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I made an inequality to solve myself, but someone pointed out my solution is wrong. I made the following inequality for myself to solve, but my friend found out an mistake:
If $a,b,c\in\mathbb{R^{+}}$ and $abc=1$, prove that $$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge\frac32$$
I tried to substitute $a=\frac xy, b= \frac yz, c= \frac zx$, but that friend found out a mistake. He wrote a solution using Muirhead’s inequality, but that is not what I want. I also tried trigonometric substitution, but it turned out that there is a mistake also. Is there any method which doesn’t use Muirhead’s inequality? Thanks for any help!
| By C-S $$\sum_{cyc}\frac{a}{b+1}=\sum_{cyc}\frac{a^2}{ab+a}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a)}.$$
Thus, it's enough to prove that
$$2(a+b+c)^2\geq3\sum_{cyc}(ab+a)$$ or
$$\sum_{cyc}(2a^2+ab)\geq3(a+b+c).$$
Now, by AM-GM $$\sum_{cyc}ab\geq3\sqrt[3]{a^2b^2c^2}=3$$ and by C-S again
$$a^2+b^2+c^2=\frac{1}{3}(1+1+1)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2.$$
Thus, it's enough to prove that
$$\frac{2}{3}(a+b+c)^2+3\geq3(a+b+c)$$ or
$$(2(a+b+c)-3)(a+b+c-3)\geq0.$$
Can you end it now?
Also, by AM-GM
$$\sum_{cyc}(2a^2+ab)\geq3\sum_{cyc}\sqrt[3]{a^2\cdot a^2\cdot bc}=3(a+b+c).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$ Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$
I tried using Induction:
For the Base Step $n=2$ we have:
$$x=\sqrt{1+\frac{1}{\sqrt{2}}}+\sqrt{1-\frac{1}{\sqrt{2}}}$$
Then we get:
$$x^2=2+\sqrt{2}\lt 4$$
So $x \lt 2$
Now Let $P(n)$ is True, We shall need to prove $P(n+1)$ is also True
We have $P(n+1)$ as:
$$\left ( 1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}+\left ( 1-\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}$$
Now i tried to use the fact that:
$$f(x)=x^{\frac{1}{x}}$$ is a Monotone Decreasing $\forall x \ge e$
Hence $\forall n \ge 3$ we have:
$$(n+1)^{\frac{1}{n+1}} \lt n^{\frac{1}{n}} \tag{2}$$ and also
$$\frac{1}{n+1} \lt \frac{1}{n} \tag{3}$$
Multiplying $(2),(3)$ We get:
$$1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1}\lt 1+\frac{n^{\frac{1}{n}}}{n}$$
Can we proceed from here?
| By the generalized binomial formula, for $x\ne0$,
$$(1+x)^{1/n}+(1-x)^{1/n}
\\=2+\frac2n\left(\frac1n-1\right)\frac{x^2}2+\frac2n\left(\frac1n-1\right)\left(\frac1n-2\right)\left(\frac1n-3\right)\frac{x^4}{4!}+\cdots\\<2.$$
That's all folks.
Alternatively, the first derivative of $(1+x)^{1/n}+(1-x)^{1/n}$ is odd and monotonic in $(-1,1)$ because the second derivative is non-negative, so it has a single maximum, at $(0,2)$.
| {
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Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$ Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$
I used the quadratic equation to get $$x=\frac{y-1\pm\sqrt{-2y-3y^2-\frac{5}{3}}}{4}$$
But I don’t see how that helps, hints and solutions would be appreciated
Taken from the 2006 IWYMIC
| Multiply out and collect terms $$8x^2+4x-4xy+2y^2+\frac{2}{3}=0.$$
Multiply by $6$ and complete the square for $x$ terms $$3(4x+1-y)^2+9y^2+6y+1=0$$
$$3(4x+1-y)^2+(3y+1)^2=0$$
Therefore $x=y=-\frac{1}{3}.$
| {
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In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time I've tried it so I'm stuck at a point...
So $$x_1+x_2+x_3 = 10$$
Subject to the condition that :
$$1\leq x_1 \leq8$$
$$1\leq x_2 \leq8$$
$$1\leq x_3 \leq8$$
As each beggar can get at maximum 8 blankets and at minimum, 1.
So the number of ways must correspond to the coefficient of $x^{10}$ in:
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)^3$
= coeff of $x^{10}$ in $x^3(1-x^7)^3(1-x)^{-3}$
= coeff of $x^{10}$ in $x^3(1-x^{21}-3x^7(1-x^7))(1-x)^{-3}$
= coeff of $x^{10}$ in $(x^3-3x^{10})(1+\binom{3}{1}x + \binom{4}{2}x^2+...+ \binom{12}{10}x^{10})$
= $\binom{9}{7} - 3 = 33$
Is this right? From here I get the answer as $\binom{9}{7} - 3 = 33$ but the answer is stated as $36$. I don't understand where I'm making a mistake
| While generating functions are all well and good to understand, it is unnecessary here and a more direct approach is usually preferred.
The technique of stars and bars leads to a direct solution of $\binom{10-1}{3-1}=\binom{9}{2}=\binom{9}{7}$ outcomes where each person receives at least one item.
| {
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If $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $ (Pythagorean triples)
Prove that, if $x$, $y$ and $z$ are integers, and if $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $.
Attempt:
$ xyz \equiv 0\pmod{60} \iff 60\mid xyz$. Now notice that $60 = 3 \cdot 4 \cdot 5 $ so the statement $3 \cdot 4 \cdot 5|xyz $ is equivalent to the original statement.
Now I'd have to prove separately that $3|xyz$, $4|xyz$, and $5|xyz$, and I'm not sure how to proceed.
At the same time, it will likely be too much work to prove each of these statements separately, so any other more concise/direct way to tackle this problem is welcome. Thanks.
| The easiest way to do this is to note:
If $m \equiv -1,0,1 \pmod 3$ then $m^2 \equiv 0,1\pmod 3$. If none of them $x, y$ or $z$ are divisible by $3$ then $x^2,y^2,z^2 \equiv 1 \pmod 3$ and $1+1\equiv 1\pmod 3$ is a contradiction. So at least one of $x,y,z$ is divisible by $3$.
Similarly if $m\equiv -2,-1,0,1,2 \pmod 5$ then $m^2 \equiv \pm 1, 0\pmod 5$. If none of $x,y,$or $z$ are divisible by $5$ then we have $\pm 1 +\pm 1 \equiv \pm 1$ which is impossible. So at least one of $x,y,z$ is divisible by $5$.
$4$ is a problem in not being prime we have $m\equiv 2\implies m^2 \equiv 0$. And it implies it's not enough to assume none of them are divisible by $4$ but that at most one is divisible by $2$.
So if $m\equiv -1,0,1,2\pmod 4$ then $m^2 \equiv 0,1$ and at most one is divisible by $2$ we can have $(0,1)+(0,1)\equiv (0,1)$ only by having one of $x$ or $y$, wolog $x$, is even and $z$ and the other odd.
But then we have $x^2 = z^2 -y^2 = (z-y)(z+y)$ Both $z-y$ and $z+y$ as sum/difference of odd numbers are even. But we can further claim that one is divisble by $4$ and the other isn't.
If $y = 2n + 1$ and $z=2m+1$ and $z -y = 2(m-n)$ with $m-n$ odd/even. Then $z+y= 2(m+n+1)$ with $m+n+1 = (m-n) + 2n + 1$ even/odd. So one is $2*odd$ and the other is $2*even = 4*something$.
So $x^2 = (z-y)(z-x)$ is divisible by $8$ and $x$ is divisible by $4$.
| {
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Why doesn't this converge to $\pi$? $\lim_{n\to\infty}\frac12\sqrt{2-2\cos( \frac{2 \pi}{n} )} \times n$ Suppose a regular polygon with $n$ side has radius (line from center to point that connect sides *I don't know how to call it) of length $r$.
From cosine law, the side would has length of $$\sqrt{2r^2-2r^2\cos( \frac{2 \pi}{n} )} $$
Perimeter would be $$\sqrt{2r^2-2r^2cos( \frac{2 \pi}{n} )} \times n $$
Circle has infinitely-many sides and circumference of $2\pi r$
So $$\lim_{n \rightarrow \infty} \sqrt{2r^2-2r^2\cos( \frac{2 \pi}{n} )} \times n= 2 \pi r$$
And
$$\lim_{n \rightarrow \infty} \frac{\sqrt{2-2\cos( \frac{2 \pi}{n} )} \times n}{2}= \pi$$
But, when I plot it on Desmos, the graph scatters at $5.9\times 10^8$ (in the picture). Can anybody give me reason why this happens?
Graph
| The limit set up is correct, indeed by standard limit we have that
$$\frac{ \sqrt{ 2-2cos( \frac{2 \pi}{n} )}n }{2}=\sqrt 2 \pi \sqrt{ \frac{1-cos( \frac{2 \pi}{n} )}{\left( \frac {2\pi} n\right)^2} }\to \pi$$
therefore the problem could be a numerical issue with desmos.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $0.9^n+0.8^n \leq 0.1$ For students who studied the logarithmic function, it is easy to solve the equation
$$0.9^n \leq 0.1$$ in $\mathbb{N}$, which has as solutions $n\geq \frac{\ln0.1}{\ln 0.9} \approx 21.85 $. That is all natural numbers starting from
$22$.
Now how can we solve the following equation
$$0.9^n+0.8^n \leq 0.1$$
| Ask a computer
Asking a numerical solver to find $n$ such that $0.8^n + 0.9^n = 0.1$, yields $n = 22.5020{\dots}$. Checking that the function decreases as $n$ increases, the solution set is $n \geq 23$.
OP has stated they do not want to use this method. But at least we know what answer we should be getting.
Basic Answer
Just start trying $n$... Clearly, when $n = 0$, the sum is $2 > 1$ and the sum decreases as $n$ increases. So, start with $n = 1$ and see when you have success. \begin{align*}
&n & &0.8^n + 0.9^n \\
&1 & & 1.7 \\
&2 & & 1.45 \\
&3 & & 1.241 \\
&4 & & 1.0657 \\
&5 & & 0.91817 \\
&\vdots & &\vdots \\
&21 & & 0.1186\dots \\
&22 & & 0.10585\dots \\
&23 & & 0.0945 \dots
\end{align*}
So the solution set in $\Bbb{N}$ is $n \geq 23$.
There are a few ways to compute less of this table.
Upper and lower bounding with binary search
Notice that $2 \cdot 0.8^n < 0.8^n + 0.9^n < 2 \cdot 0.9^n$. Solving $2 \cdot 0.8^n = 0.1$, we get $n = 13.425{\dots}$. Solving $2 \cdot 0.9^n = 0.1$, we get $n = 28.433{\dots}$. So the solution to the original equation is one of $n \geq 14$, $n \geq 15$, $\dots$, $n \geq 29$. We could check these in order (as in the table above), to find a solution.
However, we can binary search this region, which is much quicker.
\begin{align*}
&n & &0.8^n + 0.9^n \\
&14 & & 0.272\dots \\
&29 & & 0.0486\dots \\
\left\lfloor \frac{14+29}{2} \right\rfloor &= 21 & & 0.1186\dots \\
\left\lfloor \frac{21+29}{2} \right\rfloor &= 25 & & 0.07556\dots \\
\left\lfloor \frac{21+25}{2} \right\rfloor &= 23 & & 0.0945\dots \\
\left\lfloor \frac{21+23}{2} \right\rfloor &= 22 & & 0.1058\dots
\end{align*}
and we find that $n \geq 23$ is the solution.
Using solution to partial equation
You have shown that $0.9^n \leq 0.1$ when $n \geq 22$. Since $0.8^n > 0$, $0.8^n + 0.9^n > 0.9^n$, so we can shortcut the list in answer one by starting at $n = 22$, since the sum can't be smaller than $0.1$ if the larger term is not. This leads us to compute only the last two rows in the table.
| {
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Show that $(1+x)^{(1+x)}>e^x$ How can I prove that $(1+x)^{(1+x)}>e^x$ for all $x>0$?
The problem arose as I tried to prove the well-known & intuitive econometric principle that the more often you compound your interest, the more interest you ultimately get (in maths, that $\frac{d}{dn}((1+\frac{1}{n})^n)>0$ for $n>0$).
An interesting further problem is to prove that $(a+x)^{(a+x)}>e^x$ is true for all $x>0$ if and only if $a\geq1$.
| $$
(1+x)^{1+x} = 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \cdots
$$
is clearly greater than
$$
e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{24} + \cdots
$$
for $x > 0.$
| {
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Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \left(9^x+4\cdot 3^x+1\right)-(4\cdot 3^x+1)>0$$
$$a\cdot \left(9^x+4\cdot 3^x+1\right)>4\cdot 3^x+1$$
$$a>\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$$
Now if a is greater than the maximum value of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then inequality will be true for all x.
So if we can find the range of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then we can say a should be greater than the maximum value in the range.
Let's assume y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$ and substitute $3^x$ with $t$.
$$y=\frac{4t+1}{t^2+4t+1}$$
$$yt^2+4ty+y=4t+1$$
$$yt^2+4t(y-1)+y-1=0$$
We want to have real values of t satisfying the equation, so $D>=0$
$$16(y-1)^2-4*y*\left(y-1\right)>=0$$
$$4(y-1)(4y-4-y)>=0$$
$$4(y-1)(3y-4)>=0$$
$$y\in \left(-\infty,1 \right] \cup \left[\frac{4}{3},\infty\right)$$
So I am getting maximum value tending to $\infty$ for y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$
I am not able to understand where am I making mistake.
Official answer is $a\in \left[1,\infty\right) $
| I think I got the point where I did mistake, thanks to @copper.hat
It is important to note that when substituting $3^x$ with $t$, it means t will also be greater than $0$.
So we have to find the condition where at least one root is positive, so there can be multiple cases to it.
Case 1: When both the roots are positive, it means product of roots should be positive and sum of roots should be positive, so from the equation $yt^2+4t(y-1)+y-1=0$ , we can say $\frac{y-1}{y}>0$ and $\frac{-4\left(y-1\right)}{y}>0$, so $y\in \phi$.
Case 2: When one root is positive and other root is negative, it means product of roots should be negative, so from the equation $yt^2+4t(y-1)+y-1=0$, we can say $\frac{y-1}{y}<0$, so $y\in (0,1)$
Case 3: When one root is zero and other root is positive, it means product of roots should be zero and sum of roots should be positive, $\frac{y-1}{y}=0$ and $\frac{-4\left(y-1\right)}{y}>0$, so $y=1$ and $y\in \left(0,1\right)$, so $y\in \phi$
Taking union of all conditions in case $1,2,3$ , we get $y\in (0,1)$, so maximum value is tending to $1$ but not $1$. Hence $a \in \left[1,\infty\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$
for substitution $y = (x-1)^{2}$ we have:
$y^{2} - 4y + 7 = 0 $
$y_{0} = 2+i\sqrt{3}$
$y_1 = 2 - i\sqrt{3}$
and we have
$y_{0}^{1/2} + 1 = x_{0} $, $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$, $-y_{1}^{1/2} +1 = x_{3} $
I am not sure what is good results.
Please check my solution.
EDIT:
The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $
EDIT2:
I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2.
But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2.
| Answer to the revised question
You have correctly obtained $x=1\pm \sqrt{2\pm i\sqrt3}$.
By the standard method find the complex square roots. This gives one form of the full answer as $$x=1\pm \sqrt[4]7(\frac{2+\sqrt7\pm i\sqrt3}{\sqrt{14+4\sqrt7}}).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with $\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$ I'm preparing for my first exam in university (just recently enrolled in computer science) and I'm having difficulties working out this limit. I either currently lack the proper reasoning process to get it done or they haven't yet explained us all the theorems needed. I'd be really grateful if someone could point me in the right direction. Thank you!
$\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$
| Hint:
By L'Hôpital ("$\frac{0}{0}$"),
\begin{align}
\lim_{x \to 0} \frac{x^2 \sin(2x)}{\log(1 + \sin^3(3x))}
& = \lim_{x \to 0} \frac{2(x^2 \cos(2x)+ x \sin(2x))(\sin^3(3x) + 1)}{9 \cos(3x) \sin^2(3x)} \\
& = \frac{2}{9} \left( \lim_{x \to 0} \left(\frac{x}{\sin(3x)}\right)^2 + \lim_{x \to 0} \left(\frac{x \sin(2x)}{\sin^2(3x)}\right)\right) \\
&= \frac{2}{9} \left( \frac{2}{9} + \frac{1}{9}\right)
= \frac{2}{27}.
\end{align}
| {
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Range of $k$ in $x^2+y^2-6x-10y+k=0$
The circle $x^2+y^2-6x-10y+k=0$ does not touch or intersect the x-axis and the point $(1,4)$ lies inside the circle, then find the range of $k$
$$
C(3,5)\;\&\;r=\sqrt{34-k}
$$
Attempt 1
$$
d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=|3|>r=\sqrt{34-k}\implies k>25
$$
Attempt 2
$$
y=0\implies x^2-6x+k=0\implies\Delta=36-4k<0
\implies k>9
$$
The solution given in my reference is $9<k<29$. The upper limit is fine, but which case is missing in attempt 1 that I am not getting the lowest limit for $k$ there ?
| The distance from center $C(3,5)$ to the $x-$ axis is $d=5$ .
It must be greater than the radius.
thus
$$5 >\sqrt{34-k}$$
the distance from the center to the point $(1,4)$ is
$D=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{5}$
It must be smaller than the radius. thus
$$\sqrt{5}<\sqrt{34-k}$$
So
$$5<34-k<25$$
or
$$9<k<29.$$
| {
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Write the expression in terms of $\sin$ only $\sin(4x)-\cos(4x)$ I am currently taking a Precalc II (Trig) course in college. There is a question in the book that I can't figure out how to complete it. The question follows:
Write the expression in terms of sine only:
$\sin(4x)-\cos(4x)$
So far I have $A\sin(x)+B\cos(x)=k\cdot\sin(x+\theta)$
I believe I have found k: $k=\sqrt{A^2+B^2}=\sqrt{2}$
So I think it would be $\sqrt{2}\cdot\sin(4x+\theta)$ but I do not know how I would find $\theta$.
Thanks in advance for all of your help. You have no idea how much I appreciate it!
| Use the two double angle identities that you know:
\begin{align}
\sin2\theta&=2\sin\theta\cos\theta\\
\cos2\theta&=1-2\sin^2\theta
\end{align}
You will also need to know
$$\cos x=\sqrt{1-\sin^2x}$$
As such, we get
\begin{align}
\sin4x-\cos4x&=\sin2(2x)-\cos2(2x)\\
&=2\sin2x\cos2x-(1-2\sin^22x)\\
&=2(2\sin x\cos x)(1-2\sin^2x)+1+2(2\sin x\cos x)^2\\
&=2\sin x\sqrt{1-\sin^2x}(1-2\sin^2x)+1+8\sin^2x(1-\sin^2x)\\
&=2\sin x\sqrt{1-\sin^2x}-4\sin^3x\sqrt{1-\sin^2x}+1+8\sin^2x-8\sin^4x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Factor $x^{35}+x^{19}+x^{17}-x^2+1$ I tried to factor $x^{35}+x^{19}+x^{17}-x^2+1$ and I can see that $\omega$ and $\omega^2$ are two conjugate roots of $x^{35}+x^{19}+x^{17}-x^2+1$. So I divide it by $x^2+x+1$ and the factorization comes to the following
$$(x^2+x+1)(x^{33}-x^{32}+x^{30}-x^{29}+x^{27}-x^{26}+x^{24}-x^{23}+x^{21}-x^{20}+x^{18}-x^{16}+2x^{15}-x^{14}-x^{13}+2x^{12}-x^{11}-x^{10}+2x^{9}-x^{8}-x^{7}+2x^6-x^5-x^4+2x^3-x^2-x+1)$$
I couldn't go further. My question is is it end here or there is a simple way to do further?
| I think if we simply add and then subtract $x^{18}$ and x this factorization becomes easier.
$x^{35} + x^{19} + x^{17} - x^{2} +1 = x^{35}- x^{18}+ x^{17}+ x^{19}- x^{2}+x+ x^{18}- x+ 1$ = $x^{17}(x^{18}-x+1)+ x(x^{18}-x +1)+(x^{18}- x +1)$= $(x^{18}-x+1)(x^{17}+x+1)$.
Now it is easy to see that omega is a root of $x^{17}+x+1$ which means $x^{2}+x+1$ is a factor of $x^{17}+x+1$. Now applying vanishing method on $x^{17}+x+1$ we easily get
$x^{17}+x+1$ = $(x^{2}+x+1)(x^{15}-x^{14}+x^{12}-x^{11}+x^{9}-x^{8}+x^{6}-x^{5}+x^{3}-x^{2}+1$).So our required solution is
$x^{35}+x^{19}+x^{17}-x^{2}+1=(x^{18}-x+1)(x^{2}+x+1)(x^{15}-x^{14}+x^{12}-x^{11}+x^{9}-x^{8}+x^{6}-x^{5}+x^{3}-x^{2}+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Dealing with degrees in decomposition into partial fractions I had to decompose $ \frac{2x^2}{x^4-1} $ into partial fractions in order to determine its antiderivative.
