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Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ What do we do now? If we say $y=x-\dfrac{1}{x}$, we won't be able to express $\left(x^2-\dfrac{1}{x^2}\right)$ in terms of $y$ because of the minus sign as $(a-b)^2=a^2-2ab\color{red}{+b^2}$. On the other side, $$y^3=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)\\x^3-\dfrac{1}{x^3}=y^3+3y$$ I think this is a traditional issue when solving reciprocal equations, but I can't figure out how to deal with it.
The equation is:- $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Continuing the given method we get:- $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ $$\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right)-2\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ $$\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+4\right)=0$$ Thus we see $\left(x-\dfrac{1}{x}\right)=0$ is a solution. So $$x^2-1=0\\\Rightarrow x^2=1\\\Rightarrow x=\pm1$$ On the other hand:- $$\left(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+4\right)=0$$ Denoting $x+\dfrac{1}{x}$ as $a$ the equation becomes:- $$a^2-2a+2=0$$ Thus $$a = \dfrac{2 \pm \sqrt{4-4\cdot2}}{2}$$ $$\Rightarrow x+\dfrac{1}{x} = \dfrac{2 \pm 2i}{2}$$ $$\Rightarrow \dfrac{x^2+1}{x} = {1 \pm i}$$ Thus we get 2 equations:- $x^2-(1+i)x+1=0$ and $x^2-(1-i)x+1=0$ Thus $$x = \dfrac{1 \pm \sqrt{(1+i)^2-4}}{2}\\x = \dfrac{1 \pm \sqrt{(1-i)^2-4}}{2}$$ Thus the solutions to the equation are:- $\pm1$,$\dfrac{1 \pm \sqrt{(1+i)^2-4}}{2}$ and $\dfrac{1 \pm \sqrt{(1-i)^2-4}}{2}$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4522572", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 3, "answer_id": 1 }
Reciprocal sum of triangular numbers is 1. n∈N Triangular numbers set is $$\{1,3,6,10,15,21,28,36,45,55,\dots,\frac{n(n+1)}2,\dots\}$$ n≥3 I solved the equation $$\frac 1{x_1} +\frac1{x_2} +\cdots+\frac1{x_n} =1$$ for triangular numbers: $$\begin{align}n&=3,(3,3,3)\\ n&=4,(3,3,6,6)\\ n&=5,(3,6,6,6,6)\\ n&=6,(6,6,6,6,6,6)\\ n&=7,(3,6,10,10,10,10,10) \text{ and } \\ &\quad(3,3,15,15,15,15,15)\\ n&=8,(3,6,6,15,15,15,15,15)\\ n&=9,(6,6,6,6,15,15,15,15,15)\\ n&=10,(10,10,10,10,10,10,10,10,10,10)\end{align}$$ Question is, how can I find the solution of the equation for triangular numbers for other $n$?
Not an answer, just expanding on my ideas in comments. Let $x_i=\frac{m_i(m_i+1)}{2}.$ Then $\frac{1}{x_i}=\frac{2}{m_i(m_i+1)}=\frac2{m_i}-\frac2{m_i+1}.$ Then your equation can be written as: $$\sum_{i=1}^{n} \frac{1}{m_i}=\frac 12+\sum_i \frac1{m_i+1}$$ If the largest $m_i$ has $m_i+1=p^k>2$ for some prime $p,$ then the number of $i$ with $m_i=p^k-1$ must be a multiple of $p.$ This is because the right side will still have $p^k$ as a factor in the denominator if not, and it can't be a factor on the left side. If the largest $m_i$ has $m_i=p^k>2$ then the number of $m_i=p^k$ must be congruent to the number of $m_i=p^k-1$ modulo $p.$ So when the largest $x_i=10,m_i=4$ then the number of $10$s must be congruent to the number of $6$s modulo $2,$ and the number of $10$s must be a multiple of $5.$ If $m_i=5,x_i=15$ is the largest, the number of $15$s must be congruent to the number of $10$s modulo $5.$ We have $a$ $10s$ and $5b+a$ $15?$s, you get the sum of those is: $$a\left(\frac1{10}+\frac1{15}\right)+\frac{5b}{15}=\frac a6+\frac b3.$$ When $a+2b\leq 6,$ you get solutions: $((6-a-2b)\times 6,a\times 10,(a+5b)\times 15).$ This gives an answer for $n=6-a-2b+a+a+5b=6+a+3b.$ When $6-a-2b>2,$ we can replace any pairs of sixes with a $3.$ So when $a=2,b=1,$ we get another solution for $n=10,$ $x_*(3,2\times 10,7\times 15)$ or $m_*=(2,2\times 4,7\times 5).$ When the largest $m_i=6, x_i=21$ then there must be a multiple of $7$ of them. Then you get $\frac{7b}{21}=\frac b3.$ You can $\frac13$ and $\frac23$ in quite a few ways with the smaller $m_i.$ $m_i=7$ is interesting because $m_i$ and $m_i+1$ are prime powers, so if the largest $x_i=28,$ then the number of $28$s must be even and congruent to the number of $21$s modulo $7.$ If there are $a$ $21$s] and $a+7b$ $28$s, then $a+b$ must be even, the partial sum $\frac{a}{12}+\frac{b}{4}=\frac{a+3b}{12}=\frac{b+\frac{a+b}2}6$ and thus we can make thirds and sixths out of the smaller $m_i.$ But this approach will really only work for small $m_i.$ Prime powers became rarer when $m$ increases. When $m_i=14,$ neither $m_i$ nor $m_i+1$ is a prime power. You might still be able to do something with that, but it gets more complicated. Solutions for all $n>2:$ If $T(k)=\frac{k(k+1)}2,$ we have: $$\sum_{k=2}^{m-1}\frac1{T(k)}=1-\frac{2}{m},$$ we can add $m+1$ copies of $\frac1{T(m)}$ to get a total of $1,$ yielding a solution $(T(2),T(3),\dots,T(m-1),(m+1)\times T(m))$ for $n=2m-1.$ That gives a solution for all odd $n.$ For $n=11,$ this becomes $(3,6,10,15,7\times 21).$ We can get a solution for $n=2m$ by first solving for $2m-1,$ then replacing the $3$ with two $6$s. So we can find a solution for all $n>2.$ Since $\sum_{m=2}^{\infty}\frac{1}{T(m)}=1,$ there is no $n>2$ for which there no repeating values. For $n$ odd, we see there is a solution with only one repeated value. For $n=2m$ even, our solution has potentially two repeated. values, $T(3)$ and $T(m).$ There are cases when $n=T(k)$ is even where we have a solution with only one value, such as $(6,6,6,6,6,6)$ and $(10\times 10).$ But are those the only even cases where all but one value occurs only once?
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable function, what is the maximum value of $k$, provided that $b+c=a$? $1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$ In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$, $$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$ Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
Equating $g(x)$ and $g'(x)$ and using $c=a-b$ gives $$ax^2+x(b-2a)=2b-a.$$ Completing the square: $$(2ax+b-2a)^2=4a(2b-a)+(b-2a)^2=b^2+4ab.$$ We must have only one solution, that is $b=0$ or $b=-4a$. From $(2ax+b-2a)^2=0$ we get $x = 1$ or $x = 3$.
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Can we show algebraically that this is true? This problem is part of a larger proof I've been working on. Long story short, I'm trying to prove a particular explicit formula for the recursive function $$ t(p, 0) = 0 \\ t(p, k+1) = \left\lfloor \dfrac{t(p, k)}{3} \right\rfloor + p $$ where $p=2n+1, n \in \mathbb{N}$. The inductive step of proving my particular explicit formula includes the following question: Can we show algebraically that $$ \left\lfloor \dfrac{1}{3} - \dfrac{1}{3} \left\lfloor \dfrac{2n-1}{2(3^{k-1})} + \dfrac{1}{2} \right\rfloor \right\rfloor \ \ = \ \ (-1) \left\lfloor \dfrac{2n-1}{2(3^k)} + \dfrac{1}{2} \right\rfloor$$ is true for all $k,n = 1, 2, 3, ...$ ? I used Python to test it and it seems to be true, at least for all $k≤1000, \ n≤1000$. Can anyone prove this rigorously? I tried using induction on $k$ but was not successful. I did not try induction on $n$. Any help is appreciated.
Let $q = \dfrac{2n-1}{2\cdot 3^k}$, and so $\dfrac{2n-1}{2\cdot 3^{k-1}} = 3q$. Both $q$ and $3q$ are is always positive and never an integer for $k,n = 1,2,3, \ldots$. Converting the RHS into an inequality, $$\begin{array}{rcl} \left(q+\dfrac12\right) -1 <& \left\lfloor q+\dfrac12\right\rfloor &\le q+\dfrac12\\ -q+\dfrac12 >& -\left\lfloor q+\dfrac12\right\rfloor &\ge -q-\dfrac12 \end{array}$$ There is exactly one integer in this range. From the LHS, consider the content inside the outer $\lfloor \ \rfloor$, $$\begin{align*} \frac13 - \frac13\left\lfloor 3q+\frac12\right\rfloor &= -\frac13\left\lfloor-1+3q+\frac12\right\rfloor\\ &= -\frac13\left\lfloor3q-\frac12\right\rfloor \end{align*}\\ \begin{array}{rcl} \left(3q-\dfrac12\right)-1<& \left\lfloor3q-\dfrac12\right\rfloor &\le 3q-\dfrac12\\ -q+\dfrac12>& -\dfrac13\left\lfloor3q-\dfrac12\right\rfloor &\ge -q+\dfrac16\\ \end{array}$$ The middle $-\frac13\left\lfloor3q-\frac12\right\rfloor$ is a third of an integer. That means its floor satisfy a tighter inequality: $$\begin{array}{rcl} \left(-\dfrac13\left\lfloor3q-\dfrac12\right\rfloor\right) -\dfrac23 \le& \left\lfloor-\dfrac13\left\lfloor3q-\dfrac12\right\rfloor\right\rfloor &\le -\dfrac13\left\lfloor3q-\dfrac12\right\rfloor\\ \left(-q+\dfrac16\right) -\dfrac23 \le& \left\lfloor-\dfrac13\left\lfloor3q-\dfrac12\right\rfloor\right\rfloor &< -q+\dfrac12\\ -q -\dfrac12 \le& \left\lfloor-\dfrac13\left\lfloor3q-\dfrac12\right\rfloor\right\rfloor &< -q+\dfrac12\\ \end{array}$$ There is also exactly one integer in this range, and this matches the range of the RHS. So $$LHS = \left\lfloor\frac13 - \frac13\left\lfloor \dfrac{2n-1}{2\cdot 3^{k-1}}+\frac12\right\rfloor\right\rfloor = -\left\lfloor \dfrac{2n-1}{2\cdot 3^k}+\frac12\right\rfloor = RHS.$$
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions: Let $x, y, z \geq 0$ with $x+y+z=3$. Show that $$ x y+y z+z x-x y z \leq \frac{9}{4}. $$ and also find out when the equality holds. I started by plugging in $z=3-x-y$ on the LHS and got $$ 3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-(y^2+x^2)+3x-4xy+x^2y+xy^2\leq 3y-((y+x)^2)+3x-4xy+x^2y+xy^2 $$ But this got me nowhere. Then I started again with the left hand side $$ x y+y z+z x-x y z \Leftrightarrow yz(1-x)+xy+zx $$ and $x+y+z=3 \Leftrightarrow y+z-2=1-x$ so $$ yz(y+z-2)+x(y+z) $$ But this also leaves no idea. Do I have to use a known inequality?
Assume without loss of generality that $x \le 1$. Then, $$xy+yz+zx-xyz = x(y+z) + yz(1-x) \le x(y+z) + \left(\frac{y+z}{2}\right)^2(1-x).$$ This is equal to $$(3-x)\left( x + \frac{(1-x)(3-x)}{2} \right). $$ It is not too difficult to check that this is maximal when $x = 0$ (and $y=z=3/2)$, where it attains a value of $9/4$.
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Integer solution of $a^2+b^2=c^2+d^2$ that relates $a,b,c$, and $d$ explicitly. I have found that the integer solution of $a^2+b^2=c^2+d^2$ is $(a,b,c,d)=(pr+qs,ps-qr,pr-qs,ps+qr)$ for integer $p,r,q,s$. I wonder if there is an explicit relation between $a,b,c,$ and $d$? Or could you give me a hint on what topics should I learn to find this relation if it exists? Thank you for all your help.
From https://sites.google.com/site/tpiezas/003 (10. Form $mx^2+ny^2=mz^2+nt^2$-S. Realis(complete): $$ u^2+nv^2=x^2+ny^2 \\ \Leftrightarrow \small \{ u, v, x, y \}=\{ a^2-n(a-b)^2+n(a-c)^2, b^2-(a-b)^2+n(b-c)^2, a^2+nb^2-nc^2, c^2-(a-c)^2-n(b-c)^2 \}. $$ Substituting $n=1$; $$ u^2+v^2=x^2+y^2 \\ \Leftrightarrow \small \{ u, v, x, y \}=\{ a^2-(a-b)^2+(a-c)^2, b^2-(a-b)^2+(b-c)^2, a^2+b^2-c^2, c^2-(a-c)^2-(b-c)^2 \}. $$ So, this is the generalized form; $\small \{a, b, c, d\}=\{m^2-(m-n)^2+(m-l)^2, n^2-(m-n)^2+(n-l)^2, m^2+n^2-l^2, l^2-(m-l)^2-(n-l)^2\}.$
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$ Problem: Find the all possible values of $a$, such that $$4x^2-2ax+a^2-5a+4>0$$ holds $\forall x\in (0,2)$. My work: First, I rewrote the given inequality as follows: $$ \begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned} $$ Then, we have $$ \begin{aligned} 0<x<2\\ \implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned} $$ Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$. This leads, $$ \begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\ \implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned} $$ For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$. Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$. We have: $$ \begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\ \implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned} $$ Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$. Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$. This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$ Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$. Finally, we have to combine all the solution sets we get. I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
Your attempt is mostly correct, apart from a few fairly minor issues regarding an open set being used, and there's some work you didn't need to do. First, as you pointed out, the inequality can be rewritten as $$f(x) = \left(2x-\frac{a}{2}\right)^2 + \frac{3a^2}{4} - 5a + 4 \gt 0 \tag{1}\label{eq1A}$$ For any fixed value of $a$, then $\frac{3a^2}{4} - 5a + 4 = c$ is a constant. Also, with $g(x) = 2x-\frac{a}{2}$, then \eqref{eq1A} can be rewritten as $$f(x) = g^2(x) + c \gt 0 \tag{2}\label{eq2A}$$ To have \eqref{eq2A} being true $\forall \, x \in (0,2)$, with $$m = \min_{x\in[0,2]}g^2(x) \tag{3}\label{eq3A}$$ requires that $m + c \ge 0$ where $g(0) = 0 \lor g(2) = 0 \lor m \neq 0$ (i.e., points $1$ and $2$ below), and $m + c \gt 0$ otherwise (i.e., where $g(x) = 0$ for an $x \in (0,2)$, as described in point $3$ below). The first case is because the minimum value $m$ is not actually reached on the open set $(0,2)$, i.e., $g^2(x) \gt m \; \forall \; x \in (0,2)$, so $g^2(x) + c \gt 0$ means that $m + c \ge 0$. To determine this minimum value $m$ in \eqref{eq3A} involves $3$ basic cases: * *$g(x) \lt 0 \; \forall \; x \in (0,2)$, which is your case $1$, with $m$ coming from the upper limit value of $g(x)$, i.e., $m = (4 - \frac{a}{2})^2$. *$g(x) \gt 0 \; \forall \; x \in (0,2)$, which is your case $2$, with $m$ coming from the lower limit value of $g(x)$, i.e., $m = (-\frac{a}{2})^2$. *$g(x)$ having both negative and positive values (so $0$ as well), which is basically your case $3$ (except you should've used $\gt$ instead of $\ge$ to avoid overlapping with your previous cases, as well to ensure that $g(x) = 0$ for some $x \in (0,2)$), with $m = 0$. Regarding your analysis of these cases, note that $g(x)$ is a strictly increasing function, with its upper limit value of $4 - \frac{a}{2}$ being $4$ more than its lower limit value of $-\frac{a}{2}$ (and, correspondingly, the lower limit is $4$ less than the upper limit). Thus, for example, with your case $1$, having $4 - \frac{a}{2} \le 0 \; \to \; 4 \le \frac{a}{2} \; \to \; a \ge 8$, i.e., a stronger condition than your $a \ge 0$, so this didn't need to be mentioned. Further, as I explained in my point $1$, you only needed to deal with handling $4 - \frac{a}{2}$. Similarly, with your case $2$, having $a \le 0 \; \to \; -\frac{a}{2} \ge 0$ means that $4-\frac{a}{2} \ge 4$, which is a stronger condition than your $4-\frac{a}{2} \ge 0$. Finally, as you wrote, the solution is obtained by combining (more precisely, taking the union of) the sets of values of $a$ resulting from each of your $3$ cases.
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Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this: $$\int_1^{\infty} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x} } \right)dx= \frac{4}{15} \left(\sqrt{26\sqrt{2}-14 }-2\right) \approx 0.739132$$ I did a few substitutions but it didn't turn out as an easy calculation. I remember that eliminating square roots required some Euler type substitutions. Any ideas of how one can arrive at such a surprising result?
To compute \begin{align}I=\int_{0}^{1}f(x)\;dx\end{align} with $f(x)=\sqrt{\sqrt{x+1}-\sqrt{x}}$ you can use change of variables: $x=\cot^2(u)$ then $I=\int_{\pi/8}^{\pi/4} g(u) du$ where \begin{align}g(u)=2\cot(2u)\left(2+2 \cot^2 ( 2u) \right) \sqrt {{\frac {1-\cos(2u)}{\sin \left( 2\,u \right) }}}=4{\frac {\cos \left( 2\,u \right) }{ \sin^3(2u)}\sqrt {{\frac {\sin \left( u \right) }{\cos \left( u \right) }}}} \end{align}. Now another change of variables, $u=\arctan(v^2)$ makes \begin{align} I=\int_{\sqrt{\sqrt2 -1}}^{1} \frac{(1-v^4)(1+v^4)}{v^4} \;dv =-{\frac{8}{15}}+{\frac { \left( 4\,\sqrt {2}+24 \right) \sqrt {\sqrt { 2}-1}}{15}} \end{align} since $\tan\frac\pi 8=\tan\left(\frac 12 \frac \pi4\right)=\sqrt2 -1$.
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Rationalize the denominator of $\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$ Rationalize the denominator of $$\dfrac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$$ Usually we are supposed to use one of the formulas $$x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$$ I don't think they will work here. We can say $\sqrt[3]{3}=t\Rightarrow t^3=3$ and the given expression is then $$\dfrac{1}{1+t-t^2}$$ I don't see anything else. What are the available approaches?
Notice that $(1 + t - t^2)(2 + t + t^2) = 2 + 3t - t^4$. Since $t^4 = 3t$, this implies $$(1 + t - t^2)(2 + t + t^2) = 2.$$ Hence $$\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}} = \frac{1}{2}(2 + \sqrt[3]{3} + \sqrt[3]{9}).$$ EDIT: As to Mark's comment: You know it must be something of the form $a + bt + t^2$. To get rid of the linear and quadratic term, we must have $a+b=3$ and $-a+b=-1$, which easily gives $a = 2$ and $b = 1$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4530664", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0 }
The locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$. Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$. I find out that the tangents to this curve with slope $m$ has this general form: $y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}}$ The perpendicular tangent would have this form then: $y = -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}}$ I solved geometrically that the locus is the circle: $2x^2+2y^2-3\sqrt[3]{2}x=0$ (you can check it here: desmos.com/calculator/xt2xquh72l), but I couldn't solve it algebraically. I wonder if you could help me in eliminating $m$ in this system: $\begin{cases} y &= mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}} \\ y &= -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}} \\ \end{cases} $
The parametric equations for the curve $xy^2=1$ could be $\forall \ t \neq 0$: $\begin{cases} x =& \displaystyle t^2 \\ y =& \displaystyle \frac{1}{t} \\ \end{cases}$ The tangent at the point $\left( t^2, \frac{1}{t} \right)$ is: $m = \displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} = \frac{ \frac{\mathrm{d}y}{\mathrm{d}t} }{ \frac{\mathrm{d}x}{\mathrm{d}t} } = \frac{ -\frac{1}{t^2} }{ 2t } = -\frac{1}{2t^3}$ The tangent equation to the curve in the point $\left( t^2, \frac{1}{t} \right)$ passing through $(x,y)$ is then: $y - \frac{1}{t} = -\frac{1}{2t^3} \left( x-t^2 \right) \implies 2yt^3-3t^2+x=0$ Using Vieta's formulas to the previous cubic equation la anterior in $t$, their roots $t_1$, $t_2$ y $t_3$ satisfy: $\begin{cases} t_1+t_2+t_3 &= \frac{3}{2y} \\ t_1t_2+t_2t_3+t_1t_3 &= 0 \\ t_1t_2t_3 &= -\frac{x}{2y} \end{cases}$ If we want two of these tangents to be perpendicular, the condition $m_1 m_2=-1$ must be satified, where $m_1$ and $m_2$ are their slopes. Then: $\begin{array}{ll} & \displaystyle m_1 \cdot m_2 = -1 \implies \left( -\frac{1}{2t_1^3} \right) \left( -\frac{1}{2t_2^3} \right) = \frac{1}{4t_1^3t_2^3}=-1 \implies 4t_1^3t_2^3 =-1 \implies \\ & \displaystyle t_1t_2 = -\frac{1}{\sqrt[3]{4}} \end{array}$ Substituting this value in Vieta's formulas: $\begin{array}{ll} & \displaystyle \begin{cases} t_1+t_2+t_3 &= \frac{3}{2y} \\ -\frac{1}{\sqrt[3]{4}}+t_2t_3+t_1t_3 &= 0 \\ -\frac{1}{\sqrt[3]{4}} t_3 &= -\frac{x}{2y} \end{cases} \implies \begin{cases} t_1+t_2+\frac{\sqrt[3]{4}x}{2y} &= \frac{3}{2y} \\ -\frac{1}{\sqrt[3]{4}}+\frac{\sqrt[3]{4}x}{2y}t_2+\frac{\sqrt[3]{4}x}{2y}t_1 &= 0 \\ t_3 &= \frac{\sqrt[3]{4}x}{2y} \end{cases} \implies \\ & \displaystyle \begin{cases} t_1+t_2 &= \frac{3- \sqrt[3]{4}x}{2y} \\ \frac{\sqrt[3]{4}x}{2y} \left( t_1 +t_2 \right) &= \frac{1}{\sqrt[3]{4}} \\ \end{cases} \implies \frac{\sqrt[3]{4}x}{2y} \cdot \frac{3- \sqrt[3]{4}x}{2y} = \frac{1}{\sqrt[3]{4}} \implies \\ & \displaystyle 2x^2+2y^2-3 \sqrt[3]{2} x=0 \iff \left( x -\frac{3}{4}\sqrt[3]{2} \right)^2 + y^2 = \left( \frac{3}{4}\sqrt[3]{2} \right)^2 \end{array}$ Consequently, the locus is the circle centered in $\displaystyle \left( \frac{3}{4}\sqrt[3]{2},0 \right)$ and radius $\displaystyle \frac{3}{4}\sqrt[3]{2}$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4532890", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$ Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation $$f(x)f(y) = f(x+y) + f(x-y)$$ Find the value of $f(7)$. Attempt: If $x=1$ and $y=0$, we find $$f(1) f(0) = 2f(1) \implies f(0) = 2$$ If we fix $y=1$, we get the recurrence relation $$\begin{cases} f(0) = 2 \\ f(1) = 3 \\ f(x + 1) - 3f(x) + f(x - 1) = 0 & \text{for } x\ge1 \end{cases}$$ From here I can solve for $f(x)$ or sequentially compute $f(2),f(3),f(4),\ldots$ to arrive at $f(7) = 843$. Question: Is there a more elegant way of finding $f(7)$ directly from the functional equation?
