \\left( x\\right) = \\frac{1}{\\sqrt{1 - x ^{2}}}$, $\\alpha \\left( x\\right) = x$, $\\beta \\left( x\\right) = 1 - x ^{2}$, $A _{n} = \\frac{\\left( - 1\\right) ^{n}}{\\left( 2 n\\right) !!}$. The Rodrigues formula or the generating function are not efficient ways to calculate the polynomials. A better way is to use linear recurrence relations connecting $q _{n + 1}$ with $q _{n}$ and $q _{n - 1}$. (These recurrence relations can also be written out in full generality through $\\alpha \\left( x\\right)$ and $\\beta \\left( x\\right)$ but we shall save the space.) There are three computational tasks related to orthogonal polynomials: • Compute all coefficients of a given polynomial • $q_n\\left( x\\right)$ exactly. The coefficients are rational numbers, but their numerators and denominators usually grow", "a polynomial of degree $n - 1$ in $x$( the symbols $O( P)$, $\\ll P$ denote magnitudes of order $P$). At the $( n - 1)$- st step one obtains inner sum $$\\tag{2 } S( \\alpha )= \\sum _ {x= a+ 1 } ^ { {a+ } P _ {1} } e ^ {2 \\pi i \\alpha x } ,$$ where $P _ {1} \\leq P$, $\\alpha = n! \\lambda _ {1} {} \\dots \\lambda _ {n-} 1 \\alpha _ {n}$, $\\lambda _ {i} \\neq 0$. Sums of the form (2) are estimated using the inequality $$| S( \\alpha ) | \\leq \\min \\left ( P _ {1} , \\frac{1}{| \\sin \\pi \\alpha | } \\right ) ,$$ and the resulting estimate is: $$\\tag{3 } | S( f ) | ^ {\\rho _ {0} } \\leq P ^ {\\rho _ {0} - 1 } +$$ $$+ P ^ {\\rho _ {0} {- n + \\epsilon } } \\sum _ {0 < y \\leq P ^ {n-} 1 } \\min \\left ( P _ {1} , \\frac{1}{| \\sin \\pi y \\alpha _ {n} | } \\right ) .$$ The inequality (3) yields different estimates for the sum (1) in case $1/ | \\sin \\pi y \\alpha _ {n} |$ is small in comparison to $P$. These estimates depend on", "there exists a real number $c \\in \\mathbb{R}$ such that $c^2 = \\frac{\\alpha^2}{4} - \\beta$. Notice that we can rewrite the polynomial $x^2 + \\alpha x + \\beta$ as: (3) \\begin{align} \\quad x^2 + \\alpha x + \\beta = \\left ( x + \\frac{\\alpha}{2} \\right )^2 - \\left ( \\frac{\\alpha^2}{4} - \\beta \\right ) = \\left ( x + \\frac{\\alpha}{2} \\right )^2 - c^2 = \\left ( \\left ( x + \\frac{\\alpha}{2} \\right) + c \\right ) \\left ( \\left ( x + \\frac{\\alpha}{2} \\right ) - c \\right ) \\\\ = \\left ( x - \\left ( - \\frac{\\alpha}{2} - c \\right ) \\right) \\left ( x - \\left ( - \\frac{\\alpha}{2} + c \\right ) \\right ) \\end{align} • Thus we have the desired factorization with $\\lambda_1 = - \\frac{\\alpha}{2} - c$ and $\\lambda_2 = - \\frac{\\alpha}{2} + c$. $\\blacksquare$ We are now ready to look at the following theorem which gives us a second way to fac Theorem 1: Let $p(x) \\in \\wp ( \\mathbb{R} )$ be a non-constant polynomial. Then $p(x)$ has a unique factorization $p(x) = c(x - \\lambda_1)(x - \\lambda_2)...(x - \\lambda_m)(x^2 + \\alpha_1x + \\beta_1)(x^2 + \\alpha_2x + \\beta_2)...(x^2 + \\alpha_nx + \\beta_n)$ where $\\lambda_1, \\lambda_2, ..., \\lambda_m \\in \\mathbb{R}$ are the real roots of $p(x)$ and $c, \\alpha_1, \\alpha_2, ...,", "polynomial ${f(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_o}$ with integer coefficients, let ${K = \\max\\limits_{0\\le i\\le n-1}|\\frac{a_i}{a_n}|^{\\frac1{n-i}}}$. Then for any root ${\\alpha}$ of ${f}$, ${|\\alpha| < 2K}$. Proof: If ${|\\alpha| \\le K}$, we are done. Else, ${|\\alpha^n| \\le \\sum_{i=0}^{n-1}{|\\frac{a_i}{a_n}| |\\alpha^i|} \\le \\sum_{i=0}^{n-1} {K^{n-i} |\\alpha^i|} = K^n\\frac{\\left(\\frac{|\\alpha|}{K}\\right)^n-1}{\\frac{|\\alpha|}{K}-1} < K^n\\frac{\\left(\\frac{|\\alpha|}{K}\\right)^n}{\\frac{|\\alpha|}{K}-1}}$ which means that ${\\frac{|\\alpha|}{K}-1 < 1}$, i.e., ${|\\alpha| < 2K}$. $\\Box$ We can combine all these into the following result. Corollary 8 Given a polynomial ${f(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_o}$ with integer coefficients and ${\\frac{a_{n-1}}{a_n} \\ge 0}$, let ${a_{n-r}}$ be the first nonzero coefficient of ${f}$ after ${a_{n-1}}$ (i.e., ${a_m = 0}$ if ${n-r < m < n-1}$). Let ${G = \\max_{i\\le n-r}|\\frac{a_i}{a_n}|}$. Let ${K = \\max_{i\\le n-r}|\\frac{a_i}{a_n}|^{\\frac1{n-i}}}$. If there exists an integer ${m \\ge \\min(G,\\frac{1 + \\sqrt{1 + 4G}}{2},G^{1/r} + 1, 2K) + 1}$ such that ${f(m)}$ is prime, then ${f}$ is irreducible. Proof: As in Lemmas 5, 6 and 7, we can prove that either ${\\Re(\\alpha) \\le 0}$ or ${|\\alpha| \\le \\min(G,\\frac{1 + \\sqrt{1 + 4G}}{2},G^{1/r} + 1, 2K)}$. Lemma 3 then gives the result. $\\Box$ Perhaps the “generalisations” in this section, as they contain no new idea, are like the “cheap generalization” Pólya talks about: There are two kinds of generalizations. One is cheap and the other is valuable. It is", "# How can I show that $a_0x^n+…+a_n$ and $a_nx^n+…a_0$ have the same discriminant? How can I show that $a_0x^n+...+a_n$ and $a_nx^n+...a_0$ have the same discriminant? You can use two different definition of the discriminant of the polynomial $f(x)=a_nx^n+...a_0$. The first is $$D(f)=a_n^{2n-2}\\prod_{i<k}(\\alpha_i-\\alpha_j),$$ where $\\{\\alpha_i\\}_{i=1,...n}$ are the roots of the polynomial. The second is $$D(f)=(-1)^{\\frac{n(n-a)}{2}}\\frac{1}{a_n}R(f,f')$$ where $R(f,f')$ is the resultant of $f$ and $f'$. - Do you want to show it with the two definitions or only with one? – Davide Giraudo Oct 15 '12 at 21:35 Is $f'$ the formal derivative of $f$? – Alexander Gruber Oct 15 '12 at 21:37 either can be used... and $f'$ is the usual derivate – Elmo goya Oct 15 '12 at 21:44 The roots of $a_nx^n+\\cdots +a_0$ are the reciprocals of the roots of $a_0x^n+\\cdots +a_n$. Therefore $\\prod (\\alpha_i-\\alpha_j)$ is replaced with $\\prod (\\frac1{\\alpha_i}-\\frac1{\\alpha_j})=\\prod \\frac{\\alpha_j-\\alpha_i}{\\alpha_i\\alpha_j}$. The numerators yield the original discriminant (check that there is no sign change!). The denominators produce $(\\prod\\alpha_i)^{2(n-1)}=(\\frac{a_0}{a_n})^{2n-2}$, so that everything sorts out. Well, but... I can't see clearly why $\\prod \\alpha_i^{2(n-1)}=(\\frac{a_0}{a_n})^{2n-2}$ @HagenvonEitzen – Elmo goya Oct 15 '12 at 22:08 Camilo, are you familiar with $\\prod\\alpha_i=a_0/a_n$ in this context? Then raise to the power $2n-2$. – Gerry Myerson Oct 15 '12 at 22:57", "get an closed form solution, no? the order of the polynomial doesn't exceed 3, does it? i think the L4 and L5 points are the cool ones. those are the ones that are \"pits\" or \"depressions\" at a local minimum of potential energy 4. Oct 31, 2008 ### tony873004 5. Oct 31, 2008 ### D H Staff Emeritus The equation for the L2 point is $$\\frac{G(M+m)}{R^3}\\left(R\\frac{M}{M+m}+r\\right)\\,-\\,\\frac{GM}{(R+r)^2}\\,-\\,\\frac {Gm}{r^2}= 0$$ where M and m are the masses of the two massive objects (M > m), R is the distance between the two massive objects, and r, the unknown quantity, is the distance between the smaller object and the L2 point. Denoting α=m/M, x=r/R, this becomes $$(1+\\alpha)x+1-\\frac{1}{(1+x)^2}-\\frac {\\alpha}{x^2}= 0$$ $$((1+\\alpha)x+1)x^2(1+x)^2-x^2-\\alpha(1+x)^2 = 0$$ which is a fifth order polynomial. 6. Oct 31, 2008 ### Irrelativity If this is the case, the above polynomial becomes r = -R/2 if you use (r/R)^2 = approximately zero. But the solution to this question is, I believe, r = R(m/3M)^(1/3). Am I missing something here? 7. Oct 31, 2008 ### D H Staff Emeritus $r=R\\,(m/3M)^{1/3}$ is an approximate solution. How to get there from the fifth degree polynomial: First rewrite the polynomial this way: $$x^2((1+x)^3-1) -\\alpha(1+x)^2(1-x^3)= 0$$ Expanding the first term, $$x^3(3+3x+x^2) -\\alpha(1+x)^2(1-x^3)= 0$$", "+ c x + d$, then our $\\alpha$ is a factor of $\\frac{d}{a}$. In practice, Factor Theorem is used for factorising polynomials completely. So our steps in factorising a polynomial $p \\left(x\\right)$ would be 1 Check the constant term in $p \\left(x\\right)$ 2. Find its all possible factors. 3. Take one of the factors, say $\\alpha$ and replace $x$ by this factor in the given polynomial $p \\left(x\\right)$. 4. Try more factors whose number should be equal to the degree of polynomial. And you have got all the factors. Coming to your question, we have $2 \\ln \\left(4 x - 5\\right) + \\ln \\left(x + 1\\right) = 3 \\ln 3$, which can be written as $\\ln \\left\\{{\\left(4 x - 5\\right)}^{2} \\left(x + 1\\right)\\right\\} = \\ln 27$ or #(16x^2-40x+25)(x+1)}=ln27# or $16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$ $16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$ This is the same as in $\\left(b\\right)$ so we use factor theorem as follows. We should have zeros among $\\pm 2 , \\pm 1 , \\pm \\frac{1}{2} , \\pm \\frac{1}{4} , \\pm \\frac{1}{8}$.Observe that for $x = 2 , - \\frac{1}{4}$, it is $0$ and hence $x - \\left(- \\frac{1}{4}\\right)$ or $x + \\frac{1}{4}$ i.e. $\\left(4 x + 1\\right)$ and $\\left(x - 2\\right)$ are two", "+ c x + d$, then our $\\alpha$ is a factor of $\\frac{d}{a}$. In practice, Factor Theorem is used for factorising polynomials completely. So our steps in factorising a polynomial $p \\left(x\\right)$ would be 1 Check the constant term in $p \\left(x\\right)$ 2. Find its all possible factors. 3. Take one of the factors, say $\\alpha$ and replace $x$ by this factor in the given polynomial $p \\left(x\\right)$. 4. Try more factors whose number should be equal to the degree of polynomial. And you have got all the factors. Coming to your question, we have $2 \\ln \\left(4 x - 5\\right) + \\ln \\left(x + 1\\right) = 3 \\ln 3$, which can be written as $\\ln \\left\\{{\\left(4 x - 5\\right)}^{2} \\left(x + 1\\right)\\right\\} = \\ln 27$ or #(16x^2-40x+25)(x+1)}=ln27# or $16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$ $16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$ This is the same as in $\\left(b\\right)$ so we use factor theorem as follows. We should have zeros among $\\pm 2 , \\pm 1 , \\pm \\frac{1}{2} , \\pm \\frac{1}{4} , \\pm \\frac{1}{8}$.Observe that for $x = 2 , - \\frac{1}{4}$, it is $0$ and hence $x - \\left(- \\frac{1}{4}\\right)$ or $x + \\frac{1}{4}$ i.e. $\\left(4 x + 1\\right)$ and $\\left(x - 2\\right)$ are two", "see that $$\\prod_{b=1}^{2015}\\left(1+e^{\\frac{2\\pi i a b}{2015}}\\right)=\\left[\\prod_{b=1}^{\\frac{2015}{d}}\\left(1+\\exp\\left(\\frac{2\\pi i b}{\\frac{2015}{d}}\\right)\\right)\\right]^d$$ By factoring the polynomial $$x^n-1=\\prod_{b=1}^{n}\\left(x-e^{\\frac{2\\pi i b}{n}}\\right)$$we can see that $$(-1)^n-1=\\prod_{b=1}^{n}\\left(-1-e^{\\frac{2\\pi i b}{n}}\\right)=(-1)^n\\prod_{b=1}^{n}\\left(1+e^{\\frac{2\\pi i b}{n}}\\right)$$and thus $$\\prod_{b=1}^{n}\\left(1+e^{\\frac{2\\pi i b}{n}}\\right)=1-(-1)^n$$Since every divisor of $2015$ is odd, we obtain $$\\left[\\prod_{b=1}^{\\frac{2015}{d}}\\left(1+\\exp\\left(\\frac{2\\pi i b}{\\frac{2015}{d}}\\right)\\right)\\right]^d=2^d$$ Leading to the result described above. • @r9m Well, nothing intelligent, but as $2015=5*13*31$ has only $8$ divisors, it takes not to long to calculate it using $\\varphi(n)=n\\cdot\\prod_{p\\mid n}\\left(1-\\frac{1}{p}\\right)$. And thanks for the edit! :) – Redundant Aunt Dec 6 '15 at 11:53 • $\\sum\\limits_{d | n} d\\varphi (n/d)$ is convolution of two multiplicative functions, hence it's multiplicative as well. :) – r9m Dec 6 '15 at 12:07", "\\in K \\left[{X}\\right]$ such that: $f = q \\mu_\\alpha + r$ and: $\\deg r < \\deg \\mu_\\alpha$. Evaluating this expression at $\\alpha$ we find that: $f \\left({\\alpha}\\right) = q \\left({\\alpha}\\right) \\mu_\\alpha \\left({\\alpha}\\right) + r \\left({\\alpha}\\right) \\implies r \\left({\\alpha}\\right) = 0$ since $\\mu_\\alpha \\left({\\alpha}\\right) = f \\left({\\alpha}\\right) = 0$. But $\\mu_\\alpha$ has minimal degree among the non-zero polynomials that are zero at $\\alpha$. Therefore as $\\deg r < \\deg \\mu_\\alpha$ we must have $r = 0$. Therefore: $f = q \\mu_\\alpha$ That is, $\\mu_\\alpha$ divides $f$. $\\blacksquare$", "{\\left(\\frac{1}{\\alpha }-\\frac{1}{\\beta }\\right)}^{2}=\\frac{9}{4}\\phantom{\\rule{0ex}{0ex}}⇒\\frac{1}{\\alpha }-\\frac{1}{\\beta }=±\\frac{3}{2}$ #### Page No 61: By using the relationship between the zeroes of the cubic ploynomial. We have, Sum of zeroes = $\\therefore a-b+a+a+b=\\frac{-\\left(-3\\right)}{1}\\phantom{\\rule{0ex}{0ex}}⇒3a=3\\phantom{\\rule{0ex}{0ex}}⇒a=1$ Now, Product of zeros = #### Page No 63: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as ${x}^{3}-\\left(a+b+c\\right){x}^{2}+\\left(ab+bc+ca\\right)x-abc$ ...(1) Let Substituting the values in (1), we get ${x}^{3}-\\left(2-3+4\\right){x}^{2}+\\left(-6-12+8\\right)x-\\left(-24\\right)\\phantom{\\rule{0ex}{0ex}}⇒{x}^{3}-3{x}^{2}-10x+24$ #### Page No 63: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as ${x}^{3}-\\left(a+b+c\\right){x}^{2}+\\left(ab+bc+ca\\right)x-abc$ ...(1) Let Substituting the values in (1), we get ${x}^{3}-\\left(\\frac{1}{2}+1-3\\right){x}^{2}+\\left(\\frac{1}{2}-3-\\frac{3}{2}\\right)x-\\left(\\frac{-3}{2}\\right)\\phantom{\\rule{0ex}{0ex}}⇒{x}^{3}-\\left(\\frac{-3}{2}\\right){x}^{2}-4x+\\frac{3}{2}\\phantom{\\rule{0ex}{0ex}}⇒2{x}^{3}+3{x}^{2}-8x+3$ #### Page No 63: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes Therefore, the required polynomial is ${x}^{3}-5{x}^{2}-2x+24$ #### Page No 63: Quotient $q\\left(x\\right)=x-3$ Remainder $r\\left(x\\right)=7x-9$ #### Page No 63: Quotient $q\\left(x\\right)={x}^{2}+x-3$ Remainder $r\\left(x\\right)=8$ #### Page No 63: We can write and Quotient $q\\left(x\\right)=-{x}^{2}-2$ Remainder $r\\left(x\\right)=-5x+10$ #### Page No 63: Let $f\\left(x\\right)=2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$ and $g\\left(x\\right)={x}^{2}-3$ Quotient $q\\left(x\\right)=2{x}^{2}+3x+4$ Remainder $r\\left(x\\right)=0$ Since, the remainder is 0. Hence, ${x}^{2}-3$ is a", "variant when $p$ is given by a polynomial. For prime fields where $p$ has a special form, the Special-NFS algorithm may apply. Its complexity is $L_{p}[1/3, (32/9)^{1/3} = 1.526]$. For extension fields $\\mathbb{F}_{p^n}$, where $p$ is obtained by evaluating a polynomial of degree $d$ at a given value, e.g. $d=4$ for BN curves, Joux and Pierrot [PAIRING:JouPie13] proposed a dedicated polynomial selection method such that the NFS algorithm has an exptected running-time of • $L_{p^n}\\left[1/3, \\left(\\frac{d+1}{d}\\frac{64}{9}\\right)^{1/3}\\right]$ in medium characteristic, where $p = L_{p^n}[\\alpha_p, c_p]$ and $1/3 < \\alpha_p < 2/3$; • $L_{p^n}\\left[1/3, \\left(\\frac{32}{9}\\right)^{1/3}\\right]$ in large characteristic, where $p = L_{p^n}[\\alpha_p, c_p]$ and $2/3 < \\alpha_p < 1$ and moreover $d$ satisfies $d = \\frac{1}{n} \\left(\\frac{2 \\log p^n}{3 \\log \\log p^n} \\right)^{1/3}$; • $L_{p^n}\\left[1/3, \\left(\\frac{32}{9}\\right)^{1/3}\\right]$ in prime fields and tiny extension degree fields, where $p = L_{p^n}[1, c_p]$ and moreover $d$ satisfies $d = \\frac{1}{n} \\left(\\frac{2 \\log p^n}{3 \\log \\log p^n} \\right)^{1/3}$. The Kim–barbulescu method combines the TNFS method and the Joux–Pierrot method to obtain a $L_{p^n}[1/3, (32/9)^{1/3} = 1.526]$ asymptotic complexity in the medium-characteristic case. ## Discussion on key-size update This new attack reduces the asymptotic complexity of the NFS algorithm to compute DL in $\\mathbb{F}_{p^n}$, which contains the target group of pairings. The recommended sizes of $\\mathbb{F}_{p^n}$ were computed assuming that the asymptotic complexity is $L_{p^n}[1/3, 1.923]$.", "# Taylor Polynomials Find the Taylor polynomial $T_5(x)$ = of order 5 about $x = 0$ for $f(x) = \\sqrt{1+x}$. Write down the remainder term $R_5(x)$ and estimate the size of the error if $T_5(1)$ is used as an approximation to $f(1)$. My attempt at the question: I think this is the correct Taylor polynomial: $$T_5(x) = 1 + \\frac{x}2 - \\frac{x^2}8 + \\frac{x^3}{12} - \\frac{5x^4}{128}+\\frac{7x^5}{256}$$ I'm not sure if this fully answers the remainder term part: $$R_5(x) = \\frac{f^{(6)}(c)}{6!}x^6$$ Could someone please give some pointers on how to estimate the size of the error and just check if the other parts are correct. Thank you for any help. • FAQ section + Reading directions to use LaTeX to write mathematicas + using those directions in posts here = nice looking and appealing questions. – DonAntonio May 5 '13 at 10:19 • You should specify if you need the Peano (little-o reminder) or the Lagrange reminder, which is exactly the one you wrote, only with $c \\in [o,x]$. – Kore-N May 5 '13 at 10:39 • @Cornelis sorry I forgot to specify. I meant to specify Lagrange remainder – Joe S May 5 '13 at 10:46 Recall that $$(1+x)^\\alpha=1+\\sum_{k=1}^n\\frac{\\alpha(\\alpha-1)\\cdots(\\alpha-k+1)}{k!}x^k+o(x^n)$$ so take $\\alpha=\\frac{1}{2}$ and try to simplify $\\frac{\\alpha(\\alpha-1)\\cdots(\\alpha-k+1)}{k!}$ with this value. Now for the remainder: by Taylor-Lagrange formula there's" ]

[ "+0 0 38 1 Let $$x,y,z$$ be positive real numbers such that $$x + y + z = 1$$ Find the minimum value of $$\\frac{1}{x + y} + \\frac{1}{x + z} + \\frac{1}{y + z}.$$ Nov 12, 2019", "# An algebra problem by Dinesh Chavan Algebra Level 3 Let $$x,y,z$$ be positive real numbers. Given that $$xyz=1$$ Then find the minimum value of $$\\frac{1}{x^2+3x+2}+\\frac{1}{y^2+3y+2}+\\frac{1}{z^2+3z+2}$$ × Problem Loading... Note Loading... Set Loading...", "# AM-GM Application Let $x,y,z$ be positive real numbers such that $\\frac { 1 }{ x } +\\frac { 1 }{ y } +\\frac { 1 }{ z } =1$. Then what is the minimum value of $(x-1)(y-1)(z-1)?$ ×", "+0 # help 0 205 1 Let $$x,y,z$$ be positive real numbers. Find the set of all possible values of $$f(x,y,z) = \\frac{x}{x + y} + \\frac{y}{y + z} + \\frac{z}{z + x}.$$ Apr 10, 2019", "+0 # help 0 76 1 Let $$x,y,z$$ be positive real numbers. Find the set of all possible values of $$f(x,y,z) = \\frac{x}{x + y} + \\frac{y}{y + z} + \\frac{z}{z + x}.$$ Apr 10, 2019", "# Minimum Value Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that $$x^2 + y^2 + z^2 +2xyz = 4$$. Find the minimum value of $$x^2 + y^2 + z^2$$. ×", "# Math Help - x,y,z 1. ## x,y,z solve the system of equations for positive real numbers $\\frac{1}{xy}=\\frac{x}{z}+1,\\frac{1}{zy}=\\frac{y}{x }+1, \\frac{1}{xz}=\\frac{z}{y}+1$", "# An algebra problem by Dinesh Chavan Algebra Level 3 Let $x,y,z$ be positive real numbers. Given that $xyz=1$ Then find the minimum value of $\\frac{1}{x^2+3x+2}+\\frac{1}{y^2+3y+2}+\\frac{1}{z^2+3z+2}$ ×", "# Let x and y be two positive real numbers such that x + y = 1, then minimum value of:- 150 views closed Let x and y be two positive real numbers such that x + y = 1, then minimum value of:- 1/x + 1/y = (1) 2 (2) 5/2 (3) 3 (4) 4 by (40.1k points) selected The correct option is (D) 4. A.M ≥ H.M $\\frac{x + y}{2}$ ≥ $\\frac{2xy}{x+y}$ → $\\frac{2}{\\frac{1}{x} + \\frac{1}{y}}$ $\\frac{x + y}{2}$ ≥ $\\frac{2}{\\frac{1}{x} + \\frac{1}{y}}$ = (x + y) (${\\frac{1}{x} + \\frac{1}{y}}$) ≥ 4 = (${\\frac{1}{x} + \\frac{1}{y}}$) ≥ 4 (x + y = 1) Minimum Value = 4.", "# AM-GM Application Algebra Level 2 Let $x,y,z$ be positive real numbers such that $\\frac { 1 }{ x } +\\frac { 1 }{ y } +\\frac { 1 }{ z } =1$. Then what is the minimum value of $(x-1)(y-1)(z-1)?$ ×", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "# Find the Minimum Algebra Level 4 Let $$x, y, z$$, and $$a$$ be positive real numbers such that $$x + y + z + a = 1$$. What is the minimum value of $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{4}{z}+\\dfrac{16}{a}$ × Problem Loading... Note Loading... Set Loading...", "# Does Symmetricity Always Hold? Algebra Level 2 If $$x,y,z$$ are positive real numbers such that $$x+y+z=3$$, then find the minimum value of $E=\\frac { 4yz+9xz+16xy }{ xyz }.$ × Problem Loading... Note Loading... Set Loading...", "+0 # need help 0 76 2 Let x, y and z be positive real numbers such that x + y + z = 1. Find the minimum value of (x + y + z)/(xyz). Jan 25, 2022 #1 +516 +3 x + y + z = 1. Since xyz is in the denominator, we want xyz to be the largest value possible. The largest value possible of xyz is only if x = y = z, so x = y = z = 1/3. $$1\\over{({1\\over3})^2}$$ = $$(x + y + z)\\over{xyz}$$ Thus, the minimum value of $$(x + y + z)\\over{xyz}$$ is 27. Jan 25, 2022 #2 +360 0 I think you made a mistake on format you wrote $$\\frac{1}{(\\frac{1}{3})^2}$$ instead of $$\\frac{1}{(\\frac{1}{3})^3}$$ Jan 26, 2022", "+0 # Help pls :) +1 323 1 +504 Let x, y, and z be positive real numbers. Find the minimum value of $$(x + 2y + 4z) \\left( \\frac{4}{x} + \\frac{2}{y} + \\frac{1}{z} \\right)$$. Btw. I have to use the AM-GM inequality thingy. Help :) May 26, 2021", "+0 # This is very confusing 0 35 1 Let x, y, and z be nonzero real numbers, such that no two are equal, and $$x + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x}.$$ Find all possible numeric values of xyz. May 26, 2021", "# Least value Algebra Level 3 $\\frac{x^3}{x^2+y}+\\frac{y^3}{y^2+z}+\\frac{z^3}{z^2+x}$ If $x, y$, $z$ are positive real numbers with a product of 1, find the minimum value of the expression above. × Problem Loading... Note Loading... Set Loading...", "# Minimum Value If $x$ and $y$ are positive real numbers, what is the minimum value of $(2x + 3y)\\left(\\frac{8}{x} + \\frac{3}{y} \\right) ?$ ×", "9 a b c .$$Let$a, b$and$c$be positive real numbers. Show that $$(a+b+c)(a b+b c+c a) \\geq 9 a b c .$$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 If$x$and$y$are positive real numbers, what is the minimum value of $$(2 x+3 y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right) ?$$ 0 answers 57 views If$x$and$y$are positive real numbers, what is the minimum value of $$(2 x+3 y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right) ?$$If$x$and$y$are positive real numbers, what is the minimum value of $$(2 x+3 y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right) ?$$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all positive real solutions$(a, b, c, d)$to the following system of equations: $$a+b+c+d=12, a b c d=27+a b+a c+a d+b c+b d+c d .$$ 1 answer 57 views Find all positive real solutions$(a, b, c, d)$to the following system of equations: $$a+b+c+d=12, a b c d=27+a b+a c+a d+b c+b d+c d .$$Find all positive real solutions$(a, b, c, d)$to the following system of equations: $$a+b+c+d=12, a b c d=27+a b+a c+a d+b c+b d+c d .$$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all pairs of positive real numbers such that$a^{3}+b^{3}+1=3 a b$. 1 answer 74 views Find all pairs of positive real numbers such that$a^{3}+b^{3}+1=3 a b$.Find all pairs of positive real numbers such that$a^{3}+b^{3}+1=3 a b$. ... close Notice: Undefined index:", "+0 # Help 0 40 1 Let x, y, and z be nonzero real numbers, such that no two are equal, and $$x + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x}.$$Find all possible numeric values of xyz. Aug 28, 2020" ]

[ "to solve it for $c=\\sqrt{2}-1$ using AM-GM. but dont know how for its optimum value. \\begin{align*} \\frac{a_1}{a_2+a_3}+\\frac{a_2}{a_3+a_4}+\\cdots+ \\frac{a_n}{a_1+a_2} \\geq (\\sqrt{2}-1)n \\end{align*} Using Am-GM, the inequality becomes \\begin{align} \\frac{a_1+a_2+a_3}{a_2+a_3} \\cdot \\frac{a_2+a_3+a_4}{a_3+a_4}\\cdots \\frac{a_n+a_1+a_2}{a_1+a_2}\\geq (\\sqrt{2})^n \\end{align}", "[0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1 \\end{align*} and \\begin{align*} g(a_1) = &\\min_{x_1, x_2, \\cdots, x_m} \\ x_1^2 + x_2^2 + \\cdots + x_m^2\\\\ &\\mathrm{s.t.}\\quad x_i\\in [0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1. \\end{align*} It is easy to obtain $$g(a_1) = \\frac{a_1^2}{m}$$ (the minimum is attained always when $$x_1=x_2 = \\cdots = x_m = \\frac{a_1}{m}\\in [0, \\frac{1}{4}]$$). We have $$f(a_1) = \\frac{1}{16}\\lfloor 4a_1\\rfloor + \\left(a_1 - \\frac{1}{4}\\lfloor 4a_1\\rfloor\\right)^2$$ since $$h(x) = x_1^2 + x_2^2 + \\cdots + x_n^2$$ is convex, and (denote $$K = \\lfloor 4a_1\\rfloor$$) $$\\big(\\underbrace{\\frac{1}{4}, \\frac{1}{4}, \\cdots, \\frac{1}{4}}_{K}, a_1 - \\frac{1}{4}K, 0, 0, \\cdots, 0\\big)$$ majorizes $$(x_1, x_2, \\cdots, x_m)$$ with $$x_i\\in [0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1$$. iii) Clearly, (3) is true. (the \"if\" part): It is clear according to the (the \"only if\" part). This completes the proof of Lemma 1. Let us proceed. We want to find the explicit expression of $$F(a_1, a_2)$$. Let $$(x_1, \\ x_2, \\cdots, x_m)$$ be the global maximizer of the optimization problem given in (4). We claim that $$|\\{x_1, \\ x_2, \\cdots, x_m\\}|\\le 4$$ where $$|\\cdot|$$ is the cardinality of a set. Proof of the claim: Assume, for the sake of contradiction, that $$|\\{x_1, \\ x_2,", "# An Inequality for Cevians And The Ratio of Circumradius and Inradius ### Solution By the Law of Cosines, in $\\Delta XYC,$ \\displaystyle \\begin{align} XY^2 &= XC^2+YC^2-2\\cdot XC\\cdot YC\\cdot\\cos C\\\\ &\\ge 2\\cdot XC\\cdot YC\\cdot(1-\\cos C)=4\\cdot XC\\cdot YC\\cdot\\sin^2\\frac{C}{2}. \\end{align} It follows that $\\displaystyle XY\\ge 2\\sqrt{XC\\cdot YC}\\cdot\\sin\\frac{C}{2}.\\,$ Similarly, $\\displaystyle YZ\\ge 2\\sqrt{YA\\cdot ZA}\\cdot\\sin\\frac{A}{2}\\,$ and $\\displaystyle ZX\\ge 2\\sqrt{XB\\cdot ZB}\\cdot\\sin\\frac{B}{2}.\\,$ Multiplying the three we get $\\displaystyle \\small{XY\\cdot YZ\\cdot ZX\\ge 8\\sqrt{XC\\cdot XB\\cdot YC\\cdot YA\\cdot ZA\\cdot ZB}\\cdot\\sin\\frac{A}{2}\\cdot\\sin\\frac{B}{2}\\cdot\\sin\\frac{C}{2}}.$ Thus, $\\displaystyle\\small{\\frac{XB}{XY}\\cdot\\frac{YC}{YZ}\\cdot \\frac{ZA}{ZV}\\le\\sqrt{\\frac{XB}{XC}\\cdot \\frac{YC}{YA}\\cdot \\frac{ZA}{ZB}}\\cdot \\frac{\\displaystyle 1}{8\\cdot\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}}=\\frac{R}{2r}}$ because, by Ceva's theorem, $g=\\displaystyle \\frac{XB}{XC}\\cdot \\frac{YC}{YA}\\cdot \\frac{ZA}ZB.$ ### Acknowledgment The problem by Titu Andreescu has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook group and commented with a solution (above) by Marian Dinca. I am grateful to Leo Giugiuc for bringing this problem to my attention. Copyright © 1996-2018 Alexander Bogomolny 66524280", "# 2008 Indonesia MO Problems/Problem 2 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Prove that for every positive reals $x$ and $y$, $$\\frac {1}{(1 + \\sqrt {x})^{2}} + \\frac {1}{(1 + \\sqrt {y})^{2}} \\ge \\frac {2}{x + y + 2}.$$ ## Solution By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \\ge (1 + \\sqrt{x})^2$ and $(1+1)(1+y) \\ge (1 + \\sqrt{y})^2$, with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$. Because $x,y > 0$, \\begin{align*} \\frac{1}{(1 + \\sqrt{x})^2} &\\ge \\frac{1}{2+2x} \\\\ \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac{1}{2+2y} \\\\ \\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}). \\end{align*} By the AM-GM Inequality, we know that $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$. For the equality case, $\\frac{1}{1+x} = \\frac{1}{1+y}$, so $x = y$. Additionally, by the AM-GM Inequality, $\\frac12 \\cdot (\\frac{x+1}{2} + \\frac{y+1}{2}) \\ge \\sqrt{\\frac{(x+1)(y+1)}{4}}$. For the equality case, $\\frac{x+1}{2} = \\frac{y+1}{2}$, so $x = y$. Because $x,y \\ge 0$, \\begin{align*} \\frac12 \\cdot (\\frac{x+y+2}{2}) &\\ge \\frac{\\sqrt{(x+1)(y+1)}}{2} \\\\ \\frac{x+y+2}{2} &\\ge \\sqrt{(x+1)(y+1)} \\\\ \\sqrt{\\frac{1}{(x+1)(y+1)}} &\\ge \\frac{2}{x+y+2}. \\end{align*} Therefore, since $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y})$ and $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$ and $\\sqrt{\\frac{1}{(x+1)(y+1)}} \\ge \\frac{2}{x+y+2}$, we must have $\\frac{1}{(1 + \\sqrt{x})^2}", "# Solution of equation $\\frac{x\\cdot 2014^{\\frac{1}{x}}+\\frac{1}{x}\\cdot 2014^x}{2} = 2014$ Solution of equation $\\displaystyle \\frac{x\\cdot 2014^{\\frac{1}{x}}+\\frac{1}{x}\\cdot 2014^x}{2} = 2014$ $\\bf{My\\; Try::}$ Clearly Here $x>0$, Now Using $\\bf{A.M\\geq G.M}$ So Here $\\displaystyle x\\cdot 2014^{\\frac{1}{x}}>0$ and $\\displaystyle \\frac{1}{x}\\cdot 2014^x>0$ . So $\\displaystyle \\left(\\frac{x\\cdot 2014^{\\frac{1}{x}}+\\frac{1}{x}\\cdot 2014^x}{2}\\right)\\geq \\left(x\\cdot 2014^{\\frac{1}{x}}\\cdot \\frac{1}{x}\\cdot 2014^x\\right)^{\\frac{1}{2}} = \\sqrt{2014^{\\left(x+\\frac{1}{x}\\right)}} = 2014$ And equality Hold , When $\\displaystyle x = \\frac{1}{x}\\Rightarrow x= 1$ Can we solve it without Using $\\bf{A.M\\geq G.M}$ If Yes, Then please explain me. Thanks Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that $x\\ge0$, we have: \\begin{align} &\\frac{x\\cdot 2014^{\\frac{1}{x}}+\\frac{1}{x}\\cdot 2014^x}{2} = 2014 \\\\\\implies&x2014^{\\frac{1}{x}-1}+\\frac{1}{x}2014^{x-1}=2 \\\\\\implies&\\left(\\sqrt{x2014^{\\frac{1}{x}-1}}-\\sqrt{\\frac{1}{x}2014^{x-1}}\\right)^2+2\\sqrt{2014^{\\frac{1}{x}+x-2}}=2 \\\\\\implies&\\left(\\sqrt{x2014^{\\frac{1}{x}-1}}-\\sqrt{\\frac{1}{x}2014^{x-1}}\\right)^2+2\\sqrt{2014^{\\left(\\sqrt{\\frac{1}{x}}-\\sqrt{x}\\right)^2}}=2 \\\\\\implies&2\\sqrt{2014^{\\left(\\sqrt{\\frac{1}{x}}-\\sqrt{x}\\right)^2}}\\le2 \\\\\\implies&2014^{\\left(\\sqrt{\\frac{1}{x}}-\\sqrt{x}\\right)^2}\\le1 \\\\\\implies&\\left(\\sqrt{\\frac{1}{x}}-\\sqrt{x}\\right)^2\\le0 \\\\\\implies&\\left(\\sqrt{\\frac{1}{x}}-\\sqrt{x}\\right)^2=0 \\\\\\implies&\\sqrt{\\frac{1}{x}}-\\sqrt{x}=0 \\\\\\implies&\\frac{1}{x}=x \\\\\\implies&x=1 \\end{align} where we have used the fact that $x\\ge 0$ several times.", "be proven for all values of n in a variety of ways, including induction. Proofs are given on the wikipedia page. So, what my calculus student did was apply the AM-GM inequality to the case $n = 2, a = 3x$, and $b = 75/x$, which gives $15 = \\sqrt{3x \\cdot (75/x)} \\leq (3x + 75/x)/2$, or multiplying by 2: $30 \\leq 3x + 75/x$ with equality if and only if $3x = 75/x$ or $x = 5.$ Intrigued by this, I invented some more examples of optimizing with the AM-GM inequality. Problem 1 Find the minimum value of $f(x) = x^3 + \\frac{16}{x} + \\frac{32}{x^2}$ for $x > 0$. Solution By the AM-GM inequality, we have $x^3 + \\frac{16}{x} + \\frac{32}{x^2} \\geq 3 \\cdot \\sqrt[3]{x^3 \\cdot \\frac{16}{x} \\cdot \\frac{32}{x^2}} = 24$ with equality if and only if $x^3 = \\frac{16}{x} = \\frac{32}{x^2} =8$, which is achieved when $x = 2.$ So the minimum is 24. That sure beats having to find the critical numbers for that function. Problem 2 Find the minimum value of $f(x,y) = xy + \\frac{8}{x} + \\frac{1}{y}$ for $x > 0, y >0.$ Solution $xy + \\frac{8}{x} + \\frac{1}{y} \\geq 3 \\cdot \\sqrt[3]{xy \\cdot \\frac{8}{x} \\cdot \\frac{1}{y}} = 6$ with equality if and only if $xy = \\frac{8}{x} = \\frac{1}{y}=2$, which is achieved", "{y}{z}}}}+{\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}+{\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}+{\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}}{6}}\\\\&=6\\cdot {\\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\\end{aligned}}\\\\ with \\ {\\displaystyle x_{1}={\\frac {x}{y}},\\qquad x_{2}=x_{3}={\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}},\\qquad x_{4}=x_{5}=x_{6}={\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}.} Applying the AM–GM inequality for n = 6, we get {\\begin{aligned}f(x,y,z)&\\geq 6\\cdot {\\sqrt[ {6}]{{\\frac {x}{y}}\\cdot {\\frac {1}{2}}{\\sqrt {{\\frac {y}{z}}}}\\cdot {\\frac {1}{2}}{\\sqrt {{\\frac {y}{z}}}}\\cdot {\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}\\cdot {\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}\\cdot {\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}}}\\\\&=6\\cdot {\\sqrt[ {6}]{{\\frac {1}{2\\cdot 2\\cdot 3\\cdot 3\\cdot 3}}{\\frac {x}{y}}{\\frac {y}{z}}{\\frac {z}{x}}}}\\\\&=2^{{2/3}}\\cdot 3^{{1/2}}.\\end{aligned}} Further, we know that the two sides are equal exactly when all the terms of the mean are equal: $f(x,y,z)=2^{{2/3}}\\cdot 3^{{1/2}}\\text\\ when \\ {\\frac {x}{y}}={\\frac {1}{2}}{\\sqrt {{\\frac {y}{z}}}}={\\frac {1}{3}}{\\sqrt[ {3}]{{\\frac {z}{x}}}}.$ All the points (x, y, z) satisfying these conditions lie on a half-line starting at the origin and are given by ${\\displaystyle (x,y,z)={\\biggr (}t,{\\sqrt[{3}]{2}}{\\sqrt {3}}\\,t,{\\frac {3{\\sqrt {3}}}{2}}\\,t{\\biggr )}\\text with \\ t \\ >0.}$", "the ... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \\Bbb Z-module) with an action of the ring R, defined by R \\to Aut(M), but the main problem is, that not every image of \\mu is an automorphism. So, in fact, you could have \\varphi \\in End(M) \\setminus Aut(M) as an image of ... 0 Using formulae used here,$$\\frac{386}{97}=3+\\frac{95}{97}=3+\\frac1{\\frac{97}{95}}=3+\\frac1{1+\\frac2{95}}=3+\\frac1{1+\\frac1{\\frac{95}2}}=3+\\frac1{1+\\frac1{47+\\frac12}}$$So, the previous convergent of \\displaystyle \\frac{386}{97} is \\displaystyle3+\\frac1{1+\\frac1{47}}=3+\\frac{47}{48}=\\frac{191}{48}$$\\implies386\\cdot48-191\\cdot97=1\\implies ... 0 There is likely a typo in the question. Using the Euclidean algorithm, we have that: \\begin{align*} 386 &= 3(97) + 95 \\\\ 97 &= 1(95) + 2 \\\\ 95 &= 47(2) + 1 \\end{align*} Working backwards, we see that: \\begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\\\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\\\ &= 48(386) - 191(97)\\\\ ... 1 Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \\cdot p - 386\\cdot q = 1$$ Then $97^{-1} \\equiv p \\mod 386$. 1 Hint. If $n=p^k$, $k\\ge 1$, then $0=p^k\\Bbb Z/p^k\\Bbb Z<p^{k-1}\\Bbb Z/p^k\\Bbb Z<\\cdots<\\Bbb Z/p^k\\Bbb Z$ is a series of composition of length $k$. Can you generalize this to the case $n=p_1^{k_1}\\cdots p_r^{k_r}$, $k_i\\ge 1$ and $r\\ge 2$? Edit. Let's try", "- \\frac{S(0)}{6^3} - \\frac{S(2)}{6^5} -\\frac{S(4)}{6^7} \\right) &= \\frac{36}{6^2} \\cdot \\left( P - \\frac{S(0)}{6^3} - \\frac{S(2)}{6^5}\\right) - \\frac{12}{6^4} \\cdot \\left( P - \\frac{S(0)}{6^3} \\right) + \\frac{1}{6^6} \\cdot P \\end{align*}Note that $S(0) = 1, S(2) = 36$ and $S(4) = 6^4 - 2 \\cdot 6 = 1284$. Therefore, \\begin{align*} \\left(P - \\frac{1}{6^3} - \\frac{36}{6^5} -\\frac{1284}{6^7} \\right) = \\frac{36}{6^2} \\cdot \\left( P - \\frac{1}{6^3} - \\frac{36}{6^5}\\right) - \\frac{12}{6^4} \\cdot \\left( P - \\frac{1}{6^3} \\right) + \\frac{1}{6^6} \\cdot P \\end{align*}Solving for $P$, we obtain $P = \\frac{216}{431} \\implies m+n = \\boxed{647}$. -Vfire ## Solution 4 Let $A=\\frac{1}{6} \\begin{bmatrix} 5 & 1 & 0 & 0 \\\\ 4 & 1 & 1 & 0 \\\\ 4 & 1 & 0 & 1 \\\\ 0 & 0 & 0 & 0 \\\\ \\end{bmatrix}$. $A$ is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly $k$ times is $(A^k)_{1,4}$. Thus the probability that we roll the dice an odd number of times is $1-\\left(\\sum_{k=0}^\\infty A^{2k}\\right)_{1,4} = 1-\\left((I - A^2)^{-1}\\right)_{1,4} = \\frac{216}{431}$. Thus the answer is $216+431=\\boxed{647}$. ## Solution 5 quick cheat Consider it as a contest of Odd and Even. Let $P_o$ and $P_e$ be probability that Odd and Even wins, respectively. If", "Volume Inequality in Tetrahedron Solution 1 Expressing the volume in two ways, we get $abc=xbc+yca+zab.$ By the AM-GM inequality, $(xbc+yca+zab)^3\\ge 27xyz(abc)^2.$ Thus, $(abc)^3\\ge 27xyz(abc)^2,$ from which $abc\\ge 27xyz.$ Equality is attained when $M$ is the centroid of $\\Delta ABC.$ Indeed, let $O=(0,0,0),$ $A=(a,0,0),$ $B=(0,b,0)$ and $C=(0,0,c).$ Then $M=(ma,nb,pc)$ with $m+n+p=1.$ Equality holds iff $[MOAB]=[MOBC]=[MOCA]$ i.e. $mabc=nabc=pabc,$ from which $\\displaystyle m=n=p=\\frac{1}{3},$ making $M$ the centroid of $\\Delta ABC.$ Solution 2 First we prove that (*) $\\displaystyle \\frac{x}{OA}+\\frac{y}{OB}+\\frac{z}{OC}=1.$ Indeed, writing the value of the tetrahedron $OABC$ in two ways, we obtain $Vol(OABC)=Vol(MOAC)+Vol(MOAC)+Vol(MOAB),$ or, equivalently, $\\displaystyle \\frac{OB\\cdot OC\\cdot x}{6}+\\frac{OC\\cdot OA\\cdot x}{6}+\\frac{OA\\cdot OB\\cdot x}{6}=\\frac{OA\\cdot OB\\cdot OC}{6}.$ Dividing by the right-hand side proves (*). Applying to (*) the AM-GM inequality, we have $\\displaystyle 1=\\frac{x}{OA}+\\frac{y}{OB}+\\frac{z}{OC}\\ge 3\\sqrt[3]{\\frac{x}{OA}\\cdot\\frac{y}{OB}\\cdot\\frac{z}{OC}}$ which is equivalent to the require inequality. Equality occurs, iff, $\\displaystyle \\frac{x}{OA}=\\frac{y}{OB}=\\frac{z}{OC}\\;\\left(=\\frac{1}{3}\\right),$ i.e., if the tetrahedron $OABC$ is similar to the tetrahedron with the apex at $M$ and the base formed by the three projections of $M$ on the faces of $OAB.$ In other words, if the two tetrahedra are homothetic with the coefficient $\\displaystyle \\frac{1}{3}.$ This only happens when $M$ is the centroid of $\\Delta ABC.$ Solution 3 Let us choose the origin at $O$. WLOG, let $OA$, $OB$, and $OC$ be along the $X$, $Y$, and $Z$, respectively. The equation of the plane $ABC$ is $\\displaystyle", "$x - xy + y = -5$ becomes $(\\frac{\\sqrt{2}}{2}x - \\frac{\\sqrt{2}}{2}y) - (\\frac{\\sqrt{2}}{2}x - \\frac{\\sqrt{2}}{2}y) (\\frac{\\sqrt{2}}{2}x + \\frac{\\sqrt{2}}{2}y) + (\\frac{\\sqrt{2}}{2}x + \\frac{\\sqrt{2}}{2}y) = -5$ Once again this simplifies nicely to $(x' - \\sqrt{2})^2 - (y')^2 = 12$ and these two equations are solved in a fairly standard fashion via elimination i.e. $3(x' - \\sqrt{2})^2 - 3(y') ^2 = 36$ $+ (x')^2 + 3(y')^2 = 26$ $4(x')^2 - 6\\sqrt{2}x' + 6 = 62$ Solving with quadratic equation and $x' = \\frac{7\\sqrt{2}}{2}, y' = \\pm \\frac{\\sqrt{2}}{2}$ or $x' = -2\\sqrt{2}, y' = \\pm \\sqrt{6}$ If you look back at the previous solutions at first this looks odd but remember these have to be tilted back 45 degrees to give the proper solutions and sure enough for instance $(\\frac{\\sqrt{2}}{2} \\cdot \\frac{7\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2} \\cdot \\frac{7\\sqrt{2}}{2} - \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{2}}{2}) = (4,3)$ Friday, March 30, 2018 As is our routine, I started by going over the problem of the week: Find six distinct natural numbers A,B,C,D,E,F such that A + B + C = D + E + F And Once again, I had a girl write a python program to find the solution. I love all the computational computing that is occurring. I need a better way to harness this energy. This time I had her talk about", "prove that: $$\\sum_{cyc} x_1^3-\\sum_{cyc} \\frac{x_1^2x_2}{2}\\ge \\frac{n}{2}\\iff 2\\sum_{cyc} x_1^3\\ge n+\\sum_{cyc} x_1^2x_2$$ But due to AM-GM we have $\\sum_{cyc} x_1^3\\ge n\\cdot\\left(x_1^3x_2^3\\cdots x_n^3\\right)^{\\frac{1}{n}}=n$ and $\\sum_{cyc} x_1^3=\\sum_{cyc} \\frac{2x_1^3+x_2^3}{3}\\ge\\sum_{cyc} x_1^2x_2$ and adding these two yields the desired result. • Very nice! +1. You can cut down a lot of steps by using just AM-GM, then rearrangements and finally an AM-GM: $$\\sum_{cyc} \\frac{x_1^4}{x_2}-\\frac1{x_2}\\frac{x_1^4x_2^4}{(x_1^4+x_2^4)} \\ge \\sum_{cyc} \\frac{x_1^4}{x_2}-\\frac1{x_2}\\frac{x_1^4x_2^4}{2x_1^2x_2^2}= \\sum_{cyc} \\frac{x_1^4}{x_2}-\\frac{x_1^2x_2}2 \\ge \\sum_{cyc} x_1^3-\\frac{x_1^3}2 \\ge \\frac{n}2$$ – Macavity Mar 1 '15 at 12:15", "substitution of the constraint we need to minimize $$f(x,y)=\\sqrt{x^2+y^2+z^2}=\\sqrt{x^2+y^2+\\frac{1}{16x^2y^2}}$$ and we don't need Lagrange's method. In this case AM-GM is the most effective method, indeed $$\\frac{x^2+y^2+\\frac{1}{16x^2y^2} }{3}\\ge\\sqrt[3]{x^2\\cdot y^2\\cdot \\frac{1}{16x^2y^2}}\\iff x^2+y^2+\\frac{1}{16x^2y^2}\\ge3\\sqrt[3]{\\frac1{16}}=\\frac32\\sqrt[3]{\\frac1{2}}$$ and equality holds when $$x^2=y^2=\\frac{1}{16x^2y^2}\\implies x=\\pm\\sqrt[6]\\frac1{16}=\\pm\\sqrt[3]\\frac14 \\quad y=\\pm x$$ Thus the minimum distance is $$d_{min}=\\sqrt{\\frac32 \\sqrt[3]{\\frac1{2}} } =\\sqrt{\\frac32}\\sqrt[6]{\\frac1{2}}$$ attained for $$P_1=\\left(\\sqrt[3]\\frac14,\\sqrt[3]\\frac14,\\frac14\\sqrt[3]16\\right)\\quad P_2=\\left(\\sqrt[3]\\frac14,-\\sqrt[3]\\frac14,-\\frac14\\sqrt[3]16\\right)$$ $$P_3=\\left( -\\sqrt[3]\\frac14,\\sqrt[3]\\frac14,-\\frac14\\sqrt[3]16\\right)\\quad P_4=\\left( -\\sqrt[3]\\frac14,-\\sqrt[3]\\frac14,\\frac14\\sqrt[3]16\\right)$$ As an alternative, since $f(x,x)>0$ its minimum coincide with the ninimum of $$g(x,y)=f^2(x,y)=x^2+y^2+\\frac{1}{16x^2y^2}$$ thus for the critical points we have $$g_x=2 x - \\frac1{8 x^3 y^2}=0$$ $$g_y=2 y - \\frac1{8 x^2 y^3}=0$$ from which we obtain $$x=\\pm\\sqrt[6]\\frac1{16}=\\pm\\sqrt[3]\\frac14 \\quad y=\\pm x$$ • I have not done minimizing before so this is a bit strange to be. – Omari Celestine Jan 31 '18 at 13:33 • @OmariCelestine You can proceed also with the usual method but you don't need Lagrange at all. And for simplicity you can minimize $f^2$. Note that in general AM-GM inequality are very powerful and important. – user Jan 31 '18 at 13:37 • What do you bean by the usual method? – Omari Celestine Jan 31 '18 at 16:25 • By partial derivatives, I add some detail on it! – user Jan 31 '18 at 16:31 • Exactly as I was proceeding? – Omari Celestine Jan 31 '18 at 16:46 Then you have the following equations:$$2x+\\lambda\\dfrac{1}{4x^2y}=0\\\\2y+\\lambda\\dfrac{1}{4xy^2}=0\\\\2z+\\lambda=0\\\\z=\\dfrac{1}{4xy}$$or$$\\lambda=-8x^3y=-8xy^3=-2z=-\\dfrac{1}{2xy}$$since $x,y\\ne 0$ this", "# 1988 AIME Problems/Problem 8 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: $$f(x, x) = x,\\; f(x, y) = f(y, x), {\\rm \\ and\\ } (x+y)f(x, y) = yf(x, x+y).$$ Calculate $f(14,52)$. ## Solution 1 (Algebra) Let $z = x+y$. By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \\begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\\\ f(x,z) &= \\frac{z}{z-x} \\cdot f(x,z-x). \\end{align*} Using the properties of $f,$ we have \\begin{align*} f(14,52) &= \\frac{52}{38} \\cdot f(14,38) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot f(14,24) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(14,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(10,14)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(10,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(4,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot f(4,6)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(4,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(2,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot \\frac{4}{2} \\cdot f(2,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6}", "x=y=1/4 and z=1/2. We use a combination of the AM-HM and the AM-GM in that order. Starting, we have by the AM-HM $$\\frac{x + y}{2} \\ge \\frac{2}{\\frac{1}{x} + \\frac{1}{y}} = \\frac{2xy}{x + y}$$ So we have $$\\frac{x + y}{xy} \\ge \\frac{4}{x + y}$$. Multiplying by 1/z, we get $$\\frac{x + y}{xyz} \\ge \\frac{4}{(x + y)z}$$. By the AM-GM inequality, $$\\sqrt{(x + y)z} \\le \\frac{x + y + z}{2} = \\frac{1}{2}$$ meaning $$(x + y)z \\le \\frac{1}{4}$$. Hence, $$\\frac{4}{(x + y)z} \\ge 16$$. Using what we had before, we can write $$\\frac{x + y}{xy} \\ge \\frac{4}{x + y} \\ge 16$$. Equality occurs when x=y, and the minimum value is 16. If you ask me, this solution was quite interesting. I thought of using the QM-AM-GM-HM relationship, as well as the Cauchy-Schwarz Inequality, but not in this way. Thanks for your help! EnchantedLava68 Jun 30, 2021 #3 +2104 +2 YAYYY. :DDDD That is such an intresting solution. I have not yet learned about AM-GM/AM-HM, but I'm planning to do so soon through alcumus. I think the mistake I made was forgetting that x, y, and z were all positive, meaning that (x+y)/(xyz) was also positive. =^._.^= catmg Jul 1, 2021", "\\leq 1 \\\\ & 0 \\leq c \\leq d \\leq e \\leq 1 \\end{align} For fixed $c, d, e$, $h$ is minimized when $a$ and $b$ are maximized, i.e., $a = \\frac{3-2d}{3-2c}$ and $b = \\frac{3-2e}{3-2c}$. Then $h$ becomes $$h(c, d, e) = \\frac{27 - 3c - 6d - 9e}{27 - 4c - 6d - 8e}$$ same as $f(c, d, e)$. Thus the minimum value is $\\frac{12}{13}$. Summary: the greatest $k$ is $\\frac{12}{13}$. Yes, $\\color{red}{k=\\frac 17}$ works. First, $$\\frac{3}{3-2c}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3\\cdot 1+4\\cdot 1}=\\frac 13$$ Second, when $$\\frac{3a}{3-2d}\\ge \\frac{3}{3-2c}\\implies a\\ge\\frac{3-2d}{3-2c}\\ge \\frac{3-2\\cdot 1}{3-2\\cdot 0}=\\frac 13$$ we have $$\\frac{3a}{3-2d}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3a}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3a+4\\cdot 1}=1-\\frac{2}{a+2}$$$$\\ge 1-\\frac{2}{\\frac 13+2}=\\frac 17$$ Third, when $$\\frac{3b}{3-2e}\\ge\\frac{3}{3-2c}\\implies b\\ge \\frac{3-2e}{3-2c}\\ge \\frac{3-2\\cdot 1}{3-2\\cdot 0}=\\frac 13$$ we have $$\\frac{3b}{3-2e}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3b}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3\\cdot 1+4\\cdot b}=\\frac 34-\\frac{15}{4(4b+5)}$$$$\\ge \\frac 34-\\frac{15}{4(4\\cdot \\frac 13+5)}=\\frac{3}{19}$$ It follows from these that $$\\max\\left(\\frac{3}{3-2c},\\frac{3a}{3-2d},\\frac{3b}{3-2e}\\right)\\geq \\color{red}{\\frac 17}\\cdot\\frac{2+3a+4b}{9-c-2d-3e}$$ holds for any $0\\leq b\\leq a\\leq 1$ and $0\\leq c\\leq d\\leq e\\leq 1$. • Thanks. In fact, a better bound of $k=1/3$ follows from your first inequality, so the rest is not needed. The natural question would then be to ask for the best such constant $k$. I've edited the question. – pi66 Jul 13 '16 at 9:47 • @pi66: Ah, you are right. I've misunderstood something. – mathlove Jul 13 '16 at 10:28", "for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. It can be rewritten as: {\\displaystyle {\\begin{aligned}f(x,y,z)&=6\\cdot {\\frac {{\\frac {x}{y}}+{\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}+{\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}}{6}}\\\\&=6\\cdot {\\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\\end{aligned}}} with ${\\displaystyle x_{1}={\\frac {x}{y}},\\qquad x_{2}=x_{3}={\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}},\\qquad x_{4}=x_{5}=x_{6}={\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}.}$ Applying the AM–GM inequality for n = 6, we get {\\displaystyle {\\begin{aligned}f(x,y,z)&\\geq 6\\cdot {\\sqrt[{6}]{{\\frac {x}{y}}\\cdot {\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}\\cdot {\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}}}\\\\&=6\\cdot {\\sqrt[{6}]{{\\frac {1}{2\\cdot 2\\cdot 3\\cdot 3\\cdot 3}}{\\frac {x}{y}}{\\frac {y}{z}}{\\frac {z}{x}}}}\\\\&=2^{2/3}\\cdot 3^{1/2}.\\end{aligned}}} Further, we know that the two sides are equal exactly when all the terms of the mean are equal: ${\\displaystyle f(x,y,z)=2^{2/3}\\cdot 3^{1/2}\\quad {\\mbox{when}}\\quad {\\frac {x}{y}}={\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}={\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}.}$ All the points (x, y, z) satisfying these conditions lie on a half-line starting at the origin and are given by ${\\displaystyle (x,y,z)={\\biggr (}t,{\\sqrt[{3}]{2}}{\\sqrt {3}}\\,t,{\\frac {3{\\sqrt {3}}}{2}}\\,t{\\biggr )}\\quad {\\mbox{with}}\\quad t>0.}$ ## Practical applications An important practical application in financial mathematics is to computing the rate of return: the annualized return, computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect. ## Proofs of", "# 2008 Indonesia MO Problems/Problem 2 ## Problem Prove that for every positive reals $x$ and $y$, $$\\frac {1}{(1 + \\sqrt {x})^{2}} + \\frac {1}{(1 + \\sqrt {y})^{2}} \\ge \\frac {2}{x + y + 2}.$$ ## Solution 1 By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \\ge (1 + \\sqrt{x})^2$ and $(1+1)(1+y) \\ge (1 + \\sqrt{y})^2$, with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$. Because $x,y > 0$, \\begin{align*} \\frac{1}{(1 + \\sqrt{x})^2} &\\ge \\frac{1}{2+2x} \\\\ \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac{1}{2+2y} \\\\ \\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}). \\end{align*} By the AM-GM Inequality, we know that $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$. For the equality case, $\\frac{1}{1+x} = \\frac{1}{1+y}$, so $x = y$. Additionally, by the AM-GM Inequality, $\\frac12 \\cdot (\\frac{x+1}{2} + \\frac{y+1}{2}) \\ge \\sqrt{\\frac{(x+1)(y+1)}{4}}$. For the equality case, $\\frac{x+1}{2} = \\frac{y+1}{2}$, so $x = y$. Because $x,y \\ge 0$, \\begin{align*} \\frac12 \\cdot (\\frac{x+y+2}{2}) &\\ge \\frac{\\sqrt{(x+1)(y+1)}}{2} \\\\ \\frac{x+y+2}{2} &\\ge \\sqrt{(x+1)(y+1)} \\\\ \\sqrt{\\frac{1}{(x+1)(y+1)}} &\\ge \\frac{2}{x+y+2}. \\end{align*} Therefore, since $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y})$ and $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$ and $\\sqrt{\\frac{1}{(x+1)(y+1)}} \\ge \\frac{2}{x+y+2}$, we must have $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac{2}{x+y+2}$, with equality happening when $x =", "x_n \\log_2 z_n \\end{align*} Then the sum becomes: \\begin{align*} S &= \\sum_{n=1}^{\\infty} \\log_2 x_n \\log_2 y_n \\log_2 z_n \\\\ S &= \\sum_{n=1}^{\\infty} \\frac{n^3}{8^n} \\end{align*} The final step is to take iterative differences, like so: \\begin{align*} S &= \\frac{1}{8} + \\frac{8}{64} + \\frac{27}{512} + \\frac{64}{4096} + \\cdots \\\\ \\frac{1}{8}S &= \\quad \\; \\; \\; \\, \\frac{1}{64} + \\frac{8}{512} + \\frac{27}{4096} + \\cdots \\\\ \\Rightarrow \\frac{7}{8}S &= \\frac{1}{8} + \\frac{7}{64} + \\frac{19}{512} + \\frac{37}{4096} + \\cdots \\\\ \\frac{7}{64}S &= \\quad \\; \\; \\; \\, \\frac{1}{64} + \\frac{7}{512} + \\frac{19}{4096} + \\cdots \\\\ \\Rightarrow \\frac{49}{64}S &= \\frac{1}{8} + \\frac{6}{64} + \\frac{12}{512} + \\frac{18}{4096} + \\cdots \\\\ \\frac{49}{512}S &= \\quad \\; \\; \\; \\, \\frac{1}{64} + \\frac{6}{512} + \\frac{12}{4096} + \\cdots \\\\ \\Rightarrow \\frac{343}{512}S &= \\frac{1}{8} + \\frac{5}{64} + \\frac{6}{512} + \\frac{6}{4096} + \\cdots \\end{align*} Almost done. Now that the numerators are constants $(6)$ instead of cubics $(n^3)$, we can apply the formula for the sum of an infinite geometric series to get: \\begin{align*} \\frac{343}{512}S &= \\frac{1}{8} + \\frac{5}{64} + \\frac{6}{64 \\cdot 7} \\\\ \\frac{343}{512}S &= \\frac{97}{64 \\cdot 7} \\\\ S &= \\frac{776}{2401} \\end{align*} Then our answer is $\\boxed{776}$.", "to $$\\frac{\\tan t}{1+\\frac 2{\\pi} \\tan t}< t \\text{ for } t \\in \\left(0,\\frac{\\pi}{2}\\right)$$ or after rearranging $$\\tan t < \\frac t{1-\\frac 2{\\pi}t} \\text{ for } t \\in \\left(0,\\frac{\\pi}{2}\\right)$$ But this is true because for $t \\in ... 6 Note that $$(x+y)^3 = x^3+y^3+3xy(x+y).$$ Using the AM-GM inequality, we have $$xy\\sqrt[3]\\frac{x^3+y^3}{2} = \\frac{1}{x+y} \\cdot \\sqrt[3]{\\frac12 \\cdot \\left[xy(x+y)\\right]^3 \\cdot (x^3+y^3)}$$ $$\\leqslant \\frac{1}{x+y} \\cdot \\sqrt[3]{\\frac{1}{2}\\left(\\frac{3 xy(x+y) + x^3+y^3}{4}\\right)^4} = \\frac{(x+y)^3}{8}.$$ 5 Fix$n\\in\\Bbb N$. The claim is trivial in the case$k=1$, so by induction assume the claim is true for some$k<n. Then we get: \\begin{align*} \\left(1+\\frac{1}{n}\\right)^{k+1}&<\\left(1+\\frac{k}{n}+\\frac{k^2}{n^2}\\right)\\left(1+\\frac{1}{n}\\right)\\\\ &=1+\\frac{k}{n}+\\frac{k^2}{n^2}+\\frac{1}{n}+\\frac{k}{n^2}+\\frac{k^2}{n^3}\\\\ &\\overset{k<n}{&... 5 It is true that this falls under the realm of probability theory, because ifX$is a real valued random variable with density$f(x)$i.e.$P(X \\in (a,b)) = \\int_a^b f(x)dx$for all$a<b$, then :$\\int_{-\\infty}^{\\infty} f(x)dx = 1$and$f(x) \\geq 0$assert that$f$is a well-defined density and$X$a well defined random variable.$\\int_{-\\infty}^\\infty ... 4 lower bound: Notice that $$\\cos ^2 \\theta_1=\\sin^2 \\theta_2+\\sin^2 \\theta_3+\\cdots+\\sin^2 \\theta_{10}$$ similarly $$\\cos^2 \\theta_i=\\sum_{j=1 \\rightarrow 10 , j\\neq i }\\sin^2 \\theta_j$$ $$S_i=\\sum_{j=1\\rightarrow 10,j\\neq i} \\sin \\theta_i \\tag {say}$$ thus by power mean inequality $$\\cos \\theta_i\\ge \\sqrt{\\frac{S_i^2}{9}}=\\frac{S_i}{3}$$ Hence $$\\dfrac{\\cos ... 4 This is a remark on Dr. Mathva's answer, which may seem very mysterious at first but becomes more transparent when you consider a simpler problem: If a, b \\geq 0 with a^2 + b^2 = 1, prove that$$ ab+1" ]

[ "you. But what I said is still true, you can just find the hessian matrix. – Git Gud Apr 2 '14 at 11:29 $xy^2-x^2-x=x\\cdot(y^2-x-1)$, so we can make estimates based on whether $x>0$ or $x<0$: \\begin{align}x\\cdot(y^2-x-1)&\\ge\\min(\\sqrt2\\cdot(0-\\sqrt2-1),\\,-\\sqrt2\\cdot(2+\\sqrt2-1))=-2-\\sqrt2\\\\\\\\x\\cdot(y^2-x-1)&\\le\\max(x\\cdot((2-x^2)-x-1),\\,x\\cdot(0-x-1))\\\\&=\\max(-x^3-x^2+x,\\,-x^2-x)\\end{align} The minimum estimate is clear, for the maximum take the derivatives: \\begin{align}(-3x^2-x^2+x)'&=-3x^2-2x+1=(1-3x)(1+x)\\\\(-x^2-x)'&=-2x-1\\end{align} The first is $0$ when $x=\\frac13$ or $x=-1$, the second is $0$ when $x=-\\frac12$. Checking the local extremes and ends of the interval $[-\\sqrt2,\\sqrt2]$ you should get that the maximum of the second estimate higher than the maximum of the first and it's only for $x=-\\frac12$, so $$x\\cdot(y^2-x-1)\\le-(-\\!\\tfrac12)^2-(-\\!\\tfrac12)=\\tfrac14$$ So the minimum is $-2-\\sqrt2$, because it's attained for $(x,y)=(\\sqrt2,0)$ and the maximum is $\\frac14$, because it's attained for $(x,y)=(-\\!\\frac12,0)$. @JakeMitch I'm not using any advanced theory. Just simple estimates. If $x^2+y^2\\le2$, then $-\\sqrt2\\le x,y\\le\\sqrt2$. So for $x>0$ we can minimize $x\\cdot(y^2-x-1)$ by setting $x$ before the dot to be the greatest possible and $y^2-x-1$ after the dot to be least possible. Similarly the other estimate if $x<0$, then $x$ is least when $x=-\\sqrt2$ and $y^2-x-1$ is greatest when $x=-\\sqrt2$ and $y=\\sqrt2$. – user2345215 Apr 2 '14 at 12:51", "be proven for all values of n in a variety of ways, including induction. Proofs are given on the wikipedia page. So, what my calculus student did was apply the AM-GM inequality to the case $n = 2, a = 3x$, and $b = 75/x$, which gives $15 = \\sqrt{3x \\cdot (75/x)} \\leq (3x + 75/x)/2$, or multiplying by 2: $30 \\leq 3x + 75/x$ with equality if and only if $3x = 75/x$ or $x = 5.$ Intrigued by this, I invented some more examples of optimizing with the AM-GM inequality. Problem 1 Find the minimum value of $f(x) = x^3 + \\frac{16}{x} + \\frac{32}{x^2}$ for $x > 0$. Solution By the AM-GM inequality, we have $x^3 + \\frac{16}{x} + \\frac{32}{x^2} \\geq 3 \\cdot \\sqrt[3]{x^3 \\cdot \\frac{16}{x} \\cdot \\frac{32}{x^2}} = 24$ with equality if and only if $x^3 = \\frac{16}{x} = \\frac{32}{x^2} =8$, which is achieved when $x = 2.$ So the minimum is 24. That sure beats having to find the critical numbers for that function. Problem 2 Find the minimum value of $f(x,y) = xy + \\frac{8}{x} + \\frac{1}{y}$ for $x > 0, y >0.$ Solution $xy + \\frac{8}{x} + \\frac{1}{y} \\geq 3 \\cdot \\sqrt[3]{xy \\cdot \\frac{8}{x} \\cdot \\frac{1}{y}} = 6$ with equality if and only if $xy = \\frac{8}{x} = \\frac{1}{y}=2$, which is achieved", "for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. It can be rewritten as: {\\displaystyle {\\begin{aligned}f(x,y,z)&=6\\cdot {\\frac {{\\frac {x}{y}}+{\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}+{\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}+{\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}}{6}}\\\\&=6\\cdot {\\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\\end{aligned}}} with ${\\displaystyle x_{1}={\\frac {x}{y}},\\qquad x_{2}=x_{3}={\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}},\\qquad x_{4}=x_{5}=x_{6}={\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}.}$ Applying the AM–GM inequality for n = 6, we get {\\displaystyle {\\begin{aligned}f(x,y,z)&\\geq 6\\cdot {\\sqrt[{6}]{{\\frac {x}{y}}\\cdot {\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}\\cdot {\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}\\cdot {\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}}}\\\\&=6\\cdot {\\sqrt[{6}]{{\\frac {1}{2\\cdot 2\\cdot 3\\cdot 3\\cdot 3}}{\\frac {x}{y}}{\\frac {y}{z}}{\\frac {z}{x}}}}\\\\&=2^{2/3}\\cdot 3^{1/2}.\\end{aligned}}} Further, we know that the two sides are equal exactly when all the terms of the mean are equal: ${\\displaystyle f(x,y,z)=2^{2/3}\\cdot 3^{1/2}\\quad {\\mbox{when}}\\quad {\\frac {x}{y}}={\\frac {1}{2}}{\\sqrt {\\frac {y}{z}}}={\\frac {1}{3}}{\\sqrt[{3}]{\\frac {z}{x}}}.}$ All the points (x, y, z) satisfying these conditions lie on a half-line starting at the origin and are given by ${\\displaystyle (x,y,z)={\\biggr (}t,{\\sqrt[{3}]{2}}{\\sqrt {3}}\\,t,{\\frac {3{\\sqrt {3}}}{2}}\\,t{\\biggr )}\\quad {\\mbox{with}}\\quad t>0.}$ ## Practical applications An important practical application in financial mathematics is to computing the rate of return: the annualized return, computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect. ## Proofs of", "# The minimum value of $(\\frac{1}{x}-1)(\\frac{1}{y}-1)(\\frac{1}{z}-1)$ if $x+y+z=1$ $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible value of $(\\frac{1}{x}-1) (\\frac{1}{y}-1) (\\frac{1}{z}-1)$ is ? The options are: $1,4,8$ or $16$ Approach: \\begin{align*} \\left(\\frac{1}{x} -1\\right)\\left(\\frac{1}{y}-1\\right)\\left(\\frac{1}{z}-1\\right)&=\\frac{(1-x)(1-y)(1-z)}{xyz}\\\\ &=\\frac{1-(x+y+z)+(xy+yz+zx)-xyz}{xyz}\\\\ &=\\frac{1-1+(xy+yz+zx)-xyz}{xyz}\\\\ &=\\frac{xy+yz+zx}{xyz} - 1 \\end{align*} Now by applying $AM≥HM$, I got the least value of $(xy+yz+zx)/xyz$ as $9$, so I got final answer as $8$. Is it correct? - Is \"$(1/x-1)$\" supposed to be $\\frac{1}{x-1}$, or $\\frac{1}{x}-1$? – Arturo Magidin Jul 21 '12 at 4:13 The latter one sir. – Hyperbola Jul 21 '12 at 4:14 It's minimum is clearly $8$ (achieved when $x = y= z = \\frac{1}{3}$), as you have shown. But that's it. It can attain any value $\\geq 8$. – sos440 Jul 21 '12 at 4:26 Now the problem does make sense. – sos440 Jul 21 '12 at 4:27 There is no minimum possible value. The minimum $8$ is only achieved when $x=y=z$, which is ruled out (distinct positive reals). You can, however, get arbitrarily close to $8$. – Potato Jul 21 '12 at 5:19 If we put no constraint on $x$, $y$, and $z$ apart from $x$, $y$, $z$ positive and $x+y+z=1$, then indeed your calculation, and the one by Patrick Da Silva, show that the minimum value is $8$, attained at", "# 2008 Indonesia MO Problems/Problem 2 ## Problem Prove that for every positive reals $x$ and $y$, $$\\frac {1}{(1 + \\sqrt {x})^{2}} + \\frac {1}{(1 + \\sqrt {y})^{2}} \\ge \\frac {2}{x + y + 2}.$$ ## Solution 1 By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \\ge (1 + \\sqrt{x})^2$ and $(1+1)(1+y) \\ge (1 + \\sqrt{y})^2$, with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$. Because $x,y > 0$, \\begin{align*} \\frac{1}{(1 + \\sqrt{x})^2} &\\ge \\frac{1}{2+2x} \\\\ \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac{1}{2+2y} \\\\ \\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}). \\end{align*} By the AM-GM Inequality, we know that $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$. For the equality case, $\\frac{1}{1+x} = \\frac{1}{1+y}$, so $x = y$. Additionally, by the AM-GM Inequality, $\\frac12 \\cdot (\\frac{x+1}{2} + \\frac{y+1}{2}) \\ge \\sqrt{\\frac{(x+1)(y+1)}{4}}$. For the equality case, $\\frac{x+1}{2} = \\frac{y+1}{2}$, so $x = y$. Because $x,y \\ge 0$, \\begin{align*} \\frac12 \\cdot (\\frac{x+y+2}{2}) &\\ge \\frac{\\sqrt{(x+1)(y+1)}}{2} \\\\ \\frac{x+y+2}{2} &\\ge \\sqrt{(x+1)(y+1)} \\\\ \\sqrt{\\frac{1}{(x+1)(y+1)}} &\\ge \\frac{2}{x+y+2}. \\end{align*} Therefore, since $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y})$ and $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$ and $\\sqrt{\\frac{1}{(x+1)(y+1)}} \\ge \\frac{2}{x+y+2}$, we must have $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac{2}{x+y+2}$, with equality happening when $x =", "[0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1 \\end{align*} and \\begin{align*} g(a_1) = &\\min_{x_1, x_2, \\cdots, x_m} \\ x_1^2 + x_2^2 + \\cdots + x_m^2\\\\ &\\mathrm{s.t.}\\quad x_i\\in [0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1. \\end{align*} It is easy to obtain $$g(a_1) = \\frac{a_1^2}{m}$$ (the minimum is attained always when $$x_1=x_2 = \\cdots = x_m = \\frac{a_1}{m}\\in [0, \\frac{1}{4}]$$). We have $$f(a_1) = \\frac{1}{16}\\lfloor 4a_1\\rfloor + \\left(a_1 - \\frac{1}{4}\\lfloor 4a_1\\rfloor\\right)^2$$ since $$h(x) = x_1^2 + x_2^2 + \\cdots + x_n^2$$ is convex, and (denote $$K = \\lfloor 4a_1\\rfloor$$) $$\\big(\\underbrace{\\frac{1}{4}, \\frac{1}{4}, \\cdots, \\frac{1}{4}}_{K}, a_1 - \\frac{1}{4}K, 0, 0, \\cdots, 0\\big)$$ majorizes $$(x_1, x_2, \\cdots, x_m)$$ with $$x_i\\in [0, \\frac{1}{4}], \\forall i; \\ x_1 + x_2 + \\cdots + x_m = a_1$$. iii) Clearly, (3) is true. (the \"if\" part): It is clear according to the (the \"only if\" part). This completes the proof of Lemma 1. Let us proceed. We want to find the explicit expression of $$F(a_1, a_2)$$. Let $$(x_1, \\ x_2, \\cdots, x_m)$$ be the global maximizer of the optimization problem given in (4). We claim that $$|\\{x_1, \\ x_2, \\cdots, x_m\\}|\\le 4$$ where $$|\\cdot|$$ is the cardinality of a set. Proof of the claim: Assume, for the sake of contradiction, that $$|\\{x_1, \\ x_2,", "# 1988 AIME Problems/Problem 8 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: $$f(x, x) = x,\\; f(x, y) = f(y, x), {\\rm \\ and\\ } (x+y)f(x, y) = yf(x, x+y).$$ Calculate $f(14,52)$. ## Solution 1 (Algebra) Let $z = x+y$. By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \\begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\\\ f(x,z) &= \\frac{z}{z-x} \\cdot f(x,z-x). \\end{align*} Using the properties of $f,$ we have \\begin{align*} f(14,52) &= \\frac{52}{38} \\cdot f(14,38) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot f(14,24) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(14,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(10,14)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(10,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(4,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot f(4,6)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(4,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(2,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot \\frac{4}{2} \\cdot f(2,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6}", "it follows that $3n^2$ is odd. But if $3n^2$ is odd and $\\frac{m^2}{4}$ is even then we must have $\\frac{m^2}{4} \\neq 3n^2$ Therefore, there is no rational whose square is $12$. Prove Proposition 1.15. (a) We have \\begin{align} y = 1\\cdot y = \\left(\\frac{1}{x} \\cdot x\\right) \\cdot y = \\frac{1}{x} \\cdot (xy) = \\frac{1}{x} \\cdot(xz) = \\left(\\frac{1}{x} \\cdot x \\right) \\cdot z = 1 \\cdot z = z \\end{align} (b) If $xy = x$ then $xy = x \\cdot 1$, so by (a) we have $y=1$. (c) If $xy = 1$ then \\begin{align} y = 1 \\cdot y = \\left(\\frac{1}{x} \\cdot x \\right) \\cdot y = \\frac{1}{x} \\cdot (xy) = \\frac{1}{x} \\cdot 1 = \\frac{1}{x} \\end{align} (d) We have \\begin{align} \\dfrac{1}{(1/x)} = 1 \\cdot \\dfrac{1}{(1/x)} = \\left( \\frac{1}{x} \\cdot x \\right) \\cdot \\dfrac{1}{(1/x)} = \\left(\\frac{1}{x} \\cdot \\dfrac{1}{(1/x)} \\right) \\cdot x = 1 \\cdot x = x \\end{align} Let $E$ be a nonempty subset of an ordered set; suppose $\\alpha$ is a lower bound of $E$ and $\\beta$ is an upper bound of $E$. Prove that $\\alpha \\leq \\beta$. \\ Because $E$ is nonempty, there exists $x \\in E$. Since $\\alpha$ is a lower bound of $E$, we have $\\alpha \\leq x$. Since $\\beta$ is an upper bound of $E$, we have $x \\leq \\beta$. Therefore $\\alpha \\leq x", "# An Inequality for Cevians And The Ratio of Circumradius and Inradius ### Solution By the Law of Cosines, in $\\Delta XYC,$ \\displaystyle \\begin{align} XY^2 &= XC^2+YC^2-2\\cdot XC\\cdot YC\\cdot\\cos C\\\\ &\\ge 2\\cdot XC\\cdot YC\\cdot(1-\\cos C)=4\\cdot XC\\cdot YC\\cdot\\sin^2\\frac{C}{2}. \\end{align} It follows that $\\displaystyle XY\\ge 2\\sqrt{XC\\cdot YC}\\cdot\\sin\\frac{C}{2}.\\,$ Similarly, $\\displaystyle YZ\\ge 2\\sqrt{YA\\cdot ZA}\\cdot\\sin\\frac{A}{2}\\,$ and $\\displaystyle ZX\\ge 2\\sqrt{XB\\cdot ZB}\\cdot\\sin\\frac{B}{2}.\\,$ Multiplying the three we get $\\displaystyle \\small{XY\\cdot YZ\\cdot ZX\\ge 8\\sqrt{XC\\cdot XB\\cdot YC\\cdot YA\\cdot ZA\\cdot ZB}\\cdot\\sin\\frac{A}{2}\\cdot\\sin\\frac{B}{2}\\cdot\\sin\\frac{C}{2}}.$ Thus, $\\displaystyle\\small{\\frac{XB}{XY}\\cdot\\frac{YC}{YZ}\\cdot \\frac{ZA}{ZV}\\le\\sqrt{\\frac{XB}{XC}\\cdot \\frac{YC}{YA}\\cdot \\frac{ZA}{ZB}}\\cdot \\frac{\\displaystyle 1}{8\\cdot\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}}=\\frac{R}{2r}}$ because, by Ceva's theorem, $g=\\displaystyle \\frac{XB}{XC}\\cdot \\frac{YC}{YA}\\cdot \\frac{ZA}ZB.$ ### Acknowledgment The problem by Titu Andreescu has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook group and commented with a solution (above) by Marian Dinca. I am grateful to Leo Giugiuc for bringing this problem to my attention. Copyright © 1996-2018 Alexander Bogomolny 66524280", "# 2008 Indonesia MO Problems/Problem 2 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Prove that for every positive reals $x$ and $y$, $$\\frac {1}{(1 + \\sqrt {x})^{2}} + \\frac {1}{(1 + \\sqrt {y})^{2}} \\ge \\frac {2}{x + y + 2}.$$ ## Solution By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \\ge (1 + \\sqrt{x})^2$ and $(1+1)(1+y) \\ge (1 + \\sqrt{y})^2$, with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$. Because $x,y > 0$, \\begin{align*} \\frac{1}{(1 + \\sqrt{x})^2} &\\ge \\frac{1}{2+2x} \\\\ \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac{1}{2+2y} \\\\ \\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} &\\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}). \\end{align*} By the AM-GM Inequality, we know that $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$. For the equality case, $\\frac{1}{1+x} = \\frac{1}{1+y}$, so $x = y$. Additionally, by the AM-GM Inequality, $\\frac12 \\cdot (\\frac{x+1}{2} + \\frac{y+1}{2}) \\ge \\sqrt{\\frac{(x+1)(y+1)}{4}}$. For the equality case, $\\frac{x+1}{2} = \\frac{y+1}{2}$, so $x = y$. Because $x,y \\ge 0$, \\begin{align*} \\frac12 \\cdot (\\frac{x+y+2}{2}) &\\ge \\frac{\\sqrt{(x+1)(y+1)}}{2} \\\\ \\frac{x+y+2}{2} &\\ge \\sqrt{(x+1)(y+1)} \\\\ \\sqrt{\\frac{1}{(x+1)(y+1)}} &\\ge \\frac{2}{x+y+2}. \\end{align*} Therefore, since $\\frac{1}{(1 + \\sqrt{x})^2} + \\frac{1}{(1 + \\sqrt{y})^2} \\ge \\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y})$ and $\\frac12 \\cdot (\\frac{1}{1+x} + \\frac{1}{1+y}) \\ge \\sqrt{\\frac{1}{(1+x)(1+y)}}$ and $\\sqrt{\\frac{1}{(x+1)(y+1)}} \\ge \\frac{2}{x+y+2}$, we must have $\\frac{1}{(1 + \\sqrt{x})^2}", "# Minimum Value of expression 2 What is the minimum value of $$(2x+3y)(8/x + 3/y)$$ when $x$ and $y$ are positive? I have tried expanding the equation but I did not get anything I got : $$25 + \\frac{6(4y^2 +x^2)}{xy}$$ but I don't know how to answer after this. I have also tried AM-GM but I couldn't manipulate the variables to cancel out and get the inequality. This is from Sipnayan 2017 You were on the right track: \\begin{align}(2x+3y)\\left(\\frac{8}{x} + \\frac{3}{y}\\right) &= 25+24 \\frac{y}{x}+6\\frac{x}{y}\\\\ &=25+12\\left(\\frac{2y}{x}+\\frac{x}{2y}\\right) \\\\ &\\geq 25+12 \\times 2 \\\\ &=49\\end{align} and notice that $x=2,y=1$ achieves this bound, so the minimum value is $49$. The bound $\\frac{2y}{x} + \\frac{x}{2y} \\geq 2$ can be proved from AM-GM, or from letting $t = \\frac{2y}{x}$ and considering the function $f(t) = t + \\frac{1}{t}$ and using calculus. From where you have left of, For real $x-2y,$ $$(x-2y)^2\\ge0\\implies x^2+4y^2\\ge4xy$$ Alternatively, using AM-GM inequality, $$\\dfrac{x^2+4y^2}2\\ge\\sqrt{4x^2y^2}=?$$ By C-S $$(2x+3y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right)\\geq\\left(\\sqrt{2x\\cdot\\frac{8}{x}}+\\sqrt{3y\\cdot\\frac{3}{y}}\\right)^2=49.$$ The equality occurs for $x=2$ and $y=1$, which says that we got a minimal value. Done!", "\\leq 1 \\\\ & 0 \\leq c \\leq d \\leq e \\leq 1 \\end{align} For fixed $c, d, e$, $h$ is minimized when $a$ and $b$ are maximized, i.e., $a = \\frac{3-2d}{3-2c}$ and $b = \\frac{3-2e}{3-2c}$. Then $h$ becomes $$h(c, d, e) = \\frac{27 - 3c - 6d - 9e}{27 - 4c - 6d - 8e}$$ same as $f(c, d, e)$. Thus the minimum value is $\\frac{12}{13}$. Summary: the greatest $k$ is $\\frac{12}{13}$. Yes, $\\color{red}{k=\\frac 17}$ works. First, $$\\frac{3}{3-2c}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3\\cdot 1+4\\cdot 1}=\\frac 13$$ Second, when $$\\frac{3a}{3-2d}\\ge \\frac{3}{3-2c}\\implies a\\ge\\frac{3-2d}{3-2c}\\ge \\frac{3-2\\cdot 1}{3-2\\cdot 0}=\\frac 13$$ we have $$\\frac{3a}{3-2d}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3a}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3a+4\\cdot 1}=1-\\frac{2}{a+2}$$$$\\ge 1-\\frac{2}{\\frac 13+2}=\\frac 17$$ Third, when $$\\frac{3b}{3-2e}\\ge\\frac{3}{3-2c}\\implies b\\ge \\frac{3-2e}{3-2c}\\ge \\frac{3-2\\cdot 1}{3-2\\cdot 0}=\\frac 13$$ we have $$\\frac{3b}{3-2e}\\cdot\\frac{9-c-2d-3e}{2+3a+4b}\\ge \\frac{3b}{3-2\\cdot 0}\\cdot\\frac{9-1-2\\cdot 1-3\\cdot 1}{2+3\\cdot 1+4\\cdot b}=\\frac 34-\\frac{15}{4(4b+5)}$$$$\\ge \\frac 34-\\frac{15}{4(4\\cdot \\frac 13+5)}=\\frac{3}{19}$$ It follows from these that $$\\max\\left(\\frac{3}{3-2c},\\frac{3a}{3-2d},\\frac{3b}{3-2e}\\right)\\geq \\color{red}{\\frac 17}\\cdot\\frac{2+3a+4b}{9-c-2d-3e}$$ holds for any $0\\leq b\\leq a\\leq 1$ and $0\\leq c\\leq d\\leq e\\leq 1$. • Thanks. In fact, a better bound of $k=1/3$ follows from your first inequality, so the rest is not needed. The natural question would then be to ask for the best such constant $k$. I've edited the question. – pi66 Jul 13 '16 at 9:47 • @pi66: Ah, you are right. I've misunderstood something. – mathlove Jul 13 '16 at 10:28", "following your hint, let $a = \\frac{1}{x}, b = \\frac{1}{y}, c = \\frac{1}{z}$. The expression is then $\\sum_{cyc} \\frac{x^3yz}{y+z}$. Since $xyz = 1$, this is $\\sum_{cyc} \\frac{x^2}{y+z}$. Using Cauchy, this is greater than $\\frac{(x+y+z)^2}{2(x+y+z)}$. So now it is left to prove that $x+y+z \\geq 3$. This is simple using AM-GM, so $\\frac{x+y+z}{3} \\geq \\sqrt[3]{xyz}$, and we are done. $\\blacksquare$ For the second one, WLOG assume $a_1$ is the maximum and $a_n$ is the minimum. Using Cauchy, $(a_1 + a_2 + \\cdots + a_n)(\\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_n}) \\geq (\\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2)$ $(n + \\frac{1}{2})^2 \\geq (\\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2)^2$ Taking square roots of both sides, we get $n + \\frac{1}{2} \\geq \\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2$ $\\frac{5}{2} \\geq \\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}}$ Now, squaring, we get $\\frac{25}{4} \\geq \\frac{a_1}{a_n} + 2 + \\frac{a_n}{a_1}$ Multiplying both sides by $a_1 a_n$ we get $a_1^2 - \\frac{17}{4} a_1 a_n + a_n^2 \\leq 0$ Factoring... $(a_1 - 4a_n)(a_1 - \\frac{1}{4}a_n) \\leq 0$ Since $a_1$ and $a_n$ are positive reals, $(a_1 - 4a_n)$ must be nonpositive. Thus, $(a_1 - 4a_n \\leq 0 \\rightarrow a_1 \\leq 4a_n$, and we are done. - 6 years, 8 months ago Could you explain the second one as in how did you use Cauchy Schwartz in it. -", "$x=y=z=\\frac{1}{3}$. However, the problem specifies that $x$, $y$ and $z$ are distinct real numbers. If we take that constraint into account, there is no minimum. We can get arbitrarily close to $8$ (but above $8$) by choosing $x$, $y$, and $z$ distinct and close to $\\frac{1}{3}$, but we cannot attain $8$ with distinct $x$, $y$ and $z$. - Try $x = \\frac{1}{2}, \\; y = \\frac{1}{3}, \\; z = \\frac{1}{6}.$ Your product is $$(2-1)(3-1)(6-1) = 1 \\cdot 2 \\cdot 5 = 10.$$ Next, try Try $x = \\frac{4}{7}, \\; y = \\frac{2}{7}, \\; z = \\frac{1}{7}.$ Your product is $$(\\frac{7}{4}-1)(\\frac{7}{2}-1)(7-1) = \\frac{3}{4} \\cdot \\frac{5}{2} \\cdot 6 = \\frac{45}{4}.$$ Take $x = \\frac{10}{15}, \\; y = \\frac{3}{15}, \\; z = \\frac{2}{15}.$ Your product is $13.$ Take $x = \\frac{6}{9}, \\; y = \\frac{2}{9}, \\; z = \\frac{1}{9}.$ Your product is $14.$ Take $x = \\frac{15}{20}, \\; y = \\frac{3}{20}, \\; z = \\frac{2}{20}.$ Your product is $17.$ Take $x = \\frac{14}{21}, \\; y = \\frac{6}{21}, \\; z = \\frac{1}{21}.$ Your product is $25.$ Take $x = \\frac{165}{252}, \\; y = \\frac{77}{252}, \\; z = \\frac{10}{252}.$ Your product is $29.$ Take $x = \\frac{21}{28}, \\; y = \\frac{6}{28}, \\; z = \\frac{1}{28}.$ Your product is $33.$ Take $x = \\frac{65}{78}, \\; y = \\frac{10}{78}, \\; z = \\frac{3}{78}.$ Your product", "# Find the values of p so the line Question: Find the values of $p$ so the line $\\frac{1-x}{3}=\\frac{7 y-14}{2 p}=\\frac{z-3}{2}$ and $\\frac{7-7 x}{3 p}=\\frac{y-5}{1}=\\frac{6-z}{5}$ are at right angles. Solution: The given equations can be written in the standard form as $\\frac{x-1}{-3}=\\frac{y-2}{\\frac{2 p}{7}}=\\frac{z-3}{2}$ and $\\frac{x-1}{\\frac{-3 p}{7}}=\\frac{y-5}{1}=\\frac{z-6}{-5}$ The direction ratios of the lines are $-3, \\frac{2 p}{7}, 2$ and $\\frac{-3 p}{7}, 1,-5$ respectively. Two lines with direction ratios, $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$, are perpendicular to each other, if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ $\\therefore(-3) \\cdot\\left(\\frac{-3 p}{7}\\right)+\\left(\\frac{2 p}{7}\\right) \\cdot(1)+2 \\cdot(-5)=0$ $\\Rightarrow \\frac{9 p}{7}+\\frac{2 p}{7}=10$ $\\Rightarrow 11 p=70$ $\\Rightarrow p=\\frac{70}{11}$ Thus, the value of $p$ is $\\frac{70}{11}$.", "# How to find the minimum value of $\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$? Let $x,y,z\\in [1,4]$ such that $x \\geq y$ and $x \\geq z$. Find the minimum value of this expression: $$P=\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$$ - Since $x \\not= 0$, denote $y= u x$ and $z=v x$. You are, then, looking for minimum in $\\frac{1}{4} \\le u \\le 1$ and $\\frac{1}{4} \\le v \\le 1$. – Sasha Aug 26 '11 at 6:06 @Sasha: following to your suggestion, we get $P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$ What then shall we do? – KevinBui Aug 26 '11 at 9:48 or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply second partial derivative test – newbie Aug 26 '11 at 10:54 ## 2 Answers As has been mentioned in comments, we want to find the minimum value of $$P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$$ for $u,v\\in[\\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$: \\begin{align} \\frac{\\partial P}{\\partial u}&=-\\frac{3}{(2+3u)^2}+\\frac{v}{(u+v)^2}\\\\ \\frac{\\partial P}{\\partial v}&=-\\frac{u}{(u+v)^2}+\\frac{1}{(v+1)^2} \\end{align} To find an interior extremum, we need $\\frac{\\partial P}{\\partial u}=\\frac{\\partial P}{\\partial v}=0$. In that case, we need $$0=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}=-\\frac{3u}{(2+3u)^2}+\\frac{v}{(v+1)^2}$$ However, for $u,v\\in[\\frac{1}{4},1]$, we have $$\\begin{array}{c} \\frac{12}{121}\\le\\frac{3u}{(2+3u)^2}\\le\\frac{3}{25}\\\\ \\frac{4}{25}\\le\\frac{v}{(v+1)^2}\\le\\frac{1}{4} \\end{array}$$ Thus, $u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$, so there can be no interior extremum. Because $(u,v)\\cdot\\nabla P=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$ everywhere in $[\\frac{1}{4},1]\\times[\\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of", "# Calculate the minimum value of $\\frac{x^2 + 4}{y^2 + 1}$ where $1 \\le y \\le 2$ and $2y \\le xy + 2$. $$x$$ and $$y$$ are reals such that $$1 \\le y \\le 2$$ and $$2y \\le xy + 2$$. Calculate the minimum value of $$\\large \\frac{x^2 + 4}{y^2 + 1}$$ This problem is adapted from a recent competition... It really is. There must be better answers. We first need to prove that $$x \\ge 0$$. It is true that $$2y \\le xy + 2 \\iff 0 \\le \\dfrac{2(y - 1)}{y} \\le x$$ (because $$1 \\le y$$). Moreover, y \\le 2 \\implies \\left\\{ \\begin{align} y^2 &\\le 2y\\\\ xy + 2 &\\le 2x + 2 \\end{align} \\right. (x, y \\ge 0) We have that $$\\frac{x^2 + 4}{y^2 + 1} \\ge \\frac{2x + 3}{2y + 1} \\ge \\frac{2x + 3}{xy + 3} \\ge \\frac{2x + 3}{2x + 3} = 1$$ The equality sign occurs when $$x = 1$$ and $$y = 2$$. Let $$x=1$$ and $$y=2$$. Thus,we get a value $$1$$. We'll prove that it's a minimal value. Indeed, since $$x\\geq\\frac{2(y-1)}{y}\\geq0,$$ we obtain: $$\\frac{x^2+4}{y^2+1}-1=\\frac{\\left(\\frac{2(y-1)}{y}\\right)^2+4}{y^2+1}-1=\\frac{(2-y)(y^3+2y^2-3y+2)}{y^2(y^2+1)}\\geq0$$ and we are done! $$x y \\ge 2y-2\\Rightarrow x\\ge 2-\\frac {2}{\\max y}\\ge 1$$ because $$y > 0$$ now $$\\frac{\\min x^2+4}{\\max y^2+1}\\le \\frac{x^2+4}{y^2+1}$$ hence $$\\frac 55 = 1 \\le \\frac{x^2+4}{y^2+1}$$", "$x - xy + y = -5$ becomes $(\\frac{\\sqrt{2}}{2}x - \\frac{\\sqrt{2}}{2}y) - (\\frac{\\sqrt{2}}{2}x - \\frac{\\sqrt{2}}{2}y) (\\frac{\\sqrt{2}}{2}x + \\frac{\\sqrt{2}}{2}y) + (\\frac{\\sqrt{2}}{2}x + \\frac{\\sqrt{2}}{2}y) = -5$ Once again this simplifies nicely to $(x' - \\sqrt{2})^2 - (y')^2 = 12$ and these two equations are solved in a fairly standard fashion via elimination i.e. $3(x' - \\sqrt{2})^2 - 3(y') ^2 = 36$ $+ (x')^2 + 3(y')^2 = 26$ $4(x')^2 - 6\\sqrt{2}x' + 6 = 62$ Solving with quadratic equation and $x' = \\frac{7\\sqrt{2}}{2}, y' = \\pm \\frac{\\sqrt{2}}{2}$ or $x' = -2\\sqrt{2}, y' = \\pm \\sqrt{6}$ If you look back at the previous solutions at first this looks odd but remember these have to be tilted back 45 degrees to give the proper solutions and sure enough for instance $(\\frac{\\sqrt{2}}{2} \\cdot \\frac{7\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2} \\cdot \\frac{7\\sqrt{2}}{2} - \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{2}}{2}) = (4,3)$ Friday, March 30, 2018 As is our routine, I started by going over the problem of the week: Find six distinct natural numbers A,B,C,D,E,F such that A + B + C = D + E + F And Once again, I had a girl write a python program to find the solution. I love all the computational computing that is occurring. I need a better way to harness this energy. This time I had her talk about", "${\\bf\\mbox{6.1.17}}$: \\begin{align} &\\color{#c00000}{\\int_{0}^{\\infty}{\\dd x \\over 1 + x^{3}}} ={1 \\over 3}\\,{\\pi \\over \\sin\\pars{\\pi/3}} ={1 \\over 3}\\,{\\pi \\over \\root{3}/2} \\end{align} $$\\color{#00f}{\\large% \\int_{0}^{\\infty}{\\dd x \\over 1 + x^{3}} = {2\\root{3} \\over 9}\\,\\pi} \\approx 1.2092$$ - Since $$\\frac{1}{x^3+1}=\\frac{1}{(x+1)(x^2-x+1)},$$ we can find some $a,b,c \\in \\mathbb{R}$ such that $$\\frac{1}{x^3+1}=\\frac{a}{x+1}+\\frac{bx+c}{x^2-x+1}.$$ A simple computation shows that $$a=-b=\\frac{c}{2}=\\frac13,$$ i.e. \\begin{eqnarray} \\frac{1}{x^3+1}&=&\\frac13\\cdot\\frac{1}{x+1}-\\frac13\\cdot\\frac{x-2}{x^2-x+1}\\\\ &=&\\frac13\\cdot\\frac{1}{x+1}-\\frac16\\cdot\\frac{2x-1}{x^2-x+1}+\\frac12\\cdot\\frac{1}{x^2-x+1}\\\\ &=&\\frac13\\cdot\\frac{1}{x+1}-\\frac16\\cdot\\frac{2x-1}{x^2-x+1}+\\frac12\\cdot\\frac{1}{\\left(x-\\frac12\\right)^2+\\frac34}. \\end{eqnarray} It follows that \\begin{eqnarray} F(r)&=&\\int_0^r\\frac{1}{x^3+1}\\,dx=\\frac13\\ln(1+r)-\\frac16\\ln(r^2-r+1)+\\frac{1}{\\sqrt{3}}\\arctan\\frac{2r-1}{\\sqrt{3}}+\\frac{1}{\\sqrt{3}}\\arctan\\frac{1}{\\sqrt{3}}\\\\ &=&\\frac16\\ln\\frac{(r+1)^2}{r^2-r+1}+\\frac{1}{\\sqrt{3}}\\arctan\\frac{2r-1}{\\sqrt{3}}+\\frac{\\pi}{6\\sqrt{3}}\\\\ &=&\\frac16\\ln\\frac{r^2+2r+1}{r^2-r+1}+\\frac{1}{\\sqrt{3}}\\arctan\\frac{2r-1}{\\sqrt{3}}+\\frac{\\pi}{6\\sqrt{3}}. \\end{eqnarray} Thus $$\\int_0^\\infty\\frac{1}{x^3+1}\\,dx=\\lim_{r\\to\\infty}F(r)=\\frac{\\pi}{2\\sqrt{3}}+\\frac{\\pi}{6\\sqrt{3}}=\\frac{2\\pi}{3\\sqrt{3}}.$$ -", "###### back to index | new Let $x$ and $y$ be two positive real numbers. Find the maximum value of $\\frac{(3x+4y)^2}{x^2 + y^2}$. Find all positive integers $n$ and $k_i$ $(1\\le i \\le n)$ such that $$k_1 + k_2 + \\cdots + k_n = 5n-4$$ and $$\\frac{1}{k_1} + \\frac{1}{k_2} + \\cdots + \\frac{1}{k_n}=1$$ Let $a$ and $b$ be two positive real numbers. Show that if $\\frac{1}{a}+\\frac{1}{b}=1$. Prove that the following inequality holds for any positive integer $n$: $$(a+b)^n-a^n-b^n\\ge 2^{2n}-2^{n+1}$$ As shown. Let $a_1, a_2,\\cdots, a_n > 0, n\\ge 2,$ and $a_1+a_2+\\cdots+a_n=1$. Prove $$\\frac{a_1}{2-a_1} + \\frac{a_2}{2-a_2}+\\cdots+\\frac{a_n}{2-a_n}\\ge\\frac{n}{2n-1}$$ Let $a, b, c$ be the lengths of the sides of triangle $ABC$. Show that $$\\sqrt{a}(c+a-b) + \\sqrt{b}(a+b-c)+\\sqrt{c}(b+c-a)\\le\\sqrt{(a^2 + b^2 + c^2)(a+b+c)}$$" ]

[ "# Rational Zeroes Find the sum of all the rational roots of the equation $x^5-4x^4+2x^3+2x^2+x+6=0.$ ×", "# Find the Sum f(x) = square root of x , g(x) = square root of x+2 , Replace the function designators with the actual functions in .", "# Rational Zeroes Algebra Level 3 Find the sum of all the rational roots of the equation $x^5-4x^4+2x^3+2x^2+x+6=0.$ ×", "$g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved. By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\\pm1$. Clearly, these are not roots. Similary, any quadratic factor would have to be $x^2+ax\\pm1$ with $a\\in\\Bbb Z$. You might be able to exclude these manually ...", "# Rational Zeroes Algebra Level 2 Find the sum of all the rational roots of the equation $x^5-4x^4+2x^3+2x^2+x+6=0.$ ×", "# Does that sound rational to you? Find the sum of all integers $n,$ such that the equation $nx^3-(n+2)x^2-nx+(n-8)=0$ has a rational solution. ×", "Determine the sum of the cubes of the roots of the polynomial $x^6 – 4x^5 + 12x^4 + 23x^2 – 2x + 125 = 0$.", "# The $$x^\\text{th}$$ root of $$x$$ Let $$f: \\mathbb Q^+ \\to \\mathbb R^+$$, we define $$f(x) = \\sqrt[x]{x}$$. Let $$x$$ and $$y$$ be positive rational numbers such that $$x<y$$ and $$f(x) = f(y)$$. Find the sum of all possible values of $$x$$. ×", "# Find g(f(x)) f(x)=9x+4 , g(x) = square root of x , Set up the composite result function. Evaluate by substituting in the value of into . Remove parentheses. Find g(f(x)) f(x)=9x+4 , g(x) = square root of x", "= x^6 - 6x^3 - 2$ are $1, -1, 2,$ and $-2$. We find $P(1) = -7$, $P(-1)=-5$, $P(2)=14$ and $P(-2)=110$ Hence $P(x) = x^6 - 6x^3 - 2$ has no rational roots. It follows that $x = (3 + \\sqrt{11})^{\\frac 13}$ is irrational.", "# Lonely Quartic Rational Root Algebra Level 4 Prove that $f(x) = 10x^4 + 43x^3 - 68x^2 - 5x +33$ has exactly one rational root $\\alpha$. Fnd $100 \\alpha$. ×", "Problems Section: Problems 1121 - 1130 Q1121. Prove that $g(x) = 30x3 + 80x2 + 72x + 66$ is irreducible over the integers by showing it has no roots modulo $13$. What are the roots of $g(x)$ modulo $7$?", "# Thread: Factoring/Finding What X is 1. ## Factoring/Finding What X is Can someone help me with this. 432x -48x^2 + 4x^3 = 600 Can you either factor it or find the possibilities of X? 2. Originally Posted by bherrell2 Can someone help me with this. 432x -48x^2 + 4x^3 = 600 Can you either factor it or find the possibilities of X? Hi bherrell2, $432x-48x^2+4x^3=600$ Let's line everything up and simplify the equation. $4x^3-48x^2+432x-600=0$ $x^3-12x^2+108x-150=0$ This thing has no rational roots. It has one real irrational root around 1.6497 and two imaginary roots.", "# Solutions to Problems 1121–1130 Q1121. Prove that $g(x) = 30x3 + 80x2 + 72x + 66$ is irreducible over the integers by showing it has no roots modulo $13$. What are the roots of $g(x)$ modulo $7$?", "# The sum of three numbers in G.P. is $56$. If we subtract $1,7,21$ from the numbers in that order, we obtain an A.P. Find the numbers. $\\begin{array}{1 1} -8,16,-32 \\\\ 4,8,16 \\\\ 8,16,32 \\\\ 16,32,64 \\end{array}$", "# Product of Values on Roots Algebra Level 5 Denote by $x_1,\\ x_2,\\ x_3,\\ x_4$ the roots of the polynomial $f(x)=x^4-4x^3-12x^2+1$. Denote $g(x)=x^2+2$. Find the value of the product $g(x_1)g(x_2)g(x_3)g(x_4) .$ ×", "2, -2, 4, -4, 8, -8, 16, or -16. Put each of those into $-x^3+ 5x^2-14x+ 16$ to see if they make it 0. If none of them do then there are no rational roots and this cannot be factored with integer coefficients. If one does work, divide by (x- that root) to find the other factor.", "# f(f(x)) = x Algebra Level 3 Find the sum of all roots of the following equation. $(x^2 -3x +3)^2 - 3(x^2 -3x +3) = (x-3)$ ×", "# How do you find the sum of the roots of: x^2-5x+12=0? $5$ #### Explanation: The sum of roots of given quadratic equation: ${x}^{2} - 5 x + 12 = 0$ is given as -\\frac{\\text{coefficient of} \\ x}{\\text{coefficient of }\\ x^2 $= - \\setminus \\frac{\\left(- 5\\right)}{1}$ $= 5$", "= 2 - 1 - 21 - 26 - 8 = -54 \\neq 0$$ so $1$ is not a root. $$f(-1) = 2 + 1 - 21 + 26 - 8 = 0$$ Hence, $-1$ is a root, so $x - (-1) = x + 1$ is a factor. Dividing $2x^4 - x^3 - 21x^2 - 26x - 8$ by $x + 1$ yields $2x^3 - 3x^2 - 18x - 8$. Thus, $$f(x) = 2x^4 - x^3 - 21x^2 - 26x - 8 = (x + 1)(2x^3 - 3x^2 - 18x - 8)$$ Let $g(x) = 2x^3 - 3x^2 - 18x - 8$. By the Rational Roots Theorem, the possible rational roots of $2x^3 - 3x^2 - 18x - 8$ are the same as those of $f(x)$. We know $1$ is not a root of $f(x)$, so it cannot be a root of $g(x)$. $$g(-1) = -2 - 3 + 18 - 8 = 5 \\neq 0$$ so $-1$ is not a root of $g(x)$, which means it is not a double root of $f(x)$. $$g(2) = 16 - 12 - 36 - 8 = -40 \\neq 0$$ Hence, $2$ is not a root of $g(x)$, which also means it is not a root of $f(x) = (x + 1)g(x)$. $$g(-2) = -16 - 12 + 36 -" ]

[ "$g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved. By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\\pm1$. Clearly, these are not roots. Similary, any quadratic factor would have to be $x^2+ax\\pm1$ with $a\\in\\Bbb Z$. You might be able to exclude these manually ...", "1. ## how to factor---x^3+x^2-17x+15????? book says (x+5)(x-3)(x-1) Help! I do not know procedure to factor trinomial?????????? 2. ## Re: How to factor x^3+x^2-17x+15? Let $f(x)=x^3+x^2-17x+15.$ Try a number $a$ and see if $f(a)=0,$ say $a=1.$ $f(1) = 1^3+1^2-17(1)+15 = 0.$ Thus $x-1$ is a factor, i.e. $x^3+x^2-17x+15=(x-1)g(x)$ where $g(x)$ is a quadratic polynomial. Once you've found $g(x)$ you can factorize it further. 3. ## Re: How to factor x^3+x^2-17x+15? keep talking because i need more explanation???? 4. ## Re: how to factor---x^3+x^2-17x+15????? To \"factor\" a polynomial with integer coefficients typically means \"write as a product of factors with integer coefficients. For example, $x^2- 3$ can be written as $(x- \\sqrt{3})(x+ \\sqrt{3})$ but we don't normally think of that as \"factoring\". What Sylvia104 did was look for rational zeros of the polynomial. There is a \"rational root theorem\" that says \"Any rational number roots of the polynomial equation, $a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$, are of the form $\\frac{m}{n}$ with m an integer that evenly divides $a_0$ and n an integer that evenly divides $a_n$. Here, $a_n= 1$ so any rational root must have denominator 1- in other words be an integer. $a_0= 15$ so the only possible rational roots are 1, -1, 3, -3, 5, -5, 15, and -15, the factors of 15. Check each one: [tex](1)^3+", "Math With this no-prep activity, students will find actual (as opposed to possible) rational roots of polynomial functions. The rational root theorem states that if a polynomial with integer coefficients f(x)=pnxn+pn−1xn−1+⋯+p1x+p0 f(x) = p_n x^n + p_{n-1} x^{n-1} + \\cdots + p_1 x + p_0 f(x)=pn​xn+pn−1​xn−1+⋯+p1​x+p0​ has a rational root of the form r=±ab r =\\pm \\frac {a}{b}r=±ba​ with gcd⁡(a,b)=1 \\gcd (a,b)=1gcd(a,b)=1, then a∣p0 a \\vert p_0a∣p0​ and b∣pn b \\vert p_nb∣pn​. Let x2+x=n x^2 + x = nx2+x=n, where n nn is an integer. Today, they are going to play the quadratic game. Finding the rational roots (also known as rational zeroes) of a polynomial is the same as finding the rational x-intercepts. This time, the common factor on the left is q. Let’s extract it, and lump together the remaining sum as t. Again, q and p have no common factors. A series of college algebra lectures: Presenting the Rational Zero Theorem, Find all zeros for a polynomial. They also share no common factors. pn(ab)n+pn−1(ab)n−1+⋯+p1ab+p0=0. □ _\\square□​. Finding All Factors 3. f\\bigg (-\\frac {1}{2}\\bigg ) &= -\\frac {2}{8} + \\frac {7}{4} - \\frac {5}{2} + 1 = 0. □ _\\square□​. Determine the positive and negative factors of each. The Rational Roots Test: Introduction (page 1 of 2) The zero of a polynomial is an input value", "root of $g$. The resultant can be computed via the determinant Sylvester matrix, and in particular if $f$ and $g$ have integer (or rational) coefficients, then so does the polynomial you obtain from $R(f,g)$. $R(f,g)=0$ if and only if $f$ and $g$ have a zero in common. Then computing the resultant of $f(x)$ and $g(z-x)$ gives a polynomial (in $z$) with rational coefficients that has, among its roots, the root $a+b$; hence, $a+b$ is algebraic. 5. If $a$ is algebraic and $a\\neq 0$, then $\\frac{1}{a}$ is algebraic Proof. If $f(x) = x^m + a_{m-1}x^{m-1}+\\cdots + a_1x + a_0$ is a polynomial with rational coefficients and $f(a)=0$, then $$x^mf\\left(\\frac{1}{x}\\right) = 1 + a_{m-1}x + \\cdots + a_1x^{m-1} + a_0x^m$$ is a polynomial with rational coefficients that has $\\frac{1}{a}$ as a root. 6. If $a$ and $b$ are algebraic over $\\mathbb{Q}$, then $ab$ is algebraic over $\\mathbb{Q}$. Proof. If $f(x)$ has rational coefficients and $a$ as a root, and $g(x)$ has rational coefficients and $b$ as a root, then the resultant of $f(x)$ and $x^{\\deg(g)}g(\\frac{t}{x})$ is a polynomial in $t$ with rational coefficients that has $xy$ as a root. Hence $xy$ is algebraic. Thus, every complex number obtained as the result of doing a finite combination of addition, multiplication, and root extraction of algebraic numbers is algebraic. You'll note that", "We have $c+d=2a$. The prime factors of $2b$ must now be distributed between $c$ and $d$; there is only one even prime factor (2) and the rest are odd. Without loss of generality, assign the factor of 2 to $c$. No matter how the remaining odd factors are distributed, $c$ must remain even because of the 2 and $d$ must remain odd because there is no factor of 2 in it at all. Hence $c+d$ must be odd, which contradicts their sum $2a$ being even. Therefore $x^2+2ax+2b=0$ has no integer – or rational – roots for all odd integers $a,b$. In a word: Eisenstein. In a few more words, if $x^2+2ax+2b=0$ had a rational root, say $x=p/q$ with $\\gcd(p,q)=1$, then, on multiplying through by $q^2$, we have $$p^2+2apq+2bq^2=0$$ This implies $p$ is even. So writing $p=2p’$ and then dividing through by $2$, we have $$2p’^2+2ap’q+bq^2=0$$ which implies $q$ is even (since $b$ is odd). But that contradicts $\\gcd(p,q)=1$ (i.e., the fact we can write any fraction in reduced form). Thus $x^2+2ax+b=0$ cannot have a rational root if $b$ is odd. (Note, the assumption that $a$ is also odd is irrelevant to the proof.) The square root of the discriminant of the roots is $2 \\sqrt { a^2 -2b}$ This is never a perfect square as it gives 3", "# Difference between revisions of \"Rational Root Theorem\" Given a polynomial $P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \\ldots + a_1 x + a_0$ with integral coefficients, $a_n \\neq 0$. The Rational Root Theorem states that if $P(x)$ has a rational root $r = \\pm\\frac pq$ with $p, q$ relatively prime positive integers, $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$. As a consequence, every rational root of a monic polynomial with integral coefficients must be integral. This gives us a relatively quick process to find all \"nice\" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check. ## Proof Given $\\frac{p}{q}$ is a rational root of a polynomial $f(x)=a_nx^n+x_{n-1}x^{n-1}+\\cdots +a_0$, where the $a_n$'s are integers, we wish to show that $p|a_0$ and $q|a_n$. Since $\\frac{p}{q}$ is a root, $$0=a_n\\left(\\frac{p}{q}\\right)^n+\\cdots +a_0$$ Multiplying by $q^n$, we have: $$0=a_np^n+a_{n-1}p^{n-1}q+\\cdots+a_0q^n$$ Examining this in modulo $p$, we have $a_0q^n\\equiv 0\\pmod p$. As $q$ and $p$ are relatively prime, $p|a_0$. With the same logic, but with modulo $q$, we have $q|a_n$, and we are done. ## Problems ### Easy Factor the polynomial $x^3-5x^2+2x+8$. ### Intermediate Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$. Prove that $\\sqrt{2}$ is irrational, using the Rational Root Theorem.", "# Polynomials all of whose roots are rational I have two questions about the class of integer-coefficient polynomials all of whose roots are rational. I asked this at MSE, but it attracted little interest (perhaps because it is not interesting!) Q1. Is there some way to recognize such a polynomial from its coefficients $a_0, a_1, \\ldots, a_n$? I am aware of the rational-root theorem, which says that each rational root is of the form $\\pm p/q$, where $p$ is a factor of $a_0$ and $q$ a factor of $a_n$. Example. The roots of $$12544 x^5 + 24976 x^4 - 23994 x^3 - 51721 x^2 - 17080 x + 1275$$ are $$\\lbrace \\frac{3}{2}, -\\frac{5}{7}, -\\frac{5}{7}, \\frac{1}{16}, -\\frac{17}{8} \\rbrace \\;.$$ Here $a_0 = 1275 = 3 \\cdot 5 \\cdot 17$ and $a_5 = 12544 = 2^8 \\cdot 7^2$. As Mark Bennet commented at MSE, perhaps an analog of Sturm's theorem would serve. Q2. Has this class of polynomials been studied in its own right? In other words, is this class interesting? I can see it has at least a monoid structure, as the product of two such polynomials also has all rational roots. These are naive questions, well out of my expertise. Thanks in advance for educating me! - @Joseph: Why is the rational-root theorem not enough? It gives a", "1, -1, -1/18 and 1/18 Does this look correct? TFM 5. Nov 25, 2008 ### TFM Hi Tiny-Tim I had one value to be (x - 1), which I got by inserting x = 1, Have now constructed a formula in a spreadsheet which now gives me x = 1 and 3, so I have 2 factors... TFM 6. Nov 25, 2008 ### tiny-tim Well they have to multiply to 18 (or is it -18? ), so the third one is … ? 7. Nov 25, 2008 ### HallsofIvy Staff Emeritus The \"rational root theorem\", that LogicalTime mentions, says that if m/n is a rational number root of $a_px^p+ a_{p-1}x^{p-1}+ \\cdot\\cdot\\cdot+ a_0= 0$ then the numerator m must divide $a_0$ and and n must divide $a_p$. In your example, $-x^3 +10x^2 - 27x + 18$, any rational root m/n must have n dividing 1 (and so is either 1 or -1) and m dividing 18: m/n must be one of 1, -1, 2, -2, 3, -3, 6, -6, 18, or -18. It is easy to calculate that $$-(1)^3 +10(1)^2 - 27(1) + 18= 0$$ so x-1 is a factor. Divide $$-x^3 +10x^2 - 27x + 18$$ by x-1 to get the other, quadratic, factor. In this particular case you can factor that into linear factors. However, you should", "# Rational Root Theorem Given a polynomial $P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \\ldots + a_1 x + a_0$ with integral coefficients, $a_n \\neq 0$. The Rational Root Theorem states that if $P(x)$ has a rational root $r = \\pm\\frac pq$ with $p, q$ relatively prime positive integers, $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$. As a consequence, every rational root of a monic polynomial with integral coefficients must be integral. This gives us a relatively quick process to find all \"nice\" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check. ## Proof Given $\\frac{p}{q}$ is a rational root of a polynomial $f(x)=a_nx^n+x_{n-1}x^{n-1}+\\cdots +a_0$, where the $a_n$'s are integers, we wish to show that $p|a_0$ and $q|a_n$. Since $\\frac{p}{q}$ is a root, $$0=a_n\\left(\\frac{p}{q}\\right)^n+\\cdots +a_0$$ Multiplying by $q^n$, we have: $$0=a_np^n+a_{n-1}p^{n-1}q+\\cdots+a_0q^n$$ Examining this in modulo $p$, we have $a_0q^n\\equiv 0\\pmod p$. As $q$ and $p$ are relatively prime, $p|a_0$. With the same logic, but with modulo $q$, we have $q|a_n$, which completes the proof. ## Problems ### Easy 1. Factor the polynomial $x^3-5x^2+2x+8$. ### Intermediate 2. Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$. 3. Prove that $\\sqrt{2}$ is irrational, using the Rational Root Theorem. 1. $(x-4)(x-2)(x+1)$", "it factors, it must have a rational root. A polynomial has \"x- a\" as a factor if and only x= a makes the polynomial equal to 0 and the \"rational root theorem\" says that if m/n is a rational root of the equation $$a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$$ with integer coefficients, then n must divide $a_n$ and m must divide $a_0$. With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must be one of +/- 1, +/- 2, +/- 4. The only possible rational roots are +/- 1, +/- 2, +/- 4 and it is not hard to try those and see that -1 is a root and that 2 is a double root: $$x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)$$ #### craig0303 A polynomial has \"x- a\" as a factor if and only x= a makes the polynomial equal to 0 and the \"rational root theorem\" says that if m/n is a rational root of the equation $$a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$$ with integer coefficients, then n must divide $a_n$ and m must divide $a_0$. With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must", "the polynomial $f(x)$. Rational Root Theorem The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \\cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\\pm \\frac{p}{q}$. Remainder Theorem The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:", "roots and the rational root theorem cannot help us find the roots that do exist. Using a simple Tschirnhaus transformation ($t = x - \\frac{1}{3}$) and Cardano's method, I found: $x = \\frac{1}{3} + \\frac{\\sqrt[3]{46 + 6 \\sqrt{57}}}{3} + \\frac{\\sqrt[3]{46 - 6 \\sqrt{57}}}{3} \\approx 2.1304$", "The Rational Root Theorem states that if has a rational root with relatively prime positive integers, is a divisor of and is a divisor of. That means p and q share no common factors. Rational Roots Test. Rational Root Theorem Given a polynomial with integral coefficients,. Fill that space with the given pattern. Given a polynomial with integral coefficients, .The Rational Root Theorem states that if has a rational root with relatively prime positive integers, is a divisor of and is a divisor of .. As a consequence, every rational root of a monic polynomial with integral coefficients must be integral.. x5−4x4+2x3+2x2+x+6=0.x^5-4x^4+2x^3+2x^2+x+6=0.x5−4x4+2x3+2x2+x+6=0. Specifically, it describes the nature of any rational roots … That’s alot of plugging in. Hence a−ma-ma−m divides f(m)f(m)f(m). Home > Portfolio item > Definition of rational root theorem; The Rational Root Theorem says if a polynomial equation $a_n x^n + a_{n – 1} x^{n – 1} + … + a_1 x + a_0 = 0$ has rational root $\\frac{p}{q} (p, q \\in \\mathbb{Z})$ then the denominator q divides the leading coefficient and the numerator p divides $a_0$. Using this same logic, one can show that 3,5,7,...\\sqrt 3, \\sqrt 5, \\sqrt 7, ...3​,5​,7​,... are irrational, and from this one can prove that the square root of any number that is not a perfect square is irrational.", "factor of 3 in the numerator and a factor of 2 in the denominator. Find all rational zeroes of P(x)=2x4+x3−19x2−9x+9P(x) = 2x^4 + x^3 -19x^2 -9x + 9P(x)=2x4+x3−19x2−9x+9. It provides and quick and dirty test for the rationality of some expressions. Using rational root theorem, we have the following: Now, substituting these values in P(x)P(x)P(x) and checking if it equates to zero (please refer to this: Remainder Factor Theorem), we find that P(x)=0P(x) = 0P(x)=0 for the values 12,3,−3,1.\\frac{1}{2} , 3 , -3 ,1. Let’s replace all that stuff in parenthesis with an s. We don’t really care what’s in there. Take a look. Since gcd⁡(a,b)=1 \\gcd(a,b)=1gcd(a,b)=1, Euclid's lemma implies b∣pn b | p_nb∣pn​. No, this polynomial has irrational rootsC. The possibilities of p/ q, in simplest form, are . It must divide a₀: Thus, the numerator divides the constant term. Rational root There is a serum that's used to find a possible rational roots of a polynomial. The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $\\frac{p}{q}$, where p is a factor of the trailing constant and q is a factor of the leading coefficient. In this section we learn the rational root theorem for polynomial functions, also known as the rational zero theorem. Sometimes the", "# Find the roots of the cubic equation How to find the roots of this cubic equation ? $$x^3+16x-455=0$$ - Well, one of them is $\\;7\\;$ . How did I find it?! – DonAntonio Dec 12 '13 at 21:10 @DonAntonio First WA and then rational root theorem? – Git Gud Dec 12 '13 at 21:11 You knew $7^3+16(7)=455$ ? – K. Rmth Dec 12 '13 at 21:12 @GitGud What is WA? – Harald Hanche-Olsen Dec 12 '13 at 21:12 Nop, @GitGud : only the rational root theorem. Pretty easy to apply here, as $\\;455=5\\cdot7\\cdot 13\\;$ and the only one that seems to fit is $\\;7\\;$ . All the rest too small or too big – DonAntonio Dec 12 '13 at 21:12 Another solution, using algebraic manipulation : $x^3+16x=455$ $x^4+16x^2=65\\cdot 7x$, add $49x^2$ both sides, then we get $x^4+65x^2+(\\frac{65}{2})^2=49x^2+65\\cdot 7x+(\\frac{65}{2})^2$ Finally, $$x^2+\\frac{65}{2}=7x+\\frac{65}{2}$$ - Amazing. Thank you guy ! – Joel Dec 12 '13 at 21:19 @Joel You are very welcome ! – Ewin Dec 12 '13 at 21:22 Use rational root theorem to find if the cubic equation has a rational root. If it has you can easily bring it to a quadratic equation, by dividing the polynomial with $(x-x_0)$, where $x_0$ is its root. Anyway if the cubic equation doesn't have a rational root, you can use Cardano's", "factor g(x) in Q[x] which the product of (x - a_i) over some non-empty proper subset I of {1,...,n}. So to show that no such g(x) exists, one can simply work through the possibilities for I and check that the corresponding g(x) does not in fact have all its coefficients in Q[x]. Suppose, for instance that f(x) = x^4 + 2. The complex roots of f(x) are a_i = z^i * b where b is one choice of complex fourth root of 2 and z = \\sqrt{-1}. We know that there is no fourth root of 2 in Q, so f(x) cannot have a linear factor in Q[x]. The only possible irreducible monic g(x) in Q[x] which can divide f(x) without being equal to f(x) are thus of degree 2. Such a g(x) must equal (x - a_i)*(x - a_j) for some 1 <= i , j <-= 4 with i <> j. This would require a_i*a_j to be a rational number. However, the complex absolute value of |z| is 1, so |z^i * b| = |b| = the real positive fourth root of 2. So |a_i * a_j| = |a_i|*|a_j| = the real positive square root of 2. Since a_i*a_j is supposed to be rational, we would conclude that there has to be a rational square root", "$3 \\cdot 19$, so the rational root theorem guarantees that the only possible rational roots are $-57$, $-19$, $-3$, $-1$, $1$, $3$, $19$, and $57$. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. $\\square$ Example 2 Factor the polynomial $x^3-5x^2+2x+8$. Solution: After testing the divisors of 8, we find that it has roots $-1$, $2$, and $4$. Then because it has leading coefficient $1$, the factor theorem tells us that it has the factorization $(x-4)(x-2)(x+1)$. $\\square$ Example 3 Using the rational root theorem, prove that $\\sqrt{2}$ is irrational. Solution: The polynomial $x^2 - 2$ has roots $-\\sqrt{2}$ and $\\sqrt{2}$. The rational root theorem garuntees that the only possible rational roots of this polynomial are $-2, -1, 1$, and $2$. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because $\\sqrt{2}$ is a root of the polynomial, it cannot be a rational number. $\\square$", "4 x^2+ x-6$. By Integer solutions of a polynomial function theorem if $p$ has integer roots they are divisors of 6, so potential candidates for roots are $-6,-3,-2,-1,1, 2, 3$ and $6$. $$2\\cdot 1^4+ 1^3- 4\\cdot 1^2+ 1-6=-6$$ so 1 is not root. $$2\\cdot (-1)^4+ (-1)^3- 4\\cdot (-1)^2+ (-1)-6=-10$$ so -1 is not root. $$2\\cdot 2^4+ 2^3- 4\\cdot 2^2+ 2-6=20$$ so 2 is not root. $$2\\cdot (-2)^4+ (-2)^3- 4\\cdot (-2)^2+ (-2)-6=0$$ so -2 is root. Now we divide $p$ with $x+2$ and write it as $$p(x)=(x+2)(2 x^3- 3 x^2 +2 x-3).$$ Lets check if $i$ is root of $q(x):=2 x^3- 3 x^2 +2 x-3$: $$2 i^3- 3 i^2 +2 i-3=2(-i)-3(-1)+2i-3=0$$ so $i$ is root. By Irrational Root Theorem then follows that $-i$ is also root of $q$. But then by Bézout’s theorem $(x-i)$ and $(x+i)$ both divide $q$ so their product $(x-i)(x+i)=x^2+1$ also divides $q$ and we calculate $$q(x)=(x^2+1)(x-3).$$ Finally, we write $p$ as $$p(x)=(x+2)(x^2+1)(x-3)=(x+2)(x-i)(x+i)(x-3).$$ Polynomials we were factoring so far were polynomials in one variable. We can similarly factor polynomials with more variables. Example 8 Factor polynomial $p(x)=x^2 y + xy^2 + 4xy$. We see that factor $xy$ is common to all terms so we extract it: $$x^2 y + xy^2 + 4xy = xy(x + y + 4).$$ Example 9 Factor polynomial $p(x)=x^3 + x^2+ 2", "a factor of $$a_{0}$$. As the same kind of argument applies to q being a factor of $$a_{n}$$, I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in: $$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$ how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer? Last edited: #### Muzza Paragon said: I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try: Say we have a polynomial: $$F(x) = \\sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$ where all of the coefficients are integers and $$a_{0}$$ does not equal zero. I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that: $$F(\\frac{p}{q}) = 0$$ You won't be able to prove that. Exercise: find a polynomial with integer coefficients which doesn't have any rational roots. The \"rational root\" theorem that /I'm/ familiar with states that if p/q is a rational root of F and gcd(p, q) = 1,", "for saving me the trouble :-) – Jyrki Lahtonen Oct 1 '13 at 18:34 The basic idea behind the rational root theorem can be used to show that this can only factor into two quadratics if the factorization is of the form: $$x^4 - x - 1 = (x^2 + ax + 1) (x^2 + bx - 1)$$ - You can use the Rational Root Theorem to solve this problem. The possible roots under consideration are $\\pm 1$ (because the roots under consideration are $\\frac{p}{q}$, (where $p$ is the coefficient of the last term and $q$ is the coefficient of $x^4$ in $f(x)$), and neither of them are roots of $f(x)$. - this excludes that $f$ has a rational root, not that $f$ is reducible – Danae Kissinger Oct 1 '13 at 18:35 What I forgot to mention is that because $f$ does not have a rational root, it is not factorizable over $\\mathbb Q$. – Anonymous Oct 1 '13 at 18:36 Anonymous, what about $x^4 + 4 = (x^2 - 2 x + 2)(x^2 + 2 x + 2)?$ No rational roots. – Will Jagy Oct 1 '13 at 18:38 $$\\begin{eqnarray}x^4+4&=&(x^2+2i)\\cdot (x^2-2i)\\\\ &=& (x-(1-i))\\cdot (x+(1-i))\\cdot (x-(1+i))\\cdot(x+(1+i)) \\\\ &=& ((x-1)+i)\\cdot ((x-1)-i)\\cdot((x+1)-i)\\cdot((x+1)+i) \\\\ &=& ((x-1)^2+1)\\cdot((x+1)^2+1).\\end{eqnarray}$$ Reducible. – Anonymous Oct 1 '13 at 18:40 Oh I see. No... – Anonymous" ]

[ "$g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved. By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\\pm1$. Clearly, these are not roots. Similary, any quadratic factor would have to be $x^2+ax\\pm1$ with $a\\in\\Bbb Z$. You might be able to exclude these manually ...", "1. ## how to factor---x^3+x^2-17x+15????? book says (x+5)(x-3)(x-1) Help! I do not know procedure to factor trinomial?????????? 2. ## Re: How to factor x^3+x^2-17x+15? Let $f(x)=x^3+x^2-17x+15.$ Try a number $a$ and see if $f(a)=0,$ say $a=1.$ $f(1) = 1^3+1^2-17(1)+15 = 0.$ Thus $x-1$ is a factor, i.e. $x^3+x^2-17x+15=(x-1)g(x)$ where $g(x)$ is a quadratic polynomial. Once you've found $g(x)$ you can factorize it further. 3. ## Re: How to factor x^3+x^2-17x+15? keep talking because i need more explanation???? 4. ## Re: how to factor---x^3+x^2-17x+15????? To \"factor\" a polynomial with integer coefficients typically means \"write as a product of factors with integer coefficients. For example, $x^2- 3$ can be written as $(x- \\sqrt{3})(x+ \\sqrt{3})$ but we don't normally think of that as \"factoring\". What Sylvia104 did was look for rational zeros of the polynomial. There is a \"rational root theorem\" that says \"Any rational number roots of the polynomial equation, $a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$, are of the form $\\frac{m}{n}$ with m an integer that evenly divides $a_0$ and n an integer that evenly divides $a_n$. Here, $a_n= 1$ so any rational root must have denominator 1- in other words be an integer. $a_0= 15$ so the only possible rational roots are 1, -1, 3, -3, 5, -5, 15, and -15, the factors of 15. Check each one: [tex](1)^3+", "roots and the rational root theorem cannot help us find the roots that do exist. Using a simple Tschirnhaus transformation ($t = x - \\frac{1}{3}$) and Cardano's method, I found: $x = \\frac{1}{3} + \\frac{\\sqrt[3]{46 + 6 \\sqrt{57}}}{3} + \\frac{\\sqrt[3]{46 - 6 \\sqrt{57}}}{3} \\approx 2.1304$", "Math With this no-prep activity, students will find actual (as opposed to possible) rational roots of polynomial functions. The rational root theorem states that if a polynomial with integer coefficients f(x)=pnxn+pn−1xn−1+⋯+p1x+p0 f(x) = p_n x^n + p_{n-1} x^{n-1} + \\cdots + p_1 x + p_0 f(x)=pn​xn+pn−1​xn−1+⋯+p1​x+p0​ has a rational root of the form r=±ab r =\\pm \\frac {a}{b}r=±ba​ with gcd⁡(a,b)=1 \\gcd (a,b)=1gcd(a,b)=1, then a∣p0 a \\vert p_0a∣p0​ and b∣pn b \\vert p_nb∣pn​. Let x2+x=n x^2 + x = nx2+x=n, where n nn is an integer. Today, they are going to play the quadratic game. Finding the rational roots (also known as rational zeroes) of a polynomial is the same as finding the rational x-intercepts. This time, the common factor on the left is q. Let’s extract it, and lump together the remaining sum as t. Again, q and p have no common factors. A series of college algebra lectures: Presenting the Rational Zero Theorem, Find all zeros for a polynomial. They also share no common factors. pn(ab)n+pn−1(ab)n−1+⋯+p1ab+p0=0. □ _\\square□​. Finding All Factors 3. f\\bigg (-\\frac {1}{2}\\bigg ) &= -\\frac {2}{8} + \\frac {7}{4} - \\frac {5}{2} + 1 = 0. □ _\\square□​. Determine the positive and negative factors of each. The Rational Roots Test: Introduction (page 1 of 2) The zero of a polynomial is an input value", "root of $g$. The resultant can be computed via the determinant Sylvester matrix, and in particular if $f$ and $g$ have integer (or rational) coefficients, then so does the polynomial you obtain from $R(f,g)$. $R(f,g)=0$ if and only if $f$ and $g$ have a zero in common. Then computing the resultant of $f(x)$ and $g(z-x)$ gives a polynomial (in $z$) with rational coefficients that has, among its roots, the root $a+b$; hence, $a+b$ is algebraic. 5. If $a$ is algebraic and $a\\neq 0$, then $\\frac{1}{a}$ is algebraic Proof. If $f(x) = x^m + a_{m-1}x^{m-1}+\\cdots + a_1x + a_0$ is a polynomial with rational coefficients and $f(a)=0$, then $$x^mf\\left(\\frac{1}{x}\\right) = 1 + a_{m-1}x + \\cdots + a_1x^{m-1} + a_0x^m$$ is a polynomial with rational coefficients that has $\\frac{1}{a}$ as a root. 6. If $a$ and $b$ are algebraic over $\\mathbb{Q}$, then $ab$ is algebraic over $\\mathbb{Q}$. Proof. If $f(x)$ has rational coefficients and $a$ as a root, and $g(x)$ has rational coefficients and $b$ as a root, then the resultant of $f(x)$ and $x^{\\deg(g)}g(\\frac{t}{x})$ is a polynomial in $t$ with rational coefficients that has $xy$ as a root. Hence $xy$ is algebraic. Thus, every complex number obtained as the result of doing a finite combination of addition, multiplication, and root extraction of algebraic numbers is algebraic. You'll note that", "We have $c+d=2a$. The prime factors of $2b$ must now be distributed between $c$ and $d$; there is only one even prime factor (2) and the rest are odd. Without loss of generality, assign the factor of 2 to $c$. No matter how the remaining odd factors are distributed, $c$ must remain even because of the 2 and $d$ must remain odd because there is no factor of 2 in it at all. Hence $c+d$ must be odd, which contradicts their sum $2a$ being even. Therefore $x^2+2ax+2b=0$ has no integer – or rational – roots for all odd integers $a,b$. In a word: Eisenstein. In a few more words, if $x^2+2ax+2b=0$ had a rational root, say $x=p/q$ with $\\gcd(p,q)=1$, then, on multiplying through by $q^2$, we have $$p^2+2apq+2bq^2=0$$ This implies $p$ is even. So writing $p=2p’$ and then dividing through by $2$, we have $$2p’^2+2ap’q+bq^2=0$$ which implies $q$ is even (since $b$ is odd). But that contradicts $\\gcd(p,q)=1$ (i.e., the fact we can write any fraction in reduced form). Thus $x^2+2ax+b=0$ cannot have a rational root if $b$ is odd. (Note, the assumption that $a$ is also odd is irrelevant to the proof.) The square root of the discriminant of the roots is $2 \\sqrt { a^2 -2b}$ This is never a perfect square as it gives 3", "1, -1, -1/18 and 1/18 Does this look correct? TFM 5. Nov 25, 2008 ### TFM Hi Tiny-Tim I had one value to be (x - 1), which I got by inserting x = 1, Have now constructed a formula in a spreadsheet which now gives me x = 1 and 3, so I have 2 factors... TFM 6. Nov 25, 2008 ### tiny-tim Well they have to multiply to 18 (or is it -18? ), so the third one is … ? 7. Nov 25, 2008 ### HallsofIvy Staff Emeritus The \"rational root theorem\", that LogicalTime mentions, says that if m/n is a rational number root of $a_px^p+ a_{p-1}x^{p-1}+ \\cdot\\cdot\\cdot+ a_0= 0$ then the numerator m must divide $a_0$ and and n must divide $a_p$. In your example, $-x^3 +10x^2 - 27x + 18$, any rational root m/n must have n dividing 1 (and so is either 1 or -1) and m dividing 18: m/n must be one of 1, -1, 2, -2, 3, -3, 6, -6, 18, or -18. It is easy to calculate that $$-(1)^3 +10(1)^2 - 27(1) + 18= 0$$ so x-1 is a factor. Divide $$-x^3 +10x^2 - 27x + 18$$ by x-1 to get the other, quadratic, factor. In this particular case you can factor that into linear factors. However, you should", "# Rational Root Theorem Given a polynomial $P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \\ldots + a_1 x + a_0$ with integral coefficients, $a_n \\neq 0$. The Rational Root Theorem states that if $P(x)$ has a rational root $r = \\pm\\frac pq$ with $p, q$ relatively prime positive integers, $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$. As a consequence, every rational root of a monic polynomial with integral coefficients must be integral. This gives us a relatively quick process to find all \"nice\" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check. ## Proof Given $\\frac{p}{q}$ is a rational root of a polynomial $f(x)=a_nx^n+x_{n-1}x^{n-1}+\\cdots +a_0$, where the $a_n$'s are integers, we wish to show that $p|a_0$ and $q|a_n$. Since $\\frac{p}{q}$ is a root, $$0=a_n\\left(\\frac{p}{q}\\right)^n+\\cdots +a_0$$ Multiplying by $q^n$, we have: $$0=a_np^n+a_{n-1}p^{n-1}q+\\cdots+a_0q^n$$ Examining this in modulo $p$, we have $a_0q^n\\equiv 0\\pmod p$. As $q$ and $p$ are relatively prime, $p|a_0$. With the same logic, but with modulo $q$, we have $q|a_n$, which completes the proof. ## Problems ### Easy 1. Factor the polynomial $x^3-5x^2+2x+8$. ### Intermediate 2. Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$. 3. Prove that $\\sqrt{2}$ is irrational, using the Rational Root Theorem. 1. $(x-4)(x-2)(x+1)$", "it factors, it must have a rational root. A polynomial has \"x- a\" as a factor if and only x= a makes the polynomial equal to 0 and the \"rational root theorem\" says that if m/n is a rational root of the equation $$a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$$ with integer coefficients, then n must divide $a_n$ and m must divide $a_0$. With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must be one of +/- 1, +/- 2, +/- 4. The only possible rational roots are +/- 1, +/- 2, +/- 4 and it is not hard to try those and see that -1 is a root and that 2 is a double root: $$x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)$$ #### craig0303 A polynomial has \"x- a\" as a factor if and only x= a makes the polynomial equal to 0 and the \"rational root theorem\" says that if m/n is a rational root of the equation $$a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_1x+ a_0= 0$$ with integer coefficients, then n must divide $a_n$ and m must divide $a_0$. With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must", "factor of 3 in the numerator and a factor of 2 in the denominator. Find all rational zeroes of P(x)=2x4+x3−19x2−9x+9P(x) = 2x^4 + x^3 -19x^2 -9x + 9P(x)=2x4+x3−19x2−9x+9. It provides and quick and dirty test for the rationality of some expressions. Using rational root theorem, we have the following: Now, substituting these values in P(x)P(x)P(x) and checking if it equates to zero (please refer to this: Remainder Factor Theorem), we find that P(x)=0P(x) = 0P(x)=0 for the values 12,3,−3,1.\\frac{1}{2} , 3 , -3 ,1. Let’s replace all that stuff in parenthesis with an s. We don’t really care what’s in there. Take a look. Since gcd⁡(a,b)=1 \\gcd(a,b)=1gcd(a,b)=1, Euclid's lemma implies b∣pn b | p_nb∣pn​. No, this polynomial has irrational rootsC. The possibilities of p/ q, in simplest form, are . It must divide a₀: Thus, the numerator divides the constant term. Rational root There is a serum that's used to find a possible rational roots of a polynomial. The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $\\frac{p}{q}$, where p is a factor of the trailing constant and q is a factor of the leading coefficient. In this section we learn the rational root theorem for polynomial functions, also known as the rational zero theorem. Sometimes the", "# Finding the complex roots of a polynomial f knowing that it shares one root with a polynomial g So I'm asked to find the complex roots for the polynomial $f=x^4+3x-2$, knowing that it shares one of its roots with the polynomial $g=x^4+3x^3-3x+1$ My problem, however, is that when I try to find the GCD of f and g I get $(f:g)=1$, which would mean that f and g have no common roots. Am I wrong to assume this? Because else the GCD would be a grade 1 polynomial (if they share a root) Maybe I made a mistake when calculating the GCD? (I did this by hand and they were many divisions through Euclid's Algorithm, but I can't find a mistake) • Hint: $x^4+3x-2 = (x^2 + x - 1) (x^2 - x + 2)$ and $x^4+3x^3-3x +1 = (x^2+2x-1)(x^2+x-1)$ – gammatester Oct 12 '17 at 20:35 • Thanks, I'll try again later then – Francisco José Letterio Oct 12 '17 at 20:37 You made a mistake. Their GCD is $x^2 + x - 1$. Let $a$ be the common root. Therefore, $f(a)=g(a)=0$. But $f(a)=g(a)$ implies $$a^4+3a-2 = a^4+3a^3-3a+1 \\implies a^3-2a+1 = 0 \\implies (a-1)(a^2+a-1)=0.$$ Since $a=1$ is not the root of $f(x)$ and $g(x)$, we solve $$a^2+a-1=0 \\implies a = \\frac{-1\\pm\\sqrt{5}}{2}.$$ Indeed, $$f\\left(\\frac{-1\\pm\\sqrt{5}}{2}\\right)=g\\left(\\frac{-1\\pm\\sqrt{5}}{2}\\right)=0.$$ Can you find", "# Difference between revisions of \"Rational Root Theorem\" Given a polynomial $P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \\ldots + a_1 x + a_0$ with integral coefficients, $a_n \\neq 0$. The Rational Root Theorem states that if $P(x)$ has a rational root $r = \\pm\\frac pq$ with $p, q$ relatively prime positive integers, $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$. As a consequence, every rational root of a monic polynomial with integral coefficients must be integral. This gives us a relatively quick process to find all \"nice\" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check. ## Proof Given $\\frac{p}{q}$ is a rational root of a polynomial $f(x)=a_nx^n+x_{n-1}x^{n-1}+\\cdots +a_0$, where the $a_n$'s are integers, we wish to show that $p|a_0$ and $q|a_n$. Since $\\frac{p}{q}$ is a root, $$0=a_n\\left(\\frac{p}{q}\\right)^n+\\cdots +a_0$$ Multiplying by $q^n$, we have: $$0=a_np^n+a_{n-1}p^{n-1}q+\\cdots+a_0q^n$$ Examining this in modulo $p$, we have $a_0q^n\\equiv 0\\pmod p$. As $q$ and $p$ are relatively prime, $p|a_0$. With the same logic, but with modulo $q$, we have $q|a_n$, and we are done. ## Problems ### Easy Factor the polynomial $x^3-5x^2+2x+8$. ### Intermediate Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$. Prove that $\\sqrt{2}$ is irrational, using the Rational Root Theorem.", "a factor of $$a_{0}$$. As the same kind of argument applies to q being a factor of $$a_{n}$$, I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in: $$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$ how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer? Last edited: #### Muzza Paragon said: I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try: Say we have a polynomial: $$F(x) = \\sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$ where all of the coefficients are integers and $$a_{0}$$ does not equal zero. I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that: $$F(\\frac{p}{q}) = 0$$ You won't be able to prove that. Exercise: find a polynomial with integer coefficients which doesn't have any rational roots. The \"rational root\" theorem that /I'm/ familiar with states that if p/q is a rational root of F and gcd(p, q) = 1,", "the first factor has roots $a_{n-2}$. The third factor has no roots. We show that the second factor has 2 roots. Let $g_n(x)= P_{n-2}^2 - (\\sqrt{2}+1)$.We claim that the roots are $$\\pm \\sqrt{\\sqrt{\\sqrt{\\cdots \\sqrt{\\sqrt{2}+1~ (\\text{n -1 times})}\\cdots +1}+1}+1} .$$ This can be verified by factorizing and seeing that we have a expression of form $(g^2-1)^2 - b$ where $b>1$, then only the factor $g^2 - (\\sqrt{b}+1)$ can only real root in the factorization of the expression. Thus we have $a_n = a_{n-2}+2$. Thus the sequence is natural numbers. So the answer shall be then $2017$.", "When $$x = c$$, then $$f(c) = q(c) (c - c) + r = r$$. So the remainder is $$f(c)$$. This works for complex $$c$$ as well. ### Factor Theorem Version 1 (FT 1) Proposition: $$x - c \\mid f(x) \\iff f(c) = 0$$. Proof: Clearly, $$x - c \\mid f(x)$$ if and only if the remainder of $$f(x)$$ divided by $$x - c$$ is 0. By RT, the remainder of $$f(x)$$ divided by $$x - c$$ is $$f(c)$$. So $$x - c \\mid f(x) \\iff f(c) = 0$$. ### Factor Theorem Version 2 (FT 2) Proposition: $$x - c$$ is a factor of $$f(x)$$ if and only if $$c$$ is a root of $$f(x)$$. A root $$k$$ of $$f(x)$$ is a value such that $$f(k) = 0$$. This is a restatement of FT 1. So if we can find roots, we can find factors of the form $$x - c$$. This is useful because we can apply things like the quadratic formula to factor quadratic equations. The quadratic formula works even with complex numbers. Factor $$4x^2 - 8x + 5$$: The roots are $$\\frac{8 \\pm \\sqrt{8^2 - 4 \\cdot 4 \\cdot 5}}{2 \\cdot 4} = 1 \\pm \\frac{\\sqrt{-16}}{8} = 1 \\pm \\frac{1}{2}i$$. So $$4x^2 - 8x + 5 = 4 \\cdot (x - (1 + \\frac{1}{2}i))(x -", "a\\pm\\sqrt b.$ (four possibilities) Hence we get $$x^2\\mp2x\\sqrt a+a=b$$ $$x^2+a-b=\\pm2x\\sqrt a$$ $$(x^2+a-b)^2=4x^2a.$$ After getting this, I’ve tried to find a similar polynomial of smaller degree, but I can neither find it nor prove that there is no such polynomial. Can anybody help? #### Solutions Collecting From Web of \"A low-degree polynomial $g_{a,b}(x)$ which has a zero $x\\in\\mathbb N$ for any square numbers $a,b$?\" $\\def\\QQ{\\mathbb{Q}}\\def\\ZZ{\\mathbb{Z}}$I can establish the following: Theorem Let $g(a,b,x)$ be a polynomial in $\\mathbb{Q}(a,b,x)$. Suppose that $g$ has a rational root in $x$ if and only if $a$ and $b$ are rational squares. Then $\\deg_x g \\geq 4$. I am about 95% sure I can modify the proof to work with integer squares, and will come back and post such a modified version when I have the time. Suppose that $g(a,b,x) \\in \\QQ(a,b,x)$ and that, whenever $a$ and $b$ are integer squares then $g(a,b,x)$ has a rational root. (I have deliberately chosen the weakest combination of $\\QQ$ and $\\ZZ$ for my hypotheses, so the result is as strong as possible.) I will show that at least one of the following is true. 1. $g(u^2,b,x)$ has a rational root for all $(u,b) \\in \\mathbb{Q}^2$. 2. $g(a,v^2,x)$ has a rational root for all $(u,b) \\in \\mathbb{Q}^2$. 3. $g(a,a w^2,x)$ has a rational root for all $(a,w) \\in \\mathbb{Q}^2$.", "polynomial by the common denominator of it’s coefficients. For example: r(x)=\\frac{1}{3}x^2+\\frac{1}{6}x+\\frac{1}{4} \\ \\ \\ \\ /\\cdot 24 \\\\ \\Downarrow \\\\ 24\\cdot r(x)=8x^2+4x+6 Now, we can divide the new polynomial by the GCD of it’s coefficients – this yields a primitive polynomial. In our example, the GCD is 2, hence: \\frac{24}{2}\\cdot r(x)=4x^2+2x+3 And $\\frac{24}{2}=12$ is the number we were looking for. So after we understand that, let $a_g,a_h$ be the rational numbers such that $a_g\\cdot g(x), a_h\\cdot h(x)$ are polynomials with integer coefficients and are also primitive. Therefore, by Gauss’s lemma, their product: a_g g(x)\\cdot a_hh(x)=a_ga_h\\cdot g(x)h(x) \\in\\mathbb{Z}[x] Is a primitive polynomial as well. Recall that $g(x)h(x)=f(x)$ and $f$ is monic, hence the leading coefficient of $a_ga_h\\cdot g(x)h(x)$ is $a_ga_h$. But we know that this is a polynomial with integer coefficients, so $a_ga_h$ is an integer. Therefore, $a_g a_h$ is a common divisor of the coefficients of the product! Hence, it must be $\\pm 1$ (since it divides the GCD, which is one): a_ga_h=\\pm 1 \\\\ \\Downarrow \\\\ a_g=\\pm1, a_h=\\pm 1 Therefore: \\pm g(x), \\pm h(x)\\in\\mathbb{Z}[x] \\\\ \\Downarrow \\\\ g(x),h(x)\\in\\mathbb{Z}[x] As we wanted! ## Back to the cyclotomic polynomial We know that: \\Phi_n(x)=\\frac{x^n-1}{\\prod_{\\overset{d< n}{d|n}}\\Phi_d(x)} Which is the same as: \\overbrace{x^n-1}^{\\in\\mathbb{Z}[x]\\text{ and monic}}=\\overbrace{\\Phi_n(x)}^{\\in\\mathbb{Q}[x]}\\cdot \\overbrace{\\prod_{\\overset{d< n}{d|n}}\\Phi_d(x)}^{\\mathbb{Q}[x]} And this fits perfrectly to what we just proved, hence $\\Phi_n\\in\\mathbb{Z}[x]$ as we thought! ##", "# Find the roots of the cubic equation How to find the roots of this cubic equation ? $$x^3+16x-455=0$$ - Well, one of them is $\\;7\\;$ . How did I find it?! – DonAntonio Dec 12 '13 at 21:10 @DonAntonio First WA and then rational root theorem? – Git Gud Dec 12 '13 at 21:11 You knew $7^3+16(7)=455$ ? – K. Rmth Dec 12 '13 at 21:12 @GitGud What is WA? – Harald Hanche-Olsen Dec 12 '13 at 21:12 Nop, @GitGud : only the rational root theorem. Pretty easy to apply here, as $\\;455=5\\cdot7\\cdot 13\\;$ and the only one that seems to fit is $\\;7\\;$ . All the rest too small or too big – DonAntonio Dec 12 '13 at 21:12 Another solution, using algebraic manipulation : $x^3+16x=455$ $x^4+16x^2=65\\cdot 7x$, add $49x^2$ both sides, then we get $x^4+65x^2+(\\frac{65}{2})^2=49x^2+65\\cdot 7x+(\\frac{65}{2})^2$ Finally, $$x^2+\\frac{65}{2}=7x+\\frac{65}{2}$$ - Amazing. Thank you guy ! – Joel Dec 12 '13 at 21:19 @Joel You are very welcome ! – Ewin Dec 12 '13 at 21:22 Use rational root theorem to find if the cubic equation has a rational root. If it has you can easily bring it to a quadratic equation, by dividing the polynomial with $(x-x_0)$, where $x_0$ is its root. Anyway if the cubic equation doesn't have a rational root, you can use Cardano's", "$3 \\cdot 19$, so the rational root theorem guarantees that the only possible rational roots are $-57$, $-19$, $-3$, $-1$, $1$, $3$, $19$, and $57$. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. $\\square$ Example 2 Factor the polynomial $x^3-5x^2+2x+8$. Solution: After testing the divisors of 8, we find that it has roots $-1$, $2$, and $4$. Then because it has leading coefficient $1$, the factor theorem tells us that it has the factorization $(x-4)(x-2)(x+1)$. $\\square$ Example 3 Using the rational root theorem, prove that $\\sqrt{2}$ is irrational. Solution: The polynomial $x^2 - 2$ has roots $-\\sqrt{2}$ and $\\sqrt{2}$. The rational root theorem garuntees that the only possible rational roots of this polynomial are $-2, -1, 1$, and $2$. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because $\\sqrt{2}$ is a root of the polynomial, it cannot be a rational number. $\\square$", "2 We know that the roots of P are \\{1, \\ldots, 50\\} and the roots of Q are \\{-1, \\ldots, -50\\}. Therefore the roots of P \\cdot Q are \\{-50, \\ldots, -2, -1, 1, 2, \\ldots, 50\\}, which we'll call r_1, \\ldots, r_{100}. From the first of Vietas formulas we know that$$ 0 = -50 + \\ldots + -2 + -1 + 1 + 2 + \\ldots 50 = \\sum_{k = 0}^{100} r_{k} = - \\frac{... 2 The domain givesx\\geq2$and we need to solve $$\\sqrt{5x^2+27x+25}=5\\sqrt{x+1}+\\sqrt{x^2-4}$$ or $$5x^2+27x+25=25(x+1)+10\\sqrt{(x+1)(x^2-4)}+x^2-4$$ or $$2x^2+x+2=5\\sqrt{(x^2-x-2)(x+2)}$$ or $$\\frac{2x^2}{x+2}+1=5\\sqrt{\\frac{x^2}{x+2}-1}.$$ Now, take $$\\sqrt{\\frac{x^2}{x+2}-1}=t.$$ Can you end it now? 2 Yes. Write$c(h)$for the content of a polynomial in$\\Bbb Z[x]$. Let$m$and$n$be positive integers with$mf$,$ng\\in\\Bbb Z[x]$. I claim that$c(mnfg)=mn$. Certainly,$mn$divides all coefficients of$(mn)(fg)$but its leading coefficient is$mn$. Then$mn=c(mf)c(ng)$(Gauss's lemma). But$c(mf)\\mid m$as its leading coefficient is$m$, and$c(ng)\\mid n$... 2 For A, Consider $$B:=\\{z\\in \\Bbb C: f^{(n)}(z)=0\\;\\text{for some}\\; n \\in \\Bbb N \\}=\\bigcup_n\\{z\\in \\Bbb C: f^{(n)}(z)=0\\}$$ Here$B$is uncountable. That means, atleast one set in the union is uncountable. Thus,$\\exists k$so that$\\{z: f^{(k)}(z)=0\\}$is uncountable, so it has limit point in$\\Bbb C$and hence result follows! 2 If$a > 0$,$f(x)$has two sign changes, hence Descartes says there are either$2$or$0$positive real roots (counted by multiplicity). Hint:$f'(x) = 5 x^4 - 5 = 0$for$x = 1$, with$f''(1) = 20 > 0$, so$f$has a local minimum at$x=1$, and" ]

[ "In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $$30.$$ Find the sum of the four terms. (第二十一届AIME1 2003 第8题)", "The increasing sequence $$2,3,5,6,7,10,11,\\ldots$$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. (第八届AIME1990 第1题)", "The increasing sequence 1,3,4,9,10,12,13,… consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $$100^{th}$$ term of this sequenc(where 1 is the $$1^{st}$$ term, 3 is the $$2^{nd}$$ term, and so on). (第四届AIME1986 第7题)", "# How many geometric progressions? Number Theory Level 4 Find the number of $$6$$-term strictly increasing geometric progressions, such that all terms are positive integers less than $$1000.$$ ×", "# Shrajan's geometric progression Algebra Level 2 Three positive integers are in geometric progression, and have a sum of $$19$$ and a product of $$216$$. What is the least common multiple (LCM) of the three integers? This problem is posed by Shrajan V. ×", "# Infinite GP Algebra Level 2 Let $$a_1, a_2, a_3 , \\ldots$$ form an infinite geometric progression such that all of its terms are positive integers, and the product of the first 4 terms is 64. Find $$a_5$$. ×", "310 views A series $\\text{S1}$ of five positive integers is such that the third term is half the first term and the fifth term is $20$ more than the first term. In series $\\text{S2},$ the nth term defined as the difference between the $(n+1)$ term and the $n^{\\text{th}}$ term of series $\\text{S1}$, is an arithmetic progression with a common difference of $30$. Second term of $\\text{S2}$ is 1. $50$ 2. $60$ 3. $70$ 4. $\\text{None of these}$ 1 2 147 views", "# Problem #131 131 The increasing sequence $1,3,4,9,10,12,13\\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\\mbox{th}}$ term of this sequence. This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "with heights that are either increasing or decreasing. Construct a sequence of $16$ positive integers that has no increasing or decreasing subsequence of five terms. Find an increasing subsequence of maximal length and a decreasing subsequence of maximal length in the sequence $22, 5, 7, 2, 23, 10, 15, 21, 3, 17.$ Suppose that every student in a discrete mathematics class of $25$ students is a freshman, a sophomore, or a junior. Show that there are at least nine freshmen, at least nine sophomores, or at least nine juniors in the class. Show that there are either at least three freshmen, at least $19$ sophomores, or at least five juniors in the class", "# Only One Divides A sequence of numbers is formed. Each positive integer in increasing order, except 1, is 1. added to the sequence if at most one term of the sequence divides it or 2. not added to the sequence if at least two terms of the sequence divide it. The sequence begins $2,3,4,5,7,\\cdots$. Find the last three digits of the $2013^{th}$ term of the sequence. ×", "# Peculiar Sequence of Positive Integers! The sequence of increasing positive integers $${(x_n)_{n \\geq 1}}$$ is defined as follows: $$\\large{x_1 = 1}$$, the next two terms are 2 even numbers larger than the previous terms - ($$2$$ and $$4$$), the next three terms are the three smallest odd numbers that are larger than all the previous terms - ($$5, 7, 9$$), the next four terms are 4 smallest even numbers larger than all the previous terms - ($$10, 12, 14, 16$$) and so on. Find $$\\large{x_{2015}}$$. Bonus: Generalize for $$\\large{x_n}$$. ×", "# Problem #186 186 The increasing sequence $2,3,5,6,7,10,11,\\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "goes wrong if you try to use this argument to prove you can't partition the positive integers into 0 mod 2, 1 mod 4, and 3 mod 4. – Gerry Myerson May 24 '12 at 4:33 Trying an example with 1 mod 2 and 2 mod 2, I see that the terms together will cancel out the pole at -1. Andres has also mentioned that it is not possible for poles to cancel out like this when all the other terms have finite limits and one term has an infinite limit. I think I understand now. – Mark May 24 '12 at 13:53 Given an arithmetic progression $a,a+d,a+2d,\\dots$, you write down its generating function, $z^a+z^{a+d}+z^{a+2d}+\\dots$ which you recognize as a geometric series, so you sum the geometric series and get a function with certain poles. The generating function for the full set of positive integers is $z+z^2+z^3+\\dots$, which you recognnize as a geometric series, etc., etc. Partitioning the positive integers into arithmetic progressions is writing the generating function for the positive integers as the sum of generating functions for the arithmetic progressions. The desired conclusion comes from thinking about the poles on each side of the equation. This proof is due to D J Newman. - It seems that even with the more detailed answers, I'm still having", "# Geometric to Arithmetic Progression Algebra Level 3 In a sequence of four positive numbers, the first three are in geometric progression and the last three are in arithmetic progression. The first number is $$12$$ and the last number is $$\\frac{45}{2}$$. The sum of the two middle numbers can be written as $$\\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. Find $$a+b$$. ×", "# Deriving the Sum of the Arithmetic Sequence We have discussed arithmetic progression or arithmetic sequence and you have learned how to find its nth term. In the previous post, you have also learned how to find the sum of the first n positive integers. Notice that first n positive integers 1, 2, 3, 4, 5, all the way up to n is also an arithmetic sequence with first term 1 and constant difference 1. Now, the question that comes to mind is, how do we find the sum of an arithmetic sequence? Recall the method that we used in “Finding the sum of the first n positive integers.” We added the integers twice with the order of the terms reversed as shown above. Clearly, we can use this method to find the sum of the arithmetic sequence 3, 7, 11, 15, 19, 23, 27. Continue reading # Sum of the First n Positive Integers: Another Solution Late last month, we have seen how Gauss had possibly calculated the sum of the first n positive integers. In this post, we explore another solution. As we have done before, let us first start with a specific example, and then generalize later. What is the sum of the first 6 positive integers? If we let S be the sum of", "# Fun and Hard: The increasing sequence 2,3,5,6,7,10,11, ... consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. ×", "their geometric mean as $12 : 13$, then the numbers could be in the ratio: $12: 13$ $1/12: 1/13$ $4:9$ $2:3$ What is the first term of an arithmetic progression of positive integers? I. Sum of the squares of the first and second term is $116$. II. The fifth term is divisible by $7$. if the questions can be answered with the help of both the statements but not with ... . if the question can be answered with the help 0 statement II alone if the question can be answered with the help of statement I alone", "# Difference between revisions of \"1986 AIME Problems/Problem 7\" ## Problem The increasing sequence $1,3,4,9,10,12,13\\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $\\displaystyle 100^{\\mbox{th}}$ term of this sequence.", "The increasing sequence $$3, 15, 24, 48, \\ldots\\,$$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000? (第十二届AIME1994 第1题)", "# Arithmetic Mean! Algebra Level 4 Four positive integers are given.Select any three of these integers and find their A.M.,and add to this result add the fourth number.If the four answers thus got are $$17,21,23,29$$, then find the original numbers." ]

[ "implies that $3|d$, so only $9$ may work. Hence, the four terms are $18,\\ 27,\\ 36,\\ 48$, which indeed fits the given conditions. Their sum is $\\boxed{129}$. Postscript As another option, $3ad + 4d^2 = 30a + 30d$ could be rewritten as follows: $d(3a + 4d) = 30(a + d)$ $d(3a + 3d)+ d^2 = 30(a + d)$ $3d(a + d)+ d^2 = 30(a + d)$ $(3d - 30)(a + d)+ d^2 = 0$ $3(d - 10)(a + d)+ d^2 = 0$ This gives another way to prove $d<10$, and when rewritten one last time: $3(10 -d)(a + d) = d^2$ shows that $d$ must contain a factor of 3. -jackshi2006 EDIT by NealShrestha: Note that once we reach $3ad + 4d^2 = 30a + 30d$ this implies $3|d$ since all other terms are congruent to $0\\mod 3$. ## Solution 2 The sequence is of the form $a-d,$ $a,$ $a+d,$ $\\frac{(a+d)^2}{a}$. Since the first and last terms differ by 30, we have $$\\frac{(a+d)^2}{a}-a+d=30$$ $$d^2+3ad=30a$$ $$d^2+3ad-30a=0$$ $$d=\\frac{-3a + \\sqrt{9a^2+120a}}{2}.$$ Let $9a^2+120a=x^2$, where $x$ is an integer. This yields the following: $$9a^2+120a-x^2=0$$ $$a=\\frac{-120 + \\sqrt{14400+36x^2}}{18}$$ $$a=\\frac{-20 + \\sqrt{400+x^2}}{3}.$$ We then set $400+x^2=y^2$, where $y$ is an integer. Factoring using difference of squares, we have $$400=2^4 \\cdot 5^2=(y+x)(y-x).$$ Then, noticing that $y+x > y-x$, we set up several systems of equations", "10:59 Let $a,d,2m$ be the first term, the common difference, the number of terms respectively where $m\\in\\mathbb N$. This answer supposes that \"total of it's even terms $=30$\" means that $$(a+d)+(a+3d)+\\cdots +(a+(2m-1)d)=\\sum_{i=1}^{m}(a+(2i-1)d)=30,$$ i.e. $$am+2d\\cdot\\frac{m(m+1)}{2}-dm=30\\tag1$$ Also, this answer supposes that \"total of it's odd terms $=24$\" means that $$a+(a+2d)+\\cdots +(a+(2m-2)d)=\\sum_{i=1}^{m}(a+(2i-2)d)=24,$$ i.e. $$am+2d\\cdot\\frac{m(m+1)}{2}-2dm=24\\tag2$$ And we have $$|a+(2m-1)d-a|=10.5\\tag3$$ Now solve $(1)(2)(3)$ to get $a,d,2m$. From $(1)-(2)$, we have $d=\\frac 6m$. From $(3)$, we have $(2m-1)|\\frac 6m|=10.5\\Rightarrow m=4$. Finally, from $(1)$, we have $d=a=\\frac 32$.", "involving the factors of $400$. The second system we set up in this manner, $$y+x=2^3 \\cdot 5^2$$ $$y-x=2,$$ yields the solution $y=101, x=99$. Plugging back in, we get that $a=27 \\implies d=9$, so the sequence is $18,$ $27,$ $36,$ $48,$ and the answer is $\\boxed{129}.$ • Note: we do not have to check the other systems since the $x$ and $y$ values obtained via this system yield integers for $a$, $d$, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :) ## Solution 3 (Guesswork) We represent the values as $a-d$, $a$, $a+d$, and $\\frac{(a+d)^2}{a}$ Take the difference between the first and last values $$\\frac{(a+d)^2}{a}-a+d=30$$ Manipulating the values by expanding and then long division we see $$\\frac{a^2+2ad+d^2}{a}-a+d=30$$ $$\\frac{(a+2d)a+d^2}{a}-a+d=30$$ $$a+2d+\\frac{d^2}{a}-a+d=30$$ Combining like terms we get $$3d+\\frac{d^2}{a}=30$$ And looking at the values we know that since the second term must be positive (since both a and d are positive), $d$ must be a maximum of 9, which offers 9 possible values (since $d$ must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of $d=9$ at which $a=27$. The solution is thus $\\boxed{129}.$ • We could also lay down some further parameters for the value of $d$. as $a$", "on $$\\mathbb{Q}$$: $$\\frac{m}{n} < \\frac{m'}{n'}$$ if $$\\frac{m'}{n'} - \\frac{m}{n}$$ is positive (subtraction is defined by addition on the negation of the second operand). Proof of well-definedness. Assume $$\\frac{a}{b} < \\frac{c}{d}$$, $$(a, b) \\sim (a', b')$$, and $$(c, d) \\sim (c', d')$$. Since $$\\frac{a}{b} < \\frac{c}{d}$$, $$\\frac{c}{d} - \\frac{a}{b}$$ is positive, which means $$(cb - ad)bd > 0$$, so either $$cb - ad > 0$$ and $$bd > 0$$ or $$cb - ad < 0$$ and $$bd < 0$$. Since $$(a, b) \\sim (a', b')$$, $$ab' = ba'$$. Since $$(c, d) \\sim (c', d')$$, $$cd' = dc'$$. There are four cases to consider. Case 1: $$cb - ad > 0$$, $$bd > 0$$, and $$b'd' > 0$$. We have \\begin{align} & cb - ad > 0 \\\\ \\implies & cb > ad \\\\ \\implies & cbb'd' > adb'd' & \\text{(since } b'd' > 0 \\text{)} \\\\ \\implies & c'bb'd > a'dbd' & \\text{(since } cd' = dc' \\text{ and } ab' = ba' \\text{)} \\\\ \\implies & c'b' > a'd' & \\text{(since } bd > 0 \\text{)} \\\\ \\implies & c'b' - a'd' > 0 \\\\ \\implies & (c'b' - a'd')b'd' > 0 & \\text{(since } b'd' > 0 \\text{)} \\\\ \\implies & \\frac{c'b' - a'd'}{b'd'} \\text{ is positive} \\\\ \\implies & \\frac{c'}{d'} - \\frac{a'}{b'} \\text{ is positive} \\\\", "# 2003 AIME I Problems/Problem 8 ## Problem 8 In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms. ## Solution Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\\ a+d,\\ a+2d,\\ \\frac{(a + 2d)^2}{a + d}$. Since the first and fourth terms differ by $30$, we have that $\\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, $$(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).$$ This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$. Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $d > 10$ and $d < \\frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \\neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$", "# 2003 AIME I Problems/Problem 8 ## Problem 8 In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms. ## Solution Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\\ a+d,\\ a+2d,\\ \\frac{(a + 2d)^2}{a + d}$. Since the first and fourth terms differ by $30$, we have that $\\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, $$(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).$$ This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$. Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $d > 10$ and $d < \\frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \\neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$", "$a$ and $b$ are integers and $(a,b)=d$, then $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem 7. Suppose that $a\\ne 0$, $b\\ne 0$,", "the answer is $\\boxed{145}$. ~ math31415926535 ## Solution 2 (Easy Algebra) We can factor $d$ out of the last two equations. Therefore, it becomes $abc + d(bc + ac + ab) = 14$. Notice this is just $abc -4d$, since $bc + ac + ab = -4$. We now have $abc -4d = 14$ and $abcd = 30$. We then find $d$ in terms of $abc$, so $abc = \\frac{30}{d}-4d=14$. We solve for $d$ and find that it is either $\\dfrac32$ or $-5$. We can now try for these two values, and plug the rest into the equation. Thus, we have $33 + \\dfrac94 = \\dfrac{33 \\cdot 4 + 9}{4} = \\dfrac{132+9}{4} = \\dfrac{141}{4}$. We have $141 + 4 = \\boxed{145}$ and we're done. ~Arcticturn ## Solution 3 (Easy and Straightforward Algebra) $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$. Hence, $ab = 3c - 4$ Rewriting $abc+bcd+cda+dab = 14$, we get $ab(c+d) + cd(a+b) = 14$. Substitute $ab = 3c - 4$ and solving, we get, $3c^{2} - 4c - 4d - 14 = 0$ call this Equation 1 $abcd = 30$ gives $(3c-4)cd = 30$. So, $3c^{2}d - 4cd = 30$, which implies $d(3c^{2} - 4c) = 30$ or $3c^{2} - 4c = \\frac{30}{d}$ call this equation 2.", "Find this this diophantine equation the number Let $a,b$ be positive integer numbers. Find the number of pairs $(a,b)$ satisfying $$\\dfrac{ab}{1998}=\\sqrt{a^2+b^2}+a+b.$$ • I got a quick answer: $(1998.3,1998.4)$. – GAVD Aug 13 '15 at 10:12 • @MichaelGaluza: Sorry, I mean $(1998 \\times 3, 1998 \\times 4)$. – GAVD Aug 13 '15 at 10:41 At first, $\\sqrt{a^2+b^2}$ must be integer. So, $a$ and $b$ come from Pythagorean triples and $$a = d(n^2 - m^2),\\quad b=d\\cdot2nm$$ for some integers $d$, $n$, $m$ ($n > m$). Hence $\\sqrt{a^2+b^2}=d(n^2 + m^2)$, and then $$d\\frac{(n-m)(n+m)\\cdot 2nm}{1998} = n^2 + m^2 + n^2 - m^2 + 2nm=2n(n+m),$$ or $$dm(n-m)=1998\\Longrightarrow n = m + \\frac{1998}{md}.$$ Number of $(n,m)$-pairs is $$\\sum_{d\\mid 1998} \\sigma_0\\Big(\\frac{1998}{d}\\Big)=\\\\ = 16+ 8+ 12+ 6+ 8+ 4+ 4+ 8+ 2+ 4+ 6+ 3+ 4+ 2+ 2+ 1=90,$$ where $\\sigma_0(n)$ is the number of the divisors of $n$. But $(a, b)$ can repeats; I found $42$ (really $42$!) pairs ($(a,b)$ and $(b,a)$ are the same pairs, so there is $84$ solutions): $$(3997, 7988004), (3999, 2665332), (4000, 1999998), (4005, 891108), (4008, 669330), (4023, 299700), (4032, 225774), (4033, 219780), (4077, 102564), (4104, 77922), (4107, 75924), (4144, 57942), (4239, 36852), (4320, 28638), (4329, 27972), (4440, 21978), (4725, 14948), (4968, 12210), (4995, 11988), (5328, 9990), (5365, 9828), (6912, 6734), (6993, 6660), (7992, 5994), (8103, 5940), (9472, 5454), (12987,", "if we swap the position of the variables a and b), it can be expressed with the sum and product of a and b. Let c=a+b and d=ab. Then the quadratic equation $t^2-ct+d$ has two zeros, namely a and b. Thus, the discriminant $D=c^2-4d≥0$, which yields $d\\le \\frac {c^2}{4}$ ( and of course $c, d \\in\\mathbb Z$). In addition, $a^3+ab+b^3=61 \\iff a^3+b^3+ab=61$ $\\iff (a+b)^3-3(a+b)ab+ab=61$ $\\iff c^3-3cd+d=61$ $\\iff d=(c^3-61)/(3c-1)$ Since $d \\in\\mathbb Z$, $27d = \\frac{27c^3-1647}{3c-1} \\in\\mathbb Z$. But then, $\\frac{1646}{3c-1} \\in\\mathbb Z$, because $\\frac{27c^3-1647}{3c-1} = 9c^2+3c+1\\in\\mathbb Z$. This means that $3c-1$ must divide $1646=2\\cdot 823$ (823 is a prime). So $3c-1$ can be 1, 2, 823, 1646, -1, -2, -823, or -1646. Then $3c$ can be 2, 3, 824, 1647, 0, -1, -822, or -1645. Since c is an integer, the only possible values for c are c=0, 1, 549. This gives us the pair $(c,d) = (0,61), (1, -30), (549, 100528)$, and the only pair that $d\\le \\frac{c^2}{4}$ holds is $(1, -30)$. Thus, $c=1$ and $d=-30$, and it follows that a and b are the two zeros of the quadratic equation $x^2-x+30=0$. We now can conclude that the only solutions to the Diophantine equation $a^3+ab+b^3=61$ are $(a, b) = (6, -5), (-5, 6)$. - $$(a-b)(a^2+ab+b^2)=ab+61$$ $$\\implies a^2+ab+b^2 \\le \\mid ab+61\\mid$$ The negative case inside the absolute value", "# Powers and differences of positive integers Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5=b^4$, $c^3=d^2$, and $c-a=19$. Determine $d-b$. I know this question isn't particularly hard, but I've been having trouble getting any substantial ground. I've tried substituting that $c=a+19$ and $a=c-19$ into the first two equations, but it hasn't looked very useful. A path to a solution or a few hints would be great. Thanks! - We can say that $b$ divides $a$, and $d$ divides $c$. – Lee Yiyuan Jun 28 '14 at 15:13 From $a^5 = b^4 \\implies a = \\left(\\frac{b}{a}\\right)^4$, we can deduce that $\\frac{b}{a}$ is an integer. I will let $m = \\frac{b}{a}$. Similarly, from $c^3 = d^2 \\implies c = \\left(\\frac{d}{c}\\right)^2$, we can deduce that $\\frac{d}{c}$ is an integer. I will let $n = \\frac{d}{c}$. Since $a = m^4$ and $c = n^2$, we have $$n^2 - m^4 = 19$$ $$(n + m^2)(n - m^2) = 19$$ Since $19$ is a prime, we only have to check the cases for which $19 = 1\\cdot19$ and for which $19 = -1\\cdot-19$. If we let $n + m^2 = 19$ and $n - m^2 = 1$, then we can get $n = 10, m = 3$. I suppose it is trivial to show that all other cases do", "= \\frac{\\vert \\mathfrak{S}_D\\vert}{\\prod_{s\\in S} \\vert \\mathfrak{S}_{n(s,\\mathbf d)}\\vert }\\tag{2}.$$ The second inequality follows from $(1)$. On the left is what we want to find. ### The solution Since the number of elements in any symmetric group $\\mathfrak{S}_n$ is $n!$ (this often is taken as a definition of the factorial $n!$), we may immediately write down the solution from $(2)$ as $$\\vert \\mathfrak{S}_D \\cdot \\mathbf{d} \\vert = \\frac{D!}{\\prod_{s\\in S}n(s,\\mathbf d)!} = \\binom{D}{n(0, \\mathbf{d}), n(1, \\mathbf{d}), \\ldots, n(s-1, \\mathbf{d})}.$$ This is the multinomial coefficient determined by $\\mathbf d$. ### Example Let $\\mathbf{d} = (0,3,1,0)$ with $S=\\{0,1,2,3,4\\}$. Counting, \\eqalign{n(0,\\mathbf{d})=&2 \\\\ n(1,\\mathbf{d})=&1 \\\\ n(2,\\mathbf{d})=&0 \\\\ n(3,\\mathbf{d})=&1 \\\\ n(4,\\mathbf{d})=&0.} Since $D =4$, the answer is $$\\vert \\mathfrak{S}_D \\cdot \\mathbf{(0,1,3,0)} \\vert = \\binom{4}{2,1,0,1,0} = \\frac{4!}{2!1!0!1!0!}= \\frac{24}{2}=12.$$ Indeed, the twelve equivalent permutations of $\\mathbf d$ are (in an abbreviated notation, dropping parentheses and commas) \\eqalign{\\mathfrak{S}_4 \\cdot 0130=\\{ &0013,0031,0103,0130,0301,0310,\\\\&1003,1030,1300,3001,3010,3100\\ \\}.} ### Alternative solution Having seen the appearance of the multinomial coefficients, it becomes obvious the answer must be the coefficient of $\\mathbf{x}^\\mathbf{d} = x_0^{d_0}x_1^{d_1}\\cdots x_{s-1}^{d_{s-1}}$ in the multinomial expansion of $$\\left(\\mathbf{x}\\cdot \\mathbf{1}\\right)^D = (x_0+x_1+\\cdots + x_{s-1})^D = \\\\\\sum_{n_0+n_1+\\cdots+n_{s-1}=D}\\binom{D}{n_0,n_1,\\ldots, n_{s-1}}x_0^{n_0}x_1^{n_1}\\cdots x_{s-1}^{n_{s-1}},$$ because it counts the number of distinct ways of collecting $n(0,\\mathbf d)$ zeros, $n(1,\\mathbf {d})$ ones, etc, from the terms in the product and these enumerate all the distinguishable permutations of $\\mathbf{d}$. In the case of two", "2C = 3D and A, B, C, and D are positive integ [#permalink] Show Tags 26 Jul 2019, 09:02 1 The least possible value of B ? $$2C=3D$$ --> Equation (1): $$C=\\frac{3}{2}D$$ $$A+B=C$$ --> Equation (2): $$A+B=\\frac{3}{2}D$$ $$A+D=B$$ --> Equation (3): $$A-B=-D$$ Equation (2) - Equation (3) : $$2B=\\frac{5}{2}D$$ --> $$B=\\frac{5}{4}D$$ .....Equation (3) If $$B$$ and D each have to be positive integer, then we deduce that the least possible value of D is 4 from Equation (4). Consequently, the least possible value of B is $$B=\\frac{5}{4}(4) = 5$$. QUICK VERIFICATION: $$D=4;$$ $$C=\\frac{3}{2}D = 6;$$ $$B=5;$$ $$A-B=-D$$ --> $$A-5=-4$$ --> $$A=1$$ A,B,C,D are all positive integers and satisfy all known equations. Manager Joined: 24 Jun 2019 Posts: 111 Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ [#permalink] Show Tags 26 Jul 2019, 09:04 1 Quote: If A+B=C, A+D=B, 2C=3D and A, B, C, and D are positive integers, what is the least possible value of B? All variables are positive integers. We can try to determine their relations with eachother A+B=C... So, A<C & B<C........(1) A+D=B.... So, A<B & D<B...... (2) combining (1) and (2)..... A<B<C & D<B ........(3)... we dont know if D is less than A or greater than A", "### 2008 IMO SL #A5 Let $a$, $b$, $c$, $d$ be positive real numbers such that $abcd = 1$ and $a + b + c + d > \\dfrac{a}{b} + \\dfrac{b}{c} + \\dfrac{c}{d} + \\dfrac{d}{a}$. Prove that $a + b + c + d < \\dfrac{b}{a} + \\dfrac{c}{b} + \\dfrac{d}{c} + \\dfrac{a}{d}$ Solution 1: By Cauchy-Schwarz, we have $$(a+b+c+d)(ab+bc+cd+da)>\\left(\\frac ab+\\frac bc+\\frac cd+\\frac da\\right)(ab+bc+cd+da)$$$$\\ge(a+b+c+d)^2,$$which implies that $ab+bc+cd+da>a+b+c+d$. Again, by Cauchy-Schwarz, it follows that $$(a+b+c+d) \\left(\\frac ba+\\frac cb+\\frac dc+\\frac ad\\right) > \\left(\\frac ab+\\frac bc+\\frac cd+\\frac da\\right) \\left(\\frac dc+\\frac ad+\\frac ba+\\frac cb\\right)$$ $$\\ge\\left(\\sqrt{\\frac{da}{bc}}+\\sqrt{\\frac{ab}{cd}}+\\sqrt{\\frac{bc}{da}}+\\sqrt{\\frac{cd}{ab}}\\right)^2$$$$=(da+ab+bc+cd)^2 \\ge(a+b+c+d)^2,$$which implies $\\frac ba+\\frac cb+\\frac dc+\\frac ad>a+b+c+d$, as desired. $\\square$ Solution 2: Define $$S_1 = \\frac ab +\\frac bc +\\frac cd +\\frac da$$$$S_2 = \\frac ba +\\frac cb +\\frac dc +\\frac ad$$$$W =a+b+c+d$$. We claim that $3S_1+S_2 \\geq 4W$ . Clearly, this finishes because then we have $S_2 \\geq W + \\underbrace{3(W-S_1)}_{>0} >W$. To prove our claim, note that $$2 \\cdot \\frac ab + \\frac bc + \\frac ad \\overset{\\text{AM-GM}}{\\geq} 4\\sqrt [4]{\\frac {a}{d}\\cdot \\frac {a^2}{b^2} \\cdot \\frac {b}{c}} \\geq 4a$$. Cyclically summing up gives the desired claim, so we are done. $\\square$ 1. Anonymous3/02/2022 im so addicted to your blog that i need to stop going to it every 3 minutes to check for a new post", "is where $$\\frac{a^2}{3}> \\frac{a^2-81}{2} \\indent (4)$$ which simplifies to $$\\sqrt{243}>a$$ meaning $$16>a(5)$$ So now we have that $9 from $(2)$, $a<16$ from $(5)$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=\\boxed{440}$ AWD. ## Solution 3 (Calculus, not as good) Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$. Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root", "# Math Help - Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms. 1. ## Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms. Yo, some questions. As in how do u do these. Thanks. T5 = 3T2 and S6= 36, find the first three terms. also Sn = 5n2(sqaured that is) + 3n - Find the first three terms and thus find a formula for Tn Sweet as Thanks. 2. $a_5=a_2+(5-2)d$ and we know that $a_5=3a_2$ so $2a_2=3d$. And we were also given: $36=S_6=\\frac {6(a_1+a_6)}2=3((a_2-d)+(a_2+4d))=3(2a_2+3d)=18d\\rightarrow d=2$. Using this: $a_2=6$. So the first three terms: $4, 6, 8$. For the second problem: $S_n=n(3+5n)=\\frac{n(6+10n)}2=\\frac{n(6+10(n-1)+10)}2=\\frac{n(2\\cdot 8+(n-1)\\cdot 10)}2$. You can see that $a_1=8$ and $d=10$ therefore $a_n=8+(n-1)\\cdot 10$ and the first three terms: $8, 18, 28$. 3. Hello, Dev! We are expected to know these formulas: . . $n^{th}\\text{ term: }\\;T_n \\:=\\:a + (n-1)d$ . . $\\text{Sum of first }n\\text{ terms: }\\;S_n \\;=\\;\\frac{n}{2}[2a + (n-1)d]$ . . . . where: . $\\begin{Bmatrix}a &=& \\text{first term} \\\\ d &=& \\text{common diff.} \\\\ n &=& \\text{no. of terms} \\end{Bmatrix}$ $T_5 \\:=\\: 3\\!\\cdot\\!T_2\\:\\text{ and }\\:S_6\\:=\\: 36$ Find the first three terms. 2nd term: . $T_2 \\:=\\:a+d$ 5th term: . $T_5 \\:=\\:a+4d$ $T_5 = 3\\!\\cdot\\!T_2 \\quad\\Rightarrow\\quad a + 4d \\:=\\:3(a+d) \\quad\\Rightarrow\\quad d \\:=\\:2a$ .[1] $S_6", "$\\frac{29}{2}\\left(197+225\\right)\\ =\\ 6119$ Alternatively, The final term in each group is the square of the group number. In the first group 1, second group 4, ………… The final element of the 14th group is $\\left(14\\right)^2=\\ 196$, similarly for the 15th group this is : $\\left(15\\right)^2=\\ 225$ Each group contains all the consecutive elements in this range. Hence the 15th group the elements are: (197, 198, …………………………..225). This is an Arithmetic Progression with a common difference of 1 and the number of element 29. Hence the sum is given by : $\\frac{n}{2}\\cdot\\left(first\\ term\\ +last\\ term\\right)$ $\\frac{29}{2}\\cdot\\left(197+225\\right)$ 6119. Question 13: Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals a) 8 b) 12 c) 14 d) 10 Solution: Given x, y, z are three terms in an arithmetic progression. Considering x = a, y = a+d, z = a+2*d. Using the given equation x*y*z = 5*(x+y+z) a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d) =a*(a+d)*(a+2*d) = 5*(3*a+3*d) = 15*(a+d). = a*(a+2*d) = 15. Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers. The common difference is positive and greater than 2. Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1) Hence the", "Beautiful math on $\\LaTeX$ in less than 10 minutes # Arithmetic ## Basic Operators Basic operators work as expected 1 + 2 - 3 = 0 $1 + 2 - 3 = 0$ Division can be done as follows 1/2 = 1 \\div 2 = \\frac{1}{2} $1/2 = 1 \\div 2 = \\frac{1}{2}$ To typeset bigger fractions, use \\dfrac{1}{2} $\\dfrac{1}{2}$ Or use \\displaystyle. \\displaystyle \\frac{1}{3} $\\displaystyle \\frac{1}{3}$ Multiplication 2 \\cdot 3 = 2 \\times 3 $2 \\cdot 3 = 2 \\times 3$ Avoid using * for multiplication 2 * 3 $2 * 3$ Repeated decimals can be denoted by a bar \\frac{1}{3} = 0.\\bar 3 $\\frac{1}{3} = 0.\\bar 3$ ## Exponents and Indexes Exponents can by typeset with a caret 2^3 = 8 $2^3 = 8$ Use braces to group exponents … 2^{10} $2^{10}$ … to avoid 2^10 $2^10$ Subscripts, like indices, can by typeset as follows a_n = 2 \\cdot a_{n-1} $a_n = 2 \\cdot a_{n-1}$ ## Square Roots A square root can be typeset with \\sqrt{16} = 4 $\\sqrt{16} = 4$ To take the $n^\\text{th}$ root, use \\sqrt[3]{27} = 3 $\\sqrt[3]{27} = 3$ \\pm becomes a plus-minus. x^2 = 4 \\implies x = \\pm \\sqrt{4} $x^2 = 4 \\implies x = \\pm \\sqrt{4}$ Braces can be omitted if the argument is only 1 symbol \\sqrt 2", "& 22 \\\\ 23 & 24 & 25 & 26 \\\\ 27 & 28 & 29 & 30\\end{pmatrix}$ ... and so on to matrix $M_n$. Find the sum of elements of the main diagonal. We have: . $\\begin{array}{cccc} 1 & = & 1 & [1] \\\\ 2 + 5 & = & 7 & [2]\\\\ 6 + 10 + 14 & = & 30 & [3]\\\\ 15 + 20 + 25 + 30 & = & 90 & [4]\\\\ 31 + 37 + 43 + 49 + 55 & = & 215 & [5]\\end{array}$ Each is the sum of an arithmetic series. . . $\\begin{array}{cccc}\\text{In [3]:} & a = 6 & d = 4 & n = 3 \\\\ \\text{In [4]:} & a = 15 & d = 5 & n = 4 \\\\ \\text{In [5]:} & a = 31 & d = 6 & n = 5\\end{array}$ In the $n^{th}$ matrix, the common difference is $n+1$, and there are $n$ terms. We must determine the first terms of these series: $1,\\,2,\\,6,\\,15,\\,31,\\,56\\cdots$ After some algebra, we find that: . $a \\:=\\:\\frac{1}{6}(n+1)(2n^2-5n+6)$ The sum of an arithmetic series is: . $S \\;=\\;\\frac{n}{2}[2a + (n-1)d]$ So we have: . $S \\;=\\;\\frac{n}{2}\\left[2\\cdot\\frac{1}{6}(n+1)(2n^2-5n+6) + (n-1)(n+1)\\right]$ . . which simplifies to: . $\\boxed{S \\;=\\;\\frac{n}{6}(n+1)(2n^2-2n+3)}$", "$xd^2,$ so we can write that $$3x^2+2d^2=xd^2,$$ and it follows that $$3x^2-xd^2+2d^2=3x^2-\\left(d^2\\right)x+2d^2=0.$$ Now, we can treat $d$ as a constant and use the quadratic formula to get $$x=\\frac{d^2\\pm \\sqrt{d^4-4(3)(2d^2)}}{6}.$$ We can factor pull $d^2$ out of the square root to get $$x=\\frac{d^2\\pm d\\sqrt{d^2-24}}{6}.$$ Here, it is easy to test values of $d$. We find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\\sqrt{d^2-24}$ a positive integer.$^{*}$ $d=5$ gives $x=5$ and $x=\\frac{10}{3}$, but we can ignore the latter. $d=7$ gives $x=14$, as well as a fraction which we can ignore. Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\\boxed{031}$. -BorealBear, minor edit by Kinglogic $^*$To prove this, let $\\sqrt{d^2-24} = k$, then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ then remembering that $d$ and $k$ are integers see if you can figure it out. -PureSwag ## Solution 3 Proceed as in solution 2, until we reach $$3x^2+2d^2=xd^2,$$ Write $d^2=\\frac{3x^2}{x-2}$, it follows that $x-2=3k^2$ for some (positive) integer k and $k \\mid x$. Taking both sides modulo $k$, $-2 \\equiv 0 \\pmod{k}$, so $k \\mid 2 \\rightarrow k=1,2$. When $k=1$, we have $x=5$ and $d=5$. When $k=2$, we have $x=14$ and $d=7$. Summing the two cases, we have $10+21=\\framebox{031}$. -Ross Gao ## Solution" ]

[ "# 2003 AIME I Problems/Problem 8 ## Problem 8 In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms. ## Solution Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\\ a+d,\\ a+2d,\\ \\frac{(a + 2d)^2}{a + d}$. Since the first and fourth terms differ by $30$, we have that $\\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, $$(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).$$ This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$. Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $d > 10$ and $d < \\frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \\neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$", "# 2003 AIME I Problems/Problem 8 ## Problem 8 In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms. ## Solution Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\\ a+d,\\ a+2d,\\ \\frac{(a + 2d)^2}{a + d}$. Since the first and fourth terms differ by $30$, we have that $\\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, $$(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).$$ This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$. Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $d > 10$ and $d < \\frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \\neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$", "$\\frac{29}{2}\\left(197+225\\right)\\ =\\ 6119$ Alternatively, The final term in each group is the square of the group number. In the first group 1, second group 4, ………… The final element of the 14th group is $\\left(14\\right)^2=\\ 196$, similarly for the 15th group this is : $\\left(15\\right)^2=\\ 225$ Each group contains all the consecutive elements in this range. Hence the 15th group the elements are: (197, 198, …………………………..225). This is an Arithmetic Progression with a common difference of 1 and the number of element 29. Hence the sum is given by : $\\frac{n}{2}\\cdot\\left(first\\ term\\ +last\\ term\\right)$ $\\frac{29}{2}\\cdot\\left(197+225\\right)$ 6119. Question 13: Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals a) 8 b) 12 c) 14 d) 10 Solution: Given x, y, z are three terms in an arithmetic progression. Considering x = a, y = a+d, z = a+2*d. Using the given equation x*y*z = 5*(x+y+z) a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d) =a*(a+d)*(a+2*d) = 5*(3*a+3*d) = 15*(a+d). = a*(a+2*d) = 15. Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers. The common difference is positive and greater than 2. Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1) Hence the", "implies that $3|d$, so only $9$ may work. Hence, the four terms are $18,\\ 27,\\ 36,\\ 48$, which indeed fits the given conditions. Their sum is $\\boxed{129}$. Postscript As another option, $3ad + 4d^2 = 30a + 30d$ could be rewritten as follows: $d(3a + 4d) = 30(a + d)$ $d(3a + 3d)+ d^2 = 30(a + d)$ $3d(a + d)+ d^2 = 30(a + d)$ $(3d - 30)(a + d)+ d^2 = 0$ $3(d - 10)(a + d)+ d^2 = 0$ This gives another way to prove $d<10$, and when rewritten one last time: $3(10 -d)(a + d) = d^2$ shows that $d$ must contain a factor of 3. -jackshi2006 EDIT by NealShrestha: Note that once we reach $3ad + 4d^2 = 30a + 30d$ this implies $3|d$ since all other terms are congruent to $0\\mod 3$. ## Solution 2 The sequence is of the form $a-d,$ $a,$ $a+d,$ $\\frac{(a+d)^2}{a}$. Since the first and last terms differ by 30, we have $$\\frac{(a+d)^2}{a}-a+d=30$$ $$d^2+3ad=30a$$ $$d^2+3ad-30a=0$$ $$d=\\frac{-3a + \\sqrt{9a^2+120a}}{2}.$$ Let $9a^2+120a=x^2$, where $x$ is an integer. This yields the following: $$9a^2+120a-x^2=0$$ $$a=\\frac{-120 + \\sqrt{14400+36x^2}}{18}$$ $$a=\\frac{-20 + \\sqrt{400+x^2}}{3}.$$ We then set $400+x^2=y^2$, where $y$ is an integer. Factoring using difference of squares, we have $$400=2^4 \\cdot 5^2=(y+x)(y-x).$$ Then, noticing that $y+x > y-x$, we set up several systems of equations", "# Number Theory 1. (20 p.) If the corresponding terms of two arithmetic progressions are multiplied we get the sequence 1440, 1716, 1848, ... . Find the eighth term of this sequence. 2. (12 p.) Find the least positive integer $$n$$ such that when its leftmost digit is deleted, the resulting integer is equal to $$n/29$$. 3. (20 p.) Let $$0 < a < b < c < d$$ be integers such that $$a$$, $$b$$, $$c$$ is an arithmetic progression, $$b$$, $$c$$, $$d$$ is a geometric progression, and $$d - a = 30$$. Find $$a + b + c + d$$. 4. (7 p.) Let $$a$$, $$b$$, $$c$$ be positive integers forming an increasing geometric sequence such that $$b-a$$ is a square. If $$\\log_6a + \\log_6b + \\log_6c = 6$$, find $$a + b + c$$. 5. (38 p.) Let $$a,b,c$$ and $$d$$ be positive real numbers such that $$a^2+b^2-c^2-d^2=0$$ and $$a^2-b^2-c^2+d^2=\\frac {56}{53}(bc+ad)$$, Let $$M$$ be the maximum possible value of $$\\frac {ab+cd}{bc+ad}$$ ,If $$M$$ can be expressed as $$\\frac {m}{n}$$,$$(m,n)=1$$ then find $$100m+n$$ 2005-2017 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax", "$x^2+y^2-2by=0$ where $b$ must be positive $x^2+y^2-2by=0$ for any given $b \\in \\mathbb{R}$ 36 If $l=1+a+a^2+ \\dots$, $m=1+b+b^2+ \\dots$, and $n=1+c+c^2+ \\dots$, where $\\mid a \\mid <1, \\: \\mid b \\mid < 1, \\: \\mid c \\mid <1$ and $a,b,c$ are in arithmetic progression, then $l, m, n$ are in arithmetic progression geometric progression harmonic progression none of these 37 If the sum of the first $n$ terms of an arithmetic progression is $cn^2$, then the sum of squares of these $n$ terms is $\\frac{n(4n^2-1)c^2}{6}$ $\\frac{n(4n^2+1)c^2}{3}$ $\\frac{n(4n^2-1)c^2}{3}$ $\\frac{n(4n^2+1)c^2}{6}$ 1 vote The sum $\\dfrac{n}{n^2}+\\dfrac{n}{n^2+1^2}+\\dfrac{n}{n^2+2^2}+ \\cdots + \\dfrac{n}{n^2+(n-1)^2} + \\cdots \\cdots$ is $\\frac{\\pi}{4}$ $\\frac{\\pi}{8}$ $\\frac{\\pi}{6}$ $2 \\pi$ Let $y=[\\:\\log_{10}3245.7\\:]$ where $[ a ]$ denotes the greatest integer less than or equal to $a$. Then $y=0$ $y=1$ $y=2$ $y=3$ The number of integer solutions for the equation $x^2+y^2=2011$ is $0$ $1$ $2$ $3$", "a better one. – Ross Millikan Apr 8 '16 at 14:33 As some people said at the comments the arithmetic progression can also be rewritten as ## $$-3d, -2d, -d, 0, d, 2d, 3d, 4d, 5d, 6d$$ As we can see this is also an arithmetic progression. We know that $$S_n = \\frac n2\\left(2a_1 + d(n-1)\\right)$$ So I create two arithmetic progressions: $$a_1 = -3, \\quad d_a = 1$$ $$b_1 = 6, \\quad d_b = -1$$ One for last terms and one for first terms. We can say now that: $$5Sa_k = Sb_k$$ $$5\\frac n2(-6+k-1) = \\frac n2(12-(k-1))$$ We cancel out $\\frac n2$ on both sides. $$5(-6+k-1) = (12-(k-1))$$ $$5(-7+k) = 12-k+1$$ $$-35 + 5k = 12 - k +1$$ $$-48 = -6k$$ From here we find $k=8$. A trick for resolving this type of questions is to write the arithmetic progression even if you don't think you have the necessary information, as in the top. We can set $a_j=(j-4)d$ for $j\\in\\{1,2,\\cdots,10\\}$. Then $$S_{k\\text{ first}}= a_1 + a_2+ \\cdots + a_k =\\sum_{i=1}^k (i-4)d =\\left(\\frac{k(k+1)}{2}-4k\\right)d$$ and $$S_{k\\text{ last}}= \\sum_{i=11-k}^{10}(i-4)d = \\sum_{i=1}^k (7-i)d =\\left(\\frac{-k(k+1)}{2} +7k\\right)d.$$ Since $d\\ne 0$, $$\\left(\\frac{-k(k+1)}{2} +7k\\right)=5\\left(\\frac{k(k+1)}{2}-4k\\right)$$ and its roots are $k=0$ or $k=8$. Since $1\\le k\\le 10$, $k=8$. • Why is $a_n = (n-4)d$? – Pichi Wuana Apr 8 '16 at 14:22 • @PichiWuana Because you", "+0 # Algebra 0 79 3 The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D? Guest Mar 21, 2018 #2 +92206 +2 The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D? arithmetic progression A B, C= A+2(B-A) geometric progression B r=C/B=5/3 B, B(5/3), B(5/3)^2 B, 5B/3, 25B/9 so C=5B/3 D=25B/9 A, B, 5B/3, 25B/9 B - A = 5B/3 - B 2B-5B/3 = A (2-5/3)B = A A=B/3 $$A, B,C, D \\\\ \\frac{B}{3},\\;B, \\;\\frac{5B}{3},\\;\\frac{25B}{9}$$ These all have to be positive integers so B must be a multiple of 9, The smallest values are if B is 9 $$\\frac{9}{3},\\;9, \\;\\frac{45}{3},\\;\\frac{9*25}{9}\\\\ 3,\\;9, \\;15,\\;25\\\\$$ So the smallest possible value for A+B+C+D = 3+9+15+25 = 52 Melody Mar 22, 2018 Sort: #1 0 AP = 1, 3, 5...............etc. GP = 3, 5, 8 1/3.........etc. A + B + C + D =1 + 3 + 5 + 8 1/3 = 17 1/3 Guest Mar 21, 2018 #2 +92206 +2 The", "involving the factors of $400$. The second system we set up in this manner, $$y+x=2^3 \\cdot 5^2$$ $$y-x=2,$$ yields the solution $y=101, x=99$. Plugging back in, we get that $a=27 \\implies d=9$, so the sequence is $18,$ $27,$ $36,$ $48,$ and the answer is $\\boxed{129}.$ • Note: we do not have to check the other systems since the $x$ and $y$ values obtained via this system yield integers for $a$, $d$, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :) ## Solution 3 (Guesswork) We represent the values as $a-d$, $a$, $a+d$, and $\\frac{(a+d)^2}{a}$ Take the difference between the first and last values $$\\frac{(a+d)^2}{a}-a+d=30$$ Manipulating the values by expanding and then long division we see $$\\frac{a^2+2ad+d^2}{a}-a+d=30$$ $$\\frac{(a+2d)a+d^2}{a}-a+d=30$$ $$a+2d+\\frac{d^2}{a}-a+d=30$$ Combining like terms we get $$3d+\\frac{d^2}{a}=30$$ And looking at the values we know that since the second term must be positive (since both a and d are positive), $d$ must be a maximum of 9, which offers 9 possible values (since $d$ must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of $d=9$ at which $a=27$. The solution is thus $\\boxed{129}.$ • We could also lay down some further parameters for the value of $d$. as $a$", "that the three numbers $a, z$ and $b$ form a GP. The geometric mean of n numbers $a_1, a_2, \\cdots, a_n$ is given by the n-th root of the product of the given numbers. More precisely, in this case, we have GM $=\\sqrt[n]{a_1 a_2 \\cdots a_n}.$ ## Formulas of Geometric Progressions For a geometric progression (GP) having the first term $a$ and the common ratio $r$, we have the following results of the GP. • The GP has the form $a, ar, ar^2, \\cdots$ • The n-th term $a_n=ar^{n-1}$ • The sum of the first n terms is given as follows: $S$ $=\\frac{a(1-r^n)}{1-r}$ if $-11$ • The geometric mean of $a$ and $b$ is $\\sqrt{ab}$ • If the three numbers are in a GP, then we should assume the numbers as follows: $\\frac{a}{r},$ $a$ and $ar.$ ## Solved Problems of Geometric Progression Problem 1: Find a few terms of a geometric progression with first term $2$ and common ratio $\\frac{1}{3}$ Solution: The first term is $a=2$ and the common ration is $r=\\frac{1}{3}.$ So the second term $=2 \\times \\frac{1}{3}$ $=\\frac{2}{3}$, the third term $=\\frac{2}{3} \\times \\frac{1}{3}$ $=\\frac{2}{9}$, the fourth term $=\\frac{2}{9} \\times \\frac{1}{3}$ $=\\frac{2}{27}.$ So the few term of the GP are $2, \\frac{2}{3}, \\frac{2}{9}, \\frac{2}{27} \\cdots$ Share via:", "# Three integers are in the ratio 2:3:8. If 4 is added to the middle number, the resulting number is the second term of a geometric progression of which the other two integers are the first and third terms. How do you find the three integers? Nov 25, 2015 See explanation... #### Explanation: If the first term is $2 x$ then the sequence formed by adding $4$ to the middle number is: $2 x , 3 x + 4 , 8 x$ The middle of three terms of a geometric sequence is equal to the geometric mean of the preceding and following terms, so: $3 x + 4 = \\pm \\sqrt{2 x \\cdot 8 x} = \\pm \\sqrt{16 {x}^{2}} = \\pm 4 x$ If $3 x + 4 = 4 x$ then $x = 4$ and our original integers are $8$, $12$, and $32$. The geometric sequence is $8$, $16$, $32$ with common ratio $2$. If $3 x + 4 = - 4 x$ then $x = - \\frac{4}{7}$, but this is not an integer and does not give rise to integers when multiplied by $2$, $3$ and $8$. Out of curiosity let's look at this alternative non-integer solution: $2 x = - \\frac{8}{7}$ $3 x + 4 = - \\frac{12}{7} + 4 = \\frac{16}{7}$ $8 x =", "4 = 5$$ $$d=a_{3} − a_{2} = 16 − 9 = 7$$ This series is not Arithmetic Series because the common difference is not the same Continuation from the last step: $$4, 9, 16, ...$$ Here, $$r=\\frac{a_{2}}{a_{1}}=\\frac{9}{4}$$ $$r=(a_{3})/(a_{2}) = \\frac{16}{9}$$ This series is not Geometric Series because the common ratio is not the same Part d $$50, 60, 70, ...$$ Here, $$d=a_{2} − a_{1} = 60 − 50 = 10$$ $$d=a_{3} − a_{2} = 70 − 60 = 10$$ This series is Arithmetic Series with a common difference of 10 Part e $$12, 3, 18, ...$$ Here, $$r=\\frac{a_{2}}{a_{1}}=\\frac{3}{\\frac{1}{2}}=6$$ $$r=\\frac{a_{3}}{a_{2}}=\\frac{18}{3}=6$$ This series is a Geometric Series with a common ratio of 6 ### Relevant Questions Determine whether the given sequence could be geometric,arithmetic, or neither. If possoble, identify the common ratio or difference. 9, 13, 17, 21, .... a. arithmetic $$d = 4$$ b. geometric $$r = 4$$ c. geometric $$r = \\frac{1}{4}$$ d. neither Write A if the sequence is arithmetic, G if it is geometric, H if it is harmonic, F if Fibonacci, and O if it is not one of the mentioned types. Show your Solution. a. $$\\frac{1}{3}, \\frac{2}{9}, \\frac{3}{27}, \\frac{4}{81}, ...$$ b. $$3, 8, 13, 18, ..., 48$$ 1. Is the sequence $$0.3, 1.2, 2.1, 3, ...$$ arithmetic? If so find the common difference. 2.", "our four solutions are \\begin{eqnarray*} 148,~481,~814 \\\\ 185,~518,~851 \\\\ 259,~592,~925 \\\\ 296,~629,~962 \\\\ \\end{eqnarray*} If you are wondering if these numbers came out of nowhere, check out the decimal equivalents of the $27$ths. It is no coincidence that $\\displaystyle\\frac{4}{27}=0.\\overline{148}$, $\\displaystyle\\frac{13}{27}=0.\\overline{481}$, and $\\displaystyle\\frac{22}{27}=0.\\overline{814}$. #### Geometric Progression Now let's look at the last problem, again repeated for clarity: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form a geometric progression. For the geomtric sequence \\begin{eqnarray*} N_2-r\\cdot N_1&=& 100B+10C+A-r(100A+10B+C)= 0\\\\ N_3-r\\cdot N_2&=& 100C+10A+B-r(100B+10C+A)= 0 \\end{eqnarray*} There are other manipulations, but I haven't found any that are linearly independent of the two equations shown. Not only do we have two equations and four unknowns, the equations are non-linear. With some simplification, we have \\begin{eqnarray*} 100B+10C+A-r(100A+10B+C)= 0\\\\ 100C+10A+B-r(100B+10C+A)= 0 \\end{eqnarray*} Using the symbolic toolbox and rref we can simplify % a b c [ 1, 0, (r - 10)/(10*r^2 - 1)] [ 0, 1, (r*(r - 10))/(10*r^2 - 1)] Not incredibly helpful. However, we can reverse the columns of the array (now instead of A,B,C we have C,B,A) and rref gives us % c b a [ 1, 0, (10*r^2 - 1)/(r - 10)] [ 0, 1, -r] Which is a little cleaner. So where do we go from here?", "compute $S_{3n}$. Let $d\\ne 0$ be the common difference of an arithmetic sequence $\\{a_n\\}$, and positive rational number $q < 1$ be the common ratio of a geometric sequence $\\{b_n\\}$. If $a_1=d$, $b_1=d^2$, and $\\frac{a_1^2+a_2^2+a_3^2}{b_1+b_2+b_3}$ is a positive integer, what is the value of $q$? Let $S_n$ be the sum of the first $n$ terms in geometric sequence $\\{a_n\\}$. If all $a_n$ are real numbers and $S_{10}=10$, and $S_{30}=70$, compute $S_{40}$. Expanding $$\\Big(\\sqrt{x}+\\frac{1}{2\\sqrt[4]{x}}\\Big)^n$$ and arranging all the terms in descending order of $x$'s power, if the coefficients of the first three terms form an arithmetic sequence, how many terms with integer power of $x$ are there? Suppose sequence $\\{F_n\\}$ is defined as $$F_n=\\frac{1}{\\sqrt{5}}\\Big[\\Big(\\frac{1+\\sqrt{5}}{2}\\Big)^n-\\Big(\\frac{1-\\sqrt{5}}{2}\\Big)^n\\Big]$$ for all $n\\in\\mathbb{N}$. Let $$S_n=C_n^1\\cdot F_1 + C_n^2\\cdot F_2+\\cdots +C_n^n\\cdot F_n.$$ Find all positive integer $n$ such that $S_n$ is divisible by 8. Solve $\\{L_n\\}$ which is defined as $F_1=1, F_2=3$ and $F_{n+1}=F_{n}+F_{n-1}, (n = 2, 3, 4, \\cdots)$", "is 40, find the first term of the sequence? Slightly different algebraic approach......... the first, third and thirteenth terms of an arithmetic progression are in geometric progression First term = a Third Term = a+2d Thirteenth Term = a+12d They are in Geometric Progression too So $$\\frac{(a+2d)}{a} = \\frac{(a+12d)}{(a+2d)}$$ --------------------------> {When a, b, c, d are in GP then b/a = c/b = d/c.........} $$(a+2d)^2 = a(a+12d)$$ ----------> $$a^2 + 4d^2 + 4ad = a^2 + 12ad$$ -----------> d^2 = 2ad -----------> d(d-2a)=0---------> Statement 1 the sum of the fourth and seventh terms of this arithmetic progression is 40 -------> (a+3d) + (a+6d) = 40 --------> 2a + 9d = 40 --------> $$d = \\frac{40 - 2a}{9}$$ Putting $$d = \\frac{40 - 2a}{9}$$ in the Statement 1, we get $$(\\frac{40 - 2a}{9})$$$$(\\frac{40 - 2a}{9} - 2a)$$ -------------------> $$(\\frac{40 - 2a}{9})$$$$(\\frac{40 - 20a}{9})$$ ----------> $$\\frac{(40-2a)(40-20a)}{81} = 0$$ ------------------> Either (40-2a)=0 ------> a = 20 or (40-20a)=0 -----> a = 2, which is the answer _________________ Director Joined: 23 Jan 2013 Posts: 591 Schools: Cambridge'16 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 26 Oct 2014, 23:53 a1+3d+a1+6d=40 2a1+9d=40 a1=20-9d/2 all answer choices are integers, so 9d/2 should be integer less than 20 and 9d is even. Only d=4 gives 18 to get", "the sequence? Slightly different algebraic approach......... the first, third and thirteenth terms of an arithmetic progression are in geometric progression First term = a Third Term = a+2d Thirteenth Term = a+12d They are in Geometric Progression too So $$\\frac{(a+2d)}{a} = \\frac{(a+12d)}{(a+2d)}$$ --------------------------> {When a, b, c, d are in GP then b/a = c/b = d/c.........} $$(a+2d)^2 = a(a+12d)$$ ----------> $$a^2 + 4d^2 + 4ad = a^2 + 12ad$$ -----------> d^2 = 2ad -----------> d(d-2a)=0---------> Statement 1 the sum of the fourth and seventh terms of this arithmetic progression is 40 -------> (a+3d) + (a+6d) = 40 --------> 2a + 9d = 40 --------> $$d = \\frac{40 - 2a}{9}$$ Putting $$d = \\frac{40 - 2a}{9}$$ in the Statement 1, we get $$(\\frac{40 - 2a}{9})$$$$(\\frac{40 - 2a}{9} - 2a)$$ -------------------> $$(\\frac{40 - 2a}{9})$$$$(\\frac{40 - 20a}{9})$$ ----------> $$\\frac{(40-2a)(40-20a)}{81} = 0$$ ------------------> Either (40-2a)=0 ------> a = 20 or (40-20a)=0 -----> a = 2, which is the answer _________________ Director Joined: 23 Jan 2013 Posts: 598 Schools: Cambridge'16 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 26 Oct 2014, 23:53 a1+3d+a1+6d=40 2a1+9d=40 a1=20-9d/2 all answer choices are integers, so 9d/2 should be integer less than 20 and 9d is even. Only d=4 gives 18 to get 20-18=2 A VP Joined: 07 Dec 2014", "# ISI2016-MMA-14 1 vote 127 views The number of terms independent of $x$ in the binomial expansion of $\\left(3x^2 + \\dfrac{1}{x}\\right)^{10}$ is 1. $0$ 2. $1$ 3. $2$ 4. $5$ recategorized 1 vote let mth term is independent of x (here m is integer) (3x2)m x ($\\frac{1}{x}$)10-m x2m-10+m 2m-10+m=0 m=10/3 not integer Anser : A ## Related questions 1 102 views Suppose $a, b, c >0$ are in geometric progression and $a^p = b^q =c^r \\neq 1$. Which one of the following is always true? $p, q, r$ are in geometric progression $p, q, r$ are in arithmetic progression $p, q, r$ are in harmonic progression $p=q=r$ 1 vote Let $x$ and $y$ be real numbers satisfying $9x^2+16y^2=1$. Then $(x+y)$ is maximum when $y=9x/16$ $y=-9x/16$ $y=4x/3$ $y=-4x/3$ The value of the term independent of $x$ in the expansion of $(1-x)^{2}(x+\\frac{1}{x})^{7}$ is $-70$ $70$ $35$ None of these Let $(1+x)^n = C_0+C_1x+C_2x^2+ \\ldots +C_nx^n, \\: n$ being a positive integer. The value of $\\left( 1+\\frac{C_0}{C_1} \\right) \\left( 1+\\frac{C_1}{C_2} \\right) \\cdots \\left( 1+\\frac{C_{n-1}}{C_n} \\right)$ is $\\left( \\frac{n+1}{n+2} \\right) ^n$ $\\frac{n^n}{n!}$ $\\left( \\frac{n}{n+1} \\right) ^n$ $\\frac{(n+1)^n}{n!}$", "A positive integer is called tico if it is the product of three different prime numbers that add up to 74. Verify that 2014 is tico. Which y... 1. A positive integer is called tico if it is the product of three different prime numbers that add up to 74. Verify that 2014 is tico. Which year will be the next tico year? Which one will be the last tico year in history? 2. Let $ABCD$ be a trapezoid with bases $AB$ and $CD$, inscribed in a circle of center $O$. Let $P$ be the intersection of the lines $BC$ and $AD$. A circle through $O$ and $P$ intersects the segments $BC$ and $AD$ at interior points $F$ and $G$, respectively. Show that $BF=DG$. 3. Let $a$, $b$, $c$ and $d$ be real numbers such that no two of them are equal, $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}=4$ and $ac=bd$. Find the maximum possible value of $\\frac{a}{c}+\\frac{b}{d}+\\frac{c}{a}+\\frac{d}{b}.$ 4. Using squares of side 1, a stair-like figure is formed in stages following the pattern of the drawing. For example, the first stage uses 1 square, the second uses 5, etc. Determine the last stage for which the corresponding figure uses less than 2014 squares. 5. Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater", "c \\mid <1$ and $a,b,c$ are in arithmetic progression, then $l, m, n$ are in arithmetic progression geometric progression harmonic progression none of these 7 If the sum of the first $n$ terms of an arithmetic progression is $cn^2$, then the sum of squares of these $n$ terms is $\\frac{n(4n^2-1)c^2}{6}$ $\\frac{n(4n^2+1)c^2}{3}$ $\\frac{n(4n^2-1)c^2}{3}$ $\\frac{n(4n^2+1)c^2}{6}$ 8 If $^nC_{r-1}=36$, $^nC_r=84$ an $^nC_{r+1}=126$ then $r$ is equal to $1$ $2$ $3$ none of these 9 The value of $\\lambda$ such that the system of equation $\\begin{array}{} 2x & – & y & + & 2z & = & 2 \\\\ x & – & 2y & + & z & = & -4 \\\\ x & + & y & + & \\lambda z & = & 4 \\end{array}$ has no solution is $3$ $1$ $0$ $-3$ 1 vote 10 The sum $\\dfrac{n}{n^2}+\\dfrac{n}{n^2+1^2}+\\dfrac{n}{n^2+2^2}+ \\cdots + \\dfrac{n}{n^2+(n-1)^2} + \\cdots \\cdots$ is $\\frac{\\pi}{4}$ $\\frac{\\pi}{8}$ $\\frac{\\pi}{6}$ $2 \\pi$ 1 vote 11 Consider all possible words obtained by arranging all the letters of the word $\\textbf{AGAIN}$. These words are now arranged in the alphabetical order, as in a dictionary. The fiftieth word in this arrangement is $\\text{IAANG}$ $\\text{NAAGI}$ $\\text{NAAIG}$ $\\text{IAAGN}$ 12 Let $y=[\\:\\log_{10}3245.7\\:]$ where $[ a ]$ denotes the greatest integer less than or equal to $a$. Then $y=0$ $y=1$ $y=2$ $y=3$ 13 The number of", "# Number Theory 1. (9 p.) Let $$a$$, $$b$$, $$c$$ be positive integers forming an increasing geometric sequence such that $$b-a$$ is a square. If $$\\log_6a + \\log_6b + \\log_6c = 6$$, find $$a + b + c$$. 2. (3 p.) The square $$\\begin{array}{|c|c|c|} \\hline x&20&151 \\\\\\hline 38 & & \\\\ \\hline & & \\\\ \\hline\\end{array}$$ is magic, i.e. in each cell there is a number so that the sums of each row and column and of the two main diagonals are all equal. Find $$x$$. 3. (25 p.) Let $$0 < a < b < c < d$$ be integers such that $$a$$, $$b$$, $$c$$ is an arithmetic progression, $$b$$, $$c$$, $$d$$ is a geometric progression, and $$d - a = 30$$. Find $$a + b + c + d$$. 4. (15 p.) Find the least positive integer $$n$$ such that when its leftmost digit is deleted, the resulting integer is equal to $$n/29$$. 5. (46 p.) Let $$a,b,c$$ and $$d$$ be positive real numbers such that $$a^2+b^2-c^2-d^2=0$$ and $$a^2-b^2-c^2+d^2=\\frac {56}{53}(bc+ad)$$, Let $$M$$ be the maximum possible value of $$\\frac {ab+cd}{bc+ad}$$ ,If $$M$$ can be expressed as $$\\frac {m}{n}$$,$$(m,n)=1$$ then find $$100m+n$$ 2005-2020 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax" ]

[ "Coding challenge Let $a_1,a_2,\\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \\le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$? (Ans: 502.) (Ref: ACM 12A 2008.)", "+0 How 0 145 1 +133 Let $$a_1,a_2,\\ldots$$ be a sequence determined by the rule $$a_n= \\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_n=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_1 \\le 2008$$ is it true that $$a_1$$ is less than each of $$a_2,a_3,and$$ $$a_4$$? Jun 19, 2019 #1 +28176 +2 I think there are 502 that satisfy the condition. Only the odd numbers provide a2 greater than a1. There are 1004 of them. All these a2's are even, so a3 = a2/2 = (3a1 + 1)/2 which is greater than a1. The corresponding a4's must alternate even and odd. The even ones will be a3/2 = (3a1 + 1)/4 which must be less than or equal to a1 (equality when a1 = 1). The odd ones must be 3(3a1 + 1)/2 + 1 = 9a1/2 + 5/2 > a1. There are 1004/2 = 502 of these. Jun 19, 2019", "251 (C) 501 (D) 502 (E) 1004 _________________ Math Expert Joined: 02 Aug 2009 Posts: 7969 Re: Let a_1, a_2, ... be a sequence determined by the rule a_n=a_{n-1}/2 i [#permalink] ### Show Tags 27 Mar 2019, 01:34 1 1 Bunuel wrote: Let $$a_1$$, $$a_2$$, ... be a sequence determined by the rule $$a_n=\\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_n=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_1 \\leq 2008$$ is it true that $$a_1$$ is less than each of $$a_2$$, $$a_3$$, and $$a_4$$? (A) 250 (B) 251 (C) 501 (D) 502 (E) 1004 1) $$a_1$$ cannot be EVEN, otherwise it will be > $$a_2$$.. So all even 2008/2=1004 gone. 2) Now , $$a_2$$ cannot result in a multiple of 4, otherwise $$a_3$$ will become less than $$a_1$$. For this 3a+1 shouldn't be multiple of 4. So a should not leave a remainder 1 when divided by 4. That means a cannot be 1, 5, 9 and so on. Thus half the odd numbers possible ..3,7..etc D _________________ Re: Let a_1, a_2, ... be a sequence determined by the rule a_n=a_{n-1}/2 i [#permalink] 27 Mar 2019, 01:34 Display posts from previous: Sort by", "# 2010 USAJMO Problems/Problem 2 ## Problem Let $n > 1$ be an integer. Find, with proof, all sequences $x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following three properties: 1. (a). $x_1 < x_2 < \\cdots ; 2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\\ldots,n-1$; 3. (c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i+x_j = x_k$. ## Solution The sequence is $2, 4, 6, \\ldots, 2n-2$. ### Proof 1 We will prove that any sequence $x_1, \\ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done. First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true. Otherwise, for $n > 2$, we will prove by induction on $m$ that the difference $x_{n-m}", "# 2010 USAJMO Problems/Problem 2 ## Problem Let $n > 1$ be an integer. Find, with proof, all sequences $x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following three properties: 1. (a). $x_1 < x_2 < \\cdots ; 2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\\ldots,n-1$; 3. (c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i+x_j = x_k$. ## Solution The sequence is $2, 4, 6, \\ldots, 2n-2$. ### Proof 1 We will prove that any sequence $x_1, \\ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done. First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true. Otherwise, for $n > 2$, we will prove by induction on $m$ that the difference $x_{n-m}", "Let $n,a_1,a_2,\\ldots,a_n$ be integers such that $a_1a_2a_3\\cdots a_n =n$ and $a_1+a_2+a_3+\\cdots+a_n=0$. Prove that 4 divides n I need help with this problem: Let $n,a_1,a_2,\\ldots,a_n$ be integers such that $a_1a_2a_3\\cdots a_n =n$ and $a_1+a_2+a_3+\\cdots+a_n=0$. Prove that 4 divides n I know that $a_i$ divides $n$ for any $i=1,2,3,\\ldots,n$ and that $n>0$. So, there must be pair number of negative integers, but I don't know how to start the proof. I'm new at number theory. Need help please If $n$ is odd, then each of the $a_i$ are odd. Then $a_1+a_2+\\cdots+a_n$ is the sum of an odd number of odd numbers, which cannot equal zero (since zero is even)! So $n$ can't be odd. Now if $n$ is a multiple of $2$ but not $4$, exactly one of the $a_i$ must be even and all the others must be odd. (If there were two or more even factors then $a_1a_2\\cdots a_n$ would be a multiple of 4.) That means $a_1+a_2+\\cdots+a_n$ is the sum of one even number and $n-1$ odd numbers, but since $n-1$ itself is odd, the total sum must be odd and thus cannot be zero again! This leaves only the possibility that $n$ is a multiple of 4, and we are done.", "positive integers not exceeding 2n there must be an integer that divides one of the other integers. Solution 2: Write each of the $n+1$ integers $a_{1}, a_{2}, \\ldots, a_{n+1}$ as a power of 2 times an odd integer. In other words, let $a_{j} = 2^{k_{j}}q_{j}$ for $j=1, 2, 3 \\ldots n+1$, where $k_{j}$ is a nonnegative integer and $q_{j}$ is odd. The integers $q_{1}, q_{2}, \\ldots, q_{n+1}$ are all odd positive integers less than 2n. Because there are only n odd positive integers less than $2n$, it follows from the pigeonhole principle that two of the integers $q_{1}, q_{2}, \\ldots, q_{n+1}$ must be equal. Therefore, there are integers i and j such that $q_{i} = q_{j}$. Let q be the common value of $q_{i}$ and $q_{j}$. Then, $a_{i}=2^{k_{i}}q$ and $a_{j}=2^{k_{j}}q$. It follows that if $k_{i} then $a_{i}$ divides $a_{j}$. QED. A clever application of the pigeonhole principle shows the existence of an increasing or a decreasing subsequence of a certain length in a sequence of distinct integers. Some definitions will be reviewed before this application is presented. Suppose that $a_{1}, a_{2}, a_{3}, \\ldots, a_{n}$ is a sequence of real numbers. A subsequence of this sequence is a sequence of the form $a_{1_{1}}, a_{i_{2}}, \\ldots, a_{i_{n}}$, where $1 \\leq i_{1} < i_{2} < ldots < i_{n} \\leq N$. Hence,", "# Solution Rearrange the first $n$ integers $1, 2, 3, \\ldots, n$, listing them as $a_1, a_2, \\ldots a_n$. Prove that if $n$ is odd, then $(a_1 - 1)(a_2 - 2) \\cdots (a_n - n)$ must be even, regardless of how the integers were initially rearranged. We argue indirectly... Assume $(a_1 - 1)(a_2 - 2) \\cdots (a_n - n)$ is odd. Then each factor $(a_1 - 1)$, $(a_2 - 2)$, ... $(a_n - n)$ must be odd. This implies that $a_1, a_3, a_5, \\cdots, a_n$ are all even, while $a_2, a_4, \\cdots, a_{n-1}$ are all odd. Contemplating the potential use of the pigeon-hole principle (although we end up not using it after all), we count the number of odds and find there are $\\displaystyle{\\frac{(n-1)}{2}}$ of the $a_i$'s that are odd. However, the numbers $a_1, a_2, \\cdots a_n$ are precisely the numbers $1,2,3,\\ldots,n$ -- just possibly in a different order. The number of odd numbers in this last list is $\\displaystyle{\\frac{(n-1)}{2} + 1}$, so we have a contradiction. Hence, we must reject our initial assumption that the product $(a_1 - 1)(a_2 - 2) \\cdots (a_n - n)$ could be odd. Instead, it must always be even. QED.", "# Prove the formula Let be $a_1$, $a_2$, $\\ldots$, $a_{n-1}$, $a_n$, term of the arithmetics sequence. Please help me prove that $S_n=\\frac{n}{2}(a_1+a_n)$ is sum the n-th term of the given sequence. Thanks. - HINT: $a_1+a_n=a_2+a_{n-1}=\\ldots$ – Raskolnikov Oct 10 '12 at 7:57 Let be $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ real numerical sequence. That one sequence let be arithmetics sequence must that $a_2-a_1=a_3-a_2=\\cdots=d$. Note that the given sequence is $2-1=3-2=4-3=5-4=\\cdots=1$, therefore the given sequence is arithmetics sequence. Also note that $1+11=2+9=3+8=\\cdots$. So generally worth $(*)$ $\\ldots$ $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\\cdots$ Now turn prove the given formula. The given sequence is $a_1, a_2, \\ldots, a_{n-1}, a_n$. Form the sum of the given sequence, and that sum the mark $S_n$, i.e $(1)$ $\\ldots$ $S_n=a_1+a_2+\\ldots+a_{n-1}+a_n$ Because comutative low for the sum (+) is worth, have $(2)$ $\\ldots$ $S_n=a_n+a_{n-1}+\\ldots+a_2+a_1$ Equation $(1)$, $(2)$ collect side to side $S_n+S_n=a_1+a_2+\\ldots+a_{n-1}+a_n+a_n+a_{n-1}+\\ldots+a_2+a_1$ $2S_n=(a_1+a_n)+(a_2+a_{n-1})+\\cdots+(a_{n-1}+a_2)+(a_n+a_1)$ Implement the $(*)$ we have: $2S_n=(a_1+a_n)+(a_1+a_n)+\\ldots+(a_1+a_n)$ $2S_n=n(a_1+a_n)$$/$$:2$ $S_n=\\frac{n}{2}(a_1+a_n)$ - Thanky sir, for proving – user39471 Oct 10 '12 at 8:11 How did you manage to type this up in less than 3 minutes? – commenter Oct 10 '12 at 8:21 Writing the arithmetic sequence in another way, $$a_0 + (a_0 + d) + (a_0 + 2d) + ...$$ Flip it around to get another sequence that decreases to $a_0$, $$(a_0 + nd) +", "- a_{n}a_{n-1}=a_{n}a_{n-3}-a_{n-1}a_{n-2}$ This is equivalent to $a_{n-2}(a_{n+1}+a_{n-1})=a_{n}(a_{n-1}+a_{n-3})$ or $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}$ for all $n \\geq 4$. If n is even, we obtain $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}= \\ldots = \\frac{a_{3}+a_{1}}{a_{2}}=200$ while if n is odd, $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}=\\ldots=\\frac{a_{4}+a_{2}}{a_{3}}=11$ It follows that $a_{n+1} = 200a_{n}-a_{n-1}$, if n is even, and $a_{n+1}=11a_{n}-a_{n-1}$, if n is odd. An inductive argument shows that all $a_{n}$ are positive integers. More later, Nalin Pithwa # Basic Enumeration once again Prove the following identity (a): $\\sum_{k=0}^{n}\\left\\{ \\begin{array}{c} n \\\\ k \\end{array} \\right\\}x(x-1)\\ldots(x-k+1)=x^{n}$ Prove the following identity (b): $\\sum_{k=0}^{n}\\left [ \\begin{array}{c} n \\\\ k \\end{array} \\right] = x(x+1)\\ldots(x+n-1)$ Prove the following identity (c); $\\sum_{k=0}^{n}(-1)^{k}\\left\\{ \\begin{array}{c} n \\\\ k \\end{array} \\right\\}\\left[ \\begin{array}{c} k \\\\ j \\end{array} \\right]= 1$, if $j=n$ and is 0, otherwise. General Hint for all three problems: Please see Wikipedia for Stirling numbers! Hint (a): If x is an integer, both sides of the first identity count mappings of an n element set into an x-element set. Hint (b): Both sides count pairs $(\\pi, \\alpha)$ where $\\pi$ is a permutation of an n-element set S, and $\\alpha$ is a coloration of S with x colors invariant under $\\pi$. Hint (c): Combine the identities in (a) and (b). Solution (a): Suppose first that x is a positive integer. Let $|X|=x$ and $|N|=n$. The number of mappings of N into X is $x^{n}$. On", "this, but I'll mention it anyway: Pick any positive irrational $a$ you wish and any desired sequence $r_n^'$ in $[0,1).$ Then we can also pick an integer sequence $s_n$ one term at a time so that for each $n$, $|\\{s_na\\}-r_n^'| \\lt 10^{-n}$ and $\\frac{s_1}{s_n} \\lt \\frac{s_2}{s_n} \\lt \\ldots \\lt \\frac{s_{n-1}}{s_n} \\lt10^{-n}$: Suppose $s_{k-1}$ has been determined and consider in order $s_k=10^ks_{k-1},1+10^ks_{k-1},2+10^ks_{k-1},\\cdots .$ From the theorem of Weyl, you are sure to eventually arrive at a choice with $\\{s_ka\\}$ differing from $r^'_k$ by no more than $10^{-k}.$ -", "i. Example 2: Show that among any $n+1$ positive integers not exceeding 2n there must be an integer that divides one of the other integers. Solution 2: Write each of the $n+1$ integers $a_{1}, a_{2}, \\ldots, a_{n+1}$ as a power of 2 times an odd integer. In other words, let $a_{j} = 2^{k_{j}}q_{j}$ for $j=1, 2, 3 \\ldots n+1$, where $k_{j}$ is a nonnegative integer and $q_{j}$ is odd. The integers $q_{1}, q_{2}, \\ldots, q_{n+1}$ are all odd positive integers less than 2n. Because there are only n odd positive integers less than $2n$, it follows from the pigeonhole principle that two of the integers $q_{1}, q_{2}, \\ldots, q_{n+1}$ must be equal. Therefore, there are integers i and j such that $q_{i} = q_{j}$. Let q be the common value of $q_{i}$ and $q_{j}$. Then, $a_{i}=2^{k_{i}}q$ and $a_{j}=2^{k_{j}}q$. It follows that if $k_{i} then $a_{i}$ divides $a_{j}$. QED. A clever application of the pigeonhole principle shows the existence of an increasing or a decreasing subsequence of a certain length in a sequence of distinct integers. Some definitions will be reviewed before this application is presented. Suppose that $a_{1}, a_{2}, a_{3}, \\ldots, a_{n}$ is a sequence of real numbers. A subsequence of this sequence is a sequence of the form $a_{1_{1}}, a_{i_{2}}, \\ldots, a_{i_{n}}$, where $1 \\leq i_{1} <", "- 1$ of the subsets, lies in the remaining subset. 8. Let $p$ be an odd prime. Determine positive integers $x$ and $y$ for which $x \\leq y$ and $\\sqrt{2p} - \\sqrt{x} - \\sqrt{y}$ is non-negative and as small as possible. ### Sequences 1. Does there exist a sequence $F(1), F(2), F(3), \\ldots$ of non-negative integers that simultaneously satisfies the following three conditions? a) Each of the integers $0, 1, 2, \\ldots$ occurs in the sequence. b) Each positive integer occurs in the sequence infinitely often. c) For any $n \\geq 2,$ $F(F(n^{163})) = F(F(n)) + F(F(361)).$ 2. Find the maximum value of $x_{0}$ for which there exists a sequence $x_{0},x_{1}\\cdots ,x_{1995}$ of positive reals with $x_{0} = x_{1995}$, such that $x_{i - 1} + \\frac {2}{x_{i - 1}} = 2x_{i} + \\frac {1}{x_{i}},$ for all $i = 1,\\cdots ,1995$. 3. For an integer $x \\geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \\ldots$ defined by $x_0 = 1$ and $x_{n+1} = \\frac{x_n p(x_n)}{q(x_n)}$ for $n \\geq 0$. Find all $n$ such that $x_n = 1995$. 4. Suppose that", "2(n-1)]\\;=\\;n^2$ Show that the sum of the first $n$ positive even integers is $n^2+n$ We have another A.P. . $2 + 4 + 6 + \\hdots + 2n$ . . with first term $a = 2$ and common difference $d = 2.$ The sum is: . $S\\;=\\;\\frac{n}{2}[2\\cdot2 + 2(n-1)] \\;= \\;n(n + 1)$", "get that $(1*3*5*\\cdots *\\frac{l-1}{2})$ if $\\frac{l-1}{2}$ is odd integer and without $\\frac{l-1}{2}$ if its even is equal to all even numbers bigger than $\\frac{l-1}{2}$ and since the product include the odd integers that are bigger than $\\frac{l-1}{2}$ then we've got the product of all integers bigger than $\\frac{l-1}{2}$ till $l-1$ multiplied by $(\\frac{l-1}{2})!$ we arrive at $(l-1)!$. Now we arrive at $(l-1)!=2^{\\frac{l-1}{2}} (l-1)!$ but remember that sometimes $\\frac{l-1}{2}$ is even and sometimes is odd,we must multiply by $(-1)*(-1)*\\cdots *(-1)$ because we shifted the smaller odd integers to bigger even integer using (b) (modulo $l$) but i left the minus sign which now i will get back to it. When $\\frac{l-1}{2}$ is even we have $-1$ to the power $\\frac{l-1}{4}$ And When$\\frac{l-1}{2}$ is odd we have $-1$ to the power $\\frac{l+1}{4}$ Use the $\\mod 4$ to deduce that one can combine them in $(-1)^{\\frac{l-3}{4}}$. Thus completing the proof.", "selecting $$n$$ such that $$e^{-nia}$$ is near $$-1$$ for all the selected $$a$$. Lemma: For any finite set $$a_1,\\ldots,a_k$$ of odd integers and any $$\\varepsilon$$, there exists some $$n$$ such that $$|1+e^{-nia_k}| < \\varepsilon$$ for all $$k$$. Proof: By a similar argument about approximations as was previously used, we can find an integer $$n$$ that is arbitrarily close to an odd multiple of $$\\pi$$. Note that if a real number $$r$$ is within $$\\varepsilon$$ of an odd multiple of $$\\pi$$, then for any odd integer $$a$$, the value $$ar$$ is within $$a\\varepsilon$$ of an odd multiple of $$\\pi$$. Since the $$a_k$$ are fixed and finite, we may, by choosing $$n$$ sufficiently close to an odd multiple of $$\\pi$$ ensure that all the values $$na_k$$ are arbitrarily close to odd multiples of $$\\pi$$. The lemma immediately follows. To finish, we can, for any $$k$$, select $$k$$ values $$a_1,\\ldots,a_k$$ such that $$\\frac{1}{|1-e^{-ia_k}|} > a_kc$$. Using the lemma, we may then choose $$n$$ such that $$|1-e^{-ina_k}| > 1$$ for all $$k$$. The quotients $$\\frac{1-e^{-ina_k}}{1-e^{-ia_k}}$$ then all have absolute value at least $$a_kc$$ and thus $$a_k^{th}$$ Fourier coefficients of $$g_n$$ are all at least $$\\frac{4c}{\\pi}$$ in absolute value. Since there exist $$g_n$$ with arbitrarily many Fourier coefficients that are greater than some fixed lower bound, the sequence $$g_n$$ is not bounded in", "{2, 4, 6, …, $e_{max}$}) where 2 is the minimum positive even integer and where $e_{max}$ is the maximum positive even integer belonging to those sets… The positive real number, r, is determined by the algorithm for the Collatz conjecture. And therefore, its computation is generally complicated because we cannot easily compute the maximum positive even integer, $e_{max}$, in the Collatz sequence (orbit) for many odd positive integers, $n_{0} > 2^{68}$. Question: For any odd positive integer, $n_{0} > 2^{68}$ , can we approximate the maximum positive even integer, $e_{max}$, in the Collatz sequence (orbit)? The variable, k, is determined by the maximum positive even number, $e_{max}$, in the Collatz sequence: $k = \\left \\lfloor{\\frac{log(e_{max })}{log(2)}}\\right \\rfloor$. Can we refute our analysis? We assume maximum divisors, $2^{i}$, for any positive even integer. Remark: The Collatz conjecture is true for all positive integers, $n_{0} \\le 2^{68}$. For any positive odd integer, $n_{0} > 2^{68} -1$, we have $n_{1} = \\frac{3n_{0} +1}{2^{i_{1}}} = \\frac{3}{2^{i_{1}}} * r_{1}$ where $r_{1} = n_{0} +\\frac{1}{3}$. $n_{2} = \\frac{3(\\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}} = \\frac{3}{2^{i_{1}}} *\\frac{3}{2^{i_{2}}} * r_{2}$ where $r_{2} = \\frac{1}{9}(2^{i_{1}} + 9n_{0} +3)$. $n_{3} = \\frac{3(\\frac{3(\\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}}) + 1}{2^{i_{3}}} = \\frac{3}{2^{i_{1}}} *\\frac{3}{2^{i_{2}}} *\\frac{3}{2^{i_{3}}} *r_{3}$ where $r_{3} = \\frac{2^{i_{1}}}{27}*(2^{i_{2}} + 3) + n_{0} +\\frac{1}{3}$. $n_{t} = r_{t} * 3^{t}\\prod_{j=1}^{t } (\\frac{1}{2^{i_{j}}})$ where $r_{t}", "### 2011 IMO SL #A2 Determine all sequences $(x_1,x_2,\\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with$\\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1$ The answer is $\\boxed{(1,K,K,\\cdots,K) \\text{ for } K=2+3+\\cdots+2011}$. Denote the corresponding sequence value for place $n$ be $a_n$. We claim that the sequence $(a_n)$ is bounded, or in other words that there exists a constant $C$ such that $(a_n)$ is always less than it. Note that $$a_n = (x_1^n+2x_2^n + \\cdots + 2011x_{2011}^n-1)^{\\tfrac{1}{n+1}}<\\big[x_1+2x_2+\\cdots+2011x_{2011}+100\\big]^{\\tfrac{n}{n+1}}.$$ Let $$C=\\big[x_1+2x_2+\\cdots+2011x_{2011}+100\\big],$$ and then we have $$\\big[x_1+2x_2+\\cdots+2011x_{2011}+100\\big]^{\\tfrac{n}{n+1}}=C^{\\tfrac{n}{n+1}}<C,$$ so we have proven our claim. For sufficiently large primes $p$, $a_{p-1}=K$ holds true. Denote $n=p-1$. Consider the given condition $\\bmod{p}$ to obtain $$1+2+\\cdots+2011\\equiv a_{p-1}+1\\pmod p,$$ and therefore for large $p>\\max(a_n)$, we get $a_{p-1}=K$, as desired by the claim. Hence, for sufficiently large primes $p$, by dividing by $K$, we obtain $\\left(\\frac{x_1}K\\right)^{p-1}+\\cdots+2011\\left(\\frac{x_{2011}}K\\right)^{p-1}=K+\\frac1{K^{p-1}}.$ From this we get $x_i \\le K$ for all defined $i$. Considering $p \\to \\infty$, we get $(x_i/K)^{p-1}\\in\\{0,1\\}$, where equality holds true when $x_i=K$. The RHS approaches $K$, and hence we must have $x_1<K$ and $x_2=x_3 \\cdots = x_{2011}=K$. By $n=1$, we have that $(x_1-1)+K^2$ is a perfect square. However, $x_1-1<K$, so we know that $x_1=1$. We are done. $\\square$ 1. schukkayapally1/15/2022 lovely solution but get custom domain!!! the css looks great now but change", "$$a_0=5,k=4,(a_{n,k})_{n\\in\\mathbb{N}}=(5,26,7,36,9,46,12,3,16,4,1,...)$$ again we can ask about, every sequence will eventually reach at the number $$1$$? Programming for extended Collatz conjecture by Keith Matthews, including negative values, link More on extended Collatz conjecture First: Let $$a_0,t$$ be an positive integer and $$k$$ be any even positive integer, define $$a_{n+1} = \\begin{cases} (k^t+k^{t-1})a_n+k^{t-1}, & \\text{if a_n is odd} \\\\ \\lceil a_n/k\\rceil, & \\text{if a_n is even} \\end{cases}$$ Now form the sequence $$(a_{n,k,t})_{n\\in \\mathbb{N}}$$ by performing this operation repeatedly then sequence will eventually reach at the number $$1$$ Second: Let $$a_0$$ be an positive integer and $$k$$ be any even positive integer greater than $$2$$ and $$t\\in\\mathbb{Z}_{\\ge 2}$$, define $$a_{n+1} = \\begin{cases} (k^t+k^{t-2})a_n+k^{t-2}, & \\text{if a_n is odd} \\\\ \\lceil a_n/k\\rceil, & \\text{if a_n is even} \\end{cases}$$ Now form the sequence $$(a_{n,k,t})_{n\\in \\mathbb{N}}$$ by performing this operation repeatedly then sequence will eventually reach at the number $$1$$ I again extend claim on odd $$k$$ First: Let $$a_0$$ be an positive integer and $$k$$ be any odd positive integer greater than $$1$$ and $$t\\in\\mathbb{Z}_{\\ge 2}$$, define $$a_{n+1} = \\begin{cases} (k^t-k^{t-2})a_n, & \\text{if a_n is odd} \\\\ \\lfloor a_n/k\\rfloor, & \\text{if a_n is even} \\end{cases}$$ Then the sequence $$(a_{n,k,t})_{n\\in \\mathbb{N}}$$ will eventually reach at the number $$0$$ Second: Let $$a_0$$ be an positive integer and $$k$$ be any odd positive integer greater than", "Preparing NOJ # Missing Numbers 2000ms 262144K ## Description: Chouti is working on a strange math problem. There was a sequence of $n$ positive integers $x_1, x_2, \\ldots, x_n$, where $n$ is even. The sequence was very special, namely for every integer $t$ from $1$ to $n$, $x_1+x_2+...+x_t$ is a square of some integer number (that is, a perfect square). Somehow, the numbers with odd indexes turned to be missing, so he is only aware of numbers on even positions, i.e. $x_2, x_4, x_6, \\ldots, x_n$. The task for him is to restore the original sequence. Again, it's your turn to help him. The problem setter might make mistakes, so there can be no possible sequence at all. If there are several possible sequences, you can output any. ## Input: The first line contains an even number $n$ ($2 \\le n \\le 10^5$). The second line contains $\\frac{n}{2}$ positive integers $x_2, x_4, \\ldots, x_n$ ($1 \\le x_i \\le 2 \\cdot 10^5$). ## Output: If there are no possible sequence, print \"No\". Otherwise, print \"Yes\" and then $n$ positive integers $x_1, x_2, \\ldots, x_n$ ($1 \\le x_i \\le 10^{13}$), where $x_2, x_4, \\ldots, x_n$ should be same as in input data. If there are multiple answers, print any. Note, that the limit for $x_i$ is larger than for input" ]

[ "$a_3/2$ and less than $a_3$ is $\\frac23a_3$. Thus we have $a_2=2a_1$ and $a_3=3a_1$. This cannot continute to hold if we add a fourth number, so the only sets with this property are those of the form $\\{a_1,2a_1,3a_1\\}$ with $a_1\\in\\mathbb N$.", "number then $$x^2$$ is a multiple of 4, which makes $$\\frac{x^2}{2}$$ an even number and therefore $$\\frac{3x^2}{2}+1=3*even+1=even+1=odd$$. If you don't notice this, then one also do in another way. Let $$x=2k$$, for some integer k, then: A. $$\\frac{3x}{2}=\\frac{3*2k}{2}=3k$$ --> if $$k=odd$$ then $$3k=odd$$ but if $$k=even$$ then $$3k=even$$. Discard; B. $$\\frac{3x}{2}+1=\\frac{3*2k}{2}+1=3k+1$$ --> if $$k=odd$$ then $$3k+1=odd+1=even$$ but if $$k=even$$ then $$3k+1=even+1=odd$$. Discard; C. $$3x^2$$ --> easiest one as $$x=even$$ then $$3x^2=even$$, so this option is never odd. Discard; D. $$\\frac{3x^2}{2}=\\frac{3*4k^2}{2}=6k^2=even$$, so this option is never odd. Discard; E. $$\\frac{3x^2}{2}+1=\\frac{3*4k^2}{2}=6k^2+1=even+1=odd$$, thus this option is always odd. Similar question to practice: if-a-and-b-are-positive-integers-such-that-a-b-and-a-b-are-88108.html Hope it helps. _________________ Current Student Joined: 28 Mar 2012 Posts: 316 Location: India GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: If x is even integer, which of the following must be an odd [#permalink] ### Show Tags 25 Jul 2012, 20:26 Hi, Since, x is even we can assume x = 2 or x = 4, such that x/2 is both odd/even as per the value of x. Check each option with these values. We get the answer when both x=2, 4 gives an odd value. Regards, Intern Joined: 29 Nov 2012 Posts: 6 Re: If x is even integer, which of the following must be an odd", "2$$ so $$a < 2 < 4$$ so $$a < 4$$ must be true. So $$\\frac 12 < a$$ true and $$a < 4$$ is true so $$\\frac 12 < a < 4$$ must be TRUE. And $$\\frac 12 < a < 4$$ does not contradict $$1 < a < 2$$. It's just that $$1 is a stronger more precise statement. Consider this. $$a= 36$$ is an even number. $$a=36$$ is a perfect square. $$a=36$$ is a multiple of $$3$$. Fred comes and says \"$$a$$ is a multiple of $$3$$\". Then Jane comes along and says \"$$a$$ is a perfect square\". Michael says \"$$a$$ is an even number.\" Olav hears all three of these people and says \"Wait! Those are three completely different statements! So only one of them can be true! Which one of you is telling the truth and which one is lying.\" .... And they are all telling the truth. $$\\frac 12 < a < 4$$ means that it is possible that $$a$$ might be less than $$1$$ but it does not mean that it is. So $$\\frac 12 < a$$ does not contradict $$1 < a$$. This is analogous to \"$$a$$ is a multiple of $$3$$\". That means $$a$$ might be odd but it doesn't mean that it is. \"$$a$$ is a multiple of", "+ k=\\frac{k(k+1)}{2}$$ (1) $$a_{n-1}=\\frac{n(n-1)}{2}$$ is even, so $$n(n-1)$$ is divisible by 4. Hence, $$n=4k$$ or $$n-1=4k$$. If $$n=4k$$we have $$a_n=\\frac{n(n+1)}{2}=\\frac{4k(4k+1)}{2}=2k(4k+1)$$ is even. If $$n-1=4k$$ , we have $$a_n=\\frac{n(n+1)}{2}=\\frac{(4k+1)(4k+2)}{2}=(4k+1)(2k+1)$$ is odd. Insufficient. (2) $$a_{n-4}=\\frac{(n-4)(n-3)}{2}$$ is odd, so $$(n-4)(n-3)$$ is divisible by 2 and not by 4. Hence, $$n-4=4k+2$$ or $$n-3=4k+2$$. If $$n-4=4k+2 \\iff n=4k+6$$, we have $$a_n=\\frac{n(n+1)}{2}=\\frac{(4k+6)(4k+7)}{2}=(2k+3)(4k+7)$$ is an odd integer. If $$n-3=4k+2 \\iff n=4k+5$$, we have $$a_n=\\frac{n(n+1)}{2}=\\frac{(4k+5)(4k+6)}{2}=(4k+5)(2k+3)$$ is an odd integer. Sufficient. _________________ Senior SC Moderator Joined: 14 Nov 2016 Posts: 1319 Location: Malaysia For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4 [#permalink] ### Show Tags 06 Feb 2017, 17:48 Bunuel wrote: For the infinite series $$a_1$$, $$a_2$$, ..., $$a_n$$, $$a_n=a_{n−1}+n$$, and $$a_1=1$$. If n>4, is $$a_n$$ an odd integer? (1) $$a_{n−1}$$ is even (2) $$a_{n−4}$$ is odd Official solution from Veritas Prep. As with most sequence problems, your most effective strategy is typically to list out the first few terms of the sequence to see if you can establish a pattern or trend. Here you will find: First Term: 1 (Odd) Second Term: 1 + 2 = 3 (Odd) Third Term: 3 + 3 = 6 (Even) Fourth Term: 6 + 4 = 10 (Even) Fifth Term: 10 + 5 = 15 (Odd) Sixth Term: 15 + 6 = 21 (Odd) Seventh", "# Suppose that $a$ and $b$ are natural numbers such that $a^2 = b^3$. Prove that if $4$ divides $b$, then $8$ divides $a$. Proposition: Suppose that $a$ and $b$ are natural numbers such that $a^2 = b^3$. Prove that if $4$ divides $b$, then $8$ divides $a$. Proposed Proof: Assume $4$ divides $b$ then $b = ?$ for some natural number $n$. This implies... Conclude: Thus, $8$ divides $a$. Not sure how to prove this, I think a direct proof should be used? Any help would be appreciated. We can (in a unique manner) write $a=2^ru$ with $r\\in\\Bbb N_0$ and $u$ odd, similarly $b=2^sv$ with $s\\in\\Bbb N_0$ and $v$ odd. Then $$2^{2r}u^2=a^2=b^3=2^{3s}v^3$$ and (as $u^2,v^3$ are odd) conclude $2r=3s$. If $b$ is a multiple of $4$ then $s\\ge 2$, hence $2r\\ge 6$, i.e., $r\\ge 3$, i.e, $8\\mid a$. Alternatively, with direct use of the prime property $p\\mid xy\\to (p\\mid x\\lor p\\mid y)$, or more specifically: If a square of a natural number is even, then the number itself is even: If $b=4k$, then $a^2=64k^3$, so $a^2$ is even, hence $a$ iseven, say $a=2c$. So $a^2=(2c)^2=4c^2=64k^2$, or $c^2=16k^3$, which is even, so that $c$ is even, say $c=2d$: So $c^2=(2d)^2=4d^2=16k^3$, or $d^2=4k^3$, which is even, so that $d$ is even, say $d=2e$. Then $a=2c=4d=8e$ is a multiple of", "even. So $8|6n^3$ so $4|3n^3$ so $3n^2$ is even. If $n$ is odd, so is $3n^3$ so $n$ is even. But that contradicts that $\\frac mn$ was in \"lowest terms\". And .... the factor of $3$ didn't really have much to do with it. Of course, we could have used that $6$ is a multiple of $3$ rather than it was even: $6n^3 = m^3$ so $3|m^3$ if $m = 3k + i$ then $m^3 = 27k^3 + 27k^2 + 9k^3 + i^3$ and if $i = 1, 2$ then $3\\not \\mid i^3$ so $3|m$ so $27|6n^3$ and $3|2n^3$ and if $n = 3j +i$ then $2n^3 = 54j^3 + 54j^2 + 18k^3 + 2i$ and if $i= 1,2$ then $3 \\not \\mid 2i^3$ so $3|n$ and ... not in lowest terms. It... gets pretty repetitive. Best learn the PFT once and for all.", "2012, 07:57. Last edited by Bunuel on 25 Jul 2012, 08:03, edited 1 time in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 60515 Re: If x is even integer, which of the following must be an odd [#permalink] ### Show Tags 25 Jul 2012, 08:05 1 1 Stiv wrote: If x is even integer, which of the following must be an odd integer? A. $$\\frac{3x}{2}$$ B. $$\\frac{3x}{2} + 1$$ C. $$3x^2$$ D. $$\\frac{3x^2}{2}$$ E. $$\\frac{3x^2}{2} + 1$$ One can spot right away that if $$x$$ is any even number then $$x^2$$ is a multiple of 4, which makes $$\\frac{x^2}{2}$$ an even number and therefore $$\\frac{3x^2}{2}+1=3*even+1=even+1=odd$$. If you don't notice this, then one also do in another way. Let $$x=2k$$, for some integer k, then: A. $$\\frac{3x}{2}=\\frac{3*2k}{2}=3k$$ --> if $$k=odd$$ then $$3k=odd$$ but if $$k=even$$ then $$3k=even$$. Discard; B. $$\\frac{3x}{2}+1=\\frac{3*2k}{2}+1=3k+1$$ --> if $$k=odd$$ then $$3k+1=odd+1=even$$ but if $$k=even$$ then $$3k+1=even+1=odd$$. Discard; C. $$3x^2$$ --> easiest one as $$x=even$$ then $$3x^2=even$$, so this option is never odd. Discard; D. $$\\frac{3x^2}{2}=\\frac{3*4k^2}{2}=6k^2=even$$, so this option is never odd. Discard; E. $$\\frac{3x^2}{2}+1=\\frac{3*4k^2}{2}=6k^2+1=even+1=odd$$, thus this option is always odd. Similar question to practice: if-a-and-b-are-positive-integers-such-that-a-b-and-a-b-are-88108.html Hope it helps. _________________ Current Student Joined: 29 Mar 2012 Posts: 295 Location: India GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3:", "= a_1+2a_2$$ $$b_4 = b_1+2b_2$$ $$|a_1b_1+2a_2b_1-a_1b_1-2a_1b_2| = 1$$ $$2|a_2b_1-a_1b_2| = 1$$ This is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$. The answer is as follows: $$0+1+2+\\ldots+2$$ $a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$. There are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means $$\\boxed{N=197}.$$~Lopkiloinm", "can ALWAYS be expressed algebraically as $$b=\\frac{a+c}{2}$$. Now, does that mean that at leas on of them is be even? Not necessarily, consider $$a=1$$, $$b=3$$ and $$c=5$$. Of course it's also possible that $$b=even$$, for example if $$a=1$$ and $$b=7$$. Not sufficient. (2) a = b - c --> $$a+c=b$$. Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient. Why are we assuming that a and c could be even..? or could be odd? Its also possible that a is even and c is odd..in that case b can be odd Pls explain thanks It's not clear what are you trying to say here. If a is even, then abc is even. _________________ Manager Joined: 19 Aug 2016 Posts: 63 Re: The Discreet Charm of the DS [#permalink] Show Tags 18 Oct 2017, 17:50 Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c? Given: $$6a=3b=7c$$ --> least common multiple of 6, 3, and 7 is 42 hence we ca write: $$6a=3b=7c=42x$$, for some number $$x$$ --> $$a=7x$$, $$b=14x$$ and $$c=6x$$. (1) ac=6b --> $$7x*6x=6*14x$$ --> $$x^2=2x$$ --> $$x=0$$ or $$x=2$$. Not sufficient. (2) 5b=8a+4c --> $$5*14x=8*7x+4*14x$$ --> $$70x=80x$$ --> $$10x=0$$ -->", "• # question_answer The value of ${{i}^{1+3+5+...+(2n+1)}}$ is [AMU 1999] A) i if n is even, - i if n is odd B) 1 if n is even, - 1 if n is odd C) 1 if n is odd, - 1 if n is even D) i if n is even, - 1 if n is odd Let $z={{i}^{[1+3+5+....+(2n+1)]}}$ Clearly series is A.P. with common difference = 2 $\\because \\,{{T}_{n}}=2n-1$and ${{T}_{n+1}}=2n+1$ So, number of terms in A. P. $=n+1$ Now, ${{S}_{n+1}}=\\frac{n+1}{2}[2.1+(n+1-1)2]$ $\\Rightarrow {{S}_{n+1}}=\\frac{n+1}{2}[2+2n]={{(n+1)}^{2}}$ i.e. ${{i}^{{{(n+1)}^{2}}}}$ Now put $n=1,\\,2,\\,3,\\,4,\\,5,\\,.....$ $n=1,z={{i}^{4}}=1$, $n=2,\\,z={{i}^{6}}=-1$, $n=3,\\,z={{i}^{8}}=1$, $n=4,\\,z={{i}^{10}}=-1$, $n=5,\\,\\,z={{i}^{12}}=1\\,,........$", "# Difference between revisions of \"2008 AMC 12A Problems/Problem 17\" ## Problem Let $a_1,a_2,\\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \\le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$? $\\mathrm{(A)}\\ 250\\qquad\\mathrm{(B)}\\ 251\\qquad\\mathrm{(C)}\\ 501\\qquad\\mathrm{(D)}\\ 502\\qquad\\mathrm{(E)} 1004$ ## Solution All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer. • If $a_1=4n$, then $a_2=\\frac{4n}{2}=2n. • If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\\frac{12n+4}{2}=6n+2$, and $a_4=\\frac{6n+2}{2}=3n+1. • If $a_1=4n+2$, then $a_2=2n+1. • If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$. Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1. Since one fourth of the positive integers $a_1 \\le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\\frac{1}{4}\\cdot 2008 = 502 \\Rightarrow D$.", "are 2010 and 2016. So we just have to check if either $\\frac{2016}{3} - 1$ or $\\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work. Case 2: Both $a$ and $b$ are odd primes. This implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$. This leaves only $2012$ and $2016$. $2012={4}\\cdot{503}$ so either $(a+1)$ or $(b+1)$ both has a factor of $2$ or one has a factor of $4$. If it was the first case, then $a$ or $b$ will equal $1$. That means that either $(a+1)$ or $(b+1)$ has a factor of $4$. That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 \\cdot 504$ so we have $(503 + 1)(3 + 1)$. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\\boxed{\\textbf{(A)}\\ 1}$.", "we have $a=3$ if $b=6$, or $a=6$ if $b=3$. Note that this does not matter because $\\frac{1}{a}+\\frac{1}{b}$ is symmetric. I will go with $a=3$ and $b=6$, although it does not matter. Substituting into $\\frac{1}{a}+\\frac{1}{b}$ gives $\\frac{1}{3}+\\frac{1}{6}=\\frac{2}{6}+\\frac{1}{6}=\\frac{3}{6}=\\boxed{\\frac{1}{2}}$ RiemannIntegralzzz Apr 6, 2021 edited by RiemannIntegralzzz Apr 6, 2021 edited by RiemannIntegralzzz Apr 6, 2021", "# Difference between revisions of \"2008 iTest Problems/Problem 93\" ## Problem For how many positive integers $n$, $1 \\le n \\le 2008$, can the set ${1, 2, 3, . . . , 4n}$ be divided into $n$ disjoint $4$-element subsets such that every one of the $n$ subsets contains the element which is the arithmetic mean of all the elements in that subset? ## Solution Each $4$ element subset is of the form $\\left\\{a,b,c, \\frac{a+b+c}{3}\\right\\}$. The sum of the elements of this subset is $\\frac{4(a+b+c)}{3}$, which is divisible by $4$. If $n$ is odd however, then the sum of the elements $1+2+3+ \\cdots +4n$ is $\\frac{4n(4n+1)}{2} = 2(n)(4n+1)$, but $n$ and $4n+1$ are both odd, and so the sum is not divisible by $4$. Hence $n$ may not be odd. For $n=2$, we note that the construction $\\{1,5,6,8\\},\\{2,3,4,7\\}$ works. For even $n$ greater than $2$, we can divide each consecutive eight element subset using the same construction, eg, $\\{8k+1,8k+5,8k+6,8k+8\\},\\{8k+2,8k+3,8k+4,8k+7\\}$ for $0 \\le k < \\frac n8$. Hence, the answer is all even $n$, of which there are $\\boxed{1004}$.", "+0 How 0 145 1 +133 Let $$a_1,a_2,\\ldots$$ be a sequence determined by the rule $$a_n= \\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_n=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_1 \\le 2008$$ is it true that $$a_1$$ is less than each of $$a_2,a_3,and$$ $$a_4$$? Jun 19, 2019 #1 +28176 +2 I think there are 502 that satisfy the condition. Only the odd numbers provide a2 greater than a1. There are 1004 of them. All these a2's are even, so a3 = a2/2 = (3a1 + 1)/2 which is greater than a1. The corresponding a4's must alternate even and odd. The even ones will be a3/2 = (3a1 + 1)/4 which must be less than or equal to a1 (equality when a1 = 1). The odd ones must be 3(3a1 + 1)/2 + 1 = 9a1/2 + 5/2 > a1. There are 1004/2 = 502 of these. Jun 19, 2019", "× 2$$\\frac{9}{10}$$ is greater than 1 Explanation: the given mixed fractions are: 3$$\\frac{4}{5}$$ and 2$$\\frac{9}{10}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, The representation of the mixed numbers in the fraction form is: $$\\frac{19}{5}$$ and $$\\frac{29}{10}$$ So, $$\\frac{19}{5}$$ is greater than 1 $$\\frac{29}{10}$$ is greater than So, 3$$\\frac{4}{5}$$ × 2$$\\frac{9}{10}$$ is greater than 1 Hence, from the above, We can conclude that the value of 3$$\\frac{4}{5}$$ × 2$$\\frac{9}{10}$$ is greater than 1 Question 5. 8 × $$\\frac{2}{3}$$ The value of 8 is greater than $$\\frac{2}{3}$$ Explanation: The given numbers are: 8 and $$\\frac{2}{3}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\\frac{2}{3}$$ is less than 1 So, 8 is greater than $$\\frac{2}{3}$$ Hence, from the above, We can conclude that the value of 8 is greater than $$\\frac{2}{3}$$ Question 6. 1$$\\frac{7}{8}$$ × $$\\frac{1}{4}$$ The value of 1$$\\frac{7}{8}$$ × $$\\frac{1}{4}$$ is less than 1 Explanation: The given fractions are: 1$$\\frac{7}{8}$$ and $$\\frac{1}{4}$$ The representation", "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "can be written as $a^{2}+b^{2}$ ? – Richard Mar 25 '15 at 2:56 For $a,b$ in $a^2+b^2$, consider three cases: 1) Both $a,b$ are even. Then $a^2+b^2=4k^2+4k'^2$ which is of the form $4k$. 2) Both $a,b$ are odd. Then $a^2+b^2=8k+1+8k'+1=4(2k+2k')+2$ which is of the form $4k+2$. 3) $a,b$ have the deifferent parity. WLOG suppose that a is odd and b is even. Then $a^2+b^2=8k+1+4k'=4(2k+k')+1$ which is of the form $4k+1$", "red box is $$20\\%$$ bigger than the capacity of a blue box and it can hold $$60$$ books. Therefore, we want to find a number that $$20\\%$$ bigger than that number is $$60$$. Let $$x$$ be that number. Then: $$1.20×x=60$$, Divide both sides of the equation by $$1.2$$. Then: $$x=\\frac{60}{1.20}=50$$ 2- E We have two equations and three unknown variables, therefore $$x$$ cannot be obtained. 3- E $$x+2=2+2+2→x=4$$ $$y+7+3=6+5→y+10=11→y=1$$ Then, the perimeter is: $$2+6+2+5+2+3+2+7+4+1=34$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/yyyyyygj.png }$$ 4- ٍE Amount of available petrol in tank: $$60.4-4.86-23.9+11.22=42.86$$ liters 5- C Let put some values for a and b. If $$a=8$$ and $$b=2 →a×b=16 →\\frac{16}{4}=8→16$$ is divisible by $$4$$ then; A.$$2a-5b=(2×8)-(5×2)=6$$ is not divisible by $$4$$ B.$$3a-b=(3×8)-2=24-2=22$$ is not divisible by $$4$$ If $$a=10$$ and $$b=4 →a×b=40 →\\frac{40}{4}=10$$ is not divisible by $$4$$ then; C.$$2a×3b=20×(3×4)=20×12=240$$ is divisible by $$4$$ D.$$\\frac{a}{b}=\\frac{10}{4}$$ is not divisible by $$4$$ E. $$\\frac{a×b}{3}=\\frac{10×4}{3}=13.33$$ For choice C, if you try any other numbers for a and b, you will get the same result. 6- D Let’s review the choices provided: A.$$x=\\frac{1}{3}→ \\frac{6}{8}+\\frac{1}{3}=\\frac{18+8}{24}=\\frac{26}{24}≅1.083<2$$ B.$$x=\\frac{3}{5}→ \\frac{6}{8}+\\frac{3}{5}=\\frac{30+24}{40}=\\frac{54}{40}≅1.35<2$$ C.$$x=\\frac{6}{5}→ \\frac{6}{8}+\\frac{6}{5}=\\frac{30+48}{40}=\\frac{78}{40}≅1.95<2$$ D.$$x=\\frac{4}{3}→ \\frac{6}{8}+\\frac{4}{3}=\\frac{18+32}{24}=\\frac{50}{24}≅2.08>2$$ E.$$x=\\frac{2}{3}→ \\frac{6}{8}+\\frac{2}{3}=\\frac{18+16}{24}=\\frac{34}{24}≅1.41<2$$ Only choice D is correct. 7- E $$\\frac{3}{4}×20=\\frac{60}{4}=15$$ 8- B The angles on a straight line add up to $$180$$ degrees. Let’s review the choices provided: A.$$y=10→ x+25+y+2x+y=35+25+10+2(35)+10=150≠180$$ B.$$y=25→ x+25+y+2x+y=35+25+25+2(35)+25=180$$ C.$$y=30→ x+25+y+2x+y=35+25+30+2(35)+30=190≠180$$ D.$$y=40→ x+25+y+2x+y=35+25+40+2(35)+40=210≠180$$ E.$$y=50→ x+25+y+2x+y=35+25+50+2(35)+50=230≠180$$ 9- B", "sufficient. From S1 it follows that $$a + b - c$$ is an integer. Thus, $$2(a + b - c)$$ is an even integer. Statement (2) by itself is insufficient. Consider $$a = b = c = 0$$ (the answer is \"no\") and $$a = 1.25$$, $$c = -1.25$$, $$b = 0$$ (the answer is \"yes\"). I understand why answer is D but can't understand why in explanation used such example: $$a = 1.25$$, $$c = -1.25$$, $$b = 0$$ is (the answer is \"yes\") We have question \"Is $$2(a+b−c)$$ an odd?\" let's put this numbers in the question and we receive: $$2(1.25+0-1.25) = 0 = even$$ I think correct example will be $$a = \\frac{1}{4}$$ $$b = \\frac{1}{4}$$ and $$c =0$$ $$b = a - c$$ : $$\\frac{1}{4} = \\frac{1}{4} - 0$$ and $$2(a+b−c)$$: $$2(\\frac{1}{4}+\\frac{1}{4}-0) = 1 = odd$$ Am I miss something or this is misprint in explanation? _________________ Math Expert Joined: 02 Sep 2009 Posts: 44639 ### Show Tags 14 Apr 2015, 05:40 1 KUDOS Expert's post Harley1980 wrote: Bunuel wrote: Official Solution: Statement (1) by itself is sufficient. From S1 it follows that $$a + b - c$$ is an integer. Thus, $$2(a + b - c)$$ is an even integer. Statement (2) by itself is insufficient. Consider $$a = b = c =" ]

[ "+0 How 0 145 1 +133 Let $$a_1,a_2,\\ldots$$ be a sequence determined by the rule $$a_n= \\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_n=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_1 \\le 2008$$ is it true that $$a_1$$ is less than each of $$a_2,a_3,and$$ $$a_4$$? Jun 19, 2019 #1 +28176 +2 I think there are 502 that satisfy the condition. Only the odd numbers provide a2 greater than a1. There are 1004 of them. All these a2's are even, so a3 = a2/2 = (3a1 + 1)/2 which is greater than a1. The corresponding a4's must alternate even and odd. The even ones will be a3/2 = (3a1 + 1)/4 which must be less than or equal to a1 (equality when a1 = 1). The odd ones must be 3(3a1 + 1)/2 + 1 = 9a1/2 + 5/2 > a1. There are 1004/2 = 502 of these. Jun 19, 2019", "251 (C) 501 (D) 502 (E) 1004 _________________ Math Expert Joined: 02 Aug 2009 Posts: 7969 Re: Let a_1, a_2, ... be a sequence determined by the rule a_n=a_{n-1}/2 i [#permalink] ### Show Tags 27 Mar 2019, 01:34 1 1 Bunuel wrote: Let $$a_1$$, $$a_2$$, ... be a sequence determined by the rule $$a_n=\\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_n=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_1 \\leq 2008$$ is it true that $$a_1$$ is less than each of $$a_2$$, $$a_3$$, and $$a_4$$? (A) 250 (B) 251 (C) 501 (D) 502 (E) 1004 1) $$a_1$$ cannot be EVEN, otherwise it will be > $$a_2$$.. So all even 2008/2=1004 gone. 2) Now , $$a_2$$ cannot result in a multiple of 4, otherwise $$a_3$$ will become less than $$a_1$$. For this 3a+1 shouldn't be multiple of 4. So a should not leave a remainder 1 when divided by 4. That means a cannot be 1, 5, 9 and so on. Thus half the odd numbers possible ..3,7..etc D _________________ Re: Let a_1, a_2, ... be a sequence determined by the rule a_n=a_{n-1}/2 i [#permalink] 27 Mar 2019, 01:34 Display posts from previous: Sort by", "Coding challenge Let $a_1,a_2,\\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \\le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$? (Ans: 502.) (Ref: ACM 12A 2008.)", "the $a_i$ must come in weakly decreasing order. If $a_1+\\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\\dots+a_n+1+\\dots+1=2008.$$ If $a_1+\\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider $$(a_1-2)+2+a_2\\dots+a_n=2008$$ (Clearly $a_1=1$ cannot be maximal). Once you have figured out $a_1$, apply the same argument to $a_2$, $a_3$, etc. (edit: at each step above, replace $j$ trailing copies of $1$ by the integer $j$; that is, instead of using $2006+1+1$, use $2006+2$. Doing this does not change the result). This shows that each $a_i$ must be $3$ or $2$. Simplifying the terms $A_1+\\dots+A_n$ is now just adding terms of a 2 geometric series. Finally, if there are $k$ $3s$ then there are $\\frac{2008-3k}{2}$ $2s$; thus we want to maximize $$\\frac{3(3^k-1)}{2}+\\frac{3(3^k-1)}{2}\\cdot2\\cdot\\left(2^\\frac{2008-3k}{2}-1\\right)$$ From Wolframalpha this gives $668$ copies of $3$ and $2$ copies of $2$ for an answer of roughly $3\\times 10^{319}$", "# Prove the formula Let be $a_1$, $a_2$, $\\ldots$, $a_{n-1}$, $a_n$, term of the arithmetics sequence. Please help me prove that $S_n=\\frac{n}{2}(a_1+a_n)$ is sum the n-th term of the given sequence. Thanks. - HINT: $a_1+a_n=a_2+a_{n-1}=\\ldots$ – Raskolnikov Oct 10 '12 at 7:57 Let be $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ real numerical sequence. That one sequence let be arithmetics sequence must that $a_2-a_1=a_3-a_2=\\cdots=d$. Note that the given sequence is $2-1=3-2=4-3=5-4=\\cdots=1$, therefore the given sequence is arithmetics sequence. Also note that $1+11=2+9=3+8=\\cdots$. So generally worth $(*)$ $\\ldots$ $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\\cdots$ Now turn prove the given formula. The given sequence is $a_1, a_2, \\ldots, a_{n-1}, a_n$. Form the sum of the given sequence, and that sum the mark $S_n$, i.e $(1)$ $\\ldots$ $S_n=a_1+a_2+\\ldots+a_{n-1}+a_n$ Because comutative low for the sum (+) is worth, have $(2)$ $\\ldots$ $S_n=a_n+a_{n-1}+\\ldots+a_2+a_1$ Equation $(1)$, $(2)$ collect side to side $S_n+S_n=a_1+a_2+\\ldots+a_{n-1}+a_n+a_n+a_{n-1}+\\ldots+a_2+a_1$ $2S_n=(a_1+a_n)+(a_2+a_{n-1})+\\cdots+(a_{n-1}+a_2)+(a_n+a_1)$ Implement the $(*)$ we have: $2S_n=(a_1+a_n)+(a_1+a_n)+\\ldots+(a_1+a_n)$ $2S_n=n(a_1+a_n)$$/$$:2$ $S_n=\\frac{n}{2}(a_1+a_n)$ - Thanky sir, for proving – user39471 Oct 10 '12 at 8:11 How did you manage to type this up in less than 3 minutes? – commenter Oct 10 '12 at 8:21 Writing the arithmetic sequence in another way, $$a_0 + (a_0 + d) + (a_0 + 2d) + ...$$ Flip it around to get another sequence that decreases to $a_0$, $$(a_0 + nd) +", "# 2010 USAJMO Problems/Problem 2 ## Problem Let $n > 1$ be an integer. Find, with proof, all sequences $x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following three properties: 1. (a). $x_1 < x_2 < \\cdots ; 2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\\ldots,n-1$; 3. (c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i+x_j = x_k$. ## Solution The sequence is $2, 4, 6, \\ldots, 2n-2$. ### Proof 1 We will prove that any sequence $x_1, \\ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done. First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true. Otherwise, for $n > 2$, we will prove by induction on $m$ that the difference $x_{n-m}", "# 2010 USAJMO Problems/Problem 2 ## Problem Let $n > 1$ be an integer. Find, with proof, all sequences $x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following three properties: 1. (a). $x_1 < x_2 < \\cdots ; 2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\\ldots,n-1$; 3. (c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i+x_j = x_k$. ## Solution The sequence is $2, 4, 6, \\ldots, 2n-2$. ### Proof 1 We will prove that any sequence $x_1, \\ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done. First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true. Otherwise, for $n > 2$, we will prove by induction on $m$ that the difference $x_{n-m}", "# Conjecture about the powers of 2 in the prime factorizations of the series $a_n = n(n+1)(n+2)$ I found the following law and would like to know what do you think about it and if anyone can explain why this is so. Also, is this already known and proven? Consider the following series: $$1\\times2\\times3, 2\\times3\\times4, 3\\times4\\times5,\\ldots,n(n+1)(n+2)$$ or more compactly: $$a_n = n(n+1)(n+2)$$ Lets take the prime factorization for every element of $a_n$: $$2\\times3, 2^3\\times3, 2^2\\times3\\times5,2^3\\times3\\times5,\\ldots$$ Lets take all the powers of $2$ in every prime factorization in the above series and create a sequence out of them, and then group the elements of that sequence in tuples with size $4$. The result of this we will call $p_1$. $$p_1 = (1,3,2,3),(1,4,3,4),(1,3,2,3),(1,5,4,5),\\ldots$$ It seems that every $4$-tuple with odd index in $p_1$ is $(1,3,2,3)$. Lets remove all $4$-tuples with odd positions from $p_1$ and call the result $p_2$: $$p_2 = (1,4,3,4),(1,5,4,5),(1,4,3,4),(1,6,5,6),\\ldots$$ Now $p_2$ will have $(1,4,3,4)$ at every odd position. If we take the $p_3$ sequence to have only those elements of $p_2$ that are with even indices, $p_3$ will have $(1,5,4,5)$ at every odd position. And so on... $p_n$ will have $(1, n+2, n+1,n+2)$ at every odd position. • Probably shouldn't use the word partition here, because that has a very specific meaning. Maybe just \"group the elements", "< 6$. For $n > 6$, assume in order to get a contradiction that $m_k = \\frac{n(n-1)}{2}$. Note that $m_k$ is equal to $\\frac{n!}{k! 2^k(n-2k)! 1^{n-2k}}=\\frac{n!}{k! (n-2k)! 2^k}$. Thus we have $\\frac{n(n-1)}{2} = \\frac{n!}{k!(n-2k)!2^k}$ and hence $2^{k-1} = \\frac{(n-2)(n-3)\\ldots(n-2k+1)}{k!}$. Now if $k \\ge 5$, then $(n-4)(n-5)\\ldots(n-2k+1)$ is a product of at least $k$ consecutive integers and hence divisible by $k!$ (see here). Thus $2^{k-1} = \\frac{(n-2)(n-3)\\ldots(n-2k+1)}{k!} = (n-2)(n-3)\\frac{(n-4)(n-5)\\ldots(n-2k+1)}{k!}$ and hence $\\frac{(n-2)(n-3)\\ldots(n-2k+1)}{k!}$ is divisible by an odd number other than one (either $n-2$ or $n-3$). Thus $\\frac{(n-2)(n-3)\\ldots(n-2k+1)}{k!}$ cannot be a power of $2$, contradicting the above equation. When $k = 3$, the above equation becomes: $4= \\frac{(n-2)(n-3)(n-4)(n-5)}{3!} = (n-2)\\frac{(n-3)(n-4)(n-5)}{3!} = \\frac{(n-2)(n-3)(n-4)}{3!}(n-5)$. Thus if $n$ is odd, $n-2$ is an odd number other than $1$ that divides 4, a contradiction. If $n$ is even, then since $n > 6$, $n-5$ is an odd number other than $1$ that divides $4$, which is again a contradiction. Thus we cannot have any case when $m_k = \\frac{n(n-1)}{2}$. ## $S_n$ does not have an outer automorphism if $n \\neq 6$ With the above lemma we can now prove that all the automorphism of $S_n$ for $n \\neq 6$ are inner. First note that the only automorphism of $S_2 = \\{ \\text{id}, (12)\\}$ is the identity which is an inner automorphism. When $n \\ge", "July 23, 2011 ### a,b,c fraction integer Given positive integers $a,b,c$ such that $$\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}$$ is also an integer, show that $$\\frac{a^2b^2}{c},\\frac{b^2c^2}{a}, \\frac{c^2a^2}{b}$$ are all cubic numbers. ## Thursday, July 21, 2011 ### Sequence modulo 2011 A sequence of integers $a_1, \\dots, a_{2010}$ satisfy the following properties: $a_1 - 1$ is divisible by 2011 $a_k a_{k-1} - k$ is divisible by 2011 for $k = 2, 3, \\dots, 2010$ Show that $a_{2010} + 1$ is divisible by 2011. Solution First we prove the following assertion: For $k$ odd, then $a_k.2.4.6. \\cdots.(k-1) - 1.3.5.\\cdots.k$ is divisible by 2011. For $k$ even, then $a_k.1.3.5. \\cdots .(k-1) - 2.4.6.\\cdots.k$ is divisible by 2011. Proof by induction. It can easily be seen for $k=1$ it's true. For $k=2$, we have $2011 | a_2a_1 - 2 = a_2(a_1-1) + (a_2-2)$ So $2011 | a_2 - 2$ Suppose it's true for $k-1$ odd, then for $k$ even: $2011 | a_k a_{k-1} - k$ which means $$2011 | 2.4.\\cdots.(k-2) (a_k a_{k-1} - k) = 2.4.\\cdots.(k-2)a_k a_{k-1} - 2.4.\\cdots.(k-2).k$$ $$2011 | a_k (2.\\cdots.(k-2)a_{k-1} - 1.3.\\cdots.(k-1)) + (a_k.1.3.\\cdots.(k-1) - 2.\\cdots.(k-2).k)$$ And since 2011 divides the first term by induction hypothesis, then it also divides the second term, which completes the induction step. The proof for $k-1$ even and $k$ odd is similar. Now, that", "number then $$x^2$$ is a multiple of 4, which makes $$\\frac{x^2}{2}$$ an even number and therefore $$\\frac{3x^2}{2}+1=3*even+1=even+1=odd$$. If you don't notice this, then one also do in another way. Let $$x=2k$$, for some integer k, then: A. $$\\frac{3x}{2}=\\frac{3*2k}{2}=3k$$ --> if $$k=odd$$ then $$3k=odd$$ but if $$k=even$$ then $$3k=even$$. Discard; B. $$\\frac{3x}{2}+1=\\frac{3*2k}{2}+1=3k+1$$ --> if $$k=odd$$ then $$3k+1=odd+1=even$$ but if $$k=even$$ then $$3k+1=even+1=odd$$. Discard; C. $$3x^2$$ --> easiest one as $$x=even$$ then $$3x^2=even$$, so this option is never odd. Discard; D. $$\\frac{3x^2}{2}=\\frac{3*4k^2}{2}=6k^2=even$$, so this option is never odd. Discard; E. $$\\frac{3x^2}{2}+1=\\frac{3*4k^2}{2}=6k^2+1=even+1=odd$$, thus this option is always odd. Similar question to practice: if-a-and-b-are-positive-integers-such-that-a-b-and-a-b-are-88108.html Hope it helps. _________________ Current Student Joined: 28 Mar 2012 Posts: 316 Location: India GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: If x is even integer, which of the following must be an odd [#permalink] ### Show Tags 25 Jul 2012, 20:26 Hi, Since, x is even we can assume x = 2 or x = 4, such that x/2 is both odd/even as per the value of x. Check each option with these values. We get the answer when both x=2, 4 gives an odd value. Regards, Intern Joined: 29 Nov 2012 Posts: 6 Re: If x is even integer, which of the following must be an odd", "http://mathoverflow.net/questions/49977/do-there-exist-sets-of-integers-with-arbitrarily-large-upper-density-which-contai - Or just take all powers of $3$ and add to them all numbers that are congruent to $1$ modulo $3$. - It took me a bit to understand this since I thought you meant the set of sums of powers of 3 with numbers 1 mod 3. – Harry Altman Dec 17 2010 at 7:30 Much better than my construction! This also shows that the exceptional set in the question can be as large as a set without $3$-APs (if $B$ has no $3$-APs, let $A=3B \\cup (3\\mahthbb{Z}+1)$. Then no element in $3B$ is part of a $3$-AP in $A$). – Pablo Shmerkin Dec 17 2010 at 11:32 This is false. We construct $A$ inductively, so that the following holds: • $A$ contains all powers of two larger or equal than $4$ and no other even numbers. • The number of odd numbers in $A$ between $2^j$ and $2^{j+1}$ is $2^{j-2}$. • No power of two is in a 3-AP contained in $A$. We start by specifying that $4\\in A, 5\\in A, 6\\notin A,7\\notin A$. Suppose $A\\cap\\{1,\\ldots, 2^m-1\\}$ has been defined so that the above properties hold. We next define $A\\cap\\{ 2^m,\\ldots, 2^{m+1}-1\\}$ as follows: $2^m\\in A$. There are $1+2+\\ldots+2^{m-3}<2^{m-2}$ odd numbers smaller than $2^m$ in $A$; let $O_m$ be the set of all of them.", "coding this up to find such sequences for $n=7, 8, 9$ etc? – antkam Apr 16 at 14:41 • @antkam I have not, but I have some reason to believe that large such sequences exist (I'm being purposely vague here). Let me see if I can find them for the $n$ you suggest with a quick script.... – Mees de Vries Apr 16 at 14:48 If $$(a_1, \\ldots, a_n)$$ and $$(b_1, \\ldots, b_n)$$ are AASs of length $$n$$, then we claim that $$(2a_1, \\ldots, 2a_n, 2b_1 - 1, \\ldots, 2b_n-1)$$ is an AAS of length $$2n$$. Indeed it is clear that every integer $$1, \\ldots, 2n$$ appears precisely once, the even ones in the first half and the odd ones in the second half. Furthermore, suppose towards a contradiction that it contains a non-trivial arithmetic sequence. If that sequence is of the form $$(2a_i, 2a_j, x)$$ then we must have that $$x$$ is even, therefore $$x = 2a_k$$. But then $$(a_i, a_j, a_k)$$ would be an arithmetic sequence in $$(a_1, \\ldots, a_n)$$. For a similar reason,we cannot have an arithmetic sequence $$(2b_i - 1, 2b_j - 1, x)$$. Finally, a sequence $$(2a_i, 2b_j - 1, x)$$ cannot be arithmetic either, because $$x$$ would have to be even, but no even number occurs after an odd one. Note furthermore", "- a_{n}a_{n-1}=a_{n}a_{n-3}-a_{n-1}a_{n-2}$ This is equivalent to $a_{n-2}(a_{n+1}+a_{n-1})=a_{n}(a_{n-1}+a_{n-3})$ or $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}$ for all $n \\geq 4$. If n is even, we obtain $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}= \\ldots = \\frac{a_{3}+a_{1}}{a_{2}}=200$ while if n is odd, $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}=\\ldots=\\frac{a_{4}+a_{2}}{a_{3}}=11$ It follows that $a_{n+1} = 200a_{n}-a_{n-1}$, if n is even, and $a_{n+1}=11a_{n}-a_{n-1}$, if n is odd. An inductive argument shows that all $a_{n}$ are positive integers. More later, Nalin Pithwa # Basic Enumeration once again Prove the following identity (a): $\\sum_{k=0}^{n}\\left\\{ \\begin{array}{c} n \\\\ k \\end{array} \\right\\}x(x-1)\\ldots(x-k+1)=x^{n}$ Prove the following identity (b): $\\sum_{k=0}^{n}\\left [ \\begin{array}{c} n \\\\ k \\end{array} \\right] = x(x+1)\\ldots(x+n-1)$ Prove the following identity (c); $\\sum_{k=0}^{n}(-1)^{k}\\left\\{ \\begin{array}{c} n \\\\ k \\end{array} \\right\\}\\left[ \\begin{array}{c} k \\\\ j \\end{array} \\right]= 1$, if $j=n$ and is 0, otherwise. General Hint for all three problems: Please see Wikipedia for Stirling numbers! Hint (a): If x is an integer, both sides of the first identity count mappings of an n element set into an x-element set. Hint (b): Both sides count pairs $(\\pi, \\alpha)$ where $\\pi$ is a permutation of an n-element set S, and $\\alpha$ is a coloration of S with x colors invariant under $\\pi$. Hint (c): Combine the identities in (a) and (b). Solution (a): Suppose first that x is a positive integer. Let $|X|=x$ and $|N|=n$. The number of mappings of N into X is $x^{n}$. On", "# Difference between revisions of \"2008 AMC 12A Problems/Problem 17\" ## Problem Let $a_1,a_2,\\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \\le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$? $\\mathrm{(A)}\\ 250\\qquad\\mathrm{(B)}\\ 251\\qquad\\mathrm{(C)}\\ 501\\qquad\\mathrm{(D)}\\ 502\\qquad\\mathrm{(E)} 1004$ ## Solution All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer. • If $a_1=4n$, then $a_2=\\frac{4n}{2}=2n. • If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\\frac{12n+4}{2}=6n+2$, and $a_4=\\frac{6n+2}{2}=3n+1. • If $a_1=4n+2$, then $a_2=2n+1. • If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$. Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1. Since one fourth of the positive integers $a_1 \\le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\\frac{1}{4}\\cdot 2008 = 502 \\Rightarrow D$.", "## Saturday, July 23, 2011 ### a,b,c fraction integer Given positive integers $a,b,c$ such that $$\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}$$ is also an integer, show that $$\\frac{a^2b^2}{c},\\frac{b^2c^2}{a}, \\frac{c^2a^2}{b}$$ are all cubic numbers. ## Thursday, July 21, 2011 ### Sequence modulo 2011 A sequence of integers $a_1, \\dots, a_{2010}$ satisfy the following properties: $a_1 - 1$ is divisible by 2011 $a_k a_{k-1} - k$ is divisible by 2011 for $k = 2, 3, \\dots, 2010$ Show that $a_{2010} + 1$ is divisible by 2011. Solution First we prove the following assertion: For $k$ odd, then $a_k.2.4.6. \\cdots.(k-1) - 1.3.5.\\cdots.k$ is divisible by 2011. For $k$ even, then $a_k.1.3.5. \\cdots .(k-1) - 2.4.6.\\cdots.k$ is divisible by 2011. Proof by induction. It can easily be seen for $k=1$ it's true. For $k=2$, we have $2011 | a_2a_1 - 2 = a_2(a_1-1) + (a_2-2)$ So $2011 | a_2 - 2$ Suppose it's true for $k-1$ odd, then for $k$ even: $2011 | a_k a_{k-1} - k$ which means $$2011 | 2.4.\\cdots.(k-2) (a_k a_{k-1} - k) = 2.4.\\cdots.(k-2)a_k a_{k-1} - 2.4.\\cdots.(k-2).k$$ $$2011 | a_k (2.\\cdots.(k-2)a_{k-1} - 1.3.\\cdots.(k-1)) + (a_k.1.3.\\cdots.(k-1) - 2.\\cdots.(k-2).k)$$ And since 2011 divides the first term by induction hypothesis, then it also divides the second term, which completes the induction step. The proof for $k-1$ even and $k$ odd is similar.", "# Difference between revisions of \"2010 USAJMO Problems/Problem 2\" ## Problem Let $n > 1$ be an integer. Find, with proof, all sequences $x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following three properties: 1. (a). $x_1 < x_2 < \\cdots ; 2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\\ldots,n-1$; 3. (c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i+x_j = x_k$. ## Solution The sequence is $2, 4, 6, \\ldots, 2n-2$. ### Proof We will prove that any sequence $x_1, \\ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done. First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true. Otherwise, for $n > 2$, we will prove by induction on $m$ that", "# 2008 AMC 12A Problems/Problem 17 ## Problem Let $a_1,a_2,\\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \\le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$? $\\mathrm{(A)}\\ 250\\qquad\\mathrm{(B)}\\ 251\\qquad\\mathrm{(C)}\\ 501\\qquad\\mathrm{(D)}\\ 502\\qquad\\mathrm{(E)} 1004$ ## Solution All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer. • If $a_1=4n$, then $a_2=\\frac{4n}{2}=2n. • If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\\frac{12n+4}{2}=6n+2$, and $a_4=\\frac{6n+2}{2}=3n+1. • If $a_1=4n+2$, then $a_2=2n+1. • If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$. Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1. Since one fourth of the positive integers $a_1 \\le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\\frac{1}{4}\\cdot 2008 = 502 \\Rightarrow D$. 2008 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions", "# Determine positive integers If $n\\geq3$ is a positive integer, determine $n+1$ positive integers With the property that the sum of all $n$ integers from the $n+1$ numbers build the set: $$S=\\{n^2+2,n^2+4,n^2+6,...,n^2+2n-2,n^2+2n\\}.$$ For example, if $n=3$, the numbers are $3,3,5,7$. If $n=4$ the numbers are $2,4,6,6,8$. So for $n=3$ the set $S$ is $11,13,15$. We have $3+3+5=11, 3+5+7=15, 3+7+3=13$. We have four numbers and the sum of $3$ arbitrary picked numbers which I have detmined must be an element of $S$. The question is: how to detrmine $n+1$ numbers with that property, also $n$ numbers picked from the sum (arbitrary) must be an element of $S$ and all the sums of $n$ numbers must build the set $S$. - How do you get $3,3,5,7$ for $n=3$. You have to word your problem better. – user44197 Jan 10 '14 at 5:41 What is the question? – user127.0.0.1 Jan 10 '14 at 5:42 Observing the cases for smaller $n$s will give us the following numbers : When $n$ is odd, the numbers you want are $3,3,5,7,\\cdots, 2n-1,2n+1.$ When $n$ is even, the numbers you want are $2,4,6,\\cdots, n, n+2, n+2, n+4,\\cdots, 2n-2, 2n$. In the following, let us prove that these are true. In the case that $n$ is odd, you can easily see $$3+3+5+\\cdots+(2n-1)=3+\\sum_{k=1}^{n-1}(2k+1)=3+2\\cdot\\frac{n(n-1)}{2}+n-1=n^2+2,$$ $$3+5+7+\\cdots+(2n+1)=\\sum_{k=1}^{n}(2k+1)=2\\cdot\\frac{n(n+1)}{2}+n=n^2+2n.$$ Also, since the difference", "a_1,a_2,a_3,a_4\\rangle$. We know that $a_1,a_2,a_3<a_4$, so even if the first three terms are all equal, there are at least $2$ and at most $4$ different numbers in the sequence. How many such sequences are there with exactly $2$ different numbers? They’re easy to count: pick any two numbers from $\\{0,\\ldots,n\\}$, make the larger one $a_4$, and set $a_1,a_2$, and $a_3$ equal to the smaller one. This produces a sequence in $S$, and every sequence in $S$ with exactly $2$ different numbers can be formed uniquely in this way, so there are $\\binom{n+1}2$ such sequences, one for every pair of integers from the set $\\{0,\\ldots,n\\}$. To form a sequence with $3$ different members, we choose $3$ elements from $\\{0,\\ldots,n\\}$. The largest will of course be $a_4$. However, there are $6$ ways to distribute the other $2$ among $a_1,a_2$, and $a_3$. Say the two smaller numbers are $k$ and $\\ell$. There are $3$ ways to choose one of $a_1,a_2$, and $a_3$ to be the singleton and $2$ ways to decide whether it will be $k$ or $\\ell$; once one of those $3\\cdot2=6$ choices has been made, the other two of $a_1,a_2$, and $a_3$ will of course be the other one of $k$ and $\\ell$, so the entire sequence is determined. There are $\\binom{n+1}3$ ways to choose the $3$ numbers, and" ]

[ "# Give me a four, give me a two Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that: \\begin{align} xyz&=945 \\\\ x(y+1)+y(z+1)+z(x+1)&=385. \\end{align} If the minimum possible value of $$z+\\frac{y}{2}+\\frac{x}{4}$$ can be written as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a+b$$? ×", "# $$(x+a)(x+b+ci)(x+b-ci)$$ Algebra Level 2 Suppose that $$a, b, c$$ are positive reals such that \\begin{align} & x^3+7x^2+24x+18 \\\\ &= (x+a)(x+b+ci)(x+b-ci), \\end{align} where $$i$$ is the imaginary number that satisfies $$i^2=-1.$$ What is the value of $$a+b+c?$$ ×", "## Thinking Mathematically (6th Edition) $y=\\frac{C-Ax}{B}$ The given formula is: $Ax+By=C$ Simplify the given equation for y: Subtract Ax from both sides followed by dividing both sides by B. \\begin{align} & By=C-Ax \\\\ & y=\\frac{C-Ax}{B} \\\\ \\end{align} This is an equation of a straight line in standard form; Ax +By +C = 0 where A, B and C are Integers.", "+0 # PLS Help 0 60 1 Real numbers $x$ and $y$ satisfy \\begin{align*} x + xy^2 &= 250y, \\\\ x - xy^2 &= -240y. \\end{align*}Enter all possible values of $x,$ separated by commas. May 2, 2021", "This can now be evaluated by using the substitution u = x + 3, which yields $\\int\\frac{dx}{(x+3)^2 + 4} \\,=\\, \\frac{1}{2}\\arctan\\left(\\frac{x+3}{2}\\right)+C.$ ### Complex numbers Consider the expression $|z|^2 - b^*z - bz^* + c,\\,$ where z and b are complex numbers, z* and b* are the complex conjugates of z and b, respectively, and c is a real number. Using the identity |u|2 = uu* we can rewrite this as $|z-b|^2 - |b|^2 + c , \\,\\!$ which is clearly a real quantity. This is because \\begin{align} |z-b|^2 &{}= (z-b)(z-b)^*\\\\ &{}= (z-b)(z^*-b^*)\\\\ &{}= zz^* - zb^* - bz^* + bb^*\\\\ &{}= |z|^2 - zb^* - bz^* + |b|^2 . \\end{align} As another example, the expression $ax^2 + by^2 + c , \\,\\!$ where a, b, c, x, and y are real numbers, with a > 0 and b > 0, may be expressed in terms of the square of the absolute value of a complex number. Define $z = \\sqrt{a}\\,x + i \\sqrt{b} \\,y .$ Then \\begin{align} |z|^2 &{}= z z^*\\\\ &{}= (\\sqrt{a}\\,x + i \\sqrt{b}\\,y)(\\sqrt{a}\\,x - i \\sqrt{b}\\,y) \\\\ &{}= ax^2 - i\\sqrt{ab}\\,xy + i\\sqrt{ba}\\,yx - i^2by^2 \\\\ &{}= ax^2 + by^2 , \\end{align} so $ax^2 + by^2 + c = |z|^2 + c . \\,\\!$ ## Geometric perspective Consider completing the square for the equation $x^2", "## Thinking Mathematically (6th Edition) $\\left\\{ x\\ge 16 \\right\\}$. Let us suppose the number is $x$. Then, the algebraic form of the inequality is: $\\frac{3x}{4}\\,-3\\,\\ge 9$ And now to find out possible values of $x$ satisfying the inequality the inequality is solved as follows: \\begin{align} & \\frac{3x}{4}\\,-3\\,\\ge 9 \\\\ & \\frac{3x\\times 4}{4}\\,-3\\times 4\\,\\ge 9\\times 4 \\\\ & 3x\\,-12\\ge 36 \\end{align} This can be further simplified as: \\begin{align} & 3x-12+12\\ge 36+12 \\\\ & 3x\\ge 48 \\\\ & \\frac{3x}{3}\\ge \\frac{48}{3} \\\\ & x\\ge \\text{ }16 \\end{align} Hence all the real numbers greater than or equal to 16 will satisfy the condition. The set-builder form of the inequality obtained is: $\\left\\{ x\\ge 16 \\right\\}$ Therefore, the number can be represented in set builder form as $\\left\\{ x\\ge 16 \\right\\}$.", "+0 Help!!!!! +3 276 2 +730 Let x and y be real numbers whose absolute values are different and that satisfy \\begin{align*} x^3 &= 20x + 7y \\\\ y^3 &= 7x + 20y \\end{align*} Find xy. And this one If x, y, and z are real numbers for which \\begin{align*} x+y-z &= -8, \\\\ x-y+z &= 18,\\text{ and} \\\\ -x+y+z &= 30, \\\\ \\end{align*} then what is xyz? MIRB16 Sep 19, 2017 edited by MIRB16 Sep 19, 2017 Best Answer #1 +19649 +3 1. Let x and y be real numbers whose absolute values are different and that satisfy \\begin{align*} \\mathbf{ x^3 }&\\mathbf{ = 20x + 7y } \\\\ \\mathbf{ y^3 }&\\mathbf{ = 7x + 20y } \\end{align*} Find xy. $$\\begin{array}{lrcl|c|c|} &&&& I. & II.\\\\ \\hline (1) & x^3 &=& 20x + 7y \\\\ &&&& + & - \\\\ (2) & y^3 &=& 7x + 20y \\\\ \\end{array}$$ $$\\begin{array}{|rcll|} \\hline (I.) & x^3+y^3 &=& 20x + 7y +7x + 20y \\\\ & x^3+y^3 &=& 27x + 27y \\\\ & x^3+y^3 &=& 27(x+y) \\\\ &&& x^3+y^3 = (x+y)(x^2-xy+y^2) \\\\ & (x+y)(x^2-xy+y^2) &=& 27(x+y) \\\\ & \\mathbf{x^2-xy+y^2} &\\mathbf{=}& \\mathbf{27} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline (II.) & x^3-y^3 &=& 20x + 7y -(7x + 20y) \\\\ & x^3-y^3 &=& 13x -13y \\\\ & x^3-y^3 &=& 13(x-y) \\\\ &&& x^3-y^3 = (x-y)(x^2+xy+y^2)", "$y=cy_0$. Notice that \\begin{align} ax+by&=cax_0+cby_0\\\\ &=c(ax_0+by_0)\\\\ &=c \\cdot 1\\\\ &= c \\end{align} Hence, $ax+by=c$ always has a solution for every integer $c$. As for $37x+47y=103$, $x=-15$ and $y=14$ gives the smallest sum $x+y$.", "# Difference between revisions of \"2001 AIME I Problems/Problem 2\" ## Problem A finite set $\\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\\mathcal{S}\\cup\\{1\\}$ is $13$ less than the mean of $\\mathcal{S}$, and the mean of $\\mathcal{S}\\cup\\{2001\\}$ is $27$ more than the mean of $\\mathcal{S}$. Find the mean of $\\mathcal{S}$. ## Solution Let $x$ be the mean of $\\mathcal{S}$. Let $a$ be the number of elements in $\\mathcal{S}$. Then, the given tells us that $\\frac{ax+1}{a+1}=x-13$ and $\\frac{ax+2001}{a+1}=x+27$. Subtracting, we have \\begin{align*}\\frac{ax+2001}{a+1}-40=\\frac{ax+1}{a+1} \\Longrightarrow \\frac{2000}{a+1}=40 \\Longrightarrow a=49\\end{align*} We plug that into our very first formula, and get: \\begin{align*}\\frac{49x+1}{50}&=x-13 \\\\ 49x+1&=50x-650 \\\\ x&=\\boxed{651}.\\end{align*}", "# Difference between revisions of \"2014 UMO Problems/Problem 5\" ## Problem Find all positive real numbers $x, y$, and $z$ that satisfy both of the following equations. \\begin{align*} xyz & = 1\\\\ x^2 + y^2 + z^2 & = 4x\\sqrt{yz}- 2yz \\end{align*} ## Solution By AM-GM $$x^2+y^2+z^2 + 2yz\\ge x^2 + 4yz\\ge 4x\\sqrt{yz}$$ Hence, the second equation implies that $y=z$ and $x^2=4yz\\implies x=2y=2z$. Now we plug it into the first equation to get $(x,y,z) = \\left(\\sqrt[3]{4}, \\frac{\\sqrt[3]4}{2}, \\frac{\\sqrt[3]4}{2}\\right)$", "+0 # help 0 44 1 The equation $$\\frac{ax^2 - 24x + b}{x - 1} = x^2 + x$$ holds for exactly two real values of $$x,$$\\$ and their sum is 12. Find $$a - b.$$ Apr 11, 2019", "case we let one of the variables, either $$y$$ or $$z$$ be free. That is we let $$y$$ or $$z$$ be equal to some parameter $$t\\in\\mathbb{R}$$. It doesn’t matter which variable you choose to be free, though the answer will look different. We will set $$z=t$$ then we have \\begin{align*} 3y+z & =1\\\\ 3y+t & =1\\\\ 3y & =1-t\\\\ y & =\\frac{1-t}{3}. \\end{align*} From row $$1$$ and substituting $$z=t$$ and $$y=\\left(1-t\\right)/3$$ we have \\begin{align*} x+y+z & =3\\\\ x+\\frac{1-t}{3}+t & =3\\\\ 3x+1-t+3t & =9\\\\ 3x & =8-2t\\\\ x & =\\frac{8-2t}{3}. \\end{align*} So the solution here is \\begin{align*} x & =\\frac{8-2t}{3},y=\\frac{1-t}{3},z=t,\\;t\\in\\mathbb{R}. \\end{align*} This gives us an infinite number of solutions because $$t$$ can be any real number. To check the solution you can pick any value of $$t,$$determine $$x,y$$ and $$z$$ and substitute in the original equations. For example let $$t=0$$.3 Here we took $$t=0$$ because it is easy to get the corresponding $$x,\\,y$$ and $$z$$ values. But you can pick any value of $$t$$. Once you choose $$t$$, there will be a corresponding value of $$x,\\,y$$ and $$z.$$ These $$x,\\,y$$ and $$z$$ values will satisfy the original equations. Note that picking a value of $$t$$ and checking if the corresponding $$x,\\,y$$ and $$z$$ values will the original equation is not a proof that your solution is correct. A real", "z in equation (I) and we get the value of y, \\begin{align} & a\\left( \\frac{8}{a} \\right)-by-2c\\left( -\\frac{5}{c} \\right)=21 \\\\ & 8-by+10=21 \\\\ & 18-by=21 \\end{align} Now, subtract $18$ from both sides to get: \\begin{align} & 18-by-18=21-18 \\\\ & -by=3 \\end{align} By dividing both sides by -b we get, \\begin{align} & \\frac{by}{b}=-\\frac{3}{b} \\\\ & \\text{ }y=-\\frac{3}{b} \\\\ \\end{align} Hence, the values are $x=\\frac{8}{a}$, $y=-\\frac{3}{b}$, and $z=-\\frac{5}{c}$.", "# Is the question right ? Algebra Level 5 \\large \\begin{cases} \\begin{align*} x+y+z&=4 \\\\ x^2+y^2+z^2&=6 \\end{align*} \\end{cases} If $$x,y$$ and $$z$$ are real numbers that satisfy the equation above, and that they lie in the range of $$\\left [ \\frac ab, \\frac cd \\right]$$ for positive integers $$a,b,c$$ and $$d$$ such that $$\\gcd(a,b) = \\gcd(c,d) = 1$$, find the value of $$a+b+c+d$$. ×", "# Algebra experts can answer this in a second Algebra Level 2 \\begin{align*} a + b + c = 3 \\\\ \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 5 \\end{align*} If $$a,b,c$$ are nonzero real numbers satisfying the above equations, find the value of $\\frac{a}{b} + \\frac{a}{c} + \\frac{b}{a} + \\frac{b}{c} + \\frac{c}{a} + \\frac{c}{b}.$ ×", "know that from given equality real part of previous expression is zero therefore it is purely imaginary. 2. Given, $$|z - i| < 1$$ we have following \\begin{align}\\begin{aligned}|z + 12 -6i| = |(z - i) + (12 - 5i)| \\le |z - i| + |12 - 5i|\\\\\\Rightarrow |z + 12 - 6i| < 1 + 13 = 14\\end{aligned}\\end{align} 3. \\begin{align}\\begin{aligned}|z + 6| = |2z + 3| \\Rightarrow x^2 + 12x + 36 + y^2 = 4x^2 + 12x + 9 + 4y^2 \\Rightarrow 3x^2 + 3y^2 = 27\\\\\\Rightarrow x^2 + y^2 = 9 \\Rightarrow |z| = 3\\end{aligned}\\end{align} 4. Given, $$\\sqrt{a - ib} = x - iy \\Rightarrow a - ib = x^2 - y^2 - i2xy$$. Equating real and imaginary parts we have, $$a = x^2 - y^2 \\text{and} b = 2xy$$. $$\\therefore \\sqrt{a + ib} = \\sqrt{x^2 - y^2 + i2xy} = x + iy$$. 5. We will solve this problem by method of negation. Let us say there is a complex number $$y + iz$$ which is root of this equation. Therefore, this root will satisfy this equation. Hence, $\\frac{A^2}{y + iz -a} + \\frac{B^2}{y + iz -b} + ... + \\frac{H^2}{y + iz -h} = y + iz + l$ Multiplying and dividing each term with conjugate of denominator we get $\\frac{A^2(y - a -iz)}{(y", "Select Page Understand the problem Suppose that real numbers $x,y$ and $z$ satisfy the following equations: \\begin{align*} x+\\frac{y}{z} &=2,\\\\ y+\\frac{z}{x} &=2,\\\\ z+\\frac{x}{y} &=2. \\end{align*} Show that $s=x+y+z$ must be equal to $3$ or $7$. Note: It is not required to show the existence of such numbers $x,y,z$. Source of the problem Germany MO 2019, Problem 6 Topic Algebra, Simultaneous Equations 6/10 Suggested Book Challenges and Thrills of Pre College Mathematics Do you really need a hint? Try it first! Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3. Now, observe one thing that this set of equations is symmetric in (x,y,z). Observe that we are required to comment on (x+y+z). Rewriting the equations as: $xz+y = 2z, \\qquad (1)$ $xy + z = 2x, \\qquad (2)$ $yz + x = 2y \\qquad (3)$ and then summing gives us that $x+y+z = xy + yz + zx = s.$ Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let’s consider the case, where all of x,y,z is not 1. From now on we consider $x,y,z \\neq 1$. This also gives $x \\neq y \\neq z \\neq x$ Solving the first expression $x=\\frac{2z-y}{z}$ then", "of two real quadratics. To simplify the calculations, let $$\\,z=x+iy\\,$$, $$\\,\\frac{b}{a}=b'+ib''$$, $$\\,\\frac{c}{a}=c'+ic''$$, then dividing the equation by $$\\,a \\ne 0\\,$$ and isolating the real and imaginary parts gives: $$(x+iy)^2+(b'+ib'')(x+iy)+c'+ic'' = 0$$ \\iff\\;\\;\\;\\; \\begin{cases} \\begin{align} x^2 - y^2+b'x-b''y + c' &= 0 \\\\ 2xy+b'y+b''x+c'' &= 0 \\end{align} \\end{cases} For the equation to have a real root, the system must have a solution with $$\\,y=0\\,$$: \\begin{cases} \\begin{align} x^2 +b'x + c' &= 0 \\\\ b''x+c'' &= 0 \\end{align} \\end{cases} • If $$\\,b''=0\\,$$ the second equation requires $$\\,c''=0\\,$$, which is the case where $$\\frac{b}{a}, \\frac{c}{a} \\in \\mathbb R$$, so the original quadratic is equivalent to one with real coefficients. The roots will be either both complex, or both real, depending on the sign of $$\\,\\Delta'=b'^{\\,2}-4c'\\,$$. • If $$\\,b'' \\ne 0\\,$$ then the second equation gives $$\\,x=-\\frac{c''}{b''}\\,$$, and substituting back into the first equation $$\\,c''^{\\,2}+b'b''c''+b''^{\\,2}c'=0\\,$$, which is the condition for at least one real root.", "# Tanishq's roots Algebra Level 4 Real numbers $$w\\leq x\\leq y\\leq z$$ satisfy \\begin{align} w+x+y+z&=12\\\\ wx+wy+wz+xy+xz+yz&=17\\\\ wxy+wxz+wyz+xyz&=-114\\\\ wxyz&=-216. \\end{align} Consider the polynomial $g(u)= wu^3+xu^2+yu+z.$ The equation $$g(u) = 0$$ has roots $$p, q, r$$. The value of $$p^2 + q^2 + r^2$$ can be written as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are positive coprime integers. What is the value of $$a+b$$? This problem is posed by Tanishq A. ×", "+0 # Rational Function Equations +1 68 2 +999 Let $$a, b, c, x, y$$ and $$z$$ be real numbers such that $$a+x \\neq 0, b+y \\neq 0, c+z \\neq 0,$$ and $$a^2=by+cz$$ $$b^2=cz+ax$$ $$c^2=ax+by$$. Find all possible values of $$\\frac{x}{a + x} + \\frac{y}{b + y} + \\frac{z}{c + z}$$. I'm not sure where to start. Thanks in advance! Dec 23, 2019 #1 0 The possible values are -1, 1, and 2. Dec 27, 2019 #2 +107079 0 I have no idea either CoolStuffYT Dec 27, 2019" ]

[ "y=154$$ $$-2 x+7 y=19$$ Add, solve for $$x=\\frac{173}{33}$$. Instead of substituting, practice eliminating $$x$$ by scaling the first equation by 2 and the second equation by 5. $$10 x-2 y=44$$ $$-10 x+35 y=95$$ Add, solve for $$y$$. Final Answer: $$\\left(\\frac{173}{33}, \\frac{139}{33}\\right)$$ ##### Example 4 Solve the following system of equations: \\begin{aligned} 5 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y} &=11 \\\\ \\frac{1}{x}+\\frac{1}{y} &=4 \\end{aligned} The strategy of elimination still applies. You can eliminate the $$\\frac{1}{y}$$ term if the second equation is scaled by a factor of -2 $$5 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y}=11$$ $$-2 \\cdot \\frac{1}{x}-2 \\cdot \\frac{1}{y}=-8$$ Add the equations together and solve for $$x$$. \\begin{aligned}-3 \\cdot \\frac{1}{x}+0 \\cdot \\frac{1}{y} &=3 \\\\ -3 \\cdot \\frac{1}{x} &=3 \\\\ \\frac{1}{x} &=-1 \\\\ x &=-1 \\end{aligned} Substitute into the second equation and solve for $$y$$. \\begin{aligned} \\frac{1}{-1}+\\frac{1}{y} &=4 \\\\ -1+\\frac{1}{y} &=4 \\\\ \\frac{1}{y} &=5 \\\\ y &=\\frac{1}{5} \\end{aligned} The point $$\\left(-1, \\frac{1}{5}\\right)$$ is the point of intersection between these two curves. ##### Example 5 Solve the following system using elimination: \\begin{aligned} 11 \\cdot \\frac{1}{x}-5 \\cdot \\frac{1}{y} &=-38 \\\\ 9 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y} &=-25 \\end{aligned} To eliminate $$\\frac{1}{y},$$ scale the first equation by 2 and the second equation by $$5 .$$ To eliminate $$\\frac{1}{x},$$ scale the first equation by -9 and the second equation by 11 Final Answer: $$\\left(-\\frac{1}{3}, 1\\right)$$ ##### Review Solve each", "with x in the second and the third equations as we intended. Next step is to eliminate $y$ from from the third equation. First of all, we divide the second equation by $2$: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\12y-9z=-\\frac{5}{2}\\end{aligned}\\right. Next we subtract the second equation multiplied by $12$ from the third: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\21z=-\\frac{5}{2}-15=-\\frac{35}{2}\\end{aligned}\\right. As we can see, we already have our system in required form which is called triangular. What is left for us to do is to perform so called reverse course of Gauss method: first, we get $z$ from the last equation: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Then we substitute obtained value for $z$ into the second equation to obtain $y$: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}\\cdot(-\\frac{5}{6})=\\frac{5}{4}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Rearranging terms and performing required calculations we get: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y=\\frac{5}{2}\\cdot(-\\frac{5}{6})+\\frac{5}{4}=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. And the last step: we substitute values for $y,z$ into the first equation to obtain $x$: \\left\\{ \\begin{aligned}x-2(-\\frac{5}{6})+3(-\\frac{5}{6}) =\\frac{3}{2}\\\\y=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Performing necessary calculations, we finally obtain: \\left\\{ \\begin{aligned}x=\\frac{7}{3}\\\\y=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. So the answer is the following: \\left\\{ \\begin{aligned}x=\\frac{7}{3}\\\\y=z=-\\frac{5}{6}\\end{aligned}\\right. That’s it. More examples in the next section. 0 Shares Filed under Math.", "$a$ and $b$ are integers and $(a,b)=d$, then $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem 7. Suppose that $a\\ne 0$, $b\\ne 0$,", "side of the equation remains unchanged. Using Lemma 2, the following describes infinitely many solutions of the equation $1638x+357y=819$. $x=-195+357k$ and $y=897-1638k$ for all integers $k$ But the above solution set is not complete. For example, the number pair $x=-178$ and $y=819$ is also a solution to the equation $1638x+357y=819$, but is not listed above. The following theorem will give the complete solution set. Theorem 3 Let $d=\\text{GCD}(a,b)$. If the pair $x=x_0$ and $y=y_0$ is a solution to $ax+by=c$, then the complete solution set of the equation consists of all the integer pairs $(x,y)$ such that $\\displaystyle x=x_0+\\frac{b}{d} \\cdot k$ $\\displaystyle y=y_0-\\frac{a}{d} \\cdot k$ where $k$ is any integer. Proof of Theorem 3 Suppose that the pair $x=x_0$ and $y=y_0$ is a solution to $ax+by=c$. Two points to show. One is that any number pair of the above form is a solution. The other is that any solution is of the above form. To see the first, note that $\\displaystyle a(x_0+\\frac{b}{d} \\cdot k)+b(y_0-\\frac{a}{d} \\cdot k)=ax_0+by_0=c$ To see the second point, suppose that the number pair $x=x_1$ and $y=y_1$ is a solution to the equation $ax+by=c$. We have the following. $ax_0+by_0=c$ $ax_1+by_1=c$ Rearranging and dividing by $d$, we have the following. $a(x_1-x_0)=b(y_0-y_1)$ $\\displaystyle \\frac{a}{d}(x_1-x_0)=\\frac{b}{d}(y_0-y_1) \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "from twice the area of |ABC| being $ah_a = ax + by + cz$. Similarly, $\\displaystyle |PY_1Z| = \\frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\\quad \\quad (2)$ Finally, \\begin{aligned}|X_1Y_1B| &= \\left(\\frac{h_b-y}{h_b}\\right)^2|ABC|\\\\ &= \\left(1-\\frac{by}{bh_b}\\right)^2 |ABC|\\\\ &= \\left(1-\\frac{by}{2|ABC|}\\right)^2|ABC|\\\\ &= \\left(1-\\frac{by}{ax+by+cz}\\right)^2|ABC|\\\\ &= \\left(\\frac{ax +cz}{ax+by+cz}\\right)^2|ABC|. \\quad\\quad(3)\\end{aligned} Combining (1), (2) and (3), we obtain our desired answer as \\begin{aligned} \\frac{|BXPZ|}{|ABC|} &= \\frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\\\&= \\left(\\frac{ax +cz}{ax+by+cz}\\right)^2-\\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\\\&=\\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\\\ &= \\frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\\\&= \\frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\\quad\\quad(4)\\end{aligned} Similar formulas can be found for quadrilaterals $XPYC$ and $YPZA$ by permuting variables. Note that if $P$ is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables $x, y, z$. Note that (4) may also be written as $\\displaystyle \\frac{|BXPZ|}{|ABC|} = \\frac{ac(2xz + (x^2 + z^2)\\cos B)}{(ax+by+cz)^2}.\\quad\\quad(5)$ ### Special cases 1) If $\\triangle ABC$ is equilateral, $a=b=c$ and from (4) we obtain \\begin{aligned} \\frac{|BXPZ|}{|ABC|} &= \\frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\\\ &= \\frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\\\ &= \\frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\\quad\\quad(6)\\end{aligned} 2) If $P$ is at the incentre of $\\triangle ABC$, then $x = y = z = r$ (the inradius) and from (4) we", "If you find any mistakes, please make a comment! Thank you. ## Solve general linear equations with $2\\times 2$ matrix Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.8 Solution: (a) In this case the system of equations is \\begin{alignat*}{1} 0\\cdot x_1 + 0\\cdot x_2 &= 0\\\\ 0\\cdot x_1 + 0\\cdot x_2 &= 0 \\end{alignat*}Clearly any $(x_1,x_2)$ satisfies this system since $0\\cdot x=0$ $\\forall$ $x\\in F$. (b) Let $(u,v)\\in F^2$. Consider the system: \\begin{alignat*}{1} a\\cdot x_1 + b\\cdot x_2 &= u\\\\ c\\cdot x_1 + d\\cdot x_2 &= v \\end{alignat*}If $ad-bc\\not=0$ then we can solve for $x_1$ and $x_2$ explicitly as $$x_1=\\frac{du-bv}{ad-bc}\\quad x_2=\\frac{av-cu}{ad-bc}.$$Thus there's a unique solution for all $(u,v)$ and in partucular when $(u,v)=(0,0)$. (c) Assume WOLOG that $a\\not=0$. Then $ad-bc=0$ $\\Rightarrow$ $d=\\frac{cb}{a}$. Thus if we multiply the first equation by $\\frac ca$ we get the second equation. Thus the two equations are redundant and we can just consider the first one $a x_1+bx_2=0$. Then any solution is of the form $(-\\frac bay,y)$ for arbitrary $y\\in F$. Thus letting $y=1$ we get the solution $(-b/a,1)$ and the arbitrary solution is of the form $y(-b/a,1)$ as desired.", "Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem 7. Suppose that $a\\ne 0$, $b\\ne 0$, and $c$ are integers. Let $(x_0,y_0)$ be a particular solution to $ax+by=c$. Then all", "consider $$a=-A$$, $$b=1$$, $$c=-1$$, so that $$\\theta=\\sqrt[3]{c/b}=-1\\in\\Bbb Q$$, then use the transform in the affine space corresponding to $$Y=1$$: \\begin{aligned} Z &=\\frac{6x}{y-36A}\\ ,\\\\ X &=-\\frac{y+36A}{y-36A}\\ ,\\ . \\end{aligned} and introducing in $$-AZ^3+X^3=-1$$ we get $$-A\\left(\\frac{6x}{y-36A}\\right)^3 -\\left(\\frac{y+36A}{y-36A}\\right)^3 =-1\\ .$$ Now multiply with the common denominator, and reshape equivalently as: \\begin{aligned} -A\\cdot 6^3x^3 & -y^3 -3\\cdot 6^2Ay^2 -3\\cdot 6^4A^2y -6^6A^3 \\\\ = & -y^3 +3\\cdot 6^2Ay^2 -3\\cdot 6^4A^2y +6^6A^3\\ . \\end{aligned} The cancellations can now be followed with bare eyes. • thanks for your compeletly respond! – bluemath Jul 31 at 13:53", "# $cos(ax)=a\\cdot cos(x)$ Which are all the values of $a$? How do I know for which values of $$a$$ is true the following?$$\\space$$ $$cos(ax)=a\\cdot cos(x)$$ for all $$x\\in \\mathbb{R}$$ It is trivial that it is true for $$a=1$$. But are the more values of $$a$$ for which it is true? • There are no other values $a \\in \\Bbb R$ for which this holds Dec 29 '20 at 16:13 • And how can I prove it? @BenGrossmann Dec 29 '20 at 16:14 • True for all $x\\in\\mathbb R$? Then substitute $x=0$. Dec 29 '20 at 16:14 • True for all $x\\in \\mathbb{R}$ @drhab Dec 29 '20 at 16:15 • Then also for $x=0$ and substituting that we find that $1=a$. Dec 29 '20 at 16:16 Replacing with $$x=0$$ gives $$1 = a$$, so $$a=1$$ is the only value. By no means a real proof, but if you series-expand teach trig term and re-combine powers in $$x$$, it's pretty clear: \\begin{align*}\\cos\\left(ax\\right) &= a \\cos x \\\\ 1 - \\frac{a^2x^2}{2!} + \\frac{a^4x^4}{4!} - \\cdots &= a - a\\frac{x^2}{2!} + a\\frac{x^4}{4!} - \\cdots \\\\ \\left(1-a\\right) - \\frac{x^2}{2!}\\left(a^2-a\\right) + \\frac{x^4}{4!}\\left(a^4 - a\\right) - \\cdots&= 0\\end{align*} For the above to hold for any $$x$$, the terms in parentheses must each be zero. This only works with $$a=1$$. • I think your argument is", "the quadratic for $x$, and substitute the solution(s), if any, back into the equation for the line. Here's an equation which works even if the line is vertical. This is useful from a computer programming perspective. Let's set up the system of equations using standard formulae: $$\\begin{cases} (x - x_0)^2 + (y - y_0)^2 = r^2 \\\\ Ax + By + C = 0 \\end{cases}$$ where the circle with radius $$r$$ is centered at $$(x_0, y_0)$$; the line contains the points $$(x_1, y_1)$$ and $$(x_2, y_2)$$. From the line equation, we know: $$y = \\frac{-Ax-C}{B} = - \\frac{Ax+C}{B}$$ ($$B \\neq 0$$, we'll get into this in a second) Therefore, \\begin{align*} (x - x_0)^2 + \\left( - \\frac{Ax+C}{B} - y_0 \\right)^2 &= r^2 \\\\ x^2 - 2x_0x + {x_0}^2 + \\frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \\cdot \\frac{Ax+C}{B} + {y_0}^2 &= r^2 \\\\ x^2 - 2x_0x + \\frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \\cdot \\frac{Ax+C}{B} &= r^2 - {x_0}^2 - {y_0}^2 \\end{align*} Multiply both sides by $$B^2$$. \\begin{align*} B^2 x^2 - (2B^2 x_0)x + A^2 x^2 + (2AC)x + C^2 + 2By_0(Ax+C) &= B^2(r^2 - {x_0}^2 - {y_0}^2) \\\\ (A^2 + B^2)x^2 + (2AC - 2B^2 x_0)x + (2AB y_0)x &= -C^2 - 2BCy_0 + B^2(r^2 - {x_0}^2 - {y_0}^2) \\end{align*} Therefore, we have a", "Therefore, $p|a$ or $p|b$. Theorem 6. If $a$ and $b$ are integers and $(a,b)=d$, then $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem", "in two variables, and we’d like one equation in one variable. From the system, we should pick one variable (either $$x$$ or $$y$$) and choose to eliminate it (this is effectively the approach taught in most secondary school maths curricula). \\begin{align} & 12x + y = 12 \\\\ & x + 14 y = 24 \\end{align} For this system (the concrete example system), we can eliminate $$x$$ from the second system as follows. Take $$12x + y = 12$$ and divide it through by $$12$$, which gives that $$x + \\frac {y}{12} = 1$$ Then we can just subtract this equation from both sides of the other equation, which gives that $$x + 14 y - (x + \\frac {y}{12}) = 24 - 1$$ Therefore $$y \\left (14 - \\frac {1}{12}\\right ) = 23$$ So $$y = \\frac {276}{167}$$. The concrete example wasn’t too bad, so let’s try to think about a more general case, that of the system \\begin{align} & ax + by = m \\\\ & cx + dy = n \\end{align} If we proceed by using the previous method (of solving for $$y$$ by eliminating) $$x$$ we can unfortunately run into trouble! We have an $$ax$$ in the first equation, and ideally we’d like a $$-cx$$ (so that we can eliminate $$cx$$ from the other", "## Precalculus (6th Edition) Blitzer The solution of the given expression is $41$. We have to multiply both sides by $20$ to get the simplified form: \\begin{align} & 20\\cdot \\left( \\frac{x+3}{4}-\\frac{x+1}{10} \\right)=20\\cdot \\left( \\frac{x-2}{5}-1 \\right) \\\\ & 20\\cdot \\frac{x+3}{4}-20\\cdot \\frac{x+1}{10}=20\\cdot \\frac{x-2}{5}-20\\cdot 1 \\\\ & 5\\left( x+3 \\right)-2\\left( x+1 \\right)=4\\left( x-2 \\right)-20 \\\\ & 5x+15-2x-2=4x-8-20 \\end{align} And combine the terms on both sides to get the equation: $3x+13=4x-28$ And subtract $4x+13$ from both sides to get the value of x, \\begin{align} & 3x+13-\\left( 4x+13 \\right)=4x-28-\\left( 4x+13 \\right) \\\\ & 3x-4x=-28-13 \\\\ & -x=-41 \\end{align} And divide both sides by -1 to get: \\begin{align} & \\frac{-x}{-1}=\\frac{-41}{-1} \\\\ & x=41 \\end{align} Then, to check whether the solution is correct or not, put $x=41$ in the equation: \\begin{align} & \\frac{41+3}{4}-\\frac{41+1}{10}=\\frac{41-2}{5}-1 \\\\ & \\frac{44}{4}-\\frac{42}{10}=\\frac{39}{5}-1 \\end{align} And multiply both sides by 20 to get: \\begin{align} & 20\\left( \\frac{44}{4}-\\frac{42}{10} \\right)=20\\left( \\frac{39}{5}-1 \\right) \\\\ & 5\\cdot 44-2\\cdot 42=4\\cdot 39-20 \\\\ & 220-84=156-20 \\\\ & 136=136 \\end{align} Here, the left-hand side is equal to the right-hand side of the equation. When we have the left-hand side equal to the right-hand side of the equation, then the equation is satisfied. Hence, the solution of the given expression is $41$.", "$$-1 = B$$; \\begin{aligned} x^2 – 1 &= Ax(x + 1) – x – 1;\\\\ x^2 + x &= Ax(x+1); \\\\ x+1 &= A(x+1),\\end{aligned} so that $$A=1$$, and the partial fractions are: $\\frac{x-1}{x^2+1} – \\frac{2}{x+1}.$ Take as another example the fraction $\\frac{x^3-2}{(x^2+1)(x^2+2)}.$ We get \\begin{aligned} \\frac{x^3-2}{(x^2+1)(x^2+2)} &= \\frac{Ax+B}{x^2+1} + \\frac{Cx+D}{x^2+2}\\\\ &= \\frac{(Ax+B)(x^2+2)+(Cx+D)(x^2+1)}{(x^2+1)(x^2+2)}.\\end{aligned} In this case the determination of $$A$$$$B$$, $$C$$$$D$$ is not so easy. It will be simpler to proceed as follows: Since the given fraction and the fraction found by adding the partial fractions are equal, and have identical denominators, the numerators must also be identically the same. In such a case, and for such algebraical expressions as those with which we are dealing here, the coefficients of the same powers of $$x$$ are equal and of same sign. Hence, since \\begin{aligned} x^3-2 &= (Ax+B)(x^2+2) + (Cx+D)(x^2+1) \\\\ &= (A+C)x^3 + (B+D)x^2 + (2A+C)x + 2B+D,\\end{aligned} we have $$1=A+C$$;$$0=B+D$$ (the coefficient of $$x^2$$ in the left expression being zero); $$0=2A+C$$; and $$-2=2B+D$$. Here are four equations, from which we readily obtain $$A=-1$$; $$B=-2$$; $$C=2$$; $$D=0$$; so that the partial fractions are $$\\dfrac{2(x+1)}{x^2+2} – \\dfrac{x+2}{x^2+1}$$. This method can always be used; but the method shown first will be found the quickest in the case of factors in $$x$$ only. #### Case III. When, among the factors of the", "which gives us - $$A=B=1/2$$ Similarly, we can write - $$\\frac{y \\cdot dx }{(x+iy)(x-iy)} = \\frac{C\\cdot dx}{x+iy} + \\frac{D\\cdot dx}{x-iy}$$ from which we get - \\begin{aligned}y\\cdot dx &= C(x-iy)\\cdot dx + D(x+iy)\\cdot dx \\\\ &= (C + D) \\cdot x\\cdot dx + (iD - iC) \\cdot y \\cdot dx\\end{aligned} Upon comparing the coefficients of $x$ and $y$, we would obtain - $$C = i/2 \\text{ and } D= -i/2$$ Therefore, using these values, the original equation becomes - $$d\\theta = \\frac{1/2 \\cdot dy}{x+iy} + \\frac{1/2 \\cdot dy}{x-iy} - \\frac{i/2 \\cdot dx}{x+iy} + \\frac{i/2 \\cdot dx}{x-iy}$$ This form has the advantage that all the terms in the denominator are linear. We can now multiply both sides by $2i$, and collect terms with same denominators to get - \\begin{aligned}2i\\cdot d\\theta &= \\frac{dx + idy}{x+iy} - \\frac{dx - idy}{x-iy} \\\\ &= \\frac{d(x + iy)}{x+iy} - \\frac{d(x - iy)}{x-iy} \\end{aligned} We now integrate this equation, and obtain the following result - \\begin{aligned} &~~~~~~ \\int 2i\\cdot d\\theta = \\int \\frac{d(x + iy)}{x+iy} - \\int \\frac{d(x - iy)}{x-iy} \\\\ &\\Rightarrow 2i\\cdot \\theta = \\ln{(x+iy)} - \\ln{(x-iy)} + const\\end{aligned} By looking at the following diagram, we can see that when $y \\to 0$, we have $\\theta \\to 0$. This means that we can use this condition to obtain the integration constant as, $$0 = \\ln{(x)}", "desired answer as \\begin{aligned} \\frac{|BXPZ|}{|ABC|} &= \\frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\\\&= \\left(\\frac{ax +cz}{ax+by+cz}\\right)^2-\\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\\\&=\\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\\\ &= \\frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\\\&= \\frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\\quad\\quad(4)\\end{aligned} Similar formulas can be found for quadrilaterals $XPYC$ and $YPZA$ by permuting variables. Note that if $P$ is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables $x, y, z$. Note that (4) may also be written as $\\displaystyle \\frac{|BXPZ|}{|ABC|} = \\frac{ac(2xz + (x^2 + z^2)\\cos B)}{(ax+by+cz)^2}.\\quad\\quad(5)$ ### Special cases 1) If $\\triangle ABC$ is equilateral, $a=b=c$ and from (4) we obtain \\begin{aligned} \\frac{|BXPZ|}{|ABC|} &= \\frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\\\ &= \\frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\\\ &= \\frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\\quad\\quad(6)\\end{aligned} 2) If $P$ is at the incentre of $\\triangle ABC$, then $x = y = z = r$ (the inradius) and from (4) we have \\begin{aligned} \\frac{|BXPZ|}{|ABC|} &= \\frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\\\ &= \\frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\\\ &= \\frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\\quad\\quad(7)\\end{aligned} 3) If $\\triangle P$ is right-angled at $B$, then quadrilateral $BXPZ$ is a rectangle with area $xz$ and $\\triangle ABC$ has area $ac/2$", "lost here a factor 2. As an alternative by similarity of right triangles we obtain $$\\frac{d}{\\frac1b}=\\frac{\\frac1a}{\\sqrt{\\frac1{a^2}+\\frac1{b^2}}}\\implies d=\\frac{\\frac1{ab}}{\\sqrt{\\frac{a^2+b^2}{a^2b^2}}}\\implies d=\\frac1{\\sqrt{a^2+b^2}}$$ Minimum Distance Since \\begin{align} 1 &=\\left|\\,ax-by\\,\\right|\\\\ &=\\left|\\,(a,-b)\\cdot(x,y)\\,\\right|\\\\ &\\le\\left|\\,(a,-b)\\,\\right|\\left|\\,(x,y)\\,\\right| \\end{align} we have \\begin{align} \\left|\\,(x,y)\\,\\right| &\\ge\\frac1{\\left|\\,(a,-b)\\,\\right|}\\\\ &=\\frac1{\\sqrt{a^2+b^2}} \\end{align} If $(x,y)=\\frac{(a,-b)}{a^2+b^2}$, then $ax-by=1$ and $\\left|\\,(x,y)\\,\\right|=\\frac1{\\sqrt{a^2+b^2}}$ Thus, the minimum of $\\left|\\,(x,y)\\,\\right|$ is $\\frac1{\\sqrt{a^2+b^2}}$. Perpendicular Distance If $ax_1-by_1=1$ and $ax_2-by_2=1$, then $$(a,-b)\\cdot(x_1-x_2,y_1-y_2)=1-1=0$$ Thus, $(a,-b)$ is perpendicular to the line containing $(x_1,y_1)$ and $(x_2,y_2)$; i.e. the line $ax-by=1$. This means the vector from the origin to $\\frac{(a,-b)}{a^2+b^2}$ is perpendicular to the line $ax-by=1$ and the point $\\frac{(a,-b)}{a^2+b^2}$ is on the line $ax-by=1$. Thus, the perpendicular distance from the origin to the line is $$\\left|\\frac{(a,-b)}{a^2+b^2}\\right|=\\frac1{\\sqrt{a^2+b^2}}$$", "## Precalculus (6th Edition) Blitzer The simplified solution is $\\frac{1}{x}-\\frac{1}{x+1}$ and its sum is $\\frac{99}{100}$. Let us consider the rational expression. \\begin{align} & \\frac{1}{x\\left( x+1 \\right)}=\\frac{A}{x}+\\frac{B}{x+1} \\\\ & =\\frac{A\\left( x+1 \\right)+Bx}{x\\left( x+1 \\right)} \\end{align} $1=A\\left( x+1 \\right)+Bx$ (I) Substitute the value of $x=-1$ in the equation (I), \\begin{align} & 1=B\\left( -1 \\right) \\\\ & B=-1 \\end{align} Again, putting the value of $x=0$ in the equation (I), \\begin{align} & 1=A\\left( 1 \\right) \\\\ & A=1 \\end{align} Therefore, $\\frac{1}{x\\left( x+1 \\right)}=\\frac{1}{x}-\\frac{1}{x+1}$ is the simplified form of the result By using the above result to find the sum of the series, \\begin{align} & \\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}+\\cdots +\\frac{1}{99\\cdot 100} \\\\ & =\\left( \\frac{1}{1}-\\frac{1}{2} \\right)+\\left( \\frac{1}{2}-\\frac{1}{3} \\right)+\\left( \\frac{1}{3}-\\frac{1}{4} \\right)+\\cdots +\\left( \\frac{1}{99}-\\frac{1}{100} \\right) \\\\ & =\\frac{1}{1}-\\frac{1}{100} \\\\ & =\\frac{99}{100} \\end{align} Thus, the sum is $\\frac{99}{100}$.", "7. Mar 11, 2012 Mentallic I didn't have any doubts about what the OP is trying to do, just what the expression was meant to be once you raised the point. Mike, like I was saying it doesn't work in general that ax2 + b/x2 + c can be turned into a perfect square, but in this case c happened to be the right number for the job. When you get to the expression $$\\frac{25}{36}x^8+\\frac{1}{2}+\\frac{9}{100}x^{-8}$$ You should realize that it could be of the form $$\\left(ax^4+bx^{-4}\\right)^2$$ where in this case $$a=\\sqrt{\\frac{25}{36}}=\\frac{5}{6}$$ $$b=\\sqrt{\\frac{9}{100}}=\\frac{3}{10}$$ And all you'd need to do is check to see if $$2\\cdot \\frac{5}{6}\\cdot \\frac{3}{10} =\\frac{1}{2}$$", "$$i$$ measured in lumens. \\begin{aligned} \\log \\left(\\frac{i}{12}\\right) &=-0.0145 \\cdot d \\\\ \\log \\left(\\frac{i}{12}\\right) &=-0.0145 \\cdot 10 \\\\ \\log \\left(\\frac{i}{12}\\right) &=-0.145 \\\\\\left(\\frac{i}{12}\\right) &=10^{-0.145} \\\\ i &=12 \\cdot 10^{-0.145} \\approx 8.594 \\end{aligned} ### Example 4 Solve the following equation for all possible values of $$x$$. $$\\frac{e^{x}-e^{-x}}{3}=14$$ First solve for $$e^{x}$$, \\begin{aligned} \\frac{e^{x}-e^{-x}}{3} &=14 \\\\ e^{x}\\left(e^{x}-e^{-x}\\right) &=(42) e^{x} \\\\ e^{2 x}-1 &=42 e^{x} \\\\\\left(e^{x}\\right)^{2}-42 e^{x}-1 &=0 \\end{aligned} Let $$u=e^{x}$$ $$u^{2}-42 u-1=0$$ $$u=\\frac{-(-42) \\pm \\sqrt{(-42)^{2}-4 \\cdot 1 \\cdot(-1)}}{2 \\cdot 1}=\\frac{42 \\pm \\sqrt{1768}}{2} \\approx 42.023796,-0.0237960$$ Note that the negative result is extraneous because $$e^{x}$$ must be greater than zero, so you only proceed in solving for $$x$$ for one result. \\begin{aligned} e^{x} & \\approx 42.023796 \\\\ x & \\approx \\ln 42.023796 \\approx 3.738 \\end{aligned} ### Example 5 Solve the following equation for all possible values of $$x:\\left(\\log _{2} x\\right)^{2}-\\log _{2} x^{7}=-12$$. In calculus it is common to use a small substitution to simplify the problem and then substitute back later. In this case let $$u=\\log _{2} x$$. Notice that this is a quadratic problem. \\begin{aligned}\\left(\\log _{2} x\\right)^{2}-7 \\log _{2} x+12 &=0 \\\\ u^{2}-7 u+12 &=0 \\\\(u-3)(u-4) &=0 \\\\ u &=3,4 \\end{aligned} Now substitute back and solve for $$x$$ in each case. $$\\log _{2} x=3 \\leftrightarrow 2^{3}=x=8$$ $$\\log _{2} x=4 \\leftrightarrow 2^{4}=x=16$$ Review Solve each equation for $$x$$. Round each answer to three decimal" ]

[ "Therefore, $p|a$ or $p|b$. Theorem 6. If $a$ and $b$ are integers and $(a,b)=d$, then $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem", "y=154$$ $$-2 x+7 y=19$$ Add, solve for $$x=\\frac{173}{33}$$. Instead of substituting, practice eliminating $$x$$ by scaling the first equation by 2 and the second equation by 5. $$10 x-2 y=44$$ $$-10 x+35 y=95$$ Add, solve for $$y$$. Final Answer: $$\\left(\\frac{173}{33}, \\frac{139}{33}\\right)$$ ##### Example 4 Solve the following system of equations: \\begin{aligned} 5 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y} &=11 \\\\ \\frac{1}{x}+\\frac{1}{y} &=4 \\end{aligned} The strategy of elimination still applies. You can eliminate the $$\\frac{1}{y}$$ term if the second equation is scaled by a factor of -2 $$5 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y}=11$$ $$-2 \\cdot \\frac{1}{x}-2 \\cdot \\frac{1}{y}=-8$$ Add the equations together and solve for $$x$$. \\begin{aligned}-3 \\cdot \\frac{1}{x}+0 \\cdot \\frac{1}{y} &=3 \\\\ -3 \\cdot \\frac{1}{x} &=3 \\\\ \\frac{1}{x} &=-1 \\\\ x &=-1 \\end{aligned} Substitute into the second equation and solve for $$y$$. \\begin{aligned} \\frac{1}{-1}+\\frac{1}{y} &=4 \\\\ -1+\\frac{1}{y} &=4 \\\\ \\frac{1}{y} &=5 \\\\ y &=\\frac{1}{5} \\end{aligned} The point $$\\left(-1, \\frac{1}{5}\\right)$$ is the point of intersection between these two curves. ##### Example 5 Solve the following system using elimination: \\begin{aligned} 11 \\cdot \\frac{1}{x}-5 \\cdot \\frac{1}{y} &=-38 \\\\ 9 \\cdot \\frac{1}{x}+2 \\cdot \\frac{1}{y} &=-25 \\end{aligned} To eliminate $$\\frac{1}{y},$$ scale the first equation by 2 and the second equation by $$5 .$$ To eliminate $$\\frac{1}{x},$$ scale the first equation by -9 and the second equation by 11 Final Answer: $$\\left(-\\frac{1}{3}, 1\\right)$$ ##### Review Solve each", "$a$ and $b$ are integers and $(a,b)=d$, then $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem 7. Suppose that $a\\ne 0$, $b\\ne 0$,", "If you find any mistakes, please make a comment! Thank you. ## Solve general linear equations with $2\\times 2$ matrix Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.8 Solution: (a) In this case the system of equations is \\begin{alignat*}{1} 0\\cdot x_1 + 0\\cdot x_2 &= 0\\\\ 0\\cdot x_1 + 0\\cdot x_2 &= 0 \\end{alignat*}Clearly any $(x_1,x_2)$ satisfies this system since $0\\cdot x=0$ $\\forall$ $x\\in F$. (b) Let $(u,v)\\in F^2$. Consider the system: \\begin{alignat*}{1} a\\cdot x_1 + b\\cdot x_2 &= u\\\\ c\\cdot x_1 + d\\cdot x_2 &= v \\end{alignat*}If $ad-bc\\not=0$ then we can solve for $x_1$ and $x_2$ explicitly as $$x_1=\\frac{du-bv}{ad-bc}\\quad x_2=\\frac{av-cu}{ad-bc}.$$Thus there's a unique solution for all $(u,v)$ and in partucular when $(u,v)=(0,0)$. (c) Assume WOLOG that $a\\not=0$. Then $ad-bc=0$ $\\Rightarrow$ $d=\\frac{cb}{a}$. Thus if we multiply the first equation by $\\frac ca$ we get the second equation. Thus the two equations are redundant and we can just consider the first one $a x_1+bx_2=0$. Then any solution is of the form $(-\\frac bay,y)$ for arbitrary $y\\in F$. Thus letting $y=1$ we get the solution $(-b/a,1)$ and the arbitrary solution is of the form $y(-b/a,1)$ as desired.", "Proof. Since $(a,b)=d$, there exist $x,y\\in\\mathbb{Z}$ such that $ax+by=d$. Dividing the equation by $d$, we obtain $\\frac{a}{d}x+\\frac{b}{d}y=1$. By theorem 1, this implies that $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$. Example 1. Consider the equation $9x+24y=15$. Since $(9,24)=3$ and $3|15$, from theorem 1, we know that a solution exist. First, we can find a solution to $9x+24y=3$ using the Euclidean algorithm as seen before. \\begin{align*}24&=9\\cdot 2+6\\\\9&=6\\cdot 1+3\\\\6&=3\\cdot 2+0\\end{align*} Thus, \\begin{align*}3&=9-6\\cdot 1\\\\&=9-(24-9\\cdot 2)\\cdot 1\\\\&=9\\cdot 3+24\\cdot(-1)\\end{align*} Hence, $x’=3$ and $y’=-1$ is a solution to $9x+24y=3$ and thereby $x=5x’=15$ and $y=5y’=-5$ is a solution to $9x+24y=15$. Finding a solution is not a big deal. But there are other solutions. For instance, $x=-1$ and $y=1$ is also a solution to $9x+24y=15$. How do we find other solutions? We now turn our attention to this question. Suppose that $(x_0,y_0)$ is a solution to $$\\label{eq:lineqn}ax+by=c$$ Then $$\\label{eq:lineqn2}ax_0+by_0=c$$ Subtracting \\eqref{eq:lineqn2} from \\eqref{eq:lineqn}, we obtain $$\\label{eq:lineqn3}a(x-x_0)=b(y_0-y)$$ Let $d=(a,b)$. Dividing \\eqref{eq:lineqn3} by $d$, we obtain $$\\label{eq:lineqn4}\\frac{a}{d}(x-x_0)=\\frac{b}{d}(y_0-y)$$ This means that $\\frac{a}{d}|\\frac{b}{d}(y_0-y)$. Since $\\left(\\frac{a}{d},\\frac{b}{d}\\right)=1$, by theorem 2, $\\frac{a}{d}|y_0-y$ and so, $y_0-y=\\frac{a}{d}t$ for some $t\\in\\mathbb{Z}$. From \\eqref{eq:lineqn4} we also obtain $x-x_0=\\frac{b}{d}t$. Therefore, $x$ and $y$ are written as $$\\label{eq:lineqnsol}x=x_0+\\frac{b}{d}t,\\ y=y_0-\\frac{a}{d}t$$ where $t\\in\\mathbb{Z}$. Conversely, any $(x,y)$ in the form \\eqref{eq:lineqnsol} satisfies the equation \\eqref{eq:lineqn}. $$a\\left(x_0+\\frac{b}{d}t\\right)+b\\left(y_0-\\frac{a}{d}t\\right)=ax_0+by_0=c$$ Theorem 7. Suppose that $a\\ne 0$, $b\\ne 0$, and $c$ are integers. Let $(x_0,y_0)$ be a particular solution to $ax+by=c$. Then all", ":) – randomgirl Apr 24 '15 at 5:09 This can be done if and only if $5x^2+nx-13$ is a difference of two squares. $$5x^2+nx-13=\\frac15(25x^2+5nx+\\left(\\frac{n}{2}\\right)^2)-(13+\\frac{n^2}{4})$$ $$5x^2+nx-13=\\frac15(5x+\\frac{n}{2})^2-\\frac14(n^2+52).$$ Therefore, there should be an integer $m$ such that $$n^2+52=5m^2.$$ I think this will helps you. Continue from here. If a quadratic polynomial $ax^2 + bx + c$ with integer coefficients, $a \\neq 0$, has factorization $$ax^2 + bx + c = (rx + s)(tx + u) \\tag{1}$$ then \\begin{align*} a & = rt\\\\ b & = ru + st\\\\ c & = su \\end{align*} which can be demonstrated by substituting $0$, $1$, and $-1$ in equation 1, then solving the resulting system of equations. Note that $ac = (rt)(su) = rstu = (ru)(st)$, that is, $b$ is the sum of two integers whose product is $ac$. For the equation $5x^2 + nx - 13$, this means that $n$ is the sum of two integers with product $5 \\cdot -13 = -65$. Since \\begin{align*} -65 & = 1 \\cdot -65 & -65 & = -1 \\cdot 65\\\\ & = 5 \\cdot -13 & & = -5 \\cdot 13 \\end{align*} the possible integer values for $n$ are \\begin{align*} 1 + (-65) & = -64 & -1 + 65 & = 64\\\\ 5 + (-13) & = -8 & -5 + 13 & = 8", "of $ax+b+I$ Let $ax+b$ be a representative of a nonzero element of the field $\\F_3[x]/(x^2+1)$. Let $cx+d$ be its inverse. Then we have \\begin{align*} \\end{align*} since $x^2=-1$ in $\\F_3[x]/(x^2+1)$. Hence we obtain two equations \\begin{align*} \\end{align*} Since $ax+b$ is a nonzero element, at least one of $a, b$ is not zero. If $a\\neq 0$, then the first equation gives $d=-\\frac{bc}{a}. \\tag{*}$ Substituting this to the second equation, we obtain \\begin{align*} \\left(\\, \\frac{-b^2-a^2}{a} \\,\\right)c=1. \\end{align*} Observe that $a^2+b^2$ is not zero in $\\F_3$. (Since $a \\neq 0$, we have $a^2=1$. Also $b^2=0, 1$.) Hence we have \\begin{align*} c=-\\frac{a}{a^2+b^2}. \\end{align*} It follows from (*) that $d=\\frac{b}{a^2+b^2}$ Thus, if $a \\neq 0$, then the inverse element is $(ax+b)^{-1}=\\frac{1}{a^2+b^2}(-ax+b). \\tag{**}$ If $a=0$, then $b\\neq 0$ and it is clear that the inverse element of $ax+b=b$ is $1/b$. Note that the formula (**) is still true in this case. In summary, we have $(ax+b)^{-1}=\\frac{1}{a^2+b^2}(-ax+b)$ for any nonzero element $ax+b$ in the field $\\F_3[x]/(x^2+1)$. ### (c) $x$ is not a generator but $x+1$ is a generator Note that the order of $E^{\\times}$ is $8$ since $E$ is a finite field of order $9$ by part (a). We compute the powers of $x$ and obtain \\begin{align*} \\end{align*} Thus, the order of the element $x$ is $4$, hence $x$ is not a generator of the cyclic", "# 1988 AIME Problems/Problem 8 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: $$f(x, x) = x,\\; f(x, y) = f(y, x), {\\rm \\ and\\ } (x+y)f(x, y) = yf(x, x+y).$$ Calculate $f(14,52)$. ## Solution 1 (Algebra) Let $z = x+y$. By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \\begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\\\ f(x,z) &= \\frac{z}{z-x} \\cdot f(x,z-x). \\end{align*} Using the properties of $f,$ we have \\begin{align*} f(14,52) &= \\frac{52}{38} \\cdot f(14,38) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot f(14,24) \\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(14,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot f(10,14)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(10,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot f(4,10)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot f(4,6)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(4,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot f(2,4)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6} \\cdot \\frac{6}{2} \\cdot \\frac{4}{2} \\cdot f(2,2)\\\\ &= \\frac{52}{38} \\cdot \\frac{38}{24} \\cdot \\frac{24}{10} \\cdot \\frac{14}{4} \\cdot \\frac{10}{6}", "$$i$$ measured in lumens. \\begin{aligned} \\log \\left(\\frac{i}{12}\\right) &=-0.0145 \\cdot d \\\\ \\log \\left(\\frac{i}{12}\\right) &=-0.0145 \\cdot 10 \\\\ \\log \\left(\\frac{i}{12}\\right) &=-0.145 \\\\\\left(\\frac{i}{12}\\right) &=10^{-0.145} \\\\ i &=12 \\cdot 10^{-0.145} \\approx 8.594 \\end{aligned} ### Example 4 Solve the following equation for all possible values of $$x$$. $$\\frac{e^{x}-e^{-x}}{3}=14$$ First solve for $$e^{x}$$, \\begin{aligned} \\frac{e^{x}-e^{-x}}{3} &=14 \\\\ e^{x}\\left(e^{x}-e^{-x}\\right) &=(42) e^{x} \\\\ e^{2 x}-1 &=42 e^{x} \\\\\\left(e^{x}\\right)^{2}-42 e^{x}-1 &=0 \\end{aligned} Let $$u=e^{x}$$ $$u^{2}-42 u-1=0$$ $$u=\\frac{-(-42) \\pm \\sqrt{(-42)^{2}-4 \\cdot 1 \\cdot(-1)}}{2 \\cdot 1}=\\frac{42 \\pm \\sqrt{1768}}{2} \\approx 42.023796,-0.0237960$$ Note that the negative result is extraneous because $$e^{x}$$ must be greater than zero, so you only proceed in solving for $$x$$ for one result. \\begin{aligned} e^{x} & \\approx 42.023796 \\\\ x & \\approx \\ln 42.023796 \\approx 3.738 \\end{aligned} ### Example 5 Solve the following equation for all possible values of $$x:\\left(\\log _{2} x\\right)^{2}-\\log _{2} x^{7}=-12$$. In calculus it is common to use a small substitution to simplify the problem and then substitute back later. In this case let $$u=\\log _{2} x$$. Notice that this is a quadratic problem. \\begin{aligned}\\left(\\log _{2} x\\right)^{2}-7 \\log _{2} x+12 &=0 \\\\ u^{2}-7 u+12 &=0 \\\\(u-3)(u-4) &=0 \\\\ u &=3,4 \\end{aligned} Now substitute back and solve for $$x$$ in each case. $$\\log _{2} x=3 \\leftrightarrow 2^{3}=x=8$$ $$\\log _{2} x=4 \\leftrightarrow 2^{4}=x=16$$ Review Solve each equation for $$x$$. Round each answer to three decimal", "solution $x$ and $y$ Thus since $d | a$ and $d | b$, then $d | (ax + by) = c$. Now we have to prove that if $d | c$, then the equation has integral solution. Assume that $d | c$. By theorem 3 (in GCD), there exist integers $m$ and $n$ such that $d = am + bn$. And also there exists integer $k$ such that $c = dk$. Now since $c = ax + by$, we have \\begin{align} c &= dk \\\\ &= (ma + nb)k \\\\ & = a(km) + b(nk) \\end{align}. Hence a solution for the equation $ax + by = c$ is $x_0 = km$ and $y_0 = kn$. What is left to prove is that we have infinitely many solutions. Let $x = x_0 + ( \\frac{b}{d})t$ and $y = y_0 - ( \\frac{a}{d}) t$. We have to prove now that $x$ and $y$ are solutions for all integers $t$. Notice that \\begin{align} ax + by &= a(x_0 + ( \\frac{b}{d})t) + b(y_0 - ( \\frac{a}{d})t) \\\\ &= ax_0 + by_0 \\\\ & = c \\end{align}. We now show that every solution for the equation $ax + by = c$ is of the form $x = x_0 + ( \\frac{b}{d})t$ and $y = y_0 - ( \\frac{a}{d})t$. Notice that since $ax_0", "side of the equation remains unchanged. Using Lemma 2, the following describes infinitely many solutions of the equation $1638x+357y=819$. $x=-195+357k$ and $y=897-1638k$ for all integers $k$ But the above solution set is not complete. For example, the number pair $x=-178$ and $y=819$ is also a solution to the equation $1638x+357y=819$, but is not listed above. The following theorem will give the complete solution set. Theorem 3 Let $d=\\text{GCD}(a,b)$. If the pair $x=x_0$ and $y=y_0$ is a solution to $ax+by=c$, then the complete solution set of the equation consists of all the integer pairs $(x,y)$ such that $\\displaystyle x=x_0+\\frac{b}{d} \\cdot k$ $\\displaystyle y=y_0-\\frac{a}{d} \\cdot k$ where $k$ is any integer. Proof of Theorem 3 Suppose that the pair $x=x_0$ and $y=y_0$ is a solution to $ax+by=c$. Two points to show. One is that any number pair of the above form is a solution. The other is that any solution is of the above form. To see the first, note that $\\displaystyle a(x_0+\\frac{b}{d} \\cdot k)+b(y_0-\\frac{a}{d} \\cdot k)=ax_0+by_0=c$ To see the second point, suppose that the number pair $x=x_1$ and $y=y_1$ is a solution to the equation $ax+by=c$. We have the following. $ax_0+by_0=c$ $ax_1+by_1=c$ Rearranging and dividing by $d$, we have the following. $a(x_1-x_0)=b(y_0-y_1)$ $\\displaystyle \\frac{a}{d}(x_1-x_0)=\\frac{b}{d}(y_0-y_1) \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "the quadratic for $x$, and substitute the solution(s), if any, back into the equation for the line. Here's an equation which works even if the line is vertical. This is useful from a computer programming perspective. Let's set up the system of equations using standard formulae: $$\\begin{cases} (x - x_0)^2 + (y - y_0)^2 = r^2 \\\\ Ax + By + C = 0 \\end{cases}$$ where the circle with radius $$r$$ is centered at $$(x_0, y_0)$$; the line contains the points $$(x_1, y_1)$$ and $$(x_2, y_2)$$. From the line equation, we know: $$y = \\frac{-Ax-C}{B} = - \\frac{Ax+C}{B}$$ ($$B \\neq 0$$, we'll get into this in a second) Therefore, \\begin{align*} (x - x_0)^2 + \\left( - \\frac{Ax+C}{B} - y_0 \\right)^2 &= r^2 \\\\ x^2 - 2x_0x + {x_0}^2 + \\frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \\cdot \\frac{Ax+C}{B} + {y_0}^2 &= r^2 \\\\ x^2 - 2x_0x + \\frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \\cdot \\frac{Ax+C}{B} &= r^2 - {x_0}^2 - {y_0}^2 \\end{align*} Multiply both sides by $$B^2$$. \\begin{align*} B^2 x^2 - (2B^2 x_0)x + A^2 x^2 + (2AC)x + C^2 + 2By_0(Ax+C) &= B^2(r^2 - {x_0}^2 - {y_0}^2) \\\\ (A^2 + B^2)x^2 + (2AC - 2B^2 x_0)x + (2AB y_0)x &= -C^2 - 2BCy_0 + B^2(r^2 - {x_0}^2 - {y_0}^2) \\end{align*} Therefore, we have a", "us the following equations: \\begin{align} a&=-a-b-c\\\\ b&=ab+bc+ca\\\\ c&=-abc \\end{align} So let's see what happens when $c=0$; then, $a=-a-b$ and $b=ab$. Hey! This is the solution to the second degree equation! Let's keep that in mind, maybe we can prove something about that for the general case. Now we proceed to assume $c\\neq 0$. Then, $ab=-1$. Now let's see if we can express $a$ in terms of $b$ and $c$ to hopefully eliminate that variable: \\begin{align} a&=-\\tfrac12(b+c)\\\\ a&=\\frac{b-bc+1}{c}\\\\ a&=-\\frac1b \\end{align} So now we express $c$ in terms of $b$ by pairing the last equation to the first two: \\begin{align} c&=\\frac{2-b^2}{b}\\\\ c&=\\frac{-b-b^2}{1-b^2} \\end{align} And so we get $$b^4+b^3-2b^2+2=0$$ Ouch! That's ugly. But, it has a solution $b=-1$ (corresponding to $(a,b,c)=(1,-1,-1)$). The other three solutions are pretty ugly though; one real one and two complex ones (I won't work them out because they're too chaotic and seeing them won't accomplish anything. Feel free to use wolframalpha to get them though). As for our little trick with $c=0$ obtaining the solutions from the quadratic equation, that's actually quite simple to see. If $x^{n+1}+a_nx^n+\\cdots+a_0$ has roots $a_n,\\cdots,a_0$, then surely multiplying by $x$ doesn't change that: $x^{n+2}+a_nx^{n+1}+\\cdots+a_0x+0$ now also has those roots, and since we multiplied by $x$, it has an extra root $0$. As a result, we may assume the constant term is", "## Precalculus (6th Edition) Blitzer The solution of the given expression is $41$. We have to multiply both sides by $20$ to get the simplified form: \\begin{align} & 20\\cdot \\left( \\frac{x+3}{4}-\\frac{x+1}{10} \\right)=20\\cdot \\left( \\frac{x-2}{5}-1 \\right) \\\\ & 20\\cdot \\frac{x+3}{4}-20\\cdot \\frac{x+1}{10}=20\\cdot \\frac{x-2}{5}-20\\cdot 1 \\\\ & 5\\left( x+3 \\right)-2\\left( x+1 \\right)=4\\left( x-2 \\right)-20 \\\\ & 5x+15-2x-2=4x-8-20 \\end{align} And combine the terms on both sides to get the equation: $3x+13=4x-28$ And subtract $4x+13$ from both sides to get the value of x, \\begin{align} & 3x+13-\\left( 4x+13 \\right)=4x-28-\\left( 4x+13 \\right) \\\\ & 3x-4x=-28-13 \\\\ & -x=-41 \\end{align} And divide both sides by -1 to get: \\begin{align} & \\frac{-x}{-1}=\\frac{-41}{-1} \\\\ & x=41 \\end{align} Then, to check whether the solution is correct or not, put $x=41$ in the equation: \\begin{align} & \\frac{41+3}{4}-\\frac{41+1}{10}=\\frac{41-2}{5}-1 \\\\ & \\frac{44}{4}-\\frac{42}{10}=\\frac{39}{5}-1 \\end{align} And multiply both sides by 20 to get: \\begin{align} & 20\\left( \\frac{44}{4}-\\frac{42}{10} \\right)=20\\left( \\frac{39}{5}-1 \\right) \\\\ & 5\\cdot 44-2\\cdot 42=4\\cdot 39-20 \\\\ & 220-84=156-20 \\\\ & 136=136 \\end{align} Here, the left-hand side is equal to the right-hand side of the equation. When we have the left-hand side equal to the right-hand side of the equation, then the equation is satisfied. Hence, the solution of the given expression is $41$.", "and $CPD$, \\begin{align}\\tag{2}x^2+z^2=180\\end{align}. It follows that \\begin{align}\\tag{3}z^2-y^2=32\\end{align}. Since $\\sin\\alpha=\\sqrt{1-\\cos^2\\alpha}$, $$\\sqrt{1-\\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\\sqrt{1-\\frac{(x^2+z^2-260)^2}{4x^2z^2}}$$ which simplifies to $$\\frac{48^2}{y^2}=\\frac{80^2}{z^2} \\qquad \\Longrightarrow \\qquad 5y=3z.$$ Plugging this back to equations $(1)$, $(2)$, and $(3)$, it can be solved that $x=\\sqrt{130},y=3\\sqrt{2},z=5\\sqrt{2}$. Then, the area of the quadrilateral is $$x(y+z)\\sin\\alpha=\\sqrt{130}\\cdot8\\sqrt{2}\\cdot\\frac{14}{\\sqrt{260}}=\\boxed{112}$$ --Solution by MicGu ## Solution 5 As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$, but it's an AIME problem, we can take $AP=CP$, and assume the other choice will lead to the same result (which is true). From $AP=CP$, we have $[DAP]=[DCP]$, and $[BAP]=[BCP] \\implies [ABD] = [CBD]$, therefore, \\begin{align} \\nonumber \\frac 12 \\cdot AB\\cdot AD\\sin A &= \\frac 12 \\cdot BC\\cdot CD\\sin C \\\\ \\Longrightarrow \\hspace{1in} 7\\sin C &= \\sqrt{65}\\sin A \\end{align} By Law of Cosines, \\begin{align} \\nonumber 10^2+14^2-2\\cdot 10\\cdot 14\\cos C &= 10^2+4\\cdot 65-2\\cdot 10\\cdot 2\\sqrt{65}\\cos A \\\\ \\Longrightarrow \\hspace{1in} -\\frac 85 - 7\\cos C &= \\sqrt{65}\\cos A \\tag{2} \\end{align} Square $(1)$ and $(2)$, and add them, to get $$\\left(\\frac 85\\right)^2 + 2\\cdot \\frac 85 \\cdot 7\\cos C + 7^2 = 65$$ Solve, $\\cos C = 3/5 \\implies \\sin C = 4/5$, $$[ABCD] = 2[BCD] = BC\\cdot CD\\cdot \\sin C = 14\\cdot 10\\cdot \\frac 45 = \\boxed{112}$$ -Mathdummy ## Solution 6 Either $PA=PC$ or $PD=PB$. Let $PD=PB=s$. Applying Stewart's Theorem on $\\triangle ABD$ and $\\triangle BCD$, dividing by $2s$", "with x in the second and the third equations as we intended. Next step is to eliminate $y$ from from the third equation. First of all, we divide the second equation by $2$: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\12y-9z=-\\frac{5}{2}\\end{aligned}\\right. Next we subtract the second equation multiplied by $12$ from the third: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\21z=-\\frac{5}{2}-15=-\\frac{35}{2}\\end{aligned}\\right. As we can see, we already have our system in required form which is called triangular. What is left for us to do is to perform so called reverse course of Gauss method: first, we get $z$ from the last equation: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}z=\\frac{5}{4}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Then we substitute obtained value for $z$ into the second equation to obtain $y$: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y-\\frac{5}{2}\\cdot(-\\frac{5}{6})=\\frac{5}{4}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Rearranging terms and performing required calculations we get: \\left\\{ \\begin{aligned}x-2y+3z =\\frac{3}{2}\\\\y=\\frac{5}{2}\\cdot(-\\frac{5}{6})+\\frac{5}{4}=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. And the last step: we substitute values for $y,z$ into the first equation to obtain $x$: \\left\\{ \\begin{aligned}x-2(-\\frac{5}{6})+3(-\\frac{5}{6}) =\\frac{3}{2}\\\\y=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. Performing necessary calculations, we finally obtain: \\left\\{ \\begin{aligned}x=\\frac{7}{3}\\\\y=-\\frac{5}{6}\\\\z=-\\frac{5}{6}\\end{aligned}\\right. So the answer is the following: \\left\\{ \\begin{aligned}x=\\frac{7}{3}\\\\y=z=-\\frac{5}{6}\\end{aligned}\\right. That’s it. More examples in the next section. 0 Shares Filed under Math.", "to solve $\\ln x = ax^4$ is to solve $h(x) = ax^4 - \\ln x = 0$ and as $a^4 \\ge \\ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum. So $h(x) = ax^4 - \\ln x = 0$ and $h'(x) = 4ax^3 - \\frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - \\ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = \\frac 14$. So $\\frac 14 - \\ln x = 0$ and $\\ln x = \\frac 14$ and $x = e^{\\frac 14}$. Plugging that in we get $a*e - \\frac 14 = 0$. Which gives us $a = \\frac 1{4e}$ If $a =0$ then $\\ln x = ax^4 = 0$ and $x = 0$ is the onlys solution. If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a \\in (-\\infty, 0]\\cup \\{\\frac 1{4e}\\}$. Hint: Consider the function $f(x)=\\frac{\\ln(x)}{x^4}$ and use calculus. Use that $$f'(x)=\\frac{1-4\\ln(x)}{x^4}$$ • don't you mean $f'(x)=\\frac{1-4ln(x)}{x^5}$ – Mandelbrot Jul 14 '18 at 17:17 \\begin{align} \\ln x&=ax^4 \\tag{1}\\label{1} ,\\\\ -4\\ln x&=-4ax^4 ,\\\\ \\ln(x^{-4})&=-4ax^4 ,\\\\ x^{-4}\\ln(x^{-4})&=-4a . \\end{align} Let $\\ln(x^{-4})=u$, then $x^{-4}=\\exp(u)$ and we arrive at", "both sides of the equations. So the first step to do is eliminate the number $b$ from the left side of the equation. To do this, just add the opposite of $b$ which is $(- b)$ on both sides of the equation. We get \\begin{align*} ax+b+(-b)=0+(-b).\\end{align*}As $b+(-b)=0$ and $0+(-b)=-b,$ the one-step equation becomes \\begin{align*}{\\large ax=-b}.\\end{align*} As the nomber $a$ is non null, we then divide the both sides of this equation by $a$, we obtain \\begin{align*} \\frac{ax}{a}=\\frac{-b}{a}.\\end{align*} But $\\frac{ax}{a}=\\frac{a}{a} x=x$, then the solution of the equation is \\begin{align*}{\\Large x=\\frac{-b}{a}}.\\end{align*} Examples: 1- The solution of the one-step equation $3x+2=0$ is $x=\\frac{-3}{2}$. 2- The solution of the equation $\\frac{2}{3}x-4=0$ is \\begin{align*} x=\\frac{4}{\\frac{2}{3}}= 4 \\frac{3}{2}=2 \\times 3=6.\\end{align*} 3- Let’s solve the equation $5x+12=2x+9$. In this case, we are using the rule $A=B$ is the same as $A-B=0$. So our equation can be rewritten as \\begin{align*} 5x+12-(2x+9)=0.\\end{align*} On the other hand, as $-(2x+9)=-2x-9$, then $5x+12-2x-9=0$. This means that $5x-2x+12-9=0$. So $3x+3=0$. Therefore, we have turned the problem into a standard one-set equation that we can easily solve, so the solution is $x=\\frac{-3}{3}=-1$. ### The exponents of numbers (powers, subscript) are the most important part of all algebra lessons for beginners Positive integer exponents: such numbers are used to simplify the expression of a number multiplied by itself many times. For example, if", "lost here a factor 2. As an alternative by similarity of right triangles we obtain $$\\frac{d}{\\frac1b}=\\frac{\\frac1a}{\\sqrt{\\frac1{a^2}+\\frac1{b^2}}}\\implies d=\\frac{\\frac1{ab}}{\\sqrt{\\frac{a^2+b^2}{a^2b^2}}}\\implies d=\\frac1{\\sqrt{a^2+b^2}}$$ Minimum Distance Since \\begin{align} 1 &=\\left|\\,ax-by\\,\\right|\\\\ &=\\left|\\,(a,-b)\\cdot(x,y)\\,\\right|\\\\ &\\le\\left|\\,(a,-b)\\,\\right|\\left|\\,(x,y)\\,\\right| \\end{align} we have \\begin{align} \\left|\\,(x,y)\\,\\right| &\\ge\\frac1{\\left|\\,(a,-b)\\,\\right|}\\\\ &=\\frac1{\\sqrt{a^2+b^2}} \\end{align} If $(x,y)=\\frac{(a,-b)}{a^2+b^2}$, then $ax-by=1$ and $\\left|\\,(x,y)\\,\\right|=\\frac1{\\sqrt{a^2+b^2}}$ Thus, the minimum of $\\left|\\,(x,y)\\,\\right|$ is $\\frac1{\\sqrt{a^2+b^2}}$. Perpendicular Distance If $ax_1-by_1=1$ and $ax_2-by_2=1$, then $$(a,-b)\\cdot(x_1-x_2,y_1-y_2)=1-1=0$$ Thus, $(a,-b)$ is perpendicular to the line containing $(x_1,y_1)$ and $(x_2,y_2)$; i.e. the line $ax-by=1$. This means the vector from the origin to $\\frac{(a,-b)}{a^2+b^2}$ is perpendicular to the line $ax-by=1$ and the point $\\frac{(a,-b)}{a^2+b^2}$ is on the line $ax-by=1$. Thus, the perpendicular distance from the origin to the line is $$\\left|\\frac{(a,-b)}{a^2+b^2}\\right|=\\frac1{\\sqrt{a^2+b^2}}$$", "$$-1 = B$$; \\begin{aligned} x^2 – 1 &= Ax(x + 1) – x – 1;\\\\ x^2 + x &= Ax(x+1); \\\\ x+1 &= A(x+1),\\end{aligned} so that $$A=1$$, and the partial fractions are: $\\frac{x-1}{x^2+1} – \\frac{2}{x+1}.$ Take as another example the fraction $\\frac{x^3-2}{(x^2+1)(x^2+2)}.$ We get \\begin{aligned} \\frac{x^3-2}{(x^2+1)(x^2+2)} &= \\frac{Ax+B}{x^2+1} + \\frac{Cx+D}{x^2+2}\\\\ &= \\frac{(Ax+B)(x^2+2)+(Cx+D)(x^2+1)}{(x^2+1)(x^2+2)}.\\end{aligned} In this case the determination of $$A$$$$B$$, $$C$$$$D$$ is not so easy. It will be simpler to proceed as follows: Since the given fraction and the fraction found by adding the partial fractions are equal, and have identical denominators, the numerators must also be identically the same. In such a case, and for such algebraical expressions as those with which we are dealing here, the coefficients of the same powers of $$x$$ are equal and of same sign. Hence, since \\begin{aligned} x^3-2 &= (Ax+B)(x^2+2) + (Cx+D)(x^2+1) \\\\ &= (A+C)x^3 + (B+D)x^2 + (2A+C)x + 2B+D,\\end{aligned} we have $$1=A+C$$;$$0=B+D$$ (the coefficient of $$x^2$$ in the left expression being zero); $$0=2A+C$$; and $$-2=2B+D$$. Here are four equations, from which we readily obtain $$A=-1$$; $$B=-2$$; $$C=2$$; $$D=0$$; so that the partial fractions are $$\\dfrac{2(x+1)}{x^2+2} – \\dfrac{x+2}{x^2+1}$$. This method can always be used; but the method shown first will be found the quickest in the case of factors in $$x$$ only. #### Case III. When, among the factors of the" ]

[ "f(x)=x^4+15x^2-16", "let $f(x)=-2x$.", "# How do you find f(-8) given f(x)=x^2-3x? $f \\left(- 8\\right) = 88$ $f \\left(\\textcolor{red}{- 8}\\right) = {\\left(\\textcolor{red}{- 8}\\right)}^{2} - 3 \\left(\\textcolor{red}{- 8}\\right)$ $\\Rightarrow f \\left(- 8\\right) = 64 + 24 = 88$", "Let $f(x) = \\left|x^2 - 16 x + 63\\right|.$ Find the set of values of $x$ where $f$ is differentiable.", "the functions. 64. $f(x)= 1 x 2 f(x)= 1 x 2$ 65. $f(x)=5 x 2 −x+4 f(x)=5 x 2 −x+4$ 66. $f(x)=− x 2 +4x+7 f(x)=− x 2 +4x+7$ 67. $f(x)= −4 3− x 2 f(x)= −4 3− x 2$", "$3, 7, 11...$? 1. $f(n)=4n-1$ 2. $f(n)=3+2n$ 3. $f(x)=2n-3$ 4. $f(n)=4x-1$", "a different form. $f(x) = 2x^2 - 4x + 7$, $f(-3)$ Now, let’s substitute. $f(x) = 2x^2 - 4x + 7$ $f(-3) = 2 (-3)^2 - 4 (-3) + 7$ $f(-3) = 2 (9) - 4 (-3) + 7$ $f(-3) = 18 + 12 + 7$ $f(-3) = 37$", "35. $f(x)=x2−2x−2$ 36. $f(x)=x2−2x+37$", "169 views Given $f(300)=2,4771; f(304) = 2.4829; f(305) = 2.4843$ and $f(307) = 2.4871$ find $f(301)$ using Lagrange's interpolation formula. closed with the note: Numerical analysis removed from current gate CS syllabus closed | 169 views", "Evaluate the function $$f(x)=2x^2-7x+3$$ at $$x=6$$. $$f(6) = 2 \\times\\ (6)^2 - 7 \\times\\ (6) + 3$$ $$f(6) = 64 - 42 + 3$$ $$f(6) = 25$$", "formulas. I will assume the grid is evenly spaced, since that seems to give the results he obtained. WLOG, assume the $V_{N - 1}$ point is at $... 0 Apparently you want to use$x=5$. For$f(x)=x^3$, we have$f(5)=125$,$f‘(5)=75$,$f‘‘(5)=30$,$f‘‘‘(5)=6$, which does lead to$216\\$. Top 50 recent answers are included", "= 6, f(3) = 52$, and $f(4) = 310$; the sequence isn’t in OEIS, by the way. -", "If $f(x) = x^3 + 6 x^2 - 288 x + 5$, find analytically all values of $x$ for which $f'(x) = 0$. (Enter your answer as a comma separated list of numbers, e.g., -1,0,2) $x =$", "$f(x)=ax+b+(x^2+7)$. So, $f(2x^2+7)=2(ax+b)^2+7+(x^2+7)$. Now that implies $x^2+7|−14a^2+2b^2+7+4abx \\implies 7+2b^2=14a^2$. Absurd. -", "(0) and f’ (0) are given, and they are as follows: f ( 0 ) = 4 , f’ ( 0 ) = 2, f’’ ( 0 ) = 6 $f ( x ) = 4 + 2 x + \\frac { 6 }{ 2 } x ^ 2$ $f ( x ) = 4 + 2 x + 3 x ^ 2$", "27$$ $$f = X^3 + 18X^2 + 81X + 81$$ $$f = X^3 + 21X^2 + 66X + 19$$ $$f = X^3 + 27X^2 + 54X + 27$$ Additional remark: Let $$f = X^3 + 18X^2 + 81X + 27$$, then (since $$f$$ appears in the above list) we have $$O_K \\supsetneq \\mathbb{Z}[\\alpha]$$, where $$\\alpha$$ is (as usually) a zero of $$f$$. However, according to SageMath, it turns out that $$O_K = \\mathbb{Z}[\\alpha/3]$$.", "be quite handy. Let $$f(x)$$ be $x^2-3x-4=0$ We want to factorise the equation such that the equation takes the form: $(x-a)(x-b)=0\\text{, where a, b, are roots of the quadratic}$", "x ) = 2 3 x − 1$ 523. $f ( x ) = −6 f ( x ) = −6$ 524. $f ( x ) = 2 x f ( x ) = 2 x$ 525. $f ( x ) = 3 x 2 f ( x ) = 3 x 2$ 526. $f ( x ) = − 1 2 x 2 f ( x ) = − 1 2 x 2$ 527. $f ( x ) = x 2 + 2 f ( x ) = x 2 + 2$ 528. $f ( x ) = x 3 − 2 f ( x ) = x 3 − 2$ 529. $f ( x ) = x + 2 f ( x ) = x + 2$ 530. $f ( x ) = − | x | f ( x ) = − | x |$ 531. $f ( x ) = | x | + 1 f ( x ) = | x | + 1$ Read Information from a Graph of a Function In the following exercises, use the graph of the function to find its domain and range. Write the domain and range in interval notation 532. 533. 534. In the following exercises, use the graph of the function to find the indicated", "x - 17}{6 x + 1} ^ 2 = f ' \\left(x\\right)$", "2 x 2 + 8 x x 2 − 9 x + 20 f ( x ) = 2 x 2 + 8 x x 2 − 9 x + 20$ $g ( x ) = x − 5 x 2 g ( x ) = x − 5 x 2$ For the following exercises, find $R(x)=f(x)g(x)R(x)=f(x)g(x)$ where $f(x)f(x)$ and $g(x)g(x)$ are given. 67. $f ( x ) = 27 x 2 3 x − 21 f ( x ) = 27 x 2 3 x − 21$ $g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42 g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42$ 68. $f ( x ) = 24 x 2 2 x − 8 f ( x ) = 24 x 2 2 x − 8$ $g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28 g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28$ 69. $f ( x ) = 16 x 2 4 x + 36 f ( x ) = 16 x 2 4 x + 36$ $g ( x ) = 4" ]

