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Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks very much !
| For a start:
Let $z=xy+1$, then $$z(z+x+1)=a^2\implies \boxed{z^2+z(x+1)-a^2=0}$$
Since discriminant is perfect square we have $$d^2 =(x+1)^2+4a^2$$ which means that (Pythagorean triples) $$d=k\cdot (u^2+v^2),\;\; x+1 =k\cdot(u^2-v^2),\;\; a = k\cdot uv$$ for some positive integers $u,v,k$ and $\gcd(u,v)=1$. Since $x+1>0$ we have $u>v$. Then (solving boxed equation) $$xy+1 =k\cdot {v^2-u^2\pm (u^2+v^2)\over 2}$$
Since $z$ is positive we have only $+$ choise, so we have $xy+1 = kv^2$ so $y = {kv^2-1\over k(u^2-v^2)-1}$
??? Since for every integer $m$ we have $ \gcd(mk+1,k)=1$ thus $k(u^2-v^2)-1\mid k(2v^2+u^2) \implies k(u^2-v^2)-1\mid 2v^2+u^2$ and so
$$k(u^2-v^2)-1\leq 2v^2+u^2$$
So we have only finite possibilities for $k$ ...
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $f(x)=x$ is the only function satisfying certain properties
Let $\mathbb{N}$ be the set of positive integers. Prove that $f(x)=x$ is the only function from $\mathbb{N}\to\mathbb{N}$ for which $(x^2 + f(y)) | (f(x)^2 + y)$ for all $x,y \in \mathbb{N}.$
It's straightforward to show that $f(x)=x$ satisfies the required properties. Let $f$ be a function satisfying the problem conditions. Then $1+f(1) | (f(1)^2 + 1).$ Letting $a = f(1)$, we get a contradiction for $a > 1$ as $a^2 + 1\equiv 2\mod a+1\not\equiv 0\mod a+1$. Let $k\in\mathbb{N}.$ We know that $(1^2 + f(k)) | (f(1)^2 + k) = (k+1).$ So clearly $f(k) \leq k.$ $(k^2 + f(1)) | (f(k)^2 + 1).$ Thus if $f(k) < k,$ we get a contradiction. So $f(k) = k.$
Am I missing something above or is my proof correct?
| Your approach looks fine. Just for fun, here is another way to prove things.
Since $f:\mathbb{N}\to\mathbb{N}$, $x+f(x)$ is positive for all $x\ge1$, so
$$(x^2+f(y))\mid(f(x)^2+y)\implies x^2+f(y)\le f(x)^2+y\implies f(y)-y\le(f(x)-x)(f(x)+x)$$
Setting $x=y$, we find that $f(y)-y\le(f(y)-y)(f(y)+y)$, which tells us $f(y)-y\ge0$ (since $f(y)+y\gt0$ for $y\ge1$). It thus suffices to show that $f(1)=1$, since that will give us $f(y)=y$ for all $y$ from $0\le f(y)-y\le(f(1)-1)(f(1)+1)=0$.
To show $f(1)=1$, note that $(1^2+f(1))\mid(f(1)^2+1$ implies $f(1)^2+1=k(1+f(1))$ for some positive integer $k$, from which the quadratic formula tells us
$$f(1)={-k\pm\sqrt{k^2-4(1-k)}\over2}$$
But for this to be an integer, we must have $k^2-4(1-k)=(k+2)^2-8=m^2$ for some integer $m$. And the only two squares that differ by $8$ are $9$ and $1$, which tells us $k=1$, and thus $f(1)^2+1=1+f(1)$, which simplifies first to $f(1)^2=f(1)$ and then to $f(1)=1$ (since $f(1)\ge1$).
| {
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A surprising results with a interesting inequality let $x,y,z>0$ such $x+y+z=3$, Find the minimum of the value
$$F=\dfrac{1}{x}+\dfrac{1}{xy}+\dfrac{1}{xyz}$$
I use wolfapha found the surprising results $F_{min}=\dfrac{3+\sqrt{5}}{2}$
see links
maybe use AM-GM inequality,but I can't it,so How to prove $F\ge\dfrac{3+\sqrt{5}}{2}$
The question came from a middle school exam (14 age ago)
| If we allow for 1-variable calculus, then
*
*Fix $y+z = k$. Show that the minimum of $\frac{1}{y} + \frac{1}{yz} = \frac{z+1}{z(k-z) } $ subject to $ y> 0, z> 0$ has value
$\frac{1}{ (\sqrt{k+1} -1 ) ^2 }$ and occurs at $z = \sqrt{k+1} - 1$.
*So $ \frac{1}{x} + \frac{1}{xy} + \frac{1}{xyz} = \frac{1}{x} ( 1 + \frac{1}{y} + \frac{1}{yz} ) \geq \frac{1}{3-k} ( 1 + \text{ answer from above } ) $. We can verify that the minimum occurs at $ k = \frac{1 + \sqrt{5}}{2}$, and has value $\frac{ 3 + \sqrt{5} } { 2}$.
| {
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How to solve $3\sec(x)-2\cot(x)>0$ (trigonometric inequality)? I am struggling to find the solution, here is what I already tried (it's for a pre-calculus class so no calculus):
$3\sec(x)-2\cot(x)=0 \rightarrow 3(\frac{1}{\cos(x)})=2\left(\frac{\cos(x)}{\sin(x)}\right) \rightarrow 3\sin(x)=2\cos(x)^2 \rightarrow 3\sqrt{1-\cos(x)^2}=2\cos(x)^2 \rightarrow 9(1-\cos(x)^2)=4\cos(x)^4 \rightarrow 4\cos(x)^4+9\cos(x)^2-9=0$
Then I found the roots ($\pm \sqrt{3}/2$) and the respective $rad$ values for $x$ ($\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6},\frac{7\pi}{6})$, but the problem is that when I graph it I find that the functions are different, although they share the roots (but cosine one has more roots in between!) and some patterns (maximum/minimum values and the roots of the cosine function seem to have some relation to when the original one is positive or negative).
Did I do the transformations right? I though they were supposed to be equal at least on the roots, why are they related in such a weird way?
But anyway, I think my biggest problem right now is that I have no intuition for associating the roots to the original function so that I know when its positive or negative.
Edit: The solution set is for all $\mathbb{R}$.
Edit 2 : Thank you all. So I did it using $\cos(x)^2=1-\sin(x)^2$ and now the roots are right($\frac{\pi}{6},\frac{5\pi}{6}$). My only problem now is associating this with the original function. I don't know how to aproach it. Looking at the graph, the tangent and the secant parts of the function seem to be independent for some reason.
| \begin{align*}
3\sec x - 2\cot x & > 0\\
\frac{3}{\cos x} - \frac{2\cos x}{\sin x} & > 0\\
\frac{3\sin x - 2\cos^2x}{\sin x\cos x} & > 0\\
\frac{3\sin x - 2(1 - \sin^2x)}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 3\sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 4\sin x - \sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin x(\sin x + 2) - 1(\sin x + 2)}{\sin x\cos x} & > 0\\
\frac{(2\sin x - 1)(\sin x + 2)}{\sin x\cos x} & > 0
\end{align*}
The term $\sin x + 2 > 0$ for every real value of $x$. Thus, the sign of the expression depends on the terms $2\sin x - 1$ and $\sin x\cos x$.
In the interval $[0, 2\pi)$, the roots of the equation $2\sin x - 1 = 0$ are $\dfrac{\pi}{6}, \dfrac{5\pi}{6}$. By examining the sine curve, we see that in the interval $[0, 2\pi)$, $2\sin x - 1 > 0 \implies \dfrac{\pi}{6} < x < \dfrac{5\pi}{6}$.
The term $\sin x\cos x > 0$ when $\sin x$ and $\cos x$ have the same sign, which occurs in the first and third quadrants. The term $\sin x\cos x < 0$ when $\sin x$ and $\cos x$ have opposite signs, which occurs in the second and fourth quadrants.
We require that $2\sin x - 1 > 0$ and $\sin x\cos x > 0$ or that $2\sin x - 1 < 0$ and $\sin x\cos x < 0$.
In the interval $[0, 2\pi)$, $2\sin x - 1 > 0$ and $\sin x\cos x > 0$ if $\dfrac{\pi}{6} < x < \dfrac{\pi}{2}$.
In the interval $[0, 2\pi)$, $2\sin x - 1 < 0$ and $\sin x\cos x < 0$ if
$\dfrac{5\pi}{6} < x < \pi$ or $\dfrac{3\pi}{2} < x < 2\pi$.
Hence, the inequality is satisfied if
$$x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$$
or if $x$ is any angle coterminal with one of these angles.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the greatest value of $P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}$ The problem
Given that $x,y,z > 0$ and $xz-yz-yx=1$
Find the greatest value of
\begin{align} P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}\end{align}
The inequality is assymetric as you can see. I literally got nothing out of the condition $xy-yz-zx=1$. Probably the best thing I could thought of is that \begin{align}1+y^2=xz-yx-zy+y^2=(y-x)(y-z)\end{align}
But that won't be true anymore if I substitute that in for $1+x^2$ and/or $1+z^2$.
That's why I am completely stuck with this.
Any help is appreciated!
| You found $1+z^2=xy-yz−zx+z^2=(y-z)(x-z)$
Similarly for $1+x^2$ you get $(x-z)(x+y)$
And for $1+y^2$ you get $(y-z)(x+y)$
Therefore, you can replace:
\begin{align} P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}\end{align}
with
\begin{align} P = \frac{2x^2}{(x-z)(x+y)} -\frac{2y^2}{(y-z)(x+y)}+\frac{3z^2}{(y-z)(x-z)}\end{align}
With a bit of manipulation of the numerator you can have $$\frac{2x^2}{(x-z)(x+y)} \to \frac{2(x^2 + 1) - 2}{(x-z)(x+y)} \to \frac{2((x-z)(x+y)) - 2}{(x-z)(x+y)} \to 2 - \frac{2}{(x-z)(x+y)}$$
$$\frac{2y^2}{(y-z)(x+y)} \to \frac{2(y^2 + 1) - 2}{(y-z)(x+y)} \to \frac{2((y-z)(x+y)) - 2}{(y-z)(x+y)} \to 2 - \frac{2}{(y-z)(x+y)}$$
$$\frac{3z^2}{(y-z)(x-z)} \to \frac{3(z^2 + 1) - 3}{(y-z)(x-z)} \to \frac{3((y-z)(x-z)) - 3}{(y-z)(x-z)} \to 3 - \frac{3}{(y-z)(x-z)}$$
See if you can continue from here... this question is a little bit tedious and its asking you to reduce the forms to a numerical value. Note that every time you run into a stall, replace any numerical value with $(xy-yz−zx)$ until its reduced to a simple numeric or obtain a identity. For example,
$2 - \frac{2}{(y-z)(x+y)} \to 2(xy-yz−zx) - \frac{2(xy-yz−zx)}{(y-z)(x+y)}$
This way you can try to reduce your chances of running into a stall (or avoid stalling and hopefully deduce a value).
| {
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"timestamp": "2023-03-29T00:00:00",
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Verification for proof by strong induction of $a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a + 1)$ I'm trying to prove by strong induction that
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a + 1),$$
for $n \geq 1$.
By strong induction, I'd like to know if my solution is valid. What I did was:
*
*proof for $n = 1$
$$a - 1 = (a-1)(1)$$
*
*$n = k$
$$a^k - 1 = (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a + 1)$$
*proof for $n = k + 1$
We want to show that:
$$a^{k+1} - 1 = (a - 1)(a^k + a^{k-1} + a^{k-2} + a^{k-3}+···+a + 1)$$
as we know, $$a^{k+1} - 1 = a^{k+1} - a^k + a^k - 1$$
so we rewrite the statement and use that fact that for $n = k$ the property holds.
\begin{eqnarray*}
a^{k+1} - 1 &=& a^{k+1} - a^k +a^k - 1\\
&=& a^{k+1} - a^k + (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a + 1),
\end{eqnarray*}
now factorizing $a^k$ we get:
$$a^{k+1} - 1 = a^k (a-1) + (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a + 1)$$
so we have
$$a^{k+1} - 1 = (a - 1)(a^k + a^{k-1} + a^{k-2} + a^{k-3}+···+a + 1)$$
as desired.
| To prove the induction step, you should prove that
$$a^{n+1}-1=(a-1)(a^n+a^{n-1}+\ldots+a+1),$$
given that
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1).$$
Of course this follows easily from the fact that
$$(a^{n+1}-1)-(a^n-1)=a^{n+1}-a^n=(a-1)a^n.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$ Consider the following polynomial in $x$:
$$p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n.$$
I want to show that $p(x)\geq 0$ for real $x$. It is trivial to show that $p(-1) = 0$, and by graphing $p(x)$ for some $k$ we can see that $x=-1$ is in fact the global minimum of this polynomial, presumably for any $k\in \mathbb{N}^+$.
How can I show that this is in fact true?
Note that $$p(x)=f(x) - \frac{1}{2k+1}f(-x),$$ with $$f(x)=1 + x + \dots + x^{2k}=\frac{x^{2k+1} - 1}{x-1}.$$
| Changing your notation slightly, we have
$$
p_{2k}(x) = \frac{2}{2k+1}\left(b_0 + b_1 x + b_1 x^2 + \cdots + b_{2k}x^{2k}+ b_{2k+1} x^{2x+1}\right),
$$where $b_n = k$ if $n$ is even and $b_n=k+1$ if $n$ is odd. We will prove the claim by induction on $k$. The base case $n=0$ gives $p_0(x) =0$ and we also have $p_2(x) = 2/3 (x+1)^2$, which is obviously non-negative.
Suppose $p_{2k}(x) \ge 0$, with a minimum at $x=-1$. We have
$$
p_{2k+2}(x)- p_{2k}(x) = (x+1)^2\cdot \frac{2}{4k^2-1}\sum_{n=0}^{2k-2} (-1)^n\frac{(n+1)(n+2)}{2} x^n
$$Cancelling with the factor of $(x+1)^2$, the sum reduces to
$$
S_{2k}(x)=\frac{2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1}{ x + 1};$$ note that $S_{2k}(0)=1$ and $\lim_{x\to -1} S_{2k}(x) $ exists and is positive by LHR.
We claim the numerator has a unique root at $x=-1$. Indeed,
$$
2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1$$
$$ = (2k^2-k)x^{2k+1}+(4k^2-1)x^{2k}+(2k^2+k)x^{2k-1}+1,
$$
and its derivative is
$$
x^{2k-2}\bigg((2k^2-k)(2k+1)x^2 + (4k^2-1)(2k)x + (2k^2+k)(2k-1)\bigg)
$$By the quadratic formula, this has a unique double root at $x=-1$. This means the numerator of $S_{2k}$ changes sign at $x=-1$; in particular, the numerator is negative for $x<-1$ and positive for $x>-1$, implying $S_{2k}(x)>0$. In summary, $p_{2k}(x)$ is clearly zero at $x=-1$ and is increasing in $k$ for any fixed $x$, so the inequality holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}$ Simplify $$\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}$$ if $x=\dfrac{2mn}{n^2+1}$ and $m>0,n>1$.
The solution of the authors starts as follows:
$$\dfrac{\left(\sqrt{m+x}+\sqrt{m-x}\right)^2}{(\sqrt{m+x})^2-(\sqrt{m-x})^2}=\dfrac{m+x+2\sqrt{m^2-x^2}+m-x}{m+x-m+x}=\dfrac{m+\sqrt{m^2-x^2}}{x}=...$$
I don't get the idea behind this. What exactly have they done with the given expression? Thank you!
| \begin{align*}
\bigg(\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}\bigg)
\bigg(\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}+\sqrt{m-x}}\bigg)
\\
\\
=\dfrac{\big(\sqrt{m+x}+\sqrt{m-x}\big)^2}
{(m+x)-(m-x)}\\
\\
=\dfrac{2 \sqrt{m - x} \sqrt{m + x} + 2 m}
{2x}\\
\\
=\dfrac{\sqrt{m^2 - x^2} + m}
{x}
,\space x=\dfrac{2mn}{n^2+1},\space m>0,\space n>1
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to prove that $ \frac{n+1}{4n^2+3}$ is a Cauchy sequence? By definition, I need to show that for any $\epsilon \gt 0$ there exists $N \in \mathbb{N}$ such that for any $m,n \gt N$:
$ \lvert a_n - a_m \rvert \lt \epsilon$
So I write,
$ |\frac{n+1}{4n^2+3} - \frac{m+1}{4m^2+3} |\leq | \frac{n+1}{4n^2+3}| + |\frac{m+1}{4m^2+3}| = \frac{n+1}{4n^2+3} + \frac{m+1}{4m^2+3} \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
From here I try to show,
$\frac{n+1}{4n^2+3} \lt \frac{\epsilon}{2} \quad \quad \Rightarrow \quad \quad 4\epsilon n^2 -2n +3\epsilon -2 \gt 0$
$n \gt \frac{1}{4\epsilon} + \frac{1}{4\epsilon} \sqrt{1-12\epsilon^2+8\epsilon}$
Can someone please explain to me what should I do from here?
Am I in the right direction at all?
| $4n^2+3\geq 4n^2$. Therefore, $\frac{n+1}{4n^2+3} \leq \frac{2n}{4n^2}= \frac{1}{2n}$
| {
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Golden ratio in complex number squares
In the Argand diagram shown below the complex numbers
$– 1 + i, 1 + i, 1 – i, – 1 – i$ represent the vertices of a square ABCD.
The equation of its diagonal BD is $y = x$. The complex number $k + ki$
where $– 1 < k < 0$ represents the point E which is in the fourth quadrant and
lies on the line $y = x.$
EFGC is a square such that F lies on AB. The line GE meets the line CD
produced at H such that H is represented by the complex number $– 2 – i$.
(i) Show that $F=\frac{-k}{k+2}+i$
(ii) Hence show that $\frac{AF}{FB}=\varphi$
For (i), I haven't been able to arrive at the result. I've tried the following:
$M_{HG}=\frac{k+1}{k+2}$, therefore $M_{CF}=-\frac{k+2}{k+1}$
From the diagram, we know that $\Im{F}=1$. Let $f$ denote the real part. Therefore line $CF$ is given by $(y-1)=-\frac{k+2}{k+1}(x-f)$.
Similarly, line $EF$ is given by $(y-1)=-\frac{k-f}{1-k}(x-f)$
$\therefore -\frac{k-f}{1-k}=-\frac{k+2}{k+1}$, which gives $f=\frac{2k^2+2k-2}{k+1}$, which is clearly wrong. Could someone show me the correct way or at least give a hint as to the right lines to consider and if the path I'm going down is somewhat correct?