Evaluate $\pmb{f(0)}$ $f(1)=3$ and $$ f(x)f(y)=f(x+y)+f(x-y)\tag1 $$ Setting $x=1$ and $y=0$, $(1)$ gives $f(0)=2$. Determine and Solve a Recursion Setting $y=1$, $(1)$ gives $$ 3f(x)=f(x+1)+f(x-1)\tag2 $$ Solving the second order linear recurrence in $(2)$ and using $f(0)=2$ and $f(1)=3$ gives, for $x\in\mathbb{Z}$, $$ \begin{align} f(x)&=\left(\frac{3+\sqrt5}2\right)^{\large\!x}+\left(\frac{3-\sqrt5}2\right)^{\large\!x}\tag{3a}\\ &=\phi^{2x}+\phi^{-2x}\tag{3b}\\[6pt] &=L_{2x}\tag{3c} \end{align} $$ where $L_n$ is the $n^\text{th}$ Lucas Number (as Akiva Weinberger mentioned in a comment). Check the Functional Equation $\pmb{(1)}$ If $(1)$ has a solution, then it is given by $(3)$. However, to show that $(1)$ has a solution, we need to verify that $(3)$ satisfies $(1)$. $$ \begin{align} f(x)f(y) &=\left(\phi^{2x}+\phi^{-2x}\right)\left(\phi^{2y}+\phi^{-2y}\right)\tag{4a}\\[4pt] &=\phi^{2(x+y)}+\phi^{-2(x+y)}+\phi^{2(x-y)}+\phi^{-2(x-y)}\tag{4b}\\[6pt] &=f(x+y)+f(x-y)\tag{4c} \end{align} $$ So, $(1)$ is satisfied. Apply to the Question Thus, $$ \begin{align} f(7) &=L_{14}\tag{5a}\\[4pt] &=843\tag{5b} \end{align} $$
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Integration of $\int\sin^{3}2x\, dx.$ using IBP I am attempting to solve this integral: $$\int\sin^{3}2x\, dx.$$ Using an identity, I have manipulated the integral into this: $$\int\frac{1}{2}\left(1-\cos4x\right)\sin2x dx.$$ From here, using IBP, I let $u$ = $1-cos4x$ and $v'$ = sin2x However, I obtained an answer of $\cos2x+\frac{1}{2}\cos2x\cos4x\ -\ \frac{1}{6}\cos6x$, which is nowhere near the intended answer in the solutions of $\frac{1}{6}\cos^{3}2x-\frac{1}{2}\cos2x$. Any help on why my solution/method is incorrect will be appreciated.
Integration by parts is interesting as below: $$ \begin{aligned} &I=\int \sin ^3 2 x d x=-\frac{1}{2} \int \sin ^2 2 x d(\cos 2 x)\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+\frac{1}{2} \int 4 \sin 2 x \cos ^2 2 xdx\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x\left(1-\sin ^2 2 x\right) d x\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x d x-2 I\\ I&=-\frac{\cos 2 x}{6}\left(\sin ^2 2 x+2\right)+C \\(\textrm{ OR}& =\frac16 \cos^3 2x-\frac12\cos 2x+C) \end{aligned} $$
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Closed form for $\int_{0}^{\pi/2}x\cos^n x \ dx$, $n\in\mathbb{N}$ Closed form for the definite integral $$I(n)=\int_{0}^{\pi/2}x\cos^n x \ dx$$ where $n\in\mathbb{N}$. I tried using $\int_{0}^a f(x)\ dx= \int_{0}^a f(a-x)\ dx$ to get $$I(n)= \int_{0}^{\pi/2}\left(\frac{\pi}{2}-x\right)\sin^n x \ dx$$ So we have $$I(n)=\frac{\pi}{2^2}\frac{\Gamma(\frac{n+1}{2})\Gamma(1/2)}{\Gamma(\frac{n+2}{2})} -\int_{0}^{\pi/2}x\sin^n x \ dx$$
Utilize the expansions \begin{align} &\cos^{2m}x= \frac1{2^{2m}}\binom {2m}{m}+\frac1{2^{2m-1}}\sum_{k=1}^{m} \binom {2m}{m-k} \cos2kx\\ &\cos^{2m+1}x= \frac1{2^{2m}}\sum_{k=0}^{m} \binom {2m+1}{m-k} \cos(2k+1)x\\ \end{align} and the results \begin{align} &\int_0^{\pi/2}x\cos2mx\ dx= \frac{(-1)^m-1}{(2m)^2}\\ &\int_0^{\pi/2}x\cos(2m+1)x\ dx= \frac{\frac\pi2(2m+1)(-1)^m-1}{(2m+1)^2} \end{align} to obtain \begin{align} &\int_0^{\pi/2}x\cos^{2m}xdx = \frac{\pi^2}{2^{2m+3}}\binom {2m}{m}+\frac1{2^{2m-1}}\sum_{k=1}^m \binom {2m}{m-k} \frac{(-1)^{k}-1}{(2k)^2}\\ &\int_0^{\pi/2}x\cos^{2m+1}x\ dx =\frac1{2^{2m}}\sum_{k=0}^{m} \binom {2m+1}{m-k} \frac{\frac\pi2(2k+1)(-1)^k-1}{(2k+1)^2}\\ \end{align}
{ "language": "en", "url": "https://math.stackexchange.com/questions/4538742", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
Is a trig substitution the only way to solve $ \int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x $? I have an integral that looks like the following: $$ \int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x $$ I have seen a method of solving it being to substitute $x = \frac{\mathrm{tan}(u)}{\sqrt{c}}$; however, this seems somewhat sloppy to me. Is there perhaps a better way of tackling this integral?
$$ \begin{aligned} &\text { Let } y=\frac{1}{x^2} \text {, then } d x=-\frac{1}{2y^{\frac{3}{2}}} d y\\ I&=\int_{\frac{1}{a^2}}^{\frac{1}{b^2}} \frac{1}{\left(1+\frac{c}{y}\right)^{\frac{3}{2}}}\left(-\frac{d y}{2y \frac{3}{2}}\right)\\ &=\frac{1}{2} \int_{\frac{1}{b^2}}^{\frac{1}{a^2}} \frac{d y}{(y+c)^{\frac{3}{2}}}\\ &=\left[-\frac{1}{\sqrt{y+c}}\right]_{\frac{1}{b^2}}^{\frac{1}{a^2}}\\ &=\frac{b}{\sqrt{1+b^2 c}}-\frac{a}{\sqrt{1+a^2 c}} \end{aligned} $$
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$\lim\limits_{x\to0}\frac{2(\tan x-\sin x)-x^3}{x^5}=$? Original question: $$\lim_{x\to0}\frac{2(\tan x-\sin x)-x^3}{x^5}$$ What is wrong in my solution as answer is $\frac14$ not $\frac12$: $$\lim_{x\to0}\frac{2({\frac{\sin x}{\cos x}-\sin x)}-x^3}{x^5}$$ $$\lim_{x\to0}\frac{2(\sin x-\sin x\cos x)-x^3\cos x}{x^5\cos x}$$ $$\lim_{x\to0}\frac{2\sin x(1-\cos x)-x^3\cos x}{x^5\cos x}$$ $$\lim_{x\to0}\frac{(2\sin x)(2\sin^2 (\frac{x}{2}))-x^3\cos x}{x^5\cos x}$$ Dividing by $x^3$ and distributing $x$ and $x^2$ between $\sin x$ and $\sin^2 (\frac{x}{2})$ respectively, $$\lim_{x\to0}\frac{4(\frac{\sin x}{x})(\frac{\sin^2 (\frac{x}{2})}{x^2})-\cos x}{x^2\cos x}$$ $$\lim_{x\to0}\frac{(\frac{\sin x}{x})(\frac{\sin^2 (\frac{x}{2})}{\frac{x^2}{4}})-\cos x}{x^2\cos x}$$ Now I assumed $\frac{\sin x}{x}$ and $(\frac{\sin (\frac{x}{2})}{\frac{x}{2}})^2$ to be 1 as $x\to 0$ $$\lim_{x\to0}\frac{1-\cos x}{x^2\cos x}$$ $$\lim_{x\to0}\frac{2\sin^2 \frac{x}{2}}{x^2\cos x}$$ $$\lim_{x\to0}\frac{2\sin^2 \frac{x}{2}}{4(\frac{x^2}{4})\cos x}$$ Again $(\frac{\sin (\frac{x}{2})}{\frac{x}{2}})^2$ to be 1 as $x\to 0$ $$\lim_{x\to0}\frac{1}{2\cos x}$$ And now as $x\to0$ $\cos x$ will be $1$. So that's how I got $\frac12$. I am learning limits. So, correcting me will improve my misconceptions.
As Stephen pointed out in the comment, the problem is taking the limit in the third line of your computation. Using $S(x)=\sin(x)/x$, by adding $-1+1$ in the nominator, the correct expression is $$\lim_{x\rightarrow 0}\frac{S(x)S(x/2)-\cos(x)}{x^2\cos(x)} =\lim_{x\rightarrow 0}\frac{S(x)S(x/2)+\frac{1}{2}S(x/2)x^2-1}{x^2\cos(x)}\\ =\lim_{x\rightarrow 0}\frac{S(x)S(x/2)-1}{x^2\cos(x)}+\frac{1}{2},$$ and the first contribution to the limit is exactly the missing $-1/4$. If it is an option, I would recommend to apply L'Hôpital's rule, which is usually also used to obtain $S(0)=1$. Then, with $\tan'(x)=1+\tan(x)^2$, $\sin'(x)=\cos(x)$, $\cos'(x)=\sin(x)$, $(x^r)'=rx^{r-1}$ for $r\in\mathbb R\setminus\{0\}$ and the chain rule we get $$\frac{2(\tan(x)-\sin(x))-x^3}{x^5}\rightarrow \frac{2+2\tan(x)^2-2\cos(x)-3x^2}{5x^4}\rightarrow \frac{4\tan(x)+4\tan(x)^3+2\sin(x)-6x}{20x^3}\rightarrow \frac{4+16\tan(x)^2+12\tan(x)^4+2\cos(x)-6}{60x^2}\rightarrow \frac{32\tan(x)+80\tan(x)^3+48\tan(x)^5-2\sin(x)}{120x}\rightarrow \frac{32+272\tan(x)^2+480\tan(x)^4+240\tan(x)^6-2\cos(x)}{120}\rightarrow \frac{1}{4}.$$ As a more general bugfixing strategy, I recommend to plot the function in each step of your computations, for example here, from say $x=-0.05$ to $x=0.05$ (if this option is available). This helps to identify the faulty step. If you further keep track of the error terms (as illustrated above by adding $+1-1$), you can not only determine the limit, but also the asymptotical behavior as the limit is approached.
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Why am I getting $\beta=90^{\circ}$ Consider the geometry below, where the small circle is touching both semi circles of radius $5$ and the side of the square. Find the radius of the small circle. My try: $M$ and $N$ are centers of semicircles and $G$ is center of small circle indicated below.Let $r$ be the radius of small circle. We have $$\beta=\frac{3\pi}{4}-\alpha$$ So we have $$\sin \beta=\frac{\sin \alpha+\cos \alpha}{\sqrt{2}}$$ Also $$\sin \beta=\frac{r}{5-r}=\frac{\sin \alpha+\cos \alpha}{\sqrt{2}}\tag{1}$$ By cosine law $$\cos \alpha=\frac{(5-r)^2+(5 \sqrt{2})^2-(5+r)^2}{2(5 \sqrt{2})(5-r)}=\frac{5-2 r}{\sqrt{2}(5-r)}\:\Rightarrow \sin \alpha=\sqrt{\frac{2(5-r)^2-(5-2 r)^2}{2(5-r)^2}}=\frac{\sqrt{25-2r^2}}{\sqrt{2}(5-r)}$$ Hence from $(1)$ we get $$\frac{(5-2 r)+\sqrt{-2 r^2+25}}{\sqrt{2}(5-r)}=\frac{r}{5-r}$$ Solving the above equation we get $r=\frac{5}{2}$, but in that case $\beta=90^{\circ}$. Where I went wrong?
This might be easier if you notice that that when two circles are tangent, then their tangent point is co-linear with their centers. So if the lower left corner is the point $(0,0),$ then the center of your circle is $(r,y)$ where $r$ is the radius you seek, since the circle is tangent to the $y$ axis, and: $$r^2+(y-5)^2=(5-r)^2$$ and $$(r-5)^2+y^2=(5+r)^2.$$ It is not hard to solve for $r$ from here.
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Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification Solve the quartic polynomial : $$x^4+x^3-2x+1=0$$ where $x\in\Bbb C$. Algebraic, trigonometric and all possible methods are allowed. I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here. I realized there is no any rational root, by the rational root theorem. The harder part is, WolframAlpha says the factorisation over $\Bbb Q$ is impossible. Another solution method can be considered as the quasi-symmetric equations approach. (divide by $x^2$). $$x^2+\frac 1{x^2}+x-\frac 2x=0$$ But the substitution $z=x+\frac 1x$ doesn't make any sense. I want to ask the question here to find possible smarter ways to solve the quartic.
Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics: $$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \overline a)x^3 + (a\overline a + b + \overline b)x^2 + (a\overline b + \overline ab)x + b\overline b.$$ The $x^3$ coefficient is $1 = a + \overline a$, so we must have $$a = \frac12 + si$$ for some $s ∈ ℝ$. The constant term is $1 = b\overline b$, so $b$ lies on the complex unit circle, and we must have $$b = \cos θ + i \sin θ = \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2}i$$ where $t = \tan \frac θ2 ∈ ℝ$. The $x$ coefficient is now $$1 = a\overline b + \overline ab = \frac{1 - 4st - t^2}{1 + t^2},$$ so we must have $$s = -\frac t2.$$ Finally, the $x^2$ coefficient is $$0 = a\overline a + b + \overline b = \frac{t^4 - 6t^2 + 9}{4t^2 + 4} = \frac{(t^2 - 3)^2}{4t^2 + 4},$$ so $t = \sqrt3$ (or $-\sqrt3$, which would give the same solution with the quadratics swapped). Then $s = -\frac{\sqrt3}2$, $a = \frac{1 - i\sqrt3}2$, $b = \frac{-1 + i\sqrt3}2$, and the above factorization becomes $$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$
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Deriving exact value of $\sin \pi/12$ using double angle identity Deriving the exact value of $\sin \pi/12$ using double angle identity Double angle identity - $\sin 2A = 2\sin A \cos A$ So, $\sin (2 \frac{\pi}{12}) = 2 \sin \frac{\pi}{12} \cos \frac{\pi}{12} $ $\sin^2 A + \cos^2 A = 1 $ so, $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$ $\sin (2 \frac{\pi}{12}) =2 \sin \frac{\pi}{12}\sqrt{1- \sin^2 \frac{\pi}{12}} $ I square both sides of the equation to get: $(1/2)^2 = 4 \sin^2 \frac{\pi}{12} (1- \sin^2 \frac{\pi}{12})$ With this, I am moving away from finding the exact value of $\sin \pi/12$ which is $\frac{\sqrt{3}-1}{2\sqrt{2}}$
A double identity which gets you there more directly (perhaps not the one you are using) is $$\cos2x=1-2\sin^2x.$$ When $x=\pi/12$, we have \begin{align}\cos\frac\pi6&=1-2\sin^2\frac\pi{12}\\ 2\sin^2\frac\pi{12}&=1-\frac{\sqrt3}2\\ \sin^2\frac\pi{12}&=\frac12-\frac{\sqrt3}4=\frac{2-\sqrt3}4\\ \sin\frac\pi{12}&=\pm\sqrt{\frac{2-\sqrt3}4}=\pm\frac{\sqrt{2-\sqrt3}}2=\pm\frac{\sqrt3-1}{2\sqrt2}. \end{align} Since $\frac\pi{12}\in(0,\frac\pi2)$, we can reject the negative answer, leaving us with $$\sin\frac\pi{12}=\frac{\sqrt3-1}{2\sqrt2}$$ as desired.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4548727", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$ Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$. I think it might be useful to prove that f is injective and then make a substitution. It should be possible to prove $f$ is surjective too. It might also help to prove additional properties of f such as multiplicativity, that f has infinitely many fixed points, that f satisfies the functional equation $f(x+y) = f(x)+f(y)$ for all positive real numbers x,y, etc. First substituting $x=y$ gives $f(x+f(x+x^2)) = (x+1)f(x+1)-1.$ Swapping $x$ and $y$ gives $f(y+f(x+xy)) = (x+1)f(y+1)-1.$ Substituting $x=1$ into the latter equation gives $f(1+f(2)) = 2 f(2) - 1.$ In general for any positive integer $n, f(n+f(n+1)) = 2f(n+1)-1$. Suppose $f(x)=f(y).$ Then It might be useful to make the substitution $u=y+1, v= x+1$. Then we have $f(u-1 + f(u(v-1))) = uf(v) - 1$ for all $u,v > 1.$ If $f(\dfrac{x+y}2) = \dfrac{f(x)+f(y)}2$ for $x,y \in \mathbb{R}_{\ge 0}$, then letting $g(x) = f(x) - f(0)$ we get $g((x+y)/2) = (g(x)+g(y))/2$ for all $x\in \mathbb{R}^+$ so we may assume $f(0) = 0$. Then $f(x) = f(2x)/2$ for all nonnegative x and so $f(x+y) = (f(2x)+f(2y))/2 = f(x)+f(y)$, implying that $f(nx) = nf(x)$ for all rationals $n > 0$ and $x \ge 0$.
I found that the answer above had some apparent issues w.r.t OP, so I'll write an answer that should not be doubtable, hopefully. It will borrow statements from the above answer, but will try to be more precise and detailed. Let $P(x,y)$ be the assertion that $$ f(x + f(y+xy)) = (y+1)f(x+1)-1 $$ You have listed all the usual tricks that we can apply. Since we are working on a restricted domain, the possibility of substituting $0$ is out. Thankfully, because we are asked to prove $f(x)=x$ for $x>0$, we know that this function's properties, such as bijectivity, satisfying the Cauchy functional equation etc. may be sub-targets of our working. For example, if we set $x=1$ then we get $$ f(1+f(2y)) = (y+1)f(2)-1 $$ Injectivity The above statement implies injectivity. Indeed, if $f(a)=f(b)$ for $a,b>0$, then $1+f(2\frac a2) = 1+f(2 \frac b2)$. Applying $f$ to both sides (which can be done because both sides are positive as $f$ has range $\mathbb R^+$ and $\frac a2,\frac b2>0$), $$ f\left(1+f(2\frac a2)\right) = f\left(1+f(2 \frac b2)\right) $$ By the identity before this section, we get $\left(\frac{a}{2}+1\right)f(2)-1= \left(\frac b2 + 1\right)f(2)-1$, which because $f(2) \neq 0$ leads to $a = b$. Injectivity has been proven. Clever substitution Once injectivity has been proven, we may also want to try surjectivity, for example. However, prior to this, there is a very clever use for injectivity available. This is the following : If we can choose $y$ (in terms of $x$) such that the right hand side of $P(x,y)$ equals $f(z)$ for some $z$, then the left hand side of $P(x,y)$ always equals $f(z')$ for some $z'$. Therefore, $z'=z$ will be obtained by injectivity. For what value of $y$ can the right hand side $(y+1)f(x+1)-1$ be made into some $f(z)$? The value $Y = \frac 1{f(x+1)}$ fits. For starters, it's positive because $f(x+1) > 0$, hence $y>0$. Next, we observe that $$ (Y+1)f(x+1) - 1 = Y f(x+1) - 1 + f(x+1) = f(x+1) $$ Therefore, by looking at $P(x,Y)$ we get $$ f(x + f(Y+xY)) = f(x+1) $$ By injectivity, $x+f(Y+xY) = x+1$. Hence, $f(Y+xY) = 1$ for all $x>0$. Substituting $Y$ in, we have $$ f\left(\frac{x+1}{f(x+1)}\right) = 1 $$ for all $x>0$. By letting $z = x+1$, we have $$ f\left(\frac{z}{f(z)}\right) = 1 $$ for all $z>1$. This is a very useful statement. Use injectivity again! Indeed, we just saw that $f(\frac {z}{f(z)}) = 1$ for all $z>1$. That is the same as saying that $f(\frac {z}{f(z)}) = f(\frac{z'}{f(z')})$ for all $z,z'>1$. Therefore, by injectivity, $\frac{z}{f(z)} = \frac{z'}{f(z')}$ for all $z,z'$. Let's call this constant as $K$. Back to $P(x,y)$ and nearly finishing the problem To push this back into $P(x,y)$, we note that $x+1>1$ as $x>0$, therefore we obtain the analogous statement $$ f(x+f(y+xy)) = (y+1)\frac{(x+1)}{K}-1 $$ In particular, if $x>1$ then $x+f(y+xy)>1$. Therefore, $$ \frac{x+f(y+xy)}{K} = \frac{(y+1)(x+1)}{K}-1 = \frac{(y+1)(x+1)-K}{K} \\ \implies f(yx+y) = yx+y+1-K $$ Let $x=2$ now, for example, to get $$ f(3y) = 3y+1-K $$ For all $y>0$ (we never restricted the value of $y$ in our manipulations in this section!) Now, if $z=3y$ then the above statement transforms into $f(z) = z + 1-K$ for all $z>0$. All we need to do is find the value of $K$. Finding the value of $K$ Using $P(x,y)$ itself, we see that for any $x,y>0$, $$ f(x+f(y+xy)) = (y+1)f(x+1)-1 \implies (x+f(y+xy)) + 1 - K = (y+1)(x+2-K)-1 \\ \implies x+y+xy+1-K+1-K = (y+1)(x+1)+ (y+1)(1-K)-1 \\ \implies 1-2K = (y+1)(1-K)-1 \implies 2-2K = (y+1)(1-K) $$ for all $y$. Note that if $K>1$, then $f(K-1)=0$ gives a contradiction to the domain of $f$. Therefore, $K \leq 1$. However , if $K<1$ then we will get a contradiction because $y+1 = 2$ for all $y >0$ is not possible. Therefore, $K=1$ must happen. Thus, $f(z)=z$ for all $z>0$, as desired.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4549625", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 1 }
Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$ I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w} $$ and was asked to find the value $kw$. Here is my try on it: Substituting $\theta = \dfrac{\pi}{4}+x$, we get $$\begin{equation} \label{eq1} \begin{split} I &= \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{(-4x) \tan (\frac{\pi}{4}+x)}{1-\tan (\frac{\pi}{4}+x)} \, dx \\ &= -4\displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x.\dfrac{1+ \tan x}{1- \tan x}}{1- \dfrac{1+ \tan x}{1- \tan x}} \, dx \\ &= 2 \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx \end{split} \end{equation}$$ From here I am stuck on this, I don't know how to proceed from here to get answer in terms of $\pi \ln k-\frac{\pi^2}{w}$ Thank you for any help.
letting $\theta=\frac{\pi}{4}-x$ changes the integral into $$ I=2 \int_0^{\frac{\pi}{2}} \frac{x(1-\tan x)}{\tan x} d x $$ $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} d x &=\int_0^{\frac{\pi}{2}} x d(\ln (\sin x)) \\ &= \left[x \ln (\sin x) \right] _0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \ln (\sin x) d x\\ &=\frac{\pi}{2} \ln 2 \end{aligned} $$ where the last integral using the result: $\int_0^{\frac{\pi}{2}} \ln (\sin x) d x=-\frac{\pi}{2} \ln 2$. Hence $$\boxed{\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta =\pi \ln 2-\frac{\pi^2}{4}} $$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4552842", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 3, "answer_id": 2 }
Non-negative $f(n)$ satisfies $f(ab)=f(a)+f(b)$, $f(n)=0$ for $n$ a prime greater than $10$, and $f(1) For each positive integer n, a non-negative integer $f (n)$ is associated so that the following three rules are met: i) $f (ab) = f (a)+ f (b).$ ii) $f (n) = 0 $ if $n$ is a greater prime than $10$. iii) $f (1) < f (243) < f (2) < 11$ Knowing that $f(2106) < 11$, determine the value of $f(96)$. By property i), we have $f (243) = f (3^5) = 5f (3).$ Given that $0 ≤ f (1) < 5f (3) < f (2) < 11$, and that $5f (3)$ is a multiple of $5$, we have $5f (3) = 5$, that is, $f (3) = 1$. Note that $2106 = 2·3^4·13.$ So, by property i), $f(2106)=f(2)+4f(3)+f(13)=f(2)+4.$
You now know that $f(2)+4<11$ and $f(2)>5$; it then follows from $f$'s integrality that $f(2)=6$. Then $f(96)=5f(2)+f(3)=5\cdot6+1=31$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4554913", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
The solution of differential equation $2xy+6x+(x^2-4)y'=0$ When I solved the above DE, I reached to the step and confused how to continue $2x(y+3)+(x^2-4)dy/dx$ $(x^2-4) dy/dx=-2x(y+3)$ $dy/(y+3)=-2x/(x^2-4) dx$ $∫dy/(y+3)=∫-2x/(x^2-4) dx$ $\ln⁡|y+3|=-ln⁡|x^2-4|+C$ $\ln⁡|y+3|+ln⁡|x^2-4|=C$ $\ln⁡|(y+3)(x^2-4)|=C$ $|(y+3)(x^2-4)|=e^C$ then how can I get the function $y$ for the last equation and when I checked the solution by W.F, I found it as $y=c/(x^2-4)-3x^2/(x^2-4)$ plz can anyone help me to show how happed that thanks
Note from $$ \ln⁡|(y+3)(x^2-4)|=C$$ you have $$ y=-3+\frac{C}{x^2-4}=\frac{-3x^2+12+C}{x^2-4}=\frac{-3x^2}{x^2-4}+\frac{12+C}{x^2-4} $$ Now define $C_1=12+C$ and you have $$ y=\frac{-3x^2}{x^2-4}+\frac{C_1}{x^2-4} $$ which is the same as the answer.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4555213", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0 }
Why taking combinations works but conventional approach does not in this case? The question is Let two items be chosen from a lot containing 12 items of which 4 are defective. Find the probability that atleast one item is defective. Approach 1: $$ \frac{4}{12}\times\frac{8}{11}+\frac{4}{12}\times\frac{3}{11} = \frac{1}{3} $$ First term represents the case in which one item is defective and second is not defective. Here $\frac{4}{12}$ is probability choosing one item from $4$ defective items out of $12$ items. As a result, now there are $11$ items are there to choose from. $\frac{8}{11}$ is probability of choosing one from $8$ non-defective item out of $11$ items. Second term represents the case in which both items are defective. $\frac{4}{12}$ is probability of choosing one item from $4$ defective items and $\frac{3}{11}$ is probability of choosing another item from $3$ defective items. Approach 2: Total number of ways of choosing $2$ items out of $12$ $= \binom{12}{2} = 66$ Total number of ways of choosing $2$ non-defective items out of 8 $= \binom{8}{2} = 28$ Let $X$ denote number of defective items chosen. Therefore, $$ P(X = 0) = \frac{14}{33} $$ Probability of choosing atleast one defective item is denoted by $P(X \ge 1)$. Therefore, $$ \begin{split} P(X \ge 1) &= 1 - P(X = 0) \\ &= 1 - \frac{14}{33} \\ &= \frac{19}{33} \end{split} $$ Why the first approach gives a wrong answer and why second approach works?