[ "- 3x + 11$ with a remainder of $-28$ (a degree zero polynomial). $\\begin{array}{ccrcrcrcr}& & & & x^2 & - & 3x & + & 11 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x+3) & ) & x^3 & + & 0x^2 & + & 2x & + & 5 \\\\& & x^3 & + & 3x^2 & & & & \\\\& & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\& & & - & 3x^2 & + & 2x & & \\\\& & & - & 3x^2 & - & 9x & & \\\\& & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\& & & & & & 11x & + & 5 \\\\& & & & & & 11x & + & 33 \\\\& & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\& & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed to take modulo $x^3 + x + 1$. Well, dividing $x^3 + x^2$ by $x^3 + x + 1$, we see that it goes in one time with a remainder of $x^2 + x + 1$. [Note: addition and subtraction are the same in $GF(2^n)$.] $\\begin{array}{ccrcrcrcr}& & & & & & & & 1 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x^3+x+1) & ) & x^3 & + & x^2 & + & 0x & + &", "+0 # help quickly 0 48 1 Find $a$ if the remainder is constant when $10x^3-7x^2+ax+6$ is divided by $2x^2-3x+1$. Feb 21, 2020 #1 +1 Nevermind I figured it out myself We perform the polynomial division: $\\begin{array}{c|cc cc} \\multicolumn{2}{r}{5x} & +4 \\\\ \\cline{2-5} 2x^2-3x+1 & 10x^3&-7x^2&+ax&+6 \\\\ \\multicolumn{2}{r}{-10x^3} & +15x^2 & -5x \\\\ \\cline{2-4} \\multicolumn{2}{r}{0} & 8x^2 & (a-5)x & 6 \\\\ \\multicolumn{2}{r}{} & -8x^2 & +12x & -4 \\\\ \\cline{3-5} \\multicolumn{2}{r}{} & 0 & (a-5+12)x & 2 \\\\ \\end{array}$ The remainder will be constant if and only if So the answer is 7 Feb 21, 2020", "3x^2 & & & & \\\\& & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\& & & - & 3x^2 & + & 2x & & \\\\& & & - & 3x^2 & - & 9x & & \\\\& & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\& & & & & & 11x & + & 5 \\\\& & & & & & 11x & + & 33 \\\\& & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\& & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed to take modulo $x^3 + x + 1$. Well, dividing $x^3 + x^2$ by $x^3 + x + 1$, we see that it goes in one time with a remainder of $x^2 + x + 1$. [Note: addition and subtraction are the same in $GF(2^n)$.] $\\begin{array}{ccrcrcrcr}& & & & & & & & 1 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x^3+x+1) & ) & x^3 & + & x^2 & + & 0x & + & 0 \\\\& & x^3 & + & 0x^2 & + & x & + & 1 \\\\& & \\multicolumn{7}{l}{\\hrulefill} \\\\& & & & x^2 & + & x & + & 1\\\\\\end{array}$ For a reasonable way to accomplish this in the special case of our integer representations of polynomials in $GF(2^n)$,", "0... P: 492 Not relevant to the problem above, but... here is the long division for problem 4.4.2(a)! $$\\begin{equation*} \\begin{array}{rc@{}c} & \\multicolumn{2}{l}{\\, \\, \\, x^9\\,+\\,x^8\\,+\\,2x^7\\,+\\,2x^6\\,+\\,2x^5\\,+\\,2x^4\\,+\\,2x^3\\,+\\,2x^2\\,+\\,2 x\\,+\\,2} \\vspace*{0.12cm} \\\\ \\cline{2-3} \\multicolumn{1}{r}{x\\,-\\,1 \\hspace*{-4.8pt}} & \\multicolumn{1}{l}{ \\hspace*{-5.6pt} \\Big) \\hspace*{20pt} x^{10}\\,+\\,x^8} \\\\ & \\multicolumn{2}{l}{ -\\left(x^{10}\\,-\\,x^9\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{72pt}x^9\\,+\\,x^8} & \\multicolumn{2}{l}{\\hspace*{98pt}-\\left(x^9\\,-\\,x^8\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{108pt}2\\,x^8} & \\multicolumn{2}{l}{\\hspace*{134pt}-\\left(2\\,x^8\\,-\\,2\\,x^7\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{164pt}2\\,x^7} & \\multicolumn{2}{l}{\\hspace*{190pt}-\\left(2\\,x^7\\,-\\,2\\,x^6\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{220pt}2\\,x^6} & \\multicolumn{2}{l}{\\hspace*{246pt}-\\left(2\\,x^6\\,-\\,2\\,x^5\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{276pt}2\\,x^5} & \\multicolumn{2}{l}{\\hspace*{302pt}-\\left(2\\,x^5\\,-\\,2\\,x^4\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{332pt}2\\,x^4} & \\multicolumn{2}{l}{\\hspace*{358pt}-\\left(2\\,x^4\\,-\\,2\\,x^3\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{388pt}2\\,x^3} & \\multicolumn{2}{l}{\\hspace*{414pt}-\\left(2\\,x^3\\,-\\,2\\,x^2\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{444pt}2\\,x^2} & \\multicolumn{2}{l}{\\hspace*{470pt}-\\left(2\\,x^2\\,-\\,2\\,x\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{500pt}2\\,x} & \\multicolumn{2}{l}{\\hspace*{526pt}-\\left(2\\,x\\,-\\,2\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{556pt}2} \\end{array} \\end{equation*}$$ It cuts off at the end, but I think you can do the last two lines:)) P: 492 Quote by micromass 5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0... So, $$5\\,\\equiv\\,-1\\,\\left(mod\\,6\\right)$$... How do I find another root? Emeritus Sci Advisor Thanks PF Gold P: 15,871 There are only 6 elements in $$\\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots? HW Helper Thanks P: 24,975 Quote by micromass There are only 6 elements in $$\\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots? Yes, please. Check them all. You already", "useful to recall the general advice, because many calculators seem to have forgotten it and been led to calculations which were more laborious and perhaps less exact. I have determined the behaviour of the following chronometers {\\small \\begin{center} \\begin{tabular}{l@{\\,}r@{\\ }l@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,} r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r} &&&&&\\multicolumn{2}{c}{1}&\\multicolumn{2}{c}{4} &\\multicolumn{2}{c}{Breguet}&\\multicolumn{2}{c}{Barraud} &\\multicolumn{2}{c}{Kassel} \\\\ &&&h&$'$&&&&&\\multicolumn{2}{c}{3056}&\\multicolumn{2}{c}{904} &\\multicolumn{2}{c}{1252} \\\\ Greenwich&30&June& 3&22&$-8'$&$17.14''$&$+1'$&$2.37''$ \\\\ &25 &July & 2 &15 &10 &44.39 &1 &32.15 &$+30'$&$59.75''$&$+48'$&$29.20''$ &$+50'$&$29.31''$ \\\\ &28 &$''$ & 3 &13 &11 & 0.69 &1 &36.96 &30 &50.07 &48 &40.24 &50 &39.69 \\\\ & 2 &Aug. & 1 &15 &11 &28.48 &1 &44.44 &30 &31.78 &48 &58.87 &50 &52.14 \\\\ &17 &Aug. &10 &18 &12 &59.40 &2 & 6.24 &29 &35.69 &49 &57.83 &51 &38.66 \\\\ &25 &$''$ &7 &27 &13 &47.98 &2 &15.84 &29 &10.48 &50 &27.15 &52 & 2.45 \\\\ &10 &Sep. &7 &40 &15 &24.47 &2 &40.36 \\\\ \\ \\\\ Helgoland& 3&July& 3&40&$-40$&8.00&$-30$&26.84 \\\\ &22 &$''$ &12 &40 &42 & 2.02 &30 &3.89 &$-0$ &20.34 &$+16$ &47.39 &$+18$ &48.39 \\\\ & 5 &Aug. & 1 &48 &43 &18.11 &29 &43.35 & 1 &10.24 &17 &37.51 &19 &26.77 \\\\ &11 &$''$ &13 & 9 &43 &35.77 &29 &33.43 & 1 &32.75 &18 & 1.30 &19 &47.22 \\\\ &30 &$''$ &19 &30 &45 &53.08 &29 & 7.96 & 2 &40.67 &19 &17.03 &20 &47.68 \\\\ & 6 &Sep. & 3", "divide increases or diminishes by multiplication, when the ratio between the chances remains the same; so that, as in the aforesaid case, the divisor and dividend remain reciprocally in the same ratio or proportion, and the quotient remains unchanged, being the real value of the chances together. This will be better understood by the preceding example, treated as follows in three different methods, with their solution:-- \\begin{center} \\begin{tabular}{rcccrcccr} \\multicolumn{9}{c}{I.} \\\\ 6 & & & & 0 & & & & 0 \\\\ 6 & & & & 1,200 & & & & 7,200 \\\\ 4 & & & & 2,100 & & & & 8,400 \\\\ 3 & & & & 3,600 & & & & 10,800 \\\\ 2 & & & & 4,200 & & & & 8,400 \\\\ \\cline{1-1} \\cline{9-9} 21,& \\multicolumn{7}{l}{divisor} & 34,800 \\\\ \\multicolumn{9}{c}{$34,800 \\div 21 = 1,657\\ 1-7$} \\end{tabular} \\end{center} \\begin{center} \\begin{tabular}{lcccrcccr} \\multicolumn{9}{c}{II.} \\\\ 1 & & & & 0 & & & & 0 \\\\ 1 & & & & 1,200 & & & & 1,200 \\\\ $\\phantom{0}\\frac{2}{3}$ & & & & 2,100 & & & & 1,400 \\\\ $\\phantom{0}\\frac{1}{2}$ & & & & 3,600 & & & & 1,800 \\\\ $\\phantom{0}\\frac{1}{3}$ & & & & 4,200 & & & & 1,400 \\\\ \\cline{1-1} \\cline{9-9} $3\\frac{1}{2}$,& \\multicolumn{7}{l}{divisor} & 5,800 \\\\ \\multicolumn{9}{c}{$5,800 \\div", "it as: $x + 3 \\overline{\\left) x^3 + 0x^2 + 2x + 5\\right.}$ We notice that to get $(x+3) \\cdot q(x)$ to start with $x^3$, we need $q(x)$ to start with $x^2$. We then proceed to subtract $(x+3) \\cdot x^2$ and then figure out that we need a $-3x$ for the next term, and so on. We end up with $x^2 - 3x + 11$ with a remainder of $-28$ (a degree zero polynomial). $\\begin{array}{ccrcrcrcr} & & & & x^2 & - & 3x & + & 11 \\\\ & \\multicolumn{8}{l}{\\hrulefill} \\\\(x+3) & ) & x^3 & + & 0x^2 & + & 2x & + & 5 \\\\ & & x^3 & + & 3x^2 & & & & \\\\ & & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\ & & & - & 3x^2 & + & 2x & & \\\\ & & & - & 3x^2 & - & 9x & & \\\\ & & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\ & & & & & & 11x & + & 5 \\\\ & & & & & & 11x & + & 33 \\\\ & & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\ & & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed", "\\multicolumn{2}{c|}{20 } & \\multicolumn{2}{c|}{30 } & \\multicolumn{2}{c|}{45 } & \\multicolumn{2}{c|}{60 } \\\\ \\hline \\multicolumn{1}{|c|}{W} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} \\\\ \\hline \\multicolumn{1}{|c|}{4} & \\multicolumn{1}{c|}{7.77} & \\multicolumn{1}{c|}{0.92} & \\multicolumn{1}{c|}{14.23} & \\multicolumn{1}{c|}{0.72} & \\multicolumn{1}{c|}{19.23} & \\multicolumn{1}{c|}{0.48} & \\multicolumn{1}{c|}{24.10} & \\multicolumn{1}{c|}{0.18} & \\multicolumn{1}{c|}{26.61} & \\multicolumn{1}{c|}{0.01} \\\\ \\hline \\multicolumn{1}{|c|}{7} & \\multicolumn{1}{c|}{7.79} & \\multicolumn{1}{c|}{0.91} & \\multicolumn{1}{c|}{14.26} & \\multicolumn{1}{c|}{0.71} & \\multicolumn{1}{c|}{19.22} & \\multicolumn{1}{c|}{0.47} & \\multicolumn{1}{c|}{23.99} & \\multicolumn{1}{c|}{0.18} & \\multicolumn{1}{c|}{26.35} & \\multicolumn{1}{c|}{0.02} \\\\ \\hline \\end{tabular} \\caption{My caption} \\label{my-label} \\end{table} \\end{document} How can I fix this? I would like to shrink it to fit to the margins of the text. Although, if if it becomes too small I would like to have an opinion and an example of how could I display the same information but in other way. An important aspect is that I'll have 33 tables with the same problem. Each table has 18 rows in total instead of the two presented here (it was just an example). PS: This is not the document class that I'm working with. It was just an example to don't put here tons of code. In my document I have more space but the problem is the same. • What do you want to fix? Make everything smaller?", "# Remainder in polynomial division The remainder when $x^{50}$ is divided by $(x-3)(x+2)$ is of the form $ax + b$. Find the units digit of $a$. I tried to tackle the problem using the polynomial remainder theorem but got stuck as the divisor is a quadratic expression. Siong Thye Goh has a good idea but we need to work $\\bmod 50$. We can compute the following modular exponentiation using the Square and Multiply Algorithm: \\begin{align} x^{50}&=(x-3)(x+2)Q(x)+ax+b\\\\ (-2)^{50}&\\equiv24\\equiv-2a+b\\pmod{50}\\\\ 3^{50}&\\equiv49\\equiv\\phantom{-}3a+b\\pmod{50} \\end{align} Therefore, $$25\\equiv5a\\pmod{50}$$ which means that $$\\bbox[5px,border:2px solid #C0A000]{a\\equiv5\\pmod{10}}$$ Exponentiation Using The Square and Multiply Algorithm $$\\begin{array}{} &\\bmod{50}\\\\ (-2)^1&\\equiv-2\\\\ (-2)^2&\\equiv4&\\text{square}\\\\ (-2)^3&\\equiv-8&\\text{multiply}\\\\ (-2)^6&\\equiv14&\\text{square}\\\\ (-2)^{12}&\\equiv-4&\\text{square}\\\\ (-2)^{24}&\\equiv16&\\text{square}\\\\ (-2)^{25}&\\equiv-32&\\text{multiply}\\\\ (-2)^{50}&\\equiv24&\\text{square} \\end{array}$$ $$\\begin{array}{} &\\bmod{50}\\\\ 3^1&\\equiv3\\\\ 3^2&\\equiv9&\\text{square}\\\\ 3^3&\\equiv27&\\text{multiply}\\\\ 3^6&\\equiv29&\\text{square}\\\\ 3^{12}&\\equiv41&\\text{square}\\\\ 3^{24}&\\equiv31&\\text{square}\\\\ 3^{25}&\\equiv-7&\\text{multiply}\\\\ 3^{50}&\\equiv49&\\text{square} \\end{array}$$ • How did you get the last step from the one before? – tatan Oct 31 '16 at 16:57 • @robjohn very neat approach. – Siong Thye Goh Oct 31 '16 at 16:58 • @tatan $5a=50k+25$. divides by $5$. – Siong Thye Goh Oct 31 '16 at 16:59 • @SiongThyeGoh Yeah got it. Very compact and nice solution. – tatan Oct 31 '16 at 16:59 • @Rob In fact we can eliminate repeated squaring and solve it with simple mental arithmetic - see my answer. – Bill Dubuque Nov 23 '16 at 1:59 Edit: robjohn's solution is super awesome. $$x^{50}=A(x-3)(x+2)+ax+b$$ Substitute $x=-2$", "6x^4-x^2+2$,$c =-\\frac{2}{3}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$p(x) = x^4 +x^3-6x^2-7x-7$,$c=-\\sqrt{7}$\\item$p(x) = x^2-4x+1$,$c =2-\\sqrt{3}$\\label{remainderexerlast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\newpage In Exercises \\ref{factorpolyzerofirst} - \\ref{factorpolyzerolast}, you are given a polynomial and one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$x^{3} - 6x^{2} + 11x - 6, \\;\\; c = 1$\\label{factorpolyzerofirst} \\item$x^{3} - 24x^{2} + 192x - 512, \\;\\; c = 8$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$3x^{3} + 4x^{2} - x - 2, \\;\\; c = \\frac{2}{3}$\\item$2x^3-3x^2-11x+6, \\;\\; c=\\frac{1}{2}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$x^3+2x^2-3x-6, \\;\\; c = -2$\\item$2x^3-x^2-10x+5, \\;\\; c=\\frac{1}{2}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$4x^{4} - 28x^{3} + 61x^{2} - 42x + 9$,$c = \\frac{1}{2}$is a zero of multiplicity 2 \\item$x^5+2x^4-12x^3-38x^2-37x-12$,$c=-1$is a zero of multiplicity 3 \\item$125x^{5} - 275x^{4} - 2265x^{3} - 3213x^{2} - 1728x - 324$,$c = -\\frac{3}{5}$is a zero of multiplicity 3 \\item$x^{2} - 2x - 2, \\;\\; c = 1 - \\sqrt{3}$\\label{factorpolyzerolast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} In Exercises \\ref{buildapolyexerfirst} - \\ref{buildapolyexerlast}, create a polynomial$p$which has the desired characteristics. You may leave the polynomial in factored form. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item \\label{buildapolyexerfirst} \\begin{itemize} \\item The zeros of$p$are$c = \\pm 2$and$c = \\pm 1$\\item The leading term of$p(x)$is$117x^4$. \\end{itemize} \\item \\begin{itemize} \\item The zeros of$p$are$c=1$and$c = 3$\\item$c=3$is a zero of multiplicity 2. \\item The", "13x - 12 = (x - 4)(x^2 + 4x + 3) \\\\[0.5em] = (x - 4)(x^2 + 3x + x + 3) \\\\[0.5em] = (x - 4)(x(x + 3) + 1(x + 3)) \\\\[0.5em] = (x - 4)(x + 1)(x + 3)$ Hence, x3 - 13x - 12 = (x - 4)(x + 1)(x + 3). #### Question 15 Use Remainder Theorem to factorise the following polynomials completely : (i) 2x3 + x2 - 13x + 6 (ii) 3x3 + 2x2 - 19x + 6 (iii) 2x3 + 3x2 - 9x - 10 (iv) x3 + 10x2 - 37x + 26 (i) Let f(x) = 2x3 + x2 - 13x + 6 Putting, x = 2 in f(x) $f(2) = 2(2)^3 + 2^2 - 13(2) + 6 \\\\[0.5em] = 16 + 4 - 26 + 6 \\\\[0.5em] = 0$ Since, f(2) = 0 , (x - 2) is factor of f(x) by factor theorem. Dividing, f(x) by (x - 2), we get, 2x2 + 5x - 3 as quotient and remainder = 0. $\\therefore 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \\\\[0.5em] = (x - 2)(2x^2 + 6x - x - 3) \\\\[0.5em] = (x - 2)(2x(x + 3) - 1(x + 3)) \\\\[0.5em] = (x - 2)(2x - 1)(x", "0$ \\item $7x^2-4xy\\sqrt{3}+3y^2-2x-2y\\sqrt{3}-5= 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $5x^2+6xy+5y^2 - 4\\sqrt{2}x+4\\sqrt{2}y = 0$ \\item $x^2+ 2\\sqrt{3}xy+3y^2+ 2\\sqrt{3}x-2y-16 = 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $13x^2-34xy\\sqrt{3}+47y^2 - 64=0$ \\item $x^2-2\\sqrt{3} xy-y^2+8=0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $x^2-4xy+4y^2-2x\\sqrt{5}-y\\sqrt{5}=0$ \\item $8x^2+12xy+17y^2 - 20 = 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} Graph the following equations. \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{2}{1-\\cos(\\theta)}$ \\item $r = \\dfrac{3}{2 + \\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{3}{2-\\cos(\\theta)}$ \\item $r = \\dfrac{2}{1 + \\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{4}{1+3\\cos(\\theta)}$ \\item $r = \\dfrac{2}{1-2\\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{2}{1 + \\sin(\\theta - \\frac{\\pi}{3})}$ \\item $r = \\dfrac{6}{3 - \\cos\\left(\\theta + \\frac{\\pi}{4}\\right)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} The matrix $A(\\theta) = \\left[ \\begin{array}{rr} \\cos(\\theta) & -\\sin(\\theta) \\\\ \\sin(\\theta) & \\cos(\\theta) \\\\ \\end{array} \\right]$ is called a \\textbf{rotation matrix}\\index{matrix ! rotation}\\index{rotation matrix}. We've seen this matrix most recently in the proof of used in the proof of Theorem \\ref{rotatecoordinatesthm}. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item Show the matrix from Example \\ref{rotationmatrixex} in Section \\ref{MatArithmetic} is none other than $A\\left(\\frac{\\pi}{4}\\right)$. \\item Discuss with your classmates how to use $A(\\theta)$ to rotate points in the plane. \\item Using the even / odd identities for cosine and sine, show $A(\\theta)^{-1} = A(-\\theta)$. Interpret this geometrically. \\end{enumerate} \\newpage \\begin{multicols}{2}", "# Thread: Long Division Problem - 2 1. ## Long Division Problem - 2 $\\dfrac{x^{3} - 4x - 7}{x^{2}-x-6}$ First step? 2. ## Re: Long Division Problem - 2 you can always just divide $x(x^2-x-6)=x^3-x^2-6x$ $x^3-4x-7-(x^3-x^2-6x)=x^2+2x-7$ $x^2+2x-7-(x^2-x-6)=3x-1$ so your quotient is $(x+1)$ with remainder $3x-1$ 3. ## Re: Long Division Problem - 2 Hello, Jason76! $\\dfrac{x^3 - 4x - 7}{x^2-x-6}$ . . $\\begin{array}{cccccccccc} &&&&&& x & + & 1 \\\\ && --&--&--&--&--&--& -- \\\\ x^2-x-6 & | & x^3 &&& - & 4x & - & 7 \\\\ && x^3 &-& x^2 &-& 6x \\\\ && --&--&--&--&-- \\\\ &&&& x^2 &+& 2x &-& 7 \\\\ &&&& x^2 &-& x &-& 6 \\\\ &&&& --&--&--&--&-- \\\\ &&&&&& 3x &-& 1 \\end{array}$ Therefore: . $\\frac{x^3-4x-7}{x^2-x -6} \\;=\\;x+1 + \\frac{3x-1}{x^2-x-6}$", "# Multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively I want the multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively. Now I have had a search of similar questions, but I may have missed it since I am on a phone. This question should be trivial to me, but I am having trouble, so that is depressing enough, please just give me a hint. It would appear that I am dealing with: $$\\frac{7x}2\\equiv 1\\pmod 7$$ $$\\frac{7x}{3} \\equiv 2\\pmod 7$$ $$\\frac{7x}{4} \\equiv 3 \\pmod 7$$ and then I find inverses for $2,3,4$ and multiply both sides by these, then apply Chinese remainder theorem. But having the different remainders under the same modulus makes no sense, so this leads me to believe I have formulated the question wrong, and chosen the wrong modulus. Thanks • The equations you set up are wrong.Try to use the definition of (\\mod) to rewrite the equations. – rah4927 Jan 5 '15 at 5:44 • @Rah Is that in the answer the correct approach? – user142198 Jan 5 '15 at 5:45 • The first bunch is equivalent to $x\\equiv -1\\pmod{12}$. We also want $x\\equiv 0\\pmod{7}$. There will be a unique solution modulo $84$. – André Nicolas Jan 5 '15 at 5:47 Let $n$ be the number you want, then", "18.3 \\\\ 61 & 67.0 & $-0.2$ & $-0.11$ & 18.1 & $+0.1$ & $+0.18$ & $-0.08$ & $-0.04$ & 18.0 \\\\ 48 & 66.0 & $-1.2$ & $-0.69$ & 17.8 & $-0.2$ & $-0.36$ & $-0.54$ & $-0.30$ & 17.8 \\\\ 36 & 65.0 & $-2.2$ & $-1.25$ & 17.7 & $-0.3$ & $-0.53$ & $-1.00$ & $-0.56$ & 17.5 \\\\ 21 & 64.0 & $-3.2$ & $-1.83$ & 17.2 & $-0.8$ & $-1.46$ & $-1.46$ & $-0.80$ & 17.2 \\\\ \\hline\\hline & & \\multicolumn{2}{|c||}{\\ } & & \\multicolumn{3}{|c|}{Deviation from M$_s$ reckoned in} & & \\\\ & & \\multicolumn{2}{|c||}{Devation from M$_c$} & & \\multicolumn{3}{|c|}{\\ } & Smoothed & \\\\ \\cline{6-8} No.\\ of & Left & \\multicolumn{2}{|c||}{reckoned in} & Mean of & & \\multicolumn{2}{|c|}{\\ } & values & Added to \\\\ cases. & cubit. & \\multicolumn{2}{|c||}{\\ } & corresponding & & \\multicolumn{2}{|c|}{Units of Q$_s$} & multiplied & M$_s$. \\\\ & & \\multicolumn{2}{|c||}{\\ } & statures. & Inches. & \\multicolumn{2}{|c|}{\\ } & by Q$_c$. & \\\\ \\cline{3-4} \\cline{7-8} & & Inches. & Units of Q$_c$. & & & Observed. & Smoothed. & & \\\\ \\hline & inches. & & & inches. & & & & & \\\\ \\z38 & 19.25 & $+1.20$ & $+2.14$ & 70.3 & $+3.1$ & $+1.8$ & $+1.70$ & $+3.0$ & 70.2 \\\\ \\z55 &", "= 6x^4+17x^3-55x^2+16x+12$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = -3x^4-8x^3-12x^2-12x-5$ \\item $f(x) = 8x^4+50x^3+43x^2+2x-4$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^4+9x^2+20$ \\item $f(x) = x^4 + 5x^2 - 24$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^5 - x^4+7x^3-7x^2+12x-12$ \\item $f(x) = x^6-64$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^{4} - 2x^{3} + 27x^{2} - 2x + 26$ (Hint: $x = i$ is one of the zeros.) \\item $f(x) = 2x^4+5x^3+13x^2+7x+5$ (Hint: $x = -1+2i$ is a zero.) \\label{compfactpolylast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} In Exercises \\ref{buildcomppolyfirst} - \\ref{buildcompolylast}, create a polynomial $f$ with real number coefficients which has all of the desired characteristics. You may leave the polynomial in factored form. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item \\label{buildcomppolyfirst} \\begin{itemize} \\item The zeros of $f$ are $c=\\pm 1$ and $c = \\pm i$ \\item The leading term of $f(x)$ is $42x^4$ \\end{itemize} \\item \\begin{itemize} \\item $c=2i$ is a zero. \\item the point $(-1,0)$ is a local minimum on the graph of $y=f(x)$ \\item the leading term of $f(x)$ is $117x^4$ \\end{itemize} \\item \\begin{itemize} \\item The solutions to $f(x) = 0$ are $x = \\pm 2$ and $x=\\pm 7i$ \\item The leading term of $f(x)$ is $-3x^5$ \\item The point $(2,0)$ is a local maximum on the graph of $y=f(x)$. \\end{itemize} \\item \\begin{itemize} \\item $f$ is", "- 2), we get, 2x2 + 7x + 5 as quotient and remainder = 0. $\\therefore 2x^3 + 3x^2 - 9x - 10 = (x - 2)(2x^2 + 7x + 5) \\\\[0.5em] = (x - 2)(2x^2 + 5x + 2x + 5) \\\\[0.5em] = (x - 2)(x(2x + 5) + 1(2x + 5)) \\\\[0.5em] = (x - 2)(x + 1)(2x + 5)$ Hence, 2x3 + 3x2 - 9x - 10 = (x - 2)(x + 1)(2x + 5). (iv) Let f(x) = x3 + 10x2 - 37x + 26 Putting, x = 1 in f(x) $f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 \\\\[0.5em] = 1 + 10 - 37 + 26 \\\\[0.5em] = 37 - 37 \\\\[0.5em] = 0$ Since, f(1) = 0 , (x - 1) is factor of f(x) by factor theorem. Dividing, f(x) by (x - 1), we get, x2 + 11x - 26 as quotient and remainder = 0. $\\therefore x^3 + 10x^2 - 37x + 26 = (x - 1)(x^2 + 11x - 26) \\\\[0.5em] = (x - 1)(x^2 + 13x - 2x - 26) \\\\[0.5em] = (x - 1)(x(x + 13) - 2(x + 13)) \\\\[0.5em] = (x - 1)(x - 2)(x + 13)$ Hence, x3 + 10x2 - 37x + 26 = (x - 1)(x - 2)(x + 13).", "fraction is fully reduced, it is $\\frac{135}{256}$, so our answer is $135 + 256 = \\boxed{391}.$ ## Solution (casework 2) We split this into cases by making a chart. In each row, the entries $(\\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$. $$\\begin{array}{ccccccccccccl} \\multicolumn{5}{c}{R (+)\\qquad L (-)}& |&\\multicolumn{5}{c}{U (+)\\qquad D (-)}\\\\ +1&& +1&& +1&| & +1&& +1&& +1\\\\ +1&& +1&& -1& | & +1&& +1&& -1\\\\ +1&& -1&& -1& | & +1&& -1&& -1\\\\ -1 && -1&& -1& | & -1&& -1&& -1\\\\ \\\\ +1&& +1&& +1&& -1 &|& +1&& +1\\\\ +1&& +1&& -1 && -1 &|& +1 && -1\\\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\\times 2 \\text{ for symmetry by swapping }R-L\\text{ and }U-D)\\\\ \\\\ +1&& +1 &&+1 &&-1&& -1& |& +1\\\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\\times 2\\text{ symmetry})\\\\ \\\\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\\times2 \\text{ symmetry})\\\\ \\end{array}$$ Note that to account for the cases when $x=-y$, we can simply multiply the $U-D$ steps by $-1$, so for example, the first row would become $$+1 \\qquad+1\\qquad +1 \\ \\", "- 10 + 24 = 12$ $f \\left(- 1\\right) = - 1 - 3 + 10 + 24 = 30$ $f \\left(2\\right) = 8 - 12 - 20 + 24 = 0$ So $x = 2$ is a root of $f \\left(x\\right) = 0$ and $\\left(x - 2\\right)$ is a factor of $f \\left(x\\right)$ Next use, synthetic division to divide $f \\left(x\\right)$ by $\\left(x - 2\\right)$... Choose the first multiplier of $\\left(x - 2\\right)$ to match the leading term ${x}^{3}$. The multiplier we need is $\\textcolor{red}{{x}^{2}}$ ${x}^{2} \\cdot \\left(x - 2\\right) = {x}^{3} - 2 {x}^{2}$ Subtract this from ${x}^{3} - 3 {x}^{2} - 10 x + 24$ to give a remainder: $\\left({x}^{3} - 3 {x}^{2} - 10 x + 24\\right) - \\left({x}^{3} - 2 {x}^{2}\\right)$ $= - {x}^{2} - 10 x + 24$ Choose the next multiplier of $\\left(x - 2\\right)$ to match the leading term $- {x}^{2}$ of this remainder. The multiplier we need is $\\textcolor{red}{- x}$ $- x \\cdot \\left(x - 2\\right) = - {x}^{2} + 2 x$ Subtract this from the previous remainder to get a new remainder: $\\left(- {x}^{2} - 10 x + 24\\right) - \\left(- {x}^{2} + 2 x\\right)$ $= - 12 x + 24$ Choose the next multiplier of $\\left(x - 2\\right)$ to match the leading term $- 12 x$ of", "## anonymous one year ago This is the question: use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x) and express P(x) in the form d(x)*Q(x)+R(x) P(x) = x^3+4x^2-13x+60 d(x)=x+7 My answer was wrong:P(x)=(x+7)(x^2-3x+8)+6. somewhere in my divison, but I don't understand where. Help please! 1. anonymous $\\begin{array}{c|cccc} &x^2&-3x&+8 \\\\ \\hline x+7&x^3&+4x^2&-13x&+60\\\\ &-(x^3&+7x^2)\\\\ \\hline &&-3x^2\\\\ &&-(-3x^2&-21x)\\\\ \\hline &&&8x\\\\ &&&-(8x&+56)\\\\ \\hline &&&&4 \\end{array}$ Let me know if the formatting is confusing. 2. anonymous sweet! no that was perfect! everything I did was the same except for some reason I put 54 not 56 at the end. thank you! 3. anonymous You're welcome!" ]

[ "- 3x + 11$ with a remainder of $-28$ (a degree zero polynomial). $\\begin{array}{ccrcrcrcr}& & & & x^2 & - & 3x & + & 11 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x+3) & ) & x^3 & + & 0x^2 & + & 2x & + & 5 \\\\& & x^3 & + & 3x^2 & & & & \\\\& & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\& & & - & 3x^2 & + & 2x & & \\\\& & & - & 3x^2 & - & 9x & & \\\\& & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\& & & & & & 11x & + & 5 \\\\& & & & & & 11x & + & 33 \\\\& & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\& & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed to take modulo $x^3 + x + 1$. Well, dividing $x^3 + x^2$ by $x^3 + x + 1$, we see that it goes in one time with a remainder of $x^2 + x + 1$. [Note: addition and subtraction are the same in $GF(2^n)$.] $\\begin{array}{ccrcrcrcr}& & & & & & & & 1 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x^3+x+1) & ) & x^3 & + & x^2 & + & 0x & + &", "+0 # help quickly 0 48 1 Find $a$ if the remainder is constant when $10x^3-7x^2+ax+6$ is divided by $2x^2-3x+1$. Feb 21, 2020 #1 +1 Nevermind I figured it out myself We perform the polynomial division: $\\begin{array}{c|cc cc} \\multicolumn{2}{r}{5x} & +4 \\\\ \\cline{2-5} 2x^2-3x+1 & 10x^3&-7x^2&+ax&+6 \\\\ \\multicolumn{2}{r}{-10x^3} & +15x^2 & -5x \\\\ \\cline{2-4} \\multicolumn{2}{r}{0} & 8x^2 & (a-5)x & 6 \\\\ \\multicolumn{2}{r}{} & -8x^2 & +12x & -4 \\\\ \\cline{3-5} \\multicolumn{2}{r}{} & 0 & (a-5+12)x & 2 \\\\ \\end{array}$ The remainder will be constant if and only if So the answer is 7 Feb 21, 2020", "3x^2 & & & & \\\\& & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\& & & - & 3x^2 & + & 2x & & \\\\& & & - & 3x^2 & - & 9x & & \\\\& & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\& & & & & & 11x & + & 5 \\\\& & & & & & 11x & + & 33 \\\\& & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\& & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed to take modulo $x^3 + x + 1$. Well, dividing $x^3 + x^2$ by $x^3 + x + 1$, we see that it goes in one time with a remainder of $x^2 + x + 1$. [Note: addition and subtraction are the same in $GF(2^n)$.] $\\begin{array}{ccrcrcrcr}& & & & & & & & 1 \\\\& \\multicolumn{8}{l}{\\hrulefill} \\\\(x^3+x+1) & ) & x^3 & + & x^2 & + & 0x & + & 0 \\\\& & x^3 & + & 0x^2 & + & x & + & 1 \\\\& & \\multicolumn{7}{l}{\\hrulefill} \\\\& & & & x^2 & + & x & + & 1\\\\\\end{array}$ For a reasonable way to accomplish this in the special case of our integer representations of polynomials in $GF(2^n)$,", "0... P: 492 Not relevant to the problem above, but... here is the long division for problem 4.4.2(a)! $$\\begin{equation*} \\begin{array}{rc@{}c} & \\multicolumn{2}{l}{\\, \\, \\, x^9\\,+\\,x^8\\,+\\,2x^7\\,+\\,2x^6\\,+\\,2x^5\\,+\\,2x^4\\,+\\,2x^3\\,+\\,2x^2\\,+\\,2 x\\,+\\,2} \\vspace*{0.12cm} \\\\ \\cline{2-3} \\multicolumn{1}{r}{x\\,-\\,1 \\hspace*{-4.8pt}} & \\multicolumn{1}{l}{ \\hspace*{-5.6pt} \\Big) \\hspace*{20pt} x^{10}\\,+\\,x^8} \\\\ & \\multicolumn{2}{l}{ -\\left(x^{10}\\,-\\,x^9\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{72pt}x^9\\,+\\,x^8} & \\multicolumn{2}{l}{\\hspace*{98pt}-\\left(x^9\\,-\\,x^8\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{108pt}2\\,x^8} & \\multicolumn{2}{l}{\\hspace*{134pt}-\\left(2\\,x^8\\,-\\,2\\,x^7\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{164pt}2\\,x^7} & \\multicolumn{2}{l}{\\hspace*{190pt}-\\left(2\\,x^7\\,-\\,2\\,x^6\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{220pt}2\\,x^6} & \\multicolumn{2}{l}{\\hspace*{246pt}-\\left(2\\,x^6\\,-\\,2\\,x^5\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{276pt}2\\,x^5} & \\multicolumn{2}{l}{\\hspace*{302pt}-\\left(2\\,x^5\\,-\\,2\\,x^4\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{332pt}2\\,x^4} & \\multicolumn{2}{l}{\\hspace*{358pt}-\\left(2\\,x^4\\,-\\,2\\,x^3\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{388pt}2\\,x^3} & \\multicolumn{2}{l}{\\hspace*{414pt}-\\left(2\\,x^3\\,-\\,2\\,x^2\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{444pt}2\\,x^2} & \\multicolumn{2}{l}{\\hspace*{470pt}-\\left(2\\,x^2\\,-\\,2\\,x\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{500pt}2\\,x} & \\multicolumn{2}{l}{\\hspace*{526pt}-\\left(2\\,x\\,-\\,2\\right)} \\\\ \\cline{2-3} & \\multicolumn{2}{l}{\\hspace*{556pt}2} \\end{array} \\end{equation*}$$ It cuts off at the end, but I think you can do the last two lines:)) P: 492 Quote by micromass 5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0... So, $$5\\,\\equiv\\,-1\\,\\left(mod\\,6\\right)$$... How do I find another root? Emeritus Sci Advisor Thanks PF Gold P: 15,871 There are only 6 elements in $$\\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots? HW Helper Thanks P: 24,975 Quote by micromass There are only 6 elements in $$\\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots? Yes, please. Check them all. You already", "useful to recall the general advice, because many calculators seem to have forgotten it and been led to calculations which were more laborious and perhaps less exact. I have determined the behaviour of the following chronometers {\\small \\begin{center} \\begin{tabular}{l@{\\,}r@{\\ }l@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,} r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r@{\\,}r} &&&&&\\multicolumn{2}{c}{1}&\\multicolumn{2}{c}{4} &\\multicolumn{2}{c}{Breguet}&\\multicolumn{2}{c}{Barraud} &\\multicolumn{2}{c}{Kassel} \\\\ &&&h&$'$&&&&&\\multicolumn{2}{c}{3056}&\\multicolumn{2}{c}{904} &\\multicolumn{2}{c}{1252} \\\\ Greenwich&30&June& 3&22&$-8'$&$17.14''$&$+1'$&$2.37''$ \\\\ &25 &July & 2 &15 &10 &44.39 &1 &32.15 &$+30'$&$59.75''$&$+48'$&$29.20''$ &$+50'$&$29.31''$ \\\\ &28 &$''$ & 3 &13 &11 & 0.69 &1 &36.96 &30 &50.07 &48 &40.24 &50 &39.69 \\\\ & 2 &Aug. & 1 &15 &11 &28.48 &1 &44.44 &30 &31.78 &48 &58.87 &50 &52.14 \\\\ &17 &Aug. &10 &18 &12 &59.40 &2 & 6.24 &29 &35.69 &49 &57.83 &51 &38.66 \\\\ &25 &$''$ &7 &27 &13 &47.98 &2 &15.84 &29 &10.48 &50 &27.15 &52 & 2.45 \\\\ &10 &Sep. &7 &40 &15 &24.47 &2 &40.36 \\\\ \\ \\\\ Helgoland& 3&July& 3&40&$-40$&8.00&$-30$&26.84 \\\\ &22 &$''$ &12 &40 &42 & 2.02 &30 &3.89 &$-0$ &20.34 &$+16$ &47.39 &$+18$ &48.39 \\\\ & 5 &Aug. & 1 &48 &43 &18.11 &29 &43.35 & 1 &10.24 &17 &37.51 &19 &26.77 \\\\ &11 &$''$ &13 & 9 &43 &35.77 &29 &33.43 & 1 &32.75 &18 & 1.30 &19 &47.22 \\\\ &30 &$''$ &19 &30 &45 &53.08 &29 & 7.96 & 2 &40.67 &19 &17.03 &20 &47.68 \\\\ & 6 &Sep. & 3", "to take modulo $x^3 + x + 1$. Well, dividing $x^3 + x^2$ by $x^3 + x + 1$, we see that it goes in one time with a remainder of $x^2 + x + 1$. [Note: addition and subtraction are the same in $GF(2^n)$.] $\\begin{array}{ccrcrcrcr} & & & & & & & & 1 \\\\ & \\multicolumn{8}{l}{\\hrulefill} \\\\(x^3+x+1) & ) & x^3 & + & x^2 & + & 0x & + & 0 \\\\ & & x^3 & + & 0x^2 & + & x & + & 1 \\\\ & & \\multicolumn{7}{l}{\\hrulefill} \\\\ & & & & x^2 & + & x & + & 1\\\\\\end{array}$ For a reasonable way to accomplish this in the special case of our integer representations of polynomials in $GF(2^n)$, see this article about Finite Field Arithmetic and Reed Solomon Codes. In (tail-call style) Lisp, that algorithm for $GF(2^8)$ might look something like this to multiply $a$ times $b$ modulo $m$: (flet ((next-a (a) (ash a -1)) (next-b (b) (let ((overflow (plusp (logand b #x80)))) (if overflow (mod (logxor (ash b 1) m) #x100) (ash b 1))))) (labels ((mult (a b r) (cond ((zerop a) r) ((oddp a) (mult (next-a a) (next-b b) (logxor r b))) (t (mult (next-a a) (next-b b) r))))) (mult a b 0))) ### How is", "it as: $x + 3 \\overline{\\left) x^3 + 0x^2 + 2x + 5\\right.}$ We notice that to get $(x+3) \\cdot q(x)$ to start with $x^3$, we need $q(x)$ to start with $x^2$. We then proceed to subtract $(x+3) \\cdot x^2$ and then figure out that we need a $-3x$ for the next term, and so on. We end up with $x^2 - 3x + 11$ with a remainder of $-28$ (a degree zero polynomial). $\\begin{array}{ccrcrcrcr} & & & & x^2 & - & 3x & + & 11 \\\\ & \\multicolumn{8}{l}{\\hrulefill} \\\\(x+3) & ) & x^3 & + & 0x^2 & + & 2x & + & 5 \\\\ & & x^3 & + & 3x^2 & & & & \\\\ & & \\multicolumn{3}{l}{\\hrulefill} & & & & \\\\ & & & - & 3x^2 & + & 2x & & \\\\ & & & - & 3x^2 & - & 9x & & \\\\ & & & \\multicolumn{4}{l}{\\hrulefill} & & \\\\ & & & & & & 11x & + & 5 \\\\ & & & & & & 11x & + & 33 \\\\ & & & & & & \\multicolumn{3}{l}{\\hrulefill} \\\\ & & & & & & & - & 28\\end{array}$ For the example we cited earlier we had $x^3 + x^2$ which we needed", "= 6x^4+17x^3-55x^2+16x+12$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = -3x^4-8x^3-12x^2-12x-5$ \\item $f(x) = 8x^4+50x^3+43x^2+2x-4$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^4+9x^2+20$ \\item $f(x) = x^4 + 5x^2 - 24$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^5 - x^4+7x^3-7x^2+12x-12$ \\item $f(x) = x^6-64$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $f(x) = x^{4} - 2x^{3} + 27x^{2} - 2x + 26$ (Hint: $x = i$ is one of the zeros.) \\item $f(x) = 2x^4+5x^3+13x^2+7x+5$ (Hint: $x = -1+2i$ is a zero.) \\label{compfactpolylast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} In Exercises \\ref{buildcomppolyfirst} - \\ref{buildcompolylast}, create a polynomial $f$ with real number coefficients which has all of the desired characteristics. You may leave the polynomial in factored form. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item \\label{buildcomppolyfirst} \\begin{itemize} \\item The zeros of $f$ are $c=\\pm 1$ and $c = \\pm i$ \\item The leading term of $f(x)$ is $42x^4$ \\end{itemize} \\item \\begin{itemize} \\item $c=2i$ is a zero. \\item the point $(-1,0)$ is a local minimum on the graph of $y=f(x)$ \\item the leading term of $f(x)$ is $117x^4$ \\end{itemize} \\item \\begin{itemize} \\item The solutions to $f(x) = 0$ are $x = \\pm 2$ and $x=\\pm 7i$ \\item The leading term of $f(x)$ is $-3x^5$ \\item The point $(2,0)$ is a local maximum on the graph of $y=f(x)$. \\end{itemize} \\item \\begin{itemize} \\item $f$ is", "\\multicolumn{2}{c|}{20 } & \\multicolumn{2}{c|}{30 } & \\multicolumn{2}{c|}{45 } & \\multicolumn{2}{c|}{60 } \\\\ \\hline \\multicolumn{1}{|c|}{W} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} & \\multicolumn{1}{c|}{RMSE} & \\multicolumn{1}{c|}{Adj. $r^2$} \\\\ \\hline \\multicolumn{1}{|c|}{4} & \\multicolumn{1}{c|}{7.77} & \\multicolumn{1}{c|}{0.92} & \\multicolumn{1}{c|}{14.23} & \\multicolumn{1}{c|}{0.72} & \\multicolumn{1}{c|}{19.23} & \\multicolumn{1}{c|}{0.48} & \\multicolumn{1}{c|}{24.10} & \\multicolumn{1}{c|}{0.18} & \\multicolumn{1}{c|}{26.61} & \\multicolumn{1}{c|}{0.01} \\\\ \\hline \\multicolumn{1}{|c|}{7} & \\multicolumn{1}{c|}{7.79} & \\multicolumn{1}{c|}{0.91} & \\multicolumn{1}{c|}{14.26} & \\multicolumn{1}{c|}{0.71} & \\multicolumn{1}{c|}{19.22} & \\multicolumn{1}{c|}{0.47} & \\multicolumn{1}{c|}{23.99} & \\multicolumn{1}{c|}{0.18} & \\multicolumn{1}{c|}{26.35} & \\multicolumn{1}{c|}{0.02} \\\\ \\hline \\end{tabular} \\caption{My caption} \\label{my-label} \\end{table} \\end{document} How can I fix this? I would like to shrink it to fit to the margins of the text. Although, if if it becomes too small I would like to have an opinion and an example of how could I display the same information but in other way. An important aspect is that I'll have 33 tables with the same problem. Each table has 18 rows in total instead of the two presented here (it was just an example). PS: This is not the document class that I'm working with. It was just an example to don't put here tons of code. In my document I have more space but the problem is the same. • What do you want to fix? Make everything smaller?", "6x^4-x^2+2$,$c =-\\frac{2}{3}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$p(x) = x^4 +x^3-6x^2-7x-7$,$c=-\\sqrt{7}$\\item$p(x) = x^2-4x+1$,$c =2-\\sqrt{3}$\\label{remainderexerlast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\newpage In Exercises \\ref{factorpolyzerofirst} - \\ref{factorpolyzerolast}, you are given a polynomial and one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$x^{3} - 6x^{2} + 11x - 6, \\;\\; c = 1$\\label{factorpolyzerofirst} \\item$x^{3} - 24x^{2} + 192x - 512, \\;\\; c = 8$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$3x^{3} + 4x^{2} - x - 2, \\;\\; c = \\frac{2}{3}$\\item$2x^3-3x^2-11x+6, \\;\\; c=\\frac{1}{2}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$x^3+2x^2-3x-6, \\;\\; c = -2$\\item$2x^3-x^2-10x+5, \\;\\; c=\\frac{1}{2}$\\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item$4x^{4} - 28x^{3} + 61x^{2} - 42x + 9$,$c = \\frac{1}{2}$is a zero of multiplicity 2 \\item$x^5+2x^4-12x^3-38x^2-37x-12$,$c=-1$is a zero of multiplicity 3 \\item$125x^{5} - 275x^{4} - 2265x^{3} - 3213x^{2} - 1728x - 324$,$c = -\\frac{3}{5}$is a zero of multiplicity 3 \\item$x^{2} - 2x - 2, \\;\\; c = 1 - \\sqrt{3}$\\label{factorpolyzerolast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} In Exercises \\ref{buildapolyexerfirst} - \\ref{buildapolyexerlast}, create a polynomial$p$which has the desired characteristics. You may leave the polynomial in factored form. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item \\label{buildapolyexerfirst} \\begin{itemize} \\item The zeros of$p$are$c = \\pm 2$and$c = \\pm 1$\\item The leading term of$p(x)$is$117x^4$. \\end{itemize} \\item \\begin{itemize} \\item The zeros of$p$are$c=1$and$c = 3$\\item$c=3$is a zero of multiplicity 2. \\item The", "0$ \\item $7x^2-4xy\\sqrt{3}+3y^2-2x-2y\\sqrt{3}-5= 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $5x^2+6xy+5y^2 - 4\\sqrt{2}x+4\\sqrt{2}y = 0$ \\item $x^2+ 2\\sqrt{3}xy+3y^2+ 2\\sqrt{3}x-2y-16 = 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $13x^2-34xy\\sqrt{3}+47y^2 - 64=0$ \\item $x^2-2\\sqrt{3} xy-y^2+8=0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $x^2-4xy+4y^2-2x\\sqrt{5}-y\\sqrt{5}=0$ \\item $8x^2+12xy+17y^2 - 20 = 0$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} Graph the following equations. \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{2}{1-\\cos(\\theta)}$ \\item $r = \\dfrac{3}{2 + \\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{3}{2-\\cos(\\theta)}$ \\item $r = \\dfrac{2}{1 + \\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{4}{1+3\\cos(\\theta)}$ \\item $r = \\dfrac{2}{1-2\\sin(\\theta)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{2} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $r = \\dfrac{2}{1 + \\sin(\\theta - \\frac{\\pi}{3})}$ \\item $r = \\dfrac{6}{3 - \\cos\\left(\\theta + \\frac{\\pi}{4}\\right)}$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} The matrix $A(\\theta) = \\left[ \\begin{array}{rr} \\cos(\\theta) & -\\sin(\\theta) \\\\ \\sin(\\theta) & \\cos(\\theta) \\\\ \\end{array} \\right]$ is called a \\textbf{rotation matrix}\\index{matrix ! rotation}\\index{rotation matrix}. We've seen this matrix most recently in the proof of used in the proof of Theorem \\ref{rotatecoordinatesthm}. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item Show the matrix from Example \\ref{rotationmatrixex} in Section \\ref{MatArithmetic} is none other than $A\\left(\\frac{\\pi}{4}\\right)$. \\item Discuss with your classmates how to use $A(\\theta)$ to rotate points in the plane. \\item Using the even / odd identities for cosine and sine, show $A(\\theta)^{-1} = A(-\\theta)$. Interpret this geometrically. \\end{enumerate} \\newpage \\begin{multicols}{2}", "13x - 12 = (x - 4)(x^2 + 4x + 3) \\\\[0.5em] = (x - 4)(x^2 + 3x + x + 3) \\\\[0.5em] = (x - 4)(x(x + 3) + 1(x + 3)) \\\\[0.5em] = (x - 4)(x + 1)(x + 3)$ Hence, x3 - 13x - 12 = (x - 4)(x + 1)(x + 3). #### Question 15 Use Remainder Theorem to factorise the following polynomials completely : (i) 2x3 + x2 - 13x + 6 (ii) 3x3 + 2x2 - 19x + 6 (iii) 2x3 + 3x2 - 9x - 10 (iv) x3 + 10x2 - 37x + 26 (i) Let f(x) = 2x3 + x2 - 13x + 6 Putting, x = 2 in f(x) $f(2) = 2(2)^3 + 2^2 - 13(2) + 6 \\\\[0.5em] = 16 + 4 - 26 + 6 \\\\[0.5em] = 0$ Since, f(2) = 0 , (x - 2) is factor of f(x) by factor theorem. Dividing, f(x) by (x - 2), we get, 2x2 + 5x - 3 as quotient and remainder = 0. $\\therefore 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \\\\[0.5em] = (x - 2)(2x^2 + 6x - x - 3) \\\\[0.5em] = (x - 2)(2x(x + 3) - 1(x + 3)) \\\\[0.5em] = (x - 2)(2x - 1)(x", "- 2), we get, 2x2 + 7x + 5 as quotient and remainder = 0. $\\therefore 2x^3 + 3x^2 - 9x - 10 = (x - 2)(2x^2 + 7x + 5) \\\\[0.5em] = (x - 2)(2x^2 + 5x + 2x + 5) \\\\[0.5em] = (x - 2)(x(2x + 5) + 1(2x + 5)) \\\\[0.5em] = (x - 2)(x + 1)(2x + 5)$ Hence, 2x3 + 3x2 - 9x - 10 = (x - 2)(x + 1)(2x + 5). (iv) Let f(x) = x3 + 10x2 - 37x + 26 Putting, x = 1 in f(x) $f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 \\\\[0.5em] = 1 + 10 - 37 + 26 \\\\[0.5em] = 37 - 37 \\\\[0.5em] = 0$ Since, f(1) = 0 , (x - 1) is factor of f(x) by factor theorem. Dividing, f(x) by (x - 1), we get, x2 + 11x - 26 as quotient and remainder = 0. $\\therefore x^3 + 10x^2 - 37x + 26 = (x - 1)(x^2 + 11x - 26) \\\\[0.5em] = (x - 1)(x^2 + 13x - 2x - 26) \\\\[0.5em] = (x - 1)(x(x + 13) - 2(x + 13)) \\\\[0.5em] = (x - 1)(x - 2)(x + 13)$ Hence, x3 + 10x2 - 37x + 26 = (x - 1)(x - 2)(x + 13).", "t_k & \\multicolumn{1}{r}{q_k} \\\\ 0 & 1 \\\\ 1 & 0 & 4 & 732 & 153 & 1 \\\\ -4 & 1 & 1 & 612 & 120 & \\\\ \\cline{4-5} 5 & -1 & 3 & 120 & 33 & 1 \\\\ -19 & 4 & 1 & 99 & 21 & \\\\ \\cline{4-5} 24 & -5 & 1 & 21 & 12 & 1 \\\\ -43 & 9 & 1 & 12 & 9 & \\\\ \\cline{4-5} 67 & -14 & 3 & 9 & 3 & \\\\ & & & 9 & & \\\\ \\cline{4-4} & & & 0 & \\end{array}$ shows that $$3=\\gcd(153,732)=67\\cdot153-14\\cdot732$$. 9. The following computation $\\begin{array}[t]{rr@{\\qquad}r|r|r|r} s_k & t_k & \\multicolumn{1}{r}{q_k} \\\\ 0 & 1 \\\\ 1 & 0 & 2 & 6958 & 2478 & 1 \\\\ -2 & 1 & 1 & 4956 & 2002 & \\\\ \\cline{4-5} 3 & -1 & 4 & 2002 & 476 & 4 \\\\ -14 & 5 & 4 & 1904 & 392 & \\\\ \\cline{4-5} 59 & -21 & 1 & 98 & 84 & 6 \\\\ -73 & 26 & 6 & 84 & 84 & \\\\ \\cline{4-5} & & & 14 & 0 & \\end{array}$ shows that $$14=\\gcd(2478,6958)=-73\\cdot2478+26\\cdot6958$$. ### Section [sec:moregcd] 1. $$-43\\cdot133+40\\cdot143=1$$. 2. $$-512\\cdot757+319\\cdot1215=1$$. 3. Suppose $$\\sqrt{7}$$ is rational,", "- 10 + 24 = 12$ $f \\left(- 1\\right) = - 1 - 3 + 10 + 24 = 30$ $f \\left(2\\right) = 8 - 12 - 20 + 24 = 0$ So $x = 2$ is a root of $f \\left(x\\right) = 0$ and $\\left(x - 2\\right)$ is a factor of $f \\left(x\\right)$ Next use, synthetic division to divide $f \\left(x\\right)$ by $\\left(x - 2\\right)$... Choose the first multiplier of $\\left(x - 2\\right)$ to match the leading term ${x}^{3}$. The multiplier we need is $\\textcolor{red}{{x}^{2}}$ ${x}^{2} \\cdot \\left(x - 2\\right) = {x}^{3} - 2 {x}^{2}$ Subtract this from ${x}^{3} - 3 {x}^{2} - 10 x + 24$ to give a remainder: $\\left({x}^{3} - 3 {x}^{2} - 10 x + 24\\right) - \\left({x}^{3} - 2 {x}^{2}\\right)$ $= - {x}^{2} - 10 x + 24$ Choose the next multiplier of $\\left(x - 2\\right)$ to match the leading term $- {x}^{2}$ of this remainder. The multiplier we need is $\\textcolor{red}{- x}$ $- x \\cdot \\left(x - 2\\right) = - {x}^{2} + 2 x$ Subtract this from the previous remainder to get a new remainder: $\\left(- {x}^{2} - 10 x + 24\\right) - \\left(- {x}^{2} + 2 x\\right)$ $= - 12 x + 24$ Choose the next multiplier of $\\left(x - 2\\right)$ to match the leading term $- 12 x$ of", "# Remainder in polynomial division The remainder when $x^{50}$ is divided by $(x-3)(x+2)$ is of the form $ax + b$. Find the units digit of $a$. I tried to tackle the problem using the polynomial remainder theorem but got stuck as the divisor is a quadratic expression. Siong Thye Goh has a good idea but we need to work $\\bmod 50$. We can compute the following modular exponentiation using the Square and Multiply Algorithm: \\begin{align} x^{50}&=(x-3)(x+2)Q(x)+ax+b\\\\ (-2)^{50}&\\equiv24\\equiv-2a+b\\pmod{50}\\\\ 3^{50}&\\equiv49\\equiv\\phantom{-}3a+b\\pmod{50} \\end{align} Therefore, $$25\\equiv5a\\pmod{50}$$ which means that $$\\bbox[5px,border:2px solid #C0A000]{a\\equiv5\\pmod{10}}$$ Exponentiation Using The Square and Multiply Algorithm $$\\begin{array}{} &\\bmod{50}\\\\ (-2)^1&\\equiv-2\\\\ (-2)^2&\\equiv4&\\text{square}\\\\ (-2)^3&\\equiv-8&\\text{multiply}\\\\ (-2)^6&\\equiv14&\\text{square}\\\\ (-2)^{12}&\\equiv-4&\\text{square}\\\\ (-2)^{24}&\\equiv16&\\text{square}\\\\ (-2)^{25}&\\equiv-32&\\text{multiply}\\\\ (-2)^{50}&\\equiv24&\\text{square} \\end{array}$$ $$\\begin{array}{} &\\bmod{50}\\\\ 3^1&\\equiv3\\\\ 3^2&\\equiv9&\\text{square}\\\\ 3^3&\\equiv27&\\text{multiply}\\\\ 3^6&\\equiv29&\\text{square}\\\\ 3^{12}&\\equiv41&\\text{square}\\\\ 3^{24}&\\equiv31&\\text{square}\\\\ 3^{25}&\\equiv-7&\\text{multiply}\\\\ 3^{50}&\\equiv49&\\text{square} \\end{array}$$ • How did you get the last step from the one before? – tatan Oct 31 '16 at 16:57 • @robjohn very neat approach. – Siong Thye Goh Oct 31 '16 at 16:58 • @tatan $5a=50k+25$. divides by $5$. – Siong Thye Goh Oct 31 '16 at 16:59 • @SiongThyeGoh Yeah got it. Very compact and nice solution. – tatan Oct 31 '16 at 16:59 • @Rob In fact we can eliminate repeated squaring and solve it with simple mental arithmetic - see my answer. – Bill Dubuque Nov 23 '16 at 1:59 Edit: robjohn's solution is super awesome. $$x^{50}=A(x-3)(x+2)+ax+b$$ Substitute $x=-2$", "\\\\ \\cline{2-2} \\end{array} \\quad \\quad \\quad \\begin{array}{c|c|c} \\cline{2-2} & 1 & \\\\ \\hline \\multicolumn{1}{|c|}{3} & 6 & \\multicolumn{1}{c|}{2} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 4 & \\\\ \\cline{2-2} \\end{array}$ $\\displaystyle \\begin{array}{c|c|c} \\cline{2-2} & 4 & \\\\ \\hline \\multicolumn{1}{|c|}{3} & 6 & \\multicolumn{1}{c|}{2} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 1 & \\\\ \\cline{2-2} \\end{array} \\quad \\quad \\quad \\begin{array}{c|c|c} \\cline{2-2} & 1 & \\\\ \\hline \\multicolumn{1}{|c|}{3} & 6 & \\multicolumn{1}{c|}{4} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 2 & \\\\ \\cline{2-2} \\end{array} \\quad \\quad \\quad \\begin{array}{c|c|c} \\cline{2-2} & 2 & \\\\ \\hline \\multicolumn{1}{|c|}{3} & 6 & \\multicolumn{1}{c|}{4} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 1 & \\\\ \\cline{2-2} \\end{array}$ $\\displaystyle \\begin{array}{c|c|c} \\cline{2-2} & 2 & \\\\ \\hline \\multicolumn{1}{|c|}{4} & 6 & \\multicolumn{1}{c|}{1} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 3 & \\\\ \\cline{2-2} \\end{array} \\quad \\quad \\quad \\begin{array}{c|c|c} \\cline{2-2} & 3 & \\\\ \\hline \\multicolumn{1}{|c|}{4} & 6 & \\multicolumn{1}{c|}{1} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 2 & \\\\ \\cline{2-2} \\end{array} \\quad \\quad \\quad \\begin{array}{c|c|c} \\cline{2-2} & 1 & \\\\ \\hline \\multicolumn{1}{|c|}{4} & 6 & \\multicolumn{1}{c|}{2} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 3 & \\\\ \\cline{2-2} \\end{array}$ $\\displaystyle \\begin{array}{c|c|c} \\cline{2-2} & 3 & \\\\ \\hline \\multicolumn{1}{|c|}{4} & 6 & \\multicolumn{1}{c|}{2} \\\\ \\hline & 5 & \\\\ \\cline{2-2} & 1 &", "1 & 36 & 12 & \\\\ \\cline{2-3} 2 & 12 & 6 & \\\\ & 12 & & \\\\ \\cline{2-2} & 0 & \\end{array}$ Next, compute $$s_k$$ and $$t_k$$ alongside these quotients (we do not need to record the values of $$k$$): $\\begin{array}[t]{rr@{\\qquad}r|r|r|r} s_k & t_k & \\multicolumn{1}{r}{q_k} \\\\ 0 & 1 \\\\ 1 & 0 & 1 & 426 & 246 & 1 \\\\ -1 & 1 & 1 & 246 & 180 & \\\\ \\cline{4-5} 2 & -1 & 2 & 180 & 66 & 1 \\\\ -5 & 3 & 1 & 132 & 48 & \\\\ \\cline{4-5} 7 & -4 & 2 & 48 & 18 & 1 \\\\ -19 & 11 & 1 & 36 & 12 & \\\\ \\cline{4-5} 26 & -15 & 2 & 12 & 6 & \\\\ & & & 12 & & \\\\ \\cline{4-4} & & & 0 & \\end{array}$ The last nonzero remainder is the greatest common divisor, and the last linear combination gives the desired answer. We find $$\\gcd(246,426) = 6 = 26\\cdot246-15\\cdot426$$. Observe that, starting with $$k=2$$, the signs of $$s_k$$ and $$t_k$$ alternate. This provides a quick check of their signs. In addition, the signs of $$s_k$$ and $$t_k$$ are opposite for each $$k\\geq2$$. hands-on exercise $$\\PageIndex{8}\\label{he:gcd-08}$$ Use the two-column format to find", "## anonymous one year ago This is the question: use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x) and express P(x) in the form d(x)*Q(x)+R(x) P(x) = x^3+4x^2-13x+60 d(x)=x+7 My answer was wrong:P(x)=(x+7)(x^2-3x+8)+6. somewhere in my divison, but I don't understand where. Help please! 1. anonymous $\\begin{array}{c|cccc} &x^2&-3x&+8 \\\\ \\hline x+7&x^3&+4x^2&-13x&+60\\\\ &-(x^3&+7x^2)\\\\ \\hline &&-3x^2\\\\ &&-(-3x^2&-21x)\\\\ \\hline &&&8x\\\\ &&&-(8x&+56)\\\\ \\hline &&&&4 \\end{array}$ Let me know if the formatting is confusing. 2. anonymous sweet! no that was perfect! everything I did was the same except for some reason I put 54 not 56 at the end. thank you! 3. anonymous You're welcome!", "Offset} 608 & Yes \\\\\\hline 609 ROL & \\multicolumn{4}{l|}{4'h5} 610 & \\multicolumn{4}{l|}{R. Reg} 611 & \\multicolumn{3}{l|}{Cond.} 612 & \\multicolumn{21}{l|}{Operand B, truncated to low order 5 bits} 613 & \\\\\\hline 614 LOD & \\multicolumn{4}{l|}{4'h6} 615 & \\multicolumn{4}{l|}{R. Reg} 616 & \\multicolumn{3}{l|}{Cond.} 617 & \\multicolumn{21}{l|}{Operand B address} 618 & \\\\\\hline 619 STO & \\multicolumn{4}{l|}{4'h7} 620 & \\multicolumn{4}{l|}{D. Reg} 621 & \\multicolumn{3}{l|}{Cond.} 622 & \\multicolumn{21}{l|}{Operand B address} 623 & \\\\\\hline 624 SUB & \\multicolumn{4}{l|}{4'h8} 625 & \\multicolumn{4}{l|}{R. Reg} 626 & \\multicolumn{3}{l|}{Cond.} 627 32 dgisselq & \\multicolumn{21}{l|}{Operand B} 628 21 dgisselq & Yes \\\\\\hline 629 AND & \\multicolumn{4}{l|}{4'h9} 630 & \\multicolumn{4}{l|}{R. Reg} 631 & \\multicolumn{3}{l|}{Cond.} 632 & \\multicolumn{21}{l|}{Operand B} 633 & Yes \\\\\\hline 634 ADD & \\multicolumn{4}{l|}{4'ha} 635 & \\multicolumn{4}{l|}{R. Reg} 636 & \\multicolumn{3}{l|}{Cond.} 637 & \\multicolumn{21}{l|}{Operand B} 638 & Yes \\\\\\hline 639 OR & \\multicolumn{4}{l|}{4'hb} 640 & \\multicolumn{4}{l|}{R. Reg} 641 & \\multicolumn{3}{l|}{Cond.} 642 & \\multicolumn{21}{l|}{Operand B} 643 & Yes \\\\\\hline 644 XOR & \\multicolumn{4}{l|}{4'hc} 645 & \\multicolumn{4}{l|}{R. Reg} 646 & \\multicolumn{3}{l|}{Cond.} 647 & \\multicolumn{21}{l|}{Operand B} 648 & Yes \\\\\\hline 649 LSL/ASL & \\multicolumn{4}{l|}{4'hd} 650 & \\multicolumn{4}{l|}{R. Reg} 651 & \\multicolumn{3}{l|}{Cond.} 652 33 dgisselq & \\multicolumn{21}{l|}{Operand B, imm. truncated to 6 bits} 653 21 dgisselq & Yes \\\\\\hline 654 ASR & \\multicolumn{4}{l|}{4'he} 655 & \\multicolumn{4}{l|}{R. Reg} 656 & \\multicolumn{3}{l|}{Cond.} 657 33 dgisselq & \\multicolumn{21}{l|}{Operand B, imm. truncated to 6 bits} 658 21" ]

[ "+0 # Hyperbola 0 213 1 Find the distance between the foci of the hyperbola x^2/50 - y^2/22 = 1. May 26, 2021 Use the fact that $dist=2a\\sqrt{1-\\frac{b^2}{a^2}}$ and $a,b$ are given in the equation.", "Browse Questions # The distance between the foci of the hyperbola $x^2-3y^2-4x-6y-11=0$ is $\\begin {array} {1 1} (1)\\;4 & \\quad (2)\\;6 \\\\ (3)\\;8 & \\quad (4)\\;10 \\end {array}$ (3) 8", "Browse Questions # The distance between the foci of the hyperbola $x^2-3y^2-4x-6y-11=0$ is $\\begin {array} {1 1} (1)\\;4 & \\quad (2)\\;6 \\\\ (3)\\;8 & \\quad (4)\\;10 \\end {array}$ Can you answer this question? (3) 8 answered Nov 7, 2013 by", "# Recent questions tagged cl37065 Questions from: ### Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\\; 49y^2-16x^2=784$ To see more, click for the full list of questions or popular tags.", "Now, As we know the relation in a hyperbola, Hence, Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum : Exams Articles Questions", "+0 # Hyperbola +1 110 3 +30 Find the distance between the two foci of the hyperbola. $$-6x^2 + 5y^2 + 24x + 20y = 64$$ Sep 7, 2020 #1 +30 0 Someone pls help me, thanks. Sep 7, 2020 #2 +112002 +2 https://www.purplemath.com/modules/hyperbola.htm You have to start by getting it in the form $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1\\\\ (h,k) \\text{ is the centre}\\\\$$ ALSO .. Here is a youtube clip that looks fairly comprehensive.", "Find the foci and latus rectum for the hyperbola $\\large\\frac{x^2}{4}-\\frac{y^2}{5}=\\frac{1}{2}$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "+0 # The asymptotes of a hyperbola are and Also, the hyperbola passes through the point Find the distance between the foci of +1 50 3 +33 The asymptotes of a hyperbola are $$y=2x-3$$ and $$y=17-2x$$ Also, the hyperbola passes through the point $$(4,7)$$ Find the distance between the foci of the hyperbola. Once again, I would greatly appreciate any help given. Thanks! Feb 5, 2020 edited by Impasta Feb 6, 2020 #1 0 The distance between the foci is 2*sqrt(5). Feb 5, 2020 #2 +11277 +2 The asymptotes of a hyperbola are y=2x-3 and y=17-2x. Also, the hyperbola passes through the point Find the distance between the foci of the hyperbola. Once again, I would greatly appreciate any help given. Thanks! Feb 5, 2020 edited by Omi67 Feb 5, 2020 #3 +24031 +3 The asymptotes of a hyperbola are $$y=2x-3$$ and $$y=17-2x$$ Also, the hyperbola passes through the point $$(x_p,\\ y_p)$$ Find the distance between the foci of the hyperbola. $$\\text{The distance between the foci of the hyperbola \\mathbf{= \\sqrt{k}\\sqrt{5}} \\\\ with k = -(2x_p-3-y_p)(17-2x_p-y_p) }$$ The formula of the hyperbola: $$\\begin{array}{|rcll|} \\hline \\dfrac{(x-5)^2}{a^2}-\\dfrac{(y-7)^2} {b^2} &=& 1 \\\\ a &=&\\dfrac{\\sqrt{k}}{2}\\\\ b &=& \\sqrt{k} \\\\ k &=& -(2x_p-3-y_p)(17-2x_p-y_p) \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline x_p = 4,\\ y_p = 7 \\\\ \\\\ k &=& -(2x_p-3-y_p)(17-2x_p-y_p) \\\\ k &=& -(2*4-3-7)(17-2*4-7)", "hyperbolic distance$d_H(\\frac{1}{2},\\frac{1}{2}i)\\$.", "points of these Hyperbolas as you please, and thereby know that these curve lines are Hyperbolas differing little from the conical Hyperbola. And by measuring the lines C i, C k, C l, you may find other points of these Curves. For instance, when the Knives were distant from the hole in the Window ten Feet, and the Paper from the Knives nine Feet, and the Angle contained by the edges of the Knives to which the Angle A C B is equal, was subtended by a Chord which was to the Radius as 1 to 32, and the distance of the line r v from the Asymptote DE was half an Inch: I measured the lines ps, qt, rv, and found them 0'35, 0'65, 0'98 Inches respectively, and by adding to their halfs the line $\\frac{1}{2}$ mn, (which here was the 128th part of an Inch, or 0'0078 Inches) the Sums np, nq, nr, were 0'1828, 0'3328, 0'4978 Inches. I measured also the distances of the <310> brightest parts of the Fringes which run between pq and st, qr and tv, and next beyond r and v, and found them 0'5, 0'8, and 1'17 Inches. Obs. 11. The Sun shining into my darken'd Room through a small round hole made in a Plate of Lead with a", "# Math Help - Distance between vertices 1. ## Distance between vertices Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ 2. Originally Posted by viet Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ This is an hyperbola. First get it into standard form by completing the square on both the x and y terms: $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $-(9x^{2} - 18 x) + (4y^{2} + 24y) = 9$ $-9(x^2 - 2x) - 4(y^2 + 6y) = 9$ $-9 ( x^2 - 2x + 1 ) + 9 + 4(y^2 + 6y + 9) - 36 = 9$ $-9(x - 1)^2 + 4(y + 3)^2 = 36$ $-\\frac{(x - 1)^2}{4} + \\frac{(y + 3)^2}{9} = 1$ $-\\frac{(x - 1)^2}{2^2} + \\frac{(y + 3)^2}{3^2} = 1$ Can you finish the problem? -Dan 3. yeah got it thanks", "(v) Distance between foci is denoted by 2c, distance between vertices by 2a, and define the quantity b as $$b=\\sqrt{c^{2}-a^{2}} \\text { i.e., } \\sqrt{b^{2}=c^{2}-a^{2}}$$ (vi) Length of transverse axis = 2a Length of conjugate axis = 2b (vii) Eccentricity = $$e=\\frac{c}{a}(\\text { Here } c \\geq a)$$ ∴ Eccentricity is never less than 1. Standard equation of hyperbola and their graphs: Question 2. Derive the standard equation of the hyperbola in the form $$\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$$ Let F1 and F2 be the foci and ‘O’ be the midpoint of line segment F1F2. Let ‘O’ be the origin and the line through ‘O’, F1 and F2 be the x-axis. The line through O, perpendicular to the x-axis be the y-axis. Let F1= (c,0) and F2 = (-c,0) Let P = (x,y) be any point on the hyperbola such that the difference of the distances from P to the F1 and F2 be 2a. i.e., PF2-PF1=2a Using the distance formula, we have Length of latus rectum $$=\\frac{2 b^{2}}{a}$$ Note: A hyperbola in which a = b is called an equilateral hyperbola. Question 3. Define latus rectum of hyperbola. Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. Question 4. Find the coordinates of the", "#### G-GPE.A.3. Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. • No tasks yet illustrate this standard.", "Finding foci of a hyperbola It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. Solve Now What our clients say Foci Of Hyperbola foci\\:\\frac{y^2}{25}-\\frac{x^2}{9}=1; foci\\:\\frac{(x+3)^2}{25}-\\frac{(y-4)^2}{9}=1; foci\\:4x^2-9y^2-48x-72y+108=0; foci\\:x^2-y^2=1 Passing Quality", "the length of the latus rectum? 14. Find the eccentricity of the hyperbola. with foci on the x-axis if the length of its conjugate axis is ${ \\left( \\frac { 3 }{ 4 } \\right) }^{ th }$ of the length of its tranverse axis. 15. Find the equation of the hyperbola whose vertices are (0, ±7) and e = $\\frac { 4 }{ 3 }$", "the ellipse. 5. Oct 11, 2012 ### HallsofIvy The eccentricity of an ellipse is always between 0 and 1 so it cannot \"go to infinity\". As the distance between foci goes to infinity, the eccentricity goes to 1. Eccentricity 1 gives a parabola, eccentricity greater than 1 is a hyperbola.", "# Recent questions tagged exercise11-4 Questions from: ### Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\\;\\large\\frac{x^2}{16}$$-\\large\\frac{y^2}{9}$$=1$ To see more, click for the full list of questions or popular tags.", "distance between the flag posts is 8 m. Find the equation of the posts traced by the man. Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.", "Answer Comment Share Q) # The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : ( A ) $\\frac{2}{\\sqrt{3}}$ ( B ) $\\sqrt{3}$ ( C ) $\\frac{4}{3}$ ( D ) $\\frac{4}{\\sqrt{3}}$", "semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows: $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1.$ The semi-minor axis and the semi-major axis are related through the eccentricity, as follows: $b = a \\sqrt{e^2-1}.$ Note that in a hyperbola b can be larger than a. [1]" ]

[ "we see a/b = 2. We find a = 2, so 2/b = 2; thus, b = 1. The equation is: To find the foci, c2 = a2 + b2= 22 + 12 = 5 .So, c = $\\sqrt{5}$ Thus, the foci are (0, ± $\\sqrt{5}$).", "$|b|=1$. $b^2+2a+1=0\\rightarrow b^2=-1-2a$. This is a circle in the complex plane centered at $(-1,0)$ with radius $2$ since $|a|=1$. The maximum distance from the origin is $3$ at $(-3,0)$ and similarly the minimum distance is $1$ at $(1,0)$. So $1\\le |b^2|\\le 3\\rightarrow 1\\le |b|\\le \\sqrt{3}$. Both solutions give the same lower bound, $1$. So the range is $\\sqrt{3}-1=\\boxed{\\textbf{(C) }\\sqrt{3}-1}$.", "is irrelevant. Our answer is $$BC = \\sqrt {2\\left(\\frac {8\\sqrt {3}}{13}\\right)^{2}} = \\sqrt {\\frac {768}{169}}\\implies m + n = \\boxed{937}.$$ ### Solution 2 Solving for $y$ in terms of $x$ gives $y=\\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\\sqrt{4-x^2}/2)$ and $(-x,\\sqrt{4-x^2}/2)$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\\sqrt{4-x^2}/2)$ and $(0,1)$, so by the distance formula we have $$2x=\\sqrt{x^2+(1-\\sqrt{4-x^2}/2)^2}.$$ Squaring both sides and simplifying through algebra yields $x^2=192/169$, so $2x=\\sqrt{768/169}$ and the answer is $\\boxed{937}$.", "c^2=foci and because i already know foci is 2i but im a complete novice at this so i assume your right.", "27 '12 at 19:52 Let $d = a^2 + b^2$ and, $$f(d) = a^2 + b^2 + (c-1)^2 = d + \\left ( \\frac{3}{2}d -1 \\right )^2.$$ Thus, setting $f'(d) = (9/2)d-2 = 0$ gives us the critical point $d^* = 4/9$ and so, $$f(d^*) = \\frac{5}{9}.$$ Hence, the distance you're looking for is $$\\sqrt{f(d^*)} = \\frac{\\sqrt{5}}{3}.$$ -", "## College Algebra (6th Edition) Foci: $(0,2+\\sqrt{11})$ and $(0,2-\\sqrt{11})$. $\\displaystyle \\frac{(x-0)^{2}}{25}+\\frac{(y-2)^{2}}{36}=1$ (major axis vertical), $a^{2}=36,$ $a=6$ $b^{2}=25$, $b=5$ Center: $(h,k)=(0, 2)$ $c^{2}=a^{2}-b^{2}=36-25=11$ $c=\\sqrt{11}\\approx 3.317$ Foci are $c$ units above and $c$ units below the center. Foci: $(0,2+\\sqrt{11})$ and $(0,2-\\sqrt{11})$.", "$(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\\frac{1}{3}\\times120$, $\\frac{1}{3}\\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\\sqrt{7250}$ or $\\sqrt{725\\times10}$, thus $n$ is $\\boxed{725}$.", "and $$(25, -2)$$ 7. $$(5, -6)$$ and $$(3, 7)$$ 8. $$(12, -9)$$ and $$(-1, 5)$$ 9. $$(-3, 14)$$ and $$(8, 10)$$ 10. $$(-11, 3)$$ and $$(-5, 1)$$ 11. $$(5, 2)$$ and $$(11, 13)$$ 12. $$(8, 10)$$ and $$(9, -6)$$ Find the perimeter of each triangle. Round each answer to the nearest tenth. 1. $$A(3,−5),\\:B(−5,−8),\\:C(−2,7)$$ 2. $$A(5,3),\\:B(2,−7),\\:C(−1,5)$$ 3. $$A(1,2),\\:B(1,5),\\:C(4,5)$$ ## Vocabulary Term Definition Distance Formula The distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ can be defined as $$d=\\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}$$.", "\\sqrt{{9}^{2} + {2}^{2}} = \\sqrt{85}$ The foci are: $\\left(- \\sqrt{85} , 0\\right)$ and $\\left(\\sqrt{85} , 0\\right)$", "and $$b^{2} = 4$$. Thus we have: $$a^{2} - b^{2} = c^{2}$$ $$9-4 = c^{2}$$ $$5 = c^{2}$$ $$c = \\sqrt{5}$$ SInce the bigger denominator is with the y's, we find the foci by moving up and down $$\\sqrt{5}$$ units from the center. Thus we have foci at the points $$(7,3+\\sqrt{5})$$ and $$(7, 3-\\sqrt{5})$$. It's a lot of info, but hopefully it makes sense and it helps. 5. anonymous Thank you!", "the center is $(h,k) = \\boxed { (1,-2) }$ what is the foci of this equation? 7(x-2)^2 + 3(y-2)^2=21 We want to get this equation in the form $\\frac {(x - h)^2}{b^2} + \\frac {(y - k)^2}{a^2} = 1$ or $\\frac {(x - h)^2}{a^2} + \\frac {(y - k)^2}{b^2} = 1$ It's not hard to do here, just divide both sides by 21 $7(x - 2)^2 + 3(y - 2)^2 = 21$ $\\Rightarrow \\frac {(x - 2)^2}{3} + \\frac {(y - 2)^2}{7} = 1$ clearly we see that $a = \\sqrt {7}$, $b = \\sqrt {3}$, $h = 2$, $k = 2$ also, $c = \\sqrt {a^2 - b^2} = \\sqrt {7 - 3} = \\sqrt {4} = 2$ The foci are given by: $(h, k \\pm c)$ $\\Rightarrow \\mbox { Foci } = (2, 2 \\pm 2) = \\boxed {(2,4)} \\mbox { and } \\boxed {(2,0)}$ I'm beginning to feel that I'm spoiling you. Why not try the last two on your own. Tell me your solutions when you're done the vertices of this equation are: (x+3)^2- 4(y-2)^2=4 find the foci of this hyperbola: 9y^2-72y-16x^2-64x-64=0", "15$$, then $$|b – 15| = b – 15$$ but if b < 15, then$$|b – 15| = (15 - b)$$ Let’s assume the value of a to be x. Hence, $$b-a = 2x$$ and substituting the value of a, we get the value of $$b = 3x$$ Similarly, the value of $$c = 4x$$ To find As, we need to find |c|, which is the distance of c from the origin, we can say that $$|c| = 4x$$ Approach and Working Out Now, we are given the relationship between the distance of points a and b with respect to 15. However, we do not know the position of 15 with respect to a and b on the number line. So, the following cases can occur depending upon the position of 15 on the number line: o Distance between 15 and $$a = a – 15 = x – 15$$ (as a is greater than 15) o Distance between 15 and b = $$b – 15 = 3x – 15$$ (as b is greater than 15) o $$3x – 15 = 2*(x – 15)$$ {as $$|b – 15| = 2 * |a-15|$$} o $$3x – 15 = 2x – 30$$ o $$x = -15$$, which is not possible as x is the coordinate of a point that lies", "= a$, $AC = b$, $AB = c$. Note that $F$ is the incentre of $ABC$, hence the distance from $F$ to $BC$ is $r$. Now we have $$\\frac{DE}{a} = 1 - \\frac{r}{h_A} = 1 - \\frac{2S/p}{2S/a} = 1 - \\frac{a}{p} = \\frac{b + c}{p}$$ hence $$DE = \\frac{a(b+c)}{p} = \\frac{40(34 + 26)}{100} = 24.$$", "$n$, where each character is $a$, $b$, or $c$, that contain at least $2$ many $a$’s. Where does the the $c$ in \"$a$, $b$, or $c$\" come from? Question $3$: Consider a square with sides of length $17$. This square contains $n$ points. What is the minimum value of $n$ such that we can guarantee that at least two of these points have distance at most $17/\\sqrt 2$? Edit: This was originally $17\\sqrt 2$, I had made a typo. (a) 4 (b) 5 (c) 6 (d) 7 The diagonal distance of this square is $\\sqrt{{17^2} + {17^2}} = 17\\sqrt 2$. This means that even with $2$ points, this rule would still hold because even if the two points were in either corner of the square, and since the longest distance is the diagonal distance, they would always be within a distance of $17\\sqrt 2$. In this case, the lowest number to choose from is $4$, so that is the minimum -- however the answer is $5$. Where am I going wrong? • I think @Robert Z has a point below, can you recheck the Question 3 because if distance is at most $17\\sqrt{2}$ then $2$ points are enough. So it should be something else. And if the answer is $5$, it is probably $\\frac{17}{\\sqrt{2}}$. – ArsenBerk Feb 25", "## College Algebra 7th Edition $24$ The formula for the distance between two numbers is $|a-b|$. Thus, we have $|-3-21|=|-24|=24$", "Subtracting the first equation above from the second gives 2a=1, so a=1/2 and b=-1/2. It then follows that c=1. So, our answer is\\ $\\frac{1}{2} - \\left(-\\frac 12\\right) - 1 = \\boxed{0}.$ Aug 31, 2022", "notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$. after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$. Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta. Solving, we get $a=5$ and $b=-1$, so the distance is $a^2 + b^2 = \\boxed{026}$.", "y2)^2}$ and $\\sqrt{(x2 - x)^2 + (y2 - y)^2} = \\sqrt{(x1 - x2)^2 + (y1 - y2)^2}$ There will be two possible points that satisfy the the equations. Okay, that's hard. BUT if you are given one of the two points the other will be symmetrically placed on the other side of the midpoint. Ex: A=(7,3), B=(6,0), and C=(14,−1). So the midpoint of A and B is M = (6.5, 1.5). C is 14 - 6.5 = 7.5 away from M in the x direction. C is -1 - 1.5 = -2.5 away in the y directions. So the 4th point X will also be 7.5 away in the x direction and -2.5 away in the y direction. So X = (6.5 - 7.5, 1.5 - (-2.5)) = (-1, 4).", "a lot of typing... lol Good luck! 5. Originally Posted by HallsofIvy That is, $4\\sqrt{3}$ so the foci are at $(3, -1+ 4\\sqrt{3})$ and $(3, -1- 4\\sqrt{2})$. The foci are on the longer axis, here the y- axis. Dead on, except that $\\sqrt {32} = 4 \\sqrt {2}$. I'm guessing it's just a typo since you got it right in the second focus. Just thought I'd correct it so the OP doesn't get confused.", "The foci are F$= \\left(h . k + c\\right) = \\left(1 , \\sqrt{5}\\right)$ and F'$= \\left(h , k - c\\right) = \\left(1 , - \\sqrt{5}\\right)$ graph{(9x^2-18x+4y^2-27)=0 [-7.025, 7.02, -3.51, 3.51]}" ]

[ "+0 # Hyperbola 0 213 1 Find the distance between the foci of the hyperbola x^2/50 - y^2/22 = 1. May 26, 2021 Use the fact that $dist=2a\\sqrt{1-\\frac{b^2}{a^2}}$ and $a,b$ are given in the equation.", "Browse Questions # The distance between the foci of the hyperbola $x^2-3y^2-4x-6y-11=0$ is $\\begin {array} {1 1} (1)\\;4 & \\quad (2)\\;6 \\\\ (3)\\;8 & \\quad (4)\\;10 \\end {array}$ (3) 8", "Browse Questions # The distance between the foci of the hyperbola $x^2-3y^2-4x-6y-11=0$ is $\\begin {array} {1 1} (1)\\;4 & \\quad (2)\\;6 \\\\ (3)\\;8 & \\quad (4)\\;10 \\end {array}$ Can you answer this question? (3) 8 answered Nov 7, 2013 by", "# Math Help - Distance between vertices 1. ## Distance between vertices Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ 2. Originally Posted by viet Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ This is an hyperbola. First get it into standard form by completing the square on both the x and y terms: $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $-(9x^{2} - 18 x) + (4y^{2} + 24y) = 9$ $-9(x^2 - 2x) - 4(y^2 + 6y) = 9$ $-9 ( x^2 - 2x + 1 ) + 9 + 4(y^2 + 6y + 9) - 36 = 9$ $-9(x - 1)^2 + 4(y + 3)^2 = 36$ $-\\frac{(x - 1)^2}{4} + \\frac{(y + 3)^2}{9} = 1$ $-\\frac{(x - 1)^2}{2^2} + \\frac{(y + 3)^2}{3^2} = 1$ Can you finish the problem? -Dan 3. yeah got it thanks", "# Distance between vertices • July 17th 2007, 09:19 AM viet Distance between vertices Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ • July 17th 2007, 10:12 AM topsquark Quote: Originally Posted by viet Determine the distance D between the vertices of $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $D =$ This is an hyperbola. First get it into standard form by completing the square on both the x and y terms: $-9x^{2} + 18 x + 4 y^{2} + 24 y - 9 = 0$ $-(9x^{2} - 18 x) + (4y^{2} + 24y) = 9$ $-9(x^2 - 2x) - 4(y^2 + 6y) = 9$ $-9 ( x^2 - 2x + 1 ) + 9 + 4(y^2 + 6y + 9) - 36 = 9$ $-9(x - 1)^2 + 4(y + 3)^2 = 36$ $-\\frac{(x - 1)^2}{4} + \\frac{(y + 3)^2}{9} = 1$ $-\\frac{(x - 1)^2}{2^2} + \\frac{(y + 3)^2}{3^2} = 1$ Can you finish the problem? -Dan • July 17th 2007, 11:40 AM viet yeah got it thanks :)", "+0 # The asymptotes of a hyperbola are and Also, the hyperbola passes through the point Find the distance between the foci of +1 50 3 +33 The asymptotes of a hyperbola are $$y=2x-3$$ and $$y=17-2x$$ Also, the hyperbola passes through the point $$(4,7)$$ Find the distance between the foci of the hyperbola. Once again, I would greatly appreciate any help given. Thanks! Feb 5, 2020 edited by Impasta Feb 6, 2020 #1 0 The distance between the foci is 2*sqrt(5). Feb 5, 2020 #2 +11277 +2 The asymptotes of a hyperbola are y=2x-3 and y=17-2x. Also, the hyperbola passes through the point Find the distance between the foci of the hyperbola. Once again, I would greatly appreciate any help given. Thanks! Feb 5, 2020 edited by Omi67 Feb 5, 2020 #3 +24031 +3 The asymptotes of a hyperbola are $$y=2x-3$$ and $$y=17-2x$$ Also, the hyperbola passes through the point $$(x_p,\\ y_p)$$ Find the distance between the foci of the hyperbola. $$\\text{The distance between the foci of the hyperbola \\mathbf{= \\sqrt{k}\\sqrt{5}} \\\\ with k = -(2x_p-3-y_p)(17-2x_p-y_p) }$$ The formula of the hyperbola: $$\\begin{array}{|rcll|} \\hline \\dfrac{(x-5)^2}{a^2}-\\dfrac{(y-7)^2} {b^2} &=& 1 \\\\ a &=&\\dfrac{\\sqrt{k}}{2}\\\\ b &=& \\sqrt{k} \\\\ k &=& -(2x_p-3-y_p)(17-2x_p-y_p) \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline x_p = 4,\\ y_p = 7 \\\\ \\\\ k &=& -(2x_p-3-y_p)(17-2x_p-y_p) \\\\ k &=& -(2*4-3-7)(17-2*4-7)", "a hyperbola, the foci are determined to be a distance of c units from the center where c is determined by the equation $$a^{2} + b^{2} = c^{2}$$. Since the foci are 11 units away from the center, we can plug in a^2 = 4 and c^2 = 11^2 = 121 and solve for b^2. $$a^{2} +b^{2} = c^{2}$$ $$4 + b^{2} = 121$$ $$b^{2} = 117$$ Therefore the equation we're looking for is $\\frac{ y^{2} }{ 4 } - \\frac{ x^{2} }{ 117 } = 1$ Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "+0 # Hyperbola +1 110 3 +30 Find the distance between the two foci of the hyperbola. $$-6x^2 + 5y^2 + 24x + 20y = 64$$ Sep 7, 2020 #1 +30 0 Someone pls help me, thanks. Sep 7, 2020 #2 +112002 +2 https://www.purplemath.com/modules/hyperbola.htm You have to start by getting it in the form $$\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^2}=1\\\\ (h,k) \\text{ is the centre}\\\\$$ ALSO .. Here is a youtube clip that looks fairly comprehensive.", "the set of all points in a plane such that the difference of the distances from two fixed points (foci) is a constant. The foci of a hyperbola are located at:$$\\left(\\frac{c}{2},0\\right) \\text { and } \\left(-\\frac{c}{2},0\\right)$$ Where c is the distance between the foci. ### Do hyperbolas have foci? Yes, hyperbolas have two foci. The foci of a hyperbola are located at:$$\\left(\\frac{c }{2},0\\right) \\text { and } \\left(-\\frac{c}{2},0\\right)$$ Where c is the distance between the foci. ### How do you find the vertices and foci of a hyperbola? The vertices and foci of a hyperbola can be found using the equation of the hyperbola. The equation of a hyperbola is typically written in the form: $$\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1$$ Where a and b are the lengths of the semi-major and semi-minor axes, respectively. The vertices of the hyperbola are located at: $$\\left(\\pm a,0\\right) \\text { and } \\left(0,\\pm b\\right)$$ The foci of the hyperbola are located at: $$\\left(\\pm a\\sqrt{1+\\frac{b^2}{a^2}},0\\right)$$", "Another way to find the foci of a hyperbola is to use the properties of a hyperbola. A hyperbola is a conic section that is the set of all points in a plane such that the difference of the distances from two fixed points (foci) is a constant. The foci of a hyperbola are located at: $$\\left(\\frac{c}{2},0\\right) \\text { and } \\left(-\\frac{c}{2},0\\right)$$ Where c is the distance between the foci. Another way to find the foci of a hyperbola is to use the distance formula. The distance formula is:$$d=\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Where (x1,y1) and (x2,y2) are the coordinates of two points in a plane. The foci of a hyperbola are located at:$$\\left(\\frac{c}{2},0\\right) \\text { and } \\left(-\\frac{c}{2},0\\right)$$ Where c is the distance between the foci. Another way to find the foci of a hyperbola is to use the definition of a hyperbola. A hyperbola is a conic section that is the set of all points in a plane such that the difference of the distances from two fixed points (foci) is a constant. The foci of a hyperbola are located at:$$\\left(\\frac{c}{2},0\\right) \\text { and } \\left(-\\frac{c}{2},0\\right)$$ Where c is the distance between the foci. One final way to find the foci of a hyperbola is to use the properties of a hyperbola. A hyperbola is a conic section that is", "the center is $(h,k) = \\boxed { (1,-2) }$ what is the foci of this equation? 7(x-2)^2 + 3(y-2)^2=21 We want to get this equation in the form $\\frac {(x - h)^2}{b^2} + \\frac {(y - k)^2}{a^2} = 1$ or $\\frac {(x - h)^2}{a^2} + \\frac {(y - k)^2}{b^2} = 1$ It's not hard to do here, just divide both sides by 21 $7(x - 2)^2 + 3(y - 2)^2 = 21$ $\\Rightarrow \\frac {(x - 2)^2}{3} + \\frac {(y - 2)^2}{7} = 1$ clearly we see that $a = \\sqrt {7}$, $b = \\sqrt {3}$, $h = 2$, $k = 2$ also, $c = \\sqrt {a^2 - b^2} = \\sqrt {7 - 3} = \\sqrt {4} = 2$ The foci are given by: $(h, k \\pm c)$ $\\Rightarrow \\mbox { Foci } = (2, 2 \\pm 2) = \\boxed {(2,4)} \\mbox { and } \\boxed {(2,0)}$ I'm beginning to feel that I'm spoiling you. Why not try the last two on your own. Tell me your solutions when you're done the vertices of this equation are: (x+3)^2- 4(y-2)^2=4 find the foci of this hyperbola: 9y^2-72y-16x^2-64x-64=0", "(v) Distance between foci is denoted by 2c, distance between vertices by 2a, and define the quantity b as $$b=\\sqrt{c^{2}-a^{2}} \\text { i.e., } \\sqrt{b^{2}=c^{2}-a^{2}}$$ (vi) Length of transverse axis = 2a Length of conjugate axis = 2b (vii) Eccentricity = $$e=\\frac{c}{a}(\\text { Here } c \\geq a)$$ ∴ Eccentricity is never less than 1. Standard equation of hyperbola and their graphs: Question 2. Derive the standard equation of the hyperbola in the form $$\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$$ Let F1 and F2 be the foci and ‘O’ be the midpoint of line segment F1F2. Let ‘O’ be the origin and the line through ‘O’, F1 and F2 be the x-axis. The line through O, perpendicular to the x-axis be the y-axis. Let F1= (c,0) and F2 = (-c,0) Let P = (x,y) be any point on the hyperbola such that the difference of the distances from P to the F1 and F2 be 2a. i.e., PF2-PF1=2a Using the distance formula, we have Length of latus rectum $$=\\frac{2 b^{2}}{a}$$ Note: A hyperbola in which a = b is called an equilateral hyperbola. Question 3. Define latus rectum of hyperbola. Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. Question 4. Find the coordinates of the", "# Difference between Distances from Point on Hyperbola to Foci is Constant ## Theorem Let $K$ be a hyperbola. Let $F_1$ and $F_2$ be the foci of $K$. Let $P$ be an arbitrary point on $K$. Then the distance from $P$ to $F_1$ minus the distance from $P$ to $F_2$ is constant for all $P$ on $K$.", "signals travel at 960 ft/ms, find | d(P, A) - d(P, B)| The difference between the distances AP and PB is $\\frac{960\\times2640}{5280}=480$ miles. Quote: B) Find an equation for the branch of the hyperbola indicated in red in the figure, using miles as the unit of distance. The equation of a North-South opening hyperbola, centre the origin, is of the form $\\frac{y^2}{a^2}-\\frac{x^2}{b^2}=1$ and the absolute difference between the distances of any point on the hyperbola from the foci is $2a$. So here, $a = 240$. The foci are at $(0, \\pm ae)$. So $ae = 260$ $\\Rightarrow a^2e^2=260^2$ But $a^2e^2 = a^2+b^2$ $\\Rightarrow b^2= a^2e^2 - a^2$ $=260^2-240^2$ $\\Rightarrow b = 100$ Can you finish off now?", "Question # If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is A 12 B 23 C 13 D 45 Solution ## The correct option is B $$\\displaystyle \\frac{1}{\\sqrt {3}}$$$$Distance\\quad between\\quad the\\quad directrices\\quad is\\quad \\dfrac { 2a }{ e }$$$$Distance\\quad between\\quad the\\quad foci\\quad is\\quad 2ae$$$$Given\\quad :\\quad \\dfrac { 2a }{ e } =3*2ae$$$$Or,\\quad { e }^{ 2 }=\\dfrac { 1 }{ 3 }$$$$\\therefore \\quad e=\\dfrac { 1 }{ \\sqrt { 3 } }$$Option [C]Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "A hyperbola as a constant difference of distances I understand that a hyperbola can be defined as the locus of all points on a plane such that the absolute value of the difference between the distance to the foci is $2a$, the distance between the two vertices. In the simple case of a horizontal hyperbola centred on the origin, we have the following: 1. $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ 2. $c = \\sqrt{a^2 + b^2} = a\\varepsilon = a\\sqrt{1 + \\frac{b^2}{a^2}}$ The foci lie at $(\\pm c, 0)$. Now, if I'm not wrong about that, then this should be pretty basic algebra, but I can't see how to get from the above to an equation given a point $(x,y)$ describing the difference in distances to the foci as being $2a$. While I actually do care about the final result, how to get there is more important. Why do I want to know this? Well, I'd like to attempt trilateration based off differences in distance rather than fixed radii. - What is \"an equation that results in $2a$ given x and y\"? – anon Jul 5 '11 at 1:29 @anon I've reworded that bit in a way that might make more sense. – Iskar Jarak Jul 5 '11 at 2:13 We will use a little trick to avoid work.", "# Recent questions tagged cl37065 Questions from: ### Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\\; 49y^2-16x^2=784$ To see more, click for the full list of questions or popular tags.", "Browse Questions # The distance between the foci of a hyperbola is $16$ and its eccentricity is $\\sqrt 2$. Its equation is $\\begin{array}{1 1}(A)\\;x^2-y^2=32 \\\\(B)\\;\\frac{x^2}{4} -\\frac{y^2}{9}=1 \\\\(C)\\;2x^2-3y^2=7 \\\\(D)\\; none\\;of\\;these \\end{array}$ $2ae=16$ $ae=8\\qquad e=\\sqrt 2$ $a=4 \\sqrt 2$ $a^2=32$ $b^2=a^2(e^2-1)$ $\\qquad= 32 (2-1)$ $\\qquad=32$ Hence the hyperbola is $x^2-y^2=32$ Hence A is the correct answer.", "9x^2) &= 250\\\\ x^2 &= \\frac{250}{24} \\end{align*} Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \\cdot \\frac{250}{24} = \\boxed{375.}$", "# Math Help - need help with hyperbola equations 1. ## need help with hyperbola equations I am having issues with finding hyperbola equations with a translation when it gives the vertices and foci. vertices: (-5,5) (5,5) foci: (-7,5) (7,5) 2. The focal axis is horizontal, so the equation is in the form $\\frac{(x - h)^2}{a^2} - \\frac{(y - k)^2}{b^2} = 1$ The center (h, k) is in between the two vertices, so it's (0, 5). a is the distance from the center to one of the vertices, so a = 5. c is the distance from the center to one of the foci, so c = 7. The relationship between a, b, c is $c^2 = a^2 + b^2$ so solve for b: $7^2 = 5^2 + b^2$ $49 = 25 + b^2$ $24 = b^2$ $b = \\sqrt{24}$ So the equation should be $\\frac{x^2}{25} - \\frac{(y - 5)^2}{24} = 1$ Nothing to it! 01" ]

[ "u - tanp u = secp u cscp u, corresponding to the parabola x - y^2 = 1. These are analogous to the circular functions with defining equation x^2 + y^2 =1 or the hyperbolic functions, x^2-y^2 = 1. -", "# datatable.math.tanh()¶ Hyperbolic tangent of x, defined as $$\\tanh x = \\frac{\\sinh x}{\\cosh x} = \\frac{e^x-e^{-x}}{e^x+e^{-x}}$$.", "# Thread: Tangent to hyperbola 1. ## Tangent to hyperbola Prove that the lines $y=mx\\pm a\\sqrt{m^2-1}$ are tangent to the hyperbola $x^2-y^2=a^2$ for all values of m. My work: I substituted $y=mx\\pm a\\sqrt{m^2-1}$ into the equation. $x^2-m^2x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-1)=a^2$ $(1-m^2)x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-2)=0$ If $y=mx\\pm a\\sqrt{m^2-1}$ is a tangent to the hyperbola, $b^2=4ac$ $\\rightarrow(2am\\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$ I solved for m and got $\\pm 1$ So this is definitely wrong, but I don't know what is wrong. Thanks for any help! 2. Check your b^2 = 4ac calculation. 3. this is my working: $4a^2m^2(m^2-1)=-4a^2(1-m^2)(m^2-2)$ $4a^2m^4-4a^2m^2=-4a^2(3m^2-2-m^4)$ $4a^2m^4-4a^2m^2=-12a^2m^2+8a^2+4a^2m^4$ $8a^2m^2=8a^2$ $m^2=\\frac{8a^2}{8a^2}=1$ $m=\\pm 1$ 4. The last term is -a^2*m^2 + a^2 not -a^2*m^2 -a^2 5. huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. 6. Originally Posted by arze huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. The above equation becomes x^2(1-m^2) + 2amsqrt(m^2-1) -a^2*m^2 = 0 In this b^2 = 4ac. 7. oh! ok i see where i got it wrong. thanks!", "# Tanh The hyperbolic tangent is defined by the formula $$\\mathrm{tanh}(z)=\\dfrac{\\mathrm{sinh}(z)}{\\mathrm{cosh}(z)},$$ where $\\mathrm{sinh}$ is the hyperbolic sine and $\\mathrm{cosh}$ is the hyperbolic cosine.", "actually an application of the addition formula for the hyperbolic tangent, not the circular tangent. $\\tanh(x+y)=(\\tanh x+\\tanh y)/(1+\\tanh x \\tanh y)$, where the sign in the denominator is opposite to the circular version. The quantity $\\tanh^{-1}(v/c)$ is called the rapidity, and it adds linearly. – Ben Crowell Mar 11 '12 at 20:54", "Hyperbolic Tangent Calculator The calculator will find the hyperbolic tangent of the given value. The hyperbolic tangent $$y=\\tanh(x)$$$is such a function that $$y=\\frac{\\sinh(x)}{\\cosh(x)}=\\frac{e^x-e^{-x}}{e^x+e^{-x}}$$$. The domain of the hyperbolic tangent is $$(-\\infty,\\infty)$$$, the range is $$(-1,1)$$$. It is an odd function. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.", "is tangent to both the parabola y^2=4ax and the parabola x^2=4ay. The answer is x+y+a=0, but I've got no idea where that came from. Can I assume that the tangent touches the two parabola's at the same point? Help would be greatly appreciated. The tangent does not touch the parabolas (parabolae? neither of these look right. oh well) at the same point. What you can do is use the fact that the tangent will have the form y=mx+c. Then you can substitute this for y in both equations and find out what values m and c take if the equations have exactly one solution.", "Review question # Can we find the triangle areas given by a tangent? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8286 ## Question Find the equation of the tangent to the rectangular hyperbola $xy=c^2$ at the point $P(ct,c/t)$. The tangent at $P$ meets the axes of $x$ and $y$ at $L$ and $M$ respectively. $O$ is the centre of the hyperbola and $POQ$ is a diameter. The straight line $MQ$ meets the axis of $x$ in $T$. Prove that 1. the area of triangle $MOL$ is $2c^2$, 2. the area of triangle $QOT$ is $c^2/3$.", "a tangent to the hyperbola 5x2 – y2 = 5 and which passes through an external point (2, 8) is 1. a. 3x – y + 2 = 0 2. b. 3x + y – 14 = 0 3. c. 23x – 3y – 22 = 0 4. d. 3x – 23y + 178 = 0 Solution: $$\\frac{x^{2}}{1}-\\frac{y^{2}}{5}=1$$ Let the tangent of hyperbola $$y=mx\\pm \\sqrt{a^{2}m^{2}-b^{2}}$$ $$y=mx\\pm \\sqrt{m^{2}-5}$$ ……(1) Passes through (2, 8) ⇒ (8 – 2m) = $$\\pm \\sqrt{m^{2}-5}$$ On squaring both sides ⇒ (8 – 2m)2 = (m2 – 5) ⇒m = 3 or (23/3) (put in equation (1)) So equation of tangent is ⇒ $$y=3x\\pm 2$$ or $$y=\\frac{23x}{3}\\pm \\frac{22}{3}$$ ⇒ y = 3x + 2, y = 3x – 2, 3y = 23x + 22, 3 y = 23x – 22", "## Calculus 8th Edition Given: $x=y^{2}-1$ This is a parabola in the standard form as: $4.\\frac{1}{4}(x-(-1)=(y-0)^{2}$", "# Relationship between tangent, Gudermannian, and sinh The following formula holds: $$\\tan(\\mathrm{gd}(x))=\\sinh(x),$$ where $\\tan$ denotes tangent, $\\mathrm{gd}$ denotes the Gudermannian, and $\\sinh$ denotes the hyperbolic sine.", "Browse Questions # Find the equation of tangents to the hyperbola $3x^2-y^2=3$ which are perpendicular to the line $x+3y=2$. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "1, cotp u - sinp u = cscp u, and cscp u - tanp u = secp u cscp u, corresponding to the parabola x - y^2 = 1. These are analogous to the circular functions with defining equation x^2 + y^2 =1 or the hyperbolic functions, x^2-y^2 = 1.", "like a parabola ($y=x^2$)", "the equation of hyperbola. Then since we want tangents parallel to y=-3x+4 equate $\\frac{dy}{dx}=-3$", "# Hello!!! Calculus Level pending This is going to be a mixture between my 2 previous problems: Find the value for $$z$$ with $$(x , y)$$ = $$(5 , 12)$$ in the paraboloid tangent to $$x^{2} + y^{2} + z^{2} = 25$$ in its intersection points with $$z = -3$$ ×", "L*(h-h') } {r^2 + L^2 }$$ With $$\\alpha$$ the indented tangent point is easily found as $$tx = r - r*cos\\alpha$$ $$ty = h' - r*sin\\alpha$$", "the \"direction vector\" of the line. 3. So $x'=1/t, y'=t^(-1/2), z'=2t$ and evaluated at t=1, $x'=1, y'=1, z'=2$, meaning the parametric tangent line is... $x=1+ln(t), y=1+2(sqrt(t)), z=2+t^2$ ?", "Thus, the Vertex Form of the Parabola is $$y=(x-3)^2-16$$ . Solving $$(x-3)^2-16=0$$ yields the two zeros: $$x=7 , x=-1$$ . Get it now? Try the above Complete the Square Solver again.", "$$z' = (x')^2 - (y')^2$$ in the $$(x',y',z')$$ coordinate system. That is, $$z = 2xy$$ is a hyperbolic paraboloid as in Equation \\ref{Eq1.38}, but rotated $$45^\\circ$$ in the $$xy$$-plane." ]

[ "$y= x^2+ x$ which has derivative $2x_0$ when x= x_0. The slope of any tangent to that parabola must be $2x_0$ for some number $x_0$. Also, any line through (2, 3) is of the form y- 3= m(x- 2) so a line through (2, 3) tangent to the parabola is of the form $y- 3= (2x_0)(x- 2)$. Of course, the tangent line must touch the parabola at $x= x_0$, that is when $x= x_0$, $y= x_0^2+ x_0$ That is, you must have $x_0^2+ x_0- 3= 2x_0(x_0- 2)$. Solve that for $x_0$ and then write the equation of the tangent lines. (That is a parabola so there will be two solutions.) 8. Originally Posted by metallica007 Hi guys Thanks Here comes a slightly different approach to this problem: 1. A line passing through (2, -3) has the equation: $y-(-3)=m(x-2)~\\implies~y=mx-2m-3$ 3. Calculate the coordinates of the points of intersection of the parabola and the line: $x^2+x=mx-2m-3~\\implies~x^2+(1-m)x+2m+3=0$ Solve for x: $x = \\dfrac{-(1-m) \\pm \\sqrt{(1-m)^2-4 \\cdot 1 \\cdot (2m+3)}}{2}$ $x = \\dfrac{m-1}2 \\pm \\dfrac12 \\cdot \\sqrt{m^2-10m-11}$ 3. A line is tangent to the parabola if there exist only one point of intersection: the tangent point. This is only possible if the dicriminant equals zero: $m^2-10m-11=0~\\implies~m=-1~\\vee~m=11$ Plug in these values into the equation of the line at 2. 4. The x-coordinate of", "= \\pm \\sqrt {\\dfrac{64}{21}} = \\pm \\dfrac{8}{\\sqrt {21} }\\nonumber$ EXAMPLE $$\\PageIndex{5}$$ Find the points where the hyperbola $$\\dfrac{y^2}{4} - \\dfrac{x^2}{9} = 1$$ intersects the parabola $$y = 2x^2$$. Solution We can solve this system of equations by substituting $$y = 2{x^2}$$ into the hyperbola equation. $\\dfrac{(2x^2)^2}{4} - \\dfrac{x^2}{9} = 1\\nonumber$Simplify $\\dfrac{4x^4}{4} - \\dfrac{x^2}{9} = 1\\nonumber$Simplify, and multiply by 9 $9x^4 - x^2 = 9\\nonumber$Move the 9 to the left $9x^4 - x^2 - 9 = 0\\nonumber$ While this looks challenging to solve, we can think of it as a “quadratic in disguise,” since $$x^4 = (x^2)^2$$. Letting $$u = x^2$$, the equation becomes $9u^2 - u^2 - 9 = 0\\nonumber$Solve using the quadratic formula $u = \\dfrac{ - ( - 1) \\pm \\sqrt ( - 1)^2 - 4(9)( - 9) }{2(9)} = \\dfrac{1 \\pm \\sqrt {325} }{18}\\nonumber$Solve for $$x$$ $x^2 = \\dfrac{1 \\pm \\sqrt {325} }{18}\\nonumber$But $$1 - \\sqrt {325} < 0$$, so $x = \\pm \\sqrt {\\dfrac{1 + \\sqrt {325} }{18}} \\nonumber$This leads to two real solutions $x \\approx 1.028, -1.028\\nonumber$ Substituting these into $$y = 2{x^2}$$, we can find the corresponding y values. The curves intersect at the points (1.028, 2.114) and (-1.028, 2.114). Exercise $$\\PageIndex{3}$$ Find the points where the line $$y = 4x$$ intersect the ellipse $$\\dfrac{y^2}{4} - \\dfrac{x^2}{16} = 1$$ Substituting $$y =", "# How to make mx+n=0 more complicated? We all know how to solve the equation $mx+n=0$ ($m,n$ are constants).The answer is of course $-\\dfrac nm$. Then what if one day we forget about how to solve $ax+b=0$,but remember the quadratic formula for $ax^2+bx+c=0$?$x=\\frac {-b \\pm \\sqrt {b^2-4ac}}{2a}$ I have 2 ways: 1.$mx+n=0,x(mx+n)=0$.Solve for $mx^2+nx+0=0$,place $a\\rightarrow m,b\\rightarrow n,c\\rightarrow0$,we get $x=\\frac{-n\\pm\\sqrt{n^2-4m\\times0}}{2m}=\\frac{-n\\pm|n|}{2m}=\\frac{-n\\pm n}{2m}=0\\text{ or }-\\frac nm$ That $0$ is of course not a root.So the root is $-\\dfrac nm$ 2.Solve for $0x^2+mx+n=0$,place $a\\rightarrow0,b\\rightarrow m,c\\rightarrow n$. Wait,$a$ can't be $0$ ! Try using L'Hopital. If $m>0$,then $\\displaystyle\\lim_{a\\to 0}\\frac{-m-\\sqrt{m^2-4na}}{2a}$ doesn't exist. $\\lim_{a\\to 0}\\frac{-m+\\sqrt{m^2-4na}}{2a}=\\lim_{a\\to0}\\frac{\\dfrac{\\partial}{\\partial a}(-m+\\sqrt{m^2-4na}) }{\\dfrac{\\partial}{\\partial a}( 2a)}=\\lim_{a\\to0}\\dfrac{\\dfrac{-4n}{2\\sqrt{m^2-4na}}}{2}=\\frac{-n}{|m|}=-\\frac nm$ If $m<0$,then $\\displaystyle\\lim_{a\\to 0}\\frac{-m+\\sqrt{m^2-4na}}{2a}$ doesn't exist. $\\lim_{a\\to 0}\\frac{-m-\\sqrt{m^2-4na}}{2a}=\\lim_{a\\to0}\\frac{\\dfrac{\\partial}{\\partial a}(-m-\\sqrt{m^2-4na}) }{\\dfrac{\\partial}{\\partial a}( 2a)}=\\lim_{a\\to0}\\dfrac{\\dfrac{4n}{2\\sqrt{m^2-4na}}}{2}=\\frac{n}{|m|}=-\\frac nm$ Can anyone think of other ways of solving $mx+n=0$? Note by X X 2 years ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or", "\\sqrt{B^2 + 4am}}{2m}. Assuming the discriminant 0, x = -B / 2m. Thus, B = -2mx, and since \\frac{a}{x} = mx + B = mx - 2mx = -mx, we have f '(x) = m = - \\frac {a}{x^2}. ## Square Root This time, f(x) = \\sqrt{x}. Again, \\sqrt{x} = mx + B. Squaring both sides leads to a quadratic equation: m^2x^2 + (2mB - 1)x + B^2 = 0, with a solution x = \\frac{(-2mB+1) \\pm \\sqrt{4m^2B^2 - 4mB + 1 - 4m^2B^2}}{2m^2}. Simplifying we get x = \\frac{1 - 2mB}{2m^2} \\pm \\frac {\\sqrt{1 - 4mB}}{2m^2}. Since the discriminant must be zero, B = 1/4m, and therefore x = (1 - 2mB) / 2m^2 = 1 / 4m^2, and since x \\gt 0, m = 1/2\\sqrt{x}. Hence, f '(x) = 1 / 2\\sqrt{x}. ## Any Polynomial Write p(x+h) as a polynomial in h with coefficients that depend on x. The coefficient of the linear (in h) term is the derivative p'(x). For example, (x+h)^2=x^2+2xh+h^2, so ({x^2})'=2x. By the binomial formula (x+h)^n = x^n + nx^{n-1}h +(\\dots)h^2 , whence (x^n)'=nx^{n-1} . Another way to calculate p'(x) is to divide p(x)-p(a) by x-a, which is possible, because p(a)-p(a)=0, and stick x=a into the answer. For example, x^3-a^3=(x^2+xa+a^2)(x-a), so ({x^3})'=3x^2. ## Root of Degree n. To calculate the derivative of", "$x - 2 = X , y - 2 = Y$, we have, ${X}^{2} / 1 - {Y}^{2} / 9 = 1$, which represents a Hyperbola . Comparing with, ${X}^{2} / {a}^{2} - {Y}^{2} / {b}^{2} = 1$, we have, ${a}^{2} = 1 , {b}^{2} = 9$ Hence the Vertices, in $\\left(X , Y\\right)$ system, are (+-a,0)=(+-1,0) and (0,+-b)=(0,+-3) In $\\left(x , y\\right)$, they are, $x - 2 = \\pm 1 , y - 2 = 0$, i.e., $x = 2 \\pm 1 , y = 2$, or, (3,2), &, (1,2). $\\left(2 , 2 \\pm 3\\right) , i . e . , \\left(2 , 5\\right) \\mathmr{and} \\left(2 , - 1\\right)$ are also vertices. The Eccentricity $e$ is given by, ${b}^{2} = {a}^{2} \\left({e}^{2} - 1\\right) \\Rightarrow 9 = {e}^{2} - 1 \\Rightarrow {e}^{2} = 10 \\Rightarrow e = \\sqrt{10}$. Focii , in $\\left(X , Y\\right)$ system, are $\\left(\\pm a e , 0\\right) = \\left(\\pm \\sqrt{10} , 0\\right)$ $\\therefore$, In $\\left(x , y\\right)$ system, Focii are given by, $x - 2 = \\pm \\sqrt{10} , y - 2 = 0$ i.e., 3x=2+-sqrt10, y=2#. $\\therefore$ Focii, in original system, are $\\left(2 \\pm \\sqrt{10} , 2\\right)$ Asymptotes in $\\left(X , Y\\right)$ are, ${X}^{2} / 1 - {Y}^{2} / 9 = 0$, i.e., $Y = \\pm 3 X$. So, in $\\left(x ,", "\\ge \\max(0, -\\frac {c_2}{c_3})$. By squaring we added extraneous solutions.) $a^2x^4 - 2a(y-c_1)x^2 + (y-c_1)^2 = c_2 + c_3 x^2$ $a^2x^4 - [2a(y-c_1)- c_3]x^2 +[(y-c_1)^2 - c_2] = 0$ So $x^2 = \\frac { [2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$. We have some restrictions. $[2a(y-c_1)- c_3]^2 +4a^2c_2$ must be non-negative so that $4a^2c_2 \\ge - [2a(y-c_1)- c_3]^2$. And as $x^2 \\ge 0$. We must have $[2a(y-c_1)- c_3] \\ge - \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ and if $[2a(y-c_1)- c_3] < \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ we will not accept $x^2 = \\frac { [2a(y-c_1)- c_3]- \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$ as a solution. So $x = \\pm\\sqrt{\\frac { [2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}}=\\pm \\frac{\\sqrt{[2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}}{\\sqrt {2} a}$ With the restriction that everything under the radicals are positive (If you have wrong values of $y, c_1,c_2,c_3, a$ it just won't work) and the restriction that $x^2 \\ge \\max(0, -\\frac {c_2}{c_3})$. Or in other words either $x\\ge \\sqrt{\\max(0, -\\frac {c_2}{c_3})}$ or $x \\le - \\sqrt{\\max(0, -\\frac {c_2}{c_3})}$. There could be as many as $4$ solutions or as few as $0$ depending on the restrictions. Note: If $\\frac {c_2}{c_3} \\ge 0$ then this is not an issue. • wow...I did not expect line by line solutions, but thanks anyway, I will read that. – Codelearner777 Dec 18 '17 at 19:13", "\\end{pmatrix} =0$$ That is $$\\color{red}{c'}=\\frac{af^2+bg^2-2fgh}{ab-h^2}$$ The asymptotes are $$\\fbox{ax^2+2hxy+by^2+2gx+2fy+\\frac{af^2+bg^2-2fgh}{ab-h^2}=0 \\,}$$ Alternatively, using the centre of the conics $$\\left( \\frac{bg-fh}{h^2-ab}, \\frac{af-gh}{h^2-ab} \\right)$$ and the slope $m$ of an asymptote is given by $$a+2hm+bm^2=0$$ On solving, $$m=\\frac{-h \\pm \\sqrt{h^2-ab}}{b}=\\frac{a}{-h \\mp \\sqrt{h^2-ab}}$$ Therefore $$\\fbox{ y-\\frac{af-gh}{h^2-ab}= \\frac{-h \\pm \\sqrt{h^2-ab}}{b} \\left( x-\\frac{bg-fh}{h^2-ab} \\right) }$$ • Hello @Ng Chung Tak I tried to use your equation to compute the asymptotes of the equilateral hyperbola $Ax+By+xy-C=0$, but it does not work, I get $y=\\frac{-4Ax+(2B-1)AB}{4(B+x)}$, instead the right ones are $y=-A$ and $x=-B$. Can you help me to understand where I did a mistake please? – Gennaro Arguzzi May 13 at 9:16 • @GennaroArguzzi You must have made some computing error: you should get $c'=AB$, which then gives the right equations. – Aretino May 13 at 10:52 • @GennaroArguzzi Note that $$\\frac{-h \\pm \\sqrt{h^2-ab}}{b}=\\frac{a}{-h \\mp \\sqrt{h^2-ab}}$$ re-arranging the denominators and putting $a=b=0$ you may get $hx+f=0$ and $hy+g=0$ for each cases. – Ng Chung Tak May 13 at 13:34 • @GennaroArguzzi It is $2g=A$. – Aretino May 13 at 16:30 • Solve for $x$ instead of $y$. Or, better, factor the equation. – Aretino May 13 at 17:10", "# For what values of m is line y=mx tangent to the hyperbola x^2-y^2=1? Mar 31, 2018 $m = \\pm 1$ and tangents are $y = x$ and $y = - x$ #### Explanation: Put $y = m x$ in te equation of hyperbola ${x}^{2} - {y}^{2} = 1$, then ${x}^{2} - {m}^{2} {x}^{2} = 1$ or ${x}^{2} \\left(1 - {m}^{2}\\right) - 1 = 0$ the values of $x$ will give points of intersection of $y = m x$ and ${x}^{2} - {y}^{2} = 1$. But as $y = m x$ is a tangent, weshould get only one root, which would be wwhen discriminant is zero i.e. ${0}^{2} - 4 \\cdot \\left(1 - {m}^{2}\\right) \\cdot \\left(- 1\\right) = 0$ or $4 - 4 {m}^{2} = 0$ i.e. $m = \\pm 1$ and tangents are $y = x$ and $y = - x$ graph{(x^2-y^2-1)(x+y)(x-y)=0 [-10, 10, -5, 5]} Mar 31, 2018 $\\text{Slope of tangent } = \\frac{\\mathrm{dy}}{\\mathrm{dx}}$ Therefore, slope of tangent of x^2−y^2=1 => dx^2/dx−dy^2/dx=d1/dx $\\implies 2 x - 2 y \\frac{\\mathrm{dy}}{\\mathrm{dx}} = 0$ $\\implies \\frac{\\mathrm{dy}}{\\mathrm{dx}} = \\frac{x}{y}$ For what values of m is line $y = m x$ tangent to the hyperbola x^2−y^2=1? $\\implies m = \\frac{x}{y}$", "Thus $$\\boxed{f(x) = \\frac35 x^2}$$ Case $1$: Of type $x^2=4ay$. If $(10,60)$ passes through this parabola, then we get, $$100=4a (60)\\Rightarrow 4a = \\frac {5}{3}$$ The parabola is: $$x^2 =\\frac {5}{3} y$$ Case $2$: Of type $y^2=4ax$. If $(10,60)$ passes through this parabola, then we get, $$3600=4a (10) \\Rightarrow 4a =360$$ The parabola is: $$y^2=360x$$ Hope it helps. Let $f(x) = ax^2 + bx + c$ for some $a,b,c \\in \\mathbb{R}$ be the wanted parabola. Then • one root at $(0, 0) \\rightarrow f(0) = 0 = a \\cdot 0^2 + b \\cdot 0 + c \\rightarrow c = 0$ • $f(10) = 60 = a \\cdot 10^2 + b \\cdot 10 = 100a + 10b \\rightarrow 6 = 10 \\cdot a + b \\rightarrow b = 6 - 10a$ • only one root at $(0, 0) \\rightarrow \\Delta(0) = 0 = b^2-4ac = (6 - 10a)^2 - 4a\\cdot 0 = (6 - 10a)^2 = 0 \\iff 6 - 10a = 0 \\iff a = \\frac{3}{5 }$ so all the parabolae in the form $$f(x) = \\frac{3}{5} \\cdot x^2$$ satisfy the requirements. • This function will have a second root at $x=10 -\\frac6a$ (if $a\\neq 0$). I think OP wants the only root to be at $x=0$, doesn't he? – MPW Feb 20 '17 at 13:47 • yup!", "# Matrix powers and hyperbola (We're in $$\\mathbb{R}^2$$) How to find hyperbola equation, that has symmetry axis crossing (0,0) point and for $$n=1,2,\\ldots$$ points $${\\begin{pmatrix} 4 & 3 \\\\ 1 & 1 \\end{pmatrix}}^n \\begin{pmatrix} -3 \\\\ 4 \\end{pmatrix}$$ lie on that hyperbola, also how to show that all those points lie on it? By those points I mean $$\\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}$$ for $$n=1$$, $$\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$$ for $$n=2$$, etc. So far I've tried diagonalization of $$\\begin{pmatrix} 4 & 3 \\\\ 1 & 1 \\end{pmatrix}$$, but it's very messy and I don't know where it could get me to be honest. I guess that hyperbola can be rotated. Edit: So knowing that in $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ we got $$E=F=0$$ and choosing $$F=-1$$ we have $$Ax^2 + Bxy + Cy^2 = 1$$ and substituting $$x=0$$, $$y=1$$ we have $$C=1$$, again substituting $$x=3$$, $$y=1$$ we have $$B = -3A$$ and again $$x=15$$, $$y=4$$ we finally obtain $$15^2A -15\\cdot 4 \\cdot 3A = -15$$, so $$15A-12A=-1$$, thus $$A=-\\frac{1}{3}$$ and $$B=1$$, so our curve is $$-\\frac{1}{3}x^2 + xy + y^2 - 1 = 0$$ Now we can easly check what curve is it using quadratic forms. Let $$Q(X) = -\\frac{1}{3}x^2 + xy + y^2$$, then our curve is $$Q(X)", "# Thread: Tangent to hyperbola 1. ## Tangent to hyperbola Prove that the lines $y=mx\\pm a\\sqrt{m^2-1}$ are tangent to the hyperbola $x^2-y^2=a^2$ for all values of m. My work: I substituted $y=mx\\pm a\\sqrt{m^2-1}$ into the equation. $x^2-m^2x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-1)=a^2$ $(1-m^2)x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-2)=0$ If $y=mx\\pm a\\sqrt{m^2-1}$ is a tangent to the hyperbola, $b^2=4ac$ $\\rightarrow(2am\\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$ I solved for m and got $\\pm 1$ So this is definitely wrong, but I don't know what is wrong. Thanks for any help! 2. Check your b^2 = 4ac calculation. 3. this is my working: $4a^2m^2(m^2-1)=-4a^2(1-m^2)(m^2-2)$ $4a^2m^4-4a^2m^2=-4a^2(3m^2-2-m^4)$ $4a^2m^4-4a^2m^2=-12a^2m^2+8a^2+4a^2m^4$ $8a^2m^2=8a^2$ $m^2=\\frac{8a^2}{8a^2}=1$ $m=\\pm 1$ 4. The last term is -a^2*m^2 + a^2 not -a^2*m^2 -a^2 5. huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. 6. Originally Posted by arze huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. The above equation becomes x^2(1-m^2) + 2amsqrt(m^2-1) -a^2*m^2 = 0 In this b^2 = 4ac. 7. oh! ok i see where i got it wrong. thanks!", "of the systems$$ \\begin{cases} y=mx+q\\\\ x^2=-32y \\end{cases} \\qquad \\begin{cases} y=mx+q\\\\ y^2=4x \\end{cases} $$have zero discriminant. The resolvent equations are$$ x^2+32mx+32q=0 \\qquad m^2x^2+(2mq-4)x+q^2=0 $$so we get$$ \\begin{cases} 32^2m^2-4\\cdot32q=0\\\\[4px] ... 1 You can prove that the tangent line to an ellipse of equation $$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$$ at a point$(s,t)$on the ellipse is $$\\frac{sx}{a^2}+\\frac{ty}{b^2}=1$$ You can rewrite the equations of your tangent lines as $$\\frac{x}{\\sqrt{39}}+\\frac{2y}{\\sqrt{39}}=1, \\qquad \\frac{-x}{7}+\\frac{3y}{7}=1$$ Thus there must exist$(s,t)$and$(u,v)$... 0 So$\\tan\\alpha = \\frac{b}{a}$and$c^2 - a^2 = b^2$So we have:$c^2 = a^2 + b^2c^2 = a^2 + (a^2)tan^2\\alphac = \\sqrt{a^2 + (a^2)tan^2\\alpha}c = a\\sqrt{1 + tan^2\\alpha}c = a\\sqrt{sec^2\\alpha}c = (a)(sec\\alpha)\\frac{c}{a} = sec\\alpha\\epsilon = sec\\alpha$1 It likely is a conditional extremum problem and I don't know what to do next. I don't know what to do next either. So, let us take another approach. If the line$AB$is parallel to$y$-axis, we can easily see that the area of the triangle is$\\sqrt 3/2$. If the line$AB$is not parallel to$y$-axis, we can set$AB : y=ax+b$. Then, ... 0 My instinct says that this curved stretching is the wrong approach. Whatever answer you get will be useless for your CAD program. Here's a plot of the curves and the unaltered ellipse: My method would be to use your CAD program to draw a line that is tangent to the lower half of the ellipse and the blue", "side: $$y=\\pm \\sqrt{-1 \\cdot b^2}$$ $$y=\\pm bi$$ Since this problem will not have a real solution, we will not have any y-intercepts. ### A Hyperbola with y-intercepts (0,b) and (0,-b) $$\\frac{y^2}{b^2}- \\frac{x^2}{a^2}=1$$ Note: Some textbooks will keep a2 under the y2. Keeping b2 underneath y2 will make some formulas for this lesson a bit easier to remember. Our y-intercepts will occur at: (0, b) and (0, -b) This type of hyperbola opens up and down. We will not have any x-intercepts. To think about why let's plug a 0 in for y and solve for x: $$\\frac{(0)^2}{b^2}- \\frac{x^2}{a^2}=1$$ $$- \\frac{x^2}{a^2}=1$$ $$-x^2=a^2$$ Let's multiply both sides by -1: $$x^{2}=-1 \\cdot a^{2}$$ To solve for x, let's take the square root of each side: $$x=\\pm \\sqrt{-1 \\cdot a^2}$$ $$x=\\pm ai$$ Again, we know this problem will not have a real solution, therefore, we will not have any x-intercepts. To graph a hyperbola: • Locate and plot the intercepts • (a, 0) and (-a, 0) if the x2 term has a positive coefficient • (0, b) and (0, -b) if the y2 term has a positive coefficient • Find the fundamental rectangle • Endpoints: (a,b),(-a,b),(-a,-b),(a,-b) • Sketch the asymptotes • These are extended from the fundamental rectangle • y = (b/a)x • y = -(b/a)x • Sketch the graph: each branch", "and right hand side we see that we must have $xy=0$ and $x^2-y^2=4\\pm\\sqrt{15}$. But $x=0$ is not possible, because then the second of these two equations would not have a (real) solution y. Setting $y=0$ to satisfy the first equation gives the solutions $x_{1,2,3,4}=\\pm\\sqrt{4\\pm\\sqrt{15}}$ of the second equation. Hence I get $z_{1,2,3,4}=\\pm\\sqrt{4\\pm\\sqrt{15}}$ four isolated points (and not four vertical lines). 6. So what is the locus? 7. Originally Posted by Lukybear So what is the locus? As I wrote, it is just the four points $z_{1,2,3,4}=\\pm\\sqrt{4\\pm \\sqrt{15}}$ on the real axis. This agrees with the solution Soroban gave you, with the small difference that y cannot be arbitrary, but must be =0. If you take $z:= \\sqrt{4+\\sqrt{15}}+i\\cdot 2$, for example, which would lie on one of those vertical axes Soroban found, you get that the left hand side of the original equation evaluates to $z^2+1/z^2\\approx 3.90+i\\cdot 11.14$, which is rather too far away from the value of the right hand side, $8$.", "ib = \\pm i\\sqrt{-x}$. Otherwise,) Expanding, we get $(a^2-b^2) +i(2ab) = x + iy$. Equating real and imaginary parts gives us $\\begin{cases} a^2-b^2 &= x \\,\\noindent\\rule{1cm}{0.4pt} (1) \\\\ 2ab &= y \\,\\noindent\\rule{1cm}{0.4pt} (2)\\end{cases}$ Squaring, $\\begin{cases} (a^2-b^2)^2 &= x^2 \\\\ (2ab)^2 &= y^2 \\end{cases}$ Adding and using the identity $(A+B)^2 =(A-B)^2 + 4AB$ with $A=a^2, \\, B=b^2$, we get $(a^2+b^2)^2= x^2+y^2$ Since $a^2+b^2$ and $x^2+y^2$ are both positive, we can take the positive square roots of both sides to get $a^2+b^2= \\sqrt{x^2+y^2}\\noindent\\rule{1cm}{0.4pt} (3)$ $(1) + (3) \\implies 2a^2 = x + \\sqrt{x^2+y^2} \\implies a = \\pm \\sqrt{\\frac{x + \\sqrt{x^2+y^2}}{2}}$ and since $(2) \\implies b = \\frac{y}{2a},\\, a \\neq 0$, $b = \\pm \\frac{y}{2\\sqrt{\\frac{x + \\sqrt{x^2+y^2}}{2}}} = \\pm \\frac{y}{\\sqrt{2(x + \\sqrt{x^2+y^2})}}$ So cleaning up, we have $a + ib = \\pm \\sqrt{\\frac{x + \\sqrt{x^2+y^2}}{2}}\\pm i \\frac{y}{\\sqrt{2(x + \\sqrt{x^2+y^2})}}$ and we can get rid of that second $\\pm$ sign if desired: $a + ib = \\pm\\left( \\sqrt{\\frac{x + \\sqrt{x^2+y^2}}{2}} + i\\,\\mathrm{sgn}(y) \\frac{y}{\\sqrt{2(x + \\sqrt{x^2+y^2})}}\\right)$ Where $\\mathrm{sgn}\\, x$ is the sign function. $\\sqrt{x+ i y} = \\pm \\left( \\sqrt{\\frac{x+\\sqrt{x^2+y^2}}{2}} + i \\, \\text{sgn}(y) \\sqrt{\\frac{-x+\\sqrt{x^2+y^2}}{2}} \\right)$ 15. ## Re: HSC 2018 MX2 Marathon $\\sqrt{x+ i y} = \\pm \\left( \\sqrt{\\frac{x+\\sqrt{x^2+y^2}}{2}} + i \\, \\text{sgn}(y) \\sqrt{\\frac{-x+\\sqrt{x^2+y^2}}{2}} \\right)$ Woops. Forgot to rationalise the denominator. $\\frac{y}{\\sqrt{2(x + \\sqrt{x^2+y^2})}}$ $= \\sqrt{\\frac{\\frac{1}{2}y^2}{x+\\sqrt{x^2+y^2}}} \\times \\sqrt{\\frac{x-\\sqrt{x^2+y^2}}{x-\\sqrt{x^2+y^2}}} = \\sqrt{\\frac{\\frac{1}{2}y^2(x-\\sqrt{x^2+y^2})}{x^2-x^2-y^2}} = \\sqrt{\\frac{-x+\\sqrt{x^2+y^2}}{2}}$", "- M xy + y^2 \\geq 2 + M > 0.$ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$ With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get $x < My.$ The point on the hyperbola $x^2 - Mxy + y^2 = -m$ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and $$t^2 = \\frac{m}{M-2}.$$ We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$ More important than first appears, that this point is inside the unit square. We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by $$(x,y) \\mapsto (Mx - y, x)$$ gives a smaller $x+y$ value. Or, if $2x > My,$ $$(x,y) \\mapsto (y, My - x)$$ gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$ Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the", "$(x-y)^2 \\ge 0 \\implies x^2+y^2 \\ge 2xy$ $x+y=4\\implies (x+y)^2=16\\implies x^2+2xy +y^2=16\\implies 2xy=16-(x^2+y^2)$ $x^2+y^2 \\ge 16-(x^2+y^2) \\implies x^2+y^2 \\ge 8\\implies z_{min} = 8$. I solved it again in high school when quadratic equation is discussed: $x+y=4 \\implies y =4-x$ $z=x^2+y^2 \\implies z = x^2+(4-x)^2 \\implies 2x^2-8x+16-z=0$ $\\Delta = 64-4 \\cdot 2\\cdot (16-z) \\ge 0 \\implies z \\ge 8\\implies z_{min} = 8$ In my freshman calculus class, I solved it yet again: $x+y=4 \\implies y=4-x$ $z = x^2+(4-x)^2$ $\\frac{dz}{dx} = 2x+2(4-x)(-1)=2x-8+2x=4x-8$ $\\frac{dz}{dx} =0 \\implies x=2$ $\\frac{d^2 z}{dx^2} = 4 > 0 \\implies x=2, z_{min}=2^2+(4-2)^2=8$ Example-2 Find the shortest distance from the point $(1,0)$ to the parabola $y^2=4x$. We minimize $f = (x-1)^2+y^2$ where $y^2=4x$. If we eliminate $y^2$ in $f$, then $f = (x-1)^2+4x$. Solving $\\frac{df}{dx} = 2x+2=0$ gives $x=-1$, Clearly, this is not valid for it would suggest that $y^2=-4$ from $y^2=4x$, an absurdity. By Lagrange’s method, we solve $\\begin{cases} 2(x-1)-4\\lambda=0 \\\\2y\\lambda+2y = 0 \\\\y^2-4x=0\\end{cases}$ Fig. 1 The only valid solution is $x=0, y=0, k=-\\frac{1}{2}$. At $(x, y) = (0, 0), f=(0-1)^2+0^2=1$. It is the global minimum: $\\forall (x, y) \\ne (0, 0), y^2=4x \\implies x>0$. $(x-1)^2+y^2 \\overset{y^2=4x}{=}(x-1)^2+4x=x^2-2x+1+4x=x^2+2x+1\\overset{x>0}{>}1=f(0,0)$ Example-3 Find the shortest distance from the point $(a, b)$ to the line $Ax+By+C=0$. We want find a point $(x_0, y_0)$ on the line $Ax+By+C=0$ so that the distance between $(a,", "is increasing: In particular, if $\\,0 \\le x \\le y\\,$, then $\\,0 \\le \\sqrt{x} \\le \\sqrt{y}\\,$. • Renaming the square root of a square: For all real numbers $\\,x\\,$, $\\sqrt{x^2} = |x|\\,$. • The absolute value of a nonnegative number is itself: If $\\,x \\ge 0\\,$, then $\\,|x| = x\\,$. • Squared quantities are nonnegative: For all real numbers $\\,x\\,$, $\\,x^2 \\ge 0\\,$. • Solving an absolute value inequality: For all real numbers $\\,x\\,$ and for $\\,k > 0\\,$: $$\\cssId{s37}{|x| \\ge k\\ \\ \\ \\ \\text{is equivalent to}\\ \\ \\ \\ (x\\ge k\\ \\ \\text{ or }\\ \\ x\\le -k)}$$ • The square root of a product is the product of the square roots: Providing both $\\,x\\ge 0\\,$ and $\\,y\\ge 0\\,$, then $\\,\\sqrt{xy} = \\sqrt{x}\\sqrt{y}\\,$. • Solving an absolute value equation: For all real numbers $\\,x\\,$ and for $\\,k \\ge 0\\,$: $$\\cssId{s42}{|x| = k\\ \\ \\ \\ \\text{is equivalent to}\\ \\ \\ \\ x = \\pm k}$$ ## Important Information Given by the Equations of Hyperbolas Many equations easily give you lots of information about their graphs, without any memorizing! This is certainly true with hyperbolas, as follows. ## What part of the plane do hyperbolas live in? Let's rewrite and analyze the equation $\\,\\displaystyle \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1\\,$: \\displaystyle \\begin{alignat}{2} \\frac{x^2}{a^2} &= 1 + \\frac{y^2}{b^2} &\\qquad&\\text{(solve", "\\}$ and $y = 1/x$ – caverac Nov 21 '16 at 17:12 You can also directly combine the equations into complete binomial formulas $$(x+y)^2=x^2+y^2+2xy=4+2=6,\\\\ (x-y)^2=x^2+y^2-2xy=4-2=2$$ and solve the trivial linear system for each of the 4 sign combinations of the roots. • nice!$\\quad\\qquad$ – polfosol Nov 22 '16 at 5:30 The equation of the circle is $x^2 + y^2 = 4$ and the equation of the hyperbola is $xy=1$ So the point of intersection would be a common solution to $xy =1$ $x^2 + y^2 = 4$ so $y = 1/x$ $x^2 + \\frac 1{x^2} = 4$ $x^4 +1 = 4x^2$ $x^4 - 4x^2 + 1 = 0$ $x^2 = \\frac {4 \\pm \\sqrt {12}}2$ $x^2 = 2 \\pm \\sqrt 3$ $x = \\pm \\sqrt{2 \\pm \\sqrt{3}}$ $y = 1/x = \\pm \\frac 1{\\sqrt{2 \\pm \\sqrt{3}}}$ $= \\pm \\frac 1{\\sqrt{2 \\pm \\sqrt{3}}}\\frac {\\sqrt {2\\mp \\sqrt {3}}}{\\sqrt {2\\mp\\sqrt{3}}}$ $=\\pm \\frac{\\sqrt {2\\mp \\sqrt {3}}}{\\sqrt {4-3}}=\\pm {\\sqrt {2\\mp \\sqrt {3}}}$ So there are four points: $(\\sqrt{2 + \\sqrt{3}},{\\sqrt{2 - \\sqrt{3}}});(\\sqrt{2 - \\sqrt{3}},{\\sqrt{2 +\\sqrt{3}}});(-\\sqrt{2 + \\sqrt{3}},-{\\sqrt{2 - \\sqrt{3}}});(-\\sqrt{2 - \\sqrt{3}},-{\\sqrt{2 + \\sqrt{3}}});$ • You could simplify the $y$ coordinates using $1/(2\\pm\\sqrt{3}) = 2\\mp\\sqrt{3}$, which also shows the symmetry more clearly. – Klaus Draeger Nov 21 '16 at 21:22 • I could. But I personally don't like that. It treats the pm sign", "This can be simplified as $$(b^2 + m^2a^2)x^2 + (2mca^2)x + (a^2c^2 - a^2b^2) = 0\\cdots\\cdots(1)$$ Now tangent touches at a single point to the ellipse, so roots of the above quadratic equation can not be distinct. Fo this to happen, we must have discriminant equal to $0$. So we have $$(2mca^2)^2 - 4(b^2 + m^2a^2)(a^2c^2- a^2b^2)=0 \\cdots\\cdots(2)$$ Simplifying this, we have $c^2 = b^2 + m^2a^2$. Now let $(h,k)$ be an external point to the ellipse from where we draw the tangent. So $(h,k)$ must lie on the line $y = mx+c$. Therefore $k = mh+c$. This leads to $$(k-mh)^2 = c^2 = b^2+m^2a^2$$ Expanding, we get $$(h^2- a^2)m^2 - 2mhk + (k^2 - b^2) = 0$$ This is a quadratic equation in $m$, since we have two tangents from an external point to this ellipse. Now from the theory of quadratic equations, the product of roots is given by $$m_1m_2 = \\frac{k^2 - b^2}{h^2 - a^2}$$ Now in part (a), this product is equal to 1. This leads us to $h^2 - k^2 = a^2 - b^2$. So the locus of points here is $x^2 - y^2 = a^2 - b^2$, which is a hyperbola. For part (b), the product is equal to -1, which leads to $h^2 + k^2 = a^2 + b^2$. So the" ]

[ "= \\pm \\sqrt {\\dfrac{64}{21}} = \\pm \\dfrac{8}{\\sqrt {21} }\\nonumber$ EXAMPLE $$\\PageIndex{5}$$ Find the points where the hyperbola $$\\dfrac{y^2}{4} - \\dfrac{x^2}{9} = 1$$ intersects the parabola $$y = 2x^2$$. Solution We can solve this system of equations by substituting $$y = 2{x^2}$$ into the hyperbola equation. $\\dfrac{(2x^2)^2}{4} - \\dfrac{x^2}{9} = 1\\nonumber$Simplify $\\dfrac{4x^4}{4} - \\dfrac{x^2}{9} = 1\\nonumber$Simplify, and multiply by 9 $9x^4 - x^2 = 9\\nonumber$Move the 9 to the left $9x^4 - x^2 - 9 = 0\\nonumber$ While this looks challenging to solve, we can think of it as a “quadratic in disguise,” since $$x^4 = (x^2)^2$$. Letting $$u = x^2$$, the equation becomes $9u^2 - u^2 - 9 = 0\\nonumber$Solve using the quadratic formula $u = \\dfrac{ - ( - 1) \\pm \\sqrt ( - 1)^2 - 4(9)( - 9) }{2(9)} = \\dfrac{1 \\pm \\sqrt {325} }{18}\\nonumber$Solve for $$x$$ $x^2 = \\dfrac{1 \\pm \\sqrt {325} }{18}\\nonumber$But $$1 - \\sqrt {325} < 0$$, so $x = \\pm \\sqrt {\\dfrac{1 + \\sqrt {325} }{18}} \\nonumber$This leads to two real solutions $x \\approx 1.028, -1.028\\nonumber$ Substituting these into $$y = 2{x^2}$$, we can find the corresponding y values. The curves intersect at the points (1.028, 2.114) and (-1.028, 2.114). Exercise $$\\PageIndex{3}$$ Find the points where the line $$y = 4x$$ intersect the ellipse $$\\dfrac{y^2}{4} - \\dfrac{x^2}{16} = 1$$ Substituting $$y =", "# How to make mx+n=0 more complicated? We all know how to solve the equation $mx+n=0$ ($m,n$ are constants).The answer is of course $-\\dfrac nm$. Then what if one day we forget about how to solve $ax+b=0$,but remember the quadratic formula for $ax^2+bx+c=0$?$x=\\frac {-b \\pm \\sqrt {b^2-4ac}}{2a}$ I have 2 ways: 1.$mx+n=0,x(mx+n)=0$.Solve for $mx^2+nx+0=0$,place $a\\rightarrow m,b\\rightarrow n,c\\rightarrow0$,we get $x=\\frac{-n\\pm\\sqrt{n^2-4m\\times0}}{2m}=\\frac{-n\\pm|n|}{2m}=\\frac{-n\\pm n}{2m}=0\\text{ or }-\\frac nm$ That $0$ is of course not a root.So the root is $-\\dfrac nm$ 2.Solve for $0x^2+mx+n=0$,place $a\\rightarrow0,b\\rightarrow m,c\\rightarrow n$. Wait,$a$ can't be $0$ ! Try using L'Hopital. If $m>0$,then $\\displaystyle\\lim_{a\\to 0}\\frac{-m-\\sqrt{m^2-4na}}{2a}$ doesn't exist. $\\lim_{a\\to 0}\\frac{-m+\\sqrt{m^2-4na}}{2a}=\\lim_{a\\to0}\\frac{\\dfrac{\\partial}{\\partial a}(-m+\\sqrt{m^2-4na}) }{\\dfrac{\\partial}{\\partial a}( 2a)}=\\lim_{a\\to0}\\dfrac{\\dfrac{-4n}{2\\sqrt{m^2-4na}}}{2}=\\frac{-n}{|m|}=-\\frac nm$ If $m<0$,then $\\displaystyle\\lim_{a\\to 0}\\frac{-m+\\sqrt{m^2-4na}}{2a}$ doesn't exist. $\\lim_{a\\to 0}\\frac{-m-\\sqrt{m^2-4na}}{2a}=\\lim_{a\\to0}\\frac{\\dfrac{\\partial}{\\partial a}(-m-\\sqrt{m^2-4na}) }{\\dfrac{\\partial}{\\partial a}( 2a)}=\\lim_{a\\to0}\\dfrac{\\dfrac{4n}{2\\sqrt{m^2-4na}}}{2}=\\frac{n}{|m|}=-\\frac nm$ Can anyone think of other ways of solving $mx+n=0$? Note by X X 2 years ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or", "\\sqrt{B^2 + 4am}}{2m}. Assuming the discriminant 0, x = -B / 2m. Thus, B = -2mx, and since \\frac{a}{x} = mx + B = mx - 2mx = -mx, we have f '(x) = m = - \\frac {a}{x^2}. ## Square Root This time, f(x) = \\sqrt{x}. Again, \\sqrt{x} = mx + B. Squaring both sides leads to a quadratic equation: m^2x^2 + (2mB - 1)x + B^2 = 0, with a solution x = \\frac{(-2mB+1) \\pm \\sqrt{4m^2B^2 - 4mB + 1 - 4m^2B^2}}{2m^2}. Simplifying we get x = \\frac{1 - 2mB}{2m^2} \\pm \\frac {\\sqrt{1 - 4mB}}{2m^2}. Since the discriminant must be zero, B = 1/4m, and therefore x = (1 - 2mB) / 2m^2 = 1 / 4m^2, and since x \\gt 0, m = 1/2\\sqrt{x}. Hence, f '(x) = 1 / 2\\sqrt{x}. ## Any Polynomial Write p(x+h) as a polynomial in h with coefficients that depend on x. The coefficient of the linear (in h) term is the derivative p'(x). For example, (x+h)^2=x^2+2xh+h^2, so ({x^2})'=2x. By the binomial formula (x+h)^n = x^n + nx^{n-1}h +(\\dots)h^2 , whence (x^n)'=nx^{n-1} . Another way to calculate p'(x) is to divide p(x)-p(a) by x-a, which is possible, because p(a)-p(a)=0, and stick x=a into the answer. For example, x^3-a^3=(x^2+xa+a^2)(x-a), so ({x^3})'=3x^2. ## Root of Degree n. To calculate the derivative of", "# Thread: Tangent to hyperbola 1. ## Tangent to hyperbola Prove that the lines $y=mx\\pm a\\sqrt{m^2-1}$ are tangent to the hyperbola $x^2-y^2=a^2$ for all values of m. My work: I substituted $y=mx\\pm a\\sqrt{m^2-1}$ into the equation. $x^2-m^2x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-1)=a^2$ $(1-m^2)x^2\\mp 2am\\sqrt{m^2-1}x-a^2(m^2-2)=0$ If $y=mx\\pm a\\sqrt{m^2-1}$ is a tangent to the hyperbola, $b^2=4ac$ $\\rightarrow(2am\\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$ I solved for m and got $\\pm 1$ So this is definitely wrong, but I don't know what is wrong. Thanks for any help! 2. Check your b^2 = 4ac calculation. 3. this is my working: $4a^2m^2(m^2-1)=-4a^2(1-m^2)(m^2-2)$ $4a^2m^4-4a^2m^2=-4a^2(3m^2-2-m^4)$ $4a^2m^4-4a^2m^2=-12a^2m^2+8a^2+4a^2m^4$ $8a^2m^2=8a^2$ $m^2=\\frac{8a^2}{8a^2}=1$ $m=\\pm 1$ 4. The last term is -a^2*m^2 + a^2 not -a^2*m^2 -a^2 5. huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. 6. Originally Posted by arze huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result. The above equation becomes x^2(1-m^2) + 2amsqrt(m^2-1) -a^2*m^2 = 0 In this b^2 = 4ac. 7. oh! ok i see where i got it wrong. thanks!", "$y= x^2+ x$ which has derivative $2x_0$ when x= x_0. The slope of any tangent to that parabola must be $2x_0$ for some number $x_0$. Also, any line through (2, 3) is of the form y- 3= m(x- 2) so a line through (2, 3) tangent to the parabola is of the form $y- 3= (2x_0)(x- 2)$. Of course, the tangent line must touch the parabola at $x= x_0$, that is when $x= x_0$, $y= x_0^2+ x_0$ That is, you must have $x_0^2+ x_0- 3= 2x_0(x_0- 2)$. Solve that for $x_0$ and then write the equation of the tangent lines. (That is a parabola so there will be two solutions.) 8. Originally Posted by metallica007 Hi guys Thanks Here comes a slightly different approach to this problem: 1. A line passing through (2, -3) has the equation: $y-(-3)=m(x-2)~\\implies~y=mx-2m-3$ 3. Calculate the coordinates of the points of intersection of the parabola and the line: $x^2+x=mx-2m-3~\\implies~x^2+(1-m)x+2m+3=0$ Solve for x: $x = \\dfrac{-(1-m) \\pm \\sqrt{(1-m)^2-4 \\cdot 1 \\cdot (2m+3)}}{2}$ $x = \\dfrac{m-1}2 \\pm \\dfrac12 \\cdot \\sqrt{m^2-10m-11}$ 3. A line is tangent to the parabola if there exist only one point of intersection: the tangent point. This is only possible if the dicriminant equals zero: $m^2-10m-11=0~\\implies~m=-1~\\vee~m=11$ Plug in these values into the equation of the line at 2. 4. The x-coordinate of", "Thus $$\\boxed{f(x) = \\frac35 x^2}$$ Case $1$: Of type $x^2=4ay$. If $(10,60)$ passes through this parabola, then we get, $$100=4a (60)\\Rightarrow 4a = \\frac {5}{3}$$ The parabola is: $$x^2 =\\frac {5}{3} y$$ Case $2$: Of type $y^2=4ax$. If $(10,60)$ passes through this parabola, then we get, $$3600=4a (10) \\Rightarrow 4a =360$$ The parabola is: $$y^2=360x$$ Hope it helps. Let $f(x) = ax^2 + bx + c$ for some $a,b,c \\in \\mathbb{R}$ be the wanted parabola. Then • one root at $(0, 0) \\rightarrow f(0) = 0 = a \\cdot 0^2 + b \\cdot 0 + c \\rightarrow c = 0$ • $f(10) = 60 = a \\cdot 10^2 + b \\cdot 10 = 100a + 10b \\rightarrow 6 = 10 \\cdot a + b \\rightarrow b = 6 - 10a$ • only one root at $(0, 0) \\rightarrow \\Delta(0) = 0 = b^2-4ac = (6 - 10a)^2 - 4a\\cdot 0 = (6 - 10a)^2 = 0 \\iff 6 - 10a = 0 \\iff a = \\frac{3}{5 }$ so all the parabolae in the form $$f(x) = \\frac{3}{5} \\cdot x^2$$ satisfy the requirements. • This function will have a second root at $x=10 -\\frac6a$ (if $a\\neq 0$). I think OP wants the only root to be at $x=0$, doesn't he? – MPW Feb 20 '17 at 13:47 • yup!", "# For what values of m is line y=mx tangent to the hyperbola x^2-y^2=1? Mar 31, 2018 $m = \\pm 1$ and tangents are $y = x$ and $y = - x$ #### Explanation: Put $y = m x$ in te equation of hyperbola ${x}^{2} - {y}^{2} = 1$, then ${x}^{2} - {m}^{2} {x}^{2} = 1$ or ${x}^{2} \\left(1 - {m}^{2}\\right) - 1 = 0$ the values of $x$ will give points of intersection of $y = m x$ and ${x}^{2} - {y}^{2} = 1$. But as $y = m x$ is a tangent, weshould get only one root, which would be wwhen discriminant is zero i.e. ${0}^{2} - 4 \\cdot \\left(1 - {m}^{2}\\right) \\cdot \\left(- 1\\right) = 0$ or $4 - 4 {m}^{2} = 0$ i.e. $m = \\pm 1$ and tangents are $y = x$ and $y = - x$ graph{(x^2-y^2-1)(x+y)(x-y)=0 [-10, 10, -5, 5]} Mar 31, 2018 $\\text{Slope of tangent } = \\frac{\\mathrm{dy}}{\\mathrm{dx}}$ Therefore, slope of tangent of x^2−y^2=1 => dx^2/dx−dy^2/dx=d1/dx $\\implies 2 x - 2 y \\frac{\\mathrm{dy}}{\\mathrm{dx}} = 0$ $\\implies \\frac{\\mathrm{dy}}{\\mathrm{dx}} = \\frac{x}{y}$ For what values of m is line $y = m x$ tangent to the hyperbola x^2−y^2=1? $\\implies m = \\frac{x}{y}$", "\\end{pmatrix} =0$$ That is $$\\color{red}{c'}=\\frac{af^2+bg^2-2fgh}{ab-h^2}$$ The asymptotes are $$\\fbox{ax^2+2hxy+by^2+2gx+2fy+\\frac{af^2+bg^2-2fgh}{ab-h^2}=0 \\,}$$ Alternatively, using the centre of the conics $$\\left( \\frac{bg-fh}{h^2-ab}, \\frac{af-gh}{h^2-ab} \\right)$$ and the slope $m$ of an asymptote is given by $$a+2hm+bm^2=0$$ On solving, $$m=\\frac{-h \\pm \\sqrt{h^2-ab}}{b}=\\frac{a}{-h \\mp \\sqrt{h^2-ab}}$$ Therefore $$\\fbox{ y-\\frac{af-gh}{h^2-ab}= \\frac{-h \\pm \\sqrt{h^2-ab}}{b} \\left( x-\\frac{bg-fh}{h^2-ab} \\right) }$$ • Hello @Ng Chung Tak I tried to use your equation to compute the asymptotes of the equilateral hyperbola $Ax+By+xy-C=0$, but it does not work, I get $y=\\frac{-4Ax+(2B-1)AB}{4(B+x)}$, instead the right ones are $y=-A$ and $x=-B$. Can you help me to understand where I did a mistake please? – Gennaro Arguzzi May 13 at 9:16 • @GennaroArguzzi You must have made some computing error: you should get $c'=AB$, which then gives the right equations. – Aretino May 13 at 10:52 • @GennaroArguzzi Note that $$\\frac{-h \\pm \\sqrt{h^2-ab}}{b}=\\frac{a}{-h \\mp \\sqrt{h^2-ab}}$$ re-arranging the denominators and putting $a=b=0$ you may get $hx+f=0$ and $hy+g=0$ for each cases. – Ng Chung Tak May 13 at 13:34 • @GennaroArguzzi It is $2g=A$. – Aretino May 13 at 16:30 • Solve for $x$ instead of $y$. Or, better, factor the equation. – Aretino May 13 at 17:10", "has equal roots. Discriminant = 0 ⇒ (-32m2)2 – 4(m)(-32) = 0 ⇒ 1024 m4 + 128m = 0 ⇒ 128m (8m3 + 1) = 0 ⇒ 8m3 + 1 = 0 …..[∵ m ≠ 0] ⇒ m3 = $$-\\frac{1}{8}$$ ⇒ m = $$-\\frac{1}{2}$$ Substituting m = $$-\\frac{1}{2}$$ in (i), we get ⇒ $$y=-\\frac{1}{2} x+\\frac{1}{\\left(-\\frac{1}{2}\\right)}$$ ⇒ $$y=-\\frac{1}{2} x-2$$ ⇒ x + 2y + 4 = 0, which is the equation of the common tangent. Question 18. Find the equation of the locus of a point, the tangents from which to the parabola y2 = 18x are such that sum of their slopes is -3. Solution: Given equation of the parabola is y2 = 18x Comparing this equation with y2 = 4ax, we get ⇒ 4a = 18 ⇒ a = $$\\frac{9}{2}$$ Equation of tangent to the parabola y2 = 4ax having slope m is ⇒ y = mx + $$\\frac{a}{m}$$ ⇒ y = mx + $$\\frac{9}{2m}$$ ⇒ 2ym = 2xm2 + 9 ⇒ 2xm2 – 2ym + 9 = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1 + m2 = $$-\\frac{(-2 y)}{2 x}=\\frac{y}{x}$$ But, m1 + m2 = -3 $$\\frac{y}{x}$$ = -3 y = -3x, which is the required equation of locus. Question 19. The towers of a bridge,", "be taken as y – 2 = m(x – 4) = mx – 4m mx – y + (2 – 4m) = 0 ………… (1) This is a tangent to the circle x² + y² + 4x + 2y – 4 = 0 Squaring and cross-multiplication is (1 – 2m)² = (m² + 1) 4m² – 4m + 1 = m² + 1 3m² – 4m = 0 m(3m-4) = 0 m,= 0 or = $$\\frac{4}{3}$$ m = 0 Substituting in (1), equation of the tangent is y + 2 = 0 or y -2 = 0 m = $$\\frac{4}{3}$$ Substituting in (1), equation of the tangent is $$\\frac{4}{3}$$x – y + (2 – $$\\frac{16}{3}$$) = 0 ⇒ $$\\frac{4}{3}$$x – y –$$\\frac{10}{3}$$ = 0 4x – 3y – 10 = 0 Internal centre of similitude S’ divides AB internally in the ratio 3 : 1 Co-ordinates of S’ are ($$\\frac{6-2}{3+1}$$, $$\\frac{3-1}{3+1}$$x) = (1, $$\\frac{1}{2}$$) Equation of the tangent can be taken as y – $$\\frac{1}{2}$$ = m(x – 1) = mx – m mx – y + ($$\\frac{1}{2}$$ – m) = 0 ………. (2) This is a tangent to the circle x² + y² + 4x + 2y – 4 = 0 Squaring and cross-multiply is (1 – 2m)² = 4(m² + 1) 1 + 4m² – 4m", "that E( R ) = 3, and find Var( R ). If each ball is replaced before another is drawn, show that in this case E( R ) = 4. ### Solutions to Review 1 Question 1 (i) $y = f(x) = \\frac{x^2 + 14x + 50}{3(x+7)}$ $3y(x+7) = x^2 + 14x + 50$ $x^2 + (14-3y)x + 50 - 21 y = 0$ $\\text{discriminant} \\ge 0$ $(14-3y)^2 - 4(1)(50-21y) \\ge 0$ $196 - 84y + 9y^2 - 200 + 84y \\ge 0$ $9y^2 - 4 \\ge 0$ $(3y - 2)(3y + 2) \\ge 0$ $y \\le - \\frac{2}{3} \\text{~or~} y \\ge \\frac{2}{3}$ (ii) Using long division, we find that $y = \\frac{x^2 + 14x + 50}{3(x+7)} = \\frac{x}{3} + \\frac{7}{3} + \\frac{1}{3(x+7)}$ So the asymptotes are $y = \\frac{x}{3} + \\frac{7}{3}$ and $x = -7$ Question 2 (i) $x^2 - 9y^2 + 18y = 18$ $x^2 - 9(y^2 - 2y) = 18$ $x^2 - 9[(y-1)^2 - 1^2] = 18$ $x^2 - 9(y-1)^2 + 9 = 18$ $x^2 - 9(y-1)^2 = 9$ $\\frac{x^2}{9} - (y-1)^2 = 1$ This is a hyperbola with centre $(0, 1)$, asymptotes are $y = \\pm \\frac{x}{3} + 1$, and vertices $(3, 1)$ and $(-3, 1)$. $y = \\frac{1}{x^2} + 1$ is a graph with asymptotes $x = 0$ and $y=1$. Use GC to", "$x - 2 = X , y - 2 = Y$, we have, ${X}^{2} / 1 - {Y}^{2} / 9 = 1$, which represents a Hyperbola . Comparing with, ${X}^{2} / {a}^{2} - {Y}^{2} / {b}^{2} = 1$, we have, ${a}^{2} = 1 , {b}^{2} = 9$ Hence the Vertices, in $\\left(X , Y\\right)$ system, are (+-a,0)=(+-1,0) and (0,+-b)=(0,+-3) In $\\left(x , y\\right)$, they are, $x - 2 = \\pm 1 , y - 2 = 0$, i.e., $x = 2 \\pm 1 , y = 2$, or, (3,2), &, (1,2). $\\left(2 , 2 \\pm 3\\right) , i . e . , \\left(2 , 5\\right) \\mathmr{and} \\left(2 , - 1\\right)$ are also vertices. The Eccentricity $e$ is given by, ${b}^{2} = {a}^{2} \\left({e}^{2} - 1\\right) \\Rightarrow 9 = {e}^{2} - 1 \\Rightarrow {e}^{2} = 10 \\Rightarrow e = \\sqrt{10}$. Focii , in $\\left(X , Y\\right)$ system, are $\\left(\\pm a e , 0\\right) = \\left(\\pm \\sqrt{10} , 0\\right)$ $\\therefore$, In $\\left(x , y\\right)$ system, Focii are given by, $x - 2 = \\pm \\sqrt{10} , y - 2 = 0$ i.e., 3x=2+-sqrt10, y=2#. $\\therefore$ Focii, in original system, are $\\left(2 \\pm \\sqrt{10} , 2\\right)$ Asymptotes in $\\left(X , Y\\right)$ are, ${X}^{2} / 1 - {Y}^{2} / 9 = 0$, i.e., $Y = \\pm 3 X$. So, in $\\left(x ,", "one of x and y. 8. O.K..I will solve the first one for you and I am sure you will be able to pick up from there. Let $y=mx +c$ be the required line. Solving with $y^2=8x$yields $(mx+c)^2=8x$ $m^2x^2+2(mc-4)x+c^2=0$ which is a quadratic in $x$ which must have $one$ root. $Discriminant=0$ $(mc-4)^2-m^2c^2=0$ $c=\\frac{2}{m}$.....................(1) Now solving $y=mx+c$ and $xy=-1$ yields $x(mx+c)=-1$ $mx^2+cx+1=0$ which is a quadratic in $x$ which must have one root. $Discriminant=0$ $c^2-4m=0$ $c^2=4m$..........................(2) Solving (1) and (2) we get $m=1$ and $c=2$ Therefore,the equation of the common tangent is $y=x+2$. Apply the same method to other problems 9. Thanks for the help pankaj! I've tried that method on a specific example I'm working on. The two equations are the circle: $X^2 + Y^2 = 1$ and and ellipse: $((X-2)^2)/3 + Y^2 = 1$ The first equation in terms of m and c (by setting the determinant equal to zero) is $m^2 -c^2 +1 = 0$ The second equation in terms of m and c (by setting the determinant equal to zero) is $12-48mc - 12c^2 - 12m^2 = 0$ Solving these equations yields the following expression for c: $3c^4 - 2c^2 -1 = 0$ Now when I solve this in maple, and plot the resultant line, it does not seem to be tangential to the two", "Browse Questions # Find the equation of the tangent and normal to the ellipse $\\large\\frac{x^{2}}{20}+\\frac{y^{2}}{5}$$=1, Which are perpendicular to x+y+z=0 Can you answer this question? ## 1 Answer 0 votes Toolbox: • Equation of any tangent to • \\quad(i) The parabola y^2=4ax is of the form y=mx+\\large\\frac{a}{m} • \\quad(ii) The ellipse \\large\\frac{x^2}{a^2}+\\frac{y^2}{b^2}$$=1$ is of the form $y=mx\\pm\\sqrt{a^2m^2+b^2}$ • $\\quad(iii)$ The hyperbola $\\large\\frac{x^2}{a^2}-\\frac{y^2}{b^2}$$=1 is of the form y=mx\\pm\\sqrt{a^2m^2-b^2} Step 1: Tangent to \\large\\frac{x^2}{20}+\\frac{y^2}{5}$$=1$ which is $\\perp$ to the line $x+y+z=0$ Compare with $\\large\\frac{x^2}{a^2}+\\frac{y^2}{b^2}$$=1$, We have $a^2=20,b^2=5$ Any tangent i of the form $y=mx\\pm\\sqrt{a^2m^2+b^2}$ (i.e)$y=mx\\pm\\sqrt{20m^2+5}$ Step 2: The required tangents are $\\perp$ to $x+y+z=0$ whose slope is $-1$. Therefore slope of the tangent =$1=m$ The equation of the tangents are $y=x\\pm\\sqrt{20+5}$ (i.e) $y=x\\pm \\sqrt{25}$ $y=x\\pm 5$", "\\ge \\max(0, -\\frac {c_2}{c_3})$. By squaring we added extraneous solutions.) $a^2x^4 - 2a(y-c_1)x^2 + (y-c_1)^2 = c_2 + c_3 x^2$ $a^2x^4 - [2a(y-c_1)- c_3]x^2 +[(y-c_1)^2 - c_2] = 0$ So $x^2 = \\frac { [2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$. We have some restrictions. $[2a(y-c_1)- c_3]^2 +4a^2c_2$ must be non-negative so that $4a^2c_2 \\ge - [2a(y-c_1)- c_3]^2$. And as $x^2 \\ge 0$. We must have $[2a(y-c_1)- c_3] \\ge - \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ and if $[2a(y-c_1)- c_3] < \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ we will not accept $x^2 = \\frac { [2a(y-c_1)- c_3]- \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$ as a solution. So $x = \\pm\\sqrt{\\frac { [2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}}=\\pm \\frac{\\sqrt{[2a(y-c_1)- c_3]\\pm \\sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}}{\\sqrt {2} a}$ With the restriction that everything under the radicals are positive (If you have wrong values of $y, c_1,c_2,c_3, a$ it just won't work) and the restriction that $x^2 \\ge \\max(0, -\\frac {c_2}{c_3})$. Or in other words either $x\\ge \\sqrt{\\max(0, -\\frac {c_2}{c_3})}$ or $x \\le - \\sqrt{\\max(0, -\\frac {c_2}{c_3})}$. There could be as many as $4$ solutions or as few as $0$ depending on the restrictions. Note: If $\\frac {c_2}{c_3} \\ge 0$ then this is not an issue. • wow...I did not expect line by line solutions, but thanks anyway, I will read that. – Codelearner777 Dec 18 '17 at 19:13", "of the systems$$ \\begin{cases} y=mx+q\\\\ x^2=-32y \\end{cases} \\qquad \\begin{cases} y=mx+q\\\\ y^2=4x \\end{cases} $$have zero discriminant. The resolvent equations are$$ x^2+32mx+32q=0 \\qquad m^2x^2+(2mq-4)x+q^2=0 $$so we get$$ \\begin{cases} 32^2m^2-4\\cdot32q=0\\\\[4px] ... 1 You can prove that the tangent line to an ellipse of equation $$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$$ at a point$(s,t)$on the ellipse is $$\\frac{sx}{a^2}+\\frac{ty}{b^2}=1$$ You can rewrite the equations of your tangent lines as $$\\frac{x}{\\sqrt{39}}+\\frac{2y}{\\sqrt{39}}=1, \\qquad \\frac{-x}{7}+\\frac{3y}{7}=1$$ Thus there must exist$(s,t)$and$(u,v)$... 0 So$\\tan\\alpha = \\frac{b}{a}$and$c^2 - a^2 = b^2$So we have:$c^2 = a^2 + b^2c^2 = a^2 + (a^2)tan^2\\alphac = \\sqrt{a^2 + (a^2)tan^2\\alpha}c = a\\sqrt{1 + tan^2\\alpha}c = a\\sqrt{sec^2\\alpha}c = (a)(sec\\alpha)\\frac{c}{a} = sec\\alpha\\epsilon = sec\\alpha$1 It likely is a conditional extremum problem and I don't know what to do next. I don't know what to do next either. So, let us take another approach. If the line$AB$is parallel to$y$-axis, we can easily see that the area of the triangle is$\\sqrt 3/2$. If the line$AB$is not parallel to$y$-axis, we can set$AB : y=ax+b$. Then, ... 0 My instinct says that this curved stretching is the wrong approach. Whatever answer you get will be useless for your CAD program. Here's a plot of the curves and the unaltered ellipse: My method would be to use your CAD program to draw a line that is tangent to the lower half of the ellipse and the blue", "Comment Share Q) # If the line $y=mx+c$ touches the hyperbola $\\large\\frac{x^2}{a^2}+\\frac{y^2}{b^2}$$=1 then find the value of c . \\begin{array}{1 1}(a)\\;\\pm \\sqrt{a^2m^2-b^2}\\\\(b)\\;\\pm \\sqrt{a^2m^2+b^2}\\\\(c)\\;\\pm \\sqrt{b^2m^2-a^2}\\\\(d)\\;\\text{None of these}\\end{array} ## 1 Answer Comment A) Toolbox: • For equal roots of the quadratic equation Ax^2+Bx+c=0, Discriminant B^2-AC=0 Given Equation of the hyperbola is \\large\\frac{x^2}{a^2}-\\large\\frac{y^2}{b^2}$$=1$...(i) Also given that $y=mx+c$......(ii) is tangent to the hyperbola Solving (i) and (ii) we get $\\large\\frac{x^2}{a^2}-\\frac{(mx+c)^2}{b^2}$$=1$ $\\Rightarrow\\:b^2x^2-a^2(mx+c)^2=a^2b^2$ $\\Rightarrow\\:(a^2m^2-b^2)x^2+2mca^2x+a^2(b^2+c^2)=0$ If the line touches the hyperbola,then the roots of this quadratic eqn. are equal. $\\Rightarrow\\:$ The discriminant is $0.$ $B^2=4AC$ $\\Rightarrow\\:4m^2c^2a^4-4(a^2m^2-b^2)a^2(b^2+c^2)=0$ $\\Rightarrow\\:c^2=a^2m^2-b^2$ $\\Rightarrow\\:c=\\pm \\sqrt{a^2m^2-b^2}$ Hence (a) is the correct answer.", "## Calculus, 10th Edition (Anton) (a)$$x_1=4, \\qquad x_2=2$$ (b) $y$-coordinate of no point of the parabola is $10$. (c)$$x \\in \\mathbb R -(2,4)$$ (d) This function has no maximum. $(3,-1)$ is the minimum point of this function. (a) Let us find the values of $x$ for which $x^2-6x+8=0$.$$x_{1,2}=\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\\quad a=1, \\, b=-6, \\, c=8 \\\\ \\Rightarrow \\quad x_{1,2}=\\frac{6 \\pm \\sqrt{36-32}}{2} \\quad \\Rightarrow \\quad x_1=4, \\, x_2=2$$ (b) Let us find the values of $x$ for which $x^2-6x+8=-10$.$$x_{1,2}=\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\\quad a=1, \\, b=-6, \\, c=18$$The above equation has no root since$$b^2-4ac=36-4 \\cdot 18<0.$$Thus, $y$-coordinate of no point of the parabola is $10$. (c) Let us find the values of $x$ for which $x^2-6x+8 \\ge 0.$ $$\\Rightarrow \\quad (x-4)(x-2) \\ge 0 \\\\ \\Rightarrow \\quad x \\ge 4, \\, x \\ge 2 \\quad \\text{ or } \\quad x \\le 4, \\, x \\le 2 \\\\ x \\in \\mathbb R- (2,4).$$ (d) $y=x^2-6x+8=(x-4)(x-2)$ has no maximum since for example, for $x_1, x_2 \\ge 4$ and $x_1 < x_2$, we have $y_1=(x_1-4)(x_1-2) > (x_2-4)(x_2-2)$, so there is no upper bound for $y=x^2-6x+8$. However, $y=x^2-6x+8$ has minimum since it represents a parabola. Since the minimum is the midpoint of any two points on the parabola having the same $y$-coordinate, we can find the $x$-coordinate of the minimum point by means of averaging the", "Browse Questions Find the equation of the two tangents that can be drawn, from the point $(1 , 2 )$ to the hyperbola $2x^{2}-3y^{2}=6$ Toolbox: • Equation of any tangent to • $\\quad(i)$ The parabola $y^2=4ax$ is of the form $y=mx+\\large\\frac{a}{m}$ • $\\quad(ii)$ The ellipse $\\large\\frac{x^2}{a^2}+\\frac{y^2}{b^2}$$=1 is of the form y=mx\\pm\\sqrt{a^2m^2+b^2} • \\quad(iii) The hyperbola \\large\\frac{x^2}{a^2}-\\frac{y^2}{b^2}$$=1$ is of the form $y=mx\\pm\\sqrt{a^2m^2-b^2}$ Step 1: $2x^2-3y^2=6$ The above equation is divided by $6$ $\\large\\frac{x^2}{3}-\\large\\frac{y^2}{2}$$=1 Therefore a^2=3,b^2=2 Any tangent to \\large\\frac{x^2}{a^2}-\\frac{y^2}{b^2}$$=1$ is of the form $y=mx\\pm\\sqrt{3m^2-2}$ Let the required tangent to this ellipse be $y=mx+\\sqrt{3m^2-2}$ Step 2: It passes through $(1,2)$ Therefore $2=m+\\sqrt{3m^2-2}$ $(2-m)^2=3m^2-2$ $4+m^2-4m=3m^2-2$ $2m^2+4m-6=0$ $m^2+2m-3=0$ $m=\\large\\frac{-2\\pm \\sqrt{4+12}}{2}$ $\\Rightarrow\\large\\frac{ -2\\pm 4}{2}$$=-3,1$ Step 3: Using point slope form,since they pass through $(1,2)$ When $m=1$ $y-2=1(x-1)$ $x-y+1=0$ When $m=-3$ $y-2=-3(x-1)$ $(y-2)=-3x+3$ $3x+y-5=0$ Step 4: The two tangents are $x-y+1=0,3x+y-5=0$", "\\}$ and $y = 1/x$ – caverac Nov 21 '16 at 17:12 You can also directly combine the equations into complete binomial formulas $$(x+y)^2=x^2+y^2+2xy=4+2=6,\\\\ (x-y)^2=x^2+y^2-2xy=4-2=2$$ and solve the trivial linear system for each of the 4 sign combinations of the roots. • nice!$\\quad\\qquad$ – polfosol Nov 22 '16 at 5:30 The equation of the circle is $x^2 + y^2 = 4$ and the equation of the hyperbola is $xy=1$ So the point of intersection would be a common solution to $xy =1$ $x^2 + y^2 = 4$ so $y = 1/x$ $x^2 + \\frac 1{x^2} = 4$ $x^4 +1 = 4x^2$ $x^4 - 4x^2 + 1 = 0$ $x^2 = \\frac {4 \\pm \\sqrt {12}}2$ $x^2 = 2 \\pm \\sqrt 3$ $x = \\pm \\sqrt{2 \\pm \\sqrt{3}}$ $y = 1/x = \\pm \\frac 1{\\sqrt{2 \\pm \\sqrt{3}}}$ $= \\pm \\frac 1{\\sqrt{2 \\pm \\sqrt{3}}}\\frac {\\sqrt {2\\mp \\sqrt {3}}}{\\sqrt {2\\mp\\sqrt{3}}}$ $=\\pm \\frac{\\sqrt {2\\mp \\sqrt {3}}}{\\sqrt {4-3}}=\\pm {\\sqrt {2\\mp \\sqrt {3}}}$ So there are four points: $(\\sqrt{2 + \\sqrt{3}},{\\sqrt{2 - \\sqrt{3}}});(\\sqrt{2 - \\sqrt{3}},{\\sqrt{2 +\\sqrt{3}}});(-\\sqrt{2 + \\sqrt{3}},-{\\sqrt{2 - \\sqrt{3}}});(-\\sqrt{2 - \\sqrt{3}},-{\\sqrt{2 + \\sqrt{3}}});$ • You could simplify the $y$ coordinates using $1/(2\\pm\\sqrt{3}) = 2\\mp\\sqrt{3}$, which also shows the symmetry more clearly. – Klaus Draeger Nov 21 '16 at 21:22 • I could. But I personally don't like that. It treats the pm sign" ]

[ "# Points of tangency 1. Nov 17, 2004 ### ziddy83 I have the following function given, and I need to find the points of tangency... Two tangent lines to the huperbola $$9x^2 - y^2 =36$$ interesect at the y-axis. Use implicit differentiation to find the points of tangency. Ok so i multiplied both sides by dy/dx, and solved for y'...i got $$y'= \\frac {9x}{y}$$ so to find the points of tangency, do i plug in the x and y values? im a little confused on that...thanks. 2. Nov 17, 2004 ### ziddy83 Can anyone help? 3. Nov 17, 2004 ### ziddy83 anyone at all???", "the x values of the points of tangency, as given above.", "+0 # Help 0 214 0 The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\\sqrt{a}/b$, where $a$ is not divisible by the square of any prime. Find $a+b$. Mar 6, 2018", "The answer says \"points of tangency.\" The point of tangency of the line $y=1$ is $(0,1)$. – André Nicolas Aug 17 '11 at 16:21", "+0 # Help 0 72 0 The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\\sqrt{a}/b$, where $a$ is not divisible by the square of any prime. Find $a+b$. Guest Mar 6, 2018", "The tangency point Geometry Level 5 Let there be 2 graphs of 2 parabolas, $$(1): y = x^2$$, $$(2): x=y^2$$ as shown in the figure. The graph of the parabola $$(1)$$ moves along the $$y$$-axis from the bottom to the top as the parabola $$(2)$$ shifts to the right. What is the sum of coordinates of the tangency of the point? ×", "Log in or sign up to add this lesson to a Custom Course. Therefore, in our diagrams, sometimes it may appear that the tangent line touches at more than just one point. Visit the GMAT Prep: Help and Review page to learn more. Let f(x) = 12x + 11 - 7e^x. : Date: 23 October 2007: Source: Self-made This W3C-unspecified vector image was created with Inkscape. The tangency portfolio is the portfolio that one should hold. : Author: Inductiveload: Permission (Reusing this file) So, you find that the point of tangency is (2, 8); the equation of tangent line is y = 12x – 16; and the points of normalcy are approximately (–1.539, –3.645), (–0.335, –0.038), and (0.250, 0.016). 3. Remember that lines and curves in mathematics do not have a thickness to them. The x-axis intersects that parabola at a single point, labeled V. Point V is the point of tangency of the parabola and the x-axis. The police will net allow you to get close to the fence. add example. Find the length of line XY in the diagram below. For example, 0+00 indicates the beginning of the project. What is the tangency portfolio in this case? Therefore. All rights reserved. Plugging the points into y = x 3 gives you the three points: (–1.539, –3.645),", "# Least Distance Geometry Level 3 On the parabola $$y={ x }^{ 2 }$$, the point least distant from the straight line $$y=2x-1$$ is ×", "Thus, the Vertex Form of the Parabola is $$y=(x-3)^2-16$$ . Solving $$(x-3)^2-16=0$$ yields the two zeros: $$x=7 , x=-1$$ . Get it now? Try the above Complete the Square Solver again.", "# The vertex of the parabola $x^2+12x - 9y=0$ is : ( A ) $(-6, 4)$ ( B ) $(6, -4)$ ( C ) $(6,4)$ ( D ) $(-6, -4)$", "+0 +2 574 1 +598 Points $A$ and $B$ are on parabola $y=3x^2-5x-3$, and the origin is the midpoint of $\\overline{AB}$. Find the square of the length of $\\overline{AB}$. Jul 26, 2017 #1 +27475 +3 Perhaps this graph will help: . Jul 26, 2017", "## Question Find the point on the parabola y2=2x that is closest to the point (1, ). (You may round your answers to two decimal places.) (, )", "+0 # parabola 0 65 1 Find the vertex of the parabola x = -2y^2 + 12y - 15. Jun 30, 2020 We complete the square to get $$y=-2(y-3)^2+3$$. The vertex is (h, k). Therfore the vertex of the parabola $$x = -2y^2 + 12y - 15\\ \\text {is} \\ \\boxed{(3, 3)}.$$", "Z}{d y^2} = 24 - 8 = 16 > 0$ $\\text { So, the nearest point is }\\left( \\pm 2\\sqrt{2}, 4 \\right) .$ Is there an error in this question or solution? #### Video TutorialsVIEW ALL [1] Solution Find the Point on the Parabolas X2 = 2y Which is Closest to the Point (0,5) ? Concept: Graph of Maxima and Minima. S", "# How to find points of tangency on a hyperbola? If tangent lines to the hyperbola $9x^2-y^2=36 \\;$ intersect y-axis at point $(0,6)$, find the points of tangency. - The (hyperbolic-geometry) tag does not mean \"involves hyperbolas.\" :) – anon Oct 16 '11 at 9:40 This question is similar to the other question you asked. What have you tried so far in solving this one? – yunone Oct 16 '11 at 9:44 You do in the very same way I explained to you in the ellipse problem. The method works in principle for every plane algebraic curve and is very effective in the case of conics (where it reduces to an elementary question about quadratic equations): – Andrea Mori Oct 16 '11 at 9:49 @Muavia: You want a verification of what you've done without telling us what you've done? – anon Oct 16 '11 at 10:22 $\\frac{x^2}{4}-\\frac{y^2}{36}=1$ , then solve system: $\\begin{cases} y_0=kx_0+n \\\\ n^2=a^2k^2-b^2 \\end{cases}$ where $x_0=0 , y_0=6 ,a^2=4 ,b^2=36$", "crosses itself, and find the equations of the tangent line at this point. Example 4. Draw your example carefully, showing what the correct points of tangency are, and showing also the highest and lowest points, making it clear that they do not give the common tangent that we seek. So, if you have a graph with curves, like a parabola, it can have points of tangency as well. Find the length of the tangent in the circle shown below. Find the equation of the tangent line of the curve at the given point. add example. The tangent only touches the curve at one point. For example, if you put a square around a circle, then each side of the square has a … Plus, get practice tests, quizzes, and personalized coaching to help you credit-by-exam regardless of age or education level. Beschreibung Tangency Example 1.svg A diagram showing a curve (C, red), a line (L, blue) and the point of intersection point (P, black). So, the length of the tangent is 22.4 cm. 5x - xy + y^2 = 5 at (1, 1). So, let's put another point of tangency on this parabola. 16 = x 2 √16 = √x 2. x = 8. If DC = 20 inch and BC = 12 inch, calculate the radius shown", "the parabolic cylinder: x2+2ay=0.", "}^{ 3 } }{ 3 }$ about the x-axis. 42. Find the area enclosed by the parabolas 5x2-y=0 and 2x2-y+9=0.", "$y=x^2$ for $x$ to find the $y$ values and plot the points and draw your parabola.", "# Math Help - Turning point Parabolas Help 1. ## Turning point Parabolas Help How do you find the turning point of parabola: $y=2x^2+3x$ 2. Originally Posted by 22upon7 How do you find the turning point of parabola: $y=2x^2+3x$ i suppose this is for precalc? complete the square to get the parabola in its standard form, that is, the form $y = a(x - h)^2 + k$ the turning point is $(h,k)$" ]

[ "complete the square for $$x$$ and add $$9$$ to complete the square for $$y$$. It should be pretty simple from here. 4 - Since the axis of symmetry is $$x = 4$$, we can substitute that into $$y = a(x-h)^2+k$$ 5 - The simplified equation of $$y = a(x-h)^2+k$$ is $$y = a(x + 3)^2 - 2$$. We can also tell that the parabola passes through the point $$(-1,0)$$, so $$x = -1, ~ y = 0$$. Try substituting that in the equation and sees where that gets you. Hope that helped. AnxiousLlama May 28, 2021 #3 +36434 +2 1.) x = y^2 -8y + 13 re-arrange to vertex form x = (y-4)^2 -3 vertex is -3,4 May 28, 2021 #4 +36434 +2 2 ) -2x^2+8x-15 re-arrange to vertex form -2 (x^2 - 4x) -15 -2 ( x-2)^2 - 7 vertex = 2, -7 May 28, 2021 #5 +36434 +2 3) x^2 -2x + y^2 + 6y = 6 Arrange into standard circle equation form (x-1)^2 + ( y+3)^2 = 6 +1 + 9 6+ 1 + 9 = 16 = r^2 Area of a circle = pi r^2 = pi (16) = 16 pi units2 May 28, 2021 #6 +27 0 Can you answer the last two questions? Thank you so much for the other answers.... all", "Thus, the Vertex Form of the Parabola is $$y=(x-3)^2-16$$ . Solving $$(x-3)^2-16=0$$ yields the two zeros: $$x=7 , x=-1$$ . Get it now? Try the above Complete the Square Solver again.", "# What conic section has the equation x^2+y^2+12x+8y=48? Sep 13, 2014 This is an equation for a circle. You begin by reorganizing the terms of the function so that $x$ and ${x}^{2}$ are together and $y$ and ${y}^{2}$ are together. Next you will have to use the Completing the Square method. Step 1: Reorder the terms ${x}^{2} + 12 x + {y}^{2} + 8 y = 48$ Step 2: Begin Completing the square ${x}^{2} + 12 x + {y}^{2} + 8 y = 48$ ${\\left(\\frac{12}{2}\\right)}^{2} = {6}^{2} = 36$, Value to be added to complete the square ${\\left(\\frac{8}{2}\\right)}^{2} = {4}^{2} = 16$, Value to be added to complete the square ${x}^{2} + 12 x + 36 + {y}^{2} + 8 y + 16 = 48 + 36 + 16$ $\\left({x}^{2} + 12 x + 36\\right) + \\left({y}^{2} + 8 y + 16\\right) = 100$ Factor ${\\left(x + 6\\right)}^{2} + {\\left(y + 4\\right)}^{2} = 100$ Solution: Standard form of a Circle.", "Complete the square: $y=(x^2-12x +\\_\\_\\_)-\\_\\_\\_+7$ Take half the coefficient of x, square it, and add it to make a perfect square trinomial. But you have to subtract it also to keep the equation balanced. $y=x^2-12x+6^2-6^2+7$ $y=(x-6)^2-29$ The parabola opens up and has a vertex (6, -29).", "+0 # coordinates 0 21 1 In terms of pi, what is the area of the circle defined by the equation 2x^2+2y^2+10x-6y-48=0 Aug 2, 2022 #1 +2275 +2 Simplify: $$x^2 + y^2 + 5x - 3y - 24 = 0$$ Add $$1.5^2$$ and $$2.5^2$$ to both sides to complete the square: $$x^2 + y^2 + 5x - 3y - 24 + 1.5^2 + 2.5^2 = 1.5^2 + 2.5^2$$ Now, add 24 to both sides: $$x^2 + y^2 + 5x - 3y + 1.5^2 + 2.5^2 = 24 + 1.5^2 + 2.5^2$$ Complete the square on x: $$(x+2.5)^2 + y^2 - 3y - 24 = 1.5^2 + 2.5^2$$ Complete the square on y: $$(x+2.5)^2 + (y-1.5)^2 = 1.5^2 +2.5^2 + 24$$ Simplify the right-hand side: $$(x+2.5)^2 + (y - 1.5)^2 = 32.5$$ The radius of this circle is $$\\sqrt{32.5}$$, so the area is $$\\color{brown}\\boxed{32.5 \\pi}$$ Aug 2, 2022 #1 +2275 +2 Simplify: $$x^2 + y^2 + 5x - 3y - 24 = 0$$ Add $$1.5^2$$ and $$2.5^2$$ to both sides to complete the square: $$x^2 + y^2 + 5x - 3y - 24 + 1.5^2 + 2.5^2 = 1.5^2 + 2.5^2$$ Now, add 24 to both sides: $$x^2 + y^2 + 5x - 3y + 1.5^2 + 2.5^2 = 24 + 1.5^2 + 2.5^2$$ Complete the square on x:", "+0 # Help 0 177 1 Here is an equation of a parabola: y = 1/6 x^2 + 2x + 11 What are the coordinates of the vertex of the parabola? Mar 28, 2018 edited by Guest Mar 28, 2018 #1 +7348 +3 $$y=\\frac16x^2+2x+11$$ Multiply through by 6 . $$6y=x^2+12x+66$$ Subtract 66 from both sides of the equation. $$6y-66=x^2+12x$$ Add 36 to both sides to complete the square on the right side. $$6y-30=x^2+12x+36$$ Factor the right side as a perfect square trinomial. $$6y-30=(x+6)^2$$ Factor 6 out of the terms on the left side. $$6(y-5)=(x+6)^2$$ Now that the equation is in this form we can see that the vertex is (-6, 5) . Here's a graph to check it https://www.desmos.com/calculator/wrvhzfuag3 Mar 28, 2018 #1 +7348 +3 $$y=\\frac16x^2+2x+11$$ Multiply through by 6 . $$6y=x^2+12x+66$$ Subtract 66 from both sides of the equation. $$6y-66=x^2+12x$$ Add 36 to both sides to complete the square on the right side. $$6y-30=x^2+12x+36$$ Factor the right side as a perfect square trinomial. $$6y-30=(x+6)^2$$ Factor 6 out of the terms on the left side. $$6(y-5)=(x+6)^2$$ Now that the equation is in this form we can see that the vertex is (-6, 5) . Here's a graph to check it https://www.desmos.com/calculator/wrvhzfuag3 hectictar Mar 28, 2018", "square for x and y ... $x^2+2x+1+y^2+6y+9= \\frac{39}{4}+10$ $(x+1)^2 + (y+3)^2 = \\frac{79}{4}$ 4. ## Re: Circle and Radius, having to complete the square? Ah I see ok thanks! So basically it's like the tic tac toe method (if you guys heard of it), where you basically find numbers that would fit in. Ok thanks!", "– 10 y – 10 = x² + 6x To solve the given equation by using the completing the squares, We have to add 9 on both sides So, y – 10 + 9 = x² + 6x + 9 y – 1 = (x + 3)² y = (x + 3)² + 1 So, From the vertex form of the equation, The vertex point is: (-3, 1) From the above vertex point, We can observe that the x-coordinate is negative So, The parabola closes down and the y-coordinate will be the maximum value of the given equation Hence, from the above, We can conclude that The given equation has a maximum value and the maximum value of the given equation is: 1 Question 43. y = -x2 – 10x – 30 Question 44. y = -x2 + 14x – 34 The givne equation is: y = -x² + 14x – 34 Add with 34 on both sides y + 34 = -x² + 14x + 34 – 34 y + 34 = -x² + 14x y + 34 = – (x² – 14x) To solve the given equation by using the completing the squares, We have to subtract 49 on both sides So, y + 34 – 49 = – (x² – 14x + 49) y –", "- 4)^2}{(\\sqrt{27})^2} = 1$ which is the form for a hyperbola. The rest are done in a similar fashion. -Dan • Nov 12th 2007, 10:25 PM xterminal01 Thank you I appreciate it, I have tried to do the # 2 and 3 myself and this is what i came up with 2. 4x^2+9y^2+24x+18y+12=0 4x^2+9y^2+24x+18y=-12 4(x^2+6x+9)+9(y^2+2y+1)=-12+36+9 4(x+3)^2 + 9(y+1)^2 = 33 (x+3)^2 + (y+1)^2 33 33 = 1 4 9 and # 3 i guess is a parabola and this is what i got y^2+10y=2x-1 (y^2+10y+25) = 2x-1+25 (y+5)^2 = 2x + 24 (y+5)^2 = 2(x+12) Please let me know if i did the rest correct :) • Nov 12th 2007, 11:23 PM earboth Quote: Originally Posted by xterminal01 Thank you I appreciate it, I have tried to do the # 2 and 3 myself and this is what i came up with 2. 4x^2+9y^2+24x+18y+12=0 4x^2+9y^2+24x+18y=-12 4(x^2+6x+9)+9(y^2+2y+1)=-12+36+9 4(x+3)^2 + 9(y+1)^2 = 33 (x+3)^2 + (y+1)^2 33 33 = 1 4 9 (how about some squares as denominator?) and # 3 i guess is a parabola and this is what i got y^2+10y=2x-1 (y^2+10y+25) = 2x-1+25 (y+5)^2 = 2x + 24 (y+5)^2 = 2(x+12) Please let me know if i did the rest correct :) Hi, (And now it's time to learn a little bit $L_A T^E \\chi$)", "1)2 = 22 d. ($x$ - 2)2 + ($y$ - 1)2 = 22 Solution: x2 + y2 - 4x + 2y + 1 = 0 [Equation of conic section.] x2 - 4x + y2 + 2y = - 1 [Group x and y - terms.] (x2 - 4x + ?) + (y2 + 2y + ?) = - 1 (x2 - 4x + 4) + (y2 + 2y + 1) = - 1 + 4 + 1 (x - 2)2 + (y + 1)2 = (2)2 [Complete the squares.] So, the standard form of the given equation is (x - 2)2 + (y + 1)2 = (2)2, which represents a circle with center (2, - 1) and radius 2. 8. What conic section is represented by the equation 9$x$2 + 4$y$2 - 18$x$ - 8$y$ - 23 = 0? a. a circle b. a parabola c. a hyperbola d. an ellipse Solution: 9x2 + 4y2 - 18x - 8y - 23 = 0 [Equation of conic section.] 9x2 - 18x + 4y2 - 8y = 23 [Group x - terms and y - terms.] 9(x2 - 2x) + 4(y2 - 2y) = 23 9(x2 - 2x + ?) + 4(y2 - 2y + ?) = 23 9(x2 - 2x + 1) + 4(y2 - 2y + 1) =", "# Is the equation 13x² + 13y² - 26x + 52y = -78 a line, parabola, ellipse, hyperbola, or circle? Jul 25, 2018 No - it contains no real points. #### Explanation: Given: $13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$ First note that all of the coefficients are divisible by $13$, so let's divide the whole equation by $13$ to get: ${x}^{2} + {y}^{2} - 2 x + 4 y = - 6$ Adding $1 + 4 = 5$ to both sides of the equation and rearranging slightly, this becomes: ${x}^{2} - 2 x + 1 + {y}^{2} + 4 y + 4 = - 1$ That is: ${\\left(x - 1\\right)}^{2} + {\\left(y + 2\\right)}^{2} = - 1$ Note that this is always false for real values of $x$ and $y$, so this equation describes an empty set of values in the $x$ $y$ plane. Imaginary circle #### Explanation: The given equation: $13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$ $13 \\left({x}^{2} + {y}^{2} - 2 x + 4 y\\right) = - 78$ ${x}^{2} + {y}^{2} - 2 x + 4 y = - 6$ $\\left({x}^{2} - 2 x + 1\\right) + \\left({y}^{2} + 4 y + 4\\right) - 5 = - 6$ ${\\left(x -", "an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \\sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \\sqrt {25x^2-48x+36}$ = $0$. We then add $5 \\sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \\sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\\frac {48+0}{50}$. Therefore, $x = \\frac {48}{50}$. Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\\boxed {(B)4.8}$ ~Arcticturn", "the square for the $$x$$, $$y$$ and $$z$$ variables. Example 1.28 Is $$2x^{2} + 2y^{2} + 2z^{2} - 8x + 4y - 16z + 10 = 0$$ the equation of a sphere? Solution Dividing both sides of the equation by $$2$$ gives $$\\nonumber x^{2} + y^{2} + z^{2} - 4x + 2y - 8z + 5 = 0$$ $$\\nonumber (x^{2} - 4x + 4) + (y^{2} + 2y + 1) + (z^{2} - 8z + 16) + 5 - 4 - 1 - 16 = 0$$ $$\\nonumber (x - 2)^{2} + (y + 1)^{2} + (z - 4)^{2} = 16$$ which is a sphere of radius $$4$$ centered at $$(2,-1,4)$$. Example 1.29 Find the points(s) of intersection (if any) of the sphere from Example 1.28 and the line $$x = 3 + t$$, $$y = 1 + 2t$$, $$z = 3 - t$$. Solution Put the equations of the line into the equation of the sphere, which was $$(x - 2)^{2} + (y + 1)^{2} + (z - 4)^{2} = 16$$, and solve for $$t$$: \\begin{align*} \\nonumber (3 + t - 2)^{2} + (1 + 2t + 1)^{2} + (3 - t - 4)^{2} &= 16\\\\ \\nonumber (t + 1)^{2} + (2t + 2)^{2} + (-t - 1)^{2} &= 16\\\\ \\nonumber 6t^{2} + 12t -10 &= 0 \\end{align*}", "# How do I use completing the square to convert the general equation of a hyperbola to standard form? Oct 25, 2014 I can not remember how to deal with $x y$ so I will be demonstrating completing the square without one. Consider the equation below $9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$ The first thing that we should do is group our $x$s and $y$s $9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$ $\\implies \\left(9 {x}^{2} - 36 x\\right) + \\left(- 16 {y}^{2} - 96 y\\right) - 252 = 0$ Now, let's make our work easier by factoring out ${x}^{2}$'s and ${y}^{2}$'s coefficient $\\left(9 {x}^{2} - 36 x\\right) + \\left(- 16 {y}^{2} - 96 y\\right) - 252 = 0$ $\\implies 9 \\left({x}^{2} - 4 x\\right) + - 16 \\left({y}^{2} + 6 y\\right) - 252 = 0$ Before proceeding, let's recall what happens when a binomial is squared ${\\left(a x + b\\right)}^{2} = {a}^{2} {x}^{2} + 2 a b + {b}^{2}$ when a = 1, we have ${\\left(x + b\\right)}^{2} = {x}^{2} + 2 b + {b}^{2}$ Now, in our equation, $9 \\left({x}^{2} - 4 x\\right) + - 16 \\left({y}^{2} + 6 y\\right) - 252 = 0$ We want ${x}^{2} - 4 x$", "## Elementary Geometry for College Students (6th Edition) $(x,y,z) = (-3,-3,-6)$ $(x,y,z) = (3,3,6)$ $(x,y,z) = (0,0,0) + n(1,1,2) = (n, n, 2n)$ We can find the two values of $n$ such that the points on the line satisfy the equation of the sphere: $x^2+y^2+z^2 = 54$ $(n)^2+(n)^2+(2n)^2 = 54$ $6n^2 = 54$ $n^2 = 9$ $n^2-9 = 0$ $(n+3)(n-3) = 0$ $n=-3~~$ or $~~n=3$ We can find the two points: $(x,y,z) = (0,0,0) + (-3)(1,1,2) = (-3,-3,-6)$ $(x,y,z) = (0,0,0) + (3)(1,1,2) = (3,3,6)$", "for any point $$(x,y)$$ on the hyperbola. We know that the difference of these distances is $$2a$$ for the vertex $$(a,0)$$. It follows that $$d_2−d_1=2a$$ for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. \\begin{align*} d_2-d_1&=2a\\\\ \\sqrt{{(x-(-c))}^2+{(y-0)}^2}-\\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\\qquad \\text{Distance Formula}\\\\ \\sqrt{{(x+c)}^2+y^2}-\\sqrt{{(x-c)}^2+y^2}&=2a\\qquad \\text{Simplify expressions.}\\\\ \\sqrt{{(x+c)}^2+y^2}&=2a+\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Move radical to opposite side.}\\\\ {(x+c)}^2+y^2&={(2a+\\sqrt{{(x-c)}^2+y^2})}^2\\qquad \\text{Square both sides.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\\qquad \\text{Expand the squares.}\\\\ x^2+2cx+c^2+y^2&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\\qquad \\text{Expand remaining square.}\\\\ 2cx&=4a^2+4a\\sqrt{{(x-c)}^2+y^2}-2cx\\qquad \\text{Combine like terms.}\\\\ 4cx-4a^2&=4a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Isolate the radical.}\\\\ cx-a^2&=a\\sqrt{{(x-c)}^2+y^2}\\qquad \\text{Divide by 4.}\\\\ {(cx-a^2)}^2&=a^2{\\left[\\sqrt{{(x-c)}^2+y^2}\\right]}^2\\qquad \\text{Square both sides.}\\\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\\qquad \\text{Expand the squares.}\\\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\\qquad \\text{Distribute } a^2\\\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\\qquad \\text{Combine like terms.}\\\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\\qquad \\text{Rearrange terms.}\\\\ x^2(c^2-a^2)-a^2y^2&=a^2(c^2-a^2)\\qquad \\text{Factor common terms.}\\\\ x^2b^2-a^2y^2&=a^2b^2\\qquad \\text{Set } b^2=c^2−a^2\\\\. \\dfrac{x^2b^2}{a^2b^2}-\\dfrac{a^2y^2}{a^2b^2}&=\\dfrac{a^2b^2}{a^2b^2}\\qquad \\text{Divide both sides by } a^2b^2\\\\ \\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}&=1\\\\ \\end{align*} This equation defines a hyperbola centered at the origin with vertices $$(\\pm a,0)$$ and co-vertices $$(0,\\pm b)$$. STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER $$(0,0)$$ The standard form of the equation of a hyperbola with center $$(0,0)$$ and transverse axis on the $$x$$-axis is $$\\dfrac{x^2}{a^2}−\\dfrac{y^2}{b^2}=1$$ where • the length of the transverse axis is $$2a$$ • the coordinates of the vertices are $$(\\pm a,0)$$ • the length of the conjugate axis is $$2b$$", "The tangency point Geometry Level 5 Let there be 2 graphs of 2 parabolas, $$(1): y = x^2$$, $$(2): x=y^2$$ as shown in the figure. The graph of the parabola $$(1)$$ moves along the $$y$$-axis from the bottom to the top as the parabola $$(2)$$ shifts to the right. What is the sum of coordinates of the tangency of the point? ×", "origin 19. $2x−y=0,2x−y=−2,2x−y=2;2x−y=0,2x−y=−2,2x−y=2;$ three lines 21. $xx+y=−1,xx+y=0,xx+y=2xx+y=−1,xx+y=0,xx+y=2$ 23. $exy=12,exy=3exy=12,exy=3$ 25. $xy−x=−2,xy−x=0,xy−x=2xy−x=−2,xy−x=0,xy−x=2$ 27. $e−2x2=y,y=x2,y=e2x2e−2x2=y,y=x2,y=e2x2$ 29. The level curves are parabolas of the form $y=cx2−2.y=cx2−2.$ 31. $z=3+y3,z=3+y3,$ a curve in the $zy-planezy-plane$ with rulings parallel to the $x-axisx-axis$ 33. $x225+y24≤1x225+y24≤1$ 35. $x29+y24+z236<1x29+y24+z236<1$ 37. All points in $xyz-spacexyz-space$ 39. 41. 43. 45. 47. The contour lines are circles. 49. $x2+y2+z2=9,x2+y2+z2=9,$ a sphere of radius $33$ 51. $x2+y2−z2=4,x2+y2−z2=4,$ a hyperboloid of one sheet 53. $4x2+y2=1,4x2+y2=1,$ 55. $1=exy(x2+y2)1=exy(x2+y2)$ 57. $T(x,y)=kx2+y2T(x,y)=kx2+y2$ 59. $x2+y2=k40,x2+y2=k40,$ $x2+y2=k100.x2+y2=k100.$ The level curves represent circles of radii $10k/2010k/20$ and $k/10k/10$ ### Section 4.2 Exercises 61. 2.0 63. $2323$ 65. $11$ 67. $1212$ 69. $−12−12$ 71. $e−32e−32$ 73. $11.011.0$ 75. $1.01.0$ 77. The limit does not exist because when $xx$ and $yy$ both approach zero, the function approaches $ln0,ln0,$ which is undefined (approaches negative infinity). 79. every open disk centered at $(x0,y0)(x0,y0)$ contains points inside $RR$ and outside $RR$ 81. $0.00.0$ 83. $0.000.00$ 85. The limit does not exist. 87. The limit does not exist. The function approaches two different values along different paths. 89. The limit does not exist because the function approaches two different values along the paths. 91. The function $ff$ is continuous in the region $y>−x.y>−x.$ 93. The function $ff$ is continuous at all points in the $xy-planexy-plane$ except at $(0,0).(0,0).$ 95. The function is continuous at", "point P have coordinates $(x, y, z)$, A have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations$$(s-x)^2+y^2+z^2=(5\\sqrt{10})^2$$$$x^2+(s-y)^2+z^2=(5\\sqrt{5})^2$$$$x^2+y^2+(s-z)^2=(10\\sqrt{2})^2$$$$(s-x)^2+(s-y)^2+(s-z)^2=(3\\sqrt{7})^2$$These simplify into$$s^2+x^2+y^2+z^2-2sx=250$$$$s^2+x^2+y^2+z^2-2sy=125$$$$s^2+x^2+y^2+z^2-2sz=200$$$$3s^2-2s(x+y+z)+x^2+y^2+z^2=63$$Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$. Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$, so $x^2+y^2+z^2=256$. This means $PA=16$. However, we scaled down everything by 12 so our answer is $16*12=\\boxed{192}$. ~JHawk0224 Solution 2 (Solution 1 with slight simplification) Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive,$$2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.$$Subtracting the fourth equation gives,$$2(x^2 + y^2 + z^2) = 575 - 63$$$$x^2 + y^2 + z^2 = 256$$$$\\sqrt{x^2 + y^2 + z^2} = 16.$$Since point $A = (0,0,0), PA = 16$, and since we scaled the answer is $16 \\cdot 12 = \\boxed{192}$ ~Aaryabhatta1 Solution 3 Let E be the vertex of the cube such that ABED is a square. By the British Flag Theorem, we can easily we can show that$$PA^2 + PE^2 = PB^2 + PD^2$$and$$PA^2 + PG^2 = PC^2 + PE^2$$Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. Substituting in the values we know,", "anonymous one year ago Convert the equation to the standard form for a parabola by completing the square on x or y as appropriate. y2 + 2y - 2x - 3 = 0 A. (y + 1)2 = 2(x + 2) B. (y - 1)2 = -2(x + 2) C. (y + 1)2 = 2(x - 2) D. (y - 1)2 = 2(x + 2) 1. anonymous @cobra-strikes 2. anonymous when completing the square, take half of middle coefficient, square it, then add to both sides example $y^2 + 8y = 0$ $y^2 +8y + (\\frac{8}{2})^2 = (\\frac{8}{2})^2$ $y^2 +8y +16 = 16$ $(y+4)^2 = 16$" ]

[ "Log in or sign up to add this lesson to a Custom Course. Therefore, in our diagrams, sometimes it may appear that the tangent line touches at more than just one point. Visit the GMAT Prep: Help and Review page to learn more. Let f(x) = 12x + 11 - 7e^x. : Date: 23 October 2007: Source: Self-made This W3C-unspecified vector image was created with Inkscape. The tangency portfolio is the portfolio that one should hold. : Author: Inductiveload: Permission (Reusing this file) So, you find that the point of tangency is (2, 8); the equation of tangent line is y = 12x – 16; and the points of normalcy are approximately (–1.539, –3.645), (–0.335, –0.038), and (0.250, 0.016). 3. Remember that lines and curves in mathematics do not have a thickness to them. The x-axis intersects that parabola at a single point, labeled V. Point V is the point of tangency of the parabola and the x-axis. The police will net allow you to get close to the fence. add example. Find the length of line XY in the diagram below. For example, 0+00 indicates the beginning of the project. What is the tangency portfolio in this case? Therefore. All rights reserved. Plugging the points into y = x 3 gives you the three points: (–1.539, –3.645),", "The tangency point Geometry Level 5 Let there be 2 graphs of 2 parabolas, $$(1): y = x^2$$, $$(2): x=y^2$$ as shown in the figure. The graph of the parabola $$(1)$$ moves along the $$y$$-axis from the bottom to the top as the parabola $$(2)$$ shifts to the right. What is the sum of coordinates of the tangency of the point? ×", "tangency implies on $\\sqrt{(1-b) (b+1)^3}=0\\;$ hence the solutions for tangency are $b = \\pm 1$ etc. • Didn't understand why L should be a tangent and why it should tend to a Jul 15 '18 at 11:40 • Interesting method, though please explain the concepts used Jul 15 '18 at 11:40 • @Abcd The problem $\\min\\max u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = \\{u_{\\min},\\cdots,u_{\\max}\\}$. This is the idea of Lagrange multipliers method. Jul 15 '18 at 13:43", "# How to find points of tangency on a hyperbola? If tangent lines to the hyperbola $9x^2-y^2=36 \\;$ intersect y-axis at point $(0,6)$, find the points of tangency. - The (hyperbolic-geometry) tag does not mean \"involves hyperbolas.\" :) – anon Oct 16 '11 at 9:40 This question is similar to the other question you asked. What have you tried so far in solving this one? – yunone Oct 16 '11 at 9:44 You do in the very same way I explained to you in the ellipse problem. The method works in principle for every plane algebraic curve and is very effective in the case of conics (where it reduces to an elementary question about quadratic equations): – Andrea Mori Oct 16 '11 at 9:49 @Muavia: You want a verification of what you've done without telling us what you've done? – anon Oct 16 '11 at 10:22 $\\frac{x^2}{4}-\\frac{y^2}{36}=1$ , then solve system: $\\begin{cases} y_0=kx_0+n \\\\ n^2=a^2k^2-b^2 \\end{cases}$ where $x_0=0 , y_0=6 ,a^2=4 ,b^2=36$", "+0 # parabola 0 65 1 Find the vertex of the parabola x = -2y^2 + 12y - 15. Jun 30, 2020 We complete the square to get $$y=-2(y-3)^2+3$$. The vertex is (h, k). Therfore the vertex of the parabola $$x = -2y^2 + 12y - 15\\ \\text {is} \\ \\boxed{(3, 3)}.$$", "+0 +1 150 3 +409 A square is drawn such that one of its sides coincides with the line $y = 5$, and so that the endpoints of this side lie on the parabola $y = x^2 + 3x + 2$. What is the area of the square? michaelcai Jul 17, 2017 #1 +5205 +1 Let's find the two points on the parabola that touch the line y = 5 . y = x2 + 3x + 2 We want to find the x values when y is equal to 5 . 5 = x2 + 3x + 2 Subtract 5 from both sides of this equation. 0 = x2 + 3x - 3 Use the quadratic formula to find x . x = $${-3 \\pm \\sqrt{21} \\over 2}$$ So we know that the points $$({-3 + \\sqrt{21} \\over 2}\\,,\\,5)$$ and $$({-3 - \\sqrt{21} \\over 2}\\,,\\,5)$$ are the endpoints of one side of the square. Here's a graph that shows this: https://www.desmos.com/calculator/djyyxourv3 The side length of the square is the distance between these two points. side length $$={-3 + \\sqrt{21} \\over 2} - {-3 - \\sqrt{21} \\over 2} \\\\~\\\\ ={-3 + \\sqrt{21} \\,-\\,(-3-\\sqrt{21})\\over 2} \\\\~\\\\ ={-3 + \\sqrt{21} + 3+\\sqrt{21}\\over 2} \\\\~\\\\ =\\frac{2\\sqrt{21}}{2} \\\\~\\\\ =\\sqrt{21}$$ and... area of square = (side length)2 area of square = $$\\sqrt{21}^2$$ area", "Thus, the Vertex Form of the Parabola is $$y=(x-3)^2-16$$ . Solving $$(x-3)^2-16=0$$ yields the two zeros: $$x=7 , x=-1$$ . Get it now? Try the above Complete the Square Solver again.", "crosses itself, and find the equations of the tangent line at this point. Example 4. Draw your example carefully, showing what the correct points of tangency are, and showing also the highest and lowest points, making it clear that they do not give the common tangent that we seek. So, if you have a graph with curves, like a parabola, it can have points of tangency as well. Find the length of the tangent in the circle shown below. Find the equation of the tangent line of the curve at the given point. add example. The tangent only touches the curve at one point. For example, if you put a square around a circle, then each side of the square has a … Plus, get practice tests, quizzes, and personalized coaching to help you credit-by-exam regardless of age or education level. Beschreibung Tangency Example 1.svg A diagram showing a curve (C, red), a line (L, blue) and the point of intersection point (P, black). So, the length of the tangent is 22.4 cm. 5x - xy + y^2 = 5 at (1, 1). So, let's put another point of tangency on this parabola. 16 = x 2 √16 = √x 2. x = 8. If DC = 20 inch and BC = 12 inch, calculate the radius shown", "be perpendicular to the radius drawn to the points, the distance from (4,2) to either of these points will be $\\sqrt{(\\sqrt{20})^2 - 1^2} = \\sqrt{19}$. To find the points, then, we can use the intersection of the two circles $(x-4)^2 + (y-2)^2 = 19$ and $x^2 + y^2 = 1$. 3. So, if I did my math right (which may or may not be the case) (0,1) is a point of tangency, and y = 1/4x + 1 is an equation which answers the question. 4. Originally Posted by justaguy So, if I did my math right (which may or may not be the case) (0,1) is a point of tangency, and y = 1/4x + 1 is an equation which answers the question. I don't think that is right. Both points of tangency lie on the line y = 1/2 - 2x. You can use this fact in conjunction with the unit circle to produce a quadratic in x or y, which you can then use to find the actual points of tangency. 5. Yeah, that can't be right, actually. The line from (0,0) to (0,1) isn't perpendicular to the line from (0,1) to (4,2). How do you find y = 1/2 - 2x? On the plus side, I've figured out the other problem. 6. This is", "Complete the square: $y=(x^2-12x +\\_\\_\\_)-\\_\\_\\_+7$ Take half the coefficient of x, square it, and add it to make a perfect square trinomial. But you have to subtract it also to keep the equation balanced. $y=x^2-12x+6^2-6^2+7$ $y=(x-6)^2-29$ The parabola opens up and has a vertex (6, -29).", "Posted by Moo Hello, You have to found to values : h and k. First of all, you know that the summit is the point where the derivate of the parabola is 0. So calculate dy/dx and get the value for which it's 0. It'll give you the value of h. Then the parabola goes through the point (1,10). So replace x and y in y=3(x-h)^2 + k and you will get a relation between h and k. well how do i find h and k 4. Hello, polakio92! b) Square $ABCD$ is inscribed in square $EFGH$ forming four congruent right triangles. If the perimeter of the square $EFGH$ is 64 and the area of square $ABCD$ is 130, determine the values of $x\\text{ and }y.$ Code: E x A y F * - - - * - - - - - - - * | * * | | * * | x | * * | y | * * B | * *| | * * | |* * | D * * | y | * * | x | * * | | * * | * - - - - - - - * - - - * H 16-x C x G : - - - - 16 - - -", "the derivative to get the slope at that point. 1. y^3 = 8x^4 at (1, 2) 2. The 15+52.96 would indicate a point 1,552,96 feet from the beginning. You could choose any other point on the parabola as well, and draw a tangent line through that point. succeed. Create your account. Example sentences with \"point of tangency\", translation memory. Example 1: Changing a Free Entity to a Fixed Entity . Tangents can be formed around any shape, but this lesson will focus on the tangents to a circle. For example, 0+00 indicates the beginning of the project. The line intersects the curve but is not a tangent to it. All other trademarks and copyrights are the property of their respective owners. The combination in fact is the point of tangency of t For example, if you put a square around a circle, then each side of the square has a point of tangency on the circle. Draw your example carefully, showing what the correct points of tangency are, and showing also the highest and lowest points, making it clear that they do not give the common tangent that we seek. 16 = x 2 √16 = √x 2. x = 8. Another word for tangency. Plus, get practice tests, quizzes, and personalized coaching to help you To visualize", "Thread: Tangent twice -- challenge problem 1. Tangent twice -- challenge problem Find the equation of the line which is tangent to $y=-{x}^{4}+18\\,{x}^{3}-97\\,{x}^{2}+180\\,x-52$ at exactly two distinct points. Also find the points of tangency. The graph, shown with no axes and different vertical scale than horizontal scale is shown below. Note: this is just for fun, I am not looking for help. 2. Well, I tried, and after some tedious calculations, I got this... Equation of the line that contains these two points: $y = 36.46062956x + 2.620763081$ Am I right? EDIT: Points of tangency: $C_{1} = 1.450646899$ $C_{2} = 8.321649395$ 3. Originally Posted by Chop Suey Well, I tried, and after some tedious calculations, I got this... Equation of the line that contains these two points: $y = 36.46062956x + 2.620763081$ Am I right? EDIT: Points of tangency: $C_{1} = 1.450646899$ $C_{2} = 8.321649395$ Actually, the numbers should work out nicely, in fact I have chosen coefficients so the coordinates of the points of tangency are integers. So, back to the drawing board 4. Just curious, but are you aware that this exact question is posted on the AoPs forum? That is why I cannot answer it, for I would be \"cheating\" in a sense, for after seeing the solution I cannot say how much I would", "+0 # Help 0 177 1 Here is an equation of a parabola: y = 1/6 x^2 + 2x + 11 What are the coordinates of the vertex of the parabola? Mar 28, 2018 edited by Guest Mar 28, 2018 #1 +7348 +3 $$y=\\frac16x^2+2x+11$$ Multiply through by 6 . $$6y=x^2+12x+66$$ Subtract 66 from both sides of the equation. $$6y-66=x^2+12x$$ Add 36 to both sides to complete the square on the right side. $$6y-30=x^2+12x+36$$ Factor the right side as a perfect square trinomial. $$6y-30=(x+6)^2$$ Factor 6 out of the terms on the left side. $$6(y-5)=(x+6)^2$$ Now that the equation is in this form we can see that the vertex is (-6, 5) . Here's a graph to check it https://www.desmos.com/calculator/wrvhzfuag3 Mar 28, 2018 #1 +7348 +3 $$y=\\frac16x^2+2x+11$$ Multiply through by 6 . $$6y=x^2+12x+66$$ Subtract 66 from both sides of the equation. $$6y-66=x^2+12x$$ Add 36 to both sides to complete the square on the right side. $$6y-30=x^2+12x+36$$ Factor the right side as a perfect square trinomial. $$6y-30=(x+6)^2$$ Factor 6 out of the terms on the left side. $$6(y-5)=(x+6)^2$$ Now that the equation is in this form we can see that the vertex is (-6, 5) . Here's a graph to check it https://www.desmos.com/calculator/wrvhzfuag3 hectictar Mar 28, 2018", "333. Find two consecutive even numbers whose product is 360. 334. The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.) For each parabola, find which ways it opens, the axis of symmetry, the vertex, the x- and y-intercepts, and the maximum or minimum value. 335. $y = 3 x 2 + 6 x + 8 y = 3 x 2 + 6 x + 8$ 336. $y = x 2 − 4 y = x 2 − 4$ 337. $y = x 2 + 10 x + 24 y = x 2 + 10 x + 24$ 338. $y = −3 x 2 + 12 x − 8 y = −3 x 2 + 12 x − 8$ 339. $y = − x 2 − 8 x + 16 y = − x 2 − 8 x + 16$ Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry. 340. $y = 2 x 2 + 6 x + 2 y = 2 x 2 + 6 x + 2$ 341. $y = 16 x 2 + 24 x + 9 y = 16 x 2 + 24", "a point of tangency on the circle. Many translated example sentences containing \"point of tangency\" – French-English dictionary and search engine for French translations. DC 2 = 27 (10 + 27) = 27 *37. So, the radius of the circle is 21.3 inch. Thus, the value of x is 8 cm. Tangents can be formed around any shape, but this lesson will focus on the tangents to a circle. Line of the tangent is 31.61 cm might draw of this situation looks like this: get access for. Focus on the circumference of the circle OB is the point of tangency –! That lines and curves in mathematics do not have a common external point to a Custom Course it may that! To see them lines or segments are not the only objects that can be around... At any one point in a parabola, it can have points of intersection or points of tangency occur. The following diagram is an object, like a parabola are all points of tangency on.! Curve but is not a tangent to a circle = 12x + 11 -.! Property of their respective owners you can use grips to move points tangency! Filling in a and the tangent to the tangent in the point in the below. Move away design using CAD techniques this", "x and y both rational, to the equation $40x+ 72y=16x^2+79+24y^2$. Solution: First, move all terms to one side to arrive at $16x^2-40x+24y^2-72y+79=0$. Moving the constant to the right side, we find that $16x^2-40x+24y^2-72y=-79$. Completing the square in both x and y, we arrive at $16x^2-40x+25+24y^2-72y+54=-79+25+54$, or $16x^2-40x+25+24y^2-72y+54=0$. Dividing the square trinomials of x and y by the respective variables’ coefficients gives $\\frac{(x-5/4)^2}{16}+\\frac{(y-3/2)^2}{24}=0$. Here, it can be identified that the conic relation is that of a point, but this equation can also be analyzed algebraically. Any real value squared is nonnegative, so both terms on the left hand side must equal 0 for this equation to be true. The only real solution, and also the only rational solution, is $\\boxed{(5/4,3/2)}$. 7. Equilateral triangle ABC has side length $\\sqrt{3}$. D is on $\\overline{AC}$, and E and F are both on $\\overline{BC}$, between B and C. Square DEFG has side length 1. Circle O is the smallest circle that passes through G and is tangent to $\\overline{AB}$. This point of tangency is H. Find GH. Solution: Draw altitude $\\overline{AJ}$, and call its point of intersection with $\\overline{BG}$ K. Label the intersection of $\\overline{AB}$ and $\\overline{BG}$ L. See the below diagram. $\\angle A\\cong\\angle B\\cong\\angle C$, because equilateral triangles are also equiangular, and $m\\angle A=m\\angle B=m\\angle C=60^\\circ$. Since opposite sides of a square", "{x -> 11.3197 - 0.0000200237 I, y -> 15.2599 + 0.0000148534 I}}$ The question is as follows: What should I do to get the values of the $r_x$, $\\theta _1$ and $\\theta _2$? The geometries must be tangent, but I do not know how to do it ... • Can you explain what \"largest radius of the ellipse\" and \"angles of the ellipse\" mean? I see no actual ellipse in any of your images. What ellipse are you looking at? – bill s May 30 '17 at 15:34 The circle. Solve[x^2 + y^2 == 19^2, y] {{y -> -Sqrt[361 - x^2]}, {y -> Sqrt[361 - x^2]}} D[Sqrt[361 - x^2], x] -(x/Sqrt[361 - x^2]) The ellipse. With[{b = 3/2}, Simplify[Solve[x^2/a^2 + (y - 15)^2/b^2 == 1, y], a > 0]] {{y -> 15 - (3 Sqrt[a^2 - x^2])/(2*a)}, {y -> 15 + (3 Sqrt[a^2 - x^2])/(2*a)}} D[15 + (3 Sqrt[a^2 - x^2])/(2 a), x] -((3 x)/(2 a Sqrt[a^2 - x^2])) The points of tangency are the points where the two curves intersect and have the same derivatives. We make use of this to find a. Solve[ {Sqrt[361 - x^2] == 15 + (3 Sqrt[a^2 - x^2])/(2 a), x/Sqrt[361 - x^2] == (3 x)/(2 a Sqrt[a^2 - x^2])}, {a, x}] Now we can draw the figure. With[{a = 1/2 Sqrt[1/2", "$\\sqrt{170}/2$ which is slightly larger than 6.5. The distance from (5/2, 9/2) to (6, -4) is $\\sqrt{338}/2$, a little larger than 9. Finally, a problem that can be done! Geometrically, if you draw the lines from the given point tangent to the circle, the line from the given point to the center of the circle, and the line from the center of the circle to the point of tangency, you have a right triangle, the right angle being at the point of tangency. Your circle has radius $\\sqrt{170}/2$ and the distance from (6, -4) to (5/2, 9/2) is $\\sqrt{338}/2$ so you have a right triangle with hypotenuse of length $\\sqrt{338}/2$ and one leg of length $\\sqrt{170}/2$. By the Pythagorean theorem, the length of the other leg, and the distance from (8, 5) to the point of tangency, is $\\sqrt{338- 170}/2=\\sqrt{168/2}$. That means that we could find the points of tangency by striking a circle about (6, -4) with radius $\\sqrt{168}/2$. That is, the points of tangency satisfy $(x-6)^2+ (y+4)^2= 42$ and $(x- 5/2)^2+ (y- 9/2)^2= 170/4$. Solve those two equations for x and y to get the two points of tangency and, together with (6, -4), gives two points on each tangent line so you can find the equations.", "# Math Help - conics 1. ## conics Given that the parabolas x^2 = y + 2 y^2 = x + 3 meet in four points, show that these four points lie on a circle, and also on a rectangular hyperbola. 2. Originally Posted by millwallcrazy Given that the parabolas x^2 = y + 2 y^2 = x + 3 meet in four points, show that these four points lie on a circle, and also on a rectangular hyperbola. Add the two equations together: $x^2 + y^2 = y + x + 5$. Any point that satisfies the original two equations will also satisfy that new equation. But that new equation is the equation of a circle. So any point that lies on both parabolas also lies on the circle. If you subtract the equations instead of adding them, you'll get the equation of a rectangular hyperbola... ." ]

[ "= Solutions (separate by commas)...", "Write all solutions on the same line, separated by commas. ##### Question 3 Solve for the unknown: $-8x+x^2=-6-x-x^2$8x+x2=6xx2 1. Write all solutions on the same line, separated by commas. ##### Question 4 Solve the following equation: $x-\\frac{45}{x}=4$x45x=4 1. Write all solutions on the same line, separated by commas.", "by completing the square: $4x^2+11x+7=0$4x2+11x+7=0 1. Write all solutions on the same line, separated by commas.", "= 8}$ is the only solution:", "the equation as z+1=8. z+1=8 Move all terms not containing z to the right side of the equation. Subtract 1 from both sides of the equation. z=8-1 Subtract 1 from 8. z=7 z=7 Solve for z 1/2+2/(z+1)=3/4 ## Our Professionals ### Lydia Fran #### We are MathExperts Solve all your Math Problems: https://elanyachtselection.com/ Scroll to top", "solution : Find numbers in Z by divid 66.......(66=2×11×3 ) 12 solutions LOOK HERE", "# Solve the following equation by factoring z^2=8 z Solve the following equation by factoring ${z}^{2}=8z$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it likvau Concept: Factorization is the process of breaking a polynomial into product of its factors. For example: if a quadratic equation $a{x}^{2}+bx+c=0$ has roots p and q then in factor form it can be written as (x-p)(x-q)=0 Solution: ${z}^{2}=8z$ subtract 8z from both sides: ${z}^{2}-8z=8z-8z$ ${z}^{2}-8z=0$ tale out z as it is common to both: z(z-8)=0 now, the above expression shows that we have broken the given expression in to it's factors, z=0 or (z-8)=0 z=0 or z=8 therefore, by factorization the solution of the equation ${z}^{2}=8z$ is: z=0 or z =8", "$(3z^2+3a^2)=(z^2+4a^2)$ which you should have no difficulty solving. Last edited:", "# How do you solve 8(z – 3) = -24 ? Then, we get that $8 z - 24 = - 24$ We get $8 z = 0$ From here, we just divide by 8 to get the value of z, which is clearly $0$", "be : ZZ((x==6).solve(x)[0].rhs()).factor() ( 2019-03-08 18:02:18 +0200 )edit", "# How do you solve 8z+9=3? Mar 19, 2018 $z = - 0.75 \\mathmr{and} - \\frac{3}{4}$ #### Explanation: Isolate the variable $z$, then eliminate any coefficients $8 z + 9 = 3$ $8 z = - 6$ $z = - 0.75 \\mathmr{and} - \\frac{3}{4}$", "me to how they arrived at this please ? $$\\frac{z(z-1)(z+1)}{(z-3)^3}$$ The solution : ...", ")$ because of $3|15$ and $5|15$. • Note that $8$ is a solution, hence $15z+8$ is a solution for every $z\\in\\mathbb Z$ – Peter Dec 2 '15 at 14:18", "+0 # For which values of does the equation have no solution for ? Enter all the possible values of ​ separated by commas. 0 155 1 For which values of $$k$$ does the equation $$\\frac{x-1}{x-2} = \\frac{x-k}{x-6}$$ have no solution for $$x$$? Enter all the possible values of $$k,$$ separated by commas. Nov 23, 2019 #1 +108650 +1 You can see straight off that x cannot be 2 or 6. Now just for the moment you can ignore that. Try solving for x in terms of k and see what happens. The answer will present itself.", "= 8^2$$ and the solution is $$x = 8$$.", "Find all possible values of $z$ that satisfy $|z – 5| + |z + 5| = 12$ and $|z – 5| = \\operatorname{Re} z + 4$.", "solutions are as before. Worked Examples Question 1 Solve the equation $9\\left|x\\right|=4$9|x|=4. Write both solutions as equations on the same line separated by a comma. Question 2 Consider the equation $\\left|5x\\right|=15$|5x|=15. 1. On the same set of axes, graph the functions $y=\\left|5x\\right|$y=|5x| and $y=15$y=15. 2. Hence determine the solutions to the equation $\\left|5x\\right|=15$|5x|=15. Write the solutions on the same line, separated by a comma. Question 3 Solve $\\left|4x-8\\right|+1=13$|4x8|+1=13. Write both solutions as equations on the same line separated by a comma. Outcomes M7-2 Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs 91257 Apply graphical methods in solving problems", ") in the first box, and use the second box to give the spacing between solutions for any integer k. Entry tips: enter multiple answers separated by comma: Enter T by 'typing \"pi\"...", "'-8' to each side of the equation. 8 + -8 + z = -4.795831523 + -8 Combine like terms: 8 + -8 = 0 0 + z = -4.795831523 + -8 z = -4.795831523 + -8 Combine like terms: -4.795831523 + -8 = -12.795831523 z = -12.795831523 Simplifying z = -12.795831523 SolutionThe solution to the problem is based on the solutions from the subproblems. z = {-3.204168477, -12.795831523}`", "Solve the equation in the interval $[0,\\, 2 \\pi]$. If there is more than one solution write them separated by commas. Hint: To solve this problem you will have to use the quadratic formula, inverse trigonometric functions and the symmetry of the unit circle. $x=$ Your overall score for this problem is" ]

[ "find that $z= \\pm\\sqrt{3}$ are also roots of $I(z)$ (and $\\pm \\sqrt{\\frac{5}{2}}$ are not roots), so $$f(\\pm i\\sqrt{3}) = g(\\pm\\sqrt{3}) = R(z) + iI(z) = 0$$ and $f$ has two purely imaginary roots.", "and to get the $+1$ we have to also have a $+1$ in the remainder. So, our product is $$z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \\cdots + z^2) + z^2+1.$$ Then, our remainder is $z^2+1$. The remainder when dividing by $z^3-1$ must be the same when dividing by $z^2+z+1$, modulo $z^2+z+1$. So, we have that $z^2 + 1 \\equiv R(z) \\pmod{z^2+z+1}$, or $R(z) \\equiv -z\\pmod{z^2+z+1}$. This corresponds to answer choice $\\boxed{\\textbf{(A)} ~ -z}$. ~rocketsri ## Solution 2 (Complex numbers) One thing to note is that $R(z)$ takes the form of $Az + B$ for some constants A and B. Note that the roots of $z^2 + z + 1$ are part of the solutions of $z^3 -1 = 0$ They can be easily solved with roots of unity: $$z^3 = 1$$ $$z^3 = e^{i 0}$$ $$z = e^{i 0}, e^{i \\frac{2\\pi}{3}}, e^{i -\\frac{2\\pi}{3}}$$ $$\\newline$$ Obviously the right two solutions are the roots of $z^2 + z + 1 = 0$ We substitute $e^{i \\frac{2\\pi}{3}}$ into the original equation, and $z^2 + z + 1$ becomes 0. Using De Moivre's theorem, we get: $$e^{i\\frac{4042\\pi}{3}} + 1 = A \\cdot e^{i \\frac{2\\pi}{3}} + B$$ $$e^{i\\frac{4\\pi}{3}} + 1 = A \\cdot e^{i \\frac{2\\pi}{3}} + B$$ Expanding into rectangular complex number form: $$\\frac{1}{2} - \\frac{\\sqrt{3}}{2} i = (-\\frac{1}{2}A + B) + \\frac{\\sqrt{3}}{2} i", "# Thread: complex number equation 1. ## complex number equation Hi, I would like some help with this. The equation $p(z)=z^4-2z^3+4z^2-2z+3$ is given. Show that one of the roots are i. So i verified that i is indeed a root. And since complex roots are \"mirrored\", another root must be -i. The factor theorem. $(z-i)(z+i) = z^2 +1$ Then i factored the equation using polynomial division and the result is $(z^2+1)(z^2-2z+3)$ the second bracket is a simple quadratic. $\\frac{-2}{-2} \\pm \\sqrt{\\bigg(\\frac{-2}{2}\\bigg)^2 -3}$ = $1 \\pm \\sqrt{1-3}$ This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ? The correct answer is $1\\pm i\\sqrt{2}$ however, the result im getting is $1\\pm 2i$ Any ideas? 2. Originally Posted by Jones Hi, = $1 \\pm \\sqrt{1-3}$ This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ? The correct answer is $1\\pm i\\sqrt{2}$ however, the result im getting is $1\\pm 2i$ Any ideas? No, you have it right. = $1 \\pm \\sqrt{1-3} = 1 \\pm \\sqrt{-2} = 1 \\pm \\sqrt{-1*2} = 1 \\pm i\\sqrt{2}$ Just remember that the square root doesn't go away just because you have a negative number under it. In the same way $\\sqrt{8} = \\sqrt{4*2} = 2\\sqrt{2}$, right?", "that $u= z^2$. $z^2= -1+ i3^{1/2}$ so $z= \\pm\\sqrt{-1+ i 3^{1/2}}$ $z^2= -1- i3^{1/2}$ so $z= \\pm\\sqrt{-1- i3^{1/2}}$. I presume you will want to use d'Moivre's formula to take those square roots. Since $1^2+ (3^{1/2})^2= 4$ and $\\frac{3^{1/2}}{-1}= -\\sqrt{3}= arctan(\\pi/3}$, in polar form, $-1+ i3^{1/2}= 2(cos(\\pi/3)+ i sin(\\pi/3))= 2e^{i\\pi/3}$ and $-1- i3^{1/2}= 2(cos(\\pi/3)- i sin(\\pi/3))= 2e^{-i\\pi/3}$. I forgot put i, its mean -1. there are four solution.", "2x^2- 1$ so $z= \\pm\\sqrt{4x-2x^2-1}$. That z only exists when $\\frac{4- 2\\sqrt{2}}{4}< x< \\frac{4+ 2\\sqrt{2}}{4}$. In that range, the roots are x, y= 2- x and z= $\\pm\\sqrt{4x-2x^2-1}$ for every value of x. It is true that x= 1 is in that range and then y= 2-1= 1 and z= $\\sqrt{4- 2-1}= 1$ so that (1, 1, 1) is one solution. x= 1, y= 1, z= $-\\sqrt{4-2-1}= -1$ or (1, 1, -1) is also a solution. Those may be the only integer solutions. 4. Sorry, I don't really understand what do you mean by \"In that range, the roots are x, y= 2- x and z= for every value of x.\" By roots, which roots are you talking about? Please explain about the above in details and on how do you come to the conclusion that the roots are x, y= 2- x and z = . A little lost here", "- 2 = 6\\, then, by adding 2 to both sides of the equation, followed by dividing both sides by 4, we get \\log_5(x - 3) = 2\\, whence x - 3 = 5^2 = 25\\, from which we obtain x = 28.\\, Radical equation showing two ways to represent the same expression A radical equation is one that includes a radical sign, which includes square roots, \\sqrt{x}, cube roots, \\sqrt[3]{x}, and nth roots, \\sqrt[n]{x}. Recall that an nth root can be rewritten in exponential format, so that \\sqrt[n]{x} is equivalent to x^{\\frac{1}{n}}. Combined with regular exponents (powers), then \\sqrt[2]{x^3} (the square root of x cubed), can be rewritten as x^{\\frac{3}{2}}.[39] So a common form of a radical equation is a = \\sqrt[n]{x^m} (equivalent to a = x^\\frac{m}{n}) where m and n are integers. It has solution: m is odd m is even and a \\ge 0 x = \\sqrt[m]{a^n} or x = \\left(\\sqrt[m]a\\right)^n x = \\pm \\sqrt[m]{a^n} or x = \\pm \\left(\\sqrt[m]a\\right)^n For example, if: (x + 5)^{2/3} = 4,\\, then \\begin{align} x + 5 & = \\pm (\\sqrt{4})^3\\\\ x + 5 & = \\pm 8\\\\ x & = -5 \\pm 8\\\\ x & = 3,-13 \\end{align}. ### System of linear equations There are different methods to solve a system of linear equations with two variables. ####", "is where $$\\frac{a^2}{3}> \\frac{a^2-81}{2} \\indent (4)$$ which simplifies to $$\\sqrt{243}>a$$ meaning $$16>a(5)$$ So now we have that $9 from $(2)$, $a<16$ from $(5)$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=\\boxed{440}$ AWD. ## Solution 3 (Calculus, not as good) Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$. Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root", "2} = \\frac{- 1 \\pm \\sqrt{{2}^{2} - 4 \\cdot 1 \\cdot 5}}{2 \\cdot 1}$ Notice that the determinant, $\\Delta$, which is the name given to the expression that's under the square root, , is negative. $\\Delta = {b}^{2} - 4 a c$ $\\Delta = {2}^{2} - 4 \\cdot 1 \\cdot 5 = - 16$ For real numbers, you cannot take the square root of a negative number, which means that the quadratic equation has no real solutions. Its graph will not intercept the $x$-axis. However, it will have two distinct complex roots. ${x}_{1 , 2} = \\frac{- 1 \\pm \\sqrt{- 16}}{2}$ ${x}_{1 , 2} = \\frac{- 1 \\pm \\left({i}^{2} \\cdot 16\\right)}{2} = \\frac{- 1 \\pm i \\cdot \\sqrt{16}}{2}$ ${x}_{1 , 2} = \\frac{- 1 \\pm 4 i}{2}$ The two roots will thus be ${x}_{1} = \\frac{- 1 + 4 i}{2} \\text{ }$ and $\\text{ } {x}_{2} = \\frac{- 1 - 4 i}{2}$", "1+y^2+7=9,y^2=1$ So, the roots are $\\pm \\sqrt7i,1\\pm i$ Choosing $z = 1 + y i$, you obtain the formula: $$y^4 – 2 i y^3 – 9y^2 + 2 i y + 8 = 0$$ Because $y$ is real, this means that $$2 y^3 – 2y = 0$$ and $$y^4 – 9y^2 + 8 = 0$$ The first equation implies $y = 0, 1, -1$. The second equation implies that $y^2 = 8$ or $y^2 = 1$. Thus we see both $1 + i$ and $1 – i$ are solutions. Factor out the polynomial $(z – 1 + i)(z – 1 – i) = z^2 – 2z + 2$ and you’ll find the remaining roots. Maxima tells me the roots are $\\pm i \\sqrt{7}$, $1 \\pm i$. It factors as $(z^2 + 7) (z^2 – 2 z + 2)$.", "then immediately replace it as follows: $$\\sqrt{-4}=i\\sqrt{4}=i\\cdot 2=2i$$ Or in general: $$\\sqrt{-b}=i\\sqrt{b}, b>0$$ We now know how to deal with roots of negative numbers, therefore we can get back to our initial equation and finally solve it: $$x_{1,2}=\\frac{-4 \\pm \\sqrt{-4}}{2}=\\frac{-4 \\pm i\\sqrt{4}}{2}=\\frac{-4 \\pm 2i}{2}=-2 \\pm i$$ That’s it. $i$ here can be treated as some kind of a flag indicating that the number standing next to it is of different, unusual nature. ## Numbers of different nature Consider the resulting number as follows: $$-2 \\pm i=-2 \\pm 1\\cdot i$$ to indicate that $1 \\cdot i$ is also a number, only of different kind. We could go even further adopting a sign for real numbers as well: $$-2 \\pm i=-2\\cdot r \\pm 1\\cdot i$$ This makes little sense, though, as we’re already used to reals. Despite being of different nature, imaginary numbers can be added, subtracted, multiplied or whatsoever. Addition and subtraction are no challenge, let’s see: $$i-15i+3i=i(1-15+3)=-11i$$ Here we treat “$i$” as a factor which can legally be taken out front. Along with that, we can consider “$i$” as “$1$” but of different nature, subtract “$15$”, again of different nature, from it and then add “$3$”, but of different nature. Result is obviously “$-11$”, of different nature. Both ways yield the same “$-11i$”, an imaginary number. Thus, we", "(the negative root must be discarded). If $$r=1+\\sqrt{2}$$, you have $$e^{2x}-2re^x+1=0$$ hence $$e^x=r\\pm\\sqrt{r^2-1}=1+\\sqrt{2}\\pm\\sqrt{2+2\\sqrt{2}}$$ You know that the roots of the quadratic $$t^2-2rt+1=0$$ are reciprocal of one another, so the solutions are $$x=\\pm\\log(1+\\sqrt{2}+\\sqrt{2+2\\sqrt{2}})$$ From $$\\cosh^2(x) - 2\\cosh(x)-1 = 0$$, you get $$\\cosh x= 1+ \\sqrt2$$ Then, use the identity $$\\cosh^{-1}t = \\ln(t+\\sqrt{t^2-1})$$ to obtain $$x= \\pm \\cosh^{-1} (1+\\sqrt2)=\\pm\\ln (1+\\sqrt2+ \\sqrt{2\\sqrt2+2})$$", "to the equation, so that $z_0$ has two square roots: \\begin{equation*} \\pm \\sqrt{r_0}e^{i\\theta_0/2}\\text{.} \\end{equation*} For instance, $z^2 = i$ has two solutions. Since $i =1 e^{i\\pi/2}\\text{,}$ $\\sqrt{i} = \\pm e^{i\\pi/4}\\text{.}$ In Cartesian form, $\\sqrt{i} = \\pm (\\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2}i)\\text{.}$ More generally, the complex quadratic equation $\\alpha z^2 + \\beta z + \\gamma = 0$ where $\\alpha\\text{,}$ $\\beta\\text{,}$ and $\\gamma$ are complex constants, will have one or two solutions. This marks an important difference from the real case, where a quadratic equation might not have any real solutions. In both cases we may use the quadratic formula to hunt for roots, and in the complex case we have solutions \\begin{equation*} z = \\frac{-\\beta \\pm \\sqrt{\\beta^2 - 4\\alpha\\gamma}}{2\\alpha}\\text{.} \\end{equation*} For instance, $z^2 + 2z + 4 = 0$ has two solutions: \\begin{equation*} z = \\frac{-2 \\pm \\sqrt{-12}}{2} = -1 \\pm \\sqrt{3}i \\end{equation*} since $\\sqrt{-1} = i\\text{.}$ ###### Example2.4.4.Solving a quadratic equation. Consider the equation $z^2 - (3+3i)z = 2-3i\\text{.}$ To solve this equation for $z$ we first rewrite it as \\begin{equation*} z^2-(3+3i)z-(2-3i)=0\\text{.} \\end{equation*} We use the quadratic formula with $\\alpha = 1\\text{,}$ $\\beta = -(3+3i)\\text{,}$ and $\\gamma = -(2-3i)\\text{,}$ to obtain the solution(s) \\begin{align*} z \\amp = \\frac{3+3i \\pm \\sqrt{(3+3i)^2+4(2-3i)}}{2}\\\\ z \\amp = \\frac{3+3i\\pm \\sqrt{8+6i}}{2}\\text{.} \\end{align*} To determine the solutions in Cartesian form, we need to evaluate $\\sqrt{8+6i}\\text{.}$ We offer", "that each of the roots $z=1/n \\pi,\\ n=\\pm 1,\\ \\pm 2,\\ ...$ has multiplicity 3 as the first and second derivatives of $\\sin^3(1/z)$ are zero at these points, but the third derivative is non-zero at these points. (This method may also be applied to (c), which will confirm ImPerfectHacker conclusion) RonL 7. The second problem is $z\\not = 0$ We have, $\\sin^3 (1/z)=0$ Using the field property of zero, If and only if, $\\sin (1/z)=0$ We know that, $\\sin (1/z)=\\frac{e^{i/z}-e^{-i/z}}{2i}$ We require that, $e^{i/z}-e^{-i/z}=0$ Thus, $e^{i/z}=e^{-i/z}$ Since the complex exponential is no-where zero we can divde through by it obtaining a biconditional statement, $e^{2i/z}=1$ By Euler's Formula, $e^{2\\pi i}=1$ Thus, $2i/z=2\\pi i+2\\pi k$ Thus, $1/z=-\\pi+\\pi i k$ Thus, $z=\\frac{1}{\\pi i k-\\pi }$ It seems all solutions are of multiplicity of order 1.", "Originally Posted by st123 This is my first time here, so I'm not sure if this question belongs here or somewhere else. I'm actually in Year 11, but none of the pre-university topics seemed to cover this, so sorry if I posted this in the wrong place: Show that the roots of (z-1)^4 + (z+1)^4 = 0 are +/- i cot (pi/8) and +/- i cot (3pi/8) Thanks Of course you could just try substituting the purported roots into the left hand side and showing that the result is zero. CB 4. $(z - 1)^4 + (z + 1)^4 = z^4 - 4z^3 + 6z^2 - 4z + 1 + z^4 + 4z^3 + 6z^2 + 4z + 1$ $= 2z^4 + 12z^2 + 2$ $= 2(z^4 + 6z^2 + 1)$ $= 2\\left(z^4 + 6z^2 + 3^2 - 3^2 + 1\\right)$ $= 2\\left[(z^2 + 3)^2 - 8\\right]$ Therefore, if $(z - 1)^4 + (z + 1)^4 = 0$ then $2\\left[(z^2 + 3)^2 - 8\\right] = 0$ $(z^2 + 3)^2 - 8 = 0$ $(z^2 + 3)^2 = 8$ $z^2 + 3 = \\pm 2\\sqrt{2}$ $z^2 = -3 \\pm 2\\sqrt{2}$ $z = \\pm \\sqrt{-3 \\pm 2\\sqrt{2}}$. Now, you should note that since $2\\sqrt{2} < 3$, that means $-3 + 2\\sqrt{2} < 0$ and $-3 - 2\\sqrt{2} < 0$. Therefore", "Then $x^2-5y^2=-1$ looks like a Pell equation. $(x_1,y_1)=(\\pm 2,\\pm 1)$. See my answer here for how to solve Pell equations. In particular, the solution given in http://vjimc.osu.cz/history 2015 Category II Solutions Problem 2 is very similar, analogous to what I'll do here. I've also posted an analogous answer here. The solutions, in this case, are given by $$x_v=\\pm \\frac{(2+\\sqrt{5})^{2v-1}+(2-\\sqrt{5})^{2v-1}}{2}$$ $$y_v=\\pm \\frac{(2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}}{2\\sqrt{5}}$$ $v=1,2,\\ldots$ $$2\\sqrt{5}\\cdot 5^k=(2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}$$ Clearly $(2+\\sqrt{5})^{2v-1}>(2-\\sqrt{5})^{2v-1}$. If $k=0$, then $n=1$, $(m,n)=(\\pm 2,1)$. Let $k\\ge 1$. $$5^k=\\binom{2v-1}{1} 4^{v-1}+\\binom{2v-1}{3} 4^{v-2}\\cdot 5+$$ $$+\\cdots+\\binom{2v-1}{2v-1} 5^{v-1}$$ $$\\implies 2v-1\\equiv 0\\pmod{5}$$ $(2v-1)/5$ is odd. $$2\\sqrt{5}\\cdot 5\\cdot 61=(2+\\sqrt{5})^{5}-(2-\\sqrt{5})^{5}\\mid$$ $$\\mid (2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}=2\\sqrt{5}\\cdot 5^k,$$ contradiction. I think this divisibility result is true. How do I know that $$b_5:=(2+\\sqrt{5})^5+(2-\\sqrt{5})^5\\mid$$ $$\\mid (2+\\sqrt{5})^{2v-1}+(2-\\sqrt{5})^{2v-1}=:b_{2v-1}?$$ ((($:=$ denotes \"equal by definition\"))) Here I used $+$ instead of $-$. Well, the solution of $5^n=6m^2+1$ which I gave in the beginning of this answer uses this and also see this link from AoPS. I won't be self-contained here. v_Enhance said that, in this case: \"$b_{2v-1}/b_5$ is both rational and an algebraic integer.\" Kent Merryfield has a longer answer, which you should see for yourself. There are no non-trivial positive integer solutions to $m^2+1=5^n$. We can show this using fairly elementary methods. As others have noted, $n$ must be odd. If $n=2r$ then $1 = (5^r)^2 - m^2 = (5^r + m)(5^r - m)$, which is only true", "$x = 28.\\,$ Radical equation showing two ways to represent the same expression A radical equation is one that includes a radical sign, which includes square roots, $\\sqrt{x}$, cube roots, $\\sqrt[3]{x}$, and nth roots, $\\sqrt[n]{x}$. Recall that an nth root can be rewritten in exponential format, so that $\\sqrt[n]{x}$ is equivalent to $x^{\\frac{1}{n}}$. Combined with regular exponents (powers), then $\\sqrt[2]{x^3}$ (the square root of $x$ cubed), can be rewritten as $x^{\\frac{3}{2}}$.[38] So a common form of a radical equation is $a = \\sqrt[n]{x^m}$ (equivalent to $a = x^\\frac{m}{n}$) where $m$ and $n$ are integers. It has solution: $m$ is odd $m$ is even and $a \\ge 0$ $x = \\sqrt[m]{a^n}$ or $x = \\left(\\sqrt[m]a\\right)^n$ $x = \\pm \\sqrt[m]{a^n}$ or $x = \\pm \\left(\\sqrt[m]a\\right)^n$ For example, if: $(x + 5)^{2/3} = 4,\\,$ then \\begin{align} x + 5 & = \\pm (\\sqrt{4})^3\\\\ x + 5 & = \\pm 8\\\\ x & = -5 \\pm 8\\\\ x & = 3,-13 \\end{align}. ### System of linear equations There are different methods to solve a system of linear equations with two variables. #### Elimination method The solution set for the equations $x - y = -1$ and $3x + y = 9$ is the single point (2, 3). An example of solving a system of linear equations is by using the elimination method:", "0 so r2= 3 or r2= -12. The four roots to that equation are $r= \\pm\\sqrt{3}$ and $r= \\pm 2i\\sqrt{3}$. Of course, r must be positive so $r= \\sqrt{3}$.", "For the case $$(1,1,0)$$, the roots are $$\\frac{-1\\pm\\sqrt{1+4(x^4+2x^3+2x^2+x)}}{2}=\\frac{-1\\pm(2x^2+2x+1)}2=x^2+x,-x^2-x-1.$$ A solution, among many many many others:", "of $(x+2)^2=0\\!$) or two complex/imaginary roots. Only equations of this type (with real coefficients) are needed in FP1. Consider now the quadratic equation $\\mathit{z}^2+1=0\\!$. Previously, one would have been unable to solve this as it would require one to root $\\pm\\mathit{-1}\\!$. Since we now represent this value as $\\mathit{i}\\!$, this quadratic can now be solved with relative ease: Using the quadratic formula gives: \\begin{align}z & ={-b\\pm\\sqrt{b^2-4ac} \\over 2a} \\\\ & ={-0\\pm\\sqrt{0^2-4\\times1\\times1} \\over 2} \\\\ & ={-0\\pm\\sqrt{-4} \\over 2} \\\\ & ={\\pm\\sqrt{-4}\\over2} \\\\ & ={\\pm\\sqrt{4}\\sqrt{-1}\\over2} \\\\ & ={\\pm2i\\over2} \\\\ & ={\\pm i} \\\\ \\end{align} We can then use the factor theorem to prove that either $z=i\\!$ or $z=-i\\!$ are factors of $z^2+1=0\\!$. Let $P(z)=z^2+1\\!$ \\begin{align}z & =i \\Rightarrow P(z)=i^2+1\\! \\\\ & =(-1)+1\\! \\\\ & =0\\! \\\\ \\end{align} $P(i)=0\\Rightarrow i\\!$ is a factor of $z^2+1\\!$ The same result is achieved for $-i\\!$, since $\\sqrt{-1}=\\pm i\\!$. $\\Rightarrow P(z)=(z+i)(z-i)$ The same can be done with Complex Roots: Consider the quadratic equation: $z^2-14z+53\\!$ Using the quadratic formula, one can see that: \\begin{align}z & ={-b\\pm\\sqrt{b^2-4ac} \\over 2a} \\\\ & ={-(-14)\\pm\\sqrt{(-14)^2-4\\times1\\times53} \\over 2} \\\\ & ={14\\pm\\sqrt{196-212} \\over 2} \\\\ & ={14\\pm\\sqrt{-16} \\over 2} \\\\ & ={14\\pm4i\\over2} \\\\ & =7\\pm2i\\! \\\\ \\end{align} If one was to now factorise this expression: $z^2-14z+53=(z-(7+2i))(z-(7-2i))\\!$ It should be noted, once again, that $7\\pm2i\\!$ is a conjugate pair (since $7+2i\\!$ is", "9b^2 + b^4$$, i.e., $$b^4 + 9b^2 -4 = 0$$. Solve $$x^2 + 9x-4=0$$, where $$x=b^2$$. Solutions are $$x_{1,2} = \\frac{-9\\pm\\sqrt{9^2-4\\cdot 1\\cdot (-4)}}{2\\cdot 1} = \\frac{-9\\pm\\sqrt{97}}{2}$$. Solutions are $$b_{1,2,3,4} = \\pm\\sqrt{\\frac{-9\\pm\\sqrt{97}}{2}}$$. This gives the corresponding values $$a_{1,2,3,4}$$. All solution pairs are real-valued." ]

[ "and to get the $+1$ we have to also have a $+1$ in the remainder. So, our product is $$z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \\cdots + z^2) + z^2+1.$$ Then, our remainder is $z^2+1$. The remainder when dividing by $z^3-1$ must be the same when dividing by $z^2+z+1$, modulo $z^2+z+1$. So, we have that $z^2 + 1 \\equiv R(z) \\pmod{z^2+z+1}$, or $R(z) \\equiv -z\\pmod{z^2+z+1}$. This corresponds to answer choice $\\boxed{\\textbf{(A)} ~ -z}$. ~rocketsri ## Solution 2 (Complex numbers) One thing to note is that $R(z)$ takes the form of $Az + B$ for some constants A and B. Note that the roots of $z^2 + z + 1$ are part of the solutions of $z^3 -1 = 0$ They can be easily solved with roots of unity: $$z^3 = 1$$ $$z^3 = e^{i 0}$$ $$z = e^{i 0}, e^{i \\frac{2\\pi}{3}}, e^{i -\\frac{2\\pi}{3}}$$ $$\\newline$$ Obviously the right two solutions are the roots of $z^2 + z + 1 = 0$ We substitute $e^{i \\frac{2\\pi}{3}}$ into the original equation, and $z^2 + z + 1$ becomes 0. Using De Moivre's theorem, we get: $$e^{i\\frac{4042\\pi}{3}} + 1 = A \\cdot e^{i \\frac{2\\pi}{3}} + B$$ $$e^{i\\frac{4\\pi}{3}} + 1 = A \\cdot e^{i \\frac{2\\pi}{3}} + B$$ Expanding into rectangular complex number form: $$\\frac{1}{2} - \\frac{\\sqrt{3}}{2} i = (-\\frac{1}{2}A + B) + \\frac{\\sqrt{3}}{2} i", "# Math Help - Solving equations 1. ## Solving equations Solve each equation by the most suitable method. please do not use decimals in the answer. b) z(5z-1)=3z(z+3) c) x^2+6x-8=0 d) x^2 +8x/3=1 2. Originally Posted by pashah Solve each equation by the most suitable method. please do not use decimals in the answer. b) z(5z-1)=3z(z+3) c) x^2+6x-8=0 d) x^2 +8x/3=1 Hello, pashah, to b): rearrange, so that the RHS of the equation is zero. Put the factor z befor the bracket. Then you have two factors. This product can be only zero if one of the factors is zero: $z(5z-1)-3z(z+3)=0\\Longleftrightarrow z(5z-1-3z-9)=0$ $\\Longleftrightarrow z(2z-10)=0 \\Longleftrightarrow z=0\\ \\vee\\ z=5$ to c) (Are you sure that there isn't a typo? If you have +8 instead of -8 then you could use factorization) Use the formula: $x_{1,2}=-\\frac{6}{2\\cdot 1}\\pm\\sqrt{\\frac{6^2-4\\cdot 1\\cdot (-8)}{4\\cdot 1^2}}$ $\\ \\quad \\Longleftrightarrow\\ \\ x=-3+\\sqrt{17}\\vee x=-3-\\sqrt{17}$ to d) Use the formula: $x_{1,2}=-\\frac{\\frac{8}{3}}{2}\\pm\\sqrt{\\frac{\\frac{64}{9}-4\\cdot 1\\cdot (-1)}{4\\cdot 1^2}}$ $\\Longleftrightarrow$ $x=-\\frac{4}{3}+\\frac{5}{3}=\\frac{1}3}\\vee x=-\\frac{4}{3}-\\frac{5}{3}=-3$ Greetings EB 3. Hello, pashah! I assume you know how to solve a quadratic equation. . . Get all terms on the left (0 on the right). . . Factor the quadratic (if possible). . . Set each factor equal to zero and solve the resulting equations. $b)\\;\\;z(5z-1)\\:=\\:3z(z+3)$ Distribute: . $5z^2 - z\\;=\\;3x^2 + 9z$ We have: . $2z^2 - 10z\\;=\\;0$ Factor:", "# Thread: complex number equation 1. ## complex number equation Hi, I would like some help with this. The equation $p(z)=z^4-2z^3+4z^2-2z+3$ is given. Show that one of the roots are i. So i verified that i is indeed a root. And since complex roots are \"mirrored\", another root must be -i. The factor theorem. $(z-i)(z+i) = z^2 +1$ Then i factored the equation using polynomial division and the result is $(z^2+1)(z^2-2z+3)$ the second bracket is a simple quadratic. $\\frac{-2}{-2} \\pm \\sqrt{\\bigg(\\frac{-2}{2}\\bigg)^2 -3}$ = $1 \\pm \\sqrt{1-3}$ This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ? The correct answer is $1\\pm i\\sqrt{2}$ however, the result im getting is $1\\pm 2i$ Any ideas? 2. Originally Posted by Jones Hi, = $1 \\pm \\sqrt{1-3}$ This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ? The correct answer is $1\\pm i\\sqrt{2}$ however, the result im getting is $1\\pm 2i$ Any ideas? No, you have it right. = $1 \\pm \\sqrt{1-3} = 1 \\pm \\sqrt{-2} = 1 \\pm \\sqrt{-1*2} = 1 \\pm i\\sqrt{2}$ Just remember that the square root doesn't go away just because you have a negative number under it. In the same way $\\sqrt{8} = \\sqrt{4*2} = 2\\sqrt{2}$, right?", "that $u= z^2$. $z^2= -1+ i3^{1/2}$ so $z= \\pm\\sqrt{-1+ i 3^{1/2}}$ $z^2= -1- i3^{1/2}$ so $z= \\pm\\sqrt{-1- i3^{1/2}}$. I presume you will want to use d'Moivre's formula to take those square roots. Since $1^2+ (3^{1/2})^2= 4$ and $\\frac{3^{1/2}}{-1}= -\\sqrt{3}= arctan(\\pi/3}$, in polar form, $-1+ i3^{1/2}= 2(cos(\\pi/3)+ i sin(\\pi/3))= 2e^{i\\pi/3}$ and $-1- i3^{1/2}= 2(cos(\\pi/3)- i sin(\\pi/3))= 2e^{-i\\pi/3}$. I forgot put i, its mean -1. there are four solution.", "2x^2- 1$ so $z= \\pm\\sqrt{4x-2x^2-1}$. That z only exists when $\\frac{4- 2\\sqrt{2}}{4}< x< \\frac{4+ 2\\sqrt{2}}{4}$. In that range, the roots are x, y= 2- x and z= $\\pm\\sqrt{4x-2x^2-1}$ for every value of x. It is true that x= 1 is in that range and then y= 2-1= 1 and z= $\\sqrt{4- 2-1}= 1$ so that (1, 1, 1) is one solution. x= 1, y= 1, z= $-\\sqrt{4-2-1}= -1$ or (1, 1, -1) is also a solution. Those may be the only integer solutions. 4. Sorry, I don't really understand what do you mean by \"In that range, the roots are x, y= 2- x and z= for every value of x.\" By roots, which roots are you talking about? Please explain about the above in details and on how do you come to the conclusion that the roots are x, y= 2- x and z = . A little lost here", "to the equation, so that $z_0$ has two square roots: \\begin{equation*} \\pm \\sqrt{r_0}e^{i\\theta_0/2}\\text{.} \\end{equation*} For instance, $z^2 = i$ has two solutions. Since $i =1 e^{i\\pi/2}\\text{,}$ $\\sqrt{i} = \\pm e^{i\\pi/4}\\text{.}$ In Cartesian form, $\\sqrt{i} = \\pm (\\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2}i)\\text{.}$ More generally, the complex quadratic equation $\\alpha z^2 + \\beta z + \\gamma = 0$ where $\\alpha\\text{,}$ $\\beta\\text{,}$ and $\\gamma$ are complex constants, will have one or two solutions. This marks an important difference from the real case, where a quadratic equation might not have any real solutions. In both cases we may use the quadratic formula to hunt for roots, and in the complex case we have solutions \\begin{equation*} z = \\frac{-\\beta \\pm \\sqrt{\\beta^2 - 4\\alpha\\gamma}}{2\\alpha}\\text{.} \\end{equation*} For instance, $z^2 + 2z + 4 = 0$ has two solutions: \\begin{equation*} z = \\frac{-2 \\pm \\sqrt{-12}}{2} = -1 \\pm \\sqrt{3}i \\end{equation*} since $\\sqrt{-1} = i\\text{.}$ ###### Example2.4.4.Solving a quadratic equation. Consider the equation $z^2 - (3+3i)z = 2-3i\\text{.}$ To solve this equation for $z$ we first rewrite it as \\begin{equation*} z^2-(3+3i)z-(2-3i)=0\\text{.} \\end{equation*} We use the quadratic formula with $\\alpha = 1\\text{,}$ $\\beta = -(3+3i)\\text{,}$ and $\\gamma = -(2-3i)\\text{,}$ to obtain the solution(s) \\begin{align*} z \\amp = \\frac{3+3i \\pm \\sqrt{(3+3i)^2+4(2-3i)}}{2}\\\\ z \\amp = \\frac{3+3i\\pm \\sqrt{8+6i}}{2}\\text{.} \\end{align*} To determine the solutions in Cartesian form, we need to evaluate $\\sqrt{8+6i}\\text{.}$ We offer", "Then $x^2-5y^2=-1$ looks like a Pell equation. $(x_1,y_1)=(\\pm 2,\\pm 1)$. See my answer here for how to solve Pell equations. In particular, the solution given in http://vjimc.osu.cz/history 2015 Category II Solutions Problem 2 is very similar, analogous to what I'll do here. I've also posted an analogous answer here. The solutions, in this case, are given by $$x_v=\\pm \\frac{(2+\\sqrt{5})^{2v-1}+(2-\\sqrt{5})^{2v-1}}{2}$$ $$y_v=\\pm \\frac{(2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}}{2\\sqrt{5}}$$ $v=1,2,\\ldots$ $$2\\sqrt{5}\\cdot 5^k=(2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}$$ Clearly $(2+\\sqrt{5})^{2v-1}>(2-\\sqrt{5})^{2v-1}$. If $k=0$, then $n=1$, $(m,n)=(\\pm 2,1)$. Let $k\\ge 1$. $$5^k=\\binom{2v-1}{1} 4^{v-1}+\\binom{2v-1}{3} 4^{v-2}\\cdot 5+$$ $$+\\cdots+\\binom{2v-1}{2v-1} 5^{v-1}$$ $$\\implies 2v-1\\equiv 0\\pmod{5}$$ $(2v-1)/5$ is odd. $$2\\sqrt{5}\\cdot 5\\cdot 61=(2+\\sqrt{5})^{5}-(2-\\sqrt{5})^{5}\\mid$$ $$\\mid (2+\\sqrt{5})^{2v-1}-(2-\\sqrt{5})^{2v-1}=2\\sqrt{5}\\cdot 5^k,$$ contradiction. I think this divisibility result is true. How do I know that $$b_5:=(2+\\sqrt{5})^5+(2-\\sqrt{5})^5\\mid$$ $$\\mid (2+\\sqrt{5})^{2v-1}+(2-\\sqrt{5})^{2v-1}=:b_{2v-1}?$$ ((($:=$ denotes \"equal by definition\"))) Here I used $+$ instead of $-$. Well, the solution of $5^n=6m^2+1$ which I gave in the beginning of this answer uses this and also see this link from AoPS. I won't be self-contained here. v_Enhance said that, in this case: \"$b_{2v-1}/b_5$ is both rational and an algebraic integer.\" Kent Merryfield has a longer answer, which you should see for yourself. There are no non-trivial positive integer solutions to $m^2+1=5^n$. We can show this using fairly elementary methods. As others have noted, $n$ must be odd. If $n=2r$ then $1 = (5^r)^2 - m^2 = (5^r + m)(5^r - m)$, which is only true", "4/x-3 $\\frac {3}{x^2 - 9} + \\frac {4}{x - 3} = \\frac {3}{(x + 3)(x - 3)} + \\frac {4}{x - 3}$ ....the LCD is (x + 3)(x - 3) ....................... $= \\frac {3 + 4(x + 3)}{(x + 3)(x - 3)}$ ....................... $= \\frac {4x + 15}{(x + 3)(x - 3)}$ 3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor By the rational roots theorem, the possible rational roots are given by the factors of 8 are: $\\pm 1$, $\\pm 2$, $\\pm 4$, $\\pm 8$ start with $\\pm 1$ and start plugging in each of the factors of 8 into the formula, if we get zero as the result, we have a root. we realize that -1 works, since, $(-1)^3 - 5(-1)^2 + 2(-1) + 8 = 0$ since x = -1 is a root, it means x + 1 is a factor by the factor theorem 4- simplify 4x^5y^-3 / 20x^-2y $\\frac {4x^5y^{-3}}{20x^{-2}y} = \\frac {4}{20} \\cdot \\frac {x^5}{x^{-2}} \\cdot \\frac {y^{-3}}{y}$ ............. $= \\frac {1}{5} \\cdot x^{5 - (-2)} \\cdot y^{-3 - 1}$ ............. $= \\frac {1}{5} \\cdot x^{7} \\cdot y^{-4}$ ............. $= \\frac {x^7}{5y^4}$ EDIT: Thanks a lot earboth. anyway, i guess it's my own fault for going around asking a lot of questions before doing the problem", "is where $$\\frac{a^2}{3}> \\frac{a^2-81}{2} \\indent (4)$$ which simplifies to $$\\sqrt{243}>a$$ meaning $$16>a(5)$$ So now we have that $9 from $(2)$, $a<16$ from $(5)$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=\\boxed{440}$ AWD. ## Solution 3 (Calculus, not as good) Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$. Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root", "1+y^2+7=9,y^2=1$ So, the roots are $\\pm \\sqrt7i,1\\pm i$ Choosing $z = 1 + y i$, you obtain the formula: $$y^4 – 2 i y^3 – 9y^2 + 2 i y + 8 = 0$$ Because $y$ is real, this means that $$2 y^3 – 2y = 0$$ and $$y^4 – 9y^2 + 8 = 0$$ The first equation implies $y = 0, 1, -1$. The second equation implies that $y^2 = 8$ or $y^2 = 1$. Thus we see both $1 + i$ and $1 – i$ are solutions. Factor out the polynomial $(z – 1 + i)(z – 1 – i) = z^2 – 2z + 2$ and you’ll find the remaining roots. Maxima tells me the roots are $\\pm i \\sqrt{7}$, $1 \\pm i$. It factors as $(z^2 + 7) (z^2 – 2 z + 2)$.", "- 2 = 6\\, then, by adding 2 to both sides of the equation, followed by dividing both sides by 4, we get \\log_5(x - 3) = 2\\, whence x - 3 = 5^2 = 25\\, from which we obtain x = 28.\\, Radical equation showing two ways to represent the same expression A radical equation is one that includes a radical sign, which includes square roots, \\sqrt{x}, cube roots, \\sqrt[3]{x}, and nth roots, \\sqrt[n]{x}. Recall that an nth root can be rewritten in exponential format, so that \\sqrt[n]{x} is equivalent to x^{\\frac{1}{n}}. Combined with regular exponents (powers), then \\sqrt[2]{x^3} (the square root of x cubed), can be rewritten as x^{\\frac{3}{2}}.[39] So a common form of a radical equation is a = \\sqrt[n]{x^m} (equivalent to a = x^\\frac{m}{n}) where m and n are integers. It has solution: m is odd m is even and a \\ge 0 x = \\sqrt[m]{a^n} or x = \\left(\\sqrt[m]a\\right)^n x = \\pm \\sqrt[m]{a^n} or x = \\pm \\left(\\sqrt[m]a\\right)^n For example, if: (x + 5)^{2/3} = 4,\\, then \\begin{align} x + 5 & = \\pm (\\sqrt{4})^3\\\\ x + 5 & = \\pm 8\\\\ x & = -5 \\pm 8\\\\ x & = 3,-13 \\end{align}. ### System of linear equations There are different methods to solve a system of linear equations with two variables. ####", "# Background Recently, I have been factoring quadratic equation of the form $$x^2 + ax + b$$ into $$(x+y) \\cdot (x+z)$$ e.g. From $$x^2 + 3528x + 63180$$ to $$(x+3510) \\cdot (x+18)$$ Normally, it is possible to obtain $$y$$ and $$z$$ by looking into every factor of $$b$$, and find 2 of them, where there product is $$b$$ and their sum is $$a$$. However, I find this method too tedious, especially when dealing with numbers that are hard or impossible to get its factors. So, I decided to use equations to directly obtain $$y$$ and $$z$$ given $$a$$ and $$b$$. # Question According to the scenario, $$y+z=a$$ $$y \\cdot z = b$$ and so, how do I solve this system of equations, so that I get $$y = \\frac{-\\sqrt{a^2 - 4 \\cdot b} + a}{2} \\, \\text{and} \\, z = \\frac{\\sqrt{a^2 - 4 \\cdot b} + a}{2}$$ $$\\text{or} \\, y = \\frac{\\sqrt{a^2 - 4 \\cdot b} + a}{2} \\, \\text{and} \\, z = \\frac{-\\sqrt{a^2 - 4 \\cdot b} + a}{2}$$ ? If I understood the question, you want to know how to get the formulae for the roots of the second-degree polynomial equation. That isn't that hard. First, consider the equation written in the following way (this is the standard way):$$ax^2+bx+c=0$$ and let's assume $$a\\not=0$$, $$b\\not=0$$ and $$c\\not=0$$", "# Thread: [SOLVED] Solving for roots -&gt; Z^3 = -i 1. ## [SOLVED] Solving for roots -&gt; Z^3 = -i Hi all, I'm trying to work out the roots for the equation above. I've worked out one of them as being z = i, as z^3= i^3 = -i. Wolfram Alpha gives z = 1/2 (sqrt(3)-i) and z = 1/2 (-sqrt(3)-i) as the other answers, but I'm unsure how they arrived at this? Thanks. 2. Hello, Ashgasm! Tricky . . . Find all roots of: . $z^3 \\:=\\:-i$ We have: . $z^3 + i \\:=\\:0$ Since $(\\text{-}i)^3 \\:=\\:i$, we have: . $z^3 + (\\text{-}1)^3 \\;=\\;0$ . . . a sum of cubes Factor: . $\\bigg[z + (\\text{-}i)\\bigg]\\,\\bigg[z^2 -(\\text{-}1)(z) + (\\text{-}i)^2\\bigg] \\;=\\;0$ . . . . . . . . . $(z - i)\\,(z^2 + iz - 1)\\;=\\;0$ We have two equations to solve: $z - i \\:=\\:0 \\quad\\Rightarrow\\quad \\boxed{z \\:=\\:i}$ $z^2 + iz - 1 \\:=\\:0$ Quadratic Formula: . $z \\:=\\:\\frac{-i \\pm\\sqrt{i^2 - 4(1)(\\text{-}1)}}{2(1)} \\;=\\;\\boxed{\\frac{-i \\pm\\sqrt{3}}{2}}$ The three roots are: . $i,\\;\\;\\frac{\\sqrt{3}-i}{2},\\;\\;\\frac{-\\sqrt{3}-i}{2}$ 3. Originally Posted by Ashgasm Hi all, I'm trying to work out the roots for the equation above. I've worked out one of them as being z = i, as z^3= i^3 = -i. Wolfram Alpha gives z = 1/2 (sqrt(3)-i) and z = 1/2 (-sqrt(3)-i)", "side, we must add it to the other. $$x^2-14x+(\\frac{-14}{2})^2=-1+(\\frac{-14}{2})^2$$ Simplify both sides of the equation. $$x^2-14x+49=48$$ The left hand side is now a perfect-square trinomial. Let's convert it now. $$(x+\\frac{b}{2})^2=48$$ This is the form of what it will look like. Take the coefficient of the linear term, -14, and half it to get it into the desired form. $$(x-7)^2=48$$ Take the square root of both sides. $$x-7=\\pm\\sqrt{48}$$ Remember that when you take the square root of something, it results in the positive and negative answer. Add 7 to both sides. $$x=7\\pm\\sqrt{48}$$ Simplify the radical to its simplest terms. $$\\sqrt{48}=\\sqrt{16*3}=\\sqrt{16}\\sqrt{3}=4\\sqrt{3}$$ $$x=7\\pm4\\sqrt{3}$$ I'm going to split these solutions into separate ones. Ok, now we must not forget that we were solving for a--not x. Let's substitute it in. $$x_1=7+4\\sqrt{3}$$ $$x_2=7-4\\sqrt{3}$$ Substitute a^2 back into the equation. $$a^2=7+4\\sqrt{3}$$ $$a^2=7-4\\sqrt{3}$$ Take the square root of both sides. $$a=\\pm\\sqrt{7+4\\sqrt{3}}$$ $$a=\\pm\\sqrt{7-4\\sqrt{3}}$$ Now, we have to simplify those square roots. It is possible. Now, I have a clever way of simplifying these radicals so that they are nicer. I'm going to do it to both individually. $$\\sqrt{7+4\\sqrt{3}}$$ To simplify this, I'm going to add another term. What you want to do is get another term containing the square root of 3. $$\\sqrt{7+4\\sqrt{3}+\\sqrt{3}^2-3}$$ You might notice something. I've added 2 more terms. Let's simplify inside the", "# Math Help - solving radical equations directions say to find all solutions 1. (x + 4)^1\\2 + 5x(x + 4)^3/2 = 0 I don't know how to get rid of the fractions when there are two of them any help appreciated 2. Originally Posted by isundae directions say to find all solutions 1. (x + 4)^1\\2 + 5x(x + 4)^3/2 = 0 I don't know how to get rid of the fractions when there are two of them any help appreciated $(x+4)^{1/2} + 5x(x+4)^{3/2} = 0$ factor ... $(x+4)^{1/2}[1 + 5x(x+4)] = 0$ $(x+4)^{1/2}(5x^2 + 20x + 1) = 0$ set each factor equal to 0 and solve ... $x = -4$ $x = \\frac{-10 \\pm \\sqrt{95}}{5}$ 3. (x + 4)^(1/2) + 5x(x + 4)^(1+1/2) = 0, by factoring (x + 4)^(1/2). (x + 4)^(1/2)(1 + (5x)(x + 4)) = 0, (x + 4)^(1/2)(1 + 20x + 5x^2) = 0 THESE are the 3 roots: x_1 = -4, x_2 = -2 - sqrt (19/5) = -3.95 x_3 = sqrt (19/5) - 2 = -0.05 see graph,", "9b^2 + b^4$$, i.e., $$b^4 + 9b^2 -4 = 0$$. Solve $$x^2 + 9x-4=0$$, where $$x=b^2$$. Solutions are $$x_{1,2} = \\frac{-9\\pm\\sqrt{9^2-4\\cdot 1\\cdot (-4)}}{2\\cdot 1} = \\frac{-9\\pm\\sqrt{97}}{2}$$. Solutions are $$b_{1,2,3,4} = \\pm\\sqrt{\\frac{-9\\pm\\sqrt{97}}{2}}$$. This gives the corresponding values $$a_{1,2,3,4}$$. All solution pairs are real-valued.", "then immediately replace it as follows: $$\\sqrt{-4}=i\\sqrt{4}=i\\cdot 2=2i$$ Or in general: $$\\sqrt{-b}=i\\sqrt{b}, b>0$$ We now know how to deal with roots of negative numbers, therefore we can get back to our initial equation and finally solve it: $$x_{1,2}=\\frac{-4 \\pm \\sqrt{-4}}{2}=\\frac{-4 \\pm i\\sqrt{4}}{2}=\\frac{-4 \\pm 2i}{2}=-2 \\pm i$$ That’s it. $i$ here can be treated as some kind of a flag indicating that the number standing next to it is of different, unusual nature. ## Numbers of different nature Consider the resulting number as follows: $$-2 \\pm i=-2 \\pm 1\\cdot i$$ to indicate that $1 \\cdot i$ is also a number, only of different kind. We could go even further adopting a sign for real numbers as well: $$-2 \\pm i=-2\\cdot r \\pm 1\\cdot i$$ This makes little sense, though, as we’re already used to reals. Despite being of different nature, imaginary numbers can be added, subtracted, multiplied or whatsoever. Addition and subtraction are no challenge, let’s see: $$i-15i+3i=i(1-15+3)=-11i$$ Here we treat “$i$” as a factor which can legally be taken out front. Along with that, we can consider “$i$” as “$1$” but of different nature, subtract “$15$”, again of different nature, from it and then add “$3$”, but of different nature. Result is obviously “$-11$”, of different nature. Both ways yield the same “$-11i$”, an imaginary number. Thus, we", "\\sqrt{3}i) + (1 - \\sqrt{3}i)(1 + \\sqrt{3}i)$ $= z^2 - z + \\sqrt{3}iz - z - \\sqrt{3}iz + 1 + 3$ $= z^2 - 2z + 4$. Since $z^2 - 2z + 4$ is a factor, long divide $z^3 - a$ by $z^2 - 2z + 4$ and you will have the third factor. You should note that the third root will need to be real. Once you have the third factor, you can find $a$.", "# Equations- Factorization of a polynomial equation – Exercise 5604 Exercise Factor the polynomial equation $$x^4-3x^2-4=0$$ $$(x-2)(x+2)(x^2+1)=0$$ Solution First solution: $$x^4-3x^2-4=0$$ To reach a quadratic equation, we define a new variable: $$y=x^2$$ We set the new variable: $$y^2-3y-4=0$$ We got a quadratic equation. Our equation coefficients are $$a=1, b=-3, c=-4$$ We solve it with the quadratic formula. Putting the coefficients in the formula gives us $$y_{1,2}=\\frac{3\\pm \\sqrt{{(-3)}^2-4\\cdot 1\\cdot (-4)}}{2\\cdot 1}=$$ $$=\\frac{3\\pm \\sqrt{25}}{2}=$$ $$=\\frac{3\\pm 5}{2}$$ Hence, we get the solutions: $$y_1=\\frac{3+ 5}{2}=\\frac{8}{2}=4$$ $$y_2=\\frac{3- 5}{2}=\\frac{-2}{2}=-1$$ Thus, the factorizing of the quadratic equation is $$y^2-3y-4=$$ $$=(y-4)(y+1)$$ Going back to the original variable we get $$=(x^2-4)(x^2+1)$$ We break down the first factor using Short Multiplication Formulas (third formula) and get the factorizing $$=(x-2)(x+2)(x^2+1)$$ Second solution: $$x^4-3x^2-4=$$ We express the second expression as sum and we get $$=x^4-4x^2+x^2-4=$$ We extract a common factor from the first two expressions: $$=x^2(x^2-4)+(x^2-4)=$$ Once again we extract a common factor – this time the expression in parentheses: $$=(x^2-4)(x^2+1)=$$ We break down the first factor using Short Multiplication Formulas (third formula) and get the factorizing: $$=(x-2)(x+2)(x^2+1)$$ Have a question? Found a mistake? – Write a comment below! Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! Share with Friends", "2} = \\frac{- 1 \\pm \\sqrt{{2}^{2} - 4 \\cdot 1 \\cdot 5}}{2 \\cdot 1}$ Notice that the determinant, $\\Delta$, which is the name given to the expression that's under the square root, , is negative. $\\Delta = {b}^{2} - 4 a c$ $\\Delta = {2}^{2} - 4 \\cdot 1 \\cdot 5 = - 16$ For real numbers, you cannot take the square root of a negative number, which means that the quadratic equation has no real solutions. Its graph will not intercept the $x$-axis. However, it will have two distinct complex roots. ${x}_{1 , 2} = \\frac{- 1 \\pm \\sqrt{- 16}}{2}$ ${x}_{1 , 2} = \\frac{- 1 \\pm \\left({i}^{2} \\cdot 16\\right)}{2} = \\frac{- 1 \\pm i \\cdot \\sqrt{16}}{2}$ ${x}_{1 , 2} = \\frac{- 1 \\pm 4 i}{2}$ The two roots will thus be ${x}_{1} = \\frac{- 1 + 4 i}{2} \\text{ }$ and $\\text{ } {x}_{2} = \\frac{- 1 - 4 i}{2}$" ]

[ "Calculus (3rd Edition) $$A=-3$$ $$B=5$$ $\\frac{2x-3}{(x-3)(x-4)}$ = $\\frac{A}{x-3}$ + $\\frac{B}{x-4}$ Multiplying (x-3)(x-4) on both sides, we have: $2x-3=A(x-4)+B(x-3)$ If we let x=4 on both sides, we can eliminate constant A to find constant B: $8-3=A(4-4)+B(4-3)$ Thus, $B = 5$. Similarly, if we let x = 3 on both sides, we can eliminate constant B to find constant A: $6-3=A(3-4)+B(3-3)$ Thus, $3=-A$ and so $A = -3$.", "#### Problem 6P 6. Find the values of the constants $a$ and $b$ such that $$\\lim _{x \\rightarrow 0} \\frac{\\sqrt[3]{a x+b}-2}{x}=\\frac{5}{12}$$", "+0 # A and B are constants such that . Find ​. 0 56 1 A and B are constants such that $$\\frac{4x+5}{x^2+x-2}= \\frac{A}{x+2} +\\frac{B}{x-1}$$. Find $$\\frac{B}{x+1} - \\frac{A}{x-2}$$. Apr 5, 2020", "2). Find constants a, b, c, and d. This gives the following $f(2) = -4$ $f(0) = 2 \\Rightarrow d = 2$ $f'(2)=0$ Originally Posted by skeske1234 f'(x)=0 when 0=12a+4b+c You mean $f'(2)=0 \\Rightarrow 0=12a+4b+c$ Originally Posted by skeske1234 f(2)=-4 -4=24a+4b+2c+d Given $f(x) = ax^3+bx^2+cx+2$ $f(2) = 2^3a+2^2b+2c+2$ $-4 = 2^3a+2^2b+2c+2$ $-4 = 8a+4b+2c+2$", "+0 0 53 1 Find constants $A$ and $B$ such that $\\frac{x - 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$for all $x$ such that $x\\neq -1$ and \\$ Aug 29, 2021", "+0 0 31 1 +21 Find constants $A$ and $B$ such that $\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$ for all $x$ such that $x\\neq -1$ and $x\\neq 2$. Give your answer as the ordered pair $(A,B)$. Jul 18, 2020 #1 0 If you let x go infinity, then you get 0 = A + B. If you let x = 0, then you get -7/2 = A/(-2) + B. The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3). Jul 19, 2020", "that $b_1=\\frac{\\sum_i w_ix_iy_i}{\\sum_i w_ix_i^2} = \\frac{\\sum_i (w_ix_i)y_i}{\\sum_i (w_ix_i)x_i}$ Since in this case $b_1=\\frac{\\sum_i y_i}{\\sum x_i}$, you can see by inspection that the weights must be such that $w_ix_i$ is a constant.", "I get the constant $a=\\frac{3}{2}$ which is correct. However, I am completely stuck with the constant b. I am calculating using $\\begin{bmatrix}b_{1}\\\\ b_{2}\\end{bmatrix}$. But I get two exactly same equations. Thank you for your help on this one! Last edited by a moderator: May 1, 2013", "sides by $k$. Even though $C$ changes, we'll leave it as $C$ since it's a catch-all for any constant. $\\ln \\left\\mid B + k y \\right\\mid = k x + C$ $\\left\\mid B + k y \\right\\mid = {e}^{k x + C}$ Note that ${e}^{k x + C} = {e}^{k x} {e}^{C} = C {e}^{k x}$ where $C$ again represents ${e}^{C}$, some other constant. $\\left\\mid B + k y \\right\\mid = C {e}^{k x}$ The absolute value bars are just a reminder that it's possible that $B + k y < 0$. $k y = C {e}^{k x} - B$ $y = \\frac{C {e}^{k x} - B}{k}$", "+0 0 67 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19832 +1 Sep 15, 2019 #1 +19832 +1", "+0 0 47 1 Find constants A and B such that: $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x≠-1 and x≠2. Give your answer as the ordered pair (A,B). Sep 14, 2019 #1 +1 I think I saw this earlier. Check out https://web2.0calc.com/questions/help-please_7890 . Sep 14, 2019", "+ c$, where $c$ is a constant $= \\frac{1}{5} \\ln \\left(\\frac{w - 4}{w + 1}\\right) + c$, where $c$ is a constant", "+0 0 30 1 The quadratic x^2-3x+1-8x+11 can be written in the form (x+b)^2+c, where b and c are constants. What is b+c? Apr 15, 2022 $$x^2-11x+12=0 \\\\ (x-\\frac{11}{2})^2=19.25 \\\\ (x-\\frac{11}{2})^2-19.25 \\\\ b+c=\\frac{11}{2}-19.25 \\\\ b+c=\\boxed{-13.75} \\\\ \\huge{🎉}$$", "+0 0 103 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19818 0 Sep 15, 2019 #1 +19818 0", "+0 # help me! 0 38 2 Find constants A and B such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x =/= -1 and x =/= 2. Give your answer as the ordered pair (A,B). Aug 17, 2020 #1 0 Letting x go to 2, we find that A + 2B = 4. Letting x go to -1, we find 2A - B = 3. Therefore, A = 2 and B = -1. Aug 17, 2020 #2 0 its wrong :/ Aug 17, 2020", "# Find the constant of a and b Algebra ### Find the constant of a and b Greetings all, can someone explain to me how to find the constant of a and b for which $$\\frac{a}{x+b}$$+$$\\frac{2}{x-4}$$=$$\\frac{3x}{(x+b)(x-4)}$$ Thank you Guest ### Who is online Users browsing this forum: No registered users and 1 guest", "# Math Help - Determine values 1. ## Determine values Dear all, Would appreciate some guidance for this question: If q = ax + by and r = bx - ay, determine the value of: Thank you. 2. If $q=ax+by$ then would $\\frac{dq}{dx}=a$ so $\\frac{dq}{dx}\\frac{1}{y}=\\frac{1}{y}$? Is b also a constant? 3. Had edited the equation above in image format for better viewing. Thank you...", "constant expression? $$\\frac{\\left [IBr\\right ]^2}{\\left [I_{2}\\right ]\\left [Br_{2}\\right ]}$$ 1. $$I_{2}(g)+Br_{2}(g)\\rightleftharpoons IBr(g)$$ 2. $$I_{2}(g)+Br_{2}(g)\\rightleftharpoons 2 IBr(g)$$ 3. $$2 IBr(g)\\rightleftharpoons I_{2}(g)+Br_{2}(g)$$ 4. $$2 I_{2}(g)+2 Br_{2}(g)\\rightleftharpoons IBr(g)$$ 5. $$IBr(g)\\rightleftharpoons 2I_{2}(g)+2Br_{2}(g)$$ b. $$I_{2}(g)+Br_{2}(g)\\rightleftharpoons 2 IBr(g)$$", "+ B + Cx^2$ $\\Rightarrow x^2 + 75 = (A + C)x^2 + (A + B)x + B$ B is the constant term on the right, so it must be the same as the constant term on the left to have equality, thus, again, B = 75 now finish up with the method of your choice", "29, 2011 ### QuarkCharmer $$2x = A(3x+1) + B(3+x)$$ $$2x = x(3A+B) + A +3B$$ ? 5. Sep 29, 2011 ### George Jones Staff Emeritus left = right. How many x's on the left? On the right? What is the constant on the left? On the right? 6. Sep 29, 2011 ### QuarkCharmer So, The constant is A + 3B The other equation is 2=(3A+B) ? I'm guessing I system of equation these guys to find A and B now? What does the constant equal? 0? 7. Sep 29, 2011 ### QuarkCharmer That makes this I believe: $$\\frac{2x}{(3+x)(3x+1)} = \\frac{\\frac{3}{4}}{3+x} + \\frac{\\frac{-1}{4}}{3x+1}$$ Does that look correct? I can integrate those. 8. Sep 29, 2011 ### cepheid Staff Emeritus Yeah, it seems like you've got it. You can always check your answer by doing the reverse (combine the two terms on the right hand side into one fraction with a common denominator and check that the numerator simplifies to 2x)." ]

[ "and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Hello s2951 I'm not sure where you get the term $ax$ in the line that I've indicated. When you divide $4x^2-7x-15$ by $x-3$, the result is simply $4x+5$. If I may expand on acc100jt's solution a little: If $4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get: $4x^2-7x-15=Ax^2+(B-3A)x -3B$ So, when we compare coefficients: the coefficient of $x^2:4 = A$ and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Thanks, just solved and its correct. The letter \"A\" is just symbol to find the value, its redundant. 6. Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ 7. Hello Raoh Originally Posted by Raoh Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ I think what you mean is: Since $(x-3)$ is a factor, $+3$ is a root $\\Rightarrow f(3)=0$ So: $f(x) = f(x) - f(3) = 4(x^2-3^2)", "# Find the $4^{th}$ term in the expansion of $(x-2y)^{12}$ $\\begin{array}{1 1}(A)\\;-1760\\\\(B)\\;-760x^9\\\\(C)\\;-1760x^9y^3\\\\(D)\\;x^9y^3\\end{array}$ Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-......+(-1)^nnC_ra^{n-r}b^r+......nC_n(-b)^n$ $4^{th}$ term =$T_{3+1}$ in the expansion of $x+(-2y)^{12}$ $\\Rightarrow 12C_3x^{12-3}[-2y]^3$", "in $(x^2-1)^{11}(x^4+x^2+1)^5$ is to find the coefficient of $x^{-12}$ in the expansion $\\eqref{eq:expansion}$ of $\\left(x^4-\\dfrac{1}{x^2}\\right)^5\\left(x-\\dfrac{1}{x}\\right)^6$. We already know this to be $-15$. And to find the coefficient of $x^{38}$ in $(x^2-1)^{11}(x^4+x^2+1)^5$ is to find the coefficient of $x^{22}$ in the expansion $\\eqref{eq:expansion}$. This comes from $x^{20}$ in the first bracket and $x^2$ in the second, giving a coefficient of $15$.", "marks)}$ 3. Find the coefficient of $x^{2}$ in the expansion of $\\left(2 x^{2}-x-3\\right)^{6}$. $\\quad\\mbox{ (5 marks)}$ 4. Find the coefficient of $x^{2}$ and $x^{3}$ in the expansion of $\\left(x^{2}-x-2\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ 5. Evaluate the coefficients of $x^{5}$ and $x^{4}$ in the binomial expansion of $\\left(\\frac{x}{3}-3\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ Hence evaluate the coefficient of $x^{5}$ in the expansion of $\\left(\\frac{x}{3}-3\\right)^{7}(x+6)$. $\\quad\\mbox{ (5 marks)}$ 6. In the expansion of $(1+2 x)^{11}$, the coefficient of $x^{3}$ is $k$ times the coefficient of $x^{2}$. Find $k$. $\\quad\\mbox{ (5 marks)}$ 7. If the coefficient of $x^{4}$ in the expansion of $(3+2 x)^{6}$ is equal to the coefficient of $x^{4}$ in the expansion of $(2 k+3 x)^{6}$, find $k$. $\\quad\\mbox{ (5 marks)}$ 8. The ratio of the coefficient of $x^{4}$ in the expansion of $(3-2 x)^{6}$ and the coefficient of $x^{4}$ in the expansion of $(k+2 x)^{7}$ is $1: 7.$ Find the value of $k$. $\\quad\\mbox{ (5 marks)}$ 9. The first three terms in the binomial expasion of $(a+b)^{n}$, in ascending power of $b$, are denoted by $p, q$ and $r$ respectively. Show that $\\frac{q^{2}}{p r}=\\frac{2 n}{n-1}$. Given that $p=4, q=32$ and $r=96$, evaluate $n$. $\\quad\\mbox{ (5 marks)}$ 10. In the binomial expansion of $(1+x)^{n}$ the first three terms are $1+3+4+\\ldots$ Calculate the numerical values of $n$ and $x$, and the value of", "$$12-4r=0$$, that is, when $$r=3$$. Hence the constant term in the expansion of $$\\Big(x^2 + \\dfrac{1}{x^2}\\Big)^6$$ is $$\\dbinom{6}{3}= 20$$. ##### Exercise 7 Find the constant term in the expansion of $$\\Big(x + \\dfrac{1}{x^2}\\Big)^6$$. Screencast of exercise 7 #### Example Find the coefficient of $$x^5$$ in the expansion of $$(1-2x+3x^2)^5$$. #### Solution Bracket $$1-2x$$ and expand: $(1-2x+3x^2)^5=(1-2x)^5+5(1-2x)^4(3x^2)+10(1-2x)^3(3x^2)^2+\\dotsb.$ • The coefficient of $$x^5$$ in $$(1-2x)^5$$ is $$(-2)^5 = -32$$. • The coefficient of $$x^5$$ in $$5(1-2x)^4(3x^2)$$ is $$15\\tbinom{4}{3}(-2)^3 = -480$$. • The coefficient of $$x^5$$ in $$10(1-2x)^3(3x^2)^2$$ is $$90\\tbinom{3}{1}(-2) = -540$$. • The remaining terms do not contain $$x^5$$. Therefore the required coefficient is $$-32-480-540=-1052$$. #### The middle term When $$n$$ is even, there will be an odd number of terms in the expansion of $$(a+b)^n$$, and hence there will be a middle term. Let $$n=2m$$, for some positive integer $$m$$. Then, when the expansion of $$(a+b)^n$$ is arranged with terms in descending or ascending order, the middle term is $$\\dbinom{2m}{m}a^mb^m.$$ For example, the middle term of $$(a+b)^8$$ is $$\\dbinom{8}{4}a^4b^4 = 70a^4b^4.$$ When $$n$$ is odd, there will be two middle terms of $$(a+b)^n$$. Let $$n=2m+1$$, for some positive integer $$m$$. Then the middle terms are $$\\dbinom{2m+1}{m}a^{m+1}b^m \\qquad\\text{and}\\qquad \\dbinom{2m+1}{m+1}a^mb^{m+1}.$$ ##### Exercise 8 Find the middle term of $$(2x-3y)^6$$. Screencast of exercise 8 #### Greatest coefficients When $$n$$ is even, we", "of: $$(x + 2)^5$$ A $$16x^4$$ B $$80x^2$$ C $$40x^3$$ D $$40x^2$$ E $$4x^2$$ Example #2 Find each term described. 6th term in expansion of: $$(2y + 1)^5$$ A $$1$$ B $$10y$$ C $$32y^5$$ D $$5y^4$$ E $$2y^3$$ Example #3 Find each term described. 3rd term in expansion of: $$(x + 4)^4$$ A $$1024x^4$$ B $$16x$$ C $$16x^3$$ D $$96x^2$$ E $$256x$$", "marks) 11.) Consider the expansion of $\\D (x^2- 2)^5.$ (a) Write down the number of terms in this expansion. (b) The first four terms of the expansion in descending powers of $\\D x$ are $x^{10} -10x^8 + 40x^6 + Ax^4 + ...$ Find the value of A. (Total 6 marks) 12.) Given that $\\D \\left(3+\\sqrt{7}\\right)^3=p+\\sqrt{q}$ where $\\D p$ and $\\D q$ are integers, find (a) $\\D p;$ (b) $\\D q.$ (Total 6 marks) 13.) When the expression $\\D (2 + ax)^{10}$ is expanded, the coefficient of the term in $\\D x^3$ is 414720. Find the value of $\\D a.$ (Total 6 marks) 14.) Find the term containing $\\D x^3$ in the expansion of $\\D (2- 3x)^8.$ (Total 6 marks) 15.) Find the term containing $\\D x^{10}$ in the expansion of $\\D (5 + 2x^2)^7.$ (Total 6 marks) 16.) Complete the following expansion. $\\D (2 + ax)^4 = 16 + 32ax +\\cdots$ (Total 6 marks) 17.) Consider the expansion of $\\D \\left(3x^2-\\frac{1}{x}\\right)^9.$ (a) How many terms are there in this expansion? (b) Find the constant term in this expansion. (Total 6 marks) 18.) Find the coefficient of $\\D x^3$ in the expansion of $\\D (2- x)^5.$ (Total 6 marks) 19.) Use the binomial theorem to complete this expansion. $\\D (3x + 2y)^4 = 81x^4 + 216x^3 y + \\cdots$", "# Can you expand (x-1)^4 ? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 14 Aug 26, 2017 ${x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$ #### Explanation: We need to expand ${\\left(x - 1\\right)}^{4}$. Using Pascal's triangle , we know that the coefficients of each progressive terms are: $1$ $4$ $6$ $4$ $1$. We also know from the binomial theorem that when we perform an expansion of ${\\left(a + b\\right)}^{n} , n \\in {\\mathbb{Z}}^{+}$, the exponent of $a$ in each term decreases from $n$ to $0$ and the exponent of $b$ in each term increases from $0$ to $n$. In this expansion, $x$ has a positive coefficient, so the sign will stay the same, but $- 1$ will have a sign change for every alternating term. Bearing all of this in mind, ${\\left(x - 1\\right)}^{4}$ $= 1 {\\left(- 1\\right)}^{0} {x}^{4} + 4 {\\left(- 1\\right)}^{1} {x}^{3} + 6 {\\left(- 1\\right)}^{2} {x}^{2}$ $+ 4 {\\left(- 1\\right)}^{3} {x}^{1} + 1 {\\left(- 1\\right)}^{4} {x}^{0}$ $= {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$ • 12 minutes ago • 13 minutes ago • 14 minutes ago • 15 minutes ago • 4 minutes ago •", "b^{2}+4 a b^{3}+b^{4}$$ $$5$$ $$a^{4}$$ $$b^{4}$$ $$(a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}$$ $$6$$ $$a^{5}$$ $$b^{5}$$ $$(a+b)^{n}$$ $$n$$ $$a^{n}$$ $$b^{n}$$ Table 12.4.1 Notice the first and last terms show only one variable. Recall that $$a^{0}=1$$, so we could rewrite the first and last terms to include both variables. For example, we could expand $$(a+b)^{3}$$ to show each term with both variables. Generally, we don’t show the zero exponents, just as we usually write $$x$$ rather than $$1x$$. ##### Note Patterns in the Expansion of $$(a+b)^{n}$$ • The number of terms is $$n+1$$. • The first term is $$a^{n}$$ and the last term is $$b^{n}$$. • The exponents on $$a$$ decrease by one on each term going left to right. • The exponents on $$b$$ increase by one on each term going left to right. • The sum of the exponents on any term is $$n$$. Let’s look at an example to highlight the last three patterns. From the patterns we identified, we see the variables in the expansion of $$(a+b)^{n}$$, would be $$(a+b)^{n}=a^{n}+\\_\\_\\_a^{n-1}b^{1}+\\_\\_\\_a^{n-2}b^{2}+\\ldots+\\_\\_\\_a^{1}b^{n-1}+b^{n}$$. To find the coefficients of the terms, we write our expansions again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. The array to the right is called Pascal’s Triangle. Notice each number in the array", "The coefficient of $x^{13}$ in the expansion of $(1-x)^{5}(1+x+x^{2}+x^{3})^{4}$ is • A $4$ • B $-4$ • C $0$ • D none of these #### Solution Given, $(1-x)^{5}(1+x+x^2+x^3)^{4}$ $=(1-x)^{5}((1+x)(1+x^2))^{4}$ $=(1-x)(1-x^2)^{4}(1+x^2)^{4}$ $=(1-x)(1-x^4)^{4}$ $=(1-x^{4})^{4}-x(1-x^{4})^{4}$ Hence the coefficient of $x^{13}$ will be $=-[(-1)^{3}\\:^{4}C_{3}]$ $=4$", "# How do you expand (a-b)^3? Feb 22, 2017 ${a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$ #### Explanation: Use the Binomial expansion (note the exponents sum to the power in each term): ${\\left(x + y\\right)}^{3} {=}_{3} {C}_{0} {x}^{3} {y}^{0} {+}_{3} {C}_{1} {x}^{2} {y}^{1} {+}_{3} {C}_{2} {x}^{1} {y}^{2} {+}_{3} {C}_{3} {x}^{0} {y}^{3}$ Remember 3! = 3*2*1 = 6, 2! = 2*1 = 2, 1! = 1 and 0! = 1 _3C_0 = (3!)/((3-0)!(0!)) = (3!)/((3)!1) = 1 _3C_1 = (3!)/((3-1)!(1!)) = (3!)/((2)!1) = (3*2!)/(2!) = 3 _3C_2 = (3!)/((3-2)!(2!)) = (3!)/((1)!2!) = (3*2!)/(2!) = 3 _3C_3 = (3!)/((3-3)!(3!)) = (3!)/(0!*3!) = 1 Note: ${\\left(a - b\\right)}^{3} = {\\left(a + \\left(- b\\right)\\right)}^{3}$ Substitute into the Binomial expansion formula, let $x = a$ and $y = - b$: ${\\left(a - b\\right)}^{3} = {a}^{3} + 3 {a}^{2} {\\left(- b\\right)}^{1} + 3 a {\\left(- b\\right)}^{2} + {\\left(- b\\right)}^{3}$ $= {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$", "a rational expression like, $5x+6(x+4)(x+6)5x+6(x+4)(x+6)$ we would like to rewrite it as an addition of two simpler rational expressions $A(x+4)A(x+4)$ and $B(x+6)B(x+6)$ . Our goal is to find the values of A and B such that $5x+6(x+4)(x+6)=Ax+4+Bx+65x+6(x+4)(x+6)=Ax+4+Bx+6$ Find the LCD of the denominators Multiply both sides of the equation by the LCD. Distribute and cancel like terms $(x+4)(x+6)5x+6(x+4)(x+6)=Ax+4+Bx+6(x+4)(x+6)(x+4)(x+6)5x+6(x+4)(x+6)=Ax+4(x+4)(x+6)+Bx+6(x+4)(x+6)5x+16=A(x+6)+B(x+4)(x+4)(x+6)5x+6(x+4)(x+6)=Ax+4+Bx+6(x+4)(x+6)(x+4)(x+6)5x+6(x+4)(x+6)=Ax+4(x+4)(x+6)+Bx+6(x+4)(x+6)5x+16=A(x+6)+B(x+4)$ On the right side, we expand and collect terms with like terms $5x+16=A(x+6)+B(x+4)5x+16=Ax+6A+Bx+B5x+16=(A+B)x+(6A+4B)5x+16=A(x+6)+B(x+4)5x+16=Ax+6A+Bx+B5x+16=(A+B)x+(6A+4B)$ We compare the coefficients of both sides. This will give a system of two equations with two variables $5x+16=(A+B)x+(6A+4B)A+B=56A+4B=165x+16=(A+B)x+(6A+4B)A+B=56A+4B=16$ Use solving by elimination to find the values of A and B. Rewrite the original rational expression as the addition of two rational expressions with unlike denominators Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions. Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression. ### Decomposing $P( x ) Q( x ) P( x ) Q( x )$ Where Q(x) Has Only Nonrepeated Linear Factors Recall the algebra regarding adding and subtracting rational expressions. These operations", "SEARCH HOME Math Central Quandaries & Queries Question from Nick: I'm an 8th grader and am quite confused with this equation: 3x^2-19x-14=0 The answer is: -2/3 , 7 But I'm not sure how to solve for it. Hi Nick, You have equation with a quadratic expression with integer coefficients on one side and zero on the other side. Your instructions are to solve for the variable. Whenever you have this situation you should first try to factor the quadratic. So the question is, can you find integers $a, b, c$ and $d$ so that $3x^2 - 19x - 14 = (ax + b)(cx + d).$ If you knew $a, b, c$ and $d$ you could expand the right side to verify that your factoring was correct. I would start expanding the right side with $ax \\times cx = a \\times b x^2$ and this must be $3x^2$ so $a \\times c = 3.$ Since $a$ and $c$ are integers one of them must be 3 and the other 1. Thus we have $3x^2 - 19x - 14 = (3x + b)(x + d).$ I would continue expanding the right side with the constant term $b \\times d.$ This must be the constant term on the left side so $b \\times d = -14.$ This tells me two things,", "Browse Questions # Find the $4^{th}$ term in the expansion of $(x-2y)^{12}$ $\\begin{array}{1 1}(A)\\;-1760\\\\(B)\\;-760x^9\\\\(C)\\;-1760x^9y^3\\\\(D)\\;x^9y^3\\end{array}$ Can you answer this question? Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-......+(-1)^nnC_ra^{n-r}b^r+......nC_n(-b)^n$ $4^{th}$ term =$T_{3+1}$ in the expansion of $x+(-2y)^{12}$ $\\Rightarrow 12C_3x^{12-3}[-2y]^3$", "## (Math Quiz 10 – CT) In the expansion ({left( {a – 2b} right)^8}), the coefficient of the containing term ({a^4}. {b^4}) is: • Question: In the expansion $${\\left( {a – 2b} \\right)^8}$$, the coefficient of the term containing $${a^4}.{b^4}$$ is: Reference explanation: The general term in the above expansion is $${T_{k + 1}} = C_8^k. {a^{8 – k}}. {\\left( { – 2} \\right)^k}. {b^k}$$ Ask the problem to happen when k = 4. Then the coefficient of the term containing $${a^4}.{b^4}$$ is: $$C_8^4{.2^4} = 1120$$.", "$\\D x^{12}.$ (Total 6 marks) 9.) One of the terms of the expansion of $\\D (x + 2y)^{10}$ is $\\D ax^8 y^2.$ Find the value of a. (Total 6 marks) 10.) (a) Expand $\\D \\left(e+\\frac{1}{e}\\right)^4$ in terms of $\\D e.$ (b) Express $\\D \\left(e+\\frac{1}{e}\\right)^4+\\left(e-\\frac{1}{e}\\right)^4$ as the sum of three terms. (Total 6 marks) 11.) Consider the expansion of $\\D (x^2- 2)^5.$ (a) Write down the number of terms in this expansion. (b) The first four terms of the expansion in descending powers of $\\D x$ are $x^{10} -10x^8 + 40x^6 + Ax^4 + ...$ Find the value of A. (Total 6 marks) 12.) Given that $\\D \\left(3+\\sqrt{7}\\right)^3=p+\\sqrt{q}$ where $\\D p$ and $\\D q$ are integers, find (a) $\\D p;$ (b) $\\D q.$ (Total 6 marks) 13.) When the expression $\\D (2 + ax)^{10}$ is expanded, the coefficient of the term in $\\D x^3$ is 414720. Find the value of $\\D a.$ (Total 6 marks) 14.) Find the term containing $\\D x^3$ in the expansion of $\\D (2- 3x)^8.$ (Total 6 marks) 15.) Find the term containing $\\D x^{10}$ in the expansion of $\\D (5 + 2x^2)^7.$ (Total 6 marks) 16.) Complete the following expansion. $\\D (2 + ax)^4 = 16 + 32ax +\\cdots$ (Total 6 marks) 17.) Consider the expansion of $\\D \\left(3x^2-\\frac{1}{x}\\right)^9.$ (a) How many terms are", "x^{12}.$ (Total 6 marks) 9.) One of the terms of the expansion of $\\D (x + 2y)^{10}$ is $\\D ax^8 y^2.$ Find the value of a. (Total 6 marks) 10.) (a) Expand $\\D \\left(e+\\frac{1}{e}\\right)^4$ in terms of $\\D e.$ (b) Express $\\D \\left(e+\\frac{1}{e}\\right)^4+\\left(e-\\frac{1}{e}\\right)^4$ as the sum of three terms. (Total 6 marks) 11.) Consider the expansion of $\\D (x^2- 2)^5.$ (a) Write down the number of terms in this expansion. (b) The first four terms of the expansion in descending powers of $\\D x$ are $x^{10} -10x^8 + 40x^6 + Ax^4 + ...$ Find the value of A. (Total 6 marks) 12.) Given that $\\D \\left(3+\\sqrt{7}\\right)^3=p+\\sqrt{q}$ where $\\D p$ and $\\D q$ are integers, find (a) $\\D p;$ (b) $\\D q.$ (Total 6 marks) 13.) When the expression $\\D (2 + ax)^{10}$ is expanded, the coefficient of the term in $\\D x^3$ is 414720. Find the value of $\\D a.$ (Total 6 marks) 14.) Find the term containing $\\D x^3$ in the expansion of $\\D (2- 3x)^8.$ (Total 6 marks) 15.) Find the term containing $\\D x^{10}$ in the expansion of $\\D (5 + 2x^2)^7.$ (Total 6 marks) 16.) Complete the following expansion. $\\D (2 + ax)^4 = 16 + 32ax +\\cdots$ (Total 6 marks) 17.) Consider the expansion of $\\D \\left(3x^2-\\frac{1}{x}\\right)^9.$ (a) How many terms are there", "# Expansion of (x + a)(x + b)(x + c) We will discuss here about the expansion of (x + a)(x + b)(x + c). (x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)} = (x + a){x$$^{2}$$ + (b + c)x + bc} = x{x$$^{2}$$ + (b + c)x + bc} + a{x$$^{2}$$ + (b + c)x + bc} = x$$^{3}$$ + (b + c)x$$^{2}$$ + bcx + ax$$^{2}$$ + a(b + c)x + abc = x$$^{3}$$ + (a + b + c)x$$^{2}$$ + (bc + ab + ac)x + abc = x$$^{3}$$ + (a + b + c)x$$^{2}$$ + (ab + bc + ca)x + abc Therefore, (x + a)(x + b)(x + c) = x$$^{3}$$ + (Sum of the constant terms)x$$^{2}$$ + (Sum of the product of constant terms taking two at a time)x + Product of constant terms. Solved Examples on Expansion of (x + a)(x + b)(x + c) 1. Find product of (x + 1)(x + 2)(x + 3) Solution: We know that, (x + a)(x + b)(x + c) = x$$^{3}$$ + (a + b + c)x$$^{2}$$ + (ab + bc + ca)x + abc Here, a = 1, b = 2 and c = 3 Therefore, the product = x$$^{3}$$ + (1 + 2", "Question # The coefficient of ${a^3}{b^4}c$ in the expansion of ${\\left( {1 + a - b + c} \\right)^9}$ is equal to$A.{\\text{ }}\\dfrac{{9!}}{{3!6!}} \\\\ B.{\\text{ }}\\dfrac{{9!}}{{4!5!}} \\\\ C.{\\text{ }}\\dfrac{{9!}}{{3!5!}} \\\\ D.{\\text{ }}\\dfrac{{9!}}{{3!4!}} \\\\$ Hint: In order to solve such a question of expansion and finding its coefficient first assume the coefficient as some unknown variable. Use permutation to find the number of such arrangements of a, b& c in order to find the unknown variable. Given expansion is ${\\left( {1 + a - b + c} \\right)^9}$ Let the coefficient of ${a^3}{b^4}c$ is $\\lambda$ . Since the expansion has an overall power of 9. So we know that the sum of overall power of each term of expansion must be 9. In order to make the overall power of the given term as 9 let us modify the term as follows ${a^3}{b^4}c = {1^1}{a^3}{b^4}{c^1}$ Now we have 9 variables some of which are repeating in the given term $\\left\\{ {3\\left( a \\right),4\\left( b \\right),1\\left( c \\right),1\\left( 1 \\right)} \\right\\}$ Since the coefficient of each term within the expansion term is 1. So the coefficient of the given term will be the number of ways the 9 variables can be arranged. Now the number of different arrangements of $\\left\\{ {3\\left( a \\right),4\\left( b \\right),1\\left( c \\right),1\\left( 1 \\right)} \\right\\}$", "- 12xy - 6xy$ $- 12xy - \\left( {6xy} \\right) = - 18xy$ Therefore, the obtained term is $- 18xy$. (iii) $\\left( {a - b} \\right)$ from $\\left( {a + b} \\right)$ Ans: Combine the like terms (the terms which have the same variables and power) of variable $a$ and the variable $b$ then subtract the coefficients. $\\left( {a + b} \\right) - \\left( {a - b} \\right) = a + b - a + b$ $\\left( {a + b} \\right) - \\left( {a - b} \\right) = a - a + b + b$ $\\left( {a + b} \\right) - \\left( {a - b} \\right) = 2b$ Therefore, the obtained term is $2b$. (iv) $a\\left( {b - 5} \\right)$ from $b\\left( {5 - a} \\right)$ Ans: Solve the brackets then combine the like terms (the terms which have the same variables and power) of variable $ab$ and then subtract the coefficients. $b\\left( {5 - a} \\right) - a\\left( {b - 5} \\right) = 5b - ab - ab + 5a$ $b\\left( {5 - a} \\right) - a\\left( {b - 5} \\right) = 5b - 2ab + 5a$ $b\\left( {5 - a} \\right) - a\\left( {b - 5} \\right) = 5a + 5b - 2ab$ Therefore, the obtained term is $5a + 5b - 2ab$. (v) $- {m^2}" ]

[ "a and b. (ii). Hence, with the values of a and b found in (i), expand $$\\frac{ {(1 \\ + \\ ax)}^{b} }{1 \\ – \\ x}$$ as a series in ascending powers of x up to and including the term in $${\\small x^2 }$$. $${\\small 6. \\enspace}$$ Expand $${\\small {(1-2p)}^{8} }$$ up to and including the term in $${\\small p^3 }$$. Hence, find the first four terms in the expansion in ascending powers of x of $${\\small {(1 \\ – \\ 4x \\ + \\ \\frac{2}{x})}^{8} }$$. $${\\small 7. \\enspace}$$ Find the coefficient of $${\\small {x}^{2}}$$ in the expansion of $${\\small {(1 \\ + \\ 3x)}^{2}{(1 \\ – \\ 3x)}^{10} }$$. $${\\small 8. \\enspace}$$ Find the coefficient of $${ \\small x }$$ in the expansion of $${\\small {(3x \\ – \\ \\frac{2}{x})}^{9} }$$. $${\\small 9. \\enspace}$$ Find the value of a and b in the following expansion: $${\\small \\hspace{1.5em} {(a + bx)}^{9} \\ = \\ … + 672{x}^{3} + 252 {x}^{4} + … }$$ $${\\small 10.\\enspace}$$ Find the value of a and b in the expansion of $${\\small {(1 \\ + \\ ax)}^{5}{(1 \\ – \\ bx)}^{7} }$$ if the coefficients of $${ \\small x }$$ and $${\\small {x}^{2}}$$ are 3 and -9 respectively. As always, if you have any particular questions to discuss, leave it in", "and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Hello s2951 I'm not sure where you get the term $ax$ in the line that I've indicated. When you divide $4x^2-7x-15$ by $x-3$, the result is simply $4x+5$. If I may expand on acc100jt's solution a little: If $4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get: $4x^2-7x-15=Ax^2+(B-3A)x -3B$ So, when we compare coefficients: the coefficient of $x^2:4 = A$ and the constant term: $-15 = -3B \\Rightarrow B = 5$ Check with the coefficient of $x$: $B-3A = 5 -12 = -7$, which is correct So the other factor is $Ax+B = 4x+5$ Thanks, just solved and its correct. The letter \"A\" is just symbol to find the value, its redundant. 6. Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ 7. Hello Raoh Originally Posted by Raoh Here's another way,since $(x-3)$ is a factor, $-3$ is a root. We can do the following, $f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$ I think what you mean is: Since $(x-3)$ is a factor, $+3$ is a root $\\Rightarrow f(3)=0$ So: $f(x) = f(x) - f(3) = 4(x^2-3^2)", "$(3-x)^{4}$. Hence find the coefficient of $x^{2}$ in the expansion of $(1+2 x)^{5}(3-x)^{4}$. $\\mbox{ (5 marks)}$ 2.Using binomial theorem, find the coefficient of $x^{2}$ in the expansion of $\\left(3+x-2 x^{2}\\right)^{5}$. $\\mbox{ (5 marks)}$ 3.If the coefficient of $x^{2}$ in the expansion of $(4+k x)(2-x)^{6}$ is 384 , find the value of $k$ and the coefficient of $x^{3}$ in that expansion. $\\mbox{ (5 marks)}$ 4.Find the term independent of $x$ in the expansion of $\\left(5 x+\\frac{5}{x}\\right)^{6}\\left(5 x-\\frac{5}{x}\\right)^{6}$. $\\mbox{ (5 marks)}$ 5.In the expansion of $\\left(x^{2}-\\frac{2}{x}\\right)^{n}$ in descending power of $x$, the $5^{\\text {th }}$ term is independent of $x .$ Find the value of $n$ and the $4^{\\text {th }}$ term. $\\mbox{ (5 marks)}$ 6.In the binomial expansion of $\\left(1+\\frac{1}{4}\\right)^{n}$, the third term is twice the fourth term. Calculate the value of $n$. $\\mbox{ (5 marks)}$ 7.In the expansion of $\\left(1+\\frac{1}{3}\\right)^{n}$, the third term and the fourth term are in the ratio $3: 2$. Find the value of $n$ and the middle term of that expansion. $\\mbox{ (5 marks)}$ 8.If the $2^{\\text {nd }}$ and the $3^{r d}$ terms in $(a+b)^{n}$ are in the same ratio as the $3^{\\text {rd }}$ and the $4^{\\text {th }}$ terms in $(a+b)^{n+3}$, find the value of $n .$ Calculate also the middle term of $(a+b)^{n+3}$. $\\mbox{ (5 marks)}$ $\\quad\\;$$\\, 1.(1+2 x)^{5}=1+10 x+40", "Calculus (3rd Edition) $$A=-3$$ $$B=5$$ $\\frac{2x-3}{(x-3)(x-4)}$ = $\\frac{A}{x-3}$ + $\\frac{B}{x-4}$ Multiplying (x-3)(x-4) on both sides, we have: $2x-3=A(x-4)+B(x-3)$ If we let x=4 on both sides, we can eliminate constant A to find constant B: $8-3=A(4-4)+B(4-3)$ Thus, $B = 5$. Similarly, if we let x = 3 on both sides, we can eliminate constant B to find constant A: $6-3=A(3-4)+B(3-3)$ Thus, $3=-A$ and so $A = -3$.", "x^{12}.$ (Total 6 marks) 9.) One of the terms of the expansion of $\\D (x + 2y)^{10}$ is $\\D ax^8 y^2.$ Find the value of a. (Total 6 marks) 10.) (a) Expand $\\D \\left(e+\\frac{1}{e}\\right)^4$ in terms of $\\D e.$ (b) Express $\\D \\left(e+\\frac{1}{e}\\right)^4+\\left(e-\\frac{1}{e}\\right)^4$ as the sum of three terms. (Total 6 marks) 11.) Consider the expansion of $\\D (x^2- 2)^5.$ (a) Write down the number of terms in this expansion. (b) The first four terms of the expansion in descending powers of $\\D x$ are $x^{10} -10x^8 + 40x^6 + Ax^4 + ...$ Find the value of A. (Total 6 marks) 12.) Given that $\\D \\left(3+\\sqrt{7}\\right)^3=p+\\sqrt{q}$ where $\\D p$ and $\\D q$ are integers, find (a) $\\D p;$ (b) $\\D q.$ (Total 6 marks) 13.) When the expression $\\D (2 + ax)^{10}$ is expanded, the coefficient of the term in $\\D x^3$ is 414720. Find the value of $\\D a.$ (Total 6 marks) 14.) Find the term containing $\\D x^3$ in the expansion of $\\D (2- 3x)^8.$ (Total 6 marks) 15.) Find the term containing $\\D x^{10}$ in the expansion of $\\D (5 + 2x^2)^7.$ (Total 6 marks) 16.) Complete the following expansion. $\\D (2 + ax)^4 = 16 + 32ax +\\cdots$ (Total 6 marks) 17.) Consider the expansion of $\\D \\left(3x^2-\\frac{1}{x}\\right)^9.$ (a) How many terms are there", "{(1-2p)}^{8} }$$ up to and including the term in $${\\small p^3 }$$. Hence, find the first four terms in the expansion in ascending powers of x of $${\\small {(1 \\ – \\ 4x \\ + \\ \\frac{2}{x})}^{8} }$$. $${\\small 7. \\enspace}$$ Find the coefficient of $${\\small {x}^{2}}$$ in the expansion of $${\\small {(1 \\ + \\ 3x)}^{2}{(1 \\ – \\ 3x)}^{10} }$$. $${\\small 8. \\enspace}$$ Find the coefficient of $${ \\small x }$$ in the expansion of $${\\small {(3x \\ – \\ \\frac{2}{x})}^{9} }$$. $${\\small 9. \\enspace}$$ Find the value of a and b in the following expansion: $${\\small \\hspace{1.5em} {(a + bx)}^{9} \\ = \\ … + 672{x}^{3} + 252 {x}^{4} + … }$$ $${\\small 10.\\enspace}$$ Find the value of a and b in the expansion of $${\\small {(1 \\ + \\ ax)}^{5}{(1 \\ – \\ bx)}^{7} }$$ if the coefficients of $${ \\small x }$$ and $${\\small {x}^{2}}$$ are 3 and -9 respectively. As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) . 1. 1/3√1-x 2. Can you please expand this one ASAP to the power of x^3 . 3(1-x)^-1 – ((6x+1)(2x^2 – 1)^-1) 3. 11+x = (1+x)−1 = 1+((−1)x+(−1)(−2))/2!x^2+((−1)(−2)(−3))/3!x3+… = 1 – x + x2 – x3 + … – … this example question has", "$\\D x^{12}.$ (Total 6 marks) 9.) One of the terms of the expansion of $\\D (x + 2y)^{10}$ is $\\D ax^8 y^2.$ Find the value of a. (Total 6 marks) 10.) (a) Expand $\\D \\left(e+\\frac{1}{e}\\right)^4$ in terms of $\\D e.$ (b) Express $\\D \\left(e+\\frac{1}{e}\\right)^4+\\left(e-\\frac{1}{e}\\right)^4$ as the sum of three terms. (Total 6 marks) 11.) Consider the expansion of $\\D (x^2- 2)^5.$ (a) Write down the number of terms in this expansion. (b) The first four terms of the expansion in descending powers of $\\D x$ are $x^{10} -10x^8 + 40x^6 + Ax^4 + ...$ Find the value of A. (Total 6 marks) 12.) Given that $\\D \\left(3+\\sqrt{7}\\right)^3=p+\\sqrt{q}$ where $\\D p$ and $\\D q$ are integers, find (a) $\\D p;$ (b) $\\D q.$ (Total 6 marks) 13.) When the expression $\\D (2 + ax)^{10}$ is expanded, the coefficient of the term in $\\D x^3$ is 414720. Find the value of $\\D a.$ (Total 6 marks) 14.) Find the term containing $\\D x^3$ in the expansion of $\\D (2- 3x)^8.$ (Total 6 marks) 15.) Find the term containing $\\D x^{10}$ in the expansion of $\\D (5 + 2x^2)^7.$ (Total 6 marks) 16.) Complete the following expansion. $\\D (2 + ax)^4 = 16 + 32ax +\\cdots$ (Total 6 marks) 17.) Consider the expansion of $\\D \\left(3x^2-\\frac{1}{x}\\right)^9.$ (a) How many terms are", "# $(x^3 + \\frac{a}{x^2})^5 = -270$, find $a$ I am working on my scholarship exam practice and not sure how to begin. Please assume math knowledge at high school or pre-university level. Let $$a$$ be a real constant. If the constant term of $$(x^3 + \\frac{a}{x^2})^5$$ is equal to $$-270$$, then $$a=$$...... Could you please give a hint for this question? The answer provided is $$-3$$. From the binomial theorem, we know that the sum of the powers of $$x^3$$ and $$a/x^2$$ must add to $$5$$. So if our term in the expansion is $$k (x^3)^p (\\frac{a}{x^2})^q$$, $$p+q = 5$$. Moreover, we want our term to be constant (no nonzero powers of $$x$$), so we have $$3p - 2q = 0$$. Solving these two expressions, we have $$p = 2, q =3$$. So we must find $$a$$ such that the coefficient of the $$(x^3)^2 (\\frac{a}{x^2})^3$$ term is $$270$$. The binomial theorem says that the coefficient of this power in the expansion is $${5 \\choose 2} = 10$$, so in particular, the coefficient of the constant term is $$10a^3$$, which we want to equal $$270$$. Therefore, $$a = -3$$. Hint: binomial expand $$(x^3+ax^{-2})^5$$. Hint: $$(A+B)^5={A}^{5}+5\\,{A}^{4}B+10\\,{A}^{3}{B}^{2}+10\\,{A}^{2}{B}^{3}+5\\,A{B}^{4}+ {B}^{5}$$ Hint: The general term $$T_{r+1}$$ is $$\\binom5r(x^3)^{5-r}\\left(\\dfrac a{x^2}\\right)^r=\\binom5ra^rx^{3\\cdot5-3r-2r}$$ For the constant term, the exponent of $$x$$ will be $$?$$", "22, question 5) (i) Find and simplify the first three terms of the expansion, in ascending powers of $\\D x,$ of $\\D (1-4x)^5.$ [2] (ii) The first three terms in the expansion of $\\D (1-4x)^5(1+ax+bx^2)$ are $\\D 1- 23x+ 222x^2$. Find the value of each of the constants $\\D a$ and $\\D b.$ [4] 13 (CIE 2014, w, paper 11, question 6) (i) Given that the coefficient of $\\D x^2$ in the expansion of $\\D(2+ px)^6$ is 60, find the value of the positive constant $\\D p.$ [3] (ii) Using your value of $\\D p,$ find the coefficient of $\\D x^2$ in the expansion of $\\D (3- x) (2 +px)^6.$ [3] 14 (CIE 2014, w, paper 13, question 9) (a) Given that the first 3 terms in the expansion of $\\D (5-qx)^p$ are $\\D 625- 1500x +rx^2,$ find the value of each of the integers $\\D p, q$ and $\\D r.$ [5] (b) Find the value of the term that is independent of $\\D x$ in the expansion of $\\D \\left(2x+\\frac{1}{4x^3}\\right)^{12}.$ [3] 15 (CIE 2015, s, paper 11, question 3) (i) Find the first 4 terms in the expansion of $\\D (2+x^2)^6$ in ascending powers of $\\D x.$ [3] (ii) Find the term independent of $\\D x$ in the expansion of $\\D (2+x^2)^6\\left(1-\\frac{3}{x^2}\\right)^2.$ [3] 16 (CIE 2015, s,", "s, paper 12, question 4) The first 3 terms in the expansion of $\\left(3-\\frac{x}{6}\\right)^{n}$ are $81+a x+b x^{2}$. Find the value of each of the constants $n, a$ and $b .$ 26 (CIE 2017, s, paper 21, question 5) (i) Given that $a$ is a constant, expand $(2+a x)^{4}$, in ascending powers of $x$, simplifying each term of your expansion. [2] Given also that the coefficient of $x^{2}$ is equal to the coefficient of $x^{3}$, (ii) show that $a=3$, $[1]$ (iii) use your expansion to show that the value of $1.97^{4}$ is $15.1$ to 1 decimal place. 27 (CIE 2017, s, paper 23, question 6) The first three terms of the binomial expansion of $(2-a x)^{n}$ are $64-16 h x+100 h x^{2}$. Find the value of each of the integers $n, a$ and $b .$ 28 (CIE 2017, w, paper 12, question 3) (i) Find, in ascending powers of $x$, the first 3 terms in the expansion of $\\left(2-\\frac{x^{2}}{4}\\right)^{5}$ [3] (ii) Hence find the term independent of $x$ in the expansion of $\\left(2-\\frac{x^{2}}{4}\\right)^{5}\\left(\\frac{1}{x}-\\frac{3}{x^{2}}\\right)^{2}$. [3] 29 (CIE 2017, w, paper 13, question 7) (i) Find, in ascending powers of $x$, the first 3 terms in the expansion of $\\left(2-\\frac{x^{2}}{4}\\right)^{6}$. Give each term in its simplest form. [3] (ii) IIcnce find the cocfficient of $x^{2}$ in the cxpansion of $\\left(2-\\frac{x^{2}}{4}\\right)^{6}\\left(\\frac{1}{x}+x\\right)^{2}$.", "paper 22, question 5) (i) Find and simplify the first three terms of the expansion, in ascending powers of $\\D x,$ of $\\D (1-4x)^5.$ [2] (ii) The first three terms in the expansion of $\\D (1-4x)^5(1+ax+bx^2)$ are $\\D 1- 23x+ 222x^2$. Find the value of each of the constants $\\D a$ and $\\D b.$ [4] 13 (CIE 2014, w, paper 11, question 6) (i) Given that the coefficient of $\\D x^2$ in the expansion of $\\D(2+ px)^6$ is 60, find the value of the positive constant $\\D p.$ [3] (ii) Using your value of $\\D p,$ find the coefficient of $\\D x^2$ in the expansion of $\\D (3- x) (2 +px)^6.$ [3] 14 (CIE 2014, w, paper 13, question 9) (a) Given that the first 3 terms in the expansion of $\\D (5-qx)^p$ are $\\D 625- 1500x +rx^2,$ find the value of each of the integers $\\D p, q$ and $\\D r.$ [5] (b) Find the value of the term that is independent of $\\D x$ in the expansion of $\\D \\left(2x+\\frac{1}{4x^3}\\right)^{12}.$ [3] 15 (CIE 2015, s, paper 11, question 3) (i) Find the first 4 terms in the expansion of $\\D (2+x^2)^6$ in ascending powers of $\\D x.$ [3] (ii) Find the term independent of $\\D x$ in the expansion of $\\D (2+x^2)^6\\left(1-\\frac{3}{x^2}\\right)^2.$ [3] 16 (CIE 2015,", "marks)}$ 3. Find the coefficient of $x^{2}$ in the expansion of $\\left(2 x^{2}-x-3\\right)^{6}$. $\\quad\\mbox{ (5 marks)}$ 4. Find the coefficient of $x^{2}$ and $x^{3}$ in the expansion of $\\left(x^{2}-x-2\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ 5. Evaluate the coefficients of $x^{5}$ and $x^{4}$ in the binomial expansion of $\\left(\\frac{x}{3}-3\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ Hence evaluate the coefficient of $x^{5}$ in the expansion of $\\left(\\frac{x}{3}-3\\right)^{7}(x+6)$. $\\quad\\mbox{ (5 marks)}$ 6. In the expansion of $(1+2 x)^{11}$, the coefficient of $x^{3}$ is $k$ times the coefficient of $x^{2}$. Find $k$. $\\quad\\mbox{ (5 marks)}$ 7. If the coefficient of $x^{4}$ in the expansion of $(3+2 x)^{6}$ is equal to the coefficient of $x^{4}$ in the expansion of $(2 k+3 x)^{6}$, find $k$. $\\quad\\mbox{ (5 marks)}$ 8. The ratio of the coefficient of $x^{4}$ in the expansion of $(3-2 x)^{6}$ and the coefficient of $x^{4}$ in the expansion of $(k+2 x)^{7}$ is $1: 7.$ Find the value of $k$. $\\quad\\mbox{ (5 marks)}$ 9. The first three terms in the binomial expasion of $(a+b)^{n}$, in ascending power of $b$, are denoted by $p, q$ and $r$ respectively. Show that $\\frac{q^{2}}{p r}=\\frac{2 n}{n-1}$. Given that $p=4, q=32$ and $r=96$, evaluate $n$. $\\quad\\mbox{ (5 marks)}$ 10. In the binomial expansion of $(1+x)^{n}$ the first three terms are $1+3+4+\\ldots$ Calculate the numerical values of $n$ and $x$, and the value of", ", on both sides and can drop the bases to set the exponents equal to each other and solve for $r$ . $(x^3)^{4-r} (x^{-1})^r &=x^0 \\\\x^{12-3r} \\cdot x^{-r} &=x^0 \\\\x^{12-3r-r} &=x^0 \\\\x^{12-4r} &=x^0 \\\\12-4r &=0 \\\\12 &=4r \\\\r &=3$ Now, plug the value of $r$ into the rule to get the constant term in the expansion. $\\dbinom{4}{3}(4x^3)^{4-3}\\left(\\frac{1}{x}\\right)^3=4(4x^3)\\left(\\frac{1}{x}\\right)^3=16.$ ### Practice Expand the following binomials using the Binomial Theorem. 1. $(x-a)^7$ 2. $(2a+3)^4$ Find the $n^{th}$ term in the expansions of the following binomials. 1. $(7x-2)^5; \\ n = 4$ 2. $\\left(6x+\\frac{1}{2} \\right)^7; \\ n = 3$ 3. $(5-a)^9; \\ n = 7$ 4. $\\left(\\frac{2}{3}x+9y \\right)^6; \\ n = 4$ 5. Find the term with $\\ x^5$ in the expansion of $(3x-2)^7$ . 6. Find the term with $\\ y^6$ in the expansion of $(5-y)^8$ . 7. Find the term with $\\ a^3$ in the expansion of $(2a-b)^{10}$ . 8. Find the term with $\\ x^4$ in the expansion of $(8-3x)^5$ . 9. Find the constant term in the expansion of $\\left(x^2+\\frac{3}{x} \\right)^6$ . 10. Find the constant term in the expansion of $\\left(\\frac{5}{x^3}-x \\right)^8$ .", "# Find the constant term in this binomial expansion? ## ${\\left(2 {x}^{2} - \\frac{1}{x}\\right)}^{6}$ FInd the constant term in the expansion of this binomial. Apr 2, 2017 $60.$ #### Explanation: The General Term, denoted by, ${T}_{r + 1} ,$ in the Expansion of ${\\left(a + b\\right)}^{n}$ is, T_(r+1)=\"\"_nC_ra^(n-r)b^r, r=0,1,,...,n. With, a=2x^2, b=-1/x, n=6, T_(r+1)=\"\"_6C_r(2x^2)^(6-r)(-1/x)^r. =\"\"_6C_r(2)^(6-r)(-1)^r(x)^(12-2r)x^(-r) =\"\"_6C_r(2)^(6-r)(-1)^rx^(12-3r)..............(ast) For the Const. Term, the index of $x$ must be $0$. $\\therefore 12 - 3 r = 0 \\Rightarrow r = 4.$ (ast) rArr T_(4+1)=T_5=\"\"_6C_4(2)^(6-4)(-1)^4x^(12-(3)(4)), =\"\"_6C_2*2^2*(1), $= \\frac{\\left(6\\right) \\left(5\\right)}{\\left(1\\right) \\left(2\\right)} \\cdot 4.$ Hence, the desired const. term is $60$, and is the ${5}^{t h}$ term in the Expansion. Enjoy Maths.!", "+0 # i need some help with this 0 109 2 +794 Find $A$ and $B$ such that $$\\frac{4x}{x^2-8x+15} = \\frac{A}{x-3} + \\frac{B}{x-5}$$for all x besides 3 and 5. Express your answer as an ordered pair in the form $(A, B).$ Jun 20, 2020 #1 +781 +1 $$\\frac{4x}{x^2-8x+15}=\\frac{A}{x-3}+\\frac{B}{x-5}\\\\ \\frac{4x}{x^2-8x+15}=\\frac{A(x-5)}{x^2-8x+15}+\\frac{B(x-3)}{x^2-8x+15}\\\\ 4x=A(x-5)+B(x-3)\\\\ 4x=Ax-5A+Bx-3B\\\\$$ The x terms, Ax and Bx must add up to 4x so A+B=4. The constant terms, -5a and -3B must add up to 0 so -5A-3B=0. We have a system of equations: A+B=4 -5A-3B=0 Solving, we get A= -6 and B=10. Jun 20, 2020 #2 +794 -1 thanks! AnimalMaster Jun 20, 2020", "the term in $\\D x^2$ in $\\D (2 + x)^4\\left(1+\\frac{1}{x^2}\\right).$ (Total 6 marks) 3.) Find the term in $\\D x^4$ in the expansion of $\\D \\left(3x^2-\\frac{2}{x}\\right)^5$. (Total 6 marks) 4.) The fifth term in the expansion of the binomial $\\D (a + b)^n$ is given by $\\D \\iixi{10}{4}p^6(2q)^4.$ (a) Write down the value of $\\D n$. (b) Write down $\\D a$ and $\\D b,$ in terms of $\\D p$ and/or $\\D q.$ (c) Write down an expression for the sixth term in the expansion. (Total 6 marks) 5.) Let $\\D f(x) = x^3- 4x + 1.$ (a) Expand $\\D (x + h)^3.$ 6.) Find the term in $\\D x^3$ in the expansion of $\\D \\left(\\frac{2}{3}x-3\\right)^8.$ (Total 5 marks) 7.) (a) Expand $\\D (x-2)^4$ and simplify your result. (b) Find the term in $\\D x^3$ in $\\D (3x + 4)(x-2)^4.$ (Total 6 marks) 8.) Consider the expansion of the expression $\\D (x^3-3x)^6.$ (a) Write down the number of terms in this expansion. (b) Find the term in $\\D x^{12}.$ (Total 6 marks) 9.) One of the terms of the expansion of $\\D (x + 2y)^{10}$ is $\\D ax^8 y^2.$ Find the value of a. (Total 6 marks) 10.) (a) Expand $\\D \\left(e+\\frac{1}{e}\\right)^4$ in terms of $\\D e.$ (b) Express $\\D \\left(e+\\frac{1}{e}\\right)^4+\\left(e-\\frac{1}{e}\\right)^4$ as the sum of three terms. (Total 6", "division and long division for numbers: Polynomial long division doesn't involve a carry. Consider $3x(x^2+ax+b)$ what contribution do the $ax+b$ make to the leading term (the $x^3$ term) of the answer? A: they don't have any effect ... the leading term of $3x^3+ax^2+bx$ is $3x^3$, which doesn't involve an $a$ nor a $b$. – John Joy Jan 6 '15 at 18:28 As said, \"it works the same way as euclidean division works for integers\" (quoting Bernard). So, if you take $$A=\\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}$$ The division of $3x^3$ by $x^2$ (the highest terms) gives $3x$. So $$A=3x+\\Big(\\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}-3x\\Big)=3x -\\frac{11 x^2+5x+3}{x^2 + 3x + 3}=3x-B$$ Now, consider $$B=\\frac{11 x^2+5x+3}{x^2 + 3x + 3}$$ The division of $11x^2$ by $x^2$ (the highest terms) gives $11$. So $$B=\\frac{11 x^2+5x+3}{x^2 + 3x + 3}=11+\\Big(\\frac{11 x^2+5x+3}{x^2 + 3x + 3}-11\\Big)=11-\\frac{28 x+30}{x^2 + 3x + 3}=11-C$$ Now, the highest degree in the numerator of $C$ is smaller that the highest degree in its denominator so you stop and have the remainder. So, by the end $$A=\\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}=3x-11-\\frac{28 x+30}{x^2 + 3x + 3}$$ Some trick on how to do this quickly: $3x^3 - 2x^2", "? Free Version Difficult # Constant Term in an Expansion COMBIN-5X4GEF What is the constant term $c_0$ in the expansion? $$\\left(x^2+2+\\frac{1}{x^2}\\right)^4z$$ $$=c_8x^8 + c_7x^7 + \\cdots + c_0 + + c_1x\\cdots + \\frac{c_{-1}}{x}1x+\\frac{c_{-7}}{x^7} + \\frac{c_{-8}}{x^8}$$ A $16$ B $32$ C $64$ D $70$ E $256$ F none of the above", "# Constant such that $\\max\\left(\\frac{5}{5-3c},\\frac{5b}{5-3d}\\right)\\geq k\\cdot\\frac{2+3b}{5-c-2d}$ What is the greatest constant $k>0$ such that $$\\max\\left(\\frac{5}{5-3c},\\frac{5b}{5-3d}\\right)\\geq k\\cdot\\frac{2+3b}{5-c-2d}$$ for all $0\\leq b\\leq 1$ and $0\\leq c\\leq d\\leq 1$? The right-hand side looks like a weighted sum of the two terms on the left-hand side, but not quite. If we plug in $b=1$ and $c=d$, then all three terms are equal, so $k\\leq 1$. On the other hand, we have $k\\geq 3/5$. Indeed, we will show that $$\\frac{5}{5-3c}\\geq\\frac35\\cdot\\frac{2+3b}{5-c-2d}.$$ Note that $d\\leq 1$ and $2+3b\\leq 5$, so it suffices to show $$\\frac{1}{5-3c}\\geq\\frac35\\cdot\\frac{1}{3-c},$$ or $$5(3-c)\\geq 3(5-3c)$$ or $$15-5c\\geq 15-9c$$ which is true. But the bound is not tight here, since we must have $b=d=1$ and $c=0$, and we have $\\max(5, 5/2)\\geq 5/3$. (The term $\\frac{5b}{5-3d}$, which we did not use at all, is large.) Update: By dividing the cases into whether $b\\leq 3/5$ (and compare with the first term in the $\\max$ if so) or $b\\geq 3/5$ (and compare with the second term in the $\\max$ if so), we can show that $k\\geq 15/19$. Moreover, as Aravind pointed out in the comments, we have $k\\leq 15/16$. So the gap is now between $15/19$ and $15/16$. Update 2: WolframAlpha confirms that $k=15/16$ is the right answer. The question is now how to prove it: http://www.wolframalpha.com/input/?i=find+minimum+of+max(5%2F(5-3c),(5b)%2F(5-3d))*(5-c-2d)%2F(2%2B3b)+for+0%3C%3Db%3C%3D1+and+0%3C%3Dc%3C%3Dd%3C%3D1 • Does \"max\" just refer to the largest of", "5B}{x^2+11x+30} = \\frac{(A+B)x + (6A+5B)}{x^2+11x+30}. \\end{align*} For this to work out, you need $(A+B)x + (6A+5B) = -11x-60$. That means we need $A+B=-11$ (so the $x$s agree) and $6A+5B = -60$ (so the constant terms agree). That means $A=-11-B$ (from the first equation). Plugging into the second equation, we get $$-60 = 6A+5B = 6(-11-B)+5B = -66 -6B + 5B = -66-B.$$ So that means that $B=60-66 = -6$. And since $A+B=-11$, then $A=-5$. (An alternative method for finding the values of $A$ and $B$ is the Heaviside cover-up method; from the fact that $$-11x - 60 = A(x+6)+B(x+5)$$ we know that the two sides must take the same value for every value of $x$; if we plug in $x=-6$, this will \"cover up\" the $A$ on the right and we simply get $B(-6+5) = -B$; this must equal $-11(-6)-60 = 6$; so $6 = -B$, hence $B=-6$. Then plugging in $x=-5$ \"covers up\" the $B$ to give us $-11(-5)-60 = A(-5+6)$, or $-5=A$. So we obtain $A=-5$, $B=-6$, same as before.) That is, $$\\frac{-11x-60}{x^2+11x+30} = \\frac{-5}{x+5} + \\frac{-6}{x+6}.$$ Putting it all together, we have: $$\\frac{2x^2+11x}{x^2+11x+30} = 2 + \\frac{-11x-60}{x^2+11x+30} = 2 - \\frac{5}{x+5} - \\frac{6}{x+6}.$$ And <triumphant trumpet cue>ta da!</triumphant trumpet cue> Your task is done. You've decomposed the quotient into partial fractions. Caveat. You need to" ]

[ "= Solutions (separate by commas)...", "4$ and $z = 5$", "Write all solutions on the same line, separated by commas. ##### Question 3 Solve for the unknown: $-8x+x^2=-6-x-x^2$8x+x2=6xx2 1. Write all solutions on the same line, separated by commas. ##### Question 4 Solve the following equation: $x-\\frac{45}{x}=4$x45x=4 1. Write all solutions on the same line, separated by commas.", "by completing the square: $4x^2+11x+7=0$4x2+11x+7=0 1. Write all solutions on the same line, separated by commas.", "0$, so $z = 4$", "From there you can find, $d = -4$. $$( \\mathcal{P} ): x + 4y + z - 4 = 0$$", "which we detect precisely four solutions.", "= 4 \\text{ is the equation}$", "then to solve the four.", "or $$z^4$$ gives interesting results.", "can then solve. • Alternatively, since $z=0$ is not a solution we could divide by sides by $z^5,$ do a substitution $x=1+1/z$ and solve $x^5=0.$ – Ragib Zaman Aug 4 '20 at 4:23", "4$$ #4: Solutions: a) $$p = 5$$ #5: Solutions: a) $$n = \\frac{5}{3}$$", "the four solutions to the equation $4=s^2$ mod $91$.", "the known $$4$$x$$4$$-solution.", "solution : Find numbers in Z by divid 66.......(66=2×11×3 ) 12 solutions LOOK HERE", "Find the solutions of equation 2^ x equiv x^ 5 rounded to one decitrial place. Enter your answers as a comma-separated .", "for |z|$\\leq$2 $x^2+y^2$=4 What to do next?", "$4$ solutions, so the answer is $\\boxed{\\textbf{(D) } 4}$ ~IceMatrix", "Find all possible values of $z$ that satisfy $|z – 5| + |z + 5| = 12$ and $|z – 5| = \\operatorname{Re} z + 4$.", "me to how they arrived at this please ? $$\\frac{z(z-1)(z+1)}{(z-3)^3}$$ The solution : ..." ]

[ "Note that at least one of $w+3i$, $w+9i$, and $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one. Let $w=x+yi$, where $x$ and $y$ are reals. Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$, so $w=x-3i$. Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+9i}=2w-4$, so $\\overline{x+6i}=2(x-3i)-4$, implying that $x=4$, so $w=4-3i$. Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$, so $w=x-9i$. Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+3i}=2w-4$, so $\\overline{x-6i}=2(x-9i)-4$, so $x+6i=2x-4-18i$, which has no solution as $x$ is real. Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$, so $w=x$. Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$, which means that $\\overline{x+3i}=x+6i$, so $x-3i=x+6i$, which has no solutions. Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(x-(4))(x-(4+6i))(x-(4-6i))$. Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$, so we can plug in $x=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$.", "&= y\\\\ x= \\frac{29}{325} \\ \\text{and} \\ y&=\\frac{28}{65}\\\\ \\end{align*} To find the square roots of a complex number $$a+ib$$ 1. Let $$x+yi= \\sqrt{a+bi}$$ 2. Square both sides: $$(x+yi)^2=a+bi$$ 3. Expand both sides: $$x^2+2xyi-y^2=a+bi$$ 4. Equate real and imaginary parts: $$x^2-y^2=a, 2xy=b$$ 5. Solve these equations simultaneously for $$x$$ and $$y$$ ### Example 2: Find the square roots of $$3+4i$$ \\begin{align*} \\text{Let} \\ x+yi &= \\sqrt{3+4i}\\\\ (x+yi)^2&=3+4i\\\\ x^2+2xyi-y^2&=3+4i\\\\ x^2-y^2=3 , \\ xy&=2\\\\ \\text{Substitute} \\ y&= \\frac{2}{x} \\ \\text{into} \\ \\text{the first equation:}\\\\ x^2- \\frac{4}{x^2} &= 3\\\\ x^4-3x^2-4&=0\\\\ (x^2-4)(x^2+1)&=0\\\\ \\text{As} \\ x \\ \\text{cannot be imaginary,} \\ x&=±2\\\\ \\text{Therefore} \\ \\sqrt{3+4i}&=±(2+i) \\end{align*} ### Example 3: Find the square roots of $$8-6i$$ \\begin{align*} \\text{Let} \\ x+yi &= \\sqrt{8-6i}\\\\ (x+yi)^2 &= 8 -6i\\\\ x^2+2xyi – y^2&=8-6i\\\\ x^2-y^2=8, \\ xy&=-3\\\\ \\text{As} \\ x \\ \\text{cannot be an imaginary number,} \\ x&=±3\\\\ \\text{Therefore} \\sqrt{8-6i} &= ±(3-i)\\\\ \\end{align*} ## Solving Quadratic Equations in the Complex Field A quadratic equation is in the general form $$ax^2+bx+c$$. As such, there are two ways of factorising a quadratic equation. You may choose whichever method you are more comfortable with in an exam. 1. Using the quadratic formula 2. Using the completing the square method ### Example 1: i) Fully factorise $$x^2-6x+10$$ using the Quadratic Formula \\begin{align*} \\text{Consider} \\ x^2-6x+10&=0\\\\ x&= \\frac {6 \\ ± \\ \\sqrt{36-40}}{2}\\\\ &= \\frac{6", "$a = 0$ then $-b^2 = -1$ which is possible if $b = \\pm 1$. If $b = 0$ then the first equation tells us that $a^2 = -1$ which is impossible since $a$ is real number. So $z^2 = -1$ has exactly two solutions: $a = 0$ and $b = \\pm 1$ which correspond to $z = \\pm i$. 2. We repeat the same reasoning as in part (a) except that this time we are trying to solve $(a+bi)^2 = 1$. This leads to the system of equations \\begin{align} a^2 - b^2 &= 1\\\\ 2ab &= 0. \\end{align} If $a = 0$ this is not possible because $-b^2$ can not be 1. So $b = 0$ and then the first equation tells us that $a^2 = 1$ or $a = \\pm 1$. So $z^2 = 1$ has the two expected solutions and no others: $z = \\pm 1$. 3. This time we want to solve $(a+bi)^2 = i$ which is equivalent to $a^2 + 2abi + bi^2 = i$. Equating real and complex parts of the two sides as above we find \\begin{align} a^2 - b^2 &= 0\\\\ 2ab &= 1. \\end{align} The first equation tells us that $a^2 = b^2$. This means that $a = \\pm b$. If $a = b$ then the second equation gives", "by $f(x, y)$. • $\\Rightarrow$ Suppose that $n$ can be properly represented by some binary quadratic form $f(x, y) = ax^2 + bxy + cy^2$ with discriminant $d = b^2 - 4ac$. Then there exists $x_0, y_0 \\in \\mathbb{Z}$ such that $n = f(x_0, y_0)$ and $(x_0, y_0) = 1$. Since $(x_0, y_0) = 1$ there exists integers $m_1, m_2 \\in \\mathbb{Z}$ such that $m_1m_2 = 4|n|$ and $(x_0, m_1) = 1$, $(y_0, m_2) = 1$. Observe that: (4) \\begin{align} \\quad 4af(x, y) = (2ax + by)^2 - dy^2 \\end{align} • Plugging in $x = x_0$ and $y = y_0$ gives us: (5) \\begin{align} \\quad 4an = (2ax_0 + by_0)^2 - dy_0^2 \\end{align} • So $(2ax_0 + by_0)^2 \\equiv dy_0^2 \\pmod {m_2}$. Since $(y_0, m_2) = 1$, there exists $y_0^{-1} \\in \\mathbb{Z}$ such that $y_0y_0^{-1} = 1$. So: (6) \\begin{align} \\quad (y_0^{-1}[2ax_0 + by_0])^2 \\equiv d \\pmod {m_2} \\end{align} • We can similarly obtain a solution to $x^2 \\equiv d \\pmod {m_1}$. By the Chinese Remainder Theorem, there exists a unique solution modulo $m_1m_2 = 4|n|$, that is, $x^2 \\equiv d \\pmod {4|n|}$ has a solution. $\\blacksquare$ Lemma 2: Let $f(x, y)$ be a binary quadratic form. If $p$ is a prime and $p$ is represented by $f(x, y)$ then any such representation is a proper representation. •", "+ x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \\geq 0$? We reduce the problem to one in the non-negative integers. Since $x_1$ is an integer, $x_1 > 1 \\implies x_1 \\geq 2$. Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \\implies x_2 \\geq 2$ and $x_3 > 3 \\implies x_3 \\geq 4$. Let \\begin{align*} y_1 & = x_1 - 2\\\\ y_2 & = x_2 - 2\\\\ y_3 & = x_3 - 4\\\\ y_4 & = x_4 \\end{align*} Then each $y_k$, $1 \\leq k \\leq 4$, is a non-negative integer. Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields \\begin{align*} y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\\\ y_1 + y_2 + y_3 + y_4 & = 4 \\tag{3} \\end{align*} Equation 3 is an equation in the non-negative integers with $$\\binom{4 + 3}{3} = \\binom{7}{3}$$ solutions.", "&= x-y, \\\\ x^2 = x,\\qquad y^2 &= y, \\\\ \\intertext{therefore} y-2xy &= -y, \\\\ \\therefore y &=xy. \\end{align*} In other words, the $Uy$ must be included in the $Ux$ (Fig.\\ 5). Here we have assumed that the law of signs is the same as in ordinary algebra, and the result comes out correct. \\epsfig{file=boole5.eps,width=3cm,height=3.5cm,clip=} Suppose $Uz = Uxy$; then $Ux = U\\frac{1}{y}z$. How are the $Ux$'s related to the $Uy$'s and the $Uz$'s? From the diagram (in Fig.\\ 2) we see that the $Ux$'s are identical with all the $Uyz$'s together with an indefinite portion of the $U$'s, which are neither $y$ nor $z$. Boole discovered a general method for finding the meaning of any function of elementary logical symbols, which applied to the above case, is as follows. When $y$ is an elementary symbol, \\begin{align*} 1 &= y+(1-y). \\\\ \\intertext{Similarly} 1 &= z+(1-z). \\\\ \\therefore 1 &= yz+y(1-z)+(1-y)z+(1-y)(1-z), \\end{align*} which means that the $U$'s either have both qualities $y$ and $z$, or $y$ but not $z$, or $z$ but not $y$, or neither $y$ and $z$. Let $\\frac{1}{y}z = Ayz + By(1 -z) + C(1-y)z + D(1-y)(1-z),$ it is required to determine the coefficients $A$, $B$, $C$, $D$. Suppose $y = 1,$ $z = 1$; then $1 = A$. Suppose $y = 1$, $z = 0$,", "both sides: $$\\int{(x^{-1}y_2)’}dx=\\int{1-x^{-2}}dx$$ \\begin{align}x^{-1}y_2&=\\int{1dx}-\\int{x^{-2}}dx\\\\&=x+x^{-1}+c\\end{align} Hence,$$y_2=x^2+1+cx$$ Let $c=1$, we have $$y_2=x^2+x+1$$ Since we already know that $W(y_1,y_2)(x)=x^2-1\\neq 0$, Hence $y_2(x)$ is indeed another linearly independent solution. $\\\\$ $(c)$ $\\\\$ For $W(y_1,y_2)(x)=x^2-1\\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(x^2+x+1)$$ where $c_1$ and $c_2$ are some arbitrary constants. Now, $$y’=c_1+2c_2x+c_2$$ Let $x=0$ and $y=1$, $x=0$ and $y’=2$, we have $$\\cases{c_2=1\\\\c_1+c_2=2}\\implies \\cases{c_1=1\\\\c_2=1}$$ Therefore, the solution for the IVP is $$y(x)=x^2+2x+1$$", "$$then it will be of the form x^3-4x^2 + \\sqrt[3]{A^2-675}x You just need to find A such that A^2-675 is a perfect cube. Trivially A^2 = 676 will do. 1 Let x=A-15\\sqrt 3 and y=A+15\\sqrt 3, then the equation becomes$$x^{1/3}+y^{1/3}=4.$$Taking cubes of both sides gives$$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$Now since x+y=2A, and x^{1/3}+y^{1/3}=4, the equation becomes$$2A+3(xy)^{1/3}(4)=4^3,$$which gives$$6(xy)^{1/3}=32-A.$$Now, cubing and substituting for xy=A^2-15^2\\cdot 3 gives$$6^3(A^... 1 Letx,y$be the numbers$(A\\pm 15\\sqrt 3)^{1/3}$so that$x+y=4. hen we have: \\begin{aligned} 2A &= x^3+y^3\\\\ &= (x+y)^3-3xy(x+y) \\\\ &= 4^3-3xy\\cdot 4 \\\\ &=64 - 3(A^2-675)^{1/3}\\cdot 4\\ .\\text{ So:} \\\\[3mm] 6^3(A^2-675) &=(32-A)^3\\ . \\end{aligned} This gives us an equation inA$, that we may solve (with bare hands or not). sage: ... 0 In fact it is not unproblematic to work with an expression$a^x$for$x \\notin \\mathbb Z$and$a <0$. See my answer to Why$(-2)^{2.5}$isn't equal to$((-2)^{25})^{1/10}$? The \"universal definition\" would be$a^x = e^{x\\ln a}$, but is valid only for$a > 0$. Extending this to$a < 0$is possible, but involves the complex logarithm and doing so ... 0 I think it is$\\sqrt[3]{x}=x^{\\frac{1}{3}} \\ne x^{\\frac{2}{6}}$Consider this example$-1=(-1)^3=(-1)^{2.\\frac{3}{2}} \\ne ((-1)^2)^{\\frac{3}{2}}=1$The idea is that we must write the numbers we use in their irreducible forms to remove any kind of ambiguity. 2 If$n$is odd,$x^n$is an invertible function from$\\Bbb R$to itself; we denote the inverse either$\\sqrt[n]{x}$or$x^{1/n}$. This is consistent with$\\left(x^a\\right)^b=x^{ab}$. We can now uniquely define$x^{p/q}\\in\\Bbb R$for any$x\\in\\Bbb", "1$ has infinitely many solutions. Now $y=b$, $x = a -2b$ and you are done, aren't you? – Aryabhata Jan 22 at 21:16 Following Aryabhata, write it as $x^2+4xy+4y^2-3y^2=1=(x+2y)^2-3y^2$ Let us define $z=x+2y$, so this becomes $z^2-3y^2=1$. Clearly $z=1,y=0$ is a solution. Now if we have a solution $(z,y)$ we observe that $(z',y')=(2z+3y,z+2y)$ is also a solution, because \\begin {align}z'^2-3y'^2&=(2z+3y)^2-3(z+2y)^2\\\\&=4z^2+12zy+9y^2-3z^2-12yz-12y^2\\\\&=z^2-3y^2\\\\&=1 \\end{align} Given any solution we can find a larger one, so there are infinitely many. How do you know that $x$ will be an integer (or even real) solution, for a fixed $y$? This seems to be a vast oversimplification of the problem. – user61527 Feb 6 at 3:16", "# 3xy + 14x + 17y + 71 = 0 need some advice $$3xy + 14x + 17y + 71 = 0$$ Need to find both $x$ and $y$. If there was only one variable then this is easy problem. Have tried: \\begin{align}3xy &= -14x - 17y - 71 \\\\ x &= \\frac{-14x - 17y - 71}{3y}\\end{align} Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$. I don't even know how to get on right track. Please give any advice. Thanks. - The solution is not unique. – azarel Jul 26 '12 at 16:10 Solve how? One solution in $\\mathbb{Z}$ is $x = -4, y = -3$. – Cocopuffs Jul 26 '12 at 16:10 Where is the linear algebra here? – Chris Eagle Jul 26 '12 at 16:14 If $x$ and $y$ can be real numbers, with one equation in two unknowns you will have one dimension of freedom. Solving for $x$, for example, $x=- \\frac {17y+71}{3y+14}$. You can substitute in any value for $y$ you want except $\\frac {-14}3$ and find $x$. If $x$ and $y$ are integers you can use divisibility testing to restrict the options. - I think you've made an algebra error (or typo), and ended up solving for -x instead of x", "we see that y+i = m^3 – 3mn^2 + i(3m^2n – n^3). Equating real and imaginary parts, we have that \\begin{align} y &= m(m^2 – 3n^2) \\\\ 1 &= n(3m^2 – n^2). \\end{align} This second line shows that ${n \\mid 1}$. As ${n}$ is a regular integer, we see that ${n = 1}$ or ${-1}$. If ${n = 1}$, then that line becomes ${1 = (3m^2 – 1)}$, or after rearranging ${2 = 3m^2}$. This has no solutions. If ${n = -1}$, then that line becomes ${1 = -(3m^2 – 1)}$, or after rearranging ${0 = 3m^2}$. This has the solution ${m = 0}$, so that ${y+i = (-i)^3 = i}$, which means that ${y = 0}$. Then from ${y^2 + 1 = x^3}$, we see that ${x = 1}$. And so the only solution is ${(x,y) = (1,0)}$, and there are no other solutions. 4. Other Rings The Gaussian integers have many of the same properties as the regular integers, even though there are some differences. We could go further. For example, we might consider the following integer-like sets, \\mathbb{Z}(\\sqrt{d}) = { a + b \\sqrt{d} : a,b \\in \\mathbb{Z} }. One can add, subtract, and multiply these together in similar ways to how we can add, subtract, and multiply together integers, or Gaussian integers. We might", "$$z = x + iy$$, for $$x,y \\in \\mathbb{R}$$. Consequently, $$\\bar{z} = x - iy$$. Make these substitutions into your equation and isolate all of the $$x$$ and $$y$$ terms on one side, trying to make it \"look\" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative). Equate the real and imaginary parts to get a system of equations in two variables ($$x,y$$) which you can solve get your solution. Similar Exercise To Show What I Mean: Let's solve for $$z$$ with $$iz + 2\\bar{z} = 1 + 2i$$ Then, making our substitutions... \\begin{align} iz + 2\\bar{z} &= i(x + iy) + 2(x - iy) \\\\ &= ix + i^2 y + 2x - 2iy \\\\ &= ix - y + 2x - 2iy \\\\ &= (2x - y) + i(x - 2y) \\\\ \\end{align} Thus, $$(2x - y) + i(x - 2y) = 1 + 2i$$ The real part of our left side is $$2x-y$$ and the imaginary part is $$x - 2y$$. On the right, the real and imaginary parts are $$1$$ and $$2$$ respectively. Then, we get a system of equations by equating real and imaginary parts! \\begin{align} 2x - y &= 1\\\\ x - 2y &= 2\\\\ \\end{align} You can", "integer solution is $$y=0$$, with corresponding $$x=-7$$. 4. If $$x+2y=-7$$ and $$xy-3=3$$ then $$6 = xy = -(7+2y)y$$, so that $0 \\; = \\; 2y^2 + 7y + 6 \\; = \\; (2y + 3)(y + 2)$ and hence the only integer solution is $$y=-2$$, with corresponding $$x=-3$$. Thus there are four solutions: $$(3,2)$$, $$(-3,-2)$$, $$(7,0)$$ and $$(-7,0)$$. · 3 years, 9 months ago My answer is (for all integral solutions only). Ordered pairs (x, y) = (7, 0), (-7, 0), (3, 2), (-3, -2) Why is this so? We can decompose x^2 + 4y^2 = (x^2 + 4xy + 4y^2) - 4xy = (x + 2y + 2(sqrt(xy))) (x + 2y - 2 (sqrt(xy))) = 4(x^2)(y^2) - 24xy + 49 It is trivial putting xy = 0 which implies that either x or y is zero. If x is zero, then y = 7/2 or -7/2 which is not an interger. If y is zero, then x = 7 or -7. Use the discriminant and we determine that the right-hand side of the polynomial is factorable but by complex imaginary numbers which is 4(xy)^2 - 24xy + 49 = (2xy - (6 + i sqrt(13)) (2xy - (6 - i sqrt(13)) = (2xy - 6)^2 + 13. Further info: Regarding the factorized form of the right-hand side", "then these can be written as $$2k_{x}+1$$ and $$2k_{y}+1$$, respectively. Furthermore, we can write $$z_{2}=2k_{z}$$. • We then have that \\begin{align*} (2k_{x}+1)^{2}+(2k_{y}+1)^{2}&=(2k_{z})^{2} \\\\ 4k_{x}^{2} +4k_{x}+1+4k_{y}^{2}+4k_{y}+1& =4k_{z}^{2} \\\\ 2& \\equiv 0 \\text{mod} 4.\\end{align*}. • Since this is not true, we must have $$z_{2}$$ be odd and one of the $$x_{2}$$ or $$y_{2}$$ be odd. We will assume $$x_{2}$$ is odd, otherwise, we can switch the labeling of $$x_{2}$$ and $$y_{2}$$. #### Now that we have which ones are even, we rewrite the problem. • Now that we know that \\begin{align*} (z_{2})^{4}-(x_{2})^{4}&=(y_{2})^{4} \\\\ ((z_{2})^{2}-(x_{2})^{2})((z_{2})^{2}+(x_{2})^{2})&=(y_{2})^{4}.\\end{align*} • Note that if a number divides both $$(z_{2})^{2}-(x_{2})^{2}$$ and $$(z_{2})^{2}-(x_{2})^{2}$$, then it would both the sum and difference of these. Therefore, it would divide $$2(x_{2})^{2}$$ and $$2(z_{2})^{2}$$. • If this number is an odd prime, then it does not divide 2. Therefore, it divides $$x_{2}$$ and $$z_{2}$$. However, we have chosen these so that they have no common divisors. Hence, an odd prime cannot divide both of these terms. • If 4 divides both of these terms, we then have that $$(x_{2})^{2}$$ is even, which contradicts that $$(x_{2})$$ is odd. • Therefore, the greatest common divisor of $$(z_{2})^{2}-(x_{2})^{2}$$ and $$(z_{2})^{2}+(x_{2})^{2}$$ is 2. • Since $$y_{2}$$ is even, we have that $$y_{2}$$ can be written as $$2y_{3}$$ for some integer $$y_{3}$$. • Therefore, $$((z_{2})^{2}-(x_{2})^{2})*((z_{2})^{2}+(x_{2})^{2})=16(y_{3})^{4}$$. • Since", "\\le 10-z^2$ $2z^2-4z-6 \\le 0$ $z^2-2z-3 \\le 0$ $(z-3)(z+1) \\le 0$ The solution of the above inequality is $z\\in\\langle-1,3\\rangle$. EDIT: The rest of this post is my original solution, where I assumed that all 3 variables should be integers. From the first equation you get $x=4-y-2z$ and $x^2=16+y^2+4z^2-8y-16z+4yz$. By plugging this into the second equation you get \\begin{align} 16+y^2+4z^2-8y-16z+4yz+y^2+2z^2&=20\\\\ 2y^2+6z^2-8y-16z+4yz&=4\\\\ y^2+3z^2-4y-8z+2yz&=2\\\\ (y+z-2)^2+2z^2-4z&=6\\\\ (y+z-2)^2+2(z-1)^2&=8 \\end{align} So now we have a new equation $a^2+2b^2=8$. By trial and error we can find all integer solutions. (We only need to try $a=0,\\pm 1,\\pm 2$.) We get $a=0$, $b=\\pm 2$. Thus $z=1\\pm2$, i.e. $z=-1$ or $z=3$. For each $z$ we get $y=2-z$ and from the first equation we can calculate $x$. So we have: $z=-1$, $y=3$, $x=3$ $z=3$, $y=-1$, $x=-1$ Another approach would be going through representations of $20$ as sum of four squares and trying all permutations and possibilities for signs. (I guess this still can be done by hand.) We only want representations $20=x^2+y^2+z^2+w^2$, where $x+y+z+w=4$ and $z=w$. We have $20=(\\pm1)^2+(\\pm1)^2+(\\pm3)^2+(\\pm3)^2$ $20=0^2+0^2+(\\pm2)^2+(\\pm4)^2$ See wolframalpha. At the moment I don't see where AM-GM inequality can be used here. From AM-GM we get $$\\frac{x^2+y^2+z^2+z^2}4 \\ge \\left(\\frac{x+y+z+z}4\\right)^2,$$ which in this case is $\\frac{20}4\\ge 1$, which is true. So perhaps if the problem would ask about numbers different from $20$ and $4$, we could show", "is negative, then by equation (ii), x and y are of the different signs. Hence, $$\\sqrt{a + ib}$$ = $$\\pm$$ [$$\\sqrt{{(1\\over 2)}{\\sqrt{a^2 + b^2} + a}}$$ – $$i\\sqrt{{(1\\over 2)}{\\sqrt{a^2 + b^2} – a}}$$] Example : Find the square root of 7 – 24i. Solution : Let $$\\sqrt{7 – 24i}$$ = x + iy. Then, (7 – 24i) = $$(x + iy)^2$$ $$\\implies$$ 7 – 24i = $$(x^2 – y^2)$$ + 2ixy On equating real and imaginary parts, we get $$x^2 – y^2$$ = 7 ……….(i) and, 2xy = -24 …………..(ii) Now, $$(x^2 + y^2)^2$$ = $$(x^2 – y^2)^2$$ + $$4x^2y^2$$ $$\\implies$$ $$(x^2 + y^2)^2$$ = 49 + 576 = 625 $$\\implies$$ $$x^2 + y^2$$ = 25 ………..(iii) On Solving equation (i) and (iii), we get $$x^2$$ = 16 and $$y^2$$ = 9 $$\\implies$$ x = $$\\pm 4$$ and y = $$\\pm 3$$ From (ii), 2xy is negative. So, x and y are of opposite signs. $$\\therefore$$ (x = 4 and y = -3) or, (x = -4 and y = 3) Hence, $$\\sqrt{7 – 24i}$$ = $$\\pm$$ (4 – 3i)", "do it? – David K Jul 17 '15 at 12:35 • First you've write out your equation and subtract everything, then you get something that ends with $=0$ than you look for the way to write it $()()=0$ – Jan Eerland Jul 17 '15 at 12:39 • Yes, we got the \"something that ends with $=0$\" part; that was easy. It's the \"look for the way to write it $()()=0$\" that is a mystery. – David K Jul 17 '15 at 12:48 • @DavidK Posit a solution $$x^4 - 6x^3 + 12x^2 - 12x + 4 = (x^2 + ax + b)(x^2 + cx + d)$$ Matching coefficients yields \\begin{align*} a + c & = -6\\\\ ac + b + d & = 12\\\\ ad + bc & = -12\\\\ bd & = 4\\end{align*} The choice $b = 1, d = 4$ leads to the contradiction $a + c = -6$, $ac = 7$. The choice $b = 2, d = 2$ yields $a + c = -6$ and $ac = 8$, from which you can determine that $a = -4$ and $c = -2$. – N. F. Taussig Jul 18 '15 at 10:02 Given that $$(x^2+2)^2+8x^2=6x(x^2+2)$$ Put $x^2+2=y$. Then $$y^2+8x^2=6xy$$ $$8x^2-6xy+y^2=0$$ $$8x^2-4xy-2xy+y^2=0$$ $$4x(2x-y)-y(2x-y)=0$$ $$(4x-y)(2x-y)=0$$ $$(4x-x^2-2)(2x-x^2-2)=0$$ $$(x^2-4x+2)(x^2-2x+2)=0$$ Now we have $$x^2-4x+2=0$$ This is quadratic eq $$x=\\frac{-(-4) \\pm \\sqrt{(-4)^2-4(1)(2)}}{2(1)}$$", "4 < 6 \\end{align} There exists an integer $m$ such that: (10) $$-4 < 8m + 5 < 4$$ So $m = -1$. Applying $M_2$ with $m = -1$ and we get: (11) \\begin{align} \\quad f(x - y, y) &= 4(x - y)^2 + 5(x - y)y + 6y^2 \\\\ &= 4(x^2 - 2xy + y^2) + 5(xy - y^2) + 6y^2 \\\\ &= 4x^2 - 8xy + 4y^2 + 5xy - 5y^2 + 6y^2 \\\\ &= 4x^2 -3xy + 6y^2 \\end{align} We have that $-4 < -3 \\leq -4 < 6$, and so $f(x, y) = 4x^2 - 3xy + 6y^2$ is a reduced binary quadratic form that is equivalent to $6x^2 - 5xy + 4y^2$.", "f}{\\partial x}\\right)\\right) .$ Example $$\\PageIndex{7}$$: Higher order partial derivatives 1. Let $$f(x,y) = x^2y^2+\\sin(xy)$$. Find $$f_{xxy}$$ and $$f_{yxx}$$. 2. Let $$f(x,y,z) = x^3e^{xy}+\\cos(z)$$. Find $$f_{xyz}$$. SOLUTION 1. To find $$f_{xxy}$$, we first find $$f_x$$, then $$f_{xx}$$, then $$f_{xxy}$$: \\begin{align*}f_x &= 2xy^2+y\\cos(xy) \\quad\\quad f_{xx} = 2y^2-y^2\\sin(xy)\\\\f_{xxy} &= 4y-2y\\sin(xy) - xy^2\\cos(xy).\\end{align*} To find $$f_{yxx}$$, we first find $$f_y$$, then $$f_{yx}$$, then $$f_{yxx}$$: \\begin{align*}f_y &= 2x^2y+x\\cos(xy) \\quad \\quad f_{yx} = 4xy + \\cos(xy) - xy\\sin(xy)\\\\f_{yxx} &= 4y-y\\sin(xy) - \\big(y\\sin(xy) + xy^2\\cos(xy)\\big)\\\\ &= 4y-2y\\sin(xy)-xy^2\\cos(xy).\\end{align*} Note how $$f_{xxy} = f_{yxx}$$. 2. To find $$f_{xyz}$$, we find $$f_x$$, then $$f_{xy}$$, then $$f_{xyz}$$: \\begin{align*}f_x &= 3x^2e^{xy}+ x^3ye^{xy} \\quad \\quad f_{xy} = 3x^3e^{xy}+x^3e^{xy}+x^4ye^{xy} = 4x^3e^{xy}+x^4ye^{xy}\\\\ f_{xyz} &= 0.\\end{align*} In the previous example we saw that $$f_{xxy} = f_{yxx}$$; this is not a coincidence. While we do not state this as a formal theorem, as long as each partial derivative is continuous, it does not matter the order in which the partial derivatives are taken. For instance, $$f_{xxy} = f_{xyx} = f_{yxx}$$. This can be useful at times. Had we known this, the second part of Example $$\\PageIndex{7}$$ would have been much simpler to compute. Instead of computing $$f_{xyz}$$ in the $$x$$, $$y$$ then $$z$$ orders, we could have applied the $$z$$, then $$x$$ then $$y$$ order (as $$f_{xyz} = f_{zxy}$$). It is easy to see that", "+0 # Pls help thanks 0 116 2 +79 The non-negative real numbers and satisfy the two equations \\begin{align} x^2+y^2+z^2 &= 9\\\\ xy+yz+zx &= 8. \\end{align} What is the sum of $x,y,$ and $z?$ Oct 24, 2020 #1 +4 This is kind of a guess and check answer. The only way this could work is if x y and z equal 3, 0, 0 in some order, or 1, 2, 2. Since 3, 0, 0 isn't very good for the second equation, we'll go with 1, 2, 2. When you plug it in we get 1+4+4=9 for the first equation and 2+4+2=8 for the second equation. So we get the sum is 5 Oct 24, 2020 #2 +112504 +3 $$(x+y+z)^2=(x^2+y^2+z^2)+2(xy+xz+zy)\\qquad \\\\\\text{I have just expanded and simplified}\\\\ (x+y+z)^2=(9)+2(8)\\\\ (x+y+z)^2=25\\\\ x+y+z=\\pm5\\\\ \\text{but it has to be positive so}\\\\ x+y+z=5$$" ]