For (ii) I did get it through a rather convoluted way so if anyone has a quicker way that would be great to know! My way was:
$EF=F-E$
$=\frac{-k}{k+2}+i-k-ki$
$=\frac{-k}{k+2}-k+i(1-k)$
$EC=C-E$
$=1-i-k-ki$
$=1-k-i(1+k)$
Since $EFGC$ is a square, $EF=iEC$.
$\therefore \frac{-k}{k+2}-k+i(1-k)=i(1-k-i(1+k))$
$\frac{-k}{k+2}-k+i(1-k)=(1+k)+i(1-k)$
Equating the Real part gives $\frac{-k}{k+2}=(1+k)$, that is $k^2+3k+1=0$
$\therefore k=\frac{-3±\sqrt{5}}{2}$
Since $k>-1$, reject the negative solution
$AF=F-A$
$=\frac{-k}{k+2}+i-(-1+i)$
$=\frac{-k}{k+2}+1$
$FB=B-F$
$=1+i-\left(\frac{-k}{k+2}+i\right)$
$=1+\frac{k}{k+2}$
$\therefore \frac{AF}{FB}=\frac{\frac{-k}{k+2}+1}{1+\frac{k}{k+2}}$
$=\frac{1}{1+k}$
$=\frac{1}{1+\frac{-3+\sqrt{5}}{2}}$
$=\frac{1+\sqrt{5}}{2}$
$=\varphi$
| (i)
The slope of the line $EH$ is $\dfrac{k+1}{k+2}$. So, the equation of the line $CF$ is given by $y+1=-\dfrac{k+2}{k+1}(x-1)$. Since $F(f+i)$ is on this line, one has
$$1+1=-\dfrac{k+2}{k+1}(f-1)\implies f=\frac{-k}{k+2}$$
(ii)
Using the fact that $|EF|=|EC|$, one has
$$\begin{align}|EF|^2=|EC|^2&\implies (f-k)^2+(1-k)^2=(1-k)^2+(-1-k)^2
\\\\&\implies (f-k)^2=(k+1)^2
\\\\&\implies f-k=\pm (k+1)
\\\\&\implies f=2k+1,-1\end{align}$$
Since $f=\dfrac{-k}{k+2}=-1+\dfrac{2}{k+2}\gt -1$, one has
$$(f=)\ \ 2k+1=\frac{-k}{k+2}\implies k^2+3k+1=0\implies k=\frac{-3+\sqrt 5}{2}$$
from which you can show that $\dfrac{AF}{FB}=\varphi$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/4327328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the sum $\sum\limits_{k=0}^\infty (-2)^k\frac{k+2}{k+1}x^k$ $$\displaystyle\sum_{k=0}^\infty (-2)^k\dfrac{k+2}{k+1}x^k.$$
I showed that this series converges when $|x|<\dfrac{1}{2}$ because $$\displaystyle\lim_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|=2|x|.$$
Now I have to find the sum result. I tried so far trying to combine $$\ln(x+1)=\displaystyle\sum_{k=0}^\infty \dfrac{(-1)^kx^{k+1}}{k+1}$$ and $$\arctan(x)=\displaystyle\sum_{k=0}^\infty \dfrac{(-1)^kx^{2k+1}}{2k+1},$$ and trying many substitutions with multiples of $x$, $x^2$ and got close to the result, but $\dfrac{k+2}{k+1}$ confuses me.
Any help would be appreciated.
| So we want to find the value of
$$\sum_{k=0}^\infty \frac{(k+2)(-2x)^k}{k+1}$$
As you already determined, we know
$$\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^kx^{k+1}}{k+1}$$
If you multiply both sides by $x$ you get
$$x\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+2}}{k+1}$$
Now, the interesting step is to take the derivative of both sides.
$$\ln (x+1)+\frac{x}{x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-1)^kx^{k+1}$$
Now divide both sides by $x$ to get
$$\frac{1}{x}\ln (x+1)+\frac{1}{x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-x)^k$$
Substitute $2x$ into $x$ to get,
$$\ln (2x+1)+\frac{1}{2x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-2x)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Reflect a point across a line using affine transform I'm trying to reflect a point (6,12) across the line y=7 using an affine transformation. Logically, the line y=7 should be halfway the Y coordinate 12 and the Y coordinate of the answer, so (12+Y)/2=7, or 12+Y=14, or Y=2, so the answer is (6,2).
Wikipedia has examples of affine transformations at https://en.wikipedia.org/wiki/Affine_transformation#In_plane_geometry.
The approach I'm trying is 'translate Y -7', then 'reflect about X axis', then 'translate Y +7', leading me to believe I need these three transforms:
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & -7 \\
0 & 0 & 1 \\
\end{bmatrix}
.
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
.
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 7 \\
0 & 0 & 1 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & -14 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
But if I then multiply this with my point (6,12) I get the wrong answer:
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & -14 \\
0 & 0 & 1 \\
\end{bmatrix}
.
\begin{bmatrix}
6 \\
12 \\
1 \\
\end{bmatrix}
=
\begin{bmatrix}
6 \\
-26 \\
1 \\
\end{bmatrix}
$$
And if I switch around the two translation steps, the sign of the translation changes and I do get the correct answer:
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 14 \\
0 & 0 & 1 \\
\end{bmatrix}
.
\begin{bmatrix}
6 \\
12 \\
1 \\
\end{bmatrix}
=
\begin{bmatrix}
6 \\
2 \\
1 \\
\end{bmatrix}
$$
This gives the correct answer, but now I don't understand why this works, geometrically speaking! Shouldn't the positive Y transform now translate the point (2,12) to (2,19), then reflect that to (2,-19) and translate that to (2,-12)?
| To reflect about the plane $\mathbf{n} (\mathbf{r} - \mathbf{r_0} ) = 0 $, ($\mathbf{n}$ is a unit vector), the affine transformation is
$ T(\mathbf{p}) = \mathbf{r_0} + A (\mathbf{p} - \mathbf{r_0} ) $
where
$T = \mathbf{I} - 2 {\mathbf{n}\mathbf{n}}^T $
In this case, the plane is $y = 7$, so $\mathbf{n} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ and $\mathbf{r_0} = \begin{bmatrix} 0 \\ 7 \\ 0 \end{bmatrix} $
Matrix $A$ is
$A = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - 2 \begin{bmatrix} 0 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} = \begin{bmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && 1 \end{bmatrix}$
Using the formula for reflection,
$T\left(\begin{bmatrix} 6 \\ 12 \\ 0 \end{bmatrix} \right) = \begin{bmatrix} 0 \\ 7 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} 6 - 0\\ 12 - 7 \\ 0 - 0 \end{bmatrix} = \begin{bmatrix} 6 \\ 2\\ 0 \end{bmatrix} $
Note that I am using the $3D$ coordinates not homogeneous coordinates in $2D$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
:Find the area of the shaded region $ABC$ in the figure below For reference:Calculate the area of the shaded region; if: $AP = 4\sqrt2; PB = 4$ and $HN = 4$.(Answer: $\frac{32}{51}(5\sqrt2+4)$)
My progress
$HN(2r-HN)=BH^2\quad\Rightarrow{}\quad 4(2r-4)=4 \therefore R=\dfrac{5}{2}\\
S_{ABC} = \frac{a.b.c}{4R}=\frac{4.b.c}{10}\therefore \boxed{S_{ABC}= \frac{2.b.c}{5}}\\
S_{ABC}= \frac{S_{APC}}{2}\\
\frac{S_{ABC}}{S_{PAB}}=\frac{4b}{c.4\sqrt2}\implies b=\sqrt2c$
...???
By trigonometry
$S(ABC)=\frac{1}{2}S(APC)=\frac{1}{4}PA⋅PCsin(\overset{\LARGE{\frown}}{APC})=\frac{1}{4}PA⋅PC.sin(\overset{\LARGE{\frown}}{APO}+\overset{\LARGE{\frown}}{CPO})=\\
\frac{1}{4}PA⋅PC.(sin(\overset{\LARGE{\frown}}{CPO})cos(\overset{\LARGE{\frown}}{APO})+cos(\overset{\LARGE{\frown}}{CPO})sin(\overset{\LARGE{\frown}}{APO})$
...???
| Based on confirmation that $PA$ is tangent, by using power of point of $P$, $BC = 4$ and using similar triangles, $AC = AB \sqrt2$.
If $OH = x, OB = ON = 4 - x$, and applying Pythagoras in $\triangle OBH$
$x^2 + 2^2 = (4-x)^2 \implies x = OH = 3/2, OB = 5/2$
As $\angle BAC = \angle BOH = \theta $ (say)
$\cos \theta = \dfrac{3}{5}, \sin\theta = \dfrac{4}{5}$
Using law of cosine in $\triangle BAC$, $AB^2 + (AB \sqrt2)^2 - 4^2 = 2 AB \cdot AB \sqrt2 \cdot \dfrac{3}{5}$
$\implies AB^2 = \dfrac{80}{15 - 6 \sqrt2} = \dfrac{80}{51} (5 + 2 \sqrt2)$
$S_{\triangle ABC} = \dfrac 12 \cdot AB \cdot AB\sqrt2 \cdot \sin\theta $
$ = \dfrac{2\sqrt2}{5} AB^2 = \dfrac{32}{51}(5 \sqrt 2 + 4)$
If you want to avoid trigonometry, drop a perp from $C$ to $AB$ which meets $AB$ at $G$ then $\triangle ACG \sim \triangle OBH$
so, $ \displaystyle \frac{CG}{AC} = \frac{BH}{OB} \implies CG = \frac{4 \sqrt2}{5} AB$
say, $AB = y$. Then,
$ \displaystyle S_{\triangle ABC} = \frac{2 \sqrt2}{5} y^2 \tag1$
Next, equating area of $\triangle ABC$
$ \displaystyle \frac 12 AK \cdot 4 = \frac{2 \sqrt2}{5} y^2$
$ \displaystyle \implies AK = \frac{\sqrt2}{5} y^2, ~$ where $K$ is the foot of perp from $A$ to $BC$
$ \displaystyle BC = BK + CK = \sqrt{y^2 - AK^2} + \sqrt{2y^2 - AK^2} = 4$
$ \displaystyle \sqrt{y^2 - \frac{2y^4}{25}} + \sqrt{2y^2 - \frac{2y^4}{25}} = 4$
Substitute $z = y^2$ and solve for $z$.
You get two solutions and one of them can be discarded based on given dimensions.
Finally we have, $ \displaystyle z = y^2 = \frac{400+ 160 \sqrt2}{51}$
Now $(1)$ gives you the desired area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Investigate the convergence or divergence of the sequence $a_1 = \sqrt{c}, a_2 = \sqrt{c + a_1}, a_{n+1} = \sqrt{c + a_n}$ $$a_1 = \sqrt{c}, a_2 = \sqrt{c + a_1}, a_{n+1} = \sqrt{c + a_n}, c > 0$$
I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How can I do this?
| Claim: $a_n$ is increasing and bounded above by $M = \dfrac{1+\sqrt{1+4c}}{2}$. Also $\displaystyle \lim_{n\to \infty} a_n = M$
Proof: By induction on $n \ge 1$. $a_2 = \sqrt{c+\sqrt{c}}> \sqrt{c} = a_1$. Assume $a_n > a_{n-1}$, we show $a_{n+1} > a_n$. But $a_{n+1} = \sqrt{c+a_n}> \sqrt{c+a_{n-1}}= a_n$ by the inductive step. So by induction $a_{n+1} > a_n, \forall n \ge 1$. We show that $a_n < M, \forall n \ge 1$ by induction also. For $n = 1, a_1 = \sqrt{c} < M \iff \sqrt{c} < \dfrac{1+\sqrt{1+4c}}{2}\iff 2\sqrt{c} - 1 < \sqrt{1+4c}$. If $c < \dfrac{1}{4}$, we're done since $LHS < 0 < RHS$. Otherwise, $c > \dfrac{1}{4}\implies 2\sqrt{c} - 1 > 0$, and squaring both sides: $4c - 4\sqrt{c} + 1 < 1+ 4c \iff -4\sqrt{c} < 0$ is clearly true. Thus $a_1 < M$. Assume $a_n < M$, you have $a_{n+1} = \sqrt{c+a_n} < M \iff a_n < M^2 - c = \dfrac{1+2\sqrt{1+4c}+1+4c}{4} -c = \dfrac{1+\sqrt{1+4c}}{2} = M$, which is true by inductive step. So $a_n < M, \forall n \ge 1$.
Thus the claim is verified. This follows that the limit exists and call it $M$. Then $M = \sqrt{c+M}\implies M^2 = c+M\implies M^2-M -c = 0\implies M = \dfrac{1 \pm \sqrt{1+4c}}{2}$. Since $a_n > 0,\forall n \ge 1 \implies M \ge 0\implies M = \dfrac{1+\sqrt{1+4c}}{2}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When does this equality hold: $(A + B)^2 = A^2 + B^2$? Where $A$ and $B$ are matrices. Given
$A =
\begin{bmatrix}
2 & -1 \\
2 & -1
\end{bmatrix}
$
and
$
B =
\begin{bmatrix}
a & 1\\
b & -1
\end{bmatrix}
$
, what are the values of $a$ and $b$ if $(A+B)^2 = A^2 + B^2$?
I think if the equality holds then $AB + BA = 0$.
$
AB =
\begin{bmatrix}
2a-b & 3\\
2a-b & 3
\end{bmatrix}
$
and
$
BA =
\begin{bmatrix}
2a + 2 & -a-1\\
2b-2 & -b + 1
\end{bmatrix}
$
$\therefore
AB + BA =
\begin{bmatrix}
4a-b+2 & 2 - a\\
2a+b-2 & 4-b\\
\end{bmatrix}
$
I think there are no values of $a$ and $b$ for which $AB+BA = 0$.
Is this correct, or are there any mistakes in my solution?
| Yes. There are no such values.
For $AB+BA=0$, we must have $$a=2,b=4$$ and $$4a-b+2=0$$ that is simply not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $ Solve $$ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $$
using shift $x=e^{z}$ and differential operator $Dz=\frac{d}{dz}$
What does $Dz = d / dz$ mean?
I did this but I don't know how to continue. Please help.
$$ (e^{z})^{3}{y}'''+e^{z}{y}'-y = e^{z}\ln(e^{z}) \\\\(e^{3z}){y}'''+e^{z}{y}'-y = e^{z}{z}$$
And I tried $\,\,y=z^r$
$$e^{3}r(r^{2}-r-2)z^{r-3}+e^{z}rz^{r-1}-z^{r}=0$$
| Let $Y(z) = y(x) = y(e^z)$. Then Chain Rule and Product Rule give
\begin{align*}
\frac{dY}{dz} & = \frac{dy}{dx}\frac{dx}{dz} = \frac{dy}{dx}e^z = x\frac{dy}{dx} \\
\frac{d^2Y}{dz^2} & = \frac{d}{dz}\left(\frac{dy}{dx}e^z\right) = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^ze^z = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^{2z} = \frac{dY}{dz} + x^2\frac{d^2y}{dx^2} \\
\frac{d^3Y}{dz^3} & = \frac{d}{dz}\left(\frac{dY}{dz} + \frac{d^2y}{dx^2}e^{2z}\right) = \frac{d^2Y}{dz^2} + 2\frac{d^2y}{dx^2}e^{2z} + \frac{d^3y}{dx^3}e^{3z} \\
& = \frac{d^2Y}{dz^2} + 2x^2\frac{d^2y}{dx^2} + x^3\frac{d^3y}{dx^3} = \frac{d^2Y}{dz^2} + 2\left(\frac{d^2Y}{dz^2} - \frac{dY}{dz}\right) + x^3\frac{d^3y}{dx^3} \\
& = 3\frac{d^2Y}{dz^2} - 2\frac{dY}{dz} + x^3\frac{d^3y}{dx^3}.
\end{align*}
Thus, the original DE transforms to
\begin{align*}
x^3y''' + xy' - y & = x\ln x \\
Y''' - 3Y'' + 2Y' + Y' - Y & = e^zz \\
Y''' - 3Y'' + 3Y' - Y & = ze^z.
\end{align*}
**Thanks to bjorn93 for pointing out my mistake!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluating $\int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}.$ I want to evaluate the following integral ($\theta_0>0$)
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta
\end{equation*}
So I thought about $d(\cos\theta)=-\sin\theta d\theta$, then
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta = - \int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}
\end{equation*}
My question is about that step. Is that legit? I mean, that integral is doubly improper at both limits. How can I ensure that it is integrable?
| We first restrict the angle $\theta$ in the range, $0<\theta_0<\theta<\pi, $
then rewrite the integral by double-angle formula
$$1-\cos \theta=2\sin^2 \frac{\theta}{2} \text{ and } 1+\cos \theta=2\cos^2 \frac{\theta}{2} ,$$
\begin{aligned}
& I=\int_{\theta_{0}}^{\pi} \sqrt{\frac{1-\cos \theta}{\cos \theta_{0}-\cos \theta}} d \theta = \int_{\theta_{0}}^{\pi} \frac{\sqrt{2 \sin ^{2} \frac{\theta}{2}}}{\sqrt{\cos \theta_{0}-\cos \theta}} d \theta =\sqrt{2} \int_{\theta_{0}}^{\pi} \frac{\sin \frac{\theta}{2} d \theta}{\sqrt{\cos \theta_{0}-\left(2 \cos^2 \frac{\theta}{2}-1\right)}}
\end{aligned}
Putting $y=\cos \dfrac{\theta}{2}$ yields
\begin{aligned}
I &=-2 \int_{\cos \frac{\theta_{0}}{2}}^{0} \frac{d y}{\sqrt{\cos ^{2} \frac{\theta_{0}}{2}-y^{2}}} \\
&=2\left[\sin ^{-1}\left(\frac{y}{\cos \frac{\theta_{0}}{2}}\right)\right]_{0}^{\cos \frac{\theta_{0}}{2}} \\
&=2 \sin ^{-1}(1) \\
&=\pi,
\end{aligned}
which is surprisingly independent of $\theta_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Powers of a Toeplitz matrix I'm searching a closed formula to compute the powers of the following matrix
\begin{equation*}F\triangleq \begin{bmatrix}
1 & T & \frac{T^2}{2}\\
0 & 1 & T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
where $T$ is a given parameter. For example, if I'm not wrong the first 5+1 powers are
\begin{equation*}F^0\triangleq \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
\begin{equation*}F^1\triangleq \begin{bmatrix}
1 & T & \frac{T^2}{2}\\
0 & 1 & T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
\begin{equation*}F^2\triangleq \begin{bmatrix}
1 & 2T & 2T^2\\
0 & 1 & 2T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
\begin{equation*}F^3\triangleq \begin{bmatrix}
1 & 3T & \frac{9T^2}{2}\\
0 & 1 & 3T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
\begin{equation*}F^4\triangleq \begin{bmatrix}
1 & 4T & 8T^2\\
0 & 1 & 4T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
\begin{equation*}F^5\triangleq \begin{bmatrix}
1 & 5T & \frac{25T^2}{2}\\
0 & 1 & 5T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
From these results it seems to me that for any $k$ holds
\begin{equation*}F^k=\begin{bmatrix}
1 & kT & \frac{(kT)^2}{2} \\
0 & 1 & kT \\
0 & 0 & 1
\end{bmatrix}\end{equation*}
is it true? If yes, how can be proved my formula?
| Your matrix can be written as an exponential of a nilpotent matrix:
$$F=e^{TN},\qquad N=\left(\begin{array}{ccc} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right).$$
Note that $N^2=\left(\begin{array}{ccc} 0 & 0 & 1 \\
0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ and $N^3=0$. Now,
$$F^k=e^{kT N}=\mathbf 1+kTN+\frac{k^2T^2}{2}N^2,$$
which gives the result.