So, I said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}
$$
However, in the answer key, they said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x^2-1} + \frac{B}{x^2+1}
$$
Although I got the same answer, I had to do unpleasant calculations to finally get a system of four equations and four unknowns, which I had to solve as well.
What I want to know is what made them do that assumption? This isn't what we learned about decomposition into partial fractions.
I thought that maybe because in the starting fraction, the polynomial in the numerator is 2 degrees less than that of the denominator, so we must keep the same degree difference in the partial fractions.
However if we look at this example where the numerator is 5 degrees less than the denominator, this what was written in the answer key:
$$
\frac{1}{x^2(x-1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}
$$
Which is normal and compatible with what I've learned about decomposition into partial fractions.
Please can anyone help? Also please no very complex calculations because I am a biology student. Thank you.
| In your very first example, notice that the function
$$f(x)=\frac{2x^2}{x^4-1}$$
is an even function (a function that satisfies $f(-x)=f(x)$ for all $x$). So, the partial fraction descomposition
$$g(x)=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}$$
also satisfies the same. Thus
$$\begin{align}
\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}
&= \frac{a}{-x-1}+\frac{b}{-x+1}+\frac{-cx+d}{x^2+1} \\
&=-\frac{a}{x+1}-\frac{b}{x-1}+\frac{-cx+d}{x^2+1}
\end{align}$$
Can you see this implies that $c=0$ and we can consider only a one variable in the top of $(x+1)(x-1)=x^2-1$?
However, in the second example, the function is not even neither odd, so, you have to do the decomposition in the traditional way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$ If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$.
$$3\sec^4\theta-10\sec^2\theta+8=0$$
$$3\sec^4\theta-6\sec^2\theta-4\sec^2\theta+8=0$$
$$(3\sec^2\theta-4)(\sec^2\theta-2)=0$$
$$\sec^2\theta=2 \text { or } \sec^2\theta=\frac{4}{3}$$
$$1+\tan^2\theta=2 \text { or } 1+\tan^2\theta=\frac{4}{3}$$
$$\tan^2\theta=1 \text { or } \tan^2\theta=\frac{1}{3}$$
$$\begin{equation}
\tan\theta=\pm1 \text { or } \tan\theta=\pm\frac{1}{\sqrt{3}}
\end{equation}$$
So $\tan\theta$ can have four values $1,-1,\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}$.
But answer is only $1 \text{ or } \dfrac{1}{\sqrt{3}}$.
What are the things I am missing here?
| Your method is correct. Whatever source told you only the positive roots are right either made a mistake or imposed (or meant to impose) a range on $\theta$ that ensures $\tan\theta>0$, such as $\theta\in\left(0,\,\frac{\pi}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Calculate the smallest possible value to measure the ∠BAC angle Let $ABC$ be a right triangle acutangle and let $\overline{ AD}$, with D in $\overline{BC}$, be a height relative to point $A$. Let $Γ_1$
and $Γ_2$ as circumferences circumscribed to triangles $ABD$ and $ACD$, respectively. The circumference
$Γ_1$ crosses the $AC$ side at points $ A$ and $P$, while $Γ_2$ crosses the AB side at points $B$ and $Q$. Let
$X$ or line intersection point
BP with $Γ_2$ so that $P$ is between $B$ and $X$. Likewise, be $Y$ the intersection point of the line
$QC$ with $Γ_1$ so that $ Q $ is between $C$ and $Y$. Knowing that $A, X$ and $Y$
are collinear, calculate the smallest possible value to measure the $\angle{BAC}$ angle.
I think triangular inequality would help.
I'm not sure about this condition
Problem wrote:
''Let ABC be a right triangle...''
If $A=90^{\circ}$ then $P=D$ or else $D=B$ or $C$.
|
In this optimized drawing we have:
BA=BC=CY
AX=AY
$\angle BPA=90^o$
$\angle BQY=90^o$
BC=a, AC=b, AB=c
In triangle ACX, AC is diameter of circle so $\angle CXA=90^o$ and we can write:
$2AX^2=AC^2=b^2$
$AX \times YX=AX \times 2 AX=2AX^2$
⇒ $AX \times YX= b^2$
$QY \times QC = AX \times YX$
$QC=b Cos(\angle QCA)=b Sin(\angle CAQ)$
$CY=AB=BC=a$
⇒ $CY \times QY =a(a- CQ)=a(a-b Sin (\angle CAQ))=b^2$
Let $\angle CAQ=\angle CAB=\alpha$
⇒ $Sin (\alpha)= \frac{a^2-b^2}{ab}=\frac{a}{b}-\frac{b}{a}$ . . . . . . . . (1)
In triangle ABC we can write:
$Sin (\angle CBA)=Sin 2(PBA)=2Sin(90-CAB)Cos(90-CAB))= 2 Sin(CAB)Cos(CAB)=Sin 2(CAB)$
$\frac{b}{Sin (\angle CBA)}=\frac{a}{Sin (\alpha)}$
⇒ $\frac{b}{Sin 2(CAB)}=\frac{a}{Sin (CAB)}$
⇒ $\frac{b}{2 Cos (\alpha)}=a$
⇒ $\frac{b}{a}=2 Cos (\alpha)$
Putting this in relation (1) we get:
$Sin (\alpha)=\frac{1}{2 Cos(\alpha)}-2 Cos (\alpha)= \frac{1- 4 Cos^2(\alpha)}{2 Cos(\alpha)}$
⇒ $2Cos^2 (\alpha)=1-4 Cos^2(\alpha)$
⇒ $Cos^2(\alpha)=\frac{1}{6}$
⇒ $Cos (\alpha)≈ 0.4$ ⇒ $\alpha ≈ 66^o $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find real parameter $a$, such that the solution of the linear system lies in the second quadrant
For which real parameter $a$ lies the solution of the system of equations
$$\begin{aligned} \frac{x}{a+1} + \frac{y}{a-1} &= \frac{1}{a-1}\\ \frac{x}{a+1} - \frac{y}{a-1} &= \frac{1}{a+1} \end{aligned}$$
in the second quadrant?
I do not how to start to solve this system of equations. Any help?
| Subtracting the two equations you get
$$\frac{2y}{a-1} = \frac{2}{a^2-1} \Leftrightarrow y = \frac{1}{a+1}, \qquad \text{whenever $a\neq -1$}.$$
Therefore, setting $y$ in the second equation,
$$\frac{x}{a+1} = \frac{1}{a+1}+\frac{y}{a-1} = \frac{1}{a+1} + \frac{1}{a^2-1}=\frac{a}{a^2-1} \Leftrightarrow x = \frac{a}{a-1}, \qquad \text{whenever $a \neq 1$}$$
The solution is actually a point on the cartesian plane. It is the point lying on both lines defining the system of equations. So the question is: for which values of $a$ do we have a negative $x$ coordinate and a positive $y$ coordinate (the conditions that must be met, in order for a point to lie in the second quadrant)?
Well, whenever
$$y \ge 0 \implies a > -1$$
and whenever
$$x \le 0 \Leftrightarrow \frac{a}{a-1}\le 0 \Leftrightarrow 0 \le a \le 1.$$
Because $a=1$ is not allowed (the $x$ coordinate would not be defined) we have that the solution lies in the second quadrant, whenever $0 \le a < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
| After dividing by $\sqrt{x}$ you may use the substitution $\boxed{y = \sqrt{x}}$. Then, after a bit rearranging you get
$$y+1 - m = \sqrt{y^2-1}$$
Squaring removes the $y^2$ and you can solve directly for $y$:
$$y = \frac{2-2m+m^2}{2(m-1)} \Rightarrow \boxed{x = \frac{(2-2m+m^2)^2}{4(m-1)^2}}$$
Note, that there are solutions only for $m >1$, since $y\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Solving indefinite integral $\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$ How can we solve this integration?
$$\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$$
I tried to make the following substitution
$$1+x^2=w^2$$
but this substitution complicated the integral.
| Substitute $x=\sinh u$,
$$I=\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx=
\int \frac{\cosh u+\cosh^2 u}{\cosh^2 u+\cosh u - 1}du = u+I_1$$
where,
$$I_1=\int \frac{du}{\cosh^2 u+\cosh u-1}$$
Next, use the substitution $\cosh u = \frac{1+t^2}{1-t^2}$, along with $d u = \frac{2dt}{1-t^2}$,
$$I_1=2\int \frac{t^2-1}{t^4-4t^2-1}dt$$
$$=\frac{1}{\sqrt5}\int\left( \frac{\sqrt5+1}{t^2+\sqrt5-2}+\frac{\sqrt5-1}{t^2-\sqrt5-2}\right)dt$$
$$=\sqrt{\frac25}\sqrt{\sqrt5+1}\tan^{-1}\left(\sqrt{\sqrt5+2}\>t\right)
- \sqrt{\frac25}\sqrt{\sqrt5-1}\tanh^{-1}\left(\sqrt{\sqrt5-2}\>t\right)$$
Thus, the solution is,
$$I= \sinh^{-1}x +
\sqrt{\frac25(\sqrt5+1)}\tan^{-1}\left(\sqrt{\sqrt5+2}\>t\right)
-\sqrt{\frac25(\sqrt5-1)}\tanh^{-1}\left(\sqrt{\sqrt5-2}\>t\right)$$
where,
$$t= \left(\frac{\cosh u-1}{\cosh u+1}\right)^{\frac12}
= \left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)^{\frac12}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum_{j=1}^{n-1} M_j(M_{j+1} - M_j) = \frac{1}{2}\left( \sum_{j=1}^n X_j \right)^2 - \frac{n}{2}$ The following question is taken from Steven Shreve Stochastic Calculus for Finance Volume 1, question $2.5$
Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2},$ as is the probability of tail. Let $X_j = 1$ if the $j$th toss results in a head and $X_j = -1$ if the $j$th toss results in a tail. Consider the stochastic process $M_0,M_1,...$ defined by $M_0 = 0$ and
$$M_n = \sum_{j=1}^n X_j, \quad n\geq 1.$$
Define $I_0 = 0$ and
$$I_n = \sum_{j=1}^{n-1} M_j(M_{j+1} - M_j),\quad n=1,2,...$$
Show that
$$I_n = \frac{1}{2}M_n^2 - \frac{n}{2}.$$
My attempt:
$$\begin{align*}
\frac{1}{2}M_n^2 - \frac{n}{2} & = \frac{1}{2} \left( \sum_{j=1}^n X_j^2 + 2\sum_{i\neq j}X_i\cdot X_j \right) - \frac{n}{2} \\
& = \frac{1}{2} \left( n + 2\sum_{i\neq j}X_i\cdot X_j \right) - \frac{n}{2} \\
& = I_n.
\end{align*}$$
Is there an easier / shorter approach to solve this question?
| Note that since $X_j^2 = 1$ and $M_1^2 = X_1^2 = 1$,
$$I_n = \sum_{j=1}^{n-1} \frac{1}{2}(M_j+M_{j+1})(M_{j+1} - M_j) - \sum_{j=1}^{n-1} \frac{1}{2}X_{j+1}(M_{j+1} - M_j) \\= \frac{1}{2}\sum_{j=1}^{n-1}(M_{j+1}^2- M_j^2)- \frac{1}{2}\sum_{j=1}^{n-1}X_{j+1}^2 = \frac{1}{2} M_n^2 - \frac{1}{2} M_1^2 -\frac{n-1}{2} \\ = \frac{1}{2} M_n^2 - \frac{n}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Demonstrate that there are no perfect squares ending with $8$ A number n will always end in some digit of the set {$0,1,2,3,4,5,6,7,8,9$}. The last digit of $n^2$ is the last digit of its last squared digit. Like this:
$$\ldots 0^2 = \ldots 0$$
$$\ldots 1^2 = \ldots 1$$
$$\ldots 2^2 = \ldots 4$$
$$\ldots 3^2 = \ldots 9$$
$$\ldots 4^2 = \ldots 6$$
$$\ldots 5^2 = \ldots 5$$
$$\ldots 6^2 = \ldots 6$$
$$\ldots 7^2 = \ldots 9$$
$$\ldots 8^2 = \ldots 4$$
$$\ldots 9^2 = \ldots 1$$
Therefore, no perfect square ends in $8.$
I think my proof is pretty bad, is there anything more formal than that?
| It's good but you can tweak it. Notice the palindromic symmetry: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. If $n \equiv 0 \pmod{10}$, then $n^2 \equiv 0 \pmod{10}$; if $n \equiv \pm 1 \pmod{10}$, then $n^2 \equiv 1 \pmod{10}$; if $n \equiv \pm 2 \pmod{10}$, then $n^2 \equiv 4 \pmod{10}$; etc. By proving $n^2 \equiv 8 \pmod{10}$ is impossible, you've also proven it for $n^2 \equiv 2 \pmod{10}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.
| Let $I(a,b)=\int_0^{\pi/2}\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac{\pi}{2ab}$(try proving this yourself)
partially differentiating wrt $a$ and applying Leibnitz rule,
$\int_0^{\pi/2}\frac{-2a \sin^2x dx}{(a^2\sin^2x+b^2\cos^2x)^2}= \frac{- \pi}{2ba^2}$.
So $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}dx = \frac{\pi}{4a^3b}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A conjecture about the number of divisors of a natural number Conjecture:
$\tau(n)\mid\tau(n^2)\iff$ $n$ is a perfect square and $\sqrt n=p^2s$,
where $p$ is prime and $s$ is a non prime squarefree number such that $\gcd(p,s)=1$.
$\tau(n)$ is the number of factors of $n$.
See this question on MSE:
Number theory problem on divisors!
It's tested for $n<10,000$.
| If $n=p^4s^2$ with $p$ prime, $s$ is a product of $m\ge1$ distinct primes $q_1,\ldots,q_m$ and $p\nmid s$, then
$$\tau(n) =\tau(p^4)\tau(q_1^2)\cdots \tau(q_m^2)=5\cdot 3\cdots 3=5\cdot 3^m$$
and
$$\tau(n^2) =\tau(p^8)\tau(q_1^4)\cdots \tau(q_m^4)=9\cdot 5\cdots 5=3^2\cdot 5^m.$$
It follows that for such $n$, we have $\tau(n)\mid\tau(n^2)$ only if $m\le 2$.
This allows us to find an explicit counterexample:
Let $n=420^2=176400$.
Then $$\begin{align}\tau(n)&=\tau(2^4\cdot 3^2\cdot 5^2\cdot 7^2)=135\\\tau(n^2)&=\tau(2^8\cdot 3^4\cdot 5^4\cdot 7^4)=1125=8\cdot 135+45.\end{align}$$
It is also not hard to find counterexamples to the other direction:
Let $$n=29674142746122490321305600000000000000000000.$$
Then $$\begin{align}
\tau(n)=\tau(2^{40}3^{24}5^{20}7^211^213^2)&=41\cdot25\cdot 21\cdot 3^3\\
\tau(n^2)=\tau(2^{80}3^{48}5^{40}7^411^413^4)&=81\cdot49\cdot 41\cdot 5^3=35\cdot \tau(n).\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If equation rep $x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ represent pair of plane. then $k$ is The values of $k$ for which the equation
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tedious work
How do i solve it some short way Help me please
| ADDED: conclusion without the matrices:
When $k^2 - 4 k - 8 = 0,$ we get factoring over the reals because
$$ \color{magenta}{ (x+2y+z)^2 - \frac{1}{2} \left( 2y + (2-k) z \right)^2 } $$
is your quadratic form, and then we can factor as
$$ T^2 - \frac{1}{2} U^2 = \left(T + \frac{U}{ \sqrt 2}\right) \left(T - \frac{U}{ \sqrt 2}\right) $$
when $T= x+2y+z$ and $U= 2y + (2-k) z.$ Note that $2-k$ simplifies when $k$ is one of the two roots of $k^2 - 4 k - 8 = (k-2)^2 - 12.$ Thus we have either $k = 2 + 2 \sqrt 3$ or $k = 2 - 2 \sqrt 3$
ORIGINAL::
we write a quadratic form as $X^T H X,$ where capital $X$ is the column vector with elements $x,y,z$ and $X^T = (x,y,z).$ Here, $H$ is the Hessian matrix or half of that, for convenience I'm taking half this time.
The form factors only if the determinant of $H$ is zero.
$$
H =
\left(
\begin{array}{ccc}
1&2&1 \\
2&2&k \\
1&k&-5 \\
\end{array}
\right)
$$
and
$$ \det H = 8 + 4 k - k^2 $$
This becomes zero when
$$ k = 2 \pm 2 \sqrt 3 $$
To make this concrete, I will display $Q^T D Q = H,$ where $D$ is diagonal and $\det Q = 1.$
$$
\left(
\begin{array}{ccc}
1&0&0 \\
2&1&0 \\
1&\frac{2-k}{2}&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&0&0 \\
0&-2&0 \\
0&0&\frac{k^2 - 4k - 8}{2} \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&2&1 \\
0&1&\frac{2-k}{2} \\
0&0&1 \\
\end{array}
\right)=
\left(
\begin{array}{ccc}
1&2&1 \\
2&2&k \\
1&k&-5 \\
\end{array}
\right)
$$
When $k^2 - 4k - 8$ is nonzero, this expresses your quadratic form as a sum of three squares of linear terms, with coefficients.
When $k^2 - 4 k - 8 = 0,$ we get factoring over the reals because
$$ (x+2y+z)^2 - 2 \left( y + \frac{2-k}{2} z \right)^2 $$
is your quadratic form, and then we can factor
$$ V^2 - 2 W^2 = (V + W \sqrt 2)(V - W \sqrt 2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Pointwise convergence of a series of functions using the ratio test. I am asked to calculate the pointwise convergence of the series of functions
$$\sum_{n\geq 0} a_n=\sum_{n\geq 0}\frac{(2n)^{\frac{n+1}{2}}}{\sqrt{n!}}x^ne^{-nx^2}$$
Since it is a series of positive terms, I apply the ratio test and after simplifying $\frac{a_{n+1}}{a_n}$ I got to $$\frac{a_{n+1}}{a_n}=\frac{x\sqrt{2}((1+1/n)^n(1+1/n))^{1/2}}{e^{x^2}}$$
As a hint, I was told that I should use $$(1+1/n)^n\geq\frac{e}{1+1/n}$$ For all $n\geq 1$
Therefore the initial series will converge if $$1>\frac{a_{n+1}}{a_n}\geq \frac{x\sqrt{2e}}{e^{x^2}}$$
How do I finish the problem? If I had $\frac{a_{n+1}}{a_n}= \frac{x\sqrt{2e}}{e^{x^2}}$ I would say it converges for all $x$ such that $e^{x^2}<x\sqrt{2e}$.
| By ratio test we obtain
$$\left|\frac{[2(n+1)]^{\frac{n+2}{2}}}{\sqrt{(n+1)!}}\frac{\sqrt{n!}}{[2n]^{\frac{n+1}{2}}}\frac{x^{n+1}e^{-(n+1)x^2}}{x^ne^{-nx^2}}\right|=\sqrt 2|x|e^{-x^2}\left(1+\frac{1}{n}\right)^\frac{n+1}2\to \sqrt {2e}|x|e^{-x^2}\le 1$$
for $\sqrt {2e}|x|e^{-x^2}=1 \implies |x|e^{-x^2}=\frac1{\sqrt{2e}} $ we have
$$\left|\frac{2n^{\frac{n+1}{2}}}{\sqrt{n!}}x^ne^{-nx^2}\right|=\frac{(2n)^{\frac{n+1}{2}}}{\sqrt{n!}}\frac1{(2e)^\frac n2}\sim \frac{(2n)^{\frac{n+1}{2}}}{\sqrt[4]{2\pi n}n^\frac n 2}\frac{e^\frac n 2}{(2e)^\frac n2}=\frac{\sqrt{2n}}{\sqrt[4]{2\pi n}} \to \infty$$
Therefore the given series converges for any $x\in \mathbb R$ such that $\sqrt {2e}|x|e^{-x^2}<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417722",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} \equiv 9^{5} \bmod 11 = 9^{4} 9 \bmod 11, \:\: 9^{2} \equiv 4 \bmod 11 $$
$$ \implies 9^{5} \equiv 144 \bmod 11 \implies 20^{5} \equiv 1 \bmod 11 $$
Then the proof is done.
Now I will prove:
$$ 20^{15} \equiv 1 \bmod 31 $$
Notice $20^{2} \equiv 28 \bmod 31$, so
$$20 \times (20^{2})^{7} \equiv 20 \times (28)^{7} \bmod 31 \equiv 20 \times (-3)^{7} \bmod 31 \equiv -60 \times 16 \bmod 31\equiv 32 \bmod 31 $$
then the proof is done.
Also, in similar way to prove the $20^{15} \equiv 1 \bmod 61$.
Are there shorter or more efficient proof?
| $20\equiv3^2\pmod{11}$ therefore $20^{15} \equiv 3^{30}\equiv 1 \bmod11$.
$20\equiv12^2\pmod{31}$ therefore $20^{15} \equiv 12^{30}\equiv 1 \bmod31$.
$20\equiv3^4\pmod{61}$ therefore $20^{15} \equiv 3^{60}\equiv 1 \bmod61$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so there are two values
1 and 3.