In the first approach you have to count the first term twice, representing the fact that you can either pull a working item and then a defective one or the other way around. This gives $$2×\frac4{12}×\frac8{11}+\frac4{12}×\frac3{11}=\frac{19}{33}$$
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How do I write a power series from a recurrence relation? I have ODE $$y''- 2xy'+ 2ky = 0$$ From this I have found a recurrence relation $$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)}C_n$$ How would I write this out in series summation form? I am trying to relate all terms bacl to $C_0$, but am struggling to write out in summation form.
Starting with $$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n,$$ apply $n = 0,1,2, \cdots$ to determine the pattern. As seen by: \begin{align} C_{n+2} &= \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n \\ C_{2} &= \dfrac {(2k)}{(1)(2)} \, C_0 \\ C_{3} &= \dfrac {2k+2}{(2)(3)} \, C_1 \\ C_{4} &= \dfrac {2k+4}{(3)(4)} \, C_2 = \frac{(2 k)(2 k + 4)}{4!} \, C_{0} = \frac{2^2 \, k (k + 2)}{4!} \, C_{0} \\ C_{5} &= \dfrac {2k+6}{(4)(5)} \, C_3 = \dfrac {2^2 \, (k+1)(k+3)}{5!} \, C_1. \end{align} This leads to \begin{align} y(x) &= C_{0} + C_{1} \, x + C_{2} \, x^2 + \cdots \\ &= C_{0} + C_{1} \, x + \dfrac {(2k)}{(1)(2)} \, C_0 \, x^2 + \dfrac {2k+2}{(2)(3)} \, C_1 \, x^3 + \cdots \\ &= C_{0} \, \sum_{n=0}^{\infty} \frac{a_{n} \, 2^{n} \, x^{2 n}}{(2 n)!} + C_{1} \, \sum_{n=0}^{\infty} \frac{b_{n} \, 2^{n} \, x^{2 k + 1}}{(2 k + 1)!}, \end{align} where $a_{n}$ and $b_{n}$ are coefficients involving $k$. Further work shows that \begin{align} C_{2n} &= \frac{4^{n}}{(2 n)!} \, \left(\frac{k}{2}\right)_{n} \, C_{0} \\ C_{2 n +1} &= \frac{4^{n}}{(2 n +1)!} \, \left(\frac{k+1}{2}\right)_{n} \, C_{1}. \end{align} This leads to $$ y(x) = C_{0} \, \sum_{n=0}^{\infty} \frac{\left(\frac{k}{2}\right)_{n} \, (2 x)^{2n}}{(2 n)!} + \frac{C_{1}}{2} \, \sum_{n=0}^{\infty} \frac{\left(\frac{k+1}{2}\right)_{n} \, (2 x)^{2 n+1}}{(2 n+1)!}, $$ where $(a)_{n}$ is the Pochhammer symbol. Note: There seems to be an error, or missing components, with the coefficient recurrence relation. This is based upon the solution of $$ y'' - 2 \, x \, y' + 2 \, k \, y = 0$$ is $$ y(x) = c_{0} \, H_{k}(x) + c_{1} \, {}_{1}F_{1}\left(- \frac{k}{2}, \frac{1}{2}; x \right),$$ where $H_{n}(x)$ are the Hermite polynomials.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4563833", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$. Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$. Firstly, I decided to raise both sides of the equation to the power of $2$. I came at $$\dfrac{4+x}{8+x+4\sqrt{4+x}}=\dfrac{4-x}{8-x-4\sqrt{4-x}}$$ Another thing I tried is to let $\sqrt{4+x}=u\ge0$ and $\sqrt{4-x}=v\ge0$. Then $$\begin{cases}\dfrac{u}{2+u}=\dfrac{v}{2-v}\\4+x=u^2\\4-x=v^2\end{cases}$$ Adding the second to the third equation, we get $8=u^2+v^2$. And the last thing: I cross-multiplied $$2\sqrt{4+x}-\sqrt{16-x^2}=2\sqrt{4-x}+\sqrt{16-x^2}\\\sqrt{4+x}=\sqrt{4-x}+\sqrt{16-x^2}$$ Raising to the power of 2 gives $$4+x=4-x+16-x^2+2\sqrt{(4-x)(16-x^2)}\\2\sqrt{(4-x)(16-x^2)}=x^2+2x-16$$ Is there something easier?
My first instinct is to note that for $x \ge -4$ and $x \ne 0$, $$\frac{\sqrt{4+x}}{2+\sqrt{4+x}} = \frac{\sqrt{4+x}(\sqrt{4+x}-2)}{(4+x)-4} = \frac{4+x - 2\sqrt{4+x}}{x}, \tag{1}$$ and similarly, for $x \le 4$ and $x \ne 0$, $$\frac{\sqrt{4-x}}{2-\sqrt{4-x}} = \frac{\sqrt{4-x}(2+\sqrt{4-x})}{4-(4-x)} = \frac{2\sqrt{4-x} +4-x }{x}. \tag{2}$$ Therefore the original equation implies $$4+x - 2\sqrt{4+x} = 2\sqrt{4-x} + 4-x$$ when $0 < |x| \le 4$. This in turn yields $$\sqrt{4+x} + \sqrt{4-x} = x, \tag{3}$$ and now we can square both sides: $$4+x + 4-x + 2\sqrt{16-x^2} = x^2, \tag{4} $$ or $$4(16-x^2) = (x^2 - 8)^2, \tag{5} $$ hence $$x^4 - 12x^2 = 0. \tag{6}$$ This gives $x = \pm 2 \sqrt{3}$ as the only solutions satisfying the requirement that $0 < |x| \le 4$; however, the negative root, when substituted back into the original equation, is extraneous. The reason for this has to do with the first squaring step from $(3)$ to $(4)$, since the RHS of $(3)$ cannot be negative, being the sum of two (nonnegative) square roots on the LHS. Therefore the unique real-valued solution is $x = 2\sqrt{3}$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4566671", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 3, "answer_id": 0 }
Given the perimeter, find the length $\overline{\rm PQ}$ in the following setup. Let $\overline{\rm AB}$ be a diameter of a circle $\omega$ and let $C$ be a point on $\omega$, different from $A$ and $B$. The perpendicular from $C$ intersects $\overline{\rm AB}$ at $D$ and omega at $E (\ne C)$. The circle with the center at $C$ and radius $\rm CD$ intersects $\omega$ at $P$ and $Q$. If the perimeter of the $\triangle{PEQ}$ is $p$, find the length of the side $\overline{\rm PQ}$. I solved this question by taking $C$ on the diameter of $\omega$ perpendicular to $AB$. It turned out to be a very simple case where $\triangle{PEQ}$ was equilateral, and thus $\rm PQ = \frac{p}{3}$. Now this is the correct answer as the question allows us to take $C$ anywhere. But I feel that it doesn't quite grasp the essence of the question. Could anyone help solve it for the general case where $C$ lies at any random point on $\omega$?
Let the radius of the circle centered at $C$ be $r$. $$\implies \overline{\rm CP} = \overline{\rm CQ} = r$$ Therefore, in $\omega$, $\angle{CEQ} = \angle{CEP} = \theta$ And subtended by $\overline{\rm EP}, \angle{PQE} = \angle{PCE} = \delta.$ $\tag*{}$ Now in $\triangle{PEQ}$, we have: $$\frac{\sin{\delta}}{\rm PE} = \frac{\sin{2\theta}}{\rm PQ} \tag{1}$$ And in $\triangle{PEC}$, we have: $$\frac{\sin{\delta}}{\rm PE} = \frac{\sin{\theta}}{\rm CP} \tag{2}$$ From $(1)$ and $(2)$: $$\rm PQ = {\rm CP} \cdot \frac{\sin{2\theta}}{\sin{\theta}} = 2r\cos{\theta} \tag{3}$$ Now in $\triangle{CEQ}$, we have: $$\begin{align} \cos{\theta} & = \frac{{\rm QE}^2 + {\rm CE}^2 - {\rm CQ}^2}{2\cdot{\rm QE}\cdot{\rm CE}} \\ & = \frac{{\rm QE}^2 + {(2r)}^2 - {r}^2}{2\cdot{\rm QE} \cdot{2r}} \\ & = \frac{{\rm QE}^2 + 3{r}^2}{4r \cdot {\rm QE}}\end{align}$$ $$\implies {\rm QE}^2 - (4r\cos{\theta}){\rm QE} + 3r^2 = 0 \tag{4}$$ Similarly in $\triangle{CEP}$, we have: $$\begin{align} \cos{\theta} & = \frac{{\rm PE}^2 + {\rm CE}^2 - {\rm CP}^2}{2\cdot{\rm PE}\cdot{\rm CE}} \\ & = \frac{{\rm PE}^2 + {(2r)}^2 - {r}^2}{2\cdot{\rm PE} \cdot{2r}} \\ & = \frac{{\rm PE}^2 + 3{r}^2}{4r \cdot {\rm PE}}\end{align}$$ $$\implies {\rm PE}^2 -(4r\cos{\theta}){\rm PE} + 3r^2 = 0 \tag{5}$$ From $(4)$ and $(5)$ we observe that $\rm PE$ and $\rm QE$ are the two roots of the equation: $$x^2 - (4r\cos{\theta})x + 3r^2 = 0$$ Therefore, ${\rm PE} + {\rm QE} = 4r\cos{\theta} \tag{6}$ Using $(3)$ and $(6)$, we get: $$\begin{align} p & = \rm PQ + (\rm PE + \rm QE) \\ & = 2r\cos{\theta} + (4r\cos{\theta}) \\ & = 6r\cos{\theta} \end{align}$$ Therfore, $$\boxed{\rm PQ = \frac{p}{3} = \frac{1}{3}\cdot\text{Perimeter}}$$ Hence proved.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4567197", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1 }
Why is $ab+\frac{1}{2} - \frac{a+b}{2} >0$ for $0I am trying to show that the function $$ab+\frac{1}{2}-\frac{a+b}{2}$$ is positive when $0<a,b<1.$ Here is what I have done. * *The arithmetic-geometric mean inequality doesn't seem like the way to go because it implies $$ab < \sqrt{ab} \le \frac{a+b}{2}.$$ This implies $$ab+\frac{1}{2}-\frac{a+b}{2}\le \frac{1}{2}.$$ *When I fix values for $a$ (or $b$) and sketch traces of the function $ab+\frac{1}{2}-\frac{a+b}{2}$, I obtain positive values on the graph.
You have shown that $$ f(a, b) = ab+\frac{1}{2}-\frac{a+b}{2} < \frac{1}{2} $$ for $0 < a, b < 1$, because $$ ab < \sqrt{ab} \le \frac{a+b}{2}. $$ But $ f(a, b) + f(1-a, b) = \frac 12$, and therefore your upper bound also gives the desired lower bound: $$ f(a, b) = \frac 12 - \underbrace{f(1-a, b)}_{<1/2} > 0 \, . $$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4567690", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3 }
The least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$ If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function) (A)6 (B)7 (C)5 (D)4 My approach is as follow $f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$ $\alpha$ lies between 1&2 and $m$ lies between 6 and 7 hence the answer is 7. I checked it and ur us correct. But solution is required
By using the Cauchy Schwarz inequality, you can easily obtain: $$\sqrt {a^2+b^2}+\sqrt{c^2+d^2}\ge \sqrt {(a+c)^2+(b+d)^2};$$ and equality holds if $\frac{a}{b}=\frac {c}{d}$. Now, take $a=x-1$, $c=2-x$, $b=1$, $d=5$. We get: $$\sqrt {(x-1)^2+1^2}+\sqrt{(2-x)^2+5^2}\ge \sqrt {1^2+6^2}=\sqrt {37}\implies m=\sqrt {37}.$$ And, $$\frac {x-1}{1}=\frac {2-x}{5}\implies a=\frac {7}{6}.$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4570233", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1 }
Prove that: $e-\ln(10)>\sqrt 2-1.$ The author's original inequality is as follows. Prove that: $$e-\ln(10)>\sqrt 2-1$$ Is there a good approximation for $$e-\ln 10?$$ Actually, I am also wondering that, Where does $\sqrt 2-1$ come from? Maybe, there exist relevant inequality? My attempt: $$-\ln 10>\sqrt 2-1-e\\ \ln 10<e+1-\sqrt 2\\ e^{e+1-\sqrt 2}>10$$ So, can we show that $$e^x>10$$ when $x\ge e+1-\sqrt 2$? I don't have a good idea.
I hope the following will help. $$\ln10=\ln2+\ln5=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+...\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+...\right)<$$$$<2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+\frac{\left(\frac{1}{3}\right)^9}{7}+\frac{\left(\frac{1}{3}\right)^{11}}{7}+...\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}+\frac{\left(\frac{2}{3}\right)^{15}}{13}+...\right)=$$$$=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}\cdot\frac{1}{1-\left(\frac{1}{3}\right)^2}\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}\cdot\frac{1}{1-\left(\frac{2}{3}\right)^2}\right)=$$ $$=\frac{8166549323}{3546482940}.$$ Can you end it now?
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Find the flux of $\mathbf{F} = (2x-xy, -y, yz)$ out of the region W Question statement: Let $W$ be the 3D region under the graph of $f(x,y) = \exp(x^2+y^2)$ over the region in the plane defined by $1 \leq x^2 + y^2 \leq 2$. Find the flux of $\mathbf{F} = (2x-xy,-y,yz)$ out of the region W and check your answer using the divergence theorem. My attempt: The divergence theorem is the easier method. Using it, we get $$\iint_{\partial W} \mathbf{F} \cdot d\mathbf{S}=\iiint_W \nabla \cdot \mathbf{F} dV = \iiint_W 1 dV = \int_1^{\sqrt 2} \int_0^{2\pi} re^{r^2} d\theta dr = \pi(e^2-e).$$ The direct computation is more difficult and I think that there are 3 surfaces which bound the volume $W$. The first is the surface defined by $z=e^{x^2+y^2}$ over the region in the plane. I parametrise it using $\mathbf{r}(x,y) = (x,y,\exp(x^2+y^2)).$ I find it is easier to work with cylindrical polars, so I instead use $\mathbf{r}(r, \theta)=(r\cos\theta, r\sin \theta, e^{r^2}), 1 \leq r \leq \sqrt{2}, 0 \leq \theta \leq 2 \pi.$ I compute the normal $$\mathbf{N} = \mathbf{r}_r \times \mathbf{r}_{\theta} = (\cos \theta, \sin \theta, 2re^{r^2}) \times (-r\sin \theta, r \cos \theta, 0) = r(-2r \cos \theta e^{r^2}, -2r e^{r^2} \sin \theta, 1).$$ I check and this normal seems to be pointing outside the surface. Proceeding, we get $$\mathbf{F} \cdot \mathbf{N} = -r^2 e^{r^2} \sin \theta - 2e^{r^2} r^2 (2r \cos \theta - r^2 \sin \theta \cos \theta) + 2e^{r^2} r^3 \sin ^2 \theta.$$ Then $$\int_1^{\sqrt 2} \int_0^{2\pi} \mathbf{F} \cdot \mathbf{N} d\theta dr = -\pi e^2.$$ The second surface is the disk on the plane, i.e $x^2 + y^2 = r^2$ for $0 \leq r \leq \sqrt{2}.$ But there, $z = 0$ and the normal points in the $z$-direction and so the integral is $0$. The third surface is the cylinder. I parametrise using $$\mathbf{r}(\theta, z) = (\sqrt{2} \cos \theta, \sqrt{2} \sin \theta, z), 0 \leq \theta \leq 2 \pi, 0 \leq z \leq e^2.$$ Going through the steps again we get $\mathbf{N} = \mathbf{r}_{\theta} \times \mathbf{r}_z = (\sqrt{2} \cos \theta, \sqrt{2} \sin \theta, 0).$ Then we obtain that $$\mathbf{F} \cdot \mathbf{N} = 4 \cos^2 \theta - 2\sqrt{2} \cos \theta \sin \theta - 2 \sin^2 \theta.$$ So we can get that $$\int_0^{2\pi} \int_0^{e^2} \mathbf{F} \cdot \mathbf{N} dz d\theta = 2 \pi e^2.$$ But then the sum is just $\pi e^2$ instead of $\pi e^2 - \pi e$. I understand that the Divergence Theorem gives the answer quite easily, but I would be very grateful if someone could tell me where I went wrong here. Thank you very much for helping me.
There is a fourth surface: an internal cylinder with radius $1$: $$\mathbf{r}(\theta, z) = (\cos \theta,\sin \theta, z), 0 \leq \theta \leq 2 \pi, 0 \leq z \leq e.$$ Then $\mathbf{N} = \mathbf{r}_{\theta} \times \mathbf{r}_z = (\cos \theta, \sin \theta, 0)$ and $\mathbf{F} \cdot \mathbf{N} = 2 \cos^2 \theta - \cos \theta \sin \theta - \sin^2 \theta$. Hence the flux accross this fourth surface with the correct orientation, i.e. pointing towards the $z$-axis, is $$-\int_0^{2\pi} \int_0^{e} \mathbf{F} \cdot \mathbf{N}\, dz d\theta = -\pi e.$$ Finally, the total sum is $$\iint_{\partial W} \mathbf{F} \cdot d\mathbf{S}=-\pi e^2+0 +2\pi e^2-\pi e=\pi(e^2-e).$$ as we expected.
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Find a, real number for the following limit Find a, real number, such that $\displaystyle{\lim_{n\to\infty}{(n^3+an^2)^\frac{1}{3}-(n^2-an)^\frac{1}{2}}}=1$ I noted x=$(n^3+an^2)^\frac{1}{3}$ and y=$(n^2-an)^\frac{1}{2}$ I applied with the conjugate ($x^2-xy+y^2$) but I do not know how to continue. Any ideas? Thank you for your help!
Note that $$\begin{align*} x-y &= (x-n) + (n-y) \\ &= \frac{x^3-n^3}{x^2+xn+n^2} + \frac{n^2-y^2}{n+y} \\ &= \frac{an^2}{x^2+xn+n^2} + \frac{an}{n+y} \\ &= \frac{a}{(x/n)^2 + (x/n) + 1} + \frac{a}{1+(y/n)}\end{align*}$$ and therefore, as $\lim_{n\to\infty} \frac{x}{n} = \lim_{n\to\infty} \frac{y}{n} = 1$, $$\lim_{n\to\infty} (x-y) = \lim_{n\to \infty} \left(\frac{a}{(x/n)^2 + (x/n) + 1} + \frac{a}{1+(y/n)}\right) = \frac{a}{1+1+1}+\frac{a}{1+1} = \frac{5a}{6}$$ Setting $\frac{5a}{6} = 1$ yields $a=\frac{6}{5}$.
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How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? I want to solve the equation $$\frac{1}{x^2} \, \frac{d}{dx}\left(x^2\frac{dy}{dx}\right)=-y.$$ I converted it to this equation $y''+\frac{2}{x}y'+y=0$. How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? Is there any way to solve it generally?
Assume a solution with power series and derivatives $$y(x) = \sum_{n\ge0} a_n x^n \implies y'(x) = \sum_{n\ge0} (n+1) a_{n+1} x^n \implies y''(x) = \sum_{n\ge0} (n+1) (n+2) a_{n+2} x^n$$ Putting these into the ODE yields $$\begin{align*} 0 &= y'' + \frac2x y' + y \\[1ex] &= xy'' + 2y' + xy \\[1ex] &= \sum_{n\ge0} (n+1)(n+2) a_{n+2} x^{n+1} + 2 \sum_{n\ge0} (n+1) a_{n+1} x^n + \sum_{n\ge0} a_n x^{n+1} \\[1ex] &= \sum_{n\ge1} n(n+1) a_{n+1} x^n + 2 \sum_{n\ge0} (n+1)a_{n+1} x^n + \sum_{n\ge1} a_{n-1}x^n \\[1ex] &= 2a_1 + \sum_{n\ge1} \bigg((n+2)(n+1) a_{n+1} + a_{n-1}\bigg) x^n \\[1ex] \end{align*}$$ Let $a_0=A$. The coefficients are governed by the recurrence $$\begin{cases}a_0 = A \\ 2a_1 = 0 \\ (n+2)(n+1)a_{n+1} + a_{n-1} = 0 & n\ge1\end{cases} \iff \begin{cases}a_0 = A \\ a_1 = 0 \\ a_n = -\frac{a_{n-2}}{n(n+1)} & n\ge2\end{cases}$$ and we have for $k\in\Bbb N$, $$\begin{cases} a_{2k-1} = 0 \\ a_{2k} = \frac{(-1)^k}{(2k+1)!} A\end{cases}$$ so that a fundamental solution to the ODE is $$y_1(x) = \sum_{k\ge0} \frac{(-1)^k}{(2k+1)!} x^{2k} = \boxed{\frac{\sin(x)}x}$$ Reduce the order to find another solution. Let $y_2(x)=z(x)y_1(x)$. Substitute into the ODE and solve the subsequent linear equation. $$\begin{align*} 0 &= x \left(z'' y_1 + 2z'{y_1}' + z{y_1}''\right) + 2 \left(z' y_1 + z {y_1}'\right) + x z y_1 \\[1ex] &= z'' \sin(x) + 2z' \cos(x)\\[1ex] &= w' \sin(x) + 2w \cos(x) & w(x) = z'(x) \\[1ex] &= w' \sin^2(x) + w \sin(2x) \\[1ex] &= \left(w \sin^2(x)\right)' \\[2ex] \implies w(x) &= C_1 \csc^2(x) \\[2ex] \implies z(x) &= C_1 \cot(x) + C_2 \end{align*}$$ Then the other fundamental solution is $$y_2(x) = \cot(x) \cdot \frac{\sin(x)}x = \boxed{\frac{\cos(x)}x}$$
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How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general $$ I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$ by the powerful substitution $x=\frac{1-t}{1+t} .$ where $n$ is a natural number greater $1$. Let’s start with easy one \begin{aligned} I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\ &=\frac{\pi}{8} \ln 2-G\end{aligned} By my post $$I_2= \frac{\pi}{4} \ln 2-G $$ and $$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G \end{aligned} $$ $$ \begin{aligned} I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G \\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3}) \end{aligned} $$ where the last integral refers to my post. Let’s skip $I_5$ now. $$ I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\ $$ $$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$ Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$ How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
The odd cases are much harder to evaluate. Below are some of them: \begin{align} \int_0^1 \frac{\ln(1-x^5)}{1+x^2}dx= & \ \frac{9\pi}8\ln2-\frac{13}5G+\frac\pi2\ln\left(\cos\frac\pi{20}\cos\frac{3\pi}{20} \right)\\ &\>\>\>+ \frac{3\pi}{20}\ln \tan\frac{3\pi}{20} - \frac{\pi}{20}\ln \tan\frac{\pi}{20} \\ \\ \int_0^1 \frac{\ln(1-x^7)}{1+x^2}dx = & \ \frac{13\pi}8\ln2-\frac{24}7G+\frac\pi2\ln\left(\cos\frac\pi{28}\cos\frac{3\pi}{28} \cos\frac{5\pi}{28}\right)\\ &\>\>\>+ \frac{5\pi}{28}\ln \tan\frac{5\pi}{28}-\frac{3\pi}{28}\ln \tan\frac{3\pi}{28} +\frac{\pi}{28}\ln \tan\frac{\pi}{28} \\ \end{align}
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Is this limit solvable using Stolz’s theorem ? $\lim_\limits{n\to+\infty}\left(\frac12+\frac3{2^2}+\dots+\frac{2n−1}{2^n}\right)$ I would like to ask if this example is solvable using Stolz’s theorems, because I have a $2^n$ expression at the bottom and a quadratic expression at the top, so it doesn't work for me, but maybe I'm calculating wrong.