[ "&= y\\\\ x= \\frac{29}{325} \\ \\text{and} \\ y&=\\frac{28}{65}\\\\ \\end{align*} To find the square roots of a complex number $$a+ib$$ 1. Let $$x+yi= \\sqrt{a+bi}$$ 2. Square both sides: $$(x+yi)^2=a+bi$$ 3. Expand both sides: $$x^2+2xyi-y^2=a+bi$$ 4. Equate real and imaginary parts: $$x^2-y^2=a, 2xy=b$$ 5. Solve these equations simultaneously for $$x$$ and $$y$$ ### Example 2: Find the square roots of $$3+4i$$ \\begin{align*} \\text{Let} \\ x+yi &= \\sqrt{3+4i}\\\\ (x+yi)^2&=3+4i\\\\ x^2+2xyi-y^2&=3+4i\\\\ x^2-y^2=3 , \\ xy&=2\\\\ \\text{Substitute} \\ y&= \\frac{2}{x} \\ \\text{into} \\ \\text{the first equation:}\\\\ x^2- \\frac{4}{x^2} &= 3\\\\ x^4-3x^2-4&=0\\\\ (x^2-4)(x^2+1)&=0\\\\ \\text{As} \\ x \\ \\text{cannot be imaginary,} \\ x&=±2\\\\ \\text{Therefore} \\ \\sqrt{3+4i}&=±(2+i) \\end{align*} ### Example 3: Find the square roots of $$8-6i$$ \\begin{align*} \\text{Let} \\ x+yi &= \\sqrt{8-6i}\\\\ (x+yi)^2 &= 8 -6i\\\\ x^2+2xyi – y^2&=8-6i\\\\ x^2-y^2=8, \\ xy&=-3\\\\ \\text{As} \\ x \\ \\text{cannot be an imaginary number,} \\ x&=±3\\\\ \\text{Therefore} \\sqrt{8-6i} &= ±(3-i)\\\\ \\end{align*} ## Solving Quadratic Equations in the Complex Field A quadratic equation is in the general form $$ax^2+bx+c$$. As such, there are two ways of factorising a quadratic equation. You may choose whichever method you are more comfortable with in an exam. 1. Using the quadratic formula 2. Using the completing the square method ### Example 1: i) Fully factorise $$x^2-6x+10$$ using the Quadratic Formula \\begin{align*} \\text{Consider} \\ x^2-6x+10&=0\\\\ x&= \\frac {6 \\ ± \\ \\sqrt{36-40}}{2}\\\\ &= \\frac{6", "$a = 0$ then $-b^2 = -1$ which is possible if $b = \\pm 1$. If $b = 0$ then the first equation tells us that $a^2 = -1$ which is impossible since $a$ is real number. So $z^2 = -1$ has exactly two solutions: $a = 0$ and $b = \\pm 1$ which correspond to $z = \\pm i$. 2. We repeat the same reasoning as in part (a) except that this time we are trying to solve $(a+bi)^2 = 1$. This leads to the system of equations \\begin{align} a^2 - b^2 &= 1\\\\ 2ab &= 0. \\end{align} If $a = 0$ this is not possible because $-b^2$ can not be 1. So $b = 0$ and then the first equation tells us that $a^2 = 1$ or $a = \\pm 1$. So $z^2 = 1$ has the two expected solutions and no others: $z = \\pm 1$. 3. This time we want to solve $(a+bi)^2 = i$ which is equivalent to $a^2 + 2abi + bi^2 = i$. Equating real and complex parts of the two sides as above we find \\begin{align} a^2 - b^2 &= 0\\\\ 2ab &= 1. \\end{align} The first equation tells us that $a^2 = b^2$. This means that $a = \\pm b$. If $a = b$ then the second equation gives", "+ x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \\geq 0$? We reduce the problem to one in the non-negative integers. Since $x_1$ is an integer, $x_1 > 1 \\implies x_1 \\geq 2$. Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \\implies x_2 \\geq 2$ and $x_3 > 3 \\implies x_3 \\geq 4$. Let \\begin{align*} y_1 & = x_1 - 2\\\\ y_2 & = x_2 - 2\\\\ y_3 & = x_3 - 4\\\\ y_4 & = x_4 \\end{align*} Then each $y_k$, $1 \\leq k \\leq 4$, is a non-negative integer. Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields \\begin{align*} y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\\\ y_1 + y_2 + y_3 + y_4 & = 4 \\tag{3} \\end{align*} Equation 3 is an equation in the non-negative integers with $$\\binom{4 + 3}{3} = \\binom{7}{3}$$ solutions.", "Note that at least one of $w+3i$, $w+9i$, and $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one. Let $w=x+yi$, where $x$ and $y$ are reals. Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$, so $w=x-3i$. Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+9i}=2w-4$, so $\\overline{x+6i}=2(x-3i)-4$, implying that $x=4$, so $w=4-3i$. Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$, so $w=x-9i$. Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+3i}=2w-4$, so $\\overline{x-6i}=2(x-9i)-4$, so $x+6i=2x-4-18i$, which has no solution as $x$ is real. Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$, so $w=x$. Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$, which means that $\\overline{x+3i}=x+6i$, so $x-3i=x+6i$, which has no solutions. Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(x-(4))(x-(4+6i))(x-(4-6i))$. Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$, so we can plug in $x=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$.", "1$ has infinitely many solutions. Now $y=b$, $x = a -2b$ and you are done, aren't you? – Aryabhata Jan 22 at 21:16 Following Aryabhata, write it as $x^2+4xy+4y^2-3y^2=1=(x+2y)^2-3y^2$ Let us define $z=x+2y$, so this becomes $z^2-3y^2=1$. Clearly $z=1,y=0$ is a solution. Now if we have a solution $(z,y)$ we observe that $(z',y')=(2z+3y,z+2y)$ is also a solution, because \\begin {align}z'^2-3y'^2&=(2z+3y)^2-3(z+2y)^2\\\\&=4z^2+12zy+9y^2-3z^2-12yz-12y^2\\\\&=z^2-3y^2\\\\&=1 \\end{align} Given any solution we can find a larger one, so there are infinitely many. How do you know that $x$ will be an integer (or even real) solution, for a fixed $y$? This seems to be a vast oversimplification of the problem. – user61527 Feb 6 at 3:16", "$$then it will be of the form x^3-4x^2 + \\sqrt[3]{A^2-675}x You just need to find A such that A^2-675 is a perfect cube. Trivially A^2 = 676 will do. 1 Let x=A-15\\sqrt 3 and y=A+15\\sqrt 3, then the equation becomes$$x^{1/3}+y^{1/3}=4.$$Taking cubes of both sides gives$$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$Now since x+y=2A, and x^{1/3}+y^{1/3}=4, the equation becomes$$2A+3(xy)^{1/3}(4)=4^3,$$which gives$$6(xy)^{1/3}=32-A.$$Now, cubing and substituting for xy=A^2-15^2\\cdot 3 gives$$6^3(A^... 1 Letx,y$be the numbers$(A\\pm 15\\sqrt 3)^{1/3}$so that$x+y=4. hen we have: \\begin{aligned} 2A &= x^3+y^3\\\\ &= (x+y)^3-3xy(x+y) \\\\ &= 4^3-3xy\\cdot 4 \\\\ &=64 - 3(A^2-675)^{1/3}\\cdot 4\\ .\\text{ So:} \\\\[3mm] 6^3(A^2-675) &=(32-A)^3\\ . \\end{aligned} This gives us an equation inA$, that we may solve (with bare hands or not). sage: ... 0 In fact it is not unproblematic to work with an expression$a^x$for$x \\notin \\mathbb Z$and$a <0$. See my answer to Why$(-2)^{2.5}$isn't equal to$((-2)^{25})^{1/10}$? The \"universal definition\" would be$a^x = e^{x\\ln a}$, but is valid only for$a > 0$. Extending this to$a < 0$is possible, but involves the complex logarithm and doing so ... 0 I think it is$\\sqrt[3]{x}=x^{\\frac{1}{3}} \\ne x^{\\frac{2}{6}}$Consider this example$-1=(-1)^3=(-1)^{2.\\frac{3}{2}} \\ne ((-1)^2)^{\\frac{3}{2}}=1$The idea is that we must write the numbers we use in their irreducible forms to remove any kind of ambiguity. 2 If$n$is odd,$x^n$is an invertible function from$\\Bbb R$to itself; we denote the inverse either$\\sqrt[n]{x}$or$x^{1/n}$. This is consistent with$\\left(x^a\\right)^b=x^{ab}$. We can now uniquely define$x^{p/q}\\in\\Bbb R$for any$x\\in\\Bbb", "\\le 10-z^2$ $2z^2-4z-6 \\le 0$ $z^2-2z-3 \\le 0$ $(z-3)(z+1) \\le 0$ The solution of the above inequality is $z\\in\\langle-1,3\\rangle$. EDIT: The rest of this post is my original solution, where I assumed that all 3 variables should be integers. From the first equation you get $x=4-y-2z$ and $x^2=16+y^2+4z^2-8y-16z+4yz$. By plugging this into the second equation you get \\begin{align} 16+y^2+4z^2-8y-16z+4yz+y^2+2z^2&=20\\\\ 2y^2+6z^2-8y-16z+4yz&=4\\\\ y^2+3z^2-4y-8z+2yz&=2\\\\ (y+z-2)^2+2z^2-4z&=6\\\\ (y+z-2)^2+2(z-1)^2&=8 \\end{align} So now we have a new equation $a^2+2b^2=8$. By trial and error we can find all integer solutions. (We only need to try $a=0,\\pm 1,\\pm 2$.) We get $a=0$, $b=\\pm 2$. Thus $z=1\\pm2$, i.e. $z=-1$ or $z=3$. For each $z$ we get $y=2-z$ and from the first equation we can calculate $x$. So we have: $z=-1$, $y=3$, $x=3$ $z=3$, $y=-1$, $x=-1$ Another approach would be going through representations of $20$ as sum of four squares and trying all permutations and possibilities for signs. (I guess this still can be done by hand.) We only want representations $20=x^2+y^2+z^2+w^2$, where $x+y+z+w=4$ and $z=w$. We have $20=(\\pm1)^2+(\\pm1)^2+(\\pm3)^2+(\\pm3)^2$ $20=0^2+0^2+(\\pm2)^2+(\\pm4)^2$ See wolframalpha. At the moment I don't see where AM-GM inequality can be used here. From AM-GM we get $$\\frac{x^2+y^2+z^2+z^2}4 \\ge \\left(\\frac{x+y+z+z}4\\right)^2,$$ which in this case is $\\frac{20}4\\ge 1$, which is true. So perhaps if the problem would ask about numbers different from $20$ and $4$, we could show", "do it? – David K Jul 17 '15 at 12:35 • First you've write out your equation and subtract everything, then you get something that ends with $=0$ than you look for the way to write it $()()=0$ – Jan Eerland Jul 17 '15 at 12:39 • Yes, we got the \"something that ends with $=0$\" part; that was easy. It's the \"look for the way to write it $()()=0$\" that is a mystery. – David K Jul 17 '15 at 12:48 • @DavidK Posit a solution $$x^4 - 6x^3 + 12x^2 - 12x + 4 = (x^2 + ax + b)(x^2 + cx + d)$$ Matching coefficients yields \\begin{align*} a + c & = -6\\\\ ac + b + d & = 12\\\\ ad + bc & = -12\\\\ bd & = 4\\end{align*} The choice $b = 1, d = 4$ leads to the contradiction $a + c = -6$, $ac = 7$. The choice $b = 2, d = 2$ yields $a + c = -6$ and $ac = 8$, from which you can determine that $a = -4$ and $c = -2$. – N. F. Taussig Jul 18 '15 at 10:02 Given that $$(x^2+2)^2+8x^2=6x(x^2+2)$$ Put $x^2+2=y$. Then $$y^2+8x^2=6xy$$ $$8x^2-6xy+y^2=0$$ $$8x^2-4xy-2xy+y^2=0$$ $$4x(2x-y)-y(2x-y)=0$$ $$(4x-y)(2x-y)=0$$ $$(4x-x^2-2)(2x-x^2-2)=0$$ $$(x^2-4x+2)(x^2-2x+2)=0$$ Now we have $$x^2-4x+2=0$$ This is quadratic eq $$x=\\frac{-(-4) \\pm \\sqrt{(-4)^2-4(1)(2)}}{2(1)}$$", "integer solution is $$y=0$$, with corresponding $$x=-7$$. 4. If $$x+2y=-7$$ and $$xy-3=3$$ then $$6 = xy = -(7+2y)y$$, so that $0 \\; = \\; 2y^2 + 7y + 6 \\; = \\; (2y + 3)(y + 2)$ and hence the only integer solution is $$y=-2$$, with corresponding $$x=-3$$. Thus there are four solutions: $$(3,2)$$, $$(-3,-2)$$, $$(7,0)$$ and $$(-7,0)$$. · 3 years, 9 months ago My answer is (for all integral solutions only). Ordered pairs (x, y) = (7, 0), (-7, 0), (3, 2), (-3, -2) Why is this so? We can decompose x^2 + 4y^2 = (x^2 + 4xy + 4y^2) - 4xy = (x + 2y + 2(sqrt(xy))) (x + 2y - 2 (sqrt(xy))) = 4(x^2)(y^2) - 24xy + 49 It is trivial putting xy = 0 which implies that either x or y is zero. If x is zero, then y = 7/2 or -7/2 which is not an interger. If y is zero, then x = 7 or -7. Use the discriminant and we determine that the right-hand side of the polynomial is factorable but by complex imaginary numbers which is 4(xy)^2 - 24xy + 49 = (2xy - (6 + i sqrt(13)) (2xy - (6 - i sqrt(13)) = (2xy - 6)^2 + 13. Further info: Regarding the factorized form of the right-hand side", "4 < 6 \\end{align} There exists an integer $m$ such that: (10) $$-4 < 8m + 5 < 4$$ So $m = -1$. Applying $M_2$ with $m = -1$ and we get: (11) \\begin{align} \\quad f(x - y, y) &= 4(x - y)^2 + 5(x - y)y + 6y^2 \\\\ &= 4(x^2 - 2xy + y^2) + 5(xy - y^2) + 6y^2 \\\\ &= 4x^2 - 8xy + 4y^2 + 5xy - 5y^2 + 6y^2 \\\\ &= 4x^2 -3xy + 6y^2 \\end{align} We have that $-4 < -3 \\leq -4 < 6$, and so $f(x, y) = 4x^2 - 3xy + 6y^2$ is a reduced binary quadratic form that is equivalent to $6x^2 - 5xy + 4y^2$.", "# How to find a simultaneous solution to these equations? QUESTION $$x^2 + xy + yz + xz = 30$$ $$y^2 + xy + yz + xz = 15$$ $$z^2 + xy + yz + xz = 18$$ I have tried manipulating the expressions, the identity of $$(x+y+z)^2$$ but to no avail. Along with the answer, it would be great if you can explain the approach and thought process involved in dealing with such kind of questions. Hint: Let $$x+y = a, y+z = b, z + x = c$$. Rewrite the equations in term of $$a,b,c$$. The approach and thought process is by recognizing the pattern of how the LHS factorizes. • I believe that would be z+x=c. – InfiniteCool23 Oct 24 at 6:39 • Anyway, thanks for your help! – InfiniteCool23 Oct 24 at 6:40 • It works to give x=4,y=1,z=2 – InfiniteCool23 Oct 24 at 6:41 You didn't say whether $$x,y,z$$ are real or integer. But let's see what happens if we assume they're integer. We can eliminate the cross terms $$xy + yz + xz$$ by subtraction. \\begin{align} x^2-y^2 &= 15\\\\ x^2-z^2 &= 12\\\\ z^2-y^2 &= 3\\\\ \\end{align} Now factorizing, $$z^2-y^2 = (z+y)(z-y) = 3$$ So (if they're positive integers), $$z+y=3 \\, \\text{and} \\, z-y=1$$ thus $$z=2, y=1$$. If we permit negative integers, $$z=-2,", "Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers • You should try to improve your question by putting on it what you have tried so far. People here will be more interested in help you, if they see that you are interested in learn, instead of just getting the solution of this problem. – Tomás Sep 13 '13 at 13:40 • Show some effort towards your own question. Some rather boring solutions are $\\;(0,0)\\;,\\;(0,-1)\\;,\\;(-1,0)\\;,\\;(2,1)\\;,\\ldots$ – DonAntonio Sep 13 '13 at 13:43 Here's a solution for positive $x$ and $y$. I will show that the only solutions for positive $x$ and $y$ are $(x, y) = (2, 1)$ and $(11, 3)$. $x(x+1) = y(y+1)(y^2+2) =y(y^3+y^2+2y+2) =y^4+y^3+2y^2+2y$ Multiplying by 4, $(2x+1)^2-1 =4y^4+4y^3+8y^2+8y$ or $(2x+1)^2 =4y^4+4y^3+8y^2+8y+1$ My goal is to show algebraically that this polynomial in $y$ is between two consecutive squares for large enough $y$, and then examine the remaining cases. $(2y^2+y)^2 =4y^4+4y^3+y^2$. \\begin{align} (2y^2+y+1)^2 &=4y^4+4y^3+y^2 +2(2y^2+y)+1 \\\\ &=4y^4+4y^3+y^2 +4y^2+2y+1 \\\\ &=4y^4+4y^3+5y^2+2y+1 \\\\ \\end{align}. \\begin{align} (2y^2+y+2)^2 &=4y^4+4y^3+y^2 +4(2y^2+y)+4 \\\\ &=4y^4+4y^3+y^2 +8y^2+4y+4 \\\\ &=4y^4+4y^3+9y^2+4y+4 \\\\ \\end{align}. For $(2x+1)^2$ to be between these consecutive squares, we need $5y^2+2y+1 <8y^2+8y+1 <9y^2+4y+4$. The first inequality is true for $y \\ge 1$. For the second inequality to be true, we need $8y^2+8y+1 <9y^2+4y+4$ or $y^2-4y+3 > 0$ or", "we see that y+i = m^3 – 3mn^2 + i(3m^2n – n^3). Equating real and imaginary parts, we have that \\begin{align} y &= m(m^2 – 3n^2) \\\\ 1 &= n(3m^2 – n^2). \\end{align} This second line shows that ${n \\mid 1}$. As ${n}$ is a regular integer, we see that ${n = 1}$ or ${-1}$. If ${n = 1}$, then that line becomes ${1 = (3m^2 – 1)}$, or after rearranging ${2 = 3m^2}$. This has no solutions. If ${n = -1}$, then that line becomes ${1 = -(3m^2 – 1)}$, or after rearranging ${0 = 3m^2}$. This has the solution ${m = 0}$, so that ${y+i = (-i)^3 = i}$, which means that ${y = 0}$. Then from ${y^2 + 1 = x^3}$, we see that ${x = 1}$. And so the only solution is ${(x,y) = (1,0)}$, and there are no other solutions. 4. Other Rings The Gaussian integers have many of the same properties as the regular integers, even though there are some differences. We could go further. For example, we might consider the following integer-like sets, \\mathbb{Z}(\\sqrt{d}) = { a + b \\sqrt{d} : a,b \\in \\mathbb{Z} }. One can add, subtract, and multiply these together in similar ways to how we can add, subtract, and multiply together integers, or Gaussian integers. We might", "And the text in the OP would / should be improved by just adding the symbols point and comma at the many places where they are missing. And please make clear which is the question. An answer should solve (i) and (ii)? Or just give the roots of $$r=(r^4-4)^2-4\\ ?$$ – dan_fulea Oct 4 '20 at 17:00 (i) Let $$a,b$$ be such that $$P(x)=x^2+ax+b$$ has the given property. Let $$x_1,x_2$$ be the roots of $$P(x)$$. • If $$x_1^2-4=x_2^2-4$$, then we have successively $$x_1^2=x_2^2$$, $$ax_1+b=ax_2+b$$, $$a(x_1-x_2)=0$$, $$a=0$$. So $$x_1,x_2$$ are $$\\pm\\sqrt {-b}$$, and we need $$-b-4=x_1^2-4=\\pm\\sqrt b$$. So $$\\sqrt b$$ is a root of the equation $$-T^2-4=\\pm T$$ in $$T$$. There are four solutions in complex numbers, none of them is real. But i suppose we are working over $$\\Bbb R$$. So this case leads to no solution. • If $$x_1^2-4\\ne x_2^2-4$$, the remained case, then these two values are $$x_1,x_2$$ in this order or in the other one. So we have one of the cases: \\begin{aligned} x_1 &= x_1^2-4\\ ,\\\\ x_2 &= x_2^2-4\\ , \\end{aligned} \\qquad\\text{ or } \\begin{aligned} x_1 &= x_2^2-4\\ ,\\\\ x_2 &= x_1^2-4\\ . \\end{aligned} In the first case we obtain real distinct roots, they are the roots of the equation $$X^2-X-4=0\\ ,$$ and we compute them immediately, $$\\frac 12(1\\pm\\sqrt{17})$$. In the second case, we", "etc.) and simultaneous equations. let $Z = \\sqrt{4i-3}$ where Z is complex. i.e. $Z$ is of the form $Z = a + ib$, where $a$ and $b$ are REAL coefficients. (This is important later on.) Square both sides: $Z^2 = -3+4i$ Expand the standard form of $Z^2$: $(a+ib)^2 = -3+4i$ This becomes $a^2 - b^2 + 2abi = -3+4i$ The above is an identity and so the real and imaginary parts can be equated. Hence two equations are formed: $a^2 - b^2 = -3$................(1) $2abi = 4i$ which becomes $ab = 2$......(2) In (2), $b = 2/a$ so $a^2 - (2/a)^2 + 3 = 0$ i.e. $a^4 + 3a^2 - 4 = 0$ (multiplying all by $a^2$ and rearranging) which is a quadratic-styled quartic: $(a^2 + 4)(a^2 - 1) = 0.$ Solving, $a=1, -1, 2i$ and $-2i$ BUT since $a$ and $b$ are real, we discard $2i$ and $-2i$. Now we simply solve for $b$ using $b = 2/a$. Hence the solutions are $a=1, b=2$ or $a=-1, b=-2$. and so $\\sqrt{4i-3} = \\pm(1+2i)$ since we defined $Z = a+ib$. Best of luck with any future questions! Hope this helped. • Just add $around math expressions, expressed in the LaTeX formatting. – PearlSek Dec 23 '16 at 22:21 • Well that was simpler than I thought, thanks for that!", "# 3xy + 14x + 17y + 71 = 0 need some advice $$3xy + 14x + 17y + 71 = 0$$ Need to find both $x$ and $y$. If there was only one variable then this is easy problem. Have tried: \\begin{align}3xy &= -14x - 17y - 71 \\\\ x &= \\frac{-14x - 17y - 71}{3y}\\end{align} Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$. I don't even know how to get on right track. Please give any advice. Thanks. - The solution is not unique. – azarel Jul 26 '12 at 16:10 Solve how? One solution in $\\mathbb{Z}$ is $x = -4, y = -3$. – Cocopuffs Jul 26 '12 at 16:10 Where is the linear algebra here? – Chris Eagle Jul 26 '12 at 16:14 If $x$ and $y$ can be real numbers, with one equation in two unknowns you will have one dimension of freedom. Solving for $x$, for example, $x=- \\frac {17y+71}{3y+14}$. You can substitute in any value for $y$ you want except $\\frac {-14}3$ and find $x$. If $x$ and $y$ are integers you can use divisibility testing to restrict the options. - I think you've made an algebra error (or typo), and ended up solving for -x instead of x", "clearly stated in the statement. But, be careful that this could lead to fairly meaningless step if you're not observant enough a problem solver: \\begin{align*}4x^2+5xy-6y^2-4x+3y&=4x^2-4x-6y^2+3y+5xy\\\\&=4x(x-1)-3y(2y-1)+5xy\\end{align*} But whether this step is useless or useful to us depends on our algebraic manipulating skills, in fact, if you're bright, you could see if we have introduced a $2y$ and another $x$ into the factors $(x-1)$ and $(2y-1)$ respectively, we would end up having one common factor, which is $(2y+x-1)$, so that gives us the idea that we could rewrite $5xy$ so to split it into two in the following fashion: \\begin{align*}4x^2+5xy-6y^2-4x+3y&=4x^2-4x-6y^2+3y+5xy\\\\&=4x(x-1)-3y(2y-1)+8xy-3xy\\\\&=4x(2y+x-1)-3y(x+2y-1)\\\\&=(4x-3y)(2y+x-1)\\end{align*} But, we could also reach to the same conclusion even if we don't factor out $4x$ from $4x^2-4x$ and $3y$ from $-6y^2+3y$ (which the details would be discussed in the subsequent problems). Therefore both A and B are the correct answers to this second problem, and this has assured us there are many ways to skin a cat. Question 3: If you're told to substitute $x=3$ and $y=4$ simultaneously into the given equation $4x^2+5xy-6y^2-4x+3y-1=0$, what would you have observed? A. We obtained $0-1=0$, which is ridiculous. B. $x=3$ and $y=4$ are roots of $4x^2+5xy-6y^2-4x+3y-1=0$. C. $x=3$ and $x=4$ are roots of $4x^2+5xy-6y^2-4x+3y=0$. Yes, if we substitute $x=3$ and $y=4$ into the equation $4x^2+5xy-6y^2-4x+3y-1=0$, we would end up getting $0-1=0$, but", "# Ex 12.2 Q5 Algebraic-Expressions Solutions NCERT Maths Class 7 Go back to 'Ex.12.2' ## Question What should be taken away from $$3x^2 – 4y^2 + 5xy + 20$$ to obtain $$– x^2 – y^2 + 6xy + 20?$$ Video Solution Algebraic Expressions Ex 12.2 | Question 5 ## Text Solution What is known? we know that arithmetic operation will be applied. Reasoning: In this question basic concept of arithmetic operations is applied. We are given two terms and asked what should be subtracted from one term to get the required answer. For this we will subtract the given answer from the 1st term to get what should be subtracted. E.g: what should be subtracted from $$5$$ to get $$3.$$ We will subtract $$3$$ from $$5.$$ Steps: 1st term $$= 3x^2 – 4y^2 + 5xy + 20$$ Answer term $$= \\, – x^2 – y^2 + 6xy + 20$$ 2nd term = 1st term – Answer 2nd term \\begin{align}&\\!\\!\\!\\!\\!\\!= \\!3x^2\\! –\\! 4y^2 \\!+\\! 5xy \\!+ \\!20 \\!–\\! (\\!– \\!x^2\\! –\\! y^2\\! + \\!6xy \\!+ \\!20)\\\\&\\!\\!\\!\\!\\!\\!= \\!3{x^2}\\!\\!-\\!4{y^2} \\!+ \\!5xy \\!+\\! 20 \\!+\\! {x^2} \\!\\!+ \\!{y^2} \\!-\\! 6xy\\! -\\! 20\\\\&\\!\\!\\!\\!\\!\\! = \\!4{x^2} \\!-\\! 3{y^2} \\!- \\!xy\\end{align} Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and", "# 1961 IMO Problems/Problem 1 ## Problem (Hungary) Solve the system of equations: $\\begin{matrix} \\quad x + y + z \\!\\!\\! &= a \\; \\, \\\\ x^2 +y^2+z^2 \\!\\!\\! &=b^2 \\\\ \\qquad \\qquad xy \\!\\!\\! &= z^2 \\end{matrix}$ where $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x, y, z$ (the solutions of the system) are distinct positive numbers. ## Solution Note that $x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$, so the first two equations become $\\begin{matrix} \\quad (x + y) + z \\!\\!\\! &= a \\; \\; (*) \\\\ (x+y)^2 - z^2 \\!\\!\\! &=b^2 (**) \\end{matrix}$. We note that $(x+y)^2 - z^2 = \\Big[ (x+y)+z \\Big]\\Big[ (x+y)-z\\Big]$, so if $a$ equals 0, then $b$ must also equal 0. We then have $x+y = -z$; $xy = (x+y)^2$. This gives us $x^2 + xy + y^2 = 0$. Mutiplying both sides by $(x-y)$, we have $x^3 - y^3 = 0$. Since we want $x,y$ to be real, this implies $x = y$. But $x^2 + x^2 + x^2$ can only equal 0 when $x=0$ (which, in this case, implies $y,z = 0$). Hence there are no positive solutions when $a = 0$. When $a \\neq 0$, we divide $(**)$ by $(*)$ to obtain the system of", "# Find the fixed points of the two systems What are the fixed points of the two non linear systems below? \\begin{align} x(t)& = x(3-x-2y)\\\\ y(t)& = y(2-x-y) \\end{align} I know that $(0,0)$, $(0,2)$, $(3,0)$, $(1,1)$ are the fixed points. However, I don't understand some of the steps to obtain this. Any help will be appreciated. - Unless I missed something in the question, the fixed points are given by solutions of $x = x(3-x-2y)$, $y=y(2-x-y)$. If $x=0$, the first equation is trivially satisfied, and the second gives $y = y(2-y)$, or in other words $y(y-1) = 0$. Hence $(0,0)$ and $(0,1)$ are solutions. If $y=0$, the second equation is trivially satisfied, and the first gives $x=x(3-x)$ or in other words $x(2-x) = 0$. Hence $(0,0)$ and $(2,0)$ are solutions. Now suppose $x\\neq 0$ and $y\\neq 0$. Divide the equations by $x,y$ respectively to get $1 = 3-x-2y$, $1 = 2-x-y$, or equivalently $x+2y-2=0$ and $1-x-y=0$. Adding the first equation to two times the second gives $x=0$ which cannot be, hence there are no solutions with $x\\neq 0$ and $y\\neq 0$. Hence the only solutions are $(0,0), (0,1),(2,0)$. Addendum: Following Hans comments below, I will look for stationary points of the original system. This means solving $0 = x(3-x-2y)$, $0=y(2-x-y)$ instead. The same approach works: If $x=0$, then" ]

[ "Mathematics OpenStudy (anonymous): what is the smallest sloution to the equation x to the fourth - 34x to the 2 +225=0 So, you can factor this into a pair of quadratics. Basically, you have this: $x^4 - 34x^2 + 225 = 0$ Imagine that $$u = x^2$$. Then you have: $u^2 - 34u + 225 = 0$ So first you solve for $$u$$. Then, you have $$u = x^2$$, so $$x = \\sqrt{u}$$. Does that help?", "taking t = 6 and s – t = 9, so that s = 15. That gives the solution $a = 261 = 15^2+6^2$, with $a^2 = 261^2 = 189^2 + 180^2$. This certainly satisfies all the conditions (1), (2), (3), but I do not know whether it is the smallest solution apart from a = 5.", "# Pell's Equation/Examples ## Examples of Solutions of Pell's Equation ### Pell's equation: $x^2 - 13 y^2 = 1$ $x^2 - 13 y^2 = 1$ has the smallest positive integral solution: $x = 649$ $y = 180$ ### Pell's equation: $x^2 - 29 y^2 = 1$ $x^2 - 29 y^2 = 1$ has the smallest positive integral solution: $x = 9801$ $y = 1820$ ### Pell's equation: $x^2 - 61 y^2 = 1$ $x^2 - 61 y^2 = 1$ has the smallest positive integral solution: $x = 1 \\, 766 \\, 319 \\, 049$ $y = 226 \\, 153 \\, 980$", "# Find the Lowest Whole Solution What is the smallest positive integer solution for $$y$$ in the Diophantine equation, $$3x+13y=234$$? ×", "7.1.6 equations are known. There is now a known solutions to the 7.1.7 equation,(2)(M. Dodrill 1999, PowerSum), requiring an update by Guy (1994, p. 140). The smallest 7.1.8 solution is(3)(Lander et al. 1967, Ekl 1998). The smallest 7.1.9 solution is(4)(Lander et al. 1967).No solutions to the 7.2.2, 7.2.3, 7.2.4, or 7.2.5 equations are known. The smallest 7.2.6 equation is(5)(Meyrignac). The smallest 7.2.8 solution is(6)(Lander et al. 1967, Ekl 1998). A 7.2.10.10 solution is(7)(8)(Lander et al. 1967).No solutions to the 7.3.3 equation are known (Ekl 1996), nor are any to 7.3.4. The smallest 7.3.5 equations are(9)(10)No solutions are known to the 7.3.6 equation. The smallest 7.3.7 solution is(11)(Lander et al. 1967).Guy (1994, p. 140) asked if a 7.4.4 equation exists. The following.. ### Diophantine equation--6th powers The 6.1.2 equation(1)is a special case of Fermat's last theorem with , and so has no solution. No 6.1. solutions are known for (Lander et al. 1967; Guy 1994, p. 140). The smallest 6.1.7 solution is(2)(Lander et al. 1967; Ekl 1998). The smallest primitive 6.1.8 solutions are(3)(Lander et al. 1967). The smallest 6.1.9 solution is(4)(Lander et al. 1967). The smallest 6.1.10 solution is(5)(Lander et al. 1967). The smallest 6.1.11 solution is(6)(Lander et al. 1967). There is also at least one 6.1.16 identity,(7)(Martin 1893). Moessner (1959) gave solutions for 6.1.16, 6.1.18, 6.1.20, and 6.1.23", "question, how can I find the solution, where the sum of $x+y$ is the smallest? I'd appreciate any help whatsoever! Yes those are correct, you just need a bound for $k$ which can be calculated using $x \\ge 1, y\\ge 1$ or $3k-350 \\ge 1, 350-2k \\ge 1$ also, $2x+3y=350 \\implies 2(x+y)=350 -y$, to find the minimum value of $x+y$ calculate maximum value of $y$ • Thanks! This is the step I was missing. One note; we actually said that we can use $0$ as a whole number in discrete structure classes, so I have to check $x\\ge0$.. – peroxy Jan 26 '15 at 17:17 • So I found out that there are $60$ solutions, $116 \\le k \\le 175$. Do you know how I can find the right solution, where the sum of $x+y$ is the smallest? – peroxy Jan 26 '15 at 17:25", "$x^2 + 1=0$, though, i.e., just again covers the $p\\equiv1 \\pmod{4}$ case. Nov 1 '20 at 18:15 • @HarryAltman Oh, hmm, Not sure what I meant then. I'll have to think about this again. Nov 1 '20 at 19:14 This is another update of my answer, in which I also give a quick summary about solvability obstructions as suggested in Will Sawin's answer. The smallest equation with easy-to-find solution is $$0=0$$ with $$H=0$$. The smallest equations with no solutions are $$1=0$$ and $$-1=0$$ with $$H=1$$. The smallest equations with at least one variable and no solutions are $$x^2+1=0$$ and $$2x+1=0$$ (and their variants) with $$H=5$$. All equations with $$H\\leq 14$$ either have small solutions or have trivial obstructions of at least one of the following types: a) no real solutions (like $$x^2+1=0$$) or no real solution outside a region with finite number of integer points (like $$x^2-2=0$$), b) no solutions modulo some integer (like $$2x+1=0$$), or c) divisibility conditions imply at most finitely many possible solutions, and none of them works. An example is $$(x^2+2)y=1$$, where $$y$$ must be a divisor of $$1$$, but both divisors $$y=1$$ and $$y=-1$$ do not lead to a solution. Equation $$y^2=x^3-3$$ with $$H=15$$ is the smallest equation with no solutions for which the trivial reasons above do not suffice and obstructions arising", "= 0.225", "1545 + 4b = 1849 => b= 76$ too large so a+b = 36 is the only solution nice solution", "least one illustration of the numerical problem described with small $w(0)$.", "of $x(1-x)=u$ are units satisfying $a+b=1$. Start with your favorite $u$ and iterate. Bonus question: It's known that the number of solutions of the unit equation is bounded in terms of the rank of the group of units, hence the degree. What's the smallest degree of a number field where the unit equation has $n$ solutions? - Elkies's comment is here: mathoverflow.net/questions/76206/… – user2035 Oct 9 '11 at 20:53", "the equation $\\varepsilon y-F(x,y,y')$ for small $\\varepsilon$O. A. Oleinik, A. I. Zhizhina 709", "# 1.5x^2-3x+242=0 ## Simple and best practice solution for 1.5x^2-3x+242=0 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. ## Solution for 1.5x^2-3x+242=0 equation: 1.5x^2-3x+242=0 a = 1.5; b = -3; c = +242;Δ = b2-4acΔ = -32-4·1.5·242Δ = -1443Delta is less than zero, so there is no solution for the equation `", "also true that 7 is the smallest positive solution to $2343^a \\equiv a \\pmod{10000}$. I haven't thought about this long enough to know whether I should be surprised. Let $(a,x)$ denote that $x$ is the smallest positive solution to $a^x \\equiv x \\pmod{10^d}$, where $d$ is the number of digits of $x$. I found the pairs (2,36), (3,87), (4,96), (5,25), (6,56), (7,43) [which contains the last two digits of your 2343 value], (8,56), and (9,89). Solutions to $a^a \\equiv a \\pmod{10^d}$, where $d$ is the number of digits of $a$, can be found here. The sequence itself is pretty interesting. -", "given by(13)(L. Morelli 1999). No 9.2.11 solutions are known. The smallest 9.2.12 solution is(14)(Lander et al. 1967, Ekl 1998). There are no known 9.2.13 or 9.2.14 solutions.The smallest 9.2.15 solution is(15)(Lander et al. 1967).There are no known 9.3.3, 9.3.4, 9.3.5, 9.3.6, 9.3.7, or 9.3.8 solutions. The smallest 9.3.9 solution is(16)(Ekl.. ### Diophantine equation--8th powers The 8.1.2 equation(1)is a special case of Fermat's last theorem with , and so has no solution. No 8.1.3, 8.1.4, 8.1.5, 8.1.6, or 8.1.7 solutions are known. The only known 8.1.8 is(2)(S. Chase; Meyrignac). The smallest 8.1.9 is(3)(N. Kuosa). The smallest 8.1.10 is(4)(N. Kuosa, PowerSum). The smallest 8.1.11 solution is(5)(Lander et al. 1967, Ekl 1998). The smallest 8.1.12 solution is(6)(Lander et al. 1967). The general identity(7)gives a solution to the 8.1.17 equation (Lander et al. 1967).No 8.2.2, 8.2.3, 8.2.4, 8.2.5, or 8.2.6 solution is known. A single 8.2.7 solutions is known,(8)(S. Chase; Meyrignac). The smallest 8.2.8 solution is(9)The smallest 8.2.9 solution is(10)(Lander et al. 1967, Ekl 1998).No 8.3.3 or 8.3.4 solutions are known. An 8.3.5 solution is(11)(S. Chase, Meyrignac, Resta and Meyrignac 2003). No 8.3.6 solution is known. The smallest 8.3.7 solution is(12)The smallest 8.3.8 solution.. ### Diophantine equation--7th powers The 7.1.2 equation(1)is a special case of Fermat's last theorem with , and so has no solution. No solutions to the 7.1.3, 7.1.4, 7.1.5,", "less than 0. Therefore $x = 2$ is the solution", "smallest you can get is $2^{\\aleph_0}$ ($> \\aleph_0$).", "[1] 36x^2 + 225y^2 = 22500 : [1]*9 -36x^2 - 64y^2 = -16704 : [2]*-9 Add above: 161y^2 = 5796 Now solve for y, then substitute to solve for x", "Are you smarter than me? $\\mathcal Y = 2^{200} - 31 \\times 2^{192} + 2^n$ The above expression of $$\\mathcal Y$$ shows the sum and differences of different powers of $$2$$. What is the smallest value of $$n$$ such that the $$\\mathcal Y$$ is a perfect square? ×", "Find the smallest numbers $a, b$, and $c$ such that: $$a^2 = 2 b^3 = 3 c^5$$ What can you say about other solutions to this problem?" ]

[ "adding $21$ two more times gives us $19$ with no remainders. We now have $1077 + 21 + 21 + 21 = 1140$. However, this number is divisible by $120$. To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$. Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$. Adding $420$ again gives us $1980$, which is valid. However, remember that this is equal to $n + 63$, so subtracting $63$ from $1980$ gives us $1917$, which is $n$. The sum of its digits are $1 + 9 + 1 + 7 = \\boxed{\\textbf{(C) } 18}$. ~primegn ## Solution 8 (Euclidean Algorithm) By the Euclidean Algorithm, we have \\begin{alignat*}{8} \\gcd(63,n+120)&=\\hspace{1mm}&&\\gcd(63,\\phantom{ }\\underbrace{n+120-63k_1}_{(n+120) \\ \\mathrm{mod} \\ 63}\\phantom{ })&&=21, \\\\ \\gcd(n+63,120)&=&&\\gcd(\\phantom{ }\\underbrace{n+63-120k_2}_{(n+63) \\ \\mathrm{mod} \\ 120}\\phantom{ },120)&&=60. \\end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$ ## Video Solutions ### Video Solution 2 https://youtu.be/8mNMKH0T9W0 - Happytwin ### Video Solution 3 (Quick & Simple) Education The Study of Everything ### Video Solution 5 https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0", "we get $18 r 18$. By pure luck, adding $21$ two more times gives us $19$ with no remainders. We now have $1077 + 21 + 21 + 21 = 1140$. However, this number is divisible by $120$. To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$. Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$. Adding $420$ again gives us $1980$, which is valid. However, remember that this is equal to $n + 63$, so subtracting $63$ from $1980$ gives us $1917$, which is $n$. The sum of its digits are $1 + 9 + 1 + 7 = 18$. So, our answer is $\\boxed{\\textbf{(C) }18}$. ~ primegn ## Solution 11 (Euclidean Algorithm) By the Euclidean Algorithm, we have \\begin{alignat*}{8} \\gcd(63,n+120)&=\\hspace{1mm}&&\\gcd(63,\\phantom{ }\\underbrace{n+120-63k_1}_{(n+120) \\ \\mathrm{mod} \\ 63}\\phantom{ })&&=21, &&\\hspace{10mm}(1) \\\\ \\gcd(n+63,120)&=&&\\gcd(\\phantom{ }\\underbrace{n+63-120k_2}_{(n+63) \\ \\mathrm{mod} \\ 120}\\phantom{ },120)&&=60.&&\\hspace{10mm}(2) \\end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$ More generally, let $t\\in\\{1,2\\},$ so we get \\begin{align*} n+120-63k_1&=21t, &\\hspace{55.5mm}(1*) \\\\ n+63-120k_2&=60. &\\hspace{55.5mm}(2*) \\end{align*} Subtracting $(2*)$ from $(1*)$ and then simplifying give \\begin{align*} 57-63k_1+120k_2&=21t-60 \\\\ 117-63k_1+120k_2&=21t \\\\ 39-21k_1+40k_2&=7t. \\hspace{54mm}(\\bigstar) \\end{align*} Taking $(\\bigstar)$ modulo $7$ produces \\begin{align*} 4+5k_2&\\equiv0\\pmod{7} \\\\ k_2&\\equiv2\\pmod{7}. \\end{align*} Recall that $n>1000.$ From $(2*),$ it follows that $\\[1063-120k_2 from which $k_2>8.$", "$16$, respectively. $211 + 28 + 16 = 255$. Hence, the answer is $\\boxed{\\textbf{(A)}\\ 255}$. ## Solution 3 Using the closed forms for the sums, we get $m^2=n(n+1)+212$, or $m^2=n^2+n+212$. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$. Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$. Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7\\cdot11\\cdot11$. The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$, $7$ and $121$, $11$ and $77$. In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$. The sum of all possible values of $n$ is $211+28+16=\\boxed{\\textbf{(A)}\\ 255}$. ~dolphin7", ".9\\phantom{0} && \\tag1\\\\ \\underline{-7} && \\underline{-.66} \\tag2\\\\ -2 && .24 \\tag3\\\\ \\end{align} The key thing here is, what does the last line mean? On line $(1)$, the sum of the numbers in the two columns, $5$ and $0.9$, is one of the numbers we started with, $5.9$, while on line $(2)$, the sum of the numbers in the two columns, $-7$ and $-0.66$, is $-7.66$, corresponding to the operation of subtracting $7.66$. On line $(3)$, then, we must not subtract $0.24$ from $-2$ in order to obtain $-2.24$; rather, we must add the two numbers, like this: $$(-2) + 0.24 = -(2 - 0.24) = -1.76.$$ In practice, this is still too confusing, and a better method is to rearrange the original subtraction so that when you do the digit-by-digit calculation you are always doing it with the larger-magnitude number on top. So we first write $$5.9 - 7.66 = - (-5.9 + 7.66) = -(7.66 - 5.9).$$ \\begin{align} 7 &.66 \\\\ \\underline{-5} & \\underline{.9\\phantom{0}} \\\\ 1 &.76 \\end{align} Therefore $$5.9 - 7.66 = -(7.66 - 5.9) = -1.76.$$ We have $5.9$. If you subtract one, you have $4.9$. Subtract $2$, you have $3.9$. Subtract $3$, $2.9$. Naively, one might expect that subtracting $7$ therefore gives you $-2.9$. This fails, however, because this \"pattern\" breaks down once you", "and write down the resulting factorization if you find a root\". – Henning Makholm Aug 2 '18 at 9:43 1. First write $\\left(\\dfrac{ax\\phantom{+4}}{\\phantom{5}}\\right) \\left(\\dfrac{ax\\phantom{+4}}{\\phantom{5}}\\right).$ $$\\left(\\dfrac{x\\phantom{+4}}{\\phantom{5}}\\right) \\left(\\dfrac{x\\phantom{+4}}{\\phantom{5}}\\right)$$ 1. Find the product $ac$, including sign. $$ac=6$$ 1. Find the prime factorization of $ac$ using the factor tree. $$ac=6=2\\cdot 3$$ 1. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached \\sqrt{ac}. $$(1,6),(2,3),(-1,-6),(-2,-3)$$ 1. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$. $$(s,t)=(2,3)$$ 1. Write $\\left(\\dfrac{ax+s}{\\phantom{5}}\\right)\\left(\\dfrac{ax+t}{\\phantom{5}}\\right).$ $$\\left(\\dfrac{x+2}{\\phantom{5}}\\right)\\left(\\dfrac{x+3}{\\phantom{5}}\\right)$$ 1. Divide each of these binomials by its own GCF $\\left(\\dfrac{ax+s}{\\operatorname{gcf}(a,s)}\\right) \\left(\\dfrac{ax+t}{\\operatorname{gcf}(a,t)}\\right).$ Check that $$\\operatorname{gcf}(a,s)\\cdot\\operatorname{gcf}(a,t)=a.$$ $$\\left(\\dfrac{x+2}{1}\\right)\\left(\\dfrac{x+3}{1}\\right)=\\left(x+2\\right)\\left(x+3\\right)$$ In that case, as an alternative, we can proceed by completing the square as follow $$x^2+5x+6=x^2+5x+\\frac{25}{4}-\\frac{25}{4}+6=\\left(x+\\frac{5}{2}\\right)^2-\\frac{1}{4}=$$ $$=\\left[\\left(x+\\frac{5}{2}\\right)-\\frac{1}{2}\\right]\\left[\\left(x+\\frac{5}{2}\\right)+\\frac{1}{2}\\right]=\\left(x+2\\right)\\left(x+3\\right)$$ • Can you be more clear at this step $(1,6),(2,3),(-1,-6),(-2,-3)$? – Mark Aug 2 '18 at 10:12 • @C.Maxwell Those are all the possible integer pairs for the factorization of $6$. –", "\\ldots 11}^{n\\text{ digits}} \\phantom{ }) \\\\ 2c - b &= (a^2 - 9c) \\cdot (\\phantom{ }\\overbrace{11 \\ldots 11}^{n\\text{ digits}}\\phantom{ }). \\end{align*} Because $2c - b$ and $a^2 - 9c$ are constants and because there must be at least two distinct values of $n$ that satisfy, $2c - b = a^2 - 9c = 0.$ Thus, we have \\begin{align*} 2c&=b, \\\\ a^2&=9c. \\end{align*} Knowing that $a,b,$ and $c$ are single digit positive integers and that $9c$ must be a perfect square, the values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \\boxed{\\textbf{(D) } 18}.$ ~LegionOfAvatars (Solution) ~MRENTHUSIASM (Reformatting) ~ dolphin7", "2^4 +{\\phantom{2^3}}+2^2 = 64 + 32 + 16 +{\\phantom{xx}}+4,$ is represented in the bottom horizontal row and $84 = 2^6 + {\\phantom{2^5}}+ 2^4 +{\\phantom{2^3}}+2^2 = 64 +{\\phantom{32}}+ 16 +{\\phantom{xx}}+4,$ is recorded just below it in the margin. Step 3. Next, all of the counters are combined in the bottom horizontal row. Step 4. The board is abbreviated in two steps: First, from each square that had two counters after Step 3, one counter is moved to the square to the left and the other is removed. The result is shown above. Second, the abbreviation is repeated: from the square marked \"32\", one counter is moved to the left and the other is removed. The result is shown below. Step 5: Now that each square along the bottom row contains at most one counter, the sum can be read from this row as 116 + 84 = 128 + 64 + 8 = 200. Next: subtraction on the chessboard calculator # John Napier's Binary Chessboard Calculator - Subtraction Author(s): Sidney J. Kolpas and Erwin Tomash How to subtract one positive integer (the subtrahend) from another, larger positive integer (the minuend) (Gardner 1973): Step 1. Place the larger number on the bottom horizontal row. Step 2. Place the binary complement of the number to be subtracted in the margin below", "of a negative number, so $\\tfrac{225 - 15\\sqrt{229}}{2}$ is an extraneous solution. On the other hand, since we know that there is a solution to the equation and $\\tfrac{225 + 15\\sqrt{229}}{2}$ makes both sides positive in the original equation, $\\tfrac{225 + 15\\sqrt{229}}{2}$ is a valid solution. We know that $15\\sqrt{229}$ is greater than $225$, but we need to know if it’s less than $227.$ \\begin{align*} 15\\sqrt{229} &\\bigcirc 227 \\\\ 225 \\cdot 229 &\\bigcirc 227^2 \\\\ 51525 &\\bigcirc 51529 \\\\ \\end{align*} This means that $15\\sqrt{229}$ is less than $227,$ so $\\tfrac{225 + 15\\sqrt{229}}{2}$ is less than $226$. Thus, the value of the greatest integer less than or equal to $x$ is $\\boxed{225}$.", "can be squared to get 25? $$\\left( \\boxed{\\phantom{300}} \\right)^2=25$$ • There are two numbers that work, 5 and -5: $$5^2=25$$ and $$(\\text-5)^2=25$$ • If $$x+1 = 5$$, then $$x=4$$. • If $$x+1 = \\text-5$$, then $$x=\\text-6$$. This means that both $$x=4$$ and $$x=\\text-6$$ make the equation true and are solutions to the equation.", "change . . . 1. from step 0 to step 3? 2. from step 2 to step 5? 3. Conjecture about the factor from step 10 to step 13. 3. Refer to the third table. step value expression 0 1 2 3 4 5 6 2,048 1,024 512 1. Predict the value in steps 4, 5, and 6. 2. By what factor does the value change . . . 1. from step 1 to step 3? 2. from step 3 to step 5? 3. Conjecture about the factor from step 10 to step 12. 3. By what factor does the value change . . . 1. from step 0 to step 3? 2. from step 2 to step 5? 3. Conjecture about the factor from step 10 to step 13. ### 18.3: Rewriting Expressions 1. For each given expression, decide what to write in the box to create equal expressions. given expression equal expression 1 equal expression 2 $$5\\boldcdot10^8$$ $$5\\boldcdot100^\\boxed{\\phantom{3}}$$ $$5\\boldcdot\\boxed{\\phantom{3}}^2$$ $$7\\boldcdot16^9$$ $$7\\boldcdot\\boxed{\\phantom{3}}^{4\\boldcdot9}$$ $$7\\boldcdot4^\\boxed{\\phantom{3}}$$ $$(0.25)^3$$ $$(0.5)^\\boxed{\\phantom{3}}$$ $$\\boxed{\\phantom{3}}^1$$ $$3\\boldcdot(1.2)^6$$ $$3\\boldcdot1.44^\\boxed{\\phantom{3}}$$ $$3\\boldcdot1.728^\\boxed{\\phantom{3}}$$ $$6\\boldcdot0.09^{10}$$ $$6\\boldcdot\\boxed{\\phantom{3}}^5$$ $$6\\boldcdot0.3^\\boxed{\\phantom{3}}$$ 2. Write at least 3 new expressions that are equal to $$4 \\boldcdot 27^6$$.", "equations: \\begin{align}\\Box-\\Box+2&=18 \\\\ \\small{\\rm or} \\quad \\Box-\\Box+4&=18.\\end{align} Therefore, \\begin{align}\\Box-\\Box&=18-2=16 \\\\ \\small{\\rm or} \\quad \\Box-\\Box&=18-4=12.\\end{align} And finally, From the previous proof, THERE EXISTS NO SOLUTION! ### Proof: Now considering the first equation, the first box has to have an option number greater than $$16$$. The only option number like that is $$25$$. We thus have $$\\boxed{25}-\\Box=16$$ therefore $$\\Box=25-16=9$$. But $$9$$ is not an option number. That is a contradiction, so the first equation cannot exist. $$\\require{cancel}{\\xcancel {\\Box-\\Box=16}}$$ Considering the second equation, the first box needs to be greater than $$12$$. It can't be $$12$$, it has to be greater than $$12$$. Again, the only option number greater than $$12$$ is $$25$$. We thus have $$\\boxed{25}-\\Box=12$$ therefore $$\\Box=25-12=13$$. But $$13$$ is not an option number. That is a contradiction so the second equation cannot exist. $$\\require{cancel}{\\xcancel {\\Box-\\Box=12}}$$ But if both equations cannot exist, then... ...THERE IS NO SOLUTION! Therefore, Some lateral-thinking must be required, unless you do not follow by BODMAS or BEDMAS or PEDMAS. • check the tags in the question :) – Oray Sep 21 '18 at 14:19 • @Oray I did, but DEEM wrote that he/she found a solution without inverting a $6$ to a $9$, and I cannot think of anything else more lateral :P – Mr Pie Sep 21 '18 at 14:32 • @user477343 This", "# Lesson 6 Find That Factor • Let’s calculate factors that scale numbers. ### 6.1: Multiplication and Division Here are some multiplication and division equations. Write the missing pieces. The first one is completed, as an example. 1. $$6 \\div 2 = 3$$ and $$2 \\boldcdot 3 = 6$$ 2. $$20 \\div 4 = 5$$ and $$\\underline{\\hspace{.5in}}$$ 3. $$\\underline{\\hspace{.5in}}$$ and $$1.5 \\boldcdot 12 = 18$$ 4. $$9 \\div \\frac14 = 36$$ and $$\\underline{\\hspace{.5in}}$$ 5. $$12 \\div 15 = \\underline{\\hspace{.5in}}$$ and $$\\underline{\\hspace{.5in}}$$ 6. $$a \\div b = c$$ and $$\\underline{\\hspace{.5in}}$$ ### 6.2: Scaling Segments For each question, the length of the second segment (on the right) is some fraction of the length of the first segment (on the left). Complete the division and multiplication equations that relate the lengths of the segments. $$7 \\div 14 = \\frac12$$ $$14 \\boldcdot \\frac12 = 7$$ $$\\boxed{\\phantom{33}} \\div \\boxed{\\phantom{33}} = 3$$ $$\\boxed{\\phantom{33}} \\boldcdot \\boxed{\\phantom{33}} = 12$$ $$8 \\div 12 = \\boxed{\\phantom{33}}$$ $$12 \\boldcdot \\boxed{\\phantom{33}}=8$$ $$\\boxed{\\phantom{33}} \\div \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ $$\\boxed{\\phantom{33}} \\boldcdot \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ $$\\boxed{\\phantom{33}} \\div \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ $$\\boxed{\\phantom{33}} \\boldcdot \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ $$\\boxed{\\phantom{33}} \\div \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ $$\\boxed{\\phantom{33}} \\boldcdot \\boxed{\\phantom{33}}=\\boxed{\\phantom{33}}$$ ### 6.3: Medicine Wears Off Some different medications were given to patients in a clinical trial, and the amount of medication remaining in the patient’s bloodstream was measured every hour for the first three hours after the medicine was given. Here", "Thus, a = 8 and $±$3 Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5). #### Page No 267: Let the required terms be (a - d), a and (a + d). Then (a - d) + a + (a + d) = 21 ⇒ 3a = 21 ⇒ a = 7​ Also, (a - d)2 + a2 + (a + d)2​ = 165 ⇒ 3a2 + 2d2 = 165 ​(3 ⨯ 49​​ + 2 d2) = 165 ​ ⇒ 2d2 = 165 - 147 = 18 d2 = 9​ ⇒ d = $±$3 Thus, a = 7 and $±$3 Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4). #### Page No 268: Let the required angles be (a - 15)o, (a - 5)o, (a + 5)o and (a + 15)o, as the common difference is 10 (given). Then (a - 15)o + (a - 5)o + (a + 5)o + (a + 15)o = 360o ⇒ 4a = 360 ⇒ a = 90 Hence, the required angles of a quadrilateral are #### Page No 268: ${a}^{2}+9{d}^{2}-6ad+{a}^{2}+{d}^{2}-2ad+{a}^{2}+{d}^{2}+2ad+{a}^{2}+9{d}^{2}+6ad=216\\phantom{\\rule{0ex}{0ex}}⇒4{a}^{2}+20{d}^{2}=216\\phantom{\\rule{0ex}{0ex}}⇒4{\\left(7\\right)}^{2}+20{d}^{2}=216\\phantom{\\rule{0ex}{0ex}}⇒d=±1\\phantom{\\rule{0ex}{0ex}}$ #### Page No 268: Let the four parts in AP be (a − 3d), (ad), (a + d) and (a + 3d). Then, Also, $⇒960-135{d}^{2}=448-7{d}^{2}\\phantom{\\rule{0ex}{0ex}}⇒135{d}^{2}-7{d}^{2}=960-448\\phantom{\\rule{0ex}{0ex}}⇒128{d}^{2}=512\\phantom{\\rule{0ex}{0ex}}⇒{d}^{2}=4$ $⇒d=±2$ When a = 8 and d =", "winning numbers) $=\\frac{C\\left(6,6\\right)}{C\\left(53,6\\right)}\\phantom{\\rule{0ex}{0ex}}=\\frac{1}{22957480}$ b) P( Matching five from 6 numbers and 1 from 47 remaining) = $C\\left(6,5\\right)\\cdot \\frac{C\\left(47,1\\right)}{C\\left(53,6\\right)}\\phantom{\\rule{0ex}{0ex}}=6\\cdot \\frac{47}{22957480}\\phantom{\\rule{0ex}{0ex}}=\\frac{2}{162819}$ c) P(Matching four numbers) $=C\\left(6,4\\right)\\cdot \\frac{C\\left(47,2\\right)}{C\\left(53,6\\right)}\\phantom{\\rule{0ex}{0ex}}=15\\cdot \\frac{1081}{22957480}\\phantom{\\rule{0ex}{0ex}}=\\frac{3243}{4591496}$ d) P(Matching three numbers)= $C\\left(6,3\\right)\\cdot \\frac{C\\left(47,3\\right)}{C\\left(53,6\\right)}\\phantom{\\rule{0ex}{0ex}}=20\\cdot \\frac{16215}{22957480}\\phantom{\\rule{0ex}{0ex}}=\\frac{16215}{1147874}$ e) P(Matching two numbers)= $C\\left(6,2\\right)\\cdot \\frac{C\\left(47,4\\right)}{C\\left(53,6\\right)}\\phantom{\\rule{0ex}{0ex}}=15\\cdot \\frac{178365}{22957480}\\phantom{\\rule{0ex}{0ex}}=\\frac{535095}{4591496}$ We have step-by-step solutions for your answer!", "is the list of perfect squares using numbers up to 20: $\\begin{array}{llll} 1^2=1\\hspace{0.5in}&\\phantom{1}6^2=36\\hspace{0.5in}&11^2=121\\hspace{0.5in}&16^2=256 \\\\ \\\\ 2^2=4&\\phantom{1}7^2=49&12^2=144&17^2=289 \\\\ \\\\ 3^2=9&\\phantom{1}8^2=64&13^2=169&18^2=324 \\\\ \\\\ 4^2=16&\\phantom{1}9^2=81&14^2=196&19^2=361 \\\\ \\\\ 5^2=25&10^2=100&15^2=225&20^2=400 \\end{array}$ Cubes are numbers multiplied by themselves three times. A number that is being cubed is shown as having a superscript 3 attached to it. For example, 5 cubed is written as 53, which equals 5 × 5 × 5 or 125. Perfect cubes are cubes of whole numbers, such as 1, 8, 27, 64, 125, 216, and 343. They are found by cubing natural numbers. The following is the list of perfect cubes using numbers up to 20: $\\begin{array}{llll} 1^3=1\\hspace{0.5in}&\\phantom{1}6^3=216\\hspace{0.5in}&11^3=1331\\hspace{0.5in}&16^3=4096 \\\\ \\\\ 2^3=8&\\phantom{1}7^3=343&12^3=1728&17^3=4913 \\\\ \\\\ 3^3=27&\\phantom{1}8^3=512&13^3=2197&18^3=5832 \\\\ \\\\ 4^3=64&\\phantom{1}9^3=729&14^3=2744&19^3=6859 \\\\ \\\\ 5^3=125&10^3=1000&15^3=3375&20^3=8000 \\end{array}$ Percentage means parts per hundred. A percentage can be thought of as a fraction $\\dfrac{a}{b},$ where $a,$ the numerator, is the number to the left of the % sign, and $b,$ the denominator, is 100. For example: $42\\% = \\dfrac{42}{100} = 0.42.$ Absolute values. The absolute value of an expression $a,$ denoted $| a |,$ is the distance from zero of the number or operation that occurs between the absolute value signs. For example: $| -4 | = 4 \\text{ or } | -9 | = 9$ Examples of absolute values of simple operations are: $\\begin{array}{ccccc} |-8+6|=2&\\text{since}&-8+6=-2&\\text{and}&|-2|=2 \\\\ \\\\", "equations to solve for $d$. \\begin{aligned}2d + c &= 12\\\\2d + 2&= 12\\\\2d&= 10\\\\d&= 5\\end{aligned} This means that one box of donuts costs $\\$5$while a cup of coffee costs$\\$2$ at Amy’s favorite pastry shop. Practice Question 1. Which of the following shows the solution to the system of equations $\\begin{array}{ccc}3a – 4b&= \\phantom{x}18\\\\3a – 8b&= \\phantom{x}26\\end{array}$? A.$a=-2,b=\\dfrac{10}{3}$ B. $a=\\dfrac{10}{3},b=-2$ C. $a=-2,b=-\\dfrac{10}{3}$ D. $a=\\dfrac{10}{3},b=2$ 2. Which of the following shows the solution to the system of equations $\\begin{array}{ccc}4x + 5y&= \\phantom{x}4\\\\5x- 4y&= -2\\end{array}$? A. $\\left(-\\dfrac{28}{41},-\\dfrac{6}{41}\\right)$ B. $\\left(-\\dfrac{6}{41},-\\dfrac{28}{41}\\right)$ C. $\\left(\\dfrac{28}{41},\\dfrac{6}{41}\\right)$ D. $\\left(\\dfrac{6}{41},\\dfrac{28}{41}\\right)$", "# A computer science problem by Thành Đạt Lê $\\boxed{\\phantom0}\\ \\boxed{\\phantom0}\\ \\boxed{\\phantom0}\\ \\boxed{\\phantom0} \\times 3 = \\boxed{\\phantom0}\\ \\boxed{\\phantom0}\\ \\boxed{\\phantom0}\\ \\boxed{\\phantom0}\\ \\boxed{\\phantom0}$ Can we fill in the boxes with distinct digits so that the equation is satisfied? Clarification: The above equation represents a 4-digit number multiplied by 3 to produce a 5-digit number, so the leftmost box is non-zero on either side. ×", "# Calculate: 2586*3815 ## Expression: $$2586 \\cdot 3815$$ First line up the numbers vertically and match the places from the right like this. $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\end{array}$$ Now multiply the first number with the $1^{st}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 2586 with 5. Write the result 12930 at the end leaving 0 spaces to the right like this. $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\phantom{\\times99}12930\\\\\\end{array}$$ Now multiply the first number with the $2^{nd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 2586 with 1. Write the result 2586 at the end leaving 1 spaces to the right like this. $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\phantom{\\times99}12930\\\\\\phantom{\\times99}2586\\phantom{9}\\\\\\end{array}$$ Now multiply the first number with the $3^{rd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 2586 with 8. Write the result 20688 at the end leaving 2 spaces to the right like this. $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\phantom{\\times99}12930\\\\\\phantom{\\times99}2586\\phantom{9}\\\\\\phantom{\\times}20688\\phantom{99}\\\\\\end{array}$$ Now multiply the first number with the $4^{th}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 2586 with 3. Write the result 7758 at the end leaving 3 spaces to the right like this. $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\phantom{\\times99}12930\\\\\\phantom{\\times99}2586\\phantom{9}\\\\\\phantom{\\times}20688\\phantom{99}\\\\\\underline{\\phantom{\\times}7758\\phantom{999}}\\\\\\end{array}$$ $$\\begin{array}{c}\\phantom{\\times999}2586\\\\\\underline{\\times\\phantom{999}3815}\\\\\\phantom{\\times99}12930\\\\\\phantom{\\times99}2586\\phantom{9}\\\\\\phantom{\\times}20688\\phantom{99}\\\\\\underline{\\phantom{\\times}7758\\phantom{999}}\\\\\\phantom{\\times}9865590\\end{array}$$", "the correct answer is option (d). #### Question 9: If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ? (a) −5 (b) 5 (c) 3 (d) −3 (b) 5 #### Question 10: If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then (a) m = 5, n = −3 (b) m = 7, n = −18 (c) m = 17, n = −8 (d) m = 23, n = −19 (b) m = 7, n = −18 Let: $p\\left(x\\right)={x}^{3}+10{x}^{2}+mx+n\\phantom{\\rule{0ex}{0ex}}\\phantom{\\rule{0ex}{0ex}}$ Now, $x+2=0⇒x=-2$ (x + 2) is a factor of p(x). So, we have p($-$2)=0 Now, $x-1=0⇒x=1$ Also, (x $-$ 1) is a factor of p(x). We have: p(1) = 0 By substituting the value of m in (i), we get n = −18. ∴ m = 7 and n = −18 #### Question 11: 104 × 96 = ? (a) 9864 (b) 9984 (c) 9684 (d) 9884 (b) 9984 $104×96=\\left(100+4\\right)\\left(100-4\\right)$ $={100}^{2}-{4}^{2}\\phantom{\\rule{0ex}{0ex}}=\\left(10000-16\\right)\\phantom{\\rule{0ex}{0ex}}=9984$ 305 × 308 = ? (a) 94940 (b) 93840 (c) 93940 (d) 94840 (c) 93940 207 × 193 = ? (a) 39851 (b) 39951 (c) 39961 (d) 38951 (b) 39951 #### Question 14: 4a2 + b2 + 4ab + 8a + 4b + 4 = ? (a) (2a + b", "# Difference between revisions of \"2007 iTest Problems/Problem 55\" The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted. ## Problem Let $R=675$. Let $x$ be the smallest real solution of $3x^2+Rx+R=90x\\sqrt{x+1}$. Find the value of $\\lfloor x\\rfloor$. ## Solution Plugging in $R$ results in \\begin{align*} 3x^2 + 675x + 675 &= 90x\\sqrt{x+1} \\\\ x^2 + 225x + 225 &= 90x\\sqrt{x+1} \\end{align*} We can solve the equation a bit easier by letting $a = x+1$. \\begin{align*} x^2 + 225a &= 30x\\sqrt{a} \\\\ x^4 + 450x^2 a + 225^2 a^2 &= 900x^2 a \\\\ (x^2 - 225a)^2 &= 0 \\\\ x^2 - 225x - 225 &= 0 \\end{align*} Using the Quadratic Formula yields $x = \\tfrac{225 \\pm 15\\sqrt{229}}{2}$. However, we have to check for extraneous solutions and calculate the highest integer less than or equal to $x$. This can be easily done using a calculator, but in this solution, we will not use a calculator. We know that $15.1^2 = 228.01$ and $15.15^2 = 229.5225$, so $226.5 < 15\\sqrt{229} < 227.25$. That means $-1.5 < \\tfrac{225 - 15\\sqrt{229}}{2} < -1.125$. Inserting the value into the equation would result in the right hand side having a square root" ]

[ "of a negative number, so $\\tfrac{225 - 15\\sqrt{229}}{2}$ is an extraneous solution. On the other hand, since we know that there is a solution to the equation and $\\tfrac{225 + 15\\sqrt{229}}{2}$ makes both sides positive in the original equation, $\\tfrac{225 + 15\\sqrt{229}}{2}$ is a valid solution. We know that $15\\sqrt{229}$ is greater than $225$, but we need to know if it’s less than $227.$ \\begin{align*} 15\\sqrt{229} &\\bigcirc 227 \\\\ 225 \\cdot 229 &\\bigcirc 227^2 \\\\ 51525 &\\bigcirc 51529 \\\\ \\end{align*} This means that $15\\sqrt{229}$ is less than $227,$ so $\\tfrac{225 + 15\\sqrt{229}}{2}$ is less than $226$. Thus, the value of the greatest integer less than or equal to $x$ is $\\boxed{225}$.", "Mathematics OpenStudy (anonymous): what is the smallest sloution to the equation x to the fourth - 34x to the 2 +225=0 So, you can factor this into a pair of quadratics. Basically, you have this: $x^4 - 34x^2 + 225 = 0$ Imagine that $$u = x^2$$. Then you have: $u^2 - 34u + 225 = 0$ So first you solve for $$u$$. Then, you have $$u = x^2$$, so $$x = \\sqrt{u}$$. Does that help?", "# Difference between revisions of \"2007 iTest Problems/Problem 55\" The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted. ## Problem Let $R=675$. Let $x$ be the smallest real solution of $3x^2+Rx+R=90x\\sqrt{x+1}$. Find the value of $\\lfloor x\\rfloor$. ## Solution Plugging in $R$ results in \\begin{align*} 3x^2 + 675x + 675 &= 90x\\sqrt{x+1} \\\\ x^2 + 225x + 225 &= 90x\\sqrt{x+1} \\end{align*} We can solve the equation a bit easier by letting $a = x+1$. \\begin{align*} x^2 + 225a &= 30x\\sqrt{a} \\\\ x^4 + 450x^2 a + 225^2 a^2 &= 900x^2 a \\\\ (x^2 - 225a)^2 &= 0 \\\\ x^2 - 225x - 225 &= 0 \\end{align*} Using the Quadratic Formula yields $x = \\tfrac{225 \\pm 15\\sqrt{229}}{2}$. However, we have to check for extraneous solutions and calculate the highest integer less than or equal to $x$. This can be easily done using a calculator, but in this solution, we will not use a calculator. We know that $15.1^2 = 228.01$ and $15.15^2 = 229.5225$, so $226.5 < 15\\sqrt{229} < 227.25$. That means $-1.5 < \\tfrac{225 - 15\\sqrt{229}}{2} < -1.125$. Inserting the value into the equation would result in the right hand side having a square root", "The smallest solution is $m=130=2 \\cdot 5\\cdot 13$, with roots of $x^2+1 \\equiv 0 \\pmod m$ being $\\{47, 57, 73, 83\\}$ • how to derive this? – lab bhattacharjee Dec 10 '13 at 15:00 • thanks, can you tell me what was the reasoning behind that? – FranckN Dec 10 '13 at 15:00 • I didn't find a reasoning, so I did it by computer. Still manageable by hand, but not very nice. But have a look at OEIS A224770. It seems to give the possible values of $m$ (not considering the number of prime factors), though I don't know why. – Jean-Claude Arbaut Dec 10 '13 at 15:05 • it's supposed to be solvable by hand, so I think it must have certain trick to make it easier – FranckN Dec 10 '13 at 15:38", "We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube). Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$. At this moment the second photon will be at the coordinates $(12\\cdot 12,7\\cdot 12,5\\cdot 12)$. Then the distance it travelled is $\\sqrt{ (12\\cdot 12)^2 + (7\\cdot 12)^2 + (5\\cdot 12)^2 } = 12\\sqrt{12^2 + 7^2 + 5^2}=12\\sqrt{218}$. And as the factorization of $218$ is $218=2\\cdot 109$, we have $m=12$ and $n=218$, hence $m+n=\\boxed{230}$.", "taking t = 6 and s – t = 9, so that s = 15. That gives the solution $a = 261 = 15^2+6^2$, with $a^2 = 261^2 = 189^2 + 180^2$. This certainly satisfies all the conditions (1), (2), (3), but I do not know whether it is the smallest solution apart from a = 5.", "also true that 7 is the smallest positive solution to $2343^a \\equiv a \\pmod{10000}$. I haven't thought about this long enough to know whether I should be surprised. Let $(a,x)$ denote that $x$ is the smallest positive solution to $a^x \\equiv x \\pmod{10^d}$, where $d$ is the number of digits of $x$. I found the pairs (2,36), (3,87), (4,96), (5,25), (6,56), (7,43) [which contains the last two digits of your 2343 value], (8,56), and (9,89). Solutions to $a^a \\equiv a \\pmod{10^d}$, where $d$ is the number of digits of $a$, can be found here. The sequence itself is pretty interesting. -", "be any solutions at all (you can prove this by contradiction). Suppose there were two solutions $t$ and $t'$ then $a(t'-t) \\equiv 0 \\pmod {m}$, in other words if we found some minimal solution $t$, then every other solution would be of the form $t+d$ for some $ad|m$. So to compute the solutions in this case let's start by computing $\\gcd(4991,7429) = 23$ so $4991 = 217 \\cdot 23$ and $217^{-1} \\equiv 582 \\pmod {7429}$ so we should start by finding the smallest $t$ satisfying $23 t \\equiv 582 \\cdot 460 \\equiv 276 \\equiv 23 \\cdot 12 \\pmod {7429}$, which is a multiple of $23$ so we're in luck and $t=12$ is the smallest solution. Now we can just go through the divisors of $7429/23 = 323 = 17 \\cdot 19$ in order: 12+17=31 12+19=49 12+17*19=335 all are within the specified range so we have $4$ solutions. - add comment", "$16$, respectively. $211 + 28 + 16 = 255$. Hence, the answer is $\\boxed{\\textbf{(A)}\\ 255}$. ## Solution 3 Using the closed forms for the sums, we get $m^2=n(n+1)+212$, or $m^2=n^2+n+212$. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$. Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$. Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7\\cdot11\\cdot11$. The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$, $7$ and $121$, $11$ and $77$. In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$. The sum of all possible values of $n$ is $211+28+16=\\boxed{\\textbf{(A)}\\ 255}$. ~dolphin7", "# How to solve this simple equation? • January 5th 2013, 10:57 AM matte123 How to solve this simple equation? Hi, I forgot the basic rules of how to solve this equation. I know that x = 26 How would we solve this? 25.1 = 2250 + x * 10 _____________ 100 • January 5th 2013, 11:01 AM earboth Re: How to solve this simple equation? Quote: Originally Posted by matte123 Hi, I forgot the basic rules of how to solve this equation. I know that x = 26 But would we solve this? 25.1 = 2250 + x * 10 ___________ 100 1. I assume that you want to solve for x: $25.1 = \\frac{2250+x \\cdot 10}{100}$ 2. Multiply both sides by 100 to get rid of the denominator. You should get: $2510 = 2250 + 10x$ Solve for x. • January 5th 2013, 11:06 AM matte123 Re: How to solve this simple equation? Thank you earboth, Ofcourse you are right :) Then: 2510 - 2250 = 10x x = 260 /10 x = 26", "# Math Help - How to solve this simple equation? 1. ## How to solve this simple equation? Hi, I forgot the basic rules of how to solve this equation. I know that x = 26 How would we solve this? 25.1 = 2250 + x * 10 _____________ 100 2. ## Re: How to solve this simple equation? Originally Posted by matte123 Hi, I forgot the basic rules of how to solve this equation. I know that x = 26 But would we solve this? 25.1 = 2250 + x * 10 ___________ 100 1. I assume that you want to solve for x: $25.1 = \\frac{2250+x \\cdot 10}{100}$ 2. Multiply both sides by 100 to get rid of the denominator. You should get: $2510 = 2250 + 10x$ Solve for x. 3. ## Re: How to solve this simple equation? Thank you earboth, Ofcourse you are right Then: 2510 - 2250 = 10x x = 260 /10 x = 26", "is $m=130=2 \\cdot 5\\cdot 13$, with roots of $x^2+1 \\equiv 0 \\pmod m$ being $\\{47, 57, 73, 83\\}$ The smallest solution is $m=130=2 \\cdot 5\\cdot 13$, with roots of $x^2+1 \\equiv 0 \\pmod m$ being $\\{47, 57, 73, 83\\}$", "adding $21$ two more times gives us $19$ with no remainders. We now have $1077 + 21 + 21 + 21 = 1140$. However, this number is divisible by $120$. To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$. Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$. Adding $420$ again gives us $1980$, which is valid. However, remember that this is equal to $n + 63$, so subtracting $63$ from $1980$ gives us $1917$, which is $n$. The sum of its digits are $1 + 9 + 1 + 7 = \\boxed{\\textbf{(C) } 18}$. ~primegn ## Solution 8 (Euclidean Algorithm) By the Euclidean Algorithm, we have \\begin{alignat*}{8} \\gcd(63,n+120)&=\\hspace{1mm}&&\\gcd(63,\\phantom{ }\\underbrace{n+120-63k_1}_{(n+120) \\ \\mathrm{mod} \\ 63}\\phantom{ })&&=21, \\\\ \\gcd(n+63,120)&=&&\\gcd(\\phantom{ }\\underbrace{n+63-120k_2}_{(n+63) \\ \\mathrm{mod} \\ 120}\\phantom{ },120)&&=60. \\end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$ ## Video Solutions ### Video Solution 2 https://youtu.be/8mNMKH0T9W0 - Happytwin ### Video Solution 3 (Quick & Simple) Education The Study of Everything ### Video Solution 5 https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0", "with PARI/GP , for the prime $$p=29$$ the smallest solution is $$x=511417665685259113593784321$$ $$y=608113496300320416961344$$ and we have $$x\\equiv 1\\mod 837$$ • And what about the solution for $=p^2$, is that $\\equiv 2$? – Maestro13 Jun 30 '18 at 9:22 • I checked myself. The smallest solution for $=p^2$ is simply 29 times the one you found, hence one to be excluded. And $\\mod p^2-4$, it is obviously equal to $29 \\ne 2$. So not a requested solution. Or, in terms of the later added underlying question, the only way to find a solution for the second equation is, to multiply one for the first equation by $29$. – Maestro13 Jul 2 '18 at 17:16", "$(199,253)$ and $(203,248)$. We see that $252 \\cdot 198 = 49896 > 202 \\cdot 247 = 49894$. The solution is $\\boxed{896}$. ### Solution 2 We realize that drawing $x$ vertical lines and $y$ horizontal lines, the number of basic rectangles we have is $(x-1)(y-1)$. The easiest possible case to see is $223$ vertical and $223$ horizontal lines, as $(4+5)223 = 2007$. Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation $(222-4x)(222+5x)$ maximize. Expanded, this gives $-20x^{2}+222x+222^{2}$. From $-\\frac{b}{2a}$ you get that the vertex is at $x=\\frac{111}{20}$. This is not an integer though, so you see that when $x=5$, you have $-20*25+222*5+222^{2}$ and that when x=6, you have $-20*36+222*6+222^{2}$. $222 > 20*11$, so the maximum integral value for x occurs when $x=6$. Now you just evaluate $-20*36+222*6+222^{2}\\mod 1000$ which is ${896}$.", "+ \\\\ \\hline \\\\ -5 \\cdot 1 + 3 \\cdot 2 &=& 1 & \\end{matrix}$ This is the lowest value we can get for the right hand side. Since we need a right hand side of 2, we multiply by 2: $-5 \\cdot 2 + 3 \\cdot 4 = 2$ So your solution is $x=3 \\cdot 4+2=14$", "smallest solution for $$x_1$$ is 18 and for $$y_1$$ = 14, but they're both even and we have to settle for $$x_1 = 39$$ (or $$y_1 = 35$$ but that will give a larger solution). This is the result: $$31 \\times 70 = 2170$$ The solution for $$3 \\times 10$$ is rather small compared to the rectangle size and my program was able to verify (after a few weeks of calculation) that it's the minimal solution. Interestingly, the $$28 \\times 55$$ solution for $$3 \\times 5$$ is not the smallest one. The one below is smaller and is generalizable for $$3 \\times n$$ where $$n$$ is odd but not divisible by 3, but it's usually larger than the corresponding one for the previous family. $$35 \\times 38 = 1330$$ • Yup. Two down, six to go. – theonetruepath Apr 15 '18 at 10:28 • 1x2, 1x3, 1x4, 1x5 all minimal. Four more that I've found so far. – theonetruepath May 9 '18 at 10:03 • I've found seven more, including a $1 \\times 2k$ generalizable one. Will post them later today. – Glorfindel May 9 '18 at 17:12 • Your 1x6 is smaller than mine which means I have to go find a bug. 1x7 is the same, also 1x8 and 2x5. My program hasn't reached the others", "m \\\\ 19 m &= a(10^n-2) \\\\ m &= \\frac{2a(5\\cdot 10^{n-1} - 1)}{19} \\end{align} The smallest $$m$$, if one exists, requires $$a,n$$ such that 19 divides $$5\\cdot 10^{n-1}-1$$ (as 19 can't divide $$2 a$$) and the result has $$n$$-digits. It's easy to check that the smallest value of $$n$$ that satisfies the first condition is $$n=17$$. To get the smallest solution we try $$a=1$$, but this yields a value with only 16 digits. Setting $$a=2$$, however, yields the 17-digit $$m = 10526315789473684$$. The smallest solution to our puzzle is therefore the 18-digit number $$105263157894736842$$; that's surprisingly large. Numbers with this type of property are known as parasitic numbers. Later, I wondered if there were numbers with the slightly different, but equally interesting property, that moving the last digit to the front converted (\"autoconverts\") it from a value under one unit of measurement to an equivalent value under a different unit of measurement. The first one I tried was moving the last digit to the front converts from Celsius to Fahrenheit. This is a fun puzzle that eventually made its way into the New York Times. The smallest such value is 275 C, which exactly equals 527 F. What's the next smallest temperature? How about moving the first digit to the end? We'll need to use the little-known fact", "m \\\\ 19 m &= a(10^n-2) \\\\ m &= \\frac{2a(5\\cdot 10^{n-1} - 1)}{19} \\end{align} The smallest $$m$$, if one exists, requires $$a,n$$ such that 19 divides $$5\\cdot 10^{n-1}-1$$ (as 19 can't divide $$2 a$$) and the result has $$n$$-digits. It's easy to check that the smallest value of $$n$$ that satisfies the first condition is $$n=17$$. To get the smallest solution we try $$a=1$$, but this yields a value with only 16 digits. Setting $$a=2$$, however, yields the 17-digit $$m = 10526315789473684$$. The smallest solution to our puzzle is therefore the 18-digit number $$105263157894736842$$; that's surprisingly large. Numbers with this type of property are known as parasitic numbers. Later, I wondered if there were numbers with the slightly different, but equally interesting property, that moving the last digit to the front converted (\"autoconverts\") it from a value under one unit of measurement to an equivalent value under a different unit of measurement. The first one I tried was moving the last digit to the front converts from Celsius to Fahrenheit. This is a fun puzzle that eventually made its way into the New York Times. The smallest such value is 275 C, which exactly equals 527 F. What's the next smallest temperature? How about moving the first digit to the end? We'll need to use the little-known fact", "question, how can I find the solution, where the sum of $x+y$ is the smallest? I'd appreciate any help whatsoever! Yes those are correct, you just need a bound for $k$ which can be calculated using $x \\ge 1, y\\ge 1$ or $3k-350 \\ge 1, 350-2k \\ge 1$ also, $2x+3y=350 \\implies 2(x+y)=350 -y$, to find the minimum value of $x+y$ calculate maximum value of $y$ • Thanks! This is the step I was missing. One note; we actually said that we can use $0$ as a whole number in discrete structure classes, so I have to check $x\\ge0$.. – peroxy Jan 26 '15 at 17:17 • So I found out that there are $60$ solutions, $116 \\le k \\le 175$. Do you know how I can find the right solution, where the sum of $x+y$ is the smallest? – peroxy Jan 26 '15 at 17:25" ]

[ "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"E\",E,SE); label(\"F\",F,NE); label(\"P\",P,NW); label(\"Q\",Q,NE); label(\"R\",R,SE); label(\"S\",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]$ Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \\frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$. We then know that $CE = \\frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$. Since $PF+FQ=PQ=\\omega$. We can solve for $PF$ and $FQ$ in terms of $\\omega$. We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$. We can set up the proportion: $\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$. Solving for $AF$, $AF = \\frac{3}{4} PF = \\frac{3}{4}", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "# Problem #2163 2163 Square $ABCD$ has side length $s$, a circle centered at $E$ has radius $r$, and $r$ and $s$ are both rational. The circle passes through $D$, and $D$ lies on $\\overline{BE}$. Point $F$ lies on the circle, on the same side of $\\overline{BE}$ as $A$. Segment $AF$ is tangent to the circle, and $AF=\\sqrt{9+5\\sqrt{2}}$. What is $r/s$? $[asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",Ep,E); label(\"F\",F,NW); [/asy]$ $\\mathrm{(A) \\ } \\frac{1}{2}\\qquad \\mathrm{(B) \\ } \\frac{5}{9}\\qquad \\mathrm{(C) \\ } \\frac{3}{5}\\qquad \\mathrm{(D) \\ } \\frac{5}{3}\\qquad \\mathrm{(E) \\ } \\frac{9}{5}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "(1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similary, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$.", "# Problem #2136 2136 In the right triangle $\\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\\triangle DBF$ to that of $\\triangle ACE$? $\\mathrm{(A) \\ } \\frac{1}{4} \\qquad \\mathrm{(B) \\ } \\frac{9}{25} \\qquad \\mathrm{(C) \\ } \\frac{3}{8} \\qquad \\mathrm{(D) \\ } \\frac{11}{25} \\qquad \\mathrm{(E) \\ } \\frac{7}{16}$ $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",C--B,W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "#### Triangles and their classification ##### Angle Classifacation Acute Triangle All angles are less than 90° Right Triangle Has a right angle (90°) Obtuse Triangle Has an angle more than 90° ##### Side Classifacation Equilateral Triangle Three equal sides Isosceles Triangle Two equal sides Two equal angles Scalene Triangle No equal sides No equal angles This part shows what Proves ussually look like ## Proof that any nonperfect square positive integer is irrational Let us assume that $\\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\\frac{p}{q}=\\sqrt{n} \\Rightarrow \\frac{p^2}{q^2}=n \\Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\\sqrt{n}$ is irrational. ## Proof of the Pythagorean Theorem $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Proof", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "circle at $G$. Draw line $BG$. $A[\\Delta ABD] = A[\\Delta ABG] = 120$ (Since $\\square ABGD$ is cyclic) But $A[\\Delta ABE]$ is common in both with an area of 60. So, $A[\\Delta AED] = A[\\Delta BEG]$. \\therefore $A[\\Delta AED] \\cong A[\\Delta BEG]$ (SAS Congruency Theorem). In $\\Delta AED$, let $DI$ be the median of $\\Delta AED$. Which means $A[\\Delta AID] = 30 = A[\\Delta EID]$ Rotate $\\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\\frac{AE}{EF} = 2:1$. $AE$ is a radius and $EF$ is half of it implies $EF$ = $\\frac{radius}{2}$. Which means $A[\\Delta BEF] \\cong A[\\Delta DEI]$ Thus $A[\\Delta BEF] = \\boxed{\\textbf{(B) }30}$ ~phoenixfire & flamewavelight ## Solution 8 $[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(\"M\",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\\overline{AD}$ and $\\overline{CD}$, we find the area of $\\triangle ADB$ is $120$ and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The" ]

[ "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "(1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similary, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$.", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "label(\"D\",D,NNW); label(\"C\",C,SW); label(\"H\",H,SE); draw(e--H,dashed); label(\"O\",(0,0),NE); label(\"1\",(C--O),N); label(\"1\",(H--O),N); label(\"2\",(A--C),N); label(\"2\",(H--B),N); label(\"3\",(O--D),NE); label(\"3\",(O--e),NE); label(\"2\\sqrt{2}\",(D--C),W); label(\"2\\sqrt{2}\",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]$ Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\\sqrt{3^2-1^2}$ or $2\\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\\frac{(3+2+1)(2\\sqrt{2})}{2}$ or $6\\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\\frac{(1+1)(2\\sqrt{2})}{2}$ or $2\\sqrt{2}$. We also know that the area of $[OHE]=\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus the area of $[COE]=2\\sqrt{2}-\\sqrt{2}$ or $\\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\\sqrt{2}+\\sqrt{2}$ or $2\\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\\frac{2\\sqrt{2}}{6\\sqrt{2}}$ or $\\frac{1}{3}$ $\\Longrightarrow \\mathrm{(C)}$. Solution 5 We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a", "dot(\"S\", S, E); dot(\"K\",K,E); dot(\"L\",L,W); // dot(\"R\",R,NE); dot(\"Q\",Q,NW); dot(\"G\", G, SE); dot(\"Q\",Q,NW); dot(\"P\",P,NE); dot(\"M\",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label(\"x\",C+(0,0.1),N); label(\"y\",C+(-0.1,0.3),N); label(\"w\",H+(-0.03,0.1), N); label(\"z\", H+(0.06,0.12), N ); label(\"u\",T+(-0.1,-0.2), S); label(\"v\", S+(0,-0.2), S); label(\"t\",T+(0.1,-0.3), S); label(\"s\", S+(-0.08,-0.2), S); [/asy]$ Denote $\\angle{HSB}=v$, $\\angle{HTD}=u$, $\\angle{HSC}=s$, $\\angle{HTC}=t$, $\\angle{HCS}=x$, $\\angle{HCT}=y$, $\\angle{AHS}=z$, $\\angle{AHT}=w$. Since $\\angle{CHS}-\\angle{CSB} =90$ and $\\angle{CHT}-\\angle{CTD}=90$, we have $\\angle{CSA}=x+90$, $\\angle{CTA}=y+90$. Since $\\angle{CHS} = \\angle{CSB}+90$, the tangent of the circumcircle of $\\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$. Extend $CB$ to meet circumcircle of $\\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\\triangle{CTH}$ at $Q$. Then, since $\\angle{ADC}=\\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \\parallel BD$; also, $AH \\perp BD$, so $AH \\perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence, $$HP=HQ$$ We have $$u=\\angle{CHT}-90=(90-w+\\angle{DHC}) - 90$$ $$v=\\angle{CHS}-90 = (90-z+\\angle{BHC}) - 90$$ Hence, $u+v = -w-z+\\angle{DHC}+\\angle{BHC} =", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(0,10),C=(10,0),P=(0,0),B=(13.660,13.660); dot(A); label(\"A\",A,W); dot(C); label(\"C\",C,S); dot(P); label(\"P_0\",P,SW); dot(B); label(\"B\",B,NE); draw(A--C--P--A); draw(A--B--C); draw(B--P,dotted); [/asy]$ By repeatedly applying the Midpoint Formula, we can determine the coordinates of $P_1$, $P_2$, $P_3$, and so on. We can also use the Distance Formula to calculate the distance of $P_0P_1$, $P_0P_2$, and so on. The values are shown in the below table. $n$ Coordinates of $P_n$ $P_0 P_n$ 1 $(0,2a)$ $2a$ 2 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ 3 $(a-a\\sqrt{3},a-a\\sqrt{3})$ $a\\sqrt{6} - a\\sqrt{2}$ 4 $(-a+a\\sqrt{3},a+a\\sqrt{3})$ $2a\\sqrt{2}$ 5 $(2a,0)$ $2a$ 6 $(0,0)$ $0$ 7 $(0,2a)$ $2a$ 8 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ Note that the coordinates of $P_n$ as well as the distance $P_0 P_n$ cycle after $n = 6$. Thus, $P_0P_n = \\frac{\\sqrt{6} - \\sqrt{2}}{2} \\cdot P_0P_1$ if $n \\equiv 3 \\pmod{6}$, $P_0P_n = 0$ if $n \\equiv 0 \\pmod{6}$, $P_0P_n = P_0P_1$ if $n \\equiv 1, 5 \\pmod{6}$, and $P_0P_n = \\sqrt{2} \\cdot P_0P_1$ if $n \\equiv 2, 4 \\pmod{6}$.", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "\\frac{3}{2} + 39 \\cdot 4 -36}{\\sqrt{(4\\sqrt{3})^2+4^2+39^2}} = \\frac{144}{\\sqrt{1585}}$$ $$\\lfloor m+\\sqrt{n}\\rfloor = \\lfloor 144+\\sqrt{1585}\\rfloor = 183$$ Solution by SimonSun ## Solution 3 (Cosine Law & Pythagorean Bash) Diagram borrowed from Solution 1 $[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"T\",T,N);label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,NE);label(\"S\",S,NW);label(\"O\",O,SW);label(\"M\",M,NE); label(\"4\",(S+T)/2,NW);label(\"1\",(S+A)/2,NW);label(\"5\",(B+T)/2,NE);label(\"4\",(O+T)/2,W); dot(S);dot(O); [/asy]$ Apply Pythagorean Theorem on $\\bigtriangleup TOB$ yields $$BO=\\sqrt{TB^2-TO^2}=3$$ Since $\\bigtriangleup ABC$ is equilateral, we have $\\angle MOB=60^{\\circ}$ and $$BC=2BM=2(OB\\sin MOB)=3\\sqrt{3}$$ Apply Pythagorean Theorem on $\\bigtriangleup TMB$ yields $$TM=\\sqrt{TB^2-BM^2}=\\sqrt{5^2-(\\frac{3\\sqrt{3}}{2})^2}=\\frac{\\sqrt{73}}{2}$$ Apply Law of Cosines on $\\bigtriangleup TBC$ we have $$BC^2=TB^2+TC^2-2(TB)(TC)\\cos BTC$$ $$(3\\sqrt{3})^2=5^2+5^2-2(5)^2\\cos BTC$$ $$\\cos BTC=\\frac{23}{50}$$ Apply Law of Cosines on $\\bigtriangleup STB$ using the fact that $\\angle STB=\\angle BTC$ we have $$SB^2=ST^2+BT^2-2(ST)(BT)\\cos STB$$ $$SB=\\sqrt{4^2+5^2-2(4)(5)\\cos BTC}=\\frac{\\sqrt{565}}{5}$$ Apply Pythagorean Theorem on $\\bigtriangleup BSM$ yields $$SM=\\sqrt{SB^2-BM^2}=\\frac{\\sqrt{1585}}{10}$$ Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\\frac{\\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have $$TP^2=TS^2-SP^2=TM^2-PM^2$$ $$4^2-x^2=(\\frac{\\sqrt{73}}{2})^2-(\\frac{\\sqrt{1585}}{10}-x)^2$$ Cancelling out the $x^2$ term and solving gets $x=\\frac{181}{2\\sqrt{1585}}$. Finally, by Pythagorean Theorem, $$TP=\\sqrt{TS^2-SP^2}=\\frac{144}{\\sqrt{1585}}$$ so $\\lfloor m+\\sqrt{n}\\rfloor=\\boxed{183}$ ~ Nafer", "yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$: $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); label(\"P\",P,dir(235)); [/asy]$ This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "(A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\end{tikzpicture} Or did I misunderstand? How can we draw a line parallel to $AC$ from $F$ ? Isn't $F$ on $AC$ ? Or do you mean from $E$ instead of $F$ ? Because: \\begin{tikzpicture}[scale=3] \\coordinate[label=left:A] (A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0); \\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\draw (F) -- (B1); \\end{tikzpicture} Why will we not reach zero? We have a new equilateral triangle that has non-zero size. After 2 more iterations, we will have yet again a new equilateral triangle that has non-zero size, won't we? #### mathmari ##### Well-known member MHB Site Helper I thought that we get the following construction: Isn't this correct? #### Klaas van Aarsen ##### MHB Seeker Staff member I thought that we get the following construction: Isn't this correct? Ah okay. My mistake.", "Solving for $b$ in terms of $a$ yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from E to ML and label the point of intersection P: pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L) label(\"P\",P,E) label(\"$A$\",A,NW); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,NE); label(\"$E$\",E,dir(90)); label(\"$F$\",F,NE); label(\"$G$\",G,NE); label(\"$H$\",H,W); label(\"$J$\",J,S); label(\"$K$\",K,SE); label(\"$L$\",L,SE); label(\"$M$\",M,dir(90)); label(\"$N$\",N,dir(180)); (Error compiling LaTeX. label(\"P\",P,E) ^ error: could not load module '7a023ad7e27f5ed1c7813ddb2ed5dce918789de8.asy') This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$" ]

[ "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"E\",E,SE); label(\"F\",F,NE); label(\"P\",P,NW); label(\"Q\",Q,NE); label(\"R\",R,SE); label(\"S\",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]$ Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \\frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$. We then know that $CE = \\frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$. Since $PF+FQ=PQ=\\omega$. We can solve for $PF$ and $FQ$ in terms of $\\omega$. We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$. We can set up the proportion: $\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$. Solving for $AF$, $AF = \\frac{3}{4} PF = \\frac{3}{4}", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "# Problem #2136 2136 In the right triangle $\\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\\triangle DBF$ to that of $\\triangle ACE$? $\\mathrm{(A) \\ } \\frac{1}{4} \\qquad \\mathrm{(B) \\ } \\frac{9}{25} \\qquad \\mathrm{(C) \\ } \\frac{3}{8} \\qquad \\mathrm{(D) \\ } \\frac{11}{25} \\qquad \\mathrm{(E) \\ } \\frac{7}{16}$ $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",C--B,W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "#### Triangles and their classification ##### Angle Classifacation Acute Triangle All angles are less than 90° Right Triangle Has a right angle (90°) Obtuse Triangle Has an angle more than 90° ##### Side Classifacation Equilateral Triangle Three equal sides Isosceles Triangle Two equal sides Two equal angles Scalene Triangle No equal sides No equal angles This part shows what Proves ussually look like ## Proof that any nonperfect square positive integer is irrational Let us assume that $\\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\\frac{p}{q}=\\sqrt{n} \\Rightarrow \\frac{p^2}{q^2}=n \\Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\\sqrt{n}$ is irrational. ## Proof of the Pythagorean Theorem $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Proof", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "and $EF$. They are as follows:$$AG = f(x) = -\\frac{3}{5}x + 4$$$$AC = g(x) = -\\frac{4}{5}x + 4$$$$EF = h(x) = 2x - 4$$After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"P'\", (20/7,0),SSW); label(\"Q'\", (40/13,0),SSE); label(\"E'\", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy]$ Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\\frac{20}{7}$ and $\\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \\frac{40}{13} - \\frac{20}{7} = \\frac{20}{91}$. Now we can find $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F} = \\frac{\\frac{20}{91}}{2} = \\boxed{\\textbf{(D)}~\\frac{10}{91}}$ ## Solution 3 (Similar Triangles) $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"H\",H,E); [/asy]$ Extend $AG$ to intersect $CD$ at $H$. Letting $x=\\overline{HC}$, we have that $$\\triangle{HCG}\\sim\\triangle{HDA}\\implies \\dfrac{\\overline{HC}}{\\overline{CG}}=\\dfrac{\\overline{HD}}{\\overline{AD}}\\implies \\dfrac{x}{1}=\\dfrac{x+5}{4}\\implies x=\\dfrac{5}{3}.$$ Then, notice that $\\triangle{AEQ}\\sim\\triangle{HFQ}$ and $\\triangle{AEP}\\sim\\triangle{CFP}$. Thus, we see that", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "$\\overline{AB}$ and $\\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\\overline{BC}$ and $\\overline{CD}$, respectively, and points $I$ and $J$ lie on $\\overline{EH}$ so that $\\overline{FI} \\perp \\overline{EH}$ and $\\overline{GJ} \\perp \\overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?$[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot(\"A\", A, S); dot(\"B\", B, S); dot(\"C\", C, dir(90)); dot(\"D\", D, dir(90)); dot(\"E\", E, S); dot(\"F\", F, dir(0)); dot(\"G\", G, N); dot(\"H\", H, W); dot(\"I\", I, SW); dot(\"J\", J, SW); [/asy]$ $\\textbf{(A) } \\frac{7}{3} \\qquad \\textbf{(B) } 8-4\\sqrt2 \\qquad \\textbf{(C) } 1+\\sqrt2 \\qquad \\textbf{(D) } \\frac{7}{4}\\sqrt2 \\qquad \\textbf{(E) } 2\\sqrt2$ ## Problem 19 Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: $L,$ a rotation of $90^{\\circ}$ counterclockwise around the origin; $R,$ a rotation of $90^{\\circ}$ clockwise around the origin; $H,$ a reflection across the $x$-axis;", "coordinate system where $Q=(0,0)$, $P=(2,0)$ and $R=(0,2)$. In this coordinate system we have $M=(2,1)$, and the line $QM$ has the equation $2y-x=0$. As the line $RN$ is orthogonal to $QM$, it must have the equation $y+2x+q=0$ for some suitable constant $q$. As this line contains the point $R=(0,2)$, we have $q=-2$. Substituting $x=2y$ into $y+2x-2=0$, we get $y=\\frac 25$, and then $x=\\frac 45$. We can note that in $\\triangle RNQ$ $x$ is the height from $N$ onto $RQ$, hence its area is $[RNQ] = \\frac{x \\cdot RQ} 2 = \\frac{2x}2 = x = \\frac 45$, and therefore the answer is $3[RNQ]+4 = 3\\cdot \\frac 45 + 4 = \\boxed{\\frac{32}5 \\Longrightarrow B}$. ### Solution 2 Extend $RN$ to intersect $PQ$ at $O$: $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); draw(P -- (0,1)); label(\"M\",(-1,0),W); label(\"C\",(-0.1,-0.3)); label(\"A\",(-0.4,0.7)); label(\"B\",(0.7,0.4)); label(\"P\",(-1,1),NW); label(\"Q\",(1,1),NE); label(\"R\",(1,-1),SE); label(\"S\",(-1,-1),SW); label(\"N\",P,1.5*WNW); label(\"O\",(0,1),N); [/asy]$ It is now obvious that $O$ is the midpoint of $PQ$. (Imagine rotating the square $PQRS$ by $90^\\circ$ clockwise around its center. This rotation will map the segment $MQ$ to a segment that is orthogonal to $MQ$, contains $R$ and contains the midpoint of $PQ$. From $\\triangle PQM$ we can compute that $QM = \\sqrt{1^2 + 2^2} = \\sqrt 5$. Observe that $\\triangle PQM$ and $\\triangle NQO$ have the same angles and therefore they", "(1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similary, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$." ]

[ "is larger than $\\frac{1}{6}$. Therefore, we can assume that $\\frac{log_2{x}}{x}$ equals to $\\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\\frac{(2^t-3t)t}{4^t}$ is $\\textbf{(C)}\\ \\frac{1}{12}$.", "The value of the expression $\\dfrac{1}{1+ \\log_u \\: vw} + \\dfrac{1}{1+ \\log_v \\: wu} + \\dfrac{1}{1+\\log_w uv}$ is _______ 1. $-1$ 2. $0$ 3. $1$ 4. $3$", "The value of the expression $\\frac{1}{1+ \\log_u \\: vw} + \\frac{1}{1+ \\log_v \\: wu} + \\frac{1}{1+\\log_w uv}$ is _______", "Browse Questions # What is the value of $\\;$ $\\lim_{x\\to 0} log(1+x)^{\\frac{1}{x}}$? Answer:$\\;$ $\\lim_{x\\to 0} log(1+x)^{\\frac{1}{x}}$ $=e$", "What is the value of $\\frac{1}{2}$ of $\\frac{2}{3}$ of $\\frac{3}{4}$ of $\\frac{4}{5}$ of $\\frac{5}{6}$ of $\\frac{6}{7}$ of $\\frac{7}{8}$ of $\\frac{8}{9}$ of $\\frac{9}{10}$ of $1000$?", "What is the value of (1/5-1/x)/(5/x-1)? May 28, 2015 $\\textcolor{\\lim e}{- \\frac{1}{5}}$ Explanation: $\\frac{\\textcolor{red}{\\frac{1}{5} - \\frac{1}{x}}}{\\textcolor{b l u e}{\\frac{5}{x} - 1}}$ $= \\frac{\\textcolor{red}{\\frac{x - 5}{5 x}}}{\\textcolor{b l u e}{\\frac{5 - x}{x}}}$ $= \\textcolor{red}{\\frac{x - 5}{5 x}} \\cdot \\textcolor{b l u e}{\\frac{x}{5 - x}}$ $= - \\frac{1}{5}$", "Browse Questions Find the principal value of $\\cos^{-1} \\bigg( -\\frac{1}{2} \\bigg).$ Ans: $\\frac{2\\pi}{3}$", "Free Version Easy Evaluating Logs to Solve SATSTM-4EYC7T What is the approximate value of $a$ if $\\log _{4}{150}=a$ ? A $a \\approx 37.5$ B $a \\approx 3.6$ C $a \\approx 0.03$ D $a \\approx 1.4$ E $a \\approx 5.6$", "a calculator to get the exact value $\\textcolor{g r e e n}{\\overline{\\underline{| \\textcolor{w h i t e}{\\frac{a}{a}} \\textcolor{b l a c k}{\\text{pH} = 1 + \\log 4 = 1.60} \\textcolor{w h i t e}{\\frac{a}{a}} |}}}$", "known values $$t$$ = $$12540 \\text{ sec or } 209 \\text{ min}$$", "? Free Version Easy # Evaluating Logs to Solve SATSTM-4EYC7T What is the approximate value of $a$ if $\\log _{4}{150}=a$ ? A $a \\approx 37.5$ B $a \\approx 3.6$ C $a \\approx 0.03$ D $a \\approx 1.4$ E $a \\approx 5.6$", "Browse Questions # Find the principal value of $\\cos^{-1} \\bigg( -\\frac{1}{2} \\bigg).$ Ans: $\\frac{2\\pi}{3}$", "enormous values of$2^m$and$(3/2)^m$to differ by way more than 1/4, right? In short, even if$|(3/2)^{\\frac{n_k}{m_k}} - 2| \\to 0$for$\\frac{n_k}{m_k}$approximating that log-value, why does$|(3/2)^{n_k} - 2^{m_k}| \\to 0\\$? Aug 3 at 20:54 • @LukasMiaskiwskyi Ok i edited to address that. Thanks! And i agree Aug 3 at 22:55", "# Way to find the value of $a_{2011}$ Let us consider the sequence $a_1 , a_2 , a_3 ,\\ldots$ where $\\frac{1}{a_{k+1}} = \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_k}$ for $k >1$ and $a_1 = 2^{2009}$. How can I find the value of $a_{2011}$? - I changed $a_1,a_2,a_3,\\text{...}$ to $a_1,a_2,a_3,\\ldots$ and $\\frac{1}{a_2}+\\text{...}+\\frac{1}{a_k}$ to $\\frac{1}{a_2}+\\cdots+\\frac{1}{a_k}$. Those are standard usage. – Michael Hardy Nov 29 '13 at 5:23 Observe that $\\frac{1}{a_{k+1}} = \\frac{1}{a_k} + \\frac{1}{a_k}$. This will give you a simple relation between $a_k$ and $a_{k+1}$. Unfold it upto $a_1$. You have your answer.", "Browse Questions # If $a^x=b^y=c^z=d^w$, the value of $x \\bigg(\\large\\frac{1}{y}+\\frac{1}{z}+\\frac{1}{w}\\bigg)$ is $\\begin {array} {1 1} (a)\\;\\log_a(abc) & \\quad (b)\\;\\log_a(bcd) \\\\ (c)\\;\\log_b(cda) & \\quad (d)\\;\\log_c(dab) \\end {array}$ $(b)\\;\\log_a(bcd)$", "a= \\frac{1}{2}\\log_x a$", "anticipated value of $a$ is $\\frac{100}{7}=14\\frac{2}{7}$. 0 2019-05-21 02:43:17 Source", "$$I=\\int \\frac{1}{y \\ln(y)}\\,dy=\\log (\\log (y))$$ Using the last one, we could have showed that $$\\log \\left(\\frac{\\log (1+4 e)}{\\log (3)}\\right)<\\int_2^{4e} \\frac{1}{x \\ln(x+1)}\\,dx<\\log \\left(\\frac{1+\\log (4)}{\\log (2)}\\right)$$ and, surprizingly, the numerical value of the integral $(\\approx 1.02081)$ is very close to the average of the two boundind expressions $(\\approx 0.811901$ and $\\approx 1.23625)$.", "Note that $2$ is the minimum attainable value of $$\\frac{a_1}{\\sqrt{1-a_1^2}} + \\frac{a_2}{\\sqrt{1-a_2^2}}+ \\ ... \\ +\\frac{a_n}{\\sqrt{1-a_n^2}} \\ge 2$$ (take $a_1 = a_2 =\\frac 1{\\sqrt 2}$ and all other $a_i$s equal to $0$).", "P(A \\cap B)$$ (i) $$P(\\text{only A})$$ $$=$$ $$\\frac{1}{2} - \\frac{1}{6}$$ $$=$$ $$\\frac{3 - 1}{6}$$ $$=$$ $$\\frac{2}{6}$$ $$=$$ $$\\frac{1}{3}$$ Therefore, the value of $$P(\\text{only A})$$ is $$\\frac{1}{3}$$. (ii) $$P(\\text{only B})$$ $$=$$ $$\\frac{1}{3} - \\frac{1}{6}$$ $$=$$ $$\\frac{2 - 1}{6}$$ $$=$$ $$\\frac{1}{6}$$ Therefore, the value of $$P(\\text{only B})$$ is $$\\frac{1}{6}$$." ]

[ "the so-called \"change of base\" formula: $$\\log_a(x)=\\frac{\\log_b(x)}{\\log_b(a)}$$ 3. Mar 9, 2009 ### MrAnderson Thanks for the post and for the explanation. :)", "### Home > A2C > Chapter 9 > Lesson 9.1.3 > Problem9-62 9-62. 1. Without using a calculator, write the solution to each equation. Homework Help ✎ 1. 2x = 17 2. log3(x + 1) = 5 3. log3(3x) = 4 4. 4log4(x) = 7 Remember how to change the base using logarithms. $\\text{\\frac{log17}{log2}}$ 242 4 7", "# Math Help - change of base? 1. ## change of base? I am just learning Log equations How does $4^x$^2 $*2^3^x=4$ become X= $\\frac{1}{2}$, 2 2. ## Re: change of base? I assume it's $4^{x^2}\\cdot2^{3x}=4$. This is equivalent to $(2^2)^{x^2}\\cdot2^{3x}=2^2$ $2^{2x^2}\\cdot2^{3x}=2^2$ $2^{2x^2+3x}=2^2$ $2x^2+3x=2$", "# Math Help - Change of base 1. ## Change of base Hi, If anyone can help pls The formula for change of base is logbz = 1/ log z b Use this formula to solve logz5 + log5Z =2 thank you 2. Originally Posted by bungy189 Hi, If anyone can help pls The formula for change of base is logbz = 1/ log z b Use this formula to solve logz5 + log5Z =2 thank you i don't see your problem here. you were given the formula to use $\\log_z 5 + \\log_5 z = 2$ $\\Rightarrow \\frac {1}{\\log_5 z} + \\log_5 z = 2$ $\\Rightarrow \\left( \\log_5 z \\right)^2 - 2 \\log_5 z + 1 = 0$ the above is quadratic in $\\log_5 z$. now continue", "be positive integers greater than $$1$$$.$Then, by the change-of-base formula, $$\\log_b(n)=\\frac{\\log_n(n)}{\\log_n(b)}=\\frac{1}{\\log_n(b)}$$ 42) Does $$\\log_{81}(2401)=\\log_3(7)$$$?$Verify the claim algebraically.", "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "base formula. $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\,$ $\\log_{b}{m} = \\dfrac{1}{\\log_{b}{m}}$", "How do you use the Change of Base Formula and a calculator to evaluate the logarithm log_4 72? Refer to explanation Explanation: Using the change of base of formula log_(4)72=log72/log4=log(9*8)/log(2^2)=(2*log3+3*log2)/(2*log2)= 3/2+log3/log2 Now using a calculator we see that $\\log 2 = 0.69$ and $\\log 3 = 1.10$ so finally ${\\log}_{4} 72 = 3.08$", "# How do you evaluate log_7 4 using the change of base formula? May 20, 2017 $\\approx 0.7124 \\text{ to 4 dec. places}$ $\\text{Given \" log_b a\" then change of base is as follows}$ • (log_(10)a)/(log_(10)b) $\\Rightarrow {\\log}_{7} 4 = \\frac{{\\log}_{10} 4}{{\\log}_{10} 7} \\approx 0.7124$", "_{d}b}$ Further, change of base formula helps in rewriting logarithms in terms of the base log. It can be used for evaluation of log with another base than 10. Practically, you can evaluate the non-standard base by transforming it to the fractions. ### Change of Base Solve Example Question 1: Solve $\\ 2 \\log_{4} 29$ Given $\\ 2 \\log_{4} 29$ Using logarithm change of base formula, $\\ \\log_{b}x=\\frac{\\log _{d}x}{\\log _{d}b}$ $\\ 2 \\log_{4}29= 2 \\times \\frac{\\log _{10}29}{\\log _{10}4}$ $\\ 2 \\log_{4}29= 2 \\times 2.43 = 4.86$", "Change of Base Formula $x = \\log \\frac{b}{\\log} \\left(a\\right)$ In our situation, we essentially have $a = 5$ and $b = 52$. Plugging into the formula, we get $x = \\log \\frac{52}{\\log} 5$ $x \\approx 2.46$ Hope this helps!", "For the first question, follow the hint by Feldoh, and use the change of base formula; so, $$\\log_{16} 17=\\frac{\\log_4 17}{\\log_4 16}$$. Can you continue? Incidentally, you should post in the homework forums in future.", "# Rewrite in Exponential Form 2 = log base 4 of 16 For logarithmic equations, is equivalent to such that , , and . In this case, , , and . Substitute the values of , , and into the equation . Rewrite in Exponential Form 2 = log base 4 of 16 Scroll to top", "change of base formula. For example, we have that $\\log_b a = \\frac{\\log_a a}{\\log_a b} = \\frac{1}{\\log_a b}$. Using the second form of the change of base formula gives $\\log_b a^n = \\log_b a \\cdot \\log_a a^n = n \\log_b a$. One consequence of the change of base formula is that for positive constants $a, b$, the functions $f(x) = \\log_a x$ and $g(x) = \\log_b x$ differ by a constant factor, $f(x) = (\\log_a b) g(x)$ for all $x > 0$.", "# Change of Base Formula The Change of base formula helps to rewrite the logarithm in terms of another base log. Change of base formula is used in the evaluation of log and have another base than 10. $\\LARGE \\log_{b}x=\\frac{\\log _{d}x}{\\log _{d}b}$ ### Solved Examples Example 1: Solve: 2 $$\\begin{array}{l}\\log_{4} 29\\end{array}$$ Solution: Given, 2 $$\\begin{array}{l}\\log_{4} 29\\end{array}$$ Using logarithm change of base formula, $$\\begin{array}{l}log_{b} x\\end{array}$$ = $$\\begin{array}{l}\\frac{log_{d} x}{log_{d} b}\\end{array}$$ 2 $$\\begin{array}{l}\\log_{4} 29\\end{array}$$ = 2 $$\\begin{array}{l}\\times\\end{array}$$ $$\\begin{array}{l}\\frac{log_{10} 29}{log_{10} 4}\\end{array}$$ 2 $$\\begin{array}{l}\\log_{4} 29\\end{array}$$ = 2 x 2.43 = 4.86 Example 2: Simply: $$\\begin{array}{l}\\log_{32} 16\\end{array}$$ Solution: Given, $$\\begin{array}{l}\\log_{32} 16\\end{array}$$ Using change of base formula, $$\\begin{array}{l}\\log_{32} 16=\\frac{\\log_{10}16}{\\log_{10}32}\\\\=\\frac{\\log_{10}2^4}{\\log_{10}2^5}\\\\=\\frac{4\\log_{10}2}{5\\log_{10}2}\\\\=\\frac{4}{5}\\end{array}$$", "# Write in Exponential Form log base 4 of 256=4 For logarithmic equations, is equivalent to such that , , and . In this case, , , and . Substitute the values of , , and into the equation . Write in Exponential Form log base 4 of 256=4", "then $\\log_{a} a = 1$ and the above formula becomes $\\log_{b} a$ = $\\frac{1}{\\log_{a} b}$ 1. The change of base formula for converting any logarithms to common logarithms (base 10) $\\log_{b}x$ = $\\frac{\\log_{10} x}{\\log_{10} b}$ 2. The change of base formula for converting any logarithms to natural logarithms (base e) $\\log_{b}x$ = $\\frac{\\log_{e} x}{\\log_{e} b}$ ## Change of Base Examples ### Solved Examples Question 1: Use change of base formula and evaluate the following logarithm. $\\log_{6} 3$ Solution: Change of base formula $\\log_{b}x$ = $\\frac{\\log_{a} x}{\\log_{a} b}$ Given, $b = 6$ and $a = 10$ (common logarithms) $\\log_{6} 3$ = $\\frac{\\log_{10} 3}{\\log_{10} 6}$ = $0.6131472$ Question 2: Use change of base formula and evaluate the following logarithm. $\\log_{5} 8$ Solution: Change of base formula $\\log_{b}x$ = $\\frac{\\log_{a} x}{\\log_{a} b}$ Given, $b = 5$ and $a = e$ (natural logarithms) $\\log_{5} 8$ = $\\frac{\\log_{e} 8}{\\log_{e} 5}$ = $1.2920296$ Question 3: Solve $\\log_{18}60$. Solution: Change of base formula $\\log_{b}x$ = $\\frac{\\log_{a} x}{\\log_{a} b}$ Given, $b = 18$ and $a = 10$ (common logarithms) $\\log_{6} 3$ = $\\frac{\\log_{10} 60}{\\log_{10} 18}$ = $1.4165460$", "$\\log_{10} (x)$ . If you need logarithms to a base other than $e$ or 10 , for which the calculator has no buttons, you definitely need the \"change-of-base formula\" (though I've seen some recent calculators that have a \"$\\log_{\\Box} x$\" button)...", "# How do you find the value of log_6 24 using the change of base formula? Mar 7, 2018 Approximately $1.7737$ #### Explanation: Change of base formula states that: ${\\log}_{b} a = \\log \\frac{a}{\\log} b$ Here, we can substitute: ${\\log}_{6} 24$ becomes: $\\log \\frac{24}{\\log} 6$ $= \\frac{1.3802}{0.7782}$ $\\cong 1.7737$ And there we have our answer.", "that $$a^x=b$$. Hence, \\begin{align} \\log(a^x)&=\\log(b) \\\\[5pt] x\\log a &= \\log b \\\\[5pt] x &= \\frac{\\log b}{\\log a} \\, . \\end{align} Since $$x=\\log_a(b)$$, the change of base formula is valid. What this proof tells us is that the change of the base formula follows directly from the power rule: $$\\log(a^x)=x\\log a$$, which is true regardless of the base used." ]

[ "the so-called \"change of base\" formula: $$\\log_a(x)=\\frac{\\log_b(x)}{\\log_b(a)}$$ 3. Mar 9, 2009 ### MrAnderson Thanks for the post and for the explanation. :)", "change of base formula. For example, we have that $\\log_b a = \\frac{\\log_a a}{\\log_a b} = \\frac{1}{\\log_a b}$. Using the second form of the change of base formula gives $\\log_b a^n = \\log_b a \\cdot \\log_a a^n = n \\log_b a$. One consequence of the change of base formula is that for positive constants $a, b$, the functions $f(x) = \\log_a x$ and $g(x) = \\log_b x$ differ by a constant factor, $f(x) = (\\log_a b) g(x)$ for all $x > 0$.", "# How do you find the value of log_6 24 using the change of base formula? Mar 7, 2018 Approximately $1.7737$ #### Explanation: Change of base formula states that: ${\\log}_{b} a = \\log \\frac{a}{\\log} b$ Here, we can substitute: ${\\log}_{6} 24$ becomes: $\\log \\frac{24}{\\log} 6$ $= \\frac{1.3802}{0.7782}$ $\\cong 1.7737$ And there we have our answer.", "be positive integers greater than $$1$$$.$Then, by the change-of-base formula, $$\\log_b(n)=\\frac{\\log_n(n)}{\\log_n(b)}=\\frac{1}{\\log_n(b)}$$ 42) Does $$\\log_{81}(2401)=\\log_3(7)$$$?$Verify the claim algebraically.", "## Algebra 2 (1st Edition) Using the change of base formula, we find: $$\\frac{log(24/5)}{log(6)} = .875$$", "College Algebra (10th Edition) Using The Change-of-Base Formula ($\\log_a{b}=\\frac{\\log_c{b}}{\\log_c{a}}$) and then evaluating by a calculator: $\\log_{7}{62}=\\frac{\\ln{62}}{\\ln{7}}=8.031023\\approx2.121.$", "## College Algebra (11th Edition) $\\approx1.1592$ By using the change of base formula we can calculate easily the value of $\\log_{\\frac{2}{3}}\\frac{5}{8}$. $\\log_{\\frac{2}{3}}\\frac{5}{8}=\\frac{\\log_{10}\\frac{5}{8}}{\\log_{10}\\frac{2}{3}}\\approx1.1592$", "College Algebra 7th Edition $\\frac{\\log 27}{\\log 12}\\approx 1.326$ We use the change of base formula: $\\log_{12}27=\\frac{\\log 27}{\\log 12}\\approx 1.326$", "The value of the expression $\\dfrac{1}{1+ \\log_u \\: vw} + \\dfrac{1}{1+ \\log_v \\: wu} + \\dfrac{1}{1+\\log_w uv}$ is _______ 1. $-1$ 2. $0$ 3. $1$ 4. $3$", "# How do you use the Change of Base Formula and a calculator to evaluate the logarithm log_6 120? Some scientific calculators allow you to choose the base of the logarithm. However, if this is not the case, then the log function needs to be represented in terms of either base $10$ or base $e$. ${\\log}_{a} b = {\\log}_{c} \\frac{b}{\\log} _ c a$ Hence: ${\\log}_{6} 120 = {\\log}_{10} \\frac{120}{\\log} _ 10 6 = \\ln \\frac{120}{\\ln} 6$ where $\\ln$ is the natural logarithm i.e. ${\\log}_{e}$", "## donnie1999 one year ago What is the value of log3 34 rounded to the nearest hundredth 1. Nnesha do you know the change of base formula ? 2. KaiserTheSlayer 3.21? 3. KaiserTheSlayer 4. donnie1999 yes sr :D 5. confluxepic $$\\Huge log_b~^x=\\frac{log_d~^x}{log_d~^b}$$ 6. confluxepic $$\\Huge log_b~^x=\\frac{log~x}{log~b}$$", "_{d}b}$ Further, change of base formula helps in rewriting logarithms in terms of the base log. It can be used for evaluation of log with another base than 10. Practically, you can evaluate the non-standard base by transforming it to the fractions. ### Change of Base Solve Example Question 1: Solve $\\ 2 \\log_{4} 29$ Given $\\ 2 \\log_{4} 29$ Using logarithm change of base formula, $\\ \\log_{b}x=\\frac{\\log _{d}x}{\\log _{d}b}$ $\\ 2 \\log_{4}29= 2 \\times \\frac{\\log _{10}29}{\\log _{10}4}$ $\\ 2 \\log_{4}29= 2 \\times 2.43 = 4.86$", "can be done by purely following the definition. ${\\log}_{a} b = c \\text{ \" rArr } {a}^{c} = b$ Remember: \"The base stays the base and the other two change around\" ${\\log}_{14} 196 = 2 \\text{ \" rArr \" } {14}^{2} = 196$", "# Logs and the Change of Base Formula Prior to calculators, the standard way for determining a logarithm precisely was to use a look-up table for values. However, such tables were usually only printed for the common most bases, particularly 10 (log). To determine a logarithm for a different base, you needed to first convert to that base. Luckily, it’s very simple to do so. The change of base formula is: $\\log_a b = \\frac{\\log_c b}{\\log_c a}$ This works for any base, but it’s most typical to use it for 10 and e. For instance, if you want to find $$\\log_6 29$$ using a log (base 10) table, you find $$\\log 6$$ and $$\\log 29$$, then divide. That is, $\\log_6 29 = \\frac{\\log 29}{\\log 6} \\approx \\frac{1.4624}{.7782} \\approx 1.879$ Side note: You’ll notice that the log table I linked only goes up to 9.99. So how did I use it to find $$\\log 29$$? I looked up $$\\log 2.9$$ and then added 1. This is why common log tables were most common for this purpose: They’re more flexible than natural log tables. You only need the values from 1.000 to 9.999 to find the log of any positive real number. Since many calculators only have log and ln buttons, it’s useful to know this formula. Even for the", "Then do the same for log (a + b) = log(2a + x) = log( 2a.(1+x/2a) ) ≅log(2a) + x/2a + . . . Compare the results, to see in which power of x = b - a the two expressions differ. A similar approach should work for your other formulae. 5. Feb 26, 2017 ### Stephen Tashi Here's an intuitive way to look at the situation. If $A$ and $B$ are nearly equal then $A+B \\approx 2A \\approx 2B$. So a crude approximation is $f(A+B) \\approx f(2A) \\approx f(2B)$ Instead of using one of those crude approximations, we could use the \"average\" value of $f$ over the interval $[2A,2B]$. That would be $f(A+B) \\approx m = \\frac{ \\int_{2A}^{2B} f(u) du} {2B- 2A}$. Making the change of variable $2x = u$ the integration becomes $m = \\frac{ 2 \\int_A^B f(2x) dx}{2B - 2A} = \\frac{\\int_A^B f(2x) dx}{B-A}$. 6. Feb 26, 2017 ### Kumar8434 $A$ and $B$ need not be nearly equal. For example, when $A=0.4$ and $B=1.5$. Then $A+B=1.9$, $2A=0.8$ and $2B=3$. So, $A+B$ is neither approximately equal to $2A$ nor to $2B$. The formula still gives a very accurate value of $tan^{-1}1.9$. 7. Feb 26, 2017 ### Stephen Tashi However, A+B is still the average of 2A and 2B which leads to the justification for making an", "by using the following formula. $$\\log _{2}$$x=$$\\frac{\\log_{10}x}{\\log_{10}2}$$ It is based on the general formula for change of base as: $$\\log _{a}x$$=$$\\frac{\\log_{b}x}{\\log_{b}a}$$ Thus, to get the value of log base 2, we need to convert it into common logarithmic functions at base 10 by using the change of base formula. Properties of Log Base 2: Some of the log base 2 properties are as follows: • Product Rule: $$log_{2} (x \\times y)$$ = $$log_{2} x + log_{2}$$ y • Quotient Rule: $$log_{2} (\\frac{x}{y})$$ = $$log_{2} x – log_{2}$$ y • Power Rule: Raise the exponential expression to the power and multiply the exponents. Thus, $$log_{2} x^{n}$$ = $$n log_{2} x$$ • Zero Exponent Rule: $$log_{a} 1$$ = 0. • Change of Base Rule: $$log_{b} x$$ = $$\\frac {log_{2} x} { log_{2} b}$$ This formula is useful to change log 2 to log 10 also. Solved Examples for You Q.1: Find the values of following logarithm terms. 1. log 1 base 2 2. log 7 base 2 3. 3 log base 2 4. log 2 base 2 Solution: (i) log 1 base 2 = $$log_{2}$$ 1 = 0 . It is true for any base. (ii) Log7 in base 2 i.e. $$\\frac { log7}{log2}$$. It will be 2.80735492205760. (iii) 3 log base 2 i.e. $$log_{2} 3$$ = $$\\frac { log3}{log2}$$.", "## Precalculus (6th Edition) Blitzer The change-of-base property for logarithms allows us to write logarithms with base $b$ in terms of a new base $a$. Introducing base $a$, the property states that$$\\log_bM=\\frac{\\log_aM}{\\log_ab}.$$For example, $$\\log_35=\\frac{\\log_25}{\\log_23}$$", "Change of Base Formula $x = \\log \\frac{b}{\\log} \\left(a\\right)$ In our situation, we essentially have $a = 5$ and $b = 52$. Plugging into the formula, we get $x = \\log \\frac{52}{\\log} 5$ $x \\approx 2.46$ Hope this helps!", "# Difference between revisions of \"Change of base formula\" The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases. For any positive real numbers $d,a,b$ such that neither $d$ nor $b$ are $1$, we have $$\\log_b a = \\frac{\\log_d a}{\\log_d b}.$$ This allows us to rewrite a logarithm in base $b$ in terms of logarithms in any base $d$. This formula can also be written $$\\log_b a \\cdot \\log_d b = \\log_d a.$$ ### Use for computations The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that $$\\log_{\\frac{1}{4}} 32\\sqrt{2} = \\frac{\\log_2 32\\sqrt{2}}{\\log_2 \\frac{1}{4}} = \\frac{\\frac{11}{2}}{-2} = -\\frac{11}{4}.$$ The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate $\\log_{7} 19$, you would first convert it to the form $\\frac{\\log_{10}{19}}{\\log_{10}{7}}$. Then you would evaluate it using the base-10 log function on the calculator. ### Special cases and consequences Many other logarithm rules can be written in terms of the", "# How do you evaluate log_7 4 using the change of base formula? May 20, 2017 $\\approx 0.7124 \\text{ to 4 dec. places}$ $\\text{Given \" log_b a\" then change of base is as follows}$ • (log_(10)a)/(log_(10)b) $\\Rightarrow {\\log}_{7} 4 = \\frac{{\\log}_{10} 4}{{\\log}_{10} 7} \\approx 0.7124$" ]

[ "that $f(1) = 2$ and $f(xy) = f(x)f(y) - f(x + y) + 1$ for all rational numbers $x$ and $y$. #### 1 comment: 1. Let x=1; then, f(y) = f(1)*f(y) -f(x+1)+1 for all rational y, or f(y)=2f(y)-f(x+1)+1 --> f(y+1)=f(y) + 1 for all rational y. Since f(1)=2 and f(y+1)=f(y)+1 for all rational y, it follows with induction to both negative and positive integers that f(z)=z+1 for all integers z. Furthermore, it follows by induction that f(y+z)=f(y) + z for rational y and integer z. Next, suppose x= 1/w and y=w for an integer w. Then, f(1) = f(1/w)*f(w) - f(1/w + w) + 1 1=f(1/w)[w+1] - f(1/w) - w f(1/w)[w]=w+1 f(1/w)=1/w+1 - again, for all integers w. Finally, let x=m/n and y=n for some integers m and n. Then, f(m)=f(m/n)f(n) - f(m/n + n) +1 m+1 = f(m/n)[n+1] - f(m/n) - n + 1 m+n=f(m/n)(n) f(m/n)=(m+n)/n f(m/n)=1+m/n for all integers m and n; so for all rationals R, f(R)=1+R is the only function that exists to satisfy these conditions.", "Ana is investigating a function $f(n)$ defined for all positive integers $n$. She knows that $f(ab) = f(a)\\cdot f(b)$ for all relatively prime positive integers $a$ and $b$, and that $f(p+q) = f(p)+f(q)$ for all primes $p$ and $q$. Given these facts, help her find the value of $f(2012)$.", "# How Many Values Can It Attain? Algebra Level 5 Let $$f$$ be a function from the positive integers to the positive integers such that $f(m+n) + 2 > f(m) + f(f(n))$ for all positive integers $$m,n.$$ Find the last three digits of the sum of all possible values of $$f(2014).$$ Details and assumptions • The inequality is strict. • This problem is not original. • A typo has been fixed. Sorry for the inconvenience. ×", "# Finding $f(n)=f(f(n-1))+f(f(n+1))$ Determine whether a function exists from the positive integers to the positive integers which satisfies the equation: $$f(n)=f(f(n-1))+f(f(n+1))$$. My guess is that this function does not exist, as through trial and error I found that $f(n)=\\frac{n}{2}$ is a solution to the functional equation. Obviously this is not enough, but I don't know how to tackle this problem. I found that $f(n+2)-f(n)=f(f(n+3))-f(f(n-1))$, but this doesn't seem very useful to me. Can anyone help me with this problem? Here is a sketch of the proof that no such function exists. Say that such a function does exist. Take $n$ be such that $f(n)=m$ is the minimum value that $f$ takes. If $n\\neq1$, then then $m$ is expressed as the sum of two outputs of the function $f$ by the given relation. Since the outputs of $f$ are positive integers, this is impossible. If the minimum is $f(1)$ and is attained only at $1$, then take let $n_1>1$ be such that $f(n_1)$ is the minimum of the values attained by $f$ excluding $f(1)$. Then $f(n_1+1)\\geq f(n_1)>f(1)\\geq 1$, so by the minimality of $f(n_1)$, we have that $f(f(n_1+1))\\geq f(n_1)$. Again this leaves the desired relation impossible, so there is no such function. • You should rule out the case $n=1$, for which the formula does not apply, as $f(n-1)$", "# Find all functions that satisfy the following conditions Find all functions $f:\\mathbb Z\\to \\mathbb Z$ that satisfy the following conditions: (i) $f (0) = 1$ (ii) $f(f (x)) = x$ for all integers x (iii) $f(f(x + 1)+1) = x$ for all integers x How do you prove it with induction • As in my argument below, we know that $f(x)=f(x+1)+1$ for all integers $x$. Now let us prove that $f(n)=1-n$ and that $f(-n)=1+n$ for all positive integers $n\\geq 1$. The induction start is easy: we know that $f(0)=f(1)+1$ and that $f(-1)=f(0)+1$. As $f(0)=1$, it follows that $f(1)=0$ and $f(-1)=2$. If it holds for some $n\\geq 1$ that $f(n)=1-n$ and $f(-n)=1+n$, then $f(n)=f(n+1)+1$, implying $f(n+1)=f(n)-1=-n=1-(n+1)$, and $f(-(n+1))=f(-n)+1$, implying $f(-(n+1))=1+n+1=2+n=1+(n+1)$. Hence it follows that $f(n)=1-n$ and $f(-n)=1+n$ for all integers $n\\geq 1$, so that $f(x)=1-x$. – Bryder Apr 14 '14 at 19:28 Quite straightforward. $f$ is a bijection by (ii), so (ii) and (iii) imply $f(x)=f(x+1)+1$. Now $f(n)=f(n-1)-1=f(n-2)-2=f(n-3)-3=\\ldots=f(0)-n=1-n$ and $f(-n)=f(1-n)+1=\\ldots=f(0)+n=1+n$ for all integers $n>0$, so that $f(x)=1-x$ for all integers $x$. Induction is of course also applicable. By (ii), $$f(f(0))=f(1)=0$$ By (iii) $$f(f(-1+1)+1)=f(f(0)+1)=f(2)=-1$$ So $$f(f(2))=f(-1)=2$$ So $$f(f(-2+1)+1)=f(f(-1)+1)=f(3)=-2$$ So $$f(f(3))=f(-2)=3$$ So $$f(f(-3+1)+1))=f(4)=-3$$ So... • So, the solution is $f(x)=-x+1$. – user35603 Apr 14 '14 at 15:41", "# Let f be a positive integer be define recursively by $f(1)=1$ and $f(n+1)=\\sqrt{2+f(n)}$ for all integers n. Prove that $f(n) = (2^n)-1$. Let f: be a positive integer defined recursively by $f(1)=1$ and $f(n+1)=\\sqrt{2+f(n)}$ for all integers n. Prove that $f(n)<2$. I am supposed to prove this by using proof of induction. I've tested the base case of 1, and I've let $P(n)$ be the statement that $f(n)<2$. I'm not sure how to go about the rest. I can't seem to prove that it is true for the case of $n+1$. - See this closely related answer for a standard approach to these problems. – Andres Caicedo Sep 29 '13 at 18:17 Wrong title. – Did Sep 29 '13 at 18:19 Hint: $\\sqrt{2+2}=2$. ${}{}{}{}{}{}$", "x and any positive integer n. It follows that if m and n are integers (with n positive) then $e^m = f(m) = f(n\\tfrac mn) = \\bigl(f(\\tfrac mn)\\bigr)^n$. Take n'th roots to see that $e^{m/n} = f(\\tfrac mn)$. In other words, $f(r) = e^r$ for any rational number r. Finally, use continuity to deduce that the result is true for all real numbers.", "# Function from integers to integers Let $$f:\\mathbb{N} \\rightarrow \\mathbb{N}$$ be a function such that $$f(1)= 1$$, and for all positive integers $$n,$$ $$f(2n)= f(n)$$ and $$f(2n+1)= f(n) + f(n+1)$$. Find the number of odd integers $$n$$ such that $$f(n)= 2015$$. Details and assumptions • This problem is not original. ×", "# More fun in 2015, Part 39 $\\sum_{d|n}\\mu\\left(\\frac{n}{d}\\right)f(d)=n$ If $f(d)$ is an arithmetic function such that the equation above holds for all positive integers $n$, find $f(2015)$. Notation: $\\mu$ denotes the Möbius function. ×", "## IMO 2011 (Problem 5) Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$ Prove that, for all integers $m$ and $n$ with $f(m)\\leq f(n)$ , the number $f(n)$ is divisible by $f(m)$ .", "# Difference between revisions of \"2011 IMO Problems/Problem 5\" Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m - n)$. Prove that, for all integers $m$ and $n$ with $f(m) \\leq f(n)$, the number $f(n)$ is divisible by $f(m)$.", "}n\\ge 1000 \\\\ f(f(n+5)) & \\mbox{if }n<1000 \\end{cases}$ Find $f(84)$. (Source) • Let $f$ be a function with the following properties: 1. $f(n)$ is defined for every positive integer $n$; 2. $f(n)$ is an integer; 3. $f(2)=2$; 4. $f(mn)=f(m)f(n)$ for all $m$ and $n$; 5. $f(m)>f(n)$ whenever $m>n$. Prove that $f(n)=n$. (Source) • Describe all polynomials $P(x)$ such that $P(x + 1) - 1 = P(x) + P'(x)$ for all $x$.", "number $x$ we have $f(x)=f(β+(x-β))=f(β)f(x-β)=0$ which implies that $f$ is identically zero. By the continuity of $f$ we can conclude that $f(x)>0$ for all real numbers. For a positive integer $n$, $f(n)=f(1+1+...+1)=f(1)f(1)*...*f(1)=f(1)^n$. Since $f(-1)=f(1+(-2))=f(1)f(-2)=f(1)f(-1)f(-1)$, it follows that $f(-1)=f(1)^{-1}$. For a negative integer $m$, $f(m)=f(-1+(-1)+...+(-1))=f(-1)^{-m}=(f(1)^{-1})^{-m}=f(1)^m$. Since $f(0)=1=f(1)^0$, $f(n)=f(1)^n$ holds for all integers. Let $q$ and $m$ be positive integers, then $f(m)=f(1/q+1/q+...+1/q)=f(1/q)^{qm}=f(1)^m$. It follows that $f(1/q)=f(1)^{1/q}$. This will also hold for negative integers $q_1$ and $m_1$ as well. Then, $f(s/t)=f(1/t+...+1/t)=f(1/t)^s=(f(1)^{1/t})^s=f(1)^{s/t}$, for any rational number. Since the rationals are dense in the reals and $f$ is continuous, for any real number $x$ we can find a sequence $(r_n)$ that will converge to $x$ with $f(r_n)=f(1)^{r_n}$, so that $f(x)=f(1)^x$ for all real numbers. Since $f(1)>0$ and is real-valued, $\\log(f(1))=c$ exists. Then: $f(x)=f(1)^x=(\\exp(\\log(f(1))))^x=(\\exp(c))^x=(e^c)^x=e^{cx}$. Thanks again! -", "many - one for each $n$. – akkkk Nov 5 '12 at 8:30 @akkkk: $f$ does not depend on $n$, but in order to meet the conditions of the problem, such an $f$ will have particular behavior relative to the sequence of reciprocals of positive integers. Hence one method is to define what $f$ does on each interval $[1/(n+1),1/n]$, making sure the definitions agree at the endpoints. – Jonas Meyer Nov 5 '12 at 8:30", "IMO 2011 (Problem 5) Printable View • July 19th 2011, 10:06 AM FernandoRevilla IMO 2011 (Problem 5) Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$ Prove that, for all integers $m$ and $n$ with $f(m)\\leq f(n)$ , the number $f(n)$ is divisible by $f(m)$ .", "# Give it a try Let $$f(n)$$ be a function defined on the non-negative integers given the following facts: • $$f(0)=f(1)=0$$. • $$f(2)=1$$ • For $$n>2$$, $$f(n)$$ gives the smallest positive integer, which does not divide $$n$$. Let $$g(n)=f(f(f(n)))$$, find $$g(1)+g(2)+g(3)+\\cdots+g(2016)$$. ×", "Functional Equation Problems If $\\Bbb N$ denotes all positive integers. Then find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ which are strictly increasing and such for all positive integers $n$, we have: $$f(f(n)) = n+2$$ So far I know that $f(n)$ is greater than $n$ because the function is strictly increasing, but I'm not sure how to use this in order to solve the equation. • Could we not have $f(x)=x+1$, which is strictly increasing and hence $f(f(n))=f(n+1)=n+2$? – lioness99a Jun 22 '17 at 11:41 • hint: $f ( n )$ can't be greater than $n + 2$. – Mohsen Shahriari Jun 22 '17 at 11:42 • strictly increasing seems to be unnecessary – Mudream Jun 29 '17 at 5:45 • Possible duplicate of Find all functions of positive integers for $f(f(n))=n+2$ – Sil Apr 6 at 12:53 As kindly explained below strictly increasing means $f(a) > f(b)$ for $a >b.$ Let's consider $f(0)$. 1. $f(0) = 0$ is not possible because then $f^2(0) = 0 \\neq 2.$ 2. $f(0) = 1$ is possible. 3. $f(0) = m > 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) > f(0) > 1$ which implies $f(m) \\ge m + 2 > 2$ in contradiction with the assumption $f^2(0) = 2.$ Hence $f(0)", "the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that$$xf(2f(y)-x)+y^2f(2x-f(y))=\\frac{f(x)^2}{x}+f(yf(y))$$for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$. Problem 10 (woah casework, 2012 usamo/4): Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$. Problem 11 (2012 imo/4, misplaced i think): Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$, the following equality holds:$$f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).$$ Problem 12 (2011 imo/3): Let $f: \\mathbb R \\to \\mathbb R$ be a real-valued function defined on the set of real numbers that satisfies$$f(x + y) \\le yf(x) + f(f(x))$$for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \\le 0$. Problem 13 (This was the first oly FE question I solved. That's why it's last :P 2010 imo/1): Find all function $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for all $x,y\\in\\mathbb{R}$ the following equality holds $f(\\left\\lfloor x\\right\\rfloor y)=f(x)\\left\\lfloor f(y)\\right\\rfloor$ where $\\left\\lfloor a\\right\\rfloor$ is greatest integer not greater than $a.$ Nothin much ## MONSTROUS Functional Equations Ok we're not going", "# f(2014) Level pending Let $$f$$ be a function defined on non-negative integers satisfying the following conditions: $f(2n+1)=f(n)$ $f(2n)=1-f(n)$ Given that $$f(2014)$$ can be written as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. Find $$a+b$$ ×", "– gt6989b Jun 16 '15 at 21:46 Look at the function $f(x) = 2^x-x$. $f'(x)=(2^x)(\\log 2)- 1 > 0$ for $x \\geq 1$. So $f$ is increasing. Now, $f(1) = 1$. Hence, $f(m) > 0$ for every natural number $m$. More a comment than a solution, but it points to a more general result (one that I found independently and am pleased with): In Prove that $n^k < 2^n$ for all large enough $n$, I showed that if $n$ and $k$ are integers and $k≥2$ and $n≥k^2+1$, then $2^n>n^k$." ]

[ "Dear Uncle Colin, Someone gave me the puzzle: • $ff(n) = 3n$ for all real $n$, • $f(n)$ is a positive integer for all positive integer $n$ • $f(n+1) > f(n)$ for all positive integer $n$ • What is $f(13)$? Any ideas? Functions? Fun? No. Hi, FFN, This is a fun puzzle! Here’s how I tackled it: • $ff(1) = 3$, so I reckon $f(1) = 2$ and $f(2)=3$ This takes a little unpicking; $f(1)$ can’t be 1, or else $ff(1) = 1$. Could it be 3, and $f(3)=3$? again, no: $f(3) > f(2) > f(1)$. So, it has to be 2, and $f(2) = 3$. • We know that $ff(2) = 6$ and $f(2) = 3$, so $f(3) = 6$. • Now $ff(3) = 9$, so $f(6) = 9$ Let’s take stock: • $f(1) = 2$ • $f(2) = 3$ • $f(3) = 6$ • $f(6) = 9$ That means $f(4) = 7$ and $f(5) = 8$, if we want to satisfy the “increasing” condition. • We know $ff(4) = 12$ and $f(4) = 7$, so $f(7) = 12$. • Similarly, $ff(5) = 15$ and $f(5) = 8$, so $f(8) = 15$. There’s a pattern emerging! • $f(1) = 2$ • $f(2) = 3$ • $f(3) = 6$ • $f(4) = 7$ • $f(5) = 8$ •", "# Talk:2012 USAMO Problems/Problem 4 Though not as elegant as the inductive proof, my proof for this problem is quite different from the one posted in the Page, so I would like to paste it here just for reference. 1. First, $f(1)$, $f(2)$ equal to $1$ or $2$. 2. Second, suppose that for all $n$, $f(n)\\le 2$. Then for $n\\ge 3$, $$2\\ge f(n!) = f(1) + C(n!-1)\\ge -2+5C$$ , where $C\\ge 0$ is an integer. We have $5C\\le 4$ and so $C=0$ is the only possibility. Hence $f(n)!=f(n!)=f(1)$. Similar argument yields that $f(n)!=f(n!)=f(2)$, so $f$ is a constant function, which can only be $f=1$ or $f=2$. 3. From now on, suppose that there exists $n_0\\ge 3$ such that $f(n_0)>2$. 4. By $2| (n_0!-2) | f(n_0)!-f(2)$ and that $2|f(n_0)!$ we know that $2|f(2)$, or $f(2)=2$. 5. By $f(3)!=f(6)=4C+f(2)=4C+2$ we know that $f(3)\\le 3$, and by $3|f(n_0)!-f(3)$ and $3|f(n_0)!$ we know that $3|f(3)$, therefore $f(3)=3$. 6. Then by $2|f(3)-f(1)$ we know that $f(1)$ is odd, so $f(1)=1$. 7. For $n\\ge 3$, $f(n)=C_1(n-1)+1=C_2(n-2)+2$ implies that $f(n)>2$, otherwise $C_1=C_2=0$ and that $f(n)=1=2$, which is a contradiction. Since $f(n)>2$, now we have $C_2\\ge 1$, so $f(n)\\ge (n-2)+2=n$. Now we have a lower bound. What is more difficult is to find an upper bound of $f(n)$. (I know from the main Page it is", "-2$. Notice that $11^3+7<2^{11}$. It is easy to prove with induction that $11^N+7<2^{N}$ for $N\\geq 11$. Hence we mus try only with numbers in the range $0,1,2,3,4,5,6,7,8,9$. Of course all even values except $0$ need not be considered, as the numerator would be odd, and $0$ does not work. So we are left only with $1,3,5,7,9$. $n=1$ works. $n=3$ does not work as $34$ is not a multiple of $8$. $n=5$ does not work as $126$ is not a multiple of $32$ $n=7$ does not work as $344$ is not a multiple of $128$ $n=9$ does not work as $730$ is not a multiple of $512$ Let $\\displaystyle f(n)={n^3+7\\over 2^n}$. Then we have $$f(0)= 7$$ $$f(1)= 4$$ $$f(2)= 15/4$$ $$f(3)= 17/4$$ $$f(4)= 71/16$$ $$f(5)= 33/8$$ $$f(6)= 223/64$$ $$f(7)= 175/64$$ $$f(8)= 519/256$$ $$f(9)= 23/16$$ $$f(10)= 1007/1024.$$ Of these values, the only perfect square is $f(1)= 4$. Also, we easily see that $0<f(n)<1$ when $n\\ge10$ because the denominator $2^n$ grows faster than the numerator $n^3+7$ (and eventually $2^n$ grows faster than any polynomial function of $n$). Thus the only $n$ satisfying the given conditions is $n=1$.", "}n\\ge 1000 \\\\ f(f(n+5)) & \\mbox{if }n<1000 \\end{cases}$ Find $f(84)$. (Source) • Let $f$ be a function with the following properties: 1. $f(n)$ is defined for every positive integer $n$; 2. $f(n)$ is an integer; 3. $f(2)=2$; 4. $f(mn)=f(m)f(n)$ for all $m$ and $n$; 5. $f(m)>f(n)$ whenever $m>n$. Prove that $f(n)=n$. (Source) • Describe all polynomials $P(x)$ such that $P(x + 1) - 1 = P(x) + P'(x)$ for all $x$.", "$f(4)=\\#{\\mathbb{Z}_4}^*=2$, since ${\\mathbb{Z}_4}=\\{0,1,2,3\\}$ and ${\\mathbb{Z}_4}^*=\\{1,3\\}$ • $f(5)=\\#{\\mathbb{Z}_4}^*=4$, since ${\\mathbb{Z}_5}=\\{0,1,2,3,4\\}$ and ${\\mathbb{Z}_4}^*=\\{1,2,3,4\\}$ Hence $s(1)=3+4f(1)=3$ $s(2)=s(1)+4f(2)=7$ $s(3)=s(2)+4f(3)=15$ $s(4)=s(3)+4f(4)=23$ $s(5)=s(4)+4f(5)=39$ $s(n+1)=s(n)+4f(n+1)$", "n\\}$ and let us prove it for $n+1$. We place $a=1$, $b=n$, $c=-n-1$ in the original equation to obtain: $f(1)^2+n^4f(1)^2+f(n+1)^2=2n^2f(1)^2+2(n^2+1)f(n+1)f(1)$ $\\Leftrightarrow \\quad\\quad\\quad \\left(f(n+1)-(n+1)^2f(1)\\right)\\cdot \\left(f(n+1)-(n-1)^2f(1)\\right)=0.$ If $f(n+1)=(n-1)^2f(1)$ then setting $a=n+1$, $b=1-n$, and $c=-2$ in the original equation yields $2(n-1)^4f(1)^2+16f(1)^2=2\\cdot 4\\cdot 2(n-1)^2f(1)^2+2\\cdot (n-1)^4f(1)$ which implies $(n-1)^2=1$ hence $n=2$. Therefore $f(3)=f(1)$. Placing $a=1$, $b=3$, and $c=4$ into the original equation implies that $f(4)=0$ or $f(4)=4f(1)=f(2)$. If $f(4)\\neq 0$ we get $f(2)^2+f(2)^2+f(4)^2=2f(2)^2+4f(2)f(4)$ hence $f(4)=4f(2)$. We already have that $f(4)=f(2)$ and this implies that $f(2)=0$, which is impossible according to our assumption. Therefore $f(4)=0$ and the function $f$ has period $4$. Then $f(4k)=0$, $f(4k+1)=f(4k+3)=c$, and $f(4k+2)=4c$. It is easy to verify that this function satisfies the requirements of the problem. Thus the solutions are: $f(x)=cx^2$ for some $c\\in\\mathbb Z$; $f(x)=\\left\\{\\begin{array}{rl}0,& 2\\mid n,\\\\ c, & 2\\not\\mid n\\end{array}\\right.$ for some $c\\in\\mathbb Z$; and $f(x)=\\left\\{\\begin{array}{rl}0,& 4\\mid n,\\\\ c, & 2\\nmid n,\\\\ 4c, & n\\equiv 2 \\mbox{ (mod } 4)\\end{array}\\right.$ for some $c\\in\\mathbb Z$. Problem 5 Given a triangle $ABC$, assume that $\\angle C=90^{\\circ}$. Let $D$ be the foot of the perpendicular from $C$ to $AB$, and let $X$ be any point of the segment $CD$. Let $K$ and $L$ be the points on the segments $AX$ and $BX$ such that $BK=BC$ and $AL=AC$, respectively. Let $M$ be the intersection of $AL$ and $BK$. Prove that $MK=ML$.", "# If $f(1)=1$, then is it true that $f(n)=n$ for all $n \\in \\mathbb{N}\\cup\\{0\\}$. Let $f:\\mathbb{N}\\cup\\{0\\}\\to\\mathbb{N}\\cup\\{0\\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \\in\\mathbb{N}\\cup\\{0\\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. Then it's very easy to see that $f(2)=2$ and since $f(2)=2$, we can see that $f(4)=4$ and $f(5)=5$ because $5=2^2+1^2$. To see $f(3)=3$, note that $$25=f(5^2)=f(3)^2+f(4)^2 \\implies f(3)=3$$ One can even see that $f(6)=6$, $f(7)=7$ and so on. Is it generally true that $f(n)=n$ for all $n$? Edit. If anyone wondering how $f(7)=7$ here is the way. Note that $f$ satisfies $f(n^2)=f(n)^2$. One can see that $$25^{2}=24^2+7^2 \\implies 25^{2}=f(24)^{2}+f(7)^{2}$$ We have to show $f(24)=24$. For this note that $$26^2 = 24^2+10^2 \\implies f(26)^{2} = f(24)^{2}+f(10)^{2}$$ But $f(26)=26$ because $26=5^2+1^2$ and $f(10)=10$ since $10=3^{2}+1^{2}$. In short if $n=x^2+y^2$ for some $x,y$ and $f(1)=1$, then we can see that $f(n)=n$. • Why must $f(1)$ be $0$ or $1$? – copper.hat Oct 4 '17 at 4:52 • $f(0)=f(0^2+0^2)=f(0)^2+f(0)^2$ so $f(0)=0$. Next check $f(1)=f(1^2+0^2)=f(1)^2+f(0)^2=f(1)^2$. – Jorge Fernández Hidalgo Oct 4 '17 at 5:00 • @JorgeFernández: Thanks. A bit slow tonight :-(. – copper.hat Oct 4 '17 at 5:00 • how did you see $f(6)=6$? – Jorge Fernández Hidalgo Oct 4 '17 at 5:54 • Very simple. $3^2+4^2=5^2$ which gives $6^2+8^2=10^2$. $f(10)=10$", "617$. We have to find the value of $f(10)$. Let $f(10)$ = b $f(12)$ = b + 91 $f(13)$ = 91 + b + 91 = 182 + b $f(14)$ = 182+b+91+b = 273+2b $f(15)$ = 273+2b+182+b = 455+3b It has been given that 455+3b = 617 3b = 162 => b = 54 Therefore, 54 is the correct answer. Question 8: Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals Solution: Given, f(mn) = f(m)f(n) when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1 when m=1, n= 2, f(2) = f(1)*f(2) ==> f(1) = 1 when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $[f(2)]^2$ Similarly f(8) = f(4)*f(2) =$[f(2)]^3$ f(24) = 54 $[f(2)]^3$ * $[f(3)]$ = $3^3*2$ On comparing LHS and RHS, we get f(2) = 3 and f(3) = 2 Now we have to find the value of f(18) f(18) = $[f(2)]$ * $[f(3)]^2$ = 3*4=12 Question 9: Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f", "If $k$ is even $f(k)=997$ if $k$ is odd $f(k)=998$. $84$ has even parity, so $f(84)=997$. The result may be rigorously shown through induction. ## Solution 3 Assume that $f(84)$ is to be performed $n+1$ times. Then we have $$f(84)=f^{n+1}(84)=f(f^n(84+5))$$ In order to find $f(84)$, we want to know the smallest value of $$f^n(84+5)\\ge1000$$ Because then $$f(84)=f(f^n(84+5))=(f^n(84+5))-3$$ From which we'll get a numerical value for $f(84)$. Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\\frac{n}{2}$ times, the result is greater than or equal to $1000$. In this case, the value of $n$ for $f(84)$ is $916$, because $$84+\\frac{916}{2}\\cdot5-\\frac{916}{2}\\cdot3=1000\\Longrightarrow f^{916}(84+5))=1000$$ Thus $$f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\\boxed{997}$$ ~ Nafer", "$0$ for all rationals, it is identically $0$, so $$f(x^i)=f(1)^{n-i}f(x)^i$$ for all $0\\leq i\\leq n$. Originally, we had $f(1)=f(1)^n$, so $f(1)\\in\\{-1,0,1\\}$. If $f(1)=0$, we have $f(x^i)=0$, so $\\boxed{f(x)\\equiv 0}$. Otherwise, we have $$f(x^2)=f(1)^{n-2}f(x)^2=f(1)f(x)^2$$ $$f(x+y^2)=f(x)+f(y^2)=f(x)+f(1)f(y)^2.$$ If $f(1)=1$, this means $f$ is increasing, and if $f(1)=-1$ this means $f$ is decreasing. Either way, $f$ is not everywhere dense, so $f(x)=cx$ for some $c$ and all $x$. The observation that $f(1)=\\pm 1$ means $\\boxed{f(x)=x}$ and $\\boxed{f(x)=-x}$ are our only other solutions. • In $f(x^{2})=f(1)^{n-2}f(x)^{2}=f(1)f(x)^{2}$ when $f(1)\\neq 0$ how does the second equality follow? Sorry if I am missing something obvious. – Kavi Rama Murthy Jul 16 '18 at 5:46 • As $f(1)\\in\\{-1,0,1\\}$ and $f(1)\\neq 0$, we have $|f(1)|=1\\implies f(1)^2=1$; since $n$ is odd, $$f(1)^{n-2}=\\left(f(1)^2\\right)^{\\frac{n-3}{2}}f(1)=1^{\\frac{n-3}{2}}f(1)=f(1).$$ – Carl Schildkraut Jul 16 '18 at 17:02 • Nice and simpler than mine! – Sungjin Kim Jul 16 '18 at 18:27 • @KaviRamaMurthy The second equality is by $f(y^2)=f(1)f(y)^2$. – Sungjin Kim Jul 16 '18 at 18:27 This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\\in\\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you. As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\\in \\mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\\in\\mathbb{Q}$, $x\\in\\mathbb{R}$. Since $f( (x+y)^5)= f(x+y)^5", "+1 vote 50 views A function $f(x)$ satisfies $f(1)=3600$ and $f(1)+f(2)+ ... f(n) =n^2 f(n)$, for all positive integers n>1. What is the value of $f(9)$? 1. 80 2. 240 3. 200 4. 100 5. 120 f1+f2=4f2 f2=f1/3 f1+f2+f3=9f3 f1+f1/3+f3=9f3 f3=f1/6 f1+f2+f3+f4=16f4 We get f4=f1/10 So we can say fn=f1/sum of n terms therefore f9=3600/45=80", "we can cancel it to give $$f(x) = 2$$ for all $$x$$, which is a solution. Otherwise, $$f(0) = 0$$. $$P(2,2) \\implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \\implies f(2)^2 = 2f(2)$$. This gives $$f(2) = 0$$ or $$f(2) = 2$$. $$P(1,1) \\implies f(2) + f(1)^2 = 3f(1)$$. If $$f(2) = 0$$, this leads to $$f(1) = 0$$ or $$f(1) = 3$$. If $$f(2) = 2$$, this leads to $$f(1) = 1$$ or $$f(1) = 2$$. $$P(-1,1) \\implies f(-1)f(1) = 2f(-1) + f(1)$$. $$f(1) = 2$$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $$(f(1), f(2)) = (0,0), (1,2), (3,0)$$. $$P(x,1) \\implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$$. We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $$f$$ on the integers: Case 1: $$f(1) = 0,$$. Then $$P(x,1) \\implies f(x+1) = 2f(x)$$. Inducting gives $$f(x) = 0$$ for all integer $$x$$. Case 2: $$f(1) = 1$$. Then $$P(x,1) \\implies f(x+1) = f(x) + 1$$. Inducting gives $$f(x) = x$$ for all integer $$x$$. Case 3: $$f(1) = 3$$. Then $$P(x,1) \\implies f(x+1) = 3 - f(x)$$. Inducting gives $$f(x) = 3$$ if $$x$$ is odd and $$f(x) = 0$$ if $$x$$ is", "103 views The function defined for positive integers by $F(1)=1,F(2)=1,F(3)=-1$ and by identities F(2k)=F(k),F(2k+1)=F(k) for $k>=2.$The sum $F(1)+F(2)+F(3)+...+F(100)$ is________ | 103 views 0 67.? +1 May u check your question once again because according to F(3) function F(2k+1) should be equal to -F(k) not F(k). 0 34? 0 $28$ is the right 0 @Lakshman Patel RJIT is question is correct? 0 Yes ,in test series 0 It is not correct question. leave it 0 Let's start series by ignoring first 3 numbers. So F(4)=1,F(5)=1,F(6)=-1,F(7)=-1,F(8)=1,F(9)=1,F(10)=1,F(11)=1, We observe that +1 and -1 will occur simultaneously as +2,-2,+4,-4,+8,-8,+16,-16,+32 .Now total terms are 92.So,until 97 5 times -1 will be there so,32 -5=27, Now adding +1 as F(0)+F(1)+F(2)=1 So,27+1=28 by Active (1.4k points) 0 How you got this, So, until 97, 5 times -1 will be there so,32 -5=27? 0 I think from $F(1)$ $to$ $F(95)=32+1=33$ and we find the$:F(96)=F(2.48)=F(48)=F(2.24)=F(24)=F(2.12)=F(12)=F(2.6)=F(6)=F(2.3)=F(3)=-1$ similarly we can find $F(97)=-1,F(98)=-1,F(99)=-1,F(100)=-1$ So,we get $F(1)+F(2)+...........+F(95)+F(96)+F(97)+F(98)+F(99)+F(100)$ $\\Rightarrow33-1-1-1-1-1$ $\\Rightarrow 28$ +1 After 32 +1 there will be 32 -1 .But as we have reached upto 92 (+3 as I stared from 4) already and we need to go upto 97( +3) as we will have only -5. So,32-5=27. 0 Yes, see my comment is also right? 1", "# 2012 USAMO Problems/Problem 4 ## Problem Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$. ## Solution By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$. By the second condition, we have $$n\\cdot n!=(n+1)!-n! \\mid f(n+1)!-f(n)! \\qquad \\qquad (1)$$ for all positive integers $n$. Suppose that for some $n \\geq 2$ we have $f(n) = 1$. We claim that $f(k)=1$ for all $k\\ge n$. Indeed, from Equation (1) we have $f(n+1)!\\equiv 1 \\mod n\\cdot n!$, and this is only possible if $f(n+1)=1$; the claim follows by induction. We now divide into cases: Case 1: $f(1)=f(2)=1$ This gives $f(n)=1$ always from the previous claim, which is a solution. Case 2: $f(1)=2, f(2)=1$ This implies $f(n)=1$ for all $n\\ge 2$, but this does not satisfy the initial conditions. Indeed, we would have $$3-1 \\mid f(3)-f(1)$$ and so $2\\mid -1$, a contradiction. Case 3: $f(1)=1$, $f(2)=2$ We claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$. Fix $k > 1$ and suppose that $f(k)=k$. By Equation (1) we have that $$f(k+1)!", "of $3$ or we can skip to the next multiple of $13$ which is simply $13.$ From these equations we have deduced $f(0) = f(15) + 5f(12).$ Continuing we have $$f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).$$ Finally, note that $f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5$ since at any point we can either go to the next multiple of $3$ or go to the next multiple of $13$ which happens to be $39.$ Therefore $f(26) = 5.$ Similarly we find $f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4$ so the end answer is $5 \\cdot 29 + 6 \\cdot 4 = \\boxed{169}.$", "The $$\\pm$$ is used because the triples alternate: $$A>B$$ and $$A. $$\\text{With infinite integers input }(n=\\sqrt{2*m^2\\pm1}-m)\\text{ will always output an integer for some }m.$$ For all $$M\\in\\mathbb{N}$$, whenever this function yields an integer $$>0$$, we have the $$(m,n)$$ we need for a triple. The following $$19$$ were generated (in $$4$$ seconds) in a loop of $$m=1\\text{ to }16,000,000$$ and $$f(m,n)$$ shows the variable values needed to generate each triple using Euclid's formula. The $$g(n,k)$$ shows the values (generated in $$1$$ second) for an alternate set of functions in a loop of $$m=1\\text{ to }5,000,000$$. $$A=m^2-n^2\\qquad B=2mn\\qquad C=M^2+n^2$$ $$f(2,1)=g(1,1)=(3,4,5)$$ $$f(5,2)=g(2,2)=(21,20,29)$$ $$f(12,5)=g(4,5)=(119,120,169)$$ $$f(29,12)=g(9,12)=(697,696,985)$$ $$f(70,29)=g(21,29)=(4059,4060,5741)$$ $$f(169,70)=g(50,70)=(23661,23660,33461)$$ $$f(408,169)=g(120,169)=(137903,137904,195025)$$ $$f(985,408)=g(289,408)=(803761,803760,1136689)$$ $$f(2378,985)=g(697,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=g(1682,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=g(4060,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=g(9801,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=g(23661,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=g(57122,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=g(137904,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=g(332929,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=g(803761,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=g(1940450,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=g(4684660,6625109)=(211929657785303,211929657785304,299713796309065)$$ The alternate formula mentioned above is slightly faster since it deals only with the subset of triples where $$GCD(A,B,C)=(2m-1)^2,m\\in\\mathbb{N}$$. We have the needed $$(n,k)$$ values whenever the following yields a positive integer. $$k=\\sqrt{\\frac{(2n-1)^2\\pm1}{2}}$$ Having found a needed $$(n,k)$$ the following will generate a needed triple. $$A=(2n-1)^2+2(2n-1)k\\qquad B=2(2n-1)k+2k^2\\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$", "+0 # Algebra Tricky Question 0 94 1 +140 For all ordered pairs of positive integers (x,y) we define f(x,y) as follows: (a) $$f(x,1) = x.$$ (b) if $$f(x,y) = 0$$ if $$y > x.$$ (c) $$f(x + 1,y) = y[f(x,y) + f(x,y - 1)].$$ Compute f(5,5) Aug 6, 2022 #1 +2 Hi sherwyo! We are going to solve this by substituting the desired values in (c) I.e. we want f(x+1,y) = f(5,5) so x=4 and y=5: Let x=4 and y=5 $$f(5,5)=5[f(4,5)+f(4,4)]$$ by (c). Then notice: $$f(4,5)=0 \\space \\space \\text{As, y>x (5>4)}$$ by (b). So: $$f(5,5)=5f(4,4)$$ Now, we want to find $$f(4,4)$$. We have to repeat the process! let x=3, and y=4 in (c), to get f(4,4): $$f(4,4)=4[0+f(3,3)]=4f(3,3)$$ Next, we need to find $$f(3,3)$$: Let x=2, and y=3: $$f(3,3)=3f(2,2)$$ Now let's find $$f(2,2)$$: Let x=1, and y=2 $$f(2,2)=2f(1,1)=2(1)=2$$ by (a). So: $$f(3,3)=3f(2,2)=3(2)=6$$ Hence, $$f(4,4)=4f(3,3)=4(6)=24$$ Therefore, $$f(5,5)=5f(4,4)=5(24)=120$$ Extra information: Notice, this function behaves exactly like $$n!$$ for all integers n (Because it has the pattern 2,6,24,120 etc..) But, our function is of two variables, namely x and y. But $$n!$$ just relies on one variable. But, notice, this pattern of results of $$f(x,y)$$ occurs only when x=y. Therefore, $$f(x,y)=xPy$$ That is, the permutation function; which is defined as follows: $$nPk=\\frac{n!}{(n-k)!}$$ So, in this case, we know x=y: $$xPy=xPx=\\frac{x!}{0!}=x!$$, the", "cf(n).$$ We can show $g$ is in $O(f)$ using the definition of $O(f)$ by choosing positive constants for the values of $c$ and $n_0$, and showing that the property $g(n) \\le cf(n)$ holds for all values $n \\ge n_0$. To show $g$ is not in $O(f)$, we need to explain how, for any choices of $c$ and $n_0$, we can find values of $n$ that are greater than $n_0$ such that $g(n) \\le cf(n)$ does not hold. Example 7.1: $O$ Examples We now show the claimed properties are true using the formal definition. $n-7$ is in $O(n+12)$ Choose $c = 1$ and $n_0 = 1$. Then, we need to show $n-7 \\le 1(n+12)$ for all values $n \\ge 1$. This is true, since $n-7 > n+12$ for all values $n$. $n+12$ is in $O(n-7)$ Choose $c = 2$ and $n_0 = 26$. Then, we need to show $n+12 \\le 2(n-7)$ for all values $n \\ge 26$. The equation simplifies to $n + 12 \\le 2n - 14$, which simplifies to $26 \\le n$. This is trivially true for all values $n \\ge 26$. $2n$ is in $O(3n)$ Choose $c = 1$ and $n_0 = 1$. Then, $2n \\le 3n$ for all values $n \\ge 1$. $3n$ is in $O(2n)$ Choose $c = 2$ and $n_0 = 1$. Then,", "# Function 166 views The function defined for positive integers by $F(1)=1,F(2)=1,F(3)=-1$ and by identities F(2k)=F(k),F(2k+1)=F(k) for $k>=2.$The sum $F(1)+F(2)+F(3)+...+F(100)$ is________ 0 67.? 1 May u check your question once again because according to F(3) function F(2k+1) should be equal to -F(k) not F(k). 0 34? 0 $28$ is the right 0 @Lakshman Patel RJIT is question is correct? 0 Yes ,in test series 0 It is not correct question. leave it 0 Let's start series by ignoring first 3 numbers. So F(4)=1,F(5)=1,F(6)=-1,F(7)=-1,F(8)=1,F(9)=1,F(10)=1,F(11)=1, We observe that +1 and -1 will occur simultaneously as +2,-2,+4,-4,+8,-8,+16,-16,+32 .Now total terms are 92.So,until 97 5 times -1 will be there so,32 -5=27, Now adding +1 as F(0)+F(1)+F(2)=1 So,27+1=28 0 How you got this, So, until 97, 5 times -1 will be there so,32 -5=27? 0 I think from $F(1)$ $to$ $F(95)=32+1=33$ and we find the$:F(96)=F(2.48)=F(48)=F(2.24)=F(24)=F(2.12)=F(12)=F(2.6)=F(6)=F(2.3)=F(3)=-1$ similarly we can find $F(97)=-1,F(98)=-1,F(99)=-1,F(100)=-1$ So,we get $F(1)+F(2)+...........+F(95)+F(96)+F(97)+F(98)+F(99)+F(100)$ $\\Rightarrow33-1-1-1-1-1$ $\\Rightarrow 28$ 1 After 32 +1 there will be 32 -1 .But as we have reached upto 92 (+3 as I stared from 4) already and we need to go upto 97( +3) as we will have only -5. So,32-5=27. 0 Yes, see my comment is also right? ## Related questions 1 77 views The number of ways possible to form injective function from set A set B", "Yes. $\\because f(n)=f(n+5)$. Putting $n=n-5$ [where $n>5$] will yield $f(n-5)=f(n-5+5)=f(n)$ $\\text{let we have f(1) = x. Then, f(2) = f(2/2) = f(1) = x}$ $\\text{f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x }$ $\\text{f(5) = f(5+5) = f(10/2) = f(5) = y. }$ $\\text{All$N^+$except multiples of 5 are mapped to x and multiples}$ $\\text{of 5 are mapped to y so ,$\\mathbf{Answer\\space is\\space 2}$}$ edited +1 thanx @Prince Sindhiya ,the pictorial mapping clears everything., choose any number let n=17, then f(17)=f(22)=f(11)=f(16)=f(8)=f(4)=f(2)=f(1)=f(6)=f(3)=f(8)=f(4)=f(2)=f(1)=f(6)=f(3)=f(8)=f(4)=f(2)=f(1)... <this is one part> now let n=50 f(50)=f(25)=f(30)=f(15)=f(20)=f(10)=f(5)=f(10)=f(15)=f(20)=f(10)=f(5)=f(10)=f(5)=f(10)=f(5) .....<this is other part> so we can take any number and that will fall either of any cycle, these are the two types of values that Function f( ) can take. by +1 vote Let's start with the smallest number. (You can begin at any number) $f(1) = f(6) = f(3) = f(8) = f(4) = f(2) = \\color{red}{f(1)}...$ $f(2)=\\color{red}{f(1)}$ $f(3)=\\color{red}{f(1)}$ $f(4)=\\color{red}{f(1)}$ $f(5) = f(10) = \\color{blue}{f(5)}...$ $f(6)=\\color{red}{f(1)}$ $f(7) = f(12) = f(6) =\\color{red}{f(1)}$ $f(8)=\\color{red}{f(1)}$ $f(9) = f(14) = f(7)=\\color{red}{f(1)}$ $f(10)=\\color{blue}{f(5)}$ $f(11) = f(16) = f(8)=\\color{red}{f(1)}$ ... so on. We observe that all the multiples of 5 will have the value of $f(5)$ and every other number will converge to $f(1)$ ultimately. Let's assume $f(1) = i_1$ and $f(5) = i_2$ $R=\\{i∣∃j:f(j)=i\\}$ Hence," ]

[ "Dear Uncle Colin, Someone gave me the puzzle: • $ff(n) = 3n$ for all real $n$, • $f(n)$ is a positive integer for all positive integer $n$ • $f(n+1) > f(n)$ for all positive integer $n$ • What is $f(13)$? Any ideas? Functions? Fun? No. Hi, FFN, This is a fun puzzle! Here’s how I tackled it: • $ff(1) = 3$, so I reckon $f(1) = 2$ and $f(2)=3$ This takes a little unpicking; $f(1)$ can’t be 1, or else $ff(1) = 1$. Could it be 3, and $f(3)=3$? again, no: $f(3) > f(2) > f(1)$. So, it has to be 2, and $f(2) = 3$. • We know that $ff(2) = 6$ and $f(2) = 3$, so $f(3) = 6$. • Now $ff(3) = 9$, so $f(6) = 9$ Let’s take stock: • $f(1) = 2$ • $f(2) = 3$ • $f(3) = 6$ • $f(6) = 9$ That means $f(4) = 7$ and $f(5) = 8$, if we want to satisfy the “increasing” condition. • We know $ff(4) = 12$ and $f(4) = 7$, so $f(7) = 12$. • Similarly, $ff(5) = 15$ and $f(5) = 8$, so $f(8) = 15$. There’s a pattern emerging! • $f(1) = 2$ • $f(2) = 3$ • $f(3) = 6$ • $f(4) = 7$ • $f(5) = 8$ •", "# Puzzling function Let $$f$$ be a function whose domain is the set of positive integers, and for positive integers $$a$$, $$b$$ and $$n$$, if $$a + b = 2^{n}$$, then $$f(a) + f(b) = n^2$$. What is $$f(2021)$$? I started by testing values for $$a$$, $$b$$ and $$n$$, with the hope of finding a pattern, but so far I can't say I've made any headway; For $$a=b=n=1,\\ f(1)+f(1)=2^1, so f(1)=\\frac{1^2}{2}=\\frac12$$ I realize that it becomes easier to find $$f(a)$$ if $$a$$=$$b$$ So $$a = b = 2^{n-1}$$ then $$f(a) = \\frac{n^2}{2}$$ that is, $$f(2^{n-1}) = \\frac{n^2}{2}$$ At this point, I cannot see where to move forward. • Please edit to include your efforts. A natural starting point might be to find $f(k)$ for small values of $k$. – lulu Jun 24 at 16:38 • Note: may be worth remarking that $f(k)$ doesn't seem to be an integer for some small $k$. Nothing wrong with that, I suppose, but is it what you intended? – lulu Jun 24 at 16:41 • please include you efforts Jun 24 at 16:58 • In your $a=b=n=1$ example, I think you mean: $1+1=2 = 2^1$, so $f(1) + f(1) = 1^2 = 1$ and thus $f(1) = 0.5$. Jun 24 at 17:15 • Note that $2021+27=2048=2^{11}$ thus you can obtain $f(2021)$ in", "# Find all functions that satisfy the following conditions Find all functions $f:\\mathbb Z\\to \\mathbb Z$ that satisfy the following conditions: (i) $f (0) = 1$ (ii) $f(f (x)) = x$ for all integers x (iii) $f(f(x + 1)+1) = x$ for all integers x How do you prove it with induction • As in my argument below, we know that $f(x)=f(x+1)+1$ for all integers $x$. Now let us prove that $f(n)=1-n$ and that $f(-n)=1+n$ for all positive integers $n\\geq 1$. The induction start is easy: we know that $f(0)=f(1)+1$ and that $f(-1)=f(0)+1$. As $f(0)=1$, it follows that $f(1)=0$ and $f(-1)=2$. If it holds for some $n\\geq 1$ that $f(n)=1-n$ and $f(-n)=1+n$, then $f(n)=f(n+1)+1$, implying $f(n+1)=f(n)-1=-n=1-(n+1)$, and $f(-(n+1))=f(-n)+1$, implying $f(-(n+1))=1+n+1=2+n=1+(n+1)$. Hence it follows that $f(n)=1-n$ and $f(-n)=1+n$ for all integers $n\\geq 1$, so that $f(x)=1-x$. – Bryder Apr 14 '14 at 19:28 Quite straightforward. $f$ is a bijection by (ii), so (ii) and (iii) imply $f(x)=f(x+1)+1$. Now $f(n)=f(n-1)-1=f(n-2)-2=f(n-3)-3=\\ldots=f(0)-n=1-n$ and $f(-n)=f(1-n)+1=\\ldots=f(0)+n=1+n$ for all integers $n>0$, so that $f(x)=1-x$ for all integers $x$. Induction is of course also applicable. By (ii), $$f(f(0))=f(1)=0$$ By (iii) $$f(f(-1+1)+1)=f(f(0)+1)=f(2)=-1$$ So $$f(f(2))=f(-1)=2$$ So $$f(f(-2+1)+1)=f(f(-1)+1)=f(3)=-2$$ So $$f(f(3))=f(-2)=3$$ So $$f(f(-3+1)+1))=f(4)=-3$$ So... • So, the solution is $f(x)=-x+1$. – user35603 Apr 14 '14 at 15:41", "617$. We have to find the value of $f(10)$. Let $f(10)$ = b $f(12)$ = b + 91 $f(13)$ = 91 + b + 91 = 182 + b $f(14)$ = 182+b+91+b = 273+2b $f(15)$ = 273+2b+182+b = 455+3b It has been given that 455+3b = 617 3b = 162 => b = 54 Therefore, 54 is the correct answer. Question 8: Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals Solution: Given, f(mn) = f(m)f(n) when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1 when m=1, n= 2, f(2) = f(1)*f(2) ==> f(1) = 1 when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $[f(2)]^2$ Similarly f(8) = f(4)*f(2) =$[f(2)]^3$ f(24) = 54 $[f(2)]^3$ * $[f(3)]$ = $3^3*2$ On comparing LHS and RHS, we get f(2) = 3 and f(3) = 2 Now we have to find the value of f(18) f(18) = $[f(2)]$ * $[f(3)]^2$ = 3*4=12 Question 9: Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f", "that $f(1) = 2$ and $f(xy) = f(x)f(y) - f(x + y) + 1$ for all rational numbers $x$ and $y$. #### 1 comment: 1. Let x=1; then, f(y) = f(1)*f(y) -f(x+1)+1 for all rational y, or f(y)=2f(y)-f(x+1)+1 --> f(y+1)=f(y) + 1 for all rational y. Since f(1)=2 and f(y+1)=f(y)+1 for all rational y, it follows with induction to both negative and positive integers that f(z)=z+1 for all integers z. Furthermore, it follows by induction that f(y+z)=f(y) + z for rational y and integer z. Next, suppose x= 1/w and y=w for an integer w. Then, f(1) = f(1/w)*f(w) - f(1/w + w) + 1 1=f(1/w)[w+1] - f(1/w) - w f(1/w)[w]=w+1 f(1/w)=1/w+1 - again, for all integers w. Finally, let x=m/n and y=n for some integers m and n. Then, f(m)=f(m/n)f(n) - f(m/n + n) +1 m+1 = f(m/n)[n+1] - f(m/n) - n + 1 m+n=f(m/n)(n) f(m/n)=(m+n)/n f(m/n)=1+m/n for all integers m and n; so for all rationals R, f(R)=1+R is the only function that exists to satisfy these conditions.", "we can cancel it to give $$f(x) = 2$$ for all $$x$$, which is a solution. Otherwise, $$f(0) = 0$$. $$P(2,2) \\implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \\implies f(2)^2 = 2f(2)$$. This gives $$f(2) = 0$$ or $$f(2) = 2$$. $$P(1,1) \\implies f(2) + f(1)^2 = 3f(1)$$. If $$f(2) = 0$$, this leads to $$f(1) = 0$$ or $$f(1) = 3$$. If $$f(2) = 2$$, this leads to $$f(1) = 1$$ or $$f(1) = 2$$. $$P(-1,1) \\implies f(-1)f(1) = 2f(-1) + f(1)$$. $$f(1) = 2$$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $$(f(1), f(2)) = (0,0), (1,2), (3,0)$$. $$P(x,1) \\implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$$. We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $$f$$ on the integers: Case 1: $$f(1) = 0,$$. Then $$P(x,1) \\implies f(x+1) = 2f(x)$$. Inducting gives $$f(x) = 0$$ for all integer $$x$$. Case 2: $$f(1) = 1$$. Then $$P(x,1) \\implies f(x+1) = f(x) + 1$$. Inducting gives $$f(x) = x$$ for all integer $$x$$. Case 3: $$f(1) = 3$$. Then $$P(x,1) \\implies f(x+1) = 3 - f(x)$$. Inducting gives $$f(x) = 3$$ if $$x$$ is odd and $$f(x) = 0$$ if $$x$$ is", "number $x$ we have $f(x)=f(β+(x-β))=f(β)f(x-β)=0$ which implies that $f$ is identically zero. By the continuity of $f$ we can conclude that $f(x)>0$ for all real numbers. For a positive integer $n$, $f(n)=f(1+1+...+1)=f(1)f(1)*...*f(1)=f(1)^n$. Since $f(-1)=f(1+(-2))=f(1)f(-2)=f(1)f(-1)f(-1)$, it follows that $f(-1)=f(1)^{-1}$. For a negative integer $m$, $f(m)=f(-1+(-1)+...+(-1))=f(-1)^{-m}=(f(1)^{-1})^{-m}=f(1)^m$. Since $f(0)=1=f(1)^0$, $f(n)=f(1)^n$ holds for all integers. Let $q$ and $m$ be positive integers, then $f(m)=f(1/q+1/q+...+1/q)=f(1/q)^{qm}=f(1)^m$. It follows that $f(1/q)=f(1)^{1/q}$. This will also hold for negative integers $q_1$ and $m_1$ as well. Then, $f(s/t)=f(1/t+...+1/t)=f(1/t)^s=(f(1)^{1/t})^s=f(1)^{s/t}$, for any rational number. Since the rationals are dense in the reals and $f$ is continuous, for any real number $x$ we can find a sequence $(r_n)$ that will converge to $x$ with $f(r_n)=f(1)^{r_n}$, so that $f(x)=f(1)^x$ for all real numbers. Since $f(1)>0$ and is real-valued, $\\log(f(1))=c$ exists. Then: $f(x)=f(1)^x=(\\exp(\\log(f(1))))^x=(\\exp(c))^x=(e^c)^x=e^{cx}$. Thanks again! -", "# Finding $f(n)=f(f(n-1))+f(f(n+1))$ Determine whether a function exists from the positive integers to the positive integers which satisfies the equation: $$f(n)=f(f(n-1))+f(f(n+1))$$. My guess is that this function does not exist, as through trial and error I found that $f(n)=\\frac{n}{2}$ is a solution to the functional equation. Obviously this is not enough, but I don't know how to tackle this problem. I found that $f(n+2)-f(n)=f(f(n+3))-f(f(n-1))$, but this doesn't seem very useful to me. Can anyone help me with this problem? Here is a sketch of the proof that no such function exists. Say that such a function does exist. Take $n$ be such that $f(n)=m$ is the minimum value that $f$ takes. If $n\\neq1$, then then $m$ is expressed as the sum of two outputs of the function $f$ by the given relation. Since the outputs of $f$ are positive integers, this is impossible. If the minimum is $f(1)$ and is attained only at $1$, then take let $n_1>1$ be such that $f(n_1)$ is the minimum of the values attained by $f$ excluding $f(1)$. Then $f(n_1+1)\\geq f(n_1)>f(1)\\geq 1$, so by the minimality of $f(n_1)$, we have that $f(f(n_1+1))\\geq f(n_1)$. Again this leaves the desired relation impossible, so there is no such function. • You should rule out the case $n=1$, for which the formula does not apply, as $f(n-1)$", "No good. • When $n > 5$, $f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$. Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question. • We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive. • When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question. - Let $f(x)=x^3-7x^2+11x-5$. $f(1)=f(5)=0$ and so $$f(x)=(x-1)(x^2-6x+5)=(x-1)(x-1)(x-5)=(x-1)^2(x-5)$$ Then $$f(x)>0\\iff ...$$ - The polynomial was chosen so that it would factor simply. So probably a solution through factoring is expected. But let's take a look at the polynomial $n^3-7n^2+11n-5$ as if factoring might be hard. If $n\\ge 7$, then $n^3-7n^2\\ge 0$. And $11n-5\\gt 0$ for any positive integer $n$. So $n^3-7n^2+11n-5\\gt 0$ for any $n\\ge 7$. Now ask whether our polynomial is $\\gt 0$ for $n\\ge 6$. So plug in $n=6$. We get a positive result. Is our polynomial $\\gt 0$ for $n\\ge 5$? Plug in $5$. We get $0$. This is not positive. So the required number $N$ is $5$. -", "help or contributions will be greatly appreciated. Thank you! :) - You can easily narrow it down to between 18 and 27. – karakfa Mar 21 '13 at 19:51 First note that $f(f(1)) = 3 \\implies f(1) = 2$ (Why?) This gives us $f(2) = 3$. $$f(f(2)) = 3 \\times 2 = 6 \\implies f(3) = 6$$ $$f(f(3)) = 3 \\times 3 = 9 \\implies f(6) = 9$$ Since we have $$6 = f(3) < f(4) < f(5) < f(6) = 9$$ and all are positive integers, we get $f(4) = 7$ and $f(5) = 8$ (Why?). Now $$f(f(4)) = 3 \\times 4 = 12 \\implies f(7) = 12$$ $$f(f(5)) = 3 \\times 5 = 15 \\implies f(8) = 15$$ $$f(f(6)) = 3 \\times 6 = 18 \\implies f(9) = 18$$ $$f(f(7)) = 3 \\times 7 = 21 \\implies f(12) = 21$$ Again, since $$18 = f(9) < f(10) < f(11) < f(12) = 21$$ and all are positive integers, we get that $$f(10) = 19$$ EDIT If I plug in these values in OEIS, it gives this, the unique monotonic sequence of non-negative integers satisfying $a(a(n)) = 3n$. - Very nice answer. Thanks! – mathsnoob Mar 21 '13 at 19:57 Given the choices (A)12, (B)15, (C)19, (D)21, (E)30 As you need to evaluate $f(10)$ satisfying the equation", "$f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$. Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question. • We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive. • When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question. - Let $f(x)=x^3-7x^2+11x-5$. $f(1)=f(5)=0$ and so $$f(x)=(x-1)(x^2-6x+5)=(x-1)(x-1)(x-5)=(x-1)^2(x-5)$$ Then $$f(x)>0\\iff ...$$ - The polynomial was chosen so that it would factor simply. So probably a solution through factoring is expected. But let's take a look at the polynomial $n^3-7n^2+11n-5$ as if factoring might be hard. If $n\\ge 7$, then $n^3-7n^2\\ge 0$. And $11n-5\\gt 0$ for any positive integer $n$. So $n^3-7n^2+11n-5\\gt 0$ for any $n\\ge 7$. Now ask whether our polynomial is $\\gt 0$ for $n\\ge 6$. So plug in $n=6$. We get a positive result. Is our polynomial $\\gt 0$ for $n\\ge 5$? Plug in $5$. We get $0$. This is not positive. So the required number $N$ is $5$.", "}n\\ge 1000 \\\\ f(f(n+5)) & \\mbox{if }n<1000 \\end{cases}$ Find $f(84)$. (Source) • Let $f$ be a function with the following properties: 1. $f(n)$ is defined for every positive integer $n$; 2. $f(n)$ is an integer; 3. $f(2)=2$; 4. $f(mn)=f(m)f(n)$ for all $m$ and $n$; 5. $f(m)>f(n)$ whenever $m>n$. Prove that $f(n)=n$. (Source) • Describe all polynomials $P(x)$ such that $P(x + 1) - 1 = P(x) + P'(x)$ for all $x$.", "Functional Equation Problems If $\\Bbb N$ denotes all positive integers. Then find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ which are strictly increasing and such for all positive integers $n$, we have: $$f(f(n)) = n+2$$ So far I know that $f(n)$ is greater than $n$ because the function is strictly increasing, but I'm not sure how to use this in order to solve the equation. • Could we not have $f(x)=x+1$, which is strictly increasing and hence $f(f(n))=f(n+1)=n+2$? – lioness99a Jun 22 '17 at 11:41 • hint: $f ( n )$ can't be greater than $n + 2$. – Mohsen Shahriari Jun 22 '17 at 11:42 • strictly increasing seems to be unnecessary – Mudream Jun 29 '17 at 5:45 • Possible duplicate of Find all functions of positive integers for $f(f(n))=n+2$ – Sil Apr 6 at 12:53 As kindly explained below strictly increasing means $f(a) > f(b)$ for $a >b.$ Let's consider $f(0)$. 1. $f(0) = 0$ is not possible because then $f^2(0) = 0 \\neq 2.$ 2. $f(0) = 1$ is possible. 3. $f(0) = m > 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) > f(0) > 1$ which implies $f(m) \\ge m + 2 > 2$ in contradiction with the assumption $f^2(0) = 2.$ Hence $f(0)", "can show that $$f(n)\\ge (\\sqrt{2})^{n-1}$$. However, $$f$$ really shouldn't be growing that quickly, since $$f(f(n-1))=f(n+1)-f(n)<f(n+1)$$. Since $$f$$ is increasing, $$f(n-1)<n+1$$, or $$f(n)<n+2$$. Merging the two results, we have that $$(\\sqrt{2})^{n-1}\\le f(n)<n+2$$, which clearly doesn't hold for some large enough $$n$$ since the exponential function will eventually overtake the linear function. Hence we attain a contradiction, and $$f$$ does not exist. · 3 years, 2 months ago it's f(n)= n for all n>=2 · 3 years, 2 months ago Since $$f(n)$$ is defined on the positive integers and returns a positive integer for $$n \\ge 2$$, we have that operations contained within our function must always return positive integer values for any positive integer satisfying $$n \\ge$$. We know that the sum/difference of positive integers results a positive integer. The type of function that may add/subtract positive integers are polynomial functions. \"Daniel Liu\" has already shown that $$f(n)$$ is not of the form $$a_nx^{n}+a_{n-1}x^{n-1}+...+a_0$$. Now what other operations always return positive integer values for positive integers? Division won't work because if we have a rational function of the form $$\\frac{P(n)}{Q(n)}$$, where $$Q(n)$$ is not a factor of $$P(n)$$ (because then we would just get a polynomial instead of a rational function) and $$Q(n)$$ is non-constant, we will never always have a positive integer return since the denominator itself is", "Indeed, proving that $f(x)$ is odd or twice an odd is the key step. Thank you for your answer. – wythagoras May 3 '15 at 19:59 To see that such a function exists let $f$ be defined as follows, if $n=2k$ is even, let $f(n)=2f(k)$, if $n=4k+1$, let $f(n)$ be $4k+3$, and if $n=4k+3$, let $f(n)=2(4k+1)$. Then $f(f(2^j(4k+1)))=2^jf(f(4k+1))=2^jf(4k+3)=2^{j+1}(4k+1)$ as desired, and $f(f(2^j(4k+3)))=2^jf(f(4k+3))=2^jf(2(4k+1))=2^{j+1}(4k+3)$. Since any positive integer can be written in one of these forms for $j$ and $k$ nonnegative integers this function satisfies your property. In general this construction can be done in the same way if $g$ is an involution in the positive odd integers. $g$ pairs up the odd integers with each other let $S$ be a set containing one element in each pair. Then define $f(2k)=2f(k)$, and otherwise $f(n)=g(n)$ if $n\\in S$, and $f(n)=2g(n)$ if $n\\not\\in S$.", "# 2012 USAMO Problems/Problem 4 ## Problem Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$. ## Solution By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$. By the second condition, we have $$n\\cdot n!=(n+1)!-n! \\mid f(n+1)!-f(n)! \\qquad \\qquad (1)$$ for all positive integers $n$. Suppose that for some $n \\geq 2$ we have $f(n) = 1$. We claim that $f(k)=1$ for all $k\\ge n$. Indeed, from Equation (1) we have $f(n+1)!\\equiv 1 \\mod n\\cdot n!$, and this is only possible if $f(n+1)=1$; the claim follows by induction. We now divide into cases: Case 1: $f(1)=f(2)=1$ This gives $f(n)=1$ always from the previous claim, which is a solution. Case 2: $f(1)=2, f(2)=1$ This implies $f(n)=1$ for all $n\\ge 2$, but this does not satisfy the initial conditions. Indeed, we would have $$3-1 \\mid f(3)-f(1)$$ and so $2\\mid -1$, a contradiction. Case 3: $f(1)=1$, $f(2)=2$ We claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$. Fix $k > 1$ and suppose that $f(k)=k$. By Equation (1) we have that $$f(k+1)!", "$n$ that is greater than $n_0$ such that $g(n) > cf(n)$. Exercise 7.2 For each of the $g$ functions below, answer whether or not $g$ is in the set $O(n)$. Your answer should include a proof. If $g$ is in $O(n)$ you should identify values of $c$ and $n_0$ that can be selected to make the necessary inequality hold. If $g$ is not in $O(n)$ you should argue convincingly that no matter what values are chosen for $c$ and $n_0$ there are values of $n \\ge n_0$ such the inequality in the definition of $O$ does not hold. a. $g(n) = n + 5$ b. $g(n) = .01n$ c. $g(n) = 150n + \\sqrt{n}$ d. $g(n) = n^{1.5}$ e. $g(n) = n!$ Exercise 7.3 Given $f$ is some function in $O(h)$, and $g$ is some function not in $O(h)$, which of the following must always be true: a. For all positive integers $m$, $f(m) \\le g(m)$. b. For some positive integer $m$, $f(m) < g(m)$. c. For some positive integer $m_0$, and all positive integers $m > m_0$, $$f(m) < g (m).$$", "## A weird recurrence relation ### Problem 463 The function $f$ is defined for all positive integers as follows: • $f(1)=1$ • $f(3)=3$ • $f(2n)=f(n)$ • $f(4n + 1)=2f(2n + 1) - f(n)$ • $f(4n + 3)=3f(2n + 1) - 2f(n)$ The function $S(n)$ is defined as $\\sum_{i=1}^{n}f(i)$. $S(8)=22$ and $S(100)=3604$. Find $S(3^{37})$. Give the last 9 digits of your answer.", "## The function $$f$$ is defined for all positive integers $$n > 4$$ as $$f(n) = 3n - 9$$ if $$n$$ is odd and ##### This topic has expert replies Legendary Member Posts: 2276 Joined: 14 Oct 2017 Followed by:3 members ### The function $$f$$ is defined for all positive integers $$n > 4$$ as $$f(n) = 3n - 9$$ if $$n$$ is odd and by VJesus12 » Wed Feb 23, 2022 4:16 am 00:00 A B C D E ## Global Stats The function $$f$$ is defined for all positive integers $$n > 4$$ as $$f(n) = 3n - 9$$ if $$n$$ is odd and $$f(n) = 2n - 7$$ if $$n$$ is even. What is the value of the positive integer $$a?$$ (1) $$f(f(a)) = a$$ (2) $$f(f(f(a)))$$ is odd.", "interval $(1,3]$? – Siong Thye Goh Jan 22 at 10:07 The function is first defined on $$[-1,1]$$ and then extended to the whole line by adding the condition $$f(x+2)=f(x)$$ (periodocity of $$f$$). For example, if $$7 \\leq x \\leq 9$$ then $$x-8$$ lies between $$-1$$ and $$1$$ and $$f(x)$$ is defined to be $$f(x-8x)$$. [Note that $$f(x+2)=f(x)$$ for all $$x$$ implies that $$f(x+2n)=f(x)$$ for all $$x$$ and for any integer $$n$$]. • I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $f\\left(x + 2n\\right) = f\\left(x\\right)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates. – John Omielan Jan 22 at 7:43 • @JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding – Kabo Murphy Jan 22 at 7:47 • Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement \"... is defined to be $f\\left(x - 8x\\right)$\". As for your suggestion of removing" ]

[ "real numbers. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 - y^2) = x f(x) - y f(y)$ for all pairs of real numbers $x$ and $y$.", "# Functions are awesome!!! Algebra Level 4 Find the number of functions $$f : \\mathbb{R} \\rightarrow \\mathbb{R}$$ such that $f(x+y)=f(x)\\cdot f(y)\\cdot f(xy)$ for all $$x,y$$ in $$\\mathbb{R}$$. Note: $$\\mathbb{R}$$ denotes the set of real numbers. ×", "# Functions are awesome!!! -Part 3 Algebra Level 5 Find the number of functions such that $f:R➡R$ satisfying: $f(x+y)f(x-y)=(f(x)+f(y))^{2}-4x^{2}f(y)$ where $$x,y$$ belong to $$R$$ for all real numbers. ×", "# IMO Algebra Level 5 How many functions $$f: \\mathbb{R}\\rightarrow \\mathbb{R}$$ satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $$x$$ and $$y?$$ ×", "# Just Plug It In Algebra Level 5 A real function $$f : \\mathbb{R} \\to \\mathbb{R}$$ satisfies $$f(x^2 + 3x + 4) = 2x^2 + 6x + 10$$ for all real $$x$$. What is the value of $$f(0)$$? ×", "#### Function IMO ###### back to index Let $\\mathbb R$ be the set of real numbers. Determine all functions $f:\\mathbb R\\to\\mathbb R$ that satisfy the equation$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania Find all functions $f:\\mathbb Z\\rightarrow \\mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds: $f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ (Here $\\mathbb{Z}$ denotes the set of integers.) Proposed by Liam Baker, South Africa Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that, for all real numbers $x$ and $y$, $$f (f(x)f(y)) + f(x + y) = f(xy)$$", "# IMO 2017, Problem 2 Let \\mathbb{R} be the set of real numbers. Determine all functions f:\\,\\mathbb{R} \\rightarrow \\mathbb{R} such that, for any real numbers x and y, f(f(x)f(y)) + f(x+y) = f(xy). The solution can be found on a separate page.", "# Problem 40 * Find all functions f: \\mathbb R \\to \\mathbb R which satisfy f(x^3) + f(y^3) = (x + y)(f(x^2) + f(y^2) - f(xy)) for all real numbers x and y. [originally posted by und on TSR] 1 Like 2 Likes just realised I made a typo (wrote y^2 instead of y^3) but you spotted it, neat solution. 1 Like", "# 2012 IMO - Problem 4 Find all functions f:\\mathbb Z\\to\\mathbb Z such that, for all integers a , b , c with a+b+c=0 the following equality holds: f^{2}(a)+f^{2}(b)+f^{2}(c)=2f(a)f(b)+2f(b)f(c)+2f(c)f(a). Here \\mathbb Z is the set of all integers.", "# Existence of a great function Algebra Level 5 How many functions $$f : \\mathbb{R} \\to \\mathbb{R}$$ are there such that $$f(f(x)) = x^2 - 2$$ for all real $$x$$? ×", "# The Number of Functions Algebra Level 4 Find the number of functions $$f$$ from the positive reals to the positive reals such that $f(x+ f(y)) = \\frac{y}{xy+1}$ for all $$x, y > 0.$$ ×", "# Function Questions ## What is the formula to find number of one one function ? Solution : If A and B are two sets having m and n elements respectively such that m $$\\le$$ n, then the total number of one-one functions from A to B is $$^nC_m \\times m!$$ where m! is m factorial. For example, Let set A have 3 elements and set B have 4 elements, then … ## Find the domain of the function f(x) = $$1\\over x + 2$$. Solution : We have, f(x) = $$1\\over x + 2$$ Clearly f(x) assumes real values for all real values for all x except for the values of x satisfying x + 2 = 0 i.e. x = -2. Hence, Domain(f) = R – {-2} Similar Questions If y = 2[x] + 3 & y = … ## If y = 2[x] + 3 & y = 3[x – 2] + 5, then find [x + y] where [.] denotes greatest integer function. Solution : y = 3[x – 2] + 5 = 3[x] – 1 so 3[x] – 1 = 2[x] + 3 [x] = 4 $$\\implies$$ 4 $$\\le$$ x < 5 then y = 11 so x + y will lie in the interval [15, 16) so [x + y] = 15", "+0 0 208 3 The function $$f : \\mathbb{R} \\to \\mathbb{R}$$ satisfies $$f(x) f(y) = f(x + y) + xy$$ for all real numbers x and y. Find all possible functions f. I tried subsituting in simple values for x and y, but nothing works. A full explanation would be appreciated, as I am really confused! Thank you so much! Oct 24, 2021 #1 0 I checked if there were constant solutions, and there are none, so I don't think there are any other solutions. Oct 24, 2021 #2 +2401 0 I have no idea how to solve this, but I did find this. I hope this helps. :)) Let x = 0 f(0)*f(y) = f(y) + 0 f(0) = 1 =^._.^= Oct 25, 2021 #3 +117463 0 I do not know how to do it either but I would be interested in seeing the answer. Oct 25, 2021", "# A lovely function problem Algebra Level 5 A function $$f:\\mathbb{R} -\\{a_1,a_2\\} \\to \\mathbb{R},$$ where $$\\mathbb{R}$$ is the set of Real numbers, is defined by $f(x)=\\frac{Ax^2+6x-8}{A+6x-8x^2}$ How many integral values of $$A$$ exist for which $$f(x)$$ is onto? Details : $$a_1,a_2$$ are the roots of the quadratic equation $$A+6x-8x^2=0$$. ×", "$f(f(x))+af(x)=b(a+b)x.$ Problem 11 (Vietnam 2003) Let $$F$$ be the set of all functions $$f:\\mathbb{R}^+\\rightarrow\\mathbb{R}^+$$ which satisfy the inequality $$f(3x)\\geq f(f(2x))+x$$, for every positive real number $$x$$. Find the largest real number $$\\alpha$$ such that for all functions $$f\\in F$$: $$f(x)\\geq \\alpha\\cdot x$$. Problem 12 Find all functions $$f,g,h:\\mathbb{R}\\rightarrow\\mathbb{R}$$ that satisfy $f(x+y)+g(x-y)=2h(x)+2h(y).$ Problem 13 Find all functions $$f:\\mathbb{Q}\\rightarrow\\mathbb{Q}$$ for which $f(xy)=f(x)f(y)-f(x+y)+1.$ Solve the same problem for the case $$f:\\mathbb{R}\\rightarrow\\mathbb{R}$$. Problem 14 (IMO 2003, shortlist) Let $$\\mathbb{R}^+$$ denote the set of positive real numbers. Find all functions $$f:\\mathbb{R}^+\\rightarrow \\mathbb{R}^+$$ that satisfy the following conditions: • (i) $$f(xyz)+f(x)+f(y)+f(z)=f(\\sqrt{xy})f(\\sqrt{yz})f(\\sqrt{zx})$$ • (ii) $$f(x)< f(y)$$ for all $$1\\leq x< y$$. Problem 15 Find all functions $$f:[1,\\infty)\\rightarrow [1,\\infty)$$ that satisfy: • (i) $$f(x)\\leq 2(1+x)$$ for every $$x\\in [1,\\infty)$$; • (ii)$$xf(x+1)=f(x)^2-1$$ for every $$x\\in [1,\\infty)$$. Problem 16 (IMO 1999, probelm 6) Find all functions $$f:\\mathbb{R}\\rightarrow \\mathbb{R}$$ such that $f(x-f(y))=f(f(y))+xf(y)+f(x)-1.$ Problem 17 Given an integer $$n$$, let $$f:\\mathbb{R}\\rightarrow\\mathbb{R}$$ be a continuous function satisfying $$f(0)=0$$, $$f(1)=1$$, and $$f^{(n)}(x)=x$$, for every $$x\\in[0,1]$$. Prove that $$f(x)=x$$ for each $$x\\in[0,1]$$. Problem 18 Find all functions $$f: (0,+\\infty)\\rightarrow(0,+\\infty)$$ that satisfy $$f(f(x)+y)=xf(1+xy)$$ for all $$x,y\\in(0,+\\infty)$$. Problem 19 (Bulgaria 1998) Prove that there is no function $$f:\\mathbb{R}^+\\rightarrow\\mathbb{R}^+$$ such that $$f(x)^2\\geq f(x+y)(f(x)+y)$$ for every two positive real numbers $$x$$ and $$y$$. Problem 20 Let $$f:\\mathbb{N}\\rightarrow\\mathbb{N}$$ be a function satisfying $f(1)=2,\\quad f(2)=1,\\quad f(3n)=3f(n),\\quad f(3n+1)=3f(n)+2,\\quad f(3n+2)=3f(n)+1.$ Find", "# Functions Algebra Level 1 Let $f(x)=x^3$, where $x$ belongs to the real numbers. Find $f(f(x))$ at $x=2$. ×", "# Algebra of Real Function ## Algebra of Real Function Real Value Function: A function which has all real number or subset of the real number as it domain Real Valued Function: A function which has all real number or subset of the real number as it range For functions $f:X - > {\\bf{R}}$ and $g:X - > {\\bf{R}}$, we have $\\left( {f + g} \\right)\\left( x \\right) = f\\left( x \\right) + g\\left( x \\right),x \\in X$ 2. Substraction $\\left( {f - g} \\right)\\left( x \\right) = f\\left( x \\right)-g\\left( x \\right),x \\in X$ 3. Multiplication $\\left( {f.g} \\right)\\left( x \\right) = f\\left( x \\right).g\\left( x \\right),x \\in X$ 4. Multiplication by real number $\\left( {kf} \\right)\\left( x \\right) = kf\\left( x \\right),x \\in X$, where $k$ is a real number. 5. Division $\\frac{f}{g}\\left( x \\right) = \\frac{{f(x)}}{{g(x)}}$ $x \\in X$ and $g\\left( x \\right) \\ne 0$ ### Quiz Time Question 1. which is of the below relation is not a function. A) R = {(2,2),(2,4),(3,3), (4,4)} B) R = {(2,1),(3,1), (4,2)} C) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} D) None of the above Question 2. Let A = {1, 2, 3,4} and B = {5, 7}. Then possible number of relation from A to B ? A) 64 B) 256 C) 16 D) 32 Question 3.", "Find the range of each of the following functions. Question: Find the range of each of the following functions. (i) $f(x)=2-3 x, x \\in \\mathbf{R}, x>0$. (ii) $f(x)=x^{2}+2, x$, is a real number. (iii) $f(x)=x, x$ is a real number Solution: (i) $f(x)=2-3 x, x \\in \\mathbf{R}, x>0$ The values of $f(x)$ for various values of real numbers $x>0$ can be written in the tabular form as Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2. i.e., range of $f=(-\\infty, 2)$ Alter: Let $x>0$ $\\Rightarrow 3 x>0$ $\\Rightarrow 2-3 x<2$ $\\Rightarrow f(x)<2$ $\\therefore$ Range of $f=(-\\infty, 2)$ (ii) $f(x)=x^{2}+2, x$, is a real number The values of f(x) for various values of real numbers x can be written in the tabular form as Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. i.e., range of $f=[2, \\infty)$ Alter: Let x be any real number. Accordingly, $x^{2} \\geq 0$ $\\Rightarrow x^{2}+2 \\geq 0+2$ $\\Rightarrow x^{2}+2 \\geq 2$ $\\Rightarrow f(x) \\geq 2$ $\\therefore$ Range of $f=[2, \\infty)$ (iii) $f(x)=x, x$ is a real number It is clear that the range of $f$ is the set of all real numbers. $\\therefore$ Range of $f=\\mathbf{R}$", "function $f: \\mathbb{R} \\to \\mathbb{R}$ as a sum of a even and odd function by: $$f(x) = \\frac{f(x) + f(-x)}{2} + \\frac{f(x) - f(-x)}{2}$$ In many instances we try to draw analogies between operators and real/complex numbers. This often happens through the spectral theorem. -", "\\ \\ \\ \\ \\ \\ (\\because f(x) = 2-3x)$ Therefore, Range of function $f(x) = 2 -3x$ is $(-\\infty,2)$ Question:5 (ii) Find the range of each of the following functions $f ( x ) = x ^2 +2$ , x is a real number. Given function is $f ( x ) = x ^2 +2$ It is given that is a real number Now, $\\Rightarrow x^2 \\geq 0$ Add 2 on both the sides $\\Rightarrow x^2+2 \\geq 0+2$ $\\Rightarrow f(x) \\geq 2 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (\\because f(x) = x^2+2)$ Therefore, Range of function $f ( x ) = x ^2 +2$ is $[2,\\infty)$ Question:5 (iii) Find the range of each of the following functions. f (x) = x, x is a real number Given function is $f ( x ) = x$ It is given that is a real number Therefore, Range of function $f ( x ) = x$ is R ## NCERT solutions for class 11 maths chapter 2 relations and functions-Miscellaneous Exercise It is given that $f (x) = \\left\\{\\begin{matrix} x^2 & 0 \\leq x\\leq 3 \\\\ 3x &3 \\leq x \\leq 10 \\end{matrix}\\right.$ Now, $f(x) = x^2 \\ for \\ 0\\leq x\\leq 3$ And $f(x) = 3x \\ for \\ 3\\leq x\\leq" ]

[ "# Functional equation $4f(x^2+y^2)=(f(x)+f(y))^2$ Consider the following problem: Determine all functions $f:\\mathbb{N}\\rightarrow \\mathbb{N}$ that satisfy the functional equation $$4f(x^2+y^2)=(f(x)+f(y))^2.$$ So first of all plugging in $x=0=y$, we get that $4f(0)=4f(0)^2$. Hence $f(0)=0$ or $f(0)=1$. Now using $x=1$ and $y=0$, we get that $f(1)^2+(2f(0)-4)f(1)+f(0)^2=0$. Thus $f(1)=2-f(0)+2\\sqrt{1-f(0)}$. Hence if $f(0)=0$, then $f(1)=4$ and if $f(0)=1$, then $f(1)=1$. (Obviously we implicitly assumed that $f(1)\\neq 0$.) Now notice that plugging in $x=y$ yields $f(2x^2)=f(x)^2$. Let's assume that $f(0)=0$ and $f(1)=4$. Then also $4f(x^2)=f(x)^2$. From these equations we easily get that $f(2)=16, f(4)=\\frac{f(2)^2}{4}=64, f(8)=f(2)^2=256$, and so on. So we can find $f(2^n)$ using these techniques. We see that $f(2^n)=4^{n+1}$. From this one could guess that $f(x)=4x^2$ is a solution. Corr-blimey, $f(x)=4x^2$ is a solution! Now notice that $f(x)=0$ is a solution, $f(x)=1$ is also a solution and $f(x)=4x^2$ is a solution. I guess that these solutions are actually determined by there values on $0$ and $1$ for which we already found all possibilities. But how to proceed? Again assume that $f(0)=0$ and $f(1)=4$. We already know $f(2^n)$. How do we find $f(3)$? Well, notice that $4f(5)=4f(2^2+1^2)=(f(2)+f(1))^2=(16+4)^2=400$. Hence $f(5)=100=4\\cdot 5^2$. Now using the Pythagorean triple $(3,4,5)$ we find that $10000=f(5)^2=4f(5)^2=4f(3^2+4^2)=(f(3)+f(4))^2=f(3)^2+128f(3)+64^2$. Solving this equation yields $f(3)=36=4\\cdot 3^2$. Using the above procedure I can find the value of $f(n)$ for a lot of $n$, but I'm", "# 2002 USAMO Problems/Problem 4 ## Problem Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that $f(x^2 - y^2) = xf(x) - yf(y)$ for all pairs of real numbers $x$ and $y$. ## Solutions ### Solution 1 We first prove that $f$ is odd. Note that $f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$, and for nonzero $y$, $xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$, or $yf(-y) = -yf(y)$, which implies $f(-y) = -f(y)$. Therefore $f$ is odd. Henceforth, we shall assume that all variables are non-negative. If we let $y = 0$, then we obtain $f(x^2) = xf(x)$. Therefore the problem's condition becomes $f(x^2 - y^2) + f(y^2) = f(x^2)$. But for any $a,b$, we may set $x = \\sqrt{a}$, $y = \\sqrt{b}$ to obtain $f(a-b) + f(b) = f(a)$. (It is well known that the only continuous solutions to this functional equation are of the form $f(x) = kx$, but there do exist other solutions to this which are not solutions to the equation of this problem.) We may let $a = 2t$, $b = t$ to obtain $2f(t) = f(2t)$. Letting $x = t+1$ and $y = t$ in the original condition yields $\\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \\qquad", "with form $f(t) = at^2 + bt + 1$ ($+1$ since we know that $f(0) = 1$). Writing the functional equation as $(f(x) - 1)(f(y) - 1) = f(xy) - 1$, we get $(ax^2 + bx)(ay^2 + by) = (ax^2 y^2 + bxy)$, so comparing coefficients we get the following system: $a^2 = a$ $ab = 0$ $b^2 = b$ So this gives answers of $(1, 0), (0,1),$ and $(0,0)$. This gives answers of $f(x) = 1, f(x) = x+1, f(x) = x^2 + 1$. Using our initial condition $f(2) = 5$, we know that only the last equation can be correct. Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$. Now, suppose that $x_0 \\neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \\cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$ is a zero of $f(x)-1$ for all non-negative integers $k$. But $f(x)-1$ is a non-zero polynomial, and thus, can only have finitely many zeros, a contradiction. Therefore, the only zero of $f(x) - 1$ is $0$, i.e. $f(x)-1 = Cx^n$ for some constant $C$ and some non-negative integer $n$. Now, use the fact that $f(2x)-1 = 4(f(x)-1)$ to determine", "2020 at 03:32): The solution to problem 1: Let $y=f(x)=x \\log x$ for $x>0$, then $f(x)<0$ when $x<1$, $f(x)=0$ when $x=1$, and $f(x)>0$ when $x>1$. We have $f'(x)=1+\\log x$, which satisfies $f'(x)<0$ when $x<1/e$, and $f'(x)>0$ when $x>1/e$. Therefore $f(x)$ decreases as $x$ increases if $x\\in(0,1/e)$, $f(x)$ increases as $x$ increases if $x\\in(1/e, +\\infty)$, and $f(x)$ takes minimun at $x=1/e$ with value $-1/e$. Therefore if $a\\neq b$ such that $a^a=b^b$, then $f(a)=f(b)=:y$, we must have $y\\in(-1/e,0)$, $a,b\\in(0,1/e)\\cup(1/e,1)$, and • $b$ decreases and $t$ increases as $a$ increases, • either $a\\in(0,1/e),b\\in(1/e,1),t\\in(0,e^2)$, or $a\\in(1/e,1),b\\in(0,1/e),t\\in(e^2,+\\infty)$. • If $a > b$, then $1/\\sqrt{t}=\\sqrt{b^3/a} < \\sqrt{b^3/b}=b$ and hence $a+b > 2b > 2/\\sqrt{t}$. • If $a < b$, then $1/\\sqrt{t}=\\sqrt{b^3/a} > \\sqrt{b^3/b}=b$ and hence $a+b < 2b < 2/\\sqrt{t}$. The (1) comes from that since $t=4 < e^2$, we have $a < 1/e < b$, hence $a+b < 2/\\sqrt{t}=1$ and $a+b=tb^3+b>t/e^3+1/e=1/e+4/e^3$. #### Jz Pan (Nov 24 2020 at 03:42): The solution to problem 2: It's clear from the above discussion that the range of $t$ is $(e^2,+\\infty)$. Conversely, when $a$ goes up from $1/e$ to $1$ the $b$ goes down from $1/e$ to $0$, and the $t$ goes up from $e^2$ to $+\\infty$, hence given any $t>e^2$ there exist $a,b$ which satisfy the desired conditions. Since $t$ increases as $a$ increases, we only", "2017 IMO Problems/Problem 2 Let $\\mathbb{R}$ be the set of real numbers , determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$ Solution Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$. Let $f(0)=n$, so $f(n^{2})=0$. Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$. If $x+n^2=xn^2$, or $x=\\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0. This means that the only possible values of $n$ is -1,0, and 1. Go through the cases: $n=-1$ $f(x+1)=f(x)+1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=1$ $f(3)=2$ ... $f(x)=x-1$ $n=0$ $f(x)=0$ $n=1$ $f(x+1)=f(x)-1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=-1$ $f(3)=-2$ ... $f(x)=1-x$ So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.", "# IMO 2019, Problem 1 Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "# Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ Find all functions $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\\ge 0$ we have $f(a)>f(b)$ As much as I know: • putting $x=y=0$ we get $f(0)=0$. • let $x+y=0$, the we get $f(y)+f(-y)=\\frac{f(2y)}{2}$ Step 1: Prove : $f(y) = 0$ iff. $y=0$ It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$) Following proof is given by @Kyson To prove that $f(y)≠0$ for all$y<0.$ Assume there is a $y<0$ such that $f(y)=0.$ Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0$Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$ $\\$ Step 2: Prove $f(y)$ is even. Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\\frac{f(2y)}{2} = f(-y) +f(y) = \\frac{f(-2y)}{2}$ Thus $f(x)$is a even function, so $f(y)+f(-y)=2f(y)=\\frac{f(2y)}{2} \\Rightarrow f(2y)=4f(y)$ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\\infty,0)\\cup(0,\\infty)$) $\\$ Step 3: Prove:$f(y)=y^2f(1)$ By induction , find $f(ny)=n^2f(y)$ where $n \\in \\mathbb N$ Details of induction: Suppose $m\\in \\mathbb N ,m\\ge 2 ,\\ f(ny)=n^2f(y)$ holds for all $n\\le m$ Put $x=(m-1)y$ in property (a).$\\\\ \\Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\\\ \\Rightarrow f((m+1)y)=(m+1)^2f(y) \\ for\\ y\\not = 0 \\ and \\ it\\ still\\ holds\\ for\\ y=0$ Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$ Remark: Result induces $f(zy)=z^2f(y)$ where $z\\in \\mathbb Z$ , since $f$ is even. $\\$ $\\forall", "$$x\\mapsto x$$ and $$x\\mapsto -x$$. For this, assume that there are $$a,b\\in\\mathbb{R}\\backslash\\{0\\}$$ with $$f(a)=a$$ and $$f(b)=-b$$. Then $$P(a,b):\\quad f(a^2-b)=a^2+b$$ Now if $$f(a^2-b)=a^2-b$$ then $$a^2-b=a^2+b$$ and thus $$b=0$$, contradiction. But if $$f(a^2-b)=-(a^2-b)$$ then $$-(a^2-b)=a^2+b$$ and thus $$a=0$$, so again contradiction. Therefore, such $$a,b$$ don't exist, and thus either $$f(x)=x\\ \\forall x$$ or $$f(x)=-x\\ \\forall x$$, and one can verify easily that this are indeed solutions to the equation. Plugging in $$x=y=-1$$ shows that $$f(0)=0$$ and plugging in $$x=0$$ then shows that $$f(f(y))=y$$. This implies that $$f$$ is invertible, and plugging in $$y=0$$ shows that $$f(xf(x))=x^2=f(f(x^2)),$$ and hence $$xf(x)=f(x^2)$$ for all $$x\\in\\Bbb{R}$$. This shows that $$f(-x)=-f(x)$$ for all $$x\\in\\Bbb{R}$$, and that $$f(x^2+y)=xf(x)+f(y)=f(x^2)+f(y),\\tag{1}$$ for all $$x,y\\in\\Bbb{R}$$, from which it follows that $$f$$ satisfies Cauchy's functional equation $$f(x+y)=f(x)+f(y).$$ Much has been said about this functional equation, which has many pathological solutions. Note that this means $$f$$ is $$\\Bbb{Q}$$-linear, and if $$f$$ is either continuous at a point, bounded on an interval or monotonic on an interval, then $$f$$ is $$\\Bbb{R}$$-linear and so $$f(x)=cx$$ for some $$c\\in\\Bbb{R}$$. In an earlier version of this answer I rushed to the conclusion that $$f(x)=cx$$, which quickly implies that $$c=\\pm1$$ and indeed both functions $$f(x)=\\pm x$$ satisfy the functional equation. • I fail to see how you can conclude that $f(x)=\\pm x$ at the end. Apr 1,", "+ 3y^2 \\ne 0$ for any nonzero $y.$ Therefore, replacing $y \\to t/4$ in this equation, it follows that $f(t) = 0 \\implies f(2t) = 0.$ Step 5: If $f(a) = f(b) = 0$, then $f(b - a) = 0.$ $\\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \\tfrac{b - a}{2}$, so plugging $x, y$ into the given equation, we deduce that $f\\left(\\tfrac{b - a}{2}\\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$, as desired. Step 6: If $f \\not\\equiv 0$, then $f(t) = 0 \\implies t = 0.$ $\\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \\ne 0$ and $x + y = t.$ The following three facts are crucial: $\\indent$ 1. $f(y) \\ne 0.$ This is because $(x + y) - y = x$, so by Step 5, $f(y) = 0 \\implies f(x) = 0$, impossible. $\\indent$ 2. $f(x - 3y) \\ne 0.$ This is because $(x + y) - (x - 3y) = 4y$, so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \\implies f(4y) = 0 \\implies f(2y)", "# A Functional Equation from the 2012 IMO Here's Problem 4 from the 2012 IMO: Find all functions $f:\\mathbb Z\\to\\mathbb Z$ such that, for all integers $a$, $b$, $c$ with $a+b+c=0$ the following equality holds: $f^{2}(a)+f^{2}(b)+f^{2}(c)=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ Here $\\mathbb Z$ is the set of all integers. ### Solution The problem is reasonably standard. Setting first $a = b = c = 0$, we get $3f^{2}(0)=6f^{2}(0),$ implying $f(0)=0$. Next, let say $c=0$, so that $b=-a$: $f^{2}(a)+f^{2}(-a)=2f(a)f(-a)$. In other words, $(f(a) - f(-a))^{2}=0$, so that $f(a)=f(-a)$, making $f$ an even function. Using that and setting $a=b=1$ and $c=-2$ gives $2f^{2}(1) + f^{2}(2)=4f(2)f(1) + 2f^{2}(1)$, which is equivalent to $f(2)(f(2)-4f(1))=0$. Thus one of the two: either $f(2)=0$ or $f(2)=4f(1)$. We'll have to consider the two cases separately. Note that we can safely assume that $f(1)\\ne 0$, for $f(0)=0$ would imply that $f$ is identically $0$, which is an obvious solution. To see that, take successively $a=b=1$, $c=2$ which gives $f(2)=0$, and then $a=1$, $b=2$, $c=3$, to obtain $f(3)=0$, and so on. ### Case 1: $f(2)=0$ Clearly also $f(4)=0$ and $f(6)=0$, and so on. More generally, $f(2n)=0$, for all $n\\in \\mathbb Z$. This gives a good reason to suspect that this is an indication that $f$ is periodic. Let's check: $f^{2}(a+2)+f^{2}(a)+f^{2}(2)=2f(a+2)f(a)+2f(a)f(2)+2f(a+2)f(2)$. But, since $f(2)=0$, this reduces to $f^{2}(a+2)+f^{2}(a)=2f(a+2)f(a)$ or $(f(a+2)-f(a))^{2}=0$, Telling us that", "n\\}$ and let us prove it for $n+1$. We place $a=1$, $b=n$, $c=-n-1$ in the original equation to obtain: $f(1)^2+n^4f(1)^2+f(n+1)^2=2n^2f(1)^2+2(n^2+1)f(n+1)f(1)$ $\\Leftrightarrow \\quad\\quad\\quad \\left(f(n+1)-(n+1)^2f(1)\\right)\\cdot \\left(f(n+1)-(n-1)^2f(1)\\right)=0.$ If $f(n+1)=(n-1)^2f(1)$ then setting $a=n+1$, $b=1-n$, and $c=-2$ in the original equation yields $2(n-1)^4f(1)^2+16f(1)^2=2\\cdot 4\\cdot 2(n-1)^2f(1)^2+2\\cdot (n-1)^4f(1)$ which implies $(n-1)^2=1$ hence $n=2$. Therefore $f(3)=f(1)$. Placing $a=1$, $b=3$, and $c=4$ into the original equation implies that $f(4)=0$ or $f(4)=4f(1)=f(2)$. If $f(4)\\neq 0$ we get $f(2)^2+f(2)^2+f(4)^2=2f(2)^2+4f(2)f(4)$ hence $f(4)=4f(2)$. We already have that $f(4)=f(2)$ and this implies that $f(2)=0$, which is impossible according to our assumption. Therefore $f(4)=0$ and the function $f$ has period $4$. Then $f(4k)=0$, $f(4k+1)=f(4k+3)=c$, and $f(4k+2)=4c$. It is easy to verify that this function satisfies the requirements of the problem. Thus the solutions are: $f(x)=cx^2$ for some $c\\in\\mathbb Z$; $f(x)=\\left\\{\\begin{array}{rl}0,& 2\\mid n,\\\\ c, & 2\\not\\mid n\\end{array}\\right.$ for some $c\\in\\mathbb Z$; and $f(x)=\\left\\{\\begin{array}{rl}0,& 4\\mid n,\\\\ c, & 2\\nmid n,\\\\ 4c, & n\\equiv 2 \\mbox{ (mod } 4)\\end{array}\\right.$ for some $c\\in\\mathbb Z$. Problem 5 Given a triangle $ABC$, assume that $\\angle C=90^{\\circ}$. Let $D$ be the foot of the perpendicular from $C$ to $AB$, and let $X$ be any point of the segment $CD$. Let $K$ and $L$ be the points on the segments $AX$ and $BX$ such that $BK=BC$ and $AL=AC$, respectively. Let $M$ be the intersection of $AL$ and $BK$. Prove that $MK=ML$.", "function. Putting $$x=0$$ gives $$f(yf(0))=0$$ for any $$y\\in\\mathbb{R}$$ so $$f(0)=0$$. If $$f$$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $$f$$ is not zero at some point. Setting $$y=0$$, $$f(x^2)=xf(x).$$ Let $$x_0$$ be a real number for which $$f(x_0)=0$$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$x_0 f(x_0 +y) =0$$ for all $$y\\in\\mathbb{R}$$. Thus $$x_0 =0$$ since f is not identically zero. Indeed, $$f(x)=0 \\iff x=0.$$ Now setting $$y=-x$$ we have for any $$x\\in\\mathbb{R}$$, $$f(x^2-xf(x))=0$$ and so $$x^2-xf(x)=0$$ or $$f(x)=x.$$", "a solution, then the equation gives $$c^2=2c\\implies c\\in \\{0,2\\}$$. So from now on let's assume $$f$$ is non-constant. If some $$x$$ satisfies $$f(x)=0$$, then plugging that in, $$f(xy)=-f(x^2)$$, and since $$f$$ is non-constant, $$x$$ must be zero. Also, putting $$x=0$$ gives $$f(0)f(f(0)+y)=2f(0)$$, and again $$f(0)\\ne 0$$ leads to $$f$$ being constant, so in fact $$f(x)=0\\iff x=0$$. Plugging $$y=0$$ now gives $$f(x^2)=f(x)f(f(x))$$, so the equation can be written $$f(x)f(f(x)+y)=f(x)f(f(x))+f(xy)$$. Call this alternate version $$(\\star)$$. Now we claim that $$f$$ is injective. Indeed, suppose $$f(a)=f(b)$$ with $$a\\ne b$$. As seen already, $$a,b$$ are non-zero. Now substituting $$x\\mapsto a$$ and $$x\\mapsto b$$ in $$(\\star)$$, we see that all the terms except $$f(xy)$$ have the same value in both cases. Thus $$f(ay)=f(by)$$ for all reals $$y$$. This shows that $$f$$ takes the same value on any two reals with ratio $$a/b$$. Now plugging $$x=1$$ in the original equation gives $$f(1)f(f(1)+y)=f(1)+f(y)$$. Using this equation for $$y\\mapsto ay$$ and $$y\\mapsto by$$, we see that in fact $$f(f(1)+ay)=f(f(1)+by)$$, and given any $$c\\ne a/b$$, one can choose $$y$$ so that the numbers $$f(1)+ay$$ and $$f(1)+by$$ have ratio $$c$$. Thus given any ratio, one can find two reals with that ratio so that $$f$$ has the same value on them. And thus, to prove injectivity, Note that given any non-zero $$x_1,x_2$$, we can find $$c,d$$ so that", "$0$ for all rationals, it is identically $0$, so $$f(x^i)=f(1)^{n-i}f(x)^i$$ for all $0\\leq i\\leq n$. Originally, we had $f(1)=f(1)^n$, so $f(1)\\in\\{-1,0,1\\}$. If $f(1)=0$, we have $f(x^i)=0$, so $\\boxed{f(x)\\equiv 0}$. Otherwise, we have $$f(x^2)=f(1)^{n-2}f(x)^2=f(1)f(x)^2$$ $$f(x+y^2)=f(x)+f(y^2)=f(x)+f(1)f(y)^2.$$ If $f(1)=1$, this means $f$ is increasing, and if $f(1)=-1$ this means $f$ is decreasing. Either way, $f$ is not everywhere dense, so $f(x)=cx$ for some $c$ and all $x$. The observation that $f(1)=\\pm 1$ means $\\boxed{f(x)=x}$ and $\\boxed{f(x)=-x}$ are our only other solutions. • In $f(x^{2})=f(1)^{n-2}f(x)^{2}=f(1)f(x)^{2}$ when $f(1)\\neq 0$ how does the second equality follow? Sorry if I am missing something obvious. – Kavi Rama Murthy Jul 16 '18 at 5:46 • As $f(1)\\in\\{-1,0,1\\}$ and $f(1)\\neq 0$, we have $|f(1)|=1\\implies f(1)^2=1$; since $n$ is odd, $$f(1)^{n-2}=\\left(f(1)^2\\right)^{\\frac{n-3}{2}}f(1)=1^{\\frac{n-3}{2}}f(1)=f(1).$$ – Carl Schildkraut Jul 16 '18 at 17:02 • Nice and simpler than mine! – Sungjin Kim Jul 16 '18 at 18:27 • @KaviRamaMurthy The second equality is by $f(y^2)=f(1)f(y)^2$. – Sungjin Kim Jul 16 '18 at 18:27 This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\\in\\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you. As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\\in \\mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\\in\\mathbb{Q}$, $x\\in\\mathbb{R}$. Since $f( (x+y)^5)= f(x+y)^5", "f(x)=-x$. Now we'll show that $\\forall x\\in\\Bbb{R}, f(x)=-x$: Let $x\\in\\Bbb{R}$, there are three cases: $x\\in(0,\\infty)$ or $x=0$ or $x\\in(-\\infty,0)$. If $x\\in(0,\\infty)$ then we've already shown that $f(x)=-x$. If $x=0$ then $f(x)=f(0)=0=-0=-x$. If $x\\in(-\\infty, 0)$ then $-x\\in(0,\\infty)$ and so $f(-x)=-(-x)=x$ now since $f$ is odd we get that $f(-x)=-f(x)$ and so $-f(x)=x$ which implies that $f(x)=-x$. And so the only function that satisfies the conditions in the question is $f(x)=-x$ as was to be shown. Q.E.D. Select a point $x \\in \\mathbb R$. There exists $n_1$ with the property that $$f^{n_1}(x) = -x.$$ There exists $n_2$ with the property that $$f^{n_1 + n_2}(x) = f^{n_2}(f^{n_1}(x)) = f^{n_2}(-x) = -(-x) = x.$$ Repeat to get $f^{2(n_1 + n_2)}(x) = x$. The function $f$ is either increasing or decreasing. Thus $f^{2}$ is increasing. If $f^2(x) > x$ then $f^4(x) > f^2(x) > x$ and so on until you get $f^{2(n_1 + n_2)}(x) > x$. Likewise, if $f^2(x) < x$ you get $f^{2(n_1 + n_2)}(x) < x$. Either way is a contradiction. Thus $f^2(x) = x$. Any monotone function with the IVP is continuous. Can you determine all monotone continuous functions satisfying $f^2(x) = x$? • If we didn't have $f^n(x)=-x$, there would be $f(x)=x^3(x>0), f(x)=x^{1/3}(x<=0)$. – Empy2 Sep 29 '14 at 15:57", "$f(-y)f(y)=(f(y))^2$, so if $f(y) \\ne 0$, then $f(y)=f(-y)$. Furthermore, setting $y=-x$ gives us $(f(x)-x^2) \\cdot f(4x) + (f(-x)-x^2) \\cdot f(4x) = f(0) = 0$. The LHS can be factored as $(f(x)+f(-x)-2x^2) \\cdot f(4x) = 0$. Therefore, if $f(4x) \\ne 0$, then we have $f(x)+f(-x)-2x^2=0$, or $f(x)+f(-x)=2x^2$. However, since from step 2 we have that $f(x)=f(-x)$ assuming f(x) \\ne 0$, the equation becomes$2f(x)=2x^2$, so$f(x)=0, x^2$are the only solutions. ==Solution 2== Step 1:$x=y=0 \\implies f(0)=0$Step 2:$x=0 \\implies f(y)f(-y)=f(y)^{2}$. Now, assume$y \\not = 0$. Then, if$f(y)=0$, we substitute in$-y$to get$f(y)f(-y)=f(-y)^{2}$, or$f(y)=f(-y)=0$. Otherwise, we divide both sides by$f(y)$to get$f(y)=f(-y)$. If$y=0$, we obviously have$f(0)=f(0)$. Thus, the function is even. . Step 3:$y=-x \\implies 2f(4x)(f(x)-x^{2})=0$. Thus,$\\forall x$, we have$f(4x)=0$or$f(x)=x^{2}$. Step 4: We now assume$ (Error compiling LaTeX. ! Missing $inserted.)f(x) \\not = 0$,$x\\not = 0$. We have$f(\\frac{x}{4})=\\frac{x^{2}}{16}$. Now, setting$x=y=\\frac{x}{4}$, we have$f(\\frac{x}{2}=\\frac{x^{2}}{4}$or$f(\\frac{x}{2})=0$. The former implies that$f(x)=0$or$x^{2}$. The latter implies that$f(x)=0$or$f(x)=\\frac{x^{2}}{2}$. Assume the latter.$y=-2x \\implies -\\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\\frac{x^{2}}{2}$. Clearly, this implies that$f(x)$is negative for some$m$. Now, we have$f(\\frac{m}{4})=\\frac{m^{2}}{16} \\implies f(\\frac{m}{2})=0,\\frac{m^{2}}{4} \\implies f(m) \\geq 0$, which is a contradiction. Thus,$\\forall x$$(Error compiling LaTeX. ! Missing$ inserted.)f(x)=0$or$f(x)=x^{2}$. Step 5: We now assume$ (Error compiling LaTeX. ! Missing $inserted.)f(x)=0$,$f(y)=y^{2}$for some$x,y \\not = 0$. Let$m$be sufficiently large integer, let$z=|4^{m}x|$and take the absolute value of$y$(since the function is even). Choose$c$such that$3z-c=y$. Note that we have$\\frac{c}{z}$~$3$and$\\frac{y}{z}$~$0$. Note that$f(z)=0$. Now,$x=z, y=c \\implies$LHS is positive, as", "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "the case where $$0\\in\\mathbb N$$. Firstly, if $$f(x)=f(y)$$, then $$2x=f(2f(x))=f(2f(y))=2y,$$ so $$x=y$$; i.e., $$f$$ is injective. Now, consider any $$a,b,c\\in\\mathbb N$$ with $$a+b>c$$. Then $$f(f(a)+f(b))=a+b=c+(a+b-c)=f(f(c)+f(a+b-c)),$$ which implies that $$f(a)+f(b)=f(c)+f(a+b-c).$$ In particular, let $$f(1)=t$$ and $$f(2)=t+u$$. We claim that $$f(n)=t+u(n-1)$$ for all $$n$$, which can be proven by strong induction. It holds for $$n\\in\\{1,2\\}$$, and for $$n\\geq 3$$, $$f(n)+f(1)=f(2)+f(n-1),$$ which implies $$f(n)=f(n-1)+f(2)-f(1)=t+u(n-2)+t+u-t=t+u(n-1),$$ as desired. This means that $$m+n=f(f(m)+f(n))+f(2t+u(m+n-2))=t+u(2t+u(m+n-2)).$$ Since $$u\\geq 0$$, we must have $$u=1$$, since the left side is $$m+n$$ and the right side is $$u^2(m+n)$$ plus some constant independent of $$m$$ and $$n$$. Then, we need $$m+n=t+2t+m+n-2,$$ which gives $$t=1$$. As a result, the only working function is the identity. This is not an answer, mostly just results I found interesting, feel free to edit this post to complete it. I'm going to assume there's no $$0$$ in $$\\mathbb{N}$$ (if there was then the function would not be well-defined unless $$f(0)=0$$, if $$0 \\mapsto e \\not=0$$ then $$2e \\mapsto 0$$ and then $$0\\color{red}{\\mapsto} 4e \\not= e$$): $$f$$ by definition must be injective, injective basically means that $$f$$ doesn't map different elements in the domain $$(\\mathbb{N})$$ to the same element in the range/codomain $$(\\mathbb{N})$$. Say you had a function $$g$$ s.t. $$g(1)=1$$ and $$g(2)=1$$, that cannot happen if $$g$$ is injective because $$g$$ doesn't map different elements $$(1\\not=2)$$", "Thanks for reading! ## IMO 2019, Problem 1 Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "not 0, $f(2x)$ is 0 for all $x$ (including 0). Now either $f(x)$ is 0 for all $x$, or there exists some $m \\neq 0$ such that $f(m) = m^2$. Then $m$ must be odd. We can let $x = 2k$ in the original equation, and since $f(2x)$ is 0 for all $x$, stuff cancels and we get: $$y^2f(4k - f(y)) = f(yf(y))$$ [b]for $k \\neq 0$.[/b] Now, let $y = m$ and we get: $$m^2f(4k - m^2) = f(m^3)$$ Now, either both sides are 0 or both are equal to $m^6$. If both are $m^6$ then: $$m^2(4k - m^2)^2 = m^6$$ which simplifies to: $$4k - m^2 = \\pm m^2$$ Since $k \\neq 0$ and $m$ is odd, both cases are impossible, so we must have: $$m^2f(4k - m^2) = f(m^3) = 0$$ Then we can let $k$ be anything except 0, and get $f(x)$ is 0 for all $x \\equiv 3 \\pmod{4}$ except $-m^2$. Also since $x^2f(-x) = f(x)^2$, we have $f(x) = 0 \\Rightarrow f(-x) = 0$, so $f(x)$ is 0 for all $x \\equiv 1 \\pmod{4}$ except $m^2$. So $f(x)$ is 0 for all $x$ except $\\pm m^2$. Since $f(m) \\neq 0$, $m = \\pm m^2$. Squaring, $m^2 = m^4$ and dividing by $m$, $m = m^3$. Since $f(m^3) = 0$, $f(m) = 0$," ]

[ "# 2002 USAMO Problems/Problem 4 ## Problem Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that $f(x^2 - y^2) = xf(x) - yf(y)$ for all pairs of real numbers $x$ and $y$. ## Solutions ### Solution 1 We first prove that $f$ is odd. Note that $f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$, and for nonzero $y$, $xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$, or $yf(-y) = -yf(y)$, which implies $f(-y) = -f(y)$. Therefore $f$ is odd. Henceforth, we shall assume that all variables are non-negative. If we let $y = 0$, then we obtain $f(x^2) = xf(x)$. Therefore the problem's condition becomes $f(x^2 - y^2) + f(y^2) = f(x^2)$. But for any $a,b$, we may set $x = \\sqrt{a}$, $y = \\sqrt{b}$ to obtain $f(a-b) + f(b) = f(a)$. (It is well known that the only continuous solutions to this functional equation are of the form $f(x) = kx$, but there do exist other solutions to this which are not solutions to the equation of this problem.) We may let $a = 2t$, $b = t$ to obtain $2f(t) = f(2t)$. Letting $x = t+1$ and $y = t$ in the original condition yields $\\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \\qquad", "2017 IMO Problems/Problem 2 Let $\\mathbb{R}$ be the set of real numbers , determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$ Solution Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$. Let $f(0)=n$, so $f(n^{2})=0$. Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$. If $x+n^2=xn^2$, or $x=\\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0. This means that the only possible values of $n$ is -1,0, and 1. Go through the cases: $n=-1$ $f(x+1)=f(x)+1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=1$ $f(3)=2$ ... $f(x)=x-1$ $n=0$ $f(x)=0$ $n=1$ $f(x+1)=f(x)-1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=-1$ $f(3)=-2$ ... $f(x)=1-x$ So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.", "# Functional equation $4f(x^2+y^2)=(f(x)+f(y))^2$ Consider the following problem: Determine all functions $f:\\mathbb{N}\\rightarrow \\mathbb{N}$ that satisfy the functional equation $$4f(x^2+y^2)=(f(x)+f(y))^2.$$ So first of all plugging in $x=0=y$, we get that $4f(0)=4f(0)^2$. Hence $f(0)=0$ or $f(0)=1$. Now using $x=1$ and $y=0$, we get that $f(1)^2+(2f(0)-4)f(1)+f(0)^2=0$. Thus $f(1)=2-f(0)+2\\sqrt{1-f(0)}$. Hence if $f(0)=0$, then $f(1)=4$ and if $f(0)=1$, then $f(1)=1$. (Obviously we implicitly assumed that $f(1)\\neq 0$.) Now notice that plugging in $x=y$ yields $f(2x^2)=f(x)^2$. Let's assume that $f(0)=0$ and $f(1)=4$. Then also $4f(x^2)=f(x)^2$. From these equations we easily get that $f(2)=16, f(4)=\\frac{f(2)^2}{4}=64, f(8)=f(2)^2=256$, and so on. So we can find $f(2^n)$ using these techniques. We see that $f(2^n)=4^{n+1}$. From this one could guess that $f(x)=4x^2$ is a solution. Corr-blimey, $f(x)=4x^2$ is a solution! Now notice that $f(x)=0$ is a solution, $f(x)=1$ is also a solution and $f(x)=4x^2$ is a solution. I guess that these solutions are actually determined by there values on $0$ and $1$ for which we already found all possibilities. But how to proceed? Again assume that $f(0)=0$ and $f(1)=4$. We already know $f(2^n)$. How do we find $f(3)$? Well, notice that $4f(5)=4f(2^2+1^2)=(f(2)+f(1))^2=(16+4)^2=400$. Hence $f(5)=100=4\\cdot 5^2$. Now using the Pythagorean triple $(3,4,5)$ we find that $10000=f(5)^2=4f(5)^2=4f(3^2+4^2)=(f(3)+f(4))^2=f(3)^2+128f(3)+64^2$. Solving this equation yields $f(3)=36=4\\cdot 3^2$. Using the above procedure I can find the value of $f(n)$ for a lot of $n$, but I'm", "# 2001 IMO Shortlist Problems/A4 ## Problem Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, satisfying $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$. ## Solution Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\\ \\forall x$, so $f(x) = 0,\\ \\forall x$. This is a solution, so we can take it out of the way: assume $f(1)\\ne 0$. $y = 1\\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\\in\\{0,f(1)x\\}$. In particular, $f(0) = 0$. Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\\ \\forall x$. This means that $f(a),f(b)\\ne 0\\Rightarrow f\\left(\\frac ab\\right)\\ne 0\\ (*)$ ($\\frac ab$ is defined because $f(b)\\ne 0\\Rightarrow b\\ne 0$). Assume now that $x\\ne y$ and $f(x),f(y)\\ne 0$. We get $f(x) = f(1)x,\\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\\ne 0$, so $x\\ne y,\\ f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (**)$. Assume now $f(x)\\ne 0$. From $(*)$ we get $f\\left(\\frac 1x\\right)\\ne 0$, and after applying $(*)$ again to $a = x,b = \\frac 1x$ we get $f(x^2)\\ne 0\\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (\\#)$. Let $G = \\{x\\in\\mathbb R|f(x)\\ne 0\\}$. $(*)$ and $(\\#)$ simply say that $(G,\\ \\cdot)$ is a subgroup of $(\\mathbb R^{*},\\", "numerator . . . Wow! . . . How did I miss that? . . . *blush* 9. Originally Posted by linda2005 2) let $x,y,z > 0$, prove that : $\\frac{xy}{z} + \\frac{yz}{x} + \\frac{zx}{y} \\geq x + y + z$ expanding $(a-b)^2 + (b-c)^2 + (c - a)^2 \\geq 0$ gives us $a^2 + b^2 + c^2 \\geq ab + bc + ac.$ now put $a=xy, \\ b = yz, \\ c=xz$ and then divide by $xyz.$ 3) Find all function $f: \\mathbb{R} \\to \\mathbb{R}$ such that : $f\\left( xf(x) + f(y) \\right ) = \\left(f(x)\\right)^2 +y. \\ \\ \\ \\ \\ (*)$ this question is not bad! if in $(*)$ we put x = 0 we'll get $f(f(y))=(f(0))^2 + y. \\ \\ \\ \\ \\ \\ (**)$ 1) $f$ is onto: suppose $z \\in \\mathbb{R}$ is given. put $y=z - (f(0))^2.$ then by $(**): \\ f(f(y))=z.$ 2) $f(0)=0$: by 1) there exists $x$ with $f(x)=0.$ then by $(*): \\ f(f(y))=y,$ for all $y \\in \\mathbb{R}.$ let's call this result $(***).$ now $(**)$ gives us $f(0)=0.$ 3) $\\forall x \\in \\mathbb{R}: \\ (f(x))^2 = x^2$: if in $(*)$ we put y = 0, we'll get $f(xf(x))=(f(x))^2.$ again if in $(*)$ we let y = 0 and replace x with f(x) and use $(***)$ we'll get: $f(xf(x))=x^2.$ 4)", "# IMO 2019, Problem 1 Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "Z^+$, $f(y) = n$. Thus $(x,y) \\in \\mathbb Z^+ \\times \\mathbb Z^+$ and $g(x,y) = (m,n)$. Thus $g$ is onto. Thus $\\mathbb Z^+ \\times \\mathbb Z^+ \\sim \\mathbb Q \\times \\mathbb Q$. Since $\\mathbb Z^+ \\sim \\mathbb Z^+ \\times \\mathbb Z^+$ and $\\sim$ is a transitive relation, it follows that $\\mathbb Z^+ \\sim \\mathbb Q \\times \\mathbb Q$. Thus $\\mathbb Q \\times \\mathbb Q$ is denumerable. (b) From Section-6.5 Ex-6 part(b), we know that $\\mathbb Q(\\sqrt 2) = \\{ a + b \\sqrt 2 \\; \\vert \\; a \\in \\mathbb Q, b \\in \\mathbb Q \\}$. Suppose $S = \\mathbb Q(\\sqrt 2)$. Thus we can define a function $f: \\mathbb Q \\times \\mathbb Q \\to S$ such that $f(a,b) = a + b \\sqrt 2$. Clearly $f$ is onto by definition as $Ran(f) = S$. Suppose $a_1, b_1, a_2, b_2 \\in \\mathbb Q$ such that $f(a_1,b_1) = f(a_2,b_2)$. Thus $a_1 + b_1 \\sqrt 2 = a_2 + b_2 \\sqrt 2$, or $(a_1 - a_2) = \\sqrt 2 (b_2 - b_1)$. Since $\\sqrt 2 \\notin \\mathbb Q$, it follows that $a_1 = a_2$ and $b_1 = b_2$. Thus $f$ is one-to-one and onto. Thus $\\mathbb Q \\times \\mathbb Q \\sim S$. Now from part(a) of this solution, $Z^+ \\sim \\mathbb Q \\times \\mathbb Q$, and since $\\sim$ is transitive", "Z^+$, $f(y) = n$. Thus $(x,y) \\in \\mathbb Z^+ \\times \\mathbb Z^+$ and $g(x,y) = (m,n)$. Thus $g$ is onto. Thus $\\mathbb Z^+ \\times \\mathbb Z^+ \\sim \\mathbb Q \\times \\mathbb Q$. Since $\\mathbb Z^+ \\sim \\mathbb Z^+ \\times \\mathbb Z^+$ and $\\sim$ is a transitive relation, it follows that $\\mathbb Z^+ \\sim \\mathbb Q \\times \\mathbb Q$. Thus $\\mathbb Q \\times \\mathbb Q$ is denumerable. (b) From Section-6.5 Ex-6 part(b), we know that $\\mathbb Q(\\sqrt 2) = \\{ a + b \\sqrt 2 \\; \\vert \\; a \\in \\mathbb Q, b \\in \\mathbb Q \\}$. Suppose $S = \\mathbb Q(\\sqrt 2)$. Thus we can define a function $f: \\mathbb Q \\times \\mathbb Q \\to S$ such that $f(a,b) = a + b \\sqrt 2$. Clearly $f$ is onto by definition as $Ran(f) = S$. Suppose $a_1, b_1, a_2, b_2 \\in \\mathbb Q$ such that $f(a_1,b_1) = f(a_2,b_2)$. Thus $a_1 + b_1 \\sqrt 2 = a_2 + b_2 \\sqrt 2$, or $(a_1 - a_2) = \\sqrt 2 (b_2 - b_1)$. Since $\\sqrt 2 \\notin \\mathbb Q$, it follows that $a_1 = a_2$ and $b_1 = b_2$. Thus $f$ is one-to-one and onto. Thus $\\mathbb Q \\times \\mathbb Q \\sim S$. Now from part(a) of this solution, $Z^+ \\sim \\mathbb Q \\times \\mathbb Q$, and since $\\sim$ is transitive", "Thanks for reading! ## IMO 2019, Problem 1 Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "# Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt – Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\\since $f(0)=0$) Finally putting $y=(x+1)$ gives us $f(x)=0$. This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one? #### Solutions Collecting From Web of \"Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$\" You proved that $f(0)=0$, we can continue from here Suppose that there exist $k \\neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$. Now suppose that there exist no $k \\neq 0$ such that $f(k)=0$. Then plugging", "for all $x\\in\\mathbb{N}$, and b)$f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x$, $y\\in\\mathbb{N}$. 25. Consider those functions $f:\\mathbb{N}\\to\\mathbb{N}$ which satisfy the condition $f(m+n) \\geq f(m)+f(f(n))-1$ for all $m,n\\in\\mathbb{N}.$ Find all possible values of $f(2007).$ 26. Find all surjective functions $f:\\mathbb{N}\\to\\mathbb{N}$ such that for every $m,n\\in\\mathbb{N}$ and every prime $p,$ the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p.$ 27. Find all real polynomials $f$ such that $2yf(x+y)+(x-y)(f(x)+f(y)) \\geq 0\\forall x,y\\in\\mathbb{R}.$ 28. Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ with $x,y\\in\\mathbb{R}$ such that $f(x-f(y))=f(x+y)+f(y).$ 29. Show that for positive integer $n$, and for $x\\not =0$, $\\left(x^{n-1}\\sin\\dfrac{1}{x}\\right)^{(n)}=\\dfrac{(-1)^n}{x^{n+1}}\\sin\\left(\\dfrac{1}{x}+\\dfrac{n\\pi}{2}\\right).$ 30. Find all $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xy+f(x))=xf(y)+f(x)$ for every pair of real numbers $x,y$.", "to see that the function $f(x)=x$ for all $x \\in \\mathbb{R}$ and the function $f(x)=-x$ for all $x \\in \\mathbb{R}$ satisfy (1). We show that these are the only solutions. Let $a=f(0)$. Setting $x=0$ in (1) we get $$\\tag{2} f(f(y))=y+a^2 \\,.$$ Pick $b=f(-a^2)$. Then $f(b)=f(f(-a^2))=-a^2+a^2=0$. Setting $x=b$ in (1) we get $$\\tag{3} f(f(y))=y \\,.$$ By combining (2) and (3) we get $a=0$, and hence $f(0)=0$. Setting $y=0$ in (1) we get $$\\tag{4} f(xf(x))=\\left( f(x) \\right)^2 \\,.$$ Now, let $t \\in \\mathbb R$ be arbitrary. Setting $x=f(t)$ in (4), and using the fact that by (3) we have $f(x)=t$ we get $$\\tag{5} f(tf(t))=\\left( f(f(t)) \\right)^2=t^2 \\,.$$ Therefore, for all $x \\in \\mathbb R$ we have by (4) and (5) $$\\tag{6} x^2=f(xf(x))=(f(x))^2 \\,.$$ This gives that $f(x)=\\pm x$ for each $x \\in \\mathbb{R}$. Therefore, for each $x \\in \\mathbb{R}$ we have either $f(x)=x$ or $f(x)=-x$. To complete the solution we need to show that the sign is the same for all $x$. Assume by contradiction that there exist non-zero numbers $\\alpha, \\beta$ such that $f(\\alpha)=\\alpha$ and $f(\\beta)=-\\beta$. Then, setting $x=\\alpha, y=\\beta$ in (1) we get $$f \\left(\\alpha^2-\\beta \\right)= \\alpha^2+\\beta \\,,$$ and hence, by (6) $$\\left(\\alpha^2-\\beta \\right)^2=\\left(\\alpha^2+\\beta \\right)^2 \\,.$$ Expanding, we get $\\alpha^2 \\beta=0$ which contradicts the fact that $\\alpha, \\beta$ are non-zero. Since we got a contradiction, our assumption was", "= f(a,b,kx,ky) = kf(a,b,x,y)$ for any $k$ real number. Now, if either $f(a,b,a,b)$ or $f(x,y,x,y)$ is zero, the desired inequality holds trivially. So we let: $a' = a/\\sqrt{f(a,b,a,b)}, b' = b/\\sqrt{f(a,b,a,b)}$ $x' = x/\\sqrt{f(x,y,x,y)}, y' = y/\\sqrt{f(x,y,xy)}$ And, $f(a'-x',b'-y', a'-x', b'-y') \\geq 0$ $f(a'-x',b'-y',a',b') - f(a'-x',b'-y',x',y') \\geq 0$ $f(a',b',a',b') - f(x',y',a',b') - f(a',b',x',y') + f(x',y',x',y') \\geq 0$ $f(a',b',a',b') + f(x',y',x',y') \\geq 2f(a',b',x',y')$ But $f(a',b',a',b') = f(a,b,a,b) / f(a,b,a,b) = 1$, and similarly $f(x',y',x',y') = 1$. That means $f(a',b',x',y') \\leq 1$, which translates to: $f \\left(\\frac{a}{\\sqrt{f(a,b,a,b)}}, \\frac{b}{\\sqrt{f(a,b,a,b)}}, \\frac{x}{\\sqrt{f(x,y,x,y)}}, \\frac{y}{\\sqrt{f(x,y,x,y)}} \\right) \\leq 1$ $f(a,b,x,y) \\leq \\sqrt{f(a,b,a,b)} \\sqrt{f(x,y,x,y)}$ $f(a,b,x,y)^2 \\leq f(a,b,a,b) f(x,y,x,y)$ Remark: while this approach is significantly lengthier than direct expansion (see First Solution), this technique can be used for any function $f$ that satisfies the properties: $f(a,b,a,b) \\geq 0$ and $f(a,b,a,b) = 0 \\iff a=b=0$ $f(a+ka',b+kb',x,y) = f(a,b,x,y) + kf(a',b',x,y)$ $f(a,b,x,y) = f(x,y,a,b)$. Such function is usually called the \"inner product\" of vectors $(a,b)$ and $(x,y)$, and it can be generalized to $\\mathbb{R}^n$ Third Solution: Just like in the Second Solution, we can assume $P > 0$ without loss of generality. Let $f(t) = Pt^2 + 2Qt + R$. Since $PR > Q^2$, that means $f$ can be expressed as a sum of two squares: $f(t) = P(t+ C)^2 + D^2$ where: $C = Q/P, D^2 = R -", "# Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt - Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\\since $f(0)=0$) Finally putting $y=(x+1)$ gives us $f(x)=0$. This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one? You proved that $f(0)=0$, we can continue from here Suppose that there exist $k \\neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$. Now suppose that there exist no $k \\neq 0$ such that $f(k)=0$. Then plugging $x=x,y=-x$ gives $f(x^2-xf(x)) = xf(0)= 0$, which by assumption means $x^2=xf(x)$ or $f(x)=x$. So we conclude that, possible functions", "# IMO 2019, Problem 1 The International Math Olympiad 2019 had the following question: Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The reason that I decided to record this is because I thought I’d made an interesting observation that allowed me to solve the problem in only a couple of steps. However, I later realized that at least one other person has solved the problem the same way. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "B ∃a ∈ A (b = f (a)). \\nonumber$ For example, let $$X = {a,b}$$ and $$Y = {1,2,3}$$, and consider the function from $$Y$$ to $$X$$ specified by the set of ordered pairs $$\\{(1, a), (2, a), (3, b)\\}$$. This function is onto because its image, $$\\{a, b\\}$$, is equal to the range, $$X$$. However, the function from $$X$$ to $$Y$$ given by $$\\{(a,1),(b,3)\\}$$ is not onto, because its image, $$\\{1,3\\}$$, is a proper subset of its range, $$Y$$. As a further example, consider the function $$f$$ from $$\\mathbb{Z}$$ to $$\\mathbb{Z}$$ given by $$f (n) = n − 52$$. To show that $$f$$ is onto, we need to pick an arbitrary $$b$$ in the range $$\\mathbb{Z}$$ and show that there is some number $$a$$ in the domain $$\\mathbb{Z}$$ such that $$f(a) = b$$. So let $$b$$ be an arbitrary integer; we want to find an a such that $$a − 52 = b$$. Clearly this equation will be true when $$a = b + 52$$. So every element $$b$$ is the image of the number $$a = b + 52$$, and $$f$$ is therefore onto. Note that if $$f$$ had been specified to have domain $$\\mathbb{N}$$, then $$f$$ would not be onto, as for some $$b ∈ \\mathbb{Z}$$ the number $$a = b + 52$$ is not", "# Find all real real functions that satisfy the following eqation $f(x^2)+f(2y^2)=$ Find all real functions $f:\\Bbb R\\rightarrow\\Bbb R$ so that $f(x^2)+f(2y^2)=[f(x+y)+f(y)][f(x-y)+f(y)]$, for all real numbers $x$ and $y$. $f(x)=x^2$ is the only solution I think. So far I have got: $f(x^2)=[f(x)]^2+2f(x)f(0)+[f(0)]^2-f(0)$ $f(2y^2)=4[f(y)^2]-f(0)$ And by putting $x=y$ and $x=-y$ you get that $[f(x)-f(-x)][f(2x)-f(0)]=0$. If $f(2x)=f(0)$ then $f$ is a constant function of value $0$ or $1/2$, both work, if not than $f$ is an even function. Source: 3rd European mathematical cup, senior category. #### Solutions Collecting From Web of \"Find all real real functions that satisfy the following eqation $f(x^2)+f(2y^2)=$\" Very partial solution. I show that $f$ is even when $f(0)=0$. I use a comment of @soulEater: in this case we have $$(f(x+y)+f(x-y))(f(y)-f(-y))=0$$ Now suppose that there exists an $y_0$ such that $f(y_0)-f(-y_0)\\not =0$, put $L=2y_0\\not = 0$. We have $f(x+y_0)=-f(x-y_0)$ for all $x$, hence $f(x+L)=-f(x)$ and $f(x+2L)=f(x)$ for all $x$ (and $f$ is periodic). We replace $x$ by $x+L$ in the original equation $$f((x+L)^2)+f(2y^2)=(-f(x+y)+f(y))(-f(x-y)+f(y))$$ We put $x=0$ $$f(L^2)+f(2y^2)=0$$ From here, we get $f(u)=c=-f(L^2)$ for all $u\\geq 0$, and $f(u)=c$ for all $u$ by periodicity. But this contradict the existence of $y_0$. Hence we have $f(y)-f(-y)=0$ for all $y$, and $f$ is even. EDIT I suppose now that $\\displaystyle f(0)=\\frac{1}{2}$, and I suppose that $f$ is continuous. I", "# Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ Find all functions $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\\ge 0$ we have $f(a)>f(b)$ As much as I know: • putting $x=y=0$ we get $f(0)=0$. • let $x+y=0$, the we get $f(y)+f(-y)=\\frac{f(2y)}{2}$ Step 1: Prove : $f(y) = 0$ iff. $y=0$ It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$) Following proof is given by @Kyson To prove that $f(y)≠0$ for all$y<0.$ Assume there is a $y<0$ such that $f(y)=0.$ Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0$Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$ $\\$ Step 2: Prove $f(y)$ is even. Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\\frac{f(2y)}{2} = f(-y) +f(y) = \\frac{f(-2y)}{2}$ Thus $f(x)$is a even function, so $f(y)+f(-y)=2f(y)=\\frac{f(2y)}{2} \\Rightarrow f(2y)=4f(y)$ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\\infty,0)\\cup(0,\\infty)$) $\\$ Step 3: Prove:$f(y)=y^2f(1)$ By induction , find $f(ny)=n^2f(y)$ where $n \\in \\mathbb N$ Details of induction: Suppose $m\\in \\mathbb N ,m\\ge 2 ,\\ f(ny)=n^2f(y)$ holds for all $n\\le m$ Put $x=(m-1)y$ in property (a).$\\\\ \\Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\\\ \\Rightarrow f((m+1)y)=(m+1)^2f(y) \\ for\\ y\\not = 0 \\ and \\ it\\ still\\ holds\\ for\\ y=0$ Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$ Remark: Result induces $f(zy)=z^2f(y)$ where $z\\in \\mathbb Z$ , since $f$ is even. $\\$ $\\forall", "# Find $f$ such that $f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$ Denote the set of non-negative real numbers by $$\\mathbb R^+_0$$. Find all functions $$f:\\mathbb R \\rightarrow \\mathbb R_0^+$$ s.t. $$\\forall a,b,c,d\\in\\mathbb R$$ satisfying $$ab+bc+cd=0$$ we have $$f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$$ My attempt: set $$b=d=0$$ and get $$f(0)=0$$. Now set $$a=b=c=0$$ to get $$f(-d)=f(d)$$. Now we pretty much reduced this fe's domain to $$\\mathbb R^+_0\\rightarrow\\mathbb R^+_0$$. That's where I got stuck. I tried a few things but none of them gave me any progress. Any help/hints appreciated. • @AhmedS.Attaalla If $c=0$, $ab=0$. Thus $a=0$ or $b=0$. – induction601 Mar 1 at 7:11 • When $a=b$, we have $a^2 = -(a+d)c$. By letting $c=a$ and $d=-2a$, we have $\\frac{f(3a)-f(a)}{2} = f(2a)$. – induction601 Mar 1 at 7:12 • When $b=d$, we have $b(a+2c)=0$. By letting $a=-2c$, we have $f(2c+b) + f(c-b) = f(2c) + f(c+b) + f(b)$. Let $c=2b$. Then this gives us $f(5b) + f(b) = f(4b) + f(3b) + f(b)$. Thus we have $f(5b) = f(4b) + f(3b)$. – induction601 Mar 1 at 7:36 • $\\sf{f(a-b)+f(a+b)=f(2a)+f(2b)}$ – TheSimpliFire Mar 1 at 8:05 • $f(x) = kx^2$ satisfies the given conditions for any $k\\ge 0$. – FredH Mar 1 at 11:32 Let $$c \\neq 0$$. To remove our condition, we write $$d$$ in terms of the rest of the variables : $$d=\\frac{-ab-bc}{c}=-b-\\frac{ab}{c}$$ Now, we can substitute" ]

[ "+0 # Help pls 0 35 1 The points $P,$ $Q,$ and $R$ are represented by the complex numbers $z,$ $(1 + i) z,$ and $2 \\overline{z},$ respectively, where $|z| = 1.$ When $P,$ $Q$, and $R$ are not collinear, let $S$ be the fourth vertex of the parallelogram $PQSR.$ What is the maximum distance between $S$ and the origin of the complex plane? Feb 6, 2021", "+0 -1 84 2 +108 The points $P,$ $Q,$ and $R$ are represented by the complex numbers $z,$ $(1 + i) z,$ and $2 \\overline{z},$ respectively, where $|z| = 1.$ When $P,$ $Q$, and $R$ are not collinear, let $S$ be the fourth vertex of the parallelogram $PQSR.$ What is the maximum distance between $S$ and the origin of the complex plane? Oct 22, 2019 #1 +153 0 in other words, please do not post homework questions on the forum. Oct 23, 2019 edited by SoulSlayer615 Oct 23, 2019 #2 0", "Everything is Parallel Geometry Level 2 Let $$PQSR$$ be a parallelogram graphed on a coordinate plane where $$\\overline{PQ}$$ is parallel to $$\\overline{RS}$$. If the coordinates of $$P$$ are $$(0,0)$$, the coordinates of $$Q$$ are $$(3,2)$$, and the coordinates of $$S$$ are $$(4,6)$$, find the coordinates of $$R$$. If the coordinates are $$(x,y)$$, then enter $$x+y$$ as your answer. ×", "# A problem by Nit Jon Level pending Complex numbers a, b, and c are zeros of a polynomial $$P(z) = z^3 + qz + r$$, and $$|a|^2 + |b|^2 + |c|^2 = 250$$. The points corresponding to a, b, and c in the complex plane are the vertices of a right triangle with hypotenuse h. Find $$h^2$$. ×", "# Complex Numbers Geometry Level 5 Complex numbers $$a$$, $$b$$ and $$c$$ are the zeros of a polynomial $$P(z) = z^3+qz+r$$, and $$|a|^2+|b|^2+|c|^2=250$$. The points corresponding to $$a$$, $$b$$, and $$c$$ in the complex plane are the vertices of a right triangle with hypotenuse $$h$$. Find $$h^2$$. ×", "A suitable picture will be $S^1$ in the complex plane, each $n$-th root marked on it with a dot. $z^n=1$ is $(1,0)$ on the $x$-axis. From this point there are two equidistant points, $z$ and $z^{n-1}$, to its left and right on the circle. Since complex conjugation is the same as mirroring the complex number around the $x$-axis it should be clear that $\\overline{z}=z^{n-1}$. $$zw=1$$ iff $$|z||w|=1$$ and $$\\arg(z)+\\arg(w)=0\\ \\ \\ (\\!\\!\\!\\!\\!\\mod 2\\pi).$$ • Maybe I am missing something, but what about $|1||-1|=1$ and $-1\\cdot 1=-1$? – Phil-ZXX Apr 19 '14 at 11:33 • $\\arg(1)=0$, $\\arg(-1)=\\pi+2k\\pi$. – Martín-Blas Pérez Pinilla Apr 19 '14 at 11:51", "# The points $(5,0,2),\\:(2,-6,0),\\:(4,-9,6)\\:and\\:(7,-3,8)$ represent vertices of a ? $(a)\\:Square\\:\\:\\qquad\\:(b)\\:Rhombus\\:\\:\\qquad\\:(c)\\:Rectangle\\:\\:\\qquad\\:(d)\\:Parallelogram.$", "# Area of a Parallelogram Graphed on Real and Complex Plane Algebra Level 3 The Argand Diagram is a method of graphing points similar to the Cartesian coordinate system. Instead, however, points on the Argand Diagram, also called the complex coordinate plane, are represented as complex numbers in the form $$a+bi$$, where the real component $$a$$ represents the $$x$$-coordinate, and the imaginary component $$b$$ represents the $$y$$-coordinate. When $$f(x)=-x^{2}+6x-45$$ is graphed on the Cartesian coordinate plane, let the two roots be $$z_1$$ and $$z_2$$. The points $$(8, 13)$$ and $$(8, 1)$$ are graphed on the Cartesian coordinate plane, and $$z_1$$ and $$z_2$$ are graphed on the Argand Diagram. Then, they two systems are overlapped such that their origins are equal and they are proportionate (i.e. the point $$3i$$ on the Argand Diagram is equal$$(0, 3)$$ on the Cartesian coordinate plane). What is the area of the parallelogram formed by the four points? ×", "# State true or false for the following : If three complex numbers $\\;z_{1} , z_{2}\\;$ and $\\;z_{3}\\;$ are in A.P. then they lie on a circle in the complex plane . $(a)\\;True\\qquad(b)\\;false$ Explanation : False , because if $\\;z_{1} , z_{2}\\;$ and $\\;z_{3}\\;$ are in A.P. then $\\;z_{1} = \\large\\frac{z_{1}+z_{3}}{2}\\;$ => $\\;z_{2}\\;$ is the mid point of $\\;z_{1}\\;$ and $\\;z_{3}\\;$ => which implies that the points $\\;z_{1} , z_{2} , z_{3}\\;$ are collinear", "# If $P$ is the point of intersection of the line joining the points $Q(2,3,5)\\:\\:and\\:\\:R(1,-1,4)$ with the plane $5x-4y-z-1=0$. If $S$ is foot of $\\perp$ drawn from the point $T(2,1,4)$ to the line $QR$, then the length of segment $\\overrightarrow {PS}$ = ? $\\begin{array}{1 1} (a)\\:\\large\\frac{1}{\\sqrt 2}\\:\\:\\:\\qquad\\:(b)\\:\\sqrt 2\\:\\:\\:\\qquad\\:(c)\\:2\\:\\:\\:\\qquad\\:(d)\\:2\\sqrt 2. \\end{array}$", "thank you • November 1st 2010, 08:02 AM HallsofIvy Specifically, |a- b| is the distance between points in the plane, interpreted as complex numbers. So |z- a|= r is the set of points whose distance from a is r- that is, the circle with center a and radius r.", "Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then Solution: *Multiple options can be correct QUESTION: 75 Solution: Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code", "the complex plane where points are identified with complex numbers: The roots of the derivative of the polynomial $P(z)=(z-z_{1})(z-z_{2})(z-z_{3})$ are the foci of the ellipse tangent to the midpoints of $\\Delta z_{1}z_{2}z_{3}.$ ### Conic Sections > Ellipse Copyright © 1996-2018 Alexander Bogomolny 63567929 Search by google:", "# Let $P,Q,R$ and $S$ be the points on the plane with position vectors $-2 \\hat i -\\hat j ,4 \\hat i, 3 \\hat i +3 \\hat j$ and $3 \\hat i +2 \\hat j$ respectively . The quadrilateral $PQRS$ must be a $\\begin{array}{1 1} \\text{Parallelogram, which is neither rhombus nor a rectangle } \\\\ \\text{square } \\\\ \\text{rectangle , but not a square } \\\\ \\text{rhombus , but not a square } \\end{array}$", "$$z-w$$ and $$z+w$$ are the diagonals. $$2(|z|^2+|w|^2)$$ represents the sum of the squares of the sides, and $$|z+w|^2+|z-w|^2$$ represents the sum of the squares of the diagonals. The identity shows that, in any parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of the sides. ## Problems 1. Subtract the second complex number from the first. Plot them as vectors on the complex plane. 1. $$1+i$$ and $$2+i$$ 2. $$3-i$$ and $$4+i$$ 2. Find the distance between the two complex numbers on the complex plane. 1. $$3+4i$$ and $$-1-i$$ 2. $$-2-7i$$ and $$-3-2i$$ 3. $$100+100i$$ and $$-50+50i$$ 3. Show that the complex numbers $$-2-2i$$, $$1-i$$, $$2+2i$$ and $$-1+i$$ form a rhombus in the complex plane. 4. Show algebrically that $$|z+w|^2+|z-w|^2=2(|z|^2+|w|^2).$$ 5. Let $$z_1$$, $$z_2$$ and $$z_3$$ be complex numbers represented by the points $$A$$, $$B$$ and $$C$$ on the complex plane. 1. Write down the complex number representing the vectors $$\\overrightarrow{AB}$$ and $$\\overrightarrow{AC}$$. 2. Let $$D$$ be a point on the complex plane such that $$ABDC$$ is a parallogram. Show that the complex number which represents $$D$$ is $$z_2+z_3-z_1$$.", "# Finite set of non-collinear points on plane with every point having ≥ 3 equidistant neighbors? [closed] Does there exist a finite set of points on the Euclidean plane, such that: • No 3 points are collinear, and • Every one of the points has (at least) three other points in the set at the same distance from it? It seems to me that the answer should be No, but my naïve attempts to prove it have failed. • When you say “every point [P] has at least three points [Q,R,S] in the same distance”, does that just mean PQ=PR=PS, or do you also require PQ=PR=PS=P’Q’=P’R’=P’S’ for any other P’? Aug 4 at 16:53 • For instance, the 8 points in the graph of a cube, drawn with all edges of length 1? Aug 4 at 17:07 • Yes, of course, it was assumed to be on the plane. The example I wrote in answer is also a 2D projection of the 1-skeleton of a 4D polytope (the cartesian square of a triangle). Aug 4 at 19:26 • @PietroMajer's comment above also solves the generalisation for $k$ instead of $3$ points at the same distance from each point - just take a suitable projection of the $k$-dimensional hypercube graph. Aug 5 at 7:30 • It's simpler to talk", "# Parallelogram on the Coordinate Plane Geometry Level 4 A parallelogram on the coordinate plane is defined by the following lines: $\\left\\{\\begin{array}{l}y=4x+5 \\\\ y=4x+3 \\\\ y=nx+5\\\\ y=nx+3\\end{array}\\right.$ If the area of the parallelogram is $$16$$, then $$n$$ can be expressed as $$\\dfrac{p}{q}$$, where $$p,q$$ are coprime positive integers. Find the smallest possible value of $$p+q$$. ×", "22 views For any distinct points $P$ and $Q$ of the plane, we denote $m(P Q)$ the perpendicular bisector of the segment $P Q$. Let $S$ be a finite subset of the plane, with more than one element, which satisfies the following conditions: (i) If $P$ and $Q$ are distinct points of $S$, then $m(P Q)$ meets $S$. (ii) If $P_1 Q_1, P_2 Q_2$ and $P_3 Q_3$ are three distinct segments with endpoints in $S$, then no point of $S$ belongs simultaneously to the three lines $m\\left(P_1 Q_1\\right), m\\left(P_2 Q_2\\right), m\\left(P_3 Q_3\\right)$. (Mexico) | 22 views", "# Unsolved ChallengeArea of triangle #### anemone ##### MHB POTW Director Staff member Find in terms of $p,\\,q$ and $r$, a formula for the area of a triangle whose vertices are the roots of $x^3-px^2+qx-r=0$ in the complex plane.", "this line segment? 1. $-2 - 5i$ 2. $-4i$ 3. $2 + 5i$ 4. $3/2 - 92 i$ 4. Which of the listed complex numbers, when represented by a point in the complex plane, is both equidistant and collinear with the complex numbers represented by points A and K? 1. $3i$ 2. $1 + 3/2 i$ 3. $1/2 + 3i$ 4. $3 + 1/2 i$ 5. Let $z=-8+5i$. If the midpoint of the line segment created by complex numbers $z$ and $w$ is $-3+3/2 i$, what is the value of $w ?$ 1. $w = -11/2 + 13/4 i$ 2. $w = 2-2i$ 3. $w = 14 + i$ 4. $w = 4 + 7/2 i$ 6. Find the distance between $z = 6 + 7i$ and $w = -2 - 3i$ in the complex plane. 1. $4sqrt(2)$ 2. $18$ 3. $2sqrt(2)$ 4. $2sqrt(41)$ 7. Which of the following expressions represent(s) the distance between $z_1$ and $z_2$ in the complex plane? Choose all correct answers. 1. $|z_1 - z_2|$ 2. $|z_2 - z_1|$ 3. $|z_2| - |z_1|$ 4. $|z_2 + z_1|$ 8. What is the distance between the complex numbers $2-4i$ and $-3/2 - 1/2 i$ in the complex plane? 1. $7/2 sqrt(2)$ 2. $1/2sqrt(130)$ 3. $5/2 sqrt(2)$ 4. $1/8 sqrt(82)$ 9. If $z = 5 + 4i$," ]

[ "of 18 things, RHS is not GM of 18 things. I think you get $(4xy+6yz+8xz)/3\\ge((4xy)(6yz)(8xz))^{1/3}$. – Gerry Myerson Mar 6 '12 at 11:20 @GerryMyerson, you are right, i have edited the answer. – quartz Mar 6 '12 at 11:31 @vikiii RHS is $192^{1/3}*xyz^{2/3}$ and LHS is 3. – quartz Mar 6 '12 at 11:38 Of course, you then have to decide whether the bound you get is actually attained. – Gerry Myerson Mar 6 '12 at 11:49 If we want to use the AGM inequality without loss we have to \"symmetrize\" the three variables. Therefore we replace $x$, $y$, $z$ by $$x':=4x,\\quad y':=3y,\\quad z':=6z\\ .$$ The AGM inequality then says that $$\\root 3\\of {x'^2y'^2z'^2}\\leq {x'y' + y'z' + z' x'\\over 3}=4xy + 6 yz+8 zx=9\\ ,$$ i.e., $x'y'z'\\leq 27$, with equality iff $x'=y'=z'=3$. It follows that $$xyz = {x'y'z'\\over 72}\\leq{3\\over 8}$$ with equality iff $x={3\\over4}$, $y=1$, $z={1\\over2}$. - Thanks . But i want to know How did you think of x′:=4x,y′:=3y,z′:=6z . – vikiiii Mar 7 '12 at 4:40 @vikiiii: I wrote $x'=\\alpha x$, $\\ldots$, and determined $\\alpha$, $\\beta$, $\\gamma$ such that $4xy={1\\over3}x'y'$, $\\ldots\\$. – Christian Blatter Mar 7 '12 at 8:59 You can also do it by using weighted arithmetic mean and geometric mean inequality, $$\\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n} > [{x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}}]^{\\frac{1}{m_1m_1m_2m_3.....m_n}}$$", "20:52 $$|x_n| - |\\overline{x}| \\leq |x_n - \\overline{x}| < \\frac{|\\overline{x}|}{2}$$ and: $$|\\overline{x}| -|x_n| \\leq |x_n - \\overline{x}| < \\frac{|\\overline{x}|}{2}$$ by definition of $$||x| - |y||$$.Hence: $$|x_n| > |\\overline{x}| - \\frac{|\\overline{x}|}{2} = \\frac{|\\overline{x}|}{2}$$ • makes perfect sense, thanks! – Mathbeginner Feb 27 at 20:53 The 'reverse' triangle inequality states the following: for all $$x,y\\in\\mathbb{R}$$ \\begin{align*} |x-y|\\geq |x|-|y|. \\end{align*} To see this, notice that \\begin{align*} |x|-|y|=|x-y+y|-|y|\\leq |x-y|+|y|-|y|=|x-y|, \\end{align*} where the inquality follows from the normal triangle inequality. Now, if you have $$|x_n-\\bar{x}|<\\frac{|\\bar{x}|}{2}$$ then this means that \\begin{align*} |\\bar{x}|-|x_n|\\leq\\frac{|\\bar{x}|}{2}, \\end{align*} and so by adding $$|x_n|$$ to both sides and subtracting $$\\frac{|\\bar{x}|}{2}$$ the result follows.", "\\therefore |z-w|&\\le |z-x|+|x-w| \\end{align} (i.e. the case reduces to the definition of the triangle inequality with regard to complex numbers) For case two, $d(z,x)=|z-x|$ and $d(x,w)=|x|+|w|$ tell us that $z=0$. If $z=0$, then: \\begin{align} |z-w|&=|-w|=|w|\\\\ \\text{and } |z-x|+|x|+|w|&=|-x|+|x|+|w|\\\\ &=2|x|+|w|\\\\ \\text{thus } |w|&\\le 2|x|+|w|\\\\ 0 &\\le 2|x|\\\\ \\therefore |z-w| &\\le |z-x|+|x|+|w| \\end{align} Case three implies that $w=0$, so: \\begin{align} |z| &\\le |z|+|x|+|x|\\\\ 0 &\\le 2|x|\\\\ \\therefore |z-w| &\\le |z|+|w|+|x-w| \\end{align} Case four: First, let: \\begin{align} z&=a_z+b_zi\\\\ w&=a_w+b_wi\\\\ x&=a_x+b_xi \\end{align} Then, we have: \\begin{align} |z-w|&\\le |z|+2|x|+|w|\\\\ \\sqrt{(a_z-a_w)^2+(b_z-b_w)^2}&\\le \\sqrt{a_z^2+b_z^2}+2\\sqrt{a_x^2+b_x^2}+\\sqrt{a_w^2+b_w^2}\\\\ (a_z-a_w)^2+(b_z-b_w)^2 &\\le a_z^2+b_z^2+a_w^2+b_w^2+2|x|+4|x||z|+2|z||w|+4|x||w|\\\\ -2(a_za_w+b_zb_w) &\\le 2|x|+4(|x||z|+|x||w|)+2|z||w|\\\\ 0 & \\le |x|+2(|x||z|+|x||w|)+|z||w|+a_za_w+b_zb_w \\end{align} $\\arg(z)=\\arg(w)$ ( http://www.wolframalpha.com/input/?i=arg%28z%29%3Darg%28w%29 ) is the only such situation for case four, therefore the product of their real parts and their imaginary parts cannot be negative and, hence, the final inequality is true. All that is left now are the four other similar cases: \\begin{align} |z|+|w|&\\le |z-x|+|x-w|\\\\ |z|+|w|&\\le |z-x|+|x|+|w|\\\\ |z|+|w|&\\le |z|+|x|+|x-w|\\\\ |z|+|w|&\\le |z|+|x|+|x|+|w| \\end{align} Case one follows equality (namely, since neither $z$ nor $w$ are $0$, the inequality reduces to an equality because $x$ must be $0$. (The argument of $x$ cannot be equal to both $w$ and $z$. If it were, then the left side would be $|z-w|$ rather than $|z|+|w|$.)) Case four is straight to the point. It reduces to $0\\le 2|x|$. Addendum: I don't know how to prove", "how this proof came to be; what is the hidden motivation. Thank you. #### Solutions Collecting From Web of \"A proof of the Isoperimetric Inequality – how does it work?\" My attempt at an answer has two parts. First, I think the geometric mean thing slightly obscures what’s going on and contributes to your bafflement why “the $r$’s magically fall out of the equation in the last step”. A more geometric view of this step is: We know that the curve has “width” $w=2r$ (along the chosen direction). Then $A = w\\overline{h} = 2r\\overline{h}$, where $\\overline{h}$ is the “average height” of the curve (along the chosen direction). Thus we have $A + \\pi r^2 \\le lr \\;\\Leftrightarrow\\; 2r\\overline{h} + \\pi r^2 \\le lr \\;\\Leftrightarrow\\; 2\\overline{h} + \\pi r \\le l \\;\\Leftrightarrow\\; 2\\overline{h} + \\frac{\\pi}{2} w \\le l\\;.$ Now the cancellation of the $r$’s seems more natural, and the inequality gives an upper bound for the “average height” we can achieve for a given “width” $w$ and length $l$. This bound implies the isomperimetric inequality: $l^2 \\ge (2\\overline{h} + \\frac{\\pi}{2} w)^2=(2\\overline{h} – \\frac{\\pi}{2} w)^2 + 4\\pi w\\overline{h} \\ge 4\\pi w\\overline{h}=4\\pi A\\;.$ (There may be a connection here to the inequality by Bonnesen cited in Christian Blatter’s comment.) Second, regarding the proof as a whole, it seems useful to think", "$-1$ and $i$. So $z$ is on perpendicular bisector of segment between $-1$ and $i$. Hint: Let $z=a+ib$, then $|\\frac{1+z}{1-i\\overline{z}}|=\\frac{|1+z|}{|1-i\\overline{z}|}=\\frac{|(1+a)+ib|}{|(1-b)-ia|}=\\frac{\\sqrt{(1+a)^2+b^2}}{\\sqrt{(1-b)^2+a^2}}=\\frac{\\sqrt{1+a^2+2a+b^2}}{\\sqrt{1+b^2-2b+a^2}}=1$ Now this equality is true if and only if $a=-b$ which gives us that if $z=a+ib$ then $iz=ai-b=a-bi=\\overline{z}$ Take this $$|1+z|=|1-i\\bar{z}|$$ then square this babies $$(1+z)(1+\\bar{z})=(1-i\\bar{z})(1+iz)$$ $$1+z+\\bar{z}+z\\bar{z}=1-i\\bar{z}+iz+z\\bar{z}$$ so $$z+\\bar{z}=i(z-\\bar{z})$$ $$2Re(z)=i(i2Im(z)$$ therefore $$Re(z)=-Im(z)$$ Let $z=x+yi$ and then from $$|1+z|=|1-i\\bar{z}|$$ it is easy to obtain $$(x+1)^2+y^2=(-y+1)^2+x^2,$$ So $$x=-y$$ and hence $z=(1-i)x$.", "# How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying $$x + y +z + 1 =4xyz.\\tag{1}$$ Prove that $$xy + yz + zx \\geqslant x + y + z.\\tag{2}$$ I don't know how to start? - Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\\ge$G.M., we get $xyz = (x+y+z+1)/4 \\ge (xyz)^{1/4}$. Hence $xyz\\ge1$. WLOG, suppose $0<x\\le y\\le z$. There are three cases. Case 1: $x\\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\\le y< z$. Then $(2)$ also holds because $$1-xyz\\le0\\le(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz.$$ Case 3: $0<x\\le y<1<z$. Then \\begin{align*} (2)&\\Leftrightarrow z(x+y) + xy \\ge x+y+z,\\\\ &\\Leftrightarrow z(x+y-1) + xy - x-y\\ge0,\\\\ &\\Leftrightarrow z(x+y-1)-1 + (x-1)(y-1) \\ge0. \\end{align*} So it suffices to show that $z(x+y-1)-1\\ge0$. Since $x,y,z$ are positive, we must have $4xy>1$, otherwise the LHS of $(1)$ will be strictly greater than the RHS. Now $(1)$ implies that $z=(x+y+1)/(4xy-1)$. Therefore $$z(x+y-1)-1 =\\frac{(x+y+1)(x+y-1)-(4xy-1)}{4xy-1} =\\frac{(x-y)^2}{4xy-1} \\ge0$$ and we are done. From the details of above three cases, it can be shown that equality in $(2)$ holds iff $x=y=z=1$. - Firstly, I'm going to relabel $x$,$y$ and $z$ to $\\alpha$,$\\beta$ and $\\gamma$. We then consider the cubic polynomial $$f(x) = ax^3 + bx^2 + cx + d$$ Supposing this has roots", "follows that $F$ is holomorphic at each $z\\in\\mathbb{D}$. Q.E.D. (ii) Proof: Clearly, $F(0)=\\dfrac{w}{1}=w\\quad\\text{and}\\quad F(w)=\\dfrac{0}{1-\\left|w\\right|^2}=0$. Q.E.D. (ii) Proof: This is a direct consequence of part (a) since $\\left|w\\right|\\neq 1$ and $\\left|z\\right|=1$. Q.E.D. (iv) Proof: It follows that\\begin{aligned} F\\circ F(z) &= \\frac{w-\\dfrac{w-z}{1-\\overline{w}z}}{1-\\overline{w}\\dfrac{w-z}{1-\\overline{w}z}}\\\\ &= \\frac{\\dfrac{w-\\left|w\\right|^2z-w+z}{1-\\overline{w}z}}{\\dfrac{1-\\overline{w}z-\\left|w\\right|^2+\\overline{w}z}{1-\\overline{w}z}}\\\\ &= \\frac{z(1-\\left|w\\right|^2)}{1-\\left|w\\right|^2}\\\\ &= z\\quad(\\text{since w\\in\\mathbb{D}\\implies\\left|w\\right|<1}.) \\end{aligned} Thus, $F$ is bijective. Q.E.D. Status Not open for further replies.", "\\overline{G}\\\\ \\to \\frac{dY}{d \\overline{G}} = \\frac{1}{1 - C'(Y - \\overline{W})}$$ Therefore: \\begin{align*} \\frac{dY}{d \\overline{G}} - 1 &= \\frac{1}{1 - C'(Y - \\overline{W})} - 1,\\\\ &=\\frac{1 - 1 +C'(Y - \\overline{W})}{1 - C'(Y - \\overline{W})},\\\\ & = \\frac{C'(Y - \\overline{W})}{1 - C'(Y - \\overline{W})},\\\\ &= \\frac{dY}{d(- \\overline{W})} \\end{align*}", "shows that the equality holds iff $x=y=z$. Here is my approach. Using AM-GM, \\begin{align} x^2+y^2+z^2 &\\ge 3(xyz)^{2/3} \\\\ \\implies 1 &\\ge 3(xyz)^{2/3} \\\\ \\implies (xyz)^{2/3} &\\le \\dfrac{1}{3} \\\\ \\implies xyz \\le \\dfrac{1}{3\\sqrt{3}} \\tag 1 \\end{align} Again, using AM-GM, \\begin{align} \\dfrac{x^3+y^3+z^3}{3} &\\ge 3xyz \\\\ \\implies x^3+y^3+z^3-3xyz &\\ge 0 \\\\ \\implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\\ge 0\\end{align}\\tag*{} Since $$x,y,z \\in \\mathbb R^+$$, $$x+y+z \\ge 0$$. \\begin{align}\\therefore \\, x^2+y^2+z^2 -xy-yz-zx &\\ge 0 \\\\ \\implies -x^2-y^2-z^2 +xy+yz+zx &\\le 0 \\\\ \\implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\\le 0 \\\\ \\implies x^2+y^2+z^2 +2xy+2yz+2zx &\\le 3(x^2+y^2+z^2) \\\\ \\implies (x+y+z)^2 &\\le 3 \\\\ \\implies x+y+z &\\le \\sqrt{3} \\tag 2 \\end{align} Multiplying $$(1)$$ and $$(2)$$, we can get the desired result.", "z}}{{dz}} = \\frac{{dx - idy}}{{dx + idy}} = \\frac{{1 - iy'}}{{1 + iy'}}$, where$y' = \\frac{{dy}}{{dx}}$. However, alongC,$\\bar z = S(z)$, and since the derivative an analytic function is independent of the direction in which increments are taken, we have (7.4)$S'(z) = \\frac{{d\\bar z}}{{dz}} = \\frac{{1 - iy'}}{{1 + iy'}}$. Since$dx - idy = \\overline {dx + idy}$, it follows from (7.3) and (7.4) that (7.5)$|S'(z)| = 1$alongC. The equality (7.5) may also be obtained as a consequence... 11. CHAPTER 8 CONFORMAL MAPS, REFLECTIONS, AND THEIR ALGEBRA (pp. 49-88) In Chapter 6, we defined an analytic arcCas the image of the real segment$a \\le t \\le b$under a one-to-one conformal mapf. We found that if a point$z = f(t)$is in a neighborhood ofC, the reflection* ofzinCis defined by$z* = f(\\bar t)$· Since$S(z) = \\overline {z*}$, we have$S(z) = \\overline {f(\\bar t)} = \\bar f(t)$and since$t = {f^{ - 1}}(z)$, we have$S(z) = \\bar f({f^{ - 1}}(z))$. We can write this as (8.1)$S = \\bar f{f^{ - 1}}$, from which we obtain (8.1')$S' = \\bar f'{f^{ - 1}}/f'{f^{ - 1}}$. If the parametertin [0, 1] is changed by means of (8.2)$t = g(t')$,$0 \\le t' \\le 1$, wheregis real analytic on the real${t'}$axis:$\\begin{array}{l} g = \\bar g \\\\ z* = f(\\bar t \\\\ \\end{array}$, then we can compute... 12.", "points are W (5, -1), X (5, 6), Y (2, 1), Z (2, 6) Now, $$\\overline{W X}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(5 – 5)² + (1 + 6)²}$$ = 7, $$\\overline{X Y}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(5 – 2)² + (6 – 1)²}$$ = 5.8, $$\\overline{Y Z}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(2 – 2)² + (6 – 1)²}$$ = 5, $$\\overline{Z W}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(5 – 2)² + (6 + 1)²}$$ = 7.6, $$\\overline{W Y}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(5 – 2)² + (1 + 1)²}$$ = 3.6, $$\\overline{X Z}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$, = $$\\sqrt{(5 – 2)² + (6 – 6)²}$$ = 3, Hence the perimeter of the given polygon = WX + XY + YZ + ZW, = 7 + 5.8 + 5 + 7.6 = 25.4, Since all the lengths of the sides are different the area of the given polygon = $$\\frac{1}{2}$$ (d1) (d2) where d1 and d2 are the diagonals, So the area of the polygon = $$\\frac{1}{2}$$ (WY) (XZ) = $$\\frac{1}{2}$$ (3) (3.6) = 5.4, Hence from the above we can conclude that the perimeter of the given polygon is 25.4,", "the origin of the trace more clear) but that didn't get me much farther. Thanks! Update: Following a helpful comment by @whuber, I had a look at the one dimensional case in order to gain insight, using the equivalent one dimensional substitution to $$\\mathbf{y} = \\mathbf{M}^{-1}(\\mathbf{w} - \\overline{\\mathbf{w}})$$. \\begin{align} \\int \\frac{1}{2} w^2 a \\> Q(w)\\> \\mathrm{d} w &= \\int \\frac{1}{2} w^2 a \\> \\frac{\\exp(-\\frac{1}{2c}(w-\\overline{w})^2)}{(2\\pi c)^{1/2}} \\mathrm{d}w \\end{align} Now we make a substitution of $$y = \\frac{1}{c^{1/2}} (w - \\overline{w})$$. \\begin{align} & = \\int \\frac{1}{2} ( yc^{1/2} + \\overline{w})^2 a \\frac{\\exp(-\\frac{1}{2} y^2 )}{(2\\pi)^{1/2} } \\mathrm{d}y \\\\ & = \\int \\frac{1}{2} ( y^2c + 2yc^{1/2}\\overline{w} + \\overline{w}^2 ) a \\> \\frac{\\exp(-\\frac{1}{2} y^2 )}{(2\\pi)^{1/2} } \\mathrm{d} y \\\\ \\end{align} And we notice that the term with $$2yc^{1/2}\\overline{w}$$ is odd, and that $$\\overline{w}^2$$ is independent of $$y$$. \\begin{align} & =\\frac{1}{2} \\overline{w}^2 a + \\int \\frac{1}{2} y^2c a \\> \\frac{\\exp(-\\frac{1}{2} y^2 )}{(2\\pi)^{1/2} } \\mathrm{d} y \\\\ & =\\frac{1}{2} \\overline{w}^2 a + \\frac{1}{2} ca \\frac{1}{(2\\pi)^{1/2}} \\int y^2\\> \\exp(-\\frac{1}{2} y^2 ) \\mathrm{d} y \\end{align} Finally, we integrate by parts, using $$u = y$$ and $$\\frac{\\mathrm{d} v}{\\mathrm{d} y} = y \\exp(-\\frac{1}{2} y^2)$$. \\begin{align} \\int \\frac{1}{2} w^2 a \\> Q(w)\\> \\mathrm{d} w &= \\frac{1}{2}\\overline{w}^2 a + \\frac{1}{2}ca \\end{align} • First principles tell you the answer will be a bilinear function of $C$ and $A$ and a quadratic", "$a,b,c,d$ simultaneously reduces the desired expression. (This is non-trivial. It's step 3 of my proof.) Furthermore, this reduction can result in negative lengths, which adds a lot of complexity. See steps 1 and 4 of my answer. – Will Nelson Dec 8 '13 at 20:50 WLOG, assume that $$a = \\max(a, b, c, d)$$. Let $$x = a- b, \\ y = a-c, \\ z = a-d$$. Then $$x, y, z \\ge 0$$. Let $$w = b+c+d - a$$. Then $$w > 0$$ since $$a, b, c, d$$ are the sides of a quadrilateral. The inequality is written as $$\\frac{1}{4}Aw^2 + \\frac{1}{4}Bw + \\frac{1}{4}C \\ge 0$$ where \\begin{align} A &= 2\\, x^2 - x\\, y - 2\\, x\\, z + 2\\, y^2 - y\\, z + 2\\, z^2,\\\\ B &= 2\\, x^3 + 4\\, x^2\\, y - 2\\, x\\, y^2 - 4\\, x\\, y\\, z + 2\\, y^3 + 4\\, y^2\\, z - 2\\, y\\, z^2 + 2\\, z^3,\\\\ C &= (x+3z)y^3 - 4xzy^2 + (3x^3+3x^2z-3xz^2+z^3)y. \\end{align} It suffices to prove that $$A, B, C \\ge 0$$. We have \\begin{align} A = (x^2 - xy + y^2) + (y^2 - yz + z^2) + (z^2 - 2zx + x^2) \\ge 0 \\end{align} and \\begin{align} B &= 2x^3 + 3x^2y + (x^2y - 2xy^2 + y^3) - 2(2xz)y +", "# Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \\geq \\frac{1}{3}$ for $x+y+z=1$ We have $x, y, z \\in \\mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \\geq \\frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship? Hint: $(x-\\frac13)^2 + (y-\\frac13)^2 + (z-\\frac13)^2 \\geq 0$. We have $$x+y+z=1\\\\ (x+y+z)^2=1\\\\ x^2+y^2+z^2+2(xy+xz+yz)=1$$ By the rearrangement inequality, $x^2+y^2+z^2\\geq xy+xz+yz$. Inserting that gives you $$3(x^2+y^2+z^2)\\geq1$$ For any real numbers $x,y,z$, we have \\begin{align} 0\\le(x-y)^2+(y-z)^2+(z-x)^2&\\implies2(xy+yz+zx)\\le2(x^2+y^2+z^2)\\\\ &\\implies(x+y+z)^2\\le3(x^2+y^2+z^2) \\end{align} So if $x+y+z=1$, then ${1\\over3}\\le x^2+y^2+z^2$.", "# How to show a complex number inequality A classmate consulted me this problem, after a few moment's thought I found it was difficult, so I wish to try my luck here. Let $z_1,z_2,z_3,z_4\\in \\mathbb{C}$ such that $|z_1|^2+|z_2|^2=|z_3|^2+|z_4|^2=1$. How to show $$(1-|z_1|^p)^{1/p}\\le(1-|z_3|^p)^{1/p}+(1-|z_1\\overline{z}_3+z_2\\overline{z}_4|^p)^{1/p},$$ where $p\\ge2$ is an integer? - Suppose we fix $w_1 = |z_1|$ and $w_3 = |z_3|$. In order to maximize $|z_1\\overline{z}_3 + z_2\\overline{z}_4|$, we need that the arguments of $z_1\\overline{z}_3$ and $z_2\\overline{z}_4$ be the same, and in that case the value will be $|z_1||z_3| + |z_2||z_4|$. So we can rewrite the inequality in an equivalent form: $$(1-w_1^p)^{1/p} \\leq (1-w_3^p)^{1/p} + (1-(w_1w_3 + \\sqrt{(1-w_1^2)(1-w_3^2)})^p)^{1/p}.$$ Equivalently, $$|(1-w_1^p)^{1/p} - (1-w_3^p)^{1/p}| \\leq (1-(w_1w_3 + \\sqrt{(1-w_1^2)(1-w_3^2)})^p)^{1/p}.$$ Writing $w_1 = \\sin \\alpha$ and $w_3 = \\sin \\beta$, this is the same as $$|(1-(\\sin \\alpha)^p)^{1/p} - (1-(\\sin \\beta)^p)^{1/p}| \\leq (1 - (\\cos (\\alpha-\\beta))^p)^{1/p}.$$ Here $0 \\leq \\alpha,\\beta \\leq \\pi/2$. That's as far as I got. - How did you get the $+$ in the expressions on the left hand side? – Raskolnikov Jun 5 '11 at 19:25 Sorry, corrected. – Yuval Filmus Jun 5 '11 at 19:48 I think what would help is a kind of \"continuity relation\" for convex/concave functions. $f(x)=(1-x^p)^{1/p}$ for $p\\geq 2$ is a concave function. There are often inequalities of the type $|f(x)-f(y)|<f(|x-y|)$ under appropriate restrictions on $x$ and", "$\\overline{EC}$ would be $y-x$ because $\\overline{DC}$ is $y$ and $\\overline{DE}$ is $x$. $\\overline{DE}$ is $x$ because it is parallel to $\\overline{AB}$ and both are of equal length. Because of the Pythagorean theorem, we know that $(EC)^2+(BE)^2=(BC)^2$. Substituting the values we have we get $(y-x)^2+(7)^2=(x+y)^2$. Simplifying this we get $(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)$. Now we get rid of the $x^2$ and $y^2$ terms from both sides to get $(-2xy)+(49)=(2xy)$. Combining like terms we get $(49)=(4xy)$. Then we divide by $4$ to get $(12.25)=(xy)$. Now we know that $x \\times y$ (same thing as $xy$) is equal to $12.25$ which is answer choice $\\boxed{\\textbf{(B)}\\ 12.25}$. Solution By: MATHCOUNTSCMS25 P.S. I Don't Know How to Format It Properly Using $\\LaTeX$ So Could Someone Please Fix It EDIT: Fixed! (As much as my ability can) - Mliu630XYZ EDIT: Fixed Completely! - palaashgang", "MAT334-2018F > Quiz-3 Q3 TUT 0102 (1/1) Victor Ivrii: Show that the function $w = g(z) = e^{z^2}$ maps the lines $\\{x = y\\}$ and $\\{x = -y\\}$ onto the circle $\\{w\\colon |w| = 1\\}$. Show that $g$ maps each of the two pieces of the region $\\{x + iy\\colon x^2 > y^2\\}$ onto the set $\\{w\\colon |w| > 1\\}$ and each of the two pieces of the region $\\{x + iy: x^2 < y^2\\}$ onto the set $\\{w\\colon |w| < 1\\}$. Draw all regions and domains. Quentin King: Let $z = x + iy$ Then $w$ can be rewritten as: $w = e^{(x+iy)(x+iy)} \\\\ w = e^{(x^2-y^2+i2xy)} \\\\ w = e^{x^2-y^2}(\\cos(2xy) + i\\sin(2xy)) \\\\$ Now find the modulus of $w$ $|w| = |e^{x^2-y^2}\\cos(2xy) + ie^{x^2-y^2}\\sin(2xy))| \\\\ |w| = e^{x^2-y^2}\\sqrt{\\cos^2(2xy)+\\sin^2(2xy)} \\\\ |w| = e^{x^2-y^2} \\\\$ - Now note that when you consider the line ${x=y}$: $|w| = e^{x^2-x^2} \\\\ |w| = e^0 \\\\ |w| = 1$ - And similarly on the line ${x=-y}$: $|w| = e^{(-y)^2-y^2} \\\\ |w| = e^0 \\\\ |w| = 1$ - For the region $\\{x+iy : x^2 > y^2\\}$: $|w| = e^{x^2-y^2} > e^0 > 1$ - And finally for the region $\\{x+iy : x^2 < y^2\\}$: $|w| = e^{x^2-y^2} < e^0 < 1$ EDIT: I dunno why, but my drawing attachment appears sideways", "be that $W=\\bigcap_{i=1}^r N_{g_i}$ since we have shown the left hand side is contained in the right hand side and both sides have the same dimension. #### Exercise 3.6.3 Let $S$ be a set, $F$ a field, and $V(S;F)$ the space of all functions from $S$ into $F$: $$(f+g)(x)=f(x)+g(x)$$$$(cf)(x)=cf(x).$$Let $W$ be any $n$-dimensional subspace of $V(S;F)$. Show that there exist points $x_1,\\dots,x_n$ in $S$ and functions $f_1,\\dots,f_n$ in $W$ such that $f_i(x_j)=\\delta_{ij}$. Solution: I’m not sure using the double dual is really the easiest way to prove this. It can be done rather easily directly by induction on $n$ (in fact see question 121704 on math.stackexchange.com). However, since H\\&K clearly want this done with the double dual. At first glance you might try to think of $W$ as a dual on $S$ and $W^{*}$ as the double dual somehow. But that doesn’t work since $S$ is just a set. Instead I think you have to consider the double dual of $W$, $W^{**}$ to make it work. I came up with the following solution. Let $s\\in S$. We first show that the function \\begin{alignat*}{1} \\phi_s:&W\\rightarrow F\\\\ &w\\mapsto w(s) \\end{alignat*}is a linear functional on $W$ (in other words for each $s$, we have $\\phi_s\\in W^*$). Let $w_1,w_2\\in W$, $c\\in F$. Then $\\phi_s(cw_1+w_2)=(cw_1+w_2)(s)$ which by definition equals $cw_1(s)+w_2(s)$ which equals $c\\phi_s(w_1)+\\phi_s(w_2)$.", "# Not sure what to do after trying to simplify the inequality and getting nowhere? Here is the question: If $$x, y, z \\in {\\displaystyle \\mathbb {R} }$$, then prove $$x^2+y^2+z^2+4\\ge2(x+y+z)$$ Here is what I tried doing: I tried simplifying the inequality and getting all the terms on one side, like so: $$x^2+y^2+z^2+4\\ge2x+2y+2z$$ $$x^2+y^2+z^2+4-2x+2y+2z\\ge0$$ $$(x^2-2x)+(y^2-2y)+(z^2-2z)+4\\ge0$$ $$(x^2-2x)+(y^2-2y)+(z^2-2z)\\ge-4$$ Is it possible to use the fact that it is always true that $$x^2-2x\\ge-4$$, $$y^2-2y\\ge-4$$, and $$z^2-2z\\ge-4$$, to show that inequality holds? Other than that, I'm not sure what to do next. I can't think of any manipulation that might work in this case. Any help would be greatly appreciated! • $x^2-2x\\ge-1$, so $(x^2-2x)+(y^2-2y)+(z^2-2z)\\ge-3$ – J. W. Tanner Jan 14 '20 at 1:45 \\begin{align} x^2 - 2x + y^2 - 2y + z^2 - 2z + 4 = (x-1)^2 + (y-1)^2 + (z-1)^2 + 1 \\ge 1 > 0. \\end{align} If you borrow three ones, you get $$(x^2-2x)+(y^2-2y)+(z^2-2z)+4=(x-1)^2+(y-1)^2+(z-1)^2+1\\geq1.$$", "# Proving a complex inequality Let $z$ and $w$ be complex numbers with $|z|< 1, |w|<1$. Show that $|1-\\overline zw|^2-|z-w|^2=(1-|z|^2)(1-|w|^2)$, and hence prove that $|\\frac{z-w}{1-\\overline zw}|<1$ I'm having trouble getting started with the first part. I tried expanding the numbers out with their real/imaginary components but got a giant mess that I couldn't work with. Could someone point me in the right direction? Hint $$|1-\\overline zw|^2-|z-w|^2 =(1-\\overline{z}w)(1-z\\overline{w})-(z-w)(\\overline{z}-\\overline{w}).$$ • In general $|z|^2 = z\\overline{z}$. – C. Dubussy Mar 2 '16 at 11:18" ]

[ "u = (1,−1, 0), v = (3, 0,−4) and w = (0, 2, 5). A plane which contains three points will have the same coefficients as the normal vector the plane, which is also normal to any two vectors in the plane. Two vectors in the plane can be found by \\displaystyle \\begin{align*} \\overline{u\\,v} &= \\overline{u\\,O} + \\overline{O\\,v} \\\\ &= -\\overline{O\\,u} + \\overline{O\\,v} \\\\ &= \\overline{O\\,v} - \\overline{O\\,u} \\\\ &= <3, 0, 4> - <1, -1, 0> \\\\ &= <2, 1, 4>\\end{align*} and \\displaystyle \\begin{align*} \\overline{u\\,w} &= \\overline{u\\,O} + \\overline{O\\,w} \\\\ &= -\\overline{O\\,u} + \\overline{O\\,w} \\\\ &= \\overline{O\\,w} - \\overline{O\\,u} \\\\ &= <0, 2, 5> - <1, -1, 0> \\\\ &= <-1, 3, 5> \\end{align*} So now to find the normal, you need to evaluate \\displaystyle \\begin{align*} \\overline{u\\,v} \\times \\overline{u\\,w} \\end{align*}. \\displaystyle \\begin{align*} \\overline{u\\,v} \\times \\overline{u\\,w} &= \\left|\\begin{matrix}\\phantom{-}i&j&k\\\\ \\phantom{-}2&1&4 \\\\ -1&3&5\\end{matrix}\\right| \\end{align*} • Apr 28th 2012, 10:51 AM TheIntegrator Re: Vector Parametric equation Find our direction vectors u,w = (0-1, 2 - -1, 5-0) = (-1, 3, 5) u,v = (3-1, 0 - -1, -4 -0) = (2, 1, -4) Now that we found the direction vectors, we need only apply one of the three points we already have to find our equation of the plane (x, y, z) = (1, -1, 0) + r(2, 1, -4) +", "vertices (0, 0), (1, 1) and (2, 1).(8 marks) 6 (a) Obtain half range sine series for \\begin {align*} f(x) &= x, 0\\le x \\le 2 \\\\ &=4-x, 2\\le x \\le 4 \\end{align*}(6 marks) 6 (b) Prove that the transformation $w = \\dfrac {1}{z+l}$ transforms the real axis of the z-plane into a circle in the w-plane.(6 marks) 6 (c) (i) Use Stoke's Theorem to evaluate $\\int_c \\overline F. \\overline {dr}$ where $\\overline {F}=(x^3 - y^2) \\widehat{i}+2xy\\widehat{j}$ and C is the rectangle in the plane z=0, bounded by x=0, y=0, x=a and y=b.(4 marks) 6 (c) (ii) Use Gauss Divergence Theorem to evaluate. $\\iint_s \\overline F. \\widehat{n}ds \\text{ where }\\overline F= 4x\\widehat{i} + 3y\\widehat{j} - 2z\\widehat{k}$ and S is the surface bounded by x=0, y=0 and 2x+2y+z=4.(4 marks)", "# Make the complex simple Level 2 Let $$m$$ be a line in the complex plane defined by $(1-i)z+(1+i)\\overline{z} =4.$ Let $$z_1=2+2i$$ be a point in the complex plane. If the reflection of $$z_1$$ in $$m$$ is $$z_{2}$$, then compute the value of $\\overline{z_{1}}(1+i)+z_{2}(1-i).$ ×", "along respective edges. Let faces $$W$$, $$X$$, $$Y$$, $$Z$$ be opposite vertices $$O$$, $$A$$, $$B$$, $$C$$. The reader is invited to verify these straightforward expressions for edge-lengths in terms of face-areas and volume: $$(a, b, c, d, e, f) = \\frac{2}{3V}\\left( YZ\\overline{A}, ZX\\overline{B}, XY\\overline{C}, WX\\overline{D}, WY\\overline{E}, WZ\\overline{F} \\right) \\tag{\\star}$$ where $$\\overline{\\theta} := \\sin\\theta$$. So, \"all we have to do\" is determine the face-areas and volume. This is not difficult. Viewing $$X$$ as the combined shadows of $$W$$, $$Y$$, $$Z$$ (and then viewing $$Y$$ and $$Z$$ similarly) gives rise to these analogues of the $$a= b\\cos C+c\\cos B$$ relations for a triangle: \\begin{align} X &= W \\ddot D + Y \\ddot C + Z \\ddot B \\\\ Y &= X \\ddot C + W \\ddot E + Z \\ddot A \\\\ Z &= X \\ddot B + Y \\ddot A + W \\ddot F \\end{align} \\tag1 where $$\\ddot{\\theta}:=\\cos\\theta$$. This comprises a simple linear system in $$X$$, $$Y$$, $$Z$$. Solving gives \\begin{align} X \\Delta &= W \\left(\\; \\ddot A \\left(-\\ddot A \\ddot D + \\ddot B \\ddot E + \\ddot C \\ddot F\\right) + \\ddot D + \\ddot C \\ddot E + \\ddot B \\ddot F \\;\\right) \\\\ Y \\Delta &= W \\left(\\; \\ddot B \\left(\\phantom{-}\\ddot A \\ddot D - \\ddot B \\ddot E + \\ddot C \\ddot F\\right) +", "Bounds for the Sum of Distances to the Vertices Problem Consider complex numbers $a,$ $b,$ $c$ that satisfy $a+b+c=0,$ $|a|=|b|=|c|=1.$ Prove that $3\\le |z-a|+|z-b|+|z-c|\\le 4,$ for any $z,$ with $|z|\\le 1.$ Solution Let's agree that - for simplicity - for any expression $f(x),$ $\\sum f(a)$ denotes $f(a)+f(b)+f(c).$ Since $|a|=1,$ so $|\\overline{a}|=1$ and $\\overline{a}a=1,$ etc., we have, \\begin{align} \\sum |z-a| &= \\sum \\overline{a}|z-a| = \\sum |\\overline{a}z - 1|\\\\ &\\ge |(\\overline{a}z - 1)+(\\overline{b}z - 1)+(\\overline{b}z - 1)|\\\\ &= |z\\sum\\overline{a}-3|=3, \\end{align} because $\\sum\\overline{a}=\\overline{\\sum a}=0.$ Now observe that the two conditions $a+b+c=0$ and $|a|=|b|=|c|=1$ imply that points $a,b,c$ on the unit circle form an equilateral triangle. Indeed, place the origin at the center of the circle and let $\\gamma$ be the angle between $a$ and $b$ (as vectors from the origin) then, by the Law of Cosines, \\displaystyle\\begin{align} |a+b|^{2} &= |a|^{2}+|b|^{2}-2|a||b|\\cos(180^{\\circ}-\\gamma )\\\\ &= 2+2\\cos\\gamma\\\\ &= 4\\cos^{2}\\frac{\\gamma}{2}. \\end{align} To insure $a+b+c=0,$ it is necessary to have $|a+b|=1$ which is achieved only when $\\displaystyle\\cos\\frac{\\gamma}{2}=\\pm\\frac{1}{2},$ i.e., when $\\displaystyle\\frac{\\gamma}{2}=\\pm 60^{\\circ},$ or, $\\gamma =\\pm 120^{\\circ}.$ Similarly, we verify that $a$ and $b$ form angles of $120^{\\circ}$ with $c.$ For a given point (complex number) $z,$ let $pq$ be the diameter passing through $z.$ There is a real number $\\alpha ,$ $0\\le\\alpha\\le 1$ such that $z=\\alpha p+(1-\\alpha)q.$ This helps prove the right inequality $|z-a|+|z-b|+|z-c|\\le 4:$ \\begin{align} \\sum |z-a|", "operator as well as the Real part \\begin{align*} &\\Re:\\mathbb{C}->\\mathbb{R} &\\Re(z)=\\Re(x+iy)=x \\end{align*} In fact, the latter is a projection operator which orthogonally projects a complex number $x+iy$ to the $x$-axis. It has the property that repeated application of it gives the same result. \\begin{align*} \\Re(\\Re(z))=\\Re(\\Re(x+iy))=\\Re(x)=x \\end{align*} The $\\Im$ operator does not share this property. Conjugation: It's convenient to consider the easy to remember rules of complex conjugation \\begin{align*} \\overline{z_1+z_2}&=\\overline{z_1}+\\overline{z_2}\\\\ \\overline{z_1-z_2}&=\\overline{z_1}-\\overline{z_2}\\\\ \\overline{z_1\\cdot z_2}&=\\overline{z_1}\\cdot\\overline{z_2}\\\\ \\overline{\\left(\\frac{z_1}{z_2}\\right)}&= \\frac{\\overline{z_1}}{\\overline{z_2}}\\\\ \\end{align*} together with \\begin{align*} \\Re(z)&=\\frac{1}{2}\\left(z+\\overline{z}\\right)\\\\ \\Im(z)&=\\frac{1}{2i}\\left(z-\\overline{z}\\right)\\\\ \\end{align*} Complicated situations: If we consider $H(z)$ in OPs example we obtain \\begin{align*} \\Im\\left(H(z)\\right)&=\\frac{1}{2i}\\left(H(z)+\\overline{H(z)}\\right) \\end{align*} and by multiple application of the rules for complex conjugation \\begin{align*} \\overline{H(z)}&=\\overline{\\frac{z(z+1)}{(z+2)(z+3)(z+4)}}\\\\ &=\\frac{\\overline{z}(\\overline{z}+1)}{(\\overline{z}+2)(\\overline{z}+3)(\\overline{z}+4)} \\end{align*} Can you continue? • Thank you very much. Excellent and very clear explanation. – BTestQ Jan 20 '16 at 0:31 • @BTestQ: You're welcome! :-) – Markus Scheuer Jan 20 '16 at 4:26 • Could it be a mistake, that Im operator is not \"idempotent\", since, if I apply it to a, say, 5+i4, I should get number 4, and since 4 is real, next time I apply Im to 4, I get 0 ? If shown in Cartesian coordinate system, first application of Im will project to imaginary axis, and every next time, there will be no need to \"move\" dot and it will remain on the same spot", "$$z=3-2i$$ This completes the proof. Example. ## Finding the complex conjugate (2) Find the complex conjugate of $5+8i$. Solution. The complex conjugate of $5+8i$ is: $$\\overline{5+8i}=5-8i$$ # Properties of complex conjugate Theorem. ## Sum of a complex number and its complex conjugate If $z=a+bi$, then $z+\\overline{z}$ is: $$z+\\overline{z}=2a$$ This means that $z+\\overline{z}$ is a real number. Proof. If $z=a+bi$, then $z+\\overline{z}$ is: \\begin{align*} z+\\overline{z} &=a+bi+(a-bi)\\\\ &=2a \\end{align*} This completes the proof. Theorem. ## Product of a complex number and its complex conjugate If $z=a+bi$, then the product $z\\overline{z}$ is: $$z\\bar{z}=a^2+b^2$$ This means that the product of a complex number and its complex conjugate results in a purely real numberlink. This is the trick we used to get rid of the imaginary part in the denominator when dividing two complex numbers! Proof. The product $z\\bar{z}$ is: \\begin{align*} z\\bar{z}&= (a+bi)(a-bi)\\\\&= a^2-abi+abi-b^2i^2\\\\ &=a^2+b^2 \\end{align*} Here, for the second equality, we used the fact that $i^2=-1$. This completes the proof. Example. ## Multiplying a complex number and its complex conjugate Let $z=3+4i$. Compute $z\\bar{z}$. Solution. By theoremlink, we have that: \\begin{align*} z\\bar{z}&=3^2+4^2\\\\ &=25 \\end{align*} Theorem. ## Complex conjugate of a sum of two complex numbers If $z_1$ and $z_2$ are complex numbers, then the complex conjugate of $z_1+z_2$ is: $$\\overline{z_1+z_2}= \\overline{z_1}+\\overline{z_2}$$ Proof. Let $z_1$ and $z_2$ be defined as follows: \\begin{align*} z_1&=a+bi\\\\", "\\,$ $| z + w | \\leq | z | + | w | \\,$ ( triangle inequality) $| z \\cdot w | = | z | \\cdot | w | \\,$ for all complex numbers z and w. It then follows, for example, that $|1|=1$ and $|z/w|=|z|/|w|$. By defining the distance function $d(z,w)=|z-w|$ we turn the set of complex numbers into a metric space and we can therefore talk about limits and continuity. The complex conjugate of the complex number $z=x+yi$ is defined to be $x-yi$, written as $\\bar{z}$ or $z^*\\,$. As seen in the figure, $\\bar{z}$ is the \"reflection\" of z about the real axis. The following can be checked: $\\overline{z+w} = \\bar{z} + \\bar{w}$ $\\overline{z\\cdot w} = \\bar{z}\\cdot\\bar{w}$ $\\overline{(z/w)} = \\bar{z}/\\bar{w}$ $\\bar{\\bar{z}}=z$ $\\bar{z}=z$ if and only if z is real $\\bar{z}=-z$ if and only if z is purely imaginary $|z|=|\\bar{z}|$ $|z|^2 = z\\cdot\\bar{z}$ $z^{-1} = \\bar{z}\\cdot|z|^{-2}$ if z is non-zero. The latter formula is the method of choice to compute the inverse of a complex number if it is given in rectangular coordinates. That conjugation commutes with all the algebraic operations (and many functions; e.g. $\\sin\\bar z=\\overline{\\sin z}$) is rooted in the ambiguity in choice of i (−1 has two square roots). It is important to note, however, that the function $f(z) = \\bar{z}$ is", "a }$$ × $$\\overline { b }$$, $$\\overline { b }$$ × $$\\overline { c }$$, $$\\overline { c }$$ × $$\\overline { a }$$ as coterminous edges is 8 cubic units, then the volume of the parallelepiped with ($$\\overline { a }$$ × $$\\overline { b }$$) × ($$\\overline { b }$$ × $$\\overline { c }$$), ($$\\overline { b }$$ × $$\\overline { c }$$) × ($$\\overline { c }$$ × $$\\overline { a }$$) and ($$\\overline { c }$$ × $$\\overline { a }$$) × ($$\\overline { a }$$ × $$\\overline { b }$$) as coterminous edges is (a) 8 cubic units (b) 512 cubic units (c) 64 cubic units (d) 24 cubic units Solution: (c) 64 cubic units Hint: Given volume of the parallelepiped with Question 12. Consider the vectors $$\\overline { a }$$, $$\\overline { b }$$, $$\\overline { c }$$, $$\\overline { d }$$ such that ($$\\overline { a }$$ × $$\\overline { b }$$) × ($$\\overline { c }$$ × $$\\overline { d }$$) = $$\\overline { 0 }$$ Let P1 and P2 be the planes determined by the pairs of vectors $$\\overline { a }$$, $$\\overline { b }$$ and $$\\overline { c }$$, $$\\overline { d }$$ respectively. Then the angle between P1 and P2 is (a) 0° (b)", "# Complex conjugate properties Here are some complex conjugate properties and identities that are useful to know for complex numbers $$z$$ and $$w$$. Complex conjugation is distributive over addition, subtraction, multiplication and division. \\begin{align*} \\overline{z+w} &= \\overline{z} + \\overline{w} \\\\ \\overline{z-w} &= \\overline{z}-\\overline{w} \\\\ \\overline{zw} &= \\overline{z} \\, \\overline{w} \\\\ \\overline{\\left( \\frac{z}{w} \\right)} &= \\frac{\\overline{z}}{\\overline{w}}, \\text{ if w\\ne0.} \\end{align*} We can relate conjugation with the real part and the imaginary part. \\begin{align*} z + \\overline{z} &= 2\\, \\text{Re}(z) \\\\ z-\\overline{z} &= 2i \\,\\text{Im}(z). \\end{align*} We have an identity for powers of $$z$$. $$\\overline{z^n} = (\\overline{z})^n \\text{, for integer }n.$$ Lastly, the conjugate of a conjugate of a complex number $$z$$ is $$z$$. $$\\overline{\\overline{z}} = z.$$ Let’s prove a few of these. Theorem. For complex numbers $$z$$ and $$w$$, we have $$z + \\overline{z} = 2\\, \\text{Re}(z)$$. Proof. Let $$z = x+yi$$, where $$x$$ and $$y$$ are real numbrs. Then $$z + \\overline{z} = x+yi + x-yi = 2x = 2\\text{Re}(z).$$ The next one is requires a bit more working, but relies on multiplying the top and the bottom of the fraction by the conjugate of the denominator. Theorem. For complex numbers $$z$$ and $$w$$, we have $$\\overline{\\left( \\frac{z}{w} \\right)} = \\frac{\\overline{z}}{\\overline{w}}.$$ Proof. Let $$z = a + bi$$ and $$w = c+di$$. Then \\begin{align*} \\overline{\\left( \\frac{z}{w} \\right)} &=", "$$z-w$$ and $$z+w$$ are the diagonals. $$2(|z|^2+|w|^2)$$ represents the sum of the squares of the sides, and $$|z+w|^2+|z-w|^2$$ represents the sum of the squares of the diagonals. The identity shows that, in any parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of the sides. ## Problems 1. Subtract the second complex number from the first. Plot them as vectors on the complex plane. 1. $$1+i$$ and $$2+i$$ 2. $$3-i$$ and $$4+i$$ 2. Find the distance between the two complex numbers on the complex plane. 1. $$3+4i$$ and $$-1-i$$ 2. $$-2-7i$$ and $$-3-2i$$ 3. $$100+100i$$ and $$-50+50i$$ 3. Show that the complex numbers $$-2-2i$$, $$1-i$$, $$2+2i$$ and $$-1+i$$ form a rhombus in the complex plane. 4. Show algebrically that $$|z+w|^2+|z-w|^2=2(|z|^2+|w|^2).$$ 5. Let $$z_1$$, $$z_2$$ and $$z_3$$ be complex numbers represented by the points $$A$$, $$B$$ and $$C$$ on the complex plane. 1. Write down the complex number representing the vectors $$\\overrightarrow{AB}$$ and $$\\overrightarrow{AC}$$. 2. Let $$D$$ be a point on the complex plane such that $$ABDC$$ is a parallogram. Show that the complex number which represents $$D$$ is $$z_2+z_3-z_1$$.", "out, the coefficients will not be real, unless you picked 3 real roots, or two complex conjugate root and one real root. The reason why any polynomial with real coefficients of odd degree (including cubics) must have at least one real root is because the highest power term dominates when the variable $x$ gets large. Assuming the coefficient of the highest term $x^n$ is 1, since $(-x)^n=-x^n$ when $n$ is odd, then when $x\\to\\infty$ and $-\\infty$, then your polynomial will tend to $\\infty$ and $-\\infty$ respectively. Somewhere in between it must cross the $x$ axis by the intermediate value theorem, giving us a real root. The reason why complex roots come in conjugate pairs, is because complex conjugation respects addition and multiplication. More precisely, if $z=x+iy$ and $w=a+bi$, and $\\overline{z}$ represents the complex conjugate of $z$, then: \\begin{align*}\\overline{z+w}&=\\overline{x+iy+a+bi} \\\\ &=\\overline{(x+a)+(y+b)i} \\\\ &=(x+a)-(y+b)i \\\\ &=(x-yi)+(a-bi) \\\\ &=\\overline{z}+\\overline{w} \\end{align*} and \\begin{align*} \\overline{zw}&=\\overline{(x+iy)(a+bi)}\\\\ &=\\overline{(xa-yb)+(xb+ya)i} \\\\ &=(xa-yb)-(xb+ya)i \\\\ &=(x-yi)(a-bi) \\\\ &=\\overline{z}\\cdot\\overline{w} \\end{align*} Repeated application of the second identity gives $\\overline{z^n}=(\\overline{z})^n$. So suppose we know that $x=w$ is a root of a polynomial $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$$ where $a_1,\\ldots,a_n$ are real numbers. Then \\begin{align*} p(\\overline{w})&=a_n\\overline{w}^n+a_{n-1}\\overline{w}^{n-1}+\\cdots+a_1\\overline{w}+a_0 \\\\ &= a_n\\overline{w^n}+a_{n-1}\\overline{w^{n-1}}+\\cdots+a_1\\overline{w}+a_0 \\\\ &= \\overline{a_nw^n}+\\overline{a_{n-1}w^{n-1}}+\\cdots+\\overline{a_1w}+\\overline{a_0} \\\\ &= \\overline{a_nw^n+a_{n-1}w^{n-1}+\\cdots+a_1w+a_0} \\\\ &= \\overline{p(w)} \\\\ &= 0 \\end{align*} meaning that $\\overline{w}$, the complex conjugate was also a root, and therefore complex roots come", "$-1 = i^2$: \\begin{align} \\Leftrightarrow \\quad x^2 = (-1)(4) = 4 i^2 \\quad \\Rightarrow \\quad x_\\pm = \\pm 2i \\end{align} ## Geometric Representation: Complex Plane In contrast to real numbers, the complex ones have two independent components: The real part $x$ and the imaginary part $y$. While the real numbers can be represented geometrically as a line, this does not work for the complex numbers anymore. Here a two-dimensional plane is required in order to represent all possible combinations of values for $x$ and $y$. It is very intuitive to represent these components $x$ and $y$ in analogy to cartesian coordinates along a rectangular grid: Such a two-dimensional representation of the complex numbers is referred to as an Argand diagram.[3] ## Calculation Rules ### Complex Conjugate The complex conjugate $\\overline{z}$ of a complex number $z= x + iy$ is obtained by switching the sign of the imaginary part:[4] \\begin{align} \\overline{z} := x - i y \\end{align} Using the notation of complex conjugates is really helpful. For instance one can compute both the real and the imaginary part of a complex number by using its complex conjugate: \\begin{align} \\Re{z} &= \\frac{z + \\overline{z}}{2} = \\frac{x+iy + x-iy}{2} = x \\\\ \\Im{z} &= \\frac{z - \\overline{z}}{2i} = \\frac{x+iy - \\left( x-iy \\right) }{2i} = y \\end{align} Furthermore, if one defines", "Line Passing Through Two Points¶ The equation between two point $$z_1$$ and $$z_2$$ is given by the determinant $\\begin{split}\\begin{vmatrix} z & \\overline{z}&1\\\\ z_1 & \\overline{z_1}&1\\\\ z_2 & \\overline{z_2}&1 \\end{vmatrix}=0\\end{split}$ or, $\\frac{z-z_1}{\\overline{z}-\\overline{z_1}}=\\frac{z_1-z_2}{\\overline{z_1}-\\overline{z_2}}$ Please refer to determinants section and coordinate geometry books on how to establish this result. The proof is left as an exercise to the reader. The parametric form is given by $$z = iz_1 + (1 - t)z_2$$ 9.12.4. Collinear Points¶ Three points $$z_1, z_2 \\text{ and } z_3$$ are collinear if and only if $\\begin{split}\\begin{vmatrix} z_1 & \\overline{z_1}&1\\\\ z_2 & \\overline{z_2}&1\\\\ z_3 & \\overline{z_3}&1 \\end{vmatrix}=0\\end{split}$ This follows from the formula for a line passing through two points. Just substitute $$z_3$$ as a point in the equation to satisfy the equality. 9.12.5. Parallelogram¶ Four complex numbers $$A(z_1), B(z_2), C(z_3)$$ and $$D(z_4)$$ represent the vertices of a parallelogram if $$z_1 + z_3 = z_2 + z_4.$$ The diagonals $$AC$$ and $$BD$$ must bisect each other. Therefore, \\begin{align}\\begin{aligned}\\frac{1}{2}(z_1 + z_3) = \\frac{1}{2}(z_2 + z_4)\\\\\\Leftrightarrow z_1 + z_3 = z_2 + z_4\\end{aligned}\\end{align} 9.12.6. Rhombus¶ Four complex numbers $$A(z_1), B(z_2), C(z_3)$$ and $$D(z_4)$$ represent the vertices of a rhombus if $$z_1 + z_3 = z_2 + z_4$$ and $$|z_4 - z_1| = |z_2 - z_1|.$$ The diagonals $$AC$$ and $$BD$$ must bisect each other. Therefore, \\begin{align}\\begin{aligned}\\frac{1}{2}(z_1 + z_3) = \\frac{1}{2}(z_2 +", "# Make the complex simple Level pending Let $$m$$ be a line in the complex plane defined by: $(1-i)z+(1+i)\\overline{z} =4$ Let $$z_1=2+2i$$ be a point in the complex plane. If the reflection of $$z_1$$ in $$m$$ is $$z_{2}$$, then compute the value of: $\\overline{z_{1}}(1+i)+z_{2}(1-i)$ ×", "x\\geq 2y^2, \\ -\\infty is cut below by the plane $$z=0$$ and on top by the plane $$z=4-x-2y$$. These two planes intersect in the line $$\\ell: \\ x+2y=4\\ \\wedge\\ z=0\\ ,$$ creating an infinite wedge $$W$$ above $$z=0$$. Our $$D$$ then is the intersection $$P\\cap W$$. The line $$\\ell$$ intersects the parabola $$x=2y^2$$ in the points $$(8,-2)$$ and $$(2,1)$$. We therefore can write $$D$$ in the form $$D:=\\bigl\\{(x,y,z)\\bigm| -2\\leq y\\leq 1, \\>2y^2\\leq x\\leq4-2y, \\ 0\\leq z\\leq 4-x-2y\\bigr\\}\\ .$$ This indicates that we can write an integral over $$D$$ in the form $$\\int_D f(x,y,z)\\>{\\rm d}(x,y,z)=\\int_{-2}^1\\int_{2y^2}^{4-2y}\\int_0^{4-x-2y} f(x,y,z)\\>dz\\>dx\\>dy\\ .$$", "# Math Help - Please help me!! >.< 1. ## Please help me!! >.< Please teach me how to solve this question... Two complex number w and z are such that w*= z -2i and |w|^2=z+6. By eliminating z, find w in the form a+ib, where a and b are real and positive. please help me! I really have many problems with complex number!! 2. ## Re: Please help me!! >.< I assume you are using the * to represent conjugate. You should know that \\displaystyle \\begin{align*} w\\,\\overline{w} = |w|^2 \\end{align*}, so we have \\displaystyle \\begin{align*} w\\,\\overline{w} = z + 6 \\end{align*} and \\displaystyle \\begin{align*} \\overline{w} = z - 2i \\end{align*}. Subtracting the second equation from the first gives \\displaystyle \\begin{align*} w\\,\\overline{w} - \\overline{w} &= 6 + 2i \\\\ \\overline{w}\\left( w - 1 \\right) &= 6 + 2i \\\\ \\left( a - b\\,i \\right) \\left( a - 1 + b\\,i \\right) &= 6 + 2i \\\\ a(a - 1) + ab\\,i - (a - 1)b\\,i - b^2 i^2 &= 6 + 2i \\\\ a^2 - a + b^2 + b\\,i &= 6 + 2i \\\\ a^2 - a + b^2 = 6 \\textrm{ and } b &= 2 \\\\ a^2 - a + 4 &= 6 \\\\ a^2 - a - 2 &= 0 \\\\ (a - 2)(a", "$$Re(z) = Re(\\bar z)$$ • $$Im(z) =- Im(\\bar z)$$ • If $$z$$ is purely real, then $$z=\\bar z$$. For example, the complex conjugate of $$z=3$$ is $$\\bar z =3$$. • If $$z$$ is purely imaginary, then $$z=-\\bar z$$. For example, the complex conjugate of $$z=2i$$ is $$\\bar z = -3i$$. • The bar over two complex numbers with some operation in between can be distributed to each of the complex numbers. i.e., if $$z_1$$ and $$z_2$$ are any two complex numbers, then: \\begin{align} \\overline{z_1+z_2} &= \\bar{z_1}+ \\bar{z_2}\\\\[0.2cm] \\overline{z_1-z_2}&= \\bar{z_1}- \\bar{z_2}\\\\[0.2cm] \\overline{z_1 \\cdot z_2}&= \\bar{z_1} \\cdot \\bar{z_2}\\\\[0.2cm] \\overline{\\left(\\ \\dfrac{z_1}{z_2} \\right)}&= \\dfrac{ \\bar{z_1}}{ \\bar{z_2}}\\end{align} Complex Numbers grade 10 | Questions Set 1 Complex Numbers Complex Numbers", "$$\\overline{A C}$$ = P.v of C – P.v of A = i – 2 j – 3 k -(i – j + 4k) = – j – 7k Let A be the angle between $$\\overline{A B}$$ and $$\\overline{A C}$$ 3. Area of the parallelogram ### Plus Two Maths Vector Algebra Six Mark Questions and Answers Question 1. Using this figure answer the following questions. 1. Find $$\\overline{O A}$$, $$\\overline{O B}$$, $$\\overline{O C}$$ (2) 2. Find $$\\overline{O D}$$ (2) 3. Find the coordinate of the vertex D. (2) 1. $$\\overline{O A}$$ = (3 – 1)i + (-1 – 2)j + (7 – 3)k = 2i – 3j + 4k $$\\overline{O B}$$ = (2 – 1)i + (4 – 2)j +(2 – 3)k = i + 2j – k $$\\overline{O C}$$ = (4 – 1)i + (1 – 2 )j + (5 – 3 )k = 3i – j + 2 k. 2. From the figure, 3. Let the vertex of D be (x , y , z), Then, $$\\overline{O D}$$ = (x – 1)i + (y – 2)j + (z – 3)k. But we have, $$\\overline{O D}$$ = 6i – 2j + 5k = (x – 1)i + (y – 2)j +(z – 3)k x – 1 = 6 ⇒ x = 7, y – 2 = -2 ⇒", "$$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is A $$1:2:1$$ B $$1:2:3$$ C $$2:1:2$$ D $$1:1:1$$ ## Explanation Intersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\\left( {0,0} \\right)$$ and $$\\left( {4,4} \\right).$$ The graph is as shown in the figure. By symmetry, we observe $${S_1} = {S_3} = \\int\\limits_0^4 {ydx = \\int\\limits_0^4 {{{{x^2}} \\over 4}dx} }$$ $$= \\left[ {{{{x^3}} \\over {12}}} \\right]_0^4 = {{16} \\over 3}$$ sq. unit Also $${S_2} = \\int\\limits_0^4 {\\left( {2\\sqrt x - {{{x^2}} \\over 4}} \\right)} dx$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$= \\left[ {{{2{x^{{3 \\over 2}}}} \\over {{3 \\over 2}}} - {{{x^3}} \\over {12}}} \\right]_0^4$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$= {4 \\over 3} \\times 8 - {{16} \\over 3} = {{16} \\over 3}$$ $$\\therefore$$ $${S_1}:{S_2}:{S_3} = 1:1:1$$ 4 ### AIEEE 2005 The area enclosed between the curve $$y = {\\log _e}\\left( {x + e} \\right)$$ and the coordinate axes is A $$1$$ B $$2$$ C $$3$$ D $$4$$ ## Explanation The graph of the curve $$y = {\\log _e}\\left( {x + e} \\right)$$ is as shown in the fig. Required area $$A = \\int\\limits_{1 - e}^0 {ydx} = \\int\\limits_{1 - e}^0 {{{\\log }_e}} \\left( {x + e} \\right)dx$$ put $$x + e = t \\Rightarrow dx = dt$$ also At" ]

[ "is true for z= 1 and z= -3 but the complex plane is a plane including points both above and below the real line. Geometrically |z- a| can be interpreted as distance from z to a in the complex plane and so the set of points, z, satisfying |z-a|= r is the set of points that have distance r from a: a circle with radius r and center a. The set of points, z, satisfying |z+1|= |z- (-1)|= 2 is the circle with center at -1 and radius 2. The real line is a diameter of that circle and z= 1 and z= -3 are the endpoints of that diameter. If z= 2, then |z+ 1|= |2+ 1|= 3 which is outside the circle. If z= -2, then |z+ 1|= |-2+1|= 1 which in inside the circle. 4. Feb 27, 2009 ### Bacat Thank you! I understand now.", "# Too Complex to Solve! Algebra Level 3 Let $$z=a+bi$$, $$|z|=5$$, and $$b>0$$. If the distance between $$(1+2i)z^3$$ and $$z^5$$ is as large as possible, then determine $$z^4$$. Give your answer as the sum of the real and imaginary parts of $$z^4$$. ×", "How do you Plot the complex number z=-4sqrt3-4i? So plot $\\left(- 4 \\sqrt{3} , - 4\\right)$ on the rectangular graph sheet, naming y-axis", "# How do I graph the complex number 3+4i in the complex plane? Remember to use the horizontal axis to plot the REAL part and the vertical one to plot the coeficient of the immaginary part (the number with $i$). So, for a general complex number, $z = a + i b$ you have the following graph: In your case, where the number is given as: $z = 3 + 4 i$, you get:", "is the area in between them. 2) $|z-3+4i| \\leq 4$ is a circle. $|z| = |z-10|$ is the line x = 5. $|z-3+4i|\\leq4$ and $|z|\\geq |z-10|$ is the area between the circle and the line (you will need to decide which of the two possible areas it is ....) By the way, all these relations have a very simple geometric interpretation which makes it easy to get the cartesian equations .... I guess i understand what you mean. But the thing is I am very bad at inequalities and geometric. I will post my answers later to check. Thanks! 4. Do you understand that |z| is the distance from the complex number z to the number 0= 0+ 0i? From that it follows that |a- b| is the distance between the two complex numbers a and b. Saying that |z+ 3i|= |z- (-3i)|= 1 means that the distance from z to -3i is 1: z can be any number on the circle with center at -3i and radius 1. Saying that $1\\le |z+ 3i|$ means that z is on or outside that circle. Similarly |z+ 3i|= |z- (-3i)|= 4 means that the distance from z to -3i is 4: z can be any number on the circle with center at -3i and radius 4. $|z+ 3i|\\le 4$ says", "Home > English > Class 11 > Maths > Chapter > > If complex number z satisfies ... # If complex number z satisfies the condition |z-2+i|<=1, then maximum distance of origin from A(4+i(2-z)) is Updated On: 27-06-2022", "((x+2)^2) - y^2 =16 where am i going wrong |z + 2| = 4 <=> |z - (-2)| = 4. So the geometric interpretation is the set of all points in the complex plane that are a distance 4 from z = -2. This is a circle of radius 4 and centre at z = -2, that is, (-2, 0).", "# Sequence of Complex Stuff Algebra Level 4 $\\large x_1=0, \\ \\ \\text{and} \\ \\ x_{n+1}=x_n^2-i \\ \\ \\text{ for} \\ \\ n>1$ Above is the sequence $$x_1,x_2,x_3, \\ldots$$ of the complex numbers.. Find the square of the distance between $$x_{2000}$$ and $$x_{1997}$$ in the complex plane. Details and Assumptions: $$i = \\sqrt{-1}$$ is the imaginary number. ×", "# Yet another easy question Algebra Level 5 Let $$z$$ be a complex number satisfying the relation $$~~~~~ z = 2 + t + i\\sqrt{3 - t^2}.$$ Find the locus of $$z$$ in argand plane. Where $$t$$ is a real parameter and $$t^2\\leq3$$ Try more here ×", "+ iy| + |x + 1 + iy|^2 \\\\ (x-1)^2 + y^2 &= 16 - 8|x + 1 + iy| + (x+1)^2 + y^2 \\\\ 8|x + 1 + iy| &= 16 + (x+1)^2 - (x - 1)^2 \\\\ 8|x + 1 + iy| &= 16 + 4x \\\\ \\end{align} Now square both sides again and continue simplifying. 4. Feb 10, 2013 ### Ray Vickson Re: Complex analysis topic help! Need explanation!!!! For $|z-1| + |z + 1| = 4$ it is easiest to look at the geometry. The term $|z-1|$ is the distance from z to p = -1 + 0i, while $|z-1|$ is the distance from z to q = +1 + 0i. Therefore, the left-hand-side is the sum of the distances to two fixed points, and that sum must be the constant 4. Can you recognize what type of curve that would be? 5. Feb 12, 2013 ### nate9228 Re: Complex analysis topic help! Need explanation!!!! Thanks guys that definitely helps. I'm sure I'll be back with other questions soon!", "this line segment? 1. $-2 - 5i$ 2. $-4i$ 3. $2 + 5i$ 4. $3/2 - 92 i$ 4. Which of the listed complex numbers, when represented by a point in the complex plane, is both equidistant and collinear with the complex numbers represented by points A and K? 1. $3i$ 2. $1 + 3/2 i$ 3. $1/2 + 3i$ 4. $3 + 1/2 i$ 5. Let $z=-8+5i$. If the midpoint of the line segment created by complex numbers $z$ and $w$ is $-3+3/2 i$, what is the value of $w ?$ 1. $w = -11/2 + 13/4 i$ 2. $w = 2-2i$ 3. $w = 14 + i$ 4. $w = 4 + 7/2 i$ 6. Find the distance between $z = 6 + 7i$ and $w = -2 - 3i$ in the complex plane. 1. $4sqrt(2)$ 2. $18$ 3. $2sqrt(2)$ 4. $2sqrt(41)$ 7. Which of the following expressions represent(s) the distance between $z_1$ and $z_2$ in the complex plane? Choose all correct answers. 1. $|z_1 - z_2|$ 2. $|z_2 - z_1|$ 3. $|z_2| - |z_1|$ 4. $|z_2 + z_1|$ 8. What is the distance between the complex numbers $2-4i$ and $-3/2 - 1/2 i$ in the complex plane? 1. $7/2 sqrt(2)$ 2. $1/2sqrt(130)$ 3. $5/2 sqrt(2)$ 4. $1/8 sqrt(82)$ 9. If $z = 5 + 4i$,", "5 # Plot the complex numbers on the complex plane.$-3-4 i$... ## Question ###### Plot the complex numbers on the complex plane.$-3-4 i$ Plot the complex numbers on the complex plane. $-3-4 i$", "### Real(ly) Numbers If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have? ### Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. ### Quartics Investigate the graphs of y = [1 + (x - t)^2][1 + (x + t^)2] as the parameter t varies. # Three Ways ##### Stage: 5 Challenge Level: Given that $x + y = -1$ find the largest value of $xy$ 1. by coordinate geometry. 2. by calculus. 3. by algebra.", "and a in the complex plane. Saying that z satisfies $1\\le |(z+ 2i)- 0|\\le 2$ means that z+ 2i is in the \"washer\" with center 0, inner radius 1, outer radius 2. By \"writing z+ 2+ i as z+ 2i+ 2-i\" we can interpret |z+ 2+ i|= |(z+2i)- (i-2)| as the distance from the point z+2i to the point i-2. |i- 2|= $\\sqrt{5}$> 2 so i- 2 is outside that \"washer\". The shortest distance from i- 2 to the larger circle is $\\sqrt{5}- 2$ while the longest distance, to the other side of the outer circle, is $\\sqrt{5}+ 2$.", "the following formula: $Z_{x+1}=Z_{x}+C$. If the pythagorean distance of Z from 0 goes above a certain number (I used 1) then it is not in the set, move on the the next point. Color the pixel in the complex plane with something representing how many iterations the pixel took to escape. Some will never escape, others will fly out into infinity in only a couple of iterations. I always thought the set looked like a fat guy, surrounded by a nimbus of self-similar fat guys, sitting on the john. I'll be out in a minute! ### Related Posts: • No Related Posts This entry was posted in My Homeboy and tagged , , . Bookmark the permalink.", "## Geometry of complex numbers Dear professor, I was doing some revision and I came across a question I would like to ask you. Please enlighten me on it. Find the least value of $|z|$ if $|z - i| = |z - 2|.$ Hope you had a good holiday. ——————————————— Regarding your question, note that $|z-w|$ for any two complex numbers $z$ and $w$ is just the distance between them regarded as points in the plane. In particular, you should be able to draw the line (why?) defined by the condition $|z - i| = |z - 2|,$", "+0 # Help pls 0 35 1 The points $P,$ $Q,$ and $R$ are represented by the complex numbers $z,$ $(1 + i) z,$ and $2 \\overline{z},$ respectively, where $|z| = 1.$ When $P,$ $Q$, and $R$ are not collinear, let $S$ be the fourth vertex of the parallelogram $PQSR.$ What is the maximum distance between $S$ and the origin of the complex plane? Feb 6, 2021", "# Is there a proof that the distance from $0$ to $i$ in the complex plane is $1$? I was just wondering how did people know that the distance between $$0$$ and $$i$$ is $$1$$ in the complex plane, did they just assume this, is it just an axiom, or is there a proof behind it or a reason for it? • I suppose it's just a convention to put $i$ at a distance $1$ from the origin in the complex plane; but it's nice to have $|zw|=|z||w|$ where $|z|$ denotes the distance from $z$ to the origin, don't you think? Nov 3 '18 at 20:34 • In some sense, I think the ordinate of a point $z$ at complex plane is exactly its imaginary part, by definition $b\\in\\mathbb{R}$ such that $z=a+bi$ ($a\\in\\mathbb{R}$ as well). I understand the answers below, by in some sense I thinks it's a definition (maybe some answer are ciclic). I'd like to receive answer to this comment... I'm curious now. Thanks. Nov 3 '18 at 20:35 • How did people know that the distance between $0$ and $1$ is $1$ in the complex plane? Same thing. – MPW Nov 3 '18 at 20:45 • How did we know that the distance from $(0,1)$ to $(0,0)$ = $1$? Nov 3 '18 at 20:55 •", "Real(ly) Numbers If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have? Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Quartics Investigate the graphs of y = [1 + (x - t)^2][1 + (x + t^)2] as the parameter t varies. Three Ways Stage: 5 Challenge Level: Given that $x + y = -1$ find the largest value of $xy$ 1. by coordinate geometry. 2. by calculus. 3. by algebra.", "Re: Sketch the following sets in the complex plane Originally Posted by Prove It a) Completing the Square, enables you to put the circle into standard form. b) Surely you know how to evaluate the modulus of a complex number... $\\displaystyle |a + bi| = \\sqrt{a^2 + b^2}$. Ah, of course! to both Thanks a lot [SOLVED] 7. ## Re: Sketch the following sets in the complex plane Originally Posted by terrorsquid $b)~ |z+i|\\ge|z-i|$ Here is a different way to approach this question. $|z-w|$ is just the distance between $z~\\&~w$. So you want the complex numbers whose distance from $\\mathif{-i}$ is greater than or equal to the distance to $\\mathif{i}$. 8. ## Re: Sketch the following sets in the complex plane Originally Posted by Plato Here is a different way to approach this question. $|z-w|$ is just the distance between $z~\\&~w$. So you want the complex numbers whose distance from $\\mathif{-i}$ is greater than or equal to the distance to $\\mathif{i}$. Ah yeh, this is the method that someone mentioned to me but they couldn't seem to explain my question. Shouldn't it be everything below and including the $x$ axis if you want numbers whos distance between $z$ and $+i$ is $\\ge$ the distance between $z$ and $-i$, i.e. all the numbers that are closer to $-i$ (lesser" ]

[ "3i | = 5$ is the circle centered at $(-4,-3)$ of radius $5$. I claim that this circle is wholly contained inside the circle $|z| = 11$. This is evident geometrically, and to prove it observe that by the triangle inequality $|w| \\leq |w + 4 + 3i| + |-(4 + 3i)| = 5 + 5 = 10$. Equality occurs when $w = -8 - 6i$, the farther of the two points on the circle that are on the line connecting the origin to the center $(-4,-3i)$. The minimum of $11 - |w|$ on this circle occurs when $|w|$ is maximized, namely at $w = -8 - 6i$ when $|w| = 10$. So the minimum distance is given by $11 - 10 = 1$. - Let $z=a+bi$ and $w=c+di$. Now, $|z|=11\\implies \\sqrt{a^2+b^2}=11\\implies a^2+b^2=121$, so $z$ is constrained to this circle (shown in blue below). On the other hand, $|w+4+3i|=5\\implies \\sqrt{(c+4)^2+(d+3)^2}=5\\implies (c+4)^2+(d+3)^2=25$, so $w$ is constrained to this circle (in red below). The task is to minimize $|z-w|=\\sqrt{(a-c)^2+(b-d)^2}$ subject to the two constraints above. Geometrically, we are looking to find the shortest distance between any two points which lie on the blue and red circles, respectively. A little calculus reveals the minimum is $1$ and is obtained when $a=-44/5$, $b=-33/5$ and $c=-8$, $d=-6$, as shown below. - I have", "# How to find the maximum and minimum value of $\\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\\right|$ (where $|z_1|=|z_2|=|z_3|=1$)? How to find the maximum and minimum value of $\\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ? My try: \\begin{align}\\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\\right| &\\leq |z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \\\\ &\\leq (|z_1|+|z_2|)^2 + (|z_2|+|z_3|)^2 + (|z_3|+|z_1|)^2 \\\\ &\\leq 2^2+2^2+2^2 \\leq 12\\end{align} However the answer given is $8$. Where am I going wrong and how to do it correctly? • Note that these are points on the unit circle, so the max distance two points can have is the length of the diameter. However, you are dealing with 3 points. Try to find out how you would arrange them on the unit circle to maximize distance. – David Feb 17 '17 at 20:16 • @DougM I think you found the maximum of $|(z_1-z_2)|^2 + |(z_2-z_3)|^2 + |(z_3-z_1)|^2$ rather than the one given in question. I checked on Wolfram Alpha. The correct answer is 8. – user400242 Feb 17 '17 at 20:55 • We do not change the absolute value if we multiply with $\\overline z_3^2$. Therefore, $|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2| = |(\\overline z_3z_1-\\overline z_3z_2)^2+(\\overline z_3z_2-1)^2+(1-\\overline z_3z_1)^2|$. With the substitution $u_1 =\\overline z_3z_1$ and $u_2= \\overline z_3z_2$ this reduces the number of variables to 2 (which hopefully makes the problem easier). –", "is the area in between them. 2) $|z-3+4i| \\leq 4$ is a circle. $|z| = |z-10|$ is the line x = 5. $|z-3+4i|\\leq4$ and $|z|\\geq |z-10|$ is the area between the circle and the line (you will need to decide which of the two possible areas it is ....) By the way, all these relations have a very simple geometric interpretation which makes it easy to get the cartesian equations .... I guess i understand what you mean. But the thing is I am very bad at inequalities and geometric. I will post my answers later to check. Thanks! 4. Do you understand that |z| is the distance from the complex number z to the number 0= 0+ 0i? From that it follows that |a- b| is the distance between the two complex numbers a and b. Saying that |z+ 3i|= |z- (-3i)|= 1 means that the distance from z to -3i is 1: z can be any number on the circle with center at -3i and radius 1. Saying that $1\\le |z+ 3i|$ means that z is on or outside that circle. Similarly |z+ 3i|= |z- (-3i)|= 4 means that the distance from z to -3i is 4: z can be any number on the circle with center at -3i and radius 4. $|z+ 3i|\\le 4$ says", "that is where we get our x value: $(19, d)$ $(19, -d)$ $(17, -3d)$ Now, we can plug one of the first two value in as well as the last one to get the following equations: $$19^2 + d^2 = r^2$$ $$17^2 + (3d)^2 = r^2$$ Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ - therefore, we get $72 = 8d^2 \\rightarrow d^2 = 9 \\rightarrow d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\\boxed{B}$. ~Tony_Li2007 ## Solution 3 (Stewart's Theorem) $[asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label(\"O\", O, N); label(\"C\", C, SW); label(\"D\", D, SE); label(\"E\", E, SW); label(\"F\", F, SE); label(\"P\", P, SW); label(\"Q\", Q, S); [/asy]$ If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives $$19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.$$ This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem", "z^2-(4+5i)z-3+9i=0 => z=[(4+5i)+/-sqr(4+5i)^2+4(3-9i)]/2 => z=[(4+5i)+/-sqr(3+4i)]/2 => z=[(4+5i)+/-(2+i)]/2 => z1=(6+6i)/2=3+3i. Example 1. Physics. If you're using complex numbers, then every polynomial equation of degree #k# yields exactly #k# solution. Determine (24221, 122/221, Arg(2722), And Arg(21/22). Ask your question. KEAM 2016: If |z-(3/2)|=2 , then the greatest value of |z| is (A) 1 (B) 2 (C) 3 (D) 4 (E) 5. Then the minimum value of |z1 – z2| is : asked Apr 16, 2019 in Mathematics by Niharika ( 75.6k points) If z=3- 4i is turned 90^@ in anti clock direction then new position of z is. $\\begingroup$ If the series converges at $3+4i$ then it is absolutely convergent for any $|z| \\le 5$, so the radius of convergence is equal or larger than $5$. …, . rohankedia3541 is waiting for your help. z= 3-4i. If |z-3+2i|＜=4 then the difference between the greatest and the least value of |z| is : A) 2(13^1/2) B) 8 C) 4+((13)^1/2) D) (13)^1/2 The inequality |z-3+2i| 3 What is Z Array? |z−(3+4i)| ≤ 3 is the interior+boundary of a circle centre (3,4) and radius 3. z of least magnitude is where line joining O to centre meets circle. asked Jan 27, 2015 in TRIGONOMETRY by anonymous. b) If Z[K] >= R-i+1 then it is possible to extend the [L,R] interval thus we will", "draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$. Let the radius of circle $P$ be $r$. Using Distance Formula, we get the following system of three equations: $$h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2$$ By simplifying, we get $$h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1$$ By subtracting the first equation from the second and third equations, we get $$8r=-4h+12, 10r=2h+6$$ which simplifies to $$2r=3-h, 5r=h+3$$ When we add these two equations, we get $$7r=6$$ So $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ $w^5$ ## Solution 5(Inversion from Circular Inversion) Let $\\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram: $[asy] size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NE); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0),", "# Minimum value of $|z-w|$ where $z,w \\in \\mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? I was thinking about the problem: What is the minimum value of $|z-w|$ where $z,w \\in \\mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? My attempts: I notice that $|z-w| \\geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5$ gives $|w|^2+8u+6v=0$. Now i can not proceed. Can someone point me in the right direction? Thanks in advance for your time. - If you think of it geometrically, you'll get a sense of what you needed to do algebraically to actually prove what the answer is. $|z|=11$ means $z$ is a point of the circle centered in $0$ and radius $11$. $|w+4+3i|=5$ means $w$ is a point of the circle centered in $4+3i$ and radius $5$. This isn't an answer or a hint, just an observation. – Git Gud Jan 10 '13 at 17:59 Thanks a lot sir.It's been useful observation. +1 from me. – user52976 Jan 10 '13 at 18:11 Geometrically, $|z| = 11$ is the circle centered at the origin of radius $11$. If a point $w$ is inside the circle $|z| = 11$, elementary geometry shows that the the minimum distance from $w$ to this circle is $11 - |w|$ (the outward distance from $w$ to the circle). $|w + 4 +", "+0 0 92 1 Let $X$, $Y$, and $Z$ be points on a circle. Let $\\overline{XY}$ and the tangent to the circle at $Z$ intersect at $W$. If $WX = 4$, $WZ = 8$, and $\\overline{WY} \\perp \\overline{WZ}$, then find $YZ$. [asy] unitsize(2 cm); pair A, B, C, D; D = (0,1); B = dir(150); C = dir(210); A = (-sqrt(3)/2,1); draw(Circle((0,0),1)); draw(D--A--C); label(\"$W$\", A, NW); label(\"$X$\", B, SE); label(\"$Y$\", C, SW); label(\"$Z$\", D, N); label(\"$8$\", (A + D)/2, N); label(\"$4$\", (A + B)/2, W); [/asy] Feb 14, 2020", "# Difference between revisions of \"2012 AIME II Problems/Problem 6\" ## Problem 6 Let $z=a+bi$ be the complex number with $\\vert z \\vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$. ## Solution Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have $|z^3| * |z^2 - (1+2i)|$ $=|z|^3 * |z^2 - (1+2i)|$ $=125|z^2 - (1+2i)|$ Thus we only need to maximize the value of $|z^2 - (1+2i)|$. To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\\frac{-1}{\\sqrt{5}} + \\frac{-2}{\\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \\sqrt{5} (-5 - 10i)$ Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = 125$", "+0 HELPME!?!?!? 0 92 3 A circle is centered at $O.$ The tangent to the circle at $P$ is extended to $Q.$ Line segment $\\overline{QS}$ intersects the circle at $R.$ Given that $OS = 2,$ $SR = RQ = 3$, and $PQ = 6$, find the radius of the circle. [asy] unitsize(2 cm); pair A, B, C, D, E, F, G, O; A = dir(70); B = A + 1.2*dir(-20); C = dir(25); D = 2*C - B; O = (0,0); draw(Circle(O,1)); draw(A--B--D--O); dot(\"$P$\", A, N); dot(\"$Q$\", B, dir(0)); dot(\"$R$\", C, SW); dot(\"$S$\", D, NW); dot(\"$O$\", O, SW); [/asy] Aug 14, 2020 #1 0 Image Here: Aug 15, 2020 #2 0 By power of a point, the radius of the circle is 4*sqrt(7). Aug 15, 2020 #3 +110762 0 Maybe it would be in your interest to present your questions better. All those unnecessary dollar signs make it difficult to read and I feel that if you cannot be bothered to present your question properly then it is not worth my effort to attempt to answer. Aug 15, 2020", "bi. In my opinion, $\\sin$ and $\\cos$ are unchanged after increasing or decreasing an angle by $360^\\circ$ because turning something $360^\\circ$ around the origin puts it back where you started. Then z 3 = z 1z 2, where: z 3 = −8+6i = √ 100eiθ 3 θ 3 ≈ 2.498 r √ 5eiθ 1 r √ 20eiθ 2 r 10eiθ 3 ⑥ Figure 3 Applying (4) to z 1 = z 2 = −4+4i = 4 √ 2e3 4 πi (our earlier example), we get (−4+4i)2 = (4 √ 2e34πi)2 = 32e 3 2 πi = −32i. if z= 3-4i, then z 4-3z 3 +3z 2 +99z-95 is equal to ans. $\\begingroup$ If the series converges at $3+4i$ then it is absolutely convergent for any $|z| \\le 5$, so the radius of convergence is equal or larger than $5$. |z−(3+4i)| ≤ 3 is the interior+boundary of a circle centre (3,4) and radius 3. z of least magnitude is where line joining O to centre meets circle. The solution of the equation log2 x+log2(2x) = 5 is: a) x = 2; b) x = 4; c) x = 4; d) x = 1. (When taking the fifth power of a complex number, you take its magnitude to the fifth power, and multiply its argument by 5. Z^3 = -i", "line to fourth line, $$z_1+z_2\\neq4$$ since they belong to $$U$$. I think my attempt was right, but please tell me whether it is right or wrong. You can see it is injective on the unit disk directly: Suppose $$z^3-12z=z'^3-12z'$$. This means $$z^3-z'^3-12z+12z'=(z-z')(z^2+zz'+z'^2-12)=0.$$ Now, since we're n the unit disk, the triangle inequality ensures that $$\\;|z^2+zz'+z'^2|\\le |z|^2+|z||z'|+|z'|^2 \\le 3$$, so the second factor can't be $$0$$, and the only solution is $$z=z'$$. • I'm sorry, the title was wrong. It should have been $f(z)=z^2−4z$. And in the similar fashion as you did, I made some answer above. I think it's right. Thanks a lot. – Sun Joong Kim Nov 26 '19 at 14:52 • Yes it is right. You just add the details why $z_1+z_2$ can't be $4$ when $z_1,z_2$ line in the unit circle. – Bernard Nov 26 '19 at 18:37 Your argument is correct. It shows that $$f(z) = z^2 -4z$$ is in fact injective in the larger disk with radius $$2$$ centered at the origin. An alternative approach is to use that If $\\operatorname{Re}f^\\prime > 0$ on a convex domain, then $f$ is one-to-one.. Applied to $$g(z) = -f(z) = 4z - z^2$$ we have $$\\operatorname{Re} g'(z) = \\operatorname{Re} (4 - 2z) = 4 - 2 \\operatorname{Re}(z)$$ which shows that $$g$$ (and consequently, $$f$$) is injective", "that represents the slope of the radius to the tangent point. Let's start at the point $(7,3)$. If we go $4k$ units up and $3k$ units right from $(7,3)$, we would arrive at a point that's $5k$ units away. But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$. Thus, $k = \\frac{8}{5}$, and we want to travel $4\\cdot \\frac{8}{5}$ up and $3\\cdot \\frac{8}{5}$ over from the point $(7,3)$ to reach our maximum. This means the maximum value of $3x + 4y$ occurs at $(7 +3\\cdot \\frac{8}{5}, 3 + 4\\cdot \\frac{8}{5})$, which is $(\\frac{59}{5}, \\frac{47}{5})$ Plug in those values for $x$ and $y$, and you get the maximum value of $3x + 4y = 3\\cdot\\frac{59}{5} + 4\\cdot\\frac{47}{5} = \\boxed{73}$, which is option $\\boxed{B}$", "+0 0 91 2 +95 Let $$z$$ and $$w$$ be complex numbers such that $$|z + 1 + 3i| = 1$$ and $$|w - 7 - 8i| = 3.$$ Find the smallest possible value of $$|z - w|.$$ Sep 9, 2022 #1 +1 Hello WorldEndSymphony! We need to know the geometric meaning of the following: (Also, are \"hints\"). 1- $$\\left | z-(a+bi) \\right |=r$$ is a circle with centre (a,b) and radius r. 2- $$\\left |z-w \\right |$$ is the distance between z and w. And, to solve this question, we need to sketch the givens. Solution: Rewrite: $$\\left | z+1+3i \\right |=1 \\iff \\left |z-(-1-3i) \\right |=1$$ Sketch this in an Argand diagram. Next: $$\\left | w-7-8i\\right |=3 \\iff \\left | w-(7+8i) \\right |=3$$ Sketch this also in the same Argand diagram. We should get: But you might be wondering, where is z and w? What we have drawn is a \"Locus\" you see, \"z\" can be any point on the red circle (on that red line only as it is \"=\" , if for example it were $$\\left |z-(-1-3i) \\right |<1$$ \"<\" then any point inside the circle might be z, and if \">\" then outside the circle.) Similarly, for \"w\", it is any point on the blue circle (That blue lines only). So what the", "N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); [/asy]$ $C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius of $C_4$. We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\\Omega$, and all points on $\\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$, and since $\\Omega$ has radius of $6$, the resulting line must be $\\frac{36}{4} = 9$ units away from $O$. $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2,", "you can start with $|c| = c^*c$ unclear and typo: I mean $|c|^2 = c^*\\,c \\$ with $c^*$ the complex conjugate of $c$. Last edited: Apr 13, 2016 3. Apr 13, 2016 chwala Is this correct $(z-u)^2=z^2$ $(z-1+i)(z-1+i)= z^2$ $z^2-2z+2iz-2i=z^2$ $-2z+2iz-2i=0$ $z(-2+2i)=2i$ $z=(2i/-2)+(2i/2i)$ $z=-i+1$ $z=1-i$ from here the point on argand diagram is $(1,-1)$ now how do we draw the locus, why are we joining this to the origin and then we go ahead to find the bisector of this line segment from the point $(1,-1)$ to the origin. Last edited: Apr 13, 2016 4. Apr 13, 2016 chwala Is it correct to state that $|z-i|=2$ is equal to $(z-i)(z-i)=4$ $z^2-2iz-5=0$ how do we proceed from here? ok here i know that the center is $(0,1)$ how do we get here.....good evening Bvu ok we can say$(z-i)^2=2^2$ $(z-i)=2$ centre $(0,1)$ and radius of 2 units comparing with the other one i.e $(z-1+i)=0$ centre $(1,-1)$ radius 0 units? are these statements valid Last edited: Apr 13, 2016 5. Apr 13, 2016 Samy_A No. For a complex number $z$, $|z|²= \\bar z z$ or $|z|²= z^* z$, where $\\bar z$ and $z^*$ both denote the complex conjugate of $z$. 6. Apr 13, 2016 chwala sam then how do we solve$|x-2|=4?$ I think the correct step would be $(x-2)(x-2) - 16", "# Difference between revisions of \"2018 AIME II Problems/Problem 5\" ## Problem Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$. ## Solution 1 First we evaluate the magnitudes. $|xy|=80\\sqrt{17}$, $|yz|=60$, and $|zx|=24\\sqrt{17}$. Therefore, $|x^2y^2z^2|=17\\cdot80\\cdot60\\cdot24$, or $|xyz|=240\\sqrt{34}$. Divide to find that $|z|=3\\sqrt{2}$, $|x|=40\\sqrt{34}$, and $|y|=10\\sqrt{2}$. $[asy] draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); [/asy]$ This allows us to see that the argument of $y$ is $\\frac{\\pi}{4}$, and the argument of $z$ is $-\\frac{\\pi}{4}$. We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$. More division is needed to find what $x$ is. $$x=-20-12i$$ $$x+y+z=-7-5i$$ $$(-7)^2+(-5)^2=\\boxed{74}$$ $$QED\\blacksquare$$ Written by a1b2 ## Solution 2 Dividing the first equation by the second equation given, we find that $\\frac{xy}{yz}=\\frac{x}{z}=\\frac{-80-320i}{60}=-\\frac{4}{3}-\\frac{16}{3}i \\implies x=z\\left(-\\frac{4}{3}-\\frac{16}{3}i\\right)$. Substituting this into the third equation, we get $z^2=\\frac{-96+24i}{-\\frac{4}{3}-\\frac{16}{3}i}=3\\cdot \\frac{-24+6i}{-1-4i}=3\\cdot \\frac{(-24+6i)(-1+4i)}{1+16}=3\\cdot \\frac{-102i}{17}=-18i$. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative", "# Difference between revisions of \"2012 AIME II Problems/Problem 6\" ## Problem 6 Let $z=a+bi$ be the complex number with $\\vert z \\vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$. ## Solution Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have $|z^3| * |z^2 - (1+2i)|$ $=|z|^3 * |z^2 - (1+2i)|$ $=125|z^2 - (1+2i)|$ Thus we only need to maximize the value of $|z^2 - (1+2i)|$. To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\\frac{-1}{\\sqrt{5}} + \\frac{-2}{\\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \\sqrt{5} (-5 - 10i)$ Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = \\boxed{125}$ ## Solution 2 WLOG, let $z_{1}=5(\\cos{\\theta_{1}}+i\\sin{\\theta_{1}})$ and $z_{2}=1+2i=\\sqrt{5}(\\cos{\\theta_{2}+i\\sin{\\theta_{2}}})$ This means that $z_{1}^3=125(\\cos{3\\theta_{1}}+i\\sin{3\\theta_{1}})$ $z_{1}^4=625(\\cos{4\\theta_{1}}+i\\sin{4\\theta_{1}})$ Hence, this means that $z_{2}z_{1}^3=125\\sqrt{5}(\\cos({\\theta_{2}+3\\theta_{1}})+i\\sin({\\theta_{2}+3\\theta_{1}}))$ And $z_{1}^5=3125(\\cos{5\\theta_{1}}+i\\sin{5\\theta_{1}})$ Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying", "in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $\\boxed{\\mathbf{(C)}\\ 6.4}$. ### Solution 2 $[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$ Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$. Similarly, $L_3$ and $L_4$ have slope $-\\frac{1}{m}$. Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$, $L_2: y = m(x-5)$, $L_3: y = -\\frac{1}{m}(x-7)$, $L_4: y = -\\frac{1}{m}(x-13)$. Since $PQRS$ is a square, it follows that $\\Delta x$ between points $P$ and $Q$ is equal to $\\Delta y$ between points $Q$ and $R$. Our approach will be to find $\\Delta x$ and $\\Delta y$ in terms of $m$ and equate the", "# Sketching a set of $2 < |z| \\leq |z + 2| < 4$ on the complex plane. So I need to sketch $$2 < |z| \\leq |z + 2| < 4$$ on the complex plane. At first, it seamed pretty easy since I know that: • $$2 < |z|$$ is a circle with a center in point $$0,0$$ and radius = 2 • |z + 2| < 4 is a circle with a center in point $$-2,0$$ and radius = 4 BUT here comes that part $$|z| \\leq |z + 2|$$ and I actually have no idea how to connect those two. Edit - my proposed solution: $$|z| \\leq |z + 2| \\implies z^2 \\leq z^2 + 4z + 4 \\implies 0 = \\leq 4z + 4 \\implies -1 \\leq z$$ Is that correct? Is it the missing condition? • Hint. $|z|$ is the distance of $z$ from the origin. $|z+2| = |z-(-2)|$ is the distance of $z$ from $-2$. Nov 7 '20 at 2:15 • Yeah, I know. I think I have just written that. Or you mean that I can use that fact to solve my problem (I mean this part: $|z| \\leq |z + 2|$)? Nov 7 '20 at 2:18 • What points are closer to the origin than they are to $-2$? The" ]