Then $$2^x=1$$
$$x=0$$
And $$2^x=3$$
$$x=\log_2 3$$
But the answer doesn’t seem to consider the $\log_2 3$ as a viable root, and the answer is 1. Why is that the case?
| Your way is right indeed for $2^x \ge1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5+2^x-1=2^{2x}-2\cdot 2^{x} \iff2^{2x}-3 \cdot 2^x-4=0$$
for $0<2^x <1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5-2^x+1=2^{2x}-2\cdot 2^{x} \iff2^{2x}- 2^x-6=0$$
then let $2^x=t>0$ and solve keeping only the solutions which agree with the assumptions.
Note that $2^x=3$ is not valid since it was obtained under the assumption that $0<2^x <1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{2x^8}{8!}+...$
$e^x-\sin(x) = 1+\dfrac{x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{2x^7}{7!}+\dfrac{x^8}{8!}+...$
$e^x-\cos(x) = x+x^2+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{2x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$
Due to the missing term and the irregularity, I am unable to write the general formula for these series. I wish to find the general formula with compact sigma notation. Can you help with this?
| Take the series for $e^x+\sin x$ for example. This is obtained by taking the series for $e^x$, doubling the terms $\dfrac{x^n}{n!}$ for $n=4k+1$ and deleting the terms $\dfrac{x^n}{n!}$ for $n=4k+3$. So you could write it as
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$
Alternatively, let
$$i_n = \begin{cases} 1 & \text{if } n = 4k \text{ or } 4k+2 \\ 2 & \text{if } n=4k+1 \\ 0 & \text{if } n=4k+3 \end{cases}$$
so that the sequence $(i_n)_{n \ge 0}$ is given by $1,2,1,0,1,2,1,0,\dots$; then you have
$$e^x + \sin x = \sum_{n=0}^{\infty} \dfrac{i_n x^n}{n!}$$
Now note that $i_n = 1 + \sin \dfrac{n\pi}{2}$, to give you
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \left( 1 + \sin \frac{n\pi}{2} \right) \dfrac{x^n}{n!}}$$
You can do a similar trick for the other sequences you desire.
If you don't like using the sine function inside the sum, then you can notice that $i_n = 1 + \dfrac{i^n-i^{-n}}{2}$, where $i^2=-1$; this gives
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \left( 1 + \dfrac{i^n-i^{-n}}{2} \right) \dfrac{x^n}{n!}}$$
If you don't like using the sine function or complex numbers inside the sum, you can do some fiddling around with floor functions and powers of $-1$ to get an equivalent expression. For example:
$$i_n = \dfrac{1-(-1)^n}{2} \cdot (-1)^{\frac{1}{2}\left(n - \left\lfloor \frac{n}{4} \right\rfloor - 1\right)}$$
so you can write
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \dfrac{1-(-1)^n}{2} \cdot (-1)^{\frac{1}{2}\left(n - \left\lfloor \frac{n}{4} \right\rfloor - 1\right)} \cdot \dfrac{x^n}{n!}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Inequality $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\leq \frac{\sqrt{3}}{k+1}$ Hi it's related to this If $a+b+c = 3abc$ and $\frac17 \leq k \leq 7$ prove $ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $
I propose this :
Let $a,b,c>0$ and $a+b+c=abc$ and $a\geq b \geq c $ then we have :
$$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\leq \frac{\sqrt{3}}{k+1}$$
Where $k$ is a real numbers such that $8\leq k\leq \alpha $
with :
$$\alpha^6 - 12 \alpha^5 + 30 \alpha^4 - 42 \alpha^3 + 30 \alpha^2 - 12 \alpha + 1=0 $$
I try a lot of things Karamata's inequality by example .
More interesting I think we can use strong convexity . In the case of $k=8$ we have :
$$\frac{1}{8a+b}\leq \frac{1}{9}\Big(\frac{8}{9a}+\frac{1}{9b}-\frac{8}{81}\frac{(a-b)^2}{a^3}\Big)$$
And
$$\frac{1}{8b+c}\leq \frac{1}{9}\Big(\frac{8}{9b}+\frac{1}{9c}-\frac{8}{81}\frac{(b-c)^2}{b^3}\Big)$$
Remains to show with the initial conditions:
$$\frac{1}{9}\Big(\frac{8}{9b}+\frac{1}{9c}-\frac{8}{81}\frac{(b-c)^2}{b^3}\Big)+\frac{1}{9}\Big(\frac{8}{9a}+\frac{1}{9b}-\frac{8}{81}\frac{(a-b)^2}{a^3}\Big)+\frac{1}{8c+a}\leq\frac{\sqrt{3}}{9}$$
But I think it's not true and we lost some cyclicity .
So I'm really stuck .
Maybe Buffalo way's can kills it but I don't know how to use it .
Thanks a lot for sharing your time and knowledge .
| The Buffalo Way works.
After homogenization, it suffices to prove that, for $a\ge b\ge c > 0$ and $8 \le k\le \alpha$,
$$\frac{3}{(k+1)^2}\ge \frac{abc}{a+b+c}\left(\frac{1}{ka+b} + \frac{1}{kb+c} + \frac{1}{kc+a}\right)^2.$$
WLOG, assume that $c = 1$. Let $a = 1 + s + t, b = 1 + s$ for $s, t\ge 0$. It suffices to prove that,
for $s, t \ge 0$ and $8 \le k\le \alpha$,
$$\frac{3}{(k+1)^2}\ge \frac{(1+s+t)(1+s)}{3+2s+t}\left(\frac{1}{k(1+s+t)+1+s} + \frac{1}{k(1+s)+1} + \frac{1}{k+1+s+t}\right)^2.$$
After clearing the denominators, it suffices to prove that, for $s, t \ge 0$ and $8 \le k\le \alpha$,
$$q_5t^5 + q_4t^4 + q_3t^3 + q_2t^2 + q_1t + q_0 \ge 0$$
where $q_5, q_4, q_3, q_2, q_1, q_0$ are polynomials in $(s, k)$.
It is easy to prove that $q_5, q_4, q_3, q_2, q_1\ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$.
Indeed, for $8 \le k\le \alpha$, each of them can be expressed as polynomials in $s$ with non-negative coefficients (polynomials in $k$).
Thus, it suffices to prove that $q_0 \ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$.
We have
\begin{align}
q_0 &= 6 k^2 s^3+(-k^4+6 k^3+22 k^2+6 k-1) s^2\\
&\quad +(-2 k^4+14 k^3+32 k^2+14 k-2) s -(k^2-10k+1)(k+1)^2.
\end{align}
For each fixed $k$ with $8 \le k\le \alpha$, $q_0 = q_0(s)$ is a cubic function of $s$. Note that
$q_0(0) = -(k^2-10k+1)(k+1)^2 > 0$, $q_0(-\infty) = -\infty$ and $q_0(\infty) = \infty$ for $8 \le k\le \alpha$.
Also, for $8 \le k\le \alpha$, $q_0(s)$ has a discriminant
\begin{align}
\mathrm{discr}(q_0) &= 12 k^2 (k^6-12 k^5+30 k^4-42 k^3+30 k^2-12 k+1) (k+1)^2 (k-1)^4\\
&\le 0.
\end{align}
As a result, we have $q_0 \ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$.
We are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Series expansion for $\arctan(1-x)$ Series expansion for $\arctan(1-x)$
I try to expand this function into its Taylor series by means of differentiating it and then integrating it terms by terms but I fail to obtain the correct result.
The derivative of $\arctan(1-x)=-\dfrac{1}{x^2-2x+2}$. By using long division, I can obtain the series expansion for $-\dfrac{1}{x^2-2x+2}$ and it is
$-\dfrac{1}{2}-\dfrac{x}{2}-\dfrac{x^2}{4}+\dfrac{x^4}{8}+\dfrac{x^5}{8}+\dfrac{x^6}{16}...$
I integrate this series terms by terms and I obtain:
$0-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$
The series according to Wolfram is: $\dfrac{\pi}{4}-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$.
I notice that if I use the indefinite integral, there is going to be an arbitrary constant left after the integrating process. How do I obtain this $\dfrac{\pi}{4}$? Without it, is my series wrong?
Is there any other methods to expand this function into Maclaurin series without directly employing the Maclaurin formula
| For $n\in\mathbb{N}$, the derivative of $\arctan z$ is
\begin{equation*}
(\arctan z)^{(n)}
=\frac{(n-1)!}{(2z)^{n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{k}{n-k-1}\frac{(2z)^{2k}}{(1+z^2)^{k+1}}.
\end{equation*}
The function $\frac{\arctan z}{z}$ has Taylor's series expansion
\begin{equation}\label{arctan-pi-4-ser-eq}
\frac{\arctan z}{z}=\sum_{n=0}^{+\infty}(-1)^n\Bigl[\frac{\pi}{4}+T(n)\Bigr](z-1)^n,\quad |z-1|<\sqrt{2}\,,
\end{equation}
where
\begin{equation}\label{T(n)-dfn-notation}
T(n)=
\begin{cases}
0, & n=0;\\ \displaystyle
\sum_{k=1}^{n}\frac{(-1)^{k}}{2^{k/2}k}\sin\frac{3k\pi}{4}, & n\in\mathbb{N}.
\end{cases}
\end{equation}
Hence, we derive
\begin{equation}\label{arctan(1-z)-pi-4-ser-eq}
\frac{\arctan (1-z)}{1-z}=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n,\quad |z|<\sqrt{2}\,,
\end{equation}
which can be rearranged as
\begin{align}
\arctan (1-z)&=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n
-\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^{n+1}\\
&=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n
-\sum_{n=1}^{\infty}\Bigl[\frac{\pi}{4}+T(n-1)\Bigr]z^{n}\\
&=\frac{\pi}{2}+\sum_{n=1}^{\infty}[T(n)-T(n-1)]z^n\\
&=\frac{\pi}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^{n/2}n}\sin\frac{3n\pi}{4}z^n\\
&=\frac{\pi}{2}-\frac{z}{2}-\frac{z^2}{4}-\frac{z^3}{12}+\frac{z^5}{40}+\frac{z^6}{48}+\frac{z^7}{112}-\frac{z^9}{288}-\dotsm.
\end{align}
*
*Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
| Much the simplest method is to use arithmetic modulo 2,3 and 5.
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 1^{2n+1} - 1^{2n+1} - 0^{2n+1}\equiv 0 \pmod 2$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 2^{2n+1} - 0^{2n+1} - 2^{2n+1}\equiv 0 \pmod 3$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 0^{2n+1} - (-2)^{2n+1} - 2^{2n+1}\equiv 0\pmod 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality:
$$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$
and I calculated partial derivatives:
$$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$
$$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$
I noticed that:
$$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$
I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).
| We need to prove that:
$$\sum_{cyc}(4x^4+4x^3)\geq\sum_{cyc}(2+3x+3xy),$$ which is true by Muirhead.
Indeed, we need to prove that:
$$\sum_{cyc}(3x^4+3x^3+x^4+x^3)\geq\sum_{cyc}(3xy+3x+1+1).$$
Now, $$\sum_{cyc}x^4\geq\sum_{cyc}xy$$ it's
$$\sum_{cyc}x^4\geq\sum_{cyc}xy\sqrt[3]{x^2y^2z^2}$$ or
$$\sum_{cyc}x^4\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{5}{3}}z^{\frac{2}{3}},$$ which is true by Muirhead because $$\left(4,0,0\right)\succ\left(\frac{5}{3},\frac{5}{3},\frac{2}{3}\right).$$
Also, $$\sum_{cyc}x^3\geq\sum_{cyc}x$$ it's
$$\sum_{cyc}x^3\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}},$$
which is true by Muirhead again because
$$(3,0,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right).$$
Also, by AM-GM $$\sum_{cyc}x^4\geq3\sqrt[3]{x^4y^4z^4}=3=\sum_{cyc}1$$ and
$$\sum_{cyc}x^3\geq3xyz=3=\sum_{cyc}1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the limit $ \cos x / (1 - \sin x)^{2/3}$ My problem:
Find the following limit
$$\lim\limits_{x \to \pi/2} \frac{\cos x}{\left(1 - \sin x\right )^{2/3}}$$
I tried $(1 - \sin x)^2 = 1 - 2\sin x + \sin^{2}x$ but can go as far as that.
| We have that by
*
*$\cos x=\sin\left(\frac \pi 2 -x\right)$
*$\sin x=\cos\left(\frac \pi 2 -x\right)$
and $\frac \pi 2 -x =y \to 0$
$$\frac{\cos x}{\sqrt[3]{(1-\sin x)^2}} =\frac{\sin y}{\sqrt[3]{(1-\cos y)^2}}=\frac{\sin y}{y}\frac{\sqrt[3]{y^4}}{\sqrt[3]{(1-\cos y)^2}}\frac1{\sqrt[3]y} \to \pm \infty $$
indeed
*
*$\frac{\sin y}{y} \to 1$
*$\frac{\sqrt[3]{y^4}}{\sqrt[3]{(1-\cos y)^2}}=\sqrt[3]{\left(\frac{y^2}{{1-\cos y}}\right)^2}\to \sqrt[3] 2$
*$\frac1{\sqrt[3]y} \to \pm \infty$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^4 + 1020x^3 - 4x^2 + 2039x+1$ is divisible by 1019. Give all integers x in the range [1,2018] such that $x^4 + 1020x^3 - 4x^2 + 2039x+1$
is divisible by 1019.
Applying mod 1019 to the coefficients, we have
$x^4 + x^3 - 4x^2 + x + 1$ which is equal to $(x-1)^2(x^2+3x+1)$
I tried making 1 of the factors equal to a multiple of 1019, but it looks tedious and i have not found a solution other than 1020.
answer given : 1, 492, 524, 1010, 1511, 1543
| $(x-1)^2(x^2+3x+1)\equiv0\bmod1019\implies$
$ x-1\equiv0\bmod1019$ or $x^2+3x+1\equiv0\bmod1019$.
The former implies $x\equiv1\bmod 1019$; solutions in $[1,2018]$ are $1$ and $1020$.
The latter implies $x^2+3x+\dfrac94\equiv\dfrac94-1\bmod1019$, so $\left(x+\dfrac32\right)^2\equiv9\times255-1\equiv256\bmod1019$,
so $x+\dfrac32\equiv\pm16,$ so $x\equiv\pm16-3(510)\equiv492 $ or $524$;
solutions in $[1,2018]$ are $492, 524, 1511$ and $1543$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A refinement of a famous inequality on the forum . It's related to a big problem Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ .I have this (For one time I take the time to check it )
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}> \sum_{cyc}\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}\geq \frac{3\tan\Big(\frac{1}{81}\Big)}{13\tan\Big(\frac{1}{27}\Big)} $$
The difficulty exceeds the level of an Olympiad I think . Furthermore I think that we cannot use Jensen's inequality (it's not homogeneous) and Cauchy-Schwarz is really too weak .Don't tell me that $\tan(x)\geq x $ is a good approximation in this case it will be a joke .
Maybe we can prove this kind of inequality :
$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}+\frac{\tan(b^4)}{8\tan(b^3)+5\tan(a^3)}$$
But even if it's works it doesn't decide the problem .
I discouraged to use power series it's really awful.
So comments and hints are welcome but don't try it alone .
Thanks for sharing your time and knowledge.
Update :
I think it's not so hard if we remark that we have :
$$\frac{a^4}{8a^3+5b^3}>\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}$$
For $a,b>0$ and $a+b<1$
Maybe someone can prove this and prove the LHS
| The inequality
$$
f(a,b)=\frac{a^4}{8a^3+5b^3}>\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}
$$
reduces to
$$
5 a^4 \tan \left(b^3\right)-\left(8 a^3+5 b^3\right) \tan \left(a^4\right)+8 a^4 \tan \left(a^3\right) \ge 0
$$
in $$T=\{(a,b):a>0,b>0,a+b \le 1\}.$$
One way to do it is to cut $T$ into pieces and approximate $tan(x)$ with different upper and lower bounds.
For example, let
$$
T_1=\{(a,b):1/2 \ge a>0,b>0,a+b \le 1\}.
$$
Then we can use
$$
\frac{x^3}{3}+x \le \tan(x) \le 2 \frac{x^3}{3}+x
$$
for $0 <x <1/2$.
For
$$
T_2=\{(a,b):1 \ge a>9/10, 1/10>b>0,a+b \le 1\},
$$
we use
$$
\frac{b^3}{3}+b \le \tan(b^3)
$$
and (Pade Approximation)
$$
\frac{-\frac{7 (a-1)^2 \left(1+\tan ^2(1)\right)}{1+3 \tan (1)}+\frac{(a-1) \left(9-9 \tan ^2(1)+17 \tan (1)\right)}{3 (1+3 \tan (1))}+\tan (1)}{\frac{(a-1) \left(-20-54 \tan ^4(1)-36 \tan ^3(1)-74 \tan ^2(1)-36 \tan (1)\right)}{6 \left(1+3 \tan ^3(1)+\tan ^2(1)+3 \tan (1)\right)}+1} \le \tan(a^3)
$$
and
$$
\frac{\frac{1}{2} (a-1) (8-3 \tan (1))+\tan (1)}{\frac{1}{2} (a-1) (-3-8 \tan (1))+1}
\ge \tan(a^4)
$$
We can continue doing this for
$$
T_3=\{(a,b):9/10 \ge a>1/2, 1/10>b>0,a+b \le 1\},
$$
and
$$
T_4=\{(a,b):9/10 \ge a>1/2, 1/10<b,a+b \le 1\}.
$$
But a slightly quicker way is to check (rigorously) that there's no critical point inside $T_3 \cup T_4$ using IntervalRootFinding.jl. Thus we just need to verify that $f(a,b)\ge 0$ on the boundary of the $T_3$ and $T_4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| Suppose without loss of generality that $a\geqslant b$ so that $0\leqslant b \leqslant 1$ and write $a = 2- b$.
Our expression becomes
$$\sqrt{2(2-b)^2 + b + 1} + \sqrt{2b^2 - b + 3}$$
and this can be minimized for $b\in[0,1]$ with usual calculus techniques.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $x^2+px+p^2$ is a factor $(x+p)^n-x^n-p^n$, if $n$ be odd and not divisible by $3$. Question:
Prove that $x^2+px+p^2$ is a factor of $(x+p)^n-x^n-p^n$, if $n$ is odd and is not divisible by $3$.
My approach:
$$(x+p)^n-x^n-p^n=\sum_{r=0}^n\limits {n\choose r} x^{n-r}p^r-x^n-p^n$$ What can I do after that?
| First of all substitute $x$ with $py$
Observe $y^3-1=(y-1)(?)=0$
So, $y=w,w$ is a complex cube root of unity
Now $(x+p)^n-x^n-p^n=p^n((1+y)^n-1-y^n)$
Now $f(n)=(1+w)^n-1-w^n=(-w^2)^n-1-w^n$
which is $=-1-w^n-w^{2n}$ if $n$ is odd
If $3\mid n,f(n)=-1-1^n-1^{2n}=?$
Else
$w^n\ne1,f(n)=-\dfrac{1-(w^3)^n}{1-w^n}=0$
Check for $n=3m+1,3m+2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve system of equations $ y + 3x = 4x^3$ and $x + 3y = 4y^3$ Problem: In the set of real numbers find all $x,y$ that satisfy
$ y + 3x = 4x^3,x + 3y = 4y^3$.
What I don't understand, is how do I know that a condition satisfies both equations(I will show later exactly).
The solution goes like this: By addition and substraction of the equations we get the following 2 equations:
$x + y = (x + y)(x^2 − xy + y^2),x − y = 2(x − y)(x^2 + xy + y^2)$.
My question is, how do I know which conditions have to be true? According to the solution either $x+y=0$ or $x-y=0$, which I understand. My problem comes when they mention that both $(x^2 + xy + y^2)=1/2$ and $(x^2 - xy + y^2)=1$ have to be true at the same time. Why both? How do systems of equations work and how do we know that there won't be another solution if we multiply or divide the equations? Any tips for these kinds of problems are greatly appreciated.