Denote $S_{n}=\frac{1}{2}+\frac{3}{2^{2}}+\ldots+\frac{2n-1}{2^{n}}$. Firstly, observe that \begin{eqnarray*} S_{n} & = & \left\{ \frac{2}{2}+\frac{4}{2^{2}}+\frac{2n}{2^{n}}\right\} -\left\{ \frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{n}}\right\} \\ & = & 2\sum_{k=1}^{n}\frac{k}{2^{k}}-\frac{\frac{1}{2}\left[1-(\frac{1}{2})^{n}\right]}{1-\frac{1}{2}}\\ & = & 2\sum_{k=1}^{n}\frac{k}{2^{k}}-\left\{ 1-\left(\frac{1}{2}\right)^{n}\right\} . \end{eqnarray*} Now, we focus on the computation of $\sum_{k=1}^{n}\frac{k}{2^{k}}$. Let $f(x)=\sum_{k=1}^{n}x^{k}$, then $f'(x)=\sum_{k=1}^{n}kx^{k-1}$. Therefore, $\sum_{k=1}^{n}kx^{k}=xf'(x)$. On the other hand, by sum of G.P. formula (which is valid for $x\neq 1$), $f(x)=\frac{x(1-x^{n})}{1-x}$, so \begin{eqnarray*} f'(x) & = & \frac{\left(1-(n+1)x^{n}\right)(1-x)+x(1-x^{n})}{(1-x)^{2}}. \end{eqnarray*} Hence, we obtain the formula: \begin{eqnarray*} \sum_{k=1}^{n}kx^{k} & = & x\cdot\frac{\left(1-(n+1)x^{n}\right)(1-x)+x(1-x^{n})}{(1-x)^{2}}. \end{eqnarray*} Put $x=\frac{1}{2}$, then we obtain $$ \sum_{k=1}^{n}\frac{k}{2^{k}}=\frac{1}{2}\cdot\frac{\left[1-(n+1)\left(\frac{1}{2}\right)^{n}\right]\cdot\frac{1}{2}+\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\frac{1}{4}}. $$ Now, you should be able to compute the limit $\displaystyle\lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^{n}\frac{k}{2^{k}}$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4580463", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1 }
$S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} \geq \dfrac{3}{2}$ Main Problem: Let $x,y,z$ be positive real numbers and $xyz = 1$. For any $p$ real number, let's define $$ S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} $$ Prove (or disprove) that for all $p\geq 2$ values, $S_p\geq \dfrac{3}{2}$. My Efforts: It would be helpful if I explained to the readers how I created the problem and what methods I tried. * *We show that $S_4\geq \dfrac{3}{2}$ in MSE here. *Also $S_3\geq \dfrac{3}{2}$ is a problem from IMO-1995. (There are many different solution method in the link.) *For $p=2$, I proved $S_2 = \dfrac{1}{(y+z) x^2} + \dfrac{1}{(z+x) y^2} + \dfrac{1}{(x+y) z^2} \geq \dfrac{3}{2}$. Proof for $p=2$ Case: For $a,b,c$ real numbers $a^2 + b^2 + c^2 \geq ab + bc + ca$. Also $xyz=1$ is given and so we can write $(xy)^2 + (yz)^2 + (zx)^2 \geq xy\cdot yz + zx\cdot xy + yz\cdot zx = x+y+z$. By Sedrakyan's Form of Cauchy-Schwarz inequality and by the above inequality, we get $ S_2 = \dfrac{1}{(y+z) x^2} + \dfrac{1}{(z+x) y^2} + \dfrac{1}{(x+y) z^2} \\ = \dfrac{(xyz)^2}{(y+z) x^2} + \dfrac{(xyz)^2}{(z+x) y^2} + \dfrac{(xyz)^2}{(x+y)z^2} = \dfrac{(yz)^2}{(y+z)} + \dfrac{(zx)^2}{(z+x)} + \dfrac{(xy)^2}{(x+y)} \\ \geq \dfrac{\left(xy + yz + zx \right)^2}{2(x+y+z)} = \dfrac{(xy)^2 + (yz)^2 + (zx)^2 + 2(x+y+z)}{2(x+y+z)} \\ \geq \dfrac{(x+y+z)+2(x+y+z)}{2(x+y+z)}=\dfrac{3}{2}.$ $p=1$ Case: If we take $x=y=2$, $z=\dfrac{1}{4}$, we can calculate the value $S_1 = \dfrac{13}{9} < \dfrac{3}{2}$. In this case, inequality is not achieved. $p=\dfrac{3}{2}$ Case: If we take $x=y=4$, $z=\dfrac{1}{16}$, we can calculate the value $S_{3/2} = \dfrac{4}{65} + 8 > \dfrac{3}{2}$. In this case, inequality seems achieved. So, I felt that the inequality should be true when $p>1$. This is how the idea of the Main Problem came about. On the other hand, by Reza Rajaei's comment $(x,y,z) = (50,\dfrac{1}{50},1)$ and for $p=1.1$, we see that $S_{1.1}<\dfrac{3}{2}$. * *WLOG, $0 <x \leq y \leq z$. Then $x^p \leq y^p \leq z^p$ and $x+y\leq x+z \leq y+z$. I thought of using Chebyshev's sum inequality. $$ S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} \leq \dfrac{1}{3}\cdot \left( \dfrac{1}{x^p} + \dfrac{1}{y^p} + \dfrac{1}{z^p} \right) \left( \dfrac{1}{y+z} + \dfrac{1}{z+x} + \dfrac{1}{x+y} \right) .$$ This equality is correct, but unuseful unfortunalely. Because, the inequality is in the opposite direction. * *I tried homogenizing $S_p$. $$ S_p = \dfrac{(xyz)^{(p+1)/3}}{(y+z) x^p} + \dfrac{(xyz)^{(p+1)/3}}{(z+x) y^p} + \dfrac{(xyz)^{(p+1)/3}}{(x+y) z^p} $$ and now it is homogenize. I had some more futile efforts and then I got stuck. Thank you for your interest.
Oh this one is not that bad. First, substitute $x\to\frac 1x$, etc for simplicity to obtain an equivalent inequality: $$\sum\dfrac{x^{p-1}}{y+z}\geq \frac 32.$$ Then, do the natural Cauchy-Schwarz to obtain: $$\sum\dfrac{x^{p-1}}{y+z}\sum (y+z)\geq \left(\sum x^{\tfrac{p-1}{2}}\right)^2.$$ But if $p\geq 2$, then $\dfrac{p-1}{2}\geq 1$ and it's an easy exercise to prove: $$x^m+y^m+z^m\geq (xyz)^{\frac{m-1}{3}}(x+y+z)$$ when $m\geq 1.$ Combine all these together and your inequality follows. This is proved by two C-S or equivalents, so not the most strongest inequality. I would not be surprised the optimal value of such $p$ is strictly less than $2.$
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Find all the real solutions of $(1+x^2)(1+x^4)=4x^3$. As title suggests, the question is to find all the real roots to the polynomial: $$(1+x^2)(1+x^4)=4x^3$$ This problem was asked in the Kettering University Math Olympiad a few years back, it's an interesting problem with many different ways to approach it. I'm going to share my own approach here, please make sure to let me know if there are any mistakes in mine or if anything can be improved, and share your own methods too! Here's my approach for the problem: $$(1+x^2)(1+x^4)=4x^3$$ Divide by $x^3$ on both sides: $$\frac{1+x^2}{x}\cdot\frac{1+x^4}{x^2}=4$$ $$\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=4$$ Now we can set $y=x+\frac{1}{x}$, thus, $y^2-2=x^2+\frac{1}{x^2}$ $$y(y^2-2)=4$$ $$y^3-2y-4=0$$ $$y^3-8-2y+4=0$$ $$(y-2)(y^2+2y+4)-2(y-2)=0$$ $$(y-2)(y^2+2y+2)=0$$ Now obviously, since we need to find real solutions, the term $(y^2+2y+2)$ will need to be rejected, as it can easily be checked that it yields no real roots. [Alternatively, we can set $y^2+2y+2=0$ and observe that $(y+1)^2=-1$, thus this term yields no real solution] Therefore: $$y=2$$ $$x+\frac{1}{x}=2$$ $$\frac{x^2+1}{x}=2$$ $$x^2-2x+1=0$$ $$(x-1)^2=0$$ Therefore, $x=1$ is our only real solution.
After expanding, we're trying to find the roots of the polynomial $p(x)=x^6+x^4-4x^3+x^2+1$. Factoring out obvious roots one at time gives \begin{align*} p(x)&=x^6+x^4-4x^3+x^2+1\\ &=(x-1)(x^5+x^4+2x^3-2x^2-x-1)\\ &=(x-1)^2(x^4+2x^3+4x^2+2x+1)\\ &=(x-1)^2\left[(x^2+x)^2+2x^2+(x+1)^2\right] \end{align*} at which point $x=1$ is clearly the only real root.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4583516", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 2, "answer_id": 0 }
Hard triple Integral $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{2-zx^{2}-zy^{2}}dxdydz=\ln(2^{G})$ How do prove this triple integral? $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{2-zx^{2}-zy^{2}}dxdydz=\ln(2^{G})$$ where G is Catalan's constant. As my try I only reach to this hard single integral: $$\int _0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=G\ln \left(2\right)$$
Integrate by parts $$I=\int _0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx = \frac{\pi^3}{24}+\int_0^1\frac{2x\tan^{-1}x\ln\frac{1-x^2}2}{1+x^2}dx$$ Then, utilize $\int_0^1 \frac{x \tan^{-1} x \ln \left( 1-x^2\right)}{1+x^2}dx= -\frac{\pi^3}{48}-\frac{\pi}{8}\ln^2 2 +G\ln 2$ and $\int_0^1 \frac{x\tan^{-1}x}{1+x^2}dx = \frac12G-\frac\pi2\ln2$ to obtain $I= G\ln2$.
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Calculating tangent at two points $(4a,8a)$ Find the equation of the tangent to the curve $ay^2=x^3$ at the points $(4a,8a)$ I have re-arranged the equation to get $$y = \left(\frac{x^3}{a}\right)^{\frac{1}{2}}$$ Then taking its derivative I get $$\frac{dy}{dx}=\frac{3\left(\sqrt{\frac{x^3}{a}}\right)}{2x}$$, when $x=4a$, I find that $\frac{dy}{dx}=3$ and so I have tried: $$\frac{y-\left(\frac{x^3}{a}\right)^{\frac{1}{2}}}{x-4a}=3$$ However, this will not give me a simple answer which is to be $y = 3x-4a$ How do I correctly calculate at both points?
You have probably miscalculated the derivative when the chain rule was applied. Note that $\dfrac{du^\frac{1}{2}}{du}=\dfrac{1}{2}u^{-\frac{1}{2}}$. Hence, $y'=\dfrac{d}{d\dfrac{x^3}{a}}(\dfrac{x^3}{a})^\frac{1}{2}\cdot\dfrac{d}{dx}(\dfrac{x^3}{a})$ $=\dfrac{1}{2}\cdot(\dfrac{x^3}{a})^{-\frac{1}{2}}\cdot\dfrac{3x^2}{a}$ $=\dfrac{3}{2}\sqrt{\dfrac{x}{a}}$ $y'|_{x=4a}=3$ Therefore, the equation of tangent: $\dfrac{y-8a}{x-4a}=3$ $\underline{\underline{y=3x-4a}}$
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Find the remainder of 3^3^3...^3 divided by 46 Find the remainder of $3^{{{{{3^3}^3}^3}^3}^\cdots} \text{(2020 copies of 3)}$ by 46 Note that $a^{b^c}$ means $a^{(b^c)}$ not $(a^b)^c$ If there was 2 copies of 3: $3^3\equiv27 (mod 46)$. If there was 3 copies of 3: $3^{3^3}\equiv3^{27}\equiv(3^9)^3\equiv(19683)^3\equiv(41)^3\equiv68921\equiv13(mod46)$ However I can't repeat this process 2020 times. How to solve this?
Since $3$ is odd, for all odd $k$ we have $3^k$ is odd, and $3^k=3\pmod 4$. Now consider $$ x_0 = 3^{3^{3^{3^{k}}}}, x_1 = 3^{3^{3^{k}}}, x_2 = 3^{3^{k}}, x_3 = 3^{k} $$ where $k = 3^{3^{⋰^{3}}}$ is the remaining part of the exponent. We have * *$x_0$ is computed modulo $46$, and $x_0 = 3^{x_1} \pmod {46}$ *$x_1$ is computed modulo $\varphi(46)=22$, and $x_1 = 3^{x_2} \pmod {22}$ *$x_2$ is computed modulo $\varphi(\varphi(46))=10$, and $x_2 = 3^{x_3} \pmod {10}$ *$x_3$ is computed modulo $\varphi(\varphi(\varphi(46))=4$ But $k$ is odd, so $x_3 = 3^k = 3 \pmod 4$. Then $x_2 = 3^3 = 7 \pmod {10}$. Then $x_1 = 3^7 = 9 \pmod {22}$. Then $x_0 = 3^9 = 41 \pmod {46}$. So as long as the tower has more than 4 elements, the result will be $41 \pmod {46}$.
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Basic question about floor function and limit ( $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$) $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$ calculate the limit if it exists if not then prove it does not exist I tried approaching by squeeze theorem and floor function property and got $(x-2) \cdot (x+3)-1<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq (x-2) \cdot (x+3)$ but then if I calculate the limits as $x$ approaches zero I get $-7<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq-6$ which did not give me an answer according to squeeze theorem so I tried a different approach by side limits $\lim \limits_{x \to 0^+} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{0-2}\rfloor \cdot \lfloor{0+3}\rfloor = -6$ and $\lim \limits_{x \to 0^-} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{-1-2}\rfloor \cdot \lfloor{-1+3}\rfloor = -6$ so the limit exists and $L=-6$ is this correct? is there a different way? thank you !
$x+2 < \lfloor x+3\rfloor\leq x+3$ and $ x-3<\lfloor x-2 \rfloor\leq x-2$. this implies $(x+2).(x-3)<\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq (x+3).(x-2)$ so $lim_{x \to 0}(x+2).(x-3)<lim_{x \to 0}\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq lim_{x \to 0} (x+3).(x-2)$ $\Rightarrow $$lim_{x \to 0}\lfloor x+3 \rfloor.\lfloor x-2 \rfloor=-6.$ warning : the first implication is not necessary true because if the both left sides are negative and the other are positive maybe in one case we must reverse the product of sides ,but any way in all cases it will get the same limit $-6$.so try to rewrite the complete proof by yourself .
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IVP variable change. I'm working in this IVP: $$(4t^2+16t+17)y'' - 8y = 0, \quad y(-2) = 1, \quad y'(-2) = 0$$ and it's asked to do this variable change $x=t+2$. I did as asked and found the equation: $$ (4x^2 + 1)y'' - 8y = 0. $$ Changing the $t \to x$: then $ x = t + 2 $. Changing $y \to s$: then $ s = y - 1 $. With these we can change this: $ y(-2) = 1 $ to $ s(0) = 0 $ but I'm having some trouble to change the $ y'(-2) = 0 $ to $ s $ end the question. Any help is appreciated!
From $$ (4 \, t^2 + 16 \, t + 17) \, y^{''}(t) - 8 \, y^{'}(t) = 0, \quad y(-2) = 1 \quad y^{'}(-2) = 0 $$ and knowing that the transformation $t \to x-2$ needs to be performed then: $$ dt = dx \to \frac{dx}{dt} = 1$$ and \begin{align} \frac{dy}{dt} &= \frac{dx}{dt} \, \frac{dy}{dx} = \frac{dy}{dx} \\ \frac{d^2 y}{d t^2} &= \frac{d^2 y}{d x^2}. \end{align} Since $x = t + 2$ then when $t = -2$ this gives $x = 0$ such that $y(-2)=1$ becomes $y(0)=1$ when $t \to x$ and $y^{'}(-2) = 0$ becomes $y^{'}(0) = 0$. Now the differential equation becomes, in $x$ variable, $$ (4 \, x^2 + 1) \, y^{''}(x) - 8 \, y(x) = 0, \quad y(0) = 1, \quad y^{'}(0) = 0.$$ This has the solution $$ y(x) = c_{1} (4 \, x^2 + 1) + c_{2} \, ((4 \, x^2 + 1) \, \tan^{-1}(2 x) + 2 x). $$ Using $y(0) = 1$ gives $c_{1} = 1$. Using $y^{'}(0) = 0$ gives $c_{2} = 0$ and $$ y(x) = 4 \, x^2 + 1.$$ Returning to $t$ variable this gives the solution $$ y(t) = 4 \, (t + 2)^2 + 1.$$ Note: The original equation can be solved as is and gives the general solution $$ y(x) = c_{1} \, (4 \, t^2 + 16 \, t + 17) + c_{2} ((4 \, t^2 + 16 \, t + 17) \, \tan^{-1}(2 t + 2) + 2 t + 4). $$ Applying the conditions $y(-2) = 1$ and $y^{'}(-2) = 0$ gives $$ y(t) = 4 \, (t+2)^2 + 1 $$ as the solution.
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How to minimize $(x-1)^2+(y-1)^2+(z-1)^2$ under the constraint $xyz=s$? Let $s \in (0,1)$ be a parameter. Can we find an exact closed form expression for $$ F(s)=\min_{xyz=s,x,y,z>0}(x-1)^2+(y-1)^2+(z-1)^2, \tag{1} $$ and exact formulas the minimizers $x(s) \le y(s) \le z(s)$? I tried using Mathematica but failed (However I am not very skillful). Using Lagrange's multipliers, one gets $$ (x-1,y-1,z-1)=\lambda (\frac{1}{x},\frac{1}{y},\frac{1}{z}), $$ so $x,y,z$ should all satisfy the quadratic equation $t(1-t)=-\lambda$. If $t$ is a solution then so is $1-t$; thus, if we denote the solution of this equation by $a$, then $$ \{x,y,z\} \subseteq \{a,1-a \}. $$ One possible solution is the symmetric solution $x=y=z=\sqrt[3] s$. If $x,y,z$ do not all coincide, then they must take both values $a$ and $1-a$. Since we required $x,y,z>0$ this means that $a,1-a$ should be positive, so $0<a<1$. After a possible a renaming/switching, we may assume that $x=a, y=z=1-a$, so the value $1-a$ is attained twice. So, if we define, $$ G(s)=\min_{a \in (0,1),a(1-a)^{2}=s} (1-a)^2+2a^2, \tag{2} $$ then $$ F(s)=\min \{3(1-\sqrt[3]s)^2,G(s)\}. $$ Since $\max_{a \in (0,1)}a(1-a)^{2}=4/27$, it follows that the non-symmetric solution is possible only if $s \le 4/27$. The term $3(1-\sqrt[3]s)^2$ comes from comparing with the symmetric solution $x=y=s$. I am not sure how to continue, since analyzing explicitly the solutions to the quintic equation is not so nice. Motivation: This problem is a special case of the problem of finding the closest matrix to $\text{SO}_n$ with a given determinant.
You need to solve $$\nabla ((x-1)^2+(y-1)^2+(z-1)^2)=\lambda \nabla (xyz-s)$$ $$[2(x-1),2(y-1),2(z-1)]=[\lambda yz,\lambda xz, \lambda xy]$$ Equating $\lambda /2$ for the first and second terms, we have $$\frac {x-1}{yz}=\frac {y-z}{xz}$$ Note that the constraint $xyz=s $ guarantees that $x,y,z$ are all non-zero. A bit of algebra gives $$x^2-y^2=x-y$$ so $x=y$ or $x+y=1.$Similarly, by comparing first and third or second and third terms in the gradient equation, we obtain $x=z $ or $x+z=1$ and $y=z$ or $y+z=1$ respectively.If $x=y=z$ we have $x=y=z=s^{1/3}.$ If $x+y=1,x+z=1,y+z=1$ we have $x=y=z=1/2$ which is a contradiction unless $s=1/8$, if $s=1/8$ we have the previous point.If $x=y, x+z=1$, that gives $x^2z=s,x+s/x^2=1,x^3-x^2+s=0$ which is a cubic equation that we can solve by the metods of Cardano and his contemporaries.We obtain similar cubic equations for the cases $y=z,y+x=1$ and $z=x,z+y=1$. At each point $x,y,z$ we calculate $(x-1)^2+(y-1)^2+(z-1)^2$and take the minimumof these values.
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Algebraic calculation with polynomial and complex root. Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$. Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as linear combinations of $1, \alpha, \alpha^{2}$ over $\mathbb{Q}$ and conclude that $\beta$ is a zero of $f$. Is $\beta=\alpha$? I found my mistake with your help. The solution: We have $$ \beta^2=(3\alpha^{2}+4 \alpha-14)^2=9 \alpha^4 + 24 \alpha^3 - 68 \alpha^2 - 112 \alpha + 196 $$ To remove the cubic and quartic powers, we use the fact that $\alpha^3-7\alpha+7=0$ holds and gradually eliminate partial polynomials. We get $$ \beta^2=9 \alpha^4- 68 \alpha^2 + 24 \alpha^3 + 16\cdot (-7 \alpha) + 28\cdot 7 =9 \alpha^4- 68 \alpha^2 -8\cdot (-7 \alpha) \ + 4\cdot 7\alpha(9 \alpha^3+10\cdot(-7) \alpha +8\cdot 7+2\alpha)+4\cdot 7=\alpha(-5\alpha -7) + 4\cdot 7=-5\alpha^2-7\alpha+28. $$ Now we try $\beta^3$ with the same method $$ \beta^3=\beta^2\beta=(3\alpha^{2}+4 \alpha-14)^2(3\alpha^{2}+4 \alpha-14)=(-5\alpha^2-7\alpha+28)(3\alpha^{2}+4 \alpha-14)\ =-15\alpha^4+ 126 \alpha^2 -(41 \alpha^3 +30\cdot(-7\alpha) + 56\cdot 7)=-15\alpha^4+ 126 \alpha^2 -( -11\cdot(-7\alpha) + 15\cdot 7)\ =-\alpha(15\alpha^3+18\cdot(-7\alpha) +7\cdot 11) - 15\cdot 7=-\alpha(3\cdot(-7\alpha) -4\cdot7) - 15\cdot 7=21\alpha^2 +28\alpha-105. $$ And with this i get indeed $$ f(\beta)=0. $$ Thanks for your help! Maybe there are different reasonings for why $\beta\neq \alpha$?
You should review your computations, there must be some mistake in them, I tried myself to compute $\beta^{2}$ and $\beta^{3}$ and found $$ \beta^{2} = -5\alpha^{2}-7\alpha+28, $$ $$ \beta^{3} = 21\alpha^{2} +28\alpha -105 $$ Hence $$ f\left(\beta\right) = \beta^{3} - 7\beta +7 = 21\alpha^{2} +28\alpha -105 -7\left(3\alpha^{2}+4\alpha-14\right)+7 = 21\alpha^{2} +28\alpha -105 - 21\alpha^{2} -28\alpha +98 + 7 = 0. $$ I don't know where exactly is your mistake but it must be at the computation of $\beta^{3}$, also $\beta \neq \alpha$, because if we had $\beta = \alpha$ then: $$ \alpha = 3\alpha^{2}+4\alpha-14 $$ $$ \alpha^{2} = -5\alpha^{2}-7\alpha+28 $$ but these two equations have different roots, which is a contradiction.
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Is this quadratic polynomial monotone at solutions of this cubic? Let $0<s < \frac{4}{27}$. The equation $x(1-x)^{2}=s$ admits exactly two solutions in $(0,1)$: Denote by $a,b$ be these solutions, and suppose that $a<b$. Does $$ (1-a)^2+2a^2<(1-b)^2+2b^2 $$ hold? The limitation on the range of $s$ is due to $$\max_{a \in (0,1)}a(1-a)^{2}=\frac{4}{27},$$ which is attained at $a=\frac{1}{3}$. I checked this numerically. I hope for an analytic argument which does not require solving explicitly the cubic. Here is some partial information: Define $F(x)= (1-x)^2+2x^2$. Then $$ F'(x)=2(3x-1)>0 \iff x > \frac{1}{3}. $$ Thus $F|_{(0,\frac{1}{3})}$ is decreasing, while $F|_{(\frac{1}{3},1)}$ is increasing. Unfortunately, this does not seem to help. The motivation for this problem come from this question.
Using my answer to your previous question, we have (after trigonometric simplifications) $$a=\frac{4}{3} \cos ^2\left(\frac{1}{6} \left(\cos ^{-1}\left(\frac{27 s}{2}-1\right)+2 \pi \right)\right)$$ $$b=\frac{4}{3} \sin ^2\left(\frac{1}{6} \left(\cos ^{-1}\left(\frac{27 s}{2}-1\right)+\pi \right)\right)$$ Then $$(1-a)^2+2a^2=\frac{1}{3} \left(5-4 \cos \left(\frac{1}{3} \cos ^{-1}\left(1-\frac{27 s}{2}\right)\right)+2 \cos \left(\frac{2}{3} \cos ^{-1}\left(1-\frac{27 s}{2}\right)\right)\right)$$ and $$(1-b)^2+2b^2=\frac{1}{3} \left(5+4 \sin \left(\frac{1}{3} \sin ^{-1}\left(1-\frac{27 s}{2}\right)\right)-2 \cos \left(\frac{2}{3} \sin ^{-1}\left(1-\frac{27 s}{2}\right)\right)\right)$$ So, if $$\Delta=\Big[(1-b)^2+2b^2\Big]-\Big[(1-a)^2+2a^2\Big]$$ $$\Delta=\frac{8}{\sqrt{3}}\sin ^2\left(\frac{1}{6} \cos ^{-1}\left(\frac{27 s}{2}-1\right)\right) \sin \left(\frac{1}{3} \cos ^{-1}\left(\frac{27 s}{2}-1\right)\right) $$ $\Delta$ has the same sign as $\sin \left(\frac{1}{3} \cos ^{-1}\left(\frac{27 s}{2}-1\right)\right) $ which varies between $\frac{\sqrt{3}}{2}$ and $0$. So $$\color{red}{\Delta ~\geq~ 0 \quad \forall s \in \left(0,\frac 4{27}\right)}$$
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Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation. Let $z=x+yi $. Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$ And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$ So we get $$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$ Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
Hint: Write $-10+5i$ in its polar form, then recognize the exponential has period $2\pi i$. Use this fact to take the cube root of both sides and obtain all solutions. Once you have the solutions, you can find individual real and imaginary parts.