P.S. If you are uncomfortable with matrix exponentials, then simply use induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion in 0 of a series I need to find a 2 term asymptotic expansion of the serie of functions $f=\sum u_n$
where for all real $x$, $u_n(x) = 1/(n^2+x^2)$. Obviously, the function being continuous (which is easy to prove by uniform convergence over $\mathbb R$) the limit is $\pi^2/6$.
I have no idea on what to do next. I have tried to get an inequality by using serie integral comparison: if we take a real $x$, then define the function $f : t\in \mathbb R \rightarrow 1/(t^2+x^2)$, which is positive and decreasing over $\mathbb R_+$, we can get:
(skipping the integration part)
$$\arctan(x)/x -1 +1/(1+x^2)-1 \le f(x) - \pi^2/6 \le \arctan(x)/x -1$$
which gives no useful information. Any help would be appreciated.
| According to Find the sum of $\sum \frac{1}{k^2 - a^2}$ when $0<a<1$, the following identity holds
$$\sum_{n=1}^\infty \frac1{n^2-z^2}=\frac1{2z^2}-\frac{\pi\cot\,\pi z}{2z}.$$
After letting $z=ix$, we find
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\frac{\pi\coth(\pi x)}{2x}-\frac1{2x^2}.$$
Finally, after expanding the Hyperbolic Cotangent at $x=0$, we get
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\frac{\pi^2}{6}-\frac{\pi^4}{90}x^2 +O(x^4).$$
Notice that $\frac{\pi^4}{90}=\sum_{n=1}^{\infty}\frac{1}{n^4}$:
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\sum_{n=1}^\infty \frac{1}{n^2}\frac1{1+(x/n)^2}=\sum_{n=1}^\infty \frac{1}{n^2}\left(1-\frac{x^2}{n^2}+O(x^4)\right)\\=\sum_{n=1}^\infty\frac1{n^2}-\sum_{n=1}^\infty\frac1{n^4}\,x^2+O(x^4).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do next? I can't come up with anything else...
| One way to go is to use $a^3-b^3=(a-b)(a^2+ab+b^2)$.
So with $a=1$ and $b=\sqrt[3]{t}$:
$$\frac{1-t}{1-\sqrt[3]{t}} = \frac{1-(\sqrt[3]{t})^3}{1-\sqrt[3]{t}}=1^2+1.\sqrt[3]{t}+(\sqrt[3]{t})^2$$
And with $a=1$ and $b=-\sqrt[3]{t}$:
$$\frac{1+t}{1+\sqrt[3]{t}} = \frac{1-(-\sqrt[3]{t})^3}{1+\sqrt[3]{t}}=1^2-1.\sqrt[3]{t}+(-\sqrt[3]{t})^2$$
So $$A(t)=2 +2(\sqrt[3]{t})^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$ Given the real $ a $. Prove that
$$a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$$
I tried to factor it as
$$(a-1)\Bigl(a^{13}+a^5(a^2+a+1)+a\Bigr)+1$$
I think it should be written as a sum of squares.
Any idea will be appreciated.
| Notice it a bunch of $+ a^{even}$ and $-a^{odd}$ in descending order of powers so.....
Case 1: $a \le 0$ then $a^{even} \ge 0$ and $-a^{odd} \ge 0$ so $a^{14} - a^{13}+a^8-a^5+a^2 -a + 1=|a|^{14}+|a|^{13}+ |a|^8 +|a|^5 + |a|^2+|a| + 1\ge 1$.
Case 2: $a\ge 1$ then $a^{big} \ge a^{small}$ so $a^{big}-a^{small} >\ge 0$ and $(a^{14}-a^{13}) + (a^8-a^5) + (a^2 -a) + 1 \ge 1$.
Case 3: $0 < a < 1$ then $a^{big} < a^{small}$ so $-a^{big}+a^{small} > 0$ and $a^{14} + (-a^{13}+a^8) + (-a^5+a^2) + (-a + 1) > a^{14} > 0$.
And there aren't any more options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number theoretic problems leading to elliptic curves I am looking for basic number theoretic problems that can be addressed by the use of elliptic curves.
I know few of problems such as congruent number problem, or the problem of extending a Diophantine triple $\{a,b,c\}$ to a quadruple $\{a,b,c,x\} $satisfying $ax+1 = \square, ~bx+1 = \square, ~cx+1 = \square,$ where a Diophantine triple is set consisting three integers with the property that product of any two increased by one is always a perfect square e.g. $\{1,3,8\}$ and $\square$ denotes perfect square. This problem can be addressed by the elliptic curve $E: y^2 = (ax+1)(bx+1)(cx+1).$ I am looking for some more problems that can have relations with elliptic curve. What are those problems and what can be a source to understand them?
| Um, all of them? (That's an exaggeration, but only slightly.)
For example consider the $n=3$ case of Fermat's Last Theorem. You're looking for nontrivial integer solutions to $a^3 + b^3 = c^3$. Dividing by $c^3$ and setting $x = a/c$ and $y = b/c$, we get the new equation $x^3 + y^3 = 1$, where now we are looking for rational solutions instead of integer solutions (since if $a,b,c$ are integers then $x,y$ are rational numbers, and vice-versa). If you know enough general theory, you may recognize that $x^3+y^3 = 1$ is an elliptic curve. Explicitly, if we set $C : x^3 + y^3 = 1$ and $E : Y^2 - 9Y = X^3 - 27$, then the transformation
\begin{align*}
X &= \frac{3 y}{1-x} \\
Y &= \frac{9 x}{x-1}
\end{align*}
maps $C$ to $E$, and the inverse transformation is
\begin{align*}
x &= \frac{Y}{Y - 9} \\
y &= \frac{3 X}{9 - Y}
\end{align*}
mapping $E$ to $C$. Hence by finding rational points on the elliptic curve $E$, we can find rational points on $C$, and vice-versa. To find rational points on $E$, we can use the general theory of elliptic curves. For example, as told by Sagemath:
sage: E = EllipticCurve([0,0,-9,0,-27])
sage: E.rank()
0
sage: E.torsion_points()
[(0 : 1 : 0), (3 : 0 : 1), (3 : 9 : 1)]
This computation indicates that $\{\mathcal{O}, (3,0), (3,9)\}$ are the only rational points on $E$. Translating back to $C$ using the above equations, we find that $(x,y) = (1,0)$ and $(x,y) = (0,1)$ are the only rational points on $C$. So that's how you prove one of the cases of Fermat's Last Theorem using the general theory of elliptic curves.
A similar trick works with the $n=4$ case. If $a^4 + b^4 = c^2$, then divide by $b^4$ and substitute $x = a/b$ and $y = c/b^2$ to obtain $C : x^4 + 1 = y^2$. Then $C$ is an elliptic curve, birational to $E : Y^2 = X^3 - 4X$, with equations
\begin{align*}
X &= \frac{2x(y+1)}{x^3} \\
Y &= \frac{4(y+1)}{x^3} \\
x &= \frac{2X}{Y} \\
y &= \frac{2X^3-Y^2}{Y^2}
\end{align*}
mapping $C$ to $E$ and vice-versa. The only rational points on $E$ are $\{\mathcal{O}, (0,0), (\pm 2,0)\}$, from which one can deduce that $a^4+b^4=c^2$ has no nontrivial integer solutions.
Unfortunately, the elliptic curve train ends here. The $n=5$ and higher cases of Fermat's Last Theorem are not amenable to elementary elliptic curve manipulations (not counting Wiles' proof here, which of course is based on elliptic curves, but uses very advanced theory).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How is the summation being expanded? I am trying to understand summations by solving some example problems, but I could not understand how is the second to last line being expanded? I would really appreciate if you could explain me how is it being expanded.
\begin{align}
&\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=1}^{j}1 =\\
&\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}j =\\
&\sum_{i=1}^{n-1}\left(\sum_{j=1}^{n}j - \sum_{j=1}^{i}j\right) =\\
&\sum_{i=1}^{n-1}\left(\frac{n(n+1)}{2} - \frac{i(i+1)}{2}\right) =\\
&\frac{1}{2}\sum_{i=1}^{n-1}n^2+n-i^2-i =\\
&\frac{1}{2}\left((n-1)n^2 + (n-1)n - \left(\frac{n(n+1)(2n+1)}{6} - n^2\right) -
\left(\frac{n(n+1)}{2} - n\right)\right) =\\
&f(n) = \frac{n(n(n+1))}{2} - \frac{n(n+1)(2n+1)}{12} - \frac{n(n+1)}{4}
\end{align}
| I take it that what has to be explained is this (I've introduced parentheses on the left hand side for clarity):
\begin{multline*}
\sum_{i=1}^{n-1}\left(n^2 + n - i^2 - i\right) = \\
(n-1)n^2 + (n-1)n - \left(\frac{n(n+1)(2n+1)}6 - n^2\right) - \left(\frac{n(n+1)}2 - n\right).
\end{multline*}
This equation results from adding together the following four identities:
\begin{align*}
\sum_{i=1}^{n-1}n^2 & = (n - 1)n^2, \\
\sum_{i=1}^{n-1}n & = (n - 1)n, \\
\sum_{i=1}^{n-1}i^2 & = \sum_{i=1}^ni^2 - n^2 \\ & = \frac{n(n+1)(2n+1)}6 - n^2, \\
\sum_{i=1}^{n-1}i & = \sum_{i=1}^ni - n \\ & = \frac{n(n+1)}2 - n.
\end{align*}
Lines 4 and 6 follow, of course, from the familiar identities:
\begin{align*}
\sum_{i=1}^ni^2 & = \frac{n(n+1)(2n+1)}6, \\
\sum_{i=1}^ni & = \frac{n(n+1)}2.
\end{align*}
I don't know why it was done this way! It seems to me that it would have been simpler just to write:
\begin{align*}
\sum_{i=1}^{n-1}i^2 & = \frac{(n-1)n(2n-1)}6, \\
\sum_{i=1}^{n-1}i & = \frac{(n-1)n}2.
\end{align*}
(Also, in the comments, I've suggested two ways to arrive at the final answer with less calculation.)
| {
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"url": "https://math.stackexchange.com/questions/4360109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$? What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$.
Note that $f'(x)=\exp(\frac{1}{2}c^2x^2)c^2x$ and $f''(x)=\exp(\frac{1}{2}c^2x^2)(c^2x)^2+\exp(\frac{1}{2}c^2x^2)c^2$. Then $f'(0)=0$ and $f''(0)=c^2$.
Is that
$$
f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+o(x^2)=1+\frac{1}{2}c^2(\frac{1}{2}c^2x^2)^2?
$$
| Just use Taylor expansion as if it were $e^X$ where $X = \frac{1}{2}c^2x^2$ hence:
$$e^X = \sum_{k = 0}^{+\infty} \frac{X^k}{k!}$$
$$e^{\frac{1}{2}x^2x^2} = \sum_{k = 0}^{+\infty} \frac{\left(\frac{1}{2}c^2x^2\right)^k}{k!} =\sum_{k = 0}^{+\infty}\frac{1}{2^k k!} (cx)^{2k}$$
The first terms are given by
$$1+\frac{c^2 x^2}{2}+\frac{c^4 x^4}{8}+O\left(x^5\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger . I have been asked that which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger without resorting to calculator.
My Attempt:
I tried taking $f(x)$ to be
$f(x)=\sin 4x-\tan 3x$
$f'(x)=4\cos 4x-3\sec^23x=\cos 4x(4-3\sec^23x\sec 4x)$
but to no avail.
I also tried solving $\tan^2 21^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}-\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}$
but again no luck.
There doesn't appear to be a general way of doing this
| Working with radians, you want to compare
$$\sin \left(\frac{\pi }{6}-\frac{\pi}{90}\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8}-\frac{\pi }{120}\right)$$ Let $\epsilon=\frac \pi {120}$ and we shall compare
$$\sin \left(\frac{\pi }{6}-\frac 4 3\epsilon\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8}-\epsilon\right)$$ Using series around $\epsilon=0$
$$\sin \left(\frac{\pi }{6}-\frac 4 3\epsilon\right)=\frac{1}{2}-\frac{2 }{\sqrt{3}}\epsilon +O\left(\epsilon ^2\right)$$
$$\tan \left(\frac{\pi }{8}-\epsilon\right)=\sqrt 2-1+2 \left(\sqrt{2}-2\right) \epsilon +O\left(\epsilon ^2\right)$$
$$\sin \left(\frac{\pi }{6}-\frac{\pi}{90}\right)- \tan \left(\frac{\pi }{8}-\frac{\pi }{120}\right)=\left(\frac{3}{2}-\sqrt{2}\right)+\left(4-2 \sqrt{2}-\frac{2}{\sqrt{3}}\right)
\frac \pi {120} +\cdots$$ which is $0.0862$ as an approximation. The true value is $0.0856$.
No calculator used until the last line (I am too lazy and used my phone).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Proving that a given mapping is a inner product How to prove that a mapping $\langle\cdot,\cdot\rangle:\mathbb{R_{3}[x]}\times\mathbb{R_{3}[x]}\rightarrow\mathbb{R}$ is a inner product, where $\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx?$
First, for $a,b\in\mathbb{R}$ and $f,g,h\in\mathbb{R_{3}[x]}$, we have
$\langle af+bg,h\rangle=\int\limits_{-1}^{1}(af+bg)(x)h(x)dx=\int\limits_{-1}^{1}(af(x)+bg(x))h(x)dx$
$\qquad\qquad\quad\ =a\int\limits_{-1}^{1}f(x)h(x)dx+b\int\limits_{-1}^{1}g(x)h(x)dx=a\langle f,h\rangle+b\langle g,h\rangle.$
Second, for $f,g\in\mathbb{R_{3}[x]}$, we have
$\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx=\int\limits_{-1}^{1}g(x)f(x)dx=\langle g,f\rangle$.
And, third, for $f\in\mathbb{R}_3[x],$ $f\neq0,$ we have to prove $\langle f,f\rangle >0.$
I wrote $f=ax^3+bx^2+cx+d$ and I get $\langle f,f\rangle=\dfrac{2}{7}a^2+\dfrac{4}{5}ac+\dfrac{2}{3}c^2+\dfrac{2}{5}b^2+\dfrac{4}{3}bd+2d^2.$
I don't know how to prove that this is greater than 0. I tried to make a square of a binomial, but it isn't working.
| The answers given already are the most elegant, but for completeness let me give another way (continuing the approach in the post). Let $f = ax^3 +b x^2+cx +d.$ Then
\begin{align*}
\langle f, f \rangle &= \frac{2}{7} a^2 + \frac{4}{5} ac + \frac{2}{3} c^2 + \frac{2}{5} b^2 + \frac{4}{3} bd + 2d^2 \\
&= \frac{2}{3} \left( c + \frac{3a}{5}\right)^2 + \frac{8}{175} a^2 + 2 \left(d+\frac{b}{3}\right)^2 + \frac{8}{45} b^2,
\end{align*}
which is greater than zero (and equal to zero iff $f=0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a decomposition of a $2\times 2$ diagonal matrix into the sum of two diagonal matrix which satisfy some conditions on the coefficients Let $M=\begin{pmatrix} \alpha&0\\
0&\beta
\end{pmatrix} $ be an arbitrary $2\times 2$ real constant diagonal matrix with $\alpha\neq 0$ and $\beta\neq 0$. I want to find a decomposition of $M$ into the sum of two diagonal matrix $P$ and $N$, such that:
$$
P=\begin{pmatrix} a&0\\
0&b
\end{pmatrix},\,N=\begin{pmatrix} c&0\\
0&d
\end{pmatrix},a>0,b>0,c<0,d<0\mbox{ and }ad=bc.
$$
I know that, without the condition $ad=bc$, such a decomposition exists (take the example
$\begin{pmatrix} 1&0\\
0&-1
\end{pmatrix}=\begin{pmatrix} 2&0\\
0&1
\end{pmatrix}+\begin{pmatrix} -1&0\\
0&-2
\end{pmatrix}$).
I need that decomposition in ordre to guarantee that the two operators $P (\nabla \nabla\cdot)N$ and $N (\nabla \nabla\cdot)P$ commute on the space of infinitely differentiable functions of the form $u:\mathbb R^2\to \mathbb R^2;(x,y)\mapsto (u_1(x,y),u_2(x,y))$.
| If $\alpha$ and $\beta$ are both nonnegative then for any $t > 1$,
$$
\begin{pmatrix}
\alpha & 0 \\
0 & \beta
\end{pmatrix}
=
\begin{pmatrix}
t\alpha & 0 \\
0 & t\beta
\end{pmatrix}
+
\begin{pmatrix}
(1-t)\alpha & 0 \\
0 & (1-t)\beta
\end{pmatrix}.
$$
A similar argument works if both are nonpositive.