Question 2: Why can't we also try to find a solution which satisfies for example $x-y=0$ and $(x^2 − xy + y^2)=1$
| Here is a natural way to solve the system. Continue with,
$$x + y = (x + y)(x^2 − xy + y^2),\>\>\>\>\>
x − y = 2(x − y)(x^2 + xy + y^2)$$
or,
$$(x + y)(x^2 − xy + y^2-1)=(x − y)(x^2 + xy + y^2-\frac12)=0$$
So, there are four cases to be considered,
Case 1: Substitute $x-y=0$ into one of the original equations, say, $y + 3x = 4x^3$, which leads to,
$$4x(x^2-1)=0\implies x= 0, \pm 1$$
Case 2: Similarly, $x+y=0$ leads to,
$$2x(2x^2-1)=0\implies x = 0,\pm \frac 1{\sqrt2}$$
Case 3: Substitute $x^2 − xy + y^2-1=0$ into $y+3x=4x^3$, leading to
$$16x^6-28x^4+13x^2-1=0,\>\> \text{or}$$
$$(x^2-1)(16x^4-12x^2+1)=0\implies x=\pm 1,\pm\frac{\sqrt5\pm 1}{4}$$
Case 4: Similarly $x^2 + xy + y^2-\frac12=0$ leads to
$$(2x^2-1)(16x^4-12x^2+1)=0\implies x=\pm \frac1{\sqrt2},\pm\frac{\sqrt5\pm 1}{4}$$
Thus, all the real solutions are:
$(0,0), (1,1), (-1,-1),
(\frac 1{\sqrt2},-\frac 1{\sqrt2}), (-\frac 1{\sqrt2},\frac 1{\sqrt2}), $
$(\frac {\sqrt5 + 1}{4},\frac {-\sqrt5 + 1}{4}), (\frac {\sqrt5 - 1}{4},\frac {-\sqrt5 - 1}{4}),
(\frac {-\sqrt5 + 1}{4},\frac {\sqrt5 + 1}{4}), (\frac {-\sqrt5 - 1}{4},\frac {\sqrt5 - 1}{4})$
Note that the system of the two equations are implicitly of ninth order. So, it is expected to have nine pairs of solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$
and now I want to apply Cauchy-Schwarz but it is the wrong direction.
| Lemma For all $x$ we have $$\sqrt{x^2+x+1}\geq {\sqrt{3}\over 2}(x+1)$$
Proof After squaring and clearing the denominator we get $$4x^2+4x+4\geq 3(x^2+2x+1)$$
which is the same as $$x^2-2x+1\geq 0$$
Using lemma we get $$\sqrt{a^2+ab+b^2}= b\sqrt{\Big({a\over b}\Big)^2+{a\over b}+1}\geq b\cdot {\sqrt{3}\over 2}({a\over b}+1)=$$ $$={\sqrt{3}\over 2}(a+b) $$ so
$$... \geq {\sqrt{3}\over 2}(a+b) + {\sqrt{3}\over 2}(b+c)+ {\sqrt{3}\over 2}(c+a) =3\sqrt{3}$$
| {
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"url": "https://math.stackexchange.com/questions/3450882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have already tried to replace $1/x$ with $t \,\,(t\rightarrow 0)$, but the only result I get is infinity, which is incorrect.
| By Taylor expansion, we see that for $x>1$,
\begin{align*}
\dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}}-\dfrac{1}{4}\dfrac{1}{x^{4}}\leq\log\left(1+\dfrac{1}{x}\right)\leq \dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}},
\end{align*}
so
\begin{align*}
x^{2}-\dfrac{1}{2}x+\dfrac{1}{3}+1+p\left(\dfrac{1}{x}\right)\leq(x^{3}+x+1)\log\left(1+\dfrac{1}{x}\right)\leq x^{2}-\dfrac{1}{2}x+\dfrac{1}{3}+1+q\left(\dfrac{1}{x}\right),
\end{align*}
where $p$ and $q$ are polynomials.
By Squeeze Theorem, the limit goes to $-4/3$.
With L'Hospital:
\begin{align*}
&\lim_{x\rightarrow\infty}x^{2}-\dfrac{x}{2}+(x^{3}+x+1)\log\left(1+\dfrac{1}{x}\right)\\
&=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{x}-\dfrac{1}{2x^{2}}+\left(1+\dfrac{1}{x^{2}}+\dfrac{1}{x^{3}}\right)\log\left(1+\dfrac{1}{x}\right)}{\dfrac{1}{x^{3}}}\\
&=\lim_{t\rightarrow 0}\dfrac{t-\dfrac{1}{2}\cdot t^{2}+(1+t^{2}+t^{3})\log(1+t)}{t^{3}}\\
&=\lim_{t\rightarrow 0}\dfrac{1-t-(2t+3t^{2})\log(1+t)-(1+t^{2}+t^{3})\cdot\dfrac{1}{1+t}}{3t^{2}}\\
&=\lim_{t\rightarrow 0}\dfrac{1-t^{2}-(3t^{3}+5t^{2}+2t)\log(1+t)-(1+t^{2}+t^{3})}{3(t^{3}+t^{2})}\\
&=\lim_{t\rightarrow 0}\dfrac{-2t-t^{2}-(3t^{2}+5t+2)\log(1+t)}{3(t^{2}+t)}\\
&=\lim_{t\rightarrow 0}\dfrac{-2-2t-(6t+5)\log(1+t)-\dfrac{3t^{2}+5t+2}{1+t}}{3(2t+1)}\\
&=-\dfrac{4}{3}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving iff statement about polynomials I have the following question:
Let $\mathbb F$ be a field and $ x^3 +ax+b, x^2+cx-1 $ $\in \mathbb F$[x].
Prove that $(x^2+cx-1) | (x^3+ax+b) $if and only if $c=b$ and $ a= -1-b^2$.
Attempt:
If $(x^2+cx-1) | (x^3+ax+b) $ then there exists $(x+d)$ such that $(x^2+cx-1)(x+d) = (x^3+ax+b) $. Then I got $ (c+d)x^2 + (cd-1)x -d=ax+b$ and I don't know how to do the next step. Can anyone help me?
| Just equate the coefficients and eliminate $d$. We have $c+d=0, cd-1=a$ and $-d=b$. So $d=-c$. Plug this into the other equations to get $a=cd-1=-1-c^{2}$ and $b=-d=c$. Hence $b=c$ and $a =-1-b^{2}$. Now check that the converse is also true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the period of the $f(x)=\sin x +\sin3x$?
What is the period of the $f(x)=\sin x +\sin3x$?
$f(x)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos\frac{x-3x}{2}=2\sin2x\cos x=4\sin x\cos^2x\\f(x+T)=4\sin(x+T)\cos^2(x+T)=4\sin x\cos^2 x$
how can I deduct this I have no idea
| The period of $~\sin x~$ is $~2\pi~$ and that of $~\sin 3x~$ is $~\frac{2\pi}{3}~$, because $~\sin\left(3\frac{2\pi}{3}\right)=\sin 2\pi~$.
Now given that $~f(x)=\sin 3x+\sin x~$
Let $~a~$be a period $~f(x)~$, then by the definition $~f(x+a)=f(x)~$.
Here $~f(x+a)=\sin(3x+3a)+\sin(x+a)~$
From $~\sin(3x+3a)~$, we have $~a=\frac{2\pi}{3}n,~~ n$ is integer, because its the same as adding period of $~\sin 3x, ~~n$ times.
Similarly, from $~\sin(x+a)~$, we have $~a=2\pi k~$, $k$ is integer.
$~a~$ is a multiple of $~\frac{2\pi}{3}~$ and $~2\pi~$, so the period is smallest positive multiple of $~\frac{2\pi}{3}~$ and $~2\pi~$ which is $~2\pi~$, because $~2\pi=2\pi*1 (\text{multiple of}~ 2\pi), 2\pi=\frac{2\pi}{3}*3 (\text{multiple of}~ \frac{2\pi}{3})~$.
The period of $~f(x)~$, $~a=2\pi~$
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1.
2^5 ends with 2.
3^5 ends with 3.
And so on.
Do I simply add the ending digits to get my answer?
| Using Faulhaber's formula:
$$S_{99}=\frac{4a^3-a^2}{4}=a^2\cdot \frac{4a-1}{4}=\frac{9900^2}{4}\cdot \frac{2\cdot 9900-1}{3}=33\cdot 25\cdot 9900\cdot 19799,$$
where:
$$a=\frac{n(n+1)}{2}.$$
Hence, the answer is $00$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{d}{dx} \frac{1}{x+\frac{1}{x+\frac{1}{ \ddots}}}$, the derivative of an infinite continued fraction. I computed the first few derivatives in the sequence $\frac{1}{x+\frac{1}{x}}, \frac{1}{x+\frac{1}{x+\frac{1}{x}}} \dots$ and eventually lost feeling in my fingers and also realized I was seeing nothing meaningful. Let
$$f(x,t) = \overbrace{\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{\ddots}}}}^{x\text{ appearing }t \text{ times.}}$$
My question: does there exist a closed-form of the derivative of an infinite continued fraction, at least for positive arguments? That is, does $g(x)$ such that:
$$g(x)=\lim_{t \to \infty}\frac{d}{dx}f(x,t)$$
Have a closed-form expression?
| Let $y:=\frac{1}{x+\frac{1}{x+\cdots}}=\frac{1}{x+y}$ so$$y^2+xy-1=0\implies y=\frac{-x\pm\sqrt{x^2+4}}{2}.$$If $x>0$, $y>0$ so$$y=\frac{-x+\sqrt{x^2+4}}{2}\implies\frac{dy}{dx}=\frac{-1+\frac{x}{\sqrt{x^2+4}}}{2}.$$Your $f(x,\,t)$ are odd for each integer $t\ge0$, so if we are to extend this to $x\le0$ we need$$y=\frac{-x+\operatorname{sgn}x\cdot\sqrt{x^2+4}}{2},\,y^\prime=\frac{-1+\frac{|x|}{\sqrt{x^2+4}}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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calculate:$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$ calculate:$$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$$
I tried:
$$\int \frac{x+1-\frac{\sqrt{x^2+1}}{x}+\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx=\int\frac{-dx}{x}+\int \frac{\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx$$
I can not continue from here
| Let $x = \tan u $
$$\int \frac{(\tan u +1) \sec^2 u}{\sec u - \tan u ( \tan u +1 )} \; du $$ multiply with $\cos ^2 u $
$$\int \frac{(\tan u +1 )}{\cos u - \sin u ( \sin u + \cos u ) } $$
$$\int \frac{\sin u + \cos u }{\cos ^2 u - \sin u \cos u(\sin u + \cos u) } $$
$$\int \frac{2 \sqrt{2} \sin (0.25 \pi + u ) }{ 2\cos^2 u - \sin 2 u ( \sin ( 0.25 \pi +u))} \; du $$
$$\int \frac{ 2 \sqrt{2} ( \sin ( 0.25 \pi + u) ) }{\cos 2u +1 - \sin 2u ( \sin ( 0.25 \pi +u ))}$$
$\cos 2u = \cos 2 (y - 0.25 \pi) = \cos (2y - 0.5 \pi) = - \sin ( 2y) , \sin 2u = \sin(2y - 0.5 \pi ) = - \cos (2y)$ hence
$$\int \frac{2 \sqrt{2} \sin y}{- \sin 2y +1 + \cos 2y \sin y } $$
I hope this can help you. I will come back to check
| {
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Prove that $f$ is continuous in $(0,0)$ and check if $f$ differentiable in $(0,0)$ and if $f$ is uniformly continuous on $\mathbb R^2$
Let $f: \mathbb R^2 \to \mathbb R$ such that:
$$f(x,y)= \begin{cases} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1}, (x,y)\neq (0,0) \\ 0, (x,y)=(0,0) \end{cases}$$
a) Prove that $f$ is continuous in $(0,0)$.
b) Is $f$ differentiable in $(0,0)$?
c) Is $f$ is uniformly continuous on $\mathbb R^2$?
My try:
a) $f(0,0)=0$ so to prove that $f$ is continuous in $(0,0)$ we should show that $\lim _{(x,y)\to (0,0)} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1} =0$. So let's take a $(x_n,y_n)$ such that $(x_n,y_n)\to (0,0)$:
$$
\frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}-1}
\le \frac{(x_n^2+y_n^2)^3}{\sqrt{1+x_n^2+y_n^2}-1}\\
=\frac{(x_n^2+y_n^2)^3(\sqrt{1+x_n^2+y_n^2}+1)}{x_n^2+y_n^2}\\
=(x_n^2+y_n^2)^2(\sqrt{1+x_n^2+y_n^2}+1) \to (0+0)(\sqrt{1+0+0}-1)=0
$$
$$
\frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}-1} \ge \frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}}\ge \frac{(x_n+y_n)}{\sqrt{1+x_n^2+y_n^2}} \ge \frac{(x_n+y_n)}{1+x_n^2+y_n^2} \to \frac{0+0}{1+0+0}=0
$$
So from theorem of three functions $\lim _{(x,y)\to (0,0)} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1} =0$.
b) We already know that $f$ is continuous in $(0,0)$ so to prove that $f$ is differentiable in $(0,0)$ we should still show that exist linear map $A: \mathbb R^n \to \mathbb R^m$ such that: $$\lim_{||h||\to 0} \frac{||f(a+h)-f(a)-Ah||}{||h||}=0$$So:
$$
\lim_{||h||\to 0} \frac{||f(a+h)-f(a)-Ah||}{||h||}
=\lim_{||h||\to 0} \frac{|| \frac{(h_1+h_2)^3}{\sqrt{1+h_1^2+h_2^2}-1}-Ah||}{\sqrt{h_1^2+h_2^2}}\\
=\lim_{||h||\to 0} ||\frac{(h_1+h_2)^3(\sqrt{1+h_1^2+h_2^2}+1)-Ah(h_1^2+h_2^2)}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}||\\
=\lim_{||h||\to 0} ||\frac{(h_1+h_2)^3(\sqrt{1+h_1^2+h_2^2}+1)-Ah(h_1^2+h_2^2)}{(h_1^2+h_2^2)^{\frac 32}}||
$$
My questions:
In a) I would like someone to check if my solution is correct
In b) I don't know what I can do the next
| Hints:
a) No, $(x+y)^3\le (x^2+y^2)^3$ is false. Take $x=1/2,y=0$ to see this. But this is true:
$$|f(x,y)| \le \frac{(|x|+|y|)^3}{\sqrt {1+x^2+y^2}-1}=\frac{(|x|+|y|)^3}{x^2+y^2}(\sqrt {1+x^2+y^2}+1).$$
b) If $f$ is differentiable at $(0,0),$ then
$$f(x,y) = \nabla f(0,0)\cdot (x,y) + o\left(\sqrt {x^2+y^2}\right).$$
I'm getting $\nabla f(0,0)=(2,2).$
c) If $f$ were uniformly continuous, then $x\to f(x,0)$ would be uniformly continuous on $\mathbb R.$ And a uniformly continuous function on $\mathbb R$ is always bounded above by a linear function. But $f(x,0)=x(\sqrt {1+x^2}+1).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the equation of the line $ r $ Find the equation of the line $ r $ that goes through $ (1, -2,3) $, concurrent with the line
$$\begin{cases}
x=2+3t \\
y=1+2t \\
z= -1 \\
\end{cases}$$
and has orthogonal director vector $ (1, -3,1) $
Solution: line r is contained in the plane $(x-1)-3(y+2)+(z-3)=0$
$x-3y+z=10$
Next we find the intersection of the plane and the concurrent line
$2+3t-3(1+2t)-1=10$
$-3t-2=10$
$t=-4, so (-10,-7,-1)$
So line r goes through the points (1,-2,3) and (-10,-7,-1)
$\begin{cases}
x=1-11t \\
y=-2-5t \\
z=3-4t \\
\end{cases}$
How to solve without using plan concepts? (Whatever if it's parametric, symmetrical, vector representation ....)
| A straight $r$ going through point $(1,-2,3)$ has this parametric form:
$$x = 1 + a·u$$
$$y = -2 + b·u$$
$$z= 3 + c·u$$
If it has intersection with the other given line, then
$$ x = 1 + au = 2+ 3t$$
$$ y= -2 + bu = 1 +2t$$
$$ z= 3 + cu = -1$$
If it has a orthogonal director vector (1,−3,1) then the dot product with $(a,b,c)$ must be $0$
$$ 1·a -3·b + 1·c = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is there a way to guss the sign of 1 in the expressions (2^8-1)/17=15 or (3^8+1)/17 =386 I found some expressions like above that are divisible by 17 as bellow:
(4^8-1)/17 ,(5^8+1)/17 ,(6^8+1)/17 ,(7^8+1)/17 ,(8^8-1)/17 ,(9^8-1)/17,(10^8+1)/17
then I sum and get (1+2^8+3^8+...+10^8) that are divisible by 17
My question is that how can I find the sign of (1) based on 17 without testing
| We know that $a^{16}=1\mod17$ (for $17\nmid a$) by Fermat's Little Theorem. So $a^8=\pm1\bmod 17$.
It is not hard to check that $2^8=1,3^8=-1\bmod17$. Hence $4^8=8^8=1,6^8=(2^8)(3^8)=-1\bmod17$ and $9^8=(3^8)^2=1\bmod17$. Similarly $12^8=(2^8)^23^8=-1\bmod17$, so $5^8=(17-5)^8=-1\bmod17$ and $10^8=(2^8)(5^8)=-1\bmod17$. Hence $7^8=(17-10)^8=-1\bmod17$.
So we end up with $a^8=1\bmod17$ for $a=1,2,4,8,9$ and $-1$ for $a=3,5,6,7,10$. So the sum is 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$(a+bi)$ of $\frac{3+4i}{5+6i}$ and $i^{2006}$ and $\frac{1}{\frac{1}{1+i}-1}$ How can one get the Cartesian coordinate form $(a+bi)$ of the following complex numbers?
$$\frac{3+4i}{5+6i}$$
$$i^{2006}$$
$$\frac{1}{\frac{1}{1+i}-1}$$
Regarding $\frac{3+4i}{5+6i}$ I tried expanding it with $\frac{5+6i}{5+6i}$ and got $\frac{-9+38i}{-9+60i}$, but that doesn't get me anywhere.
Regarding $i^{2006}$ I have $i^{2006} = -1$ (because $2006$ is an even number and $i^2 = -1$). The Cartesian coordinate form would then be $-1 + 0i$. So the real part is $-1$ and the imaginary is $0i$ is that correct?
Regarding $\frac{1}{\frac{1}{1+i}-1}$ I tried expanding the denominator with $\frac{1+i}{1+i}$ and got $\frac{1}{\frac{1}{1+i}-\frac{1+i}{1+i}} = \frac{\frac{1}{1}}{\frac{1+i}{i^2}} = \frac{-1}{1+i}$, but how do I continue from there?
| $$ \frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} $$
$$\frac{z_1}{z_2}=(\frac{r_1}{r_2})e^{i(\theta_1-\theta_2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
${2n \choose 1} + {2n \choose 3} + ... {2n \choose 2n-1}$ $(1) \ \ \ \ {2n \choose 1} + {2n \choose 3} + ... {2n \choose 2n-1} = ?$
I know that $\sum_{k=0}^n {n \choose k} = 2^n$ and it is pretty easy to obtain, as it's an amout of all possible sets we can get from ${1,2,...,n}$. There is also a formula for $k = k + at$ $\forall_{a,t} \in R$, but it's pretty complicated, so I thought of figuring it by myself.
An assumption I made is that I can also express $(1)$ a bit differently:
$\sum_{k=0}^n {n \choose k} = \sum_{k=0}^n {n \choose 2k-1} + \sum_{k=0}^n {n \choose 2k} = A + B$. Let's take an example: $n = 4$.
What I get is:
$A = {8 \choose 1} + {8 \choose 3} + {8 \choose 5} + {8 \choose 7}$
$B = {8 \choose 0} + {8 \choose 2} + {8 \choose 4} + {8 \choose 6} + {8 \choose 8}$
There comes a confusing part that I'm not sure at all, but let $A$ be an amount of all sets with an odd amount of elements, and same for $B$, but now we think of an even amount of sets with an even amount of elements. I thought:
$A + B + x_4 = 2^4 + 2^5 +x_4 = 2^8$
Then: $ x_4 = 2^4 \cdot 13$. Ok, kinda interesting due to the fact we get $2^4$. Having it done for a few more examples:
$x_2 = 2^2 \cdot 1$
$x_3 = 2^3 \cdot 5$
$x_4 = 2^4 \cdot 13$
$x_5 = 2^5 \cdot 29$
$x_6 = 2^6 \cdot 61$
$x_{2n} = 2^{2n} \cdot a_{2n}$
An intresting part is a series of these numbers $(5,13,29,61,...)$. It seems, that every following number $a_{2i} = a_{2i-1} + (n - 5) \cdot 8$. So if I knew whether the formula is true $\forall_{2n \in R}$ I could obtain, for example, $\sum_{k=1}^n {2n \choose 2k-1}$. But... is it true at all?
TL;DR
What is an elegant way to obtain a sum ${100 \choose 1} + {100 \choose 3} + ... + {100 \choose 99}?$ Of course there is a formula I mentioned above:
But it's complicated and doesn't seem to be any useful during a test.
| We have the following binomial expansion:
$$(1+1)^{2n}=\binom{2n}{0}+\binom{2n}{1}+\dots+\binom{2n}{2n}$$
and
$$(1-1)^{2n}=\binom{2n}{0}-\binom{2n}{1}\pm\dots+\binom{2n}{2n}.$$
Summing both gives
$$2^{2n}=2\binom{2n}{0}+2\binom{2n}{2}+\dots+2\binom{2n}{2n},$$
So dividing throughout by $2$ we get
$$2^{2n-1}=\binom{2n}{0}+\binom{2n}{2}+\dots+\binom{2n}{2n}.$$
Of course, by subtracting this from the first equation (the binomial expansion of $(1+1)^{2n}$) we have
$$2^{2n-1}=2^{2n}-2^{2n-1}=\binom{2n}{1}+\binom{2n}{3}+\dots+\binom{2n}{2n-1}.$$
The key trick here was to consider two different binomial expansions, one of which has alternating signs, and then exploit cancellation on every other term to isolate the terms we want to look at in the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find area bounded by $y=\frac 3x, y=\frac 5x, y=3x, y=6x$ Let S be the area of the region bounded by the curves
$$y=\frac 3x,\>\>\> y=\frac 5x,\>\>\> y=3x,\>\>\> y=6x$$
Need to find $S$.
The coordinates of the vertices of the resulting figure were found. The problem with the transition from a double integral to a repeated one.