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Intersection of an elliptic paraboloid and a sphere Problem: Prove that sphere $x^2+y^2+z^2=50z$ and elliptic paraboloid $\frac{x^2}{25}+\frac{y^2}{16}=2z$ intersects in two circles and find the radius and center of these circles. I figured out that intersection of them lies in two planes $3y\pm4z$. Since every plane section of a sphere is a circle, we got what we wanted to prove (am I right?) After that I decided to express $y$ in terms of $z$ and substitute y into the equation of the sphere to try find center and radius of circles. After various transformations, I got this equation: $z^2-18z-\frac{9x^2}{25}=0$. And here I stuck and don't understand how to get to the radius and center.
$x^2 + y^2 + z^2 = 50 z $ and $ \dfrac{x^2}{25} + \dfrac{y^2}{16} = 2 z $ Multiplying the second equation by $25$ and subtracting, we get $ y^2 \bigg(1 - \dfrac{25}{16} \bigg) + z^2 = 0 $ Multiply through by $16$: $ - 9 y^2 + 16 z^2 = 0 $ which is, as you said in the question statment, the equation of the two planes $ (4 z - 3 y) (4 z + 3 y) = 0 $ So the two planes are $4z - 3y = 0 $ and $ 4z + 3 y = 0 $ The intersection of a plane as stated is a circle, whose center is the foot of the perpendicular from the center of the sphere to the plane. From $x^2 + y^2 + z^2 = 50 z $ we get $ x^2 + y^2 + (z - 25)^2 = 625 = 25^2 $ Hence the center of the sphere is $(0, 0, 25) $. The normal to the first plane is $(0, -3, 4) $. Therefore, the foot of the perpendicular to the plane from the center of the sphere is $ C = (0, 0, 25) + t (0, -3, 4) $ $C$ lies in the plane, so $(0, -3, 4) \cdot (0, -3 t, 25 + 4 t) = 0 $ From which $ t (9 + 16) = -100 $, therefore, $t = -4 $ which implies that $ C = (0, 0, 25) - 4 (0, -3, 4) = (0, 12, 9 ) $ The distance between $C$ and the center of the sphere is $ d = | t | \| (0, -3, 4 ) \| = 4 (5) = 20 $ Hence the radius of the circle is $\sqrt{ 25^2 - 20^2} = 15$ For the second plane $ 4 z + 3 y = 0 $, the normal to the plane is $(0, 3, 4) $ $ C = (0, 0, 25) + t (0, 3, 4) $ $C$ lies in the plane, so $(0, 3, 4) \cdot (0, 3 t, 25 + 4 t) = 0 $ From which $ t (9 + 16) = -100 $, therefore, $t = -4 $ which implies that $ C = (0, 0, 25) - 4 (0, 3, 4) = (0, -12, 9 ) $ The distance between $C$ and the center of the sphere is $ d = | t | \| (0, 3, 4 ) \| = 4 (5) = 20 $ Hence the radius of the second circle is $\sqrt{ 25^2 - 20^2} = 15$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4602745", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$? If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ? The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$ How to proceed?
You used the quadratic formula incorrectly. $$\begin{align*}20x^2 + 13x - 15 = 0 \implies x &= \frac{-(13) \pm \sqrt{(13)^2 - 4(20)(-15)}}{2(20)}\\ &= \frac{-13 \pm \sqrt{169 + 1200}}{40} \\ &= \frac{-13 \pm \sqrt{1369}}{40} \\ &= \frac{-13 \pm 37}{40} \\ &= \frac{24}{40} \text{ or } - \frac{50}{40} \\ &= \frac 3 5 \text{ or } - \frac 5 4 \\ \end{align*}$$ Therefore $$ 20 x^2 + 13 x - 15 = 20 \left( x - \frac 3 5 \right) \left( x + \frac 5 4 \right)$$ Bring a factor of $5$ into the first parenthetical and $4$ into the second, each from the $20$, to get the desired form and find your answer.
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Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$ My attempt: $$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$ $$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1)-(x^2-2x)}{x^2-3x+1}} \ \ = \ \ \frac{x-2}{\frac{x^3-3x^2+x-2x^2+6x-2-x^2+2x}{x^2-3x+1}} $$ $$ =\frac{(x-2)(x^2-3x+1)}{x^3-6x^2+9x-2} \ \ = \ \ \frac{x^3-3x^2+x-2x^2+6x-2}{x^3-6x^2+9x-2} $$ $$ =\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2} $$ But the answer is given as $ \ \frac{x^2-3x+1}{x^2-4x+1} \ $, so I went wrong somewhere but can't see it. Any help is appreciated thanks.
In the comments I explain how your answer is fine, just not simplified yet. The way I would have simplifed this though avoids that issue. And you may like to study this technique. $$\begin{align} \frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}} &=\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}\cdot\frac{x-2}{x-2}}\\ &=\frac{x-2}{x-2-\frac{x(x-2)}{x(x-2)-(x-1)}}\\ &=\frac{x-2}{x-2-\frac{x(x-2)}{x^2-3x+1}}\cdot\frac{x^2-3x+1}{x^2-3x+1}\\ &=\frac{(x-2)\left(x^2-3x+1\right)}{(x-2)\left(x^2-3x+1\right)-x(x-2)} \end{align}$$ All those $(x-2)$ cancel now, before you would expend effort multiplying out. $$\frac{x^2-3x+1}{x^2-3x+1-x}$$
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Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that $$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$ is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$. I first computed the derivative, but I don't see why it must be positive: $$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$ Any ideas? Thanks a lot for your help. Case $p=3/4$. Then $$f(x)=(x+1)^{1/4}x^{1/4}(\sqrt{x+1}-\sqrt{x})=\frac{(x+1)^{1/4}x^{1/4}}{\sqrt{x+1}+\sqrt{x}}$$ It suffices to show that $\ln f(x)$ is strictly increasing. We have $$\ln f(x) =\frac{1}{4} \ln(x+1)+\frac{1}{4} \ln(x)-\ln(\sqrt{x+1}+\sqrt{x})$$ Taking derivative w.r.t. $x$ we get $$\frac{d \ln f(x) }{d x}=\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x}-\frac{1}{\sqrt{x+1}+\sqrt{x}}(\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}})$$ Hence $\frac{d \ln f(x) }{d x}>0$ is equivalent to $$(\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x})(\sqrt{x+1}+\sqrt{x})>\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}}$$ which is equivalent to $$\frac{1}{4}\frac{\sqrt{x}}{x+1}+\frac{1}{4}\frac{\sqrt{x+1}}{x}>\frac{1}{4}\frac{1}{\sqrt{x+1}}+\frac{1}{4}\frac{1}{\sqrt{x}}$$ or $$\frac{\sqrt{x+1}-\sqrt{x}}{x}>\frac{\sqrt{x+1}-\sqrt{x}}{x+1}$$ which holds. Hence $f(x)$ is strictly increasing.
As zwim just a comment : Starting with your derivative and using the lemma 7.1 (see the reference) we have : $$\frac{px}{x+1}\left(\left(1-p\right)^{2}+\frac{x+1}{x}p(2-p)-\frac{x+1}{x}p(1-p)\ln\frac{x+1}{x}\right)\leq p(x+1)^{p-1}x^{1-p}$$ Or : $$\frac{p(1-p)^{2}x}{x+1}+p^{2}(2-p)-p^{2}(1-p)\ln\frac{x+1}{x}\le p(x+1)^{p-1}x^{1-p}$$ In the lemma we have : $$a=\frac{x+1}{x},c=p$$ We can do it for the four partial expression . I think there is some simplifications at the end . Reference : VASILE CIRTOAJE, PROOFS OF THREE OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS, Journal of Nonlinear Sciences and Applications, 4 (2011), no. 2, 130-137 Another way : Considering $a>0$ fixed and the function for $0.5\leq x\leq 1$ : $$g(x)=(a+1)^{x-1}a^{1-x}-(a+1)^{x}a^{-x}+(a+1)^{-x}a^{x}-(a+1)^{1-x}a^{x-1}$$ $$g'(x)=-a^{\left(-x-1\right)}(a+1)^{\left(x-1\right)}(a^{\left(2x+1\right)}+a^{2x}-a(a+1)^{\left(2x\right)})(\log(a)-\log(a+1))\leq 0$$ So the function in $x$ is decreasing so $$f(1)\leq f(x)$$ Now consider your derivative : $$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$ We have : $$\frac{d f(x)}{d x}=p\left((x+1)^{p-1}x^{1-p}-(x+1)^{p}x^{-p}\right)+(x+1)^{p}x^{-p}+p\left((x+1)^{-p}x^{p}-(x+1)^{1-p}x^{p-1}\right)-(x+1)^{-p}x^{p}$$ Or : $$\frac{d f(x)}{d x}=p\left((x+1)^{p-1}x^{1-p}-(x+1)^{p}x^{-p}+(x+1)^{-p}x^{p}-(x+1)^{1-p}x^{p-1}\right)+(x+1)^{p}x^{-p}-(x+1)^{-p}x^{p}$$ Now use $g(x)$ and another function like $g(x)$ to conclude partially. In fact I show that $f'(x)\leq 0$ for $0<p\leq 1/2$
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Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as: $\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $ Is there another simpler method to evaluate the integral? Your comments and alternative methods are highly appreciated.
As a heuristic calclulation which can be proved later: Let $I(a) = \int^{2\pi}_{0}\frac{dx}{1+a\cos(x)}$. Using the geometric series: \begin{align*} \frac{1}{1+a\cos(x)} = \sum^{\infty}_{n=1} (-1)^na^n\cos^n(x) \end{align*} We have: \begin{align*} I(a) = \sum^{\infty}_{n=1} (-1)^na^n \int^{2\pi}_{0} \cos^n(x)dx \end{align*} Consider: \begin{align*} \int\cos^n(x)dx &= \cos^{n-1}(x)\sin(x) + (n-1) \int\cos^{n-2}(x)\sin^2(x)dx \end{align*} This means $\int^{2\pi}_{0}\cos^n(x)dx = \frac{(n-1)}{n}\int^{2\pi}_{0 }\cos^{n-2}(x)dx$. By iterating you would find the odd terms zero and the series expansion to be: \begin{align*} I(a) = (2\pi)\left(1 + \frac{1}{2}a^2 + \frac{3}{8}a^4 ...\right) \end{align*} which would be the expansion of $2\pi(\sqrt{1-a^2})^{-1}$ I think we can justify using the uniform convergence of the geometric series when $|a|<1$ that we can switch summation and integration
{ "language": "en", "url": "https://math.stackexchange.com/questions/4605188", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 6, "answer_id": 2 }
Is it possible to adjust the sizes of square and circle inside a unit circle so the combined area is half of that unit circle. A square and a circle fits together inside an unit circle symmetrically. Is it possible to adjust the sizes of circle and the square so that their combined area is half of the area of that unit circle ? I tried it like this : As the circle is of unit radius so radius is equal to 1 and so its area is irrational. Now let the length of the square be $a$ and that radius of circle be $r$ and now the question becomes is it possible such that $a^2$+$πr^2$ = $π/2$ since the combined area is half of the unit circle. There I got stuck as RHS is irrational . Any help would be appreciated.
Trying to Proving the Existence ( Answering the Question in the title ) : Where-ever we take the Centre $X$ of the Inner Circle , we can get the Combined total Area by Drawing the Square beside it. We can move the Point $X$ from right Extreme to left Extreme. In the Image , we can Draw the largest Square ( Blue ) inside the Circle & then Draw the Small Circle ( Blue ) beside it. Here $X$ is right Extreme. Area of outer Circle is $\pi \approx 3.14$ Here , the Combined Blue total Area is more than half the outer Circle. Diagonal of Square is $2$ , Area is $2$ , Combined Area will be a little more than $2$. Now , the Purple Circle can be almost the whole Outer Circle, with a very small Purple Square. Here $X$ is left Extreme. Hence , Combined Purple total Area is more than half the outer Circle. Combined Area varies continuously from more than half to some minimum to more than half. Is there a Point $X$ where Combined Area is Exactly half ? If YES , Where is that Point $X$ ? We can figure that out by making a Equation. Let the Inner Circle have Diameter $x$ , & the Square have Side $y$. Combined Area is $ y^2 + \pi x^2 / 4 = \pi / 2 $ [[ EQ1 ]] The left most Part $ z = 1 - \sqrt {1-y^2/4} $ Hence the Outer Circle Diameter is $ 2 = x+y+z $ $ 2 = x + y + 1 - \sqrt {1-y^2/4} $ $ x+y-1 = \sqrt {1-y^2/4} $ $ (x+y-1)^2 = 1-y^2/4 $ [[ EQ2 ]] Solving [[ EQ1 ]] & [[ EQ2 ]] will give the necessary Point $X$. Value will be cumbersome , involving square root , $\pi$ , square , Etc. $ y^2 + \pi x^2 / 4 = \pi / 2 $ $ (x+y-1)^2 = 1-y^2/4 $ $ y^2 = \pi / 2 - \pi x^2 / 4 $ $ y = \sqrt { \pi / 2 - \pi x^2 / 4 } $ $ (x+\sqrt { \pi / 2 - \pi x^2 / 4 }-1)^2 = 1-\sqrt { ( \pi / 2 - \pi x^2 / 4 ) }^2/4 $ $ (x+\sqrt { \pi / 2 - \pi x^2 / 4 }-1)^2 = 1- ( \pi / 2 - \pi x^2 / 4 )/4 $ $ (x-1)^2+ ( \pi / 2 - \pi x^2 / 4 ) + 2(x-1)\sqrt { \pi / 2 - \pi x^2 / 4 } = 1- ( \pi / 2 - \pi x^2 / 4 )/4 $ $ 2(x-1)\sqrt { \pi / 2 - \pi x^2 / 4 } = -(x-1)^2 - ( \pi / 2 - \pi x^2 / 4 ) + 1- ( \pi / 2 - \pi x^2 / 4 )/4 $ $ 4(x-1)^2 ( \pi / 2 - \pi x^2 / 4 ) = [ -(x-1)^2 - \pi / 2 + \pi x^2 / 4 + 1 - \pi / 8 + \pi x^2 / 16 ] ^2 $ We can use Wolfram to get the answer $x ≈ 0.995126$ & hence $y ≈ 0.890525$ . We can then verify this : Combined Area is $ 0.890525^2 + \pi 0.995126 / 4 = 1.57460...$ which is quite close to half the Outer Circle Area.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4607126", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Indefinite integral $\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $ involving $\mathrm{sgn}()$ function How to solve integrals like $$\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $$ If we factor out $x^2$ from denominator it become $=\int \frac{x}{\left| x \right|} \frac{x}{\sqrt{1 + x^2}} \, dx $ $=\int \mathrm{sgn}(x) \frac{x}{\sqrt{1 + x^2}} \, dx $ What to do next ? can't we do this like $=\int \frac{x}{x} \frac{x}{\sqrt{1 + x^2}} \, dx $ $=\int \frac{x}{\sqrt{1 + x^2}} \, dx $ it becomes an easy integral with substitution now. Is it wrong ? So my quetion is that can we write $\sqrt{x^2}$ as x in this case ?
To avoid the sign function, integrate as follows \begin{align} \int \frac{x^2}{\sqrt{x^2 + x^4}}dx =\int \frac{\sqrt{x^2 + x^4}}{x^2}\overset{ibp}{dx} - \int \frac{1}{\sqrt{x^2 + x^4}} dx =\frac{\sqrt{x^2 + x^4}}{x} \end{align}
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Where is error in this solution to $ \sqrt{x} + \sqrt{-x} = 2 $? Given the equation: $$ \sqrt{x} + \sqrt{-x} = 2 $$ The solutions are $x = \pm 2i$. This can be seen via Wolfram Alpha $$ \left( \sqrt{x} + \sqrt{-x} \right)^2 = 2^2 $$ $$ \sqrt{x}^2 + 2\sqrt{x}\sqrt{-x} + \sqrt{-x}^2 = 4 $$ $$ x + 2\sqrt{x}\sqrt{-x} - x = 4 $$ $$ \sqrt{-x^2} = 2 $$ $$ -x^2 = 4 $$ $$ x = \pm 2i $$ However, my approach to the problem only found the positive value to this equation. $$ \sqrt{x} + \sqrt{-x} = 2 $$ $$ \sqrt{x} + i\sqrt{x} = 2 $$ $$ (1+i) \sqrt{x} = 2 $$ $$ (1-i)(1+i) \sqrt{x} = 2(1-i) $$ $$ 2 \sqrt{x} = 2 (1-i) $$ $$ x = (1-i)^2 $$ $$ x = 1^2 - 2i + (-i)^2 = -2i $$ Where did I make a mistake here that resulted in me only getting one of the two solutions to this equation?
Squaring twice, $$ x +(-x) + 2 \sqrt{x}\sqrt{-x} =4 \to -x^2=4$$ and solutions of quadratic equation $$ x^2+4 = 0$$ should be two in number $$ x= \pm \sqrt {2} i~. $$
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Solving $\tan\beta\sin\gamma-\tan\alpha\sec\beta\cos\gamma=b/a$, $\tan\alpha\tan\beta\sin\gamma+\sec\beta\cos\gamma=c/a$ for $\beta$ and $\gamma$ I am trying to solve the following tricky system of two trigonometric equations. $\alpha, a, b, c$ are all given (and $a$ is nonzero) so I am trying to solve for $\beta$ and $\gamma$. $$\tan{\beta}\sin{\gamma}-\tan{\alpha}\sec{\beta}\cos{\gamma}=\frac{b}{a}$$ $$\tan{\alpha}\tan{\beta}\sin{\gamma}+\sec{\beta}\cos{\gamma}=\frac{c}{a}$$ If this is too difficult to solve by hand, are there any pieces of software/programming libraries that could help me solve this? I plugged it into woflramalpha and got a horrendous ~34 line equation for $\gamma$. I am hoping that a cleaner solution exists after some simplification.
In order to see more symmetry, multiply both equations $$ \left\{ \begin{aligned} \tan\beta \sin\gamma - \tan\alpha \sec\beta \cos\gamma &= \frac{b}{a} \\[3pt] \tan\alpha \tan\beta \sin\gamma + \sec\beta \cos\gamma &= \frac{c}{a} \end{aligned} \right. $$ by $\cos \alpha$ to get $$ \left\{ \begin{aligned} \cos\alpha \tan\beta \sin\gamma - \sin\alpha \sec\beta \cos\gamma &= \frac{b}{a} \\[3pt] \sin\alpha \tan\beta \sin\gamma + \cos\alpha \sec\beta \cos\gamma &= \frac{c}{a} \end{aligned} \right. $$ which is of the form $$ \left\{ \begin{aligned} U \cos\alpha - V \sin\alpha &= \frac{b}{a} \\[3pt] U \sin\alpha + V \cos\alpha &= \frac{c}{a} \end{aligned} \right. $$ for $U = \tan\beta \sin\gamma$ and $V = \sec\beta \cos\gamma$. This is rotation transformation of coordinates $(U, V)$ which we can invert: $$ \left\{ \begin{aligned} U &= \frac{b}{a} \cos\alpha + \frac{c}{a} \sin\alpha \\[3pt] V &= -\frac{b}{a} \sin\alpha + \frac{c}{a} \cos\alpha \end{aligned} \right. $$ Now, we need to solve the equations for $(U, V)$ in terms of $(T, C) = (\tan\beta, \cos\gamma)$. Squaring gives us $$ \left\{ \begin{aligned} U^2 &= T^2 (1 - C^2) \\ V^2 &= (1 + T^2) C^2 \end{aligned} \right. $$ This is a system of quadratic equations in $T^2$ and $C^2$, so it has at most $4$ solutions.
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Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$. My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$ Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
The given inequality can be transformed to various equivalent forms: $$\begin{align} 9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9 &\iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3} \iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3\\ &\iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}\end{align}$$ Notice $3 > 2\sqrt{2} \sim 2.828 > e \sim 2.718 $ and the function $x^{\frac1x}$ is strictly decreasing for $x > e$ (standard calculus exercise). The rightmost "inequality" is true and hence the original "inequality" $9^{\sqrt{2}} < \sqrt{2}^9$ is true.
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I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me. $x,y,z>0$. Prove that:$$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$ My solution:$$\frac{x+y}{2}≥\frac{2xy}{x+y}$$ $$\frac{y+z}{2}≥\frac{2yz}{y+z}$$ $$\frac{x+z}{2}≥\frac{2xz}{x+z}$$ Summing up these inequalities $$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me.
This can even be shown without the help of any theorems. Note that $4 x y = (x+y)^2 - (x-y)^2$, likewise for the other terms. Substituting this into the question gives $$ \frac{(x+y)^2 - (x-y)^2}{2 (x+y)} + \frac{(y+z)^2 - (y-z)^2}{2 (y+z)} + \frac{(x+z)^2 - (x-z)^2}{2 (x+z)} \\ ≤ \frac12 ((x+y) + (y+z) + (x+z)) $$ or $$ - \frac{(x-y)^2}{2 (x+y)} - \frac{(y-z)^2}{2 (y+z)} - \frac{ (x-z)^2}{2 (x+z)} ≤ 0 $$ which is obviously true. It also shows that equality will only be obtained for $x=y=z$. $\qquad \Box$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4612207", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 0 }
Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions, $$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$ For the integral $J$, we are going to transforms it into a Beta function by letting $y=\frac{1}{1+x^n}$. $$\displaystyle \begin{aligned}J & =\frac{1}{n} \int_0^1 y^{-\frac{2}{n}}(1-y)^{\frac{2}{n}-1} d y \\& =\frac{1}{n} B\left(-\frac{2}{n}+1, \frac{2}{n}\right) \\& =\frac{\pi}{n} \csc \left(\frac{2 \pi}{n}\right) \quad \textrm{ (By Euler Reflection Formula)}\end{aligned}\tag*{} $$ Next, we are going to evaluate the integral $ \displaystyle K=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $ by the theorem $ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $ We first split the integral into two integrals $$ \displaystyle \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$$ Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x}$ m, we have $$\displaystyle \displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $$ Putting back yields $ \begin{aligned}\displaystyle K&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\ & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\ & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right] \end{aligned}\tag*{} $ Modifying yields $$\displaystyle \begin{aligned} K&=\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\& =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right)\end{aligned} \tag*{} $$ By the Theorem, $$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $$ where $ \displaystyle z\notin Z,$ we have $$ \displaystyle \displaystyle K=\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n} =J\tag*{} $$ We can now conclude that $\displaystyle \boxed{I=J-K=0 }\tag*{} $ Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$
Here is maybe the simplest proof\begin{align} &\int_0^{\infty} \frac{x^a-2 x+1}{x^{2 a}-1} d x\\ =& \int_0^{\infty} \frac{x^a-x}{x^{2 a}-1} d x - \int_0^{\infty} \frac{x-1}{x^{2 a}-1}\overset{x\to 1/x}{ d x}\\ =& \int_0^{\infty} \frac{x^{2a-3}-x}{x^{2 a}-1} d x - \int_0^{\infty} \frac{x^{2a-2}-x^a}{x^{2 a}-1}\overset{x\to x^2}{ d x}\\ =& \int_0^{\infty} \frac{x^{2a-3}-x}{x^{2 a}-1} d x -\int_0^{\infty} \frac{2(x^{4a-3}-x^{2a+1})}{(x^{2 a}-1)(x^{2 a}+1)}d x\\ =& \int_0^{\infty} \frac{x^{2a-3}-x}{x^{2 a}+1}\overset{x\to 1/x}{d x}=0 \end{align} Note that $a$ need not be an integer.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4614562", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "14", "answer_count": 3, "answer_id": 1 }
Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$. How can I approach this? (The maximum value of the objective function $x^2 + y^2$ seems to be 2. If I start out by assuming this I am able to show that the function cannot exceed this value. But this is kind of doing it in reverse as I start by evaluating the function at $y=-x=1.$ This also doesn't prove that $y=-x$ is the only optimal solution. How would I do it without "guessing" the solution first?)