It's not hard to show there is no solution for
$$
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Calculating a limit with exponents There is a limit I want to calculate, and I've calculated it as follows:
$$\lim_{x \to \infty} \frac{7^{-x+1}-2\cdot 5^{-x}}{3^{-x}-7^{-x}} = \lim_{x \to \infty} \frac{\frac{7}{7^x}-\frac{2}{5^x}}{\frac{1}{3^x}-\frac{1}{7^x}}$$
If I now multiply the denominator and numerator by $3^x$, I get the following:
$$\lim_{x \to \infty} \frac{7\cdot\frac{3^x}{7^x}-2\cdot\frac{3^x}{5^x}}{1-\frac{3^x}{7^x}} = \frac{0}{1}.$$
Is it correct to assume that the fractions in the numerator will limit to $0$ and not to infinity (because they are multiplied by $7$ and $2$ respectively)?
Does this way of thinking apply to all limits of such format (except for those that limit to $e$) or are there exceptions?
| Your answer is correct. In order to be sure it's nice to write :
$$\lim_{x \to +\infty} \dfrac{7 \dfrac{3^x}{7^x} - 2 \dfrac{3^x}{5^x}}{1 - \dfrac{3^x}{7^x}} = \dfrac{7 \times 0 - 2 \times 0}{1 - 0} = \dfrac{0}{1} = 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to use some kind of factorisation. I tried for like an hour but all in vain.
Then, I tried to solve the problem by multiplying both sides by 2 and then squaring both sides. The equation became too complicated.
By using some hit and trial, I get to know that $x = 1$ satisfies the equation. But what about complex solutions. So this method is also of no use.
I am sure this question has to be solved using some special equality which I'm unaware of. I want a method so that, I could obtain all the possible values. Can anyone help me or just give some hints?
| There is only a single real solution:
For $x\geq -1$ :
$$\left[\sqrt{3} \sqrt{1+x} - \sqrt{2} \sqrt{2+x}\right]^2 + \left[(1+x) - \sqrt{2} \sqrt{1+x^2}\right]^2 \geq 0$$
$$\left[7 + 5 x - 2 \sqrt{6(1+x)(2+x)} \right] + \left[ 3 + 2 x + 3 x^2 - 2 (1+x) \sqrt{2(1+x^2)} \right] \geq 0$$
$$ 10 + 7x + 3 x^2 \geq 2 \sqrt{6(1+x)(2+x)} + 2 (1+x) \sqrt{2(1+x^2)}$$
Both terms in the first line are zero if and only if $x=1$.
For $x\leq -2$ additional minus signs are required within both square roots of the first term, but no additional solution exists.
For $-2<x<-1$ the term $\sqrt{6(1+x)(2+x)}$ results in complex numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Deriving the formula for $\cos^{-1}x+\cos^{-1}y$ I am trying to prove the following:
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\ge0\\2\pi-cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\le0\end{cases}$$
Let $\cos^{-1}x=A, \cos^{-1}y=B.$ Since $-1\le x,y\le1\implies0\le A,B\le\pi\implies0\le A+B\le2\pi$
$\cos(A+B)=\cos A\cos B-\sin A\sin B=xy-\sqrt{1-x^2}\sqrt{1-y^2}$ (since the sine function is positive in $1$st and $2$nd quadrants.)
Therefore, $\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})=\cos^{-1}(\cos(A+B))=\begin{cases}A+B;&0\le A+B\le\pi\implies0\le A,B\le\frac\pi2\\2\pi-(A+B);&\pi\le A+B\le2\pi\implies\frac\pi2\le A,B\le\pi\end{cases}$
From this, how do we conclude about $x+y$?
| In first case you found that
$$0\leq A+B\leq \pi$$ which becomes
$$0\leq\cos^{-1}x+\cos^{-1}y\leq \pi$$
Note the first inequality above is true for all $x,y\in[-1,1]$
Now the second inequality becomes $$\cos^{-1}x+\cos^{-1}y\leq \pi$$
$$\cos^{-1}x\leq\pi-\cos^{-1}y$$
$$\cos^{-1}x\leq\cos^{-1}(-y)$$
Now $f(x)=\cos^{-1}x$ is strictly decreasing function , therefore $$\cos^{-1}x\leq\cos^{-1}(-y)\implies x\geq-y\implies x+y\geq0$$
because for a decreasing function $$f(x)\leq f(y)\iff x\geq y$$
Similarly second case could be proven.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula? Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
| Along the line of @robjohn's solution, consider the change of variables $u=\frac{1}{1+x^2}$. Then
$$du=-2 u^2\Big(\frac{1}{u}-1\Big)^{1/2} dx=-\frac12 u^{3/2}(1-u)^{-1/2}dx$$
and so
$$\int^\infty_0\frac{dx}{(1+x^2)^n}=\frac12\int^1_0u^{n-\tfrac12-1}(1-u)^{\tfrac12-1}\,du=\frac12B\big(n-\tfrac12,\frac12\big)=\frac12\frac{\Gamma(n-\tfrac12)\Gamma(\tfrac12)}{\Gamma(n)}$$
The identity of the OP follows from properties of the Gamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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Rolling dice game - who first will roll number 3 Let's consider very easy game with players A and B - they roll a dice starting with player A. If any of players roll a three, then he wins. I want to calculate probability that player B wins.
Intuition
Intuition is that $P(\textrm{player B wins}) < P(\textrm{player A wins})$
because they have even chances on winning, and player A starts, so player A has one more roll, therefore bigger chance to win. In other words player A is one roll ahead of player B so what should hold is that:
$$P(\textrm{player A wins}) = P(\textrm{player B wins}) + \frac 16$$
Out of this we can already calculate desire probability $P(\textrm{player B wins}) = \frac{5}{12}$
Normal approach
I want to calculate this normally (without any tricks) to compare the results. Please see the probability tree that I've created:
Out of this tree we can see that:
$$P(\textrm{B won}) = \frac{5}{6} \cdot \frac 1 6 + (\frac{5}{6})^2 \cdot \frac{5}{6} \cdot \frac 1 6 + (\frac{5}{6})^4 \cdot \frac 1 6 + ... = \sum_{n = 0}^\infty (\frac 5 6)^{2n}\frac{5}{6}\frac{1}{6} = $$
$$= \sum_{n = 0}^\infty(\frac{25}{36})^n\frac{5}{6}\cdot \frac 1 6 = \frac{1}{1 - \frac{25}{36}} \cdot \frac{5}{36} = \frac{36}{11} \cdot \frac{5}{36} = \frac{5}{11}$$
Question
As you can see those two probabilities differ. Second result also matches our intuition that $P(\textrm{player B wins}) < P(\textrm{Player A wins})$ but I want to ask you - which result is correct and where is the mistake with the wrong one?
| Denoting the ultimate probability of $A$ winning by $a$,
either $A$ wins immediately, or both fail and we are back to square $1$
So we have $a = \frac16 + \frac{5}6\frac56\cdot{a}$
which yields $a = \frac6{11},\;\; b = \frac5{11}$
You should be able to now spot the error in your first formulation
| {
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$\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? How to calculate integral $\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? I got this integral by using Abel-Plana formula on series $\sum_{n=0}^\infty \frac{1}{(n+1)^2}$. This integral can be splitted into two integrals with bounds from 0 to 1 and from 1 to infinity and the both integrals converge, so does the sum. I checked with WolframAlpha and the value of the integral is $\frac{-9 + \pi^2}{24}$, but I don't know how to compute it. Also, I tried to write $\frac{2xdx}{(1+x^2)^2}=d\frac{1}{x^2+1}$ and then tried to use partial integration, but didn't succeded.
Any help is welcome. Thanks in advance.
| Recall Binet's second $\ln \Gamma$ formula:
$$\int_{0}^{\infty} \frac{\arctan (t z)}{e^{2\pi t}-1} \, dt = \frac{1}{2} \ln \Gamma \left(\frac{1}{z}\right) + \frac{1}{2}\left(\frac{1}{z}-\frac{1}{2}\right) \ln (z) + \frac{1}{2z} - \frac{1}{4} \ln (2 \pi)$$
Consider the following integral:
$$\int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz = \frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} + \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)}$$
Since $$\frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} = \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right)$$
$$\implies \int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \int_{0}^{\infty} \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right) \, dt+ \int_{0}^{\infty} \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)} \, dt$$
Using Binet's formula we determine then:
$$\int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \frac{1}{4z}-\frac{1}{8}\ln(2\pi z)-\frac{\psi\left(\frac{1}{z}\right)}{4z}+\frac{1}{4}\ln\left(\Gamma\left(\frac{1}{z}\right)\right)-\frac{1}{8}$$
where $\psi$ is the digamma function.
Taking the derivative with respect to $z$ then the limit as $z \to 1$ we determine:
$$\boxed{\int_{0}^{\infty} \frac{x}{(e^{2 \pi x}-1)(x^2+1)^2} \, dx = \frac{\pi^2}{24} - \frac{3}{8}}$$
| {
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If a, b, c are positive numbers then prove that $\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq a+b+c$.
If a, b, c are positive numbers then prove that $\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq a+b+c$.
I tried using AM-GM inequality, $$\frac{a+b}{2}\geq \sqrt{ab}$$ $$a^2+b^2+2ab\geq 4ab$$ $$ a^2+b^2\geq 2ab$$
Dividing both sides by $a+b$, $$\frac{a^2+b^2}{a+b}\geq \frac{2ab}{a+b}$$
Similarly, $$\frac{b^2+c^2}{b+c}\geq \frac{2bc}{b+c}$$ $$\frac{c^2+a^2}{c+a}\geq \frac{2ca}{c+a}$$
Adding all, we get, $$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq \frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ca}{c+a}$$
Now, I have no idea what to do. Can we simplify RHS to get $a+b+c$ or is there a different method?
| Hint: Use $\dfrac{x^2+y^2}{x+y} \ge \dfrac{x+y}{2}$.
| {
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Integrating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without trig sub So I am evaluating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without using trig sub integrals. So far I have
$$u=\sqrt{3-x^2}, x^2=3-u^2,du=-\frac{x}{\sqrt{3-x^2}}dx, dx = -\frac{\sqrt{3-x^2}}{x}$$
So rewriting I get
$$\int \frac{1}{x\sqrt{3-x^2}}dx=-\int\frac{1}{3-u^2}du$$
Then I use partial fraction decamp to find that both variables are $\frac{1}{2\sqrt{3}}$ so I get
$$-\int\frac{1}{3-u^2}du=-\frac{1}{2\sqrt{3}}\bigg[\int \frac{1}{\sqrt{3}+u}du+\frac{1}{\sqrt{3}-u}du\bigg]=-\frac{1}{2\sqrt{3}}\bigg( \ln \bigg\vert \sqrt{3} +\sqrt{3-x^2}\bigg \vert - \ln \bigg \vert \sqrt{3}-\sqrt{3-x^2} \bigg \vert \bigg)+C$$
But this isn't the answer, where am I going wrong?
| Your answer looks fine to me. To double check, you can differentiate your answer, or confer with wolfram, which gives
$$\int \frac{1}{x\sqrt {3-x^2}}=-\frac{1}{\sqrt 3}\tanh^{-1}\sqrt{1-x^2/3}+C,$$
and using the identity
$$\tanh^{-1}t=\frac{1}{2}\ln \left(\frac{1+t}{1-t}\right),t\in(-1,1)$$
we have
$$\begin{align}-\frac{1}{\sqrt 3}\tanh^{-1}\sqrt{1-x^2/3}+C&=-\frac{1}{2\sqrt 3}\ln \left(\frac{1+\sqrt{1-x^2/3}}{1-\sqrt{1-x^2/3}}\right)+C\\
&=-\frac{1}{2\sqrt 3}\ln \left(\frac{\sqrt 3+\sqrt{3-x^2}}{\sqrt 3-\sqrt{3-x^2}}\right)+C\\
&=-\frac{1}{2\sqrt 3}\left(\ln \left(\sqrt 3+\sqrt{3-x^2}\right)-\ln \left(\sqrt 3-\sqrt{3-x^2}\right)\right)+C
\end{align}.$$
| {
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If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
I found the similar question in Quora. There was slightly a mistake.
$$(\tan\theta+\sin\theta)^2-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\tan\theta\sin\theta-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{\tan^2\theta-\sin^2\theta}-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
I just said that "we have to take $4\sqrt{m(\tan\theta-\sin\theta)}=4\sqrt{mn}$ to satisfy $\tan\theta-\sin\theta=n$ but I don't like it, I just want direct derivation."
And in quora they just took $4\sqrt{m}$ common and wrote $$(\tan\theta-\sin\theta)^2+4\sqrt{m}(\sqrt{\tan\theta-\sin\theta-n})-n^2=0$$
which is totally wrong
cause $\sqrt{\tan\theta-\sin\theta-n}\neq\sqrt{\tan\theta-\sin\theta}-\sqrt{n}$
| Let's call $x=\tan\theta-\sin\theta$ then
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
becomes
\begin{align}
x^2+4\sqrt{m}\sqrt{x}-n^2-4\sqrt{m}\sqrt{n}&=0\\
(x^2-n^2)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(x-n)(x+n)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(\sqrt{x}-\sqrt{n})(\sqrt{x}+\sqrt{n})(x+n)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(\sqrt{x}-\sqrt{n})\Big((\sqrt{x}+\sqrt{n})(x+n)+4\sqrt{m}\Big)&=0
\end{align}
and since the second term is strictly positive $\sqrt{x}=\sqrt{n}\iff x=n$ as we wanted.
| {
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Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simplified to:
$$|z|^3z^4=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
$$|z|^7(\cos{4\alpha} + i\sin{4\alpha})=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
From here I thought just to compare $|z|^7 = 8\sqrt{2}$ and the sine, cosine part.
Is this the correct way to go about it or could it be done using some simpler method?
| Let $z=r*e^{i\theta}$, So here we have,
$$\bar{z}*z^2*|z|=8\sqrt{2}(-sin\frac{\pi}{5}-icos\frac{\pi}{5})^8$$
$$(e^{-i\theta})(e^{i\theta})^2(r)=8\sqrt{2}(-i(cos\frac{\pi}{5}-isin\frac{\pi}{5}))^8$$
$$(e^{-i\theta})(e^{-\theta^2})(r)=8\sqrt{2}*1*(cos(-\frac{\pi}{5})+isin(-\frac{\pi}{5}))^8$$
$$(cos\theta-isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}(e^{-i\frac{\pi}{5}})^8$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}e^{-i\frac{8\pi}{5}}$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}(cos(\frac{8\pi}{5})-isin(\frac{8\pi}{5}))$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})-i8\sqrt{2}sin(\frac{8\pi}{5})$$
So now we have segregated the real and complex parts,
So now, $(cos\theta)(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})$ and $(sin\theta)(e^{-\theta^2})(r)=8\sqrt{2}sin(\frac{8\pi}{5})$
On dividing Both Equations,
$$cot\theta=cot(\frac{8\pi}{5})$$
So from here we get $\theta=\frac{8\pi}{5}+2n$,where n is any integer.
So Now,
$$(cos(\frac{8\pi}{5}+2n))(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})$$
so both cosines will cancel out
$$(e^{-\theta^2})(r)=8\sqrt{2}$$
$$(r)=8\sqrt{2}(e^{\frac{8\pi}{5}^2})$$
Hence, $$z=8\sqrt{2}(e^{\frac{8\pi}{5}^2})(e^{i(\frac{8\pi}{5}+2n)})$$ where n is any positive integer.
| {
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"source": "stackexchange",
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Trigonometric Integrals and Hypergeometric function I'm dealing with these two integrals:
$$I_1=\int_{-\pi}^{\pi} \frac{\cos(x)\cos(nx)}{(1+e \cos(x))^3} \mathrm{d}x, \quad I_2=\int_{-\pi}^{\pi} \frac{\sin(x)\sin(nx)}{(1+e \cos(x))^3}\mathrm{d}x$$
is there a way to reduce them to a Hypergeometric form or to solve them analytically?
| From this answer we know that
$$
\int_0^\pi\frac{\cos(mx)}{p-q\cos (x)}\ \mathrm{d}x=\frac{\pi}{\sqrt{p^2-q^2}}\left(\frac{p-\sqrt{p^2-q^2}}{q}\right)^m\quad\hbox{for}\quad |q|<p \tag{1}
$$
We'll come back to this identity to solve both integrals.
For $I_1$ first notice that
$$
\frac{\partial}{\partial p}\left(\frac{\partial}{\partial q}\frac{\cos(mx)}{p-q\cos (x)} \right) = \frac{\partial}{\partial p}\left(\frac{\cos(x)\cos(mx)}{\left(p-q\cos (x)\right)^2 }\right) = -2 \frac{\cos(x)\cos(mx)}{\left(p-q\cos (x)\right)^3}
$$
So using Feynman's trick, differentiating both sides of $(1)$ first with respect to $q$ and then $p$, we get
\begin{align}
-2 \int_0^\pi\frac{\cos(x)\cos(mx)}{\left(p-q\cos (x)\right)^3} \ \mathrm{d}x & =-\frac{\pi \left(p - \sqrt{p^2 -q^2}\right)^m\left(m^2 p\left(p^2 -q^2\right) + m\left(p^2 + 2q^2\right) \sqrt{p^2 - q^2}+ 3pq^2 \right) }{q^{m+1}\left(p^2 - q^2 \right)^{\frac{5}{2}}}
\end{align}
Now, since $I_1$ has an even integrand, we know that $\int_{-\pi}^{\pi} = 2 \int_{0}^{\pi}$. Substituting $p=1$ and $q = - \xi$ (which holds since you established that the parameter $0<\xi<1$) we can combine all the above to get
\begin{align}
\boxed{\int_{-\pi}^\pi\frac{\cos(x)\cos(nx)}{\left(1+ \xi\cos (x)\right)^3} \ \mathrm{d}x = - \frac{\pi \left( \sqrt{1 -\xi^2}-1\right)^n\left(n^2 \left(1 -\xi^2\right) + n\left(1 + 2\xi^2\right) \sqrt{1 - \xi^2}+ 3\xi^2 \right) }{\xi^{n+1}\left(1 - \xi^2 \right)^{\frac{5}{2}}}}
\end{align}
valid for $0<|\xi|<1$.