There's another idea. Make a replacement
\begin{cases}
\xi=xy \\ \eta=\frac{y}{x}
\end{cases}
Вut none led to the correct answer.
| 1) Subtract the two integrals of $5/x$ and $3/x$ in the range of $a$ which is the solution of $\frac{3}{x}=6x$ (the intersection of two lines) to $b$ which is the solution of $\frac{5}{x}=3x$.
2) One would obtain $area=[ln(2x)]_{a}^{b}$.
3)Now, to subtract the area of the two hyperbolic triangles, simply obtain $c$, the solution of $\frac{5}{x}=6x$ and $d$, the solution of $\frac{3}{x}=3x$ and integrate the function $5/x$ from a to c minus the integral of function $6x$ over the same range. Do the same for the other two terms ($3/x$ and $3x$) over range $b$ to $d$. (${a,b,c,d}\in R^+$)
$a=\frac{\sqrt{2}}{2},b=\sqrt{\frac{5}{3}},c=\sqrt{\frac{5}{6}},d=1$
Now, $S = area - (\int_{a}^{c}\frac{5}{x} - 6x dx ) - (\int_{b}^{d}\frac{3}{x} - 3x dx ) = [ln(2x)]_{a}^{b} - [ln(5x)-3x^2]_{a}^{c}-[ln(3x)-1.5x^2]_{b}^{d}$
Since $\ln(a)+\ln(b)=\ln(ab), \ln(a)-\ln(b)=\ln(a/b)$ then the above can be written as $\ln(\frac{b}{cd})+3c^{2}-3a^{2}+1.5d^{2}-1.5b^{2}=\ln(\sqrt{2})$
Since it is a hyperbola (mirror image on 4th quadrant), multiply the area between the two functions $5/x$ and $3/x$ by two. So, $2\ln(\sqrt{2})=\ln2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding sum of series $\sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$
what i try
Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$
$\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Rightarrow \sum^{\infty}_{k=0}x^{k+1}=\frac{x}{1-x}=\frac{1-(1-x)}{1-x}$
differentiating both side
$\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$
How do i solve it help me please
| Another way:
Let $f(m)=\dfrac{a+bm+cm^2}{2^m}$
and $\dfrac{(k+1)(k+2)}{2^k}=f(k)-f(k+1)$
so that $$\sum_{k=0}^n(f(k)-f(k+1))=f(0)-f(n+1)$$
$$\implies2(k+1)(k+2)=2(a+bk+ck^2)-[a+b(k+1)+c(k+1)^2]$$
$$\iff4+6k+2k^2=k^2(2c-c)+k(2b-b-2c)+2a-a-b-c$$
Compare the constants and the coefficients of $k,k^2$ to find $a,b,c$
Now we can prove $$\lim_{n\to\infty}f(n+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\
\end{align}
But I stuck at this point because how to solve $\frac{1}{D(D+4i)}x^2?$ The solution provided this
\begin{align}
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{24}[6x^2\cos 2x+x(8x^2-3)\sin 2x]
\end{align}
I still can't figure out how they do this. Any help will be appreciated.
Update: Using @Isham answer I tired
\begin{align}
&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac{x^2}{2}-\frac{x}{4}\right)\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos 2x-\frac{x}{4}2\cos 2x\right]
\end{align}
But I think I am again Lost.
| $$E=\frac 1 {D+4i}=\frac 1 {4i}\frac 1 {(1+D/4i)}$$
$$E=\frac 1 {4i}({1+D/4i})^{-1}$$
Now you can use this ( after a certain number the derivative is zero for $x^2$:
$$\frac 1 {1-x}=\sum_{n=0}^{\infty} x^n$$
So we have that:
$$E=\frac 1 {4i}(1-\frac D {4i}-\frac {D^2}{16}+........)$$
Apply the operator on $x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^3+3y-xy^\prime=0$
Solve $x^3+3y-xy^\prime=0$
Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$
Let $v=\frac{y}{x}$ then $y^\prime=v^\prime x+v$
then $x^2+3v=v^\prime x+v$
then $v^\prime=x+2\frac{v}{x}$
Used integrating factor $e^{\int -2/x dx}=e^{-x^2}$
then $e^{-x^2}v=\int e^{-x^2}\cdot x dx$
But when I solve this $e^{-x^2}v=\frac{-1}{2}e^{-x^2}$
$v=\frac{-1}{2}$
But this isn't the right answer.
|
Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$
From this step, multiply by negative one to form
$$y'-\frac{3}{x}y=x^2$$
therefore the integrating factor is
$$\mu(x)=\text{exp}\left(-{\int {\frac{3}{x}}dx}\right)=x^{-3}$$
hence the equation can be written as
$$\frac{d}{dx}\left(\frac{y}{x^3}\right)=\frac{1}{x}$$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the recurrence relation solution. $a_n$ = $a_{n-1} + 3n - 5$ $a_n$ = $a_{n-1} + 3n - 5$ $,$ $a_0 = 7$
So far what I got is:
$a_0 = 7$
$a_1 = 7+3(1)-5$
$a_2 = 7+3(1)-5 + 3(2)-5$
$a_3 = 7+3(1)-5 + 3(2)-5 + 3(3)-5$
$a_4 = 7+3(1)-5 + 3(2)-5 + 3(3)-5 + 3(4)-5$
This is where I'm stuck I'm having trouble finding a pattern to the relation.
At first I tried solving this as a quadratic sequence and got $a_n = \frac{3}{2}{n^2} - \frac{7}2n + 7$ which works but on a test I solved a recurrence relation problem as a quadratic sequence got the right answer but got a 0 on the problem because the professor wanted us to use forward or backward substitution which is what I am attempting above (forward substitution).
| By telescoping sum, $a_n-7=a_n-a_0=\sum\limits_{k=1}^n(a_k-a_{k-1})=\sum\limits_{k=1}^n(3k-5)=3\sum\limits_{k=1}^n k-\sum\limits_{k=1}^n 5=3\frac{n(n+1)}2-5n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating $\frac1{a+b\cos(x)}$ using $e^{ix}$ In calculus class our teacher demonstrated us the evaluation of the definite integral $\int_0^\pi \dfrac{1}{a+b\cos(x)}dx=\dfrac{\pi}{\sqrt{a^2-b^2}}$, for $a\ne 0, b \ne 0, |\dfrac ba \lt 1|$, but there's a part I could not grasp.
Our teacher started out by setting up an equality: $bz^2+2az+b=b(z-\alpha)(z-\beta)$, where $\alpha, \beta$ are the roots of the expression on the LHS in $\Bbb C$, and $\alpha$ takes on the value with a negative sign before the radical, while $\beta$ is the other one. Letting $z=e^{ix}$, the integrand is transformed into the following form:
$\dfrac{1}{\sqrt{a^2-b^2}}\left(\dfrac{\alpha}{\alpha-e^{ix}}-\dfrac{\beta}{e^{ix}-\beta}\right)=\dfrac{1}{\sqrt{a^2-b^2}}\left(\dfrac{1}{\alpha e^{-ix}-1}+\dfrac{1}{1-\beta e^{-ix}}\right)$
$=\dfrac{1}{\sqrt{a^2-b^2}}\sum_{n=0}^{\infty}(\beta^n-\alpha^n)e^{-inx}=\dfrac{1}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}(\beta^n-\alpha^n)(\cos(nx)-i\sin(nx))$
But then $\int_0^\pi \cos(nx)-i\sin(nx) dx=\dfrac 1n(\sin(nx)+i\cos(nx))|_0^\pi=\dfrac in(\cos(n\pi)-1)$ and I was completely lost on how the next part was carried out. For if we substitute that back into the infinite sum, we get that integrated version of that expression would be $\dfrac{1}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}(\beta^n-\alpha^n)\dfrac in(\cos(n\pi)-1)$
$=\dfrac{1}{\sqrt{a^2-b^2}}\sum_{k=1}^{\infty}(\beta^{2k-1}-\alpha^{2k-1})\dfrac {i}{2k-1}(\cos((2k-1)\pi)-1)$
$=\dfrac{1}{\sqrt{a^2-b^2}}\sum_{k=1}^{\infty}(\beta^{2k-1}-\alpha^{2k-1})\dfrac {-2i}{2k-1}$
Yet I fail to see how this could lead to the supposed final result, especially with that $i$ there.
| A simpler method:
Let us recall the fact that when $A, B, C\in{\mathbb R}$ and $\Delta=B^2-4 A C < 0$ then
\begin{equation}
\int \frac{d t}{A t^2 + B t + C} = \frac{2}{\sqrt{-\Delta}} \arctan\left(\frac{2 A t + B}{\sqrt{- \Delta}}\right)
\end{equation}
The substitution $t = \tan(x/2)$ in the original integral gives
\begin{equation}
I = \int_0^\pi\frac{1}{a + b \cos(x)} d x= \int_0^{+\infty}\frac{1}{a + b\frac{1-t^2}{1+t^2}}\times \frac{2}{1 + t^2} dt
=\int_0^{+\infty}\frac{2}{(a-b)t^2+(a+b)} d t
\end{equation}
Here $\Delta = -4 (a-b)(a+b) = -4(a^2-b^2)<0$, hence
\begin{equation}
I = \left[ \frac{2}{\sqrt{a^2-b^2}}\arctan\left(\frac{(a-b)t}{\sqrt{a^2-b^2}}\right)\right]_0^{+\infty} = \frac{\pi}{\sqrt{a^2-b^2}}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Count a triple integral on a sphere $\int\int\int_{x^2+y^2+z^2 \le 1} |x+y|e^{2xy-z^2}dxdydz$
I have tried to change coordinates to spherical, but it didn't help. Can you give me a hint, please?
| Use the substitution
$$\begin{cases} u = \frac{x+y}{\sqrt{2}} \\ v = \frac{x-y}{\sqrt{2}} \\ w = z\\ \end{cases} \implies u^2-v^2 = 2xy$$
(chosen so that the Jacobian is $1$) to get the following integral
$$\sqrt{2}\iiint_{u^2+v^2+w^2\leq 1} |u|e^{u^2-v^2-w^2}\:du \:dv \:dw = \sqrt{2} \iint_{v^2+w^2\leq 1} e^{1-2(v^2+w^2)}\: dv \:dw$$
by using symmetry (the integrand is an even function of $u$, so double the value and then integrate from $u=0$ to $u=\sqrt{1-v^2-w^2}$). Then convert to "polar":
$$ = 2\sqrt{2}\pi e \int_0^1 r\:e^{-2r^2}dr = \frac{\pi(e^2-1)}{e\sqrt{2}}$$
$\textbf{EDIT}$: Alternatively, at the $uvw$ stage, convert to spherical coordinates with $u$ as the "$z$". The integrand becomes
$$ r^3|\cos\theta|\sin\theta e^{r^2(\cos^2\theta-\sin^2\theta)} = \frac{1}{2}r^3\sin2\theta e^{r^2\cos2\theta}$$
so it becomes clear which order of integration is easiest. Using symmetry, we only have to integrate the "upper" half to avoid the absolute value.
$$= 2\sqrt{2}\int_0^{2\pi} \int_0^1 \int_0^{\frac{\pi}{2}}\frac{1}{2}r^3\sin2\theta e^{r^2\cos2\theta} \:d\theta \:dr \:d\phi = \pi\sqrt{2}\int_0^1 -re^{r^2\cos2\theta}\Biggr|_0^{\frac{\pi}{2}}dr$$
$$=\pi\sqrt{2}\int_0^1 2r\sinh(r^2)dr = \pi \sqrt{2}\cosh(r^2)\Biggr|_0^1 = \pi\sqrt{2}(\cosh(1)-1)$$
which is the same thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
| Another way of doing it. Using the result that,
$$\sqrt{1+\frac{n}{m}}-\sqrt{1-\frac{n}{m}} ≥ \frac{n}{m}$$
$$\sqrt{1+\frac{3}{32}}-\sqrt{1-\frac{3}{32}} >\frac{3}{32}$$
$$\sqrt{1+\frac{3}{32}} > \frac{3}{32}+\sqrt{1-\frac{3}{32}}$$
$$8\sqrt{1+\frac{3}{32}} > \frac{3}{4}+8\sqrt{1-\frac{3}{32}}$$
$$\sqrt{70} > \frac{3}{4}+\sqrt{57}\tag{1}$$
Assuming $a<b$;
$$\sqrt{7}+\sqrt{10}<\sqrt{3}+\sqrt{19}$$
$$17+2\sqrt{70}<22+2\sqrt{57}$$
$$\sqrt{70}<\frac{5}{2}+\sqrt{57}$$
By Using (1);
$$\sqrt{70}<\frac{5}{2}+\sqrt{70}-\frac{3}{4}$$
$$0<1+\frac{3}{4}$$
Hence true, so $a<b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebyshev polynomials)
Thanks.
| $$
\alpha = \frac{\omega+\omega^4}{2}\\
\alpha^2 = \frac{\omega^2 + \omega^3 + 2}{4}\\
2 \alpha + 4 \alpha^2 = \omega+\omega^4 + \omega^2 + \omega^3 + 2\\
4 \alpha^2 + 2 \alpha -1 = 0\\
$$
Do $\alpha^n$ enough times that you see all the powers of $\omega^k$. Then you can see how to combine them that gives you the minimal polynomial for $\omega$.
| {
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} |
Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y\cos y + \sin y + \cos y + 1)$
$= 2(\sin y\cos y + \sin y + \cos y + \sin^2 y + \cos^2 y) (\sin^2 y + \cos^2 y = 1)$
$= 2(\sin^2 y + \sin y\cos y + \sin y + \cos^2 y + \cos y)$
I got stuck here. I do not know what to do from here.
I have tried and tried several days even contacted friends but all to no avail.
| If you have difficulty finding clever groupings, you could brute-force the problem by expanding everything and seeing what happens:
Writing $c := \cos y$ and $s := \sin y$ to save typing, we have
$$\require{cancel}\begin{align}
2(s+1)(c+1) &= (s+c+1)^2 \\
2(s c + s + c + 1) &= s^2 + c^2 + 1 + 2 s c + 2 s + 2 c \\
\cancel{2s c} + \cancel{2s} + \cancel{2c} + 2 &= s^2 + c^2 + 1 + \cancel{2 s c} + \cancel{2 s} + \cancel{2 c} \\
2-1 &= s^2 + c^2 \\
1 &= \sin^2 y + \cos^2 y \quad\checkmark
\end{align}$$
Since the steps are reversible, the identity is verified. $\square$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Harmonic series bounded above by $\sqrt{n}$ Question:
(a) Prove that $$\sqrt{n+1}-\sqrt{n} > \frac{1}{2\sqrt{n+1}}$$
(b) Prove by induction, for $n \geqslant 7$, that $$1 + \frac{1}{2} + \frac13 + \cdots + \frac1n<\sqrt{n}$$
| Answer:
(a) Rationalising the numerator, we have:
\begin{align}
\sqrt{n+1}-\sqrt{n} &= \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} \\
&=\frac{1}{\sqrt{n+1}+\sqrt{n}} \\
&> \frac{1}{2\sqrt{n+1}} \qquad (\text{since $\sqrt{n}<\sqrt{n+1}$})
\end{align}
(b) The base case is satisfied, since $H_7 = 2.592\ldots < 2.645\ldots = \sqrt{7}$. Assuming true for $n=k$, we will show it to be true for $n=k+1$. Now,
\begin{align}
1 + \frac{1}{2} + \frac13 + \cdots + \frac1k + \frac{1}{k+1} &< \sqrt{k} + \frac{1}{k+1} \qquad (\text{by induction hypothesis})\\
&< \sqrt{k+1} - \frac{1}{2\sqrt{k+1}} + \frac{1}{k+1} \qquad (\text{by part (a)}) \\
&= \sqrt{k+1} - \frac{\sqrt{k+1} - 2}{2(k+1)} \\
&< \sqrt{k+1} \qquad (\text{since $\sqrt{k+1}>2$ for $k \geqslant 7$})
\end{align}
Therefore, by the principle of mathematical induction, the claim is true for all integers $n \geqslant 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
| Using a technique similar to Liouville's theorem, $\sqrt{3}$ is a root of $P_2(x)=x^2-3$. Then, for any $\frac{m}{n}$ we have an $\varepsilon$ in between $\sqrt{3}$ and $\frac{m}{n}$ such that (this is MVT)
$$\left|P_2\left(\frac{m}{n}\right)\right|=
\left|P_2(\sqrt{3})-P_2\left(\frac{m}{n}\right)\right|=
|P_2'(\varepsilon)|\cdot \left|\sqrt{3}-\frac{m}{n}\right|$$
or
$$\left|\sqrt{3}-\frac{m}{n}\right|=
\left|\frac{m^2-3n^2}{2\varepsilon \cdot n^2}\right|\geq
\frac{1}{2\left|\varepsilon\right| \cdot n^2}\tag{1}$$
Now, if $\frac{m}{n}<\varepsilon<\sqrt{3}$ then $(1)$ becomes $\frac{1}{2\left|\varepsilon\right| \cdot n^2}>\frac{1}{2\sqrt{3}n^2}>\frac{1}{5n^2}$ and we are done.
If $\sqrt{3}<\varepsilon<\frac{m}{n}<\frac{5}{2}$ then $2\varepsilon<5$ and $(1)$ becomes $\frac{1}{2\left|\varepsilon\right| \cdot n^2}>\frac{1}{5n^2}$. So, we are done.
If $\sqrt{3}<\frac{5}{2}<\varepsilon<\frac{m}{n}$ then
$$\left|\frac{m}{n}-\sqrt{3}\right|>
\left|\frac{5}{2}-\sqrt{3}\right|=
\frac{\frac{25}{4}-3}{\frac{5}{2}+\sqrt{3}}=
\frac{13}{10+ 4\sqrt{3}}>
\frac{1}{5\cdot 1^2}\geq
\frac{1}{5\cdot n^2}$$
for all $n\geq1$.
| {
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"url": "https://math.stackexchange.com/questions/3475708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $\cos(A+B+C)=\cos A\cos B\cos C\neq 0$, then evaluate $\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$
If
$$\cos(A+B+C)=\cos A\cos B\cos C, \quad\text{with}\;A,B,C\neq \frac{k\pi}{2}$$
then
$$\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$$ is what integer?
What I tried:
$$\begin{align}
\cos(A+B+C)
&=\cos A \cos B \cos C - \sin A \sin B \sin C - \sin C \sin A \cos B - \sin B \sin C \cos A \\
&=\cos A\cos B\cos C\left(1-\tan A\tan B-\tan B\tan C-\tan C\tan A\right)
\end{align}$$
How do i solve it Help me please
| Using $\sin(2x)=2\sin(x)\cos(x)$ along with $\sin(A+B)=\sin(\pi-C)=\sin(C)$ and $\cos(A)\cos(B)\cos(C)=\cos(A+B+C)=\cos(\pi)=-1$ should give the value as $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $f'\left(\frac13\right)$ does not exist? Consider the function $f(x)=\begin{cases}
x^2\left|\cos\dfrac{\pi}{2x}\right|, &x\ne 0\\
0, & x=0
\end{cases}$
$$f(x)=\begin{cases}
x^2\cos\dfrac{\pi}{2x}, &\dfrac{\pi}{2x}\in\left[2n\pi-\dfrac{\pi}{2},2n\pi+\dfrac{\pi}{2}\right]\\
-x^2\cos\dfrac{\pi}{2x}, &\dfrac{\pi}{2x}\in\left(2n\pi+\dfrac{\pi}{2},2n\pi+\dfrac{3\pi}{2}\right)\\
0, & x=0
\end{cases}$$
$$f(x)=\begin{cases}
x^2\cos\dfrac{\pi}{2x}, &x\in\left[\dfrac{1}{4n+1},\dfrac{1}{4n-1}\right]\\
-x^2\cos\dfrac{\pi}{2x}, &x\in\left(\dfrac{1}{4n+3},\dfrac{1}{4n+1}\right)\\
0, & x=0
\end{cases}$$
$$f'\left(\dfrac{1}{3}\right)=\lim_{h\to0}\dfrac{\left(\dfrac{1}{3}+h\right)^2\cos\left(\dfrac{3\pi}{2(1+3h)}\right)}{h}$$
$$f'\left(\dfrac{1}{3}\right)=\lim_{h\to0}\dfrac{\left(\dfrac{1}{h}+9h+6\right)\cos\left(\dfrac{3\pi}{2(1+3h)}\right)}{9}$$
$$f'\left(\dfrac{1}{3}\right)=\lim_{h\to0}\dfrac{\cos\left(\dfrac{3\pi}{2(1+3h)}\right)}{9h}+\lim_{h\to0}h\cos\left(\dfrac{3\pi}{2(1+3h)}\right)+\dfrac{2}{3}\lim_{h\to0}\cos\left(\dfrac{3\pi}{2(1+3h)}\right)$$
$$f'\left(\dfrac{1}{3}\right)=\lim_{h\to0}\dfrac{\cos\left(\dfrac{3\pi}{2(1+3h)}\right)}{9h}$$
Assuming $\dfrac{1}{1+3h}=y$, as $h\rightarrow 0, \dfrac{1}{1+3h}\rightarrow 1,h=\dfrac{1-y}{3y}$
$$f'\left(\dfrac{1}{3}\right)=\lim_{y\to1}\dfrac{y\cos\left(\dfrac{3\pi y}{2}\right)}{3(1-y)}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ determinant form
$$f'\left(\dfrac{1}{3}\right)=\lim_{y\to1}\dfrac{-y\sin\left(\dfrac{3\pi y}{2}\right)\dfrac{3\pi}{2}+\cos\left(\dfrac{3\pi y}{2}\right)}{-3}$$
$$f'\left(\dfrac{1}{3}\right)=-\dfrac{\pi}{2}$$
It seems function is differentiable at $\dfrac{1}{3}$, what am I missing here?
| For a derivative to exist, the right hand derivative must be equal to the left hand derivative.