[ "is the area in between them. 2) $|z-3+4i| \\leq 4$ is a circle. $|z| = |z-10|$ is the line x = 5. $|z-3+4i|\\leq4$ and $|z|\\geq |z-10|$ is the area between the circle and the line (you will need to decide which of the two possible areas it is ....) By the way, all these relations have a very simple geometric interpretation which makes it easy to get the cartesian equations .... I guess i understand what you mean. But the thing is I am very bad at inequalities and geometric. I will post my answers later to check. Thanks! 4. Do you understand that |z| is the distance from the complex number z to the number 0= 0+ 0i? From that it follows that |a- b| is the distance between the two complex numbers a and b. Saying that |z+ 3i|= |z- (-3i)|= 1 means that the distance from z to -3i is 1: z can be any number on the circle with center at -3i and radius 1. Saying that $1\\le |z+ 3i|$ means that z is on or outside that circle. Similarly |z+ 3i|= |z- (-3i)|= 4 means that the distance from z to -3i is 4: z can be any number on the circle with center at -3i and radius 4. $|z+ 3i|\\le 4$ says", "# Difference between revisions of \"2012 AIME II Problems/Problem 6\" ## Problem 6 Let $z=a+bi$ be the complex number with $\\vert z \\vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$. ## Solution Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have $|z^3| * |z^2 - (1+2i)|$ $=|z|^3 * |z^2 - (1+2i)|$ $=125|z^2 - (1+2i)|$ Thus we only need to maximize the value of $|z^2 - (1+2i)|$. To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\\frac{-1}{\\sqrt{5}} + \\frac{-2}{\\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \\sqrt{5} (-5 - 10i)$ Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = 125$", "3i | = 5$ is the circle centered at $(-4,-3)$ of radius $5$. I claim that this circle is wholly contained inside the circle $|z| = 11$. This is evident geometrically, and to prove it observe that by the triangle inequality $|w| \\leq |w + 4 + 3i| + |-(4 + 3i)| = 5 + 5 = 10$. Equality occurs when $w = -8 - 6i$, the farther of the two points on the circle that are on the line connecting the origin to the center $(-4,-3i)$. The minimum of $11 - |w|$ on this circle occurs when $|w|$ is maximized, namely at $w = -8 - 6i$ when $|w| = 10$. So the minimum distance is given by $11 - 10 = 1$. - Let $z=a+bi$ and $w=c+di$. Now, $|z|=11\\implies \\sqrt{a^2+b^2}=11\\implies a^2+b^2=121$, so $z$ is constrained to this circle (shown in blue below). On the other hand, $|w+4+3i|=5\\implies \\sqrt{(c+4)^2+(d+3)^2}=5\\implies (c+4)^2+(d+3)^2=25$, so $w$ is constrained to this circle (in red below). The task is to minimize $|z-w|=\\sqrt{(a-c)^2+(b-d)^2}$ subject to the two constraints above. Geometrically, we are looking to find the shortest distance between any two points which lie on the blue and red circles, respectively. A little calculus reveals the minimum is $1$ and is obtained when $a=-44/5$, $b=-33/5$ and $c=-8$, $d=-6$, as shown below. - I have", "z^2-(4+5i)z-3+9i=0 => z=[(4+5i)+/-sqr(4+5i)^2+4(3-9i)]/2 => z=[(4+5i)+/-sqr(3+4i)]/2 => z=[(4+5i)+/-(2+i)]/2 => z1=(6+6i)/2=3+3i. Example 1. Physics. If you're using complex numbers, then every polynomial equation of degree #k# yields exactly #k# solution. Determine (24221, 122/221, Arg(2722), And Arg(21/22). Ask your question. KEAM 2016: If |z-(3/2)|=2 , then the greatest value of |z| is (A) 1 (B) 2 (C) 3 (D) 4 (E) 5. Then the minimum value of |z1 – z2| is : asked Apr 16, 2019 in Mathematics by Niharika ( 75.6k points) If z=3- 4i is turned 90^@ in anti clock direction then new position of z is. $\\begingroup$ If the series converges at $3+4i$ then it is absolutely convergent for any $|z| \\le 5$, so the radius of convergence is equal or larger than $5$. …, . rohankedia3541 is waiting for your help. z= 3-4i. If |z-3+2i|＜=4 then the difference between the greatest and the least value of |z| is : A) 2(13^1/2) B) 8 C) 4+((13)^1/2) D) (13)^1/2 The inequality |z-3+2i| 3 What is Z Array? |z−(3+4i)| ≤ 3 is the interior+boundary of a circle centre (3,4) and radius 3. z of least magnitude is where line joining O to centre meets circle. asked Jan 27, 2015 in TRIGONOMETRY by anonymous. b) If Z[K] >= R-i+1 then it is possible to extend the [L,R] interval thus we will", "+0 0 91 2 +95 Let $$z$$ and $$w$$ be complex numbers such that $$|z + 1 + 3i| = 1$$ and $$|w - 7 - 8i| = 3.$$ Find the smallest possible value of $$|z - w|.$$ Sep 9, 2022 #1 +1 Hello WorldEndSymphony! We need to know the geometric meaning of the following: (Also, are \"hints\"). 1- $$\\left | z-(a+bi) \\right |=r$$ is a circle with centre (a,b) and radius r. 2- $$\\left |z-w \\right |$$ is the distance between z and w. And, to solve this question, we need to sketch the givens. Solution: Rewrite: $$\\left | z+1+3i \\right |=1 \\iff \\left |z-(-1-3i) \\right |=1$$ Sketch this in an Argand diagram. Next: $$\\left | w-7-8i\\right |=3 \\iff \\left | w-(7+8i) \\right |=3$$ Sketch this also in the same Argand diagram. We should get: But you might be wondering, where is z and w? What we have drawn is a \"Locus\" you see, \"z\" can be any point on the red circle (on that red line only as it is \"=\" , if for example it were $$\\left |z-(-1-3i) \\right |<1$$ \"<\" then any point inside the circle might be z, and if \">\" then outside the circle.) Similarly, for \"w\", it is any point on the blue circle (That blue lines only). So what the", "# Sketching a set of $2 < |z| \\leq |z + 2| < 4$ on the complex plane. So I need to sketch $$2 < |z| \\leq |z + 2| < 4$$ on the complex plane. At first, it seamed pretty easy since I know that: • $$2 < |z|$$ is a circle with a center in point $$0,0$$ and radius = 2 • |z + 2| < 4 is a circle with a center in point $$-2,0$$ and radius = 4 BUT here comes that part $$|z| \\leq |z + 2|$$ and I actually have no idea how to connect those two. Edit - my proposed solution: $$|z| \\leq |z + 2| \\implies z^2 \\leq z^2 + 4z + 4 \\implies 0 = \\leq 4z + 4 \\implies -1 \\leq z$$ Is that correct? Is it the missing condition? • Hint. $|z|$ is the distance of $z$ from the origin. $|z+2| = |z-(-2)|$ is the distance of $z$ from $-2$. Nov 7 '20 at 2:15 • Yeah, I know. I think I have just written that. Or you mean that I can use that fact to solve my problem (I mean this part: $|z| \\leq |z + 2|$)? Nov 7 '20 at 2:18 • What points are closer to the origin than they are to $-2$? The", "# Minimum value of $|z-w|$ where $z,w \\in \\mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? I was thinking about the problem: What is the minimum value of $|z-w|$ where $z,w \\in \\mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? My attempts: I notice that $|z-w| \\geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5$ gives $|w|^2+8u+6v=0$. Now i can not proceed. Can someone point me in the right direction? Thanks in advance for your time. - If you think of it geometrically, you'll get a sense of what you needed to do algebraically to actually prove what the answer is. $|z|=11$ means $z$ is a point of the circle centered in $0$ and radius $11$. $|w+4+3i|=5$ means $w$ is a point of the circle centered in $4+3i$ and radius $5$. This isn't an answer or a hint, just an observation. – Git Gud Jan 10 '13 at 17:59 Thanks a lot sir.It's been useful observation. +1 from me. – user52976 Jan 10 '13 at 18:11 Geometrically, $|z| = 11$ is the circle centered at the origin of radius $11$. If a point $w$ is inside the circle $|z| = 11$, elementary geometry shows that the the minimum distance from $w$ to this circle is $11 - |w|$ (the outward distance from $w$ to the circle). $|w + 4 +", "draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$. Let the radius of circle $P$ be $r$. Using Distance Formula, we get the following system of three equations: $$h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2$$ By simplifying, we get $$h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1$$ By subtracting the first equation from the second and third equations, we get $$8r=-4h+12, 10r=2h+6$$ which simplifies to $$2r=3-h, 5r=h+3$$ When we add these two equations, we get $$7r=6$$ So $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ $w^5$ ## Solution 5(Inversion from Circular Inversion) Let $\\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram: $[asy] size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NE); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0),", "that is where we get our x value: $(19, d)$ $(19, -d)$ $(17, -3d)$ Now, we can plug one of the first two value in as well as the last one to get the following equations: $$19^2 + d^2 = r^2$$ $$17^2 + (3d)^2 = r^2$$ Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ - therefore, we get $72 = 8d^2 \\rightarrow d^2 = 9 \\rightarrow d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\\boxed{B}$. ~Tony_Li2007 ## Solution 3 (Stewart's Theorem) $[asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label(\"O\", O, N); label(\"C\", C, SW); label(\"D\", D, SE); label(\"E\", E, SW); label(\"F\", F, SE); label(\"P\", P, SW); label(\"Q\", Q, S); [/asy]$ If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives $$19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.$$ This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem", "this line segment? 1. $-2 - 5i$ 2. $-4i$ 3. $2 + 5i$ 4. $3/2 - 92 i$ 4. Which of the listed complex numbers, when represented by a point in the complex plane, is both equidistant and collinear with the complex numbers represented by points A and K? 1. $3i$ 2. $1 + 3/2 i$ 3. $1/2 + 3i$ 4. $3 + 1/2 i$ 5. Let $z=-8+5i$. If the midpoint of the line segment created by complex numbers $z$ and $w$ is $-3+3/2 i$, what is the value of $w ?$ 1. $w = -11/2 + 13/4 i$ 2. $w = 2-2i$ 3. $w = 14 + i$ 4. $w = 4 + 7/2 i$ 6. Find the distance between $z = 6 + 7i$ and $w = -2 - 3i$ in the complex plane. 1. $4sqrt(2)$ 2. $18$ 3. $2sqrt(2)$ 4. $2sqrt(41)$ 7. Which of the following expressions represent(s) the distance between $z_1$ and $z_2$ in the complex plane? Choose all correct answers. 1. $|z_1 - z_2|$ 2. $|z_2 - z_1|$ 3. $|z_2| - |z_1|$ 4. $|z_2 + z_1|$ 8. What is the distance between the complex numbers $2-4i$ and $-3/2 - 1/2 i$ in the complex plane? 1. $7/2 sqrt(2)$ 2. $1/2sqrt(130)$ 3. $5/2 sqrt(2)$ 4. $1/8 sqrt(82)$ 9. If $z = 5 + 4i$,", "4i|=4. Then the minimum value of |z1 -z2| is: (2019) (a) 0 (b) √2 (c) 1 (d) 2 Ans. a |Z1| = 9, |z2 - 3 - 4i| = 4 z1 lies on a circle with centre C1(0, 0) and radius r1 = 9 z2 lies on a circle with centre C2(3, 4) and radius r2 = 4 So, minimum value of |z1 -z2| is zero at point of contact (i.e. A) Q.12. If then (1 + iz + z5 + iz8)9 is equal to: (a) 0 (b) 1 (c) (-1 + 2i)9 (d) -1 Ans. d where ω is imaginary cube root of unity. Q.13. All the points in the set lie on a: (2019) (a) straight line whose slope is 1. (b) circle whose radius is 1. (c) circle whose radius is √2 . (d) straight line whose slope is -1. Ans. b Let z∈S then Since, z is a complex number and let z = x + iy Then, (by rationalisation) Then compare both sides ...(1) ...(2) Now squaring and adding equations (1) and (2) Q.14. Let z ∈ C be such that |z| < 1. If ω = then: (2019) (a) 5 Re (ω) > 4 (b) 4 Im (ω) > 5 (c) 5 Re (ω) >1 (d) 5 Im (ω) < 1 Ans.", "# Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value I'm having a hard time solving following question: Determine the complex number Z that satisfy $$|z-3-3i|=1$$ and that has maximum absolute value. Z should be written on the form z=x+iy. I have determined with the help of the triangle inequality that $$|z|=1+\\sqrt{18}$$. This is the point where i run into problem. I don't know how to determine z on the form $$z=x+iy$$, using the information above. If someone could give me a hint i would be very thankful. • This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin. – saulspatz Nov 15 '18 at 16:58 Think of the complex numbers as a real coordinate plane. The equation $$|z-3-3i|=1$$ is basically a circle of radius $$1$$ with the center at $$(3, 3)$$. What point on the circumference is farthest away from the origin? Any number with magnitude $$1$$ can be written as $$e^{it}$$ for real $$t$$, so your $$z$$ must be of the form $$z= z_0+e^{it}$$ where $$z_0=3+3i$$. Note that we can rewrite $$z_0$$ as $$3\\sqrt{2}e^{i\\pi/4}$$. Let $$\\mathbf v_0$$ denote the vector from $$0$$ to $$z_0$$, and $$\\mathbf v_1$$ the vector from $$0$$ to $$e^{it}$$. You are looking to maximize $$|\\mathbf v_0+\\mathbf v_1|$$. The vector $$\\mathbf v_0$$ always points", "and a in the complex plane. Saying that z satisfies $1\\le |(z+ 2i)- 0|\\le 2$ means that z+ 2i is in the \"washer\" with center 0, inner radius 1, outer radius 2. By \"writing z+ 2+ i as z+ 2i+ 2-i\" we can interpret |z+ 2+ i|= |(z+2i)- (i-2)| as the distance from the point z+2i to the point i-2. |i- 2|= $\\sqrt{5}$> 2 so i- 2 is outside that \"washer\". The shortest distance from i- 2 to the larger circle is $\\sqrt{5}- 2$ while the longest distance, to the other side of the outer circle, is $\\sqrt{5}+ 2$.", "Samy_A So $z=12 + 16i$. You are looking for the point $z$ that satisfies $|z-25i| \\leq 15$ and with the lowest possible argument. $z$ has to be in (or on) the circle with centre at (0,25) and with radius 15. You will get the lowest argument with the tangent from the origin to the circle. That gives you the point S (or $z$ as complex number). As POS is a right triangle, you easily get that r=20. And then you can compute $\\arcsin \\phi$ from the same triangle POS. (Sorry for the clumsy drawing, never used that whiteboard before. P is supposed to be the centre of the circle.) Last edited: Dec 11, 2015 10. Dec 11, 2015 ### Staff: Mentor Keep in mind that |z - 25i| represents the distance from a complex number z to the complex number 25i. You should be able to solve this problem simply by drawing a diagram. 11. Dec 11, 2015 ### Samy_A Deleted as irrelevant. Last edited: Dec 11, 2015", "# How do you sketch the set of all the complex numbers z such that |z- 3- 4i| = 2? Dec 23, 2015 This will be a circle of radius $2$ centred on $3 + 4 i$ #### Explanation: graph{((x-3)^2+(y-4)^2 - 4)((x-3)^2+(y-4)^2-0.003) = 0 [-9.5, 10.5, -1.36, 8.64]} If ${z}_{1}$ and ${z}_{2}$ are Complex numbers, then $\\left\\mid {z}_{1} - {z}_{2} \\right\\mid$ is the distance between ${z}_{1}$ and ${z}_{2}$. So we can read $\\left\\mid z - 3 - 4 i \\right\\mid = 2$ as \"the distance between $z$ and $3 + 4 i$ is $2$\".", "when it comes to defining $|z|$ for some $z = a+bi$ (especially as it relates to multiplicative inverses of complex numbers as previously computed). Recalling the distance formula, $d = \\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$ (derived from the Pythagorean theorem) that calculates the distance $d$ between two points $(x_0,y_0)$ and $(x_1,y_1)$ on a coordinate plane -- we see that for points $(a,b)$ and $(0,0)$, we have $d = \\sqrt{(a-0)^2 + (b-0)^2} = \\sqrt{a^2 + b^2}$. As such, we define $|a + bi|$ to equal $\\sqrt{a^2 + b^2}$, calling this either the absolute value, magnitude, or modulus of $a + bi$, depending on the circumstances at hand -- the last being a dimunitive form of the Latin modus, which means \"measure\". While it may appear unrelated for a brief moment -- let us consider the set of complex values of unit magnitude (i.e., complex values $z = a + bi$ with $|z| = 1$). As the set of all points equidistant from a given point in a plane forms a circle, and the corresponding ordered pairs $(a,b)$ for all such $z$ are at unit distance from $(0,0)$, these $z$ must exist on a circle of radius $1$ centered at the origin. Let us call this the unit circle in the complex plane. Note that we can scale", "+0 HELPME!?!?!? 0 92 3 A circle is centered at $O.$ The tangent to the circle at $P$ is extended to $Q.$ Line segment $\\overline{QS}$ intersects the circle at $R.$ Given that $OS = 2,$ $SR = RQ = 3$, and $PQ = 6$, find the radius of the circle. [asy] unitsize(2 cm); pair A, B, C, D, E, F, G, O; A = dir(70); B = A + 1.2*dir(-20); C = dir(25); D = 2*C - B; O = (0,0); draw(Circle(O,1)); draw(A--B--D--O); dot(\"$P$\", A, N); dot(\"$Q$\", B, dir(0)); dot(\"$R$\", C, SW); dot(\"$S$\", D, NW); dot(\"$O$\", O, SW); [/asy] Aug 14, 2020 #1 0 Image Here: Aug 15, 2020 #2 0 By power of a point, the radius of the circle is 4*sqrt(7). Aug 15, 2020 #3 +110762 0 Maybe it would be in your interest to present your questions better. All those unnecessary dollar signs make it difficult to read and I feel that if you cannot be bothered to present your question properly then it is not worth my effort to attempt to answer. Aug 15, 2020", "If a complex number z lies in the interior or on the boundary of a circle of radius as 3 and centre at (0, –4) then greatest and least value of |z + 1| are- (a) $3+\\sqrt{17},\\sqrt{17}-3$ (b) 6, 1 (c) $\\sqrt{17},1$ (d) 3, 1 Sol: Greatest and least value of |z + 1| means maximum and minimum distance of circle from the point (–1, 0). In a circle, greatest and least distance of it from any point is along the normal. ∴ $\\,\\,\\,\\,Greatest\\,dis\\tan ce\\,=\\,3+\\sqrt{{{1}^{2}}+{{4}^{2}}}=3+\\sqrt{17}$ Least distance = $\\sqrt{{{1}^{2}}+{{4}^{2}}}-3=\\sqrt{17}-3$ Illustration: Find the equation of the circle for which arg $\\left( \\frac{z-6-2i}{z-2-2i} \\right)=\\pi /4.$ Sol: $\\arg \\left( \\frac{z-6-2i}{z-2-2i} \\right)=\\pi /4$ represent a major arc of circle of which Line joining (6, 2)) and (2, 2) is a chord that subtends an angle $\\frac{\\pi }{4}$ at circumference. Clearly AB is parallel to real (x) axis, M is mid-point, M ≡ (4, 2), OM = AM = 2 ∴ $\\,\\,O=(4,4)\\,and\\,O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}}=2\\sqrt{2}$ Equation of required circle is $|z-4-4i|=2\\sqrt{2}$ Illustration: if |z| > 3, prove that the least value of $\\left| z+\\frac{1}{z} \\right|\\,is\\,\\frac{8}{3}$ Sol: Here $\\left| z+\\frac{1}{z} \\right|\\,\\underline{>}\\,|z|-\\frac{1}{|z|}$ Now |z| > 3 ∴ $\\,\\,\\,\\,\\frac{1}{|z|}\\underline{<}\\frac{1}{3}\\,or\\,-\\frac{1}{|z|}\\underline{>}-\\frac{1}{3}$ (i) $|z|-\\frac{1}{|z|}\\underline{>}3-\\frac{1}{3}=\\frac{8}{3}$ Hence from (i) and (ii), we get $\\left| z+\\frac{1}{z} \\right|\\underline{>}\\frac{8}{3}$ (ii) ∴ $\\,\\,\\,Least\\,value\\,is\\frac{8}{3}$ Illustration: If z1, z2, z3 are non-zero complex numbers such that ${{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0\\,and\\,z_{1}^{-1}+z_{2}^{-1}+z_{3}^{-1}=0$ then prove that", "each midpoint of a side of $S$, segments are drawn to the two closest vertices of $S'$. The result is a four-pointed starlike figure inscribed in $S$. The star figure is cut out and then folded to form a pyramid with base $S'$. Find the volume of this pyramid. $[asy] pair S1 = (20, 20), S2 = (-20, 20), S3 = (-20, -20), S4 = (20, -20); pair M1 = (S1+S2)/2, M2 = (S2+S3)/2, M3=(S3+S4)/2, M4=(S4+S1)/2; pair Sp1 = (7.5, 7.5), Sp2=(-7.5, 7.5), Sp3 = (-7.5, -7.5), Sp4 = (7.5, -7.5); draw(S1--S2--S3--S4--cycle); draw(Sp1--Sp2--Sp3--Sp4--cycle); draw(Sp1--M1--Sp2--M2--Sp3--M3--Sp4--M4--cycle); [/asy]$ ## Problem 6 Let $z=a+bi$ be the complex number with $\\vert z \\vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$. ## Problem 7 Let $S$ be the increasing sequence of positive integers whose binary representation has exactly $8$ ones. Let $N$ be the 1000th number in $S$. Find the remainder when $N$ is divided by $1000$. ## Problem 8 The complex numbers $z$ and $w$ satisfy the system $$z + \\frac{20i}w = 5+i$$ $$w+\\frac{12i}z = -4+10i$$ Find the smallest possible value of $\\vert zw\\vert^2$. ## Problem 9 Let $x$ and $y$ be real numbers such that $\\frac{\\sin x}{\\sin y} = 3$ and $\\frac{\\cos x}{\\cos y} = \\frac12$.", "# If $z$ is a complex number such that $\\frac{z-i}{z-1}$ is purely imaginary, find the minimum value of $|z-(2+2i)|$. This is a question from an entrance examination. Let $z$ be a complex number such that $\\dfrac{z-i}{z-1}$ is purely imaginary. Find the minimum value of $|z-(2+2i)|$. My attempt:Since $\\dfrac{z-i}{z-1}$ is purely imaginary, $$\\dfrac{z-i}{z-1}+\\overline{\\left(\\dfrac{z-i}{z-1}\\right)}=0$$ This reduces to $$(z-i)(\\bar z-1)+(\\bar z+i)(z-1)=0$$ Assuming my calculations are correct, I get the following result: $$|z|^2=\\Re(z)+\\Im(z)$$ This represents the locus of $z$ on the Argand Plane. Thus, the minimum value of $|z-(2+2i)|$ will be the shortest distance between any point $z$ lying on $|z|^2=\\Re(z)+\\Im(z)$ and the point $(2,2)$ on the Argand Plane. This is where I am getting stuck. I am not able to recognize the locus represented by $|z|^2=\\Re(z)+\\Im(z)$. If hypothetically it had been a straight line, the the minimum value of $|z-(2+2i)|$ would have been the perpendicular distance of $(2,2)$ from the hypothetical line represented by $|z|^2=\\Re(z)+\\Im(z)$. Could somebody please help me identify the locus represented by $|z|^2=\\Re(z)+\\Im(z)$ ($\\textbf{ preferably without }$ first setting $z=x+iy$ and hence using normal Coordinate Geometry to recognize the locus)? Also could I please be told how to proceed with solving this problem? Many thanks in anticipation! • Could be easier to start with $\\,(z-i)/(z-1)=ix\\,$ where $x \\in \\mathbb{R} \\setminus \\{0\\}$. – dxiv Apr 30 '18 at 19:44" ]

[ "• # question_answer The minimum value of $3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$ is A) 6 B) 15 C) 24 D) none of these Using $A.M.\\ge G.M$ $\\Rightarrow$$\\frac{1}{2}(3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta )\\ge 6$ $\\Rightarrow$$3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$has minimum value 12.", "and minimum should occur. The maximum value of $\\sin(\\theta)$ occurs at $\\theta = 2n \\pi + \\dfrac{\\pi}{2}$ and the minimum value of $\\sin(\\theta)$ occurs at $\\theta = 2n \\pi - \\dfrac{\\pi}{2}$ and it is at these points the tangent line will be horizontal. Hence, the tangent line to $f(x)$ is horizontal at $$x + \\dfrac{\\pi}4 = 2n \\pi \\pm \\dfrac{\\pi}2$$ -", "## Trigonometry (11th Edition) Clone On the graph, we can see that the maximum value of $\\theta$ occurs when $x \\approx 1.4$ We can graph the following function: $\\theta = tan^{-1}(\\frac{x}{x^2+2})$ On the graph, we can see that the maximum value of $\\theta$ occurs when $x \\approx 1.4$", "lies between $-4$ and $10$ 39. If $m\\tan(\\theta - 30^\\circ) = n\\tan(\\theta + 120^\\circ),$ show that $\\cos2\\theta = \\frac{m + n}{2(m - n)}$ 40. if $\\alpha + \\beta = \\theta$ and $\\tan\\alpha:\\tan\\beta = x:y,$ prove that $\\sin(\\alpha - \\beta) = \\frac{x - y}{x + y}\\sin\\theta$ 41. Find the maximum and minimum value of $7\\cos\\theta + 24\\sin\\theta$ 42. Show that $\\sin 100^\\circ - \\sin 10^\\circ$ is positive.", "# Is there a better way to find the minimum value of $\\frac{1}{\\sin \\theta \\cos \\theta \\left ( \\sin \\theta + \\cos \\theta \\right )}$? To find the minimum value of $$\\frac{1}{\\sin \\theta \\cos \\theta \\left ( \\sin \\theta + \\cos \\theta \\right )}$$ I can see that I can convert it to $$\\frac{\\sqrt{2}}{\\sin (2 \\theta) \\sin \\left ( \\theta + \\frac{\\pi}{4} \\right )}$$ And from here I can convert it to a pair of cosecant functions and then proceed with the usual product rule, solving that out with zero etc. Once that is all done, we can find that the function attains a minimum at $\\theta= \\frac{\\pi}{4}$ and this minimum value is $\\sqrt{2}$. However, I was wondering if there was perhaps a better way of finding the minimum value of this function that does NOT require heavy amounts of algebra bashing. Perhaps by some means of observation. So for example, one could find the maximum of $\\sin \\theta$ by realising that it oscillates between $1$ and $-1$. Could a similar idea be used to find the minimum of this function? • What if $\\theta \\to 0$ from the left? Wouldn't this tend to $-\\infty$ and have no minimum? – jameselmore Sep 29 '15 at 16:24 • if f(x) has an extremum at x = x0, then 1/f(x)", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "# Finding minimum and maximum value of complex number Let a be a positive real number and let $M_a =\\{z \\in C^*: |z+\\frac{1}{z}|=a\\}$ Find the minimum and maximum value of $|z|$ when z$\\in M_a$ • What is $\\mathbb{C}^*$? – copper.hat Feb 22 '13 at 5:52 • $C*$ = non zero complex number – Sachin Sharmaa Feb 22 '13 at 6:19 Let $z=r(\\cos \\theta+i\\sin \\theta)$ $z^{-1}=r^{-1}(\\cos \\theta-i\\sin \\theta)$ We have , $|z+\\frac{1}{z}|=|(r+1/r)\\cos \\theta +(r-1/r)i\\sin\\theta|$ So we have, $\\displaystyle|(r+1/r)\\cos \\theta +(r-1/r)i\\sin\\theta|=\\sqrt{(r^2+1/r^2)+2\\cos 2\\theta}$ Thus we have , $(r^2+1/r^2)+2\\cos 2\\theta=a^2$ Let $r^2=x$, $\\Rightarrow x^2+1=x(a^2-2\\cos2\\theta)$ $\\Rightarrow x^2-x(a^2-2\\cos2\\theta)+1=0$ $\\displaystyle x=\\frac{(a^2-2\\cos 2\\theta)+\\sqrt{(a^2-2\\cos 2\\theta)^2-4}}{2}$ and $\\displaystyle x=\\frac{(a^2-2\\cos 2\\theta)-\\sqrt{(a^2-2\\cos 2\\theta)^2-4}}{2}$ Clearly x attains its maximum value at $2\\theta=0$ using the first expression, The value is $\\displaystyle \\frac{(a^2+2)+\\sqrt{(a^2+2)^2-4}}{2}$ So the maximum value of $r=\\sqrt{\\displaystyle \\frac{(a^2+2)+\\sqrt{(a^2+2)^2-4}}{2}}$ This on simplification turns out to be, $\\displaystyle r=\\frac{a+\\sqrt{a^2+4}}{2}$", "Find the minimum and maximum values of $y = \\sqrt{8} \\theta - \\sqrt{4} \\sec \\theta$ on the interval $[ 0, \\frac {\\pi}{3}]$ $f_{\\text{min}}=$ $f_{\\text{max}}=$", "# Find the maximum and minimum value of $\\sin ^4 \\theta + \\cos ^4 \\theta$? Please suggest suitable approach for this problem - In which interval? It's periodic with period $\\frac{\\pi}{2}$, so it will have a lot of maxima and minima. But if you're restricting to $\\left[0,\\frac{\\pi}{2}\\right]$, then it has one minimum and two maxima at the ends. – Ｊ. Ｍ. Nov 6 '10 at 9:59 I solved it the answer is 1 and 1/2. – Trewick Marian Nov 6 '10 at 10:10 Setting $x=\\sin^2\\theta$ this reduces to finding the extrema of $x^2+(1-x)^2$ for $x\\in[0,1]$. – Robin Chapman Nov 6 '10 at 10:26 Note that $$1 = (\\sin^2\\theta + \\cos^2\\theta)^2$$ and use $\\sin2\\theta = 2\\sin\\theta\\cos\\theta.$", "× # Minimum Value I recently solved a trigonometric question which asked for the minimum value of the $$(\\sin\\theta + \\csc\\theta)^{2}+(\\cos\\theta +\\sec\\theta)^{2}$$.I found there ar two answers, which one of them is correct? A)8 B)9 Note by Puneet Pinku 3 weeks, 1 day ago Sort by: Hint: Expand and simplify the expression by using $$\\sin^2 A + \\cos^2 A = 1$$. Then apply double angle formula $$\\sin(2A) = 2\\sin A \\cos A$$. · 3 weeks ago", "to $\\tan \\theta = \\frac{1}{\\mu}$ using some simple trig you get $\\sin \\theta = \\frac{1}{\\sqrt{\\mu^2 + 1}}$ and $\\cos \\theta = \\frac{\\mu}{\\sqrt{\\mu^2 + 1}}$ put this into the formula (with your value of $/mu$) I get the minimum value of $F = 0.148 mg$. you should be able to notice that F will be a maximum when $\\theta = \\frac{\\pi}{2}$ in which case $F = \\mu mg$ Edit: F is actually maximum in the case when $\\theta = 0$ in this case you would just be pulling the sledge vertically • Mar 22nd 2008, 06:02 PM pakman134 hm... I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm..... • Mar 22nd 2008, 06:10 PM bobak Quote: Originally Posted by pakman134 I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm..... Well the maximum value", "Browse Questions The maximum and minimum values of $7\\cos\\theta+24\\sin\\theta$ are equal to $(a)\\;25\\;and\\;-25\\qquad(b)\\;24\\;and\\;-24\\qquad(c)\\;5\\;and\\;-5\\qquad(d)\\;None \\;of\\;these$ We know that the maximum and minimum values of $a\\cos\\theta+b\\cos\\theta$ are $\\sqrt{a^2+b^2}$ and $-\\sqrt{a^2+b^2}$ respectively. Hence,the maximum and minimum values of $7\\cos\\theta+24\\sin\\theta$ are $\\sqrt{7^2+{24}^2}=25$ and $-\\sqrt{7^2+{24}^2}=-25$ respectively. Hence (a) is the correct answer.", "the sine is at its maximum value of $1$. This means a maximum is at $\\theta+\\phi = \\pi / 2$ so $\\theta=\\pi / 2 - \\phi$ The wave then drops through zero to its minimum value of $-1$ when the angle equals $3 \\pi / 2$. This means a minimum is at $\\theta+\\phi = 3\\pi / 2$ so $\\theta=3\\pi / 2-\\phi$ There is an infinite number of maxima and minima. It is normal to give values in the range $0 ≤ \\theta ≤ 2\\pi$. Maxima are at $\\theta=2n\\pi+\\phi$ Minima are at $\\theta=(2n+1)\\pi+\\phi$ It is normal to give values of $\\theta$ in the range $0 ≤ \\theta ≤ 2\\pi$. Example 8.5: Find the value and position of the first maximum and minimum of$y= 4sin\\theta+3cos \\theta + 3$ Your browser does not support the HTML5 canvas tag. $r$ $=$ $\\sqrt{3^2 + 4^2}=5$. and $\\phi$ $=$ $tan^{-1}(3/4) = 0.64$ radians Maximum $=$ $5 + 3=8$ Minimum $=$ $-5 + 3=-2$ Maximum $=8$ and is at $\\theta+0.64$ $=$ $\\pi/2$ so $\\theta_{max}$ $=$ $\\pi/2 - 0.64=0.93$ radians Minimum $=-2$ and is at $\\theta+0.64$ $=$ $3 \\pi/2$ so $\\theta_{min}$ $=$ $3 \\pi/2-0.64=4.07$ radians", "# What is the maximum value of 1sec θ? Question: What is the maximum value of $\\frac{1}{\\sec \\theta}$ ? Solution: The maximum value of $\\frac{1}{\\sec \\theta}$ is 1 because the maximum value of $\\cos \\theta$ is 1 that is $\\frac{1}{\\sec \\theta}=\\cos \\theta$ $\\frac{1}{\\sec \\theta}=1$", "# Minimum value of $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2}$ Minimum value of $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2}$ Find the minimum value of $h(\\theta)$ $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2} = 5 \\sin (\\theta + 53.13) + \\sqrt{2}$ Minimum value - $5\\sin (\\theta + 53.13) + \\sqrt{2} = -5$ Therefore min value is = $-5/5 - \\sqrt{2}$ Why am I wrong ? And how should I do this question.. • I would say that the standard way is finding the derivative of the function and solving for $h'(\\theta) = 0$. But algebraic manipulation can work too. Numerically, I find the minimum to be $\\approx -3.5858$. – Matti P. Feb 19 '18 at 7:13 • @user175089...Find h'(θ) and h''(θ)...Take h'(θ)=0...For the value of θ for which h''(θ)>0 ,gives the minimum value... – Nehal Samee Feb 19 '18 at 7:14 • You made a sign error. $-5\\color{red}+\\sqrt2$. And why do you divide by $5$ ?? – Yves Daoust Feb 19 '18 at 7:14 Minimum is attained when $\\sin (\\theta + 53.13) = -1$ that is $$h_{min}=h(3\\pi/2+k\\pi)=5 \\sin (3\\pi/2+k\\pi) + \\sqrt{2}=-5+\\sqrt{2}$$ for the same reason the maximum is attained when $\\sin (\\theta + 53.13) = 1$. Hint You must note that the range of $a\\sin\\theta \\pm b\\cos\\theta$ is $\\left[ -\\sqrt {a^2+b^2},", "# Maximum value of the given expression Assuming $\\theta \\in [\\frac{-5\\pi}{12},\\frac{-\\pi}{3}]$, find the maximum value of $$\\frac{\\tan(\\theta+\\frac{2\\pi}{3})-\\tan(\\theta+\\frac{\\pi}{6})+\\cos(\\theta+\\frac{\\pi}{6})}{\\sqrt{3}}$$ I know we can differentiate and then put $f'(x)=0$ to get the critical points, but I strongly believe that's very hefty. The range of $\\theta$ increases the trouble. Can anyone give a clear method that applies easily? $\\tan(\\theta + \\frac{2\\pi}{3})-\\tan(\\theta + \\frac{\\pi}{6})$ is in that range always greater than $0$ and rising, just look at the singularities and the period of both tangents, also the cosine rises constantly in the given range if you look at its period, so the maximum of the function is actually at the maximum of the range at $\\theta = -\\frac{\\pi}{3}$, evaluate that in the function to obtain the maximum value. • just noticed-$tan(\\theta+\\frac{2\\pi}{3})=-cot(\\theta+\\frac{\\pi}{6})$, so now if we assume $x=-tan(\\theta+\\frac{\\pi}{6})$, then x is positive. Eventually greatest value occurs at $\\theta=\\frac{-\\pi}{3}$. Thanks! – Ashish Gaurav Feb 19 '13 at 19:33", "+0 # Help 0 89 1 Let a and b be real numbers. Find the maximum value of $$a \\cos \\theta + b \\sin \\theta$$ in terms of a and b. Aug 31, 2019 $$a \\cos(\\theta) + b \\sin(\\theta) = \\\\ \\sqrt{a^2 + b^2}\\sin(\\theta + \\phi),~\\phi = \\tan^{-1}\\left(b,a\\right)\\\\ -1 \\leq \\sin(\\theta+\\phi) \\leq 1\\\\ \\text{The max value is thus \\sqrt{a^2+b^2}}$$", "\\:=\\:R\\cos(\\theta + \\alpha)$ Therefore: . $\\boxed{R \\:=\\:\\sqrt{13}}$ Quote: Show that: . $\\alpha \\,\\approx\\, 33.7^o$ Since $\\sin\\alpha \\,=\\,\\tfrac{2}{\\sqrt{13}}$, then: . $\\alpha \\:=\\:\\sin^{-1}\\left(\\tfrac{2}{\\sqrt{13}}\\right) \\quad\\Rightarrow\\quad \\boxed{\\alpha\\;\\approx\\; 33.7^o}$ Quote: Hence, find the maximum value of $3\\cos\\theta - 2\\sin\\theta$ and find a positive value of $\\theta$ at which the maximum value occurs. The maximum of: . $3\\cos\\theta - 2\\sin\\theta \\:=\\:\\sqrt{13}\\,\\cos(\\theta + \\alpha)$ occurs when $\\cos(\\theta + \\alpha) \\:=\\:1$ . . Then the maximum value is $\\boxed{\\sqrt{13}}$ If $\\cos(\\theta + \\alpha) \\:=\\:1$, then: . $\\theta + \\alpha \\:=\\:360^o$ Therefore: . $\\theta \\:=\\:360^o - \\alpha \\:=\\:360^o - 33.7^o \\quad\\Rightarrow\\quad\\boxed{\\theta\\;=\\;326.3^o}$", "sin \\theta =-1$ • b. $2cos \\theta - sin \\theta =2$ • c. $3cos \\theta - 4sin \\theta =-1$ • d. $\\sqrt{3}cos \\theta + sin \\theta=1$ A combined wave is made from two or more sine and cosine waves. Once a combined wave has been calculated we would like to know the position and magnitude of the turning points, that is the points at which the combined wave is at a maximum or a minimum. Consider $y= a cos⁡ \\theta + bsin \\theta + c$. Cosines and sines vary between $+1$ and $-1$ so the maximum value is equal to $\\sqrt{a^2+b^2}+c$ and the minimum value is equal to $-\\sqrt{a^2+b^2}+c$. Maximum = $\\sqrt{a^2+b^2}+c$ Minimum = $-\\sqrt{a^2+b^2}+c$ When the angle equals zero the cosine is at its maximum value of $1$. This means a maximum is at $\\theta-\\phi = 0$ so $\\theta=\\phi)$ The wave then drops through zero to its minimum value of $-1$ when the angle equals $\\pi$. This means a minimum is at $\\theta-\\phi = \\pi$ so $\\theta=\\pi+\\phi$ There is an infinite number of maxima and minima but it is normal to give values in the range $0 ≤ \\theta ≤ 2\\pi$. Maxima are at $\\theta=2n\\pi+\\phi$ Minima are at $\\theta=(2n+1)\\pi+\\phi$ It is normal to give values of $\\theta$ in the range $0 ≤ \\theta ≤ 2\\pi$. Example 8.5:", "Courses Courses for Kids Free study material Free LIVE classes More # Show that max and min values of $8\\cos \\theta - 15\\sin \\theta$ are 17 and -17 respectively. Last updated date: 21st Mar 2023 Total views: 307.2k Views today: 6.86k Verified 307.2k+ views Hint: Here, we will use the extreme values of the form $a\\cos \\theta + b\\sin \\theta$ to find the max and min values. Given, $8\\cos \\theta - 15\\sin \\theta \\to (1)$ Let us compare the equation (1) with $a\\cos \\theta + b\\sin \\theta$, we get $a = 8,b = - 15$ As, we know the maximum and minimum values of $a\\cos \\theta + b\\sin \\theta$ are $\\sqrt {{a^2} + {b^2}}$ and -$\\sqrt {{a^2} + {b^2}}$respectively. Therefore, substituting the values of a and b, we get $\\Rightarrow \\max = \\sqrt {{a^2} + {b^2}} = \\sqrt {{8^2} + {{( - 15)}^2}} = \\sqrt {64 + 225} = \\sqrt {289} = 17 \\\\ \\Rightarrow \\min = - \\sqrt {{a^2} + {b^2}} = - \\sqrt {{8^2} + {{( - 15)}^2}} = - \\sqrt {64 + 225} = - \\sqrt {289} = - 17 \\\\$ Hence, the maximum value of $8\\cos \\theta - 15\\sin \\theta$ is 17 and minimum value of $8\\cos \\theta - 15\\sin \\theta$ is -17. Note: The maximum and minimum of the $a\\cos \\theta + b\\sin" ]

[ "# Find the minimum value of $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta$ Find the minimum value of $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta$ $a.)\\ 1 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ b.)\\ 3 \\\\ c.)\\ 5 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ d.)\\ 7$ $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta \\\\ =\\sin^{2} \\theta +\\dfrac{1}{\\sin^{2} \\theta }+\\cos^{2} \\theta+\\dfrac{1}{\\cos^{2} \\theta }+\\tan^{2} \\theta+\\dfrac{1}{\\tan^{2} \\theta } \\\\ \\color{blue}{\\text{By using the AM-GM inequlity}} \\\\ \\color{blue}{x+\\dfrac{1}{x} \\geq 2} \\\\ =2+2+2=6$ Which is not in options. But I am not sure if I can use that $AM-GM$ inequality in this case. I look for a short and simple way . I have studied maths upnto $12$th grade . • The AM-GM inequality is useful, but you cannot assume that $\\sin^{2} \\theta +\\dfrac{1}{\\sin^{2} \\theta }$,$\\cos^{2} \\theta+\\dfrac{1}{\\cos^{2} \\theta }$, and $\\tan^{2} \\theta+\\dfrac{1}{\\tan^{2} \\theta }$ all are minimized at the same value of $\\theta$. In fact they are not. – David K Jan 20 '16 at 22:54 Hint: You can use the Pythagorean identities $\\color{blue}{\\sin^2\\theta+\\cos^2\\theta=1}$, $\\color{blue}{\\sec^2 \\theta=\\tan^2 \\theta+1}$ and $\\color{blue}{\\csc^2\\theta=\\cot^2 \\theta+1}$, giving us \\begin{align}\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta&=3+2\\tan^2\\theta+2\\cot^2\\theta\\\\ &=3+2\\left(\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}\\right) \\end{align} • is the answer $7$. – R K Jan 20 '16 at 22:49 • @RK: Yes, it", "$k=\\boxed{\\sqrt{26}}$ is the highest possible value. ### Fourth Solution By AM-QM, $\\frac{2x + 3y}{13} = \\frac{4 \\cdot \\frac{x}{2} + 9 \\cdot \\frac{y}{3}}{13} \\le \\sqrt{\\frac{4 \\cdot \\left(\\frac{x}{2}\\right)^2 + 9 \\left(\\frac{y}{3}\\right)^2}{13}} = \\sqrt{\\frac{2}{13}}$, so $2x + 3y \\le \\sqrt{26}$, equality when $\\frac x2 = \\frac y3$.", "# How to correctly find the value of theta for which $\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta}$ is minimum? I was solving a question which required me to find the value of $$\\theta$$ for which the expression $$\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta}$$ has its minimum value. Given that, $$a=3\\sqrt3, b=1$$. Note: It was a fault from my end for not including the values of $$a,b$$ which were given in the question. I apologize for that. Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following: ### Using AM-GM inequality $$\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta} \\ge 2\\sqrt{\\frac{ab}{\\cos\\theta\\sin\\theta}}\\\\ \\implies \\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta} \\ge 2\\sqrt{\\frac{2ab}{\\sin2\\theta}}$$ Since, we're minising the given expression, therefore $$\\sin2\\theta$$ should have maximum value, i.e. $$\\sin2\\theta=1$$. Therefore, $$\\theta=\\dfrac{\\pi}{4}$$. ### Using calculus $$\\text{Let }f(\\theta)=a\\sec\\theta+b\\csc\\theta \\\\ \\therefore f'(\\theta)=a\\sec\\theta\\tan\\theta-b\\csc\\theta\\cot\\theta \\\\ \\text{For mininma, }f(\\theta)=0, \\implies a\\sec\\theta\\tan\\theta=b\\csc\\theta\\cot\\theta \\\\ \\text{or, }\\tan^3\\theta=\\frac{b}{a}=\\frac{1}{3\\sqrt3} \\\\ \\implies \\theta=\\frac{\\pi}{6}$$ Why does solving this problem using derivatives yield $$\\theta=\\dfrac{\\pi}{6}$$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus? • Please show your work solving using derivatives; doesn't $\\theta$ depend on $a$ and $b$? Apr 8, 2019 at 14:11 • @J.W.Tanner Done! Apr 8, 2019 at 14:28 • Thanks for clarifying; you have the correct answer using calculus now Apr 8, 2019 at 14:37 • Simply, you cannot apply AM-GM inequality here. This applies to means. You", "# trig equality - stick • February 28th 2009, 03:02 PM TYTY trig equality - stick $\\frac{(\\sqrt{5}tan \\theta)^2*\\sqrt{5}sec^2 \\theta}{\\sqrt{5 +5tan^2\\theta}}$ is supposed to equal $5tan^2\\theta$ it's the part in the middle that's screwing me :o • February 28th 2009, 03:52 PM skeeter did you post the numerator correctly? here's what can be done with the denominator ... $\\sqrt{5 + 5\\tan^2{\\theta}} $ $\\sqrt{5(1 + \\tan^2{\\theta})} $ $\\sqrt{5\\sec^2{\\theta}}$", "• # question_answer The minimum value of $3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$ is A) 6 B) 15 C) 24 D) none of these Using $A.M.\\ge G.M$ $\\Rightarrow$$\\frac{1}{2}(3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta )\\ge 6$ $\\Rightarrow$$3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$has minimum value 12.", "## 2bornot2b Group Title Evaluate the integral $\\int\\frac{1}{(x^2\\sqrt{x^2-16)}}dx$ using the substitution $x=4\\sec\\theta$ This question is originally posted by SteelSeries 2 years ago 2 years ago 1. SteelSeries I wonder if many people know Calc II here... 2. Mr.Math $$x=4\\sec\\theta \\implies dx=4\\tan\\theta \\sec\\theta d\\theta$$, substituting that we get: $\\int \\frac{4\\tan\\theta \\sec\\theta}{16\\sec^2\\theta\\sqrt{16\\sec^2\\theta-16}}d\\theta=\\int \\frac{4\\tan\\theta}{16\\sec\\theta\\cdot 4\\tan\\theta}=\\frac{1}{16}\\int \\cos\\theta=\\cdots$ 3. Mr.Math Don't forget to substitute back for $$x$$. 4. myininaya $dx=4sec( \\theta) tan( \\theta) d \\theta$ $\\int\\limits_{}^{}\\frac{ 4 \\sec(\\theta) \\tan(\\theta) d \\theta}{4^2 \\sec^2(\\theta) \\sqrt{4 ^2 \\sec^2(\\theta)-16}}$ $\\frac{4}{4^2}\\int\\limits_{}^{}\\frac{\\tan(\\theta) d \\theta}{\\sqrt{16} \\sec(\\theta) \\sqrt{\\sec^2(\\theta)-1}}$ $\\frac{1}{4} \\cdot \\frac{1}{\\sqrt{16}} \\int\\limits_{}^{} \\frac{\\tan(\\theta) d \\theta}{\\sec(\\theta) \\sqrt{\\tan^2(\\theta)}}$ $\\text{ Assume } \\tan(\\theta)>0 =>|\\tan(\\theta)|=\\tan(\\theta)$ So we have $\\frac{1}{16}\\int\\limits_{}^{}\\cos(\\theta) d \\theta= \\frac{1}{16}\\sin(\\theta)+C$ Recall the sub: $x=4 \\sec(\\theta) => \\sec(\\theta)=\\frac{x}{4} (=\\frac{hyp}{adj})$ So opposite=... $opp=\\sqrt{x^2-4^2}=\\sqrt{x^2-16}$ So we have the answer is $\\frac{1}{16} \\frac{\\sqrt{x^2-16}}{x}+C$ 5. SteelSeries Woah 6. Mr.Math |dw:1328289496907:dw| 7. myininaya |dw:1328289591112:dw| 8. Mr.Math Right sqrt{x^2-16}. 9. SteelSeries Wow, thanks guys, that really helps a lot! 10. 2bornot2b So @steelseries, remember the tricks :) to increase your probability of getting answers. OK? 11. SteelSeries Yeah, will do, thanks Captain! 12. Mr.Math What are those tricks? We want to know them too :-P 13. 2bornot2b If so then here is the real story. http://openstudy.com/users/steelseries#/updates/4f2c1192e4b0e1bedb5d1d04 14. saifoo.khan", "(1) we get $$x\\cdot(2b-x)=a^2 \\leftrightarrow -x^2+2bx-a^2=0 \\leftrightarrow x=-\\frac{4b^2\\pm\\sqrt{4b^2-4a^2}}{2}$$ Thus not even a calculation of gm from am is possible, but for every positive $(a;b)$ am, gm pair we can find the $(x,y)$ real number pair, whose am and gm are the given values! -", "# Trigonometric inequality with Tangent If $a \\in (\\pi/2, \\pi)$, then $\\sqrt{x^2+x}+\\frac{\\tan^2a}{\\sqrt{x^2+x}}$ is always greater than or equal to (A) $2\\tan a$ (B) $1$ (C) $-1$ (D) $-2\\tan a$ I know that the $\\tan(\\cdot)$ function takes its minimum value as $0$. Tried to use this. But I'm not able to eliminate $x$ from my answer. Hint: For positive numbers $a$ and $b$, $$a+b \\ge 2\\sqrt{ab}.$$ Hint: use the AM-GM inequality, then consider the sign of $\\tan a\\,$: $$\\require{cancel} \\dfrac{\\sqrt{x^2+x} + \\dfrac{\\tan^2 a}{\\sqrt{x^2+x}}}{2} \\ge \\sqrt{\\cancel{\\sqrt{x^2+x}} \\cdot \\dfrac{\\tan^2 a}{\\cancel{\\sqrt{x^2+x}}}} = |\\tan a|$$", "# Thread: Minimum value of trigonometric expression? 1. ## Minimum value of trigonometric expression? The minimum value of 27^cos x + 81^sin x is? 2. Does using AM-GM inequality help? 3. Hello, Originally Posted by fardeen_gen Does using AM-GM inequality help? Yes ! $27=3^3$ and $81=3^4$ Therefore $27^{\\cos(x)}+81^{\\sin(x)}=3^{3 \\cos(x)}+3^{4 \\sin(x)}$ By AM-GM : $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$ So you just have to find the minimum of $3 \\cos(x)+4 \\sin(x)$ 4. Originally Posted by fardeen_gen The minimum value of 27^cos x + 81^sin x is? This is periodic with perion $2 \\pi$, so sketch the curve. This has a minimum near $x=4$. The minimum is a root of: $f(x)=\\ln(81) \\cos(x) e^{\\ln(81) \\sin(x)}-\\ln(27) \\sin(x) e^{\\ln(27) \\cos(x)}$ Running the bisection method over this gives the minimum occurs at $x \\approx 3.86147$, where the function is very close to $0.141494$ . RonL 5. Originally Posted by Moo Hello, Yes ! $27=3^3$ and $81=3^4$ Therefore $27^{\\cos(x)}+81^{\\sin(x)}=3^{3 \\cos(x)}+3^{4 \\sin(x)}$ By AM-GM : $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$ So you just have to find the minimum of $3 \\cos(x)+4 \\sin(x)$ And how exactly do you show that the minimum of a lower bound is equal to the minimum of the function? Note a numerical solution to the first problem is not equal to the function value at a", "we have $\\frac{|a|+|c|}{2}-\\sqrt|ac|>0$ Looking at the AM-GM inequality, we want a and c to be as far apart as possible. So |a| has to be 5 and |c| has to be 0 or vice versa. Applying this, we have $\\frac{|a|+|c|}{2}-\\sqrt|ac|<\\frac{5}{2}$", "# Using AM-GM Inequality (Arithmetic mean- Geometric Mean) Find the minimum value of $2x^2+\\frac{1}{x^4}$ Problem: The problem is to find the minimum value of $2x^2+\\frac{1}{x^4}$ and though you can use any method in finding the answer, you probably want to use AM-GM method My thought process before getting stuck In this case I know that according to the AM-GM theorem/equation $\\frac{a_1+a_2+...a_n}{n}\\ge {\\root n\\of {a_1*a_2....a_n}}$. But then i do not now how to use the AM GM method in finding the minimum value • Hint: $2x^2+\\frac{1}{x^4} = x^2+x^2+\\frac{1}{x^4}$. Jul 7 '16 at 7:36 $2x^2+\\frac{1}{x^4}=x^2+x^2+\\frac{1}{x^4} \\geq 3 \\sqrt[3]{(x^2)(x^2)\\left(\\frac{1}{x^4}\\right)}=3$ The equation $$\\frac{a+b+c}2\\left[(b-c)^2+(c-a)^2+(a-b)^2\\right]=a^3+b^3+c^3-3abc$$ shows that equality in the AGM $$\\sqrt[\\large3]{xyz}\\le\\frac{x+y+z}3$$ occurs only when $x=y=z$. This in conjunction with Jason M's observation that the AGM applies to this question as $$3\\le x^2+x^2+\\frac1{x^4}$$ shows that the minimum is $3$ and it is attained when $x=1$.", "## 2bornot2b Group Title Evaluate the integral $\\int\\frac{1}{(x^2\\sqrt{x^2-16)}}dx$ using the substitution $x=4\\sec\\theta$ This question is originally posted by SteelSeries 2 years ago 2 years ago 1. SteelSeries Group Title I wonder if many people know Calc II here... 2. Mr.Math Group Title $$x=4\\sec\\theta \\implies dx=4\\tan\\theta \\sec\\theta d\\theta$$, substituting that we get: $\\int \\frac{4\\tan\\theta \\sec\\theta}{16\\sec^2\\theta\\sqrt{16\\sec^2\\theta-16}}d\\theta=\\int \\frac{4\\tan\\theta}{16\\sec\\theta\\cdot 4\\tan\\theta}=\\frac{1}{16}\\int \\cos\\theta=\\cdots$ 3. Mr.Math Group Title Don't forget to substitute back for $$x$$. 4. myininaya Group Title $dx=4sec( \\theta) tan( \\theta) d \\theta$ $\\int\\limits_{}^{}\\frac{ 4 \\sec(\\theta) \\tan(\\theta) d \\theta}{4^2 \\sec^2(\\theta) \\sqrt{4 ^2 \\sec^2(\\theta)-16}}$ $\\frac{4}{4^2}\\int\\limits_{}^{}\\frac{\\tan(\\theta) d \\theta}{\\sqrt{16} \\sec(\\theta) \\sqrt{\\sec^2(\\theta)-1}}$ $\\frac{1}{4} \\cdot \\frac{1}{\\sqrt{16}} \\int\\limits_{}^{} \\frac{\\tan(\\theta) d \\theta}{\\sec(\\theta) \\sqrt{\\tan^2(\\theta)}}$ $\\text{ Assume } \\tan(\\theta)>0 =>|\\tan(\\theta)|=\\tan(\\theta)$ So we have $\\frac{1}{16}\\int\\limits_{}^{}\\cos(\\theta) d \\theta= \\frac{1}{16}\\sin(\\theta)+C$ Recall the sub: $x=4 \\sec(\\theta) => \\sec(\\theta)=\\frac{x}{4} (=\\frac{hyp}{adj})$ So opposite=... $opp=\\sqrt{x^2-4^2}=\\sqrt{x^2-16}$ So we have the answer is $\\frac{1}{16} \\frac{\\sqrt{x^2-16}}{x}+C$ 5. SteelSeries Group Title Woah 6. Mr.Math Group Title |dw:1328289496907:dw| 7. myininaya Group Title |dw:1328289591112:dw| 8. Mr.Math Group Title Right sqrt{x^2-16}. 9. SteelSeries Group Title Wow, thanks guys, that really helps a lot! 10. 2bornot2b Group Title So @steelseries, remember the tricks :) to increase your probability of getting answers. OK? 11. SteelSeries Group Title Yeah, will do, thanks Captain! 12. Mr.Math Group Title What are those tricks? We want to know them too :-P 13. 2bornot2b Group Title If so then here is the real", "chain rule, $\\frac{d}{dt}(\\tan\\theta)=(\\sec^2\\theta)\\left(\\frac {d\\theta}{dt}\\right)=(1+\\tan^2\\theta)\\frac{d\\theta }{dt}$; and you can find $\\frac{d}{dt}\\left(\\frac{\\dot{y}}{\\dot{x}}\\right)$ via the quotient rule. Then replace $\\tan\\theta$ on the left hand side with $\\frac{\\dot{y}}{\\dot{x}}$ (since those are equal), then solve the equation for $\\frac{d\\theta}{dt}$ in terms of $\\dot{x}$, $\\dot{y}$, $\\ddot{x}$, and $\\ddot{y}$. --Kevin C. 4. Thank you for the help!", "Browse Questions # If $\\tan (\\pi \\cos \\theta)=\\cot (\\pi \\sin \\theta)$ then a value of $\\cos \\bigg( \\theta-\\large\\frac{\\pi}{4}\\bigg)$ among the following is : $(a)\\;\\frac{1}{2\\sqrt 2} \\quad (b)\\;\\frac{1}{\\sqrt 2} \\quad (c)\\;\\frac{1}{2} \\quad (d)\\;\\frac{1}{4}$ $(a)\\;\\frac{1}{2\\sqrt 2}$", "\\pm z_{\\alpha/2}\\sqrt{\\frac{\\hat{\\pi}\\cdot (1-\\hat{\\pi})}{n}}=0.514 \\pm 0.052$$ where, $$z_{\\alpha/2}=1.96$$ The 1-proportion z-interval is (0.4623, 0.5655).", "# If u=sqrt(a^2 cos^2 theta + b^2sin^2theta)+sqrt(a^2 sin^2 theta + b^2 cos^2 theta), then the difference between the maximum and minimum values of u^2 is given by : (a) (a-b)^2 (b) 2sqrt(a^2+b^2) (c) (a+b)^2 (d) 2(a^2+b^2) Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free,", "2\\theta>1$ which would contradict $w\\geq h$. Comparing $f(x,y)$ in the two cases, we find that the maximum is: • $(x,y)=h/2\\cdot\\left(1/\\sin\\theta,1/\\cos\\theta\\right)$ when $w/h\\geq 1/\\sin 2\\theta$ • $(x,y) = 1/\\cos 2\\theta \\cdot \\left(w\\cos\\theta - h\\sin\\theta, h\\cos\\theta - w\\sin\\theta\\right)$ when $w/h\\leq 1/\\sin 2\\theta$ When $\\theta=\\pi/4$, the maximum is $x=y=h/2$.", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "1. ## differentiate! 1) $f(x)=(7x+\\sqrt{x^2+6})^4$ 2) $f(x)=tan^3 (4x)$ 3) $g(x)=(\\frac{x-2}{2x+2})^9$ 4) $p(x)= [3x(5x-1)^5]^4$ I tried to work out all 4 of these but I keep getting the wrong answer. 2. Originally Posted by yoman360 1) $f(x)=(7x+\\sqrt{x^2+6})^4$ 2) $f(x)=tan^3 (4x)$ 3) $g(x)=(\\frac{x-2}{2x+2})^9$ 4) $p(x)= [3x(5x-1)^5]^4$ I tried to work out all 4 of these but I keep getting the wrong answer. all involve the chain rule ... 1) $f'(x) = 4(7x + \\sqrt{x^2+6})^3 \\cdot \\left(7 + \\frac{x}{\\sqrt{x^2+6}}\\right)$ 2) $f'(x) = 3\\tan^2(4x) \\cdot \\sec^2(4x) \\cdot 4$ 3) $g'(x) = 9\\left(\\frac{x-2}{2x+2}\\right)^8 \\cdot \\frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}$ 4) $p'(x)= 4[3x(5x-1)^5]^3 \\cdot [3x \\cdot 5(5x-1)^4 \\cdot 5 + (5x-1)^5 \\cdot 3]$ you can clean up the algebra. 3. Originally Posted by skeeter all involve the chain rule ... 1) $f'(x) = 4(7x + \\sqrt{x^2+6})^3 \\cdot \\left(7 + \\frac{x}{\\sqrt{x^2+6}}\\right)$ 2) $f'(x) = 3\\tan^2(4x) \\cdot \\sec^2(4x) \\cdot 4$ 3) $g'(x) = 9\\left(\\frac{x-2}{2x+2}\\right)^8 \\cdot \\frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}$ 4) $p'(x)= 4[3x(5x-1)^5]^3 \\cdot [3x \\cdot 5(5x-1)^4 \\cdot 5 + (5x-1)^5 \\cdot 3]$ you can clean up the algebra. Thanks I kinda don't understand the chain rule but as I do more problems it might make more sense.", "Sep 27, 2015 ### terryds $n=\\frac{R_1^2+2R_1R_2+R_2^2}{R_1R_2} = 2 + \\frac{R_1^2}{R_1R_2} + \\frac{R_2^2}{R_1R_2}$ The AM GM of function y= $\\frac{R_1^2}{R_1R_2} + \\frac{R_2^2}{R_1R_2}$ is $\\frac{R_1^2+R_2^2}{R_1R_2} \\geq 2\\sqrt{\\frac{R_1^2 R_2^2}{R_1^2 R_2^2}}$ $\\frac{R_1^2+R_2^2}{R_1R_2} \\geq 2$ The minimum value of this function is 2 So, the minimum value of n is 2+2 = 4... Thanks a lot for your help !" ]

[ "# How to correctly find the value of theta for which $\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta}$ is minimum? I was solving a question which required me to find the value of $$\\theta$$ for which the expression $$\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta}$$ has its minimum value. Given that, $$a=3\\sqrt3, b=1$$. Note: It was a fault from my end for not including the values of $$a,b$$ which were given in the question. I apologize for that. Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following: ### Using AM-GM inequality $$\\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta} \\ge 2\\sqrt{\\frac{ab}{\\cos\\theta\\sin\\theta}}\\\\ \\implies \\frac{a}{\\cos\\theta}+\\frac{b}{\\sin\\theta} \\ge 2\\sqrt{\\frac{2ab}{\\sin2\\theta}}$$ Since, we're minising the given expression, therefore $$\\sin2\\theta$$ should have maximum value, i.e. $$\\sin2\\theta=1$$. Therefore, $$\\theta=\\dfrac{\\pi}{4}$$. ### Using calculus $$\\text{Let }f(\\theta)=a\\sec\\theta+b\\csc\\theta \\\\ \\therefore f'(\\theta)=a\\sec\\theta\\tan\\theta-b\\csc\\theta\\cot\\theta \\\\ \\text{For mininma, }f(\\theta)=0, \\implies a\\sec\\theta\\tan\\theta=b\\csc\\theta\\cot\\theta \\\\ \\text{or, }\\tan^3\\theta=\\frac{b}{a}=\\frac{1}{3\\sqrt3} \\\\ \\implies \\theta=\\frac{\\pi}{6}$$ Why does solving this problem using derivatives yield $$\\theta=\\dfrac{\\pi}{6}$$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus? • Please show your work solving using derivatives; doesn't $\\theta$ depend on $a$ and $b$? Apr 8, 2019 at 14:11 • @J.W.Tanner Done! Apr 8, 2019 at 14:28 • Thanks for clarifying; you have the correct answer using calculus now Apr 8, 2019 at 14:37 • Simply, you cannot apply AM-GM inequality here. This applies to means. You", "# Find the minimum value of $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta$ Find the minimum value of $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta$ $a.)\\ 1 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ b.)\\ 3 \\\\ c.)\\ 5 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ d.)\\ 7$ $\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta \\\\ =\\sin^{2} \\theta +\\dfrac{1}{\\sin^{2} \\theta }+\\cos^{2} \\theta+\\dfrac{1}{\\cos^{2} \\theta }+\\tan^{2} \\theta+\\dfrac{1}{\\tan^{2} \\theta } \\\\ \\color{blue}{\\text{By using the AM-GM inequlity}} \\\\ \\color{blue}{x+\\dfrac{1}{x} \\geq 2} \\\\ =2+2+2=6$ Which is not in options. But I am not sure if I can use that $AM-GM$ inequality in this case. I look for a short and simple way . I have studied maths upnto $12$th grade . • The AM-GM inequality is useful, but you cannot assume that $\\sin^{2} \\theta +\\dfrac{1}{\\sin^{2} \\theta }$,$\\cos^{2} \\theta+\\dfrac{1}{\\cos^{2} \\theta }$, and $\\tan^{2} \\theta+\\dfrac{1}{\\tan^{2} \\theta }$ all are minimized at the same value of $\\theta$. In fact they are not. – David K Jan 20 '16 at 22:54 Hint: You can use the Pythagorean identities $\\color{blue}{\\sin^2\\theta+\\cos^2\\theta=1}$, $\\color{blue}{\\sec^2 \\theta=\\tan^2 \\theta+1}$ and $\\color{blue}{\\csc^2\\theta=\\cot^2 \\theta+1}$, giving us \\begin{align}\\sin^{2} \\theta +\\cos^{2} \\theta+\\sec^{2} \\theta+\\csc^{2} \\theta+\\tan^{2} \\theta+\\cot^{2} \\theta&=3+2\\tan^2\\theta+2\\cot^2\\theta\\\\ &=3+2\\left(\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}\\right) \\end{align} • is the answer $7$. – R K Jan 20 '16 at 22:49 • @RK: Yes, it", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "• # question_answer The minimum value of $3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$ is A) 6 B) 15 C) 24 D) none of these Using $A.M.\\ge G.M$ $\\Rightarrow$$\\frac{1}{2}(3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta )\\ge 6$ $\\Rightarrow$$3{{\\tan }^{2}}\\theta +12{{\\cot }^{2}}\\theta$has minimum value 12.", "\\:=\\:R\\cos(\\theta + \\alpha)$ Therefore: . $\\boxed{R \\:=\\:\\sqrt{13}}$ Quote: Show that: . $\\alpha \\,\\approx\\, 33.7^o$ Since $\\sin\\alpha \\,=\\,\\tfrac{2}{\\sqrt{13}}$, then: . $\\alpha \\:=\\:\\sin^{-1}\\left(\\tfrac{2}{\\sqrt{13}}\\right) \\quad\\Rightarrow\\quad \\boxed{\\alpha\\;\\approx\\; 33.7^o}$ Quote: Hence, find the maximum value of $3\\cos\\theta - 2\\sin\\theta$ and find a positive value of $\\theta$ at which the maximum value occurs. The maximum of: . $3\\cos\\theta - 2\\sin\\theta \\:=\\:\\sqrt{13}\\,\\cos(\\theta + \\alpha)$ occurs when $\\cos(\\theta + \\alpha) \\:=\\:1$ . . Then the maximum value is $\\boxed{\\sqrt{13}}$ If $\\cos(\\theta + \\alpha) \\:=\\:1$, then: . $\\theta + \\alpha \\:=\\:360^o$ Therefore: . $\\theta \\:=\\:360^o - \\alpha \\:=\\:360^o - 33.7^o \\quad\\Rightarrow\\quad\\boxed{\\theta\\;=\\;326.3^o}$", "and minimum should occur. The maximum value of $\\sin(\\theta)$ occurs at $\\theta = 2n \\pi + \\dfrac{\\pi}{2}$ and the minimum value of $\\sin(\\theta)$ occurs at $\\theta = 2n \\pi - \\dfrac{\\pi}{2}$ and it is at these points the tangent line will be horizontal. Hence, the tangent line to $f(x)$ is horizontal at $$x + \\dfrac{\\pi}4 = 2n \\pi \\pm \\dfrac{\\pi}2$$ -", "lies between $-4$ and $10$ 39. If $m\\tan(\\theta - 30^\\circ) = n\\tan(\\theta + 120^\\circ),$ show that $\\cos2\\theta = \\frac{m + n}{2(m - n)}$ 40. if $\\alpha + \\beta = \\theta$ and $\\tan\\alpha:\\tan\\beta = x:y,$ prove that $\\sin(\\alpha - \\beta) = \\frac{x - y}{x + y}\\sin\\theta$ 41. Find the maximum and minimum value of $7\\cos\\theta + 24\\sin\\theta$ 42. Show that $\\sin 100^\\circ - \\sin 10^\\circ$ is positive.", "the sine is at its maximum value of $1$. This means a maximum is at $\\theta+\\phi = \\pi / 2$ so $\\theta=\\pi / 2 - \\phi$ The wave then drops through zero to its minimum value of $-1$ when the angle equals $3 \\pi / 2$. This means a minimum is at $\\theta+\\phi = 3\\pi / 2$ so $\\theta=3\\pi / 2-\\phi$ There is an infinite number of maxima and minima. It is normal to give values in the range $0 ≤ \\theta ≤ 2\\pi$. Maxima are at $\\theta=2n\\pi+\\phi$ Minima are at $\\theta=(2n+1)\\pi+\\phi$ It is normal to give values of $\\theta$ in the range $0 ≤ \\theta ≤ 2\\pi$. Example 8.5: Find the value and position of the first maximum and minimum of$y= 4sin\\theta+3cos \\theta + 3$ Your browser does not support the HTML5 canvas tag. $r$ $=$ $\\sqrt{3^2 + 4^2}=5$. and $\\phi$ $=$ $tan^{-1}(3/4) = 0.64$ radians Maximum $=$ $5 + 3=8$ Minimum $=$ $-5 + 3=-2$ Maximum $=8$ and is at $\\theta+0.64$ $=$ $\\pi/2$ so $\\theta_{max}$ $=$ $\\pi/2 - 0.64=0.93$ radians Minimum $=-2$ and is at $\\theta+0.64$ $=$ $3 \\pi/2$ so $\\theta_{min}$ $=$ $3 \\pi/2-0.64=4.07$ radians", "# Finding minimum and maximum value of complex number Let a be a positive real number and let $M_a =\\{z \\in C^*: |z+\\frac{1}{z}|=a\\}$ Find the minimum and maximum value of $|z|$ when z$\\in M_a$ • What is $\\mathbb{C}^*$? – copper.hat Feb 22 '13 at 5:52 • $C*$ = non zero complex number – Sachin Sharmaa Feb 22 '13 at 6:19 Let $z=r(\\cos \\theta+i\\sin \\theta)$ $z^{-1}=r^{-1}(\\cos \\theta-i\\sin \\theta)$ We have , $|z+\\frac{1}{z}|=|(r+1/r)\\cos \\theta +(r-1/r)i\\sin\\theta|$ So we have, $\\displaystyle|(r+1/r)\\cos \\theta +(r-1/r)i\\sin\\theta|=\\sqrt{(r^2+1/r^2)+2\\cos 2\\theta}$ Thus we have , $(r^2+1/r^2)+2\\cos 2\\theta=a^2$ Let $r^2=x$, $\\Rightarrow x^2+1=x(a^2-2\\cos2\\theta)$ $\\Rightarrow x^2-x(a^2-2\\cos2\\theta)+1=0$ $\\displaystyle x=\\frac{(a^2-2\\cos 2\\theta)+\\sqrt{(a^2-2\\cos 2\\theta)^2-4}}{2}$ and $\\displaystyle x=\\frac{(a^2-2\\cos 2\\theta)-\\sqrt{(a^2-2\\cos 2\\theta)^2-4}}{2}$ Clearly x attains its maximum value at $2\\theta=0$ using the first expression, The value is $\\displaystyle \\frac{(a^2+2)+\\sqrt{(a^2+2)^2-4}}{2}$ So the maximum value of $r=\\sqrt{\\displaystyle \\frac{(a^2+2)+\\sqrt{(a^2+2)^2-4}}{2}}$ This on simplification turns out to be, $\\displaystyle r=\\frac{a+\\sqrt{a^2+4}}{2}$", "1. ## minimum positive value. If $a,b$ are positive quantities and $a\\geq b$.Then find Minimum positive value of $asec\\theta-btan\\theta$ 2. $\\displaystyle a \\sec \\theta \\tan \\theta - b \\sec^{2} \\theta = 0$ $\\displaystyle \\frac{a}{b} = \\frac{\\sec \\theta}{\\tan \\theta}$ $\\displaystyle \\frac{a}{b} = \\frac{1}{\\sin \\theta}$ $\\displaystyle \\frac{b}{a} = \\sin \\theta$ $\\displaystyle \\theta = \\arcsin \\frac {b}{a} + 2 \\pi n, \\ n = 0, \\ \\pm 1, \\pm2, ...$ 3. but ans is in terms of $a$ and $b$ i.e $asec\\theta-btan\\theta \\geq \\sqrt{a^2-b^2}$ 4. Hello, jacks! but ans is in terms of $\\,a$ and $\\,b$ $a\\sec\\theta-b\\tan\\theta \\:\\geq\\: \\sqrt{a^2-b^2}$ Random Variable did the hard work for you . . . $\\text{We have: }\\:f(\\theta) \\:=\\:a\\sec\\theta - b\\tan\\theta$ $\\text{He found that }\\,f'(\\theta) = 0\\,\\text{ when }\\,\\sin\\theta \\,=\\,\\dfrac{b}{a}$ . . $\\text{Hence: }\\;\\sec\\theta \\,=\\,\\dfrac{a}{\\sqrt{a^2-b^2}},\\;\\;\\tan\\theta \\,=\\,\\dfrac{b}{\\sqrt{a^2-b^2}}$ Substitute into the function: . . $\\displaystyle a\\sec\\theta - b\\tan\\theta \\;\\ge\\;a\\frac{a}{\\sqrt{a^2-b^2}} - b\\frac{b}{\\sqrt{a^2-b^2}}$ . . . . . . . . . . . . . . $\\displaystyle =\\;\\frac{a^2}{\\sqrt{a^2-b^2}} - \\frac{b^2}{\\sqrt{a^2-b^2}}$ . . . . . . . . . . . . . . $\\displaystyle =\\;\\frac{a^2-b^2}{\\sqrt{a^2-b^2}} \\;=\\;\\sqrt{a^2-b^2}$ $\\text{Therefore: }\\;a\\sec\\theta - b\\tan\\theta \\;\\ge \\;\\sqrt{a^2+b^2}$ 5. Thanks Random variable and Sorobon.", "# Find minimum value of $\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$ Find minimum value of $$\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$$ I know this question has already answered here Then minimum value of $\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha}$ but my answer is coming different. i applied AM-GM directly to two fractions and by changing terms of sec and tan into sin and cos and simplifying a little we get that $$\\frac{\\sec^4 \\alpha}{\\tan^2 \\beta}+\\frac{\\sec^4 \\beta}{\\tan^2 \\alpha} \\geq\\frac2{\\cos\\alpha \\cos\\beta \\sin\\alpha \\sin\\beta}\\geq2$$ but minimum value is coming 8 ??? • Your bound $2/(\\cos\\alpha\\sin\\alpha\\cos\\beta\\sin\\beta)\\geq 2$ is too weak. You can tighten that to $\\geq 8$ by using double angle formula for sine. Jul 23 '20 at 3:29 • Right, so the expression is always $\\ge2$. But $2$ isn't the minimum value, since there are no $\\alpha$ and $\\beta$ making the expression $2$ (since actually it's always $\\ge8$). Jul 23 '20 at 3:29 • @AnginaSeng but how we know that there are no $α$ and $β$ making the expression $2$ , and how we get to know that 8 is correct minimum in the same sense ? Jul 23 '20 at 3:31 • Jul 23 '20 at 3:31 • @AnginaSeng thanks,but i have already seen it and also i have mention same link in my question,but i am asking how we know that 2 is", "# Thread: Minimum value of trigonometric expression? 1. ## Minimum value of trigonometric expression? The minimum value of 27^cos x + 81^sin x is? 2. Does using AM-GM inequality help? 3. Hello, Originally Posted by fardeen_gen Does using AM-GM inequality help? Yes ! $27=3^3$ and $81=3^4$ Therefore $27^{\\cos(x)}+81^{\\sin(x)}=3^{3 \\cos(x)}+3^{4 \\sin(x)}$ By AM-GM : $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$ So you just have to find the minimum of $3 \\cos(x)+4 \\sin(x)$ 4. Originally Posted by fardeen_gen The minimum value of 27^cos x + 81^sin x is? This is periodic with perion $2 \\pi$, so sketch the curve. This has a minimum near $x=4$. The minimum is a root of: $f(x)=\\ln(81) \\cos(x) e^{\\ln(81) \\sin(x)}-\\ln(27) \\sin(x) e^{\\ln(27) \\cos(x)}$ Running the bisection method over this gives the minimum occurs at $x \\approx 3.86147$, where the function is very close to $0.141494$ . RonL 5. Originally Posted by Moo Hello, Yes ! $27=3^3$ and $81=3^4$ Therefore $27^{\\cos(x)}+81^{\\sin(x)}=3^{3 \\cos(x)}+3^{4 \\sin(x)}$ By AM-GM : $3^{3 \\cos(x)}+3^{4 \\sin(x)} \\ge 2 \\sqrt{3^{3 \\cos(x)+4 \\sin(x)}}$ So you just have to find the minimum of $3 \\cos(x)+4 \\sin(x)$ And how exactly do you show that the minimum of a lower bound is equal to the minimum of the function? Note a numerical solution to the first problem is not equal to the function value at a", "Comment Share Q) Find the maximum and minimum values of the following expression $3\\cos\\theta+5\\sin (\\theta-\\large\\frac{\\pi}{6})$ $\\begin{array}{1 1}(A)\\;Y_{max}=\\sqrt {19},Y_{min}=-\\sqrt{19}\\\\(B)\\;Y_{max}=\\sqrt {17},Y_{min}=-\\sqrt{17}\\\\(C)\\;Y_{max}=\\sqrt {13},Y_{min}=-\\sqrt{13}\\\\(D)\\;Y_{max}=\\sqrt {21},Y_{min}=-\\sqrt{21}\\end{array}$", "# Minimum value of $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2}$ Minimum value of $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2}$ Find the minimum value of $h(\\theta)$ $h(\\theta)= 3 \\sin \\theta - 4\\cos \\theta + \\sqrt{2} = 5 \\sin (\\theta + 53.13) + \\sqrt{2}$ Minimum value - $5\\sin (\\theta + 53.13) + \\sqrt{2} = -5$ Therefore min value is = $-5/5 - \\sqrt{2}$ Why am I wrong ? And how should I do this question.. • I would say that the standard way is finding the derivative of the function and solving for $h'(\\theta) = 0$. But algebraic manipulation can work too. Numerically, I find the minimum to be $\\approx -3.5858$. – Matti P. Feb 19 '18 at 7:13 • @user175089...Find h'(θ) and h''(θ)...Take h'(θ)=0...For the value of θ for which h''(θ)>0 ,gives the minimum value... – Nehal Samee Feb 19 '18 at 7:14 • You made a sign error. $-5\\color{red}+\\sqrt2$. And why do you divide by $5$ ?? – Yves Daoust Feb 19 '18 at 7:14 Minimum is attained when $\\sin (\\theta + 53.13) = -1$ that is $$h_{min}=h(3\\pi/2+k\\pi)=5 \\sin (3\\pi/2+k\\pi) + \\sqrt{2}=-5+\\sqrt{2}$$ for the same reason the maximum is attained when $\\sin (\\theta + 53.13) = 1$. Hint You must note that the range of $a\\sin\\theta \\pm b\\cos\\theta$ is $\\left[ -\\sqrt {a^2+b^2},", "Share Find the Maximum and Minimum Values of Y = Tan X − 2 X . - Mathematics Question Find the maximum and minimum values of y = tan $x - 2x$ . Solution $\\text { Given }: f\\left( x \\right) = y = \\tan x - 2x$ $\\Rightarrow f'\\left( x \\right) = \\sec^2 x - 2$ $\\text { For a local maxima or local minima, we must have }$ $f'\\left( x \\right) = 0$ $\\Rightarrow \\sec^2 x - 2 = 0$ $\\Rightarrow \\sec^2 x = 2$ $\\Rightarrow \\sec x = \\pm \\sqrt{2}$ $\\Rightarrow x = \\frac{\\pi}{4} \\text { and } \\frac{3\\pi}{4}$ $\\text { Thus, x }= \\frac{\\pi}{4} \\text { and }x = \\frac{3\\pi}{4}\\text { are the possible points of local maxima or a local minima } .$ $\\text { Now,}$ $f''\\left( x \\right) = 2 \\sec^2 x \\tan x$ $\\text { At }x = \\frac{\\pi}{4}:$ $f''\\left( \\frac{\\pi}{4} \\right) = 2 \\sec^2 \\left( \\frac{\\pi}{4} \\right) \\tan \\left( \\frac{\\pi}{4} \\right) = 4 > 0$ $\\text { So, }x = \\frac{\\pi}{4} \\text { is a point of local minimum } .$ $\\text { The local minimum value is given by }$ $f\\left( \\frac{\\pi}{4} \\right) = \\tan\\left( \\frac{\\pi}{4} \\right) - 2 \\times \\frac{\\pi}{4} = 1 - \\frac{\\pi}{2}$ $\\text { At x} = \\frac{3\\pi}{4}:$ $f''\\left( \\frac{3\\pi}{4} \\right) = 2 \\sec^2 \\left( \\frac{3\\pi}{4}", "# How do you find the average value of f(x)=x/sqrt(x^2+1), 0<=x<=4? Feb 17, 2017 The average value of $f \\left(x\\right)$ on 0 ≤ x ≤ 4 is $\\frac{1}{4} \\left(\\sqrt{17} - 1\\right)$, which is approximately $0.78$ #### Explanation: The average value of a continuous function $f \\left(x\\right)$ on $\\left[a , b\\right]$, is given by the formula $A = \\frac{1}{b - a} {\\int}_{a}^{b} f \\left(x\\right) \\mathrm{dx}$. $A = \\frac{1}{4 - 0} {\\int}_{0}^{4} \\frac{x}{\\sqrt{{x}^{2} + 1}} \\mathrm{dx}$ $A = \\frac{1}{4} {\\int}_{0}^{4} \\frac{x}{\\sqrt{{x}^{2} + 1}} \\mathrm{dx}$ This now becomes a trig substitution problem. Let $x = \\tan \\theta$. Then $\\mathrm{dx} = {\\sec}^{2} \\theta d \\theta$. Adjust the bounds of integration accordingly. $A = \\frac{1}{4} {\\int}_{0}^{\\tan 4} \\tan \\frac{\\theta}{\\sqrt{{\\tan}^{2} \\theta + 1}} \\cdot {\\sec}^{2} \\theta d \\theta$ Use the pythagorean identity ${\\tan}^{2} \\alpha + 1 = {\\sec}^{2} \\alpha$. $A = \\frac{1}{4} {\\int}_{0}^{\\tan 4} \\tan \\frac{\\theta}{\\sqrt{{\\sec}^{2} \\theta}} \\cdot {\\sec}^{2} \\theta d \\theta$ $A = \\frac{1}{4} {\\int}_{0}^{\\tan 4} \\tan \\frac{\\theta}{\\sec} \\theta \\cdot {\\sec}^{2} \\theta$ $A = \\frac{1}{4} {\\int}_{0}^{\\tan 4} \\tan \\theta \\sec \\theta$ This is a known integral. $A = \\frac{1}{4} {\\left[\\sec \\theta\\right]}_{0}^{\\tan 4}$ We know from our initial substitution that $\\frac{x}{1} = \\tan \\theta$. This means that $\\sqrt{{x}^{2} + 1}$ is the hypotenuse. That's to say $\\sec \\theta = \\sqrt{{x}^{2} + 1}$. $A = \\frac{1}{4} {\\left[\\sqrt{{x}^{2} + 1}\\right]}_{0}^{4}$ Evaluate this using the 2nd fundamental", "Let M and m Question: Let $\\mathrm{M}$ and $\\mathrm{m}$ respectively be the maximum and minimum values of the function $f(x)=\\tan ^{-1}(\\sin x+\\cos x)$ in $\\left[0, \\frac{\\pi}{2}\\right]$, Then the value of $\\tan (\\mathrm{M}-\\mathrm{m})$ is equal to: 1. $2+\\sqrt{3}$ 2. $2-\\sqrt{3}$ 3. $3+2 \\sqrt{2}$ 4. $3-2 \\sqrt{2}$ Correct Option: , 4 Solution: Let $g(x)=\\sin x+\\cos x=\\sqrt{2} \\sin \\left(x+\\frac{\\pi}{4}\\right)$ $\\mathrm{g}(\\mathrm{x}) \\in[1, \\sqrt{2}]$ for $\\mathrm{x} \\in[0, \\pi / 2]$ $f(x)=\\tan ^{-1}(\\sin x+\\cos x) \\in\\left[\\frac{\\pi}{4}, \\tan ^{-1} \\sqrt{2}\\right]$ $\\tan \\left(\\tan ^{-1} \\sqrt{2}-\\frac{\\pi}{4}\\right)=\\frac{\\sqrt{2}-1}{1+\\sqrt{2}} \\times \\frac{\\sqrt{2}-1}{\\sqrt{2}-1}=3-2 \\sqrt{2}$", "Problem 60 Suppose that $\\tan \\theta=2$ and $0<\\theta<\\pi / 2$ (a) Compute $\\sin \\theta$ and $\\cos \\theta$ (b) Using the values obtained in part (a), make the substitutions $$x=X \\cos \\theta-Y \\sin \\theta \\quad \\text { and } \\quad y=X \\sin \\theta+Y \\cos \\theta$$ in the expression $7 x^{2}-8 x y+y^{2},$ and simplify the result. Check back soon! ### Problem 61 In this exercise, we are going to find the minimum value of the function $$f(t)=\\tan ^{2} t+9 \\cot ^{2} t \\quad 0<t<\\frac{\\pi}{2}$$ (a) Set your calculator in the radian mode and complete the table. Round the values you obtain to two decimal places. \\begin{tabular}{cccccccc} $t$ & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 & 1.2 & 1.4 \\\\ \\hline$f(t)$ & & & & & & & \\\\\\hline\\end{tabular} (b) Of the seven outputs you calculated in part (a), which is the smallest? What is the corresponding input? (c) Prove that $\\tan ^{2} t+9 \\cot ^{2} t=(\\tan t-3 \\cot t)^{2}+6$ (d) Use the identity in part (c) to explain why $\\tan ^{2} t+9 \\cot ^{2} t \\geq 6$ (e) The inequality in part (d) tells us that $f(t)$ is never less than 6. Furthermore, in view of part (c), $f(t)$ will equal 6 when $\\tan t-3 \\cot t=0 .$ From this last equation, show that $\\tan ^{2}", "$cosec\\theta$ is equal to a) $\\frac{5}{3}$ b) $2$ c) $\\frac{1}{2}$ d) $4$ Question 10: The value of $\\frac{1}{1 + tan^2\\theta}$ + $\\frac{1}{1 + cot^2\\theta}$ is a) 1 b) 2 c) $\\frac{1}{2}$ d) $\\frac{1}{4}$ Taking $cos\\theta$ outside in numerator and in denominator and making $tan\\theta$ hence eq will be $(\\frac{5tan\\theta – 3}{5tan\\theta + 3})$ As it is given that $5tan\\theta$ = 4 after putting values and solving we will get the equation reduced to 1/7. $4sec^2\\theta+9cosec^2\\theta$ or $4+4tan^2\\theta+9+9cot^2\\theta$ or $13+4tan^2\\theta+9cot^2\\theta$ or $13+4tan^2\\theta+\\frac{9}{tan^2\\theta}$ or $13-12+(2tan\\theta+\\frac{3}{tan\\theta})^2$ (eq. (1) ) or now above expression to be minimum, equation $(2tan\\theta+\\frac{3}{tan\\theta})^2$ should be minimum. So applying $A.M.\\geq G.M.$ $\\frac{(2tan\\theta +\\frac{3}{tan\\theta})}{2} \\geq \\sqrt{6}$ or ${(2tan\\theta+\\frac{3}{tan\\theta})}=2\\sqrt{6}$ ( for value to be minimum) After putting above value in eq.(1) , we will get least value of expression as 25. $x=cosec\\theta – sin\\theta=\\frac{cos^2\\theta}{sin\\theta}=cot\\theta cos\\theta$ Similarly $y=tan\\theta sin\\theta$ $xy=sin\\theta cos\\theta$ $x^2+y^2+3=(sec^2\\theta +cosec^2\\theta )$ Now putting above values in given equation, and after solving it will be reduced to 1 $2y=tan\\theta$ $x=2ycosec\\theta$ Hence value of $x^2 – 4y^2$ = $4y^2(cosec^2\\theta – 1)$ or $tan^2\\theta cot^2\\theta$ = 1 Given equation can be written as $(cos^4\\theta + cos^2\\theta)^3 -1$ as $sin\\theta + sin^2\\theta = 1$ or $sin\\theta = cos^2\\theta$ putting above value in given equation it will be $(sin^2\\theta + sin\\theta)^3 -1 = 0$ $\\frac{sin\\theta}{1-cos\\theta} – \\frac{1}{sin\\theta}$ or $\\frac{cos\\theta – cos^2\\theta}{(1-cos\\theta)sin\\theta}$ =", "# Is there a better way to find the minimum value of $\\frac{1}{\\sin \\theta \\cos \\theta \\left ( \\sin \\theta + \\cos \\theta \\right )}$? To find the minimum value of $$\\frac{1}{\\sin \\theta \\cos \\theta \\left ( \\sin \\theta + \\cos \\theta \\right )}$$ I can see that I can convert it to $$\\frac{\\sqrt{2}}{\\sin (2 \\theta) \\sin \\left ( \\theta + \\frac{\\pi}{4} \\right )}$$ And from here I can convert it to a pair of cosecant functions and then proceed with the usual product rule, solving that out with zero etc. Once that is all done, we can find that the function attains a minimum at $\\theta= \\frac{\\pi}{4}$ and this minimum value is $\\sqrt{2}$. However, I was wondering if there was perhaps a better way of finding the minimum value of this function that does NOT require heavy amounts of algebra bashing. Perhaps by some means of observation. So for example, one could find the maximum of $\\sin \\theta$ by realising that it oscillates between $1$ and $-1$. Could a similar idea be used to find the minimum of this function? • What if $\\theta \\to 0$ from the left? Wouldn't this tend to $-\\infty$ and have no minimum? – jameselmore Sep 29 '15 at 16:24 • if f(x) has an extremum at x = x0, then 1/f(x)" ]

[ "# Compute the sum k=1 100000 1/k and sum k=1 1000000 1/k. Are there any differences? Compute the sum k=1 100000 1/k^2 and sum k=1 1000000 1/k^2. Are there any differences? Mauricio Mathis 2022-07-27 Answered Compute the sum $\\sum _{k=1}^{10000}\\frac{1}{k}$ and $\\sum _{k=1}^{100000}\\frac{1}{k}$. Are there any differences? Compute the sum $\\sum _{k=1}^{100000}\\frac{1}{{k}^{2}}$ and $\\sum _{k=1}^{1000000}\\frac{1}{{k}^{2}}$. Are there any differences? You can still ask an expert for help • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Jaylene Tyler We are taking k = 1 to 10000 for the first one and k = 1 to 100000 for the second one. So, yes, definitely, for the first one, we have : 1/1 + 1/2 + 1/3 + .... 1/10000 Second one would be 1/1 + 1/2 + 1/3 + .... 1/10000 + 1/10001 + 1/10002 + .... 1/99999 + 1/100000 So, yes, the second one would be slightly larger but yes, they will be different We have step-by-step solutions for your answer!", "# Modulus 1009 # 2 $\\frac{(1009^2-2)!}{1009^{1008}}$ Find the remainder when the number above is divided by 1009. ×", "to $N=1000$, we get Pretty good.", "1000 } ℓ\\\\ =39.5ℓ$$ Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "sum of all the odd numbers lying between 1 and 1000 is $S_{n}=83667$. Hence proved", "### Home > CCA2 > Chapter 10 > Lesson 10.1.4 > Problem10-63 10-63. Use any method to find the sum of the integers from 100 through 1000. Describe your strategy. Homework Help ✎ 1000 + 100 = 1100 There are 901 integers between 100 and 1000, inclusive. Substitute these numbers into the formula: $\\frac{901(1000+100)}{2}$ 495,550", "1001, 1002. For all nontrivial divisors $d_1|1001, d_2|1002$ we can write: $n(1001) = d_1(\\frac{1000!}{d_1}+ \\frac{1001}{d_1})$ and $n(1002) = d_2(\\frac{1000!}{d_2}+ \\frac{1002}{d_2})$. This is because $d_1,d_2 < 1000$ hence $d_1,d_2|1000!$.", "# Coding challenge Write some code that sums the first $1000$ consecutive integers. Show your result equals $S_n=\\frac{1}{2}n(a_1+a_n)$, for $n=1000$ ($a_1=1$ and $a_{1000}=1000$). Does this hold for other values of $n$? Say, the first 100 integers?", "\\sum_{n=2}^{\\infty} \\frac{n}{n-1}$$", "it. Nevertheless, as mathematicians, we tend to be very fancy or \"elegant\". So, let's generalize this finding: We want to add all the integers from $1$ to $n$. In other words, we want to know what$$\\sum_{i=1}^{n}i=1+2+3+\\cdots+n$$adds up to. Straight from Gauss' problem, it's clear that he was adding to $100$ by multiplying $101$ by $50$. In other words, if we let $n=100$, then he just multiplied $n+1$ by $n\\div2$. That's it! We just generalized the problem and can now claim that:$$\\sum_{i=1}^{n}i=\\frac{n(n+1)}{2}.$$From this expression, it follows that the sum of all the numbers from $1$ to $1,234,567$ is$$\\frac{1,234,567(1,234,568)}{2}=762,078,456,028.$$I bet noticing this pattern saved our little fella some playground time.", "1. ## Evaluate this summation Evaluate this summation $\\sum^{99}_{n=0}2^n(100C(n+1))$ 100C(n+1) is meant to be combinatorics C(100,n+1) 2. Here is a hint: $\\displaystyle 3^{100}=(2+1)^{100}=\\sum\\limits_{n = 0}^{100} {\\binom{100}{n}2^n}$.", "100 + 1)}{6}$$ = $$\\frac{100 × 101 × 201}{6}$$ = $$\\frac{2030100}{6}$$ = 338350 Arithmetic Progression From Sum of the Squares of First n Natural Numbers to HOME PAGE", "# Coding challenge Write some code that sums the first $1000$ consecutive integers. Show your result equals $S_n=\\frac{1}{2}n(a_1+a_n)$, for $n=1000$ ($a_1=1$ and $a_{1000}=1000$). Does this hold for other values of $n$? Say, the first 100 integers? Show a friend, family member, or teacher what you've done!", "a percent, we get this lower bound on n: $$\\frac{\\mathrm{ln} \\frac{2}{.05}}{2(0.001)^2} \\approx 9.22 \\times 10^5 \\leq n$$ The number might seem discouragingly large, but generating a million characters is feasible on modern computers; just write code to do it!", "number then $\\sum_{n=0}^N a_n \\cos \\frac{\\pi n t}{2}$ will approximate $f(t)$. Take something like $N=50,100,200$ and see how the curves approximate the function. 3. Hi, Thanks. I get that part, what i'm stuck with is actually coming up with the Matlab code to get the coefficients and then plotting the approximation.", "k^2 = {n+1 \\choose 2} = \\sum_{k=1}^n \\; k = \\frac{(n+1) \\; n}{2}$$ -", "## Algebra and Trigonometry 10th Edition $S_{100}=10000$ Sequence of the positive odd integers: 1, 3, 5, 7, ... That is, $a_1=1$ and $d=2$ $a_n=a_1+(n-1)d$ $a_{100}=1+(100-1)2=1+(99)2=1+198=199$ $S_n=\\frac{n}{2}(a_1+a_n)$ $S_{100}=\\frac{100}{2}(1+199)=50(200)=10000$", "# 900 Followers Problem! $\\frac{m^3-n^3}{m^2+n^2-mn}$ Find the sum of all prime numbers less than 900 that can be expressed as the above fraction where $m$ and $n$ are positive integers. × Problem Loading... Note Loading... Set Loading...", "geometric, it shouldn't be hard to compute the sum. • July 14th 2012, 01:30 AM lord287 Re: I am having problems related to series and sequences and here are those questions Quote: Originally Posted by richard1234 Q2: Series obviously converges. Also, note that $\\frac{2}{10} + \\frac{4}{1000} + \\frac{6}{100000} + \\dots = (\\frac{2}{10} + \\frac{2}{1000} + \\dots) + (\\frac{2}{1000} + \\frac{2}{100000} + \\dots) + \\dots$ Factor stuff out. $= (\\frac{2}{10} + \\frac{2}{1000} + \\dots)(1 + \\frac{1}{100} + \\frac{1}{10000} + \\dots)$ Both series are geometric, it shouldn't be hard to compute the sum. Didn't understand how did u factor that stuff out. I had been trying to factor it but could not and I am not understanding how come the way u have written it is the factor of that equation?? • July 14th 2012, 01:31 AM lord287 Re: I am having problems related to series and sequences and here are those questions Quote: Originally Posted by richard1234 Q3 should be pretty simple. What are all the cubes from 1 to 10000, and how do you sum the first n cubes? I'm still figuring out Q1 but it probably involves AM-GM inequality. Oh I am sorry I wrote wrongly the real question was sum of cubes between 100 and 10000, not 1 to 10000", "## Discrete Mathematics with Applications 4th Edition The sum of the first n integers is given by: 1 + 2 + ··· + n = $\\frac{n(n+1)}{2}$ 3 + 4 + 5 + 6 + ··· + 1000 = (1 + 2 + 3 + 4 + ··· +1000) − (1 + 2) = $\\frac{1000\\times(1001)}{2}$ -3 =500500 - 3 = 500,497" ]

[ "write $$\\frac1{(3n−2)(3n+1)} = \\frac 13\\frac{(3n+1) - (3n-2)}{(3n−2)(3n+1)} =$$ and most terms telescope.", "# Difference between revisions of \"2002 AIME II Problems/Problem 6\" ## Problem Find the integer that is closest to $1000\\sum_{n=3}^{10000}\\frac1{n^2-4}$. ## Solution You know that $\\frac{4}{n^2 - 4} = \\frac{1}{n-2} - \\frac{1}{n + 2}$. So if you pull the $\\frac{1}{4}$ out of the summation, you get $250\\sum_{n=3}^{10,000} (\\frac{1}{n-2} - \\frac{1}{n + 2})$. Now that telescopes, leaving you with: $250 (1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} - \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000}) = 250 + 125 + 83.3 + 62.5 - 250 (- \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ $250(-\\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ is not enough to bring $520.8$ lower than $520.5$ so the answer is $\\fbox{521}$", "100 terms of $\\sum \\frac{1}{n^2 + 3n + 2}.$ Solution: First we can factor the denominator, which will allow us to decompose this rational function into two: $$\\frac{1}{n^2 + 3n + 2} = \\frac{1}{(n + 1)(n + 2)}$$ Now let's decompose: \\begin{align} \\frac{1}{(n + 1)(n + 2)} &= \\frac{A}{n + 1} + \\frac{B}{n + 2}\\\\[5pt] 1 &= A(n + 2) + B(n + 1)\\\\[5pt] \\text{Let }\\; n &= -1, \\; \\text{ then } \\; A = 1 \\\\[5pt] \\text{Let }\\; n &= -2, \\; \\text{ then } \\; B = -1, \\; \\text{ so} \\\\[5pt] \\frac{1}{(n + 1)(n + 2)} &= \\frac{1}{n + 1} - \\frac{1}{n + 2} \\end{align} Now we'll telescope the series: \\begin{align} S &= \\left( \\frac{1}{2} - \\frac{1}{3} \\right) + \\left( \\frac{1}{3} - \\frac{1}{4} \\right) + \\left( \\frac{1}{4} - \\frac{1}{5} \\right) + \\dots + \\left( \\frac{1}{101} - \\frac{1}{102} \\right)\\\\[5pt] &= \\frac{1}{2} + \\left( \\frac{1}{3} - \\frac{1}{3} \\right) + \\left( \\frac{1}{4} - \\frac{1}{4} \\right) + \\left( \\frac{1}{5} - \\frac{1}{5} \\right) + \\dots - \\frac{1}{102}\\\\[5pt] &= \\frac{1}{2} - \\frac{1}{102} = \\frac{51 - 1}{102} = \\frac{25}{51} \\approx \\frac{1}{2} \\end{align} It's easy to see that the sum of this series as n → ∞ is ½. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images", "\\sum_{i=1}^3 \\sum_{j=1}^5 y_{ij} \\right)^2}{N_{total}} = \\frac{(\\mbox{Total of all observations})^2}{N_{total}} = \\frac{(108.1)^2}{15} = 779.041", "that $$1+\\frac{1}{i^{2}}+\\frac{1}{(i+1)^{2}}=\\frac{i^{2}(i+1)^{2}+(i+1)^{2}+i^{2}}{i^{2}(i+1)^{2}}$$ $$=\\frac{i^{4}+2i^{3}+3i^{2}+2i+1}{i^{2}(i+1)^{2}}=\\frac{(i^{2}+i+1)^{2}}{i^{2}(i+1)^{2}}.$$ Hence our sum is $$\\sum_{i=1}^{69}\\frac{i^{2}+i+1}{i(i+1)}=\\sum_{i=1}^{69}\\left(\\frac{i^{2}+2i+1}{i(i+1)}-\\frac{i}{i(i+1)}\\right)$$ $$=\\sum_{i=1}^{69}\\frac{i+1}{i}-\\frac{1}{i+1}=69+\\sum_{i=1}^{69}\\left(\\frac{1}{i}-\\frac{1}{i+1}\\right).$$ The right hand side telescopes, so we conclude that the answer is $$69+1-\\frac{1}{70}=70-\\frac{1}{70}=\\frac{4899}{70}.$$", "understand where you're trying to put your parentheses (since you have$\\frac1{p_n}$on the LHS I presume it's trying to be 'captured' by the sum), then the sum should be easily relatable to the Mertens constant :$\\sum(\\log\\log p_{n+1}-\\log\\log p_n)$telescopes, so if you sum from$n=1$to$n=K$your partial sum will be just$\\log\\log ... Top 50 recent answers are included", "I'd like to see if there's another way of showing the relationship without knowing that it's a Riemann sum for the above reason. – Jam Nov 21 '19 at 19:55 • Approximate $\\frac{1}{k^2}$ with something that telescopes, like $\\frac{1}{k^2 - \\frac{1}{4}}$ or $\\frac{1}{k(k\\pm 1)}$ and show the difference doesn't matter. – Daniel Fischer Nov 21 '19 at 19:57 • @DanielFischer Thank you Daniel. – Jam Nov 21 '19 at 19:59 A more elementary approach. $$\\frac{1}{2n}=\\sum_{i=1}^{n}\\frac{1}{(n+i)(n+i-1)}$$ because the sum telescopes to $$\\frac{1}{n}-\\frac{1}{2n}.$$ So: \\begin{align}\\frac{1}{2n}-\\sum_{i=1}^{n}\\frac{1}{\\left(n+i\\right)^{2}}&=\\sum_{i=1}^{n}\\left(\\frac{1}{(n+i)(n+i-1)}-\\frac{1}{(n+i)^2}\\right)\\\\ &=\\sum_{i=1}^{n}\\frac{1}{(n+i)^2(n+i-1)}\\tag{1}\\\\&<\\sum_{i=1}^{n}\\frac{1}{n^3}\\\\&=\\frac{1}{n^2} \\end{align} Also, the value at (1) is positive. So we have: $$0<\\frac{1}{2}-n\\sum_{i=1}^{n}\\frac{1}{(n+i)^2}<\\frac{1}{n}$$ and hence$$n\\sum_{i=1}^{n}\\frac{1}{(n+i)^2}\\to\\frac{1}{2}$$ We have that $$\\sum_{i=1}^{n}\\frac{1}{\\left(n+i\\right)^{2}}=\\sum_{i=n+1}^{2n}\\frac{1}{i^{2}}=\\sum_{i=1}^{2n}\\frac{1}{i^{2}}-\\sum_{i=1}^{n}\\frac{1}{i^{2}}$$ then we can use the result indicated here • Perfect. This is what I was looking for. Thank you, user. – Jam Nov 21 '19 at 20:01 • The Euler–Maclaurin one? That's kind of weird if you want to avoid integrals. – Daniel Fischer Nov 21 '19 at 20:01 • @Jam Nice to may help you! – user Nov 21 '19 at 20:03 • @DanielFischer You're right; I hadn't thought of that. – Jam Nov 21 '19 at 20:05 Yet another way: since $$\\frac{1}{x^2}$$ is convex on $$\\mathbb{R}^+$$, the Hermite-Hadamard inequality ensures $$\\frac{1}{2n+1}-\\frac{1}{(2n+1)(4n-1)}=\\int_{n+1/2}^{2n-1/2}\\frac{dx}{x^2}\\geq\\sum_{k=1}^{n}\\frac{1}{(n+k)^2}\\geq \\int_{n+1}^{2n}\\frac{dx}{x^2}=\\frac{1}{2n}-\\frac{1}{n(n+1)}.$$ • Sure, but that assumes the integral of $1/x^2$ and, as the question points out, if you know that, you can just", "$$\\mathrm{R}\\left(\\frac{1}{\\mathrm{n}_1^2}-\\frac{1}{\\mathrm{n}_2^2}\\right)$$ For second member of Lyman senes, $$\\frac{1}{5400}$$ = $$R\\left(\\frac{1}{1^2}-\\frac{1}{3^2}\\right)$$ ⇒ $$\\frac{1}{5400}$$ = $$\\frac{8 \\mathrm{R}}{9}$$ —-> (1) For first member of Lyman series, $$\\frac{1}{\\lambda^1}$$ = $$\\mathrm{R}\\left(\\frac{1}{1^2}-\\frac{1}{2^2}\\right)$$ $$\\frac{1}{\\lambda^1}$$ = $$\\frac{3 R}{4}$$ —–> (2) $$\\frac{(1)}{(2)}$$ ⇒ $$\\frac{\\lambda^1}{5400}$$ = $$\\frac{8 R}{9} \\times \\frac{4}{3 R}$$ ∴ λ’ = $$\\frac{32}{27}$$ × 5400 = 6400A. Question 11. Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer senes limit. Given : R = 10970000m-1. Solution: $$\\frac{1}{\\lambda}$$ = $$\\mathrm{R}\\left(\\frac{1}{\\mathrm{n}_1^2}-\\frac{1}{\\mathrm{n}_2^2}\\right)$$ R = 10970000 = 1.097 × 107 ms-1 For Balmer senes limit n1 = 2 and n2 = ∞ $$\\frac{1}{\\lambda}$$ = $$\\mathrm{R}\\left(\\frac{1}{2^2}-\\frac{1}{\\infty}\\right)$$ ⇒ $$\\frac{1}{\\lambda}$$ = $$\\frac{R}{4}$$ λ = $$\\frac{4}{\\mathrm{R}}$$ = $$\\frac{4}{1.097 \\times 10^7}$$ = 3646.3A Question 12. Using the Rydberg formula, calcûlate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum. Solution: Question 1. Choose the correct alternative from the clues given at the end of the each statement: a) The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model. (much greater than / no different from / much less than.) b) In the ground state of ……. electrons are in stable equilibrium, while in …… electrons always experience a net force. (Thomson’s model / Rutherford’s model.) c) A classical atom based on …… is doomed to collapse. (Thomson’s", "and arrive at $$\\frac{1}{k^2-1}=\\frac{-\\frac{1}{2}}{k+1}+\\frac{\\frac{1}{2}}{k-1}=\\frac{1}{2}\\left(\\frac{1}{k-1}-\\frac{1}{k+1}\\right).$$ Taking the partial sum $$\\sum_{4 \\le k \\le m}\\frac{1}{k^2-1}=\\frac{1}{2}\\sum_{4 \\le k \\le m}\\frac{1}{k-1}-\\frac{1}{k+1}.$$ The sum telescopes, so we see that $$\\sum_{4 \\le k \\le m}\\frac{1}{k+1}-\\frac{1}{k-1}=\\color{red}{\\frac{1}{3}}-\\color{green}{\\frac{1}{5}}+\\color{red}{\\frac{1}{4}}-\\frac{1}{6}+\\color{green}{\\frac{1}{5}}-\\frac{1}{7}+\\dots+\\color{green}{\\frac{1}{m-1}}-\\frac{1}{m-3}+\\color{red}{\\frac{1}{m}}-\\frac{1}{m-2}+\\color{red}{\\frac{1}{m+1}}-\\color{green}{\\frac{1}{m-1}}=\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{m}+\\frac{1}{m+1}.$$ Taking the limit, we have $$\\lim_{m \\to \\infty}\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{m}+\\frac{1}{m+1}=\\frac{1}{3}+\\frac{1}{4}+\\lim_{m \\to \\infty}\\frac{1}{m}+\\frac{1}{m+1}=\\frac{1}{3}+\\frac{1}{4}=\\frac{7}{12}.$$ Recalling the factor of $\\frac{1}{2}$, we arrive at $\\sum_{4 \\le k \\le \\infty}\\frac{1}{k^2-1}=\\frac{1}{2}\\cdot\\frac{7}{12}=\\frac{7}{24}$. - thanks a lot.It is crystal clear now. – learner Dec 2 '12 at 7:42 @learner, I'm glad! I love colors! – 000 Dec 2 '12 at 8:36 I love the use of colors! I would recommend something other than red and green though, considering red-green is the most common type of colorblindness. – Antonio Vargas Dec 2 '12 at 23:58 @AntonioVargas, LOL. What luck I have! I will keep that in mind. :) – 000 Dec 3 '12 at 0:51", "\\sbrackets {f(k)} - \\sum _{k=1}^{n} \\sbrackets {f(k)} \\end{align} • Example 3.2.1 Find a closed-form (read: nice) formula for the sum $$\\sum _{1 \\leqq r \\leqq n} \\left [ (r^2 + r - 1)(r-1)! \\right ]$$ The first step (assuming that we wish to solve this using a telescoping series) is to split the expression summand into two, which will telescope. The source I originally found this question from did first give the necessary identity which very much helps, but we can also work this out ourselves. Let us first be very clear about what our goal is; we guess that the sum will telescope, and assuming that it does we must find some way of exploiting this. Specifically, we want to split the summand into two parts (which telescope). It would be lovely if we could break $$r^2 + r -1$$ into two parts which telescope. todo: why isn’t this possible The other thing we can do is this \\begin{align} r^2 (r-1)! - r(r-1)! - (r-1)! \\\\ & = (r-1)!(r^2 - r) - (r-1)! \\\\ & = (r-1)!(r)(r+1) - (r-1)! \\\\ & = (r+1)! - (r-1)! \\end{align} Therefore, \\begin{align} \\sum _{1 \\leqq r \\leqq n} \\left [(r^2 + r - 1)(r-1)!\\right ] & = \\sum _{1 \\leqq r \\leqq n} (r+1)! - \\sum _{1 \\leqq r \\leqq n}", "# Telescope Summation Algebra Level 4 $\\sum_{n=1}^{20} \\dfrac{8n}{4n^4+1}= \\dfrac85 + \\dfrac{16}{65} + \\dfrac{24}{325} + \\cdots + \\dfrac{160}{640001}$ Let $$x$$ denote the value of the summation above. Find $$841x$$. ×", "# Science:Math Exam Resources/Courses/MATH101/April 2018/Question 08 (ii)/Solution 1 We start by splitting the series ${\\displaystyle \\sum _{n=2}^{\\infty }\\left({\\frac {(-3)^{n}}{2^{2n}}}+{\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)=\\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{2^{2n}}}+\\sum _{n=2}^{\\infty }\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right).}$ Now the second series will telescope. Observe that for any positive integer ${\\textstyle N}$, we have ${\\displaystyle \\sum _{n=2}^{N}\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)={\\bigg (}{\\frac {1}{\\sqrt {4}}}-{\\frac {1}{\\sqrt {3}}}{\\bigg )}+{\\bigg (}{\\frac {1}{\\sqrt {5}}}-{\\frac {1}{\\sqrt {4}}}{\\bigg )}+\\cdots +{\\bigg (}{\\frac {1}{\\sqrt {N+2}}}-{\\frac {1}{\\sqrt {N+1}}}{\\bigg )}=-{\\frac {1}{\\sqrt {3}}}+{\\frac {1}{\\sqrt {N+2}}},}$ so we have ${\\displaystyle \\lim _{N\\to \\infty }\\sum _{n=2}^{N}\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)=\\lim _{N\\to \\infty }{\\bigg [}-{\\frac {1}{\\sqrt {3}}}+{\\frac {1}{\\sqrt {N+2}}}{\\bigg ]}=-{\\frac {1}{\\sqrt {3}}}.}$ It remains to calculate the series ${\\displaystyle \\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{2^{2n}}}=\\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{(2^{2})^{n}}}=\\sum _{n=2}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}=-1+{\\frac {3}{4}}+\\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}=-{\\frac {1}{4}}+\\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}.}$ Now the series is a geometric series, so using the formula for the geometric series we get that ${\\displaystyle \\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}={\\frac {1}{1+{\\frac {3}{4}}}}={\\frac {4}{7}}.}$ It follows that ${\\displaystyle \\sum _{n=2}^{\\infty }\\left({\\frac {(-3)^{n}}{2^{2n}}}+{\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)={\\frac {4}{7}}-{\\frac {1}{4}}-{\\frac {1}{\\sqrt {3}}}={\\frac {9}{28}}-{\\frac {1}{\\sqrt {3}}}.}$ Answer: The correct answer is ${\\textstyle {\\color {blue}{\\frac {9}{28}}-{\\frac {1}{\\sqrt {3}}}}}$.", "covering one slit. To subtract the contributions from the 1000 slits and the single slit, you have to take into account the relative intensity, as it differs by a factor of 1000. 6. May 29, 2014 unscientific I subtracted the angular displacement: $$u = u_0 \\left[e^{i(kz-\\omega t)} e^{i(\\frac{N}{2}\\delta - \\frac{\\delta}{2})} \\frac{sin \\left(\\frac{N\\delta}{2}\\right)}{sin \\left( \\frac{\\delta}{2}\\right)}\\space - \\space e^{i(\\frac{N\\delta}{2})}\\right]$$ There should be a prefactor of $\\frac{1}{r}$, but we asume that the distances are large so $r$ doesn't vary with angle. 7. May 29, 2014 Staff: Mentor There should be a prefactor of 1000 as the first comes from 1000 slits and the second just from one. 8. May 29, 2014 unscientific So It should be: $$u = u_0 \\left[e^{i(kz-\\omega t)} e^{i(\\frac{N}{2}\\delta - \\frac{\\delta}{2})} \\frac{sin \\left(\\frac{N\\delta}{2}\\right)}{sin \\left( \\frac{\\delta}{2}\\right)}\\space - \\frac{1}{1000}\\space e^{i(\\frac{N\\delta}{2})}\\right]$$ $$= I = I_0 \\left[ \\frac{sin^2 \\left( \\frac{N\\delta}{2} \\right)}{sin^2\\left(\\frac{\\delta}{2}\\right)} \\space - 2\\frac{1}{10^6} cos\\left(\\frac{\\delta (2N+1)}{2})\\right) \\frac{sin \\left( \\frac{N\\delta}{2} \\right)}{sin \\left(\\frac{\\delta}{2}\\right)} + \\frac{1}{10^6} \\right]$$ Last edited: May 29, 2014 9. May 29, 2014 Staff: Mentor How did you expand the small term, where does the sine come from and why does it have an N? The single slit does not \"know\" about the total number of slits. I would expect a slight shift in the position of the minima - when the sine is zero, the cosine is not zero and the", "# Difference between revisions of \"2002 AIME II Problems/Problem 6\" ## Problem Find the integer that is closest to $1000\\sum_{n=3}^{10000}\\frac1{n^2-4}$. ## Solution You know that $\\frac{4}{n^2 - 4} = \\frac{1}{n-2} - \\frac{1}{n + 2}$. So if you pull the $\\frac{1}{4}$ out of the summation, you get $250\\sum_{n=3}^{10,000} (\\frac{1}{n-2} - \\frac{1}{n + 2})$. Now that telescopes, leaving you with: $250 (1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} - \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000}) = 250 + 125 + 83.3 + 62.5 - 250 (- \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ $250(-\\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ is not enough to bring $520.8$ lower than $520.5$ so the answer is $\\fbox{521}$ ## See also 2002 AIME II (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions Invalid username Login to AoPS", "are really the same. I think we composed them at about the same time too. Oct 21, 2011 at 19:39 • That's beautiful,I never realized that latex could be so colorful.Aha,wish I could give you another +1 for \"high-definition\"! Oct 21, 2011 at 20:01 • $\\color{Blue}{+}\\color{Red}{1}$. :) Oct 22, 2011 at 0:29 By multiplying to get a common denominator we have that $$1+\\frac{1}{i^{2}}+\\frac{1}{(i+1)^{2}}=\\frac{i^{2}(i+1)^{2}+(i+1)^{2}+i^{2}}{i^{2}(i+1)^{2}}$$ $$=\\frac{i^{4}+2i^{3}+3i^{2}+2i+1}{i^{2}(i+1)^{2}}=\\frac{(i^{2}+i+1)^{2}}{i^{2}(i+1)^{2}}.$$ Hence our sum is $$\\sum_{i=1}^{69}\\frac{i^{2}+i+1}{i(i+1)}=\\sum_{i=1}^{69}\\left(\\frac{i^{2}+2i+1}{i(i+1)}-\\frac{i}{i(i+1)}\\right)$$ $$=\\sum_{i=1}^{69}\\frac{i+1}{i}-\\frac{1}{i+1}=69+\\sum_{i=1}^{69}\\left(\\frac{1}{i}-\\frac{1}{i+1}\\right).$$ The right hand side telescopes, so we conclude that the answer is $$69+1-\\frac{1}{70}=70-\\frac{1}{70}=\\frac{4899}{70}.$$ This is a late response (since I joined three weeks ago) but here is another method. Write \\begin{align} 1 + \\frac{1}{i^2} + \\frac{1}{(i + 1)^2} &= 1 + \\frac{2}{i(i+1)} + \\left(\\frac{1}{i^2} - \\frac{2}{i(i+1)} + \\frac{1}{(i+1)^2}\\right)\\\\ &= 1 + 2\\left(\\frac{1}{i} - \\frac{1}{i+1}\\right) + \\left(\\frac{1}{i} - \\frac{1}{i+1}\\right)^2\\\\ &= \\left(1 + \\frac{1}{i} - \\frac{1}{i+1}\\right)^2. \\end{align} Then $$\\sum_{i = 1}^{69} \\sqrt{1 + \\frac{1}{i} + \\frac{1}{i^2}} = \\sum_{i = 1}^{69} \\left(1 + \\frac{1}{i} - \\frac{1}{i + 1}\\right) = 69 + \\sum_{i = 1}^{69} \\left(\\frac{1}{i} - \\frac{1}{i+1}\\right) = 70 - \\frac{1}{70} = \\frac{4899}{70}.$$", "\\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{2n-1} + \\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2n} - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots +\\frac{1}{2n}\\phantom{=} \\end{align*} - @GreenBeans, clearly this is the answer you are looking for. – leo Mar 1 '12 at 3:36 @leo You spoiled it. GreenBeans, this are the droids you are looking for. – Pedro Tamaroff Mar 1 '12 at 4:14 This sequence has popped up many times in SE. You can write it as $$s_n = \\sum_{k=1}^{n} \\frac{1}{n+k}$$ Note that the partial sums of the series for $\\log 2$ are \\eqalign{ & {p_1} = 1 \\cr & {p_2} = 1 - \\frac{1}{2} = \\frac{1}{2} \\cr & {p_3} = \\frac{1}{2} + \\frac{1}{3} \\cr & {p_4} = \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} = \\frac{1}{3} + \\frac{1}{4} \\cr & {p_5} = \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} \\cr & {p_6} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} \\cr & {p_7} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} \\cr & {p_8} = \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\cr} I guess that with Henning's answer and this, you can probably relate the $s_n$ with", "# series..diverging, converging • March 3rd 2009, 08:46 AM champrock series..diverging, converging Q2004. Can you help me with this question.. http://img401.imageshack.us/img401/8...09111306pm.png • March 3rd 2009, 10:15 AM TheEmptySet Quote: Originally Posted by champrock Q2004. Can you help me with this question.. http://img401.imageshack.us/img401/8...09111306pm.png Note that $\\frac{1}{n(n+1)}=\\frac{1}{n}-\\frac{1}{n+1}$ so this telescopes to 1. So it is none of these. • March 3rd 2009, 10:25 AM mateoc15 This seems like a good candidate for \"plug and chug\". Just try a few values and see what happens. You'll figure it out. • March 3rd 2009, 05:36 PM champrock Quote: Originally Posted by TheEmptySet Note that $\\frac{1}{n(n+1)}=\\frac{1}{n}-\\frac{1}{n+1}$ so this telescopes to 1. So it is none of these. i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity. So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be \"some number between 0 and 1\" ? i am confused on this part. • March 4th 2009, 03:58 PM TheEmptySet Quote: Originally Posted by champrock i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity. So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be \"some number between 0 and 1\" ? i am confused on this part. $\\sum_{n=1}^{\\infty}\\frac{1}{n(n+1)}=\\sum_{n=1}^{\\i nfty}\\frac{1}{n}-\\frac{1}{n+1}=\\sum_{n=1}^{\\infty}\\frac{1}{n}-\\sum_{n=1}^{\\infty}\\frac{1}{n+1}=$ Now lets", "answers are really the same. I think we composed them at about the same time too. – Eric Naslund Oct 21 '11 at 19:39 That's beautiful,I never realized that latex could be so colorful.Aha,wish I could give you another +1 for \"high-definition\"! – VelvetThunder Oct 21 '11 at 20:01 $\\color{Blue}{+}\\color{Red}{1}$. :) – Srivatsan Oct 22 '11 at 0:29 By multiplying to get a common denominator we have that $$1+\\frac{1}{i^{2}}+\\frac{1}{(i+1)^{2}}=\\frac{i^{2}(i+1)^{2}+(i+1)^{2}+i^{2}}{i^{2}(i+1)^{2}}$$ $$=\\frac{i^{4}+2i^{3}+3i^{2}+2i+1}{i^{2}(i+1)^{2}}=\\frac{(i^{2}+i+1)^{2}}{i^{2}(i+1)^{2}}.$$ Hence our sum is $$\\sum_{i=1}^{69}\\frac{i^{2}+i+1}{i(i+1)}=\\sum_{i=1}^{69}\\left(\\frac{i^{2}+2i+1}{i(i+1)}-\\frac{i}{i(i+1)}\\right)$$ $$=\\sum_{i=1}^{69}\\frac{i+1}{i}-\\frac{1}{i+1}=69+\\sum_{i=1}^{69}\\left(\\frac{1}{i}-\\frac{1}{i+1}\\right).$$ The right hand side telescopes, so we conclude that the answer is $$69+1-\\frac{1}{70}=70-\\frac{1}{70}=\\frac{4899}{70}.$$ - This is a late response (since I joined three weeks ago) but here is another method. Write \\begin{align} 1 + \\frac{1}{i^2} + \\frac{1}{(i + 1)^2} &= 1 + \\frac{2}{i(i+1)} + \\left(\\frac{1}{i^2} - \\frac{2}{i(i+1)} + \\frac{1}{(i+1)^2}\\right)\\\\ &= 1 + 2\\left(\\frac{1}{i} - \\frac{1}{i+1}\\right) + \\left(\\frac{1}{i} - \\frac{1}{i+1}\\right)^2\\\\ &= \\left(1 + \\frac{1}{i} - \\frac{1}{i+1}\\right)^2. \\end{align} Then $$\\sum_{i = 1}^{69} \\sqrt{1 + \\frac{1}{i} + \\frac{1}{i^2}} = \\sum_{i = 1}^{69} \\left(1 + \\frac{1}{i} - \\frac{1}{i + 1}\\right) = 69 + \\sum_{i = 1}^{69} \\left(\\frac{1}{i} - \\frac{1}{i+1}\\right) = 70 - \\frac{1}{70} = \\frac{4899}{70}.$$ -", "# Math Help - Series... 1. ## Series... Find the sum of: 1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2)) 2. Originally Posted by bearej50 Find the sum of: 1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2)) Telescopic series: $\\frac{1}{n(n+1)}=\\frac{1}{n}-\\frac{1}{n+1}\\Longrightarrow \\frac{1}{n(n+1)(n+2)}=\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)\\frac{1}{n+2}=$ $\\frac{1}{n(n+2)}$ $-\\frac{1}{(n+1)(n+2)}$ $=\\frac{1}{2}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right)-\\left(\\frac{1}{n+1}-\\frac{1}{n+2}\\right)$ . I think you can take it from here Tonio 3. note that in tonio's solution the first term telescopes not so fast than the second term which does faster. another way on doing this is to see that $\\frac{1}{n(n+1)(n+2)}=\\frac{1}{2}\\cdot \\frac{n+2-n}{n(n+1)(n+2)}=\\frac{1}{2}\\left( \\frac{1}{n(n+1)}-\\frac{1}{(n+1)(n+2)} \\right),$ and this one telescopes easily. 4. Originally Posted by Krizalid note that in tonio's solution the first term telescopes not so fast than the second term which does faster. another way on doing this is to see that $\\frac{1}{n(n+1)(n+2)}=\\frac{1}{2}\\cdot \\frac{n+2-n}{n(n+1)(n+2)}=\\frac{1}{2}\\left( \\frac{1}{n(n+1)}-\\frac{1}{(n+1)(n+2)} \\right),$ and this one telescopes easily. Much better than my solution. Tonio", "since the sum telescopes. • Got it, thanks. – rain Feb 11 '18 at 0:56 $S_n(x)=1+x+x^2+x^3+ . . .x^n=1+x+x^2+x^3+ . . .x^n +x^{n+1}-x^{n+1}=1-x^{n+1} + x(1 +x+x^2+x^3 . . .+x^n)=1-x^{n+1} +x S_(n)$ ⇒ $(1-x)S_n(x)=1-x^{n+1}$ ⇒ $S_n(x)=\\frac{1-x^{n+1}}{1-x}$" ]

[ "write $$\\frac1{(3n−2)(3n+1)} = \\frac 13\\frac{(3n+1) - (3n-2)}{(3n−2)(3n+1)} =$$ and most terms telescope.", "$$\\mathrm{R}\\left(\\frac{1}{\\mathrm{n}_1^2}-\\frac{1}{\\mathrm{n}_2^2}\\right)$$ For second member of Lyman senes, $$\\frac{1}{5400}$$ = $$R\\left(\\frac{1}{1^2}-\\frac{1}{3^2}\\right)$$ ⇒ $$\\frac{1}{5400}$$ = $$\\frac{8 \\mathrm{R}}{9}$$ —-> (1) For first member of Lyman series, $$\\frac{1}{\\lambda^1}$$ = $$\\mathrm{R}\\left(\\frac{1}{1^2}-\\frac{1}{2^2}\\right)$$ $$\\frac{1}{\\lambda^1}$$ = $$\\frac{3 R}{4}$$ —–> (2) $$\\frac{(1)}{(2)}$$ ⇒ $$\\frac{\\lambda^1}{5400}$$ = $$\\frac{8 R}{9} \\times \\frac{4}{3 R}$$ ∴ λ’ = $$\\frac{32}{27}$$ × 5400 = 6400A. Question 11. Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer senes limit. Given : R = 10970000m-1. Solution: $$\\frac{1}{\\lambda}$$ = $$\\mathrm{R}\\left(\\frac{1}{\\mathrm{n}_1^2}-\\frac{1}{\\mathrm{n}_2^2}\\right)$$ R = 10970000 = 1.097 × 107 ms-1 For Balmer senes limit n1 = 2 and n2 = ∞ $$\\frac{1}{\\lambda}$$ = $$\\mathrm{R}\\left(\\frac{1}{2^2}-\\frac{1}{\\infty}\\right)$$ ⇒ $$\\frac{1}{\\lambda}$$ = $$\\frac{R}{4}$$ λ = $$\\frac{4}{\\mathrm{R}}$$ = $$\\frac{4}{1.097 \\times 10^7}$$ = 3646.3A Question 12. Using the Rydberg formula, calcûlate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum. Solution: Question 1. Choose the correct alternative from the clues given at the end of the each statement: a) The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model. (much greater than / no different from / much less than.) b) In the ground state of ……. electrons are in stable equilibrium, while in …… electrons always experience a net force. (Thomson’s model / Rutherford’s model.) c) A classical atom based on …… is doomed to collapse. (Thomson’s", "# Difference between revisions of \"2002 AIME II Problems/Problem 6\" ## Problem Find the integer that is closest to $1000\\sum_{n=3}^{10000}\\frac1{n^2-4}$. ## Solution You know that $\\frac{4}{n^2 - 4} = \\frac{1}{n-2} - \\frac{1}{n + 2}$. So if you pull the $\\frac{1}{4}$ out of the summation, you get $250\\sum_{n=3}^{10,000} (\\frac{1}{n-2} - \\frac{1}{n + 2})$. Now that telescopes, leaving you with: $250 (1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} - \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000}) = 250 + 125 + 83.3 + 62.5 - 250 (- \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ $250(-\\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ is not enough to bring $520.8$ lower than $520.5$ so the answer is $\\fbox{521}$", "100 terms of $\\sum \\frac{1}{n^2 + 3n + 2}.$ Solution: First we can factor the denominator, which will allow us to decompose this rational function into two: $$\\frac{1}{n^2 + 3n + 2} = \\frac{1}{(n + 1)(n + 2)}$$ Now let's decompose: \\begin{align} \\frac{1}{(n + 1)(n + 2)} &= \\frac{A}{n + 1} + \\frac{B}{n + 2}\\\\[5pt] 1 &= A(n + 2) + B(n + 1)\\\\[5pt] \\text{Let }\\; n &= -1, \\; \\text{ then } \\; A = 1 \\\\[5pt] \\text{Let }\\; n &= -2, \\; \\text{ then } \\; B = -1, \\; \\text{ so} \\\\[5pt] \\frac{1}{(n + 1)(n + 2)} &= \\frac{1}{n + 1} - \\frac{1}{n + 2} \\end{align} Now we'll telescope the series: \\begin{align} S &= \\left( \\frac{1}{2} - \\frac{1}{3} \\right) + \\left( \\frac{1}{3} - \\frac{1}{4} \\right) + \\left( \\frac{1}{4} - \\frac{1}{5} \\right) + \\dots + \\left( \\frac{1}{101} - \\frac{1}{102} \\right)\\\\[5pt] &= \\frac{1}{2} + \\left( \\frac{1}{3} - \\frac{1}{3} \\right) + \\left( \\frac{1}{4} - \\frac{1}{4} \\right) + \\left( \\frac{1}{5} - \\frac{1}{5} \\right) + \\dots - \\frac{1}{102}\\\\[5pt] &= \\frac{1}{2} - \\frac{1}{102} = \\frac{51 - 1}{102} = \\frac{25}{51} \\approx \\frac{1}{2} \\end{align} It's easy to see that the sum of this series as n → ∞ is ½. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images", "\\sum_{i=1}^3 \\sum_{j=1}^5 y_{ij} \\right)^2}{N_{total}} = \\frac{(\\mbox{Total of all observations})^2}{N_{total}} = \\frac{(108.1)^2}{15} = 779.041", "that $$1+\\frac{1}{i^{2}}+\\frac{1}{(i+1)^{2}}=\\frac{i^{2}(i+1)^{2}+(i+1)^{2}+i^{2}}{i^{2}(i+1)^{2}}$$ $$=\\frac{i^{4}+2i^{3}+3i^{2}+2i+1}{i^{2}(i+1)^{2}}=\\frac{(i^{2}+i+1)^{2}}{i^{2}(i+1)^{2}}.$$ Hence our sum is $$\\sum_{i=1}^{69}\\frac{i^{2}+i+1}{i(i+1)}=\\sum_{i=1}^{69}\\left(\\frac{i^{2}+2i+1}{i(i+1)}-\\frac{i}{i(i+1)}\\right)$$ $$=\\sum_{i=1}^{69}\\frac{i+1}{i}-\\frac{1}{i+1}=69+\\sum_{i=1}^{69}\\left(\\frac{1}{i}-\\frac{1}{i+1}\\right).$$ The right hand side telescopes, so we conclude that the answer is $$69+1-\\frac{1}{70}=70-\\frac{1}{70}=\\frac{4899}{70}.$$", "\\frac{1}{(i+1)(i+2) \\ldots (i + k)}\\text{?}$ The first sum is familiar enough to me – writing $\\displaystyle \\frac{1}{(i+1)(i+2)}$ as $\\displaystyle \\frac{1}{i+1} - \\frac{1}{i+2}$ the sum telescopes: \\begin{aligned} \\sum_{i=0}^{\\infty} \\frac{1}{(i+1)(i+2)} &= \\lim_{N \\rightarrow \\infty} \\sum_{i=0}^N \\frac{1}{(i+1)(i+2)}\\\\ &= \\lim_{N \\rightarrow \\infty} \\sum_{i=0}^N \\left( \\frac{1}{i+1} - \\frac{1}{i+2}\\right )\\\\ &= \\lim_{N \\rightarrow \\infty} \\sum_{i=0}^N \\frac{1}{i+1} - \\sum_{i=1}^{N+1} \\frac{1}{i+1}\\\\ &= \\lim_{N \\rightarrow \\infty} \\frac{1}{2} - \\frac{1}{N+2} \\\\ &= \\frac{1}{2}.\\end{aligned} To do the next sum I thought one could use the partial fraction expansion $\\displaystyle \\frac{1}{(i+1)(i+2)(i+3)} = \\frac{1/2}{i+1} - \\frac{1}{i+2} + \\frac{1/2}{i+3}$ and continue as in the previous case. While this is doable, it is not easily generalisable. It is simpler to write $\\displaystyle \\frac{1}{(i+1)(i+2)(i+3)} = \\frac{1}{2}\\left( \\frac{1}{(i+1)(i+2)} - \\frac{1}{(i+2)(i+3)} \\right).$ In the general case, we write \\begin{aligned} \\frac{1}{(i+1)(i+2) \\ldots (i + k)} &= \\frac{1}{(i+2)\\ldots (i+k-1)} \\left( \\frac{1}{(i+1)(i+k)} \\right)\\\\ &= \\frac{1}{(i+2)\\ldots (i+k-1)} \\cdot \\frac{1}{(k-1)}\\left( \\frac{1}{(i+1)} - \\frac{1}{(i+k)} \\right)\\\\ &= \\frac{1}{k-1} \\left( \\frac{1}{(i+1)(i+2)\\ldots (i+k-1)} - \\frac{1}{(i+2)(i+3)\\ldots (i+k)} \\right), \\end{aligned} \\begin{aligned} \\sum_{i=0}^{\\infty} \\frac{1}{(i+1)(i+2) \\ldots (i + k)} &= \\lim_{N \\rightarrow \\infty} \\sum_{i=0}^N \\frac{1}{(i+1)(i+2) \\ldots (i + k)}\\\\ &= \\lim_{N \\rightarrow \\infty} \\frac{1}{k-1} \\sum_{i=0}^N \\left( \\frac{1}{(i+1)(i+2)\\ldots (i+k-1)} - \\frac{1}{(i+2)(i+3)\\ldots (i+k)} \\right) \\\\ &= \\lim_{N \\rightarrow \\infty} \\frac{1}{k-1} \\left( \\sum_{i=0}^N \\frac{1}{(i+1)(i+2) \\ldots (i+k-1)} - \\sum_{i=1}^{N+1} \\frac{1}{(i+1)(i+2) \\ldots (i + k - 1)} \\right) \\\\ &= \\lim_{N \\rightarrow \\infty} \\frac{1}{k-1} \\left( \\frac{1}{1 \\times 2 \\times \\ldots \\times (k-1) } -", "# Difference between revisions of \"2002 AIME II Problems/Problem 6\" ## Problem Find the integer that is closest to $1000\\sum_{n=3}^{10000}\\frac1{n^2-4}$. ## Solution You know that $\\frac{4}{n^2 - 4} = \\frac{1}{n-2} - \\frac{1}{n + 2}$. So if you pull the $\\frac{1}{4}$ out of the summation, you get $250\\sum_{n=3}^{10,000} (\\frac{1}{n-2} - \\frac{1}{n + 2})$. Now that telescopes, leaving you with: $250 (1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} - \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000}) = 250 + 125 + 83.3 + 62.5 - 250 (- \\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ $250(-\\frac{1}{9997} - \\frac{1}{9998} - \\frac{1}{9999} - \\frac{1}{10000})$ is not enough to bring $520.8$ lower than $520.5$ so the answer is $\\fbox{521}$ ## See also 2002 AIME II (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions Invalid username Login to AoPS", "and arrive at $$\\frac{1}{k^2-1}=\\frac{-\\frac{1}{2}}{k+1}+\\frac{\\frac{1}{2}}{k-1}=\\frac{1}{2}\\left(\\frac{1}{k-1}-\\frac{1}{k+1}\\right).$$ Taking the partial sum $$\\sum_{4 \\le k \\le m}\\frac{1}{k^2-1}=\\frac{1}{2}\\sum_{4 \\le k \\le m}\\frac{1}{k-1}-\\frac{1}{k+1}.$$ The sum telescopes, so we see that $$\\sum_{4 \\le k \\le m}\\frac{1}{k+1}-\\frac{1}{k-1}=\\color{red}{\\frac{1}{3}}-\\color{green}{\\frac{1}{5}}+\\color{red}{\\frac{1}{4}}-\\frac{1}{6}+\\color{green}{\\frac{1}{5}}-\\frac{1}{7}+\\dots+\\color{green}{\\frac{1}{m-1}}-\\frac{1}{m-3}+\\color{red}{\\frac{1}{m}}-\\frac{1}{m-2}+\\color{red}{\\frac{1}{m+1}}-\\color{green}{\\frac{1}{m-1}}=\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{m}+\\frac{1}{m+1}.$$ Taking the limit, we have $$\\lim_{m \\to \\infty}\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{m}+\\frac{1}{m+1}=\\frac{1}{3}+\\frac{1}{4}+\\lim_{m \\to \\infty}\\frac{1}{m}+\\frac{1}{m+1}=\\frac{1}{3}+\\frac{1}{4}=\\frac{7}{12}.$$ Recalling the factor of $\\frac{1}{2}$, we arrive at $\\sum_{4 \\le k \\le \\infty}\\frac{1}{k^2-1}=\\frac{1}{2}\\cdot\\frac{7}{12}=\\frac{7}{24}$. - thanks a lot.It is crystal clear now. – learner Dec 2 '12 at 7:42 @learner, I'm glad! I love colors! – 000 Dec 2 '12 at 8:36 I love the use of colors! I would recommend something other than red and green though, considering red-green is the most common type of colorblindness. – Antonio Vargas Dec 2 '12 at 23:58 @AntonioVargas, LOL. What luck I have! I will keep that in mind. :) – 000 Dec 3 '12 at 0:51", "# Science:Math Exam Resources/Courses/MATH101/April 2018/Question 08 (ii)/Solution 1 We start by splitting the series ${\\displaystyle \\sum _{n=2}^{\\infty }\\left({\\frac {(-3)^{n}}{2^{2n}}}+{\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)=\\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{2^{2n}}}+\\sum _{n=2}^{\\infty }\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right).}$ Now the second series will telescope. Observe that for any positive integer ${\\textstyle N}$, we have ${\\displaystyle \\sum _{n=2}^{N}\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)={\\bigg (}{\\frac {1}{\\sqrt {4}}}-{\\frac {1}{\\sqrt {3}}}{\\bigg )}+{\\bigg (}{\\frac {1}{\\sqrt {5}}}-{\\frac {1}{\\sqrt {4}}}{\\bigg )}+\\cdots +{\\bigg (}{\\frac {1}{\\sqrt {N+2}}}-{\\frac {1}{\\sqrt {N+1}}}{\\bigg )}=-{\\frac {1}{\\sqrt {3}}}+{\\frac {1}{\\sqrt {N+2}}},}$ so we have ${\\displaystyle \\lim _{N\\to \\infty }\\sum _{n=2}^{N}\\left({\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)=\\lim _{N\\to \\infty }{\\bigg [}-{\\frac {1}{\\sqrt {3}}}+{\\frac {1}{\\sqrt {N+2}}}{\\bigg ]}=-{\\frac {1}{\\sqrt {3}}}.}$ It remains to calculate the series ${\\displaystyle \\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{2^{2n}}}=\\sum _{n=2}^{\\infty }{\\frac {(-3)^{n}}{(2^{2})^{n}}}=\\sum _{n=2}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}=-1+{\\frac {3}{4}}+\\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}=-{\\frac {1}{4}}+\\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}.}$ Now the series is a geometric series, so using the formula for the geometric series we get that ${\\displaystyle \\sum _{n=0}^{\\infty }{\\bigg (}{\\frac {-3}{4}}{\\bigg )}^{n}={\\frac {1}{1+{\\frac {3}{4}}}}={\\frac {4}{7}}.}$ It follows that ${\\displaystyle \\sum _{n=2}^{\\infty }\\left({\\frac {(-3)^{n}}{2^{2n}}}+{\\frac {1}{\\sqrt {n+2}}}-{\\frac {1}{\\sqrt {n+1}}}\\right)={\\frac {4}{7}}-{\\frac {1}{4}}-{\\frac {1}{\\sqrt {3}}}={\\frac {9}{28}}-{\\frac {1}{\\sqrt {3}}}.}$ Answer: The correct answer is ${\\textstyle {\\color {blue}{\\frac {9}{28}}-{\\frac {1}{\\sqrt {3}}}}}$.", "\\tanh(c_2x) \\ \\ \\text{ with c_1 = \\frac 1 2 and c_2 = \\log(3) } \\end{align} The last formula is the hyperbolic tangent of a multiple of $x$, so it's symmetric with respect to the origin. From this, it follows that the $f(x) + f(-x) = 0$, which, after the transformation, means $g(x) + g(1-x) = 1$. From this the claim follows. \\begin{align} &g(t)=\\frac {3^t}{3^t+3^{\\frac 12}}=\\frac {3^{t-\\frac12}}{3^{t-\\frac 12}+1}\\\\ \\Rightarrow \\qquad &g\\left(\\frac12+u\\right)+g\\left(\\frac12-u\\right)=\\frac {3^u}{3^u+1}+\\frac {3^{-u}}{3^{-u}+1}=\\frac {3^u}{3^u+1}+\\frac {1}{1+3^u}=1\\\\\\\\ \\sum_{r=1}^{2014}g\\left(\\frac r{2015}\\right) &=\\sum_{r=1}^{1007}g\\left(\\frac r{2015}\\right)+\\sum_{r=1008}^{2014}g\\left(\\frac r{2015}\\right)\\\\ &=\\sum_{i=1}^{1007}g\\left(\\frac{1007\\frac12-(i-\\frac12)}{2015}\\right)+g\\left(\\frac{1007\\frac12+(i-\\frac12)}{2015}\\right)\\\\ &=\\sum_{i=1}^{1007}g\\left(\\frac12-\\frac{i-\\frac12}{2015}\\right)+g\\left(\\frac12+\\frac{i-\\frac12}{2015}\\right)\\\\ &=\\sum_{i=1}^{1007}1\\\\ &=1007\\qquad\\blacksquare\\end{align} Notice that $g (x) = g (1 - x)$. Indeed $$g(x)+g(1-x) = \\frac {1} {1 + 3^{1/2-x}} + \\frac {1} {1+3^{x-1/2}} = 1$$. Then we conclude by $$\\sum_{k = 1}^{2014} g\\left(\\frac{k}{2015}\\right) = \\sum_{k = 1}^{1007} \\left ( g\\left(\\frac{k}{2015} \\right ) + g\\left(\\frac{2015 - k}{2015} \\right ) \\right ) = 1007.$$", "the same, we just write ${\\displaystyle s\\left(\\int _{1}^{2}{\\frac {1}{x^{s+1}}}dx+\\int _{2}^{3}{\\frac {2}{x^{s+1}}}dx+\\dots \\right)}$(1) Now we exploit the Fundamental Theorem of Calculus, computing ${\\displaystyle \\int _{n}^{n+1}{\\frac {n}{x^{s+1}}}dx=n\\left[-{\\frac {x^{-s}}{s}}\\right]_{n}^{n+1}=n\\left(-{\\frac {(n+1)^{-s}}{s}}+{\\frac {n^{-s}}{s}}\\right).}$ So, the summation in Equation 1 can, more explicitly be written as ${\\displaystyle s\\sum _{n=1}^{\\infty }n\\left(-{\\frac {(n+1)^{-s}}{s}}+{\\frac {n^{-s}}{s}}\\right)=\\sum _{n=1}^{\\infty }\\left({\\frac {n}{n^{s}}}-{\\frac {n}{(n+1)^{s}}}\\right)}$ However, grouping common denominators, we observe that the sum partially telescopes to yield more simply ${\\displaystyle \\sum _{n=1}^{\\infty }{\\frac {1}{n^{s}}}=\\zeta (s).}$ ### b Having now proved Part a it suffices to show that ${\\displaystyle s\\int _{1}^{\\infty }{\\frac {[x]}{x^{s+1}}}dx={\\frac {s}{s-1}}-s\\int _{1}^{\\infty }{\\frac {x-[x]}{x^{s+1}}}dx.}$ By the Fundamental Theorem of Calculus we have ${\\displaystyle \\int _{1}^{\\infty }{\\frac {1}{x^{s}}}dx={\\frac {1}{s-1}}}$ So \\begin{eqnarray} \\int_1^\\infty \\frac{x}{x^{s+1}} dx&=&\\frac{1}{s-1}\\\\ \\Rightarrow s \\int_1^\\infty \\frac{x}{x^{s+1}} dx&=&\\frac{s}{s-1}\\\\ \\Rightarrow s \\int_1^\\infty \\left( \\frac{x-[x]}{x^{s+1}} + \\frac{[x]}{x^{s+1}} \\right) dx&=&\\frac{s}{s-1}\\\\ \\Rightarrow s \\int_1^\\infty \\left( \\frac{x-[x]}{x^{s+1}} + \\frac{[x]}{x^{s+1}} \\right) dx&=&\\frac{s}{s-1}\\\\ \\Rightarrow s \\int_1^\\infty \\frac{[x]}{x^{s+1}} dx &=&\\frac{s}{s-1} - s \\int_1^\\infty \\frac{x-[x]}{x^{s+1}} dx\\\\ \\end{eqnarray*} as desired\\ end part b It remains now to show that the integral in Part \\ref{2} converges. Since for ${\\displaystyle x\\in (1,\\infty )[itex]wehave[itex]0\\leq {\\frac {x-[x]}{x^{s+1}}}\\leq {\\frac {1}{x^{s+1}}}}$ we know that ${\\displaystyle \\int _{1}^{\\infty }{\\frac {x-[x]}{x^{s+1}}}dx}$ converges if and only if ${\\displaystyle \\int _{1}^{\\infty }{\\frac {1}{x^{s+1}}}dx}$ converges. However, ${\\displaystyle \\int _{1}^{\\infty }{\\frac {1}{x^{s+1}}}dx}$ converges by the integral test (Problem 8) since we have already shown that the sequence ${\\displaystyle \\sum _{x=1}^{\\infty }{\\frac {1}{x^{s+1}}}}$", "are close to the end. Express what you got as $$\\left(\\frac{1/2}{n}-\\frac{1/2}{n+1}\\right)-\\left(\\frac{1/2}{n+1}-\\frac{1/2}{n+2}\\right).$$ It looks a little better as $$\\frac{1/2}{n(n+1)}-\\frac{1/2}{(n+1)(n+2)}.$$ Now add up, and (in either version) watch almost all the terms cancel. Hint: \\begin{align} \\sum_{n=1}^\\infty\\frac1{n(n+1)(n+2)} &=\\frac12\\sum_{n=1}^\\infty\\left[\\frac1{n(n+1)}-\\frac1{(n+1)(n+2)}\\right]\\\\ &=\\frac12\\sum_{n=1}^\\infty\\frac1{n(n+1)}-\\frac12\\sum_{n=2}^\\infty\\frac1{n(n+1)}\\\\ \\end{align} More Generally \\begin{align} &\\frac1{n(n+1)\\dots(n+k-1)}-\\frac1{(n+1)(n+2)\\dots(n+k)}\\\\ &=\\frac1{(n+1)(n+2)\\dots(n+k-1)}\\left(\\frac1n-\\frac1{n+k}\\right)\\\\ &=\\frac{k}{n(n+1)\\dots(n+k)} \\end{align} Therefore, $$\\frac1{n(n+1)\\dots(n+k)}=\\frac1k\\left[\\frac1{n(n+1)\\dots(n+k-1)}-\\frac1{(n+1)(n+2)\\dots(n+k)}\\right]$$", "of thinking about it can also be made clear by drawing a suitably shaped figure. If you split the area into vertical strips of unit width, you see that the total area is the sum on Brian's rightmost column. OTOH if you split it into horizontal strips of heights $2^{-n}$ and width $n$ you see that the total area is the original sum. – Jyrki Lahtonen Jul 11 '13 at 19:14 Hint: \\begin{align} \\frac12+\\frac14+\\frac18+\\frac1{16}+\\dots&=\\color{red}{1}\\\\ \\frac14+\\frac18+\\frac1{16}+\\dots&=\\color{red}{\\frac12}\\\\ \\frac18+\\frac1{16}+\\dots&=\\color{red}{\\frac14}\\\\ \\frac1{16}+\\dots&=\\color{red}{\\frac18}\\\\ \\hline \\frac12+\\frac24+\\frac38+\\frac4{16}+\\dots&=2 \\end{align} • Thanks for your support. :+) – Mikasa Jul 13 '13 at 19:04 This is the derivative of a geometric series: Let $$f(x)=\\sum_{n=0}^\\infty x^n.$$ Then (by taking derivative summandwise) $$f'(x) = \\sum_{n=1}^\\infty nx^{n-1}.$$ Since $f(x)=\\frac1{1-x}$ if $|x|<1$, we have $f'(x)=\\frac1{(1-x)^2}$. Your sum is just $\\frac12f'(\\frac12)$. Consider the power series $$f(x)=\\sum_{n=0}^\\infty\\frac{x^n}{2^n}=\\sum_{n=0}\\left(\\frac x2\\right)^n=\\sum_{n=0}^\\infty t^n=\\frac{1}{1-t}=\\frac{1}{1-\\frac x2}=\\frac{2}{2-x}$$ Then we have that $f(1)=2$. We also have that $$f'(x)=\\sum_{n=0}^\\infty\\frac{nx^{n-1}}{2^n}$$ But we also have that $$\\left(\\frac2{2-x}\\right)'=\\frac2{(2-x)^2}$$ Therefore $f'(1)=2$ as wanted. • I'll be glad to hear what's wrong in this answer. – Asaf Karagila Jul 11 '13 at 20:08 The method that says $$\\sum_n nx^n = x \\sum_n \\frac{d}{dx} x^n,\\text{ etc, etc.,}$$ has already been mentioned. Here's another way: $$\\begin{array}{rrrrrrrrrrrrrrrr} 1/2 \\\\[8pt] {}+ 1/4 & {}+ 1/4 \\\\[8pt] {}+ 1/8 & {}+1/8 & {}+1/8 \\\\[8pt] {}+\\cdots{} \\end{array}$$ Now find the sum of the entries in", "I'd like to see if there's another way of showing the relationship without knowing that it's a Riemann sum for the above reason. – Jam Nov 21 '19 at 19:55 • Approximate $\\frac{1}{k^2}$ with something that telescopes, like $\\frac{1}{k^2 - \\frac{1}{4}}$ or $\\frac{1}{k(k\\pm 1)}$ and show the difference doesn't matter. – Daniel Fischer Nov 21 '19 at 19:57 • @DanielFischer Thank you Daniel. – Jam Nov 21 '19 at 19:59 A more elementary approach. $$\\frac{1}{2n}=\\sum_{i=1}^{n}\\frac{1}{(n+i)(n+i-1)}$$ because the sum telescopes to $$\\frac{1}{n}-\\frac{1}{2n}.$$ So: \\begin{align}\\frac{1}{2n}-\\sum_{i=1}^{n}\\frac{1}{\\left(n+i\\right)^{2}}&=\\sum_{i=1}^{n}\\left(\\frac{1}{(n+i)(n+i-1)}-\\frac{1}{(n+i)^2}\\right)\\\\ &=\\sum_{i=1}^{n}\\frac{1}{(n+i)^2(n+i-1)}\\tag{1}\\\\&<\\sum_{i=1}^{n}\\frac{1}{n^3}\\\\&=\\frac{1}{n^2} \\end{align} Also, the value at (1) is positive. So we have: $$0<\\frac{1}{2}-n\\sum_{i=1}^{n}\\frac{1}{(n+i)^2}<\\frac{1}{n}$$ and hence$$n\\sum_{i=1}^{n}\\frac{1}{(n+i)^2}\\to\\frac{1}{2}$$ We have that $$\\sum_{i=1}^{n}\\frac{1}{\\left(n+i\\right)^{2}}=\\sum_{i=n+1}^{2n}\\frac{1}{i^{2}}=\\sum_{i=1}^{2n}\\frac{1}{i^{2}}-\\sum_{i=1}^{n}\\frac{1}{i^{2}}$$ then we can use the result indicated here • Perfect. This is what I was looking for. Thank you, user. – Jam Nov 21 '19 at 20:01 • The Euler–Maclaurin one? That's kind of weird if you want to avoid integrals. – Daniel Fischer Nov 21 '19 at 20:01 • @Jam Nice to may help you! – user Nov 21 '19 at 20:03 • @DanielFischer You're right; I hadn't thought of that. – Jam Nov 21 '19 at 20:05 Yet another way: since $$\\frac{1}{x^2}$$ is convex on $$\\mathbb{R}^+$$, the Hermite-Hadamard inequality ensures $$\\frac{1}{2n+1}-\\frac{1}{(2n+1)(4n-1)}=\\int_{n+1/2}^{2n-1/2}\\frac{dx}{x^2}\\geq\\sum_{k=1}^{n}\\frac{1}{(n+k)^2}\\geq \\int_{n+1}^{2n}\\frac{dx}{x^2}=\\frac{1}{2n}-\\frac{1}{n(n+1)}.$$ • Sure, but that assumes the integral of $1/x^2$ and, as the question points out, if you know that, you can just", "\\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{2n-1} + \\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2n} - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots +\\frac{1}{2n}\\phantom{=} \\end{align*} - @GreenBeans, clearly this is the answer you are looking for. – leo Mar 1 '12 at 3:36 @leo You spoiled it. GreenBeans, this are the droids you are looking for. – Pedro Tamaroff Mar 1 '12 at 4:14 This sequence has popped up many times in SE. You can write it as $$s_n = \\sum_{k=1}^{n} \\frac{1}{n+k}$$ Note that the partial sums of the series for $\\log 2$ are \\eqalign{ & {p_1} = 1 \\cr & {p_2} = 1 - \\frac{1}{2} = \\frac{1}{2} \\cr & {p_3} = \\frac{1}{2} + \\frac{1}{3} \\cr & {p_4} = \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} = \\frac{1}{3} + \\frac{1}{4} \\cr & {p_5} = \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} \\cr & {p_6} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} \\cr & {p_7} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} \\cr & {p_8} = \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\cr} I guess that with Henning's answer and this, you can probably relate the $s_n$ with", "I remember we did this in class but I couldn't find it in my notebook. I just found it in my other notebook. It was driving me crazy. Thank you so much for your assistance. I appreciate it. Melissa No problem. In case you are still interested, here is how to do it with infinite series: $n=0.0\\overline{1485}=0.01485+\\frac{0.01485}{10000} +\\frac{0.01485}{10000^2}+\\cdots=\\sum_{n=0}^\\infty \\frac{0.01485}{10000^n}$ $=\\frac{1485}{10^5}\\sum_{n=0}^\\infty\\left(\\frac{1}{ 10^{4n}}\\right)=\\frac{1485}{10^5}\\sum_{n=0}^\\infty \\left(\\frac{1}{10^4}\\right)^n$ $=\\frac{1485}{10^5}\\left(\\frac1{1-\\frac1{10^4}}\\right)=\\frac1{\\left(\\frac{9999}{10^4 }\\right)}$ $=\\frac{1485}{10^5}\\left(\\frac{10^4}{9999}\\right)=\\ frac{1485}{99990}=\\frac{3^3\\cdot5\\cdot11}{2\\cdot3^ 2\\cdot5\\cdot11\\cdot101}=\\frac3{202}$", "understand where you're trying to put your parentheses (since you have$\\frac1{p_n}$on the LHS I presume it's trying to be 'captured' by the sum), then the sum should be easily relatable to the Mertens constant :$\\sum(\\log\\log p_{n+1}-\\log\\log p_n)$telescopes, so if you sum from$n=1$to$n=K$your partial sum will be just$\\log\\log ... Top 50 recent answers are included", "1-\\frac{ 1 }{ 10 } } \\\\ &= \\frac{\\left(\\frac{7}{10}\\right)}{\\left(\\frac{9}{10}\\right)} \\\\ &= \\left(\\frac{7}{10}\\right)\\left(\\frac{10}{9}\\right)\\\\ &= \\frac { 7 }{ 9 } \\end{align} } The formula works for any repeating term. Some more examples are: \\displaystyle{ \\begin{align} 0.123412341234 \\cdots &= \\frac { a }{ 1-r } \\\\ &= \\frac { \\frac{1234}{ 10000 } }{ 1-\\frac{ 1 }{ 10000 } } \\\\ &= \\frac{\\left(\\frac{1234}{ 10000 }\\right)}{\\left(\\frac{9999}{10000}\\right)} \\\\ &= \\left(\\frac{1234}{ 10000 }\\right)\\left(\\frac{10000}{9999}\\right)\\\\ &= \\frac { 1234 }{ 9999 } \\end{align} } \\displaystyle{ \\begin{align} 0.0909090909 \\cdots &= \\frac { a }{ 1-r } \\\\ &= \\frac { \\frac{9}{ 100 } }{ 1-\\frac{ 1 }{ 100 } } \\\\ &= \\frac{\\left(\\frac{9}{ 100 }\\right)}{\\left(\\frac{99}{100}\\right)} \\\\ &= \\left(\\frac{9}{100}\\right)\\left(\\frac{100}{99}\\right)\\\\ &= \\frac { 9 }{ 99 } \\\\ &= \\frac{1}{11} \\end{align} } \\displaystyle{ \\begin{align} 0.143814381438 \\cdots &= \\frac { a }{ 1-r } \\\\ &= \\frac { \\frac{1438}{ 10000 } }{ 1-\\frac{ 1 }{ 10000 } } \\\\ &= \\frac{\\left(\\frac{1438}{ 10000 }\\right)}{\\left(\\frac{9999}{10000}\\right)} \\\\ &= \\left(\\frac{1438}{ 10000 }\\right)\\left(\\frac{10000}{9999}\\right)\\\\ &= \\frac { 1438 }{ 9999 } \\end{align} } \\displaystyle{ \\begin{align} 0.9999 \\cdots &= \\frac { a }{ 1-r } \\\\ &= \\frac { \\frac{ 9 }{ 10 } }{ 1-\\frac{ 1 }{ 10 } } \\\\ &= \\frac{\\left(\\frac{9}{10}\\right)}{\\left(\\frac{9}{10}\\right)} \\\\ &= \\left(\\frac{9}{10}\\right)\\left(\\frac{10}{9}\\right)\\\\ &= \\frac { 9 }{ 9 } \\\\ &= 1 \\end{align} } That is, a repeating decimal with a repeating", ") \\right ]^{2}$ = $\\frac{1}{n}\\sum_{i = 1}^{n} x_{i}^{2} – \\frac{1}{n}\\sum_{i = 1}^{n}2\\left ( \\frac{n + 1}{2} \\right )x_{i} + \\frac{1}{n}\\sum_{i = 1}^{n}\\left ( \\frac{n + 1}{2} \\right )^{2}$ = $\\frac{1}{n}\\frac{n\\left ( n + 1 \\right )\\left ( 2n + 1 \\right )}{6} – \\left ( \\frac{n + 1}{n} \\right )\\left [ \\frac{n\\left ( n + 1 \\right )}{2} \\right ] + \\frac{\\left ( n + 1 \\right )^{2}}{4n} \\times n$ = $\\frac{\\left ( n + 1 \\right )\\left ( 2n + 1 \\right )}{6} – \\frac{\\left (n + 1 \\right )^{2}}{2} + \\frac{\\left ( n + 1 \\right )^{2}}{4}$ = $\\frac{\\left ( n + 1 \\right )\\left ( 2n + 1 \\right )}{6} – \\frac{\\left ( n + 1 \\right )^{2}}{4}$ = $\\left ( n + 1 \\right )\\left [\\frac{ 4n + 2 – 3n – 3 }{12}\\right ]$ = $\\frac{\\left ( n + 1 \\right )\\left ( n – 1 \\right )}{12}$ = $\\frac{n^{2} – 1}{12}$ Q3. Calculate the mean and variance for the first 10 multiples of 3. Sol: The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Number of observations, n = 10 Mean, $\\bar{x} = \\frac{\\sum_{i = 1}^{10}}{10}$ = $\\frac{165}{10}$ = 16.5 The given table is obtained, $x_{i}$ $(x_{i} – \\bar{x})$ $(x_{i} – \\bar{x})^{2}$ 3 -13.5" ]

[ "+0 # Pls help +1 32 1 Find the number of real solutions to \\[(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \\dots + x^2 + 1) = 2006x^{2005}.\\] Feb 13, 2021", "# Solution for boxes Find the number of real solutions of the equation $\\frac{1}{[x]}+\\frac{1}{[2x]}= x-[x]+\\frac{1}{3}.$ Note: 1.$$[x]$$ stands for the greatest integer function. 2. This problem is not original. ×", "# Math Help - find number of solutions of x 1. ## find number of solutions of x find number of solutions of x ..where x belongs to real number for which $64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$ 2. Try factoring each number as a multiple of 8. You might get some cancellation there.", "# find number of solutions of x #### banku12 find number of solutions of x ..where x belongs to real number for which $$\\displaystyle 64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$$ #### Ackbeet MHF Hall of Honor Try factoring each number as a multiple of 8. You might get some cancellation there.", "get one real solution. $$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$ $$x = \\frac{-4 \\pm \\sqrt{0}}{2 \\times 2}$$ $$x = \\frac{-4}{4}$$ $$x = -1$$", "# find number of solutions of x • July 8th 2010, 09:38 AM banku12 find number of solutions of x find number of solutions of x ..where x belongs to real number for which $64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$ • July 8th 2010, 12:13 PM Ackbeet Try factoring each number as a multiple of 8. You might get some cancellation there.", ") - 2C) Find total number of real s... Question # ) - 2C) Find total number of real solutions to the equation 52 + 122 = 132 15(S+ JEE/Engineering Exams Maths Solution 97 4.0 (1 ratings) here, given ( 5^{2^{2}+12^{2}=13^{2^{2}}} ) [ begin{array}{l}text { We know, } 5^{2}+12^{2}=13^{21} therefore frac{z^{2}=2}{17=pm sqrt{21}} text { Ary } text { total no of real son = two(2) }end{array} ]", "+0 # Math Question +1 81 2 +119 Let $$r$$ be the positive real solution to $$x^3 + \\frac{2}{5} x - 1 = 0.$$ Find the exact numerical value of $$r^2 + 2r^5 + 3r^8 + 4r^{11} + \\dotsb.$$ Thank You! Jan 5, 2020 edited by NewMember Jan 5, 2020 #1 0 From the formula for a geometric series, the sum is 19/3. Jan 5, 2020 #2 +24054 +2 Let $$r$$ be the positive real solution to $$x^3 + \\dfrac{2}{5} x - 1 = 0$$. Find the exact numerical value of $$r^2 + 2r^5 + 3r^8 + 4r^{11} + \\dotsb$$ . Jan 6, 2020", "### Home > A2C > Chapter 9 > Lesson 9.2.3 > Problem9-110 9-110. $x=\\frac{-b\\pm\\sqrt{b^{2}-4(5)(20)}}{2(5)}$ The equation will have real solutions as long as the value under the radical is greater than or equal to zero.", "29 views What is the number of solutions of: $x=\\frac{x^{2}}{50}-\\cos \\frac{x}{2}+2$ in $[0,10]$ ? 1. $0$ 2. $1$ 3. $2$ 4. $\\infty$", "The number of distinct solutions of the equation Question: The number of distinct solutions of the equation $\\log _{\\frac{1}{2}}|\\sin x|=2-\\log _{\\frac{1}{2}}|\\cos x|$ in the interval $[0,2 \\pi]$, is___________. Solution:", "# How Many Solutions? For real numbers $a, b,$ and $c$ such that $a \\ne \\frac{c-b}{2},$ how many real values of $x$ satisfy the equation $(x-a)(x-c)=(x+b)(a-x)?$ ×", "1. Find all real solutions of the equation ${\\large (x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1}.$ 2. Find all real solutions of the equation ${\\large (2-x^2)^{(x^2 - 3\\sqrt{2}x + 4)} = 1}.$", "# 200 Followers Problem! $\\frac{r^2}{s}+200=\\frac{s^2}{r}+200^2$ In the equation above, $r$ and $s$ are both rational numbers and their difference is an integer. What is the number of solutions to the equation? ×", "another point is that, there are infinite numbers of solutions to the equation: $$\\frac{-16}{25} = \\sin (2 \\theta)$$, not just one as Swapnil mentioned.", "# Solutions to Problems 1161-1170 Q1161. Find all values of $x$ (real number) satisfying $$\\frac{x-1}{2004} + \\frac{x-3}{2002} + \\frac{x-5}{2000} + \\cdots + \\frac{x-2003}{2} = \\cdots$$", "× Weird Equation Find all real solutions of : $$\\sqrt{x-\\frac{1}{x}}+\\sqrt{1-\\frac{1}{x}}=x$$. This Problem is not original. Note by Lawrence Bush 2 years, 9 months ago", "= 2\\sqrt{AB} \\cosh\\left(\\gamma x + \\frac{1}{2}\\log\\left(\\frac{A}{B}\\right) \\right).$$ Notice that, in order to have well defined real solution, then you must pay care to the sign of $A$ and $B$. Indeed, they must be both positive or both negative. Also, $A=0$ and/or $B = 0$ are forbidden.", "# The number of distinct solutions Question: The number of distinct solutions of the equation, $\\log _{1 / 2}|\\sin x|=2-\\log _{1 / 2}|\\cos x|$ in the interval $[0,2 \\pi]$, is_________. Solution: $\\log _{1 / 2}|\\sin x|=2-\\log _{1 / 2}|\\cos x|$ $\\Rightarrow \\log _{1 / 2}|\\sin x \\cos x|=2$ $\\Rightarrow|\\sin x \\cos x|=\\frac{1}{4}$ $\\Rightarrow \\quad \\sin 2 x=\\pm \\frac{1}{2}$", "# Thread: Find the Real Solutions 1. ## Find the Real Solutions I need to find the real solutions to a simple equation: $\\sqrt{x+5}=x$ $x+5=x^2$ $x^2-x-5=0$ $x=\\frac{1\\pm \\sqrt{21}}{2}$ It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution. I need to find the real solutions to a simple equation: $\\sqrt{x+5}=x$ $x+5=x^2$ $x^2-x-5=0$ $x=\\frac{1\\pm \\sqrt{21}}{2}$ It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution. From the first equation you know that $\\sqrt{x+5}$ is a positive real number. Therefore x must be a positive real number. But your last equation has also a negative number which can't be a solution of the original equation. 3. ## Only one solution There is only one real solution: 2.79129. In your first step, you squared both sides. You forgot that it is \"plus or minus\" though, when you do that. Make sense?" ]

[ "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = $1008-x$ , then you", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2,N); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+N)/2,W); (Error compiling LaTeX. 228aa9a09ec6a18b4a0ab11366b7b8f2ec44b852.asy: 27.6: no matching function 'label(string, pair[], pair)') Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow h^2 = (504 + x)(504 - x)\\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$.", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "\\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(N+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow h^2 = (504 + x)(504 - x)\\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$.", "get $$t=\\frac{160}{3}.$$ Finally, the sum of the numerator and denominator is $160+3=\\boxed{163}.$ ### Solution 2 Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$. Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively. $[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,50)); label(\"A\",A,SW); label(\"B\",B,NNW); label(\"C\",C,S); label(\"D\",D,NE); label(\"P\",P,S); label(\"Q\",Q,NE); label(\"O\",O,W); label(\"X\",X,ESE); [/asy]$ From the problem statement, $AB=3t$, and $CD=t$. Since $\\Delta ABX \\sim \\Delta CDX$, $CX=AC\\cdot\\left(\\frac{CD}{AB-CD}\\right)=200\\cdot\\left(\\frac{t}{3t-t}\\right)=100$. Since $PC=100$, $PX=200$. So, $\\tan(\\angle OXP)=\\frac{OP}{PX}=\\frac{50}{200}=\\frac{1}{4}$. Since circle $O$ is tangent to $BX$ and $AX$, $OX$ is the angle bisector of $\\angle BXA$. Thus, $\\tan(\\angle BXA)=\\tan(2\\angle OXP)=\\frac{2\\tan(\\angle OXP)}{1- \\tan^2(\\angle OXP)} = \\frac{2\\cdot \\left(\\frac{1}{4}\\right)}{1-\\left(\\frac{1}{4}\\right)^2}=\\frac{8}{15}$. Therefore, $t = CD = CX\\cdot\\tan(\\angle BXA) = 100 \\cdot \\frac{8}{15} = \\frac{160}{3}$, and the answer is $\\boxed{163}$. ### Solution 3 $[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O--Q--R--cycle); draw(Circle(O,50)); label(\"A\",A,SW); label(\"B\",B,NNW); label(\"C\",(200,-205),S); label(\"D\",D,NE); label(\"P\",(100,-205),S); label(\"Q\",Q,NE); label(\"O\",O,SW); label(\"R\",R,NE); label(\"S\",S,W); [/asy]$ Let $t$ be the time they walk.", "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "2, 3, 1, 1, 3, 1, 7, 2] println(collect(ContinuedFraction(1414214, 1000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] println(collect(ContinuedFraction(14142136, 10000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2] println(collect(ContinuedFraction(13 // 11))) # => [1, 5, 2] println(collect(ContinuedFraction(√2), 20)) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2] ## Kotlin // version 1.1.2 // compile with -Xcoroutines=enable flag from command line import kotlin.coroutines.experimental.buildSequence fun r2cf(frac: Pair<Int, Int>) = buildSequence { var num = frac.first var den = frac.second while (Math.abs(den) != 0) { val div = num / den val rem = num % den num = den den = rem yield(div) } } fun iterate(seq: Sequence<Int>) { for (i in seq) print(\"$i \") println() } fun main(args: Array<String>) { val fracs = arrayOf(1 to 2, 3 to 1, 23 to 8, 13 to 11, 22 to 7, -151 to 77) for (frac in fracs) { print(\"${\"%4d\".format(frac.first)} /${\"%-2d\".format(frac.second)} = \") iterate(r2cf(frac)) } val root2 = arrayOf(14142 to 10000, 141421 to 100000, 1414214 to 1000000, 14142136 to 10000000) println(\"\\nSqrt(2) ->\") for (frac in root2) { print(\"${\"%8d\".format(frac.first)} /${\"%-8d\".format(frac.second)} = \") iterate(r2cf(frac)) } val pi = arrayOf(31 to 10, 314", "yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$: $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); label(\"P\",P,dir(235)); [/asy]$ This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "by a ratio of $\\frac{BC}{AD} = \\frac{125}{251}$. Since the homothety carries the midpoint of $\\overline{BC}$, $M$, to the midpoint of $\\overline{AD}$, which is $N$, then $E,M,N$ are collinear. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M[0]+H)/2,(1,0)); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, because $\\triangle RPM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED", "\\frac{3}{2} + 39 \\cdot 4 -36}{\\sqrt{(4\\sqrt{3})^2+4^2+39^2}} = \\frac{144}{\\sqrt{1585}}$$ $$\\lfloor m+\\sqrt{n}\\rfloor = \\lfloor 144+\\sqrt{1585}\\rfloor = 183$$ Solution by SimonSun ## Solution 3 (Cosine Law & Pythagorean Bash) Diagram borrowed from Solution 1 $[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"T\",T,N);label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,NE);label(\"S\",S,NW);label(\"O\",O,SW);label(\"M\",M,NE); label(\"4\",(S+T)/2,NW);label(\"1\",(S+A)/2,NW);label(\"5\",(B+T)/2,NE);label(\"4\",(O+T)/2,W); dot(S);dot(O); [/asy]$ Apply Pythagorean Theorem on $\\bigtriangleup TOB$ yields $$BO=\\sqrt{TB^2-TO^2}=3$$ Since $\\bigtriangleup ABC$ is equilateral, we have $\\angle MOB=60^{\\circ}$ and $$BC=2BM=2(OB\\sin MOB)=3\\sqrt{3}$$ Apply Pythagorean Theorem on $\\bigtriangleup TMB$ yields $$TM=\\sqrt{TB^2-BM^2}=\\sqrt{5^2-(\\frac{3\\sqrt{3}}{2})^2}=\\frac{\\sqrt{73}}{2}$$ Apply Law of Cosines on $\\bigtriangleup TBC$ we have $$BC^2=TB^2+TC^2-2(TB)(TC)\\cos BTC$$ $$(3\\sqrt{3})^2=5^2+5^2-2(5)^2\\cos BTC$$ $$\\cos BTC=\\frac{23}{50}$$ Apply Law of Cosines on $\\bigtriangleup STB$ using the fact that $\\angle STB=\\angle BTC$ we have $$SB^2=ST^2+BT^2-2(ST)(BT)\\cos STB$$ $$SB=\\sqrt{4^2+5^2-2(4)(5)\\cos BTC}=\\frac{\\sqrt{565}}{5}$$ Apply Pythagorean Theorem on $\\bigtriangleup BSM$ yields $$SM=\\sqrt{SB^2-BM^2}=\\frac{\\sqrt{1585}}{10}$$ Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\\frac{\\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have $$TP^2=TS^2-SP^2=TM^2-PM^2$$ $$4^2-x^2=(\\frac{\\sqrt{73}}{2})^2-(\\frac{\\sqrt{1585}}{10}-x)^2$$ Cancelling out the $x^2$ term and solving gets $x=\\frac{181}{2\\sqrt{1585}}$. Finally, by Pythagorean Theorem, $$TP=\\sqrt{TS^2-SP^2}=\\frac{144}{\\sqrt{1585}}$$ so $\\lfloor m+\\sqrt{n}\\rfloor=\\boxed{183}$ ~ Nafer", "draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"A\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"B\",(168/5, 24/5),NE); dot((24,0)); label(\"C\",(24,0),NW); dot((40, 0)); label(\"D\",(40, 0),NE); dot((24, 12)); label(\"E\",(24, 12),NE); dot((24, -12)); label(\"F\",(24, -12),SE); dot((54/5, 72/5)); label(\"G\",(54/5, 72/5),NW); dot((0, 0)); label(\"O_1\",(0, 0),S); dot((30, 0)); label(\"O_2\",(30, 0),S); [/asy]$ $r_1 = O_1A = 24$$r_2 = O_2B = 6$$AG = BO_2 = r_2 = 6$$O_1G = r_1 - r_2 = 24 - 6 = 18$$O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$$\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$$\\frac{O_2D}{30} = \\frac{6}{18}$$O_2D = 10$ $CD = O_2D + r_1 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{\\textbf{192}}$ ~isabelchen ## Solution 2 Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim", "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$" ]

[ "three real solutions for $x$, so we are done and the answer is $\\boxed{\\textbf{(E) }a>\\frac12}$. ~ ccx09 ## Solution 8 Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$, which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$. This further simplifies to $x^2(1 + x^2 - 2a) = 0$. Thus, either $x^2 = 0$, or $x^2 - 2a + 1 = 0$. Since we care about $a$, we consider the second case. Then, we solve in terms of $a$ giving $a = \\frac{x^2}{2} + \\frac{1}{2}$. We see that in order to find the range in which $a$ lies, we must find the vertex of the that equation, which turns out to be $(0, \\frac{1}{2})$, so we know that the minimum is $\\frac{1}{2}$, which further implies that $\\boxed{\\textbf{(E) }a > \\frac{1}{2}}$ ## Solution 9 Now, let's graph these two equations. We want the blue parabola to be inside this red circle. $[asy] import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*\"a\",(0,5),NE); dot(scale(.7)*\"-a\",(0,-5),N); dot(scale(.7)*\"a\",(5,0),NE); dot(scale(.7)*\"-a\",(-5,0),SE); [/asy]$ Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\\pm\\sqrt{2a-1}$. Therefore, $2a-1>0$, which means", "draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); label(\"$y = mx$\", (8,12*sqrt(221)/49*8), N); dot(\"$(9,6)$\", (9,6), NE); dot(\"$(kr,r)$\", O[1], N); dot(P,red); [/asy] Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\\frac{|r - krm|}{\\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \\neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \\neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \\frac{2k}{k^2 - 1} = \\frac{2 \\sqrt{\\frac{117}{68}}}{\\frac{117}{68} - 1} = \\boxed{\\frac{12 \\sqrt{221}}{49}}.$$", "2, 3, 1, 1, 3, 1, 7, 2] println(collect(ContinuedFraction(1414214, 1000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] println(collect(ContinuedFraction(14142136, 10000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2] println(collect(ContinuedFraction(13 // 11))) # => [1, 5, 2] println(collect(ContinuedFraction(√2), 20)) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2] ## Kotlin // version 1.1.2 // compile with -Xcoroutines=enable flag from command line import kotlin.coroutines.experimental.buildSequence fun r2cf(frac: Pair<Int, Int>) = buildSequence { var num = frac.first var den = frac.second while (Math.abs(den) != 0) { val div = num / den val rem = num % den num = den den = rem yield(div) } } fun iterate(seq: Sequence<Int>) { for (i in seq) print(\"$i \") println() } fun main(args: Array<String>) { val fracs = arrayOf(1 to 2, 3 to 1, 23 to 8, 13 to 11, 22 to 7, -151 to 77) for (frac in fracs) { print(\"${\"%4d\".format(frac.first)} /${\"%-2d\".format(frac.second)} = \") iterate(r2cf(frac)) } val root2 = arrayOf(14142 to 10000, 141421 to 100000, 1414214 to 1000000, 14142136 to 10000000) println(\"\\nSqrt(2) ->\") for (frac in root2) { print(\"${\"%8d\".format(frac.first)} /${\"%-8d\".format(frac.second)} = \") iterate(r2cf(frac)) } val pi = arrayOf(31 to 10, 314", "a slope of $\\frac{4}{3}$. Since we know $m=\\frac{4}{3}$ , we can see that the line rises by $\\frac{8}{5}$ and moves to the right by $\\frac{6}{5}$ to meet the hypotenuse. (Let $2 = 5x$ and the rise be $4x$ and the run be $3x$ and then solve.) Therefore, line $\\l$ intersects the hypotenuse at the point $(x+\\frac{6}{5}, x+\\frac{8}{5})$. Plugging into the equation for the hypotenuse we have $x=\\frac{2}{7}$ , and after a bit of computation we get $\\boxed{\\textbf{(D) } \\frac{145}{147}}$ ## Solution 7(Slightly different from first solution): Same drawing as before: $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"2\", (.8,1), E); label(\"S\", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); draw((0.3,0.3)--(4,0), dashed); draw((0.3,0.3)--(0,3), dashed); label(\"\\small{a}\", (0.15,0.3), N); label(\"\\small{a}\", (0.3,0.15), E); [/asy]$ Let's assign $a$ as the side length of box S. We then get each of the smaller triangle areas. The sum of all the triangular areas(not including the box) is equal to $\\frac{(3-a) \\cdot a}{2} + \\frac{(4-a) \\cdot a}{2} + \\frac{5 \\cdot 2}{2} = \\frac{3 \\cdot 4}{2} - a^2$ You can solve for $a=\\frac{2}{7}$ Then, the ratio would be $1-\\dfrac{{\\frac{2}{7}}^2}{6}$ which is equal to $\\boxed{\\textbf{(D) } \\frac{145}{147}}$ ~Starshooter11 ## Solution 8(Similar but cleaner than first solution): Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone", "\\times \\frac{8}{\\sqrt{3}} + 8 \\times \\frac{10}{\\sqrt{3}} = \\frac{144}{\\sqrt{3}} = 48\\sqrt{3}$. Thus, $m+n = \\boxed{051}$. Solution 2 $[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label(\"A (0,0)\",(0,0),S);label(\"B (b,2)\",(10/sqrt(3),2),SE);label(\"C\",(18/sqrt(3),6),E);label(\"D\",(10/sqrt(3),10),N);label(\"E\",(0,8),NW);label(\"F\",(-8/sqrt(3),4),W); [/asy]$ From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6. $[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label(\"A (0,0)\",(0,0),SE);label(\"B (b,2)\",(10/sqrt(3),2),SE);label(\"C\",(18/sqrt(3),6),E);label(\"D\",(10/sqrt(3),10),N);label(\"E\",(0,8),NW);label(\"F\",(-8/sqrt(3),4),W); xaxis(\"x\");yaxis(\"y\"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label(\"\\theta\",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]$ Let the angle between the $x$-axis and segment $AB$ be $\\theta$, as shown above. Thus, as $\\angle FAB=120^\\circ$, the angle between the $x$-axis and segment $AF$ is $60-\\theta$, so $\\sin{(60-\\theta)}=2\\sin{\\theta}$. Expanding, we have $\\sin{60}\\cos{\\theta}-\\cos{60}\\sin{\\theta}=\\frac{\\sqrt{3}\\cos{\\theta}}{2}-\\frac{\\sin{\\theta}}{2}=2\\sin{\\theta}$ Isolating $\\sin{\\theta}$ we see that $\\frac{\\sqrt{3}\\cos{\\theta}}{2}=\\frac{5\\sin{\\theta}}{2}$, or $\\cos{\\theta}=\\frac{5}{\\sqrt{3}}\\sin{\\theta}$. Using the fact that $\\sin^2{\\theta}+\\cos^2{\\theta}=1$, we have $\\frac{28}{3}\\sin^2{\\theta}=1$, or $\\sin{\\theta}=\\sqrt{\\frac{3}{28}}$. Letting the side length of the hexagon be $y$, we have $\\frac{2}{y}=\\sqrt{\\frac{3}{28}}$. After simplification we find that that $y=\\frac{4\\sqrt{21}}{3}$. In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$, hence $b=10\\sqrt{3}/3$. Also, if $C=(c,6)$, then $y^2=BC^2=4^2+(c-b)^2$, hence $c-b=8\\sqrt{3}/3,$ and thus $c=18\\sqrt{3}/3$. Using similar methods (or symmetry), we determine that $D=(10\\sqrt{3}/3,10)$, $E=(0,8)$, and $F=(-8\\sqrt{3}/3,4)$. By the Shoelace theorem, $$[ABCDEF]=\\frac12\\left|\\begin{array}{cc} 0&0\\\\ 10\\sqrt{3}/3&2\\\\ 18\\sqrt{3}/3&6\\\\ 10\\sqrt{3}/3&10\\\\ 0&8\\\\ -8\\sqrt{3}/3&4\\\\ 0&0\\\\ \\end{array}\\right|=\\frac12|60+180+80-36-60-(-64)|\\sqrt{3}/3=48\\sqrt{3}.$$ Hence the answer is $\\boxed{51}$. 2003 AIME II (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 •", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "119 views If $x_{0} = 1, x_{1} = 2$ and $x_{n+2} = \\dfrac{1 + x_{n+1}}{x_{n}}, n = 0, 1, 2, 3, \\dots ,$ then $x_{2021}$ is equal to 1. $1$ 2. $3$ 3. $4$ 4. $2$ Given that, $x_{0}=1, x_{1}=2, x_{n+2} = \\frac{1+x_{n+1}}{x_{n}}; n=0,1,2,3,\\dots$ Now, • $x_{0}=1$ • $x_{1}=2$ • $x_{2}=\\frac{1+2}{1} = 3$ • $x_{3}=\\frac{1+3}{2} = 2$ • $x_{4}=\\frac{1+2}{3} = 1$ • $x_{5}=\\frac{1+1}{2} = 1$ • $x_{6}=\\frac{1+1}{1} = 2$ • $x_{7}=\\frac{1+2}{1} = 3$ • $x_{8}=\\frac{1+3}{2} = 2$ • $x_{9}=\\frac{1+2}{3} = 1$ • $\\vdots \\quad \\vdots \\quad \\vdots \\quad \\vdots$ We can see the pattern, $1,2,3,2,{\\color{Red} {1}},1,2,3,2,{\\color{Red} {1}},\\dots$ Every $5^{\\text{th}}$ multiple is $1.$ So, $x_{2020}=1.$ $\\therefore$ The value of $x_{2021} = 2.$ Correct Answer $:\\text{D}$ 10.3k points 1 vote 1 308 views", "1.1.5 As illustrated below, the scale need not always be one unit. In the first number line, each tick mark represents two units. In the second, each tick mark represents $$\\frac{1}{7}$$: Figure 1.1.6 The graph of each real number is shown as a dot at the appropriate point on the number line. A partial graph of the set of integers $$\\mathbb{Z}$$, follows: Figure 1.1.7 Example $$\\PageIndex{3}$$: Graph the following set of real numbers: $$\\{−\\frac{5}{2}, 0, \\frac{3}{2}, 2\\}$$. Solution: Graph the numbers on a number line with a scale where each tick mark represents $$\\frac{1}{2}$$ unit. Figure 1.1.8 The opposite38 of any real number a is −a. Opposite real numbers are the same distance from the origin on a number line, but their graphs lie on opposite sides of the origin and the numbers have opposite signs. Figure 1.1.9 Given the integer $$−7$$, the integer the same distance from the origin and with the opposite sign is $$+7$$, or just $$7$$. Figure 1.1.10 Therefore, we say that the opposite of $$−7$$ is $$−(−7) = 7$$. This idea leads to what is often referred to as the double-negative property39. For any real number a, $$−(−a)=a$$ Example $$\\PageIndex{4}$$: Calculate: $$−(−(−\\frac{3}{8})).$$ Solution: Here we apply the double-negative within the innermost parentheses first. $$−(\\color{Cerulean}{−(−\\frac{3}{8})})=−(\\frac{3}{8})$$ $$=−\\frac{3}{8}$$ $$−\\frac{3}{8}$$ In general, an odd number of sequential", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "set coefficients equal to each other: $a=\\frac{2a}{e^i}-\\frac{i}{2}$ $b=2be^{i}+\\frac{i}{2}$Thus,$a=\\frac{ie^i}{4-2e^i}$and $b=\\frac{i}{2-4e^i}$, so the following is one possible solution to $F_n$: $F_n=\\frac{ie^i}{4-2e^i}e^{in}+\\frac{i}{2-4e^i}e^{-in}$This provides with one particular solution. However, we then need to find the particular solution matching our initial condition. For this, we simply solve for the general solution to$F_n=2F_{n-1}$and tack that on as a new term to our solution: $F_n=\\frac{ie^i}{4-2e^i}e^{in}+\\frac{i}{2-4e^i}e^{-in}+c2^n$Now, knowing$F_0$, plug $n=0$ to solve for $c$ and you have your equation.Here is a Python implementation of the above: >>> from math import cos, sin >>> def exp(imaginary): return cos(imaginary)+1j*sin(imaginary) >>> def func(initial, n): ... c = initial-((1j*exp(1))/(4-2*exp(1))+1j/(2-4*exp(1))) ... return c*2**n+1j*exp(1)/(4-2*exp(1))*exp(n)+1j/(2-4*exp(1))*exp(-n) ... >>> func(10, 0) (10+0j) >>> func(10, 1) (20.8414709848079+0j) >>> func(10, 2) (42.59223939644147+0j) >>> 2*func(10, 1)+sin(2) (42.59223939644148+0j) `As you can see, I tried the formula assuming $F_0=10$ and was able to verify that $2F_1+\\sin(2)=F_2$, which shows that the equation works. » 3 months ago, # | +7 Well, $Cg(n)=2Cn-6C$ is clearly a linear function, so let's assume that $F(n)$ is also a linear function $mn+b$. Then, we get: $mn+b=2Cn-6C+m(n-1)+b+m(n-2)+b+m(n-3)+b$ $mn+b=(2C+3m)n-6C-6m+3b$ $0=(2C+2m)n-6C-6m+2b$Thus,$2C+2m=0$, so $m=-C$, and $-6C-6m+2b=-6C+6C+2b=2b=0$, so $b=0$. Thus, we know that $F(n)=-Cn$ is at least one solution to the linear recurrence. However, this might not meet the initial conditions required by $F(1),F(2),F(3)$.Thus, we need to solve the characteristic equation of the linear recurrence. After doing this, we find the", "\\frac{8}{\\sqrt{3}} + 8 \\times \\frac{10}{\\sqrt{3}} = \\frac{144}{\\sqrt{3}} = 48\\sqrt{3}$. Thus, $m+n = \\boxed{051}$. ## Solution 2 $[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label(\"A (0,0)\",(0,0),S);label(\"B (b,2)\",(10/sqrt(3),2),SE);label(\"C\",(18/sqrt(3),6),E);label(\"D\",(10/sqrt(3),10),N);label(\"E\",(0,8),NW);label(\"F\",(-8/sqrt(3),4),W); [/asy]$ From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6. $[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label(\"A (0,0)\",(0,0),SE);label(\"B (b,2)\",(10/sqrt(3),2),SE);label(\"C\",(18/sqrt(3),6),E);label(\"D\",(10/sqrt(3),10),N);label(\"E\",(0,8),NW);label(\"F\",(-8/sqrt(3),4),W); xaxis(\"x\");yaxis(\"y\"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label(\"\\theta\",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]$ Let the angle between the $x$-axis and segment $AB$ be $\\theta$, as shown above. Thus, as $\\angle FAB=120^\\circ$, the angle between the $x$-axis and segment $AF$ is $60-\\theta$, so $\\sin{(60-\\theta)}=2\\sin{\\theta}$. Expanding, we have $\\sin{60}\\cos{\\theta}-\\cos{60}\\sin{\\theta}=\\frac{\\sqrt{3}\\cos{\\theta}}{2}-\\frac{\\sin{\\theta}}{2}=2\\sin{\\theta}$ Isolating $\\sin{\\theta}$ we see that $\\frac{\\sqrt{3}\\cos{\\theta}}{2}=\\frac{5\\sin{\\theta}}{2}$, or $\\cos{\\theta}=\\frac{5}{\\sqrt{3}}\\sin{\\theta}$. Using the fact that $\\sin^2{\\theta}+\\cos^2{\\theta}=1$, we have $\\frac{28}{3}\\sin^2{\\theta}=1$, or $\\sin{\\theta}=\\sqrt{\\frac{3}{28}}$. Letting the side length of the hexagon be $y$, we have $\\frac{2}{y}=\\sqrt{\\frac{3}{28}}$. After simplification we find that that $y=\\frac{4\\sqrt{21}}{3}$. In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$, hence $b=10\\sqrt{3}/3$. Also, if $C=(c,6)$, then $y^2=BC^2=4^2+(c-b)^2$, hence $c-b=8\\sqrt{3}/3,$ and thus $c=18\\sqrt{3}/3$. Using similar methods (or symmetry), we determine that $D=(10\\sqrt{3}/3,10)$, $E=(0,8)$, and $F=(-8\\sqrt{3}/3,4)$. By the Shoelace theorem, $$[ABCDEF]=\\frac12\\left|\\begin{array}{cc} 0&0\\\\ 10\\sqrt{3}/3&2\\\\ 18\\sqrt{3}/3&6\\\\ 10\\sqrt{3}/3&10\\\\ 0&8\\\\ -8\\sqrt{3}/3&4\\\\ 0&0\\\\ \\end{array}\\right|=\\frac12|60+180+80-36-60-(-64)|\\sqrt{3}/3=48\\sqrt{3}.$$ Hence the answer is $\\boxed{51}$. ### Note By symmetry the area of $ABCDEF$ is twice the area of $ABCF$. Therefore, you only need to calculate the coordinates of", "= 2 and y = 5, then what is the value of \\frac{x^4+2y^2}{6} ? Level 1 Algebra We have $\\frac{x^4 + 2y^2}{6} = \\frac{2^4 + 2(5^2)}{6} = \\frac{16+2(25)}{6} = \\frac{16+50}{6} = \\frac{66}{6} = \\boxed{11}.$ The sequence of integers in the row of squares and in each of the two columns of squares form three distinct arithmetic sequences. What is the value of N? [asy] unitsize(0.35inch); draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((3,0)--(3,1)); draw((4,0)--(4,1)); draw((5,0)--(5,1)); draw((6,0)--(6,1)); draw((6,2)--(7,2)--(7,-4)--(6,-4)--cycle); draw((6,-1)--(7,-1)); draw((6,-2)--(7,-2)); draw((6,-3)--(7,-3)); draw((3,0)--(4,0)--(4,-3)--(3,-3)--cycle); draw((3,-1)--(4,-1)); draw((3,-2)--(4,-2)); label(\"21\",(0.5,0.8),S); label(\"14\",(3.5,-1.2),S); label(\"18\",(3.5,-2.2),S); label(\"N\",(6.5,1.8),S); label(\"-17\",(6.5,-3.2),S); [/asy] Level 2 Algebra Since 18 - 14 = 4, the common difference in the first column of squares is 4, so the number above 14 is 14 - 4 = 10, and the number above 10 is 10 - 4 = 6. This is also the fourth number in the row, so the common difference in the row is (6 - 21)/3 = -5. Then the seventh (and last) number in the row is 21 - 5 \\cdot 6 = -9. In the second column, the common difference is [(-17) - (-9)]/4 = -2, so N = -9 - (-2) = \\boxed{-7}. Tim wants to invest some money in a bank which compounds quarterly with an annual interest rate of 7\\%. To the nearest dollar, how much money should he invest if he", "# Math Help - Really confusing? 1. ## Really confusing? use (1/x) - (1/x+1) to find the following sum. (1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100) show/explain how to get the answer the answer is 99/100 if that helps 2. Originally Posted by yoman360 use (1/x) - (1/x+1) to find the following sum. (1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100) show/explain how to get the answer the answer is 99/100 if that helps When you expand the sum, something very nice will happen... $\\sum_{x=1}^{99}\\left(\\frac{1}{x}-\\frac{1}{x+1}\\right)=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\dots+\\left(\\frac{1}{99}-\\frac{1}{100}\\right)$ $=1-\\frac{1}{100}=\\boxed{\\frac{99}{100}}$ Does this make sense? 3. Originally Posted by Chris L T521 When you expand the sum, something very nice will happen... $\\sum_{x=1}^{99}{\\left(\\frac{1}{x}-\\frac{1}{x+1}\\right)}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\dots+\\left(\\frac{1}{99}-\\frac{1}{100}\\right)$ $=1-\\frac{1}{100}=\\boxed{\\frac{99}{100}}$ Does this make sense? Note that your original series is $\\frac{1}{1 \\times 2} + \\frac{1}{2 \\times 3} + \\frac{1}{3 \\times 4} + \\dots + \\frac{1}{99 \\times 100} = \\sum_{i = 1}^{99}{\\frac{1}{x(x + 1)}}$ Also note that $\\frac{1}{x} - \\frac{1}{x + 1} = \\frac{1}{x}\\times\\frac{x + 1}{x + 1} - \\frac{1}{x + 1} \\times \\frac{x}{x}$ $= \\frac{x + 1}{x(x + 1)} - \\frac{x}{x( x + 1)}$ $= \\frac{x + 1 - x}{x(x + 1)}$ $= \\frac{1}{x(x + 1)}$. So $\\sum_{i = 1}^{99}{\\frac{1}{x(x + 1)}} = \\sum_{i = 1}^{99}{\\left(\\frac{1}{x} - \\frac{1}{x + 1}\\right)}$. Then refer to Chris' post. 4. Originally Posted by Chris L T521 When you expand the sum, something", "$$50$$ seconds. (Animation will only show in the HTML version) Source code used for the above ________________________________________________________________________________________ ##### 3.2.2.7 [264] Specific example of the above problem number 264 Problem 8.3.5 from Richard Haberman applied partial differential equations book, 5th edition, but added specific values for the parameters in order to do animations Solve for $$u(r,t)$$ $u_t= k (u_rr+1/r u_r) + f(r,t)$ Inside the circle ($$r<a$$) with $$u=0$$ at $$r=a$$ and initially $$u=0$$. Let $$k=\\frac {1}{100},a=2$$. Let $$f(r,t)=r t e^{-t}$$ Mathematica $\\left \\{\\left \\{u(r,t)\\to \\underset {K[1]=1}{\\overset {\\infty }{\\sum }}\\frac {1600 e^{-\\frac {1}{400} t \\left (j_{0,K[1]}\\right ){}^2} J_0\\left (\\frac {1}{2} r j_{0,K[1]}\\right ) \\left (e^{\\frac {1}{400} t \\left (j_{0,K[1]}\\right ){}^2-t} \\left (t \\left (j_{0,K[1]}\\right ){}^2-400 (t+1)\\right )+400\\right ) \\, _1F_2\\left (\\frac {3}{2};1,\\frac {5}{2};-\\frac {1}{4} \\left (j_{0,K[1]}\\right ){}^2\\right )}{3 J_1\\left (j_{0,K[1]}\\right ){}^2 \\left (\\left (j_{0,K[1]}\\right ){}^2-400\\right ){}^2}\\right \\}\\right \\}$ Maple sol=() Hand solution The basic solution for this type of PDE was already given in problem 3.2.2.5 on page 1231 as\\begin {align*} u\\left ( r,t\\right ) & =\\sum _{n=1}^{\\infty }a_{n}\\left ( t\\right ) \\operatorname {BesselJ}\\left ( 0,\\frac {\\Lambda _{n}}{a}r\\right ) \\\\ & =\\sum _{n=1}^{\\infty }\\left ( \\int _{0}^{t}b_{n}\\left ( \\tau \\right ) e^{k\\left ( \\frac {\\Lambda _{n}}{a}\\right ) ^{2}\\tau }d\\tau \\right ) e^{-k\\left ( \\frac {\\Lambda _{n}}{a}\\right ) ^{2}t}\\operatorname {BesselJ}\\left ( 0,\\frac {\\Lambda _{n}}{a}r\\right ) \\end {align*} Where$b_{n}\\left ( t\\right )", "Simplify. $$4x = 6 \\tag{8.3.80}$$ Divide by 4. $$\\frac{4x}{\\textcolor{red}{4}} = \\frac{6}{\\textcolor{red}{4}} \\tag{8.3.81}$$ Simplify. $$x = \\frac{3}{2} \\tag{8.3.82}$$ Check: Let x = $$\\frac{3}{2}$$. $$\\begin{split} \\frac{1}{2} (6x - 2) &= 5 - x \\\\ \\frac{1}{2} \\left(6 \\cdot \\textcolor{red}{\\dfrac{3}{2}} - 2 \\right) &\\stackrel{?}{=} 5 - \\textcolor{red}{\\frac{3}{2}} \\\\ \\frac{1}{2} (9 - 2) &\\stackrel{?}{=} \\frac{10}{2} - \\frac{3}{2} \\\\ \\frac{1}{2} (7) &\\stackrel{?}{=} \\frac{7}{2} \\\\ \\frac{7}{2} &= \\frac{7}{2}\\; \\checkmark \\end{split}$$ Exercise 8.69: Solve: $$\\frac{1}{3}$$(6u + 3) = 7 − u. Exercise 8.70: Solve: $$\\frac{2}{3}$$(9x − 12) = 8 + 2x. In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations. Example 8.36: Solve: 0.24(100x + 5) = 0.4(30x + 15). ##### Solution Distribute. $$24x + 1.2 = 12x + 6 \\tag{8.3.83}$$ Subtract 12x to get all the x s to the left. $$24x + 1.2 \\textcolor{red}{-12x} = 12x + 6 \\textcolor{red}{-12x} \\tag{8.3.84}$$ Simplify. $$12x + 1.2 = 6 \\tag{8.3.85}$$ Subtract 1.2 to get the constants to the right. $$12x + 1.2 \\textcolor{red}{-1.2} = 6 \\textcolor{red}{-1.2} \\tag{8.3.86}$$ Simplify. $$12x = 4.8 \\tag{8.3.87}$$ Divide. $$\\frac{12x}{\\textcolor{red}{12}} = \\frac{4.8}{\\textcolor{red}{12}} \\tag{8.3.88}$$ Simplify. $$x = 0.4 \\tag{8.3.89}$$ Check: Let x = 0.4. $$\\begin{split} 0.24(100x + 5) &= 0.4(30x + 15) \\\\ 0.24[100(\\textcolor{red}{0.4}) + 5] &\\stackrel{?}{=} 0.4[30(\\textcolor{red}{0.4}) + 15] \\\\ 0.24(40 + 5) &\\stackrel{?}{=} 0.4(12 + 15) \\\\ 0.24(45) &\\stackrel{?}{=} 0.4(27) \\\\", "- 2$ $\\frac{2 x}{\\textcolor{red}{2}} = - \\frac{2}{\\textcolor{red}{2}}$ $x = - 1$ or $\\left(- 1 , 1\\right)$ We can next plot the two points on the coordinate plane and draw a line through the two points: graph{(2x + y - 3)(2x + y + 1)(x^2+(y+1)^2-0.075)((x+1)^2+(y - 1)^2-0.075)=0 [-20, 20, -10, 10]} From the graphs of the lines it appears the lines are parallel so there are no points in common so there is no solution. Or, the solution is the null or empty set: $\\left\\{\\emptyset\\right\\}$", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou", "the lower bound yields $\\frac{19}{28}-\\frac{2}{3}=\\frac{1}{84}$. This is answer $1+84=$ $\\boxed{\\textbf{(E)} ~85}$. ~theAJL ### Diagram $[asy] /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label(\"m=2/3\", (32,20)); label(\"m=19/28\", (32.3,20.8)); [/asy]$ ## Solution 3 An alternative approach with the same methodology as Solution 1 can be done using Pick's Theorem. Wikipedia page: https://en.wikipedia.org/wiki/Pick%27s_theorem It's a formula to find the amount of lattice points strictly inside a polygon. Approximation of the lower bound is still necessary. ## Video Solution , Very Easy https://youtu.be/PC8fIZzICFg ~hippopotamus1 (Video solution is in Chinese) ~jhu08 ~Interstigation", "in Exercises $23-28$ with its graph in $(a)-(f)$ below and on the next page. $$\\frac{x^{2}}{16}+\\frac{y^{2}}{25}+\\frac{z^{2}}{4}=1$$ Check back soon! ### Problem 28 Match each equation in Exercises $23-28$ with its graph in $(a)-(f)$ below and on the next page. $$z=5\\left(x^{2}+y^{2}\\right)^{-1 / 2}$$ Sid W. University of Louisville ### Problem 29 Let $f(x, y)=4 x^{2}-2 y^{2},$ and find the following. (a) $\\frac{f(x+h, y)-f(x, y)}{1}$ (b) $\\frac{f(x, y+h)-f(x, y)}{h}$ (c) $\\lim _{h \\rightarrow 0} \\frac{f(x+h, y)-f(x, y)}{h}$ (d) $\\lim _{h \\rightarrow 0} \\frac{f(x, y+h)-f(x, y)}{h}$ Jonathon B. ### Problem 30 Let $f(x, y)=5 x^{3}+3 y^{2},$ and find the following. (a) $\\frac{f(x+h, y)-f(x, y)}{h}$ (b) $\\frac{f(x, y+h)-f(x, y)}{h}$ (c) $\\lim _{h \\rightarrow 0} \\frac{f(x+h, y)-f(x, y)}{h}$ (d) $\\lim _{h \\rightarrow 0} \\frac{f(x, y+h)-f(x, y)}{h}$ Sid W. University of Louisville ### Problem 31 Let $f(x, y)=x y e^{x^{2}+y^{2}}$ . Use a graphing calculator or spreadsheet to find each of the following and give a geometric interpretation of the results. (Hint: First factor $e^{2}$ from the limit and then evaluate the quotient at smaller and smaller values of $h . )$ (a) $\\lim _{h \\rightarrow 0} \\frac{f(1+h, 1)-f(1,1)}{h}$ (b) $\\lim _{h \\rightarrow 0} \\frac{f(1,1+h)-f(1,1)}{h}$ Ethan S. ### Problem 32 The following table provides values of the function $f(x, y) .$ However, because of potential errors in measurement, the functional values may be slightly inaccurate." ]

[ "# Let's FInd Some Real Values ! Level pending How many real values of $$x$$ which satisfy $$$(x+9)^6 + 2=(x^2+ 18)^6$$$ ? ×", "plug in any three numbers there instead of 5, 10, and 6 and it works because the real line is a (very simple) vector space of sorts. What you do to test this is simple. If I want to test the distributive property on real numbers, I go to what I just wrote and say, hm, is (5+10)x6 -> 15x6 really equal to 5x6 + 10x6? At that point, you just evaluate both sides until you get something that's obviously (un)equal. 5. Jul 19, 2012 ### g.lemaitre ok, i got it.", "back to index | new Find all the real values of $x$ that satistify: $$\\sqrt{3x^2 + 1} + \\sqrt{x} - 2x - 1=0$$ Find all the real values of $x$ that satistify: $$\\sqrt{3x^2 + 1} - 2\\sqrt{x} + x - 1=0$$ Find all the real values of $x$ that satistify: $$\\sqrt{3x^2 + 1} - 2\\sqrt{x} - x + 1=0$$", "Let $$t$$ be a real number satisfying $$t^3-3t^2-9t+c = 0$$, then find the sum of all possible integer values of $$c$$ for which $$x+\\dfrac 1x = t$$ has 6 real and distinct roots.", "+0 Find the positive real value of t that satisfies ​. 0 283 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "+0 # Find the positive real value of t that satisfies ​. 0 35 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "# How Many X Can You Find? Find the largest possible value of $$n$$, such that there exists $$n$$ real numbers $$x_1,...,x_n$$ which satisfy $$|x_i-x_j|> \\frac{1}{100}(1+x_ix_j)$$ for all values of $$i \\neq j$$. ×", "$m\\leq x\\leq m+1$ inductively. The function $h(x)=f(-m+1-x)$ can be treated in the same manner and that allows us to prove $f(x)=0$ for every real number $x$.", "+0 # Find all real values of t that satisfy the equation 0 97 1 Find all real values of t that satisfy the equation t^4 = 324. If you find more than one, then list your values in increasing order, separated by commas. Nov 1, 2020 $$\\sqrt[4]{324}$$ = - +4.24264", "# Math Help - find number of solutions of x 1. ## find number of solutions of x find number of solutions of x ..where x belongs to real number for which $64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$ 2. Try factoring each number as a multiple of 8. You might get some cancellation there.", "b}Show that inf[a,b) =0 and sup[a,6) = b. Consider real numbers a and b where a < b. We define [a,6] = {r € R:a < x < b}, [a,6) = {2 € R:a < x < b}, (a,6) = {r € R:a < = < b}, (a,6] = {x € R:a < x < b} Show that inf[a,b) =0 and sup[a,6) = b....", "Question # The sum of all real values of $$x$$ satisfying the equation $$(x^{2} - 5x + 5)^{x^{2} + 4x - 60} = 1$$ is: A 3 B 4 C 6 D 5 Solution ## The correct option is A $$3$$$$(x^2-5x+5)^{x^2+4x-60}=1$$First possibility is $$x^2+4x-60 = 0$$$$x=-10,6$$Seond possibility is $$x^2-5x+5=1$$$$x=1 ,4$$Third Possibility$$x^2-5x+5=-1$$$$x=2,3$$ substitute $$x=3$$ it will give odd number in $$x^2+4x-60$$So $$x=2$$ is only acceptable.Sum of the integers $$=-10+4+1+6+2=3$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "Each solution to $x^{2} + 5x + 8 = 0$ can be written in the form $x = a + b i$, where $a$ and $b$ are real numbers. What is $a + b^{2}$?", "× # Can you rigidly prove it? Nonnegative real numbers satisfy $$a^5+b^5 \\leq 1$$ and $$c^5+d^5 \\leq 1$$. Prove that $$a^2c^3+b^2d^3 \\leq 1$$. Note by Finn Hulse 3 years, 2 months ago", "# Find the Value of X in the Following: 5^(2x+3)=1 - Mathematics Find the value of x in the following: 5^(2x+3)=1 #### Solution Given 5^(2x+3)=1 5^(2x+3)=5^0 On equating the exponents we get ⇒ 2x + 3 = 0 ⇒ 2x = -3 rArr x = (-3)/2 Concept: Laws of Exponents for Real Numbers Is there an error in this question or solution? #### APPEARS IN RD Sharma Mathematics for Class 9 Chapter 2 Exponents of Real Numbers Exercise 2.2 | Q 10.7 | Page 26 Share", "# Find $$x+y$$ Level pending Let $$x,y$$ be real numbers such that $$(x+y)(x+1)(y+1)=2$$ and $$x^3+y^3=1$$. Find the value of $$x+y$$.", "+0 # Help 0 78 1 Let $S$ be the set of all nonzero real numbers. Let $f : S \\to S$ be a function such that $f(x) + f(y) = f(xyf(x + y))$for all $x,$ $y \\in S$ such that $x + y \\neq 0.$ Let $n$ be the number of possible values of $f(4),$ and let $s$ be the sum of all possible values of $f(4).$ Find $n \\times s.$ May 6, 2021", "number line. Represent sqrt3.5, sqrt9.4, sqrt10.5 on the real number line. #### Page 0 Visualise 2.665 on the number line, using successive magnification. Visualise the representation of 5.3bar7` on the number line upto 5 decimal places, that is upto 5.37777. S", "and then divide them by 2 Sp we get, = ( 2 / 5 ) + 2, = ( 2 / 5 ) + ( 2 /1 ). Taking L.C.M of 5 and 1 as 5, we first make the denominator same For this, we multiply the numerator and the denominator of 2 by 5, we get, = ( 2 / 5 ) + 2*5/5 == ( 2 / 5 ) + ( 10 / 5 ) = ( 2 + 10 ) / 5 = 12 / 5 Now dividing the above number by 2, we get the real number which lies between = ( 2 / 5 ) and 2 = ( 12 / 5 ) ÷ 2 = (6 / 5 ) Now if we proceed in the same way we observe that there exist a real number between 2/5 and 6 / 5. Also there exist a real number between 6/5 and 2. This check can go up to an endless process and we will find a new number every time. Thus this list is endless and so there are infinite numbers between the given two numbers. This property of real numbers, where we can find infinite real numbers between any two real numbers is called the Density Property of Real Numbers.", "If x=y and x+2, then y+2. Example 2: If m=5 and m-2, then 5-2. Example 3: If x=3 and 5+x=8, then 5+3=8. You might also be interested in: Classifying Real Numbers Properties of Real Numbers" ]

[ "is that $c$ is an integer. $$\\left( {k \\over 10}+n+c \\right)^2 -n^2 \\le m \\lt \\left({{k+1} \\over {10}}+n+c \\right)^2-n^2$$ Factoring gets, $${{k^2} \\over {100}}+{{k \\cdot n} \\over 5}+{{c\\cdot (k+10n)} \\over 5}+c^2 \\le m \\lt {{k^2} \\over {100}}+{{k \\cdot (10n+1)} \\over {50}}+{n \\over 5}+{{c \\cdot (k+10n+1)} \\over 5}+{1 \\over {100}}+c^2$$ Note that these are linear functions. This inequality explains everything given that, $p=0$ and we need to find the occurrence of a $k$ digit. For instance, the slope of the lines increases whenever we increase $k$. This explains the behavior you observed when conducting your simulations. Proof that solutions for integer $m$ exist: To prove that this inequality always contains an integer $m$, find the width of the interval given by the inequality. The width $w$ is, $$w={c \\over 5} + {k \\over {50}}+{{20n+1} \\over 100}$$ By the Pigeon Hole Principle, if the width $w$ is greater than or equal to $1$, the values that $m$ can take on must include an integer. If we set $w=$ and solve for $n$ we get, $$n \\gt {{99-20 \\cdot c -2 \\cdot k} \\over 20}$$ So any $n$ greater than the value on the right, will create an interval of solutions $m$ that contains an integer. • Are you aware that this answer is very similar to your previous one? –", "get the result. The original inequality must be very loose indeed! 4. Originally Posted by tonio $5^{k+1}+5=5\\cdot 5^k+5=4\\cdot 5^k+\\left(5^k+5)<4\\cdot 5^k+5^{k+1}<5\\cdot 5^k+5^{k+1}=2\\cdot 5^{k+1}<5\\cdot 5^{k+1}=5^{k+2}$ Tonio Thank you soo much this was great help.. im just a tad confused how the 4x5^k and the 2x5^k+1 got there. 5. Thread closed due to this member deleting questions after getting help.", "so that $$a\\cdot a= b\\cdot b =1 +2^{2}+3^{2}+4^{2}=30$$ $$\\Rightarrow \\|a\\|=\\sqrt{30} =\\| b \\| \\mbox{ and } \\ \\big(\\|a\\|+\\|b\\|\\big)^{2}=(2\\sqrt{30})^{2}=120\\, .$$ Since $$a+b= \\begin{pmatrix}5\\\\5\\\\5\\\\5\\end{pmatrix}\\, , \\quad$$ we have $$\\|a+b\\|^{2}=5^{2}+5^{2}+5^{2}+5^{2}=100<120=\\big(\\|a\\|+\\|b\\|\\big)^{2}$$ as predicted by the triangle inequality. Notice also that $$a\\cdot b=1.4+2.3+3.2+4.1=20< \\sqrt{30}.\\sqrt{30}=30=\\|a\\|\\, \\|b\\|$$ in accordance with the Cauchy--Schwarz inequality.", "## Intermediate Algebra (6th Edition) $[0,4]$ We are given that $x^{2}-4x\\leq0$. To solve, we can first solve the related equation, $x^{2}-4x=0$. $x^{2}-4x=0$ Factor an x out of both terms on the left side. $x(x-4)=0$ Therefore, $x=0$ or $x=4$. We can now test numbers around these values by plugging them into the equation. $(-1)^{2}-4(-1)=1+4=5\\gt0$ $(1)^{2}-4(1)=1-4=-3\\lt0$ $(5)^{2}-4(5)=25-20=5\\gt5$ Therefore, we know that values of x between and including 0 and 4 satisfy the given inequality.", "Nicolas Jun 15 '13 at 19:30 Excellent work: You can conclude, since you have shown $$3^{3(k+1)+1}\\;\\; <\\;\\; 3^3 \\cdot 2^{5n+6} \\;\\;{\\color{blue}{\\bf <}}\\;\\; 2^{5(k+1)+6}$$ or simply, $$3^{3(k+1)+1}\\; {\\color{blue}{\\bf <}}\\; 2^{5(k+1)+6}$$ as desired. The \"blue\" strict inequality is all you need. You have shown, prior to your conclusion, that $$3^{3(k+1)+1}\\;\\; <\\;\\; 3^3 \\cdot 2^{5n+6}$$ and $$3^3\\cdot 2^{5n+6} \\;\\;<\\;\\; 2^{5(k+1)+6}$$ - But is $3^3∗2^{5n+6}=2^{5(k+1)+6}$? How do i show it? – Maximus Programus Jun 15 '13 at 19:27 You showed it when you expanded: $$3^3 \\cdot 2^{5n+6} = 27 \\cdot 2^{5n+6}$$ $$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \\cdot 2^{5n+6} = 32 \\cdot 2^{5n+6}$$ – amWhy Jun 15 '13 at 19:31 After your first invokation of the inductive hypothesis you reach $3^3 \\cdot 3^{3k+1} < 3^3 \\cdot 2^{5k+6}$ So why not just direction show that $3^3 < 2^5$, completing the proof? $3^3 \\cdot 2^{5k+6} < 2^5 \\cdot 2^{5k+6} = 2^{5(k+1)+6}$ - It’s fine. You can compress it to a single chain of inequalities: $$3^{3(k+1)+1}=3^3\\cdot3^{3k+1}=27\\cdot3^{3k+1}<32\\cdot3^{3k+1}<32\\cdot2^{5k+6}=2^5\\cdot2^{5k+6}=2^{5(k+1)+6}\\;.$$ -", "factors x and x - 5 need to have different signs for the product to be negative such that: `x gt 0` `x - 5 lt 0 =gt x lt 5` or `x lt 0` `x - 5 gt 0 =gt x gt 5` Since the intersection of the intervals `(-oo,0)` and `(5,oo)` yields `O/` , hence the common values for x that satisfy the inequality `x(x - 5)lt0` are in interval `(0,5).` Approved by eNotes Editorial Team", "06:20 PM Ackbeet Wow. Amazing how many of those inequalities are not tight, and yet you still get the result. The original inequality must be very loose indeed! • July 12th 2010, 06:36 PM jess0517 Quote: Originally Posted by tonio $5^{k+1}+5=5\\cdot 5^k+5=4\\cdot 5^k+\\left(5^k+5)<4\\cdot 5^k+5^{k+1}<5\\cdot 5^k+5^{k+1}=2\\cdot 5^{k+1}<5\\cdot 5^{k+1}=5^{k+2}$ Tonio Thank you soo much this was great help.. im just a tad confused how the 4x5^k and the 2x5^k+1 got there. • July 14th 2010, 02:24 AM mr fantastic Thread closed due to this member deleting questions after getting help.", "+0 # What is the value of $c$ for which the inequality $x\\cdot(3x+1) 0 184 1 What is the value of$c$for which the inequality$x\\cdot(3x+1) Oct 10, 2021 #1 +13724 +1 What is the value of c for which the inequality \\$x\\cdot(3x+1) Hello Guest! $$x\\cdot(3x+1)$$ What is the inequality called and where is c? ! Oct 10, 2021", "then it is impossible to go any further as 1 is the smallest value. Therefore 11 is the smallest possible answer for n. Conclusion: n=17, 15 and 11 are the only possible solutions. Here is Mariana's solution 1.Solution with consecutive numbers . The sum of integers 1 to n is: $1 + 2 + 3 + \\cdots + n = {n(n+1)\\over 2}.$ Let the three numbers removed be $x$, $x+1$ and $x+2$ which add up to $3x+3$. Then $${{n(n+1)\\over 2} -(3x+3)\\over n-3}=7.5 \\quad (1)$$ This simplifies to: $${n^2-14n+39\\over 6} = x. \\quad (2)$$ The smallest value of the first of the three consecutive numbers is 1 and the largest value is $(n-2)$. Therefore $1\\leq x \\leq n-2.$ Combining equation (2) with this inequality: $$1 \\leq {n^2 - 14n +39\\over 6} \\leq n-2.$$ This simplifies to $6 \\leq n^2-14n+39 \\leq 6n-12.$ So $$0\\leq n^2 -14n + 33 =(n - 11)(n - 3)$$ and $$n^2-20n + 51 =(n-3)(n-17)\\leq 0.$$ From the first inequality $n\\leq 3$ or $n\\geq 11$ and from the second inequality $3\\leq n \\leq 17.$ However, $n=3$ is not a valid solution because, after removing three consecutive numbers, there would not be any numbers left to yield an average of 7.5. Therefore, $11\\leq n \\leq 17.$ Using equation (2) again, the following table contains the values of $x$", "accepted answer scores 10. – user59671 Mar 12 '13 at 11:27 Note that from Cauchy-Schwarz, we have $$(x_1 + x_2 + \\cdots + x_n) \\leq \\sqrt{n} \\sqrt{x_1^2 + x_2^2 + \\cdots + x_n^2}$$ and equality holds iff $x_1=x_2 = \\cdots = x_n$. Hence, we need $$a^2 \\leq nb$$ In general, if strict inequality holds there are infinitely many solutions. If the inequality is violated, then there is no solution. If equality holds, there is only one unique solution. For $(n,a,b) = (20,200,2000)$, we have $a^2 = 200^2 = 40000 = 20 \\times 2000 = nb$. Hence, for this specific case, there is only one solution, viz, $$x_1 = x_2 = x_3 \\cdots = x_n = 10$$ -", "## Precalculus (6th Edition) Option $A$ The given inequality is \"$-2$ is less than $x$, and $x$ is less than or equal to $6$.\" The numbers included in this set are the numbers between $-2$ and $6$, including $6$. The equivalent interval notation of the given inequality will have: $\\bf\\text{lower limit:}$ $-2$, but a parenthesis will be used since $-2$ is not included $\\bf\\text{upper limit:}$ $6$, a bracket will be used since $6$ is included Therefore, the interval is: $\\color{blue}{(-2, 6]}$", "prime factorize $20^9$ as $2^{18} \\cdot 5^9$. Denote $a_1$ as $2^{b_1} \\cdot 5^{c_1}$, $a_2$ as $2^{b_2} \\cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \\cdot 5^{c_3}$. In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\\le b_2\\le b_3$, and $c_1\\le c_2\\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor. We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\\le b_1. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\\dbinom{21}{3}$. The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is $$\\frac{\\dbinom{21}{3} \\cdot \\dbinom{12}{3}}{190^3}.$$ Simplification gives $\\frac{77}{1805}$, and therefore the answer is $\\boxed{077}$. -molocyxu ## Solution 2 Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we must have $b_1", "2018-04 An inequality Let $$x_1,x_2,\\ldots,x_n$$ be reals such that $$x_1+x_2+\\cdots+x_n=n$$ and $$x_1^2+x_2^2+\\cdots +x_n^2=n+1$$. What is the maximum of $$x_1x_2+x_2x_3+x_3x_4+\\cdots + x_{n-1}x_n+x_nx_1$$? GD Star Rating loading... 2018-04 An inequality, 3.1 out of 5 based on 7 ratings", "W); [/asy]$ First, prime factorize $20^9$ as $2^{18} \\cdot 5^9$. Denote $a_1$ as $2^{b_1} \\cdot 5^{c_1}$, $a_2$ as $2^{b_2} \\cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \\cdot 5^{c_3}$. In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\\le b_2\\le b_3$, and $c_1\\le c_2\\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor. We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\\le b_1. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\\dbinom{21}{3}$. The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is $$\\frac{\\dbinom{21}{3} \\cdot \\dbinom{12}{3}}{190^3}.$$ Simplification gives $\\frac{77}{1805}$, and therefore the answer is $\\boxed{077}$. -molocyxu ## Solution 2 Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we", "2^{n-1}+3\\cdot 2^{n-3}+3 &< 2^{n-1}+2^2\\cdot 2^{n-3}+3\\\\ &= 2^{n-1}+2^{n-1}+3\\\\ &=2^n + 3\\;; \\end{align*} the only problem is that extra $+3$. But since we used a pretty estimate when we replaced $3$ by $4$ in the second term of $(2)$, and since that almost worked, we can reasonably hope that in fact $3\\cdot 2^{n-3}+3\\le 2^{n-1}$, which would be enough to make $2^{n-1}+3\\cdot 2^{n-3}+3\\le 2^n$. The inequality $3\\cdot 2^{n-3}+3\\le 2^{n-1}$ can be written $3+\\dfrac3{2^{n-3}}\\le 4$, and you’re assuming that $n\\ge 8$, so ...", "(x) > 0$ for all x in $(-∞, -4) or (0, ∞)$. Conclusion: Thus, the interval notation of the given inequality is $(-∞, -4) ∪ (0, ∞)$ and the graph of the solution set on a number line is shown as follows:", "## Precalculus (6th Edition) Option $D$ The given inequality is \"$6$ is less than or equal to $x$\". This is equivalent to \"$x$ is greater than or equal to $6$. The numbers included in this set are the numbers that are greater than or equal to $6$. The equivalent interval notation of the given inequality will have: $\\bf\\text{lower limit:}$ $6$, a bracket will be used since $6$ is included $\\bf\\text{upper limit:}$ none, so positive infinity will be used Therefore, the interval is: $\\color{blue}{[6, +\\infty)}$", "# 2018-04 An inequality Let $$x_1,x_2,\\ldots,x_n$$ be reals such that $$x_1+x_2+\\cdots+x_n=n$$ and $$x_1^2+x_2^2+\\cdots +x_n^2=n+1$$. What is the maximum of $$x_1x_2+x_2x_3+x_3x_4+\\cdots + x_{n-1}x_n+x_nx_1$$? GD Star Rating", "# A Muirhead Like Inequality I am looking for a proof of the inequality as follow: Let $n$ be an integer number $n \\ge 2$ and $x_1, \\cdots, x_n$ and $y_1,\\cdots, y_n$ are nonegative real numbers such that $(x_1,\\cdots, x_n)$ majorizes $(y_1,\\cdots, y_n)$; Let $0 \\leq a_1, a_2,\\cdots,a_n \\leq 1$ then $$\\sum_{\\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\\cdots {x_n}^{a_n}\\leq \\sum_{\\text{sym}} {y_1}^{a_1}{y_2}^{a_2}\\cdots {y_n}^{a_n}$$ Note: The inequality above is not Muirhead's Inequality. Example: Let $0 \\leq a_i \\leq 1$ then • $4^{a_1}1^{a_2}+ 4^{a_2}1^{a_1} \\le 3^{a_1}2^{a_2}+ 3^{a_2}2^{a_1}$ • $5^{a_1}5^{a_2}2^{a_3}+5^{a_1}5^{a_3}2^{a_2}+5^{a_2}5^{a_1}2^{a_3}+5^{a_2}5^{a_3}2^{a_1}+5^{a_3}5^{a_1}2^{a_2}+5^{a_3}5^{a_2}2^{a_1} \\leq 4.5^{a_1}4^{a_2}3.5^{a_3}+4.5^{a_1}4^{a_3}3.5^{a_2}+4.5^{a_2}4^{a_1}3.5^{a_3}+4.5^{a_2}4^{a_3}3.5^{a_1}+4.5^{a_3}4^{a_1}3.5^{a_2}+4.5^{a_3}4^{a_2}3.5^{a_1}$ For $x=(x_1,\\dots,x_n)\\in(0,\\infty)$, let \\begin{equation*} f(x):=2\\sum_{\\si\\in S_n}\\prod_{k=1}^n x_k^{a_{\\si(k)}}, \\end{equation*} where $S_n$ is the set of all permutations of the set $\\{1,\\dots,n\\}$. By the symmetry and the Schur--Ostrowski criterion, it suffices to show that \\begin{equation*} x_1\\le x_2\\overset{\\text{(?)}}\\implies f_1(x)-f_2(x)\\ge0, \\tag{1} \\end{equation*} where $f_j(x)$ is the partial derivative of $f(x)$ in $x_j$. Consider the bijection $S_n\\ni\\si\\leftrightarrow\\tsi\\in S_n$ defined by the formula \\begin{equation*} \\tsi(k):= \\left\\{ \\begin{aligned} \\si(2)&\\text{ if }k=1,\\\\ \\si(1)&\\text{ if }k=2,\\\\ \\si(k)&\\text { otherwise}. \\end{aligned} \\right. \\end{equation*} Then \\begin{align*} f(x)&:=\\sum_{\\si\\in S_n}x_1^{a_{\\si(1)}}x_2^{a_{\\si(2)}}\\prod_{k=3}^n x_k^{a_{\\si(k)}} + \\sum_{\\tsi\\in S_n}x_1^{a_{\\tsi(1)}}x_2^{a_{\\tsi(2)}}\\prod_{k=3}^n x_k^{a_{\\tsi(k)}} =\\sum_{\\si\\in S_n}Q P, \\tag{2} \\end{align*} where \\begin{equation*} Q:=Q_\\si(x):=x_1^{b_1}x_2^{b_2}+x_1^{b_2}x_2^{b_1},\\quad P:=P_\\si(x):=\\prod_{k=3}^n x_k^{b_k}, \\end{equation*} \\begin{equation*} b_k:=b_{\\si;k}:=a_{\\si(k)}. \\end{equation*} Denoting by $Q_j$ the partial derivative of $Q$ in $x_j$, we have \\begin{equation*} Q_1-Q_2=b_1 (x_1 x_2)^{b_1-1}(x_2^{b_2-b_1+1}-x_1^{b_2-b_1+1}) +b_2 (x_1 x_2)^{b_2-1}(x_2^{b_1-b_2+1}-x_1^{b_1-b_2+1})\\ge0. \\end{equation*} Now noting that $P$ does not depend on $(x_1,x_2)$, we see that (1) follows from (2), as desired.", "then the last value is certainly less than $2\\cdot 10^i$. But we can divide this by $10^{2i}$ to get $t^2\\leq t_i^2/10^{2i}\\lt t^2+ 2\\cdot10^{-i}$ - or in other words, $0.201120122013 \\leq t_i^2/10^{2i} \\le 0.201120122013+2\\cdot10^{-i}$ - and all we have to do to get this to match up with our original inequality is to take an $i$ such that $2\\cdot 10^{-i}$ is even less than $10^{-12}$ - for instance, $i=13$ will do. This gives us an answer that $t_{13} = 4484641814159$ squares to $t_{13}^2=20112012201303326692877281$. -" ]

[ "## Precalculus (6th Edition) Blitzer $5|x|\\sqrt{5}$ RECALL: For any non-negative real numbers $a$ and $b$, $\\sqrt{ab} = \\sqrt{a}\\cdot \\sqrt{b}$ and $\\sqrt{a} \\cdot \\sqrt{b} = \\sqrt{ab}$. Factor the radicand so that one factor is a perfect square to obtain $\\sqrt{9x^2 \\cdot 5}$. Use the rule above to obtain $\\sqrt{25x^2} \\cdot \\sqrt{5} \\\\=\\sqrt{(5x)^2} \\cdot \\sqrt{5}.$ Note that the variable $x$ represents any real number. This means that $x$ could be negative. Thus, to simplify the expression, use the rule $\\sqrt{a^2}=|a|$ to obtain $|5x|\\sqrt{5} \\\\=5|x|\\sqrt{5}.$", "get the result. The original inequality must be very loose indeed! 4. Originally Posted by tonio $5^{k+1}+5=5\\cdot 5^k+5=4\\cdot 5^k+\\left(5^k+5)<4\\cdot 5^k+5^{k+1}<5\\cdot 5^k+5^{k+1}=2\\cdot 5^{k+1}<5\\cdot 5^{k+1}=5^{k+2}$ Tonio Thank you soo much this was great help.. im just a tad confused how the 4x5^k and the 2x5^k+1 got there. 5. Thread closed due to this member deleting questions after getting help.", "# How do you solve the inequality x^2+2x>=24? May 9, 2017 By trial and error. Your answer is with the exception of -6<x<4, all x values provide solution. #### Explanation: x must be a real number and let me try +3 provide the solution. There are two numbers provide the solution (one is positive and the other one is negative): ${3}^{2} + 2 \\cdot 3 = 15$ therefore it does not provide solution. Now let me try +4 ${4}^{2} + 2 \\cdot 4 = 24$ Yes. This is one answer. Therefore x must be greater than or equal to +4. On the negative side, let me try -3 first: ${\\left(- 3\\right)}^{2} - 2 \\cdot 3 = 2$ does not satisfy the requested condition. Try -4 now: ${\\left(- 4\\right)}^{2} - 2 \\cdot 4 = 8$ does not satisfy the requested condition. Try -5 now: ${\\left(- 5\\right)}^{2} - 2 \\cdot 5 = 15$ does not satisfy the requested condition. Try -6 now: ${\\left(- 6\\right)}^{2} - 2 \\cdot 6 = 24$ Yes it satisfies. Any number less than -6 (including -6) will satisfy the given. For instance (-8): ${\\left(- 8\\right)}^{2} - 2 \\cdot 8 = 48$ which is greater than 24. Now your solution is x must be greater than or equal to 4 or x must be less than or", "in the inequality $$25 - 4n < 0$$. Adding $$4n$$and dividing by 4, we find $$n>6.25$$. The probability that I chose one of the numbers 7, 8, 9, or 10 is $$\\boxed{\\frac{2}{5}}$$ . Aug 11, 2018 #5 +809 +2 Thank you, both CPhill and azsun for wonderful solutions! Aug 11, 2018 #6 +98173 +1 OK, mathtoo !!!!! CPhill Aug 11, 2018", "Given real number $x_1, x_2, \\ldots, x_n$ such that $x_1 + x_2 + \\ldots +x_n>0$, must this other inequality be true? Given real numbers $x_1, x_2, \\ldots, x_n$ such that $$x_1 + x_2 + \\ldots + x_n >0$$ with the condition that $\\max\\{|x_i|\\}_{i=1}^n = \\max\\{x_i\\}_{i=1}^n$, must it be true that $${x_1}^m + {x_2}^m + \\ldots + {x_n}^m > 0$$ for all natural numbers $m$? If so, can you prove it? I don't know how to approach the problem. Is there any generalization of the power-mean inequality to real numbers that might help? The inequality clearly holds true for all even $m$ from the Trivial Inequality, but the case for odd $m$ feels difficult. Thanks, I appreciate any help! • Well, no: Consider $x_1=3,\\ x_2=x_3=x_4=\\frac12,\\ x_5=-1,\\ x_6=-3$. Then the sum for odd values of $m$ is $\\frac{3}{2^m}-1$. – user228113 Oct 30 '16 at 15:48", "+0 # help again -1 179 1 +389 Both $x$ and $y$ are positive real numbers, and the point $(x,y)$ lies on or above both of the lines having equations $2x+5y = 10$ and $3x+4y = 12$. What is the least possible value of $8x+13y$? Mar 19, 2020 #1 +260 0 With the info, we try to sum the equation to something useful. We turn the lines into inequalities. Adding the first inequality to two times the second gets us $$8x+13y=(2x+5y)+2(3x+4y)\\geq 10+2\\cdot 12 = \\boxed{34}$$ credit to aops for the solution Apr 6, 2020", "06:20 PM Ackbeet Wow. Amazing how many of those inequalities are not tight, and yet you still get the result. The original inequality must be very loose indeed! • July 12th 2010, 06:36 PM jess0517 Quote: Originally Posted by tonio $5^{k+1}+5=5\\cdot 5^k+5=4\\cdot 5^k+\\left(5^k+5)<4\\cdot 5^k+5^{k+1}<5\\cdot 5^k+5^{k+1}=2\\cdot 5^{k+1}<5\\cdot 5^{k+1}=5^{k+2}$ Tonio Thank you soo much this was great help.. im just a tad confused how the 4x5^k and the 2x5^k+1 got there. • July 14th 2010, 02:24 AM mr fantastic Thread closed due to this member deleting questions after getting help.", "+0 # Inequality 0 27 1 Find the set of all real numbers x so that x^2 <= 3. May 14, 2022 #1 +13581 +1 The set of all real numbers x so that $$(x^2\\leq3)\\ is\\ \\color{blue}x\\in\\mathbb R\\ |\\ -\\sqrt{3}\\leq x\\leq +\\sqrt{3}.$$ ! May 14, 2022 edited by asinus May 14, 2022", "Consider the inequality $9^x - a \\cdot 3^x - a + 3 \\leq 0$ where $a$ is a real parameter independent to $x$. If this inequality has at least one positive solution, then find the range of $a$.", "# Classical Inequalities addicts can do this. Part 1 Algebra Level 5 \\begin{cases}\\begin{align} x_{1} + x_{2} + x_{3} + \\cdots + x_{n} &= 2016 \\\\\\\\ x_{1}^{4} + x_{2}^{4} + x_{3}^{4} + \\cdots + x_{n}^{4} &= 2016\\times 512 \\end{align}\\end{cases} Let $$x_1, x_2, \\ldots , x_n$$ be real numbers. Find the smallest possible value of $$n$$ such that there exists real solutions to the equations above. For more problems like this, try answering this set . ×", "+0 # What is the value of $c$ for which the inequality $x\\cdot(3x+1) 0 184 1 What is the value of$c$for which the inequality$x\\cdot(3x+1) Oct 10, 2021 #1 +13724 +1 What is the value of c for which the inequality \\$x\\cdot(3x+1) Hello Guest! $$x\\cdot(3x+1)$$ What is the inequality called and where is c? ! Oct 10, 2021", "Trivial 6.0.34992.1 x6 - 3x5 + 5x3 - 3x + 1 $$-\\,2^{4}\\cdot 3^{7}$$ $S_3$ (as 6T2) Trivial 6.0.35099.1 x6 + x4 + x2 - x + 1 $$-\\,35099$$ $S_6$ (as 6T16) Trivial Next", "# [Solutions] Baltic Way Mathematical Competition 2014 1. Show that $\\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\cos(2^2\\cdot 56^{\\circ})\\cdot . . . \\cdot \\cos(2^{23}\\cdot 56^{\\circ}) = \\frac{1}{2^{24}} .$ 2. Let $a_0, a_1, . . . , a_N$ be real numbers satisfying $a_0 = a_N = 0$ and $a_{i+1} - 2a_i + a_{i-1} = a^2_i$ for $i = 1, 2, . . . , N - 1.$ Prove that $a_i\\leq 0$ for $i = 1, 2, . . . , N- 1.$ 3. Positive real numbers $a, b, c$ satisfy $\\frac{1}{a} +\\frac{1}{b} +\\frac{1}{c} = 3.$ Prove the inequality $\\frac{1}{\\sqrt{a^3+ b}}+\\frac{1}{\\sqrt{b^3 + c}}+\\frac{1}{\\sqrt{c^3 + a}}\\leq \\frac{3}{\\sqrt{2}}.$ 4. Find all functions $f$ defined on all real numbers and taking real values such that $f(f(y)) + f(x - y) = f(xf(y) - x),$ for all real numbers $x, y.$ 5. Given positive real numbers $a, b, c, d$ that satisfy equalities $a^2 + d^2 - ad = b^2 + c^2 + bc, \\ \\ a^2 + b^2 = c^2 + d^2$ find all possible values of the expression $$\\frac{ab+cd}{ad+bc}.$$ 6. In how many ways can we paint $16$ seats in a row, each red or green, in such a way that the number of consecutive seats painted in the same colour is always odd? 7. Let $p_1, p_2, . . . , p_{30}$", "Nicolas Jun 15 '13 at 19:30 Excellent work: You can conclude, since you have shown $$3^{3(k+1)+1}\\;\\; <\\;\\; 3^3 \\cdot 2^{5n+6} \\;\\;{\\color{blue}{\\bf <}}\\;\\; 2^{5(k+1)+6}$$ or simply, $$3^{3(k+1)+1}\\; {\\color{blue}{\\bf <}}\\; 2^{5(k+1)+6}$$ as desired. The \"blue\" strict inequality is all you need. You have shown, prior to your conclusion, that $$3^{3(k+1)+1}\\;\\; <\\;\\; 3^3 \\cdot 2^{5n+6}$$ and $$3^3\\cdot 2^{5n+6} \\;\\;<\\;\\; 2^{5(k+1)+6}$$ - But is $3^3∗2^{5n+6}=2^{5(k+1)+6}$? How do i show it? – Maximus Programus Jun 15 '13 at 19:27 You showed it when you expanded: $$3^3 \\cdot 2^{5n+6} = 27 \\cdot 2^{5n+6}$$ $$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \\cdot 2^{5n+6} = 32 \\cdot 2^{5n+6}$$ – amWhy Jun 15 '13 at 19:31 After your first invokation of the inductive hypothesis you reach $3^3 \\cdot 3^{3k+1} < 3^3 \\cdot 2^{5k+6}$ So why not just direction show that $3^3 < 2^5$, completing the proof? $3^3 \\cdot 2^{5k+6} < 2^5 \\cdot 2^{5k+6} = 2^{5(k+1)+6}$ - It’s fine. You can compress it to a single chain of inequalities: $$3^{3(k+1)+1}=3^3\\cdot3^{3k+1}=27\\cdot3^{3k+1}<32\\cdot3^{3k+1}<32\\cdot2^{5k+6}=2^5\\cdot2^{5k+6}=2^{5(k+1)+6}\\;.$$ -", "Inequalities finding the set of solutions Find the set of solutions to this inequality? $|x − 3| + |x − 6| < 5$ I have been taught to do it by treating $x$ in $3$ separate cases however I am not getting the correct answer. The answer is 'The set of real numbers $x$ such that $2 < x < 7.$ I am getting The set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$. Method Case 1 $x<3$ $|x-3|= -(x-3)$ $|x-6|= -(x-6)$ $-(x-3)-(x-6)<5$ $-2x+9<5$ $-2x<-4$ $2x>4$ $x>2$ Because you have assumed $x<3$ the solutions is the intersection so $2<x<3$. Case 2: $3<x<6$ $|x-3|= (x-3)$ $|x-6|= -(x-6)$ $(x-3)-(x-6)<5$ $3<5$ No solutions Case 3: $x>6$ $|x-3|= (x-3)$ $|x-6|= (x-6)$ $(x-3)+(x-6)<5$ $2x-9<14$ $x<7$ Assumed $x>6$ therefore $6<x<7$. So joint solutions the set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$. Where am I going wrong?? Case 2: 3 < x < 6 |x-3|= (x-3) |x-6|= -(x-6) (x-3)-(x-6)<5 3<5 3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5 Finally we have |x − 3| + |x − 6| < 5 $\\Leftrightarrow$ 2 < x < 7", "so that $$a\\cdot a= b\\cdot b =1 +2^{2}+3^{2}+4^{2}=30$$ $$\\Rightarrow \\|a\\|=\\sqrt{30} =\\| b \\| \\mbox{ and } \\ \\big(\\|a\\|+\\|b\\|\\big)^{2}=(2\\sqrt{30})^{2}=120\\, .$$ Since $$a+b= \\begin{pmatrix}5\\\\5\\\\5\\\\5\\end{pmatrix}\\, , \\quad$$ we have $$\\|a+b\\|^{2}=5^{2}+5^{2}+5^{2}+5^{2}=100<120=\\big(\\|a\\|+\\|b\\|\\big)^{2}$$ as predicted by the triangle inequality. Notice also that $$a\\cdot b=1.4+2.3+3.2+4.1=20< \\sqrt{30}.\\sqrt{30}=30=\\|a\\|\\, \\|b\\|$$ in accordance with the Cauchy--Schwarz inequality.", "# 2018-04 An inequality Let $$x_1,x_2,\\ldots,x_n$$ be reals such that $$x_1+x_2+\\cdots+x_n=n$$ and $$x_1^2+x_2^2+\\cdots +x_n^2=n+1$$. What is the maximum of $$x_1x_2+x_2x_3+x_3x_4+\\cdots + x_{n-1}x_n+x_nx_1$$? GD Star Rating", "Consider the inequality $9^{x}-a\\cdot 3^{x}-a+3\\leq 0$ ,where $a$ is a real parameter. The given inequality has at least one negative solution for $a \\in$", "# A Muirhead Like Inequality I am looking for a proof of the inequality as follow: Let $n$ be an integer number $n \\ge 2$ and $x_1, \\cdots, x_n$ and $y_1,\\cdots, y_n$ are nonegative real numbers such that $(x_1,\\cdots, x_n)$ majorizes $(y_1,\\cdots, y_n)$; Let $0 \\leq a_1, a_2,\\cdots,a_n \\leq 1$ then $$\\sum_{\\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\\cdots {x_n}^{a_n}\\leq \\sum_{\\text{sym}} {y_1}^{a_1}{y_2}^{a_2}\\cdots {y_n}^{a_n}$$ Note: The inequality above is not Muirhead's Inequality. Example: Let $0 \\leq a_i \\leq 1$ then • $4^{a_1}1^{a_2}+ 4^{a_2}1^{a_1} \\le 3^{a_1}2^{a_2}+ 3^{a_2}2^{a_1}$ • $5^{a_1}5^{a_2}2^{a_3}+5^{a_1}5^{a_3}2^{a_2}+5^{a_2}5^{a_1}2^{a_3}+5^{a_2}5^{a_3}2^{a_1}+5^{a_3}5^{a_1}2^{a_2}+5^{a_3}5^{a_2}2^{a_1} \\leq 4.5^{a_1}4^{a_2}3.5^{a_3}+4.5^{a_1}4^{a_3}3.5^{a_2}+4.5^{a_2}4^{a_1}3.5^{a_3}+4.5^{a_2}4^{a_3}3.5^{a_1}+4.5^{a_3}4^{a_1}3.5^{a_2}+4.5^{a_3}4^{a_2}3.5^{a_1}$ For $x=(x_1,\\dots,x_n)\\in(0,\\infty)$, let \\begin{equation*} f(x):=2\\sum_{\\si\\in S_n}\\prod_{k=1}^n x_k^{a_{\\si(k)}}, \\end{equation*} where $S_n$ is the set of all permutations of the set $\\{1,\\dots,n\\}$. By the symmetry and the Schur--Ostrowski criterion, it suffices to show that \\begin{equation*} x_1\\le x_2\\overset{\\text{(?)}}\\implies f_1(x)-f_2(x)\\ge0, \\tag{1} \\end{equation*} where $f_j(x)$ is the partial derivative of $f(x)$ in $x_j$. Consider the bijection $S_n\\ni\\si\\leftrightarrow\\tsi\\in S_n$ defined by the formula \\begin{equation*} \\tsi(k):= \\left\\{ \\begin{aligned} \\si(2)&\\text{ if }k=1,\\\\ \\si(1)&\\text{ if }k=2,\\\\ \\si(k)&\\text { otherwise}. \\end{aligned} \\right. \\end{equation*} Then \\begin{align*} f(x)&:=\\sum_{\\si\\in S_n}x_1^{a_{\\si(1)}}x_2^{a_{\\si(2)}}\\prod_{k=3}^n x_k^{a_{\\si(k)}} + \\sum_{\\tsi\\in S_n}x_1^{a_{\\tsi(1)}}x_2^{a_{\\tsi(2)}}\\prod_{k=3}^n x_k^{a_{\\tsi(k)}} =\\sum_{\\si\\in S_n}Q P, \\tag{2} \\end{align*} where \\begin{equation*} Q:=Q_\\si(x):=x_1^{b_1}x_2^{b_2}+x_1^{b_2}x_2^{b_1},\\quad P:=P_\\si(x):=\\prod_{k=3}^n x_k^{b_k}, \\end{equation*} \\begin{equation*} b_k:=b_{\\si;k}:=a_{\\si(k)}. \\end{equation*} Denoting by $Q_j$ the partial derivative of $Q$ in $x_j$, we have \\begin{equation*} Q_1-Q_2=b_1 (x_1 x_2)^{b_1-1}(x_2^{b_2-b_1+1}-x_1^{b_2-b_1+1}) +b_2 (x_1 x_2)^{b_2-1}(x_2^{b_1-b_2+1}-x_1^{b_1-b_2+1})\\ge0. \\end{equation*} Now noting that $P$ does not depend on $(x_1,x_2)$, we see that (1) follows from (2), as desired.", "# How to solve this problem using Jensen's inequality? This is the task to find the minimum value of $x^5 + y^5 + z^5 - 5xyz$ where $x,y$ and $z$ are any positive numbers. I know that I may use this inequality: $$(t_1\\cdot t_2\\cdot t_3\\cdots t_n)^{\\frac{1}{n}} \\leq \\frac{t_1+t_2+t_3+\\cdots +t_n}{n}$$ By the inequality we have that $\\frac{x^5 + y^5 + z^5 + 1 + 1}{5} \\ge \\sqrt[5]{x^5y^5z^5\\cdot1 \\cdot 1}$. Hence $x^5 + y^5 + z^5 + 2 \\ge 5xyz$ $$x^5 + y^5 + z^5 - 5xyz = x^5 + y^5 + z^5 + 2 - 5xyz - 2 \\ge 5xyz - 5xyz - 2 = -2$$ It's obtained for $x=y=z$ Let $xyz=t^3$. Hence, by AM-GM (which is just Jensen!) $x^5+y^5+z^5-5xyz\\geq3t^5-5t^3\\geq-2,$ where the last inequality it's AM-GM again: $3t^5+2\\geq5\\sqrt[5]{t^{15}}=5t^3$" ]

[ "If $$a$$ and $$b$$ are positive, and $$a+b=1$$, then what is the maximum value of $${ (\\sqrt { 4a+1 } +\\sqrt { 4b+1 } ) }^{ 2 }$$.", "therefore the lowest value abc can have is 8. If, however, a, b, and c are the lowest possible real numbers, this is a very different problem. From the condition by AM-GM we obtain: $$\\prod\\limits_{cyc}\\frac{a}{1+a}=\\prod\\limits_{cyc}\\left(\\frac{1}{1+b}+\\frac{1}{1+c}\\right)\\geq\\prod\\limits_{cyc}\\frac{2}{\\sqrt{(1+b)(1+c)}}=8\\prod\\limits_{cyc}\\frac{1}{1+a}$$ Id est, $abc\\geq8$.", "# Maximum? This shouldn't be tough Let $a, b, c, d \\in \\left[\\frac{1}{2}, 2\\right]$. Suppose $abcd = 1$. Find the maximum possible value of $\\left(a + \\dfrac{1}{b}\\right)\\left(b + \\dfrac{1}{c}\\right)\\left(c + \\dfrac{1}{d}\\right)\\left(d + \\dfrac{1}{a}\\right)$. ×", "find values of a,b,c and d such that $g(x)=ax^3+bx^2+cx+d$ has a local maximum at $(2,4)$ and a local minimum at (0,0) 2. Well you know $g(2)=4$ and $g(0)=0 \\implies d = 0$ you can also use $g'(2)=g'(0)=0$", "If $$a$$ $$b$$ and $$c$$ are positive real numbers with $$a+b+c = 6$$ find the maximum value of $$(a+ \\sqrt{bc})(b+\\sqrt{ac})(c+\\sqrt{ab}) + abc$$", ",13 ,12 ,11 ,10 ,9 ,8 ,7 ,6 ,5 ,4 ,3 ,2 ,1 \\}$ maximum value for $A=30$ and $1 \\leq B \\leq 29$ from left to right. – Ahmad Jul 10 '17 at 23:28", "# Maximum value of absolute sum in a polynomial Suppose that for any $$-1\\leq x\\leq 1$$, $$|ax^2+bx+c|\\leq 1$$. Find the maximum possible value of $$|a|+|b|+|c|$$. My attempts: It is suffices to find the maximum value of $$|a|+|b|+|c|$$ if for any $$-1\\leq x\\leq 1$$, $$(ax^2+bx+c)^2\\leq 1$$. I inserted $$x=-1, x=0, x=1$$, and I got that $$c\\leq 1, a+c\\leq 1, a+b+c\\leq 1$$. How to continue? Or maybe it is not ppssoble to solve the problem with my attempts? Any solution? • Almost identical question: math.stackexchange.com/q/314686/42969 – the same method can be used to solve this question. Mar 3, 2019 at 13:06 • @Phillips Hubert In which year it was posted in Qvant? Mar 3, 2019 at 13:10 Hint: We have $$-1\\le a+b+c\\le 1$$ and $$|c|\\le 1$$ and $$-1\\le a-b+c\\le 1$$ Multiplying $$-1\\le a+b+c\\le 1$$ by $$(-1)$$ we get $$-1\\le -a-b-c\\le 1$$ adding this to $$-1\\le a+b+c\\le 1$$ we get $$-2\\le -2b\\le 2$$ so $$|b|\\le 1$$ The last $$|a|$$ is for you! Second solution: Plugging $$x=0,x=-1,x=1$$ we get $$|c|\\le \\epsilon,|a+b+c|\\le \\epsilon,|a-b+c|\\le \\epsilon$$ we get $$-2\\epsilon \\le -2c\\le 2\\epsilon$$ $$-\\epsilon \\le a+b+c\\le \\epsilon$$ $$-\\epsilon\\le a-b+c\\le \\epsilon$$ doing the same like above and using that $$|a|+|b|=|a-b|$$ we get $$|a|+|b|+|c|\\le 4\\epsilon$$. • @Sonnhard How did you use $-1\\leq a-b+c\\leq1$ and $|b|\\leq1$ in your solution? Also, what is your answer? Also, if you'll see some", "# Calculate the maximum value of $|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$ where $a^2 + b^2 + c^2 = 1$. $$a, b, c$$ are reals such that $$a^2 + b^2 + c^2 = 1$$. Calculate the maximum value of $$\\large |ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$$ This problem appeared in a 9th-grade book about elementary ways to prove inequalities. And I struggled hard with this problem in particular. (I'm a student in 10th grade now.) I'm still questioning if they should delete this example (not an actual exercise), (and they needn't, but the writers of the book should suggest this problem to the authors of a different book for further education in Mathematics). Furthermore, there are problems which are marked difficult or challenging. This one doesn't (What are the attributes of a hard problem that this one doesn't have?). It is unusual. In retrospect, I can't solve it. We have that $$|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)| = |(a + b + c)(a - b)(b - c)(c - a)|$$ And I'm done, factorizing the expression is all I got. The maximum value is $$\\dfrac{9\\sqrt 2}{32}$$, occurs when $$(a, b, c) = \\left(\\dfrac{2 - 3\\sqrt 2}{4\\sqrt 3}, \\dfrac{1}{2\\sqrt 3}, \\dfrac{2 + 3\\sqrt 2}{4\\sqrt 3}\\right)$$", "Say we have 8 numbers, ${a,b,c,d,e,f,g,h}$ such that $abcd=4$ and $efgh=9$. Find the minimum value of $(ae)^2 + (bf)^2 + (cg)^2 + (dh)^2$.", "# Maximum? This shouldn't be tough Algebra Level 5 Let $$a, b, c, d \\in \\left[\\frac{1}{2}, 2\\right]$$. Suppose $$abcd = 1$$. Find the maximum possible value of $$\\left(a + \\dfrac{1}{b}\\right)\\left(b + \\dfrac{1}{c}\\right)\\left(c + \\dfrac{1}{d}\\right)\\left(d + \\dfrac{1}{a}\\right)$$. ×", "# Does Vieta's Help In Any Way? For non negative integers $a$ and $b$ what is the maximum value that $b$ could be such that a value for $a$ can be found such that $ab-(a+b)=1$? Problem Credit: Problem courtesy of my son, David! :) ×", "# What is the maximal value of the expression $\\overline{abc}-(a^3+b^3+c^3)?$ Let $\\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\\overline{abc}-(a^3+b^3+c^3)?$ Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand. You can write the whole thing as $100a + 10b + c - a^3 - b^3 - c^3$. Rearranging the terms gives $(100a - a^3) + (10b - b^3) + (c - c^3)$, which you can solve one summand at a time: You maximize $100a - a^3$ for $a = 6$. You maximize $10b - b^3$ for $b = 2$. You maximize $c - c^3$ for $c = 0$ or $1$. You can write $\\overline{abc} = 100a + 10 b + c$. Then $$\\overline{abc} - (a^3 + b^3 + c^3) = a(100-a^2) + b(10 - b^2) + c(1-c^2).$$ You just need to find the digits $a,b,c$ that maximize each term individually.", "maximal value of $$a$$ which is $$0.2$$. Therefore, $$-a c^3 + 15 a + c\\ge0.2(15-27)+2.6>0$$ Therefore, we can assume that $$0=a\\le b\\le c$$. Now $$f(a,b,c)=b^3c^6\\le2^6\\left(\\frac{3\\times b+6\\times0.5c}{9}\\right)^9=64$$ and it is proved.", "# How find the maximum value of $|bc|$ Question: Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \\leq 1$ holds true for any complex number $z, |z| \\leq 1$. Find the maximum value of $|bc|$ It is said this is answer is $$|bc|\\le \\dfrac{3\\sqrt{3}}{16}$$ My idea: let $z=1$,then $$|a+b+c|\\le 1$$ let $z=-1$,then $$|a-b+c|\\le 1$$ let $z=0$, then $$|c|\\le 1$$ let $|\\theta|=1$,then we have $$|a(z/\\theta)^2+b(z/\\theta)+c|\\le 1\\Longrightarrow |az^2+b\\theta z+c\\theta^2|\\le 1$$ and only this can't solve this problem,Thank you • What value of a b c leads to your equality case? – Calvin Lin May 8 '14 at 5:15 You can assume that $a,b,c$ are reals and $b,c\\ge0$. For $c$ this is obvious. For $b$ you can get this substituting $\\theta z$ for $z$, as you wrote in the question. Finally, you can replace the polynomial by $\\frac{(az^2+bz+c)+(\\bar{a}z^2+bz+c)}2$. At the point $e^{it}$ we have $$1 \\ge |ae^{2it}+be^{it}+c|^2 = |a+be^{-it}+ce^{-2it}|^2 \\\\ = \\big(a+b\\cos t+c\\cos 2t\\big)^2 + \\big(b\\sin t+c\\sin 2t\\big)^2 \\\\ = \\big(a+b\\cos t+c\\cos 2t\\big)^2 + \\big(b\\sin t-c\\sin 2t\\big)^2 + 4bc \\sin t \\sin 2t \\\\ \\ge 4bc \\sin t \\sin 2t.$$ The maximum of $\\sin t \\sin 2t$ is $\\frac{16}{3\\sqrt3}$ (at $t=\\arcsin\\frac{\\sqrt2}{\\sqrt3}$ where $\\cos t=\\frac1{\\sqrt3}$, $\\sin t=\\frac{\\sqrt2}{\\sqrt3}$, $\\cos 2t=-\\frac13$ and $\\sin 2t=\\frac{2\\sqrt2}{3}$). So, at this point we get the bound $bc\\le \\frac{3\\sqrt3}{16}$. To preserve equality in the extreme case,", "Let $a,b,c> 0,abc=1$ Let the minimum of P is M $P= \\frac{a^2}{\\sqrt{2+2ab}}+\\frac{b^2}{\\sqrt{2+2bc}}+\\frac{c^2}{\\sqrt{2+2ca}}$ If $M=\\frac{m}{n}$ Find the value of m+n if $$\\gcd{m,n}=1$$", "Thus, to find what the minimum value for the desired expression is, we have to calculate the minimum of -a-b-c, and the minimum of ab+ac+bc. Since a, b, and c are negative, we cannot use the AM-GM inequality for 3 variables on it, but after multiplying each of the variables by -1, making them positive, we have $$\\frac{-a-b-c}{3}≥ \\sqrt[3]{-abc}$$, simplifying to $$-(a+b+c)≥9$$, so the minimum value of -(a+b+c) is 9. Since a, b, c are negative, ab, ca, and bc must all be positive, hence allowing for an easy AM-GM application. $$\\frac{ab+ac+bc}{3}≥\\sqrt[3]{abc^2}$$, simplifying to $$ab+ac+bc≥ 27$$. Both -(a+b+c) and ab+ac+bc reach the minimum value when the equality condition of the AM-GM inequality is met, namely, when -a=-b=-c, or when a=b=c, and when ab=ac=bc. Both of these simplify to a=b=c. Since abc=-27, a=b=c=-3. Thus, the minimum value of −abc+ab+ac+bc−a−b−c+1 occurs when a=b=c=-3, rendering the minimum value of (1-a)(1-b)(1-c) as 64, as you assumed.", "+0 # Help 0 89 1 Let a and b be real numbers. Find the maximum value of $$a \\cos \\theta + b \\sin \\theta$$ in terms of a and b. Aug 31, 2019 $$a \\cos(\\theta) + b \\sin(\\theta) = \\\\ \\sqrt{a^2 + b^2}\\sin(\\theta + \\phi),~\\phi = \\tan^{-1}\\left(b,a\\right)\\\\ -1 \\leq \\sin(\\theta+\\phi) \\leq 1\\\\ \\text{The max value is thus \\sqrt{a^2+b^2}}$$", "+ b + c = 15, find the maximum value of abc? Solution: AM ≥ GM $\\frac{{a + b + c}}{3} \\ge \\sqrt[3]{{a \\times b \\times c}}$ Or $\\frac{{15}}{3} \\ge \\sqrt[3]{{a \\times b \\times c}}$ Or, $abc \\le {5^3}$. Hence the maximum value of abc = 125, when a= b = c = 5. Example: For positive real number x, y, and z, If xyz = 64, find the minimum value of x+y+z? Solution: Clearly, the sum is minimum if all the variables are equal, i.e. x = y = z = 4. Hence the minimum value of x + y + z = 4+4+4 = 12. Example: Let a, b and c be nonnegative integers such that a + b+ c = 15. What is the maximum value of a.b.c + a.b + b.c + c.a? Solution: (a+1) (b+1)(c+1) = a.b.c + a.b + b.c + c.a + a + b + c + 1 = a.b.c + a.b + b.c + c.a + 16 Applying AM-GM $\\frac{{{\\rm{(a + 1) }} + {\\rm{(b + 1)}} + {\\rm{(c + 1) }}}}{3} \\ge \\sqrt[3]{{{\\rm{(a + 1) (b + 1)(c + 1) }}}}$ Thus (a+1) (b+1)(c+1) is maximum at 216, which occurs when a = b =c = 5. Therefore a.b.c + a.b + b.c + c.a = 200. ###", "# Find the maximum of the $a+ab+abc+abcd$ Let $m$ be any give postive ream numbers,and $a+b+c+d=m$,where $a,b,c,d\\ge 0$ Find the maximum of the value $$a+ab+abc+abcd$$ try when $m=1$,we have $a+ab+abc+abcd\\le a(b+1)(c+1)(d+1)\\le\\left(\\dfrac{a+b+c+d+3}{4}\\right)^4=1$ when $a=1,b=c=d=0$ when $m=3$, I got the $$a+ab+abc+abcd=a+ab(1+c+cd)\\le a+ab(1+c+d+cd)=a+ab(1+c)(1+d)\\le a+a\\left(\\dfrac{b+1+c+1+d}{3}\\right)^3=a+a\\cdot\\left(\\dfrac{5-a}{3}\\right)^3=4-\\dfrac{1}{27}(a-2)^2(a^2-11a+27)\\le 4$$ when $a=2,b=1,c=d=0$ for other any postive $m$,How to find it? partial derivative? • Yes, the Lagrange multipliers method helps, but the answer depend on $m$ . – Michael Rozenberg Aug 22 '17 at 13:01 • Form a Lagrangian with multiplier $\\lambda$. The first order conditions give $$abc=ab(1+d)=a(1+c+cd)=1+b(1+c+cd)=\\lambda.$$ From here $c=1+d$ and $b=(1+c^2)/c$ while $a = m-b-c-d$, so we may express $a,b,d$ as explicit functions of $c$. This leaves just a univariate optimization over $d$. – encore Aug 22 '17 at 13:01 • I suspect that (a) with $0\\le m \\le 1$ you will achieve the maximum with $b=c=d=0$; (b) with $1\\le m \\le 3$ you will achieve the maximum with $c=d=0$ and $a=b+1$; (c) with $3\\le m \\le \\frac{11}2$ you will achieve the maximum with $d=0$ and $b=c+1$ and $a=b+\\frac{1}{b}$; and (d) $\\frac{11}2 \\le m$ you will achieve the maximum with $c=d+1$ and $b=c+\\frac{1}{c}$ and $a=b+\\frac{1}{bc}$ – Henry Aug 22 '17 at 13:02 • Yes, the answer will naturally depend on $m$. – encore Aug 22 '17 at 13:03 This seems to me a classical problem", "\\leq (a^{2}+b^{2})^{1/2} (c^{2}+d^{2})^{1/2}$ with $c=d=1$. If you restrict $x_i,y_i, i=1.2$ to psoitive values then the maximum is $2$ and it is attained when $x_2=1,x_1=0, y_1=1,y_2=0$/. Sep 21 '19 at 0:03" ]

[ "get $$a+b+c\\leq\\sqrt{\\frac{3}{2}}.$$ $$a+b+c=\\sqrt{\\sum_{cyc}(a^2+2ab)}\\leq\\sqrt{\\sum_{cyc}\\left(a^2+\\frac{1}{2}a^2+\\frac{3}{2}ab\\right)}\\leq\\sqrt{\\frac{3}{2}}.$$ The equality occurs for $$a=b=c$$ and $$\\sum\\limits_{cyc}(a^2+ab)=1,$$ which says that we got a maximal value. $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2=6(a^2+b^2+c^2+ab+bc+ca)$$ and so $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2\\le 6$$ Each of $$(a-b)^2,(b-c)^2,(c-a)^2$$ is non-negative and so $$4(a+b+c)^2\\le 6.$$ Therefore $$a+b+c\\le \\sqrt \\frac{3}{2}$$.", "\\neq 0$, the global maximum $w_{max}$ is $w_2$. Solving $w = w_2$ for $a$: $\\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \\frac{(b+1)(2\\sqrt{b(b+1)}e^2+2be^2+e^2-2\\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$, we have $a = \\sqrt{1+ \\frac{1}{b}}$. Therefore, $e^2-2be-2e+b+1 \\neq 0 \\implies w$ attains maximum at $a = \\sqrt{1+ \\frac{1}{b}}$. In fact, $w$ attains maxima at $a = \\sqrt{1+\\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below: Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have $e_1= -\\sqrt{b(b+1)}+b+1$ or $e_2 = \\sqrt{b(b+1)} + b + 1$. Only $e_1$ is valid (see Exercise-2), When $e = e_1$, $w(\\sqrt{1+\\frac{1}{b}} )- w(a) = - \\frac{(b+1)g(a)}{4 \\sqrt{b(b+1)} (\\sqrt{b(b+1)}+ab) (a\\sqrt{b(b+1)}+b+1)}\\quad(8)$ where $g(a) = (2a^2b+4ab+2b+2a+2)\\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$ Solve quadratic equation $g(a) = 0$ for $a$ yields $a = \\sqrt{1+\\frac{1}{b}}$. The coefficient of $a^2$ in $g(a)$ is $2b\\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3). The implication is that $g(a)$ is a negative quantity when $a \\neq \\sqrt{1 + \\frac{1}{b}}$. Hence, (8) is a positive quantity, i.e., $e^2-2be-2e+b+1 = 0, a \\neq \\sqrt{1+\\frac{1}{b}} \\implies w(\\sqrt{1+\\frac{1}{b}})-w(a) > 0$ We therefore conclude $\\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \\sqrt{1+\\frac{1}{b}}$. Fig. 2 Exercise-1 Prove:$00 \\implies$ $be^2-2b\\sqrt{b(b+1)}e-2b^2e-2be+2b^2\\sqrt{b(b+1)}+2b\\sqrt{b(b+1)}+2b^3+3b^2+b > 0$ Exercise-2 Prove: $b > 0 \\implies 0 <-\\sqrt{b(b+1)} + b +1 <1$ Exercise-3 Prove: $b > 0 \\implies 2b\\sqrt{b(b+1)}-2b^2-b < 0$ # Viva Rocketry! Part 2 Fig. 1 A rocket with $n$ stages is a composition of $n$ single", "# Maximum value of $(ab)(a+b)$ given $a^2 + b^2 = 100$ Can anyone help me with finding the maximum value of $$y = {(ab)(a+b)}$$ With that condition that $$100 = a^2 + b^2$$ and both $a,b$ are positive numbers. Any help appreciated. Thanks. • Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Apr 1 '16 at 9:17 • Use \"Lagrange multipliers\" – SomeOne Apr 1 '16 at 9:17 • The answer might be an irrational one. – TheRandomGuy Apr 1 '16 at 10:00 Since $a,b>0$ and hence $y>0$ you can equivalently maximize $y^2$ which you can write by the given constraint as $$y^2=(ab)^2(a+b)^2=(ab)^2(100+2ab)$$ so that $y^2$ (and hence $y$) is maximized whenever $ab$ is maximized. By the AM-GM inequality $$ab=\\sqrt{(ab)^2}\\overset{AM-GM}\\le\\frac{a^2+b^2}{2}=\\frac{100}{2}=50$$ with equality if $a=b$, see here. Hence $a=b=\\sqrt{50}$ with $$y_{\\max}=\\sqrt{50}^2(\\sqrt{50}+\\sqrt{50})=100\\sqrt{50}$$ To use the method of Lagrange multipliers you can write this problem as \\begin{align}\\max_{a,b} {f(a,b)}&=(ab)(a+b)\\\\\\mbox{s.t. } g(a,b)&=a^2+b^2-100=0\\end{align} The Lagrange function is $$\\mathcal L(a,b,λ)=f(a,b)-λg(a,b)$$ with partial derivatives \\begin{align}\\frac{\\partial \\mathcal L}{\\partial a}&=2ab+b^2-2λa\\\\\\frac{\\partial \\mathcal L}{\\partial b}&=2ab+a^2-2λb\\\\\\frac{\\partial \\mathcal L}{\\partial λ}&=-α^2-b^2+100\\end{align} If $(a_0,b_0)$ is a solution to the problem then there exists $λ_0$ such that $(a_0,b_0,λ_0)$ is a stationary point", "maximum $w_{max}$ is $w_2$. Solving $w = w_2$ for $a$: $\\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \\frac{(b+1)(2\\sqrt{b(b+1)}e^2+2be^2+e^2-2\\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$, we have $a = \\sqrt{1+ \\frac{1}{b}}$. Therefore, $e^2-2be-2e+b+1 \\neq 0 \\implies w$ attains maximum at $a = \\sqrt{1+ \\frac{1}{b}}$. In fact, $w$ attains maxima at $a = \\sqrt{1+\\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below: Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have $e_1= -\\sqrt{b(b+1)}+b+1$ or $e_2 = \\sqrt{b(b+1)} + b + 1$. Only $e_1$ is valid (see Exercise-2), When $e = e_1$, $w(\\sqrt{1+\\frac{1}{b}} )- w(a) = - \\frac{(b+1)g(a)}{4 \\sqrt{b(b+1)} (\\sqrt{b(b+1)}+ab) (a\\sqrt{b(b+1)}+b+1)}\\quad(8)$ where $g(a) = (2a^2b+4ab+2b+2a+2)\\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$ Solve quadratic equation $g(a) = 0$ for $a$ yields $a = \\sqrt{1+\\frac{1}{b}}$. The coefficient of $a^2$ in $g(a)$ is $2b\\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3). The implication is that $g(a)$ is a negative quantity when $a \\neq \\sqrt{1 + \\frac{1}{b}}$. Hence, (8) is a positive quantity, i.e., $e^2-2be-2e+b+1 = 0, a \\neq \\sqrt{1+\\frac{1}{b}} \\implies w(\\sqrt{1+\\frac{1}{b}})-w(a) > 0$ We therefore conclude $\\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \\sqrt{1+\\frac{1}{b}}$. Fig. 2 Exercise-1 Prove:$00 \\implies$ $be^2-2b\\sqrt{b(b+1)}e-2b^2e-2be+2b^2\\sqrt{b(b+1)}+2b\\sqrt{b(b+1)}+2b^3+3b^2+b > 0$ Exercise-2 Prove: $b > 0 \\implies 0 <-\\sqrt{b(b+1)} + b +1 <1$ Exercise-3 Prove: $b > 0 \\implies 2b\\sqrt{b(b+1)}-2b^2-b < 0$ # Prelude to Taylor’s theorem As an application of derivative, we may prove the Binomial theorem that concerns the expansion of $(1+x)^n$", "abc=1)$$\\frac{b+c}{\\sqrt{a}}+\\frac{c+a}{\\sqrt{b}}+\\frac{a+b}{\\sqrt{c}} \\ge \\sqrt{a}+\\sqrt{b}+\\sqrt{c}+3\\sqrt[6]{abc}$$Since equality is attainable for a=b=c, this would show that the desired k is 3. Observe that, in virtue of AM-GM$$\\frac{\\frac{a}{\\sqrt{b}}+\\frac{b}{\\sqrt{a}}+\\frac{b}{\\sqrt{c}... 3 When we saw this problem years ago, the idea was to involve Beatty sequences. IIRC, We couldn't solve it without using Beatty sequences (or its ideas in some form). Here's how the wishful thinking went: Let the sequence be $a_n$. Define $b_n = a_n + n$, which is the complement. IF (and that's a big wishful if) $a_n = \\lfloor \\alpha \\times n \\rfloor$, then $... 2 Comments only: Since you have already known the answer, the following is just hindsight. First of all we know that$\\{ a_n \\}$and$\\{ b_n = a_n+n \\}$is a partition of$\\mathbb N$. On the other hand if a formula of$\\{ a_n \\}$ensures that$\\{a_n\\}$and$\\{a_n+n\\}$partition$\\mathbb N$then we are done (Edit: not quite, we still need$a_{n-1}+n-1>a_n$, ... 3 An alternative way: By the rearrangement inequality we have $$\\frac{b}{\\sqrt{a}}+\\frac{c}{\\sqrt{a}}+\\frac{c}{\\sqrt{b}}+\\frac{a}{\\sqrt{b}}+\\frac{a}{\\sqrt{c}}+\\frac{b}{\\sqrt{c}}\\geq \\frac{a}{\\sqrt{a}}+\\frac{a}{\\sqrt{a}}+\\frac{b}{\\sqrt{b}}+\\frac{b}{\\sqrt{b}}+\\frac{c}{\\sqrt{c}}+\\frac{c}{\\sqrt{c}}=2(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})$$ Now proceed by AM-GM as in the ... 1 By the CRT it’s enough to figure out$F_{2020}$mod$8$and$125$. It’s easy to check that six applications of the linear recurrence relation turn$(0,1)=(F_0,F_1)$into$(F_6,F_7)=(8,13)=5 \\cdot (0,1)$mod$8$. As the multiplicative order of$5$mod$8$is two, it follows that$F_n$mod$8$has a period of$12$. Now note that$2020$is congruent to$4$mod$... 6 By AM-GM $$\\frac{b+c}{\\sqrt{a}}+\\frac{c+a}{\\sqrt{b}}+\\frac{a+b}{\\sqrt{c}} \\ge 2(\\sqrt{\\frac{bc}{a}} + \\sqrt{\\frac{ac}{b}} + \\sqrt{\\frac{ab}{c}} )=2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$$ Now using $x^2+y^2+z^2\\ge", "# If $ab+bc+ca+abc=4$ then $\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\leq3$, via AM-GM Suppose, for positive reals $a$, $b$, $c$, that $$ab+bc+ca+abc=4$$ Prove that $$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\leq3$$ I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \\geq 3\\qquad\\text{and}\\qquad abc \\leq1$$ The exact equation is as follows $$1=\\frac{ab+bc+ca+abc}{4}\\ge\\sqrt[4]{(abc)^3}\\implies 1\\ge abc$$ However, I didn't manage to get any further than this after applying AM-GM to several other inequalities. I'd like a solution for this that utilizes AM-GM only, as I'm very new to inequalities. • Something is missing. If $a=b=c=-2$ the assumption $ab+bc+ca+abc=4$ holds, but the sum of those square roots is $6>3$. Did you forget to include the assumption that $a,b,c$ are positive? Exactly how did you apply AM-GM to the equation? Or, is something else missing given that you refer to the first inequality. – Jyrki Lahtonen Oct 15 '17 at 5:52 • i forgot to write abc are positive reals and i meant first equality. – John Doe 1234 Oct 15 '17 at 6:00 • If the condition is $ab+bc+ca=\\mbox{const}$, the function $\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}$ has a maximum. If the condition is $abc=\\mbox{const}$, the function has a minimum. How do AM-GM equations know at what combination coefficients the minimum-maximum sign changes? – Zhuoran He Oct 15 '17 at 6:12 • It's the same as this math.stackexchange.com/questions/1180443/…, just the condition is", "0 & 1 \\\\ 1 & 0 & 1 \\\\ x_A & \\sqrt{1 - x_A^2} & 1 \\end{pmatrix}\\right| = \\sqrt{1 - x_A^2}\\,.$$ Well, this function has a single critical point respect to which it assumes the maximum value: $$x_A = 0\\,, \\; \\; \\; f(x_A) = 1\\,.$$ 2 triangles Considering two triangles of vertices respectively: $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; A\\left(x_A, \\; \\sqrt{1 - x_A^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; B\\left(x_B, \\; \\sqrt{1 - x_B^2}\\right)$$ with $$-1 \\le x_A \\le 0$$ and $$0 \\le x_B \\le 1$$, the area of the region defined by them is equal to the sum of the areas of the two triangles minus the area of the intersection triangle: $$f(x_A,\\,x_B) := \\sqrt{1 - x_A^2} + \\sqrt{1 - x_B^2} - \\frac{2\\,\\sqrt{1 + x_A}\\,\\sqrt{1 - x_B}}{\\sqrt{1 + x_A}\\,\\sqrt{1 + x_B} + \\sqrt{1 - x_A}\\,\\sqrt{1 - x_B}}\\,.$$ Well, this function has a single critical point respect to which it assumes the maximum value: $$\\left(x_A,\\,x_B\\right) = \\left(-\\frac{\\sqrt{3}-1}{2}\\,,\\frac{\\sqrt{3}-1}{2}\\right), \\; \\; \\; f(x_A,\\,x_B) = \\sqrt{6\\sqrt{3} - 9}\\,.$$ 3 triangles Considering three triangles of vertices respectively: $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; A\\left(x_A, \\; \\sqrt{1 - x_A^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; B\\left(x_B, \\; \\sqrt{1 - x_B^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; C\\left(x_C, \\; \\sqrt{1 -", "# Inequality question: If $a + b + c =1$, what is the minimum value of $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. I've tried AM-HM but it gave $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq 9$ which gives $\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} + 2 (\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca} )\\geq 81$ $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$ = $\\frac{c}{abc} + \\frac{a}{abc} + \\frac{b}{abc}$ = ${\\frac{1}{abc}}$ We know that AM${\\ge}$GM ${\\frac{a+b+c}{3}}$ ${\\ge} \\sqrt[3]{ abc}$ ${\\frac{1}{3}}$ ${\\ge} \\sqrt[3]{ abc}$ ${\\frac{1}{27}}$ ${\\ge} { abc}$ ${\\frac{1}{abc}}$ ${\\ge} 27$ so minimum value is 27 This minimum value is achieved when a=b=c=${\\frac{1}{3}}$ Assuming $a,b,c$ are positive reals, $$\\sum\\limits_{\\textrm{cyc}}\\frac 1{ab}=\\frac 1{abc}\\sum\\limits_{\\textrm{cyc}}a=\\frac 1{abc}\\geq \\frac 1{\\left(\\frac{a+b+c}3\\right)^3}=27$$ with equality when $a=b=c=1/3$ giving $3\\cdot\\dfrac 1{(1/3)^3}=27$ where the penultimate step follows from the AM-GM inequality By Holder $$\\sum_{cyc}\\frac{1}{ab}=\\sum_{cyc}\\frac{1}{ab}\\sum_{cyc}a\\sum_{cyc}b\\geq\\left(\\sum_{cyc}\\sqrt[3]{\\frac{1}{ab}\\cdot a\\cdot b}\\right)^3=\\left(\\sum_{cyc}1\\right)^3=27.$$ The equality occurs for $a=b=c=\\frac{1}{3}$, which says that $27$ is a minimal value.", "# If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\\over{ab}}$ If $a^2+b^2=1$, where $a>0$ and $b>0$, then find the minimum value of $a+b+{1\\over{ab}}$ This can be easily done by calculas but is there any way to do do this by algebra #### Solutions Collecting From Web of \"If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\\over{ab}}$\" Edit: The last line is incorrect. In light of @Lemur’s answer, it will suffice to show that $$a + b \\geq \\sqrt{2}$$ where $a,b > 0, a^2 + b^2 = 1$ is enforced. To see this, note that by AMGM, $ab \\leq \\frac{a^2 + b^2}{2} = \\frac{1}{2}$, and so $$(a + b)^2 = a^2 + b^2 + 2 ab \\geq 1 + 2 ab \\geq 1 + 1 = 2$$ Thus $a + b \\geq \\sqrt{2}$. Another way is to use AM-GM while preserving the point of equality, i.e.: \\begin{align} a+b+\\frac1{ab} &= a+b+\\frac1{2 \\sqrt2 ab} + \\left(1-\\frac1{2\\sqrt2}\\right)\\frac1{ab} \\\\ &\\geqslant \\frac3{\\sqrt2}+\\left(1-\\frac1{2\\sqrt2}\\right)\\frac2{a^2+ b^2} \\\\ &= \\frac3{\\sqrt2}+\\left(1-\\frac1{2\\sqrt2}\\right)\\cdot 2 = 2+\\sqrt{2} \\end{align} $$a + b + \\frac{1}{ab} = a + b + \\frac{ a^2 + b^2}{ab} \\geq a + b + 2$$ I have used AM-GM ineq: $$\\frac{a^2 + b^2}{2} \\geq ab$$ Remark: IT is still left to show that $a+b \\geq \\sqrt{2}$ constrained to $a^2 + b^2 = 1$. See A", "# minimum value of $a^2 + b^2 + \\tfrac{1}{(a + b)^2}$ Let $$a$$ and $$b$$ be positive real numbers. Find the minimum value of $$a^2 + b^2 + \\frac{1}{(a + b)^2}.$$ So I really have no idea how to start with this one. I've tried using AM-GM to try and cancel out the $$(a+b)^2$$: $$\\frac{(a+b)^2-2ab+\\tfrac{1}{(a+b)^2}}{3} \\geq \\sqrt[3]{-2ab}$$ $$a^2+b^2+\\frac{1}{(a+b)^2} \\geq -3\\sqrt[3]{2ab}$$ However this doesn't seem much better.. • Unless I’m missing something, you can trivially bound the expression from below by $0$, which is better than the bound you have given. Mar 30, 2021 at 23:11 • Plugging in $(1,1)$ gives $2.25$, and plugging in $(.5, .5)$ gives $1.5$, and it is bounded below by $0$ Mar 30, 2021 at 23:14 Well, $$(a+b)^2=a^2+b^2+\\color{green}{2ab}\\leq 2(a^2+b^2)$$ (by AM-GM) which in turn gives $$\\frac{1}{(a+b)^2}\\geq\\frac{1}{2(a^2+b^2)}$$ which gives $$a^2+b^2+\\frac{1}{(a+b)^2}\\geq a^2+b^2+\\frac{1}{2(a^2+b^2)}$$ So now the question is reduced to (setting $$t=a^2+b^2>0$$) : $$\\min_{t\\in\\Bbb R^+}(t+\\frac{1}{2t})=\\sqrt2$$ Explaining achievability per Clement's comment. Equality for $$a^2+b^2\\geq2ab$$ as seen in the first line (and there on) happens $$\\iff a=b$$. This is not disrupted later (we did not, for example, divide through by $$a-b$$) so the minimum is attainable. Indeed, the $$t$$ which minimises the given expression is $$t=2^{-\\frac12}\\implies a=b=2^{-\\frac 34}$$. • (+1) To be complete, though, the answer should show that this is achievable (i.e., the inequalities are tight at the", "as the parabola is squeezed between them. $f(0) = 1$ = minimum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(1) = f(-1) = \\sqrt{2}$ = maximum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(0)[1 - (-1)] < \\int_{-1}^{1} \\sqrt{x^2+1} \\, dx < f(1)[1 - (-1] $ $2 < \\int_{-1}^{1} \\sqrt{x^2+1} \\, dx < 2\\sqrt{2} $ 7. Originally Posted by skeeter $f(0) = 1$ = minimum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(1) = f(-1) = \\sqrt{2}$ = maximum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ Sorry but why did you sub in 0 and 1 into f ? 8. Originally Posted by SyNtHeSiS Sorry but why did you sub in 0 and 1 into f ? Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that mf (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(ba) and M(ba), it follows that $m(b-a) \\le \\int_a^b f(x) \\, dx \\le M(b-a)$ if you require further explanation, see your instructor.", "Try $x=-1$. By the way, $$\\frac{\\sqrt{2x^2+1}}{x}=\\sqrt{2+\\frac{1}{x^2}}$$ for $x>0$ and $$\\frac{\\sqrt{2x^2+1}}{x}=-\\sqrt{2+\\frac{1}{x^2}}$$ for $x<0$. The both equalities we can write so: $$\\sqrt{2+\\frac{1}{x^2}}=\\frac{\\sqrt{2x^2+1}}{\\sqrt{x^2}}=\\frac{\\sqrt{2x^2+1}}{|x|}$$ 1) $x>0,$ the equality is OK. 2) $x < 0$, LHS is negative, RHS positive. LHS: $\\dfrac{\\sqrt{2x^2 +1} }{x}=$ $\\dfrac {\\sqrt{x^2(2+\\dfrac{1}{x^2})}}{x} =$ $\\dfrac{|x|\\sqrt{2+\\dfrac{1}{x^2}}}{x} =$ $\\dfrac{|x|}{x}\\sqrt{2+\\dfrac{1}{x^2}}.$ Note: $\\dfrac{|x|}{x} = 1$ for $x>0;$ $\\dfrac{|x|}{x} = -1$ for $x<0.$", "Tag Info 10 Due to symmetry, assume wlog that $a\\geqslant b\\geqslant c\\geqslant d\\geqslant 0$. Since I will use this fact later, I am proving it before the cases: Lemma: We have that $2\\geqslant a+d$. In fact, this follows from Cauchy-Schwarz: $$2= 2\\sqrt{\\frac{a^2+b^2+c^2+d^2}{3}}\\geqslant 2\\sqrt{\\frac{a^2+3d^2}{3}}=\\sqrt{1+\\frac13}\\cdot \\sqrt{a^2+3d^2}\\geqslant a+d... 3 Since a^2+b^2+c^2+d^2 = 3 defines a compact set in \\mathbb{R}^4, f(a,b,c,d)=abcd+3-a-b-c-d will have a global minimum and maximum over that set, that will occur (can be easily justified) in critical points of the Lagrangian$$ L(a,b,c,d,\\lambda) = abcd+3-a-b-c-d -\\lambda(a^2+b^2+c^2+d^2-3) $$So, just compute the critical points, compute the value of ... 1 For any a, b > 0, x=a, y=-b gives \\max(f(a+b), f(a-b)) | \\min(af(-b)+bf(a), -ab). Since f takes positive values, \\max(f(a + b), f(a-b)) | ab. For a=b=1 this shows that f(0)=f(2)=1. For a=b this shows that f(2a) | a^2, and for b=1 this shows that \\max(f(a+1), f(a-1))|a. Combined together these imply that f(x) = 1 for all even x \\... 2 Elementary solution: Suppose n=t-1, putting in we get: A=n^3-3n+1=t^3-3t^2+3 For t=3m we get: A=27m^3-27+3=3[9(m^3-m^2)+1] We can assume m^3-m^2=k so 9k+1|A Then n=3m-1 For example: m=2, \\rightarrow:, p=37, n=5 m=3, \\rightarrow:, p=163, n=8 m=4, \\rightarrow:, p=433, n=11 Hence values of n make an arithmetic progression . 5 Hint: x^3 - 3x + 1 is the minimal polynomial of \\zeta_9 + \\zeta_9^{-1} over", "x_C^2}\\right)$$ with $$-1 \\le x_A \\le -\\frac{1}{3}$$, $$-\\frac{1}{3} \\le x_B \\le \\frac{1}{3}$$ and $$\\frac{1}{3} \\le x_C \\le 1$$, the area of the region defined by them is equal to the sum of the areas of the three triangles minus the area of the two triangles intersection: $$\\small f(x_A,\\,x_B,\\,x_C) := \\sqrt{1 - x_A^2} + \\sqrt{1 - x_B^2} + \\sqrt{1 - x_C^2} \\\\ \\small - \\frac{2\\,\\sqrt{1 + x_A}\\,\\sqrt{1 - x_B}}{\\sqrt{1 + x_A}\\,\\sqrt{1 + x_B} + \\sqrt{1 - x_A}\\,\\sqrt{1 - x_B}} - \\frac{2\\,\\sqrt{1 + x_B}\\,\\sqrt{1 - x_C}}{\\sqrt{1 + x_B}\\,\\sqrt{1 + x_C} + \\sqrt{1 - x_B}\\,\\sqrt{1 - x_C}}\\,.$$ Well, this function has a single critical point respect to which it assumes the maximum value: $$\\small \\left(x_A,\\,x_B,\\,x_C\\right) = \\left(-\\frac{\\sqrt[3]{3\\sqrt{33}+17}-\\sqrt[3]{3\\sqrt{33}-17}-1}{3},\\,0,\\,\\frac{\\sqrt[3]{3\\sqrt{33}+17}-\\sqrt[3]{3\\sqrt{33}-17}-1}{3}\\right), \\\\ \\small f(x_A,\\,x_B,\\,x_C) = 2\\sqrt[3]{\\frac{11\\sqrt{33}+63}{9}} - 2\\sqrt[3]{\\frac{11\\sqrt{33}-63}{9}} - 3\\,.$$ n triangles Based on what has been studied so far, both the algorithm to be followed and the fact that at the computational level a numerical rather than analytical approach is preferable should be clear. In particular, in Wolfram Mathematica, writing: nmax = 30; frames = {}; area = {}; vertices = {}; For[n = 1, n <= nmax, n++, fct = 0; bc = {}; var = {}; For[i = 1, i <= n, i++, j = ToExpression[StringJoin[\"a\", ToString[i]]]; k = ToExpression[StringJoin[\"a\", ToString[i + 1]]]; l = ToExpression[StringJoin[ToString[-1 + 2 (i - 1)/n, InputForm],", "# Maximum value of $a+b+c$ in an inequality Given that $$a$$, $$b$$ and $$c$$ are real positive numbers, find the maximum possible value of $$a+b+c$$, if $$a^2+b^2+c^2+ab+ac+bc\\le1.$$ From the AM-GM theorem, I have $$a^2+b^2+c^2+ab+ac+bc\\geq 6\\sqrt[6]{a^4b^4c^4} = 6\\sqrt[3]{a^2b^2c^2} \\\\ 6\\sqrt[3]{a^2b^2c^2} \\le1 \\\\ a^2b^2c^2 \\le \\frac{1}{216} \\\\ abc \\le \\frac{\\sqrt{6}}{36}$$ However, I don't know where to go from here. \\begin{align}& a^2+b^2+c^2+ab+ac+bc\\le1 \\\\ & (a+b+c)^2 -(ab+bc+ac) \\le 1 \\\\ & (a+b+ c) \\le \\sqrt{1+(ab+bc+ac)} \\quad \\quad \\color{red}{\\text{ (1.)}} \\end{align} Also note that : $$\\frac {(ab+bc+ac)}{3} \\ge \\sqrt[3]{a^2b^2c^2}$$ Assuming you are correct : $${(ab+bc+ac)} \\ge 3 \\cdot\\frac16 \\implies (ab+bc+ac) \\ge \\frac12$$ Hence from $$\\color{red}{\\text{ (1.)}}$$: $$(a+b+c) \\le \\sqrt{1+\\frac12} \\implies \\boxed {\\color {blue}{(a+b+c) \\le \\sqrt {\\frac32}}}$$ • The equality $ab+ac+bc\\geq\\frac{1}{2}$ is wrong. Try $a=1$ and $b=c\\rightarrow0^+$. – Michael Rozenberg Oct 20 at 17:29 Hint: the inequality is equivalent to $$(a+b)^{2} + (b+c)^{2} + (c+a)^{2} \\leq 2.$$ Now put $$x = b+c, y = c+a, z = a +b$$ and then we need to find maximum of $$a + b + c = \\frac{1}{2}(x+y+z)$$ under the condition $$x^{2} + y^{2} + z^{2} \\leq 2$$ and you may use Cauchy-Schwartz inequality. Note that you need to check $$x + y >z, y +z>x, z+x>y$$ for $$(x, y, z)$$ that gives maximum. Notice that $$ab+bc+ca\\leq\\frac{(a+b+c)^2}{3}.$$ Hence \\begin{align} a^2+b^2+c^2+ab+bc+ca&=(a+b+c)^2-(ab+bc+ca)\\\\&\\geq (a+b+c)^2-\\frac{(a+b+c)^2}{3}\\\\&=\\frac{2}{3}(a+b+c)^2. \\end{align} That is $$\\frac{2}{3}(a+b+c)^2\\leq 1,$$ which we", "(3) is valid (see Exercise-1) Hence $\\frac{m_1}{m_2} = -\\frac{(\\sqrt{b}\\sqrt{b+1}-b-1)P}{\\sqrt{b^2+b}P-b P} = \\frac{\\sqrt{b+1}}{\\sqrt{b}} = \\sqrt{1+\\frac{1}{b}}$ The entire process is captured in Fig. 2. Fig. 2 Exercise-1 Given $b>0, 00$, prove: 1. $- \\frac{(\\sqrt{b}\\sqrt{b+1}-b-1)P}{b}>0$ 2. $\\frac{\\sqrt{b^2+b}P-bP}{b}>0$ Exercise-2 From (1), prove the extreme value attained under (2) is a maximum.", "# Can Mathematica Resolve this inequality? I want Mathematica to Indicate this inequality is (hopefully) true for all values that meet the initial conditions: a+b+c=1 and a,b,c are positive real numbers Sqrt[ab]/Sqrt[c + ab] + Sqrt[bc]/Sqrt[a + bc] + Sqrt[ac]/Sqrt[b + ac] <= 1.5 I tried this: Resolve[ForAll[{a, b, c}, a > 0 && b > 0 && c > 0 && a + b + c == 1, Sqrt[ab]/Sqrt[c + ab] + Sqrt[bc]/Sqrt[a + bc] + Sqrt[ac]/Sqrt[b + ac] <= 1.5], Reals] But it keeps on Evaluating.Is there something that i did wrong or Mathematica is unable to resolve this?Please consider that im new to Mathematica. • @Cesareo: This results in {0.,{a->0.333333,b->0.333333,c->0.333333}}. so this is not it. Apr 23 '20 at 17:36 • This result means that $1.5-\\frac{\\sqrt{a b}}{\\sqrt{a b+c}}-\\frac{\\sqrt{a c}}{\\sqrt{a c+b}}-\\frac{\\sqrt{b c}}{\\sqrt{a+b c}}\\ge 0$ and the equality is attained at $a=b=c=\\frac 13$ Apr 23 '20 at 17:45 • @Cesareo: You deleted your comment with your suggestion to try Minimize[{1.5-Sqrt[a*b]/Sqrt[c + a*b] + Sqrt[b*c]/Sqrt[a + b*c] + Sqrt[a*c]/Sqrt[b + a*c] ,a > 0 && b > 0 && c > 0 && a + b + c == 1},{a,b,c}]. You are not right that its result does the job. Upgrade your math. Apr 23 '20 at 18:00 • I deleted it because it wasn't understood. No need", "+1} ge 3 / 2$$ Attempt: Note that from AM-Gm $$frac {a} {b ^ {2} + 1} + frac {b} {c ^ {2} +1} + frac {c} {a ^ {2} +1} ge 3 frac { sqrt (3) {abc}} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}}$$ Now AM-GM again $$a ^ {2} + b ^ {2} + c ^ {2} + 3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} … (1)$$ Then $$a + b + c = 3 ge 3 sqrt (3) {abc} implies 1 ge sqrt (3) {abc}$$, Also $$a ^ {2} + b ^ {2} + c ^ {2} ge 3 sqrt (3) {(abc) ^ {2}}$$ times $$1 ge sqrt (3) {abc}$$ and will get $$a ^ {2} + b ^ {2} + c ^ {2} ge 3 abc … (2)$$ subtract $$(1)$$ With $$(2)$$ and get $$3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} – 3 abc$$ $$3 + 3 abc ge sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}$$ $$frac {3abc} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}} ge 1 – frac { 3}", "Let $a_{1}>0,a_{2}>0$ and $a_{n}=\\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\\{ a_{n}\\}$ converges to $\\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\\{ a_{n}\\}$ converges to $\\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$. My attempt: \\begin{align} a_{n} &= \\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}} \\\\ &= \\frac{2}{\\dfrac{1}{a_{n-1}}+\\dfrac{1}{a_{n-2}}} \\\\ & \\le \\frac{1}{\\sqrt{a_{n-1}a_{n-2}}} \\end{align} I used AM- GM inequality here. I am not able to proceed further. How to solve the question? Please help me. Hint: let $b_n=1/a_n$ then $b_{n+1}=(b_n+b_{n-1})/2 \\iff (b_{n+1}-b_n)=-(b_n-b_{n-1})/2$. Then: $$b_{n+1}-b_n=\\frac{-1}{2}(b_n-b_{n-1})=\\left(\\frac{-1}{2}\\right)^2(b_{n-1}-b_{n-2}) = \\cdots = \\left(\\frac{-1}{2}\\right)^{n-1}(b_{2}-b_{1})$$ Next, telescope: $$b_{n+1} = b_n + \\left(\\frac{-1}{2}\\right)^{n-1}(b_{2}-b_{1}) = b_{n-1} + \\left(\\left(\\frac{-1}{2}\\right)^{n-1}+\\left(\\frac{-1}{2}\\right)^{n-2}\\right)(b_2-b_1) = \\;\\cdots$$ • Can you please expand the answer? I am not able to think. – Unknown x Aug 10 '17 at 4:39 • I got the answer. Your hint is really helpful – Unknown x Aug 10 '17 at 5:13 • @M.Narayanan Thank you, and glad you worked it out. (I have always believed that a hint that you turn into a proof is more helpful and better remembered than a complete solution upfront.) – dxiv Aug 10 '17 at 5:19 You are close. From $a_{n} =\\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}$, $\\dfrac1{a_{n}} =\\dfrac{a_{n-1}+a_{n-2}}{2a_{n-1}a_{n-2}} =\\dfrac12(\\frac1{a_{n-2}}+\\frac1{a_{n-1}})$. Letting $\\dfrac1{a_n} =b_n$, this becomes $b_n =\\frac12(b_{n-1}+b_{n-2})$. You should be able to solve this. (I think I did in one of my answers.) Following in the footsteps of @martycohen elsewhere on this page, a generalized version of the sequence in the OP can be constructed. Thus,", "MNP]\\le \\frac{1}{4}[\\Delta ABC].$ This follows from the inequality for $a,b,c\\gt 0:$ $(a+b)(b+c)(c+a)\\ge 8abc,$ with equality only when $a=b=c.$ Indeed, by the AM-GM inequality $\\displaystyle\\frac{a+b}{2}\\frac{b+c}{2}\\frac{c+a}{2}\\ge \\sqrt{ab}\\sqrt{bc}\\sqrt{ac}=abc.$ It follows that the only time when the area of a cevian triangle equals $\\displaystyle\\frac{1}{4}[\\Delta ABC]$ is when $D=(1:1:1),\\;$ i.e., when $D$ is the centroid of $\\Delta ABC.$" ]

[ "\\neq 0$, the global maximum $w_{max}$ is $w_2$. Solving $w = w_2$ for $a$: $\\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \\frac{(b+1)(2\\sqrt{b(b+1)}e^2+2be^2+e^2-2\\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$, we have $a = \\sqrt{1+ \\frac{1}{b}}$. Therefore, $e^2-2be-2e+b+1 \\neq 0 \\implies w$ attains maximum at $a = \\sqrt{1+ \\frac{1}{b}}$. In fact, $w$ attains maxima at $a = \\sqrt{1+\\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below: Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have $e_1= -\\sqrt{b(b+1)}+b+1$ or $e_2 = \\sqrt{b(b+1)} + b + 1$. Only $e_1$ is valid (see Exercise-2), When $e = e_1$, $w(\\sqrt{1+\\frac{1}{b}} )- w(a) = - \\frac{(b+1)g(a)}{4 \\sqrt{b(b+1)} (\\sqrt{b(b+1)}+ab) (a\\sqrt{b(b+1)}+b+1)}\\quad(8)$ where $g(a) = (2a^2b+4ab+2b+2a+2)\\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$ Solve quadratic equation $g(a) = 0$ for $a$ yields $a = \\sqrt{1+\\frac{1}{b}}$. The coefficient of $a^2$ in $g(a)$ is $2b\\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3). The implication is that $g(a)$ is a negative quantity when $a \\neq \\sqrt{1 + \\frac{1}{b}}$. Hence, (8) is a positive quantity, i.e., $e^2-2be-2e+b+1 = 0, a \\neq \\sqrt{1+\\frac{1}{b}} \\implies w(\\sqrt{1+\\frac{1}{b}})-w(a) > 0$ We therefore conclude $\\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \\sqrt{1+\\frac{1}{b}}$. Fig. 2 Exercise-1 Prove:$00 \\implies$ $be^2-2b\\sqrt{b(b+1)}e-2b^2e-2be+2b^2\\sqrt{b(b+1)}+2b\\sqrt{b(b+1)}+2b^3+3b^2+b > 0$ Exercise-2 Prove: $b > 0 \\implies 0 <-\\sqrt{b(b+1)} + b +1 <1$ Exercise-3 Prove: $b > 0 \\implies 2b\\sqrt{b(b+1)}-2b^2-b < 0$ # Viva Rocketry! Part 2 Fig. 1 A rocket with $n$ stages is a composition of $n$ single", "maximum $w_{max}$ is $w_2$. Solving $w = w_2$ for $a$: $\\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \\frac{(b+1)(2\\sqrt{b(b+1)}e^2+2be^2+e^2-2\\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$, we have $a = \\sqrt{1+ \\frac{1}{b}}$. Therefore, $e^2-2be-2e+b+1 \\neq 0 \\implies w$ attains maximum at $a = \\sqrt{1+ \\frac{1}{b}}$. In fact, $w$ attains maxima at $a = \\sqrt{1+\\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below: Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have $e_1= -\\sqrt{b(b+1)}+b+1$ or $e_2 = \\sqrt{b(b+1)} + b + 1$. Only $e_1$ is valid (see Exercise-2), When $e = e_1$, $w(\\sqrt{1+\\frac{1}{b}} )- w(a) = - \\frac{(b+1)g(a)}{4 \\sqrt{b(b+1)} (\\sqrt{b(b+1)}+ab) (a\\sqrt{b(b+1)}+b+1)}\\quad(8)$ where $g(a) = (2a^2b+4ab+2b+2a+2)\\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$ Solve quadratic equation $g(a) = 0$ for $a$ yields $a = \\sqrt{1+\\frac{1}{b}}$. The coefficient of $a^2$ in $g(a)$ is $2b\\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3). The implication is that $g(a)$ is a negative quantity when $a \\neq \\sqrt{1 + \\frac{1}{b}}$. Hence, (8) is a positive quantity, i.e., $e^2-2be-2e+b+1 = 0, a \\neq \\sqrt{1+\\frac{1}{b}} \\implies w(\\sqrt{1+\\frac{1}{b}})-w(a) > 0$ We therefore conclude $\\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \\sqrt{1+\\frac{1}{b}}$. Fig. 2 Exercise-1 Prove:$00 \\implies$ $be^2-2b\\sqrt{b(b+1)}e-2b^2e-2be+2b^2\\sqrt{b(b+1)}+2b\\sqrt{b(b+1)}+2b^3+3b^2+b > 0$ Exercise-2 Prove: $b > 0 \\implies 0 <-\\sqrt{b(b+1)} + b +1 <1$ Exercise-3 Prove: $b > 0 \\implies 2b\\sqrt{b(b+1)}-2b^2-b < 0$ # Prelude to Taylor’s theorem As an application of derivative, we may prove the Binomial theorem that concerns the expansion of $(1+x)^n$", "# If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\\over{ab}}$ If $a^2+b^2=1$, where $a>0$ and $b>0$, then find the minimum value of $a+b+{1\\over{ab}}$ This can be easily done by calculas but is there any way to do do this by algebra #### Solutions Collecting From Web of \"If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\\over{ab}}$\" Edit: The last line is incorrect. In light of @Lemur’s answer, it will suffice to show that $$a + b \\geq \\sqrt{2}$$ where $a,b > 0, a^2 + b^2 = 1$ is enforced. To see this, note that by AMGM, $ab \\leq \\frac{a^2 + b^2}{2} = \\frac{1}{2}$, and so $$(a + b)^2 = a^2 + b^2 + 2 ab \\geq 1 + 2 ab \\geq 1 + 1 = 2$$ Thus $a + b \\geq \\sqrt{2}$. Another way is to use AM-GM while preserving the point of equality, i.e.: \\begin{align} a+b+\\frac1{ab} &= a+b+\\frac1{2 \\sqrt2 ab} + \\left(1-\\frac1{2\\sqrt2}\\right)\\frac1{ab} \\\\ &\\geqslant \\frac3{\\sqrt2}+\\left(1-\\frac1{2\\sqrt2}\\right)\\frac2{a^2+ b^2} \\\\ &= \\frac3{\\sqrt2}+\\left(1-\\frac1{2\\sqrt2}\\right)\\cdot 2 = 2+\\sqrt{2} \\end{align} $$a + b + \\frac{1}{ab} = a + b + \\frac{ a^2 + b^2}{ab} \\geq a + b + 2$$ I have used AM-GM ineq: $$\\frac{a^2 + b^2}{2} \\geq ab$$ Remark: IT is still left to show that $a+b \\geq \\sqrt{2}$ constrained to $a^2 + b^2 = 1$. See A", "get $$a+b+c\\leq\\sqrt{\\frac{3}{2}}.$$ $$a+b+c=\\sqrt{\\sum_{cyc}(a^2+2ab)}\\leq\\sqrt{\\sum_{cyc}\\left(a^2+\\frac{1}{2}a^2+\\frac{3}{2}ab\\right)}\\leq\\sqrt{\\frac{3}{2}}.$$ The equality occurs for $$a=b=c$$ and $$\\sum\\limits_{cyc}(a^2+ab)=1,$$ which says that we got a maximal value. $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2=6(a^2+b^2+c^2+ab+bc+ca)$$ and so $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2\\le 6$$ Each of $$(a-b)^2,(b-c)^2,(c-a)^2$$ is non-negative and so $$4(a+b+c)^2\\le 6.$$ Therefore $$a+b+c\\le \\sqrt \\frac{3}{2}$$.", "# Maximum value of $(ab)(a+b)$ given $a^2 + b^2 = 100$ Can anyone help me with finding the maximum value of $$y = {(ab)(a+b)}$$ With that condition that $$100 = a^2 + b^2$$ and both $a,b$ are positive numbers. Any help appreciated. Thanks. • Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Apr 1 '16 at 9:17 • Use \"Lagrange multipliers\" – SomeOne Apr 1 '16 at 9:17 • The answer might be an irrational one. – TheRandomGuy Apr 1 '16 at 10:00 Since $a,b>0$ and hence $y>0$ you can equivalently maximize $y^2$ which you can write by the given constraint as $$y^2=(ab)^2(a+b)^2=(ab)^2(100+2ab)$$ so that $y^2$ (and hence $y$) is maximized whenever $ab$ is maximized. By the AM-GM inequality $$ab=\\sqrt{(ab)^2}\\overset{AM-GM}\\le\\frac{a^2+b^2}{2}=\\frac{100}{2}=50$$ with equality if $a=b$, see here. Hence $a=b=\\sqrt{50}$ with $$y_{\\max}=\\sqrt{50}^2(\\sqrt{50}+\\sqrt{50})=100\\sqrt{50}$$ To use the method of Lagrange multipliers you can write this problem as \\begin{align}\\max_{a,b} {f(a,b)}&=(ab)(a+b)\\\\\\mbox{s.t. } g(a,b)&=a^2+b^2-100=0\\end{align} The Lagrange function is $$\\mathcal L(a,b,λ)=f(a,b)-λg(a,b)$$ with partial derivatives \\begin{align}\\frac{\\partial \\mathcal L}{\\partial a}&=2ab+b^2-2λa\\\\\\frac{\\partial \\mathcal L}{\\partial b}&=2ab+a^2-2λb\\\\\\frac{\\partial \\mathcal L}{\\partial λ}&=-α^2-b^2+100\\end{align} If $(a_0,b_0)$ is a solution to the problem then there exists $λ_0$ such that $(a_0,b_0,λ_0)$ is a stationary point", "# minimum value of $a^2 + b^2 + \\tfrac{1}{(a + b)^2}$ Let $$a$$ and $$b$$ be positive real numbers. Find the minimum value of $$a^2 + b^2 + \\frac{1}{(a + b)^2}.$$ So I really have no idea how to start with this one. I've tried using AM-GM to try and cancel out the $$(a+b)^2$$: $$\\frac{(a+b)^2-2ab+\\tfrac{1}{(a+b)^2}}{3} \\geq \\sqrt[3]{-2ab}$$ $$a^2+b^2+\\frac{1}{(a+b)^2} \\geq -3\\sqrt[3]{2ab}$$ However this doesn't seem much better.. • Unless I’m missing something, you can trivially bound the expression from below by $0$, which is better than the bound you have given. Mar 30, 2021 at 23:11 • Plugging in $(1,1)$ gives $2.25$, and plugging in $(.5, .5)$ gives $1.5$, and it is bounded below by $0$ Mar 30, 2021 at 23:14 Well, $$(a+b)^2=a^2+b^2+\\color{green}{2ab}\\leq 2(a^2+b^2)$$ (by AM-GM) which in turn gives $$\\frac{1}{(a+b)^2}\\geq\\frac{1}{2(a^2+b^2)}$$ which gives $$a^2+b^2+\\frac{1}{(a+b)^2}\\geq a^2+b^2+\\frac{1}{2(a^2+b^2)}$$ So now the question is reduced to (setting $$t=a^2+b^2>0$$) : $$\\min_{t\\in\\Bbb R^+}(t+\\frac{1}{2t})=\\sqrt2$$ Explaining achievability per Clement's comment. Equality for $$a^2+b^2\\geq2ab$$ as seen in the first line (and there on) happens $$\\iff a=b$$. This is not disrupted later (we did not, for example, divide through by $$a-b$$) so the minimum is attainable. Indeed, the $$t$$ which minimises the given expression is $$t=2^{-\\frac12}\\implies a=b=2^{-\\frac 34}$$. • (+1) To be complete, though, the answer should show that this is achievable (i.e., the inequalities are tight at the", "# Inequality question: If $a + b + c =1$, what is the minimum value of $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. I've tried AM-HM but it gave $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq 9$ which gives $\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} + 2 (\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca} )\\geq 81$ $\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$ = $\\frac{c}{abc} + \\frac{a}{abc} + \\frac{b}{abc}$ = ${\\frac{1}{abc}}$ We know that AM${\\ge}$GM ${\\frac{a+b+c}{3}}$ ${\\ge} \\sqrt[3]{ abc}$ ${\\frac{1}{3}}$ ${\\ge} \\sqrt[3]{ abc}$ ${\\frac{1}{27}}$ ${\\ge} { abc}$ ${\\frac{1}{abc}}$ ${\\ge} 27$ so minimum value is 27 This minimum value is achieved when a=b=c=${\\frac{1}{3}}$ Assuming $a,b,c$ are positive reals, $$\\sum\\limits_{\\textrm{cyc}}\\frac 1{ab}=\\frac 1{abc}\\sum\\limits_{\\textrm{cyc}}a=\\frac 1{abc}\\geq \\frac 1{\\left(\\frac{a+b+c}3\\right)^3}=27$$ with equality when $a=b=c=1/3$ giving $3\\cdot\\dfrac 1{(1/3)^3}=27$ where the penultimate step follows from the AM-GM inequality By Holder $$\\sum_{cyc}\\frac{1}{ab}=\\sum_{cyc}\\frac{1}{ab}\\sum_{cyc}a\\sum_{cyc}b\\geq\\left(\\sum_{cyc}\\sqrt[3]{\\frac{1}{ab}\\cdot a\\cdot b}\\right)^3=\\left(\\sum_{cyc}1\\right)^3=27.$$ The equality occurs for $a=b=c=\\frac{1}{3}$, which says that $27$ is a minimal value.", "# How can I get exact maximal value of this expression? I am trying to find the exact maximum value of the expression $$E= \\sqrt{5 a^2+a (4 b-2 c)+2b^2+4 b c+5 c^2}+\\sqrt{2a^2+a (2 b+2 c)+2 b^2-2 bc+2 c^2}+\\sqrt{26 a^2+a (10c-2 b)+26 b^2+10 b c+2 c^2} ,$$ where $$a^2 + b^2 + c^2 = 1$$ and $$a, b, c > 0$$. I know that, the answer is $$4\\sqrt{3} + \\sqrt{6} .$$ When I tried NMaximize[{(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] + Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]), a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0}, {a, b, c}] {9.37769, {a -> 0.801784, b -> 0.534523, c -> 0.267261}} When I tried, Maximize[{(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] + Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]), a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0}, {a, b, c}] It's take about 3 minutes, I could't get the answer. How can I get exact maximize value of that expression? Writing: rad1 = Sqrt[5", "0 & 1 \\\\ 1 & 0 & 1 \\\\ x_A & \\sqrt{1 - x_A^2} & 1 \\end{pmatrix}\\right| = \\sqrt{1 - x_A^2}\\,.$$ Well, this function has a single critical point respect to which it assumes the maximum value: $$x_A = 0\\,, \\; \\; \\; f(x_A) = 1\\,.$$ 2 triangles Considering two triangles of vertices respectively: $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; A\\left(x_A, \\; \\sqrt{1 - x_A^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; B\\left(x_B, \\; \\sqrt{1 - x_B^2}\\right)$$ with $$-1 \\le x_A \\le 0$$ and $$0 \\le x_B \\le 1$$, the area of the region defined by them is equal to the sum of the areas of the two triangles minus the area of the intersection triangle: $$f(x_A,\\,x_B) := \\sqrt{1 - x_A^2} + \\sqrt{1 - x_B^2} - \\frac{2\\,\\sqrt{1 + x_A}\\,\\sqrt{1 - x_B}}{\\sqrt{1 + x_A}\\,\\sqrt{1 + x_B} + \\sqrt{1 - x_A}\\,\\sqrt{1 - x_B}}\\,.$$ Well, this function has a single critical point respect to which it assumes the maximum value: $$\\left(x_A,\\,x_B\\right) = \\left(-\\frac{\\sqrt{3}-1}{2}\\,,\\frac{\\sqrt{3}-1}{2}\\right), \\; \\; \\; f(x_A,\\,x_B) = \\sqrt{6\\sqrt{3} - 9}\\,.$$ 3 triangles Considering three triangles of vertices respectively: $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; A\\left(x_A, \\; \\sqrt{1 - x_A^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; B\\left(x_B, \\; \\sqrt{1 - x_B^2}\\right)$$ $$P(-1,\\,0)\\,, \\; \\; \\; Q(1,\\,0)\\,, \\; \\; \\; C\\left(x_C, \\; \\sqrt{1 -", "# If $ab+bc+ca+abc=4$ then $\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\leq3$, via AM-GM Suppose, for positive reals $a$, $b$, $c$, that $$ab+bc+ca+abc=4$$ Prove that $$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\leq3$$ I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \\geq 3\\qquad\\text{and}\\qquad abc \\leq1$$ The exact equation is as follows $$1=\\frac{ab+bc+ca+abc}{4}\\ge\\sqrt[4]{(abc)^3}\\implies 1\\ge abc$$ However, I didn't manage to get any further than this after applying AM-GM to several other inequalities. I'd like a solution for this that utilizes AM-GM only, as I'm very new to inequalities. • Something is missing. If $a=b=c=-2$ the assumption $ab+bc+ca+abc=4$ holds, but the sum of those square roots is $6>3$. Did you forget to include the assumption that $a,b,c$ are positive? Exactly how did you apply AM-GM to the equation? Or, is something else missing given that you refer to the first inequality. – Jyrki Lahtonen Oct 15 '17 at 5:52 • i forgot to write abc are positive reals and i meant first equality. – John Doe 1234 Oct 15 '17 at 6:00 • If the condition is $ab+bc+ca=\\mbox{const}$, the function $\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}$ has a maximum. If the condition is $abc=\\mbox{const}$, the function has a minimum. How do AM-GM equations know at what combination coefficients the minimum-maximum sign changes? – Zhuoran He Oct 15 '17 at 6:12 • It's the same as this math.stackexchange.com/questions/1180443/…, just the condition is", "affect the arguments. However, I know that for $a\\le b\\le c$ the inequality is weak, because if $a\\le b\\le c$ then $f(a,b,c) \\le f(b,a,c)$ (this can easily be verified). In other words, if the inequality is true for $b\\le a \\le c$ (or $a\\ge b \\ge c$), then it is also true for $a\\le b\\le c$, but the reverse direction is not true. – Khue Apr 2 at 21:44 One more remark: you found the maximum at $y=0^+$ and $x\\approx 0.1547$, but these values clearly violate the constraint $x\\leq y \\leq 1/2$. By the way, you might be interested in the closed-form values of the numerical values that you found. Here they are: $$x=\\frac{2+2\\sqrt{3}}{3+2\\sqrt{3}} \\approx 0.845, \\quad y =\\frac{1}{3+2\\sqrt{3}} \\approx 0.1547$$ and the maximum value is $$\\frac{2\\sqrt{3}-2}{\\sqrt{2\\sqrt{3}}}+\\sqrt{-1+\\frac{2}{\\sqrt{3}}} \\approx 1.179.$$ These values were totally found by hand (+ a pen and a scrap paper :D). – Khue Apr 2 at 22:05 As I said the inequality is not symmetric. It is only cyclic, not symmetric (clearly $f(a,b,c)\\neq f(b,a,c)$), thus if we assume $a\\le b\\le c$ then generality is lost. Your arguments above even prove that. Indeed, according to your arguments, if $a\\le b\\le c$ then the maximum value of $f(a,b,c)$ is only $1.115$, i.e. $f(a,b,c)$ will never reach its true maximum value $1.179$, which can be achieved only in", "Find the minimum value of $\\sum_{cyc}\\frac{1}{\\sqrt{a^2 - ab + 3b^2 + 1}}$ where $a, b, c > 0$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\le 3$. $$a$$, $$b$$ and $$c$$ are positives such that $$\\dfrac{1}{a} + \\dfrac{1}{b} + \\dfrac{1}{c} \\le 3$$. Calculate the maximum value of $$\\large \\frac{1}{\\sqrt{a^2 - ab + 3b^2 + 1}} + \\frac{1}{\\sqrt{b^2 - bc + 3c^2 + 1}} + \\frac{1}{\\sqrt{c^2 - ca + 3a^2 + 1}}$$ I want to ask if there are any other solutions that are more practical, please. This was taken directly from an exam I recently took. I have posted a solution below if anyone to check out. We can use also the following way. By C-S and Tangent Line method we obtain: $$\\sum_{cyc}\\frac{1}{\\sqrt{a^2-ab+3b^2+1}}=\\sum_{cyc}\\frac{2}{\\sqrt{(3+1)(a^2-ab+3b^2+1}}\\leq$$ $$\\leq2\\sum_{cyc}\\frac{1}{\\sqrt{3(a^2-ab+3b^2)}+1}\\leq\\frac{2}{(3+1)^2}\\sum_{cyc}\\left(\\frac{3^2}{\\sqrt{a^2-ab+3b^2}}+\\frac{1^2}{1}\\right)=$$ $$=\\frac{3}{8}+\\frac{9}{8}\\sum_{cyc}\\left(\\frac{1}{\\sqrt{3(a^2-ab+3b^2)}}-\\frac{1}{18a}-\\frac{5}{18b}\\right)+\\frac{9}{8}\\sum_{cyc}\\left(\\frac{1}{18a}+\\frac{5}{18b}\\right)\\leq$$ $$\\leq\\frac{3}{8}+\\frac{3}{8}\\sum_{cyc}\\frac{1}{a}\\leq\\frac{3}{8}+\\frac{9}{8}=\\frac{3}{2}.$$ The equality occurs for $$a=b=c=1,$$ which says that we got a maximal value. I got $$\\frac{1}{\\sqrt{3(a^2-ab+3b^2)}}\\leq\\frac{1}{18a}+\\frac{5}{18b}$$ by the following way. Let $$a=xb$$. Now, we'll choose $$k$$ and $$m$$ such that a graph of $$g(x)=\\frac{k}{x}+m$$ is a tangent to the graph $$f(x)=\\frac{1}{\\sqrt{3(x^2-x+3)}}$$ for $$x=1$$, for which we need $$f(1)=g(1)$$ and $$f'(1)=g'(1).$$ After solving of this system we obtain $$k=\\frac{1}{18}$$ and $$m=\\frac{5}{18}.$$ Now, easy to show that the inequality $$\\frac{1}{\\sqrt{3(x^2-x+3)}}\\leq\\frac{1}{18x}+\\frac{5}{18}$$ is true for all $$x>0$$. • What is the Tangent Line Method? I have never heard of that. – Lê Thành Đạt Jun 3 '19", "# Maximum value of $a+b+c$ in an inequality Given that $$a$$, $$b$$ and $$c$$ are real positive numbers, find the maximum possible value of $$a+b+c$$, if $$a^2+b^2+c^2+ab+ac+bc\\le1.$$ From the AM-GM theorem, I have $$a^2+b^2+c^2+ab+ac+bc\\geq 6\\sqrt[6]{a^4b^4c^4} = 6\\sqrt[3]{a^2b^2c^2} \\\\ 6\\sqrt[3]{a^2b^2c^2} \\le1 \\\\ a^2b^2c^2 \\le \\frac{1}{216} \\\\ abc \\le \\frac{\\sqrt{6}}{36}$$ However, I don't know where to go from here. \\begin{align}& a^2+b^2+c^2+ab+ac+bc\\le1 \\\\ & (a+b+c)^2 -(ab+bc+ac) \\le 1 \\\\ & (a+b+ c) \\le \\sqrt{1+(ab+bc+ac)} \\quad \\quad \\color{red}{\\text{ (1.)}} \\end{align} Also note that : $$\\frac {(ab+bc+ac)}{3} \\ge \\sqrt[3]{a^2b^2c^2}$$ Assuming you are correct : $${(ab+bc+ac)} \\ge 3 \\cdot\\frac16 \\implies (ab+bc+ac) \\ge \\frac12$$ Hence from $$\\color{red}{\\text{ (1.)}}$$: $$(a+b+c) \\le \\sqrt{1+\\frac12} \\implies \\boxed {\\color {blue}{(a+b+c) \\le \\sqrt {\\frac32}}}$$ • The equality $ab+ac+bc\\geq\\frac{1}{2}$ is wrong. Try $a=1$ and $b=c\\rightarrow0^+$. – Michael Rozenberg Oct 20 at 17:29 Hint: the inequality is equivalent to $$(a+b)^{2} + (b+c)^{2} + (c+a)^{2} \\leq 2.$$ Now put $$x = b+c, y = c+a, z = a +b$$ and then we need to find maximum of $$a + b + c = \\frac{1}{2}(x+y+z)$$ under the condition $$x^{2} + y^{2} + z^{2} \\leq 2$$ and you may use Cauchy-Schwartz inequality. Note that you need to check $$x + y >z, y +z>x, z+x>y$$ for $$(x, y, z)$$ that gives maximum. Notice that $$ab+bc+ca\\leq\\frac{(a+b+c)^2}{3}.$$ Hence \\begin{align} a^2+b^2+c^2+ab+bc+ca&=(a+b+c)^2-(ab+bc+ca)\\\\&\\geq (a+b+c)^2-\\frac{(a+b+c)^2}{3}\\\\&=\\frac{2}{3}(a+b+c)^2. \\end{align} That is $$\\frac{2}{3}(a+b+c)^2\\leq 1,$$ which we", "+1} ge 3 / 2$$ Attempt: Note that from AM-Gm $$frac {a} {b ^ {2} + 1} + frac {b} {c ^ {2} +1} + frac {c} {a ^ {2} +1} ge 3 frac { sqrt (3) {abc}} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}}$$ Now AM-GM again $$a ^ {2} + b ^ {2} + c ^ {2} + 3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} … (1)$$ Then $$a + b + c = 3 ge 3 sqrt (3) {abc} implies 1 ge sqrt (3) {abc}$$, Also $$a ^ {2} + b ^ {2} + c ^ {2} ge 3 sqrt (3) {(abc) ^ {2}}$$ times $$1 ge sqrt (3) {abc}$$ and will get $$a ^ {2} + b ^ {2} + c ^ {2} ge 3 abc … (2)$$ subtract $$(1)$$ With $$(2)$$ and get $$3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} – 3 abc$$ $$3 + 3 abc ge sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}$$ $$frac {3abc} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}} ge 1 – frac { 3}", "first post? 7. Alpha does not change anything, it is just the difference between the LHS and the RHS 8. Originally Posted by mestmoha Alpha does not change anything, it is just the difference between the LHS and the RHS I think I found the solution. Please, check it for me: $\\frac{\\sin A}{a}=\\frac{\\sin B}{b}=\\frac{\\sin C}{c}$ $\\frac{{\\sin}^2 A}{{a}^2}=\\frac{\\sin B \\sin C}{bc}$ $\\sin A + \\sin B \\sin C= \\Sin A (1 + \\frac{bc}{{a}^2} \\sin A)$ For any A, and a $\\sin A + \\sin B \\sin C$ is maximum if and only if b=c (you can use the area= 1/2 bc sin A to approach it). hence for a triangle with b=c: $\\sin A + \\sin B \\sin C= \\sin A + \\frac{1}{2} \\cos A +\\frac{1}{2}$ if: $x=\\sin A \\quad then \\quad f(x) =x + \\frac{1}{2}\\sqrt{1-x^2} + \\frac {1} {2}$ by using the derrivative of f(x), and setting f'(x)=0 we get: $x=\\sin A= \\frac{2}{\\sqrt{5}}$ Then the maximum value of f(x) $\\sin A + \\sin B \\sin C \\quad$) is $\\frac{1+\\sqrt{5}}{2}$ now we proved that for a triangle ABC $\\sin A + \\sin B \\ sin C \\leq \\frac{1 + \\sqrt{5}}{2}$ Let's consider a function g(x) such that: $g(x)=-x^2 + x +1$ The roots of g(x) are : $\\frac{1-\\sqrt{5}}{2} \\quad and \\quad \\frac{1+\\sqrt{5}} {2}$ Hence: $for \\quad every \\quad x", "# Find the minimum value of $\\sqrt{\\frac{x^2}{y^2} + \\frac{y^2}{x^2} + 2} + \\frac{\\sqrt{xy}}{x + y}$ where $x, y > 0$. Given that $$x$$ and $$y$$ are positives, find the minimum value of $$\\large \\sqrt{\\frac{x^2}{y^2} + \\frac{y^2}{x^2} + 2} + \\frac{\\sqrt{xy}}{x + y}$$ We have that $$\\sqrt{\\dfrac{x^2}{y^2} + \\dfrac{y^2}{x^2} + 2} = \\sqrt{\\left(\\dfrac{x}{y}\\right)^2 + \\left(\\dfrac{y}{x}\\right)^2 + 2 \\cdot \\dfrac{x}{y} \\dfrac{y}{x}}$$ $$= \\sqrt{\\left(\\dfrac{x}{y} + \\dfrac{y}{x}\\right)^2} = \\dfrac{x}{y} + \\dfrac{y}{x} = \\dfrac{x^2 + y^2}{xy}$$. Furthermore, $$2(x^2 + y^2) \\ge (x + y)^2$$ $$\\implies \\dfrac{x^2 + y^2}{xy} \\ge \\dfrac{(x + y)^2}{2xy} \\implies \\sqrt{\\dfrac{x^2}{y^2} + \\dfrac{y^2}{x^2} + 2} \\ge \\dfrac{(x + y)^2}{2xy}$$. Using the AM-GM inequality, we have that $$\\frac{(x + y)^2}{2xy} + 2 \\cdot \\frac{\\sqrt{xy}}{2(x + y)} \\ge 3 \\cdot \\sqrt[3]{\\frac{(x + y)^2}{2xy} \\left[\\frac{\\sqrt{xy}}{2(x + y)}\\right]^2} = \\frac{3}{2}$$ $$\\implies \\sqrt{\\dfrac{x^2}{y^2} + \\dfrac{y^2}{x^2} + 2} + \\dfrac{\\sqrt{xy}}{x + y} \\ge \\dfrac{3}{2}$$. But the equality sign happens when $$x = y = 0$$, which contradicts the fact that $$x, y > 0$$. • Equality happens when $x=y$, not necessarily $=0$ – Hagen von Eitzen May 30 '19 at 10:35 • No, if you solve all of the condition, you have $x = y$ and $x + y - \\sqrt{xy} = 0$, which is only possible if $x = y = 0$. – Lê Thành Đạt May 30 '19 at 10:37 • $\\sqrt{\\frac{x^2}{y^2}+\\frac{y^2}{x^2}+2}\\ge \\sqrt{2+2}=2$. $\\frac{3}{2}$", "abc=1)$$\\frac{b+c}{\\sqrt{a}}+\\frac{c+a}{\\sqrt{b}}+\\frac{a+b}{\\sqrt{c}} \\ge \\sqrt{a}+\\sqrt{b}+\\sqrt{c}+3\\sqrt[6]{abc}$$Since equality is attainable for a=b=c, this would show that the desired k is 3. Observe that, in virtue of AM-GM$$\\frac{\\frac{a}{\\sqrt{b}}+\\frac{b}{\\sqrt{a}}+\\frac{b}{\\sqrt{c}... 3 When we saw this problem years ago, the idea was to involve Beatty sequences. IIRC, We couldn't solve it without using Beatty sequences (or its ideas in some form). Here's how the wishful thinking went: Let the sequence be $a_n$. Define $b_n = a_n + n$, which is the complement. IF (and that's a big wishful if) $a_n = \\lfloor \\alpha \\times n \\rfloor$, then $... 2 Comments only: Since you have already known the answer, the following is just hindsight. First of all we know that$\\{ a_n \\}$and$\\{ b_n = a_n+n \\}$is a partition of$\\mathbb N$. On the other hand if a formula of$\\{ a_n \\}$ensures that$\\{a_n\\}$and$\\{a_n+n\\}$partition$\\mathbb N$then we are done (Edit: not quite, we still need$a_{n-1}+n-1>a_n$, ... 3 An alternative way: By the rearrangement inequality we have $$\\frac{b}{\\sqrt{a}}+\\frac{c}{\\sqrt{a}}+\\frac{c}{\\sqrt{b}}+\\frac{a}{\\sqrt{b}}+\\frac{a}{\\sqrt{c}}+\\frac{b}{\\sqrt{c}}\\geq \\frac{a}{\\sqrt{a}}+\\frac{a}{\\sqrt{a}}+\\frac{b}{\\sqrt{b}}+\\frac{b}{\\sqrt{b}}+\\frac{c}{\\sqrt{c}}+\\frac{c}{\\sqrt{c}}=2(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})$$ Now proceed by AM-GM as in the ... 1 By the CRT it’s enough to figure out$F_{2020}$mod$8$and$125$. It’s easy to check that six applications of the linear recurrence relation turn$(0,1)=(F_0,F_1)$into$(F_6,F_7)=(8,13)=5 \\cdot (0,1)$mod$8$. As the multiplicative order of$5$mod$8$is two, it follows that$F_n$mod$8$has a period of$12$. Now note that$2020$is congruent to$4$mod$... 6 By AM-GM $$\\frac{b+c}{\\sqrt{a}}+\\frac{c+a}{\\sqrt{b}}+\\frac{a+b}{\\sqrt{c}} \\ge 2(\\sqrt{\\frac{bc}{a}} + \\sqrt{\\frac{ac}{b}} + \\sqrt{\\frac{ab}{c}} )=2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$$ Now using $x^2+y^2+z^2\\ge", "\\dfrac{f(a) +f(b)+f(c)}{3}$ $\\Rightarrow 6 \\leq a\\sqrt{a+3} +b\\sqrt{b+3} +c\\sqrt{c+3}$, as required First set up a function $$f(a,b,c)=\\sqrt{a+3} a+\\sqrt{b+3} b+\\sqrt{c+3} c.$$ Then, set $a=3-b-c$ to eliminate $a$ and get: $$f(b,c)=-\\sqrt{-b-c+6} (b+c-3)+\\sqrt{b+3} b+c \\sqrt{c+3},$$ then find the minimum of the function $f(b,c)$, which is $f(b=1,c=1)=6$. Therefore, we have $$f \\ge 6$$ as requested.", "# Limit of a sequence. • Oct 18th 2008, 12:55 PM Showcase_22 Limit of a sequence. I wasn't sure which section this belonged to so I decided to post it here. Quote: Use a method to approximate $\\sqrt 2$ using a sequence of rational numbers. Do the same for $\\sqrt 3.$ I started it like this: $a_1, \\ a_2, \\ a_3, \\ ........... \\ a_n \\ ...... \\ L$ $\\frac{1}{a+a_1}+b, \\ \\frac{1}{a+a_2}+b \\ ............\\ \\frac{1}{a+a_n}+b \\ ..... \\ \\frac{1}{a+L}+b$ where L is the limit. Both sequences must have the same limit: $\\frac{1}{a+L}+b=L$ $\\frac{1+ba+bL}{a+L}=L$ $1+ba+bL=L^2+aL$ $L^2+aL-bL-1-ba=0$ $L^2+L(a-b)-1-ab=0$ Since we want L= $\\sqrt 2$: $f(\\sqrt 2)=2+ \\sqrt {2} (a-b)-1-ab=0$ Since we want this to be 0 a=b since the middle term cannot exist. $1-b^2=0$ $b=1 \\ or \\ -1$ I want the positive $\\sqrt 2$ so b=1. Therefore the sequence becomes $a_{n+1}=\\frac{1}{1+a_n}+1$. This sequence tends to $\\sqrt 2$. Using the same method to find one that tends to $\\sqrt3$ gives: $\\frac{1}{a+L}+b=L$ $L^2+L(a-b)-1-ab=0$ Once again, the central term cannot be present at the end so a=b: $f(\\sqrt3)=3-1-b^2=0$ $b^2=2$ $b=\\sqrt 2$ But b can only be a rational number so it doesn't make sense. I think i'm supposed to use the first answer to find the second one except the two sequences are entirely different! So am I on the right lines?", "as the parabola is squeezed between them. $f(0) = 1$ = minimum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(1) = f(-1) = \\sqrt{2}$ = maximum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(0)[1 - (-1)] < \\int_{-1}^{1} \\sqrt{x^2+1} \\, dx < f(1)[1 - (-1] $ $2 < \\int_{-1}^{1} \\sqrt{x^2+1} \\, dx < 2\\sqrt{2} $ 7. Originally Posted by skeeter $f(0) = 1$ = minimum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ $f(1) = f(-1) = \\sqrt{2}$ = maximum value of $y = \\sqrt{x^2+1}$ on $[-1,1]$ Sorry but why did you sub in 0 and 1 into f ? 8. Originally Posted by SyNtHeSiS Sorry but why did you sub in 0 and 1 into f ? Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that mf (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(ba) and M(ba), it follows that $m(b-a) \\le \\int_a^b f(x) \\, dx \\le M(b-a)$ if you require further explanation, see your instructor." ]

[ "# Minimum Value Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that $$x^2 + y^2 + z^2 +2xyz = 4$$. Find the minimum value of $$x^2 + y^2 + z^2$$. ×", "A nice use Algebra Level 4 For positive real numbers $x,y,$ and $z$ such that $xyz=32,$ what is the minimum value of $x^{2} + 4xy + 4y^{2} + 2z^{2}?$ × Problem Loading... Note Loading... Set Loading...", "# An algebra problem by Dinesh Chavan Algebra Level 3 Let $$x,y,z$$ be positive real numbers. Given that $$xyz=1$$ Then find the minimum value of $$\\frac{1}{x^2+3x+2}+\\frac{1}{y^2+3y+2}+\\frac{1}{z^2+3z+2}$$ × Problem Loading... Note Loading... Set Loading...", "# Maximize This Algebra Level 4 Positive reals $x$, $y$ and $z$ are such that $x+y+z=xyz$. Find the maximum value of $\\large (x-1)(y-1)(z-1)$ Given the answer in 4 decimal places. ×", "+0 0 38 1 Let $$x,y,z$$ be positive real numbers such that $$x + y + z = 1$$ Find the minimum value of $$\\frac{1}{x + y} + \\frac{1}{x + z} + \\frac{1}{y + z}.$$ Nov 12, 2019", "# A nice use Algebra Level 4 For positive real numbers $$x,y,$$ and $$z$$ such that $$xyz=32,$$ what is the minimum value of $x^{2} + 4xy + 4y^{2} + 2z^{2}?$ ×", "How can you find a maximum for $f(x,y,z) = xyz + z$ where z is unconstrained. I mean z could be any positive real number or zero right? then $f(x,y, \\infty) = \\infty$.", "# An algebra problem by Dinesh Chavan Algebra Level 3 Let $x,y,z$ be positive real numbers. Given that $xyz=1$ Then find the minimum value of $\\frac{1}{x^2+3x+2}+\\frac{1}{y^2+3y+2}+\\frac{1}{z^2+3z+2}$ ×", "• zmudz Find the minimum of $$x-y$$ among all ordered pairs of real numbers $$(x, y)$$, $$x$$ and $$y$$between 0 and 1, where there exists a real number $$a \\neq 1$$ such that $$\\log_{x}a + \\log_{y}a = 4\\log_{xy}a.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Find the Minimum Algebra Level 4 Let $$x, y, z$$, and $$a$$ be positive real numbers such that $$x + y + z + a = 1$$. What is the minimum value of $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{4}{z}+\\dfrac{16}{a}$ × Problem Loading... Note Loading... Set Loading...", "# Minimum of sum of squares Algebra Level pending Let $$x$$, $$y$$ and $$z$$ be positive real numbers such that $$x+2y+z=1$$. What is the minimum value of $${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$$?", "• zmudz Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of $$9x+12y+8z.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# AM-GM Application Let $x,y,z$ be positive real numbers such that $\\frac { 1 }{ x } +\\frac { 1 }{ y } +\\frac { 1 }{ z } =1$. Then what is the minimum value of $(x-1)(y-1)(z-1)?$ ×", "show that this value of $z$ is a minimum. To check whether $z$ has a minimum when $x=2$ and $y=\\dfrac{1}{2}$, we can look at the second derivative of $z$ with respect to $x$. A positive second derivative corresponds to a local minimum. From $\\eqref{eq:2}$ we obtain $z''(x) = 160x^{-6},$ which gives $z''(2)=2.5>0$ (or we could note that $z''(x)$ is positive for all values of $x$). Therefore the value of $z$ when $x=2$ and $y=\\dfrac{1}{2}$ is indeed a minimum.", "Algebra Level 4 Let $$x,y,z$$ be positive real numbers satisfying $$x,y,z > 1$$. It is true that $\\log_y(xz)\\cdot\\log_z(xy) \\ge k$ for some real number $$k$$. Find $$k$$ accurate to 3 decimal places. Details and Assumptions $$k$$ may possibly be an integer. ×", "# How to calculate minimum! Algebra Level 4 If $$x,y$$ and $$z$$ are positive reals, find the minimum value of $$\\dfrac{x^4 +y^4+z^4}{xyz}$$. ×", "# AM-GM Application Algebra Level 2 Let $x,y,z$ be positive real numbers such that $\\frac { 1 }{ x } +\\frac { 1 }{ y } +\\frac { 1 }{ z } =1$. Then what is the minimum value of $(x-1)(y-1)(z-1)?$ ×", "## Algebra 2 (1st Edition) This function is in the form $y=a(x-h)^2 +k$. Thus, we see that k will be the critical value. If $a$ is positive, k will be the minimum, while if $a$ is negative, k will be the maximum. Using this, we see that the given function has a minimum of 130.", "Algebra Level 5 Let x,y,z be positive reals for which: $$\\sum_{cyc} (xy)^{2}=6xyz$$ Find the minimum value of: $$\\sum_{cyc} \\sqrt{\\frac{x}{x+yz}}$$ If the minimum value can be expressed in the form a$$\\sqrt{b}$$ + c, then find a + b + c ×", "Find the Minimum Algebra Level 4 Let $$x, y, z$$, and $$a$$ be positive real numbers such that $$x + y + z + a = 1$$. What is the minimum value of $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{4}{z}+\\dfrac{16}{a}$ ×" ]

[ "shows that the equality holds iff $x=y=z$. Here is my approach. Using AM-GM, \\begin{align} x^2+y^2+z^2 &\\ge 3(xyz)^{2/3} \\\\ \\implies 1 &\\ge 3(xyz)^{2/3} \\\\ \\implies (xyz)^{2/3} &\\le \\dfrac{1}{3} \\\\ \\implies xyz \\le \\dfrac{1}{3\\sqrt{3}} \\tag 1 \\end{align} Again, using AM-GM, \\begin{align} \\dfrac{x^3+y^3+z^3}{3} &\\ge 3xyz \\\\ \\implies x^3+y^3+z^3-3xyz &\\ge 0 \\\\ \\implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\\ge 0\\end{align}\\tag*{} Since $$x,y,z \\in \\mathbb R^+$$, $$x+y+z \\ge 0$$. \\begin{align}\\therefore \\, x^2+y^2+z^2 -xy-yz-zx &\\ge 0 \\\\ \\implies -x^2-y^2-z^2 +xy+yz+zx &\\le 0 \\\\ \\implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\\le 0 \\\\ \\implies x^2+y^2+z^2 +2xy+2yz+2zx &\\le 3(x^2+y^2+z^2) \\\\ \\implies (x+y+z)^2 &\\le 3 \\\\ \\implies x+y+z &\\le \\sqrt{3} \\tag 2 \\end{align} Multiplying $$(1)$$ and $$(2)$$, we can get the desired result.", "\\ge \\dfrac{7}{4}$$. To sum it up, $$\\sqrt{\\dfrac{x^2}{y^2} + \\dfrac{y^2}{x^2} + 2} + \\dfrac{\\sqrt{xy}}{x + y} \\ge \\dfrac{3}{4} + \\dfrac{7}{4} = \\dfrac{5}{2}$$. The equality sign happens when \\left\\{ \\begin{align*} (x + y)^2 = 2(x^2 + y^2)\\\\ \\dfrac{(x + y)^2}{16xy} = \\dfrac{\\sqrt{xy}}{x + y}\\\\ (x + y)^2 = 4xy \\end{align*}\\\\ \\right.\\implies x = y. Are there any other solutions that are more practical? Assume $$xy= 1\\,\\therefore\\,{y}'= -\\,\\frac{y}{x}$$ $$\\therefore\\,(\\,x- y\\,)(\\,x- 1\\,)\\geqq 0$$ $$\\because\\,x\\geqq y \\tag{ASSUME!ing}$$ We let $$w(\\,x\\,)= \\frac{x}{y}+ \\frac{y}{x}+ \\frac{\\sqrt{xy}}{x+ y}- \\frac{5}{2}= (\\,x+ y\\,)^{\\,2}+ \\frac{1}{x+ y}- \\frac{9}{2}$$ $$\\therefore\\,{w}'(\\,x\\,)= \\frac{(\\,{y}'+ 1\\,)[\\,2(\\,x+ y\\,)^{\\,3}- 1]}{(\\,x+ y\\,)^{\\,2}}$$ $$\\therefore\\,{w}'(\\,x\\,)= \\frac{(\\,x- y\\,)[\\,2(\\,x+ y\\,)^{\\,3}- 1]}{x(\\,x+ y\\,)^{\\,2}}$$ Using above that I found $$\\therefore\\,(\\,x- 1\\,){w}'(\\,x\\,)\\geqq 0$$ $$\\therefore\\,w(\\,x\\,)\\geqq w(\\,1\\,)= 0$$ $$\\because\\,(\\,x- 1\\,){w}'(\\,x\\,)= w(\\,x\\,)- w(\\,1\\,) \\tag{tangent equation}$$", "# Minimum value of $(x + 2y)(y + 2z)(xz + 1)$ when $xyz=1$ Let $$x,$$ $$y,$$ and $$z$$ be positive real numbers such that $$xyz = 1.$$ Find the minimum value of $$(x + 2y)(y + 2z)(xz + 1).$$ I am pretty sure this problem either uses AM-GM or Rearrangement Inequality, but I don't know how to solve it. Can anyone give me a hint? • I solved your problem by another way. If you want to see my solution, show please your attempts. May 15 '20 at 6:13 ALERT: this is my solution, not a hint. $$\\underbrace{(x+2y)}_{2AM \\ge 2GM} \\ \\ \\underbrace{(y+2z)}_{2AM \\ge 2GM} \\ \\ \\underbrace{(xz+1)}_{2AM \\ge 2GM}\\ge 2\\sqrt {x \\cdot2y} \\cdot 2\\sqrt {y \\cdot2z} \\cdot 2 \\sqrt{xz} = 8 \\sqrt{4x^2y^2z^2}=16xyz=16$$ Using the AM-GM inequality we have that : $$(x+2y)\\geq 2 \\sqrt{2xy}$$ $$(y+2z)\\geq 2 \\sqrt{2yz}$$ $$(xz+1)\\geq 2 \\sqrt{xz}$$ Then $$(x+2y) (y+2z) (xz+1) \\geq 8 \\sqrt{4 (xyz)^2}\\implies (x+2y)(y+2z)(xz+1) \\geq 16$$ • Well, you basically repeated what I just said :) May 15 '20 at 5:15 • I think we answered almost simultaneously, I didn't see that it already had an answer when I posted mine, when I started writing the solution there wasn't an answer May 15 '20 at 5:18 • Oh, alright :), sorry May 15 '20 at 5:19", "# Minimum Value of expression 2 What is the minimum value of $$(2x+3y)(8/x + 3/y)$$ when $x$ and $y$ are positive? I have tried expanding the equation but I did not get anything I got : $$25 + \\frac{6(4y^2 +x^2)}{xy}$$ but I don't know how to answer after this. I have also tried AM-GM but I couldn't manipulate the variables to cancel out and get the inequality. This is from Sipnayan 2017 You were on the right track: \\begin{align}(2x+3y)\\left(\\frac{8}{x} + \\frac{3}{y}\\right) &= 25+24 \\frac{y}{x}+6\\frac{x}{y}\\\\ &=25+12\\left(\\frac{2y}{x}+\\frac{x}{2y}\\right) \\\\ &\\geq 25+12 \\times 2 \\\\ &=49\\end{align} and notice that $x=2,y=1$ achieves this bound, so the minimum value is $49$. The bound $\\frac{2y}{x} + \\frac{x}{2y} \\geq 2$ can be proved from AM-GM, or from letting $t = \\frac{2y}{x}$ and considering the function $f(t) = t + \\frac{1}{t}$ and using calculus. From where you have left of, For real $x-2y,$ $$(x-2y)^2\\ge0\\implies x^2+4y^2\\ge4xy$$ Alternatively, using AM-GM inequality, $$\\dfrac{x^2+4y^2}2\\ge\\sqrt{4x^2y^2}=?$$ By C-S $$(2x+3y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right)\\geq\\left(\\sqrt{2x\\cdot\\frac{8}{x}}+\\sqrt{3y\\cdot\\frac{3}{y}}\\right)^2=49.$$ The equality occurs for $x=2$ and $y=1$, which says that we got a minimal value. Done!", "# Maximum value of $(ab)(a+b)$ given $a^2 + b^2 = 100$ Can anyone help me with finding the maximum value of $$y = {(ab)(a+b)}$$ With that condition that $$100 = a^2 + b^2$$ and both $a,b$ are positive numbers. Any help appreciated. Thanks. • Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Apr 1 '16 at 9:17 • Use \"Lagrange multipliers\" – SomeOne Apr 1 '16 at 9:17 • The answer might be an irrational one. – TheRandomGuy Apr 1 '16 at 10:00 Since $a,b>0$ and hence $y>0$ you can equivalently maximize $y^2$ which you can write by the given constraint as $$y^2=(ab)^2(a+b)^2=(ab)^2(100+2ab)$$ so that $y^2$ (and hence $y$) is maximized whenever $ab$ is maximized. By the AM-GM inequality $$ab=\\sqrt{(ab)^2}\\overset{AM-GM}\\le\\frac{a^2+b^2}{2}=\\frac{100}{2}=50$$ with equality if $a=b$, see here. Hence $a=b=\\sqrt{50}$ with $$y_{\\max}=\\sqrt{50}^2(\\sqrt{50}+\\sqrt{50})=100\\sqrt{50}$$ To use the method of Lagrange multipliers you can write this problem as \\begin{align}\\max_{a,b} {f(a,b)}&=(ab)(a+b)\\\\\\mbox{s.t. } g(a,b)&=a^2+b^2-100=0\\end{align} The Lagrange function is $$\\mathcal L(a,b,λ)=f(a,b)-λg(a,b)$$ with partial derivatives \\begin{align}\\frac{\\partial \\mathcal L}{\\partial a}&=2ab+b^2-2λa\\\\\\frac{\\partial \\mathcal L}{\\partial b}&=2ab+a^2-2λb\\\\\\frac{\\partial \\mathcal L}{\\partial λ}&=-α^2-b^2+100\\end{align} If $(a_0,b_0)$ is a solution to the problem then there exists $λ_0$ such that $(a_0,b_0,λ_0)$ is a stationary point", "# Tag Info ## Hot answers tagged lagrange-multiplier 7 HINT: Use $$\\begin{cases} f_x = \\lambda g_x \\\\ f_y = \\lambda g_y \\\\ xy = 3\\end{cases}$$ Note that $f_x = 2x, f_y = 2y, g_x = y, g_y = x$. 2 Here's another approach for fun. We can use the AM-GM mean: $$\\sqrt{ab}\\leq \\frac{a+b}{2},\\quad \\text{for all}\\ a,b\\geq 0$$ where equality holds if and only if $a=b$. Use $a=x^2,\\ b=y^2$ to get: $$\\sqrt{x^2y^2}\\leq \\frac{x^2+y^2}{2},$$ and our minimum will be where $x=y$. This implies $x=y=\\pm\\sqrt{3}$ and thus: $$2\\sqrt{(xy)^2} = 2|xy| = 6\\leq x^2+y^2,$$ ... 2 $A = x^{4} + y^{4} - 4b^{2}xy + 2b^4$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2x^{2}b^2 + 2y^2b^2 - 4b^{2}xy$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x^2 + y^2 -2xy)$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x - y)^2$ Which is positive as a sum of squares 2 Hints: let $f(x,y) = x^4+y^4-4b^2xy$. Show that the critical points (gradient equals zero) of $f$ are $(b,b)$ and $(-b,-b)$, show that they are local minima, and note that $f$ at these points takes the value $-2b^4$. 2 you must solve the system $$2x+\\lambda(1+2x)=0$$ $$2y+\\lambda(1+2y)=0$$ $$x+y+x^2+y^2=12$$ easy you will get $$x=y=-3,\\lambda=-\\frac{6}{5}$$ or $$x=y=2,\\lambda=-\\frac{4}{5}$$ 2 Your solution is correct, except the last part. You should get four minimizing", "for any values of $(b_1, b_2, b_3)$, so we can choose $(\\sqrt{y+z}, \\sqrt{z+x}, \\sqrt{x+y})$, which makes $( \\spadesuit )$ simplify to: \\begin{align} S &\\geq \\dfrac{1}{2}(x+y+z) \\tag{from the above equation} \\\\[3mm] &\\geq \\dfrac{1}{2} \\cdot 3 \\sqrt[3]{xyz} \\tag{AM-GM} \\\\[3mm] &= \\dfrac{3}{2} \\tag{since xyz=1} \\end{align} $\\square$ That will do it for this post. In part 2, I will show the proof of the AM-GM inequality, along with a few more examples of its use in miscellaneous problems. Thanks for reading, and post any questions in the comments section.", "at this point not available to you. Or else we could let the sides be $x$, $y$, and $z$. Then $2xy+2yz+2xz=6a^2$. (i) Keeping $y$ fixed, use the calculus to show that for maximum volume we must have $x=z$. (ii) Now allow $y$ to vary, and use exactly the argument we gave above to conclude that $x=y=a$. But there is an interesting purely algebraic way to handle the problem, the famous Arithmetic Mean/Geometric Mean Inequality, which has a purely algebraic proof. AM-GM (in the case of $3$ variables) says that if $p$, $q$, and $r$ are positive, then $$\\frac{p+q+r}{3} \\ge \\sqrt[3]{pqr},$$ with equality precisely when $p=q=r$. To apply this result, let $p=xy$, $q=yz$, and $r=zx$. We know that $2xy+2yz+2zx=6a^2$, so $\\frac{p+q+r}{3}=a^2$. Using AM-GM, we find that $$a^2 \\ge \\sqrt[3]{pqr}=\\sqrt{3}{x^2y^2z^2}=(xyz)^{2/3},$$ with equality only when $x=y=z$. This says that $xyz \\le a^3$, with equality only when $x=y=z$. So the volume is always $\\le a^3$, and it is $a^3$ only when the three sides are equal. - It's very nice to see a 80k+ user reacts against edits. – Gigili Aug 6 '12 at 22:52 What is that supposed to mean, @Gigili? – Potato Aug 6 '12 at 23:38 @Potato: This is the precise definition of it. – Gigili Aug 6 '12 at 23:44", "2 terms as you have no information on if $f''$ exists), for any $x\\in [0,2]$ we have \\begin{align}f(x) &= f(0) + f'(c)x \\\\ f(x) &= f(2) + f'(d)(x-2)\\end{align} for some $c,d \\in [0,2]$. The first one shows $$1-x \\leq f(x) \\leq 1 + x \\qquad (*)$$ and the second shows $$1 +(x-2) \\leq f(x) \\leq ... 3 Note that 2ab > -ab \\Rightarrow a^2+b^2+2ab > a^2 + b^2 -ab \\Rightarrow (a+b)^2 > a^2 + b^2 -ab \\Rightarrow a+b > \\sqrt{a^2 + b^2 -ab} I was able to do this by squaring the original inequality I was supposed to prove, and then I worked backwards. 3 3ab>0 Hence, 2ab>-ab Hence, a^2+2ab+b^2>a^2+b^2-ab Hence a+b>\\sqrt{a^2+b^2-ab} because a^2+b^2-ab>a^2+b^2-2ab=(a-b)^2>0 2 Using AM-GM, we have:$$4x^2+9y^2 \\ge 12xy3y^2+12z^2 \\ge 12yz18z^2+2x^2\\ge 12zx$$Combining and using the constraint,$$6x^2+12y^2+30z^2 \\ge 12(xy+yz+zx) \\implies x^2+2y^2+5z^2 \\ge 2$$Equality is when x:y:z = 3:2:1 and satisfying the constraint, i.e. when x = \\frac3{\\sqrt{11}}, y = \\frac2{\\sqrt{11}}, z = \\frac1{\\sqrt{11}}. In general for ... 2$$|x^2-4|<2 \\\\ \\implies -2<x^2-4<2 \\\\ \\implies 2<x^2<6 \\\\ \\implies \\sqrt{2}<\\pm x<\\sqrt{6} \\\\ \\implies \\sqrt{2}<x<\\sqrt{6},\\ \\ -\\sqrt{2}>x>-\\sqrt{6}$$2 You want to show that$$2-\\frac{1}{\\sqrt{n}}+\\frac1{(n+1)^4}\\le 2-\\frac1{\\sqrt{n+1}}.\\tag{1}$$If (1) is true, then using induction hypothesis one gets$$1+\\frac1{2^4}+\\cdots+\\frac{1}{n^4}+\\frac{1}{(n+1)^4}\\le 2-\\frac1{\\sqrt{n}}+\\frac1{(n+1)^4}\\le 2-\\frac1{\\sqrt{n+1}},$$and we are done by induction. Showing (1) should be easy since ... 2 A symmetrical approach:$$\\begin{align}x^2+2x+1\\quad &=(x+1)^2&&=2(x^2+1)-(x^2+1-2x)\\\\ \\\\ x^2+1+\\underbrace{2x}_{>0}\\quad &=(x+1)^2&&=2(x^2+1)-\\underbrace{(x-1)^2}_{\\geq 0}\\\\ x^2+1\\quad &<(x+1)^2&&\\leq 2(x^2+1)\\qquad \\text{QED} \\blacksquare \\end{align}$$2", "# What is the maximum value of $\\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem: Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\\sqrt6 xy + 4yz$ I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$. • use the Lagrange Multiplier methode – Dr. Sonnhard Graubner Jan 9 '16 at 22:47 • I get $\\sqrt{\\frac{11}{2}}$ at $\\left(\\sqrt{\\frac{3}{22}},\\sqrt{\\frac{11}{22}},\\sqrt{\\frac{8}{22}}\\right)$. – CommonerG Jan 9 '16 at 23:20 Let $a,b\\in\\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\\in\\mathbb{R}$). By Cauchy-Schwarz, $axy+byz=(ax+bz)y\\leq \\left(\\sqrt{a^2+b^2}\\sqrt{x^2+z^2}\\right)|y|$. By AM-GM, $\\sqrt{x^2+z^2}\\,|y|\\leq \\frac{\\left(x^2+z^2\\right)+y^2}{2}=\\frac{1}{2}$. That is, $axy+byz\\leq \\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$. (Similarly, we also have the lower bound $axy+byz\\geq -\\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},-\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$.) • Pretty neat. Thank you. – Ashish Gupta Jan 9 '16 at 23:03 • I think your $y-$ and $z-$coördinates should be reversed in your solutions. – Théophile Jan 10 '16 at 0:15 • @Théophile Thanks, fixed. – Batominovski Jan 10 '16 at 0:52 We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\\sqrt{6}xy \\leq A^2x^2+B^2y^2$, and $4yz \\leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \\dfrac{\\sqrt{6}}{2},", "rigorously afterwards whether the tentative results actually make sense. It seems that he is doing this the hard way. If you know that $A^3 = 2I$, then any infinite sum of powers of $A$, if it converges, should reduce to a sum of the form $xI + yA + zA^2.$ So it seems reasonable to at least try to solve $$(xI + yA + zA^2)(A^2 - 2A + 2I) = I$$ for $x, y,$ and $z$. \\begin{align} I &= (xI + yA + zA^2)(2I - 2A + A^2)\\\\ &= 2xI +(-2x + 2y)A + (x - 2y + 2z)A^2 + (y - 2z)A^3 + zA^4\\\\ &= 2xI +(-2x + 2y)A + (x - 2y + 2z)A^2 + (2y - 4z)I + 2zA\\\\ &= (2x + 2y - 4z)I + (-2x + 2y + 2z)A + (x - 2y + 2z)A^2 \\end{align} Solve \\begin{align} 2x + 2y - 4z &= 1 \\\\ -2x + 2y + 2z &= 0 \\\\ x - 2y + 2z &= 0 \\end{align} and you get $x = \\frac 25,\\, y = \\frac 3{10},\\, z = \\frac 1{10}$ The basic rule of thumb for such problems is that as long as all of the components of your equations are invertible and any multiplications used will commute, things will work just fine (this is the", "$x^2 - 2xy - y^2 = [x - (1+\\sqrt{2})y][x - (1-\\sqrt{2})y].$ For the other case, $$x^2 + (2y)x + (-y^2) = 0\\to \\\\ x= \\frac{-2y \\pm \\sqrt{4y^2 - 4(1)(-y^2)}}{2} \\\\ =-y \\pm\\sqrt{2}y = (1 \\pm \\sqrt{2})y.$$ So $x^2 + 2xy - y^2 = [x - (1-\\sqrt{2})y][x - (1+\\sqrt{2})y].$ There is no simple factorization of $x^2+2xy-y^2$ nor $x^2-2xy-y^2$, although you can write: \\begin{align}x^2+2xy-y^2 &= (x+y)^2-2y^2 \\\\&= \\left(x+y(1+\\sqrt{2})\\right)\\left(x+y(1-\\sqrt{2})\\right) \\end{align} and similarly: $$x^2-2xy-y^2 = \\left(x-y(1+\\sqrt{2})\\right)\\left(x-y(1-\\sqrt{2})\\right)$$ This is another way of solving this problem using Completing the Square method. $x^2+2xy-y^2=??$ First we have, $x^2+2xy-y^2=0$ $x^2+2xy=y^2,$ By adding both sides by $y^2$ to make it perfect square then, $x^2+2xy+y^2=y^2+y^2$ $(x+y)^2=2y^2$ $x+y=[2^(1/2)]y$ $x=(+ or -)[2^(1/2)]y-y$ $x=[2^(1/2)]y-y ; x=-[2^(1/2)]y-y$ $x=[2^(1/2)-1]y ; x=-[2^(1/2)-1]y$ therefore we have, ${x-[2^(1/2)-1]y}{x+[2^(1/2)-1]y}$ By solving $x^2-2xy-y^2=??$ Try to solve this by relying with this pattern. Thank you. • I don't think the answer is complete. Also please check the formatting – Shailesh Mar 31 '16 at 11:01 You can factor a quadratic trinomial $$ax^2+bxy+cy^2$$ by finding the roots of $$\\frac{ax^2+bxy+cy^2}{y^2}=a\\left(\\frac xy\\right)^2+b\\left(\\frac xy\\right)+c=0.$$ Then $$ax^2+bx+c=y^2a\\left(\\frac xy-r_0\\right)\\left(\\frac xy-r_1\\right)=a\\left(x-r_0y\\right)\\left(x-r_1y\\right).$$", "# Not sure what to do after trying to simplify the inequality and getting nowhere? Here is the question: If $$x, y, z \\in {\\displaystyle \\mathbb {R} }$$, then prove $$x^2+y^2+z^2+4\\ge2(x+y+z)$$ Here is what I tried doing: I tried simplifying the inequality and getting all the terms on one side, like so: $$x^2+y^2+z^2+4\\ge2x+2y+2z$$ $$x^2+y^2+z^2+4-2x+2y+2z\\ge0$$ $$(x^2-2x)+(y^2-2y)+(z^2-2z)+4\\ge0$$ $$(x^2-2x)+(y^2-2y)+(z^2-2z)\\ge-4$$ Is it possible to use the fact that it is always true that $$x^2-2x\\ge-4$$, $$y^2-2y\\ge-4$$, and $$z^2-2z\\ge-4$$, to show that inequality holds? Other than that, I'm not sure what to do next. I can't think of any manipulation that might work in this case. Any help would be greatly appreciated! • $x^2-2x\\ge-1$, so $(x^2-2x)+(y^2-2y)+(z^2-2z)\\ge-3$ – J. W. Tanner Jan 14 '20 at 1:45 \\begin{align} x^2 - 2x + y^2 - 2y + z^2 - 2z + 4 = (x-1)^2 + (y-1)^2 + (z-1)^2 + 1 \\ge 1 > 0. \\end{align} If you borrow three ones, you get $$(x^2-2x)+(y^2-2y)+(z^2-2z)+4=(x-1)^2+(y-1)^2+(z-1)^2+1\\geq1.$$", "we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x \\qquad(R_1^{[i]}),$$ for every integer $$i$$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. \\qquad (R_2^{[i]})$$ If we now conjugate $$(R_3)$$ by $$y$$, the left-hand side becomes \\begin{align}yzxz^{-1}y^{-1} &= z^2y\\cdot x \\cdot y^{-1}z^{-2}\\\\ & = z^2y^{-1}xz^{-2}\\\\ & = z^2 y^{-1}z^{-2}\\cdot z^2xz^{-2}\\\\ & = y^{-1}z^2\\cdot x^4 \\end{align} (the first equality is a double use of the relation $$yz = z^2y$$, a reformulation of $$(R_2)$$ ; the second uses $$(R_1^{[1]})$$ and the last uses the inverse of $$(R_2^{[2]})$$ and $$R_3$$ twice). On the other hand, the left side becomes \\begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \\\\ &= y^{-1}xy^{-1}x \\\\ &= y^{-3}x^2 \\end{align} (the first equality uses the $$(R_1^{[1]})$$ twice, the last uses the inverse of $$(R_1)$$). Put together, we have proven $$y^{-1}z^2x^4 = y^{-3}x^2$$, which gives $$z^2 = y^{-2}x^{-2}.\\qquad (R^*)$$ If we conjugate $$y$$ by $$z^{-2}$$, we now get on the one hand \\begin{align}z^{-2}y z^2 &= x^2 y^2 \\cdot y \\cdot y^{-2} x^{-2} \\\\ &= x^2 y x^{-2} \\\\ &= y^4 \\end{align} (the first equality uses $$(R^*)$$ twice, the last uses $$(R_1)$$ twice.) But, on the other hand, $$z^{-2}yz^2 = z^2 y$$ because of $$(R_2^{[-2]}$$). So we finally get $$z^2 y = y^4$$, which translates to $$z^2 = y^3.$$ This proves", "$x+y = 2\\sqrt 6$. $(x-y)^2 = (x^2+y^2) - 2xy = 16 - 8 = 8$. So $x+y = 2\\sqrt 2$. So \\begin{align} (x+y)+(x-y) &= 2\\sqrt 6 + 2 \\sqrt 2 \\\\ 2x &= 2\\sqrt 6 + 2 \\sqrt 2 \\\\ x &= \\sqrt 6 + \\sqrt 2 \\end{align} Since $x$ can also be negative, then $x = \\pm(\\sqrt 6 + \\sqrt 2)$ We compute $(x + (\\sqrt 6 + \\sqrt 2))(x - (\\sqrt 6 + \\sqrt 2)) = x^2 - 8 + 4\\sqrt 3$ We find that $x^4- 8\\sqrt{3}x^2 - 16 = (x^2 - 4\\sqrt 3 - 8)(x^2 - 4\\sqrt 3 + 8)$ It only seems proper to check that $(\\sqrt 6 - \\sqrt 2)^2 = 8 - 4\\sqrt 3$ So the roots are $$\\{\\pm(\\sqrt 6 + \\sqrt 2), \\pm i(\\sqrt 6 - \\sqrt 2) \\}$$", "# Difference between revisions of \"2001 IMO Shortlist Problems/N2\" ## Problem Consider the system $x + y = z + u,$ $2xy & = zu.$ (Error compiling LaTeX. ! Misplaced alignment tab character &.) Find the greatest value of the real constant $m$ such that $m \\leq x/y$ for any positive integer solution $(x,y,z,u)$ of the system, with $x \\geq y$. ## Solution First consider the real solutions to the system. We have by AM-GM that $\\frac{z+u}{2}\\ge\\sqrt{zu}$ and substituting we get $\\frac{x+y}{2}\\ge\\sqrt{2xy}$. Squaring and simplifying and dividing by $y^2$, we get the inequality $r^2-6r+1\\ge0$, where $r=\\frac{x}{y}$. Then $r^2-6r+9\\ge8$, so $r\\ge3+2\\sqrt2$ or $r\\le3-2\\sqrt2$. Since $r\\ge1$, we discard the second inequality and have that $3+2\\sqrt2$ is a lower bound for $r$ This bound is also attainable for real values when $z=u$. Since $\\mathbb{Q+}$ is dense, it is always possible to assign rational values to $x, y, z,$ and $w$ so that $r$ approaches $3+2\\sqrt2$, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of $r$. Thus, $\\boxed{m=3+2\\sqrt2}$.", "# Difference between revisions of \"2014 UMO Problems/Problem 5\" ## Problem Find all positive real numbers $x, y$, and $z$ that satisfy both of the following equations. \\begin{align*} xyz & = 1\\\\ x^2 + y^2 + z^2 & = 4x\\sqrt{yz}- 2yz \\end{align*} ## Solution By AM-GM $$x^2+y^2+z^2 + 2yz\\ge x^2 + 4yz\\ge 4x\\sqrt{yz}$$ Hence, the second equation implies that $y=z$ and $x^2=4yz\\implies x=2y=2z$. Now we plug it into the first equation to get $(x,y,z) = \\left(\\sqrt[3]{4}, \\frac{\\sqrt[3]4}{2}, \\frac{\\sqrt[3]4}{2}\\right)$", "# Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\\in$ $\\mathbb{R}$ if \\begin{align} x^2+x-1=y \\\\ y^2+y-1=z\\\\ z^2+z-1=x \\end{align} My Try: if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions. if $x \\ne y \\ne z$ Then we have \\begin{align} x(x+1)=y+1 \\\\ y(y+1)=z+1\\\\ z(z+1)=x+1 \\end{align} Multiplying all we get $$xyz=1 \\tag{1}$$ and adding all we get $$x^2+y^2+z^2=3 \\tag{2}$$ Now from Original Equations \\begin{align} x^2=y+1-x\\\\ y^2=z+1-y\\\\ z^2=x+1-z \\end{align} Multiplying all and Using $(1)$ we get $$(y+1-x)(z+1-y)(x+1-z)=1$$ $\\implies$ $$xy+yz+zx-3=(x-y)(y-z)(z-x) \\tag{3}$$ I am unable to proceed further.. #### Solutions Collecting From Web of \"Solve the System of Equations in Real $x$,$y$ and $z$\" using the RMS GM inequality $$\\sqrt{\\frac{x^2+y^2+z^2}{3}}\\geq\\sqrt[3]{xyz}$$ with equality if and only if $x=y=z$, plugging in your values for $xyz$ and$x^2+y^2+z^2$ we get that $$\\sqrt{\\frac{3}{3}}\\geq\\sqrt[3]{1}$$ $$1=1$$ thus the only possible solutons are those you already stated. No solutions exist for real, distinct $(x,y,z)$. Writing $f(x)=x^2+x-1$, the existence of such a solution would mean that $f$ has a real point of least period 3; that is, $f^3(x)=x$ for some real $x$ with $f(x)\\neq1$ i.e. $x\\neq \\pm 1$. Sarkovsky’s theorem then implies that $f$ has points of arbitrary least period. In particular, there would exist $x\\in\\mathbb{R}-\\{-1,1\\}$ such that $f^2(x)=x$. But $$f^2(x) – x = (x^2+x-1)^2+(x^2+x-1)-1-x = x^4+2x^3-2x-1 = (x-1)(x+1)^3.$$ and so there", "\\end{align} We know that each of the two areas equals $x$. For notational simplicity, temporarily write \"$y$\" for \"$\\sqrt{x}$\", so that $x = y^2$. \\begin{align} g^2 &= x = y^2 &(1) \\\\ 4 w^2 + 4 g w &= x = y^2 &(2) \\end{align} From equation $(1)$, we get that $g = y$. Substituting that into $(2)$ gives $$4 w^2 + 4 w y = y^2 \\qquad \\to \\qquad 4 w^2 + 4 w y - y^2 = 0$$ Now, simply solve the quadratic equation for $w$: $$w = \\frac{- 4 y \\pm \\sqrt{(4y)^2-4\\cdot 4\\cdot(-y^2)}}{2\\cdot 4} = \\frac{-4y \\pm \\sqrt{16y^2+16y^2}}{8} = \\frac{-4y\\pm 4y\\sqrt{2}}{8} = \\frac{y}{2}\\left(-1\\pm\\sqrt{2}\\right)$$ We (presumably) want the positive root, so (since $\\sqrt{2} > 1$) take \"$\\pm$\" to be \"$+$\"; also, replace \"$y$\" with its defined value, \"$\\sqrt{x}$\": $$w = \\frac{\\sqrt{x}}{2}\\left(\\sqrt{2} - 1\\right)$$ - thanks very much for your help – dato datuashvili Aug 9 '13 at 7:57", "to solve it for $c=\\sqrt{2}-1$ using AM-GM. but dont know how for its optimum value. \\begin{align*} \\frac{a_1}{a_2+a_3}+\\frac{a_2}{a_3+a_4}+\\cdots+ \\frac{a_n}{a_1+a_2} \\geq (\\sqrt{2}-1)n \\end{align*} Using Am-GM, the inequality becomes \\begin{align} \\frac{a_1+a_2+a_3}{a_2+a_3} \\cdot \\frac{a_2+a_3+a_4}{a_3+a_4}\\cdots \\frac{a_n+a_1+a_2}{a_1+a_2}\\geq (\\sqrt{2})^n \\end{align}" ]

[ "Given that $x, y, z$ are positive real numbers satisfying $xyz = 32$, find the minimum value of \\begin{align*} x^{2} + 4xy + 4y^{2} + 2z^{2} \\end{align*} ### Solution Using $AM \\geq GM$ as all the quantities involved are positive, we have \\begin{align*} \\frac{x^{2} + 2xy + 2xy + 4y^{2} + z^{2} + z^{2}}{6} &\\geq& (x^{2} \\cdot 2xy \\cdot 2xy \\cdot 4y^{2} \\cdot z^{2} \\cdot z^{2})^{\\frac{1}{6}} = 16 \\ \\implies x^{2} + 4xy + 4y^{2} + 2z^{2} &\\geq& 96 \\end{align*} Therefore, the minimum value of $x^{2} + 4xy + 4y^{2} + 2z^{2}$ is $96$.", "# AM-GM & Minimization Proof [duplicate] I want to prove that for all $$x, y > 0$$, $$\\cfrac{x+y}{2} \\geq \\sqrt{xy}$$ Particularly, I want to show that the minimum of $$(x+y)/2$$ is exactly $$\\sqrt{xy}$$. This is my attempt: $$\\textbf{Proof}$$ (Contradiction). Assume if $$x, y > 0$$, then $$(x+y)/2 < \\sqrt{xy}$$. But \\begin{align*} x+y &< 2\\sqrt{xy} \\\\ x^2+2xy+y^2 &< 4xy \\\\ x^2-2xy+y^2 &< 0 \\\\ (x-y)^2 &< 0 \\end{align*} gives us a contradiction since for all $$x, y > 0$$, we know $$(x-y)^2 > 0$$. Therefore there exists no positive reals $$x$$ and $$y$$ such that $$(x+y)/2 < \\sqrt{xy}$$, or $$\\min\\bigg(\\cfrac{x+y}{2}\\bigg) = \\sqrt{xy}$$. $$\\qquad \\square$$ • No. The minimum can not be an \"expression\". It must be only a finite number. Indeed, you can also write $$x+y+1≥3\\sqrt [3]{xy}\\implies \\frac {x+y}{2}≥\\frac {3\\sqrt [3]{xy}-1}{2}$$ Therefore, the minimum can not be considered as a non-constant expression. Jan 13 at 4:17 • Your claim about the minimum is false. With $x, y > 0$ the LHS has no minimum and gets arbitrarily close to $0$. Jan 13 at 4:37 Instead of talking about minimum, we may say that the inequality $$\\frac{x+y}{2} \\geq \\sqrt{xy}$$ holds for any $$x, y>0$$ and the equality holds for $$x=y.$$ We may prove the inequality directly by starting with \\begin{aligned} & x+y-2 \\sqrt{x y}=(\\sqrt{x}-\\sqrt{y})^2 \\geqslant 0 \\\\ \\Rightarrow \\quad &", "# If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is $\\bf{My\\; Try::}$ Here I have Used $\\bf{A.M\\geq G.M}$ Inequality So $$\\displaystyle \\frac{x^2+4xy+4y^2+4z^2}{4}\\geq \\left(x^3\\cdot y^3\\cdot z^2\\right)^{\\frac{1}{4}}\\;,$$ But I did not How can I solve it Help me, Thanks Applying the AM-GM is the right strategy, but you need to do it a bit differently. $$x^2+4xy+4y^2+4z^2$$ $$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$ $$\\ge 6\\sqrt[6]{x^2 \\cdot 2xy \\cdot 2xy \\cdot 4y^2 \\cdot 2z^2 \\cdot 2z^2}$$ $$= 6\\sqrt[6]{64x^4y^4z^4}$$ $$= 12(xyz)^{2/3}$$ $$= 12 \\cdot 32^{2/3}$$ Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$. Let $f: (x,y,z) \\mapsto x^{2} + 4xy + 4y^{2} + 4z^{2}$ on $\\mathbb{R}^{3}$; let $g: (x,y,z) \\mapsto xyz - 32$ on $\\mathbb{R}^{3}_{++}$; and let $(x,y,z) \\in g^{(-1)}\\{ 0 \\}$ such that $f(x,y,z)$ is an extremum. Then there is some $\\lambda \\in \\mathbb{R}$ such that $$\\nabla f(x,y,z) = (2x+4y, 4x+8y, 8z) = \\lambda \\nabla g = \\lambda (yz, xz, xy).$$ And we can find the candidate possible extrema of $f|_{g^{(-1)}\\{ 0 \\}}$, inspecting them.", "# Minimum value of $(x + 2y)(y + 2z)(xz + 1)$ when $xyz=1$ Let $$x,$$ $$y,$$ and $$z$$ be positive real numbers such that $$xyz = 1.$$ Find the minimum value of $$(x + 2y)(y + 2z)(xz + 1).$$ I am pretty sure this problem either uses AM-GM or Rearrangement Inequality, but I don't know how to solve it. Can anyone give me a hint? • I solved your problem by another way. If you want to see my solution, show please your attempts. May 15 '20 at 6:13 ALERT: this is my solution, not a hint. $$\\underbrace{(x+2y)}_{2AM \\ge 2GM} \\ \\ \\underbrace{(y+2z)}_{2AM \\ge 2GM} \\ \\ \\underbrace{(xz+1)}_{2AM \\ge 2GM}\\ge 2\\sqrt {x \\cdot2y} \\cdot 2\\sqrt {y \\cdot2z} \\cdot 2 \\sqrt{xz} = 8 \\sqrt{4x^2y^2z^2}=16xyz=16$$ Using the AM-GM inequality we have that : $$(x+2y)\\geq 2 \\sqrt{2xy}$$ $$(y+2z)\\geq 2 \\sqrt{2yz}$$ $$(xz+1)\\geq 2 \\sqrt{xz}$$ Then $$(x+2y) (y+2z) (xz+1) \\geq 8 \\sqrt{4 (xyz)^2}\\implies (x+2y)(y+2z)(xz+1) \\geq 16$$ • Well, you basically repeated what I just said :) May 15 '20 at 5:15 • I think we answered almost simultaneously, I didn't see that it already had an answer when I posted mine, when I started writing the solution there wasn't an answer May 15 '20 at 5:18 • Oh, alright :), sorry May 15 '20 at 5:19", "# If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is $\\bf{My\\; Try::}$ Here I have Used $\\bf{A.M\\geq G.M}$ Inequality So $$\\displaystyle \\frac{x^2+4xy+4y^2+4z^2}{4}\\geq \\left(x^3\\cdot y^3\\cdot z^2\\right)^{\\frac{1}{4}}\\;,$$ But I did not How can I solve it Help me, Thanks ## 2 Answers Applying the AM-GM is the right strategy, but you need to do it a bit differently. $$x^2+4xy+4y^2+4z^2$$ $$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$ $$\\ge 6\\sqrt[6]{x^2 \\cdot 2xy \\cdot 2xy \\cdot 4y^2 \\cdot 2z^2 \\cdot 2z^2}$$ $$= 6\\sqrt[6]{64x^4y^4z^4}$$ $$= 12(xyz)^{2/3}$$ $$= 12 \\cdot 32^{2/3}$$ Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$. Lagrange's theorem is helpful. Let $f: (x,y,z) \\mapsto x^{2} + 4xy + 4y^{2} + 4z^{2}$ on $\\mathbb{R}^{3}$; let $g: (x,y,z) \\mapsto xyz - 32$ on $\\mathbb{R}^{3}_{++}$; and let $(x,y,z) \\in g^{(-1)}\\{ 0 \\}$ such that $f(x,y,z)$ is an extremum. Then there is some $\\lambda \\in \\mathbb{R}$ such that $$\\nabla f(x,y,z) = (2x+4y, 4x+8y, 8z) = \\lambda \\nabla g = \\lambda (yz, xz, xy).$$ And we can find the candidate possible extrema of $f|_{g^{(-1)}\\{ 0 \\}}$, inspecting them. • The tag is (algebra-precalculus), though. Perhaps something different is expected. – Brian Tung Aug 13 '15 at 6:55", "# Difference between revisions of \"2014 UMO Problems/Problem 5\" ## Problem Find all positive real numbers $x, y$, and $z$ that satisfy both of the following equations. \\begin{align*} xyz & = 1\\\\ x^2 + y^2 + z^2 & = 4x\\sqrt{yz}- 2yz \\end{align*} ## Solution By AM-GM $$x^2+y^2+z^2 + 2yz\\ge x^2 + 4yz\\ge 4x\\sqrt{yz}$$ Hence, the second equation implies that $y=z$ and $x^2=4yz\\implies x=2y=2z$. Now we plug it into the first equation to get $(x,y,z) = \\left(\\sqrt[3]{4}, \\frac{\\sqrt[3]4}{2}, \\frac{\\sqrt[3]4}{2}\\right)$", "Thursday, June 4, 2015 Second Method of Solving IMO Optimization Contest Problem: Find minimum $xy$ Second Method of Solving IMO Optimization Contest Problem: Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers. The second method is the solution provided by a well-known mathematics retired professor from the University in the U.K. He is a great and keen observer. He first noticed that the minimum value of $xy$ must be positive, because if $xy\\leqslant 0$, from the given equation $xy+xz+yz=4$, then we have: $(x+y)z\\geqslant 4$ So $(x+y)^2\\geqslant\\dfrac{16}{z^2}$, and from the other given equation where $x^2+y^2+z^2=7$: $7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \\geqslant (x+y)^2 \\geqslant \\frac{16}{z^2}.$ Thus $z^2 + \\dfrac{16}{z^2} \\leqslant 7$. AM-GM inequality formula tells us that cannot happen, because the minimum value of $z^2 + \\dfrac{16}{z^2}\\ge 2\\sqrt{\\left(z^2\\cdot \\dfrac{16}{z^2} \\right)}=2(4)=8$, occurring when $z^2 = 4$. So we may assume that $xy>0$. Let $u = \\sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But again from the AM-GM inequality we have: $vz\\leqslant\\bigl(\\frac12(v+z)\\bigr)^2 = \\frac{15}4.$ Therefore $vz+u^2=4$ $u^2=4-vz$ $xy = u^2 \\geqslant 4-\\frac{15}4 = \\frac14$. To see that the minimum value $xy=\\frac14$ is attained, check that if $x= \\frac14(\\sqrt{15}", "+0 0 95 1 Assuming $$x$$, $$y$$, and $$z$$ are positive real numbers satisfying: \\begin{align*} xy-z&=15, \\\\ xz-y&=0, \\text{ and} \\\\ yz-x&=0, \\end{align*} then, what is the value of $$xyz$$? Jan 4, 2020 #1 +42 +2 Hello! Notice our equations 2 and 3 are equivalent. xy-y=yz-x y(z+1)=x(z+1) x=y Now we can rewrite our 2nd equation as x(z-1)=0 First we note that x cannot =0, as x,y, and z are all positive. Thus, z=1. We can plug this in, giving us x^2-1=15. Hence, x=4 (x cannot be -4) Therefore, x=4, y=4, and z=1. Thus, our answer is 4*4*1= 16. Jan 4, 2020", "# If $x,y,z$ are positive real number number, Then minimum value of $\\frac{x^4+y^4+z^2}{xyz}$ If $x,y,z$ are positive real number number, Then minimum value of $\\displaystyle \\frac{x^4+y^4+z^2}{xyz}$ $\\bf{My\\; Try::}$ Given $x,y,z>0.$ So Using $\\bf{A.M\\geq G.M\\;,}$ We get $$\\displaystyle x^4+y^4\\geq 2x^2y^2$$ and then Using $$2x^2y^2+z^2\\geq 2\\sqrt{2}|xyz|\\geq 2\\sqrt{2}xyz$$ So we get $$x^4+y^4+z^4\\geq 2x^2y^2+z^2 \\geq 2\\sqrt{2}xyz$$ and equality hold when $x^4=y^4$ and $2x^2y^2=z^2$ So we get $x=y$ and $z=\\sqrt{2}x^2$, Now Minimum occur at $\\left(x,x,\\sqrt{2}z^2\\right)$ So at that point we get $\\displaystyle \\left[\\frac{x^4+y^4+z^4}{xyz}\\right]_{\\bf{\\min}} = \\sqrt{2}+2x^2$ and Answer given is $2\\sqrt{2}$, I did not understand how this can happen olz explain me, Thanks My Question is • You found that $x^4+y^4+z^2\\ge 2\\sqrt{2}xyz$, thus $\\displaystyle \\frac{x^4+y^4+z^2}{xyz}\\ge \\frac{2\\sqrt{2}xyz}{xyz}=2\\sqrt{2}$. – Galc127 Feb 18 '16 at 12:59 • Please fix the equation in title : $z^2$ and not $z^4$. I was lost ! Cheers. – Claude Leibovici Feb 18 '16 at 13:03 $$x^4+y^4+z^2=x^4+x^4+(\\sqrt{2}x^2)^2=4x^4\\\\ xyz=xx\\sqrt{2}x^2=\\sqrt{2}x^4$$ We have, with AM-GM, that \\begin{align} x^4+y^4+z^2&=4\\cdot\\frac{x^4+y^4+\\tfrac12z^2+\\tfrac12z^2}{4}\\\\ &\\geq 4\\sqrt[4]{x^4y^4z^4\\cdot\\tfrac14}\\\\ &=xyz\\cdot 2\\sqrt{2} \\end{align} So that $$\\frac{x^4+y^4+z^2}{xyz}\\geq 2\\sqrt{2}$$ With equality iff $x^4=y^4=\\frac12z^2$, so for example $x=y=1$ and $z=\\sqrt{2}$.", "+0 # intermediate algebra 0 75 1 Let x, y and z be positive real numbers such that x + y + z = 1. Find the maximum value of xyz. Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ MaxWong Jan 28, 2021", "# Give me a four, give me a two Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that: \\begin{align} xyz&=945 \\\\ x(y+1)+y(z+1)+z(x+1)&=385. \\end{align} If the minimum possible value of $$z+\\frac{y}{2}+\\frac{x}{4}$$ can be written as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a+b$$? ×", "+0 # Assuming $x$, $y$, and $z$ are positive real numbers satisfying ​then, what is the value of $xyz$? 0 48 2 +389 Assuming $x$, $y$, and $z$ are positive real numbers satisfying:\\begin{align*} xy-z&=15, \\\\ xz-y&=0, \\text{ and} \\\\ yz-x&=0, \\end{align*}then, what is the value of $xyz$? May 4, 2020 #1 +1498 +1 Haven't done these in some time! So. Bullets 2 and 3 tell us that xz = y, and yz = x. xz = y yz = x xz + yz = x+y = z(x + y) z = (x+y)/(x+y) = 1! Ayyyy! xy - 1 = 15 xy = 16 16*1 = 16 If you don't understand anything feel free to ask. Of course idk if this is the right answer but i'm reasonably sure that this is. May 4, 2020 #2 +109519 -1", "# Prove that for all positive real numbers $x$ and $y$, $(x+y)(\\frac1x+\\frac1y)\\ge4$. Prove that for all positive real numbers $$x$$ and $$y$$, $$(x+y)(\\frac1x+\\frac1y)\\ge4$$. Any help would be appreciated thanks. • What have you tried? Jan 21 '19 at 4:17 • I tried expanding the left side but I do not know how to show that the left sider is also greater than 4, only equal to. – macy Jan 21 '19 at 4:19 • Use $AM \\geq HM$ Jan 21 '19 at 5:43 To begin with, notice that \\begin{align*} (x+y)\\left(\\frac{1}{x} + \\frac{1}{y}\\right) = 2 + \\frac{x}{y} + \\frac{y}{x} \\end{align*} According to the inequality $$AM\\geq GM$$, we have \\begin{align*} \\frac{x}{y} + \\frac{y}{x} \\geq 2\\sqrt{\\frac{x}{y}\\cdot\\frac{y}{x}} = 2 \\end{align*} from whence we obtain the claimed result. EDIT Due to JavaMan's observation, I provide you another approach to solve the second part. Indeed, for any $$a\\in\\textbf{R}_{>0}$$, we have $$(a-1)^{2}\\geq 0 \\Leftrightarrow a^{2} - 2a + 1 \\geq 0 \\Leftrightarrow a^{2} + 1 \\geq 2a\\Leftrightarrow a + \\frac{1}{a} \\geq 2$$ In the present case, it suffices to take $$a = x/y$$. • AM-GM is too much here. Simple algebra will do it. Jan 21 '19 at 4:21 • If you take a=x/y, doesn't the rhs still remain as 2? – macy Jan 21 '19 at 15:57 • Don't forget to add the previous", "# How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying $$x + y +z + 1 =4xyz.\\tag{1}$$ Prove that $$xy + yz + zx \\geqslant x + y + z.\\tag{2}$$ I don't know how to start? - Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\\ge$G.M., we get $xyz = (x+y+z+1)/4 \\ge (xyz)^{1/4}$. Hence $xyz\\ge1$. WLOG, suppose $0<x\\le y\\le z$. There are three cases. Case 1: $x\\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\\le y< z$. Then $(2)$ also holds because $$1-xyz\\le0\\le(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz.$$ Case 3: $0<x\\le y<1<z$. Then \\begin{align*} (2)&\\Leftrightarrow z(x+y) + xy \\ge x+y+z,\\\\ &\\Leftrightarrow z(x+y-1) + xy - x-y\\ge0,\\\\ &\\Leftrightarrow z(x+y-1)-1 + (x-1)(y-1) \\ge0. \\end{align*} So it suffices to show that $z(x+y-1)-1\\ge0$. Since $x,y,z$ are positive, we must have $4xy>1$, otherwise the LHS of $(1)$ will be strictly greater than the RHS. Now $(1)$ implies that $z=(x+y+1)/(4xy-1)$. Therefore $$z(x+y-1)-1 =\\frac{(x+y+1)(x+y-1)-(4xy-1)}{4xy-1} =\\frac{(x-y)^2}{4xy-1} \\ge0$$ and we are done. From the details of above three cases, it can be shown that equality in $(2)$ holds iff $x=y=z=1$. - Firstly, I'm going to relabel $x$,$y$ and $z$ to $\\alpha$,$\\beta$ and $\\gamma$. We then consider the cubic polynomial $$f(x) = ax^3 + bx^2 + cx + d$$ Supposing this has roots", "shows that the equality holds iff $x=y=z$. Here is my approach. Using AM-GM, \\begin{align} x^2+y^2+z^2 &\\ge 3(xyz)^{2/3} \\\\ \\implies 1 &\\ge 3(xyz)^{2/3} \\\\ \\implies (xyz)^{2/3} &\\le \\dfrac{1}{3} \\\\ \\implies xyz \\le \\dfrac{1}{3\\sqrt{3}} \\tag 1 \\end{align} Again, using AM-GM, \\begin{align} \\dfrac{x^3+y^3+z^3}{3} &\\ge 3xyz \\\\ \\implies x^3+y^3+z^3-3xyz &\\ge 0 \\\\ \\implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\\ge 0\\end{align}\\tag*{} Since $$x,y,z \\in \\mathbb R^+$$, $$x+y+z \\ge 0$$. \\begin{align}\\therefore \\, x^2+y^2+z^2 -xy-yz-zx &\\ge 0 \\\\ \\implies -x^2-y^2-z^2 +xy+yz+zx &\\le 0 \\\\ \\implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\\le 0 \\\\ \\implies x^2+y^2+z^2 +2xy+2yz+2zx &\\le 3(x^2+y^2+z^2) \\\\ \\implies (x+y+z)^2 &\\le 3 \\\\ \\implies x+y+z &\\le \\sqrt{3} \\tag 2 \\end{align} Multiplying $$(1)$$ and $$(2)$$, we can get the desired result.", "# Minimum Value of expression 2 What is the minimum value of $$(2x+3y)(8/x + 3/y)$$ when $x$ and $y$ are positive? I have tried expanding the equation but I did not get anything I got : $$25 + \\frac{6(4y^2 +x^2)}{xy}$$ but I don't know how to answer after this. I have also tried AM-GM but I couldn't manipulate the variables to cancel out and get the inequality. This is from Sipnayan 2017 You were on the right track: \\begin{align}(2x+3y)\\left(\\frac{8}{x} + \\frac{3}{y}\\right) &= 25+24 \\frac{y}{x}+6\\frac{x}{y}\\\\ &=25+12\\left(\\frac{2y}{x}+\\frac{x}{2y}\\right) \\\\ &\\geq 25+12 \\times 2 \\\\ &=49\\end{align} and notice that $x=2,y=1$ achieves this bound, so the minimum value is $49$. The bound $\\frac{2y}{x} + \\frac{x}{2y} \\geq 2$ can be proved from AM-GM, or from letting $t = \\frac{2y}{x}$ and considering the function $f(t) = t + \\frac{1}{t}$ and using calculus. From where you have left of, For real $x-2y,$ $$(x-2y)^2\\ge0\\implies x^2+4y^2\\ge4xy$$ Alternatively, using AM-GM inequality, $$\\dfrac{x^2+4y^2}2\\ge\\sqrt{4x^2y^2}=?$$ By C-S $$(2x+3y)\\left(\\frac{8}{x}+\\frac{3}{y}\\right)\\geq\\left(\\sqrt{2x\\cdot\\frac{8}{x}}+\\sqrt{3y\\cdot\\frac{3}{y}}\\right)^2=49.$$ The equality occurs for $x=2$ and $y=1$, which says that we got a minimal value. Done!", "# Solve for positive reals Solve for positive reals $x,y$ $(x+y)(1+\\frac{1}{xy})+4=2(\\sqrt{2x+1}+\\sqrt{2y+1})$ I started by accumulating the terms of $x$ and then used AM-GM inequality but unsuccessfully.... • Are you interested in the answer or the steps of the solution? – Paul Sundheim Aug 15 '14 at 18:28 • Steps of course... – Satvik Mashkaria Aug 15 '14 at 18:29 • You cannot really use AM-GM inequality, since you wish to solve an equality (though in some cases it would work). That would yield a solution only in some cases. – Theon Alexander Aug 15 '14 at 18:37 First: \\begin{align} (x+y)\\left(1+\\frac{1}{xy}\\right)+4 &= x+y+\\frac{1}y+\\frac{1}x +4\\\\ &= \\left(x+2+\\frac{1}{x}\\right)+\\left(y+2+\\frac{1}{y}\\right) \\\\ &=\\frac{1}{x} \\left(x+1\\right)^2 + \\frac{1}{y}\\left(y+1\\right)^2 \\end{align} So, define $$f(x)=\\frac{1}{x}\\left(x+1\\right)^2-2\\sqrt{2x+1}$$ Then you want to find two positive values $x,y$ so that $f(x)=-f(y)$. Can $f(u)$ ever be negative or zero? Letting $g(x)=xf(x) = (x+1)^2-2x\\sqrt{2x+1}$, you need to solve $g(x)\\leq 0$. So $(x+1)^4 \\leq 4x^2(2x+1)$ or $$x^4-4x^3+2x^2+4x+1\\leq 0$$ $x^4-4x^3+2x^2+4x+1$ factors (via Wolfram Alpha) as $(x^2-2x-1)^2$. So there are no negative values for $g$ and only one positive zero at $x=1+\\sqrt 2$, and therefore a solution of $x=y=1+\\sqrt 2$ to your original equation, and this is the only positive real solution. Let's see. Symmetric polynomials in $x,y$ are always polynomials in $x+y, xy$. The same goes with rational functions. The problem here is that we have radicals,", "# What is the maximum value of $\\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem: Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\\sqrt6 xy + 4yz$ I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$. • use the Lagrange Multiplier methode – Dr. Sonnhard Graubner Jan 9 '16 at 22:47 • I get $\\sqrt{\\frac{11}{2}}$ at $\\left(\\sqrt{\\frac{3}{22}},\\sqrt{\\frac{11}{22}},\\sqrt{\\frac{8}{22}}\\right)$. – CommonerG Jan 9 '16 at 23:20 Let $a,b\\in\\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\\in\\mathbb{R}$). By Cauchy-Schwarz, $axy+byz=(ax+bz)y\\leq \\left(\\sqrt{a^2+b^2}\\sqrt{x^2+z^2}\\right)|y|$. By AM-GM, $\\sqrt{x^2+z^2}\\,|y|\\leq \\frac{\\left(x^2+z^2\\right)+y^2}{2}=\\frac{1}{2}$. That is, $axy+byz\\leq \\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$. (Similarly, we also have the lower bound $axy+byz\\geq -\\frac{\\sqrt{a^2+b^2}}{2}$. The equality holds iff $\\left(x,y,z\\right)=\\pm\\left(\\frac{a}{\\sqrt{2\\left(a^2+b^2\\right)}},-\\frac{1}{\\sqrt{2}},\\frac{b}{\\sqrt{2\\left(a^2+b^2\\right)}}\\right)$.) • Pretty neat. Thank you. – Ashish Gupta Jan 9 '16 at 23:03 • I think your $y-$ and $z-$coördinates should be reversed in your solutions. – Théophile Jan 10 '16 at 0:15 • @Théophile Thanks, fixed. – Batominovski Jan 10 '16 at 0:52 We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\\sqrt{6}xy \\leq A^2x^2+B^2y^2$, and $4yz \\leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \\dfrac{\\sqrt{6}}{2},", "Let $x, y, z$ be positive real numbers such that $x^{2} + y^{2} + z^{2} = 1$. Prove that $x^{2}yz + xy^{2}z + xyz^{2} \\leq \\frac{1}{3}$. ## Solution Using Muirhead’s inequality, we have \\begin{align} 2(x^{2}yz + xy^{2}z + xyz^{2}) \\leq 2(x^{4}+y^{4}+z^{4}) \\label{y2002r1p1e1}\\\\ 2(x^{2}yz + xy^{2}z + xyz^{2}) \\leq 2x^{2}y^{2}+2x^{2}z^{2}+2y^{2}z^{2} \\label{y2002r1p1e2} \\end{align} We have $\\ref{y2002r1p1e1}$ because (4,0,0) majorizes (2,1,1). We have $\\ref{y2002r1p1e2}$ because (2,2,0) majorizes (2,1,1). Adding $\\ref{y2002r1p1e1}$ and $\\ref{y2002r1p1e2}$, we get \\begin{align} 3(x^{2}yz + xy^{2}z + xyz^{2}) \\leq (x^{2}+y^{2}+z^{2})^{2} = 1 \\\\ \\end{align}", "given positive real numbers x,y how to demonstrate that is defined as the maximum of x and y? Is it true that given positive real numbers $x,y$, then we have that $$\\sqrt{x^2 + y^2} \\geq \\max\\{ x, y \\}$$ I cant find a counter-example although it seems it is true... Any comments? - This is indeed true and follows directly from the fact that $y^2,x^2\\ge0$ hence $|x|\\sqrt{1+(y/x)^2}\\ge |x|$ (conversely the same is true for $y$) hence the result – b00n heT Feb 2 '14 at 10:57 why didn't you accept my answer? (I was the first to post...) – el.Salvador Feb 2 '14 at 11:19 I did it now! thank you very much for your help! – Marvin Gaye Feb 2 '14 at 12:22 We know that $${x^2} + \\underbrace {{y^2}}_{ \\ge 0} \\ge {x^2}$$ $$\\underbrace {{x^2}}_{ \\ge 0} + {y^2} \\ge {y^2}$$ So, taking the square root on both sides of each expression, we get $$\\sqrt {{x^2} + {y^2}} \\ge \\left| x \\right| \\ge x$$ $$\\sqrt {{x^2} + {y^2}} \\ge \\left| y \\right| \\ge y$$ Thus $$\\sqrt {{x^2} + {y^2}} \\ge \\max \\left\\{ {x,y} \\right\\}.$$ - Yes. To see this, note that $$\\tag{1}\\sqrt{x^2+y^2}\\geq \\sqrt{x^2}=x$$ since $x$ is positive. Similarly, we have $$\\tag{2}\\sqrt{x^2+y^2}\\geq \\sqrt{y^2}=y$$ since $y$ is positive. Combining $(1)$ and $(2)$, we have $$\\sqrt{x^2+y^2}\\geq\\max\\{x,y\\}.$$ -" ]