Substitutions work much faster . Let $\thinspace x=a+b,\thinspace y=a-b$, then you have : $$\begin{align}&x^2+y^2=1-xy\\ \iff &3a^2+b^2=1\end{align}$$ This leads to the following : $\underline{\rm {Global\thinspace\thinspace maximum}}$ $$\begin{align}x^2+y^2&=2(a^2+b^2)\\ &=2(a^2+1-3a^2)\\ &=2-4a^2\leq 2\end{align}$$ Equality occurs iff, when $a=0,\thinspace |b|=1$, which corresponds to $\left(x,y\right)=\left(\pm 1,\mp 1\right)\thinspace . $ $\underline {\rm {Global\thinspace\thinspace minimum }}$ $$\begin{align}x^2+y^2&=2(a^2+b^2)\\ &=2\left(\frac {1-b^2}{3}+b^2\right)\\ &=\frac 23\left(2b^2+1\right)\geq\frac 23\end{align}$$ Equality occurs iff, when $b=0,\thinspace |a|=\frac {\sqrt 3}{3}$, which corresponds to $\left(x,y\right)=\left(\pm \frac {\sqrt 3}{3},\pm \frac {\sqrt 3}{3}\right).$ This completes the answer.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4616478", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 7, "answer_id": 4 }
Do 1.$\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$ 2. $\sum_{n=1}^\infty(e^{\cos n/n}−\cos\frac{1}n)$ converge? Does $$ 1. \hspace{8mm} \sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n} $$ and $$ 2. \hspace{8mm} \sum_{n=1}^{\infty} \left(e^\frac{\cos{n}}{n} - \cos\frac{1}{n}\right) $$ converge? In $\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$, I allocate two sequences of 1. cos n is bounded and $\frac{1}{n^\frac{2}{3} + (-1)^n}$, where $(-1)^n$ irrelevant next to $n^\frac{2}{3}$ for large and, so I get sequence $\frac{1}{n^\frac{2}{3}}$, this sequence is monotonically decreases to $0$. On the basis of Dirichlet, the series converges. Is it so? But I don't have any ideas about the second number..
For the first series, even though $(-1)^n$ is neglible compared with $n^{2/3}$ as $n$ gets large, the monotone hypothesis of the Dirichlet test is not met and it cannot be applied directly. However, we have $$\sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}= \sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}\frac{n^\frac{2}{3} - (-1)^n}{n^\frac{2}{3} - (-1)^n} = \underbrace{\sum_{n=2}^m \frac{n^{\frac{2}{3}}\cos{n}}{n^\frac{4}{3} -1}}_{A_m}- \underbrace{\sum_{n=2}^m \frac{(-1)^n\cos{n}}{n^\frac{4}{3} -1}}_{B_m}$$ Now we can apply the Dirichlet test to show that the sequences of partial sums $A_m$ and $B_m$ on the RHS converges. To handle $B_m$, note that $(-1)^n \cos n = \cos n\pi\cos n = \frac{1}{2}[\cos (n(\pi+1)) + \cos (n(\pi-1))]$, and use the fact that $\sum_{n=1}^m \cos nx$ is bounded for all $m$ and fixed $x$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4625043", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
$\int_0^1\frac{1}{7^{[1/x]}}dx$ $$\int_0^1\frac{1}{7^{[1/x]}}dx$$ Where $[x]$ is the floor function now as the exponent is always natural, i converted it to an infinite sum $$\sum\limits_{k=1}^{\infty} \frac{1}{7^{[1/k]}}$$ Which is an infinite GP, whose sum is 6. However, this is wrong, why is this wrong?
Note that when $x\in\left(\frac{1}{n+1},\frac{1}{n}\right],$ for some $n\in\mathbb{N},$ we have that $\lfloor\frac{1}{x}\rfloor=n.$ This observation motivates me to split up the given integral as such: $$\begin{align*} \int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\int_\frac{1}{2}^1\frac{dx}{7}+\int_\frac{1}{3}^\frac{1}{2}\frac{dx}{7^2}+\int_\frac{1}{4}^\frac{1}{3}\frac{dx}{7^3}+\ldots\\ &=\left(\frac{1}{7}+\frac{1}{2\times7^2}+\frac{1}{3\times7^3}+\ldots\right)-\left(\frac{1}{2\times7}+\frac{1}{3\times7^2}+\frac{1}{4\times7^3}+\ldots\right)\\ &=\boxed{1+6\ln{\left(\frac{6}{7}\right)}} \end{align*}$$ I used the series expansions for $\ln(1+x)$ and $\ln(1-x)$ in the last step. Note: Just so there is no confusion, I have more explicitly performed the calculation. From the observation, we get that $$\begin{align*} \int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\sum_{i\geq1}\int_\frac{1}{i+1}^\frac{1}{i}\frac{dx}{7^i}\\ &=\sum_{i\geq1}\left(\frac{1}{i\times7^i}-\frac{1}{(i+1)\times7^i}\right)\\ &=\sum_{i\geq1}\frac{1}{i\times7^i}-\sum_{i\geq1}\frac{1}{(i+1)\times7^i}\\ &=-\ln{\left(1-\frac{1}{7}\right)}-7\left(-\ln{\left(1-\frac{1}{7}\right)-\frac{1}{7}}\right)\\ &=\boxed{1+6\ln{\left(\frac{6}{7}\right)}} \end{align*}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4628117", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$. I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$. In both cases, squares are always congruent to $0$ or $1$. So \begin{align*}n^2 &\equiv_3 0, 1 \\ \implies (n-1)^2 &\equiv_3 n-2, n-1\end{align*} Given $-1\equiv_3 2$, then $(n-1)^2\equiv_3 n-2$ or $(n-1)^2\equiv_3 n-1$. But now I'm stuck here. I know the polynomial is made up of a product of 3 consecutive numbers, so either two of them are even and one odd, or two of them are odd and one even. When I made the assumption $n$ is even, then $n-1$ is odd and $n-2$ is even, eventually landing at $n(n-2)(n-1)^2 \equiv_4 n(n-2)$ but so far nothing helpful has come up. What am I missing?
If $n$ is even, $(n-2)$ is also even, and $n(n-2)$ is divisible by 4 If $n$ is odd, $(n-1)$ is even, and $(n-1)^2$ is divisible by 4.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4629961", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0 }
Transformation of Variables (Finding The Distribution) Let $V \sim f(v) = Tv^2,$ such that $-2<v<1$ and $T$ is a constant. I want to know the distribution of $W=V^2$ and the value $T$. My attempt is that $V = \pm \sqrt{ W }.$ We know that $$\int_{-2}^1 Kx^2 = 1 \quad \Rightarrow 3K =1, \quad K = 1/3$$ Then $V\sim f(v) = \frac{1}{3}v^2, \ -2<v<1$ and $$F_V(v) = \int_{-2}^y \frac{1}{3} v^2 dy + \int_{y}^1 \frac{1}{3} v^2 dy = \frac{y^3+8}{9}+\frac{1-y^3}{9} = 1$$ Then, \begin{align*} F_V (W \le w) & = P(V \le \pm \sqrt{w})\\ & = P(-\sqrt{w} \le V \le \sqrt{w})\\ & = F_V(\sqrt{w}) - F_V(-\sqrt{w})\\ f_V(w) & = \frac12 f_V (w^{-1/2}) - \frac12 f_V (-w^{-1/2})\\ & = \frac12\left(\frac13(w^{-1}) \right) + \frac12\left(\frac13(w^{-1}) \right)\\ & \stackrel{?}{=} \frac{1}{3}w \end{align*} Could that be wrong?
The value of the constant is right. I continue later with your approach. First of all we need the cdf of $V$, since we work with inequalities. For this purpose we integrate the pdf from $-2$ (lower bound) to $v$: $$\int_{-2}^v \frac13t^2 \ dt=\left[\frac19t^3\right]_{-2}^v=\frac{v^3}{9}+\frac{8}{9}$$ Thus the cdf is $$F_V(v)=\begin{cases} 0, \ v\leq -2 \\ \frac{v^3}{9}+\frac{8}{9}, \ -2< v \leq 1 \\ 1, \ v>1 \end{cases}$$ Since the interval is not symmetric around 0 we make a case distinction: First case: $v\in [-1,1) \Rightarrow w\in [0,1)$. Then we have $v=w^{1/2} $ for $v\in [0,1)$ and $v=-w^{1/2}$ for $v\in [-1,0)$ Second case: $v\in [-2,-1] \Rightarrow w\in [1,4]$ At the first case we can use your interval: $$F_V(\sqrt{w}) - F_V(-\sqrt{w})=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{\left(-w^{1/2}\right)^3}{9}+\frac{8}{9}\right)$$ $$F_W(w)=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{-w^{3/2}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac{8}{9}+\frac{w^{\frac32}}{9}-\frac89=\frac{2}{9}w^{\frac32}$$ At the second case $V$ is negative. Thus we have $P((-V)^2<w)=P(-V<w^{\frac12})=P(V>-w^{\frac12})=1-F\left(-w^{\frac12} \right)$ Inserting the value into the cdf results in $$F_W(w)=1-\left(\frac{\left(-w^{\frac12}\right)^3}{9}+\frac{8}{9}\right)=1-\left(\frac{-w^{\frac32}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac19$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4631927", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Proof of n-times matrix multiplication I want to proof the following: $$ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n= \left[ {\begin{array}{cc} 1 & 0 \\ 2^n-1 & 2^n \\ \end{array} } \right] $$ I betitle the matrix on the left with $A$ and the matrix on the right with $B$. I am not finding the right idea. I started with stating that the resulting matrix of $ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n $ has to be a $2\times2$ matrix because of the definition of matrix multiplication. Then I wrote the following $$ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n= \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]\times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]\times\dots\times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 1\times 1+0\times 1 & 1\times 0 + 0\times 2 \\ 1\times 1 + 2\times 1 & 1\times 0 + 2\times 2 \\ \end{array} } \right]\times\dots\times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 1 & 0 \\ 3 & 4 \\ \end{array} } \right]\times\dots\times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]$$ My idea behind this was that this should show that the first line will never change, because we always just multiply with the same matrix again. I think this is quite easy to see. For the second line it's in my opinion quite easy to see that the second number ($b_{22}$) is always the old second number ($a_{22}$) times 2 and because the first number with wich we started ($a_{22}$) is the number 2 it's just always powers of 2. And the first number ($b_{21}$) is always the old first number ($a_{21}$) plus the old second number ($a_{22}$). And because the first number with wich we started ($a_{21}$) is on less than the second number with wich we started ($a_{22}$) it's always one less than the power of 2. I have problems with explaining my proof and I am not quite sure if it's correct, so can maybe someone recommend a more exact way of proofing this, because I don't know if my sentences are exact enough. Induction Try Thanks to the comment from @Jaap Sherphuis I now tried to proof the statement with induction and started with the basecase $n=1$. $$ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^1=\left[ {\begin{array}{cc} 1 & 0 \\ 2^1-1 & 2^1 \\ \end{array} } \right]$$ This is obviously true. Now I take $ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n= \left[ {\begin{array}{cc} 1 & 0 \\ 2^n-1 & 2^n \\ \end{array} } \right] $ as given and try to show with it the statement for $n=n+1$. $$ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^{n+1}= \left[ {\begin{array}{cc} 1 & 0 \\ 2^{n+1}-1 & 2^{n+1} \\ \end{array} } \right]\Leftrightarrow \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n \times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 1 & 0 \\ 2^n-1 & 2^n \\ \end{array} } \right] \times \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right] $$ Because $ \left[ {\begin{array}{cc} 1 & 0 \\ 1 & 2 \\ \end{array} } \right]^n= \left[ {\begin{array}{cc} 1 & 0 \\ 2^n-1 & 2^n \\ \end{array} } \right] $ is true the whole statement is true for $n=n+1$ and thus the whole statement is true for every $n\in\mathbb{N}$. Is this the correct approach?
The matrix can be represented as $$I+P,\quad P=\begin{pmatrix}0&0\\ 1&1\end{pmatrix}$$ We have $P^2=P.$ Thus $$(I+P)^n=I+\sum_{k=1}^n{n\choose k}P^k=I+(2^n-1)P$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/4634252", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Solution of complex differential equation Finding solution of differential equation $\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$ Where $y=f(x)$ and $f(0)=0$ and $f(1)=1$ I have write $\displaystyle \int \frac{xdy-ydx}{x^2}=-\int \frac{1}{2x^2}(xdx-ydy)\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}$ $\displaystyle \int d\bigg(\frac{y}{x}\bigg)=-\int \frac{1}{x^2}d(x^2-y^2)\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}$ But I did not understand what I do with $x^2$ term in denominator so that right side become integrable. Please have a look.
$$\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$$ $$2x^2d\dfrac yx=-\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}d(x^2-y^2)$$ Substitute $x^2-y^2=v$ and $u=\dfrac yx$: $$2x^2du=-\sqrt{\frac{v+1}{v}}dv$$ For the $x^2$ factor we have: $$v=x^2-y^2=x^2(1-u^2)$$ $$ \implies {x^2}=\dfrac v{1-u^2}$$ The DE becomes: $$\dfrac {2v} {u^2-1}du=\sqrt{\frac{v+1}{v}}dv$$ The DE is separable but not sure if it's easy to integrate. Try to substitute $w=\dfrac 1 v$. Or maybe it's better to substitute $w^2=1+\dfrac 1v$ and $2wdw=-\dfrac {d v}{v^2}$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4637107", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 1, "answer_id": 0 }
$2\sqrt3 h\geq$ perimeter Problem: Let $ABC$ be an acute triangle and $h$ be the longest height. Show that $2\sqrt3h\geq$ the perimeter of $ABC$. I have a solution using geometric arguments, where I reduce the problem into isoceles triangle and bash out the length. Here is my problem: When I tried to use algebraic approaches, I stuck at the following: $$3(b+c-a)(c+a-b)(a+b-c)\geq c^2(a+b+c)$$ where $a,b,c$ is the side length with $c$ the minimum. I wonder if this is doable, and if so, how? Thanks a lot.
Not entirely complete but even I've tried and I am also stuck at an inequality, I took $a$ as the smallest side $h = \frac{2\triangle}{a} \ \ \ and \ \ 4R\triangle = abc\\ h = \frac{bc}{2R}\\ h = \frac{bc\ sina}{a}\\ $ According to the question $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\sqrt3h > 2s \\\\ \ \ \ \ \ \frac{\sqrt3bcsina}{a} > s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin(a) > \frac{as}{\sqrt3 bc}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin^2(a) > \frac{a^2s^2}{3b^2c^2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 - cos^2(a) > \frac{a^2s^2}{3b^2c^2}\\ 1 - \frac{(b^2 + c^2 - a^2)^2}{4b^2c^2} > \frac{a^2s^2}{3b^2c^2}\\ 12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > 4a^2s^2 \\ \ \ \ \ \ \ \ \ \ \ 12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > a^2(a+b+c)^2 $$ Simplyfing it is $6\displaystyle \sum_{cyc}a^2 b^2 > 3(\displaystyle \sum_{cyc}a^4) + a^2(\displaystyle \sum_{cyc}a^2) + 2a^2\displaystyle \sum_{cyc}(ab)$ Objective is to prove $a<b$ and $a<c$ I suppose but I've had no luck so far, I'll comment if I've managed to find something.
{ "language": "en", "url": "https://math.stackexchange.com/questions/4639497", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 5, "answer_id": 4 }
Taylor expansion of $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$ Let $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$, for $x>0.$ Show that for all $n\in \mathbb N$, the function $g(x)=f(x)+\ln(x)$ has an expansion of order $n$. Comptue for $n=8$. I have tried to develop the term $\frac{1}{\sqrt{t^3+x^3}}$ in the neihborhood of $0$ but in vain. Any help is really appreciated.
Assume for a moment that $0 < x < 1$, and write \begin{align*} f(x) &= \int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}} - \int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}. \end{align*} The first integral can be handled by substituting $t = x u^{1/3}$: \begin{align*} \int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}} &= \frac{1}{3\sqrt{x}} \int_{0}^{\infty} \frac{u^{1/3-1}}{(1 + u)^{1/2}} \, \mathrm{d}u \\ &= \frac{1}{3\sqrt{x}} B\left(\frac{1}{3}, \frac{1}{6} \right) = \frac{\Gamma(4/3)\Gamma(1/6)}{\Gamma(1/2)\sqrt{x}}. \end{align*} The second integral can be integrated term-by-term using the binomial series: \begin{align*} \int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}} &= \int_{1}^{\infty} \frac{\mathrm{d}t}{t^{3/2} \sqrt{1 + (x/t)^3}} \\ &= \int_{1}^{\infty} \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{t^{3n+3/2}} \\ &= \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{3n+1/2} \\ &= 2 - \frac{x^3}{7} + \frac{3 x^6}{52} - \frac{5 x^9}{152} + \cdots. \end{align*} Therefore, we get $$ f(x) = \frac{\Gamma(4/3)\Gamma(1/6)}{\Gamma(1/2)\sqrt{x}} - \left( \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{3n+1/2} \right). $$ This series is the same as what Robert Israel found.
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Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to $$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$ However, I am not sure how to prove the above or if expansion and rearrangement are even correct steps.
Since $f(x)=x^\frac{3}{2}$ is convex in $\mathbb{R}^+$ (its second derivative is positive), we have $$\left(\frac{x^2+y^2}{2}\right)^{3/2}=f\left(\frac{x^2+y^2}{2}\right)\leq\frac{f(x^2)+f(y^2)}{2}=\frac{x^3+y^3}{2}.$$ by the definition of convexity. Taking the square of both sides gives the result.
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Show that the Maclaurin expansion of exact ODE general solution has the same form as the power series solution So, given the ODE $$ y''+2y'+y=0 $$ I have found a power series solution, coefficient recurrence relation and the general solution in terms of elementary functions: $$ \begin{aligned} a_{n+2} &= -\frac{2a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}, \: \: n=0,1,2,... \\ \\ y(x) &= \sum_{n=0}^{\infty}a_nx^n\\ &= a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)\\ \\ &= C_1e^{-x} + C_2xe^{-x}\\ \\ &= C_1\sum_{n=0}^{\infty}\frac{(-x)^n}{n!} + C_2x\sum_{n=0}^{\infty}\frac{(-x)^n}{n!} \end{aligned} $$ Now, I want to confirm my power series solution by showing that the Maclaurin expansion of the exact general solution has the same form as the power series, and find relations for $C_1$ and $C_2$ in terms of $a_0$ and $a_1$. I.e. find A, B, C, D such that $$ \begin{align} C_1 &= Aa_0 + Ba_1\\ C_2 &= Ca_0 + Da_1 \end{align} $$ I'm really struggling with this.. Equating coefficients feels like it's leading me nowhere and I don't know what else to try. Edit: my original ODE power series solution $$ \begin{align} y = \sum_{n = 0}^{\infty}a_nx^n, \:\: y' = \sum_{n = 0}^{\infty}na_nx^{n-1}, \:\: y'' = \sum_{n = 0}^{\infty}(n-1)na_nx^{n-2} \end{align} $$ So the ODE becomes $$ \begin{align} &\sum_{n = 0}^{\infty}(n-1)na_nx^{n-2} + 2\sum_{n = 0}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0 \\ \implies & \sum_{n = 2}^{\infty}(n-1)na_nx^{n-2} + 2\sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0\\ \implies & \sum_{i=0}^{\infty}(i+1)(i+2)a_{i+2}x^{i} + 2\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k} + \sum_{n = 0}^{\infty}a_nx^n = 0\\ \implies & \sum_{n=0}^{\infty}[(n+1)(n+2)a_{n+2}+(n+1)2a_{n+1} + a_n]x^n = 0\\ \end{align} $$ Comparing Coefficients leads to the recurrence relation: $$ \begin{align} (n+1)(n+2)a_{n+2}+(n+1)2a_{n+1} + a_n &= 0\\ \implies a_{n+2} &= -\frac{2a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}, \: \: n=0,1,2,... \end{align} $$ Which gives the solution: $$ y = a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...) $$ Apparently this is incorrect but I am really struggling to see a mistake after re-doing this
$$y = a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)$$ $$y(x)=a_0y_1(x)+a_1y_2(x)$$ There is no mistake in your answer. The recurrence relation is correct and your final answer $y(x)$ is also correct. Your calculations are correct. Now let's check that $y_1(x)$ is indeed a solution of the DE. To do this I will use the Sigma ntation ($\sum $) but you can use another notation if you prefer : $$ \begin {align} y_1(x) &= a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...) \\ &= a_0(1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{3}{4!}x^4 +...) \\ &= a_0\left ( 1-\sum_{n=2}^{\infty}\frac{(n-1)}{n!}(-x)^n\right )\\ &= a_0\left ( 1-\sum_{n=2}^{\infty}\frac{n}{n!}(-x)^n +\sum_{n=2}^{\infty}\frac{(-x)^n}{n!}\right ) \\ &= a_0\left ( -\sum_{n=2}^{\infty}\frac{n}{n!}(-x)^n +e^{-x} +x\right )\\ &= a_0\left ( -\sum_{n=2}^{\infty}\frac{(-x)^n}{(n-1)!} +e^{-x} +x\right ) \\ &= a_0\left ( -\sum_{m=1}^{\infty}\frac{(-x)^{m+1}}{m!} +e^{-x} +x\right ) \\ &= a_0\left ( -\sum_{m=0}^{\infty}\frac{(-x)^{m+1}}{m!} +e^{-x}\right ) \\ &= a_0\left ( x\sum_{m=0}^{\infty}\frac{(-x)^m}{m!} +e^{-x}\right )\\ y_1(x) &= a_0\left (xe^{-x}+e^{-x}\right ) \\ \end{align} $$ This is an obvious solution of the original differential equation. $$y''+2y'+y=0$$ $$\implies y(x)=(C_1+C_2x)e^{-x}$$ You already have the solution: $$y_2(x)= a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)=a_1xe^{-x}$$ $$y(x)=a_0y_1(x)+a_1y_2(x)$$ $$y(x)=a_0(xe^{-x}+e^{-x})+a_1xe^{-x}$$ $$y(x)=xe^{-x}(a_0+a_1)+a_0e^{-x}$$ So that $C_1=a_0$ and $C_2=a_1+a_0$. Your answer is perfectly fine.
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Range of $f(x)=x \sqrt{1-x^2}$ I have to find the range of $f(x)=x\sqrt{1-x^2}$ on the interval $[-1,1]$. I have done so by setting $x=\sin\theta$ and thus finding it to be $[-0.5,0.5]$. Let $x=\sinθ$. Then, for $x\in[-1,1]$ we get that $θ \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, $f(x)$ becomes: $f(\theta)=\sin\theta \sqrt{1-(\sin\theta)^2}= \sin\theta |\cos\theta|$. Since for $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ we have that $cos\theta \geq 0$: $f(θ)=\sinθ\cosθ=\frac{1}{2}\sin2θ$. Which has maximum value $\frac{1}{2}$ when $\theta = \frac{\pi}{4}$ and minimum value $-\frac{1}{2}$ when $\theta = -\frac{π}{4}$. When $\theta = \frac{\pi}{4}$ we have $x=\frac{\sqrt{2}}{2}$ and when $\theta = -\frac{\pi}{4}$ we have $x=-\frac{\sqrt{2}}{2}$, which are the maximum and minimum positions respectively. So, $f(\frac{\sqrt{2}}{2})=\frac{1}{2}$ and $f(-\frac{\sqrt{2}}{2})=-\frac{1}{2}$. Thus the range of $f(x)$ is $[-\frac{1}{2},\frac{1}{2}]$. I know that it can be found with derivatives as well. I was wondering how can quadratic theory can be used to find the Range.
You could reason like this: $f(x)$ will have its maximum value at the same point as $g(x) = (f(x))^2 = x^2 - x^4.$ This is quadratic in $x^2$ and an even function. Since $z-z^2$ takes its maximum value when $z=1/2$ (by finding the vertex of the parabola) then we know $g(x)$, and hence $f(x)$ takes its maximum value when $x^2=1/2$ or $x=\pm 1/\sqrt{2}.$ And $f(1/\sqrt{2}) = 1/2.$ To get the lower end of the range, you can argue by symmetry.