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)-f(x-h)}{h}$$
Also the limits for defining functions will be $[\frac{1}{4n+1},\frac{1}{4n-1}] \text{ and }(\frac{1}{4n+3},\frac{1}{4n+1})$
For $x=1/3$, the left hand derivative will use the last part of the equation while the right handed derivative will use the second part of the equation. You just evaluated the second part. Also evaluate $\lim_{h\to0}\frac{f(x)-f(x-h)}{h}$ using the definition for $f(1/3-h)=-(1/3-h)^2\cos(\frac{\pi}{2(1/3-h)})$and you'll see they're not equal
| {
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"source": "stackexchange",
"question_score": "2",
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How does $49\cdot\left(\frac{1}{7}\right)^{x}-\left(\frac{1}{7}\right)^x$ become $\left(\frac{1}{7}\right)^{x}\left(49-1\right)$? Rewrite $\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^x$ as $A\cdot\left(\frac{1}{7}\right)^x$ is a question I got.
Now the answer to A is 48, but I don't really get how they got to the answer:
$$\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^x = \frac{\left(\frac{1}{7}\right)^{x}}{\left(\frac{1}{7}\right)^{2}}-\left(\frac{1}{7}\right)^x$$
$$ = 49\cdot\left(\frac{1}{7}\right)^{x}-\left(\frac{1}{7}\right)^x$$
so far this makes sense, multiplying by the reciprocal of $\frac{1}{\left(\frac{1}{7}\right)^2}$ ends up in $\left(\frac{7}{1}\right)^2$ I suppose, so 49. But the next step does not make any sense to me:
$$ = \left(\frac{1}{7}\right)^{x}\left(49-1\right)$$
How do we end up with 49 minus 1? Or do I have to treat this expression the same as $10x - x$ which is the same as $10x - 1x$, so we end up with $9x$. And in this specific case the $x$'s are the fractions to the $x$ power?
| Note that, for all $a$ and $b$, due to the distributive property, you have
$$(a-1)b = ab - b \tag{1}\label{eq1A}$$
Thus, with $a = 49$ and $b = \left(\frac{1}{7}\right)^x$, you get what you're asking about, i.e.,
$$49\left(\frac{1}{7}\right)^x - \left(\frac{1}{7}\right)^x = (49 - 1)\left(\frac{1}{7}\right)^x \tag{2}\label{eq2A}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest positive number k For any positive number $n$, let $a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$ ($2$ appear $n$) and let $k$ is positive number such that $\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$ for any positive number $n$, then find the smallest positive number $k$.
I have $a_1=\sqrt 2$ and $a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N $
Consider
$a_1=\sqrt 2 \lt 2$
$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$
$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$
Use Mathematical Induction, I conclude $\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$
Thus, $3-a_{n+1}\gt1$ and $7-a_{n}\gt 5$
Since $k\in\mathbb N$, I have $\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$
Hence, $3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$
Therefore $k=5$
Please check my solution, Is it correct?, Thank you
| You have given a lower & upper bound for $a_n$. With this, you've determined a possible value for $k$, but you haven't shown it's necessarily the smallest such value of $k$.
Instead, note that $a_{n+1} = \sqrt{2 + a_n}$. Thus, you have
$$\begin{equation}\begin{aligned}
\frac{3-a_{n+1}}{7-a_n} & = \frac{3-\sqrt{2 + a_n}}{7-a_n} \\
& = \frac{(3-\sqrt{2 + a_n})(3 + \sqrt{2 + a_n})}{(7-a_n)(3 + \sqrt{2 + a_n})} \\
& = \frac{9-(2 + a_n)}{(7-a_n)(3 + \sqrt{2 + a_n})} \\
& = \frac{7 - a_n}{(7-a_n)(3 + \sqrt{2 + a_n})} \\
& = \frac{1}{3 + \sqrt{2 + a_n}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, if $L$ is the supremum of $a_n$, then $k = 3 + \sqrt{2 + L}$. To determine $L$, you can prove that $a_n$ is a strictly increasing sequence (I'll leave it to you fill in the details, such as what is shown in Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$), with an upper bound as you've shown, so it must converge to limiting value of its supremum. To determine this value, use
$$\begin{equation}\begin{aligned}
L & = \sqrt{2 + L} \\
L^2 & = 2 + L \\
L^2 - L - 2 & = 0 \\
(L - 2)(L + 1) & = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, since $L \gt 0$, you have $L = 2$, as you surmised. You also have
$$k = 3 + \sqrt{2 + 2} = 5 \tag{3}\label{eq3A}$$
which matches what you got.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of $a+b+c+d$ with following conditions?
$$
\begin{split}
a^2 &+b &+c &+d&=10\\
a &+b^2 &+c &+d&=12\\
a &+b &+c^2 &+d&=16\\
a &+b &+c &+d^2&=22\\
a &+b &+c &+d&=?
\end{split}
$$
By adding these equations I get that $3(a+b+c+d)=60-(a^2+b^2+c^2+d^2)$.
So can I get some hint that how can I get the value of $(a^2+b^2+c^2+d^2)$ (or relate it with the equation) or my approach is wrong and there are some easy way to solve this problem?
| Subtract the first equation from the second to get $b^2 + a - a^2 - b = 2$, which becomes $2 = (b - a)(b + a - 1)$.
Assuming that $a,b,c,d$ are all positive integers, then $b - a = 1$ and $b + a - 1 = 2$ OR $b - a = 2$ and $b + a - 1 = 1$. The restriction that all four numbers are positive integers gives $a = 1$ and $b = 2$.
Substituting these values into the first equation gives $c + d = 7$ and into the fourth equation gives $c + d^2 = 19$. Solving simultaneously gives $c = 3, d = 4$.
$\therefore a + b + c + d = 1 + 2 + 3 + 4 = 10$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find $\int\frac{x^2-1}{x^4+x^2+1}dx$
Find
$$
\int\frac{x^2-1}{x^4+x^2+1}dx
$$
$x^4+x^2+1>0$, so not possible to factor, so I guess there is no direct way or through partial fraction decomposition.
$$
\int\frac{x^2-1}{x^4+x^2+1}dx=\int\frac{(x^2-1)^2}{(x^2)^3-1}dx\\
$$
I have no clue of how to solve it, is there any substitution that I can give to $x$ so that it becomes simple ?
| At the bottom, write $ x^2 + 1/x^2 = (x + 1/x)^2 -2$. Then assume $ x+ 1/x = t$. Then you're done.
Let me right the detailed solution. Divide the numerator and denominator of the integrand by $x^2.$ Then we have $$ \frac{1-1/x^2}{(x+1/x)^2 - 1}$$. Then do the substitution business. Then just integrate the following nice cute function: $$ \frac{1}{t^2 -1}$$. After integrating, we have the following function $$ 1/2 ln|\frac{t-1}{t+1}|$$. QED
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I calculate the limit $\lim\limits_{n \to \infty} \frac1n\ln \left( \frac{2x^n}{x^n+1} \right)$. I have the following limit to find:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg)$$
Where $n \in \mathbb{N}^*$ and $x \in (0, \infty)$.
I almost got it. For $x > 1$, I observed that:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n(1 + \frac{1}{x^n})} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2}{1+\frac{1}{x^n}} \bigg)$$
Because $x>1$, we have that $x^n \rightarrow \infty$ as $n \rightarrow \infty$, so that means that we have:
$$\dfrac{1}{\infty} \cdot \ln \bigg ( \dfrac{2}{1+\frac{1}{\infty}} \bigg ) = 0 \cdot \ln 2 = 0$$
The problem I have is in calculating for $x \in (0, 1]$. If we have that $x \in (0, 1]$ that means $x^n \rightarrow 0$ as $n \to \infty$, so:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n + 1} \bigg ) = \lim\limits_{n \to \infty} \dfrac{\ln \bigg( \dfrac{2x^n}{x^n + 1}\bigg )}{n} $$
And I tried using L'Hospital and after a lot of calculation I ended up with
$$\ln x \lim\limits_{n \to \infty} \dfrac{x^n + 1}{x^n}$$
which is
$$\ln x\cdot \dfrac{1}{0}$$
And this is my problem. Maybe I applied L'Hospital incorrectly or something, I'm not sure. Long story short, I do not know how to calculate the following limit:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n+1} \bigg )$$
when $x \in (0, 1]$.
| The term reduces to
\begin{align*}
\dfrac{1}{n}\log\left(2-\dfrac{2}{x^{n}+1}\right)&=\dfrac{1}{n}\log 2+\dfrac{1}{n}\log\left(1-\dfrac{1}{x^{n}+1}\right).
\end{align*}
We do the L'Hopital to the second term, it becomes
\begin{align*}
&\lim_{n\rightarrow\infty}\dfrac{\dfrac{1}{1-\dfrac{1}{x^{n}+1}}\dfrac{1}{(x^{n}+1)^{2}}x^{n}\log x}{1}\\
&=\lim_{n\rightarrow\infty}\dfrac{x^{n}+1}{x^{n}}\dfrac{1}{(x^{n}+1)^{2}}x^{n}\log x\\
&=\lim_{n\rightarrow\infty}(x^{n}+1)^{-1}\log x\\
&=\log x.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 5,
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} |
Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results.
Getting two different answers when tried using two different methods:-
M-$1$:
$$\int \dfrac{5-x}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx$$
$$\dfrac{1}{2}\int\dfrac{-2x+7}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(5-x\right)\left(x-2\right)}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{-x^2+7x-10}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(\dfrac{3}{2}\right)^2-\left(x-\dfrac{7}{2}\right)^2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{x-\dfrac{7}{2}}{\dfrac{3}{2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$
M-$2$:
$$x=5\sin^2\theta+2\cos^2\theta$$
$$dx=\left(10\sin\theta\cos\theta-4\cos\theta\sin\theta\right) \, d\theta$$
$$\int \sqrt{\dfrac{5\cos^2\theta-2\cos^2\theta}{5\sin^2\theta-2\sin^2\theta}}\cdot6\sin\theta\cos\theta \,d\theta$$
$$6\int \cos^2\theta \,d\theta$$
$$\int 3\left(1+\cos2\theta\right) \,d\theta$$
$$3\left(\theta+\dfrac{\sin2\theta}{2}\right)$$
$$3\theta+\dfrac{3}{2}\sin2\theta$$
$$x=5\sin^2\theta+2-2\sin^2\theta$$
$$\sin^{-1}\sqrt{\dfrac{x-2}{3}}=\theta$$
$$\cos2\theta=1-2\sin^2\theta$$
$$\cos2\theta=1-2\cdot\dfrac{x-2}{3}$$
$$\cos2\theta=\dfrac{7-2x}{3}$$
$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\dfrac{3}{2}\cdot\dfrac{\sqrt{9-(49+4x^2-28x)}}{3}$$
$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\sqrt{\left(5-x\right)\left(x-2\right)}$$
In first and second method I am getting the different results of $\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$ and $3\sin^{-1}\sqrt{\dfrac{x-2}{3}}$ respectively. I checked that these are not inter-convertible. Why am I getting this difference?
| Both your answers are correct. Your two functions differ by $3\pi/4$, so they have the same derivative.
Let $b=\sqrt{\frac{x-2}3}$. For your integral to make sense you need $0\leq b\leq1$ (this comes from $2\leq x\leq5$). Let $a=\arcsin b$. Note $0\leq\arcsin b\leq \tfrac\pi2$, and so the sine is injective when applied to $2a-\tfrac\pi2$. Then
\begin{align}
\sin\left(2a-\tfrac\pi2\right)
&=-\cos(2a)=-(1-2\sin^2a)=2b^2-1.
\end{align}
So
\begin{align}
2\arcsin\sqrt{\frac{x-2}3}\ -\frac\pi2
&=2\arcsin b -\frac\pi2=2a-\frac\pi2\\ \ \\
&=\arcsin(2b^2-1)\\ \ \\
&=\arcsin\frac{2x-7}3.
\end{align}
Then, multiplying by $3/2$,
$$
3\arcsin\sqrt{\frac{x-2}3}\ -\frac{3\pi}4=\frac32\,\arcsin\frac{2x-7}3.
$$
| {
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"answer_count": 2,
"answer_id": 0
} |
Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron whose vertices are the given points:
$$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$
Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
| your tetrahedron is also a pyramid. with the volume of $\frac{1}{3}\cdot \frac{2\cdot 2}{2}\cdot 2=\frac{4}{3}$
| {
"language": "en",
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"answer_id": 2
} |
On sums such as $\sum_{k=0}^\infty \binom{2k}{k}\frac{1}{8^k}=\sqrt{2}$ This identity is a special case of a more general formula found here, and known at least since 1972 (published in Abramowitz and Stegun 1972, p. 555.) Many remarkable series involving the inverse of binomial coefficients are known and listed in the same source, see also here for a way to derive them.
However, there are plenty of interesting series involving binomial coefficients (not their inverse) that are not listed in the references that I checked. For instance:
$$\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)$$
$$\sum_{k=0}^\infty \binom{4k}{2k}\frac{1}{32^k}= \sin\frac{\pi}{8}+\cos\frac{\pi}{8}$$
Another example is:
Question
How do you prove these results? I found them using WolframAlpha symbolic calculator, see here for an example. One of these results is proved here, but I am looking for a proof that would apply to a broad class of such series.
Background
The reason that I am interested in powers of two ($8^{-k}, 32^{-k}$) is because I am looking for series converging to irrational numbers, with each term being a fraction: the denominator is a power of two, and the numerator is an integer (a binomial coefficient in this case.) The goal is to gain some insights in the binary digit distribution of numbers such as $\sqrt{2}$.
For instance, a result that could be useful for me is the following:
$$\sqrt{2} = \lim_{n\rightarrow\infty}\frac{P_n}{8^n} = \lim_{n\rightarrow\infty}\frac{1}{8^n}\sum_{k=0}^n 8^{n-k}\binom{2k}{k},$$
with $P_n$ being an integer.
| The Lagrange Inversion Formula provides an appropriate method to derive
\begin{align*}
\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)\tag{1}
\end{align*}.
Let a formal power series $w=w(t)$ be implicitely defined by a relation $w=t\Phi(w)$, where $\Phi(t)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:
$$[t^n]w(t)^k=\frac{k}{n}[t^{n-k}]\Phi(t)^n$$
A variation stated as formula $G6$ in Lagrange Inversion: when and how by R. Sprugnoli (etal) is:
Let $F(t)$ be any formal power series and $w=t\Phi(w)$ as before, then the following is valid:
\begin{align*}
[t^n]F(t)\Phi(t)^n=\left[\left.\frac{F(w)}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\tag{2}
\end{align*}
Note: The notation $[\left.f(w)\right|w=g(t)]$ is a linearization of $\left.f(w)\right|_{w=g(t)}$ and denotes the substitution of $g(t)$ to every occurrence of $w$ in $f(w)$ (that is, $f(g(t))$). In particular, $w=t\Phi(w)$ is to be solved in $w=w(t)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.
In order to prove (1) we set $F(t)=1$ and $\Phi(t)=(1+t)^3$. We then have
$$t\Phi'(w)=3t(1+w)^2=\frac{3t\Phi(w)}{1+w}=\frac{3w}{1+w}$$
It follows:
\begin{align*}
\binom{3n}{n}&=[t^n]F(t)\Phi(t)^n=[t^n](1+t)^{3n}\\
&=[t^n]\left[\left.\frac{1}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\\
&=[t^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=t\Phi(w)\right]\\
&=[t^n]\left[\left.\frac{1+w}{1-2w}\right|w=t\Phi(w)\right]\\
\end{align*}
Let
\begin{align*}
A(t):=\sum_{n\ge0}\binom{3n}{n}t^n=\left.\frac{1+w}{1-2w}\right|_{w=t\Phi(w)}
\end{align*}
Expressing $A(t)=\frac{1+w}{1-2w}$ in terms of $w$, we get
$$w=\frac{A(t)-1}{2A(t)+1}$$
Since $w=t\Phi(w)=t(1+w)^3$, we obtain
\begin{align*}
\frac{A(t)-1}{2A(t)+1}=t\left(1+\frac{A(t)-1}{2A(t)+1}\right)^3
\end{align*}
which simplifies to:
\begin{align*}
(4-27t)A(t)^3-3A(t)-1=0\tag{3}
\end{align*}
In order to get the RHS of $(1)$ we first analyse the structure of (3) which is
$$f(t)A(t)^3-3A(t)=1$$
with $f(t)$ linear and observe a similarity of this structure with the identity
$$4\cos^3{t}-3\cos{t}=\cos{3t}$$
Thus we use the ansatz:
\begin{align*}
A(t) := \frac{2\cos\left(g(t)\right)}{\sqrt{4-27t}}\tag{4}
\end{align*}
We see
\begin{align*}
(4-27t)A(t)^3-3A(t)
&=\frac{8\cos^3\left(g(t)\right)}{\sqrt{4-27t}}-\frac{6\cos\left(g(t)\right)}{\sqrt{4-27t}}=\\
&=\frac{2\cos\left(3g(t)\right)}{\sqrt{4-27t}}\\
&=1
\end{align*}
Now, since
\begin{align*}
2\cos\left(3g(t)\right)&=\sqrt{4-27t}\\
4\cos^2\left(3g(t)\right)&=4-27t\\
\sin^2\left(3g(t)\right)&=\frac{27}{4}t\\
\end{align*}
we get
\begin{align*}
g(t)&=\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\tag{5}\\
\end{align*}
We finally conclude from (4) and (5)
\begin{align*}
\color{blue}{\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}}
&=\left.\frac{2\cos(g(t))}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\
&=\left.\frac{2\cos\left(\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\right)}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\
&\,\,\color{blue}{=\frac{4\sqrt{10}}{5}\cos\left(\frac{1}{3}\arcsin\left( \frac{3\sqrt{6}}{8}\right)\right)}
\end{align*}
and the claim follows.
| {
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"answer_count": 3,
"answer_id": 0
} |
Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method with convex function but no result:
Since $f(x)=\frac{1}{x}$ is a convex function, by Jensen we obtain:
$$\frac{1}{x+y+z}\sum_{cyc}\frac{x^3}{x^2+y^2}=\sum_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{1}{\frac{x^2+y^2}{x^2}}\right)\geq$$
$$\geq\frac{1}{\sum\limits_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{x^2+y^2}{x^2}\right)}=\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}.$$
Thus, it's enough to prove that
$$\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}\geq\frac{1}{2}$$ or
$$x+y+z\geq\sum_{cyc}\frac{y^2}{x},$$ which is wrong.
thanks
| We apply AM-GM Inequality with $x^2 + y^2 \geq 2xy$.
\begin{align*}
&\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \\
&= \dfrac{x^3 + xy^2}{x^2+y^2}+\dfrac{y^3 + yz^2}{y^2+z^2}+\dfrac{z^3 + zx^2}{z^2+x^2} - \left(\dfrac{xy^2}{x^2+y^2}+\dfrac{yz^2}{y^2+z^2}+\dfrac{zx^2}{z^2+x^2}\right) \\
&= x + y + z - \left(\dfrac{xy^2}{x^2+y^2}+\dfrac{yz^2}{y^2+z^2}+\dfrac{zx^2}{z^2+x^2}\right) \\
&\geq x + y + z - \left(\dfrac{xy^2}{2xy}+\dfrac{yz^2}{2yz}+\dfrac{zx^2}{2xz}\right) \\
&= x + y + z - \left(\frac{x}{2} + \frac{y}{2} + \frac{z}{2}\right) \\
&= \frac{x + y + z}{2}
\end{align*}
| {
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"answer_count": 3,
"answer_id": 1
} |
Formula for probability of sums of rolling 2 six-sided dice I'm learning about expected probability and have got stuck.
In summing the outcomes of rolling 2 six-sided dice, how is the probability for each event given by $p_i = \frac{1}{36} (6 - abs(7 - i))p$ please?
Obviously I can check by counting instances of each sum in a multiplication grid, but I would like some intuition for how the formula works.
| The best way is indeed to look at the table of the outcomes. Each outcome has a probability of $\frac1{36}$.
$$\begin{array}{|c|c|c|c|c|c|}\hline \text{die 1 / die 2 } & 1 &2 &3 &4 &5 &6 \\ \hline\hline \hline 1 & 2&3 &4 &5 &6 &7 \\ \hline 2 & 3 &4 &5 &6 &7&8 \\ \hline 3 &4 &5 &6 &7&8&9 \\ \hline 4 &5 &6 &7&8&9&10 \\ \hline 5 &6 &7&8&9&10 &11 \\ \hline 6 &7&8&9&10 &11 &12 \\ \hline\end{array}$$
If I use your notation I have:
$p_2=1, p_3=2, p_4=3,p_5=4,p_6=5,p_7=6,p_8=5,p_9=4,p_{10}=3,p_{11}=2,p_{12}=1$
We see a maximum is at $p_7=6$. Right and left from $p_7$ the number of values decrease by 1. So we subtract the absolute value $6-|A|$.