[ "by completing the square: $4x^2+11x+7=0$4x2+11x+7=0 1. Write all solutions on the same line, separated by commas.", "JoyLoveReynolds: Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. Find the solution set. The solution set for 7q2 − 28 = 0 is { }. (Separate the solutions with a comma) JoyLoveReynolds: JoyLoveReynolds: @theDeviliscoming @ThatCreepyBoy theDeviliscoming: $7q ^{2}-28=0$ you would add 28 to each side to make it $7q ^{2}=28$ then you would divide both sides by 7 to get $q ^{2}=4$ and then you would find the squareroot of each side to get $\\sqrt{q}=\\sqrt{4}$ the squarroot of 4 is 2 so q=2 theDeviliscoming: *squareroot JoyLoveReynolds: so its 2? theDeviliscoming: yes", "Write all solutions on the same line, separated by commas. ##### Question 3 Solve for the unknown: $-8x+x^2=-6-x-x^2$8x+x2=6xx2 1. Write all solutions on the same line, separated by commas. ##### Question 4 Solve the following equation: $x-\\frac{45}{x}=4$x45x=4 1. Write all solutions on the same line, separated by commas.", "Find the solutions of equation 2^ x equiv x^ 5 rounded to one decitrial place. Enter your answers as a comma-separated .", "+0 # Find all real values of t that satisfy the equation 0 97 1 Find all real values of t that satisfy the equation t^4 = 324. If you find more than one, then list your values in increasing order, separated by commas. Nov 1, 2020 $$\\sqrt[4]{324}$$ = - +4.24264", "### Home > INT3 > Chapter 2 > Lesson 2.2.3 > Problem2-86 2-86. Solve $\\sqrt{x+2}=8$ and check your solution. Square both sides. $x+2=64$ $x=62$ Be sure to show a check for your solution.", "the solutions of the equation. Two real solutions, two imaginary solutions, or one real solution. 2t^2-4t=0 7. Math/College Algebra 1.)Complete the square to find all real solutions of equation 3 x^2 + 4 x - 1 = 0. X1= X2= 2.)\\sqrt{7 x - 9} + 6 = 3x there are two solutions 3.)4 <= 5 x - 4 <10 4.)(x/4x-16)-7 = (1/x-4) 5.) (1/R) = (1/R1)+ (1/R2) for R1 8. algebra If (x-3) is a factor of 3x^3+bx^2-17x-12. a. Find the value of b. b.Hence , find the roots fo the equation 3x^3+bx^2-17x-12=0. 9. Math Find all real solutions of equation 2x^2+1-4=0 Does the equation have real solutions? 10. Pre-Cal Find all Real solutions of the Equation: (ENter as comma sparated answers, if there are no real solutions then just leave as NRS) x^(4/3)-9x^(2/3)+14=0 x=? More Similar Questions", "+0 # help people 0 67 1 Find all solutions to the equation $x^2 - 7x = 30$. If you find more than one, then list the values separated by commas. If the solutions are not real, then they should be written in $a + bi$ form. Jan 23, 2021 #1 0 x2 -7x - 30 = 0 Use Quadratic Formula a = 1 b = -7 c = -30 $$x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}$$ you will find x = 10 and -3 (x-10)(x+3) = x^2 -7x-30 check ! Jan 23, 2021", "/ 121m / 141 m /... ##### Consider f(x) = 5x3 31x2_ 29x + 7 . Find the solutions of each of the following equations_ (a) f(x) = 0 (b) f(x 1)=0 (c) f(x + 3) = 0 (d) f(3x) = 0(a) The solutions of flx) = 0 are x = (Type an integer : simplified fraction_ Use comma to separate answers as needed ) (b) The solutions of f(x- 1) =0 are X= (Type an integer Or simplified fraction Use comma to separale answers as needed ) (c) The solutions of f(x + 3) = 0 are x = (Type an integer or simplified fraction Use comma t0 separale answers Consider f(x) = 5x3 31x2_ 29x + 7 . Find the solutions of each of the following equations_ (a) f(x) = 0 (b) f(x 1)=0 (c) f(x + 3) = 0 (d) f(3x) = 0 (a) The solutions of flx) = 0 are x = (Type an integer : simplified fraction_ Use comma to separate answers as needed ) (b) The solutions of f(x- 1) =0 ... ##### What is the distance between (3, 2) and (8, 8)? What is the distance between (3, 2) and (8, 8)?... ##### 2. Data was collected in order to fit inear regression model [ Y = po + p *X", "# Find all real solutions of the equation. (Enter your answers as a comma-separated list.) 4x3 − 20x2 = 0 Question Find all real solutions of the equation. (Enter your answers as a comma-separated list.) 4x3 − 20x2 = 0", "+0 0 156 3 +11 Find all solutions of the equation and express them in the form a + bi. (Enter your answers as a comma-separated list. Simplify your answer completely.) x2 + 16x + 65 = 0 Apr 8, 2020 #1 0 By the quadratic formula, the solutions are -8 - 2i, -8 + 2i. Apr 8, 2020 #2 +10785 +1 Find all solutions of the equation and express them in the form a + bi. (Enter your answers as a comma-separated list. Simplify your answer completely.) x2 + 16x + 65 = 0 Hello Guest! $$x^2 + 16x + 65 = 0\\\\ x=-8\\pm\\sqrt{8^2-65}\\\\ x=-8\\pm\\sqrt{-1}$$ $$\\mathbb L=\\{-8+i\\ ,\\ -8-i\\}$$ ! Apr 8, 2020 #3 0 thaaaaannnkkkkssss Guest Apr 8, 2020", "(Hint: use find and string slicing). Show/Hide Solution 1. Calculate the zeros of the equation $$ax^2-b = 0$$ where a = 10 and b = 1. Hint: use math.sqrt or ** 0.5. Finally check that substituting the obtained value of x in the equation gives zero. Show/Hide Solution", "+0 0 345 1 Find all solutions to the equation $\\sqrt{3x+6}=x+2$. If there are multiple solutions, order them from least to greatest, separated by comma(s). Aug 17, 2017 #1 +22188 +1 Find all solutions to the equation \\sqrt{3x+6}=x+2 ($$\\sqrt{3x+6}=x+2$$). If there are multiple solutions, order them from least to greatest, separated by comma(s). $$\\begin{array}{|rcll|} \\hline \\sqrt{3x+6} &=& x+2 \\quad & | \\quad \\text{square both sides} \\\\ 3x+6 &=& (x+2)^2 \\\\ 3x+6 &=& x^2+4x+4 \\\\ x^2 +x -2 &=& 0 \\\\ (x+2)(x-1) &=& 0 \\\\\\\\ x_1 &=& -2 \\\\ x_2 &=& 1 \\\\ \\hline \\end{array}$$ Solution Set: {-2,1} Proof: $$x=-2:\\qquad \\sqrt{3\\cdot(-2)+6} = -2+2\\quad \\checkmark \\\\ x=1: \\qquad \\sqrt{3\\cdot(1)+6} = 1+2 \\quad \\checkmark$$ Aug 17, 2017 #1 +22188 +1 Find all solutions to the equation \\sqrt{3x+6}=x+2 ($$\\sqrt{3x+6}=x+2$$). If there are multiple solutions, order them from least to greatest, separated by comma(s). $$\\begin{array}{|rcll|} \\hline \\sqrt{3x+6} &=& x+2 \\quad & | \\quad \\text{square both sides} \\\\ 3x+6 &=& (x+2)^2 \\\\ 3x+6 &=& x^2+4x+4 \\\\ x^2 +x -2 &=& 0 \\\\ (x+2)(x-1) &=& 0 \\\\\\\\ x_1 &=& -2 \\\\ x_2 &=& 1 \\\\ \\hline \\end{array}$$ Solution Set: {-2,1} Proof: $$x=-2:\\qquad \\sqrt{3\\cdot(-2)+6} = -2+2\\quad \\checkmark \\\\ x=1: \\qquad \\sqrt{3\\cdot(1)+6} = 1+2 \\quad \\checkmark$$ heureka Aug 17, 2017", "# finding solutions • April 27th 2009, 06:21 PM jumpman23 finding solutions find all the solutions between 0 and 2pi 4csc^2 (x/2) - 7 csc (x/2)-2 = 0 • April 27th 2009, 06:43 PM Reckoner Quote: Originally Posted by jumpman23 find all the solutions between 0 and 2pi 4csc^2 (x/2) - 7 csc (x/2)-2 = 0 This is a quadratic in $\\csc\\frac x2.$ Factor, and then set each factor equal to 0. Of the two resulting equations, only one has a real solution.", "+0 # Math Help 0 92 2 +169 Given that x is an integer such thatx * √x - 5x - 9 * √x = 35 , find x. matth99 Jul 2, 2018 #1 +1 The solution to this is quite involved: Solve for x: -9 sqrt(x) - 5 x + x^(3/2) = 35 Subtract 35 from both sides: -35 - 9 sqrt(x) - 5 x + x^(3/2) = 0 Simplify and substitute y = sqrt(x). -35 - 9 sqrt(x) - 5 x + x^(3/2) = -35 - 9 sqrt(x) - 5 (sqrt(x))^2 + (sqrt(x))^3 = y^3 - 5 y^2 - 9 y - 35: y^3 - 5 y^2 - 9 y - 35 = 0 The left hand side factors into a product with two terms: (y - 7) (y^2 + 2 y + 5) = 0 Split into two equations: y - 7 = 0 or y^2 + 2 y + 5 = 0 y = 7 or y^2 + 2 y + 5 = 0 Substitute back for y = sqrt(x): sqrt(x) = 7 or y^2 + 2 y + 5 = 0 Raise both sides to the power of two: x = 49 or y^2 + 2 y + 5 = 0 Subtract 5 from both sides: x = 49 or y^2 + 2", "1. ## finding solutions find all the solutions between 0 and 2pi 4csc^2 (x/2) - 7 csc (x/2)-2 = 0 2. Originally Posted by jumpman23 find all the solutions between 0 and 2pi 4csc^2 (x/2) - 7 csc (x/2)-2 = 0 This is a quadratic in $\\csc\\frac x2.$ Factor, and then set each factor equal to 0. Of the two resulting equations, only one has a real solution.", "+0 # For which values of does the equation have no solution for ? Enter all the possible values of ​ separated by commas. 0 155 1 For which values of $$k$$ does the equation $$\\frac{x-1}{x-2} = \\frac{x-k}{x-6}$$ have no solution for $$x$$? Enter all the possible values of $$k,$$ separated by commas. Nov 23, 2019 #1 +108650 +1 You can see straight off that x cannot be 2 or 6. Now just for the moment you can ignore that. Try solving for x in terms of k and see what happens. The answer will present itself.", "+0 # help algebra 0 128 1 Find all solutions to the equation 6x^2 - 31x + 46 = 0. If you find more than one, then list the values separated by commas. If the solutions are not real, then they should be written in a + bi form. Jan 18, 2022 #1 +2437 +1 Use the quadratic formula $${-b \\pm \\sqrt{b^2-4ac} \\over 2a}$$ You will get two imaginary solutions: 31/12 + (sqrt143i)/12 and: 31/12 - (sqrt143i)/12 Jan 18, 2022", "+0 0 96 1 Find all solutions to the equation x^2 - 7x = 30 + 5x. If you find more than one, then list the values separated by commas. If the solutions are not real, then they should be written in a + bi form. Jun 4, 2022 #1 +2448 0 Simplifying into a quadratic, we have: $$x^2 - 12x - 30 = 0$$. Now, to solve, use the quadratic formula: $$x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}$$, where $$a = 1$$$$b = -12$$, and $$c = -30$$ Can you take it from here? Jun 4, 2022", "Enable text based alternatives for graph display and drawing entry Try Another Version of This Question Use the square root property to determine all real solutions for each of the following equations. 5x^2-900 = 0 x= 5j^2+400 = 0 j= Give exact solutions (don't use decimals), and separate multiple solutions with commas. If there are no real solutions, type DNE. Get help:" ]

[ "multiplying both sides of the above equation. We get, $48 + {y^2} = 19y$ ${y^2} - 19y + 48 = 0$ Now we had to solve the above quadratic equation of y to get the required value of y. So, splitting the middle term of the above quadratic equation. We get, ${y^2} - 16y - 3y + 48 = 0$ $y\\left( {y - 16} \\right) - 3\\left( {y - 16} \\right) = 0$ Now taking factor (y – 16) common from the above equation. We get, $\\left( {y - 16} \\right)\\left( {y - 3} \\right) = 0$ So, y = 16 or y = 3 Putting the value of y in equation 6. We get, When y = 16 Then, $x = \\dfrac{{48}}{y} = \\dfrac{{48}}{{16}} = 3$ When y = 3 Then, $x = \\dfrac{{48}}{y} = \\dfrac{{48}}{3} = 16$ So, the values of x and y are x = 16, y = 3 or x = 3, y = 16. Now as we know that the square root of $19 + 8\\sqrt 3$ is $\\sqrt x + \\sqrt y$. So, putting the value of x and y in $\\sqrt x + \\sqrt y$. We get, $\\sqrt x + \\sqrt y = \\sqrt {16} + \\sqrt 3$ Now as we know that the square root of 16 is 4.", "Adding each side of $$-6x – 2y = 12$$ to the corresponding side of $$6x+4y=10$$ gives $$2y=22$$ or $$y=11$$. Finally, substituting $$11$$ for $$y$$ in $$6x+4y=10$$ gives $$6x+4(11)=10$$ or $$x=-\\frac{17}{3}$$. 49- Choice B is correct $$x_{1,2} =\\frac{-b ± \\sqrt{b^2-4ac}}{2a}$$ , $$ax^2 + bx + c = 0, 4x^2 + 14x + 6 = 0$$ ⇒ then: $$a = 4$$, $$b = 14$$ and $$c = 6$$, $$x = \\frac{-14 +\\sqrt{14^2 – 4 .4 .6}}{2.4} = – \\frac{1}{2}, x =\\frac{-14 – \\sqrt{14^2 – 4 .4 .6}}{2.4} = – 3$$ 50- Choice A is correct Use Pythagorean theorem: $$a^2+b^2=c^2→s^2+h^2=(5s)^2→s^2+h^2=25s^2$$ Subtracting s^2 from both sides gives: $$h^2=24s^2$$ Square roots of both sides: $$h=\\sqrt{24s^2}=\\sqrt{4×6×s^2} =\\sqrt{4}×\\sqrt{6}×\\sqrt{s^2 }=2×s×\\sqrt{6}=2s\\sqrt{6}$$ 51- Choice D is correct Let $$x$$ be the number of purple marbles. Let’s review the choices provided: A. $$\\frac{1}{10}$$, if the probability of choosing a purple marble is one out of ten, then: Probability $$=\\frac{number \\space of \\space desired \\space outcomes}{number \\space of \\space total \\space outcomes}=\\frac{x}{20+30+40+x}=\\frac{1}{10}$$ Use cross multiplication and solve for $$x$$. $$10x=90+x→9x=90→x=9$$ Since, number of purple marbles can be 9, then, choice be the probability of randomly selecting a purple marble from the bag. Use same method for other choices. B. $$\\frac{1}{4}$$ $$\\frac{x}{20+30+40+x}=\\frac{1}{4}→4x=90+x→3x=90→x=30$$ C. $$\\frac{2}{5}$$ $$\\frac{x}{20+30+40+x}=\\frac{2}{5}→5x=180+2x→3x=180→x=60$$ D. $$\\frac{7}{15}$$ $$\\frac{x}{20+30+40+x}=\\frac{7}{15}→15x=630+7x→8x=630→x=78.75$$ Number of purple marbles cannot be a decimal. 52- Choice C is correct", "$x_1=289$ gives $y_1=\\sqrt{289\\cdot 144}=17\\cdot 12 = 204$, hence $(x,y)=(577,408)$, and the answer is $x+y=\\boxed{985}$. Solution 2 We quickly find the first solution, $(x,y)=(3,2)$. Factoring, we get $$(3-2\\sqrt{2})(3+2\\sqrt{2})=1$$ We can square both sides to get $$(3-2\\sqrt{2})^2(3+2\\sqrt{2})^2=1^2 \\Rightarrow (17-12\\sqrt{2})(17+12\\sqrt{2})=1$$ So $(x,y)=(17,12)$ is another solution. This gives us a way to generate whatever solutions we want to the equation. Raising the first equation to the fourth power gives us $$(577-408\\sqrt{2})(577+408\\sqrt{2})=1$$ The answer is $577+408=\\boxed{985}$.", "That's right. When you solve the two equations you get x=1 and y=1. 11. ok but i still don't understand how you came up with the two equations from the equation $x^3+3x^2y\\sqrt2+3xy^2\\sqrt2+y^3\\sqrt2$. i subsitutuded but it was unsuccessful. 12. When you arrange the equation into the form $x^3+6xy^2+(2y^3+3x^2y)\\sqrt{2}=7+5\\sqrt{2}$ The coefficient of $\\sqrt{2}$ on the left hand side is $2y^3+3x^2y$ and the coefficient of the $\\sqrt{2}$ on the right hand side of the equation is $5$ So one equation to solve is $2y^3+3x^2y=5$. Then the next equation is what is not the coefficient of $\\sqrt{2}$ on the left, equal to what is not the coeffient of $\\sqrt{2}$ on the right: $x^3+6xy^2=7$ 13. ok, firstly i understand how you got the two equations but the part that is confusing me is how did you get this equation: $x^3+3x^2y\\sqrt2+3xy^2\\sqrt2+y^3\\sqrt2$ to this equation: $x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$. 14. As posted above: $x+y\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ $\\Rightarrow (x+y\\sqrt{2})^3=7+5\\sqrt{2}$ (from cubing both sides) $\\Rightarrow x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$ 15. Originally Posted by Stroodle For the first question: $x+y\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ $\\Rightarrow (x+y\\sqrt{2})^3=7+5\\sqrt{2}$ $\\Rightarrow x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$ So from this you can now get two equations to solve simultaneously: $x^3+6xy^2=7$ and $2y^3+3x^2y=5$ Hence $x=1$ and $y=1$ $\\Rightarrow 1+\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ Quick question? How would you go about solving the 2 equations if it would not have been as easy as spotting that x=1 and y=1 works. Would you sub y=mx?", "get: ⇒ $$\\rm \\frac{336}{\\sqrt x}=\\frac{112}{\\sqrt y}$$ ⇒ $$\\rm 3{\\sqrt y}=\\sqrt x$$ ... (3) Squaring both sides, we get: ⇒ 9y = x ... (4) Substituting values from equations (3) and (4) in equation (1) [or equation (2)], we get: ⇒ $$\\rm 9y\\left(3\\sqrt y\\right) + y\\sqrt y=\\frac{336}{3\\sqrt y}$$ ⇒ 81y2 + 3y2 = 336 ⇒ 84y2 = 336 ⇒ y2 = 4 Since x and y are positive real numbers, we must have y = 2, and using equation (4) we get x = 18. ∴ x + y = 20. # How many prime numbers are there between 1 and 50? 1. 17 2. 15 3. 14 4. 16 5. None of the above Option 2 : 15 ## Detailed Solution Calculation: Prime numbers between 1 and 50 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 Total number of prime numbers = 15 ∴ There are 15 prime numbers between 1 and 50 # What will be the remainder when 340 is divided by 4? 1. 1 2. 2 3. 3 4. 0 Option 1 : 1 ## Detailed Solution Given: Number = 340 Divided by = 4 Calculations: ⇒ 340/4 ⇒ (32)20/4 ⇒ (9)20/4 ⇒ (1)20 ⇒ 1 ∴ When 340 is divided by 4 the remainder will be", "it is equal to 64. Therefore, I can take the cube root of both sides of the equation. The cube root of x3, $$\\sqrt[3]{x^{3}}$$, is x because x3 = x ∙ x ∙ x. The cube root of 64, $$\\sqrt [ 3 ]{ 64 }$$, is 4 because 64 = 4 ∙ 4 ∙ 4. Therefore, x = 4. Exercise 7. Find the positive value of x that makes the equation true: x2 = 256-1. x2 = 256-1 $$\\sqrt{x^{2}}$$ = $$\\sqrt{256^{-1}}$$ x = $$\\sqrt{256^{-1}}$$ x = $$\\sqrt{\\frac{1}{256}}$$ x = $$\\frac{1}{16}$$ x = 16-1 Check: (16-1)2 = 256-1 16-2 = 256-1 $$\\frac{1}{16^{2}}$$ = 256-1 $$\\frac{1}{256}$$ = 256-1 256-1 = 256-1 Exercise 8. Find the positive value of x that makes the equation true: x3 = 343-1. x3 = 343-1 $$\\sqrt[3]{x^{3}}$$ = $$\\sqrt[3]{343^{-1}}$$ x = $$\\sqrt[3]{343^{-1}}$$ x = $$\\sqrt[3]{\\frac{1}{343}}$$ x = $$\\frac{1}{7}$$ x = 7-1 Check: (7-1)3 = 343-1 73 = 343-1 ($$\\frac{1}{7}$$)3 = 343-1 $$\\frac{1}{343}$$ = 343-1 343-1 = 343-1 Exercise 9. Is 6 a solution to the equation x2 – 4 = 5x? Explain why or why not. 62 – 4 = 5(6) 36 – 4 = 30 32 ≠ 30 No, 6 is not a solution to the equation x2 – 4 = 5x. When the number is substituted into the equation and simplified, the left side", "Browse Questions # Solve the equation $sin^{-1}6x+sin^{-1}6\\sqrt 3 x = -\\frac{\\pi}{2}.$ Can you answer this question? Toolbox: • $sin^{-1}x+sin^{-1}y=sin^{-1}\\: x\\sqrt{1-y^2}+y\\sqrt{1-x^2}$ • $sin(-\\frac{\\pi}{2})=-1$ By taking 6x in the place of x and $6\\sqrt3x$ in the place of y in the formula of $sin^{-1}x+sin^{-1}y$ we get $sin^{-1}6x+sin^{-1}6\\sqrt3x=$ $sin^{-1} \\bigg[ 6x \\sqrt{1-108x^2}+6\\sqrt 3 x\\sqrt{1-36x^2} \\bigg]$$= -\\frac{\\pi}{2}$ Taking sin on both the sides $\\Rightarrow\\: 6x\\sqrt{1-108x^2}+6\\sqrt 3 x\\sqrt{1-36x^2}=sin(-\\frac{\\pi}{2})=-1$ We know that $sin(-\\frac{\\pi}{2})=-1$ $\\Rightarrow 6\\sqrt 3 x \\sqrt{1-36x^2}=-1-6x\\sqrt{1-108x^2}$ Squaring on both the sides we get $\\Rightarrow 108x^2(1-36x^2)=1+36x^2(1-108x^2)+12x\\sqrt{1-108x^2}$ $\\Rightarrow\\: 72x^2-1=12x\\sqrt{1-108x^2}$ Again squaring both the sides we get $5184x^4+1-144x^2=144x^2(1-108x^2)=144x^2-15552x^4$ answered Feb 28, 2013 edited Mar 21, 2013", "# If x and y are positive real numbers satisfying the system of equations $$\\rm x^2 + y\\sqrt{xy}=336$$ and $$\\rm y^2 + x\\sqrt{xy} = 112$$, then x + y is: This question was previously asked in NIMCET 2014 Official Paper View all NIMCET Papers > 1. $$\\sqrt{448}$$ 2. $$\\sqrt{224}$$ 3. 20 4. 40 Option 3 : 20 Free NIMCET 2020 Official Paper 1342 120 Questions 480 Marks 120 Mins ## Detailed Solution Concept: Simultaneous equations of degree not equal to 1 (not linear) can be solved comparatively easily by elimination/substitution. Calculation: It is given that $$\\rm x^2 + y\\sqrt{xy}=336$$. Dividing by $$\\rm \\sqrt x$$, we get: ⇒ $$\\rm x\\sqrt x + y\\sqrt y=\\frac{336}{\\sqrt x}$$ ... (1) It is also given that $$\\rm y^2 + x\\sqrt{xy} = 112$$. Dividing by $$\\rm \\sqrt y$$, we get: ⇒ $$\\rm y\\sqrt y + x\\sqrt x=\\frac{112}{\\sqrt y}$$ ... (2) From equations (1) and (2), we get: ⇒ $$\\rm \\frac{336}{\\sqrt x}=\\frac{112}{\\sqrt y}$$ ⇒ $$\\rm 3{\\sqrt y}=\\sqrt x$$ ... (3) Squaring both sides, we get: ⇒ 9y = x ... (4) Substituting values from equations (3) and (4) in equation (1) [or equation (2)], we get: ⇒ $$\\rm 9y\\left(3\\sqrt y\\right) + y\\sqrt y=\\frac{336}{3\\sqrt y}$$ ⇒ 81y2 + 3y2 = 336 ⇒ 84y2 = 336 ⇒ y2 = 4 Since x and y are positive real", "is $$\\frac{7}{12}$$, this gives $$(\\frac{12}{7})×\\frac{7}{12} y=\\frac{3}{5}×(\\frac{12}{7})$$, which simplifies to $$y=\\frac{36}{35}$$. 27- Choice A is correct Subtracting $$3x$$ and adding 2 to both sides of $$3x+6≥5x-2$$ gives $$8≥ 2x$$ and $$4≥x$$. Therefore, $$x$$ is a solution to $$3x+6≥5x-2$$ if and only if $$x$$ is less than or equal to 4 and $$x$$ is NOT a solution to $$3x+6≥5x-2$$ if and only if $$x$$ is greater than 4. Of the choices given, only 6 is greater than 4 and, therefore, cannot be a value of $$x$$. 28- Choice A is correct Given the two equations, substitute the numerical value of a into the second equation to solve for $$x$$. $$a=\\sqrt{5}, 6a=\\sqrt{12x}$$ Substituting the numerical value for a into the equation with $$x$$ is as follows. $$6(\\sqrt{5})=\\sqrt{12x}$$, From here, distribute the 4. $$6\\sqrt{5}=\\sqrt{12x}$$ Now square both side of the equation. $$(6\\sqrt{5})^2=(\\sqrt{12x})^2$$ Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. $$6^2 \\sqrt{5}^2=12x, 36(5)=12x, 180=12x, x=15$$ 29- Choice B is correct First square both sides of the equation to get $$-4m+5=m^2$$ Subtracting both sides by $$-4m+5$$ gives us the equation $$m^2+4m-5=0$$ Here you can solve the quadratic equation by factoring to get $$(m-1)(m+5)=0$$ For the expression $$(m-1)(m+5)$$ to equal zero, $$m=1$$ or $$m=-5$$ ## The Absolute Best Book to Ace the SAT", "-2: 3x^2- y- 2x= -2. That was the mistaken equation I got before. It should be $3x^2- y- 2x= -2$ So $3x^2- y- 2x= 2$ and y= 2x+1. Replace y in the first equation by 2x+1 and we have $3x^2- 2x-1- 2x= -2$ or $3x^2- 4x+ 1= 0$. By the quadratic formula, the roots of that are [tex]\\frac{4\\pm\\sqrt{16- 9}}{6}= \\frac{4\\pm\\sqrt{7}}{6}[/itex]. Since 7 is just a little less than 9, $\\sqrt{7}$ is just a little less than 3. If we were to take the positive, we would have $4+ \\sqrt{7}$ a little bit more than 7 and $(4+ \\sqrt{7})/6$ a little bit larger than 1. If we take the negative, we have $4- \\sqrt{7}$ a little smaller than 1 and $(4-\\sqrt{7})/6$ between 0 and 1. In order that (x,y) be between (0,1) and (1,3) x must be between 0 and 1 and so $x= (4- \\sqrt{7})/6$. Then, of course, $y= 2x+ 1= (4- \\sqrt{7}}/3+ 1= (7- \\sqrt{7})/3$ which is a little larger than 1 and so between 1 and 3.", "inches. 11- C Multiplying each side of $$\\frac{4}{x}=\\frac{12}{x-8}$$ by $$x(x-8)$$ gives $$4(x-8)=12(x)$$, distributing the 4 over the values within the parentheses yields $$x-8=3x$$ or $$x=-4$$. Therefore, the value of $$\\frac{x}{2}=\\frac{-4}{2}=-2$$. 12- D Given the two equations, substitute the numerical value of a into the second equation to solve for $$x$$. $$a=\\sqrt{3}, 4a=\\sqrt{4x}$$ Substituting the numerical value for a into the equation with $$x$$ is as follows. $$4(\\sqrt{3})=\\sqrt{4x}$$, From here, distribute the $$4$$. $$4\\sqrt{3}=\\sqrt{4x}$$ Now square both side of the equation. $$(4\\sqrt{3})^2=(\\sqrt{4x})^2$$ Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. $$4^2 \\sqrt{3}^2=4x, 16(3)=4x, 48=4x, x=12$$ 13- D In the figure angle A is labeled $$(3x-2)$$ and it measures $$37$$. Thus, $$3x-2=37$$ and $$3x=39$$ or $$x=13$$. That means that angle B, which is labeled $$(5x)$$, must measure $$5×13=65$$. Since the three angles of a triangle must add up to $$180, 37+65+y-8=180$$, then: $$y+94=180→y=180-94=86$$ 14- C Let $$x$$ be all expenses, then $$\\frac{22}{100} x=660 →x=\\frac{100×660}{22}=3,000$$ Mr. Jones spent for his rent: $$\\frac{ 27}{100}×3,000=810$$ 15- A The amount of money for $$x$$ bookshelf is: $$200x$$, Then, the total cost of all bookshelves is equal to: $$200x+600$$, The total cost, in dollar, per bookshelf is: $$\\frac{Total \\ cost}{number \\ of \\ items}=\\frac{200x+600}{x}$$ 16- B It is given that g(5)=4. Therefore, to find", "Math Help - solve by factoring/give an example please 1. solve by factoring/give an example please w2-169=0 36a2= 169 15t2-11t-12= 0 2. Please use the ^ sign for powers. As you have it written, the first one means w times 2. But anyway, for the first one. Notice the difference of two squares?. $(w)^{2}-(13)^{2}$. Now, can you factor using the proper form?. (x+y)(x-y). 3. $36a^2=169$ Take the square root of both sides. $\\sqrt{36a^2}\\ = \\sqrt 169$ $6a=13$ $a= \\frac{13}{6}$ For the other one, you can factor it. $15t^2-11t-12 = (3t-4)(5t+3)$ And from there, it's easy to solve. 4. Originally Posted by quarks $36a^2=169$ Take the square root of both sides. $\\sqrt{36a^2}\\ = \\sqrt 169$ So far, so good. But $\\sqrt{36a^2} = \\sqrt{169}$ implies $6a = \\pm 13$ $a = \\pm \\frac{13}{6}$ -Dan 5. Yeah. Thanks. =) My mistake.", "$x=-13\\sqrt{11}$ or $x=-\\sqrt{11}$ The approximate value of $\\sqrt{11} = 3$ Then, $x=-39$ or $x=-3$ II: $y^2-22\\sqrt{3}y+360=0$ $y^2-20\\sqrt{3}y-12\\sqrt{3}y+360=0$ $y(y-20\\sqrt{3})-12\\sqrt{3}(y-20\\sqrt{3})=0$ $(y-20\\sqrt{3})(y-12\\sqrt{3})=0$ $y=20\\sqrt{3}$ or $y=12\\sqrt{3}$ The approximate value of $\\sqrt{3}=1$ Then, $x=20$ or $x=12$ Comparing x and y, Both the x values are negative and both the y values are positive. Therefore, x is less than y. I: $x^3-128=1727872$ $x^3 = 1728000$ $x=120$ II: $\\sqrt{3}{y^2} = \\dfrac{\\sqrt{2}{y^3}}{121^\\frac{5}{6}}$ $y^\\frac{2}{3} = \\dfrac{y^\\frac{3}{2}}{121^\\frac{5}{5}}$ $y^{\\frac{3}{2}-\\frac{2}{3}} = 121^\\frac{5}{6}$ $y^\\frac{5}{6}=121^\\frac{5}{6}$ $y = 121$ Comparing x and y, $120 < 121$. Therefore, x is less than y. I: $x^2-x-812=0$ $x^2-29x+28x-812=0$ $x(x-29)+28(x-29)=0$ $(x-29)(x+28)=0$ $x=29$ or $x=-28$ II: $y^2+y-1332=0$ $y^2+37y-36y-1332=0$ $y(y+37)-36(y+37)=0$ $(y+37)(y-36)=0$ $y=-37$ or $y=36$ Comparing x and y, $29>-37$ $29<36$ $-28>-37$ $-29<36$ Therefore, The relationship between x and y cannot be established. I: $x^2+0.25x-60=0$ $x^2+\\dfrac{x}{4}-60=0$ $4x^2+x-240=0$ $4x^2+16x-15x-240=0$ $4x(x+16)-15(x+16)=0$ $(x+16)(4x-15)=0$ $x=-16$ or $x=\\dfrac{15}{4}$ II: $y^2-0.33y-8=0$ $y^2-\\dfrac{y}{3}-8=0$ $3y^2-y-24=0$ $3y^2-9y+8y-24=0$ $3y(y-3)+8(y-3)=0$ $(y-3)(3y+8)=0$ $y=3$ or $y=\\dfrac{-8}{3}$ Comparing x and y $-16<3$ $-16$\\dfrac{15}{4}>3\\dfrac{15}{4}>\\dfrac{-8}{3}$Therefore, The relationship between x and y cannot be established. 8) Answer (A) I:$\\sqrt{x+14}+\\sqrt{841} = \\sqrt{1369}\\sqrt{x+14}+29=37\\sqrt{x+14}=8x+14=64x=40$II:$y^2+0.5y-60=02y^2+y-120=02y^2+16y-15y-120=02y(y+8)-15(y+8)=0(y+8)(2y-15)=0y=-8$or$y=\\dfrac{15}{2}=7.5$Comparing x and y$40 > -840 > 7.5$Therefore, x is greater than y. 9) Answer (E) I:$x^2-16\\sqrt{5}x+300=0x^2-10\\sqrt{5}x-6\\sqrt{5}x+300=0x(x-10\\sqrt{5})-6\\sqrt{5}(x-10\\sqrt{5})=0(x-10\\sqrt{5})(x-6\\sqrt{5})=0x=10\\sqrt{5}$or$x=6\\sqrt{5}$II:$y^2-31\\sqrt{5}y+750=0y^2-25\\sqrt{5}y-6\\sqrt{5}y+750=0y(y-25\\sqrt{5})-6\\sqrt{5}(y-25\\sqrt{5})=0(y-25\\sqrt{5})(y-6\\sqrt{5})=0y=25\\sqrt{5}$or$y=6\\sqrt{5}$Comparing x and y,$10\\sqrt{5} < 25\\sqrt{5}10\\sqrt{5} > 6\\sqrt{5}6\\sqrt{5} < 25\\sqrt{5}6\\sqrt{5}=6\\sqrt{5}$Therefore, The relationship between x and y cannot be determined. 10) Answer (B) I:$6\\sqrt{x}+\\dfrac{5}{\\sqrt{x}} = \\sqrt{x}\\dfrac{6x+5}{\\sqrt{x}} = \\sqrt{x}6x+5=x5x=-5x=-1$II:$\\dfrac{2^\\frac{5}{9}}{\\sqrt[3]{y}} = y^\\frac{2}{9}2^\\frac{5}{9} = y^\\frac{2}{9} \\times y^\\frac{1}{3}2^\\frac{5}{9} = y^\\frac{5}{9}y=2$By comparing x and y,$-1<2$Therefore, x is", "another pair of the equations we have, AO =BO sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9-y)^2) Squaring on both sides of the equation we have, (5 - x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2 25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y 6x + 2y = 4 3x + y = 2 Equating another pair of the equations we have, AO = CO sqrt((5 -x)^2 + (-8 - y)^2) = sqrt((2 -x)^2 + (1 - y)^2) 25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 -4x + 1 + y^2 - 2y` 6x - 8y = 84 x - 3y = 14 Now we have two equations for ‘x’ and ‘y’, which are 3x + y = 2 x - 3y = 14 From the second equation we have y = -3x + 2. Substituting this value of ‘y’ in the first equation we have, x - 3(-3x + 2) = 14 x + 9x - 6 = 14 10x = 20 x = 2 Therefore the value of ‘y’ is, y = -3x + 2 = -3(2) + 2 y = -4 Hence the", "+0 # Math Help 0 92 2 +169 Given that x is an integer such thatx * √x - 5x - 9 * √x = 35 , find x. matth99 Jul 2, 2018 #1 +1 The solution to this is quite involved: Solve for x: -9 sqrt(x) - 5 x + x^(3/2) = 35 Subtract 35 from both sides: -35 - 9 sqrt(x) - 5 x + x^(3/2) = 0 Simplify and substitute y = sqrt(x). -35 - 9 sqrt(x) - 5 x + x^(3/2) = -35 - 9 sqrt(x) - 5 (sqrt(x))^2 + (sqrt(x))^3 = y^3 - 5 y^2 - 9 y - 35: y^3 - 5 y^2 - 9 y - 35 = 0 The left hand side factors into a product with two terms: (y - 7) (y^2 + 2 y + 5) = 0 Split into two equations: y - 7 = 0 or y^2 + 2 y + 5 = 0 y = 7 or y^2 + 2 y + 5 = 0 Substitute back for y = sqrt(x): sqrt(x) = 7 or y^2 + 2 y + 5 = 0 Raise both sides to the power of two: x = 49 or y^2 + 2 y + 5 = 0 Subtract 5 from both sides: x = 49 or y^2 + 2", "= 1331 – 1 Question 3. Determine the positive value of x that makes the equation true, and then explain how you solved the equation. $$\\frac{x^{9}}{x^{7}}$$ – 49 = 0 $$\\frac{x^{9}}{x^{7}}$$ – 49 = 0 x2 – 49 = 0 x2 – 49 + 49 = 0 + 49 x2 = 49 $$\\sqrt{x^{2}}$$ = $$\\sqrt{49}$$ x = 7 Check: 72 – 49 = 0 49 – 49 = 0 0 = 0 To solve the equation, I first had to simplify the expression $$\\frac{x^{9}}{x^{7}}$$ to x2. Next, I used the properties of equality to transform the equation into x2 = 49. Finally, I had to take the square root of both sides of the equation to solve for x. Question 4. Determine the positive value of x that makes the equation true. (8x)2 = 1 (8x)2 = 1 64x2 = 1 $$\\sqrt{64^{2}}$$ = $$\\sqrt{1}$$ 8x = 1 $$\\frac{8x}{8}$$ = $$\\frac{1}{8}$$ x = $$\\frac{1}{8}$$ Check: (8($$\\frac{1}{8}$$))2 = 1 12 = 1 1 = 1 Question 5. (9$$\\sqrt{x}$$)2 – 43x = 76 (9$$\\sqrt{x}$$)2 – 43x = 76 92 (√x)2 – 43x = 76 81x – 43x = 76 38x = 76 $$\\frac{38x}{38}$$ = $$\\frac{76}{38}$$ x = 2 Check: (9($$\\sqrt{2}$$))2 – 43(2) = 76 92 ($$\\sqrt{2}$$)2 – 86 = 76 81(2) – 86 = 76 162 – 86 = 76 76", "distributed on the circle $$R = \\{(x, y) \\in \\R^2: x^2 + y^2 \\lt 36\\}$$. The conditional PDF of $$X$$ given $$Y = y$$ is denoted $$x \\mapsto g(x \\mid y)$$. The conditional PDF of $$Y$$ given $$X = x$$ is denoted $$y \\mapsto h(y \\mid x)$$. • For $$y \\in (-6, 6)$$, $$g(x \\mid y) = \\frac{1}{12}$$ for $$x \\in (-6, 6)$$. • For $$x \\in (-6, 6)$$, $$h(y \\mid x) = \\frac{1}{12}$$ for $$y \\in (-6, 6)$$. • $$X$$, $$Y$$ are independent. • For $$y \\in (-6, 6)$$, $$g(x \\mid y) = \\frac{1}{6 - y}$$ for $$x \\in (y, 6)$$ • For $$x \\in (-6, 6)$$, $$h(y \\mid x) = \\frac{1}{x + 6}$$ for $$y \\in (-6, x)$$ • $$X$$, $$Y$$ are dependent. • For $$y \\in (-6, 6)$$, $$g(x \\mid y) = \\frac{1}{2 \\sqrt{36 - y^2}}$$ for $$x \\in \\left(-\\sqrt{36 - y^2}, \\sqrt{36 - y^2}\\right)$$ • For $$x \\in (-6, 6)$$, $$g(x \\mid y) = \\frac{1}{2 \\sqrt{36 - x^2}}$$ for $$y \\in \\left(-\\sqrt{36 - x^2}, \\sqrt{36 - x^2}\\right)$$ • $$X$$, $$Y$$ are dependent. In the bivariate uniform experiment, run the simulation 1000 times in each of the following cases. Watch the points in the scatter plot and the graphs of the marginal distributions. 1. square 2. triangle 3. circle Suppose that $$(X, Y, Z)$$ is", "$y \\leq 5$ 2. Solve the equation : $3(6x+2)=2x$ 1. $x=- \\frac{3}{8}$ 2. $x=\\frac{3}{8}$ 3. $x=\\frac{8}{3}$ 4. $x=-\\frac{8}{3}$ 5. $x=-3$ 3. What is the multiplicative inverse of $-7$? 1. $\\frac{1}{7}$ 2. $-\\frac{1}{7}$ 3. $0$ 4. $7$ 5. $-7$ 4. Simplify : $\\sqrt{169}.\\sqrt{81}.\\sqrt{196}$ 1. $13689$ 2. $21294$ 3. $1638$ 4. $1764$ 5. $12356$ 5. Factor: $b^{2}-121$ 1. $(b-11)(b+11)$ 2. $(b^{2}-11)(b^{2}+11)$ 3. $(b^{2}+11)(b^{2}+11)$ 4. $(b-11)(b-11)$ 5. $(b+11)(b+11)$ 6. Solve: $1+\\frac{1}{x}=\\frac{12}{x^{2}}$ 1. $x=-4$ or $x=3$ 2. $x= 4$ or $x=-3$ 3. $x=-4$ or $x=-4$ 4. $x=-3$ or $x=3$ 5. $x= 6$ or $x=-2$ 7. Factor the following equation: $a^{2}-2a-35$ 1. $(a-7)(a+5)$ 2. $(a+7)(a+5)$ 3. $(a-5)(a-7)$ 4. $(a-1)(a+35)$ 5. $(a-35)(a+1)$ 8. Solve the following inequality: $9(6-2x) \\leq -12$ 1. $x\\geq 11$ 2. $x \\leq \\frac{11}{3}$ 3. $x \\geq \\frac{11}{3}$ 4. $x\\geq$ 3 5. $x \\leq$ -11 9. Determine the x-intercepts of the following parabola: $y=3x^{2}+x-14$ 1. $(2,0)$ and $(-\\frac{7}{3},0)$ 2. $(2,0)$ and $(-3,0)$ 3. $(-\\frac{3}{7},0)$ and $(-4,0)$ 4. $(0,2)$ and $(0,-\\frac{7}{3})$ 5. $(0,-\\frac{7}{3})$ and $(2,0)$ 10. A Graph of $x^{2} + y^{2} = 9$ is a 1. Parabola 2. Hyperbola 3. Circle 4. Line 5. ellipse ## Friday, June 24, 2011 ### CLEP College Mathematics Practice Questions - 1 1. Simplify the following expression : $8x+3-3x-7$ 1. $5x-4$ 2. $5x+4$ 3. $11x-4$ 4. $10x-4$ 5. $-5x-4$ 2. Determine the mean of the", "# Find all solutions to the non-linear systems: 4sqrtx + sqrty = 34 and sqrtx + 4sqrty = 16 ? Dec 5, 2017 x=64 y=4 #### Explanation: if $\\sqrt{x} = a$ $\\sqrt{y} = b$ then $4 a + b = 34$ $a + 4 b = 16$ multyply the first equation by -1; the second by 4 and add them together. Then you get $15 b = 64 - 34$ $b = \\frac{30}{15} = 2$ Now. Let's go back to find a. 4a+b=34 so after substitating b=2 you get: $a = \\frac{32}{4} = 8$ then back: $\\sqrt{x} = a$ $\\sqrt{x} = 8$ $x = {8}^{2} = 64$ $\\sqrt{y} = b$ $\\sqrt{y} = 2$ $y = {2}^{2} = 4$ the answers are $\\therefore x = 64 \\mathmr{and} y = 4$ #### Explanation: the equations are $4 \\sqrt{x} + \\sqrt{y} = 34 \\to \\left(1\\right)$ $\\sqrt{x} + 4 \\sqrt{y} = 16 \\to \\left(2\\right)$ multiplying $\\left(2\\right) b y - 4$ we get $- 4 \\sqrt{x} - 16 \\sqrt{y} = - 64 \\to \\left(3\\right)$ solving $\\left(2\\right) + \\left(3\\right)$ $4 \\sqrt{x} + \\sqrt{y} - 4 \\sqrt{x} - 16 \\sqrt{y} = - 64 + 34 \\Rightarrow - 15 \\sqrt{y} = - 30 \\Rightarrow \\sqrt{y} = 2 \\Rightarrow y = 4$ put $y = 4$ in $\\left(1\\right)$ we get $4 \\sqrt{x} + 2 = 34 \\Rightarrow", "How do you solve sqrt(8x)+1=65 and check the solution? May 23, 2017 See a solution process below: Explanation: First, subtract $\\textcolor{red}{1}$ from each side of the equation to isolate the radical while keeping the equation balanced: $\\sqrt{8 x} + 1 - \\textcolor{red}{1} = 65 - \\textcolor{red}{1}$ $\\sqrt{8 x} + 0 = 64$ $\\sqrt{8 x} = 64$ Next, square each side of the equation to remove the radical while keeping the equation balanced: ${\\left(\\sqrt{8 x}\\right)}^{2} = {64}^{2}$ $8 x = 4096$ Now, divide each side of the equation by $\\textcolor{red}{8}$ to solve for $x$ while keeping the equation balanced: $\\frac{8 x}{\\textcolor{red}{8}} = \\frac{4096}{\\textcolor{red}{8}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{8}}} x}{\\cancel{\\textcolor{red}{8}}} = 512$ $x = 512$ To check the solution substitute $\\textcolor{red}{512}$ for $\\textcolor{red}{x}$ and calculate each side of the equation to ensure both sides are equal: $\\sqrt{8 \\textcolor{red}{x}} + 1 = 65$ becomes: $\\sqrt{8 \\cdot \\textcolor{red}{512}} + 1 = 65$ $\\sqrt{4096} + 1 = 65$ $64 + 1 = 65$ $65 + 65$ Both sides of the equation are equal therefore the solution is valid." ]

[ "it is equal to 64. Therefore, I can take the cube root of both sides of the equation. The cube root of x3, $$\\sqrt[3]{x^{3}}$$, is x because x3 = x ∙ x ∙ x. The cube root of 64, $$\\sqrt [ 3 ]{ 64 }$$, is 4 because 64 = 4 ∙ 4 ∙ 4. Therefore, x = 4. Exercise 7. Find the positive value of x that makes the equation true: x2 = 256-1. x2 = 256-1 $$\\sqrt{x^{2}}$$ = $$\\sqrt{256^{-1}}$$ x = $$\\sqrt{256^{-1}}$$ x = $$\\sqrt{\\frac{1}{256}}$$ x = $$\\frac{1}{16}$$ x = 16-1 Check: (16-1)2 = 256-1 16-2 = 256-1 $$\\frac{1}{16^{2}}$$ = 256-1 $$\\frac{1}{256}$$ = 256-1 256-1 = 256-1 Exercise 8. Find the positive value of x that makes the equation true: x3 = 343-1. x3 = 343-1 $$\\sqrt[3]{x^{3}}$$ = $$\\sqrt[3]{343^{-1}}$$ x = $$\\sqrt[3]{343^{-1}}$$ x = $$\\sqrt[3]{\\frac{1}{343}}$$ x = $$\\frac{1}{7}$$ x = 7-1 Check: (7-1)3 = 343-1 73 = 343-1 ($$\\frac{1}{7}$$)3 = 343-1 $$\\frac{1}{343}$$ = 343-1 343-1 = 343-1 Exercise 9. Is 6 a solution to the equation x2 – 4 = 5x? Explain why or why not. 62 – 4 = 5(6) 36 – 4 = 30 32 ≠ 30 No, 6 is not a solution to the equation x2 – 4 = 5x. When the number is substituted into the equation and simplified, the left side", "How do you solve sqrt(8x)+1=65 and check the solution? May 23, 2017 See a solution process below: Explanation: First, subtract $\\textcolor{red}{1}$ from each side of the equation to isolate the radical while keeping the equation balanced: $\\sqrt{8 x} + 1 - \\textcolor{red}{1} = 65 - \\textcolor{red}{1}$ $\\sqrt{8 x} + 0 = 64$ $\\sqrt{8 x} = 64$ Next, square each side of the equation to remove the radical while keeping the equation balanced: ${\\left(\\sqrt{8 x}\\right)}^{2} = {64}^{2}$ $8 x = 4096$ Now, divide each side of the equation by $\\textcolor{red}{8}$ to solve for $x$ while keeping the equation balanced: $\\frac{8 x}{\\textcolor{red}{8}} = \\frac{4096}{\\textcolor{red}{8}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{8}}} x}{\\cancel{\\textcolor{red}{8}}} = 512$ $x = 512$ To check the solution substitute $\\textcolor{red}{512}$ for $\\textcolor{red}{x}$ and calculate each side of the equation to ensure both sides are equal: $\\sqrt{8 \\textcolor{red}{x}} + 1 = 65$ becomes: $\\sqrt{8 \\cdot \\textcolor{red}{512}} + 1 = 65$ $\\sqrt{4096} + 1 = 65$ $64 + 1 = 65$ $65 + 65$ Both sides of the equation are equal therefore the solution is valid.", "side, we must add it to the other. $$x^2-14x+(\\frac{-14}{2})^2=-1+(\\frac{-14}{2})^2$$ Simplify both sides of the equation. $$x^2-14x+49=48$$ The left hand side is now a perfect-square trinomial. Let's convert it now. $$(x+\\frac{b}{2})^2=48$$ This is the form of what it will look like. Take the coefficient of the linear term, -14, and half it to get it into the desired form. $$(x-7)^2=48$$ Take the square root of both sides. $$x-7=\\pm\\sqrt{48}$$ Remember that when you take the square root of something, it results in the positive and negative answer. Add 7 to both sides. $$x=7\\pm\\sqrt{48}$$ Simplify the radical to its simplest terms. $$\\sqrt{48}=\\sqrt{16*3}=\\sqrt{16}\\sqrt{3}=4\\sqrt{3}$$ $$x=7\\pm4\\sqrt{3}$$ I'm going to split these solutions into separate ones. Ok, now we must not forget that we were solving for a--not x. Let's substitute it in. $$x_1=7+4\\sqrt{3}$$ $$x_2=7-4\\sqrt{3}$$ Substitute a^2 back into the equation. $$a^2=7+4\\sqrt{3}$$ $$a^2=7-4\\sqrt{3}$$ Take the square root of both sides. $$a=\\pm\\sqrt{7+4\\sqrt{3}}$$ $$a=\\pm\\sqrt{7-4\\sqrt{3}}$$ Now, we have to simplify those square roots. It is possible. Now, I have a clever way of simplifying these radicals so that they are nicer. I'm going to do it to both individually. $$\\sqrt{7+4\\sqrt{3}}$$ To simplify this, I'm going to add another term. What you want to do is get another term containing the square root of 3. $$\\sqrt{7+4\\sqrt{3}+\\sqrt{3}^2-3}$$ You might notice something. I've added 2 more terms. Let's simplify inside the", "multiplying both sides of the above equation. We get, $48 + {y^2} = 19y$ ${y^2} - 19y + 48 = 0$ Now we had to solve the above quadratic equation of y to get the required value of y. So, splitting the middle term of the above quadratic equation. We get, ${y^2} - 16y - 3y + 48 = 0$ $y\\left( {y - 16} \\right) - 3\\left( {y - 16} \\right) = 0$ Now taking factor (y – 16) common from the above equation. We get, $\\left( {y - 16} \\right)\\left( {y - 3} \\right) = 0$ So, y = 16 or y = 3 Putting the value of y in equation 6. We get, When y = 16 Then, $x = \\dfrac{{48}}{y} = \\dfrac{{48}}{{16}} = 3$ When y = 3 Then, $x = \\dfrac{{48}}{y} = \\dfrac{{48}}{3} = 16$ So, the values of x and y are x = 16, y = 3 or x = 3, y = 16. Now as we know that the square root of $19 + 8\\sqrt 3$ is $\\sqrt x + \\sqrt y$. So, putting the value of x and y in $\\sqrt x + \\sqrt y$. We get, $\\sqrt x + \\sqrt y = \\sqrt {16} + \\sqrt 3$ Now as we know that the square root of 16 is 4.", "# inverse of function Thanks for the help! Here is the solution.. i had a problem: $$f(x)=\\frac{(\\sqrt x+8)}{(5-\\sqrt x)}$$ i had to find the inverse, so lets begin... 1) first i write in terms of $y$ $$y=\\frac{(\\sqrt x+8)}{(5-\\sqrt x)}$$ 2) now try to get $x$ by itself $$(5-\\sqrt x)y=\\sqrt x+8$$ 3)Distribute y along the left hand side both sides $$(5y-y\\sqrt x)=\\sqrt x+8$$ 4)subtract 8 and from both sides and add $$y\\sqrt x$$ to both sides $$5y-8=y\\sqrt{x} + \\sqrt x$$ 5) factor right hand side $$5y-8=\\sqrt{x} (y + 1)$$ 6)divide out $$(y+1)$$ from right side $$((5y-8)/(y+1))=\\sqrt x$$ 7) square both sides leaving $$x=((5y-8)/(y+1))^2$$ 8) so the solution is $$f^-1(x)=((5x-8)/(x+1))^2$$ - please type the equations in \\$\\$. thx- – Seyhmus Güngören Feb 12 '13 at 0:28 Step 3) is invalid. You should distribute $y$ across the sum in the left hand side first. Then collect all terms containing \"$\\sqrt x$\" on one side, and factor it out. – David Mitra Feb 12 '13 at 0:29 Ditto for step 4). – Gerry Myerson Feb 12 '13 at 0:32 -", "= 1331 – 1 Question 3. Determine the positive value of x that makes the equation true, and then explain how you solved the equation. $$\\frac{x^{9}}{x^{7}}$$ – 49 = 0 $$\\frac{x^{9}}{x^{7}}$$ – 49 = 0 x2 – 49 = 0 x2 – 49 + 49 = 0 + 49 x2 = 49 $$\\sqrt{x^{2}}$$ = $$\\sqrt{49}$$ x = 7 Check: 72 – 49 = 0 49 – 49 = 0 0 = 0 To solve the equation, I first had to simplify the expression $$\\frac{x^{9}}{x^{7}}$$ to x2. Next, I used the properties of equality to transform the equation into x2 = 49. Finally, I had to take the square root of both sides of the equation to solve for x. Question 4. Determine the positive value of x that makes the equation true. (8x)2 = 1 (8x)2 = 1 64x2 = 1 $$\\sqrt{64^{2}}$$ = $$\\sqrt{1}$$ 8x = 1 $$\\frac{8x}{8}$$ = $$\\frac{1}{8}$$ x = $$\\frac{1}{8}$$ Check: (8($$\\frac{1}{8}$$))2 = 1 12 = 1 1 = 1 Question 5. (9$$\\sqrt{x}$$)2 – 43x = 76 (9$$\\sqrt{x}$$)2 – 43x = 76 92 (√x)2 – 43x = 76 81x – 43x = 76 38x = 76 $$\\frac{38x}{38}$$ = $$\\frac{76}{38}$$ x = 2 Check: (9($$\\sqrt{2}$$))2 – 43(2) = 76 92 ($$\\sqrt{2}$$)2 – 86 = 76 81(2) – 86 = 76 162 – 86 = 76 76", "That's right. When you solve the two equations you get x=1 and y=1. 11. ok but i still don't understand how you came up with the two equations from the equation $x^3+3x^2y\\sqrt2+3xy^2\\sqrt2+y^3\\sqrt2$. i subsitutuded but it was unsuccessful. 12. When you arrange the equation into the form $x^3+6xy^2+(2y^3+3x^2y)\\sqrt{2}=7+5\\sqrt{2}$ The coefficient of $\\sqrt{2}$ on the left hand side is $2y^3+3x^2y$ and the coefficient of the $\\sqrt{2}$ on the right hand side of the equation is $5$ So one equation to solve is $2y^3+3x^2y=5$. Then the next equation is what is not the coefficient of $\\sqrt{2}$ on the left, equal to what is not the coeffient of $\\sqrt{2}$ on the right: $x^3+6xy^2=7$ 13. ok, firstly i understand how you got the two equations but the part that is confusing me is how did you get this equation: $x^3+3x^2y\\sqrt2+3xy^2\\sqrt2+y^3\\sqrt2$ to this equation: $x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$. 14. As posted above: $x+y\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ $\\Rightarrow (x+y\\sqrt{2})^3=7+5\\sqrt{2}$ (from cubing both sides) $\\Rightarrow x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$ 15. Originally Posted by Stroodle For the first question: $x+y\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ $\\Rightarrow (x+y\\sqrt{2})^3=7+5\\sqrt{2}$ $\\Rightarrow x^3+2\\sqrt{2}y^3+3\\sqrt{2}x^2y+6xy^2=7+5\\sqrt{2}$ So from this you can now get two equations to solve simultaneously: $x^3+6xy^2=7$ and $2y^3+3x^2y=5$ Hence $x=1$ and $y=1$ $\\Rightarrow 1+\\sqrt{2}=(7+5\\sqrt{2})^{\\frac{1}{3}}$ Quick question? How would you go about solving the 2 equations if it would not have been as easy as spotting that x=1 and y=1 works. Would you sub y=mx?", "Math Help - solve by factoring/give an example please 1. solve by factoring/give an example please w2-169=0 36a2= 169 15t2-11t-12= 0 2. Please use the ^ sign for powers. As you have it written, the first one means w times 2. But anyway, for the first one. Notice the difference of two squares?. $(w)^{2}-(13)^{2}$. Now, can you factor using the proper form?. (x+y)(x-y). 3. $36a^2=169$ Take the square root of both sides. $\\sqrt{36a^2}\\ = \\sqrt 169$ $6a=13$ $a= \\frac{13}{6}$ For the other one, you can factor it. $15t^2-11t-12 = (3t-4)(5t+3)$ And from there, it's easy to solve. 4. Originally Posted by quarks $36a^2=169$ Take the square root of both sides. $\\sqrt{36a^2}\\ = \\sqrt 169$ So far, so good. But $\\sqrt{36a^2} = \\sqrt{169}$ implies $6a = \\pm 13$ $a = \\pm \\frac{13}{6}$ -Dan 5. Yeah. Thanks. =) My mistake.", "$x_1=289$ gives $y_1=\\sqrt{289\\cdot 144}=17\\cdot 12 = 204$, hence $(x,y)=(577,408)$, and the answer is $x+y=\\boxed{985}$. Solution 2 We quickly find the first solution, $(x,y)=(3,2)$. Factoring, we get $$(3-2\\sqrt{2})(3+2\\sqrt{2})=1$$ We can square both sides to get $$(3-2\\sqrt{2})^2(3+2\\sqrt{2})^2=1^2 \\Rightarrow (17-12\\sqrt{2})(17+12\\sqrt{2})=1$$ So $(x,y)=(17,12)$ is another solution. This gives us a way to generate whatever solutions we want to the equation. Raising the first equation to the fourth power gives us $$(577-408\\sqrt{2})(577+408\\sqrt{2})=1$$ The answer is $577+408=\\boxed{985}$.", "# Find all solutions to the non-linear systems: 4sqrtx + sqrty = 34 and sqrtx + 4sqrty = 16 ? Dec 5, 2017 x=64 y=4 #### Explanation: if $\\sqrt{x} = a$ $\\sqrt{y} = b$ then $4 a + b = 34$ $a + 4 b = 16$ multyply the first equation by -1; the second by 4 and add them together. Then you get $15 b = 64 - 34$ $b = \\frac{30}{15} = 2$ Now. Let's go back to find a. 4a+b=34 so after substitating b=2 you get: $a = \\frac{32}{4} = 8$ then back: $\\sqrt{x} = a$ $\\sqrt{x} = 8$ $x = {8}^{2} = 64$ $\\sqrt{y} = b$ $\\sqrt{y} = 2$ $y = {2}^{2} = 4$ the answers are $\\therefore x = 64 \\mathmr{and} y = 4$ #### Explanation: the equations are $4 \\sqrt{x} + \\sqrt{y} = 34 \\to \\left(1\\right)$ $\\sqrt{x} + 4 \\sqrt{y} = 16 \\to \\left(2\\right)$ multiplying $\\left(2\\right) b y - 4$ we get $- 4 \\sqrt{x} - 16 \\sqrt{y} = - 64 \\to \\left(3\\right)$ solving $\\left(2\\right) + \\left(3\\right)$ $4 \\sqrt{x} + \\sqrt{y} - 4 \\sqrt{x} - 16 \\sqrt{y} = - 64 + 34 \\Rightarrow - 15 \\sqrt{y} = - 30 \\Rightarrow \\sqrt{y} = 2 \\Rightarrow y = 4$ put $y = 4$ in $\\left(1\\right)$ we get $4 \\sqrt{x} + 2 = 34 \\Rightarrow", "6M) or (2W 5M) or (3W 4M) ∴ The possible ways are (1) (8) + (4) (28) + (6) (56) + (4) (70) = 8 + 112 + 336 + 280 = 736 ways [OR] (b) Evaluate $$\\lim _{x \\rightarrow 0} \\frac{\\sqrt{1+\\sin x}-\\sqrt{1-\\sin x}}{\\tan x}$$ Solution: Question 46 (a). If y = $$\\frac{\\sin ^{-1} x}{\\sqrt{1-x^{2}}}$$ show that (1 – x²)y2 – 3xy1 – y = 0 Solution: y = $$\\frac{\\sin ^{-1} x}{\\sqrt{1-x^{2}}}$$ ⇒ y $$\\sqrt{1-x^2}$$ = sin-1 x differentiating w.r.to x we get multiplying both sides by $$\\sqrt{1-x^2}$$ we get -xy + (1 – x²)y1 = 1 differentiating both sides again w.r.to x. – [xy-1 +y(1)] + (1 – x²) (y2) + y1(-2x) = 0 (i.e.) -xy1 – y + (1 – x²)y2 – 2xy1 = 0 (1 – x²)y2 – 3xy1 – y = 0 [OR] (b) If y = cos (m sin-1 x), prove that (1 – x²) y3 – 3xy2 + (m² – 1) y1 = 0 Solution: We have y = cos (m sin-1 x) y1 = sin(m sin-1 x) $$\\frac{m}{\\sqrt{1-x^2}}$$ $$y_{1}^{2}$$ = sin² (m sin-1 x) $$\\frac{m^2}{(1-x^2)}$$ This implies (1 – x²) $$y_{1}^{2}$$ = m² sin² (m sin-1 x) = m² [1 – cos² (m sin-1 x)] This is, (1 – x²) $$y_{1}^{2}$$ = m² (1 – y²). Again differentiating, (1 – x²)", "get: ⇒ $$\\rm \\frac{336}{\\sqrt x}=\\frac{112}{\\sqrt y}$$ ⇒ $$\\rm 3{\\sqrt y}=\\sqrt x$$ ... (3) Squaring both sides, we get: ⇒ 9y = x ... (4) Substituting values from equations (3) and (4) in equation (1) [or equation (2)], we get: ⇒ $$\\rm 9y\\left(3\\sqrt y\\right) + y\\sqrt y=\\frac{336}{3\\sqrt y}$$ ⇒ 81y2 + 3y2 = 336 ⇒ 84y2 = 336 ⇒ y2 = 4 Since x and y are positive real numbers, we must have y = 2, and using equation (4) we get x = 18. ∴ x + y = 20. # How many prime numbers are there between 1 and 50? 1. 17 2. 15 3. 14 4. 16 5. None of the above Option 2 : 15 ## Detailed Solution Calculation: Prime numbers between 1 and 50 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 Total number of prime numbers = 15 ∴ There are 15 prime numbers between 1 and 50 # What will be the remainder when 340 is divided by 4? 1. 1 2. 2 3. 3 4. 0 Option 1 : 1 ## Detailed Solution Given: Number = 340 Divided by = 4 Calculations: ⇒ 340/4 ⇒ (32)20/4 ⇒ (9)20/4 ⇒ (1)20 ⇒ 1 ∴ When 340 is divided by 4 the remainder will be", "for `y in [3,+oo)` , such that: `y - 2 + y - 3 = 1 => 2y = 6 => y = 3 in [3,+oo)` You need to solve for `x` the following inequality, such that: `2<=sqrt(x - 2) <= 3 => 4<= x - 2 = 9 => 6 <= x <= 11` Hence, evaluating the solutions to the given equation yields that the equation holds for all real values `x in [6,11].` Approved by eNotes Editorial Team The root of the equation `(x+2-4(x-2)^(1/2))^(1/2)+(x+7-6(x-2)^(1/2))^(1/2)-1=0` has to be determined. `(x+2-4(x-2)^(1/2))^(1/2)+(x+7-6(x-2)^(1/2))^(1/2)-1=0` => `sqrt(x-4*sqrt(x-2)+2) + sqrt(x-6*sqrt(x-2)+7) = 1` Take the square of both the sides `2*sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)` +`2*x-10*sqrt(x-2)+9` = 1 `sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)` +`x-5*sqrt(x-2)+4` = 0 `x-5*sqrt(x-2)+4` = `-sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)` Take the square of both the sides `x^2+33*x+(-10*x-40)*sqrt(x-2)-34` = `x^2+33*x+(-10*x-40)*sqrt(x-2)-34 ` => 0 = 0 This equation is true for all values of x Approved by eNotes Editorial Team", "appears in the equation. Example 1 Find the real solutions of the equation $\\sqrt{2x-1}=5$. Solution Since the radical expression is already isolated, we can just square both sides of the equation in order to eliminate the radical sign: $\\left(\\sqrt{2x-1}\\right)^2=5^2$ $\\text{Remember that} \\ \\sqrt{a^2}=a \\ \\text{so the equation simplifies to:} && 2x-1& =25\\\\\\text{Add one to both sides:} && 2x& =26\\\\\\text{Divide both sides by 2:} &&& \\underline{\\underline{x=13}}$ Finally we need to plug the solution in the original equation to see if it is a valid solution. $\\sqrt{2x-1}=\\sqrt{2(13)-1}=\\sqrt{26-1}=\\sqrt{25}=5$ The solution checks out. Example 2 Find the real solutions of $\\sqrt[3]{3-7x}-3=0$. Solution $\\text{We isolate the radical on one side of the equation:} && \\sqrt[3]{3-7x}& =3\\\\\\text{Raise each side of the equation to the third power:} && \\left(\\sqrt[3]{3-7x}\\right)^3& =3^3\\\\\\text{Simplify:} && 3-7x& =27\\\\\\text{Subtract 3 from each side:} && -7x& =24\\\\\\text{Divide both sides by -7:} &&& \\underline{\\underline{x=-\\frac{24}{7}}}$ Check: $\\sqrt[3]{3-7x}-3=\\sqrt[3]{3-7 \\left(-\\frac{24}{7}\\right)}-3=\\sqrt[3]{3+24}-3=\\sqrt[3]{27}-3=3-3=0$. The solution checks out. Example 3 Find the real solutions of $\\sqrt{10-x^2}-x=2$. Solution $\\text{We isolate the radical on one side of the equation:} && \\sqrt{10-x^2}& =2+x\\\\\\text{Square each side of the equation:} && \\left(\\sqrt{10-x^2}\\right)^2& =(2+x)^2\\\\\\text{Simplify:} && 10-x^2& =4+4x+x^2\\\\\\text{Move all terms to one side of the equation:} && 0& =2x^2+4x-6\\\\\\text{Solve using the quadratic formula:} && x& =\\frac{-4 \\pm \\sqrt{4^2-4(2)(-6)}}{4}\\\\\\text{Simplify:} && x& =\\frac{-4 \\pm \\sqrt{64}}{4}\\\\\\text{Re-write} \\ \\sqrt{24} \\ \\text{in simplest form:} && x& =\\frac{-4 \\pm 8}{4}\\\\\\text{Reduce all terms by", "# Math Help - Solve for X(2) 1. ## Solve for X(2) $0=-\\frac{\\sqrt{x+2}(x-7)}{2}+12$ 2. Originally Posted by Punch $0=-\\frac{\\sqrt{x+2}(x-7)}{2}+12$ The method here is the same as before. If we multiply by 2 throughout: Again $x > -2$ $0 = -(x-7)\\sqrt{x+2} + 24$ $(x-7)\\sqrt{x+2} = 24$ square both sides $(x-7)^2(\\sqrt{x+2})^2 = 24^2$ Simplify and solve for x. Remember to check for extraneous solutions 3. I did it like this, however won't we get a power of 3 when we expand? 4. Originally Posted by Punch I did it like this, however won't we get a power of 3 when we expand? Yep (x-7)^2(x+2) (x+2)(x^2-14x+49) = x^3-14x^2+49x+2x^2-28x+98 Collect like terms: x^3-12x^2+21x+98 x^3-12x^2+21x-478=0 Solve for x, very tricky: http://www.wolframalpha.com/input/?i=x^3-12x^2%2B21x-478%3D0 5. For general solution of the cubic, you might Google \"Cardano\". The history is interesting also. 6. Originally Posted by e^(i*pi) The method here is the same as before. If we multiply by 2 throughout: Again $x > -2$ $0 = -(x-7)\\sqrt{x+2} + 24$ $(x-7)\\sqrt{x+2} = 24$ square both sides $(x-7)^2(\\sqrt{x+2})^2 = 24^2$ Simplify and solve for x. Remember to check for extraneous solutions Just a correction. here $x \\geq -2$", "Browse Questions # Solve the equation $sin^{-1}6x+sin^{-1}6\\sqrt 3 x = -\\frac{\\pi}{2}.$ Can you answer this question? Toolbox: • $sin^{-1}x+sin^{-1}y=sin^{-1}\\: x\\sqrt{1-y^2}+y\\sqrt{1-x^2}$ • $sin(-\\frac{\\pi}{2})=-1$ By taking 6x in the place of x and $6\\sqrt3x$ in the place of y in the formula of $sin^{-1}x+sin^{-1}y$ we get $sin^{-1}6x+sin^{-1}6\\sqrt3x=$ $sin^{-1} \\bigg[ 6x \\sqrt{1-108x^2}+6\\sqrt 3 x\\sqrt{1-36x^2} \\bigg]$$= -\\frac{\\pi}{2}$ Taking sin on both the sides $\\Rightarrow\\: 6x\\sqrt{1-108x^2}+6\\sqrt 3 x\\sqrt{1-36x^2}=sin(-\\frac{\\pi}{2})=-1$ We know that $sin(-\\frac{\\pi}{2})=-1$ $\\Rightarrow 6\\sqrt 3 x \\sqrt{1-36x^2}=-1-6x\\sqrt{1-108x^2}$ Squaring on both the sides we get $\\Rightarrow 108x^2(1-36x^2)=1+36x^2(1-108x^2)+12x\\sqrt{1-108x^2}$ $\\Rightarrow\\: 72x^2-1=12x\\sqrt{1-108x^2}$ Again squaring both the sides we get $5184x^4+1-144x^2=144x^2(1-108x^2)=144x^2-15552x^4$ answered Feb 28, 2013 edited Mar 21, 2013", "and (2) and get: $$\\sin^{8}\\frac{\\pi}{12}+\\cos^{8}\\frac{\\pi}{12}\\sim 75802\\cdot 10^{-5} \\sim\\frac{97}{128}$$ (3) Daniel. Last edited: Jan 5, 2005 9. Jan 5, 2005 ### Hurkyl Staff Emeritus Let s = sin^2 x, c = cos^2 x be a solution to the equation. As Zurtex Started, this equation is a quartic in s: 97/128 = s^4 + (1-s)^4 The symmetry suggests making the substitution s = t + 1/2 Note that t is in the range -1/2 <= t <= 1/2, if we assume to look for real solutions. 97/128 = (1/2 + t)^4 + (1/2 - t)^4 = 2t^4 + 3t^2 + 1/8 = 2t^4 + 3t^2 + 16/128 so 2t^4 + 3t^2 - 81/128 = 0 yielding the solution: t^2 = -3/4 +/- (1/4) sqrt(9 + 81/16) = -3/4 +/- (1/4) sqrt(225/16) = -3/4 +/- (1/4) (15/4) = -12/16 +/- 15/16 = (-12 +/- 15) / 16 t^2 = -27 / 16 or 3 / 16 However, we know that 0 <= t^2 <= 9/4, so we have the single solution: t^2 = 3/16 so t = sqrt(3) / 4 s = t + 1/2 = (2 + sqrt(3)) / 4 Now, s = sin^2 x, so we have: sin x = (1/2) sqrt(2 + sqrt(3)) Where both square roots could be either positive or negative. For some reason, I feel", "multiplied by 4, satisfies the equation 52 + (4$$\\sqrt{x}$$)2 = 112. 52 + (4$$\\sqrt{x}$$)2 = 112 25 + 42 ($$\\sqrt{x}$$)2 = 121 25 – 25 + 42 ($$\\sqrt{x}$$)2 = 121 – 25 16x = 96 $$\\frac{16x}{16}$$ = $$\\frac{96}{16}$$ x = 6 The value of x is 6. Check: 52 + (4$$\\sqrt{6}$$)2 = 112 25 + 16(6) = 121 25 + 96 = 121 121 = 121 ### Eureka Math Grade 8 Module 7 Lesson 5 Problem Set Answer Key Find the positive value of x that makes each equation true, and then verify your solution is correct. Question 1. x2 (x + 7) = $$\\frac{1}{2}$$ (14x2 + 16) x2 (x + 7) = $$\\frac{1}{2}$$ (14x2 + 16) x3 + 7x2 = 7x2 + 8 x3 + 7x2 – 7x2 = 7x2 – 7x2 + 8 x3 = 8 $$\\sqrt[3]{x^{3}}$$ = $$\\sqrt [ 3 ]{ 8 }$$ x = 2 Check: 22 (2 + 7) = $$\\frac{1}{2}$$ (14(22 ) + 16) 4(9) = $$\\frac{1}{2}$$ (56 + 16) 36 = $$\\frac{1}{2}$$ (72) 36 = 36 Question 2. x3 = 1331 – 1 x3 = 1331 – 1 $$\\sqrt[3]{x^{3}}$$ = $$\\sqrt[3]{1331^{ – 1}}$$ x = $$\\sqrt[3]{\\frac{1}{1331}}$$ x = $$\\sqrt[3]{\\frac{1}{11^{3}}}$$ x = $$\\frac{1}{11}$$ Check: ($$\\left(\\frac{1}{11}\\right)^{3}$$)3 = 1331 – 1 $$\\frac{1}{11^{3}}$$ = 1331 – 1 $$\\frac{1}{1331}$$ = 1331 – 1 1331 – 1", "is $$\\frac{7}{12}$$, this gives $$(\\frac{12}{7})×\\frac{7}{12} y=\\frac{3}{5}×(\\frac{12}{7})$$, which simplifies to $$y=\\frac{36}{35}$$. 27- Choice A is correct Subtracting $$3x$$ and adding 2 to both sides of $$3x+6≥5x-2$$ gives $$8≥ 2x$$ and $$4≥x$$. Therefore, $$x$$ is a solution to $$3x+6≥5x-2$$ if and only if $$x$$ is less than or equal to 4 and $$x$$ is NOT a solution to $$3x+6≥5x-2$$ if and only if $$x$$ is greater than 4. Of the choices given, only 6 is greater than 4 and, therefore, cannot be a value of $$x$$. 28- Choice A is correct Given the two equations, substitute the numerical value of a into the second equation to solve for $$x$$. $$a=\\sqrt{5}, 6a=\\sqrt{12x}$$ Substituting the numerical value for a into the equation with $$x$$ is as follows. $$6(\\sqrt{5})=\\sqrt{12x}$$, From here, distribute the 4. $$6\\sqrt{5}=\\sqrt{12x}$$ Now square both side of the equation. $$(6\\sqrt{5})^2=(\\sqrt{12x})^2$$ Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. $$6^2 \\sqrt{5}^2=12x, 36(5)=12x, 180=12x, x=15$$ 29- Choice B is correct First square both sides of the equation to get $$-4m+5=m^2$$ Subtracting both sides by $$-4m+5$$ gives us the equation $$m^2+4m-5=0$$ Here you can solve the quadratic equation by factoring to get $$(m-1)(m+5)=0$$ For the expression $$(m-1)(m+5)$$ to equal zero, $$m=1$$ or $$m=-5$$ ## The Absolute Best Book to Ace the SAT", "into 372 to obtain 3. Write 3 as the next number in the quotient and also place 3 next to 124 in the divisor. Multiply 1243 by 3 and place the product under 3729. Subtract. The remainder is zero. $\\therefore$ $\\sqrt { 3881.29 }$ = 62.3 Example 2: Find the square root of 6432.04. Solution : In the above example, since 16 does not divide 3, put a zero both in the root (quotient) and the divisor and bring down the next pair 04 also. Example 3: Find the value of sf3136 and use it to find the value of $\\sqrt { 31.36 }$ + $\\sqrt { 0.3136 }$ Solution : $\\sqrt { 31.36 }$ = 5.6 $\\therefore$ $\\sqrt { 31.36 }$ = 5.6 and $\\sqrt { 0.3136 }$ = 0.56 $\\therefore$ $\\sqrt { 31.36 }$ + $\\sqrt { 0.3136 }$ = 5.6 + 0.56 = 6.16 We can also work out as under $\\sqrt { 31.36 }$ + $\\sqrt { 0.3136 }$ $\\sqrt { \\frac { 3136 } { 100 } }$ + $\\sqrt { \\frac { 3136 } { 10000 } }$ $\\frac { \\sqrt { 3136 } } { 10 }$ + $\\frac { \\sqrt { 3136 } } { 100 }$ = $\\frac {56 }{ 10 }$$\\frac { 56 }{ 100 }$ = 5.6" ]

[ "Browse Questions # What is the range of the function $f(x) = \\large\\frac{1}{3 - \\sin x}$? $\\bigg[ \\large\\frac{1}{4}$$, \\large\\frac{1}{2} \\bigg]$", "Consider the function $f(x)$, for which $f(0) = 3$ and $f'(0) = -7$. Find $h'(0)$ for the function $h(x) = \\frac{1}{f(x)}$. $h'(0)$ =", "+0 0 112 1 +979 Find the range of the function $$h(x) = \\frac{5x^2+20x+23}{x^2+4x+5}$$ Dec 14, 2020 $$3\\le \\:f\\left(x\\right)<5$$", "# What is the range of the function $f(x) = \\large\\frac{1}{3 - \\sin x}$? ## 1 Answer $\\bigg[ \\large\\frac{1}{4}$$, \\large\\frac{1}{2} \\bigg]$ Hence (C) is the correct answer. answered Jun 29, 2014", "are in the range of the function $f(x) = \\frac{4x^2+75}{2x^2+3}$?", "# What is the range of the function h(x) = 1 /(x^2 - 9)? $y \\setminus \\ge q - \\frac{1}{9}$, $y \\ne 0$ $y$ is in the range if and only if the polynomial $p \\left(x\\right) = y {x}^{2} - y 9 - 1 = 0$ has a root, and it has a root if and only if $9 + \\frac{1}{y} \\ge q 0$ and $y \\ne 0$, so $y \\setminus \\ge q - \\frac{1}{9}$, $y \\ne 0$", "What is the range of the function Question: What is the range of the function $f(x)=\\frac{|x-1|}{x-1} ?$ Solution: $f(x)=\\frac{|x-1|}{x-1}=\\frac{\\pm(x-1)}{x-1}=\\pm 1$ Range of $f=\\{-1,1\\}$", "# Math Help - help finding the limit.. 1. ## help finding the limit.. i need to find the limit as x -> negative infinity for the function (sqrt(x^2-2x))/(2x+7) 2. Multiply by $\\frac{\\tfrac{1}{x}}{\\tfrac{1}{x}}$ to get $\\frac{\\sqrt{x^2-2x}}{2x+7} \\cdot \\frac{\\tfrac{1}{x}}{\\tfrac{1}{x}}$ $= \\frac{\\sqrt{\\frac{x^2-2x}{x^2}}}{2 + \\frac{7}{x}}$ $= \\frac{\\sqrt{1-\\frac{2}{x}}}{2 + \\frac{7}{x}}$", "+0 # Help pls -1 124 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020", "The range of the function f(x) Question: The range of the function $f(x)=\\frac{x}{|x|}$ is (a) R − {0} (b) R − {−1, 1} (c) {−1, 1} (d) None of these Solution: (c) {−1, 1} $f(x)=\\frac{x}{|x|}$ Let $y=\\frac{x}{|x|}$ For $x>0,|x|=x$ $\\Rightarrow y=\\frac{x}{x}=1$ For $x<0,=-x$ $\\Rightarrow y=\\frac{x}{-x}=-1$ Thus, r ange of $f(x)$ is $\\{-1,1\\}$.", "Thread: Find the range of a function 1. Find the range of a function How do you algebraicly find the range of a function. I can tell what the range is supposed to be by looking at the graph, but I cannot show it using algebra. 2. Originally Posted by redier How do you algebraicly find the range of a function. I can tell what the range is supposed to be by looking at the graph, but I cannot show it using algebra. Excellent question. It used to bother me too, until I developed a trick. Consider a function, $y=f(x)$ The range are all the values of $y$ such that there exists an $x$ in the domain such as, $y=f(x)$. --- Example, Find the range of, $y=\\frac{1}{1+x^2}$ The domain is all real numbers. Thus, for what values of $y$ does the equation have a solution. Say, $y=\\frac{1}{1+x^2}$ Then, $\\frac{1}{y}=1+x^2$ Thus, $\\frac{1}{y}-1=x^2$ Note this equation can only have a solution when the Left Hand side is non-negative. That is, $\\frac{1}{y}-1\\geq 0$. Thus, $\\frac{1}{y}\\geq 1$ If you solve this inequality (details omitted) you have that, $0 --- As an excersce, Find the range of, $y=\\sin x+\\cos x$", "+0 # Help pls -1 61 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020", "# How do you evaluate the function h(x)=7-2/3x for h(15)? $h \\left(15\\right) = 7 - \\frac{2}{3} \\left(15\\right)$ $h \\left(15\\right) = - 3$", "# The range of the function f : R –{–2) → R given by Question: The range of the function $f: R-\\{-2) \\rightarrow R$ given by $f(x)=\\frac{x+2}{|x+2|}$ is _________. Solution: Given: $f(x)=\\frac{x+2}{|x+2|}$ $f(x)=\\frac{x+2}{|x+2|}$ $= \\begin{cases}\\frac{x+2}{x+2} & , x+2 \\geq 0 \\\\ \\frac{x+2}{-(x+2)} & , x+2<0\\end{cases}$ $= \\begin{cases}1 & , x+2 \\geq 0 \\\\ -1 & , x+2<0\\end{cases}$ To find the range, we find the real values of y obtained. Hence, the range of the function $f: R-\\{-2) \\rightarrow R$ given by $f(x)=\\frac{x+2}{|x+2|}$ is $\\{-1,1\\}$", "# What is the range of the function 2 / x^2? R=(0,+oo) If $f \\left(x\\right) = \\frac{2}{x} ^ 2$ We know that $\\frac{2}{x} ^ 2 > 0 \\implies f \\left(x\\right) > 0$ so the range is : R=(0,+oo)", "# The range of the function f(x) Question: The range of the function $f(x)=\\frac{|x-4|}{x-4}$ is ______ . Solution: $f(x)=\\frac{|x-4|}{x-4}$ $\\frac{-\\frac{(x-4)}{x-4} \\frac{(x-4)}{x-4}}{4}$ $= \\begin{cases}\\frac{x-4}{x-4} & ; x \\geq 4 \\\\ -\\frac{(x-4)}{x-4} & ; x<+4\\end{cases}$ $f(x)= \\begin{cases}1 & ; x \\geq 4 \\\\ -1 & ; x<4\\end{cases}$ Range of f(x) is {−1, 1}", "figure out what's the range for $r$, which is a function in $\\theta$. – Jack Mar 21 '12 at 0:08 According to Maple, the answer is $$\\frac12\\,\\ln \\left( \\sqrt {2}-1 \\right) -\\frac32\\,\\ln \\left( 1+\\sqrt {2} \\right) -\\frac32\\,\\ln \\left( -1+\\sqrt {10} \\right) -\\ln \\left( -3+\\sqrt {10} \\right) +\\frac32\\,\\ln \\left( 1+\\sqrt {10} \\right)$$ This agrees with Raymond's answer. -", "1. Function Let the function h be defined by h(x)=14+ $\\frac{x^2}{4}$. If h(2m)=9m, what is one possible value of m? 2. Hello, fabxx! Let: . $h(x)\\:=\\:14+ \\frac{x^2}{4}$. If $h(2m) \\:{\\color{red}=}\\:9m$, what is one possible value of $m$? Since $h(2m) \\:=\\:14 + \\frac{(2m)^2}{4} \\:=\\:14 + m^2$ . . we have: . $14 + m^2 \\:=\\:9m \\quad\\Rightarrow\\quad m^2 - 9m + 14 \\:=\\:0$ Factor:. . $(m - 2)(m - 7) \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ m \\;=\\;2,\\;7}$", "## anonymous 3 years ago find the range of function (i)cos x/3 1. anonymous $(ii) \\frac{ x+2 }{ |x+2| }$ 2. anonymous (1) it is a cosine function so range always lies [-1,1] for all values of x (2) {-1,1} for all values of x except x=-2 for which function is not defined:)", "above discussed above to find the range of the given function. Let y = f ( x ). Then, we have y = $\\frac{x-2}{3-x}$ ⇒ y ( 3 – x ) = x – 2 ⇒ 3 y – x y = x – 2 ⇒ x ( y + 1 ) = 3 y + 2 ⇒ x = $\\frac{3 y+2}{y+1}$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) We can see from equation ( 1 ) that x assumes real values of all y except y + 1 = 0. This means that y = – 1 Hence, the range of the function f ( x ) = $\\frac{x-2}{3-x}$ will be Range ( f ) = R – { – 1 } Example 2 Find the domain and range of the function f ( x ) = $\\frac{1}{2-sin 3 x}$ Solution We have been given the function f ( x ) = $\\frac{1}{2-sin 3 x}$ and we need to find its domain and range. Let us find them one by one. Domain of f We know that – 1 sin 3 x 1 for all x ∈ R ⇒ – 1" ]

[ "\\cup (0, \\infty)$, basically ur function can take on every value except 0.", "## Precalculus (6th Edition) Blitzer $(-\\infty, -3) \\cup (-3, 5) \\cup (5, +\\infty)$ Factor the denominator to obtain: $g(x) = \\dfrac{3}{(x-5)(x+3)}$ The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominator of the given function equal to zero. These numbers are $5$ and $-3$. Thus, the value of $x$ can be any real number except $5$ and $-3$. Therefore, the domain is $(-\\infty, -3) \\cup (-3, 5) \\cup (5, +\\infty)$.", "## Precalculus (6th Edition) $(-\\infty, 8) \\cup (8, +\\infty)$ The denominator is not allowed to be equal to zero since dividing zero to ny number is undefined. Since $8$ will make the denominator equal to zero, then the value of $x$ is not allowed to be $8$. The function is defined for all real numbers except $8$. Thus, the function's domain is: $(-\\infty, 8) \\cup (8, +\\infty)$.", "f =(-$\\infty$ , 2 ) $\\\\$ (ii) f ( x )= x $^2$+ 2, x , is a real number $\\\\$ The values of f (x ) for various of real numbers x can be written in the tabular form as $\\\\$ $\\begin{array}{|c|c|} \\hline x&0&\\pm0.3&\\pm0.8&\\pm1&\\pm2&\\pm 3&... \\\\ \\hline f(x)&2&2.09&2.64&3&6&11&...\\\\ \\hline \\end{array}$ $\\\\$ Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. i.e., range of f=[2, $\\infty$] $\\\\$ Alter: $\\\\$ Let x be any real number. Accordingly, $\\\\$ x $^2$$\\leq$ 0 $\\\\$ $\\\\$ x $^2$2 + 0$\\leq$ 2 $\\\\$ x $^2$ 2 $\\leq$ 2 $\\\\$ f ( x ) $\\leq$ 2 $\\\\$ Range of f =[ 2, $\\infty$] $\\\\$ (iii) f ( x )= x , x is a real number $\\\\$ It is clear that the range of f is the set of all real numbers. $\\\\$ Range of f = R .", "when $x$ is between -2 and -1 it is not defined, or there are no $x-$ values. Mathematically, this would be written: $x \\in(-\\infty, -2]\\cup(-1, \\infty)$ . The $\\cup$ symbol means “union.” In words, the domain is “all real numbers except those between -2 and -1.” Notice that -2 is included in the domain because the dot at -2 is closed. To find the range, we need to look at the possible $y-$ values. Changing our viewpoint to look at the $y-$ axis, at first glance, it looks like the function is not defined from 1 to -3. However, upon further investigation, the branch on the left does pass through the yellow region, where we though the function was not defined. This means that the function is defined between 1 and -3 and thus for all real numbers. However, below -3, there are no $y-$ values. The range is $y \\in [-3, \\infty)$ . Intro Problem Revisit The function represented by this situation can be written as $y = 150 + 5x$ , where x is the number of sales you make. You can't make a negative number of sales, so the least amount of sales you can make is zero. To find the maximum number of sales before you reach the cap, we must plug in$250 for", "# Math Help - Find range of values 1. ## Find range of values find the range of values for k for which the function $\\frac{x^2-1}{(x-2)(x+k)}$ where x is real, takes all real values. I don't know how to start. Any pointers? Thanks 2. The question asks you for what values of X, would F(x) exist. Or what values of X would F(x) not exist. Hint: C/0 doesn't exist (where C is an arbitrary constant). 3. so x cannot be 2 or -k. i dont know what else the answer is $|k|\\leq 1$ 4. Sorry, I misread the question. I thought you were supposed to find values for x instead of k. So same as idea as before. We know that k cannot be -x and we also know that x can't be 2. Therefore, k cannot be -2 (I think). 5. ok got that, but how to continue confuses me 6. What part are you confused about? You know that k can't be -x, and when x = 2, k can't be -2. 7. yes, i meant how to find the values of k other than that it isn't -2. 8. The domain for is k $(- \\infty, -x)\\cup(-x, \\infty)$ The domain of x is $(- \\infty, -2)\\cup(-2, \\infty)$", "# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set: 11 $(-\\infty, -1) \\cup (-1, 1) \\cup (1, +\\infty)$ #### Work Step by Step The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. The first denominator, $x^2+1$, is nonzero for all real numbers. However, the second denominator is equal to zero when $x=1$ or $x=-1$. Thus, the value of $x$ can be any real number except $1$ and $-1$. Therefore, the domain is $(-\\infty, -1) \\cup (-1, 1) \\cup (1, +\\infty)$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "# Chapter 3 - Polynomial and Rational Functions - 3.5 Rational Functions: Graphs, Applications, and Models - 3.5 Exercises: 1 The domain of the function is $(-\\infty, 0) \\cup (0, \\infty)$. The range of the function is $(-\\infty, 0) \\cup (0, \\infty)$. #### Work Step by Step The denominator is not allowed to be positive since dividing $0$ to any number gives an undefined expression. This means the value of $x$ cannot be zero. Thus, the domain is any real number except zero. In interval notation, the domain is $(-\\infty, 0) \\cup (0, \\infty)$. The range is the set that contains all the possible values of $f(x)$ (or $y$). Note that the value of $\\dfrac{1}{x}$ will never be equal to zero since when any non-zero number is divided to $1$, the quotient is always a non-zero number. Thus, the range is any real number except zero. In interval notation, the range is $(-\\infty, 0) \\cup (0, \\infty)$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "18x + 55), whose discriminants are 49^2 and 63^2. Both cubic have all real roots, so we can assume ... 3 The function g(x)=\\cos x-kx is the sum of two strictly decreasing functions in [0,\\pi/2], so it can not have more than one zero. On the other hand, g(0)=1>0 and g(\\pi/2)=-k\\pi/2<0. By Bolzano's theorem, it must have one zero. 3 For this to even make sense, x must be in (-\\infty,-1]\\cup[1,\\infty). If it's in [1,\\infty), then$$\\sqrt{x^2-1}>x\\iff x^2-1>x^2, which is impossible. If, on the other hand, $x$ is in $(-\\infty,-1]$, it's trivially true as the left side can't be negative. Only top voted, non community-wiki answers of a minimum length are eligible", "## Precalculus (6th Edition) Blitzer Published by Pearson # Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set - Page 258: 10 #### Answer $(-\\infty, -8) \\cup (-8, 10) \\cup (10, +\\infty)$ #### Work Step by Step The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. These numbers are $-8$ for the first denominator and $10$ for the second denominator. Thus, the value of $x$ can be any real number except $-8$ and $10$. Therefore, the domain is $(-\\infty, -8) \\cup (-8, 10) \\cup (10, +\\infty)$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "$(x-2)$ from the numerator like you said but in this case you have always that $x \\neq 2$ because you consider the limit so canceling the term is a valid step to get the function without the disconuity.", "the restriction in the domain. Range: Just like the function $y = 1/x$ in Part 2, we can have any value for $h(x)$, but it’s impossible for it to be $0$ no matter how small $x-3$ is. So, the range is the set of real numbers except 0 or $(- \\infty, 0) \\cup (0, \\infty)$ 4. $p(x) = \\frac{5}{2x + 5}$ Domain: The denominator of the equation of the function $p$ must not equal to $0$, so we find the value that will make it $0$. That is, we equate $2x + 5$ to $0$. Now, $2x + 5 = 0$ means $x = -5/2$. Therefore, $x$ must not equal $-5/2$. So, the domain of the function is the set of real numbers except $x = -5/2$ or $(- \\infty, -5/2) \\cup (-5/2, \\infty)$. Range: Left as an exercise 5. $q(x) = \\sqrt {x - 5}$ Domain: We have learned that the value inside the radical must be positive or 0, so, $\\sqrt {x-5} \\geq 0$ Squaring both sides, $x - 5 \\geq 0$ $x \\geq 5$. So, the domain of the function $q$ is the set of real numbers greater than or equal to 5. That is $[5, \\infty)$ in interval notation. Range: The value inside the radical sign must be greater than or equal to 0,", "## Precalculus (6th Edition) Blitzer $(-\\infty, -2) \\cup (-2, 2) \\cup (2, +\\infty)$ The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. The first denominator, $x^2+4$, is nonzero for all real numbers. However, the second denominator is equal to zero when $x=-2$ and $x=2$. Thus, the value of $x$ can be any real number except $-2$ and $2$. Therefore, the domain is $(-\\infty, -2) \\cup (-2, 2) \\cup (2, +\\infty)$.", "us a real number. It is undefined. We are dealing with only real numbers. $D = \\{x | x \\in \\mathbb{R}; x \\ne 0\\} \\\\[3ex] D = (-\\infty, 0) \\bigcup (0, \\infty) \\\\[3ex]$ To find the range, let us assume the output is $0$ $y = \\dfrac{3}{x} \\\\[5ex] 0 = \\dfrac{3}{x} \\\\[5ex] \\dfrac{3}{x} = 0 \\\\[5ex] 3 = x * 0 \\\\[3ex] 3 = 0? \\:This\\: is\\: a\\: contradiction \\\\[3ex]$ $0$ can never be an output. Any other number can be an output. $R = \\{y | y \\in \\mathbb{R}; y \\ne 0\\} \\\\[3ex] R = (-\\infty, 0) \\bigcup (0, \\infty)$ (19.) $xy = -7$ $xy = -7 \\\\[3ex] y = -\\dfrac{7}{x} \\\\[5ex]$ It is a function (rational function). The denominators of rational functions must not be zero. We cannot divide anything by nothing. Dividing anything (something or nothing) by nothing does not give us a real number. It is undefined. We are dealing with only real numbers. $D = \\{x | x \\in \\mathbb{R}; x \\ne 0\\} \\\\[3ex] D = (-\\infty, 0) \\bigcup (0, \\infty) \\\\[3ex]$ To find the range, let us assume the output is $0$ $y = -\\dfrac{7}{x} \\\\[5ex] 0 = -\\dfrac{7}{x} \\\\[5ex] -\\dfrac{7}{x} = 0 \\\\[5ex] -7 = x * 0 \\\\[3ex] -7 = 0? \\:This\\: is\\: a\\: contradiction \\\\[3ex]$ $0$ can never be", "In order to be able to cancel with the 4y- 3 which is a factor $8y^2- 2y- 3= (2y+ 1)(4y- 3)$, Soroban factored out a \"-1\" writing it as -(4y- 3).", "< q < 7, what is the maximum possible value of p2 + pq + q2? 13. Inequalities - Integer Solutions For how many integer values does the following inequality hold good? (x + 2) (x + 4) (x + 6)........(x + 100) < 0. 14. Maximum possible value If a, b, c are integers such that – 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc? 15. Properties of Inequalities Solve x2 - |x + 3| + x > 0 1. x $\\in (-\\infty,-1] \\cup [\\sqrt{3}, 3)$ 2. x $\\in (-\\infty,-3] \\cup [\\sqrt{3}, \\infty)$ 3. x $\\in (-4,-3) \\cup (4, \\infty)$ 4. x $\\in (-8,-3] \\cup [2, \\infty)$ • Properties of Inequalities • Medium 16. Properties of Inequalities Find range of f(x) = x2 – 6x + 14 1. $(-\\infty, 8)$ 2. $(-\\infty, 100)$ 3. $(-\\infty, 45)$ 4. $(5, \\infty)$ • Properties of Inequalities • Medium 17. Properties of Inequalities Solve :$\\frac{(x – 4) (x+3)}{(x + 4) ( x +5)} > 0$ 1. $(-\\infty,-5) \\cup (4, \\infty)$ 2. $(-\\infty,-3] \\cup [3, \\infty)$ 3. $(-\\infty,-5] \\cup [1, \\infty)$ 4. $(-\\infty,-1] \\cup [2, \\infty)$ • Properties of Inequalities • Medium 18. Modulus - Tricky Consider three distinct positive integers a, b, c all less than", "## Precalculus (6th Edition) The domain of the function is $(-\\infty, 0) \\cup (0, \\infty)$. The range of the function is $(-\\infty, 0) \\cup (0, \\infty)$. The denominator is not allowed to be positive since dividing $0$ to any number gives an undefined expression. This means the value of $x$ cannot be zero. Thus, the domain is any real number except zero. In interval notation, the domain is $(-\\infty, 0) \\cup (0, \\infty)$. The range is the set that contains all the possible values of $f(x)$ (or $y$). Note that the value of $\\dfrac{1}{x}$ will never be equal to zero since when any non-zero number is divided to $1$, the quotient is always a non-zero number. Thus, the range is any real number except zero. In interval notation, the range is $(-\\infty, 0) \\cup (0, \\infty)$.", "the function is the set of real numbers greater than or equal to $-3$ or $[-3, \\infty)$ in interval notation. 2. $g(x) = |x - 3| + 4$ Domain: The function is defined for any real number $x$, so the domain of $f$ is the set of real numbers. Range: The minimum value of any number inside the absolute value sign is $0$ (here, that happens at $g(3)$). So adding $4$ to $g(3)$ will make the minimum value of the range $g(3) + 0 = 0 + 4 = 4$. So the range is the set of real numbers greater than or equal to $4$ or $[4, \\infty)$ in interval notation. Division by 0 3. $h(x) = \\frac{1}{x-3}$ Domain: The denominator of the equation of the function $h$ must not equal to $0$, so we find the value that will make it $0$. That is, we equate $x - 3$ to $0$. Now, $x - 3 = 0$ gives us $x = 3$. Therefore, $x$ must not equal to $3$. So, the domain of the function $h$ is the set of real numbers except $x \\neq 3$ or $(- \\infty, 3) \\cup (3, \\infty)$. Notice, that to get the restrictions in the domain we equate the denominator to $0$ and solve for $x$. The value of $x$ is", "Precalculus (6th Edition) The given relation defines $y$ as a function of $x$. domain: $(-\\infty, 3) \\cup (3, +\\infty)$ range: $(-\\infty, 0) \\cup (0, +\\infty)$ The given equation above will give only one value of $y$ for every value of $x$ within its domain. This means that each $x$ is paired with only one value of $y$. Thus, the given relation defines $y$ as a function of $x$. The denominator is not allowed to be zero. Since $3$ will make the given expression's denominator equal to zero, then $x$ cannot be $3$. Thus, the domain is $(-\\infty, 3) \\cup (3, +\\infty)$. Note that when $2$ is divided by any non-zero number, the quotient will never be equal to zero. Thus, the value of $y$ can be any real number except zero. In interval notation, the range is $(-\\infty, 0) \\cup (0, +\\infty)$.", "Trigonometry (11th Edition) Clone Domain: $(-\\infty, +\\infty)$ Range: $(-\\infty, -2] \\cup [2, +\\infty)$ RECALL: The cosecant function $y=\\csc{x}$ has: (1) $(-\\infty, +\\infty)$ (the set of real numbers) as its domain; and (2) $(-\\infty, -1] \\cup [1, +\\infty)$ as its range. The value of the function is never in the interval $(-1, 1)$. Thus, the domain of the given function is also the set of real numbers. The given function has $2$ as a multiplier outside the cosecant function. The value of the given function is will never be between $-2$ and $2$ so its range is $(-\\infty, -2] \\cup [2, +\\infty)$." ]

[ "$x=0$ or $h(x)=3$ there is a contradiction, so we know that $h(x)\\neq 3$. - Assuming you mean that $h(x)=\\frac{(3x+2)(x-2)}{x(x-2)}$, the function is the same as $\\frac{3x+2}{x}$, which is $3+\\frac2x$ except at $x=2$, where $y$ would otherwise be $4$. Such a function is a transform of $y=\\frac1x$, which is one to one and only has one missing point from its domain, which is the limit of the function as $x\\to\\infty$. In this case, that is $3$. So both $3$ and $4$ are missing from the range. The range is $(-\\infty,3)\\cup(3,4)\\cup(4\\infty)$. -", "# Range of this function $$f(x) = \\frac{1 - x}{\\sqrt{5 + 7x - x^{2}}}$$ I know that the range of that function is $$(-\\infty, 0) U (0, +\\infty)$$ But how do I get it? I'm having difficulties in isolating x. Can you explain it to me? Thank you. :) HallsofIvy Homework Helper You don't get that range- it's wrong. It is easy to see that $$f(1)= \\frac{1- 1}{\\sqrt{5+ 7- 1}}= \\frac{0}{\\sqrt{11}}= 0$$ so 0 certainly is in the range. Since you have determined what x can be, what values of f do those values of x give? Moreover, for sufficiently large |x|, the denominator is not real. HallsofIvy Right. $5+ 7x- x^2= 0$ when $x= (7\\pm\\sqrt{49+ 20})/2$ so that the domain of the function is only from $(7- \\sqrt{69})/2$ to $7+\\sqrt{69})/2$, or from between -1 and 0 to between 7 and 8. The function goes to infinity as x approaches the end points. The lower bound on the range will occur at the minimum of the function.", "mention it? The range is $$(0,\\infty) \\cup (-\\infty, -\\frac 1 3]$$. To see this write the range as $$\\{\\frac 1 {t-3}: t \\geq 0, t \\neq 3\\}$$. Find $$\\{\\frac 1 {t-3}:0 \\leq t < 3\\}$$ and $$\\{\\frac 1 {t-3}: 3 < t <\\infty)\\}$$ separately. These can be written as $$\\{\\frac 1 s:-3 \\leq s < 0\\}$$ and $$\\{\\frac 1 s: 0 < s <\\infty)\\}$$. Can you compute the range now? I took a closer look at the method for determining the range of a composite function described on the U of T website and I came up with the same answer as the one already posted (unfortunately I can't fully follow the reasoning in that solution, even though it seems to be correct). The method described on the U of T website states that to find the range of a composite function, we must first find the range of the outer function. Then we determine the range of the inner function, which we can also regard as the domain of the outer function in the composition, and we check whether this domain eliminates any of the values from the range of the outer function. Below is the application of this method to the given problem. Given $$f(x) = \\dfrac{1}{x - 3}$$ and $$g(x) = \\sqrt{x}$$, we want to", "\\infty )- \\{ 2 \\} = [-2, 2) \\cup (2, \\infty )$$(Note that for $x \\in \\mathbb R - \\{ 2 \\}$ the fraction under the root can be written as $\\frac{(x+2)(x-2)}{x-2}=x+2$). To find the range of $H(x)= \\sqrt{ \\frac{x^2 -4}{x-2}}= \\sqrt{x+2}$ (for $x \\in D_H$), one should note that the range of $f_1(x)= \\sqrt{x}$ is $[0, \\infty )$ and the range of $f_2(x)=x+2$ is $\\mathbb R$. Thus, the range of $H$ is$$R_H= [0, \\infty ) - \\{ 2 \\}= [0, 2) \\cup (2, \\infty )$$(Note that one must exclude the value of $\\sqrt{x+2}$ for $x=2$ since $x=2$ is not in the domain of $H$).", "# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set: 16 $(-\\infty, 2) \\cup (2, \\frac{10}{3}) \\cup (\\frac{10}{3}, +\\infty)$. #### Work Step by Step The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. The expression $\\dfrac{4}{x-2}$ is undefined when $x=2$. To find the value of $x$ that will make the denominator $\\dfrac{4}{x-2} -3$ equal to $0$, equate the denominator to zero then solve the equation to obtain: $\\dfrac{4}{x-2} - 3 = 0 \\\\\\dfrac{4}{x-2} = 3$ Cross-multiply: $\\\\4 = 3(x-2) \\\\4 = 3x - 6 \\\\4+6 = 3x \\\\10 = 3x \\\\\\frac{10}{3} = x$ Thus, the value of $x$ can be any real number except $2$ and $\\frac{10}{3}$. Therefore, the domain is $(-\\infty, 2) \\cup (2, \\frac{10}{3}) \\cup (\\frac{10}{3}, +\\infty)$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "= (- \\infty,3) \\cup (3,\\infty)$ (see graph below). Now, let's discuss ranges: Note, as $x$ grows increasingly large, (very very large: $\\to infty$), or increasingly small ($\\to \\infty$), $f(x) \\to 1$ in each case. That is, we have a horizontal asymptote at $y = 1$ So, $\\text{Range}(f) = (- \\infty,1) \\cup (1,\\infty)$, and you can view that below: graph of $f(x) = \\dfrac{x-1}{x-3}$, a hyperbola. Now, if you want to determine $\\text{Range}(y)$, we need to omit $f(2) = 0$ from $\\text{Range}(f)$ because, as we noted above, $2 \\notin \\text{Domain}(y)$ and because $f$ is a one-to-one function (which means that $f$ does not attain the value $0$ anywhere else other than at $x = 2$, and it is undefined at x = 2). Therefore, $\\text{Range}(y) = (- \\infty,0) \\cup (0,1) \\cup (1,\\infty)$. - thanks..... but if no factors cancel out then? – user60576 Feb 1 '13 at 17:23 thanks..... but if no factors cancel out then? like if it is.....y=((x^2)-3x+2)/((x^2)-7x+12) – user60576 Feb 1 '13 at 17:30 Then you check where the denominator is $0$, where it will be undefined, if there is such a value(s). In this case $x^2 - 7x + 12 = (x - 3)(x - 4)$ and so the denominator will be $0$, and hence undefined, at $x = 3$ and at $x = 4$", "such that – 3 < p < 4, – 8 < q < 7, what is the maximum possible value of p2 + pq + q2? 13. ### Inequalities - Integer Solutions For how many integer values does the following inequality hold good? (x + 2) (x + 4) (x + 6)........(x + 100) < 0. 14. ### Maximum possible value If a, b, c are integers such that – 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc? 15. ### Properties of Inequalities Solve x2 - |x + 3| + x > 0 1. x $\\in (-\\infty,-1] \\cup [\\sqrt{3}, 3)$ 2. x $\\in (-\\infty,-3] \\cup [\\sqrt{3}, \\infty)$ 3. x $\\in (-4,-3) \\cup (4, \\infty)$ 4. x $\\in (-8,-3] \\cup [2, \\infty)$ • Properties of Inequalities • Medium 16. ### Properties of Inequalities Find range of f(x) = x2 – 6x + 14 1. $(-\\infty, 8)$ 2. $(-\\infty, 100)$ 3. $(-\\infty, 45)$ 4. $(5, \\infty)$ • Properties of Inequalities • Medium 17. ### Properties of Inequalities Solve :$\\frac{(x – 4) (x+3)}{(x + 4) ( x +5)} > 0$ 1. $(-\\infty,-5] \\cup [4, \\infty)$ 2. $(-\\infty,-3] \\cup [3, \\infty)$ 3. $(-\\infty,-5] \\cup [1, \\infty)$ 4. $(-\\infty,-1] \\cup [2, \\infty)$ • Properties of Inequalities • Medium", "< q < 7, what is the maximum possible value of p2 + pq + q2? 13. Inequalities - Integer Solutions For how many integer values does the following inequality hold good? (x + 2) (x + 4) (x + 6)........(x + 100) < 0. 14. Maximum possible value If a, b, c are integers such that – 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc? 15. Properties of Inequalities Solve x2 - |x + 3| + x > 0 1. x $\\in (-\\infty,-1] \\cup [\\sqrt{3}, 3)$ 2. x $\\in (-\\infty,-3] \\cup [\\sqrt{3}, \\infty)$ 3. x $\\in (-4,-3) \\cup (4, \\infty)$ 4. x $\\in (-8,-3] \\cup [2, \\infty)$ • Properties of Inequalities • Medium 16. Properties of Inequalities Find range of f(x) = x2 – 6x + 14 1. $(-\\infty, 8)$ 2. $(-\\infty, 100)$ 3. $(-\\infty, 45)$ 4. $(5, \\infty)$ • Properties of Inequalities • Medium 17. Properties of Inequalities Solve :$\\frac{(x – 4) (x+3)}{(x + 4) ( x +5)} > 0$ 1. $(-\\infty,-5) \\cup (4, \\infty)$ 2. $(-\\infty,-3] \\cup [3, \\infty)$ 3. $(-\\infty,-5] \\cup [1, \\infty)$ 4. $(-\\infty,-1] \\cup [2, \\infty)$ • Properties of Inequalities • Medium 18. Modulus - Tricky Consider three distinct positive integers a, b, c all less than", "# Find all possible values which following expression can take $\\sqrt{x^2-7x+6}$ Given expression is: $\\sqrt{x^2-7x+6}$. Now i first find values of x for which this is a valid question by putting guy inside square root equals to greater than 0. I get $x \\in[-\\infty,1]\\cup[6,\\infty]$. Now i completed the square and i got $$\\sqrt{(x-\\frac{7}{2})^2-\\frac{25}{4}}$$. Now from here i calculated range as x $\\in$ $[0,\\infty]$. Now i have taken intersection of values of obtained and i got answer to be $[0,1]\\cup[6,\\infty]$. but my textbook says answer to be $[0,\\infty]$. Where is my mistake? Thanks • Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer Oct 28 '16 at 17:30 You want to see what are the real numbers $c$ such that the equation $$\\sqrt{x^2-7x+6}=c$$ has a solution. As you observe, $c\\ge0$ by definition of square root. Now we can square, getting $x^2-7x+6=c^2$ or $$x^2-7x+6-c^2=0$$ which has a solution if and only if its discriminant is $\\ge0$. This amounts to $$49-4(6-c^2)\\ge0$$ which is true for every $c$. So the range is $[0,\\infty)$. With completion of the square it's the same: the equation is $$\\left(x-\\frac{7}{2}\\right)^{\\!2}=c^2+\\frac{25}{4}$$ which is solvable for every $c$. You are right that the function is only defined over $(-\\infty,1]\\cup[6,\\infty)$, but this doesn't affect the range. Clearly the polynomial under the radical can", "of $(-\\infty, -5) \\cup \\(-5, -3) \\cup (-3, -2) \\cup (-2, 0) \\cup (0, 5) \\cup (5, 7) \\cup (7, 8) \\cup (8, 10) \\cup (10, \\infty) = S$. Now consider any $(x \\in (-\\infty, 0)$. We see that the denominator > 0. So, the numerator must also be > 0, So only consider odd power terms in numerator, since even power gives positive no matter what. So take $(x-8)(x+3)^3(x+5) > 0$, since we see $(x-8) < 0 for x \\in (-\\infty, 0)$, it must be that $(x+3)^3(x+5) < 0$(negative * negative = positive) . Which means either $x - 3 < 0 AND x + 5 > 0$ OR $x - 3 > 0 AND x + 5 < 0$ which means $x < 3 AND x > -5$ OR $x > 3 AND x < -5$., the second is impossible, x cannot be > 3 and < -5 at the same time.. So x \\in $(-5, 3) \\cap S$ Now we solved what happens when $x \\in (-\\infty, 0))$ we got for $x \\in (-5, -3) \\cap S = (-5, -3)$ the inequality holds. Now assume $x \\in (0, \\infty)$ We only need to consider $\\frac{x-8}{5-x}^3$ because every other term evaluates to positive. So for $\\frac{x-8}{5-x}^3 > 0$ One option is for both num and denom", "+0 +1 157 4 The problem is: Find the set of values $$a$$ for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. I got $$[-1 , ∞)$$ by finding x in terms of y and a, then finding when that x value would be real. May 31, 2021 #1 +1 [-1.inf) is correct. Good job! May 31, 2021 #2 0 Oh, it was incorrect for some reason. Guest May 31, 2021 #3 0 I got it right :) Guest May 31, 2021 #4 +13045 +1 Find the set of values a for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. Hello Guest! $$a\\in \\mathbb R\\ |\\ a\\neq -1$$ ! May 31, 2021", "# Math Help - Find range of values 1. ## Find range of values find the range of values for k for which the function $\\frac{x^2-1}{(x-2)(x+k)}$ where x is real, takes all real values. I don't know how to start. Any pointers? Thanks 2. The question asks you for what values of X, would F(x) exist. Or what values of X would F(x) not exist. Hint: C/0 doesn't exist (where C is an arbitrary constant). 3. so x cannot be 2 or -k. i dont know what else the answer is $|k|\\leq 1$ 4. Sorry, I misread the question. I thought you were supposed to find values for x instead of k. So same as idea as before. We know that k cannot be -x and we also know that x can't be 2. Therefore, k cannot be -2 (I think). 5. ok got that, but how to continue confuses me 6. What part are you confused about? You know that k can't be -x, and when x = 2, k can't be -2. 7. yes, i meant how to find the values of k other than that it isn't -2. 8. The domain for is k $(- \\infty, -x)\\cup(-x, \\infty)$ The domain of x is $(- \\infty, -2)\\cup(-2, \\infty)$", "## Thomas' Calculus 13th Edition Published by Pearson # Chapter 1: Functions - Section 1.1 - Functions and Their Graphs - Exercises 1.1 - Page 11: 5 #### Answer The domain is $(-\\infty,3)\\cup(3,\\infty)$ The range is $(-\\infty,0)\\cup(0,\\infty)$ #### Work Step by Step $f(t)=\\frac{4}{3-t}$ $3-t\\neq0$ - Because we can't divide by 0 $t \\neq -3$ The domain is $(-\\infty,3)\\cup(3,\\infty)$ When $t<3$, we know that $3-t>0$ and therefore $\\frac{4}{3-t}>0$. When $t>3$, we know that $3-t<0$ and therefore $\\frac{4}{3-t}<0$. This means $t$ can be any real number except $0$, which makes the range $(-\\infty,0)\\cup(0,\\infty)$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "# Range of this function 1. Aug 31, 2010 ### yik-boh $$f(x) = \\frac{1 - x}{\\sqrt{5 + 7x - x^{2}}}$$ I know that the range of that function is $$(-\\infty, 0) U (0, +\\infty)$$ But how do I get it? I'm having difficulties in isolating x. Can you explain it to me? Thank you. :) 2. Sep 1, 2010 ### HallsofIvy You don't get that range- it's wrong. It is easy to see that $$f(1)= \\frac{1- 1}{\\sqrt{5+ 7- 1}}= \\frac{0}{\\sqrt{11}}= 0$$ so 0 certainly is in the range. Since you have determined what x can be, what values of f do those values of x give? 3. Sep 1, 2010 Moreover, for sufficiently large |x|, the denominator is not real. 4. Sep 2, 2010 ### HallsofIvy Right. $5+ 7x- x^2= 0$ when $x= (7\\pm\\sqrt{49+ 20})/2$ so that the domain of the function is only from $(7- \\sqrt{69})/2$ to $7+\\sqrt{69})/2$, or from between -1 and 0 to between 7 and 8. The function goes to infinity as x approaches the end points. The lower bound on the range will occur at the minimum of the function. 5. Sep 2, 2010 ### uart Assuming we're talking about real functions then the domain is [itex](7-\\sqrt{69})/2 < x < (7+\\sqrt{69})/2[/tex] and the range is all real [itex]y[/tex]. Halls, note that the function", "us a real number. It is undefined. We are dealing with only real numbers. $D = \\{x | x \\in \\mathbb{R}; x \\ne 0\\} \\\\[3ex] D = (-\\infty, 0) \\bigcup (0, \\infty) \\\\[3ex]$ To find the range, let us assume the output is $0$ $y = \\dfrac{3}{x} \\\\[5ex] 0 = \\dfrac{3}{x} \\\\[5ex] \\dfrac{3}{x} = 0 \\\\[5ex] 3 = x * 0 \\\\[3ex] 3 = 0? \\:This\\: is\\: a\\: contradiction \\\\[3ex]$ $0$ can never be an output. Any other number can be an output. $R = \\{y | y \\in \\mathbb{R}; y \\ne 0\\} \\\\[3ex] R = (-\\infty, 0) \\bigcup (0, \\infty)$ (19.) $xy = -7$ $xy = -7 \\\\[3ex] y = -\\dfrac{7}{x} \\\\[5ex]$ It is a function (rational function). The denominators of rational functions must not be zero. We cannot divide anything by nothing. Dividing anything (something or nothing) by nothing does not give us a real number. It is undefined. We are dealing with only real numbers. $D = \\{x | x \\in \\mathbb{R}; x \\ne 0\\} \\\\[3ex] D = (-\\infty, 0) \\bigcup (0, \\infty) \\\\[3ex]$ To find the range, let us assume the output is $0$ $y = -\\dfrac{7}{x} \\\\[5ex] 0 = -\\dfrac{7}{x} \\\\[5ex] -\\dfrac{7}{x} = 0 \\\\[5ex] -7 = x * 0 \\\\[3ex] -7 = 0? \\:This\\: is\\: a\\: contradiction \\\\[3ex]$ $0$ can never be", "f =(-$\\infty$ , 2 ) $\\\\$ (ii) f ( x )= x $^2$+ 2, x , is a real number $\\\\$ The values of f (x ) for various of real numbers x can be written in the tabular form as $\\\\$ $\\begin{array}{|c|c|} \\hline x&0&\\pm0.3&\\pm0.8&\\pm1&\\pm2&\\pm 3&... \\\\ \\hline f(x)&2&2.09&2.64&3&6&11&...\\\\ \\hline \\end{array}$ $\\\\$ Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. i.e., range of f=[2, $\\infty$] $\\\\$ Alter: $\\\\$ Let x be any real number. Accordingly, $\\\\$ x $^2$$\\leq$ 0 $\\\\$ $\\\\$ x $^2$2 + 0$\\leq$ 2 $\\\\$ x $^2$ 2 $\\leq$ 2 $\\\\$ f ( x ) $\\leq$ 2 $\\\\$ Range of f =[ 2, $\\infty$] $\\\\$ (iii) f ( x )= x , x is a real number $\\\\$ It is clear that the range of f is the set of all real numbers. $\\\\$ Range of f = R .", "real root function $\\sqrt{f}$ is the set of real numbers, $\\mathbb R$, minus those points which make $f$ negative. So the domain of $H(x)= \\sqrt{ \\frac{x^2 -4}{x-2}}$ is$$D_H=[-2, \\infty )- \\{ 2 \\} = [-2, 2) \\cup (2, \\infty )$$(Note that for $x \\in \\mathbb R - \\{ 2 \\}$ the fraction under the root can be written as $\\frac{(x+2)(x-2)}{x-2}=x+2$). To find the range of $H(x)= \\sqrt{ \\frac{x^2 -4}{x-2}}= \\sqrt{x+2}$ (for $x \\in D_H$), one should note that the range of $f_1(x)= \\sqrt{x}$ is $[0, \\infty )$ and the range of $f_2(x)=x+2$ is $\\mathbb R$. Thus, the range of $H$ is$$R_H= [0, \\infty ) - \\{ 2 \\}= [0, 2) \\cup (2, \\infty )$$(Note that one must exclude the value of $\\sqrt{x+2}$ for $x=2$ since $x=2$ is not in the domain of $H$). After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "the function is the set of real numbers greater than or equal to $-3$ or $[-3, \\infty)$ in interval notation. 2. $g(x) = |x - 3| + 4$ Domain: The function is defined for any real number $x$, so the domain of $f$ is the set of real numbers. Range: The minimum value of any number inside the absolute value sign is $0$ (here, that happens at $g(3)$). So adding $4$ to $g(3)$ will make the minimum value of the range $g(3) + 0 = 0 + 4 = 4$. So the range is the set of real numbers greater than or equal to $4$ or $[4, \\infty)$ in interval notation. Division by 0 3. $h(x) = \\frac{1}{x-3}$ Domain: The denominator of the equation of the function $h$ must not equal to $0$, so we find the value that will make it $0$. That is, we equate $x - 3$ to $0$. Now, $x - 3 = 0$ gives us $x = 3$. Therefore, $x$ must not equal to $3$. So, the domain of the function $h$ is the set of real numbers except $x \\neq 3$ or $(- \\infty, 3) \\cup (3, \\infty)$. Notice, that to get the restrictions in the domain we equate the denominator to $0$ and solve for $x$. The value of $x$ is", "all real numbers <= -100 Is this correct? No. in fact, $25(-7x-4)^{64}\\geq 0$ for all $x\\in\\mathbb{R}$. - But the correct answer is quite easy to find. Because the range of $y=-7x-4$ is $(-\\infty,+\\infty)$ and because the exponent 64 is even (and the factor 25 is $> 0$), the range of this function must be $[0,+\\infty)$ 3. y = x^2 / (x^2-16) How would I do this one? Thanks a bunch This one is more complicated. First the domain is $\\mathbb{R}\\backslash\\{-4,+4\\}$, because for $x=\\pm 4$ a division by zero would result. As to the range: consider first the case that $-4. In this case, the numerator is always $\\geq 0$ ( $=0$ for $x=0$ only), while the denominator is always $<0$ and tends to 0 from the left (negative side) as x approaches -4 from the right or +4 from the left. So for $x\\in (-4,+4)$, this function assumes all values in $(-\\infty,0]$. Now consider the second case, namely that $x<-4$ or $+4. In this case, the numerator $x^2$ is always positive and larger than the the denominator $x^2-16$, and hence the function values are $\\geq 1$. From $y=\\frac{x^2}{x^2-16}=\\frac{1}{1-\\frac{16}{x^2}}$ we conclude that for $x\\rightarrow\\pm \\infty$ the function values get closer and closer to +1 (from above). To sum up: the range is $(-\\infty,0]\\cup (1,+\\infty)=\\mathbb{R}\\backslash (0,1]$. 3. #2. - How", "The range of f(x)= Question: The range of $f(x)=\\frac{1}{1-2 \\cos x}$ is (a) [1/3, 1] (b) [−1, 1/3] (c) (−∞, −1) ∪ [1/3, ∞) (d) [−1/3, 1] Solution: We know that −1 ≤ cosx ≤ 1 for all x ∈ R. Now, $-1 \\leq \\cos x \\leq 1$ $\\Rightarrow-1 \\leq-\\cos x \\leq 1$ $\\Rightarrow-2 \\leq-2 \\cos x \\leq 2$ $\\Rightarrow-1 \\leq 1-2 \\cos x \\leq 3$ (Adding 1 to each term) But, $\\cos x \\neq \\frac{1}{2}$ $\\Rightarrow 1-2 \\cos x \\in[-1,3]-\\{0\\}$ $\\Rightarrow \\frac{1}{1-2 \\cos x} \\in(-\\infty,-1] \\cup\\left[\\frac{1}{3}, \\infty\\right)$ $\\therefore$ Range of $f(x)=(-\\infty,-1] \\cup\\left[\\frac{1}{3}, \\infty\\right)$ Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as \"(−∞, −1] ∪ [1/3, ∞)\"." ]

[ "or $$(\\sqrt[n]{x})^m$$ Example: Find the value of $$8^{\\frac{5}{3}}$$ and $$256^{\\frac{3}{4}}$$ Solution: 1. $$8^{\\frac{7}{3}} = (\\sqrt[3]{8})^7 = (\\sqrt[3]{2^3})^7 = 2^7 = 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 = 128$$ 2. $$256^{\\frac{3}{4}} = (\\sqrt[4]{256})^3 = (\\sqrt[4]{4^4})^3 = 4^3 = 4 \\times 4 \\times 4 = 64$$", "Step 2 put the values, to get $$16\\times10^{-2}=\\frac{1}{2}\\times9.8\\times t^{2}$$ $$\\Rightarrow t^{2}=\\frac{2\\times16\\times10^{-2}}{9.8}$$ $$\\Rightarrow t=0.18s$$ So, the time taken for the bill to fall beyond her grasp is 0.18s. And as the average reaction time is 0.25s, so the friend would never catch it.", "{3}{2^2} \\approx 0.477 - 2 \\times 0.301$. It is useful to know what $\\log 2, \\log 3$ is approximately equal to. After the first year, the value is $\\frac{3}{4}\\times 900$. After another year, it is $\\frac{3}{4}\\times\\frac{3}{4}\\times 900$. Continuing this way, one sees that after eight years, the value is $(\\frac{3}{4})^8\\times 900$.", "unknown value ...\"), which is rather different to a situation where $\\theta^* \\in \\{\\frac{1}{4}, \\frac{3}{4}, \\frac{1}{2}\\}$, a finite set. In this situation, we take each of the possible values in the set in turn and say \"if $\\theta$ was this value, what is the likelihood\": (Coin 1) $L_3(\\frac{1}{4} = f(x_1, x_2, x_3;\\theta = \\frac{1}{4}) = f(1,0,0; \\theta = \\frac{1}{4}) = \\frac{1}{4} \\times (1-\\frac{1}{4}) \\times (1-\\frac{1}{4}) = \\frac{1}{4} \\times \\frac{3}{4} \\times \\frac{3}{4} = \\frac{9}{64}$ (Coin 2) $L_3(\\frac{3}{4} = f(x_1, x_2, x_3;\\theta = \\frac{3}{4}) = f(1,0,0; \\theta = \\frac{3}{4}) = \\frac{3}{4} \\times (1-\\frac{3}{4}) \\times (1-\\frac{3}{4}) = \\frac{3}{4} \\times \\frac{1}{4} \\times \\frac{1}{4} = \\frac{3}{64}$ (Coin 3) $L_3(\\frac{1}{2} = f(x_1, x_2, x_3;\\theta = \\frac{1}{2}) = f(1,0,0; \\theta = \\frac{1}{2}) = \\frac{1}{2} \\times (1-\\frac{1}{2}) \\times (1-\\frac{1}{2}) = \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{8} = \\frac{8}{64}$ So, the \"most likely\" estimate following the three tosses on the basis of the maximum likelihood principle is $\\widehat{\\theta}_3 = \\frac{1}{4}$ (Coin 1). ## Maximum Likelihood and the $Exponential(\\lambda)$ RV The $Exponential$ is parameterised by $\\lambda$. We have seen that, for a given $\\lambda \\in (0,\\infty)$, an $Exponential(\\lambda)$ random variable has the following PDF $f$ and DF $F$: $f(x;\\lambda) = \\lambda e^{-\\lambda x}$ $F(x;\\lambda) = 1 - e^{-\\lambda x}$ You try in class On paper, try to work out $f(x_1,x_2,\\ldots,x_n; \\lambda)$, the joint density of $X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim}", "• # question_answer What is the value of $\\frac{\\mathbf{1}}{\\mathbf{2}}\\mathbf{lo}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{36-21o}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{3+lo}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{15?}$ A) 2 B) 3 C) 1D) 0 (c): $\\frac{1}{2}{{\\log }_{10}}36-2lo{{g}_{10}}3+lo{{g}_{10}}15$ $=lo{{g}_{10}}{{36}^{1/2}}-lo{{g}_{10}}{{3}^{2}}+lo{{g}_{10}}15$ $=lo{{g}_{10}}6-lo{{g}_{10}}9+lo{{g}_{10}}15$ $={{\\log }_{10}}\\frac{6\\times 15}{9}={{\\log }_{10}}\\frac{90}{9}={{\\log }_{10}}10=1$", "# How fast can you find the value? $\\frac {2.3^3 - 0.027}{2.3^2 + 0.69 + 0.09}$ Quick! Calculate the expression above as fast as possible! Time is of the essence! ×", "that at least two passwords hash to the same value is given by: $\\displaystyle 1-\\frac{N\\times N-1\\times N-2\\times \\ldots \\times N-k+1}{N^k}$ Since md5 hash function has $N=2^{128}$ possible values, the probability that two passwords hash to the same value is $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k}$ We want to compute for k so that this probability is at least 50%. $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} \\ge 0.5$ which is equivalent to $\\displaystyle \\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} < 0.5$ Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us. We can write the probability in the following way: $\\displaystyle 1-\\frac{N}{N}\\times\\frac{N-1}{N}\\times\\frac{N-2}{N}\\times\\frac{N-3}{N}\\times\\ldots\\times\\frac{N-k+1}{N}$ $= \\displaystyle 1-\\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ Since N is large, the quantities $\\displaystyle \\frac{1}{N}, \\frac{2}{N}, \\frac{3}{N}, \\frac{k-1}{N}$ are very small. Because of this, we can use the approximation $e^{-x} \\approx 1-x$ The above approximation comes from the Taylor expansion of $e^{-x}$: $\\displaystyle e^{-x} = 1 - x + \\frac{x^2}{2!} - \\frac{x^3}{3!} + \\frac{x^4}{4!} \\ldots$ If $x$ is small, the higher order terms like $x^2, x^3, x^4, \\ldots$ vanish. Using this approximation, we can write the probability as: $\\displaystyle \\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ $\\displaystyle = e^{-\\frac{1}{N}}\\cdot e^{-\\frac{2}{N}}\\cdot e^{-\\frac{3}{N}}\\cdot \\ldots\\cdot e^{-\\frac{k-1}{N}}$ $\\displaystyle = e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}}$ Since $\\displaystyle", "• # question_answer Simplify the given expression and find the value of the expression. $\\sqrt{\\frac{~0.256\\times 0.081\\times 4.356}{1.5625\\times 0.0121\\times 129.6\\times 64}}$ A) 10.96 B) 0.024 C) 2.196 D) 4.096 (a): $\\sqrt{256}=16$ $\\sqrt{81}=9$ $\\sqrt{4356}=66$ $\\sqrt{15625}=125$ $\\sqrt{121}=11$ $\\sqrt{1296}=36$ Expression, $\\sqrt{\\frac{256\\times {{10}^{-}}^{3}\\times 81\\times {{10}^{-3}}\\times 4356\\times {{10}^{-3}}}{15625\\times {{10}^{-4}}\\times 121\\times {{10}^{-}}^{4}\\times 1296\\times {{10}^{-}}^{1}\\times 64}}$ $\\sqrt{\\frac{16\\times {{9}^{2}}\\times {{66}^{2}}\\times \\bcancel{{{10}^{-9}}}}{{{125}^{2}}\\times {{11}^{2}}\\times {{36}^{2}}\\times {{8}^{2}}\\times \\bcancel{{{10}^{-9}}}}}$ $=\\frac{16\\times 9\\times 66}{125\\times 11\\times 36\\times 8}=0.024$", "\\times \\frac{{75}}{{25}} = 20$$", "} I = \\frac{P \\times n \\times (n+1)}{2} \\times \\frac{r}{100} \\times \\frac{1}{12}$ $\\displaystyle 8325 = \\frac{P \\times 36 \\times 37}{2} \\times \\frac{7.5}{100} \\times \\frac{1}{12}$ $\\displaystyle \\Rightarrow P = \\frac{8325 \\times 2 \\times 100 \\times 12}{36 \\times 37 \\times 7.5} = 2000 \\text{ Rs.}$ (ii) Maturity Value $\\displaystyle = P \\times n + I = 2000 \\times 36 + 8325 = 80325 \\text{ Rs.}$", "# A Lot of Calculation? Level pending What is the value of $810000-\\frac{810000 \\times 810001+1}{810901}?$ ×", "by $$(8!\\times2!\\times2!\\times2!)^2$$. Final answer: $$\\frac{64!}{32!\\times(8!\\times2!\\times2!\\times2!)^2} = 4634726695587809641192045982323285670400000$$ Fun fact: $$1.46966\\times10^{35}$$ years would contain these many seconds.", "What is the value of $\\frac{1}{2}$ of $\\frac{2}{3}$ of $\\frac{3}{4}$ of $\\frac{4}{5}$ of $\\frac{5}{6}$ of $\\frac{6}{7}$ of $\\frac{7}{8}$ of $\\frac{8}{9}$ of $\\frac{9}{10}$ of $1000$?", "$\\frac{3\\sqrt{12}}{2\\sqrt{28}}\\div \\frac{2\\sqrt{21}}{\\sqrt{98}}$ is approximately equal to : [A]1.0727 [B]1.6026 [C]1.0606 [D]1.6007 1.0606 We have to find the approx value of $\\frac{3\\sqrt{12}}{2\\sqrt{28}}\\div \\frac{2\\sqrt{21}}{\\sqrt{98}}$ $\\frac{3\\sqrt{12}}{2\\sqrt{28}}\\div \\frac{2\\sqrt{21}}{\\sqrt{98}}$ $=\\frac{3\\sqrt{12}}{2\\sqrt{28}}\\times \\frac{\\sqrt{98}}{2\\sqrt{21}}$ $= \\frac{3\\times 2\\times \\sqrt{3}}{2\\times 2\\times \\sqrt{7}}\\times \\frac{7\\times \\sqrt{2}}{2\\times \\sqrt{3}\\times \\sqrt{7}}$ $= \\frac{3\\sqrt{2}}{4} = \\frac{3\\times 1.414}{4} = 1.0605\\approx 1.0606$ Hence option [C] is correct answer.", "$P_{1}=\\frac{1622\\times&space;25}{26}=15600$ $16224=P_{2}(1+\\frac{4}{100})^{2}$ $P_{2}=15000$ Total value = 16244+15600+15000 = 46,844", "=\\frac{T_1-T_2}{T_1}\\times 100$$ $$Percentage \\ change \\ in\\ the\\ tool \\ life =\\frac{1.07166 \\ C\\ -\\ 0.18944\\ C}{1.07166\\ C}\\times 100$$ 82.32 %", "given a hash function $f(x)$, the probability that at least two passwords hash to the same value is given by: $\\displaystyle 1-\\frac{N\\times N-1\\times N-2\\times \\ldots \\times N-k+1}{N^k}$ Since md5 hash function has $N=2^{128}$ possible values, the probability that two passwords hash to the same value is $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k}$ We want to compute for k so that this probability is at least 50%. $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} \\ge 0.5$ which is equivalent to $\\displaystyle \\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} < 0.5$ Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us. We can write the probability in the following way: $\\displaystyle 1-\\frac{N}{N}\\times\\frac{N-1}{N}\\times\\frac{N-2}{N}\\times\\frac{N-3}{N}\\times\\ldots\\times\\frac{N-k+1}{N}$ $= \\displaystyle 1-\\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ Since N is large, the quantities $\\displaystyle \\frac{1}{N}, \\frac{2}{N}, \\frac{3}{N}, \\frac{k-1}{N}$ are very small. Because of this, we can use the approximation $e^{-x} \\approx 1-x$ The above approximation comes from the Taylor expansion of $e^{-x}$: $\\displaystyle e^{-x} = 1 - x + \\frac{x^2}{2!} - \\frac{x^3}{3!} + \\frac{x^4}{4!} \\ldots$ If $x$ is small, the higher order terms like $x^2, x^3, x^4, \\ldots$ vanish. Using this approximation, we can write the probability as: $\\displaystyle \\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ $\\displaystyle = e^{-\\frac{1}{N}}\\cdot e^{-\\frac{2}{N}}\\cdot e^{-\\frac{3}{N}}\\cdot \\ldots\\cdot e^{-\\frac{k-1}{N}}$", "what is the value of $$\\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ? A. 2 B. 4 C. 8 D. 16 E. 32 $$\\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html _________________ Kudos [?]: 129394 [1], given: 12197 Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] 13 Nov 2013, 01:21 Display posts from previous: Sort by", "# What is the value of (-1/4-1/2)div(-4/7)? Jun 8, 2018 $\\frac{21}{16}$ #### Explanation: $\\left(- \\frac{1}{4} - \\frac{1}{2}\\right)$ = $\\frac{- 1 - 2}{4}$ = $- \\frac{3}{4}$ $- \\frac{3}{4} \\div \\left(- \\frac{4}{7}\\right)$ which is equal to $- \\frac{3}{4} \\times \\left(- \\frac{7}{4}\\right)$ $- \\frac{3}{4} \\times \\left(- \\frac{7}{4}\\right)$ = $\\frac{21}{16}$", "is larger than $\\frac{1}{6}$. Therefore, we can assume that $\\frac{log_2{x}}{x}$ equals to $\\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\\frac{(2^t-3t)t}{4^t}$ is $\\textbf{(C)}\\ \\frac{1}{12}$." ]

[ "$$\\left( {n - r + 1} \\right)$$. Let us take this product all the way from n to 1. We do this by multiplying the existing product with \\begin{align}&\\left( {n - r} \\right) \\times \\left( {n - r - 1} \\right) \\times \\left( {n - r - 2} \\right) \\times ... \\times 1\\\\&{\\rm{or}},\\quad\\left( {n - r} \\right)!,\\end{align} and dividing by this same expression, as follows: \\begin{align}&^n{P_r} \\;= \\left( n \\right) \\times \\left( {n - 1} \\right) \\times ... \\times \\left( {n - r + 1} \\right)\\\\&\\qquad = \\frac{{\\left( n \\right) \\times \\left( {n - 1} \\right) \\times ... \\times \\left( {n - r + 1} \\right) \\times \\left( {n - r} \\right)!}}{{\\left( {n - r} \\right)!}}\\end{align} Now, in the numerator, the product goes all the way from n to 1. Thus, $^n{P_r} = \\frac{{n!}}{{\\left( {n - r} \\right)!}}$ Some examples: \\begin{align}&^5{P_2} = \\frac{{5!}}{{\\left( {5 - 2} \\right)!}} = \\frac{{5!}}{{3!}} = \\frac{{120}}{6} = 20\\\\&^6{P_5} = \\frac{{6!}}{{\\left( {6 - 5} \\right)!}} = \\frac{{6!}}{{1!}} = \\frac{{120}}{1} = 120\\\\&^7{P_2} = \\frac{{7!}}{{\\left( {7 - 2} \\right)!}} = \\frac{{7!}}{{5!}} = \\frac{{7 \\times 6 \\times 5!}}{{5!}} = 42\\end{align} Let us talk about an interesting result which follows from this discussion. We have seen that the number of permutations of n objects, taken all together, is $$n!$$. Thus, \\begin{align}&^n{P_n} = n!\\quad\\Rightarrow \\,\\,\\,\\frac{{n!}}{{\\left( {n - n} \\right)!}}", "with $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right)$$”; while the right-hand side is “make $$\\frac{a}{b}$$, but in­stead of start­ing out with $$1$$, start with $$\\frac{c}{d}$$; then do the same but start with $$\\frac{e}{f}$$; and then put the two to­gether”. If we draw out di­a­grams for the right-hand side, and put them next to each other, we get the di­a­gram for the left-hand side. (As an ex­er­cise, you should do this for some spe­cific val­ues of $$a,b,c,d,e,f$$.) We’ll prove it now us­ing the in­stant rules. The left-hand side is $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right) \\times \\frac{a}{b} = \\frac{c \\times f + d \\times e}{d \\times f} \\times \\frac{a}{b} = \\frac{(c \\times f + d \\times e) \\times a}{(d \\times f) \\times b}$$ which is $$\\frac{c \\times f \\times a + d \\times e \\times a}{d \\times f \\times b}$$. The right-hand side is $$\\left(\\frac{a}{b} \\times \\frac{c}{d}\\right) + \\left(\\frac{a}{b} \\times \\frac{e}{f}\\right) = \\frac{a \\times c}{b \\times d} + \\frac{a \\times e}{b \\times f} = \\frac{(a \\times c) \\times (b \\times f) + (b \\times d) \\times (a \\times e)}{(b \\times d) \\times (b \\times f)} = \\frac{(a \\times c \\times b \\times f) + (b \\times d \\times a \\times e)}{(b \\times d \\times b \\times f)}$$ No­tice that ev­ery­thing on the top has a $$b$$ in it some­where, so we can use this same “fil­ter­ing-through”", "## Precalculus (6th Edition) Blitzer The value of $\\frac{n!}{\\left( n-r \\right)!\\times r!}$ for $n=8$ and $r=3$ is $56$. The given expression is $\\frac{n!}{\\left( n-r \\right)!\\times r!}$ Put $n=8$ and $r=3$ in the expression $\\frac{n!}{\\left( n-r \\right)!\\times r!}$ and simplify, \\begin{align} & \\frac{n!}{\\left( n-r \\right)!\\times r!}=\\frac{8!}{\\left( 8-3 \\right)!\\times 3!} \\\\ & =\\frac{8!}{5!\\times 3!} \\end{align} It is known that \\begin{align} & n!=n\\times \\left( n-1 \\right)\\times \\left( n-2 \\right)\\times \\left( n-3 \\right)..........3\\times 2\\times 1 \\\\ & =n\\times \\left( n-1 \\right)! \\end{align} \\begin{align} & n!=\\frac{8\\times 7\\times 6\\times 5!}{5!\\times 3\\times 2\\times 1} \\\\ & =\\frac{8\\times 7\\times 6}{3\\times 2\\times 1} \\\\ & =56 \\end{align} Thus, the value of $\\frac{n!}{\\left( n-r \\right)!\\times r!}$ for $n=8$ and $r=3$ is $56$.", "product is, $4 \\times 7p = 4 \\times 7 \\times p$ $4 \\times 7p = 28p$ iii. $- 4p$ and $7p$ Ans: The required product is, $- 4p \\times 7p = \\left( { - 4} \\right) \\times p \\times 7 \\times p$ $- 4p \\times 7p = \\left( { - 4 \\times 7} \\right) \\times \\left( {p \\times p} \\right)$ $- 4p \\times 7p = - 28{p^2}$ iii. $- 4p$ and $7pq$ Ans: The required product is, $- 4p \\times 7pq = \\left( { - 4} \\right) \\times p \\times 7 \\times p \\times q$ $- 4p \\times 7pq = \\left( { - 4 \\times 7} \\right) \\times \\left( {p \\times p} \\right) \\times q$ $- 4p \\times 7pq = - 28{p^2}q$ iv. $4{p^3}$ and $- 3p$ Ans: The required product is, $4{p^3} \\times \\left( { - 3p} \\right) = 4 \\times {p^3} \\times \\left( { - 3} \\right) \\times p$ $4{p^3} \\times \\left( { - 3p} \\right) = \\left( {4 \\times - 3} \\right) \\times \\left( {{p^3} \\times p} \\right)$ $4{p^3} \\times \\left( { - 3p} \\right) = - 12{p^4}$ v. $4p$ and $0$ Ans: The required product is, $4p \\times \\left( 0 \\right) = 4 \\times p \\times 0$ $4p \\times \\left( 0 \\right) = 0$ 2. Find the areas of rectangles with the following pairs", "## Precalculus (6th Edition) Blitzer The value of $\\frac{n!}{\\left( n-r \\right)!}$ for $n=20$ and $r=3$ is $6840$. The given expression is $\\frac{n!}{\\left( n-r \\right)!}$. Put $n=20$ and $r=3$ in the expression $\\frac{n!}{\\left( n-r \\right)!}$ and simplify, \\begin{align} & \\frac{n!}{\\left( n-r \\right)!}=\\frac{20!}{\\left( 20-3 \\right)!} \\\\ & =\\frac{20!}{17!} \\end{align} It is know that \\begin{align} & n!=n\\times \\left( n-1 \\right)\\times \\left( n-2 \\right)\\times \\left( n-3 \\right)..........3\\times 2\\times 1 \\\\ & =n\\times \\left( n-1 \\right)! \\end{align} \\begin{align} & n!=\\frac{20\\times 19\\times 18\\times 17!}{17!} \\\\ & =20\\times 19\\times 18 \\\\ & =6,840 \\end{align} Thus, the value of $\\frac{n!}{\\left( n-r \\right)!}$ for $n=20$ and $r=3$ is $6840$.", "• # question_answer12) Find the product: (i) $({{a}^{2}})\\times (2{{a}^{22}})\\times (4{{a}^{26}})$ (ii) $\\left( \\frac{2}{3}xy \\right)\\times \\left( \\frac{-9}{10}{{x}^{2}}{{y}^{2}} \\right)$ (iii) $\\left( -\\frac{10}{3}\\,p{{q}^{3}} \\right)\\,\\times \\left( \\frac{6}{5}\\,{{p}^{3}}q \\right)$ (iv) $x\\times {{x}^{2}}\\times {{x}^{3}}\\times {{x}^{4}}$. $({{a}^{2}})\\times (2{{a}^{22}})\\times (4{{a}^{26}})$ $({{a}^{2}})\\times (2{{a}^{22}})\\times (4{{a}^{26}})$ $=(2\\times 4)\\times ({{a}^{2}}\\times {{a}^{22}}\\times {{a}^{26}})$ $=8\\times {{a}^{50}}=8{{a}^{50}}$. (ii) $\\left( \\frac{2}{3}xy \\right)\\times \\,\\left( -\\frac{9}{10}{{x}^{2}}{{y}^{2}} \\right)$ $\\left( \\frac{2}{3}xy \\right)\\times \\,\\left( -\\frac{9}{10}{{x}^{2}}{{y}^{2}} \\right)$ $=\\left\\{ \\frac{2}{3}\\times \\,\\left( -\\frac{9}{10} \\right) \\right\\}\\times (x\\times {{x}^{2}})\\times (y\\times {{y}^{2}})$ $=-\\frac{3}{5}\\,{{x}^{3}}{{y}^{3}}$. (iii) $\\left( -\\frac{10}{3}p{{q}^{3}} \\right)\\times \\,\\left( \\frac{6}{5}{{p}^{3}}q \\right)$ $\\left( -\\frac{10}{3}p{{q}^{3}} \\right)\\times \\,\\left( \\frac{6}{5}{{p}^{3}}q \\right)$ $=\\left\\{ \\left( -\\frac{10}{3} \\right)\\times \\frac{6}{5} \\right\\}\\times (p\\times {{p}^{3}})\\times ({{p}^{3}}\\times q)$ $=-4{{p}^{4}}{{q}^{4}}$. (iv) $x\\times {{x}^{2}}\\times {{x}^{3}}\\times {{x}^{4}}$ $x\\times {{x}^{2}}\\times {{x}^{3}}\\times {{x}^{4}}$$={{x}^{1}}\\times {{x}^{2}}\\times {{x}^{3}}\\times {{x}^{4}}$ $={{x}^{1+2+3+4}}$ $={{x}^{10}}$.", "Find each of the following product: Question: Find each of the following product: $\\left(\\frac{7}{9} a b^{2}\\right) \\times\\left(\\frac{15}{7} a c^{2} b\\right) \\times\\left(-\\frac{3}{5} a^{2} c\\right)$ Solution: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \\times a^{n}=a^{m+n}$. We have: $\\left(\\frac{7}{9} a b^{2}\\right) \\times\\left(\\frac{15}{7} a c^{2} b\\right) \\times\\left(-\\frac{3}{5} a^{2} c\\right)$ $=\\left\\{\\frac{7}{9} \\times \\frac{15}{7} \\times\\left(-\\frac{3}{5}\\right)\\right\\} \\times\\left(a \\times a \\times a^{2}\\right) \\times\\left(b^{2} \\times b\\right) \\times\\left(c^{2} \\times c\\right)$ $=-a^{4} b^{3} c^{3}$ Thus, the answer is $-a^{4} b^{3} c^{3}$.", "10^{-1})^2}{(2\\times 10^{-5})^2}\\right)\\right]\\:=\\:80$, will work. But you can't use braces { } to do the same job as parentheses ( ). They have completely different functions in TeX. For the multiplication sign, use \\times. Notice also that if an exponent consists of more than one symbol then you need to enclose the symbols in braces. Otherwise only the first symbol will be superscripted. For example, 10^-1 gives $10^-1$, but 10^{-1} gives $10^{-1}$. 7. Thanks very very much! that is a massive help. 8. I'll try this again: I hope i get it right! $10\\:log.\\left[\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)^3\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)\\:=\\:80$ \\:=\\:80 9. Why did the last part mess up with the \\: & 80?? 10. ## Just testing I'll try again: I hope i get better this time?! 10\\:log.\\:\\left[\\left(\\frac{(2\\times 10^{-1})^3}{2\\times10^{-5})^3}\\right)\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}){2\\times10^{-5})\\right)\\:=120 11. Originally Posted by Paul46 Why did the last part mess up with the \\: & 80?? Probably because what you wrote was \"$$10\\:log.\\left[\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)^3\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)\\:=\\:80$$\\:=\\:80\", with an additional \"\\:=\\:80\" at the end. While I'm about it, here are a couple of extra comments. You don't normally put a full stop after \"log\", and in TeX you should write \"\\log\". The backslash converts the letters from italic to roman, which looks better (the same applies to other common functions: \\cos, \\sin, \\exp, ...). Also, you shouldn't normally need to put extra space in math expressions by inserting", "Find the following product: Question: Find the following product: $\\left(-\\frac{7}{4} a b^{2} c-\\frac{6}{25} a^{2} c^{2}\\right)\\left(-50 a^{2} b^{2} c^{2}\\right)$ Solution: To find the product, we will use distributive law as follows: $\\left(-\\frac{7}{4} a b^{2} c-\\frac{6}{25} a^{2} c^{2}\\right)\\left(-50 a^{2} b^{2} c^{2}\\right)$ $=\\left\\{\\left(-\\frac{7}{4} a b^{2} c\\right)\\left(-50 a^{2} b^{2} c^{2}\\right)\\right\\}-\\left\\{\\left(\\frac{6}{25} a^{2} c^{2}\\right)\\left(-50 a^{2} b^{2} c^{2}\\right)\\right\\}$ $=\\left\\{\\left\\{-\\frac{7}{4} \\times(-50)\\right\\}\\left(a \\times a^{2}\\right) \\times\\left(b^{2} \\times b^{2}\\right) \\times\\left(c \\times c^{2}\\right)\\right\\}-\\left\\{\\left(\\frac{6}{25}\\right)(-50)\\left(a^{2} \\times a^{2}\\right) \\times\\left(b^{2}\\right) \\times\\left(c^{2} \\times c^{2}\\right)\\right\\}$ $=\\left\\{-\\frac{7}{4} \\times(-50)\\right\\}\\left(a^{1+2} b^{2+2} c^{1+2}\\right)-\\left\\{\\left(\\frac{6}{25}\\right)(-50)\\left(a^{2+2} b^{2} c^{2+2}\\right)\\right\\}$ $=\\frac{175}{2} a^{3} b^{4} c^{3}-\\left(-12 a^{4} b^{2} c^{4}\\right)$ $=\\frac{175}{2} a^{3} b^{4} c^{3}+12 a^{4} b^{2} c^{4}$ Thus, the answer is $\\frac{175}{2} a^{3} b^{4} c^{3}+12 a^{4} b^{2} c^{4}$.", "product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times {\\left( 2x \\right)}^{2} \\times 8y+3 \\times {\\left( 2x \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( 8y \\right)}^{2} \\times 2x+3 \\times {\\left( 8y \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( -4z \\right)}^{2} \\times 2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times {\\left( 2x \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( 8y \\right)}^{2} \\times 2x+3 \\times {\\left( 8y \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( -4z \\right)}^{2} \\times 2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( 8y \\right)}^{2} \\times 2x+3 \\times {\\left( 8y \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( -4z \\right)}^{2} \\times 2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times 64{y}^{2} \\times 2x+3 \\times {\\left( 8y \\right)}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( -4z \\right)}^{2} \\times", "1. $\\displaystyle \\alpha$ 2. ## Just testing $\\displaystyle 10\\:log\\:\\left[\\frac{\\left(2\\times10^{-1}}\\right)^3{\\left(2\\times10^{-5}}\\right)^3\\right]\\:=30\\:log\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\:120\\:$ 3. ## Debugging LaTeX Originally Posted by Paul46 $\\displaystyle 10\\:log\\:\\left[\\frac{\\left(2\\times10^{-1}}\\right)^3{\\left(2\\times10^{-5}}\\right)^3\\right]\\:=30\\:log\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\:120\\:$ The LaTeX compiler used by this Forum is very unforgiving and very uncommunicative. So, what do you do when after entering an impressive formula like $$10\\:log\\:\\left[\\frac{\\left(2\\times10^{-1}}\\right)^3{\\left(2\\times10^{-5}}\\right)^3\\right]\\:=30\\:log\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\:120\\:$$ in LaTeX, you hit the Preview button and all that happens is that you get the message [LaTeX Error: Syntax error]? Here are a few hints on how to debug your LaTeX. The first and most important one is to narrow down where the errors are occurring. Split the formula into two at some convenient break point like an = sign, so that it becomes two separate formulas: $$10\\:log\\:\\left[\\frac{\\left(2\\times10^{-1}}\\right)^3{\\left(2\\times10^{-5}}\\right)^3\\right]\\:$$ $$=30\\:log\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\:120\\:$$. Now hit the Preview button and you see [LaTeX Error: Syntax error] [LaTeX Error: Syntax error]. Bad news. That means there is at least one error on both sides of the equation. But at least we are narrowing down where the errors might lie. Take the second of the two parts of the equation: $$=30\\:log\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\:120\\:$$. Can we subdivide it further? Not easily, because one thing that the LaTeX compiler is very picky about is that a command like \"\\left(\" must always be matched by a \"\\right)\". So we can't just split this formula down the middle. What we can do is to replace", "6 | Page 14 Find each of the following product: $\\left( \\frac{- 24}{25} x^3 z \\right) \\times \\left( - \\frac{15}{16}x z^2 y \\right)$ Ex. 6.30 | Q 7 | Page 14 Find each of the following product: $\\left( - \\frac{1}{27} a^2 b^2 \\right) \\times \\left( \\frac{9}{2} a^3 b^2 c^2 \\right)$ Ex. 6.30 | Q 8 | Page 14 Find each of the following product: $( - 7xy) \\times \\left( \\frac{1}{4} x^2 yz \\right)$ Ex. 6.30 | Q 9 | Page 14 Find each of the following product: (7ab) × (−5ab2c) × (6abc2) Ex. 6.30 | Q 10 | Page 14 Find each of the following product: (−5a) × (−10a2) × (−2a3) Ex. 6.30 | Q 11 | Page 14 Find each of the following product: (−4x2) × (−6xy2) × (−3yz2) Ex. 6.30 | Q 12 | Page 14 Find each of the following product: $\\left( - \\frac{2}{7} a^4 \\right) \\times \\left( - \\frac{3}{4} a^2 b \\right) \\times \\left( - \\frac{14}{5} b^2 \\right)$ Ex. 6.30 | Q 13 | Page 14 Find each of the following product: $\\left( \\frac{7}{9}a b^2 \\right) \\times \\left( \\frac{15}{7}a c^2 b \\right) \\times \\left( - \\frac{3}{5} a^2 c \\right)$ Ex. 6.30 | Q 14 | Page 14 Find each of the following product: $\\left( \\frac{4}{3} u^2 vw \\right) \\times \\left( - 5uv w^2 \\right)", "is the divisor. Mixed Number A mixed number is a number made up of a whole number and a fraction such as $4 \\frac{3}{5}$ . Reciprocal The reciprocal of a number is the inverse of that number. If $\\frac{a}{b}$ is a nonzero number, then $\\frac{b}{a}$ is its reciprocal. The product of a number and its reciprocal is one. Quotient The quotient is the answer of a division problem. ### Guided Practice 1. $\\left(\\frac{7}{10}\\right) \\div \\left ( \\frac{5}{6} \\right )=?$ 2. $\\left(6 \\frac{2}{5}\\right) \\div \\left(1 \\frac{2}{3}\\right)=?$ 3. How many pieces of plywood 0.375 in. thick are in a stack of 30 in. high? 1. $\\left(\\frac{7}{10}\\right) \\div \\left(\\frac{5}{6}\\right)$ $=\\frac{7}{10} \\times \\frac{6}{5}$ $=\\frac{7 \\times 6}{10 \\times 5}$ $=\\frac{42}{50}=\\frac{21}{25}$ 2. $\\left(6\\frac{2}{5}\\right) \\div \\left(1\\frac{2}{3}\\right)$ $=\\left(\\frac{32}{5}\\right) \\div \\left(\\frac{5}{3}\\right)$ $=\\left(\\frac{32}{5}\\right) \\times \\left(\\frac{3}{5}\\right)$ $=\\frac{32 \\times 3}{5 \\times 5}$ $& =\\frac{96}{25}\\\\& =3\\frac{21}{25}$ 3. To determine the number of pieces of plywood in the stack, divide the thickness of one piece into the height of the pile. $& \\overset{ \\qquad \\qquad \\ 80}{\\underset{\\ \\ \\longrightarrow}{0.375} \\overline{ ) {\\underset{\\ \\ \\longrightarrow}{30.000}}}}\\\\& \\qquad \\underline{- 3000\\;}\\\\& \\qquad \\qquad \\quad 0\\\\& \\qquad \\ \\underline{- \\;\\;\\;\\;\\;\\;\\;0}\\\\& \\qquad \\qquad \\quad 0$ There are 80 pieces of plywood in the pile. ### Practice Find each quotient or product. 1. $(+14)\\div (+2)$ 2. $(-14) \\div (+2)$ 3. $(-9)\\div (-3)$ 4. $(+16) \\div (+4)$ 5. $(+25) \\div", "20 Observe the following pattern $1^2 = \\frac{1}{6}\\left[ 1 \\times \\left( 1 + 1 \\right) \\times \\left( 2 \\times 1 + 1 \\right) \\right]$ $1^2 + 2^2 = \\frac{1}{6}\\left[ 2 \\times \\left( 2 + 1 \\right) \\times \\left( 2 \\times 2 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 = \\frac{1}{6}\\left[ 3 \\times \\left( 3 + 1 \\right) \\times \\left( 2 \\times 3 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 + 4^2 = \\frac{1}{6}\\left[ 4 \\times \\left( 4 + 1 \\right) \\times \\left( 2 \\times 4 + 1 \\right) \\right]$ and find the values : 12 + 22 + 32 + 4+ ... + 102 Ex. 3.20 | Q 10.2 | Page 20 Observe the following pattern $1^2 = \\frac{1}{6}\\left[ 1 \\times \\left( 1 + 1 \\right) \\times \\left( 2 \\times 1 + 1 \\right) \\right]$ $1^2 + 2^2 = \\frac{1}{6}\\left[ 2 \\times \\left( 2 + 1 \\right) \\times \\left( 2 \\times 2 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 = \\frac{1}{6}\\left[ 3 \\times \\left( 3 + 1 \\right) \\times \\left( 2 \\times 3 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 + 4^2 = \\frac{1}{6}\\left[ 4 \\times \\left( 4 + 1 \\right) \\times \\left( 2 \\times 4 + 1 \\right) \\right]$ and find the values : 52 + 62 + 72", "do most questions like this in your head. Consider the following solution. \\begin{align*}7.5\\times3.9&=(0.1\\times75)\\times(0.1\\times39)\\\\\\\\&=0.01\\times75\\times39\\\\\\\\&=75\\%\\,\\,\\text{of}\\,\\,39\\quad(\\text{as}\\,\\,\\%\\,\\,\\text{means}\\,\\,1/100=0.01)\\tag{1}\\\\\\\\&=\\frac34\\times39\\quad\\small{\\left[=\\frac{4-1}4\\times39=\\left(\\frac44-\\frac14\\right)\\times39\\right]}\\tag2\\\\\\\\&=\\left(1-\\frac14\\right)\\times39\\quad\\small{\\left[=1\\times39-\\frac14\\times39\\right]}\\\\\\\\&=39-\\frac{39}4\\\\\\\\&=39-\\frac{36+3}4\\quad\\small{\\left[=39-\\left(\\frac{36}4+\\frac34\\right)\\right]}\\\\\\\\&=39-\\left(9+\\frac34\\right)\\\\\\\\&=39-9-\\frac34\\\\\\\\&=30-0.75\\\\\\\\&=29.25\\end{align*} $$(1)$$ As an aside, it may be of interest to mention that $$75\\%\\times39=39\\%\\times75$$. $$(2)$$ Comparing the length of this solution to Martin's, it shows the importance of choosing whether to express $$3=4-1$$ (as in this answer), or $$39=40-1$$ which yields a quicker solution. $$\\left\\lfloor \\frac{\\left(x+y\\right)^2}{4} \\right\\rfloor - \\left\\lfloor \\frac{\\left(x-y\\right)^2}{4} \\right\\rfloor = xy.$$ This answer uses the floor function (which essentially means to get rid of the decimal). You can then quickly consult a table to find 0.25*x^2. X, 0.25*x^2 1, 0.25 2, 1 3, (9/4) 4, 4 5, (25/4) 6, 9 7, (49/4) 8, 16 9, (81/4) 10, 25 11, (121/4) 12, 36 13, (169/4) 14, 49 15, (225/4) • Can you please expand on this a little by editing your answer to perform the computation in the question? – Xander Henderson Apr 21 at 17:47 This is similar to Xander Henderson's approach, but the calculations have been made slightly easier by using two FOIL multiplications instead of just one: $$3.9 \\times (3.8 + 3.7)$$ $$=3.9 \\times 3.8 + 3.9 \\cdot 3.7$$ $$=(4 - 0.1)(4-0.2)+(4-0.1)(4-0.3)$$ $$=16-0.8-0.4+0.02+16-1.2-0.4+0.03$$ $$=16+16-0.8-1.2-0.4-0.4+0.02+0.03$$ $$=(16+16)-(0.8+1.2)-(0.4+0.4)+(0.02-0.03)$$ $$=32-2-0.8+0.05$$ $$=29.2+0.05$$ $$=29.25$$", "What are all the factor pairs of 32? Then teach the underlying concepts Don't copy without citing sources preview ? Explanation Explain in detail... Explanation: I want someone to double check my answer 8 Tony B Share May 22, 2016 1, 32 2, 16 4, 8 Explanation: Assumption: Looking for whole number factors. If you include fractional or decimal factors then there are infinitely many factors. One way is to use a factor tree and then use that to build up different combinations. From this we observe that we have the following: $\\left[1\\right] \\times \\left[32\\right] \\to 1 , 32$ $\\left[2\\right] \\times \\left[2 \\times 2 \\times 2 \\times 2\\right] \\to 2 , 16$ $\\left[2 \\times 2\\right] \\times \\left[2 \\times 2 \\times 2\\right] \\to 4 , 8$ $\\left[2 \\times 2 \\times 2\\right] \\times \\left[2 \\times 2\\right] \\to 8 , 4 \\text{ \"larr \"repeated pair}$ $\\left[2 \\times 2 \\times 2 \\times 2\\right] \\times \\left[2\\right] \\to 16 , 2 \\text{ \"larr \"repeated pair}$ $\\left[32\\right] \\times \\left[1\\right] \\to 32 , 1 \\text{ \"larr \"repeated pair}$ • 7 minutes ago • 7 minutes ago • 7 minutes ago • 8 minutes ago • 16 seconds ago • 41 seconds ago • A minute ago • 2 minutes ago • 4 minutes ago • 4 minutes ago • 7 minutes ago • 7 minutes ago", "(2×109)4= 1.6×1037 Question 3. $$\\frac{\\left(1.2 \\times 10^{4}\\right)+\\left(2 \\times 10^{4}\\right)+\\left(2.8 \\times 10^{4}\\right)}{3}$$= 2×104 Question 4. $$\\frac{7 \\times 10^{15}}{14 \\times 10^{9}}$$ = 5×105 Question 5. $$\\frac{4 \\times 10^{2}}{2 \\times 10^{8}}$$= 2×10-6 Question 6. $$\\frac{\\left(7 \\times 10^{9}\\right)+\\left(6 \\times 10^{9}\\right)}{2}$$=", "symmetry. Thus $$p(2,3)=p(1,2)=p(0,1)=\\frac 12$$ It follows that: $$p(2,3)=\\frac 12\\times \\left(1+0\\right)=\\frac 12$$ $$p(2,2)=\\frac 12\\times \\left(1+\\frac 12\\right)=\\frac 34$$ $$p(2,1)=\\frac 12 \\times \\left(1+\\frac 34\\right)=\\frac 78$$ $$p(2,0)=\\frac 12 \\times \\left(1+\\frac 78\\right)=\\frac {15}{16}$$ $$p(1,3)=\\frac 12\\times \\left(\\frac 12+0\\right)=\\frac 14$$ $$p(1,2)=\\frac 12 \\times \\left(\\frac 34+\\frac 14\\right)=\\frac 12$$ $$p(1,1)=\\frac 12 \\times \\left(\\frac 78 + \\frac 12\\right)=\\frac {11}{16}$$ $$p(1,0)=\\frac 12 \\times \\left(\\frac {15}{16}+\\frac {11}{16}\\right)=\\frac {26}{32}$$ Finally we get $$p(0,0)=\\frac 12 \\times \\left( \\frac {26}{32}+\\frac 12\\right)=\\boxed {\\frac {42}{64}}$$ Note: the above is straight forward but somewhat error prone. I advise checking. • I had a feeling we had crossed paths before... It's the rectangle around the answer... What do you see as a discrepancy with my answer here? – Antoni Parellada Dec 13 '16 at 1:12 • @AntoniParellada this problem is nearly identical to the baseball problem, though I did not approach it that way. As trueblueanil points out, you can do it simply by asking whether Heads comes up at least three times in the first six tosses. After all, if we see at least three $H's$ then Heads wins, if we don't then we must have had at least four $T's$ so Tails wins. Much better approach than mine. My method works best for problems in which we might return to these states infinitely often, but that isn't the case here. Of course, the answer comes", "2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times 64{y}^{2} \\times 2x+3 \\times 64{y}^{2} \\times \\left( -4 \\right)z+3 \\times {\\left( -4z \\right)}^{2} \\times 2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times 64{y}^{2} \\times 2x+3 \\times 64{y}^{2} \\times \\left( -4 \\right)z+3 \\times 16{z}^{2} \\times 2x+3 \\times {\\left( -4z \\right)}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ To raise a product to a power, raise each factor to that power $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times 64{y}^{2} \\times 2x+3 \\times 64{y}^{2} \\times \\left( -4 \\right)z+3 \\times 16{z}^{2} \\times 2x+3 \\times 16{z}^{2} \\times 8y+6 \\times 2x \\times 8y \\times \\left( -4 \\right)z$ Multiplying an odd number of negative terms makes the product negative $8{x}^{3}+512{y}^{3}-64{z}^{3}+3 \\times 4{x}^{2} \\times 8y+3 \\times 4{x}^{2} \\times \\left( -4 \\right)z+3 \\times 64{y}^{2} \\times 2x+3 \\times 64{y}^{2} \\times \\left( -4 \\right)z+3 \\times 16{z}^{2} \\times 2x+3 \\times 16{z}^{2} \\times 8y-6 \\times 2x \\times 8y \\times 4z$ Calculate", "# Find what a Infinite product number approaches Find what the following number approaches: $$\\frac{6}{5}\\times\\frac{26}{25}\\times\\frac{626}{625}\\times\\frac{390626}{390625}\\times...$$ What I found is that each fraction can be simplified as:$$1+\\frac{1}{5^n}$$ Where $$n$$ is $$1,2,4,8$$ and so on (instead of $$1,2,3,4,\\ldots$$) What should I do next? Note: Please use junior high school math if possible • @Damien Thanks for the notice. – Cyh1368 Jan 27 at 9:39 • or do u mean the reciprocal of what you have written? – Aatmaj Jan 27 at 9:41 • Hint: Notice that the product is in fact $$\\sum_{i=0}^\\infty 5^{-i}$$ you can observe it by just expand the product. Take one term in each product and you will see that. – macton Jan 27 at 9:42 • The same idea math.stackexchange.com/q/622321/399263 – zwim Jan 27 at 11:27 You have $$\\left(1+\\frac{1}{5^1}\\right)\\left(1+\\frac{1}{5^2}\\right)\\left(1+\\frac{1}{5^4}\\right)\\left(1+\\frac{1}{5^8}\\right)\\cdots$$ $$=\\left(1+\\frac{1}{5^1}+\\frac{1}{5^2}+\\frac{1}{5^3}\\right)\\left(1+\\frac{1}{5^4}\\right)\\left(1+\\frac{1}{5^8}\\right)\\cdots$$ $$=\\left(1+\\frac{1}{5^1}+\\frac{1}{5^2}+\\frac{1}{5^3}+\\frac{1}{5^4}+\\frac{1}{5^5}+\\frac{1}{5^6}+\\frac{1}{5^7}\\right)\\left(1+\\frac{1}{5^8}\\right)\\cdots$$ $$=1+\\frac{1}{5^1}+\\frac{1}{5^2}+\\frac{1}{5^3}+\\frac{1}{5^4}+\\frac{1}{5^5}+\\frac{1}{5^6}+\\frac{1}{5^7}+\\frac{1}{5^8}+\\cdots$$ $$= \\frac{5}{5-1} \\\\= 1+\\frac{1}{4}$$ Put $$x=\\frac{1}{5}$$ then we have $$\\lim_{n \\to \\infty}(1+x)(1+x^2)(1+x^4)(1+x^8)..(1+x^{2^n})$$ now multiply $$1-x$$ in numerator and denominator and use $$(1-x^{2^k})(1+x^{2^k})=1-x^{2^{k+1}}$$ repeatedly and see the magic!" ]

[ "or $$(\\sqrt[n]{x})^m$$ Example: Find the value of $$8^{\\frac{5}{3}}$$ and $$256^{\\frac{3}{4}}$$ Solution: 1. $$8^{\\frac{7}{3}} = (\\sqrt[3]{8})^7 = (\\sqrt[3]{2^3})^7 = 2^7 = 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 = 128$$ 2. $$256^{\\frac{3}{4}} = (\\sqrt[4]{256})^3 = (\\sqrt[4]{4^4})^3 = 4^3 = 4 \\times 4 \\times 4 = 64$$", "with $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right)$$”; while the right-hand side is “make $$\\frac{a}{b}$$, but in­stead of start­ing out with $$1$$, start with $$\\frac{c}{d}$$; then do the same but start with $$\\frac{e}{f}$$; and then put the two to­gether”. If we draw out di­a­grams for the right-hand side, and put them next to each other, we get the di­a­gram for the left-hand side. (As an ex­er­cise, you should do this for some spe­cific val­ues of $$a,b,c,d,e,f$$.) We’ll prove it now us­ing the in­stant rules. The left-hand side is $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right) \\times \\frac{a}{b} = \\frac{c \\times f + d \\times e}{d \\times f} \\times \\frac{a}{b} = \\frac{(c \\times f + d \\times e) \\times a}{(d \\times f) \\times b}$$ which is $$\\frac{c \\times f \\times a + d \\times e \\times a}{d \\times f \\times b}$$. The right-hand side is $$\\left(\\frac{a}{b} \\times \\frac{c}{d}\\right) + \\left(\\frac{a}{b} \\times \\frac{e}{f}\\right) = \\frac{a \\times c}{b \\times d} + \\frac{a \\times e}{b \\times f} = \\frac{(a \\times c) \\times (b \\times f) + (b \\times d) \\times (a \\times e)}{(b \\times d) \\times (b \\times f)} = \\frac{(a \\times c \\times b \\times f) + (b \\times d \\times a \\times e)}{(b \\times d \\times b \\times f)}$$ No­tice that ev­ery­thing on the top has a $$b$$ in it some­where, so we can use this same “fil­ter­ing-through”", "that at least two passwords hash to the same value is given by: $\\displaystyle 1-\\frac{N\\times N-1\\times N-2\\times \\ldots \\times N-k+1}{N^k}$ Since md5 hash function has $N=2^{128}$ possible values, the probability that two passwords hash to the same value is $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k}$ We want to compute for k so that this probability is at least 50%. $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} \\ge 0.5$ which is equivalent to $\\displaystyle \\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} < 0.5$ Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us. We can write the probability in the following way: $\\displaystyle 1-\\frac{N}{N}\\times\\frac{N-1}{N}\\times\\frac{N-2}{N}\\times\\frac{N-3}{N}\\times\\ldots\\times\\frac{N-k+1}{N}$ $= \\displaystyle 1-\\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ Since N is large, the quantities $\\displaystyle \\frac{1}{N}, \\frac{2}{N}, \\frac{3}{N}, \\frac{k-1}{N}$ are very small. Because of this, we can use the approximation $e^{-x} \\approx 1-x$ The above approximation comes from the Taylor expansion of $e^{-x}$: $\\displaystyle e^{-x} = 1 - x + \\frac{x^2}{2!} - \\frac{x^3}{3!} + \\frac{x^4}{4!} \\ldots$ If $x$ is small, the higher order terms like $x^2, x^3, x^4, \\ldots$ vanish. Using this approximation, we can write the probability as: $\\displaystyle \\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ $\\displaystyle = e^{-\\frac{1}{N}}\\cdot e^{-\\frac{2}{N}}\\cdot e^{-\\frac{3}{N}}\\cdot \\ldots\\cdot e^{-\\frac{k-1}{N}}$ $\\displaystyle = e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}}$ Since $\\displaystyle", "$\\mathbf{1}_X(x) = \\begin{cases} 1 & x \\in X\\\\ 0 & x \\notin X \\end{cases}$ where $$\\in$$ means in and $$\\notin$$ means not in, and so equals one if the $$j^\\mathrm{th}$$ nearest neighbour is a member of the $$i^\\mathrm{th}$$ cluster and zero otherwise. For example, if we were to choose the weights $w = \\left[\\tfrac12, \\tfrac14, \\tfrac18, \\tfrac1{16}, \\tfrac1{32}\\right]$ then we'd have $\\sum_{j=0}^4 w_j = \\tfrac{31}{32}$ and \\begin{align*} p_0 &= \\frac{1 \\times \\tfrac12 + 0 \\times \\tfrac14 + 1 \\times \\tfrac18 + 1 \\times \\tfrac1{16} + 0 \\times \\tfrac1{32}}{\\tfrac{31}{32}} = \\frac{1 \\times 16 + 0 \\times 8 + 1 \\times 4 + 1 \\times 2 + 0 \\times 1}{31} = \\frac{22}{31}\\\\ p_1 &= \\frac{0 \\times \\tfrac12 + 0 \\times \\tfrac14 + 0 \\times \\tfrac18 + 0 \\times \\tfrac1{16} + 1 \\times \\tfrac1{32}}{\\tfrac{31}{32}} = \\frac{0 \\times 16 + 0 \\times 8 + 0 \\times 4 + 0 \\times 2 + 1 \\times 1}{31} = \\frac1{31}\\\\ p_2 &= \\frac{0 \\times \\tfrac12 + 1 \\times \\tfrac14 + 0 \\times \\tfrac18 + 0 \\times \\tfrac1{16} + 0 \\times \\tfrac1{32}}{\\tfrac{31}{32}} = \\frac{0 \\times 16 + 1 \\times 8 + 0 \\times 4 + 0 \\times 2 + 0 \\times 1}{31} = \\frac8{31} \\end{align*} Both variants of the algorithm are implemented by the ak.knnClasses function in listing 3. Listing 3: ak.knnClasses ak.knnClasses = function(candidates, arg1,", "1,657 views What is the value of $x$ when $81\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144?$ 1. $1$ 2. $-1$ 3. $-2$ 4. $\\text{Can not be determined}$ Migrated from GO Civil 3 years ago by Arjun 1 comment Easiest way is to substitute value, If you try to smiplify it you may end up something like $(\\frac{16}{9})$ $^{x+1}$ = 1 , only -1 stastify $81\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144$ $\\implies9^{2}\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144$ $\\implies9^{2}\\times\\left (\\frac{4}{5} \\right )^{2(x+2)}\\div\\left (\\frac{3}{5} \\right )^{2(x+2)}=12^{2}$ $\\implies\\left [ 9\\times\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]^{2}=12^{2}$ $\\implies\\left [ 9\\times\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]=12$ $\\implies \\left [\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]=\\frac{4}{3}$ $\\implies\\frac{\\left(\\frac{4}{5}\\right)^{x+2}}{\\left(\\frac{3}{5}\\right)^{x+2}} = \\frac{4}{3}$ $\\implies\\left(\\frac{4}{3}\\right)^{x+2}= \\left(\\frac{4}{3}\\right)^{1}$ Compare both side and we get $x+2=1$ $\\implies x=-1$ So, correct answer is option (B). \\begin{align} 81\\times\\left( \\frac{16}{25} \\right)^{x+2}\\div\\left( \\frac{3}{5} \\right)^{2x+4}&=144\\\\ \\Rightarrow \\left( \\frac{16}{25} \\right)^{x+2}\\times\\left( \\frac{5}{3} \\right)^{2(x+2)}&=\\frac{144}{81}\\\\ \\Rightarrow \\left( \\frac{16}{25} \\right)^{x+2}\\times\\left( \\frac{25}{9} \\right)^{x+2}&=\\frac{16\\times9}{9\\times9}\\\\ \\Rightarrow \\left( \\frac{16}{25}\\times\\frac{25}{9} \\right)^{x+2}&=\\frac{16}{9}\\\\ \\Rightarrow \\left( \\frac{16}{9} \\right)^{x+2}&=\\frac{16}{9}\\\\ \\Rightarrow x+2&=1\\\\ \\therefore x&=-1\\end{align} So the correct answer is B. option b is right Option B is right. by", "10^{-1})^2}{(2\\times 10^{-5})^2}\\right)\\right]\\:=\\:80$, will work. But you can't use braces { } to do the same job as parentheses ( ). They have completely different functions in TeX. For the multiplication sign, use \\times. Notice also that if an exponent consists of more than one symbol then you need to enclose the symbols in braces. Otherwise only the first symbol will be superscripted. For example, 10^-1 gives $10^-1$, but 10^{-1} gives $10^{-1}$. 7. Thanks very very much! that is a massive help. 8. I'll try this again: I hope i get it right! $10\\:log.\\left[\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)^3\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)\\:=\\:80$ \\:=\\:80 9. Why did the last part mess up with the \\: & 80?? 10. ## Just testing I'll try again: I hope i get better this time?! 10\\:log.\\:\\left[\\left(\\frac{(2\\times 10^{-1})^3}{2\\times10^{-5})^3}\\right)\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}){2\\times10^{-5})\\right)\\:=120 11. Originally Posted by Paul46 Why did the last part mess up with the \\: & 80?? Probably because what you wrote was \"$$10\\:log.\\left[\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)^3\\right]\\:=\\:30\\:log.\\left(\\frac{2\\times10^{-1}}{2\\times10^{-5}}\\right)\\:=\\:80$$\\:=\\:80\", with an additional \"\\:=\\:80\" at the end. While I'm about it, here are a couple of extra comments. You don't normally put a full stop after \"log\", and in TeX you should write \"\\log\". The backslash converts the letters from italic to roman, which looks better (the same applies to other common functions: \\cos, \\sin, \\exp, ...). Also, you shouldn't normally need to put extra space in math expressions by inserting", "on each table. They can be split only by the outcome on A because it is equally likely to be odd or even. Some of the rows and columns are repeated. A simpler version of the tables is shown below, with fewer rows and columns - but the ratio between the odd and even outcomes on each dice is preserved. Using the second pair of tables, there are 18 possible outcomes. 9 of them are 'O'. So the probability of getting an odd sum is $\\frac12$. Using a tree diagram The tree diagram below shows the possible outcomes. The outcomes where the sum is an odd number are indicated with a green arrow. So the total probability is: $$\\left(\\frac12\\times\\frac23\\times\\frac13\\right)+\\left(\\frac12\\times\\frac13\\times\\frac23\\right)+\\left(\\frac12\\times\\frac23\\times\\frac23\\right)+\\left(\\frac12\\times\\frac13\\times\\frac13\\right)\\\\ = 2\\times\\left(\\frac12\\times\\frac23\\times\\frac13\\right)+\\left(\\frac12\\times\\frac23\\times\\frac23\\right)+\\left(\\frac12\\times\\frac13\\times\\frac13\\right)\\\\ = 2\\times\\frac{1\\times2\\times1}{2\\times3\\times3}+\\frac{1\\times2\\times2}{2\\times3\\times3}+\\frac{1\\times1\\times1}{2\\times3\\times3}\\\\ =\\frac{2\\times2}{18}+\\frac{4}{18}+\\frac{1}{18}\\\\ =\\frac{4+4+1}{18}\\\\ =\\frac9{18}\\\\ =\\frac12$$ Using combinations of products To add three numbers and get an odd number, the numbers must have all been odd, or two evens and an odd. The probability that all of the numbers are odd is $\\dfrac12\\times\\dfrac23\\times\\dfrac13=\\dfrac{1}{9}$. There are three ways of getting two even numbers and an odd number. The odd number could be on any of the three dice. Die A odd: $\\dfrac12\\times\\dfrac13\\times\\dfrac23=\\dfrac19$ Die B odd: $\\dfrac12\\times\\dfrac23\\times\\dfrac23=\\dfrac29$ Die C odd: $\\dfrac12\\times\\dfrac13\\times\\dfrac13=\\dfrac1{18}$ So the probability of getting an odd sum is: $$\\begin{split}\\frac19+\\frac19+\\frac29+\\frac1{18}&=\\frac2{18}+\\frac2{18}+\\frac4{18}+\\frac1{18}\\\\&=\\frac9{18}=\\frac12\\end{split}$$ You can find more short problems, arranged by", "}\\times \\left( 100+\\frac { 12 }{ 2 } \\right) \\%$$ Solving this gives $${ F }_{ 5 }=\\frac { 110 }{ 106\\% } =103.77\\%$$ This means we have to purchase $$103.77\\%\\times 100=103.77$$ face value of bond 5 At 18 months, only bonds 4 and 5 make a payment. We can therefore obtain the value of $${ F }_{ 4 }$$ as follows: $$10={ F }_{ 2 }\\times 0+{ F }_{ 3 }\\times 0+{ F }_{ 4 }\\times \\left( 100+\\frac { 10 }{ 2 } \\right) \\%+103.77\\times \\frac { 12\\% }{ 2 }$$ Solving this gives $${ F }_{ 4 }=\\frac { 10-103.77 \\times 0.06 }{ 1.05 } =3.59\\%$$ This, again, means we have to purchase$3.59 face value of bond 4 To solve for $${ F }_{ 3 }$$, $$10={ F }_{ 2 }\\times 0+{ F }_{ 3 }\\times \\left( 100+\\frac { 24 }{ 2 } \\right) \\%+3.59\\times \\frac { 10\\% }{ 2 } +103.77\\times \\frac { 12\\% }{ 2 }$$ $${ F }_{ 3 }=\\frac { 10-0.18-6.23 }{ 1.12 } =3.21\\%$$ Similarly, $$10={ F }_{ 2 }\\times \\left( 100+\\frac { 14 }{ 2 } \\right)\\% +3.21\\times \\frac { 24\\% }{ 2 } +3.59\\times \\frac { 10\\% }{ 2 } +103.77\\times \\frac { 12\\% }{ 2 }$$ $${ F }_{ 2 }=\\frac { 10-0.39-0.18-6.23 }{ 1.07 } =2.99\\%$$ Cash", "46 views What is the value of $x$ when $81\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144?$ 1. $1$ 2. $-1$ 3. $-2$ 4. $\\text{Can not be determined}$ edited | 46 views Migrated from GO Civil 2 months ago by Arjun 0 Easiest way is to substitute value, If you try to smiplify it you may end up something like $(\\frac{16}{9})$ $^{x+1}$ = 1 , only -1 stastify +1 vote $81\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144$ $\\implies9^{2}\\times\\left (\\frac{16}{25} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{2x+4}=144$ $\\implies9^{2}\\times\\left (\\frac{4}{5} \\right )^{2(x+2)}\\div\\left (\\frac{3}{5} \\right )^{2(x+2)}=12^{2}$ $\\implies\\left [ 9\\times\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]^{2}=12^{2}$ $\\implies\\left [ 9\\times\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]=12$ $\\implies \\left [\\left (\\frac{4}{5} \\right )^{x+2}\\div\\left (\\frac{3}{5} \\right )^{x+2}\\right ]=\\frac{4}{3}$ $\\implies\\frac{\\left(\\frac{4}{5}\\right)^{x+2}}{\\left(\\frac{3}{5}\\right)^{x+2}} = \\frac{4}{3}$ $\\implies\\left(\\frac{4}{3}\\right)^{x+2}= \\left(\\frac{4}{3}\\right)^{1}$ Compare both side and we get $x+2=1$ $\\implies x=-1$ So, correct answer is option (B). by Boss (45.8k points) selected by option b is right by Boss (34.4k points)", "20 Observe the following pattern $1^2 = \\frac{1}{6}\\left[ 1 \\times \\left( 1 + 1 \\right) \\times \\left( 2 \\times 1 + 1 \\right) \\right]$ $1^2 + 2^2 = \\frac{1}{6}\\left[ 2 \\times \\left( 2 + 1 \\right) \\times \\left( 2 \\times 2 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 = \\frac{1}{6}\\left[ 3 \\times \\left( 3 + 1 \\right) \\times \\left( 2 \\times 3 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 + 4^2 = \\frac{1}{6}\\left[ 4 \\times \\left( 4 + 1 \\right) \\times \\left( 2 \\times 4 + 1 \\right) \\right]$ and find the values : 12 + 22 + 32 + 4+ ... + 102 Ex. 3.20 | Q 10.2 | Page 20 Observe the following pattern $1^2 = \\frac{1}{6}\\left[ 1 \\times \\left( 1 + 1 \\right) \\times \\left( 2 \\times 1 + 1 \\right) \\right]$ $1^2 + 2^2 = \\frac{1}{6}\\left[ 2 \\times \\left( 2 + 1 \\right) \\times \\left( 2 \\times 2 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 = \\frac{1}{6}\\left[ 3 \\times \\left( 3 + 1 \\right) \\times \\left( 2 \\times 3 + 1 \\right) \\right]$ $1^2 + 2^2 + 3^2 + 4^2 = \\frac{1}{6}\\left[ 4 \\times \\left( 4 + 1 \\right) \\times \\left( 2 \\times 4 + 1 \\right) \\right]$ and find the values : 52 + 62 + 72", "hash to the same value is $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k}$ We want to compute for k so that this probability is at least 50%. $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} \\ge 0.5$ which is equivalent to $\\displaystyle \\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} < 0.5$ Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us. We can write the probability in the following way: $\\displaystyle 1-\\frac{N}{N}\\times\\frac{N-1}{N}\\times\\frac{N-2}{N}\\times\\frac{N-3}{N}\\times\\ldots\\times\\frac{N-k+1}{N}$ $= \\displaystyle 1-\\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ Since N is large, the quantities $\\displaystyle \\frac{1}{N}, \\frac{2}{N}, \\frac{3}{N}, \\frac{k-1}{N}$ are very small. Because of this, we can use the approximation $e^{-x} \\approx 1-x$ The above approximation comes from the Taylor expansion of $e^{-x}$: $\\displaystyle e^{-x} = 1 - x + \\frac{x^2}{2!} - \\frac{x^3}{3!} + \\frac{x^4}{4!} \\ldots$ If $x$ is small, the higher order terms like $x^2, x^3, x^4, \\ldots$ vanish. Using this approximation, we can write the probability as: $\\displaystyle \\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ $\\displaystyle = e^{-\\frac{1}{N}}\\cdot e^{-\\frac{2}{N}}\\cdot e^{-\\frac{3}{N}}\\cdot \\ldots\\cdot e^{-\\frac{k-1}{N}}$ $\\displaystyle = e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}}$ Since $\\displaystyle \\sum_1^n j = 1+2+3+4+ \\ldots + n = \\frac{n(n+1)}{2}$ we have $e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}} = e^{-k\\cdot (k-1)/2N }$ Computing for k Let’s compute k so that $\\displaystyle e^{-k\\cdot (k-1)/2N} < 0.5$ Taking the", "hash to the same value is $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k}$ We want to compute for k so that this probability is at least 50%. $\\displaystyle 1-\\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} \\ge 0.5$ which is equivalent to $\\displaystyle \\frac{2^{128}\\times 2^{128}-1\\times 2^{128}-2\\times \\ldots \\times 2^{128}-k+1}{(2^{128})^k} < 0.5$ Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us. We can write the probability in the following way: $\\displaystyle 1-\\frac{N}{N}\\times\\frac{N-1}{N}\\times\\frac{N-2}{N}\\times\\frac{N-3}{N}\\times\\ldots\\times\\frac{N-k+1}{N}$ $= \\displaystyle 1-\\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ Since N is large, the quantities $\\displaystyle \\frac{1}{N}, \\frac{2}{N}, \\frac{3}{N}, \\frac{k-1}{N}$ are very small. Because of this, we can use the approximation $e^{-x} \\approx 1-x$ The above approximation comes from the Taylor expansion of $e^{-x}$: $\\displaystyle e^{-x} = 1 - x + \\frac{x^2}{2!} - \\frac{x^3}{3!} + \\frac{x^4}{4!} \\ldots$ If $x$ is small, the higher order terms like $x^2, x^3, x^4, \\ldots$ vanish. Using this approximation, we can write the probability as: $\\displaystyle \\frac{N}{N}\\times (1-\\frac{1}{N})\\times (1-\\frac{2}{N})\\times (1-\\frac{3}{N}) \\times\\ldots\\times (1-\\frac{k-1}{N})$ $\\displaystyle = e^{-\\frac{1}{N}}\\cdot e^{-\\frac{2}{N}}\\cdot e^{-\\frac{3}{N}}\\cdot \\ldots\\cdot e^{-\\frac{k-1}{N}}$ $\\displaystyle = e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}}$ Since $\\displaystyle \\sum_1^n j = 1+2+3+4+ \\ldots + n = \\frac{n(n+1)}{2}$ we have $e^{-\\frac{1+2+3+4+\\ldots + k-1}{N}} = e^{-k\\cdot (k-1)/2N }$ Computing for k Let’s compute k so that $\\displaystyle e^{-k\\cdot (k-1)/2N} < 0.5$ Taking the", "at 19:55 • If $100 = 7\\times 14 + 2$ then $2 = 100 - 7\\times 14$. So $2\\times \\frac {999,999}7 = (100-7\\times 14)\\times \\frac {999999}7 = (100 - 7\\times 14)\\times \\frac {10^6 -1}7$ Now just expand $(100-7\\times 14)\\times \\frac{10^6 - 1}7 = 100\\times \\frac {10^6-1}7 - 7\\times 14\\times \\frac {10^6-1}7= 100\\times \\frac {10^6-1}7 +-14\\times (10^6 -1) = 100\\times \\frac {10^6-1}7 - 14\\times 10^6 + 14$. ... Now $100\\times {10^6-1}7$ will add two zeros to $142857$ to get $14285700$. And $-14\\times 10^6=-14000000$ removes the leading $14$ to get $285700$ and $+14$ adds it back to the other side: $285714$. Sep 22 at 20:06 • $1999998 = 2\\cdot 999,999 \\ne 2\\cdot \\frac {999999}7=285714$. Why did you do anything with $2\\times 999999$ that was not part of the question and never part of my solution? Sep 22 at 20:08 • It $2 = 100 - 7\\times 14$ (which it does) then $$2 \\times ANYTHING = (100 -7\\times 14)ANYTHING = 100\\times ANYTHING - 7\\times 14\\times ANYTHING$$. And if $$ANYTHING = 142857 = \\frac {1000000 - 1}7$$ then $$2\\times 142857 = 100\\times 142857 - 7\\times 14\\times\\frac {1000000 -1}7 = 100\\times 142857 - 14\\times 1000000 + 14$$. $$=\\color{green}{14}\\color{red}{2857}\\color{purple}{00} - \\color{green}{14}\\color{red}{0000}\\color{purple}{00} + \\color{purple}{14}= \\color{red}{2857}\\color{purple}{00}$$That's all. Sep 22 at 20:11 • Yes I got confused, sorry about that – Jim Sep 23 at 20:50", "# Find the value of (2sqrt6)^6? Aug 16, 2017 ${\\left(2 \\sqrt{6}\\right)}^{6} = 13824$ #### Explanation: ${\\left(2 \\sqrt{6}\\right)}^{6}$ = ${2}^{6} \\times {\\left(\\sqrt{6}\\right)}^{6}$ = $\\left(2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2\\right) \\times \\left(\\sqrt{6 \\times 6 \\times 6 \\times 6 \\times 6 \\times 6}\\right)$ = $64 \\times \\sqrt{\\underline{6 \\times 6} \\times \\underline{6 \\times 6} \\times \\underline{6 \\times 6}}$ = $64 \\times 6 \\times 6 \\times 6$ = $64 \\times 216$ = $13824$ Aug 16, 2017 $13 , 824$ #### Explanation: Use the laws of indices. ${\\left(2 \\sqrt{6}\\right)}^{6} = {2}^{6} \\times {\\left(\\sqrt{6}\\right)}^{6} \\text{ } \\leftarrow \\left({\\left(a b\\right)}^{m} = {a}^{m} {b}^{m}\\right)$ $= {2}^{6} \\times {\\left(6\\right)}^{\\frac{6}{2}} \\text{ } \\leftarrow \\sqrt{x} = {x}^{\\frac{1}{2}}$ $= {2}^{6} \\times {6}^{3}$ $= \\left(2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2\\right) \\times \\left(6 \\times 6 \\times 6\\right)$ It helps to know the powers less than 1000 by heart. $= 64 \\times 216$ $= 13 , 824$ Jun 21, 2018 $13824$ #### Explanation: ${\\left(2 \\sqrt{6}\\right)}^{6}$ $\\therefore = {\\left(2 \\times {6}^{\\frac{1}{2}}\\right)}^{6}$ $\\therefore = {2}^{6} \\times {6}^{\\frac{6}{2}}$ $\\therefore = 64 \\times {6}^{3}$ $\\therefore = 64 \\times 216$ $\\therefore = 13824$", "do most questions like this in your head. Consider the following solution. \\begin{align*}7.5\\times3.9&=(0.1\\times75)\\times(0.1\\times39)\\\\\\\\&=0.01\\times75\\times39\\\\\\\\&=75\\%\\,\\,\\text{of}\\,\\,39\\quad(\\text{as}\\,\\,\\%\\,\\,\\text{means}\\,\\,1/100=0.01)\\tag{1}\\\\\\\\&=\\frac34\\times39\\quad\\small{\\left[=\\frac{4-1}4\\times39=\\left(\\frac44-\\frac14\\right)\\times39\\right]}\\tag2\\\\\\\\&=\\left(1-\\frac14\\right)\\times39\\quad\\small{\\left[=1\\times39-\\frac14\\times39\\right]}\\\\\\\\&=39-\\frac{39}4\\\\\\\\&=39-\\frac{36+3}4\\quad\\small{\\left[=39-\\left(\\frac{36}4+\\frac34\\right)\\right]}\\\\\\\\&=39-\\left(9+\\frac34\\right)\\\\\\\\&=39-9-\\frac34\\\\\\\\&=30-0.75\\\\\\\\&=29.25\\end{align*} $$(1)$$ As an aside, it may be of interest to mention that $$75\\%\\times39=39\\%\\times75$$. $$(2)$$ Comparing the length of this solution to Martin's, it shows the importance of choosing whether to express $$3=4-1$$ (as in this answer), or $$39=40-1$$ which yields a quicker solution. $$\\left\\lfloor \\frac{\\left(x+y\\right)^2}{4} \\right\\rfloor - \\left\\lfloor \\frac{\\left(x-y\\right)^2}{4} \\right\\rfloor = xy.$$ This answer uses the floor function (which essentially means to get rid of the decimal). You can then quickly consult a table to find 0.25*x^2. X, 0.25*x^2 1, 0.25 2, 1 3, (9/4) 4, 4 5, (25/4) 6, 9 7, (49/4) 8, 16 9, (81/4) 10, 25 11, (121/4) 12, 36 13, (169/4) 14, 49 15, (225/4) • Can you please expand on this a little by editing your answer to perform the computation in the question? – Xander Henderson Apr 21 at 17:47 This is similar to Xander Henderson's approach, but the calculations have been made slightly easier by using two FOIL multiplications instead of just one: $$3.9 \\times (3.8 + 3.7)$$ $$=3.9 \\times 3.8 + 3.9 \\cdot 3.7$$ $$=(4 - 0.1)(4-0.2)+(4-0.1)(4-0.3)$$ $$=16-0.8-0.4+0.02+16-1.2-0.4+0.03$$ $$=16+16-0.8-1.2-0.4-0.4+0.02+0.03$$ $$=(16+16)-(0.8+1.2)-(0.4+0.4)+(0.02-0.03)$$ $$=32-2-0.8+0.05$$ $$=29.2+0.05$$ $$=29.25$$", "than RRR. You can fix it by making a much larger sample space taking in more detailed outcomes. An easier way to solve the problem would be to use a tree. The first level of the tree would be rolling the first die, etc. – Arby Mar 15 '17 at 12:28 • @Arby Thanks for your input. I have a very rusty background in probability, so I wasn't really sure if I was doing the right thing. – RoadRunner Mar 15 '17 at 12:32 • In addition, there's a fourth action you need to take into account, that you pick one of the three dice at the end. If you use a tree, it will actually have 4 levels. – Arby Mar 15 '17 at 12:34 It looks as if all $6+6+6=18$ faces are equally likely to be chosen in the end, and $2+1+3=6$ are green, making the answer $$\\frac{6}{18}=\\frac13$$ If you want a longer answer based on your sample space, try $$\\frac26 \\times \\frac16\\times\\frac36\\times \\frac33 + \\frac26 \\times \\frac56\\times\\frac36\\times \\frac13 +\\frac26 \\times \\frac16\\times\\frac36\\times \\frac23 + \\frac26 \\times \\frac56\\times\\frac36\\times \\frac23 \\\\ + \\frac46 \\times \\frac56\\times\\frac36\\times \\frac03 + \\frac46 \\times \\frac16\\times\\frac36\\times \\frac23 + \\frac46 \\times \\frac16\\times\\frac36\\times \\frac13 + \\frac46 \\times \\frac56\\times\\frac36\\times \\frac13$$ but it is a little more complicated • It may be obvious to you (and possibly to", "$C(12,4)=C(12,8)$C(12,4)=C(12,8) and $C(7,1)=C(7,6)$C(7,1)=C(7,6). There is a nice symmetry with this expression. Since $0!=1$0!=1 and $1!=1$1!=1, we have by definition that $C\\left(n,0\\right)=\\frac{n!}{0!\\times\\left(n-0\\right)!}=\\frac{n!}{1\\times n!}=1$C(n,0)=n!0!×(n0)!=n!1×n!=1 For example, $C\\left(5,0\\right)=1$C(5,0)=1 and $C\\left(12,0\\right)=1$C(12,0)=1. Finally we can form a relationship between $C(n,r)$C(n,r) and the previous $P(n,r)$P(n,r) as follows: $C\\left(n,r\\right)$C(n,r) $=$= $\\frac{n!}{r!\\times\\left(n-r\\right)!}$n!r!×(n−r)!​ $=$= $\\frac{n\\times\\left(n-1\\right)\\times\\left(n-2\\right)\\times...\\times\\left(n-r+1\\right)\\times\\left(n-r\\right)!}{r!\\times\\left(n-r\\right)!}$n×(n−1)×(n−2)×...×(n−r+1)×(n−r)!r!×(n−r)!​ $=$= $\\frac{n\\times\\left(n-1\\right)\\times\\left(n-2\\right)\\times...\\times\\left(n-r+1\\right)}{r!}$n×(n−1)×(n−2)×...×(n−r+1)r!​ $=$= $\\frac{P\\left(n,r\\right)}{r!}$P(n,r)r!​ So $C\\left(n,r\\right)=\\frac{P\\left(n,r\\right)}{r!}$C(n,r)=P(n,r)r! and this is another important mathematical result in the study of probability. Another type of notation that is used in combinations is nC it means the same as $C(n,r)$C(n,r). We can also write it as $\\binom{n}{r}$(nr) #### Worked Examples ##### QUESTION 1 Evaluate $\\nCr{7}{2}$7C2. ##### QUESTION 2 A boss wants to select one group of $4$4 people from his $28$28 staff. How many different groups are possible? ### Outcomes #### M8-3 Use permutations and combinations", "• # question_answer Simplify the given expression and find the value of the expression. $\\sqrt{\\frac{~0.256\\times 0.081\\times 4.356}{1.5625\\times 0.0121\\times 129.6\\times 64}}$ A) 10.96 B) 0.024 C) 2.196 D) 4.096 (a): $\\sqrt{256}=16$ $\\sqrt{81}=9$ $\\sqrt{4356}=66$ $\\sqrt{15625}=125$ $\\sqrt{121}=11$ $\\sqrt{1296}=36$ Expression, $\\sqrt{\\frac{256\\times {{10}^{-}}^{3}\\times 81\\times {{10}^{-3}}\\times 4356\\times {{10}^{-3}}}{15625\\times {{10}^{-4}}\\times 121\\times {{10}^{-}}^{4}\\times 1296\\times {{10}^{-}}^{1}\\times 64}}$ $\\sqrt{\\frac{16\\times {{9}^{2}}\\times {{66}^{2}}\\times \\bcancel{{{10}^{-9}}}}{{{125}^{2}}\\times {{11}^{2}}\\times {{36}^{2}}\\times {{8}^{2}}\\times \\bcancel{{{10}^{-9}}}}}$ $=\\frac{16\\times 9\\times 66}{125\\times 11\\times 36\\times 8}=0.024$", "\\times 6 \\times 3^{5}}$ Solution: $=\\sqrt{(0.5)^{2} \\times 0.5 \\times 3 \\times 2 \\times 3^{5}}$ $=\\sqrt{(0.5)^{2} \\times 0.5 \\times 2 \\times 3 \\times 3^{5}}$ $=\\sqrt{(0.5)^{2} \\times 1.0 \\times 3^{6}}[0.5 \\times 2=1.0]$ $=\\sqrt{(0.5)^{2} \\times 1 \\times 3^{6}}=0.5 \\times 3^{3}$ $=0.5 \\times 27=13.5$ (iii) $\\sqrt{\\left(5+2 \\frac{21}{25}\\right) \\times \\frac{0.69}{16}}$ Solution: $=\\sqrt{\\left(5+\\frac{71}{25}\\right) \\times \\frac{0.169}{1600}}=\\sqrt{\\frac{196}{25} \\times \\frac{169}{1600}}$ $=\\sqrt{\\frac{14 \\times 14}{5 \\times 5} \\times \\frac{13 \\times 13}{40 \\times 40}}=\\frac{14 \\times 13}{5 \\times 40}$ $=\\frac{7 \\times 13}{5 \\times 20}=\\frac{91}{100}=0.91$ (iv) $\\sqrt{5\\left(2 \\frac{3}{4}-\\frac{3}{10}\\right)}$ Solution : $\\sqrt{5\\left(2 \\frac{3}{4}-\\frac{3}{10}\\right)}=\\sqrt{5\\left(\\frac{11}{4}-\\frac{3}{10}\\right)}$ $=\\sqrt{5\\left(\\frac{55-6}{20}\\right)}=\\sqrt{5\\left(\\frac{49}{20}\\right)}$ $=\\sqrt{\\frac{5 \\times 49}{20}}=\\sqrt{\\frac{49}{4}}=\\sqrt{\\frac{7 \\times 7}{2 \\times 2}}$ $=\\frac{7}{2}=3 \\frac{1}{2}$ (v) $\\sqrt{248+\\sqrt{52+\\sqrt{144}}}$ Solution: $=\\sqrt{248+\\sqrt{52+12}} (\\sqrt{144}=12)$ $=\\sqrt{248+\\sqrt{64}}=(\\sqrt{248+8} ( \\sqrt{64}=8)$ $=\\sqrt{256}=16 (\\sqrt{256}=\\sqrt{16 \\times 16}=16)$ Question 8. A man, after a tour, finds that he had spent day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all 1,296? Solution: Let the number of days he had spent =x Number of rupees spent in each day =x Total money spent $=x \\times x=x^{2}=1,296(\\text { given })$ $x=\\sqrt{1296}$ $\\Rightarrow x=\\sqrt{4 \\times 4 \\times 9 \\times 9}$ $x=4 \\times 9=36$ Hence required number of days =36 Question 9. Out of 745 students, maximum is to be arranged in the school field for a PT. display, such that the number of rows is equal to the number of columns. Find the", "the correct option. The value of $$\\frac{3.157\\times4126\\times3.198}{63.972\\times2835.121}$$ is closet to: A2 B0.002 C0.02 D0.2 Explanation: Here is no explanation for this answer Workspace NA SHSTTON 2 Solv. Corr. 1 Solv. In. Corr. 3 Attempted 0 M:0 S Avg. Time 66 / 70 Choose the correct option. $$\\frac{0.0203\\times2.92}{0.0073\\times14.5\\times0.7}=?$$ A3.25 B0.8 C1.45 D2.4 Explanation: $$\\frac{0.0203\\times2.92}{0.0073\\times14.5\\times0.7}=\\frac{203\\times292}{73\\times145\\times7}=\\frac{4}{7}=0.8$$ Workspace NA SHSTTON 3 Solv. Corr. 1 Solv. In. Corr. 4 Attempted 0 M:0 S Avg. Time 67 / 70 Choose the correct option. Simplify: $$\\frac{0.2\\times0.2+0.2\\times0.02}{0.044}$$ A2 B0.004 C0.4 D1 Explanation: Given Expression=$$\\frac{0.2\\times0.2+0.2\\times0.02}{0.044}=\\frac{0.2\\times0.22}{0.044}=\\frac{0.044}{0.044}=1$$ Workspace NA SHSTTON 1 Solv. Corr. 3 Solv. In. Corr. 4 Attempted 0 M:0 S Avg. Time 68 / 70 Choose the correct option. $$\\frac{(36.54)^2-(3.46)^2}{?}=40$$ A330.8 B3.308 C4 D33.08 Explanation: Here is no explanation for this answer Workspace NA SHSTTON 1 Solv. Corr. 2 Solv. In. Corr. 3 Attempted 0 M:0 S Avg. Time 69 / 70 Choose the correct option. $$(7.5\\times7.5+37.5+2.5\\times2.5)$$ is equal to: A100 B30 C60 D80 Explanation: Here is no explanation for this answer Workspace NA SHSTTON 3 Solv. Corr. 2 Solv. In. Corr. 5 Attempted 0 M:0 S Avg. Time 70 / 70 Choose the correct option. What decimal of an hour is a second A0.0028 B0.0027 C0.0026 D0.0025 E0.0024 Explanation: 1/(60*60) = 1/3600 = .0027 Workspace ## Quantitative Aptitude Decimal Fraction Questions and Answers pdf At Quantitative" ]

[ "• zmudz Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of $$9x+12y+8z.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Maximum value Calculus Level 3 If $$x,y$$ are real numbers where $$x+y = 1$$, determine the maximum value of $$(x^3+1)(y^3+1)$$. ×", "Inspired by Brian Charlesworth $x$ and $y$ are non- negative integers such that $2x+y=10$ Find the sum of maximum and minimum values of $(x+y)$. ×", "Practice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.", "# Find the radical maximum value Algebra Level 3 $\\large \\sqrt{3x+1} + \\sqrt{3y+3} +\\sqrt{3z+5}$ Suppose $$x,y,z$$ are non-negative real numbers such that $$x+y+z=1$$, then find the maximum value of the expression above. ###### For more problems, click here. × Problem Loading... Note Loading... Set Loading...", "# Yet another question 8 Calculus Level 4 Find the minimum value of $$\\max( |y^2- xy| )$$ given that $$x$$ is any real number and $$0\\leq y \\leq 1$$. Try more here ×", "# Tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ . Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "$${a,b,c}$$ are nonnegative real numbers. Find the minimum value of $$\\dfrac{a^2}{bc} + \\dfrac{b^2}{ac} + \\dfrac{c^2}{ab} + 3(\\dfrac{a+b}{c} + \\dfrac{a+c}{b} + \\dfrac{b+c}{a} )$$", "# Find the maximum Algebra Level pending $$x$$, $$y$$ and $$z$$ are non-negative real numbers so that $$x+y+z=3$$. What is the maximum value of $$x^{2} y^{2} z$$?", "# Very Very Weird Inequality Algebra Level 4 Let $x$, $y$ and $z$ be nonnegative real numbers such that $x+y+z=5$. Find the maximum value of $xy^2 + yz^2 + 2xyz.$ × Problem Loading... Note Loading... Set Loading...", "# Maximum Possible Value Algebra Level 3 $$x, y$$ and $$z$$ are non-negative real numbers such that $$x+y+z=1$$. What is the maximum possible value of $$x+y^2+z^3$$? ×", "# Very Very Weird Inequality Algebra Level 4 Let $$x$$, $$y$$ and $$z$$ be nonnegative real numbers such that $$x+y+z=5$$. Find the maximum value of $xy^2 + yz^2 + 2xyz.$ ×", "How can you find a maximum for $f(x,y,z) = xyz + z$ where z is unconstrained. I mean z could be any positive real number or zero right? then $f(x,y, \\infty) = \\infty$.", "# $$x^2+y^2+z^2=32$$ Algebra Level 2 If real numbers $$x$$, $$y$$ and $$z$$ satisfy $x+y+z=8, x^2+y^2+z^2=32,$ what is the maximum value of $$z?$$ ×", "Great and tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "Maximum value Algebra Level 4 Let $x$ and $y$ be positive real numbers satisfying $x+y=1$, find the maximum value of $36(x^4y + xy^4)$. ×", "# Integers here! Max? Let $$x,y,z$$ be three non-negative integers such that $$x+y+z=10$$. The maximum possible value of $$xyz+xy+yz+zx$$ is: ×", "# Zero uh? No Algebra Level 5 Suppose $$x, y, z$$ are non-negative real numbers with $$x+y+z=3$$. Find the maximum value of: $\\large P=(x-y)(y-z)(z-x)$ ×", "# Strange Suppose $$x$$ and $$y$$ are real numbers such that $$x^2$$ and $$y^2$$ are positive integers. Find the maximum value of $$x^2-xy$$ if $$\\left(3x^2-y^2\\right)^2+\\left(x^2+y^2\\right)^2=72$$. ### This is a part of the Set. × Problem Loading... Note Loading... Set Loading...", "Don't try x = y = z Algebra Level 4 Let $x, y, z$ be real numbers such that $x^{2}+y^{2}+z^{2}=1$. Let the maximum possible value of $\\sqrt {6} xy+4yz$ be $A$.Find $A^{2}$. ×" ]

[ "real number is nonnegative, we know that the sum is greater than or equal to $0$. Equality holds when the value inside the parhentheses is equal to $0$. We find that $$(x,y) = (5,-3)$$ and the sum we are looking for is $$5+(-3)=2 \\implies \\boxed{\\textbf{(B)}}.$$ - Honestly", "G$, so $[z, y] = 1$ as $Q \\cap N = 1$ This shows that $Z(P) \\le Z(Q)$, equality follow from the fact that they have the same order.", "# Properties of Equality The following are the properties of equality for real numbers . Some textbooks list just a few of them, others list them all. These are the logical rules which allow you to balance, manipulate, and solve equations. PROPERTIES OF EQUALITY Reflexive Property For all real numbers $x$ , $x=x$ . A number equals itself. These three properties define an equivalence relation Symmetric Property For all real numbers $x\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}y$ , if $x=y$ , then $y=x$ . Order of equality does not matter. Transitive Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , if $x=y$ and $y=z$ , then $x=z$ . Two numbers equal to the same number are equal to each other. Addition Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , if $x=y$ , then $x+z=y+z$ . These properties allow you to balance and solve equations involving real numbers Subtraction Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , if $x=y$ , then $x-z=y-z$ . Multiplication Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , if $x=y$ , then $xz=yz$ . Division Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , if $x=y$ , and $z\\ne 0$ , then $\\frac{x}{z}=\\frac{y}{z}$ . Substitution Property For all real numbers $x\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}y$ , if $x=y$ , then $y$ can be substituted for $x$ in any expression. Distributive Property For all real numbers $x,y,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}z$ , $x\\left(y+z\\right)=xy+xz$ For more, see", "(2)$$ The equality occurs when $x=y=z=2$. Using C-S: $$(x^2+y^2+z^2)(y^2+z^2+x^2)\\ge (xy+yz+zx)^2 \\Rightarrow x^2+y^2+z^2\\ge 12 \\ \\ \\ \\ (3)$$ The equality occurs when $x=y=z=2$. From the first equation: $$(x+y+z)^2=24+x^2+y^2+z^2\\ge 36 \\Rightarrow x+y+z\\ge 6 \\ \\ \\ \\ \\ (4)$$ From $(2)$ and $(4)$, and then from the second equation we get: $$x+y+z=6, \\\\ xyz=8.$$ So, we get: $$x=y=z=2.$$", "× # Equality Problem $$x^{p} + y^{q} = y^{r} + z^{p} = z^{q} + x^{r}$$, x, y and z are positive reals, p, q, and r are positive integers. Find all values of x,y,z,p,q,r. General solutions are acceptable (i.e. x=y=z, p=q=r). How would I go about solving this? Note by Phil Peters 2 years, 6 months ago", "# Tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ . Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "$X$ and $Y$ are two positive real numbers such that $2X+Y \\leq 6$ and $X + 2Y \\leq 8$. For which of the following values of $(X,Y)$ the function $f(X,Y)=3X + 6Y$ will give maximum value ? 1. $\\left(\\dfrac{4}{3} , \\dfrac{10}{3}\\right) \\\\$ 2. $\\left(\\dfrac{8}{3} , \\dfrac{20}{3}\\right) \\\\$ 3. $\\left(\\dfrac{8}{3} , \\dfrac{10}{3}\\right) \\\\$ 4. $\\left(\\dfrac{4}{3} , \\dfrac{20}{3}\\right)$ edited", "Great and tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "equality (as in the beginning) we continue: $e^{-z}-e^z=e^{\\overline{z}}-e^{-\\overline{z}}\\Longleftrightarrow$ putting $z=x+iy\\,,\\,\\,x,y\\in\\mathbb{R}$ and comparing real and imaginary parts I get $x=0$ as unique condition... Tonio", "Strange Suppose $$x$$ and $$y$$ are real numbers such that $$x^2$$ and $$y^2$$ are positive integers. Find the maximum value of $$x^2-xy$$ if $$\\left(3x^2-y^2\\right)^2+\\left(x^2+y^2\\right)^2=72$$. ×", "+ 1 = y$ and $y = 4x + 5$ then if we apply the transitive property of equality, we get: $x + 1 = 4x + 5$ 11. Michele_Laino ok! 12. Michele_Laino", "## Monday Math 67 Suppose that x, y, and z are positive real numbers with x+y+z=1. Then find (without calculus) the minimum value of . Here, we can use the AM-GM inequality: and , but , so , and so and so , with equality when .", "a + 0i. Therefore, every real number is a complex number number. Hence, R $$\\subset$$ C, where R is the set of all real numbers. Equality of Complex Numbers Two Complex numbers $$z_1$$ = $$a_1 + ib_1$$ and $$z_2$$ = $$a_2 + ib_2$$ are equal if $$a_1$$ = $$a_2$$ and $$b_1$$ = $$b_2$$ i.e. $$Re(z_1)$$ = $$Re(z_2)$$ and $$Im(z_1)$$ = $$Im(z_2)$$ Example : If $$z_1$$ = 2 – iy and $$z_2$$ = x + 3i are equal, find x and y. Solution : We have, $$z_1$$ = $$z_2$$ $$\\implies$$ 2 – iy = x + 3i $$\\implies$$ 2 = x and -y = 3 $$\\implies$$ x = 2 and y = -3", "Practice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.", "# Equality of complex numbers Theorem. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. In other words, if $$z = a+bi$$ and $$w = c+di$$, where $$a,b,c,d$$ are real, and $$z = w$$, then $$a = c$$ and $$b = d$$. Let’s prove this fact. Proof. It’s clear that if $$a = c$$ and $$b = d$$, then $$a + bi = c + di$$. If $$a + bi = c + di$$, then $$a-c = (d-b)i$$. Squaring both sides and collecting the terms gives us $$(a-c)^2 + (d-b)^2 = 0.$$ Now recall that the square of any real number is greater than or equal to zero. Thus, this is only possible if $$a-c = d-b = 0$$, which means that $$a = c$$ and $$b = d$$. ## Problems 1. Find $$x$$ and $$y$$, given that they are real numbers. 1. $$4+5i = x+yi$$ 2. $$-2 -3i = x+(y-1)i$$ 3. $$2+i = (x-1) + (2y+3)i$$ 4. $$1+i =2x + yi + 4$$ 5. $$2 + 3i = x + y + (x-y)i$$", "$1$. Hence their sum is at least the distance $|z-1|$ between $z$ and $1$ (and the equality holds if and only if $z$ is real nonnegative).", "Question 72 # If x and y are non-negative integers such that $$x + 9 = z$$, $$y + 1 = z$$ and $$x + y < z + 5$$, then the maximum possible value of $$2x + y$$ equals Solution We can write x=z-9 and y=z-1 Now we have x+y< z+5 Substituting we get z-9+z-1<z+5 or z<15 Hence the maximum possible value of z is 14 Maximum value of \"x\" is 5 and maximum value of \"y\" is 13 Now 2x+y = 10+13=23", "+0 0 251 1 +159 x, y, and z are all positive real numbers. x+y+z = 1 prove: x^2/(y+2z) + y^2/(z+2x) + z^2/(z+2y) is greater or equal to 1/3 This has to be proved using the Cauchy inequality. Any help is appreciated!! May 24, 2019 $$\\displaystyle 1=[x+y+z]=[(x/k_{1})k_{1}+(y/k_{2})k_{2}+(z/k_{3})k_{3}]$$ Apply Cauchy-Schwarz to that and then deduce appropriate forms for $$\\displaystyle k_{1},k_{2}\\text{ and }k_{3}.$$", "a,b $${\\in}$$ W then a – b $${\\in}$$ W and ab $${\\in}$$ W. Let $$S= \\frac{m}{2^n} \\mid$$ m and n are integers} (A) neither S nor T is a ring; (B) S is a ring and T is not; (C) T is a ring and S is not; (D) both S and T are rings. 42. For a real number a, define $${a^+} = max{a,0}$$ . for example $${2^+}$$ = 2, $${(-3)^+}$$ = 0. Then for two real numbers a and b, the equality $$(ab)^{+} = (a^{+})(b^{+})$$ holds if and only if (A) both a and b are positive;(B) a and b have the same sign;(C) a=b=0;(D) at least one of a and b is greater than or equals to 0. 43. For any real number x, let [x] denote the largest integer less than or equal to x and <x> = x – [x], that is, the fractional part of x. For arbitrary real numbers x,y and z only one of the following statement is correct. Which one is it? (A) [x+y+z] = [x]+[y]+[z] (B) [x+y+z] = [x+y] + [z] = [x] + [y+z] = [x+z] + [y] (C) < x+y+z > = y+z – [y+z]+ <x> (D) [x+y+z] = [x+y] + [z+ <y+x>]. 44. Suppose that $${x_1}$$, …. $${x_n}$$ (n>2) are real numbers such that $${x_i}$$", "do you prove the addition of inequalites property for real numbers: If x < y and w < z, then x + w < y + z, for all real numbers w,x,y,z. Thanks We have $y = x+(y-x) = x+r_1$ and $z = w+(z-w) = w+r_2$ where $r_1,r_2>0$ Thus $y+z = x+w+r_1+r_2 > x+w$ I believe this reduces your problem to $a,b>0 \\implies a+b > 0$ 6. Thanks for the help, everyone. 7. One of the defining properties of an \"ordered field\" is that if a< b then, for any number c, a+ c< b+ c. Now, if you have x< y and w< z, then you can start with x+ w< y+ w from the property I just stated. If then from w< z, w+ y< z+ y. Finally, you use the \"transitivity\" of inequality, \"if a< b and b< c then a< c\" to argue that since x+ w< y+ w and w+ y< z+ y, you get x+ w< z+ y= y+ z." ]

[ "# Tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ . Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "Practice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.", "Great and tough Inequality Level pending Let x,y,z be real numbers in $]-1,1[$ such that : $x+y+z=0$ Find the maximum value of : $\\sqrt{x+1+\\frac{y^2}{6}}+\\sqrt{y+1+\\frac{z^2}{6}}+\\sqrt{z+1+\\frac{x^2}{6}}$ ×", "real number is nonnegative, we know that the sum is greater than or equal to $0$. Equality holds when the value inside the parhentheses is equal to $0$. We find that $$(x,y) = (5,-3)$$ and the sum we are looking for is $$5+(-3)=2 \\implies \\boxed{\\textbf{(B)}}.$$ - Honestly", "+0 # help 0 73 1 If $$y = -x^2 + 5$$ and x is a real number, then what is the maximum value possible for y? Jul 25, 2019 #1 +5798 +2 $$-x^2 \\leq 0\\\\ -x^2+5 \\leq 5\\\\ y\\leq 5$$ . Jul 25, 2019", "Question 72 # If x and y are non-negative integers such that $$x + 9 = z$$, $$y + 1 = z$$ and $$x + y < z + 5$$, then the maximum possible value of $$2x + y$$ equals Solution We can write x=z-9 and y=z-1 Now we have x+y< z+5 Substituting we get z-9+z-1<z+5 or z<15 Hence the maximum possible value of z is 14 Maximum value of \"x\" is 5 and maximum value of \"y\" is 13 Now 2x+y = 10+13=23", "+0 # intermediate algebra 0 75 1 Let x, y and z be positive real numbers such that x + y + z = 1. Find the maximum value of xyz. Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ Jan 28, 2021 #1 +8456 +2 x + y + z = 1. $$\\dfrac{x + y + z}3 = \\dfrac13$$. By AM-GM inequality, $$\\dfrac{x + y +z}3 \\geq \\sqrt[3]{xyz}\\\\ \\sqrt[3]{xyz} \\leq\\dfrac13\\\\ xyz \\leq \\dfrac1{27}$$ Maximum value = $$\\dfrac1{27}$$ MaxWong Jan 28, 2021", "How can you find a maximum for $f(x,y,z) = xyz + z$ where z is unconstrained. I mean z could be any positive real number or zero right? then $f(x,y, \\infty) = \\infty$.", "+0 # help 0 242 1 If $$y = -x^2 + 5$$ and x is a real number, then what is the maximum value possible for y? Jul 25, 2019 $$-x^2 \\leq 0\\\\ -x^2+5 \\leq 5\\\\ y\\leq 5$$", "+0 # Inequality 0 48 2 If w, x, y, and z are positive real numbers such that w + 2x + 3y + 4z = 8, then what is the maximum value of wxyz? Aug 14, 2021 #1 0 The minimum value is 9/25. Aug 14, 2021 #2 +436 0 By AM-GM, $$\\frac{w+2x+3y+4z}{4} \\ge \\sqrt[4]{24wxyz}\\\\ 2\\ge\\sqrt[4]{24wxyz}\\\\ 16\\ge24wxyz\\\\ \\frac{2}{3}\\ge wxyz$$ so the maximum value is $$\\boxed{\\frac{2}{3}}$$ for completeness, equality occurs when the terms are equal: $$w=2, x=1, y=\\frac{2}{3}, z=\\frac{1}{2}$$ Aug 21, 2021", "1. Show that 1 8 P. 2. Use Axiom C to show that every nonempty set of real numbers with a lower bound has a greatest lower bound. 3. Prove Proposition 1 using Axiom C. 4. If x and y are two real numbers, we define max (x, y) to be x if x > y and y if y > x. We often denote max (x, y) by 'xvj;'. Similarly, we define min (x, y) to be the smaller of x and y, and denote it by 'xa/. Show that a. (xaj/)az = xa(j;az). b. xAy + xvy = x + y. c. (-x)a(-j;)= -(xvy). d. xvj; + z = (x + z)v(]/+z). e. z(x v y) = (zx) v (zy) if z > 0. 5. We define |x| to be x if x > 0 and to be — x if x < 0. Show that a. \\xy\\ = \\x\\ \\y\\. b. \\x + y\\ <\\x\\ + \\y\\. C. | X | = X V (— x). d. xvy = ±(x + y + \\x — y\\). e. if — y < x < y, then | x | < y. 2 The Natural and Rational Numbers as Subsets of R We have adopted the procedure of taking the natural numbers for granted and of using", "+0 # Inequalities 0 98 1 +179 Let x and y be real numbers such that $$x - \\sqrt{x + 6} = \\sqrt{y + 6} - y$$. Let m be the minimum value of x+y, and M be the maximum value of x+y. Enter the ordered pair (m, M). Thanks for any tips, hints, or answers you provide, anything is appreciated. Jun 7, 2021 #1 +2205 +2 I tried, but couldn't figure out the answer. :(( x - sqrt(x+6) = sqrt(y+6) - y x + y = sqrt(y+6) + sqrt(x+6) I'm not sure if this is true, but I feel like the maximum value of x+y would be when x = y. So x = y = 3, making x + y = 6. =^._.^= Jun 7, 2021", "# Very Very Weird Inequality Algebra Level 4 Let $$x$$, $$y$$ and $$z$$ be nonnegative real numbers such that $$x+y+z=5$$. Find the maximum value of $xy^2 + yz^2 + 2xyz.$ ×", "Strange Suppose $$x$$ and $$y$$ are real numbers such that $$x^2$$ and $$y^2$$ are positive integers. Find the maximum value of $$x^2-xy$$ if $$\\left(3x^2-y^2\\right)^2+\\left(x^2+y^2\\right)^2=72$$. ×", "## $v w x y z = v + w + x + y + z$ Find the maximum possible value of $x$ where $v , w , x , y , z > 0$ Apr 4, 2018 Maximum for natural numbers is $x = 5$ #### Explanation: Given: $v w x y z = v + w + x + y + z$ For solutions in natural numbers, we can look at possible combinations of small values. Note that: • If $v = w = x = y = z = 1$ then $v w x y z = 1 < 5 = v + w + x + y + z$ • If $v = w = x = y = z = 2$ then $v w x y z = 32 > 10 = v + w + x + y + z$ The minimum possible values of $v , w , y , z$ are $v = w = y = z = 1$, resulting in: $x = x + 4 \\text{ }$ which has no solutions. Considering the case where exactly one of $v , w , y , z > 1$, and putting $v = 2$, $w = y = z = 1$ we get: $2 x = x + 5 \\text{", "• zmudz Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of $$9x+12y+8z.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Maximum Sum Value Given 10 non-negative reals such that $$a_i \\leq i$$ and $$(1-a_1)(2-a_2)\\ldots(9-a_9)(10-a_{10}) \\geq 1$$, what is the maximum value of $$a_1 + a_2 + \\cdots + a_9 + a_{10}$$? ×", "# Very Very Weird Inequality Algebra Level 4 Let $x$, $y$ and $z$ be nonnegative real numbers such that $x+y+z=5$. Find the maximum value of $xy^2 + yz^2 + 2xyz.$ × Problem Loading... Note Loading... Set Loading...", "###### Waste less time on Facebook — follow Brilliant. × Back to all chapters # Power Mean Inequalities This chain of inequalities forms the foundation for many other classical inequalities. See how the four common \"means\" - arithmetic, geometric, harmonic, and quadratic - relate to each other. # AM-GM Suppose the average of $$5$$ positive numbers is equal to $$2.$$ What is the maximum possible product of the numbers? Suppose $$xyz = 8$$ with $$x, y, z > 0.$$ What is the minimum value of $$x + y + z$$? What is the largest $$n$$ for which $x^3 + y^3 + z^3 \\geq nxyz$ holds for all $$x, y, z > 0$$? What is the largest $$n$$ for which $(x + y)(y + z)(x + z) \\geq nxyz$ holds for all positive $$x, y, z$$? Suppose $$2x + 3y + z = 18.$$ Find the maximum value of $$xyz$$ for $$x, y, z > 0.$$ × Problem Loading... Note Loading... Set Loading...", "For real numbers, $x$ and $y$, with $y= 3x^{2}+3x+1$, the maximum and minimum value of $y$ for $x\\in [-2,0]$ are respectively,_____. 1. $7$ and $\\frac{1}{4}$ 2. $7$ and $1$ 3. $-2$ and $\\frac{-1}{2}$ 4. $1$ and $\\frac{1}{4}$" ]

[ "+0 Find the positive real value of t that satisfies ​. 0 283 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "+0 # Find the positive real value of t that satisfies ​. 0 35 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "if I can find some inequality to complete the proof, but I'm a bit in over my head and thought some expert help would be in order. Thank you! You're working too hard. Let $$a=\\sqrt[n]{x}$$ and $$b=\\sqrt[n]{y}$$. Then $$ab$$ is positive and $$(ab)^n=a^nb^n=xy$$ By uniqueness of positive $$n$$-th roots, it follows that $$\\sqrt[n]{x}\\sqrt[n]{y}=ab=\\sqrt[n]{xy}$$", "Proof of an expression involving absolute value 1. Dec 17, 2011 mindauggas 1. The problem statement, all variables and given/known data Prove: |ab|=|a||b|. (Hint: Consider the cases separately for various signs of a and b.) 2. Relevant equations I think it asks me to prove it only from the definition of absolute value (if it is possible). 3. The attempt at a solution |ab|= $\\frac{(ab), if (ab)\\geq0}{-(ab), if (ab)<0}$ I do the same with |a||b| .... but does it constitute a proof? I actually have no idea how to prove such a thing. Last edited: Dec 17, 2011 2. Dec 17, 2011 mindauggas Sry for the interruption, i found the answers in this forum ... 3. Dec 17, 2011 The1337gamer Could you say this: abs(ab) = positive sqrt((ab)^2) = positive sqrt((a^2)(b^2)) = positive sqrt(a^2)*positive sqrt(b^2) = abs(a)*abs(b) 4. Dec 17, 2011 Frogeyedpeas So you've got the problem solved or do you need more into it? 5. Dec 17, 2011 mindauggas Yeah I found the proof in the other threads", "Find the positive value of $x$ that satisfies $x = 1.900 \\sin(x)$. Give the answer to four places of accuracy. $x\\approx$ Remember to calculate the trig functions in radian mode.", "If $$a$$ $$b$$ and $$c$$ are positive real numbers with $$a+b+c = 6$$ find the maximum value of $$(a+ \\sqrt{bc})(b+\\sqrt{ac})(c+\\sqrt{ab}) + abc$$", "formula applied to a variable t times the variables x in the vector. i.e. use the fact that a quadratic equation has a solution if and only if the discriminant b^2 -4ac is non negative. Actually with your formula, the sum norm, it is even easier to prove your request. indeed it seems obvious from the properties of absolute value. try it and see. of course your homework is now 3 months overdue so you are not reading this anymore. 6. Dec 14, 2010 ### tahawalu |ab|>=ab 2|ab|>=2ab ||a|+|b||>=|a+b| Hope this helps:shy:", "$(2)$, and $(3)$ also satisfies condition $(4)$.", "If b is positive, is ab positive? 1. a^2b>0 2. a^2+b=13 7 19 Jul 2010, 06:22 Display posts from previous: Sort by", "? Free Version Easy # Odd Power and Root if the Radicand is Positive ALGEBR-XPBCV2 If $x$ is positive, what is the value of $\\sqrt[7]{x^7}$? A $x$ B $-x$ C $7$ D $-7$", "05 Jun 2015, 05:23 If b is positive, is ab positive? 1. a^2b>0 2. a^2+b=13 7 17 Jul 2010, 11:19 1 What is the value of (a^-2)(b^-3)? (1) (a^-3)(b^-2) = 36^-1 4 08 May 2010, 08:53 9 Is a^2*b > 0? 19 17 Jun 2007, 20:58 Display posts from previous: Sort by", "solution, if $ab$ IS negative, is ${a=0,b=0}$", "check that the function $f(k) = (b-k)(k+a)$ satisfies this recursion, and hence that $E_0(T) = ab$.", "< b$, then $a' = (b-a)$ and $b' = a$ also satisfy it and have $a' \\le b' < b$, so that the max absolute value of the two items in the pair decreases What the heck...let's check that: we want to show that $(b' - a') (b' + a') - (a'b' -1)$ is zero. So compute \\begin{align} (b' - a') (b' + a') - (a'b' -1) &=(a - (b-a)) (a + (b-a)) - (a(b-a) -1)\\\\ &=(2a - b)) (b)) - (a(b-a) -1)\\\\ &=2ab - b^2 - ab + a^2 +1 \\\\ &=ab - b^2 + a^2 +1 \\end{align} which is $0$ because $a$ and $b$ satisfy the relation, which expanded out says that $b^2 - a^2 - ab + 1 = 0$. Case 1b: if $b < a < 0$, then $b' = b-a$ and $a' = b$ do as well, and $b' < a' < 0$, and $|b'| = |a| < |b|$. Proof: exactly the same as before. Once again, the max absolute value of the two items in the pair decreases. Case 1c: $b$ and $a$ have opposite signs. If $b$ is positive, then $a$ is negative, and $|a| > |b|$. If $b$ is negative, then $a$ is positive, so $b-a$ is negative, so $b+a$ is positive, and once again $|a| > |b|$. Again, by", "# Case Case Case Let $$a,b,c$$ be positive integers such that $$1<a<b<c$$. It is known that only 1 triplet ($$a,b,c$$) satisfies $$abc|(ab-1)(bc-1)(ca-1)$$. Find the value of $$a+b+c$$. ×", "+0 Find the positive real value of t that satisfies ​ 0 265 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$ Mar 21, 2019 #1 +6192 +2 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$ . Mar 21, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$", "# ABC soup Let $$a$$, $$b$$, $$c$$ be positive integers satisfying $$ab+bc = 518$$ and $$ab-ac=360$$. Find the largest possible value of the product $$abc$$. ×", "Institutions: Global |ALU | TUT | UCT | UJ | UNISA | UZ | Wits +1 vote 13 views Prove that $\\sqrt[n]{a} \\sqrt[n]{b} = \\sqrt[n]{ab}$ | 13 views Proof If we multiply the expression on the left hand side by itself $n$ times, we have $(\\sqrt[n]{a} \\sqrt[n]{b})^n = (\\sqrt[n]{a})^n (\\sqrt[n]{b})^n = ab$. On the other hand, a positive number ab has only a single positive root, $\\sqrt[n]{ab}$. Therefore, $\\sqrt[n]{a} \\sqrt[n]{b} = \\sqrt[n]{ ab}$. This completes our proof. by Diamond (55,292 points)", "Check that the product of two \"positive\" numbers is positive. Same deal here: check that no matter what combination of (i), (ii), (iii)$a + b\\sqrt{d}$and$a' + b'\\sqrt{d}$could satisfy, their product satisfies one of them. 3) Check that every element$a + b\\sqrt{d}$of$\\mathbb{Z}[\\sqrt{d}]$satisfies precisely one of the following:$a + b \\sqrt{d} = 0$(i.e.$a = b = 0$),$a + b \\sqrt{d}$is \"positive\" or$-(a + b \\sqrt{d}) = -a - b \\sqrt{d}$is \"positive\". As$0 = 0 + 0 \\sqrt{d}$does not satisfy the definition of \"positive\", it suffices to show the following two statements: (a) if$a + b \\sqrt{d}$is non-zero and not positive, then$-a -b\\sqrt{d}$is positive (b) if$a + b \\sqrt{d}$is positive, then$-a -b\\sqrt{d}$is not positive Let's start with (a): Suppose$a + b\\sqrt{d} \\neq 0$satisfies none of (i), (ii), (iii). i.e. the three following statements hold: (i)'$a \\leq 0$or$b \\leq 0$(ii)'$a \\leq 0$,$b > 0$or$a^2 \\leq b^2 d$(iii)'$a > 0$,$b \\leq 0$or$a^2 \\geq b^2 d$But this means the following hold (i)''$-a \\geq 0$or$-b \\geq 0$(ii)''$-a \\geq 0$,$-b < 0$or$a^2 \\leq b^2 d$(iii)''$-a < 0$,$-b \\geq 0$or$a^2 \\geq b^2 d$Now$a + b\\sqrt{d} \\neq 0$combined with (i)'' implies$-a > 0$or$-b > 0$. If$-a > 0$and$-b > 0$, then$-a - b\\sqrt{d}$satisfies (i) and we're done. Else$-a \\leq 0$or$-b \\leq 0$. If$-a \\leq 0$, then$-b > 0$(see the paragraph above starting with \"Now\"), and now by (ii)'',$a^2", "# Whats the absolute value of abs(-(-5)-(-3))? $= | + 5 + 3 | = | 8 | = 8$ $8$ is already positive, the brackets have no job here." ]

[ "\\;=\\; ab - (a+b) + 1 \\;=\\; -6 - 4 + 1\\quad\\Rightarrow\\quad \\boxed{AB\\:=\\:-9}$ Therefore, the equation is: . $\\boxed{x^2 -2x - 9\\:=\\:0}$", "\\dfrac{a}{b}=\\dfrac{3+2\\sqrt{2}}{3-2\\sqrt{2}} \\\\ \\end{align} Hence, we have proved that the ratio of the two numbers is $\\dfrac{3+2\\sqrt{2}}{3-2\\sqrt{2}}$. Note: There is a possibility that one may write $t$ as our answer in a hurry to solve the question. But since $t=\\sqrt{\\dfrac{a}{b}}$ and we are required to find the $\\dfrac{a}{b}$ in the question, we have to perform squaring and then answer the question.", "in eqn (1),we get b=6ab(3+22)ab$b = 6\\sqrt{ab} – (3 + 2\\sqrt{2})\\sqrt{ab}$ => b=(322)ab$b = (3 – 2\\sqrt{2})\\sqrt{ab}$ ab=(3+22)ab(322)ab=3+22322$\\frac{a}{b} = \\frac{(3 + 2\\sqrt{2})\\sqrt{ab}}{(3 – 2\\sqrt{2})\\sqrt{ab}} = \\frac{3 + 2\\sqrt{2}}{3 – 2\\sqrt{2}}$ Hence, required ratio = (3+22):(322)$\\left ( 3 + 2\\sqrt{2} \\right ) : \\left ( 3 – 2\\sqrt{2} \\right )$ Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are A±(A+G)(AG)$A \\pm \\sqrt{(A + G)(A – G)}$ Soln: Given G and A are A.M and G.M Assume the numbers be a and b. AM=A=a+b2(1)$AM = A = \\frac{a + b}{2} ……… (1)$ GM=G=ab(2)$GM = G = \\sqrt{ab} ……… (2)$ From Eqn(1) and Eqn (2), we get a + b = 2A ……. (3) ab = G2 ……. (4) Putting the value of a and b from (3) and (4) (a – b)2 = (a + b)2 – 4ab, we get (a – b)2 = 4A2 – 4G2 = 4(A2 – G2) (ab)=2(A+G)(AG)(5)$(a – b) = 2\\sqrt{(A + G)(A – G)} …… (5)$ From eqn (3) and (5), we get 2a=2A+2(A+G)(AG)$2a = 2A + 2\\sqrt{(A + G)(A – G)}$ => a=A+(A+G)(AG)$a = A + \\sqrt{(A + G)(A – G)}$ Putting the value of a in eqn(3), we get b=2AA(A+G)(AG)=A(A+G)(AG)$b = 2A – A – \\sqrt{(A + G)(A – G)} = A", "= 3\\sqrt{\\frac{b-a}{2}}$ We want the sum of $a$, $b$, $c$, and $d$. We can now rewrite this $a + b + 12\\sqrt{\\frac{b-a}{2}}$ We are told that $a$, $b$, $c$, and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this $\\frac{b-a}{2} = k^2$ where $k$ is some positive integer. Rearranging, we get $b = a + 2k^2$ Now we can rewrite the sum as $a + a + 2k^2 +12k$. Since both $a$ and $k$ must be at least $1$, the minimum value is $1 + 1 + 2(1)^2 + 12(1) = 16 \\Rightarrow \\boxed{\\text{D}}$.", "of positive and/or negative numbers are positive or negative. For complex numbers we can not make these assumptions. But we can assume $|ab|=|a||b|$ and so $|\\sqrt {a}||\\sqrt {b}|=|\\sqrt {ab}|$. So $\\sqrt {-4}\\sqrt {-9} =\\pm 2i * \\pm 3i = 6i^2= -6$. But it doesn't equal $\\sqrt {-4*-9}=\\sqrt {36}=6$. • Unless you choose a particular branch cut, the expression is two-valued and, of course, the values are $6$ and $-6$. – egreg Aug 11 '16 at 20:35 • Yeah, I was assuming we would pick a branch cut (do you mean either 2i, 3i or -2,-3i but not 2i, -3i or -2i, 3i?) but I didn't want to get bogged down in minutia. I was hoping no one would notice and I wouldn't have to explain. – fleablood Aug 11 '16 at 20:39 • Somebody did. ;-) – egreg Aug 11 '16 at 20:43 Things to understand: 1. $\\sqrt a \\times \\sqrt b =\\sqrt ab$ is only valid when $a,b\\geq 0$ 2. The most common mistake I noticed in this thread is like $\\sqrt{25} = \\pm 5$ or $\\sqrt{-4}=\\pm 2i$. T. Bongers has mentioned it in his comment as well. The correct way to think is the equation $x^2=25$ has two solutions $x=\\pm\\sqrt{25}=\\pm 5$ or the equation $t^2=-4$ has two solutions $t=\\pm\\sqrt{-4}=\\pm 2i$. But nevertheless, $\\sqrt{25}$ is just $5$", "## Algebra 2 (1st Edition) $y=313\\cdot\\sqrt2^x$ Let $y=ab^x$. Then two of our equations are: $313=ab^0$ and $616=ab^2$. If we divide the second equation by the first one we get: $2=b^2$, thus because $b$ is positive: $b=\\sqrt2$. Also $b^0=1$, thus $a=313$. Thus $y=313\\cdot\\sqrt2^x$", "4}.$ also since $b < c$ and by (2) we have $\\sqrt{b}(\\sqrt{b}-\\sqrt{c})=3-c,$ and $\\sqrt{c}(\\sqrt{c}-\\sqrt{b})=3-b,$ we must have $\\boxed{c > 3}$ and $\\boxed{b < 3}.$ finally since c > 3, we'll get from (1) that $c=3+\\sqrt{ab}.$ thus: $6=a+b+c=a+b+\\sqrt{ab}+3.$ hence $a+b+\\sqrt{ab}=3.$ but: $3a < a+b+\\sqrt{ab} < 3b,$ because $a thus $\\boxed{a< 1}$ and $\\boxed{b>1}$ and we are done. Also,find the range of $a^3+b^3+c^3.$ since $abc=a(a-3)^2$ and $0 < a < 1,$ it's clear that $0 < abc < 4.$ but this is not enough to claim that the range of $abc$ is $(0,4)$ because it might be smaller. in fact we'll prove that the range is indeed the interval $(0,4)$: let $abc=k.$ the cubic $x^3 -6x^2+9x - k=0$ must have 3 distinct real roots $a,b,c,$ which happens iff $\\Delta,$ the discriminant of the cubic, is positive. a simple calculation shows that $\\Delta=27k(4-k). \\ \\Box$ 4. ok, the second part of the question was to find the range of $a^3 + b^3 + c^3$ and, for some strange reasons, i thought it was to find the range of $abc.$ however, since $a^3+b^3+c^3=54+3abc,$ as ADARSH showed, and the range of $abc,$ as i showed, is (0,4), the range of $a^3+b^3+c^3$ will be $(54, 66).$ 5. Thanks NCA The value of $abc$ can also be found by the fact that if $f(x)=0$ will", "# If $a^{1/a}=b^{1/b}=c^{1/c}$ and $a^{bc}+b^{ac}+c^{ab}=729$, find the value of $a^{1/a}$ If $$a^{1/a}=b^{1/b}=c^{1/c}$$ and $$a^{bc}+b^{ac}+c^{ab}=729$$, which of the following equals to $$a^{1/a}$$? 1. $$\\sqrt[abc]{81}$$ 2. $$\\sqrt{2}$$ 3. $$\\sqrt[abc]{27}$$ 4. $$\\sqrt[abc]{9}$$ This question is from the book, Mathematics, Class 9 (The IIT Foundation Series) , page number 1.25, question number 58. My attempts to solve this question have failed several times. However, I did find the value of $$a^{1/a}$$ but not in the correct format. Below is my method to do so. $$a^{1/a}=b^{1/b}=c^{1/c}$$ $$\\Rightarrow \\sqrt[a]{a}=\\sqrt[b]{b}=\\sqrt[c]{c}$$ $$\\Rightarrow (\\sqrt[a]{a})^{abc}=(\\sqrt[b]{b})^{abc}=(\\sqrt[c]{c})^{abc}$$ $$\\Rightarrow a^{bc}=b^{ac}=c^{ab}$$ Here I conclude our first equation, $$a^{bc}=b^{ac}=c^{ab}$$. Moving on to the next equation, we have: $$a^{bc}+b^{ac}+c^{ab}=729$$ $$\\Rightarrow a^{bc}+b^{ac}+c^{ab}=729$$ $$\\Rightarrow a^{bc}+a^{bc}+a^{bc}=729$$ $$\\Rightarrow 3a^{bc}=729$$ $$\\Rightarrow a^{bc}=243$$ $$\\Rightarrow (a^{bc})^{1/abc}=243^{1/abc}$$ $$\\Rightarrow a^{1/a}=\\sqrt[abc]{243}$$ $$\\Rightarrow a^{1/a}=\\sqrt[abc]{3^5}$$ Here I finally find the value of $$a^{1/a}$$ as $$\\sqrt[abc]{3^5}$$. However, none of the options match with my result. Please help me to solve the question completely. Thanks! • You've shown that none of the options can be correct. May 4 at 6:09 • @TobyMak If that’s the case, thanks for notifying. I had been smashing my head for an incorrect question. Also, thanks for your time on this question. Your presence was really helpful! May 4 at 6:14 I think the question is wrong. Your math is correct that means that none of the options are right and since that", "- 3 \\sqrt{2}$ #### Explanation: $a + b = 22 \\implies b = 22 - a$ $a b = 103$ $a \\cdot \\left(22 - a\\right) = 103$ $- {a}^{2} + 22 a = 103$ ${a}^{2} - 22 a + 103 = 0$ We have ${\\left(a - 11\\right)}^{2} - 121 + 103 = {\\left(a - 11\\right)}^{2} - 18$ $a - 11 = \\pm \\sqrt{18} = \\pm 3 \\sqrt{2}$ $a = 11 + 3 \\sqrt{2}$ $b = 11 - 3 \\sqrt{2}$", "\\rightarrow \\boxed{(n\\sqrt{2},n,n)}$ for all positive $n.$", "Question 18 # If the sum of squares of two numbers is 97, then which one of the following cannot be their product? Solution Let 'a' and 'b' are those two numbers. $$\\Rightarrow$$ $$a^2+b^2 = 97$$ $$\\Rightarrow$$ $$a^2+b^2-2ab = 97-2ab$$ $$\\Rightarrow$$ $$(a-b)^2 = 97-2ab$$ We know that $$(a-b)^2$$ $$\\geq$$ 0 $$\\Rightarrow$$ 97-2ab $$\\geq$$ 0 $$\\Rightarrow$$ ab $$\\leq$$ 48.5 Hence, ab $$\\neq$$ 64. Therefore, option D is the correct answer.", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", ". . $A - B \\;=\\;\\pm\\sqrt{B^2+4A} \\mp \\sqrt{A^2+4B}$ . . . . . . $A - B \\;=\\;\\pm\\left(\\sqrt{B^2+4A} - \\sqrt{A^2+4B}\\right)$ Square both sides: . . $A^2 - 2AB + B^2 \\;=\\;B^2 + 4A - 2\\sqrt{A^2+4B}\\sqrt{B^2+4A} + A^2 + 4B$ . . . . $\\sqrt{A^2+4B}\\sqrt{B^2+4A} \\;=\\;2A + 2B + AB$ Square both sides: . . $(A^2+4B)(B^2+4) \\;=\\;(2A+2B+ AB)^2$ . . $A^2+B^2 + 4A^3 + 4B^3 + 16AB \\;=\\;4A^2 + 8AB + 4A^2B + 4B^2 + 4AB^2 + A^2B^2$ This simplifies to: . $A^3 - A^2B - AB^2 + B^3 \\;=\\;A^2 - 2AB + B^2$ . . Factor: . $A^2(A-B) - B^2(A-B) \\;=\\;(A-B)^2 \\quad\\Rightarrow\\quad (A-B)(A^2-B^2) \\;=\\;(A-B)^2$ . . Factor: . $(A-B)(A-B)(A+B) \\:=\\:(A-B)^2 \\quad\\Rightarrow\\quad (A-B)^2(A+B) \\:=\\:(A-B)^2$ Since $A \\neq B$, we can divide by $(A-B)^2\\!:\\;\\;\\boxed{A + B \\;=\\;1}$ Can you finish it? 4. F1 = 1/2 f2 = 1/2", "= 6\\sqrt{ab} – (3 + 2\\sqrt{2})\\sqrt{ab}$ => $b = (3 – 2\\sqrt{2})\\sqrt{ab}$ $\\frac{a}{b} = \\frac{(3 + 2\\sqrt{2})\\sqrt{ab}}{(3 – 2\\sqrt{2})\\sqrt{ab}} = \\frac{3 + 2\\sqrt{2}}{3 – 2\\sqrt{2}}$ Hence, required ratio = $\\left ( 3 + 2\\sqrt{2} \\right ) : \\left ( 3 – 2\\sqrt{2} \\right )$ Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are $A \\pm \\sqrt{(A + G)(A – G)}$ Soln: Given G and A are A.M and G.M Assume the numbers be a and b. $AM = A = \\frac{a + b}{2} ……… (1)$ $GM = G = \\sqrt{ab} ……… (2)$ From Eqn(1) and Eqn (2), we get a + b = 2A ……. (3) ab = G2 ……. (4) Putting the value of a and b from (3) and (4) (a – b)2 = (a + b)2 – 4ab, we get (a – b)2 = 4A2 – 4G2 = 4(A2 – G2) $(a – b) = 2\\sqrt{(A + G)(A – G)} …… (5)$ From eqn (3) and (5), we get $2a = 2A + 2\\sqrt{(A + G)(A – G)}$ => $a = A + \\sqrt{(A + G)(A – G)}$ Putting the value of a in eqn(3), we get $b = 2A – A – \\sqrt{(A + G)(A – G)} = A – \\sqrt{(A + G)(A –", "$5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$. Use your previously obtained equation to plug in for $a$ and solve for $b$, which should yield $5\\sqrt{3}$. $a$ is then $\\frac{21}{5}\\sqrt{3}$. Multiplying $a$ and $b$ yields $\\boxed{315}$.", "as well 37. Write $\\eps=a_1+b_1\\sqrt d$ with $a_1,b_1\\in\\Z^+$. We can always get positive coefficients by appropriately choosing one of the four fundamental units. Now let $\\eps^k:=a_k+b_k\\sqrt d$ be the positive powers of $\\eps$ and note that $b_k=a_1b_{k-1}+b_1a_{k-1}$ so the sequence $\\{b_k\\}$ is increasing. Thus, if you want to find a fundamental unit, just guess and check. Start with $b_1=1$ and check to see if $db_1^2\\pm1$ is a perfect square. If not, move on to $b_1=2$ and repeat. Once you’ve found a value that works, write $nb_1^2\\pm1=a_1^2$ and your fundamental unit is $a_1+b_1\\sqrt d$. Example Let $d=11$. If we take $b_1=1$, then $11b_1^2\\pm1=\\{10,12\\}$ so no good. If we take $b_1=2$, then $11b_1^2\\pm1=\\{45,43\\}$ so still no luck. Now we try $b_1=3$ to get $11b_1^2\\pm1=\\{100,98\\}$ and we have a winner. Our fundamental unit is $10+3\\sqrt{11}$. Indeed, $10^2-11*3^2=1$ is a solution to Pell’s equation. Example Now, take $d=2$ instead. If we let $b_1=1$, then $2b_1^2\\pm1=\\{1,3\\}$ so our fundamental unit is $\\eps=1+\\sqrt 2$. However, this has norm $1-2=-1$ so it’s not a solution to Pell’s equation. In cases like this, we instead focus our attention on $\\eps^2=3+2\\sqrt2$ and use this to generate solutions. I’d like to say that’s everything, but I’ve left a few loose ends. These include what to do if $d$ isn’t square free, and what about the case where $d\\equiv1\\pmod4$", "how did you go from $(a+b)^3+c^3−3ab(a+b+c)$ to $(a+b+c)((a+b)^2−(a+b)c+c^2−3ab)$ in the second line? – MattMath Aug 8 '17 at 20:22 • @MattMath I used $x^3+y^3=(x+y)(x^2-xy+y^2)$. – Michael Rozenberg Aug 9 '17 at 2:30 Hint: Let $\\sqrt[3]{m+3}=a$ and $\\sqrt[3]{m-3}=b$ $a-b=?$ $a^3-b^3=?$ Can you find $ab$ and $a^3+b^3=(a+b)\\{(a-b)^2+ab\\}$ Go ahead and cube: $$m+9=27+27\\sqrt[3]{m-9}+9\\sqrt[3]{(m-9)^2}+m-9.$$ Denote: $\\sqrt[3]{m-9}=t$ to get: $$t^2+3t+1=0 \\Rightarrow t=\\frac{-3\\pm \\sqrt{5}}{2}$$ Now: $$m-9=t^3$$ will produce the answers.", "+ \\sqrt{x^2 - 56}$. We can find $AT_2$ by essentially coming in from the other way. Since $AB = 27$, we also know that $AT_3 = 27 - \\sqrt{x^2 - 56}$. By POP, we know that $AT_2 = AT_3$, so $1 + \\sqrt{x^2 - 56} = 27 - \\sqrt{x^2 - 56}$. Let $\\sqrt{x^2 - 56} = A$, for simplicity. We can change the equation into $1 + A = 27 - A$, which we find $A$ to be $13$. Therefore, $\\sqrt{x^2 - 56} = 13$, which further implies that $x^2 - 56 = 169$. After simplifying, we find $x^2 = 225$, so $x = \\boxed{\\textbf{(A) } 15}$ ~EricShi1685", "2: $AB = 16$, then $r = 3\\sqrt{3}$ and $BQ = 7$.", "Courses Courses for Kids Free study material Free LIVE classes More # The sum of the two numbers is 6 times their geometric mean. Show that the two numbers are in the ratio $\\left( 3+2\\sqrt{2} \\right):\\left( 3-2\\sqrt{2} \\right)$. Last updated date: 19th Mar 2023 Total views: 305.1k Views today: 3.84k Verified 305.1k+ views Hint: Consider two numbers and give each number a variable, say a and b. For these two numbers, the geometric mean is given by $\\sqrt{ab}$. In the question, we are given that the sum of the two numbers is 6 times the geometric mean of the two numbers. Use this to find the value of the ratio of these two numbers. Before proceeding with the question, we must know the formula that will be required to solve this question. For any two numbers a and b, the geometric mean of these two numbers is given by, $\\sqrt{ab}$ . . . . . . . . . . . . (1) In this question, it is given that the sum of the two numbers is 6 times their geometric mean and we are required to find the ratio of these two numbers. Let us assume that these two numbers are a and b. From (1), the geometric mean of these two numbers is equal to $\\sqrt{ab}$." ]

[ "+0 # Algebra 2 help 0 126 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$. Oct 10, 2019 #1 +6179 +1 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$ . Oct 10, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$", "+0 # Algebra 2 help 0 57 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$. Oct 10, 2019 #1 +6045 +1 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$ . Oct 10, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\~\\\\ |6-4i| = \\sqrt{36+16} = \\sqrt{52}=2\\sqrt{13}\\\\~\\\\ |t+2i\\sqrt{3}| = \\dfrac{26}{2\\sqrt{13}} = \\sqrt{13}\\\\~\\\\ t^2 + 4\\cdot 3 = 13\\\\ t=1$$", "all this explicitly by taking $$\\omega = -\\frac{1}{2} + i \\frac{\\sqrt{3}}{2} .$$ 19. Dec 9, 2017 ### eatsleepmath It's easy to simplify this to $(a+b)=\\sqrt{ab}$. But extending AM-GM to include negative numbers (which is fine since this setup won't invalidate the underlying stipulations) tells us that $(a+b) \\geq 2\\sqrt{ab}$. Substitution gives $\\sqrt{ab} \\geq 2\\sqrt{ab}$. The only time this is true is the equality case, i.e. $\\sqrt{ab}=0$. Thus at least one of $a$ and $b$ is zero. Solving for the other gives $a^2+a(0)+(0)^2=0$ so the other is zero as well. 20. Dec 10, 2017 ### Staff: Mentor Notwithstanding what someone on Reddit has to say, this is not a good proof. For one thing, it looks like you are concluding that $a^2 + b^2 = -ab$. You should not conclude this -- you are assuming it is true. Since you are assuming that $a^2 + ab + b^2 = 0$, it follows trivially that $a^2 + b^2 = -ab$, regardless of whether a and b are positive, negative, of mixed signs, whatever. The two equations are equivalent. You don't need any machinery to get from one to the other, apart from adding equal quantities to both sides of either equation. If a and b are both positive or both negative, then the equation $a^2 + b^2 = -ab$ can't", "+0 Find the positive real value of t that satisfies ​ 0 265 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$ Mar 21, 2019 #1 +6192 +2 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$ . Mar 21, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$", "part d) as follows: We want to find a positive value for $a$ such that: $\\frac{a}{(t+1)^2}>\\frac{1}{t^2+1}$ for all $t$ in the given domain. This means we want: $(a-1)t^2-2t+(a-1)>0$ This means the discriminant must be negative: $4-4(a-1)^2<0$ $0 which implies: $2. Now, since: $\\lim_{x\\to\\infty}\\left[a\\int_0^x\\frac{1}{(t+1)^2}\\,dt \\right]=a$ we may state: $0. Excellent!", "+0 # Find the positive real value of t that satisfies ​ 0 84 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}| |6-4i| = 26$$ Mar 21, 2019 #1 +5172 +2 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$ . Mar 21, 2019 $$|t+2i\\sqrt{3}||6-4i|=26\\\\ |t+2i\\sqrt{3}|^2|6-4i|^2=676\\\\ (t^2+12)(36+16)=676\\\\ t^2 + 12 = 13\\\\ t^2 = 1\\\\ t = \\pm 1\\\\ \\text{The positive value of }t \\text{ satisfying above is }1$$", "< b$, then $a' = (b-a)$ and $b' = a$ also satisfy it and have $a' \\le b' < b$, so that the max absolute value of the two items in the pair decreases What the heck...let's check that: we want to show that $(b' - a') (b' + a') - (a'b' -1)$ is zero. So compute \\begin{align} (b' - a') (b' + a') - (a'b' -1) &=(a - (b-a)) (a + (b-a)) - (a(b-a) -1)\\\\ &=(2a - b)) (b)) - (a(b-a) -1)\\\\ &=2ab - b^2 - ab + a^2 +1 \\\\ &=ab - b^2 + a^2 +1 \\end{align} which is $0$ because $a$ and $b$ satisfy the relation, which expanded out says that $b^2 - a^2 - ab + 1 = 0$. Case 1b: if $b < a < 0$, then $b' = b-a$ and $a' = b$ do as well, and $b' < a' < 0$, and $|b'| = |a| < |b|$. Proof: exactly the same as before. Once again, the max absolute value of the two items in the pair decreases. Case 1c: $b$ and $a$ have opposite signs. If $b$ is positive, then $a$ is negative, and $|a| > |b|$. If $b$ is negative, then $a$ is positive, so $b-a$ is negative, so $b+a$ is positive, and once again $|a| > |b|$. Again, by", "$a$, $b$ and $c$, with $c$ being the length of the hypotenuse, $a^2 + b^2 = c^2.$ In particular, with $a = x$, $b = y$, and $c = (x+1)$, we obtain $x^2 + y^2 = (x+1)^2$. Expanding and rearranging gives $y^2 = 2x + 1$ so that $y = \\sqrt{2x+1}.$ (As $y$ is the length of a side, it can’t be negative, so we take the positive square root here.) Hence find a member of $T$ for which $y > 10$. Since $y > 10$, we know that $2x + 1 = y^2 > 100$. We are thus looking for a positive integer $x$ such that $2x + 1$ is a square number that exceeds $100$. Since $11^2 = 121 = 2 \\times 60 + 1$, one suitable member of $T$ has $x = 60$ and $y = 11$. In fact, $11$ isn’t special here; any odd number larger than $10$ will fit the bill.", "$$a$$ and $$b$$ are positive, $$\\frac{a}{b}$$ is what percent of $$ab$$? (1) $$|b - 2| = 4$$ (2) $$b^2 - 3b - 18 = 0$$ The answer is relatively very simple. Point to remember a and b are positive. The question is asking $$a/b$$= x% of $$ab$$, find x. We can further simplify - expression $$a/b = x/100 * ab$$ ----> $$100/b^2$$ = x All we have to know is b. So we can rephrase the question WHAT IS THE VALUE OF b? 1) $$|b - 2| = 4$$ b = 6 or b = -2 But since b>0, b=6 Suff 2) $$b^2 - 3b - 18 = 0$$ Solving we will get b =6 or -4 Again b>0, hence b = 6 Suff option D winner _________________ Sortem sternit fortem! Retired Moderator Joined: 19 Mar 2014 Posts: 969 Location: India Concentration: Finance, Entrepreneurship GPA: 3.5 ### Show Tags 06 Jul 2017, 13:30 Bunuel wrote: If $$a$$ and $$b$$ are positive, $$\\frac{a}{b}$$ is what percent of $$ab$$? (1) $$|b - 2| = 4$$ (2) $$b^2 - 3b - 18 = 0$$ We are looking for the value of $$\\frac{1}{b^2} * 100$$ (1) $$|b - 2| = 4$$ Gives two values b = 6 & -2 as b is positive b = 6 Hence (1) =====> is SUFFICIENT (2)", "If $$a$$ $$b$$ and $$c$$ are positive real numbers with $$a+b+c = 6$$ find the maximum value of $$(a+ \\sqrt{bc})(b+\\sqrt{ac})(c+\\sqrt{ab}) + abc$$", "ab where a and b are numbers such that a > 0, b > 0. Also, ( -1) x (-a) = - (-a) = a 2. Negative number multiplied by a positive number gives a negative number. That is, (-a) x (-b) = -(ab) Or, (-b) x (a) = -(ab) Also, (-1) x (+a) = - (+a) = -a Or, (+1) x (-a) = + (-a) = -a 3. Positive number multiplied by a positive number gives a positive number. That is, (a) x (b) = ab Or, (b) x (a) = ba Or, (1) x (+a) = + (+a) = +a Now, let us consider the expression given in the question as $A=\\left( -8 \\right)+\\left( -9 \\right)+\\left( +17 \\right)$ According to BODMAS, first we have to solve the bracket of above equation, Also, we know that +(-a) = -a, +(+a) = +a and +(-a) = -a. By applying these in above expression, we get A = - 8 – 9 + 17 Since, we know that – a – b = - (a+b) Therefore, we get – 8 – 9 = – (8+9) = – 17 By putting the value in the above expression, we get, A = – 17 + 17 A = 0 Therefore, we get A = (-8) + (-9) + (+17) equal to", "$$a=0$$ (and $$b=\\text{any value}$$, including zero) so in this case $$ab=0\\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient. (2) $$ab^2=b$$ $$ab^2-b=0$$; $$b(ab-1)=0$$: either $$b=0$$ (and $$a=\\text{any value}$$, including zero) so in this case $$ab=0 \\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient. (1)+(2) either $$a=b=0$$, so in this case $$ab=0 \\neq 1$$ and the answer to the question is NO, OR $$ab=1$$ and the answer to the question is YES. Two different answers, not sufficient. When we consider two statements together we should take the values which satisfy both statements. For this question $$ab=1$$ satisfies both statement, but $$a=b=0$$ also satisfies both statements. So what you call \"common solution\" for this question is: $$ab=1$$ OR $$ab=0\\neq{1}$$. _________________ Intern Joined: 23 Jan 2014 Posts: 27 Schools: Sloan '19 (D) ### Show Tags 15 Aug 2016, 21:45 Sorry Bunuel, but if b=0, it will not satisfy the statement 1, will it? Bunuel wrote: svijayaug12 wrote: Hi Bunuel, I have a question based on the solution that you have mentioned below: Let's say a question in DS is as follows? If x is a positive integer, is x=4? Stmt -1) X is a root of the quadratic equation - x^2 - 12x+12=0 Stmt -2) X is a root of the quadratic equation - x^2 - 9x+20=0 Now solving each", "b = 9 value of ab = 13 × 9 ⇒ ab = 117 ∴ The value of ab is 117. # For what value of 'a', does the inequality 9a - a2$$\\leq$$ 17a + 15 holds 1. -2 2. -5 3. -6 4. All the above Option 4 : All the above ## Detailed Solution Calculation: Given: 9a - a2 $$\\leq$$ 17a + 15 On rearranging -a2 + 9a $$\\leq$$ 17a + 15 Shifting the sign a2 - 9a $$≥$$ - 17a - 15 a2 - 9a + 17a + 15 $$≥$$ 0 a2 + 8a + 15 $$≥$$ 0 a2 + 5a + 3a + 15 $$≥$$ 0 a(a+ 5) + 3(a + 5) $$≥$$ 0 (a + 3)(a + 5) $$≥$$ 0 So, -3 and -5 are the roots of the equation. Now look at the diagram given below, We see that all the numbers less than -5 and all the numbers greater than -3 will give us positive result. While numbers between -5 and -3 will give us negative results So, all the above value holds for the above equation. 1. B 2. C 3. A 4. D Option 1 : B -4 = -7 + 3x ⇒ 3x = 7 – 4 ⇒ x = 3/3 ∴ x = 1 # If $${\\left(", "$$t$$ must be positive, we can see that $$t = \\boxed{9}.$$ . Dec 30, 2017", "if any of A or B alone seems impossible, you can pick the other answer choice and it would be correct every single time. Make sure all of this only happens only if you correctly sense a c-trap, but I hope such situation never occur to you. All the best! Director Joined: 19 Oct 2018 Posts: 503 Location: India Re: If x is a positive integer, what is the value of (x + 24)^(1/2) - x [#permalink] ### Show Tags 28 May 2019, 20:46 Statement 1 - Clearly insufficient, x can be any number that is perfect square. Statement 2- Same explanation as statement 1 Combining both statement- Let $$\\sqrt{x+24}$$=a and $$\\sqrt{x}$$=b x+24=a^2 and x=b^2 a^2-b^2=24 (a+b)(a-b)=24 Possible Solutions 1. a+b=12 and a-b=2 a=7,b=5 2. a+b=6 and a-b=4 a=5 and b=1 Insufficient. Bunuel wrote: If x is a positive integer, what is the value of $$\\sqrt{x+24}-\\sqrt{x}$$? (1) $$\\sqrt{x}$$ is an integer (2) $$\\sqrt{x+24}$$ is an integer Re: If x is a positive integer, what is the value of (x + 24)^(1/2) - x [#permalink] 28 May 2019, 20:46 Display posts from previous: Sort by", "both positive integers such that a < b, which of the fo [#permalink] ### Show Tags 27 Jul 2018, 04:31 1 dave13 wrote: PKN wrote: Bunuel wrote: If a and b are both positive integers such that a < b, which of the following must be true? (A) $$a < \\sqrt{ab} <b$$ (B) $$\\sqrt{a} < \\sqrt{ab} < \\sqrt{b}$$ (C) $$\\sqrt{a} < \\sqrt{b} < \\sqrt{ab}$$ (D) $$\\sqrt{ab}<a<b$$ (D) $$a<\\sqrt{ab}<\\sqrt{b}$$ Let a=4 & b=16 (a<b) (A) $$a < \\sqrt{ab} <b$$: 4<8<16 (True) (B) $$\\sqrt{a} < \\sqrt{ab} < \\sqrt{b}$$: 2<8<4 (False) (C) $$\\sqrt{a} < \\sqrt{b} < \\sqrt{ab}$$: 2<4<8 (False) (D) $$\\sqrt{ab}<a<b$$: 8<4<16 (False) (E) $$a<\\sqrt{ab}<\\sqrt{b}$$: 2<8<4 (False) Ans. (A) P.S.- Perfect square are chosen for checking purpose since square roots are involved in the answer choices. PKN hey there, can you please explain how can option C be wrong. 2<4<8 if A=2 B= 4 o correct 2<4 or a<b so whats wring with it ? thanks and have a great day !:-) Hi dave13 , Have a great day!! First of all, this is a MUST BE TRUE question. We have to find out the correct relationship among a,b, a√a,b√b, and ab−−√ab in terms of magnitude. And that relationship MUST HOLD TRUE at all the possible positive integer values of a and b, available in the universe. (with the given limitation:-", "+0 Find the positive real value of t that satisfies ​. 0 283 1 Find the positive real value of t that satisfies $$|t+2i\\sqrt{3}||6-4i|=26$$. Mar 24, 2019", "$s_3\\:=\\:-(-1) - 3^2 \\:=\\:-8$ Hence $s_4$ is the negative of $s_3$. minus $4^2.$ . . $s_4\\:=\\:-(-8) - 4^2\\:=\\:-8$ Therefore: $s_5$ is the negative of $s_4$, minus $5^2.$ . . $s_5\\:=\\:-(-8) - 5^2\\:=\\:\\boxed{-17}$ 5. Yes thanks that is the answer I got.", "that a & b are positive integers, find the value of a & b. Ans.: (From text book): a = 2, b = 5 $(a+\\sqrt{3})(b-\\sqrt{3})=7+3\\sqrt{3}$ $(ab-3)+(-a+b)\\sqrt3=7+3\\sqrt{3}$ So $ab-3=7~\\&~-a+b=3.$ Solve for $a~\\&~b.$ 4. ## Re: Surds From above, then we let b, -a are root of a quadric equation s.t. -ab=-10, b+(-a)=3 ie, equation is x^2-3x-10=0 (x+2)(x-5)=0 then (a=2,b=5) or (a=5,b=2) by putting to original equation, answer is found 5. ## Re: Surds Originally Posted by GrigOrig99 I'm having a bit of trouble with this one. Can anyone help me out? Many thanks. Q. Given that $(a+\\sqrt{3})(b-\\sqrt{3})=7+3\\sqrt{3}$ & that a & b are positive integers, find the value of a & b. Attempt: $ab-a\\sqrt{3}+b\\sqrt{3}=7+3\\sqrt{3}$ There's an error in your first line. You have \"FOI\" but not \"L\"! $(-\\sqrt{3})(\\sqrt{3}= -3$ so this should be $ab- a\\sqrt{3}+ b\\sqrt{3}- 3= 7+ 3\\sqrt{3}$ You have ab- 3= 7 and b- a= 3 => $ab-a\\sqrt{3}=3\\sqrt{3}-b\\sqrt{3}+10$ => $a(b-\\sqrt{3})=3\\sqrt{3}-b\\sqrt{3}+10$ => $a=\\frac{3\\sqrt{3}-b\\sqrt{3}+10}{b-\\sqrt{3}}$ => $a=\\frac{\\sqrt{3}(3-b)+10}{b-\\sqrt{3}}$ => $a=\\frac{\\sqrt{3}(3-b)+10}{b-\\sqrt{3}}.\\frac{b+\\sqrt{3}}{b+\\sqrt{3}}$ => $a=\\frac{3b\\sqrt{3}-b^2\\sqrt{3}+10b+9-3b+10\\sqrt{3}}{b^2-3}$... Ans.: (From text book): a = 2, b = 5 a= -5, b= -2 is also a solution. 6. ## Re: Surds Great. Thanks guys.", "4}.$ also since $b < c$ and by (2) we have $\\sqrt{b}(\\sqrt{b}-\\sqrt{c})=3-c,$ and $\\sqrt{c}(\\sqrt{c}-\\sqrt{b})=3-b,$ we must have $\\boxed{c > 3}$ and $\\boxed{b < 3}.$ finally since c > 3, we'll get from (1) that $c=3+\\sqrt{ab}.$ thus: $6=a+b+c=a+b+\\sqrt{ab}+3.$ hence $a+b+\\sqrt{ab}=3.$ but: $3a < a+b+\\sqrt{ab} < 3b,$ because $a thus $\\boxed{a< 1}$ and $\\boxed{b>1}$ and we are done. Also,find the range of $a^3+b^3+c^3.$ since $abc=a(a-3)^2$ and $0 < a < 1,$ it's clear that $0 < abc < 4.$ but this is not enough to claim that the range of $abc$ is $(0,4)$ because it might be smaller. in fact we'll prove that the range is indeed the interval $(0,4)$: let $abc=k.$ the cubic $x^3 -6x^2+9x - k=0$ must have 3 distinct real roots $a,b,c,$ which happens iff $\\Delta,$ the discriminant of the cubic, is positive. a simple calculation shows that $\\Delta=27k(4-k). \\ \\Box$ 4. ok, the second part of the question was to find the range of $a^3 + b^3 + c^3$ and, for some strange reasons, i thought it was to find the range of $abc.$ however, since $a^3+b^3+c^3=54+3abc,$ as ADARSH showed, and the range of $abc,$ as i showed, is (0,4), the range of $a^3+b^3+c^3$ will be $(54, 66).$ 5. Thanks NCA The value of $abc$ can also be found by the fact that if $f(x)=0$ will" ]

[ "3 ( p 1 ) / 2 many solutions of type I. 3. From (2), S 11 has 3 ( 11 1 ) / 2 = 15 solutions of type I. By Lemma 3.1, S p = 1 8 ( 11 1 ) 11 2 + 3 ( 1 ) 11 + 1 2 = 15 . Thus, S 11 has no solutions of type II.□ Next, we focus on 3Q-partitions of 0 in r p . We derive formulas α p , β p , and relationships between them. We examine an example first. ## Example 3.7 Consider p = 11 . r 11 = { 1 , 4 , 9 , 5 = 4 2 , 3 = 5 2 } . From Proposition 3.6, X 2 ( 11 ) = . X ( 11 ) = { { 1 , 1 , 9 } , { 1 , 5 , 5 } , { 3 , 3 , 5 } , { 3 , 4 , 4 } , { 4 , 9 , 9 } } = X 1 ( 11 ) . Each above partition produces three solutions of type I in S 11 . The set S 11 is given as follows: S 11 = { ( 1 , 1 , 9", "s^2 + t^2 = 1$ $S_3 = r^3 + s^3 + t^3 = 33$ $S_4 = r^4 + s^4 + t^4 = -127$ Which gives us our desired solutions, -127 and 1.", "be to do 3r^2=12 , thus r=2 yup, that's what i said. the equation $3r^2~r' = 12~r'$ must hold. so just solve for r", "2,544 views Let $r$ be a root of the equation $x^{2} + 2x + 6 = 0.$ Then the value of the expression $(r+2) (r+3) (r+4) (r+5)$ is 1. $51$ 2. $-51$ 3. $126$ 4. $-126$ Since $r$ is a root of the given equation $\\text{}x^2+2x+6=0, \\\\$ So, $r$ will satisfy this equation. Hence, we have $r^2+2r+6=0$ $\\Rightarrow \\boxed{r^{2} + 2r = -6}$ The value of expression $(r+2)(r+3)(r+4)(r+5)\\\\ = (r^2+5r+6)(r^2+9r+20)\\\\ = (r^2+2r+6+3r)(r^2+2r+6+14+7r)\\\\ =(3r)(7r+14)\\\\= 21(r^2+2r)\\\\ = 21 \\times -6 \\\\ = \\color{blue}{-126}.$ Realized the importance of practice and smart approach to a problem. 😍", "HomeEnglishClass 10MathsChapterHeart Of Algebra 3(r+2s)=r+s <br> If (r, s) is ... # 3(r+2s)=r+s <br> If (r, s) is a solution to the equation above and sne0, what is the value of (r)/(s)? Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, Text Solution -(5)/(2)-(2)/(5)(2)/(3)(3)/(2) A", "= 1$, and $r_Q(4) = 3$, corresondign respectively to the solutions $(0, -1, 0, -1)$ and $(-1, 0, -1, 0), (0, -1, -1, -1), (0, 0, 0, -1)$.</p>", "determine the value of $x$. $3 x - 4 \\left(- \\frac{1}{2}\\right) = 11$ $3 x + 2 = 11 \\implies x = \\frac{11 - 2}{3} = \\textcolor{g r e e n}{3}$ The solutions to this system of equations are $\\left\\{\\begin{matrix}x = 3 \\\\ y = - \\frac{1}{2}\\end{matrix}\\right.$", "for $R_{11}$ and $R_{22}$.", "=4a^2(5+4k^2)$ must be a square of a rational so $5+4k^2$ must be a square of a rational. Multiplying out, we want solutions to $5 p^2 + 4q^2 = r^2$. I'll leave it at this, since I have to go. - add comment", "solutions exist. For example $r_i=0$ for all $i \\neq 1,2$ and set $r_1=1$ and $r_2=-2$.", "# How do you solve abs(-2r-1)=11? Oct 30, 2017 $r = - 6 , r = 5$ #### Explanation: This is an absolute value equation meaning that we need to split this into two equations; the original equation and the opposite of it: $- 2 r - 1 = 11$ and $- 2 r - 1 = - 11$ Notice that the $11$ became $- 11$. Now we can solve this by simplifying. Let's simplify the first one: $- 2 r - 1 = 11$ $- 2 r = 12$ $r = - 6$ Second one: $- 2 r - 1 = - 11$ $- 2 r = - 10$ $r = 5$ Now let's check our work by plugging what we got for r back into the original equation $| - 2 r - 1 | = 11$. $| - 2 \\left(- 6\\right) - 1 | = 11$ $| 12 - 1 | = 11$ $11 = 11$ :) $| - 2 \\left(5\\right) - 1 | = 11$ $| - 10 - 1 | = 11$ $| - 11 | = 11$ $11 = 11$ :) So our two solutions are $r = - 6$ and $r = 5$.", "{ 2x-1 }{ 3 }$ > 1, x ∈ R and represent the solution set on the number line. $\\frac { 3x }{ 5 } -\\frac { 2x-1 }{ 3 }$ > 1 => 9x – (10x – 5) > 15 => 9x – 10x + 5 > 15 => – x > 15 – 5 => – x > 10 => x < – 10 x ∈ R. .’. Solution set = {x : x ∈R, x < – 10} Solution set on the number line #### Question 23 Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000) – 3 ≤ 3 – 2x < 9 – 3 ≤ 3 – 2x and 3 – 2x < 9 2x ≤ 3 + 3 and – 2x < 9 – 3 2x ≤ 6 and – 2x < 6 => x ≤ 3 and – x < 3 => x ≤ – 3 and – 3 < x – 3 < x ≤ 3. Solution set= {x : x ∈ R, – 3 < x ≤ 3) Solution on number line #### Question 24 Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003) 2 ≤ 2x", "we get $(r - 2)(r^2 + r + 1)$", "us that: $P_1 + 3 = 0$ $P_2 + 3P_1 + 8 = 0$ $P_3 + 3P_2 + 4P_1 - 24 = 0$ $P_4 + 3P_3 + 4P_2 - 8P_1 = 0$ Solving, first for $P_1$, and then for the other variables, yields, $P_1 = r + s + t = -3$ $P_2 = r^2 + s^2 + t^2 = 1$ $P_3 = r^3 + s^3 + t^3 = 33$ $P_4 = r^4 + s^4 + t^4 = -127$ Which gives us our desired solutions, $\\boxed{1}$ and $\\boxed{-127}$. 2019 AMC 12A #17", "r^2$ and $n = s^2$. Now we have $a = r^2 - s^2$, $b=2rs$, and $c = r^2 + s^2$. To recap: if $(a,b,c)$ is a solution, and if $a$, $b$ and $c$ are coprime, then they must be of the form $a=r^2-s^2$, $b = 2rs$, $c = r^2 + s^2$, for some positive integers $r$ and $s$ with $r > s$. So all the primitive solutions must be of this form. But does every triple of this form give a solution? Let’s see. We can just do the algebra. We have $a^2 + b^2 = (r^2-s^2)^2 + (2rs)^2 = r^4 - 2r^2s^2+s^4+4r^2s^2=r^4+2r^2s^2+s^4=(r^2+s^2)^2=c^2$, so we do indeed always get a solution. We do not necessarily get a primitive solution (where the side lengths are coprime), though. For example, $r=3$ and $s=1$ gives $a=8$, $b=6$, $c=10$. But let’s be a bit more careful. We know that with our primitive solution we had $a$ and $c$ odd, so one of $m$ and $n$ must be odd and the other even. The same must apply to $r$ and $s$ We also saw that we must have $m$ and $n$ coprime, so $r$ and $s$ must be coprime. If we impose those conditions (which would rule out $r=3$, $s=1$), do we necessarily get a primitive solution? It turns out that we do.", "we use it to minus off each equation in the system. We can get $$11a_i = \\frac {9 + 99 + 999 + ... + 10^n –1}{n–11} – 10^i +1$$ for $$i = 1, 2, 3, ... , n$$. Then, we can easily get the solutions for the system. When n = 11, $$LHS = (11–11)(a_1 + a_2 + a_3 + ... + a_{11}) = 0$$ $$RHS = 9 + 99 + 999 + ... + 10^{11} –1$$ Clearly $$0 \\neq 9 + 99 + 999 + ... + 10^{11} –1$$ so $$LHS \\neq RHS$$ and 11 is the only number which the system doesn't admit a solution. Staff - 5 years, 10 months ago Solution A - Imagine, given some $$n$$, how to actually solve the system of equations. One of the easiest ways would be to add up all the equations and divide by $$n-11$$. This will result in an equation with a coefficient of one for each variable. From there, we could take this new equation and subtract the first equation to obtain an equation with only $$a_1$$, which can be easily solved to find the value of $$a_1$$. Similarly, we can subtract the second equation to get the value of $$a_2$$, and so on. Using this method, we can solve this system for any", "## Algebra 2 (1st Edition) $|8r+1|=3r$ implies $8r+1=3r\\\\1=-5r\\\\r=-0.2$ or $8r+1=-3r\\\\1=-11r\\\\r=\\frac{-1}{11}$. Here $r=-0.2$ and $r=\\frac{-1}{11}$ are extraneous solutions. Thus we have no solutions.", "don't seem to have seen problems like this you are probably in basic calculus. Since you must have $S= \\pi rl= \\pi r\\sqrt{r^2+ h^2}= 4\\sqrt{3\\pi}$. Squaring both sides of that $\\pi^2r^2(r^2+ h^2)= 48\\pi$. You can solve that for h and replace h in V= \\pi r^2h/3 by that to get a problem in just the one variable r. Use whatever methods you know to find the largest possible value of that function.", "Select Page Question: Let $$C \\subset \\mathbb{ZxZ}$$ be the set of integer pairs $$(a,b)$$ for which the three complex roots $$r_1,r_2,r_3$$ of the polynomial $$p(x)=x^3-2x^2+ax-b$$ satisfy $$r_1^3+r_2^3+r_3^3=0$$. Then the cardinality of $$C$$ is A. $$\\infty$$ B. 0 C. 1 D. $$1<|C|<\\infty$$ Discussion: We have $$r_1+r_2+r_3=-(-2)=2$$ $$r_1r_2+r_2r_3+r_3r_1=a$$ and $$r_1r_2r_3=-(-b)=b$$ Also, of the top of our head, we can think of one identity involving the quantities $$r_1^3+r_2^3+r_3^3=0$$ and the three mentioned just above. Let’s apply that: $$r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))$$ …$$…(1)$$ Also, note that $$r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)$$ So $$r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a$$. So by equation $$(1)$$, $$0-3b=2(4-2a-a)$$ or, $$6a-3b=8$$. Now, we notice the strangest thing about the above equation: $$3|(6a-3b)$$ but $$3$$ does not divide $$8$$. So, we get a contradiction. Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some $$a,b$$. So, the ANSWER: $$|C|=0$$.", "Math Help - solve in R 1. solve in R solve in R : 2. Start by using double angle formula to change the denominators. $\\sin 2x = 2\\sin x \\cos x$" ]

[ "## Vieta's Formulas For a polynomial of the form \\begin{align*} f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \\end{align*} with roots $$r_1,r_2,r_3,...r_n$$, Vieta's formulas state that: \\begin{align*} r_1+r_2+r_3+...+r_n=-\\frac{a_{n-1}}{a_n} \\\\ r_1r_2 +r_1r_3+...+r_{n-1}r_n=\\frac{a_{n-2}}{a_n} \\\\ r_1r_2r_3+r_1r_2r_4+...+r_{n-2}r_{n-1}r_n=-\\frac{a_{n-3}}{a_n} \\\\ \\vdots \\\\ r_1r_2r_3...r_n=(-1)^n\\frac{a_0}{a_n} \\\\ \\end{align*} Generalization: When the $$n$$ roots are taken in groups of $$k$$ (i.e. $$r_1+r_2+...+r_n$$ is taken in groups of $$1$$ and $$r_1r_2...r_n$$ is taken in groups of $$n$$), this is equivalent to \\begin{align*} (-1)^k\\frac{a_{n-k}}{a_n} \\end{align*} source", "# Difference between revisions of \"2008 AIME II Problems/Problem 7\" ## Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Solution 1 By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is $$(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).$$ Also by Vieta's formulas we have $$rst = -rs(r + s) = -\\dfrac{2008}{8} = -251$$ so negating both sides and multiplying through by 3 gives our answer of $\\boxed{753}.$ ## Solution 2 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ ## Solution 3 Vieta's formulas gives $r + s + t = 0$.", "2008 AIME II Problems/Problem 7 Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. Solution 1 By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is $$(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).$$ Also by Vieta's formulas we have $$rst = -rs(r + s) = -\\dfrac{2008}{8} = -251$$ so negating both sides and multiplying through by 3 gives our answer of $\\boxed{753}.$ Solution 2 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ Solution 3 Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3", "# 2008 AIME II Problems/Problem 7 ## Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Solution ### Solution 1 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ ### Solution 2 Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3 + 1001r + 2008 = 0\\Longleftrightarrow - 8r^3 = 1001r + 2008$, and the same can be done with $s,\\ t$. Therefore, we have \\begin{align*}8\\{(r + s)^3 + (s + t)^3 + (t + r)^3\\} &= - 8(r^3 + s^3 + t^3)\\\\ &= 1001(r + s + t) + 2008\\cdot 3 = 3\\cdot 2008\\end{align*}yielding the answer $753$.", "330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$. From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \\frac{5r+13}{2}$. We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \\frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \\boxed{420}$. ## See also 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •", "+0 # HELP HELP NOWW 0 259 5 Let $r$ and $s$ be the roots of $3x^2 + 4x + 12 = 0.$ Find $r^2 + s^2.$ Apr 25, 2020 #1 +21959 +1 Apr 25, 2020 #2 0 where????? Apr 25, 2020 #3 +1 $$3x^2+4x+12=0$$ Is a quadratic equation of the form $$ax^2+bx+c=0$$ where r and s are roots By Vieta's formulas: $$r+s=-\\frac{b}{a}$$ Thus $$r+s=-\\frac{4}{3}$$ $$r*s=\\frac{c}{a}$$ Thus $$rs=4$$ Now notice that: $$r^2+s^2=(r+s)^2-2rs$$ Substitute $$(-\\frac{4}{3})^2-2(4)=-\\frac{56}{9}=-6.22$$ Apr 25, 2020 #4 0 it is wrong????? Apr 26, 2020 #5 +112000 0 Good, I hope it is, you are too lazy to even lay your question out properly. And too lazy to search old answers. I would have preferred that you were not answered at all but if the answer is wrong that is the next best thing. (No disrespect intended to answering guest, I know you are just trying to help) Melody Apr 26, 2020", "\\begin{align} f(x_2) = \\sqrt{3}\\, x^2_2 + 2 \\, x_2 + \\sqrt{3} - \\frac{1}{r_1} &=0\\\\ \\sqrt{3}\\big((x_2 + x_3)^2 - 2\\, x_2x_3\\big) + 2 \\, {x_2\\, x_3} &= \\frac{1}{r_1}\\\\ f(x_3) = \\sqrt{3}\\, x^2_3 + 2 \\, x_3 + \\sqrt{3} - \\frac{1}{r_1} &=0 \\end{align} or in another form \\begin{align} f(x_2) = \\sqrt{3}\\, x^2_2 + 2 \\, x_2 + \\sqrt{3} - \\frac{1}{r_1} &=0\\\\ \\sqrt{3}(x_2 + x_3)^2 + (2- 2\\sqrt{3})\\, x_2 \\, x_3 &= \\frac{1}{r_1}\\\\ f(x_3) = \\sqrt{3}\\, x^2_3 + 2 \\, x_3 + \\sqrt{3} - \\frac{1}{r_1} &=0 \\end{align} Case1: Assume $x_2$ and $x_3$ are the two roots of $f$ (could be that $x_2\\neq x_3$ or could be that $x_2=x_3$ is a double root). By Vieta's formulas $$x_2+x_3 = - \\, \\frac{2}{\\sqrt{3}} \\,\\,\\,\\,\\, x_2\\, x_3 = 1- \\frac{1}{\\sqrt{3} \\, r_1}$$ so if we plug them in the second equaiton $$\\frac{4 \\sqrt{3}}{3} + (2- 2\\sqrt{3}) - (2- 2\\sqrt{3})\\frac{1}{\\sqrt{3} \\, r_1} = \\frac{1}{r_1}$$ $$2 - \\frac{2 \\sqrt{3}}{3} + \\left(2 - \\frac{2 \\sqrt{3}}{3}\\right)\\frac{1}{ r_1} = \\frac{1}{r_1}$$ $$1 + \\frac{1}{r_1}=\\frac{3}{6-2\\sqrt{3}} \\, \\frac{1}{r_1}$$ $$1 = \\frac{3}{6-2\\sqrt{3}} \\, \\frac{1}{r_1} - \\frac{1}{r_1} = \\left( \\frac{3}{6-2\\sqrt{3}} - 1\\right) \\, \\frac{1}{r_1}$$ when you solve for $r_1$ one gets $$r_1 = \\frac{\\sqrt{3}-1}{4}.$$ The solutions $x_2$ and $x_3$ are different, one of them is, let's say $x_2$ is $1$ and the other, $x_3$, is greater than $1$ in absolute value, which leads to $r_3>1$.", "= \\boxed{420}$. ## Solution 3 By Vieta's, we know that the sum of roots of $p(x)$ is $0$. Therefore, the roots of $p$ are $r, s, -r-s$. By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$. Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$. Since $p(x)$ and $q(x)$ have the same coefficient for $x$, we can go ahead and match those up to get \\begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\\\ 0 &= -13 - 5r + 2s \\\\ s &= \\frac{5r + 13}{2} \\end{align*} At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$. Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$. Therefore, $q(s) = 240$. If we plug that into our expression, we get that \\begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &=", "Vieta's Formulas – IMT DeCal # Vieta’s Formulas by Suraj Rampure Vieta’s formulas give us a way to interpret a polynomial in standard form, e.g. $p(x) = 4x^2 + 3x + 3$, in terms of its roots, without having to find the roots specifically. While there are more practical applications to this, the derivation of Vieta’s formulas prove to be an interesting combinatoral problem. ## Naive Approach, using the Quadratic Formula Let $p(x) = ax^2 + bx + c$. We know that $p(x)$ will have two roots, i.e. solutions to $p(x) = 0$ (they might be complex and/or equal). Suppose $p(x)$'s roots are $r_1$ and $r_2$, and suppose we want to find the sum ($r_1 + r_2$) and product ($r_1 r_2$) of these roots. One way to do this would be to use the quadratic formula to solve for both roots, and simplify. $r_1, r_2 = \\frac{-b + \\sqrt{b^2 - 4ac}}{2a}, \\frac{-b - \\sqrt{b^2 - 4ac}}{2a}$ Then: $r_1 + r_2 = \\frac{-b + \\sqrt{b^2 - 4ac} - b - \\sqrt{b^2 - 4ac}}{2a} = -\\frac{2b}{2a} = -\\frac{b}{a}$ $r_1r_2 = \\left( \\frac{-b + \\sqrt{b^2 - 4ac}}{2a} \\right)\\left( \\frac{-b - \\sqrt{b^2 - 4ac}}{2a} \\right) = \\frac{b^2 - (b^2 - 4ac)}{4a^2} = \\frac{4ac}{4a^2} = \\frac{c}{a}$ We’ve found that the sum of the roots is $-\\frac{b}{a}$, and the product of the roots is", ".$ Solution: The expression is equal to $$r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$$. By Vieta's Formula, we know that $r_1+r_2+r_3=-\\frac{-11}{5}=\\frac{11}{5}, r_1r_2+r_2r_3+r_3r_1=\\frac{7}{5}, r_1r_2r_3 = - \\frac{3}{5},$ so $$r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \\frac{11}{5}+2\\times \\frac{7}{5} = \\frac{25}{5}=5$$. Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are. 3. Let $$r_1, r_2$$ and $$r_3$$ be the roots of the polynomial $$5x^3 -11x^2+7x+3$$. Evaluate $$r_1^3+r_2^3+r_3^3.$$ Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that $r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right].$ Now we only need to know how to calculate $$r_1^2+r_2^2+r_3^2$$. Again, from factorization, we have that $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$ so this allows us to conclude that $\\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \\left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3\\times (-\\frac{3}{5}) +(\\frac{11}{5}) \\left[ \\left(\\frac{11}{5} \\right)^2-3\\times \\frac{7}{5} \\right] \\\\ & = -\\frac{49}{125}.\\\\ \\end{array}$ 4. [IMO shortlist] Determine all real values of the parameter $$a$$ for which the equation $16x^4-ax^3+(2a+17)x^2-ax+16 = 0$ has exactly four distinct real roots that form a geometric progression.", "r_2(1+r_3+r_1) + r_3(1+r_1+r_2) .$ Solution: The expression is equal to $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$. By Vieta's Formula, we know that $r_1+r_2+r_3=-\\frac{-11}{5}=\\frac{11}{5}, r_1r_2+r_2r_3+r_3r_1=\\frac{7}{5}, r_1r_2r_3 = - \\frac{3}{5},$ so $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \\frac{11}{5}+2\\times \\frac{7}{5} = \\frac{25}{5}=5$. Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are. ### 3. Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1^3+r_2^3+r_3^3.$ Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that $r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right].$ Now we only need to know how to calculate $r_1^2+r_2^2+r_3^2$. Again, from factorization, we have that $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$ so this allows us to conclude that $\\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \\left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3\\times (-\\frac{3}{5}) +(\\frac{11}{5}) \\left[ \\left(\\frac{11}{5} \\right)^2-3\\times \\frac{7}{5} \\right] \\\\ & = -\\frac{49}{125}.\\\\ \\end{array}$ ### 4. [IMO shortlist] Determine all real values of the parameter $a$ for which the equation $16x^4-ax^3+(2a+17)x^2-ax+16 = 0$ has exactly four distinct real roots", "roots, $$r = 0.366 = \\frac{1}{2}(\\sqrt{3}-1)$$ and $$r= 1$$. Let $$r_0$$ be the smaller of these roots. Suppose we could show that $$\\frac{b_{n-1}^2}{b_n} = r_0$$. Then we can show by induction that $$b_n=r_0 s^{2^n}$$ for some value of $$s$$. Suppose this is true for some value $$n$$. Then \\begin{align} b_{n+1} &= \\frac{b_{n}^2}{r_0} = \\frac{\\left(r_0 s^{2^n}\\right)^2}{r_0} = r_0 s^{2^{n+1}}. \\end{align} In fact, we won't quite show this; what we will show is that $$\\frac{b_{n-1}^2}{b_n}$$ goes to $$r_0$$ as $$n$$ goes to $$\\infty$$. With some analysis (given in an appendix), we can complete the proof that $$b \\approx r_0 s^{2^n}$$ for some value of $$s$$. The actual value of $$s$$ depends on the initial conditions, and the only way I know to get it is computationally. Maple seems to say that $$s$$ is around 2.280142. If you start with $$a_1=2$$, $$a_2=7$$, Maple gives $$s \\approx$$ 4.189425. Why does $$\\frac{b_{n-1}^2}{b_n}$$ go to $$r_0$$? Take the recurrence, and divide $$b_{n-1}^2$$ by both sides. This gives $$\\frac{b_{n-1}^2}{b_n} = \\frac{1}{3-2\\left(\\frac{b_{n-2}^2}{b_{n-1}}\\right)^2},$$ or, letting $$t_n = b_{n-1}^2/b_n$$, $$t_n=\\frac{1}{3-2 t_{n-1}^2}.$$ This has three fixed points, but the only attractive point is $$r_0$$, and $$t_n$$ approaches this point from any initial value except the two other fixed points. Appendix: Here is a sketch of the argument that if $$\\lim_{n\\rightarrow \\infty} t_n = r_0$$, then $$s$$ exists. We'll consider", "are also roots of $f(x)$. Let these roots be $u,v$. We get the system \\begin{align*} au^{17} + bu^{16} + 1 &= 0, \\\\ av^{17} + bv^{16} + 1 &= 0. \\end{align*} If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get \\begin{align*} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\\\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \\end{align*} Now subtracting, we get $$a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \\implies a = \\frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.$$ By Vieta's, $uv=-1$ so the denominator becomes $u-v$. By difference of squares and dividing out $u-v$ we get $$a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).$$ A simple exercise of Vieta's gets us $a= \\boxed{987}.$ ~bobthegod78", "Vieta’s formulas for degree 3. However, at this point, you should try and derive these on your own, then come back here to check your work. ## Derivation of Vieta’s Formulas for Polynomials of Degrees 3, 4 Let’s start with cubic polynomials. Assume $p(x)$ has roots $r_1, r_2, r_3$, and has the form $p(x) = ax^3 + bx^2 + cx + d$. Then: \\begin{aligned} p(x) &= a(x-r_1)(x-r_2)(x-r_3) \\\\ &= a \\left[(x^3 - (r_1 + r_2 + r_3)x + (r_1r_2 + r_1r_3 + r_2r_3)x - r_1r_2r_3\\right] \\\\ &= ax^3 - a(r_1 + r_2 + r_3)x^2 + a(r_1r_2 + r_1r_3 + r_2r_3)x - ar_1r_2r_3 \\ \\end{aligned} $\\implies r_1 + r_2 + r_3 = -\\frac{b}{a}$ $\\implies r_1r_2 + r_1r_3 + r_2r_3 = \\frac{c}{a}$ $\\implies r_1r_2r_3 = -\\frac{d}{a}$ We could follow the same process to find the formulas for degrees 4, 5 and so on and so forth. However, there’s a pattern here. When multiplying out $(x - r_1)(x - r_2)(x - r_3)$, we know our product must contain every combination of one term from the first bracket, one term from the second bracket and one term from the third bracket. Then, our choices are the following: • We can choose $3$ $x$s and no roots, yielding $x^3$ • We can choose $2$ $x$s and one root, yielding $(-r_1 - r_2 -", "+0 # help 0 94 2 Let a, b, c be the roots of x^3 - 17x - 19. Find a^3 + b^3 + c^3. Nov 27, 2019 #1 +4330 +2 This involves factoring from Vieta's Formulas: The first step is to break $$a^3+b^3+c^3$$ into $$3r_1r_2r_3+(r_1+r_2+r_3)[r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1)$$. The confusing part might be trying to find $$r_1^2+r_2^2+r_3^2$$, yet we know that is equal to $$(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).$$ This can be better written as $$3r_1r_2r_3+(r_1+r_2+r_3)[(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)].$$ Remember that $$r_1, r_2$$, and $$r_3$$ are the roots of the polynomial which is an expression. Note that $$r_1+r_2+r_3=\\frac{-b}{a}$$ , $$r_1r_2+r_2r_3+r_3r_1=\\frac{c}{a}$$, and $$r_1r_2r_3=\\frac{-d}{a}.$$ Try to plug the values in, and be careful and don't forget the $$x^2$$ term. Nov 27, 2019 #2 +23841 +3 Let $$a$$, $$b$$, $$c$$ be the roots of $$x^3 - 17x - 19$$. Find $$a^3 + b^3 + c^3$$. $$\\begin{array}{|rcll|} \\hline && x^3 - 17x - 19 \\\\ &=& x^3 + {\\color{red}0}x^2 {\\color{green}-17}x {\\color{blue}-19} \\\\\\\\ \\hline \\mathbf{\\text{vieta:}}\\\\ {\\color{red}0} &=& -(a+b+c) \\\\ \\mathbf{a+b+c} &=& \\mathbf{0} \\\\\\\\ {\\color{green}-17} &=& ab+ac+bc \\\\ \\mathbf{ab+ac+bc} &=& \\mathbf{-17} \\\\\\\\ {\\color{blue}-19} &=& -abc \\\\ \\mathbf{abc} &=& \\mathbf{19} \\\\ \\hline \\end{array}$$ Formula: $$\\begin{array}{|rcll|} \\hline a^3+b^3+c^3 &=& 3\\underbrace{abc}_{=19}+\\underbrace{(a+b+c)}_{=0}\\left(a^2+b^2+c^2-(ab+bc+ca)\\right) \\\\ a^3+b^3+c^3 &=& 3\\times 19 \\\\ \\mathbf{a^3+b^3+c^3} &=& \\mathbf{57} \\\\ \\hline \\end{array}$$ Nov 28, 2019", "}$$ in $$f\\left( x \\right)$$ or even simpler, $${ a }_{ n }$$ is the leading coefficient of $$f\\left( x \\right)$$. Example Lets have an explicit example where $$m=3$$ and $$f\\left( x \\right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120$$ So $$\\left( { r }_{ 1 }+3 \\right) \\left( { r }_{ 2 }+3 \\right) \\left( { r }_{ 3 }+3 \\right) ={ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27$$ Then using Vieta's, $${ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27=-\\frac { -120 }{ 2 } +3\\times \\frac { -14 }{ 2 } -9\\times \\frac { 12 }{ 2 } +27\\\\ =12$$ and indeed $${ \\left( -1 \\right) }^{ 3 }\\frac { f\\left( -3 \\right) }{ 2 } =12$$ and also the roots of $$f\\left( x \\right) =2{ x }^{ 3 }+12{ x }^{", "should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 The roots of $p(x)$ are $r$, $s$, and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$). Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$, as they too sum to $0$. Then: $a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and $a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$. From these equations, we can write that $$rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a$$ and simplifying gives $$2s-5r-13=0 \\Rightarrow s = \\frac{5r+13}{2}.$$ We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get $$rs(r+s) = b$$ $$(r+4)(s-3)(r+s+1)=b + 240.$$ Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding, simplifying, substituting $s = \\frac{5r+13}{2}$, and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$, or $b = (1)(9)(1 + 9) = 90$. The answer is $|-330|+|90|", "+0 # Let r and s be the roots of 3x^2 + 4x + 12 = 0. Find r^2 + s^2. 0 66 1 +202 Let r and s be the roots of 3x^2 + 4x + 12 = 0. Find r^2 + s^2. Aug 4, 2020 #1 +169 +3 $$x = {-4 \\pm \\sqrt{4^2-4(3)(12)} \\over 2(3)}\\\\ x = {-4 \\pm \\sqrt{-128} \\over 6}\\\\ x = {-4 \\pm 3\\sqrt{-2} \\over 6}\\\\ ({-4 + 3\\sqrt{-2} \\over 6})^2+({-4 - 3\\sqrt{-2} \\over 6})^2\\\\ {(-4 + 3\\sqrt{-2})^2 \\over 6^2}+{(-4 - 3\\sqrt{-2})^2 \\over 6^2}\\\\ {(-4 + 3\\sqrt{-2})^2 \\over 36}+{(-4 - 3\\sqrt{-2})^2 \\over 36}\\\\ {(-4 + 3\\sqrt{-2})^2+(-4 - 3\\sqrt{-2})^2 \\over 36}\\\\$$ Well this is kind of bashy, lOl. $$r + s = -\\frac{4}{3} \\text{ due to vieta}\\\\ (r + s)^2 = \\frac{16}{9} \\text{ because we square -4}\\\\ r^2+s^2+2rs=\\frac{16}9\\\\ rs=4 \\text{ due to vieta}\\\\ r^2 + s^2 = \\frac{16}{9} - 2(4) = \\frac{16}{9}-8=-6\\frac{2}{9}$$ Someone check this Aug 4, 2020", "\\begin{align*} x+\\left(2x^2-1\\right)+\\left(4x^3-3x\\right)&=-\\frac12 \\\\ 4x^3+2x^2-2x-\\frac12&=0 \\\\ x^3+\\frac12 x^2 - \\frac12 x - \\frac18 &= 0. \\end{align*} It follows that $(a,b,c)=\\left(\\frac12,-\\frac12,-\\frac18\\right),$ and $abc=\\boxed{\\textbf{(D) }\\frac{1}{32}}.$ ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) ### Solution 4.2 (Vieta's Formulas--Explains Solution 1 Using De Moivre's Theorem) Let $z=e^{\\frac{2\\pi i}{7}}.$ Since $z$ is a $7$th root of unity, $z^7=1.$ Geometrically, it follows that $$\\begin{array}{ccccc} \\cos{\\frac{2\\pi}{7}} &=& \\frac{z+z^6}{2} &=& \\frac{z+z^{-1}}{2} \\\\ [2ex] \\cos{\\frac{4\\pi}{7}} &=& \\frac{z^2+z^5}{2} &=& \\frac{z^2+z^{-2}}{2} \\\\ [2ex] \\cos{\\frac{7\\pi}{7}} &=& \\frac{z^3+z^4}{2} &=& \\frac{z^3+z^{-3}}{2} \\end{array}$$ Recall that $\\sum_{k=0}^{6}z^k=0$ (so that $\\sum_{k=1}^{6}z^k=-1$), and let $(r,s,t)=\\left(\\cos{\\frac{2\\pi}{7}},\\cos{\\frac{4\\pi}{7}},\\cos{\\frac{6\\pi}{7}}\\right).$ By Vieta's Formulas and the results above, the answer is \\begin{align*} abc&=[-(r+s+t)](rs+st+tr)(-rst) \\\\ &=(r+s+t)(rs+st+tr)(rst) \\\\ &=\\left(\\frac{\\sum_{k=1}^{6}z^k}{2}\\right)\\left(\\frac{2\\sum_{k=1}^{6}z^k}{4}\\right)\\left(\\frac{1+\\sum_{k=0}^{6}z^k}{8}\\right) \\\\ &=\\frac{1}{32}\\left(\\sum_{k=1}^{6}z^k\\right)\\left(\\sum_{k=1}^{6}z^k\\right)\\left(1+\\sum_{k=0}^{6}z^k\\right) \\\\ &=\\frac{1}{32}(-1)(-1)(1) \\\\ &=\\boxed{\\textbf{(D) }\\frac{1}{32}}. \\end{align*} ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) ~ pi_is_3.14", "w^3 + 2g w^2 + g^2 w + \\frac{1-g^4}{4k} = 0. \\end{align} This cubic equation has three roots; call them $w_1, w_2, w_3$. Since $s$ is one root, fix $w_1=s$ without loss of generality. Now Vieta's formulas give \\begin{align} -2g = s + w_2 + w_3, && g^2 = sw_2+ sw_3 + w_2w_3, && \\frac{g^4-1}{4k} = sw_2w_3. \\end{align} Combining the first two relations implies \\begin{align*} 4g^2 = (-2g)^2 = (s + w_2 + w_3)^2 = s^2 + w_2^2 + w_3^2 + 2(sw_2+ sw_3 + w_2w_3) = s^2 + w_2^2 + w_3^2 + 2g^2, \\end{align*} which gives $$2g^2-s^2 = w_2^2+w_3^2. \\tag{2}$$ A cubic function either has three real roots, or it has one real root and two nonreal complex conjugate roots. Since $w_1=s$ is real, $w_2$ and $w_3$ are either both real or they are nonreal complex conjugates. Since we have fixed $k$, and are seeking an integer solution with $1 < g = r-s$, we may take $w_2,w_3 \\in \\mathbb{R}$, where [at least] one of them is a second solution of ($\\star$) for our fixed quotient $k$. Now (2) implies $2g^2-s^2 = w_2^2+w_3^2 > 0$, yielding \\begin{equation*} s^2 < 2g^2 = 2(r-s)^2. \\end{equation*} Therefore \\begin{equation*} \\biggl(\\frac{s}{r-s}\\biggr)^{\\!2} < 2 \\qquad\\implies\\qquad \\frac{r}{s} > \\frac{1+\\sqrt{2}}{\\sqrt{2}} > \\frac{3}{2}, \\end{equation*} and hence $2r > 3s$ as claimed. • Without proving the $w_2,w_3 \\not\\in\\mathbb{R}$" ]

[ "2008 AIME II Problems/Problem 7 Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. Solution 1 By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is $$(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).$$ Also by Vieta's formulas we have $$rst = -rs(r + s) = -\\dfrac{2008}{8} = -251$$ so negating both sides and multiplying through by 3 gives our answer of $\\boxed{753}.$ Solution 2 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ Solution 3 Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3", "# Difference between revisions of \"2008 AIME II Problems/Problem 7\" ## Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Solution 1 By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is $$(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).$$ Also by Vieta's formulas we have $$rst = -rs(r + s) = -\\dfrac{2008}{8} = -251$$ so negating both sides and multiplying through by 3 gives our answer of $\\boxed{753}.$ ## Solution 2 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ ## Solution 3 Vieta's formulas gives $r + s + t = 0$.", "## Vieta's Formulas For a polynomial of the form \\begin{align*} f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \\end{align*} with roots $$r_1,r_2,r_3,...r_n$$, Vieta's formulas state that: \\begin{align*} r_1+r_2+r_3+...+r_n=-\\frac{a_{n-1}}{a_n} \\\\ r_1r_2 +r_1r_3+...+r_{n-1}r_n=\\frac{a_{n-2}}{a_n} \\\\ r_1r_2r_3+r_1r_2r_4+...+r_{n-2}r_{n-1}r_n=-\\frac{a_{n-3}}{a_n} \\\\ \\vdots \\\\ r_1r_2r_3...r_n=(-1)^n\\frac{a_0}{a_n} \\\\ \\end{align*} Generalization: When the $$n$$ roots are taken in groups of $$k$$ (i.e. $$r_1+r_2+...+r_n$$ is taken in groups of $$1$$ and $$r_1r_2...r_n$$ is taken in groups of $$n$$), this is equivalent to \\begin{align*} (-1)^k\\frac{a_{n-k}}{a_n} \\end{align*} source", "330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$. From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \\frac{5r+13}{2}$. We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \\frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \\boxed{420}$. ## See also 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •", "# 2008 AIME II Problems/Problem 7 ## Problem Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Solution ### Solution 1 By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization $$r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0$$ we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \\boxed{753}.$ ### Solution 2 Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3 + 1001r + 2008 = 0\\Longleftrightarrow - 8r^3 = 1001r + 2008$, and the same can be done with $s,\\ t$. Therefore, we have \\begin{align*}8\\{(r + s)^3 + (s + t)^3 + (t + r)^3\\} &= - 8(r^3 + s^3 + t^3)\\\\ &= 1001(r + s + t) + 2008\\cdot 3 = 3\\cdot 2008\\end{align*}yielding the answer $753$.", ".$ Solution: The expression is equal to $$r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$$. By Vieta's Formula, we know that $r_1+r_2+r_3=-\\frac{-11}{5}=\\frac{11}{5}, r_1r_2+r_2r_3+r_3r_1=\\frac{7}{5}, r_1r_2r_3 = - \\frac{3}{5},$ so $$r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \\frac{11}{5}+2\\times \\frac{7}{5} = \\frac{25}{5}=5$$. Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are. 3. Let $$r_1, r_2$$ and $$r_3$$ be the roots of the polynomial $$5x^3 -11x^2+7x+3$$. Evaluate $$r_1^3+r_2^3+r_3^3.$$ Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that $r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right].$ Now we only need to know how to calculate $$r_1^2+r_2^2+r_3^2$$. Again, from factorization, we have that $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$ so this allows us to conclude that $\\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \\left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3\\times (-\\frac{3}{5}) +(\\frac{11}{5}) \\left[ \\left(\\frac{11}{5} \\right)^2-3\\times \\frac{7}{5} \\right] \\\\ & = -\\frac{49}{125}.\\\\ \\end{array}$ 4. [IMO shortlist] Determine all real values of the parameter $$a$$ for which the equation $16x^4-ax^3+(2a+17)x^2-ax+16 = 0$ has exactly four distinct real roots that form a geometric progression.", "r_2(1+r_3+r_1) + r_3(1+r_1+r_2) .$ Solution: The expression is equal to $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$. By Vieta's Formula, we know that $r_1+r_2+r_3=-\\frac{-11}{5}=\\frac{11}{5}, r_1r_2+r_2r_3+r_3r_1=\\frac{7}{5}, r_1r_2r_3 = - \\frac{3}{5},$ so $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \\frac{11}{5}+2\\times \\frac{7}{5} = \\frac{25}{5}=5$. Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are. ### 3. Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1^3+r_2^3+r_3^3.$ Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that $r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right].$ Now we only need to know how to calculate $r_1^2+r_2^2+r_3^2$. Again, from factorization, we have that $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$ so this allows us to conclude that $\\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \\left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \\right] \\\\ & =3\\times (-\\frac{3}{5}) +(\\frac{11}{5}) \\left[ \\left(\\frac{11}{5} \\right)^2-3\\times \\frac{7}{5} \\right] \\\\ & = -\\frac{49}{125}.\\\\ \\end{array}$ ### 4. [IMO shortlist] Determine all real values of the parameter $a$ for which the equation $16x^4-ax^3+(2a+17)x^2-ax+16 = 0$ has exactly four distinct real roots", "should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 The roots of $p(x)$ are $r$, $s$, and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$). Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$, as they too sum to $0$. Then: $a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and $a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$. From these equations, we can write that $$rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a$$ and simplifying gives $$2s-5r-13=0 \\Rightarrow s = \\frac{5r+13}{2}.$$ We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get $$rs(r+s) = b$$ $$(r+4)(s-3)(r+s+1)=b + 240.$$ Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding, simplifying, substituting $s = \\frac{5r+13}{2}$, and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$, or $b = (1)(9)(1 + 9) = 90$. The answer is $|-330|+|90|", "= \\boxed{420}$. ## Solution 3 By Vieta's, we know that the sum of roots of $p(x)$ is $0$. Therefore, the roots of $p$ are $r, s, -r-s$. By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$. Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$. Since $p(x)$ and $q(x)$ have the same coefficient for $x$, we can go ahead and match those up to get \\begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\\\ 0 &= -13 - 5r + 2s \\\\ s &= \\frac{5r + 13}{2} \\end{align*} At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$. Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$. Therefore, $q(s) = 240$. If we plug that into our expression, we get that \\begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &=", "We are trying to find $-(r^3 + s^3 + t^3)$. Substituting: $$-(r^3 + s^3 + t^3) = -3srt = \\frac{-2008*3}{8} = \\boxed{753}.$$ ## Solution 5 Write $(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)$ and let $f(x)=8x^3+1001x+2008$. Then $$f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.$$ Solving for $r^3+s^3+t^3$ and negating the result yields the answer $\\boxed{753}.$ ## Solution 6 Here by Vieta's formulas: $r+s+t = 0$ --(1) $rst = \\frac{-2008}{8} = -251$ --(2) By the factorisation formula: Let $a = r+s$, $b = s+t$, $c = t+r$, $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$ (By (1)) So $$a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \\boxed{753}.$$ ## Solution 7 Let's construct a polynomial with the roots $(r+s), (s+t),$ and $(t+r)$. sum of the roots: $=2(r+s+t)=2\\cdot0=0$ pairwise product of the roots: $(r+s)(s+t)+(s+t)(t+r)+(t+r)(r+s)=r^2+s^2+t^2+3(rs+st+tr)$ $=(r+s+t)^2+rs+st+tr=0+\\frac{1001}{8}$ product of the roots: $(r+s)(s+t)(t+r)=r^2t+r^2s+s^2r+s^2t+t^2r+t^2s+3rst$ $=(rs+st+tr)(r+s+t)-3rst+2rst=-rst=-\\frac{2008}{8}$ thus, the polynomial we get is $x^3+\\frac{1001}{8}x+-\\frac{2008}{8}=0$ as $(r+s), (s+t),$ and $(t+r)$ are roots of this polynomial, we know that (using power reduction) $(r+s)^3+\\frac{1001}{8}(r+s)-\\frac{2008}{8}=0$ $(s+t)^3+\\frac{1001}{8}(s+t)-\\frac{2008}{8}=0$ $(t+r)^3+\\frac{1001}{8}(t+r)-\\frac{2008}{8}=0$ adding all of the equations up, we see that $(r+s)^3+(s+t)^3+(t+r)^3=3\\cdot\\frac{2008}{8}-\\frac{1001}{8}(2r+2s+2t)=251(3)+0=\\boxed{753}$", "Vieta's Formulas – IMT DeCal # Vieta’s Formulas by Suraj Rampure Vieta’s formulas give us a way to interpret a polynomial in standard form, e.g. $p(x) = 4x^2 + 3x + 3$, in terms of its roots, without having to find the roots specifically. While there are more practical applications to this, the derivation of Vieta’s formulas prove to be an interesting combinatoral problem. ## Naive Approach, using the Quadratic Formula Let $p(x) = ax^2 + bx + c$. We know that $p(x)$ will have two roots, i.e. solutions to $p(x) = 0$ (they might be complex and/or equal). Suppose $p(x)$'s roots are $r_1$ and $r_2$, and suppose we want to find the sum ($r_1 + r_2$) and product ($r_1 r_2$) of these roots. One way to do this would be to use the quadratic formula to solve for both roots, and simplify. $r_1, r_2 = \\frac{-b + \\sqrt{b^2 - 4ac}}{2a}, \\frac{-b - \\sqrt{b^2 - 4ac}}{2a}$ Then: $r_1 + r_2 = \\frac{-b + \\sqrt{b^2 - 4ac} - b - \\sqrt{b^2 - 4ac}}{2a} = -\\frac{2b}{2a} = -\\frac{b}{a}$ $r_1r_2 = \\left( \\frac{-b + \\sqrt{b^2 - 4ac}}{2a} \\right)\\left( \\frac{-b - \\sqrt{b^2 - 4ac}}{2a} \\right) = \\frac{b^2 - (b^2 - 4ac)}{4a^2} = \\frac{4ac}{4a^2} = \\frac{c}{a}$ We’ve found that the sum of the roots is $-\\frac{b}{a}$, and the product of the roots is", "Vieta’s formulas for degree 3. However, at this point, you should try and derive these on your own, then come back here to check your work. ## Derivation of Vieta’s Formulas for Polynomials of Degrees 3, 4 Let’s start with cubic polynomials. Assume $p(x)$ has roots $r_1, r_2, r_3$, and has the form $p(x) = ax^3 + bx^2 + cx + d$. Then: \\begin{aligned} p(x) &= a(x-r_1)(x-r_2)(x-r_3) \\\\ &= a \\left[(x^3 - (r_1 + r_2 + r_3)x + (r_1r_2 + r_1r_3 + r_2r_3)x - r_1r_2r_3\\right] \\\\ &= ax^3 - a(r_1 + r_2 + r_3)x^2 + a(r_1r_2 + r_1r_3 + r_2r_3)x - ar_1r_2r_3 \\ \\end{aligned} $\\implies r_1 + r_2 + r_3 = -\\frac{b}{a}$ $\\implies r_1r_2 + r_1r_3 + r_2r_3 = \\frac{c}{a}$ $\\implies r_1r_2r_3 = -\\frac{d}{a}$ We could follow the same process to find the formulas for degrees 4, 5 and so on and so forth. However, there’s a pattern here. When multiplying out $(x - r_1)(x - r_2)(x - r_3)$, we know our product must contain every combination of one term from the first bracket, one term from the second bracket and one term from the third bracket. Then, our choices are the following: • We can choose $3$ $x$s and no roots, yielding $x^3$ • We can choose $2$ $x$s and one root, yielding $(-r_1 - r_2 -", "roots, $$r = 0.366 = \\frac{1}{2}(\\sqrt{3}-1)$$ and $$r= 1$$. Let $$r_0$$ be the smaller of these roots. Suppose we could show that $$\\frac{b_{n-1}^2}{b_n} = r_0$$. Then we can show by induction that $$b_n=r_0 s^{2^n}$$ for some value of $$s$$. Suppose this is true for some value $$n$$. Then \\begin{align} b_{n+1} &= \\frac{b_{n}^2}{r_0} = \\frac{\\left(r_0 s^{2^n}\\right)^2}{r_0} = r_0 s^{2^{n+1}}. \\end{align} In fact, we won't quite show this; what we will show is that $$\\frac{b_{n-1}^2}{b_n}$$ goes to $$r_0$$ as $$n$$ goes to $$\\infty$$. With some analysis (given in an appendix), we can complete the proof that $$b \\approx r_0 s^{2^n}$$ for some value of $$s$$. The actual value of $$s$$ depends on the initial conditions, and the only way I know to get it is computationally. Maple seems to say that $$s$$ is around 2.280142. If you start with $$a_1=2$$, $$a_2=7$$, Maple gives $$s \\approx$$ 4.189425. Why does $$\\frac{b_{n-1}^2}{b_n}$$ go to $$r_0$$? Take the recurrence, and divide $$b_{n-1}^2$$ by both sides. This gives $$\\frac{b_{n-1}^2}{b_n} = \\frac{1}{3-2\\left(\\frac{b_{n-2}^2}{b_{n-1}}\\right)^2},$$ or, letting $$t_n = b_{n-1}^2/b_n$$, $$t_n=\\frac{1}{3-2 t_{n-1}^2}.$$ This has three fixed points, but the only attractive point is $$r_0$$, and $$t_n$$ approaches this point from any initial value except the two other fixed points. Appendix: Here is a sketch of the argument that if $$\\lim_{n\\rightarrow \\infty} t_n = r_0$$, then $$s$$ exists. We'll consider", "a= -89$$ and finally, $$1+\\frac{ac}{10} = b = 1+\\frac{-89 \\cdot 900}{10} = b = -8009$$. We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$. We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \\boxed{\\textbf{(C)}\\,-7007}$. ## Solution 8 Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$. Let $r_4$ be the additional root of $f(x)$. Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$, we obtain the following: \\begin{align*} r_1+r_2+r_3&=-a \\\\ r_1+r_2+r_3+r_4&=-1 \\end{align*} Thus $r_4=a-1$. Now applying Vieta's formulas on the constant term of $g(x)$, the linear term of $g(x)$, and the linear term of $f(x)$, we obtain: \\begin{align*} r_1r_2r_3 & = -10\\\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\\\ \\end{align*} Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain: $$-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100$$ It follows that $r_4=-90$. But $r_4=a-1$ so $a=-89$ Now we can factor $f(x)$ in terms of $g(x)$ as $$f(x)=(x-r_4)g(x)=(x+90)g(x)$$ Then $f(1)=91g(1)$ and $$g(1)=1^3-89\\cdot 1^2+1+10=-77$$ Hence $f(1)=91\\cdot(-77)=\\boxed{\\textbf{(C)}\\,-7007}$. ## Solution 9 (Risky) Let the roots of $g(x)$ be $r_1$, $r_2$, and $r_3$. Let the roots of $f(x)$ be $r_1$, $r_2$, $r_3$, and $r_4$. From Vieta's, we have: \\begin{align*} r_1+r_2+r_3=-a \\\\ r_1+r_2+r_3+r_4=-1 \\\\ r_4=a-1 \\end{align*} The", "+0 # HELP HELP NOWW 0 259 5 Let $r$ and $s$ be the roots of $3x^2 + 4x + 12 = 0.$ Find $r^2 + s^2.$ Apr 25, 2020 #1 +21959 +1 Apr 25, 2020 #2 0 where????? Apr 25, 2020 #3 +1 $$3x^2+4x+12=0$$ Is a quadratic equation of the form $$ax^2+bx+c=0$$ where r and s are roots By Vieta's formulas: $$r+s=-\\frac{b}{a}$$ Thus $$r+s=-\\frac{4}{3}$$ $$r*s=\\frac{c}{a}$$ Thus $$rs=4$$ Now notice that: $$r^2+s^2=(r+s)^2-2rs$$ Substitute $$(-\\frac{4}{3})^2-2(4)=-\\frac{56}{9}=-6.22$$ Apr 25, 2020 #4 0 it is wrong????? Apr 26, 2020 #5 +112000 0 Good, I hope it is, you are too lazy to even lay your question out properly. And too lazy to search old answers. I would have preferred that you were not answered at all but if the answer is wrong that is the next best thing. (No disrespect intended to answering guest, I know you are just trying to help) Melody Apr 26, 2020", "a handful of sign tests. – Yves Daoust Jul 30 '15 at 14:06 I believe that Vieta's Formulae can help here. First, you divide the whole equation by $$a$$. We can redefine variables $$b, c, d,$$ and $$e$$ as $$\\frac{b}{a}, \\frac{c}{a}, \\frac{d}{a},$$ and $$\\frac{e}{a}$$ respectively. Basically, you expand the following, where $$p, q, r,$$ and $$s$$ are roots: $$(x-p)(x-q)(x-r)(x-s)$$ And get: \\begin{align}b &= -(p+q+r+s) \\\\ c &= pq + pr + ps + qr + qs + rs \\\\ d &= -(pqr + pqs + prs + qrs)\\\\ e &= pqrs\\end{align} You now have $$4$$ variables and $$4$$ equations, so it is now a matter of solving this system. Admittedly, it is rather tedious, and requires quite a lot of work.", "+0 # help 0 94 2 Let a, b, c be the roots of x^3 - 17x - 19. Find a^3 + b^3 + c^3. Nov 27, 2019 #1 +4330 +2 This involves factoring from Vieta's Formulas: The first step is to break $$a^3+b^3+c^3$$ into $$3r_1r_2r_3+(r_1+r_2+r_3)[r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1)$$. The confusing part might be trying to find $$r_1^2+r_2^2+r_3^2$$, yet we know that is equal to $$(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).$$ This can be better written as $$3r_1r_2r_3+(r_1+r_2+r_3)[(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)].$$ Remember that $$r_1, r_2$$, and $$r_3$$ are the roots of the polynomial which is an expression. Note that $$r_1+r_2+r_3=\\frac{-b}{a}$$ , $$r_1r_2+r_2r_3+r_3r_1=\\frac{c}{a}$$, and $$r_1r_2r_3=\\frac{-d}{a}.$$ Try to plug the values in, and be careful and don't forget the $$x^2$$ term. Nov 27, 2019 #2 +23841 +3 Let $$a$$, $$b$$, $$c$$ be the roots of $$x^3 - 17x - 19$$. Find $$a^3 + b^3 + c^3$$. $$\\begin{array}{|rcll|} \\hline && x^3 - 17x - 19 \\\\ &=& x^3 + {\\color{red}0}x^2 {\\color{green}-17}x {\\color{blue}-19} \\\\\\\\ \\hline \\mathbf{\\text{vieta:}}\\\\ {\\color{red}0} &=& -(a+b+c) \\\\ \\mathbf{a+b+c} &=& \\mathbf{0} \\\\\\\\ {\\color{green}-17} &=& ab+ac+bc \\\\ \\mathbf{ab+ac+bc} &=& \\mathbf{-17} \\\\\\\\ {\\color{blue}-19} &=& -abc \\\\ \\mathbf{abc} &=& \\mathbf{19} \\\\ \\hline \\end{array}$$ Formula: $$\\begin{array}{|rcll|} \\hline a^3+b^3+c^3 &=& 3\\underbrace{abc}_{=19}+\\underbrace{(a+b+c)}_{=0}\\left(a^2+b^2+c^2-(ab+bc+ca)\\right) \\\\ a^3+b^3+c^3 &=& 3\\times 19 \\\\ \\mathbf{a^3+b^3+c^3} &=& \\mathbf{57} \\\\ \\hline \\end{array}$$ Nov 28, 2019", "+ \\ r_2^2r_3 \\ + \\ r_1^2r_3 \\ + \\ r_1r_2^2 \\ + \\ r_2r_3^2 \\ )$$ and $$r_1^2r_2 \\ + \\ r_1r_3^2 \\ + \\ r_2^2r_3 \\ + \\ r_1^2r_3 \\ + \\ r_1r_2^2 \\ + \\ r_2r_3^2$$ $$= \\ \\ (r_1 + r_2 + r_3) · (r_1r_2 \\ + \\ r_1r_3 \\ + \\ r_2r_3) \\ - \\ 3r_1r_2r_3 \\ \\ ,$$ we obtain $$s_1^3 \\ \\ = \\ \\ (r_1 + r_2 + r_3)^3 \\ - \\ \\left(3 + \\frac32 \\right)·( \\ r_1^2r_2 \\ + \\ r_1r_3^2 \\ + \\ r_2^2r_3 \\ + \\ r_1^2r_3 \\ + \\ r_1r_2^2 \\ + \\ r_2r_3^2 \\ )$$ $$= \\ \\ (r_1 + r_2 + r_3)^3 \\ - \\ \\frac92 ·[ \\ (r_1 + r_2 + r_3) · (r_1r_2 \\ + \\ r_1r_3 \\ + \\ r_2r_3) \\ - \\ 3r_1r_2r_3 \\ ] \\ \\ .$$ Now we can apply the Viete relations to $$\\ f(x) \\$$ to evaluate this as $$\\ 3^3 \\ - \\ \\frac92 ·[ \\ 3 · (-4) \\ - \\ 3·(-4) \\ ]$$ $$= \\ 27 - 0 \\ \\ .$$ The real part of the sum of the coefficients of $$\\ g(x) \\$$ is then $$\\ 1 - 27 \\ = \\ -26 \\ \\ .$$", "theorem is actually just a special case of Vieta’s formulas, when all roots are the same! For example, suppose $n = 4$, $r_i = c$ for all $i$ and the leading coefficient is 1. Then: $p(x) = (x-c)^4 = {4 \\choose 0}x^4 - {4 \\choose 1} x^3c + {4 \\choose 2}x^2c^2 - {4 \\choose 3}xc^3 + {4 \\choose 4}c^4$ Using Vieta’s formulas for $n = 4$: \\begin{aligned} a_3 &= -(r_1 + r_2 + r_3 + r_4) = -4c = {4 \\choose 1}(-c)^1 \\\\ a_2 &= r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 = 6c^2 = {4 \\choose 2}(-c)^2 \\\\ a_1 &= -(r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4) = -4c^3 = {4 \\choose 3}(-c)^3 \\\\ a_0 &= r_1r_2r_3r_4 = c^4 = {4 \\choose 4}(-c)^4 \\end{aligned} ## Symmetric Sums Vieta’s formulas give us expressions for the sum of the roots and product of the roots of polynomials. However, we are also given expressions for sums of the form $r_1r_2 + r_1r_3 + r_2r_3$, for instance. A symmetric sum is an sum in which the value does not change when the order of variables are swapped. For example, $f(a, b, c) = ab + ac + bc$ is symmetric, because in any permutation of our variables $a, b, c$, the value stays the same: \\begin{aligned} f(a,", "\\begin{align} f(x_2) = \\sqrt{3}\\, x^2_2 + 2 \\, x_2 + \\sqrt{3} - \\frac{1}{r_1} &=0\\\\ \\sqrt{3}\\big((x_2 + x_3)^2 - 2\\, x_2x_3\\big) + 2 \\, {x_2\\, x_3} &= \\frac{1}{r_1}\\\\ f(x_3) = \\sqrt{3}\\, x^2_3 + 2 \\, x_3 + \\sqrt{3} - \\frac{1}{r_1} &=0 \\end{align} or in another form \\begin{align} f(x_2) = \\sqrt{3}\\, x^2_2 + 2 \\, x_2 + \\sqrt{3} - \\frac{1}{r_1} &=0\\\\ \\sqrt{3}(x_2 + x_3)^2 + (2- 2\\sqrt{3})\\, x_2 \\, x_3 &= \\frac{1}{r_1}\\\\ f(x_3) = \\sqrt{3}\\, x^2_3 + 2 \\, x_3 + \\sqrt{3} - \\frac{1}{r_1} &=0 \\end{align} Case1: Assume $x_2$ and $x_3$ are the two roots of $f$ (could be that $x_2\\neq x_3$ or could be that $x_2=x_3$ is a double root). By Vieta's formulas $$x_2+x_3 = - \\, \\frac{2}{\\sqrt{3}} \\,\\,\\,\\,\\, x_2\\, x_3 = 1- \\frac{1}{\\sqrt{3} \\, r_1}$$ so if we plug them in the second equaiton $$\\frac{4 \\sqrt{3}}{3} + (2- 2\\sqrt{3}) - (2- 2\\sqrt{3})\\frac{1}{\\sqrt{3} \\, r_1} = \\frac{1}{r_1}$$ $$2 - \\frac{2 \\sqrt{3}}{3} + \\left(2 - \\frac{2 \\sqrt{3}}{3}\\right)\\frac{1}{ r_1} = \\frac{1}{r_1}$$ $$1 + \\frac{1}{r_1}=\\frac{3}{6-2\\sqrt{3}} \\, \\frac{1}{r_1}$$ $$1 = \\frac{3}{6-2\\sqrt{3}} \\, \\frac{1}{r_1} - \\frac{1}{r_1} = \\left( \\frac{3}{6-2\\sqrt{3}} - 1\\right) \\, \\frac{1}{r_1}$$ when you solve for $r_1$ one gets $$r_1 = \\frac{\\sqrt{3}-1}{4}.$$ The solutions $x_2$ and $x_3$ are different, one of them is, let's say $x_2$ is $1$ and the other, $x_3$, is greater than $1$ in absolute value, which leads to $r_3>1$." ]

[ "Let the function $f(x)$ have the property that $f'(x) = \\frac{x+2}{x-3}$. If $g(x) = f(x^2)$ find $g'(x)$. $g'(x) =$", "Example 3: $m =\\frac 1 2 ; b = 1: f(x) = \\frac 1 2x + 1$ Each arrow passes through a single point, which is labeled $F = [1/2,1]$. The point $F$ completely determines the function $f$. Given a point / number, $x$, on the source line, there is a unique arrow passing through $F$ meeting the target line at a unique point / number, $\\frac 1 2x + 1$, which corresponds to the linear function's value for the point/number, $x$.", "+0 # Help 0 71 1 Define the function $f(x)=\\frac{a}{1-x}$. If $f(-1)=f^{-1}(4a+1)$, find the product of all possible values of $a$. Oct 26, 2019 #1 +2552 0 Stop posting in unclear latex, actually convert to this site's latex so people can see more clearly. $$f(x)=\\frac{a}{1-x}$$ $$f(-1)=f^{-1}(4a+1)$$ . Oct 26, 2019", "Function $f(x) = \\frac{1}{x + b} + a$ with $a,b \\in \\mathbb R$ in an orthonormed system Click to show the fonction $f(x) = \\frac{1}{x + b} + a$ with a = and b =", "of the function. $$f(x)=\\frac{2 x^{3}-5}{x^{2}+x-6}$$...", "## College Algebra 7th Edition We are given: $f(x)=\\displaystyle \\frac{1}{2}(3x-1)$ $f(x)=\\frac{3}{2}x-\\frac{1}{2}$ Since the function has the form $f(x)=ax+b$, it is linear.", "The function $f(x) = \\frac{x - 2}{x^2 - 3x + 2}$ is not defined for x = 2. Find a value to be given to f(2) that will make f continuous at 2. 2. $f(x)=\\dfrac{(x-2)}{(x-2)(x-1)}$", "theorem to the function $$f(x)=\\frac{1-x^4}7$$ on the interval $[0,1]$.", "$$f_1(\\frac a b) = \\frac {2a}b$$ is a well-defined function, because applying it to two different representation of the same fraction gives the same result (e.g. $$f_1(\\frac 1 2)=f_1(\\frac 2 4)$$). $$f_2(\\frac a b) = a+b$$, on the other hand, is not a well-defined function, because $$f_2(\\frac 1 2)=3$$, while $$f_2(\\frac 2 4)=6$$. You've said that you've defined $$exp$$ recursively, and implicit in saying that this is a definition is saying that the function is well-defined.", "## Intermediate Algebra (6th Edition) We are given that $f(x)=(\\frac{1}{2})^{-x}$. In order to find out which graph matches this function, we can find the value of $f(1)$. $f(1)=(\\frac{1}{2})^{-1}=2^{1}=2$ Therefore, $f(x)$ matches Graph B, as it is the only graph that contains the ordered pair $(1,2)$.", "# Function Help • April 23rd 2014, 02:12 AM LRNAB Function Help $f(x) = \\frac{x + 3}{2}$ Given that $f(-9) + f(t) = A + Bt$, find the values of A and B. A is $-\\frac{3}{2}$ B is $\\frac{1}{2}$", "whole f(z) function, ie. ((z^2 + 2z + 5) / (iz+1)) 4. Feb 16, 2015 ### RUber This would be either the product rule or the quotient rule. $[\\frac{g(z)}{f(z)}]'= \\frac{fg'-gf'}{[f(z)]^2}$", "# IMO 2019, Problem 1 Find all functions $f:\\Bbb{Z}\\to \\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$. The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$. Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.", "The function f is defined by f(x) = ~$-\\frac{1}{x}$~ for all nonzero numbers x. If f(a) = ~$-\\frac{1}{2}$~ and f(ab) = ~$\\frac{1}{6}$~ , then b = 3 ~$\\frac{1}{3}$~ ~$-\\frac{1}{3}$~ -3 -12", "Browse Questions # Let f, g be two functions defined by $f(x)=\\frac{x}{x+1},g(x)=\\frac{x}{1-x},\\;then\\;find\\;(fog)^{-1}(x)$ Can you answer this question?", "Consider the function $f(x)$, for which $f(0) = 3$ and $f'(0) = -7$. Find $h'(0)$ for the function $h(x) = \\frac{1}{f(x)}$. $h'(0)$ =", "The function f is defined by ~$f\\left ( x \\right )=-\\frac{1}{x}$~ for all nonzero numbers x. If ~$f\\left ( a \\right )=-\\frac{1}{2}$~ and ~$f\\left ( ab \\right )=\\frac{1}{6}$~, then b = 3 ~$\\frac{1}{3}$~ ~$-\\frac{1}{3}$~ -3 -12", "# Algebra of Function Algebra Level 3 If we have $f(x)+xf(1-x)=\\frac{x}{x-1}$ and $f(2)=\\frac{a}{b},$ then determine $$a^2+b^2+\\frac{b}{a}$$. ×", "linear For any functions f and g and any real numbers a and b, the derivative of the function h(x) = af(x) + bg(x) with respect to x is : h'(x) = a f'(x) + b g'(x). In this is written as: : \\frac = a\\frac +b\\frac. Special cases include: * ''The ''constant factor rule'' :(af)' = af' * ''The ''sum rule'' :(f + g)' = f' + g' * ''The subtraction rule'' :(f - g)' = f' - g'. The product rule For the functions ''f'' and ''g'', the derivative of the function ''h''(''x'') = ''f''(''x'') ''g''(''x'') with respect to ''x'' is : h' ... [...More Info...] [...Related Items...]", "Calculus 8th Edition Calculate $f(t)$ and $f(-t)$ $f(t)=\\frac{t}{t^{2}+1}$ $f(-t)=\\frac{-t}{(-t)^{2}+1}=-\\frac{t}{t^{2}+1}$ therefore we can see that the function f is an odd function that satisfies $f(-x)=-f(x)$ for every x in the domain." ]

[ "Algebra # Vieta's Formula Problem Solving Suppose one of the roots of the quadratic equation $x^2+6ax+b=0$ is $\\sqrt{3+\\sqrt{8}}.$ If $a$ and $b$ are rational numbers, what is the value of $15ab?$ What is the sum of the two roots of the following equation: $(3x-1355)^2+4(3x-1355)+5=0?$ $A$ and $B$ are the two roots of the quadratic equation $f(x)=0$. If $A+B=6$, what is the sum of the two roots of $f(2x-21)=0$? Let $a$ and $b$ be the roots of the quadratic polynomial $x^2-x-5=0.$ Let $f(x)$ be a cubic polynomial such that $f(a)=a, f(b)=b,$ $f(a+b)=a+b,$ and $f(ab)=-155.$ What is the value of $f(6)$? Find the sum of all real solutions to the equation $6 \\cdot 4^x - 2057 \\cdot 2^x + 3072 = 0.$ ×", "b^2c + abc + c^2a + c^2b + abc - abc =$ $a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc =$ $(a + b + c)(ab + bc + ac) - abc = (-3)(4) - 11 = -23$. The value of $t$ is the negation of this, which is $-(-23) = \\boxed{023}$ ~Extremelysupercooldude ## Solution 3 We have that $x^3+3x^2+4x-11=0$ for roots $a,b,c.$ In the second cubic function $x^3+rx^2+sx+t=0,$ the roots are $a+b,b+c,c+a.$ By Vieta's formulae, we see that $t= -(a+b)(b+c)(a+c).$ As we know that the sum of the roots of the first polynomial, $a+b+c$ is $-3$ by applying Vieta's again. Using this fact, we can rewrite $t$ as $-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).$ Seeing this, we can find the value of the product of the roots by applying this to the first equation. This can be done by setting $x'=a+3,$ so, from this, we see that we should plug in $x'-3$ for $x$ in $x^3+3x^2+4x-11=0$. After simplifying, we get that the polynomial is $x'^3 - 6x'^2 + 13x' - 23 = 0.$ Given that the product of the roots of this equation is equivalent to our desired value for $t$, we can apply Vieta's formulae for a third time to find that $t = -(\\frac{-23}{1}) = \\boxed{023}.$ ~MathWhiz35", "+0 # Help. Thank you. +3 75 2 +294 Let a and b be the solutions of the equation $$2x^2-10x+5=0$$. What is the value of $$(2a-3)(4b-6)$$? Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +3 Thank you! calvinbun Feb 23, 2021", "# Difference between revisions of \"1984 USAMO Problems/Problem 1\" ## Problem In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$. ## Solution Let the four roots be a, b, c, and d, so that ab=-32. From here we show two methods; the second is more slick, but harder to see. Solution #1 Using Vieta's formulas, we have: \\begin{align*} a+b+c+d &= 18, ab+ac+ad+bc+bd+cd &= k, abc+abd+acd+bcd &= -200, abcd &= -1984. \\en... From the last of these equations, we see that cd = \\frac{abcd}{ab} = \\frac{-1984}{-32} = 62. Thus, the second equation becomes -32+ac+ad+bc+bd+62=k, and so ac+ad+bc+bd=k-30. The key insight is now to factor the left-hand side as a product of two binomials: (a+b)(c+d)=k-30, so that we now only need to determine a+b and c+d rather than all four of a,b,c,d. Let p=a+b and q=c+d. Plugging our known values for ab and cd into the third Vieta equation, -200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b), we have -200 = -32(c+d) + 62(a+b) = 62p-32q. Moreover, the first Vieta equation, a+b+c+d=18, gives p+q=18. Thus we have two linear equations in p and q, which we solve to obtain p=4 and q=14. Therefore, we have (\\underbrace{a+b}_4)(\\underbrace{c+d}_{14}) = k-30, yielding k=4\\cdot", "+0 # help 0 123 2 Let $$a,b,c$$ and $$d$$ be the roots of $$x^4 + 8x^3 + 9x^2 + 5x + 4 = 0.$$ Find the value of $$\\frac{1}{ab} + \\frac{1}{ac} + \\frac{1}{ad} + \\frac{1}{bc} + \\frac{1}{bd} + \\frac{1}{cd}.$$ Apr 16, 2019 #1 +1 Try making the expression you're trying to find into one big fraction. Then use Vieta's Formulas and plug coefficients in. Apr 17, 2019 #2 +7711 +2 Let P be the value of the required expression. $$P = \\dfrac{cd+bd+bc+ad+ac+ab}{abcd} = \\dfrac{\\dfrac{9}{1}}{\\dfrac{4}{1}} = \\dfrac{9}{4}$$ . Apr 18, 2019", "$f(a,b)$ is defined as a quotient, it won't be defined if the denominator is zero, i.e., if $$a+b-3ab=0$$ However, this can't happen since $a,b\\in \\Bbb Z$ (why?), so the value of $f$ is defined everywhere and therefore $f$ is a function. • But how do we know that condition is 2 is satisfied rather than eye-balling it or depending upon intuition? – Benedict Voltaire May 3 '15 at 14:23 • It rests on the well definition of basic operations as sum, multiplication and division: If $(a,b)=(c,d)$, then $ab=cd$, $a+b=c+d$, $a+b-3ab=c+d-3cd$ and $\\frac{a+b}{a+b-3ab}=\\frac{c+d}{c+d-3cd}$ so if you have not any issue in assuming those operations are well defined, $f$ will be well defined. – Daniel May 3 '15 at 14:27 • Okay, I see that. Just to make sure that I understand why $a+b-3ab=0$ will never be true when $a,b \\mathbb{Z^*}$, is because it turns out that if we solve the equation for $b$, say, we have the quotient $b=\\frac{a}{3a-1}$. But how do we know that there does not exist a value for $a$ such that the fraction may be reduced in such a way as to make $b \\in \\mathbb{Z}$ ? – Benedict Voltaire May 3 '15 at 14:35 • @BenedictVoltaire I think you're concerned about 2 because you're confusing this with functions of the type $\\Bbb Q\\to \\text{[some", "$$\\Delta < 0$$, then the equation has no solution. Here t is unknown. --------------- $$v_f ^ 2 = v_i^2 + 2ad$$ Isolate d by adding $=v_i^2$ to both sides. $$v_f ^ 2 - v_i^2 = 2ad$$ To isolate d, divide both sides by 2a: $$\\frac{v_f ^ 2 - v_i^2}{2a} = \\frac{2ad}{2a}$$ $$\\frac{v_f ^ 2 - v_i^2}{2a} = d$$ Viet Dao, VietDao29 said: $$d = v_it + \\frac{1}{2}at^2$$ And you want to solve for t. Rearrange it as: $$\\frac{1}{2}at^2 + v_it - d = 0$$. And it's a quadratic equation. $$ax^2 + bx + c = 0$$, a, b, c are already known, x is the unknown. And its solution can be found by letting: $$\\Delta = b ^ 2 - 4ac$$ If $$\\Delta \\geq 0$$, then the equation has solution(s): $$x = \\frac{-b \\pm \\sqrt{\\Delta}}{2a}$$ If $$\\Delta < 0$$, then the equation has no solution. Here t is unknown. --------------- $$v_f ^ 2 = v_i^2 + 2ad$$ Isolate d by adding $=v_i^2$ to both sides. $$v_f ^ 2 - v_i^2 = 2ad$$ To isolate d, divide both sides by 2a: $$\\frac{v_f ^ 2 - v_i^2}{2a} = \\frac{2ad}{2a}$$ $$\\frac{v_f ^ 2 - v_i^2}{2a} = d$$ Viet Dao, careful with the signs. VietDao29 Homework Helper Whoops, stupid typo... :grumpy:", "## Being Receptive- A Quadratic Example A few days ago, I presented some students with this problem from the 2003 AMC 10A (problem #5): Let $d$ and $e$ denote the solutions of $2x^2 + 3x - 5 = 0$. What is the value of $(d-1)(e-1)$? Almost as soon as the problem was stated, the students began working out the solution by applying the quadratic formula to solve for $d$ and $e$, and then evaluating the desired formula by substituting their found values for $d$ and $e$. I picked this problem, however, because it invites many different approaches. Before addressing the most direct way, I want to describe a way to solve this problem using Vieta’s formulas. If you have a quadratic equation of the form $ax^2 + bx + c = 0$ and $r_1$ and $r_2$ are the roots, Vieta’s formulas tell us that $r_1 + r_2 = -\\frac{b}{a}$ and $r_1r_2 = \\frac{c}{a}$. This is readily verified by comparing coefficients in the equality $ax^2 + bx + c = a(x-r_1)(x-r_2)$. Applying Vieta’s formulas to the specific problem, we can read off from the coefficients that $d + e = -1.5$ and $de = -2.5$. We can then evaluate $(d-1)(e-1)$ without ever applying the quadratic formula: $(d-1)(e-1) = de - (d+e) + 1 = -2.5 + 1.5 + 1", "# 1977 USAMO Problems/Problem 3 ## Problem If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. ## Solution This problem needs a solution. If you have a solution for it, please help us out by adding it. a,b,c,d are roots of equation $x^4+x^3-1=0$ then by vietas relation $$ab +bc+cd+da+ac+bd= 0$$ let us suppose $ab,bc,cd,da,ac,bd$ are roots of $x^6+x^4+x^3-x^2-1=0$. then sum of roots $= ab +bc+cd+da+ac+bd=c/a = -b/a=0$ sum taken two at a time $= ab\\times bc + bc\\times ca +..........=c/a=1$ similarly we prove for the roots taken three four five and six at a time to prove $ab,bc,cd,da,ac,bd$ are roots of second equation Given the roots $a,b,c,d$ of the equation $x^{4}+x^{3}-1=0$. First, $a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1$. Then $cd=-\\frac{1}{ab}$ and $c+d=-1-(a+b)$. Remains $ab+(a+b)(c+d)+cd = 0$ or $ab+(a+b)[-1-(a+b)]-\\frac{1}{ab}=0$. Let $a+b=s$ and $ab=p$, so $p+s(-1-s)-\\frac{1}{p}=0$(1). Second, $a$ is a root, $a^{4}+a^{3}=1$ and $b$ is a root, $b^{4}+b^{3}=1$. Multiplying: $a^{3}b^{3}(a+1)(b+1)=1$ or $p^{3}(p+s+1)=1$. Solving $s= \\frac{1-p^{4}-p^{3}}{p^{3}}$. In (1): $\\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0$. $p^{8}+p^{5}-2p^{4}-p^{3}+1=0$ or $(p-1)(p+1)(p^{6}+p^{4}+p^{3}-p^{2}-1)= 0$. Conclusion: $p =ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.", "$f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \\neq 0$ ? It is known that $f(5) = – 3f(2)$. and that 3 is a root of $f(x) = 0$. What is the other root of f(x) = 0? [CAT 2008] a) -7 b) – 4 c) 2 d) 6 e) cannot be determined Solution: f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c f(2) = 4a + 2b + c f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c => 37a + 11b + 4c = 0 –> (1) 4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2) From (1) and (2), a – b = 0 => a = b => c = -12a The equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$ => -4 is a root of the equation. Question 3: Let $f(x)\\neq0$ for any ‘x’ be a function satisfying $f(x)f(y) = f(xy)$ for all real x, y. If $f(2) = 4$, then what is the value of $f(\\frac{1}{2})$? a) 0 b) 1/4 c) 1/2 d) 1 e) cannot be determined Solution: $f(1)^2$ = f(1) => f(1) = 1", "search and it basically states that the sum of the two roots is $$\\frac{-b}{a}$$, and the product of the two roots is $$\\frac{c}{a}$$, correct? So using that explanation, it means that since we are looking for the sign of the products of the roots, we need to find the value of $$\\frac{c}{a}$$ and therefore, $$\\frac{(x_1 * x_2)}{a}$$. Correct? We aren't given any information about a so how can we assume whether a is positive or negative? EDIT: Are there similar problems that I can apply this theorem to? Thanks! $$a$$ there is the coefficient of $$x^2$$ ($$ax^2+bx+c=0$$), hence for $$x^2+bx+c=0$$ it equals to 1 ($$1*x^2+bx+c=0$$). Questions involving Viete's theorem to practice: in-the-equation-x-2-bx-12-0-x-is-a-variable-and-b-is-a-109771.html if-x-2-3-is-one-factor-of-the-equation-x-2-4-3-x-160524.html if-x-2-12x-k-0-is-x-155465.html in-the-equation-ax-2-bx-c-0-a-b-and-c-are-constants-148766.html new-algebra-set-149349-80.html#p1200987 if-q-is-one-root-of-the-equation-x-2-18x-11c-0-where-141199.html if-f-x-5x-2-and-g-x-x-2-12x-85-what-is-the-sum-of-all-85989.html if-4-is-one-solution-of-the-equation-x2-3x-k-10-where-139119.html Theory on Algebra: algebra-101576.html DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29 PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50 Special algebra set: new-algebra-set-149349.html Hope this helps. _________________ Senior Manager Joined: 03 Aug 2011 Posts: 280 GMAT 1: 640 Q44 V34 GMAT 2: 700 Q42 V44 GMAT 3: 680 Q44 V39 GMAT 4: 740 Q49 V41 GPA: 3.7 WE: Project Management (Energy and Utilities) Followers: 12 Kudos [?]: 63 [0], given: 913 Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags 16 Aug 2014, 08:19 Hi guys, I couldn't solve", "Home · Algebra · Quadratic equations. Vieta's formulas The solutions $x_1$ and $x_2$ of the quadratic equation $ax^2+bx+c=0$ can be found by special Vieta’s formulas. $$x_1 + x_2 = -\\frac{b}{a}, \\ \\ x_1 \\cdot x_2 = \\frac{c}{a}$$ 2020-11-29", "+ (k+1)/2$. That is, you are given a statement of the form: $$a \\ge b$$ and are trying to establish a statement of the form $$a\\cdot c \\ge d$$ So you need to establish $b \\ge \\frac dc$, that is, you need to establish: $$1 + \\frac k2 \\ge \\dfrac{1 + \\dfrac{k + 1}{2}}{1 + \\dfrac 12}$$ Should be straightforward. Continue expanding the product. $$(1 + \\frac{1}{2})^{k+1} =(1 + \\frac{1}{2})^k \\cdot (1 + \\frac{1}{2}) \\ge (1+\\frac{k}2)(1 + \\frac{1}{2}) = 1+\\frac{k}2 + \\frac12+\\frac{k}{4}>1+\\frac{k+1}{2}$$", "have real roots so, $(b-3)^{2}-4(b^{2}-3b+3) \\ge 0$ Simplifying the quadratic inequality above we obtain, $-3(b-1)^{2} \\ge 0$ Solving the quadratic inequality we observe that the only solution for b is, $\\boxed{b=1}$. Hence, solving for $a$ using $(3)$ we obtain, $a = \\frac{-1(1-3) \\pm \\sqrt{-3(1-1)^{2}}}{2} =\\frac{2}{2}$ $\\boxed{a=1}$ Therefore, $c$ can be easily calculated using equation $(1)$, $c = 3 - 1 -1 = 1$ The only real solution to this question is: # $\\boxed{ (a,b,c) = (1,1,1) }$ - 6 years ago Log in to reply Nice. Is there an easier way to get at this? Staff - 6 years ago Log in to reply $a+b+c=3\\\\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=3\\\\ which\\quad implies,{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca=0\\\\ =>\\frac { 1 }{ 2 } (({ a }^{ 2 }-2ab+{ b }^{ 2 })+({ a }^{ 2 }-2ac+{ c }^{ 2 })+({ b }^{ 2 }-2bc+{ c }^{ 2 }))=0\\\\ =>{ (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (a-c) }^{ 2 }=0\\\\ =>a=b=c=1$ - 6 years ago Log in to reply Incredible! Nice solution. Clean, short and clear. - 6 years ago Log in to reply thanks! - 6 years ago Log in to reply that's what i approached good job buddy !!! -", "# MIE 2015/Day 1/Problem 2 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem 2 The polynomial $x^3+ax^2+bx+c$ has real roots $\\alpha$, $-\\alpha$ and $\\frac{1}{\\alpha}$. Thus the value of the sum of $b+c^2+ac+\\frac{b}{c^2}$ is: (a) $-2$ (b) $-1$ (c) $0$ (d) $1$ (e) $2$ ## Solution By Girard's relations (also called Vieta's formulas or Newton's identities), $\\begin{cases}a=-\\frac{1}{\\alpha}\\\\b=-\\alpha^2\\\\c=\\alpha\\end{cases}$ $b+c^2+ac+\\frac{b}{c^2}$ $-\\alpha^2+\\alpha^2-\\frac{1}{\\alpha}\\cdot\\alpha+\\frac{-\\alpha^2}{\\alpha^2}=\\boxed{-2}~~~~\\boxed{\\text{A}}$", "# Difference between revisions of \"Mock AIME 3 Pre 2005 Problems/Problem 3\" ## Problem A function $f(x)$ is defined for all real numbers $x$. For all non-zero values $x$, we have $$2f\\left(x\\right) + f\\left(\\frac{1}{x}\\right) = 5x + 4$$ Let $S$ denote the sum of all of the values of $x$ for which $f(x) = 2004$. Compute the integer nearest to $S$. ## Solution Substituting $\\frac{1}{x}$, we have $$2f\\left(\\frac 1x\\right) + f\\left(x\\right) = \\frac{5}{x} + 4$$ This gives us two equations, which we can eliminate $f\\left(\\frac 1x\\right)$ from (the first equation multiplied by two, subtracting the second): \\begin{align*} 3f(x) &= 10x + 4 - \\frac 5x \\\\ 0 &= x^2 - \\frac{3 \\times 2004 - 4}{10}x + \\frac 52\\end{align*} Clearly, the discriminant of the quadratic equation $\\Delta > 0$, so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the $x$ term, so our answer is $\\left[\\frac{3 \\times 2004 - 4}{10}\\right] = \\boxed{601}$.", "text. I proceeded to solve the first part of the question as follows. I need to find $p$such that $f(x+p)=f(x)$. Thus, $f(x+p)=a(x+p)+b=ax+b=f(x) \\tex{ or } ax+ap+b=ax$. This implies $ap=0$ or, since $a\\neq0\\tex{, } p=0$. Hence, there is no p which satisfies the given definition. For the solution to the quadratic, I took a similar approach. Again, I need to find $p$such that $f(x+p)=f(x)$. Thus, $f(x+p)=a(x+p)^2+b(x+p)+c=ax^2+bx+c=f(x)$ or $ax^2+2xpa+ap^2+bx+bp+c=ax^2+bx+c$. This implies $ap^2+p(2xa+b)=0$ or, since $a\\neq0\\tex{, } p(p+2x+\\frac{b}{a})=0$. Thus $p=\\left\\{0,-2x-\\frac{b}{a}\\right\\}$. The first solution for $p$ is identical to the solution in the previous problem. However, the second solution for $p$ is a function of $x$. Is it because of the fact that $p$ depends on $x$, it is a function and not a number, that the second solution fails to give a period? Or is my interpretation of the definition incorrect? It makes sense that the period cannot be dependent on $x$, but I'm trying to make my solution fit the definition. Thus, I'm looking for the reason why $f(x)=a^2+bx+c$ has no period. Any assistance is appreciated. Last edited: Aug 10, 2008 20. ### temurman of no wordsRegistered Senior Member Messages: 1,330 Remember that the relation should hold for each x. This means that if you get a polynomial expression in x then all the coefficients for the powers of x", "# $x_1^3+3x_2+3x_3$ in terms of the roots $x_1$, $x_2$, $x_3$ of $x^3−3x−15=0$ Let $$x_1$$, $$x_2$$, $$x_3$$ be the roots of $$x^3−3x−15=0$$. Find $$x_1^3+3x_2+3x_3$$. I tried solving the problem using formulas from Vieta's theorem, but I was unable to find any plausible ways to calculate the end result. Does anyone know how to do this? $$3(x_1+x_2+x_3)=0$$ since the polynomial has no $$x^2$$ term. Thus, $$x_1^3+3x_2+3x_3=x_1^3-3x_1=15.$$ Let $$a,b,c$$ be the roots of $$x^3 -3x -15=0$$ Let $$I=a^3 + 3b + 3c$$ We have $$a^3 = 3a + 15$$ Since $$a+b+c=- \\frac{a_2}{a_3} = 0$$ (Vieta's formulas), $$I= 3a + 15 + 3b + 3c = 15 + 3(a+b+c) = 15$$", "some $j. In the former case, this can happen in $\\tbinom 32=3$ ways, and in the latter case, this can happen in $3^2=9$ ways. Hence, \\begin{align*} 19=3\\sum_{i=1}^{673} z_i^2+9\\sum_{1\\le j and the requested sum is $343+9=\\boxed{352}$. (Solution by TheUltimate123) ## Solution 4 Let $$(f(x))^3=x^{2019}+20x^{2018}+19x^{2017}+g(x).$$ Therefore, $f(x)=(x-z_{1})(x-z_{1})\\cdots (x-z_{673})$. This is also equivalent to $$f(x)=x^{673}+ax^{672}+bx^{671}+h(x)$$ for some real coefficients $a$ and $b$ and some polynomial $h(x)$ with degree $670$. We can see that the big summation expression is simply summing the product of the roots of $f(x)$ taken two at a time. By Vieta's, this is just the coefficient $b$. The first three terms of $(f(x))^3$ can be bashed in terms of $a$ and $b$ to get $$20 = 3a$$ $$19 = 3a^2+3b$$ Thus, $a=\\frac{20}{3}$ and $b=\\frac{1}{3}\\left(19-3\\left(\\frac{20}{3} \\right)^2 \\right)$. That is $|b|=\\frac{343}{9}=\\frac{m}{n}$. $m+n=343+9=\\boxed{352}$ ~jakeg314", "and $$x=6$$. If $$3x-8=7$$, we have $$3x= 15$$, and $$x=5$$. If $$4x - 5 - 3a = 15 + 9a$$, we have $$4x = 20 + 12a$$, and consequently $$x=5 + 3a$$. 578 When the first equation has the form $$\\dfrac{x}{a}=b$$, we multiply both sides by $$a$$, in order to have $$x = ab$$. But if it is $$\\dfrac{x}{a} + b - c = d$$, we must first make $$\\dfrac{x}{a} = d- b + c$$; after which we find $x=(d -b + c)a=ad-ab + ac.$ Let $$\\frac{1}{2}x-3=4$$, then $$\\frac{1}{2}x=7$$, and $$x= 14$$. Let $$\\frac{1}{3}x - 1 + 2a = 3 + a$$, then $$\\frac{1}{3}x = 4 - a$$, and $$x=12-3a$$. Let $$\\dfrac{x}{a-1}-1=a$$, then $$\\dfrac{x}{a-1} =a + 1$$, and $$x = a^2 - 1$$. 579 When we have arrived· at such an equation as $$\\dfrac{ax}{b} = c$$, we first multiply by $$b$$, in order to have $$ax =bc$$, and then dividing by $$a$$, we find $$x = \\dfrac{bc}{a}$$. If $$\\dfrac{ax}{b}- c =d$$, we begin by giving the equation this form, $$\\dfrac{ax}{b} = d + c$$; after which, we obtain the value of $$ax = bd + bc$$, and then that of $$x = \\dfrac{bd+bc}{a}$$. Let $$\\frac{2}{3}x - 4 = 1$$, then $$\\frac{2}{3}x = 5$$, and $$2x = 15$$; whence $$x=\\frac{15}{2}=7\\frac{1}{2}$$. If $$\\frac{3}{4}x+\\frac{1}{2}=5$$, we have $$\\frac{3}{4}x =5 -\\frac{1}{2} =\\frac{9}{2}$$; whence" ]