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How to solve this series $f(n) = f(n/2) + n$? Can I solve this as: $f(n) = f(n/2) + n$ let find, $$f(n/2) = f(n/2/2) + n/2\\ f(n/2) = f(n/4) + n/2$$ Now, $$\begin{split} f(n) &= f(n/4) + n/2 + n\\ f(n) &= f(n/8) + n/4 + n/2 + n \end{split}$$ hence so on. $$\vdots$$ Now, $n = 2^i$. $$\begin{split} f(2^i) &= f(2^i/2^i) + 2^i/2^{i-1} + 2^i/2^{i-2} + \cdots + 2^i\\ f(2^i) &= f(1) + 2^1 + 2^2 + \cdots + 2^i\\ f(2^i) &= 2^0 + 2^1 + 2^2 + \cdots + 2^i\\ f(2^i) &= 2^{i+1} - 1\\ f(2^i) &= 2^i\cdot 2^1 - 1\\ f(n) &= 2n - 1\\ f(2^k) &= f(2^{k-k}) + k\\ f(2^k) &= f(2^0) + k\\ \cdots &= f(1) + k\\ \cdots &= 1 + k \end{split}$$ As we know $$\begin{split} n &= 2^k\\ \log (n) &= k \log(2)\\ k &= \log (n) / \log(2)\\ k &= \log_2 (n)\\ f(n) &= \log_2(n) + 1 \end{split}$$
You’ve just given a recursion. A recursion needs have a base to be well defined. The base here could be the set of all $x \in (0.5, 1]$ (then any $x>0$ can be uniquely written as $2^kx_0$ for some $k\in\mathbb Z$ and $x_0\in (0.5,1]$). So choose for one such $x_0$ a $y_0$ and define $f(x_0) := y_0$. Then suppose $n = 2^ky_0$. This gives the recursion $$ f_0(0) = y_0 $$ $$ f_0(k) = f_0(k-1) + 2^ky_0 $$ Thus $$ f_0(k) = y_0\left(\sum_{i=0}^k 2^k\right) = y_0(2^{k+1}-1)$$ If $n = y_02^k$ then $k = \log_2 n - \log_2{y_0}$. Also as $y_0\in(0.5,1]$ we have $\log_2y_0 \in (-1, 0]$. So $k = \lfloor \log_2 n \rfloor$, $y_0 = 2^{\{\log_2 n\}}$ (where $\{x\} := x - \lfloor x \rfloor$ is the fractional part). Thus given any function $g:(0.5, 1]\to\mathbb R$ and the condition $f|{(0.5,1]} = g$ the solution to your recursion will be given by $$ f(n) = g(2^{\{\log_2 n\}})(2^{\lfloor \log_2 n\rfloor + 1} - 1) $$ or alternatively using $g^*:(-1:0]\to\mathbb R$ with $g^*(x) := g(2^x)$ we get $$ f(n) = g^*(\{\log_2 n\})(2^{\lfloor \log_2 n\rfloor + 1} - 1) $$
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Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?
A complete answer using the disk method would be the following. If you revolve $y=\sqrt{r^2-x^2}$ about the $x$-axis (and form a solid) you get the volume of a sphere. Form a disk with height $f(x)$, and find its area. The area of the red disk above is $\pi r^2$, or $\pi f(x)^2$, or we could say $\pi \sqrt{r^2-x^2}^2=\pi \left (r^2-x^2\right )$ at any point $x$ between $x=-r$ and $x=r$. To find the volume of a sphere of radius $r$ then, just add up the areas of infinitesimally thin disks on as $x$ goes from $-r$ to $r$. To work it out:\begin{align*}\int \limits _{-r}^r\pi \sqrt{r^2-x^2}^2\,dx & =2\pi \int \limits _0^rr^2-x^2\,dx \\ & =2\pi \left (\left .r^2x-\frac{1}{3}x^3\right )\right |_0^r \\ & =2\pi \left (r^3-\frac{1}{3}r^3\right ) \\ & =\frac{4\pi}{3}r^3. \end{align*}
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Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$ doesn't converge, on the other hand it grows very slowly? I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.
The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know. I also like the following argument. I'm not sure what students who are new to the topic will think about it. Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way: $$1 + \frac{1}{2} > \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$$ $$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$ $$\frac{1}{5} + \frac{1}{6} > \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$ Continuing in this way, we get $S > S$, a contradiction.
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Perfect numbers, the pattern continues The well known formula for perfect numbers is $$ P_n=2^{n-1}(2^{n}-1). $$ This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$: $$ 496=1+2+4+8+16+31+62+124+248 $$ one can see that the first pattern is a sequence of powers of $2$ that stops at $16$, the second pattern starts with a prime number, in this case $31$, the rest of them are multiples of $31$, e.g. $2\cdot 31, 4\cdot 31$ and $8\cdot 31$. But I found that the pattern proceeds after $31$, for example $31=2^5-2^0$, $62=2^6-2^1$, $124=2^7-2^2$ and finally $248=2^8-2^3$, so the perfect number can be written as $$ 496=1+2+4+8+16+(2^5-2^0)+(2^6-2^1)+(2^7-2^2)+(2^8-2^3) $$ or $$ 496=(1+2+4+8+16+32+64+128+256) -(1+2+4+8). $$ So the formula follows very naturally from this. I've searched but didn't find this formulation anywhere. Well, is this something new? Has anyone seen this somewhere?
What you have discovered is the special case $\rm\ a = 2,\ m = n-1\ $ of the following simple identity $\rm\displaystyle\ a^{m}\:\frac{a^n-1}{a-1}\: =\: \frac{a^{m+n}-1}{a-1} - \frac{a^m-1}{a-1}\: =\: 1+a+\cdots+a^{m+n-1} - (1+a+\cdots+a^{m-1})$ For example, it yields $\ 11111000 = 11111111 - 111\ $ for $\rm\ a = 10,\ m = 3,\ n = 5\:$. However, except in the special case when the left hand side is an even perfect number, the right hand side is not the (rearranged) sum of the divisors of the left hand side, so it bears little relation to perfect numbers.
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us?
The most recent issue of The American Mathematical Monthly (August-September 2011, pp. 641-643) has a new proof by Luigi Pace based on elementary probability. Here's the argument. Let $X_1$ and $X_2$ be independent, identically distributed standard half-Cauchy random variables. Thus their common pdf is $p(x) = \frac{2}{\pi (1+x^2)}$ for $x > 0$. Let $Y = X_1/X_2$. Then the pdf of $Y$ is, for $y > 0$, $$p_Y(y) = \int_0^{\infty} x p_{X_1} (xy) p_{X_2}(x) dx = \frac{4}{\pi^2} \int_0^\infty \frac{x}{(1+x^2 y^2)(1+x^2)}dx$$ $$=\frac{2}{\pi^2 (y^2-1)} \left[\log \left( \frac{1+x^2 y^2}{1+x^2}\right) \right]_{x=0}^{\infty} = \frac{2}{\pi^2} \frac{\log(y^2)}{y^2-1} = \frac{4}{\pi^2} \frac{\log(y)}{y^2-1}.$$ Since $X_1$ and $X_2$ are equally likely to be the larger of the two, we have $P(Y < 1) = 1/2$. Thus $$\frac{1}{2} = \int_0^1 \frac{4}{\pi^2} \frac{\log(y)}{y^2-1} dy.$$ This is equivalent to $$\frac{\pi^2}{8} = \int_0^1 \frac{-\log(y)}{1-y^2} dy = -\int_0^1 \log(y) (1+y^2+y^4 + \cdots) dy = \sum_{k=0}^\infty \frac{1}{(2k+1)^2},$$ which, as others have pointed out, implies $\zeta(2) = \pi^2/6$.
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What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.) Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$. Expand $\sin(A+B)$ and do the manipulations to get $2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$ which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$. Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$. So the triangle is a right-angled triangle.
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How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$ how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$ Thank you
Summation by parts gives that with the choice $a_k=2^k, b_k=k$ we have $$ A_k = 2^1+\ldots+2^k = 2^{k+1}-2 $$ and $$ \sum_{k=1}^{n}a_k b_k = A_n b_n - \sum_{k=1}^{n-1} A_k $$ so: $$ \sum_{k=1}^{n} k 2^k = (2^{n+1}-2)n-\sum_{k=1}^{n-1}(2^{k+1}-2) = (2^{n+1}-2)n+2(n-1)-2(2^n-2)$$ and the RHS simplifies to $(n-1) 2^{n+1}+2$ as wanted.
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Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem: $$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$
Hint: $$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$ Hint2: $$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$
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Computing the sum $\sum \frac{1}{n (2n-1)}$ I was asked to sum the given series: * *$\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n\cdot (2n-1)}=\frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty$ Here i workout the details. \begin{align*} \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n-1)} &= \frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty \\ &= 1 + \Bigl( \frac{1}{2} - \frac{1}{3}\Bigr) + \frac{1}{2} \Bigl(\frac{1}{3} - \frac{1}{5}\Bigr) + \frac{1}{3} \Bigl(\frac{1}{4}-\frac{1}{7}\Bigr) + \cdots \infty \\ &= 1 + \biggl[ \frac{1}{1 \cdot 2 } + \frac{1}{2 \cdot 3} + \cdots \infty\biggr] - \biggl[ \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \cdots \infty \biggr] \\ &= 2 - \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n+1)} \end{align*} Now for summing that second sum note that $$-\log(1-x^{2}) =-\biggl[x^{2} + \frac{x^{4}}{2} + \frac{x^{6}}{3} + \cdots \infty\biggr]$$ Have i done this correctly or is there some error in computing the sum. Now what i need to evaluate is the integral $$\int\limits_{0}^{1} \log(1-x^{2}) \ \rm{dx}$$ Can anyone tell me how to evaluate this integral. I tried by using Integration by parts but that didn't actually work out. Moreover, i would also be interested to see if there are other ways of solving this sum.
We have, $$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)} = 2 \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right)$$ The RHS has the alternating harmonic series, and its value is $\ln 2$. So your final answer would be $2 \ln 2$
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How to find the sum of the following series How can I find the sum of the following series? $$ \sum_{n=0}^{+\infty}\frac{n^2}{2^n} $$ I know that it converges, and Wolfram Alpha tells me that its sum is 6 . Which technique should I use to prove that the sum is 6?
You can also apply the formula for the sum of a geometric series three times in a row : $$\begin{array}{rcl} \sum_{n=1}^\infty \frac{n^2}{2^n} &=& \sum_{n=1}^\infty (2n-1) \sum_{k=n}^\infty \frac{1}{2^k} = \sum_{n=1}^\infty \frac{2n-1}{2^{n-1}} \\ &=& \sum_{k=0}^\infty \frac{1}{2^k} + \sum_{n=1}^\infty 2 \sum_{k=n}^\infty \frac{1}{2^k} \\ &=& 2 + \sum_{n=1}^\infty \frac{2}{2^{n-1}} \\ &= & 2 + 2 \cdot 2 = 6 \end{array}$$ Also, if cutting up those geometric series is too hard, there is an easier way to do the same thing : $$\begin{array}{rcl} (1-\frac{1}{2})^3 \cdot \sum_{n=1}^\infty \frac{n^2}{2^n} & = & (1-\frac{1}{2})^2 \cdot \sum_{n=1}^\infty \frac{n^2-(n-1)^2}{2^n} = (1-\frac{1}{2})^2 \cdot \sum_{n=1}^\infty \frac{2n-1}{2^n} \\ &=& (1-\frac{1}{2}) \cdot \left(\frac{1}{2} + \sum_{n=2}^\infty \frac{(2n-1)-(2n-3)}{2^n} \right) \\ &=& (1-\frac{1}{2}) \cdot \left(\frac{1}{2} + \sum_{n=2}^\infty \frac{2}{2^n} \right) \\ &=& \frac{1}{4} + \frac{2}{4} = \frac{3}{4} = \left(1-\frac{1}{2} \right)^3 \cdot 6 \end{array}$$
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Minimum and maximum There are five real numbers $a,b,c,d,e$ such that $$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$ How can we find the maximum and minimum possible values of any one of the numbers ? Source
I just wanted to point out that if you don't want to appeal to symmetry there is a Lagrange multiplier argument which isn't too bad: Plugging in the first equation into the second, the last 4 variables satisfy the equation $$(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2 = 10$$ Suppose you want to maximize $b$ for example. Then the gradient of $(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2$ has to be a multiple of the gradient of $b$, which means there is some $\lambda$ such that $$-2(7 - b - c - d - e) + 2b = \lambda$$ $$-2(7 - b - c - d - e) + 2c = 0$$ $$-2(7 - b - c - d - e) + 2d = 0$$ $$-2(7 - b - c - d - e) + 2e = 0$$ But the last three equations are the same as $-2a + 2c = 0$, $-2a + 2d = 0$, and $-2a + 2e = 0$, which lead to $a = c = d = e$. Thus the $4$ variables other than $b$ are the same and one can now proceed as in Sivaram's argument.
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
Just to give a different approach, we have $$\begin{align} n^7-n&=n(n^6-1)\\ &=n(n^3-1)(n^3+1)\\ &=n(n-1)(n^2+n+1)(n+1)(n^2-n+1)\\ &=n(n-1)(n+1)(n^2+n-6+7)(n^2-n-6+7)\\ &=n(n-1)(n+1)\Big((n-2)(n+3)(n+2)(n-3)+7\cdot\text{quadratic in }n\Big) \end{align}$$ where the quadratic in $n$, to be explicit, is $2(n^2-6)+7$. What's important, though, is that $n(n-1)$, being the product of two consecutive integers, is always divisible by $2$, $n(n-1)(n+1)$, being the product of three consecutive integers, is always divisible by $3$, and $n(n-1)(n+1)(n-2)(n+3)(n+2)(n-3)$, being the product of seven consecutive numbers, is always divisible by $7$.
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Proof by contradiction: $r - \frac{1}{r} =5\Longrightarrow r$ is irrational? Prove that any positive real number $r$ satisfying: $r - \frac{1}{r} = 5$ must be irrational. Using the contradiction that the equation must be rational, we set $r= a/b$, where a,b are positive integers and substitute: $\begin{align*} &\frac{a}{b} - \frac{1}{a/b}\\ &\frac{a}{b} - \frac{b}{a}\\ &\frac{a^2}{ab} - \frac{b^2}{ab}\\ &\frac{a^2-b^2}{ab} \end{align*}$ I am unsure what to do next?
Since you have already seen answers which complete your proof, here is another proof using continued fractions. If $r$ was rational, it will have a finite continued fraction, say $[a_1, a_2, \dots, a_n]$, but that can be extended to $[5, a_1, a_2, \dots , a_n]$ and $[5,5,a_1, a_2, \dots a_n]$ etc, because $$ r = 5 + \frac{1}{r} = 5 + \frac{1}{[a_1, a_2, \dots, a_n]} = [5, a_1, a_2, \dots, a_n]$$ In fact, you can easily give the infinite continued fraction of $r$ $$[5,5,5,5,\dots]$$
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Prove that if $a^{k} \equiv b^{k} \pmod m $ and $a^{k+1} \equiv b^{k+1} \pmod m $ and $\gcd( a, m ) = 1$ then $a \equiv b \pmod m $ My attempt: Since $a^{k} \equiv b^{k}( \text{mod}\ \ m ) \implies m|( a^{k} - b^{k} )$ and $a^{k+1} \equiv b^{k+1}( \text{mod}\ \ m ) \implies m|( a^{k+1} - b^{k+1} ) $ Using binomial identity, we have: $$a^{k} - b^{k} = ( a - b )( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$$ $$a^{k + 1} - b^{k + 1} = ( a - b )( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$$ Now there are two cases: 1. If $m|(a - b)$, we're done. 2. Else $m|( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$ and $m|( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$ And I was stuck from here, since I could not deduce anything from these two observations. I still have $(a, m) = 1$, and I guess this condition is used to prevent $m$ divides by the two right hand side above. A hint would be greatly appreciated. Thanks, Chan
We have $$m|(a^{k}+a^{k-1}b+a^{k-2}b^2 + \cdots + ab^{k-1}+b^k)$$ $$-t(a^{k-1}+a^{k-2}b+a^{k-3}b^2+ \cdots + ab^{k-2} + b^{k-1}) = a^k$$ which is impossible.
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How does summation formula work with floor function? Prove that if a and b are relatively prime, then $$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \frac{(a - 1)(b - 1)}{2}$$ My attempt was: We have: $$\sum_{i=1}^{n-1} i = \frac{n(n - 1)}{2}$$ Then, $$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \left\lfloor \frac{a(a - 1)b}{2a}\right\rfloor$$ Could I apply the summation formula for floor function like above? Am I in the right track? Thanks, Chan
Notice for each $n$ $$\left\lfloor\frac{nb}{a}\right\rfloor=\frac{nb}{a}-\frac{r_n}{a}$$ where $r_n\equiv nb \pmod a$ and $0\leq r_n< a$. Then since $a$ and $b$ are relatively prime, $nb\equiv mb \pmod a$ implies $n\equiv m \pmod a$ so that we conclude each remainder $r_n$ is unique. But, we know $r_n\in \{1,2,\dots,a-1\}$ for each $n$, and since there are $a-1$ different values for $n$ we see that $r_n$ takes on each of these values. Hence $$\sum_{i=1}^{a-1}r_i=\sum_{i=1}^{a-1}i=\frac{a(a-1)}{2}.$$ Then our original sum is $$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \sum_{n=1}^{a-1} \frac{nb}{a}-\sum_{n=1}^{a-1}\frac{r_n}{a}=\frac{a(a - 1)b}{2a}-\frac{a(a-1)}{2a}=\frac{(a - 1)(b-1)}{2}$$ as desired.
{ "language": "en", "url": "https://math.stackexchange.com/questions/24184", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "7", "answer_count": 3, "answer_id": 1 }
Proving the bound on a recurrence relation I am trying to prove the recurrence $2T(n-1) + 1$ has the bound $\theta(2^{n})$. $T(1) = \theta (1)$ My attempted solution: \begin{align*} T(n) &= 2T(n-1) + 1 \\ &= 2 \{ 2T(n-2) + 1 \} + 1 \\ &= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\ & \ldots \\ &= 2^{i}T(n-i)+\sum_{i=0}^{k} 2^{i} + 1. \end{align*} I then proceed to substitute $n-1$ for $i$ which gives me \begin{align*} & 2^{n-1}T(n-(n-1)) + \sum_{i=0}^{n-1} 2^{i} + 1 \\ &= 2^{n-1}T(n-n+1) + \sum_{i=0}^{n-1} 2^{i} + \sum_{i=0}^{n-1} 1 \\ &= 2^{n-1}T(1) + 2^{n-1+1} + n - 1 \\ &= 2^{n-1}\theta(1) + 2^{n} + n - 1\\ \end{align*} At this point I can see that $2^{n}$ is the dominating factor, however in the solution hints of my book it states that I should use $i = n$. However when I do that I get $T(0)$. I suspect there is something wrong with my substitution/unraveling step, however I can't seem to figure it out. Any help would be appreciated
Maybe here? $$\begin{align*} T(n) &= 2T(n-1) + 1 \\ &= 2 \{ 2T(n-2) + 1 \} + 1 \\ &= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\ & \ldots \\ &= 2^{i}T(n-i)+\sum_{k=1}^{i-1} 2^{k} + 1. \end{align*}$$ Also, $$ \begin{align*} & 2^{n-1}T(n-(n-1)) + \sum_{i=1}^{n-2} 2^{i} + 1 \\ &= 2^{n-1}T(n-n+1) + \sum_{i=1}^{n-2} 2^{i} + 1 \\ &= 2^{n-1}T(1) + 2^{n-1} -1 \\ &= 2^{n-1}\theta(1) + 2^{n-1} - 1\\ \end{align*}$$
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For a fixed positive integer n, show that the determinant below is divisible by n For a fixed positive integer n, if $D = \left|\begin{array}{ccc} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{array} \right|$ show that $\left(\dfrac{D}{(n!)^{3}} - 4 \right)$ is divisible by $n$. Any ideas on how to go about solving this?? Thank You in advance.
$$\begin{eqnarray*} D &=&% \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)!% \end{vmatrix} \\ &=&% \begin{vmatrix} n! & (n+1)n! & (n+2)(n+1)n! \\ (n+1)n! & (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! \\ (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! & (n+4)(n+3)(n+2)(n+1)n!% \end{vmatrix} \\ &=&n!^{3}(n+1)^{2}(n+2)\underset{\text{This determinant is 2 (see below)}}{\underbrace{% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \\ 1 & (n+2) & (n+3)(n+2) \\ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% }} \\ &=&n!^{3}\left( 2n^{3}+8n^{2}+10n+4\right) \end{eqnarray*}$$ I have used the following property repeatedly: "If $B$ results from $A$ by multiplying one row or column with a number $c$, then $\det(B) = c \cdot \det(A)$". (Wikipedia). $n!$ is a factor of rows 1, 2 and 3. $n+1$ is a factor of rows 2 and 3. $n+2$ is a factor of row 3. Thus $$\frac{D}{n!^{3}}-4=2n^{3}+8n^{2}+10n=(2n^{2}+8n+10)n.$$ Evaluation of the last determinant expanding it by the minors of column 1. $% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \\ 1 & (n+2) & (n+3)(n+2) \\ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% $ $=% \begin{vmatrix} (n+2) & (n+3)(n+2) \\ (n+3) & (n+4)(n+3)% \end{vmatrix}% -% \begin{vmatrix} (n+1) & (n+2)(n+1) \\ (n+3) & (n+4)(n+3)% \end{vmatrix}% +% \begin{vmatrix} (n+1) & (n+2)(n+1) \\ (n+2) & (n+3)(n+2)% \end{vmatrix}% $ $=(n+2)(n+3)% \begin{vmatrix} 1 & (n+3) \\ 1 & (n+4)% \end{vmatrix}% -(n+1)(n+3)% \begin{vmatrix} 1 & (n+2) \\ 1 & (n+4)% \end{vmatrix}% $ $+(n+1)(n+2)% \begin{vmatrix} 1 & (n+2) \\ 1 & (n+3)% \end{vmatrix}% $ $=(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)$ $=\left( n^{2}+5n+6\right) -\left( 2n^{2}+8n+6\right) +\left( n^{2}+3n+2\right) =2$
{ "language": "en", "url": "https://math.stackexchange.com/questions/24976", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus: $|z-1| = 2|z+1|$ that is, where the modulus of $z -1$ is twice the modulus of $z+1$ I've solved this problem algebraically (by letting $z=x+iy$) as follows: $\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$ $(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$ $x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$ $3x^2 + 10x + 3y^2 = -3$ $x^2 + \frac{10}{3}x + y^2 = -1$ $(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$ therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle. However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?
View the point $z$ as a vector in $\mathbb R^2$. Without doing any calculations, we know that $\lVert z - (1,0) \rVert^2$ and $\lVert z - (-1,0) \rVert^2$ are of the form $z^Tz + b^T z + c$ for some $b$ and $c$. So the equation of the locus is $$(z^Tz + b_1^Tz + c_1) = 4(z^Tz + b_2^Tz + c_2),$$ or $$3z^Tz + b_3^Tz + c_3 = 0,$$ which is the equation of a circle. In general, the form $x^TAx + b^Tx + c$ defines a quadric, and if you know something about the properties of the matrix $A$, it can tell you what the shape of the quadric is. In this case, $A$ is a multiple of the identity, so it's a circle/sphere/hypersphere.
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Simple 4-cycle permutation I call a 4-cycle permutation simple if I can write it as $(i,i+1,i+2,i+3)$ so $(2,3,4,5)$ is a simple 4-cycle but $(1,3,4,5)$ is not. I want to write $(1,2,3,5)$ as a product of simple 4-cycles. So this is what I did: $$ (1,2,3,5)=(1,2)(1,3)(1,5) $$ but $$\begin{align} (1,3)&=(2,3)(1,2)(2,3)\\ (1,5)&=(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5) \end{align}$$ So now $$(1,2,3,5)=(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$ Can you please give me a hint on how I can express $$(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$ as a product of simple 4-cycles. Note: We do permutation multiplication from left to right.
One way would be to write $(1,2,3,5) = (4,5)(1,2,3,4)(4,5)$ and try to write $(4,5)$ as a simple $4$-cycle instead of trying to do so for all the 2-cycles you came up with.
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Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Can someone point me to a proof, or explain if it's a simple answer? What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series. A lot of the information I find when looking for that seems to point back to arctan.
The derivative of the arc tangent is $$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$ From the formula for geometric series (see for example this answer for a proof) shows that $$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$ Plugging in $-x^2$ for $y$, we get that $$\begin{align*} \frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\ &= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\ &= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots \end{align*}$$ provided that $|-x^2| \lt 1$; that is, provided $|x|\lt 1$. All the computations below are done under this hypothesis (see comments at the end). So we have that: $$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$ Because this is a Taylor series, it can be integrated term by term. That is, up to a constant, we have: $$\begin{align*} \arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\ &= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\ &= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\ &= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1}}{2n+1}\right) + C\\ &= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} +\cdots\right). \end{align*}$$ Evaluating at $x=0$ gives $0 = \arctan(0) = C$, so we get $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \cdots,\qquad\text{if }|x|\lt 1.$$ the equality you ask about. Note however that this does not hold for all $x$: it certainly works if $|x|\lt 1$, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is $$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.$$ Using the Ratio Test, we have that $$\begin{align*} \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3}}{2n+3}\quad}{\frac{|x|^{2n+1}}{2n+1}}\\ &= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3}}{(2n+3)|x|^{2n+1}}\\ &= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\ &= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\ &= |x|^2. \end{align*}$$ By the Ratio Test, the series converges absolutely if $|x|^2\lt 1$ (that is, if $|x|\lt 1$) and diverges if $|x|\gt 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is $1$, and the equality is valid for $x\in [-1,1]$ only (that is, if $|x|\leq 1$; we gained two points in the process). However, the arc tangent has a nice property, namely that $$\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan(x),$$ So, given a value of $x$ with $|x|\gt 1$, you can use this identity to compute $\arctan(x)$ by computing $\arctan(\frac{1}{x})$ instead, and for this argument the series is valid.