For $I_2$, integrating by parts gives
$$
I_2 = \int_{-\pi}^{\pi} \sin(nx)\left[\frac{\partial}{\partial x} \ \frac{1}{\color{blue}{2}\xi(1+\xi \cos(x))^2} \right] \ \mathrm{d}x = - \frac{n}{\xi} \int_{\color{blue}{0}}^{\pi} \frac{\cos(nx)}{(1+\xi \cos(x))^2} \ \mathrm{d}x
$$
since $\sin(nx)=0$ at $x = \pm \pi$. Differentiating $(1)$ with respect to $p$ gives
$$
-\int_0^\pi\frac{\cos(mx)}{(p-q\cos (x))^2}\ \mathrm{d}x = -\frac{\pi \left(\frac{p - \sqrt{p^2 - q^2}}{q}\right)^m \left(p + m \sqrt{p^2 - q^2}\right)}{\left(p^2 - q^2\right)^{\frac{3}{2}}}
$$
And again substituting $p=1$ and $q = - \xi$, we can conclude that
$$\boxed{
\int_{-\pi}^{\pi} \frac{\sin(x)\sin(nx)}{(1+\xi \cos(x))^3}\ \mathrm{d}x = -\frac{n \pi \left( \sqrt{1 - \xi^2}-1\right)^n \left(1 + n \sqrt{1 - \xi^2}\right)}{\xi^{n+1} \left(1 - \xi^2\right)^{\frac{3}{2}}} }
$$
also valid for $0<|\xi|<1$.
| {
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Matching pairs of gloves A closet contains $13$ different pairs of gloves. The 26 gloves are randomly arranged into $13$ pairs.
(a) Find the probability that all left-hand gloves are paired with right-hand gloves (not necessarily matching).
(b) Find the probability that all of the gloves are arranged so that each left-hand glove is paired with its matching right-hand glove.
Updated:
To deal with (a). First I get the number of ways to arrange 26 gloves into pairs ($n=13$), the order does not matter: $$\frac{(2n)!}{2^{n}}\cdot\frac{1}{n!}$$
After that, I calculate the number of ways in which all left-hand gloves are paired with right-hand gloves (not fully matching): $n!=13!$
So the probability of (a) equals: $$n!\cdot\frac{(2^{n})n!}{(2n)!}$$
And the probability of (b) equals (only one case when all gloves are fully matching): $$\frac{(2^{n})n!}{(2n)!}$$
Is my approach correct?
| Your solutions are correct.
There is another way to count the number of elements in the sample space. Suppose we have $n$ pairs of gloves in the closet. Grab any of the $2n$ gloves in the closet. It does not matter which glove we choose. There are $2n - 1$ gloves left in the closet. Pair one of them with the first glove you select. Set that pair aside. Take any of the remaining $2n - 2$ gloves out of the closet. Again, it does which glove we choose. Pair any of the remaining $2n - 3$ gloves left in the closet with that glove. Iterate the process. The number of ways of forming $n$ pairs of gloves is
$$(2n - 1)!! = (2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)$$
where $k!!$ is referred to as the double factorial.
We can verify that this expression is equivalent to your expression as follows. Based partly on work you did in the comments, you found that the number of ways to place the gloves in pairs is
$$\frac{1}{n!2^n}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \cdots \binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{(2n)!}{n!2^n}$$
Observe that
\begin{align*}
\frac{(2n)!}{n!2^n} & = \frac{(2n)(2n - 1)(2n - 2)(2n - 3) \cdots (4)(3)(2)(1)}{n(n - 1)(n - 2)(n - 3) \cdots (3)(2)(1)(2^n)}\\
& = \frac{(2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)(n!)}{n!}\\
& = (2n - 1)(2n - 3)(2n - 5) \cdots 5 \cdot 3 \cdot 1\\
& = (2n - 1)!!
\end{align*}
where we first cancel a $2$ from each even factor to obtain the expression in the second line, then cancel $n!$ to obtain the expression in the third line.
In this particular problem, $n = 13$, so the number of elements in our sample space is
$$25!! = 25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
Find the probability that all left-hand gloves are paired with right-hand gloves (not necessarily matching).
If we place the right-hand gloves in a line (in any order), there are $n$ left-hand gloves which can be matched with the first right-hand glove, $n - 1$ left-hand gloves which can be matched with the second right-hand glove, $n - 2$ left-hand gloves which can be matched with the third right-hand glove, and so forth. Hence, there are $n!$ favorable cases. Thus, the probability that each left-hand glove is matched with a right-hand glove is
$$\frac{n!}{(2n - 1)!!}$$
In this particular problem, we obtain
$$\frac{13!}{25!!} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}$$
Find the probability that all of the gloves are arranged so that each left-hand glove is paired with its matching right-hand glove.
There is only one favorable case. Hence, the probability is
$$\frac{1}{(2n - 1)!!}$$
In this particular problem, we obtain
$$\frac{1}{25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}$$
| {
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$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.
| We have $e^x > 1+x+x^2/2+x^3/6$, so we show that the right side is bigger than $x^2+x$ for positive $x$. Let
$$f(x) = 1+x+x^2/2+x^3/6-(x^2+x) = x^3/6-x^2/2+1.$$
Find the minimum value of $f(x)$ using usual calculus methods:
$$f'(x) = x^2/2-x = 0$$
gives one positive critical point $x=2$. $f''(x) = x-1$ and $f''(2) >0$ so the critical point is a minimum. Since $f(2) = 1/3>0$ the minimum value of $f$ on $x>0$ is positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$
The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve for $A+B$ (what we want).
How can I do that?
| Let $x = \sqrt[3]{40 + 11\sqrt{13}} + \sqrt[3]{40 - 11\sqrt{13}}$. Then we have that
\begin{align*}
x^{3} = 80 + 9x & \Longleftrightarrow x^{3} - 9x - 80 = 0\\\\
& \Longleftrightarrow (x^{3} - 25x) + (16x - 80) = 0\\\\
& \Longleftrightarrow x(x^{2} - 25) + 16(x - 5) = 0\\\\
& \Longleftrightarrow x(x + 5)(x - 5) + 16(x - 5) = 0\\\\
& \Longleftrightarrow (x^{2} + 5x + 16)(x - 5) = 0\\\\
& \Longleftrightarrow x = 5
\end{align*}
and we are done.
Hopefully this helps !
| {
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Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ using AM-GM inequality, }\space\space t_n\ge (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\le t_n\le \dfrac{1}{n}(1+\cdots+1)=\dfrac{n}{n}=1$
$(1\cdot2\cdot 3\cdots n)^{\dfrac{1}{n}}=(1)^{\frac{1}{n}}\cdot (2)^{\frac{1}{n}}\cdot (3)^{\frac{1}{n}}\cdots\cdot (n)^{\frac{1}{n}}\stackrel{n\to\infty}{\to}1$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\to1$
Hence, by sandwich theorem $t_n\to1$
Is this correct?
| Note that your sequence can be written as follows
$$t_n = \frac{1}{n}\sum_{j=1}^n\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/2}}dt\; + \frac{1}{\sqrt{n+1}}.$$
Clearly the last term of the summation vanishes as $n\to+\infty$ while the first term is the Cesaro sum of the sequence
$$a_j:=\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/2}}dt.$$
By a direct computation you can verify that $\lim_{j\to+\infty}a_j=0$ and by Cesaro theorem this means that $1/n\sum_{j=1}^n a_j$ converges to the same limit as $n\to+\infty$ and this shows that $t_n\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How can I find the equation of a parabola with focus and directrix? Focus: (3,5)
Directrix: $y=-x-1$
I'm comfortable with questions where the directrix is simply $y=3$ or something that is constant (i.e. a horizontal line).
I know that I need to find the distance between a general point (x,y) and (3,5) as well as the distance between (x,y) and the line $y=-x-1$
The distance between (x,y) and (3,5) is $\sqrt{(x-3)^2+(y-5)^2}$
I see online that the formula for the distance from a line to a point is given by
$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$
I want to know if there is some easier way for me to calculate this distance? Do I just need to memorize this?
Also, using this formula, is my calculation correct?
Distance from $(x,y)$ to $y=-x-1$:
$$x+y+1=0$$
$$ \frac{|x+y+1|}{\sqrt{1^2+1^2}}$$
$$\sqrt{(x-3)^2+(y-5)^2} = \frac{|x+y+1|}{\sqrt{2}}$$
$$(x-3)^2+(y-5)^2=\frac{(x+y+1)^2}{2}$$
Then from here I just expand everything out and solve for $x$:
$$2(x^2-6x+9+y^2-10y+25)=x^2+y^2+2xy+2x+2y+1$$
$$2x^2+2y^2-12x-20y+68=x^2+y^2+2xy+2x+2y+1$$
$$x^2=-67+14x+22y+2xy-y^2$$
Not exactly sure what to do from here?
| First, you can find the vertex of the parabola. It lies half way between the focus and the directrix.
The axis of the parabola passes through the focus, and is perpendicular to the directrix. The directrix is
$x + y + 1 = 0$
So the unit vector along the axis is
$u_1 = [1, 1]^T / \sqrt{2} $
The distance between the focus and the directrix is
$2 p = \dfrac{|3 + 5 + 1 | }{\sqrt{2} } = \dfrac{9}{\sqrt{2}}$
Therefore the vertex of the parabola is
$ V = (3, 5) - \dfrac{9}{2 \sqrt{2}} (1, 1)/ \sqrt{2} = (0.75, 2.75) $
$p = \dfrac{9}{2 \sqrt{2}} $ is the focal length.
Rotate $u_1$ by ($90^\circ$ clockwise) to generate the other axis
$ u_2 = (1, -1) / \sqrt{2} $
Now in the coordinate refernce frame specified by $u_1, u_2$ the equation of the parabola is
$ y' = \dfrac{1}{4 p} x'^2 $
where $x'$ is along $u_2$ and $y'$ is along $u_1$
Hence
$ y' = \dfrac{\sqrt{2}}{18} x'^2 $
Finally to get the equation in the $xy$ frame, we note that
$(x, y) = V + [u_2, u_1] (x', y') $
Thus
$ (x', y') = [u_2,u_1]^{-1} ( (x, y) - V ) $
Now:
$[u_2, u_1]^{-1} = \sqrt{2} \begin{bmatrix} 1 && 1\\-1&&1 \end{bmatrix}^{-1} = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} $
Writing $x', y'$ explicitly in terms of $x, y$, we have
$x' = \dfrac{1}{\sqrt{2}} (x - y + 2) $
and
$y' = \dfrac{1}{\sqrt{2}} ( x + y - 3.5 ) $
Plug these two expressions into
$ y' = \dfrac{\sqrt{2}}{18} x'^2 $
You get
$ 9 (x + y - 3.5) = \dfrac{1}{2} ( x- y + 2 )^2 $
Therefore,
$ 18 (x + y - 3.5) = x^2 + y^2 + 4 + 4 x - 4 y - 2 xy $
and finally,
$ x^2 - 2 xy + y^2 - 14 x - 22 y + 67 = 0 $
which is exactly what you got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor Series of $(\frac{z}{1 - z})^2$ around $z = i$ I am having some trouble trying to find the Taylor series of $f(z) = \left(\frac{z}{1 - z}\right)^2$ around $z = i$. I have started a little bit, but unsure how to complete. Below is my attempt:
Since $f(z) = \left(\frac{z}{1 - z}\right)^2$, we are able to rewrite this as:
\begin{equation*}
f(z) = \left(\frac{1}{1 - z} - 1\right)^2 = \dfrac{1}{(1 - z)^2} - \dfrac{2}{1 - z} + 1
\end{equation*}
Observe that $2\frac{1}{1 - z}$ is the Taylor series:
\begin{equation*}
2\frac{1}{1 - z} = 2\sum_{n = 0}^{\infty} \frac{(z - i)^n}{(1 - i)^{n + 1}}
\end{equation*}
around $z = i$
Also observe that the Taylor series of $\frac{1}{(1 - z)^2}$ is the Taylor series:
\begin{equation*}
\frac{1}{(1 - z)^2} = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}}
\end{equation*}
around $z = i$
Then we have that:
\begin{equation*}
f(z) = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1
\end{equation*}
I am not sure how else to proceed from here, and would like some assistance. Thank you.
| Observe that for
\begin{equation*}
f(z) = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1
\end{equation*}
you may reshift the indices of the first summation so that
\begin{equation*}
f(z) = \sum_{n = 1}^{\infty} \dfrac{(z - i)^{n}}{(1 - i)^{n + 1}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1.
\end{equation*}
From here it is just trivial simplifications, namely taking out one term in the second summation and then combining them again.
$$ f(z) = 1 - \frac{2}{1-i} - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}}\\
\implies
f(z) = -\frac{1+i}{1-i} - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}} \\
\implies f(z) = -i - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}}$$
If you wanted to you could shift it once more to start at $ n = 0$ as that's usually the convention.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What type of product? $\mathbf{a} \circledast \mathbf{b}:= \mathbf{a}\mathbf{b}^T - \mathbf{b}\mathbf{a}^T$ What do you know about the names and properties of the products of the following definitions?
\begin{align}
&\mathbf{a,b}\in\mathbb{R}^N, \mathbf{C}\in\mathbb{R}^{N\times N}\\
&\mathbf{C} = \mathbf{a} \circledast \mathbf{b}:= \mathbf{a}\mathbf{b}^T - \mathbf{b}\mathbf{a}^T
\end{align}
I found some propeties.
\begin{align}
& \mathbf{b} \circledast \mathbf{a} = - \mathbf{a} \circledast \mathbf{b} = (\mathbf{a} \circledast \mathbf{b})^T\\
& \mathbf{a} \circledast (\beta \mathbf{b} + \gamma \mathbf{c}) = \beta (\mathbf{a} \circledast \mathbf{b}) + \gamma (\mathbf{a} \circledast \mathbf{c})
\end{align}
In addition, let $\mathbf{x} = \alpha \mathbf{a} + \beta \mathbf{b}, \mathbf{A} = [\mathbf{a}~\mathbf{b}] \in \mathbb{R}^{N \times 2}$,
\begin{align}
& \mathbf{x}^T (\mathbf{a} \circledast \mathbf{b}) \mathbf{x} = 0 \\
&\|(\mathbf{a} \circledast \mathbf{b}) \mathbf{x}\|_2 = \sqrt{\det \mathbf{A}^T\mathbf{A}} \| \mathbf{x} \|_2
\end{align}
therefore, $\mathbf{R} = \frac{1}{\sqrt{\det \mathbf{A}^T\mathbf{A}}}\mathbf{a} \circledast \mathbf{b}$ is $\pi/2$ [rad] rotation matrix for $\mathbf{x} \in \mathrm{range}\mathbf{A}$.
Do you know the name of this product?
Can you find any other properties about this product?
We now understand that this product is closely related to the wedge product.
How can I make this product more interesting as wedge product?
For example, the wedge product can be stacked many times.
\begin{align}
a \wedge b \wedge c
\end{align}
But my product cannot be stacked.
| This looks like the representation of the wedge product $a\wedge b$ as a skew-symmetric matrix. (Note that $\mathbf C$ is skew-symmetric.) If $a=\sum a_ie_i$ and $b=\sum b_je_j$, then $a\wedge b = \sum\limits_{i<j} (a_ib_j-a_jb_i)e_i\wedge e_j$.
(You might read more about exterior algebra.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{-\infty}^\infty \frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$ I recently attempted to evaluate the following integral
$$\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$$
I started by inserting a parameter, $t$
$$F(t)=\int_{-\infty}^\infty\frac{\ln{(tx^4+x^2+t)}}{x^4+1}dx$$
Where F(0) is the following
$$F(0)=\int_{-\infty}^\infty\frac{\ln{(x^2)}}{x^4+1}dx=2\int_0^\infty\frac{\ln{(x^2)}}{x^4+1}dx=4\int_0^\infty\frac{\ln x}{x^4+1}dx$$
We can evaluate this using a common integral from complex analysis and taking the derivative using Leibniz’s rule.
$$\int_0^\infty\frac{x^m}{x^n+1}dx=\frac{1}{m+1}\int_0^\infty\frac{(m+1)x^m}{(x^{m+1})^\frac{n}{m+1}+1}dx=\frac{1}{m+1}\int_0^\infty\frac{du}{x^\frac{n}{m+1}+1}$$
$$=\frac{1}{m+1}\frac{\pi}{\frac{n}{m+1}\sin{\frac{\pi}{\frac{n}{m+1}}}}=\frac{\pi}{n\sin{\frac{\pi(m+1)}{n}}}=\frac{\pi}{n}\csc{\frac{\pi(m+1)}{n}}$$
$$\int_0^\infty\frac{\ln{x}}{x^n+1}dx=\frac{d}{dm}\int_0^\infty\frac{x^m}{x^n+1}dx\Big|_{m=0}$$
$$=\frac{\pi}{n}\frac{d}{dm}\csc{\frac{\pi(m+1)}{n}}\Big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi(m+1)}{n}}\cot{\frac{\pi(m+1)}{n}}\big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi}{n}}\cot{\frac{\pi}{n}}$$
Therefore
$$F(0)=4\int_0^\infty\frac{\ln{x}}{x^4+1}dx=-\frac{\pi^2\sqrt 2}{4}=-\frac{\pi^2}{2\sqrt 2}$$
Now that we found F(0), we can start applying Feynman’s trick.
$$F’(t)=\int_{-\infty}^{\infty}\frac{dx}{tx^4+x^2+t}$$
Using a formula I derived we can continue
$$\int_{-\infty}^\infty\frac{dx}{ax^4+bx^2+c}=\frac{\pi}{\sqrt{c}\sqrt{b+2\sqrt{ac}}}$$
$$F’(t)=\frac{\pi}{\sqrt{t}\sqrt{1+2t}}$$
Integrating both sides
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}+C$$
Set $t=0$
$$C=F(0)=-\frac{\pi^2}{2\sqrt2}$$
Therefore
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}-\frac{\pi^2}{2\sqrt2}$$
$$I=\pi\sqrt2\ln{(\sqrt2+\sqrt3)}-\frac{\pi^2}{2\sqrt2}$$
WolframAlpha confirms it numerically
I am not satisfied with this solution. I am curious as to what other solutions there might be. How else can we solve this integral?
| Utilize the integral
$\int_{-\infty}^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{2\pi}b \ln(a+b)$, along with the shorthands $p=e^{i\frac\pi6}$ and $q= e^{-i\frac\pi4} $
\begin{align}
&\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx\\
=& \int_{-\infty}^\infty\frac{\ln{(x^2+p^2) (x^2+\bar{p}^2)}}{(x^2+q^2)( x^2+\bar{q}^2) }dx\\
=& \ \Im \int_{-\infty}^\infty
\frac{\ln(x^2+p^2)}{x^2+{q}^2} + \frac{\ln(x^2+\bar{p}^2)}{x^2+{q}^2} \ dx\\
=& \ \Im \frac{2\pi}{q}\left[\ln (p+q)+\ln(\bar p+q)\right]\\
= &\ \sqrt2\pi\left( \ln(\sqrt2+\sqrt3)-\frac\pi4\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work because I lose equality case.
| We have $|z^2 -1|\cdot |z-1|^2 = \lvert p(z) \rvert$ with $p(z) = (z^2-1)(z-1)^2$. Since $p(z)$ is holomorphic, $ \lvert p(z) \rvert$ attains its maximum on the closed disk $D^2 = \{ z \in \mathbb C \mid \lvert z \rvert \le 1\}$ on the boundary $\partial D^2 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. This follows from the maximum modulus principle.