[ "$\\frac{3x - 2}{x+1}$ b) $f:\\ x\\ \\rightarrow\\ 2x\\ +\\ 3e^{x}$ Solution: To find the value of $f(-2)$ for these functions, we should put $x$ = $-2$ in each function and evaluate. a) $f(x)$ = $\\frac{3x - 2}{x+1}$ Putting $x$ = $-2$, $f(-2)$ = $\\frac{3\\times (-2) - 2}{-2+1}$ = $\\frac{-8}{-1}$ = $8$. b) $f:\\ x\\ \\rightarrow\\ 2x\\ +\\ 3e^{x}$ This can be written as $f(x)$ = $2x\\ +\\ 3e^{x}$ Putting $x$ = $-2$, $f(-2)$ = $2\\ \\times\\ -\\ 2\\ +\\ 3e^{-2}$ = $-3.59399415029$ Example 2: Match the functions in column A to their equivalent functions in column B. A B $f(x)$ = $\\frac{3x-4y}{3y+2x}$ $f(x)$ = $e^{3x} - 2x$ $f$ = ${(x, y)|y = 3x^3 - 2x^2 -1}$ $f\\overset{x}{\\rightarrow}(2x+3logx)$ $f:\\ x\\ \\rightarrow\\ 2x\\ +\\ 3logx$ $f\\overset{x,y}{\\rightarrow}$$(\\frac{3x-4y}{3y+2x})$ $f\\overset{x}{\\rightarrow}(e^{3x}-2x)$ $y$ = $3x^3 - 2x^2 -1$ Solution: The functions equivalent to each other are given here. $A1\\ \\rightarrow\\ B3$ $A2\\ \\rightarrow\\ B4$ $A3\\ \\rightarrow\\ B2$ $A4\\ \\rightarrow\\ B1$", "# Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question: If $$\\alpha$$ and $$\\beta$$ are the roots of the equation $$ax^2+bx+c=0$$... Then find the roots of the equation $$ax^2-bx(x-1)+c(x-1)^2=0$$ My Attempt: $$(a-b+c)x^2+(b-2c)x+c=0$$ Now $$\\text{Sum of roots}=\\dfrac{-b}{a} = \\dfrac{2c-b}{a-b+c}$$ And $$\\text{Product of roots}=\\dfrac{c}{a}=\\dfrac{c}{a-b+c}$$ But I don't seem to be going anywhere with the way I'm proceeding We look at the equation $$ax^2-bx(x-1)+c(x-1)^2=0.\\tag{1}$$ If $a\\ne 0$, then $1$ is not a solution, so Equation (1) is equivalent to $$a\\left(\\frac{x}{x-1}\\right)^2-b\\frac{x}{x-1}+c=0.\\tag{2}$$ If we put $y=-\\frac{x}{x-1}$ then the solutions of (2) are the solutions of $ay^2+by+c=0$. It follows that $$\\frac{-x}{x-1}=\\alpha\\quad\\text{or}\\quad \\frac{-x}{x-1}=\\beta.$$ Now we can easily solve for $x$. hint: $\\dfrac{2c-b}{a-b+c} = \\dfrac{2\\dfrac{c}{a}-\\dfrac{b}{a}}{1-\\dfrac{b}{a}+\\dfrac{c}{a}}=\\dfrac{2\\alpha\\cdot \\beta+(\\alpha+\\beta)}{1+\\alpha+\\beta+\\alpha\\cdot \\beta}$, can you continue? • Yeah...Thanks a lot! :) Jul 6, 2015 at 4:02 Let $\\alpha = \\dfrac{-b+\\sqrt{b^2-4ac}}{2a}$ and $\\beta = \\dfrac{-b-\\sqrt{b^2-4ac}}{2a}$. Note that $\\alpha + \\beta = -\\dfrac{b}{a}$ and $\\alpha\\beta = \\dfrac{c}{a}$. We want to calculate: $$\\dfrac{2c-b\\pm \\sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)}$$ In terms of $\\alpha$ and $\\beta$. $$\\dfrac{2c-b\\pm \\sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} = \\dfrac{2c-b\\pm \\sqrt{b^2-4bc+4c^2 -4ac + 4bc -4c^2}}{2(a-b+c)} =$$ $$= \\dfrac{a}{a-b+c}\\cdot\\dfrac{2c-b\\pm \\sqrt{b^2-4ac}}{2a} =\\dfrac{\\frac{c}{a}}{1-\\frac{b}{a}+\\frac{c}{a}} +\\dfrac{1}{1-\\frac{b}{a}+\\frac{c}{a}}\\cdot\\dfrac{-b\\pm \\sqrt{b^2-4ac}}{2a} =$$ $$= \\dfrac{\\alpha\\beta}{1+\\alpha+\\beta + \\alpha\\beta} +\\dfrac{1}{1+\\alpha+\\beta + \\alpha\\beta}\\cdot\\dfrac{-b\\pm \\sqrt{b^2-4ac}}{2a}$$ So the roots are: $$\\dfrac{\\alpha\\beta + \\alpha}{1+\\alpha+\\beta + \\alpha\\beta}\\qquad \\dfrac{\\alpha\\beta + \\beta}{1+\\alpha+\\beta + \\alpha\\beta}$$ Simplifying: $$\\dfrac{\\alpha}{\\alpha+1} \\qquad \\dfrac{\\beta}{\\beta+1}$$ • Sweet stuff. +1 Jul 6, 2015 at 9:53", "therefore be expressed as a function ##f## of variables ##a, b## alone. ##(a+1)(b+1)(c+1) \\equiv f(a, b) = (a+1)(b+1)\\left(\\dfrac{a+b+2}{ab-1} + 1\\right)## ##= (a+1)(b+1)\\dfrac{(a+1)(b+1)}{ab-1} = \\dfrac{(a+1)^2(b+1)^2}{ab-1}## (Eq. 2) Since ##a, b > 0##, it follows that the numerator of (Eq. 1) is positive. Since we also need ##c > 0##, we also need denominator in that equation to be positive and hence we must have ##ab > 1##. To find the minimum value of ##f(a, b)## for a fixed value of ##b##, find the partial derivative of this function w.r.t ##a## and equate to 0.##\\dfrac {\\partial f} {\\partial a} = 0 \\equiv \\dfrac{(b+1)^2 (a+1) (ab - 2 - b)}{(ab-1)^2} = 0 \\Rightarrow a = \\dfrac{2+b}{b}## (Eq. 3) That the above is a minimum, not a maximum, can be proven by looking at the sign of ##\\dfrac{\\partial^2 f} {\\partial a^2}## at ##a = \\frac{2+b}{b}##. ##\\dfrac{\\partial^2 f} {\\partial a^2} = \\dfrac{(b+1)^2}{(ab-1)^2} \\left( \\dfrac{-2b(a+1)(ab - 2 - b)}{ab - 1} + (2ab - 2) \\right)##, which when evaluated at ##a = \\dfrac{2+b}{b}## is equal to ##\\dfrac{1}{b+1} \\left( -2(2+2b)(2+b-2-b) + 2(b+1)^2 \\right) = 2(b+1)##. Since this value is positive for any positive value of ##b## (in fact it is positive for ##b > -1##), ##a = \\frac{2+b}{b}## must correspond to a local minimum, not a maximum. ##f(\\frac{2+b}{b}, b)## is therefore minimum value of ##f(a,", "$y=a+bx$ C: $y=ax+(b+2a)$ D: $y=bx+(a+2b)$ 7. If the line $y=2x-1$ is the equation of the slant asymptote for the rational function $\\frac{2x^2+bx+5}{x-1}$, the value of $b$ is $\\cdots\\cdots$ A: $-3$ B: $-1$ C: $3$ D: $1$ 8. Find the equation of the slant asymptote of the rational function $y=\\frac{5x^2+2x+1}{3x+7}$ A: $y=\\frac{5x}{3}-\\frac{29}{3}$ B: $y=\\frac{5x}{2}-\\frac{29}{9}$ C: $y=\\frac{5x}{2}-\\frac{35}{9}$ D: $y=\\frac{5x}{3}-\\frac{29}{9}$ 9. Find the equation of the slant asymptote of the rational function $y=\\frac{x^2-4x+4}{x-3}$ A: $y=x-1$ B: $y=1-x$ C: $y=x+1$ D: $y=x-2$ 10. Find the equation of the slant asymptote of the rational function $y=\\frac{x^2-a^2}{x-b}$ A: $y=x+a$ B: $y=x-a$ C: $y=x-b$ D: $y=x+b$", "## weak formulation multiply a function $v(x)$ on both sides of the original equation. \\begin{aligned}&-\\frac{d}{dx}\\left( c(x)\\frac{du(x)}{dx} \\right)=f(x),\\quad a<x<b\\\\ \\Rightarrow &-\\frac{d}{dx}\\left( c(x)\\frac{du(x)}{dx} \\right)v(x)=f(x)v(x), \\quad a<x<b \\\\ \\Rightarrow &-\\int_a^b\\frac{d}{dx}\\left( c(x)\\frac{du(x)}{dx} \\right)v(x)=\\int_a^b f(x)v(x).\\end{aligned} $u(x)$ is called a trial function and $v(x)$ is called a test funcion. using integeration by part, we obtain \\begin{aligned}&\\int_a^b \\frac{d}{dx}\\left(c(x)\\frac{du(x)}{dx}\\right)v(x)dx\\\\=&\\int_a^b(cu')'v\\quad dx\\\\=&cu'v|_a^b-\\int_a^bcu'dv\\end{aligned} then $$-c(b)u'(b)v(b)+c(a)u'(a)v(a)+\\int_a^bcu'dv=\\int_a^b fv dx$$ since the solution at $x=a$ and $x=b$are given by $u(a)=g_a,u(b)=g_b$, then we can choose the test function $v(x)$ such that $v(a)=v(b)=0$ hence $$\\int_a^bcu'v'dx=\\int_a^b fv dx$$ • Weak formulation: find $u\\in H^1(I)$ such that$$\\int_a^bcu'v'\\;dx=\\int_a^bfv\\;dx$$for any $v\\in H_0^1(I)$where $I=(a,b)$ Let $a(u,v)=\\int_a^bcu'v'\\;dx$ and $(f,v)=\\int_a^bfv\\;dx$. • Weak formulation: find $u\\in H^1(I)$ such that$$a(u,v)=(f,v)$$for any $v\\in H_0^1(I)$where $I=(a,b)$ Assume there is a finite dimensional subspace $U_h \\sub H^1(a,b).$ Define $U_{h0}$ to be the space which consists of the functions of $U_h$ with value 0 on the Dirichlet boundary. • Galerkin formulation: find $u_h \\in U_h$ such that $$a(u_h,v_h)=(f,v_h)$$ for any $v_h\\in U_{h0}$ ## Sobolve spaces • support • compactly support • $C_0^{\\infty}(I)$ is the set of all functions that are infinitely differentiable on $I$ and compactly supported in $I$ • for $v \\in C_0^\\infty$ ,we have $v(a)=v(b)=0$ $$\\int_a^bu'v'dx=-\\int_a^b fv dx \\tag{1}$$ • weak derivative $$\\int_a^b wv \\;dx=-\\int_a^buv'\\;dx$$ $u'$不存在，$w$替代$u'$，$w$叫做弱导数(weak dericative) • $L^2 space$ $$L^2(I)=\\left\\{v: I\\to R:\\int_a^bv^2dx<\\infty\\right\\}$$ where $I=(a,b)$ • $H^1 space$ $$H^1(I)=\\left\\{v\\in L^2(I):v'\\in L^2(I)\\right\\}$$ where $I=(a,b)，v'$ is weak", "# 1977 USAMO Problems/Problem 3 ## Problem If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. ## Solution This problem needs a solution. If you have a solution for it, please help us out by adding it. a,b,c,d are roots of equation $x^4+x^3-1=0$ then by vietas relation $$ab +bc+cd+da+ac+bd= 0$$ let us suppose $ab,bc,cd,da,ac,bd$ are roots of $x^6+x^4+x^3-x^2-1=0$. then sum of roots $= ab +bc+cd+da+ac+bd=c/a = -b/a=0$ sum taken two at a time $= ab\\times bc + bc\\times ca +..........=c/a=1$ similarly we prove for the roots taken three four five and six at a time to prove $ab,bc,cd,da,ac,bd$ are roots of second equation Given the roots $a,b,c,d$ of the equation $x^{4}+x^{3}-1=0$. First, $a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1$. Then $cd=-\\frac{1}{ab}$ and $c+d=-1-(a+b)$. Remains $ab+(a+b)(c+d)+cd = 0$ or $ab+(a+b)[-1-(a+b)]-\\frac{1}{ab}=0$. Let $a+b=s$ and $ab=p$, so $p+s(-1-s)-\\frac{1}{p}=0$(1). Second, $a$ is a root, $a^{4}+a^{3}=1$ and $b$ is a root, $b^{4}+b^{3}=1$. Multiplying: $a^{3}b^{3}(a+1)(b+1)=1$ or $p^{3}(p+s+1)=1$. Solving $s= \\frac{1-p^{4}-p^{3}}{p^{3}}$. In (1): $\\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0$. $p^{8}+p^{5}-2p^{4}-p^{3}+1=0$ or $(p-1)(p+1)(p^{6}+p^{4}+p^{3}-p^{2}-1)= 0$. Conclusion: $p =ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.", "[ ]: f in SR Then we can ask to perform operations, like expanding the formula we entered above: In [ ]: expand(f) We can also try to check whether two formulas are equivalent: In [ ]: t1 = (a^2+2*a*b+b^2)/(a+b); t1 In [ ]: t2 = a+b; t2 In [ ]: t1 == t2 In [ ]: bool(t1 == t2) This is not restricted to algebraic equations: In [ ]: bool(sin(a)^2 + cos(a)^2 == 1) #### Exercise Try to check the following identities, some of which are true and some of which are false. Which one does SageMath get wrong? $a \\leq \\max(a,b) \\\\ \\exp(a) \\cdot \\exp(b) = \\exp(a+b) \\\\ \\frac{1}{\\frac{1}{a}-\\frac{1}{b}} = \\frac{a \\cdot b}{a - b} \\\\ \\frac{1}{2}(a+b) \\leq \\max(a,b) \\\\ (a+b+c)^2 = a^2 + b^2 + c^2 + 2 \\cdot(ab + ac + bc) \\\\ \\sin\\left(\\frac{\\pi}{4}\\right) = \\frac{1}{\\sqrt{2}}$ Solution (uncomment to see) ### Substitution We can make variable substitutions as follows: In [ ]: var('a,b,c,d') f = a^2 + b*c; f In [ ]: f.subs({a:b, c:b+1}) Note that the argument of f.subs is a dictionary, sending variables appearing in f to their new values. Alternatively, we can treat f as if it was a function, and call it (but specifying the names of the variables as follows): In [ ]: f(a=2,b=3,c=4) In [ ]: f(a=d,b=d,c=d) ###", "with zero and form an equation. Then we solve the equation. The roots of an equation are the roots of a function. Suppose the given polynomial is f(x)=2x+1 and we have to find the zero of the polynomial. Now equating the function with zero we get, $2x+1=0$ or, $2x=-1$ or, $x=- \\frac{1}{2}$ Therefore the zero of the polynomial 2x+1 is $x=- \\frac{1}{2}$. Example: Find the root of the function $\\frac{x}{a}-\\frac{x}{b}-a+b$. First, we equate the function with zero and form an equation. Then we solve the equation and find x. $\\frac{x}{a}-\\frac{x}{b}-a+b=0$ or, $\\frac{x}{a}-\\frac{x}{b}=a-b$ or, $\\frac{bx-ax}{ab}=a-b$ or, $\\frac{x(b-a)}{ab}=-\\left ( b-a \\right )$ or, $x(b-a)=-ab\\left ( b-a \\right )$ or, $x=\\frac{-ab(b-a)}{(b-a)}$ canceling (b-a) we get, or, $x=-ab$ There the zeros or roots of a function is $-ab$. ### How to find the zeros of a function on a graph This method is the easiest way to find the zeros of a function. In this method, we have to find where the graph of a function cut or touch the x-axis (i.e., the x-intercept). The points where the graph cut or touch the x-axis are the zeros of a function. Now look at the examples given below for better understanding Question: How to find the zeros of a function on a graph y=x. Here the graph of the function y=x cut the", "Vieta's Formulas – IMT DeCal # Vieta’s Formulas by Suraj Rampure Vieta’s formulas give us a way to interpret a polynomial in standard form, e.g. $p(x) = 4x^2 + 3x + 3$, in terms of its roots, without having to find the roots specifically. While there are more practical applications to this, the derivation of Vieta’s formulas prove to be an interesting combinatoral problem. ## Naive Approach, using the Quadratic Formula Let $p(x) = ax^2 + bx + c$. We know that $p(x)$ will have two roots, i.e. solutions to $p(x) = 0$ (they might be complex and/or equal). Suppose $p(x)$'s roots are $r_1$ and $r_2$, and suppose we want to find the sum ($r_1 + r_2$) and product ($r_1 r_2$) of these roots. One way to do this would be to use the quadratic formula to solve for both roots, and simplify. $r_1, r_2 = \\frac{-b + \\sqrt{b^2 - 4ac}}{2a}, \\frac{-b - \\sqrt{b^2 - 4ac}}{2a}$ Then: $r_1 + r_2 = \\frac{-b + \\sqrt{b^2 - 4ac} - b - \\sqrt{b^2 - 4ac}}{2a} = -\\frac{2b}{2a} = -\\frac{b}{a}$ $r_1r_2 = \\left( \\frac{-b + \\sqrt{b^2 - 4ac}}{2a} \\right)\\left( \\frac{-b - \\sqrt{b^2 - 4ac}}{2a} \\right) = \\frac{b^2 - (b^2 - 4ac)}{4a^2} = \\frac{4ac}{4a^2} = \\frac{c}{a}$ We’ve found that the sum of the roots is $-\\frac{b}{a}$, and the product of the roots is", "order recurrence relations. Define $b_0=0$ and $b_n=b_{n-1}+a$ for $n\\geq 1$. In the same way define $c_0=1$ and $c_{n}=c_{n-1}+a$ for $n\\geq 1$. Now we get $F_{2n}=b_n$ and $F_{2n+1}=c_n$, the first couple of terms of the generating function look like $$F_0 + F_{1} x + F_{2} x^2 + F_{3} x^3 + \\cdots = b_0 + c_0 x + b_1 x^2 + c_1 x^3 + \\dots.$$ If you let $B(x), C(x)$ and $F(x)$ correspond to the generating functions of $b_n,c_n$ and $F_n$ respectively you see that $$F(x) = B(x^2) + x\\cdot C(x^2).$$ You can easily solve $b_n= a\\cdot n$ from which you should be able to get $$B(x)= \\frac{a x}{(x-1)^2}.$$ The other recurrence relation can just as easily be solved as $c_n=1+a\\cdot n.$ So we get $$C(x) = (1 + x^2 + x^3 + \\cdots) + B(x) = \\frac{1}{1-x} + \\frac{a x}{(x-1)^2}.$$ Now put it all together $$F(x) = \\frac{a x^2}{(x^2-1)^2} + x\\left(\\frac{1}{1-x^2} + \\frac{a x^2}{(x^2-1)^2}\\right).$$ This can be simplified to $$F(x)=\\frac{x \\left((a-1) x+1\\right)}{(x-1)^2 (x+1)}.$$", "# Difference between revisions of \"1997 AIME Problems/Problem 12\" ## Problem The function $f$ defined by $f(x)= \\frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\\frac{-d}{c}$. Find the unique number that is not in the range of $f$. ## Solution ### Solution 1 First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have $\\frac {a\\frac {ax + b}{cx + d} + b}{c\\frac {ax + b}{cx + d} + d} = x$, which reduces to $$\\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\\frac {px + q}{rx + s} = x.$$ In order for this fraction to reduce to $x$, we must have $q = r = 0$ and $p = s\\not = 0$. From $c(a + d) = b(a + d) = 0$, we get $a = - d$ or $b = c = 0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$, so $f(x) = \\frac {ax + b}{cx - a}$. The only value that is not in the range of this function is $\\frac {a}{c}$. To find $\\frac {a}{c}$, we use the two values of the", "# Difference between revisions of \"1997 AIME Problems/Problem 12\" ## Problem The function $f$ defined by $f(x)= \\frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\\frac{-d}{c}$. Find the unique number that is not in the range of $f$. ## Solution ### Solution 1 First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have $\\frac {a\\frac {ax + b}{cx + d} + b}{c\\frac {ax + b}{cx + d} + d} = x$, which reduces to $$\\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\\frac {px + q}{rx + s} = x.$$ In order for this fraction to reduce to $x$, we must have $q = r = 0$ and $p = s\\not = 0$. From $c(a + d) = b(a + d) = 0$, we get $a = - d$ or $b = c = 0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$, so $f(x) = \\frac {ax + b}{cx - a}$. The only value that is not in the range of this function is $\\frac {a}{c}$. To find $\\frac {a}{c}$, we use the two values of the", "+0 # Help. Thank you. +3 75 2 +294 Let a and b be the solutions of the equation $$2x^2-10x+5=0$$. What is the value of $$(2a-3)(4b-6)$$? Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ Feb 20, 2021 #1 +944 +2 Once again, we can use Vieta's equations. The sum of the roots is $$-\\frac{b}{a} = -\\frac{-10}{2} = 5$$. The product of the roots is $$\\frac{c}{a} = \\frac{5}{2}$$. Now, we can simplify: $$(2a-3)(4b-6) = 8ab - 12(a+b) + 18 = 8(\\frac{5}{2}) - 12(5) + 18 = \\boxed{-22}$$ CubeyThePenguin Feb 20, 2021 #2 +294 +3 Thank you! calvinbun Feb 23, 2021", "months ago The function $M(a,b)$ is minimized when both of its arguments equal each other. That is: $3a^2 + 2b = 3b^2 + 2a \\Rightarrow 3a^2 - 3b^2 + 2b - 2a = 0 \\Rightarrow 3(a+b)(a-b) - 2(a-b) = 0 \\Rightarrow (a-b)[3(a+b) - 2] = 0$ or when: $a = b$ (i), $a+b = \\frac{2}{3}$ (ii). Substituting (i) into either argument results in the quadratic equation: $f(a) = 3a^2 + 2a$ whose first & second derivavtives yield: $f'(a) = 6a + 2 = 0 \\Rightarrow a = -\\frac{1}{3}$; $f''(-\\frac{1}{3}) = 6 > 0$ (hence, a global minimum) which produces the critical pair $P_{1}(a,b) = (-\\frac{1}{3}, -\\frac{1}{3})$. Likewise, substituting (ii) produces: $g(a) = 3a^2 + 2(\\frac{2}{3} - a) = 3a^2 - 2a + \\frac{4}{3}$; $g'(a) = 6a - 2 = 0 \\Rightarrow a = \\frac{1}{3}$; $g''(\\frac{1}{3}) = 6 > 0$ (also a global minimum) and the critical pair $P_{2}(a,b) = (\\frac{1}{3}, \\frac{1}{3})$. Checking both critical pairs against $M(a,b)$ now results in: $M(-\\frac{1}{3}, -\\frac{1}{3}) = 3(-\\frac{1}{3})^2 + 2(-\\frac{1}{3}) = \\boxed{-\\frac{1}{3}}$. $M(\\frac{1}{3}, \\frac{1}{3}) = 3(\\frac{1}{3})^2 + 2(\\frac{1}{3}) = \\boxed{1}$. Since $M_{P_{1}} \\le M_{P_{2}}$, $\\boxed{a = b = -\\frac{1}{3}}$ are the critical values. Now this is a solution I can understand ! However, the rather bold statement at the beginning surely needs some explaining. i.e. \"The function M(a,b) is minimized when both", "# Difference between revisions of \"1997 AIME Problems/Problem 12\" ## Problem The function $f$ defined by $f(x)= \\frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\\frac{-d}{c}$. Find the unique number that is not in the range of $f$. ## Solution ### Solution 1 First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have ${\\frac {a\\frac {ax + b}{cx + d} + b}{c\\frac {ax + b}{cx + d} + d} = x}$, which reduces to ${\\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = \\frac {ex + f}{gx + h} = x}$. In order for this fraction to reduce to $x$, we must have $f = g = 0$ and $e = h\\not = 0$. From $c(a + d) = b(a + d) = 0$, we get $a = - d$ or $b = c = 0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$, so $f(x) = \\frac {ax + b}{cx - a}$. The only value that is not in the range of this function is $\\lim_{x\\to - d/c}f(x) = \\frac {a}{c}$. To find $\\frac {a}{c}$, we use", "## Calculus: Early Transcendentals 8th Edition The function $y=Ax^2+Bx+C$ would satisfy the differential equation if $A=-\\frac{1}{2}$, $B=-\\frac{1}{2}$ and $C=-\\frac{3}{4}$ $$y''+y'-2y=x^2$$ Consider the function $y=Ax^2+Bx+C$ - First derivative: $y'=2Ax+B$ - Second derivative: $y''=2A$ To find $A, B, C$, we need to apply the above function and derivatives to the differential equation: $$2A+(2Ax+B)-(2Ax^2+2Bx+2C)=x^2$$ $$2A+2Ax+B-2Ax^2-2Bx-2C=x^2$$ $$(-2Ax^2)+(2A-2B)x+(2A+B-2C)=x^2+0x+0$$ Now we see that, to satisfy the differential equation, the followings must be satisfied: 1) $-2A=1$ So, $A=-\\frac{1}{2}$ 2) $(2A-2B)=0$. Therefore, $A-B=0$. So, $B=A=-\\frac{1}{2}$ 3) $2A+B-2C=0$. Therefore, $C=A+\\frac{B}{2}=-\\frac{1}{2}-\\frac{1}{4}=-\\frac{3}{4}$ These are the values of A, B and C we need to find.", "# Difference between revisions of \"Mock AIME 3 Pre 2005 Problems/Problem 3\" ## Problem A function $f(x)$ is defined for all real numbers $x$. For all non-zero values $x$, we have $$2f\\left(x\\right) + f\\left(\\frac{1}{x}\\right) = 5x + 4$$ Let $S$ denote the sum of all of the values of $x$ for which $f(x) = 2004$. Compute the integer nearest to $S$. ## Solution Substituting $\\frac{1}{x}$, we have $$2f\\left(\\frac 1x\\right) + f\\left(x\\right) = \\frac{5}{x} + 4$$ This gives us two equations, which we can eliminate $f\\left(\\frac 1x\\right)$ from (the first equation multiplied by two, subtracting the second): \\begin{align*} 3f(x) &= 10x + 4 - \\frac 5x \\\\ 0 &= x^2 - \\frac{3 \\times 2004 - 4}{10}x + \\frac 52\\end{align*} Clearly, the discriminant of the quadratic equation $\\Delta > 0$, so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the $x$ term, so our answer is $\\left[\\frac{3 \\times 2004 - 4}{10}\\right] = \\boxed{601}$.", "the two values of the function given to us. We get $2(97)a + b = 97^2c$ and $2(19)a + b = 19^2c$. Subtracting the second equation from the first will eliminate $b$, and this results in $2(97 - 19)a = (97^2 - 19^2)c$, so ${\\frac {a}{c} = \\frac {(97 - 19)(97 + 19)}{2(97 - 19)} = \\boxed{058}$ (Error compiling LaTeX. ! Missing } inserted.). ### Solution 2 First, we note that $e = \\frac ac$ is the horizontal asymptote of the function, and since this is a linear function over a linear function, the unique number not in the range of $f$ will be $e$. $\\frac{ax+b}{cx+d} = \\frac{b-\\frac{cd}{a}}{cx+d} + \\frac{a}{c}$. Without loss of generality, let $c=1$, so the function becomes $\\frac{b- \\frac{d}{a}}{x+d} + e$. (Considering $\\infty$ as a limit) By the given, $f(f(\\infty)) = \\infty$. $\\lim_{x \\rightarrow \\infty} f(x) = e$, so $f(e) = \\infty$. $f(x) \\rightarrow \\infty$ as $x$ reaches the vertical asymptote, which is at $-\\frac{d}{c} = -d$. Hence $e = -d$. Substituting the givens, we get $\\begin{eqnarray*} 17 &=& \\frac{b - \\frac da}{17 - e} + e\\\\ 97 &=& \\frac{b - \\frac da}{97 - e} + e\\\\ b - \\frac da &=& (17 - e)^2 = (97 - e)^2\\\\ 17 - e &=& \\pm (97 - e) \\end{eqnarray*}$ Clearly we can discard the positive root,", "\\frac{-3 \\pm \\sqrt{8}}{2}$. ## The discriminant of the quadratic equation The discriminant of the quadratic equation $ax^2 + bx + c$ is the number $D$: $$D=b^2 – 4ac.$$ Then the solutions of the quadratic equation written by using the discriminant are: $$x_{1,2}= \\frac{-b \\pm \\sqrt{D}}{2a}.$$ If $\\ D > 0$, then the quadratic equation thus has two distinct real solutions, if $\\ D < 0$, the quadratic equation has two solutions that are complex conjugates , and if $\\ D = 0$ the quadratic equation has one real solution of multiplicity two. ## Vieta’s formulas French mathematician François Viète also studied quadratic equation and came to an important relation between quadratic equation and system of two equations with two unknowns. Those unknowns are the solutions to observed quadratic equation. The solutions $x_1$ and $x_2$ of the quadratic equation $\\ ax^2 + bx + c = 0$ meet the criteria of the Vieta’s formulas: $$x_1 + x_2 = -\\frac{b}{a},$$ $$x_1 \\cdot x_2 = \\frac{c}{a}.$$ Example 9. If $\\ x = 3$ is one solution to the equation $\\ x^2 + bx + 8 = 0$, what is the other solution and what is $b$? Solution: According to the Vieta’s formulas we find that: $$x_1 + x_2 = -\\frac{b}{1}$$ $$x_1 \\cdot x_2 = \\frac{8}{1}.$$ Since we already have one solution", "is on the graph. [3] There is a critical point at $x = 2.$ Since $f(0) = 2:\\;\\;a\\cdot0^3 + b\\cdot0^2 + c\\cdot0 + d\\:=\\:2\\quad\\Rightarrow\\quad d = 2$ . . The function (so far) is: . $f(x)\\;=\\;ax^3 + bx^2 + cx + 2$ Since $f(-2) = 0:\\;\\;a(-2)^3 + b(-2)^2 + c(-2) + 2\\;=\\;0$ . . $-8a + 4b - 2c + 2\\:=\\:0\\quad\\Rightarrow\\quad 8a - 4b + 2c\\:=\\:2$ [1] The derivative is: . $f'(x) \\:=\\:3ax^2 + 2bx + c$ Since $f'(2) = 0:\\;\\;3a\\cdot2^2 + 2b\\cdot2 + c\\:=\\:0\\quad\\Rightarrow\\quad 12a + 4b + c \\:=\\:0$ [2] Add [1] and [2]: . $20a + 3c\\:=\\:2\\quad\\Rightarrow\\quad c\\,=\\,\\frac{2-20a}{3}$ Substitute into [2]: . $12a + 4b + \\left(\\frac{2 - 20a}{3}\\right)\\:=\\:0\\quad\\Rightarrow\\quad b = \\frac{-8a - 1}{6}$ Therefore: . $\\boxed{f(x) \\;= \\;ax^3 - \\left(\\frac{8a + 1}{6}\\right)\\!x^2 + \\left(\\frac{2 - 20a}{3}\\right)\\!x \\,+ \\,2\\,}$ And, if there is no more information, that's the best we can do. Edit: corrected that silly \"3\" . . . Thanks for the heads-up! 5. Originally Posted by Soroban . . $-8a + 4b - 2c + 2\\:=\\:0\\quad\\Rightarrow\\quad 8a - 3b + 2c\\:=\\:2$ [1] wouldn't $-8a + 4b - 2c + 2\\:=\\:0\\quad\\Rightarrow\\quad8a - 4b + 2c \\:=\\:2$ instead of $8a - 3b + 2c\\:=\\:2$ 6. ## thanks thanks for checking my answer I reworked it and got the right answer." ]

[ "Determine coefficients $a$ and $b$ such that $p(x)=x^{2}+ax+b$ satisfies $p(1)=-3$ and $p'(1)=-4$. $a =$ $b =$", "###### back to index | new Let polynomials $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ satisfy: $$P(x^5) + xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$$ Prove: $(x-1) | P(x)$ Determine all polynomials such that $P(0) = 0$ and $P(x^2 + 1) = P(x)^2 + 1$. Prove there cannot exist a $998$-degree polynomial with real number coefficients $P(x)$ such that $$[P(x)]^2-1=P(x^2+1)$$ holds for any $x\\in\\mathbb{C}$.", "# Write the standard form of a cubic polynomial with real coefficients. Question: Write the standard form of a cubic polynomial with real coefficients. Solution: The most general form of a cubic polynomial with coefficients as real numbers is of the form $f(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real number and $a \\neq 0$", "$P(x)$ will be $>0$ for all real $x$ if $c-b^2>0\\iff c>b^2$ Conversely, if $c>b^2, P(x)=(x+b)^2+c-b^2>0$ for all real $x$", "# Let P denote the set of all polynomials in the real variable x which varies over the interval [0, 1]. What is the closure of P in C[0, 1] Let P denote the set of all polynomials in the real variable x which varies over the interval [0, 1]. What is the closure of P in C[0, 1] (with its usual supnorm topology)", "+ x^2 + 1$ is 4? Degree of a polynomial $P(x)$ is the highest power of $x$ in $P(x)$. Therefore, $x^3 + x^{a-4} + x^2 + 1$, $x^{a-4}$ = $x^4$ $a-4$ = $4$, $a$ = $4 + 4$ = $8$ Consider, $P(x)$ = $x^2 – 3x + 2$, Put $x$ = $3$ in $P(x)$ which gives, $P(3)$ = $9 – 9 + 2$ = $2$ Replace $x$ by 2 in the polynomial $x^2 – 3x + 2$, which gives $P(2)$ = $4 – 6 + 2$ = $0$. Similarly, value of $x^2 – 3x + 2$ at $x$ = $0$ is, $P(0)$ = $0 – 0 + 2$ = $2$ In general; if P(x) is a polynomial in x and k is any real number, then value of $P(k)$ at $x$ = $k$ is denoted by $P(k)$ is found by replacing $x$ by $k$ in $P(x)$. In the polynomial $x^2 – 3x + 2$, Replacing $x$ by 1 gives, $P(1)$ = $1 – 3 + 2$ = $0$ Similarly, replacing $x$ by 2 gives, $P(2)$ = $4 – 6 + 2$ = $0$ For a polynomial $P(x)$, real number k is said to be zero of polynomial $P(x)$, if $P(k)$ = $0$. Therefore, 1 and 2 are the zeros of polynomial $x^2 – 3x + 2$. Consider, $P(x)$", "## Fall 2015, problem 3 For a fixed $N \\in \\mathbb{N}$, find all the real polynomials $P(x)$ such that $(P \\circ P)(x) = (P(x))^N$.", "# Solve a Polynomial in R A polynomial p(x) is an expression of the form: $$p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + … + a_nx^n$$ Where n is any non-negative integer. ## Solve a polynomial p(x) in R To solve the equation $$p(x) = 0$$ in R, we can use the function: polyroot. For example, let’s solve the equation: $$p(x) = -\\sqrt{2} – x + x^2 + 0.4x^3 = 0$$ # we use the coefficients of p(x) as inputs to polyroot solutions = polyroot(c(-sqrt(2), -1, 1, 0.4)) print(solutions) # outputs: 1.3390011-0i -0.8977101+0i -2.9412910-0i ## Check if the solutions are real numbers So, we can see that the solutions of p(x) are all real numbers (associated with 0i), but we can check for this in R: # extracting the imaginary parts of the solutions Im(solutions) # outputs: -3.333339e-18 6.981698e-18 -3.648359e-18 # Which are numbers very close to 0, # To get them to be exactly 0, we need to round our answer: round(Im(solutions), 5) # outputs: 0 0 0 # Finally, the following code outputs: # TRUE if the solutions are real # FALSE otherwise all(round(Im(solutions), 5) == 0) # outputs: TRUE # With the help of this code, # we can remove the imaginary parts of the solutions # if they are real numbers if(all(round(Im(solutions),", "+ n^3r^2)$. See Michael Sagraloff and Kurt Mehlhorn, Computing Real Roots of Real Polynomials. – Matt F. Nov 19 '18 at 1:45 • Here is the link for the paper mentioned by Matt's comment: people.mpi-inf.mpg.de/~msagralo/RealRootComputation.pdf – Ali Enayat Nov 19 '18 at 21:44", "Let $n$ be a positive integer. Given that the polynomial $P(x)$ satisfies the relation $2 P(x+1) – P(x) = x(x-1)(x-2)\\cdots(x-n)$ for all $x$, find the value of $P(0)$.", "for every real $x$, then $f(x)\\geqslant0$ for every real $x$. It follows that Fact 2: If $f(x)$ is a polynomial with real coefficients, and $f(x) +f'(x) \\geqslant0$ for every real $x$, then $f(x)\\geqslant0$ for every real $x$. To prove Fact 2, let $g(x) = f(-x)$. Then $g'(x) = -f'(-x)$ and so $g(x) - g'(x) = f(-x) + f'(-x) \\geqslant0$ for every real $x$. It follows from Fact 1 that $g(x)\\geqslant0$ for every real $x$. Therefore $f(x)\\geqslant0$ for every real $x$. Write $D= \\frac d{dx}$ for the operator of differentiation, and $I$ for the identity operator. Then those two Facts can be written as Fact 1: If $f(x)$ is a polynomial with real coefficients, and $(I-D)f(x) \\geqslant0$ for every real $x$, then $f(x)\\geqslant0$ for every real $x$; Fact 2: If $f(x)$ is a polynomial with real coefficients, and $(I+D)f(x) \\geqslant0$ for every real $x$, then $f(x)\\geqslant0$ for every real $x$. Now let $p(x)$ be a polynomial with real coefficients such that $p(x) - p'(x) - p''(x) + p'''(x) \\geqslant0$ for every real $x$. That condition says that $(I-D - D^2 + D^3)p(x)\\geqslant0$ for every real $x$. But $I-D-D^2+D^3 = (I-D)(I-D^2)$. Therefore $(I-D - D^2 + D^3)p(x) = (I-D)\\bigl((I-D^2)p(x)\\bigr) \\geqslant0$ for every real $x$. It follows from Fact 1 that $(I-D^2)p(x) \\geqslant0$ for every real $x$. But $I-D^2 = (I-D)(I+D)$.", "can be factored as a product of linear and quadratic factors. If $p$ is a real polynomial and $p(a+ib) = 0$, then $p(a-ib) = 0$, so $p(z)$ is divisible by $(z-(a+ib))(z-(a-ib)) = z^2-2az+(a^2+b^2)$.", "# $P(x)$ with coefficient $\\pm1$ Find all polynomial $P(x)$ with coefficient $\\pm1$ and have all real roots. My attempted work : Let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ $-P(x) = -a_nx^n - a_{n-1}x^{n-1} - ... - a_1x - a_0$ $P(x)$ have all real roots $\\Leftrightarrow -P(x)$ have all real roots. WLOG, $a_n = 1$ $P(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ Let $r_1, r_2, ..., r_n$ be roots of the equation. $P(x) = (x-r_1)(x-r_2)...(x-r_n)$", "suppose that $p(x)$ is a polynomial of odd degree. If the leading coefficient of $p(x)$ is positive, then $$\\lim_{x\\to\\infty}p(x)=\\infty\\quad\\text{ and }\\quad\\lim_{x\\to -\\infty}p(x)=-\\infty\\;,$$ while if the leading coefficient is negative, then $$\\lim_{x\\to\\infty}p(x)=-\\infty\\quad\\text{ and }\\quad\\lim_{x\\to -\\infty}p(x)=\\infty\\;.$$ In either case we can find real numbers $a$ and $b$ such that $p(a)<0$ and $p(b)>0$, and the intermediate value theorem then ensures that $p(x)$ has a zero somewhere between $a$ and $b$. -", "× # Polynomial problem Let p(x)= $$a$$$$x^{2}$$ + $$b$$$$x$$ + $$c$$ be such that p(x) takes real values for real values of $$x$$ and non-real values for non-real values of $$x$$. Prove that $$a$$=0. I tried it out taking $$a$$,$$b$$ and $$c$$ to be complex values and then implemented the conditions given in the problem which only got me to a weird expression. Any correct approach ? Note by Nishant Sharma 4 years, 1 month ago Sort by: Hint: first prove that a, b, c are real. For this it suffices to use the fact that p(-1), p(0), p(1) are real. Specifically, express a, b & c in terms of these three values. Next, suppose a>0. Find a sufficiently small M<0 such that p(x) = M has no real roots, or equivalently, the discriminant for $$ax^2 + bx + (c-M)=0$$ is negative. Do something similar for a<0. · 4 years, 1 month ago Could not follow your second argument. Please explain a bit further. · 4 years, 1 month ago", "# Please explain Illustration: The quadratic polynomial p(x) has the following properties : p(x) can be positive or zero for all real numbers p(1) = 0 and p(2) = 2 Then find the quadratic polynomial. Sol: p(x) is positive or zero for all real numbers. Also p(1) = 0 Then we have p(x) = k(x-1)2, where k > 0 Now p(2) = 2$⇒$ k=2∴ p(x) = 2(x-1)2 Dear student Regards • 3 there is nothing to be explained in this question its too simple • -16 What are you looking for?", "The graph below is a polynomial function in the form $f(x) = (x-a)^2(x-b)(x-c).$ Find suitable unique real numbers $a$, $b$, and $c$ that describe the graph. Answer: $a =$ , $b =$ , and $c =$ Note: You can click on the graph to enlarge the image.", "$(m,n,p)$. In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{i\\frac{2\\pi}{3}}$ and $re^{i\\frac{4\\pi}{3}}$ being conjugates and since $m+n+p = 5$, (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\\boxed{\\textbf{(C) } 2}$. ~Murtagh FORMAT IT PROPERLY KID", "real coefficients and no real roots of the form $x^4+A x^2+ B$ can be factored as follows ( $p,s$ real ) $\\left(x^2-s x + p\\right)\\left(x^2+s\\text{ }x +p\\right)$ multply out to get $x^4+\\left(2p-s^2\\right)x^2+p^2$ and compare coefficients and note that $B$ must be positive otherwise the polynomial has real roots", "56 views Consider $p(s)=s^{3}+ a_{2}s^{2}+a_{1}s+a_{0}$ with all real coefficients. It is known that its derivatives ${p}'(s)$ has no real roots. The number of real roots of $p(s)$ is 1. $0$ 2. $1$ 3. $2$ 4. $3$" ]

[ "8y - 1 \\end{align*} $$2(x- 2y)(x^2 + xy + y^2)$$ \\begin{align*} 2(x- 2y)(x^2 + xy + y^2) &= 2(x^3 + x^2y + xy^2 - 2x^2y - 2xy^2 - y^3) \\\\ &= 2(x^3 - x^2y - xy^2 - y^3) \\\\ &=2x^3 - 2x^2y - 2xy^2 - 2y^3 \\end{align*} $$3(a-3b)(a^2 + 3ab - b^2)$$ \\begin{align*} 3(a-3b)(a^2 + 3ab - b^2) &= 3(a^3 + 3a^2b - ab^2 - 3a^2b - 9ab^2 + 3b^3) \\\\ &=3(a^3 - 10ab^{2} + 3b^3) \\\\ &= 3a^3- 30ab^{2} + 9b^3 \\end{align*} $$(2a- b)(2a + b)(2a^2 - 3ab + b^2)$$ \\begin{align*} (2a- b)(2a + b)(2a^2 - 3ab + b^2) &= (4a^2 - b^2)(2a^2 - 3ab + b^2) \\\\ &= 8a^4 - 12a^3b + 4a^2b^2 - 2a^2b^2 + 3ab^3 - b^4 \\\\ &= 8a^4 - 12a^3b + 2a^2b^2 + 3ab^3 - b^4 \\end{align*} $$2(3x + y)(3x - y) - (3x -y)^2$$ \\begin{align*} 2(3x + y)(3x - y) - (3x -y)^2 &= 2(9x^2 - y^2) - 9x^2 + 6xy - y^2 \\\\ &= 18x^2 - 2y^2 - 9x^2 + 6xy - y^2 \\\\ &= 9x^2 + 6xy - 3y^2 \\end{align*} $$(x+y)(x-3y) + (2x - y)^2$$ \\begin{align*} (x+y)(x-3y) + (2x - y)^2 &= x^2 - 3xy +xy - 3y^2 + 4x^2 - 4xy + y^2 \\\\ &=5x^2 - 6xy - 2y^2 \\end{align*} $$\\left(\\dfrac{x}{3} - \\dfrac{3}{x}\\right)\\left(\\dfrac{x}{4} + \\dfrac{4}{x}\\right)$$ \\begin{align*} \\left(\\frac{x}{3} -", "and $$\\ P\\left(\\left.\\xi_j=0\\,\\right|N=n\\right)=1-p\\$$—that is, if $$\\ i_1,i_2, \\dots, i_n\\$$ is a sequence of ones and zeros containing $$\\ m\\$$ ones and $$\\ n-m\\$$ zeros, then \\begin{align} P\\left(\\xi_1=i_1, \\xi_2=i_2, \\dots, \\xi_n=i_n |\\,N=n\\right)&=\\prod_{j=1}^nP\\left(\\xi_j=i_j|\\,N=n\\right)\\\\ &=p^m(1-p)^{n-m} \\end{align} Therefore \\begin{align} P \\left(\\xi_1+ \\xi_2+ \\dots +\\xi_n=m \\,|\\,N=n\\right)&=\\sum_{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n:\\\\ i_1+i_2+ \\dots+i_n=m} p^m(1-p)^{n-m}\\\\ &={n\\choose m} p^m(1-p)^{n-m}\\ , \\end{align} because $$\\ \\left|\\left\\{i_1,i_2, \\dots, i_n\\in \\{0,1\\}^n\\,|\\, i_1+i_2+ \\dots+i_n=m\\right\\}\\right|= {n\\choose m}\\$$.", "\\hat{\\vc{e}}_2 &= b_{12} \\vc{e}_1 + b_{22} \\vc{e}_2 + \\dots + b_{n,2} \\vc{e}_n, \\\\ &\\dots \\\\ \\hat{\\vc{e}}_n &= b_{1,n} \\vc{e}_1 + b_{2,n} \\vc{e}_2 + \\dots + b_{n,n} \\vc{e}_n, \\end{align} (6.79) where a particular vector $\\vc{v}$ has the following two representations \\begin{align} \\vc{v} &= v_1\\vc{e}_1 + v_2\\vc{e}_2 + v_3\\vc{e}_3+\\dots+v_n\\vc{e}_n = \\\\ &= \\hat{v}_1\\hat{\\vc{e}}_1 + \\hat{v}_2\\hat{\\vc{e}}_2 + \\hat{v}_3\\hat{\\vc{e}}_3+\\dots+\\hat{v}_n\\hat{\\vc{e}}_n, \\end{align} (6.80) and let $\\mx{B}$ be the matrix with $\\hat{\\vc{e}}_i$ as column vectors, then it holds that \\begin{align} \\vc{v} = \\mx{B}\\hat{\\vc{v}} = \\begin{pmatrix} b_{11} & b_{12} & \\dots & b_{1,n} \\\\ b_{21} & b_{22} & \\dots & b_{2,n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b_{n,1} & b_{n,2} & \\dots & b_{n,n} \\end{pmatrix} \\vc{\\hat{v}}, \\end{align} (6.81) where $\\vc{v}=(v_1,v_2,v_3,\\dots,v_n)$ and $\\hat{\\vc{v}}=(\\hat{v}_1,\\hat{v}_2,\\hat{v}_3,\\dots,\\hat{v}_n)$. We start by rewriting the representation of $\\vc{v}$ as \\begin{align} \\vc{v} &= \\sum_{i=1}^n x_i\\vc{e}_i \\\\ &= \\sum_{i=1}^n \\hat{x}_i\\hat{\\vc{e}}_i, \\end{align} (6.82) and $\\hat{\\vc{e}}_j$ as \\begin{align} \\hat{\\vc{e}}_j = \\sum_{i=1}^n b_{ij}\\vc{e}_i, \\end{align} (6.83) and then insert Equation (6.83) into line 2 of Equation (6.82), which results in \\begin{align} \\vc{v} = \\sum_{j=1}^n \\hat{x}_j\\hat{\\vc{e}}_j = \\sum_{j=1}^n \\hat{x}_j \\Biggl( \\sum_{i=1}^n b_{ij}\\vc{e}_i \\Biggr) = \\sum_{i=1}^n \\Biggl( \\sum_{j=1}^n b_{ij} \\hat{x}_j \\Biggr) \\vc{e}_i. \\end{align} (6.84) Since there is only a single set of coordinates (see Theorem 2.5 for the three-dimensional case) for a vector in a basis, and we have the following two ways of representing $\\vc{v}$, \\begin{align} \\vc{v} &=", "+ 2} + \\dots + (x + 1)^{n + 2(m - 1)} = \\dfrac{(x + 1)^n((x + 1)^{2m} - 1)}{(x + 1)^2 - 1} \\;,$$ and $${n - 1 \\choose n - 1} + {n + 1 \\choose n - 1} + {n + 3 \\choose n - 1} + {n + 5 \\choose n - 1} + \\dots + {n + 2m - 3 \\choose n - 1}$$ is the coefficient of $x^{n - 1}$ in $$(x + 1)^{n - 1} + (x + 1)^{n + 1} + \\dots + (x + 1)^{n + 2m - 3} = \\dfrac{(x + 1)^{n - 1}((x + 1)^{2m} - 1)}{(x + 1)^2 - 1} \\;.$$ Now \\begin{align*}&{n \\choose n - 1} - {n - 1 \\choose n - 1} + {n + 2 \\choose n - 1} - {n + 1 \\choose n - 1} +\\binom{n+4}{n-1}-\\binom{n+3}{n-1}+\\dots \\\\ &\\dots + {n + 2(m - 1) \\choose n - 1} - {n + 2m - 3 \\choose n - 1} \\end{align*} is the coefficient of $x^{n - 1}$ in \\begin{align*}&\\dfrac{(x + 1)^n((x + 1)^{2m} - 1)}{(x + 1)^2 - 1} - \\dfrac{(x + 1)^{n - 1}((x + 1)^{2m} - 1)}{(x + 1)^2 - 1}=\\\\ &\\dfrac{x(x + 1)^{n - 1}((x + 1)^{2m} - 1)}{(x + 1)^2 - 1}=\\\\ &x\\Big((x + 1)^{n - 1}", "10)\\\\ & = 2(x + 1)(x + 10) \\end{align*} $$6 a^{2} + 14 a + 8$$ \\begin{align*} 6 a^{2} + 14 a + 8 & = \\text{2} (3 a^{2} + 7 a + 4)\\\\ & = \\text{2} \\left(a + 1 \\right) \\left( 3 a + 4 \\right) \\end{align*} $$6 v^{2} - 27 v + 27$$ \\begin{align*} 6 v^{2} - 27 v + 27 & = \\text{3} (2 v^{2} - 9 v + 9)\\\\ & = \\text{3} \\left(2 v - 3 \\right) \\left( v - 3 \\right) \\end{align*} $$6 g^{2} - 15 g - 9$$ \\begin{align*} 6 g^{2} - 15 g - 9 & = \\text{3} (2 g^{2} - 5 g - 3)\\\\ & = \\text{3} \\left(g - 3 \\right) \\left( 2 g + 1 \\right) \\end{align*} $$3x^{2} + 19x + 6$$ $3x^{2} + 19x + 6 = (3x + 1)(x + 6)$ $$3x^{2} + 17x - 6$$ $3x^{2} + 17x - 6 = (3x - 1)(x + 6)$ $$7x^{2} - 6x - 1$$ $7x^{2} - 6x - 1 = (7x + 1)(x - 1)$ $$6x^{2} - 15x - 9$$ \\begin{align*} 6x^{2} - 15x - 9 & = 3(2x^{2} - 5x - 3)\\\\ & = 3(2x + 1)(x - 3) \\end{align*} $$a^2 - 7ab + 12b$$ $a^2 - 7ab + 12b^2 = (a - 4b)(a -3b)$ $$3a^2 + 5ab", "b _ i \\right).$ 2. $(a _ 0, \\dots, a _ {2 ^ N - 1}) \\bigoplus (b _ 0 - 1, \\dots, b _ {2 ^ N - 1}) = (c _ 0 - a _ 0, \\dots, c _ {2 ^ N - 1} - a _ {2 ^ N - 1}).$ 3. \\begin{align} (a _ 0 + a _ 1, \\dots, a _ {2 ^ N - 2} + a _ {2 ^ N - 1}) \\bigoplus (b _ 0 + b _ 1, \\dots, b _ {2 ^ N - 2} + b _ {2 ^ N - 1}) &= (c _ 0 + c _ 1, \\dots, c _ {2 ^ N - 2} + c _ {2 ^ N - 1}), \\\\ (a _ 0 - a _ 1, \\dots, a _ {2 ^ N - 2} - a _ {2 ^ N - 1}) \\bigoplus (b _ 0 - b _ 1, \\dots, b _ {2 ^ N - 2} - b _ {2 ^ N - 1}) &= (c _ 0 - c _ 1, \\dots, c _ {2 ^ N - 2} - c _ {2 ^ N - 1}) \\end{align} Proof. 1. and 2. are straightforward. To see the first line of 3., Let $a _ i ‘", "$$x$$: $$16\\left(x+1\\right)={x}^{2}\\left(x+1\\right)$$ \\begin{align*} 16\\left(x+1\\right) &= {x}^{2}\\left(x+1\\right) \\\\ 16x + 16 &= {x}^{3} + {x}^{2} \\\\ 0 &= {x}^{3} + {x}^{2} - 16x - 16 \\\\ \\text{Let } a(x) &= {x}^{3} + {x}^{2} - 16x - 16 \\\\ a(-1) &= (-1)^{3} + (-1)^{2} - 16(-1) - 16 \\\\ &= -1 + 1 +16 -16 \\\\ &= 0 \\\\ \\therefore a(x) &= (x + 1)(x^{2} - 16) \\\\ &= (x + 1)(x - 4)(x +4) \\\\ \\therefore 0 &= (x + 1)(x - 4)(x +4) \\\\ \\therefore x = -1 &\\text{ or } x = 4 \\text{ or } x= -4 \\end{align*} Show that $$x-2$$ is a factor of $$3{x}^{3}-11{x}^{2}+12x-4$$. \\begin{align*} \\text{Let } a(x) &= 3{x}^{3}-11{x}^{2}+12x-4 \\\\ a(2) &= 3(2)^{3}-11(2)^{2}+12(2)-4 \\\\ &= 24 - 44 +24 - 4 \\\\ &= 0 \\\\ \\therefore (x-2) &\\text{ is a factor of } a(x) \\end{align*} Hence, by factorising completely, solve the equation: $$3{x}^{3}-11{x}^{2}+12x-4=0$$ \\begin{align*} 3{x}^{3}-11{x}^{2}+12x-4 &= 0 \\\\ (x -2)(3x^{2} - 5x + 2) &= 0 \\\\ \\therefore (x - 2)(3x - 2)(x - 1) &= 0 \\\\ \\therefore x = 2 &\\text{ or } x = \\frac{2}{3} \\text{ or } x = 1 \\end{align*} $$2{x}^{3}-{x}^{2}-2x+2=Q\\left(x\\right).\\left(2x-1\\right)+R$$ for all values of $$x$$. What is the value of $$R$$? \\begin{align*} \\text{Let } a(x) &= 2{x}^{3}-{x}^{2}-2x+2 \\\\ R = a \\left( \\frac{1}{2} \\right) &= 2\\left( \\frac{1}{2}", ". \\left . \\begin{align} N_1(x_1) & = 1 \\\\ N_1(x_2) & = 0 \\\\ N_1(x_3) & = 0 \\\\ N_1(x_4) & = 0 \\\\ \\tfrac{}{} \\\\ N_2(x_1) & = 0 \\\\ N_2(x_2) & = 1 \\\\ N_2(x_3) & = 0 \\\\ N_2(x_4) & = 0 \\\\ \\tfrac{}{} \\\\ N_3(x_1) & = 0 \\\\ N_3(x_2) & = 0 \\\\ N_3(x_3) & = 1 \\\\ N_3(x_4) & = 0 \\\\ \\tfrac{}{} \\\\ N_4(x_1) & = 0 \\\\ N_4(x_2) & = 0 \\\\ N_4(x_3) & = 0 \\\\ N_4(x_4) & = 1 \\\\ \\end{align} \\right \\} \\Rightarrow \\begin{align} \\tfrac{}{} a_{11} + a_{12} x_1 + a_{13} x_1^2 + a_{14} x_1^3 & = 1 \\\\ \\tfrac{}{} a_{11} + a_{12} x_2 + a_{13} x_2^2 + a_{14} x_2^3 & = 0 \\\\ \\tfrac{}{} a_{11} + a_{12} x_3 + a_{13} x_3^3 + a_{14} x_3^3 & = 0 \\\\ \\tfrac{}{} a_{11} + a_{12} x_4 + a_{13} x_4^3 + a_{14} x_4^3 & = 0 \\\\ \\tfrac{}{} \\\\ \\tfrac{}{} a_{21} + a_{22} x_1 + a_{23} x_1^2 + a_{24} x_1^3 & = 0 \\\\ \\tfrac{}{} a_{21} + a_{22} x_2 + a_{23} x_2^2 + a_{24} x_2^3 & = 1 \\\\ \\tfrac{}{} a_{21} + a_{22} x_3 + a_{23} x_3^2 + a_{24} x_3^3 & = 0 \\\\ \\tfrac{}{} a_{21} + a_{22} x_4 + a_{23} x_4^2 + a_{24} x_4^3 & = 0 \\\\ \\tfrac{}{}", "'18 at 22:30 The OP is equivalent to: given \\begin{align} P(x)&=\\sum_{i=0}^{3n} a_i x^i \\tag{1}\\label{1} ,\\\\ P(3i-2)&=1,\\quad i=1,\\cdots,n \\tag{2}\\label{2} ;\\\\ P(3i-1)&=0,\\quad i=1,\\cdots,n \\tag{3}\\label{3} ;\\\\ P(3i)&=2,\\quad i=1,\\cdots,n \\tag{4}\\label{4} ;\\\\ P(3n+1)&=730 \\tag{5}\\label{5} ;\\\\ P(0)&=2 \\tag{6}\\label{6} , \\end{align} determine $n$. Conditions \\eqref{2}-\\eqref{5} define the system of $(3n+1)\\times(3n+1)$ linear equations, \\begin{align} Au&=v \\tag{7}\\label{7} , \\end{align} where $A$ is a special case of the $(3n+1)\\times(3n+1)$ Vandermonde matrix \\begin{align} A&=\\begin{bmatrix} 1&1&1&\\dots &1 \\\\ 1&2^1&2^2&\\dots &2^{3n} \\\\ 1&3^1&3^2&\\dots &3^{3n} \\\\ \\vdots &\\vdots &\\vdots &\\ddots &\\vdots \\\\ 1&(3n+1)^1&(3n+1)^2&\\dots &(3n+1)^{3n}\\\\ \\end{bmatrix} ,\\\\ \\text{or }\\quad A_{ij}&=(i+1)^{j} ,\\quad i,j=0,\\cdots,3n . \\end{align} $u$ is the vector of coefficients $a_i$ \\begin{align} u&=[a_0,\\cdots,a_{3n}]^{\\mathsf{T}} , \\end{align} and the right-hand side of \\eqref{7} is a vector of the form \\begin{align} v&=[\\underbrace{1,0,2,1,0,2,\\cdots,1,0,2}_{3n},730]^{\\mathsf{T}} ,\\\\ \\text{or }\\quad v_{3n}&=730 ,\\\\ v_{i}&= 2-(i+1\\mod 3) ,\\quad i=0,\\cdots,3n-1 . \\end{align} It follows from the condition \\eqref{6} that \\begin{align} a_0&=2 , \\end{align} For any $n$ all the coefficients $a_i$ of $P(x)$, including $a_0$, can be found as the solution of \\eqref{7}. A quick test reveals that $a_0=2$ for $n=4$. This is a log of Maxima session for the test: (%i1) a[i,j]:=(i+1)^j$(%i2) v[i]:=2-mod(i+1,3)$ (%i3) n:4$(%i4) A:apply('matrix,makelist(makelist(a[i,j],j,0,3*n),i,0,3*n))$ (%i5) V:makelist(v[i],i,0,3*n)$(%i6) V[3*n+1]:730$ (%i7) V; (%o7) [1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 730] (%i8) a0=invert(A)[1] . V; (%o8) a0 = 2 Let $\\omega:=\\exp\\left(\\frac{2\\pi\\text{i}}{3}\\right)=\\frac{-1+\\sqrt{-3}}{2}$. Define $$\\tau(z):=\\left(\\frac{1-\\omega^2}{3}\\right)z^2+\\left(\\frac{1-\\omega}{3}\\right)z+1\\text{ for all }z\\in\\mathbb{C}\\,.$$ Note that, for each", "both sides : \\begin{align} \\\\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\\\ x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\\\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\\\ \\end{align} factor $(x-1)$ and you're done ! • Your sums should end in $\\dots+x^2+x$, not $\\dots+x+x$. – Joonas Ilmavirta Aug 17 '14 at 10:10 • Ahh yes ! will fix it thanks :) – AgentS Aug 17 '14 at 10:11 Use the formula for sum of a geometric series: $$1+x+\\ldots +x^{n-2}+x^{n-1}=\\frac{x^n-1}{x-1}$$ • Isn't geometric series formula derived from what is to be proven here? – zbrojny120 Apr 20 at 15:14 You may just open hooks in right half of expression: $$(x - 1)(x^{n - 1} + x^{n - 2} + \\dots + x + 1) = x * (x^{n - 1} + x^{n - 2} + \\dots + x + 1) - 1 * (x^{n - 1} + x^{n - 2} + \\dots + x + 1) = (x^n + x^{n - 1} + \\dots + x^2 + x) - (x^{n - 1} + x^{n - 2} + \\dots + x + 1) = x^ n - 1$$ No problems) $$(x−1)(x_n−1+x_n−2+...+x+1)=x^n+x^{n-1}+.....+x^2+x-[x^{n-1}+x^{n-2}+....+x+1]=x^n-1$$", "as $$f_1$$ with different initial conditions. From $$g(n) = f_1(n-2) + f_1(n-4) + g(n-1)$$, we get the infinite series $$g(n) = f_1(n-2) + f_1(n-3) + 2f_1(n-4) + 2f_1(n-5) + \\dotsb$$, and this gives us a recurrence \\begin{align} f_1(n) &= f_1(n-1) + f_2(n-1) + 2g(n-1) \\\\ &= f_1(n-1) + f_1(n-3) + 2g(n-1) \\\\ &= f_1(n-1) + 3f_1(n-3) + 2f_1(n-4) + 4f_1(n-5) + 4f_1(n-6) + \\dotsb \\\\ \\end{align} Subtracting $$f_1(n-1)$$ from $$f_1(n)$$, we get $$f_1(n) - f_1(n-1) = \\color{red}{(f_1(n-1) + 3f_1(n-3) + 2f_1(n-4) + 4f_1(n-5) + 4f_1(n-6) + \\dotsb )} - \\color{blue}{(f_1(n-2) + 3f_1(n-4) + 2f_1(n-5) + 4f_1(n-6) + 4f_1(n-7) + \\dotsb )}$$ and most of the red terms cancel with blue terms, leaving us with $$f_1(n) - f_1(n-1) = f_1(n-1) - f_1(n-2) + 3f_1(n-3) - f_1(n-4) + 2f_1(n-5)$$ or $$f_1(n) = 2f_1(n-1) - f_1(n-2) + 3f_1(n-3) - f_1(n-4) + 2f_1(n-5)$$. As observed earlier, this also means that the recurrence $$f(n) = 2f(n-1) - f(n-2) + 3f(n-3) - f(n-4) + 2f(n-5)$$ holds. The first few nonzero terms of the sequence are $$1, 1, 2, 5, 10, 22, 49, 105, 227, 494, 1071, \\dots$$, as computed by @PeterKagey in the comments and in the upcoming OEIS listing. • I submitted this sequence as a draft to the OEIS. Once approved, it will be here. – Peter Kagey Nov 7 at 21:05", "y^3 + 3y^2z + z^3 + (y^2z - 2yz^2 + z^3)\\nonumber\\\\ &\\ge 2x^3 + 3x^2y - 2(x^2+z^2)y + y^3 + 3y^2z + z^3\\nonumber\\\\ &= 2x^3 + x^2y + y^3 + 2y^2z + (y^2z + z^3 - 2yz^2)\\nonumber\\\\ &\\ge 0. \\end{align} And $$C\\ge 0$$ follows from \\begin{align} &4(x+3z)(3x^3+3x^2z-3xz^2+z^3) - (4zx)^2\\nonumber\\\\ =\\ & 12x^4+48x^3z+8x^2z^2-32xz^3+12z^4\\nonumber\\\\ =\\ & (12x^4 - 16x^3z + 24x^2z^2) + 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4\\\\ \\ge \\ & 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4 \\nonumber \\\\ = \\ & 4z(4x+3z)(2x-z)^2\\nonumber\\\\ \\ge \\ &0. \\end{align} We are done. use the ptolemy inequality ! $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\\ge 0$$\\Longrightarrow 2(ab^3+bc^3+cd^3+da^3)\\ge$$b^2\\cdot{a}(b+c)+c^2\\cdot{b}(c+d)+d^2\\cdot{c}(a+d)+a^2\\cdot{d}(a+b)$$\\Longrightarrow 2(ab^3+bc^3+cd^3+da^3)\\ge$$b^2\\cdot{a}l_{1}+c^2\\cdot{b}l_{2}+d^2\\cdot{c}l_{1}+a^2\\cdot{d}l_{2}$$\\Longrightarrow 8abcd\\ge$$b^2\\cdot{a}l_{1}+c^2\\cdot{b}l_{2}+d^2\\cdot{c}l_{1}+a^2\\cdot{d}l_{2}$$\\Longrightarrow 8\\ge$$bl_{1}/cd+cl_{2}/ad+dl_{1}/ab+al_{2}/bc$ if $bl_{1}/cd,cl_{2}/ad,dl_{1}/ab,al_{2}/bc\\ge2$ then, $l_{1}l_{2}abcd\\ge bc\\cdot(cd)^2+ad\\cdot(ab)^2+ab\\cdot(bc)^2+cd\\cdot(ad)^2$ $\\Longrightarrow$$ac+bd\\ge 4\\sqrt{abcd}$$\\Longrightarrow$$l_{1}l_{2}\\ge ac+bd$ which contradict with the ptolemy inequality ! Let $$\\;$$ $$LHS=f(a,b,c,d)$$. Note that $$f(a,b,c,d)>f(a-k,b-k,c-k,d-k)$$ for $$0. So, WLOG we can take $$d=0$$ to prove the inequality. In that case we should show $$ab^2(b-c)+bc^3\\ge0$$. If $$b\\ge c$$ obviously we are done. For $$c\\ge b \\ge a$$ and $$c\\ge a \\ge b$$ cases arranging the inequality shows us $$ab^3+bc(c^2-ab)\\ge0$$. Now, $$a\\ge c \\ge b$$ case remained. We will use the inequality $$a\\ge c$$. Let's arrange the inequality again. Then, $$ab^2(b-c)+bc^3\\ge cb^2(b-c)+bc^3=b^3c-b^2c^2+bc^3\\ge0$$ $$\\;$$ (by AM-GM $$b^3c+bc^3\\ge 2b^2c^2$$ )", "$n=2$. \\begin{aligned}c_a(-3,3)&={3 \\choose 1}(x)-{3 \\choose 2}(x^2+2x)+{3 \\choose 3}(x^3+2x^2+3x)\\\\&=3x+3(x^2+2x)+x^3+2x^2+3x\\\\&=x^3-x^2\\\\&=b_3\\\\\\end{aligned} For $n\\geq 4$, \\begin{aligned}c_a(-3,n)=&\\sum^{3}_{k=0}(-1)^{3-k}{3 \\choose 3-k}a_{n-3+k}\\\\=&-{3 \\choose 0}\\sum^{n-4}_{i=0}(i+1)x^{n-i-3}+{3 \\choose 1}\\sum^{n-3}_{i=0}(i+1)x^{n-i-1}\\\\&-{3 \\choose 2}\\sum^{n-2}_{i=0}(i+1)x^{n-i-1}+{3 \\choose 3}\\sum^{n-1}_{i=0}(i+1)x^{n-i}\\\\=&-\\sum^{n-1}_{i=3}(i-2)x^{n-i}+3\\sum^{n-1}_{i=2}(i-1)x^{n-i}-3\\sum^{n-1}_{i=1}ix^{n-i}\\\\ &\\quad+\\sum^{n-1}_{i=0}(i+1)x^{n-i}\\\\=&-\\sum^{n-1}_{i=3}(i-2)x^{n-i}+3\\sum^{n-1}_{i=3}(i-1)x^{n-i}-3\\sum^{n-1}_{i=3}ix^{n-i}\\\\&+\\sum^{n-1}_{i=3}(i+1)x^{n-i}+3x^{n-2}-3x^{n-1}-6x^{n-2}+x^n+2x^{n-1}+3x^{n-2}\\\\=&\\sum^{n-1}_{i=3}[-(i-2)+3(i-1)-3i(i+1)]x^{n-i}+x^n-x^{n-1}\\\\=&\\, x^n-x^{n-1}\\\\=&\\, b_n\\\\\\end{aligned} Therefore,$(a_n)\\sim_{\\Sigma}(b_n)$. #### Chains of Sequences with Leading Zeros Adding zeros to the beginning of a sequence adds columns of zeros to be beginning of the sequences summation chain. #### Lemma. Let c be a nontrivial $\\Sigma$-chain in $M$. Let $N=\\min\\{n \\in \\mathbb{N}:c(0,n)\\neq 0\\}$. For every integer, m, and for every natural number, $n, $c(m,n)=0$. #### Proposition. Let M be a $\\mathbb{Z}$-Module. Let c be a nontrivial $\\Sigma$-chain in $M$. Let the natural number $N=\\min\\{n \\in \\mathbb{N}:c(0,n)\\neq 0\\}$. Let $d:\\mathbb{Z}\\times\\mathbb{N}\\to M$ be defined such that $d(m,n) = c(m,n+N+1)$ for every $m\\in \\mathbb{Z}$ and $n\\in \\mathbb{N}$. Then d is a $\\Sigma$-chain. #### Corollary Let M be a $\\mathbb{Z}$-Module, and let $(a_n)$ and $(b_n)$ be sequences in M, such that for a positive integer k, $b_{n+k}=a_n$ for all natural numbers n and $b_n=0$ when $n\\leq k$. Let $c_a=\\Phi_{\\Sigma}((a_n))$ and $c_b=\\Phi_{\\Sigma}((a_b))$. For every integer m and natural number $n$, $c_b(m,n+k)=c_a(m,n)$, and $c_b(m,n)=0$ when $n< k$. ### Conclusion Editor’s note: This paper represents the second installment of the concept of summation chains of complex valued sequences. Here, a new way to determine the relationship between two sequences was defined. The concept of sum-relation was defined, and a theorem", "- 12b^2$$ $3a^2 + 5ab - 12b^2 = (3a - 4b )(a + 3b)$ $$98x^{4} + 14x^{2} - 4$$ \\begin{align*} 98x^4 + 14x^2 - 4 &= 2(49x^4 - 7x^2 - 2) \\\\ &=2((7x+2)(7x-1)) \\end{align*} $$(x-2)^2 - 7(x-2) + 12$$ \\begin{align*} (x-2)^2 - 7(x-2) + 12 &=((x-2)-4)((x-2)-3) \\\\ &=(x-6)(x-5) \\end{align*} $$(a-2)^2 - 4(a-2) -5$$ \\begin{align*} (a-2)^2 - 4(a-2) -5 &= ((a-2)-5)((a-2)+1) \\\\ =&(a-7)(a-1) \\end{align*} $$(y+3)^2 - 3(y+3) - 18$$ \\begin{align*} (y + 3)^2 - 3(y + 3) - 18 &= ((y + 3) - 6)((y + 3) + 3) \\\\ &=(y - 3)(y + 6) \\end{align*} $$3(b^2 + 5b) +12$$ \\begin{align*} 3(b^2 + 5b) +12 &= 3(b^2 + 5b) +3(4) \\\\ &=3(b^2 + 5b + 4) \\\\ &= 3(b + 4)(b+1) \\end{align*} $$6(a^2 +3a) -168$$ \\begin{align*} 6(a^2 +3a) -168 &= 6(a^2 + 3a) - 6(28) \\\\ &= 6(a^2 + 3a - 28) \\\\ &=6(a + 7)(a -4) \\end{align*} ### Sum and difference of two cubes (EMAP) We now look at two special results obtained from multiplying a binomial and a trinomial: Sum of two cubes: \\begin{align*} \\left(x + y\\right)\\left({x}^{2} - xy + {y}^{2}\\right) & = x\\left({x}^{2} - xy + {y}^{2}\\right) + y\\left({x}^{2} - xy + {y}^{2}\\right) \\\\ & = \\left[x\\left({x}^{2}\\right) + x\\left(-xy\\right) + x\\left({y}^{2}\\right)\\right] + \\left[y\\left({x}^{2}\\right) + y\\left(-xy\\right) + y\\left({y}^{2}\\right)\\right]\\\\ & = {x}^{3} - {x}^{2}y + x{y}^{2} + {x}^{2}y", "(A_1 \\Delta \\dots A_n) \\Delta A_{n+1} \\right) \\\\ & = P \\left( A_1 \\Delta \\dots A_n \\right) + P(A_{n+1}) - 2 P \\left( (A_1 \\Delta \\dots A_n) \\cap A_{n+1} \\right) \\\\ & = \\sum_{i=1}^{n+1} P(A_i) - 2 \\sum_{i<j}^{n} P(A_i \\cap A_j) + 4 \\sum_{i<j<k}^{n} P(A_i \\cap A_j \\cap A_k) + \\dots + (-2)^{n-1}P(A_1 \\cap \\dots \\cap A_n) - 2 P \\left( (A_1 \\Delta \\dots A_n) \\cap A_{n+1} \\right) \\end{align} Where at first we applied the step we saw for $n=2$ and then used our inductive step. Finally, recall that intersection distributes over symmetric difference. With that, we can use our inductive step to address this last term. \\begin{align} P \\left( (A_1 \\Delta \\dots A_n) \\cap A_{n+1} \\right) & = P \\left( (A_1 \\cap A_{n+1}) \\Delta \\dots \\Delta (\\dots A_n \\cap A_{n+1}) \\right) \\\\ & = \\sum_{i=1}^{n} P(A_i \\cup A_{n+1}) - 2 \\sum_{i<j} P(A_i \\cap A_j \\cap A_{n+1}) + \\dots + (-2)^{(n-1)-1}P(A_1 \\cap \\dots \\cap A_{n+1}) \\end{align} Thus, $P(A_1 \\Delta \\dots \\Delta A_n) = \\sum_{i=1}^{n} P(A_i) - 2 \\sum_{i<j} P (A_i \\cap A_j) + 4 \\sum_{i<j<k} P(A_i \\cap A_j \\cap A_k) + \\dots + (-2)^{n-1} P(A_1 \\cap \\dots \\cap A_n).$", "+ (x + 1)^{n + 1} + ... + (x + 1)^{n + 2m - 3}\\Big)\\;, \\end{align*} which is $${n - 1 \\choose n - 2} + {n + 1 \\choose n - 2} + {n + 3 \\choose n - 2} + {n + 5 \\choose n - 2} + \\dots + {n + 2m - 3 \\choose n - 2} \\;.$$ So \\begin{align*} &{n \\choose n - 1} + {n + 2 \\choose n - 1} + {n + 4 \\choose n - 1} + {n + 6 \\choose n - 1} + \\dots + {n + 2(m - 1) \\choose n - 1} =\\\\ &\\left({n - 1 \\choose n - 1} + {n + 1 \\choose n - 1} + {n + 3 \\choose n - 1} + {n + 5 \\choose n - 1} + \\dots + {n + 2m - 3 \\choose n - 1}\\right)+\\\\ &\\qquad+\\left({n - 1 \\choose n - 2} + {n + 1 \\choose n - 2} + {n + 3 \\choose n - 2} + {n + 5 \\choose n - 2} + \\dots + {n + 2m - 3 \\choose n - 2}\\right)\\;. \\end{align*} But $${n - 1 \\choose n - 2} + {n + 1 \\choose n - 2} + {n + 3 \\choose n - 2}", "0:35 Let, $$S(n)=a^4+(a+d)^4+\\dots+(a+nd)^4\\tag1$$ Then we can write $$S(n) =a^4+(a^4+4a^3d+6a^2d^2+4ad^3+d^4)+\\dots+(a^4+4a^3dn+6a^2d^2n^2+4ad^3n^3+d^4n^4)\\\\ =(n+1)a^4+4a^3d(1+2+\\dots+n)+6a^2d^2(1^2+2^2+\\dots+n^2)+4ad^3(1^3+2^3+\\dots+n^3)+d^4(1^4+2^4+\\dots+n^4)$$ Using Faulhaber's formula for $F(x) = 1^x+2^x+\\dots+n^x$ we get $$S(n)=(n+1)a^4+4a^3d \\cdot F(1)+6a^2d^2 \\cdot F(2)+4ad^3\\cdot F(3)+d^4 \\cdot F(4)$$ Explicitly $$S(n)=(n+1)\\big(a^4+2a^3dn+a^2d^2n(2n+1)+ad^3n^2(n+1)+\\tfrac{1}{30}d^4n(2n+1)(3n^2+3n-1)\\big)\\tag2$$ In general, let $$s(n)=a^x+(a+d)^x+\\dots+(a+nd)^x\\\\ =a^x+(a^x+C^1_xa^{x-1}d+\\dots+d^x)+\\dots+\\big(a^x+C_x^1a^{x-1}dn+\\dots+(dn)^x\\big)\\\\ =(n+1)a^x+C_x^1a^{x-1}d(1+2+\\dots+n)+C_x^2a^{x-2}d^2(1^2+2^2+\\dots+n^2)+\\dots+d^x(1^x+2^x+\\dots+n^x)\\\\ =(n+1)a^x+C_x^1a^{x-1}d\\cdot F(1)+C_x^2a^{x-2}d^2\\cdot F(2)+\\dots+d^x\\cdot F(x)$$ So $$s(n)=(n+1)a^x+\\sum_{i=1}^x C_x^ia^{x-i}d^iF(i)$$ • Thanks! Very nice, using a general approach. Gave $+1$. (I've edited the LaTex format of your answer a bit so it will look clearer.) I've verified your formula $(2)$ gives the correct result when $a,d,n = 29,2,63$. – Tito Piezas III Dec 24 '14 at 16:57", "\\in K$. Then \\begin{align*} f(\\mathbf{v})&=f(c_1\\mathbf{v}_1+\\cdots+c_n \\mathbf{v}_n)\\\\ &=c_1f(\\mathbf{v}_1)+\\cdots+c_n f(\\mathbf{v}_n)\\\\ &=c_1 0+\\dots +c_n 0=0. \\end{align*} Thus $f(\\mathbf{v})=0$ for any $\\mathbf{v} \\in V$, hence $f=0$ and $\\phi$ is injective. ### To show that $\\phi$ is surjective, let $(\\mathbf{u}_1, \\dots, \\mathbf{u}_n)\\in V^{\\oplus n}$ be an arbitrary vector in $V^{\\oplus n}$. Using the basis $\\mathbf{v}_1, \\dots, \\mathbf{v}_n$, we have linear combinations $\\mathbf{u}_i=c_{1i}\\mathbf{v}_1+\\cdots+c_{ni} \\mathbf{v}_n$ for some $c_{1i}, \\dots, c_{ni} \\in K$ for $1\\leq i \\leq n$. Let $f \\in \\End(V)$ be a linear transformation whose matrix representation with respect to the basis $B:=\\{\\mathbf{v}_1, \\dots, \\mathbf{v}_n\\}$ is given by $[f]_B=\\begin{bmatrix} c_{11} & c_{12} & \\dots & c_{1n} \\\\ c_{21} &c_{22} & \\dots & c_{2n} \\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\ c_{n1} & c_{n2} & \\dots & c_{nn} \\end{bmatrix}.$ Since $[f]_B=[\\,[f(\\mathbf{v}_1)]_B \\dots, [f(\\mathbf{v}_n)]_B\\,]$, we have $[f(\\mathbf{v}_i)]_B=\\begin{bmatrix} c_{1i} \\\\ c_{2i} \\\\ \\vdots \\\\ c_{ni} \\end{bmatrix}.$ Namely, we have $f(\\mathbf{v}_i)=c_{1i}\\mathbf{v}_1+\\cdots+c_{ni} \\mathbf{v}_n=\\mathbf{u}_i.$ Therefore $\\phi(f)= (\\mathbf{u}_1, \\dots, \\mathbf{u}_n)\\in V^{\\oplus n}$ and $\\phi$ is surjective. In conclusion, $\\phi$ is a bijective homomorphism from $\\End(V)$ to $V^{\\oplus n}$. Thus $\\End(V)$ and $V^{\\oplus n}$ are isomorphic. Let $V$ be the set of all $n \\times n$ diagonal matrices whose traces are zero. That is, \\begin{equation*} V:=\\left\\{...", "$t_2>t_3>0, u_2>u_3>0$ with $h_t\\ne h_u$ where $h_\\alpha=\\alpha_2\\otimes \\alpha_3$. Let $\\alpha_1=-\\alpha_2-\\alpha_3$ and \\begin{align} a_1&=t_3u_3-t_1u_1\\notag\\\\ a_2&=t_1u_1-t_2u_2\\notag\\\\ a_3&=t_2u_2-t_3u_3\\notag\\\\ b_1&=t_2u_3-t_1u_1\\notag\\\\ b_2&=t_1u_1-t_3u_2\\notag\\\\ b_3&=t_3u_2-t_2u_3\\notag\\\\ a_{7-i}&=-a_i\\tag{asym}\\label{asym}\\\\ b_{7-i}&=-b_i\\tag{bsym}\\label{bsym} \\end{align} Then, for $j=0,\\dots,5$ and $n=6$, \\begin{align} \\sum_{i=1}^n a_i^j=\\sum_{i=1}^n b_i^j=\\Sigma_j,\\tag{te}\\label{te} \\end{align} say. $\\Sigma_0=n=6$, and \\ref{asym} and \\ref{bsym} ensure $\\Sigma_1=\\Sigma_3=\\Sigma_5=0$. Moreover, \\begin{align} \\Sigma_2=4h_t h_u; \\Sigma_4=4(h_t h_u)^2 \\implies\\frac{\\Sigma_4}{\\Sigma_2^2}&=\\frac14.\\tag{reg}\\label{reg} \\end{align} \\eqref{reg} is not a necessary precondition for \\eqref{te}. For example, $$a_{1\\dots3}=25, 23, 6; b_{1\\dots3}=27, 19, 10$$ satisfy \\eqref{te} with $\\Sigma_2=1190$ and $\\Sigma_4=671762$. However, \\eqref{reg} is satisfied for all of the first 18 examples (ordered by increasing $\\Sigma_2$) and most of the first few thousand examples. $$a_2\\otimes a_3=b_2\\otimes b_3=h_t h_u.$$ Such an equality implies that, for each of the above $\\otimes$ formulas in turn, its operands may be used as the values of $t_2, t_3$ with another pair of values for $u_2, u_3$ to produce a collection of up to four sets of six values each, where all those sets of six have the same $\\Sigma_2$ and the same $\\Sigma_4$, so any two of those sets of six satisfy \\eqref{te} as the $a_i$ and $b_i$. For example, $$\\begin{array}{rrrrrrrrrrrrr} t_2&t_3&u_2&u_3&h_t&h_u&h_t h_u&a_1&a_2&a_3&b_1&b_2&b_3\\\\ 2&1&3&1&7&13&91&-11&\\color{red}6&\\color{red}5&-10&\\color{blue}9&\\color{blue}1\\\\ \\color{red}6&\\color{red}5&3&2&91&19&1729&-45&37&8&-43&40&3\\\\ \\color{blue}9&\\color{blue}1&3&2&91&19&1729&-48&23&25&-32&47&-15 \\end{array}$$ \\eqref{te} may also be stated as \\begin{align} \\sum_i f(a_i)=\\sum_i f(b_i)\\tag{poly}\\label{poly} \\end{align} for $f=x\\mapsto x^j$ for $j=0,\\dots,k$. This implies that \\eqref{poly} is true for $f$ being any polynomial function of degree $\\leqslant k$. In particular, we may take $f=x\\mapsto", "2x^3 - 3x^2 - x^3 - 2x^2 + 3x - 6x^2 - 12x + 18 \\\\ & = x^4 + x^3 - 11x^2 - 9x + 18 \\end{align*} $$(a + 2)^2 - (2a - 4)^2$$ \\begin{align*} (a + 2)^2 - (2a - 4)^2 &= a^2 + 4a + 4 - (4a^2 - 16a + 16) \\\\ &= a^2 + 4a + 4 - 4a^2 + 16a - 16 \\\\ &= -3a^2 + 20a - 12 \\end{align*} Expand the following products: $$(2x + 3)^2 - (x-2)^2$$ \\begin{align*} (2x + 3)^2 - (x-2)^2 &= 4x^2 + 12x + 9 - (x^2 - 4x + 4) \\\\ &= 4x^2 + 12x + 9 - x^2 + 4x - 4 \\\\ &= 3x^2 + 16x + 5 \\end{align*} $$(2a^2 - a - 1 )(a^2 + 3a + 2)$$ \\begin{align*} (2a^2 - a - 1 )(a^2 + 3a + 2) &= 2a^4 + 6a^3 + 4a^2 - a^3 - 3a^2 - 2a - a^2 - 3a - 2 \\\\ &= 2a^4 + 5a^3 - 5a - 2 \\end{align*} $$(y^2 + 4y - 1)(1 - 4y - y^2)$$ \\begin{align*} (y^2 + 4y - 1)(1 - 4y - y^2) &= y^2 - 4y^3 - y^4 + 4y - 16y^2 - 4y^3 - 1 + 4y + y^2 \\\\ &= -y^4 - 8y^3 - 14y^2 +" ]

[ "and $$g$$. Polynomials should be denoted only using Capital letters. The symbol to denote polynomial is $$P(x)$$ if the equation has $$x$$ as variable. Example: \\begin{align*} P(x) = 2x^2 -x+3 \\end{align*} The $$x$$ in $$P(x)$$ represents the variable $$x$$. If the variable of the equation is $$y$$, the symbol to denote the polynomial is $$P(y)$$ Example: \\begin{align*} P(y) = 2y^2 -y+3 \\end{align*} The $$y$$ in $$P(y)$$ represents the variable $$y$$. Example 1: It is given that $$P(x) = 2x^2 - x + 3$$ and \\begin{align*} Q(x) = x^3 + 2x -1 \\end{align*} Evaluate $$P(2) +Q(-1)$$. Solution: To find $$P(2)$$, we need to replace the $$x$$ with $$2$$. \\begin{align*} P(2) = 2(2)^2 - 2 +3 \\end{align*} To find $$Q(-1)$$, we need to replace the $$x$$ with $$-1$$. \\begin{align*} Q(-1) = (-1)^3 + 2(-1) -1 \\end{align*} Adding both of the above equations. \\begin{align*} P(2) +Q(-1) = [\\;2(2)^2 - 2 +3\\;] \\;\\;+\\;\\;[\\;(-1)^3 + 2(-1) -1\\;] \\end{align*} Solving the above will give the answer. $$=5$$ Therefore, the answer is $$5$$. Example 2: Given that \\begin{align*} x^3 - 5x+3 = (Ax+3)(x-1)(x-2) \\;+\\; B(x-1) \\;+\\; C \\end{align*} for all real values of $$x$$, find the values of $$A,B$$ and $$C$$. Solution: In the question, it is given as “for all real values of $$x$$” which means we can substitute any value for $$x$$ and", "# Find new polynomial coefficients using Vieta's formulas $p(x)=x^3+ax^2+bx+c$ has roots $x_1,x_2,x_3$. Find $a,b,c$ for $q(x)$ which has roots at $x_1+x_2$, $x_1+x_3$ and $x_2+x_3$. I know that I'm supposed to use Vieta's formulas where $$x_1+x_2+x_3=-a$$ $$x_1x_2+x_1x_3+x_2x_3=b$$ $$x_1x_2x_3=-c$$ but I do not know how to tackle this problem. Any solutions or tips? First I want to mention that you should obtain that $a$ is the component of your second order term, hier $a$ should be replaced by the third order term, which is $1$, so you should obtain that \\begin{align} &x_1+x_2+x_3=-a,\\\\ &x_1x_2+x_2x_3+x_3x_1=b,\\\\ &x_1x_2x_3=-c. \\end{align} Now let us suppose that $q(x)=x^3+a'x^2+b'x+c'$, which has roots $x_1+x_2,x_2+x_3,x_3+x_1$. Then again from Vieta's formula \\begin{align} -a'&=2(x_1+x_2+x_3)=-2a,\\\\ b'&=(x_1+x_2)(x_2+x_3)+(x_2+x_3)(x_3+x_1)+(x_3+x_1)(x_1+x_2)\\\\ &=(x_1x_2+x_2x_3+x_3x_1)+(x_1+x_2+x_3)^2,\\\\ &=b+a^2,\\\\ -c'&=(x_1+x_2)(x_2+x_3)(x_3+x_1)\\\\ &=-(a+x_1)(a+x_2)(a+x_3)\\\\ &=-(a^3+(x_1+x_2+x_3)a^2+(x_1x_2+x_2x_3+x_3x_1)a+x_1x_2x_3)\\\\ &=c-ab. \\end{align} So the new polynimil is $q(x)=x^3+2ax^2+(b+a^2)x+ab-c$. • I do not really get the first step in -c'. How do you include a in that part? – Oscar3 Jul 18 '17 at 12:32 • Since $x_1+x_2+x_3=-a$, you can replace $x_1+x_2=-a-x_3$ and so on. – Student Jul 18 '17 at 12:36 • @Winther yes, thanks^^ – Student Jul 18 '17 at 12:39 • And hence, $-2ax^2$ – Oscar3 Jul 18 '17 at 13:04 Given that $x_1,x_2,x_3$ are the roots of the polynomial $$x^3+ax^2+bx+c$$ then by Vieta's formulas $$x_1+x_2+x_3=-a,\\;x_1x_2+x_2x_3+x_1x_3=b,\\;x_1x_2x_3=-c.$$ Again by Vieta's formulas the polynomial with roots $x_1+x_2$, $x_2+x_3$, $x_3+x_1$ is equal to", "polynomials of higher degree. In this situation, we define polynomials \\begin{align} P_3(x) & = P_2(x) + c_3(x − a)^3 \\\\ P_4(x) & = P_3(x) + c_4(x − a)^4 \\\\ P_5(x) & = P_4(x) + c_5(x − a)^5 \\end{align} and so on, with the general one being $P_n(x) = P_{n−1}(x) + c_n(x − a)^n.$ The defining property of these polynomials is that for each $$n$$, $$P_n(x)$$ must have its value and all its first n derivatives agree with those of $$f$$ at $$x = a$$. In other words we require that $P^{(k)}_n (a) = f^{(k)}(a)$ for all $$k$$ from 0 to $$n$$. To see the conditions under which this happens, suppose $P_n(x) = c_0 + c_1(x − a) + c_2(x − a)^2 + \\ldots + c_n(x − a)^n .$ Then \\begin{align} P^{(0)}_n (a) & = c_0 \\\\ P^{(1)}_n (a) & = c_1 \\\\ P^{(2)}_n (a) &= 2c_2 \\\\ P^{(3)}_n (a) &= (2)(3)c_3 \\\\ P^{(4)}_n (a) &= (2)(3)(4)c_4 \\\\ P^{(5)}_n (a) & = (2)(3)(4)(5)c_5 \\end {align} and, in general, $P^{(k)}_n (a) = (2)(3)(4) \\ldots (k − 1)(k)c_k = k!c_k .$ So having $P^{(k)}_n (a) = f^{(k)} (a)$ means that $k!c_k = f^{(k)} (a)$ and therefore $c_k = \\dfrac{f^{(k)}(a)}{k!}$ for each value of $$k$$. The expression for $$c_k$$ we have found the formula for the degree $$n$$ polynomial approximation of $$f$$", "19:57 • No, I don't think it has to include all polynomials of degree $3$ or less. Just all polynomials $p(x)$ of degree $3$ or less such that $p(1) = p(-2)$. May 24 at 21:05 If $$p(x) = a_0+a_1x+a_2x^2+a_3x^3$$ is a such polynomial, then \\begin{align} a_0+a_1+a_2+a_3 &= p(1) \\\\ &= p(-2) = a_0-2a_1+4a_2-8a_3, \\end{align} which is equivalent to $$3a_1-3a_2+9a_3=0$$, or $$a_2=a_1+3a_3$$. Hence \\begin{align} p(x) &= a_0+a_1x+(a_1+3a_3)x^2+a_3x^3 \\\\ &= a_0 \\color{red}1 + a_1\\color{red}{(x+x^2)} + a_3 \\color{red}{(3x^2+x^3)}. \\end{align} Thus, we see that every polynomial $$p(x)$$ with $$p(1)=p(-2)$$ is a linear combination of the polynomials $$1$$, $$x+x^2$$ and $$3x^2+x^3$$. One can easily verify that these two indeed form a basis for that vector space (i.e. they are linearly independent). • For anyone wondering the so-referred to easy verification, your best bet is probably to use repeated differentiation. May 25 at 3:58 Using @Jair Taylor's hint: If $$p(x)$$ is a polynomial that is in $$V$$, we can \"shift\" the function down so that $$p(1)$$ and $$p(-2)$$, would be zeroes, i.e. $$p(x) - p(1)$$ or equivalently $$p(x) - p(-2)$$. This can then be factored into the form: $$q(x)(x - 1)(x + 2)$$ so then $$p(x) = q(x)(x - 1)(x + 2) + p(1)$$, which is a set of polynomials (of degree 3 or less) spanned by: $$\\{(x - 1)(x + 2), x(x -", "# Third degree Polynomial has the properties Let $p(x)$ be a polynomial of degree $3$ with real coefficients. Which of the following is possible ? a) $p(x)$ has no real roots b) $p(x)$ has exactly two real roots c) $p(1)=-1, \\; p(2)=1,\\; p(3)=11,\\; p(4)=35$ d) $i-1$ and $i+1$ are roots of $p(x)$, where $i=\\sqrt{-1}$ Every odd degree real polynomial has at least one real root, so a) is false b) is false as $x^3-1$ serve an example d) is also false as complex roots are occur in pairs. so c) is true. My try for c): take $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ where $a_3 \\neq 0$ $p(1)=-1 \\Rightarrow a_0+a_1+a_2+a_3=-1$ $p(2)=1 \\Rightarrow a_0+2a_1+2^2a_2+2^3a_3=1$ $p(3)=11 \\Rightarrow a_0+3a_1+3^2a_2+3^3a_3=11$ $p(4)=35 \\Rightarrow a_0+4a_1+4^2a_2+4^3a_3=35$ So this can be written as \\begin{align*}\\left[\\begin{matrix}1&1 &1 &1\\\\1&2 & 2^2 & 2^3\\\\ 1 & 3 & 3^2 &3^3\\\\ 1 &4 &4^2 &4^3\\end{matrix}\\right]\\left[\\begin{matrix}a_0\\\\a_1\\\\a_2\\\\a_3\\end{matrix}\\right] &= \\left[\\begin{matrix}-1\\\\1\\\\11\\\\35\\end{matrix}\\right] \\end{align*} Since the coefficient matrix is Vandermonde, so the determinant is not zero and hence this system has a unique solution. So such a unique polynomial exist. Is my argument correct? or any other method to show this? • Have you solved this system? – Dr. Sonnhard Graubner Apr 13 '18 at 14:12 • For $b$, the question asks what is possible, not inevitable. Therefore showing a single example for which the property does not hold does not show", "polynomial with coefficients all $1$ and taken $\\pmod n$ if the exponents cover $0,1,…,n-1$, you can rest assured it is divisible by $(x^n-1)/(x-1)$. May not be easy to recognise always, but it is a simple enough test. +1 Sep 7, 2020 at 14:02 Hmm. I'd look at it and say \"There's no rational root\" because $$x = \\pm 1$$ doesn't work. So there's some irrational root, $$\\alpha$$, and two complex-conjugate pairs. So there's no nice obvious linear factor I can write down. Then I'd say \"Maybe there's a quadratic factor.\" I can assume it's monic, so I'm looking to write $$x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r).$$ from which I can expand to get $$x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br$$ if I've done the algebra right. Equating coefficients I see that \\begin{align} 0 &= a + p\\\\ 0 &= q + ap + b\\\\ 1 &= ra + bq\\\\ 1 &= br \\end{align} so $$p = -a$$, and $$r = \\frac1b$$,and these equations become \\begin{align} 0 &= a + -a\\\\ 0 &= q - a^2 + b\\\\ 1 &= \\frac1b a + bq\\\\ 1 &= b(1/b) \\end{align} which simplify down to", "# Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less ## Problem 137 Let $P_n(\\R)$ be the vector space over $\\R$ consisting of all degree $n$ or less real coefficient polynomials. Let $U=\\{ p(x) \\in P_n(\\R) \\mid p(1)=0\\}$ be a subspace of $P_n(\\R)$. Find a basis for $U$ and determine the dimension of $U$. ## Solution. Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$ be an element in $U$. Since $p(1)=0$, we have $a_n+a_{n-1}+\\cdots+a_1+a_0=0$. Thus we have \\begin{align*} p(x)&=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x-(a_n+a_{n-1}+\\cdots+a_1)\\\\ &=a_n(x^n-1)+a_{n-1}(x^{n-1}-1)+\\cdots a_1(x-1). \\end{align*} Let us define $q_n(x)=x^n-1, q_{n-1}(x)=x^{n-1}-1, \\dots, q_1(x)=x-1 \\in U.$ Then the above computation shows that we have $p(x)=a_nq_n(x)+a_{n-1}q_{n-1}(x)+\\cdots a_1q_1(x).$ In other words, any element $p(x)\\in U$ is a linear combination of $q_1(x), \\dots, q_n(x)$. Hence $B:=\\{q_1(x), \\dots, q_n(x)\\}$ is a spanning set for $U$. We show that $B$ is a linearly independent set. Consider a linear combination $c_1q_1(x)+\\cdots+c_n q_n(x)=\\theta(x):=0x^{n}+0x^{n-1}+\\cdots+0x+0.$ Then we have \\begin{align*} c_nx^n+c_{n-1}x^{n-1}+\\cdots+c_1x-(c_1+c_2+\\cdots+c_n)=0. \\end{align*} Comparing the coefficients of both sides, we obtain $c_1=c_2=\\cdots=c_n=0$ and thus $q_1(x), \\dots, q_n(x)$ are linearly independent, and hence $B=\\{q_1(x), \\dots, q_n(x)\\}$ is a basis for the subspace $U$. Since the dimension is the number of vectors in a basis, the dimension of $U$ is $n$. ### More from my site • A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors", "Coefficients of product of polynomials of degree 1 I have the following equations: \\begin{aligned} (x+x_1)(x+x_2)&=x^2+(x_1+x_2)x+x_1x_2\\\\ (x+x_1)(x+x_2)(x+x_3)&= x^3+(x_1+x_2+x_3)x^2\\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ +(x_1x_2+x_1x_3+x_2x_3)x+x_1x_2x_3\\\\ (x+x_1)(x+x_2)(x+x_3)(x+x_4)&=x^4+(x_1+x_2+x_3+x_4)x^3\\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ +(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_3+x_3x_4)x^2\\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ +(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)x\\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ +x_1x_2x_3x_4\\\\ &\\ \\ \\vdots\\\\ (x+x_1)(x+x_2)\\cdots(x+x_{n-1})(x+x_n)&=c_nx^n+c_{n-1}x^{n-1}+\\cdots+c_1x+c_0 \\end{aligned} Where $x_1,\\dots,x_n$ are known. My question is if there is a general formula to calculate all of the coefficients $c_i$. At the moment I only know that $c_n=1$ and $c_0=\\prod\\limits^n_{i=1}x_i$. Thanks in advance. Yes, in fact. The generating function of the elementary symmetric polynomials in $n$ variables is $$S_n(x)=\\prod_{i=1}^n{(1+x_ix)}=\\sum_{i=0}^n{e_i(x_1,\\ldots,x_n)x^i}$$ where $$e_0(x_1,\\ldots,x_n)=1$$ and $$e_i(x_1,\\ldots,x_n)=\\sum_{1\\leq p_1<\\cdots<p_i\\leq n}{x_{p_1}\\cdots x_{p_i}}$$ It's simply the sum over all subsets of the variables with $i$ elements of the product of the variables in the subset. This is not quite your product. Your product is $$x^nS_n\\left(\\frac1x\\right)$$ Thus the coefficient of $x^{n-i}$ is $e_i$.", "with coefficients in $$\\mathbb R$$, and let $$X=\\{\\text{functions}\\,\\mathbb{R}[X]\\to\\mathbb{R}[X]\\}$$, so every element of $$X$$ is a function that maps polynomials to polynomials. Let $$*$$ be composition of functions in $$X$$, i.e, if $$a,b\\in X$$, then $$a*b=a\\circ b$$. Clearly $$*$$ is associative and \\begin{align} e:\\mathbb{R}[X]&\\to\\mathbb{R}[X]\\\\ P(X)&\\mapsto P(X) \\end{align} is the identity element of $$(X,*)$$. We give two elements $$a,b\\in X$$ such that $$a*b=e$$ yet $$b*a\\neq e$$, which shows that $$a$$ and $$b$$ don't have inverses. Let \\begin{align} a:\\mathbb{R}[X]&\\to\\mathbb{R}[X]\\\\ \\sum_{i=0}^n\\alpha_i X^i&\\mapsto\\sum_{i=1}^n\\alpha_{i}X^{i-1}. \\end{align} What $$a$$ does is that it removes the constant term of a polynomial and reduces the degree of the other terms by $$1$$. One can define it by the following formula for $$P(X)$$ nonconstant polynomial: $$a(P(X))=\\dfrac{P(X)-P(0)}{X}$$ and with $$a(c(X))=0$$ for constant polynomials. Examples: $$a(5+2X+7X^2-3X^3)=2+7X-3X^2;$$ $$a(3-2X+9X^3)=-2+9X^2;$$ $$a(\\sqrt{2}X-3X^3)=\\sqrt{2}-3X^2.$$ Now let \\begin{align} b:\\mathbb{R}[X]&\\to\\mathbb{R}[X]\\\\ P(X)&\\mapsto XP(X). \\end{align} $$b$$ simply multiplies the polynomial by $$X$$, which has the opposite effect on the degrees compared to $$a$$. For example, $$b(\\sqrt{2}-3X^2)=\\sqrt{2}X-3X^3$$. Now I claim that $$a*b=e$$. Indeed, if $$P(X)$$ is a polynomial, $$b(P(X))=XP(X)$$. What $$b$$ did is that it added $$1$$ degree to every term. Now what would $$a(b(P(X))$$ be? $$a$$ removes the constant term and then substracts the degree of the other terms by $$1$$. Since $$XP(X)$$ has no constant term, all $$a$$ does is substract by $$1$$ the degree of all the other terms,", "Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less Problem 137 Let $P_n(\\R)$ be the vector space over $\\R$ consisting of all degree $n$ or less real coefficient polynomials. Let $U=\\{ p(x) \\in P_n(\\R) \\mid p(1)=0\\}$ be a subspace of $P_n(\\R)$. Find a basis for $U$ and determine the dimension of $U$. Solution. Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$ be an element in $U$. Since $p(1)=0$, we have $a_n+a_{n-1}+\\cdots+a_1+a_0=0$. Thus we have \\begin{align*} p(x)&=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x-(a_n+a_{n-1}+\\cdots+a_1)\\\\ &=a_n(x^n-1)+a_{n-1}(x^{n-1}-1)+\\cdots a_1(x-1). \\end{align*} Let us define $q_n(x)=x^n-1, q_{n-1}(x)=x^{n-1}-1, \\dots, q_1(x)=x-1 \\in U.$ Then the above computation shows that we have $p(x)=a_nq_n(x)+a_{n-1}q_{n-1}(x)+\\cdots a_1q_1(x).$ In other words, any element $p(x)\\in U$ is a linear combination of $q_1(x), \\dots, q_n(x)$. Hence $B:=\\{q_1(x), \\dots, q_n(x)\\}$ is a spanning set for $U$. We show that $B$ is a linearly independent set. Consider a linear combination $c_1q_1(x)+\\cdots+c_n q_n(x)=\\theta(x):=0x^{n}+0x^{n-1}+\\cdots+0x+0.$ Then we have \\begin{align*} c_nx^n+c_{n-1}x^{n-1}+\\cdots+c_1x-(c_1+c_2+\\cdots+c_n)=0. \\end{align*} Comparing the coefficients of both sides, we obtain $c_1=c_2=\\cdots=c_n=0$ and thus $q_1(x), \\dots, q_n(x)$ are linearly independent, and hence $B=\\{q_1(x), \\dots, q_n(x)\\}$ is a basis for the subspace $U$. Since the dimension is the number of vectors in a basis, the dimension of $U$ is $n$. • A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors Show that the set $S=\\{1, 1-x, 3+4x+x^2\\}$ is", "a_1 x_{m-2} + \\dots + a_{m-1} = \\sum_{j=0}^{m-1}a_{m-1-j} x^j;$$ then \\begin{align*} \\sum_{i=0}^n q_{m-1}(i) &= \\sum_{i=0}^n \\sum_{j=0}^{m-1} a_{m-1-j} i^j\\\\ &= \\sum_{j=0}^{m-1}\\sum_{i=0}^n a_{m-1-j} i^j\\\\ &= \\sum_{j=0}^{m-1} a_{m-1-j} \\sum_{i=0}^n i^j\\\\ &= \\sum_{j=0}^{m-1} a_{m-1-j} \\left(\\frac{n^{j+1}}{j+1} + p_j(n) \\right) \\end{align*} is a polynomial in $n$ of degree at most $m$, which I’ll call $r_m(n)$. Substitute this into $(*)$: $\\displaystyle(n+1)^{m+1} = (m+1)\\sum_{i=0}^n i^m + r_m(n)$, so \\begin{align*} (m+1)\\sum_{i=0}^n i^m &= (n+1)^{m+1} - r_m(n)\\\\ &= n^{m+1} + s_m(n) - r_m(n), \\end{align*} where $s_m$ is a polynomial of degree $m$. Here again I’ve used the binomial theorem. Now $s_m(n) - r_m(n)$ is clearly a polynomial in $n$ of degree at most $m$, so if we set $$p_m(n) = \\frac{1}{m+1}(s_m(n) - r_m(n),$$ then $p_m$ is a polynomial of degree at most $m$, and $$\\sum_{i=0}^n i^m = \\frac{n^{m+1}}{m+1} + p_m(n),$$ as desired. - Suppose we know that $$\\sum_{i=1}^n\\;i^k=\\frac{n^{k+1}}{k+1}+P_k(n)$$ where $P_k$ is a polynomial of degree $k$. Sum in $n$ to get \\begin{align} \\sum_{n=1}^m\\;\\sum_{i=1}^n\\;i^k&=\\sum_{n=1}^m\\left(\\frac{n^{k+1}}{k+1}+P_k(n)\\right)\\\\ \\sum_{i=1}^m\\;\\sum_{n=i}^m\\;i^k&=\\frac{1}{k+1}\\sum_{n=1}^m\\;n^{k+1}+Q_{k+1}(m)\\\\ \\sum_{i=1}^m\\;(m-i+1)i^k&=\\frac{1}{k+1}\\sum_{n=1}^m\\;n^{k+1}+Q_{k+1}(m)\\\\ (m+1)\\left(\\frac{m^{k+1}}{k+1}+P_k(m)\\right)-\\sum_{i=1}^m\\;i^{k+1}&=\\frac{1}{k+1}\\sum_{n=1}^m\\;n^{k+1}+Q_{k+1}(m)\\\\ (m+1)\\left(\\frac{m^{k+1}}{k+1}+P_k(m)\\right)-Q_{k+1}(m)&=\\frac{k+2}{k+1}\\sum_{n=1}^m\\;n^{k+1}\\\\ \\frac{m^{k+2}}{k+2}+P_{k+1}(m)&=\\sum_{n=1}^m\\;n^{k+1} \\end{align} - One approach would be to say \"We assume that $\\sum_{i=1}^n i^k = \\frac{n^{k+1}}{k+1} +$ some terms which are $O(n^k)$ or smaller. Now we try to add $(n+1)^k$. You then use the binomial expansion to show that $\\frac{(n+1)^{k+1}}{k+1} = \\frac{n^{k+1}}{k+1} + (n+1)^k$ + terms which are $O(n^{k-1})$ or smaller.\" The problem with this approach as stated is that we", "the same derivative, for all derivatives. What’s the form of a polynomial? It looks something like this: $p(x) = a + bx + cx^2 + dx^3 + ex^4 + \\ldots$ 0th derivative We need $$p(0) = f(0)$$. When $$x = 0$$ all terms of the polynomial $$p$$ go to zero, other than the first. In other words: \\begin{align*} f(0) &= p(0) \\\\ &= a + bx + cx^2 + dx^3 + ex^4 + \\ldots |_{x=0} \\\\ &= a \\end{align*} We’ve discovered our first coefficient in $$p$$: $$p(x) = f(0) + bx + cx^2 + dx^3 + ex^4 + \\ldots$$ 1st derivative We need $$p'(0) = f'(0)$$. \\begin{align*} f'(0) &= p'(0) \\\\ &= 0 + b + 2cx + 3dx^2 + 4ex^3 + \\ldots |_{x=0} \\\\ &= b \\end{align*} We’ve discovered our second coefficient in $$p$$: $$p(x) = f(0) + f'(0)x + cx^2 + dx^3 + ex^4 + \\ldots$$ 2nd derivative We need $$p''(0) = f''(0)$$. \\begin{align*} f''(0) &= p''(0) \\\\ &= 0 + 0 + 2c + 3 \\cdot 2dx + 4 \\cdot 3ex^2 + \\ldots |_{x=0} \\\\ &= 2c \\end{align*} We’ve discovered our third coefficient in $$p$$: $$p(x) = f(0) + f'(0)x + \\frac{f''(0)}{2}x^2 + dx^3 + ex^4 + \\ldots$$ 3rd derivative We need $$p'''(0) = f'''(0)$$. \\begin{align*} f'''(0) &= p'''(0) \\\\ &= 0 + 0 +", "Q(n)2^{n}$ for some polynomial $Q$? There is no unbound $k$ on the left side. – Thomas Andrews Dec 14 '12 at 21:12 Specifically, all polynomials, $P(n,k)$ can be written as $$P(n,k)=\\sum_{i=0}^d P_i(n) \\binom {k}{i}$$ For some sequence of polynomials $P_i$. Then $$Q(n)=\\sum_{i=0}^d P_i(n) \\binom{n}{i} 2^{-i}$$ – Thomas Andrews Dec 14 '12 at 21:21 $$(N-2n)^2 = (N-n)^2 + n^2 - 2n(N-n)$$ Hence, \\begin{align} S_N(n) & = \\dfrac{(N-2n)^2}{n!(N-n)!}\\\\ & = \\dfrac{(N-n)^2 + n^2 - 2n(N-n)}{n!(N-n)!}\\\\ & = \\dfrac{N-n}{n!(N-n-1)!} + \\dfrac{n}{(n-1)! (N-n)!} - 2 \\dfrac1{(n-1)!(N-n-1)!}\\\\ & = \\dfrac1{n!(N-n-2)!} + \\dfrac1{n!(N-n-1)!} + \\dfrac1{(n-2)!(N-n)!} + \\dfrac1{(n-1)!(N-n)!} - \\dfrac2{(n-1)!(N-n-1)!} \\end{align} Now note that from the binomial expansion of $2^{N-k-m}$, we have that $$\\sum_{n=\\min(k,a)}^{\\max(N-m,b)} \\dfrac1{(n-k)!(N-n-m)!} = \\dfrac{2^{N-k-m}}{(N-k-m)!}$$ Hence, $$\\sum_{n=0}^{n=N} S_N(n) = \\dfrac{2^{N-2}}{(N-2)!} + \\dfrac{2^{N-1}}{(N-1)!} + \\dfrac{2^{N-2}}{(N-2)!} + \\dfrac{2^{N-1}}{(N-1)!} -2 \\dfrac{2^{N-2}}{(N-2)!} = \\dfrac{2^N}{(N-1)!}$$ - Note that $\\sum_{n=0}^{N}\\frac{x^{2n}}{n!(N-n)!}=\\frac{(1+x^2)^N}{N!}$. Thus: $$\\sum_{n=0}^{N}\\frac{x^{2n-N}}{n!(N-n)!}=\\frac{x^{-N}(1+x^2)^N}{N!}$$ Now diffrentiate with repect to $x$ to get: $$\\sum_{n=0}^{N}\\frac{(2n-N)x^{2n-N-1}}{n!(N-n)!}=\\frac{d}{dx}[\\frac{x^{-N}(1+x^2)^N}{N!}]$$ Multiply by $x$ to get: $$\\sum_{n=0}^{N}\\frac{(2n-N)x^{2n-N}}{n!(N-n)!}=x\\frac{d}{dx}[\\frac{x^{-N}(1+x^2)^N}{N!}]$$ Diffrentiate with respect to $x$ to get: $$\\sum_{n=0}^{N}\\frac{(2n-N)^2x^{2n-N-1}}{n!(N-n)!}=\\frac{d}{dx}[x\\frac{d}{dx}[\\frac{x^{-N}(1+x^2)^N}{N!}]]$$ Now let $x=1$ to get your sum on the LHS. - I think the last line should say $x=1$? – joriki Dec 14 '12 at 20:32 Thanks. I will edit it. – Amr Dec 14 '12 at 20:33 \\begin{align} \\sum_{n=0}^N\\frac{(N-2n)^2}{n!(N-n)!} &= \\sum_{n=0}^N\\frac{((N-n)-n)^2}{n!(N-n)!} \\\\ &= \\sum_{n=0}^N\\frac{(N-n)^2}{n!(N-n)!}+\\sum_{n=0}^N\\frac{n^2}{n!(N-n)!}-2\\sum_{n=0}^N\\frac{(N-n)n}{n!(N-n)!} \\\\ &= \\sum_{n=0}^N\\frac{n^2}{n!(N-n)!}+\\sum_{n=0}^N\\frac{n^2}{n!(N-n)!}-2\\sum_{n=1}^{N-1}\\frac1{(n-1)!(N-n-1)!} \\\\ &= 2\\sum_{n=0}^N\\frac{n^2}{n!(N-n)!}-2\\sum_{n=0}^{N-2}\\frac1{n!(N-2-n)!}\\;. \\end{align} Now consider $$\\sum_{n=0}^m\\binom mnq^n=(1+q)^m$$ and thus $$\\sum_{n=0}^m\\frac{q^n}{n!(m-n)!}=\\frac{(1+q)^m}{m!}\\;.$$ Applying $q\\partial/\\partial q$ twice yields", "The next step is to set up the matrix. Suppose that our numerator polynomial is $p(x)=a_0+a_1x+\\dots+a_mx^m$ and that our denominator polynomial is $q(x)=b_0+b_1x+\\dots+b_mx^m$. Then we have the following equations, writing $n$ for $m+k+1$: \\begin{align*} p(x_1)-y_1q(x_1)&=a_0+a_1x_1+\\dots +a_mx_1^m-y_1(b_0+b_1x_1+\\dots +b_kx_1^k)=0\\\\ p(x_2)-y_2q(x_2)&=a_0+a_1x_2+\\dots +a_mx_2^m-y_2(b_0+b_1x_2+\\dots +b_kx_2^k)=0\\\\ &\\;\\;\\vdots\\\\ p(x_n)-y_nq(x_n)&=a_0+a_1x_n+\\dots +a_mx_n^m-y_n(b_0+b_1x_n+\\dots +b_kx_n^k)=0 \\end{align*} This can be written as $$\\begin{pmatrix} 1&x_1&\\dots&x_1^m&-y_1&-y_1x_1&\\dots&-y_1x_1^k\\\\ 1&x_2&\\dots&x_2^m&-y_2&-y_2x_2&\\dots&-y_2x_2^k\\\\ \\vdots\\\\ 1&x_n&\\dots&x_n^m&-y_n&-y_nx_n&\\dots&-y_nx_2^k \\end{pmatrix} \\begin{pmatrix} a_0\\\\a_1\\\\\\vdots\\\\a_m\\\\b_0\\\\b_1\\\\\\vdots\\\\b_k \\end{pmatrix}=\\mathbf 0$$ We generate this matrix columnwise: <<RINTERP Implementation>>= tempvec: List F := [1 for i in 1..(m+k+1)] collist: List List F := cons(tempvec, [(tempvec := [tempvec.i * xlist.i _ for i in 1..(m+k+1)]) _ for j in 1..max(m,k)]) collist := append([collist.j for j in 1..(m+1)], _ [[- collist.j.i * ylist.i for i in 1..(m+k+1)] _ for j in 1..(k+1)]) @ Now we can solve the system: <<RINTERP Implementation>>= res: List Vector F := nullSpace((transpose matrix collist) _ ::Matrix F) @ Note that it may happen that the system has several solutions. In this case, some of the data points may not be interpolated correctly. However, the solution is often still useful, thus we do not signal an error. <<RINTERP Implementation>>= if #res~=1 then output(\"Warning: unattainable points!\" _ ::OutputForm)$OutputPackage @ In this situation, all the solutions will be equivalent, thus we can always simply take the first one: <<RINTERP Implementation>>= reslist: List List P := _ [[monomial((res.1).(i+1),i) for i in", "& 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & \\dots \\\\ \\hline 1 & c & b & a & c^2 & bc & ac & b^2 & ab & a^2 & c^3 & bc^2 & ac^2 & cb^2 & abc & a^2c & \\dots \\end{array}$$ I ran out of space to write all the terms for the third degree, but you should be able to notice a pattern here: each degree group contains a two-dimensional indexing subproblem so we can just add the formula from before to the index of the first term in the group. Meanwhile, the index of the first term is the arithmetic sum of the arithmetic sum up to the degree of the group, which is $$i_3(p_a,p_b,p_c) = \\frac{(p_a+p_b+p_c)(p_a+p_b+p_c+1)}{3\\cdot2}+i_2(p_a,p_b)$$ Now that we know how to deduce the indexing formula, we can write the general indexing formula for the $$n$$ dimensonal polynomial $$i_n(p_1,\\dots,p_n) = \\sum_{k=1}^n \\binom{(\\sum_{l=1}^{k}p_l)+k-1}{k}$$ where $$p_1,p_2\\dots$$ is taken to mean $$p_a,p_b,\\dots$$ i.e. the order of the term of each dimension. Just thought I would post an answer to this question I had since I found solutions. First, the number of coefficients for a polynomial basis complete up to degree $$p$$ in $$n$$ dimensions, designated $$N_p^n$$, is: \\begin{align} N_p^n", "identifies the degree of the polynomial. The term \\begin{align*}5x^4\\end{align*} is the highest degree so the degree of the polynomial is 4. The answer is that the polynomial is of the fourth degree. #### Example 4 Name the degree of the expression \\begin{align*}6y^3+3xy+9\\end{align*}. First, look at the degrees for each term in the expression. \\begin{align*}6y^3\\end{align*} has a degree of 3 \\begin{align*}3xy\\end{align*} has a degree of 2 \\begin{align*}9\\end{align*} has a degree of 0 Next, the highest degree identifies the degree of the polynomial. The term \\begin{align*}6y^3\\end{align*} is the highest degree so the degree of the polynomial is 3. The answer is that the polynomial is of the third degree. #### Example 5 Write the following polynomial in standard form and identify the degree of the polynomial. \\begin{align*}7x^2+3x-2x^4+8x^6-7\\end{align*} First, look at the degrees for each term in the expression. \\begin{align*}7x^2\\end{align*} has a degree of 2 \\begin{align*}3x\\end{align*} has a degree of 1 \\begin{align*}-2x^4\\end{align*} has a degree of 4 \\begin{align*}8x^6\\end{align*} has a degree of 6 \\begin{align*}-7\\end{align*} has a degree of 0 Next, write this polynomial in order by degree, highest to lowest \\begin{align*}8x^6-2x^4+7x^2+3x-7\\end{align*} Then, the highest degree identifies the degree of the polynomial. The term \\begin{align*}8x^6\\end{align*} is the highest degree so the degree of the polynomial is 6. The answer is \\begin{align*}8x^6-2x^4+7x^2+3x-7\\end{align*} and the polynomial is of the sixth degree. ### Review Write", "of degree 0. Hence $$0 \\in P_{3}$$. Furhtermore, \\begin{align*} &p(x)+0 \\\\ &=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+0 \\\\ &=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+(p_{0}+0) \\\\ &=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0} \\\\ &=p(x). \\end{align*} Therefore, there is a additive identity for polynomials (the constant 0) of degree 3 or less. Let $$s(x)=(-p_{3})x^{3}+(-p_{2})x^{2}+(-p_{1})x+(-p_{0})$$. Note that $$-p_{i} \\in \\mathbb{R}$$ since $$p_{i} \\in \\mathbb{R}$$ for $$i \\in \\mathbb{Z}$$ and $$0 \\leq i \\leq 3$$. Therefore, $$s(x) \\in P_{3}$$. Furthermore, \\begin{align*} &p(x)+s(x)\\\\ &=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+(-p_{3})x^{3}+(-p_{2})x^{2}+(-p_{1})x+(-p_{0}) \\\\ &=(p_{3}-p_{3})x^{3}+(p_{2}-p_{2})x^{2}+(p_{1}-p_{1})x+(p_{0}-p_{0}) \\\\ &=0x^{3}+0x^{2}+0x+0 \\\\ &=0. \\end{align*} Hence, $$s(x)$$ is the additive inverse of $$p(x)$$. Therefore, every polynomial has a $$-p(x)$$ such that $$p(x)+(-p(x))=0$$. Scalar Multiplication Closure Here we note that \\begin{align*} cp(x)&=c(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\\\ &=cp_{3}x^{3}+cp_{2}x^{2}+cp_{1}x+cp_{0}. \\end{align*} Furthermore, $$c p_{i} \\in \\mathbb{R}$$ since $$c, p_{i} \\in \\mathbb{R}$$ for all $$i \\in \\mathbb{K}$$ with $$0 \\leq i \\leq 3$$. Hence, $$cp(x) \\in P_{3}$$. Distributive We now have that \\begin{align*} (c+d)p(x)&=(c+d)(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\\\ &=(c+d)p_{3}x_{3}+(c+d)p_{2}x^{2}+(c+d)p_{1}x+(c+d)p_{0} \\\\ &=(cp_{3}+dp_{3})x^{3}+(cp_{2}+dp_{2})x^{2}+(cp_{1}+dp_{1})x+cp_{0}+dp_{0} \\\\ &=cp_{3}x^{3}+cp_{2}x^{2}+cp_{1}x+cp_{0}+dp_{3}x^{3}+dp_{2}x^{2}+dp_{1}x+dp_{0} \\\\ &=cp(x)+dp(x). \\end{align*} Hence, you can distribute across scalar addition. Associative Here, we see that \\begin{align*} (cd)p(x)&=(cd)(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\\\ &=(cd)p_{3}x_{3}+(cd)p_{2}x^{2}+(cd)p_{1}x+(cd)p_{0} \\\\ &=c(dp_{3})x^{3}+c(dp_{2})x^{2}+c(dp_{1})x+c(dp_{0}) \\\\ &=c((dp_{3})x^{3}+(dp_{2})x^{2}+(dp_{1})x+(dp_{0})) \\\\ &=c(dp(x)). \\end{align*} Therefore, scalar multiplication is associative. Identity Note that $$1$$ is a real number. Furthermore, we have that \\begin{align*} 1*p(x)&=1*(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\\\ &=1*p_{3}x^{3}+1*p_{2}x^{2}+1*p_{1}x+1*p_{0} \\\\ &=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0} \\\\ &=p(x). \\end{align*} Therefore, $$1$$ acts as an identity for scalar multiplication in $$P_{3}$$. Conclusion We have now shown that the set of polynomials of degree 3 or less satisfied all", "2. $[2^{2^{n-1}}]^2 = [\\frac{3}{2}2^{2^{n-1}}]$ 3. $x^2 + x$ takes precisely the values $0, 1, \\dotsc, [2^{2^n-1}-1]$ as $x$ varies in $[2^{2^n}]$ The lexicographically earliest irreducible polynomial over $[2^{2^n}]$ is $x^2 + x = [2^{2^n-1}]$, because $x^2 = \\alpha$ always has a root in finite field of characteristic $2$, and $x^2 + x = \\alpha$ has a root for earlier $\\alpha$ according to statement 3. We know by Theorem 5 that $[2^{2^n}]$ is a root of $x^2 + x = [2^{2^n-1}]$, hence $$\\textstyle [2^{2^n}]^2 = [2^{2^n}] + [2^{2^n-1}] = [2^{2^n} + 2^{2^n-1}] = [\\frac{3}{2}2^{2^n}].$$ We obtain the field $[2^{2^{n+1}}]$ as a vector space over $[2^{2^n}]$ with typical element $X = [2^{2^n}]x + y$. We examine the polynomial \\begin{align}X^2 + X &= ([2^{2^n}]x + y)^2 + [2^{2^n}]x + y \\\\ &= [2^{2^n}]^2 x^2 + y^2 + [2^{2^n}]x + y \\\\ &= [2^{2^n}](x^2 + x) + ([2^{2^n-1}]x^2 + y^2 + y).\\end{align} By induction, $x^2 + x$ can take any value in $[2^{2^n-1}]$. Note that $x^2 + x$ remains unchanged when we replace $x$ by $x + 1$. The same is true for $y^2 + y$. It follows that $[2^{2^n-1}]x^2 + y^2 + y$ can be made to take any value in $[2^{2^n}]$ without affecting the value of $x^2 + x$. This implies that the values of $X^2 + X$ can be", "x^n \\}$. Consider a linear combination of these vectors: (3) \\begin{align} a_01 + a_1(1 - x) + a_2(1 - x^2) + ... + a_n(1 - x^n) \\\\ = a_0 + a_1 - a_1x + a_2 - a_2x + ... + a_n - a_nx^n \\\\ = (a_0 + a_1 + a_2 ... + a_n) + (-a_1)x + (-a_2)x^2 + ... + (-a_n)x^n \\end{align} It should be clear that any polynomial of degree less than or equal to $n$ can be written in this form. For example, the polynomial $p(x) = 3 + 2x^2 + 5x^4$ can be written in the form by letting $a_0 = 10$, $-a_2 = 2$ and letting $-a_4 = 5$ and all other $a_i = 0$ (4) \\begin{align} p(x) = (10 + (-2) + (-5)) + 2x^2 + 5x^4 \\\\ p(x) = 3 + 2x^2 + 5x^4 \\end{align}", "# Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form $\\sum_{k=0}^a\\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\\epsilon_k=\\pm1$. I discovered a relatively simple algorithm to generate these patterns: simply take $(n+2^p-1)^p$ and subtract $(n+2^p-2)^p$ (using $n\\to n-1$) to get a polynomial of degree $p-1$. Take this difference and subtract $(n+2^p-3)^p-(n+2^p-4)^p$ (using $n\\to n-2$) to get a polynomial of degree $p-2$. Repeat this process until $n^p$ is reach. The first few examples of this are \\begin{align} n^0&=1\\\\ (n+1)^1-n^1&=1\\\\ (n+3)^2-(n+2)^2-(n+1)^2+n^2&=4\\\\ (n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3&=48 \\end{align} Upon doing this for the next several powers and checking OEIS, it would appear the constant corresponding to the power $p$ is $$\\large C_p=2^{\\frac{p(p-1)}2} p!$$ However, this is an observation only, and I have no idea how to go about proving this. The only thing I notice is that $\\frac{p(p-1)}{2}=\\sum\\limits_{k=1}^{p-1}k$, but I don't know how to use this fact. Does any one know how to prove this observation? • $f(n) = \\displaystyle\\sum_{k = 0}^{a}\\epsilon_k(n+k)^p$ is a polynomial with degree at most $p$. So if you can show that $f(n) = C$ for $p+1$ distinct integers $n$, then $f(n)$ must be constant. Dec 24, 2014 at 7:21 • I already know that it's constant, the nature of the algorithm described above reduces the degree" ]

[ "+0 # Complex Numbers 0 115 1 Determine the complex number z satisfying the equation $2z - 3 \\overline{z} = -7 + 2i$. Note that denotes the conjugate of z. Jan 14, 2022", "+0 # Complex Numbers 0 110 1 Determine the complex number z satisfying the equation $2z - 3i \\overline{z} = -7 + 2i$. Note that $\\overline{z}$ denotes the conjugate of z. Oct 25, 2021", "+0 # Complex Numbers 0 31 1 Determine the complex number z satisfying the equation $2z - 3 \\overline{z} = -7 + 2i$. Note that denotes the conjugate of z. Jan 14, 2022 #1 +2287 0 Jan 15, 2022", "+0 # Complex Numbers 0 63 1 Determine the complex number z satisfying the equation $2z - 3i \\overline{z} = -7 + 2i$. Note that $\\overline{z}$ denotes the conjugate of z. Oct 25, 2021 #1 0 z = 8 - i Dec 7, 2021", "# Think and solve Algebra Level 5 For distinct positive integers $p, q$, how many complex numbers $z$ satisfy the equation $z ^ p = \\bar{z} ^ q ?$ Note: $\\bar{z}$ is the complex conjugate of $z.$ ×", "How to Find the Conjugate of a Complex Number Here you will learn how to find the conjugate of a complex number and properties of conjugate with examples. Let’s begin – How to Find the Conjugate of a Complex Number Let z = a + ib be a complex number. Then the conjugate of z is denoted by $$\\bar{z}$$ and is equal to a – ib. Thus, z = a + ib $$\\implies$$ $$\\bar{z}$$ = a – ib It follows from this definition that the conjugate of a complex number is obtained by replacing i by -i. For Example : If z = 3 + 4i, then $$\\bar{z}$$ = 3 – 4i. Properties of Conjugate If $$z$$, $$z_1$$, $$z_2$$ are complex numbers, then (i) $$\\bar{\\bar{z}}$$ = z (ii) z + $$\\bar{z}$$ = 2 Re (z) (iii) z – $$\\bar{z}$$ = 2i Im (z) (iv) z = $$\\bar{z}$$ $$\\iff$$ z is purely real (v) z + $$\\bar{z}$$ = 0 $$\\implies$$ z is purely imaginary (vi) z$$\\bar{z}$$ = $$[Re (z)]^2$$ + $$[Im (z)]^2$$ (vii) $$\\bar{z_1 + z_2}$$ = $$\\bar{z_1}$$ + $$\\bar{z_2}$$ (viii) $$\\bar{z_1 – z_2}$$ = $$\\bar{z_1}$$ – $$\\bar{z_2}$$ (ix) $$\\bar{z_1z_2}$$ = $$\\bar{z_1}$$ $$\\bar{z_2}$$ (x) $$\\bar{z_1\\over z_2}$$ = $$\\bar{z_1}\\over \\bar{z_2}$$ Example : Multiply 3 – 2i by its conjugate. Solution : The conjugate of 3 – 2i is 3 +", "$\\bar{z}$ be a complex conjugate of $z$. Which of the following formulas is correct. $z\\bar{z} = |z|$ $z\\bar{z} = |z|^2$ $z\\bar{z} = z^2$ • Q11: 3 pts Let $\\bar{z}$ be a complex conjugate of $z$, than $\\frac{z - \\bar{z}}{2i}$ is real part of $z$ imaginary part of $z$ none of these", "# Complex Number -A problem on conjugate |$z_1$|=2,|$z_2$|=3,|$z_3$|=4 and |$2z_1+3z_2+4z_3$|=9 then the absolute value of $8z_2z_3+27z_3z_1+64z_1z_2$ must be equal to? ($z_1,z_2,z_3$ are complex numbers) I tried manipulating with the conjugates and stuff...but not being able to figure out the right technique to solve..help please!! HINT: Try to multiply the expression $|2\\bar{z_1}+3\\bar{z_2}+4\\bar{z_3}|$ by the product $|z_1z_2z_3|$. What is the value of $|2\\bar{z_1}+3\\bar{z_2}+4\\bar{z_3}|$? Given expression equals $|z_1z_2z_3||8/z_1+27/z_2+64/z_3|$ =$|z1||z2||z3||2z_1+3z_2+4z_3|$=2.3.4.9=216", "# Question #bd452 Dec 26, 2016 3. #### Explanation: Yes you are correct that the conjugate of $3 + 2 i$ is $3 - 2 i$ The problem is to find the sum of $5 - 3 i$ and $3 - 2 i$ , which is $5 - 3 i + 3 - 2 i$ $\\implies 8 - 5 i$ Hence, option 3 is the correct option", "you did not need to specify a second argument in the call to conj (alternatively, you could have passed the empty list: []). Now suppose that b is complex-valued with complex conjugate b_bar. The conjugate of a+I*b is a-I*b_bar: > $\\mathrm{conj}\\left(a+Ib,\\left[\\left[b,\\mathrm{b_bar}\\right]\\right]\\right)$ ${a}{-}{I}{}{\\mathrm{b_bar}}$ (2) Now suppose that both a and b are complex-valued. Compute the complex conjugate of a+I*b: > $\\mathrm{conj}\\left(a+Ib,\\left[\\left[a,\\mathrm{a_bar}\\right],\\left[b,\\mathrm{b_bar}\\right]\\right]\\right)$ ${\\mathrm{a_bar}}{-}{I}{}{\\mathrm{b_bar}}$ (3)", "Complex numbers daster Could someone please show me how: $$z_1\\bar{z_2} + \\bar{z_1}z_2 = 2\\:Re(z_1\\bar{z_2})$$ where $\\bar{z}$ is the conjugate of $z$.", "# Complex numbers 1. Dec 7, 2004 ### daster Could someone please show me how: $$z_1\\bar{z_2} + \\bar{z_1}z_2 = 2\\:Re(z_1\\bar{z_2})$$ where $\\bar{z}$ is the conjugate of $z$. 2. Dec 7, 2004 ### daster Nevermind. I got it.", "26th 2010, 12:40 PM running-gag Just for fun you can solve it another way $(2+i)z+(3-2i) \\bar{z}=32$ Take the conjugate of both sides $(2-i)\\bar{z}+(3+2i) z=32$ This leads to the following system $(2+i)z+(3-2i) \\bar{z}=32$ $(3+2i) z + (2-i)\\bar{z}=32$ that you can solve for z by eliminating $\\bar{z}$ • March 26th 2010, 04:45 PM Fandango1992 Thanks a lot for your help! Now I can move to question two, haha. Good times...", "write the quotient of two complex numbers, allowing us to do division: z zw zw = = 2 w ww w EXAMPLE 1.1 Find the complex conjugate, sum, product, and quotient of the complex numbers z = 2 − 3i w = 1+ i SOLUTION To find the complex conjugate of each complex number we let i → − i . Hence z = 2 − 3i = 2 + i 3 w = 1+ i = 1− i CHAPTER 1 Complex Numbers 7 The sum of the two complex numbers is formed by adding the real and imaginary parts, respectively: z + w = ( 2 − 3i ) + (1 + i ) = ( 2 + 1) + i ( −3 + 1) = 3 − 2i We can form the product as follows: zw = ( 2 − 3i )(1 + i ) = 2 − 3i + 2i − 3i 2 = ( 2 + 3) + i ( −3 + 2 ) = 5−i Finally, we use the complex conjugate of w to form the quotient: z z ⎛ w ⎞ ( 2 − 3i ) (1 − i ) 2 − 3i − 2i + 3i 2 2 − 3 − 5i −1 − 5i = ⎜ ⎟= = = =", "## anonymous one year ago Determine the conjugate of the complex number -4 + 2016i $-4 \\mathbf{\\color{red}-} 2016i$", "# Conjugate Complex Numbers Definition of conjugate complex numbers: In any two complex numbers, if only the sign of the imaginary part differ then, they are known as complex conjugate of each other. Conjugate of a complex number z = a + ib, denoted by $$\\bar{z}$$, is defined as $$\\bar{z}$$ = a - ib i.e., $$\\overline{a + ib}$$ = a - ib. For example, (i) Conjugate of z$$_{1}$$ = 5 + 4i is $$\\bar{z_{1}}$$ = 5 - 4i (ii) Conjugate of z$$_{2}$$ = - 8 - i is $$\\bar{z_{2}}$$ = - 8 + i (iii) conjugate of z$$_{3}$$ = 9i is $$\\bar{z_{3}}$$ = - 9i. (iv) $$\\overline{6 + 7i}$$ = 6 - 7i, $$\\overline{6 - 7i}$$ = 6 + 7i (v) $$\\overline{-6 - 13i}$$ = -6 + 13i, $$\\overline{-6 + 13i}$$ = -6 - 13i Properties of conjugate of a complex number: If z, z$$_{1}$$ and z$$_{2}$$ are complex number, then (i) $$\\bar{(\\bar{z})}$$ = z Or, If $$\\bar{z}$$ be the conjugate of z then $$\\bar{\\bar{z}}$$ = z. Proof: Let z = a + ib where x and y are real and i = √-1. Then by definition, (conjugate of z) = $$\\bar{z}$$ = a - ib. Therefore, (conjugate of $$\\bar{z}$$) = $$\\bar{\\bar{z}}$$ = a + ib = z. Proved. (ii) $$\\bar{z_{1} + z_{2}}$$ = $$\\bar{z_{1}}$$ + $$\\bar{z_{2}}$$ Proof:", "Re z = 2 and Im z = √2 ii) √3/ 5 i Solution : Let z = √3/5 i = 0 + √3/5 i Re z = 0 and Im z √3/5 3) Find a and b such that 2a + i 4b and 2i represents the same complex-numbers. Solution : 2a + i 4b = 2i 2a + i 4b = 0 + 2i By the definition of equality, 2a = 0 ⇒ a = 0 And 4b = 2 ⇒ b = 2/4 = ½ ∴ a = 0 and b = ½ 4) Find the conjugates of 2 + 5i and √5 Solution : By definition, the complex conjugate is obtained by reversing the sign of the imaginary part of the complex-numbers. Hence the required conjugates are 2 – 5i and √5 (as there is no imaginary part)", "Algebra # Complex Numbers: Level 3 Challenges If the square of $$x$$ and the square root of $$x$$ are equal to each other, but not equal to $$x$$, then what is the value of $$x^2+x$$? Let $$z=x+iy$$ be a complex number where $$x$$ and $$y$$ are integers. Find the area of the rectangle whose vertices are the roots of the equation $\\large z \\bar{z}^3+\\bar{z}z^3=350.$ Details: $$\\bar{z}=x-iy$$ and $$i=\\sqrt{-1}$$. How many complex numbers are the conjugate of their own cube? The complex numbers $$p$$ and $$q$$ satisfy $$p^3=5+i\\sqrt{2}$$ and $$q^3=5-i\\sqrt{2}$$. Find the only possible integer value of $$p+q$$. $\\large e ^{ i\\theta }e^{ 2i\\theta} e^{ 3i\\theta }...\\large e^{ ni\\theta } = 1$ Find the value of $$\\theta$$ satisfying the above equation for $$m\\in\\mathbb{N}$$. ×", "number z = 3~-~4i is.! Is ¯z = 3 is ¯z = 3 is ¯z = 3 z ¯ = 3 is =. ] or z\\ [ conjugate ] gives the complex tangent bundle of \\$ \\mathbb { }!", "Do they represent conic sections? Algebra Level 4 $\\begin{cases} |z+\\bar{z}|+|z-\\bar{z}|=2 \\\\ |iz-1|+|z-i|=2\\end{cases}$ Find the complex number $$z$$ which does not satisfy the above equations simultaneously. ×" ]

[ "etc.) and simultaneous equations. let $Z = \\sqrt{4i-3}$ where Z is complex. i.e. $Z$ is of the form $Z = a + ib$, where $a$ and $b$ are REAL coefficients. (This is important later on.) Square both sides: $Z^2 = -3+4i$ Expand the standard form of $Z^2$: $(a+ib)^2 = -3+4i$ This becomes $a^2 - b^2 + 2abi = -3+4i$ The above is an identity and so the real and imaginary parts can be equated. Hence two equations are formed: $a^2 - b^2 = -3$................(1) $2abi = 4i$ which becomes $ab = 2$......(2) In (2), $b = 2/a$ so $a^2 - (2/a)^2 + 3 = 0$ i.e. $a^4 + 3a^2 - 4 = 0$ (multiplying all by $a^2$ and rearranging) which is a quadratic-styled quartic: $(a^2 + 4)(a^2 - 1) = 0.$ Solving, $a=1, -1, 2i$ and $-2i$ BUT since $a$ and $b$ are real, we discard $2i$ and $-2i$. Now we simply solve for $b$ using $b = 2/a$. Hence the solutions are $a=1, b=2$ or $a=-1, b=-2$. and so $\\sqrt{4i-3} = \\pm(1+2i)$ since we defined $Z = a+ib$. Best of luck with any future questions! Hope this helped. • Just add $around math expressions, expressed in the LaTeX formatting. – PearlSek Dec 23 '16 at 22:21 • Well that was simpler than I thought, thanks for that!", "$\\bar{z} = \\bar{a + bi} = a - bi.$ We also notice that $z\\bar{z} = (a + bi)(a - bi) = a^{2} + b^{2},$ and so the product of a complex number with its conjugate is a real number. In fact, $z\\bar{z} = a^{2} + b^{2} = |z|^{2},$, and so $|z| = \\sqrt{z\\bar{z}}$ Exercise $$\\PageIndex{4}$$ Let $$w = 2 + 3i$$ and $$z = -1 + 5i$$ 1. Find $$\\bar{w}$$ and $$\\bar{z}$$. 2. Compute $$|w|$$ and $$|z|$$. 3. Compute $$w\\bar{w}$$ and $$z\\bar{z}$$. 4. What is $$\\bar{z}$$ if $$z$$ is a real number? 1. Using the definition of the conjugate of a complex number we find that $$\\bar{w} = 2 - 3i$$ and $$\\bar{z} = -1 - 5i$$. 2. Using the definition of the norm of a complex number we find that $$|w| = \\sqrt{2^{2} + 3^{2}} = \\sqrt{13}$$ and $$|z| = \\sqrt{(-1)^{2} + 5^{2}} = \\sqrt{26}$$. 3. Using the definition of the product of complex numbers we find that $w\\bar{w} = (2 + 3i)(2 - 3i) = 4 + 9 = 13$ $z\\bar{z} = (-1 + 5i)(-1 - 5i) = 1 + 25 = 26$ 4. Let $$z = a + 0i$$ for some $$a \\in \\mathbb{R}$$. Then $$\\bar{z} = a - 0i$$. Thus, $$\\bar{z} = z$$ when $$z \\in \\mathbb{R}$$. Summary In this section, we studied the", "some complex number that solves $x^3=-1$, then we can find it by solving the $a$ and $b$ that will make the following equation true: $(a + bi)^3 = -1$ You are going to solve that equation now. When you find $a$ and $b$, you will have found the answers. Stop now and make sure you understand how I have set up this problem, before you go on to solve it. 3. What is $(a + bi)^3?$ Multiply it out. 4. Now, rearrange your answer so that you have collected all the real terms together and all the imaginary terms together. Now, we are trying to solve the equation $(a+bi)^3=-1$. So take the formula you just generated in number $4$, and set it equal to $-1$. This will give you two equations: one where you set the real part on the left equal to the real part on the right, and one where you set the imaginary part on the left equal to the imaginary part on the right. 5. Write down both equations. OK. If you did everything right, one of your two equations factors as $b(3a^2-b^2)=0$. If one of your two equations doesn’t factor that way, go back—something went wrong! If it did, then let’s move on from there. As you know, we now have two things", "and express absolute values using conjugation. $$\\exists z\\in\\mathbb C\\,\\forall k\\in\\{1,2,3\\}:\\quad 1=(z-z_k)(\\bar z-\\bar z_k)=z\\bar z-z\\bar z_k-\\bar zz_k+z_k\\bar z_k$$ Now we can treat $a=-\\bar z$ and $b=-z$ as two distinct variables and obtain $$ab + az_k + b\\bar z_k + \\lvert z_k\\rvert^2 - 1 = 0$$ Three conditions for two variables, which is still too much. So let's treat $c=ab$ as a third variable. $$az_k + b\\bar z_k + c = 1 - \\lvert z_k\\rvert^2$$ So now we have a $3\\times 3$ system of equations, and once we have solved that, we can check whether $c=ab$ holds. I guess that if it does, then $a=\\bar b$ will hold as well. To verify that guess: For reasons of symmetry $c$ is real. Therefore $az_k$ and $b\\bar z_k$ must have opposite imaginary parts. Furthermore, $az_k\\cdot b\\bar z_k=c\\lvert z_k\\rvert^2\\in\\mathbb R$ so they must have opposite argument as well. If two numbers have opposite imaginary part and opposite argument, then they are conjugate to one another. Unfortunately solving a system of three linear equations is still harder than what I'd have hoped for. One big win is that this approach is very symmetric, and can be done symbolically as well. Unfortunately, the result still looks evil: $$z_1^2\\bar z_1^2z_2\\bar z_2 - z_1\\bar z_1^2z_2^2\\bar z_2 - z_1^2\\bar z_1z_2\\bar z_2^2 + z_1\\bar z_1z_2^2\\bar z_2^2 - z_1^2\\bar z_1^2\\bar", "is also \"unique\" that is, if z= a1+ b1i and the same z= a2+ b2i, we must have a1= a2 and b1= b2. Now, in this problem, attempting to find (2+ 6i)1/3, they are suggesting that you write the answer as \"a+ bi\", find (a+ bi)3 and set the \"real\" and \"imaginary\" parts equal. For example, (a+ bi)2= (a+ bi)(a+ bi)= a2+ 2abi+ b2i2= (a2- b2+ 3abi. If we wanted to find (1+ i)1/2, we could now set s2- b2= 1 and ab= 1, giving two equations to solve for a and b. Yes, that is write. By any definition of \"complex\" number, we can write any complex number in the form z= a+ bi for some real numbers \"a\" and \"b\". That form is also \"unique\" that is, if z= a1+ b1i and the same z= a2+ b2i, we must have a1= a2 and b1= b2. Now, in this problem, attempting to find (2+ 6i)1/3, they are suggesting that you write the answer as \"a+ bi\", find (a+ bi)3 and set the \"real\" and \"imaginary\" parts equal. For example, (a+ bi)2= (a+ bi)(a+ bi)= a2+ 2abi+ b2i2= (a2- b2+ 3abi. If we wanted to find (1+ i)1/2, we could now set s2- b2= 1 and ab= 1, giving two equations to solve for a and b. Thanks", "+0 # Find all complex solutions such that z^4 = -4 Note: All solutions should be expressed in the form a+bi, where a and b are real numbers help! 0 116 1 Find all complex solutions such that z^4 = -4 Note: All solutions should be expressed in the form a+bi, where a and b are real numbers. I took a square root on both sides, and then got z^2 = +/- 2i (sorry, don't know how to use latex) from there, I know I have to take a square root on both sides again, and I get the square root of 2, but I really don't know the way to take the square root of i, which is where I need help at. If you're kind enough to help, it would be great! May 26, 2022 #1 +26321 +2 Find all complex solutions such that z^4 = -4 Note: All solutions should be expressed in the form a+bi, where a and b are real numbers help! $$\\huge{z^2 = \\pm 2i}$$ $$\\begin{array}{|rcll|} \\hline z^2 &=& \\pm 2i \\quad | \\quad z = a+bi \\\\ \\left(a+bi\\right)^2 &=& \\pm 2i \\\\ a^2+2abi+b^2i^2 &=& \\pm 2i \\quad | \\quad i^2 = -1 \\\\ a^2+2abi-b^2 &=& \\pm 2i \\\\ \\mathbf{a^2-b^2+2abi} &=& \\mathbf{\\pm 2i} \\\\ \\begin{array}{|rcll|} \\hline a^2-b^2+2abi &=& \\mathbf{+2i}\\\\ a^2-b^2+2abi &=& 0+2i", "$d=-kb$. So $c+di=k(a-bi)$. To get an imaginary number Geometrically, there are two ways of looking at it. Since multiplication corresponds to a rotation, we could consider the angle that $z_1$ needs to be rotated by to reach the imaginary axis: Here, the horizontal distance from $0+0i$ to $z_1$ is the same as the vertical distance from $0+0i$ to $z_2$, and vice versa. This is why the angle between $z_1$'s line and the imaginary axis is the same as the angle between the real axis and $z_2$'s line. Alternatively, we could use the result for real products: find $z_2$ such that the product is real, and then multiply by $i$. In this case, to get a real number, $z_1=1+3i$ could be multiplied by $z_2=1-3i$ to get a real number, so to get an imaginary number, is should be multiplied by $iz_2=i(1-3i)=3+i$, shows as $z_4$ below. $iz_2$ is the image of $z_2$ under a rotation $90^\\circ$ anticlockwise about the origin: Algebraically, we can use this second idea. Since we know that multiplying $a+bi$ by real multiples of $a-bi$ gives us real numbers, we know that multiplying $a+bi$ by imaginary multiples of $a-bi$ gives us imaginary numbers! That is the same as multiplying by real multiples of $i(a-bi)=ai-bi^2=b+ai$ - swapping the coefficients $a$ and $b$. So multiplying $a+bi$ by $k(b+ai)$ for", ") – (2+ i) ( $z-\\bar{z}$ ) + 14i = 0, then $z\\bar{z}$ is equal to 1) 3 2) 8 3) 10 4) none of these Solution: Given $(3+i)(z+\\bar{z})-(2+i)(z-\\bar{z})+14i = 0$ ..(i) Suppose z = x+iy $\\bar{z}$ = x-iy $z+\\bar{z}$ = 2x $z-\\bar{z}$ = 2iy Substituting above values in (i) (3+i)2x-(2+i)2iy+14i = 0 6x+2ix-4iy+2y+14i = 0 (6x+2y)+(2x-4y+14)i = 0 Comparing real part 6x+2y = 0 6x = -2y y = 6x/-2 y = -3x Comparing imaginary part 2x-4y+14 = 0 Put y = -3x in above equation 2x+12x+14 = 0 14x = -14 x = -1 So y = 3 Now z = -1+3i $\\bar{z}$ = -1-3i $z\\bar{z}$ = (-1+3i)(-1-3i) = 1+9 = 10 Hence, option (3) is the answer. Related video", "$z_2$ horizontally gives other real sums: $(-1-3i)+(4+3i)=3+0i$ $(-3-3i)+(1+3i)=-2+0i$ For these last three examples, $z_1=a_1-3i$ and $z_2=a_2+3i$ for some real numbers $a_1$ and $a_2$. We can check this will always work: $$z_1+z_2=(a_1-3i)+(a_2+3i)=a_1+a_2-3i+3i=(a_1+a_2)+0i$$ What if we had another pair in the form $a_1+bi$ and $a_2-bi$ (for some real number $b$)? Then $$z_1+z_2=(a_1+bi)+(a_2-bi)=a_1+a_2+bi-bi=(a_1+a_2)+0i$$ So whenever $z_1=a_1+bi$ and $z_2=a_2-bi$, $z_1+z_2$ will be real. How it works geometrically When you add $z_2$ to $z_1$, you add the real parts, $a_1+a_2$, so $z_1$ is moved by $a_2$ along the real axis. The imaginary parts are also added, $b_1+b_2$, so $z_1$ is moved by $b_2$ along the imaginary axis. This is the same as a translation by the vector from $0+0i$ to $z_2$: $(1+2i)+(1-2i)=2+0i$ $(4+4i)+(-2-4i)=2+0i$ We can also prove algebraically that $z_1+z_2$ is only ever a real number if $z_1=a_1+bi$ and $z_2=a_2-bi$. To do this, we should begin by letting $z_1$ and $z_2$ be any complex numbers, so $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$. Then: $$z_1+z_2=(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$$ This is real only if the imaginary part is equal to $0i$, so $(b_1+b_2)i=0i$, so $b_1+b_2=0$, which means $b_2=-b_1$. So $z_1=a_1+bi$ and $z_2=a_2-bi$. Adding two complex numbers to get an imaginary number is very similar. Here are some examples: $(2+i)+(-2+3i)=0+4i$ $(3+2i)+(-3+i)=0+3i$ $(3-4i)+(-3-i)=0-5i$ Similarly, notice that if $z_1=a+b_1i$ and $z_2=-a+b_2i$ then $z_1+z_2=0+(b_1+b_2)i$, which is imaginary (if $b_1+b_2=0$, then $z_1+z_2=0+0i$, which is real -", "+0 # Ok. I think this might be calculus. real hard. +13 106 7 +1132 All help is greatly appreciated for this problem. Let $$z = a + bi,$$ where a and b are positive real numbers. If $$z^3 + |z|^2 + z = 0,$$ then enter the ordered pair $$(a,b).$$ Jul 8, 2022 #1 -1 (a,b) = (1,sqrt(3)). Jul 8, 2022 #2 +1132 +10 No explanation, and it's wrong. nerdiest Jul 8, 2022 #3 +117849 +2 Hi Nerdiest. I just worked out each bit seperately and then put it altogether. I have not presented it becasue it is a bit long and I no doubt made careless errors But I will walk you through it $$z=a+bi\\\\ z^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=etc\\\\ |z|^2=a^2+b^2$$ The real part is 0 and the imaginary part is 0 Solve simultaneously. Jul 9, 2022 #4 +1 Hi. Given: $$z=a+bi$$ So: $$z^3=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i$$ Given: $$z^3+\\left | z\\right |^2+z=0$$ Substitute $$z,z^3$$, and we know: $$\\left |z \\right |^2=\\sqrt{a^2+b^2}^2=a^2+b^2$$ $$a^3+3a^2bi-3ab^2-b^3i+(a^2+b^2)+a+bi=0$$ We equate real parts = 0 and imaginary parts = 0 (As the right-hand side is zero, this means both the real and imaginary parts are zero.) (Real parts without the \"i\" and imaginary parts with the \"i\"). So: Set $$Re=0:$$ $$a^3-3ab^2+a^2+b^2+a=0$$ (1) Set $$Im=0:$$ $$3a^2b-b^3+b=0$$ (2) Given b is not equal to zero so, divide (2) by b: $$3a^2-b^2+1=0$$ Let's", "+0 # Precalc 0 42 5 +163 There are four complex numbers such that $$z \\overline{z}^3 + \\overline{z} z^3 = 350,$$ and both the real and imaginary parts of $$z$$ are integers. These four complex numbers are plotted in the complex plane. Find the area of the quadrilateral formed by the four complex numbers as vertices. Jan 1, 2020 #1 +1 The four complex nubmers are $$\\pm 5 \\pm 4i$$, so the area is 2*5*2*4 = 80. Jan 1, 2020 #2 +163 +1 The complex numbers are all different though So I don't think this is correct. Not that I can do better or anything. Keep up the good work! Jan 1, 2020 #3 +106887 +1 $$z\\bar{z}^3+\\bar{z}z^3=350\\\\ (a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\\\ (a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\\\ (a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\\\ (a^2+b^2)[(2a^2-2b^2)]=350\\\\ (a^2+b^2)(a^2-b^2)=175\\\\ a^4-b^4=175$$ So the solution are ( $$z=+4\\pm3i\\\\ z=-4\\pm3i$$ So assuming i have made no silly mistakes the area is 48 units squared coding: z\\bar{z}^3+\\bar{z}z^3=350\\\\ (a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\\\ (a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\\\ (a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\\\ (a^2+b^2)[(2a^2-2b^2)]=350\\\\ (a^2+b^2)(a^2-b^2)=175\\\\ a^4-b^4=175 Jan 6, 2020 #4 +163 +1 Good Job!!! Melody, you are correct. I also got 48. A bit late on your response, but I answered the problem before seeing you post this. Jan 7, 2020 #5 +106887 0 Good work EpicWater. It is always good when you can solve it by yourself! Melody Jan 7, 2020", "# Algebraic Form of Complex Numbers Complex number z=(a;b) can be represented in the form z=a+bi. This is called algebraic form of complex number. In the representation z=a+bi, i^2=-1 and i is called imaginary unit. a is real part (denoted by Re z) and b is imaginary part (denoted by Im z) of complex number. If imaginary part of complex number a+bi is not 0 (b!=0) then such number is called imaginary, for example 1+bi. If a=0 and b!=0 then complex number is called purely imaginary. For example, 3i is purely imaginary. If b=0 then we obtain real number a. If we are given complex number z=a+bi then number a-bi is called complex conjugate of z. It is denoted by bar(z) or z^(**). For example complex conjugate of 2+3i is 2-3i. Sum and product of two complex conjugates are real numbers. Indeed, if z=a+bi then bar(z)=a-bi , so z+bar(z)=a+bi+a-bi=2a. Also, zbar(z)=(a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2-b^2(-1)=a^2+b^2. Absolute value (or modulus) of complex number z is real number r=|z|=sqrt(zbar(z))=sqrt(a^2+b^2). Conjugate complex numbers have same absolute values, i.e. |a+bi|=|a-bi|.", "$(a+bi)^2(a+bi)=2a \\Rightarrow (a^3-3ab^2)+(3a^2b-b^3)i=2a$ So , you should solve following system of equations : $\\begin{cases} a^3-3ab^2=2a \\\\ 3a^2b-b^3=0 \\end{cases}$ - I thought there was a way around this, so as to not to have to expand the $(a+bi)$ binomial (which I mistakenly thought would be a pain), however this seems to do the trick! – Milosz Wielondek Mar 14 '12 at 13:16 A hint: From your equation it follows that $z^3$ is real. Now write $z=r\\,e^{i\\phi}$ and draw your conclusions about possible $\\phi$'s and $r$'s. There will be several cases. - I did write it like that, but I missed the fact that you can figure out $\\operatorname{Im} z$ from the argument. Thanks! – Milosz Wielondek Mar 14 '12 at 15:45", "of the problem. It also Pythagorean triples. $$u^2+v^2=\\sqrt{a^2+b^2}$$ The decision boils down to this primitive idea. $$a=2uv$$ $$b=u^2-v^2$$ – individ Oct 8 '15 at 4:48 As soon as you get to general $$z^3 = a + b i$$ this may be impossible. In solving a cubic $z^3 + p z + q = 0$ with, say, rational coefficients, one may use Cardano's formula. If there is only one real (irrational) root and two complex conjugate roots, then this works in the sense of being able to separately calculate real and imaginary parts using real square and cube roots. However, if there are three real irrational roots, a situation called CASUS IRREDUCIBILIS, then Cardano's formula is just a sum of cube roots of complex numbers $a+bi,$ with no way to separate real and imaginary parts. The terms are summed in a way that guarantees real answers, but that is not satisfying. Now, if you begin with general $z^3 = a + b i,$ say with $a,b \\in \\mathbb Q,$ and carefully write out equations for the real and imaginary parts of this $z,$ you get cubics with three real roots, where you cannot separate parts for those...it is all pretty circular, and hopeless. - Hi Will, indeed I knew about the casus irreducibilis which might one of the key", "$r^4e^ = 1.$ This means that $r^4 = 1,$ so $r=1$ since $r$ is a real number and $r geq 0$. In addition $4 heta$ must be an integer multiple of $2pi$. Choosing multiples of 0, 1, 2, and 3 gives $heta= 0, frac<2>, pi, frac<3pi><2>$. These correspond to the four numbers 1, $i$, -1, and $-i$, as pictured below: The fourth roots of 1 divide the unit circle evenly into four sections. Solution: 3 Solving equations 1. Writing $(a+bi)^3=(a^3-3ab^2)+(3a^2b-b^3)i=1+0i$ and equating real and imaginary parts of $(a+bi)^3=1$ gives $a^3 - 3ab^2 = 1$ and $3a^2b - b^3 = 0$ Factoring the second equation as $b(3a^2-b^2)=0$, we see that either $b=0$ or $3a^2=b^2$. If $b=0$, then $a^3=1$, giving the obvious cube root of 1. If $b eq 0$, then $3a^2=b^2$, and substituting this into $a^3-3ab^2=1$ gives $a^3-9a^3=1$, so $a=-frac<1><2>$, and then $b=pmfrac><2>$. 2. Similarly, if we write $(a+bi)^4=((a^2-b^2)+2abi)^2= (a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)i$ then equating imaginary parts in $(a+bi)^4=1$, gives $(4a^3b-4ab^3)=0.$ Factoring the left-hand side as $4ab(a^2-b^2)=0$, we see that any 4th root of 1 must have $a=0$, $b=0$, or $a^2=b^2$. If $a=0$, then the equation $(bi)^4=1$ quickly gives $b=pm 1$. Similarly, if $b=0$, then the equation $a^4=1$ again gives $a=pm 1$ (remember that $a$ and $b$ are real). Finally, if $a=pm b$, then equating the real parts of $(apm ai)^4=1$ gives", "$|b|^2-|a|^2=b_1^2+b_2^2-a_1^2-a_2^2\\neq 0$, and we get $$z_1=\\frac{c_1(a_1-b_1)-c_2(-a_2+b_2)}{|b|^2-|a|^2}$$ $$z_2=\\frac{(a_1+b_1)c_2-(a_2+b_2)c_1}{|b|^2-|a|^2}$$ The conclusion is, the system has exactly one solution only if $|a|\\neq |b|$. Hope this helps you! An approach that does not involve using the \"expanded\" form of a complex number. Let $w$ be a different solution than $z$, such that: $aw+b \\bar w +c=0$ and $az + b \\bar z + c= 0$. By subtracting we get: $a(z-w)+b(\\bar z- \\bar w)=0$ $⇒ a(z-w)=-b(\\bar z- \\bar w)$ (1) The conjucate of this expression must also equal zero, thus: $\\bar a(\\bar z- \\bar w)=-\\bar b(z-w)$ (2) By multyplying (1) and (2) we get: $a \\bar a(z-w)(\\bar z- \\bar w)=b\\bar b (z-w)(\\bar z- \\bar w) ⇒ a \\bar a=b\\bar b$ and finally we get: $|a|=|b|$ • Thank you very much, I must say I much prefer this way. – GRS Jan 17 '16 at 19:02 It's easier than that. What you wrote is a linear system of two equations (collect the real and imaginary terms into separate equations): \\begin{align} a_1 z_1-a_2 z_2 + b_1 z_1+b_2 z_2&=-c_1\\\\ a_2 z_1+a_1 z_2 + b_2 z_1-b_1 z_2&=-c_2 \\end{align} This has a single solution, if the rank of the matrix on the left is full (its determinant nonzero) - that's basics of linear algebra. If the determinant is zero, then you'll have infinite amount of solutions or no", "bi is an imaginary number if b = 0, and (a + bi )(a + bi ) = a2 + b2 is a real number. So for a complex number z = a + bi , z z = a2 + b2 and thus we can define the modulus of z to be z z = a2 + b2 , which we denote by | z |.Example 6.9 Let z1 = -2 + 3 i and z2 = 3 + 4 i . Find z1 + z2 , z1 - z2 , z1 z2 , z1 / z2 , | z1 |, and | z2 |.Solution: Using our rules and definitions, we have:z1 + z2 = (-2 + 3 i ) + (3 + 4 i ) z1 - z2 = (-2 + 3 i ) - (3 + 4 i ) z1 z2 = (-2 + 3 i ) (3 + 4 i )= -18 + i = -5 - i = 1 + 7i= ((-2)(3) - (3)(4)) + ((-2)(4) + (3)(3)) i-2 + 3 i z1 = z2 3 + 4i (-2)(3) + (3)(4) + ((3)(3) - (-2)(4)) i = 32 + 42 17 6 + i = 25 25 | z1 | = = | z2 | = = 5(-2)2 + 32 13 32 +", "# Solving Complex Number Equation Let $z = a + ib , z^* = a - ib$ I need to find all possible solutions to $2z=(z^*)^2$ $(z^*)^2 = (a^2 - b^2) -i2ab$ $2z = 2a + i2b$ $\\implies 2a + i2b = a^2 -b^2 -ia2b$ $\\implies a^2 - b^2 -i2ab - 2a - i2b = 0$ $\\implies a(a -i2b - 2) - b(b + i2) = 0$ I found solution to the above equation $z = 0$. However, I do not know how to find the rest of the solutions from here. - Alternatively, you could use polar coordinates. – 1015 Mar 25 '13 at 16:14 You should separate the real and imaginary parts of your equation, keeping in mind that $a,b$ are real. So from $a^2-b^2-2iab-2a-2ib=0$ you go to $$a^2-b^2-2a=0\\\\-2iab-2ib=0\\\\ab+b=0$$ and solve the first and third as a pair of simultaneous equations. @0xFF, do you know, $a+ib=c+id\\iff a=c,b=d$. Now $0=0+i0$ i.e., $c=d=0$ – lab bhattacharjee Mar 25 '13 at 16:30 @oxFF: two complex numbers are equal precisely when the real and imaginary parts are both equal. In particular, the origin of the complex plane is $0+0i$. So if you wanted to solve $z^2=i$, you could write $z=a+bi, z^2=a^2+2abi-b^2=i$, so $a^2-b^2=0, 2ab=1$ and solve these two. – Ross Millikan Mar 25 '13 at 16:30", "# Proving if $z$ is an n'th root, $\\bar z$ is also an n'th root Let $$n>0$$ be an even number, and let $$z$$ be an $$n$$'th root of a real number. Is $$\\bar z$$ also an $$n$$'th root of this number? My answer is yes. The way I solved this was to consider a complex number $$z = a+bi$$ on the (polar) form $$z = re^{i\\theta}$$. Solving $$z^n = re^{i\\theta}$$, I get $$z = \\sqrt[n] r e^{{(i\\theta+2\\pi m)}/n}$$, $$0 \\leq m \\leq n-1$$. Since $$n>0$$ is an even number, we will always have an even number of solutions. If we then draw the solutions in the complex plane, we will always have that for each solution that $$z$$ that has an angle $$\\phi=\\theta$$, we will always find $$\\bar z$$, also a solution with $$\\phi=-\\theta$$ on the complex plane. However, my answer is more a geometric interpreration/realization than a rigorious algebraic proof. I wanted to try to prove this in a more general algebraic form: My attempt was to assume $$z=a+bi$$ is an $$n$$'th root of a real number. We now want to show that $$z=a-bi$$ is also an $$n$$'th root of the same real number. I tried setting up $$(a+bi)^2 = a^2-b^2+2iab$$ and $$(a-bi)^2 = a^2-b^2-2iab$$, noticing that the real part of these numbers are always the", "# On complex numbers and absolute values Exercise 1.31 of Analysis by Apostol states: Given three complex numbers $z_1,z_2,z_3$ such that $|z_1| = |z_2| = |z_3| = 1$ and $z_1 + z_2 +z_3 = 0$. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with centre at the origin. Solution: It is clear that the three numbers are vertices of a triangle inscribed in the unit circle with centre in the origin. It remains to show that $|z_1-z_2| = |z_2 -z_3| = |z_3 - z_1|$ Note that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$ ... I have a problem noticing that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$ can anybody help me out? By direct computation, $$|2z_1+z_3|^2=(2z_1+z_3)(2\\bar{z}_1+\\bar{z}_3)=4|z_1|^2+2z_3\\bar{z}_1+2z_1\\bar{z}_3+|z_3|^2\\\\ =5+2z_3\\bar{z}_1+2z_1\\bar{z}_3.$$ Similarly, $$|z_1+2z_3|^2=(z_1+2z_3)(\\bar{z}_1+2\\bar{z}_3)=|z_1|^2+2z_3\\bar{z}_1+2z_1\\bar{z}_3+4|z_3|^2\\\\ =5+2z_3\\bar{z}_1+2z_1\\bar{z}_3.$$ So the equality you seek follows from $|z_1|=|z_3|$ rather than $z_1+z_2+z_3=0$." ]

[ "and express absolute values using conjugation. $$\\exists z\\in\\mathbb C\\,\\forall k\\in\\{1,2,3\\}:\\quad 1=(z-z_k)(\\bar z-\\bar z_k)=z\\bar z-z\\bar z_k-\\bar zz_k+z_k\\bar z_k$$ Now we can treat $a=-\\bar z$ and $b=-z$ as two distinct variables and obtain $$ab + az_k + b\\bar z_k + \\lvert z_k\\rvert^2 - 1 = 0$$ Three conditions for two variables, which is still too much. So let's treat $c=ab$ as a third variable. $$az_k + b\\bar z_k + c = 1 - \\lvert z_k\\rvert^2$$ So now we have a $3\\times 3$ system of equations, and once we have solved that, we can check whether $c=ab$ holds. I guess that if it does, then $a=\\bar b$ will hold as well. To verify that guess: For reasons of symmetry $c$ is real. Therefore $az_k$ and $b\\bar z_k$ must have opposite imaginary parts. Furthermore, $az_k\\cdot b\\bar z_k=c\\lvert z_k\\rvert^2\\in\\mathbb R$ so they must have opposite argument as well. If two numbers have opposite imaginary part and opposite argument, then they are conjugate to one another. Unfortunately solving a system of three linear equations is still harder than what I'd have hoped for. One big win is that this approach is very symmetric, and can be done symbolically as well. Unfortunately, the result still looks evil: $$z_1^2\\bar z_1^2z_2\\bar z_2 - z_1\\bar z_1^2z_2^2\\bar z_2 - z_1^2\\bar z_1z_2\\bar z_2^2 + z_1\\bar z_1z_2^2\\bar z_2^2 - z_1^2\\bar z_1^2\\bar", "The Complex Conjugate of a Complex Number # The Complex Conjugate of a Complex Number Recall that a complex number is in the form $z = a + bi$ where $a, b \\in \\mathbb{R}$, then the real part of $z$ is $\\Re (z) = a$ and the imaginary part of $z$ is $\\Im (z) = b$. Therefore we have that $z = \\Re (z) + \\Im (z) i$. We will now defined a special number call the complex conjugate of $z$ and look at its interesting properties. This will become useful when diving further into linear algebra. Definition: If $z = a + bi = \\Re(z) + \\Im(z) i$ is a complex number, then the Complex Conjugate of $z$ denoted $\\bar{z} = a - bi = \\Re(z) - \\Im(z) i$. For example, consider the complex number $z = 5 + 2i$. Then the complex conjugate of $z$ is $\\bar{z} = 5 - 2i$. Another example is the complex number $z = i - 1$ whose complex conjugate is $\\bar{z} = -1 - i$. Theorem 1: Let $z, z' \\in \\mathbb{C}$ be complex numbers $z = a + bi$ and $z' = a' + b'i$. Then: a) $z + \\bar{z} = 2 \\Re (z)$. b) $z - \\bar{z} = 2 \\Im (z) i$. c) $\\bar{\\bar{z}} = z$. d) $\\overline{z", "$\\bar{z} = \\bar{a + bi} = a - bi.$ We also notice that $z\\bar{z} = (a + bi)(a - bi) = a^{2} + b^{2},$ and so the product of a complex number with its conjugate is a real number. In fact, $z\\bar{z} = a^{2} + b^{2} = |z|^{2},$, and so $|z| = \\sqrt{z\\bar{z}}$ Exercise $$\\PageIndex{4}$$ Let $$w = 2 + 3i$$ and $$z = -1 + 5i$$ 1. Find $$\\bar{w}$$ and $$\\bar{z}$$. 2. Compute $$|w|$$ and $$|z|$$. 3. Compute $$w\\bar{w}$$ and $$z\\bar{z}$$. 4. What is $$\\bar{z}$$ if $$z$$ is a real number? 1. Using the definition of the conjugate of a complex number we find that $$\\bar{w} = 2 - 3i$$ and $$\\bar{z} = -1 - 5i$$. 2. Using the definition of the norm of a complex number we find that $$|w| = \\sqrt{2^{2} + 3^{2}} = \\sqrt{13}$$ and $$|z| = \\sqrt{(-1)^{2} + 5^{2}} = \\sqrt{26}$$. 3. Using the definition of the product of complex numbers we find that $w\\bar{w} = (2 + 3i)(2 - 3i) = 4 + 9 = 13$ $z\\bar{z} = (-1 + 5i)(-1 - 5i) = 1 + 25 = 26$ 4. Let $$z = a + 0i$$ for some $$a \\in \\mathbb{R}$$. Then $$\\bar{z} = a - 0i$$. Thus, $$\\bar{z} = z$$ when $$z \\in \\mathbb{R}$$. Summary In this section, we studied the", "The complex conjugate of $$x-iy$$ is $$x+iy$$. ### Complex Conjugate Examples Complex number $$z$$ Its conjugate $$\\bar z$$ 2 + 3i 2 - 3i -1 - 5i -1 + 5i $$i-\\dfrac{1}{2}$$ $$-i-\\dfrac{1}{2}$$ Important Notes 1. The process of finding the complex conjugate in math is NOT just changing the middle sign always, but changing the sign of the imaginary part. Observe the last example of the above table for the same. 2. The sum of a complex number and its conjugate is twice the real part of the complex number. $z+\\bar{z}=(x+ iy)+(x- iy)=2 x=2{Re}(z)$ 3. The difference between a complex number and its conjugate is twice the imaginary part of the complex number. $z-\\bar{z}=(x+ iy)-(x- iy)=2(iy)=2 {Im}(z)$ ## Properties of Complex Conjugates We know that $$z$$ and $$\\bar z$$ are conjugate pairs of complex numbers. Here are the properties of complex conjugates. • $$Re(z) = Re(\\bar z)$$ • $$Im(z) =- Im(\\bar z)$$ • If $$z$$ is purely real, then $$z=\\bar z$$. For example, the complex conjugate of $$z=3$$ is $$\\bar z =3$$. • If $$z$$ is purely imaginary, then $$z=-\\bar z$$. For example, the complex conjugate of $$z=3i$$ is $$\\bar z = -3i$$. • The bar over two complex numbers with some operation in between them can be distributed to each of the complex numbers. That is, if", "conjugate complex values, so set $y=a+bi, z=a-bi$ with $x, a$ and $b$ real. This gives \\begin{align} &x+2a=4 \\\\ &x^2+2a^2-2b^2=4 \\\\ &x^3+2a^3-6ab^2=4 \\end{align} which is actually easier than the real version. You can solve the second for $b^2$, insert it into the third along with $x=4-2a$ and get a cubic that will have $a=1$ as a solution. I tried playing with the symmetric polynomials, but didn't find any slick approach. -", "# Solving complex equation How do i further solve the following complex equation: $$z\\cdot \\bar{z} + z + \\bar{z} + i\\cdot z - \\overline{i \\cdot z} = 9 + 4i$$ $$a^{2} - b^{2} + 2a - 2b = 9 + 4i$$ How do i solve from here on ? - Do you mean $z\\bar{z}+z+\\bar{z}+iz-\\overline{iz}=9+4i$? – André Nicolas Aug 29 '12 at 14:29 Yes my latex is bad i didnt know how to do that negate sign. – Sterling Duchess Aug 29 '12 at 14:30 To solve the equation $$z\\bar{z}+z+\\bar{z}+iz-\\overline{iz}=9+4i,$$ we can proceed much as you did. Let $z=a+ib$. Then $z\\bar{z}=(a+ib)(a-ib)=a^2+b^2$. We have $iz=-b+ia$, so its conjugate is $-b-ia$. Our expression is therefore equal to $$a^2+b^2+(a+ib)+(a-ib)+(-b+ia)-(-b-ia),$$ which simplifies to $a^2+b^2+2a+2ia$. This is $9+4i$ precisely if the imaginary parts match and the real parts match. We end up with the equations $2a=4$ and $a^2+b^2+2a=9$. Now $a$ and then $b$ are easy to find. - To finish the job, a = 2 and b = 1 :) – Martin Aug 29 '12 at 14:44 @Martin: Well, there is also $b=-1$. – André Nicolas Aug 29 '12 at 14:46 So $$2ai = 4i, a^{2} + b^{2} + 2a = 9$$ Would then a = 2i and i put this a in to $$a^2{2} + b^{2} + 2a = 9$$ to solve for", "# Conjugate Complex Numbers Definition of conjugate complex numbers: In any two complex numbers, if only the sign of the imaginary part differ then, they are known as complex conjugate of each other. Conjugate of a complex number z = a + ib, denoted by $$\\bar{z}$$, is defined as $$\\bar{z}$$ = a - ib i.e., $$\\overline{a + ib}$$ = a - ib. For example, (i) Conjugate of z$$_{1}$$ = 5 + 4i is $$\\bar{z_{1}}$$ = 5 - 4i (ii) Conjugate of z$$_{2}$$ = - 8 - i is $$\\bar{z_{2}}$$ = - 8 + i (iii) conjugate of z$$_{3}$$ = 9i is $$\\bar{z_{3}}$$ = - 9i. (iv) $$\\overline{6 + 7i}$$ = 6 - 7i, $$\\overline{6 - 7i}$$ = 6 + 7i (v) $$\\overline{-6 - 13i}$$ = -6 + 13i, $$\\overline{-6 + 13i}$$ = -6 - 13i Properties of conjugate of a complex number: If z, z$$_{1}$$ and z$$_{2}$$ are complex number, then (i) $$\\bar{(\\bar{z})}$$ = z Or, If $$\\bar{z}$$ be the conjugate of z then $$\\bar{\\bar{z}}$$ = z. Proof: Let z = a + ib where x and y are real and i = √-1. Then by definition, (conjugate of z) = $$\\bar{z}$$ = a - ib. Therefore, (conjugate of $$\\bar{z}$$) = $$\\bar{\\bar{z}}$$ = a + ib = z. Proved. (ii) $$\\bar{z_{1} + z_{2}}$$ = $$\\bar{z_{1}}$$ + $$\\bar{z_{2}}$$ Proof:", "0$ Firstly let us remark: The equation $z^4 + z^3 + z^2 + z + 1 = 0$ has no real solutions and since the polynomial has real coefficients the 4 non-real solutions come in conjugate pairs. Now the conjugate pair part is important since this means that if $\\alpha$ is a root then so is $\\bar{\\alpha}$ and the polynomial $z^4 + z^3 + z^2 + z + 1$ must contain the factor $(z - \\alpha)(z - \\bar{\\alpha}) = z^2 - (\\alpha + \\bar{\\alpha})z + |\\alpha|^2$ Since the roots $\\alpha$ all lie on the unit circle we have $|\\alpha|= 1$ and further we note that $\\alpha + \\bar{\\alpha}$ is twice the real part of $\\alpha$ which is precisely the quantity we want to determine. So what we really need to find is the coefficient $a$ and $b$ in the factorization $z^4 + z^3 + z^2 + z + 1 = (z^2- az + 1)(z^2 - bz + 1)$. Computing the product we get $(z^2- az + 1)(z^2 - bz + 1) = z^4 - (a + b)z^3 + (2 + ab)z^2 - (a + b)z + 1$ And then by comparing coefficients we get the system of equations $\\begin{cases}a + b = -1 & \\\\ ab = -1\\end{cases}$ Oh look! It’s a good old sum-product system of equations", "0$ Firstly let us remark: The equation $z^4 + z^3 + z^2 + z + 1 = 0$ has no real solutions and since the polynomial has real coefficients the 4 non-real solutions come in conjugate pairs. Now the conjugate pair part is important since this means that if $\\alpha$ is a root then so is $\\bar{\\alpha}$ and the polynomial $z^4 + z^3 + z^2 + z + 1$ must contain the factor $(z - \\alpha)(z - \\bar{\\alpha}) = z^2 - (\\alpha + \\bar{\\alpha})z + |\\alpha|^2$ Since the roots $\\alpha$ all lie on the unit circle we have $|\\alpha|= 1$ and further we note that $\\alpha + \\bar{\\alpha}$ is twice the real part of $\\alpha$ which is precisely the quantity we want to determine. So what we really need to find is the coefficient $a$ and $b$ in the factorization $z^4 + z^3 + z^2 + z + 1 = (z^2- az + 1)(z^2 - bz + 1)$. Computing the product we get $(z^2- az + 1)(z^2 - bz + 1) = z^4 - (a + b)z^3 + (2 + ab)z^2 - (a + b)z + 1$ And then by comparing coefficients we get the system of equations $\\begin{cases}a + b = -1 & \\\\ ab = -1\\end{cases}$ Oh look! It’s a good old sum-product system of equations", "# On complex numbers and absolute values Exercise 1.31 of Analysis by Apostol states: Given three complex numbers $z_1,z_2,z_3$ such that $|z_1| = |z_2| = |z_3| = 1$ and $z_1 + z_2 +z_3 = 0$. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with centre at the origin. Solution: It is clear that the three numbers are vertices of a triangle inscribed in the unit circle with centre in the origin. It remains to show that $|z_1-z_2| = |z_2 -z_3| = |z_3 - z_1|$ Note that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$ ... I have a problem noticing that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$ can anybody help me out? By direct computation, $$|2z_1+z_3|^2=(2z_1+z_3)(2\\bar{z}_1+\\bar{z}_3)=4|z_1|^2+2z_3\\bar{z}_1+2z_1\\bar{z}_3+|z_3|^2\\\\ =5+2z_3\\bar{z}_1+2z_1\\bar{z}_3.$$ Similarly, $$|z_1+2z_3|^2=(z_1+2z_3)(\\bar{z}_1+2\\bar{z}_3)=|z_1|^2+2z_3\\bar{z}_1+2z_1\\bar{z}_3+4|z_3|^2\\\\ =5+2z_3\\bar{z}_1+2z_1\\bar{z}_3.$$ So the equality you seek follows from $|z_1|=|z_3|$ rather than $z_1+z_2+z_3=0$.", "the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :) – Laura Salas Dec 29 '18 at 21:26 Another approach is to take the complex conjugate of your equation: $$3\\overline z-iz=4-i.$$ You now have two equations for $$z$$ and $$\\overline z$$. Now eliminate $$\\overline z$$ from them and solve for $$z$$. • !! Also works, thank you. This is a shorter way to do it. – Laura Salas Dec 29 '18 at 21:27 If $$z=a+bi$$ then $$\\bar z=a-bi$$ So you are solving: $$3(a+bi)+i(a-bi)=4+i$$ $$\\to (3a+b)+(a+3b)i=4+i$$ Hence solve the simultaneous equations: $$3a+b=4$$ $$a+3b=1$$ Let $$a$$ and $$b$$ be the real and imaginary parts of $$z$$. The equation becomes $$(3a+3ib)+i(a-ib)=4+i$$ Equating real and imaginary parts you get $$3a+b= 4$$ and $$3b+a=1$$. Now you should be able to discover that $$a=\\frac {11} 8$$ and $$b =-\\frac 1 8$$, so $$z=\\frac {11} 8-i\\frac 1 8$$. • It’s $3a + 3ib$ in the first bracket of your first equation. – Live Free or π Hard Dec 28 '18 at 0:45 • @LiveFreeorπHard That was a typo. I had used $3a+3ib$", "How to Find the Conjugate of a Complex Number Here you will learn how to find the conjugate of a complex number and properties of conjugate with examples. Let’s begin – How to Find the Conjugate of a Complex Number Let z = a + ib be a complex number. Then the conjugate of z is denoted by $$\\bar{z}$$ and is equal to a – ib. Thus, z = a + ib $$\\implies$$ $$\\bar{z}$$ = a – ib It follows from this definition that the conjugate of a complex number is obtained by replacing i by -i. For Example : If z = 3 + 4i, then $$\\bar{z}$$ = 3 – 4i. Properties of Conjugate If $$z$$, $$z_1$$, $$z_2$$ are complex numbers, then (i) $$\\bar{\\bar{z}}$$ = z (ii) z + $$\\bar{z}$$ = 2 Re (z) (iii) z – $$\\bar{z}$$ = 2i Im (z) (iv) z = $$\\bar{z}$$ $$\\iff$$ z is purely real (v) z + $$\\bar{z}$$ = 0 $$\\implies$$ z is purely imaginary (vi) z$$\\bar{z}$$ = $$[Re (z)]^2$$ + $$[Im (z)]^2$$ (vii) $$\\bar{z_1 + z_2}$$ = $$\\bar{z_1}$$ + $$\\bar{z_2}$$ (viii) $$\\bar{z_1 – z_2}$$ = $$\\bar{z_1}$$ – $$\\bar{z_2}$$ (ix) $$\\bar{z_1z_2}$$ = $$\\bar{z_1}$$ $$\\bar{z_2}$$ (x) $$\\bar{z_1\\over z_2}$$ = $$\\bar{z_1}\\over \\bar{z_2}$$ Example : Multiply 3 – 2i by its conjugate. Solution : The conjugate of 3 – 2i is 3 +", "(the real part $a$ or the imaginary part $bi$) can be completely independent of each other. As a result a complex number can represent in condensed form 2 numbers. Moreover this also means that a complex number to be completely determined each of the dimensions needs to be determined as well. On the other hand, from every complex number $z=a+bi$ (along with its complex conjugate $\\bar{z}=a-bi$), one can calculate 2 real numbers ($a$, $b$) as: $$a = (z + \\bar{z})/2$$ $$b = (z - \\bar{z})/{2i}$$ Since $a$ and $b$ can be completely independent of each other, so can $z$ and $\\bar{z}$. There is a complete symmetry of representation (if such a term can be used). This means that in QFT (for example), instead of doing variations on the $a$, $b$ real fields, one can equivalently (by the same token) do variations on the $z$, $\\bar{z}$ complex fields and so on. UPDATE: To get into the abstract mathematics a little more. Complex conjugation is (the natural) automorphism of the field of complex numbers. Furthermore the complex conjugate of a complex number $z$ can not be derived from any analytic function of $z$ (roughly meaning rational functions of $z$ and power series). This further makes the complex conjugate $\\bar{z}$ natural candidate for treating as separate field. Quiz: How many components", "# Solving a complex number equation with both $z$ and its conjugate $\\bar z$ Determine all possible values of $$z\\in\\mathbb{C}$$ that satisfy the equation $$4z = \\overline{z}^2$$. Where $$\\overline{z}$$ represents the complex conjugate. (Hint: There are $$4$$ solutions.) ### Observations If we had $$4z=z^2$$, that would be an easy quadratic equation, with solutions $$0,4$$. And if it was $$4\\bar z = \\bar z^2$$, then after substitution $$\\zeta=\\bar z$$ we have a quadratic equation again. But this equation has both $$z$$ and $$\\bar z$$. I'm not sure how to solve these types of problems. Any tips or how to do these would be great thanks! Hint: use the polar representation $z = r e^{i\\theta}$. There is the obvious solution $z=0$. From now on, assume $z\\ne 0$. Taking norms, we find that $|z|=4$. Let $z=4e^{i\\theta}$. Then we want $e^{i\\theta}=e^{-2i\\theta}$. We leave the rest to you. Denote $z=a+bi$ with $a,b\\in\\mathbb{R}$, then your equation says: $$4(a+bi)=(a-bi)^2.$$ Which means that $$4a+4bi=(a^2-b^2)-2abi.$$ Now the above equality of complex numbers is a system of equations of real numbers: $$\\begin{cases} 4a=a^2-b^2\\\\ 4b=-2ab \\end{cases}.$$ Solve it and you'll find the solutions. • Solving the above system I got the $4$ solutions: $z=-2\\pm\\sqrt{12}i$, $z=0$, or $z=4$. – DKal Mar 25, 2014 at 6:53 • I got the first two solutions, but how did you get z=0 or z=4?", "of $$x+iy$$ is $$x-iy$$. This means that it either goes from positive to negative or from negative to positive. &= -6 -4i \\end{align}. The notation for the complex conjugate of $$z$$ is either $$\\bar z$$ or $$z^*$$.The complex conjugate has the same real part as $$z$$ and the same imaginary part but with the opposite sign. Can we help Emma find the complex conjugate of $$4 z_{1}-2 i z_{2}$$ given that $$z_{1}=2-3 i$$ and $$z_{2}=-4-7 i$$? In the same way, if $$z$$ lies in quadrant II, can you think in which quadrant does $$\\bar z$$ lie? The complex conjugate has the same real component a a, but has opposite sign for the imaginary component imaginary part of a complex If z=x+iyz=x+iy is a complex number, then the complex conjugate, denoted by ¯¯¯zz¯ or z∗z∗, is x−iyx−iy. Hide Ads About Ads. These complex numbers are a pair of complex conjugates. The bar over two complex numbers with some operation in between can be distributed to each of the complex numbers. This consists of changing the sign of the imaginary part of a complex number. The complex conjugate of a + bi is a – bi , and similarly the complex conjugate of a – bi is a + bi . number formulas. If $$z$$ is purely real, then $$z=\\bar", "z_3) + z_2 Im(\\bar{z_3} z_1 ) + z_3 Im(\\bar{z_1} z_2)$ $\\large \\frac{1}{2 i} (z_1 (\\bar{z_2} z_3) – z_2 \\bar{z_3}) + z_2 (\\bar{z_3} z_1) – z_3 \\bar{z_1}) + z_3 (\\bar{z_1} z_2) – z_1 \\bar{z_2}))$ $\\large \\frac{1}{2 i} (z_1 \\bar{z_2}z_3 – z_1 z_2 \\bar{z_3} + z_2 \\bar{z_3} z_1 – z_2 z_3 \\bar{z_1} + z_3 \\bar{z_1} z_2 -z_3 z_1 \\bar{z_2})$ = 0 Illustration : Consider a quadratic equation az2 + bz + c = 0 where a , b , c , are complex numbers. Find the condition that the equation has (i) one purely imaginary root (ii) one purely real root (iii) two purely imaginary roots (iv) two purely real roots Illustration : Let z1 , z2 , z3 be three distinct complex numbers satisfying | z1 − 1| = | z2 − 1| = | z3 −1|. Let A, B, and C be the points represented in the Argand plane corresponding to z1 , z2 and z3 respectively. Prove that z1 + z2 + z3 = 3 if and only if ΔABC is an equilateral triangle. Solution: |z1 − 1| = |z2 − 1| =|z3 −1| ⇒ The point corresponding to 1(say P) is equidistant from the points A , B and C. ⇒ P is the circumcentre of the ΔABC Now if z1 + z2 + z3 =", "Using $z_1$ and $z_2$ from above we get $$z_1-z_2=(2-4)+(5-2)i=-2+3i$$ ## 17.4 Product of Complex Numbers The product of two complex numbers can be found by multiplying the parts of one number by the parts of the other. Using $z_1$ and $z_2$ from above we get: $z_1 \\times z_2$ $=$ $(2+5i)(4+2i)$ $=$ $2 \\times 4 + 2 \\times 2i + 5i \\times 4 + 5i \\times 2i$ $=$ $8 + 4i + 20i + 10i^2$ Remember that $i=\\sqrt{-1}$ so $i^2=-1$ so $z_1 \\times z_2$ $=$ $8 + 24i - 10$ $=$ $-2 + 24i$ Another, and easier, way to find the product of two complex numbers is to put the numbers in polar form. Imagine we have $z_1 = r_1(cos \\theta + isin \\theta)$ and $z_2 = r_2(cos \\phi + isin \\phi)$ then $$z_1 z_2 = r_1 r_2(cos(\\theta + \\phi) + isin( \\theta + \\phi))$$ or $$z_1 z_2 = r_1 r_2 \\angle (\\theta + \\phi)$$ ## 17.5 The Complex Conjugate You probably already know the identity $(a+b)(a-b)=a^2-b^2$ With a small change we can write $(a+ib)(a-ib)=a^2-(ib)^2$. We also know $i = \\sqrt{-1}$ so $i^2 = -1$ which means $(a+ib)(a-ib)=a^2+b^2$. The complex conjugate is the quantity that creates a real number from a complex number. If $z=a+ib$ then the complex conjugate is $z^*=a-ib$ and if $z=a-ib$ then the complex conjugate is", "this correct? I recognized that since there's a 3, this means the other root must be a conjugate, hence $(z-(3+7i))(z-(3-7i))$ $(z-3)^2-(7i)^2 =0$ $z^2+6z+58=0$ $3z^2+18z+174=0$ I would take a more general approach to this. You are given that: $3x^2 + Ax + B = 3(x - z)(x - \\overline{z})$ Where $z = 3 + 7i$. Try expressing $A, B$ in terms of $z$ in general first. And see what you get. #### jisbon I would take a more general approach to this. You are given that: 3x2+Ax+B=3(x−z)(x−¯¯¯z)3x2+Ax+B=3(x−z)(x−z¯) Where z=3+7iz=3+7i. Try expressing A,BA,B in terms of zz in general first. And see what you get. I got: 3x2+Ax+B=3(x2−x¯¯¯z−xz+z¯¯¯z)3x2+Ax+B=3(x2−xz¯−xz+zz¯) Ax+B=−(3¯¯¯z+z)x+z¯¯¯zAx+B=−(3z¯+z)x+zz¯ SoA=−(3¯¯¯z+z)A=−(3z¯+z) and B=z¯¯¯zB=zz¯ #### PeroK Homework Helper Gold Member 2018 Award I got: $3x^2+Ax+B = 3 (x^2 -x\\overline{z}-xz+z\\overline{z})$ $Ax+B= -(3\\overline{z}+z)x+z\\overline{z}$ So$A =-(3\\overline{z}+z)$ and $B= z\\overline{z}$ That's not quite right. You need to be more careful. #### HallsofIvy Homework Helper I am puzzled by your statement \"since there's a 3, this means the other root must be a conjugate\". The other root must be the complex conjugate of 3+ 7i, 3- 7i, because all coefficients are real. Because the leading coefficient is 3, the polynomial must be 3(x-(3+7i))(x- (3- 7i))= 3((x-3)- 7i)((x-3)+ 7i)= 3((x- 3)^2+ 49)= 3(x^2- 6x+ 58)= 3x^2- 18x+ 174. #### jisbon That's not quite right. You need to be", "= 4+6i$ \\kern-7ex \\begin{aligned} (1+2i)(3+4i)\\ &= 1(3+4i)+2i(3+4i) = 3+4i+6i+8i^2 \\\\ &= (3 - 8) + 10 i = -5+10i \\\\[5pt] 5(3+4i)\\ &= 15+20i\\\\[5pt] (-1)(c+di)\\ &= -c -di \\end{aligned} The conjugate of $z = a+bi$ is $\\bar{z} = a-bi$. Reflection in real axis. We'll use this for division of complex numbers in a moment. Theorem (Properties of conjugates): Let $w$ and $z$ be complex numbers. Then: 1. $\\bar{\\bar{z}} = \\query{z}$ 2. $\\overline{w+z} = \\query{\\bar{w} + \\bar{z}}$ 3. $\\overline{w z} = \\query{\\bar{w} \\bar{z}}$ (typo in text) (good exercise) 4. If $z \\neq 0$, then $\\overline{w/z} = \\query{\\bar{w} / \\bar{z}}$ (see below for division) 5. $z$ is real if and only if $\\bar{z} = \\query{z}$ The absolute value or modulus $|z|$ of $z = a+bi$ is $$\\kern-4ex |z| = |a+bi| = \\sqrt{a^2+b^2}, \\qtext{the distance from the origin.}$$ Examples: $|3| = 3,\\, |-3| = 3,\\, |\\pm i| = 1,\\, |3+4i| = \\sqrt{3^2+4^2} = 5$. Note that $$\\kern-7ex z \\bar{z} = (a+bi)(a-bi) = a^2 -abi+abi-b^2 i^2 = a^2 + b^2 = |z|^2$$ This means that for $z \\neq 0$ $$\\kern-4ex \\frac{z \\bar{z}}{|z|^2} = 1 \\qtext{so} z^{-1} = \\frac{\\bar{z}}{|z|^2}$$ This can be used to compute quotients of complex numbers: $$\\kern-4ex \\frac{w}{z} = \\frac{w}{z} \\frac{\\bar{z}}{\\bar{z}} = \\frac{w \\bar{z}}{|z|^2}.$$ Example: $$\\kern-8ex \\frac{-1+2i}{3+4i} = \\frac{-1+2i}{3+4i} \\frac{3-4i}{3-4i} = \\frac{5+10i}{3^2+4^2} = \\frac{5+10i}{25} = \\frac{1}{5} + \\frac{2}{5}i$$ Theorem (Properties of", "# Algebraic Form of Complex Numbers Complex number z=(a;b) can be represented in the form z=a+bi. This is called algebraic form of complex number. In the representation z=a+bi, i^2=-1 and i is called imaginary unit. a is real part (denoted by Re z) and b is imaginary part (denoted by Im z) of complex number. If imaginary part of complex number a+bi is not 0 (b!=0) then such number is called imaginary, for example 1+bi. If a=0 and b!=0 then complex number is called purely imaginary. For example, 3i is purely imaginary. If b=0 then we obtain real number a. If we are given complex number z=a+bi then number a-bi is called complex conjugate of z. It is denoted by bar(z) or z^(**). For example complex conjugate of 2+3i is 2-3i. Sum and product of two complex conjugates are real numbers. Indeed, if z=a+bi then bar(z)=a-bi , so z+bar(z)=a+bi+a-bi=2a. Also, zbar(z)=(a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2-b^2(-1)=a^2+b^2. Absolute value (or modulus) of complex number z is real number r=|z|=sqrt(zbar(z))=sqrt(a^2+b^2). Conjugate complex numbers have same absolute values, i.e. |a+bi|=|a-bi|." ]

[ "+0 Help! 0 106 2 +803 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Lightning Jul 6, 2018 #1 0 Guest Jul 6, 2018 #2 +20108 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "+0 # Help! 0 219 2 +1110 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Jul 6, 2018 #1 0 Jul 6, 2018 #2 +22188 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "answers 83 views Define the operation $*$ by $a * b = 4a−3b+ab$ for all $a, b ∈ R$. For what real numbers y does $12 = 3 * y$?Define the operation by $a &nbsp; b = 4a−3b+ab$ for all $a, b ∈ R$. For what real numbers y does $12 = 3 y$? ... close", "The operation Δ is defined for all nonzero x and y by x Δ y ~$=x+\\frac{x}{y}$~ . If a > o, then 1 Δ (1 Δ a) = a a+1 ~$\\frac{a}{a+1}$~ ~$\\frac{a+2}{a+1}$~ ~$\\frac{2a+1}{a+1}$~", "the indicated operation. $\\frac{5}{4} \\div \\frac{15}{8}$...", "+0 # Operation +1 39 1 For any three real numbers a, b, and c, with $b \\neq -c$, the operation P is defined by $P(a,b,c) = \\frac{a}{b + c}.$ What is P(P(1,2,3),P(2,3,1),P(3,1,2))? Apr 6, 2021 #1 +485 +1 $P(1,2,3)=\\frac{1}{2+3}=\\frac{1}{5}$ $P(2,3,1)=\\frac{2}{3+1}=\\frac{2}{4}=\\frac{1}{2}$ $P(3,1,2)=\\frac{3}{1+2}=\\frac{3}{3}=1$ $P(\\frac{1}{5},\\frac{1}{2},1)=\\frac{\\frac{1}{5}}{\\frac{1}{2}+1}=\\frac{1}{\\frac{15}{2}}=\\boxed{\\frac{2}{15}}$ Apr 6, 2021", "something like the operation $|a - b|$.", "An operation @ is defined by the equation a@b = (a - b) / (a 10 04 Mar 2007, 16:49 Display posts from previous: Sort by", "08:04 Applying the operation as described, we get $$(4#)# = (\\frac{1}{4}+4)# = \\frac{17}{4}# = \\frac{4}{17}+\\frac{17}{4} = \\frac{508}{68} = 4.48 < 4.5$$ Re: The operation # is defined by x# [#permalink] 28 Sep 2017, 08:04 Display posts from previous: Sort by", "+0 # Functional Operations +1 36 1 +193 The operation $*$ is defined for non-zero $a$ and $b$ as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 13$ and $a \\times b = 25$, what is the value of $a*b$? Express your answer as a common fraction. Feb 14, 2021 #1 +283 +2 Since we know that a x b = 25, we can put 1/a + 1/b as (b + a) / 25. Then we know that a + b = 13, so 1/a + 1/b would equal 13/25, which is your answer :) Feb 14, 2021 #1 +283 +2 Since we know that a x b = 25, we can put 1/a + 1/b as (b + a) / 25. Then we know that a + b = 13, so 1/a + 1/b would equal 13/25, which is your answer :) Logarhythm Feb 14, 2021", "operation is $b^2$? Nov 23 '15 at 4:22 • Is it correct that answer is $b*2^{b/2}$? Nov 23 '15 at 4:28 • You can neglect the factor $b$. When you expand $2^\\frac{b}{2}$ into a power series, all powers of $b$ will occur. Multiplying the power series with $b$ will not change the overall picture (just increase all powers by $1$. Nov 23 '15 at 8:11", "of c? The operation  is defined by r s=12 $r^{-1}$(s+3) for all positive numbers r and s. If x 3=18, what is the value of x? The operation is defined by x✸y=$\\frac{x^{2}}{y}$+$\\frac{x}{y}$ for all numbers x and y, where y≠0. What is the value of (9✸ (-9))+((-9) ✸9)? 25000 +道题目 174本备考书籍", "that a ≠ -b. If a ≠ -c and a@c = 0, then c = (A) -a (B) -1/a (C) 0 (D) 1/a (E) a Math Expert Joined: 02 Sep 2009 Posts: 60734 Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink] ### Show Tags 13 Dec 2012, 10:16 3 2 An operation @ is defined by the equation a@b = (a - b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c and a@c = 0, then c = (A) -a (B) -1/a (C) 0 (D) 1/a (E) a Given that $$a@b = \\frac{a - b}{a + b}$$, thus $$a@c = \\frac{a - c}{a + c}$$. Also given that $$a@c = \\frac{a - c}{a + c}=0$$ --> $$a-c=0$$ --> $$a=c$$. _________________ ##### General Discussion Manager Joined: 21 Oct 2013 Posts: 175 Location: Germany GMAT 1: 660 Q45 V36 GPA: 3.51 Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink] ### Show Tags 24 Nov 2013, 12:04 1 Bunuel wrote: An operation @ is defined by the equation a@b = (a - b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c", "that a ≠ -b. If a ≠ -c and a@c = 0, then c = (A) -a (B) -1/a (C) 0 (D) 1/a (E) a Given that $$a@b = \\frac{a - b}{a + b}$$, thus $$a@c = \\frac{a - c}{a + c}$$. Also given that $$a@c = \\frac{a - c}{a + c}=0$$ --> $$a-c=0$$ --> $$a=c$$. Hi Bunnuel, I've never senn such a task before in my life. Could you please explain what the @ means or what the whole expression \"operation @\" means? or do you recommend any books/links for this ? (I already worked through MGMAT Foundations of Math but really NEVER seen this before) Math Expert Joined: 02 Sep 2009 Posts: 27512 Followers: 4319 Kudos [?]: 42453 [0], given: 6029 Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink] 24 Nov 2013, 12:47 Expert's post unceldolan wrote: Bunuel wrote: An operation @ is defined by the equation a@b = (a - b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c and a@c = 0, then c = (A) -a (B) -1/a (C) 0 (D) 1/a (E) a Given that $$a@b = \\frac{a - b}{a + b}$$, thus $$a@c = \\frac{a - c}{a + c}$$. Also given", "1, 4]$. In a single operation with $i = 1$, we set $a_1 = \\min(\\frac{0+1-2}{2}, 0)$ and $a_2 = \\max(\\frac{0+1+2}{2}, 1)$. So, after a single operation with $i = 1$, $a$ becomes equal to $[-\\frac{1}{2}, \\frac{3}{2}, 4]$. We can show that no operation has any effect on this array, so $F(a, b) = -\\frac{1}{2}$.", "# What operation is denoted by this subscript notation? What operation is denoted here? $$11_3$$ I believe the answer to this is 4. $11_3$ is the number $4$ written in base $3$: $$11_3 = 1\\cdot 3^1+1\\cdot 3^0 = 3+1=4$$ A small nitpick: it is not technically an operation, but a notation. $11_3=4_{10}$.", "# #4 - Operations with Fractions ### Sample Question: 4. $$\\frac{1}{2} + \\left (\\frac{2}{3} \\div \\frac{3}{4}\\right ) - \\left (\\frac{4}{5} \\times \\frac{5}{6}\\right )=$$ ? A. $$\\frac{1}{16}$$ B. $$\\frac{17}{27}$$ C. $$\\frac{13}{18}$$ D. $$\\frac{7}{9}$$ E. $$\\frac{5}{6}$$", "# Operations with rational numbers Keywords: $\\frac{a}{b}+\\frac{c}{d}=\\frac{ad+bc}{bd}$ Addition $\\frac{a}{b}-\\frac{c}{d}=\\frac{ad-bc}{bd}$ Subtraction $\\frac{a}{b}·\\frac{c}{d}=\\frac{ac}{bd}$ Multiplication $\\frac{a}{b}:\\frac{c}{d}=\\frac{ad}{bc}$ Division", "N. 13. (a) The operation (*) is defined on the set of real numbers, R, by $$x * y = \\frac{x + y}{2}, x, y \\in R$$. (i) Evaluate $$3 * \\frac{2}{5}$$. (ii) If $$8 * y = 8\\frac{1}{4}$$, find the value of y. (b) In $$\\Delta ABC, \\overline{AB} = \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}$$ and $$\\overline{AC} = \\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix}$$. If P is the midpoint of $$\\overline{AB}$$, express $$\\overline{CP}$$ as a column vector. Subscribe Notify of", "Last updated: 25 July 2022 Algebra Arithmetic Operations $$a(b+c) = ab + ac$$ $${{a \\over b} + {c \\over d}} = {{ad + bc} \\over bd}$$ $${{a + c} \\over b} = {{a \\over b} + {c \\over b}}$$ {{a \\over b } \\over {c \\over d}} = {{a \\over b} \\times {d \\over c}}…" ]

[ "that it is \"proven\" that $$1 + 3 + 9 + 27 + ... = -\\frac{1}{2}$$ - I think this should have been included for the premise in the video: Let $x:=1-1+1-1+1-\\dots$, then we have $$x=1-(1-1+1-1+\\dots)=1-x$$ so $2x=1,\\ \\ x=1/2$. -", "Multiply $x$ by $10$, and subtract: $10x = 9.999\\dots$ $x = 0.999\\dots$ $\\therefore (10-1)x = 9$ $x =1$ $0.999\\dots = 1$ Previous Post", "of the second sequence is $\\frac{99,100}{100} = \\boxed{991, A}\\, .$", "1 = 0.$ Hence, we can write $(z – x^2)(z – x^6)\\dots (z – x^{358}) = z^{90} + 1$, and so[\\frac{1}{M} = \\dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \\dots |1 – x^{358}| = \\dfrac{1}{2^{90}} |1^{90} + 1| = \\dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.", "x_i$ for $n = 1, 2, \\dots , 10$. Should this be: $s_n = \\sum_{i=1}^{n} x_i$ for $n = 1, 2, \\dots , 10$? • Dec 27th 2008, 06:27 AM awkward Quote: Originally Posted by Grandad Should this be: $s_n = \\sum_{i=1}^{n} x_i$ for $n = 1, 2, \\dots , 10$?", "range $\\{2,3,\\dots,2^n\\}$. But this can be proved using Chebyshev's theorem.", "\\dotsb )$$ where $$|X_n|=1$$ for $$n\\gg 0$$.", "\\dotsc + 1 = \\frac{1}{2}n(n - 1)$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910}$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "+0 # The numbers \\$1,\\$ \\$2,\\$ \\$\\dots,\\$ \\$10\\$ are to be entered into the 10 boxes shown below, so that each number is used exactly once: \\[P = (\\squar 0 33 1 The numbers \\$1,\\$ \\$2,\\$ \\$\\dots,\\$ \\$10\\$ are to be entered into the 10 boxes shown below, so that each number is used exactly once: \\[P = (\\square + \\square + \\square + \\square + \\square)(\\square + \\square + \\square + \\square + \\square).\\]What is the maximum value of \\$P\\$? What is the minimum value of \\$P\\$? May 4, 2020", "$= \\frac{431}{900}$. Therefore $0.478888\\dots = \\frac{431}{900}$.", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "+0 0 64 1 +95 Let $$(a_1,b_1), (a_2,b_2), \\dots, (a_n,b_n)$$ be all the ordered pairs $$(a,b)$$ of complex numbers with $$a^2+b^2\\neq 0, a+\\frac{10b}{a^2+b^2}=5, \\text{ and } b+\\frac{10a}{a^2+b^2}=4.$$ Find $$a_1 + b_1 + a_2 + b_2 + \\dots + a_n + b_n.$$ Sep 17, 2022", "know the value of $x(1)$ since I haven't opened boxes $1,101,201,\\dots$ yet. What I can do though is let $y(1) = max(x(2),x(3),...)+1$ be larger than all the observed numbers. It just so happened that $x(4)=7$ was the biggest, so $y(1)=8$. It's finally time to open most of the remaining boxes. I'll open all the boxes in my sequence $1, 101, 201, \\dots$, starting with the box given by $y(1)=8$: box $801$. (Since $y(1)$ has to be at least $1$, this strategy always leaves at least box $001$ closed.) Let's see what I got: • 1 -> ??? • 101 -> ??? • ... • 701 -> ??? • 801 -> 7pi+sqrt(3) • 901 -> 2 • 1001 -> 8 • 1101 -> 4 • etc. Seeing those last digits, I know enough to figure out which group it belongs to: the group with representative (e,2,7,1,8,2,8,1,8,2,8,4,6,...). I now know enough to make my guess. I'm going to use the representative, and pick the box given by $y(1)-1=7$, which is the maximum value of $\\{x(2),x(3),\\dots\\}$ (which we said was $x(4)$ in this example). In this case, the seventh entry is $1$ (note that we're zero-indexing so that the first entry goes with box $001$). So I'll guess that box $701$ contains $1$. Of course, mathematician #$n$ will do exactly the", "box. $(a_1) \\to (1,2, \\dots r) =r$ ways $(a_2) \\to (1,2, \\dots r)=r$ ways . . . $(a_n) \\to (1,2, \\dots r)= r$ ways $(a_i) \\to (1,2, \\dots r)$ represents the possible slots for $a_i$. -", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "location: 43. Two players play a game: the first player thinks of $n$ integers ${x}_{1}$, ${x}_{2}$, $\\dots$, ${x}_{n}$, each with one digit, and the second player selects some numbers ${a}_{1}$, ${a}_{2}$, $\\dots$, ${a}_{n}$ and asks what is the value of the sum ${a}_{1}{x}_{1}+{a}_{2}{x}_{2}+\\dots +{a}_{n}{x}_{n}$. What is the minimum number of questions used by the second player to find the integers ${x}_{1}$, ${x}_{2}$, $\\dots$, ${x}_{n}$? Solution. We are going to prove that the second player needs only one question to find the integers ${x}_{1}$, ${x}_{2}$, $\\dots$, ${x}_{n}$. Indeed, let him choose ${a}_{1}=100$, ${a}_{2}={100}^{2}$, $\\dots$, ${a}_{n}={100}^{n}$ and ask for the value of the sum ${S}_{n}=100{x}_{1}+{100}^{2}{x}_{2}+\\dots +{100}^{n}{x}_{n} .$ Note that $\\begin{array}{cc}|\\hfill & \\frac{100{x}_{1}+{100}^{2}{x}_{2}+\\dots +{100}^{n-1}{x}_{n-1}}{{100}^{n}}|\\hfill \\\\ \\multicolumn{0}{c}{}& \\le \\frac{100‖{x}_{1}‖+{100}^{2}‖{x}_{2}‖+\\dots +{100}^{n-1}‖{x}_{n-1}‖}{{100}^{n}}\\hfill \\\\ \\multicolumn{0}{c}{}& \\le \\frac{9\\left(100+{100}^{2}+\\dots +{100}^{n-1}\\right)}{{100}^{n}}\\hfill \\\\ \\multicolumn{0}{c}{}& <\\frac{{10}^{2n-1}}{{10}^{2n}}=\\frac{1}{10} .\\hfill \\end{array}$ Hence $|\\frac{{S}_{n}}{{100}^{n}}-{x}_{n}|=\\frac{100{x}_{1}+{100}^{2}{x}_{2}+\\dots +{100}^{n-1}{x}_{n-1}}{{100}^{n}}<\\frac{1}{10} ,$ and ${x}_{n}$ can be obtained. Now, we can find the sum ${S}_{n-1}={S}_{n}-{100}^{n}{x}_{n}=100{x}_{1}+{100}^{2}{x}_{2}+\\dots +{100}^{n-1}{x}_{n-1}$ and similarly obtain ${x}_{n-1}$. The procedure continues until all the numbers are found. 44. Find the permutation $\\left\\{{a}_{1},{a}_{2},\\dots ,{a}_{n}\\right\\}$ of the set $\\left\\{1,2,\\dots ,n\\right\\}$ for which the sum $S=‖{a}_{2}-{a}_{1}‖+‖{a}_{3}-{a}_{2}‖+\\dots +‖{a}_{n}-{a}_{n-1}‖$ has maximum value. Solution. Let $a=\\left({a}_{1},{a}_{2},\\dots ,{a}_{n}\\right)$ be a permutation of $\\left\\{1,2,\\dots ,n\\right\\}$ and define $f\\left(a\\right)=\\sum _{k=1}^{n-1}‖{a}_{k+1}-{a}_{k}‖+‖{a}_{n}-{a}_{1}‖ .$ With ${a}_{n+1}={a}_{1}$ and ${\\epsilon }_{0}={\\epsilon }_{n}$, we find that $f\\left(a\\right)=\\sum _{k=1}^{n}{\\epsilon }_{k}\\left({a}_{k+1}-{a}_{k}\\right)=\\sum _{k=1}^{n}\\left({\\epsilon }_{k-1}-{\\epsilon }_{k}\\right){a}_{k}$ where ${\\epsilon }_{1}$, ${\\epsilon }_{2}$, $\\dots$, ${\\epsilon }_{n}$ are all equal to" ]

[ "+0 Help! 0 106 2 +803 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Lightning Jul 6, 2018 #1 0 Guest Jul 6, 2018 #2 +20108 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "+0 # Help! 0 239 1 +973 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Jul 7, 2018 #1 +10104 +2 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Jul 7, 2018 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction.", "+0 # Help! 0 219 2 +1110 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Jul 6, 2018 #1 0 Jul 6, 2018 #2 +22188 +1 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$?", "The operation Δ is defined for all nonzero x and y by x Δ y ~$=x+\\frac{x}{y}$~ . If a > o, then 1 Δ (1 Δ a) = a a+1 ~$\\frac{a}{a+1}$~ ~$\\frac{a+2}{a+1}$~ ~$\\frac{2a+1}{a+1}$~", "+0 # Help! 0 99 1 +803 For any three real numbers $a$, $b$, and $c$, with $b \\neq c$, the operation $\\P$ is defined by $\\P(a, b, c) = \\frac{a}{b-c}.$What is $\\P\\left(\\P(1,2,3), \\P(2,3,1), \\P(3,1,2)\\right)$? Lightning Jul 6, 2018 #1 +963 +1 $$\\text{For any three real numbers a, b, and c, with } b \\neq c,\\\\\\text{the operation P is defined by } P(a, b, c) = \\frac{a}{b-c}.\\\\ \\text{What is } P\\left(P(1,2,3), P(2,3,1), P(3,1,2)\\right)?\\\\ P(1,2,3)=\\frac1{2-3}=-1\\\\ P(2,3,1)=\\frac2{3-1}=1\\\\ P(3,1,2)=\\frac3{1-2}=-3\\\\ P(-1,1,-3)=\\frac{-1}{1-(-3)}=-\\frac14\\\\ \\text{I hope this helped,}\\\\ \\text{Gavin.}$$ GYanggg Jul 6, 2018", "have $$10x=10\\left(\\frac{9}{10}+\\frac{9}{100}+\\frac{9}{1000}\\right)=\\frac{9}{1}+\\frac{9}{10}+\\frac{9}{100}=9.99.$$ But what about non-terminating numbers? It would seem that if $x=0.999\\dotsc$ then you can just write $$x=\\frac{9}{10}+\\frac{9}{100}+\\frac{9}{1000}+\\dotsb,$$ but how do you calculate this infinite sum? It is not obvious that you can do it because for example the sum $$1-1+1-1+\\dotsb$$ does not have an obvious or intuitive value. For example, it could be zero: $$1-1+1-1+\\dotsb = (1-1)+(1-1)+\\dotsb=0+0+\\dotsb = 0,$$ or it could be one: $$1-1+1-1+\\dots = 1+(-1+1)+(-1+1)+\\dotsb = 1+0+0+\\dotsb = 1$$ or something else. Now, once you have a definition for calculating (some) infinite sums, you must prove that with that definition $$10\\left(\\frac{9}{10}+\\frac{9}{100}+\\frac{9}{1000}\\dotsb\\right)=\\frac{9}{1}+\\frac{9}{10}+\\frac{9}{100}+\\dotsb=9.999\\dotsc$$ So, there is quite a bit of stuff hidden (or forgotten) in the first proof. The above examples about $1-1+1-1+\\dotsb$ being zero or one shows that with infinite sums you must prove even basic \"intuitive facts\" because they might not actually hold! If with your definition of infinite sums you can prove $$1-1+1-1+\\dotsb = 1-(1-1+1-1+\\dotsb),$$ then your second proof is also valid. It also shows that whatever definition for $1-1+1-1+\\dotsb$ you come up with, if it has the above property, then it must give this sum the value 1/2. However, while the standard definition of infinite sums does give a value for $9/10+9/100+\\dotsb$ and has the properties we want, it does not give any value to $1-1+1-1+\\dotsb$, so you cannot use that, as", "is defined by the equation $$x Ф y = \\frac{x^2}{4} - xy + y^2$$ for all numbers x and y. What is the value of 186 Ф 91? A) 1 B) 4 C) 8 D) 16 E) 24 *kudos for all correct solutions $$x$$ $$Ф$$ $$y = \\frac{x^2}{4} - xy + y^2$$ $$=> \\frac{(x^2 - 4xy + 4y^2)}{4}$$ $$x$$ $$Ф$$ $$y = \\frac{(x - 2y)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{(186 - 2*91)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{(186 - 182)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{4^2}{4}$$ $$186$$ $$Ф$$ $$91 = 4$$ _________________ Please Press \"+1 Kudos\" to appreciate. Manager Joined: 22 May 2017 Posts: 114 An operation Ф is defined by the equation x Ф y = x²/4 [#permalink] ### Show Tags 02 Aug 2017, 19:31 1 Good question to be done mentally xФy=(x/2−y)^2=(93−91)^2=4 Posted from my mobile device Posted from my mobile device GMAT Club Legend Joined: 12 Sep 2015 Posts: 4226 Re: An operation Ф is defined by the equation x Ф y = x²/4 [#permalink] ### Show Tags 03 Aug 2017, 07:40 Top Contributor 1 GMATPrepNow wrote: An operation Ф is defined by the equation $$x Ф y = \\frac{x^2}{4} - xy + y^2$$ for all numbers x and y. What is the value of 186 Ф 91? A) 1 B) 4 C) 8 D) 16 E)", "+0 # Functional Operations +1 36 1 +193 The operation $*$ is defined for non-zero $a$ and $b$ as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 13$ and $a \\times b = 25$, what is the value of $a*b$? Express your answer as a common fraction. Feb 14, 2021 #1 +283 +2 Since we know that a x b = 25, we can put 1/a + 1/b as (b + a) / 25. Then we know that a + b = 13, so 1/a + 1/b would equal 13/25, which is your answer :) Feb 14, 2021 #1 +283 +2 Since we know that a x b = 25, we can put 1/a + 1/b as (b + a) / 25. Then we know that a + b = 13, so 1/a + 1/b would equal 13/25, which is your answer :) Logarhythm Feb 14, 2021", "+0 # Operation +1 39 1 For any three real numbers a, b, and c, with $b \\neq -c$, the operation P is defined by $P(a,b,c) = \\frac{a}{b + c}.$ What is P(P(1,2,3),P(2,3,1),P(3,1,2))? Apr 6, 2021 #1 +485 +1 $P(1,2,3)=\\frac{1}{2+3}=\\frac{1}{5}$ $P(2,3,1)=\\frac{2}{3+1}=\\frac{2}{4}=\\frac{1}{2}$ $P(3,1,2)=\\frac{3}{1+2}=\\frac{3}{3}=1$ $P(\\frac{1}{5},\\frac{1}{2},1)=\\frac{\\frac{1}{5}}{\\frac{1}{2}+1}=\\frac{1}{\\frac{15}{2}}=\\boxed{\\frac{2}{15}}$ Apr 6, 2021", "Sep 2009 Posts: 60694 Re: The operation # is defined for all nonzero x and y by x # y [#permalink] ### Show Tags 22 Oct 2013, 10:14 3 5 Bunuel wrote: kartboybo wrote: The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) = A. a B. a+1 C. a/(a+1) D. (a+2)/(a+1) E. (2a+1)/(a+1) $$1#(1#a) = 1#(1+\\frac{1}{a}) = 1#(\\frac{a+1}{a})= 1+ \\frac{1}{(\\frac{a+1}{a})}=1+\\frac{a}{a+1}=\\frac{a+1+a}{a+1}=\\frac{2a+1}{a+1}$$. Similar questions to practice: the-operation-is-defined-by-x-y-1-x-1-y-for-all-67650.html gmat-prep-question-101282.html if-the-operation-is-defined-by-x-y-xy-1-2-for-all-144272.html an-operation-is-defined-by-the-equation-a-b-a-b-a-144074.html let-denote-a-mathematical-operation-is-it-true-that-x-y-131347.html for-any-operation-that-acts-on-two-numbers-x-and-y-the-127024.html the-operation-is-defined-for-all-nonzero-numbers-a-and-b-106495.html the-operation-x-n-for-all-positive-integers-greater-than-99064.html for-all-integers-x-and-y-the-operation-is-defined-by-x-y-86208.html Hope it helps. _________________ ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 60694 Re: The operation # is defined for all nonzero x and y by x # y [#permalink] ### Show Tags 22 Oct 2013, 10:07 1 4 kartboybo wrote: The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) = A. a B. a+1 C. a/(a+1) D. (a+2)/(a+1) E. (2a+1)/(a+1) $$1#(1#a) = 1#(1+\\frac{1}{a}) = 1#(\\frac{a+1}{a})= 1+ \\frac{1}{(\\frac{a+1}{a})}=1+\\frac{a}{a+1}=\\frac{a+1+a}{a+1}=\\frac{2a+1}{a+1}$$. _________________ Manager Joined: 24 Apr 2013 Posts: 50 Location: United States Re: The operation # is defined for all nonzero x and y by x # y [#permalink] ### Show Tags 22 Oct 2013, 10:13 kartboybo wrote: The operation # is defined for all nonzero x and y by x#y", "08:04 Applying the operation as described, we get $$(4#)# = (\\frac{1}{4}+4)# = \\frac{17}{4}# = \\frac{4}{17}+\\frac{17}{4} = \\frac{508}{68} = 4.48 < 4.5$$ Re: The operation # is defined by x# [#permalink] 28 Sep 2017, 08:04 Display posts from previous: Sort by", "N. 13. (a) The operation (*) is defined on the set of real numbers, R, by $$x * y = \\frac{x + y}{2}, x, y \\in R$$. (i) Evaluate $$3 * \\frac{2}{5}$$. (ii) If $$8 * y = 8\\frac{1}{4}$$, find the value of y. (b) In $$\\Delta ABC, \\overline{AB} = \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}$$ and $$\\overline{AC} = \\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix}$$. If P is the midpoint of $$\\overline{AB}$$, express $$\\overline{CP}$$ as a column vector. Subscribe Notify of", "+0 # Help! 0 114 1 +705 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Lightning Jul 7, 2018 ### Best Answer #1 +9558 +2 The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Omi67 Jul 7, 2018 #1 +9558 +2 Best Answer The operation $*$ is defined for non-zero integers as follows: $a * b = \\frac{1}{a} + \\frac{1}{b}$. If $a+b = 9$ and $a \\times b = 20$, what is the value of $a*b$? Express your answer as a common fraction. Omi67 Jul 7, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website. For more information: our cookie policy and privacy policy.", "1, 4]$. In a single operation with $i = 1$, we set $a_1 = \\min(\\frac{0+1-2}{2}, 0)$ and $a_2 = \\max(\\frac{0+1+2}{2}, 1)$. So, after a single operation with $i = 1$, $a$ becomes equal to $[-\\frac{1}{2}, \\frac{3}{2}, 4]$. We can show that no operation has any effect on this array, so $F(a, b) = -\\frac{1}{2}$.", "operation Ф is defined by the equation x Ф y = x²/4 [#permalink] ### Show Tags 02 Aug 2017, 11:34 2 1 GMATPrepNow wrote: An operation Ф is defined by the equation $$x Ф y = \\frac{x^2}{4} - xy + y^2$$ for all numbers x and y. What is the value of 186 Ф 91? A) 1 B) 4 C) 8 D) 16 E) 24 *kudos for all correct solutions $$x$$ $$Ф$$ $$y = \\frac{x^2}{4} - xy + y^2$$ $$=> \\frac{(x^2 - 4xy + 4y^2)}{4}$$ $$x$$ $$Ф$$ $$y = \\frac{(x - 2y)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{(186 - 2*91)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{(186 - 182)^2}{4}$$ $$186$$ $$Ф$$ $$91 = \\frac{4^2}{4}$$ $$186$$ $$Ф$$ $$91 = 4$$ _________________ Please Press \"+1 Kudos\" to appreciate. Manager Joined: 22 May 2017 Posts: 118 An operation Ф is defined by the equation x Ф y = x²/4 [#permalink] ### Show Tags 02 Aug 2017, 19:31 1 Good question to be done mentally xФy=(x/2−y)^2=(93−91)^2=4 Posted from my mobile device Posted from my mobile device CEO Joined: 12 Sep 2015 Posts: 2587 Re: An operation Ф is defined by the equation x Ф y = x²/4 [#permalink] ### Show Tags 03 Aug 2017, 07:40 Top Contributor 1 GMATPrepNow wrote: An operation Ф is defined by the equation $$x Ф y = \\frac{x^2}{4} - xy", "answers 83 views Define the operation $*$ by $a * b = 4a−3b+ab$ for all $a, b ∈ R$. For what real numbers y does $12 = 3 * y$?Define the operation by $a &nbsp; b = 4a−3b+ab$ for all $a, b ∈ R$. For what real numbers y does $12 = 3 y$? ... close", "are those appropriate for binary operation symbols. $\\qquad$ – Michael Hardy Jun 20 '16 at 23:35 • @Michael Hardy: Thanks; figured there was something better than what I wrote... – Chas Brown Jun 21 '16 at 5:04 I think I have an answer. Look at $\\frac{1}{3}$ in binary format and you will notice that it is $0.\\overline{01}$. This tells us something, namely that $[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to: $$\\sum \\limits_{i=0}^{50} [2^{2i}/3]+\\sum \\limits_{i=0}^{49} [2^{2i+1}/3]=\\sum \\limits_{i=0}^{49} [2^{2i}/3]+\\sum \\limits_{i=0}^{49} [2^{2i+1}/3]+[2^{100}/3]$$ $$\\sum \\limits_{i=0}^{49} [2^{2i}/3]+\\sum \\limits_{i=0}^{49} [2^{2i+1}/3]+[2^{100}/3]=\\sum(4^i-1)+[2^{100}/3]=\\frac{4^{50}-151}{3}+[2^{100}/3]$$ Edit 1: Added after I knew what to shoot for: I suppose we can simplify this by noting that $4 \\bmod 3 = 1$ thus $[2^{2i}/3] = \\frac{2^{2i}-1}{3}$. This would give us: $$\\frac{4^{50}-151}{3}+2^{100}/3 - 1/3 = \\frac{2^{101}-152}{3}$$ Edit 2: Going for a more general solution (boring but useful I guess) Consider $m=2n+1$ odd: $$\\sum \\limits_{i=0} ^{2n+1} \\lfloor\\frac{2^i}{3}\\rfloor = \\sum \\limits_{i=0} ^n (4^i-1) = \\frac{4^{n+1}-3n-4}{3}=\\frac{2^{m+1}-3\\frac{m-1}{2}-4}{3} =\\frac{2^{m+2}-3 m-5}{6}$$ For $m=2n$ even: $$\\sum \\limits_{i=0} ^{2n} \\lfloor\\frac{2^i}{3}\\rfloor = \\sum \\limits_{i=0} ^{n-1} (4^i-1) + \\lfloor \\frac{4^n}{3} \\rfloor = \\frac{2^{2n+1}-3n-2}{3}=\\frac{2^{m+1}-\\frac{3m}{2}-2}{3}=\\frac{2^{m+2}-3m-4}{6}$$ We can then rewrite this as: $$\\frac{2^{m+3}-6m-9+(-1)^{m}}{12}$$ • the answer is actually (2^(101) - 152) / 3 – uzumaki Jun 20 '16 at 23:21 • @uzumaki It's very likely that that answer and the one given are the same. – MCT Jun 20 '16 at", "= $$\\frac{16b}{12b + 16 – 12b} = \\frac{16b}{16} = b$$ Thus, fog = IR – $$\\left \\{- \\frac{4}{3} \\right \\}$$ and gof = IRange f Therefore we have, the inverse of the function g is the map f: Range g → R – $$\\left \\{- \\frac{4}{3} \\right \\}$$, which will be given by g(b) = $$\\frac{4b}{4 – 3b}$$ Hence we know, Answer is b. #### EXERCISE – 1.4 Q-1: Check whether each of the following definitions gives a binary operation or not. In each of the event, that * will not be a binary operation, give explanation for this condition. (i) On Z+, define * by x * y = x – y (ii) On Z+, define * by x * y = xy (iii) On R, define * by x * y = xy2 (iv) On Z+, define * by x * y = |x – y| (v) On Z +, define * by x * y = x (i) On Z+, * will be defined by x * y = x – y It’s not a binary operation, Such that the image of (2, 3) under * is 2 * 3 = 2 – 3 = -1 $$\\notin \\; Z^{+}$$ (ii) On Z+, * will be defined by x * y = xy. So for every", "+0 # The operation is defined for non-zero a andb as follows: If a+b=13 and , what is the value of ​?Express +1 90 2 +591 The operation $$*$$ is defined for non-zero a andb as follows: $$a * b = \\frac{1}{a} + \\frac{1}{b}.$$ If a+b=13 and $$a \\times b=25$$ , what is the value of $$a*b$$?Express your answer as a common fraction. ant101 Aug 28, 2018 #1 +3270 +2 We know that: $$\\frac{1}{a}+\\frac{1}{b}=a*b$$ . If we multiply the denominator by b and a, respectively, we get: $$\\frac{1b}{ab}+\\frac{1a}{ba}=a*b$$ . Suddenly, we go back to the problem, and it says: $$ab=25.$$ So, we have:$$\\frac{1a}{25}+\\frac{1b}{25}=\\frac{1a+1b}{25}.$$ We also know that $$a+b=13$$ , so $$\\frac{1a+1b}{25}=\\frac{a+b}{25}=\\boxed{\\frac{13}{25}}.$$ tertre Aug 28, 2018 #1 +3270 +2 We know that: $$\\frac{1}{a}+\\frac{1}{b}=a*b$$ . If we multiply the denominator by b and a, respectively, we get: $$\\frac{1b}{ab}+\\frac{1a}{ba}=a*b$$ . Suddenly, we go back to the problem, and it says: $$ab=25.$$ So, we have:$$\\frac{1a}{25}+\\frac{1b}{25}=\\frac{1a+1b}{25}.$$ We also know that $$a+b=13$$ , so $$\\frac{1a+1b}{25}=\\frac{a+b}{25}=\\boxed{\\frac{13}{25}}.$$ tertre Aug 28, 2018 #2 +20147 +2 The operation $$*$$ is defined for non-zero a andb as follows: $$a * b = \\frac{1}{a} + \\frac{1}{b}.$$ If $$a+b=13$$ and $$a \\times b=25$$ what is the value of $$a*b$$ ? Express your answer as a common fraction. $$\\begin{array}{|rcll|} \\hline a+b &=& 13 \\\\ a\\times b &=& 25 \\\\\\\\ \\dfrac{a+b}{a\\times b}&=& \\dfrac{13}{25}", "operation is $b^2$? Nov 23 '15 at 4:22 • Is it correct that answer is $b*2^{b/2}$? Nov 23 '15 at 4:28 • You can neglect the factor $b$. When you expand $2^\\frac{b}{2}$ into a power series, all powers of $b$ will occur. Multiplying the power series with $b$ will not change the overall picture (just increase all powers by $1$. Nov 23 '15 at 8:11" ]

[ "least 440. TQ.. You want $\\sum_{i=1}^n(2i+1)\\geq440$ Or $2\\frac{n(n+1)}{2}+n\\geq440$. Solve the inequality.", "## Thinking Mathematically (6th Edition) $\\left\\{ x\\ge 16 \\right\\}$. Let us suppose the number is $x$. Then, the algebraic form of the inequality is: $\\frac{3x}{4}\\,-3\\,\\ge 9$ And now to find out possible values of $x$ satisfying the inequality the inequality is solved as follows: \\begin{align} & \\frac{3x}{4}\\,-3\\,\\ge 9 \\\\ & \\frac{3x\\times 4}{4}\\,-3\\times 4\\,\\ge 9\\times 4 \\\\ & 3x\\,-12\\ge 36 \\end{align} This can be further simplified as: \\begin{align} & 3x-12+12\\ge 36+12 \\\\ & 3x\\ge 48 \\\\ & \\frac{3x}{3}\\ge \\frac{48}{3} \\\\ & x\\ge \\text{ }16 \\end{align} Hence all the real numbers greater than or equal to 16 will satisfy the condition. The set-builder form of the inequality obtained is: $\\left\\{ x\\ge 16 \\right\\}$ Therefore, the number can be represented in set builder form as $\\left\\{ x\\ge 16 \\right\\}$.", "## anonymous 4 years ago SOMEBODY PLZ HELP ME!!!!!!!!1 SOLVE EACH INEQUALITY AND GIVE THE SOLUTION IN INTERVAL NOTATION. 4+4X/3 IS LESS THAN 6 1. lgbasallote $\\frac{4+4x}{3} < 6$ IS THAT YOUR QUESTION??? 2. anonymous its 4 + 4x over 3 3. anonymous the first four is not over the 3 4. lgbasallote $4 + \\frac {4x}3 < 6$ ??? 5. anonymous yes", "How do you solve #3/4x + 1/4y > 1? To solve the inequality $\\frac{3}{4} x + \\frac{1}{4} y > 1$, implies identifying the region in the coordinate plane in which the inequality holds good. To do this, it is first required to graph the line $\\frac{3}{4} x + \\frac{1}{4} y = 1$. Then, to identify the region consider some point, say the origin (0,0) and see if it satisfies the inequality. In this case it is found that this does not satisfy the inequality, because it results in 0>1, which is not true.", "## anonymous 5 years ago I need to graph the inequality of a plane for: 2x+3y≤6 First you solve for $$y$$ $$y\\leq -\\frac{2}{3}x+2$$ Then draw the line $$y=-\\frac{2}{3}x+2$$ The region described by the inequality is exactly the region below the line.", "\\frac{x -3}{x -4} = \\frac{x + 2}{x + 6} 1.10mins Q2f Solve the inequality. \\displaystyle \\frac{4}{x - 3} < 1 1.40mins Q4a Solve the inequality. \\displaystyle \\frac{7}{x+ 1} > 7 1.08mins Q4b Solve the inequality. \\displaystyle \\frac{5}{x+ 4} \\leq \\frac{2}{x + 1} 2.11mins Q4c Solve the inequality. \\displaystyle \\frac{(x-2)(x+1)^2}{(x - 4)(x+5)} \\geq 0 1.27mins Q4d Solve the inequality. \\displaystyle \\frac{x^2-16}{x^2 -4x - 5} > 0 1.19mins Q4e Solve the inequality. \\displaystyle \\frac{x-2}{x} > \\frac{x -4}{x - 6} 2.02mins Q4f Solve each this inequality. \\displaystyle \\frac{x^2 + 9x + 14}{x^2 -6x + 5} > 0 3.04mins Q5a Solve each this inequality. \\displaystyle \\frac{2x^2 + 5x -3}{x^2 + 8x + 16} < 0 1.41mins Q5b Solve each this inequality. \\displaystyle \\frac{x^2 - 3x - 4}{x^2 + 11x + 30} \\leq 0 1.44mins Q5c Solve each this inequality. \\displaystyle \\frac{3x^2 - 8x + 4}{2x^2 - 9x - 5} \\geq 0 2.04mins Q5d Write a rational equation that cannot have x = 3 or x = -5 as a solution. 0.59mins Q6 Solve \\frac{x}{x + 1} < \\frac{2x}{x - 2} by graphing the functions f(x) = \\frac{x}{x + 1} and g(x) = \\frac{2x}{x -2} with or without using technology. Determine the points of intersection and when f(x) < g(x). 5.22mins Q7 Solve \\displaystyle \\frac{x}{x -3} > \\frac{3x}{x + 5} by graphing two", "what integer amounts will satisfy the conditions. 4. Jan 27, 2007 ### arildno Note that $$0\\leq{C}=100-\\frac{2}{2}A-\\frac{15}{2}A=\\frac{200-17A}{2}<30$$ Thus, you have the inequality: 0<200-17A<60", "Abby A. # Solve the inequality 19y<13, and write the solution in interval notation. Solve the inequality 19y<13, and write the solution in interval notation.", "have \\begin{align} e^x \\leq T_{2n}(x) \\end{align} for all $n$ where \\begin{align} T_n(x) = 1+x+\\frac{x}{2!}+\\ldots +\\frac{x^n}{n!}. \\end{align} Combining both result yields the desired inequality for all $r>0$ and $0<x<1$.", "+ 130 – 130 = 210 – 130 8d = 80 To solve above multiplication equation we need to divide both sides of the equation by 8 to find d. 8d/8 = 80/8 d = 10 The value of d is equal to 10. Question 14. 33 + $$\\frac{1}{3}$$c ≤ 66 Given inequality is 33 + $$\\frac{1}{3}$$c ≤ 66 33 + $$\\frac{1}{3}$$c – 33 ≤ 66 – 33 $$\\frac{1}{3}$$c ≤ 33 $$\\frac{1}{3}$$c x 3 ≤ 33 x 3", "$$P(X \\geq a) \\leq \\frac{E(x)}{a}$$ Chebyshev’s inequality guarantees that $$P(|X-\\mu| \\geq c) = P((X-\\mu)^2 \\geq c^2) \\geq \\frac{\\sigma^2}{c^2}$$ Built with Jekyll on the Swiss theme.", "\\frac{x-q}{x+q}\\right)^m = \\left(\\frac{2x}{x+q} \\right)^x -1 .$$ This inequality looks more tractable. However, even here, all the classical techniques as CS, AM-GM, concavity/convexity seem to fail.", "## College Algebra (6th Edition) $[59,95]$ Use the given information in the problem to create an inequality based on the original equation. Then multiply $\\frac{9}{5}$, and add $32$ on all three sides to solve for $F$. $C=\\frac{5}{9}(F-32)$ $15\\leq C\\leq35$ $15\\leq\\frac{5}{9}(F-32)\\leq35$ $27\\leq(F-32)\\leq63$ $59\\leq F\\leq95$", "# Logarithmic inequality? Find the set of values of k for which any solution of the inequality $\\frac{\\log_{2}(x^2 - 5x + 6)}{\\log_{2}(2x)}< 1$ is also a solution of the inequality $x^2 - 7kx + 2k - 6\\leq0$", "## Thinking Mathematically (6th Edition) $\\left\\{ x\\le 50 \\right\\}$. Let us suppose the number is $x$. Then, the algebraic form of the inequality is: $\\frac{3x}{5}\\,+4\\,\\le 34$ And now to find out possible values of $x$ satisfying the inequality the inequality is solved as follows: \\begin{align} & \\frac{3x}{5}\\,+4\\,\\le 34 \\\\ & \\frac{3x\\times 5}{5}\\,+4\\times 5\\,\\le 34\\times 5 \\\\ & 3x\\,+20\\le 170 \\end{align} This can be further simplified as: \\begin{align} & 3x\\le 150 \\\\ & \\frac{3x}{3}\\le \\frac{150}{3} \\\\ & x\\le \\text{ }50 \\end{align} Hence all the real numbers less than or equal to 50 will satisfy the condition. The set-builder form of the inequality obtained is: $\\left\\{ x\\,x\\le \\,50 \\right\\}$ Therefore, the number can be represented in set builder form as $\\left\\{ x\\le 50 \\right\\}$.", "# How do you solve the inequality 9x^2+16>=24x and write your answer in interval notation? May 25, 2017 Solution: $x \\in \\mathbb{R}$ , In Interval notation: $\\left(- \\infty , \\infty\\right)$ #### Explanation: $9 {x}^{2} + 16 \\ge 24 x \\mathmr{and} 9 {x}^{2} + 16 - 24 x \\ge 0 \\mathmr{and} 9 {x}^{2} - 24 x + 16 \\ge 0$ or ${\\left(3 x - 4\\right)}^{2} \\ge 0 \\mathmr{and} \\left(3 x - 4\\right) \\left(3 x - 4\\right) \\ge 0$. Critical point is $x = \\frac{4}{3}$ At x =4/3 ; (3x-4)(3x-4) =0.On either side of x=4/3 ; (3x-4)(3x-4) >0 :. x in RR , In Interval notation: $\\left(- \\infty , \\infty\\right)$ Solution: $x \\in \\mathbb{R}$ , In Interval notation: $\\left(- \\infty , \\infty\\right)$ graph{9x^2-24x+16 [-10, 10, -5, 5]} [Ans]", "inequality for $\\triangle XYZ$ as $\\frac{40\\sqrt{53}}{11} \\approx 26.5 > 7 + 13 = 20$. But the former does thus our answer is $\\textbf{(E)}\\ \\boxed{\\sqrt{30}}$. ~Aaryabhatta1", "Solve the inequality. 2. Hence, plot the inequality $8x-11>5x+4$8x11>5x+4 on the number line below. ##### Question 2 Consider the inequality $\\frac{9x}{16}+3>\\frac{x}{16}+2$9x16+3>x16+2. 1. Solve the inequality. 2. Hence, plot the inequality $\\frac{9x}{16}+3>\\frac{x}{16}+2$9x16+3>x16+2 on the number line below. ##### Question 3 Consider the inequality $\\frac{4x-3}{3}+\\frac{x-39}{2}\\le\\frac{2x+5}{5}$4x33+x3922x+55. 1. Solve the inequality. 2. Hence, plot the inequality $\\frac{4x-3}{3}+\\frac{x-39}{2}\\le\\frac{2x+5}{5}$4x33+x3922x+55 on the number line below.", "## Algebra 2 Common Core $-4 \\le x \\le 2$ Refer to the graph below. Solve each inequality: \\begin{align*} 3x &\\ge -12\\\\\\\\ \\frac{3x}{3} &\\ge \\frac{-12}{3}\\\\\\\\ x &\\ge -4 \\end{align*} $$\\text{and}$$ \\begin{align*} 8x&\\le 16\\\\\\\\ \\frac{8x}{8} &\\le \\frac{16}{8}\\\\\\\\ x&\\le 2 \\end{align*} Graph each inequality to obtain the one in the attached image below. Since the compound inequality involves the conjunction $\\text{AND}$, then the solutions to the inequality are numbers common to both graphs, which are the numbers in the interval $-4 \\le x \\le 2$.", "since $$\\sum_{cyc} (-\\frac{1}{36} a^2 + \\frac{25}{36} b^2) = \\frac{24}{36}\\sum_{cyc} a^2 = \\frac{2}{3} (a^2+b^2+c^2)$$ ) Indeed, by AP-GP, we have $$41 (a^3 + b^3+b^3) \\ge 41 \\cdot 3 ab^2$$ and $$b^3 + a^2b \\ge 2 ab^2$$ Summing up, we have $$41a^3 + 83b^3 + a^2 b \\ge 125 ab^2$$ which is equivalent to: $$36(a^3 + 3b^3) \\ge (5a + b)(-a^2 + 25b^2)$$ Dividing $36(5a+b)$ from both sides, the desired inequality is proved. --Lightest 15:31, 7 May 2012 (EDT)" ]

[ "at 8:56 As I wrote in the comments, $$g(1+x)=\\sqrt{2+x+\\sqrt{3-2x-8x^2}}\\;,\\tag{1}$$ so the first step is to ensure that $3-2x-8x^2\\ge 0$. This quadratic happens to factor: $3-2x-8x^2=(3+4x)(1-2x)$, so it’s equal to $0$ when $x=-\\frac34$ and when $x=\\frac12$. To see where it’s positive you can check to see where $3+4x$ and $1-2x$ have the same sign, but it’s easier to realize that the graph of $y=3-2x-8x^2$ is a parabola opening down, so it lies above the $x$-axis between the roots $-\\frac34$ and $\\frac12$. Thus, $3-2x-8x^2\\ge 0$ when $-\\frac34\\le x\\le\\frac12$. (You must have made a sign error in the calculation in your comment.) Okay, now we know that the domain is at most the interval $\\left[-\\frac34,\\frac12\\right]$, but there might be values of $x$ in that interval that make $$2+x+\\sqrt{3-2x-8x^2}\\tag{2}$$ negative. Are there? At $x=-\\frac34$, $(2)$ is equal to $2-\\frac34=\\frac54$, which is certainly not negative. As $x$ increases from $-\\frac34$ to $\\frac12$, $2+x$ also increases, so it won’t be a problem: it will always be at least $\\frac54$. What about $\\sqrt{3-2x-8x^2}$? On the interval $\\left[-\\frac34,\\frac12\\right]$ it’s always at least $0$, so it won’t be a problem either, and we can be sure that $(2)$ is always at least $\\frac54$ when $-\\frac34\\le x\\le\\frac12$. In particular, it won’t be negative, so we can take its square root. Thus, the domain of $g(1+x)$ is the whole", "values. Case 1 : $x\\ge\\frac{40}{3}$ we get 3x-20 +3x-40 =20 6x=80 x=$\\frac{80}{6}$=$\\frac{40}{3}$=13.33 Case 2 :$\\frac{20}{3}\\le\\ x<\\frac{40}{3}\\$ we get 3x-20+40-3x =20 we get 20=20 So we get x$\\in\\ \\left[\\frac{20}{3},\\frac{40}{3}\\right]$ Case 3 $x<\\frac{20}{3}$ we get 20-3x+40-3x =20 40=6x x=$\\frac{20}{3}$ but this is not possible so we get from case 1,2 and 3 $\\frac{20}{3}\\le\\ x\\le\\frac{40}{3}$ Now looking at options we can say only option C satisfies for all x . Hence 7<x<12. $f(x) = \\frac{x^2 + 2x – 15}{x^2 – 7x – 18}$<0 $\\frac{\\left(x+5\\right)\\left(x-3\\right)}{\\left(x-9\\right)\\left(x+2\\right)}<0$ We have four inflection points -5, -2, 3, and 9. For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive. When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again. We are left with the range (-5,-2) and (3,9) where the expression will be negative. We have $\\mid n – 60 \\mid < \\mid n – 100 \\mid < \\mid n – 20 \\mid$. Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number. Example: The absolute difference of", "and open dots (representing values that are not part of the solution) at x = 2, -1, and 4: Now, each of the factors with an even power will always be positive (except where it is zero, which we have covered), so we can now ignore it. The factors with odd powers will change sign at the dots corresponding to each of them; so the shortcut Doctor Rick mentioned is to first find the sign when x is far to the right, like this: $$\\frac{x(x+5)^{2016}(x-3)^{2017}(6-x) ^{1231}}{(x-2)^{10000}(x+1)^{2015}(4-x)^{242}} = \\frac{(+)(+)^{2016}(+)^{2017}(-) ^{1231}}{(+)^{10000}(+)^{2015}(-)^{242}} = \\frac{(+)(+)(+)(-) }{(+)(+)(+)} = \\frac{-}{+} = -$$ So we can mark the sign as negative at the right, and then work to the left, changing the sign at each of the odd factors, namely -1, 0, 3, and 6: The same can be determined (more slowly, but perhaps more safely) by repeating what we did above for each interval separately, listing the sign of each factor, and so finding the sign of the entire expression. Now, we mark the intervals on which the sign is positive; then the dots and segments together constitute the solution of the inequality: And there is the solution Arsh got before: $$\\{-5\\}\\cup(-1, 0]\\cup[3, 4)\\cup(4, 6]$$ At the start it was mentioned that this could be obtained by taking some of the steps required for", "three terms at the chosen value of $$p\\text{.}$$ \\begin{equation*} \\fp(p) = \\frac{8}{3} \\cdot \\underbrace{p^{-\\frac{1}{3}}}_{\\lt 0}\\cdot \\underbrace{(p-1)}_{\\lt 0}\\underbrace{(p+1)}_{\\gt 0}\\text{.} \\end{equation*} We have a “negative × negative × positive” giving a positive number; $$f$$ is increasing on $$(-1,0)\\text{.}$$ Interval 3: $$(0,1)$$ We do a similar sign analysis as before, using $$p$$ in $$(0,1)\\text{.}$$ \\begin{equation*} \\fp(p) = \\frac{8}{3} \\cdot \\underbrace{p^{-\\frac{1}{3}}}_{\\gt 0}\\cdot \\underbrace{(p-1)}_{\\lt 0}\\underbrace{(p+1)}_{\\gt 0}\\text{.} \\end{equation*} We have two positive factors and one negative factor; $$\\fp(p)\\lt 0$$ and so $$f$$ is decreasing on $$(0,1)\\text{.}$$ Interval 4: $$(1,\\infty)$$ Similar work to that done for the other three intervals shows that $$\\fp(x)\\gt 0$$ on $$(1,\\infty)\\text{,}$$ so $$f$$ is increasing on this interval. We conclude by stating that $$f$$ is increasing on the intervals $$(-1,0)$$ and $$(1,\\infty)$$ and decreasing on the intervals $$(-\\infty,-1)$$ and $$(0,1)\\text{.}$$ The sign of $$\\fp$$ changes from negative to positive around $$x=-1$$ and $$x=1\\text{,}$$ meaning by Theorem 3.3.12 that $$f(-1)$$ and $$f(1)$$ are relative minima of $$f\\text{.}$$ As the sign of $$\\fp$$ changes from positive to negative at $$x=0\\text{,}$$ we have a relative maximum at $$f(0)\\text{.}$$ Figure 3.3.21 shows a graph of $$f\\text{,}$$ confirming our result. We also graph $$\\fp\\text{,}$$ highlighting once more that $$f$$ is increasing when $$\\fp\\gt 0$$ and is decreasing when $$\\fp\\lt 0\\text{.}$$ We have seen how the first derivative of a function helps determine when the graph", "only conclude that $g$ is always increasing for $x \\lt \\frac{1}{b}$ and decreasing for $x \\gt \\frac{1}{b}\\text{,}$ but also that $g$ has a global maximum at $\\big(\\frac{1}{b}, g\\big(\\frac{1}{b}\\big)\\big)$ and no local minimum. The first derivative sign chart for $g$ is shown below in Figure3.41. We turn next to analyzing the concavity of $g\\text{.}$ With $g'(x) = ae^{-bx}-abxe^{-bx}\\text{,}$ we differentiate to find that \\begin{equation*} g''(x) = ae^{-bx}(-b)-ab\\left(e^{-bx}+xe^{-bx}(-b)\\right)\\text{.} \\end{equation*} Combining like terms and factoring, we now have \\begin{equation*} g''(x) = ab^2xe^{-bx} - 2abe^{-bx} = abe^{-bx}(bx - 2)\\text{.} \\end{equation*} We observe that $abe^{-bx}$ is always positive, and thus the sign of $g''$ depends on the sign of $(bx-2)\\text{,}$ which is zero when $x = \\frac{2}{b}\\text{.}$ Since $b$ is positive, the value of $(bx-2)$ is negative for $x \\lt \\frac{2}{b}$ and positive for $x \\gt \\frac{2}{b}\\text{.}$ The sign chart for $g''$ is shown below in Figure3.42. Thus, $g$ is concave down for all $x \\lt \\frac{2}{b}$ and concave up for all $x \\gt \\frac{2}{b}\\text{.}$ Finally, we analyze the long-term behavior of $g$ by considering two limits. First, we note that \\begin{equation*} \\lim_{x \\to \\infty} g(x) = \\lim_{x \\to \\infty} axe^{-bx} = \\lim_{x \\to \\infty} \\frac{ax}{e^{bx}}\\text{.} \\end{equation*} This limit has indeterminate form $\\frac{\\infty}{\\infty}\\text{,}$ so with our current tools, we can not evaluate the limit algebraically. We can, however, reason that $ax$ is linear", "# Question #bf7f4 Dec 3, 2017 That is correct. Furthermore, the function increases for $x \\ge \\frac{1}{2}$ #### Explanation: A function is increasing if $\\frac{\\mathrm{dy}}{\\mathrm{dx}} > 0$. $\\frac{\\mathrm{dy}}{\\mathrm{dx}} > 0$ $8 x - \\frac{1}{x} ^ 2 > 0$ Since ${x}^{2} > 0$, multiplying it to both sides of the inequality retains the sign. $8 {x}^{3} - 1 > 0$ $8 {x}^{3} > 1$ ${x}^{3} > \\frac{1}{8}$ Since the cube root is an increasing function, taking the cube root on both sides does not change the sign too. $x > \\frac{1}{2}$ The point $\\left(\\frac{1}{2} , 3\\right)$ is included as well.", "3\\right)$ satisfies $\\frac{3}{x + 3} > \\frac{4 x}{x - 2}$. For the case of $x \\setminus \\in \\left(- 3 , 2\\right)$, $x + 3$ is positive while $x - 2$ is negative. $3 \\left(x - 2\\right) < 4 x \\left(x + 3\\right)$ The sign flips as the product of a positive and a negative expressions is negative. Completing the square, $0 < {\\left(x + \\frac{9}{8}\\right)}^{2} + \\frac{15}{64}$ Since the RHS is always $\\ge \\frac{15}{64} > 0$, all real value of $x$ satisfies $3 \\left(x - 2\\right) > 4 x \\left(x + 3\\right)$. Therefore, all values of $x \\setminus \\in \\left(- 3 , 2\\right)$ satisfies $\\frac{3}{x + 3} > \\frac{4 x}{x - 2}$. For the case of $x \\setminus \\in \\left(2 , \\infty\\right)$, both $x + 3$ and $x - 2$ are negative. $3 \\left(x - 2\\right) > 4 x \\left(x + 3\\right)$ The sign remains the same as the product of 2 negative expressions is positive. Completing the square, $0 > {\\left(x + \\frac{9}{8}\\right)}^{2} + \\frac{15}{64}$ Since the RHS is always $\\ge \\frac{15}{64}$, no real value of $x$ satisfies $3 \\left(x - 2\\right) > 4 x \\left(x + 3\\right)$. Therefore, no value of $x \\setminus \\in \\left(2 , \\infty\\right)$ satisfies $\\frac{3}{x + 3} > \\frac{4 x}{x - 2}$. Combining the solutions on all 3 intervals, we get $\\frac{3}{x", "in an interval, then $$f$$ is decreasing $$(\\searrow)$$ on that interval. (c) If $$f'(x)=0$$ for every $$x$$ in an interval, then $$f$$ is constant on that interval. In applying this theorem, we need to determine the sign of $$f'(x)$$. If $$f'(x)$$ is a polynomial, this may be conveniently done by breaking it up into factors and determining the sign of each factor. A factor of the form $$x-a$$ is negative when $$x<a$$ and is positive when $$x>a$$. Suppose, then, we wish to determine the sign of $(x+2)(x-1)(x-3).$ 1. When $$-\\infty<x<-2$$, all three factors are negative, and thus the product is negative $(-)(-)(-)=(-).$ 2. When $$-2<x<1$$, the first factor is positive, and the other factors are negative. Therefore, the product is positive $(+)(-)(-)=(+).$ 3. When $$1<x<3$$, the first two factors are positive, but the last factor is negative. Therefore, the product is negative $(+)(+)(-)=(-).$ 4. When $$3<x<+\\infty$$, all three factors are positive, and the product is positive too $(+)(+)(+)=(+).$ Obviously the product is zero at $$-2,1$$, and $$3$$. We may summarize these into the following “sign table.” $$x$$ $$-\\infty$$ $$-2$$ 1 3 $$+\\infty$$ sign of $$(x+2)$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$ $$+$$ $$+$$ sign of $$(x-1)$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$ sign of $$(x-3)$$ $$-$$ $$-$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$ $$\\therefore$$ sign of product $$-$$ $$0$$", "be the product of the base and the previous output, regardless of the value of$\\,a.$ Notice from the table that • the output values are positive for all values of $x;$ • as$\\,x\\,$increases, the output values increase without bound; and • as$\\,x\\,$decreases, the output values grow smaller, approaching zero. (Figure) shows the exponential growth function $\\,f\\left(x\\right)={2}^{x}.$ Figure 1. Notice that the graph gets close to the x-axis, but never touches it. The domain of$\\,f\\left(x\\right)={2}^{x}\\,$is all real numbers, the range is$\\,\\left(0,\\infty \\right),$ and the horizontal asymptote is$\\,y=0.$ To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form$\\,f\\left(x\\right)={b}^{x}\\,$whose base is between zero and one. We’ll use the function$\\,g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}.\\,$Observe how the output values in (Figure) change as the input increases by$\\,1.$ $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g(x)=\\left(\\frac{1}{2}\\right)^{x}$ $8$ $4$ $2$ $1$ $\\frac{1}{2}$ $\\frac{1}{4}$ $\\frac{1}{8}$ Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio$\\,\\frac{1}{2}.$ Notice from the table that • the output values are positive for all values of$\\,x;$ • as$\\,x\\,$increases, the output values grow smaller, approaching zero; and • as$\\,x\\,$decreases, the output values grow without bound. (Figure) shows the exponential decay function,$\\,g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}.$ Figure 2. The domain of$\\,g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}\\,$is all real numbers,", "is changed. We see that in the expression $\\frac{-2(2x-1)}{(x-2)(x-4)}$ degrees of the points $2$, $4$ and $\\frac{1}{2}$ they are odds because $$\\frac{-2(2x-1)}{(x-2)(x-4)}=\\frac{-2(2x-1)^1}{(x-2)^1(x-4)^1},$$ which says that all signs are changed and we got the answer. Another example. Let, we need to solve $$\\frac{(x-2)^2}{(x-1)(x-3)}\\geq0.$$ The series of the signs is $+$, $-$, $-$, $+$ (draw it!) and we got the answer: $$(-\\infty,1)\\cup(3,+\\infty)\\cup\\{2\\}$$ • What do you mean by the \"intervals method\"? – ಠ ಠ Sep 6 '18 at 14:11 • @ಠ ಠ I added something. See now. – Michael Rozenberg Sep 6 '18 at 15:10 I have another way which is an illustrating good one for you. I started from @Michael point. Try to find what values of $x$, make that fraction negative. Be careful of $x=2$ and $x=4$. Look at this graphic table. I hope you find the whole story which both of us are trying to tell you:", "as the hand is lying along the \"x\" axis, going through zero to a maximum negative value when the hand is lying along the negative \"y\" axis. It would then begin to increase until it returned to a maximum value when the rotating hand was again lying along the positive \"x\" axis. ### Practice Graph each of the following functions. 1. $f(x)=\\cos(x)$ . 2. $h(x)=\\cos(2x)$ . 3. $k(x)=\\cos(2x+\\pi)$ . 4. $m(x)=-2\\cos(2x+\\pi)$ . 5. $g(x)=-2\\cos(2x+\\pi)+1$ . 6. $f(x)=\\sec(x)$ . 7. $h(x)=\\sec(3x)$ . 8. $k(x)=\\sec(3x+\\pi)$ . 9. $m(x)=2\\sec(3x+\\pi)$ . 10. $g(x)=3+2\\sec(3x+\\pi)$ . 11. $h(x)=\\cos(\\frac{x}{2})$ . 12. $k(x)=\\cos(\\frac{x}{2}+\\frac{\\pi}{2})$ . 13. $m(x)=2\\cos(\\frac{x}{2}+\\frac{\\pi}{2})$ . 14. $g(x)=2\\cos(\\frac{x}{2}+\\frac{\\pi}{2})-3$ . 15. $h(x)=\\sec(\\frac{x}{4})$ . 16. $k(x)=\\sec(\\frac{x}{4}+\\frac{3\\pi}{2})$ . 17. $m(x)=-3\\sec(\\frac{x}{4}+\\frac{3\\pi}{2})$ . 18. $g(x)=2-3\\sec(\\frac{x}{4}+\\frac{3\\pi}{2})$ . ### Vocabulary Language: English Circular Function Circular Function A circular function is a function measured by examining the angle of rotation around the coordinate plane. Sep 26, 2012 Feb 26, 2015", "sign without ever being $0$, as it is undefined at least at $1$ and $\\frac 43$. As I said in my earlier comment, the entire function is undefined between those values (unless you meant to take absolute value inside the log. – lulu Apr 27 '16 at 23:43 • So the function is increasing in (-∞, 1) and (4/3, ∞)? – Phil Apr 27 '16 at 23:46 First, notice that x is not defined at $x=1$ and $x=\\frac{4}{3}$. Set $y'=0$ $3+\\frac{1}{3x^2-7x+4}=0 \\Rightarrow 3x^2-7x+4=-\\frac{1}{3} \\Rightarrow 3x^2-7x+\\frac{13}{3}=0$ $(-7)^2-4*3*\\frac{13}{3}=-3<0$, so it doesn't have a solution. However, we were assuming $3x^2-7x+4 \\neq 0$ before. Now consider the case that $3x^2-7x+4=0$, and we get $x=1$ or $x=\\frac{4}{3}$. Neither of the two points fall in the range of x, so we don't have to consider these two cases. We can see that $y'$ is always greater than $0$, which means y is monotonically increasing. And since it is monotonically increasing, it is neither convex nor concave.", "# Evaluate: 8 x ^ 2+2 x-3 = 0 ## Expression: $$8 x ^ { 2 } + 2 x - 3 = 0$$ To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $8x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved. $$a+b=2$$ $$ab=8\\left(-3\\right)=-24$$ Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-24$. $$-1,24$$ $$-2,12$$ $$-3,8$$ $$-4,6$$ Calculate the sum for each pair. $$-1+24=23$$ $$-2+12=10$$ $$-3+8=5$$ $$-4+6=2$$ The solution is the pair that gives sum $2$. $$a=-4$$ $$b=6$$ Rewrite $8x^{2}+2x-3$ as $\\left(8x^{2}-4x\\right)+\\left(6x-3\\right)$. $$\\left(8x^{2}-4x\\right)+\\left(6x-3\\right)$$ Factor out $4x$ in the first and $3$ in the second group. $$4x\\left(2x-1\\right)+3\\left(2x-1\\right)$$ Factor out common term $2x-1$ by using distributive property. $$\\left(2x-1\\right)\\left(4x+3\\right)$$ To find equation solutions, solve $2x-1=0$ and $4x+3=0$. $$x=\\frac{1}{2}$$ $$x=-\\frac{3}{4}$$ Random Posts Random Articles", "$${2^{ - 2}} = \\frac{1}{4} = 0.25$$ $$- 3$$ $${2^{ - 3}} = \\frac{1}{8} = 0.125$$ $$- 5$$ $${2^{ - 5}} = \\frac{1}{{32}} \\approx 0.03$$ $$- 10$$ $${2^{ - 10}} = \\frac{1}{{1024}} \\approx 0.001$$ The trend is once again clear: as decreases from 0 towards negative infinity, f decreases from 1 towards 0. We can say that as $$x \\to - \\infty$$,$$f \\to 0$$. Note that no matter how large a negative value x takes, f will never become exactly 0,though it will keep coming closer and closer to 0. The following figure shows the graph of $$f\\left( x \\right) = {2^x}$$ plotted using a plotter: Note the spectacular rate of increase towards the positive side. This is understandable: for every unit change in x,the function value doubles. For example, in going from $$x = 5$$ to$$x = 6$$ , the input variable changes by only 1 unit, but the output changes by 32. On the negative side, note the slow decay towards zero. This is shown in more detail in the following figure: The curve never actually touches the x-axis,but it does keep on coming closer and closer towards the x-axis as we move towards negative infinity. Clearly, the range of $$f\\left( x \\right) = {2^x}$$ is $$\\left( {0,\\infty } \\right)$$. The number 0 is not included", "you multiplied both sides by (3x + 5), you forgot to consider the case when (3x + 5) < 0 – user137481 Sep 1 '14 at 23:32 If you multiply by a positive number the $\\geq$ stays but when you multiply by a negative number sign changes to $\\leq$.So you can take 2 cases when $3x+5\\geq0$ and $3x+5<0$.Anyway it's better to go $$\\frac{4x-5}{3x+5}-3\\geq 0\\\\\\frac{4x-5-9x-15}{3x+5}\\geq0\\\\\\frac{-5x-20}{3x+5}\\geq0$$ For $x>-\\frac{5}{3}$: $$4x-5 \\geq 3(3x+5) \\Rightarrow 4x-5 \\geq 9x+15 \\Rightarrow 5x \\geq -20 \\Rightarrow x \\geq -4$$ For $x<-\\frac{5}{3}$: $$4x-5 \\leq 3(3x+5) \\Rightarrow 4x-5 \\leq 9x+15 \\Rightarrow \\dots$$", "get (1)-(2): $$2\\left(\\sqrt[4]{\\frac{x^4}{3}+4}-\\sqrt[4]{\\frac{y^4}{3}+4}\\right)=y-x.\\tag{3}$$ Notice that the function $x\\mapsto\\sqrt[4]{\\frac{x^4}{3}+4}$ is increasing. So for $x>y$ the LHS is positive and the RHS is negative. Similarly, for $y<x$ the signs of the two expressions are opposite. So we can only find a real solution for $x=y$, which gives us $$2\\sqrt[4]{\\frac{x^4}{3}+4}=1+\\sqrt{\\frac{3}{2}}x\\tag{4}$$ From this we get $$16\\left(\\frac{x^4}{3}+4\\right)=\\left(1+\\sqrt{\\frac{3}{2}}x\\right)^4\\tag{5}$$ This is a quartic equation. In theory, this can be done by hand, but it will be very probably quite messy. WolframAlpha returns this. (One of the solution, according to WA, is $\\sqrt6$.) -", "+ 2x ≥ 2 6 – x ≥ 2 – x ≥ 2 – 6 –x ≥ –4 Since x is negative, we multiply both sides by -1 & change the signs (– 1) × (–x) ≤ (– 1) × (–4) x ≤ 4 Hence x is a real number which is less than or equal to 4, Thus, x ∈. First go to the Algebra Calculator main page. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. 36(x3)2 36 ( x 3) 2. Get free shipping on qualified 6 x 3 Sheds products or Buy Online Pick Up in Store today in the Storage & Organization Department. The graphs of these are therefore the same except when x =2. 6 x 2/3 = (6x2)/3. Precalculus. Two numbers r and s sum up to -3 exactly when the average of the two numbers is \\frac{1}{2}*-3 = -\\frac{3}{2}. −3(x+ 2) = 6 - 3 ( x + 2) = 6. Goto windows settings > Update & Security check the box that says \"For developers\" 4. This video helps students review some of the fundamentals of . Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Multiply both sides of the equation", "can thus rewrite the quadratic as $6 \\left(x - \\frac{3}{2}\\right) \\left(x + \\frac{2}{3}\\right) = 0$ So, you need this expression to be negative, which implies that $\\left(x - \\frac{3}{2}\\right)$ must be positive whenever $\\left(x + \\frac{2}{3}\\right)$ is negative, and vice versa. For $x > - \\frac{2}{3}$ and $x < \\frac{3}{2}$ you get $\\left\\{\\begin{matrix}x - \\frac{3}{2} < 0 \\\\ x + \\frac{2}{3} > 0\\end{matrix}\\right. \\implies \\left(x - \\frac{3}{2}\\right) \\left(x + \\frac{2}{3}\\right) < 0$ Any value of $x > \\frac{3}{2}$ will make both terms positive, and any value of $x < - \\frac{2}{3}$ will make both terms negative. Keeping in mind that you also need $x \\ne 0$, the solution set for this inequality will be $x \\in \\left(- \\frac{2}{3} , 0\\right) \\cup \\left(0 , \\frac{3}{2}\\right)$.", "## Calculus (3rd Edition) $f(x)$ increasing on $\\left( -\\infty,-3\\right )\\cup (3,\\infty)$ and decreasing on $(-3,-\\sqrt{3})\\cup ( 0,\\sqrt{3})\\cup (\\sqrt{3}, 3)$ $f(x)$ has a local maximum at $x=-3$ and a local minimum at $x=3$. Given $$y=\\frac{x^{3}}{x^{2}-3}$$ Since \\begin{align*} f'(x) &=\\frac{\\frac{d}{dx}\\left(x^3\\right)\\left(x^2-3\\right)-\\frac{d}{dx}\\left(x^2-3\\right)x^3}{\\left(x^2-3\\right)^2}\\\\ &=\\frac{x^4-9x^2}{\\left(x^2-3\\right)^2} \\end{align*} Then $f'(x)=0$ for $x^4-9x^2= 0$. Hence, $$x=0,\\ \\ \\ \\ x=-3,\\ \\ \\ \\ x=3 ,\\ \\ \\ \\ x=\\pm \\sqrt{3}$$ Choose $x=-1,\\ \\ x=1$: \\begin{align*} f'(-4)&> 0\\\\ f'(-2)&<0\\\\ f'(-1) &<0\\\\ f'(2)&<0\\\\ f'(4)&> \\end{align*} Then $f(x)$ is increasing on $\\left( -\\infty,-3\\right )\\cup (3,\\infty)$ and decreasing on $(-3,-\\sqrt{3})\\cup ( 0,\\sqrt{3})\\cup (\\sqrt{3}, 3)$. Hence, $f(x)$ has a local maximum at $x=-3$ and a local minimum at $x=3$.", "or negative value. Right off the bat, we can immediately see that the values $x=-3, 1$ satisfy the 'equals' part of the $\\ge$ inequality. But we'll try values that are just above/below the three critical values. $x=-3.1$: $$\\frac{\\left(-\\right)\\left(-\\right)}{\\left(-\\right)} \\lt 0$$ $x=-2.9$: $$\\frac{\\left(+\\right)\\left(-\\right)}{\\left(-\\right)} \\gt 0$$ $x=-0.1$: $$\\frac{\\left(+\\right)\\left(-\\right)}{\\left(-\\right)} \\gt 0$$ $x=0.1$: $$\\frac{\\left(+\\right)\\left(-\\right)}{\\left(+\\right)} \\lt 0$$ $x=0.9$: $$\\frac{\\left(+\\right)\\left(-\\right)}{\\left(+\\right)} \\lt 0$$ $x=1.1$: $$\\frac{\\left(+\\right)\\left(+\\right)}{\\left(+\\right)} \\gt 0$$ So from that we can see that $x+2\\ge \\frac{3}{x}$ for $-3 \\le x \\lt 0$ and $x \\ge 1$. ## Modulus inequalities In FP2 you also need to solve inequalities where either or both of the expressions contain a modulus sign. You must then consider the cases when the terms inside the modulus signs are positive and negative. Recall that when drawing a graph with modulus signs, the graph of $|f\\left(x\\right)|$ is the graph of $f\\left(x\\right)$ with the negative values of $f\\left(x\\right)$ reflected across the $x$-axis, and that the graph of $f\\left(|x|\\right)$ is the graph of $f\\left(x\\right)$ with all values of $f\\left(x\\right)$ for $x \\gt 0$ reflected across the $y$-axis, i.e. $f\\left(-x\\right)=f\\left(x\\right)$. Also recall that if $k$ is a constant, $k|x|=|kx|$. ### Example Q) Solve $\\frac{3}{|x|} \\gt -2 - x$ graphically. A) Firstly draw the graphs of $\\frac{3}{|x|}$ and $-2 - x$. Note that $\\frac{3}{|x|} = \\left|\\frac{3}{x}\\right|$. Clearly $\\frac{3}{|x|}$ is bigger than $-2 - x$ to the right" ]

[ "and open dots (representing values that are not part of the solution) at x = 2, -1, and 4: Now, each of the factors with an even power will always be positive (except where it is zero, which we have covered), so we can now ignore it. The factors with odd powers will change sign at the dots corresponding to each of them; so the shortcut Doctor Rick mentioned is to first find the sign when x is far to the right, like this: $$\\frac{x(x+5)^{2016}(x-3)^{2017}(6-x) ^{1231}}{(x-2)^{10000}(x+1)^{2015}(4-x)^{242}} = \\frac{(+)(+)^{2016}(+)^{2017}(-) ^{1231}}{(+)^{10000}(+)^{2015}(-)^{242}} = \\frac{(+)(+)(+)(-) }{(+)(+)(+)} = \\frac{-}{+} = -$$ So we can mark the sign as negative at the right, and then work to the left, changing the sign at each of the odd factors, namely -1, 0, 3, and 6: The same can be determined (more slowly, but perhaps more safely) by repeating what we did above for each interval separately, listing the sign of each factor, and so finding the sign of the entire expression. Now, we mark the intervals on which the sign is positive; then the dots and segments together constitute the solution of the inequality: And there is the solution Arsh got before: $$\\{-5\\}\\cup(-1, 0]\\cup[3, 4)\\cup(4, 6]$$ At the start it was mentioned that this could be obtained by taking some of the steps required for", "# Question #bf7f4 Dec 3, 2017 That is correct. Furthermore, the function increases for $x \\ge \\frac{1}{2}$ #### Explanation: A function is increasing if $\\frac{\\mathrm{dy}}{\\mathrm{dx}} > 0$. $\\frac{\\mathrm{dy}}{\\mathrm{dx}} > 0$ $8 x - \\frac{1}{x} ^ 2 > 0$ Since ${x}^{2} > 0$, multiplying it to both sides of the inequality retains the sign. $8 {x}^{3} - 1 > 0$ $8 {x}^{3} > 1$ ${x}^{3} > \\frac{1}{8}$ Since the cube root is an increasing function, taking the cube root on both sides does not change the sign too. $x > \\frac{1}{2}$ The point $\\left(\\frac{1}{2} , 3\\right)$ is included as well.", "at 8:56 As I wrote in the comments, $$g(1+x)=\\sqrt{2+x+\\sqrt{3-2x-8x^2}}\\;,\\tag{1}$$ so the first step is to ensure that $3-2x-8x^2\\ge 0$. This quadratic happens to factor: $3-2x-8x^2=(3+4x)(1-2x)$, so it’s equal to $0$ when $x=-\\frac34$ and when $x=\\frac12$. To see where it’s positive you can check to see where $3+4x$ and $1-2x$ have the same sign, but it’s easier to realize that the graph of $y=3-2x-8x^2$ is a parabola opening down, so it lies above the $x$-axis between the roots $-\\frac34$ and $\\frac12$. Thus, $3-2x-8x^2\\ge 0$ when $-\\frac34\\le x\\le\\frac12$. (You must have made a sign error in the calculation in your comment.) Okay, now we know that the domain is at most the interval $\\left[-\\frac34,\\frac12\\right]$, but there might be values of $x$ in that interval that make $$2+x+\\sqrt{3-2x-8x^2}\\tag{2}$$ negative. Are there? At $x=-\\frac34$, $(2)$ is equal to $2-\\frac34=\\frac54$, which is certainly not negative. As $x$ increases from $-\\frac34$ to $\\frac12$, $2+x$ also increases, so it won’t be a problem: it will always be at least $\\frac54$. What about $\\sqrt{3-2x-8x^2}$? On the interval $\\left[-\\frac34,\\frac12\\right]$ it’s always at least $0$, so it won’t be a problem either, and we can be sure that $(2)$ is always at least $\\frac54$ when $-\\frac34\\le x\\le\\frac12$. In particular, it won’t be negative, so we can take its square root. Thus, the domain of $g(1+x)$ is the whole", "can thus rewrite the quadratic as $6 \\left(x - \\frac{3}{2}\\right) \\left(x + \\frac{2}{3}\\right) = 0$ So, you need this expression to be negative, which implies that $\\left(x - \\frac{3}{2}\\right)$ must be positive whenever $\\left(x + \\frac{2}{3}\\right)$ is negative, and vice versa. For $x > - \\frac{2}{3}$ and $x < \\frac{3}{2}$ you get $\\left\\{\\begin{matrix}x - \\frac{3}{2} < 0 \\\\ x + \\frac{2}{3} > 0\\end{matrix}\\right. \\implies \\left(x - \\frac{3}{2}\\right) \\left(x + \\frac{2}{3}\\right) < 0$ Any value of $x > \\frac{3}{2}$ will make both terms positive, and any value of $x < - \\frac{2}{3}$ will make both terms negative. Keeping in mind that you also need $x \\ne 0$, the solution set for this inequality will be $x \\in \\left(- \\frac{2}{3} , 0\\right) \\cup \\left(0 , \\frac{3}{2}\\right)$.", "by \"x\" and not flipping the inequality sign, you were assuming that x was positive! 5. Originally Posted by mark1950 This is how I did it. $x - 2 > \\frac{x + 4}{x}$ $x^2 - 2x > x + 4$ $x^2 - 3x - 4 > 0$ $(x - 4)(x + 1) > 0$ Since the expression is > 0, hence, the x values are above the x -axis, right? And since the graph is a \"smiling\" graph, x > 4 or x < -1. That's how I got it. You made a mistake here . You CANNOT multiply x-2 with x because u don know whether x is positive or negative . If its positive , the sign is not affected but if its negative , the sign will be affected. So start like this : $(x-2)-\\frac{x+4}{x}>0$ $ \\frac{x(x-2)}{x}-\\frac{x+4}{x}>0 $ 6. Oh! No wonder...I got the previous question wrong too. Wow. This is a new concept for me. So, I can't multiply unknowns in inequalities. Thanks, guys! 7. hi $\\frac{{x}^{2}-3\\,x-4}{x} > 0$ $(x \\neq 0)$ solve ${x}^{2}-3\\,x-4$", "at 0:25 When you have the form $\\frac{a}{b} < 0$ you know there are only two options: $a < 0$, $b>0$ and $a>0$, $b<0$, because $a$ and $b$ need to differ in sign for the fraction to be negative. First case: $$2x+1 < 0 \\implies x < -\\frac12 \\\\ (x-1)(x+2) > 0 \\implies x < ...$$ and then repeat for the other case - Thanks for keeping it simple. – Kat Oct 18 '13 at 21:37 $\\frac{1}{x-1}<-\\frac{1}{x+2} \\rightarrow \\frac{1}{x-1}+\\frac{1}{x+2}<0 \\rightarrow$ $\\frac{(x+2)+(x-1)}{(x+2)(x-1)} <0 \\rightarrow \\frac{(x+2)+(x-1)}{(x+2)(x-1)}((x+2)(x-1)^2<0 \\rightarrow (x+2+x-1)(x+2)(x+1)<0$ for $x$ different to $-2,1$. (-2 and 1 arent soultions cause they are undefined.) using the fact that rational functions are continuous wherever they are defined and the intermediate value theorem: $(x+2+x-1)(x+2)(x-1)=2(x+\\frac{1}{2})(x+2)(x+1)<0$ Since the polynomial 2(x+\\frac{3}{2})(x+2)(x+1) has three roots:(each with multiplicity 1) $-\\frac{1}{2},-1,-2$ and -1000 is a solution to the inequality we can know that the solution set is $(-\\infty,-2)\\cup (-1,-\\frac{1}{2})$ - how did you get $((x+2)(x+1))^2$? – Kat Oct 18 '13 at 0:48 i multiply both sides by that (its standard in this type of problems) its usefull because since squares of real numbers are non negative im preserving the sign, and therefore its also gonna be positive unless its 0. But it can only be 0 if x is a root of the polynomial. – dREaM Oct 18 '13 at", "will handle it the same way. You can assume that the odd power is equal to 1 and proceed as usual. This is so because the sign of (x – a) will be the same as the sign of $$(x – a)^3$$. e.g. $$(x – 4)(x – 2)(x + 1)^3 < 0$$ The expression will be negative in the range $$x < -1$$ or $$2 < x < 4$$. Now let’s look at a question involving both these complications. Question: Find the range of x for which the given inequality holds. $$\\frac{(2x – 5) * (6 – x)^3}{x^2} < 0$$ Solution: I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors. $$\\frac{2(x – \\frac{5}{2}) * (-1)(x – 6)^3}{x^2}< 0$$ (take 2 common) $$\\frac{2(x – \\frac{5}{2}) * (x – 6)^3}{(x – 0)^2} > 0$$ (multiply both sides by -1) Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power. We want to find the range where the expression is positive. The required range is $$x < \\frac{5}{2}$$ or $$x > 6$$. But we are missing something here. $$x < \\frac{5}{2}$$ implies that all values less than 5/2 are acceptable but note that x cannot", "# How do you solve (2x-1)(x+2)(x-4)<=0? Oct 13, 2015 The inequality holds for $x \\setminus \\le - 2$ and $\\frac{1}{2} \\setminus \\le x \\setminus \\le 4$. #### Explanation: Keep in mind the rule for multiplying: • Positive times positive is positive; • Positive times negative is negative; • Negative times positive is negative; • Negative times negative is positive. Combining these rules, you can clearly see that a product of factors is negative if and only if there is an odd number of negative factors (in this case, one or all three). Now, $2 x - 1$ is positive if and only if $x > \\frac{1}{2}$, $x + 2$ is positive if and only if $x > - 2$, and $x - 4$ is positive if and only if $x > 4$. So, the important points are $- 2$, $\\frac{1}{2}$ and $4$. Before $- 2$, all three factors are negative, so the product is negative. Between $- 2$ and $\\frac{1}{2}$, $x + 2$ is positive and the other two are negative, so the product is positive. Between $\\frac{1}{2}$ and $4$, only $x - 4$ is negative, and the other two are positive, so the product is negative. After $4$, all the factors are positive, so the product is positive.", "comprising 30 GMAT level questions in inequalities. Explanatory answers and answer key are provided for all questions in the test. Here is a typical solved example from this chapter. ### Sample Question Find the range of values of x for which $$frac {1} {x - 2}$ > -2? #### Explanatory Answer There is a linear expression in 'x' in the denominator. Let us eliminate the expression in x in the denominator by multiplying both sides of the inequality by$x - 2)2 in the numerator. (x - 2)2 is the square of a number and is therefore, always positive for real values of x. Multiplying both sides of an inequality by a positive number will not affect the sign of the inequality. (x - 2)2 * $$frac {1} {x - 2}$ >$x - 2)2 * (-2) (x - 2) > -2(x - 2)2 (x - 2) > -2(x2 - 4x + 4) (x - 2) > -2x2 + 8x - 8 0 > -2x2 + 7x - 6 Or 2x2 - 7x + 6 > 0 Factorize the expression: 2x2 - 4x - 3x + 6 > 0 (2x - 3) (x - 2) > 0 The range of values of x that satisfy the above inequality are ${$left$ {{$infty },{\\frac {3} {2}}} \\right$}) $\\cup$ $$2,{\\infty }$) In other", "question. Hi Banuel, When i factorsie 15x^2 - X-2 =0 i got only only 2 roots : -1/3 and 2/5. ( I cross multiplied the denominator x so it become \"0\" on the RHS.) So i got only 3 intervals. How did you got 3 roots , especially 0 as a root. please provide me the insight and tell me where did i went wrong. By the way thank you very much for your links to learn inequalities concepts. It's very useful. Regards _________________ Kabilan.K Kudos is a boost to participate actively and contribute more to the forum Math Expert Joined: 02 Sep 2009 Posts: 31287 Followers: 5351 Kudos [?]: 62237 [0], given: 9445 Re: Range for variable x in a given inequality [#permalink] 10 May 2013, 04:50 Expert's post kabilank87 wrote: Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x - (2/x) > 1? A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5 $$15x-\\frac{2}{x}>1$$ ---> $$\\frac{15x^2-x-2}{x}>0$$ --> $$\\frac{(3x+1)(5x-2)}{x}>0$$ -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: $$x<-\\frac{1}{3}$$, $$-\\frac{1}{3}<x<0$$, $$0<x<\\frac{2}{5}$$ and $$x>\\frac{2}{5}$$ --> in rightmost range expression is positive: so -+-+ --> $$-\\frac{1}{3}<x<0$$,", "# How do you resolve the following inequality: ((x^2-4)(3x-6))/(x-7)>0? Apr 19, 2016 $x \\in \\left(- \\infty , - 2\\right) \\cup \\left(7 , \\infty\\right)$ #### Explanation: If we factor the numerator of the left hand side completely, we obtain $\\frac{3 \\left(x + 2\\right) {\\left(x - 2\\right)}^{2}}{x - 7} > 0$ We wish to identify the intervals on which the left hand side is positive. As the sign can only change at points where the left hand side evaluates to $0$ (a factor of the number is $0$) or is undefined (a factor of the denominator is $0$), we will check the intervals with endpoints at those values. We can also simplify the task slightly by noting that ${\\left(x - 2\\right)}^{2} > 0$ for all $x \\ne 2$, and so that factor will never change the sign of the expression. Proceeding: On $\\left(- \\infty , - 2\\right)$ we have $x + 2 < 0$ and $x - 7 < 0$, meaning we have a $\\frac{-}{-} = +$ case, and so $\\frac{3 \\left(x + 2\\right) {\\left(x - 2\\right)}^{2}}{x - 7} > 0$. On $\\left(- 2 , 7\\right)$ we have $x + 2 > 0$ and $x - 7 < 0$, meaning we have a $\\frac{+}{-} = -$ case, and so $\\frac{3 \\left(x + 2\\right) {\\left(x - 2\\right)}^{2}}{x - 7} < 0$", "2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy$$ Quantity B: $${ (x-y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-2xy$$ X is negative & y is positive. Therefore, xy is negative. $${ x }^{ 2 }+{ y }^{ 2 }-2xy>{ x }^{ 2 }+{ y }^{ 2 }+2xy\\\\ { (x-y) }^{ 2 }>{ (x+y) }^{ 2 }$$ Therefore, Quantity B is greater. Problem 13) If y<0 & 4x>y, which of the following could be equal to $$\\frac { x }{ y }$$ ? a) 0 b) $$\\frac { 1 }{ 4 }$$ c) $$\\frac { 1 }{ 2 }$$ d) 1 e) 4 Sol. 4x>y (divide both sides by y) $$\\frac { 4x }{ y } <1$$ Remember to switch the direction of the inequality sign when multiplying or dividing by a negative Quantity. $$\\frac { 4x }{ y } <1$$ (Now divide both sides by 4) $$\\frac { x }{ y } <\\frac { 1 }{ 4 }$$ The only answer choice less than $$\\frac { 1 }{ 4 }$$ is 0. Problem 14) If |1-x|=6 & |2y-6|=10, which of the following could be the value of xy? Indicate all such value? a) -40 b) -10 c) -14 d) 56 Sol. Remember that to solve the absolute value equation, First set the", "on (from) both the sides of inequalities then the sign of inequality does not changes. • When both the sides of the inequality is multiplies or divided by the same positive number then the sign of inequality does not changes. • Whereas, if both the sides are multiplied or divided by the same negative number then the sign of inequality gets reversed. Calculation: Given: $$\\frac{{2x\\; + \\;1}}{{7x\\; - \\;1}} > 5\\;and\\;\\frac{{x\\; + \\;7}}{{x\\; - \\;8}} > 2$$ Let’s first solve: $$\\frac{{2x\\; + \\;1}}{{7x\\;-\\;1}} > 5$$ $$⇒ \\frac{{2x\\; + \\;1}}{{7x\\;-\\;1}} - 5 > 0$$ ⇒ x ∈ (1/7, 6/35) ---------(1) Now, let’s solve: $$\\frac{{x\\; + \\;7}}{{x\\; - \\;8}} > 2$$ $$⇒ \\frac{{x\\; + \\;7}}{{x\\; - \\;8}} - 2 > 0$$ ⇒ x ∈ (8, 23) ---------(2) So, from (1) and (2) we get, no such value of x exist which satisfies both the inequalities. #### Inequalities Question 11 Solution of (log3x)2 < log3x4 will be 1. 1 < x < 81 2. 1 > x > 81 3. 1 ≤ x ≤ 81 4. 1 ≥ x ≥ 81 Option 1 : 1 < x < 81 #### Inequalities Question 11 Detailed Solution Formula used: logam = m loga 0 < a < 1 ⇒ logx < log Calculation: Given that, (log3x)2 < log3x4 ∵ logam = m loga", "Solution Given: An inequality,$$\\frac{(x+6) (3x-4)^3}{(x+2)^2}$$>0 To find: The range of x, that satisfies the above inequality Approach and Working: The first step is to multiply the inequality by $$(x+2)^2$$ Thus, we get, $$\\frac{(x+2)^2 (x+6) (3x-4)^3}{(x+2)^2} >0$$ Which implies, $$(x + 6) * (3x - 4)^3 > 0$$ and the denominator, x + 2 ≠ 0, implies, x≠ -2 This can be written as $$(x + 6) * (3x - 4) * (3x - 4)^2 > 0$$ Since, $$(3x-4)^2$$, is always positive, expect for x = 4/3 We can write the inequality as, (x + 6) * (3x - 4) > 0 The zero points of this inequality are, x = -6 and x = 4/3 Plot these points on the number line. Since, 4/3 > -6, the inequality will be positive, for all the points to the right of 4/3, on the number line And, it is negative, in the region, between -6 and 4/3 It is again positive for all the values less than -6 Therefore, the range of x is x > 4/3 and x < -6 and x ≠ -2 Hence, the correct answer is option D. _________________ Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of", "May 2013, 03:17 Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x - (2/x) > 1? A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5 $$15x-\\frac{2}{x}>1$$ ---> $$\\frac{15x^2-x-2}{x}>0$$ --> $$\\frac{(3x+1)(5x-2)}{x}>0$$ -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: $$x<-\\frac{1}{3}$$, $$-\\frac{1}{3}<x<0$$, $$0<x<\\frac{2}{5}$$ and $$x>\\frac{2}{5}$$ --> in rightmost range expression is positive: so -+-+ --> $$-\\frac{1}{3}<x<0$$, and $$x>\\frac{2}{5}$$. Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve $$15x-\\frac{2}{x}>1$$ for the true ranges then the answer is D but if the question asks for which of the following ranges inequality $$15x-\\frac{2}{x}>1$$ holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486 Not a good", "values. Case 1 : $x\\ge\\frac{40}{3}$ we get 3x-20 +3x-40 =20 6x=80 x=$\\frac{80}{6}$=$\\frac{40}{3}$=13.33 Case 2 :$\\frac{20}{3}\\le\\ x<\\frac{40}{3}\\$ we get 3x-20+40-3x =20 we get 20=20 So we get x$\\in\\ \\left[\\frac{20}{3},\\frac{40}{3}\\right]$ Case 3 $x<\\frac{20}{3}$ we get 20-3x+40-3x =20 40=6x x=$\\frac{20}{3}$ but this is not possible so we get from case 1,2 and 3 $\\frac{20}{3}\\le\\ x\\le\\frac{40}{3}$ Now looking at options we can say only option C satisfies for all x . Hence 7<x<12. $f(x) = \\frac{x^2 + 2x – 15}{x^2 – 7x – 18}$<0 $\\frac{\\left(x+5\\right)\\left(x-3\\right)}{\\left(x-9\\right)\\left(x+2\\right)}<0$ We have four inflection points -5, -2, 3, and 9. For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive. When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again. We are left with the range (-5,-2) and (3,9) where the expression will be negative. We have $\\mid n – 60 \\mid < \\mid n – 100 \\mid < \\mid n – 20 \\mid$. Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number. Example: The absolute difference of", "\\left(\\frac{(n-2)(n+3) + \\frac{n+3}{2} + \\frac{n+3}{2} + 4}{n+1}\\right)^{n+1} = \\left(n+1\\right)^{n+1}$$ The inequality to be shown is $$(n+1)^{n+1}\\geqslant(n+3)^n,$$ for every positive integer $n$. Introduce the function $u$ defined by $$u(x)=(x+1)\\log(x+1)-x\\log(x+3),$$ then standard computations yield $$u'(x)=\\frac3{3+x}+\\log\\left(\\frac{1+x}{3+x}\\right),\\qquad u''(x)=\\frac{3-x}{(1+x)(3+x)^2},$$ hence $u'$ increases on $(0,3)$ and decreases on $(3,+\\infty)$. Since $u'(1)=\\frac34-\\log2\\gt0$ and $u'(+\\infty)=0$, $u$ is increasing on $(1,+\\infty)$. Since $u(1)=0$, this yields $u(n)\\geqslant0$ for every positive integer, QED. Edit: Equivalently, one wants to prove that $(k+2)^{k-1}\\leqslant k^k$ for every $k\\geqslant2$, that is, $k+2\\geqslant\\left(1+\\frac2k\\right)^k$. If one knows that the RHS is increasing and converges to $\\mathrm e^2\\lt8$, this yields the result for every $k\\geqslant6$. The cases $2\\leqslant k\\leqslant5$ can be checked manually. • Re Bernoulli's inequality $(1+x)^r\\gt1+rx$, recall that it is valid for every nonnegative integer $r$ and every real number $x\\geqslant-1$. Using it when $r$ is not an integer can lead to chaos. – Did Jan 17 '13 at 18:56", "# Help with inecuation inverse Hello! a)$\\frac{1}{2}<\\frac{5}{3x+42}<\\frac{3}{2}$ b)$\\frac{1}{2}<\\frac{3x+1}{3x+4}<\\frac{3}{2}$ c)$\\frac{1}{2}<\\frac{3x+1}{3x+4}<\\frac{2}{3}$ Can you tell me how I can resolve it? In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2, but when I pass 2x-1 to multipliy with 1 \"(3<1·2x-1)\" Why I have not change \"<\" by \">\" The denominator is negative, when is negative I thought I had change it. The solution set is $(-\\infty,\\frac{1}{2})\\cup(2,+\\infty)$ for x> 2 not to have to change the sign to move the denominator to multiply by 1. Thank you very much for your help :) Last edited: robert2734 If x<-4/3 then (3x+1)/(3x+4) is always greater than one. When x=0, (3x+1)/(3x+4)=1/4. When x=infinity, (3x+1)/(3x+4)=1. There's some region, while x is positive, that the expression is within your region of interest. Staff Emeritus Homework Helper Gold Member Hello! a)$\\frac{1}{2}<\\frac{5}{3x+42}<\\frac{3}{2}$ b)$\\frac{1}{2}<\\frac{3x+1}{3x+4}<\\frac{3}{2}$ c)$\\frac{1}{2}<\\frac{3x+1}{3x+4}<\\frac{2}{3}$ Can you tell me how I can resolve it? In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2, but when I pass 2x-1 to multiply with 1 \"(3<1·2x-1)\" Why I have not change \"<\" by \">\" The denominator is negative, when is negative I thought I had change it. The solution set is $(-\\infty,\\frac{1}{2})\\cup(2,+\\infty)$ for x> 2 not to have to change the sign to move the denominator to", "it inverts the sign. Manager Joined: 16 Jun 2010 Posts: 118 Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink] ### Show Tags 03 Oct 2010, 12:26 Thanks for the lightning quick reply, I was not aware of this. Manager Joined: 16 Jun 2010 Posts: 118 Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink] ### Show Tags 03 Oct 2010, 12:45 Bunuel wrote: One could also do as follows: $$15x-\\frac{2}{x}>1$$ --> $$\\frac{15x^2-x-2}{x}>0$$ --> $$\\frac{(x+\\frac{1}{3})(x-\\frac{2}{5})}{x}>0$$: Both denominator and nominator positive --> $$x>\\frac{2}{5}$$; Both denominator and nominator negative --> $$-\\frac{1}{3}<x<0$$; So inequality holds true in the ranges: $$-\\frac{1}{3}<x<0$$ and $$x>\\frac{2}{5}$$. Bunuel you are awesome Manager Joined: 25 Jun 2010 Posts: 71 Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink] ### Show Tags 03 Oct 2010, 18:00 Nice alternate solution. $$15x^2-x-2 -> x^2 - x/15 - 2/15 -> x^2 + x/3 - 2x/5 - 2/15 -> (x+1/3)(x-2/5)$$ This works well, however I would suggest to rely on quadratic equation formula: $$-b(+-)\\sqrt{D}/2a$$, as you don't need to think, how to factorize an equation? Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect", "07:24 Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x - (2/x) > 1? A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5 $$15x-\\frac{2}{x}>1$$ ---> $$\\frac{15x^2-x-2}{x}>0$$ --> $$\\frac{(3x+1)(5x-2)}{x}>0$$ -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: $$x<-\\frac{1}{3}$$, $$-\\frac{1}{3}<x<0$$, $$0<x<\\frac{2}{5}$$ and $$x>\\frac{2}{5}$$ --> in rightmost range expression is positive: so -+-+ --> $$-\\frac{1}{3}<x<0$$, and $$x>\\frac{2}{5}$$. Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve $$15x-\\frac{2}{x}>1$$ for the true ranges then the answer is D but if the question asks for which of the following ranges inequality $$15x-\\frac{2}{x}>1$$ holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486 Not a good question. Could" ]

[ "real numbers. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 - y^2) = x f(x) - y f(y)$ for all pairs of real numbers $x$ and $y$.", "for all $x\\in\\mathbb{N}$, and b)$f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x$, $y\\in\\mathbb{N}$. 25. Consider those functions $f:\\mathbb{N}\\to\\mathbb{N}$ which satisfy the condition $f(m+n) \\geq f(m)+f(f(n))-1$ for all $m,n\\in\\mathbb{N}.$ Find all possible values of $f(2007).$ 26. Find all surjective functions $f:\\mathbb{N}\\to\\mathbb{N}$ such that for every $m,n\\in\\mathbb{N}$ and every prime $p,$ the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p.$ 27. Find all real polynomials $f$ such that $2yf(x+y)+(x-y)(f(x)+f(y)) \\geq 0\\forall x,y\\in\\mathbb{R}.$ 28. Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ with $x,y\\in\\mathbb{R}$ such that $f(x-f(y))=f(x+y)+f(y).$ 29. Show that for positive integer $n$, and for $x\\not =0$, $\\left(x^{n-1}\\sin\\dfrac{1}{x}\\right)^{(n)}=\\dfrac{(-1)^n}{x^{n+1}}\\sin\\left(\\dfrac{1}{x}+\\dfrac{n\\pi}{2}\\right).$ 30. Find all $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xy+f(x))=xf(y)+f(x)$ for every pair of real numbers $x,y$.", "# IMO 2017, Problem 2 Let \\mathbb{R} be the set of real numbers. Determine all functions f:\\,\\mathbb{R} \\rightarrow \\mathbb{R} such that, for any real numbers x and y, f(f(x)f(y)) + f(x+y) = f(xy). The solution can be found on a separate page.", "2017 IMO Problems/Problem 2 Let $\\mathbb{R}$ be the set of real numbers , determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$ Solution Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$. Let $f(0)=n$, so $f(n^{2})=0$. Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$. If $x+n^2=xn^2$, or $x=\\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0. This means that the only possible values of $n$ is -1,0, and 1. Go through the cases: $n=-1$ $f(x+1)=f(x)+1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=1$ $f(3)=2$ ... $f(x)=x-1$ $n=0$ $f(x)=0$ $n=1$ $f(x+1)=f(x)-1$ $f(1)=0$ (Based on $f(n^2)=0$) $f(2)=-1$ $f(3)=-2$ ... $f(x)=1-x$ So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.", "+0 # Functions help +1 47 3 +46 Let be S the set of all nonzero real numbers. The function $$f : S \\to \\mathbb{R}$$ satisfies the following two properties: (i) First, $$f\\left( \\frac{1}{x} \\right) = xf(x)$$ for all $$x \\in S.$$ (ii) Second, $$f \\left( \\frac{1}{x} \\right) + f \\left( \\frac{1}{y} \\right) = 1 + f \\left( \\frac{1}{x + y} \\right)$$ for all $$x \\in S$$ and $$y \\in S$$ such that $$x + y \\in S.$$ Let n be the number of possible values of $$f(1),$$ and let s be the sum of all possible values of $$f(1).$$ Find $$n \\times s.$$ Apr 4, 2022 #1 -1 ns = 10 Apr 5, 2022 #2 +46 +1 Sorry, $$n\\times s \\neq 10.$$ SpaceXGeek Apr 5, 2022 #3 +46 0 Hello? I don't know how to bump this without doing this, sorry. Apr 6, 2022", "+0 # Help 0 78 1 Let $S$ be the set of all nonzero real numbers. Let $f : S \\to S$ be a function such that $f(x) + f(y) = f(xyf(x + y))$for all $x,$ $y \\in S$ such that $x + y \\neq 0.$ Let $n$ be the number of possible values of $f(4),$ and let $s$ be the sum of all possible values of $f(4).$ Find $n \\times s.$ May 6, 2021", "\\in \\mathbb{R}$, $y \\neq 0$, $f(2x)=af(x)+bx$ and $f(x)f(y)=f(xy)+f\\left(\\dfrac{x}{y}\\right).$ 11. Prove that for all integers $a > 1$ and $b > 1$ there exists a function $f$ from the positive integers to the positive integers such that $f(a\\cdot f(n))=b\\cdot n$ for all $n$ positive integer. 12. Find all functions $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(xf(y)+f(x)) = 2f(x)+xy$ for every reals $x,y$. 13. Let $f(x)$ be a real-valued function defined on the positive reals such that (1)If $x < y$, then $f(x) < f(y)$, (2)$f\\left(\\dfrac{2xy}{x+y}\\right) \\geq\\dfrac{f(x) + f(y)}{2}$ for all $x,y>0$. Show that $f(x) < 0$ for some value of $x$. 14. Define $f$ on the positive integers by $f(n) = k^2 + k + 1$, where $n=2^k(2l+1)$ for some $k,l$ nonnegative integers. Find the smallest $n$ such that $f(1) + f(2) + ... + f(n) \\geq 123456.$ 15. Find all functions $f:\\mathbb{N}\\to\\mathbb{N}$ satisfying, for all $x\\in\\mathbb{N}$, $f(2f(x)) = x + 1998.$ 16. a) Show that there are no functions $f,g: \\mathbb{R}\\to\\mathbb{R}$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x\\in\\mathbb{R}$. b)Let $S$ be the set of all real numbers greater than $1$. Show that there are functions $f,g:S\\to S$ satsfying the condition above. 17. Let $f(x)= x^2-C$ where $C$ is a rational constant. Show that exists only finitely many rationals $x$ such that $\\{x,f(x),f(f(x)),\\cdots\\}$ is finite. 18. Find", "#### Function IMO ###### back to index Let $\\mathbb R$ be the set of real numbers. Determine all functions $f:\\mathbb R\\to\\mathbb R$ that satisfy the equation$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania Find all functions $f:\\mathbb Z\\rightarrow \\mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds: $f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ (Here $\\mathbb{Z}$ denotes the set of integers.) Proposed by Liam Baker, South Africa Let $\\mathbb{R}$ be the set of real numbers. Determine all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that, for all real numbers $x$ and $y$, $$f (f(x)f(y)) + f(x + y) = f(xy)$$", "\\in \\mathbb{Z} : n \\le i \\le m\\}$$. If $$a_i$$ is a real number assigned to each integer $$i \\in X$$, then we have $\\sum_{i=n}^{m}a_i = \\sum_{i \\in X}a_i$ 5. Disjoint sets. Let $$X$$, $$Y$$ be disjoint finite sets ($$X \\cap Y = \\emptyset$$), and $$f: X \\cup Y \\rightarrow \\mathbb{R}$$ is a function. Then we have $\\sum_{z \\in X \\cup Y}f(z) = \\left( \\sum_{x \\in X} f(x) \\right) + \\left( \\sum_{y \\in Y} f(y) \\right)$ 6. Linearity, part I. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ and $$g: X \\rightarrow \\mathbb{R}$$ be functions. Then we have $\\sum_{x \\in X}(f(x) + g(x)) = \\sum_{x \\in X}f(x) + \\sum_{x \\in X}g(x)$ 7. Linearity, part II. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ be a function, and let $$c$$ be a real number. Then $\\sum_{x \\in X}cf(x) = c\\sum_{x \\in X}f(x)$ 8. Monotonicity. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ and $$g: X \\rightarrow \\mathbb{R}$$ be functions such that $$f(x) \\le g(x) \\text{ for all } x \\in X$$. Then $\\sum_{x \\in X}f(x) \\le \\sum_{x \\in X}g(x)$ 9. Triangle inequality. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ be a function, then $|\\sum_{x \\in X}f(x)| \\le \\sum_{x \\in X}|f(x)|$ ### The", "f(n+1)-1$$ So $f(n+1) = n+2$, thus for $n \\in \\mathbb{N}$ we have $f(n) = n+1$. For the other way, let $x=0$ and $y=n-1$ then $$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$ So we see that if $f(y) =0$ for some $y \\in \\mathbb{Z}$, we have $f(x) = x+1$. So we can now assume $f(x) \\neq 0$ for all $x \\in \\mathbb{Z}$. If there exists an $y \\in \\mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$. Suppose there exists an $y \\in \\mathbb{Z}$ such that $f(y)=y$, then with $x=y$ $$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$ So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \\ge 0$ that $$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$ So we see that $f(y+n) = y$ for every $n \\in \\mathbb{N}$. Now, if $y>0$, we would find with $x=2y$ $$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$ So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$. Now, we have for all $x \\in \\mathbb{Z}$ $$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$ And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\\ge 0$ we have $$f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$ Then", "# Algebra of Real Function ## Algebra of Real Function Real Value Function: A function which has all real number or subset of the real number as it domain Real Valued Function: A function which has all real number or subset of the real number as it range For functions $f:X - > {\\bf{R}}$ and $g:X - > {\\bf{R}}$, we have $\\left( {f + g} \\right)\\left( x \\right) = f\\left( x \\right) + g\\left( x \\right),x \\in X$ 2. Substraction $\\left( {f - g} \\right)\\left( x \\right) = f\\left( x \\right)-g\\left( x \\right),x \\in X$ 3. Multiplication $\\left( {f.g} \\right)\\left( x \\right) = f\\left( x \\right).g\\left( x \\right),x \\in X$ 4. Multiplication by real number $\\left( {kf} \\right)\\left( x \\right) = kf\\left( x \\right),x \\in X$, where $k$ is a real number. 5. Division $\\frac{f}{g}\\left( x \\right) = \\frac{{f(x)}}{{g(x)}}$ $x \\in X$ and $g\\left( x \\right) \\ne 0$ ### Quiz Time Question 1. which is of the below relation is not a function. A) R = {(2,2),(2,4),(3,3), (4,4)} B) R = {(2,1),(3,1), (4,2)} C) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} D) None of the above Question 2. Let A = {1, 2, 3,4} and B = {5, 7}. Then possible number of relation from A to B ? A) 64 B) 256 C) 16 D) 32 Question 3.", "i \\le m\\}$$. If $$a_i$$ is a real number assigned to each integer $$i \\in X$$, then we have $\\sum_{i=n}^{m}a_i = \\sum_{i \\in X}a_i$ 5. [...]. Let $$X$$, $$Y$$ be disjoint finite sets ($$X \\cap Y = \\emptyset$$), and $$f: X \\cup Y \\rightarrow \\mathbb{R}$$ is a function. Then we have $\\sum_{z \\in X \\cup Y}f(z) = \\left( \\sum_{x \\in X} f(x) \\right) + \\left( \\sum_{y \\in Y} f(y) \\right)$ 6. [...], part I. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ and $$g: X \\rightarrow \\mathbb{R}$$ be functions. Then we have $\\sum_{x \\in X}(f(x) + g(x)) = \\sum_{x \\in X}f(x) + \\sum_{x \\in X}g(x)$ 7. [...], part II. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ be a function, and let $$c$$ be a real number. Then $\\sum_{x \\in X}cf(x) = c\\sum_{x \\in X}f(x)$ 8. [...]. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ and $$g: X \\rightarrow \\mathbb{R}$$ be functions such that $$f(x) \\le g(x) \\text{ for all } x \\in X$$. Then $\\sum_{x \\in X}f(x) \\le \\sum_{x \\in X}g(x)$ 9. [...]. Let $$X$$ be a finite set, and let $$f: X \\rightarrow \\mathbb{R}$$ be a function, then $|\\sum_{x \\in X}f(x)| \\le \\sum_{x \\in X}|f(x)|$", "the positive reals such that (1)If $x < y$, then $f(x) < f(y)$, (2)$f\\left(\\dfrac{2xy}{x+y}\\right) \\geq\\dfrac{f(x) + f(y)}{2}$ for all $x,y>0$. Show that $f(x) < 0$ for some value of $x$. 14. Define $f$ on the positive integers by $f(n) = k^2 + k + 1$, where $n=2^k(2l+1)$ for some $k,l$ nonnegative integers. Find the smallest $n$ such that $f(1) + f(2) + ... + f(n) \\geq 123456.$ 15. Find all functions $f:\\mathbb{N}\\to\\mathbb{N}$ satisfying, for all $x\\in\\mathbb{N}$, $f(2f(x)) = x + 1998.$ 16. a) Show that there are no functions $f,g: \\mathbb{R}\\to\\mathbb{R}$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x\\in\\mathbb{R}$. b)Let $S$ be the set of all real numbers greater than $1$. Show that there are functions $f,g:S\\to S$ satsfying the condition above. 17. Let $f(x)= x^2-C$ where $C$ is a rational constant. Show that exists only finitely many rationals $x$ such that $\\{x,f(x),f(f(x)),\\cdots\\}$ is finite. 18. Find all real-valued functions on the positive integers such that $f(x + 1019) = f(x)$ for all $x$, and $f(xy) = f(x)f(y)$for all $x,y$. 19. Find at least one function $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(0)=0$ and $f(2x+1) = 3f(x) + 5$ for any real $x$. 20. Let $f(x)=\\dfrac{ax+b}{cx+d}$, $F_n(x)=f(f(f\\cdots (f(x))\\cdots))$ (with $n\\ f's$). Suppose that $f(0)\\not =0$, $f(f(0))\\not =0$, and for some $n$ we have $F_n(0)=0$, show that $F_n(x)=x$", "(for any valid $x$). 21. Find all functions $f: \\mathbb{Z}\\rightarrow\\mathbb{Z}$ such that for all $x,y \\in \\mathbb{Z}$ $f(x-y+f(y))=f(x)+f(y).$ 22. Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $\\forall x,y,z\\in\\mathbb{R}$ we have: If $x^3+f(y) \\cdot x+f(z)=0,$ then $f(x)^3+y \\cdot f(x)+z=0.$ 23. Let $S\\subseteq\\mathbb{R}$ be a set of real numbers. We say that a pair $(f, g)$ of functions from $S$ into $S$ is a Spanish Couple on $S$, if they satisfy the following conditions (i) Both functions are strictly increasing, i.e. $f(x) < f(y)$ and $g(x) < g(y)$ for all $x$, $y\\in S$ with $x < y$; (ii) The inequality $f\\left(g\\left(g\\left(x\\right)\\right)\\right) < g\\left(f\\left(x\\right)\\right)$ holds for all $x\\in S$. Decide whether there exists a Spanish Couple a)On the set $S=\\mathbb{N}$ of positive integers; b)On the set $S=\\{a-\\dfrac{1}{b}: a, b\\in\\mathbb{N}\\}$. 24. For every $n\\in\\mathbb{N}$ let $d(n)$ denote the number of (positive) divisors of $n$. Find all functions $f:\\mathbb{N}\\to\\mathbb{N}$ with the following properties: a)$d\\left(f(x)\\right)=x$ for all $x\\in\\mathbb{N}$, and b)$f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x$, $y\\in\\mathbb{N}$. 25. Consider those functions $f:\\mathbb{N}\\to\\mathbb{N}$ which satisfy the condition $f(m+n) \\geq f(m)+f(f(n))-1$ for all $m,n\\in\\mathbb{N}.$ Find all possible values of $f(2007).$ 26. Find all surjective functions $f:\\mathbb{N}\\to\\mathbb{N}$ such that for every $m,n\\in\\mathbb{N}$ and every prime $p,$ the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p.$ 27. Find all real", "the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that$$xf(2f(y)-x)+y^2f(2x-f(y))=\\frac{f(x)^2}{x}+f(yf(y))$$for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$. Problem 10 (woah casework, 2012 usamo/4): Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$. Problem 11 (2012 imo/4, misplaced i think): Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$, the following equality holds:$$f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).$$ Problem 12 (2011 imo/3): Let $f: \\mathbb R \\to \\mathbb R$ be a real-valued function defined on the set of real numbers that satisfies$$f(x + y) \\le yf(x) + f(f(x))$$for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \\le 0$. Problem 13 (This was the first oly FE question I solved. That's why it's last :P 2010 imo/1): Find all function $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for all $x,y\\in\\mathbb{R}$ the following equality holds $f(\\left\\lfloor x\\right\\rfloor y)=f(x)\\left\\lfloor f(y)\\right\\rfloor$ where $\\left\\lfloor a\\right\\rfloor$ is greatest integer not greater than $a.$ Nothin much ## MONSTROUS Functional Equations Ok we're not going", "n)=f(m) \\cdot f(n)$ for every $m, n \\in A$ wi... Let $$S=\\{1,2,3,4,5,6\\}$$. Then the number of one-one functions $$f: \\mathrm{S} \\rightarrow \\mathrm{P}(\\mathrm{S})$$, where $$\\mathrm{P}(\\mathrm{S})$$... Suppose $$f$$ is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x,y \\in N$$ and $$f(1) = {1 \\over 5}$$. If $$\\sum\\limits_{n = 1}^m {{{f(n)... For some a, b, c$$\\in\\mathbb{N}$$, let$$f(x) = ax - 3$$and$$\\mathrm{g(x)=x^b+c,x\\in\\mathbb{R}}$$. If$${(fog)^{ - 1}}(x) = {\\left( {{{x - 7} \\over... For $$\\mathrm{p}, \\mathrm{q} \\in \\mathbf{R}$$, consider the real valued function $$f(x)=(x-\\mathrm{p})^{2}-\\mathrm{q}, x \\in \\mathbf{R}$$ and $$\\mathr... The number of functions$$f$$, from the set$$\\mathrm{A}=\\left\\{x \\in \\mathbf{N}: x^{2}-10 x+9 \\leq 0\\right\\}$$to the set$$\\mathrm{B}=\\left\\{\\mathrm... Let $$f(x)=2 x^{2}-x-1$$ and $$\\mathrm{S}=\\{n \\in \\mathbb{Z}:|f(n)| \\leq 800\\}$$. Then, the value of $$\\sum\\limits_{n \\in S} f(n)$$ is equal to _____... Let $$f(x)$$ be a quadratic polynomial with leading coefficient 1 such that $$f(0)=p, p \\neq 0$$, and $$f(1)=\\frac{1}{3}$$. If the equations $$f(x)=0... The sum of the maximum and minimum values of the function$$f(x)=|5 x-7|+\\left[x^{2}+2 x\\right]$$in the interval$$\\left[\\frac{5}{4}, 2\\right]$$, whe... Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If$$f(g(x)) = 8{x^2} - 2x$$and$$g(f(x)) = 4{x^2} + 6x + 1$$, then the val... Let c, k$$\\in$$R. If$$f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$$and$$f(x + y) = f(x) + f(y) - xy$$, for all x, y$$\\in$$R, then", "\\mathbb{R}x \\times (y + z) = (x \\times y) + (x \\times z)$$ 7. $$\\forall x \\in \\mathbb{R}x + 0 = x = 0 + x$$ 8. $$\\forall x \\in \\mathbb{R}x \\times 1 = x = 1 \\times x$$ 9. Every real number has an additive inverse 10. Every nonzero real number has a multiplicative inverse 11. $$\\forall x \\in \\mathbb{R}x \\leq x$$ 12. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$x \\leq y$$ and $$y \\leq x$$ then $$x = y$$ 13. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}x \\leq y$$ or $$y \\leq x$$ 14. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}\\forall z \\in \\mathbb{R}$$ if $$x \\leq y$$ and $$y \\leq z$$ then $$x \\leq z$$ 15. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}\\forall z \\in \\mathbb{R}$$ if $$y \\leq z$$ then $$x + y \\leq x + z$$ 16. $$\\forall x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$x \\nleq 0$$ and $$y \\nleq 0$$ then $$x \\times y \\nleq 0$$ 17. For any subset of the set of all real number such that it's nonempty and its complement is nonempty and for any real number in the subset, all real numbers less than or equal to it are in the subset, $$\\exists x \\in \\mathbb{R}\\forall y \\in \\mathbb{R}$$ if $$y \\neq x$$ then", "3. Let $n$ be a positive integer. Find the largest nonnegative real number $f(n)$ (depending on $n$) with the following property: whenever $a_1,a_2,...,a_n$ are real numbers such that $a_1+a_2+\\cdots +a_n$ is an integer, there exists some $i$ such that $\\left|a_i-\\dfrac{1}{2}\\right|\\ge f(n)$. 4. Let ${\\bf R}$ denote the set of all real numbers. Find all functions $f$ from ${\\bf R}$ to ${\\bf R}$ satisfying (i)There are only finitely many $s$ in ${\\bf R}$ such that $f(s)=0$, And (ii)$f(x^4+y)=x^3f(x)+f(f(y))$ for all $x,y$ in ${\\bf R}$. 5. Find all $a\\in\\mathbb{R}$ for which there exists a non-constant function $f: (0,1]\\rightarrow\\mathbb{R}$ such that $a+f(x+y-xy)+f(x)f(y)\\leq f(x)+f(y)$ for all $x,y\\in(0,1].$ 6. Consider function $f: \\mathbb{R}\\to\\mathbb{R}$ which satisfies the conditions for any mutually distinct real numbers $a,b,c,d$ satisfying $\\dfrac{a-b}{b-c}+\\dfrac{a-d}{d-c}=0$, $f(a),f(b),f(c),f(d)$ are mutully different and $\\dfrac{f(a)-f(b)}{f(b)-f(c)}+\\dfrac{f(a)-f(d)}{f(d)-f(c)}=0.$ Prove that function $f$ is linear. 7. Find all complex polynomial $P(x)$ such that for any three integers $a,b,c$ satisfying $a+b+c\\not=0,$ $\\dfrac{P(a)+P(b)+P(c)}{a+b+c}$ is an integer. 8. Find all functions $f:\\mathbb{Q}^{+}\\to\\mathbb{Q}^{+}$ such that $f(x)+f(y)+2xyf(xy)=\\dfrac{f(xy)}{f(x+y)}\\forall x,y\\in\\mathbb{Q}^+.$ 9. Let $\\alpha$ be given positive real number, find all the functions $f:\\mathbb{N}^{+}\\to\\mathbb{R}$ such that $f(k + m) = f(k) + f(m)$ holds for any positive integers $k$, $m$ satisfying $\\alpha m \\leq k \\leq (\\alpha + 1)m$. 10. Given non-zero reals $a$, $b$, find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that for every $x, y", "# IMO 2011 (Problem 3) Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a real-valued function defined on the set of real numbers that satisfies $f (x + y) \\leq yf (x) + f (f (x))$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x\\leq 0$.", "# [Shortlists] International Mathematical Olympiad 2005 ### Algebra 1. Find all pairs of integers $a,b$ for which there exists a polynomial $P(x) \\in \\mathbb{Z}[X]$ such that product $(x^2+ax+b)\\cdot P(x)$ is a polynomial of a form $x^n+c_{n-1}x^{n-1}+\\cdots+c_1x+c_0$ where each of $c_0,c_1,\\ldots,c_{n-1}$ is equal to $1$ or $-1$. 2. We denote by $\\mathbb{R}^+$ the set of all positive real numbers. Find all functions $f: \\mathbb R^ + \\rightarrow\\mathbb R^ +$ which have the property $f(x)f(y)=2f(x+yf(x))$ for all positive real numbers $x$ and $y$. 3. Four real numbers $p,q,r,s$ satisfy $$p+q+r+s = 9$$ and $$p^{2}+q^{2}+r^{2}+s^{2}=21.$$ Prove that there exists a permutation $\\left(a,b,c,d\\right)$ of $\\left(p,q,r,s\\right)$ such that $ab-cd \\geq 2$. 4. Find all functions $f: \\mathbb{R}\\to\\mathbb{R}$ such that $$f(x+y)+f(x)f(y)=f(xy)+2xy+1$$ for all real numbers $x$ and $y$. 5. Let $x,y,z$ be three positive reals such that $xyz\\geq 1$. Prove that $\\frac { x^5-x^2 }{x^5+y^2+z^2} + \\frac {y^5-y^2}{x^2+y^5+z^2} + \\frac {z^5-z^2}{x^2+y^2+z^5} \\geq 0 .$ ### Combinatorics 1. A house has an even number of lamps distributed among its rooms in such a way that there are at least three lamps in every room. Each lamp shares a switch with exactly one other lamp, not necessarily from the same room. Each change in the switch shared by two lamps changes their states simultaneously. Prove that for every initial state of the lamps there exists a" ]

[ "# Solve the following functional equation $f(xf(y))+f(yf(x))=2xy$ Find all function $f:\\mathbb{R}\\rightarrow \\mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$. By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$. • $f(xf(x))=x^2=f(-xf(-x))$ and therefore $xf(x)=-xf(-x)$. Hence $f(x)=-f(-x)$. – Pp.. Jan 6 '15 at 19:47 • @Pp : That is definition of all $odd functions$ – Narasimham Jan 6 '15 at 20:14 • @Narasimham Yes, it is a property that $f$ must have. – Pp.. Jan 6 '15 at 20:15 Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\\pm 1$ Let $f(1)=1$. $x\\rightarrow xf(x), y\\rightarrow \\frac{1}{x} \\Rightarrow f(xf(x)f(\\frac{1}{x}))+f(x)=2f(x) \\rightarrow f(x)f(\\frac{1}{x})=1$. Putting now $x\\rightarrow \\frac{1}{x}, y \\rightarrow 1$ we get $\\frac{1}{f(x)}+ \\frac{1}{f(f(x))}=\\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$. This is all I have found so far: Just keep feeding the snake with its own tail. As you noted $f(f(1))=1$. So with $x=y=f(1)$ we then see that $1=f(f(1))=(f(1))^2$ so that $f(1)=\\pm 1$. As you almost correctly noted, we have for $y=f(1)$ that $$f(x)+f(f(x)f(1))=2xf(1)$$ which indeed shows that $f$ is injective. With that observation, the comment by Pp.. works to conclude", "# 2001 IMO Shortlist Problems/A4 ## Problem Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, satisfying $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$. ## Solution Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\\ \\forall x$, so $f(x) = 0,\\ \\forall x$. This is a solution, so we can take it out of the way: assume $f(1)\\ne 0$. $y = 1\\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\\in\\{0,f(1)x\\}$. In particular, $f(0) = 0$. Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\\ \\forall x$. This means that $f(a),f(b)\\ne 0\\Rightarrow f\\left(\\frac ab\\right)\\ne 0\\ (*)$ ($\\frac ab$ is defined because $f(b)\\ne 0\\Rightarrow b\\ne 0$). Assume now that $x\\ne y$ and $f(x),f(y)\\ne 0$. We get $f(x) = f(1)x,\\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\\ne 0$, so $x\\ne y,\\ f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (**)$. Assume now $f(x)\\ne 0$. From $(*)$ we get $f\\left(\\frac 1x\\right)\\ne 0$, and after applying $(*)$ again to $a = x,b = \\frac 1x$ we get $f(x^2)\\ne 0\\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (\\#)$. Let $G = \\{x\\in\\mathbb R|f(x)\\ne 0\\}$. $(*)$ and $(\\#)$ simply say that $(G,\\ \\cdot)$ is a subgroup of $(\\mathbb R^{*},\\", "of the problem which appeared here in different times, i.e. the version posted before the OP edited his post and the current one. First I want to solve the problem as it is stated by the OP. Problem I $$f(y+f(x))=f(x)f(y)+f(f(x))+f(y)-xy.$$ As already noticed, setting $y=0$, one arrives to $$f(f(x))=f(0)f(x)+f(f(x))+f(0)$$ which leads to $f(x)\\equiv -1$, which is not a solution of the problem, or $f(0)=0$. Incidentally, even though it is not crucial for the solution, notice that if we assume that there is an $\\bar x$ such that $f(\\bar x)=0$, setting $x=\\bar x$ we infer that $f(y)=f(y)-\\bar xy$. In particular, setting $y=1$, we conclude $\\bar x=0$. Next, plug inthe equation $f(y)$ instead of $y$ to get $$f(f(y)+f(x))=f(x)f(f(y))+f(f(x))+f(f(y))-xf(y).$$ Interchange the roles of $x$ and $y$ to obtain similarly $$f(f(x)+f(y))=f(f(x))f(y)+f(f(y))+f(f(x))-yf(x).$$ Compare these two equations to conclude that $$\\frac{f(f(x))+x}{f(x)}=\\frac{f(f(y))+y}{f(y)}=k,$$ where $k$ is a suitable constant. Substitute in the original equation to get $$f(y+f(x))=f(x)f(y)-x+kf(x)+f(y)-xy.$$ Denote $a:=f(-1)$ and choose $y=-1$ to obtain $$f(f(x)-1)=(a+k)f(x)+a$$ Edit Thanks to Dejan who pointed out a flaw in my proof Plug in the original equation $f(y)-1$ instead of $y$ to obtain \\begin{aligned}f(f(y)+f(x)-1)&=f(x)[(a+k)f(y)+a]-x+kf(x)-x(f(y)-1)\\\\&=(a+k)(f(x)+f(y))-xf(y).\\end{aligned} Exchange $x$ and $y$ to obtain $$f(f(y)+f(x)-1)=(a+k)(f(x)+f(y))-yf(x).$$ Compare these two equations to deduce that $$\\frac{f(x)}{x}=\\frac{f(y)}{y}=a.$$ Substitute in the original equation to find out that the admissible values for $a$ are just $\\pm 1$ i.e. $$f(x)=\\pm x.$$ Problem", "# Functional equation $f(f(f(x)f(y)))=f(x)f(y^2)$ Find all functions $f: \\mathbb R_{>0} \\rightarrow \\mathbb R_{>0}$ such that $f(f(f(x)f(y)))=f(x)f(y^2)$ for all $x, y \\in \\mathbb R_{>0}$. • f(x) = k gives an entire family of functions. – Chris Cudmore Jul 22 '16 at 20:02 • @ChrisCudmore That doesn't work for every $k$. – wythagoras Jul 22 '16 at 20:03 $f(x)f(1)=f(f(f(x)f(1)))=f(f(f(1)f(x)))=f(1)f(x^2)$. Therefore $f(x)=f(x^2)$ for all $x$. Now $f(x)^2=f(x)f(x^2)=f(f(f(x)f(x)))=f(f(f(x)))$. So $f(f(x))=f(f(x)^2)=f(f(f(f(x))))=f(f(x))^2$. Therefore $f(f(x))=1$ for all $x$. But now $f(x)^2=f(f(f(x)))=1$, so $f(x)=1$. The only solution is the constant function $f(x)=1$. • That's doing my brain in... how do you get $f(f(f(f(x)))) = f(f(x))^2$? I just about get every other part though - nice solution! – Shuri2060 Jul 22 '16 at 21:21 • With $f(z)^2=f(f(f(z)))$, substitute $z=f(x)$. – f'' Jul 22 '16 at 21:29 • Thanks. Out of interest, is the solution to $f(x) = f(x^2)$ a constant in general (out of context of the question)? – Shuri2060 Jul 22 '16 at 21:33 • If $f$ must be continuous over $\\mathbb{R}^+$, then yes, because $\\lim_{x\\rightarrow1}f(x)$ would not exist if it was not constant. – f'' Jul 22 '16 at 21:41 • most appropriate username for this solution – elias Jul 22 '16 at 22:03 Extremely-partial initial observations: The LHS is symmetrical in x,y so the RHS is too so $f(x)f(y^2)=f(x^2)f(y)$ so $f(x^2)/f(x)$ is constant; say", "# 2014 USAJMO Problems/Problem 3 ## Problem Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that $$xf(2f(y)-x)+y^2f(2x-f(y))=\\frac{f(x)^2}{x}+f(yf(y))$$ for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$. ## Solution Let's assume $f(0)\\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get $$2f(0)^2=f(2f(0))^2/2f(0)+f(0)$$ $$2f(0)^2(2f(0)-1)=f(2f(0))^2$$ This means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\\neq 0, y=0$ to get $$xf(-x)=\\frac{f(x)^2}{x} \\implies x^2f(-x)=f(x)^2.$$ Similarly, $$x^2f(x)=f(-x)^2.$$ From these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ Both of these are valid solutions on their own, so let's see if there are any solutions combining the two. Let's say we can find $f(x)=x^2, f(y)=0,$ and $x,y\\neq 0.$ Then $$xf(-x)+y^2f(2x)=f(x)^2/x.$$ $$y^2f(2x)=x-x^3.$$ If $f(2x)=4x^2,$ then $y^2=\\frac{x-x^3}{4x^2}=\\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields $$0=2f(0)+1f(2)=0+f(1)=1,$$ which is impossible. So $f(2)=f(-2)=4$ and there are no solutions \"combining\" $f(x)=x^2$ and $f(x)=0.$ Therefore our only solutions are $\\boxed{f(x)=0}$ and $\\boxed{f(x)=x^2.}$", "# Solve the functional equation $f(xf(y)+x)=xy+f(x)$ Find all $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ so that $f(xf(y)+x)=xy+f(x)$. If you put $x=1$ it's easy to prove that f is injective. Now putting $y=0$ you can get that $f(0)=0$. $y=\\frac{-f(x)}{x}$ gives us $f(\\frac{-f(x)}{x})=-1$ • I tried putting $y=x=1$ and got $f(f(1)+1)=1+f(1)$ from which I guessed that $f(x)=x$. This seems to fit the identity you give but doesn't seem like a strong enough proof. – Mufasa Jan 18 '15 at 14:10 • It's a step in the right direction. Actually, if you put $f(x)=ax$ you get that $a=\\pm 1$, so $f(x)=\\pm x$ are for now the only slutions I have, but are not poven. – HeatTheIce Jan 18 '15 at 14:12 Set $x = 1$ and $y = t - f(1)$, then $f(1+f(t - f(1))) = t$, for all $t \\in \\mathbb{R}$, hence $f$ is a surjective function. So, $\\exists\\, y_1 \\in \\mathbb{R}$, s.t., $f(y_1) = 0$ Then, $$f(x) = f(x+f(y_1)x) = xy_1+f(x) \\implies xy_1 = 0, \\forall x \\in \\mathbb{R} \\implies y_1 = 0$$ and $\\exists\\, y_2 \\in \\mathbb{R}$, s.t., $f(y_2) = -1$ Then, $$0 = f(0) = f(x+xf(y_2)) = xy_2 + f(x)$$ i.e., $f(x) = -y_2x$ for $x \\in \\mathbb{R}$. Plugging it back in the functional equation we may verify $y_2 = \\pm 1$. Let f(1)=b. Putting now $x=1,y=-b$ we get that $f(-b)=-1$ Now", "one-one. Find the inverse of the function f: [- 1, 1] → Range f. [Hint : For y ∈ R Range f, y = f(x) = $$\\frac{x}{x+2}$$, for some x in [- 1, 1] i.e., x = $$\\frac{2 y}{1-y}$$] Solution. f: [- 1, 1] → R, is given as f(x) = $$\\frac{x}{x+2}$$ Let f(ix) = f(y). ⇒ $$\\frac{x}{x+2}=\\frac{y}{y+2}$$ ⇒ 2x = 2y ⇒ x = y ∴ f is one-one function. It is clear that f: [- 1,1] Range f is onto. ∴ f: [- 1, 1] → Range f is one-one and onto and therefore, the inverse of the function : f: [- 1, 1] → Range f exists. Let g: Range f → [- 1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since, f: [- 1, 1] → Range f is onto , we have y = f(x) for some x ∈ [- 1, 1] ⇒ y = $$\\frac{x}{x+2}$$ ⇒ xy + 2y = x ⇒ x(1 – y) = 2y ⇒ x = $$\\frac{2 y}{1-y}$$, y ≠ 1 Now, let us define g: Range f → [- 1, 1] as g(y) = $$\\frac{2 y}{1-y}$$, y ≠ 1. Now, (gof)(x) = g(f(x)) = g($$\\frac{x}{x+2}$$) = $$\\frac{2\\left(\\frac{x}{x+2}\\right)}{1-\\frac{x}{x+2}}$$ = $$\\frac{2 x}{x+2-x}=\\frac{2 x}{2}$$ = x (fog)(y) = f(g(y)) = f($$\\frac{2 y}{1-y}$$)", "then it is easy to show that $f(x) = 0$ for all $x$. So suppose $f(x) \\neq 0$ for any $x$. Now $f(x) = (f(x/2))^2 \\gt 0$. Thus it makes sense to talk about $g(x) = \\log f(x)$. This satisfies $g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation and continuity at $0$ implies $g(x) ... 0 Let$x=y=0$then$f(0)=f(0)^2$if$f(0)=0$. Let$y=0$we have$f(x)=f(x)f(0)=0$Hence we have a function :$f \\equiv 0$. If$f(0)=1$Let$y=-x$we have$1=f(0)=f(x)f(-x)$. I can't prove$f(x)>0$. If$f(x)>0$then let$g(x)=\\ln(f(x))$and$g(0)=0$. Hence,$g(x+y)=g(x)+g(y)$thus,$\\frac{g(x+y)-g(y)}{x}=\\frac{g(x)}{x}=g'(c)$with$c \\in (x,x+y)$... 1 From @Haderlump Let's define$x_i$on$x$being$b$as$x^b$. Since $$P(x_j,j \\ne i) = P(x^0) + P(x^1)$$ We can do $$P(x_i = 1 | x_j,j \\ne i) = \\frac{P(x^1)}{P(x^0) + P(x^1)}$$ It's better to see$P(x^1)$as divider, and cleaner to get rid of the +1 in$1/1 + e^{-a_i}$, so we flip the division and subtract one, $$1 / P(x_i = 1 | x_j,j \\ne i) = ... 1 The continued fraction representation x=[a_0;a_1,a_2,\\ldots] is unique when x is irrational. If x is rational, there are exactly two continued fraction representations: x=[a_0;a_1,\\cdots,a_n]=[a_0,a_1,\\cdots,(a_{n}-1),1], so it's easy to see the question-mark function is indeed independent of continued fraction representation. 2 Start from the fact that, for each i, the density of the full random vector (X_j)_j can be factored as$$P(X_i=x_i,\\hat X_i=\\hat x_i)=h(\\hat x_i)\\cdot\\mathrm e^{a_i(x_i,\\hat x_i)x_i},\\qquad$$for some positive function h whose exact value is irrelevant, with$$\\hat X_i=(X_j)_{j\\ne i},\\qquad\\hat", "# How can one show that $f(0)=0$ for $f$ satisfying certain conditions? Given the functional equation $$f(x+(1+x)f(y))=y+(1+y)f(x)$$ Such that $f:(-1,\\infty) \\to (-1,\\infty)$ and the function $g(x):=\\frac{f(x)}{x}$ is strictly increasing in $I=(-1,0)\\cup(0,+\\infty),$ how can one show that for every $t \\in I$ we have $f(t)\\neq t$? Plugging $x=y=0,$ that makes $f(f(0))=f(0).$ Does that mean that $f(0)=0$? Making $x=y,$ $f(x+(1+x)f(x))=x+(1+x)f(x)$ so we make $x+(1+x)f(x)=t$ that makes $f(t)=t$ but we can't make $t=0$ because $t$ is not in $I$? So do I have to use the fact that $\\frac{f(x)}{x}$ is increasing? Then after all this work we have to deduce that $f(0)=0$ and $f(x)=\\frac{-x}{x+1}$ for $x \\in (-1,\\infty).$ • Do $F$ and $f$ denote the same function here? – JimmyK4542 Dec 15 '15 at 23:20 • @JimmyK4542 fixed – user233658 Dec 15 '15 at 23:21 • To answer your question in line 5, $f(f(0)) = f(0)$ does not imply that $f(0) = 0$ unless you have other facts. For instance, $f(x) = 1$ satisfies $f(f(0)) = f(0)$ but not $f(0) = 0$. – JimmyK4542 Dec 15 '15 at 23:28 • it was me, and I'm not familiar with these notations, so I guess not only me.... but then you have empty set, perhaps you want union $\\cup$? – Michael Medvinsky Dec 15 '15 at 23:32 • Is $g$ increasing on $I$", "If f is a function satisfying such that , find the value of n. Question: If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \\in \\mathrm{N}$ such that $f(1)=3$ and $\\sum_{x=1}^{n} f(x)=120$, find the value of $n$. Solution: It is given that, $f(x+y)=f(x) \\times f(y)$ for all $x, y \\in \\mathrm{N}$ $f(1)=3$ Taking $x=y=1$ in (1), we obtain $f(1+1)=f(2)=f(1) f(1)=3 \\times 3=9$ Similarly, $f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \\times 9=27$ $f(4)=f(1+3)=f(1) f(3)=3 \\times 27=81$ $\\therefore f(1), f(2), f(3), \\ldots$, that is $3,9,27, \\ldots$, forms a G.P. with both the first term and common ratio equal to 3 . It is known that, $S_{n}=\\frac{a\\left(r^{n}-1\\right)}{r-1}$ It is given that, $\\sum_{x=1}^{n} f(x)=120$ $\\therefore 120=\\frac{3\\left(3^{n}-1\\right)}{3-1}$ $\\Rightarrow 120=\\frac{3}{2}\\left(3^{n}-1\\right)$ $\\Rightarrow 3^{n}-1=80$ $\\Rightarrow 3^{n}=81=3^{4}$ $\\therefore n=4$ Thus, the value of $n$ is 4 .", "From the given condition $$f(\\frac{x+y}{2}) = [f(x) f(1+\\frac{y}{x})]/2$$ so $$x \\neq 0$$ Here put y=0. So, $$f(\\frac{x}{2})=\\frac{f(x)f(1)}{2}$$ ie, $$2f(x)=f(2x)f(1)$$ Now $$f'(x) = \\lim_{h\\rightarrow 0} \\frac{f(x+h) - f(x)}{h}$$ $$= \\lim_{h\\rightarrow 0} \\frac{f(\\frac{2x+2h}{2}) - f(x)}{h}$$ $$= \\lim_{h\\rightarrow 0} \\frac{f(2x)f(1+\\frac{h}{x}) - 2f(x)}{2h}$$ Since $$2f(x)=f(2x)f(1)$$ we have, $$=(\\frac{f(2x)}{2x}) \\lim_{h\\rightarrow 0} \\frac{f(1+\\frac{h}{x}) - f(1)}{\\frac{h}{x}}$$ So, we have $$f'(x) = \\frac{f(2x)}{2x} f'(1)$$ Since $$2f(x)=f(2x)f(1)$$ and $$f(1)=f'(1)$$ $$f'(x)=\\frac{f(x)}{x}$$ $$f(x)=|cx|$$ Hence, by substituting back into the original condition, the only solutions are $$f(x) = |x|$$ and $$f(x) = 0$$ Sci Advisor HW Helper P: 9,396 Your 'solution' somhow omits the identity function and appears to claim that |x| is everywhere differentiable, and makes untold assumptions about f HW Helper PF Gold P: 1,197 That's why I changed the question, so that f(x) is differentiable everywhere except x=0 was given Quote by siddharth So you are given that f(x) is differentiable at all points except x=0. May I know what untold assumptions are being made? Sci Advisor HW Helper P: 9,396 Firstly I must admit I've given up on rereading the new assumptions you keep making, and secondly you conclude f(x)=|x| is a solution. That is not an everywhere differentiable function, and you *do not conclude* that f(x)=x is a solution. Read your own proof and its conclusion that states the solutions are f(x)=|x| and f(x)=0! HW", "[y_0x_0]^2) \\\\ &= f([x_0y_0]^2) - f(x_0y_0^2x_0) - f(y_0x_0^2y_0) + f([y_0x_0]^2) \\\\ &= [f(x_0y_0)]^2 - f(x_0)^2f(y_0^2) - f(y_0)^2f(x_0^2) + [f(y_0x_0)]^2 \\\\ &= \\left [ \\frac{1}{2} \\right ]^2 - 0 - f(y_0)^2[f(x_0)]^2 + \\left [ \\frac{1}{2} \\right ]^2 \\\\ &= \\frac{1}{4} - 0 - 0 + \\frac{1}{4} \\\\ &= \\frac{1}{2} \\end{align} • So we have arrived at a contradiction. So the assumption that $f(xy) \\neq f(yx)$ is false. Thus $f(xy) = f(yx)$ for all $x, y \\in \\mathfrak{A}$. So by $(\\star)$ we have that for all $x, y \\in \\mathfrak{A}$: (17) \\begin{align} \\quad 2f(xy) = f(xy) + f(yx) = f(xy + yx) = 2f(x)f(y) \\end{align} • Thus for all $x, y \\in \\mathfrak{A}$, $f(xy) = f(x)f(y)$. Thus $f$ is multiplicative. $\\blacksquare$ Observe the equality $(*)$ in the above proof. Note that if $\\mathfrak{A}$ is a commutative Banach algebra then the Jordan functional $f$ being multiplicative comes immediately since then $xy = yx$, so $f(xy + yx) = f(2xy) = 2f(xy)$, so $2f(xy) = 2f(x)f(y)$, i.e., $f(xy) = f(x)f(y)$ for all $x, y \\in \\mathfrak{A}$.", "# For $f(x)f(y)+f(3/x) f(3/y)=2f(xy)$ choose correct choices Consider $f:\\mathbb{R}^+ \\to \\mathbb{R}$ such that $f(3)=1$ and $$f(x)f(y)+f(3/x) f(3/y)=2f(xy)\\;\\;\\;\\;\\;\\;\\forall x,y \\in \\mathbb{R}^+$$ then choose the correct option(s): (A) $f(2014)+f(2015)-f(2010)=100$ (B) $f$ is an even function (C) $\\frac{f(100)}{f(10)+f(90)}=\\frac{1}{2}$ (D) $f$ is a periodic function By putting $x=1$ in given equation to get $f(1)=1$ Generally I have been taught to solve such question using first principle of differentiation but here I am not able to deal with the second term of L.H.S. in given equation. May someone provide some help? If we write $y=3/x$ then we get $$2f(x)f({3\\over x}) = 2f(3)=2$$ so we have for all $x>0$: $$f\\big({3\\over x}\\big) = {1\\over f(x)}\\;\\;\\;(*)$$ Pluging this in original equation we get $$f(x)f(y) +{1\\over f(x)f(y)} = 2f(xy)$$ which is valid for all $x,y>0$. Put now $x=y$ we get $$f(x)^2 +{1\\over f(x)^2} = 2f(x^2)$$ By AG-GM we have: $f(x)^2 +{1\\over f(x)^2}\\geq 2$, so $f(x^2)\\geq 1$ for all $x>0$ so $f(x)\\geq 1$ for all $x>0$. Finally, useing $(*)$ we get $$f\\big({3\\over x}\\big) \\leq 1$$ for all $x>0$. But since $x\\mapsto {3\\over x}$ is bijective on $\\mathbb{R}^+$ we have $f(x) \\leq 1$ for all $x>0$. So we conclude $f(x)=1$ for all $x> 0$, so the answer is $(C)$. • Double-check your inequality directions from \"Finally\" onwards. – J.G. Apr 17 '18 at 21:03 • You don't need the", "$f(x^3) = f(x^3+0) =_{(1)} x^2f(x)$, and $f(x^3) = f(0+x^3) =_{(1)} xf(x^2)$. That is, $f(x^3) = x^2f(x) = xf(x^2).$ (2) When $x=0$ equation (2) tells us that $f(0)=0$; when $x \\ne 0$ we can divide through by $x$ to get $xf(x) = f(x^2)$. Thus, for all real $x$, $f(x^2)=xf(x).$ (3) Furthermore, equation (2), when put back into (1), tells us that $$f\\left(x^3+y^3\\right) = x^2f(x) + yf(y^2) =_{(2)} f(x^3)+f(y^3)$$ for all real numbers $x$ and $y$. It follows that for any reals $a$ and $b$ we can set $x=a^{1/3}$ and $b=y^{1/3}$, and deduce that $f(a+b) = f(a) + f(b).$ (4) We recognize (4) to be Cauchy's functional equation, which has infinitely many other solutions in addition to the linear solutions $f(x)=mx$. Several of our correspondents felt obliged to make further assumptions (such as continuity) to eliminate the unwanted nonlinear solutions. The majority of our solvers saw, however, that equation (3) already does the job without the help of further assumptions! Note on the one hand that for all $x$, $$f\\left((x+1)^2\\right) =_{(3)} (x+1)f(x+1) =_{(4)} (x+1)(f(x)+f(1)) = xf(x)+xf(1)+f(x)+f(1).$$ On the other hand, \\begin{eqnarray*} f\\left((x+1)^2\\right) = f\\left((x^2+x)+(x+1)\\right) &=_{(4)}& f\\left((x^2+x)\\right) + f(x+1)\\\\ &=_{(4)}& f(x^2) + f(x) + f(x) + f(1)\\\\ & =_{(3)}& xf(x) + f(x) + f(x) + f(1). \\end{eqnarray*} It follows that $$xf(x)+xf(1)+f(x)+f(1) = xf(x) + f(x) + f(x) + f(1),$$ whence, $f(x)", "1=0$.) We set $f(0)=0.$ Indeed, we show first show that the function satisfies the conditions. First, we see that $f(x)$ does not vanish for non-zero $x$. We claim that if $x,y \\in K_g, f(x)/f(y)= (-n+1)^{v_p(\\frac{x}{y})}$. This is obvious by substituting the definitions of $of f(x)$ and $f(y)$, and then LHS$=(-n+1)^{v_p(\\frac{x}{\\kappa_g})-v_p(\\frac{y}{\\kappa_g})}=(-n+1)^{v_p(\\frac{x}{y})}=$RHS. Now, we verify the conditions. Condition 1 is already checked. If $x=0$, $f(x)+f(2x)+...+f(nx)=0+0+...+0=0$. If $x \\ne 0$, let $x \\in K_g$. For $jx$ with $j\\ne p$, we have $v_p(\\frac {jx}{x})=v_p(j)=0$, and thus $f(jx)=f(x)$. Since $p\\leq n < 2p$, there are $(n-1)$ such $j'$s. $f(px)=(-n+1)^{v_p(\\frac{px}{x})}f(x)=(-n+1)f(x)$. Thus $f(x)+f(2x)+...+f(nx)= (n-1)f(x)+(-n+1)f(x)=0$. We have found the desired function, for each of $n\\in \\mathbb N^+,n\\ge 2$. Q.E.D. Comment: to be written later. Standard IMO # IMO 2011 Q3 $f: \\mathbb {R} \\to \\mathbb {R}$ satisfies $f(x+y) \\leq yf(x)+f(f(x))$ for all real numbers $x$ and $y$. Show that $f(x)=0$ for all $x \\leq 0$. First, set $y=0$ gives us inequality $f(f(x)) \\geq f(x)$ for all $x\\in \\mathbb R$. Let the inequality be denoted by $P(x,y)$. $P(x, f(y)-x)$ gives us $f(f(y)) \\leq (f(y)-x)f(x) + f(f (x))$ for all $x,\\,y \\in \\mathbb R$. Similarly, we have $f(f(x)) \\leq (f(x)-y)f(y) + f(f(y))$, by switching $x$ and $y$. Adding the two inequalities up, we get $2f(x)f(y) \\geq xf(x) + yf(y),\\, \\forall x,\\,y \\in \\mathbb R$. Let $y=2f(x)$ in this", "Find all functions $f : [0, \\infty ) \\to \\mathbb{R},$ that is, an expression for $f(x),$ such that for all $x,y \\in [0,\\infty)$ we have $f(x+2y)+2f^2(y)+x = f^2(x+2y)+2f(y)-\\frac{1}{4}.$ Determine the value of $f$ for some simple cases. What about the value of $f(0)?$ Why not try getting an equation of $f$ with a single variable? … by substituting a suitable value for $y.$ By setting $x=y=0,$ \\begin{aligned} f(0) + 2f^2(0) &= f^2(0) + 2f(0) - \\frac{1}{4} \\\\ f^2(0) - f(0) + \\frac{1}{4} &= 0 \\\\ \\big(f(0) - \\frac{1}{2}\\big)^2 &= 0, \\\\ \\end{aligned} which gives us $f(0) = \\frac{1}{2}.$ Now, we can obtain $f(x)$ by setting $y=0:$ \\begin{aligned} f(x)+2f^2(0)+x &= f^2(x)+2f(0)-\\frac{1}{4} \\\\ f^2(x)-f(x)+\\frac{1}{4}-x &= 0.\\\\ \\end{aligned} Solving this quadratic equation gives us $f(x) = \\frac{1}{2} \\pm \\sqrt{x}.$", "# Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ Find all functions $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\\ge 0$ we have $f(a)>f(b)$ As much as I know: • putting $x=y=0$ we get $f(0)=0$. • let $x+y=0$, the we get $f(y)+f(-y)=\\frac{f(2y)}{2}$ Step 1: Prove : $f(y) = 0$ iff. $y=0$ It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$) Following proof is given by @Kyson To prove that $f(y)≠0$ for all$y<0.$ Assume there is a $y<0$ such that $f(y)=0.$ Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0$Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$ $\\$ Step 2: Prove $f(y)$ is even. Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\\frac{f(2y)}{2} = f(-y) +f(y) = \\frac{f(-2y)}{2}$ Thus $f(x)$is a even function, so $f(y)+f(-y)=2f(y)=\\frac{f(2y)}{2} \\Rightarrow f(2y)=4f(y)$ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\\infty,0)\\cup(0,\\infty)$) $\\$ Step 3: Prove:$f(y)=y^2f(1)$ By induction , find $f(ny)=n^2f(y)$ where $n \\in \\mathbb N$ Details of induction: Suppose $m\\in \\mathbb N ,m\\ge 2 ,\\ f(ny)=n^2f(y)$ holds for all $n\\le m$ Put $x=(m-1)y$ in property (a).$\\\\ \\Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\\\ \\Rightarrow f((m+1)y)=(m+1)^2f(y) \\ for\\ y\\not = 0 \\ and \\ it\\ still\\ holds\\ for\\ y=0$ Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$ Remark: Result induces $f(zy)=z^2f(y)$ where $z\\in \\mathbb Z$ , since $f$ is even. $\\$ $\\forall", "$x\\to0$ so that we get $f(1)=yf(y)=1$ whence $f(x)=1/x$. Checking this solution, we get $$f(1+xf(y))=yf(y+x)\\Rightarrow \\frac1{1+x/y}=\\frac y{y+x},$$ which is true, so $f(x)=1/x$ is a solution. • If we fix some $y\\ne1$ such that $f(y)\\ne1$ (which exists, because $f(x)=1$ is not a solution) and choose $x$ so that $1+xf(y)=y+x$, then $$x=\\frac{y-1}{f(y)-1}\\Rightarrow f\\Big(\\!\\frac{yf(y)-1}{f(y)-1}\\!\\Big)=yf\\Big(\\!\\frac{yf(y)-1}{f(y)-1}\\!\\Big)\\Rightarrow f\\Big(\\!\\frac{yf(y)-1}{f(y)-1}\\!\\Big)=0\\notin\\mathbb R^+,$$ which is a contradiction, so either $x=\\frac{y-1}{f(y)-1}\\le0$ or $y+x=\\frac{yf(y)-1}{f(y)-1}\\le0$. The second condition implies the first, which is equivalent to $$f(y)\\ge1\\mbox{ if }y\\le1\\mbox{ and }f(y)\\le1\\mbox{ if }y\\ge1.$$ • If we suppose $a:=\\max(f(1),f(1)^{-1})\\ne1$, then $f(1+ax)=f(1+x)$ (setting $y=1$). Moreover, this generalizes to $$yf(x+y)=f(1+xf(y))=f(1+axf(y))=yf(ax+y)\\Rightarrow f(y+x)=f(y+ax),$$ so letting $y=1-x$, we get $a=f((a-1)x+1)$, so $f(x)=f(1)$ on $[1,a)$ (assuming $a>1$), and by $f(1+ax)=f(1+x)$, it is also constant on $[1,a^2)$, $[1,a^3)$, etc. so that (by induction) it is constant for all $x\\ge1$. But then, choosing $y=2$, $f(1+xf(2))=2f(x+2)\\Rightarrow$ $f(1)=2f(1)$, since both arguments are greater than $1$, and this is a contradiction. Thus, $f(1)=1$. • If we assume $f$ is continuous at $1$, then if we let $x\\to0$, $$\\lim_{x\\to0}f(y+x)=\\lim_{x\\to0}\\frac{f(1+xf(y))}y=\\frac{f(1)}y=\\frac1y$$ which implies that $f$ is continuous almost everywhere (discontinuous on a nowhere-dense set) and equals $1/x$ where it is continuous. For the remaining points, let $x$ be small enough that $f(1+xf(y))=\\frac1{1+xf(y)}$ and $f(y+x)=\\frac1{y+x}$. Then $\\frac1{1+xf(y)}=\\frac y{y+x}$ so that $f(y)=1/y$ on $(0,\\infty)$. In summary: $f(x)=1/x$ is a solution, and $\\operatorname{sgn}(y-1)=\\operatorname{sgn}(1-f(y))$. Other than that, it is known that any", "we obtain: $$f(x)=x^2f\\left(\\frac{1}{x}\\right)=x^2f\\left(\\frac{1}{x}-1+1\\right)=x^2\\left(f\\left(\\frac{1}{x}-1\\right)+1\\right)=$$ $$=x^2+x^2f\\left(\\frac{1-x}{x}\\right)=x^2+x^2\\cdot\\frac{f\\left(\\frac{x}{1-x}\\right)}{\\frac{x^2}{(x-1)^2}}=x^2+(x-1)^2f\\left(\\frac{x}{1-x}\\right)=$$ $$=x^2+(x-1)^2f\\left(\\frac{1}{1-x}-1\\right)=x^2+(x-1)^2\\left(f\\left(\\frac{1}{1-x}\\right)-1\\right)=$$ $$=x^2-(x-1)^2+(x-1)^2\\cdot\\frac{f(1-x)}{(1-x)^2}=2x-1-f(x-1).$$ Thus, $$f(x)+f(x-1)=2x-1$$ or $$f(x)+f(x+1)=2x+1,$$ which gives $$f(x)+f(x)+1=2x+1$$ or $$f(x)=x.$$ Now, show that $$f(0)=0$$ and $$f(1)=1.$$", "# Find all functions such that $f(x^2+y)=xf(x)+f(y)$ Find all functions (over real numbers) such that $$f(x^2+y)=xf(x)+f(y).$$ My idea: Put $x=1$. Therefore the given equation becomes $f(y+1)=f(y)+f(1)$. It is in the Cauchy's first form, so we get $f(x)=cx$. It also satisfies that given functional equation. • That's not in Cauchy's form. – Saad Jun 16 '18 at 6:21 • @Alex Francisco $f(x+y)=f(x)+f(y)$ is a Cauchy form! And its solution is $f(x)=cx$ – Sufaid Saleel Jun 16 '18 at 6:23 • What you derived is $f(y+1)=f(y)+f(1)$. – Saad Jun 16 '18 at 6:24 • No, you are not right. If $g$ is a function of period $1$ with $g(0)=0,$ then $f(x)=g(x) +cx$ satisfies $f(x+1)=f(x)+f(1).$ – Thomas Andrews Jun 16 '18 at 6:24 • Is there any statement of continuity? – DanielV Jun 16 '18 at 6:25 Here is a proof that $f$ does satisfy Cauchy's functional equation. For now, I do not have a complete answer, unless some sort of continuity conditions is given. Let $P(x,y)$ denote the condition that $f\\left(x^2+y\\right)=x\\,f(x)+f(y)$. Then, $P(1,0)$ implies that $f(0)=0$. Now, $P(x,0)$ implies that $f\\left(x^2\\right)=x\\,f(x)$ for all $x\\in\\mathbb{R}$. This shows that $f\\left(x^2+y\\right)=f\\left(x^2\\right)+f(y)$ for all $x,y\\in\\mathbb{R}$. Next, $P\\left(x,-x^2\\right)$ leads to $f\\left(-x^2\\right)=-f\\left(x^2\\right)$, and consequently, $f(-x)=-f(x)$ for every $x\\in\\mathbb{R}$. From this, it can be easily shown that $f$ satisfies Cauchy's functional equation. • $$( x + 1" ]

[ "# 2001 IMO Shortlist Problems/A4 ## Problem Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, satisfying $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$. ## Solution Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\\ \\forall x$, so $f(x) = 0,\\ \\forall x$. This is a solution, so we can take it out of the way: assume $f(1)\\ne 0$. $y = 1\\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\\in\\{0,f(1)x\\}$. In particular, $f(0) = 0$. Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\\ \\forall x$. This means that $f(a),f(b)\\ne 0\\Rightarrow f\\left(\\frac ab\\right)\\ne 0\\ (*)$ ($\\frac ab$ is defined because $f(b)\\ne 0\\Rightarrow b\\ne 0$). Assume now that $x\\ne y$ and $f(x),f(y)\\ne 0$. We get $f(x) = f(1)x,\\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\\ne 0$, so $x\\ne y,\\ f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (**)$. Assume now $f(x)\\ne 0$. From $(*)$ we get $f\\left(\\frac 1x\\right)\\ne 0$, and after applying $(*)$ again to $a = x,b = \\frac 1x$ we get $f(x^2)\\ne 0\\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\\ne 0\\Rightarrow f(xy)\\ne 0\\ (\\#)$. Let $G = \\{x\\in\\mathbb R|f(x)\\ne 0\\}$. $(*)$ and $(\\#)$ simply say that $(G,\\ \\cdot)$ is a subgroup of $(\\mathbb R^{*},\\", "# Functional equation $f(f(f(x)f(y)))=f(x)f(y^2)$ Find all functions $f: \\mathbb R_{>0} \\rightarrow \\mathbb R_{>0}$ such that $f(f(f(x)f(y)))=f(x)f(y^2)$ for all $x, y \\in \\mathbb R_{>0}$. • f(x) = k gives an entire family of functions. – Chris Cudmore Jul 22 '16 at 20:02 • @ChrisCudmore That doesn't work for every $k$. – wythagoras Jul 22 '16 at 20:03 $f(x)f(1)=f(f(f(x)f(1)))=f(f(f(1)f(x)))=f(1)f(x^2)$. Therefore $f(x)=f(x^2)$ for all $x$. Now $f(x)^2=f(x)f(x^2)=f(f(f(x)f(x)))=f(f(f(x)))$. So $f(f(x))=f(f(x)^2)=f(f(f(f(x))))=f(f(x))^2$. Therefore $f(f(x))=1$ for all $x$. But now $f(x)^2=f(f(f(x)))=1$, so $f(x)=1$. The only solution is the constant function $f(x)=1$. • That's doing my brain in... how do you get $f(f(f(f(x)))) = f(f(x))^2$? I just about get every other part though - nice solution! – Shuri2060 Jul 22 '16 at 21:21 • With $f(z)^2=f(f(f(z)))$, substitute $z=f(x)$. – f'' Jul 22 '16 at 21:29 • Thanks. Out of interest, is the solution to $f(x) = f(x^2)$ a constant in general (out of context of the question)? – Shuri2060 Jul 22 '16 at 21:33 • If $f$ must be continuous over $\\mathbb{R}^+$, then yes, because $\\lim_{x\\rightarrow1}f(x)$ would not exist if it was not constant. – f'' Jul 22 '16 at 21:41 • most appropriate username for this solution – elias Jul 22 '16 at 22:03 Extremely-partial initial observations: The LHS is symmetrical in x,y so the RHS is too so $f(x)f(y^2)=f(x^2)f(y)$ so $f(x^2)/f(x)$ is constant; say", "# Solve the following functional equation $f(xf(y))+f(yf(x))=2xy$ Find all function $f:\\mathbb{R}\\rightarrow \\mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$. By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$. • $f(xf(x))=x^2=f(-xf(-x))$ and therefore $xf(x)=-xf(-x)$. Hence $f(x)=-f(-x)$. – Pp.. Jan 6 '15 at 19:47 • @Pp : That is definition of all $odd functions$ – Narasimham Jan 6 '15 at 20:14 • @Narasimham Yes, it is a property that $f$ must have. – Pp.. Jan 6 '15 at 20:15 Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\\pm 1$ Let $f(1)=1$. $x\\rightarrow xf(x), y\\rightarrow \\frac{1}{x} \\Rightarrow f(xf(x)f(\\frac{1}{x}))+f(x)=2f(x) \\rightarrow f(x)f(\\frac{1}{x})=1$. Putting now $x\\rightarrow \\frac{1}{x}, y \\rightarrow 1$ we get $\\frac{1}{f(x)}+ \\frac{1}{f(f(x))}=\\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$. This is all I have found so far: Just keep feeding the snake with its own tail. As you noted $f(f(1))=1$. So with $x=y=f(1)$ we then see that $1=f(f(1))=(f(1))^2$ so that $f(1)=\\pm 1$. As you almost correctly noted, we have for $y=f(1)$ that $$f(x)+f(f(x)f(1))=2xf(1)$$ which indeed shows that $f$ is injective. With that observation, the comment by Pp.. works to conclude", "# 2014 USAJMO Problems/Problem 3 ## Problem Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that $$xf(2f(y)-x)+y^2f(2x-f(y))=\\frac{f(x)^2}{x}+f(yf(y))$$ for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$. ## Solution Let's assume $f(0)\\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get $$2f(0)^2=f(2f(0))^2/2f(0)+f(0)$$ $$2f(0)^2(2f(0)-1)=f(2f(0))^2$$ This means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\\neq 0, y=0$ to get $$xf(-x)=\\frac{f(x)^2}{x} \\implies x^2f(-x)=f(x)^2.$$ Similarly, $$x^2f(x)=f(-x)^2.$$ From these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ Both of these are valid solutions on their own, so let's see if there are any solutions combining the two. Let's say we can find $f(x)=x^2, f(y)=0,$ and $x,y\\neq 0.$ Then $$xf(-x)+y^2f(2x)=f(x)^2/x.$$ $$y^2f(2x)=x-x^3.$$ If $f(2x)=4x^2,$ then $y^2=\\frac{x-x^3}{4x^2}=\\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields $$0=2f(0)+1f(2)=0+f(1)=1,$$ which is impossible. So $f(2)=f(-2)=4$ and there are no solutions \"combining\" $f(x)=x^2$ and $f(x)=0.$ Therefore our only solutions are $\\boxed{f(x)=0}$ and $\\boxed{f(x)=x^2.}$", "# Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ Find all functions $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\\ge 0$ we have $f(a)>f(b)$ As much as I know: • putting $x=y=0$ we get $f(0)=0$. • let $x+y=0$, the we get $f(y)+f(-y)=\\frac{f(2y)}{2}$ Step 1: Prove : $f(y) = 0$ iff. $y=0$ It's easy to see $f(y)>0$ for $y>0$ (from porperty (b) with value $a=y,b=0$) Following proof is given by @Kyson To prove that $f(y)≠0$ for all$y<0.$ Assume there is a $y<0$ such that $f(y)=0.$ Choose $x=−2y>0$ and hence $f(x)>0$ and $f(x+y)=f(−y)>0$Then $[f(x)+f(y)][f(x+2y)+f(y)]=0$ while $f(x+y)2+f(y)f(2y)>0.$ $\\$ Step 2: Prove $f(y)$ is even. Put $x+y=0$ in equation given to get:$f(y)+f(-y)=\\frac{f(2y)}{2}$ for all y. $f(y)+f(-y)=\\frac{f(2y)}{2} = f(-y) +f(y) = \\frac{f(-2y)}{2}$ Thus $f(x)$is a even function, so $f(y)+f(-y)=2f(y)=\\frac{f(2y)}{2} \\Rightarrow f(2y)=4f(y)$ (Since $f(x)$ is even ,property (b) actually shows that $f(y) > 0$ on $(-\\infty,0)\\cup(0,\\infty)$) $\\$ Step 3: Prove:$f(y)=y^2f(1)$ By induction , find $f(ny)=n^2f(y)$ where $n \\in \\mathbb N$ Details of induction: Suppose $m\\in \\mathbb N ,m\\ge 2 ,\\ f(ny)=n^2f(y)$ holds for all $n\\le m$ Put $x=(m-1)y$ in property (a).$\\\\ \\Rightarrow ((m-1)^2+1)f(y)f((m+1)y)=(m^4+4-(m-1)^2-1)f^2(y) \\\\ \\Rightarrow f((m+1)y)=(m+1)^2f(y) \\ for\\ y\\not = 0 \\ and \\ it\\ still\\ holds\\ for\\ y=0$ Hint: $(m+1)^2 ((m-1)^2 +1) = (m^2-1)^2+(m+1)^2=m^4 -m^2+2 =m^4+4-(m-1)^2-1$ Remark: Result induces $f(zy)=z^2f(y)$ where $z\\in \\mathbb Z$ , since $f$ is even. $\\$ $\\forall", "If f is a function satisfying such that , find the value of n. Question: If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \\in \\mathrm{N}$ such that $f(1)=3$ and $\\sum_{x=1}^{n} f(x)=120$, find the value of $n$. Solution: It is given that, $f(x+y)=f(x) \\times f(y)$ for all $x, y \\in \\mathrm{N}$ $f(1)=3$ Taking $x=y=1$ in (1), we obtain $f(1+1)=f(2)=f(1) f(1)=3 \\times 3=9$ Similarly, $f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \\times 9=27$ $f(4)=f(1+3)=f(1) f(3)=3 \\times 27=81$ $\\therefore f(1), f(2), f(3), \\ldots$, that is $3,9,27, \\ldots$, forms a G.P. with both the first term and common ratio equal to 3 . It is known that, $S_{n}=\\frac{a\\left(r^{n}-1\\right)}{r-1}$ It is given that, $\\sum_{x=1}^{n} f(x)=120$ $\\therefore 120=\\frac{3\\left(3^{n}-1\\right)}{3-1}$ $\\Rightarrow 120=\\frac{3}{2}\\left(3^{n}-1\\right)$ $\\Rightarrow 3^{n}-1=80$ $\\Rightarrow 3^{n}=81=3^{4}$ $\\therefore n=4$ Thus, the value of $n$ is 4 .", "# Solve the functional equation $f(xf(y)+x)=xy+f(x)$ Find all $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ so that $f(xf(y)+x)=xy+f(x)$. If you put $x=1$ it's easy to prove that f is injective. Now putting $y=0$ you can get that $f(0)=0$. $y=\\frac{-f(x)}{x}$ gives us $f(\\frac{-f(x)}{x})=-1$ • I tried putting $y=x=1$ and got $f(f(1)+1)=1+f(1)$ from which I guessed that $f(x)=x$. This seems to fit the identity you give but doesn't seem like a strong enough proof. – Mufasa Jan 18 '15 at 14:10 • It's a step in the right direction. Actually, if you put $f(x)=ax$ you get that $a=\\pm 1$, so $f(x)=\\pm x$ are for now the only slutions I have, but are not poven. – HeatTheIce Jan 18 '15 at 14:12 Set $x = 1$ and $y = t - f(1)$, then $f(1+f(t - f(1))) = t$, for all $t \\in \\mathbb{R}$, hence $f$ is a surjective function. So, $\\exists\\, y_1 \\in \\mathbb{R}$, s.t., $f(y_1) = 0$ Then, $$f(x) = f(x+f(y_1)x) = xy_1+f(x) \\implies xy_1 = 0, \\forall x \\in \\mathbb{R} \\implies y_1 = 0$$ and $\\exists\\, y_2 \\in \\mathbb{R}$, s.t., $f(y_2) = -1$ Then, $$0 = f(0) = f(x+xf(y_2)) = xy_2 + f(x)$$ i.e., $f(x) = -y_2x$ for $x \\in \\mathbb{R}$. Plugging it back in the functional equation we may verify $y_2 = \\pm 1$. Let f(1)=b. Putting now $x=1,y=-b$ we get that $f(-b)=-1$ Now", "Find all continuous functions $f$ : $\\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying: $f(xy)+f(x+y)=f(xy+x)+f(y)\\quad\\forall x,y \\in \\mathbb{R}$ Find all continuous functions $f$ : $\\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying: $$f(xy)+f(x+y)=f(xy+x)+f(y)$$ $\\forall x,y \\in \\mathbb{R}$ I have tried that : $P(y;x)-P(x;y)$: $f(xy+y)+f(x)=f(xy+x)+f(y)$ $P(x;1)$: $f(x)+f(x+1)=f(2x)+f(1)$ $P(x+1;1)$:$f(x+1)+f(x+2)=f(2x+2)+f(1)$ So $f(x+2)-f(x)=f(2x+2)-f(2x) , \\forall x \\in \\mathbb{R}$ let $g(x)=f(x+2)-f(x)$ We have $g(2x)=g(x)=g(\\frac{x}{2})=...=g(0)=c$ because $g$ is the continous functions. To here, I have no idea to solve the problem. - Let's denote the functional equation by $(*)$. Note that if $f(x)$ satisfies to $(*)$ then so do the functions of the form $af(x)+b$, for any real numbers $a$ and $b$. So we can assume that $f(0)=0$ by changing the function linearly, if needed. Now, as you got, we have $f(x+2)-f(x)=g(x)=g(0)=f(2)-f(0)=f(2)$. On the other hand, if we take $y=2$ in $(*)$, we'll get $f(2x)+f(x+2)=f(3x)+f(2)$, and if we substitute the value of $f(2)$ from the above equation, we'll get the following $f(2x)+f(x+2)=f(3x)+f(x+2)-f(x) \\implies f(2x)+f(x)=f(3x)$. Now substitute $x+1$ for $x$: $f(3x+3)=f(3(x+1))=f(2(x+1))+f(x+1)=f(2x+2)+f(x+1)=f(2x)+f(2)+f(x+1)=f(2x)+f(x)-f(x)+f(2)+f(x+1)=f(3x)+f(x+1)-f(x)+f(2)$. So, we'll have $f(x+1)-f(x)+f(2)=f(3x+3)-f(3x)=f(3x+1)+f(2)-f(3x) \\implies f(x+1)-f(x)=f(3x+1)-f(3x)$. Therefore, in the same way as you did, we can get $f(x+1)-f(x)=f(1)-f(0)=f(1)$ using the continuity of the function $f$. Using simple induction, we'll get $(1)\\space f(x+n)-f(x)=nf(1)$ for any integer $n$. In particular, we have $f(n)=nf(1)$ for all $n\\in\\mathbb Z$. If we plug in $y=n$ in $(*)$ assuming $n$ is an integer, we get $f(nx)+f(x+n)=f((n+1)x)+f(n)\\implies f((n+1)x)=f(nx)+f(x)$. Therefore", "Find all functions $f : [0, \\infty ) \\to \\mathbb{R},$ that is, an expression for $f(x),$ such that for all $x,y \\in [0,\\infty)$ we have $f(x+2y)+2f^2(y)+x = f^2(x+2y)+2f(y)-\\frac{1}{4}.$ Determine the value of $f$ for some simple cases. What about the value of $f(0)?$ Why not try getting an equation of $f$ with a single variable? … by substituting a suitable value for $y.$ By setting $x=y=0,$ \\begin{aligned} f(0) + 2f^2(0) &= f^2(0) + 2f(0) - \\frac{1}{4} \\\\ f^2(0) - f(0) + \\frac{1}{4} &= 0 \\\\ \\big(f(0) - \\frac{1}{2}\\big)^2 &= 0, \\\\ \\end{aligned} which gives us $f(0) = \\frac{1}{2}.$ Now, we can obtain $f(x)$ by setting $y=0:$ \\begin{aligned} f(x)+2f^2(0)+x &= f^2(x)+2f(0)-\\frac{1}{4} \\\\ f^2(x)-f(x)+\\frac{1}{4}-x &= 0.\\\\ \\end{aligned} Solving this quadratic equation gives us $f(x) = \\frac{1}{2} \\pm \\sqrt{x}.$", "× # Problem 5! IMO 2015 Let $$\\mathbb{R}$$ be the set of real numbers. Determine all functions $$f:\\mathbb{R}\\to\\mathbb{R}$$ that satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $$x$$ and $$y$$. ##### This is part of the set IMO 2015 Note by Sualeh Asif 1 year, 10 months ago Sort by: Let $$P(x,y)$$ be the above FE. $P(0,0)\\implies f(f(0))+f(0)=f(0)\\implies f(f(0))=0$ $P(0, f(0))\\implies 2f(0)=f(f(0))+f(0)^2\\implies f(0)^2-2f(0)=0\\implies f(0)=0, 2$ Case 1: $$f(0)=0$$. Then $P(0, y)\\implies f(f(y)) + f(0)=f(y)+yf(0)\\implies f(f(y))=f(y)$ Since the range of $$f:\\mathbb{R}\\to\\mathbb{R}$$ is $$\\mathbb{R}$$, then $$\\boxed{f(x)=x}$$. Case 2: $$f(0)=2$$. Then let $$g(x)=f(x)-2$$ which changes the FE to $P(x):= g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y$ [I'll finish later] · 1 year, 10 months ago It does look like $$f(x) = 2-x$$ satisfies the functional equation. My guess would be $$f(x) = x$$ or $$f(x) = 2-x$$, but I do agree with the comment that case 1 still has some work to be done to get to $$f(x) = x$$. · 1 year, 10 months ago This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution. · 1 year, 10 months ago Daniel Liu The problem is one of those bash types on the IMO where if you keep bashing long enough you do end up on a solution! You were on the right track! Here", "# Functional equation $4f(x^2+y^2)=(f(x)+f(y))^2$ Consider the following problem: Determine all functions $f:\\mathbb{N}\\rightarrow \\mathbb{N}$ that satisfy the functional equation $$4f(x^2+y^2)=(f(x)+f(y))^2.$$ So first of all plugging in $x=0=y$, we get that $4f(0)=4f(0)^2$. Hence $f(0)=0$ or $f(0)=1$. Now using $x=1$ and $y=0$, we get that $f(1)^2+(2f(0)-4)f(1)+f(0)^2=0$. Thus $f(1)=2-f(0)+2\\sqrt{1-f(0)}$. Hence if $f(0)=0$, then $f(1)=4$ and if $f(0)=1$, then $f(1)=1$. (Obviously we implicitly assumed that $f(1)\\neq 0$.) Now notice that plugging in $x=y$ yields $f(2x^2)=f(x)^2$. Let's assume that $f(0)=0$ and $f(1)=4$. Then also $4f(x^2)=f(x)^2$. From these equations we easily get that $f(2)=16, f(4)=\\frac{f(2)^2}{4}=64, f(8)=f(2)^2=256$, and so on. So we can find $f(2^n)$ using these techniques. We see that $f(2^n)=4^{n+1}$. From this one could guess that $f(x)=4x^2$ is a solution. Corr-blimey, $f(x)=4x^2$ is a solution! Now notice that $f(x)=0$ is a solution, $f(x)=1$ is also a solution and $f(x)=4x^2$ is a solution. I guess that these solutions are actually determined by there values on $0$ and $1$ for which we already found all possibilities. But how to proceed? Again assume that $f(0)=0$ and $f(1)=4$. We already know $f(2^n)$. How do we find $f(3)$? Well, notice that $4f(5)=4f(2^2+1^2)=(f(2)+f(1))^2=(16+4)^2=400$. Hence $f(5)=100=4\\cdot 5^2$. Now using the Pythagorean triple $(3,4,5)$ we find that $10000=f(5)^2=4f(5)^2=4f(3^2+4^2)=(f(3)+f(4))^2=f(3)^2+128f(3)+64^2$. Solving this equation yields $f(3)=36=4\\cdot 3^2$. Using the above procedure I can find the value of $f(n)$ for a lot of $n$, but I'm", "# Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt – Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\\since $f(0)=0$) Finally putting $y=(x+1)$ gives us $f(x)=0$. This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one? #### Solutions Collecting From Web of \"Find all real functions $f:\\mathbb {R} \\to \\mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$\" You proved that $f(0)=0$, we can continue from here Suppose that there exist $k \\neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$. Now suppose that there exist no $k \\neq 0$ such that $f(k)=0$. Then plugging", "### Home > PC > Chapter 7 > Lesson 7.2.2 > Problem7-57 7-57. If $y$ varies inversely as $x + 6$ and $y = 1$ when $x = 1$, complete the following problems. 1. Express y as a function of $x$, $y = f\\left(x\\right)$. If $y$ varies inversely as $x$, then $y=\\frac{k}{x}$. $y=\\frac{k}{x+6}$ Use the fact that $y = 1$ when $x = 1$ to solve for $k$. Substitute $k$ back into your original equation. To write $y$ as a function of $x$, change $y =$ to $f\\left(x\\right) =$. 2. Find $f\\left(−3\\right)$, $f\\left(0\\right)$, $f(\\frac { 1 } { 3 })$, and $f(\\frac { 1 } { a })$. $f\\left ( \\frac{1}{a} \\right )=\\frac{7a}{1+6a}$", "× # Problem 5! IMO 2015 Let $$\\mathbb{R}$$ be the set of real numbers. Determine all functions $$f:\\mathbb{R}\\to\\mathbb{R}$$ that satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $$x$$ and $$y$$. ##### This is part of the set IMO 2015 Note by Sualeh Asif 2 years, 2 months ago Sort by: Let $$P(x,y)$$ be the above FE. $P(0,0)\\implies f(f(0))+f(0)=f(0)\\implies f(f(0))=0$ $P(0, f(0))\\implies 2f(0)=f(f(0))+f(0)^2\\implies f(0)^2-2f(0)=0\\implies f(0)=0, 2$ Case 1: $$f(0)=0$$. Then $P(0, y)\\implies f(f(y)) + f(0)=f(y)+yf(0)\\implies f(f(y))=f(y)$ Since the range of $$f:\\mathbb{R}\\to\\mathbb{R}$$ is $$\\mathbb{R}$$, then $$\\boxed{f(x)=x}$$. Case 2: $$f(0)=2$$. Then let $$g(x)=f(x)-2$$ which changes the FE to $P(x):= g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y$ [I'll finish later] · 2 years, 2 months ago It does look like $$f(x) = 2-x$$ satisfies the functional equation. My guess would be $$f(x) = x$$ or $$f(x) = 2-x$$, but I do agree with the comment that case 1 still has some work to be done to get to $$f(x) = x$$. · 2 years, 2 months ago This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution. · 2 years, 2 months ago Daniel Liu The problem is one of those bash types on the IMO where if you keep bashing long enough you do end up on a solution! You were on the right track! Here", "# Functional equation : Find all pairs of functions $f,g:\\mathbb{R}→\\mathbb{R}$ Find all pairs of functions $f,g : \\mathbb{R}\\rightarrow \\mathbb{R}$ such that (a) if $x < y$, then $f(x) < f(y)$; (b) for all $x,y \\in \\mathbb{R}$, $f(xy) = g(y)f(x) + f(y)$. My work : Let $P(x,y) : f(xy) = g(y)f(x) + f(y)$ $x<y \\leftrightarrow f(x) < f(y)$ $P(x,1) : f(1)(1-g(1)) = f(1)$ $g(1) = 1 \\rightarrow f(xy) = f(x) + f(y)$ so $f(1) = f(1) + f(1) \\rightarrow f(1) = 0$ $P(x,1) :$ $f(1) = 0 \\rightarrow f(x) = g(1)f(x)f(1)$ so $f(x)(1-g(1)) = 0 \\rightarrow g(1) = 1$ The answer is, $(f,g)$ satisfies $g=\\text{sgn}(x)\\exp({b\\log|x|})$ and $f(x)=a(g(x)-1)$ for some real numbers $a>0$ and $b>0$. It is easy to see that all such $(f,g)$ satisfy the conditions. We will show that they are the only possibilities. The proof is the following: Take $y=1$, we get $$f(x)=g(1)f(x)+f(1).$$ Since $f$ is not constant, $f(1)=0$, $g(1)=1$. Take $y=0$, $$f(0)=g(0)f(x)+f(0).$$ Hence $g(0)=0$. Take $x=0$ $$f(0)=g(y)f(0)+f(y),$$ Hence $$f(x)=a(g(x)-1),$$ here we assume $f(0)=-a.$ Of course $a>-f(1)=0$ by assumption, and $g$ is monotonic increasing as $f$ is. Use this equation to the original one, we get $$g(xy)=g(x)g(y).$$ Take $x=y=-1$, note that $g$ is monotonic, hence $g(-1)=1$. Take $y=-1$, we know that $g$ is an odd function. It is enough to show that $g=\\exp({b\\log x})$ for $x>0$. This", "function. Putting $$x=0$$ gives $$f(yf(0))=0$$ for any $$y\\in\\mathbb{R}$$ so $$f(0)=0$$. If $$f$$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $$f$$ is not zero at some point. Setting $$y=0$$, $$f(x^2)=xf(x).$$ Let $$x_0$$ be a real number for which $$f(x_0)=0$$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$x_0 f(x_0 +y) =0$$ for all $$y\\in\\mathbb{R}$$. Thus $$x_0 =0$$ since f is not identically zero. Indeed, $$f(x)=0 \\iff x=0.$$ Now setting $$y=-x$$ we have for any $$x\\in\\mathbb{R}$$, $$f(x^2-xf(x))=0$$ and so $$x^2-xf(x)=0$$ or $$f(x)=x.$$", "# For $f(x)f(y)+f(3/x) f(3/y)=2f(xy)$ choose correct choices Consider $f:\\mathbb{R}^+ \\to \\mathbb{R}$ such that $f(3)=1$ and $$f(x)f(y)+f(3/x) f(3/y)=2f(xy)\\;\\;\\;\\;\\;\\;\\forall x,y \\in \\mathbb{R}^+$$ then choose the correct option(s): (A) $f(2014)+f(2015)-f(2010)=100$ (B) $f$ is an even function (C) $\\frac{f(100)}{f(10)+f(90)}=\\frac{1}{2}$ (D) $f$ is a periodic function By putting $x=1$ in given equation to get $f(1)=1$ Generally I have been taught to solve such question using first principle of differentiation but here I am not able to deal with the second term of L.H.S. in given equation. May someone provide some help? If we write $y=3/x$ then we get $$2f(x)f({3\\over x}) = 2f(3)=2$$ so we have for all $x>0$: $$f\\big({3\\over x}\\big) = {1\\over f(x)}\\;\\;\\;(*)$$ Pluging this in original equation we get $$f(x)f(y) +{1\\over f(x)f(y)} = 2f(xy)$$ which is valid for all $x,y>0$. Put now $x=y$ we get $$f(x)^2 +{1\\over f(x)^2} = 2f(x^2)$$ By AG-GM we have: $f(x)^2 +{1\\over f(x)^2}\\geq 2$, so $f(x^2)\\geq 1$ for all $x>0$ so $f(x)\\geq 1$ for all $x>0$. Finally, useing $(*)$ we get $$f\\big({3\\over x}\\big) \\leq 1$$ for all $x>0$. But since $x\\mapsto {3\\over x}$ is bijective on $\\mathbb{R}^+$ we have $f(x) \\leq 1$ for all $x>0$. So we conclude $f(x)=1$ for all $x> 0$, so the answer is $(C)$. • Double-check your inequality directions from \"Finally\" onwards. – J.G. Apr 17 '18 at 21:03 • You don't need the", "of the problem which appeared here in different times, i.e. the version posted before the OP edited his post and the current one. First I want to solve the problem as it is stated by the OP. Problem I $$f(y+f(x))=f(x)f(y)+f(f(x))+f(y)-xy.$$ As already noticed, setting $y=0$, one arrives to $$f(f(x))=f(0)f(x)+f(f(x))+f(0)$$ which leads to $f(x)\\equiv -1$, which is not a solution of the problem, or $f(0)=0$. Incidentally, even though it is not crucial for the solution, notice that if we assume that there is an $\\bar x$ such that $f(\\bar x)=0$, setting $x=\\bar x$ we infer that $f(y)=f(y)-\\bar xy$. In particular, setting $y=1$, we conclude $\\bar x=0$. Next, plug inthe equation $f(y)$ instead of $y$ to get $$f(f(y)+f(x))=f(x)f(f(y))+f(f(x))+f(f(y))-xf(y).$$ Interchange the roles of $x$ and $y$ to obtain similarly $$f(f(x)+f(y))=f(f(x))f(y)+f(f(y))+f(f(x))-yf(x).$$ Compare these two equations to conclude that $$\\frac{f(f(x))+x}{f(x)}=\\frac{f(f(y))+y}{f(y)}=k,$$ where $k$ is a suitable constant. Substitute in the original equation to get $$f(y+f(x))=f(x)f(y)-x+kf(x)+f(y)-xy.$$ Denote $a:=f(-1)$ and choose $y=-1$ to obtain $$f(f(x)-1)=(a+k)f(x)+a$$ Edit Thanks to Dejan who pointed out a flaw in my proof Plug in the original equation $f(y)-1$ instead of $y$ to obtain \\begin{aligned}f(f(y)+f(x)-1)&=f(x)[(a+k)f(y)+a]-x+kf(x)-x(f(y)-1)\\\\&=(a+k)(f(x)+f(y))-xf(y).\\end{aligned} Exchange $x$ and $y$ to obtain $$f(f(y)+f(x)-1)=(a+k)(f(x)+f(y))-yf(x).$$ Compare these two equations to deduce that $$\\frac{f(x)}{x}=\\frac{f(y)}{y}=a.$$ Substitute in the original equation to find out that the admissible values for $a$ are just $\\pm 1$ i.e. $$f(x)=\\pm x.$$ Problem", "Solving functional equations can be very difficult but there are some common methods of solving them. A discussion of involutary functions is useful. For example, consider the function $f\\left(x\\right) = frac\\left\\{1\\right\\}\\left\\{x\\right\\}$. Then consider f(f(x)) = x, if we continue the pattern we end up with x for an even number of compositions and f(x) for an odd number. This same idea applies to many other functions, i.e. $f\\left(x\\right) = frac\\left\\{1\\right\\}\\left\\{1-x\\right\\}, f\\left(x\\right) = 1-x$ and many others. Example 1: Solve $f\\left(x+y\\right)^2 = f\\left(x\\right)^2 + f\\left(y\\right)^2$ for all $x,y in mathbb\\left\\{R\\right\\},$ assuming f is a real-valued function. Let $x=y=0$: $f\\left(0\\right)^2=f\\left(0\\right)^2+f\\left(0\\right)^2$. So $f\\left(0\\right)^2=0$ and $f\\left(0\\right)=0$. Now, let $y=-x$: $f\\left(x-x\\right)^2=f\\left(x\\right)^2+f\\left(-x\\right)^2$ $f\\left(0\\right)^2=f\\left(x\\right)^2+f\\left(-x\\right)^2$ $0=f\\left(x\\right)^2+f\\left(-x\\right)^2$ A square of a real number is nonnegative, and a sum of nonnegative numbers is zero iff both numbers are 0. So $f\\left(x\\right)^2=0$ for all x and $f\\left(x\\right)=0$ is the only solution.", "# Math Help - Derivative Problem 1. ## Derivative Problem Suppose $f$ is a function that satisifies the equation $f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $x,y$. Suppose that $\\lim_{x->0}\\frac{f(x)}{x}=1$. Find $f(0),f'(0),f'(x)$. I star by writing $x+y=0,x=-y, y=-x$ $f(0)=f(x)+f(-x)+x^3-x^3$ $=f(x)+f(-x)$ So now I have: $f(x)=f(0)-f(-x)$ $f(0)=f(0)-f(0)=0$ For $f'(0)$: $f'(0)=\\lim_{x->0}\\frac{f(x)-f(0)}{x-0}$ $=\\lim_{x->0}\\frac{f(x)}{x}=1$ So I got $f'(0)=1,f(0)=0$. I can't seem to get started on finding $f'(x)$. Suppose $f$ is a function that satisifies the equation $f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $x,y$. Suppose that $\\lim_{x->0}\\frac{f(x)}{x}=1$. Find $f(0),f'(0),f'(x)$. I star by writing $x+y=0,x=-y, y=-x$ $f(0)=f(x)+f(-x)+x^3-x^3$ $=f(x)+f(-x)$ So now I have: $f(x)=f(0)-f(-x)$ $f(0)=f(0)-f(0)=0$ For $f'(0)$: $f'(0)=\\lim_{x->0}\\frac{f(x)-f(0)}{x-0}$ $=\\lim_{x->0}\\frac{f(x)}{x}=1$ So I got $f'(0)=1,f(0)=0$. I can't seem to get started on finding $f'(x)$. to get $f(0) = 0$ it's easier to put $y = 0.$ to find $f'(x)$: $f'(x)=\\lim_{y \\to 0} \\frac{f(x+y)-f(x)}{y}=\\lim_{y\\to0} \\left(\\frac{f(y)}{y} + x^2 + xy \\right)=x^2 + 1.$ this also gives you $f(x)=\\frac{x^3}{3} + x.$" ]