The value of this has to be 6 if $i=7$. Thus $|A|=0$
The value of this has to be 5 if $i=8$. Thus $|A|=1$
The value of this has to be 5 if $i=6$. Thus $|A|=1$
This should be enough to evaluate the expression for $|A|$: $|7-i|$
So in total the probability that the sum of two dice is $i$ is $\frac1{36}\cdot (6-|7-i|)$
| {
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If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamma^3})=16$$
Here's what I have tried,
By Vieta's rule
$\alpha+\beta+\gamma=\frac{-1}{2}\text{. ...........}(1)$
$\alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{2}\text{. ...........}(2)$
$\alpha \beta \gamma=\frac{-1}{2}\text{. ...........}(3)$
Squaring $(1)$,
$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\alpha\gamma)=\frac{1}{4}\text{. ...........}(4)$
From $(2)$,
$\alpha^2+\beta^2+\gamma^2=\frac{-3}{4}\text{. ...........}(5)$
Putting the roots and adding these equations,
$2\alpha^3+\alpha^2+\alpha+1=0$
$2\beta^3+\beta^2+\beta+1=0$
$2\gamma^3+\gamma^2+\gamma+1=0$
We get,
$2(\alpha^3+\beta^3+\gamma^3)+(\alpha^2+\beta^2+\gamma^2)+(\alpha+\beta+\gamma)+3=0$
Putting the values,
$2(\alpha^3+\beta^3+\gamma^3)+\frac{-3}{4}+\frac{-1}{2}+3=0$
$(\alpha^3+\beta^3+\gamma^3)=\frac{-7}{8}$
Then I divided $2x^3+x^2+x+1=0$ by $x$ and I found out $(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma})$ the same way I found out $(\alpha^3+\beta^3+\gamma^3)$. Likewise doing the same, finding $\sum\frac{1}{\alpha^2}$ then $\sum\frac{1}{\alpha^3}$ I found out
$\sum\frac{1}{\alpha^3}=-4$
But still I'm far from the answer, also the $-$ sign is creating problems.
Any help would be highly appreciated
| Notice that if we multiplly the polynomial with $x-1$ we get $$x^3(x-1)+(x-1)(x^3+x^2+x+1)=0$$
so $$2x^4-x^3-1=0\implies \boxed{{1\over x^3}= 2x-1}$$
So your expression is $$E=(2\beta + 2\gamma -2\alpha -1)(2\beta -2\gamma +2\alpha -1)(-2\beta + 2\gamma +2\alpha -1)$$
Now notice that $$\beta + \gamma +\alpha =-{1\over 2}$$ so \begin{align}E &= -8(2\alpha +1)(2\beta +1)(2\gamma +1)\\
&=-64 (\alpha +{1\over 2})(\beta +{1\over 2})(\gamma +{1\over 2}) \\
&=-32\cdot p(-{1\over 2})=-16\end{align}
| {
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How do you calculate the following limit of a standard Brownian motion $lim_{n\rightarrow \infty} E\left[\sum_{k=1}^{n}\left(B(\frac{kt}{n})-B(\frac{(k-1)t}{n}) \right)^2\right]$, where $\{B(t); t\geq 0\}$ is a standard Brownian motion.
Here is my attempt to solve it. I'd appreciate any feedback on it:
Recall that a standard Brownian motion has stationary increments and thus, $B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)$ ~$ N(0, t/n)$ . Further, by the CLT we get that $\frac{B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)}{\sqrt{\frac{t}{n}}}$ ~ $N(0,1)$. So, if we let for each n $Z_{n,k}= \frac{B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)}{\sqrt{\frac{t}{n}}}$, then $\{Z_{n,k}\}$ is a sequence of i.i.d N(0,1) random variables. Furthermore, by the WLLN we have that $E[\frac{1}{n}\sum_{k=1}^nZ^2_{n,k}]\rightarrow 1$ as $n\rightarrow \infty$. Therefore, $$lim_{n\rightarrow \infty} E\left[\sum_{k=1}^{n}\left(B(\frac{kt}{n})-B(\frac{(k-1)t}{n}) \right)^2\right] = lim_{n\rightarrow \infty} E \left[\sum_{k=1}^n \frac{t}{n}Z^2_{n,k} \right]=t\cdot lim_{n\rightarrow}E[\frac{1}{n}\sum_{k=1}^nZ^2_{n,k}]=t$$
| By linearity of expectation and expanding the quadratic term, we have
\begin{align}
\mathbb E\left[ \sum_{k=1}^n \left(B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right)\right)^2\right] &= \sum_{k=1}^n \mathbb E\left[B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right) \right]^2\\
&= \sum_{k=1}^n\mathbb E\left[B\left(\frac{kt}n\right)^2 - 2 B\left(\frac{kt}n\right)B\left(\frac{(k-1)t}n\right) + B\left(\frac{(k-1)t}n\right)^2 \right]\\
&= \sum_{k=1}^n\left(\mathrm{Var}\left(B\left(\frac{kt}n\right)\right) - 2\mathrm{Cov}\left((B\left(\frac{kt}n\right),\left(\frac{(k-1)t}n\right)\right) + \mathrm{Var}\left(B\left(\frac{(k-1)t}n\right)\right)\right)\\
&= \sum_{k=1}^n \mathrm{Var}\left(B\left(\frac{kt}n\right)\right) - 2\sum_{k=1}^n \mathrm{Cov}\left((B\left(\frac{kt}n\right),\left(\frac{(k-1)t}n\right)\right) + \sum_{k=1}^n \mathrm{Var}\left(B\left(\frac{(k-1)t}n\right)\right)\\
&= \sum_{k=1}^n \frac{kt}n - 2 \sum_{k=1}^n \frac{(k-1)t}n +\sum_{k=1}^n \frac{(k-1)t}n\\
&= \frac tn\left(\sum_{k=1}^n k - 2\sum_{k=1}^n (k-1) + \sum_{k=1}^n (k-1) \right)\\
&= \frac tn\left(\sum_{k=1}^n k - \sum_{k=1}^n (k-1) \right)\\
&= t.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3493596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the correlation coefficient between $2$ variables Value $A$ is the product of $2$ integers $X$ and $Y$. Each of the last two integers takes values from the set {$1,2,5$} with probabilities: $0.2$, $0.5$, $0.3$. Determine the correlation coefficient between $X$ and $A$.
Any ideas on how to solve this? I would appreciate any help.
| We have
\begin{align}
\mathbb P(X = 1) &= \frac15\\
\mathbb P(X = 2) &= \frac12\\
\mathbb P(X = 5) &= \frac3{10}\\
\mathbb P(A = 1) &= \frac1{25}\\
\mathbb P(A = 2) &= \frac15\\
\mathbb P(A = 4) &= \frac14\\
\mathbb P(A = 5) &= \frac3{25}\\
\mathbb P(A = 10) &= \frac3{10}\\
\mathbb P(A = 25) &= \frac9{100},
\end{align}
and so
$$
\mathbb E[X] = 1\cdot\frac15 + 2\cdot\frac12+5\cdot\frac3{10} = \frac{27}{10},
$$
$$
\mathbb E[X^2] = 1^2\cdot\frac15 + 2^2\cdot\frac12+5^2\cdot\frac3{10} = \frac{97}{10}
$$
and
$$
\mathrm{Var}(X) = \mathbb E[X^2] - \mathbb E[X]^2 = \frac{97}{10} - \frac{729}{100} = \frac{241}{100}.
$$
Similarly,
$$
\mathbb E[A] = \frac{729}{100}, \quad \mathbb E[A^2] = \frac{9409}{100},\quad \mathrm{Var}(A) = \frac{9409}{100} - \left(\frac{729}{100}\right)^2 = \frac{409459}{100000}.
$$
Moreover,
\begin{align}
\mathbb E[XA] &= \mathbb E[X^2Y]\\
&= \mathbb E[X^2]\mathbb E[Y]\\
&= \frac{241}{100}\cdot \frac{27}{10}\\
&= \frac{6507}{1000}.
\end{align}
It follows that
\begin{align}
\rho(X,A) &= \frac{\mathrm{Cov}(X,A)}{\sigma(X)\sigma(A)}\\
&= \frac{\mathbb E[XA] - \mathbb E[X]\mathbb E[A]}{\sqrt{\mathrm{Var}(X)}\sqrt{\mathrm{Var}(A)}}\\
&=\frac{\frac{6507}{1000} - \frac{27}{10}\cdot\frac{9409}{100}}{\sqrt{\frac{241}{100}}\sqrt{\frac{409459}{100000}}}\\
\end{align}
There is an error in my computations as this value is not between $-1$ and $1$, but the method should be correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why must absolute value be used in this equation? I have a simple equation:
$(x-2)^2 < 3$
My first solution was:
$x-2 < \sqrt{3}$
But this gives me only:
$x < \sqrt{3} + 2$
Which is only one solution, so it's not enough. I've figured it out that I must use:
$|x-2| < \sqrt{3}$
Then, solutions are:
$-\sqrt{3} + 2 < x < \sqrt{3} + 2$
My question is how can I know that here:
$(x-2)^2 < 3$
I must use absolute value after square rooting both sides of inequality? Like: $|x-2| < \sqrt{3}$
| If you are not comfortable using $|x-2|$ then you can divide your derivation into two cases as follows:
Case 1: $x-2\ge 0$ :
$(x-2)^2 < 3 \\ \Rightarrow 0 \le x-2 < \sqrt 3 \\ \Rightarrow 2 \le x < 2+\sqrt{3}$
Case 2: $x-2< 0$ :
$(x-2)^2 < 3 \\ \Rightarrow -\sqrt 3 < x-2 < 0\\ \Rightarrow 2-\sqrt 3 < x < 2$
Taking the union of the intervals from the two cases, we get $2-\sqrt{3}<x<2+\sqrt{3}$. The case by case approach is useful for more complex inequalities e.g. $2 < (x-2)^2<3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $\tan 70^\circ - \sec 10^\circ$ have the exact value of $\sqrt{3}$? Recently, while trying to solve a problem posed by the Youtube video here, I found the following relation:
$$\tan 70^\circ - \sec 10^\circ = \sqrt{3}$$
This relation is exact, and that can be proved by combining the purely geometric arguments presented in the video linked above and some basic trigonometry.
But as far as I know, the trigonometric ratios of $10^\circ$ and $70^\circ$ cannot be expressed in exact form (using only square roots). In fact, Gauss proved that the sine or cosine of any angle ${360^\circ}\over n$ cannot be expressed in terms of fractions and square roots unless $n = 2^m\cdot\prod p_i$ where every $p_i$ is a Fermat prime: neither $10^\circ$ nor $70^\circ$ is even constructible.
So how does the stated relation hold? I have been unable to prove why it should ... any insights appreciated!
| Great answer by Michael! Since his answer was a bit terse and I had to spend some time figuring it out, here is an expansion. Since $\tan 60^\circ = \sqrt{3}$, and because $\sec 10^\circ = \frac {1} {\cos 10^\circ}$ by definition, we can say that we need to prove
$$
\frac {1} {\cos 10^\circ} = \tan 70^\circ - \tan 60^\circ
$$
Since $\sin \theta = ( \cos 90^\circ - \theta )$, and since $\sin 30^\circ = {1 \over 2}$, the RHS further reduces as follows:
$$
= \frac {\sin 70^\circ} {\cos 70^\circ} - \frac {\sin 60^\circ} {\cos 60^\circ} = \frac {\cos 20^\circ} {\sin 20^\circ} - \frac {\cos 30^\circ} {\sin 30^\circ}
= \frac {\sin 30^\circ \cdot \cos 20^\circ - \cos 30^\circ \cdot \sin 20^\circ} {\sin 20^\circ \cdot \sin 30^\circ}
= \frac {\sin (30^\circ - 20^\circ)} {\sin 20^\circ \cdot \sin 30^\circ}
= \frac {\sin 10^\circ} {\sin 20^\circ \cdot \sin 30^\circ}
=2\cdot\frac{\sin 10^\circ} {\sin 20^\circ} = 2\cdot\frac{\sin 10^\circ} {2\cdot \sin 10^\circ \cdot \cos 10^\circ}
=\frac {1} {\cos 10^\circ}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=5a$$
I could substitute the value of a in the above equation, but that gives me a biquadractic equation in b, and I don’t think it’s supposed to go that way. What am I doing wrong?
| From the first equation, $\alpha^2=\frac{5}{a},2\alpha=\frac{2b}{a}$ and we obtain $\alpha =\frac{5}{b}$.
From the second equation
$\alpha + \beta=2b, \alpha\beta=-10$. Then $\beta=-2b$ and $\frac{5}{b}=4b$ i.e.
$b^2=\frac{5}{4}.$
Then $\alpha^2+\beta^2=(\alpha+\beta)^2 -2\alpha\beta=4b^2+20=25.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim\limits_{n \to \infty} \int\limits_0^n \frac1{1 + n^2 \cos^2 x} dx$. I have to find the limit:
$$\lim\limits_{n \to \infty} \displaystyle\int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx$$
How should I approach this?
I kept looking for some appropriate bounds (for the Squeeze Theorem) that I could use to determine the the limit, but I didn't come up with anything useful.
| Let
$$ S_n=\int_0^\pi\frac{1}{1+n^2\cos^2x}\,dx $$
Then
$$\lim_{n\to\infty}S_n=0 $$
Assume that
$$ L=\lim_{n\to \infty} \int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx $$
Then
$$ \left\lfloor\frac{n}{\pi} \right\rfloor S_n\le L\le\left(\left\lfloor\frac{n}{\pi} \right\rfloor+1\right)S_n $$
So
$$ L=\lim_{n\to\infty}\left\lfloor\frac{n}{\pi} \right\rfloor S_n $$
So you need to find
$$ \lim_{n\to\infty}\left\lfloor\frac{n}{\pi} \right\rfloor S_n $$
Using an $\arctan$ substitution gives $S_n=\frac{\pi}{\sqrt{n^2+1}}$ which gives $L=1$.
Addendum: Because of the discontinuity at $\frac{\pi}{2}$ and given the symmetry of the function, it will be better to express
$$ S_n=2\int_0^\frac{\pi}{2} \frac{dx}{1+n^2\cos^2x}$$
Then
\begin{eqnarray}
S_n&=&2\int_0^\frac{\pi}{2} \frac{dx}{\sin^2x+\cos^2x+n^2\cos^2x}\\
&=&2\int_0^\frac{\pi}{2} \frac{dx}{\sin^2x+(1+n^2)\cos^2x}\\
&=&2\int_0^\frac{\pi}{2} \frac{\sec^2x\,dx}{\tan^2x+(\sqrt{1+n^2})^2}\\
&=&2\int_0^\infty \frac{du}{u^2+(\sqrt{1+n^2})^2}\quad\text{where }u=\tan x\\
&=&\frac{2}{\sqrt{n^2+1}}\left[\arctan\left(\frac{u}{\sqrt{n^2+1}}\right)\right]_0^\infty\\
&=&\frac{2}{\sqrt{n^2+1}}\cdot\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt{n^2+1}}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Linear transformation with respect to basis problem Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $T(\begin{bmatrix} 3 \\ 1 \end{bmatrix}) = \begin{bmatrix} 1 \\2 \end{bmatrix}$ and $T(\begin{bmatrix} -1 \\ 0 \end{bmatrix}) = \begin{bmatrix} 1 \\1 \end{bmatrix}$. Find the matrix $A$ representing $T$.
I understand that to approach this problem, I have to view $\begin{bmatrix} 3 \\1 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ as a basis, $B = \{ v_1, v_2 \}$, where $[v1 v2]$ is the transition matrix from $[x]B$ to $x$. How do I use $\begin{bmatrix} 1\\2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\1 \end{bmatrix}$? I'm unclear on their connection to the basis vectors $v_1$ and $v_2$.
| $\begin {bmatrix} 1\\0 \end{bmatrix} = - \begin {bmatrix} -1\\0 \end{bmatrix}$
$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = -T(\begin {bmatrix} -1\\0 \end{bmatrix}) = -\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} -1\\-1 \end{bmatrix}$
$\begin {bmatrix} 0\\1 \end{bmatrix} = \begin {bmatrix} 3\\1 \end{bmatrix} + 3\begin {bmatrix} -1\\0 \end{bmatrix}$
$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = T(\begin {bmatrix} 3\\1 \end{bmatrix}) + 3T(\begin {bmatrix} -1\\0 \end{bmatrix}) = \begin {bmatrix} 1\\2 \end{bmatrix} + 3\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} 4\\5 \end{bmatrix}$
$T = \begin {bmatrix} -1&4\\-1&5 \end{bmatrix}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
| Let $a+b+c=p$, then consider a function $f(a)=\dfrac{a}{(p-a)^2}$, then $f''(a)=\dfrac{2(a+p)}{(p-a)^4} >0$. So from the Jensen's in equality it follows that
$$\frac{f(a)+f(b)+f(c)}{3} \ge f\left[\frac{a+b+c}{3}\right]$$
So we get $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+c)^2} \ge 3*\frac{p/3}{(p-p/3)^2}=\frac{9}{4(a+b+c)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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} |
An old and interesting problem in combinatorics from Russia Mathematics Olympiad Can the numbers from $1$ to $81$ be written on a $9 \times 9$ board, so that the sum of the numbers in each $3\times 3$ square is the same?
I believe I have not made much progress and am missing the key insight here. Any advice?
| I believe if you take the following matrix:
$$\left[\begin{array}{ccccccccc}0&3&6&0&3&6&0&3&6\\1&4&7&1&4&7&1&4&7\\2&5&8&2&5&8&2&5&8\\3&6&0&3&6&0&3&6&0\\4&7&1&4&7&1&4&7&1\\5&8&2&5&8&2&5&8&2\\6&0&3&6&0&3&6&0&3\\7&1&4&7&1&4&7&1&4\\8&2&5&8&2&5&8&2&5\end{array}\right]$$
it already satisfies the conditions, and so does the transpose $A^T$:
$$A^T=\left[\begin{array}{ccccccccc}0&1&2&3&4&5&6&7&8\\etc.\end{array}\right]$$
and so does an affine combination $B=A+9A^T+1$, where "$1$" is the matrix filled with "all ones". One should verify that the resulting matrix $B$ has all the different values from $1$ to $81$. It begins something like this:
$$B=\left[\begin{array}{ccccccccc}1&13&25&28&40&52&55&67&79\\etc.\end{array}\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{Q}^{2\times2}, f=X^2+\alpha \in \mathbb{Q}[X].$
Consider $$f(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}+\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\ \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} + \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\\begin{pmatrix} a^2+bc + \alpha & ab+bd \\ ac+cd & bc+d^2 + \alpha \end{pmatrix}.$$
How do I get to the claim from here on?
Thanks in advance!
| Pick some $X$ that maps some vector $v$ to another (independent) vector $w$ and that $w$ maps to $-\alpha v$.
*
*What does that imply for $X^2$?
*How many different $X$ can you obtain this way? (Say, let $v=e_1$, then how many choices for $w$ are possible?)
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the positive sequence $a_{n+1} = \sqrt{1+\frac{a^2_n}{4}} $ is strictly increasing for $0 \leq a_0<\frac{2}{\sqrt{3}}$ My attempt:
$a_{n+1} - a_n= \sqrt{1+\frac{a^2_n}{4}} -a_n > \frac{a_n}{2}-a_n = -\frac{1}{2}a_n$, which doesn't tell me any thing. How do I prove that this sequence is strickly increasing? Thank you!
| $a_n^2=1+\frac{a_{n-1}^2}{4}=\frac{1}{4}\left(4+a_{n-1}^2\right)=\frac{1}{4^2}\left(4^2+1\cdot4+a_{n-2}^2\right)=\frac{1}{4^3}\left(4^3+4^2+4+a_{n-3}^2\right)$
Hence, a recursive formula can be written as follows: $$ a_{n+1}^2=\frac{1}{4^n}\left(4^n+4^{n-1}+...4+a_1\right)=\frac{4}{3}\cdot\frac{4^n-1}{4^n}+\frac{a_1^2}{4^n}$$.
$\therefore\lim\limits_{n\to\infty}a_n=\frac{2}{\sqrt3}$ irrespective of the choice of $a_1$.
Also $a_{n+1}^2-a_n^2=\frac{4-3a_n^2}{4}>0 \forall a_n<\frac{2}{\sqrt3}$.
Can you do the rest?
| {
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$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=b=c$ , $a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$ , hence $a^3 + b^3 + c^3 = 3abc$.
But I was wondering if $a^3 + b^3 + c^3 = 3abc$ can be true even when none of the above two relations are true. Please guide me toward a solution.
| Without loss of generality, suppose that $a\ge b$ and $a\ge c$. Then $$a^2+b^2+c^2-ab-bc-ca=(a-b)(a-c)+(b-c)^2\ge 0.$$
Equality can only occur if $a=b=c$ and so there are no other solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can any polynomials in the rational field be decomposed like this I' ve learned that the following examples can be used to decompose a
factor in this way:
x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) +
1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 -
3/16 (5 a^2 - 12 a) + 1/16 (12 a^3 - 5 a^4) +
1/8 (12 a^2 - 5 a^3) - (9 a)/4) x^2 + ((5 a^4)/16 + a^3/4 + (
5 a^2)/16 + 1/16 (12 a - 5 a^2) + 1/16 (12 a^3 - 5 a^4) + a/2 +
1/4 (12 - 5 a) - 3) x + a x^3 + a/4 + 1/4 (5 a - 12) + x^4 - 2)
Can any polynomials f(x) in the field of rational numbers be factorized into the form of (x - a) g (x, a)?