Each $z$ with $\lvert z \rvert = 1$ has the form
$$z = x \pm i \sqrt{1-x^2} .$$
We get
$$z^2 -1 = x^2 -(1-x^2) \pm 2i x \sqrt{1-x^2} -1 = 2(x^2 -1 \pm ix \sqrt{1-x^2}), $$
$$\lvert z^2 -1 \rvert = 2\sqrt{(x^2-1)^2 + x^2(1-x^2)} = 2\sqrt{1-x^2},$$
$$z -1 = x -1 \pm i \sqrt{1-x^2} ,$$
$$|z-1|^2 = (x-1)^2 + 1- x^2 = 2(1-x) .$$
Thus
$$\lvert p(z) \rvert = 4\sqrt{1-x^2}(1-x) .$$
It therefore suffices to find the maximum of $q(x) = \sqrt{1-x^2}(1-x)$ for $x \in[-1,1]$. Since $q(x)$ is differentiable in $(-1,1)$, the maximum is either attained at one of the boundary points $x = \pm 1$ of the interval $[-1,1]$ or at a point $x \in (-1,1)$ with $q'(x) = 0$. Since $q(-1)= q(1) = 0$ and $q(0) = 1$, we have to find $x$ with $q'(x) = 0$. We get
$$q'(x) = \frac 1 2 (1-x^2)^{-1/2}(-2x)(1-x) + \sqrt{1-x^2}(-1) = \frac{2x^2 -x -1}{\sqrt{1-x^2}} .$$
We have $2x^2 -x -1 = 0$ for $x = \frac 1 4 \pm \frac 3 4$, i.e. for $x = 1$ and for $x = -\frac 1 2$. But $x =1$ is not contained in $(-1,1)$ (in fact $q'(x)$ is not defined for $x = 1$), thus the maximum is attained at $x = -\frac 1 2$.
We therefore get
$$\lvert p(z) \rvert \le 4q(-\frac 1 2) = 3 \sqrt 3 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x$ I was wondering if it is possible to compute this integral in closed form:
$$
\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x
$$
I tried making a substitution with $s = x^2$ and also tried evaluating it in Wolfram Alpha, but was not successful.
| Under substitution $\frac{x}{\sqrt{2}} \to x$ we get that your integral is equal to $I =2 \sqrt{2} \int_0^{\infty} \frac{x^4}{(x^2 + \xi)^2}e^{-x^2}\mathrm{d}x$ where $\xi = \frac{a}{2}$. Additionally, since $ \frac{x^4}{(x^2 + \xi)^2} = 1 - \frac{2\xi}{x^2 + \xi} + \frac{\xi^2}{\left(x^2 + \xi \right)^2}$, the problem reduces to solving
$$
I_n = \int_0^{\infty} \frac{e^{-x^2}}{(x^2 + \xi)^n}\, \mathrm{d}x,\ \ \quad \text{for}\quad n=1,2
$$
For the first integral we get
\begin{align*}
I_1 & =\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x\\ & = \int_0^{\infty}\int_0^{\infty} e^{-x^2}e^{-y\left(x^2+ \xi\right) } \, \mathrm{d}y \, \mathrm{d}x \\
& = \int_0^{\infty}\int_0^{\infty} e^{-x^2(y+1)}e^{-y \xi } \, \mathrm{d}x \, \mathrm{d}y
\end{align*}
Introducing the change of coordinate system $x = \sqrt{\xi}\frac{u}{v}$ and $ y = \frac{v^2}{\xi} -1$ gives $v^2 -\xi= \xi y$ and $u^2 = x^2 (y+1)$ and thus
\begin{align*}
I_1 &= \int_{\sqrt{\xi}}^{\infty} \int_{0}^{\infty} e^{-u^2}e^{-v^2 +\xi} \begin{vmatrix} \sqrt{\xi}\frac{1}{v} & -\sqrt{\xi}\frac{u}{v^2} \\ 0 & \frac{2v}{\xi}\end{vmatrix}\, \mathrm{d}u \, \mathrm{d}v\\
& = \frac{2}{\sqrt{\xi}} e^{\xi} \int_{0}^{\infty}e^{-u^2}\mathrm{d}u \int_{\sqrt{\xi}}^{\infty}e^{-v^2}\mathrm{d}v\\
& = \frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right)
\end{align*}
recalling that $\int_0^{\infty} e^{-t^2}\mathrm{d}t = \frac{\sqrt{\pi}}{2}$ and that $\int_z^{\infty} e^{-t^2}\mathrm{d}t = \frac{\sqrt{\pi}}{2}\mathrm{erfc}(z)$ by the definition of the complementary error function. Thus
$$\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x =\frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right) \qquad \text{for} \quad \xi >0 $$
For the second integral we'll use Feynman's trick. Since $\frac{\partial}{\partial \xi} \frac{e^{-x^2}}{x^2 + \xi} = - \frac{e^{-x^2}}{\left(x^2 + \xi\right)^2}$ we see that differentiating the previous result we can obtain $I_2$:
$$
-I_2 = \int_0^{\infty}\frac{\partial}{\partial \xi} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}\xi}\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}\xi}\frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right)
$$
By the F.T.C. we know that $\frac{\mathrm{d}}{\mathrm{d}z}\mathrm{erfc}(z) = - \frac{2}{\sqrt{\pi}}e^{-z^2}$ using the definition of the complementary error function. Taking the derivative we get that $\frac{\mathrm{d}}{\mathrm{d}\xi} \frac{e^\xi}{\sqrt{\xi}}\mathrm{erfc}\left(\sqrt{\xi}\right) = \frac{e^\xi (2\xi-1)}{2\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right) - \frac{1}{\sqrt{\pi}\xi} $ which allows us to conclude
$$\int_0^{\infty} \frac{e^{-x^2}}{\left(x^2 + \xi\right)^2}\, \mathrm{d}x =\frac{\sqrt{\pi}}{2\xi} - \frac{\pi e^\xi (2\xi-1)}{4\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right) \qquad \text{for} \quad \xi >0 $$
Finally, combining all of the above we can evaluate the original integral to be
\begin{align*}
I & = 2 \sqrt{2} \int_0^{\infty} \frac{x^4}{(x^2 + \xi)^2}e^{-x^2}\mathrm{d}x \\
& = 2 \sqrt{2} \left[\int_0^{\infty} e^{-x^2}\mathrm{d}x - 2\xi \int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi} \mathrm{d}x + \xi^2\int_0^{\infty} \frac{e^{-x^2}}{\left(x^2 + \xi \right)^2}\mathrm{d}x \right]\\
& = 2 \sqrt{2} \left[\frac{\sqrt{\pi}}{2} - 2\xi \frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right) + \xi^2\left( \frac{\sqrt{\pi}}{2\xi} - \frac{\pi e^\xi (2\xi-1)}{4\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right)\right) \right]\\
& = (\xi+1)\sqrt{2\pi} - (2\xi+3) \pi e^\xi \sqrt{\frac{\xi}{2}}\mathrm{erfc}\left(\sqrt{\xi}\right)
\end{align*}
and recalling that $ \xi = \frac{a}{2}$ we get the final result
$$
\boxed{\int_{-\infty}^{\infty} \left(\frac{x^2}{x^2+a} \right)^2 e^{-\frac{x^2}{2}} \, \mathrm{d}x =\frac{1}{2}\left[(a+2)\sqrt{2\pi} - \sqrt{a}(a+3) \pi e^\frac{a}{2} \mathrm{erfc}\left(\sqrt{\frac{a}{2}}\right) \right] \qquad \text{for} \quad a>0}
$$
verifying WA's result posted in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the probability density function of the random variable Let $X$ be a random variable with
$$f(x) = \frac{1}{2 \theta} , -\theta < x < \theta$$
Let $Y=\frac{1}{X^2}$.Then what is the probability density function of $Y$?
Case $1:$ let $y < 0$. Then $P(\{Y \le y\}) = 0$
Case $2$: let $0 \le y < \theta$.
Then $P(Y \le y) = P(\frac{1}{X^2} \le y) = 1 - P(\frac{1}{X^2} \ge y) = 1 - P(X^2 \le y) = 1- P(-\sqrt{y} \le X \le \sqrt{y}) = \int_{-\theta}^{y}f(y)dy - \int_{-\theta}^{-y}f(y)dy$
Case $3:$ $y \ge \theta$
$P(Y \le y) =1$
The on differentiating the function $F(y)$ we get the probability density function.
Is my approach correct?
| $\begin{aligned}
F_Y(y) &= P(Y \le y) \\
&= P \left(\frac{1}{X^2} \le y\right) \\
&= 1 - P \left(X^2 \le \frac 1y \right) \\
&= 1- P \left(- \frac 1 {\sqrt{y}} \le X \le \frac 1 {\sqrt{y}}\right) \\
&= 1 - \frac{1}{\theta \cdot \sqrt y} \\
\end{aligned}$
Now differentiating it gives you the density function of $~Y$ and as $ Y = \dfrac 1 {X^2}~, ~$ support of $Y$ is $ ~\displaystyle \frac 1 {\theta^2} \lt y \lt \infty$
Alternatively, applying Jacobian method -
$\begin{aligned}
f_Y(y) &= \frac{f_X \left( 1 / {\sqrt y}\right)}{|dy / dx|_{x = 1 / \sqrt y}} + \frac{f_X \left(- 1 / {\sqrt y}\right)}{|dy / dx|_{x = - 1 / \sqrt y}} \\
&= \frac {1}{2\theta} \cdot \frac{1}{2 y^{3/2}} + \frac {1}{2\theta} \cdot \frac{1}{2 y^{3/2}} \\
\text {So, } f_Y(y) &= \frac{1}{2 \theta \cdot y^{3/2}}~, ~\frac{1}{\theta^2} \lt \ y \lt \infty
\end{aligned}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the right hand side into $ab(a-b)$ and equated with the expansion earlier, giving the following inequality: $a^2+ab+b^2 \geq ab$. However by AM-GM, it follows that $a^2+b^2 \geq 2ab$ which my statement above does not follow. Help would be much appreciated.
| Given that $a > b > 0$, we take
$$
S = a^3 - b^3, \ \ T = a^2 b - b^2 a = a b (a - b)
$$
Note that we can write
$$
S = a^3 - b^3 = (a - b) (a^2 + a b + b^2)
$$
Thus,
$$
S - T = (a - b) (a^2 + a b + b^2) - a b (a - b) = (a - b)(a^2 + b^2) > 0
$$
since $a > b > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the probability density function of the random variable $\frac{1}{X}$? How to find the probability density function of the random variable $\frac{1}{X}$?
Let $X$ be a random variable with pdf
$f_X(x)= \begin{cases} 0 ; x \le 0 \\ \frac{1}{2} ; 0 < x \le 1 \\ \frac{1}{2x^2} ; 1 < x < \infty \end{cases}$
I was trying to avoid using formula. So we see that
$P(Y \le y) = 1 - P(X \le y)$
Case$1$: if $y \le 0$
Then $F(y) = 1 - P[X \le y] = 1$
Case $2:$ if $0 < y \le 1$
Then $F(y) = 1 - P(X \le y) = 1 - (0 + \int_{0}^{y}\frac{1}{2}dy) = 1-\frac{y}{2}$
Case $3:$ if $1 < y < \infty$
Then $F(y) = 1 - P(X \le y) = 1 - (\frac{1}{2} + \int_{1}^{y}\frac{1}{y^2}dy) = \frac{1}{2}+\frac{1}{y}-1$
Then on differentiating we can get the probability density function.
I thibk that the distribution function that i got is not correct . Can someone help me out please
| $X$ is a positive r.v. and so is $Y$. Hence, $P(Y\leq y)=0$ for $y \leq 0$.
Also, $P(Y \leq y)=\frac y 2$ for $0<y \leq 1$ and $P(Y \leq y)=1-\frac 1 {2y}$ for $y >1$.
[For $0<y\leq 1$ we have $P(Y \leq y)=P(\frac 1 X \leq y)=P(X\geq \frac 1 y)=\int_{1/y}^{\infty} \frac 1 {2x^{2}}dx=\frac y 2$. I will leave the case $y>1$ to you].
| {
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Can we prove the inequality without opening the parentheses? $(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$$
I tried to use the following inequalities:
$$x^2+y^2+z^2≥xy+yz+xz$$
and The Cauchy–Schwarz inequality:
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
$$x^2+y^2+z^2≥x+y+z$$
I also tried
$$x+y+z ≥3\\ x^2+y^2+z^2≥3\\ xy+yz+xz≥3$$
But, I couldn't make progress.
I want to solve this inequality without expansion. That is, without opening the parentheses. Is this possible?
| Rewrite the left hand side as follows:
$$
(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)=
\\
=\left(\frac{2}{3}(x+y+z)(xy+yz+zx)\right)\cdot(x^2+y^2+z^2)+
\\
+\left(\frac{1}{3}(x+y+z)(x^2+y^2+z^2)\right)\cdot(xy+yz+zx).
$$
Can you continue now?
| {
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Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to expand the numerator (Harmonic numbers) in terms of integral, to get:
$\displaystyle \tag*{} \sum \limits_{n=1}^{\infty} \frac{(-1)^n}{n2^n\binom{2n}{n}} \int _{0}^{1} \frac{x^n - x^{2n}}{1-x} \ \mathrm dx$
And I found that with the help of series expansion of $\sin^{-1}(x)$ and subsituting $x = i \sqrt{x} /8 $ where $i^2=1$
$\displaystyle \tag*{} -2(\sinh^{-1} (\sqrt{x}/8))^2 = \sum \limits_{n=1}^{\infty} \frac{(-1)^nx^n}{n^22^n \binom{2n}{n}} $
But this has $n^2$ in the denominator, which makes it complicated. EDIT: we can eliminate $n^2$ by differentiating and multiplying by $x$ as mentioned in the comments. But now, how can we solve our sum since $H_{2n}-H_n$ is numerator?
And I have the general formula for generating sum:
$ \displaystyle \tag*{} \sum \limits _{n=1}^{\infty} \frac{x^n}{n^y \binom{2n}{n}}$
And this doesn't have $n$ in the denominator and also it has closed-form $\forall \ y \geq 2$
Maybe if there is a way of expressing the denominator in the form integral, the sum can be changed in evaluating the double integral. I think there are other easy ways (such as using Hypergeometric functions)? Any help would be appreciated.
EDIT 2:
From the help of comments and a quora user,
$\DeclareMathOperator{\arcsinh}{arcsinh}$
By @Bertrand87, we have:
$\displaystyle \tag{1} H_{k} - H_{2k} + \ln (2) = \int _{0}^{1} \frac{x^{2k}}{1+x} \ \mathrm dx$
To make use of this, we express our sum as follows:
$\displaystyle \tag*{} S = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}} + \sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}}$
We know
$\displaystyle \tag*{} 2\arcsin^2(x) = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k}}{k^2 \binom{2k}{k}}$
We differentiate both sides w.r.t $x$ both sides,
$\displaystyle \tag*{} \frac{2 \arcsin(x)}{\sqrt{1-x^2}} = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k-1}}{k \binom{2k}{k}}$
Now, we multiply both sides by $(2x)$ and define $x:= ix/ \sqrt{8}$ to get:
$\displaystyle \tag{2} \frac{-2x \arcsinh (x/ \sqrt {8})}{\sqrt{8}\sqrt{1+x^2/8}} = \sum \limits_{k=1}^{\infty} \frac{(-1)^k x^{2k}}{k2^k \binom{2k}{k}}$
We now multiply both sides by $-1/(1+x)$ and integrate from $0$ to $1$ and arrive at:
$\displaystyle \tag*{} \frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}}$
Similarly, from $(2)$ if we let $x=1$ and multiply both sides by $\ln 2$, it yields:
$\displaystyle \tag*{} \frac{-2 \arcsinh (1/ \sqrt{8}) \ln 2}{ \sqrt{8} \sqrt{1 + 1/8}} =\sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}} \approx -0.1601$
Now, our only problem is to evaluate the integral:
$\displaystyle \tag*{} \boxed{\frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx} $
Can anyone help me with this integral?
| $$\color{royalblue}{\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n})}{n2^n \binom{2n}{n}} = \frac{\pi ^2}{36}-\frac{\log ^2(2)}{3}}$$
Proof sketch: Suffices to find $S=\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n2^n \binom{2n}{n}}$. Note that
$$\int_0^1 \frac{x^n(1-x)^n}{x} \log x dx = \frac{1}{n\binom{2n}{n}} (H_{n-1}-H_{2n})$$
so $$S = \int_0^1 \frac{(1-x)\log x}{1+\frac{1}{2}x(1-x)} dx$$
whose antiderivative can be expressed via dilogarithm $\text{Li}_2$, using special value of $\text{Li}_2(1/2)$ completes the proof.
As direct generalizations:
$$\small\begin{aligned}\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^22^n \binom{2n}{n}} &= -\frac{\zeta (3)}{8}-\frac{1}{6} \log ^3(2)+\frac{1}{12} \pi ^2 \log (2) \\ \sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^32^n \binom{2n}{n}} &= -2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13}{8} \zeta (3) \log (2)+\frac{19 \pi ^4}{720}-\frac{1}{24} \log ^4(2)+\frac{1}{24} \pi ^2 \log ^2(2) \end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}}$ Original Question:
Use the proof of Thm 7.2.4 given above to show that if $n \geq 2$, then $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}} < \frac{e}{4^{\frac{1}{n}}}$.
Thm 7.2.4: $\displaystyle\lim_{n\rightarrow \infty} \dfrac{n}{(n!)^\frac{1}{n}} = e$
Proof: Let $a_n = \dfrac{n^n}{n!}$, i.e. $\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n+1}{n}\right)^n \rightarrow e$
Source: Limits - A New Apporach to Real Analysis - Alan F. Beadon - Undergraduate Texts in Mathematics - Springer
In particular, I have already proved the 2nd inequality. However, IDK how to solve the first inequality.