[ "minimum value of$$ a+\\frac{1}{b(a-b)} ? $$Let a&gt;b be positive real numbers. What is the minimum value of$$ a+\\frac{1}{b(a-b)} ? ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$.Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. ...", "therefore the lowest value abc can have is 8. If, however, a, b, and c are the lowest possible real numbers, this is a very different problem. From the condition by AM-GM we obtain: $$\\prod\\limits_{cyc}\\frac{a}{1+a}=\\prod\\limits_{cyc}\\left(\\frac{1}{1+b}+\\frac{1}{1+c}\\right)\\geq\\prod\\limits_{cyc}\\frac{2}{\\sqrt{(1+b)(1+c)}}=8\\prod\\limits_{cyc}\\frac{1}{1+a}$$ Id est, $abc\\geq8$.", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "+0 # hard algebra 0 35 1 Let a, b, c be nonzero real numbers. Find all possible values of $$\\displaystyle \\frac{a}{|a|} + \\frac{b}{|b|} + \\frac{c}{|c|} + \\frac{abc}{|abc|}$$ May 14, 2020 #1 +20906 0 Each one of the individual terms will have either a value of 1 or -1. If a, b, and c are all positive, then so is abc, so the sum will be 1 + 1 + 1 + 1 = 4 If one of a, b, and c is negative, then so is abc, so the sum will be 0. If two of a, b, and c are negative, then abc is positive, so the sum will be zero. If all three are negative, then so is abc, so the sum will be -4. The possible values are -4, 0, or 4. May 14, 2020", "# Distinct real numbers Let $$a, b, c$$ be distinct non-zero real numbers such that \\begin{align} a + \\dfrac{1}{b} = b+ \\dfrac{1}{c} = c + \\dfrac{1}{a}.\\end{align} Then, which of following is a possible value for $$abc?$$ ×", "## find all possible values Let $a,b,c,d$ positive real numbers such that $a+b+c+d=12$ , and $abcd =27+ab+ac+ad+bc+bd+cd$ find all possible values of $a,b,c,d$", "# Algebra Problem Algebra Level 3 Let $a,b$ and $c$ be non-zero distinct real numbers satisfying $a+\\frac{1}{b}=b+\\frac{1}{c}=c+\\frac{1}{a}.$ Find $|abc|$. ×", "# Strange numbers Algebra Level 5 Let $a$, $b$, and $c$ be non-zero real numbers such that $(ab+bc+ca)^3 = abc (a+b+c)^3.$ Which of the following is definitely true about $a$, $b$ and $c$? Relevant wikis: ×", "+C.$ Theorem 9 Assume that $$k> 1$$ is a positive integer and $$b$$ a non-zero real number. Then: $\\int\\frac{x}{\\left( x^2+b^2\\right)^k}\\,dx=\\frac{\\left( x^2+b^2\\right)^{1-k}}{2 (1-k)} +C.$ 2005-2017 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax", "# find all possible values Let $a,b,c,d$ positive real numbers such that $a+b+c+d=12$ , and $abcd =27+ab+ac+ad+bc+bd+cd$ $a,b,c,d$", "# The answer is not 0 Let $a,b,c$ be non-negative reals. Find the maximum value of $k$ such that $(a+b+c)\\left(\\frac { 1 }{ a } +\\frac { 1 }{ b } +\\frac { 1 }{ c } \\right)+\\frac { k(ab+bc+ca) }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \\ge 9+k.$ Write your answer to 3 decimal places. ×", "\\right.$$ For all $n \\ge 0$. Determine all values of $a_0$ for which there is a number $A$ such that $a_n = A$ for infinitely many values of $n$. Find all numbers $n \\ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and$a_{i}a_{i+1} + 1 = a_{i+2}$for $i = 1, 2, ..., n.$", "+0 # Complex +2 155 1 +140 Let $a$ and $b$ be nonzero complex numbers such that $|a| = |b| = |a + b|.$Find the sum of all possible values of $\\frac{a}{b}.$ Jun 13, 2019", "# Let * be a binary operation on the set of all nonzero real numbers, defined Question: Let * be a binary operation on the set of all nonzero real numbers, defined by $a * b=\\frac{a b}{5}$ Find the value of x given that2 * (x * 5) = 10. Solution: To find: value of $x$ Given: $a^{*} b=\\frac{a b}{5}$ $\\Rightarrow x^{* 5}=\\frac{5 x}{5}=x$ Now $(2 * x)=\\frac{2 x}{5}$ $\\Rightarrow \\frac{2 x}{5}=10 \\Rightarrow x=25$", "# Algebra posted by on . If a, b and c are non-zero reals such that a + b + c = 11 and \\frac {1}{a} + \\frac {1}{b} +\\frac {1}{c} = 0, what is the value of a^2 + b^2 + c^2? • Algebra - , stupid", "Mathematics # Assume all variables and groupings are nonzero real numbers. Write expressions with only positive exponents. ab -4 c -3 the -4 and -3 are exponents. $ab^{-4}c^{-3}=\\\\ \\frac{a}{b^4c^3}$", "# The answer is not 1/9 If $a, b, c$ are non-negative real numbers, what is the maximum value of $\\frac{ ab^2 + bc^2 + ca^2 } { ( a + b + c) ^ 3 } ?$ ×", "# Minimum Value Let $$a,b,c$$ be positive real numbers such that $$abc=1$$. The minimum value of $\\dfrac{1}{a^3(b+c)}+\\dfrac{1}{b^3(a+c)}+\\dfrac{1}{c^3(a+b)}$ can be written as $$\\dfrac{m}{n}$$. Find $$m+n$$ This is not an original problem ×", "+0 # Let a, b and c be complex numbers such that |a|=|b|=|c|=1 and 0 32 2 Let a, b and c be complex numbers such that |a|=|b|=|c|=1 and $$\\frac{a^2}{bc} + \\frac{b^2}{ac} + \\frac{c^2}{ab} = -1.$$ Find all possible values of |a+b+c| Enter all the possible values, separated by commas. May 2, 2022 #1 0 The possible values of |a + b + c| are 1, sqrt(3), and 3. May 2, 2022 #2 +1", "$${a,b,c}$$ are nonnegative real numbers. Find the minimum value of $$\\dfrac{a^2}{bc} + \\dfrac{b^2}{ac} + \\dfrac{c^2}{ab} + 3(\\dfrac{a+b}{c} + \\dfrac{a+c}{b} + \\dfrac{b+c}{a} )$$" ]

[ "9$$ Equation 1: $abc(a+b+c) = 100a + 10b + c$ Note that $$a+b+c$$ is present on both sides and rewrite the equation a bit: \\begin{align} abc(a+b+c) &= 99a+9b+(a+b+c)\\\\ abc &= \\frac{9(11a+b)+(a+b+c)}{(a+b+c)}\\\\ abc &= \\frac{9(11a+b)}{(a+b+c)} + 1\\end{align} Equation 2: $abc-1 = \\frac{9(11a+b)}{a+b+c}$ Since we are dealing exclusively with positive non-zero integers, both $$abc-1$$ and $$\\frac{9(11a+b)}{a+b+c}$$ must both be integral. This means that the term $$a+b+c$$ must be a factor of $$9(11a+b)$$. Because $$9(11a+b)$$ is > 1 by definition 1, the unique factorization theorem states that it is a product of primes and the factorization is unique. Thus the factors are $$3, 3^2$$ and $$11a + b$$. We won't tackle the determination of if $$11a + b$$ is prime, but will instead show a contradiction results: \\begin{align} a+b+c &= 11a+b\\\\ c &= 10a\\end{align}Which contradicts definition 1. Next we examine if $$a+b+c=3$$. This implies that $$a=1$$, $$b=1$$ and $$c=1$$, which leads to a contradiction of equation 1. This leaves the $$a+b+c=9$$ possibility. Substituting into equation 2 produces: \\begin{align} abc-1 &= \\frac{9(11a+b)}{(a+b+c)}\\\\ abc-1 &= \\frac{9(11a+b)}{9}\\\\ abc-1 &= 11a+b\\end{align} Now we solve for $$c$$: \\begin{align} abc &= 11a+b+1\\\\ c &= \\frac{11a+b+1}{ab}\\\\ c &= \\frac{11a}{ab}+\\frac{b}{ab}+\\frac1{ab}\\\\ c &= \\frac{11}b+\\frac{b+1}{ab}\\\\ c &= \\frac{11}b+\\frac{\\frac{b+1}a}b\\\\ c &= \\frac{11+\\frac{b+1}a}b\\\\\\end{align} Since we are dealing only with non-zero positive integers, this means b must evenly divide $$11+\\frac{b+1}a$$. Therefore, there exists some", "on! This is certainly easier to solve. At this point though, it isn’t clear how to proceed. But if you assume for a while that n is a constant, then we notice that the equation becomes a conic section on the variables a and b, and it turns out that such equations are usually reducible to one of the following: • A linear diophantine equation • Generalized Pell equation • Factorization problem Let’s see what we got by assuming n is constant and throwing all “constants” to the right side: \\begin{align*} ab &= A(a+n) + B(b+n) + C\\\\\\ ab - Aa - Bb &= (A+B)n + C \\end{align*} Here, we can try to “complete the factorization”: \\begin{align*} ab - Aa - Bb &= (A+B)n + C\\\\\\ ab - Aa - Bb+AB &= (A+B)n + AB + C \\\\\\ (a-B)(b-A) &= (A+B)n + AB + C \\end{align*} This means we have reduced the problem to factorizing all numbers of the form (A+B)n + AB + C, where n \\le N # Enumerating factors Now we’re really getting somewhere. We want to enumerate the triples (a,b,n) satisfying all the following things: • (a-B)(b-A) = (A+B)n + AB + C. • a+n \\le N and b+n \\le N. • (a\\land b) = (a\\land n) = (b\\land n) = 0. The remaining", "Since the inequality is cyclic, it seems natural to take these products in a way that preserves the cyclic nature. For instance, we multiply $$ab$$ with $$(a + b - c)$$ because $$a$$ and $$b$$ have the same sign in $$(a + b - c)$$: \\begin{align*} ab( a + b - c) &= a^2b +ab^2 -abc \\ge 0, \\\\ ab (-a + b +c) &= -abc + b^2c + bc^2 \\ge 0, \\\\ ca (a - b + c) &=a^2c -abc + ac^2 \\ge 0. \\end{align*} We see the $$-3abc$$ in the sum of these products. Examining the other terms in $ab(a + b - c) = a^2b + ab^2 - abc,$ we notice that $$a^2b + ab^2$$ are factors that would pop out of $$(a - b)^2(a + b - c)$$. Expanding the squared parts of expressions like $$(a-b)^2(a+b-c)$$ we get \\begin{align*} (a - b)^2(a + b - c) &= a^2(a + b - c) - 2ab(a + b - c) + b^2(a + b - c) \\ge 0, \\\\ (b - c)^2(-a + b + c) &= b^2(-a + b + c) - 2bc(-a + b + c) + c^2(-a + b + c) \\ge 0, \\\\ (c - a)^2(a - b + c) &= c^2(a - b + c) - 2ca(a - b + c)", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "# Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$ a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor. b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$ So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to use this to hence factor the expression. Also I was reading a solution which factorised the expression without using part a) but how does the second last line turn into the last line? \\begin{align}&\\color{white}=(bc^3−cb^3)−(ac^3−ca^3)+(ab^3−ba^3)\\\\ &=bc(c^2−b^2)−ac(c^2−a^2)+ab(b^2−a^2)\\\\ &=(c−b)(c+b)bc−ac(c−a)(c+a)+ab(b−a)(b+a)\\\\ &=−(a−c)(b−c)(a−b)(a+b+c)\\end{align} • The second turns into the third by the factorization of each respective difference of squares. Apr 19, 2017 at 4:36 • Related: 1, 2, 3, 4, 5, 6. Apr 19, 2017 at 5:38 • I'm just curious, where is this problem taken from? Apr 19, 2017 at 6:44 • @amWhy This is not a duplicate. The other question doesn't ask how to do this in a way that uses the factorization of $a^3 + b^3 + c^3 - 3abc$. Apr 19, 2017 at 19:29 The polynomial $$f(a,b,c)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$$ is an alternating function of $a$, $b$, $c$. That is it changes sign whenever you swap two variables: $$f(a,b,c)=-f(b,a,c)=-f(c,b,a)=-f(a,c,b).$$ This means that $a-b$, $a-c$ and $b-c$ are all factors (if a polynomial changes sign when you swap $a$ and $b$ then $a-b$ is a factor). Therefore $$f(a,b,c)=(a-b)(a-c)(b-c)g(a,b,c)$$ for", "have helped. another way could be to take partial derivatives w.r.t a, b, c and then compare... did this help?? 3. correct factor is (b-c)(a-c)(a-b)(a+b+c) you got one wrong (c-a) should be (a-c) 4. Originally Posted by skoker correct factor is (b-c)(a-c)(a-b)(a+b+c) you got one wrong (c-a) should be (a-c) that's right thanks! how did you factorize it? check my post too. 5. Hello, mathguy80! This one requires some serious factoring . . . Prove that (b - c) is a factor of: .a^3(b - c) + b^3(c - a) + c^3(a - b) Factor the expression completely. We have: . a^3(b - c) + b^3 c - ab^3 + ac^3 - bc^3 . . = .a^3(b - c) + b^3 c - bc^3 - ab^3 + ac^3 . . = . a^3(b - c) + bc(b^2 - c^2) - a(b^3 - c^3) . . = . a^3(b - c) + bc(b - c)(b + c) - a(b - c)(b^2 + bc + c^2) . . = . (b - c) [a^3 + bc(b + c) - a(b^2 + bc + c^2)] . . = . (b - c) [a^3 + b^2 c + bc^2 - ab^2 - abc - ac^2] . . = . (b - c) [a^3 - ab^2 - abc + b^2 c - ac^2", "A= & \\frac{(a+abc)}{2}-\\frac{(b+c)}{2}=\\frac{1}{2}(a-b-c+abc) \\\\ & B= & \\frac{(b+abc)}{2}-\\frac{(a+c)}{2}=\\frac{1}{2}(-a+b-c+abc) \\\\ & C= & \\frac{(c+abc)}{2}-\\frac{(a+b)}{2}=\\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! where $a\\,\\!$, $b\\,\\!$, $c\\,\\!$ and $abc\\,\\!$ are the treatments included in the ${2}^{3-1}\\,\\!$ design. Similarly, the two factor interactions can also be obtained as: \\begin{align} & BC= & \\frac{(a+abc)}{2}-\\frac{(b+c)}{2}=\\frac{1}{2}(a-b-c+abc) \\\\ & AC= & \\frac{(b+abc)}{2}-\\frac{(a+c)}{2}=\\frac{1}{2}(-a+b-c+abc) \\\\ & AB= & \\frac{(c+abc)}{2}-\\frac{(a+b)}{2}=\\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! The equations for $A\\,\\!$ and $BC\\,\\!$ above result in the same effect values showing that effects $A\\,\\!$ and $BC\\,\\!$ are confounded in the present ${2}^{3-1}\\,\\!$ design. Thus, the quantity, $\\tfrac{1}{2}(a-b-c+abc),\\,\\!$ estimates $A+BC\\,\\!$ (i.e., both the main effect $A\\,\\!$ and the two-factor interaction $BC\\,\\!$). The effects, $A\\,\\!$ and $BC,\\,\\!$ are called aliases. From the remaining equations given above, it can be seen that the other aliases for this design are $B\\,\\!$ and $AC\\,\\!$, and $C\\,\\!$ and $AB\\,\\!$. Therefore, the equations to calculate the effects in the present ${2}^{3-1}\\,\\!$ design can be written as follows: \\begin{align} & A+BC= & \\frac{1}{2}(a-b-c+abc) \\\\ & B+AC= & \\frac{1}{2}(-a+b-c+abc) \\\\ & C+AB= & \\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! ### Calculation of Aliases Aliases for a fractional factorial design can be obtained using the defining relation for the design. The defining relation for the present ${2}^{3-1}\\,\\!$ design is: $I=ABC\\,\\!$ Multiplying both sides of the previous equation by the main effect, $A,\\,\\!$ gives the alias effect of $A\\,\\!$ : \\begin{align} & A\\cdot I= & A\\cdot", "(2) \\begin{align} (aC)(bC)=abC=ab(b^{-1}a^{-1}ba)C\\\\ =(abb^{-1}a^{-1}baC=baC=(bC)(aC). \\end{align} Furthermore if $N$ is a normal subgroup of $G$ and $G/N$ is abelian, then $(a^{-1}N)(b^{-1}N)=(b^{-1}N)(a^{-1}N)$; that is, $aba^{-1}b^{-1}N=N$, so $aba^{-1}b^{-1} \\in N$, and $C \\leq N$. Finally, if $C \\leq N$, then (3) \\begin{align} (aN)(bN)=abN=ab(b^{-1}a^{-1}ba)N\\\\ =(abb^{-1}a^{-1})baN=baN=(bN)(aN). \\end{align}", ". . . $(b+c)\\big[a^2 + ab + ac + bc\\big] \\;=\\;0$ . . . . . . . . . . $(b+c)\\big[a(a+b) + c(a+b)\\big] \\;=\\;0$ Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \\;=\\;0$ We see that the product of three factors equals zero. Hence, at least one of the factors must be zero. Therefore: . $\\begin{Bmatrix}a+b \\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}b \\\\ & \\text{or} \\\\ b+c \\:=\\:0 & \\Rightarrow & b \\:=\\:\\text{-}c \\\\ & \\text{or} \\\\ a+c\\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}c \\end{Bmatrix}$ 3. ## Re: Opposite number proof Originally Posted by Soroban Hello, DIOGYK! I think I got it . . . We have: . $\\frac{1}{a+b+c} \\;=\\;\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ . . . . . . . . $\\frac{1}{a+b+c} \\;=\\;\\frac{bc + ac + ab}{abc}$ . . . . . . . . . . . . $abc \\;=\\;(a+b+c)(bc+ac+ab)$ . . . . . . . . . . . . $abc \\;=\\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$ . . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \\;=\\;0$ n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \\;=\\;0$ . . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \\;=\\;0$ Factor: .", "# Prove that $512^3 + 675^3 + 720^3$ is a composite number We have to prove that the number $$N=512^3 + 675^3 + 720^3$$ is composite. I tried to use the identity $(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ hoping to take out some common factors from the R.H.S. but it didn't work. I also used $a^3+b^3=(a+b)(a^2-ab+b^2)$ (in all possible combos) and tried to combine with $c^3$ but that too didn't work. I nearly spent about 5 hours struggling with the question but no result :( Let $a=512, b=675, c=720$. Now the number looks like $a^3+b^3+c^3$ but we require a sort of $3abc$ term to resolve it into factors. First we factorize $a,b,c$ into prime factors. So, $a=2^9, b=3^3\\times 5^2, c=2^4\\times 3^2\\times 5$. Now it can be seen that $2c^2=3ab$. Hence, $a^3+b^3+c^3=a^3+b^3-c^3+2c^2c$ and thus the problem can be solved. It can then be seen that the given number is composite.", "= (a_1x + b_1)(a_2x + b_2)...(a_nx + b_n)...$, where none of these factors are repeated and none of these factors are a constant multiple of one another, that is for a constant C, $(a_ix + b_i) ≠ C(a_jx + b_j)$ for all factors. Hence, the partial fraction decomposition of $\\frac{R(x)}{Q(x)}$ will be: (5) \\begin{align} \\quad \\frac{R(x)}{Q(x)} = \\frac{A_1}{(a_1x + b_1)} + \\frac{A_2}{(a_2x + b_2)} + ... + \\frac{A_n}{(a_nx + b_n)} + ... \\end{align} ### Example 1 Integrate $f(x) = \\frac{2x}{x^2 - x - 2}$. By factoring the denominator of $f$, we get two distinct linear factors, namely $(x - 2)$ and $(x + 1)$. We hence know that for some $A$ and $B$: (6) \\begin{align} \\frac{2x}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1} \\\\ \\frac{2x}{x^2 - x - 2} = \\frac{A(x + 1) + B(x - 2)}{(x - 2)(x + 1)} \\end{align} Hence it follows that $2x = A(x + 1) + B(x - 2)$. Hence, we can now choose any $x$ and solve for $A$ and $B$. When $x = 2$, then we get that $4 = 3A$, or more appropriately $A = \\frac{4}{3}$. When $x = -1$, we get that $-2 = -3B$, or rather $B = \\frac{2}{3}$. Hence it follows that our partial fraction decomposition is: (7) \\begin{align} \\frac{2x}{x^2 -x", "the following factorization: $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Proof 1: Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra $P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc$. Now, once again basic algebra says that as a, b, c are roots/solutions of the above: $P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0$ $P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0$ $P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc$=0\\$ $0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc$ So, we get $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Also, the above formula can be written as $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$ Proof 2: Consider the following determinant D: $\\left| \\begin{array}{ccc} a & b & c\\\\c & a & b\\\\ b & c & a \\end{array} \\right|$ On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following: $D = \\left| \\begin{array}{ccc} a+b+c & b & c \\\\a+b+c & a & b\\\\a+b+c & c & a \\end{array} \\right|$. Note that columns 2 and 3 of the three by three determinant do not change. On expanding the original determinant D, we get $D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)$ $D= a^{3}-abc-bac+b^{3}+c^{3}-cab$ $D= a^{3}+b^{3}+c^{3}-3abc$ Whereas we get from the other transformed but equal D: $D =(a+b+c) \\left| \\begin{array}{ccc} 1 & b & c \\\\ 1 & a & b \\\\ 1 & c & a \\end{array}\\right|$ $D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))$ So, that we again get $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Proof 3: Now, let us consider $E=a^{2}+b^{2}+c^{2}-ab-bc-ca$ as a quadratic in a with b and c as parameters.", "\\\\ 0 &\\ge& (bc-1)+(ac-1)+(ab-1) \\\\ 0 &\\ge& bc+ac+ab-3 \\\\ 3 &\\ge& bc+ac+ab \\\\ \\end{array}$ 11. experimentX Group Title Here's a short proof hint ... i had written earlier. \\begin{align*} 3(a^2+b^2+c^2) & = a^2 + b^2 + c^2 + (a^2+b^2) + (b^2 + c^2) + (c^2+a^2) \\\\ & \\ge a^2 + b^2 +c^2 + 2 a b + 2 b c + 2 c a \\\\ & = (a+b+c)^2 \\\\ & = \\frac{2(a^2 + b^2 +c^2)}{2} + 2 a b + 2 b c + 2 ca \\\\ & \\ge \\frac{ 2 a b + 2 b c + 2 ca }{2} + 2 a b + 2 b c + 2 ca \\\\ & = 3(a+b+c) \\end{align*} 12. experimentX Group Title the other hint would to be make use of AM-GM inequality for 3 variables. 13. experimentX Group Title $3(a^2+b^2+c^2) \\ge (a+b+c)^2 \\\\ 3 = 3 \\sqrt[3]{abc} \\le (a+b+c)$ Dividing will do the job!! 14. sauravshakya Group Title (a+b+c)/3 >= (abc)^1/3 a+b+c>=3 Now, a^2+b^2+c^2=(a+b+c)^2 -2ab -2bc-ca SO, a^2+b^2+c^2>=(a+b+c)^2 .....i Since, a+b+c>=3 (a+b+c)^2 >=a+b+c ....ii Thus, from i and ii, a^2+b^2+c^2 >=a+b+c", "# Let $a,b,c > 1$, Show that $abc + \\frac 1 {a}+\\frac 1 {b} + \\frac 1 {c} > a+ b+ c +\\frac 1 {abc}$ I'm confused how to prove this inequality, I tried by multiplying both sides by abc to get ride of fraction which got me $abc² + bc+ac+ab -(a+c+b)abc-1>0<=> abc²+bc+ac+ab-ab²-bc²-ca²-1 > 0$ I still can't see any way to to transform this inequality. When i looked at the book solution i found that they come up with formula without explaining how they got there. here's book solution of the inequality $(a-\\frac 1 {b})(b-\\frac 1 {c})(c-\\frac 1 {a}) > 0$ What do i need to know so i can make this kind of transformation to the inequality. which got me $$\\;\\ldots \\;\\; abc²+bc+ac+ab-ab²-bc²-ca²-1 > 0$$ You made a couple of mistakes in the intermediate calculations, which is fairly obvious since the expression you got in the end is not symmetric in $$\\,a,b,c\\,$$, as it should be. What you would actually get, instead, is: $$a^2b^2c^2 + ab+bc+ca - a^2bc-ab^2c-abc^2-1 \\gt 0 \\;\\;\\iff\\;\\; (a b - 1) (a c - 1) (b c - 1) \\gt 0$$ [ EDIT ] The factorization is easier to \"see\" by defining $$\\,u=bc, v=ac, w=ab\\,$$, then: \\begin{align} a^2b^2c^2 + ab+bc+ca - a^2bc-ab^2c-abc^2-1 &= uvw - uv-vw-wu+u+v+w -1 \\\\ &= (u-1)(v-1)(w-1) \\end{align} •", "= a_0 \\\\ \\quad b_1 + b_2 = a_1 \\\\ \\quad b_2 - b_1 = a_2 \\end{align} The third equation can be rewritten as $b_2 = a_2 + b_1$, and plugging this into the second equation gives us that $2b_1 + a_2 = a_1$ so $b_1 = \\frac{a_1 - a_2}{2}$. Plugging this back into the third equation gives us $b_2 = a_2 + \\frac{a_1 - a_2}{2} = \\frac{2a_2 + a_1 - a_2}{2} = \\frac{a_1 + a_2}{2}$, and so we have that: (3) \\begin{align} \\quad b_0 = a_0 \\\\ \\quad b_1 = \\frac{a_1 - a_2}{2} \\\\ \\quad b_2 = \\frac{a_1 + a_2}{2} \\end{align} Therefore $\\{ 1, x - x^2, x + x^2 \\}$ spans $\\wp_2 (\\mathbb{R})$. Now for $c_0, c_1, c_2 \\in \\mathbb{F}$, consider the following vector equation: (4) \\begin{align} \\quad c_0(1) + c_1(x - x^2) + c_2(x + x^2) = 0 \\\\ \\quad c_0 + (c_1 + c_2)x + (c_2 - c_1)x^2 = 0 \\end{align} We thus get that: (5) \\begin{align} \\quad c_0 = 0 \\\\ \\quad c_1 + c_2 = 0 \\\\ \\quad c_2 - c_1 = 0 \\end{align} The third equation implies that $c_1 = c_2$ and plugging this into the second equation gives us $2c_1 = 0$ so $c_1 = 0$ and thus $c_2 = 0$. The first equation says that $c_0 = 0$, so", "+0 # help +1 86 2 +160 n Part a of this problem, you found the values of integers that satisfied the system of equations. In this part of the problem you will write up a full solution that describes how to solve this problem. Four positive integers $a$, $b$, $c$, and $d$ satisfy \\begin{align*} ab + a + b &= 524, \\\\ bc + b + c &= 146, \\\\ cd + c + d &= 104, \\\\ \\end{align*} and $abcd=8!$. What are $a$, $b$, $c$, and $d?$ #1 +867 +4 First of all, we have the equations: \\begin{align*} ab + a + b &= 524, \\\\ bc + b + c &= 146, \\\\ cd + c + d &= 104, \\\\ \\end{align*} and abcd=8!. From the first equation, we have: $$ab + a + b = 524.$$ After factoring out the b, we have: $$b(1 + a) = 524 - a.$$ Solving for b, we have: $$b = \\frac{524 -a}{1 + a}$$ From the second equation, we are going to try to write c in the form of a. $$bc + b + c = 146.$$ Factoring out the b, we have: $$b(c+1)+c=146.$$ Solving for b, we have: $$b=\\frac{146-c}{1+c}.$$ Since b = b, we have: $$\\frac{146-c}{1+c}=\\frac{524 -a}{1 + a}.$$After cross-multiplying, we end up with: $$(1+c)(524-a)=(1+a)(146-c).$$", "3a^b + 3ab^2 \\;= \\;-3c^2 - 3c - 1$ Factor: . $a^3 + b^3 + c^3 + 3ab\\underbrace{(a + b)} \\;= \\;-3c^2 - 3c - 1$ . . . . . $a^3 + b^2 + c^3 + 3ab(-c - 1) \\;=\\;-3c^2 - 3x - 1$ . . . . . . $a^3 + b^3 + c^3 - 3abc - 3ab \\;= \\;-3c^2 - 3c - 1$ . . . . . . . . . . . $a^3 + b^3 + c^3 - 3ab \\;= \\;3ab - 3c^2 - 3c - 1$ Hence: . $\\boxed{Numerator \\;= \\;3ab - 3c^2 - 3c - 1}$ Square [1]: . $(-c-1)^2\\;=\\;(a + b)^2$ . . . . . . . . $c^2 + 2c + 1 \\;= \\;a^2 + 2ab + b^2$ . . . . . . . . . $-a^2 - b^2 \\;= \\; 2ab - c^2 - 2c - 1$ Add $ab + bc + ac - c^2$ to both sides: . . . $(ab + bc + ac - c^2) - a^2 - b^2 \\;= \\;(ab + bc + ac - c^2)$ $+ 2ab - c^2 - 2c - 1$ . . . . $ab + bc + ac - a^2 - b^2 - c^2 \\;= \\;3ab + ac + bc - 2c^2 - 2c - 1$ We", "obtained as explained in Half-fraction Designs as $A=-BC\\,\\!$, $B=-AC\\,\\!$ and $C=-AB\\,\\!$. The effects for this design can be calculated as: \\begin{align} & A-BC= & \\frac{1}{2}(ab+ac-(1)-bc) \\\\ & B-AC= & \\frac{1}{2}(ab-ac+(1)-bc) \\\\ & C-AB= & \\frac{1}{2}(-ab+ac-(1)+bc) \\end{align}\\,\\! These equations can be combined with the equations for (A+BC), (B+AC) and (C+AB) to obtain the de-aliased main effects and two factor interactions. For example, adding equations (A+BC) and (A-BC) returns the main effect $A\\,\\!$. \\begin{align} & 2A= & \\frac{1}{2}(a-b-c+abc)+ \\\\ & & \\frac{1}{2}(ab+ac-(1)-bc) \\end{align}\\,\\! The process of augmenting a fractional factorial design by a second fraction of the same size by simply reversing the signs (of all effect columns except $I\\,\\!$) is called folding over. The combined design is referred to as a fold-over design. ## Quarter and Smaller Fraction Designs At times, the number of runs even for a half-fraction design are very large. In these cases, smaller fractions are used. A quarter-fraction design, denoted as ${2}^{k-2}\\,\\!$, consists of a fourth of the runs of the full factorial design. Quarter-fraction designs require two defining relations. The first defining relation returns the half-fraction or the ${2}^{k-1}\\,\\!$ design. The second defining relation selects half of the runs of the ${2}^{k-1}\\,\\!$ design to give the quarter-fraction. For example, consider the ${2}^{4}\\,\\!$ design. To obtain a ${2}^{4-2}\\,\\!$ design from this design, first a half-fraction", "see that $-1 = \\frac{2b^3}{3b^2 - 1}.$ Rearranging this gives $2b^3 + 3b^2 - 1 = 0$, which can be factorised as $(b+1)^2 (2b-1) = 0$. Since $b$ can’t be positive, we conclude that $b=-1$. Then using part (iv), we obtain $c=2$, and from part (i), we see that $m=3b^2 - 1 = 2$. So we can now calculate \\begin{align*} \\text{Area of R} &{}= \\int_b^c (m(x+1) - (x^3 - x)) \\, dx \\\\ &{}= \\int_{-1}^2 -x^3 + 3x + 2 \\, dx \\\\ &{}= \\left[ -\\frac{x^4}{4} + \\frac{3x^2}{2} + 2x \\right]_{-1}^2 \\\\ &{}= 6-\\left(-\\frac{3}{4}\\right)\\\\ &{}= \\frac{27}{4}. \\end{align*}", "# How I can find the value of $abc$ using the given equations? If I have been given the value of \\begin{align*} a+b+c&= 1\\\\ a^2+b^2+c^2&=9\\\\ a^3+b^3+c^3 &= 1 \\end{align*} Using this I can get the value of $$ab+bc+ca$$ How i can find the value of $abc$ using the given equations? I just need a hint. I have tried by squaring the equations. But could not get it. - You can also cube your first equation, and using $ab+bc+ac$ you found, you should be able to find $abc$. – Lazar Ljubenović Mar 15 '12 at 16:17 Here is a solution I like, but way overkill to write as an answer. Let $A$ be any $3x3$ matrix which has eigenvalues $a,b,c$. Then you are given $tr(A), tr(A^2), tr(A^3)$ and need to find $\\det A$. Write the characteristic equation of $A$ and apply trace on it.... – N. S. Mar 15 '12 at 17:09 I think you have got enough answers,you should go ahead and chose whichever is your acceptable answer – Jeremy Carlos Mar 15 '12 at 17:11 You can get a term involving $abc$ by cubing $a+b+c$: \\begin{align*} (a+b+c)^3 &= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3\\\\ &= a^3+3a^2b+3ab^2 + b^3 + 3a^2c+\\color{blue}{6abc} + 3b^2c + 3ac^2 + 3bc^2 + c^3. \\end{align*} Now use the other information you have" ]

[ "9$$ Equation 1: $abc(a+b+c) = 100a + 10b + c$ Note that $$a+b+c$$ is present on both sides and rewrite the equation a bit: \\begin{align} abc(a+b+c) &= 99a+9b+(a+b+c)\\\\ abc &= \\frac{9(11a+b)+(a+b+c)}{(a+b+c)}\\\\ abc &= \\frac{9(11a+b)}{(a+b+c)} + 1\\end{align} Equation 2: $abc-1 = \\frac{9(11a+b)}{a+b+c}$ Since we are dealing exclusively with positive non-zero integers, both $$abc-1$$ and $$\\frac{9(11a+b)}{a+b+c}$$ must both be integral. This means that the term $$a+b+c$$ must be a factor of $$9(11a+b)$$. Because $$9(11a+b)$$ is > 1 by definition 1, the unique factorization theorem states that it is a product of primes and the factorization is unique. Thus the factors are $$3, 3^2$$ and $$11a + b$$. We won't tackle the determination of if $$11a + b$$ is prime, but will instead show a contradiction results: \\begin{align} a+b+c &= 11a+b\\\\ c &= 10a\\end{align}Which contradicts definition 1. Next we examine if $$a+b+c=3$$. This implies that $$a=1$$, $$b=1$$ and $$c=1$$, which leads to a contradiction of equation 1. This leaves the $$a+b+c=9$$ possibility. Substituting into equation 2 produces: \\begin{align} abc-1 &= \\frac{9(11a+b)}{(a+b+c)}\\\\ abc-1 &= \\frac{9(11a+b)}{9}\\\\ abc-1 &= 11a+b\\end{align} Now we solve for $$c$$: \\begin{align} abc &= 11a+b+1\\\\ c &= \\frac{11a+b+1}{ab}\\\\ c &= \\frac{11a}{ab}+\\frac{b}{ab}+\\frac1{ab}\\\\ c &= \\frac{11}b+\\frac{b+1}{ab}\\\\ c &= \\frac{11}b+\\frac{\\frac{b+1}a}b\\\\ c &= \\frac{11+\\frac{b+1}a}b\\\\\\end{align} Since we are dealing only with non-zero positive integers, this means b must evenly divide $$11+\\frac{b+1}a$$. Therefore, there exists some", "# Let a,b,c be real numbers such that a^3+b^3+c^3=1. Find that value of a(ab+c^2)+b(bc+a^2)+c(ca+b^a). 8 years ago $(a+b+c)^{3}=$ $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}-6abc=3a(ab+c^{2})+3b(bc+a^{2})+3c(ca+b^{2})$ $\\frac{(a+b+c)^{3}-a^{3}-b^{3}-c^{3}-6abc}{3}=a(ab+c^{2})+b(bc+a^{2})+c(ca+b^{2})$", ". . . $(b+c)\\big[a^2 + ab + ac + bc\\big] \\;=\\;0$ . . . . . . . . . . $(b+c)\\big[a(a+b) + c(a+b)\\big] \\;=\\;0$ Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \\;=\\;0$ We see that the product of three factors equals zero. Hence, at least one of the factors must be zero. Therefore: . $\\begin{Bmatrix}a+b \\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}b \\\\ & \\text{or} \\\\ b+c \\:=\\:0 & \\Rightarrow & b \\:=\\:\\text{-}c \\\\ & \\text{or} \\\\ a+c\\:=\\:0 & \\Rightarrow & a \\:=\\:\\text{-}c \\end{Bmatrix}$ 3. ## Re: Opposite number proof Originally Posted by Soroban Hello, DIOGYK! I think I got it . . . We have: . $\\frac{1}{a+b+c} \\;=\\;\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ . . . . . . . . $\\frac{1}{a+b+c} \\;=\\;\\frac{bc + ac + ab}{abc}$ . . . . . . . . . . . . $abc \\;=\\;(a+b+c)(bc+ac+ab)$ . . . . . . . . . . . . $abc \\;=\\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$ . . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \\;=\\;0$ n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \\;=\\;0$ . . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \\;=\\;0$ Factor: .", "$a+c-b = 23$ and $b+c-a = 1571$ are all primes. And in that case $d = 1594.$ Week 3 • Problem (posted September 17th) Now an algebra problem for the third week. Let $a$, $b$ and $c$ be distinct non-zero real numbers such that $$\\frac{1-a^3}{a} = \\frac{1-b^3}{b}= \\frac{1-c^3}{c}.$$ Determine all possible values of $a^3 + b^3 +c^3$. • Solution (posted September 24th) The third week problem came from Round 2 of the 2007 Alberta High School Mathematics Competition. We give solutions from two of the contestants that were featured in the report on the competition. First Solution by Brett Baek From $\\frac{1-a^3}{a}= \\frac{1-b^3}{b}$, we have $b-a^3 b = a -a b^{3}$. Hence $a-b =ab(b^2-a^2)$ so that $ab(a+b) = -1.$ Similarly, $bc(b+c) = -1.$ From these two equations, we have $a^2 -c^2 = bc -ab$ or $(a+c)(a-c) = -b(a-c).$ Since $a \\neq c$, we have $a+b+c = 0$. Hence $abc(a+b) = -c = a+b.$ Since $a+b+c =0$ but $c \\neq 0,$ $a+b \\neq 0$ and we have $abc =1.$ Now $$0 = (a+b+c)^3 = a^3 + 3 a^2(b+c) +3a (b+c)^2 + (b+c)^3 = (a^3 +b^3 + c^3) + 3(a+b+c)(bc+ca+ab)-3abc$$ $$= (a^3 +b^3 +c^3) + 0 - 3.$$ It follows that the only possible value of $a^3+b^3+c^3$ is $3.$ Alternate Solution by Jerry Lo For those with more knowledge of", "# If three positive real numbers a, b, c are in A.P. and abc = 4, Question: If three positive real numbers abc are in A.P. and abc = 4, then the minimum possible value of b is_________. Solution: Let us suppose three positive real numbers a, b and c are in A.P and abc = 4 Since arithmetic mean ≥ geometric mean i. e $\\frac{a+b+c}{3} \\geq \\sqrt[3]{a b c}$ i. e $\\frac{a+b+c}{3} \\geq \\sqrt[3]{4}$ (given) Since $b=\\frac{a+c}{2}$ $2 b=a+c$ i. e $\\frac{2 b+b}{3} \\geq(4)^{\\frac{1}{3}}$ i.e $\\frac{3 b}{3} \\geq(4)^{\\frac{1}{3}}$ i. e $b \\geq(2)^{\\frac{2}{3}}$ i.e minimum possible value of $b$ is $(2)^{\\frac{2}{3}}$.", "+0 # help +1 86 2 +160 n Part a of this problem, you found the values of integers that satisfied the system of equations. In this part of the problem you will write up a full solution that describes how to solve this problem. Four positive integers $a$, $b$, $c$, and $d$ satisfy \\begin{align*} ab + a + b &= 524, \\\\ bc + b + c &= 146, \\\\ cd + c + d &= 104, \\\\ \\end{align*} and $abcd=8!$. What are $a$, $b$, $c$, and $d?$ #1 +867 +4 First of all, we have the equations: \\begin{align*} ab + a + b &= 524, \\\\ bc + b + c &= 146, \\\\ cd + c + d &= 104, \\\\ \\end{align*} and abcd=8!. From the first equation, we have: $$ab + a + b = 524.$$ After factoring out the b, we have: $$b(1 + a) = 524 - a.$$ Solving for b, we have: $$b = \\frac{524 -a}{1 + a}$$ From the second equation, we are going to try to write c in the form of a. $$bc + b + c = 146.$$ Factoring out the b, we have: $$b(c+1)+c=146.$$ Solving for b, we have: $$b=\\frac{146-c}{1+c}.$$ Since b = b, we have: $$\\frac{146-c}{1+c}=\\frac{524 -a}{1 + a}.$$After cross-multiplying, we end up with: $$(1+c)(524-a)=(1+a)(146-c).$$", "# $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$? The polynomial is allowed to have repeated roots. - [There was a mistake in the first version of this answer that caused one solution to go missing.] The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have \\begin{align} a&=-a-b-c\\;,\\\\ b&=ab+bc+ca\\;,\\\\ c&=-abc\\;. \\end{align} If $c=0$, this becomes \\begin{align} a&=-a-b\\;,\\\\ b&=ab\\;,\\\\ \\end{align} and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$. If $c\\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields $$c=-2a+\\frac1a\\;,$$ and then the second equation becomes $$-\\frac1a=-1+2-\\frac1{a^2}-2a^2+1\\;.$$ Multiplying through by $a^2$ yields $$2a^4-2a^2-a+1=0\\;.$$ The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields $$2a^3+2a^2-1=0\\;,$$ which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$. Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively. - Not much familiar with solutions of biquadratic equations...\"no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions\" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can", "# 2021 AIME II Problems/Problem 7 ## Problem Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \\begin{align*} a + b &= -3, \\\\ ab + bc + ca &= -4, \\\\ abc + bcd + cda + dab &= 14, \\\\ abcd &= 30. \\end{align*} There exist relatively prime positive integers $m$ and $n$ such that $$a^2 + b^2 + c^2 + d^2 = \\frac{m}{n}.$$Find $m + n$. ## Solution 1 From the fourth equation we get $d=\\frac{30}{abc}.$ substitute this into the third equation and you get $abc + \\frac{30(ab + bc + ca)}{abc} = abc - \\frac{120}{abc} = 14$. Hence $(abc)^2 - 14(abc)-120 = 0$. Solving we get $abc = -6$ or $abc = 20$. From the first and second equation we get $ab + bc + ca = ab-3c = -4 \\Longrightarrow ab = 3c-4$, if $abc=-6$, substituting we get $c(3c-4)=-6$. If you try solving this you see that this does not have real solutions in $c$, so $abc$ must be $20$. So $d=\\frac{3}{2}$. Since $c(3c-4)=20$, $c=-2$ or $c=\\frac{10}{3}$. If $c=\\frac{10}{3}$, then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$. Since you already know $d=\\frac{3}{2}$ and $c=-2$, so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\\frac{141}{4}$. So", "+ 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0$$ Take 3ab common in the numerator. $$\\frac{3(ab) * [(ab)^2 + 3(ab) - 18]}{(a-1)(a + 2)} = 0$$ Note that $$(ab)^2 + 3ab - 18$$ is a quadratic which we can split into factors just like we do for $$x^2 + 3x - 18$$. $$x^2 + 3x - 18 = (x + 6)(x - 3)$$ so $$(ab)^2 + 3ab - 18 = (ab + 6)*(ab - 3)$$ $$\\frac{3(ab) * [(ab + 6)*(ab - 3)]}{(a-1)(a + 2)} = 0$$ Now, when will this fraction be 0? When one of the factors in the numerator is 0. (Note that denominator cannot be 0. So a cannot the 1 or -2.) So either ab = 0 (a and b are non zero so not possible) or (ab + 6) = 0 i.e. ab = -6 or (ab - 3) = 0 i.e. ab = 3 So, we get that one of these two must hold. Either ab = -6 or ab = 3. Now let's look at the possible values of b. Can b be 2? If b = 2, a = -3 (in this case, ab = -6. Satisfies) Can b be 3? If b = 3, a is -2 or 1. Both values are not possible so b cannot be 3. Can", "# Prove that $512^3 + 675^3 + 720^3$ is a composite number We have to prove that the number $$N=512^3 + 675^3 + 720^3$$ is composite. I tried to use the identity $(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ hoping to take out some common factors from the R.H.S. but it didn't work. I also used $a^3+b^3=(a+b)(a^2-ab+b^2)$ (in all possible combos) and tried to combine with $c^3$ but that too didn't work. I nearly spent about 5 hours struggling with the question but no result :( Let $a=512, b=675, c=720$. Now the number looks like $a^3+b^3+c^3$ but we require a sort of $3abc$ term to resolve it into factors. First we factorize $a,b,c$ into prime factors. So, $a=2^9, b=3^3\\times 5^2, c=2^4\\times 3^2\\times 5$. Now it can be seen that $2c^2=3ab$. Hence, $a^3+b^3+c^3=a^3+b^3-c^3+2c^2c$ and thus the problem can be solved. It can then be seen that the given number is composite.", "A= & \\frac{(a+abc)}{2}-\\frac{(b+c)}{2}=\\frac{1}{2}(a-b-c+abc) \\\\ & B= & \\frac{(b+abc)}{2}-\\frac{(a+c)}{2}=\\frac{1}{2}(-a+b-c+abc) \\\\ & C= & \\frac{(c+abc)}{2}-\\frac{(a+b)}{2}=\\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! where $a\\,\\!$, $b\\,\\!$, $c\\,\\!$ and $abc\\,\\!$ are the treatments included in the ${2}^{3-1}\\,\\!$ design. Similarly, the two factor interactions can also be obtained as: \\begin{align} & BC= & \\frac{(a+abc)}{2}-\\frac{(b+c)}{2}=\\frac{1}{2}(a-b-c+abc) \\\\ & AC= & \\frac{(b+abc)}{2}-\\frac{(a+c)}{2}=\\frac{1}{2}(-a+b-c+abc) \\\\ & AB= & \\frac{(c+abc)}{2}-\\frac{(a+b)}{2}=\\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! The equations for $A\\,\\!$ and $BC\\,\\!$ above result in the same effect values showing that effects $A\\,\\!$ and $BC\\,\\!$ are confounded in the present ${2}^{3-1}\\,\\!$ design. Thus, the quantity, $\\tfrac{1}{2}(a-b-c+abc),\\,\\!$ estimates $A+BC\\,\\!$ (i.e., both the main effect $A\\,\\!$ and the two-factor interaction $BC\\,\\!$). The effects, $A\\,\\!$ and $BC,\\,\\!$ are called aliases. From the remaining equations given above, it can be seen that the other aliases for this design are $B\\,\\!$ and $AC\\,\\!$, and $C\\,\\!$ and $AB\\,\\!$. Therefore, the equations to calculate the effects in the present ${2}^{3-1}\\,\\!$ design can be written as follows: \\begin{align} & A+BC= & \\frac{1}{2}(a-b-c+abc) \\\\ & B+AC= & \\frac{1}{2}(-a+b-c+abc) \\\\ & C+AB= & \\frac{1}{2}(-a-b+c+abc) \\end{align}\\,\\! ### Calculation of Aliases Aliases for a fractional factorial design can be obtained using the defining relation for the design. The defining relation for the present ${2}^{3-1}\\,\\!$ design is: $I=ABC\\,\\!$ Multiplying both sides of the previous equation by the main effect, $A,\\,\\!$ gives the alias effect of $A\\,\\!$ : \\begin{align} & A\\cdot I= & A\\cdot", "know \\begin{align*} a+b+c &= 1/5\\\\ ab+bc+ca &= -2/5\\\\ abc &= -3/5 \\end{align*} which are enough to establish the things you're asked to establish, with some manipulation. For example, \\begin{align*} 1/a + 1/b + 1/c &= bc/abc + ca / abc + ab / abc\\\\ &= (ab+bc+bc) / abc \\end{align*} and $$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$$ The sum of cubes is left as an exercise. - Hey thanks! This looks a bit more friendly. I will have a go at it myself. – Magpie Aug 6 '12 at 15:10 What Aakash has shown is in fact the traditional way to derive the Vieta formulae. – J. M. Aug 6 '12 at 15:19 ok thanks for letting me know. – Magpie Aug 6 '12 at 17:06 This is by no means \"precalculus\", and no calculator will help you to see that. Also, your $y = 0$ is nonsensical, except if you want to state the problem as \"$a, b, c$ are roots to $y(x) = 0$, where $y(t) = 5t^3 ...$\" A simple application of Viete's relations and Newton's formulae solves most of such problems. For instance, $$\\sum \\frac{1}{a} = \\frac{ab + bc + ca}{abc} = \\frac{-2/5}{-3/5} = \\frac{2}{3}$$ $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2\\sum ab = \\frac{1}{25} - \\frac{2(-2)}{5} = \\frac{21}{25}.$$ For", "the following factorization: $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Proof 1: Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra $P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc$. Now, once again basic algebra says that as a, b, c are roots/solutions of the above: $P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0$ $P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0$ $P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc$=0\\$ $0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc$ So, we get $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Also, the above formula can be written as $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$ Proof 2: Consider the following determinant D: $\\left| \\begin{array}{ccc} a & b & c\\\\c & a & b\\\\ b & c & a \\end{array} \\right|$ On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following: $D = \\left| \\begin{array}{ccc} a+b+c & b & c \\\\a+b+c & a & b\\\\a+b+c & c & a \\end{array} \\right|$. Note that columns 2 and 3 of the three by three determinant do not change. On expanding the original determinant D, we get $D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)$ $D= a^{3}-abc-bac+b^{3}+c^{3}-cab$ $D= a^{3}+b^{3}+c^{3}-3abc$ Whereas we get from the other transformed but equal D: $D =(a+b+c) \\left| \\begin{array}{ccc} 1 & b & c \\\\ 1 & a & b \\\\ 1 & c & a \\end{array}\\right|$ $D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))$ So, that we again get $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ Proof 3: Now, let us consider $E=a^{2}+b^{2}+c^{2}-ab-bc-ca$ as a quadratic in a with b and c as parameters.", "$a - (-b) = a + b$. Note that when $n$ is even, $(-b)^n + b^n = 2b^n$, rather than 0, so this gives us no useful information. ## Vieta's/Newton Factorizations These factorizations are useful for problems that could otherwise be solved by Newton sums or problems that give a polynomial and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere. • $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ • $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$ • $(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$ • $(a+b+c)^7=a^7+b^7+c^7+7(a+b)(b+c)(c+a)((a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c))$ ## Practice Problems • Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$. • Prove that $2222^{5555} + 5555^{2222}$ is divisible by 7. - USSR Problem Book • Factor $(x-y)^3 + (y-z)^3 + (z-x)^3$. • Factor $x^4 + 1$ into two polynomials with real coefficients. • Given that $a+b+c=0$, prove that $abc=\\dfrac{a^3+b^3+c^3}{3}$.", "solutions. Here $n=3$ so, let's take $m=0,1,2$. $m=0\\implies \\cos\\frac{\\pi}{3}+i\\sin \\frac{\\pi}{3}=\\frac{1+i\\sqrt3}{2}$ $m=1\\implies \\cos \\pi+i\\sin\\pi=-1$ $m=2\\implies \\cos\\frac{5\\pi}{3}+i\\sin \\frac{5\\pi}{3}=\\frac{1-i\\sqrt3}{2}$ - I think the second factor in your first line should be $x^2-x+1$. – Daryl Sep 8 '12 at 12:32 @Daryl, thanks for identifying the lapsus calami. – lab bhattacharjee Sep 8 '12 at 12:34 Hi, where did you got $x^3+1=(x+1)(x^2-x+1)$? I mean also the site, because I found $x^n-1=(x-1)(x^{n-1}+ \\dots +1)$ - planetmath.org/encyclopedia/PrimeFactorsOfXn1.html. – alvoutila Sep 8 '12 at 19:02 $x^3+y^3=(x)^3+(y)^3=(x+y)^3-3xy(x+y)=(x+y)((x+y)^2-3xy)=(x+y)(x^2+y^2-xy)$ – lab bhattacharjee Sep 9 '12 at 3:39 Let $x=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^3 = -1$. Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$ We can separate the real and imaginary parts of this equation: \\begin{align} a^3 -3ab^2 & = -1 \\\\ 3a^2b - b^3 & = 0 \\end{align} Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution $$x =-1.$$ So suppose $b\\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \\pm a\\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$a^3 - 9a^3 = -1$$ so $a = \\frac12.$ Since $b = \\pm a\\sqrt3$, we have $b=\\pm\\frac{\\sqrt3}2$. This gives us the other two solutions, namely $$x=\\frac12 \\pm\\!\\frac{\\sqrt3}2i.$$ - Set $\\displaystyle x=re^{i \\theta}$. So $\\displaystyle r^3e^{i3\\theta}= x^3= -1= e^{i \\pi}$, hence $r^3=1$ and $3", "c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc Substituting the given values of a, b ,c in the identity , we get a3 + b3 + c3 = ($\\frac{1}{2}$ +$\\frac{1}{3}$$\\frac{5}{6}$)( a2 + b2 + c2 – ab – bc – ca) + 3abc by applying least common multiple method a3 + b3 + c3 = $(\\frac{1*6}{2*6} + \\frac{1*4}{3*4} – \\frac{5*2}{6*2})$( a2 + b2 + c2 – ab – bc – ca) + 3abc a3 + b3 + c3 = $(\\frac{6}{12} + \\frac{4}{12} – \\frac{10}{12})$( a2 + b2 + c2 – ab – bc – ca) + 3abc a3 + b3 + c3 = 0( a2 + b2 + c2 – ab – bc – ca) + 3abc a3 + b3 + c3 = 3abc ($\\frac{1}{2}$)3 + ($\\frac{1}{3}$)3 – ($\\frac{-5}{6}$)3 = 3*$\\frac{1}{2}$* $\\frac{1}{3}$* $\\frac{-5}{6}$ = $\\frac{1}{2}$ * $\\frac{-5}{6}$ = $\\frac{-5}{12}$ Therefore, the value of $(\\frac{1}{2})^{3} + (\\frac{1}{3})^{3} – (\\frac{5}{6})^{3}$ is $\\frac{-5}{12}$ (d) (0.2)3 – (0.3)3 + (0.1)3 As we know that, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) Using the above identity, consider the value of a = 0.2, b = 0.3, c = 0.1 a3 + b3 + c3 – 3abc", "$r^4e^ = 1.$ This means that $r^4 = 1,$ so $r=1$ since $r$ is a real number and $r geq 0$. In addition $4 heta$ must be an integer multiple of $2pi$. Choosing multiples of 0, 1, 2, and 3 gives $heta= 0, frac<2>, pi, frac<3pi><2>$. These correspond to the four numbers 1, $i$, -1, and $-i$, as pictured below: The fourth roots of 1 divide the unit circle evenly into four sections. Solution: 3 Solving equations 1. Writing $(a+bi)^3=(a^3-3ab^2)+(3a^2b-b^3)i=1+0i$ and equating real and imaginary parts of $(a+bi)^3=1$ gives $a^3 - 3ab^2 = 1$ and $3a^2b - b^3 = 0$ Factoring the second equation as $b(3a^2-b^2)=0$, we see that either $b=0$ or $3a^2=b^2$. If $b=0$, then $a^3=1$, giving the obvious cube root of 1. If $b eq 0$, then $3a^2=b^2$, and substituting this into $a^3-3ab^2=1$ gives $a^3-9a^3=1$, so $a=-frac<1><2>$, and then $b=pmfrac><2>$. 2. Similarly, if we write $(a+bi)^4=((a^2-b^2)+2abi)^2= (a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)i$ then equating imaginary parts in $(a+bi)^4=1$, gives $(4a^3b-4ab^3)=0.$ Factoring the left-hand side as $4ab(a^2-b^2)=0$, we see that any 4th root of 1 must have $a=0$, $b=0$, or $a^2=b^2$. If $a=0$, then the equation $(bi)^4=1$ quickly gives $b=pm 1$. Similarly, if $b=0$, then the equation $a^4=1$ again gives $a=pm 1$ (remember that $a$ and $b$ are real). Finally, if $a=pm b$, then equating the real parts of $(apm ai)^4=1$ gives", "# value of $abc$ in algebraic expression If $a+b+c=0$ and $\\displaystyle \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} = 3$. where $a,b,c$ are non zero real numbers then $ab(a+b)+bc(b+c)+ca(c+a)$ is Attempt: from $a+b+c=0$ we have to find value of $-3abc$ and from $\\displaystyle \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} = 3$ we have $\\displaystyle \\frac{ab+bc+ca}{abc} = 3$ could some help how to get value of $abc,$ thanks We have $$(a+b+c)^2 =a^2+b^2+c^2+2 (ab+bc+ca) =0$$ $$\\Rightarrow ab+bc+ca = -0.5 (a^2+b^2+c^2)$$ We already know $ab + bc + ca =3abc$, so hope you can take it from here. On a side note, observe that $$(a+b+c)^3 =a^3+b^3+c^3+ (a+b+c)(ab+bc+ca) \\Rightarrow a^3+b^3+c^3=0$$ • i have got $$\\frac{3a^3}{3a-1}$$ what should be the result? – Dr. Sonnhard Graubner Feb 24 '17 at 10:33", "a^2 \\\\ 3n &=& a^2 - 1 \\\\ n &=& \\frac{a^2-1}{3}\\\\ \\hline n+1 &=& \\frac{a^2-1}{3} +1 \\\\ n+1 &=& \\frac{a^2-1+3}{3}\\\\ n+1 &=& \\frac{a^2+2}{3}\\\\ \\end{array} }$$ Because $$\\frac{a^2+2}{3}$$ is a integer then $$a^2+2$$ is divisible by 3, then $$a^2$$ is not divisible by 3 and also $$a$$ is not divisible by 3, because if 3 is not a prime factor in $$a^2$$ a perfect spuare, then 3 is not a prime factor in $$a$$ Two numbers are not divisible by 3. It is $$3b +1$$ and $$3b + 2$$ 1. We substitute $$a = 3b+1$$ $$\\small{ \\begin{array}{rcl} n+1 &=& \\frac{a^2+2}{3}\\qquad \\text{substitute }\\ a = 3b+1\\\\ n+1 &=& \\frac{(3b+1)^2+2}{3} \\\\ n+1 &=& \\frac{9b^2+6b+1+2}{3} \\\\ n+1 &=& \\frac{9b^2+6b+3}{3} \\\\ n+1 &=& 3b^2+2b+1 \\\\ n+1 &=& b^2 + b^2 + b^2 +2b+1 \\\\ n+1 &=& b^2 + b^2 + (b+1)^2\\\\ \\end{array} }$$ So n + 1 is the sum of three perfect squares and $$b = \\frac{a-1}{3}$$. 2. We substitute $$a = 3b+2$$ $$\\small{ \\begin{array}{rcl} n+1 &=& \\frac{a^2+2}{3}\\qquad \\text{substitute }\\ a = 3b+2\\\\ n+1 &=& \\frac{(3b+2)^2+2}{3} \\\\ n+1 &=& \\frac{9b^2+12b+4+2}{3} \\\\ n+1 &=& \\frac{9b^2+12b+6}{3} \\\\ n+1 &=& 3b^2+4b+2 \\\\ n+1 &=& b^2 + b^2 + b^2 +2b+2b+1+1 \\\\ n+1 &=& b^2 + b^2+2b+1 + b^2 +2b+1 \\\\ n+1 &=& b^2 +(b+1)^2+(b+1)^2\\\\ \\end{array} }$$ So n + 1 is the sum of three", "are (x1, y1),(x2, y2) and (x3 y3) is = $$\\frac{1}{2}$$ab[1(r – r2) – 1 (r – r2) + 1 (r3 – r3)] = $$\\frac{1}{2}$$ab[r – r2 – r + r2 + 0] = $$\\frac{1}{2}$$ab × 0 = 0 ∴ The points (x1, y1),(x2, y2) and (x3 y3) are collinear. Question 19. If denotes the greatest integer less than or equal to the real number under consideration and – 1 ≤ x< 0, 0 ≤ y < 1, 1 ≤ z < 2 , then the value of the determinant (1) (2) (3) (4) (1) Explaination: Question 20. If a ≠ b, b, c satisfy $$\\left| \\begin{matrix} a & 2b & 2c \\\\ 3 & b & c \\\\ 4 & a & b \\end{matrix} \\right|$$ = 0 then abc = (1) a + b + c (2) 0 (3) b3 (4) ab + bc (3) b3 Explaination: (a – 6) (b2 – ac) = 0 Since a ≠ 6 , we have a – 6 ≠ 0 ∴ b2 – ac = 0 b2 = ac b3 = abc Question 21. If A = $$\\left| \\begin{matrix} -1 & 2 & 4 \\\\ 3 & 1 & 0 \\\\ -2 & 4 & 2 \\end{matrix} \\right|$$ and B = $$\\left| \\begin{matrix} -2 & 4 & 2 \\\\ 6 &" ]

[ "PolynomialAndEquation FunctionProperty AMC10/12 2016 Problem - 2887 A binary operation $\\diamondsuit$ has the properties that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = (a\\,\\diamondsuit \\,b)\\cdot c$ and that $a\\,\\diamondsuit \\,a=1$ for all nonzero real numbers $a, b,$ and $c$. (Here $\\cdot$ represents multiplication). The solution to the equation $2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100$ can be written as $\\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q?$", "= a\\ \\diamondsuit\\ ( a\\ \\diamondsuit\\ a) = ( a\\ \\diamondsuit\\ a) \\cdot a = a.$$ Hence, $( a\\ \\diamondsuit\\ b) \\cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\\ \\diamondsuit\\ b) = \\frac{a}{b}.$ Hence, the given equation becomes $\\frac{2016}{\\frac{6}{x}} = 100$. Solving yields $x=\\frac{100}{336} = \\frac{25}{84},$ so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ### Solution 3 One way to eliminate the $\\diamondsuit$ in this equation is to make $a = b$ so that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = c$. In this case, we can make $b = 2016$. $$2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100\\implies (2016\\, \\diamondsuit\\, 6) \\cdot x = 100$$ By multiplying both sides by $\\frac{6}{x}$, we get: $$(2016\\, \\diamondsuit\\, 6) \\cdot 6 = \\frac{600}{x}\\implies 2016 \\, \\diamondsuit\\, (6\\, \\diamondsuit\\, 6) = \\frac{600}{x}$$ Because $6\\, \\diamondsuit\\, 6 = 2016\\, \\diamondsuit\\, 2016 = 1:$ $$2016 \\, \\diamondsuit\\, (2016\\, \\diamondsuit\\, 2016) = \\frac{600}{x}\\implies (2016\\, \\diamondsuit\\, 2016) \\cdot 2016 = \\frac{600}{x}\\implies 2016 = \\frac{600}{x}$$ Therefore, $x = \\frac{600}{2016} = \\frac{25}{84}$, so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ~ pi_is_3.14", "way to express $$110$$ as the product of at least $$3$$ numbers is $$2 \\cdot 5 \\cdot 11.$$ This means that $$110n^3$$ has no other prime factors. Then the exponents must be $$1, 4,$$ and $$10$$ from the above formula. Let $$2^3$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$2^9,$$ which makes $$110n^3$$ have a factor of $$2^{10}.$$ Then let $$5$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$5^3,$$ which means that $$110n^3$$ has a factor of $$5^4.$$ We do not have to alter the number of $$11$$s since it already has $$1$$ as its exponent. This means that we can let $$n = 2^3 \\cdot 5.$$ Then $81n^4 = 3^4 \\cdot 2^12 \\cdot 5^4.$ This number has $5 \\cdot 13 \\cdot 5 = 325$ factors. Thus, the correct answer is D. 23. A binary operation $$\\diamondsuit$$ has the properties that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = (a\\,\\diamondsuit \\,b)\\cdot c$ and that $$a\\,\\diamondsuit \\,a=1$$ for all nonzero real numbers $$a, b,$$ and $$c.$$ (Here $$\\cdot$$ represents multiplication). The solution to the equation $2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100$ can be written as $$\\frac{p}{q},$$ where $$p$$ and $$q$$ are relatively prime positive integers. What is $$p+q?$$ $$109$$ $$201$$ $$301$$ $$3049$$ $$33,601$$ ###### Solution(s): Let us see what properties we can gather from the given conditions.", "of the Bell polynomial Bn,k(x1,x2,...) when all xs are equal to 1 is a Stirling number of the second kind: $B_{n,k}(1,1,\\dots)=S(n,k)=\\left\\{{n\\atop k}\\right\\}.$ The sum $\\sum_{k=1}^n B_{n,k}(1,1,1,\\dots) = \\sum_{k=1}^n \\left\\{{n\\atop k}\\right\\}$ is the nth Bell number, which is the number of partitions of a set of size n. ### Convolution identity For sequences xn, yn, n = 1, 2, ..., define a sort of convolution by: $(x \\diamondsuit y)_n = \\sum_{j=1}^{n-1} {n \\choose j} x_j y_{n-j}$. Note that the bounds of summation are 1 and n − 1, not 0 and n . Let $x_n^{k\\diamondsuit}\\,$ be the nth term of the sequence $\\displaystyle\\underbrace{x\\diamondsuit\\cdots\\diamondsuit x}_{k\\ \\mathrm{factors}}.\\,$ Then $B_{n,k}(x_1,\\dots,x_{n-k+1}) = {x_{n}^{k\\diamondsuit} \\over k!}.\\,$ For example, let us compute $B_{4,3}(x_1,x_2)$. We have $x = ( x_1 \\ , \\ x_2 \\ , \\ x_3 \\ , \\ x_4 \\ , \\dots )$ $x \\diamondsuit x = ( 0,\\ 2 x_1^2 \\ ,\\ 6 x_1 x_2 \\ , \\ 8 x_1 x_3 + 6 x_2^2 \\ , \\dots )$ $x \\diamondsuit x \\diamondsuit x = ( 0 \\ ,\\ 0 \\ , \\ 6 x_1^3 \\ , \\ 36 x_1^2 x_2 \\ , \\dots )$ and thus, $B_{4,3}(x_1,x_2) = \\frac{ ( x \\diamondsuit x \\diamondsuit x)_4 }{3!} = 6 x_1^2 x_2.$ ## Applications of Bell polynomials ### Faà di Bruno's formula Faà di", "and $u+v$ is the concatenation of $u$ and $v$ in decimal 4. and $x \\diamond y$ is the operation of interest then $$x \\diamond y = b(d(x) + d(y))$$ for all binary strings $x$ and $y$. I'm not sure if that's what you're looking for, though. -", "value of $a$, then binary search needs $lg~A$ operations Note that $b~lg~a = lg~n$, that is $$lg~A = \\frac{lg~n}{b}$$ When summing up, $$\\sum lg~A = lg~n \\cdot (\\frac{1}{1} + \\frac{1}{2} + … + \\frac{1}{B}) = lg~n \\cdot lg~B = lg~n \\cdot lg~lg~n$$ In other words, all the operations for binary search is $O(lg~n \\cdot lg~lg~n)$ Consider the operation of $a^b$, it is $O(lg~n \\cdot (lg~lg~n)^2)$ finally. ps: All the lg are base 2.", "/ $S^3$ / $S^7$ binary operation is the product of unit norm real numbers / complex numbers / quaternions / octonions respectively – Grigory M Dec 26 '14 at 14:41 The operation on $S^1$ is addition of angles (or complex multiplication of numbers of modulus 1). The one on $S^3$ is unit-quaternion multiplication. The one on $S^7$ is unit-\"octonion\" multiplication. (On $S^0 = \\{-1, 1\\}$, it's mutliplication of real numbers of absolute value 1, of course.) • Thank you. For some reason, I thought of $\\mathbb S^n$ as $\\{x \\in \\mathbb R^n | |x|=1\\}$ and not $\\{x \\in \\mathbb R^{n+1} | |x|=1\\}$ – slinshady Dec 26 '14 at 14:46", "3. $(0,1,1,1,1,0)$ 4. $(0,0,0,1,1,0)$ ##### Exercise5 Consider the linear code defined the generator matrix\\\\ $G=\\left( \\begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 1 \\\\ \\end{array} \\right)$ 1. What size blocks does this code encode and what is the length of the code words? 2. What are the code words for this code? 3. With this code, can you detect single bit errors? Can you correct all, some, or no single bit errors? # Exercises16.1.6Exercises for Section 16.1 ##### Exercise1 Review the definition of rings to show that the following are rings. The operations involved are the usual operations defined on the sets. Which of these rings are commutative? Which are rings with unity? For the rings with unity, determine the unity and all units. 1. $[\\mathbb{Z};+,\\cdot ]$ 2. $[\\mathbb{C};+,\\cdot ]$ 3. $[\\mathbb{Q};+,\\cdot ]$ 4. $\\left[M_{2\\times 2}(\\mathbb{R}),+, \\cdot \\right]$ 5. $\\left[\\mathbb{Z}_2,+_2,\\times_2\\right]$ ##### Exercise3 Show that the following pairs of rings are not isomorphic: 1. $[\\mathbb{Z};+,\\cdot ]$ and $\\left[M_{2\\times 2}(\\mathbb{Z}),+,\\cdot \\right]$ 2. $[3\\mathbb{Z};+, \\cdot ]$ and $[4\\mathbb{Z};+, \\cdot ]$. ##### Exercise5 1. Show that $3\\mathbb{Z}$ is a subring of the ring $[\\mathbb{Z}, +, \\cdot]$ 2. Find all subrings of $\\mathbb{Z}_8$. 3. Find all subrings of $\\mathbb{Z}_2 \\times \\mathbb{Z}_2$. ##### Exercise7 1. Determine all solutions of the equation $x^2 - 5x + 6", "many linked operations which, while mainly done for the purpose of doing so, are perhaps not too overly frivolous. Here is one: Given a finite set$S$with$k$elements, there are$k^{k^2}$binary operations and each operation$\\diamond$can be represented by its Cayley Table the$k \\times k$square with$i,j$entry$i \\diamond j$(assuming a known order). This single operation defines a quasigroup quasi-group if the table is a Latin Square, each symbol appears once in each row and column. Equivalently (and without an agreed order), for every$a,b$there are unique$x,y$with$x \\diamond a=b$and$a \\diamond y=b.$This defines two operations$y=a \\backslash b$and$x=a x=b / b$a$ with $$b=a \\diamond (a \\backslash b)=a \\backslash (a \\diamond b)=(b /a) \\diamond a=(b \\diamond a) / a.$$ This equational definition of a quasi-group via 3 operations is available for infinite quasi-groups as well. If we introduce the projections $k$ is a prime power (a\\pi_1b=a$and perhaps only$a\\pi_2b=b$then ) it is possible we can write the previous equations as $$a\\pi_2b=(a\\pi_1b)\\diamond(a\\backslash b)=\\cdots$$ Two Latin Squares (given by quasigroups$(S,\\cdot)$and$(S,\\diamond)$) are orthogonal if the map$(x,y) \\to have (x\\cdot y,x\\diamond y)$is a bijection from$k-1$S \\times S$ to itself. This It is not known how many pairwise orthogonal quasi-groups (aka Mutually Orthogonal Latin Squares in that any two give all MOLS) are possible on a set of size $k^2$ k.$Certainly no more than$k-1$and this is possible ordered pairs when combined position by position(and", "represent a set of ur-elements, and addition represent a set of sets, so the first table above has $a+b+c+abc$ as its left column and $1+ab+ac+ad$ as its second column, the $1$ being the product of members of the empty set. If, further, we represent the whole table as a fraction with the odd column as the numerator and the even column as the denominator, then the operation of adding the new element is this: $$\\frac{\\frac{a+b+c+abc}{1+ab+ac+bc}+d}{1+\\frac{a+b+c+abc}{1+ab+ac+bc}d} = \\frac{a+b+c+d+abc+abd+acd+bcd}{1+ab+ac+ad+bc+bd+cd+abcd}.$$ So we're working with a binary operation: $a\\,\\diamondsuit\\, b=\\dfrac{a+b}{1+ab}$. When the set we start with is empty, the the numerator is the sum of $0$ terms and is $0$ and the denominator is the product of $0$ terms and $\\text{is }1$. (And of course $\\tanh(x+y)=(\\tanh x)\\,\\diamondsuit\\,(\\tanh y)$, and there's also an application to something more like ordinary trigonometric functions.) QUESTIONS: • Does this algebraic way of looking at this shed light on the combinatorics, or vice-versa? • What is the combinatorial meaning of the fact that the binary operation is associative? • I tried this with the even terms in the numerator and odd in the denominator: $a\\,\\spadesuit\\, b$. We get $(a\\,\\spadesuit\\,b)\\,\\spadesuit\\,c =a\\,\\diamondsuit\\,b\\,\\diamondsuit\\,c$; and $(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_n)\\,\\spadesuit\\, a_{n+1} = 1/(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_{n+1})$ --- So if you apply the $\\spadesuit$ operation instead of $\\diamondsuit$ each time a new element is added, then you alternate between odd-in-the-numerator-and-even-in-the-denominator", "of Computation, April 1968, 22, #102, 423–427), the answer is yes. On page 423 they define a function$N_k(h)$, which is the count of 1 bits in the$h$th position of the sequence$n^k$over one of ... 1 Set$X_i$to be the 0/1 random variable denoting whether two binary strings differ in position$i$. These$L$random variables are independent, and each 0/1 with equal probability. By linearity of expectation,$E(X_1+X_2+\\cdots+X_L)=E(X_1)+E(X_2)+\\cdots+E(X_L)=1/2+1/2+\\cdots+1/2=L/2$. 1 A binary operation is any operation acting with two arguments, irrespective of the base, or indeed, whether they are numbers or not. Adding two decimal numbers is a binary operation:$15 + 13 = 28$. Here, the addition operation takes two numbers,$15$and$13$, and acts on them. In some notation systems, we can make this more clear by writing it as$(+\\ 15\\ ... 3 The answer is that \"it depends\". If your binary numbers are integers, then the first result is correct: $$630=48\\cdot13+6,$$ so the integer part of the quotient is, indeed, 48 and the remainder is six. The link that you gave specifically states that it is doing polynomial division. What that means is that it interprets the individual bits as coefficients ... 1 Absolutely you should get the correct result. Note that both the first and second subtraction they subtract $1001_2-1101_2=100_2$ They must be working in the field of two elements, not in $\\Bbb", "# rig A rig $(R,+,\\cdot)$ is a set $R$ together with two binary operations $+:R^{2}\\to R:(a,b)\\mapsto a+b$ and $\\cdot:R^{2}\\to R:(a,b)\\mapsto ab$, such that both $(R,+)$ and $(R,\\cdot)$ are monoids, where $\\cdot$ distributes over $+$. That is if $\\{a,b,c,d\\}\\subset R$ then $(a+b)(c+d)=ac+ad+bc+bd$. The natural numbers with ordinary addition and multiplication $(\\mathbf{N},+,\\cdot)$ is a rig. A rig $(R,+,\\cdot)$ is a ring if $(R,+)$ is a group. The integers with ordinary addition and multiplication $(\\mathbf{Z},+,\\cdot)$ is a ring. Title rig Rig 2013-03-22 14:44:29 2013-03-22 14:44:29 HkBst (6197) HkBst (6197) 8 HkBst (6197) Definition msc 20M99 semigroup ring", "A binary operation can be looked at as a function $\\cdot : G \\times G \\to G$. For example, the binary operation in the example you saw, $a * b = a^b$ can be looked at as $* : \\mathbb{N} \\times \\mathbb{N} \\to \\mathbb{N}, ~ *(a,b) = a^b$ Another example for a binary operation: Take $\\mathbb{Z}_n = \\{0,1,...,n-1 \\}$ and define $a * b = a + b ~ (mod ~ n)$.", "\\mathbf{abs}\\ {-20} = 20 unary Floor/integer part operator $$\\mathbf{floor}\\ 2.5 = 2$$ \\mathbf{floor}\\ 2.5 = 2 unary Integer division operator $$5\\ \\mathbf{div}\\ 2 = 2$$ 5\\ \\mathbf{div}\\ 2 = 2 binary Remainder operator $$5\\ \\mathbf{rem}\\ {-2} = 1$$ 5\\ \\mathbf{rem}\\ {-2} = 1 binary Modulus operator $$5\\ \\mathbf{mod}\\ {-2} = -1$$ 5\\ \\mathbf{mod}\\ {-2} = -1 binary Equality $$a = b$$ a = b binary Inequality $$a \\neq b$$ a \\neq b binary Less than $$a > b$$ a > b binary Less than or equal $$a \\ge b$$ a \\ge b binary More than $$a < b$$ a < b binary More than or equal $$a \\le b$$ a \\le b binary ## Bags¶ Concept Math LaTeX Notes Bag $$[\\![ a,b,b,c,d ]\\!]$$ [\\![ a,b,b,c,d ]\\!] Members can repeat, unordered, also \\llbracket and \\rrbracket from package stmaryrd look better Count operator $$\\mathbf{count}[\\![ a,b,b,c,d ]\\!] b = 2$$ \\mathbf{count}[\\![ a,b,b,c,d ]\\!] b = 2 How many times element $$b$$ occurs in bag Bag cardinality $$\\mathbf{card}(B)$$ \\mathbf{card}(B) Number of members of $$B$$ Bag union (1) $$[\\![ a,b,b,c,d ]\\!] \\cup [\\![ a,b,c,c,d ]\\!] = [\\![ a,b,b,c,c,d ]\\!]$$ [\\![ a,b,b,c,d ]\\!] \\cup [\\![ a,b,c,c,d ]\\!] = [\\![ a,b,b,c,c,d ]\\!] Maximum number an element occurs Bag union (2) $$[\\![ a,b,b,c,d ]\\!] \\cup [\\![ a,b,c,c,d ]\\!] = [\\![ a,a,b,b,b,c,c,c,d,c ]\\!]$$ [\\![ a,b,b,c,d ]\\!] \\cup", "# Cyclic Distribution on the reals? Do there exist binary operators *, **, and *** on the real numbers, such that * distributes over **, ** distributes over ***, *** distributes over *, but not vice versa? - Likely there are such over a countable, perhaps even small finite set. Do you require anything else of the operators? Gerhard \"Ask Me About System Design\" Paseman, 2012.09.08 – Gerhard Paseman Sep 9 '12 at 4:11 These conditions can be expressed by first-order formulas, hence if such operators exist on some infinite set (or on arbitrarily large finite sets), then they exist on every infinite set. – Emil Jeřábek Sep 9 '12 at 9:27 What have you tried? – Trevor Wilson Sep 9 '12 at 16:46 I shall show that we can have commutative associative binary operations that satisfy the required properties. Without loss of generality, we can replace $\\mathbb{R}$ by any other infinite set. Consider the binary operations $+,\\cdot,\\uparrow$ on the interval [1,\\infty] where $+,\\cdot$ are just normal addition and multiplication and $x\\uparrow y=\\infty$ for all $x,y$. Every elementary school student should know the distributive property for addition and multiplication $(a+b)\\cdot c=a\\cdot c+b\\cdot c$. Furthermore, we have $$(a\\cdot b)\\uparrow c=\\infty=(a\\uparrow c)\\cdot(b\\uparrow c)=\\infty$$ and $$(a\\uparrow b)+c=\\infty+c=\\infty=(a+c)\\uparrow(b+c).$$ On the other hand, the average kindergartner can verify that $(1\\cdot 2)+3=5\\neq 20=(1+3)\\cdot(2+3)$. Therefore", "9 '14 at 2:08 It can't be $o(\\log N)$ for any finite set of binary operations. For a set of operations of size $k$, you can only have of order $k^N$ legal strings, so you can't represent more numbers than that. - Not all of the operations are binary. For example, $$4!!!!\\cdots !!!$$ for any number of factorials, can be written with one four – Zubin Mukerjee Jan 11 at 16:50", "b$$ or $$(a \\bigcirc b) \\bigcirc b = a \\bigcirc (b \\bigcirc b)$$ • flexibility: $$(a \\bigcirc b) \\bigcirc a = a \\bigcirc (b \\bigcirc a)$$ Distributivity of a binary operator $$\\diamond$$ over operator $$\\bigcirc$$ is more involved: it can be left-distributed: $$a \\diamond (b \\bigcirc c) = (a \\diamond b) \\bigcirc (a \\diamond c)$$ or right-distributed: $$(a \\bigcirc b) \\diamond c = (a \\diamond c) \\bigcirc (b\\diamond c)$$ Division is right-distributive over add, not left-distributive. Distributivity of multiply over add is at play in rings and fields. But the converse does not apply. There are mathematical structures where two binary operators distribute over each other, like Boolean or switching algebras. Apart from that, I cannot remember of standard cases of multiply-like $$\\ast$$ or $$\\cdot$$ that would distribute. However, I will details sme specific cases, ending with the Hilbert transform and Bedrosian theorem. In the case of multiplication and convolution, or $$a$$ scalar, this is associative, as : $$a.[b(t)\\cdot c(t)] = [a.b(t)]\\cdot c(t)= b(t)\\cdot [a.c(t)]$$ and $$a.[b(t)\\ast c(t)] = [a.b(t)]\\ast c(t)= b(t)\\ast [a.c(t)]$$ You see that you cannot directly\"distribute it\", as this would become \"bilinear\". Convolution and point-wise product are linear. So, if you consider that $$c(t)$$ (or $$b(t)$$ by symmetry), there exists a linear operator $$\\Lambda$$ such that: $$\\Lambda(c(t)) = a(t)\\ast[b(t)\\cdot c(t)]$$ Whether $$\\Lambda$$ has a simple", "comment, the following sentence (found on the linked page) is a bit sloppy. \"Examples of binary operation on $A$ from $A\\times A$ to $A$ include addition $(+)$, subtraction $(-)$, multiplication $(\\times)$ and division $(\\div)$.\" They probably should have said something along the lines of: Addition $(+)$, subtraction $(-)$, multiplication $(\\times)$ and division $(\\div)$ are examples of binary operations (for the appropriate choice of set $A$ in each case). A binary operation on a non-empty set $A$ is a function $f : A\\times A \\to A$, so technically the set $A$ is specified implicitly by $f$; however, the words addition, subtraction, multiplication, and division do not implicitly specify a particular set. - @Adeeb: After you ask a question here, if you get an acceptable answer, you should \"accept\" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, How does accept rate work?. – Michael Albanese Feb 20 '13 at 7:04 @Adeeb: The above comment is just general advice as I see you have not accepted any of the answers to", "of operation can be better explained with an example. Take $n = 151$. Its binary representation is $10010111_2$. This binary representation can be turned into a sum of powers of two: $$\\begin{array}{rcl} 151 & = & 1 \\cdot 2^7 + 0 \\cdot 2^6 + 0 \\cdot 2^5 + 1 \\cdot 2^4 + 0 \\cdot 2^3 + 1 \\cdot 2^2 + 1 \\cdot 2^1 + 1 \\cdot 2^0 \\\\ & = & 2^7 + 2^4 + 2^2 + 2^1 + 2^0 \\end{array}$$ (We have taken each binary digit of $n$ and multiplied it by a power of two.) In view of this, we can write: $$151 \\cdot P = 2^7 P + 2^4 P + 2^2 P + 2^1 P + 2^0 P$$ What the double and add algorithm tells us to do is: • Take $P$. • Double it, so that we get $2P$. • Add $2P$ to $P$ (in order to get the result of $2^1P + 2^0P$). • Double $2P$, so that we get $2^2P$. • Add it to our result (so that we get $2^2P + 2^1P + 2^0P$). • Double $2^2P$ to get $2^3P$. • Don't perform any addition involving $2^3P$. • Double $2^3P$ to get $2^4P$. • Add it to our result (so that we get $2^4P + 2^2P + 2^1P + 2^0P$).", "# q-Powerseries and binary strings Consider the infinite product $(1+qz)(1+qz^2)(1+qz^4)(1+qz^8)\\cdots$, and write this as $\\sum\\limits_{\\nu \\geqslant 0} q^{B_\\nu}z^\\nu$. Relate $B_\\nu$ to the binary expression of $\\nu$. The sequence starts off as: $1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,\\ldots$. The sequence $B_n$ is exactly the sum of 1s appearing in the binary expression of $n$ (one needs $q$ not to be a root of unity in order for $B_n$ to be well-defined)." ]

[ "= a\\ \\diamondsuit\\ ( a\\ \\diamondsuit\\ a) = ( a\\ \\diamondsuit\\ a) \\cdot a = a.$$ Hence, $( a\\ \\diamondsuit\\ b) \\cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\\ \\diamondsuit\\ b) = \\frac{a}{b}.$ Hence, the given equation becomes $\\frac{2016}{\\frac{6}{x}} = 100$. Solving yields $x=\\frac{100}{336} = \\frac{25}{84},$ so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ### Solution 3 One way to eliminate the $\\diamondsuit$ in this equation is to make $a = b$ so that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = c$. In this case, we can make $b = 2016$. $$2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100\\implies (2016\\, \\diamondsuit\\, 6) \\cdot x = 100$$ By multiplying both sides by $\\frac{6}{x}$, we get: $$(2016\\, \\diamondsuit\\, 6) \\cdot 6 = \\frac{600}{x}\\implies 2016 \\, \\diamondsuit\\, (6\\, \\diamondsuit\\, 6) = \\frac{600}{x}$$ Because $6\\, \\diamondsuit\\, 6 = 2016\\, \\diamondsuit\\, 2016 = 1:$ $$2016 \\, \\diamondsuit\\, (2016\\, \\diamondsuit\\, 2016) = \\frac{600}{x}\\implies (2016\\, \\diamondsuit\\, 2016) \\cdot 2016 = \\frac{600}{x}\\implies 2016 = \\frac{600}{x}$$ Therefore, $x = \\frac{600}{2016} = \\frac{25}{84}$, so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ~ pi_is_3.14", "of sets, so the first table above has $a+b+c+abc$ as its left column and $1+ab+ac+ad$ as its second column, the $1$ being the product of members of the empty set. If, further, we represent the whole table as a fraction with the odd column as the numerator and the even column as the denominator, then the operation of adding the new element is this: $$\\frac{\\frac{a+b+c+abc}{1+ab+ac+bc}+d}{1+\\frac{a+b+c+abc}{1+ab+ac+bc}d} = \\frac{a+b+c+d+abc+abd+acd+bcd}{1+ab+ac+ad+bc+bd+cd+abcd}.$$ So we're working with a binary operation: $a\\,\\diamondsuit\\, b=\\dfrac{a+b}{1+ab}$. When the set we start with is empty, the the numerator is the sum of $0$ terms and is $0$ and the denominator is the product of $0$ terms and $\\text{is }1$. QUESTIONS: • Does this algebraic way of looking at this shed light on the combinatorics, or vice-versa? • What is the combinatorial meaning of the fact that the binary operation is associative? • I tried this with the even terms in the numerator and odd in the denominator: $a\\,\\spadesuit\\, b$. We get $(a\\,\\spadesuit\\,b)\\,\\spadesuit\\,c =a\\,\\diamondsuit\\,b\\,\\diamondsuit\\,c$; and $(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_n)\\,\\spadesuit\\, a_{n+1} = 1/(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_{n+1})$ --- So if you apply the $\\spadesuit$ operation instead of $\\diamondsuit$ each time a new element is added, then you alternate between odd-in-the-numerator-and-even-in-the-denominator and even-in-the-numerator-and-odd-in-the-denominator. Does the alternation have some combinatorial significance? • The $\\diamond$ operator sure looks a lot like the formula for $\\tan(a\\pm b)$, which makes things even more", "\\sum_{k=0}^N \\dfrac{(-1)^k \\diamondsuit_k + (-1)^{-k-1} \\diamondsuit_{-k-1}}{2k+1} = \\sum_{k=0}^N \\dfrac{(-1)^k \\left(\\diamondsuit_k - \\diamondsuit_{-k-1} \\right)}{2k+1}$$ Now for $k \\geq 0$ \\begin{align} \\left(\\diamondsuit_k - \\diamondsuit_{-k-1} \\right) & = (-1)^{k+1} \\sum_{j=N-k +1}^{N+k +1} \\dfrac{(-1)^j}{j} - (-1)^{-k} \\sum_{j=N-(k+1) +1}^{N+(k+1) +1} \\dfrac{(-1)^j}{j}\\\\ & = 2 \\cdot (-1)^{k+1} \\cdot \\sum_{j=N-k+1}^{N+ k +1} \\dfrac{(-1)^j}{j} + (-1)^{N+1} \\left( \\dfrac1{N+k+2} + \\dfrac1{N-k}\\right) \\end{align} Hence, $$\\left \\vert \\diamondsuit_k - \\diamondsuit_{-k-1} \\right \\vert = \\mathcal{O} \\left( \\dfrac1N\\right)$$ $$\\left \\vert (\\heartsuit) \\right \\vert \\leq \\sum_{k=0}^N \\dfrac1{2k+1} \\mathcal{O}(1/N) = \\mathcal{O}(\\log(2N+1)/N) \\to 0$$ Hence, $$\\left(\\sum_{k=-N-1}^{k=N} \\dfrac{(-1)^k}{(2k+1)} \\right)^2 - \\sum_{k=-N-1}^{k=N} \\dfrac{1}{(2k+1)(2k+1)} \\to 0$$ Hence, $$\\left(\\sum_{k=-\\infty}^{\\infty} \\dfrac{(-1)^k}{(2k+1)} \\right)^2 = \\sum_{k=-\\infty}^{\\infty} \\dfrac{1}{(2k+1)(2k+1)} = 2 \\sum_{k=0}^{\\infty} \\dfrac{1}{(2k+1)^2} = 2 \\cdot \\dfrac34 \\cdot \\zeta(2)$$ Hence,$$\\boxed{\\zeta(2) = \\dfrac23 \\cdot \\dfrac{\\pi^2}4 = \\dfrac{\\pi^2}6}$$ • This is a great answer! Quite $(\\heartsuit)$ly. – sai Mar 9 '13 at 23:37 The idea that Leibniz's series $$\\frac \\pi 4 = \\sum_{i = 0}^\\infty \\frac{(-1)^i}{2i + 1} \\tag{1}$$ could be used to prove Euler's $$\\zeta(2) = \\frac{\\pi^2}{6} = \\sum_{i=1}^\\infty \\frac{1}{i^2} \\tag{2}$$ seems even more tantalizing when you consider that $(2)$ is equivalent to $$\\frac{\\pi^2}{8} = \\sum_{i=0}^\\infty \\frac{1}{(2i+1)^2} \\tag{3}.$$ The equivalence between $(2)$ and $(3)$ is clear from $$\\frac{3}{4}\\zeta(2) = \\sum_{i=1}^\\infty \\frac{1}{i^2}- \\sum_{i=1}^\\infty \\frac{1}{(2i)^2}= \\sum_{i=0}^\\infty \\frac{1}{(2i+1)^2}.$$ Compare $(1)$ and $(3)$! My goodness.", "way to express $$110$$ as the product of at least $$3$$ numbers is $$2 \\cdot 5 \\cdot 11.$$ This means that $$110n^3$$ has no other prime factors. Then the exponents must be $$1, 4,$$ and $$10$$ from the above formula. Let $$2^3$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$2^9,$$ which makes $$110n^3$$ have a factor of $$2^{10}.$$ Then let $$5$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$5^3,$$ which means that $$110n^3$$ has a factor of $$5^4.$$ We do not have to alter the number of $$11$$s since it already has $$1$$ as its exponent. This means that we can let $$n = 2^3 \\cdot 5.$$ Then $81n^4 = 3^4 \\cdot 2^12 \\cdot 5^4.$ This number has $5 \\cdot 13 \\cdot 5 = 325$ factors. Thus, the correct answer is D. 23. A binary operation $$\\diamondsuit$$ has the properties that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = (a\\,\\diamondsuit \\,b)\\cdot c$ and that $$a\\,\\diamondsuit \\,a=1$$ for all nonzero real numbers $$a, b,$$ and $$c.$$ (Here $$\\cdot$$ represents multiplication). The solution to the equation $2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100$ can be written as $$\\frac{p}{q},$$ where $$p$$ and $$q$$ are relatively prime positive integers. What is $$p+q?$$ $$109$$ $$201$$ $$301$$ $$3049$$ $$33,601$$ ###### Solution(s): Let us see what properties we can gather from the given conditions.", "to 2 days from now, so I could give users a chance. #### Opalg ##### MHB Oldtimer Staff member Without using the square root button on a calculator, determine which side has a larger value: $$\\sqrt{2} \\ + \\sqrt{5} \\ + \\sqrt{11}\\$$ versus $$\\ \\ 7$$ In the inequality $\\sqrt{2} + \\sqrt{5} + \\sqrt{11} \\; \\diamondsuit\\; 7$, you have to decide whether the $\\diamondsuit$ symbol should be < or >. Start by subtracting $\\sqrt2$ from both sides: $\\sqrt{5} + \\sqrt{11} \\ \\diamondsuit\\ 7 - \\sqrt{2}.$ Now square both sides: $16 + 2\\sqrt{55} \\ \\diamondsuit\\ 51 - 14\\sqrt2$. Thus $2\\sqrt{55} + 14\\sqrt2 \\ \\diamondsuit\\ 35$. Now square both sides again: $612 + 56\\sqrt{110} \\ \\diamondsuit\\ 1225$, and therefore $56\\sqrt{110} \\ \\diamondsuit\\ 613$. So far, that has scarcely even needed a calculator. The last step is to square both sides again, and for that you do need the calculator, to get $344960 \\ \\diamondsuit\\ 375769$, from which it is clear that $\\diamondsuit$ has to be <. Last edited: #### CaptainBlack ##### Well-known member So far, that has scarcely even needed a calculator. The last step is to square both sides again, and for that you do need the calculator, to get $344960 \\ \\diamondsuit\\ 375769$, from which it is clear that $\\diamondsuit$ has to be <. It might be convenient", "to show that this relation implies $$(a^2+d^2)^3(b+c)^3+(b^2+c^2)^3(a+d)^3\\geq\\frac{(a+b)^3(a+d)^3(b+c)^3}{8}+\\frac{(c+d)^3(a+d)^3(b+c)^3}{8}$$ which in turn implies $$\\left(\\frac{a^2+d^2}{a+d}\\right)^3+\\left(\\frac{b^2+c^2}{b+c}\\right)^3\\geq \\left(\\frac{a+b}{2}\\right)^3+\\left(\\frac{c+d}{2}\\right)^3,$$ as desired. - Proof by Texas Hold'em :) – user13838 Sep 15 '11 at 15:14 $\\clubsuit$, $\\heartsuit$, $\\spadesuit$, $\\diamondsuit$ are the best symbols in Latex. They combine two of my great passions: maths and Hold'em :) – uforoboa Sep 15 '11 at 15:53 Thank you very much – Lina Sep 16 '11 at 14:17", "as follows: $\\displaystyle \\heartsuit 2=1020 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit 2=2020 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit 2=3020 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit 2=4020$ $\\displaystyle \\heartsuit 3=1030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit 3=2030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit 3=3030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit 3=4030$ $\\displaystyle \\heartsuit 4=1040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit 4=2040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit 4=3040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit 4=4040$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\", "\\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\displaystyle \\heartsuit Q=1120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit Q=2120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit Q=3120 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit Q=4120$ $\\displaystyle \\heartsuit K=1130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit K=2130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit K=3130 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit K=4130$ $\\displaystyle \\heartsuit A=1140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit A=2140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit A=3140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit A=4140$ ___________________________________________________________________________________________________________________ The card numbers need to be encrypted before they can be passed between the two players. Here’s how it works. Both players agree to choose a large prime number $p$. This number $p$ needs to be", "\\ \\ \\ \\ \\ \\ \\ \\clubsuit 2=4020$ $\\displaystyle \\heartsuit 3=1030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit 3=2030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit 3=3030 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit 3=4030$ $\\displaystyle \\heartsuit 4=1040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit 4=2040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit 4=3040 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit 4=4040$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "### 2016 AMC 10B Exam Problems Scroll down and press Start to try the exam! Or, go to the printable PDF, answer key, or solutions. Used with permission of the Mathematical Association of America. Time Left: 1:15:00 1. What is the value of $\\dfrac{2a^{-1}+\\frac{a^{-1}}{2}}{a}$ when $$a= \\tfrac{1}{2}?$$ $$\\ 1$$ $$\\ 2$$ $$\\ \\dfrac{5}{2}$$ $$\\ 10$$ $$\\ 20$$ ###### Solution(s): The expression is equivalent to $2a^{-2}+\\frac{a^{-2}}{2} = 2.5 (a^{-1})^2.$ Then $$a^{-1}$$ is equal to $$\\dfrac{1}{\\frac 12} = 2,$$ so our expression is equal to $$2.5\\cdot 2^2 = 10.$$ Thus, the correct answer is D. 2. If $$n\\heartsuit m=n^3m^2,$$ what is $$\\frac{2\\heartsuit 4}{4\\heartsuit 2}?$$ $$\\ \\dfrac{1}{4}$$ $$\\ \\dfrac{1}{2}$$ $$\\ 1$$ $$\\ 2$$ $$\\ 4$$ ###### Solution(s): For any nonzero $$a,b,$$ we have $\\dfrac{a\\heartsuit b}{b\\heartsuit a} = \\dfrac{a^3b^2}{b^3 a^2} = \\dfrac ab .$ Using $$a=2,b=4,$$ we get that $$\\frac{ 2}{4} = \\frac 12.$$ Thus, the correct answer is B. 3. Let $$x=-2016.$$ What is the value of $\\Bigg\\vert\\Big\\vert |x|-x\\Big\\vert-|x|\\Bigg\\vert-x?$ $$\\ -2016$$ $$\\ 0$$ $$\\ 2016$$ $$\\ 4032$$ $$\\ 6048$$ ###### Solution(s): Observe that: \\begin{align*} &\\Bigg\\vert\\Big\\vert |x|-x\\Big\\vert-|x|\\Bigg\\vert-x\\\\ &=\\Bigg\\vert \\Big\\vert -x-x\\Big\\vert-|x|\\Bigg\\vert-x \\\\ &=\\Bigg\\vert\\Big\\vert -2x\\Big\\vert-|x|\\Bigg\\vert-x \\end{align*} since $$x < 0$$ implies that $$-x = |x|.$$ Substituting values, we can see that $|4032-2016|-(-2016)$$= 2016 +2016$$= 4032.$ Thus, the correct answer is D. 4. Zoey read $$15$$ books, one at a time. The first book took", "\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\displaystyle \\heartsuit Q=1120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit Q=2120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit Q=3120 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit Q=4120$ $\\displaystyle \\heartsuit K=1130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit K=2130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit K=3130 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit K=4130$ $\\displaystyle \\heartsuit A=1140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit A=2140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit A=3140 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit A=4140$ ___________________________________________________________________________________________________________________ The", "& x - \\frac{x+2}{2} \\\\ \\hline \\text{second} & \\frac{x-2}{2} & \\frac{x-2}{4} + 1 & \\frac{x-2}{2} - \\frac{x+2}{4} \\\\ \\hline \\text{third} & \\frac{x-6}{4} & \\frac{x-6}{8} + 1 & \\frac{x-6}{4} - \\frac{x+2}{8} \\\\ \\hline \\text{fourth} & \\frac{x-14}{8} & \\frac{x-14}{16} + 1 & \\frac{x-14}{8} - \\frac{x+2}{16} \\\\ \\hline \\end{array}$$ The value of the last entry of the table must be zero by hypothesis (that is how many diamonds did the fifth thief find!). We can then solve that equation to find the solution: $$\\frac{x-14}{8} = \\frac{x+2}{16}.$$ This gives the amount of diamonds in the original pile. What is the value of $$x$$? ### Backwards method The backwards method starts by realizing that the fourth thief finds only two diamonds in the pile (since “one half of the diamonds plus one” is all the diamonds present in the store). Algebraically, this reads $$x_4 = x_4/2 +1$$, where $$x_4$$ is the amount of diamonds found by the fourth thief. How many diamonds did the third thief find in the store? (let’s label that value $$x_3$$) We know that, after taking half of them plus one, there remained only 2. The only possibility is thus $$x_3 = 6$$. Again, algebraically this is the solution to the equation $$x_3 - (x_3/2 + 1) = 2.$$ Continue in this fashion, finding the values of $$x_2$$ and", "+0 # Help! -1 728 3 +1206 Let $x\\mathbin{\\spadesuit}y = x^2/y$ for all $x$ and $y$ such that $y\\neq 0$. Find all values of $a$ such that $a\\mathbin{\\spadesuit} 3 = 9$. List the values you find in increasing order, separated by commas. Jul 14, 2018 #1 +147 0 Let $$x\\mathbin{\\spadesuit}y = x^2/y$$ for all $$x$$ and $$y$$ such that $$y\\neq 0$$. Find all values of $$a$$ such that $$a\\mathbin{\\spadesuit} 3 = 9$$. List the values you find in increasing order, separated by commas. Since $$a\\mathbin{\\spadesuit} 3 = 9$$, we know that $$\\frac{a^2}{3^2}=9$$. Simplifying this gives us $$\\frac{a^2}{9}=9$$. When we multiply by 9 on both sides, we get $$a^2=81$$. So, $$a=-9,9.$$ . Jul 14, 2018 #2 +1206 0 YOur woring, but thatnks for trying. It's -3\\sqrt{3},3\\sqrt{3}. Lightning Jul 14, 2018 #3 +7683 +1 $$a\\spadesuit 3 = 9\\\\\\dfrac{a^2}{3} = 9\\\\a^2 = 27\\\\a = \\pm\\sqrt{27} = \\pm3\\sqrt{3}$$ Which means a = -3sqrt(3) or a = 3sqrt(3) . Jul 15, 2018 edited by MaxWong Jul 15, 2018", "since then $\\rm\\,1 < [R[z]:R]\\mid [C:R]=p,\\,$ so $\\rm\\,[R[z]:R] = p.\\,$ This is true, e.g. for $\\rm\\,R = \\Bbb R,\\ C = \\Bbb C$. I assume that $C$ wants to be $\\Bbb C$ and $z\\in\\Bbb C\\setminus\\Bbb R$. (Else, if $z\\in R$, then for all $p\\in R[x]$, we have $p(z)\\in R$ whatever ring $R$ is.) Let's assume then that $z=a+bi$ for $a,b\\in\\Bbb R$ and $b\\ne 0$. Then $$g(x):=\\frac1b\\cdot (x-a)$$ and for arbitrary $r+si\\in\\Bbb C$, consider $p(x):=r+s\\cdot g(x)\\ \\in \\Bbb R[x]$, then, as $g(z)=i$, we have $p(z)=r+s\\cdot i$. Note $\\:z\\in\\Bbb C\\setminus\\Bbb R\\:\\Rightarrow\\:\\Bbb R[z] = \\Bbb C\\:$ follows immediately from $\\rm\\,[\\Bbb C:\\Bbb R] = 2\\,$ is prime, see my answer. – Math Gems Mar 6 '13 at 18:30", "zero is zero. lemma (in ring0) Ring_ZF_1_L6: assumes A1: $$x\\in R$$ shows $$0 \\cdot x = 0$$, $$x\\cdot 0 = 0$$proof let $$a = x\\cdot 1$$ let $$b = x\\cdot 0$$ let $$c = 1 \\cdot x$$ let $$d = 0 \\cdot x$$ from A1 have $$a + b = x\\cdot (1 + 0 )$$, $$c + d = (1 + 0 )\\cdot x$$ using Ring_ZF_1_L2, ring_oper_distr moreover have $$x\\cdot (1 + 0 ) = a$$, $$(1 + 0 )\\cdot x = c$$ using Ring_ZF_1_L2, Ring_ZF_1_L3 ultimately have $$a + b = a$$ and T1: $$c + d = c$$ moreover from A1 have $$a \\in R$$, $$b \\in R$$ and T2: $$c \\in R$$, $$d \\in R$$ using Ring_ZF_1_L2, Ring_ZF_1_L4 ultimately have $$b = 0$$ using ring_cancel_add moreover from T2, T1 have $$d = 0$$ using ring_cancel_add ultimately show $$x\\cdot 0 = 0$$, $$0 \\cdot x = 0$$ qed Negative can be pulled out of a product. lemma (in ring0) Ring_ZF_1_L7: assumes A1: $$a\\in R$$, $$b\\in R$$ shows $$( - a)\\cdot b = - (a\\cdot b)$$, $$a\\cdot ( - b) = - (a\\cdot b)$$, $$( - a)\\cdot b = a\\cdot ( - b)$$proof from A1 have I: $$a\\cdot b \\in R$$, $$( - a) \\in R$$, $$(( - a)\\cdot b) \\in R$$, $$( - b) \\in R$$,", "for small $n$, we need some polynomials $B_{n,k}$ in variables $f_j$ and $g_j$ ($j$-th derivative of $f$ and $j$). We do so by applying the $\\diamond$ operator to $a=(f_1,f_2,\\ldots)$ and $b=(g_1,g_2,\\ldots)$. \\begin{array}{rlllll} a^{1\\diamond}&=\\left(f_1,\\right.&f_2,&f_3,&f_4,&\\left.\\ldots\\right)\\\\ b^{1\\diamond}&=\\left(g_1,\\right.&g_2,&g_3,&g_4,&\\left.\\ldots\\right)\\\\ \\\\ a^{2\\diamond}&=\\left(0,\\right.&2f_1^2,&6f_1f_2,&8f_1f_3+6f_2^2,&\\left.\\ldots\\right)\\\\ a^{1\\diamond}\\diamond b^{1\\diamond}&=\\left(0,\\right.&2f_1g_1,&3f_1g_2+3f_2g_1,& 4f_1g_3+6f_2g_2+4f_3g_1,&\\left.\\ldots\\right)\\\\ b^{2\\diamond}&=\\left(0,\\right.&2g_1^2,&6g_1g_2,&8g_1g_3+6g_2^2,&\\left.\\ldots\\right)\\\\ \\\\ a^{3\\diamond}&=\\left(0,\\right.&0,&6f_1^3,&36f_1^2f_2,&\\left.\\ldots\\right)\\\\ a^{2\\diamond}\\diamond b^{1\\diamond}&=\\left(0,\\right.&0,&6f_1^2g_1,&12f_1^2g_2+24f_1f_2g_1,&\\left.\\ldots\\right)\\\\ a^{1\\diamond}\\diamond b^{2\\diamond}&=\\left(0,\\right.&0,&6f_1g_1^2,&24f_1g_1g_2+12f_2g_1^2,&\\left.\\ldots\\right)\\\\ b^{3\\diamond}&=\\left(0,\\right.&0,&6g_1^3,&36g_1^2g_2,&\\left.\\ldots\\right)\\\\ \\end{array} Case $n=2$: Each of the three sums of the identity is calculated separately. \\begin{align*} \\sum_{k=1}^{2}&B_{2,k}^{f\\cdot(\\ln \\circ g)}\\\\ &=B_{2,1}^{f\\cdot(\\ln \\circ g)}+B_{2,2}^{f\\cdot(\\ln \\circ g)}\\\\ &=\\frac{d^2}{{dx}^2}\\left(f (\\ln \\circ g)\\right)+\\left(\\frac{d}{dx}\\left(f(\\ln \\circ g)\\right)\\right)^2\\\\ &=\\left(f_2(\\ln \\circ g)+2f_1\\frac{g_1}{g}+f\\frac{g_2}{g}-f\\frac{g_1^2}{g^2}\\right)+\\left(f_1(\\ln \\circ g)+f\\frac{g_1}{g}\\right)^2\\\\ &=\\left(f_2(\\ln \\circ g)+2f_1\\frac{g_1}{g}+f\\frac{g_2}{g}-f\\frac{g_1^2}{g^2}\\right)+\\left(f_1^2(\\ln \\circ g)^2+2ff_1(\\ln \\circ g)\\frac{g_1}{g}+f^2\\frac{g_1^2}{g^2}\\right)\\\\ \\\\ \\sum_{k=1}^{2}&(\\ln \\circ g)^kB_{2,k}^f\\\\ &=(\\ln \\circ g)B_{2,1}^f+(\\ln \\circ g)^2B_{2,2}^f\\\\ &=(\\ln \\circ g)f_2+(\\ln \\circ g)^2f_1^2\\\\ \\\\ \\sum_{k=1}^{2}&\\sum_{m=0}^{2-k}\\sum_{j=0}^m\\binom{m}{j} \\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!}\\\\ &=\\sum_{k=1}^{2}\\frac{1}{g^k}\\sum_{m=0}^{2-k}\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!} \\sum_{j=0}^m\\binom{m}{j}\\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\\\ &=\\frac{1}{g}\\sum_{m=0}^1\\frac{\\left(a^{m_\\diamond}\\diamond b^{1_{\\diamond}}\\right)_2}{m!1!} \\sum_{j=0}^m\\binom{m}{j}(\\ln \\circ g)^{m-j}\\frac{d^j}{d(f)^j}(f)\\\\ \\sum_{j=0}^m\\binom{m}{j}(\\ln \\circ g)^{m-j}\\frac{d^j}{d(f)^j}\\left(f(f-1)\\right)\\\\ &=\\frac{1}{g}\\left[\\frac{\\left(b^{1_\\diamond}\\right)_2}{1!}\\binom{0}{0}f+\\frac{\\left(a^{1_\\diamond}\\diamond b^{1_\\diamond}\\right)_2}{1!1!}\\left(\\binom{1}{0}(\\ln \\circ g)f+\\binom{1}{1}\\frac{d}{d(f)}f\\right)\\right]\\\\ &=\\frac{1}{g}\\left[g_2f+2f_1g_1\\left((\\ln \\circ g) f+1\\right)\\right]+\\frac{1}{g^2}\\left[g_1^2f\\left(f-1\\right)\\right] \\end{align*} Comparison of the results of these sums shows the validity of the claim: \\begin{align*} \\sum_{k=1}^{2}B_{2,k}^{f\\cdot(\\ln \\circ g)}&=\\sum_{k=1}^{2}(\\ln \\circ g)^kB_{2,k}^f\\\\ \\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!}\\\\ \\end{align*} Case $n=3$: Each of the three sums of the identity is calculated separately. \\begin{align*} \\sum_{k=1}^{3}&B_{3,k}^{f\\cdot(\\ln \\circ g)}\\\\ &=B_{3,1}^{f\\cdot(\\ln \\circ g)}+B_{3,2}^{f\\cdot(\\ln \\circ g)}+B_{3,3}^{f\\cdot(\\ln \\circ g)}\\\\ &=\\frac{d^3}{{dx}^3}\\left(f(\\ln \\circ g)\\right)+3\\frac{d}{dx}\\left(f(\\ln \\circ g)\\right)\\frac{d^2}{{dx}^2}\\left(f(\\ln \\circ g)\\right) +\\left(\\frac{d}{dx}(\\ln \\circ g)\\right)^3\\\\ &=\\left(f_3(\\ln \\circ g)+3f_2\\frac{g_1}{g}+3f_1\\frac{g_2}{g}-3f_1\\frac{g_1^2}{g^2}+f\\frac{g_3}{g} +2f\\frac{g_1^3}{g^3}-3f\\frac{g_1g_2}{g^2}\\right)\\\\ &=\\left(f_3(\\ln \\circ g)+3f_2\\frac{g_1}{g}+3f_1\\frac{g_2}{g}-3f_1\\frac{g_1^2}{g^2}+f\\frac{g_3}{g} +2f\\frac{g_1^3}{g^3}-3f\\frac{g_1g_2}{g^2}\\right)\\\\ &\\qquad+\\left(3f_1f_2(\\ln \\circ g)^2+3ff_2(\\ln \\circ g)\\frac{g_1}{g}+6f_1^2(\\ln \\circ g)\\frac{g_1}{g}+6ff_1\\frac{g_1^2}{g^2}\\right.\\\\ &\\qquad+\\left(f_1^3(\\ln \\circ g)^3+3ff_1^2(\\ln \\circ g)^2\\frac{g_1}{g}+3f^2f_1(\\ln \\circ g)\\frac{g_1^2}{g^2}+f^3\\frac{g_1^3}{g^3}\\right) \\\\ \\\\ \\sum_{k=1}^{3}&(\\ln \\circ g)^kB_{3,k}^f\\\\ &=(\\ln \\circ g)B_{3,1}^f+(\\ln", "drawn from Bag 1 and a red is returned to Bag 1. . . [2] A black is drawn from Bag 1 and a black is returned to Bag 1. [1] A red is drawn from Bag 1: . $P(R_1) = \\frac{3}{8}$ . . .Bag 2 now contains:7 red and 2 black discs: . . .A red is drawn from Bag 2: . $P(R_2) \\:=\\:\\frac{7}{9}$ . . . . . $P(R_1R_2) \\:=\\:\\frac{3}{8}\\cdot \\frac{7}{9}$ [2] A black is drawn from Bag 1: $P(B_1) = \\frac{5}{8}$ . . .Bag 2 now contains: 6 red and 3 black discs. . . .A black is drawn from Bag 2: . $P(B_2) = \\frac{3}{9}$ . . . . . $P(B_1B_2) \\:=\\:\\frac{5}{8}\\cdot\\frac{3}{9}$ Therefore: . $P(R_1R_2\\text{ or }R_1B_2) \\;=\\;\\frac{3}{8}\\cdot\\frac{7}{9} + \\frac{5}{8}\\cdot\\frac{3}{9} \\;=\\;\\frac{7}{24} + \\frac{5}{24} \\;=\\;\\frac{12}{24} \\;=\\;\\dfrac{1}{2}$", "diamonds, but still all had en equal share. Her Prerogative Thoria changed her mind. She withdrew her suggestion of dividing by family. She was cool with the way things are now. Final Count How many diamonds did each girl end up with, and how many girls shared in the division? Another solution: Let $$G$$ be the number of girls in the village and $$d$$ be the number of Diamonds. Then $$(G-2)$$ is the number of families, Equation we have: $$\\frac d{G-2}-\\frac dG=5$$ So $$2d=5G(G-2)$$ As $$2|5G(G-2)$$, so $$2|G$$, so let $$G=2G'$$. $$\\begin{split}2d&=20G'(G'-1)\\\\d&=20\\cdot\\frac{G'(G'-1)}2\\end{split}$$ As $$\\frac{G'(G'-1)}2$$ is an integer, $$20|d$$. Checking values of $$d$$, we get $$(G,d)=(4,20),(6,60)$$ But after a girl left the share, the share is still equal, so $$(G,d)=(6,60)$$ is the valid solution. In conclusion There are $$6-1=5$$ girls in the share with $$\\frac{60}5=12$$ diamonds per girl. • @Kwan, and Strack - it's solutions like yours that you both posted that motivates me to spend my free time posting puzzles for my kids to study and learn. Screen shots are taken and studied and discussed in group, and among fellow programming students as well. Thank you for contributing not only the answer, but a lesson as well. Apr 10, 2020 at 14:32 There are: a total of 60 diamonds, and in the end 5 girls shared them, each", "\\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\displaystyle \\heartsuit Q=1120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit Q=2120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit Q=3120 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit Q=4120$ $\\displaystyle \\heartsuit K=1130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit K=2130 \\ \\ \\ \\ \\ \\ \\", "\\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\cdots$ $\\displaystyle \\heartsuit Q=1120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit Q=2120 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\spadesuit Q=3120 \\ \\ \\ \\ \\ \\ \\ \\ \\clubsuit Q=4120$ $\\displaystyle \\heartsuit K=1130 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\diamondsuit K=2130 \\ \\ \\ \\ \\ \\ \\" ]

[ "= a\\ \\diamondsuit\\ ( a\\ \\diamondsuit\\ a) = ( a\\ \\diamondsuit\\ a) \\cdot a = a.$$ Hence, $( a\\ \\diamondsuit\\ b) \\cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\\ \\diamondsuit\\ b) = \\frac{a}{b}.$ Hence, the given equation becomes $\\frac{2016}{\\frac{6}{x}} = 100$. Solving yields $x=\\frac{100}{336} = \\frac{25}{84},$ so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ### Solution 3 One way to eliminate the $\\diamondsuit$ in this equation is to make $a = b$ so that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = c$. In this case, we can make $b = 2016$. $$2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100\\implies (2016\\, \\diamondsuit\\, 6) \\cdot x = 100$$ By multiplying both sides by $\\frac{6}{x}$, we get: $$(2016\\, \\diamondsuit\\, 6) \\cdot 6 = \\frac{600}{x}\\implies 2016 \\, \\diamondsuit\\, (6\\, \\diamondsuit\\, 6) = \\frac{600}{x}$$ Because $6\\, \\diamondsuit\\, 6 = 2016\\, \\diamondsuit\\, 2016 = 1:$ $$2016 \\, \\diamondsuit\\, (2016\\, \\diamondsuit\\, 2016) = \\frac{600}{x}\\implies (2016\\, \\diamondsuit\\, 2016) \\cdot 2016 = \\frac{600}{x}\\implies 2016 = \\frac{600}{x}$$ Therefore, $x = \\frac{600}{2016} = \\frac{25}{84}$, so the answer is $25 + 84 = \\boxed{\\textbf{(A) }109.}$ ~ pi_is_3.14", "PolynomialAndEquation FunctionProperty AMC10/12 2016 Problem - 2887 A binary operation $\\diamondsuit$ has the properties that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = (a\\,\\diamondsuit \\,b)\\cdot c$ and that $a\\,\\diamondsuit \\,a=1$ for all nonzero real numbers $a, b,$ and $c$. (Here $\\cdot$ represents multiplication). The solution to the equation $2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100$ can be written as $\\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q?$", "way to express $$110$$ as the product of at least $$3$$ numbers is $$2 \\cdot 5 \\cdot 11.$$ This means that $$110n^3$$ has no other prime factors. Then the exponents must be $$1, 4,$$ and $$10$$ from the above formula. Let $$2^3$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$2^9,$$ which makes $$110n^3$$ have a factor of $$2^{10}.$$ Then let $$5$$ be a factor of $$n.$$ Then $$n^3$$ has a factor of $$5^3,$$ which means that $$110n^3$$ has a factor of $$5^4.$$ We do not have to alter the number of $$11$$s since it already has $$1$$ as its exponent. This means that we can let $$n = 2^3 \\cdot 5.$$ Then $81n^4 = 3^4 \\cdot 2^12 \\cdot 5^4.$ This number has $5 \\cdot 13 \\cdot 5 = 325$ factors. Thus, the correct answer is D. 23. A binary operation $$\\diamondsuit$$ has the properties that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = (a\\,\\diamondsuit \\,b)\\cdot c$ and that $$a\\,\\diamondsuit \\,a=1$$ for all nonzero real numbers $$a, b,$$ and $$c.$$ (Here $$\\cdot$$ represents multiplication). The solution to the equation $2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100$ can be written as $$\\frac{p}{q},$$ where $$p$$ and $$q$$ are relatively prime positive integers. What is $$p+q?$$ $$109$$ $$201$$ $$301$$ $$3049$$ $$33,601$$ ###### Solution(s): Let us see what properties we can gather from the given conditions.", "of sets, so the first table above has $a+b+c+abc$ as its left column and $1+ab+ac+ad$ as its second column, the $1$ being the product of members of the empty set. If, further, we represent the whole table as a fraction with the odd column as the numerator and the even column as the denominator, then the operation of adding the new element is this: $$\\frac{\\frac{a+b+c+abc}{1+ab+ac+bc}+d}{1+\\frac{a+b+c+abc}{1+ab+ac+bc}d} = \\frac{a+b+c+d+abc+abd+acd+bcd}{1+ab+ac+ad+bc+bd+cd+abcd}.$$ So we're working with a binary operation: $a\\,\\diamondsuit\\, b=\\dfrac{a+b}{1+ab}$. When the set we start with is empty, the the numerator is the sum of $0$ terms and is $0$ and the denominator is the product of $0$ terms and $\\text{is }1$. QUESTIONS: • Does this algebraic way of looking at this shed light on the combinatorics, or vice-versa? • What is the combinatorial meaning of the fact that the binary operation is associative? • I tried this with the even terms in the numerator and odd in the denominator: $a\\,\\spadesuit\\, b$. We get $(a\\,\\spadesuit\\,b)\\,\\spadesuit\\,c =a\\,\\diamondsuit\\,b\\,\\diamondsuit\\,c$; and $(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_n)\\,\\spadesuit\\, a_{n+1} = 1/(a_1\\,\\diamondsuit\\,\\cdots\\,\\diamondsuit\\,a_{n+1})$ --- So if you apply the $\\spadesuit$ operation instead of $\\diamondsuit$ each time a new element is added, then you alternate between odd-in-the-numerator-and-even-in-the-denominator and even-in-the-numerator-and-odd-in-the-denominator. Does the alternation have some combinatorial significance? • The $\\diamond$ operator sure looks a lot like the formula for $\\tan(a\\pm b)$, which makes things even more", "### 2016 AMC 10B Exam Problems Scroll down and press Start to try the exam! Or, go to the printable PDF, answer key, or solutions. Used with permission of the Mathematical Association of America. Time Left: 1:15:00 1. What is the value of $\\dfrac{2a^{-1}+\\frac{a^{-1}}{2}}{a}$ when $$a= \\tfrac{1}{2}?$$ $$\\ 1$$ $$\\ 2$$ $$\\ \\dfrac{5}{2}$$ $$\\ 10$$ $$\\ 20$$ ###### Solution(s): The expression is equivalent to $2a^{-2}+\\frac{a^{-2}}{2} = 2.5 (a^{-1})^2.$ Then $$a^{-1}$$ is equal to $$\\dfrac{1}{\\frac 12} = 2,$$ so our expression is equal to $$2.5\\cdot 2^2 = 10.$$ Thus, the correct answer is D. 2. If $$n\\heartsuit m=n^3m^2,$$ what is $$\\frac{2\\heartsuit 4}{4\\heartsuit 2}?$$ $$\\ \\dfrac{1}{4}$$ $$\\ \\dfrac{1}{2}$$ $$\\ 1$$ $$\\ 2$$ $$\\ 4$$ ###### Solution(s): For any nonzero $$a,b,$$ we have $\\dfrac{a\\heartsuit b}{b\\heartsuit a} = \\dfrac{a^3b^2}{b^3 a^2} = \\dfrac ab .$ Using $$a=2,b=4,$$ we get that $$\\frac{ 2}{4} = \\frac 12.$$ Thus, the correct answer is B. 3. Let $$x=-2016.$$ What is the value of $\\Bigg\\vert\\Big\\vert |x|-x\\Big\\vert-|x|\\Bigg\\vert-x?$ $$\\ -2016$$ $$\\ 0$$ $$\\ 2016$$ $$\\ 4032$$ $$\\ 6048$$ ###### Solution(s): Observe that: \\begin{align*} &\\Bigg\\vert\\Big\\vert |x|-x\\Big\\vert-|x|\\Bigg\\vert-x\\\\ &=\\Bigg\\vert \\Big\\vert -x-x\\Big\\vert-|x|\\Bigg\\vert-x \\\\ &=\\Bigg\\vert\\Big\\vert -2x\\Big\\vert-|x|\\Bigg\\vert-x \\end{align*} since $$x < 0$$ implies that $$-x = |x|.$$ Substituting values, we can see that $|4032-2016|-(-2016)$$= 2016 +2016$$= 4032.$ Thus, the correct answer is D. 4. Zoey read $$15$$ books, one at a time. The first book took", "b$$ or $$(a \\bigcirc b) \\bigcirc b = a \\bigcirc (b \\bigcirc b)$$ • flexibility: $$(a \\bigcirc b) \\bigcirc a = a \\bigcirc (b \\bigcirc a)$$ Distributivity of a binary operator $$\\diamond$$ over operator $$\\bigcirc$$ is more involved: it can be left-distributed: $$a \\diamond (b \\bigcirc c) = (a \\diamond b) \\bigcirc (a \\diamond c)$$ or right-distributed: $$(a \\bigcirc b) \\diamond c = (a \\diamond c) \\bigcirc (b\\diamond c)$$ Division is right-distributive over add, not left-distributive. Distributivity of multiply over add is at play in rings and fields. But the converse does not apply. There are mathematical structures where two binary operators distribute over each other, like Boolean or switching algebras. Apart from that, I cannot remember of standard cases of multiply-like $$\\ast$$ or $$\\cdot$$ that would distribute. However, I will details sme specific cases, ending with the Hilbert transform and Bedrosian theorem. In the case of multiplication and convolution, or $$a$$ scalar, this is associative, as : $$a.[b(t)\\cdot c(t)] = [a.b(t)]\\cdot c(t)= b(t)\\cdot [a.c(t)]$$ and $$a.[b(t)\\ast c(t)] = [a.b(t)]\\ast c(t)= b(t)\\ast [a.c(t)]$$ You see that you cannot directly\"distribute it\", as this would become \"bilinear\". Convolution and point-wise product are linear. So, if you consider that $$c(t)$$ (or $$b(t)$$ by symmetry), there exists a linear operator $$\\Lambda$$ such that: $$\\Lambda(c(t)) = a(t)\\ast[b(t)\\cdot c(t)]$$ Whether $$\\Lambda$$ has a simple", "for small $n$, we need some polynomials $B_{n,k}$ in variables $f_j$ and $g_j$ ($j$-th derivative of $f$ and $j$). We do so by applying the $\\diamond$ operator to $a=(f_1,f_2,\\ldots)$ and $b=(g_1,g_2,\\ldots)$. \\begin{array}{rlllll} a^{1\\diamond}&=\\left(f_1,\\right.&f_2,&f_3,&f_4,&\\left.\\ldots\\right)\\\\ b^{1\\diamond}&=\\left(g_1,\\right.&g_2,&g_3,&g_4,&\\left.\\ldots\\right)\\\\ \\\\ a^{2\\diamond}&=\\left(0,\\right.&2f_1^2,&6f_1f_2,&8f_1f_3+6f_2^2,&\\left.\\ldots\\right)\\\\ a^{1\\diamond}\\diamond b^{1\\diamond}&=\\left(0,\\right.&2f_1g_1,&3f_1g_2+3f_2g_1,& 4f_1g_3+6f_2g_2+4f_3g_1,&\\left.\\ldots\\right)\\\\ b^{2\\diamond}&=\\left(0,\\right.&2g_1^2,&6g_1g_2,&8g_1g_3+6g_2^2,&\\left.\\ldots\\right)\\\\ \\\\ a^{3\\diamond}&=\\left(0,\\right.&0,&6f_1^3,&36f_1^2f_2,&\\left.\\ldots\\right)\\\\ a^{2\\diamond}\\diamond b^{1\\diamond}&=\\left(0,\\right.&0,&6f_1^2g_1,&12f_1^2g_2+24f_1f_2g_1,&\\left.\\ldots\\right)\\\\ a^{1\\diamond}\\diamond b^{2\\diamond}&=\\left(0,\\right.&0,&6f_1g_1^2,&24f_1g_1g_2+12f_2g_1^2,&\\left.\\ldots\\right)\\\\ b^{3\\diamond}&=\\left(0,\\right.&0,&6g_1^3,&36g_1^2g_2,&\\left.\\ldots\\right)\\\\ \\end{array} Case $n=2$: Each of the three sums of the identity is calculated separately. \\begin{align*} \\sum_{k=1}^{2}&B_{2,k}^{f\\cdot(\\ln \\circ g)}\\\\ &=B_{2,1}^{f\\cdot(\\ln \\circ g)}+B_{2,2}^{f\\cdot(\\ln \\circ g)}\\\\ &=\\frac{d^2}{{dx}^2}\\left(f (\\ln \\circ g)\\right)+\\left(\\frac{d}{dx}\\left(f(\\ln \\circ g)\\right)\\right)^2\\\\ &=\\left(f_2(\\ln \\circ g)+2f_1\\frac{g_1}{g}+f\\frac{g_2}{g}-f\\frac{g_1^2}{g^2}\\right)+\\left(f_1(\\ln \\circ g)+f\\frac{g_1}{g}\\right)^2\\\\ &=\\left(f_2(\\ln \\circ g)+2f_1\\frac{g_1}{g}+f\\frac{g_2}{g}-f\\frac{g_1^2}{g^2}\\right)+\\left(f_1^2(\\ln \\circ g)^2+2ff_1(\\ln \\circ g)\\frac{g_1}{g}+f^2\\frac{g_1^2}{g^2}\\right)\\\\ \\\\ \\sum_{k=1}^{2}&(\\ln \\circ g)^kB_{2,k}^f\\\\ &=(\\ln \\circ g)B_{2,1}^f+(\\ln \\circ g)^2B_{2,2}^f\\\\ &=(\\ln \\circ g)f_2+(\\ln \\circ g)^2f_1^2\\\\ \\\\ \\sum_{k=1}^{2}&\\sum_{m=0}^{2-k}\\sum_{j=0}^m\\binom{m}{j} \\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!}\\\\ &=\\sum_{k=1}^{2}\\frac{1}{g^k}\\sum_{m=0}^{2-k}\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!} \\sum_{j=0}^m\\binom{m}{j}\\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\\\ &=\\frac{1}{g}\\sum_{m=0}^1\\frac{\\left(a^{m_\\diamond}\\diamond b^{1_{\\diamond}}\\right)_2}{m!1!} \\sum_{j=0}^m\\binom{m}{j}(\\ln \\circ g)^{m-j}\\frac{d^j}{d(f)^j}(f)\\\\ \\sum_{j=0}^m\\binom{m}{j}(\\ln \\circ g)^{m-j}\\frac{d^j}{d(f)^j}\\left(f(f-1)\\right)\\\\ &=\\frac{1}{g}\\left[\\frac{\\left(b^{1_\\diamond}\\right)_2}{1!}\\binom{0}{0}f+\\frac{\\left(a^{1_\\diamond}\\diamond b^{1_\\diamond}\\right)_2}{1!1!}\\left(\\binom{1}{0}(\\ln \\circ g)f+\\binom{1}{1}\\frac{d}{d(f)}f\\right)\\right]\\\\ &=\\frac{1}{g}\\left[g_2f+2f_1g_1\\left((\\ln \\circ g) f+1\\right)\\right]+\\frac{1}{g^2}\\left[g_1^2f\\left(f-1\\right)\\right] \\end{align*} Comparison of the results of these sums shows the validity of the claim: \\begin{align*} \\sum_{k=1}^{2}B_{2,k}^{f\\cdot(\\ln \\circ g)}&=\\sum_{k=1}^{2}(\\ln \\circ g)^kB_{2,k}^f\\\\ \\frac{(\\ln \\circ g)^{m-j}}{g^k}\\frac{d^j}{d(f)^j}[\\left(f\\right)_k]\\frac{\\left(a^{m_\\diamond}\\diamond b^{k_{\\diamond}}\\right)_2}{m!k!}\\\\ \\end{align*} Case $n=3$: Each of the three sums of the identity is calculated separately. \\begin{align*} \\sum_{k=1}^{3}&B_{3,k}^{f\\cdot(\\ln \\circ g)}\\\\ &=B_{3,1}^{f\\cdot(\\ln \\circ g)}+B_{3,2}^{f\\cdot(\\ln \\circ g)}+B_{3,3}^{f\\cdot(\\ln \\circ g)}\\\\ &=\\frac{d^3}{{dx}^3}\\left(f(\\ln \\circ g)\\right)+3\\frac{d}{dx}\\left(f(\\ln \\circ g)\\right)\\frac{d^2}{{dx}^2}\\left(f(\\ln \\circ g)\\right) +\\left(\\frac{d}{dx}(\\ln \\circ g)\\right)^3\\\\ &=\\left(f_3(\\ln \\circ g)+3f_2\\frac{g_1}{g}+3f_1\\frac{g_2}{g}-3f_1\\frac{g_1^2}{g^2}+f\\frac{g_3}{g} +2f\\frac{g_1^3}{g^3}-3f\\frac{g_1g_2}{g^2}\\right)\\\\ &=\\left(f_3(\\ln \\circ g)+3f_2\\frac{g_1}{g}+3f_1\\frac{g_2}{g}-3f_1\\frac{g_1^2}{g^2}+f\\frac{g_3}{g} +2f\\frac{g_1^3}{g^3}-3f\\frac{g_1g_2}{g^2}\\right)\\\\ &\\qquad+\\left(3f_1f_2(\\ln \\circ g)^2+3ff_2(\\ln \\circ g)\\frac{g_1}{g}+6f_1^2(\\ln \\circ g)\\frac{g_1}{g}+6ff_1\\frac{g_1^2}{g^2}\\right.\\\\ &\\qquad+\\left(f_1^3(\\ln \\circ g)^3+3ff_1^2(\\ln \\circ g)^2\\frac{g_1}{g}+3f^2f_1(\\ln \\circ g)\\frac{g_1^2}{g^2}+f^3\\frac{g_1^3}{g^3}\\right) \\\\ \\\\ \\sum_{k=1}^{3}&(\\ln \\circ g)^kB_{3,k}^f\\\\ &=(\\ln \\circ g)B_{3,1}^f+(\\ln", "= (g^{\\diamond 4}(x) + 1)$ $g^{\\diamond 1}(x) = (g^{\\diamond 5}(x) + 1)$ $g^{\\diamond 1}(x) = (g^{\\diamond 7}(x) + 1)$ $g^{\\diamond 1}(x) = (g^{\\diamond 1729}(x) + 1)$ ... till n approaching infinity !!! simultaneous ! it seems f and g are not functions , the diamond is not a function nor operator nor composition and none of them are finite !?!? i prefer different definitions and equations. once again your first post contains mistakes. all the rest that follows is even more paradoxal. no offense , but i prefer you look at it first ... mathematics requires thinking. at least twice. regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 11/16/2011, 12:11 AM (This post was last modified: 11/16/2011, 12:22 AM by JmsNxn.) you made a very major mistake... uhh $f^{\\diamond - n}(f^{\\diamond n + 1}(x)) \\neq f^{\\diamond 1}(x)$ this is exactly the assumption I was describing doesn't work. $(f^{\\diamond n} \\circ f^{\\diamond m})(x) \\neq f^{\\diamond m +n}(x)$ instead, we'd need to create a new operator $(f^{\\diamond n} \\diamond f^{\\diamond m})(x) = f^{\\diamond m + n}(x)$ the diamond operator is NOT composition. And the diamond superscript is an index of superfunctions. $f^{\\diamond n}(x) \\neq (f_1 \\circ f_2 \\circ f_3 ... f_n)(x) = f^{\\circ n}(x)$ instead, it's defined by $f(x) = f^{\\diamond 1}(g^{\\diamond 1}(x) + 1)$", "+0 # Help! -1 728 3 +1206 Let $x\\mathbin{\\spadesuit}y = x^2/y$ for all $x$ and $y$ such that $y\\neq 0$. Find all values of $a$ such that $a\\mathbin{\\spadesuit} 3 = 9$. List the values you find in increasing order, separated by commas. Jul 14, 2018 #1 +147 0 Let $$x\\mathbin{\\spadesuit}y = x^2/y$$ for all $$x$$ and $$y$$ such that $$y\\neq 0$$. Find all values of $$a$$ such that $$a\\mathbin{\\spadesuit} 3 = 9$$. List the values you find in increasing order, separated by commas. Since $$a\\mathbin{\\spadesuit} 3 = 9$$, we know that $$\\frac{a^2}{3^2}=9$$. Simplifying this gives us $$\\frac{a^2}{9}=9$$. When we multiply by 9 on both sides, we get $$a^2=81$$. So, $$a=-9,9.$$ . Jul 14, 2018 #2 +1206 0 YOur woring, but thatnks for trying. It's -3\\sqrt{3},3\\sqrt{3}. Lightning Jul 14, 2018 #3 +7683 +1 $$a\\spadesuit 3 = 9\\\\\\dfrac{a^2}{3} = 9\\\\a^2 = 27\\\\a = \\pm\\sqrt{27} = \\pm3\\sqrt{3}$$ Which means a = -3sqrt(3) or a = 3sqrt(3) . Jul 15, 2018 edited by MaxWong Jul 15, 2018", "Wrote: now I've found a way to generalize this to what I am labeling \"meta-superfunctions\". These have to do with iteration but in a very different complicated light. consider the definition: $f^{\\diamond n}(g^{\\diamond n}(x)) = g^{\\diamond n}(f^{\\diamond n}(x)) = x$ $f^{\\diamond n}(x) = f^{\\diamond n+1}(g^{\\diamond n+1}(x) + 1)$ NO I DONT CONSIDER THE DEFINITION : the very first equation seems wrong from first glance already ( and im eating chocolate ! ) take the last equation take $f^{\\diamond -n}$ on both sides => $x = f^{\\diamond 1}(g^{\\diamond n+1}(x) + 1)$ ( name this equation 3) this gives the same result as if we had used $g^{\\diamond n}$ on both sides. hence together with the first equation we conclude f and g are inverse under the diamond operator. also now consider equation 3 and keep in mind all 3 equations and f and g are inverses then : take $g^{\\diamond 1}$ on both sides : => $g^{\\diamond 1}(x) = (g^{\\diamond n+1}(x) + 1)$ for all n !!!! this looks really bad. for small n we get ( n = 0 ) $g^{\\diamond 1}(x) = (g^{\\diamond 1}(x) + 1)$ which seems paradoxal , and actually , it is. for larger n we get e.g. $g^{\\diamond 1}(x) = (g^{\\diamond 2}(x) + 1)$ $g^{\\diamond 1}(x) = (g^{\\diamond 3}(x) + 1)$ $g^{\\diamond 1}(x)", "(1998), 1253-1283. • There is also a follow-up paper with improved asymptotic running time and simpler treatment: D. J. Bernstein, H. W. Lenstra Jr. and J. Pila, Detecting perfect powers by factoring into coprimes, Math. Comp. 76 (2007), 385–388. – Erick Wong Jan 15 '14 at 2:56 I found an interesting and elegant solution in the paper: On the implementation of AKS class primality test, by R.Crandall and J.Papadopoulos, 18 Mar 2003. Somehow, I can show that the binary search algorithm is $O(lg~n \\cdot (lg~lg~n)^2)$. Firstly, $a^b = n$, there is $b<lg~n$. Binary Search Algorithm: For each $b$, we use binary search to find $a$. Each time the computation of $a^b$ cost $lg~b = lg~lg~n$ operations by using fast exponentiation. Therefore, the remaining issue is the range of $a$. If $A$ is the maximal possible value of $a$, then binary search needs $lg~A$ operations Note that $b~lg~a = lg~n$, that is $$lg~A = \\frac{lg~n}{b}$$ When summing up, $$\\sum lg~A = lg~n \\cdot (\\frac{1}{1} + \\frac{1}{2} + ... + \\frac{1}{B}) = lg~n \\cdot lg~B = lg~n \\cdot lg~lg~n$$ In other words, all the operations for binary search is $O(lg~n \\cdot lg~lg~n)$ Consider the operation of $a^b$, it is $O(lg~n \\cdot (lg~lg~n)^2)$ finally. ps: All the lg are base 2. Python code: #--- a^n --------------------------------------- def fast_exponentation(a, n): ans =", "many linked operations which, while mainly done for the purpose of doing so, are perhaps not too overly frivolous. Here is one: Given a finite set$S$with$k$elements, there are$k^{k^2}$binary operations and each operation$\\diamond$can be represented by its Cayley Table the$k \\times k$square with$i,j$entry$i \\diamond j$(assuming a known order). This single operation defines a quasigroup quasi-group if the table is a Latin Square, each symbol appears once in each row and column. Equivalently (and without an agreed order), for every$a,b$there are unique$x,y$with$x \\diamond a=b$and$a \\diamond y=b.$This defines two operations$y=a \\backslash b$and$x=a x=b / b$a$ with $$b=a \\diamond (a \\backslash b)=a \\backslash (a \\diamond b)=(b /a) \\diamond a=(b \\diamond a) / a.$$ This equational definition of a quasi-group via 3 operations is available for infinite quasi-groups as well. If we introduce the projections $k$ is a prime power (a\\pi_1b=a$and perhaps only$a\\pi_2b=b$then ) it is possible we can write the previous equations as $$a\\pi_2b=(a\\pi_1b)\\diamond(a\\backslash b)=\\cdots$$ Two Latin Squares (given by quasigroups$(S,\\cdot)$and$(S,\\diamond)$) are orthogonal if the map$(x,y) \\to have (x\\cdot y,x\\diamond y)$is a bijection from$k-1$S \\times S$ to itself. This It is not known how many pairwise orthogonal quasi-groups (aka Mutually Orthogonal Latin Squares in that any two give all MOLS) are possible on a set of size $k^2$ k.$Certainly no more than$k-1$and this is possible ordered pairs when combined position by position(and", "\\in \\mathcal F_a$, we do actually get $f_{a,b}(S) \\in \\mathcal F_b$. – Misha Lavrov Apr 2 '17 at 6:04 • Please check my work if it's correct or not. Thanks. – carat Apr 2 '17 at 7:43 • To prove $f$ is surjective, you shouldn't be choosing $l$; that's fixed when you define $f$. \"Choose $a_1, a_2, \\dots, a_k$ such that ....\" would technically make that sentence correct, but not a very good explanation: it's better to say: \"Choose $a_1 = b_1 - l(a-b) \\bmod p$\" and so on, then explain that this makes $f(\\{a_1,\\dots,a_k\\}) = \\{b_1, \\dots, b_k\\}$. – Misha Lavrov Apr 2 '17 at 14:40 Using the Polya Enumeration Theorem we deploy the unlabled set operator $\\mathfrak{P}$ and get for the generating function $$Z(P_k)(z+z^2+\\cdots+z^p)$$ for a closed form of $$\\frac{1}{p} \\sum_{r=0}^{p-1} \\left. Z(P_k)(z+z^2+\\cdots+z^p) \\right|_{z=\\zeta_p^r}$$ where $\\zeta_p = \\exp(2\\pi i/p).$ The exponential formula tells us the OGF of the cycle index for the set operator $Z(P_k)$ which is $$Z(P_k) = [w^k] \\exp\\left(\\sum_{l\\ge 1} (-1)^{l-1} a_l \\frac{w^l}{l}\\right)$$ Doing the substitution we find $$[w^k] \\frac{1}{p} \\sum_{r=0}^{p-1} \\left. \\exp\\left(\\sum_{l\\ge 1} (-1)^{l-1} \\left(\\sum_{q=1}^p z^{ql} \\right) \\frac{w^l}{l}\\right) \\right|_{z=\\zeta_p^r} \\\\ = [w^k] \\frac{1}{p} \\sum_{r=0}^{p-1} \\exp\\left(\\sum_{l\\ge 1} (-1)^{l-1} \\left(\\sum_{q=1}^p (\\zeta_p^{rl})^{q} \\right) \\frac{w^l}{l}\\right).$$ With $p$ prime the innermost sum of powers of $\\zeta_p$ is $p$ if $p|rl$ and zero otherwise. We thus have two contributions,", "\\frac{PA}{PB} = \\frac{AC}{AB}$$. So we replace in $$(\\diamondsuit)$$ the fraction $$AC:AB$$ by $$PA:PB$$. We can finally show $$(\\diamond)$$: $$\\frac{AC}{AB}\\cdot \\frac{B'C}{BC}\\cdot \\frac{\\sin y}{\\sin x} = \\frac{PA}{PB}\\cdot \\frac{1}{\\sin \\widehat{B'BC}}\\cdot \\frac{PO\\cdot OB\\;\\sin y}{PO\\cdot OA\\;\\sin x} = \\frac{PA}{PB}\\cdot \\frac{1}{\\sin \\widehat{B'BC}}\\cdot \\frac {\\operatorname{Area}(POB)} {\\operatorname{Area}(POA)} \\\\ = \\frac{PA}{PB}\\cdot \\frac{1}{\\sin \\widehat{B'BC}}\\cdot \\frac {PB\\cdot BO\\;\\sin\\widehat{PBO}} {PA\\cdot AO\\;\\sin\\widehat{PAO}} = \\frac {\\sin \\widehat{PBO}} {\\sin \\widehat{B'BC}} =1\\ .$$ $$\\square$$ This concludes the proof. Corollary: This concludes the problem in the OP, since $$\\widehat{BAE} = \\widehat{BAY} = \\widehat{AYZ} = \\widehat{ACZ} = \\widehat{ACD} \\ .$$ $$\\square$$ Unfortunately i could not make the proof more structural. My hope was that it should be a bridge using results like Pascal's theorem and / or configurations of the shape Pappus, Desargue... As bonus, here are some colinearities that can be established using Pascal's theorem. The problem in the above tempting picture is that at the place where the point $$D$$ is placed, there are in fact three points, • the point $$D:=AB'\\cap BX$$, • and the point $$D_1:=CZ\\cap BX$$, • and the point $$D_2:=CZ\\cap AB'$$. And we need the equality of two of them. To see colinearities from the picture, apply Pascal's theorem for instance for... • $$AABXA'B'$$, giving the colinearity of $$D^*$$, $$\\infty$$, $$D$$, as mentioned above, • $$XYZCAB$$, giving the colinearity of $$F:=XY\\cap CA$$, $$\\infty$$, $$D_1$$, • $$AA'XBB'Y$$, giving the colinearity of", "then $s_a = s_b = \\frac{a+b}{2} = 1008$. Then, applying the moves $$m_1 = (1, 2015), m_2 = (2, 2014), \\dots, m_{1007} = (1007, 1009),$$ we get $s_1 = s_2 = \\cdots = s_{2015} = 1008$. Otherwise, suppose $a+b\\ne 2016$. Then consider the following $3$ moves: $$m_{-1} = (2016-a, 2016-b), m_{-2} = (a, 2016-a), m_{-3} = (b, 2016-b)$$ We have $m_{-1}$ makes $s_{2016-a} = s_{2016-b} = 2016 - \\frac{a+b}{2}$. So, $m_{-2}$ makes $$s_a = s_{2016-a} = \\frac{2016 - \\frac{a+b}{2} + \\frac{a+b}{2}}{2} = 1008$$ Similarly for $m_{-3}$ with $s_b$ and $s_{2016-b}$. Then, finishing up with the moves $$m_1 = (1, 2015), m_2 = (2, 2014), \\dots, m_{1007} = (1007, 1009),$$ we get $s_1 = s_2 = \\cdots = s_{2015} = 1008$. (95% Sure) Doesn't this solution fail for Case $2$ when $a$ or $b$ is $1008$? (Because $2016-1008 = 1008$ so $m_{-1}$ is basically non-existent.) - ike.chen ## Solution 3 (INCORRECT) Let the set be ${a_1,a_2,\\dots ,a_{2015}}$, where all the terms are nonnegative. Note that the sum of all the terms in this sequence will always be the same after any amount of moves. To prove this, let $i,j$ be integers with $1\\le i, and we have $a_i+a_j = \\frac{a_i+a_j}{2}+\\frac{a_i+a_j}{2}$. Also, $a_ia_j \\le (\\frac{a_i+a_j}{2})(\\frac{a_i+a_j}{2})$ by AM-GM, so the product of all the terms will not decrease after any number", "a\\cdot b$$, $$a \\neq a\\cdot b$$ using assms , PositiveSet_def , OrdRing_ZF_1_L16 , OrdRing_ZF_1_L3 , Ring_ZF_1_L12C If the right hand side of an inequality is positive we can multiply it by a number that is greater than one. lemma (in ring1) OrdRing_ZF_3_L17: assumes A1: $$\\text{HasNoZeroDivs}(R,A,M)$$ and A2: $$b\\in R_+$$ and A3: $$a\\leq b$$ and A4: $$1 \\lt c$$ shows $$a \\lt b\\cdot c$$proof from A1, A2, A4 have $$b \\lt b\\cdot c$$ using OrdRing_ZF_3_L16 with A3 show $$a \\lt b\\cdot c$$ by (rule ring_strict_ord_transit ) qed We can multiply a right hand side of an inequality between positive numbers by a number that is greater than one. lemma (in ring1) OrdRing_ZF_3_L18: assumes A1: $$\\text{HasNoZeroDivs}(R,A,M)$$ and A2: $$a \\in R_+$$ and A3: $$a\\leq b$$ and A4: $$1 \\lt c$$ shows $$a \\lt b\\cdot c$$proof from A2, A3 have $$b \\in R_+$$ using OrdRing_ZF_3_L7 with A1, A3, A4 show $$a \\lt b\\cdot c$$ using OrdRing_ZF_3_L17 qed In ordered rings with no zero divisors if at least one of $$a,b$$ is not zero, then $$0 < a^2+b^2$$, in particular $$a^2+b^2\\neq 0$$. lemma (in ring1) OrdRing_ZF_3_L19: assumes A1: $$\\text{HasNoZeroDivs}(R,A,M)$$ and A2: $$a\\in R$$, $$b\\in R$$ and A3: $$a \\neq 0 \\vee b \\neq 0$$ shows $$0 \\lt a^2 + b^2$$proof { assume $$a \\neq 0$$ with A1, A2 have $$0 \\leq a^2$$,", "diamonds, but still all had en equal share. Her Prerogative Thoria changed her mind. She withdrew her suggestion of dividing by family. She was cool with the way things are now. Final Count How many diamonds did each girl end up with, and how many girls shared in the division? Another solution: Let $$G$$ be the number of girls in the village and $$d$$ be the number of Diamonds. Then $$(G-2)$$ is the number of families, Equation we have: $$\\frac d{G-2}-\\frac dG=5$$ So $$2d=5G(G-2)$$ As $$2|5G(G-2)$$, so $$2|G$$, so let $$G=2G'$$. $$\\begin{split}2d&=20G'(G'-1)\\\\d&=20\\cdot\\frac{G'(G'-1)}2\\end{split}$$ As $$\\frac{G'(G'-1)}2$$ is an integer, $$20|d$$. Checking values of $$d$$, we get $$(G,d)=(4,20),(6,60)$$ But after a girl left the share, the share is still equal, so $$(G,d)=(6,60)$$ is the valid solution. In conclusion There are $$6-1=5$$ girls in the share with $$\\frac{60}5=12$$ diamonds per girl. • @Kwan, and Strack - it's solutions like yours that you both posted that motivates me to spend my free time posting puzzles for my kids to study and learn. Screen shots are taken and studied and discussed in group, and among fellow programming students as well. Thank you for contributing not only the answer, but a lesson as well. Apr 10, 2020 at 14:32 There are: a total of 60 diamonds, and in the end 5 girls shared them, each", "propose to decrease this cost by derandomizing'' the composition: instead of feeding into $f \\diamond g$ independent inputs $x_{1},\\ldots,x_{m}$, we generate $x_{1},\\ldots,x_{m}$ using a shorter seed. We show that this idea can be realized in the particular setting of the composition of functions and universal relations. To this end, we provide two different techniques for achieving such a derandomization: a technique based on averaging samplers, and a technique based on Reed-Solomon codes. Changes to previous version: Fixed a few typos. ### Paper: TR17-146 | 1st October 2017 14:58 #### On Derandomized Composition of Boolean Functions TR17-146 Authors: Or Meir Publication: 1st October 2017 19:04 Keywords: Abstract: The composition of two Boolean functions $f:\\left\\{0,1\\right\\}^{m}\\to\\left\\{0,1\\right\\}$, $g:\\left\\{0,1\\right\\}^{n}\\to\\left\\{0,1\\right\\}$ is the function $f \\diamond g$ that takes as inputs $m$ strings $x_{1},\\ldots,x_{m}\\in\\left\\{0,1\\right\\}^{n}$ and computes $(f \\diamond g)(x_{1},\\ldots,x_{m})=f\\left(g(x_{1}),\\ldots,g(x_{m})\\right).$ This operation has been used several times for amplifying different hardness measures of $f$ and $g$. This comes at a cost: the function $f \\diamond g$ has input length $m\\cdot n$ rather than $m$ or $n$, which is a bottleneck for some applications. In this paper, we propose to decrease this cost by derandomizing'' the composition: instead of feeding into $f \\diamond g$ independent inputs $x_{1},\\ldots,x_{m}$, we generate $x_{1},\\ldots,x_{m}$ using a shorter seed. We show that this idea can be realized in the particular setting of", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}", "zero is zero. lemma (in ring0) Ring_ZF_1_L6: assumes A1: $$x\\in R$$ shows $$0 \\cdot x = 0$$, $$x\\cdot 0 = 0$$proof let $$a = x\\cdot 1$$ let $$b = x\\cdot 0$$ let $$c = 1 \\cdot x$$ let $$d = 0 \\cdot x$$ from A1 have $$a + b = x\\cdot (1 + 0 )$$, $$c + d = (1 + 0 )\\cdot x$$ using Ring_ZF_1_L2, ring_oper_distr moreover have $$x\\cdot (1 + 0 ) = a$$, $$(1 + 0 )\\cdot x = c$$ using Ring_ZF_1_L2, Ring_ZF_1_L3 ultimately have $$a + b = a$$ and T1: $$c + d = c$$ moreover from A1 have $$a \\in R$$, $$b \\in R$$ and T2: $$c \\in R$$, $$d \\in R$$ using Ring_ZF_1_L2, Ring_ZF_1_L4 ultimately have $$b = 0$$ using ring_cancel_add moreover from T2, T1 have $$d = 0$$ using ring_cancel_add ultimately show $$x\\cdot 0 = 0$$, $$0 \\cdot x = 0$$ qed Negative can be pulled out of a product. lemma (in ring0) Ring_ZF_1_L7: assumes A1: $$a\\in R$$, $$b\\in R$$ shows $$( - a)\\cdot b = - (a\\cdot b)$$, $$a\\cdot ( - b) = - (a\\cdot b)$$, $$( - a)\\cdot b = a\\cdot ( - b)$$proof from A1 have I: $$a\\cdot b \\in R$$, $$( - a) \\in R$$, $$(( - a)\\cdot b) \\in R$$, $$( - b) \\in R$$," ]

[ "# That's got to be a fibonacci number. find the no.of ordered pair of integers $$(a,b)$$ such that - $$a^{2} + b^{2} = 1126478964579$$ also note they can be negative. ×", "That's got to be a fibonacci number. find the no.of ordered pair of integers $(a,b)$ such that - $a^{2} + b^{2} = 1126478964579$ also note they can be negative. × Problem Loading... Note Loading... Set Loading...", "# That's got to be a fibonacci number. Number Theory Level 4 find the no.of ordered pair of integers $$(a,b)$$ such that - $$a^{2} + b^{2} = 1126478964579$$ also note they can be negative. ×", "# Fibonacci Decimals Algebra Level 4 $\\displaystyle \\frac{1}{10^0} + \\frac{1}{10^1} + \\frac{2}{10^2} + \\frac{3}{10^3} + \\frac{5}{10^4} + \\frac{8}{10^5} + \\frac{13}{10^6} + \\cdots$ If the value of the sum above can be expressed as $$\\dfrac mn$$ for positive coprime integers $$m$$ and $$n$$, find $$\\dfrac{n + 1}{\\sqrt{m}}$$. ×", "# Sreejato's ordered pairs For positive integers $$x$$, let $$f(x)$$ be the number of ordered pairs of positive integers $$(a, b)$$ such that $\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{x}.$ Find the smallest possible value of $$N$$ such that $$N$$ is a prime power and $$f(N)= 13$$. This problem is posed by Sreejato B. Details and assumptions A prime power is of the form $$p^n$$, where $$p$$ is a prime number and $$n$$ is a positive integer. ×", "\\sum_{k\\ge 0} z^{2k} \\frac{1}{(1-z)^{k+1}} = \\frac{1}{1-z}\\frac{1}{1-z^2/(1-z)} = \\frac{1}{1-z-z^2}.$$ This is immediately seen to be the OGF of $F_{n+1}$ ($n+1$-th Fibonacci number) and we are done. The same trick was used here at this MSE link. -", "Fibonacci numbers – gcd (Part 1) Find all ordered pairs $(m, \\; n)$ of positive integers for which there’s an integer $x$ satisfying the equation: $gcd \\,(F_m, \\, F_n) \\, x^2 \\; - \\; (F_m + F_n) \\, x \\; + \\; F_{gcd \\,(m, \\,n)} \\; = \\; 0$ Next blog: The gcd of any two Fibonacci numbers is also a Fibonacci number", "# Fibonacci is Life Discrete Mathematics Level 5 There are $$S$$ positive integers $$n$$ satisfying the following: • The digits of $$n$$ are composed of 1,3 and 4. • The sum of the digits of $$n$$ is 42. Find $$\\sqrt {S}$$. ×", "# Fib Let $$f_n$$ be the $$n$$th Fibonacci number such that $$f_1=f_2=1$$ and $$f_k=f_{k-1}+f_{k-2}$$ for $$k \\ge 3$$. Is it possible that $f_x+f_{x+1}+f_{x+2}+f_{x+3}+f_{x+4}+f_{x+5}+f_{x+6}+f_{x+7}=f_{y}$ where $$x$$ and $$y$$ are positive integers? ×", "# Pretty cool Fibonacci Let $${ F }_{ n }$$ be the $$n$$th Fibonacci number. Find the first four digits of the number of digits of $${ F }_{ p }$$, where $$p={ 10 }^{ 12 }$$. Details: $${ F }_{ n }={ F }_{ n-1 }+{ F }_{ n-2 }$$ and $${ F }_{ 0 }=0$$ and $${ F }_{ 1 }=1$$. ×", "# If $m$ and $n$ are positive integers, then $(F_m,F_n)=F_{(m,n)}$. Edit: The $F$'s are Fibonacci numbers. I need an idea on how to show the following: If $m$ and $n$ are positive integers, then $(F_m,F_n)=F_{(m,n)}$. I believe that using the fact that $F_{m+n}=F_mF_{n+1}+F_nF_{m-1}$ could come in handy. Moreover, Euclid's algorithm may as well be needed. But I am not certain, as there may be better methods to achieve this. What is $f_m$, $f_n$? – Daan Michiels Apr 25 '12 at 18:16 Hint. $F_{kn}$ is divisible by $F_n$ – Arturo Magidin Apr 25 '12 at 18:22 Josué: The proof is induction on $n+m$, so this is inductive hypothesis you can assume. – sdcvvc Apr 25 '12 at 18:40", "2], {i, 35}], {j, 35}]; Notice the smallest positive integer $a$ for which there is a smaller number $b$ so that the Euclidean algorithm applied to $a$ and $b$ stops in $n$ steps is the $(n+2)$nd Fibonacci number. Further, the corresponding $b$ is the $(n+1)$st Fibonacci number.", "# Minimizing Fibb Author: mathforces Problem has been solved: 120 times Русский язык | English Language Suppose that $F_1=F_2=1$ и $F_{n+2}=F_{n+1}+F_n$, for all positive integer $n$. Let $g(n)$ be the least possible value of $| a_1F_1 + a_2 F_2+....+a_nF_n |$, where $a_i=\\pm 1$, for all $i=1,2,...,n$. What is the value of $g(1)+g(2)+...+g(2020)$?", "F_{n + 1})$ for which $F_{n + 1} < N$. • For each pair, find integers $s$ and $t$ such that $sF_n + tF_{n + 1} = 1$ using the algorithm from Bézout’s Lemma above. • Let $k = \\lceil\\frac{sN}{F_{n + 1}}\\rceil - 1$. • Let $T_1 = sN - F_{n + 1} k$ and $T_2 = tN + F_n k$. If $T_1$ and $T_2$ are positive, then store these values, along with $n$. • Output the values of $T_1$ and $T_2$ that correspond to the largest value of $n$ for which these values were positive. This is the solution which we submitted. It was accepted, and ran in time, but we can make a few improvements. When calculating the integers $s$ and $t$ such that $sF_{n}+tF_{n+1}=1$, for example, it is not necessary for the recursion to run to the base case. If $(s, t)$ are the required integers for $(F_{n - 1}, F_n)$ (which we have already computed), then the integers for the pair $(F_n, F_{n + 1})$ are $(t - s, s)$. It is also possible to find the required integers $s$ and $t$ explicitly. Cassini’s Identity tells us that $F_{n + 1}F_{n - 1} - F_n^2 = (-1)^n$, so that one possible solution is $s = -F_n$ and $t = F_{n - 1}$ if $n$", "Fibonacci Power! How many values of $$k$$ are there such that $$F_k$$ is of the form $$2^n$$? Note that both $$k$$ and $$n$$ are non-negative integers. Also $$F_0=0,F_1=1$$ and $$F_m=F_{m-1}+F_{m-2}$$ × Problem Loading... Note Loading... Set Loading...", "# IMO 1993 b2 proof \"Let $$\\mathbb{N}=\\{ 1,2,3,\\ldots \\}$$. Determine if there exists a strictly increasing function $$f:\\mathbb{N}\\to\\mathbb{N}$$ such that 1) $$f(1)=2$$ 2) $$f(f(n))=f(n)+n$$ for all $$n$$.\" Of the solutions I've seen online, one such $$f(n)$$ that should work is $$f(n)=\\left[\\phi n + \\frac{1}{2}\\right]$$ where $$\\phi\\approx 1.618$$ is the golden ratio and the square brackets denote the integral part. It is easy to see that $$f(1)=2$$ and that $$f(n)$$ is strictly increasing - however the proof that this satisfies condition 2 seems to be incorrect or unclear in the sources I have found. A clear proof that this $$f(n)$$ satisfies 2) would be appreciated. • I think if you use condition 2 repeatedly, you will find that for the n'th fibonacci number $F_n$, our $f$ must give $f(F_n)=F_{n+1}$. – Merk Zockerborg Apr 28 at 15:41 • Can you show the proofs that you claim are incorrect or unclear? Perhaps they can be clarified. – kccu Apr 28 at 15:49 Claim: $$f(n)=\\left[\\phi n + \\frac{1}{2}\\right]$$ satisfies $$f(f(n))=f(n)+n$$. Now $$f(f(n))-f(n)-n=\\left[\\phi\\left[\\phi n+\\frac{1}{2}\\right] +\\frac{1}{2}\\right]-\\left[\\phi n +\\frac{1}{2}\\right]-n$$, so it suffices to show $$X(n)=\\phi\\left[\\phi n + \\frac{1}{2}\\right]+\\frac{1}{2}-\\left[\\phi n +\\frac{1}{2}\\right]-n=(\\phi-1)\\left[\\phi n +\\frac{1}{2}\\right]+\\frac{1}{2}-n$$ satisfies $$0\\leq X(n) < 1$$. For the lower bound, $$\\left[\\phi n +\\frac{1}{2}\\right]\\geq \\phi n -\\frac{1}{2}$$, giving $$X(n)\\geq \\phi^2n-\\frac{\\phi}{2}-\\phi n+1-n.$$ But $$\\phi=\\frac{1+\\sqrt{5}}{2}$$, so $$n(\\phi^2-\\phi-1)=0$$. Hence $$X(n)\\geq 1-\\frac{\\phi}{2}>0.$$ For the upper bound, $$\\left[\\phi n +\\frac{1}{2}\\right]", "the Fibonacci numbers and the Lucas numbers. Define the Fibonacci number $F_n$ and the Lucas number $L_n$ as $F_1 = F_2 = 1$ and, for $n > 2$, $F_{n} = F_{n-1} + F_{n-2}$ and $L_1 = L$, $L_2=3$ and, for $n > 2$, $L_{n} = L_{n-1} + L_{n-2}$ respectively. Then, $\\exp (L_n+\\frac{L_{2n}x^2}{2}+\\frac{L_{3n}x^3}{3} + ...) =\\frac{1}{F_n}(F_n+F_{2n}x+F_{3n}x^2 + ...)$ Gem 3. Maximum value of a monic polynomial Ramachandra and Balasubramanian (and possibly Chebyshev) If $f(x)$ is a polynomial having integral coefficients and the coefficient of the highest order term is 1 (monic polynomial) then, on the interval $-2 \\le x \\le 2$, $f(x)$ must attain a magnitude of at least 2: i.e. $\\max_{-2 \\le x \\le 2}|f(x)| \\ge 2$.", "# Does this really have a closed form? Algebra Level 4 $\\large 22 \\sum_{n=0}^\\infty \\dfrac{F_n}{4^n}$ Let $$F_n$$ denote the $$n^\\text{th}$$ Fibonacci number, where $$F_0 = 0, F_1 = 1$$ and $$F_n = F_{n-1} + F_{n-2}$$ for $$n=2,3,4,\\ldots$$. Compute the expression above. ×", "# Fibonacci is Life There are $$S$$ positive integers $$n$$ satisfying the following: • The digits of $$n$$ are composed of 1,3 and 4. • The sum of the digits of $$n$$ is 42. Find $$\\sqrt {S}$$. ×", "and $f_{44}$. Exercise. Write out the first 50 Fibonacci numbers. Exercise. Show that $f_{n+3}+f_n=2f_{n+2}$ whenever $n$ is a positive integer. Exercise. Show that $f_{n+3}-f_n=2f_{n+1}$ whenever $n$ is a positive integer. Exercise. Show that $f_{n-2}+f_{n+2}=3f_n$ whenever $n$ is a positive integer with $n\\geq 2.$ Exercise. Show that $f_{2n}=f_{n}^2+2f_{n-1}f_n$ whenever $n$ is a positive integer with $n\\geq 2.$ Exercise. Show that $f_{2n+1}=f_{n+1}^2+f_{n}^2$ whenever $n$ is a positive integer. Exercise. Show that $f_{2n}=f_{n+1}^2-f_{n-1}^2$ whenever $n$ is a positive integer. Exercise. Show that for all positive integers $n,$ $$f_1f_2 + \\left(f_2f_3+f_3f_4\\right)\\cdots +\\left(f_{2n-2}f_{2n-1}+f_{2n-1}f_{2n}\\right)=\\left(f_{2n}\\right)^2.$$ Exercise. Show that $f_{n+2}^2-f_{n+1}^2=f_n f_{n+3},$ whenever $n$ is a positive integer greater than 1. ## More Exercises on the Fibonacci Numbers You may also attempt to solve the following exercises. Note that we have also introduced Lucas numbers in this section, so feel free to look it up. Exercise. Show that $f_{n+1}f_n-f_{n-1}f_{n-2}=f_{2n-1}$ whenever $n$ is a positive integer greater than 2. Exercise. Show that $f_1 f_2+f_2f_3+\\cdots +f_{2n-1}f_{2n}=f_{2n}^2$ whenever $n$ is a positive integer. Exercise. Prove that $f_{m+n}=f_m f_{n+1}+f_n f_{m-1}$ whenever $n$ is a positive integer. Exercise. Prove that $f_n > \\left(\\frac{5}{3}\\right)^{n-1}$, for all $n\\geq 2$. Exercise. Show that for $n\\geq 2$, $$f_n=\\frac{f_{n-1}+\\sqrt{5 f_{n-1}^2+4(-1)^n}}{2}.$$ Notice that this formula gives $f_n$ in terms of one predecessor rather than two predecessors. Exercise. Let $F=\\begin{bmatrix}1 & 1 \\\\ 1 & 0\\end{bmatrix}$. Show that $$F^n=\\begin{bmatrix}f_{n+1}" ]

[ "\\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac{-3x^{5}} {x^{5}} &= -3 \\cdot \\frac{ x^{5}} {x^{5}} \\\\ &= -3 \\cdot x^{5-5} \\\\ &= -3 \\cdot x^{0} \\\\ &= -3 \\cdot 1 \\\\ &= -3 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac{(j^{2}k)^{4}} {(j^{2}k) \\cdot (j^{2}k)^{3}} &= \\frac{(j^{2}k)^{4}} {(j^{2}k)^{1+3}} \\\\ &= \\frac{(j^{2}k)^{4}} {(j^{2}k)^{4}} \\\\ &= (j^{2}k)^{4-4} \\\\ &= (j^{2}k)^{0} \\\\ &= 1 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac {5(rs^{2})^{2}} {(rs^{2})^{2}} &= 5(rs^{2})^{2-2} \\\\ &= 5(rs^{2})^{0} \\\\ &= 5 \\cdot 1 \\\\ &= 5 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} ## 1.2.5 Using the Negative Rule of Exponents Another useful result occurs if we relax the condition that $m>n$ in the quotient rule even further. For example, can we simplify $\\frac { h^{3}} {h^{5}}$? When $m<n$—that is, where the difference $m-n$ is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, $\\frac { h^{3}} {h^{5}}$. \\begin {align*} \\frac { h^{3}} {h^{5}} &= \\frac {h \\cdot h \\cdot h} {h \\cdot h \\cdot h \\cdot h \\cdot h} \\\\ &= \\frac { \\cancel{h} \\cdot", "- 3 \\cdot 5^2)} {(3^2 - 2^2)} = \\frac{(2 \\cdot 16 - 3 \\cdot 25)} {(9 - 4)} = \\frac{(32 - 75)} {5} = \\frac{-43} {5}\\end{align*} (c) \\begin{align*} \\left (\\frac{3^3} {2^2}\\right )^{-2} \\cdot \\frac{3} {4} = \\left (\\frac{2^2} {3^3}\\right )^2 \\cdot \\frac{3} {4} = \\frac{2^4} {3^6} \\cdot \\frac{3} {4} = \\frac{2^4} {3^6} \\cdot \\frac{3} {2^2} = \\frac{2^2} {3^5} = \\frac{4} {243}\\end{align*} Example 8 Evaluate the following expressions for \\begin{align*}x=2, y=-1, z=3\\end{align*}. (a) \\begin{align*} 2x^2 - 3y^3 + 4z\\end{align*} (b) \\begin{align*} (x^2 - y^2)^2\\end{align*} (c) \\begin{align*} \\left (\\frac{3x^2 y^5} {4z}\\right )^{-2}\\end{align*} Solution (a) \\begin{align*}2x^2 -3y^3 +4z=2 \\cdot 2^2 -3\\cdot (-1)^3 +4 \\cdot 3=2\\cdot 4-3\\cdot (-1)+4\\cdot 3=8+3+12=23\\end{align*} (b) \\begin{align*}(x^2 - y^2)^2=(2^2-(-1)^2)^2=(4-1)^2=3^2=9\\end{align*} (c) \\begin{align*} \\left (\\frac{3x^2 - y^5} {4z}\\right )^{-2} = \\left (\\frac{3 \\cdot 2^2 \\cdot (-1)^5} {4 \\cdot 3}\\right )^{-2} = \\left (\\frac{3 \\cdot 4 \\cdot (-1)} {12}\\right )^{-2} = \\left (\\frac{-12} {12}\\right )^{-2} = \\left (\\frac{-1} {1}\\right )^{-2} = \\left (\\frac{1} {-1}\\right )^2 = (-1)^2 = 1\\end{align*} ## Review Questions Simplify the following expressions, be sure that there aren't any negative exponents in the answer. 1. \\begin{align*}x^{-1} \\cdot y^2\\end{align*} 2. \\begin{align*}x^{-4}\\end{align*} 3. \\begin{align*} \\frac{x^{-3}} {x^{-7}}\\end{align*} 4. \\begin{align*} \\frac{x^{-3} y^{-5}} {z^{-7}}\\end{align*} 5. \\begin{align*}\\left ( x^{\\frac{1}{2}}y^{-\\frac{2}{3}} \\right ) \\left ( x^2 y^{\\frac{1}{3}} \\right )\\end{align*} 6. \\begin{align*}\\left (\\frac{a} {b}\\right )^{-2}\\end{align*} 7. \\begin{align*}(3a^{-2} b^2 c^3)^3\\end{align*} 8. \\begin{align*} x^{-3} \\cdot x^3\\end{align*} Simplify the following", "\\end{aligned} \\\\ f_{(1,0,1)} &= \\frac{1}{8} \\begin{aligned}[t] \\bigl[&f_{(1,0,1)}-f_{(1,0,1)}\\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\\cdot{(-1)} \\\\ &-f_{(1,0,1)}\\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\\cdot{(-1)}+f_{(1,0,1)}\\bigr] = f_{(1,0,1)} \\end{aligned} \\\\ f_{(1,1,0)} &= \\frac{1}{8} \\begin{aligned}[t] \\bigl[&f_{(1,1,0)}-f_{(1,1,0)}\\cdot{(-1)}-f_{(1,1,0)}\\cdot{(-1)}+f_{(1,1,0)} \\\\ &+f_{(1,1,0)} -f_{(1,1,0)}\\cdot{(-1)}-f_{(1,1,0)}\\cdot{(-1)}+f_{(1,1,0)}\\bigr] = f_{(1,1,0)} \\end{aligned} \\\\ f_{(1,1,1)} &= \\frac{1}{8} \\begin{aligned}[t] \\bigl[&f_{(1,1,1)}-f_{(1,1,1)}\\cdot{(-1)}-f_{(1,1,1)}\\cdot{(-1)}+f_{(1,1,1)} \\\\ & -f_{(1,1,1)}\\cdot{(-1)} +f_{(1,1,1)}+f_{(1,1,1)}-f_{(1,1,1)}\\cdot{(-1)}\\bigr] = f_{(1,1,1)} \\end{aligned} \\end{align*} \\end{document} • I cannot thank you enough for the solution provided. +1 – Maths Apr 1 at 11:45 • Now If I may ask, how can I apply a similar trick to the tables in the question? – Maths Apr 1 at 11:47 • Please post another question instead of highjacking, and note that tabulars does not belong in math mode, use array instead. – daleif Apr 1 at 11:56", "5\\cdot ( \\frac{4}{35}+\\frac{15}{35} - f_{\\le50} ) &=& 2\\cdot ( \\frac{16}{35}+f_{\\le50} ) \\\\ 5\\cdot ( \\frac{19}{35} - f_{\\le50} ) &=& 2\\cdot ( \\frac{16}{35}+f_{\\le50} ) \\\\ \\frac{95}{35} - 5\\cdot f_{\\le50} &=& \\frac{32}{35}+2\\cdot f_{\\le50} \\\\ 7\\cdot f_{\\le50} &=& \\frac{95}{35} - \\frac{32}{35} \\\\ 7\\cdot f_{\\le50} &=& \\frac{63}{35} \\\\ f_{\\le50} &=& \\frac{63}{35\\cdot 7 } \\\\ \\text{female}_{\\le50} &=& \\frac{9}{35} \\end{array}$$ $$\\begin{array}{rcl} \\hline \\text{continuation}\\\\ \\hline \\\\ \\frac{ \\text{female}_{\\le50}}{f+m} &=& \\frac{9}{35} \\\\ \\text{female}_{\\le50} &=& \\frac{9}{35}\\cdot (f+m) \\quad | \\quad :f \\\\ \\frac{ \\text{female}_{\\le50}}{f} &=& \\frac{9}{35}\\cdot (1+\\frac{m}{f}) \\\\\\\\ \\boxed{~ \\begin{array}{rcl} \\frac{m}{m+f} &=& \\frac{4}{7} \\\\ \\frac{f}{m+f} &=& 1- \\frac{4}{7} = \\frac{3}{7}\\\\ \\frac{ \\frac{m}{m+f} } { \\frac{f}{m+f} } &=& \\frac{ \\frac{4}{7} } { \\frac{3}{7} }\\\\ \\frac{ m } { f } &=& \\frac43 \\end{array} ~}\\\\\\\\ \\frac{ \\text{female}_{\\le50}}{f} &=& \\frac{9}{35}\\cdot (1+ \\frac43) \\\\ \\frac{ \\text{female}_{\\le50}}{f} &=& \\frac{9}{35}\\cdot (\\frac73) \\\\ \\frac{ \\text{female}_{\\le50}}{f} &=& \\frac{9\\cdot 7}{35\\cdot 3} \\\\ \\frac{ \\text{female}_{\\le50}}{f} &=& \\frac{3}{5} \\\\ \\end{array}$$ heureka Jan 8, 2016 #23 +2493 0 i think guest s answer is best Solveit Jan 8, 2016 ### 32 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "sizes is \\begin{align}\\label{eq:diff} s_{i+1} - s_i &= \\frac{i+1}{100}\\cdot\\prod_{j=0}^{i} \\frac{100 - j}{100} - \\frac{i}{100}\\cdot\\prod_{j=0}^{i-1} \\frac{100 - j}{100} \\\\ &= \\frac{i+1}{100}\\cdot\\frac{100 - i}{100}\\cdot\\prod_{j=0}^{i-1} \\frac{100 - j}{100} - \\frac{i}{100}\\cdot\\prod_{j=0}^{i-1} \\frac{100 - j}{100} \\\\ &= \\left(\\frac{i+1}{100}\\cdot\\frac{100 - i}{100} - \\frac{i}{100}\\right)\\prod_{j=0}^{i-1} \\frac{100 - j}{100} \\end{align} For the first few guests, sizes are increasing (you can check this easily from the table above) but at some point this trend must be reversed. We are searching for the guest with the largest piece of pie, so we will look for the first instance of $$s_{i+1} - s_i$$ being negative. Because $$\\prod_{j=0}^{i-1} \\frac{100 - j}{100}$$ is always $$> 0$$ this will occur when the term in parentheses becomes negative: \\begin{align} & &\\frac{i+1}{100}\\cdot\\frac{100 - i}{100} - \\frac{i}{100} &< 0 \\\\ &\\iff &(i+1)(100-i) - 100\\,i &< 0\\\\ &\\iff &100\\,i -i² + 100 - i - 100\\,i &< 0 \\\\ &\\iff &i² + i - 100 &> 0 \\end{align} This inequality holds true for all $$i \\geq 10$$. And thus the guest with the largest piece of pie is number $$10$$. With the help of R, we can also solve the puzzle rather more quickly by brute force. In keeping with our variable names from above: i <- 1:100 # guest numbers p <- i/100 # percentages r <- c(1, cumprod(1 - p))[i] # rests before guests are served", "1 &= 2(m-2)! C_2 \\\\ C_2 &= \\frac{1}{2(m-2)!} = \\frac{m(m-1)}{2(m!)} \\\\ C_2 &= \\frac{ \\binom {m}{2} } {m!} \\\\ \\end{align*} Now lets consider the last term. We set $$x = -m$$ and we get: \\begin{align*} 1 &= C_m(-m)(-m+1)(-m+2) \\cdots (x+m -1) \\\\ C_m &= \\frac{(-1)^{m}}{m!} \\\\ \\end{align*} Now lets consider one of the middle terms. We set $$x = -k$$ where $$0 <= k <= m$$ and we get: \\begin{align*} 1 &= C_k(-k)(-k+1)(-k+2) \\cdots (-1) (1)(2) \\cdots (-k + m - 1) \\\\ 1 &= {-1}^k C_k(k-1)(k-2) \\cdots (1)(2) \\cdots (-k + m - 1) \\\\ C_k &= \\frac{ 1 }{ {(-1)}^k C_k(k-1)(k-2) \\cdots (1)(2) \\cdots (-k + m - 1) } \\\\ C_k &= \\frac{ k! }{ {(-1)}^k C_k(k-1)(k-2) \\cdots (1)m! } \\\\ C_k &= \\frac{ \\binom {m}{k} }{ {(-1)}^k m! } \\\\ \\end{align*} Hence the answer is: $$\\sum_{k=0}^{k=m} \\left( \\frac{ \\binom {m}{k} }{ {(-1)}^k m! }\\right) \\ln{|x+k|} + C$$ • What's your question? – user376343 May 19 at 19:05 • @user376343 - From the opening of the post, \"I am hoping somebody can check my work.\" – Eevee Trainer May 19 at 19:07 • It seems it is right, but the powers of $-1$ must be in parentheses. ... and it is usual to put them in numerators rather than in denominators. – user376343 May 19", "mass): \\begin{align} \\label{massengesetz} & L + S = 100 \\text{ %} \\\\[5px] \\end{align} The two equations can now be inserted into each other and solved for the values $$L$$ or $$S$$: \\begin{align} \\underline{L\\cdot a = S \\cdot b} ~~~\\text{and}~~~\\underline{S = 100 \\text{ %} -L} ~~~\\text{:} \\\\[5px] \\end{align} \\begin{align} L \\cdot a &= (100 \\text{ %} – L) \\cdot b \\\\[5px] L \\cdot a &= 100 \\text{ %} \\cdot b – L \\cdot b \\\\[5px] L \\cdot a + L \\cdot b &= 100 \\text{ %} \\cdot b \\\\[5px] L \\cdot (a+b) &= 100 \\text{ %} \\cdot b \\\\[5px] \\end{align} \\begin{align} \\label{Sm} \\boxed{L = \\frac{b}{a+b} \\cdot 100 \\text{ %}} \\\\[5px] \\end{align} The phase fraction of the solid solution is obtained in the same way: \\begin{align} \\underline{L \\cdot a = S \\cdot b} ~~~\\text{and}~~~\\underline{L = 100 \\text{ %} -S} ~~~\\text{:} \\\\[5px] \\end{align} \\begin{align} (100 \\text{ %} – S) \\cdot a &= S \\cdot b \\\\[5px] 100 \\text{ %} \\cdot a – S \\cdot a &= S \\cdot b \\\\[5px] 100 \\text{ %} \\cdot a &= S \\cdot b + S \\cdot a \\\\[5px] 100 \\text{ %} \\cdot a &= S \\cdot (a+b) \\\\[5px] \\end{align} \\begin{align} \\label{Mk} \\boxed{S = \\frac{a}{a+b} \\cdot 100 \\text{ %}} \\\\[5px] \\end{align} Finally, the results can be interpreted as follows. The fraction of a phase is always", "= \\frac{1}{1 + l} = \\frac{1 + k'}{2}$$ Again \\begin{align}\\left(\\frac{ll'}{kk'}\\right)^{2} &= \\left(\\frac{2\\sqrt{k'}}{1 + k'}\\cdot\\frac{1 - k'}{1 + k'}\\cdot\\frac{1}{kk'}\\right)^{2}\\notag\\\\ &= \\frac{4(1 - k')^{2}}{k'(1 + k')^{4}(1 - k'^{2})} = \\frac{4(1 - k')}{k'(1 + k')^{5}}\\notag\\end{align} Therefore \\begin{align}&\\frac{1}{6}\\frac{d}{dk}\\left\\{\\log\\left(\\frac{ll'}{kk'}\\right)\\right\\} + \\frac{1}{2}\\frac{d}{dk}\\left\\{\\log\\left(\\frac{L}{K}\\right)\\right\\}\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,= \\frac{1}{12}\\frac{d}{dk}\\left\\{\\log\\left(\\frac{ll'}{kk'}\\right)^{2}\\right\\} + \\frac{1}{2}\\frac{d}{dk}\\left\\{\\log\\left(\\frac{L}{K}\\right)\\right\\}\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,= \\frac{1}{12}\\frac{dk'}{dk}\\frac{d}{dk'}\\left\\{\\log\\left(\\frac{4(1 - k')}{k'(1 + k')^{5}}\\right)\\right\\} + \\frac{1}{2}\\frac{dk'}{dk}\\frac{d}{dk'}\\left\\{\\log\\left(\\frac{1 + k'}{2}\\right)\\right\\}\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,= \\frac{1}{12}\\frac{dk'}{dk}\\left(-\\frac{1}{k'} - \\frac{1}{1 - k'} - \\frac{5}{1 + k'} + \\frac{6}{1 +k'}\\right)\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,= \\frac{1}{12}\\frac{-k}{k'}\\left(-\\frac{1}{k'} - \\frac{1}{1 - k'} + \\frac{1}{1 + k'}\\right)\\notag\\\\ &\\,\\,\\,\\,\\,\\,\\,\\,= \\frac{1}{12}\\cdot\\frac{1 + k'^{2}}{kk'^{2}}\\notag\\end{align} and thus \\begin{align}2P(q^{2}) - P(q) &= \\frac{2K^{2}}{\\pi^{2}}(1 + k'^{2})\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\cdot\\frac{1 + k'^{2}}{1 + k'}\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\cdot\\frac{1 + l^{2}}{1 + l}\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\cdot\\frac{1 + l^{2} + k'(1 + l) - 1 + l}{1 + l}\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\{k' + l\\}\\notag\\end{align} For $n = 3$ and above it is best to recast the equation $(2)$ in a simpler form. Let us also use the notation $m = K/L$ for Ramanujan's multiplier. Then we have $$\\frac{dl}{dk} = \\frac{nll'^{2}}{m^{2}kk'^{2}}$$ and \\begin{align}nP(q^{n}) - P(q) &= \\frac{4K^{2}}{\\pi^{2}}(kk'^{2})\\frac{d}{dk}\\left\\{\\log\\left(\\frac{ll'}{kk'}\\right)\\right\\} - \\frac{12K^{2}}{\\pi^{2}}(kk'^{2})\\frac{d}{dk}(\\log m)\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\left(\\frac{mkk'^{2}}{ll'}\\frac{d}{dk}(ll') - \\frac{mkk'^{2}}{kk'}\\frac{d}{dk}(kk') - 3kk'^{2}\\frac{dm}{dk}\\right)\\notag\\\\ &= \\frac{4KL}{\\pi^{2}}\\left(\\frac{mkk'^{2}}{ll'}\\frac{dl}{dk}\\frac{1 - 2l^{2}}{l'} - m(1 - 2k^{2}) - 3kk'^{2}\\frac{dm}{dk}\\right)\\notag\\\\ \\Rightarrow nP(q^{n}) - P(q) &= \\frac{4KL}{\\pi^{2}}\\left((1 - 2l^{2})\\frac{n}{m} - (1 - 2k^{2})m - 3kk'^{2}\\frac{dm}{dk}\\right)\\tag{6}\\end{align} If $n = 3$ then we have \\begin{align}l^{2} &= \\beta = \\frac{(m - 1)^{3}(m + 3)}{16m}\\notag\\\\ k^{2} &= \\alpha = \\frac{(m - 1)(m + 3)^{3}}{16m^{3}}\\notag\\\\ l'^{2} &=", "(using the product rule) we get (107)\\begin{align}\\begin{aligned}\\K uu &= \\Id \\eta \\diff{G_{i, 1}}{u_j} + \\diff{\\eta}{u_j} G_{i, 1},\\\\\\K uv &= \\Id \\eta \\diff{G_{i, 1}}{v_j} + \\diff{\\eta}{v_j} G_{i, 1},\\\\\\K vu &= \\Id \\eta \\diff{G_{i, 2}}{u_j} + \\diff{\\eta}{u_j} G_{i, 2},\\\\\\K vv &= \\Id \\eta \\diff{G_{i, 2}}{v_j} + \\diff{\\eta}{v_j} G_{i, 2}.\\end{aligned}\\end{align} The derivatives of $$\\eta$$ are computed using the chain rule: \\begin{align}\\begin{aligned}\\diff{\\eta}{u_j} &= \\diff{\\eta}{\\gamma}\\, \\diff{\\gamma}{u_j},\\\\\\diff{\\eta}{v_j} &= \\diff{\\eta}{\\gamma}\\, \\diff{\\gamma}{v_j}.\\end{aligned}\\end{align} Taking derivatives of $$\\eta$$ and $$\\gamma$$ (91) gives (108)\\begin{align}\\begin{aligned}\\diff{\\eta}{\\gamma} &= \\frac{B}{2} \\cdot \\exponent \\cdot \\p{\\gamma + \\frac{\\varepsilon_{0}}{2} }^{\\exponent-1}\\\\&= \\eta \\cdot \\exponent \\cdot \\p{\\gamma + \\frac{\\varepsilon_{0}}{2} }^{-1}.\\end{aligned}\\end{align} and (109)\\begin{align}\\begin{aligned}\\diff{\\gamma}{u_j} &= 2u_x \\diff{u_x}{u_j} + 2v_y \\diff{v_y}{u_j} + \\diff{u_x}{u_j}v_y + u_x\\diff{v_y}{u_j} +\\frac 14 \\cdot 2 \\p{u_y + v_x}\\p{\\diff{u_y}{u_j} + \\diff{v_x}{u_j} }\\\\&+ \\frac 14 \\cdot 2 u_z \\diff{u_z}{u_j} + \\frac 14 \\cdot 2 v_z \\diff{v_z}{u_j}\\\\&= 2u_x\\dphi[j]{x} + v_y\\dphi[j]{x} + \\frac 12 \\dphi[j]{y}\\p{u_y + v_x} + \\frac 12 u_z \\dphi[j]{z},\\\\\\diff{\\gamma}{v_j} &= 2u_x \\diff{u_x}{v_j} + 2v_y \\diff{v_y}{v_j} + \\diff{u_x}{v_j}v_y + u_x\\diff{v_y}{v_j} +\\frac 14 \\cdot 2 \\p{u_y + v_x}\\p{\\diff{u_y}{v_j} + \\diff{v_x}{v_j} }\\\\&+ \\frac 14 \\cdot 2 u_z \\diff{u_z}{v_j} + \\frac 14 \\cdot 2 v_z \\diff{v_z}{v_j}\\\\&= 2v_y\\dphi[j]{y} + u_x\\dphi[j]{y} + \\frac 12 \\dphi[j]{x}\\p{u_y + v_x} + \\frac 12 v_z \\dphi[j]{z}.\\end{aligned}\\end{align} The derivatives of $$G_{\\cdot, \\cdot}$$ are (110)\\begin{align}\\begin{aligned}\\diff{G_{i, 1}}{u_j} &= 4\\dpsi{x}\\dphi[j]{x} + \\dpsi{y}\\dphi[j]{y} + \\dpsi{z}\\dphi[j]{z},\\\\\\diff{G_{i, 1}}{v_j} &= 2\\dpsi{x}\\dphi[j]{y} + \\dpsi{y}\\dphi[j]{x},\\\\\\diff{G_{i, 2}}{u_j} &= 2\\dpsi{y}\\dphi[j]{x} + \\dpsi{x}\\dphi[j]{y},\\\\\\diff{G_{i, 2}}{v_j} &= 4\\dpsi{y}\\dphi[j]{y} + \\dpsi{x}\\dphi[j]{x} + \\dpsi{z}\\dphi[j]{z}.\\end{aligned}\\end{align} To", "3 \\cdot 2 \\cdot a_3 \\\\ f''''(c) &= 4 \\cdot 3 \\cdot 2 \\cdot a_4 \\end{align} re-arranging, we get \\begin{align} a_3 &= \\frac{f'''(c)}{3 \\cdot 2} \\\\ a_4 &= \\frac{f''''(c)}{4 \\cdot 3 \\cdot 2} \\end{align} Therefore, the general pattern is \\begin{align} a_0 &= f(c) = \\frac{f(c)}{0!} \\\\ a_1 &= f'(c) = \\frac{f'(c)}{1!} \\\\ a_2 &= \\frac{f''(c)}{2} = \\frac{f''(c)}{2!} \\\\ a_3 &= \\frac{f'''(c)}{3 \\cdot 2} = \\frac{f'''(c)}{3!} \\\\ a_4 &= \\frac{f''''(c)}{4 \\cdot 3 \\cdot 2} = \\frac{f''''(c)}{4!} \\\\ &\\vdots \\\\ a_n &= \\frac{f^{(n)}(c)}{n!} \\end{align} So, the power series representation of $$f(x)$$ is \\begin{align} f(x) &= \\frac{f(c)}{0!} + \\frac{f'(c)}{1!}(x - c) + \\frac{f''(c)}{2!}(x-c)^2 + \\frac{f'''(c)}{3!}(x-c)^3 + \\cdots \\\\ &= \\sum_{n=0}^\\infty \\frac{f^{(n)}(c)}{n!} (x-c)^n \\end{align}", "terms: \\begin{align*} f(x_0 + h, y_0) - f(x_0,y_0) &= h \\cdot \\frac{\\partial f}{\\partial x}(b_1,y_0)\\\\ f(x_0 + h, y_0 + k) - f(x_0 + h,y_0) &= k \\cdot \\frac{\\partial f}{\\partial y} (x_0+h,b_2) \\end{align*} for some $b_1 \\in (x_0,x_0 + h)$ and $b_2 \\in (y_0, y_0 + k)$. Therefore, \\begin{align*} & \\frac{\\left| f(x_0+h,y_0+k)-f(x_0,y_0)-h\\frac{\\partial f}{\\partial x}(x_0,y_0)-k\\frac{\\partial f}{\\partial y}(x_0,y_0) \\right|}{\\| (h,k)\\|}\\\\ ={} & \\frac{\\left| h\\left( \\frac{\\partial f}{\\partial x}(b_1,y_0)-\\frac{\\partial f}{\\partial x}(x_0,y_0) \\right) - k \\left( \\frac{\\partial f}{\\partial y}(x_0+h,b_2) - \\frac{\\partial f}{\\partial y}(x_0,y_0) \\right) \\right|}{\\| (h,k)\\|}\\\\ \\leq{} & \\frac{| h |}{\\| (h,k) \\|} \\cdot \\left| \\frac{\\partial f}{\\partial x}(b_1,y_0)-\\frac{\\partial f}{\\partial x}(x_0,y_0) \\right| + \\frac{|k |}{\\| (h,k) \\|} \\cdot \\left| \\frac{\\partial f}{\\partial y}(x_0+h,b_2) - \\frac{\\partial f}{\\partial y}(x_0,y_0) \\right| \\\\ \\leq{} & \\left| \\frac{\\partial f}{\\partial x}(b_1,y_0)-\\frac{\\partial f}{\\partial x}(x_0,y_0) \\right| + \\left| \\frac{\\partial f}{\\partial y}(x_0+h,b_2) - \\frac{\\partial f}{\\partial y}(x_0,y_0) \\right| \\end{align*} since $| h | / \\| (h,k) \\|$ and $| k | / \\| (h,k) \\|$ are less than or equal to $1$ for all $(h,k) \\neq (0,0)$. Now, taking $\\lim_{\\|(h,k)\\| \\to 0}$, we get $0$, because $(b_1,y_0) \\to (x_0,y_0)$ and $(x_0+h,b_2) \\to (x_0,y_0)$ as $(h,k) \\to (0,0)$, and the partial derivatives are continuous at $(x_0,y_0)$. This is where we use the continuity of the partial derivatives at $(x_0,y_0)$.", "\\frac{272}{675} \\cdot \\frac{1}{1 - 16 z} + 2 \\cdot \\frac{1}{1 - 2 z} + \\frac{675 t_0 - 1558}{675} \\cdot \\frac{1}{1 - z}$$ Now use the generalized binomial theorem: $$(1 + u)^{-m} = \\sum_{k \\ge 0} \\binom{-m}{n} u^k = \\sum_{k \\ge 0} (-1)^k \\binom{k + m - 1}{m - 1} u^k$$ in particular: $$\\binom{-2}{k} = (-1)^k(k + 1)$$ to finish this off: \\begin{align} t_k &= \\frac{32}{45} (k + 1) \\cdot 16^k - \\frac{272}{675} \\cdot 16^k + 2 \\cdot 2^k + \\frac{675 t_0 - 1558}{675} \\\\ &= \\frac{32}{45} \\cdot k \\cdot 16^k + \\frac{208}{675} \\cdot 16^k + 2 \\cdot 2^k + \\frac{675 t_0 - 1558}{675} \\\\ t(n) &= \\frac{32}{45} n^2 \\log_4 n - \\frac{272}{675} n^2 + 2 \\sqrt{n} + \\frac{675 t_0 - 1558}{675} \\end{align}", "\\cdots k^3+(k+1)^3=\\left(\\frac{(k+1)(k+2)}{2} \\right)^2=(1+2+3+ \\cdots k+(k+1))^2\\end{align*} \\begin{align*}\\therefore\\end{align*} The symbol \\begin{align*}\\therefore\\end{align*} is one of many indicators like QED that follow a proof to tell the reader that the proof is complete. #### Example 5 Prove the following statement using induction: For \\begin{align*}n \\ge 1, \\ 1+2+3+4+ \\cdots + n=\\frac{n(n+1)}{2}\\end{align*} Base Case: \\begin{align*}1=\\frac{1(1+1)}{2}=1 \\cdot \\frac{2}{2}=1\\end{align*} Inductive Hypothesis: \\begin{align*}1+2+3+4+ \\cdots +k=\\frac{k(k+1)}{2}\\end{align*} Proof: Start with what you know and work to showing it true for \\begin{align*}k+1\\end{align*}. Inductive Hypothesis: \\begin{align*}1+2+3+4+ \\cdots +k=\\frac{k(k+1)}{2}\\end{align*} Add \\begin{align*}k+1\\end{align*} to both sides: \\begin{align*}1+2+3+4+ \\cdots + k+(k+1)=\\frac{k(k+1)}{2}+(k+1)\\end{align*} Find a common denominator for the right side: \\begin{align*}1+2+3+4+ \\cdots +k+(k+1)=\\frac{k^2+k}{2}+\\frac{2k+2}{2}\\end{align*} Simplify the right side: \\begin{align*}1+2+3+4+ \\cdots +k+(k+1)=\\frac{k^2+3k+2}{2}\\end{align*} Factor the numerator of the right side: \\begin{align*}1+2+3+4+ \\cdots +k+(k+1)=\\frac{(k+1)(k+2)}{2}\\end{align*} Rewrite the right side: \\begin{align*}1+2+3+4+ \\cdots +k+(k+1)=\\frac{(k+1)((k+1)+1)}{2}\\end{align*} \\begin{align*}\\therefore\\end{align*} ### Review For each of the following statements: a) show the base case is true; b) state the inductive hypothesis; c) state what you are trying to prove in the inductive step/proof. Do not prove yet. 1. For \\begin{align*}n \\ge 5, \\ 4n < 2^n\\end{align*}. 2. For \\begin{align*}n \\ge 1, \\ 8^n-3^n\\end{align*} is divisible by 5. 3. For \\begin{align*}n \\ge 1, \\ 7^n-1\\end{align*} is divisible by 6. 4. For \\begin{align*}n \\ge 2, \\ n^2 \\ge 2n\\end{align*}. 5. For \\begin{align*}n \\ge 1, \\ 4^n+5\\end{align*} is divisible by 3. 6. For \\begin{align*}n \\ge 1, \\ 0^2+1^2+ \\cdots + n^2=\\frac{n(n+1)(2n+1)}{6}\\end{align*}", "K_{a_1} &= \\frac{c(\\ce{H3+O})\\cdot c(\\ce{H2A-}) }{c(\\ce{H3A})} \\tag4\\\\ K_{a_2} &= \\frac{c(\\ce{H3+O})\\cdot c(\\ce{HA^{2-}})}{c(\\ce{H2A-})} \\tag5\\\\ \\end{align} We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels. \\begin{align} K_{a_1} \\cdot K_{a_2} &= \\frac{c(\\ce{H3+O})\\cdot c(\\ce{H2A-}) }{c(\\ce{H3A})} \\cdot \\frac{c(\\ce{H3+O})\\cdot c(\\ce{HA^{2-}})}{c(\\ce{H2A-})} \\\\ K_{a_1} \\cdot K_{a_2} &= c^2(\\ce{H3+O})\\cdot \\frac{c(\\ce{HA^{2-}})}{c(\\ce{H3A})} \\tag7\\\\ \\end{align} Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate $$c(\\ce{HA^{2-}})=c(\\ce{H3A})\\tag8$$ and rewrite $(7)$ as \\begin{align} c^2(\\ce{H3+O}) &= K_{a_1} \\cdot K_{a_2}\\\\ c(\\ce{H3+O}) &= \\sqrt{K_{a_1} \\cdot K_{a_2}}\\\\ \\ce{pH} &= \\frac12\\left(\\mathrm{p}K_{a_1}+\\mathrm{p}K_{a_2}\\right). \\end{align}", "; \\frac{1}{2} -\\frac{1}{2} }. This gives \\label{eqn:qmLecture16:1000} 2 \\ket{ \\frac{3}{2} \\frac{-1}{2} } = \\sqrt{\\frac{2}{3}} \\lr{ \\sqrt{ 2 } \\ket{ 1 -1 ; \\frac{1}{2} \\frac{1}{2} } + \\ket{ 1 0 ; \\frac{1}{2} \\frac{-1}{2} } } + \\inv{\\sqrt{3}} \\sqrt{ 2 } \\ket{ 1 0 ; \\frac{1}{2} -\\frac{1}{2} }, or \\label{eqn:qmLecture16:1001} \\boxed{ \\ket{ \\frac{3}{2} \\frac{-1}{2} } = \\inv{\\sqrt{ 3 }} \\ket{ 1 -1 ; \\frac{1}{2} \\frac{1}{2} } + \\sqrt{\\frac{2}{3}} \\ket{ 1 0 ; \\frac{1}{2} \\frac{-1}{2} }. } There’s one more possible state with total angular momentum $$\\frac{3}{2}$$. This time \\label{eqn:qmLecture16:1060} \\begin{aligned} \\hat{L}_{-}^{\\textrm{tot}} \\ket{ \\frac{3}{2} \\frac{-1}{2} } &= \\sqrt{ 1 \\times 3 } \\ket{ \\frac{3}{2} \\frac{-3}{2} } \\\\ &= \\inv{\\sqrt{ 3 }} \\hat{L}_{-}^{(2)} \\ket{ 1 -1 ; \\frac{1}{2} \\frac{1}{2} } + \\sqrt{\\frac{2}{3}} \\hat{L}_{-}^{(1)} \\ket{ 1 0 ; \\frac{1}{2} \\frac{-1}{2} } \\\\ &= \\inv{\\sqrt{ 3 }} \\sqrt{1 \\times 1 } \\ket{ 1 -1 ; \\frac{1}{2} \\frac{-1}{2} } + \\sqrt{\\frac{2}{3}} \\sqrt{ 1 \\times 2 } \\ket{ 1 -1 ; \\frac{1}{2} \\frac{-1}{2} }, \\end{aligned} or \\label{eqn:qmLecture16:1080} \\boxed{ \\ket{ \\frac{3}{2} \\frac{-3}{2} } = \\ket{ 1 -1 ; \\frac{1}{2} \\frac{-1}{2} }. } The $$\\ket{ \\frac{1}{2} \\frac{1}{2} }$$ state is constructed as normal to $$\\ket{ \\frac{3}{2} \\frac{1}{2} }$$, or \\label{eqn:qmLecture17:1120} \\boxed{ \\ket{ \\frac{1}{2} \\frac{1}{2} } = \\sqrt{\\frac{1}{3}} \\ket{ 1 0 ; \\frac{1}{2} \\frac{1}{2} } \\sqrt{ \\frac{2}{3} } \\ket{ 1 1 ; \\frac{1}{2} -\\frac{1}{2} }, } and $$\\ket{ \\frac{1}{2}", "\\tag{2.101} \\frac{dv_x}{dt}&= -\\rho _f\\frac{A}{2m}v_r^2(C_D\\cdot \\cos(\\phi)+C_L\\sin(\\phi)) \\\\ \\frac{dv_y}{dt}&=\\rho _f\\frac{A}{2m}v_r^2(C_L\\cdot \\cos(\\phi)-C_D\\sin(\\phi))-g \\tag{2.102} \\end{align} From the figure we have \\begin{align*} \\cos (\\phi)&=\\frac{v_{rx}}{v_r} \\\\ \\sin(\\phi)&=\\frac{v_{ry}}{v_r} \\end{align*} We assume that the particle is a sphere, such that $$C=\\rho _f\\frac{A}{2m}=\\frac{3\\rho_f}{4\\rho_kd}$$ as in (2.6.4 Example: Sphere in free fall). Here $$d$$ is the diameter of the sphere and $$\\rho_k$$ the density of the sphere. Now (2.101) and (2.102) become \\begin{align} \\tag{2.103} \\frac{dv_x}{dt} &= -C\\cdot v_r(C_D\\cdot v_{rx}+C_L\\cdot v_{ry})\\\\ \\frac{dv_y}{dt} &= C\\cdot v_r(C_L\\cdot v_{rx}-C_D\\cdot v_{ry})- \\tag{2.104}g \\end{align} With $$\\frac{dx}{dt}=v_x$$ and $$\\frac{dy}{dt}=v_y$$ we get a system of 1st order equations as follows, \\begin{align} &\\frac{dx}{dt}=v_x \\nonumber \\\\ & \\frac{dy}{dt}=v_y \\nonumber\\\\ & \\frac{dv_x}{dt} = -C\\cdot v_r(C_D\\cdot v_{rx}+C_L\\cdot v_{ry}) \\tag{2.105}\\\\ &\\frac{dv_y}{dt} = C\\cdot v_r(C_L\\cdot v_{rx}-C_D\\cdot v_{ry})-g \\nonumber \\end{align} Introducing the notation $$x=y_1$$, $$y=y_2$$, $$v_x=y_3$$, $$v_y=y_4$$, we get \\begin{align} &\\frac{dy_1}{dt}=y_3\\nonumber\\\\ & \\frac{dy_2}{dt}=y_4\\nonumber\\\\ & \\frac{dy_3}{dt} = -C\\cdot v_r(C_D\\cdot v_{rx}+C_L\\cdot v_{ry}) \\tag{2.106}\\\\ &\\frac{dy_4}{dt} = C\\cdot v_r(C_L\\cdot v_{rx}-C_D\\cdot v_{ry})-g\\nonumber \\end{align} Here we have $$v_{rx}=v_x-v_{fx}=y_3-v_{fx},\\ v_{ry}=v_y-v_{fy}=y_4-v_{fy},\\\\ v_r=\\sqrt{v_{rx}^2+v_{ry}^2}$$ Initial conditions for $$t=0$$ are \\begin{align*} y_1&=y_2=0 \\\\ y_3&=v_0\\cos(\\alpha)\\\\ y_4&=v_0\\sin(\\alpha) \\end{align*} Let's first look at the case of a smooth ball. We use the following data (which are the data for a golf ball): $$\\begin{equation*} \\tag{2.107} \\text{Diameter } d = 41 \\text{mm},\\text{ mass } m = 46\\text{g which gives } \\rho_k=\\frac{6m}{\\pi d^3} = 1275 \\text{kg/m}^3 \\end{equation*}$$ We use the initial velocity $$v_0=50$$ m/s and solve (2.106) using the Runge-Kutta", "I_S(x,y) = \\begin{cases} 0 & c_{max}(x,y) = 0 \\\\ \\frac{\\Delta(x,y)}{c_{max}(x,y)} & c_{max}(x,y) \\neq 0 \\end{cases} \\end{aligned} \\begin{aligned} I_V(x,y) = 25.5 \\cdot c_{max}(x,y) \\end{aligned} ##### HSV to RGB Auxiliary variables: \\begin{aligned} T(x,y) = \\frac{I_S(x,y)}{100} + \\frac{I_V(x,y)}{100} \\end{aligned} \\begin{aligned} t(x,y) = \\frac{I_V(x,y)}{100} - T(x,y) \\end{aligned} \\begin{aligned} q(x,y) = T(x,y) \\cdot [1 - |(\\frac{I_H(x,y)}{60} mod 2) - 1|] \\end{aligned} Final calculation: 1. If IH(x,y) is between 0º and 60º \\begin{aligned} I_R(x,y) = (T(x,y)+t(x,y)) \\cdot 255 && I_G(x,y) = (q(x,y) + t(x,y)) \\cdot 255 && I_B(x,y) = t(x,y) \\cdot 255 \\end{aligned} 2. If IH(x,y) is between 60º and 120º \\begin{aligned} I_R(x,y) = (q(x,y) + t(x,y)) \\cdot 255 && I_G(x,y) = (T(x,y) + t(x,y)) \\cdot 255 && I_B(x,y) = t(x,y) \\cdot 255 \\end{aligned} 3. If IH(x,y) is between 120º and 180º \\begin{aligned} I_R(x,y) = t(x,y) \\cdot 255 && I_G(x,y) = (T(x,y) + t(x,y)) \\cdot 255 && I_B(x,y) = (q(x,y) + t(x,y)) \\cdot 255 \\end{aligned} 4. If IH(x,y) is between 180º and 240º \\begin{aligned} I_R(x,y) = t(x,y) \\cdot 255 && I_G(x,y) = (q(x,y) + t(x,y)) \\cdot 255 && I_B(x,y) = (T(x,y) + t(x,y)) \\cdot 255 \\end{aligned} 5. If IH(x,y) is between 240º and 300º \\begin{aligned} I_R(x,y) = (q(x,y) + t(x,y)) \\cdot 255 && I_G(x,y) = t(x,y) \\cdot 255 && I_B(x,y) = (T(x,y) + t(x,y)) \\cdot 255 \\end{aligned} 6. If IH(x,y) is between 300º and", "c_k!\\, p_k!}\\cdot \\prod_{k=2}^n \\left(\\frac{(k-1)!}{2}\\right)^{c_k} \\prod_{k=2}^n \\left(\\frac{k!}{2}\\right)^{p_k} \\\\ &= \\frac{n!}{c_1!} \\prod_{k=2}^n \\frac{1}{k!^{c_k}\\, k!^{p_k}\\, c_k!\\, p_k!}\\cdot \\left(\\frac{(k-1)!}{2}\\right)^{c_k} \\left(\\frac{k!}{2}\\right)^{p_k} \\\\ &= \\frac{n!}{c_1!} \\prod_{k=2}^n \\frac{1}{k^{c_k}\\,c_k!\\, p_k!}\\cdot \\left(\\frac{1}{2}\\right)^{c_k} \\left(\\frac{1}{2}\\right)^{p_k} \\\\ &= \\frac{n!}{c_1!} \\prod_{k=2}^n \\frac{1}{(2k)^{c_k}\\,2^{p_k} \\,c_k!\\, p_k!} \\\\ &= \\frac{n!}{c_1! \\displaystyle\\prod_{k=2}^n (2k)^{c_k}\\,2^{p_k} \\,c_k!\\, p_k!} \\end{aligned} We now have a simple formula for the answer! This can be computed in O(n) time by precomputing some values such as factorials and powers of two. Just remember to reduce all intermediate values modulo 10^9 + 7! The overall running time is just O(n^2), dominated by the traversal of the graph (and taking complements). O(n^2) ### AUTHOR’S AND TESTER’S SOLUTIONS: [setter][333] [tester][444] [editorialist][555] [333]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server. [444]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server. [555]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server.", "= \\tau_D = \\tau_E = 0 \\\\ H_A: & \\text{ not } h_0 \\end{align} We can perform an F test. \\begin{align} \\bar{y}_{++} & = \\text{(10*12.05 + 10*11.02 + 10*10.27 + 10*9.27 + 10*12.17)/50} \\\\ & = 10.956 \\\\ \\\\ SSModel & = \\sum_{i=1}^{t} \\sum_{j=1}^{n_i} (\\bar{y}_{i+} - \\bar{y}_{++})^2 \\\\ & = 10 (9.27\\, -10.956)^2+10 (10.27\\, -10.956)^2+10 (11.02\\, -10.956)^2+10 (12.05\\, -10.956)^2+10 (12.17\\, -10.956)^2 \\\\ & = 59.8792 \\\\ \\\\ MSModel & = \\frac{ 59.8792 }{ 5 - 1 }\\\\ & = 14.9698 \\\\ \\\\ SSE & = \\sum_{i=1}^{t} \\sum_{j=1}^{n_i} (y_{ij} - \\bar{y}_{i+})^2 \\\\ & = \\sum_{i=1}^{t} \\sum_{j=1}^{n_i} s_i^2 (n_i - 1) \\\\ & = 9 \\cdot 0.83^2 + 9 \\cdot 1.12^2 + 9 \\cdot 1.03^2 + 9 \\cdot 1.16^2 + 9 \\cdot 0.79^2 \\\\ & = 44.7651 \\\\ \\\\ MSE & = \\frac{ 44.7651 }{ 50 - 5 } \\\\ & = 0.99478 \\\\ \\\\ F & = \\frac{ MSModel } { MSE } \\\\ & = 15.0484 \\end{align} We can compare this to $F^4_{45, 0.05} = 2.56873$. Thus, using an $\\alpha = 0.05$ we have sufficient evidence to reject the null in favor of the alternative that not all the treatment effects are 0. ii After carrying out all pairwise comparisons at a familywise error rate $\\alpha = 0.05$, identify which differences are significant. Be clear about which multiple", "twice: \\begin{align} ((f\\cdot g)\\cdot h)^{(n)} &=\\sum_{k=0}^n\\binom{n}{k}(f\\cdot g)^{(k)}h^{(n-k)}\\\\ &=\\sum_{k=0}^n\\binom{n}{k}\\sum_{j=0}^k\\binom{k}{j}f^{(j)}g^{(k-j)}h^{(n-k)}\\\\ &=\\sum_{k=0}^n\\sum_{j=0}^k\\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\\\ &=\\sum_{j=0}^n\\sum_{k=j}^n\\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\\\ &=\\sum_{j=0}^n\\sum_{k=0}^{n-j}\\frac{n!}{j!k!(n-k-j)!}f^{(j)}g^{(k)}h^{(n-k-j)}\\\\ &=\\sum_{\\substack{i+j+k=n\\\\i,j,k\\ge0}}\\frac{n!}{i!j!k!}f^{(j)}g^{(k)}h^{(i)} \\end{align} - This was added to Shuchang's answer before I finished writing. – robjohn Dec 23 '13 at 15:23" ]

[ "easily verified: the recurrences become \\left\\{\\begin{align*} F_{2n}&=F_{2n-2}+F_{2n-1}\\\\ F_{2n+1}&=F_{2n-1}+F_{2n}\\;, \\end{align*}\\right. which reduce to the single familiar Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$, and the initial values $F_0=b_0$ and $F_1=a_0$ are correct. The problem is now reduced to standard results about the Fibonacci sequence. In particular, if $\\varphi=\\frac12\\left(1+\\sqrt5\\right)$, it’s well known that $F_n$ is the integer nearest $\\frac{\\varphi^n}{\\sqrt5}$, so $a_n$ is the integer nearest $\\frac{\\varphi^{2n+1}}{\\sqrt5}$. Equivalently, $$a_n=\\left\\lfloor\\frac{\\varphi^{2n+1}}{\\sqrt5}+\\frac12\\right\\rfloor\\;.$$ Hagen von Eitzen’s closed form is easily derived from this and the observation that $$\\varphi^2=\\frac{3+\\sqrt5}2\\;.$$", "backwards to get $s_0 = b - 3$, so: \\begin{align} S(z) = \\frac{(b - 3) + b z + b z^2 + b z^3}{1 - z^2 - z^4} \\end{align} The generating function for the Fibonacci numbers $F_n$, defined by: $$F_{n + 2} = F_{n + 1} + F_n \\qquad F_0 = 0, F_1 = 1$$ is: $$F(z) = \\sum_{n \\ge 0} F_n z^n = \\frac{z}{1 - z- z^2}$$ For Fibonacci numbers shifted by one: $$\\sum_{n \\ge 0} F_{n + 1} z^n = \\frac{1}{1 - z - z^2}$$ so our mystery generating function is: \\begin{align} S(z) &= ((b - 3) + b z + b z^2 + b z^3) \\sum_{n \\ge 0} F_{n + 1} z^2 \\\\ [z^{2 n}] S(z) &= (b - 3) F_{n + 1} + b F_{n - 1} \\\\ [z^{2 n + 1}] S(z) &= b F_{n + 1} + b F_{n - 1} \\end{align} Need to find which one satisfies this. For $n = 1$ the second option gives: $$2015 = 2015 (F_2 + F_0)$$ while the first one gives for $n = 4$: $$2015 = (290 - 3) F_5 + 290 F_3$$ so it looks like the largest value is $b = 290$ (checked up to $n = 18$).", "\\\\ &= \\frac{1 - z}{1 - z - z^2} \\end{align*} Consider the Fibonacci sequence, defined by $$F_{n + 2} = F_{n + 1} + F_n$$ with $$F_0 = 0, F_1 = 1$$. It's generating function is: \\begin{align*} F(z) &= \\sum_{n \\ge 0} F_n z^n = \\frac{z}{1 - z - z^2} \\end{align*} But also: \\begin{align*} \\sum_{n \\ge 0} F_{n + 1} z^n &= \\frac{F(z) - F_0}{z} = \\frac{1}{1 - z - z^2} \\end{align*} We see that our generating function is: \\begin{align*} \\frac{1 - z}{1 - z - z^2} &= \\sum_{n \\ge 0} (F_{n + 1} - F_n) z^n \\\\ &= \\sum_{n \\ge 0} F_{n - 1} z^n \\end{align*} So the number of sequences of length $$n$$ is $$F_{n - 1}$$, where we extend to $$F_{-1} = 1$$ (as it should be to get one sequence of length 0).", "\\quad \\sum_{k=0}^{\\left \\lfloor \\frac{n}{2} \\right \\rfloor} \\binom{n-k}{k} = \\binom{n}{0} + \\binom{n-1}{1} + ... + \\binom{n-\\left \\lfloor \\frac{n}{2} \\right \\rfloor}{\\left \\lfloor \\frac{n}{2} \\right \\rfloor} \\end{align} Therefore a general formula for the Fibonacci sequence for $n = 0, 1, 2, ...$ is given by: (4) \\begin{align} \\quad \\{ f_n \\} = \\left \\{ \\sum_{k=0}^{\\left \\lfloor \\frac{n}{2} \\right \\rfloor} \\binom{n-k}{k} \\right \\} = \\left \\{ \\sum_{k=0}^{\\left \\lfloor \\frac{n}{2} \\right \\rfloor} \\frac{(n-k)!}{k!(n - 2k)!} \\right \\} \\end{align} Let $f : \\{0, 1, 2, ... \\} \\to \\mathbb{Z}$ be the function defined by the formula $f(n) = \\sum_{k=0}^{\\left \\lfloor \\frac{n}{2} \\right \\rfloor} \\frac{(n-k)!}{k!(n - 2k)!}$. Then $f(n)$ gives us the $(n+1)^{\\mathrm{st}}$ number in the Fibonacci sequence. For example, suppose that we wanted to find the $14^{\\mathrm{th}}$ number in the Fibonacci sequence. Then we need to compute $f(13)$ which is: (5) \\begin{align} \\quad f(19) = \\sum_{k=0}^{\\left \\lfloor \\frac{13}{2} \\right \\rfloor} \\frac{(19-k)!}{k!(19 - 2k)!} \\\\ \\quad = \\sum_{k=0}^{6} \\frac{(13-k)!}{k!(13-2k)!} \\\\ \\quad = \\frac{13!}{0! \\cdot 13!} + \\frac{12!}{1! \\cdot 11!} + \\frac{11!}{2! \\cdot 9!} + \\frac{10!}{3! \\cdot 7!} + \\frac{9!}{4! \\cdot 5!} + \\frac{8!}{5! \\cdot 3!} + \\frac{7!}{6! \\cdot 1!} \\\\ \\quad = 1 + 12 + 55 + 120 + 126 + 56 + 7 \\quad = 377 \\end{align}", "# Induction technique on the fibonacci sequence [duplicate] This question already has an answer here: $$F_n=\\frac{(\\frac{1+\\sqrt{5}}{2})^{n}-(\\frac{1-\\sqrt{5}}{2})^{n}}{\\sqrt{5}}$$ Proof: Let n=1 thus, \\begin{align*} F_1&=\\frac{(\\frac{1+\\sqrt{5}}{2})^{1}-(\\frac{1-\\sqrt{5}}{2})^{1}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{1+\\sqrt{5}}{2}-\\frac{1-\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{1+\\sqrt{5}-1+\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{\\sqrt{5}+\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{2\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{{\\sqrt{5}}}{\\sqrt{5}}\\\\ &=1 \\end{align*} Suppose $$F_k=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k}}{\\sqrt{5}}$$ Also $F(k+1)=F(k)+F(k-1)$ \\begin{align*} F(k)+F(k-1)&=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k}}{\\sqrt{5}}+\\frac{(\\frac{1+\\sqrt{5}}{2})^{k-1}-(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}+(\\frac{1+\\sqrt{5}}{2})^{k-1}-(\\frac{1-\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{1+\\sqrt{5}}{2})+1)(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{1-\\sqrt{5}}{2})+1)(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{3+\\sqrt{5}}{2}))(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{3-\\sqrt{5}}{2}))(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{1+\\sqrt{5}}{2})^{2})(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{1-\\sqrt{5}}{2})^{2})(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k+1}-(\\frac{1-\\sqrt{5}}{2})^{k+1}}{\\sqrt{5}}\\hfill \\end{align*} I was told there was a mistake but I feel this is right can I get clarification on my proof if it seems right? ## marked as duplicate by Dietrich Burde, Community♦Oct 28 '16 at 15:27 • You should have two bases cases $F_1$ and $F_2$ for the inductive step to work on. – Ian Miller Oct 28 '16 at 15:21 • In your initial case you only proved $F_1$. In your inductive step you find $F_{k+1}$ using both $F_k$ and $F_{k-1}$. If we use the inductive step to prove $F_3$ it relies upon both $F_2$ and $F_1$ being true but you haven't established that $F_2$ is true. – Ian Miller Oct 28 '16 at 15:25 • Also instead of doing $F_2$ you could do $F_0$ which is slightly easier arithmetically. Then $F_2$ is done inductively from $F_1$ and $F_0$. – Ian Miller Oct 28 '16 at 15:28", "20:24 Letting $a_n$ denote the sum in question you can show $a_n = a_{n-1} + a_{n-2} + F_{n+4}$ where $F_k$ is the $k$-th Fibonacci number. Maybe this helps. – Cocopuffs Sep 5 '12 at 20:50 @HagenvonEitzen : Thanks for pointing out those links. I checked them and tried to solve it according to one of those methods. I edited the question with my results. However, I got stuck after a certain point. Please do check it out and help me to complete it. – Arjun Datta Sep 6 '12 at 2:52 Using the closed form for the Fibonacci numbers, $$F_n=\\frac{\\phi^n-(-1/\\phi)^n}{\\sqrt{5}}\\tag{1}$$ and Lucas numbers $$L_n=\\phi^n+(-1/\\phi)^n\\tag{2}$$ we get \\begin{align} F_iF_{n-i} &=\\frac{\\phi^i-(-1/\\phi)^i}{\\sqrt{5}}\\frac{\\phi^{n-i}-(-1/\\phi)^{n-i}}{\\sqrt{5}}\\\\ &=\\frac{\\phi^n+(-1/\\phi)^n-(-1)^i\\left(\\phi^{n-2i}+(-1/\\phi)^{n-2i}\\right)}{5}\\\\ &=\\frac{L_{n}-(-1)^iL_{n-2i}}{5}\\tag{3} \\end{align} To sum Lucas numbers, we use $(2)$ and the formula for the sum of a geometric series: \\begin{align} \\sum_{i=j}^{k-1}F_iF_{n-i} &=\\sum_{i=j}^{k-1}\\frac{L_{n+5}-(-1)^iL_{n-2i}}{5}\\\\ &=\\color{#C00000}{\\frac{k-j}{5}L_{n}}\\\\ &\\color{#00A000}{-\\frac15\\sum_{i=j}^{k-1}\\phi^{n}\\left(-1/\\phi^2\\right)^i}\\\\ &\\color{#0000FF}{-\\frac15\\sum_{i=j}^{k-1}(-1/\\phi)^{n}\\left(-\\phi^2\\right)^i}\\\\ &=\\color{#C00000}{\\frac{k-j}{5}L_{n}}\\\\ &\\color{#00A000}{-\\frac{\\phi^{n}}{5}\\frac{\\left(-1/\\phi^2\\right)^k-\\left(-1/\\phi^2\\right)^j}{\\left(-1/\\phi^2\\right)-1}}\\\\ &\\color{#0000FF}{-\\frac{(-1/\\phi)^{n}}{5}\\frac{\\left(-\\phi^2\\right)^k-\\left(-\\phi^2\\right)^j}{\\left(-\\phi^2\\right)-1}}\\\\ &=\\color{#C00000}{\\frac{k-j}{5}L_{n}}\\\\ &\\color{#00A000}{+\\frac15\\frac{-(-1)^{n-k}(-1/\\phi)^{2k-n-1}-(-1)^{j}\\phi^{n-2j+1}}{\\phi+1/\\phi}}\\\\ &\\color{#0000FF}{+\\frac15\\frac{+(-1)^{n-k}\\phi^{2k-n-1}+(-1)^{j}(-1/\\phi)^{n-2j+1}}{\\phi+1/\\phi}}\\\\ &=\\frac{k-j}{5}L_{n}+\\frac15\\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\\right)\\tag{4}\\\\ &=\\frac{k-j}{5}(F_{n-1}+F_{n+1})+\\frac15\\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\\right)\\tag{5} \\end{align} Applying $(4)$ to the current problem yields $$\\sum_{k=4}^{n+1}F_kF_{n+5-k}=\\frac{n-2}{5}(F_{n+6}+F_{n+4})-\\frac25F_{n-2}\\tag{5}$$", "for all $$k$$. So let us isolate in the denominator of $$(2)$$ an expression involving only two Fibonacci numbers, my choice being $$F_k$$ and $$F_{k+1}$$. We split for this $$5n^2 = (5n^2-1)+\\color{blue}1$$ and form thus from $$\\color{blue}{1}\\cdot F_k+F_{k-1}$$ one $$F_{k+1}$$. The other Fibonacci numbers can the be easily linearly written in terms of $$F_k$$ and $$F_{k+1}$$ (using small Fibonacci numbers as coefficiente), e.g. $$F_{k+3}=2F_{k+1}+F_k$$, so we have to show: $$\\sum_{n\\ge0} (-1)^n \\frac {(5n^2+4n+1)F_k + (5n+3)F_{k+1}} {{2n \\choose n}\\cdot2(2n+1)} = \\color{red}{F_{k+1}}\\qquad(3) \\ .$$ We have thus to show that the coefficients of $$F_k,F_{k+1}$$ extracted from the expression in the L.H.S. are $$0,2$$, as in the R.H.S. We completely avoid hypergeometric sums for this, so let us finally start the answer. We use the notations $$a_n= \\frac 1{\\binom{2n}n}\\ ,\\qquad b_n= \\frac n{\\binom{2n}n}\\ ,$$ and compute simply \\begin{aligned} a_n+a_{n+1} &= \\frac 1{\\binom{2n}n} \\left[1+\\frac 1{\\ \\frac{(2n+1)(2n+2)}{n+1)(n+1)}\\ }\\right] = \\frac {5n+3}{\\binom{2n}n \\cdot2(2n+1)} \\ , \\\\ b_n+b_{n+1} &= \\frac 1{\\binom{2n}n} \\left[n+\\frac {n+1}{\\ \\frac{(2n+1)(2n+2)}{n+1)(n+1)}\\ }\\right] = \\frac {5n^2+4n+1}{\\binom{2n}n \\cdot2(2n+1)} \\ , \\end{aligned} and no the first relation already realizes the telescope for the coefficient of $$F_{k+1}$$, for short this is $$\\sum_{n\\ge 0}(-1)^n(a_n+a_{n+1}) = a_0=1$$. We simimilarly use the second relation to see that the coefficient of $$F_k$$ is $$b_0=0$$. Done. $$\\square$$ If the question insists to have the values for $$A,B,C$$, then let us some", "# A Problem with the Generating Function of Fibonacci. So basically I want to find the closed form of $G_n = \\sum_{k = 1}^n \\binom{n+k - 1}{2k-1}$. After checking for $n = 1,2,3,4$ the values are $1, 3, 8, 21$ respectively. I claim that it is $F_{2n}$ where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$. My idea was to prove this using generating functions. We have that $\\binom{n+k-1}{2k-1}$ is the $n$th coefficient of $X^k \\sum_{i \\geq 0} \\binom{i + 2k - 1}{i} X^i = \\frac{X^k}{(1-X)^{2k}}$ so the $$\\sum_{n \\geq 0} G_n X^n = \\sum_{k = 1}^n \\frac{X^k}{(1-X)^{2k}}$$ Now we want to find what $F_0 + F_2X + F_4X^2 + ...$ is and we know that $F_0 + F_1X + ... = \\frac{X}{1-X-X^2}$. So \\begin{align*} F_0 + F_2X^2 + F_4X^4 + ... & = \\frac{1}{2} \\left ( \\frac{X}{1-X-X^2} + \\frac{-X}{1 + X - X^2} \\right ) \\\\ & = \\frac{X^2}{(1-X^2)^2 - X^2} \\end{align*} which means $$F_0 + F_2X + F_4X^2 + ... = \\frac{X}{(1-X)^2 - X}.$$ So if suffices to show \\begin{align*} \\frac{X}{(1-X)^2 - X} & = \\sum_{k = 1}^n \\frac{X^k}{(1-X)^{2k}} \\\\ \\frac{1}{(1-X)^2 - X} & = \\sum_{k = 1}^n \\frac{X^{k-1}}{(1-X)^{2k}} \\\\ \\frac{(1-X)^{2n}}{(1-X)^2 - X} & = ((1-X)^2)^{n-1} + ((1-X)^2)^{n-2}X + ... + X^{n-1} \\\\ \\frac{(1-X)^{2n}}{(1-X)^2 - X} & = \\frac{(1-X)^{2n} - X^n}{(1-X)^2 -", "= (-A\\phi)\\phi^{-k} + (-B\\psi)\\psi^{-k}$$ Here, $\\phi$ is the golden ratio $\\frac{1 + \\sqrt{5}}{2}$ and $\\psi = -\\phi^{-1}$ is its conjugate. We then have an easy description of the asymptotic behavior of the $w$ language: it runs in $\\Theta(\\phi^n)$. In fact, if you expand everything out, you'll find that $$w_k = \\frac{\\phi^{k} - \\psi^{k}}{\\sqrt{5}} = \\left\\lceil \\frac{\\phi^{k}}{\\sqrt{5}} \\right\\rceil$$ There's also an intricate connection to another common combinatorial class. This is just the Fibonacci numbers! Now, suppose you have $w_k$, which counts the number of strings of size $k$ that starts and ends with $k$ (and also contains no $aa$ substrings), how can we build a string that can start or end with an $a$? Well, it's simple: an admissible string is either in $w$ (starts and ends with $b$), or it is $a w$ (starts with $a$), or it is $w a$ (ends with $a$), or it is $a w a$ (starts and ends with $a$). Therefore: $$f(n) = w_n + w_{n-2} + 2 * w_{n - 1}$$ Recall that $w_n$ is the fibonacci sequence, so $w_{n-1} + w_{n-2} = w_n$, which means that \\begin{align*} f(n) &= (w_n + w_{n-1}) + (w_{n-2} + w_{n-1}) \\\\ &= w_{n+1} + w_{n} \\\\ &= w_{n+2} \\end{align*} Therefore, $f(n) = \\text{fib}(n + 2) = \\left\\lceil \\frac{\\phi^{n+2}}{\\sqrt{5}} \\right\\rceil$ Now you probably don't have to", "# Math Help - Fibonacci Proof II 1. ## Fibonacci Proof II (n choose 0)*F(0) + (n choose 1)*F(1) + ... + (n choose n)*F(n)=F(2n) I'm beginning to think induction is a bad approach... 2. Originally Posted by veronicak5678 (n choose 0)*F(0) + (n choose 1)*F(1) + ... + (n choose n)*F(n)=F(2n) I'm beginning to think induction is a bad approach... Let $\\displaystyle \\varphi=\\frac{1+\\sqrt{5}}{2}$, then a common fact (Binet's formula) says that $\\displaystyle F_n=\\frac{\\varphi^n-\\left(1-\\varphi\\right)^n}{\\sqrt{5}}$. Thus, \\displaystyle \\begin{aligned}\\sum_{k=0}^{n}{n\\choose k}F(k) &=\\frac{1}{\\sqrt{5}}\\sum_{k=0}^{n}{n\\choose k}\\left(\\varphi^k-\\left(1-\\varphi\\right)^k\\right)\\\\ &= \\frac{1}{\\sqrt{5}}\\left[\\sum_{k=0}^{n}{n\\choose k}\\varphi^k-\\sum_{k=0}^{n}{n\\choose k}\\left(1-\\varphi\\right)^k\\right]\\\\ &= \\frac{1}{\\sqrt{5}}\\left[\\left(1+\\varphi\\right)^n-\\left(2-\\varphi\\right)^n\\right]\\end{aligned} That said, a quick check shows that $\\varphi^2=1+\\varphi$ and $\\displaystyle \\left(1-\\varphi\\right)^2=1-2\\varphi+\\varphi^2=1-\\varphi+(\\varphi^2-\\varphi)=1-\\varphi+1=2-\\varphi$. So that the above shows \\displaystyle \\begin{aligned}\\sum_{k=0}^{n}{n\\choose k}F(k) &= \\frac{1}{\\sqrt{5}}\\left[\\left(1+\\varphi\\right)^n-\\left(2-\\varphi\\right)^n\\right]\\\\ &= \\frac{1}{\\sqrt{5}}\\left[\\left(\\varphi^2\\right)^n-\\left(\\left(1-\\varphi\\right)^2\\right)^n\\right]\\\\ &= \\frac{1}{\\sqrt{5}}\\left[\\varphi^{2n}-\\left(1-\\varphi\\right)^{2n}\\right]\\\\ &= F(2n)\\end{aligned} 3. A couple more proofs here. A more general form would be: $f_{n\\cdot m} = \\sum_{k=0}^n{\\binom{n}{k}\\cdot f_k \\cdot f_m^k\\cdot f^{n-k}_{m-1}}$ ( setting $m = 2$ you get the particular case above).", "5 } }{ 2^2 \\sqrt{5} } \\\\ &= \\frac{ 4 \\sqrt{5} }{ 2^2 \\sqrt{5} } \\\\ &= 1, \\end{aligned} \\label{eqn:fibonacci:200} \\begin{aligned} F_3 &= \\frac{ 1 + 3 \\sqrt{5} + 3 (5) + \\sqrt{5} 5 – \\lr{ 1 – 3 \\sqrt{5} + 3(5) – \\sqrt{5} 5 } }{ 2^3 \\sqrt{5} } \\\\ &= \\frac{ 2 \\lr{ 3 \\sqrt{5} + \\sqrt{5} 5 } }{ 2^3 \\sqrt{5} } \\\\ &= \\frac{ 3 + 5 }{ 2^2 } \\\\ &= 2. \\end{aligned} In the general case, we have \\label{eqn:fibonacci:220} \\begin{aligned} 2^n \\sqrt{5} F_n &= \\sum_{k = 0}^n \\binom{n}{k} {\\sqrt{5}}^k \\sum_{k = 0}^n \\binom{n}{k} (-\\sqrt{5})^k \\\\ &= 2 \\sum_{1 \\le k \\le n, \\mbox{$k$ is odd}} \\binom{n}{k} (\\sqrt{5})^k \\\\ &= 2 \\sqrt{5} \\sum_{m = 0}^{\\lfloor (n-1)/2 \\rfloor} \\binom{n}{2 m + 1} 5^m, \\end{aligned} so (for any $$n > 0$$), \\label{eqn:fibonacci:240} F_n = \\inv{2^{n-1}} \\sum_{m = 0}^{\\lfloor (n-1)/2 \\rfloor } \\binom{n}{2 m + 1} 5^m. Since only the odd powers of $$\\sqrt{5}$$ in the binomial expansions survive, the root in the basement is obliterated every time, leaving only integers upstairs, and a power of two factor downstairs. It is still somewhat remarkable seeming that there is always a perfect cancellation of all the factors of two in the basement.", "# Sum of inverse of Fibonacci numbers If $F(n)$ is the nth Fibonacci number, How can I prove that: $$\\sum_{i=1}^{\\infty} \\frac{1}{F(i)}\\approx 3.36\\, .$$ \\begin{align} F_n &=\\frac{\\phi^n-(-1/\\phi)^n}{\\sqrt5}\\\\ &=\\frac{\\phi^n}{\\sqrt5}\\left(1-\\left(-\\frac1{\\phi^2}\\right)^n\\right) \\end{align} Therefore, \\begin{align} \\sum_{n=1}^\\infty\\frac1{F_n} &=\\sum_{n=1}^\\infty\\frac{\\sqrt5}{\\phi^n}\\left(1+\\left(-\\frac1{\\phi^2}\\right)^n+\\left(-\\frac1{\\phi^2}\\right)^{2n}+\\left(-\\frac1{\\phi^2}\\right)^{3n}+\\cdots\\right)\\\\ &=\\sqrt5\\left(\\frac1{\\phi-1}-\\frac1{\\phi^3+1}+\\frac1{\\phi^5-1}-\\cdots\\right)\\\\ &=\\sqrt5\\sum_{k=0}^\\infty\\frac{(-1)^k}{\\phi^{2k+1}-(-1)^k} \\end{align} Since $\\frac{\\sqrt5}{\\phi^{19}+1}=0.0002392$, $$\\sqrt5\\sum_{k=0}^8\\frac{(-1)^k}{\\phi^{2k+1}-(-1)^k}=3.3600587$$ is less than $0.0002392$ too high. • It is good approach for approximation of exact value of this summation. Thanks – Amin235 Mar 24 '16 at 9:46 • +1 This is one of the good ways how to estimate the error. – yo' Mar 24 '16 at 10:01 • @Amin23 : Yes. That bound is zero, uniformly in $n$. en.wikipedia.org/wiki/… – Eric Towers Mar 24 '16 at 12:42 • Is there a bound or function based on parameter $n$ for this approximation $\\mid F(n)-\\frac{\\phi^n-{(-1/\\phi)}}{\\sqrt{5}} \\mid$ – Amin235 Mar 24 '16 at 12:47 • As $n$ gets larger, that error tends to $\\frac1{\\phi\\sqrt5}$, it does not vanish. – robjohn Mar 24 '16 at 14:34 Since $F(n) \\approx \\frac{\\phi^n}{\\sqrt 5}$ for large enough $n$, you may use that as an approximation, which give you a geometric series. For instance, if we use the approximation from the fifth term on, we get $$\\frac{1}{1} + \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\sum_{n = 4}^\\infty \\frac{\\sqrt 5}{\\phi^n} \\approx 3.3612$$ To get a feel for how accurate this is, $F(5) = 5$, while $\\frac{\\phi^5}{\\sqrt 5} \\approx", "Fibonacci sequence $$F_n=K_1\\phi^n+K_2\\left(\\frac{-1}{\\phi}\\right)^n$$ with $$F_0= 0=K_1+K_2$$ and $$F_1=1=K_1 \\phi -\\frac{K_2}{\\phi}$$ which results in $$K_2=-K_1$$ et $$1=K_1 (\\phi+\\frac{1}{\\phi})=K_1(\\frac{1+\\sqrt{5}}{2}-\\frac{1-\\sqrt{5}}{2})=K_1\\sqrt{5}$$. The golden ratio can be derived from the Fibonacci sequence. The ratio sequence of two successive integers in this sequence: $(\\frac{F_{n+1}}{F_{n}})_{n\\ge 1}$ the first 9 terms are $$1, 2, \\frac{3}{2}, \\frac{5}{3},\\frac{8}{5},\\frac{13}{8},\\frac{21}{13},\\frac{34}{21}$$. The golden ratio is the limit of this new sequence; it is hence the limit of the ratios of two consecutive terms of the Fibonacci sequence. Proof: In accordance with the formula above, as $$\\frac{1}{\\phi}<1$$, for a sufficiently large $$n$$, $$\\left(\\frac{1}{\\phi}\\right)^n$$ becomes very small and $$F_n\\simeq\\frac{1}{\\sqrt{5}}\\phi^n$$ hence $$\\frac{F_{n+1}}{F_n}$$ tends towards $$\\frac{\\frac{1}{\\sqrt{5}}\\phi^{n+1}}{\\frac{1}{\\sqrt{5}}\\phi^n}=\\phi$$. 3. Golden spirals The largest possible square within a golden rectangle is, once again, a golden rectangle. Proof: Start with a golden rectangle whose length is $$L$$ and width is $$\\ell$$, or $$\\frac{L}{\\ell}=\\phi$$. Remove a square from side $$\\ell$$. What remains is a rectangle whose length is $$\\ell$$ and width is $$L-\\ell$$. It is a golden rectangle $\\frac{\\ell}{L-\\ell}=\\frac{1}{\\left(\\frac{L-\\ell}{\\ell}\\right)}=\\frac{1}{\\phi-1}=\\phi$ since $$\\phi(\\phi-1)=\\phi^2-\\phi=1$$. To obtain a golden spiral, draw quarter circles in each successive square derived from the golden rectangles. Pages: 1 2 3", "F_k &= \\frac 1{\\sqrt{5}}\\left[\\phi^k-(-\\frac 1\\phi)^k \\right]\\\\ F_{k-1} &= \\frac 1{\\sqrt{5}}\\left[\\phi^{k-1}-(-\\frac 1\\phi)^{k-1} \\right]\\\\ &=\\frac 1{\\sqrt{5}} \\left[\\phi^k \\cdot \\frac 1\\phi -(-\\frac 1\\phi)^k \\cdot (-\\phi)\\right]\\\\ \\end{align} Hence, \\begin{align} F_{k+1}&=F_{k}+F_{k-1}\\\\ &=\\frac 1{\\sqrt{5}} \\left[\\phi^k \\cdot \\left( 1+\\frac 1\\phi \\right) -(-\\frac 1\\phi)^k \\cdot \\left( 1-\\phi \\right)\\right]\\\\ &=\\frac 1{\\sqrt{5}} \\left[\\phi^k \\cdot \\phi -(-\\frac 1\\phi)^k \\cdot \\left( -\\frac 1\\phi \\right)\\right]\\\\ &=\\frac 1{\\sqrt{5}} \\left[\\phi^{k+1}-(-\\frac 1\\phi)^{k+1} \\right] \\end{align} i.e. if formula is true for $n=k-1$ and $n=k$, it is also true for $n=k+1$. For $n=0$ and $n=1$, $F_0=0$ and $F_1=1$ respectively. Hence $F_2=F_0+F_1=1$. It can easily be shown that the formula is true for $n=2$. Hence, by induction, formula is true for all positive integer $n\\geq 2$. • You're welcome :) – hypergeometric Aug 12 '14 at 14:36 Proof: Let n=1 thus, \\begin{align*} F_1&=\\frac{(\\frac{1+\\sqrt{5}}{2})^{1}-(\\frac{1-\\sqrt{5}}{2})^{1}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{1+\\sqrt{5}}{2}-\\frac{1-\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{1+\\sqrt{5}-1+\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{\\sqrt{5}+\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{\\frac{2\\sqrt{5}}{2}}{\\sqrt{5}}\\\\ &=\\frac{{\\sqrt{5}}}{\\sqrt{5}}\\\\ &=1 \\end{align*} Suppose $$F_k=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k}}{\\sqrt{5}}$$ Also $F(k+1)=F(k)+F(k-1)$ \\begin{align*} F(k)+F(k-1)&=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k}}{\\sqrt{5}}+\\frac{(\\frac{1+\\sqrt{5}}{2})^{k-1}-(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k}+(\\frac{1+\\sqrt{5}}{2})^{k-1}-(\\frac{1-\\sqrt{5}}{2})^{k}-(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{1+\\sqrt{5}}{2})+1)(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{1-\\sqrt{5}}{2})+1)(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{3+\\sqrt{5}}{2}))(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{3-\\sqrt{5}}{2}))(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{((\\frac{1+\\sqrt{5}}{2})^{2})(\\frac{1+\\sqrt{5}}{2})^{k-1}-((\\frac{1-\\sqrt{5}}{2})^{2})(\\frac{1-\\sqrt{5}}{2})^{k-1}}{\\sqrt{5}}\\\\ &=\\frac{(\\frac{1+\\sqrt{5}}{2})^{k+1}-(\\frac{1-\\sqrt{5}}{2})^{k+1}}{\\sqrt{5}}\\hfill \\end{align*} • This is a good proof but you should have two bases cases $F_1$ and $F_0$ for the inductive step to work. I.e. at the moment you can not prove $F_2$ is true as it relies upon $F_1$ and $F_0$ being true. – Ian Miller Oct 28 '16 at 15:32 • @IanMiller Is that way the inductive hypothesis has an \"also\", because it is two inductive hypotheses? – GFauxPas Oct 28 '16 at 16:26 By", "# Looking for $\\sum_{n=1}^{\\infty}\\frac{1}{F_{n}F_{n+2}}$ I am looking for the sum of the series: $$\\sum_{n=1}^{\\infty}\\frac{1}{F_{n}F_{n+2}}$$ where $F_{n}$ is the $n$-th Fibonacci number. I was thinking about splitting the fraction into 2 like in the case of $\\dfrac{1}{n(n+1)}$, but do not see how it should work in this particular case. Thank you. \\begin{align}\\sum_{n=1}^{N}\\frac{1}{f(n)f(n+2)} &=\\sum_{n=1}^{N}\\frac{\\color{red} {f(n+1)}}{f(n)\\color{red} {f(n+1)}f(n+2)}\\\\ &=\\sum_{n=1}^{N}\\frac{f(n+2)-f(n)}{f(n)f(n+1)f(n+2)}\\\\ &=\\sum_{n=1}^{N}\\frac{1}{f(n)f(n+1)}-\\frac{1}{f(n+1)f(n+2)}\\\\ &=\\frac{1}{f(1)f(2)}-\\frac{1}{f(N+1)f(N+2)}\\end{align}.", "small values of a: • For an odd prime p ≠ 3, $\\left(\\frac{3}{p}\\right) = (-1)^{\\big\\lfloor \\frac{p+1}{6}\\big\\rfloor} =\\begin{cases} 1 & \\mbox{ if }p \\equiv 1\\mbox{ or }11 \\pmod{12} \\\\ -1 & \\mbox{ if }p \\equiv 5\\mbox{ or }7 \\pmod{12}. \\end{cases}$ • For an odd prime p ≠ 5, $\\left(\\frac{5}{p}\\right) =(-1)^{\\big\\lfloor \\frac{p+2}{5}\\big \\rfloor} =\\begin{cases} 1 & \\mbox{ if }p \\equiv 1\\mbox{ or }4 \\pmod5 \\\\ -1 & \\mbox{ if }p \\equiv 2\\mbox{ or }3 \\pmod5. \\end{cases}$ • The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... are defined by the recurrence F1 = F2 = 1, Fn+1 = Fn + Fn−1. If p is a prime number then $F_{p-\\left(\\frac{p}{5}\\right)} \\equiv 0 \\pmod p, \\qquad F_{p} \\equiv \\left(\\frac{p}{5}\\right) \\pmod p.$ • For example, \\begin{align} (\\tfrac{2}{5}) &= -1, && F_3 = 2, F_2 = 1,\\\\ (\\tfrac{3}{5}) &= -1, && F_4 = 3, F_3 = 2,\\\\ (\\tfrac{5}{5}) &= 0, && F_5 = 5, \\\\ (\\tfrac{7}{5}) &= -1, && F_8 = 21, F_7 = 13,\\\\ (\\tfrac{11}{5})&= 1, && F_{10} = 55, F_{11} = 89. \\end{align} • This result comes from the theory of Lucas sequences, which are used in primality testing.[4] See Wall–Sun–Sun prime. ## Legendre symbol and quadratic reciprocity Let p and q be odd primes. Using the Legendre symbol, the quadratic reciprocity law can be stated concisely:", "step: use the inductive hypothesis to show that the statement is true for the k + 1th step. In the case of the integer sum formula, we would prove the following: assuming that the sum of the first k positive integers is k((1 + k)/2) , the sum of the first k +1 positive integers is ((k + 1) (1 + (k + 1)/2) . Carrying out this kind of proof requires that you perform each of these steps., For the third step in particular you must rely on your algebra skills. #### Example A Prove: The sum of the first n positive integers is n(1+n)2\\begin{align*}\\frac{n(1 + n)} {2}\\end{align*} Solution Use the three steps of mathematical induction: Step 1) The base case: If n = 1, the entire sequence is just 1 and therefore the sum is 1. Also, n(1+n)2=1(1+1)2=22=1\\begin{align*}\\frac{n(1 + n)} {2} = \\frac{1(1 + 1)} {2} = \\frac{2} {2} = 1\\end{align*}. This establishes the base case. Step 2) Assume that the sum of the first k\\begin{align*}k\\end{align*} positive integers is k(1+k)2\\begin{align*}\\frac{k(1 + k)} {2}\\end{align*} . In other words, assume that 1+2+3++k=k(1+k)2\\begin{align*}1+2+3+\\cdots+k=\\frac{k(1+k)}{2}\\end{align*}. Step 3) We must show that the sum of the first k+1\\begin{align*}k + 1\\end{align*} positive integers is (k+1)(1+(k+1))2\\begin{align*}\\frac{(k +1) (1 + (k + 1))} {2}\\end{align*}. In other words, we must show that 1+2+3++k+(k+1)=(k+1)(1+(k+1))2\\begin{align*}1+2+3+\\cdots+k+(k+1)=\\frac{(k+1)(1+(k+1))}{2}\\end{align*}. There are two", "&= \\frac{(r + m - 2)!}{r! (m - 2)!} + \\frac{(r + m - 2)!}{(r - 1)! (m - 1)!} \\\\ &= \\frac{(r + m - 2)!}{(r - 1)! (m - 2)!} \\left(\\frac{1}{r} + \\frac{1}{m - 1}\\right) \\\\ &= \\frac{(r + m - 2)!}{(r - 1)! (m - 2)!} \\cdot \\frac{r + m - 1}{r(m - 1)} \\\\ &= \\frac{(r + m - 1)!}{r! (m - 1)!} \\\\ &= \\binom{r + m - 1}{m - 1} \\end{align*} If $$g(r, m) = \\binom{n}{k}$$, then \\begin{align*} \\binom{r + m - 1}{m - 1} &= \\binom{n}{k} \\\\ r = n - k, m &= k + 1 \\end{align*} For $$n \\lt k$$, $$\\binom{n}{k} = g(n - k, k + 1)$$. For $$n = k$$, $$\\binom{n}{k} = 1 = g(1, 1)$$. #### Susam Pal from cotpi added: Let the number in the $$r$$th row and $$c$$th column of the grid be $$f(r, c)$$. We can show that every row contains a strictly monotonically increasing sequence of positive integers. This can be done by proving that $$f(r, c) \\lt f(r, c + 1)$$ for integers $$r \\geqslant 1$$ and $$c \\geqslant 1$$. This is clearly true for the first row since $$f(r, c) = c$$ and $$f(r, c + 1) = c + 1$$. If we assume that this is true for $$(r", "- 1 = (n - 1)(n + 1)$ to find something like \\begin{align*} \\left(1 - \\frac 1 {2^2}\\right)&\\left(1 - \\frac 1 {3^2}\\right) \\cdots \\left(1 - \\frac 1 {10000^2}\\right) \\\\ & =\\left(\\frac 1 2 \\cdot\\frac 3 2\\right) \\left(\\frac 2 3 \\cdot\\frac 4 3\\right) \\left(\\frac 3 4 \\cdot \\frac 5 4\\right) \\left(\\frac 4 5 \\cdot \\frac 6 5 \\right)\\cdots \\left(\\frac{10000 - 1}{10000} \\cdot \\frac{10000+1}{10000}\\right) \\end{align*} Convince yourself that everything cancels (i.e. the product telescopes), except for a factor of $1/2$, and the final factor of $10001/10000$. - Thanks! I should have gone into that direction more deeply but anyway this is a lesson that I should always remember: \"Try all the pistes even if they seem irrelevant\" – LoveFood Mar 7 '14 at 18:35 \\begin{align} \\prod_{k=2}^n\\left(1-\\frac1{k^2}\\right) &=\\prod_{k=2}^n\\frac{(k-1)(k+1)}{k^2}\\\\ &=\\prod_{k=2}^n\\frac{k-1}{k}\\hspace{1.2cm}\\prod_{k=3}^{n+1}\\frac{k}{k-1}\\\\ &=\\frac12\\left(\\prod_{k=3}^n\\frac{k-1}{k}\\right)\\frac{n+1}{n}\\left(\\prod_{k=3}^n\\frac{k}{k-1}\\right)\\\\ &=\\frac12\\frac{n+1}{n}\\\\ &=\\frac{n+1}{2n} \\end{align} - Thank you man, but is there some trick that doesn't use the pi notation ($\\prod$) like Gauss' trick to evaluate that product? I will be very thankful if you help me. – LoveFood Mar 10 '14 at 11:45 @LoveFood: don't fear the $\\prod$ notation; it is very useful for writing products of a sequence in a compact way. I am not sure which trick of Gauss you mean. – robjohn Mar 10 '14 at 12:00 $$\\left(\\frac{3}{2}\\frac{4}{3}\\frac{5}{4}\\cdots\\frac{n}{n-1}\\frac{n+1}{n}\\right)\\cdot\\left(\\frac{1}{2}\\frac{2}{3}\\frac{3}{4}\\cdots\\frac{n-2}{n-1}\\frac{n-1}{n}\\right)$$ After mass cancellations, pull the $$\\frac{n+1}{2}\\text{ and }\\frac{1}{n}$$ $$\\frac{n+1}{2}\\cdot\\frac{1}{n}$$ Substitute the", "\\alpha a + \\beta b \\\\ &= \\alpha \\lr{ a – b } \\\\ &= 1, \\end{aligned} and \\label{eqn:fibonacci:340} \\begin{aligned} F_n &= F_{n-1} + F_{n-2} \\\\ &= \\alpha \\lr{ a^{n-1} + a^{n-2} } -\\alpha \\lr{ b^{n-1} + b^{n-2} } \\\\ &= \\alpha a^{n-2} \\lr{ 1 + a } -\\alpha b^{n-2} \\lr{ 1 + b }, \\end{aligned} so \\label{eqn:fibonacci:360} \\begin{aligned} a^2 &= a + 1 \\\\ b^2 &= b + 1. \\end{aligned} If we complete the square we find \\label{eqn:fibonacci:380} \\lr{ a – \\inv{2} }^2 = 1 + \\inv{4} = \\frac{5}{4}, or \\label{eqn:fibonacci:400} a, b = \\inv{2} \\pm \\frac{\\sqrt{5}}{2}. Out pop the golden ratio and it’s complement. Clearly we need to pick alternate roots for $$a$$ and $$b$$ or else we’d have zero for every value of $$n > 0$$. Suppose we pick the positive root for $$a$$, then to find the scaling constant $$\\alpha$$, we just compute \\label{eqn:fibonacci:420} \\begin{aligned} 1 &= \\alpha \\lr{ \\frac{ 1 + \\sqrt{5}}{2} – \\frac{ 1 – \\sqrt{5} }{2} } \\\\ &= \\alpha \\sqrt{5}, \\end{aligned} so our system \\ref{eqn:fibonacci:280} has the solution: \\label{eqn:fibonacci:520} \\begin{aligned} a &= \\frac{1 + \\sqrt{5}}{2} \\\\ b &= \\frac{1 – \\sqrt{5}}{2} \\\\ \\alpha &= \\inv{\\sqrt{5}} \\\\ \\beta &= -\\inv{\\sqrt{5}}. \\end{aligned} We now see a path that will systematically lead us from the Fibonacci difference equation to the final result, and" ]

[ "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]", "should be transported from godown B to shop E and F, respectively. ∴ (60 – x) ≥ 0, i.e., x ≤ 60, (50 – y) ≥ 0 Z = 6x + 3y + 2.50(100 – x – y) + 4(60 – x) + 2(50 – y) + 3{40 – [100 – (x + y)]} = 6x + 3y + 150 – 2.50x – 2.50y + 240 – 4x + 100 – 2y + 120 – 300 + 3x + 3y = 2.50x + 1.5y + 410 So, our L.P.P. is to minimise Z = 2.5x + 1.5y + 410 subject to constraints x ≤ 60, y ≤ 50, x + y ≥ 60, x + y ≤ 100, x, y ≥ 0. Plot the graphs of equations x = 60, y = 50, x + y = – 60, x + y = 100 on a graph paper. The region ABCD shows shaded area in the figure which satisfies the inequalities x ≤ 60, y ≤ 50, x + y ≥ 60, x + y ≤ 100 and x, y ≥ 0. We calculate values of Z at points A, B, C and D which are the corner points of the region ABCD. At A(10, 50), Z = 2.5x + 1.5y + 410 = 2.5 × 10 +", "graph(f,-10,10,n=200,operator..) where n=200 is the number of sample points, and operator.. means curvy connecting that is necessary in this case. // http://asymptote.ualberta.ca/ unitsize(8mm,2cm); size(8cm); import graph; real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(f,-10,10,n=200,operator..),red); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,1.5,Arrow(TeXHead)); label(\"The graph of $y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\",point(N),2N); label(\"$-2$\",(0,-2),SW); shipout(bbox(5mm,invisible)); Appendix The function f(x) is asymptotic to cos(x) when |x| is big. unitsize(1cm,4cm); size(8cm); import graph; import math; //for drawline real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(cos,-20,20,n=200,operator..),lightblue+1pt); draw(graph(f,-20,20,n=200,operator..),red+.5pt); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,2.5,Arrow(TeXHead)); drawline((0,1),(1,1),gray+.2pt); drawline((0,-1),(1,-1),gray+.2pt); drawline((0,-2),(1,-2),gray+.2pt); label(scale(.7)*\"$1$\",(0,1),NE); label(scale(.6)*\"$-1$\",(0,-1),SW); label(scale(.6)*\"$-2$\",(0,-2),SW); label(\"$\\color{red}y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\"+\" and \"+\"$\\color{blue}y=\\cos x$\",point(N),2N); shipout(bbox(5mm,invisible)); • Maybe return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x));. Oct 17 at 16:37 • @NguyenVanChi1998 Yes, that is more concise syntax real f(real x) { return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x)); }; Oct 17 at 16:41", "You weren't shown how to graph these problems? Constraint [4] puts us in Quandrant 1. For $[1]\\;3x + 2y \\:\\leq \\:120$, graph the line: . $3x + 2y \\:=\\:120$ It has intercepts: (40, 0), (0, 60) Draw the line and shade the region below it. For $[2]\\;x + 4y \\:\\leq \\:80$ graph the line: . $x + 4y \\:=\\:80$ It has intercepts: (80, 0), (0, 20) Draw the line and shade the region below it. The shaded region looks like this . . . Code: | 60 * |* | * | * | * | * 20 o * |:::* * |:::::::o |::::::::* * |:::::::::* * |::::::::::* * - - o - - - - - o - - - - - * - - | 40 80 Constraint [3] is useless. The line $2x + y \\:=\\:100$ has intercepts (50, 0), (0, 100) . . and falls outside of the shaded region. We are concerned with the vertices of the shaded region. We can see three of them: .(0, 0), (40, 0), (0, 20) We find the fourth by locating the intersection of the two slanted lines. We solve the system: . $\\begin{array}{ccc}3x + 2y &=&120 \\\\ x + 4y &=& 80\\end{array}\\;\\text{ and get: }\\32,\\:12)\" alt=\"\\begin{array}{ccc}3x + 2y &=&120 \\\\ x + 4y &=& 80\\end{array}\\;\\text{ and", "> x) = 1 – P (X < x). Figure 5.9 Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = $1 20 1 20$, 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. Figure 5.10 $P(2.3 Try It 5.1 Consider the function f(x) = $1 8 1 8$ for 0 ≤ x ≤ 8. Draw the graph of f(x) and find P(2.5 < x < 7.5). Order a print copy As an Amazon Associate we earn from qualifying purchases.", "figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ /* dots and labels */ label(\"f\",(-6.12,-0.12),NE * labelscalefactor); label(\"a = 2\",(1.46,0.24),NE * labelscalefactor,zzttqq); dot((-3.14,0),dotstyle); label(\"A\",(-3.06,0.12),NE * labelscalefactor); dot((2.2,2.04),dotstyle); dot((4.5,2.08),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ Here is the Asymptote code produced: (with some spacing added) /* Geogebra to Asymptote conversion, documentation at userscripts.org/scripts/show/72997 */ import graph; size(14.26cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.22,xmax = 8.04,ymin = -4.18,ymax = 4.76; /* image dimensions */ pen fueaev = rgb(0.96,0.92,0.9); pen zzttqq = rgb(0.6,0.2,0); real f2 (real x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); filldraw((-3.14,0)--(0,0)--(0,3.14)--(-3.14,pi )--cycle, fueaev, zzttqq); Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real", "# Difference between revisions of \"1982 AHSME Problems/Problem 25\" ## Problem The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad \\text{(D)}\\frac{21}{32}\\qquad \\text{(E)}\\frac{3}{4}$ ## Solutions ### Solution 1 $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); label(\"C_3\", (17,7), SW); label(\"C_2\", (17,11), SW); label(\"C_1\", (17,15), SW); label(\"C_0\", (17,19), SW); filldraw((17,7)--(18,7)--(18,8)--(17,8)--cycle, black); filldraw((17,11)--(18,11)--(18,12)--(17,12)--cycle, black); filldraw((17,15)--(18,15)--(18,16)--(17,16)--cycle, black); filldraw((17,19)--(18,19)--(18,20)--(17,20)--cycle, black); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ The probability that the student passes", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "that graphs those $3$ equations for some constant $a$. WLOG, we can set $a=1$. To make our life easier, we can also focus entirely on the first quadrant, since all $4$ graphs have different shapes in the first quadrant. Therefore, $x \\geq 0$ and $y \\geq 0$. First, we have $|x| + |y| = 1$. Since $x \\geq 0$ and $y \\geq 0$, this becomes $x + y = 1$. After a bit of rearranging, we get $$y = -x + 1$$. $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); label(\"X\", (-28,0), S); label(\"Y\", (-40,12), E); [/asy]$ This eliminates $III$, which does not have a diagonal line in the first quadrant. Next, we have $\\sqrt{2(x^{2}+y^{2})} = 1$. Squaring both sides and dividing by $2$, we get $$x^2+y^2=\\frac{\\sqrt{2}}{2}^2$$ Note that this describes a circle with the center $(0,0)$ and radius $\\frac{\\sqrt{2}}{2}$. We can find that $(\\frac{1}{2}, \\frac{1}{2})$ is on the circle, which is also on the line $y = -x + 1$. Therefore, the circle intersects the line at that point. We can graph this as follows: $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); draw(arc((-40,0),4.25, 0, 90)); label(\"X\", (-28,0), S); label(\"Y\", (-40,12), E); [/asy]$ This eliminates $I$, which does not include the intersection point listed above. Finally, we have $2\\mbox{Max}(|x|, |y|) = 1$. Rearranging and substituting $x$ for $|x|$ and", "between the x-axis and the graph of f over the interval $[−1,2][−1,2]$ using rectangles. For the rectangles, use the square units, and approximate both above and below the lines. Use geometry to find the exact answer. For the following exercises, consider the function $f(x)=1−x2.f(x)=1−x2.$ (Hint: This is the upper half of a circle of radius 1 positioned at $(0,0).)(0,0).)$ 26. Sketch the graph of f over the interval $[−1,1].[−1,1].$ 27. Use the preceding exercise to find the aproximate area between the x-axis and the graph of f over the interval $[−1,1][−1,1]$ using rectangles. For the rectangles, use squares 0.4 by 0.4 units, and approximate both above and below the lines. Use geometry to find the exact answer. For the following exercises, consider the function $f(x)=−x2+1.f(x)=−x2+1.$ 28. Sketch the graph of f over the interval $[−1,1].[−1,1].$ 29. Approximate the area of the region between the x-axis and the graph of f over the interval $[−1,1].[−1,1].$ Order a print copy As an Amazon Associate we earn from qualifying purchases.", "O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"O\", O, ENE); label(\"A\", A, NW); label(\"B\", B, W); label(\"C\", C, E); label(\"H\", H, E); label(\"H'\", K, NE); label(\"X\", X, W); label(\"Y\", Y, NE); label(\"O'\", P, E); label(\"M\", M, NE); label(\"L\", L, NE); label(\"D\", D, NNE); label(\"2\", X -- H, NW); label(\"3\", H -- A, SW); label(\"6\", H -- Y, NW); label(\"R\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy]$ Diagram not to scale. We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\\angle HBC = 90 - \\angle C = \\angle H'AC = \\angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\\triangle BHC$ is the image of circle $O$ over line", "refer to the following diagram. Note that the area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of", "methods. From $x \\geq 0,\\;y \\geq 0$, we are in Quadrant 1. Graph the line: . $2x + y \\:=\\:10$ It has intercepts: $(5,0),\\:(0,10)$ Draw the line and shade the region above the line. Graph the line: . $-3x + 2y \\:=\\:6$ It has intercepts: $(-2,0),\\;(0,3)$ Draw the line and shade the region below the line. Graph the line: . $x + y \\:=\\:6$ It has intercepts: $(6,0),\\;(0,6)$ Draw the line and shade the region above the line. Code: | 10 * |* | * | * | * | * | * | * 6 * * * | * * *:::: | * * *:::::::: | * o:::::::::::: | * *::::::::::: | * * *:::::::::: 3 * * *::::::::: * | * *:::::::: * | **::::::: * | o:::::: * | **:::: - * - - - - - - - - - + - - - - - - - - - * o - - -2 | 5 6 The shaded region has vertices: . $(6,0),\\;(4,2),\\;(2,6)$ Test those in the $Z$-function for minimum $Z.$", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We", "draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\\frac{1^2\\sqrt{3}}{4}\\implies\\frac{1\\sqrt{3}}{4}$, and $\\frac{2^2\\sqrt{3}}{4}\\implies\\frac{4\\sqrt{3}}{4}\\implies\\sqrt{3}$ and the area of the large equilateral triangle is $\\frac{7^2\\sqrt{3}}{4}\\implies\\frac{49\\sqrt{3}}{4}$ so the area of the hexagon would be $\\frac{49\\sqrt{3}}{4}-\\frac{\\sqrt {3}}{4}-\\frac{\\sqrt{3}}{4}-\\sqrt{3}\\implies\\boxed{\\frac{43\\sqrt{3}}{4}}$ Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can", "now taking shape our x and y axis are still a little lacking. Lets improve them by adding ticks and arrow heads: 1. \\begin{tikzpicture}[scale = 2] 3. \\def\\ANG{40} % the angle 4. \\draw[step=0.2, style=gridstyle1] (-1,-1) grid (1,1); 5. \\draw [->] (-1.2,0) — (1.2,0); \\draw [->] (0,-1.2) — (0,1.2); 6. \\draw (0,0) circle (\\R); 7. \\filldraw[fill=green!20!white, draw=green!50!black] 8. (0,0) — (0.3,0) arc (0:\\ANG:0.3) — cycle; 9. \\foreach \\x in {-1,-0.9, …, 1}{ 10. \\draw (\\x,-0.05) — (\\x,0.05); 11. } 12. \\foreach \\y in {-1,-0.9, …,1}{ 13. \\draw (-0.05,\\y) — (0.05,\\y); 14. } 15. \\end{tikzpicture} • The arrow heads for the x and y axis prove to be quite easy to add. Just include the [->] option before after the \\draw. Arrows heads can also be added to arcs and can have a head on both sides by using []. • To add ticks to our axis the for‘ loop is introduced. This allows us to define a symbol (in this case \\x), which we iterate over values given in the braces: {}. We could list the values separated by commas but in this case it is easier to specify it by {start, start+step…, end}. The statement(s) in the loop are then repeated for each \\x. • The y-ticks are produced in a similar way. Now we might like to", "# Difference between revisions of \"2004 AMC 10A Problems/Problem 21\" ## Problem Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\\pi$ radians is $180$ degrees.) $[asy] size(85); fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7)); fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7)); fill((-20,0)..(0,20)--(0,-20)..cycle,white); fill((20,0)..(0,20)--(0,-20)..cycle,white); fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7)); fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7)); fill((0,10)..(-10,0)--(10,0)..cycle,white); fill((0,-10)..(-10,0)--(10,0)..cycle,white); fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7)); fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7)); draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),20),linewidth(0.7)); draw(Circle((0,0),30),linewidth(0.7)); draw((-28,-21)--(28,21),linewidth(0.7)); draw((-28,21)--(28,-21),linewidth(0.7));[/asy]$ $\\mathrm{(A) \\ } \\frac{\\pi}{8} \\qquad \\mathrm{(B) \\ } \\frac{\\pi}{7} \\qquad \\mathrm{(C) \\ } \\frac{\\pi}{6} \\qquad \\mathrm{(D) \\ } \\frac{\\pi}{5} \\qquad \\mathrm{(E) \\ } \\frac{\\pi}{4}$ ## Solution 1 Let the area of the shaded region be $S$, the area of the unshaded region be $U$, and the acute angle that is formed by the two lines be $\\theta$. We can set up two equations between $S$ and $U$: $S+U=9\\pi$ $S=\\dfrac{8}{13}U$ Thus $\\dfrac{21}{13}U=9\\pi$, and $U=\\dfrac{39\\pi}{7}$, and thus $S=\\dfrac{8}{13}\\cdot \\dfrac{39\\pi}{7}=\\dfrac{24\\pi}{7}$. Now we can make a formula for the area of the shaded region in terms of $\\theta$: $\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot (4\\pi-\\pi)+\\dfrac{2\\theta}{2\\pi}(9\\pi-4\\pi)=\\theta +3\\pi-3\\theta+5\\theta=3\\theta+3\\pi=\\dfrac{24\\pi}{7}$ Thus $3\\theta=\\dfrac{3\\pi}{7}\\Rightarrow \\theta=\\dfrac{\\pi}{7}\\Rightarrow\\boxed{\\mathrm{(B)}\\ \\frac{\\pi}{7}}$ ## Solution 2 As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $\\theta$. $\\implies\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot", "# Difference between revisions of \"1998 AJHSME Problems/Problem 23\" ## Problem If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle? $[asy] unitsize(10); draw((0,0)--(12,0)--(6,6sqrt(3))--cycle); draw((15,0)--(27,0)--(21,6sqrt(3))--cycle); fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black); draw((30,0)--(42,0)--(36,6sqrt(3))--cycle); fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black); fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black); fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black); draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black); fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black); fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black); fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black); fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black); fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black); [/asy]$ $\\text{(A)}\\ \\frac{3}{8}\\qquad\\text{(B)}\\ \\frac{5}{27}\\qquad\\text{(C)}\\ \\frac{7}{16}\\qquad\\text{(D)}\\ \\frac{9}{16}\\qquad\\text{(E)}\\ \\frac{11}{45}$ ## Solution All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern: $0, 1, 3, 6, ...$ which is the pattern of \"triangular numbers\". Each time, the number $1, 2, 3, 4, 5...$ is added to the previous term. Thus, the first eight terms are: $0, 1, 3, 6, 10, 15, 21, 28$ In the eighth diagram, there will be $28$ shaded triangles. The total number of small triangles follows the pattern: $1, 4, 9, 16, ...$ which is the pattern of \"square numbers\". Thus, the eighth triangle will be divided into $8^2 = 64$ small triangles in total. The ratio of shaded to total triangles will be the fraction of the whole figure that's shaded, since all triangles are congruent. Thus, the answer is $\\frac{28}{64} = \\frac{7}{16}$, and the correct choice is $\\boxed{C}$ 1998 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 •", "(x) < 3:42 \u0018;1:5\u0001 ;1] ;0:12 < f (x) < 1:46 \u0018;1\u0001 ;0:5] ;1:53 < f (x) < 0:01 \u0018;0:5\u0001 0] ;1:52 < f (x) < ;0:95 \u00180\u0001 0:5] ;1:52 < f (x) < ;1:37 \u00180:5\u0001 1] ;1:49 < f (x) < ;0:83 \u00181\u0001 1:5] ;0:95 < f (x) < 0:22 \u00181:5\u0001 2] 0:08 < f (x) < 1:76 f\u00181\u0001 1:5]\u0001 \u00181:5\u0001 2]\u0001 \u00182\u0001 2:5]\u0001 \u00182:5\u0001 3]\u0001 \u00183\u0001 3:5]g f\u0018;0:5\u0001 0]\u0001 \u00180\u0001 0:5]\u0001 \u00180:5\u0001 1]\u0001 \u00181\u0001 1:5]g f\u0018;2\u0001 ;1:5]\u0001 \u0018;1:5\u0001 ;1]\u0001 \u0018;1\u0001 0]\u0001 \u00180\u0001 1:5]g f\u0018;2\u0001 ;1:5]\u0001 \u0018;1:5\u0001 ;1]\u0001 \u0018;1\u0001 ;0:5]g f\u00181\u0001 1:5]\u0001 \u00181:5\u0001 2]\u0001 \u00182\u0001 2:5]\u0001 \u00182:5\u0001 3]\u0001 \u00183\u0001 3:5]g f\u0018;0:5\u0001 0]\u0001 \u00180\u0001 0:5]\u0001 \u00180:5\u0001 1]\u0001 \u00181\u0001 1:5]g f\u0018;2\u0001 ;1:5]\u0001 \u0018;1:5\u0001 ;1]\u0001 \u0018;1\u0001 0]\u0001 \u00180\u0001 1:5]g f\u00180\u0001 0:5]\u0001 \u00180:5\u0001 1]\u0001 \u00181\u0001 1:5]\u0001 \u00181:5\u0001 2]g Table 2.8: Edges and Vertices for the graphs of X = \u0018;2\u0001 2] and Y = \u0018;2\u0001 4]. what was obtained with intervals of unit length. In fact, one can obtain as good an approximation as one likes by choosing the edge lengths su\u0017ciently small. In Figure 2.9 one sees the graph of the multivalued map when the lengths of the edges is 0:1. 2.2.2 Constructing Chain Maps In the previous section we considered the problem of approximating maps from one interval to another. Of course the goal of this course is to" ]

[ "to be zero). Rotate the figure by moving the blue point on the slider to the left to see that, since it is a graph in two dimensions, the graph is really a line. And, in three dimensions, the graph of $x=3$ is a plane. A vertical line or a plane. The graph of $x=3$ would be a line, if it were an equation in two dimensions, in terms of the variables $x$ and $y$, for example. However, here it is an equation in three dimensions, in terms of the variables $x$, $y$, and $z$, so that its graph is really a plane. Rotate the graph with you mouse to see this fact. Since the equation does not depend on $y$ or $z$, the plane is parallel to both the $y$ and the $z$ axis. You can press Home to restore the original view.", "Math Insight Applet: A vertical line or a plane The graph of $x=3$ would be a line, if it were an equation in two dimensions, in terms of the variables $x$ and $y$, for example. However, here it is an equation in three dimensions, in terms of the variables $x$, $y$, and $z$, so that its graph is really a plane. Rotate the graph with you mouse to see this fact. Since the equation does not depend on $y$ or $z$, the plane is parallel to both the $y$ and the $z$ axis. You can press Home to restore the original view.", "20:50 If $k > 1$ (resp. $0 < k < 1$) then it stretches (resp. shrinks) the plot in the $y$-axis direction. If $k$ is negative then it's the same $+$ mirroring about $x$-axis. – user2468 Aug 20 '12 at 21:02 For a positive value $k$, the graph of $y = kf(x)$ is obtained from the graph of $y=f(x)$ by scaling in the vertical direction. In coordinates, if $(x_0, y_0)$ is on the graph of $y=f(x)$, then that point moves to $(x_0, ky_0)$ on the graph of $y=kf(x)$. Any point on the $x$-axis remains fixed (as then $y=0$). When $k > 1$, the graph is stretched vertically (away from the $x$-axis), while if $0 < k < 1$, the graph is compressed vertically (toward the $x$-axis). Another way to think about this: The graph of $y=kf(x)$ is exactly the same as the graph of $y=f(x)$ if the $y$-axis is scaled so that $1$ is relabeled $k$, $2 \\mapsto 2k$, $-1 \\mapsto -k$, etc.", "## Precalculus (6th Edition) (a) $f(2)=-3$ (b) $f(-1)=2$ (a) The value of $f(2)$ can be found by looking for the value of $y$ on the graph when $x=2$. The graph shows that when $x=2$, the corresponding value of $y$ is $-3$. Thus, $f(2) =-3$. (b) The value of $f(-1)$ can be found by looking for the value of $y$ on the graph when $x=-1$. The graph shows that when $x=-1$, the corresponding value of $y$ is $2$. Thus, $f(-1) =2$.", "3, whereas the graph of $f(x)=x2/3f(x)=x2/3$ is the graph of $y=x2y=x2$ compressed vertically by a factor of $33$ (Figure 1.25). Figure 1.25 (a) If $c>1,c>1,$ the graph of $y=cf(x)y=cf(x)$ is a vertical stretch of the graph of $y=f(x).y=f(x).$ (b) If $0 the graph of $y=cf(x)y=cf(x)$ is a vertical compression of the graph of $y=f(x).y=f(x).$ The horizontal scaling of a function occurs if we multiply the inputs $xx$ by the same positive constant. For $c>0,c>0,$ the graph of the function $f(cx)f(cx)$ is the graph of $f(x)f(x)$ scaled horizontally by a factor of $c.c.$ If $c>1,c>1,$ the graph of $f(cx)f(cx)$ is the graph of $f(x)f(x)$ compressed horizontally. If $0 the graph of $f(cx)f(cx)$ is the graph of $f(x)f(x)$ stretched horizontally. For example, consider the function $f(x)=2xf(x)=2x$ and evaluate $ff$ at $x/2.x/2.$ Since $f(x/2)=x,f(x/2)=x,$ the graph of $f(x)=2xf(x)=2x$ is the graph of $y=xy=x$ compressed horizontally. The graph of $y=x/2y=x/2$ is a horizontal stretch of the graph of $y=xy=x$ (Figure 1.26). Figure 1.26 (a) If $c>1,c>1,$ the graph of $y=f(cx)y=f(cx)$ is a horizontal compression of the graph of $y=f(x).y=f(x).$ (b) If $0 the graph of $y=f(cx)y=f(cx)$ is a horizontal stretch of the graph of $y=f(x).y=f(x).$ We have explored what happens to the graph of a function $ff$ when we multiply $ff$ by a constant $c>0c>0$ to get a new function $cf(x).cf(x).$ We", "$f(x)= (x-2) 2 f(x)= (x-2) 2$ 39. $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41. $f(x)=− x f(x)=− x$ 43. $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45. $f(x)= −x +1 f(x)= −x +1$ 47. even 49. odd 51. even 53. The graph of $g g$ is a vertical reflection (across the $x x$-axis) of the graph of $f. f.$ 55. The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57. The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59. The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61. The graph of $g g$ is a horizontal reflection across the $y y$-axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63. $g(x)=|−4x| g(x)=|−4x|$ 65. $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67. $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69. The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units. 71. The graph of $f(x)=|x| f(x)=|x|$ is", "moves to $(3k,y(k))$ in the stretched graph, so if the old graph is $y = f(x)$ and the new graph is $y = F(x)$, you can see how $F(x)$ must be related to the function $f(x)$. You have it partly right, but not completely right. For 3: a point $(k,f(k)$ in the old graph moves to $(k-1, f(k)$ in the new graph. From that, you can get the equation $y = F(x)$ of the new graph in terms of the old graph $y = f(x).$ 3. Mar 9, 2017 ### Staff: Mentor I agree complete with what Ray already said: you do not have an equation here. You should have an equation here. Specifically, $y = 9x^2 \\sin(3x)$. Again, this should be an equation. Also, you wrote \"stretching vertically by a factor 3\" -- the stretch is supposed to be horizontally by a factor of 3. You need to be working with $y=x^2\\sin(3x)$ Should be an equation. Start with $y = x^2\\sin(3x)$, and shift it to the left by 1 unit. Aside from being incorrect, you didn't really simplify it by expanding $(x + 1)^2$.", "$f(x)= x+3 −1 f(x)= x+3 −1$ 37. $f(x)= (x-2) 2 f(x)= (x-2) 2$ 39. $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41. $f(x)=− x f(x)=− x$ 43. $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45. $f(x)= −x +1 f(x)= −x +1$ 47. even 49. odd 51. even 53. The graph of $g g$ is a vertical reflection (across the $x x$ -axis) of the graph of $f. f.$ 55. The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57. The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59. The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61. The graph of $g g$ is a horizontal reflection across the $y y$ -axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63. $g(x)=|−4x| g(x)=|−4x|$ 65. $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67. $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69. The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down", "# Math Insight ### A line or a plane or a point? #### Example 1: $y=3x-2$ Is the graph of $y=3x-2$ the graph of a line or a plane or a point? You might recognize this as a familiar form of the equation of a line with slope 3 and $y$-intercept $-2$, as in the following graph. However, if we think $y=3x-2$ is a line, we have implicitly assumed that the equation $y=3x-2$ is an equation in two dimensions, i.e., in the two variables $x$ and $y$. It could actually be an equation in three dimensions, say with the variables $x$, $y$, and $z$. Although the $z$ doesn't appear in the equation, we could think of the equation as having a $z$ with a coefficient of zero. We could write the equation as $y=3x+0z-2$. Here's what plotting the equation $y=3x-2$ in a three-dimensional coordinate system looks like. An angled line or a plane. The graph of the equation $y=3x-2$ looks like a line, which it would be if were an equation in two dimensions, i.e., in the $xy$-plane. However, rotate the graph with the mouse to give you a new perspective on the graph. Since in this case, the graph is really in three dimensions, the graph of the equation becomes a plane. The equation just happens to", "both nonnegative. The other quadrants are mirror images of that (why?). 4. $|f(x,y)|=1$ exactly if $f(x,y)=1$ or $f(x,y)=-1$. As in (2), draw these cases separately. - Interesting indeed. Serves a lot. – azetina Dec 22 '12 at 17:21 $y^2 = x^2$ So, for example, for any given x: $(x, x), (-x, -x), (-x, x), (x, -x)$ together represent all solutions to the equation above: each representing a line originating at the origin, and extending orthogonally, one in each quadrant. But this can be represented by the lines: • $y = x\\quad$ $(\\iff -y = -x);\\;$ slope = $\\;1$ • $y = -x\\;$ $(\\iff -y = \\;\\;x);\\;$ slope = $-1$ $(1)$ The graph of the above is then two perpendicular lines intersecting at the origin, one with slope $1$, the other with slope $-1$: $\\quad\\quad\\quad$ • Note that the graph of $y^2 = x^2$ is also the graph of the equation $|y| = |x|$. • The graph of $y = |x|$ is the portion of the graph $(1)$ at and above the x-axis; • the graph of $|y| = x$ is the portion of the graph $(1)$ at and to the right of the y-axis. $(2)$ The graph of $|x| + |y| = 1$ is the graph of $y = |x| - 1$ and $y = 1 - |x|$", "of $f$. 59. The graph of $g$ is a horizontal stretch by a factor of 3 of the graph of $f$. 61. The graph of $g$ is a horizontal reflection across the $y$ -axis and a vertical stretch by a factor of 3 of the graph of $f$. 63. $g\\left(x\\right)=|-4x|$ 65. $g\\left(x\\right)=\\frac{1}{3{\\left(x+2\\right)}^{2}}-3$ 67. $g\\left(x\\right)=\\frac{1}{2}{\\left(x - 5\\right)}^{2}+1$ 69. The graph of the function $f\\left(x\\right)={x}^{2}$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units. 71. The graph of $f\\left(x\\right)=|x|$ is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up. 73. The graph of the function $f\\left(x\\right)={x}^{3}$ is compressed vertically by a factor of $\\frac{1}{2}$. 75. The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units. 77. The graph of $f\\left(x\\right)=\\sqrt{x}$ is shifted right 4 units and then reflected across the vertical line $x=4$. 79. 81.", "# Stretches of Graphs ## Stretch Rule 1 For $y=pf(x)$, $p \\gt 0$, the effect of $p$ is to vertically stretch the graph by a factor by $p$. • If $p \\gt 1$, it moves points of $y=f(x)$ further away from the $x$-axis. • If $0 \\lt p \\lt 1$, it moves points of $y=f(x)$ closer to the $x$-axis. ## Stretch Rule 2 For $y=f(kx)$, $k \\gt 0$, the effect of $k$ is to horizontally compress the graph by a factor of $k$. • If $k \\gt 1$, it moves points of $y=f(x)$ further away from the $x$-axis. • If $0 \\lt k \\lt 1$, it moves points of $y=f(x)$ closer to the $x$-axis. ### Example 1 Given that the point $(-2,1)$ lies on $y=f(x)$, find the corresponding point on the image function $y=3f(2x)$. $2x$ means that the graph is to horizontally compress by a factor of $2$. Thus the $x$-value $-2$ transforms to $-1$. $3f(2x)$ means that the graph is to vertically stretch by a factor of $3$. Thus the $y$-value $1$ transforms to $3$. Therefore the point $(-2,1)$ transforms $(-1,3)$. ### Example 2 Find the point which is moved to the point $(-7,3)$ under the transformation $y=3f(2x)$. $2x$ means that the graph is to horizontally compress by a factor of $2$. Thus the $x$-value $-7$ was", "(x-2) 2$ 39 . $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41 . $f(x)=− x f(x)=− x$ 43 . $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45 . $f(x)= −x +1 f(x)= −x +1$ 47 . even 49 . odd 51 . even 53 . The graph of $g g$ is a vertical reflection (across the $x x$ -axis) of the graph of $f. f.$ 55 . The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57 . The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59 . The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61 . The graph of $g g$ is a horizontal reflection across the $y y$ -axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63 . $g(x)=|−4x| g(x)=|−4x|$ 65 . $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67 . $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69 . The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor", "not depend on the variable $z$, so the plane is parallel to the $z$-axis. Before rotating, the graph looks like the previous one, as the $z$ dimension is hidden. The fact that the graph is a plane doesn't become obvious until you rotate the axis. We conclude that we don't know what the graph of $y=3x-2$ is unless we specify the number of dimensions. If we want to specify a line, one way is to use a parametrization of a line. The parametrization of a line has the advantage that it gives you a line no matter how many dimensions you are working in. #### Example 2: $x=3$ Is the graph of $x=3$ the graph of a line or a plane or a point? Just like above, the answer will depend on the dimensions we are working in. In one dimension, $x=3$ is just a point, as in a point on the number line. In two dimensions, the graph is a line, as you can see by rotating the below graph. A point or a line. The graph of the equation $x=3$ could be a point, if it were an equation in one variable. But, in this case, it is an equation in two dimensions, in the variables $x$ and $y$ (where the coefficient of $y$ just happens", "of the graph of $f$... The fact is, a value of $f$ doesn't determine the corresponding value of $f'$ and we shouldn't expect the graph of $f'$ to emerge from shifting the graph of $f$, stretching or shrinking, or flipping, etc. Example. Can we do this more strategically, with fewer points to test? We first look at the points with a horizontal tangent, i.e., for those $x$'s where $f'(x) = 0$. These will typically separate intervals on which $f' > 0$ or $f' < 0$. To see which one, we look at the directions of the graph of $f$, up or down. We also notice the flattening of the graph on the far right; the result is also flattening of the graph of the derivative -- toward $0$. $\\square$ Example (sin). Let's apply this method to the graph of $y=\\sin x$. It is easy: The resulting graph of the derivative looks like that of $y=\\cos x$! We will show below that this isn't a coincidence. $\\square$ Let's summarize. As $\\Delta x$ is approaching zero, there are more and more nodes in the partition and more and more points found on the graph of the function. As a result, there are more and more values of the difference quotient found. Eventually, the former points form the graph of the", "Once you have this marginal density you can combine it with the joint density to arrive at the conditional: $$f_{X|Y}(x, y) = \\frac{f_{XY}(x, y)}{f_{Y}(y)} .$$ If you carry out these calculations you should get the answer you gave. This is the kind of problem where it helps tremendously to have a visual image of the meaning of all those mathematical formulas in your book or notes. • The graph of the pdf $f_X(u)$ of one random variable $X$ is a curve that lies (on or) above the $u$ axis, and the area under the curve is $1$. This is what all that gobbledygook $f_X(u) \\geq 0$ and $\\int_{-\\infty}^\\infty f_X(u)\\, \\mathrm du =1$. \"But, but, ...., \" you splutter, \"the pdf of $X$ is $f_X(x)$, not $f_X(u)$ and so what you are saying makes no sense!\" OK, sketch the graph of $f_X(x)$ on axes labeled $x$ and $f_X(x)$, and also the graph of $f_X(u)$ on axes labeled $u$ and $f_X(u)$. Apart from the labels on the axes, is there any difference between the two curves? • Similarly, the graph of the joint pdf $f_{X,Y}(u,v)$ of two random variables $X$ and $Y$ is a surface that lies (on or) above the $u$-$v$ plane, and the volume of the solid bounded by the surface and the $u$-$v$ plane is $1$. This", "and horizontally compressed by a factor of $1 4 1 4$ . 64. The graph of $f(x)= x f(x)= x$ is reflected over the $x x$ -axis and horizontally stretched by a factor of 2. 65. The graph of $f(x)= 1 x 2 f(x)= 1 x 2$ is vertically compressed by a factor of $1 3 , 1 3 ,$ then shifted to the left 2 units and down 3 units. 66. The graph of $f(x)= 1 x f(x)= 1 x$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. 67. The graph of $f(x)= x 2 f(x)= x 2$ is vertically compressed by a factor of $1 2 , 1 2 ,$ then shifted to the right 5 units and up 1 unit. 68. The graph of $f(x)= x 2 f(x)= x 2$ is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 69. $g(x)=4 (x+1) 2 −5 g(x)=4 (x+1) 2 −5$ 70. $g(x)=5 (x+3) 2 −2 g(x)=5 (x+3) 2 −2$ 71. $h(x)=−2|x−4|+3 h(x)=−2|x−4|+3$ 72. $k(x)=−3 x −1 k(x)=−3 x −1$ 73. $m(x)= 1 2 x", "$f$ by one unit in the negative $y$ direction. 1. between the graphs of $f$ and $fg$. The graph of $fg$ is a translation of $f$ by one unit in the positive $x$ direction. 1. Express $(1-x)(x-3)$ in the form $a-(x-b)^2$, where $a$ and $b$ are constants. We expand $(1-x)(x-3)$ to obtain $-x^2+4x-3$. Completing the square, we find that $(1-x)(x-3)=-x^2+4x-3=-(x-2)^2+4-3=1-(x-2)^2.$ We could also do this by expanding both expressions and solving for $a$ and $b$ after comparing coefficients. State the coordinates of the maximum point on the graph of $y=(1-x)(x-3)$, and state what symmetry the graph possesses. From the rewriting of the equation $y=(1-x)(x-3)$ in the form $y=1-(x-2)^2$, we see that the maximum point on the graph occurs when $y=1$ because the expression $-(x-2)^2$ must be negative or zero. The expression $-(x-2)^2$ is $0$ when $x=2$ so the maximum point has coordinates $(2,1)$. The graph possesses line symmetry about $x=2$. Sketch the graph of $y=e^{(1-x)(x-3)}$. So the graph • has a maximum at $x=2$, $y=e$; • has symmetry about $x=2$; • has its $y$-intercept at $e^{-3}$; • sees $y\\rightarrow0$ as $x\\rightarrow\\pm\\infty$. The sketch is", "the graph of f opens indefinitely to the right and upward. Consequently, when the points on this half of the graph of f are projected onto the x-axis, the “shadow” or projection extends indefinitely to the right. Consequently, the entire x-axis lies in “shadow,” making the domain of f to be $\\text { Domain of } f=(-\\infty, \\infty)=\\{x : x \\in \\mathbb{R}\\}$ To determine the range of f, we must project all points on the graph of f onto the y-axis. This projection is indicated by the red “shadow” (or thicker line if you are viewing this in black and white) shown on the y-axis in Figure $$\\PageIndex{13}$$(b). Two important points need to made about this “shadow” or projection. Figure $$\\PageIndex{13}$$. Determining the range from the graph of f. 1. The graph of f passes through the origin (the point (0, 0)). This is the lowest point on the graph and hence its shadow is the endpoint on the low end of the shaded region on the y-axis. 2. The arrows at the end of each half of the graph of f indicate that the graph opens upward indefinitely. Hence, when points on the graph of f are projected onto the y-axis, the “shadow” or projection extends upward indefinitely. This is indicated by an arrow on the upper", "Sketch a graph of $y = f''(x)\\text{;}$ label it appropriately. 4. Is the slope of the tangent line to $y = f(x)$ increasing, decreasing, or neither when $x = 2\\text{?}$ Explain. 5. Sketch a possible graph of $y = f(x)$ near $x = 2\\text{.}$ Include a sketch of $y=L(x)$ (found in part (a)). Explain how you know the graph of $y = f(x)$ looks like you have drawn it. 6. Does your estimate in (b) over- or under-estimate the true value of $f(2.07)\\text{?}$ Why? Hint 1. Find the value of $f'(2)$ from the given graph of $y=f'(x)\\text{.}$ 2. Remember that $f(2.07) \\approx L(2.07)\\text{.}$ 3. $f''(x)$ is the derivative of $f'(x)\\text{.}$ 4. Is $f'$ increasing, decreasing, or neither when $x = 2\\text{?}$ 5. Draw $y = L(x)$ first. Then think about options for $f$ relative to the graph of $L\\text{.}$ 6. Does the tangent line at $x=2$ lie above or below the graph of $y = f(x)$ at $x=2.07\\text{?}$ 1. $L(x) = -1 + 2(x-2)\\text{.}$ 2. $f(2.07) \\approx L(2.07) = -0.86\\text{.}$ 3. See the rightmost image in parte below. 4. Neither. 5. The leftmost image below shows a possible graph of $y = f(x)$ near $x = 2\\text{,}$ along with the tangent line $y = L(x)$ through $(2, f(2))\\text{.}$ 6. It is an overestimate. Solution 1. Since $f(2) =" ]

[ "$f(x)= x+3 −1 f(x)= x+3 −1$ 37. $f(x)= (x-2) 2 f(x)= (x-2) 2$ 39. $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41. $f(x)=− x f(x)=− x$ 43. $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45. $f(x)= −x +1 f(x)= −x +1$ 47. even 49. odd 51. even 53. The graph of $g g$ is a vertical reflection (across the $x x$ -axis) of the graph of $f. f.$ 55. The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57. The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59. The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61. The graph of $g g$ is a horizontal reflection across the $y y$ -axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63. $g(x)=|−4x| g(x)=|−4x|$ 65. $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67. $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69. The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down", "39. $y=|x|+1 y=|x|+1$ For the following exercises, graph the given functions by hand. 40. $y=| x |−2 y=| x |−2$ 41. $y=−| x | y=−| x |$ 42. $y=−| x |−2 y=−| x |−2$ 43. $y=−| x−3 |−2 y=−| x−3 |−2$ 44. $f(x)=−|x−1|−2 f(x)=−|x−1|−2$ 45. $f(x)=−|x+3|+4 f(x)=−|x+3|+4$ 46. $f(x)=2|x+3|+1 f(x)=2|x+3|+1$ 47. $f(x)=3| x−2 |+3 f(x)=3| x−2 |+3$ 48. $f(x)=| 2x−4 |−3 f(x)=| 2x−4 |−3$ 49. $f( x )=| 3x+9 |+2 f( x )=| 3x+9 |+2$ 50. $f(x)=−| x−1 |−3 f(x)=−| x−1 |−3$ 51. $f(x)=−| x+4 |−3 f(x)=−| x+4 |−3$ 52. $f(x)= 1 2 | x+4 |−3 f(x)= 1 2 | x+4 |−3$ #### Technology 53. Use a graphing utility to graph $f(x)=10|x−2| f(x)=10|x−2|$ on the viewing window $[ 0,4 ]. [ 0,4 ].$ Identify the corresponding range. Show the graph. 54. Use a graphing utility to graph$f(x)=−100|x|+100 f(x)=−100|x|+100$on the viewing window$[ −5,5 ]. [ −5,5 ].$Identify the corresponding range. Show the graph. For the following exercises, graph each function using a graphing utility. Specify the viewing window. 55. $f(x)=−0.1| 0.1(0.2−x) |+0.3 f(x)=−0.1| 0.1(0.2−x) |+0.3$ 56. $f(x)=4× 10 9 | x−(5× 10 9 ) |+2× 10 9 f(x)=4× 10 9 | x−(5× 10 9 ) |+2× 10 9$ #### Extensions For the following exercises, solve the inequality. 57. $|−2x− 2 3 (x+1)|+3>−1 |−2x− 2 3 (x+1)|+3>−1$ 58. If possible,", "(x-2) 2$ 39 . $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41 . $f(x)=− x f(x)=− x$ 43 . $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45 . $f(x)= −x +1 f(x)= −x +1$ 47 . even 49 . odd 51 . even 53 . The graph of $g g$ is a vertical reflection (across the $x x$ -axis) of the graph of $f. f.$ 55 . The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57 . The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59 . The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61 . The graph of $g g$ is a horizontal reflection across the $y y$ -axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63 . $g(x)=|−4x| g(x)=|−4x|$ 65 . $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67 . $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69 . The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor", "$f(x)= (x-2) 2 f(x)= (x-2) 2$ 39. $f(x)=|x+3|−2 f(x)=|x+3|−2$ 41. $f(x)=− x f(x)=− x$ 43. $f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2$ 45. $f(x)= −x +1 f(x)= −x +1$ 47. even 49. odd 51. even 53. The graph of $g g$ is a vertical reflection (across the $x x$-axis) of the graph of $f. f.$ 55. The graph of $g g$ is a vertical stretch by a factor of 4 of the graph of $f. f.$ 57. The graph of $g g$ is a horizontal compression by a factor of $1 5 1 5$ of the graph of $f. f.$ 59. The graph of $g g$ is a horizontal stretch by a factor of 3 of the graph of $f. f.$ 61. The graph of $g g$ is a horizontal reflection across the $y y$-axis and a vertical stretch by a factor of 3 of the graph of $f. f.$ 63. $g(x)=|−4x| g(x)=|−4x|$ 65. $g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3$ 67. $g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1$ 69. The graph of the function $f(x)= x 2 f(x)= x 2$ is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units. 71. The graph of $f(x)=|x| f(x)=|x|$ is", "[−5,∞).$ 25. Domain is $( −∞,∞ ). ( −∞,∞ ).$ Range is $[−12,∞). [−12,∞).$ 27. $f(x)= x 2 +4x+3 f(x)= x 2 +4x+3$ 29. $f(x)= x 2 -4x+7 f(x)= x 2 -4x+7$ 31. $f(x)= -149 x 2 +649x +8949 f(x)= -149 x 2 +649x +8949$ 33. $f(x)= x 2 -2x+1 f(x)= x 2 -2x+1$ 35. Vertex: (3, −10), axis of symmetry: x = 3, intercepts: $(3+10,0)(3+10,0)$ and $(3-10,0)(3-10,0)$ 37. Vertex: $(32,-12) (32,-12)$, axis of symmetry: $x=32 x=32$, intercept: $( 3+23 2 , 0) ( 3+23 2 , 0)$ and $( 3-23 2 , 0) ( 3-23 2 , 0)$ 39. 41. $f(x)= x 2 +2x+3 f(x)= x 2 +2x+3$ 43. $f(x)=-3 x 2 −6x−1 f(x)=-3 x 2 −6x−1$ 45. $f(x)=-14 x 2 −x+2 f(x)=-14 x 2 −x+2$ 47. $f(x)= x 2 +2x+1 f(x)= x 2 +2x+1$ 49. $f(x)= - x 2 +2x f(x)= - x 2 +2x$ 50. $f(x)=2 x 2 f(x)=2 x 2$ 53. The graph is shifted to the right or left (a horizontal shift). 55. The suspension bridge has 1,000 feet distance from the center. 57. Domain is $(−∞,∞). (−∞,∞).$ Range is $(-∞,2]. (-∞,2].$ 59. Domain: $(-∞,∞) (-∞,∞)$ ; range: $[100,∞) [100,∞)$ 61. $f(x)=2 x 2 +2 f(x)=2 x 2 +2$ 63. $f(x)=- x 2 −2 f(x)=- x 2 −2$ 65. $f(x)=3 x 2 +6x-15", "of 4 of the graph of $$f$$. Exercise 1.5.56 $$g(x)=6f(x)$$ Exercise 1.5.57 $$g(x)=f(5x)$$ The graph of $$g$$ is a horizontal compression by a factor of 15 of the graph of $$f$$. Exercise 1.5.58 $$g(x)=f(2x)$$ Exercise 1.5.59 $$g(x)=f(\\frac{1}{3}x)$$ The graph of $$g$$ is a horizontal stretch by a factor of 3 of the graph of $$f$$. Exercise 1.5.60 $$g(x)=f(\\frac{1}{5}x)$$ Exercise 1.5.61 $$g(x)=3f(−x)$$ The graph of $$g$$ is a horizontal reflection across the y-axis and a vertical stretch by a factor of 3 of the graph of $$f$$. Exercise 1.5.62 $$g(x)=−f(3x)$$ For the following exercises, write a formula for the function $$g$$ that results when the graph of a given toolkit function is transformed as described. Exercise 1.5.63 The graph of $$f(x)=|x|$$ is reflected over the y-axis and horizontally compressed by a factor of $$\\frac{1}{4}$$ . $$g(x)=|−4x|$$ Exercise 1.5.64 The graph of $$f(x)=\\sqrt{x}$$ is reflected over the x-axis and horizontally stretched by a factor of 2. Exercise 1.5.65 The graph of $$f(x)=\\dfrac{1}{x^2}$$ is vertically compressed by a factor of $$\\dfrac{1}{3}$$, then shifted to the left 2 units and down 3 units. $$g(x)=\\dfrac{1}{3(x+2)^2}−3$$ Exercise 1.5.66 The graph of $$f(x)=\\dfrac{1}{x}$$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. Exercise 1.5.67 The graph of $$f(x)=x^2$$ is vertically compressed by a factor of", "graph(f,-10,10,n=200,operator..) where n=200 is the number of sample points, and operator.. means curvy connecting that is necessary in this case. // http://asymptote.ualberta.ca/ unitsize(8mm,2cm); size(8cm); import graph; real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(f,-10,10,n=200,operator..),red); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,1.5,Arrow(TeXHead)); label(\"The graph of $y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\",point(N),2N); label(\"$-2$\",(0,-2),SW); shipout(bbox(5mm,invisible)); Appendix The function f(x) is asymptotic to cos(x) when |x| is big. unitsize(1cm,4cm); size(8cm); import graph; import math; //for drawline real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(cos,-20,20,n=200,operator..),lightblue+1pt); draw(graph(f,-20,20,n=200,operator..),red+.5pt); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,2.5,Arrow(TeXHead)); drawline((0,1),(1,1),gray+.2pt); drawline((0,-1),(1,-1),gray+.2pt); drawline((0,-2),(1,-2),gray+.2pt); label(scale(.7)*\"$1$\",(0,1),NE); label(scale(.6)*\"$-1$\",(0,-1),SW); label(scale(.6)*\"$-2$\",(0,-2),SW); label(\"$\\color{red}y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\"+\" and \"+\"$\\color{blue}y=\\cos x$\",point(N),2N); shipout(bbox(5mm,invisible)); • Maybe return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x));. Oct 17 at 16:37 • @NguyenVanChi1998 Yes, that is more concise syntax real f(real x) { return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x)); }; Oct 17 at 16:41", "and horizontally compressed by a factor of $1 4 1 4$ . 64. The graph of $f(x)= x f(x)= x$ is reflected over the $x x$ -axis and horizontally stretched by a factor of 2. 65. The graph of $f(x)= 1 x 2 f(x)= 1 x 2$ is vertically compressed by a factor of $1 3 , 1 3 ,$ then shifted to the left 2 units and down 3 units. 66. The graph of $f(x)= 1 x f(x)= 1 x$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. 67. The graph of $f(x)= x 2 f(x)= x 2$ is vertically compressed by a factor of $1 2 , 1 2 ,$ then shifted to the right 5 units and up 1 unit. 68. The graph of $f(x)= x 2 f(x)= x 2$ is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 69. $g(x)=4 (x+1) 2 −5 g(x)=4 (x+1) 2 −5$ 70. $g(x)=5 (x+3) 2 −2 g(x)=5 (x+3) 2 −2$ 71. $h(x)=−2|x−4|+3 h(x)=−2|x−4|+3$ 72. $k(x)=−3 x −1 k(x)=−3 x −1$ 73. $m(x)= 1 2 x", "y if and only if it belongs to the shaded triangle bounded by the lines x=y, y=1, and x=0, the area of which is 1/2. The ratio of the area of the triangle to the area of the rectangle is \\frac{1/2}{4} = \\boxed{\\frac{1}{8}}. [asy] draw((-1,0)--(5,0),Arrow); draw((0,-1)--(0,2),Arrow); for (int i=1; i<5; ++i) { draw((i,-0.3)--(i,0.3)); } fill((0,0)--(0,1)--(1,1)--cycle,gray(0.7)); draw((-0.3,1)--(0.3,1)); draw((4,0)--(4,1)--(0,1),linewidth(0.7)); draw((-0.5,-0.5)--(1.8,1.8),dashed); [/asy]$$ Mellie Apr 27, 2015 #3 +17711 +10 Best Answer You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... ) If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values. Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2. geno3141 Apr 27, 2015 #4 +80983 +5 I get something a little different here than geno.....see the following pic........ All points inside triangle ADF will have x values less than their associated y values. And the area of this triangle = 1/2 sq units And the area of the whole rectangle = (4)(1) = 4 sq units So....the probability", "+0 # A few questions! -1 715 2 +1206 1. The graph of $y=f(x)$ is shown in red below. For how many values of $c$ is $f(c) = -1$? (Assume grid lines are spaced 1 unit apart.) 2. The graph of $y=f(x)$ is shown in red below. Assuming the graph consists of three line segments, what is $f(-2)$? (Assume grid lines are spaced 1 unit apart.) 3. The graph of $y = h(x)$ is shown in red below. Compute $h(h(7))$. (Assume grid lines are spaced 1 unit apart.) 4. The graph of $y=f(x)$ is shown below in red. Given that $f$ is invertible, find $f^{-1}(4)$. (Assume grid lines are spaced 1 unit apart.) 5. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So \"(1,-3); (2,3); (5,-7)\" - without the quotes - is a valid answer format.) 6. Let $a$ and $b$ be real numbers, where $a < b$, and let $A = (a,a^2)$ and $B = (b,b^2)$. The line $$$\\overline{AB}$$$ (meaning the unique line that contains the point $A$ and the point $B$) has $x$-intercept $(-3/2,0)$ and $y$-intercept $(0,3)$. Find $a$ and $b$. Express your answer as the", "of the graph of $$f$$. Exercise 1.5.56 $$g(x)=6f(x)$$ Exercise 1.5.57 $$g(x)=f(5x)$$ Solution The graph of $$g$$ is a horizontal compression by a factor of 15 of the graph of $$f$$. Exercise 1.5.58 $$g(x)=f(2x)$$ Exercise 1.5.59 $$g(x)=f(\\frac{1}{3}x)$$ Solution The graph of $$g$$ is a horizontal stretch by a factor of 3 of the graph of $$f$$. Exercise 1.5.60 $$g(x)=f(\\frac{1}{5}x)$$ Exercise 1.5.61 $$g(x)=3f(−x)$$ Solution The graph of $$g$$ is a horizontal reflection across the y-axis and a vertical stretch by a factor of 3 of the graph of $$f$$. Exercise 1.5.62 $$g(x)=−f(3x)$$ For the following exercises, write a formula for the function $$g$$ that results when the graph of a given toolkit function is transformed as described. Exercise 1.5.63 The graph of $$f(x)=|x|$$ is reflected over the y-axis and horizontally compressed by a factor of $$\\frac{1}{4}$$ . Solution $$g(x)=|−4x|$$ Exercise 1.5.64 The graph of $$f(x)=\\sqrt{x}$$ is reflected over the x-axis and horizontally stretched by a factor of 2. Exercise 1.5.65 The graph of $$f(x)=\\dfrac{1}{x^2}$$ is vertically compressed by a factor of $$\\dfrac{1}{3}$$, then shifted to the left 2 units and down 3 units. Solution $$g(x)=\\dfrac{1}{3(x+2)^2}−3$$ Exercise 1.5.66 The graph of $$f(x)=\\dfrac{1}{x}$$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. Exercise 1.5.67 The graph of $$f(x)=x^2$$ is vertically compressed by", "You weren't shown how to graph these problems? Constraint [4] puts us in Quandrant 1. For $[1]\\;3x + 2y \\:\\leq \\:120$, graph the line: . $3x + 2y \\:=\\:120$ It has intercepts: (40, 0), (0, 60) Draw the line and shade the region below it. For $[2]\\;x + 4y \\:\\leq \\:80$ graph the line: . $x + 4y \\:=\\:80$ It has intercepts: (80, 0), (0, 20) Draw the line and shade the region below it. The shaded region looks like this . . . Code: | 60 * |* | * | * | * | * 20 o * |:::* * |:::::::o |::::::::* * |:::::::::* * |::::::::::* * - - o - - - - - o - - - - - * - - | 40 80 Constraint [3] is useless. The line $2x + y \\:=\\:100$ has intercepts (50, 0), (0, 100) . . and falls outside of the shaded region. We are concerned with the vertices of the shaded region. We can see three of them: .(0, 0), (40, 0), (0, 20) We find the fourth by locating the intersection of the two slanted lines. We solve the system: . $\\begin{array}{ccc}3x + 2y &=&120 \\\\ x + 4y &=& 80\\end{array}\\;\\text{ and get: }\\32,\\:12)\" alt=\"\\begin{array}{ccc}3x + 2y &=&120 \\\\ x + 4y &=& 80\\end{array}\\;\\text{ and", "is $[-5543,5543]$, but does not show the graph near $4$ when the vertical scale is $[-5542,5542]$. See (G) and (H) below. All nine points above with $y$-values less than $5543$ are plotted in red in both (G) and (H). Why does JSXGraph ‘connect the dots’ in (G), but not in (H)? In other words, why is the graph near $x = 4$ shown in (G), but not in (H)? Evidently JSXGraph uses some method to decide when it is possible to display a meaningful graph. Perhaps the interested reader can study the code to see precisely what is happening here! (G) viewing window for $y$ : $[-5543,5543]$ (H) viewing window for $y$ : $[-5542,5542]$ (I) viewing window for $y$ : $[-27,27]$ (J) viewing window for $y$ : $[-26,26]$ • The last two viewing windows above illustrate the graph ‘starting’ to disappear near $x = -2$ (I), and then disappearing completely (J). In both cases, points $(-2.045,-1660.48)$ and $(-1.955,1175.93)$ are far, far below and above the current viewing windows. The point of this optional section is to illustrate how difficult it can be to get a meaningful viewing window for a graph, and also to illustrate some of the misleading things that can occur when you use technology to produce graphs. Move on to the next section: Solve an", "the graph is defined; range=4 means that the graph will be drawn in the x-coordinate range of -2 to 2. let x = -range/2; -range/2=-2. The value of x varies from -2 to 2 and the graph is drawn. document.write('<svg width=400 height=400> <svg width=400 height=400> sets the size of the canvas to 400 pixels. <line x1=0 y1=200 x2=400 y2=200 stroke=\"black\"/> <line x1=200 y1=0 x2=200 y2=400 stroke=\"black\"/>'); A straight line on the x and y axes is drawn. for(let i=0; i < 400; i++) { The curve consists of 400 short lines. The for statement iterates over a short line 400 times. document.write('<g transform=\"translate(200,200)scale('+(400/range)+', '+(-400/range)+')\"><line x1='+x+' y1='+f(x)+' x2='+(x+range/400)+' y2='+f(x+range/400)+' stroke=\"blue\" stroke-width=\"'+range/400*2+'\" /></g>'); The curve $y=f(x)$ is drawn. translate(200,200) sets the center coordinates of the canvas to (0,0). scale() sets the scale of the canvas. The direction of y-axis is opposite in math graphs and canvases. When the graph is drawn, you’ll find the graph upside-down. So, when the y-axis is set to a negative value, the direction of the y-axis is what you want it to be. <line> draws a straight line with two coordinates. Imagine 400 points on a curve. A short straight line is drawn between one point on the curve and its neighbors. document.write('<g transform=\"translate(200,200)scale('+(400/range)+', '+(-400/range)+')\"><line x1='+x+' y1='+fd(x)+' x2='+(x+range/400)+' y2='+fd(x+range/400)+' stroke=\"green\" stroke-width=\"'+range/400*2+'\" /></g>'); A tangent line", "RN and LN for $N=1,10,100.N=1,10,100.$ 51. [T] $y=x4−5x2+4y=x4−5x2+4$ on the interval $[−2,2],[−2,2],$ which has an exact area of $32153215$ 52. [T] $y=lnxy=lnx$ on the interval $[1,2],[1,2],$ which has an exact area of $2ln(2)−12ln(2)−1$ 53. Explain why, if $f(a)≥0f(a)≥0$ and f is increasing on $[a,b],[a,b],$ that the left endpoint estimate is a lower bound for the area below the graph of f on $[a,b].[a,b].$ 54. Explain why, if $f(b)≥0f(b)≥0$ and f is decreasing on $[a,b],[a,b],$ that the left endpoint estimate is an upper bound for the area below the graph of f on $[a,b].[a,b].$ 55. Show that, in general, $RN−LN=(b−a)×f(b)−f(a)N.RN−LN=(b−a)×f(b)−f(a)N.$ 56. Explain why, if f is increasing on $[a,b],[a,b],$ the error between either LN or RN and the area A below the graph of f is at most $(b−a)f(b)−f(a)N.(b−a)f(b)−f(a)N.$ 57. For each of the three graphs: 1. Obtain a lower bound $L(A)L(A)$ for the area enclosed by the curve by adding the areas of the squares enclosed completely by the curve. 2. Obtain an upper bound $U(A)U(A)$ for the area by adding to $L(A)L(A)$ the areas $B(A)B(A)$ of the squares enclosed partially by the curve. 58. In the previous exercise, explain why $L(A)L(A)$ gets no smaller while $U(A)U(A)$ gets no larger as the squares are subdivided into four boxes of equal area. 59. A unit circle is made up", "3(3t + 2s = 42.50) ➝ 9t + 6s = 127.50 2(5t + 3s = 67.50) ➝ 10t + 6s = 135 10t + 6s – 9t – 6s = 135 – 127.50 t = 7.5 The cost of each T-shirt is \\$7.5. Question 11. The red figure is congruent to the blue figure. Which of the following is a sequence of rigid motions between the figures? F.Translate the red triangle 6 units left and then 4 units down. G. Reflect the red triangle in the x-axis, and then translate 4 units down. H. Reflect the red triangle in the y-axis, and then translate 4 units down. I. Rotate the red triangle 180° clockwise about the origin. I. Rotate the red triangle 180° clockwise about the origin. Explanation: The vertices of the red triangle is (1, 1), (4, 1), (3, 4) The vertices of the blue triangle are (-1, -3), (-4, -3), (-3, 0) Question 12. Which of the following is true about the graph of the linear equation y = 2? A. The graph is a vertical line that passes through (2, 0). B. The graph is a vertical line that passes through (0, 2). C. The graph is a horizontal line that passes through (2, 0). D. The graph is a horizontal line that passes through", "> x) = 1 – P (X < x). Figure 5.9 Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = $1 20 1 20$, 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. Figure 5.10 $P(2.3 Try It 5.1 Consider the function f(x) = $1 8 1 8$ for 0 ≤ x ≤ 8. Draw the graph of f(x) and find P(2.5 < x < 7.5). Order a print copy As an Amazon Associate we earn from qualifying purchases.", "units 11. $$f(x + 2) + 1 = \\sqrt{x + 2} + 1$$ 13. $$f(x - 3) - 4 = \\dfrac{1}{x - 3} - 4$$ 15. $$g(x) = f(x - 1)$$, $$h(x) = f(x) + 1$$ 17. 19. 21. 23. 25. $$y = |x - 3| - 2$$ 27. $$y = \\sqrt{x + 3} - 1$$ 29. $$y = -\\sqrt{x}$$ 31. 33. a. $$-f(-x) = -6^{-x}$$ b. $$-f(x + 2) - 3 = -6^{x + 2} - 3$$ 35. $$y = -(x + 1)^2 + 2$$ 37. $$y = \\sqrt{-x} + 1$$ 39. a. Even b. Neither c. Odd 41. Reflect $$f(x)$$ about the $$x$$-axis 43. Vertically stretch $$y$$ values by 4 45. Horizontally compress $$x$$ values by 1/5 47. Horizontally stretch $$x$$ values by 3 49. Reflect $$f(x)$$ about the $$y$$-axis and vertically stretch $$y$$ values by 3 51. $$f(-4x) = |-4x|$$ 53. $$\\dfrac{1}{3} f(x + 2) - 3 = \\dfrac{1}{3(x + 2)^2} - 3$$ 55. $$f(2(x - 5)) + 1 = (2 (x - 5))^2 + 1$$ 57. Horizontal shift left 1 unit, vertical stretch $$y$$ values by 4, vertical shift down 5 units becomes 59. Horizontal shift right 4 units, vertical stretch $$y$$ values by 2, reflect over $$x$$ axis, vertically shift up 3 units. becomes 61. Vertically compress $$y$$ values by 1/2 becomes 63.", "need a graph that covers these three points. We also need to remember the hole, vertical asymptotes, and intercepts. Here is a first attempt: f(x)=(x^3 - x^2 - 14*x + 24)/(x^4 + 9*x^3 + 18*x^2 - 32*x - 96) plot(f,xmin=-5,xmax=20,ymin=-10,ymax=30) So this picture is not very clear. We need to make xmin smaller so we can see more of the branch on the left. It would also be nice to extend the range on the y-axis. The vertical asymptotes at $x=-4$ and $x=-3$ are marked with solid lines, but it would be nice to make them dashed instead. We also should mark the hole at $x=2$. We can see the local minimum at $x\\approx-3.48$, but the local maximum at $x\\approx9.48$ is not clear from the graph. We also can't see the inflection point at $x\\approx15.97$. The x-intercept at $x=3$ is also not obvious. If we want to see all these features, we'll need a really big picture (which won't fit on our screen). So we will have to produce a second graph that reveals the local max, inflection point, and x-intercept. Here's another graph of the overall picture, and then we'll do a second plot that zooms in on the positive x-axis. plot(f,xmin=-15,xmax=20,ymin=-20,ymax=40,detect_poles=True,figsize=5)+point((2,-1/30),size=15,faceted=True,color='white',markeredgecolor='blue')+line([(-4,-25),(-4,45)],linestyle='dashed')+line([(-3,-25),(-3,45)],linestyle='dashed')+point([(-3.48,25.96),(9.48,0.0385),(15.97,0.0342),(3,0)],color='black',size=20) This is a good picture of the overall shape. Note: I do not expect you", "20:50 If $k > 1$ (resp. $0 < k < 1$) then it stretches (resp. shrinks) the plot in the $y$-axis direction. If $k$ is negative then it's the same $+$ mirroring about $x$-axis. – user2468 Aug 20 '12 at 21:02 For a positive value $k$, the graph of $y = kf(x)$ is obtained from the graph of $y=f(x)$ by scaling in the vertical direction. In coordinates, if $(x_0, y_0)$ is on the graph of $y=f(x)$, then that point moves to $(x_0, ky_0)$ on the graph of $y=kf(x)$. Any point on the $x$-axis remains fixed (as then $y=0$). When $k > 1$, the graph is stretched vertically (away from the $x$-axis), while if $0 < k < 1$, the graph is compressed vertically (toward the $x$-axis). Another way to think about this: The graph of $y=kf(x)$ is exactly the same as the graph of $y=f(x)$ if the $y$-axis is scaled so that $1$ is relabeled $k$, $2 \\mapsto 2k$, $-1 \\mapsto -k$, etc." ]

[ "# Freaky Functions Algebra Level 5 Suppose $f(x)$ and $g(x)$ are non-zero polynomials with real coefficients, such that $f(g(x))=f(x)\\cdot g(x).$ If $g(2)=37,$ what is $g(3)?$ Details and assumptions The zero polynomial is a polynomial that is identically 0, i.e. $f=0$. A non-zero polynomial is a polynomial that is not the zero polynomial. Equivalently, there is a value $\\alpha$ such that $f(\\alpha) \\neq 0$. A non-zero polynomial is allowed to have specific values that evaluate to 0. E.g. $f(x) = x-1$ is a non-zero polynomial even though $f(1) = 0$. ×", "# Polynomials Find polynomials $f(x), g(x)$ and $h(x)$ such that for all $x$, $F(x) = |f(x)|-|g(x)|+h(x) = \\begin{cases} -1 \\ \\ \\ \\ \\ \\ \\ \\ \\text{if} \\ x < -1 \\\\ 3x+2 \\ \\ \\ \\ \\text{if} \\ -1 \\leq x \\leq 0 \\\\ -2x+2 \\ \\ \\text{if} \\ x>0 \\end{cases}$ So it seems that the functions should have a simple form since $F(x) = -1$ for all $x<-1$. Also do we know that these functions are unique? How do we know that such functions exist?", "reduced numerator of $f-g$ is the zero polynomial. That means the unreduced numerator of $f-g$ is the zero polynomial, potentially with some holes at noninteger $x$-values. So $f-g=0$ except at finitely many $x$ where $f-g$ is undefined. At these $x$-values, either $f$ or $g$ is undefined. So $f=g$ up to where their nonintegral holes are. -", "hints. Suppose $g=\\sum_{i=0}^{d}b_iX^i\\in R[X]$ is nonzero and $fg=0$ 1. Show that we may assume $b_0\\neq 0$. 2. Write down the equations that express that the coefficient of $1, X, X^2, X^3$ in $fg$ are zero. 3. Determine how $b_0$ may be used to create elements that annihilate $a_0,a_1, a_2, a_3$. 4. Now take a guess at what nonzero element $b$ might do the job.", "if you think of polynomials as functions, and the roots as those inputs for which the function is 0, you should be able to see why. Thinking of $g$ as a function, you will see that, for any $x$, the number $g(x)$ is simply 6 times the number $f(x)$. What is six times 0?", "non constant polynomials $f\\left( t \\right),g\\left( t \\right) \\in {\\Bbb C}\\left( t \\right)$ such that $F(f(t),g(t))=0$ -", "# Prove there is no nonzero polynomial g(t) such that g(D)=0 Let $$V=P(R)$$(a set of polynomials having infinite degrees) and this is an infinite dimensional vector space. Let $$D:V \\rightarrow V$$ be the linear operator defined by $$D(f(x))=f'(x)$$. Prove there is no nonzero polynomial g(t) such that g(D)=0 I want to show that linear operators on infinite dimensional vector spaces do not always have minimal polynomials. That means I want to show that there exists no g(D) such that g(D)=0. But I am not sure how to argue this. Any help is appreciated. I am wondering does Cayley-Hamilton theorem help with this (but that is for finite dimensional). Assume $$g(D)=0$$ where $$g(x)=\\sum_{k=0}^n a_kx^k$$. Let $$k$$ be minimal with $$a_k\\ne 0$$. Then let $$f(x)=x^k$$ and observe that $$g(D)(f)=k!a_k\\ne0$$. • can you explain why it is $k! a_k$? I am not getting it. Apr 22, 2020 at 5:22", "## Divisibility for polynomials This post provides the details for the Divisibility lesson. Let $F$ be a field and $f,g \\in F[X]$. Definition 1 $f$ divides $g$, written $f|g$, if $g = hf$ for some $h ∈ F[X]$. We also say that $g$ is a multiple of $f$. If $\\deg(f) \\ge 1$, we say that $f$ is prime or…", "Let $f \\in F[X]$ be an irreducible polynomial where F has characteristic $p > 0$. Express $f(X) = g(X^{p^m})$ where $m \\in N$ is a large as possible. Show that g is irreducible and separable.", "+0 -1 33 1 +181 1. We have that $$3 \\cdot f(x) + 4 \\cdot g(x) = h(x)$$where f(x),g(x),and h(x) are all polynomials in x If the degree of f(x) is 8 and the degree of h(x) is 9, then what is the minimum possible degree of g(x) 2. Let f, g, and h be polynomials such that $$h(x) = f(x)\\cdot g(x)$$. If the constant term of f(x) is -4 and the constant term of h(x) is 3, what is g(0)? Mar 18, 2020 edited by qwertyzz Mar 18, 2020", "of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $\\deg r<\\deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.", "# finding the order of the polynomials in a polynomial groups under addition My question is: if G:= Group of polynomials under addition with elements from $$\\{0,1,2,3,4,5,6,7,8,9\\}$$ $$f(x)=7x^2+5^x+4$$ and $$g(x)=4x^2+8^x+6$$ ,$$f(x)+g(x)=x^2+3x$$ what is the order of $$f(x)$$,$$g(x)$$,$$f(x)+g(x)$$ i find that the answers are $$10$$,$$5$$ and $$10$$ . Am i correct?? (i noticed that zero polynomial is the identity here) then if $$h(x)=a_nx^n+...+a_0$$ belongs to $$G$$ what is the order of it? given that $$gcd(a_1,a_2,,,a_n)=1$$, $$gcd(a_1,a_2,,,,a_n)=2$$, $$gcd(a_1,a_2,,,a_n)=5$$, $$gcd(a_1,a_2,,,a_n)=10$$ Here i do not understand how to proceeded ?", "may assume that $f$ and $g$ do not have the monomial term $z^2$ in common. Also, note that since $f,g \\in I$, the following terms cannot be nonzero in $f,g: 1,x,y,z,xy,x^2.$ Since we must have polynomials $a,b$ such that $z^2 - x^2y = af + bg$, somehow a nonzero $z^2$ must appear. It cannot come about from a $1$ or $z$ in $f,g$, and so one of these must contain the term $z^2$. WLOG $f$ does, and so $a$ must have nonzero constant term. Now choose polynomials $p,q,r$ with $f = p(z^2 - x^2y) + q(x^3 - yz) + r(y^2 - xz)$. Since $z^2$ appears in $f$ as a nonzero term, the constant term of $p$ must be nonzero, and the $-x^2y$ term from $p(z^2 - x^2y)$ cannot be cancelled from any other term from the RHS of the equation for $f$ above. Hence $x^2y$ is a nonzero term of $f$. Now, write $x^3 - yz = cf + dg$ for some polynomials $c,d$. Then one of the constant terms of $c$ or $d$ must be nonzero and the term $yz$ appear in $f$ or $g$. Suppose that the constant term of $c$ is nonzero. Then the $x^2y$ term of $f$ appears and must be cancelled by some term of $dg$. Hence $g$ must contain one of the", "as a root, is divisible by $x^2-2$. In our current case, this means that the polynomials $f$ and $g$ have a common root in $\\alpha$ in $\\overline{K}$ and so both $f$ and $g$ have the minimal polynomial of $\\alpha$ over $K$ as a divisor. In particular, this minimal polynomial lies in $K[x]$.", "Polynomials with integer coefficients Algebra Level 4 $$f(x)$$ and $$g(x)$$ are non-constant polynomials with integer coefficients such that $f(x) \\times g(x) = x^5 - x^2 - x - 1.$ What is the value of $$\\lvert f(3) + g(3) \\rvert$$? Details and assumptions You may use the fact that $$f(3) \\times g(3) = 230$$. ×", "The following are equivalent: 1. There is a short exact sequence $$0 \\to G(p, \\lambda) \\to G(p, \\nu) \\to G(p, \\mu) \\to 0$$ 2. The Hall polynomial $g_{\\lambda \\mu}^{\\nu}(p)$ is nonzero. 3. The Littlewood-Richarsdon coefficient $c_{\\lambda \\mu}^{\\nu}$ is nonzero. 4. Anyone of the zillion equivalent conditions that you can find in, for example, Knutson and Tao or Zelevinsky For further discussion of this, see Fulton, especially Section 2. -", "$x\\ne0$ you have $f(x)=g(x)$ and hence $f(x)/x=g(x)/x$. Both those polynomials are continuous, so taking $x\\to0$ you have them also equal at $x=0$ and now you apply your inductive hypothesis. – almagest May 5 '16 at 8:52 Let $f(x)=p(x)-q(x)$. $f(x)$ is also a polynomial with real coefficients. From the question: $$\\forall x\\in\\mathbb R: f(x)=0$$", "+0 # help 0 248 1 Let $$f(x) = \\frac{x^3 - 1}{x - 1}$$. If $$g(x)$$ is a polynomial in $$x$$ such that $$f(g(x)) = f(x)f(x-1)$$ for all $$x \\neq 1,2$$, then find $$g(x)$$. Apr 14, 2019", "# if $f \\in A[x]$ is a zero divisor, then there exists $a ≠ 0$ in $A$ such that $af = 0$. [duplicate] The title of the question indicates what I am attempting to prove, that if $f$ is a member of a polynomial ring over a commutative ring with identity, and $f$ is a zero divisor, then there exists a nonzero element of A such that $af = 0$. I have begun the proof by obtaining the existence of a polynomial $g$ of least degree such that $g$ is nonzero and $fg = 0$; since $f$ is a zero divisor, such a $g$ must exist. I want to use induction to show that $a_{n-i}g = 0$ for $i = 0, 1, \\ldots, n$, and I have finished the base case $a_ng = 0$. I am having trouble coming up with exactly which term of $fg$ I should use for the inductive case. My induction begins by supposing that for some $i \\geq 0$, $a_{n-k}g=0$ for $0 \\leq k \\leq i$, and I want to show that $a_{n-(i+1)}g=0$. I believe I can do this by examining a particular term of $fg$ and showing that it is in fact equal to both $a_{n-(i+1)}$ times some coefficient of $g$ and $0$, and then using the same kind of logic I", "nonzero only if $x \\geq 0$. In this setup, $f(x)$ represents the coefficient of $t^x$. Addition of polynomials becomes pointwise addition of functions. Now here’s the good part: polynomial multiplication becomes convolution: if $f,g:\\Z\\to\\R$ are two polynomials then $$(fg)(y) = \\sum_{x\\in \\Z} f(y-x)g(x).$$ For a finite group $G$ then, complex representations of $G$ correspond to representations of the algebra of complex-valued functions on $G$ where the multiplication is the convolution product. # Check out the new post series page Posted by Jason Polak on 15. April 2017 · Write a comment · Categories: math I’ve made a small organizational change to this blog. I realize it might not have been easy to find series of posts on one topic, or the “post series”. They used to be on a page called “features” but I think that was too obscure. Now there’s the page “post series”, which is found above on the header. Along with the current series I have going on Non-unique Factorisation, it also lists five posts on Waldhausen categories. Perhaps the most exciting in my mind is the six posts of “Wild Spectral Sequences”, which give little toy problems that can be solved using spectral sequences. I came up with all the proofs myself, and some of them might be published nowhere else. Though I admit" ]

[ "is decided that, if important two factor interactions are found to be present, additional experiment trials may be conducted to separate the aliased effects. The performance of the fuel cone is measured on a scale of 1 to 15. In DOE++, the design for this experiment is set up using the properties shown in the following figure. The Fraction Generators for the design, $E=ABC\\,\\!$ and $F=BCD\\,\\!$, are the same as the defaults used in DOE++. The resulting ${2}^{6-2}\\,\\!$ design and the corresponding response values are shown in the following two figures. Design properties for the experiment in the example. Experiment design for the example. The complete alias structure for the 2 $_{\\text{IV}}^{6-2}\\,\\!$ design is shown next. $I=ABCE=ADEF=BCDF\\,\\!$ \\begin{align} & A= & BCE=DEF=ABCDF \\\\ & B= & ACE=CDF=ABDEF \\\\ & C= & ABE=BDF=ACDEF \\\\ & D= & AEF=BCF=ABCDE \\\\ & E= & ABC=ADF=BCDEF \\\\ & F= & ADE=BCD=ABCEF \\end{align}\\,\\! \\begin{align} & AB= & CE=ACDF=BDEF \\\\ & AC= & BE=ABDF=CDEF \\\\ & AD= & EF=ABCF=BCDE \\\\ & AE= & BC=DF=ABCDEF \\\\ & AF= & DE=ABCD=BCEF \\\\ & BD= & CF=ABEF=ACDE \\\\ & BF= & CD=ABDE=ACEF \\end{align}\\,\\! \\begin{align} & ABD= & ACF=BEF=CDE \\\\ & ABF= & ACD=BDE=CEF \\end{align}\\,\\! In DOE++, the alias structure is displayed in the Design Summary and as part of the Design Evaluation result, as shown next: The", "iff $m\\cdot n=0$. Using the scalar product we are able to once more reformulate our statement 1. $(x - z)(x - z)=(y - w)(y - w)$ iff $(a - c)(b-d)=0$. 2. $(x - z)(y - w)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$. Substituting $x$, $y$, $z$, $w$ from the definition reduces the above to 1. $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ iff $(a - c)(b-d)=0$. 2. $(a+b - c-d)(b+c-d-a)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$. All that remains is pure algebra. For example, $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ without the squares that are certainly the same on both sides, appears as $2(ab-ac-ad-bc-bd+cd)=2(bc-bd-ab-cd-ac+ad)$, which is equivalent to $(a - c)(b-d)=0$. Similarly, written explicitly, $(a+b - c-d)(b+c-d-a)=0$ gives \\begin{align}ab+ac & -ad-aa +bb+bc-bd-ab \\\\& -bc-cc+cd+ac -bd-cd+dd+ad \\\\& =2ac-aa+bb-2bd-cc+dd=0.\\end{align} which is equivalent to $(a-c)(a-c)=(b-d)(b-d)$. ### Related materialRead more... • Classification of Quadrilateral • Bimedians in a Quadrilateral • A Property of Circumscribed Quadrilaterals • Area of a Bicentric Quadrilateral", "already have from earlier that $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ and so: (3) \\begin{align} \\quad T(u + v) = T((b_1v_1 + b_2v_2 + ... + b_nv_n) + (a_1v_1 + a_2v_2 + ... + a_nv_n)) = T([b_1 + a_1]v_1 + [b_2 + a_2]v_2 + ... + [b_n + a_n]v_n) \\\\ \\quad = [b_1 + a_1]w_1 + [b_2 + a_2]w_2 + ... + [b_n + a_n]w_n = (b_1w_1 + b_2w_2 + ... + b_nw_n) + (a_1w_1 + a_2w_2 + ... + a_nw_n) \\\\ \\quad = T(u) + T(v) \\end{align} • Thus the additivity property holds. Now we will show that the homogeneity property holds. Let $c \\in \\mathbb{F}$ and let $v \\in V$ we such that $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ (like from earlier). Then we have that: (4) \\begin{align} \\quad T(cv) = T(c(a_1v_1 + a_2v_2 + ... + a_nv_n)) = T(ca_1v_1 + ca_2v_2 + ... + ca_nv_n) \\\\ \\quad = (ca_1)w_1 + (ca_2)w_2 + ... + (ca_n)w_n = c(a_1w_1 + a_2w_2 + ... + a_nw_n) = cT(v) \\end{align} • Therefore the homogeneity property holds and indeed $T$ is a linear map. • We now need to show that this linear map is unique. Notice that the homogeneity and additivity of $T$ implies that $T$ is uniquely determined by the vectors in", "+ g(b) + 2ab So, g(x) = x^2 does not have property 1 because 2ab≠0 (except when a=0 and/or b=0. Good! 5. ## Re: Property f(a + b) Originally Posted by Debsta Good! That's two for two. 6. ## Re: Property f(a + b) For $a=b=0$, \\begin{align}f(a+b) &= f(a)+f(b) \\\\ \\implies f(0) &= f(0)+f(0) \\\\ \\implies f(0) &= 0 \\end{align} This doesn't differentiate between the functions $f(x)=2x$ and $g(x)=x^2$, but it does eliminate vast quantities of functions. Also, for $b=-a$ \\begin{align}f(a+b) &= f(a)+f(b) \\\\ \\implies f(0) &= f(a) + f(-a) \\\\ \\implies f(-a) &= -f(a) &\\text{for all values of $a$} \\end{align} So $f(x)$ is recibo an odd function. This is a reason that $g(x)$ doesn't satisfy the functional equation, because $g(x)=x^2$ is even. Also \\begin{align}f(a+b) &= f(a)+f(b) \\\\ b=a \\implies f(2a) &= f(a) + f(a) \\\\ \\implies f(2a) &= 2f(a) \\\\ b=2a \\implies f(3a) = f(a) + f(2a) \\\\ \\implies f(3a) &= 3f(a) \\\\ &\\vdots \\\\ \\implies f(na) &= nf(a) & \\text{for all $a$ and all $n \\in \\mathbb N$} \\end{align} Also, if $f$ is differentiable, holding $a$ constant and differentiating with respect to $b$ we have \\begin{align}f(a+b) &= f(a) + f(b) \\\\ f'(a+b) &= f'(b) &\\text{for all $a$}\\end{align} So the derivative of $f$ is constant, and thus $f(x) = c_1x + c_2$, but since $f(0) = 0$", "If: Let $$g(x)$$ be as given. For any $$v_1, v_2 \\in V$$ and $$t \\in [0, 1],$$ we have \\begin{align*} g(tv_1 + (1 - t)v_2) &= f(tv_1 + (1 - t)v_2) + c \\newline &= tf(v_1) + (1 - t)f(v_2) + c \\newline &= t(f(v_1) + c) + (1 - t)(f(v_2) + c) \\newline &= t\\cdot g(v_1) + (1 - t)\\cdot g(v_2). \\end{align*} Only if: Let $$g(x)$$ be affine. Define $$f(x) = g(x) - g(\\mathbf 0).$$ We must show that $$f(x)$$ is linear; i.e., homogeneous and additive. To show homogeneity, we have for any $$c \\in [0, 1]$$ \\begin{align*} f(cx) &= f(cx + (1 - c)\\mathbf 0) \\newline &= g(cx + (1 - c)\\mathbf 0) - g(\\mathbf 0) \\newline &= c\\cdot g(x) + (1 - c)\\cdot g(\\mathbf 0) - g(\\mathbf 0) \\newline &= c\\cdot g(x) - g(\\mathbf 0) = c \\cdot f(x). \\end{align*} For $$c > 1$$ we have $$c^{-1} \\in (0, 1).$$ Therefore \\begin{align*} f(cx) &= f(c^2x/c) = \\dfrac1c f(c^2 x). \\end{align*} Rewriting, we have $$cf(cx) = f(c^2x).$$ Setting $$y = cx,$$ we obtain the desired. It remains to show that $$f$$ is homogeneous for a negative constant. We have \\begin{align*} f(-x) &= g(-x) - g(\\mathbf 0) \\newline &= g(-x) - g\\left(\\dfrac12 x + \\dfrac12 (-x)\\right) \\newline &= g(-x) - \\dfrac12 g(x) - \\dfrac12 g(-x) \\newline &=", "there are two solutions in this case: $$ex+fy+g=0\\quad\\text{or}\\quad y=-kx+C,$$ where $k$ is defined as above, in terms of $a, b, c, e, f, g.$ If you have an initial value problem, and your initial value is not on the line $ex+fy+g=0,$ then you will have to choose the other solution in order to take advantage of the arbitrary constant. And that wraps up this case fairly well. Case 2. The two lines $ax+by+c=0$ and $ex+fy+g=0$ are parallel and non-intersecting. Algebraically, we must have a non-zero $k$ such that \\begin{align*} a&=ke\\\\ b&=kf\\\\ c&\\not=kg. \\end{align*} We can rewrite the DE as \\begin{align*} (ax+by+c)\\,dx+(ex+fy+g)\\,dy&=0 \\\\ (kex+kfy+c)\\,dx+(ex+fy+g)\\,dy&=0 \\\\ k(ex+fy+c/k)\\,dx+(ex+fy+g)\\,dy&=0. \\end{align*} Now we let $u=ex+fy,$ with $du=e\\,dx+f\\,dy$ and $du-e\\,dx=f\\,dy,$ as before. Substituting into the DE yields \\begin{align*} k(u+c/k)\\,dx+(u+g)(du-e\\,dx)&=0\\\\ k\\,u\\,dx+c\\,dx+u\\,du-e\\,u\\,dx+g\\,du-eg\\,dx&=0\\\\ (ku+c-eu-eg)\\,dx+(u+g)\\,du&=0\\\\ (u(k-e)+c-eg)\\,dx+(u+g)\\,du&=0. \\end{align*} From here, you can see that the DE is separable! There are way too many subcases and possibilities to consider, so I will stop here for Case 2. Case 3: The two lines $ax+by+c=0$ and $ex+fy+g=0$ intersect at one point, call it $(x_0,y_0).$ Where is that point? We solve the two equations simultaneously: \\begin{align*} ax_0+by_0&=-c\\\\ ex_0+fy_0&=-g\\\\ x_0&=\\frac{\\left|\\begin{matrix}-c&b\\\\-g&f\\end{matrix}\\right|}{\\left|\\begin{matrix}a&b\\\\e&f\\end{matrix}\\right|}\\\\ &=\\frac{bg-cf}{af-be}\\\\ y_0&=\\frac{\\left|\\begin{matrix}a&-c\\\\e&-g\\end{matrix}\\right|}{\\left|\\begin{matrix}a&b\\\\e&f\\end{matrix}\\right|}\\\\ &=\\frac{ce-ag}{af-be}. \\end{align*} Naturally, this is only possible if $af-be\\not=0,$ the condition for intersecting lines in the first place. The idea here is to substitute a translation $t=x-x_0$ and $s=y-y_0.$ This would mean", "You can draw diagram easily) $$x= FI = EI = GE = FG$$ (since sides of the rhombus are equal) $$EH = GH-GE = a-x$$ $\\triangle EHI$ is right angle triangle. That means \\begin{align*}(EI)^2 &= (EH)^2 + (IH)^2\\\\ X^2 &= (a-x)^2 + b^2\\end{align*} From above, we will get $x = \\dfrac{a^2 + b^2}{2a}$ Draw rectangle $CEDF$ such that $GE\\perp FC$ and $ED\\perp FI$. $EF$ is the diagonal of rectangle $CEDF$ and also $ED = HI = b$ $$FD = FI-DI = FI-(EH)$$ (since $DI = EH$) $$x - a + x = 2x - a = \\frac{a^2 + b^2}{a} - a = \\frac{b^2}{a}$$ $\\triangle EDF$ is right angle triangle. That means \\begin{align*}y^2 &= (EF)^2 = (ED)^2 + (FD)^2\\\\ &= b^2 + \\left({\\frac{b^2}{a^2}}\\right)^2\\\\ &= \\frac{b^2(a^2 + b^2)}{a^2} \\end{align*} then you can easily get value of y. - Look at the triangle $FGJ$. It's a right-triangle, you know one side, you can express another in terms of $a$ and $x$, and the hypotenuse in terms of $x$. You should be able to use that to get an expression for $x$ in terms of $a$ and $b$. - I don't know - there isn't any $D$ anywhere that I can see. – Gerry Myerson May 14 '12 at 4:56 Since its a rhombus, GE=EI+FI=GF $GF=\\sqrt{b^2+(a-x)^2}=GE=x$ this equation gives you x Finding", "tells us that every single term in $$d(a,b,c)$$ has to be degree 4, this tells us that the polynomial $$p(a,b,c,)$$ actually has to be linear: \\begin{align} p(a,b,c)=\\alpha a+\\beta b+\\gamma c\\text{, for some }\\alpha,\\beta,\\gamma\\in\\mathbb{R}.\\end{align} Okay, so it's lookin' preeeetty good...we're getting a good handle-hold on what $$d(a,b,c)$$ looks like. Thanks to observation 3, we know that $$d(a,b,c)=-d(b,a,c)$$ and so: \\begin{align} (a-b)(b-c)(c-a)p(a,b,c)&=-(b-a)(a-c)(c-b)p(b,a,c)\\\\&=(a-b)(b-c)(c-a)p(a,b,c)\\\\ \\Rightarrow p(a,b,c)&=p(b,a,c)\\\\ \\Rightarrow \\alpha a +\\beta b+\\gamma c &= \\alpha b+\\beta a+\\gamma c.\\end{align} Since $$a$$ and $$b$$ are independent variables, this means that $$\\alpha=\\beta$$. By the same logic applied to $$a$$ and $$c$$, we get that: \\begin{align} \\alpha=\\beta=\\gamma.\\end{align} Soooo, we've just shown that the determinant may be written as \\begin{align} d(a,b,c)=\\alpha (a-b)(b-c)(c-a)(a+b+c),\\end{align} and all we need to do now, is to show that $$\\alpha=1$$. One way to do this is to substitute nice numbers into $$a,b,c$$ and to work it out. In fact, if we just set $$a=0$$ we see that \\begin{align}d(0,b,c)=\\left| \\begin{array}{ccc} 1 & 1 & 1 \\\\ 0 & b & c \\\\ 0 & b^3 & c^3\\end{array}\\right|=\\left|\\begin{array}{cc}b & c\\\\ b^3 & c^3\\end{array}\\right|=bc(c-b)(b+c).\\end{align} Since we know that \\begin{align}d(0,b,c)=\\alpha bc(c-b)(b+c),\\end{align} this gives us the desired conclusion that $$\\alpha=1$$. AND WE'RE DONE! Now, this technique can also be used to do the other determinant in this question. It's a little bit trickier though...and you'll need to show, using", "The claim is that the top coefficient $$b_m$$ will suffice. • Write the product: \\begin{align*} 0 = f(x)g(x) = (a_0 + \\cdots + a_{n-1}x^{n-1} + a_n x^n) (b_0 + \\cdots + b_{m-1}x^{m-1} + b_m x^m) .\\end{align*} • Equating coefficients, the coefficient for $$x^{m+n}$$ must be zero, so (importantly) $$a_n b_m = 0$$. • Since $$a_n b_m=0$$, consider $$a_ng(x)$$. This has degree $$d_1 \\leq m-1$$ but satisfies $$a_ng(x)f(x) = a_n(g(x)f(x)) = 0$$, so by minimality $$a_ng(x) = 0$$. • This forces $$a_n b_0 = \\cdots = a_n b_{m-1} = 0$$, so $$a_n$$ annihilates all of the $$b_k$$. • Now consider the coefficient of $$x^{m+n-1}$$, given by $$a_{n-1}b_m + a_{n}b_{m-1}$$. • The second term $$a_n b_{m-1}=0$$ since $$a_n$$ annihilates all $$b_k$$, so (importantly) $$a_{n-1} b_m = 0$$. • Considering now $$a_{n-1}g(x)$$: • The same argument shows this has degree $$d_2 \\leq m-1$$ but $$a_{n-1}g(x)f(x) = 0$$, so $$a_{n-1}g(x) = 0$$. • So $$a_{n-1}$$ annihilates all $$b_k$$, and allowing this process to continue inductively. • For good measure, the coefficient of $$x^{m+n-2}$$ is $$a_{n-2}b_m + a_{n-1}b_{m-1} + a_{n}b_{m-2}$$. • Note that $$a_n, a_{n-1}$$ annihilate all $$b_k$$, so (importantly) $$a_{n-2} b_m=0$$, and so on. • Thus $$a_k b_m = 0$$ for all $$0\\leq k \\leq n$$, and by linearity and commutativity, we have \\begin{align*} b_m f(x) = b_m \\sum_{k=0}^n a_k x^k =", "+ (ad + bc)i \\in \\mathbb{C} \\end{align} • Field Axiom 7 (Commutativity of Multiplication): (7) \\begin{align} \\quad zw = (ac - bd) + (ad + bc)i = (ca - db) + (da + cb)i = wz \\end{align} • Field Axiom 8 (Associativity of Multiplication): (8) \\begin{align} \\quad z(wv) &= z[(ce-df) + (cf+de)i] \\\\ &= [(ace)-(adf)]-[(bcf) - (bde)] + [((acf)+(ade)) + (bce) - (bdf)]i \\end{align} • And also: (9) \\begin{align} \\quad (zw)v &= [(ac - bd) + (ad + bc)i]v \\\\ &= [(ace) - (bde)] - [(adf) + (bcf)] + [((acf) - (bdf)) + ((ade) + (bce))]i \\end{align} • It's not hard to see that these two equations are equal, so $(zw)v = z(wv)$. • Field Axiom 9 (Existence of a Multiplicative Identity): (10) \\begin{align} \\quad z \\cdot 1 = (a + bi) \\cdot 1 = a + bi = z \\end{align} • Field Axiom 10 (Existence of Multiplicative Inverses for Nonzero Elements): For any $z \\in \\mathbb{C}$, $n \\neq 0$, let: (11) \\begin{align} \\quad z^{-1} = \\frac{a}{a^2 + b^2} - \\frac{b}{a^2 + b^2}i \\end{align} • Then: (12) \\begin{align} \\quad zz^{-1} &= (a + bi) \\frac{a - bi}{a^2 + b^2} \\\\ \\quad &= \\frac{(a + bi)(a - bi)}{a^2 + b^2} \\\\ \\quad &= \\frac{a^2 + b^2}{a^2 + b^2} \\\\ \\quad &= 1 \\end{align} • Since $\\mathbb{C}$ satisfies all of", "n+r_1)(x + n+r_2)\\cdots(x + n+r_i) + l(x + n+s_1)(x + n+s_2)\\cdots(x + n+s_j)$$ Observation 2.2 (Linear terms are absorbent.) For a quadratic $f(x) = a + bx + cx^2$, if the leading coefficient divides the constant term, we can decompose $f$ into polynomials with integer roots. Let $a = cpq$ and rewrite $f$ as \\begin{align} f(x) & = a + bx + cx^2 \\\\ & = cpq + bx + cx^2 \\\\ & = c(x^2 + pq) + bx \\\\ & = c(x + p)(x + q) - c(p+q)x + bx \\\\ f(x) & = c(x + p)(x + q) + (b - c[p+q])x \\end{align} Returning now to our initial motivation. If $f(x) = a + bx + cx^2$ and $\\gcd(b, c) \\mid a$, we can factor out $\\gcd(b, c)$ from each term, so after pulling out a constant we may assume that $\\gcd(b, c) = 1$. Now we can find an $n$ such that $c \\mid f(n)$, by considering the quadratic modulo $c$. \\begin{align} f(n) & \\equiv 0 \\pmod c \\\\ a + bn + cn^2 & \\equiv 0 \\pmod c \\\\ bn & \\equiv -a \\pmod c \\\\ \\end{align} where $b$ is invertible since $\\gcd(b, c) = 1$. If we \"shift\" $f(x)$ by $n$ using observation (2.1), then the constant term $f(n)$ will be divisible by", "d_0. \\end{align*} where the highest degree coefficent might also be zero. Evaluating \\eqref{eq:polycommfactor} and expanding we get $$n+m$$ homogeneous linear equations in $$n+m$$ quantities: \\begin{align} c_0 a_0 & & & & = & d_0 b_0 & & & \\nonumber \\\\ c_0 a_1 + & c_1 a_0 & & & = & d_0 b_1 + & d_1 b_0 & & \\nonumber \\\\ c_0 a_2 + & c_1 a_1 + & c_1 a_1 & & = & d_0 b_2 + & d_1 b_1 + & d_1 b_1 & \\nonumber \\\\ \\ddots & \\ddots & \\ddots & & & & & & \\nonumber \\\\ & & c_{m-2} a_n + & c_{m-1} a_{n-1} & = & & & d_{n-2} b_m + & d_{n-1} b_{m-1} \\nonumber \\\\ & & & c_{m-1}a_n & = & & & & d_{n-1} b_m. \\label{eq:sylveqn} \\end{align} For a solution to exist, the determinant of the Sylvester matrix should vanish: $$R = \\begin{vmatrix} a_0 & a_1 & \\cdots & a_n & & & & \\\\ & a_0 & a_1 & \\cdots & a_n & & & \\\\ & & \\cdots & \\cdots & & & & \\\\ & & & a_0 & a_1 & \\cdots & a_n & \\\\ b_0 & b_1 & \\cdots & \\cdots & b_m & & & \\\\ & b_0 & b_1 & \\cdots &", "$(C)$ in $H.$ Then $EG=FH.$ \\begin{align} EN^{2} &= CE^{2}-CN^{2} & (\\mbox{from right}\\space\\Delta CEN) \\\\ &= AC^{2}-AE^{2}-CN^{2} & (\\mbox{from right}\\space\\Delta ACE) \\\\ &= AC^{2}-R_{A}^{2}-R_{C}^{2} & (\\mbox{radii of}\\space (A)\\mbox{and}\\space (C)) \\\\ &= AC^{2}-R_{C}^{2}-R_{A}^{2} & \\\\ &= AC^{2}-CF^{2}-AM^{2} & \\\\ &= AF^{2}-AM^{2} & (\\mbox{from right}\\space\\Delta ACF) \\\\ &= FM^{2} & (\\mbox{from right}\\space\\Delta AFM). \\end{align} Thus, $EN=FM.$ Now, by the Power of a Point theorem, $FM^{2}=FE\\cdot FG$ and also $EN^{2}=EF\\cdot EH.$ It follows that $FE\\cdot FG=EF\\cdot EH$ and, therefore, $FG=EH.$ Finally (and here is the only difference with the first case), $EG=EH-GH=FG-GH=FH,$ as required. The Two Cases Together We found that the identity $EG=FH$ holds in the two cases. It is interesting to see whether the length of the segments is the same in both cases. In fact, there is an equality. (Now that both cases need to be depicted in the same diagram, the symbols for the points in the second case are decorated with an apostrophe: By the construction, $\\angle AEC=\\angle AF'C=\\angle AFC,$ implying that points $E,F,F'$ lie on the circle with diameter $AC.$ Further $\\angle CEF=\\angle CEF'$ since both those angles are subtended by equal chords $(CF=CF').$ Also, since $EN$ and $EN'$ are two tangents from $E$ to $(C),$ $\\angle CEN=\\angle LEN';$ it follows that $\\angle FEN'=\\angle F'EN$ such that the chords $FH$ and $F'H'$ are equally inclined to", "col}{\\rm .}\\,{\\rm of}\\;{\\bf A}_{\\,h} ( - {\\bf a} - {\\bf b})$$ and from here we can establish various relations, like for instance $$0^{\\, - {\\bf a} - {\\bf b}} = {\\bf A}_{\\,h} ( - {\\bf a} - {\\bf b})\\,0^{\\,{\\bf 0}_{\\,h} } = {\\bf A}_{\\,h} ( - {\\bf a} - {\\bf b})\\,\\;{\\bf A}_{\\,h} ( - {\\bf a})^{\\, - \\,{\\bf 1}} \\;0^{\\, - {\\bf a}}$$ But again, the last matrix product does not translate in general into a simple expression. To add to the discussion I will investigate if there are transformations that leave this star product unchanged. I will just look at the 2D case here. Define the general vectors $u, v$ and matrix $T$ as follows. $$u=\\pmatrix{u_1\\\\u_2} \\quad v=\\pmatrix{v_1\\\\v_2}\\quad T=\\pmatrix{a&b\\\\c&d}$$ So we will look for a $T$ such that $Tu\\star Tv=u\\star v$. Finally define $s_1=u_1+v_1,s_2=u_2+v_2$ \\begin{align} u\\star v&=(u_1+v_1)(u_2+v_2)=s_1s_2\\\\\\\\ Tu\\star Tv&=(au_1+bu_2+av_1+bv_2)(cu_1+du_2+cv_1+dv_2)\\\\ &=acs_1^2+bds_2^2+(ad+bc)s_1s_2 \\end{align} From this we can conclude that $ac=bd=0$ and $(ad+bc)=1$. This is a bad sign because it seems like $T=0$ is the only solution but the split complex numbers come to the rescue. The split complex numbers are defined in a similar way to the regular complex numbers by $j^2=1$. In the split complex numbers you can define the numbers $e,e^*$ by: $$e=\\frac{1+j}{2}\\quad e^*=\\frac{1-j}{2}$$ These have the useful properties that $e^2=e,\\ e^{*2}=e^*$ and $e e^*=0$. From $ac=bd=0$", "of the table entries is the product of the polynomials. Thus \\begin{align} (a+b+c)(w+x+y+z) = {} & aw + ax + ay + az \\\\ & {} + bw + bx + by + bz \\\\ & {} + cw + cx + cy + cz . \\end{align} Similarly, to multiply $(ax^2+bx+c)(dx^3+ex^2+fx+g),$ one writes the same table $\\begin{matrix} \\times & d & e & f & g \\\\ a & ad & ae & af & ag \\\\ b & bd & be & bf & bg \\\\ c & cd & ce & cf & cg \\end{matrix}$ and sums along antidiagonals: \\begin{align}(ax^2&+bx+c)(dx^3+ex^2+fx+g)\\\\ &= adx^5 + (ae+bd)x^4 + (af+be+cd)x^3 + (ag+bf+ce)x^2+(bg+cf)x+cg.\\end{align} Generalizations The FOIL rule cannot be directly applied to expanding products with more than two multiplicands, or multiplicands with more than two summands. However, applying the associative law and recursive foiling allows one to expand such products. For instance, \\begin{align}(a+b+c+d)(x+y+z+w)=&\\,((a+b)+(c+d))((x+y)+(z+w)) \\\\ =&\\,(a+b)(x+y)+(a+b)(z+w) \\\\ &\\,{}+(c+d)(x+y)+(c+d)(z+w) \\\\ =&\\,ax+ay+bx+by+az+aw+bz+bw \\\\ &\\,{}+cx+cy+dx+dy+cz+cw+dz+dw. \\end{align} Alternate methods based on distributing forgo the use of the FOIL rule, but may be easier to remember and apply. For example, \\begin{align} (a+b+c+d)(x+y+z+w)&=(a+(b+c+d))(x+y+z+w) \\\\ &=a(x+y+z+w)+(b+c+d)(x+y+z+w) \\\\ &=a(x+y+z+w)+(b+(c+d))(x+y+z+w) \\\\ &=a(x+y+z+w)+b(x+y+z+w) \\\\ &\\qquad +(c+d)(x+y+z+w) \\\\ &=a(x+y+z+w)+b(x+y+z+w) \\\\ &\\qquad +c(x+y+z+w)+d(x+y+z+w) \\\\ &=ax+ay+az+aw+bx+by+bz+bw \\\\ &\\qquad +cx+cy+cz+cw+dx+dy+dz+dw. \\end{align} References • Steege, Ray; Bailey, Kerry (1997), Schaum's Outline of Theory and Problems of Intermediate", "+ C' + C'$ $A' + C' + B + B'B$ $A' + C' + B$ 3). $[(AB’)+C’D]’$ $(AB')' (C'D)'$ $(A'+B)(C+D')$ $A'C + A'D' + CB + BD'$ 4). $[A+B(C’+D)]’$ $A'B'+(C'+D)'$ $A'B' + CD'$ 5). $(A \\oplus BC)+BD+ACD$ $(A'BC + AB'C') + BD + ACD$ $A'BC + AB'C' + BD + ACD$ $A'B + A'C + AB'C' + BD + ACD$ - Improve the formatting like this – Semsem Mar 15 at 3:55 I fixed the formatting. I tried to fix XOR, but I don't know how to insert the XOR symbol. – user3000964 Mar 15 at 4:13 use this \\oplus for $\\oplus$ and \\veebar for $\\veebar$ – Semsem Mar 15 at 4:27 My solution for 1) \\begin{align} & (A+B)(C+B)(D'+B)(ACD'+E) \\\\ & (AC+B)(ACD'+D'E+BE) \\\\ & (ACD'+BE) \\end{align} Rather than multiplying out everything and simplifying at the end, I have simplified intermediate factors to reduce the length of my calculation. Simplification rules: $$x + x = x$$ $$x + xy = x$$ $$x x = x$$ $$x x' = \\text{false}$$ $$x + x' = \\text{true}$$ $$x + x'y = x + y$$ It helps to keep factor literals in alphabetical order. My solution for 2) \\begin{align} & (A'+B+C')(A'+C'+D)(B'+D') \\\\ & (A'+BD+C')(B'+D') \\\\ & (A'B'+A'D'+B'C'+C'D') \\end{align} My solution for 3) \\begin{align} & [(AB′)+C′D]′ \\\\ & (AB')'(C'D')' \\\\ & (A'+B)(C+D) \\\\", "segment $$AB$$ has two mid points $$‘C’$$ and $$‘D’$$. Let assume $$C$$ is mid-point of $$AB$$ $AC = BC$ Adding $$AC$$ on both sides, we get \\begin{align} AC+AC & = BC+AC \\\\ 2 AC & = AB \\\\ AC& = \\frac { 1 } { 2 } AB \\ldots \\ldots ( 1 ) \\end{align} Let us consider a point $$D$$ lying on $$AB,$$ Let assume that $$D$$ be another mid-point of $$AB.$$ Therefore $$AD = BD$$ Adding equal length $$AD$$ on both the sides, we get \\begin{align} AD+AD & = BD+AD \\\\ 2 AD & = AB \\\\ AD & = \\frac { 1 } { 2 } AB \\ldots \\ldots ( 2 ) \\end{align} From equations ($$1$$) and ($$2$$), we can conclude that $$AC = AD$$ • $$C$$ Coincides with $$D.$$ • Axiom $$4$$: Things which coincide with one another are equal to one another. • A line segment has only one midpoint. ## Question 6 In the given figure, if $$AC = BD$$ then prove that $$AB = CD.$$ ### Solution Reasoning: We know that when equals are subtracted from equals, the remainders are equal. Steps: \\begin{align} AC & = BD \\\\ AB + BC & = BC + CD \\end{align} [ Point $$B$$ lies between $$A$$ and $$C$$ point $$C$$ lies between $$B$$ and $$D$$.]", "vector} && (af)(x)=af(x)=af(-x)=(af)(-x) \\\\ \\text{comm. add} && (f+g)(x)=f(x) +g(x)=g(x)+f(x)=(g+f)(x) \\end{aligned} Asso. Add \\begin{aligned} (f+(g+h))(x)&=f(x)+(g+h)(x)=f(x)+(g(x)+h(x)) =(f(x)+g(x))+h(x)=(f+g)(x)+h(x)\\\\&=((f+g)+h)(x) \\end{aligned} Let $0_v(x)=0$ be function where $\\forall x \\in R$ $0_v(x)=0$.$$(f+0_v)(x)=f(x)+0_v(x)=f(x)+0_v=f(x)=(f)(x)$$ \\begin{align} \\text{existence additive inver.} && (f+-1f)(x)=f(x)+-1f(x)=f(x)*(1-1)=f(x)*(0_v)=0_v \\\\ \\text{Neutral elem. mult.} && (1f)(x)=1f(x)=f(x)=(f)(x) \\\\ \\text{ass.__} && ((ab)f)(x)=(ab)(f(x))=a(bf(x))=(a(bf))x \\\\ \\text{Dist 1} && (a(f+g))(x)=a(f(x)+g(x))=af(x)+bg(x)=(af)(x)+(ag)(x) \\\\ \\text{Dist 2} && ((a+b)f)(x)=(a+b)f(x)=af(x)+bf(x)=(af)(x)+(bf)(x) \\end{align} Just reversed engineered the answer. Not sure if it actually holds. The middle part formatting of the attempt is a little chaotic. I'll try to make it prettier. Appreciate constructive critique. • Looks fine. Maybe you should have observed that $0_v$ is even. And maybe (?) many of the verifications can be omitted, since an awful lot of properties are inherited from the mother vector space of all functions from the reals to the reals. Perhaps that has already been proved to be a vector space. – André Nicolas Jan 24 '16 at 18:14 • @AndréNicolas, yes, I think you only need evenness for the closure axioms. The remaining properties are inherited from the more general space of functions on the reals. – Sherif F. Jan 24 '16 at 18:57 • I do understand to check for 4 prop closed add, closed scalar mult, existence of zero, existinence of additive inverse. Like subring test. It will save time but it is on the next section", "constrained to move in a circle that passes through point $O$, then point $Q$ will trace out a straight line. See below for proof. Image 18 We know that $AB = CD, BC = AD$ As a result, $BD \\parallel AC$ Draw line $OQ \\parallel AC$, intersecting $AD$ at point $P$. Consequently, points $O,P,Q$ are collinear Construct rectangle $EFCA$ \\begin{align} AC \\cdot BD & = EF \\cdot BD \\\\ & = (ED + EB) \\cdot (ED - EB) \\\\ & = (ED)^2 - (EB)^2 \\\\ \\end{align} For $\\begin{array}{lcl} (ED)^2 + (AE)^2 & = & (AD)^2 \\\\ (EB)^2 + (AE)^2 & = & (AB)^2 \\end{array}$, We then have $AC \\cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2$. Further, due to $\\frac{OP}{BD} = m, \\frac{OQ}{AC} = 1-m$ where $0 We have \\begin{align} OP \\cdot OQ & = m(1-m)BD \\cdot AC\\\\ & = m(1-m)((AD)^2 - (AB)^2) \\end{align} ## Other Straight Line Mechanism Image 19 Image 20 Image 21 Bryant, & Sangwin, 2008, p.44 There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to Image 19. Consider two circles $C_1$ and $C_2$ with radius having the relation $2r_2=r_1$. We roll $C_2$ inside $C_1$ without slipping as show in Image 20. Then the arch lengths $r_1\\beta = r_2\\alpha$. Voila! $\\alpha = 2\\beta$", "of $$H,F,G,I$$ is the negative golden ratio: $$\\frac{HG\\cdot FI}{FG\\cdot HI}=-\\frac{1+\\sqrt{5}}{2} = -\\phi$$ The original version is the limit of the general version as $$I$$ goes to infinity. But the general version is projectively invariant, so we can prove it using coordinates where $$A$$ is at vertical infinity, the conic is $$y=x^2$$, and the tangent at $$A$$ is the line at infinity. Then the first coordinates are simple: \\begin{align} A &= \\infty(0,1)\\\\ B &= (b,\\ b^2)\\\\ C &= (c,\\ c^2)\\\\ A' &= ((b+c)/2,\\ bc)\\\\ B' &= \\infty(1,2c)\\\\ C' &= \\infty(1,2b)\\\\ D &= (c,\\ (b-c)^2+c^2)\\\\ E &= (b,\\ (b-c)^2+b^2)\\\\ F &= ((1-\\phi)b+\\phi c,\\ ((1-\\phi)b+\\phi c)^2)\\\\ \\end{align} and we can stop computing coordinates there. The graphic below shows the case $$b=-1$$, $$c=2$$. (There is also another solution for $$F$$, not mentioned in the original question, which leads to switching $$\\phi$$ and $$1-\\phi$$ in the above and the below.) We want to evaluate the cross ratio $$(HG\\cdot FI)/(FG \\cdot HI)$$. Since $$I$$ is a point at infinity, the $$FI$$ and $$HI$$ terms will cancel. Then we can compute the directed quotient $$HG/FG$$ from the $$x-$$coordinates, which are $$c$$ and $$b$$ for $$G$$ and $$H$$. This gives us the desired result: $$\\frac{HG\\cdot FI}{FG \\cdot HI}=\\frac{HG}{FG}=\\frac{b-c}{(b-c)(1-\\phi)}=-\\phi$$" ]