Besides a, other coefficients of g (x, a) should also be in the rational number field.
In the case of $x^5-5 x+12$, we can know the algebraic relations of his five roots (The letter a is a root of equation $x^5−5x+12$):
$x^5-5 x+12=(x-a) \left(x-\frac{1}{8} \left(-a^4-a^3-a^2-2 \sqrt{2}
\sqrt{3 a^3-2 a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8}
\left(-a^4-a^3-a^2+2 \sqrt{2} \sqrt{3 a^3-2
a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8} \left(a^4+a^3+a^2-2
\sqrt{2} \sqrt{a^4+a^2+6 a-8}-3 a-4\right)\right)
\left(x-\frac{1}{8} \left(a^4+a^3+a^2+2 \sqrt{2} \sqrt{a^4+a^2+6
a-8}-3 a-4\right)\right)$
I used the function in this link to find the relationship between a polynomial Galois group and a root set.
| Every (monic) polynomial $f(x)\in \Bbb{Q}[x]$ factorizes into $$f(x) = \prod_{j=1}^{\deg(f)} (x-\beta_j)=\prod_{j=1}^{\deg(f)} (x-g_j(\alpha)), \qquad g_j\in \Bbb{Q}[x]$$
$\beta_j$ are the roots, they are algebraic numbers, they generate $f$'s splitting field $$K=\Bbb{Q}(\beta_1,\ldots,\beta_{\deg(f)})$$
The primitive element theorem gives $$K=\Bbb{Q}(\alpha)=\Bbb{Q}[\alpha]$$
Thus each $\beta_j$ is a polynomial in $\alpha$.
The main problem is that in general $\alpha$ is not defined by some $n$-th roots, when it is we say that $f$ is radical. Every polynomial of degree $\le 4$ is radical. The unsovable quintic theorem shows that most polynomials of degree $\ge 5$ are not radical.
For non-radical polynomials, $\alpha$ (as well as the roots of $f$) is only defined by its minimal polynomial, we don't have any simpler expression for it.
For your quintic polynomial, you have exploited the fact that $[\Bbb{Q}(\beta_1,\beta_j):\Bbb{Q}(\beta_1)]\le 4$ to obtain that $\beta_j$ is radical in $\beta_1$, obtaining $$f(x) = \prod_{j=1}^{\deg(f)} (x-f_j(\beta_1))$$ where $f_j$ is an expression with a variable $t$, some rational numbers, additions, multiplications and $n$-th roots.
| {
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"url": "https://math.stackexchange.com/questions/3520447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this would eventually simplify into the original circle equation? I was trying to solve this problem:
The point $A$ has coordinates $(5, 16)$ and the point $B$ has coordinates $(-4,4)$. The variable $P$ has coordinate $(x,y)$ and moves on a path such that $ AP = 2BP$. Show that the Cartesian equation of the path of the $P$ is:
$$(x+7)^2 +y^2 = 100\tag{*}$$
So, what I did is to found a point on the circular path, and shows that the relationship holds. However, the actual answer is to let:
$$(x-5)^2 +(y-16)^2 = 4(x+4)^2 +4(y-4)^2$$
This also just means $AP=2BP$.
However, it will simplify to the original circle equation $(*)$. I literally don't know why. Even if $AP=2BP$ , why would this simplify to $(*)$. What's the mechanism behind this.
Thank you very much for you guys reply.
| I'll suppose you use the usual euclidian distance between two points. Then the distance between $A$ and $P$ is
$$d(A, P) =\sqrt{(x-5)^2+(y-16)^2}$$
Same goes for the distance between $B$ and $P$.
$$d(B, P)=\sqrt{(x-(-4))^2+(y-4)^2}$$
Since $d(A, P)=D(B, P)$, we have
$$\sqrt{(x-5)^2+(y-16)^2} =2\times\sqrt{(x+4)^2+(y-4)^2}$$
Squaring each side
$$(x-5)^2 +(y-16)^2 = 4\left((x+4)^2 +(y-4)^2\right)$$
Developping each squares
$$x^2-10x+25+y^2-32y+256 = 4x^2+32x+64+4y^2-32y+64$$
Bring all variables to the right, the rest to the left.
$$153=3x^2+42x+3y^2$$
Since the coefficients of $x^2$ and $y^2$ are the same, it is a circle. Divide every term by $3$.
$$51=x^2+14x+y^2$$
Complete the square for $x$
$$51+49=x^2+14x+49+y^2$$
$$100=(x+7)^2+y^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$I_n=\int_0^1{\frac{x^n}{x^n+1}}$. Prove $I_{n+1} \le I_n$ for any $n \in \mathbb N$ $$I_n=\int_0^1{\frac{x^n}{x^n+1}}$$
Prove $\lim_{n\to\infty}{I_n} = 0$
Here is what I tried.
First, I rewrite $I_n$.
$$I_n=\int_0^1{1-\frac{1}{x^n+1}}=1 - \int_0^1{\frac{1}{x^n+1}}$$
Now the limit becomes:
$$L=1-\lim_{n\to\infty}\int_0^1{\frac{1}{x^n+1}}$$
Next I try to solve the limit of the integral using Squeeze Theorem, with no success.
Using the fact $0\le x \le 1$ I get to the following double inequality:
$$\int_0^1\frac{1}{x^{n-1}+1} \le \int_0^1\frac{1}{x^{n}+1} \le 1 $$
$$I_{n-1} \le I_n \le 1$$
I don't know what to do next, I would greatly appreciate some help.
| $I_n
=\int_0^1{\frac{x^n}{x^n+1}}dx
=\int_0^1{\frac{x^n+1-1}{x^n+1}}dx
=1-\int_0^1{\frac{1}{x^n+1}}dx
=1-J_n$.
Need to show that
$J_{n+1} \ge J_n
$.
$\begin{array}\\
J_{n+1} - J_n
&=\int_0^1{\frac{1}{x^{n+1}+1}}dx-\int_0^1{\frac{1}{x^n+1}}dx\\
&=\int_0^1(\frac{1}{x^{n+1}+1}-\frac{1}{x^n+1})dx\\
&=\int_0^1\frac{x^n+1-(x^{n+1}+1)}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^{n+1}+1)(x^n+1)}dx\\
&\gt 0\\
\end{array}
$
More directly:
$\begin{array}\\
I_{n} - I_{n+1}
&=\int_0^1{\frac{x^n}{x^n+1}}dx-\int_0^1{\frac{x^{n+1}}{x^{n+1}+1}}dx\\
&=\int_0^1(\frac{x^n}{x^n+1}-\frac{x^{n+1}}{x^{n+1}+1})dx\\
&=\int_0^1\frac{x^n(x^{n+1}+1)-x^{n+1}(x^n+1)}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^{2n+1}+x^n-x^{2n+1}-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^n+1)(x^{n+1}+1)}dx\\
&\gt 0\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/3526292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Special Eigenvalues of a Matrix in Strang p.368 This question arises from Strang's Linear Algebra p.368.
It concerns the matrix
$$A = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 \\
\end{array}
\right)$$.
Some straightforward computation shows that
$$AA^T = \left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 \\
\end{array}
\right) \ \text{ and } \ (AA^T)^{-1} = \left(
\begin{array}{cccc}
2 & -1 & 0 & 0 \\
-1 & 2 & -1 & 0 \\
0 & -1 & 2 & -1 \\
0 & 0 & -1 & 1 \\
\end{array}
\right)$$
Now the textbook claims that all eigenvalues of $(AA^T)^{-1}$ have the form $2-2\cos(\theta)$.
Is it straightforward from the context?
| If we're looking for the fact that each eigenvalue can be expressed as $2 - 2 \cos \theta$ for some $\theta$, which is to say that the eigenvalues fall on the interval $[0,4]$, then there is indeed a straightforward way to confirm that this is the case.
It suffices to note that the eigenvalues are real because $(AA^T)^{-1}$ is symmetric, and that they must satisfy $|\lambda - 2| \leq 2$ either by the Gershgorin disk theorem or by considering either the $p=1$ or $p=\infty$ operator $p$-norm of $(AA^T)^{-1} - 2I$.
If we're looking for the exact expression for the eigenvalues, then no, the fact is not quite straightforward. However, we can see this relatively quickly if we correctly "guess" the eigenvectors of this matrix. In particular, every eigenvector has the form
$$
v = \left(\sin (\theta),
\sin \left( 2\theta\right),
\sin \left( 3\theta \right),
\sin \left( 4\theta\right)
\right)^T
$$
For $\theta = \frac{\pi k}{5}$ with $k = 1,\dots,4$. It now suffices to verify that for these $\theta$, we have
$$
(AA^T)^{-1}v = \left[2 - 2 \cos(\theta)\right] v.
$$
Let $\theta$ be arbitrary, let $v$ be the vector above, and let $y = (y_1,\dots,y_4)^T$ be the vector $(AA^T)^{-1}v$. We would like to find the $\theta$ for which $y$ is a multiple of $v$. We compute as follows:
$$
y_1 = 2 \sin \theta - \sin(2 \theta) = 2 \sin\theta - 2 \sin\theta \cos \theta = (2 - 2 \cos \theta)\sin \theta = (2 - 2 \cos \theta) v_1
$$
The computations for $y_2$ and $y_3$ are similar. Using the the sum to product identity, we have
$$
\begin{align}
y_2 &= 2 \sin(2 \theta) - (\sin \theta + \sin (3 \theta))
\\ & =
4 \sin \theta \cos \theta - 2\sin\left( \frac{3 \theta + \theta}{2}\right) \cos \left( \frac{3 \theta - \theta}{2}\right)
\\ & =
(2 - 2 \cos \theta)\sin(2 \theta) = (2 - 2\cos \theta) v_2
\end{align}
$$
In general, we have
$$
2 \sin(k \theta) - (\sin ((k-1)\theta) + \sin ((k+1) \theta)) =
(2 - 2 \cos \theta)\sin(k \theta).
$$
So, we can similarly say that $y_3 = (2 - 2\cos \theta) v_3$. If you prefer, we could also have handled $y_1$ with this trick, since
$$
2 \sin \theta - \sin(2 \theta) = 2 \sin (1 \cdot \theta) - (\sin(0 \cdot \theta) + \sin (2 \cdot \theta)).
$$
So far, we have not deduced any conditions on $\theta$. For the last entry, we reapply the above computational trick to find that
$$
\begin{align}
y_4 = 2 \sin(4 \theta) - \sin (3 \theta) &=
[2 \sin(4 \theta) - (\sin (3 \theta) + \sin(5\theta))] + \sin (5 \theta)
\\ & =
(2 - 2\cos \theta)\sin(4 \theta) + \sin(5 \theta)
= (2 - 2\cos \theta)v_4 + \sin(5 \theta).
\end{align}
$$
So, we see that $y$ will be a multiple of $v$ (which is to say that $v$ will be an eigenvector) if and only if $\sin(5\theta) = 0$, which leads us to the characterization above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\frac{ 1- \sin\frac{5\pi}{18}}{\sqrt{3} \sin \frac{5\pi}{18}}= \tan\frac{\pi}{18} $ It's easy to verify on WolframAlpha, but I have difficulty deriving it.
It's easy to see
$$
\tan\left(\frac{\pi}{18}\right)=\frac{\sqrt{3}-\tan(5/18 \pi)}{1+\sqrt{3}\tan(5/18 \pi)}
$$
| Evaluate,
$$\frac1{\sin\frac{5\pi}{18}}-\sqrt3 \tan\frac{\pi}{18}
=\frac{\cos\frac{\pi}{18}-\sqrt3 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}}{\sin\frac{5\pi}{18}\cos\frac{\pi}{18}}\tag 1$$
Examine the numerator,
$$\cos\frac{\pi}{18}-\frac{\sqrt3}2\cdot2 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}$$
$$= \cos\frac{\pi}{18}-\cos\frac\pi6(\cos\frac{2\pi}{9}-\cos\frac{\pi}{3})$$
$$= \cos\frac{\pi}{18}-\frac12(\cos\frac{\pi}{18} + \cos\frac{7\pi}{18})+\cos\frac{\pi}{6}\cos\frac{\pi}{3}$$
$$= \frac12(\cos\frac{\pi}{18} - \cos\frac{7\pi}{18})+\frac12\cos\frac{\pi}{6}$$
$$= \sin\frac{\pi}{6} \sin\frac{2\pi}{9}+\frac12\cos\frac{\pi}{6}$$
$$= \frac12( \sin\frac{2\pi}{9}+\sin\frac{\pi}{3})
=\sin\frac{5\pi}{18}\cos\frac{\pi}{18}$$
Substitute the result for the numerator into (1) to have
$$\frac1{\sin\frac{5\pi}{18}}-\sqrt3 \tan\frac{\pi}{18}= 1$$
Then, rearrange to obtain,
$$\frac{ 1- \sin\frac{5\pi}{18}}{\sqrt{3} \sin\frac{5\pi}{18}}= \tan\frac{\pi}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Factor pairs problem Is it possible to find a number $N$ such that it has two factor pairs {a,b}, {c,d} such that (a+1)(b+1)=(c+1)(d+1)
My intuition tells me this is impossible in the case where a,b,c,d are positive integers. However I’m curious if this is still true for negative $N$ too. For negative $N$, assume $a$ and $c$ are negative.
| Let $a, b, c, d, N \in \mathbb{Z}\setminus\{0\}$ such that $N = ab = cd$. Suppose that $(a+1)(b+1) = (c+1)(d+1)$.
Expanding $(a+1)(b+1) = (c+1)(d+1)$ we get that $a+b = c + d$.
Now substituting $b = \frac{N}{a}$ and $d = \frac{N}{c}$ we get $a+\frac{N}{a} = c + \frac{N}{c}$.
Rearranging we have $(a-c)(ac-N)=0$ and using that $N =ab$ we get $(a-c)(c-b) = 0$.
Hence the only solution is $\{a, b\} = \{c, d\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of
sin(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ...
arcsin(x) = x + $\frac{1}{2}\cdot\frac{x^3}{3}$ + ($\frac{1}{2}\cdot\frac{3}{4}$)$\cdot\frac{x^5}{5}$ + ...
However, I do not know how to prove this using there series...Could anyone have some ideas?
Thank you!
| Hint: A nicer representation of the arcsin series is
$$\arcsin(x) = \sum_{n = 0}^\infty \frac{1}{4^n} {2n \choose n} \cdot \frac{x^{2n + 1}}{2n + 1}$$
Since both $\sin(x)$ and $\arcsin(x)$ are both odd, we can look at the coefficients on all the odd-degree terms. Can you prove that for all $n \geq 1$,
$$\frac{1}{4^n} {2n \choose n} \cdot \frac{1}{2n + 1} \geq \frac{1}{(2n + 1)!} \iff \frac{1}{4^n} {2n \choose n} \geq \frac{1}{(2n)!}$$
This would prove all coefficients of terms with degree greater than $1$ in the expansion of $\arcsin(x) + \sin(x)$ are positive, and thus $\sin(x) + \arcsin(x) = 2x + C$, where $C$ is strictly positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Finding power series representation of the fucntion $f(x) = \frac{1+x^2}{1-x^2}$. So I want to find the power series representation of the function $f(x) = \frac{1+x^2}{1-x^2}. $ So how do I solve this?
I can start by reducing the function to the known form of geometric series but I have a summation at the numerator. How do I overcome that?
$\frac{1+x^2}{1-x^2} = \frac{(1+x)^2 -2x}{1-x^2} = \frac{1+x}{1-x} - \frac{2x }{1-x^2}$, now second part I can solve, it would be $2x \sum_{n =0}^{\infty} x^{2n}.$ How do I go about the first part? Please give some hint. Thank You.
| I would go straight using with $X=x^2$
$${1\over 1-X}=\sum_{n=0}^\infty X^n$$
This leads to
$${1+x^2\over 1-x^2}=\left(1+x^2\right)\sum_{n=0}^\infty x^{2n}$$
Now rearranging the sum
$${1+x^2\over 1-x^2}=1+2\sum_{n=1}^\infty x^{2n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'+\frac{4}{x}v=\frac{-10}{x^2}$$
this is now a first order linear ODE where:
$$y(x)=\frac{1}{u(x)}\int u(x)q(x)$$
$u(x)=e^{4\int\frac{1}{x}}=x^4$
$q(x) = \frac{-10}{x^2}$
$$\frac{1}{x^4}\int x^4 \frac{-10}{x^2}=\frac{1}{x^4}\frac{-10x^{2+1}}{2+1}+C$$
which leaves us with :
$$\frac{1}{y^2} = \frac{-10}{3x}+x^{-4}C$$
naturally
$$y^2= \frac{1}{\frac{-10}{3x}+x^{-4}C}$$
The book states the answer as being:
$$y^2= \frac{x}{2+Cx^5}$$
| You switched one sign too many in
$$
-\frac12v'+\frac2xv=\frac5{x^2}
$$
Then
$$
\left(\frac{v}{x^4}\right)'=\frac{v'}{x^4}-\frac{4v}{x^5}=-\frac{10}{x^6}
\implies
\frac{v}{x^4}=\frac2{x^5}+C
$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.
I can't use polar coordinates. I decided to used Lagrange multipliers.
Problem: after using Lagrange multipliers I got this system:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a
\end{array}
\right.
$$
And then I add one more equation to the system: $5=(x-4)^2+(y-2)^2$ and $45=(x-4)^2+(y-2)^2$, so I have to find the $a$ that satisfies both system of equations:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
5=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
and
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
45=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
The thing is that I can't resolve those systems, I already tried a lot of ways but I never get a result. I would really appreciate some help.
| The point $(4,2)$ lies on the line $y=\frac{1}{2}x$ which passes through the center of the circle.
Now the minimum and maximum is happened in the two intersection points of the line and the circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$ I have to solve the following integral:
$$\int\frac{1}{(x^4+1)^2}dx$$
I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex roots and got stuck. Is there a trick for this kind of integral or some kind of helpful substitution? Thanks.
EDIT:
I did the following:
Let $x^2=\tan\theta$, then $x = \sqrt{\tan\theta}$ and $dx=\frac{\sec^2\theta}{2x}d\theta$
Then:
$$I=\int\frac{1}{(x^4+1)^2}dx = \int\frac{1}{(\tan^2\theta+1)^2} \frac{\sec^2\theta}{2x}d\theta=\int\frac{1}{\sec^4\theta} \frac{\sec^2\theta}{2x}d\theta$$
$$I=\frac{1}{2}\int{\frac{1}{\sec^2\theta \sqrt{\tan\theta}}}d\theta$$.
After this I don't know how to proceed.
| We first decrease the power 2 using integration by parts.
$$\begin{aligned} \int \frac{d x}{\left(1+x^{4}\right)^{2}} &=-\frac{1}{4}\int \frac{1}{x^{3}} d\left(\frac{1}{1+x^{4}}\right) \\ &=-\frac{1}{4}\left[\frac{1}{x^{3}\left(1+x^{4}\right)}+3 \int \frac{d x}{x^{4}\left(1+x^{4}\right)}\right] \\ &=-\frac{1}{4}\left[\frac{1}{x^{3}\left(1+x^{4}\right)}-\frac{1}{x^{3}}-3\int \frac{d x}{1+x^{4}}\right] \\ &=\frac{x}{4\left(1+x^{4}\right)}+\frac{3}{4} \int \frac{d x}{1+x^{4}} \end{aligned}$$
By my post,
$$\int \frac { d x } { x ^ { 4 } + 1 } = \frac { 1 } { 4 \sqrt { 2 } } \left[ 2 \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 } x } \right) + \ln \left| \frac { x ^ { 2 } + \sqrt { 2 } x + 1 } { x ^ { 2 } - \sqrt { 2 } x + 1 } \right|\right] + C,$$
we get$$
\int \frac{d x}{\left(1+x^{4}\right)^{2}}=\frac{x}{4\left(1+x^{4}\right)}+\frac{3}{16 \sqrt{2}}\left[2 \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2} x}\right)+\ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|\right]+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $ (x+y)$ is even, then $(x−y)$ is even, for integers. Did I prove correctly? is this a direct proof? For $x+y$ to be even, either $x$ and $y$ are both even, or $x$ and $y$ are both odd.
If $x$ and $y$ are both even we obtain: $x=2k$ and $y=2j$.
substituting into $x-y$ we get $2k-2j$.
$2(k-j)$ is even, so we have proved the first case.
Now if $x$ and $y$ are both odd, we obtain: $x=2k+1$ and $y=2j+1$.
substituting into $x-y$ we get $(2k+1)-(2j+1)$,
$2k+1-2j-1= 2k-2j= 2(k-j)$ which is even, and we have proved the remaining case.
| Note that if the difference between $p^2$ and $q^2$ is a multiple of $4,$ then $p$ and $q$ must be of equal parity, for otherwise, we have wlog that $(2m+1)^2-(2n)^2=(2m+1-2n)(2m+1+2n)=(2M+1)(2N+1),$ which is never even.
Now since $(x+y)^2-(x-y)^2=4xy,$ it follows that $x+y$ and $x-y$ are of equal parity, and in particular if one of them is even then the other must be as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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