I can show that $\frac{(n+1)^{(n+1)}}{n!}>e^n$, but I don't know the further steps. Any hint? Thanks.
| You can use induction. Check the case $n=2$ by hand. Then
\begin{align*}
\frac{{(n + 1)^{n + 1} }}{{(n + 1)!}} & = \frac{{n^n }}{{n!}}\frac{{(n + 1)^n }}{{n^n }} > \left( {\frac{n}{{n + 1}}} \right)^n \frac{{e^n }}{{n + 1}}\frac{{(n + 1)^n }}{{n^n }} = \frac{{e^n }}{{n + 1}} \\ &= \left( {\frac{{n + 1}}{{n + 2}}} \right)^{n + 1} \frac{{e^{n + 1} }}{{n + 2}}\frac{1}{e}\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2} > \left( {\frac{{n + 1}}{{n + 2}}} \right)^{n + 1} \frac{{e^{n + 1} }}{{n + 2}}.
\end{align*}
In the last step we used $e < \left( {1 + \frac{1}{m}} \right)^{m + 1}$.
| {
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Maximize $z$ over $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$ Suppose that $x$, $y$, and $z$ are real numbers such that $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$, but this doesn't say anything. I also tried some different ways to apply Cauchy, but they all didn't do much.
Thanks in advance!!
| Hint: Since you tried using CS inequality, here is a way:
$$(1+1)\cdot(x^2+y^2)\geqslant (x+y)^2 \implies 2\cdot(6-z^2)\geqslant (3-z)^2 \\ \implies 1+\sqrt2 \geqslant z \geqslant 1-\sqrt2$$
| {
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The number of solutions $(x,y)$ of the congruence $n \equiv X^2-XY+Y^2$ (mod $p^\alpha$). Is there general formula of the solutions of the congruence?
\begin{equation}
n\equiv X^2-XY+Y^2 \pmod r,
\end{equation}
where $n\in\Bbb Z$ and $r\in\Bbb N$.
If we define an arithmetic function (two variables or one variable with fixed $n$)
\begin{equation}
N(n,r):=|\{(x,y)\in\Bbb Z^2:x^2-xy+y^2\equiv n\pmod r,0\leq x,y< r\}|.
\end{equation}
I know that such function is multiplicative in $r$.
I have no approach for this question. Please give me a hint to solve this problem, or let me know that related something. Thanks.
| So far, just an answer when $r=3^a.$
When $r$ is odd, this can be multiplied by $4$ to get $$(2X-Y)^2+3Y^2\equiv 4n\pmod r$$
Then solutions to $Z^2+3Y^2\equiv 4n$ are in $1-1$ correspondence with your congruence.
If $r=3^a,$ then:
*
*If $n\equiv 1\pmod 3,$ there is two solutions for every $Y, $ for $N(n,3^a)=2r$ solutions total.
*If $n\equiv 2\pmod 3,$ then there are no solutions, so $N(n,3^a)=0.$
*What if $n\equiv0\pmod3?$ If $a=1,$ then $N(n,3)=3.$ If $a\geq2,$ then $3\mid Z.$ Then the number of solutions is three times the number of solutions to $$3Z_1^2+Y^2\equiv 4n/3\pmod{3^{a-1}},$$ since there are three $Y$ modulo $3^{a+1}$ for every $Y$ modulo $3^{a}.$ So $N(3n,3^{a+1})=3N(n,3^a)$
So, if $r=3^a,$ $n=3^kn’,$ with $\gcd(n’,3)=1,$ then:
$$N(n,3^a)=\begin{cases}3^a&k\geq a\\2\cdot 3^a&k<a,n’\equiv 1\pmod 3\\0&k<a,n’\equiv 2\pmod 3\end{cases}$$
Not sure what to do with the case $r=2^a.$
Nor yet the case $r=p^a$ when $p>3.$
| {
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prove that the sum of the elements in two subsets is the same
Eight consecutive positive integers are partitioned into two subsets such that the sum of the squares in each subset is the same. Prove that the sum of the elements in each subset is also the same, assuming that the smallest element is at least 56.
The method below is mostly a brute force approach.
Is there a better method than brute force?
For a set $S, $ let $f(S)$ denote the sum of the squares of its elements and let $g(S)$ denote the sum of the elements of S.
Let the integers be $x + 1,\cdots, x + 8$. The sum of the squares in each subset must be $\frac{1}2 \sum_{i=1}^8 (x+i)^2 = \frac{1}2 (8x^2 + 72x + 204) = 4x^2 + 36x+102.$
First we need to show that the partitions must in fact have the same number of elements. If one partition has more than $4$ elements, then its sum of squares will be at least $(x+1)^2 + (x+2)^2 + (x+3)^2 + (x+4)^2 + (x+5)^2 = 5x^2 + 30x + 55,$ which is too large (for $x\ge 11,$ since we would have $x^2 - 6x - 47 > 0$). So no subset can have more than 4 elements, implying than both subsets must have exactly 4 elements.
One of the subsets, say $S_1$, must contain $x+8$, and so the sum of squares of the remaining three elements in this case must be $3x^2 + 20 x +38.$ Let $S_2$ be the other subset. Suppose this subset also contains $x+7$. Then the remaining two elements have a sum of squares equal to $2x^2 + 6x -11.$ The minimum possible sum of squares of two elements is $(x+1)^2 + (x+2)^2 = 2x^2 + 6x + 5$, so we cannot have $x+7$ in $S_1$. If $x+6$ is in $S_1$, then $f(S_1\backslash \{x+8, x+6\}) = 2x^2 +8x+2$. The only two possible remaining elements are $x+1$ and $x+2$, as $(x+1)^2 + (x+3)^2 > 2x^2 + 8x+2$, and is the second smallest sum of squares of elements. But then $f(S_1) < 2X^2 + 8x+2$, a contradiction.
So suppose $x+5\in S_1$. $f(S_1\backslash \{x+8,x+5\} ) = 2x^2 + 10x + 13,$ and in this case one can verify that $S_1 = \{x+8, x+5, x+2, x+3\}$, since all other choices of two remaining elements give either a smaller or larger sum of squares than $2x^2 + 10x+13$. In this case, $g(S_1) = 4x+18$, so $g(S_1) = g(S_2)$.
We know $x+7$ and $x+6$ are in $S_2 $. So we need to find the remaining two elements of $S_2$. If $x+5$ is in $S_2$, then the square of the remaining element in $S_2$ must be $ 4x^2 + 36x + 102 - ((x+7)^2 + (x+6)^2 + (x+5)^2) = x^2 - 8,$ which is clearly impossible.
So the only possibility is $S_1 = \{x+8, x+5, x+2, x+3\}$.
| Here is a counterexample, if the two partition sets don't have to have the same number of elements, using $\{2,3,4,5,6,7,8,9\}$:
$2^2+3^2+4^2+7^2+8^2=142=5^2+6^2+9^2$.
But $2+3+4+7+8=24$ and $5+6+9=20$.
| {
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Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in the numerator and no square in the denominator. I am unsuccesful even in using $\arcsin^2x$ expansion.
However, we know a well known result:
$\displaystyle \tag*{} \sum \limits_{k=1}^{\infty} \frac{\binom{2n}{n}}{k^24^k} = \frac{\pi^2}{6} - 2\ln^2(2)$
So this somehows tells (?) maybe we can decompose our sum into two parts with one being Basel sum. Any help would be appreciated, thanks.
| Note that $$\frac{1+(-1)^k}{2} = \begin{cases}1 &\text{if $k$ is even}\\ 0 &\text{if $k$ is odd}\end{cases}$$
implies that $$\sum_k a_{2k} = \sum_k \frac{1+(-1)^k}{2} a_k.$$ Taking $a_k = \frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}}$ yields
\begin{align}
\sum_{k=1}^\infty \frac{\binom{4k}{2k}}{k^2 16^k}
&= \sum_{k=1}^\infty \frac{1+(-1)^k}{2}\frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}} \\
&= 2\sum_{k=1}^\infty (1+(-1)^k)\frac{\binom{2k}{k}}{k^2 4^k} \\
&= 2\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 4^k} + 2\sum_{k=1}^\infty (-1)^k\frac{\binom{2k}{k}}{k^2 4^k} \\
\end{align}
| {
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$x^4+y^4+z^4=\frac{m}{n}$, find $m+n$. $x^4+y^4+z^4$=$m\over n$
x, y, z are all real numbers,
satisfying $xy+yz+zx=1$ and $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$
m, n are positive integers and their greatest common divisor is 1. Calculate m+n.
My thinkings so far are as follows:
*
*to operate such that we can arrange 5,12,13 into some kind of pythagorean triplets.
*to do whole square of x+1/x or x^2 + 1/x^2 such that
the third term is a constant and can be moved to other side.
| Taking resultants of the polynomials given by
$$
f=5\left(x+\frac{1}{x}\right)-12\left(y+\frac{1}{y}\right),g=5\left(x+\frac{1}{x}\right)-13\left(z+\frac{1}{z}\right), h=xy+yz+zx-1,
$$
set to zero, i.e., with
\begin{align*}
0 & = 5x^2y - 12xy^2 - 12x + 5y,\\
0 & = 5x^2z - 13xz^2 - 13x + 5z,\\
0 & = xy + xz + yz - 1
\end{align*}
we obtain the linear equation $13z=12y+25x$, and then by substituting $z$ and taking resultants again,
$$
(x,y,z)=\left(\frac{1}{5},\frac{2}{3},1\right),\; \left(-\frac{1}{5},-\frac{2}{3},-1\right).
$$
Hence we have
$$
x^4+y^4+z^4=\frac{60706}{50625}.
$$
So we have $(m,n)=(60706,50625)$, so that
$$
m+n=111331.
$$
Note: Over the complex numbers we obtain two additional solutions, namely
$$
(x,y,z)=(i,-i,i),(-i,i,-i).
$$
But then $x^4+y^4+z^4=3$, so that $m+n=4$.
| {
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What is the formula for higher integration of Lambert´s W function? First I will present the notation for higher integration, which I copied from Danya Rose
\begin{align}
J_x^n(f(x)) = \int ...\int f(x) dx^n
\end{align}
Here are the higher integrations of the principial branch of the Lambert W function.
\begin{align}
J_x^1(W(x)) &=\frac{x}{W(x)}\left( W^2(x) -W(x) + 1\right) \\
J_x^2(W(x)) &=\frac{x^2}{W^2(x)}\left(\frac{1}{2} W^3(x) - \frac{3}{4}W^2(x) +\frac{3}{4}W(x) + \frac{1}{8}\right) \\
J_x^3(W(x)) &=\frac{x^3}{W^3(x)}\left(\frac{1}{6} W^4(x) -\frac{11}{36} W^3(x) + \frac{11}{36}W^2(x) +\frac{19}{216}W(x) + \frac{1}{81}\right) \\
J_x^4(W(x)) &=\frac{x^4}{W^4(x)}\left(\frac{1}{24} W^5(x) - \frac{25}{288} W^4(x) +\frac{25}{288} W^3(x) + \frac{115}{3456}W^2(x) +\frac{175}{20736}W(x) + \frac{1}{1024}\right)
\end{align}
I checked the correctness of these expressions on Wolfram|Alpha. Constant terms are not important in this question, so I will not mention them. Then I made a hypothesis
\begin{align}
J^k_x(W(x)) = \frac{x^k}{W^k(x)} \mathcal{J}_{-k}(W(x))
\label{100}
\end{align}
where $\mathcal{J}_{-k}(W(x))$ is $\mathcal{J}$-polynomial.
$\mathcal{J}$-polynomial
Definition
I defined the $\mathcal{J}$-polynomial using $j$-coefficients as
\begin{align}
\mathcal{J}_{-n}(x) = \sum_{k=0}^{n+1}x^k j(n+1 - k, -n )
\end{align}
for $n\geq 1$. I will immediately explain why $-n$ is there. The coefficients that define the $\mathcal{J}$-polynomial are in such an ugly form, because my stupidity defined the coefficients in an inelegant way. $\mathcal{J}$-polynomials also occur in expressions for higher derivates of the principal branch of the Lambert W function. So I will give a complete definition of this polynomial.
\begin{align}
\mathcal{J}_n(x) = \left\{
\begin{array}{ll}
(-1)^{n-1}\sum_{k=0}^{n}x^k j( n - k-1, n) &\mbox{if} \ n\geq 1, n \in \mathbb{Z} \\
1 & \mbox{if} \ n=0 \\
\sum_{k=0}^{|n|+1}x^k j(|n|+1 - k, n) & \mbox{if} \ n \leq -1, n \in \mathbb{Z}
\end{array}
\right.
\end{align}
Coefficients
Here is table of coefficients j(k, n).
(I apologize for the poor quality)
Certain patterns can be noticed from this table
\begin{align}
j(1, n) &= - j(2, n) < 0 & \ \mbox{for} \ n \leq -1\\
j(0, n) &= \frac{1}{|n|!} & \ \mbox{for} \ n \leq -1\\
j(1, n) & = -\frac{H_{|n|}}{|n|!} & \ \mbox{for} \ n \leq -1\\
\end{align}
\begin{align}
j(0, n) &= (n-1)! & \ \mbox{for} \ n \geq 1\\
j(1, n) &= 2(n-1)(n-1)! & \ \mbox{for} \ n \geq 1\\
j(n, n) & =(n-1)^{(n-2)} & \ \mbox{for} \ n \geq 0\\
\mathcal{J}_n(x) &= \frac{(x+1)\mathcal{J}_n^{'}(x) - \mathcal{J}_{n+1}(x)}{1 - n x - 3n} & \ \mbox{for} \ n \geq 2 .
\end{align}
I did not discover the last formula, but found it in this article on page 340. An explicit formula for j-coefficients for $\mathcal{J}$-polynomials of higher derivatives is given in this
paper on page 121.
\begin{align}
j(n-k-1, n) =\sum_{m=0}^k \frac{1}{m!} \binom{2n-1}{k-m} \sum_{q=0}^m \binom{m}{q} (-1)^q (q+n)^{m+n-1}
\end{align}
for $n \geq 1$.
Explicit formula for the $\mathcal{J}$-polynomial
First I used the power expansion of the Lambert W function, which converges for $|x|<\frac{1}{e}$.
\begin{align}
W_0(z) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}x^n
\label{101}
\end{align}
then I took the nth integration of this expansion.
\begin{align}
J_x^n(W_0) = \sum_{k=0}^\infty \frac{(-k)^{k-1}}{k!} \frac{k!x^{k+n}}{(k+n)!} = \sum_{k=0}^\infty \frac{(-k)^{k-1}x^{k+n}}{(k+n)!}
\end{align}
I used the hypothesis I mentioned here
\begin{align}
J^k_x(W(x)) = \frac{x^k}{W^k(x)} \mathcal{J}_{-k}(W(x))
\end{align}
I substituted into the equation and got formula
\begin{align}
\frac{x^n}{W^n(x)} \mathcal{J}_{-n}(W(x)) = \sum_{k=0}^\infty \frac{(-k)^{k-1}x^{k+n}}{(k+n)!}
\end{align}
so
\begin{align}
\mathcal{J}_{-n}(W(x)) = \frac{W^n(x)}{x^n}\sum_{k=0}^\infty \frac{(-k)^{k-1}x^{k+n}}{(k+n)!}
\end{align}
This unfortunately only holds for $|x|<\frac{1}{e}$.
Final formula
The partial formula for higher integration of the Lambert W function is
\begin{align}
J^n_x(W(x)) = \sum_{k=0}^\infty \frac{(-k)^{k-1}x^{k+n}}{(k+n)!}
\end{align}
for $|x|<\frac{1}{e}$.
Questions
What is the formula for j-coefficients $j(k, n)$ with $n \leq -1$?
Is there a formula for the nth integration of the Lambert W function that holds to all real numbers?
Thank you for your help.
| This is not an answer.
First remark
$$I_n=\int \big[W(x)\big]^n \,dx=(-1)^n \Big[\Gamma (n+1,-W(x))-\Gamma (n+2,-W(x))\Big]$$
Second remark
I shall use $W=W(x)$ to have more room.
$$J_n=\int J_{n-1} \,dx \qquad \text{with} \qquad J_1=\frac x{W}\big[W^2-W+1 \big]$$
Define
$$K_n=\frac{W^n \,x^{-n}\, J_n-\,n^{-(n+1)}}{W}=\frac {P_n(W)}{a_n}$$ where tha $a_n$ for the sequence
$$\{1,4,216,20736\}$$ and the first $P_n(W)$ are
$$\left(
\begin{array}{cc}
n & P_n(W) \\
1 & W-1 \\
2 & 2 W^2-3 W+3 \\
3 & 36 W^3-66 W^2+66 W+19 \\
4 & 864 W^4-1800 W^3+1800 W^2+690 W+175
\end{array}
\right)$$
| {
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"question_score": "2",
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"answer_id": 0
} |
Consider the recursively defined sequence $x_{1} =0$ $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=(1/2) + x_{2n}$ I've posted this question before, but there was a typo which messed up my search for a solution
$x_{1} =0$, $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=1/2 + x_{2n}$
1)Which is the limit of the sequences $x_{2n}$ and $x_{2n+1}$?
I thought I could substitute $$\lim_{n\rightarrow\infty}x_{2n-1} = \lim_{n\rightarrow\infty}x_{2n+1}$$ but then I got stuck and quite don't really know what to do
2)Does $x_{n}$ converges? If yes to which point?
Thanks in advance.
| First, you can substitute $x_{2n}$ in $x_{2n+1} = \frac{1}{2} + x_{2n}$ with $x_{2n} = \frac{1}{2} x_{2n-1} $ to get $x_{2n+1} = \frac{1}{2} + \frac{1}{2}x_{2n-1}$.
Define another sequence $y_n = x_{2n+1}$, so $y_0 = x_1 = 0$ and $y_n = x_{2n+1} = \frac{1}{2} + \frac{1}{2} x_{2(n-1) + 1} = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of these reccurent equation is $y_n = 1 - \frac{1}{2^n}$, so $x_{2n+1} = 1 - \frac{1}{2^n}$.
From $x_{2n} = \frac{1}{2} x_{2n-1}$ it's now obvious that $x_{2n} = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2^{n-1}} = \frac{1}{2} - \frac{1}{2^n}$, so
$\lim_{n\to\infty} x_{2n} = \frac{1}{2}$ and $\lim_{n\to\infty} x_{2n+1} = 1$. As these limits aren't equal, the sequence $x_n$ does not converge.
| {
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"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n \to \infty}a_n$ where $a_1=1$ and $a_{n+1}=a_n+\frac{1}{2^na_n}$. There are some attempts as follows.