[ "is decided that, if important two factor interactions are found to be present, additional experiment trials may be conducted to separate the aliased effects. The performance of the fuel cone is measured on a scale of 1 to 15. In DOE++, the design for this experiment is set up using the properties shown in the following figure. The Fraction Generators for the design, $E=ABC\\,\\!$ and $F=BCD\\,\\!$, are the same as the defaults used in DOE++. The resulting ${2}^{6-2}\\,\\!$ design and the corresponding response values are shown in the following two figures. Design properties for the experiment in the example. Experiment design for the example. The complete alias structure for the 2 $_{\\text{IV}}^{6-2}\\,\\!$ design is shown next. $I=ABCE=ADEF=BCDF\\,\\!$ \\begin{align} & A= & BCE=DEF=ABCDF \\\\ & B= & ACE=CDF=ABDEF \\\\ & C= & ABE=BDF=ACDEF \\\\ & D= & AEF=BCF=ABCDE \\\\ & E= & ABC=ADF=BCDEF \\\\ & F= & ADE=BCD=ABCEF \\end{align}\\,\\! \\begin{align} & AB= & CE=ACDF=BDEF \\\\ & AC= & BE=ABDF=CDEF \\\\ & AD= & EF=ABCF=BCDE \\\\ & AE= & BC=DF=ABCDEF \\\\ & AF= & DE=ABCD=BCEF \\\\ & BD= & CF=ABEF=ACDE \\\\ & BF= & CD=ABDE=ACEF \\end{align}\\,\\! \\begin{align} & ABD= & ACF=BEF=CDE \\\\ & ABF= & ACD=BDE=CEF \\end{align}\\,\\! In DOE++, the alias structure is displayed in the Design Summary and as part of the Design Evaluation result, as shown next: The", "iff $m\\cdot n=0$. Using the scalar product we are able to once more reformulate our statement 1. $(x - z)(x - z)=(y - w)(y - w)$ iff $(a - c)(b-d)=0$. 2. $(x - z)(y - w)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$. Substituting $x$, $y$, $z$, $w$ from the definition reduces the above to 1. $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ iff $(a - c)(b-d)=0$. 2. $(a+b - c-d)(b+c-d-a)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$. All that remains is pure algebra. For example, $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ without the squares that are certainly the same on both sides, appears as $2(ab-ac-ad-bc-bd+cd)=2(bc-bd-ab-cd-ac+ad)$, which is equivalent to $(a - c)(b-d)=0$. Similarly, written explicitly, $(a+b - c-d)(b+c-d-a)=0$ gives \\begin{align}ab+ac & -ad-aa +bb+bc-bd-ab \\\\& -bc-cc+cd+ac -bd-cd+dd+ad \\\\& =2ac-aa+bb-2bd-cc+dd=0.\\end{align} which is equivalent to $(a-c)(a-c)=(b-d)(b-d)$. ### Related materialRead more... • Classification of Quadrilateral • Bimedians in a Quadrilateral • A Property of Circumscribed Quadrilaterals • Area of a Bicentric Quadrilateral", "$\\gcd(f, g) = 1$. For example, consider the field of real numbers and the following polynomials: (1) \\begin{align} \\quad f(x) = x^2 -2x + 1 \\\\ \\quad g(x) = x^2 - 1 \\end{align} Note that $(x - 1) | f(x)$ and $(x - 1) | g(x)$. Furthermore, $(x - 1)$ is a monic polynomial and any other divisor of $f$ and $g$ is a divisor of $(x - 1)$. Therefore: (2) \\begin{align} \\quad \\gcd (x^2 -2x + 1, x^2 - 1) = x - 1 \\end{align} We now look at an important theorem which allows us to write the greatest common divisor of two polynomials $f$ and $g$ in a special form. Theorem 1: Let $(F, +, \\cdot)$ be a field and let $f, g \\in F[x]$ with $f(x) \\neq 0$ and $g(x) \\neq 0$. Then there exists polynomials $a, b \\in F[x]$ such that if $\\gcd (f, g) = d$ then $d(x) = a(x)f(x) + b(x)g(x)$. • Proof: Let $f, g \\in F[x]$ with $f(x) \\neq 0$ and $g(x) \\neq 0$. Consider the following set of polynomials over $F$: (3) \\begin{align} \\quad I = \\{ a(x)f(x) + b(x)g(x) : a, b \\in F[x] \\} \\end{align} • Note that $I$ contains a nonzero polynomial. Furthermore, it's easy to see that if $s, t \\in I$ then $(s +", "If: Let $$g(x)$$ be as given. For any $$v_1, v_2 \\in V$$ and $$t \\in [0, 1],$$ we have \\begin{align*} g(tv_1 + (1 - t)v_2) &= f(tv_1 + (1 - t)v_2) + c \\newline &= tf(v_1) + (1 - t)f(v_2) + c \\newline &= t(f(v_1) + c) + (1 - t)(f(v_2) + c) \\newline &= t\\cdot g(v_1) + (1 - t)\\cdot g(v_2). \\end{align*} Only if: Let $$g(x)$$ be affine. Define $$f(x) = g(x) - g(\\mathbf 0).$$ We must show that $$f(x)$$ is linear; i.e., homogeneous and additive. To show homogeneity, we have for any $$c \\in [0, 1]$$ \\begin{align*} f(cx) &= f(cx + (1 - c)\\mathbf 0) \\newline &= g(cx + (1 - c)\\mathbf 0) - g(\\mathbf 0) \\newline &= c\\cdot g(x) + (1 - c)\\cdot g(\\mathbf 0) - g(\\mathbf 0) \\newline &= c\\cdot g(x) - g(\\mathbf 0) = c \\cdot f(x). \\end{align*} For $$c > 1$$ we have $$c^{-1} \\in (0, 1).$$ Therefore \\begin{align*} f(cx) &= f(c^2x/c) = \\dfrac1c f(c^2 x). \\end{align*} Rewriting, we have $$cf(cx) = f(c^2x).$$ Setting $$y = cx,$$ we obtain the desired. It remains to show that $$f$$ is homogeneous for a negative constant. We have \\begin{align*} f(-x) &= g(-x) - g(\\mathbf 0) \\newline &= g(-x) - g\\left(\\dfrac12 x + \\dfrac12 (-x)\\right) \\newline &= g(-x) - \\dfrac12 g(x) - \\dfrac12 g(-x) \\newline &=", "n+r_1)(x + n+r_2)\\cdots(x + n+r_i) + l(x + n+s_1)(x + n+s_2)\\cdots(x + n+s_j)$$ Observation 2.2 (Linear terms are absorbent.) For a quadratic $f(x) = a + bx + cx^2$, if the leading coefficient divides the constant term, we can decompose $f$ into polynomials with integer roots. Let $a = cpq$ and rewrite $f$ as \\begin{align} f(x) & = a + bx + cx^2 \\\\ & = cpq + bx + cx^2 \\\\ & = c(x^2 + pq) + bx \\\\ & = c(x + p)(x + q) - c(p+q)x + bx \\\\ f(x) & = c(x + p)(x + q) + (b - c[p+q])x \\end{align} Returning now to our initial motivation. If $f(x) = a + bx + cx^2$ and $\\gcd(b, c) \\mid a$, we can factor out $\\gcd(b, c)$ from each term, so after pulling out a constant we may assume that $\\gcd(b, c) = 1$. Now we can find an $n$ such that $c \\mid f(n)$, by considering the quadratic modulo $c$. \\begin{align} f(n) & \\equiv 0 \\pmod c \\\\ a + bn + cn^2 & \\equiv 0 \\pmod c \\\\ bn & \\equiv -a \\pmod c \\\\ \\end{align} where $b$ is invertible since $\\gcd(b, c) = 1$. If we \"shift\" $f(x)$ by $n$ using observation (2.1), then the constant term $f(n)$ will be divisible by", "You can draw diagram easily) $$x= FI = EI = GE = FG$$ (since sides of the rhombus are equal) $$EH = GH-GE = a-x$$ $\\triangle EHI$ is right angle triangle. That means \\begin{align*}(EI)^2 &= (EH)^2 + (IH)^2\\\\ X^2 &= (a-x)^2 + b^2\\end{align*} From above, we will get $x = \\dfrac{a^2 + b^2}{2a}$ Draw rectangle $CEDF$ such that $GE\\perp FC$ and $ED\\perp FI$. $EF$ is the diagonal of rectangle $CEDF$ and also $ED = HI = b$ $$FD = FI-DI = FI-(EH)$$ (since $DI = EH$) $$x - a + x = 2x - a = \\frac{a^2 + b^2}{a} - a = \\frac{b^2}{a}$$ $\\triangle EDF$ is right angle triangle. That means \\begin{align*}y^2 &= (EF)^2 = (ED)^2 + (FD)^2\\\\ &= b^2 + \\left({\\frac{b^2}{a^2}}\\right)^2\\\\ &= \\frac{b^2(a^2 + b^2)}{a^2} \\end{align*} then you can easily get value of y. - Look at the triangle $FGJ$. It's a right-triangle, you know one side, you can express another in terms of $a$ and $x$, and the hypotenuse in terms of $x$. You should be able to use that to get an expression for $x$ in terms of $a$ and $b$. - I don't know - there isn't any $D$ anywhere that I can see. – Gerry Myerson May 14 '12 at 4:56 Since its a rhombus, GE=EI+FI=GF $GF=\\sqrt{b^2+(a-x)^2}=GE=x$ this equation gives you x Finding", "of the table entries is the product of the polynomials. Thus \\begin{align} (a+b+c)(w+x+y+z) = {} & aw + ax + ay + az \\\\ & {} + bw + bx + by + bz \\\\ & {} + cw + cx + cy + cz . \\end{align} Similarly, to multiply $(ax^2+bx+c)(dx^3+ex^2+fx+g),$ one writes the same table $\\begin{matrix} \\times & d & e & f & g \\\\ a & ad & ae & af & ag \\\\ b & bd & be & bf & bg \\\\ c & cd & ce & cf & cg \\end{matrix}$ and sums along antidiagonals: \\begin{align}(ax^2&+bx+c)(dx^3+ex^2+fx+g)\\\\ &= adx^5 + (ae+bd)x^4 + (af+be+cd)x^3 + (ag+bf+ce)x^2+(bg+cf)x+cg.\\end{align} Generalizations The FOIL rule cannot be directly applied to expanding products with more than two multiplicands, or multiplicands with more than two summands. However, applying the associative law and recursive foiling allows one to expand such products. For instance, \\begin{align}(a+b+c+d)(x+y+z+w)=&\\,((a+b)+(c+d))((x+y)+(z+w)) \\\\ =&\\,(a+b)(x+y)+(a+b)(z+w) \\\\ &\\,{}+(c+d)(x+y)+(c+d)(z+w) \\\\ =&\\,ax+ay+bx+by+az+aw+bz+bw \\\\ &\\,{}+cx+cy+dx+dy+cz+cw+dz+dw. \\end{align} Alternate methods based on distributing forgo the use of the FOIL rule, but may be easier to remember and apply. For example, \\begin{align} (a+b+c+d)(x+y+z+w)&=(a+(b+c+d))(x+y+z+w) \\\\ &=a(x+y+z+w)+(b+c+d)(x+y+z+w) \\\\ &=a(x+y+z+w)+(b+(c+d))(x+y+z+w) \\\\ &=a(x+y+z+w)+b(x+y+z+w) \\\\ &\\qquad +(c+d)(x+y+z+w) \\\\ &=a(x+y+z+w)+b(x+y+z+w) \\\\ &\\qquad +c(x+y+z+w)+d(x+y+z+w) \\\\ &=ax+ay+az+aw+bx+by+bz+bw \\\\ &\\qquad +cx+cy+cz+cw+dx+dy+dz+dw. \\end{align} References • Steege, Ray; Bailey, Kerry (1997), Schaum's Outline of Theory and Problems of Intermediate", "to do it systematically. First take $a\\,\\;$ and multiply it with $(d + e + f)\\,\\;$. You get $(ad + ae + af)\\,\\;$. Then multiply $b\\,\\;$ by $(d + e + f)\\,\\;$. You get $(bd + be + bf)\\,\\;$. Finally, multiply $c\\,\\;$ by $(d + e + f)\\,\\;$. You get $(cd + ce + cf)\\,\\;$. Add all three of these results and you get $(ad + ae + af)+(bd + be + bf)+(cd + ce + cf)\\,\\;$. Take out the parenthesis and you get $ad + ae + af+bd + be + bf+cd + ce + cf\\,\\;$.", "m/s$ $x = v_1' =\\ ?$ --- $b = m_2 = 0.25\\ kg$ $f = v_2 = -0.8\\ m/s$ $y = v_2' =\\ ?$ CoM: $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$ $ad + bf = ax + by$ $x = \\frac{b(f-y)}{a}+d \\hspace{35pt} (1)$ KE: $m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2$ $ad^2 + bf^2 = ax^2 + by^2$ $ad^2 + bf^2 = a\\left(\\frac{b(f-y)}{a}+d\\right)^2 + by^2\\hspace{35pt}$ <-- Replace by $(1)$ here. $ad^2 + bf^2 = a\\left(\\frac{b^2(f^2-2fy+y^2)}{a^2} + \\frac{2bd(f-y)}{a}+d^2\\right) + {by^2}$ $\\underline{ad^2} + bf^2 = \\frac{b^2f^2 - 2b^2fy + b^2y^2}{a} + 2bdf - 2bdy+ \\underline{ad^2} + by^2$ $0 = b^2f^2-2b^2fy + b^2y^2 + 2abdf - 2abdy+ aby^2 - abf^2$ At this point I found the coefficients of the polynomial separately: $(b^2 + ab)y^2 = (0.25^2+0.2\\cdot0.25)y^2 = 0.1125y^2$ --- $-(2b^2f+2abd)y = -(2\\cdot0.25^2\\cdot(-0.8)+2\\cdot0.2\\cdot0.25\\cdot1.5)y= -0.05y$ --- $b^2f^2 + 2abdf - abf^2 = 0.25^2\\cdot0.8^2+2\\cdot0.2\\cdot0.25\\cdot1.5\\cdot(-0.8)-0.2\\cdot0.25\\cdot0.8^2= 0.04-0.12-0.032=-0.112$ Then: $0 = 0.1125y^2-0.05y-0.112$ $y= \\frac{0.05 \\pm \\sqrt{0.05^2-4\\cdot0.1125\\cdot(-0.112)}}{0.225}$ $y= \\frac{0.05 \\pm \\sqrt{0.0529}}{0.225}$ $y_1 = 1.24\\ m/s$ $y_2 = -0.8\\ m/s$ $y_2$ turns out to be the velocity before the collision, so $v_2' = 1.24\\ m/s$. Then $v_1'$ is found to be $-1.06\\ m/s$. These values agree with a CoM of $0.1\\ kg\\dot\\ m/s$ and a Co(KE) of $0.305\\ J$. When vectors lie right in the x or y axis it's easy to spot their sign, but I have", "calculate the coefficients $a_j$. This is quick and simple and it will generate the correct values of the $a_j$ if $f(n)$ is indeed a polynomial. But it won't provide any justification that $f(n)$ really is a polynomial of degree $d + 1$ i.e. that $f(n) = g(n)$ for all non-negative integers $n$. To be more rigorous, we are going to calculate the coefficients a different way... Let $h(i)$ be the expression in the summation i.e. $h(i) = i^2 \\tag{3}$ [Note that you can generalize this method using other polynomial functions for $h$.] We are going to choose values for the $a_j$ coefficients so that: \\begin{align*} g(i) - g(i-1) &= h(i) & \\forall i \\in \\mathbb{N} & \\tag{4} \\\\ \\end{align*} Providing we can find such values, this will enable us to prove that \\begin{align*} g(n) &= f(n) & \\forall n \\in \\mathbb{N} \\end{align*} This is because: \\begin{align*} f(n) &= \\sum_{{i}={1}}^{n}h(i) \\\\ &= \\sum_{{i}={1}}^{n}[g(i) - g(i-1)] \\\\ &= [g(n) - g(n-1)] + [g(n-1) - g(n-2)] + ... + [g(2) - g(1)] + [g(1) - g(0)] \\\\ &= g(n) - g(0) & \\text{because all other terms cancel out} \\\\ &= g(n) & \\text{since } g(0) = 0 \\text{ by equation 2} \\end{align*} There's some tedious and error-prone algebraic manipulation involved in calculating $g(i) - g(i-1)$. Instead of doing it by hand,", "+ g(b) + 2ab So, g(x) = x^2 does not have property 1 because 2ab≠0 (except when a=0 and/or b=0. Good! 5. ## Re: Property f(a + b) Originally Posted by Debsta Good! That's two for two. 6. ## Re: Property f(a + b) For $a=b=0$, \\begin{align}f(a+b) &= f(a)+f(b) \\\\ \\implies f(0) &= f(0)+f(0) \\\\ \\implies f(0) &= 0 \\end{align} This doesn't differentiate between the functions $f(x)=2x$ and $g(x)=x^2$, but it does eliminate vast quantities of functions. Also, for $b=-a$ \\begin{align}f(a+b) &= f(a)+f(b) \\\\ \\implies f(0) &= f(a) + f(-a) \\\\ \\implies f(-a) &= -f(a) &\\text{for all values of $a$} \\end{align} So $f(x)$ is recibo an odd function. This is a reason that $g(x)$ doesn't satisfy the functional equation, because $g(x)=x^2$ is even. Also \\begin{align}f(a+b) &= f(a)+f(b) \\\\ b=a \\implies f(2a) &= f(a) + f(a) \\\\ \\implies f(2a) &= 2f(a) \\\\ b=2a \\implies f(3a) = f(a) + f(2a) \\\\ \\implies f(3a) &= 3f(a) \\\\ &\\vdots \\\\ \\implies f(na) &= nf(a) & \\text{for all $a$ and all $n \\in \\mathbb N$} \\end{align} Also, if $f$ is differentiable, holding $a$ constant and differentiating with respect to $b$ we have \\begin{align}f(a+b) &= f(a) + f(b) \\\\ f'(a+b) &= f'(b) &\\text{for all $a$}\\end{align} So the derivative of $f$ is constant, and thus $f(x) = c_1x + c_2$, but since $f(0) = 0$", "+ (ad + bc)i \\in \\mathbb{C} \\end{align} • Field Axiom 7 (Commutativity of Multiplication): (7) \\begin{align} \\quad zw = (ac - bd) + (ad + bc)i = (ca - db) + (da + cb)i = wz \\end{align} • Field Axiom 8 (Associativity of Multiplication): (8) \\begin{align} \\quad z(wv) &= z[(ce-df) + (cf+de)i] \\\\ &= [(ace)-(adf)]-[(bcf) - (bde)] + [((acf)+(ade)) + (bce) - (bdf)]i \\end{align} • And also: (9) \\begin{align} \\quad (zw)v &= [(ac - bd) + (ad + bc)i]v \\\\ &= [(ace) - (bde)] - [(adf) + (bcf)] + [((acf) - (bdf)) + ((ade) + (bce))]i \\end{align} • It's not hard to see that these two equations are equal, so $(zw)v = z(wv)$. • Field Axiom 9 (Existence of a Multiplicative Identity): (10) \\begin{align} \\quad z \\cdot 1 = (a + bi) \\cdot 1 = a + bi = z \\end{align} • Field Axiom 10 (Existence of Multiplicative Inverses for Nonzero Elements): For any $z \\in \\mathbb{C}$, $n \\neq 0$, let: (11) \\begin{align} \\quad z^{-1} = \\frac{a}{a^2 + b^2} - \\frac{b}{a^2 + b^2}i \\end{align} • Then: (12) \\begin{align} \\quad zz^{-1} &= (a + bi) \\frac{a - bi}{a^2 + b^2} \\\\ \\quad &= \\frac{(a + bi)(a - bi)}{a^2 + b^2} \\\\ \\quad &= \\frac{a^2 + b^2}{a^2 + b^2} \\\\ \\quad &= 1 \\end{align} • Since $\\mathbb{C}$ satisfies all of", "\\bmod{4}$, which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even. Let $M = (c+a)/2$ and $N = (c-a)/2$. Then \\begin{align} a &= M - N\\tag{6}\\\\ c &= M + N\\tag{7}\\\\ b^2 &= 4MN\\tag{8} \\end{align} Thus, we have that $M \\gt N \\gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $\\gcd(M,N)$ divides $a$, $b$, and $c$; thus, $\\gcd(M,N) = 1$. Since $b^2 = 4MN$ and $\\gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. $(4)\\Rightarrow(5):$ Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied: \\begin{align} a^2 + b^2 &= (m^2 - n^2)^2 + (2mn)^2\\\\ &= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\\\\ &= m^4 + 2 m^2 n^2 + n^4\\\\ &= (m^2 + n^2)^2\\\\ &= c^2\\tag{9} \\end{align} Furthermore, $a$ and $b$ are relatively prime since \\begin{align} \\gcd(a,b) &= \\gcd(m^2-n^2,2mn)\\\\ &\\:\\mid\\:\\gcd(m-n,2) \\gcd(m-n,m) \\gcd(m-n,n)\\\\ &\\times\\gcd(m+n,2) \\gcd(m+n,m) \\gcd(m+n,n)\\\\ &=\\gcd(m+n,2)^2 \\gcd(n,m)^4\\\\ &= 1\\tag{10} \\end{align}\\\\ $\\square$ -", "g & h & i \\end{bmatrix} \\\\[5ex]$ This implies that: $a_{11} = a \\\\[3ex] a_{12} = b \\\\[3ex] a_{13} = c \\\\[3ex] a_{21} = d \\\\[3ex] a_{22} = e \\\\[3ex] a_{23} = f \\\\[3ex] a_{31} = g \\\\[3ex] a_{32} = h \\\\[3ex] a_{33} = i \\\\[3ex] minor\\:\\: A = \\begin{bmatrix} ei - hf & di - gf & dh - ge \\\\[3ex] bi - hc & ai - gc & ah - gb \\\\[3ex] bf - ec & af - dc & ae - db \\end{bmatrix} \\\\[10ex] cofactor\\:\\: A = \\begin{bmatrix} ei - hf & -(di - gf) & dh - ge \\\\[3ex] -(bi - hc) & ai - gc & -(ah - gb) \\\\[3ex] bf - ec & -(af - dc) & ae - db \\end{bmatrix} \\\\[3ex]$ Can we make the Cofactor easier? $-(di - gf) = -di + gf = gf - di \\\\[3ex] -(bi - hc) = -bi + hc = hc - bi \\\\[3ex] -(ah - gb) = -ah + gb = gb - ah \\\\[3ex] -(af - dc) = -af + dc = dc - af \\\\[3ex]$ Therefore, $cofactor\\:\\: A = \\begin{bmatrix} ei - hf & gf - di & dh - ge \\\\[3ex] hc - bi & ai - gc & gb - ah \\\\[3ex] bf - ec & dc - af", "The claim is that the top coefficient $$b_m$$ will suffice. • Write the product: \\begin{align*} 0 = f(x)g(x) = (a_0 + \\cdots + a_{n-1}x^{n-1} + a_n x^n) (b_0 + \\cdots + b_{m-1}x^{m-1} + b_m x^m) .\\end{align*} • Equating coefficients, the coefficient for $$x^{m+n}$$ must be zero, so (importantly) $$a_n b_m = 0$$. • Since $$a_n b_m=0$$, consider $$a_ng(x)$$. This has degree $$d_1 \\leq m-1$$ but satisfies $$a_ng(x)f(x) = a_n(g(x)f(x)) = 0$$, so by minimality $$a_ng(x) = 0$$. • This forces $$a_n b_0 = \\cdots = a_n b_{m-1} = 0$$, so $$a_n$$ annihilates all of the $$b_k$$. • Now consider the coefficient of $$x^{m+n-1}$$, given by $$a_{n-1}b_m + a_{n}b_{m-1}$$. • The second term $$a_n b_{m-1}=0$$ since $$a_n$$ annihilates all $$b_k$$, so (importantly) $$a_{n-1} b_m = 0$$. • Considering now $$a_{n-1}g(x)$$: • The same argument shows this has degree $$d_2 \\leq m-1$$ but $$a_{n-1}g(x)f(x) = 0$$, so $$a_{n-1}g(x) = 0$$. • So $$a_{n-1}$$ annihilates all $$b_k$$, and allowing this process to continue inductively. • For good measure, the coefficient of $$x^{m+n-2}$$ is $$a_{n-2}b_m + a_{n-1}b_{m-1} + a_{n}b_{m-2}$$. • Note that $$a_n, a_{n-1}$$ annihilate all $$b_k$$, so (importantly) $$a_{n-2} b_m=0$$, and so on. • Thus $$a_k b_m = 0$$ for all $$0\\leq k \\leq n$$, and by linearity and commutativity, we have \\begin{align*} b_m f(x) = b_m \\sum_{k=0}^n a_k x^k =", "then, is to translate the input of our original problem (evaluations of $P$ on all $n^{\\text{th}}$-ROUs) into the inputs of our new, smaller, problems (evaluations of $P_0$ and $P_1$ on all $(n/2)^{\\text{th}}$-ROUs). We know (from the FFT algorithm) that polynomials $P_0,P_1$ of degree $<n/2$ exist such that $P(x)=P_0(x^2)+x\\cdot P_1(x^2)$. Now, let $g$ be a $n^\\text{th}$-PROU. And let $i$ be an integer in the range $\\left[0,n/2\\right)$. We have the following two equations obtain by setting $x=g^i$ and $x=g^{n/2+i}$ in the equation above: \\begin{align} P(g^i) &= P_0(g^{2i}) + g^i\\cdot P_1(g^{2i}) \\\\ P(g^{n/2+i}) &= P_0(g^{2(n/2 +i)}) + g^{n/2+i}\\cdot P_1(g^{2(n/2+i)}) \\\\ \\end{align} Arranging the second equation a little bit we get: \\begin{align} P(g^i) &= P_0(g^{2i}) + g^i\\cdot P_1(g^{2i}) \\\\ P(g^{n/2+i}) &= P_0(g^{n}g^{2i}) + g^{n/2}g^i\\cdot P_1(g^{n}g^{2i}) \\\\ \\end{align} Now, recall that $g$ is $n^\\text{th}$-PROU so $g^n=1$ and $g^{n/2}=-1$. Substituting these two identities in the second equation, we get: \\begin{align} P(g^i) &= P_0((g^2)^i)) + g^i\\cdot P_1((g^2)^i)) \\\\ P(g^{n/2+i}) &= P_0((g^2)^i) - g^i\\cdot P_1((g^2)^i) \\\\ \\end{align} This is looking good! Let's see what we have here: - $P(g^i)$ and $P(g^{n/2+i})$ are known to us, since we are given an array with the evaluations of $P$. - $g^i$ is known to us. - $P_0((g^2)^i)$ and $P_1((g^2)^i)$ are unknown to us, when taken over all $i$ values in the range $\\left[0,n/2\\right)$ these are exactly the evaluations of", "d_0. \\end{align*} where the highest degree coefficent might also be zero. Evaluating \\eqref{eq:polycommfactor} and expanding we get $$n+m$$ homogeneous linear equations in $$n+m$$ quantities: \\begin{align} c_0 a_0 & & & & = & d_0 b_0 & & & \\nonumber \\\\ c_0 a_1 + & c_1 a_0 & & & = & d_0 b_1 + & d_1 b_0 & & \\nonumber \\\\ c_0 a_2 + & c_1 a_1 + & c_1 a_1 & & = & d_0 b_2 + & d_1 b_1 + & d_1 b_1 & \\nonumber \\\\ \\ddots & \\ddots & \\ddots & & & & & & \\nonumber \\\\ & & c_{m-2} a_n + & c_{m-1} a_{n-1} & = & & & d_{n-2} b_m + & d_{n-1} b_{m-1} \\nonumber \\\\ & & & c_{m-1}a_n & = & & & & d_{n-1} b_m. \\label{eq:sylveqn} \\end{align} For a solution to exist, the determinant of the Sylvester matrix should vanish: $$R = \\begin{vmatrix} a_0 & a_1 & \\cdots & a_n & & & & \\\\ & a_0 & a_1 & \\cdots & a_n & & & \\\\ & & \\cdots & \\cdots & & & & \\\\ & & & a_0 & a_1 & \\cdots & a_n & \\\\ b_0 & b_1 & \\cdots & \\cdots & b_m & & & \\\\ & b_0 & b_1 & \\cdots &", "there are two solutions in this case: $$ex+fy+g=0\\quad\\text{or}\\quad y=-kx+C,$$ where $k$ is defined as above, in terms of $a, b, c, e, f, g.$ If you have an initial value problem, and your initial value is not on the line $ex+fy+g=0,$ then you will have to choose the other solution in order to take advantage of the arbitrary constant. And that wraps up this case fairly well. Case 2. The two lines $ax+by+c=0$ and $ex+fy+g=0$ are parallel and non-intersecting. Algebraically, we must have a non-zero $k$ such that \\begin{align*} a&=ke\\\\ b&=kf\\\\ c&\\not=kg. \\end{align*} We can rewrite the DE as \\begin{align*} (ax+by+c)\\,dx+(ex+fy+g)\\,dy&=0 \\\\ (kex+kfy+c)\\,dx+(ex+fy+g)\\,dy&=0 \\\\ k(ex+fy+c/k)\\,dx+(ex+fy+g)\\,dy&=0. \\end{align*} Now we let $u=ex+fy,$ with $du=e\\,dx+f\\,dy$ and $du-e\\,dx=f\\,dy,$ as before. Substituting into the DE yields \\begin{align*} k(u+c/k)\\,dx+(u+g)(du-e\\,dx)&=0\\\\ k\\,u\\,dx+c\\,dx+u\\,du-e\\,u\\,dx+g\\,du-eg\\,dx&=0\\\\ (ku+c-eu-eg)\\,dx+(u+g)\\,du&=0\\\\ (u(k-e)+c-eg)\\,dx+(u+g)\\,du&=0. \\end{align*} From here, you can see that the DE is separable! There are way too many subcases and possibilities to consider, so I will stop here for Case 2. Case 3: The two lines $ax+by+c=0$ and $ex+fy+g=0$ intersect at one point, call it $(x_0,y_0).$ Where is that point? We solve the two equations simultaneously: \\begin{align*} ax_0+by_0&=-c\\\\ ex_0+fy_0&=-g\\\\ x_0&=\\frac{\\left|\\begin{matrix}-c&b\\\\-g&f\\end{matrix}\\right|}{\\left|\\begin{matrix}a&b\\\\e&f\\end{matrix}\\right|}\\\\ &=\\frac{bg-cf}{af-be}\\\\ y_0&=\\frac{\\left|\\begin{matrix}a&-c\\\\e&-g\\end{matrix}\\right|}{\\left|\\begin{matrix}a&b\\\\e&f\\end{matrix}\\right|}\\\\ &=\\frac{ce-ag}{af-be}. \\end{align*} Naturally, this is only possible if $af-be\\not=0,$ the condition for intersecting lines in the first place. The idea here is to substitute a translation $t=x-x_0$ and $s=y-y_0.$ This would mean", "</mrow> <mi>2a</mi> </mfrac> </mrow> 1. 1 The square root formula is too simple, it’s not what I’d consider “complicated”. What I was thinking about was aligned equations, using other fonts or with symbols not included in the default. Take for example \\[fa] n >= 0: \\fCcn(n)\\fP ~\\[==]~ C sub n = { ( 2 n ) ! } over { ( n + 1 ) ! cdot n ! } or .EQ L wp(\\fC\\(dq df := df * x; x := x - 2\\(dq\\fI, I) .EN .EQ I \\(== ~ wp(\\fC\\(dq df := df * x; x := x - 2\\(dq\\fI, df cdot x !! = n !! ~ \\(AN ~ x >= 0) .EN .EQ I \\(== ~ wp(\\fC\\(dq df := df * x\\(dq\\fI, df cdot ( x - 2 ) !! = n !! ~ \\(AN ~ x - 2 >= 0) .EN .EQ I \\(== ~ wp(\\fC\\(dq\\(dq\\fI, df cdot x cdot ( x - 2 ) !! = n !! ~ \\(AN ~ x - 2 >= 0) .EN .EQ I \\(== ~ df cdot x cdot ( x - 2 ) !! = n !! ~ \\(AN ~ x >= 2 .EN both which I remember took a white to correctly typeset. 2. 2 Eqn IMO has a cleaner syntax compared to main stream", "A and B in the B direction: AC = AG + CG AB = AF - BF I verified it with my Mathematica calculation: A = {0,0}, B = {1,0}, C = (1+2a, (1+2a)*(2t)/(1-t2)} where a is arbitrary and t = tan((BAC)/2) AB = 1, AC = (1+2a)*(1+t2)/(1-t2) BD = CD = (1/2)*BC = sqrt(t2 + 4t2*a + (1+t2)2*a2)/(1-t2) D = (1+a, (1+2a)*(t)/(1-t2)} E = (1 + a*(1+t2))/(1 - t2) * {1,t} AE = (1+(1+t2)*a)*(1+t2)/(1-t2) BE = CE = sqrt(1+t2)*sqrt(t2 + 4t2*a + (1+t2)2*a2)/(1-t2) F = (1+(1+t2)*a)/(1-t2) * {1,0} G = (1+(1+t2)*a)/(1-t2) * {1-t2,2t}/(1+t2 AF = AG = (1+(1+t2)*a)/(1-t2) EF = EG = (1+(1+t2)*a) * t/(1-t2) BF = CG = (t2+(1+t2)*a) /(1-t2) AB = AF - BF AC = AG + CG Here, |t| < 1 and a > 0, so that the distance values come out positive, so they don't have to have reversed signs to be absolute values. lpetrich Contributor Here are some more problems: For integers a, b, c, d: a*b - 2*c*d = 3 a*c + b*d = 1 For nonnegative integers a, b, m, n: 5m + 5n = a2 + b2 For positive integers a, b, c, n: a*b*c - 1 = n * (a-1) * (b-1) * (c-1) Swammerdami Staff member AB = AF - BF AC = AG + CG" ]

[ "# Yet another question 8 Calculus Level 4 Find the minimum value of $$\\max( |y^2- xy| )$$ given that $$x$$ is any real number and $$0\\leq y \\leq 1$$. Try more here ×", "of the minimum value of $\\sqrt{9+x^{2}}+\\sqrt{(4-x)^{2}+(3-y)^{2}}+\\sqrt{4+y^{2}}$ as $x$ and $y$ range over all real numbers. ×", "• zmudz Find the minimum of $$x-y$$ among all ordered pairs of real numbers $$(x, y)$$, $$x$$ and $$y$$between 0 and 1, where there exists a real number $$a \\neq 1$$ such that $$\\log_{x}a + \\log_{y}a = 4\\log_{xy}a.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Minimize this Algebra Level 5 For real numbers $x,y$ and $z$, find the minimum value of $2x^2+5y^2+2z^2+4xy-4yz-2z-2x.$ ×", "# Min value Algebra Level 3 Find the minimum value of $4x^2-6x+1$ where $$x$$ ranges across all real numbers. ×", "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "+0 0 145 1 +102 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "# Minimum Hui Algebra Level 4 As $x, y, z$ ranges over all possible real numbers, what is the minimum value of $3x^2+12y^2+27z^2-4xy-12yz-6xz-8y-24z+100 ?$ This problem has been proposed by Hui. ×", "Inspired by Brian Charlesworth $x$ and $y$ are non- negative integers such that $2x+y=10$ Find the sum of maximum and minimum values of $(x+y)$. ×", "# Maximum value Calculus Level 3 If $$x,y$$ are real numbers where $$x+y = 1$$, determine the maximum value of $$(x^3+1)(y^3+1)$$. ×", "# Inspired by Rohit Kumar Algebra Level 2 As $x$ ranges over all real values, what is the minimum of $x^2 + ( x + 1) ^2 ?$ ×", "Minimum of the real valued function $f(x)=(x-1)^{2/3}$ occurs at $x$ equal to 1. $-\\infty$ 2. $0$ 3. $1$ 4. $\\infty$", "• zmudz Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of $$9x+12y+8z.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Minimum Hui Algebra Level 4 As $$x, y, z$$ ranges over all possible real numbers, what is the minimum value of $3x^2+12y^2+27z^2-4xy-12yz-6xz-8y-24z+100 ?$ This problem is proposed by Hui. ×", "# Minimum Value If $x$ and $y$ are positive real numbers, what is the minimum value of $(2x + 3y)\\left(\\frac{8}{x} + \\frac{3}{y} \\right) ?$ ×", "# Max value Algebra Level 2 Find the maximum value for $$y$$, as $$x$$ ranges over all real values, in the equation $y=\\frac{x}{x^2+1}.$ ×", "# Find Zi Song's minimum Algebra Level 3 Let $a$ be the minimum of $x(x - 4) - 2y(x - y)$ as $x$ and $y$ range over all real numbers. What is the value of $\\vert a \\vert$? This problem is posed by Zi Song Y. ×", "# Minimum Value Algebra Level 5 Let $$x,y,z$$ be positive real numbers such that $$x^2 + y^2 + z^2 +2xyz = 4$$. Find the minimum value of $$x^2 + y^2 + z^2$$. ×", "Given that, $$x^4y^3+x^3y^2+x^2y+x$$=139655 such that $$x , y$$ are Natural Numbers. Find the minimum values of $$x , y$$ which satisfy above condition , if,$$3x+2y=27$$. Then find $$x+y$$.", "# Find the minimum value! Algebra Level 3 Let $$N$$ be the minimum value of $6x^2+7x+5,$ where $$x$$ ranges across all real numbers. Find the value of $$\\lfloor N \\rfloor$$. ×" ]

[ "# How do you complete the square for y=x^2 - 9x? $y = {\\left(x - \\frac{9}{2}\\right)}^{2} - \\frac{81}{4}$ ${\\left(x + a\\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ => ${x}^{2} + b x = {\\left(x + \\frac{b}{2}\\right)}^{2} - {b}^{2} / 4$ Therefore substituting $b = - 9$ we have: $y = {x}^{2} - 9 x = {\\left(x - \\frac{9}{2}\\right)}^{2} - \\frac{81}{4}$", "way, and probably simplest: From x+ y= 1, y= 1- x. Putting that into [tex]x^3+ y^3+ xy[/itex] as Prove It suggested, we have $x^3+ (1- x)^3+ x(1- x)= x^3+ 1- 3x+ 3x^2- x^3+ x- x^2= 2x^2- 2x+ 1$. We can write that as $2(x^2- x)+ 1$ and, completing the square, $2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2$. Since a square is never negative, that will have its minimum value $2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2$ when x= 1/2.", "$0$ Complete the square to find the minimum or maximum value. $x^2-8x+6$ 1. $-22$ 2. $11/2$ 3. $16/3$ 4. $-10$ $x^2-3x+2$ 1. $-1/4$ 2. $-5$ 3. $-7/5$ 4. $17/4$", "2x+ 1$. We can write that as $2(x^2- x)+ 1$ and, completing the square, $2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2$. Since a square is never negative, that will have its minimum value $2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2$ when x= 1/2.", "(x- a/2)^2+ b- \\frac{a^2}{4}[/tex] Since a square is always positive, that has its minimum value, $b- \\frac{a^2}{4}$ when x= a/2. Your are told that x= a/2= -2 and $y= b- \\frac{a^2}{4}= 5$, the same equations Captain Black got. The question is in the calculus sub-forum. CB", "## Calculus (3rd Edition) $y=0$ Given $$y=x^{2}-6 x+9$$ So, we have \\begin{aligned} y&=x^{2}-6 x+9\\\\ &=(x-3)(x-3)\\\\ &=(x-3)^2 \\\\ \\end{aligned} The smallest value for a square power is zero. So, we see that $y=0$ is the minimum, when $x=3$.", "9. the minimum value of $(3x^2-12x+5)^2$ is 0. $ 3x^2-12x+5 = 0 $ completing the square is messy ... use the quadratic formula ... $ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a} $", "# Solving for Maximum / Minimum (Complete the Square) 1. 1 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=x^2+6x+2$$ For extra review on completing the square, click here 2. 2 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=2x^2+20x+8$$ 3. 3 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-3x^2+18x+1$$ 4. 4 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-4x^2+5x-3$$ 5. 5 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=0.5x^2+3x+6$$ 6. 6 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=\\frac{2}{3}x^2+8x+5$$ 7. 7 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-\\frac{1}{3}x^2-3x+4$$ 8. 8 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-0.2x^2+0.5x-7$$ For extra review on completing the square, click here $$e=mc^2$$", "\\frac{1}{2} $ Completing the square gives $ \\left( \\sin^2 \\theta - \\frac{1}{2}\\right)^2 + \\frac{1}{4} \\ge \\frac{1}{4} $ which obviously is true. 5. Assume, without loss of generality, that $x^2\\ge 1 \\ge y^2$. $x^2+y^2 > 2 \\Leftrightarrow x^2-1 > 1-y^2$, where $x^2-1 \\ge 0$ and $1-y^2 \\le 0$, so $x^2(x^2-1)>y^2(1-y^2)$, and thus $x^4+y^4 > x^2+y^2$.", "-3\\left(x^2+4x+\\frac{7}{3} \\right) $ Now complete the square as shown previously inside the brackets.", "# How do you complete the square for y=x^2 - 9x? ##### 1 Answer Jul 10, 2015 $y = {\\left(x - \\frac{9}{2}\\right)}^{2} - \\frac{81}{4}$ #### Explanation: ${\\left(x + a\\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ => ${x}^{2} + b x = {\\left(x + \\frac{b}{2}\\right)}^{2} - {b}^{2} / 4$ Therefore substituting $b = - 9$ we have: $y = {x}^{2} - 9 x = {\\left(x - \\frac{9}{2}\\right)}^{2} - \\frac{81}{4}$", "2 \\| x \\|_V \\| y \\|_V$$ Finally, this gives us $$\\| x + y \\|_V^2 \\leq \\|x \\|_V^2 + \\| y \\|_V^2 + 2 \\| x \\|_V \\| y \\|_V = (\\| x \\|_V + \\| y \\|_V)^2$$ so taking away the square gives us $$\\| x + y \\|_V \\leq \\| x \\|_V + \\| y \\|_V$$, as required.", "$(x+ a)^2= x^2+ 2ax+ a^2$. Comparing that to $x^2+ bx$ we see that we need 2a= b so that a= b/2 and then $a^2= \\frac{b^2}{4}$. To make that a perfect square, we need to add $\\frac{b^2}{4}$ and, of course, subtract it: $x^2+ bx= x^2+ bx+ \\frac{b^2}{4}- \\frac{b^2}{4}= \\left(x- \\frac{b}{2}\\right)^2- \\frac{b^2}{4}$", "is never negative since it is the square of a real number. We have $(x-4)^2 = 0$ only when $x - 4 = 0$. So the expression $(x-4)^2$ has a minimum value of 0 when $x = 4$. This means that the expression $(x-4)^2 + 6$ has a minimum value when $x = 4$ and that minimum value is 6. The structure of the expression $(x-4)^2 + 6$ was vital in determining its minimum value as it allows us to focus on the simpler expression $(x-4)^2$ and use the fact that the only real number whose square equals 0 is 0. 2. In order to write $x^2 - 6x - 3$ as a perfect square plus a number, we focus on the first two terms $x^2 - 6x$. To write $x^2 - 6x$ as a perfect square plus a number note that, for any number $a$, $$(x+a)^2 = x^2 + 2ax +a^2.$$ We have $-6x$ in the expression $x^2 - 6x$ so this means we want $2a = -6$ or $a = -3$. With this choice of $a$ we find $$x^2 - 6x - 3 = (x-3)^2 -12.$$ As in part (a), the minimum value will occur when $(x-3)^2 = 0$ or $x = 3$. When $x = 3$ we see that the minimum value is -12. 3.", "# How do you find the max and min of y=-x^2+6 by completing the square? May 9, 2015 \"Completing the squares\" is a bit unusual for the given equation: $y = - {x}^{2} + 6$ since in \"completed square form\" it would be: $y = \\left(- 1\\right) \\left({x}^{2} - 0 x + 0\\right) + 6$ or $y = \\left(- 1\\right) {\\left(x - 0\\right)}^{2} + 6$ This is also the vertex form; so the vertex is at $\\left(0 , 6\\right)$ Because of the $\\left(- 1\\right)$ factor we know that the parabola opens downward $\\rightarrow$ the vertex is at a maximum. As the magnitude of $x$ increases the value of $y$ decreases without bound. So the maximum value is $6$ and the minimum is $- \\infty$", "9 = x$$ - You have $$\\sqrt{x} = \\frac{3\\sqrt{3}}{\\sqrt{3}}.$$ Note here that $x$ must be positive. You can now simply square both sides: $$x = (\\sqrt{x})^2 = \\left(\\frac{3\\sqrt{3}}{\\sqrt{3}}\\right)^2.$$ Left for you to do is to simplify things. -", "are $\\left(\\dfrac{3}{2}, \\dfrac{35}{4}\\right)$. The difference here is that this is a negative quadratic, so the graph is going to have a maximum value rather than a minimum. In terms of completing the square, all we have to do is take a factor of $-1$ out of the quadratic $-x^2 + 10x - 3 = -(x^2 - 10x + 3)$, and then complete the square on the inside of the bracket as normal. So, half of $-10$ is $-5$, therefore $-(x^2 - 10x + 3) = -\\left[(x - 5)^2 + 3 - (-5)^2\\right].$ This simplifies to $-\\left[(x - 5)^2 - 22\\right] = -(x - 5)^2 + 22$. Here we can see why it’s a maximum point: the first part of the expression has a minus sign in front of it, meaning it is the negative of a square so must always be negative. So, the maximum exists where $-(x-5)^2$ is zero, which means that coordinates of the maximum point (and thus, the turning point) are $(5, 22)$. To complete the square on this, we first take a factor of $2$ out of the whole quadratic: $y=2(x^2+10x+7)$ Now, we complete the square on the inside section. Half of $10$ is $5$, so we get $y=2\\left[(x+5)^2+7-25\\right]=2\\left[(x+5)^2-18\\right]=2(x+5)^2-36$ Having the extra number in front of the bracket doesn’t actually change anything. We’re still", "square which is limited straight: $x=0$, $y=0$, $x=1$, $y=1$. $\\int_0^1\\int_0^1 f(x,y) dy dx$", "value is $y=0$y=0 • $x$x value corresponding to minimum $y$y value: we would need to substitute $x=3$x=3 to get $y=0$y=0 • Vertex: $\\left(3,0\\right)$(3,0) $y=\\left(x-3\\right)^2$y=(x3)2 is the result of a $3$3 unit translation of $y=x^2$y=x2 to the right. 5) $y=\\left(x+3\\right)^2$y=(x+3)2 (concave up so $y$y has a minimum value at vertex) • Minimum value of $y$y: $y$y is still the result of squaring a value so the smallest possible value is $y=0$y=0 • $x$x value corresponding to minimum $y$y value: here we would need to substitute $x=-3$x=3 to get $y=0$y=0 • Vertex: $\\left(-3,0\\right)$(3,0) $y=\\left(x+3\\right)^2$y=(x+3)2 is the result of $3$3 unit translation of $y=x^2$y=x2 to the left. 6) $y=-\\left(x-4\\right)^2$y=(x4)2 (concave down so $y$y has a maximum value at vertex) • Maximum value of $y$y: $y$y is the result of squaring a value, and then multiplying that square value by $-1$1, so the largest possible value is $y=0$y=0 • $x$x value corresponding to maximum $y$y value: here we would need to substitute $x=4$x=4 to get $y=0$y=0 • Vertex: $\\left(4,0\\right)$(4,0) $y=-\\left(x-4\\right)^2$y=(x4)2 is the result of a $4$4 unit translation of $y=x^2$y=x2 to the right. Now that we know how to represent vertical and horizontal translations algebraically, we can summarise the key concepts in the form $y=a\\left(x-h\\right)^2+k$y=a(xh)2+k, where: • $a$a determines concavity • $k$k represents the vertical shift upwards or downwards • $h$h represents", "\\left(x- \\frac{1}{2}\\right)^2 -\\frac{1}{4} \\right) \\left( \\left(x+\\frac{1}{2} \\right)^2 +\\frac{3}{4}\\right) (x^8 +1) + 1$$ From there, expand the first two brackets as such: $$f(x) = \\left[ ( x-\\frac{1}{2})^2 \\left( (x+\\frac{1}{2})^2 + \\frac{3}{4} \\right) - \\frac{1}{4} \\left( (x+\\frac{1}{2})^2 + \\frac{3}{4} \\right) \\right] (x^8+1) + 1$$ To get the minimum value of the function, we first must find the minimum value of the terms in the square brackets; this means we want the first term to be as small as possible, and our second completed square to be as large as possible. For the positive term, the minimum value of the perfect square in [0,1] is 1/4, whilst the minimum value of the completed square is 1. Hence, their product can never be less than 1/4. For the negative term, we maximize the completed square in [0,1] to to see our negative term has at most a magnitude of 3/4. Hence, the minimum value of the square brackets in -1/2. Since we want to subtract as much as possible, maximize the degree 8 term in (0,1). That term will always be less than 2. Hence we will always subtract less than 1 from the remaining constant term if x (0,1), and if we let x=0 and x=1 into f(x) we see that f(x) =1. And so f(x) will always be greater than" ]

[ "If x is a real number, Then what would be the minimum value of $\\left ( x^{2}-x-1 \\right )$: [A] $\\frac{3}{4}$ [B] 0 [C] 1 [D] $\\frac{1}{4}$", "$0$ Complete the square to find the minimum or maximum value. $x^2-8x+6$ 1. $-22$ 2. $11/2$ 3. $16/3$ 4. $-10$ $x^2-3x+2$ 1. $-1/4$ 2. $-5$ 3. $-7/5$ 4. $17/4$", "# How do you find the minimum and maximum value of y=(x-6)^2+3? $x = 6$ is the minimum and we don't have a maximum First we have a square and the plus three. If x is a real number, its square must be non-negative. So, taking this and the fact that three is constant, the minimum value happens when $x - 6 = 0 \\implies x = 6$", "## Calculus (3rd Edition) $y=0$ Given $$y=x^{2}-6 x+9$$ So, we have \\begin{aligned} y&=x^{2}-6 x+9\\\\ &=(x-3)(x-3)\\\\ &=(x-3)^2 \\\\ \\end{aligned} The smallest value for a square power is zero. So, we see that $y=0$ is the minimum, when $x=3$.", "# Minimum Value If $x$ and $y$ are positive real numbers, what is the minimum value of $(2x + 3y)\\left(\\frac{8}{x} + \\frac{3}{y} \\right) ?$ ×", "9. the minimum value of $(3x^2-12x+5)^2$ is 0. $ 3x^2-12x+5 = 0 $ completing the square is messy ... use the quadratic formula ... $ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a} $", "is never negative since it is the square of a real number. We have $(x-4)^2 = 0$ only when $x - 4 = 0$. So the expression $(x-4)^2$ has a minimum value of 0 when $x = 4$. This means that the expression $(x-4)^2 + 6$ has a minimum value when $x = 4$ and that minimum value is 6. The structure of the expression $(x-4)^2 + 6$ was vital in determining its minimum value as it allows us to focus on the simpler expression $(x-4)^2$ and use the fact that the only real number whose square equals 0 is 0. 2. In order to write $x^2 - 6x - 3$ as a perfect square plus a number, we focus on the first two terms $x^2 - 6x$. To write $x^2 - 6x$ as a perfect square plus a number note that, for any number $a$, $$(x+a)^2 = x^2 + 2ax +a^2.$$ We have $-6x$ in the expression $x^2 - 6x$ so this means we want $2a = -6$ or $a = -3$. With this choice of $a$ we find $$x^2 - 6x - 3 = (x-3)^2 -12.$$ As in part (a), the minimum value will occur when $(x-3)^2 = 0$ or $x = 3$. When $x = 3$ we see that the minimum value is -12. 3.", "+0 # Minimum problem 0 87 1 Find the minimum value of 2x^2 + 2xy + y^2 - 2x + 2y + 4 - x^2 + 4x over all real numbers $x$ and $y.$ Jun 25, 2022 #1 +2448 0 Rearrange as: $$(y^2 + 2y) +(x^2 +2x+ 2xy) + 4$$ Consider this as 2 separate equations ($$x^2 +2x+ 2xy$$ and $$y^2 + 2y$$) and a constant. Our goal is to minimize the equations. Start with $$y^2 + 2y$$. The minimum value of this is -1, and it occurs when $$y = {-b \\over 2a} = {-2 \\over 2} = -1$$ Now, subbing in $$y = -1$$ into the second equation gives us $$x^2 + 2x - 2x = x^2$$ The minimum value of this, quite obviously, is 0, and it occurs when $$x = 0$$. This means the minimum value is: $$0 - 1 + 4 = \\color{brown}\\boxed{3}$$ Jun 26, 2022", "# The minimal Algebra Level 4 Find the square of the minimum value of $\\sqrt{9+x^{2}}+\\sqrt{(4-x)^{2}+(3-y)^{2}}+\\sqrt{4+y^{2}}$ as $x$ and $y$ range over all real numbers. ×", "2x+ 1$. We can write that as $2(x^2- x)+ 1$ and, completing the square, $2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2$. Since a square is never negative, that will have its minimum value $2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2$ when x= 1/2.", "Practice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.", "problem by testing values or use the following properties: For a given sum of two numbers, the sum of even powers of these numbers is minimized, when they are equal. Simply put if it's given that $$x + y = 10$$, then $$x^{(even \\ integer)} + y^{(even \\ integer)}$$ is minimum when $$x = y = 5$$. So, the minimum value of $$x^2 + y^2$$, given that $$x + y = 10$$ is $$5^2 + 5^2 = 50$$. For a given sum of two numbers their product is maximized when they are equal. Simply put if it's given that $$x + y = 10$$, then xy is maximized when $$x = y = 5$$. So, the maximum value of xy, given that $$x + y = 10$$ is 5*5 = 25. (1) $$a + b = 4$$. According to the first property above. $$a^4 + b^4$$ will be minimized given that $$a + b = 4$$, when $$a = b = 2$$. In this case $$a^4 + b^4 = 2^4 + 2^4 = 32$$. So, the minumum possible value of $$a^4 + b^4$$ is 32 (not less than 32). Sufficient. (2) $$a^2 + b^2 = 8$$ You can use the same property (which would be easier) or square and test the second property: $$a^4 + 2a^2b^2 + b^4", "+0 # algebra 0 55 1 +73 Let $$x$$ and $$y$$ be real numbers such that $$0 \\le x \\le 1$$ and $$0 \\le y \\le 1.$$ Find the maximum value of $$x^2 y - xy^2$$. any tips / hints / solutions would be greatly appreciated! :) Feb 4, 2021 #1 +73 +1 the answer was $$1 \\over 4$$! to anyone who has also been trying to tackle this problem: you could factor the expression like so: $$xy(x-y)$$ then, since you are trying to find the maximum value, you can assume that $$x \\ge y$$. by AM-GM, $$y(x - y) \\le \\frac{x^2}{4}$$, and then continue from there:) Feb 4, 2021", "+0 # help 0 151 3 Find the minimum value of $$x^2 + 2xy + 3y^2 - 6x - 2y,$$ over all real numbers $$x$$ and $$y.$$ Apr 16, 2019 #1 +191 +1 Try to rewrite the expression as the sum of squares plus a number. Is the answer -11? Inspiration: https://www.quora.com/If-x-and-y-are-real-numbers-then-what-is-the-minimum-value-of-x-2-+4xy+6y-2-4y+4 Don't take my word for it, though. Apr 16, 2019 #2 +191 0 I think it's x=-4, y=1. Apr 16, 2019 #3 +7711 +1 $$x^2+2xy+3y^2 -6x-2y\\\\ =(x+y)^2 + 2y^2-2y-6x\\\\ =(x+y)^2 + 2y(y-1) - 6x\\\\ \\boxed{\\text{Minimum occurs when x = -y.}}\\\\ = 2y^2-2y+6y\\\\ = 2(y^2+2y)\\\\ =2((y+1)^2-1)\\\\ =2(y+1)^2 - 2\\\\ \\text{Minimum value} = -2.$$ . Apr 18, 2019", "+0 # Help pls :) +1 323 1 +504 Let x, y, and z be positive real numbers. Find the minimum value of $$(x + 2y + 4z) \\left( \\frac{4}{x} + \\frac{2}{y} + \\frac{1}{z} \\right)$$. Btw. I have to use the AM-GM inequality thingy. Help :) May 26, 2021", "# How do you find the max and min of y=-x^2-8x+2 by completing the square? May 31, 2015 $- {x}^{2} - 8 x + 2 = - \\left({x}^{2} + 8 x - 2\\right)$ $= - \\left({\\left(x + 4\\right)}^{2} - 16 - 2\\right)$ $= - \\left({\\left(x + 4\\right)}^{2} - 18\\right)$ $= 18 - {\\left(x + 4\\right)}^{2}$ Being the square of a real number ${\\left(x + 4\\right)}^{2} \\ge 0$ So the maximum value $18$ occurs when ${\\left(x + 4\\right)}^{2} = 0$ (that is when $x = - 4$) The value is unbounded in the other direction. $y = - {x}^{2} - 8 x + 2 \\to - \\infty$ as $x \\to \\infty$ or $x \\to - \\infty$", "# Solving for Maximum / Minimum (Complete the Square) 1. 1 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=x^2+6x+2$$ For extra review on completing the square, click here 2. 2 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=2x^2+20x+8$$ 3. 3 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-3x^2+18x+1$$ 4. 4 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-4x^2+5x-3$$ 5. 5 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=0.5x^2+3x+6$$ 6. 6 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=\\frac{2}{3}x^2+8x+5$$ 7. 7 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-\\frac{1}{3}x^2-3x+4$$ 8. 8 Find the maximum or minimum value and the value of $x$ when it occurs. $$y=-0.2x^2+0.5x-7$$ For extra review on completing the square, click here $$e=mc^2$$", "+0 # Inequalities 0 100 1 +179 Let x and y be positive real numbers. Find the minimum value of $$\\left( x + \\frac{1}{y} \\right) \\left( x + \\frac{1}{y} - 2018 \\right) + \\left( y + \\frac{1}{x} \\right) \\left( y + \\frac{1}{x} - 2018 \\right).$$ I really do not know where to start without involving huge expressions. Naturally, however, if you could write $$y+\\frac{1}{x}$$ in terms of $$x+\\frac{1}{y}$$, or vice versa, that would really help. Thanks for any replies! Jun 20, 2021", "# Find the maximum or minimum value of the quadratic function. Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum. $y=2x^2-4x+7$ $x^2$ has a coefficient of $2$, how should I complete the squares? $y=2x^2-4x+7$ therefore $y=2(x^2-2x+7/2)$ completing the square now gives $y=2[(x-1)^2-1+7/2]$ which rearranges to $y=2(x-1)^2+5$ For completing the square, I always divide by a number that ensures I have $1$ as the coefficient of $x^2$. To get the minimum value there are two options: 1) By looking at the equation after we have completed the square we can see that we will always have the $+5$ term, so we need to minimize the $2(x-1)^2$ term. Because it's square it will never be negative, however, we can make it equal to zero. So, $2(x-1)^2=0$, therefore $(x-1)^2=0$ and $x-1=0$ so $x=1$ is the minimum point. 2) Using differentiation. $y'=4x-4$ then set this equal to zero so $4x-4=0$, $4x=4$ and $x=1$. • Hey I have a prob in the same topic, would you help me out? – Kiara Feb 25 '14 at 19:12 • @Kiara of course! – The Ref Feb 25 '14 at 19:16 • There is another similar question in my book, $y=2x^2-5x-1$, I did it in the same", "Find the minimum value of the quantity where $a , b , c$ are real positive numbers. Find the minimum value of the quantity where $a , b , c$ are real positive numbers. $$\\left(\\frac{a^2 + 3a + 1}{a}\\right) \\left(\\frac{b^2 +3b + 1}{b}\\right)\\left(\\frac{c^2 + 3c + 1}{c}\\right)$$ I think the to get the answer we need to use $A.M.\\ge G.M.$ How i can achieve this? You can use the fact that $x + \\frac{1}{x} \\ge 2$, which can be proved using $\\text{AM} \\ge \\text{GM}$, or just completing the square." ]

[ "# How do you find the solution to the quadratic equation x^2-3x=10? $y = {x}^{2} - 3 x - 10$ = (x - p)(x - q).", "A cubic polynomial $$p(x)$$ is such that $$p(1)=1$$, $$p(2)=2$$, $$p(3)=3$$ and $$p(4)=5$$. Then the value of $$p(6)$$ is:", "# Math Help - discrete math help, please 1. ## discrete math help, please Prove that there is no quadratic polynomial p(x)=ax squared+bx+c, where a is not equal to 0 with p(-1)=2, p(0)=1, p(1)=3, p(2)=10 2. Originally Posted by mtaylor98212 Prove that there is no quadratic polynomial p(x)=ax squared+bx+c, where a is not equal to 0 with p(-1)=2, p(0)=1, p(1)=3, p(2)=10 Use the first three values to solve for the coefficients, then substiture $x=2$ into the resulting quadratic to show that $p(2) \\ne 10$. CB", "$\\rm\\,p\\,$ and $\\rm\\,-q,\\,$ we know the coefficients of the quadratic polynomial having $\\rm\\,p\\,$ and $\\rm\\,-q\\,$ as roots, namely $$\\rm (x-p)(x+q)\\, =\\, x^2 - (\\color{#C00}{p\\!-\\!q})\\, x \\color{#0A0}{-pq}\\ =\\ x^2 - k\\, x -n$$ Therefore, solving $\\rm\\ x^2 - k\\, x -n\\, =\\, 0\\$ yields the roots $\\rm\\:p\\:$ and $\\rm\\:-q.$ - Note that $p=k+q$, so $n=q(k+q)$ or $q^2+kq-n=0$. This is a quadratic in $q$. -", "# Please explain Illustration: The quadratic polynomial p(x) has the following properties : p(x) can be positive or zero for all real numbers p(1) = 0 and p(2) = 2 Then find the quadratic polynomial. Sol: p(x) is positive or zero for all real numbers. Also p(1) = 0 Then we have p(x) = k(x-1)2, where k > 0 Now p(2) = 2$⇒$ k=2∴ p(x) = 2(x-1)2 Dear student Regards • 3 there is nothing to be explained in this question its too simple • -16 What are you looking for?", "be a cubic polynomial. Since $$P'(x)$$ is a quadratic polynomial with two roots $$x=1$$ and $$x=3$$, we have $$P'(x)=3a(x-1)(x-3)=3ax^2-12ax+9a$$ for some constant $$a$$ (the factor $$3$$ is there simply to make the integration easier). Thus, $$P(x)=ax^3-6ax^2+9ax+c$$ for some constant $$c$$. Since $$P(1)=6$$ and $$P(3)=2$$, we get $$4a+c=6\\text{ and }c=2\\,,$$ so $$a=1$$ and $$c=2$$. That is, $$P(x)=x^3-6x^2+9x+2\\,,$$ with $$P'(x)=3(x-1)(x-3)=3x^2-12x+9\\text{ and }P''(x)=6(x-2)\\,.$$ This makes $$P''(1)=-6<0$$ and $$P''(3)=6>0$$, so $$x=1$$ and $$x=3$$ are the local maximum and the local minimum, respectively. Hence, $$P(x)$$ is indeed the desired polynomial.", "in quadratic equation quadratic function we can say that polynomial P ( x =. A polynomial like the one in the form of an equation is given quadratic...", "× # APPROACH! Let p(x) and q(x) be two quadratic polynomials with integer coefficients.Suppose they have a non-rational zero in common. Show that p(x) = r * q(x) for some rational number r. Note by Raven Herd 2 years, 3 months ago Sort by: It's intriguing you because it is not true. For example, take $$p(x) = (2x-1) ^2$$ and $$q(x) = (2x-1) ( 3x-1 )$$. Staff · 2 years, 3 months ago", "# The quadratic polynomial p(x) has the following properties:p(x)geq0 for all real numbers, p(1)=0 and p(2)=2 . Find the value of p(3) is__________. Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free,", "that $y=1$ is a root of this polynomial and factor it out to get $(y - 1) (y^2 + 4 y + 1).$ The quadratic $(y^2 + 4 y + 1)$ has no real roots, so our only solutions are $x= \\pm 1,$ which gives $p \\leq -1$ or $p\\geq1$.", "+0 0 104 1 Let p(x) be a quadratic polynomial with integer coefficients which has $$3+\\sqrt{7}$$ as a root. Compute $$\\frac{p(2)}{p(3)}$$. Jul 21, 2019", "# Please solve this question of quadratic-: Q.1 A quadratic polynomial f (x)= x2 + ax +b is formed with one of its zeros being $\\frac{4+3\\sqrt{3}}{2+\\sqrt{3}}$ where a and b are integers. Also g (x) = x4 + 2x3 -10x2 + 4x- 10 is a biquadratic polynomial such that g $\\left(\\frac{4+3\\sqrt{3}}{2+\\sqrt{3}}\\right)=\\mathrm{c}\\sqrt{3}+\\mathrm{b}$ where c and d are also integers. Find the values of a. b. c and d. • 1 What are you looking for?", "The polynomial of degree 4, $P(x)$ has a root of multiplicity 2 at $x=4$ and roots of multiplicity 1 at $x=0$ and $x=-2$. It goes through the point $(5,17.5)$. Find a formula for $P(x)$. $P(x) =$ , Your overall score for this problem is", "}$ 16. Find a quadratic equation whose roots are sin 15and cos 15o", "+0 # help algebra 0 16 1 Given that the polynomial x^2 - 15x + t = 0 has only positive integer roots, find the average of all distinct possible values of t. Aug 1, 2022 #1 +2275 +2 The roots of the quadratic are $${15 \\pm \\sqrt{225-4t} \\over 2}$$ Note that $$\\sqrt {225 - 4t}$$ must be an odd, positive integer. To solve, set $$\\sqrt{225 - 4t}$$ equal to 13, 11, 9, 7, 5, 3, and 1. Now, solve for t in all cases and find the average. Aug 2, 2022", "Q # Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : (i) p(x) = x ^ 3 - 3x ^2 + 5x – 3 Q1 (1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : $(i) p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2$", "and deduce an expression for $P$ by solving a quadratic. Now substitute $pt$ for $x$ and $1-p$ for $y$.", "10 | QUADRATIC EQUATIONS Find the value of p, for which one root of the quadratic equation px^2-14x + 8 = 0 is 6 times the other. Latest from Doubtnut", "## Sir pls explain me with steps all the tick mark sums Q. Factorize each of the following quadratic polynomials by using the method of completing the square: (1). ${p}^{2}+6p+8$ (2). ${q}^{2}-10q+21$ (3). $4{y}^{2}+12y+5$ (4). ${p}^{2}+6p-16$ (5). ${x}^{2}+12x+20$ What are you looking for?", "1. ## function Given P(x) = x^3 - 3x^2 - 5x a.evaluate P9x) for each integer value of x from -3 through 5 b. Find all zeros of P(x) d. Prove that 5 is an upper bound on the zeros of P(x) 2. ## Re: function a. Do you expect us do this for you? b. Part (a) gives you one zero; use polynomial long division to get a quadratic equation, which gives you the other two zeros. c. Estimate the value of square roots from (b). E.g., $\\sqrt{29}<\\sqrt{36}=6$. 3. ## Re: function Originally Posted by doctorbleachers Given P(x) = x^3 - 3x^2 - 5x a.evaluate P9x) for each integer value of x from -3 through 5 Do you not understand that this is integer arithmetic? Can you not multiply, add, and subtract integers? b. Find all zeros of P(x) You can factor an \"x\" out to P(x) easily, leaving a quadratic term. Do you know the quadratic formula? d. Prove that 5 is an upper bound on the zeros of P(x) You just need an estimate of the square root that shows up in (b). (What happened to (c)?)" ]

[ "such that $A(x)=B(x)Q(x)+R(x)$. The integer division $a=bq+r$ is a particular case. – Américo Tavares Jun 12 '13 at 23:06 By expanding and grouping like terms together, we obtain: \\begin{align*} 2x^3+3x^2-14x-5&=(ax+b)(x+3)(x+1) +C\\\\ 2x^3+3x^2-14x-5&=(ax+b)(x^2+4x+3) +C\\\\ 2x^3+3x^2-14x-5&=(ax)(x^2+4x+3)+b(x^2+4x+3) +C\\\\ 2x^3+3x^2-14x-5&=(ax^3+4ax^2+3ax)+(bx^2+4bx+3b) +C\\\\ 2x^3+3x^2-14x-5&=(a)x^3+(4a+b)x^2+(3a+4b)x+(3b+C) \\end{align*} Now let's compare coefficients. In particular, focus on the coefficients of $x^3$, $x^2$, and $x^0$ (the constant term). From this, we obtain: \\begin{align*} x^3 &: \\boxed{2=a} \\\\ x^2 &: 3 = 4a+b \\implies b = 3-4a=3-4(2) \\implies \\boxed{b=-5} \\\\ x^0 &: -5=3b+C \\implies C=-5-3b=-5-3(-5) \\implies \\boxed{C=10} \\end{align*} - So the coefficient of $x$ is $3a+4b=6-20/3=-2/3$ which is quite different from $-14$. – egreg Jun 12 '13 at 21:57 @egreg Sorry, didn't see the edit. For $x^1$, we now have $3a+4b=3(2)+4(-5)=-14$, as desired. – Adriano Jun 12 '13 at 21:59 I wanted only to warn you that the method was unsufficient; after the edit it works. – egreg Jun 12 '13 at 22:01 Hint: look at the coefficients at $1$ and $x^3$. - can you explain what you mean by at 1 ? – peter_gent Jun 12 '13 at 21:52 @peter_gent I meant the term with no $x$ (or, as if you wish, $x^0$). $3b$ and $-5$ in your case. – TZakrevskiy Jun 12 '13 at 21:53 if you compare x^3 must you also compare x^2 for A? – peter_gent", "+0 # Suppose $P(x)=ax^2+bx+c$ and $P(1)=1,P(2)=2$, and $P(3)=4$. Find $a - b - c$. +1 195 3 +588 Suppose P(x)=ax^2+bx+c and P(1)=1,P(2)=2, and P(3)=4. Find a - b - c. Thank you Aug 29, 2022 #1 +2540 0 Substituting the info gives us a system with 3 equations: $$1 = a + b + c$$ $$2 = 4a + 2b + c$$ $$3 = 16a + 4b + c$$ Subtracting the first equation from the second and the second from the third gives us another system: $$3a + b = 1$$ $$12a + 2b = 1$$ Solving this system gives us $$a = -{1 \\over 6}$$ and $$b = 1 {1 \\over 2}$$ Substituting this into the first equation gives us $$c = -{1 \\over 3}$$, so $$a - b - c = -{1 \\over 6} - {3 \\over 2} + {1 \\over 3} = \\color{brown}\\boxed{-1 {1 \\over 3}}$$ Aug 30, 2022 #2 +588 +1 That seems to be wrong.... Aug 31, 2022 #3 +588 +1 We translate the given information about P into information about a, b and c. \\begin{align*} P(1) &= 1 \\implies a+b+c =1,\\\\ P(2) &= 2 \\implies 4a+2b+c =2,\\\\ P(3) &= 4 \\implies 9a+3b+c =4. \\end{align*} Subtracting the first equation from the second and the second equation from the third gives \\begin{align*}3a+b&=1,\\\\ 5a+b&=2. \\end{align*}", "of $2$, so it is $1$ or $2$. I didnt find a way to continue with the $d\\mid 2ab$.... • Well done! To use $2ab$ you could have done $d\\mid 2b(a+b)-2ab=2a^2$, and the rest follows. Apr 17 '14 at 8:35 • I improved the formatting using $\\LaTeX$. You can click the edit button to see what I exactly did, and perhaps try to do this yourself next time. Questions and answers without $\\LaTeX$ tend to get downvoted. Apr 17 '14 at 8:44 • thanks!Ill try to use Latex. – Jam Apr 17 '14 at 10:41 Suppose $p$ is a prime that divides $a+b$ and $a^2+b^2$, then $p$ also divides $2ab$ and therefore it divides $2$ or $a$ or $b$, but if $p$ divides $a$ and $a+b$... Hint $\\,\\ d=(a\\!+\\!b,a^2\\!+b^2)\\mid 2a^2,2ab,2b^2\\Rightarrow\\,d\\mid(2a^2,2ab,2b^2)= 2(a,b)^2 = 2\\$ by $\\ (a,b)=1$ Remark $\\$ More generally one easily proves $\\ (a^2\\!+b^2,a\\!+\\!b) = (2(a,b)^2,a\\!+\\!b)$ Bezout based approach Since $(a,b)=1$, there are $x,y$ so that $ax+by=1$. Then \\begin{align} 1 &=(ax+by)^3\\\\ &=a^2\\color{#C00000}{(ax^3+3bx^2y)}+b^2\\color{#00A000}{(3axy^2+by^3)} \\end{align} Then, since $$(a^2+b^2)+(a+b)(a-b)=2a^2$$ and $$(a^2+b^2)-(a+b)(a-b)=2b^2$$ we have \\begin{align} &(a^2+b^2)\\left[\\color{#C00000}{(ax^3+3bx^2y)}+\\color{#00A000}{(3axy^2+by^3)}\\right]\\\\ +&(a+b)(a-b)\\left[\\color{#C00000}{(ax^3+3bx^2y)}-\\color{#00A000}{(3axy^2+by^3)}\\right]\\\\ =&2a^2\\color{#C00000}{(ax^3+3bx^2y)}+2b^2\\color{#00A000}{(3axy^2+by^3)}\\\\[6pt] =&2 \\end{align} Therefore, $(a+b,a^2+b^2)\\mid2$.", "8) = x^2(x-2) + 2x(x-2) + 4(x-2) = (x-2)(x^2+2x+4)$ (I choose $2x^2$ and $4x$ because I want every $(\\cdot)$-term to give an $x-2$ as a factor) – Winther Aug 21 '14 at 23:23 The problem is with the third step. Generally, when we talk about factoring by grouping, we want the stuff in brackets to match exactly, so that we can factor them out. For example: $$7x(Ax + B) - 3(Ax + B) = (7x - 3)(Ax + B)$$ If we really want to factor the given binomial by grouping (not recommended!), then notice that: \\begin{align*} x^3 - 8 &= (x^3 + 2x^2 + 4x) + (-2x^2 - 4x - 8) \\\\ &= x(x^2 + 2x + 4) - 2(x^2 + 2x + 4) \\\\ &= (x - 2)(x^2 + 2x + 4) \\\\ \\end{align*} I admit that the first step is not very intuitive. • Sorry, how did you realize that $x^3 - 8 = (x^3 + 2x^2 + 4x) + (-2x^2 - 4x - 8)$? – Zol Tun Kul Aug 21 '14 at 22:49 • Well it's easy to see that the equation is true, since the extra terms just cancel out. But as I noted, figuring out how to add zero in a clever way is certainly nontrivial. Instead, you should probably be using something", "Precalculus (6th Edition) Blitzer The values of m and b are $-4$ and $3$ respectively. We have the function $f\\left( x \\right)=mx+b$. To solve $f\\left( -2 \\right),$ put the value of $x=-2$ in the equation: $m\\left( -2 \\right)+b=11$ It gives: $-2m+b=11$ (I) Then solve $f\\left( 3 \\right),$ put the value of $x=3$ in the equation: $m\\left( 3 \\right)+b=-9$ It gives: $3m+b=-9$ (II) And subtract the equation (II) from equation (I): \\begin{align} & -2m+b-\\left( 3m+b \\right)=11-\\left( -9 \\right) \\\\ & -2m+b-3m-b=11+9 \\\\ & -5m=20 \\\\ & m=-4 \\end{align} Substitute the value of m in equation (I) and get: \\begin{align} & -2\\left( -4 \\right)+b=11 \\\\ & 8+b=11 \\\\ & b=11-8 \\\\ & b=3 \\end{align} Hence, the values of m and b are -4 and 3 respectively.", "x, \\begin{align} & ax-b\\left( \\frac{5}{b} \\right)+2c\\left( \\frac{5}{c} \\right)=-4 \\\\ & ax-5+10=-4 \\\\ & ax+5=-4 \\end{align} Now subtract 5 from both sides, we get, \\begin{align} & ax+5-5=-4-5 \\\\ & ax=-9 \\end{align} Now, dividing both sides by a to get, \\begin{align} & \\frac{ax}{a}=-\\frac{9}{a} \\\\ & \\text{ }x=-\\frac{9}{a} \\\\ \\end{align} Hence, the values are $x=-\\frac{9}{a}$, $y=\\frac{5}{b}$, and $z=\\frac{5}{c}$.", "'18 at 15:52 Let $x_1, x_2, x_3$ be the roots. We have $f(x) = (x - x_1)(x - x_2)(x-x_3)$. so $x_1 = x_2 = x_3 = 0$. Hint: Let the three real roots of $f(x)$ be $r,s,t$ (not necessarily distinct). Then, we may write $$f(x)=(x-r)(x-s)(x-t)$$ Expand this, and set equal to $x^3+ax^2+bx+c$, and continue... Alternatively, note that for $z=a+bi\\in \\mathbb Z, \\bar{z}=a-bi \\in \\mathbb Z$, the norm is: $$|z|=\\sqrt{z\\cdot \\bar{z}}=\\sqrt{a^2+b^2}.$$ So: $$|f(i)|=|c-a+(b-1)i|=\\sqrt{(c-a)^2+(b-1)^2}=1 \\Rightarrow (c-a)^2+b^2-2b=0 \\tag{1}$$ Let $x_1,x_2,x_3$ be the roots of $f(x)=0 \\iff x^3+ax^2+bx+c=0.$ By the Vieta's formulas: \\begin{align}\\begin{cases} x_1+x_2+x_3&=-a\\\\ x_1x_2+x_1x_3+x_2x_3&=b\\\\ x_1x_2x_3&=-c\\end{cases} \\tag{2}\\end{align} Plug $(2)$ to $(1)$: $$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \\iff \\\\ x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \\iff \\\\ x_1=x_2=x_3=0.$$", "Intermediate Algebra (6th Edition) $f(c)=c^2-3c$ Substituting $x=c$ in $f(x)=x^2-3x$, then, \\begin{align*} f(c)=c^2-3c .\\end{align*}", "# Find values of the constants in the following identity: x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D Another question on identities: $$x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D$$ How can I find the coefficients for this? I've got as far as multiplying out the brackets to get: $$x^4+Ax^3 + 5x^2 + x + 3 = (x^4-x^3+Bx^2+4x^2-4x+4B)+Cx+ D$$ It would be useful to get a hint at least on where to go next. - You're almost to the final solution. Just combine all coefficients of various powers or $x$ and compare each of them individually: $$x^4+Ax^3 + 5x^2 + x + 3 = x^4-x^3+ (B+4)x^2+(C-4)x+4B+ D$$ You'll get \\begin{align} A &= -1 \\\\ B &= +5 - 4 \\\\ &= 1 \\\\ C &= +1 +4 \\\\ &= 5 \\\\ D &= 3 - 4B \\\\ &= -1 \\end{align} ### Edit To explain a little as OP mentioned: The process is as simple as taking everything onto one side of $=$: $$(x^4 - x^4) + (A x^3 + x^3) + (5 x^2 - (B + 4) x^2 ) + (x - (C - 4)x ) + \\left(3 - (4B + D)\\right) = 0 \\\\ \\implies (A + 1) x^3 + (1 - B) x^2 + (4 - C) x + (3 - 4B", "1990 AIME Problems/Problem 15 Problem Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \\begin{align*} ax + by &= 3, \\\\ ax^2 + by^2 &= 7, \\\\ ax^3 + by^3 &= 16, \\\\ ax^4 + by^4 &= 42. \\end{align*} Solution 1 Set $S = (x + y)$ and $P = xy$. Then the relationship $$(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})$$ can be exploited: $\\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\\\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\\end{eqnarray*}$ Therefore: $\\begin{eqnarray*}7S & = & 16 + 3P \\\\ 16S & = & 42 + 7P\\end{eqnarray*}$ Consequently, $S = - 14$ and $P = - 38$. Finally: $\\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\\\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\\\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\\\ ax^5 + by^5 & = & \\boxed{020}\\end{eqnarray*}$ Solution 2 A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$, where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$", "Proof: Since $a(x)b(x) \\equiv a(x)c(x) \\pmod {p(x)}$ we have that: (7) \\begin{align} \\quad p(x) | (a(x)b(x) - a(x)c(x)) = a(x)[b(x) - c(x)] \\end{align} • Since $\\gcd (a, p) = 1$ this means that $p(x) | (b(x) - c(x))$ so $b(x) \\equiv c(x) \\pmod {p(x)}$. $\\blacksquare$", "8y - 1 \\end{align*} $$2(x- 2y)(x^2 + xy + y^2)$$ \\begin{align*} 2(x- 2y)(x^2 + xy + y^2) &= 2(x^3 + x^2y + xy^2 - 2x^2y - 2xy^2 - y^3) \\\\ &= 2(x^3 - x^2y - xy^2 - y^3) \\\\ &=2x^3 - 2x^2y - 2xy^2 - 2y^3 \\end{align*} $$3(a-3b)(a^2 + 3ab - b^2)$$ \\begin{align*} 3(a-3b)(a^2 + 3ab - b^2) &= 3(a^3 + 3a^2b - ab^2 - 3a^2b - 9ab^2 + 3b^3) \\\\ &=3(a^3 - 10ab^{2} + 3b^3) \\\\ &= 3a^3- 30ab^{2} + 9b^3 \\end{align*} $$(2a- b)(2a + b)(2a^2 - 3ab + b^2)$$ \\begin{align*} (2a- b)(2a + b)(2a^2 - 3ab + b^2) &= (4a^2 - b^2)(2a^2 - 3ab + b^2) \\\\ &= 8a^4 - 12a^3b + 4a^2b^2 - 2a^2b^2 + 3ab^3 - b^4 \\\\ &= 8a^4 - 12a^3b + 2a^2b^2 + 3ab^3 - b^4 \\end{align*} $$2(3x + y)(3x - y) - (3x -y)^2$$ \\begin{align*} 2(3x + y)(3x - y) - (3x -y)^2 &= 2(9x^2 - y^2) - 9x^2 + 6xy - y^2 \\\\ &= 18x^2 - 2y^2 - 9x^2 + 6xy - y^2 \\\\ &= 9x^2 + 6xy - 3y^2 \\end{align*} $$(x+y)(x-3y) + (2x - y)^2$$ \\begin{align*} (x+y)(x-3y) + (2x - y)^2 &= x^2 - 3xy +xy - 3y^2 + 4x^2 - 4xy + y^2 \\\\ &=5x^2 - 6xy - 2y^2 \\end{align*} $$\\left(\\dfrac{x}{3} - \\dfrac{3}{x}\\right)\\left(\\dfrac{x}{4} + \\dfrac{4}{x}\\right)$$ \\begin{align*} \\left(\\frac{x}{3} -", "of $$a$$ and $$c$$. Therefore, there are two possible options. Option 1 Option 2 $$\\left(x-1\\right)\\left(3x+1\\right)$$ $$\\left(x+1\\right)\\left(3x-1\\right)$$ $$3{x}^{2}-2x-1$$ $$3{x}^{2}+2x-1$$ ### Check that the solution is correct by multiplying the factors \\begin{align*} \\left(x + 1\\right)\\left(3x - 1\\right) & = 3{x}^{2} - x + 3x - 1\\\\ & = 3{x}^{2} + 2x - 1 \\end{align*} $$3{x}^{2} + 2x - 1 = \\left(x + 1\\right)\\left(3x - 1\\right)$$ Exercise 1.8 Factorise the following: $$x^{2} + 8x + 15$$ $x^{2} + 8x + 15 = (x + 5)(x + 3)$ $$x^{2} + 9x + 8$$ $x^{2} + 9x + 8 = (x + 8)(x + 1)$ $$x^{2} + 12x + 36$$ \\begin{align*} x^{2} + 12x + 36 & = (x + 6)(x + 6) \\\\ & = (x + 6)^{2} \\end{align*} $$2h^{2}+5h-3$$ $2h^{2}+5h-3 = (h+3)(2h-1)$ $$3x^{2}+4x+1$$ $3x^{2}+4x+1 = (x+1)(3x+1)$ $$3s^{2}+s-10$$ $3s^{2}+s-10 = (s+2)(3s-5)$ $$x^{2} - 2x - 15$$ $x^{2} - 2x - 15 = (x + 3)(x - 5)$ $$x^{2} + 2x - 3$$ $x^{2} + 2x - 3 = (x + 3)(x - 1)$ $$x^{2} + x - 20$$ $x^{2} + x - 20 = (x + 5)(x - 4)$ $$x^{2} - x - 20$$ $x^{2} - x - 20 = (x - 5)(x + 4)$ $$2x^{2} - 22x + 20$$ \\begin{align*} 2x^{2} + 22x + 20 & = 2(x^{2} + 11x +", "< 2(5)m - 6 < 5 \\quad \\Leftrightarrow \\quad -5 < 10m - 6 < 5 \\end{align} So $m = 1$. Therefore using the matrix $M_2 = \\begin{bmatrix} 1 & m \\\\ 0 & 1 \\end{bmatrix} = \\begin{bmatrix} 1 & 1 \\\\ 0 & 1 \\end{bmatrix}$ to get: (3) \\begin{align} \\quad f(x, y) = g(x + y, y) = 5(x + y)^2 - 6(x + y)y +6y^2 = 5x^2 + 10xy + 5y^2 - 6xy - 6y^2 + 6y^2 = 5x^2 + 4xy + 5y^2 \\end{align} Observe that $f(x, y)$ is such that $0 < 4 < 5 = 5$, and so $f(x, y)$ is equivalent to the reduced binary quadratic form $g(x, y) = 5x^2 + 4xy + 5y^2$. ## Example 2 Reduce the binary quadratic form $h(x, y) = 11x^2 - 2xy + 2y^2$. We have that $a = 11$, $b = -2$, and $c = 2$. So $-11 < -2 \\leq 11 \\not < 2$. We apply the matrix $M_1 = \\begin{bmatrix} 0 & -1\\\\ 1 & 0 \\end{bmatrix}$ to get: (4) \\begin{align} \\quad g(x, y) = h(-y, x) = 11(-y)^2 - 2(-y)(x) + 2(x)^2 = 2x^2 + 2xy + 11y^2 \\end{align} Observe that $g(x, y)$ is such that $-2 < 2 \\leq 2 < 11$. Therefore $g(x, y)$ is a reduced binary quadratic form", "- 2)\\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, &= (x + 1)(x - 2)\\end{align} $$\\therefore\\!{x^3}\\!\\!- \\!2{x^2}\\!\\!-\\!x\\!+\\!2\\!=\\!(x\\!\\!-\\!1)\\!(x\\!-\\!2)(x\\!+\\!\\!1)$$ Method 2: \\begin{align}{x^3}\\!\\!-\\!2{x^2}\\!\\!-\\!x\\!+\\!2\\!&=\\!({x^3}\\!\\!-\\!2{x^2})\\!-\\!(x\\!\\!-\\!\\!2)\\\\ &= {x^2}(x - 2)\\!-\\!1(x\\! -\\!\\! 2)\\\\ &= (x - 2)({x^2} - 1)\\\\ &= (x\\!-\\!2)(x\\!+\\!1)(x\\!-\\!\\! 1) \\\\&\\left( \\begin{array}{l}{\\text{By using }}{a^2} - {b^2}\\\\ = (a + b)(a - b)\\end{array} \\right)\\end{align} (ii) Let $$p(x) = {x^3} - 3{x^2} - 9x - 5$$ By the factor theorem we know that $$x-a$$ is a factor of $$p(x)$$ if $$p(a) = 0.$$ We shall find a factor of $$p(x)$$ by using some trial value of $$x,$$ say $$x = 1.$$ \\begin{align}p(1) &= {(1)^3} - 3{(1)^2} - 9(1) - 5 \\\\ &= 1 - 3 - 9 - 5\\\\ &= - 16 \\ne 0\\end{align} Since the remainder of $$p(1) \\ne 0$$ , by factor theorem we can say $$x=1$$ is not a factor of $$p(x) = {x^3} - 3{x^2} - 9x - 5.$$ Now say $$x = -1.$$ \\begin{align} p(\\!-\\!1) &\\!=\\!\\!{(\\!-\\!1)^3}\\!-\\!\\!3{( -\\!1)^2}\\!\\!-\\!\\! 9(\\!-\\!1)\\!-\\!\\!5 \\\\ &= - 1 - 3 + 9 - 5\\\\ &= - 9 + 9 = 0\\end{align} Since the remainder of $$p( - 1) = 0$$ , by factor theorem we can say $$x=-1$$ is a factor of $$p(x) = {x^3} - 3{x^2} - 9x - 5.$$ Now dividing $$p(x)$$ by $$x+1$$using long division. Hence $${x^3} -\\!\\!3{x^2}\\!\\!-\\!\\!9x\\!-\\!\\!5\\!\\!=\\!\\!(x\\!+\\!1)\\!({x^2}\\!-\\!4x\\!-\\!5)$$ Now taking $${x^2} - 4x - 5$$ , find $$2$$ numbers", "(x+y)(x-y) \\\\ x^3 - y^3 &= (x-y)(x^2+xy+y^2) \\\\ x^3 + y^3 &= (x+y)(x^2-xy+y^2) \\\\ x^4 - y^4 &= (x^2-y^2)(x^2+y^2) \\end{align} ## Application and Extensions ### The identity $$4(x+7)(2x-1)=Ax^2+Bx+C$$ holds for all real values of $$x$$. What is $$A+B+C$$? Multiplying out the left side of the identity we get $4(x+7)(2x-1)=8x^2+52x-28.$ This expression must be equal to the right hand side of the identity, therefore $8x^2+52x-28=Ax^2+Bx+C,$ so $$A=8$$, $$B=52$$, and $$C=-28$$, which gives us $$A+B+C=8+52-28=32$$. $$_\\square$$ ### If $$A+B=8$$ and $$AB=13$$, what is $$A^3+B^3$$? While you could solve for $$A$$ and $$B$$, a more elegant solution exploits the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ which can be rewritten as $x^3+y^3 = (x+y)^3-3xy(x+y).$ Substituting in $$A$$ and $$B$$ for $$x$$ and $$y$$ we get \\begin{align} A^3+B^3 &= (A+B)^3-3AB(A+B) \\\\ &= (8)^3-3(13)(8) \\\\ &= 512-312 \\\\ &= 200 \\quad_\\square \\end{align} Note by Arron Kau 3 years, 4 months ago Sort by: Somes part of the text bug it operations i cant see it? · 3 years, 3 months ago good explanation · 4 months, 2 weeks ago can anyone give me some notes of lessons lines and angles and triangles with inequalities and with sums · 2 years, 11 months ago Why minus to minus is not plus · 2 years, 10 months ago your expaination is good", "but we can use our knowledge of combinations to solve this. $\\editable{B}=\\editable{C}$B=C because of of the symmetry. $\\editable{B}$B also equals the value of $\\nCr{9}{4}$9C4 and $\\editable{C}$C$=$=$\\nCr{9}{5}$9C5, but we also know that $\\nCr{9}{4}=\\nCr{9}{5}$9C4=9C5 (confirming what we already knew from symmetry that the values will be the same). $\\editable{B}$B $=$= $\\nCr{9}{4}$9C4 $=126$=126 Thus both $\\editable{B}$B and $\\editable{C}=126$C=126. ## Pascal's triangle and binomial distributions $(a+b)^0$(a+b)0 $=$= $1$1 $(a+b)^1$(a+b)1 $=$= $a+b$a+b $(a+b)^2$(a+b)2 $=$= $a^2+2ab+b^2$a2+2ab+b2 $(a+b)^3$(a+b)3 $=$= $a^3+3a^2b+3ab^2+b^3$a3+3a2b+3ab2+b3 $(a+b)^4$(a+b)4 $=$= $a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4 $(a+b)^5$(a+b)5 $=$= $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5 Consider the distributions above of $(a+b)^n$(a+b)n. Particularly note the following patterns. • For each distribution to the power $n$n, there are $n+1$n+1 elements. • For each term, the sum of the exponents is $n$n. • Powers of $a$a decrease from left to right, from $n$n down to $0$0. • Powers of $b$b increase from left to right, from $0$0 up to $n$n • The coefficients start at $1$1, end at $1$1, and are the terms of the relevant row from Pascals triangle. #### Worked example ##### Example 2 What are the coefficients for the distribution of $(a+1)^7$(a+1)7, and then write out the full distribution. So we can see that we will have $n+1=8$n+1=8 terms. We can refer to the relevant row in Pascal's triangle, specifically this row This shows us that the coefficients will be $1,7,21,35,35,21,7,1$1,7,21,35,35,21,7,1. Thus the full", "+ aX_2 + 1$, and observing that $a + b^*a = b^*a$, we obtain the equivalent system \\begin{align*} X_1 &= (a^*b + a)X_2 + b^*(aX_1 + aX_2 + 1) = b^*aX_1 + (a^*b + b^*a)X_2 + b^* \\\\ X_2 &= (a+b)X_1 + b(aX_1 + aX_2 + 1) + 1 = (a + b + ba)X_1 + baX_2 + b + 1\\\\ X_3 &= aX_1 + aX_2 + 1 \\end{align*} We deduce from the second equation $$X_2 = (ba)^*((a + b + ba)X_1 + b + 1)$$ and replacing $X_2$ by its value in the first equation, we obtain \\begin{align*} X_1 &= b^*aX_1 + (a^*b + b^*a)(ba)^*((a + b + ba)X_1 + b + 1) + b^*\\\\ &= (b^*a + (a^*b + b^*a)(ba)^*(a + b + ba))X_1 + (a^*b + b^*a)(ba)^*(b + 1) + b^* \\end{align*} Finally, the language recognised by the automaton is $$X_1 = \\bigl(b^*a + (a^*b + b^*a)(ba)^*(a + b + ba)\\bigr)^*[(a^*b + b^*a)(ba)^*(b + 1) + b^*]$$ since $1$ is the unique initial state. Now, if you want to implement this algorithm, you may end up with more complicated regular expressions. For instance, the sentence \"observing that $a + b^*a = b^*a$\" might be difficult to implement. Other useful simplifications like $a^* + 1 = a^*$ or even $K + K = K$ or $K", "hinted at: p(x) = (x - 1)2(x - 2)3(ax + b) therefore: p(0) = (-1)2(-2)3(b) = -8b since p(0) = 2, this becomes: 2 = -8b b = -1/4 now let's compute p'(x): p'(x) = [(x - 1)2(x - 2)3]'(ax + b) + (x - 1)2(x - 2)3(a) = [(x - 1)2(3(x - 2)2) + 2(x - 1)(x - 2)3](ax + b) + (x - 1)2(x - 2)3(a) = [(x - 1)(3) + (2)(x - 2)](x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a) = [3x - 3 + 2x - 4](x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a) = (5x - 7)(x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a) = [(5x - 7)(ax + b) + a(x - 1)(x - 2)](x - 1)(x - 2)2 = [5ax2 + (5b - 7a)x - 7b + ax2 - 3ax + 2a](x - 1)(x - 2)2 = (6ax2 + (5b - 10a)x + (2a - 7b))(x - 1)(x - 2)2 therefore: p'(0) = (2a - 7b)(-1)(-2)2 = (2a - 7b)(-4) = 28b - 8a now we know that b = -1/4, so p'(0) = -7 - 8a. since p'(0) = 1, we have: 1 = -7 - 8a 8 = -8a a = -1 so p(x)", "\\end{align*} The special cases $k=2$ and $k=3$ are probably familiar: $$a^2-b^2=(a-b)(a+b)$$ and $$a^3-b^3=(a-b)\\left(a^2+ab+b^2\\right)\\;.$$ -" ]

[ "is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\\mathbb Q}[x]$, then it has a zero in ${\\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \\begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \\end{align*} where each factor is in ${\\mathbb Z}[x]$ by Gauss's Lemma. Hence, \\begin{align*} a + c & = - 2\\\\ ac + b + d & = 0\\\\ ad + bc & = 1\\\\ bd & = 1. \\end{align*} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \\begin{equation*}ad + bc = b( a + c ) = 1.\\end{equation*} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\\mathbb Q}$. ##### Example17.18 The polynomial \\begin{equation*}f(x) =", "is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\\mathbb Q}[x]$, then it has a zero in ${\\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \\begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \\end{align*} where each factor is in ${\\mathbb Z}[x]$ by Gauss's Lemma. Hence, \\begin{align*} a + c & = - 2\\\\ ac + b + d & = 0\\\\ ad + bc & = 1\\\\ bd & = 1. \\end{align*} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \\begin{equation*}ad + bc = b( a + c ) = 1.\\end{equation*} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\\mathbb Q}$. ##### Example17.18 The polynomial \\begin{equation*}f(x) =", "be a cubic polynomial. Since $$P'(x)$$ is a quadratic polynomial with two roots $$x=1$$ and $$x=3$$, we have $$P'(x)=3a(x-1)(x-3)=3ax^2-12ax+9a$$ for some constant $$a$$ (the factor $$3$$ is there simply to make the integration easier). Thus, $$P(x)=ax^3-6ax^2+9ax+c$$ for some constant $$c$$. Since $$P(1)=6$$ and $$P(3)=2$$, we get $$4a+c=6\\text{ and }c=2\\,,$$ so $$a=1$$ and $$c=2$$. That is, $$P(x)=x^3-6x^2+9x+2\\,,$$ with $$P'(x)=3(x-1)(x-3)=3x^2-12x+9\\text{ and }P''(x)=6(x-2)\\,.$$ This makes $$P''(1)=-6<0$$ and $$P''(3)=6>0$$, so $$x=1$$ and $$x=3$$ are the local maximum and the local minimum, respectively. Hence, $$P(x)$$ is indeed the desired polynomial.", "the $$x$$-axis are the roots of the zeros of the quadratic polynomial. Important Notes 1. Standard form of quadratic polynomial: $$p(x)=ax^2+bx+c$$, $$a\\neq0$$ 2. The curve of the quadratic polynomial is in the form of a parabola. 3. The roots of a quadratic polynomial are the zeros of the quadratic polynomial. 4. If $$\\alpha$$ and $$\\beta$$ are the two zeros of a quadratic polynomial, then the quadratic polynomial is given by $$x^2-(\\alpha+\\beta)x+\\alpha\\beta$$ 5. A quadratic polynomial has at most two zeros. ## Solved Examples Example 1 Ellie knows that the zeros of a quadratic polynomial are -2 and -7 Can you find the quadratic polynomial whose zeros are -2 and -7? ### Solution The zeros of the quadratic polynomial are -2 and -7 Let $$a=-2$$ and $$b=-7$$. Then, $$a+b=-9$$ and $$a\\times b=14$$ So, if $$a$$ and $$b$$ are the zeros of the quadratic polynomial, the quadratic polynomial is given by $$x^2-(a+b)x+ab$$. \\begin{align}x^2-(a+b)x+ab&=x^2-(-9)x+14\\\\&=x^2+9x+14\\end{align} So, the quadratic polynomial is $$x^2+9x+14$$. Example 2 Find the zeros of the quadratic polynomial $$p(x)=x^2-5$$ and verify the relationship between the coefficients of the polynomial and its zeros. Solution Recall the algebraic identity: $$a^-b^2=(a+b)(a-b)$$ Using this identity, we get $$p(x)=x^2-5=(x+\\sqrt{5})(x-\\sqrt{5})$$. So, the value of polynomial is zero when $$x=\\sqrt{5},-\\sqrt{5}$$. These are the roots of quadratic polynomial. \\begin{align}\\text{Sum of zeros}&=\\sqrt{5}-\\sqrt{5}\\\\&=0\\\\&=-\\dfrac{\\text{Coefficient of $$x$$}}{\\text{Coefficient of $$x^2$$}}\\end{align} \\begin{align}\\text{Product of zeros}&=-\\sqrt{5}\\times\\sqrt{5}\\\\&=-5\\\\&=-\\dfrac{5}{1}\\\\&=-\\dfrac{\\text{Constant", "+0 # Polynomial 0 140 1 Let P(x) be a cubic polynomial such that P(0) = -3 and P(1) = 4. When P(x) is divided by x^2 + x + 1, the remainder is 2x + 3. What is the quotient when P(x) is divided by x^2 + x + 1? Jun 11, 2021 #1 +287 +2 Since $P(x)$ is cubic and $x^2+x+1$ is quadratic, the quotient must be linear. Let the quotient be $ax+b$ for some constants $a$ and $b$. Then $$P(x) = (x^2+x+1)(ax+b) + 2x+3.$$ We are given that $P(0)= -3$, so $$(0^2+0+1)(a(0)+b) + 2(0)+3 = -3$$ that is, $$b+3=-3,$$ giving $b=-6$. We are also given that $P(1) = 4$, so $$(1^2+1+1)(a(1)-6) + 2(1)+3 = 4$$ that is, $$3(a-6)+5=4$$ giving $a=\\frac{17}{3}$. Therefore, the quotient is $\\frac{17}{3}x-6$. Jun 11, 2021", "(or the vertex), so it’s given by $$h=-b/{2a}$$. • $k$ is the $y$-coordinate of the vertex, so it’s given by plugging $h$ back into the original expression: $k=ah^2+bh+c$. This is very much like the process of completing the square that you performed in Questions compSquareQn and compSquareQn2, when $a=1$ and $b=2m$. We’ll start by rewriting $x^2+4x+3$ in vertex form. What are the values of $a, h, k$ for this polynomial? Use scratch paper. What is the vertex form of $x^2+4x+3$? Check your answer to the last question, by taking the polynomial in vertex form you just found, expanding it, and simplifying the result. If you’ve done all your algebra correctly, you should get back the original polynomial $x^2+4x+3$. We can simplify the formula $k=ah^2+bh+c$. Multiplying both sides of $$h=-b/{2a}$$ by $-2a$, we get $-2ah=b$. Plugging this equation for $b$ into the one for $k$, we get: $$\\cl\"tight\"{\\table k, =, ah^2, +, bh, +, c; , =, ah^2, -, 2ah(h), +, c; , =, ah^2, -, 2ah^2, +, c; , =, \\colspan 3 -ah^2, +, c}$$ Use the formula $k=-ah^2+c$ to compute $k$ for the quadratic polynomial $x^2+4x+3$. Remember that you already determined that $$h=-4/{2(1)}=-2$$ for this polynomial. Does this match the value of $k$ you found for that polynomial in vertForm? We’ll check algebraically that $a(x-h)^2+k = ax^2+bx+c$.", "Let $$a$$ and $$b$$ be the roots of the quadratic polynomial $$x^2-x-4=0.$$ Let $$f(x)$$ be a cubic polynomial such that $$f(a)=a, f(b)=b,$$ $$f(a+b)=a+b,$$ and $$f(ab)=-84.$$ What is the value of $$f(6)$$?", "Dan Nov 18, 2016 at 19:01 • @DanB You're welcome. Nov 18, 2016 at 19:02 To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\\tag1$$ Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$. In other words, we have \\begin{align*} & r+s=-b\\\\ & rs=c\\end{align*}\\tag2 for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$ X(X+1)(X+3) Therefore (X+1)(X+3) = X^2+4X+3. Therefore multiplying this by X you get X^3+4X^2+3X. So yes well done that is correct. Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$ Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on. Suppose we have a quadratic $p(x)$ that factorizes as: $$p(x) = (x + s)(x + t)$$ where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get: $$p(x) = x^2 + sx + tx + st = x^2", "5,058 views A polynomial $p(x)$ is such that $p(0) = 5, p(1) = 4, p(2) = 9$ and $p(3) = 20$. The minimum degree it should have is 1. $1$ 2. $2$ 3. $3$ 4. $4$ ### First read Sachin Mittal sir's comment that is written below the best answer, too good approach. Although best answer that is written by Arjun sir is also good but it is like formal approach. Lets take $p(x) = ax +b$ Now, $p(0) = 5 \\implies b = 5.$ $p(1) = 4 \\implies a + b = 4, a = -1$ $p(2) = 9 \\implies 2a + b = 9 \\implies -2 + 5 = 9$, which is false. So, degree $1$ is not possible. Let $p(x) =$ $ax^{2}$ $+$ $bx$ $+$ $c$ $p(0) = 5 \\implies c = 5$ $p(1) = 4 \\implies a + b + c = 4 \\implies a + b = -1 \\quad \\to(1)$ $p(2) = 9 \\implies 4a + 2b + c = 9 \\implies 2a + b = 2 \\quad \\to (2)$ $(2) - (1) \\implies a = 3, b = -1 - 3 = -4$ $p(3) = 20 \\implies 9a + 3b + c = 20, 27 -12 + 5 = 20$, equation holds. So, degree $2$ also will suffice. Correct Answer: $B$ by", "If a = 4, then b =1, c =1 Therefore, the quadratic polynomial is $4x^{2} + x+1$. (vi) 4,1 Let the polynomial be $ax^{2}+ bx + c$. $\\alpha + \\beta = 4=\\frac{4}{1}= \\frac{-b}{a}$ $\\alpha\\beta = 1=\\frac{1}{1}= \\frac{c}{a}$ If a = 1, then b =-4, c =1 Therefore, the quadratic polynomial is $4x^{2} + x+1$.", "Polynomial? A polynomial of the form $$p(x)=ax^2+bx+c$$, where $$a \\neq 0$$ is called a quadratic polynomial. ### Zeros of Quadratic Polynomial: Definition real number $$\"k\"$$ of a quadratic polynomial $$p(x)$$ is 0 if $$p(k)=0$$ . Consider the polynomial: $$p(x)=x^2-3x-4$$ Let's find the value of the polynomial at $$x=-1$$ by substituting -1 for $$x$$ in $$p(x)$$. \\begin{align}p(-1)&=(-1)^2-3(-1)-4\\\\&=1+3-4\\\\&=0\\end{align} So, here $$x=-1$$ is a zero of the quadratic polynomial $$p(x)=x^2-3x-4$$. ## How To Find Zeros of a Quadratic Polynomial? There are two methods to find the zeros of a quadratic polynomial: Let us consider the quadratic polynomial $$2x^2+x-3=0$$ Let’s find the determinant to find the nature of the solution. \\begin{align}b^2-4ac&=1^2-4\\times 2\\times (-3)\\\\&=1+24\\\\&=25>0\\end{align} This means that the quadratic polynomial has two real and distinct zeros. Just plug in the values $$a=2$$, $$b=1$$ and $$c=-3$$ in the quadratic formula. \\begin{align}x&=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\\\&=\\frac{-1\\pm \\sqrt{1^2-4\\times 2\\times (-3)}}{2\\times2}\\\\&=\\frac{-1\\pm \\sqrt{25}}{4}\\\\&=\\frac{-1\\pm 5}{4}\\\\&=-\\frac{3}{2},1\\end{align} \\begin{align} x=-\\frac{3}{2},1 \\end{align} ### Factorization Let’s solve this quadratic polynomial by factorization method. Here, we need to find $$a$$ and $$b$$ such that $$a+b=1$$ and $$ab=-6$$ Considering this, we see that the numbers are $$a=3$$ and $$b=-2$$ \\begin{align}2x^2+x-3&=0\\\\2x^2+3x-2x-3&=0\\end{align} Take the common factors out from the first two terms and last two terms. So, this quadratic polynomial factoring gives: \\begin{align}x(2x+3)-(2x+3)&=0\\2x+3)(x-1)&=0\\end{align} When the product of two numbers is \\(0, then either of the numbers are $$0$$ This means $$2x+3=0$$", "## anonymous 4 years ago Just another problem: A quadratic polynomial $$P(x)$$ is such that $$P(x)$$ never takes any negative values and $$P(0)=8$$ and the $$P(8)=0$$. Find $$P(-4)$$ Genre: algebra pre-calculus Rating: Easy 1. Zarkon 18 2. anonymous @zarkon: could you please explain with steps 3. anonymous It is just too easy for Zarkon :) 4. Zarkon you did clasify this as easy :) 5. Mr.Math Let $$P(x)=ax^2+bx+c$$, then $$P(0)=8 \\implies c=8$$ and $$P(8)=0 \\implies 8^2a+8b+8=0 \\implies b=-8a-1$$. Now, $$P(x)=ax^2-(8a+1)x+8\\ge 0 \\implies (x-8)(ax-1)\\ge 0 \\implies a=\\frac{1}{8}.$$ Thus $$P(-4)=\\frac{1}{8}(-4)^2-2(-4)+8=2+8+8=18.$$ 6. Zarkon i used the fact that (8,0) must be the vertex to get the 2nd equation 7. Zarkon used the vertex formula 8. Mr.Math Oh that's better I think. 9. ash2326 Let the quadratic polynomial be $P(x)=ax^2+bx+c$ as it's given it doesn't take negative values D<0 $b^2-4ac<0$ let's substitute x=0 P(0)=8 c=8 P(8)=64a+8b+8=0 or $8a+b+1=0$ we have b^2-4ac<0 or b^2-32a<0 or b^2<32 a we have 8a=-1-b b^2<-4-4b b^2+4b+4<0 (b+2)^<0 so b is a complex no. of the form =-2-xi now let x be 1 so $P(x)=\\frac{-(1+i)}{8} x^2+(-2+i)x+8$ Could anyone point out my mistake?? 10. Zarkon you way is good 11. anonymous Seems I am only one who used Maxima-minima. 12. anonymous maxima-minima : u mean by taking the second derivatives of the eqs and substituting as rt-s^2 ?? 13.", "# Math Help - discrete math help, please 1. ## discrete math help, please Prove that there is no quadratic polynomial p(x)=ax squared+bx+c, where a is not equal to 0 with p(-1)=2, p(0)=1, p(1)=3, p(2)=10 2. Originally Posted by mtaylor98212 Prove that there is no quadratic polynomial p(x)=ax squared+bx+c, where a is not equal to 0 with p(-1)=2, p(0)=1, p(1)=3, p(2)=10 Use the first three values to solve for the coefficients, then substiture $x=2$ into the resulting quadratic to show that $p(2) \\ne 10$. CB", "# Find $a$ and $b$ in the given cubic polynomial Find $a$ and $b$ such that $x+1$ and $x+2$ are factors of the polynomials $x^3+ax^2-bx+10$. Here I am not sure that how can I obtain the value of $a$ and $b$, I tried to multiply $x+1$ and $x+2$ to obtain a quadratic equation by which I divided the obtained quadratic polynomial with the above given cubic polynomial, But it didn't worked. How can I overcome my answer. - Let $x+c$ be the 3rd/last factor as the expression is of degree three. So, $$(x+c)(x+1)(x+2)=x^3+ax^2-bx+10$$ But, $$(x+c)(x+1)(x+2)=(x+c)(x^2+3x+2)=x^3+x^2(c+3)+x(3c+2)+2c$$ So, $$x^3+x^2(c+3)+x(3c+2)+2c=x^3+ax^2-bx+10$$ Comparing the coefficients of the different powers of $x,$ $2c=10\\implies c=5$ $a=c+3=5+3=8$ $-b=3c+2=3\\cdot5+2=17,b=-17$ - If $x+1$ is a factor of $p(x) = x^3+ax^2-bx+10$, Then Sub. $x=-1$ in $p(x) = 0$ we Get $p(-1)=0\\Leftrightarrow -1+a+b+10=0\\Leftrightarrow a+b=-9$ Similarly If $x+2$ is a factor of $p(x) = x^3+ax^2-bx+10$, Then Sub. $x=-2$ in $p(x) = 0$ we Get $P(-2)=0\\Leftrightarrow -8+4a+2b+10=0\\Leftrightarrow 2a+b=-1$ After Solving these two equations, We Get $a=8$ and $b=-17$ - Hint $\\ \\ x-c\\,$ divides $\\, f(x) \\iff f(c) = 0\\$ (Factor Theorem) - If $P(x)$ is our polynomial, then $P(-1)=0$, and $P(-2)=0$. Substitute. You get two linear equations in two unknowns. Solve. -", "coefficients. Here the polynomial is $x^4+ 3x^2+ 6x+ 10$. 10 factors as (1)(10) or (2)(5). So we might try $(x^2+ ax+ 1)(x^2+ bx+ 10)= x^2(x^2+ bx+ 10)+ax(x^2+ bx+ 10)+ 1(x^2+ bx+ 10)= x^4+ (a+ b)x^3+ (10+ ab+ 1)x^2+ (10a+ b)x+ 10$. We want that to equal to $x^4+ 3x^2+ 6x+ 10$ so we must have a+ b= 0, 11+ ab= 3, and 10a+ b= 6. From a+ b= 0, b= -a, of course. Then 10a+ b= 10a- a= 9a= 6 so that a= 2/3 not an integer! Further, if a= 2/3, b must equal -2/3 so 11+ ab becomes 11- 4/9= 95/9, not 3. If we try $(x^2+ ax+ 2)(x^2+ bx+ 5)= x^2(x^2+ bx+ 5)+ ax(x^2+ bx+ 5)+ 2(x^2+ bx+ 5)= x^4+ bx^3+ 5x^2+ ax^3+ abx^2+ 5ax+ 2x^2+ 2bx+ 10= x^4+ (a+b)x^3+ (5+ab+ 2)x^2+ (5a+ 2b)x+ 10$. We want that to equal to $x^4+ 3x^2+ 6x+ 10$ so we must have a+ b= 0, 5+ ab= 3, and 5a+2b= 6. Again, a+ b= 0 gives b= -a so 5a+ 2b= 5a- 2a= 3a= 6 so a= 2 and b= -2. Then 7+ ab= 7- 4= 3! Yes, $(x^2+ 2x+ 2)(x^2- 2x+ 5)= x^4+ 3x^w+ 6x+ 10$. Of course, there are other possibilities: $(x^2+ ax- 1)(x^2+ bx- 10)$ or $(x^2+ ax- 2)(x^2+ bx- 5)$ or a linear term times", "polynomial with coefficients all $1$ and taken $\\pmod n$ if the exponents cover $0,1,…,n-1$, you can rest assured it is divisible by $(x^n-1)/(x-1)$. May not be easy to recognise always, but it is a simple enough test. +1 Sep 7, 2020 at 14:02 Hmm. I'd look at it and say \"There's no rational root\" because $$x = \\pm 1$$ doesn't work. So there's some irrational root, $$\\alpha$$, and two complex-conjugate pairs. So there's no nice obvious linear factor I can write down. Then I'd say \"Maybe there's a quadratic factor.\" I can assume it's monic, so I'm looking to write $$x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r).$$ from which I can expand to get $$x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br$$ if I've done the algebra right. Equating coefficients I see that \\begin{align} 0 &= a + p\\\\ 0 &= q + ap + b\\\\ 1 &= ra + bq\\\\ 1 &= br \\end{align} so $$p = -a$$, and $$r = \\frac1b$$,and these equations become \\begin{align} 0 &= a + -a\\\\ 0 &= q - a^2 + b\\\\ 1 &= \\frac1b a + bq\\\\ 1 &= b(1/b) \\end{align} which simplify down to", "$3^2=9$ and $9|180$ $p=5$ works. For the last one Since both 2 and 5 are relatively prime the theorem does not apply directly. As suggested by the hint: $p(x)=x^2+2x-5 \\implies p(x+1)=(x+1)^2+2(x+1)-5=x^2+4x-2$ Now if we let $p=2$ the polynomial $p(x+1)$ is irreducible. So this implies that $p(x)$ is irreducible. Lets see why. Suppose that $p(x+1)$ is irreduicble but $p(x)$ is then $p(x)=(ax+b)(cx+d)$ for $a,b,c,d \\in \\mathbb{Q}$, but then $p(x+1)=[a(x+1)+d][c(x+1)+d]=(ax+a+d)(cx+c+d)$ but now $p(x+1)$ is reducible and this is a contradiction.", "8th April Morning Slot Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$-$ 1 and it leaves remainder 6 when divided by x + 1; then : A p(2) = 11 B p(2) = 19 C p($-$ 2) = 19 D p($-$ 2) = 11 ## Explanation Let, P(x) = ax2 + bx + c As, P(0) = 1, $\\therefore\\,\\,\\,$ a(0)2 + b(0) + c = 1 $\\Rightarrow $$\\,\\,\\, c = 1 \\therefore\\,\\,\\, P(x) = ax2 + bx + 1 If P(x) is divided by x - 1, remainder = 4 \\Rightarrow$$\\,\\,\\,$ P$\\left( 1 \\right) = 4$ $\\therefore\\,\\,\\,$ a + b + 1 = 4 . . . . . (1) If P(x) is divided by x + 1, remainder = 6 $\\Rightarrow$$\\,\\,\\,$ P($-$ 1) = 6 $\\therefore\\,\\,\\,$ a $-$ b + 1 = 6 . . . .(2) By solving (1) and (2) we get, a = 4, and b = $-$1 $\\therefore\\,\\,\\,$ P(x) = 4x2 $-$ x + 1 P(2) = 4(2)2 $-$ 2 + 1 = 15 P($-$ 2) = 4 ($-$2)2 $-$ ($-$ 2) + 1 = 19", "So it is reducible. • (d) $x^4 +10 x^2 + 1$ in $\\ZZ[x]$. Solution. Assume it has a linear factor. The only possible integer solutions of this polynomial are $\\pm 1$ but neither is a actual solution. Now assume it can factor into two quadratic terms $x^4+10x^2+1 = (x^2+ax+b)(x^2+cx+d)$. Multiply out, \\begin{align*} x^4 + (a+c)x^3 +(ac+b+d)x^2 + (ad+bc)x + bd = x^4 +1. \\end{align*} Matching the coefficients we get \\begin{align*} \\left\\{ \\begin{array}{rcl} a+c &=& 0 \\\\ ac+b+d &=&10 \\\\ ad+bc &=&0 \\\\ bd &=& 1. \\end{array}\\right. \\implies \\left\\{ \\begin{array}{rcl} b &=& \\pm 1 \\\\ d &=& \\pm 1. \\\\ \\end{array} \\right. \\end{align*} On the other hand, from the second equation, $ac = 8$ or $12$. But $a = -c$, we get $-a^2 = 8$ or $12$. A contradiction. Therefore $x^4+10^2 + 1$ is irreducible in $\\ZZ[x]$. Problem 14. Let $p(x) = x^4 - 4x^2 +8x +2$. Show it is irreducible over quadratic field $F=\\QQ(\\sqrt{-2}) = \\{a + b \\sqrt{-2} \\mid a, b \\in \\QQ\\}$. Proof. Note $\\ZZ[\\sqrt{-2}]$ is an integral domain; its field of fraction is $\\QQ(\\sqrt{-2})$. It suffices to show $p(x)$ is irreducible over $\\ZZ[\\sqrt{-2}]$. First assume $p(x)$ has a linear factor $x - r$. Then $r \\mid 2$ in $\\ZZ[\\sqrt{-2}]$. Use the norm $N(r) = N(a+b\\sqrt{-2}) = a^2 + 2b^2$, $N(r) \\mid 4$. Solving $a^2 +", "# Factoring a quadratic polynomial (absolute beginner level), are both answers correct? I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial: $$x^2 + 3x - 10$$ And I'm given the task of finding the values of $a$ and $b$ in: $$(x + a) (x + b)$$ Obviously the answer is: $$(x + 5)(x - 2)$$ However the answer can be also: $$(x - 2) (x + 5)$$ I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$. Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct: Answer $1$ $a = -2$ $b = 5$ or Answer $2$ $a = 5$ $b = -2$ I'm sure this is a completely obvious question, but I'm just a beginner in this. • Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 at 8:18 • They are both valid answers, since the order of the factors doesn't matter. – Ludvig Lindström Aug 23 '18 at 8:18 • You teacher should have stated what context they want you" ]

[ "$\\sum_{k=0}^{\\infty}ar^k=\\frac{a}{1-r}$", "Show that $\\sum_{k = 0}^{\\infty} k^2x^k = x(1 + x)/(1 - x)^3$ for $% $. From chapter text $\\sum_{k = 0}^{\\infty} kx^k = \\frac x {(1 - x)^2}$ (Eqn. A.8). By differentiating this with respect to $x$, we get: Multiplying both sides by $x$, we get:", "x^2}+\\sum_{k=3}^{+\\infty} \\frac{x^{-k}}{k!}$$ You can then multiply by $x²$ and write the apply the limit. The limit goes to $+\\infty$.", "\\infty }\\sum_{k=1}^{N}f(a+k\\frac{b-a}{N})(\\frac{b-a}{N})$", "## How to compute this series: $\\sum_{k=0}^\\infty \\frac{C_k}{2^{2k+1}}$ How to compute this series: $$\\sum_{k=0}^\\infty \\frac{C_k}{2^{2k+1}}$$ where $$C_k$$ is the catalan number: $$C_k=\\frac{1}{k+1}{2k \\choose k}$$. (Further, is there any general method to treat this question with general $$C_k$$?)", "\\sum_{k=1}^{\\infty} \\frac{(-1)^{k-1}}{(2k)!} \\zeta (2k)$ (3) For $k\\ge 2$ is $1 < \\zeta (k) < 2$ so that the series in second term of (3) converges... and converges also very quickly! ... Kind regards $\\chi$ $\\sigma$", "we can combine the sums let k=n $1+\\sum_{n=1}^{\\infty}\\frac{1}{n+1}-\\sum_{n=1}^{\\infty}\\frac{1}{n+1}=1+\\sum_{n=1}^{\\in fty}\\left(\\frac{1}{n+1}-\\frac{1}{n-1} \\right)=1+\\sum_{n=1}^{\\infty}0=1$", "get $\\sum\\limits_{k = 1}^\\infty {kr^{k} } = \\frac{r}{{\\left( {1 - r} \\right)^2 }}$. In this question $r=\\frac{-1}{8}$.", "P(t)^2}} = \\sum_{j=0}^\\infty \\dfrac{(2j)! 4^{-j}}{j!^2} k^{2j} P(t)^{2j+1}$$", "B_n=\\sum_{k=m}^n \\frac{(-1)^{k+1}}{k}. Since ... 4 You can prove that for all x\\in(-1,1),$$\\log(1+x)=\\sum_{k=1}^\\infty \\frac{(-1)^{k+1}}{k}x^k\\$ and use Abel theorem. Top 50 recent answers are included", "\\sum_{k=1}^{n}\\left(\\frac{[kx]}{k}\\right)$", "\\mathcal{O}(1)$ we arrive at: $$\\sum_{k=1}^\\infty (-1)^k \\frac{\\log(k)}{k} = \\gamma \\log(2) - \\frac{\\log^2(2)}{2}$$", "k^2 = {n+1 \\choose 2} = \\sum_{k=1}^n \\; k = \\frac{(n+1) \\; n}{2}$$ -", "= 1$ and hence $\\sum_{k=1}^\\infty \\|x_k\\| < r \\sum_{k=1}^\\infty \\frac{1}{2^k} = r$. Hope this helps. -", "\\frac{2^{n+1}-1}{n+1}=\\sum_{k=0}^{n}\\binom{n}{k}\\int_{0}^{1}x^kdx=\\sum_{k=0}^{n}\\frac{\\binom{n}{k}}{k+1}. \\tag {2}$$ Likewise, take derivatives of $(1)$ w.r.t. $x$ to get $7)$ and $8)$.", "problem can be reducted to: $$\\sum_{k=0}^{\\infty} -\\frac{8}{(4k+1)(4k+3)(4k+5)}$$", "\\infty} \\sum_{k \\mathop = 1}^m a_{\\pi \\left({k}\\right)} = \\sum_{k \\mathop = 1}^\\infty a_{\\pi \\left({k}\\right)}$ $\\blacksquare$", "that $$\\sum_{k=0}^{\\infty} (k+1)|f_n(k)|^2 \\leq R \\quad\\forall n\\in \\mathbb{N}$$ For each $m\\in \\mathbb{N}$, this implies $$\\sum_{k=0}^m (k+1)|f_n(k)|^2 \\leq R$$ ...", "$$\\sum_{k=1}^n \\sigma_k^2 \\to \\infty$$", "\\sum_{k = 0}^{\\infty} \\frac{\\Bigl(\\Psi \\Bigl(\\frac{1}{2} + k\\Bigr) + \\gamma + 2 \\operatorname{ln} (2)\\Bigr) a^{k}}{(1 + 2 k) k!} x + \\operatorname{O} \\bigl(x^{2}\\bigr)$$ -" ]

[ "6(3)2 + 4(3) + 1 + …. + 4n3 + 6n2 + 4n + 1 (n + 1)4 – 14 = 4(13 + 23 + 33 +…+ n3) + 6(12 + 22 + 32 + …+ n2) + 4(1 + 2 + 3 +…+ n) + n $n^4 + 4n^3 + 6n^2 + 4n + 1 – 1 =4\\sum_{k=1}^{n}k^3 + 6\\sum_{k=1}^{n}k^2 + 4\\sum_{k=1}^{n}k+n$ We know that, $\\sum_{k=1}^{n}k=\\frac{n(n+1)}{2}$ And $\\sum_{k=1}^{n}k^2=\\frac{n(n+1)(2n+1)}{6}$ Thus, $n^4 + 4n^3 + 6n^2 + 4n = 4\\sum_{k=1}^{n}k^3 + 6[\\frac{n(n+1)(2n+1)}{6}]+4[\\frac{n(n+1)}{2}]+n$ By rearranging the terms, $4\\sum_{k=1}^{n}k^3 = n^4 + 4n^3 + 6n^2 + 4n – 6[\\frac{n(n+1)(2n+1)}{6}]-4[\\frac{n(n+1)}{2}]-n\\\\\\sum_{k=1}^{n}k^3 = \\frac{1}{4}[n^4 + 4n^3 + 6n^2 + 4n – n (2n^2 + 3n + 1) – 2n(n + 1) – n]$ = (1/4) [n4 + 4n3 + 6n2 + 4n – 2n3 – 3n2 – n – 2n2 – 2n – n] = (1/4) [n4 + 2n3 + n2] = (1/4)[n2(n2 + 2n + 1)] = (1/4)[n2(n + 1)2] Therefore, the sum of cubes of first n natural numbers = [n(n + 1)]2/4 Also, get the Sum of cubes formula here. ### Solved Example Question: Find the sum to n terms of the series: 2 + 5 + 14 + 41 +…. Solution: 2 + 5 + 14 + 41 +…. The difference between two consecutive terms of this series is: 3,", "1. ## Induction - Summation $3+6+9+...+3n=\\frac{n}{2}(3n+3)$ 2. When n=1: $3=\\frac{1}{2}(3+3)=3$ Hence true for n=1. Assume true for n=k: $ 3+6+9+...+3k=\\frac{k}{2}(3k+3) $ ...and i'll leave it up to you to prove it for n=k+1! 3. Originally Posted by Showcase_22 ...and i'll leave it up to you to prove it for n=k+1! Thats where I'm stuck... 4. Originally Posted by nerdzor Thats where I'm stuck... What have you tried? How far have you gotten? You plugged \"k + 1\" into the formula: . . . . . $3\\, +\\, 6\\, +\\, 9\\, +\\, ...\\, +\\, 3k\\, +\\, 3(k\\, +\\, 1)$ You broke off the \"just k\" part so you could apply the assumption: . . . . . $\\left(3\\, +\\, 6\\, +\\, 9\\, +\\, ...\\, +\\, 3k\\right)\\, +\\, 3k\\, +\\, 3$ . . . . . $\\frac{k}{2}\\left(3k\\, +\\, 3\\right)\\, +\\, 3k\\, +\\, 3\\, =\\, \\frac{k(3k\\, +\\, 3)}{2}\\, +\\, \\frac{6k\\, +\\, 6}{2}$ . . . . . $=\\, \\frac{3k^2\\, +\\, 3k\\, +\\, 6k\\, +\\, 2}{2}\\, =\\, \\frac{3k^2\\, +\\, 9k\\, +\\, 6}{2}$ . . . . . $=\\, \\frac{3(k^2\\, +\\, 3k\\, +\\, 2)}{2}\\, =\\, \\frac{3(k\\, +\\, 2)(k\\, +\\, 1)}{2}$ . . . . . $=\\, \\frac{3(k\\, +\\, 1)((k\\, +\\, 1)\\, +\\, 1)}{2}$ Where did you go from here?", "then obviously for $k\\ge1$ we have that $3k^2+3k+1<3k^2+3k^2+k^2=7k^2.$ So you want an upperbound making the nominator larger, well in this case we ended upp with $\\frac{7}{k^4}$ The books answer is $1$. How do I go from $\\frac{7}{k^4}$ to $1$? Also my root test turned up like this, first time so no $\\Rightarrow$ $3^k-2^k = 3^k(1-\\frac{2^k}{3^k})$ $\\lim_{x_\\to \\infty} [3^k(1-\\frac{2^k}{3^k})]^{1/k}$ $ =\\lim_{x_\\to \\infty} 3(1-\\frac{2^k}{3^k})^{1/k}$ And here I´m stuck If I multiply by $\\frac{1}{3^k}$ i get $[\\frac{1}{3^k}-\\frac{2^k}{3^{2k}}]$ so that didn´t work But hey still have other algebraic idéas. Lets multiply by $\\sqrt{3^k}$ ..and that gives me the original expression to the k:th root multiplied by 3...well....nja that doesnt help....(exept that I understand that if $(1-\\frac{2^k}{3^k})^{1/k}<\\frac{1}{3}$ then the limit is smaller then 1 and the series converges. Well great thats what I want to find out by still, im back where I started....lets try something else..and here I´m stuck.... 6. $3^k - 2^k = 3^k(1 - \\frac{2^k}{3^k} ) > \\frac{3^k}{2}$ for all sufficiently large $k$. Now compare your series to $\\sum \\frac{2}{3^k}$", "with the non-homogeneous recurrence: (1) $S_{n+1}=S_{n}+\\frac{n(n+1)}{2}+1$ where $S_0=0$ (2) $S_{n+2}=S_{n+1}+\\frac{(n+1)(n+2)}{2}+1$ Subtracting (1) from (2) yields: (3) $S_{n+2}=2S_{n+1}-S_{n}+n+1$ (4) $S_{n+3}=2S_{n+2}-S_{n+1}+(n+1)+1$ Subtracting (3) from (4) yields: (5) $S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+1$ (6) $S_{n+4}=3S_{n+3}-3S_{n+2}+S_{n+1}+1$ Subtracting (5) from (6) yields: $S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}$ Now we have a homogeneous recurrence whose associated characteristic equation is: $r^4-4r^3+6r^2-4r+1=0$ $(r-1)^4=0$ We have the root r = 1 of multiplcity 4, thus, the closed-form for $S_n$ will take the form: $S_n=k_1+k_2n+k_3n^2+k_4n^3$ $S_0=0\\:\\therefore\\:k_1=0$ Computing: $S_1=1$ $S_2=3$ $S_3=7$ We now have the 3X3 system: $k_2+k_3+k_4=1$ $2k_2+4k_3+8k_4=3$ $3k_2+9k_3+27k_4=7$ and we find: $k_2=\\frac{5}{6}$ $k_3=0$ $k_4=\\frac{1}{6}$ Therefore, we have: $S_n=\\frac{5n}{6}+\\frac{n^3}{6}=\\frac{n\\left(n^2+5 \\right)}{6}$ Thus: $n\\left(n^2+5\\right)=6S_n$ From (1), noting that the product of two consecutive integers is always even, we know $S_n$ is an integer. • December 24th 2011, 02:30 PM Deveno Re: Induction - Divisibility so far, i count 4 ways to skin this cat. outstanding! • December 24th 2011, 04:53 PM Wilmer Re: Induction - Divisibility https://oeis.org/ enter sequence# A004006 • December 24th 2011, 07:19 PM MarkFL Re: Induction - Divisibility In a method very similar to that of Also sprach Zarathustra, we know: $(n-1)n(n+1)$ is divisible by 6, and so must: $(n-1)n(n+1)+6n$ be divisible by 6. $(n-1)n(n+1)+6n=n\\left(n^2-1\\right)+6n=n\\left(n^2-1+6\\right)=n\\left(n^2+5\\right)$ • December 24th 2011, 08:04 PM Prove It Re: Induction - Divisibility Quote: Originally Posted by MarkFL2 In a method very similar to that of Also sprach Zarathustra, we know:", "= 3m^2 + 7m + 4$ • If $n = 6m+5$ then $p_{123}(n) = (3m^2 + 7m + 4) + (m+1) = 3m^2 + 8m + 5.$ This can be written as $\\frac{n^2}{12} + \\frac{n}{2} + c$ where $c$ is $1$, $\\frac5{12}$, $\\frac23$, $\\frac34$, $\\frac23$, or $\\frac5{12}$ depending on whether $n$ modulo $6$ is $0$, $1$, $2$, $3$, $4$ or $5$ respectively. Although all this looks very ad hoc, I believe it can be generalised to any general set $\\{a_1, a_2, \\dots, a_k\\}$. You'll still have to take cases and evaluate sums about $k$ times, and you'll get a polynomial of degree $k-1$, and (I guess) at most $\\mathrm{lcm}(a_1, \\dots, a_k)$ cases. I don't think this is necessarily messier to do by hand than actually carrying out the generating functions -> partial fractions -> … approach to completion. - Denote by $N_3=\\{1,2,3\\}\\,$ then your question can be reformulated: Find the number of partitions of natural number n in parts over set $N_3$ that number we denote $p(n,N_3)$, $n=6k+l$, where $k\\in \\{0,1,2,...\\}$,and $l\\in \\{0,1,2,3,4,5\\}\\,$ Using recurrences (no generating functions) I find that exact solution is. [1] $p(6k,N_3)=3k^2+3k+1$ [2] $p(6k+1,N_3)=3k^2+4k+1$ [3] $p(6k+2,N_3)=3k^2+5k+2$ [4] $p(6k+3,N_3)=3k^2+6k+3$ [5] $p(6k+4,N_3)=3k^2+7k+4$ [6] $p(6k+5,N_3)=3k^2+8k+5$ To explain each formula is needed a lot of introduction example: $p(10,N_3)=14$ because [1,1,1,1,1,1,1,1,1,1] [1,1,1,1,1,1,1,1,2] [1,1,1,1,1,1,1,3] [1,1,1,1,1,1,2,2] [1,1,1,1,1,2,3] [1,1,1,1,2,2,2] [1,1,2,2,2,2] [1,1,1,1,3,3] [1,1,1,2,2,3]", "This gives that we have to show (for $$k \\leq n/2$$): $$\\frac{\\frac {x_1^3}{b}+\\frac {x_2^3}{b}+\\ldots+ \\frac {x_{n-k}^3}{b} + \\frac {x_{n-k+1}^3}{a} + \\ldots + \\frac {x_n^3}{a}}{x_1^2+x_2^2+\\ldots+x_n^2} \\leq \\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2} \\leq \\frac {a^4+b^4}{ab (a^2+b^2)}$$ Now in the LHS, we have that $$\\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2}$$ is $$1$$ for $$k=0$$ and maximal for $$k = n/2$$ where we have $$\\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2} \\quad {{(k = n/2)}\\atop {=}} \\quad \\frac {a^4+b^4}{ab (a^2+b^2)}$$ The same method can be applied for $$k \\geq n/2$$ where again the LHS is $$1$$ for $$k=n$$. For odd $$n$$ we have to take extra care of the central term with $$i = (n+1)/2$$. This proves the claim.", "of order k. I can not check if it's still a polynomial for $$k = 7$$. $$k = 0: 1$$ $$k = 1: n-1$$ $$k = 2: n ^ 2-3n + 2$$ $$k = 3: n ^ 3-5n ^ 2 + 8n-5$$ $$k = 4: n ^ 4- frac {15} 2n ^ 3 + frac {29} 2n ^ 2 + 3n-17$$ $$k = 5: n ^ 5- frac {65} 6n ^ 4 + frac {173} 6n ^ 3 + frac {148} 3n ^ 2- frac {862} 3n + 265$$ $$k = 6: n ^ 6- frac {883} {60} n ^ 5 + frac {157} 3n ^ 4 + frac {2155} {12} n ^ 3- frac {4570} 3n ^ 2+ frac {42767} {15} n-967$$ $$k = 7: frac {1679} {1680} n ^ 7- frac {2765} {144} n ^ 6 + frac {21541} {240} n ^ 5 + frac {69163} {144} n ^ 4- frac {717821} {120} n ^ 3 + frac {1462277} {72} n ^ 2- frac {10388033} {420} n + 5037$$ If I adjust the $$n$$ to one where the first positive value occurs, it is: $$k = 0: 1$$ $$k = 1: n$$ $$k = 2: n ^ 2 + n$$ $$k = 3: n ^ 3 + n ^ 2-1$$ $$k = 4: n ^ 4", "Step 1) Base case: $1^3 = 1$ $\\frac{1^2(1 + 1)^2} {4} = \\frac{2^2} {4} = 1$ Step 2) Assume that $1^3 + 2^3 + 3^3 + ... + k^3 = \\frac{k^2(k + 1)^2} {4}$ Step 3) Show that $1^3 + 2^3 + 3^3 + ... + k^3 + (k + 1)^3 = \\frac{(k + 1)^2((k + 1) + 1)^2} {4}$ First, note that $\\frac{(k + 1)^2((k + 1) + 1)^2} {4} = \\frac{(k + 1)^2(k + 2)^2} {4}$ Now we have: $1^3 + 2^3 + 3^3 + ... + k^3 + (k + 1)^3$ $= \\frac{k^2(k + 1)^2} {4} + (k + 1)^3$ $= \\frac{k^2(k + 1)^2 + 4(k + 1)^3} {4}$ $= \\frac{(k + 1)^2 \\left[k^2 + 4(k + 1) \\right]} {4} = \\frac{(k + 1)^2 \\left[k^2 + 4k + 4 \\right]} {4} = \\frac{(k + 1)^2(x + 2)^2} {4}$ 3) Prove that $1 + 4 + 7 + \\cdots + (3n-2) = \\frac{3n^2-n}{2}$ Step 1) Base case: $1 = 1$ $\\frac{3(1)^2 - 1} {2} = \\frac{2} {2} = 1$ Step 2) Inductive hypothesis: assume that $1 + 4 + 7 + ... + (3k - 2) = \\frac{3k^2 - k} {2}$ Step 3) Show that $1 + 4 + 7 + ... + (3(k + 1) - 2) = \\frac{3(k + 1)^2 - (k + 1)}", "$n = k$ $\\Rightarrow$ $1.3 + 2.4 + 3.5 + \\ldots + k(k+2) =$ $\\frac{1}{6}$ $k(k+1)(2k+7)$ … … … … … i) Now we have to show that $P(n)$ is true for $n = k+1$ $1.3 + 2.4 + 3.5 + \\ldots + k(k+2) + ( k+1)(k+3)$ $=$ $\\frac{1}{6}$ $k(k+1)(2k+7) + ( k+1)(k+3)$ $= (k+1) \\Big[$ $\\frac{k}{6}$ $(2k+7)+(k+3) \\Big]$ $= (k+1) \\Big[$ $\\frac{2k^2+13k+18}{6}$ $\\Big]$ $=$ $\\frac{1}{6}$ $(k+1) (2k+9)(k+2)$ $=$ $\\frac{1}{6}$ $(k+1) [( k+1)+1] [2(k+1)+7]$ $\\Rightarrow P(n)$ is true for $n = k+1$ Hence by the principle of mathematical induction $1.3 + 2.4 + 3.5 + \\ldots + n(n+2) =$ $\\frac{1}{6}$ $n(n+1)(2n+7)$ is true for all $n \\in N$ $\\\\$ Question 13: $1.3 + 3.5 + 5.7 + \\ldots + (2n-1)(2n+1) =$ $\\frac{n(4n^2+6n-1)}{3}$ Let $P(n) :$ $1.3 + 3.5 + 5.7 + \\ldots + (2n-1)(2n+1) =$ $\\frac{n(4n^2+6n-1)}{3}$ For $n = 1$, L.H.S $= 3$ R.H.S $= 3$ $\\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$ Let $P(n)$ is true for $n = k$ $\\Rightarrow$ $1.3 + 3.5 + 5.7 + \\ldots + (2k-1)(2k+1) =$ $\\frac{k(4k^2+6k-1)}{3}$ … … … … … i) Now we have to show that $P(n)$ is true for $n = k+1$ $1.3 + 3.5 + 5.7 + \\ldots + (2k-1)(2k+1) + [2(k+1)-1][2(k+1)+1]$ $=$ $\\frac{k(4k^2+6k-1)}{3}$ $+ (2k+1)(2k+3)$ $=$ $\\frac{4k^3+6k^2-k+ 12k^2+24k+9}{3}$ $=$ $\\frac{4k^3+18k^2+23k+9}{3}$ $=$", "When $$n=2$$, this means $$a^2+b^2+c^2=k_2s_2$$ for some constant $$k_2$$. Setting $$(a,b,c)=(2,-1,-1)$$, we see $$k_2=-2$$. When $$n=3$$, this means $$a^3+b^3+c^3=k_3s_3$$ for some constant $$k_3$$. Setting $$a=2,b=c=-1$$, we get $$k_3=3$$. Similarly $$a^5+b^5+c^5=k_5s_2s_3$$, and again we use $$(a,b,c)=(2,-1,-1)$$ to get that $$k_5=-5$$. Likewise, $$a^7+b^7+c^7=7s_2^2s_3$$. Similar formula under the same condition: $$\\frac{a^2+b^2+c^2}{2}\\frac{a^3+b^3+c^3}{3} = \\frac{a^5+b^5+c^5}{5}\\\\ \\frac{a^2+b^2+c^2}{2}\\frac{a^5+b^5+c^5}{5} = \\frac{a^7+b^7+c^7}{7}$$ More generally, if $$a,b,c\\in\\mathbb Z$$ with $$a+b+c=0$$ and $$n$$ is relatively prime to $$6$$ then $$a^n+b^n+c^n$$ is divisible by $$s_2s_3=(ab+ac+bc)abc$$. There's actually an expression for $$a^n+b^n+c^n$$ with explicit coefficients: $$a^n+b^n+c^n = \\sum_{i+2j+3k=n} (-1)^j\\frac{n}{i+j+k}\\binom{i+j+k}{i,j,k} s_1^is_2^js_3^k\\tag{1}$$ I confess I found that formula when looking at Fermat's Last many many years ago. It occurred to me at the time that Fermat can be expresses, for odd $$n$$, as Given odd positive integer $$n>1$$. Then for integers $$a,b,c$$, $$a^n+b^n+c^n=0$$ if and only if $$a+b+c=0$$ and $$abc=0$$. which seemed to imply it is a question about symmetric polynomials. • Nice, but what does $\\binom{i+j+k}{i,j,k}$ mean? – vesszabo Feb 28 '16 at 14:14 • It's a multinomial. @vesszabo – Thomas Andrews Feb 28 '16 at 18:39 $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$ The key here is the identity (for all $n\\in\\mathbb Z_{\\ge 3}$): $$a^n+b^n+c^n=(a+b+c)\\left(a^{n-1}+b^{n-1}+c^{n-1}\\right)-$$ $$-(ab+bc+ca)\\left(a^{n-2}+b^{n-2}+c^{n-2}\\right)+abc\\left(a^{n-3}+b^{n-3}+c^{n-3}\\right)$$ Therefore: $$a^3+b^3+c^3=3abc\\\\ a^4+b^4+c^4=2(ab+bc+ca)^2\\\\ a^5+b^5+c^5=-5abc(ab+bc+ca)\\\\a^7+b^7+c^7=7abc(ab+bc+ca)^2$$ Therefore: $$\\frac{a^3+b^3+c^3}{3}\\frac{a^7+b^7+c^7}{7} = \\left(\\frac{a^5+b^5+c^5}{5}\\right)^2\\\\\\frac{a^2+b^2+c^2}{2}\\frac{a^3+b^3+c^3}{3}=\\frac{a^5+b^5+c^5}{5}\\\\ \\frac{a^2+b^2+c^2}{2}\\frac{a^5+b^5+c^5}{5}=\\frac{a^7+b^7+c^7}{7}\\\\$$ If we remember the well-known formulas $(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$ $(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy", "has parametric solution: $$a^2+2b^2=3c^2$$ $$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$$ For $$k=4$$ we get: $$(a,b,c)=(5,13,11)$$ If we use Pythagorean triples we have: $$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$$ $$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$$ Here $$a=m^2+n^2$$, $$c=m^2-n^2$$ Now we rearrange the equation as: $$a^2-c^2=2(c^2-b^2)$$ Comparing these relations means: $$c^2-b^2=2m^2n^2$$ $$2m^2n^2=(m^2-n^2)^2-b^2$$ $$b^2=m^4+n^4-4m^2n^2$$ For example with $$m=2$$ and $$n=1$$ we have: $$a=2^2+1^2=5$$, $$c=2^2-1^2=3$$ and: $$b^2=2^4+1^4-4\\times2^2\\times1^2=1$$ or $$b=1$$ and we have: $$5^2+2\\times1^2=3\\times3^2$$ This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $$n^2$$: $$(5n)^2+2(n)^2=(3n)^2$$ $$a+b+c<=25000000$$ $$5n+n+3n=9n<=25000000$$ $$n<=2777777$$ This gives a set of solutions. Also $$a=b=c=1$$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $$b$$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions. $$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$ $$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$ I don't think there is a solution. I have solved many such problems for $$n$$ in terms of $$m$$ and sometimes other variables. Whenever I found an integer value for $$n$$, I had the $$m,n$$ for a Pythagorean triple.", "express this as $\\sum_{k=1}^{n} (k^3 - (k-1)^3)$ We can set these sums equal to each other, since $\\sum_{k=1}^{n} k^3 - \\sum_{k=0}^{n} k^3 = \\sum_{k=1}^{n} (k^3 - (k-1)^3)$. So now: $$\\sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3$$ Now $(k^3 - (k-1)^3) = 3k^2 -3k + 1$ so: $$3\\sum_{k=1}^{n}k^2 - 3\\sum_{k=1}^{n}k + \\sum_{k=1}^{n}1 = n^3$$ This is looking more like it. So now we have: $3\\sum_{k=1}^{n}k^2 = n^3 - n + \\frac{3n}{2}(n+1)$ since we can substitute the sum for $k$ and $1$ summed $n$ times is trivial. So then $\\sum_{k=1}^{n}k^2 = \\frac{n}{6}(2n^2 - 2 + 3(n+1)) = \\frac{n}{6}(2n^2 + 3n + 1)$. The quadratic $2n^2 + 3n + 1 = (2n + 1)(n+1)$ which gives us, after all that: $$\\sum_{k=1}^{n}k^2 = \\frac{n}{6}(n+1)(2n+1)$$ Finally for $\\sum_{k=1}^{n} k^3$. I'll adapt the proofwiki bottom proof for this - we know that $\\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \\ldots + (n-1)^3 + n^3 = \\sum_{k=0}^{n-1}(n-k)^3$. Expanding that last sum, $\\sum_{k=1}^{n} k^3 = \\sum_{k=0}^{n-1}(n^3 - k^3 + 3nk^2 - 3n^2k)$. The intermediate step here took me a moment to notice. Observe that $\\sum_{k=0}^{n-1}(n^3 - k^3 + 3nk^2 - 3n^2k) = n^3\\sum_{k=0}^{n-1}1 - \\sum_{k=0}^{n-1}k^3 + 3n\\sum_{k=0}^{n-1}k^2 -3n^2\\sum_{k=0}^{n-1}k$. So now: $$\\sum_{k=1}^{n} k^3 = n^4 + \\frac{3n^2}{6}(n+1)(2n+1) - \\frac{3n^3}{2}(n+1) - \\sum_{k=0}^{n-1}k^3$$ Which simplifies to $\\sum_{k=1}^{n} k^3 = n^4 + \\frac{n^2}{2}(2n^2 +3n +1) - \\frac{3n^3}{2}(n+1) - \\sum_{k=0}^{n-1}k^3$ and again", "= (n(2n - 1)(2n + 1))/3 P: 1,004 ## Proof by Induction $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \\frac{n(2n - 1)(2n + 1)}{3} = \\frac{4n^3 - n}{3}$$ So for $$k = n + 1$$: $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 + (2n + 1)^2 = \\frac{(n + 1)(2n + 1)(2n + 3)}{3}$$ $$\\underline{1^2 + 3^2 + 5^2 + ... + (2n - 1)^2} + (2n + 1)^2 = \\frac{4n^3 + 12n^2 + 11n + 3}{3}$$ We already know it's true for $$n$$ so you can replace the underlined part: $$\\frac{4n^3 - n}{3} + (2n + 1)^2 = \\frac{4n^3 - n}{3} + 4n^2 + 4n + 1 = \\frac{4n^3 + 12n^2 + 11n + 3}{3}$$ Multiply by 3: $$4n^3 - n + 12n^2 + 12n + 3 = 4n^3 + 12n^2 + 11n + 3$$ QED. P: 403 I have seen a beautiful geometrical way of deriving the sum of cubes of numbers, i.e. 1^3 + 2^3 + ... + n^3 = (1+2+...n)^2. I wonder if there are simple ways of deriving $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \\frac{n(2n - 1)(2n + 1)}{3}$$ P: 1,004 You can prove this: $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \\frac{n(2n", "+ (k - 1)d] + (a + kd)$ $= \\frac{2ak}{2} + \\frac{k(k - 1)d}{2} + \\frac{2a}{2} + \\frac{2kd}{2}$ $= \\frac{1}{2}[2ak + k(k - 1)d + 2a + 2kd]$ $= \\frac{1}{2}[2ak + 2a + (k^2 - k + 2k)d]$ $= \\frac{1}{2}[2a(k + 1) + (k^2 + k)d]$ $= \\frac{1}{2}[2a(k + 1) + k(k + 1)d]$ $= \\frac{k + 1}{2}[2a + kd]$ $= RHS$. Q.E.D.", "easily be wrongly applied to infinite sums to show things like 0=1. Was just being annoyingly rigorous and hoping OP would catch on and realize either you must show it $\\to 0$ or do it your way with only summing to N terms. 6. You are right to be rigorous ! 7. Hi Guys, This is the part I need explained:- (k-1)/2^(k+1) = k/2^k-k+1/2^(k+1). 8. $\\frac{k}{2^k} = \\frac{2k}{2^{k+1}}$ (multiplying numerator and denominator by 2) $\\frac{k}{2^k}-\\frac{k+1}{2^{k+1}} = \\frac{2k}{2^{k+1}}-\\frac{k+1}{2^{k+1}} = \\frac{2k-(k+1)}{2^{k+1}} = \\frac{k-1}{2^{k+1}}$ 9. How would we handle k-1/3^(k+1)? 2k/3^(K+1) won't break down into k/3^k. If we try the following: 3k/3^(k+1) - 2k+1/3^(k+1) = k/3^k - 2k+1/3^(k+1). Which is no longer a telescoping series ie. the middle bits don't cancel. 10. $\\frac{k-1}{3^{k+1}} = \\frac{2k-1}{4 \\cdot 3^k} - \\frac{2(k+1)-1}{4 \\cdot 3^{k+1}}$", "1 + 1$ (one example is $k_1 = 2$, $k_2 =1$, $k_3 = 1$, $k_4 = 0$); and finally there is exactly one way to use the partition $1 + 1 + 1 + 1$, namely $k_1 = 1$, $k_2 = 1$, $k_3 = 1$, $k_4 = 1$. Thus there are $35$ ways to achieve the sum $k = 4$ by applying the partitions of $4$ to the values of $k_1$, $k_2$, $k_3$ and $k_4$, so there should be $35$ different types of terms in the reduced expression. These will be $\\frac{1}{4!} \\bigg[\\partial_{1111} v_1^4 + \\partial_{2222} v_2^4 + \\partial_{3333} v_3^4 + \\partial_{4444} v_4^4 \\ +$ $4 \\partial_{1112} v_1^3 v_2 + 4 \\partial_{1222} v_1 v_2^3 + 4 \\partial_{2223} v_2^3 v_3 \\ +$ $4 \\partial_{3444} v_3 v_4^3 + 4 \\partial_{1113} v_1^3 v_3 + 4 \\partial_{1333} v_1 v_3^3 \\ +$ $4 \\partial_{2333} v_2 v_3^3 + 4 \\partial_{2224} v_2^3 v_4 + 4 \\partial_{1114} v_1^3 v_4 \\ +$ $4 \\partial_{1444} v_1 v_4^3 + 4 \\partial_{3334} v_3^3 v_4 + 4 \\partial_{2444} v_2 v_4^3 \\ +$ $6 \\partial_{1122} v_1^2 v_2^2 + 6 \\partial_{2233} v_2^2 v_3^2 + 6 \\partial_{1133} v_1^2 v_3^2 \\ +$ $6 \\partial_{2244} v_2^2 v_4^2 + 6 \\partial_{1144} v_1^2 v_4^2 + 6 \\partial_{3344} v_3^2 v_4^2 \\ +$ $12 \\partial_{1123} v_1^2 v_2 v_3 + 12 \\partial_{1124} v_1^2 v_2 v_4 + 12 \\partial_{1134} v_1^2 v_3 v_4", "2k + 4k + 4 = 2k2 + 6k + 4 = 2(k + 1)(k + 2) P(k + 1) is also true. ∴ By Mathematical Induction, P(n) for all value n ∈ N. Question 4. 1 + 4 + 7 + ……. + (3n – 2) = $$\\frac{n(3 n-1)}{2}$$ for all n ∈ N. Solution: Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = $$\\frac{n(3 n-1)}{2}$$ Put n = 1, LHS = 1 RHS = $$\\frac{1(3-1)}{2}$$ = 1 ∴ P(1) is true. Assume P(k) is true for n = k P(k): 1 + 4 + 7 + ……. + (3k – 2) = $$\\frac{k(3 k-1)}{2}$$ To prove P(k + 1) is true, i.e., to prove 1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = $$\\frac{(k+1)(3(k+1)-1)}{2}$$ 1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = $$\\frac{(k+1)(3 k+2)}{2}$$ 1 + 4 + 7 + …… + (3k + 1) = $$\\frac{(k+1)(3 k+2)}{2}$$ 1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = $$\\frac{k(3 k-1)}{2}$$ + (3k + 1) = $$\\frac{k(3 k-1)+2(3 k+1)}{2}$$ = $$\\frac{3 k^{2}-k+6 k+2}{2}$$ = $$\\frac{3 k^{2}+5 k+2}{2}$$ = $$\\frac{(k+1)(3 k+2)}{2}$$ ∴ P(k + 1) is true", "if we place summation for EQ(1), we get the following {EQ(2)}: $$\\sum_{K=1}^N [ K^3 - (K-1)^3 ] = \\sum_{K=1}^N 3*K^2 + \\sum_{K=1}^N K - \\sum_{K=1}^N 1$$ Now, the first part is equal to K^2. $$\\sum_{K=1}^N [ K^3 - (K-1)^3 ] = N^3$$ As Σ (K^3 - (K-1)^3) = 1^3 - 0^3 + 2^3 - 1^3 + ... + (N^3) - (N-1)^3 = N^3 (as all other terms are cancelled) $$N^3 = \\sum_{K=1}^N 3*K^2 + \\sum_{K=1}^N K - \\sum_{K=1}^N 1$$ $$N^3 = 3 * F_2(N) + F_1(N) + N$$ where F_2(N) = 1^2 + 2^2 + ... + N^2 (our focus) F_1(N) = 1 + 2 + ... + N = N * (N+1) / 2 $$N^3 = 3 * F_2(N) + N*(N+1)/2 - N$$ $$F_2(N) = ( N^3 - N*(N+1)/2 + N )/3$$ $$F_2(N) = ( 2N^3 + N^2 + 3N ) / 6$$ $$F_2(N) = N * (N+1) * (2N+1) / 6$$ Hence, we have arrived at our equation that is: F_2(N) = N * (N+1) * (2*N+1) / 6 With this, we can find both parts instantly and calculate the difference. Hence, the two equations are: $$F_1(N) = N * (N+1) / 2$$ $$F_2(N) = N * (N+1) * (2*N+1) / 6$$ With this, our pseudocode will be: int difference(N) { int sum_1 = N *", "+ 12 \\partial_{1124} v_1^2 v_2 v_4 + 12 \\partial_{1134} v_1^2 v_3 v_4 \\ +$ $12 \\partial_{1223} v_1 v_2^2 v_3 + 12 \\partial_{1224} v_1 v_2^2 v_4 + 12 \\partial_{1334} v_1 v_3^2 v_4 \\ +$ $12 \\partial_{1344} v_1 v_3 v_4^2 + 12 \\partial_{1244} v_1 v_2 v_4^2 + 12 \\partial_{1233} v_1 v_2 v_3^2 \\ +$ $12 \\partial_{2334} v_2 v_3^2 v_4 + 12 \\partial_{2234} v_2^2 v_3 v_4 + 12 \\partial_{2344} v_2 v_3 v_4^2 \\ +$ $24 \\partial_{1234} v_1 v_2 v_3 v_4 \\bigg]$ The coefficient on each of these terms is the relevant multinomial coefficient, and note that these coefficients add up to $256$. The following are examples of how I calculated the multinomial coefficients in this quadvariate case: For the term involving $\\partial_{1112}$ we have $k_1 = 3$, $k_2 = 1$, $k_3 = 0$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{3! 1! 0! 0!} = \\frac{24}{6} = 4$ For the term involving $\\partial_{1122}$ we have $k_1 = 2$, $k_2 = 2$, $k_3 = 0$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{2! 2! 0! 0!} = \\frac{24}{4} = 6$ For the term involving $\\partial_{1123}$ we have $k_1 = 2$, $k_2 = 1$, $k_3 = 1$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{2! 1! 1! 0!} = \\frac{24}{2} = 12$ Finally, for the term involving", "then distribute and simplify $acd + add – bcc – bcd = acd – bcd \\\\ \\Rightarrow add – bcc = 0 \\Rightarrow ad^2 = bc^2$ However, in all cases where a misadded fraction sum yields the correct answer, at least one of the fractions can be simplified. It will never be the case, that is, that both $$a$$ and $$c$$ and $$b$$ and $$d$$ are coprime. We can see this fairly simply. Since $$bc^2 = ad^2$$, either $$a$$ or $$d$$ must be a multiple of $$c$$. Formally stated (where $$k$$ is some integer): $bc^2 = ad^2 \\Rightarrow bc^2 = (kcd^2 \\vee a(kc)^2)$ If $$a = kc$$, then $$\\frac{a}{c} = \\frac{kc}{c}$$, which can clearly be reduced. If $$d = kc$$ then $$bc^2 = ak^2c^2$$ and $$b = ak^2$$ and so $$\\frac{b}{d} = \\frac{ak^2}{kc}$$, which can clearly be reduced. There is a third possibility, where $$c = mn$$ and where $$a = km$$ and $$d = ln$$, but this leads to both fractions being reducible. In this case, $$bc^2 = ad^2 \\Rightarrow bm^2n^2= kml^2n^2$$$$\\Rightarrow bm = kl^2 \\Rightarrow b = kl^2 \\div m$$. This requires that: (1) $$k$$ is a multiple of $$m$$, (2) $$l$$ is a multiple of $$m$$, or (3) $$kl$$ is a multiple of $$m$$ (that is, all of the factors of $$k$$ and $$l$$, taken" ]

[ "6(3)2 + 4(3) + 1 + …. + 4n3 + 6n2 + 4n + 1 (n + 1)4 – 14 = 4(13 + 23 + 33 +…+ n3) + 6(12 + 22 + 32 + …+ n2) + 4(1 + 2 + 3 +…+ n) + n $n^4 + 4n^3 + 6n^2 + 4n + 1 – 1 =4\\sum_{k=1}^{n}k^3 + 6\\sum_{k=1}^{n}k^2 + 4\\sum_{k=1}^{n}k+n$ We know that, $\\sum_{k=1}^{n}k=\\frac{n(n+1)}{2}$ And $\\sum_{k=1}^{n}k^2=\\frac{n(n+1)(2n+1)}{6}$ Thus, $n^4 + 4n^3 + 6n^2 + 4n = 4\\sum_{k=1}^{n}k^3 + 6[\\frac{n(n+1)(2n+1)}{6}]+4[\\frac{n(n+1)}{2}]+n$ By rearranging the terms, $4\\sum_{k=1}^{n}k^3 = n^4 + 4n^3 + 6n^2 + 4n – 6[\\frac{n(n+1)(2n+1)}{6}]-4[\\frac{n(n+1)}{2}]-n\\\\\\sum_{k=1}^{n}k^3 = \\frac{1}{4}[n^4 + 4n^3 + 6n^2 + 4n – n (2n^2 + 3n + 1) – 2n(n + 1) – n]$ = (1/4) [n4 + 4n3 + 6n2 + 4n – 2n3 – 3n2 – n – 2n2 – 2n – n] = (1/4) [n4 + 2n3 + n2] = (1/4)[n2(n2 + 2n + 1)] = (1/4)[n2(n + 1)2] Therefore, the sum of cubes of first n natural numbers = [n(n + 1)]2/4 Also, get the Sum of cubes formula here. ### Solved Example Question: Find the sum to n terms of the series: 2 + 5 + 14 + 41 +…. Solution: 2 + 5 + 14 + 41 +…. The difference between two consecutive terms of this series is: 3,", "then obviously for $k\\ge1$ we have that $3k^2+3k+1<3k^2+3k^2+k^2=7k^2.$ So you want an upperbound making the nominator larger, well in this case we ended upp with $\\frac{7}{k^4}$ The books answer is $1$. How do I go from $\\frac{7}{k^4}$ to $1$? Also my root test turned up like this, first time so no $\\Rightarrow$ $3^k-2^k = 3^k(1-\\frac{2^k}{3^k})$ $\\lim_{x_\\to \\infty} [3^k(1-\\frac{2^k}{3^k})]^{1/k}$ $ =\\lim_{x_\\to \\infty} 3(1-\\frac{2^k}{3^k})^{1/k}$ And here I´m stuck If I multiply by $\\frac{1}{3^k}$ i get $[\\frac{1}{3^k}-\\frac{2^k}{3^{2k}}]$ so that didn´t work But hey still have other algebraic idéas. Lets multiply by $\\sqrt{3^k}$ ..and that gives me the original expression to the k:th root multiplied by 3...well....nja that doesnt help....(exept that I understand that if $(1-\\frac{2^k}{3^k})^{1/k}<\\frac{1}{3}$ then the limit is smaller then 1 and the series converges. Well great thats what I want to find out by still, im back where I started....lets try something else..and here I´m stuck.... 6. $3^k - 2^k = 3^k(1 - \\frac{2^k}{3^k} ) > \\frac{3^k}{2}$ for all sufficiently large $k$. Now compare your series to $\\sum \\frac{2}{3^k}$", "Also I think this is an overcount, since it includes several possibilities where $b = c$ (which occurs whenever $b = (100-a)/2$). – Vik78 Apr 2 '17 at 9:36 • Oh yes you are right my bad. I'll edit to fix this but either way I found your solution much more elegant and has a result upvoted. :) – Ziad Fakhoury Apr 2 '17 at 9:39 $a<b<c$ can be translated by $b=a+x$ and $c=b+y=a+(x+y)$, where $a,x,y>0$. $$a + b + c = 100\\iff a+a+x+a+x+y=100 \\iff3a+2x+y=100$$ Let $a_k$ the number of solutions of the diophantine equation $3a+2x+y=k$, where $a,x,y>0$. $$\\sum_{d=0}^\\infty a_k t^k=\\sum_{k=0}^\\infty(\\sum_{3a+2x+y=k\\\\a,x,y>0}t^k)=(\\sum_{k=1}^\\infty t^{3k})(\\sum_{k=1}^\\infty t^{2k})(\\sum_{k=1}^\\infty t^{k})=\\frac{t^6}{(1-t^3)(1-t^2)(1-t)}$$ We will calculate explecitely $a_k$ now : $$\\frac{t^6}{(1-t^3)(1-t^2)(1-t)}\\\\=\\frac{t + 2}{9 (t^2 + t + 1))} - \\frac{89}{72 (t - 1)} + \\frac{1}{8 (t + 1)} - \\frac{3}{4 (t - 1)^2} -\\frac{1}{6 (t - 1)^3}-1\\\\=\\sum_{n=1}^∞ t^n\\frac{1}{72} (47 + 9 (-1)^n + 8 e^{-(2 i n π)/3} + 8 e^{(2 i n π)/3} + 6 n^2-36n)\\\\=\\sum_{n=1}^∞ t^n\\frac{1}{72} (47 + 9 (-1)^n + 16\\cos(2\\pi \\frac{n}{3}) + 6 n^2-36n)$$ So : $$a_{100}=\\frac{1}{72} (47+9 -8 +60000-3600)=784$$ By stars and bars there are ${99\\choose2}$ nonnegative solutions to the equation $x_1+x_2+x_3=97$. Among these there are $3\\cdot49$ having two equal $x_i\\in[0\\ ..\\ 48]$, so that there are $${1\\over6}\\left({99\\choose2}-3\\cdot49\\right)=784$$ solutions satisfying $x_1<x_2<x_3$. For each of these the numbers $a:=1+x_1$, $b:=1+x_2$, $c:=1+x_3$ form an", "= 2f(n-1)+n^2$ starting at $f(0)=0$. It is not difficult to find the values of this for small $n$, namely 0, 1, 6, 21, 58, 141, 318, 685, 1434, 2949, 5998,... and it becomes clear that this is a little less than $6\\times 2^n$ much as you might expect from the question. It is not difficult then to take differences (or looking up OEIS A047520, which I considered 11 years ago) and suspect it is likely to satisfy $f(n) = 6\\times 2^n -n^2-4n-6$, and this can easily be proved by induction. Then all you have to do is divide by $2^n$ and take the limit. - Consider the series $$\\sum_{k=0}^\\infty\\;\\frac{x^k}{2^k}=\\frac{1}{1-x/2}$$ Take a derivative $$\\sum_{k=0}^\\infty\\;k\\frac{x^{k-1}}{2^k}=\\frac{1/2}{(1-x/2)^2}\\tag{1}$$ Take another derivative $$\\sum_{k=0}^\\infty\\;k(k-1)\\frac{x^{k-2}}{2^k}=\\frac{1/2}{(1-x/2)^3}\\tag{2}$$ Adding $(1)$ and $(2)$ at $x=1$, we get $$\\sum_{k=0}^\\infty\\;\\frac{k^2}{2^k}=6$$ - I just noticed the pre-calculus tag. Derivatives may not be a good way to go, then. – robjohn Sep 7 '11 at 0:18", "# Chaitanya's Random Pages ## November 22, 2014 ### A cute sum of Ramanujan Filed under: mathematics — ckrao @ 3:09 am Here is a beautiful sum I found in [1], apparently due to Ramanujan. $\\displaystyle \\frac{1}{1^3\\times 2^3} + \\frac{1}{2^3 \\times 3^3} + \\frac{1}{3^3 \\times 4^3} + \\cdots = \\sum_{k=1}^{\\infty} \\frac{1}{k^3(k+1)^3} = 10-\\pi^2\\quad\\quad(1)$ Note that the result also demonstrates that $\\pi^2$ is slightly less than 10, an alternative to the approaches in [2]. Many of his results require advanced number theory to prove, but this one is not too tricky, as long as we know the following similarly attractive result that I had previously mentioned in this blog post. $\\displaystyle \\sum_{k=1}^{\\infty} \\frac{1}{k^2} = \\frac{\\pi^2}{6}\\quad\\quad(2)$ To prove (1), the idea is to write $\\frac{1}{k^3(k+1)^3}$ as a sum of partial fractions and then sum a telescoping series. You might wish to try this yourself before reading further. We write $\\displaystyle \\frac{1}{k^3(k+1)^3} = \\frac{a(k)}{k^3} + \\frac{b(k)}{(k+1)^3} = \\frac{a(k)(k+1)^3 + b(k)k^3}{k^3(k+1)^3},\\quad\\quad(3)$ where $a(k)$ and $b(k)$ are quadratic polynomials. One way of finding $a(k)$ and $b(k)$ would be to compare coefficients of 1, $k$, $k^2$, $k^3$ and solve a system of equations. Another approach, the Extended Euclidean Algorithm, does so via finding the greatest common divisor of $k^3$ and $(k+1)^3$. The following manipulations verify that the greatest common divisor of $k^3$ and $(k+1)^3$", "1 + 1$ (one example is $k_1 = 2$, $k_2 =1$, $k_3 = 1$, $k_4 = 0$); and finally there is exactly one way to use the partition $1 + 1 + 1 + 1$, namely $k_1 = 1$, $k_2 = 1$, $k_3 = 1$, $k_4 = 1$. Thus there are $35$ ways to achieve the sum $k = 4$ by applying the partitions of $4$ to the values of $k_1$, $k_2$, $k_3$ and $k_4$, so there should be $35$ different types of terms in the reduced expression. These will be $\\frac{1}{4!} \\bigg[\\partial_{1111} v_1^4 + \\partial_{2222} v_2^4 + \\partial_{3333} v_3^4 + \\partial_{4444} v_4^4 \\ +$ $4 \\partial_{1112} v_1^3 v_2 + 4 \\partial_{1222} v_1 v_2^3 + 4 \\partial_{2223} v_2^3 v_3 \\ +$ $4 \\partial_{3444} v_3 v_4^3 + 4 \\partial_{1113} v_1^3 v_3 + 4 \\partial_{1333} v_1 v_3^3 \\ +$ $4 \\partial_{2333} v_2 v_3^3 + 4 \\partial_{2224} v_2^3 v_4 + 4 \\partial_{1114} v_1^3 v_4 \\ +$ $4 \\partial_{1444} v_1 v_4^3 + 4 \\partial_{3334} v_3^3 v_4 + 4 \\partial_{2444} v_2 v_4^3 \\ +$ $6 \\partial_{1122} v_1^2 v_2^2 + 6 \\partial_{2233} v_2^2 v_3^2 + 6 \\partial_{1133} v_1^2 v_3^2 \\ +$ $6 \\partial_{2244} v_2^2 v_4^2 + 6 \\partial_{1144} v_1^2 v_4^2 + 6 \\partial_{3344} v_3^2 v_4^2 \\ +$ $12 \\partial_{1123} v_1^2 v_2 v_3 + 12 \\partial_{1124} v_1^2 v_2 v_4 + 12 \\partial_{1134} v_1^2 v_3 v_4", "= 3m^2 + 7m + 4$ • If $n = 6m+5$ then $p_{123}(n) = (3m^2 + 7m + 4) + (m+1) = 3m^2 + 8m + 5.$ This can be written as $\\frac{n^2}{12} + \\frac{n}{2} + c$ where $c$ is $1$, $\\frac5{12}$, $\\frac23$, $\\frac34$, $\\frac23$, or $\\frac5{12}$ depending on whether $n$ modulo $6$ is $0$, $1$, $2$, $3$, $4$ or $5$ respectively. Although all this looks very ad hoc, I believe it can be generalised to any general set $\\{a_1, a_2, \\dots, a_k\\}$. You'll still have to take cases and evaluate sums about $k$ times, and you'll get a polynomial of degree $k-1$, and (I guess) at most $\\mathrm{lcm}(a_1, \\dots, a_k)$ cases. I don't think this is necessarily messier to do by hand than actually carrying out the generating functions -> partial fractions -> … approach to completion. - Denote by $N_3=\\{1,2,3\\}\\,$ then your question can be reformulated: Find the number of partitions of natural number n in parts over set $N_3$ that number we denote $p(n,N_3)$, $n=6k+l$, where $k\\in \\{0,1,2,...\\}$,and $l\\in \\{0,1,2,3,4,5\\}\\,$ Using recurrences (no generating functions) I find that exact solution is. [1] $p(6k,N_3)=3k^2+3k+1$ [2] $p(6k+1,N_3)=3k^2+4k+1$ [3] $p(6k+2,N_3)=3k^2+5k+2$ [4] $p(6k+3,N_3)=3k^2+6k+3$ [5] $p(6k+4,N_3)=3k^2+7k+4$ [6] $p(6k+5,N_3)=3k^2+8k+5$ To explain each formula is needed a lot of introduction example: $p(10,N_3)=14$ because [1,1,1,1,1,1,1,1,1,1] [1,1,1,1,1,1,1,1,2] [1,1,1,1,1,1,1,3] [1,1,1,1,1,1,2,2] [1,1,1,1,1,2,3] [1,1,1,1,2,2,2] [1,1,2,2,2,2] [1,1,1,1,3,3] [1,1,1,2,2,3]", "with the non-homogeneous recurrence: (1) $S_{n+1}=S_{n}+\\frac{n(n+1)}{2}+1$ where $S_0=0$ (2) $S_{n+2}=S_{n+1}+\\frac{(n+1)(n+2)}{2}+1$ Subtracting (1) from (2) yields: (3) $S_{n+2}=2S_{n+1}-S_{n}+n+1$ (4) $S_{n+3}=2S_{n+2}-S_{n+1}+(n+1)+1$ Subtracting (3) from (4) yields: (5) $S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+1$ (6) $S_{n+4}=3S_{n+3}-3S_{n+2}+S_{n+1}+1$ Subtracting (5) from (6) yields: $S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}$ Now we have a homogeneous recurrence whose associated characteristic equation is: $r^4-4r^3+6r^2-4r+1=0$ $(r-1)^4=0$ We have the root r = 1 of multiplcity 4, thus, the closed-form for $S_n$ will take the form: $S_n=k_1+k_2n+k_3n^2+k_4n^3$ $S_0=0\\:\\therefore\\:k_1=0$ Computing: $S_1=1$ $S_2=3$ $S_3=7$ We now have the 3X3 system: $k_2+k_3+k_4=1$ $2k_2+4k_3+8k_4=3$ $3k_2+9k_3+27k_4=7$ and we find: $k_2=\\frac{5}{6}$ $k_3=0$ $k_4=\\frac{1}{6}$ Therefore, we have: $S_n=\\frac{5n}{6}+\\frac{n^3}{6}=\\frac{n\\left(n^2+5 \\right)}{6}$ Thus: $n\\left(n^2+5\\right)=6S_n$ From (1), noting that the product of two consecutive integers is always even, we know $S_n$ is an integer. • December 24th 2011, 02:30 PM Deveno Re: Induction - Divisibility so far, i count 4 ways to skin this cat. outstanding! • December 24th 2011, 04:53 PM Wilmer Re: Induction - Divisibility https://oeis.org/ enter sequence# A004006 • December 24th 2011, 07:19 PM MarkFL Re: Induction - Divisibility In a method very similar to that of Also sprach Zarathustra, we know: $(n-1)n(n+1)$ is divisible by 6, and so must: $(n-1)n(n+1)+6n$ be divisible by 6. $(n-1)n(n+1)+6n=n\\left(n^2-1\\right)+6n=n\\left(n^2-1+6\\right)=n\\left(n^2+5\\right)$ • December 24th 2011, 08:04 PM Prove It Re: Induction - Divisibility Quote: Originally Posted by MarkFL2 In a method very similar to that of Also sprach Zarathustra, we know:", "equation $a_n=\\frac{2}{n}a_{n-1},n\\gt1$ that $l=\\lim_{n\\to\\infty}a_n=(\\lim_{n\\to\\infty}\\frac{2}{n})(\\lim_{n\\to\\infty}a_{n-1})=0\\times l=0$. We first see that $(n-1)! > 2^n$ , for $n=6$ , moreover $k≥6$ and $(k-1)!>2^k$ implies $k! > k(2^k) > 2(2^k)=2^(k+1)$ i.e., $((k+1)-1)! > 2^(k+1)$ ; hence $(n-1)! > 2^n$ is true for all $n≥6$ , by principle of mathematical induction ; so that $n!> n (2^n)$ is true for all $n ≥ 6$ i.e., $2^n /n! < 1/n$ is true for all $n ≥6$ and since $n→∞$ implies $n≥6$ and $1/n→0$ and since $2^n / n!$ is never negative , we conclude $2^n / n!$ tends to 0 as $n→∞$", "+ (k - 1)d] + (a + kd)$ $= \\frac{2ak}{2} + \\frac{k(k - 1)d}{2} + \\frac{2a}{2} + \\frac{2kd}{2}$ $= \\frac{1}{2}[2ak + k(k - 1)d + 2a + 2kd]$ $= \\frac{1}{2}[2ak + 2a + (k^2 - k + 2k)d]$ $= \\frac{1}{2}[2a(k + 1) + (k^2 + k)d]$ $= \\frac{1}{2}[2a(k + 1) + k(k + 1)d]$ $= \\frac{k + 1}{2}[2a + kd]$ $= RHS$. Q.E.D.", "+ 12 \\partial_{1124} v_1^2 v_2 v_4 + 12 \\partial_{1134} v_1^2 v_3 v_4 \\ +$ $12 \\partial_{1223} v_1 v_2^2 v_3 + 12 \\partial_{1224} v_1 v_2^2 v_4 + 12 \\partial_{1334} v_1 v_3^2 v_4 \\ +$ $12 \\partial_{1344} v_1 v_3 v_4^2 + 12 \\partial_{1244} v_1 v_2 v_4^2 + 12 \\partial_{1233} v_1 v_2 v_3^2 \\ +$ $12 \\partial_{2334} v_2 v_3^2 v_4 + 12 \\partial_{2234} v_2^2 v_3 v_4 + 12 \\partial_{2344} v_2 v_3 v_4^2 \\ +$ $24 \\partial_{1234} v_1 v_2 v_3 v_4 \\bigg]$ The coefficient on each of these terms is the relevant multinomial coefficient, and note that these coefficients add up to $256$. The following are examples of how I calculated the multinomial coefficients in this quadvariate case: For the term involving $\\partial_{1112}$ we have $k_1 = 3$, $k_2 = 1$, $k_3 = 0$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{3! 1! 0! 0!} = \\frac{24}{6} = 4$ For the term involving $\\partial_{1122}$ we have $k_1 = 2$, $k_2 = 2$, $k_3 = 0$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{2! 2! 0! 0!} = \\frac{24}{4} = 6$ For the term involving $\\partial_{1123}$ we have $k_1 = 2$, $k_2 = 1$, $k_3 = 1$, $k_4 = 0$, so $\\binom{k}{k_1, k_2, k_3, k_4} = \\frac{4!}{2! 1! 1! 0!} = \\frac{24}{2} = 12$ Finally, for the term involving", "# Show that $\\sum_{k=0}^{\\infty}\\frac{x^{3k}}{(3k)!} = \\frac{1}{3} e^x + \\frac{2}{3} e^{-\\frac{x}{2}} \\cos\\left(\\frac{\\sqrt{3}}{2} x\\right)$ Is there anyone who knows, and want to help, how to show that this is true $\\sum_{k=0}^{\\infty}\\frac{x^{3k}}{(3k)!} = \\frac{1}{3} e^x + \\frac{2}{3} e^{-\\frac{x}{2}} \\cos\\left(\\frac{\\sqrt{3}}{2} x\\right)$ ? I know that $\\sum_{k=0}^{\\infty}\\frac{x^{k}}{k!}= e^x$, but how to use it I don't know. I can't find my lecture notes so it is like that. • Use $\\cos y=\\frac12(e^y+e^{-y})$. – Angina Seng Jan 5 '18 at 16:55 • @LordSharktheUnknown you forgot the $i$'s – Shashi Jan 5 '18 at 16:56 • @LordSharktheUnknown that looks more like $\\cosh(x)$ – Henry Jan 5 '18 at 16:57 Outline: Differentiate three times and notice that the series satisfies $y'''=y$. Then solve that differential equation, and match with the conditions $y(0)=1$ and $y'(0)=y''(0)=0$, that you get by looking at the coefficients of the series. • $\\sum_{k=0}^{\\infty}[\\frac{1}{x^3 (3k-3)!}-\\frac{1}{(3k)!} ]x^{3k} -1 = 0$ is this in the right direction? This is what I get when trying to solve y'''-y=0 – 123 Jan 5 '18 at 17:30 • @potete I think using the characteristic polynomial will be better than that. Just solve $\\lambda^3-1=0$ you will get three different lambda's and your solution will be $$y=Ae^{\\lambda_1 x}+Be^{\\lambda_2 x} +Ce^{\\lambda_3 x}$$ don't forget to make your solution real. Using series is circular. – Shashi Jan 5 '18 at 17:45 Outline: Let", "of order k. I can not check if it's still a polynomial for $$k = 7$$. $$k = 0: 1$$ $$k = 1: n-1$$ $$k = 2: n ^ 2-3n + 2$$ $$k = 3: n ^ 3-5n ^ 2 + 8n-5$$ $$k = 4: n ^ 4- frac {15} 2n ^ 3 + frac {29} 2n ^ 2 + 3n-17$$ $$k = 5: n ^ 5- frac {65} 6n ^ 4 + frac {173} 6n ^ 3 + frac {148} 3n ^ 2- frac {862} 3n + 265$$ $$k = 6: n ^ 6- frac {883} {60} n ^ 5 + frac {157} 3n ^ 4 + frac {2155} {12} n ^ 3- frac {4570} 3n ^ 2+ frac {42767} {15} n-967$$ $$k = 7: frac {1679} {1680} n ^ 7- frac {2765} {144} n ^ 6 + frac {21541} {240} n ^ 5 + frac {69163} {144} n ^ 4- frac {717821} {120} n ^ 3 + frac {1462277} {72} n ^ 2- frac {10388033} {420} n + 5037$$ If I adjust the $$n$$ to one where the first positive value occurs, it is: $$k = 0: 1$$ $$k = 1: n$$ $$k = 2: n ^ 2 + n$$ $$k = 3: n ^ 3 + n ^ 2-1$$ $$k = 4: n ^ 4", "if we place summation for EQ(1), we get the following {EQ(2)}: $$\\sum_{K=1}^N [ K^3 - (K-1)^3 ] = \\sum_{K=1}^N 3*K^2 + \\sum_{K=1}^N K - \\sum_{K=1}^N 1$$ Now, the first part is equal to K^2. $$\\sum_{K=1}^N [ K^3 - (K-1)^3 ] = N^3$$ As Σ (K^3 - (K-1)^3) = 1^3 - 0^3 + 2^3 - 1^3 + ... + (N^3) - (N-1)^3 = N^3 (as all other terms are cancelled) $$N^3 = \\sum_{K=1}^N 3*K^2 + \\sum_{K=1}^N K - \\sum_{K=1}^N 1$$ $$N^3 = 3 * F_2(N) + F_1(N) + N$$ where F_2(N) = 1^2 + 2^2 + ... + N^2 (our focus) F_1(N) = 1 + 2 + ... + N = N * (N+1) / 2 $$N^3 = 3 * F_2(N) + N*(N+1)/2 - N$$ $$F_2(N) = ( N^3 - N*(N+1)/2 + N )/3$$ $$F_2(N) = ( 2N^3 + N^2 + 3N ) / 6$$ $$F_2(N) = N * (N+1) * (2N+1) / 6$$ Hence, we have arrived at our equation that is: F_2(N) = N * (N+1) * (2*N+1) / 6 With this, we can find both parts instantly and calculate the difference. Hence, the two equations are: $$F_1(N) = N * (N+1) / 2$$ $$F_2(N) = N * (N+1) * (2*N+1) / 6$$ With this, our pseudocode will be: int difference(N) { int sum_1 = N *", "to $\\sum_{k=1}^{n} k^3 = n^4 + \\frac{n^2}{2}((2n^2 +3n +1) - 3n^3(n+1)) - \\sum_{k=0}^{n-1}k^3$ and again to: $\\sum_{k=1}^{n} k^3 = n^4 + \\frac{1}{2}(2n^4 + 3n^3 + n^2 - 3n^4 -3n^3) - \\sum_{k=0}^{n-1}k^3$. This we can reduce to: $\\sum_{k=1}^{n} k^3 = n^4 + \\frac{1}{2}(n^2-n^4) - \\sum_{k=0}^{n-1}k^3$ which can be shortened yet further: $\\sum_{k=1}^{n} k^3 = \\frac{1}{2}(n^4 + n^2) - \\sum_{k=0}^{n-1}k^3$. The final step, then, is to add $\\sum_{k=1}^{n} k^3$ to both sides? Why? Well, we observed a bit further back that $\\sum_{k=1}^{n} k^3 - \\sum_{k=0}^{n} k^3 = n^3$ so here $2\\sum_{k=1}^{n} k^3 = \\frac{1}{2}(n^4 + n^2) + \\sum_{k=1}^{n} k^3 - \\sum_{k=0}^{n-1}k^3$ gives us $2\\sum_{k=1}^{n} k^3 = \\frac{1}{2}(n^4 + n^2) + n^3$. We now simply need to reduce this algebraically - $2\\sum_{k=1}^{n} k^3 = \\frac{1}{2}(n^4 + 2n^3 + n^2)$ combines the $n^3$, then factor out $n^2$: $2\\sum_{k=1}^{n} k^3 = \\frac{n^2}{2}(n^2 + 2n + 1)$. Finally, we observe that $n^2 + 2n + 1 = (n+1)^2$ and that we can divide through by two to give: $$\\sum_{k=1}^{n} k^3 = \\frac{n^2}{4}(n+1)^2$$ That's enough fun for one evening", "# Calculating $\\sum_{k=1}^\\infty \\frac{k^2}{2^k}=\\frac{1}{2}+\\frac{4}{4}+\\frac{9}{8}+\\frac{16}{16}+\\frac{25}{32}+\\cdots+\\frac{k^2}{2^k}+\\cdots$ [duplicate] I want to know the value of $$\\sum_{k=1}^\\infty \\frac{k^2}{2^k}=\\frac{1}{2}+\\frac{4}{4}+\\frac{9}{8}+\\frac{16}{16}+\\frac{25}{32}+\\cdots+\\frac{k^2}{2^k}+\\cdots$$ I added up to $k=50$ and got the value $5.999999999997597$, so it seems that it converges to $6.$ But, I don't know how to get the exact value. Is there any other simple method to calculate it? ## marked as duplicate by Martin R, Hans Lundmark, Lord Shark the Unknown, Claude Leibovici, draks ...May 29 '17 at 8:42 If we start with the power series $$\\sum_{k=0}^{\\infty}x^k=\\frac{1}{1-x}$$ (valid for $|x|<1$) and differentiate then multiply by $x$, we get $$\\sum_{k=1}^{\\infty}kx^k=\\frac{x}{(1-x)^2}$$ If we once again differentiate then multiply by $x$, the result is $$\\sum_{k=1}^{\\infty}k^2x^k=\\frac{x(x+1)}{(1-x)^3}$$ and setting $x=\\frac{1}{2}$ shows that $$\\sum_{k=1}^{\\infty}k^22^{-k}=\\frac{\\frac{3}{4}}{\\frac{1}{8}}=6$$ as you guessed. Start with the geometric series $\\frac{1}{1-x} = \\sum_i x^i$. Differentiate it once to get $\\frac{d}{dx} \\left[ \\frac{1}{1-x} \\right] = \\frac{d}{dx} \\sum_i x^i = \\sum_i i x^{i-1}$. Differentiate again to get $\\frac{d^2}{dx^2} \\left[ \\frac{1}{1-x} \\right] = \\frac{d^2}{dx^2} \\sum_i x^i = \\sum_i i (i-1) x^{i-2}$. Now, plug in $x= \\frac{1}{2}$ and adjust indices in teh summations. Alternatively, note that if $T = \\sum_{k \\geq 0}k^2 2^{-k}$ then $$T = 2T-T = \\sum_{k\\geq 0}\\frac{2k+1}{2^k} = 2\\sum_{k \\geq 0}\\frac{k}{2^k} + \\sum_{k \\geq 0}2^{-k}$$ But we can let $S= \\sum_{k \\geq 0}k2^{-k}$ and note that $$S = 2S-S = \\sum_{k\\geq 0}2^{-k} = 2.$$ Hence $T = 2\\cdot 2 +", "This gives that we have to show (for $$k \\leq n/2$$): $$\\frac{\\frac {x_1^3}{b}+\\frac {x_2^3}{b}+\\ldots+ \\frac {x_{n-k}^3}{b} + \\frac {x_{n-k+1}^3}{a} + \\ldots + \\frac {x_n^3}{a}}{x_1^2+x_2^2+\\ldots+x_n^2} \\leq \\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2} \\leq \\frac {a^4+b^4}{ab (a^2+b^2)}$$ Now in the LHS, we have that $$\\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2}$$ is $$1$$ for $$k=0$$ and maximal for $$k = n/2$$ where we have $$\\frac{k \\frac {a^3}{b}+ (n - 2k)\\frac {b^3}{b}+ k \\frac {b^3}{a} }{(n-k) b^2 + k a^2} \\quad {{(k = n/2)}\\atop {=}} \\quad \\frac {a^4+b^4}{ab (a^2+b^2)}$$ The same method can be applied for $$k \\geq n/2$$ where again the LHS is $$1$$ for $$k=n$$. For odd $$n$$ we have to take extra care of the central term with $$i = (n+1)/2$$. This proves the claim.", "# Find $n$ and $k$ if $\\:\\binom{n\\:}{k-1}=2002\\:\\:\\:\\binom{n\\:}{k}=3003\\:\\:$ $$\\:\\binom{n\\:}{k-1}=2002\\:\\:\\:\\binom{n\\:}{k}=3003\\:\\:$$ What are the values for n and k? My initial idea was to divide those two: $$\\frac{\\binom{n\\:}{k-1}}{\\binom{n\\:}{k}}=\\frac{2002}{3003}\\:\\:$$ $$\\frac{\\frac{n!}{\\left(k-1\\right)!\\left(n+1-k\\right)!}}{\\frac{n!}{k!\\left(n-k\\right)!}}\\:=\\frac{2}{3}$$ $$\\frac{k}{n+1-k}\\:=\\frac{2}{3}$$ After that sum them up: $$\\binom{n\\:}{k-1}+\\binom{n\\:}{k}=\\binom{n+1\\:}{k}=5005$$ And now devide the third with the first term and solve everything. However I would get both times $5k = 2n+2$ and that's the problem. • Well k is even and n = 4 mod 5 And you have $\\frac {n!}{(k-1)!(n-k)!} = 1001 = 7.11.13$ You have enough for some real educated guesses. n = 14 k=6,works. – fleablood Nov 18 '15 at 19:10 You have already found that $\\frac{k}{n-k+1}=\\frac23$ from which it follows that $$\\frac32=\\frac{n-k+1}{k}=\\frac{n+1}{k}-1,$$ so $n=\\tfrac52k-1$, which also shows that $k$ is a multiple of $2$ and that $n+1$ is a multiple of $5$. Let $k:=2m$ so that $n=5m-1$. Plugging this back in to $\\tbinom{n+1}{k}=5005$ yields $$5005=\\binom{n+1}{k}=\\frac{(n+1)!}{(n-k+1)!k!}=\\frac{(5m)!}{(3m)!(2m)!}.$$ You could note that $n\\geq13$ because both $2002$ and $3003$ are multiples of $13$, and hence that $m\\geq3$. Either way, trying small values of $m$ shows that $m=3$ works. • The part I like is that $$\\:\\binom{n\\:}{k-2}=1001.$$ I'm not sure that ever happens again, three consecutive entries $A,2A,3A$ in a row of Pascal's triangle – Will Jagy Nov 18 '15 at 18:51 • @WillJagy that's an interesting question, actually :) – Servaes Nov 18 '15 at", "c = 1$ We then find the general terms corresponding to each of these two triplets: \\begin{aligned} t_{4, 0, 4} &= (-1)^0 \\frac{8!}{4!0!4!} 3^0 4^4 x^4 = 17920x^4 \\\\ t_{6, 1, 1} &= (-1)^1 \\frac{8!}{6!1!1!}3^14^1x^4 = - 336x^4 \\end{aligned} Then, in the final expansion, the term containing $x^4$ is $t_{4, 0, 4} + t_{6, 1,1} = 17920x^4 - 336x^4 = 17584x^4$, and the coefficient of $x^4$ is $17584$. Generalization Note that can write the expansion of a trinomial raised to an exponent as a sum, though it is slightly less meaningful: $(x + y + z)^n = \\sum_{a, b, c : a + b + c = n} \\frac{n!}{a!b!c!}x^ay^bz^c$ ### The Multinomial Theorem We can generalize this even further. Let’s introduce the multinomial coefficient: ${n \\choose k_1, k_2, … k_m} = \\frac{n!}{k_1!k_2! … k_m!}$ Under the assumption $k_1 + k_2 + … + k_m = n$, this term is the number of ways to select one subset of size $k_1$, one subset of size $k_2$, … and one subset of size $k_m$ from a group of $n$ items. Notice that with $m = 2$, this is the same as the binomial coefficient that we’re used to. Then: $(x_1 + x_2 + … + x_m)^n = \\sum_{k_1 + k_2 + … + k_m = n} {n \\choose k_1, k_2,", "$f(x)$ mentioned in PWMs post are $\\operatorname{Re}\\{F[0]\\} = \\pi[2n]$ and $\\operatorname{Im}\\{F[0]\\} = 0$ and so their contribution to the sums $R(2n)$ and $\\pi[2n]$ is $\\pi^2[2n]/(2n)$ respectively, which is approximately $(2n/\\log(2n))^2/(2n) = 2n/(\\log^2(2n))$ which is equivalent to the bulk rate mentioned in Charles reply. My intuitive guess for a lower bound for the number of Goldbach partitions would be more like $\\pi^2[2n]/(2n)$. For those number crunchers out there I have included a number of further interrelated relationships between the number of goldbach partitions R(2n) upto 2n , the number of odd primes π(2n) upto 2n, and the number of twin primes T(2n) upto 2n . $$R(2n)=\\frac{1}{2n}\\sum_{k=0}^{k=2n-1}F^{2}(k)$$ $$R(2n-2)=\\nabla+\\frac{1}{2n}\\sum_{k=0}^{k=2n-1}F^{2}(k)e^{-j2π.2k/2n}$$ $$R(2n+2)=\\frac{1}{2n}\\sum_{k=0}^{k=2n-1}F^{2}(k)e^{j2π.2k/2n}$$ $$π(2n)=\\frac{1}{2n}\\sum_{k=0}^{k=2n-1}F(k).F^{*}(k)$$ $$T(2n)=\\frac{1}{2n}\\sum_{k=0}^{k=2n-1}F(k).F^{*}(k)e^{-j2π.2k/2n}$$ where $$\\nabla=-1$$ if 2n-1 is an odd prime otherwise $$\\nabla=0$$ and where F(k) is the discrete Fourier transform of the prime number function f(x) $$F(k)=\\sum_{p\\, odd\\, primes=3}^{p<2n}f(x).e^{-j2π pk/2n}\\quad where\\: k=0,1,2....,2n-1$$ where f(x)=1 if x is an odd prime otherwise f(x)=0 for x=0,1,2...,2n-1. I have source code for these if any one is interested. Email address [email protected] There is indeed a simple relationship between the total number of primes pi(2n) upto 2n and the total number of goldbach partitions R(2n) for that even number. $$R(2n)=\\frac{1}{2n}\\sum_{k=0}^{2k-1}a_{k}^{2}-b_{k}^{2}$$ and $$π[2n]=\\frac{1}{2n}\\sum_{k=0}^{2k-1}a_{k}^{2}+b_{k}^{2}$$ where a = Real{ F[k] } and b = Imag { F[k] } and F[k] is the Fourier transform of the prime number" ]

[ "+ a^2)$ In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \\dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$. $$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \\dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \\dots +a_{0})$$ $$= a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \\dots + a_{1}(x-r)$$ Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that $$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \\dots + r^{n-1})$$ and so $$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \\dots + r^{n-1}) + a_{n-1}(x^{n-2} + \\dots +r^{n-1}) + \\dots + a_1)$$ Once we know a root, we can also try using Polynomial Long Division to get the other factor. For cubics, the roots can be found without the need to guess. Check this out: Cardano's Method. Does that help? - en.wikipedia.org/wiki/Factor_theorem might also be useful. – Hans Lundmark Sep 28 '10 at 20:12 @Hans: Right, I will edit that into the answer. Thanks. – Aryabhata Sep 28 '10 at 20:13 Cardano is a bit of a sledgehammer here; the only other thing I'll note is that synthetic division may be more convenient than long division, depending on the user. – Ｊ. Ｍ. Sep 28 '10 at 23:58 @J.M: Right, just wanted", "coefficients of powers of x, viz, $r1+r2=(-b/a)$. so you will get that maximum of f occurs at $x=(-b/2a)$. so use $f(-b/2a)=18$. from this you will get one more relation in a and b. knowing that $a+b+10=0$ your problem can be solved.", "+ \\dots + x^k$ is the minimal polynomial of $\\alpha$ over $F$, then $g$ divides $f$, and $F = \\mathbb{Q}(a_0, a_{1}, \\dots)$. The tiny bit is in step 2, because you will be (implicitly, perhaps) using elements of the Galois group to get all the roots. PS By the way, the above is straight from the proof that a simple extension has a finite number of intermediate fields. -", "\\dots ,k \\}$$ for some $$k \\in \\mathbb{N}$$ A naive attempt was to define $$f$$ by the positiveness of a polynomial of degree $$n$$ but the problem is that it is unknown whether computing the roots of a polynomial of degree $$n$$ can be done in polynomial time. If it can be done in polynomial time then $$f(S)$$ can be done in $$\\theta(n) \\neq \\theta(|S| \\cdot n)$$ worst case time by summing the size of ranges of positiveness. Is there any better approach?", "has no multiple roots. Now, let $g(x)=Irr(\\alpha,B).$ Clearly $g(x)|f(x)$ in $B[x]$ (as $f(\\alpha)=0$ and $f\\in K[x]\\subset B[x]$ and $g$ is the polynomial of minimum degree in $B[x]$ with this property), so if $f$ has no multiple roots, nor has $g$. Your question simply follows from the fact that $B[x]$ is a UFD.", "coefficients has exactly n roots, counting multiplicities. Proof: Let $f(z)=a_0+a_1z+a_2z^2+\\dots+a_nz^n$. Chose $R\\gg 1$ sufficiently large so that on the circle $|z|=R$, \\begin{aligned} |a_0+a_1z+a_2z^2+\\dots+a_{n-1}z^{n-1}|&\\leq|a_0|+|a_1|R+|a_2|R^2+\\dots+|a_{n-1}|R^{n-1}\\\\ &<(\\sum_{i=0}^{n-1}|a_i|)R^{n-1}\\\\ &<|a_n|R^n\\\\ &=|a_nz^n| \\end{aligned} Since $a_nz^n$ has $n$ roots inside the circle, $f$ also has $n$ roots in the circle, by Rouche’s Theorem. Since $R$ can be arbitrarily large, this proves the Fundamental Theorem of Algebra.", "\\dots, n$. Then every element of order $\\le n$ is a zero of the polynomial $x^N-1=0$. Or else, more wastefully, you can use the polynomial $\\prod_{k=1}^n (x^k-1)$. - Hint: by pigeonholing, if there are infinitely many elements of order $\\le n$ then for some $k\\le n$ there are infinitely many elements of order $k,\\:$ hence $x^k - 1$ has infinitely many roots $\\:\\Rightarrow\\Leftarrow$ -", "case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots. • For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case. Jul 23 '15 at 20:40 First, notice that if any differentiable function $f(x)$ has $r$ real (distinct) roots, then $f'(x)$ must have at least $r-1$ distinct roots. This follows from the following: Suppose that $a_1<...<a_r$ are (distinct) roots from $f(x)$ and $i \\in \\{1,...,r-1\\}$. Then consider every interval of the form $[a_i,a_{i+1}]$. Since this interval is compact, any function on this interval achieves a maximum/minimum. Each minimum/maximum (depending on whether the function $f$ is dipping below or above the real line) corresponds to a root of $f'(x)$. Since we have $r -1$ such intervals, we have at least $r-1$ roots. Since all polynomials are differentiable, we notice that if $f(x)$ has $r$ (distinct) roots, then $f'(x)$ has at least $r - 1$ distinct roots. Your question isn't the clearest, but maybe someone else can come up with a better question", "other roots between $$\\beta$$ and $$\\gamma$$ and hence $$g\"(x)$$ will have a root between $$\\alpha$$ and $$\\gamma.$$ 4. If this degree four polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and hird root. The location of the extrema does not change with $$a$$, only their $$y$$ values. Thus we are looking for $$a$$ such that the two local minima have negative $$y$$ values and the local maximum has positive $$y$$ value. Locate the local extrema as roots of the derivative $$4x^3 - 12x^2 - 16x$$ which has roots as $$-1, 0, 4$$ $$\\Rightarrow a = 0, a = 3, a = 128$$ $$\\therefore 0\\le a \\le 3$$ 5. This problem is similar to 226 and can be solved similarly. 6. $$f'(x) = 5(x - 1)^4 + 7(x + 2)^6 + 9(7x - 5)^8$$ can never be zero for any real value of $$x$$ therefore the given equation can have only one real real root at maximum. 7. Let $$x + 2 = t^2$$ then $$\\sqrt{2(t^2 + 1)} - t = 3 \\Rightarrow 2t^2 + 2 = 9 + 6t + t^2$$ $$\\Rightarrow t = 7, -1$$ Now it is trivial to calculate the values. 308 through 317 have", "Jan 29 '18 at 3:30 • yes sorry, i mean at most three real roots – Alb P Jan 29 '18 at 3:32 • Ok, good, just checking. You just say things like \"there must be at least 3 real roots\" a few times. – jgon Jan 29 '18 at 3:33 • yeah, just edited the question to fix that up – Alb P Jan 29 '18 at 3:34 • Do you know Descartes' rule of signs? en.wikipedia.org/wiki/Descartes%27_rule_of_signs – kimchi lover Jan 29 '18 at 3:36 We can actually do this directly. Note that between each pair of adjacent but distinct real roots, the polynomial must have a local maximum or a local minimum, since it has to turn around to get back to zero. Thus between each pair of adjacent and distinct real roots, the derivative of $f$ has a real root. So let's look at $f'(x)=5x^4-1$. We can see immediately that this has the four fourth roots of $\\frac{1}{5}$ as its roots: $\\frac{1}{\\sqrt[4]{5}}$, $\\frac{i}{\\sqrt[4]{5}}$, $\\frac{-1}{\\sqrt[4]{5}}$, $\\frac{i}{\\sqrt[4]{5}}$. Since only two of these are real, $f$ can only have at most three distinct real roots. All that remains is to check that $f$ has no repeated roots. We can do this by checking that $f$ and $f'$ are relatively prime. Dividing $f$ by $f'$, we get $f(x)=f'(x)(\\frac{1}{5}x) -\\frac{4}{5}x", "} f(x).$$ • Theorem Let $F$ be a field. Let $f(x) \\in F[x]$ with degree $n$ and suppose $n \\geq 1$. Then $f(x)$ has at most $n$ roots in $F$. Next section on Irreducible polynomials Back to Chapter 2: Polynomials", "# Looking for group of polynomials with only real roots Assume $P_\\mathbb R$ is the set of all polynomials which have only real coefficients and only real roots. Define $0$ as a polynomial with infinitely many real roots and all other constant polynomials as having no roots, thus no complex roots either. Say we want to find an operation $\\star$ such that $(P_\\mathbb R,\\star)$ is a group. Let's start looking with some obvious ones. If we select $\\star = +,$ ie. addition of polynomials, we immediately see that $(P_\\mathbb R,+)$ is not a group. Take as an example $p_1 = 1+x$ and $p_2 = x^2.$ Then clearly $p_i \\in P_\\mathbb R$ but $p_1+p_2 = 1+ x+ x^2$ has no real roots. However, construction $(P_\\mathbb R,+)$ has an identity element $0$ and every member has an inverse, namely inverse of $p$ is $-p$. It also has associativity. Thus $(P_\\mathbb R,+)$ is a groupoid. If we select $\\star = \\times$, meaning multiplication of polynomials, then clearly $(P_\\mathbb R,\\times)$ is closed. If $p_1$ has roots $r_1, r_2 \\dots r_n$ and $p_2$ has roots $R_1, R_2 \\dots R_k$, then $p_1 p_2$ has roots $r_1, r_2 \\dots r_n, R_1, R_2 \\dots R_k$. These obviously need not be distinct. So multiplication seems promising. We also have an identity element, namely constant polynomial $1$. $$1\\times", "# Minimum number of real roots 1. Oct 16, 2013 ### utkarshakash 1. The problem statement, all variables and given/known data Let f(x) be a non-constant twice differentiable function defined on R such that f(x)=f(1-x) and f'(1/4) =0 then what is the minimum number of real roots of the equation (f\"(x))^2+f'(x)f'''(x)=0. 3. The attempt at a solution f'(x)=-f'(1-x) f\"(x)=f\"(1-x) f'''(x)=-f'''(1-x) putting x = 1/4 in the first eqn gives f'(3/4)=0 putting x=1/4 in the given equation (f\"(x))^2+f'(x)f'''(x)=0 gives f\"(1/4)=0. ∴f\"(3/4)=0. So I have got a total of 2 roots. But there are some more which I can't find. 2. Oct 17, 2013 ### Saitama Observe that the given equation is $$\\frac{d}{dx}(f'(x)\\cdot f''(x))=0$$ See if that helps. 3. Oct 17, 2013 ### utkarshakash Let y= f'(x)f\"(x) According to Rolle's Theorem there must lie a root of dy/dx between 1/4 and 3/4. But what about other roots? 4. Oct 17, 2013 ### Saitama I am unsure how to proceed but how about x=1/2? At how many points is f'(x) zero? At how many points is f''(x) zero? Last edited: Oct 17, 2013 5. Oct 17, 2013 ### pasmith You need to find the known zeroes of $y = f'f''$, at which point you can apply Rolle's theorem to get the known zeroes of $y'$, which is what you're interested in.", "What is the maximum number of real roots a polynomial of any degree can have? Is the following reasoning correct? According to the complex conjugate root theorem, the number of complex roots of a polynomial is always equal to its degree. Since odd degree polynomials have a maximum of 2 turning points, they can have a maximum of 3 real roots. And since even degree polynomials have a maximum of 1 turning point, they can have a maximum of 3 real roots. Therefore, the maximum number of a real roots a polynomial of any degree can have is 3, all other roots are non-real. Actually the maximum number of turning points of a degree $n$ polynomial is $n-1$ and the one in the picture indeed has four of them, and five real roots. Generalizing, if $n$ is any positive integer, the polynomial $$(x-1)(x-2)\\dots(x-n)$$ has $n$ distinct roots.", "### Home > INT3 > Chapter 8 > Lesson 8.3.1 > Problem8-102 8-102. Each graph below represents a polynomial function. For each function, state its minimum degree. Then state the number of real and complex roots it has. Example: The graph of $y = f\\left(x\\right)$ at right has degree $3$. It has three real roots and no complex roots. 1. Degree $3$ There are two distinct real roots, one of which is repeated. 1. Degree $4$ There are two real roots and two complex roots.", "113 views Let $f$ be a polynomial of degree $n \\geq 3$ all of whose roots are non-positive real numbers. Suppose that $f(1)=1$. What is the maximum possible value of $f^{\\prime}(1)?$ 1. $1$ 2. $n$ 3. $n+1$ 4. $\\frac{n(n+1)}{2}$ 5. $f^{\\prime}(1)$ can be arbitrarily large given only the constraints in the question", "of the second root found, $\\dots ,{\\mathbf{z}}\\left[n-{n}_{R}-1\\right]\\mathbf{.}\\mathbf{re}$ and ${\\mathbf{z}}\\left[n-{n}_{R}-1\\right]\\mathbf{.}\\mathbf{im}$ contain the real and imaginary parts of the ${n}_{R}$th root found. After the failure has occurred, the remaining $\\left({n}_{R}\\right)$ elements of z have identical real and imaginary parts, equal to the large relative number $-1.0/\\left({\\mathbf{nag_real_safe_small_number}}×\\sqrt{2}\\right)$. ## 9 Example This example finds the roots of the fifth degree polynomial $z5 + 2⁢z4 + 3⁢z3 + 4⁢z2 + 5z + 6=0 .$ ### 9.1 Program Text Program Text (c02agce.c) ### 9.2 Program Data Program Data (c02agce.d) ### 9.3 Program Results Program Results (c02agce.r)", "$f(x)$, a polynomial of degree $n$, can have at most $n$ roots. -", "there can be at most three, because it's a cubic polynomial in $\\rho$. No more than two of those can correspond to maxima of the original log likelihood, because you can't have three local maxima in a row not separated by other critical points. Thus Lehmann is perfectly correct. – whuber Mar 4 '16 at 20:37 The question concerns maximizing the likelihood, which is a function of the sole parameter $\\rho \\in [-1,1]$, by finding its critical points. Among them are zeros of the derivative of the log likelihood. In this case, the derivative of the log likelihood is a rational function of $\\rho$: the numerator is a cubic polynomial and the denominator is a quartic polynomial (equal to $(1-\\rho^2)^2$). Since any cubic polynomial may have up to three real roots, how can we be assured there is a unique maximum of the likelihood? Well, we cannot: as Lehmann points out, there may be two local maxima. But no more than that. The reason is that whenever there are two (anywhere among the real numbers, not just in the range $[-1,1]$) then--because any polynomial is differentiable--there must be a local minimum lying between them. That local minimum would correspond to a third root, and that's all there can be. This argument can be appreciated by looking at graphs", "k[x]$. Assume $f(t) = 0$ for all $t \\in k$. Assume to the contrary that $f$ is not the zero polynomial. Then $f$ is a polynomial of degree $n \\geq 1$. Choose distinct elements $z_1, z_2, z_3, \\dots, z_{n+1}$ of $k$. By repeated use of the linear factor theorem, we know that \\$f(x) = (x-z_1)(x-z_2)(... Only top voted, non community-wiki answers of a minimum length are eligible" ]

[ "are considering polynomial with real coefficients. In that case, for $$n=3$$, we can't have 1 repeated imaginary (conjugate) and a real root and the number of possibilities is only $$4$$. In general when the degree is $$n$$, we can have $$k=n-2j$$ real roots with $$j=0,\\dots,\\lfloor n/2\\rfloor$$ with non-negative multiplicities $$m_1,m_2,\\dots,m_k$$ such that $$m_1+2m_2+3m_3+\\dots+km_k=k.$$ The number of non-negative integer solutions of this equation is $$p(k)$$ the number of partitions of $$k$$. Hence the total number of cases is given by the formula $$\\sum_{j=0}^{\\lfloor n/2\\rfloor}p(j)\\cdot p(n-2j)$$ which gives the sequence A002513: $$1, 3, 4, 9, 12, 23, 31, 54, 73, 118,\\dots$$.", "-2$, $n >3$ the example I needed? Because $f_n$ has real and nonreal roots. – user893458 May 11 '14 at 20:48 • Yes it is, but do check everything, because when you present it as a solution, then YOU are making that claim. So you need to recall a tool proving that it is irreducible, and also justify the presence of real an non-real roots. Sounds like you are well on your way! – Jyrki Lahtonen May 11 '14 at 20:51 • For irreducible I would say: Eisenstein's criterion for $p=2$. – user893458 May 11 '14 at 20:53 • Correct. Well done :-) – Jyrki Lahtonen May 11 '14 at 20:54 • The real root is $x=2^{(1/n)}$ and the complex roots are some $n$th roots of unity multiplied by $x$. – user893458 May 11 '14 at 20:55 As $g$ has nonreal roots, we see that complex conjugation $\\psi\\colon z\\mapsto \\bar z$ is a nontrivial automorphism of $K$. Let $\\alpha$ be a real root and $\\beta$ a complex root. By transitivity of the Galois group, there exists an automorphism $\\phi$ with $\\phi(\\alpha)=\\beta$. Then $\\psi(\\phi(\\alpha))=\\overline\\beta$, but $\\phi(\\psi(\\alpha)=\\beta$.", "# Compute: $\\lim_{n\\to\\infty} ({x_{2}}^n+{x_{3}}^n)$ Let be $x_{1}, x_{2}, x_{3}$ the roots of $x^3-x^2-1=0$. If $x_{1}$ is the real root of the equation, then calculate: $$\\lim_{n\\to\\infty} ({x_{2}}^n+{x_{3}}^n)$$ I wonder if this limit can be computed without being necessary to know the exact values of $x_{2}$ and $x_{3}$. - We will prove that the magnitude of the complex roots is less than $1$. Let $f(x) = x^3-x^2-1$. Then $f(1) = -1 <0$ and $f(2) = 3>0$. Hence, there is a real root, $x_1$, in the interval $(1,2)$. You can prove that this is the only real root and the other two have to be complex. This can be done by looking at the derivative in the interval $(1,2)$ which will turn out to be positive and a local maximum occurs at $x=0$ and a local minimum occurs at $x= \\dfrac23$. These guarantee that the remaining two roots are complex. The complex roots $x_2$ and $x_3$ can be written as $x_2 = r e^{i \\theta}$ and $x_3 = r e^{- i \\theta}$ since the complex roots occur in conjugate pairs. But the product of the roots is $1$ i.e. $r^2 x_1 = 1 \\implies r^2 = \\dfrac1{x_1} < 1$. The complex roots must have magnitude less than $1$. Hence, $$x_2^n + x_3^n = 2r^n \\cos(n \\theta)$$ Hence, $$\\lim_{n \\to \\infty}\\left( x_2^n", "asked for a method of generating examples. – Travis Apr 16 '16 at 14:25 • @Travis. OK Thanks. I agree with your argument and your computation of 2922. I also have the number of such cubics with coefficients in -n, ... , n as: 0, 6, 32, 104, 248, 484, 850, 1356, 2046, 2922 for n=1,..,10. – Geoffrey Critzer Apr 16 '16 at 16:04 If $$x^3+ax^2+bx+c$$ (with integers $a,b,c$) has a rational root, then it is in fact an integer root and is a (positive or negative) divisor of $c$ (rational root theorem). In particular, if you let $c=\\pm1$, you need only ensure that $\\pm1$ are no roots, i.e., that $a+b+c+1\\ne 0$ and $-1+a-b+c\\ne 0$. If $c$ is $\\pm$ a prime there are only a few more conditions and if $c$ is composite, the situation is still not difficult to handle as long as we know the factorization of $c$. However, in order to exclude non-real roots, we have to do one additional check: The cubic above will have a pair of conjugate complex roots iff its discriminant $$\\Delta=a^2b^2-4b^3-4a^3c-27c^2+18abc$$ is negative. Using these two tests it is not too hard to determine whether a given cubic has only real-irrational roots. • But he presumably wants (1) all roots real as well as (2) no roots rational. – almagest", "1. The factorization of the second polynomial is ${\\displaystyle f(-x)=-(x-1)^{2}(x+1),\\,}$ So here, the roots are 1 (twice) and −1, the negation of the roots of the original polynomial. ## Nonreal roots Any nth degree polynomial has exactly n roots in the complex plane, if counted according to multiplicity. So if f(x) is a polynomial which does not have a root at 0 (which can be determined by inspection) then the minimum number of nonreal roots is equal to ${\\displaystyle n-(p+q),\\,}$ where p denotes the maximum number of positive roots, q denotes the maximum number of negative roots (both of which can be found using Descartes' rule of signs), and n denotes the degree of the equation. ### Example: zero coefficients, nonreal roots The polynomial ${\\displaystyle f(x)=x^{3}-1\\,,}$ has one sign change, so the maximum number of positive real roots is 1. From ${\\displaystyle f(-x)=-x^{3}-1\\,,}$ we can tell that the polynomial has no negative real roots. So the minimum number of nonreal roots is ${\\displaystyle 3-(1+0)=2\\,.}$ Since nonreal roots of a polynomial with real coefficients must occur in conjugate pairs, we can see that x3 − 1 has exactly 2 nonreal roots and 1 real (and positive) root. ## Special case The subtraction of only multiples of 2 from the maximal number of positive roots occurs because the polynomial may", "# $2P(x)=Q(x)+R(x)$ and $Q(x),R(x)$ have $n$ real coefficients both. Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients can be written as the average of two monic polynomials of degree n with n real roots. This is USAMO 2002 problem 3. The result is trivial for $n=1$, so assume $n>1$. Let $g(x)=(x-2)(x-4)...(x-2n+2)$. We claim that $q(x)=x^n-kg(x),r(x)=2p(x)-q(x)$ works for sufficiently large $k$. The values $g(1),g(3), ... , g(2n-1)$ alternate in sign and have magnitude at least 1. Take $k$ so that $x^n<k/3$ and $|p(x)-x^n|<k/3$ for all $1\\le x\\le 2n-1$. Then if follows that $q(1),q(3),...,q(2n-1)$ also alternate in sign, and similarly $r(1),r(3),...,r(2n-1)$ alternate in sign. So $q(x),r(x)$ each have at least $n-1$ real roots. But polynomials with real coefficients must have an even number of non-real roots, so $q(x),r(x)$ each have $n$ real roots.", "non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$ It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\\lim_{x\\to\\pm\\infty} q(x)=+\\infty$ guarantee two more real zeros. You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each. First observe that $x=0$ is a root for: $f(x)=x^5 − 1102x^4 − 2015x$ Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root. third, count negative real roots by counting sign changes in $f(-x)$ where: $f(-x)=-x^5 - 1102x^4 + 2015x$ here we have the signs: (--)(-+), so we have 1 negative root. From the above, we have 3 real roots for $f(x)$. $$x^5 − 1102 \\cdot x^4 − 2015 \\cdot x = 0$$ This factors into, $$\\left( x^4 − 1102x^3 − 2015 \\right) \\cdot x = 0$$ Therefore $x=0$ is a root. Keep simplifying, $$x^4 − 1102x^3 − 2015=0$$ Set this equal to $f(x)$, $$f(x)=x^4 − 1102x^3 − 2015$$ $f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between", "square root, and in every case & was within a factor 3 of the minimum. It is noticeable that in these tests Q,, was usually acceptably small (less than 100, say); the variation of (Y, as measured by (Y,,/(Y,,,~,, was at times very large, however, indicating the value of using Algorithm SQRT. There is no reason to expect the a,,,,,,-values in the four tables to be of similar size, and in fact the ones in Table 4 are noticeably larger than those in the other tables. A partial explanation for this is afforded by the expression (4.6), from which it may be concluded that if (for the block Rii with TABLE 3 REALUPPERTRIANGULAR,POSITIVEEIGENVALUESa Proportionwith x “All square roots real. Maximum 9.1 1.1 x 10’” 4.3 x 109 1 rglO6 100% 2% 6% &=a 11,111. : 100 < X Q 1066 18% 26% 160% REAL SQUARE ROOTS 427 TABLE 4 REALUPPERQUASITRIANGULAR\" Proportion with Maximum x 9.3 x 10’ %in 1.2 x 108 1.2x 105 2.16 GaX a,nax/GX, ‘/%nin X 1oo~r~1ooo 28% 24% 12% 44% “Only real square roots computed. eigenvalue X in the real Schur decomposition of A) (Y= Re XII2 is small relative to IIRiill, then there is the possibility that the real square roots * Tii will have large elements and hence that a(T) will be large.", "so the maximum number of positive real roots is 1. From f(-x) = -x^3-1\\, , we can tell that the polynomial has no negative real roots. So the minimum number of nonreal roots is 3 - (1+0) = 2 \\, . Since nonreal roots of a polynomial with real coefficients must occur in conjugate pairs, we can see that x3 - 1 has exactly 2 nonreal roots and 1 real (and positive) root. ## Special case The subtraction of only multiples of 2 from the maximal number of positive roots occurs because the polynomial may have nonreal roots, which always come in pairs since the rule applies to polynomials whose coefficients are real. Thus if the polynomial is known to have all real roots, this rule allows one to find the exact number of positive and negative roots. Since it is easy to determine the multiplicity of zero as a root, the sign of all roots can be determined in this case. ## Generalizations If the real polynomial P has k real positive roots counted with multiplicity, then for every a > 0 there are at least k changes of sign in the sequence of coefficients of the Taylor series of the function eaxP(x). For a sufficiently large, there are exactly k such changes of sign.[1][2] In the", "Select Page The equation $x^3 + 10x^2 - 100x + 1729$ has at least one complex root α such that |α| > 12. False ** Discussion: A fun fact : 1729 is the Ramanujan Number; it is the smallest number expressible as the sum of two cubes in two different ways We conduct normal extrema tests. First derivative is $3x^2 + 20x - 100$. Critical points are (zeroes of first derivative) are -10 and 20/6. Second Derivative (6x + 20) test tells that -10 is the (x coordinate of ) maxima and 20/6 is minima. Since these maxima and minima turns out to be global extremas, we check f(20/6) and find that it is positive. Hence the function has exactly one real root smaller than -10 (other wise the global minima would have been negative). Product of the roots is – 1729. Since the coefficients are all real we conclude the complex roots must have occurred in conjugates. Let $\\alpha$ be a complex root and t be the real root. Then $|\\alpha|^2 t = -1729$ . We wish to show that $| \\alpha | > 12 \\implies | \\alpha |^2 > 144 \\implies | \\alpha |^2 t < 144t$ since t is negative. This says that -13 < t but plugging in -13 in f(x) we see that", "other roots between $$\\beta$$ and $$\\gamma$$ and hence $$g\"(x)$$ will have a root between $$\\alpha$$ and $$\\gamma.$$ 4. If this degree four polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and hird root. The location of the extrema does not change with $$a$$, only their $$y$$ values. Thus we are looking for $$a$$ such that the two local minima have negative $$y$$ values and the local maximum has positive $$y$$ value. Locate the local extrema as roots of the derivative $$4x^3 - 12x^2 - 16x$$ which has roots as $$-1, 0, 4$$ $$\\Rightarrow a = 0, a = 3, a = 128$$ $$\\therefore 0\\le a \\le 3$$ 5. This problem is similar to 226 and can be solved similarly. 6. $$f'(x) = 5(x - 1)^4 + 7(x + 2)^6 + 9(7x - 5)^8$$ can never be zero for any real value of $$x$$ therefore the given equation can have only one real real root at maximum. 7. Let $$x + 2 = t^2$$ then $$\\sqrt{2(t^2 + 1)} - t = 3 \\Rightarrow 2t^2 + 2 = 9 + 6t + t^2$$ $$\\Rightarrow t = 7, -1$$ Now it is trivial to calculate the values. 308 through 317 have", "# Minimum Monic Algebra Level 5 $$f(x)$$ is a monic quadratic polynomial with positive real roots, and satisfies $$f(1) = 441$$. What is the minimum value of $$f(0)$$? Details and assumptions A monic polynomial has a leading coefficient of 1, i.e. $$f(x) =x^n + \\ldots$$ for some non-negative integer $$n$$. The roots of the polynomial may be repeated real roots. ×", "result should depend on the roots of the components. Would also be nice if the group had some nontrivial subgroups, for example, polynomials with rational coefficients. – Valtteri Jun 14 '15 at 22:04 • If you change the condition to \"only real roots and real poles\", then you can just allow division as well as multiplication, but you have to exclude 0. – Matt Samuel Jun 14 '15 at 22:10 • I changed the question to a hopefully more constructive one – Valtteri Jun 14 '15 at 22:44 A non-zero polynomial with all real roots corresponds uniquely to a finite (possibly empty) set of real numbers (the roots) tagged with positive integers (the order of vanishing), together with a non-zero real number (the leading coefficient). That is, $$c\\prod_{k=1}^{N} (x - a_{k})^{m_{k}} \\leftrightarrow\\bigl\\{(a_{1}, m_{1}), \\dots, (a_{N}, m_{N}), c\\bigr\\}.$$ Alternatively, the data on the right may be viewed as a non-zero real number $c$ together with an \"order-of-vanishing\" function $m$ from the reals to the non-negative integers that is non-zero at only finitely many points. (In the notation of the preceding equation, $m(a_{k}) = m_{k}$ for $k = 1, \\dots, N$, and $m(x) = 0$ for all other real $x$.) The product operation has the pleasant interpretation of multiplying the leading coefficients and adding the order-of-vanishing functions. It should be", "- i$ satisfies the rest of conditions, but the only roots are $\\pm i\\,$. – dxiv Jun 11 '17 at 5:19 • @dxiv: Yes, I missed that. So a corrected statement might be something like this: If $P$ has degree at least one, and at least one nonzero real coefficient, then if some of the non-real roots of $P$ are grouped in complex conjugate pairs, at least one non-real root is not paired. Ugh! – quasi Jun 11 '17 at 5:33 As pointed out in the comments you have an issue with multiplying by scalars. A way to remedy this is to assume your polynomial is monic, that is $a_n = 1$. Then the statement would be: Given any monic polynomial $p$ we can represent it $p = \\prod (x - z_i)$. Then splitting into real roots and complex conjugates we have $p = \\prod (x - a_i) \\prod (x - z_i) (x - \\bar{z_i})$. It is an easy calculation to check that $(x- z_i)(x-\\bar{z_i})$ has all real coefficients for any complex $z_i$.", "or non-Real. The equation $0 = {x}^{3} + 450 {x}^{2} - 1350$ is of degree $3$, hence has $3$ Complex roots counting multiplicity. That's all the fundamental theorem of algebra will tell you. Beyond this, since this polynomial has Real coefficients, any non-Real Complex roots will occur in conjugate pairs. So we can deduce that our polynomial will have at least one Real root. It so happens that all of these $3$ roots are Real in this example, being approximately $- 450$, $- 1.74$ and $1.73$ To show that all of the roots are Real, let $f \\left(x\\right) = {x}^{3} + 450 {x}^{2} - 1350$ and evaluate $f \\left(x\\right)$ for a few values: $f \\left(- 450\\right) = - {450}^{3} + {450}^{3} - 1350 = - 1350 < 0$ $f \\left(- 2\\right) = - 8 + 450 \\cdot 4 - 1350 = - 8 + 1800 - 1350 = 442 > 0$ $f \\left(0\\right) = 0 + 0 - 1350 = - 1350 < 0$ $f \\left(2\\right) = 8 + 450 \\cdot 4 - 1350 = 8 + 1800 - 1350 = 458 > 0$ So $f \\left(x\\right)$ crosses the $x$ axis $3$ times and $f \\left(x\\right) = 0$ has $3$ distinct Real roots.", "your question about the image $F[B_b(\\epsilon)]$, though. – Eric Wofsey Jul 8 '18 at 6:58 As I commented, your function $F$ is not well-defined. However, your question still makes sense without it. You can ask, given a multiset of nonzero complex numbers $c_1,\\dots,c_n$ invariant under conjugation and $\\epsilon>0$, does there exist $\\epsilon'>0$ such that for any $r_1,\\dots,r_n\\in\\mathbb{R}$ with $|r_k-|c_k||<\\epsilon'$ for each $k$, there exist $d_1,\\dots,d_n\\in\\mathbb{C}$ also invariant under conjugation such that $|d_k|=r_k$ for each $i$ and coefficients of the monic polynomial of with the $d_k$ as roots are within $\\epsilon$ of the coefficients of the monic polynomial with the $c_k$ as roots. (The invariance under conjugation condition here corresponds to the requirement that the polynomial has real coefficients.) The answer to this question is no. For instance, let $n=2$, and $(c_1,c_2)=(i,-i)$ and consider $r_1=1+\\epsilon'/2$ and $r_2=1$. Notice then that if $|d_k|=r_k$ for $k=1,2$, then $d_1$ and $d_2$ cannot be conjugates of each other, and so they must both be real in order to be invariant under conjugation. It follows that there are no such $d_1$ and $d_2$ that give rise to a polynomial close to the polynomial $t^2+1$ with roots $c_1$ and $c_2$. Or, in more concrete terms, there is no polynomial with real coefficients close to $t^2+1$ whose roots have distinct absolute values, since any nearby polynomial", "# proving the existence of a complex root If $x^5+ax^4+ bx^3+cx^2+dx+e$ where $a,b,c,d,e \\in {\\bf R}$ and $2a^2< 5b$ then the polynomial has at least one non-real root. We have $-a = x_1 + \\dots + x_5$ and $b = x_1 x_2 + x_1 x_3 + \\dots + x_4 x_5$. #### Solutions Collecting From Web of \"proving the existence of a complex root\" We will prove the slightly stronger statement Claim: Let $$p(x) = x^n + ax^{n-1} + bx^{n-2} + \\cdots + c$$ be a polynomial with real coefficients. If $n=2,3,4,5$ and $2a^2 < 5b$ then $p$ has at least one non-real zero. If $n \\geq 6$ then $p$ may have all real zeros regardless of whether $2a^2 < 5b$ or $2a^2 \\geq 5b$. For example, the polynomial $$\\left(x – \\sqrt{\\frac{2}{n(n-1)}}\\right)^n = x^n – \\sqrt{\\frac{2n}{n-1}} x^{n-1} + x^{n-2} + \\cdots + \\left(\\frac{2}{n(n-1)}\\right)^n$$ has all real zeros with $2a^2 < 5b$. (In fact $p$ has at least two real zeros since they will come in complex conjugate pairs.) As noted in the question, if $x_1,\\ldots,x_n$ are the zeros of $p$ then $$a = -\\sum_{k=1}^{n}x_k \\qquad \\text{and} \\qquad b = \\sum_{1 \\leq j < k \\leq n} x_j x_k,$$ so that \\begin{align} 2a^2 &= 2\\sum_{k=1}^{n} x_k^2 + 4\\sum_{1 \\leq j < k \\leq n} x_j x_k \\\\ &= 2\\sum_{k=1}^{n} x_k^2", "More fun in 2016, Part 3 Algebra Level 5 Let $$f(x)=x^{2016}\\pm x^{2015}\\pm \\ldots \\pm x \\pm 1$$ be a monic polynomial whose non-leading coefficients are either 1 or $$-1$$. If $$f(x)$$ has no real roots, what is the maximal number of coefficients $$-1$$ the polynomial $$f(x)$$ can have? ×", "has no real roots and hence the roots must be a complex conjugate pair. • Thanks. Will the ''conclusion'' hold: if we let the roots to be identical real numbers, i.e., $c=(\\beta, \\beta, \\dots, \\beta)$? Here by conclusion I mean: an open cube $(|\\beta| - \\varepsilon', |\\beta| + \\varepsilon' )^n$ such that all the conjugate invariant subset is contained in the image of $F[B_b(\\varepsilon)]$? – user1101010 Jul 8 '18 at 16:04 • Yes, it is true if the roots are all real. This follows immediately from the fact that the function taking a collection of roots to the coefficients of the corresponding polynomial is continuous (since it's just given by the elementary symmetric polynomials). So, just pick a collection of real roots near $\\beta$ with the desired absolute values, and if they are sufficiently near $\\beta$, the resulting polynomial will be sufficiently near the initial polynomial. – Eric Wofsey Jul 8 '18 at 16:30", "centered at the real axis and of radius less than $2^{-L}$ together with positive integers $\\mu_{1},\\ldots,\\mu_{s}$ such that each disk $\\Delta_{i}$ contains exactly $\\mu_{i}$ roots of $f$ counted with multiplicity. In addition, it is ensured that each real root of $f$ is contained in one of the disks. If $f$ has only simple real roots, our algorithm can also be used to isolate all real roots. The bit complexity of our algorithm is polynomial in $k$ and $\\log n$, and near-linear in $L$ and $\\tau$, where $2^{-\\tau}$ and $2^{\\tau}$ constitute lower and upper bounds on the absolute values of the non-zero coefficients of $f$, and $n$ is the degree of $f$. For root isolation, the bit complexity is polynomial in $k$ and $\\log n$, and near-linear in $\\tau$ and $\\log\\sigma^{-1}$, where $\\sigma$ denotes the separation of the real roots.}, } Endnote %0 Report %A Jindal, Gorav %A Sagraloff, Michael %+ Algorithms and Complexity, MPI for Informatics, Max Planck Society Algorithms and Complexity, MPI for Informatics, Max Planck Society %T Efficiently Computing Real Roots of Sparse Polynomials : %G eng %U http://hdl.handle.net/11858/00-001M-0000-002D-8AD1-7 %U http://arxiv.org/abs/1704.06979 %D 2017 %X We propose an efficient algorithm to compute the real roots of a sparse polynomial $f\\in\\mathbb{R}[x]$ having $k$ non-zero real-valued coefficients. It is assumed that arbitrarily good approximations of the non-zero coefficients are" ]

[ "are considering polynomial with real coefficients. In that case, for $$n=3$$, we can't have 1 repeated imaginary (conjugate) and a real root and the number of possibilities is only $$4$$. In general when the degree is $$n$$, we can have $$k=n-2j$$ real roots with $$j=0,\\dots,\\lfloor n/2\\rfloor$$ with non-negative multiplicities $$m_1,m_2,\\dots,m_k$$ such that $$m_1+2m_2+3m_3+\\dots+km_k=k.$$ The number of non-negative integer solutions of this equation is $$p(k)$$ the number of partitions of $$k$$. Hence the total number of cases is given by the formula $$\\sum_{j=0}^{\\lfloor n/2\\rfloor}p(j)\\cdot p(n-2j)$$ which gives the sequence A002513: $$1, 3, 4, 9, 12, 23, 31, 54, 73, 118,\\dots$$.", "# Is it possible to scale roots such that half of the coefficients of two real monic polynomials agree? Suppose we have two real monic polynomials \\begin{align*} & p_1(t) = t^n + a_{n-1} t^{n-1} + \\dots + a_0, \\\\ & p_2(t) = t^n + b_{n-1} t^{n-1} + \\dots + b_0. \\end{align*} Let $\\alpha = (\\alpha_1, \\dots, \\alpha_n) \\subset \\mathbb C$ and $\\beta = (\\beta_1, \\dots, \\beta_n) \\subset \\mathbb C$ be the roots of $p_1$ and $p_2$ respectively and assume no root in $\\alpha$ and $\\beta$ equal $0$. Necessarily, $\\alpha$ and $\\beta$ must be invariant under conjugation since the coefficients are real. Does there exist nonzero real numbers $r_1, r_2$ such that the monic polynomials $\\hat{p}_1(t)$ with roots $r_1 \\alpha= (r_1 \\alpha_1, \\dots, r_1 \\alpha_n)$ and $\\hat{p}_2(t)$ with roots $r_2 \\beta = (r_2 \\beta_1, \\dots, r_2 \\beta_2)$ have either $a_{2j} = b_{2j}$ for $j =0, 1, \\dots, [\\frac{n}{2}]$ or $a_{2j+1} = b_{2j+1}$ for $j = 0, 1, \\dots, [\\frac{n-1}{2}]$ where $[s]$ denotes the greatest integer less than $s \\in \\mathbb R$? Quadratic case seems always true. Is there a way to see the existence or nonexistence for higher orders? It is not possible in general for $\\,n \\gt 2\\,.$ By Vieta's relations: $$\\hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\\ldots +r_1^{n-1}a_1t + r_1^na_0 \\\\ \\hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\\\$$ For", "# $2P(x)=Q(x)+R(x)$ and $Q(x),R(x)$ have $n$ real coefficients both. Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients can be written as the average of two monic polynomials of degree n with n real roots. This is USAMO 2002 problem 3. The result is trivial for $n=1$, so assume $n>1$. Let $g(x)=(x-2)(x-4)...(x-2n+2)$. We claim that $q(x)=x^n-kg(x),r(x)=2p(x)-q(x)$ works for sufficiently large $k$. The values $g(1),g(3), ... , g(2n-1)$ alternate in sign and have magnitude at least 1. Take $k$ so that $x^n<k/3$ and $|p(x)-x^n|<k/3$ for all $1\\le x\\le 2n-1$. Then if follows that $q(1),q(3),...,q(2n-1)$ also alternate in sign, and similarly $r(1),r(3),...,r(2n-1)$ alternate in sign. So $q(x),r(x)$ each have at least $n-1$ real roots. But polynomials with real coefficients must have an even number of non-real roots, so $q(x),r(x)$ each have $n$ real roots.", "# Compute: $\\lim_{n\\to\\infty} ({x_{2}}^n+{x_{3}}^n)$ Let be $x_{1}, x_{2}, x_{3}$ the roots of $x^3-x^2-1=0$. If $x_{1}$ is the real root of the equation, then calculate: $$\\lim_{n\\to\\infty} ({x_{2}}^n+{x_{3}}^n)$$ I wonder if this limit can be computed without being necessary to know the exact values of $x_{2}$ and $x_{3}$. - We will prove that the magnitude of the complex roots is less than $1$. Let $f(x) = x^3-x^2-1$. Then $f(1) = -1 <0$ and $f(2) = 3>0$. Hence, there is a real root, $x_1$, in the interval $(1,2)$. You can prove that this is the only real root and the other two have to be complex. This can be done by looking at the derivative in the interval $(1,2)$ which will turn out to be positive and a local maximum occurs at $x=0$ and a local minimum occurs at $x= \\dfrac23$. These guarantee that the remaining two roots are complex. The complex roots $x_2$ and $x_3$ can be written as $x_2 = r e^{i \\theta}$ and $x_3 = r e^{- i \\theta}$ since the complex roots occur in conjugate pairs. But the product of the roots is $1$ i.e. $r^2 x_1 = 1 \\implies r^2 = \\dfrac1{x_1} < 1$. The complex roots must have magnitude less than $1$. Hence, $$x_2^n + x_3^n = 2r^n \\cos(n \\theta)$$ Hence, $$\\lim_{n \\to \\infty}\\left( x_2^n", "# Do roots of a monic polynomial contain an open cube? Let us consider monic polynomials of $n^{\\text{th}}$ degree $p(t) = t^n + a_{n-1} t^{n-1} + \\dots + a_1 t + a_0$ with real coefficients. We know the roots are continuous functions of the coefficients. Let us define a map $F: \\mathbb R^n \\to \\mathbb R_+^n$ by \\begin{align*} F: (a_{n-1}, \\dots, a_0) \\mapsto (r_1, \\dots, r_n) \\mapsto (|r_1|, \\dots, |r_n|), \\end{align*} where $r_j = r_j(a_{n-1}, \\dots, a_0)$ are roots of the polynomial and $|\\cdot|$ denotes the modulus of a complex number. Clearly $F$ is a continuous function. Let us take the Euclidean norm on $\\mathbb R^n$. Suppose for some fixed coefficients $b = (b_{n-1}, \\dots, b_0)$, the roots are all nonzero and let $(c_1, \\dots, c_n)$ be the roots. if we take an open ball $B_b(\\varepsilon)$ around $b$, will the image $F[ B_b(\\varepsilon) ]$ contain a cube, i.e., $(|c_1|-\\varepsilon', |c_1|+\\varepsilon') \\times (|c_2|-\\varepsilon', |c_2|+\\varepsilon') \\times \\dots \\times (|c_n|-\\varepsilon', |c_n|+\\varepsilon')$ for some possible smaller $\\varepsilon'$? • Your map $F$ is not well-defined: the roots are only defined as an unordered collection, not as an ordered $n$-tuple. It would be well-defined (and continuous) if you changed the codomain to $\\mathbb{R}^n_+/S_n$. – Eric Wofsey Jul 8 '18 at 6:57 • You could still consider your $F$ as a \"multi-valued function\" and ask", "your question about the image $F[B_b(\\epsilon)]$, though. – Eric Wofsey Jul 8 '18 at 6:58 As I commented, your function $F$ is not well-defined. However, your question still makes sense without it. You can ask, given a multiset of nonzero complex numbers $c_1,\\dots,c_n$ invariant under conjugation and $\\epsilon>0$, does there exist $\\epsilon'>0$ such that for any $r_1,\\dots,r_n\\in\\mathbb{R}$ with $|r_k-|c_k||<\\epsilon'$ for each $k$, there exist $d_1,\\dots,d_n\\in\\mathbb{C}$ also invariant under conjugation such that $|d_k|=r_k$ for each $i$ and coefficients of the monic polynomial of with the $d_k$ as roots are within $\\epsilon$ of the coefficients of the monic polynomial with the $c_k$ as roots. (The invariance under conjugation condition here corresponds to the requirement that the polynomial has real coefficients.) The answer to this question is no. For instance, let $n=2$, and $(c_1,c_2)=(i,-i)$ and consider $r_1=1+\\epsilon'/2$ and $r_2=1$. Notice then that if $|d_k|=r_k$ for $k=1,2$, then $d_1$ and $d_2$ cannot be conjugates of each other, and so they must both be real in order to be invariant under conjugation. It follows that there are no such $d_1$ and $d_2$ that give rise to a polynomial close to the polynomial $t^2+1$ with roots $c_1$ and $c_2$. Or, in more concrete terms, there is no polynomial with real coefficients close to $t^2+1$ whose roots have distinct absolute values, since any nearby polynomial", "# $p_n(x)=p_{n-1}(x)+p_{n-1}^{\\prime}(x)$, then all the roots of $p_k(x)$ are real $p_0(x)=a_mx^m+a_{m-1}x^{m-1}+\\dotsb+a_1x+a_0(a_m,\\dotsc,a_1,a_0\\in\\Bbb R)$ is a polynomial, and $$p_n(x)=p_{n-1}(x)+p_{n-1}^{\\prime}(x),\\qquad n=1,2,\\dotsc$$ then, there exist $N\\in\\Bbb N$, such that $\\forall n \\geq N$, all the roots of $p_n(x)$ are real I think, let $f_n(x)=e^xp_n(x)$, then $$f_n(x)=\\frac {d(f_{n-1}(x))}{dx}$$ we get that $$f_n(x)=\\frac{d^n\\left( e^xp_0(x)\\right)}{dx^n}$$ so, If there exists a $p_N(x)$ such that $p_N(x)$ has $m$ real roots, then $\\forall n\\geq N$, $p_n(x)$ has $m$ real roots. I'd be grateful for any help • One point of interest, either $p_n(x)$ or $p_n'(x)$ has at least one real root for all $n$. You can use this to guarantee one real root for some $n$ in $p_n(x)$. – abiessu Jun 3 '14 at 19:47 • @user90041 Why? If it's constant, $p_n$ is also constant and has no zeros (or identically zero) and the claim still true (all zeroes real), right? – ploosu2 Jun 3 '14 at 20:13 • Another way to look at this is to notice that $p_n(x)=\\sum_{i=0}^n{n\\choose i}p_0^{(i)}(x)$ for $p^{(k)}(x)$ the $k$th derivative of $p$ at $x$. – abiessu Jun 3 '14 at 20:53 • @Edwin I believe $p_0(x)$ is just some arbitrary polynomial, and is taken here to be of degree $m$. – Bob Pego Jun 6 '14 at 21:17 • If $p(x)=\\prod_i(x+d_i)$ with $d_1<d_2<d_3<\\cdots$, and $q(x)=p(x)+p'(x)$, then we easily see that $q(-d_1)>0$, $q(-d_2)<0$,", "other roots between $$\\beta$$ and $$\\gamma$$ and hence $$g\"(x)$$ will have a root between $$\\alpha$$ and $$\\gamma.$$ 4. If this degree four polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and hird root. The location of the extrema does not change with $$a$$, only their $$y$$ values. Thus we are looking for $$a$$ such that the two local minima have negative $$y$$ values and the local maximum has positive $$y$$ value. Locate the local extrema as roots of the derivative $$4x^3 - 12x^2 - 16x$$ which has roots as $$-1, 0, 4$$ $$\\Rightarrow a = 0, a = 3, a = 128$$ $$\\therefore 0\\le a \\le 3$$ 5. This problem is similar to 226 and can be solved similarly. 6. $$f'(x) = 5(x - 1)^4 + 7(x + 2)^6 + 9(7x - 5)^8$$ can never be zero for any real value of $$x$$ therefore the given equation can have only one real real root at maximum. 7. Let $$x + 2 = t^2$$ then $$\\sqrt{2(t^2 + 1)} - t = 3 \\Rightarrow 2t^2 + 2 = 9 + 6t + t^2$$ $$\\Rightarrow t = 7, -1$$ Now it is trivial to calculate the values. 308 through 317 have", "asked for a method of generating examples. – Travis Apr 16 '16 at 14:25 • @Travis. OK Thanks. I agree with your argument and your computation of 2922. I also have the number of such cubics with coefficients in -n, ... , n as: 0, 6, 32, 104, 248, 484, 850, 1356, 2046, 2922 for n=1,..,10. – Geoffrey Critzer Apr 16 '16 at 16:04 If $$x^3+ax^2+bx+c$$ (with integers $a,b,c$) has a rational root, then it is in fact an integer root and is a (positive or negative) divisor of $c$ (rational root theorem). In particular, if you let $c=\\pm1$, you need only ensure that $\\pm1$ are no roots, i.e., that $a+b+c+1\\ne 0$ and $-1+a-b+c\\ne 0$. If $c$ is $\\pm$ a prime there are only a few more conditions and if $c$ is composite, the situation is still not difficult to handle as long as we know the factorization of $c$. However, in order to exclude non-real roots, we have to do one additional check: The cubic above will have a pair of conjugate complex roots iff its discriminant $$\\Delta=a^2b^2-4b^3-4a^3c-27c^2+18abc$$ is negative. Using these two tests it is not too hard to determine whether a given cubic has only real-irrational roots. • But he presumably wants (1) all roots real as well as (2) no roots rational. – almagest", "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # (a) Show that a polynomial of degree $3$ has at most three real roots.(b) Show that a polynomial of degree $n$ has at most $n$ real roots. ## (a) Suppose that a cubic polynomial $P(x)$ has roots $a_{1}<a_{2}<a_{3}<a_{4},$ so $P\\left(a_{1}\\right)=P\\left(a_{2}\\right)=P\\left(a_{3}\\right)=P\\left(a_{4}\\right)$By Rolle's Theorem there are numbers $c_{1}, c_{2}, c_{3}$ with $a_{1}<c_{1}<a_{2}, a_{2}<c_{2}<a_{3}$ and $a_{3}<c_{3}<a_{4}$ and$P^{\\prime}\\left(c_{1}\\right)=P^{\\prime}\\left(c_{2}\\right)=P^{\\prime}\\left(c_{3}\\right)=0 .$ Thus, the second-degree polynomial $P^{\\prime}(x)$ has three distinct real roots, which isimpossible.(b) We prove by induction that a polynomial of degree $n$ has at most $n$ real roots. This is certainly true for $n=1$. Supposethat the result is true for all polynomials of degree $n$ and let $P(x)$ be a polynomial of degree $n+1$. Suppose that $P(x)$ hasmore than $n+1$ real roots, say $a_{1}<a_{2}<a_{3}<\\dots<a_{n+1}<a_{n+2} .$ Then $P\\left(a_{1}\\right)=P\\left(a_{2}\\right)=\\dots=P\\left(a_{n+2}\\right)=0$By Rolle's Theorem there are real numbers $c_{1}, \\ldots, c_{n+1}$ with $a_{1}<c_{1}<a_{2}, \\ldots, a_{n+1}<c_{n+1}<a_{n+2}$ and$P^{\\prime}\\left(c_{1}\\right)=\\cdots=P^{\\prime}\\left(c_{n+1}\\right)=0 .$ Thus, the $n$ th degree polynomial $P^{\\prime}(x)$ has at least $n+1$ roots. This contradiction showsthat $P(x)$ has at most $n+1$ real roots. Derivatives Differentiation Volume ### Discussion You must be signed in to discuss. ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp ### Video Transcript it's s o We're being asked to show that upon your", "1. The factorization of the second polynomial is ${\\displaystyle f(-x)=-(x-1)^{2}(x+1),\\,}$ So here, the roots are 1 (twice) and −1, the negation of the roots of the original polynomial. ## Nonreal roots Any nth degree polynomial has exactly n roots in the complex plane, if counted according to multiplicity. So if f(x) is a polynomial which does not have a root at 0 (which can be determined by inspection) then the minimum number of nonreal roots is equal to ${\\displaystyle n-(p+q),\\,}$ where p denotes the maximum number of positive roots, q denotes the maximum number of negative roots (both of which can be found using Descartes' rule of signs), and n denotes the degree of the equation. ### Example: zero coefficients, nonreal roots The polynomial ${\\displaystyle f(x)=x^{3}-1\\,,}$ has one sign change, so the maximum number of positive real roots is 1. From ${\\displaystyle f(-x)=-x^{3}-1\\,,}$ we can tell that the polynomial has no negative real roots. So the minimum number of nonreal roots is ${\\displaystyle 3-(1+0)=2\\,.}$ Since nonreal roots of a polynomial with real coefficients must occur in conjugate pairs, we can see that x3 − 1 has exactly 2 nonreal roots and 1 real (and positive) root. ## Special case The subtraction of only multiples of 2 from the maximal number of positive roots occurs because the polynomial may", "-2$, $n >3$ the example I needed? Because $f_n$ has real and nonreal roots. – user893458 May 11 '14 at 20:48 • Yes it is, but do check everything, because when you present it as a solution, then YOU are making that claim. So you need to recall a tool proving that it is irreducible, and also justify the presence of real an non-real roots. Sounds like you are well on your way! – Jyrki Lahtonen May 11 '14 at 20:51 • For irreducible I would say: Eisenstein's criterion for $p=2$. – user893458 May 11 '14 at 20:53 • Correct. Well done :-) – Jyrki Lahtonen May 11 '14 at 20:54 • The real root is $x=2^{(1/n)}$ and the complex roots are some $n$th roots of unity multiplied by $x$. – user893458 May 11 '14 at 20:55 As $g$ has nonreal roots, we see that complex conjugation $\\psi\\colon z\\mapsto \\bar z$ is a nontrivial automorphism of $K$. Let $\\alpha$ be a real root and $\\beta$ a complex root. By transitivity of the Galois group, there exists an automorphism $\\phi$ with $\\phi(\\alpha)=\\beta$. Then $\\psi(\\phi(\\alpha))=\\overline\\beta$, but $\\phi(\\psi(\\alpha)=\\beta$.", "+0 # Cubic Polynomial. Help thanks. 0 202 2 +22 All of the roots of the cubic polynomial $$f(x) = 3x^3+bx^2+cx+d$$ are negative real numbers. If f(0) =81, find the smallest possible value of f(1). Prove that your answer is the minimum. Jun 4, 2021 #1 +179 0 Here is my take on this question, I could be wrong (inequalities are tricky for me). From the statement f(0)=81, we have d=81 This, f(x) becomes 3x^3+bx^2+cx+81. Let the roots be r, s, and t. Thus, the product rst is equal to -27 by Vieta's formulas. b=-(r+s+t), which is a positive number, given that r, s, and t are all negative. Hence, to minimize b, we want to maximize r+s+t, knowing that they are negative. We seek to relate rst to r+s+t through inequality, and the AM-GM works for this. The problem arises though, that the AM-GM only works for positive numbers, and r, s, and t are each negative. Luckily, by setting -r, -s, and -t into the equation, we have all positive inputs, hence allowing for the AM-GM to come into use. The AM-GM for 3 variables gives $$\\frac{-r-s-t}{3}≥ \\sqrt[3]{-rst}$$, simplifying to $$-(r+s+t)≥ 3\\sqrt[3]{27}$$ or $$r+s+t≤-9$$ from this, the maximum value of r+s+t is -9, so the minimum value for b is 9. This can only be achieved", "Un­for­tu­nately, as was poin­ted out earli­er, this prop­erty does not im­ply $$P_3$$. The case of $$\\bigl\\lfloor \\frac 3k X\\bigr\\rfloor$$ is more chal­len­ging, since for each root with pos­it­ive real part, one must find a neg­at­ive root that com­pensates for it. It is not simply a mat­ter of prov­ing some prop­erty for each root, as was the case for $$P_1$$ and $$P_2$$. All we can of­fer at this point is spec­u­la­tion based on com­pu­ta­tions of roots in spe­cial cases. If $$f$$ is $$P_3$$ but not $$P_2$$, $$f$$ will have com­plex roots of pos­it­ive real part. For a neg­at­ive real root $$w$$ and a con­jug­ate pair of roots $$z,\\bar z$$ with pos­it­ive real part, let $\\langle z,w\\rangle=(u-w)(u-z)(u-\\bar z).$ For this cu­bic poly­no­mi­al to have pos­it­ive coef­fi­cients, it is ne­ces­sary and suf­fi­cient that $z+\\bar z\\leq -w\\leq\\frac{z \\bar z}{z+\\bar z}.$ Here is what seems to be the case. Let $$w_1,w_2,\\dots$$ be the real roots of $$f$$ ordered in in­creas­ing or­der of their ab­so­lute val­ues, and $$z_1,z_2,\\dots$$ be the roots with pos­it­ive real and ima­gin­ary parts, ordered in in­creas­ing or­der of the real parts. As­sume that $$X$$ is SR and takes val­ues $$0,1,\\dots,n$$. Let $$f$$ be the pgf of $$\\bigl\\lfloor\\frac 34X\\bigr\\rfloor$$. The de­gree of $$f$$ is $$d=\\bigl\\lfloor\\frac34 n\\bigr\\rfloor$$. Then $$f$$ ap­pears to be $$P_3$$ in gen­er­al. The num­ber of $$z_i$$’s and the num­ber", "# Counting roots of a multivariate polynomial over a finite field How many roots can there be of a polynomial $f \\in K[x_1, x_2, \\dots , x_n]$ where $K$ is a finite field and the maximum exponent of $x_i$ in any term is $m$ for all $i$, assuming not all coefficients are zero? I found this related question, but I'm really interested in the worst case; I won't have any \"singularity condition\", necessarily. Here's an idea: View $f$ as a polynomial over $(K(x_2, \\dots , x_n))[x_1]$, where $K(x_2, \\dots , x_n)$ is the field of fractions of $K[x_2, \\dots , x_n]$. At least one coefficient of $f$ when viewed this way is non-zero. Now $f$ has at most $m$ roots, since its single indeterminate, $x_1$, has exponent at most $m$. For each of these $m$ roots $y_{1,1}, y_{1,2}, \\dots, y_{1,m}$ and any $x_2, \\dots, x_n$, $f(y_{1,j}, x_2, \\dots, x_n) = 0$. There are $m k^{n-1}$ tuples of this form, where $k$ is the size of $K$. Additionally, for each $v \\in K$, $f(v, x_2, \\dots, x_n)$ is a polynomial in $n-1$ variables. For the $v$s that are not roots of $f$ when viewed as above, $f(v, x_2, \\dots, x_n)$ has at least one non-zero coefficient. This sets up an inductive equation: $$c(n) \\leq \\begin{cases} m & n =", "Jan 29 '18 at 3:30 • yes sorry, i mean at most three real roots – Alb P Jan 29 '18 at 3:32 • Ok, good, just checking. You just say things like \"there must be at least 3 real roots\" a few times. – jgon Jan 29 '18 at 3:33 • yeah, just edited the question to fix that up – Alb P Jan 29 '18 at 3:34 • Do you know Descartes' rule of signs? en.wikipedia.org/wiki/Descartes%27_rule_of_signs – kimchi lover Jan 29 '18 at 3:36 We can actually do this directly. Note that between each pair of adjacent but distinct real roots, the polynomial must have a local maximum or a local minimum, since it has to turn around to get back to zero. Thus between each pair of adjacent and distinct real roots, the derivative of $f$ has a real root. So let's look at $f'(x)=5x^4-1$. We can see immediately that this has the four fourth roots of $\\frac{1}{5}$ as its roots: $\\frac{1}{\\sqrt[4]{5}}$, $\\frac{i}{\\sqrt[4]{5}}$, $\\frac{-1}{\\sqrt[4]{5}}$, $\\frac{i}{\\sqrt[4]{5}}$. Since only two of these are real, $f$ can only have at most three distinct real roots. All that remains is to check that $f$ has no repeated roots. We can do this by checking that $f$ and $f'$ are relatively prime. Dividing $f$ by $f'$, we get $f(x)=f'(x)(\\frac{1}{5}x) -\\frac{4}{5}x", "+ a^2)$ In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \\dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$. $$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \\dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \\dots +a_{0})$$ $$= a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \\dots + a_{1}(x-r)$$ Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that $$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \\dots + r^{n-1})$$ and so $$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \\dots + r^{n-1}) + a_{n-1}(x^{n-2} + \\dots +r^{n-1}) + \\dots + a_1)$$ Once we know a root, we can also try using Polynomial Long Division to get the other factor. For cubics, the roots can be found without the need to guess. Check this out: Cardano's Method. Does that help? - en.wikipedia.org/wiki/Factor_theorem might also be useful. – Hans Lundmark Sep 28 '10 at 20:12 @Hans: Right, I will edit that into the answer. Thanks. – Aryabhata Sep 28 '10 at 20:13 Cardano is a bit of a sledgehammer here; the only other thing I'll note is that synthetic division may be more convenient than long division, depending on the user. – Ｊ. Ｍ. Sep 28 '10 at 23:58 @J.M: Right, just wanted", "# Minimum Monic Algebra Level 5 $$f(x)$$ is a monic quadratic polynomial with positive real roots, and satisfies $$f(1) = 441$$. What is the minimum value of $$f(0)$$? Details and assumptions A monic polynomial has a leading coefficient of 1, i.e. $$f(x) =x^n + \\ldots$$ for some non-negative integer $$n$$. The roots of the polynomial may be repeated real roots. ×", "most,$P-P'$has degree$n$, so it has$n$distincts roots at most. Since$P(x_i)=y_i=P'(x_i)$for$i\\in\\{1,\\dots,n\\}$,$P'(\\beta) \\neq P(\\beta)=0$, so$P'$cannot have$\\beta$has a root. 2 If$F$is a field, then, in$F[x]$, if$f$divides$g$, then so does$\\alpha f$for any non-zero$\\alpha \\in F$. Requiring the gcd to be monic makes it unique. In your example with$F = \\mathbb{R}$,$6x$or$72x$would be just as good values for$\\gcd(5x^2, 25x)\\$ if we didn't adopt this convention (not all authors do). If the ring of coefficients is not a ... Only top voted, non community-wiki answers of a minimum length are eligible", "result should depend on the roots of the components. Would also be nice if the group had some nontrivial subgroups, for example, polynomials with rational coefficients. – Valtteri Jun 14 '15 at 22:04 • If you change the condition to \"only real roots and real poles\", then you can just allow division as well as multiplication, but you have to exclude 0. – Matt Samuel Jun 14 '15 at 22:10 • I changed the question to a hopefully more constructive one – Valtteri Jun 14 '15 at 22:44 A non-zero polynomial with all real roots corresponds uniquely to a finite (possibly empty) set of real numbers (the roots) tagged with positive integers (the order of vanishing), together with a non-zero real number (the leading coefficient). That is, $$c\\prod_{k=1}^{N} (x - a_{k})^{m_{k}} \\leftrightarrow\\bigl\\{(a_{1}, m_{1}), \\dots, (a_{N}, m_{N}), c\\bigr\\}.$$ Alternatively, the data on the right may be viewed as a non-zero real number $c$ together with an \"order-of-vanishing\" function $m$ from the reals to the non-negative integers that is non-zero at only finitely many points. (In the notation of the preceding equation, $m(a_{k}) = m_{k}$ for $k = 1, \\dots, N$, and $m(x) = 0$ for all other real $x$.) The product operation has the pleasant interpretation of multiplying the leading coefficients and adding the order-of-vanishing functions. It should be" ]

[ "operators, we simply write the functions with the operator and simplify. Given two functions $$f$$ and $$g$$, we can define four new functions: $$(f+g)(x)=f(x)+g(x)$$ Sum $$(f−g)(x)=f(x)−g(x)$$ Difference $$(f·g)(x)=f(x)g(x)$$ Product $$(\\frac{f}{g})(x)=\\frac{f(x)}{g(x)}$$ for$$g(x)≠0$$ Quotient Exercise: 1) Combining Functions Using Mathematical Operations Given the functions $$f(x)=2x−3$$ and $$g(x)=x^2−1$$, find each of the following functions and state its domain. 1. $$(f+g)(x)$$ 2. $$(f−g)(x)$$ 3. $$(f·g)(x)$$ 4. $$(\\frac{f}{g})(x)$$ Solution: 1. $$(f+g)(x)=(2x−3)+(x^2−1)=x^2+2x−4.$$ The domain of this function is the interval $$(−∞,∞)$$. 2.$$(f−g)(x)=(2x−3)−(x^2−1)=−x^2+2x−2.$$ The domain of this function is the interval $$(−∞,∞)$$. 3. $$(f·g)(x)=(2x−3)(x^2−1)=2x^3−3x^2−2x+3.$$ The domain of this function is the interval $$(−∞,∞)$$. 4. $$(\\frac{f}{g})(x)=\\frac{2x−3}{x^2−1}$$. The domain of this function is {$$x∣∣x≠±1$$}. 2) For $$f(x)=x^2+3$$ and $$g(x)=2x−5$$, find $$(f/g)(x)$$ and state its domain. Solution: $$(\\frac{f}{g})(x)=\\frac{x^2+3}{2x−5}$$. The domain is {$$x|x≠\\frac{5}{2}$$}. ### Function Composition When we compose functions, we take a function of a function. For example, suppose the temperature $$T$$ on a given day is described as a function of time $$t$$ (measured in hours after midnight) as in Table. Suppose the cost $$C$$, to heat or cool a building for 1 hour, can be described as a function of the temperature $$T$$. Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating $$C(T(t))$$. We have defined a new function, denoted $$C∘T$$, which is", "# Question #f89eb ##### 1 Answer Jan 12, 2017 In interval notation it is $\\left(- \\infty , 3\\right]$ #### Explanation: The domain of the given function would be all x for which $6 - 2 x \\ge 0$ That means $6 \\ge 2 x$ or $x \\le 3$ In interval notation it is $\\left(- \\infty , 3\\right]$", "Thread: State the domain of this graph using interval notation 1. State the domain of this graph using interval notation The picture of the graph is in the attachment. Hello I keep trying to put the domain of this graph into interval notation, but I keep getting it wrong. Can someone help me? 2. Re: State the domain of this graph using interval notation $-5 \\le x < 3$ or $x \\in [-5,3)$", "+0 # function 0 36 1 What is the domain of the function $$f(x) = \\frac{(2x-3)(2x+5)}{(3x-9)(3x+8)}~?$$ Express your answer as an interval or as a union of intervals. Aug 18, 2021", "of f on the same domain of f if and only if G is of the form G(x)= F(x) + c, where c is a constant. Proof : If G(x) = F(x) + c and F '(x) = f(x) and 'c' is a constant then G'(x) = $\\frac{\\text{d}}{\\text{d}x}[F(x) +c]$ = F' (x) + 0 G'(x) = f(x)", "The domain of the function is . Note: Write your answer in interval notation. If the answer includes more than one interval write the intervals separated by the union symbol, U. If the answer involves $-\\infty$, input -infinity ; if the answer involves $\\infty$, input infinity . Your overall score for this problem is", "sign is + for x<0 & - for x>0...so next is i need to find the domain for the function for the interval value...right?", "+0 # Stuck on these questions 0 128 1 +62 1.Let $f(x) = \\frac{2}{\\sqrt{x}}$. What is the domain of $f$? Express your answer with interval notation.\\ 2.Let $f(x) = \\frac{3x - 7}{x + 1}.$Find the domain of $f$. Give your answer using interval notation. Mar 1, 2019", "## Calculus with Applications (10th Edition) Interval notation uses a combination of [ ] and () to describe the domain [ ] are the equivalent of $\\leq$ and $\\geq$ ( ) are the equivalent of $\\lt$ and $\\gt$ Thus the domain is written as (-2,5] in interval notation", "## Discrete Mathematics with Applications 4th Edition $g(x)=\\frac{2x^{3}+2x}{x^{2}+1}=\\frac{(2x)(x^{2}+1)}{x^{2}+1}=2x=f(x)$ Yes, the functions $f$ and $g$ are equivalent. Note that $x^{2}+1\\ne0$; therefore, $g$ has the same domain as $f$.", "# How do you find the domain of the function in interval notation for g(x)= sqrtx/(x^2+x-90)? Apr 7, 2015 $g \\left(x\\right) = \\frac{\\sqrt{x}}{{x}^{2} + x - 90}$ = sqrt(x)/((x+10)(x-9) Restrictions based by the numerator $x \\ge 0$ Restrictions place by denominator $x \\ne - 10$ (this is eliminated by the restriction of the numerator) and $x \\ne 9$ In interval notation Domain of $g \\left(x\\right) = \\left\\{\\left[0 , 9\\right) , \\left(9 , + \\infty\\right]\\right\\}$", "## Calculus 8th Edition Published by Cengage # Chapter 2 - Derivatives - 2.2 The Derivative as a Function - 2.2 Exercises: 24 #### Answer $g'(t) = \\frac{-1}{2}t^{\\frac{-3}{2}}$ Domain: t $\\gt$ 0 #### Work Step by Step $g(t) = \\frac{1}{\\sqrt t} = t^{-1/2}$ $g'(t) = (-1/2)t^{(-1/2) - 1}$ $g'(t) = \\frac{-1}{2}t^{\\frac{-3}{2}}$ The domain is t $\\gt$ 0 because for any negative t value or if t = 0, then g'(t) would be undefined After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "# Question #f89eb In interval notation it is $\\left(- \\infty , 3\\right]$ The domain of the given function would be all x for which $6 - 2 x \\ge 0$ That means $6 \\ge 2 x$ or $x \\le 3$ In interval notation it is $\\left(- \\infty , 3\\right]$", "of the function $$g(t) = |3t - 4|$$ is $$\\frac{4}{3}$$.", "can be closed interval $$[a,0]$$ or $$[0,b]$$ that has zero as endpoint. E.g. the function $$f(x) = \\frac{1}{x}$$ is not defined near the origin of the real line, but the function $$g(x) = \\frac{1}{x + 1}$$ is.", "### Home > AC > Chapter 12 > Lesson 12.4.4 > Problem12-104 12-104. Graph the function $f(x)=\\frac{5}{x-2}$. Use inequalities to describe its domain and range. Domain: $x>2$ and $x<2$ Range: $y>0$ and $y<0$", "### Home > CALC > Chapter 1 > Lesson 1.3.3 > Problem1-129 1-129. Given: $g ( x ) = \\frac { 1 } { x ^ { 2 } - x }$ 1. Find the domain of $g(x)$. Solve for all values for which the denominator is zero. This will make the function undefined. That is, solve $x^2 - x = 0$. 1. Solve for $x$ if $g(x) = 0.5$. 1. Explain why $g(x)$ does not have an inverse function. Recall that to create the inverse, reflect the original function over the line $y = x$.", "\\frac{5}{3}$. Both are in the domain of $f$ (Domain of $f = \\mathbb{R}$), so they are the critical numbers for $f$.", "≤ 10 The domain of r (x) is: -10 ≤ x ≤ 10 The range of r (x) is: 0 ≤y ≤ 2 The range of f (x) is: 0 ≤ y ≤10 Question 36. Graph f(x) = | x – 2 | + 4 and g(x) = | 3x – 2 | + 4. Compare the graph of g to the graph of f. Answer: The given functions are: f (x) = | x – 2 | + 4 g (x) = | 3x – 2 | + 4 Hence, The representation of f (x) and g (x) in the same coordinate plane is: Hence, from the above, We can conclude that g (x) translates 5 units away from f (x) on the x-axis Question 37. Let g(x) = $$\\frac{1}{3}$$ | x – 1 | – 2. (a) Describe the transformations from the graph of f(x) = | x | to the graph of g. Answer: The given functions are: f (x) = | x | g (x) = $$\\frac{1}{3}$$ | x – 1 | – 2 Hence, The representation of f (x) and g (x) in the coordinate plane is: Hence, from the above, We can conclude that g (x) translates 2 units away from f (x) on the y-axis (b) Graph g. Answer: It is", "# How to write domain with asymptotes To write the domain in interval notation, we need to use three intervals and use the vertical asymptotes for endpoints of the intervals. Domain: {eq}(-\\infty, -2)\\cup (-2, 1)\\cup (1,\\infty) {/eq} • Get detailed step-by-step resolutions" ]

[ "## Precalculus (6th Edition) Blitzer $(-\\infty, -3) \\cup (-3, 5) \\cup (5, +\\infty)$ Factor the denominator to obtain: $g(x) = \\dfrac{3}{(x-5)(x+3)}$ The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominator of the given function equal to zero. These numbers are $5$ and $-3$. Thus, the value of $x$ can be any real number except $5$ and $-3$. Therefore, the domain is $(-\\infty, -3) \\cup (-3, 5) \\cup (5, +\\infty)$.", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\\color{blue}{(-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)}$ RECALL: The domain of the sum, difference, and product of $f(x)$ and $g(x)$ is the common elements of the domains of the two functions. The domain of the respective functions are: For $f(x)$, $-4.5$ will make the denominator zero so its domain is: $(-\\infty, -4.5) \\cup (-4.5, +\\infty)$ $g(x)$ is defined for all real numbers except $1$ (as it makes the denominator zero) so its domain is: $(-\\infty,1) \\cup (1, +\\infty)$. Note that: $[(-\\infty, -4.5) \\cup(-4.5, +\\infty)] \\cap [(-\\infty, 1) \\cup (1, +\\infty)] \\\\= (-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)$. Thus, the domain of the sum, difference, and product of $f(x)$ and $g(x)$ is: $\\color{blue}{(-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)}$", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\\color{blue}{(-\\infty, 0.25) \\cup (0.25, 3) \\cup (3, +\\infty)}$ Find the domain of each function: For $f(x)$, the value of $x$ can be any real number except 3 since it will make the denominator zero, which is not allowed. Thus, domain of $f(x)$ is $(-\\infty, 3) \\cup (3, +\\infty)$ For $f(x)$, the value of $x$ can be any real number except $0.25$ since it will make the denominator zero, which is not allowed. Thus, domain of $f(x)$ is $(-\\infty, 0.25) \\cup (0.25, +\\infty)$ RECALL: The domain of the sum, difference, and product of $f(x)$ and $g(x)$ is the common elements of the domains of the two functions. Note that: $[(-\\infty, 3) \\cup(3, +\\infty)] \\cap [(-\\infty, 0.25) \\cup (0.25, +\\infty)] \\\\= (-\\infty, 0.25) \\cup (0.25, 3) \\cup (3, +\\infty)$. Thus, the domain of the sum, difference, and product of $f(x)$ and $g(x)$ is: $\\color{blue}{(-\\infty, 0.25) \\cup (0.25, 3) \\cup (3, +\\infty)}$", "## Calculus with Applications (10th Edition) $(-\\infty,6)\\cup(-6,6)\\cup(6,\\infty)$ The domain is the set of all possible values of x for which the value f(x) exists (is defined). $f(x)$ is defined only for those x which do not produce 0 in the denominator. So, excluded from the domain of real numbers are those x for which $x^{2}-36=0$ $x^{2}=36$ $x=\\pm 6$ In interval notation, the domain is $(-\\infty,6)\\cup(-6,6)\\cup(6,\\infty)$", "and so on. In general, $g_{n}$ would be define on $\\{-n,-n+1,\\ldots,-1\\}\\cup[0,\\infty)$, by $g(x) = f(x)$ if $x\\geq 0$ and $g(x)=k$ if $x\\lt n$. As I noted in the comments, your formulas didn't really say that; instead, they only specified $g_1$ at $-1$, and then said, for example, that $g_2(x-1)=k$. That is at best confusing. What is $g_2(-0.5)$? According to this, I have to think of $-0.5$ as $0.5 - 1$, and then it's $k$, and ... Well, a bit of a confusing issue arises... In any case, whether this works as an answer or not depends on whether you are assuming that your functions need to be defined on the same set or not. Normally, we would be looking for functions $g$ such that $f(x)=g(|x|)$ and we want both $f$ and $g$ to have the same domain. Remember that two functions are equal if and only if they have the same domain, the same codomain, and the same value at every element of the domain. So even if you could have set up the induction properly to get the functions you wanted (or if you wanted to define $g_1$ on $[-1,0)$, then $g_2$ extended to all of $[-2,0)$, and so on sothat $g_n$ was defined on $[-n,\\infty)$) it still would not give a good answer to the problem", "$$f$$, we proceed as we did in Section 1.4: we find the zeros of the denominator and exclude them from the domain. Setting $$x+1=0$$ results in $$x=-1$$. Hence, our domain is $$(-\\infty, -1) \\cup (-1,\\infty)$$. The expression $$f(x)$$ is already in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done. 2. Proceeding as before, we determine the domain of $$g$$ by solving $$x+1=0$$. As before, we find the domain of $$g$$ is $$(-\\infty, -1) \\cup (-1,\\infty)$$. To write $$g(x)$$ in the form requested, we need to get a common denominator $\\begin{array}{rclclcl} g(x) & = & 2 - \\dfrac{3}{x+1} & = & \\dfrac{2}{1} - \\dfrac{3}{x+1} & = & \\dfrac{(2)(x+1)}{(1)(x+1)} - \\dfrac{3}{x+1} \\\\ & = & \\dfrac{(2x+2) - 3}{x+1} & = & \\dfrac{2x-1}{x+1} & & \\\\ \\end{array}$ This formula is now completely simplified. 3. The denominators in the formula for $$h(x)$$ are both $$x^2-1$$ whose zeros are $$x = \\pm 1$$. As a result, the domain of $$h$$ is $$(-\\infty, -1) \\cup (-1,1) \\cup (1, \\infty)$$. We now proceed to simplify $$h(x)$$. Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor. $\\begin{array}{rclcl} h(x) & = & \\dfrac{2x^2-1}{x^2-1} - \\dfrac{3x-2}{x^2-1}", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\\color{blue}{(-\\infty, -5) \\cup (-5, +\\infty)}$ RECALL: The domain of the sum, difference, and product of $f(x)$ and $g(x)$ is the common elements of the domains of the two functions. The domain of the respective functions are: $f(x)$ is defined for all real numbers except $-5$ (since 5 will make the denominator zero) so its domain is: $(-\\infty, 5) \\cup (5, +\\infty)$ $g(x)$ is defined for all real numbers so its domain is: $(-\\infty, +\\infty)$ Note that: $[(-\\infty, -5) \\cup(-5, +\\infty)] \\cap (-\\infty, +\\infty) \\\\= (-\\infty, -5) \\cup (-5, +\\infty)$. Thus, the domain of the sum, difference, and product of $f(x)$ and $g(x)$ is: $\\color{blue}{(-\\infty, -5) \\cup (-5, +\\infty)}$", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\\color{blue}{(-\\infty, -2.5) \\cup (-2.5, -1) \\cup (-1, +\\infty)}$ Find the domain of each function: For $f(x)$, the value of $x$ cannot be $-1$ since it will make the denominator zero. . Thus, the domain of $f(x)$ is $(-\\infty, -1) \\cup (-1, +\\infty)$. For $g(x)$, the value of $x$ can be any real number. Thus, the domain of $g(x)$ is $(-\\infty, +\\infty)$. RECALL: The domain of the $(f/g)(x)$ is the common elements of the domains of $f(x)$ and $g(x)$ excluding $x$ values for which $g(x)=0$. Note that when $g(-2.5)=0$. Thus, the domain of $(f/g)(x)$ is the set of all real numbers except $-1$ and $-2.5$. In interval notation, the domain of $f/g$ is: $\\color{blue}{(-\\infty, -2.5) \\cup (-2.5, -1) \\cup (-1, +\\infty)}$", "$G(x)={{x}^{2}}+2x+5$. • The definition of $G$ gives the value at each element of the domain by a formula. The value at $3$, for example, is $G(3)=3^2+2\\cdot3+5=20$. • The definition of $G$ does not specify the domain. The convention in the case of functions defined on the real numbers by a formula is to take the domain to be all real numbers at which the formula is defined. In this case, that is every real number, so the domain is $\\mathbb{R}$. • The definition does not specify the codomain, either. However, must include all real numbers greater than or equal to 4. (Why?) ### What the specification means • The specification guarantees that a function satisfies all five of the properties listed. • The specification does not define a mathematical structure in the way mathematical structures have been defined in the past: In particular, it does not require a function to be one or more sets with structure. • Even so, it is useful to have the specification, because: Many mathematical definitions introduce extraneous technical elements ## History Until late in the nineteenth century, functions were usually thought of as defined by formulas (including infinite series). Problems arose in the theory of harmonic analysis which made mathematicians require a more general notion of function. They came up with the", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Published by Pearson # Chapter 7 - Functions and Graphs - 7.4 The Algebra of Functions - 7.4 Exercise Set: 63 #### Answer $\\color{blue}{(-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)}$ #### Work Step by Step RECALL: The domain of the sum, difference, and product of $f(x)$ and $g(x)$ is the common elements of the domains of the two functions. The domain of the respective functions are: For $f(x)$, $-4.5$ will make the denominator zero so its domain is: $(-\\infty, -4.5) \\cup (-4.5, +\\infty)$ $g(x)$ is defined for all real numbers except $1$ (as it makes the denominator zero) so its domain is: $(-\\infty,1) \\cup (1, +\\infty)$. Note that: $[(-\\infty, -4.5) \\cup(-4.5, +\\infty)] \\cap [(-\\infty, 1) \\cup (1, +\\infty)] \\\\= (-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)$. Thus, the domain of the sum, difference, and product of $f(x)$ and $g(x)$ is: $\\color{blue}{(-\\infty, -4.5) \\cup (-4.5, 1) \\cup (1, +\\infty)}$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "$$g$$ is $$(-\\infty, 0) \\cup (0, \\infty)$$. Since $$f(x) = 6x^2-2x$$ is valid for all real numbers, we have no further restrictions. Thus the domain of $$g-f$$ matches the domain of $$g$$, namely, $$(-\\infty, 0) \\cup (0, \\infty)$$. A second method is to analyze the formula for $$(g-f)(x)$$ \\textit{before simplifying} and look for the usual domain issues. In this case, $(g-f)(x) = g(x) - f(x) = \\left(3-\\dfrac{1}{x}\\right) - \\left(6x^2 - 2x\\right),$ so we find, as before, the domain is $$(-\\infty, 0) \\cup (0, \\infty)$$. Moving along, we need to simplify a formula for $$(g-f)(x)$$. In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get $\\begin{array}{rclr} (g-f)(x) & = & g(x) - f(x) & \\\\ [5pt] & = & \\left(3-\\dfrac{1}{x}\\right) - \\left(6x^2 - 2x\\right) &\\\\ & = & 3 - \\dfrac{1}{x} - 6x^2 + 2x & \\\\ [10pt] & = & \\dfrac{3x}{x} - \\dfrac{1}{x} - \\dfrac{6x^3}{x} + \\dfrac{2x^2}{x} & \\text{get common denominators} \\\\ & = & \\dfrac{3x - 1 - 6x^3 - 2x^2}{x} & \\\\ & = & \\dfrac{-6x^3-2x^2+3x-1}{x} & \\\\ \\end{array}$ 4. \\item As in the previous example, we have two ways to approach finding the domain of $$\\frac{g}{f}$$. First, we can find the domain of $$g$$ and $$f$$ separately, and find the intersection of these two sets. In addition,", "defined there). Then saying that assuming you have defined $g_n$ with a domain of, say, $[-n,\\infty)$, you define $g_{n+1}$ on $[-n-1,\\infty)$ by $$g_{n+1}(x) = \\left\\{\\begin{array}{ll} g_n(x) &\\mbox{if x\\in[-n,\\infty).}\\\\ \\mbox{whatever} &\\mbox{if x\\in [-n-1,-n);} \\end{array}\\right.$$ This would indeed give an inductive definition for your $g_n$, defined on $[-n,\\infty)$, each extending the previous one. - Thank you a lot for your time, I learned a lot reading your posts. – user5501 Feb 6 '11 at 1:44 I think that you meant $[-2,0)$ at the fifth paragraph, and $[-n-1,-n)$ in your definition of $g_{n+1}$. Please correct me if I'm wrong in either case. – user5501 Feb 6 '11 at 1:45 @LovreP: Yes on both; thanks. – Arturo Magidin Feb 6 '11 at 1:49 It is enough to observe that $g$ can have any value whatsoever for $x < 0$. Since there is an infinite number of values for each $f(x)$ for $x < 0$, we're done. (you really don't need to do it any more difficult than that.) - Close. It is not that there are many values of x<0, but many possible values of g(x) for x<0. Imagine that something in the problem forced g(x) to be 0 for x<0. You still have lots of x, but only one g. – Ross Millikan Feb 6 '11 at 1:00 @Frederik: Actually, you", "# What is the domain of g(x)=x^3=1? Nov 25, 2017 See explanation. #### Explanation: I assume that there is a typo in the equation and the second equality sign should be either + or - sign. If the above assumption is correct then (no matter if it is + or -) then the function is a polynomial, so its domain is the whole $\\mathbb{R}$ set: ## $D = \\mathbb{R}$ Generally to find the domain of a function you need to look for any values which can be excluded from the domain (i.e. the values for which the function's value is undefined). Such numbers can be found if the function's formula has: • variable in the denominator - then you have to exclude those values of $x$ for which denominator becomes zero • variable under square root sign (or more generally root of an even degree) - this expression can only be calculated if the expression is non negative (zero or positive) • logarithms - these can only be calculated for positive values.", "in the chapter on examples of functions. • The definition of $F$ says \"$F$ is defined on the set $\\left\\{1,\\,2,\\,3,\\,6 \\right\\}$\". That phrase means that the domain is that set. • The value of $F$ at each element of the domain is given explicitly. The value at 3, for example, is 2, because the definition says that $F(2) = 3$. No other reason needs to be given. Mathematical definitions can be arbitrary. • The codomain of $F$ is not specified, but must include the set $\\{1,2,3\\}$. The codomain of a function is often not specified when it is not important — which is most of the time in freshman calculus (for example). #### A real-valued function Let $G$ be the real-valued function defined by the formula $G(x)={{x}^{2}}+2x+5$. • The definition of $G$ gives the value at each element of the domain by a formula. The value at $3$, for example, is $G(3)=3^2+2\\cdot3+5=20$. • The definition of $G$ does not specify the domain. The convention in the case of functions defined on the real numbers by a formula is to take the domain to be all real numbers at which the formula is defined. In this case, that is every real number, so the domain is $\\mathbb{R}$. • The definition does not specify the codomain, either. However, must include", "per minute 47 . approximately –0.6 milligrams per day ### 3.4 Section Exercises 1 . Find the numbers that make the function in the denominator $g g$ equal to zero, and check for any other domain restrictions on $f f$ and $g, g,$ such as an even-indexed root or zeros in the denominator. 3 . Yes. Sample answer: Let Then $f(g(x))=f(x−1)=(x−1)+1=x f(g(x))=f(x−1)=(x−1)+1=x$ and $g(f(x))=g(x+1)=(x+1)−1=x. g(f(x))=g(x+1)=(x+1)−1=x.$ So $f∘g=g∘f. f∘g=g∘f.$ 5 . $(f+g)( x )=2x+6, (f+g)( x )=2x+6,$ domain: $(−∞,∞) (−∞,∞)$ $(f−g)( x )=2 x 2 +2x−6, (f−g)( x )=2 x 2 +2x−6,$ domain: $(−∞,∞) (−∞,∞)$ $(fg)( x )=− x 4 −2 x 3 +6 x 2 +12x, (fg)( x )=− x 4 −2 x 3 +6 x 2 +12x,$ domain: $(−∞,∞) (−∞,∞)$ $( f g )( x )= x 2 +2x 6− x 2 , ( f g )( x )= x 2 +2x 6− x 2 ,$ domain: $(−∞,− 6 )∪(− 6 , 6 )∪( 6 ,∞) (−∞,− 6 )∪(− 6 , 6 )∪( 6 ,∞)$ 7 . $(f+g)( x )= 4 x 3 +8 x 2 +1 2x , (f+g)( x )= 4 x 3 +8 x 2 +1 2x ,$ domain: $(−∞,0)∪(0,∞) (−∞,0)∪(0,∞)$ $(f−g)( x )= 4 x 3 +8 x 2 −1 2x , (f−g)( x )= 4 x 3 +8 x 2 −1 2x", "could be input into$\\sin(1/x)$, since$x=0$is the only number excluded from the domain.The takeaway is that, while it is possible to design funky and unusual variations of these typical function families, we can always guarantee continuity of these function types for any value of$x$that could be input into it to obtain a real answer (i.e. the domain). ## Mr. Math Makes It Mean The last consideration about continuity that we need to make is the case where some function$h(x)$is defined as some combination of two (or possibly more) other functions, say$f(x)$and$g(x)$. This could happen by adding, subtracting, multiplying, or dividing$f$and$g$, or it could be a result of composition, e.g.$h(x) = f(g(x))$. In any case, there is nothing too surprising happening here. Definition:Let$h(x)$be comprised of two functions,$f(x)$and$g(x)$, either by arithmetic or composition. It follows that$h(x)$will be continuous everywhere that$f(x)$and$g(x)$are both continuous. For function division, such as$h(x) = f(x) \\, / \\, g(x)$for example, we must also exclude$x$values that make$g(x)=0$, for$h(x) = f(x) \\, / \\, g(x)$would then be undefined. Similarly, we must exclude any$x$values that make$g(x)$equal to values that are not in the domain of$f(x)$for a composition of the form$f\\big( g(x) \\big)$, as to avoid causing domain issues with$f$. It is not a surprising fact that domain and continuity go hand-in-hand, but it's worth mentioning explicitly for clarity, and teachers", "## Precalculus (6th Edition) Blitzer $(-\\infty, -8) \\cup (-8, 10) \\cup (10, +\\infty)$ The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. These numbers are $-8$ for the first denominator and $10$ for the second denominator. Thus, the value of $x$ can be any real number except $-8$ and $10$. Therefore, the domain is $(-\\infty, -8) \\cup (-8, 10) \\cup (10, +\\infty)$.", "functions. • The definition of $F$ says \"$F$ is defined on the set $\\left\\{1,\\,2,\\,3,\\,6 \\right\\}$\". That phrase means that the domain is that set. • The value of $F$ at each element of the domain is given explicitly. The value at 3, for example, is 2, because the definition says that $F(2) = 3$. No other reason needs to be given. Mathematical definitions can be arbitrary. • The codomain of $F$ is not specified, but must include the set $\\{1,2,3\\}$. The codomain of a function is often not specified when it is not important — which is most of the time in freshman calculus (for example). #### A real-valued function Let $G$ be the real-valued function defined by the formula $G(x)={{x}^{2}}+2x+5$. • The definition of $G$ gives the value at each element of the domain by a formula. The value at $3$, for example, is $G(3)=3^2+2\\cdot3+5=20$. • The definition of $G$ does not specify the domain. The convention in the case of functions defined on the real numbers by a formula is to take the domain to be all real numbers at which the formula is defined. In this case, that is every real number, so the domain is $\\mathbb{R}$. • The definition does not specify the codomain, either. However, must include all real numbers greater than or", "composed with g.\" The formula for $f o g$ is written $\\left(f o g\\right) \\left(x\\right)$. The domain and range for the functions are $f : A \\to B$ and $g : B \\to C$", "## Precalculus (6th Edition) Blitzer $(-\\infty, 0) \\cup (0, 4) \\cup (4, +\\infty)$. The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. The expression $\\dfrac{4}{x}$ is undefined when $x=0$. The denominator $\\dfrac{4}{x} -1$ is undefined when $x=4$. Thus, the value of $x$ can be any real number except $0$ and $4$. Therefore, the domain is $(-\\infty, 0) \\cup (0, 4) \\cup (4, +\\infty)$." ]

[ "### Home > CALC > Chapter 1 > Lesson 1.2.2 > Problem1-40 1-40. State the domain for each of the functions below. 1. $f ( x ) = \\frac { x } { x ^ { 2 } + 1 }$ We all know that $0$ is excluded from the denominator of a fraction. What value of $x$, if any, would make the denominator $x^2 + 1 = 0$? 1. $g ( x ) = \\frac { 1 } { x } - \\frac { x } { x + 1 }$ Consider the domain of each part of the subtraction problem separately. Since $g(x)$ is the difference between $\\frac{1}{x}$ and $\\frac{x}{x+1}$, both excluded values must be excluded from $g(x)$. 1. $h ( x ) = \\sqrt { x ^ { 2 } - 9 }$ Typically, a square root function has a domain of $x ≥ 0$, but in this case, $x$ is squared... meaning both positive or negative values will yield positive outputs. However, do not neglect to consider the $-9$. What values of $x$, both positive and negative, will make $x^2 - 9 > 0$? $x ≤ − 3\\text{ U } x ≥ 3$ 1. $k ( x ) = \\frac { \\operatorname { log } ( x - 3 ) } { \\sqrt {", "## Calculus 8th Edition Published by Cengage # Chapter 2 - Derivatives - 2.2 The Derivative as a Function - 2.2 Exercises: 24 #### Answer $g'(t) = \\frac{-1}{2}t^{\\frac{-3}{2}}$ Domain: t $\\gt$ 0 #### Work Step by Step $g(t) = \\frac{1}{\\sqrt t} = t^{-1/2}$ $g'(t) = (-1/2)t^{(-1/2) - 1}$ $g'(t) = \\frac{-1}{2}t^{\\frac{-3}{2}}$ The domain is t $\\gt$ 0 because for any negative t value or if t = 0, then g'(t) would be undefined After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "operators, we simply write the functions with the operator and simplify. Given two functions $$f$$ and $$g$$, we can define four new functions: $$(f+g)(x)=f(x)+g(x)$$ Sum $$(f−g)(x)=f(x)−g(x)$$ Difference $$(f·g)(x)=f(x)g(x)$$ Product $$(\\frac{f}{g})(x)=\\frac{f(x)}{g(x)}$$ for$$g(x)≠0$$ Quotient Exercise: 1) Combining Functions Using Mathematical Operations Given the functions $$f(x)=2x−3$$ and $$g(x)=x^2−1$$, find each of the following functions and state its domain. 1. $$(f+g)(x)$$ 2. $$(f−g)(x)$$ 3. $$(f·g)(x)$$ 4. $$(\\frac{f}{g})(x)$$ Solution: 1. $$(f+g)(x)=(2x−3)+(x^2−1)=x^2+2x−4.$$ The domain of this function is the interval $$(−∞,∞)$$. 2.$$(f−g)(x)=(2x−3)−(x^2−1)=−x^2+2x−2.$$ The domain of this function is the interval $$(−∞,∞)$$. 3. $$(f·g)(x)=(2x−3)(x^2−1)=2x^3−3x^2−2x+3.$$ The domain of this function is the interval $$(−∞,∞)$$. 4. $$(\\frac{f}{g})(x)=\\frac{2x−3}{x^2−1}$$. The domain of this function is {$$x∣∣x≠±1$$}. 2) For $$f(x)=x^2+3$$ and $$g(x)=2x−5$$, find $$(f/g)(x)$$ and state its domain. Solution: $$(\\frac{f}{g})(x)=\\frac{x^2+3}{2x−5}$$. The domain is {$$x|x≠\\frac{5}{2}$$}. ### Function Composition When we compose functions, we take a function of a function. For example, suppose the temperature $$T$$ on a given day is described as a function of time $$t$$ (measured in hours after midnight) as in Table. Suppose the cost $$C$$, to heat or cool a building for 1 hour, can be described as a function of the temperature $$T$$. Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating $$C(T(t))$$. We have defined a new function, denoted $$C∘T$$, which is", "composed with g.\" The formula for $f o g$ is written $\\left(f o g\\right) \\left(x\\right)$. The domain and range for the functions are $f : A \\to B$ and $g : B \\to C$", "8) (4, 5) (4, 6) (4, 7) (4, 8)} ∴ A × C ⊂ B × D Hence it is verified Question 8. If f(x) = $$\\frac{x-1}{x+1}, x \\neq 1$$ Show that f(f(x)) = – $$\\frac { 1 }{ x }$$, Provided x ≠ 0. Question 9. The functions f and g are defined by f{x) = 6x + 8; g(x) = $$\\frac { x-2 }{ 3 }$$ (i) Calculate the value of gg $\\frac { 1 }{ 2 }$ (a) Write an expression for gf (x) in its simplest form. f(x) = 6x + 8 ; g(x) = $$\\frac { x-2 }{ 3 }$$ Question 10. Write the domain of the following real functions (i) f (x) = $$\\frac { 2x+1 }{ x-9 }$$ If the denominator vanishes when x = 9 So f(x) is not defined at x = 9 ∴ Domain is x ∈ [R – {9}] (ii) if p(x) = $$=\\frac{-5}{4 x^{2}+1}$$ p(x) is defined for all values of x. So domain is x ∈ R. (iii) g(x) = $$\\sqrt { x-2 }$$ When x < 2 g(x) becomes complex. But given “g” is real valued function. So x > 2 Domain x ∈ (2, α) (iv) h (x) = x + 6 For all values of x, h(x) is defined. Hence domain is", "the function. Keep in mind that the function is indeed defined at $x=\\frac{1}{3}$. Roots are undefined when the radicand is negative right? It is still defined at $x=\\frac{1}{3}$! It just gives a value of 0. On top of that, we know that basic root functions are continuous over their domains. Here is a quick way to find the domain AND interval of continuity. We have the function $f(x)=(3x-1)^\\frac{1}{4}$ We know that the radicand must be greater than zero right? It also can be equal to 0, because the $f(x)=0\\ at\\ x=\\frac{1}{3}$. So the domain and interval of continuity lies in the inequality; $3x-1 \\geq 0$ $3x \\geq 1$ $x \\geq \\frac{1}{3}$ Just to add on interval notation, we use the closed bracket on \\frac{1}{3} because the point is included in the interval. $[\\frac{1}{3}, \\infty)$ is the same as $x \\geq \\frac{1}{3}$ $(\\frac{1}{3}, \\infty)$ is the same as $x > \\frac{1}{3}$ Does this help clarify? 5. Yah its clearer now. So when I find the domain of the function, I should pick an internal point within the domain and substitute that point for x in the function to prove my point? 6. Originally Posted by VitaX Yah its clearer now. So when I find the domain of the function, I should pick an internal point within the domain and", "h$ since the domains are different: the domain of the first is $\\mathbb{R}^2$ and the domain of the second is $\\mathbb{R}$. In these circumstances, writing $$\\frac{df}{dx} = \\frac{\\partial f}{\\partial x} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dx}$$ is a notational abuse (one of the many to which the Liebnitz notation lends itself).", "the derivative $$f'(x) = \\frac{1}{x+1}$$ has a domain of $$(-\\infty,-1) \\cup (-1,\\infty)$$ or $$x | x \\neq -1$$ which lies within the range of $$(-\\infty,0) \\cup (0,\\infty)$$ or $$y | y \\neq 0$$", "domain of the composite function. Since $( g \\circ f )(x)$ or g(f(x)) is written as $g(\\frac{1}{x+2})$, we want f(x) to be in the domain of g(x). That is, g (can take on any value except 3). Therefore, we need to know when the function f(x) gives us a value of 3. To know that, we solve algebraically: $\\frac{1}{x+2} = 3$ $3x+6 = 1$ $x = \\frac{-5}{3}$ We get f(x) = 3 when $x = \\frac{-5}{3}$. It implies that the domain of $latex( g \\circ f )(x)$ will be all real numbers except -2 and $\\frac{-5}{3}$. Note: • Always remember that it is important to know the domain of the starting function to be able to determine which values can be allowed in the domain of the composite function. • Solve algebraically to calculate the restricted values of the composite function. Part 2 to come… How clear is this post?", "of f on the same domain of f if and only if G is of the form G(x)= F(x) + c, where c is a constant. Proof : If G(x) = F(x) + c and F '(x) = f(x) and 'c' is a constant then G'(x) = $\\frac{\\text{d}}{\\text{d}x}[F(x) +c]$ = F' (x) + 0 G'(x) = f(x)", "Precalculus (6th Edition) Blitzer The domain of the function $g\\left( x \\right)=\\frac{4}{x-7}$ is $\\left( -\\infty,7 \\right)\\cup \\left( 7,\\infty \\right)$. Consider the given function $g\\left( x \\right)=\\frac{4}{x-7}$. We can see that this function contains division and division by 0 is not defined, excluding those values of $x$ from the domain that cause the denominator to be zero. Thus, set the denominator equal to 0, that is, $x-7=0$. This means that $x=7$. Therefore, the domain of the given function is the set of all real numbers excluding the number 7. Thus, the domain of the function $g\\left( x \\right)=\\frac{4}{x-7}$ is $\\left( -\\infty,7 \\right)\\cup \\left( 7,\\infty \\right)$.", "we will get: $~~~~~~~~~~~~~$ x = $y^{\\frac 13}$ This gives a function g:Y →X. This new function g can be defined as $~~~~~~~~~~~~~$ g(y) = $y^{\\frac 13}$ This function g is the inverse of the function f since it’s domain is same as the range of the function f.Since g(y) = $y ^{\\frac 12}$ , representing the independent variable with x , we get g(x) = $x^{\\frac 13}$ = $f^{-1}(x)$..", "can be defined. Although it looks like g(f(x)) can be defined at x = 0, you should remember that you will not get through f(x) in the first place. This is because f(0) cannot be defined. Now, work out the domain of g(f(x)) and convince yourself that the domain is all real numbers except x = 0 and x = -1 (c) We will start by evaluating the composite function f(f(x)). $f(x + \\frac{1}{x}) = x + \\frac{1}{x} + \\frac{1}{x + \\frac{1}{x}}$ Make sure you work it out and get to the point where: $f(x + \\frac{1}{x}) = \\frac{x^4 + 3x^2 + 1}{x(x^2+1)}$ Here, you can see that x = 0 is excluded from the domain of the composite function (since f(x) is the starting function). Let’s solve algebraically to see if x = 0 is the only restricted value on the domain. Remember, we do not want f(0), so we need to find when f(x) gives us 0. $x+\\frac{1}{x}=0$ $x^2+1=0$ $x = \\sqrt{-1}$ But the domain of the square root is $x\\ge 0$. Thus, we can conclude that the domain of the composite function is all real numbers except for x = 0. (d) Finally, we will look at g(g(x)), which is a bit similar to part (b). Evaluating the composite function gives us: $g(\\frac{x+8}{x+2}) = \\frac{\\frac{x+8}{x+2}", "$G(x)={{x}^{2}}+2x+5$. • The definition of $G$ gives the value at each element of the domain by a formula. The value at $3$, for example, is $G(3)=3^2+2\\cdot3+5=20$. • The definition of $G$ does not specify the domain. The convention in the case of functions defined on the real numbers by a formula is to take the domain to be all real numbers at which the formula is defined. In this case, that is every real number, so the domain is $\\mathbb{R}$. • The definition does not specify the codomain, either. However, must include all real numbers greater than or equal to 4. (Why?) ### What the specification means • The specification guarantees that a function satisfies all five of the properties listed. • The specification does not define a mathematical structure in the way mathematical structures have been defined in the past: In particular, it does not require a function to be one or more sets with structure. • Even so, it is useful to have the specification, because: Many mathematical definitions introduce extraneous technical elements ## History Until late in the nineteenth century, functions were usually thought of as defined by formulas (including infinite series). Problems arose in the theory of harmonic analysis which made mathematicians require a more general notion of function. They came up with the", "# What is the domain of g(x)=x^3=1? Nov 25, 2017 See explanation. #### Explanation: I assume that there is a typo in the equation and the second equality sign should be either + or - sign. If the above assumption is correct then (no matter if it is + or -) then the function is a polynomial, so its domain is the whole $\\mathbb{R}$ set: ## $D = \\mathbb{R}$ Generally to find the domain of a function you need to look for any values which can be excluded from the domain (i.e. the values for which the function's value is undefined). Such numbers can be found if the function's formula has: • variable in the denominator - then you have to exclude those values of $x$ for which denominator becomes zero • variable under square root sign (or more generally root of an even degree) - this expression can only be calculated if the expression is non negative (zero or positive) • logarithms - these can only be calculated for positive values.", "# How do you find the domain of the function g(x)=7/(10-3x)? $g \\left(x\\right) = \\frac{7}{10 - 3 x}$ is defined for all Real values of $x$ except $x = \\frac{10}{3}$ (which would result in an attempt to divide by zero). Therefore the domain of $g \\left(x\\right)$ is $\\left[- \\infty , \\frac{10}{3}\\right) \\bigcup \\left(\\frac{10}{3} , + \\infty\\right]$ That is the union of all values less that $\\frac{10}{3}$ with all values greater than $\\frac{10}{3}$", "8. Originally Posted by colby2152 Wouldn't the range be: $[0,1]$ because the function $\\frac{x-2}{x+6}$ has an infinite limit of one giving it a horizontal asymptote of $y=1$. Let $g(x) = \\frac{x - 2}{x + 6} = \\frac{x + 6 - 8}{x + 6} = 1 - \\frac{8}{x + 6}$. g has a range of (-oo, 1) U (1, +oo). Let $f(x) = \\sqrt{x}$. The maximal domain of f is [0, +oo). Therefore $y = \\sqrt{\\frac{x-2}{x+6}} = f(g(x))$ is only defined when the range of g is resticted to lie in [0, oo). Under this restriction, the 'maximal input' for f will be [0, 1) U (1, +oo) and so the 'maximal output' of f will be [0, 1) U (1, +oo). Therefore the range of $y = \\sqrt{x-2}{x+6}$ is [0, 1) U (1, +oo). 9. Originally Posted by mr fantastic Let $g(x) = \\frac{x - 2}{x + 6} = \\frac{x + 6 - 8}{x + 6} = 1 - \\frac{8}{x + 6}$. g has a range of (-oo, 1) U (1, +oo). Let $f(x) = \\sqrt{x}$. The maximal domain of f is [0, +oo). Therefore $y = \\sqrt{\\frac{x-2}{x+6}} = f(g(x))$ is only defined when the range of g is resticted to lie in [0, oo). Under this restriction, the 'maximal input' for f will be [0, 1) U", "+ 8}{\\frac{x+8}{x+2} +2}$ And, finally resulting in: Work this out $g(g(x)) = \\frac{3x+8}{x+4}$ Here, you see that we have a problem similar to the one in part (b). x = -2 looks like it can be included because the function looks fine. However, you should remember that since x = -2 is not defined for g(x) (which is the starting function here), it should be excluded from the domain of the composite function. Looking at the simplified equation, it is easy to see that x = -4 is also excluded from the domain. Therefore, the domain of g(g(x)) is all real numbers except -2 and -4. How clear is this post?", "## Discrete Mathematics with Applications 4th Edition $g(x)=\\frac{2x^{3}+2x}{x^{2}+1}=\\frac{(2x)(x^{2}+1)}{x^{2}+1}=2x=f(x)$ Yes, the functions $f$ and $g$ are equivalent. Note that $x^{2}+1\\ne0$; therefore, $g$ has the same domain as $f$.", "## Calculus: Early Transcendentals 8th Edition $G(x)$ is continuous on $(-\\infty,\\frac{-1}{2})\\cup(\\frac{-1}{2},1)\\cup(1,\\infty)$ $G(x)=\\frac{x^2+1}{2x^2-x-1}$ $G(x)=\\frac{x^2+1}{(2x+1)(x-1)}$ (factorize the denominator) 1) Find the domain of $G(x)$ $G(x)$ is defined only when $(2x+1)(x-1)\\ne0$ $G(x)$ is defined only when $(2x+1)\\ne0$ and $(x-1)\\ne0$ $G(x)$ is defined only when $x\\ne\\frac{-1}{2}$ and $x\\ne1$ Therefore, the domain of $G(x)$ is $(-\\infty,\\frac{-1}{2})\\cup(\\frac{-1}{2},1)\\cup(1,\\infty)$ 2) Since $G(x)$ is a rational function, according to Theorem 5, $G(x)$ is continuous on its demain. Therefore, $G(x)$ is continuous on $(-\\infty,\\frac{-1}{2})\\cup(\\frac{-1}{2},1)\\cup(1,\\infty)$" ]

[ "# Functions and Logarithms Algebra Level 4 \\begin{aligned} f(x) & = & 10^{10x} \\\\ g(x) & = & \\log_{10} \\left ( \\frac x {10} \\right ) \\\\ h_1 (x) & = & g(f(x)) \\\\ h_n (x) & = & h_1 (h_{n-1} (x)) \\\\ \\end{aligned} Denote the set functions above. Evaluate the sum of digits of $h_{2011} (1)$. Note- This is not original. Adapted from a Math contest. ×", "is a function $h = (b, \\oplus)$ of the form \\begin{align} h\\; 0 &= b \\\\ h\\; (n + 1) &= n \\oplus (h\\; n) \\end{align} For example, the factorial function has $b = 1$ and $n \\oplus m = (n + 1) \\times m$. For lists, a paramorphism would be a function $h$ of the form \\begin{align} h\\; \\textsf{nil} &= b \\\\ h\\; (\\textsf{cons}\\; a\\; b) &= a \\oplus (b, h\\; b) \\end{align}", "Algebra # Function Arithmetic Suppose that \\begin{aligned} f(x) &=4x^2-9,\\\\ g(x) &=3x+7,\\\\ h(x) &=f(x)+g(x). \\end{aligned} What is the value of $h(3)$? Consider the functions \\begin{aligned} f(x)&=-5x+7,\\\\ g(x)&=-2x-3. \\end{aligned} What is the value of $f(6) \\cdot g(1)$? Given the two functions $f(x) = 2x + 1, \\quad g(x) = -x - 2,$ if $h(x) = f(x) + g(x), \\quad k(x) = f(x) - g(x),$ what is the value of $h(2) \\cdot k(1)?$ Consider the functions \\begin{aligned} f(x) &= 3x + 1 \\\\ g(x) &= x + \\frac{1}{3}. \\end{aligned} If \\begin{aligned} h(x) &= f(x) + g(x) \\\\ k(x) &= \\frac{g(x)}{f(x)}, \\end{aligned} what is the value of $x$ satisfying $h(x) - k(x) = 5?$ Suppose that \\begin{aligned} f(x)&=-6x+7,\\\\ g(x)&=-3x-3. \\end{aligned} What is the value of $f(6) \\cdot g(2)$? ×", "# Problem #160 160 The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: \\begin{align*} f(x,x) &=x, \\\\ f(x,y) &=f(y,x), \\quad \\text{and} \\\\ (x + y) f(x,y) &= yf(x,x + y). \\end{align*} Calculate $f(14,52)$. This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "do not apply. #### Case 1: Positive bases Let us seek to define $x^a$ from first principles for arbitrary $x$ positive and $a$ real. We can be motivated by the continuity of the function $x^t$ as a function of $t$ (where $x$ is fixed) and think of $x^a$ as the limit of values $x^r$ as rational numbers $r$ approach $a$. To me it is easier to proceed via the logarithm function which may be defined as $\\displaystyle \\log(x) := \\int_1^x \\frac{1}{t}\\ \\text{d}t\\quad x > 0.\\quad\\quad(9)$ Being the area under a continuous positive-valued function, this function is continuous and monotonically increasing. It is also apparent that $\\log(1) = 0$. From the definition the important identity $\\log (xy) = \\log(x) + \\log(y)$ (where $x, y > 0$) can be derived via a change of variable $u=t/x$ as follows: \\begin{aligned} \\log (xy) &= \\int_1^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&=\\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_x^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{ux}x\\ \\text{d}u\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{u}\\ \\text{d}u\\\\&=\\log(x) + \\log(y). \\quad \\quad (10)\\end{aligned} By repeated use of this rule, $\\log(x^n) = n \\log(x)$ for positive integers $n$. As $\\log(2) > 0$, $\\log(2^n) = n \\log 2$ grows without bound as $n$ increases, which shows that the log function is unbounded above. It is also unbounded below due to the relationship $\\log(1/x) = -\\log(x)$", "limit of values $x^r$ as rational numbers $r$ approach $a$. To me it is easier to proceed via the logarithm function which may be defined as $\\displaystyle \\log(x) := \\int_1^x \\frac{1}{t}\\ \\text{d}t\\quad x > 0.\\quad\\quad(9)$ Being the area under a continuous positive-valued function, this function is continuous and monotonically increasing. It is also apparent that $\\log(1) = 0$. From the definition the important identity $\\log (xy) = \\log(x) + \\log(y)$ (where $x, y > 0$) can be derived via a change of variable $u=t/x$ as follows: \\begin{aligned} \\log (xy) &= \\int_1^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&=\\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_x^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{ux}x\\ \\text{d}u\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{u}\\ \\text{d}u\\\\&=\\log(x) + \\log(y). \\quad \\quad (10)\\end{aligned} By repeated use of this rule, $\\log(x^n) = n \\log(x)$ for positive integers $n$. As $\\log(2) > 0$, $\\log(2^n) = n \\log 2$ grows without bound as $n$ increases, which shows that the log function is unbounded above. It is also unbounded below due to the relationship $\\log(1/x) = -\\log(x)$ (which follows from $0 = \\log (1) = \\log(x.(1/x)) = \\log x + \\log (1/x)$). Hence the log function is monotonic, continuous and has range $(-\\infty, \\infty)$. It thus has a continuous inverse which is how we may define the exponential function: $\\exp(x)$ is the unique positive value $y$ satisfying", "at the very end of Friday’s lecture that if we define a function: $2^{\\Box}:\\mathbb{R}\\rightarrow \\mathbb{R}^+$, by $2^{\\Box}(x)=2^x$, then we define $\\log_2:\\mathbb{R}^+\\rightarrow \\mathbb{R}$ as an inverse function: $\\log_2:=(2^{\\Box})^{-1}$. ## Week 10 We will introduce more and study the properties of uses of logarithms. ### Recent Comments J.P. McCarthy on MATH6040: Winter 2019, Week… Student on MATH6040: Winter 2019, Week… J.P. McCarthy on MATH7019: Winter 2019, Week… Student on MATH7019: Winter 2019, Week… J.P. McCarthy on MATH6040: Winter 2019, Week…", "# How do you evaluate the function H(x) = 18 – x, when x=-6? You simply replace the $x$ in the function by the value at which it is evaluated. So $H \\left(- 6\\right) = 18 - \\left(- 6\\right) = 18 + 6 = 24$.", "# Barnes G at positive integer The following formula holds: $$G(n) = \\left\\{ \\begin{array}{ll} 0&\\quad n=-1,-2,\\ldots \\\\ \\displaystyle\\prod_{k=0}^{n-2} k!&\\quad n=0,1,2,\\ldots, \\end{array} \\right.$$ where $G$ denotes the Barnes G function and $i!$ denotes the factorial.", "### Home > CALC > Chapter 2 > Lesson 2.1.3 > Problem2-35 2-35. Construct a composite function $k(x)$ using$f ( x ) = \\sqrt { x }$ and $g(x) =\\text{log }x$ if $k ( x ) = \\frac { 1 } { 2 } \\operatorname { log } x$. Since $\\text{log}x^a = a\\text{log}x$ (power log of logarithms).", "Which of the twelve basic functions are bounded above? Apr 27, 2015 The Sine function: $f \\left(x\\right) = \\sin \\left(x\\right)$ The Cosine function: $f \\left(x\\right) = \\cos \\left(x\\right)$ and The Logistic function: $f \\left(x\\right) = \\frac{1}{1 - {e}^{- x}}$ are the only function of the \"Basic Twelve Functions\" which are bounded above.", "# How to find the inverse function of h(x)= 2^x ? Apr 10, 2018 $h \\left(x\\right) = {\\log}_{2} \\left(x\\right)$ #### Explanation: The find the inverse of an invertible function $y = f \\left(x\\right)$, swap $y$ and $x$ and solve for $y$. We have $h \\left(x\\right) = y = {2}^{x}$. Swapping $x$ and $y$ gives $x = {2}^{y}$. We now wish to solve for $y$. Take the log with base 2 of both sides of the equation to free the $y$ variable. $x = {2}^{y}$ ${\\log}_{2} \\left(x\\right) = {\\log}_{2} \\left({2}^{y}\\right) = y$ Thus, $y = h \\left(x\\right) = {\\log}_{2} \\left(x\\right)$. See that we used the fact that ${\\log}_{n} \\left({n}^{x}\\right) = x$.", "power. For $a > 0$ and relatively prime positive integers $p, q$ we define $a^{p/q}$ to be the unique positive number $y$ such that $y^q = a^p$. (That the $q$‘th root exists follows from the fact that the function $x^q$ has an inverse for $x > 0$). From (12) we then have $\\displaystyle (\\exp(x))^p = \\exp(px) = \\exp((px/q).q) = \\exp(px/q)^q.$ Setting $\\exp(x)$ to $a$ and $\\exp(px/q)$ to $y$ we have thus shown $a^p = y^q$, so $\\exp(x)^{p/q} = \\exp(px/q)$. Therefore (12) is satisfied when $n$ is positive rational. Finally we use $\\exp(-x) = 1/\\exp(x)$ (which can be proved using (11)) to extend (12) to negative rationals: $\\displaystyle \\exp(nx) = 1/\\exp(-nx) = 1/\\exp(x)^{-n} = \\exp(x)^n.$ We are finally ready to define $x^a$ for arbitrary $x$ positive and $a$ real relying on continuity. We simply extend (12) replacing $\\exp(x)$ with $x$ (also replacing $x$ with $\\log(x)$) and $n$ with real values $a$: $\\displaystyle x^a := \\exp(a \\log(x)), \\quad x > 0, a \\in \\mathbb{R}.\\quad\\quad(13)$ From this definition it is quick to verify laws (1) and (2): \\begin{aligned} x^a x^b &= \\exp(a \\log(x))\\exp(b \\log (x))\\\\ &= \\exp(a \\log x + b\\log x)\\\\ &= \\exp((a+b)\\log x)\\\\ &= x^{a+b}\\quad\\quad(14)\\\\\\text{and }\\ (x^a)^b &= \\exp(b \\log(x^a))\\\\ &= \\exp(b a \\log(x))\\\\ &= x^{ab}\\quad\\quad(15)\\end{aligned} Law (3) is immediate if either $x$ or $y$ is equal to 1.", "functions(2) • Apr 10th 2011, 03:00 AM Punch functions(2) given the functions $g(x)=e^x-2, x=(a,infinity)$ $ h(x)=ln(lnx), x>1 $ Find the smallest value of $a$ in exact form such that the composite function $hg$ exists. • Apr 10th 2011, 03:15 AM Prove It Assuming that you are trying to find $\\displaystyle h\\left(g(x)\\right)$... $\\displaystyle h\\left(g(x)\\right) = \\ln{\\left[\\ln{\\left(e^x - 2\\right)}\\right]}$. A logarithm is only defined for positive values of the independent variable, so the innermost logarithm can only be evaluated where $\\displaystyle e^x - 2 > 0$. But since another logarithm will be taken, you can only accept those values for which the innermost logarithm is positive. So where $\\displaystyle e^x - 2 > 1$. Solve for $\\displaystyle x$.", "& =\\left[ (152.64)-(96.96) \\right]+1 \\\\ & =[55.6]+1 \\\\ & =55+1 \\\\ & =56 \\\\ \\end{align} Hence the number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ is equal to 56 Note: The characteristic of the logarithm of a positive number is positive and it is one less than the number of digits in the number. Hence we use find number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ by taking greatest integer of log of $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ and adding 1.", "with $x$. Then translate the remaining phrase back into symbols and set it equal to $f(x)$. \\begin{aligned}{\\color{#0047af}\\text{one divided by }x} \\text{ or } {\\color{#0047af} \\text{the reciprocal of }x}&=\\frac{1}{x}\\\\ f(x) &=\\frac{1}{x}\\end{aligned} Solution Check that the functions can be composed into $h(x)$. $h(x)=\\frac{1}{x+3}$ Remember that: \\begin{aligned}f(x)&= \\frac{1}{x} \\\\ g(x)&=x+3\\end{aligned} Use the values to check the functions: $f(g(x)) = \\frac{1}{x+3} = h(x)$ So, the given function can be decomposed as $f(g(x))$. ### Associative Property Since composition of functions is associative, the way in which three or more functions are grouped in a composite function does not change the result. Composition of functions is associative, which means that the way in which functions are grouped when they are composed does not matter. $(f \\circ (g \\circ h))(x)=((f \\circ g) \\circ h)(x)$ Step-By-Step Example Confirming That Composition Is Associative Consider the given values for $f(x)$, $g(x)$, and $h(x)$: \\begin{aligned}f(x)&=x+6\\\\g(x)&=8x\\\\h(x)&=\\frac{x}{2}\\end{aligned} Identify $(f \\circ (g \\circ h))(x)$ and $((f \\circ g) \\circ h)(x)$, and then determine if they are equivalent. Step 1 Identify the composition of the two inside functions of $(f \\circ (g \\circ h))(x)$. \\begin{aligned}g(h(x))&=g\\left(\\frac{x}{2}\\right)\\\\&=8\\left(\\frac{x}{2}\\right)\\\\&=4x\\end{aligned} Step 2 Use the result to find $(f \\circ (g \\circ h))(x)$. \\begin{aligned}(f \\circ(g(h(x)))&=f(4x)\\\\&=(4x)+6\\\\&=4x+6\\end{aligned} Step 3 Identify the composition of the two inside functions of $((f \\circ g) \\circ h)(x)$. \\begin{aligned}f(g(x))&=f(8x)\\\\&=(8x)+6\\\\&=8x+6\\end{aligned} Step 4 Use the result to find", "x$ on the other hand is defined only over positive reals, and since it is always increasing from left to right, at some point the value of the function will exceed 5. Specifically, this will occur for $x>32$. Likewise, $y=\\log_2 x$ will be less than -5 for $x<\\frac{1}{32}$. And it’s apparent that in the interval $\\frac{1}{32} the log function is going to intersect the sine function as it goes up and down on its periodic path. How many times does it intersect? AIME writers like to catch students on “off by one” errors, so take care to examine what is happening close to $x=\\frac{1}{32}$ and $x=32$ before entering your answer!", "(x+1))}+\\dfrac{1}{(\\log (x+1))(\\log (x+2))}+ \\cdots\\right)\\\\ f(x)&= \\frac{1+f(x+1)}{\\log(x)}\\\\ \\implies f(x+1)&=\\log(x) f(x)-1 \\end{align} So we now have a recurrence relation for the function, and we are seeking $f(2)$. By repeatedly reducing $f(x+n+1)$, we also obtain $$f(x+n)=\\left(\\prod_{k=0}^{n-1}\\log(x+k)\\right) f(x)-\\sum_{k=0}^{n-1}\\prod_{i=1}^{n-1}\\log(x+i)$$ So if we can come up with $f(x)$ for some integer $x$, we can come up with $f(x)$ for any integer $x$. I can't seem to come up with a value for anything though, so hopefully someone else can.", "= a \\end{align*} • This function isn't quite like the ones we saw earlier — it's a function of $t$ rather than $x\\text{.}$ Recall that a function is a rule which assigns to each input value an output value. So far, we have usually called the input value $x\\text{.}$ But this “$x$” is just a dummy variable representing a generic input value. There is nothing wrong with calling a generic input value $t$ instead. Indeed, from time to time you will see functions that are not written as formulas involving $x\\text{,}$ but instead are written as formulas in $t$ (for example representing time — see Section 1.2), or $z$ (for example representing height), or other symbols. • So let us write the definition of the derivative \\begin{align*} f'(a) &= \\lim_{h \\to 0} \\frac{f(a+h)-f(a)}{h} \\end{align*} and then translate it to the function names and variables at hand: \\begin{align*} h'(a) &= \\lim_{h \\to 0} \\frac{h(a+h)-h(a)}{h} \\end{align*} • But there is a problem — “$h$” plays two roles here — it is both the function name and the small quantity that is going to zero in our limit. It is extremely dangerous to have a symbol represent two different things in a single computation. We need to change one of them. So let's rename the small quantity that is going to", "on Wikipedia. Basically, you have a function $f(x)$; now define \\begin{align} f_\\mathrm e(x) &= \\frac{1}{2}[f(x) + f(-x)] \\\\ f_\\mathrm o(x) &= \\frac{1}{2}[f(x) - f(-x)... Only top voted, non community-wiki answers of a minimum length are eligible" ]

[ "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "# How to find the value of $h(99)$ in the function? If $$h(x) + h(x+1) = 2x^2$$ and $$h(33) = 99$$ What will be the value of $h(99)$? - $h(33)+h(34)=2*33^2$ so $h(34)=...$. Then, continue for $f(35)$ and so on, try to find a logic behind it and you'll get $h(99)$ quite fast. – barto Jan 26 '13 at 19:34 $$h(34)=2\\times 33^2-h(33)$$ $$h(35)=2\\times 34^2-h(34)=2\\times 34^2-2\\times 33^2+h(33)$$ $$\\dots$$ $$h(99)=2\\times 98^2 - 2\\times 97^2 + 2\\times 96^2 -\\cdots + 2\\times 34^2 - 2\\times 33^2 +h(33)\\\\=2(98^2-97^2+96^2-\\cdots+34^2-33^2)+99\\\\ =2(98+97+\\cdots +34+33)+99\\\\ =8745$$", "got a= 1, b=-28 $\\boxed {\\therefore p(x)=x^4+2x^3+x^2-28x-60 }$ continue....", "by (3.55555555556) to get x 100/3.55555555556=x 28.125=x x=28.125 ### now we have: 36 is 28.125% of 128 Step-by-step explanation:", "= \\ln(2) = 0.693147 \\dots\\, .$", "a + 99d = (29/400) + 99(-1/400) = (29/400) – (99/400) = (-70/400) =(-7/40) Thus, the 100th term of the H.P = 1/(100th term of the A.P) = (-40/7) Therefore, the 100th term of the H.P is (-40/7).", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "l u e}{8} + 24$ =$113$", "/ 109 + {y}^{2} / 100 = 1$", "2*4*6....*94*96 = H(94)*96 H(94) + H(96) = H(94) + H(94)*96 = H(94) { 1 + 96 } = H(94) * 97 = 2*4*.....*94*97 Hence greatest prime factor is 97 _________________ Consistency is the Key Manager Joined: 05 Oct 2017 Posts: 66 Location: India Schools: GWU '21, Isenberg'21 Re: For every positive even integer n, the function h(n) is defined to be [#permalink] ### Show Tags 01 Jan 2019, 07:52 This is how i have solved. h(94) = 2*4*6*8........94 = 2^47 ( 1 * 2 * 3 * 4 ......47) .........(i) h(96) = 2*4*6*8........96 =2^48 ( 1 * 2 * 3 * 4 ......48 ) .........(ii) adding above two equation and taking 2^47 ( 1 * 2 * 3 * 4 ......47) common , we get -->2^47 ( 1 * 2 * 3 * 4 ......47) (1 + $$2*48$$) --> 2^47 ( 1 * 2 * 3 * 4 ......47) (97) Hope it helps. _________________ The Path is the Goal WTF is “that” doing in my sentence? By VerbalNinja Re: For every positive even integer n, the function h(n) is defined to be &nbs [#permalink] 01 Jan 2019, 07:52 Display posts from previous: Sort by", "So $M=7$ and we have $x=[s]\\,[e]\\,[M]=0\\,1001\\,111$", "divides $n$. Therefore, $B\\cap C = \\{35, 70, \\dots, 140\\}$.", "$m+n=2+89=\\boxed{091}$.", "the answer is $960-2=\\boxed{958}$. -brainiacmaniac31", "# What is the measurement 1042 L rounded off to two significant digits? $1000 L$ 2 s.f. means the first two digits of a number, i.e. $\\textcolor{red}{23} 9.35$ becomes $\\textcolor{red}{24} 0$ after rounding. $000 \\textcolor{red}{16} .345$ becomes $\\textcolor{red}{16}$ when rounded. Finally, $0.00 \\textcolor{red}{18} 734$ becomes $0.00 \\textcolor{red}{19}$ when rounded.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "{O}_{2}$ $t i m e C {O}_{2} = 112.569978 h o u r s$ Since there are only 2 significant figures the answer would be $110 h o u r s$", "$0 = a$, that means $x+32 = 1001$, and so $x = 1001 - 32 = 969$. So, $9$ + $6$ + $9$ = $\\boxed{\\textbf{(E)}\\ 24}$. ~songmath20 ## Video Solution ~Education, the Study of Everything" ]

[ "at the very end of Friday’s lecture that if we define a function: $2^{\\Box}:\\mathbb{R}\\rightarrow \\mathbb{R}^+$, by $2^{\\Box}(x)=2^x$, then we define $\\log_2:\\mathbb{R}^+\\rightarrow \\mathbb{R}$ as an inverse function: $\\log_2:=(2^{\\Box})^{-1}$. ## Week 10 We will introduce more and study the properties of uses of logarithms. ### Recent Comments J.P. McCarthy on MATH6040: Winter 2019, Week… Student on MATH6040: Winter 2019, Week… J.P. McCarthy on MATH7019: Winter 2019, Week… Student on MATH7019: Winter 2019, Week… J.P. McCarthy on MATH6040: Winter 2019, Week…", "& =\\left[ (152.64)-(96.96) \\right]+1 \\\\ & =[55.6]+1 \\\\ & =55+1 \\\\ & =56 \\\\ \\end{align} Hence the number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ is equal to 56 Note: The characteristic of the logarithm of a positive number is positive and it is one less than the number of digits in the number. Hence we use find number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ by taking greatest integer of log of $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ and adding 1.", "# How do you evaluate the function H(x) = 18 – x, when x=-6? You simply replace the $x$ in the function by the value at which it is evaluated. So $H \\left(- 6\\right) = 18 - \\left(- 6\\right) = 18 + 6 = 24$.", "# Functions and Logarithms Algebra Level 4 \\begin{aligned} f(x) & = & 10^{10x} \\\\ g(x) & = & \\log_{10} \\left ( \\frac x {10} \\right ) \\\\ h_1 (x) & = & g(f(x)) \\\\ h_n (x) & = & h_1 (h_{n-1} (x)) \\\\ \\end{aligned} Denote the set functions above. Evaluate the sum of digits of $h_{2011} (1)$. Note- This is not original. Adapted from a Math contest. ×", "+0 0 88 1 1. The function $h(x)$ is defined as: $h(x) = \\left\\{ \\begin{array}{cl} \\lfloor 4x \\rfloor & \\text{if } x \\le \\pi, \\\\ 3-x & \\text{if }\\pi < x \\le 5.2, \\\\ x^2& \\text{if }5.2< x. \\end{array}\\right.$ Find $h(h(\\sqrt{2}))$. 2. Find constants $A$ and $B$ such that $\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$ for all $x$ such that $x\\neq -1$ and $x\\neq 2$. Give your answer as the ordered pair $(A,B)$. 3. a) Suppose that $|a - b| + |b - c| + |c - a| = 20.$ What is the maximum possible value of $|a - b|$? b) Suppose that $|a - b| + |b - c| + |c - d| + \\dots + |m-n| + |n-o| + \\cdots+ |x - y| + |y - z| + |z - a| = 20.$ What is the maximum possible value of $|a - n|$? Guest Apr 22, 2018 Sort:", "# Problem #160 160 The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: \\begin{align*} f(x,x) &=x, \\\\ f(x,y) &=f(y,x), \\quad \\text{and} \\\\ (x + y) f(x,y) &= yf(x,x + y). \\end{align*} Calculate $f(14,52)$. This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "2*4*6....*94*96 = H(94)*96 H(94) + H(96) = H(94) + H(94)*96 = H(94) { 1 + 96 } = H(94) * 97 = 2*4*.....*94*97 Hence greatest prime factor is 97 _________________ Consistency is the Key Manager Joined: 05 Oct 2017 Posts: 66 Location: India Schools: GWU '21, Isenberg'21 Re: For every positive even integer n, the function h(n) is defined to be [#permalink] ### Show Tags 01 Jan 2019, 07:52 This is how i have solved. h(94) = 2*4*6*8........94 = 2^47 ( 1 * 2 * 3 * 4 ......47) .........(i) h(96) = 2*4*6*8........96 =2^48 ( 1 * 2 * 3 * 4 ......48 ) .........(ii) adding above two equation and taking 2^47 ( 1 * 2 * 3 * 4 ......47) common , we get -->2^47 ( 1 * 2 * 3 * 4 ......47) (1 + $$2*48$$) --> 2^47 ( 1 * 2 * 3 * 4 ......47) (97) Hope it helps. _________________ The Path is the Goal WTF is “that” doing in my sentence? By VerbalNinja Re: For every positive even integer n, the function h(n) is defined to be &nbs [#permalink] 01 Jan 2019, 07:52 Display posts from previous: Sort by", "We then define the logarithm of $c$ base $a$ as follows: $\\log_a(c)=b$ I.e. the logarithm of $c$ base $a$ is the power to which $a$ must be raised in order for it to equal $c$. Some examples include: \\begin{aligned} 10^3 = 1000 &\\longleftrightarrow \\log_{10}(1000) = 3 \\\\ \\\\2^{10} = 1024 &\\longleftrightarrow \\log_2(1024) = 10 \\\\ \\\\7^8 = 5,764,801 &\\longleftrightarrow \\log_7(5,764,801) = 8 \\end{aligned} It should be noted that, though I am using Natural Numbers in the examples above, the definition of logarithms extends to fractions (aka Rational Numbers) and even – as we will see in a bit – numbers that cannot be expressed as fractions (aka Irrational Numbers) We can now see why tables of logarithms used to be helpful. Supposing we wanted to multiply two largish numbers, say $123456789$ and $987654321$. Then: \\begin{aligned}&\\log_{10}(123456789)\\approx 8.091515\\hspace{1mm}&\\Rightarrow\\hspace{1mm}&10^{8.091515}\\approx 123456789 \\\\ \\\\&\\log_{10}(987654321)\\approx 8.994605\\hspace{1mm}&\\Rightarrow\\hspace{1mm}&10^{8.994605}\\approx 987654321\\end{aligned} Then we have that: $123456789\\times 987654321 \\approx 10^{8.091515} \\times 10^{8.994605}$ and by $(1)$ above, we have: $123456789\\times 987654321 \\approx 10^{8.091515 + 8.994605}$ Algorithmically, we take the logarithms of each multiplicand, add these and then raise 10 to the power of the resulting number. In this case: $10^{17.08612}\\approx 1.21933\\times 10^{17}$ Which is a reasonable approximation to the actual answer, probably close enough for most practical purposes. As adding the logarithms is easier then multiplying two nine digit numbers", "with real values $a$: $\\displaystyle x^a := \\exp(a \\log(x)), \\quad x > 0, a \\in \\mathbb{R}.\\quad\\quad(13)$ From this definition it is quick to verify laws (1) and (2): \\begin{aligned} x^a x^b &= \\exp(a \\log(x))\\exp(b \\log (x))\\\\ &= \\exp(a \\log x + b\\log x)\\\\ &= \\exp((a+b)\\log x)\\\\ &= x^{a+b}\\quad\\quad(14)\\\\\\text{and }\\ (x^a)^b &= \\exp(b \\log(x^a))\\\\ &= \\exp(b a \\log(x))\\\\ &= x^{ab}\\quad\\quad(15)\\end{aligned} Law (3) is immediate if either $x$ or $y$ is equal to 1. Otherwise, we can find $c$ so that $y = x^c$ (choose $c = \\log(y)/\\log(x)$) and then we have from (1) and (2) the following: \\begin{aligned} (xy)^a &= (x.x^c)^a\\\\ &= (x^{1+c})^a\\\\ &= x^{a + ac}\\\\ &= x^a x^{ac}\\\\ &= x^a (x^c)^a\\\\ &= x^a y^a.\\quad\\quad (16) \\end{aligned} #### Case 2: Negative bases When the base $x$ is negative we can define $x^a$ by the real number $(-1)^a(-x)^a$ provided $(-1)^a$ exists. This will be the case if $a$ is rational and has odd-valued denominator (square roots of -1 are not real-valued nor are irrational powers of -1). Writing $a = p/q$ where $p$ and $q$ are relatively prime ($q$ odd), we then have $(-1)^{p/q} = (-1)^p$. Since odd denominators are preserved under addition, law (1) can be shown to hold. To check law (3) we consider the cases of both $x,y$ being negative or only one (say $y$) being", "limit of values $x^r$ as rational numbers $r$ approach $a$. To me it is easier to proceed via the logarithm function which may be defined as $\\displaystyle \\log(x) := \\int_1^x \\frac{1}{t}\\ \\text{d}t\\quad x > 0.\\quad\\quad(9)$ Being the area under a continuous positive-valued function, this function is continuous and monotonically increasing. It is also apparent that $\\log(1) = 0$. From the definition the important identity $\\log (xy) = \\log(x) + \\log(y)$ (where $x, y > 0$) can be derived via a change of variable $u=t/x$ as follows: \\begin{aligned} \\log (xy) &= \\int_1^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&=\\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_x^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{ux}x\\ \\text{d}u\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{u}\\ \\text{d}u\\\\&=\\log(x) + \\log(y). \\quad \\quad (10)\\end{aligned} By repeated use of this rule, $\\log(x^n) = n \\log(x)$ for positive integers $n$. As $\\log(2) > 0$, $\\log(2^n) = n \\log 2$ grows without bound as $n$ increases, which shows that the log function is unbounded above. It is also unbounded below due to the relationship $\\log(1/x) = -\\log(x)$ (which follows from $0 = \\log (1) = \\log(x.(1/x)) = \\log x + \\log (1/x)$). Hence the log function is monotonic, continuous and has range $(-\\infty, \\infty)$. It thus has a continuous inverse which is how we may define the exponential function: $\\exp(x)$ is the unique positive value $y$ satisfying", "## Algebra and Trigonometry 10th Edition We are given the function: $h(x)=[[x+3]]$ denotes the greatest integer less than or equal to $x+3$. a) $h(-2)=[[-2+3]]=[[1]]=1$ b) $h\\left(\\dfrac{1}{2}\\right)=[[0.5+3]]=[[3.5]]=3$ c) $h(4.2)=[[4.2+3]]=[[7.2]]=7$ d) $h\\left(-21.6\\right)=[[-21.6+3]]=[[-18.6]]=-19$", "+0 # Help This HW is way to hard, URGENT +1 297 1 +24 1)$$Compute \\lfloor \\log_3 1000 \\rfloor.$$ 2)Among the following functions, let a be the number of functions that are monotonically increasing, let b be the number of functions that are monotonically decreasing, and let c be the number of functions that are neither monotonically increasing nor monotonically decreasing. Enter your answer in the form \"a,b,c\". \\begin{align*} y &= x^4 \\\\ y &= \\log_{\\frac{1}{2}} x \\\\ y &= x^2 - 2x \\\\ y &= \\frac{1}{x^2} \\\\ y &= -\\frac{1}{\\sqrt{x}} \\\\ y &= \\sqrt[3]{x + 4} \\\\ y &= 11^{-x} \\\\ y &= 5 - 3x \\end{align*} 3)Among the following functions, let a be the number of functions that are even, let b be the number of functions that are odd, and let c be the number of functions that are neither even nor odd. Enter your answer in the form \"a,b,c\". \\begin{align*} y &= \\frac{x}{x^2 - 1} \\\\ y &= \\frac{x}{x - 1} \\\\ y &= 2x - 11 \\\\ y &= \\log_3 (x^2 - 1) \\\\ y &= 2^{-x} \\\\ y &= 6x^5 + x + \\frac{2}{x^3} \\\\ y &= \\sqrt[3]{x} \\\\ y &= \\lfloor x \\rfloor \\end{align*} May 12, 2019 #1 0 1. The answer is 18. 2. The answer is (3,1,4). 3. The", "the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify. \\begin{aligned} f(2) &=2^{2}+3(2)-4 \\ &=4+6-4 \\ &=6 \\end{aligned} b. In this case, the input value is a letter so we cannot simplify the answer any further. $$f(a)=a^{2}+3 a-4$$ c. With an input value of $a+h$, we must use the distributive property. \\begin{aligned} f(a+h) &=(a+h)^{2}+3(a+h)-4 \\ &=a^{2}+2 a h+h^{2}+3 a+3 h-4 \\end{aligned} ## 微积分网课代修|预备微积分代写precalculus辅导|Evaluation of Functions in Algebraic Forms Evaluate $f(x)=x^{2}+3 x-4$ 在: C。 $a+h$ d。 $\\frac{f(a+h)-f(a)}{h}$ $$f(2)=2^{2}+3(2)-4 \\quad=4+6-4=6$$ $$f(a)=a^{2}+3 a-4$$ C。输入值为 $a+h$ ，我们必须使用分配属性。 $$f(a+h)=(a+h)^{2}+3(a+h)-4 \\quad=a^{2}+2 a h+h^{2}+3 a+3 h-4$$", "# I call it the Germain function $g(n)= \\begin{cases} 1, \\quad \\quad\\quad \\quad\\quad\\quad\\quad\\quad\\quad \\quad n = 2^0, 2^1, 2^1, 2^2, \\ldots \\\\ \\displaystyle \\prod_{\\text{odd prime }p} \\left( \\dfrac{p-1}2 \\right)^{v_p (n) }, \\quad \\text{otherwise} \\end{cases}$ Let $$g(n)$$ be defined as above, where $$v_p(n)$$ denotes the highest power of prime $$p$$ that divides $$n$$. Let $$M$$ be the set of all rational numbers $$\\mu$$ for which there is at least one positive integer n, such that $$\\mu=\\log _{ n }{ g(n) }$$. In other words: $$M=Q\\cap \\left\\{ \\log _{ n }{ g(n) } :\\quad n\\in \\mathbb N \\right\\}$$, wher $$Q$$ is the set of all rational numbers and $$N$$ the set of all positive integers. What is the cardinality of $$M$$? ×", "$log_64 x=2$ 2. $log_2 64=x$ 3. $log_2 x=64$ 4. $log_x 64=2$ Which of the following is the best definition of a parabola? 1. A U-shaped curve. 2. For a given point and line not through the point, the set of all points equidistant from this given point and line. 3. The object contained by a focal point and directrix when shown in a graphic form. 4. An equation with an $x^2$ term. Grade 11 Functions and Relations CCSS: HSF-IF.A.1 Which of the following relations is NOT a function? 1. ${(0, 1), (1, 0), (2, 1), (3, 1), (4, 2)}$ 2. ${(7, 4), (4, 9), (-3, 1), (1, 7), (2, 8)}$ 3. ${(1, 4), (3, 2), (5, 2), (1, -8), (6, 7)}$ 4. All of the above are functions. 5. None of the above are functions. Grade 11 Functions and Relations CCSS: HSF-IF.A.1 What is the range of the function? {(5,1), (6,2), (7,3)} 1. {1, 2, 3} 2. {5, 6, 7} 3. {1, 2, 3, 5, 6, 7} 4. {1, 7} Which is standard form of a quadratic function? 1. $y = mx+b$ 2. $y = x^3$ 3. $y= ax^2 + bx + c$ 4. $y=x^2$ Add and simplify. $(x+2)/(x-1)+(x-3)/(x+1)$ 1. $(x^2-3x+1)/(x^2-x+4)$ 2. $(2x-1)/(2x)$ 3. $(x^2-x-6)/(x-1)$ 4. $(2x^2-x+5)/((x-1)(x+1))$", "= a$ and $log_{70}5 = b$, then the value of $log_{70}392$ is: a) 3 (1 + b – a) b) 3 – 3b – a c) 3 ( 1 – ab) d) 3 – 3a – b Question 87: If $\\log_{24}{6}=x$ $\\log_{12}{54}=y$ then a) $y=\\frac{(4+x)}{(x-3)}$ b) $y=\\frac{(3-x)}{(x+2)}$ c) $x=\\frac{(3-y)}{(y-2)}$ d) $x=\\frac{(y+2)}{(8-y)}$ Question 88: Two functions A(x) and B(x) are such that $4A^{2}(x)-2B(x)B(-x)=B^{2}(x)+B^{2}(-x)$. If $A(4) = 24$ what is the value of $A(-4)$? a) 24 b) -24 c) 32 d) Cannot be determined Question 89: If the domain of the function f(x) is [-100,700]. Two functions g and h are defined such that g(y) = f($2y^2+5$) and h(z) = g($z^2$). Find the number of integral values for which h(z) is defined. a) 6 b) 9 c) 4 d) 2 Question 90: If a function F($x$) is defined such that F($x$)+F($x-1$)=$x^2$ and F(10)=2019. Then the value of F(52). Question 91: A quadratic function f(x) has a value 5 at x=-2. If the maximum value of the function is 8 at x=-1, then f(-3) is a) 3 b) -6 c) -4 d) 5 Question 92: Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f", "do not apply. #### Case 1: Positive bases Let us seek to define $x^a$ from first principles for arbitrary $x$ positive and $a$ real. We can be motivated by the continuity of the function $x^t$ as a function of $t$ (where $x$ is fixed) and think of $x^a$ as the limit of values $x^r$ as rational numbers $r$ approach $a$. To me it is easier to proceed via the logarithm function which may be defined as $\\displaystyle \\log(x) := \\int_1^x \\frac{1}{t}\\ \\text{d}t\\quad x > 0.\\quad\\quad(9)$ Being the area under a continuous positive-valued function, this function is continuous and monotonically increasing. It is also apparent that $\\log(1) = 0$. From the definition the important identity $\\log (xy) = \\log(x) + \\log(y)$ (where $x, y > 0$) can be derived via a change of variable $u=t/x$ as follows: \\begin{aligned} \\log (xy) &= \\int_1^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&=\\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_x^{xy} \\frac{1}{t}\\ \\text{d}t\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{ux}x\\ \\text{d}u\\\\&= \\int_1^{x} \\frac{1}{t}\\ \\text{d}t + \\int_1^{y} \\frac{1}{u}\\ \\text{d}u\\\\&=\\log(x) + \\log(y). \\quad \\quad (10)\\end{aligned} By repeated use of this rule, $\\log(x^n) = n \\log(x)$ for positive integers $n$. As $\\log(2) > 0$, $\\log(2^n) = n \\log 2$ grows without bound as $n$ increases, which shows that the log function is unbounded above. It is also unbounded below due to the relationship $\\log(1/x) = -\\log(x)$", "nearly identical. Here's a generalized root function $$h,$$ which can be instantiated whenever $$f$$ is strictly increasing. \\begin{align} & g(x, y) = \\begin{cases} g(x + 1, y) & \\text{if }f(x) \\lt y \\\\[0.5em] x & \\text{if }f(x) \\ge y \\end{cases} \\\\[1em] & h(y) = g(0, y) \\end{align} For example, suppose we want to find the integer part of the square root of 11. Then $$f(x) = x^2,$$ and our computation proceeds as follows: \\begin{align} h(11) &= g(0, 11) \\\\[0.25em] &= g(1, 11) \\\\[0.25em] &= \\ldots \\\\[0.25em] &= g(4, 11) \\\\[0.25em] &= 4 - 1 \\\\[0.25em] &= 3 \\end{align} def f(x): return x ** 2 def g(x, y): if f(x) < y: return g(x + 1, y) elif f(x) == y: return x else: return x - 1 def h(y): return g(0, y) for i in range(0, 10): print(str(i) + ' ' + str(h(i)))", "power. For $a > 0$ and relatively prime positive integers $p, q$ we define $a^{p/q}$ to be the unique positive number $y$ such that $y^q = a^p$. (That the $q$‘th root exists follows from the fact that the function $x^q$ has an inverse for $x > 0$). From (12) we then have $\\displaystyle (\\exp(x))^p = \\exp(px) = \\exp((px/q).q) = \\exp(px/q)^q.$ Setting $\\exp(x)$ to $a$ and $\\exp(px/q)$ to $y$ we have thus shown $a^p = y^q$, so $\\exp(x)^{p/q} = \\exp(px/q)$. Therefore (12) is satisfied when $n$ is positive rational. Finally we use $\\exp(-x) = 1/\\exp(x)$ (which can be proved using (11)) to extend (12) to negative rationals: $\\displaystyle \\exp(nx) = 1/\\exp(-nx) = 1/\\exp(x)^{-n} = \\exp(x)^n.$ We are finally ready to define $x^a$ for arbitrary $x$ positive and $a$ real relying on continuity. We simply extend (12) replacing $\\exp(x)$ with $x$ (also replacing $x$ with $\\log(x)$) and $n$ with real values $a$: $\\displaystyle x^a := \\exp(a \\log(x)), \\quad x > 0, a \\in \\mathbb{R}.\\quad\\quad(13)$ From this definition it is quick to verify laws (1) and (2): \\begin{aligned} x^a x^b &= \\exp(a \\log(x))\\exp(b \\log (x))\\\\ &= \\exp(a \\log x + b\\log x)\\\\ &= \\exp((a+b)\\log x)\\\\ &= x^{a+b}\\quad\\quad(14)\\\\\\text{and }\\ (x^a)^b &= \\exp(b \\log(x^a))\\\\ &= \\exp(b a \\log(x))\\\\ &= x^{ab}\\quad\\quad(15)\\end{aligned} Law (3) is immediate if either $x$ or $y$ is equal to 1.", "a function of $x$. Include at least the interval $[-5,5]$ for $x$-values. However, each $x$ does determine a unique value for $y$, and there are mathematical procedures by which $y$ can be found to any desired accuracy. Solve a = 2 - b for a. can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. With an input value of $a+h$, we must use the distributive property. Ask students to come up with a rule for the table. Invented in the early 1600s century by John Napier, log tables were a crucial tool for every mathematician for over 350 years. Problem 4. That's going to be the output of that function. Given a square root equation, the student will solve the equation using tables or graphs - connecting the two methods of solution. \\begin{align}h\\left(p\\right)&={p}^{2}+2p \\\\ h\\left(4\\right)&={\\left(4\\right)}^{2}+2\\left(4\\right) \\\\ &=16+8 \\\\ &=24 \\end{align}. It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula. Moving horizontally along the line $y=4$, we locate two points of the curve with output value $4:$ $\\left(-1,4\\right)$ and $\\left(3,4\\right)$. The tabular form for function $P$ seems ideally suited to this function, more so than writing it in paragraph or function form. As we saw above," ]

[ "+0 # Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, 0 126 1 +979 Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. Enter the most specific answer. $$\\left(\\frac x2 - 3\\right)^2 + y^2 = 10$$ Nov 30, 2020 #1 +133 +1 I would say the graph is an ellipse. Here is the graph: Nov 30, 2020", "Is g... ##### Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola.$3 x^{2}=27-3 y^{2}$ Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $3 x^{2}=27-3 y^{2}$...", "following as circle, parabola, ellipse, or hyperbola: $$x^2 + 4y^2 - 16 = 0.$$ 28. Solution ellipse 29. Identify the following as circle, parabola, ellipse, or hyperbola: $$x + 2y^2 - 8y + 4 = 0.$$ 30. Solution parabola", "the equation x^2-y^2+8x=16 is a parabola, circle, ellipse, or hyperbola and how do you graph it? How do you identity if the equation x^2-y^2+8x=16 is a parabola, circle, ellipse, or hyperbola and how do you graph it?...", "1. If , the equation represents an Ellipse, a Circle (degenerate Ellipse), a Point (degenerate Circle), or has no graph. 2. If , the equation represents a Hyperbola or pair of intersecting lines (degenerate Hyperbola). 3. If , the equation represents a Parabola, a Line (degenerate Parabola), a pair of Parallel lines (degenerate Parabola), or has no graph.", "the Graph at More Than One Point • Graph of the parabola f\\left( x \\right) = {x^2} - 2. • Graph of absolute value function f\\left( x \\right) = \\left| x \\right| • Graph of semi-circle f\\left( x \\right) = \\sqrt {7 - {x^2}} You might also be interested in: Vertical Line Test", "# Identify the graph of the given nondegenerate conic sections: 3x^2 + 2y^2 + 12x - 4y + 2 = 0. Identify the graph of the given nondegenerate conic sections: $3{x}^{2}+2{y}^{2}+12x-4y+2=0$. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it insonsipthinye Step 1 A nondegenerate conic section of the form, $A{x}^{2}+B{y}^{2}+Cx+Dy+E=0$ Where A and B both are not zero. 1) A circle if $A=B$ 2) A parabola if $AB=0$ 3) An ellipse if A is not equal to 4) A hyperbola if $AB<0$ Step 2 Given that, $3{x}^{2}+2{y}^{2}+12x-4y+2=0$ Here So A and B are not equal and $AB=\\left(3\\right)\\left(2\\right)=6>0$ Hence the graph of the equation is ellipse.", "Example. Solve the following system graphically: {(x^2+y^2=25),(xy=12):}. Graph of the equation x^2+y^2=25 is a circle with center at the origin and radius equal 5. Graph of the equation xy=12 is a hyperbola y=12/x.", "### Home > CCA2 > Chapter 8 > Lesson 8.1.1 > Problem8-10 8-10. Describe the possible numbers of intersections for each of the following pairs of graphs. Sketch a graph for each possibility. For example, a circle could intersect a line twice, once, or not at all. Your solution to each part should include all of the possibilities and a sketched example of each one. 8-10 HW eTool (Desmos). 1. Two different lines. $0,1,\\ \\text{or}\\ ∞$ 1. A line and a parabola. $0, 1,\\ \\text{or}\\ 2$ 1. Two different parabolas. 1. A parabola and a circle.", "the distances $\\overline{MF_1} + \\overline{MF_2}$ is close to $2a$. You may do the above activity for different values of $a, b, h$ and $k$ and as many points as needed to better understand the definition of an ellipse. 2 - Use the values of $a$ and $b$ to find $c$ using the relationship between $a, b$ and $c$ given by $a^2 = b^2 + c^2$. Use the above results to find the coordinates of $F_1, F_2, V_1$ and $V_2$ and check the results graphically. 3 - Set $a, b, h$ and $k$ to some values. Find the x and y intercepts and check your results graphically. 4 - Exercise: Show by algebraic calculations that the following equation $\\dfrac{(x + 2)^2}{5} + 5(y-3)^2 = 5$ is that of an ellipse and find the center, foci and vertices of the ellipse given by the equation then use the app to graph it and check your answers. If needed, Free graph paper is available. ## More References and Links to Topics Related to the Equation of the Ellipse Ellipse Similar tutorials on circle , Parabola and the hyperbola can be found in this site.", "Browse Questions # If the three distinct lines $\\;x+2ay+a=0\\;,x+3by+b=0\\;$ and $\\;x+4ay+a=0\\;$ are concurrent , then the point (a,b) lines on a : $(a)\\;circle\\qquad(b)\\;hyperbola\\qquad(c)\\;straight\\;line\\qquad(d)\\;parabola$", "+0 # help 0 145 1 Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. $x^2 + 2y^2 - 6x - 20y + 59 = 12$ Dec 9, 2018 #1 +1 It is an ellispe because putting it into conic form we have $$(x^2-6x+9)+2(y^2-10y+25)=12$$ $$(x-3)^2+2(y-5)^2=12$$ $$\\frac{(x-3)^2}{12}+\\frac{(y-5)^2}{6}=1$$ you can also put it into desmos to see the graph of it Dec 9, 2018 #1 +1 It is an ellispe because putting it into conic form we have $$(x^2-6x+9)+2(y^2-10y+25)=12$$ $$(x-3)^2+2(y-5)^2=12$$ $$\\frac{(x-3)^2}{12}+\\frac{(y-5)^2}{6}=1$$ you can also put it into desmos to see the graph of it Guest Dec 9, 2018", "section. $$e =0$$ gives a circle, $$0 \\lt e \\lt 1$$ gives an ellipse, $$e=1$$ gives a parabola, and $$e \\gt 1$$ gives a hyperbola.", "• Loser66 $$\\sqrt{(x+3)^2+(y-2)^2}=\\sqrt{(x-3)^2+y^2}$$ In the xy plane, the set of points whose coordinates satisfy the equation above is?? 1) A line 2) A circle 3) An ellipse 4) A parabola 5) One branch of a hyperbola Please, help Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "Comment Share Q) # The equation $16x^2 + y^2 + 8xy - 74x -78 y +212 = 0$ represents : ( A ) an ellipse ( B ) a hyperbola ( C ) a parabola ( D ) a circle", "the turning point. The $x$-intercepts Once you have the graph, you can turn your attention to answering the question: solve ${x}^{2} - 2 x - 24 = 0$ from a graph. If you compare the equations: $y = {x}^{2} - 2 x - 24 \\text{ and } 0 = {x}^{2} - 2 x - 24$, you will realise that $y = 0$ The question being asked is \"Where does the parabola cross the $x$-axis?\" You can read these values as the $x$-intercepts from the graph. These are seen to be $x = - 4 \\mathmr{and} x = 6$ graph{y= x^2 -2x -24 [-10, 10, -5, 5]}", "Lia M. # Identify this equation without completing the square. Identify this equation without completing the square. x2 + y2 - 8x + 4y = 0 The equation defines a circle. The equation defines an ellipse. The equation defines a hyperbola. The equation defines a parabola.", "+0 Determine whether the equation represents a circle, a point, or have no graph a) 1 – x 2 – y 2 = 0 0 150 1 Determine whether the equation represents a circle, a point, or have no graph a) 1 – x2 – y2 = 0 b) x2+ y2 + x –y = 0 Oct 29, 2021 #1 +13718 +1 Determine whether the equation represents a circle, a point, or have no graph Hello Guest! a) 1 – x^2 – y^2 = 0 $$y=\\pm \\sqrt{1-x^2}$$ is the equation of a circle. b) $$x^2+ y^2 + x –y = 0$$ $$y=\\frac{1}{2}(1\\pm \\sqrt{-4x^2-4x+1})\\ |\\ wolfram\\ alpha$$ is the equation of a circle. ! Oct 29, 2021", "the below with a center (j, k) instead of (0, 0), replace each x term with (x-j) and each y term with (y-k) circle: ellipse: parabola. Writing an equation to describe the relationship between the number of scoops in an ice cream cone and the price. Research writing on algebraic formula Rated 5/5 based on 11 review 2018.", "a hyperbola. 59. Given $k{x}^{2}+8xy+8{y}^{2}-12x+16y+18=0$, find $k$ for the graph to be a parabola. 60. Given $6{x}^{2}+12xy+k{y}^{2}+16x+10y+4=0$, find $k$ for the graph to be an ellipse." ]

[ "expression without too much computation? 10. Dec 4, 2007 ### Ben Niehoff \"Completing the cube\" doesn't work, unfortunately. For a quadratic, all you have to do to complete the square is add a constant. But a cubic has too many variables to do that. Try it. 11. Dec 4, 2007 ### Ben Niehoff But wait! This particular cubic can be completed by adding a constant! Well, then. To the OP: Remember that $$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ Now, write $$y^3 + 6y^2 + 12y = x - 7$$ and see what you can do. 12. Dec 4, 2007 ### Big-T In general yes, but x^3 + 6x^2 + 12x + 7 = (x+2)^3-1. So one would think that the coefficients 6 and 12 were not chosen arbitrarily? Last edited: Dec 4, 2007 13. Dec 4, 2007 ### agentnan Thank you ALL so much...you have definitely sent me in the correct direction particularly Big-T and Ben Niehoff! I will finish my paper right now! Thanks again...Nan Last edited: Dec 4, 2007", "Try getting rid of the xy term first... by completing the square! (Hint: instead of looking at (x-a)², try looking at (x-ay)²) 3. Aug 23, 2006 ### GregA By completing the square to eliminate xy I get $$(x+2y)^2 +y^2-6y-4x+7$$...I can go further and wrap up y to get $$(x+2y)^2 + (y-3)^2 -4x - 2$$ but I cannot see any useful clues...(apart from what I discovered from the other factorisation), nor can I see a way to force exact values upon x or y 4. Aug 24, 2006 ### Hurkyl Staff Emeritus You can still complete the square to get rid of the x term. (Doing so will introduce another y term, but you already know how to get rid of that) 5. Aug 24, 2006 ### GregA Aha! $$x^2 + 4xy + 5y^2 -4x -6y + 7 =$$ $$(x +2y -2)^2 + y^2 +2y +3 =$$ $$(x +2y -2)^2 +(y+1)^2 +2$$ Thanks for steering me in the right direction Hurkyl!", "equation (1 )and (2) we should get, $$x\\equiv 2z \\implies x+z \\equiv 0 \\pmod{3}$$. $$\\implies x+z \\equiv y+w \\equiv 0 \\pmod{3}$$. Similarly we have $x-z \\equiv y-w \\equiv 0\\pmod{3}$. So , only possibility is $x=y=z=w=3$. So there will exactly one unit square covered exactly once. Contradiction ! And we are done. ~ftheftics", "added the missing term (actually, subtracting is more useful here): $1 - 2\\sqrt{n} x + nx^2 = (1 - \\sqrt{n}x)^2$. Now this is a square, so it is nonnegative. What can you conclude? 5. Nov 4, 2013 ### Lee33 Thanks, jbunniii for the hint!", "2 \\| x \\|_V \\| y \\|_V$$ Finally, this gives us $$\\| x + y \\|_V^2 \\leq \\|x \\|_V^2 + \\| y \\|_V^2 + 2 \\| x \\|_V \\| y \\|_V = (\\| x \\|_V + \\| y \\|_V)^2$$ so taking away the square gives us $$\\| x + y \\|_V \\leq \\| x \\|_V + \\| y \\|_V$$, as required.", "complete the square as $(x-y)^2$ and not $(x-y+1)^2$ or something like that? – Archer Oct 17 '17 at 18:59 • @Abcd 1. Trial and error; 2. Rigorous Practice. – SchrodingersCat Oct 17 '17 at 19:11", "we can split the left-hand term as a difference of squares again; we can repeat as needed until we have a sum with odd outer exponents. Therefore $g(y)\\mid f_n(y)$ and thus $g(x)\\mid f_n(x)$.", "2 added 12 characters in body I was asked (by myself) to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary than I thought (or am I being silly?). Does anybody know it, and can give an answer or a reference to it? Of course, I'm quite sure it should fit within a larger theory in combinatorics or in probability, but an elementary answer would be appreciated. Let $S$ be a (say open) subset of a square $[0,1]^2$ with Lebesgue measure $|S|>1/2$. Then, there exists a rectangle with a vertex on the diagonal, and the other three vertices in $S$ (in other words, there are three points of $S$ of the form $(x,y)$, $(x,z)$ and $(y,z)\\$). The constant $1/2$ cannot be lowered, as the example of the subset $S^* :=(0,1/2)\\times(1/2,1)\\cup(1/2,1)\\times(0,1/2)$ shows (for any three points $x,y,z$ in $[0,1]$, at least 2 of them are both either smaller or larger than $1/2$, so the corresponding pair is not in $S^*$. 1 # A puzzle about finding three points $(x,y)$, $(x,z)$ and $(y,z)$ in a subset of a square. I was asked to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary", "For $x=10$ we get $y=0$ Exists in table (hence it is a complete square) So corresponding point is: $(10,0)$ So all the affine points are $(0,1)$ $(0,10)$ $(1,0)$ $(2,2)$$(2,9)$ $(9,2)$ $(9,9)$ $(10,0)$", "$-1$ case never has solutions, so we can focus on the $+1$ case. We can see that $(n^2-2)y^2+1 = n^2y^2 - 2y^2+1$ is less than $n^2y^2$ when $y\\geq 1$ as we assumed. We can also check that $$(n^2-2)y^2+1 > (ny-1)^2$$ by writing a sequence of equivalent inequalities until we see one to be true. Expanding gives:$$n^2y^2 - 2y^2 + 1 > n^2y^2 - 2ny + 1 .$$ Subtracting common terms leaves $$-2y^2 > - 2ny$$ and that is equivalent to $y<n$ which is assumed. Thus if $1\\leq y<n$ then $(n^2-2)y^2+1$ lies between two consecutive squares, and thus can not be a perfect square itself.", "side is $\\dfrac{6x^3+13x^2+9x+5}{3x+2}=\\ldots$ instead? If we compare this fraction to the one considered previously we notice that the only difference is the constant term in the cubic expression, which is now $5$ instead of $2$. If we were to set up a multiplication grid to tackle this problem it might still look like the one above. • Which part of the multiplication grid tells us about the constant term in the cubic expression? • At what stage did we choose to complete this box in the previous problem? Do we have any choice about the order in which it is completed? Can you use a similar approach to divide $4x^4+3x^3+2x+1$ by $x^2+x+2$? Are you convinced that taking this approach gives you the same result as other methods?", "# Why does $\\left(\\frac b2\\right)^2$ “geometrically complete the square?” I was just reading this MathisFun article on completing the square. It states that geometry can help complete the square. It starts off with a square and a rectangle (pictures come from link): Then, it cuts $b$ in half, and moves it under the $x^2$ square: Now, the square is \"nearly completed\", but it has this part that completes the square that equals $\\left(\\frac b2 \\right)^2$ (circled in blue): My question is, where did that part come from, and why does it equal $\\left(\\frac b2 \\right)^2$. The article doesn't give me a reason and there are no other sources as to why. • can you expand $(x+b/2)^2$ please ? and the target is to find $C$ and $\\rho$ such that $x^2 + bc + C = (x+\\rho)^2$ – reuns Apr 21 '16 at 21:33 • The picture describes the situation perfectly. What is unclear ? The green part is needed to complete the big square, and it is obvious that the sides have length $b/2$ – Peter Apr 21 '16 at 21:35 • After you add the horizontal and vertical rectangles to the $x$ by $x$ square (shown in your first figure), notice that there is a gap (see the arrow in your first figure) that is a square", "(n-9)^2$$ or $n = 18$. For $1 \\leq n < 12$, it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$ We conclude that the answer is $1 + 9 + 10 + 18 = \\boxed{38}.$ ## Solution 4: Graphing If we graphed $$y = \\sqrt{x^2-19x+99}$$ we would see that only four values of x return integer values of y: 10, 9, 1, 18. Thus, the answer if $\\boxed{38}$.", "# Least Squares Approx. If I was using a least squares approximation of the form $y = A_1 + A_2\\sin(wx) + A_3\\cos(wx)$, would you be minimising the function $\\sum_{i=0}^n (y_i - (A_1 + A_2\\sin(wx) + A_3\\cos(wx))^2$ ? I've never tried this for periodic data before! - That's generally the way it's done in the \"least squares\" sense. Except that in your error sum, replace $x$ with $x_i$ ;) – Emily Feb 22 '13 at 15:22 I realised afterwards, and the 0 by a 1. :D – Sanya Feb 22 '13 at 15:31", "the numbers. In $$x^2+21=10x$$ the square of half of $$10x$$ is the square of $${1\\over 2}(x^2+21)$$. This is $$25x^2$$. You subtract from it the product of one of them by the other $$\\left[21x^2\\right],$$ leaving a square of the difference between one of them and half of their sum. Subtracting $$21x^2$$ from $$25x^2$$ leaves $$4x^2$$, which by rule (2) is $$\\left({1\\over 2}|p-q|\\right)^2$$. But by rule (3) this is equal to $$\\left|{1\\over 2}s-p\\right|^2$$. So $$4x^2$$ is $$\\left|{1\\over 2}s-p\\right|^2$$. Because $$p$$ can be either $$x^2$$ or 21, $$4x^2$$ is either $$|5x-x^2|^2$$ or $$|5x-21|^2$$. And because we do not know which of $${1\\over 2}s$$ ($$5x$$) or $$p$$ (21 or $$x^2$$) is greater, $$\\sqrt{4x^2}$$ might be $$x^2-5x$$, $$21-5x$$, $$5x-x^2$$, or $$5x-21$$. He considers the four cases in order: If you take its root to get things $$\\left[2x\\right],$$ and you add it to half of their sum, which is half of the things which are in the equation $$\\left[5x\\right],$$ and you equate that with the māl, it results in the first type $$\\left[7x=x^2\\right].$$ And if you equate it with the number [21] it gives the third type $$\\left[7x=21\\right].$$ And if you subtract [from the $$5x$$] a root of half of their remembered sum $$\\left[2x\\right]$$ and you equate it with the māl, it likewise results in the first type $$\\left[3x=x^2\\right].$$ And if you equate it", "the same arguments we get (x^3-2x^2+2x-1)=(x-1)(x^2-x+1). Combining our equation is (x-1)^2(x^2-x+1) and completing the square gives (x-1)^2((x-1/2)^2+3/4) which is clearly nonnegative. This technique was never taught to me at school. I don't know how many people catch on to this; it was just one of the things I've always done for fast computations. 7. Oct 25, 2007 ### cristo Staff Emeritus Your method is just dividing one polynomial into another (i.e. divide (x-1) into the original polynomial). Instead of writing it down in the long division type way, you've just written it down in words, but it's basically exactly the same method. 8. Oct 25, 2007 ### ZioX I realize this. But the point is you can do it all in your head. Anyways, that's why I called it 'division'. 9. Oct 25, 2007 ### dynamicsolo The completion of the square is really what cinches the proof. (And spares one the nuisance of having to show that x^2 - x >= -1 .) Once again, though, we've been terribly nice and solved the problem. I wonder if the OP has even seen any of this yet... 10. Oct 25, 2007 ### Dick The OP is a Homework Helper. I suspect he's just on vacation or something. I'm not worried about him. 11. Oct 25, 2007 ### dynamicsolo Sorry,", "# Lesson 15: Completing the square for x2+bx+c Introduction Completing the square is quite simply ingenious and you'll find it incredibly useful later on in your studies. Again, we're just going to be changing the form of an expression, this time from $$~x^2+bx+c~$$ to $$~(x+p)^2+q~$$. If you're thinking of taking advanced level maths, you want to be able to do this in your sleep! SECTION A Let's say you're asked to write $$~x^2+10x+22~$$ in the form $$~(x+p)^2+q~$$. The method You need three values: $$⊘~$$ Half the coefficient of $$~x~$$ ⇒ $$~+5~$$ $$■~$$ $$~(+5)^2=25~$$ $$□~$$ The constant term ⇒ $$~+22~$$ Which you arrange like this: $$~(x~~~~~~⊘)^2-■~~~~~~□$$ In this case, $$~(x+5)^2-25+22~$$ Which simplifies to $$~(x+5)^2-3~$$ so $$~p=5~$$ and $$~q=-3~$$ Important The sign in front of $$~■~$$ is ALWAYS a minus sign. Sometimes you're asked to state the values of $$~p~$$ and $$~q~$$ with your answer. We'll do that here just to get you into the habit! They're quite special values actually when it comes to sketching graphs. Anyway, back to it ... Another example Express $$~x^2-8x+21~$$ in the form $$~(x+p)^2+q~$$ Half the coefficient of $$~x~$$ is $$~-4~$$ $$~(-4)^2=16~$$ The constant term is $$~+21~$$ So we have $$~(x-4)^2-16+21~$$ Which simplifies to $$~(x-4)^2+5~$$ where $$~p=-4~$$ and $$~q=5~$$ # Practise to master SECTION A Write these expressions in the form $$~(x+p)^2+q~$$ and state", "# Why does completing the square give you the minimum point? Say we have an equation:$y=$ ${x^2} + 2x + 1$ Completing the square we get: \\eqalign{ & y={x^2} + 2x + 1 \\cr & = {(x + 1)^2} - {(1)^2} + 1 \\cr & = {(x + 1)^2} \\cr} The minimum point of this parabola is (-1,0) What I would like to know is how/why does putting a quadratic equation in completing the square form give you the minimum point of a parabola? What is it about this form that corresponds to give you the minimum point? I hope i've made myself clear, if not please ask me to make myself clearer. Thank you! • No matter what $x$ is, $(x + 1)^2 \\geq 0$. The smallest it can ever get is $0$, which happens when $x = -1$. – littleO Oct 3 '13 at 20:29 The expression $(x+a)^2+b$ attains its minimum at $x=-a$ because $(x+a)^2\\ge 0$ for all real values of $x$ with equality if and only if $x=-a$.", "# Thread: Completing the square/Quadratics 1. ## Completing the square/Quadratics Find the positive constant a and b such that x^4+9/x^4 = (x^2-a/x^2)^2 + b for all non-zero values of x. Hence write down or obtain otherwise, the least possible value of x^4+9/x^4 for real values of x. Please show a step by step approach to the problem. Thanks 2. Originally Posted by umeirhuss Find the positive constant a and b such that x^4+9/x^4 = (x^2-a/x^2)^2 + b for all non-zero values of x. Hence write down or obtain otherwise, the least possible value of x^4+9/x^4 for real values of x. Please show a step by step approach to the problem. Thanks Have you tried expanding the RHS and see how it can be compared to be the LHS? 3. in other words, what is $(x^2- a/x^2)^2$? How does that differ from $(x^4+ 9/x^4)$? 4. i try to but i got stuck and don't know where to go... i got a=3 but no answer for b!! I guess I'm doing something wrong. your help will be much appreciated. Thanks 5. Originally Posted by HallsofIvy in other words, what is $(x^2- a/x^2)^2$? How does that differ from $(x^4+ 9/x^4)$? what about the value of b?", "# What do you need to add to complete the square for x^2+8x? Apr 1, 2018 16 #### Explanation: Since the $8 x$ is positive, the equation ${\\left(x + y\\right)}^{2}$ must have $y$ as a positive number. Let's see this! An easy way to complete the square if you have the constants for ${x}^{2}$ and $x$ but not the non-variable number is to divide the $x$ constant, and then square it. Just like this: $8 x$/$2$$= 4$. ${4}^{2} = 16$. The last number must be 16. Let's check it. ${\\left(x + 4\\right)}^{2} = {x}^{2} + 8 x + \\textcolor{red}{16}$ The answer is $16$. Yay!" ]

[ "Graph $$x^2+4y^2-2x+24y+33 = 0$$. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. Solution Since we have a sum of squares and the squared terms have unequal coefficients, it's a good bet we have an ellipse on our hands.\\footnote{The equation of a parabola has only one squared variable and the equation of a circle has two squared variables with identical coefficients. We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to $$1$$. $\\begin{array}{rclr} x^2+4y^2-2x+24y+33 & = & 0 & \\\\ x^2-2x+4y^2+24y & = & -33 & \\\\ x^2 - 2x + 4\\left(y^2+6y\\right) & = & - 33 & \\\\ \\left(x^2 - 2x +1\\right) + 4\\left(y^2+6y+9\\right) & = & - 33 + 1 + 4(9)& \\\\ (x-1)^2 + 4(y+3)^2 & = & 4 & \\\\[5pt] \\dfrac{(x-1)^2 + 4(y+3)^2}{4} & = & \\dfrac{4}{4} & \\\\[10pt] \\dfrac{(x-1)^2}{4} + (y+3)^2 & = & 1 & \\\\[10pt] \\dfrac{(x-1)^2}{4} + \\dfrac{(y+3)^2}{1} & = & 1 & \\\\ \\end{array}$ Now that this equation is in the standard form of Equation \\ref{standardellipse}, we see that $$x-h$$ is $$x-1$$ so $$h = 1$$, and $$y-k$$ is $$y+3$$ so $$k = -3$$. Hence, our ellipse is centered at $$(1,-3)$$. We see that $$a^2 = 4$$ so", "square or differentiate the parabola. 8. Oct 5, 2014 ### Staff: Mentor Pretty much. Different equations cause different behaviors in the graphs, though. Here are a few examples. 1. y = x2 + 2 -- shift the graph of y = x2 up by two units. There is no need to factor the right side. 2. y = (x - 3)2 -- shift the graph of y = x2 right by three units. Vertex will be at (3, 0). The right side is already factored. 3. y = -x2 + 2x -- Can't tell until it's factored For x-intercepts, set -x2 + 2x = 0, so -x(x - 2) = 0, so x = 0 or x = 2 To find vertex, complete the square on the right side -x2 + 2x = -(x2 - 2x) = -(x2 - 2x + 1) + 1 = (x - 1)2 + 1 This represents a shift to the right by one unit, and a shift up by one unit, so the vertext is at (1, 1). 9. Oct 5, 2014 ### Staff: Mentor Yes and thanks. I absent-mindedly was looking at Dick's work to find the vertex. Obviously if x(x + 3) = 0, the solutions are x = 0 and x = -3. I edited my earlier post to", "# Identify the graph of the given nondegenerate conic sections: 3x^2 + 2y^2 + 12x - 4y + 2 = 0. Identify the graph of the given nondegenerate conic sections: $3{x}^{2}+2{y}^{2}+12x-4y+2=0$. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it insonsipthinye Step 1 A nondegenerate conic section of the form, $A{x}^{2}+B{y}^{2}+Cx+Dy+E=0$ Where A and B both are not zero. 1) A circle if $A=B$ 2) A parabola if $AB=0$ 3) An ellipse if A is not equal to 4) A hyperbola if $AB<0$ Step 2 Given that, $3{x}^{2}+2{y}^{2}+12x-4y+2=0$ Here So A and B are not equal and $AB=\\left(3\\right)\\left(2\\right)=6>0$ Hence the graph of the equation is ellipse.", "# What is the vertex form of y=x^2-2x+6? Mar 2, 2018 In vertex form, the parabola's equation is $y = {\\left(x - 1\\right)}^{2} + 5$. #### Explanation: To convert a parabola in standard form to vertex form, you have to make a squared binomial term (i.e. ${\\left(x - 1\\right)}^{2}$ or ${\\left(x + 6\\right)}^{2}$). These squared binomial terms -- take ${\\left(x - 1\\right)}^{2}$, for example -- (almost) always expand to have ${x}^{2}$, $x$, and constant terms. ${\\left(x - 1\\right)}^{2}$ expands to be ${x}^{2} - 2 x + 1$. In our parabola: $y = {x}^{2} - 2 x + 6$ We have a part that looks similar to the expression we wrote before: ${x}^{2} - 2 x + 1$. If we rewrite our parabola, we can \"undo\" this squared binomial term, like this: $y = {x}^{2} - 2 x + 6$ $\\textcolor{w h i t e}{y} = \\textcolor{red}{{x}^{2} - 2 x + 1} + 5$ $\\textcolor{w h i t e}{y} = \\textcolor{red}{{\\left(x - 1\\right)}^{2}} + 5$ This is our parabola in vertex form. Here's its graph: graph{(x-1)^2+5 [-12, 13.7, 0, 13.12]}", "# How to graph a parabola y=(x-2)^2-3? Jun 18, 2015 The graph has the same shape as y = x^2, but there are some shifts #### Explanation: Replacing $x$ with $x - 2$ makes $x = 2$ act in the new equation just like $x = 0$ did in the old one. (That is where I would find ${0}^{2}$.) That shifts the graph $2$ to the right. Compare $y = {x}^{2}$ and $y + 3 = {\\left(x - 2\\right)}^{2}$ Replacing $x$ with $x + 2$ moves the graph $2$ in the positive $x$ direction (2 to the right.) What do we expect to happen when we replace $y$ with $y + 3$? If you said \"move the graph $3$ in the negative $y$ direction (3 downward)\", then you are right! A different way of thinking about it: After we find the square, what do we do? In $y = {x}^{2}$ we're done, that is the $y$ value. In $y = {\\left(x - 2\\right)}^{2}$, after we square, we are done, that is the $y$ value. In $y = {\\left(x - 2\\right)}^{2} + 3$, after we square, we still need to subtract 3 from the number, that moves us down 3. The vertex of $y = {x}^{2}$ is the point $\\left(0 , 0\\right)$ The vertex of $y = {\\left(x -", "1)2 = 22 d. ($x$ - 2)2 + ($y$ - 1)2 = 22 Solution: x2 + y2 - 4x + 2y + 1 = 0 [Equation of conic section.] x2 - 4x + y2 + 2y = - 1 [Group x and y - terms.] (x2 - 4x + ?) + (y2 + 2y + ?) = - 1 (x2 - 4x + 4) + (y2 + 2y + 1) = - 1 + 4 + 1 (x - 2)2 + (y + 1)2 = (2)2 [Complete the squares.] So, the standard form of the given equation is (x - 2)2 + (y + 1)2 = (2)2, which represents a circle with center (2, - 1) and radius 2. 8. What conic section is represented by the equation 9$x$2 + 4$y$2 - 18$x$ - 8$y$ - 23 = 0? a. a circle b. a parabola c. a hyperbola d. an ellipse Solution: 9x2 + 4y2 - 18x - 8y - 23 = 0 [Equation of conic section.] 9x2 - 18x + 4y2 - 8y = 23 [Group x - terms and y - terms.] 9(x2 - 2x) + 4(y2 - 2y) = 23 9(x2 - 2x + ?) + 4(y2 - 2y + ?) = 23 9(x2 - 2x + 1) + 4(y2 - 2y + 1) =", "to be a hyperbola. However, when we complete squares, we find that the equation becomes $2\\left(x-3\\right)^2-3\\left(y-5\\right)^2=0$2(x3)23(y5)2=0. Without going into the algebra, the left hand side can be factorised into the difference of two squares and each factor can be brought to zero, so that the equation represents two straight lines with equations given by $\\sqrt{2}\\left(x-3\\right)-\\sqrt{3}\\left(y-5\\right)=0$2(x3)3(y5)=0 and $\\sqrt{2}\\left(x-3\\right)+\\sqrt{3}\\left(y-5\\right)=0$2(x3)+3(y5)=0 as shown here: Thus it is a necessary, but not sufficient condition that $AB$AB to be negative for the equation to represent a hyperbola. #### Mathspace Worked Examples ##### Question 1 What is the graph of a conic section represented by the equation $x^2-10x-y=0$x210xy=0? 1. Parabola A Ellipse B Hyperbola C Circle D Parabola A Ellipse B Hyperbola C Circle D ##### Question 2 What type of graph does the equation $4x^2+32x-9y^2+54y-53=0$4x2+32x9y2+54y53=0 represent? 1. Circle A Ellipse B Parabola C Hyperbola D Circle A Ellipse B Parabola C Hyperbola D ##### Question 3 What type of graph does the conic section represented by the equation $\\frac{x^2}{9}+\\frac{y^2}{4}=1$x29+y24=1 have? 1. Hyperbola A Ellipse B Parabola C Circle D Hyperbola A Ellipse B Parabola C Circle D ### Outcomes #### M8-1 Apply the geometry of conic sections #### 91573 Apply the geometry of conic sections in solving problems", "333. Find two consecutive even numbers whose product is 360. 334. The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.) For each parabola, find which ways it opens, the axis of symmetry, the vertex, the x- and y-intercepts, and the maximum or minimum value. 335. $y = 3 x 2 + 6 x + 8 y = 3 x 2 + 6 x + 8$ 336. $y = x 2 − 4 y = x 2 − 4$ 337. $y = x 2 + 10 x + 24 y = x 2 + 10 x + 24$ 338. $y = −3 x 2 + 12 x − 8 y = −3 x 2 + 12 x − 8$ 339. $y = − x 2 − 8 x + 16 y = − x 2 − 8 x + 16$ Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry. 340. $y = 2 x 2 + 6 x + 2 y = 2 x 2 + 6 x + 2$ 341. $y = 16 x 2 + 24 x + 9 y = 16 x 2 + 24", "and remove the lower half. To draw “–sinx”, we take image of y=sinx in x-axis. From the graph, we see that |sinx| is greater than “-sinx” in (0,π). Note that end points are not included. The domain is written with general notation as : $x\\in \\left(2n\\pi ,\\left(2n+1\\right)\\pi \\right)$ Problem : Determine graphically the points where graphs of $|y|={\\mathrm{log}}_{e}|x|$ and ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ intersect each other. Solution : The function $|y|={\\mathrm{log}}_{e}|x|$ is obtained by transforming $y=\\mathrm{log}{}_{e}x$ . To draw $y={\\mathrm{log}}_{e}|x|$ , we need to remove left half (but here there is no left half) and take image of right half in y-axis. To draw $|y|=\\mathrm{log}{}_{e}|x|$ , we transform the graph of $y=\\mathrm{log}{}_{e}|x|$ . For this, we remove the lower half and take image of upper half in x-axis. On the other hand, ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ is a circle with center at 1,0 having radius of 2 units. Finally, superposing two graphs, we determine the intersection points. Clearly, there are three intersection points as shown by solid circles. ## Exercises Draw the graph of function given by : $f\\left(x\\right)=\\frac{1}{\\left[x\\right]-1}$ Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw 1/x-1. It is a hyperbola shifted right by 1 unit. Its center is (1,0). Remove left half and take the image of right half in y-axis. 2. Draw the graph of", "and remove the lower half. To draw “–sinx”, we take image of y=sinx in x-axis. From the graph, we see that |sinx| is greater than “-sinx” in (0,π). Note that end points are not included. The domain is written with general notation as : $x\\in \\left(2n\\pi ,\\left(2n+1\\right)\\pi \\right)$ Problem : Determine graphically the points where graphs of $|y|={\\mathrm{log}}_{e}|x|$ and ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ intersect each other. Solution : The function $|y|={\\mathrm{log}}_{e}|x|$ is obtained by transforming $y=\\mathrm{log}{}_{e}x$ . To draw $y={\\mathrm{log}}_{e}|x|$ , we need to remove left half (but here there is no left half) and take image of right half in y-axis. To draw $|y|=\\mathrm{log}{}_{e}|x|$ , we transform the graph of $y=\\mathrm{log}{}_{e}|x|$ . For this, we remove the lower half and take image of upper half in x-axis. On the other hand, ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ is a circle with center at 1,0 having radius of 2 units. Finally, superposing two graphs, we determine the intersection points. Clearly, there are three intersection points as shown by solid circles. ## Exercises Draw the graph of function given by : $f\\left(x\\right)=\\frac{1}{\\left[x\\right]-1}$ Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw 1/x-1. It is a hyperbola shifted right by 1 unit. Its center is (1,0). Remove left half and take the image of right half in y-axis. 2. Draw the graph of", "and remove the lower half. To draw “–sinx”, we take image of y=sinx in x-axis. From the graph, we see that |sinx| is greater than “-sinx” in (0,π). Note that end points are not included. The domain is written with general notation as : $x\\in \\left(2n\\pi ,\\left(2n+1\\right)\\pi \\right)$ Problem : Determine graphically the points where graphs of $|y|={\\mathrm{log}}_{e}|x|$ and ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ intersect each other. Solution : The function $|y|={\\mathrm{log}}_{e}|x|$ is obtained by transforming $y=\\mathrm{log}{}_{e}x$ . To draw $y={\\mathrm{log}}_{e}|x|$ , we need to remove left half (but here there is no left half) and take image of right half in y-axis. To draw $|y|=\\mathrm{log}{}_{e}|x|$ , we transform the graph of $y=\\mathrm{log}{}_{e}|x|$ . For this, we remove the lower half and take image of upper half in x-axis. On the other hand, ${\\left(x-1\\right)}^{2}+{y}^{2}-4=0$ is a circle with center at 1,0 having radius of 2 units. Finally, superposing two graphs, we determine the intersection points. Clearly, there are three intersection points as shown by solid circles. ## Exercises Draw the graph of function given by : $f\\left(x\\right)=\\frac{1}{\\left[x\\right]-1}$ Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw 1/x-1. It is a hyperbola shifted right by 1 unit. Its center is (1,0). Remove left half and take the image of right half in y-axis. 2. Draw the graph of", "### Home > PC > Chapter 7 > Lesson 7.2.4 > Problem7-81 7-81. Complete the square (if necessary) to find the center and radius of the circle. Then sketch the graph. 1. $x^{2} + y^{2} − 10x + 8y + 5 = 0$ Regroup with $x'\\text{s}$ and y's together and the constant on the other side. $\\left(x^{2} − 10x \\:+\\: ?\\right) + \\left(y^{2} + 8y \\:+\\: ?\\right) = − 5$ Complete the square finding the ?'s and adding the same amount on both sides. $\\left(x^{2} − 10x + 25\\right) + \\left(y^{2} + 8y + 16\\right) = − 5 + 25 + 16$ Simplify and graph. $\\left(x − 5\\right)^{2} + \\left(y + 4\\right)^{2} = 36$ 1. $x^{2} + y^{2} − 8x + 6y − 56 ≤ 0$ Follow the same thinking process as part (a).", "+0 # Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, 0 126 1 +979 Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. Enter the most specific answer. $$\\left(\\frac x2 - 3\\right)^2 + y^2 = 10$$ Nov 30, 2020 #1 +133 +1 I would say the graph is an ellipse. Here is the graph: Nov 30, 2020", "If ax^2 +2hxy+by^2=0 be the two sides of a ||gm and px+qy=1 be its one diagonal then prove that the other diagonal is y(bp-hq)=x(aq-hp)? • If ax^2 +2hxy+by^2=0 be the two sides of a ||gm and px+qy=1 be its one diagonal then prove that the other diagonal is y(bp-hq)=x(aq-hp)? The equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ represents a pair of parallel lines. Prove that the equation of the line mid way between the two parallel lines ... Positive: 65 % In a rectangular hyperbola x 2 – y 2 = a 2 , prove ... of the hyperbola x 2 - y 2 = 9, then the ... squares of its distances from the other two sides ... Positive: 62 % More resources ... circle of the ellipse. If $$a>b,$$ then $$x^2+y^2 ... ax^{2}+2hxy+by^{2}+2gx+2fy+c=0,$$ then for ... passing through one of the focii. There are two ... Positive: 65 % Question bank for Class 10 - Optional Mathematics ... second degree ax 2 + 2hxy + by 2 = 0 always ... of its two diameters are 2x - y = 5 and x ... Positive: 60 % MATHEMATICS SEMESTER SYSTEM MATHEMATICS July, ... one from each topic, ... Angle between two lines given by ax2 + 2hxy + by2 = 0. $$ax^2+2hxy+by^2=0$$ is $$\\dfrac{x^2-y^2}{a-b ... and$$Q(x,y)$$be its image about x-axis, then$$x=\\alpha", "Then determine whether the function is even, odd, or neither. Question 47. h(x) = $$\\frac{6}{x^{2}+1}$$ Answer: Question 48. f(x) = $$\\frac{2 x^{2}}{x^{2}-9}$$ Answer: Question 49. y = $$\\frac{x^{3}}{3 x^{2}+x^{4}}$$ Answer: Question 50. f(x) = $$\\frac{4 x^{2}}{2 x^{3}-x}$$ Answer: Question 51. MAKING AN ARGUMENT Your friend claims it is possible for a rational function to have two vertical asymptotes. Is your friend correct? Justify your answer. Answer: Question 52. HOW DO YOU SEE IT? Use the graph of f to determine the equations of the asymptotes. Explain. Answer: Question 53. DRAWING CONCLUSIONS In what line(s) is the graph of y = $$\\frac{1}{x}$$ symmetric? What does this symmetry tell you about the inverse of the function f(x) = $$\\frac{1}{x}$$? Answer: Question 54. THOUGHT PROVOKING There are four basic types of conic sections: parabola, circle, ellipse, and hyperbola. Each of these can be represented by the intersection of a double-napped cone and a plane. The intersections for a parabola, circle, and ellipse are shown below.Sketch the intersection for a hyperbola. Answer: Question 55. REASONING The graph of the rational function f is a hyperbola. The asymptotes of the graph of f intersect at (3, 2). The point (2, 1) is on the graph. Find another point on the graph. Explain your reasoning. Answer: Question 56. ABSTRACT REASONING Describe the intervals where", "intervals where $${{x}^{2}}>4$$, we need to see where the curved graph (parabola) is greater or higher than the line. We can see that this is $$\\displaystyle \\left( {-\\infty ,-2} \\right)\\cup \\left( {2,\\infty } \\right)$$. (Remember that our answer will be in intervals along the $$x$$-axis.) $$2{{x}^{2}}-7x\\le -3$$ Let’s do the same here by putting $$2{{x}^{2}}-7x$$ in $${{Y}_{1}}=$$ and –3 in $${{Y}_{2}}=$$, and graph. You can use the TRACE button (then arrows) to move closer to each intersection before finding that intersection. Then can use the Intercept function to find the points of intersection: Push “2nd TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER, ENTER, ENTER. Since we want the intervals where $$2{{x}^{2}}-7x\\le -3$$, we need to see where the curved graph (parabola) is less than or lower than the line. We can see that this is $$\\displaystyle \\left[ {.5\\,\\,,3} \\right]$$. (Remember that our answer will be in intervals along the $$x$$-axis.) # Solving Algebraically, including Completing the Square When solving algebraically, take the square root of each side, but we need to worry about the inequality sign. Break the equation into two equations like we did in the Solving Absolute Value Inequalities section (one with a plus, one with a minus), but", "### Home > PC > Chapter 7 > Lesson 7.2.2 > Problem7-51 7-51. Completing the square is also possible when the coefficient of the $x^{2}$ term is not $1$. Work your way through this problem to see how. \\begin{align*}y&=3x^2 -18x+1\\\\y-1&=3(x^2 -6x)\\end{align*} 1. What number needs to be added to both sides of the equation so that the expression inside the parentheses is a perfect square? Hint: $9$ is the wrong answer. The outside $3$ will distribute with whatever is inside the ( )'s. 2. Finish completing the square and find the vertex of the parabola. $y − 1 = 3\\left(x^{2} − 6x + 9\\right) − 27$", "The equation $$x^2=6y^2+8$$ is equivalent to \\begin{gather*} x^2-6y^2=8\\qquad \\text{ or }\\qquad \\frac{x^2}{\\left( \\sqrt{8} \\right)^2} - \\frac{y^2}{\\left( \\frac{2}{\\sqrt{3}} \\right)^2}=1 \\end{gather*} The graph is a hyperbola that opens left and right. 2. The equation $$x^2=\\dfrac{4-y^2}{2}$$ is equivalent to The graph is an ellipse with major axis on the $$y$$-axis because $$2\\gt\\sqrt{2} \\text{.}$$ Write each equation in standard form and describe its graph. 1. $$4y^2=x^2-8$$ Standard form: $$=1$$ The graph is • a circle • an ellipse • a hyperbola • a parabola centered at the point , with vertices . There • are • are not asymptotes: $$y=\\pm$$. 2. $$4x^2+y=0$$ Standard form: $$y=$$ The graph is • a circle • an ellipse • a hyperbola • a parabola with vertex at the point , opening • left • right • upward • downward . $$\\frac{x^{2}}{8}-\\frac{y^{2}}{2}$$ $$\\text{a hyperbola}$$ $$\\left(0,0\\right)$$ $$\\left(-2\\sqrt{2},0\\right), \\left(2\\sqrt{2},0\\right)$$ $$\\text{are}$$ $$\\frac{1}{2}x$$ $$-4x^{2}$$ $$\\text{a parabola}$$ $$\\left(0,0\\right)$$ $$\\text{downward}$$ Solution. 1. $$\\dfrac{x^2}{8}-\\dfrac{y^2}{2}=1 \\text{.}$$ A hyperbola centered at the origin with vertices $$\\left(2\\sqrt{2},0 \\right)$$ and $$\\left(-2\\sqrt{2},0 \\right) \\text{,}$$ and asymptotes $$y = \\dfrac{1}{2}x$$ and $$y = -\\dfrac{1}{2} x\\text{.}$$ 2. $$y=-4x^2\\text{.}$$ A parabola opening downward with vertex at the origin. ### Subsection9.4.3Translated Hyperbolas The standard form for the equations of hyperbolas centered at the point $$(h,k)$$ can be derived using the distance formula and the definition of hyperbola. #### Hyperbolas. The equation for a", "+0 # help 0 145 1 Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. $x^2 + 2y^2 - 6x - 20y + 59 = 12$ Dec 9, 2018 #1 +1 It is an ellispe because putting it into conic form we have $$(x^2-6x+9)+2(y^2-10y+25)=12$$ $$(x-3)^2+2(y-5)^2=12$$ $$\\frac{(x-3)^2}{12}+\\frac{(y-5)^2}{6}=1$$ you can also put it into desmos to see the graph of it Dec 9, 2018 #1 +1 It is an ellispe because putting it into conic form we have $$(x^2-6x+9)+2(y^2-10y+25)=12$$ $$(x-3)^2+2(y-5)^2=12$$ $$\\frac{(x-3)^2}{12}+\\frac{(y-5)^2}{6}=1$$ you can also put it into desmos to see the graph of it Guest Dec 9, 2018", "that $(xa-e)^2 +(y-0)^2 = e^2\\left(x-\\dfrac ae \\right)^2$ expanding gives us x2 − 2aex + a2e2 + y2 = e2x2 − 2aex + a2 factorizing gives us (e2 − 1)x2y2 = a2(e2 − 1) Hence$\\dfrac{x^2}{a^2} - \\dfrac{y^2}{a^2(e^2-1)} = 1$ Since a2(e2 − 1) > 0 we may replace it by b2, where b is a constant, to obtain the standard form of the hyperbola: $\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1$(24) i As with the ellipse it follows from the symmetry that there is a second focus and a second directrix which will produce the same curve. The two foci are the points F(ae, 0) and F′(−ae, 0) and the two directrices have equations x = a/e and x = −a/e. Notice from Figures 4d and 15 that the hyperbola has y two distinct branches: this is quite different from an ellipse which is always a closed curve. We can readily verify for Equation 24 that when y = 0, x = ± a (giving the points A′ and A in Figure 15), but there is no value of y for which x = 0; indeed, since Equation 24 can be rearranged to give the equation $\\dfrac{y^2}{b^2} = \\dfrac{x^2-a^2}{a^2}$ in which the left–hand side is positive, it follows that x2a2, so there is no part of the curve in the band" ]