{ "language": "en", "url": "https://math.stackexchange.com/questions/29649", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "41", "answer_count": 10, "answer_id": 2 }
How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$? How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$? The book solution $$\left(\dfrac{27}{101}\right) = \left(\dfrac{3}{101}\right)^3 = \left(\dfrac{101}{3}\right)^3 = (-1)^3 = -1$$ My solution $$\left(\dfrac{27}{101}\right) = \left(\dfrac{101}{27}\right) = \left(\dfrac{20}{27}\right) = \left(\dfrac{2^2}{27}\right) \cdot \left(\dfrac{5}{27}\right)$$ $$= (-1) \cdot \left(\dfrac{27}{5}\right) = (-1) \cdot \left(\dfrac{2}{5}\right) = (-1) \cdot (-1) = 1.$$ Whenever I encounter $\left(\dfrac{2^b}{p}\right)$, I use the formula $$(-1)^{\frac{p^2 - 1}{8}}$$ I guess mine was wrong, but I couldn't figure out where? Any idea? Thank you,
$\big(\frac{4}{27}\big) = +1,$ not $-1$. You can use the formula $\big( \frac {2^b}{m} \big) = (-1)^{(m^2-1)/8}$ only when $b$ is odd. When $b$ is even, $2^b$ is a square so the value is $+1$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/31200", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
inverse of a matrix What is the inverse of the following matrix ? $$ \begin{bmatrix} \binom{N}{0} &\binom{N+1}{0} &... &\binom{2N-1}{0} \\ \binom{N}{1} &\binom{N+1}{1} &... &\binom{2N-1}{1}\\ ...& ... &... &... \\ \binom{N}{N-1} &\binom{N+1}{N-1} &... &\binom{2N-1}{N-1} \end{bmatrix} $$
Your matrix, call it $A_N$, can be factorized as $B_N C_N$, where $$B_N = \left[ \binom{N}{i-j} \right]_{i,j=0,\dots,N-1}$$ and $$C_N = \left[ \binom{j}{j-i} \right]_{i,j=0,\dots,N-1}.$$ For example, for $N=5$ this will be $$ \left( \begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 5 & 6 & 7 & 8 & 9 \\ 10 & 15 & 21 & 28 & 36 \\ 10 & 20 & 35 & 56 & 84 \\ 5 & 15 & 35 & 70 & 126 \end{array} \right) = \left( \begin{array}{lllll} 1 & 0 & 0 & 0 & 0 \\ 5 & 1 & 0 & 0 & 0 \\ 10 & 5 & 1 & 0 & 0 \\ 10 & 10 & 5 & 1 & 0 \\ 5 & 10 & 10 & 5 & 1 \end{array} \right) \left( \begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 & 6 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right). $$ The inverses of these factors are $$B_N^{-1} = \left[ (-1)^{i+j} \binom{N-1+i-j}{N-1} \right]_{i,j=0,\dots,N-1}$$ and $$C_N^{-1} = \left[ (-1)^{i+j} \binom{j}{j-i} \right]_{i,j=0,\dots,N-1}.$$ Then $A_N^{-1}=C_N^{-1} B_N^{-1}$ of course, but I don't know if that simplifies to something nice. (I didn't find any hits in OEIS.) In any case, the fact that $B_N$ and $C_N$ have determinant one (since they are triangular with ones on the diagonal) explains why $A_N^{-1}$ has integer entries. Edit: Here's the picture promised in my comment. It illustrates the situation for $N=5$. Entry $(i,j)$ (counting from zero) in the matrix $A_5 = B_5 C_5$ equals the number of paths through the directed graph from source number $i$ on the left to sink number $j$ on the right. Incidentally, the network shows that $B_N$ can be factorized further, just for fun: $$ B_5 = \left( \begin{array}{lllll} 1 & 0 & 0 & 0 & 0 \\ 4 & 1 & 0 & 0 & 0 \\ 6 & 4 & 1 & 0 & 0 \\ 4 & 6 & 4 & 1 & 0 \\ 1 & 4 & 6 & 4 & 1 \end{array} \right) \left( \begin{array}{lllll} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 \\ 1 & 4 & 6 & 4 & 1 \end{array} \right). $$ (The first factor corresponds to the first four blue arrows, and the second factor to the next four blue arrows.)
{ "language": "en", "url": "https://math.stackexchange.com/questions/33086", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 1, "answer_id": 0 }
How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series $$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$ My attempt at a solution - I saw that I could rewrite it as $$\frac{1}{5}\left(1 - \frac{4}{10} \left( 1 - \frac{7}{15} \left(\cdots \left(1 - \frac{3k - 2}{5k}\left( \cdots \right)\right)\right.\right.\right.$$ and that $\frac{3k - 2}{5k} \to \frac{3}{5}$ as $k$ grows larger. Using this I thought it might converge to $\frac{1}{8}$, but I was wrong, the initial terms deviate significantly from $\frac{3}{5}$. According to Wolfram Alpha it converges to $1-\frac{\sqrt[3]{5}}{2}$. How can I get that ?
Here is a way that uses the Gamma function, and that was strangely the first thing to come to mind. (Warning!: It is perhaps more complicated than necessary.) Let $I$ denote your infinite series. We can write this as $$I=\sum_{n=0}^{\infty}(-1)^{n}\prod_{m=0}^{n}\frac{3m+1}{5m+5}$$ which becomes $$I=\sum_{n=0}^{\infty}\left((-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)\right).$$ Now since $$\Gamma\left(\frac{1}{3}\right)\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)=\Gamma\left(n+\frac{4}{3}\right)$$ we have $$I=\frac{1}{\Gamma\left(1/3\right)}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\Gamma\left(n+\frac{4}{3}\right).$$ By the definition of $\Gamma(s)$ we have $$\Gamma\left(n+\frac{4}{3}\right)=\int_{0}^{\infty}t^{n+\frac{1}{3}}e^{-t}dt,$$ which implies that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}dt$$ by switching the order of summation and integration. As $$1-\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n}t^{n}}{5^{n}n!}=e^{-\frac{3t}{5}}$$ it follows that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\left(1-e^{-\frac{3t}{5}}\right)dt=1-\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt$$ Setting $u=\frac{8}{5}t$ we get $$\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt=\frac{1}{\Gamma\left(1/3\right)}\left(\frac{5}{8}\right)^{\frac{1}{3}}\int_{0}^{\infty}u^{-\frac{2}{3}}e^{-u}du=\frac{\left(5\right)^{\frac{1}{3}}}{2}$$and hence $$I=1-\frac{\left(5\right)^{\frac{1}{3}}}{2}$$ as desired. I hope that helps,
{ "language": "en", "url": "https://math.stackexchange.com/questions/34671", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "20", "answer_count": 3, "answer_id": 1 }
Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is Write each of the following as a decimal and a fraction: (iii) $66\frac{2}{3}$% (iv) $16\frac{2}{3}$% For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$. For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$. What I don't understand is a part of the model answer for this exercise. They say $66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$ and $16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$ I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator. Could you please explain?
The recurring decimal $0.\overline{a_1\ldots a_n}$ is equal to $$\frac{a_1\cdots a_n}{10^n-1}.$$ E.g., $x=0.\overline{285} = 0.285285285\cdots$, then $$x = \frac{285}{10^3-1} = \frac{285}{999}.$$ That is, you get the periodic portion divided by a number that consists of as many $9$s as the length of the periodic portion. There are many ways of seeing this; one is using geometric series. Another is to use some manipulations: if $$x = 0.\overline{a_1\ldots a_n}$$ then $$10^nx = a_1\ldots a_n . \overline{a_1\cdots a_n}$$ so $$(10^n-1)x = 10^n x - x = a_1\cdots a_n.$$ The first "model solution" is using this: since $x = 0.\overline{6}$, then $x = \frac{6}{9}$ (the period has length $1$, so you get a single $9$ in the denominator. When the periodic decimal does not start right after the decimal point, you need to shift it a bit first. So for example, if you had $$ x = 0.1\overline{285} = 0.1285285285\ldots,$$ then first we take $10 x = 1.\overline{285}$, then proceed as before: $$\begin{align*} 10^3(10 x) &= 1285.\overline{285}\\ 10x &= 1.\overline{285}\\ 10x(10^3-1) &= 1284\\ x(9990)&= 1284\\ x &= \frac{1284}{9990}. \end{align*}$$ The second model solution uses this method. Added. For the series method, in case anyone is interested, suppose that $x$ is of the form $x=0.\overline{a_1\cdots a_n}$. This means, explicitly, that $$ x = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{(10^n)^k} = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{10^{nk}} = a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}}.$$ This is a geometric series, with initial term $\frac{1}{10^{n}}$ and common ratio $\frac{1}{10^n}$, so it converges. A geometric series with initial term $a$ and common ratio $r$, $|r|\lt 1$, converges to $$\frac{a}{1 - r},$$ so we have $$\begin{align*} x &= a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}} \\ &= a_1\cdots a_n\left(\frac{\frac{1}{10^n}}{1 - \frac{1}{10^n}} \right)\\ &= a_1\cdots a_n\left(\frac{\quad\frac{1}{10^n}\quad}{\quad\frac{10^n-1}{10^n}\quad}\right)\\ &= a_1\cdots a_n\left(\frac{1}{10^n-1}\right) = \frac{a_1\cdots a_n}{10^n-1}\\ &= \frac{a_1\cdots a_n}{\underbrace{9\cdots 9}_{n\text{ digits}}}. \end{align*}$$ And similarly if you have to "shift" the decimal before you get to the period; you simply add enough $0$s to the $9$s in the denominator to account for the shift.
{ "language": "en", "url": "https://math.stackexchange.com/questions/35631", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 2 }
Monotonicity of $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$ Define $f: \mathbb{N} \rightarrow \mathbb{R}$ as $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$. I was wondering how to tell if $f$ is a increasing or decreasing function? Thanks and regards!
A formal proof would be $$ \begin{align} f(n+1) - f(n) &= \frac{1}{n+1} \sum_{i=1}^{n} \frac{1}{i} - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} \\ &= \frac{n}{n+1}\frac{1}{n} \left(\sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n}\right) - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} \\ &= (\frac{n}{n+1} - 1)\frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n(n+1)} \\ & = \frac{1}{n(n+1)} \left(1 - \sum_{i=1}^{n-1} \frac{1}{i}\right) < 0 \end{align} $$ for $\forall n \geq 3$
{ "language": "en", "url": "https://math.stackexchange.com/questions/36483", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 2 }
How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Given a polynomial such as $X^4 + 4X^3 + 6X^2 + 4X + 1,$ where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory?
Such a polynomial with symmetric coefficients is called a palindromic polynomial. If $x_0$ is a zero of a palindromic polynomial than $\frac{1}{x_0}$ is a zero , too. If we divide the polynomial $$Ax^4+Bx^3+Cx^2+Bx+A=0$$ by $A$ we get $$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$ $$(x^4+1)+a(x^3+x)+bx^2=0$$ $$x^2(x^2+\frac{1}{x^2})+ax^2(x+\frac{1}{x})+bx^2=0$$ $$x^2\left((x^2+\frac{1}{x^2})+a(x+\frac{1}{x})+b\right)=0$$ because $x \ne0$. For the same reason we can divide the equation by $x^2$ to get $$(x^2+\frac{1}{x^2})+a(x+\frac{1}{x})+b=0 \tag{2}$$ We set $$y=x+\frac{1}{x} \tag{3}$$ and by squaring we get $$y^2=x^2+\frac{1}{x^2}+2$$ If we substitute this in $(2)$ we get a quadratic equation of $y$ $$(y^2-2)+ay+b=0 \tag{4}$$ So $(1)$ can be solved by first solving the quadratic equation $(4)$ and then the quadratic equation $(3)$ In a similar manner the equation $$x^4+ax^3+bx^2-ax+1=0 \tag{5}$$ can be solved: $$(x^2+\frac{1}{x^2})+a(x-\frac{1}{x})+b=0$$ We set $$y=x-\frac{1}{x} \tag{6}$$ and now get $$y^2=x^2+\frac{1}{x^2}-2$$ and so $$(y^2+2)+ay+b=0 \tag{7}$$ So here we have to solve the quadratic equations $(7)$ and $(6)$ to get the solution of $(5)$. An palindromic equation of degree $3$ can be solved by a different tick: We transform $$x^3+ax^2+ax+1=0 \tag{8}$$ to $$(x^3+1)+a(x^2+x)=0$$ $$(x+1)\left((x^2-x+1)+ax\right)=0$$ So one solution is $$x=-1$$ This can also be seen by plugging $-1$ into $(8)$. The two other solutions are the solutions of the quadratic equation $$(x^2-x+1)+ax=0$$ Finally the equation $$x^3+ax^2-ax-1=0$$ can be transformed to $$(x^3-1)+a(x^2-x)=0$$ and therefore to $$(x-1)\left((x^2+x+1)+ax\right)=0$$ An palindromic equation of 6th or 9th degree again can be solved in two steps by solving first an equation of $y$ of degree $3$ or $4$ and then solve the equation $$y=x+\frac{1}{x}$$ The palindromic equation of degree $5$ has the solution $x=-1$ and can be reduced to an equation of degree $$. The polynomial equations up to degree $4$ can be solved by radicals, so the palindromic polynomial equations of degree $5,6,8$ can be solved by radicals, too.
{ "language": "en", "url": "https://math.stackexchange.com/questions/40864", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "12", "answer_count": 6, "answer_id": 2 }
Primes of the form $x^2 +ny^2$ where swapping $x$ and $y$ still gives a prime I am studying primes of the form $x^2+ny^2$, and i was wondering if there are any known results about the primes of this form such that when you swap $x$ and $y$ you also get a prime. ie for $y^2+nx^2$ you get another prime, I have found quite a lot of these primes and was wondering if there might be infinitely many of these pairs of primes. For $n=2$ I have found for example: $11, 43, 59, 67, 83, 107, 139, 163, 179, 211, \ldots$, all of these primes can be written as $x^2+2y^2$ and for these values of $x$ and $y$, $y^2+2x^2$ is also a prime. Thank you!
This could make a nice trick question: Prove or disprove if $p=x^2+2 y^2$ is a prime which is $3 \mod 8$, then $y^2+2x^2$ is prime. The answer is "disprove". The first counter-example I can find is $p=131$, which is $9^2+2 \times 5^2$, where $5^2+2 \times 9^2 = 187=11 \times 17$. As I will explain below, there is a good reason where there are no significantly smaller counter-examples. First of all, let me explain the $3 \mod 8$ business. An odd prime is of the form $x^2+2 y^2$ if and only if it is $1$ or $3 \mod 8$. If $p$ is $1 \mod 8$, then $y$ is even and $y^2+2x^2$ is not prime. So the only primes we need to care about are the ones which are $3 \mod 8$. Let $z=y^2+2x^2$ and let's think about what a prime divisor $q$ of $z$ could look like. $z$ is odd, so $q$ isn't $2$. Also, $x^2+2y^2$ is divisible by $3$ if and only if $y^2+2 x^2$ is, so $q$ isn't $3$. Since $x^2+2y^2$ is prime, $x$ and $y$ are relatively prime. So $q$ divides neither $x$ nor $y$ and we can change $y^2+2x^2 \equiv 0 \mod q$ into $y^2 \equiv -2x^2 \mod q$. So $-2$ is a square modulo $q$, meaning that $q$ is $1$ or $3$ mod $8$. To give a counter-example, $z$ must be a number which is $3 \mod 8$, not divisible by $3$, and has all prime factors equal to $1$ or $3 \mod 8$. The first such number is $187$ and, sure enough, that gives a counter-example. In fact, it gives two: $187$ is also $13^2+2 \times 3^2$, and $3^2+2 \times 13^2 = 347$, which is prime. Some other values worth trying are $17 \times 19 = 323$, $11 \times 41 = 451$ and $17 \times 43 = 731$. Here is the experiment I would do if I wanted to pursue this further: Generate a collection of random pairs $(x,y)$ with $x$ and $y$ both odd, relatively prime, and exactly one of $(x,y)$ divisible by $3$. (Just generate random integers between, say $1$ and $10^4$ with the right properties modulo $2$ and $3$, and reject pairs which have a common factor.) Compute the proportion of pairs for which $x^2+2y^2$ is prime, call this proportion $p$, and the proportion of pairs for which $x^2+2 y^2$ and $y^2+2 x^2$ are both prime, call that proportion $q$. I would bet $q$ is pretty close to $p^2$. If so, I think that the obvious explanation covers all there is to see here.
{ "language": "en", "url": "https://math.stackexchange.com/questions/42829", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "17", "answer_count": 1, "answer_id": 0 }
How can I compute the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$? I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.
You can also do it by partial fraction decomposition. We have $$ \frac{2\sqrt{2}}{1+t^4} = \frac{t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{t - \sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$ We have $$\int_0^\infty \frac{dt}{t^2 \pm \sqrt{2}t + 1} = \int_0^\infty \frac{dt}{(t \pm \sqrt{2}/2)^2 + 1/2} = \int_0^\infty \frac{2dt}{(\sqrt{2} t \pm 1)^2 + 1} = \sqrt{2} \arctan (\sqrt{2}t \pm 1) \big|_0^\infty.$$ Continuing, we get $$\frac{\sqrt{2}\pi}{2} - \sqrt{2} \arctan (\pm 1) = \frac{\sqrt{2}\pi}{2} \mp \frac{\sqrt{2}}{4} = \frac{\sqrt{2}\pi (2 \mp 1)}{4}.$$ Next, we have $$ \int_0^\infty \frac{(2t \pm \sqrt{2}) dt}{t^2 \pm \sqrt{2}t + 1} = \log (t^2 \pm \sqrt{2}t + 1) \big|_0^\infty. $$ The integral doesn't converge, but we can consider instead $$ \int_0^\infty \frac{(2t + \sqrt{2}) dt}{t^2 + \sqrt{2}t + 1} - \frac{(2t - \sqrt{2}) dt}{t^2 - \sqrt{2}t + 1} = \log \frac{t^2 + \sqrt{2}t + 1}{t^2 - \sqrt{2}t + 1} \big|_0^\infty = 0. $$ Therefore we rewrite our initial expansion $$ \frac{4\sqrt{2}}{1+t^4} = \frac{2t + 2\sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - 2\sqrt{2}}{t^2 - \sqrt{2}t + 1} = \frac{2t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - \sqrt{2}}{t^2 - \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 + \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$ Integrating, we get $$ \int_0^\infty \frac{4\sqrt{2}}{1+t^4} = \sqrt{2} \frac{\sqrt{2} \pi (2-1)}{4} + \sqrt{2} \frac{\sqrt{2} \pi (2+1)}{4} = \frac{\pi}{2} + \frac{3\pi}{2} = 2\pi. $$ Therefore the integral we want is $$ \int_0^\infty \frac{1}{1 + t^4} = \frac{2\pi}{4\sqrt{2}} = \frac{\sqrt{2}\pi}{4}. $$
{ "language": "en", "url": "https://math.stackexchange.com/questions/43457", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "20", "answer_count": 7, "answer_id": 3 }
Effect of adding a constant to both Numerator and Denominator I was reading a text book and came across the following: If a ratio $a/b$ is given such that $a \gt b$, and given $x$ is a positive integer, then $$\frac{a+x}{b+x} \lt\frac{a}{b}\quad\text{and}\quad \frac{a-x}{b-x}\gt \frac{a}{b}.$$ If a ratio $a/b$ is given such that $a \lt b$, $x$ a positive integer, then $$\frac{a+x}{b+x}\gt \frac{a}{b}\quad\text{and}\quad \frac{a-x}{b-x}\lt \frac{a}{b}.$$ I am looking for more of a logical deduction on why the above statements are true (than a mathematical "proof"). I also understand that I can always check the authenticity by assigning some values to a and b variables. Can someone please provide a logical explanation for the above? Thanks in advance!
For the first question. If $x>0$, we have: $$a>b\Rightarrow ax>bx\Rightarrow ab+ax>ab+bx\Rightarrow a(b+x)>b(a+x)$$ $$\Rightarrow \frac{a}{b}>\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}<% \frac{a}{b}.$$ Then $-x<0$. We thus have $$a>b\Rightarrow -ax<-bx\Rightarrow ab-ax<ab-bx\Rightarrow a(b-x)<b(a-x)$$ $$\Rightarrow \frac{a}{b}<\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}>% \frac{a}{b}.$$ For the second question. If $x>0$, we have: $$a<b\Rightarrow ax<bx\Rightarrow ab+ax<ab+bx\Rightarrow a(b+x)<b(a+x)$$ $$\Rightarrow \frac{a}{b}<\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}>% \frac{a}{b}.$$ Then $-x<0$. We thus have $$a<b\Rightarrow -ax>-bx\Rightarrow ab-ax>ab-bx\Rightarrow a(b-x)>b(a-x)$$ $$\Rightarrow \frac{a}{b}>\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}<% \frac{a}{b}.$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/46156", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "9", "answer_count": 8, "answer_id": 4 }
Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$ I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the $\sin^2\theta + \cos^2\theta = 1$ stuff pretty well though. For example just knowing the above how do I express $\cot(2a)$ in terms of $\cot a$? That is one of my problems and I seem to get stuck half way through.
Three examples of algebraic derivations of trigonometric identities as an application of the addition and subtraction formulas. 1. Example on how to deduce the logarithmic transformation formulas (sum to product formulas) from $$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (1)$$ $$\sin (a-b)=\sin a\cos b-\sin b\cos a.\qquad (2)$$ If you write $$\left\{ \begin{array}{c} a=\frac{p+q}{2}, \\ b=\frac{p-q}{2},\end{array}\right. \Leftrightarrow \left\{ \begin{array}{c} a+b=p, \\ a-b=q,\end{array}% \right. $$ you get $$\sin (a+b)+\sin (a-b)=2\sin a\cos b,$$ $$\sin (a+b)-\sin (a-b)=2\sin b\cos a,$$ and thus $$\sin p+\sin q=2\sin \frac{p+q}{2}\cos \frac{p-q}{2},\qquad (3)$$ $$\sin p-\sin q=2\sin \frac{p-q}{2}\cos \frac{p+q}{2}.\qquad (4)$$ You can use $(3)$ to solve the equation $$\sin (5x)+\sin x=\sin (3x)$$ that appeared in my exam in 1968. (See a comment of mine to this post ). 2. As for the example in your question we present the following derivation. From $$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (5)$$ $$\cos (a+b)=\cos a\cos b-\sin a\sin b,\qquad (6)$$ we get $$\cot (a+b)=\frac{\cos (a+b)}{\sin (a+b)}=\frac{\cos a\cos b-\sin a\sin b}{\sin a\cos b+\sin b\cos a},$$ or dividing the numerator and denominator by $\sin a\cos b$ $$\begin{eqnarray*} \cot (a+b) &=&\dfrac{\dfrac{\cos a\cos b-\sin a\sin b}{\sin a\cos b}}{\dfrac{% \sin a\cos b+\sin b\cos a}{\sin a\cos b}}=\dfrac{\dfrac{\cos a\cos b}{\sin a\cos b}-\dfrac{\sin a\sin b}{\sin a\cos b}}{\dfrac{\sin a\cos b}{\sin a\cos b}% +\dfrac{\sin b\cos a}{\sin a\cos b}} \\ &=&\dfrac{\dfrac{\cos a}{\sin a}-\dfrac{\sin b}{\cos b}}{1+\dfrac{\sin b\cos a}{% \sin a\cos b}}=\dfrac{\cot a-\tan b}{1+\tan b\cot a}=\dfrac{\cot a-\dfrac{1}{% \cot b}}{1+\dfrac{\cot a}{\cot b}} \\ &=&\dfrac{\dfrac{\cot a\cot b-1}{\cot b}}{\dfrac{\cot b+\cot a}{\cot b}}=\dfrac{% \cot a\cot b-1}{\cot b+\cot a},\qquad (7) \end{eqnarray*}$$ a particular case of which (for $a=b$) is given by $$\cot (2a)=\dfrac{\cot ^{2}a-1}{2\cot a}.\qquad (8)$$ 3. The summation and subtraction formula for the tangent. From $$\sin (a\pm b)=\sin a\sin b\pm \sin b\cos a$$ we get $$\tan a\pm \tan b=\frac{\sin a}{\cos a}\pm \dfrac{\sin b}{\cos b}=\frac{\sin a\cos b\pm \sin b\cos a}{\cos a\cos b}=\frac{\sin (a\pm b)}{\cos a\cos b}.\qquad (9)$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/48938", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "12", "answer_count": 5, "answer_id": 0 }
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end. Prove that, $$ \dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2} $$
Using the identities$$1+\cos A=2 \cos ^{2} \frac{A}{2}, 1-\sin A=2 \sin ^{2} \frac{1}{2} , \sin A=2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}, $$ We have $$ \begin{aligned} \frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=\frac{2 \sin ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}} \\ &=\frac{\sin \frac{A}{2}\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)}{\cot \frac{A}{2}\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)} \\ &=\tan \frac{A}{2} \end{aligned} $$
{ "language": "en", "url": "https://math.stackexchange.com/questions/50093", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "13", "answer_count": 8, "answer_id": 7 }