Obviously, $\{a_n\}$ is increasing. Thus $a_n\ge a_1=1$. Therefore
$$a_{n+1}-a_n=\frac{1}{2^na_n}\le \frac{1}{2^n},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\le a_1+\sum_{k=1}^{n-1}\frac{1}{2^k}=2-\frac{1}{2^{n-1}}.$$
Hence, $a_n\le 2$. Similarily,
$$a_{n+1}-a_n=\frac{1}{2^na_n}\ge \frac{1}{2^{n+1}},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\ge a_1+\sum_{k=1}^{n-1}\frac{1}{2^{k+1}}=\frac{3}{2}-\frac{1}{2^{n}}.$$
Put the both aspects together, we have
$$\frac{3}{2}-\frac{1}{2^n}\le a_n\le 2-\frac{1}{2^{n-1}}.$$
But this does not satisfy the applying conditions of the squeeze theorem. Any other solutions?
| When written under this form : $\quad 2^na_{n+1}a_n=2^n{a_n}^2+1$
We see that we can get rid of the $2^n$ term by setting $\ b_n=\sqrt{2^n}\,a_n\implies \frac 1{\sqrt{2}}b_{n+1}b_n={b_n}^2+1$
And we reduced the study to $$b_{n+1}=\sqrt{2}\left(b_n+\frac 1{b_n}\right)$$
Which according to this question Study convergence of $x_{n+1} = a\left(x_n + {1\over x_n}\right)$, for $x_1 = a$ and $a \in (0, 1)$ is divergent since $\sqrt{2}\ge 1$, which was expected provided $a_n$ has a limit (it is cauchy from $a_{n+1}-a_n<2^{-n}$ Proving a Sequence Converges - Cauchy?).
Maybe you can try your way in finding an asymptotic developpement for $b_n$ from there; but I fear a closed form would be difficult, despite its apparent simplicity it is uncooperative...
| {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Show that $E(X_n)=0$ but a.s. The sum of $\frac{1}{n}(X_1+X_2+…X_n)\rightarrow -1$ as $n\rightarrow\infty$ If $X_1,X_2,…$ is a sequence of independent random variables such that,
\begin{cases}
n^2-1 & \text{with probability } \frac{1}{n^2} \\
-1 & \text{with probability } 1-\frac{1}{n^2}
\end{cases}
How do I show that for every $n$ we have $E(X_n)=0$, but almost surely we have $\frac{X_1+X_2+\dots+X_n}{n}\longrightarrow-1$ as $n\rightarrow\infty$.
This is what I've managed to do.
First, let's calculate the expectation of $X_n$.
\begin{align*}
E(X_n) = (n^2-1)\frac{1}{n^2}+(-1)(1-\frac{1}{n^2}) &= 1-\frac{1}{n^2}-1+\frac{1}{n^2} \\
&= 0
\end{align*}
Given the $P(X_n\in\{n^2-1,-1\})=1$ and $P(X_n=n^2-1)=\frac{1}{n^2}$ and $P(X_n=-1)=1-\frac{1}{n^2}$
\begin{align*}
&\sum^\infty_{n=1}P(X_n=n^2-1) = \sum^\infty_{n=1}\frac{1}{n^2}\rightarrow\frac{\pi^2}{6} \\
&\sum^\infty_{n=1}P(X_n=-1) = \sum^\infty_{n=1}1-\frac{1}{n^2}\rightarrow\infty
\end{align*}
Now, $P(X_n=-1)\longrightarrow1$ as $n\rightarrow\infty$ by Borel Cantelli, since $P(\limsup(X_n=n^2-1))\rightarrow0$. Following this,
\begin{align}
\lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_k)=\lim_{n\rightarrow\infty}\big(\sum^{n^*}_{k=1}\frac{X_k}{n}+\sum^{n}_{k=n^*}\frac{-1}{n} \big) \longrightarrow -1
\end{align}
where, $n^*$ is such that $X_n=-1$ for $n>n^*$ and $n^*>0$.
\
\
As $n\rightarrow\infty, \sum^{n^*}_{k=1}\frac{X_k}{n}\rightarrow0$ and $\sum^{n}_{k=n^*}\frac{-1}{n}=\frac{n^*}{n}-1$. Therefore,
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_n)\longrightarrow -1
\end{align*}
| Use the Borel-Cantelli lemma to conclude that $X_n=-1$ for all but finitely many $n$.
| {
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"source": "stackexchange",
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} |
Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{\sin x} dx = 7~\zeta(3) ,\text { equivalently }\ \int_{0}^{1}\frac{x - x^{2} }{ \sin(\pi x)}dx = 7\frac{\zeta (3)}{\pi^{3}},$$ has been solved here.)
| A more general solution:
I’ll state the following propositions that can be trivially proven using induction, sums of geometric series, second derivatives and the use of $\tan^2 x=\sec^2 x-1$.
For $n\in\mathbb{N}$
$$\sec^{2n}(x)=\frac{(-1)^n \, 2^{2n}}{(2n-1)!} \sum_{k=1}^{\infty} (-1)^k e^{2ixk} \prod_{r=0}^{2n-2} (k-n+r+1)$$
$$\sec^{2n+1}(x)=\frac{(-1)^{n-1} \, 2^{2n+1}}{(2n)!} \sum_{k=1}^{\infty} (-1)^k e^{ix(2k-1)} \prod_{r=0}^{2n-1} (k-n+r)$$
Letting $x\mapsto \frac{\pi}{2}-x$, we recover representations of powers of $\csc x$.
To illustrate an example:
$$\csc^2 (x)= -4\sum_{k=1}^{\infty} (-1)^k k e^{2i\left(\frac{\pi}{2}-x\right)k}$$
Let’s use this to integrate
$$\begin{align} \int_{0}^{\pi} x^2 (\pi-x)^2 \csc^2 (x)\,dx &= -4\sum_{k=1}^{\infty} (-1)^k k \int_{0}^{\pi} x^2(\pi^2-x)^2 e^{2i\left(\frac{\pi}{2}-x\right)k}\,dx\\&=-4\sum_{k=1}^{\infty} (-1)^k k \cdot \left(-\frac{3\pi \, (-1)^k}{2k^4}\right)\\&=6\pi \sum_{k=1}^{\infty}\frac{1}{k^3}\\&=6\pi \zeta(3)\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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} |
find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt{(5y+2)(y-2)}+1$$
then $f(x,y)=x+\frac{1}{2y}=\frac{2xy+1}{2y}=\frac{\sqrt{(5y+2)(y-2)}+2}{2y}=g(y)$,$$\therefore g'(y) = \frac{2 + 2 y - \sqrt{-4 - 8 y + 5 y^2}}{y^2 \sqrt{-4 - 8 y + 5 y^2}}$$
and $2+2y = \sqrt{5y^2-8y-4} \Rightarrow y = 8+6\sqrt{2}$, with the help of graphing calculator ,i know this is the maximum point. $$\therefore f_{\text{max}}(x,y)=g(8+6\sqrt{2})=\frac{3\sqrt{2}}{2}-1$$
Is there any other way to work out it?
| there is another way to solve it, I copy it from internet, but I don't know how to find this method.
first to deformation the algebraic expression: $$(2xy-1)^2=(5y+2)(y-2)\iff (2xy-1)^2 = 9y^2-4y^2-8y-4\iff (2xy-1)^2+(2y+2)^2=9y^2$$
then we have $$\left(2x-\frac{1}{y}\right)^2+\left(2+\frac{1}{y}\right)^2=9$$
and $$x+\frac{1}{2y}=\frac{1}{2}\left(2x+\frac{1}{y}\right)=\frac{1}{2}\left(2x-\frac{1}{y}+\frac{2}{y}+2\right)-1 \leq \frac{1}{2}\sqrt{2\left[\left(2x-\frac{1}{y}\right)^2+\left(\frac{2}{y}+2\right)^2\right]}-1=\frac{1}{2}\sqrt{2\times 9}-1=\frac{3\sqrt{2}}{2}-1$$
| {
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"source": "stackexchange",
"question_score": "2",
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Maclaurin series of $1- \cos^{2/3} x$ has all coefficients positive Experimenting with WA I noticed that the function $1- \cos^{\frac{2}{3}}x$ has the Maclaurin expansion with all coefficients positive ( works for any exponent in $[0, \frac{2}{3}]$). A trivial conclusion from this is $|\cos x|\le 1$, but it implies more than that, for instance see this. Maybe some ''natural'' proofs are available. Thank you for your interest!
Note: an attempt used a differential equation satisfied by the function. But the answer by @metamorphy just solved it the right way.
$\bf{Added:}$ Some comments about series with positive coefficients.
By $P$ we denote a series with positive coefficients ( no free term),
*
*If $a>0$ then $\frac{1}{(1-P)^a} = 1+P$ (moreover, the positive expression on RHS is a polynomial in $a$ with positive coefficients
*If $0<a < 1$ then $(1-P)^a = 1-P$. Similarly the expressions for $a = \frac{t}{t+1}$ are positive in $t$.
2'. If $1-f= P$ then $1- f^{a} =P$ for any $0 < a < 1$, and similar with above.
$\bf{Added:}$ It turns out that the function $\cos^{2/3} x$ has a continued fraction (an $S$-fraction, from Stieltjes) that is "positive" ( similar to the continued fraction for $\tan x$). This is a stronger statement than the one before. Maybe there is some approach using hypergeometric functions.
| Start with the positive series
$$\csc x = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{15120} x^5 + \cdots $$
Take the square and obtain the positive series
$$\csc^2 x =\frac{1}{x^2} + \frac{1}{3} + \frac{1}{15} x^2 + \frac{2}{189} x^4 + \cdots $$
Now take the derivative and get the series
$$- \frac{2\cos x}{\sin^3 x} = -\frac{2}{x^3} + \frac{2}{15} x + \frac{8}{189} x^3 + \cdots $$
where all the coefficients after the first are positive. We conclude that the series
$$ 1 - \frac{x^3 \cos x}{ \sin ^3 x}=\frac{1}{15}x^4 + \frac{4}{189}x^6 + \frac{1}{225}x^8 + \cdots $$
is positive. Now, since $(1-t)^{-1/3}$ is a positive series in $t$, if we substitute for $t$ a positive series in $x$ without free term we get again a positive series in $x$. We conclude that
$$\frac{\sin x}{x \cos^{1/3} x}= 1 + \frac{1}{45}x^4 + \frac{4}{567}x^6 + \frac{1}{405} x^8+ \cdots $$
is a positive series. Now multiply by $\frac{2}{3} x$ and integrate and get the positive $1- \cos^{2/3} x$ ( all coefficients are, except perhaps the free term, which turns out to be $0$).
$\bf{Note:}$ In fact we can prove that the series for
$$\sin x - \frac{x^3 \cos x}{ \sin^2 x}$$
is positive. For this, we start with the expansion
$$\csc x= x^{-1} + \frac{1}{6} x + \frac{7}{360}x^3 + \frac{31}{15120} x^5 + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2(2^{2n-1} -1) B_{2n}}{(2n)!}x^{2n-1}$$
Taking the derivative with respect to $x$ we get the expansion of $\frac{\cos x}{\sin^2 x}$. From here we conclude that $\sin x- \frac{x^3 \cos x}{\sin^2 x}$ is positive. Multiplying by the positive $\csc x$ we conclude that $1 - \frac{x^3 \cos x}{\sin^3 x}$ is positive.
| {
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"question_score": "17",
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Problem regarding the distance between a circle and a parabola The question is this, copy pasted here:
Consider the circle $C$ whose equation is
$$(x - 2) ^ 2 + (y - 8) ^ 2 = 1$$
and the parabola $P$ with the equation
$$y^ 2 =4x.$$
I have no idea how to model the question, I tried showing that the $x-$coordinates of the two points have to be the same but that seems to be wrong too? Can anyone help?
| Let $(x,y) $ be on the parbola, then
$ y^2 = 4 x$
The normal vector is $[ 4, -2 y] $ , while the vector from $(2, 8)$ to $(x,y)$ is $[ x - 2, y - 8 ]$, and we want these two vectors to be aligned; so we can use the determinant and set it equal to zero, namely,
$4 (y-8) + 2 y (x - 2) = 0 $
We have to solve the above equation together with $ y^2 = 4 x $
Hence,
$ 4 ( y - 8) + 2 y (y^2/4 - 2) = 0 $
From which
$ -64 + y^3 = 0$
Whose solution is $ y = 4 $, and therefore, $ x = y^2 /4 = 4 $
This means that the shortest distance between the circle and the parabola is
$ d = \sqrt{ (2 - 4)^2 + (8 - 4)^2 } - 1 = \sqrt{20} - 1 = 2 \sqrt{5} - 1 $
| {
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"source": "stackexchange",
"question_score": "1",
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What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be minimum, it gave me an absurd relation like $-6=6$.
How can we solve this? Why is the differentiation method not working here? Please help !!!
$f(x)= \frac{x^2+3x-6}{x^2+3x+6}$
$f'(x)= \frac{(x^2+3x+6)(2x+3)-(x^2+3x-6)(2x+3)}{(x^2+3x+6)^2}$
Then equated this to $0$ to find the value of $x$.
Thanks in advance !!!
| Notice that we have $2x+3=0$ and you should check that $x=-\frac32$ is a minimum.
$$f'(x) = \frac{12(2x+3)}{(x^2+3x+6)^2}$$ shows that $x=-\frac32$ is the global minimum as the gradient is negative for $x<-\frac32$ and then positive for $x>-\frac32$.
| {
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$a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$ sequence
Sequence $\{a_n\},n\in\mathbb N_+$ with all terms positive integers satisfy $a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$. Find $\{a_n\}$.
I suppose all terms of $\{a_n\}$ are $1$. This problem makes me think of a lot of conclusions, including
*
*$n\in\mathbb{Z}_+$ can be written as the sum of two squares as long as for every prime $p\equiv3\pmod4$ there's $2\mid V_p(n)$.
*$(m^2+n^2)^2=(m^2-n^2)^2+(2mn)^2$.
*$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2$.
Perhaps we can let the first non-$1$ term of the sequence be $a_s$ and derive a contradiction?
| We claim that all $a_{n}$ are $1$, we will use proof by contradiction to make our claim clear.
Suppose $\{ a_{n}\}$ is some sequence which is not the trivial sequence composed of $1$'s and satisfies your requirements.
$\textbf{Let}$ $v$ $\textbf{be the smallest positive integer so that}$ $a_{v} \neq 1$. Note that $a_{1} = 1$ as $a_{1} = a_{1}^2$ and $a_{1} \geq 1$, and $a_{2} = a_{1}^2 = 1.$ Thus $v \geq 3$. Note that $v$ is $\textbf{not}$ a sum of two positive squares, because if we have $v = c^2 +d^2$ for $c,d \in \mathbb{N}_{+}$ then $a_{v} = a_{c}a_{d} = 1$. $\textbf{Nor}$ is $v$ a square as if $v = j^2$ then $a_{v} = a_{j} = 1$.
$\textbf{Lemma:}$ For each $w \in \mathbb{N}$, there exists $\alpha_{w} \in \{1,3,5\}$ so that $\frac{w-5\alpha_{w}}{3}$ and $\frac{w+4\alpha_{w}}{3}$ are integers and
$w^2+(\frac{w-5\alpha_{w}}{3})^2 = (w-\alpha_{w})^2+(\frac{w+4\alpha_{w}}{3})^2$
In the above lemma note that when $w \geq 10$ we have $|\frac{w-5\alpha_{w}}{3}|\leq w$, $0 < (w-\alpha_{w}) < w,$ and $\frac{w+4\alpha_{w}}{3}$.
Due to the above lemma, if $v \geq 11$ then there are two cases;
$\textbf{Case 1:}$ $\frac{v-5\alpha_{v}}{3} = 0$,
In this case one has that $v^2 = (v-\alpha_{v})^2+(\frac{v+4\alpha_{v}}{3})^2$.
Thus $$a_{v^2} = a_{v-\alpha_{v}}a_{\frac{v+4\alpha_{v}}{3}} = 1,$$ but $a_{v^2} = a_{v} > 1$ which is impossible. Hence this case is impossible.
$\textbf{Case 2:}$ $\frac{v-5\alpha_{v}}{3} \neq 0$.
In this case, note that $$a_{v^2+(\frac{v-5\alpha_{v}}{3})^2} = a_{v}a_{\frac{v-5\alpha_{v}}{3}} = a_{v} > 1$$
but also
$$a_{v^2+(\frac{v-5\alpha_{v}}{3})^2} = a_{(v-\alpha_{v})^2+(\frac{v+4\alpha_{v}}{3})^2} = 1$$
which is impossible. Thus this case is impossible.
Hence we know that $3 \leq v \leq 10$. Now it is time for brute force...
$v \neq 3$ as $a_{9+16} = a_{3}a_{4} = a_{3}a_{2} = a_{3}$ and $a_{25} = a_{5} = a_{1+4} = a_{1}a_{2} = 1$ thus $a_{3} = 1$
$v \neq 4$ as $4$ is a square (see above)
$v \neq 5$ as $5 = 1+2^2$ (see above)
$v \neq 6$. To see this note that $6^2 +8^2 = 100$. Thus $a_{100} = a_{6}a_{8}=a_{6}(a_{2})^2 = a_{6}$ and $a_{100} = a_{10} = a_{9+1} = a_{3}a_{1} = 1$. Thus $a_{6} = a_{1}$.
$v \neq 7$. To see this note that $7^2+1^2 = 50 = 5^2+5^2$. Thus $$1 = a_{5}^2 = a_{50} = a_{7}$$
$v \neq 8$ as $8 = 4+4$
$v \neq 9$ as $9$ is a square.
$v \neq 10$ as $10 = 9+1$.
$\textbf{Thus there is no smallest integer}$ $v$ $\textbf{such that}$ $a_{v} \neq 1$. This is a contradiction towards our assumption.
| {
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Number of solution of equation $16 \sin^3x=14+(\sin x+7)^{\frac{1}{3}}$ in $[0,4\pi]$. Number of solutions of equation $16 \sin^3x=14+(\sin x+7)^{\dfrac{1}{3}}$ in $[0,4\pi]$.
My thinking:
*
*Directly satisfy $\sin x=1$.
*I thought of doing $f(x)=f^{-1}(x)\;$ but couldn't proceed further because I can't think of a function on the left side whose inverse is the right side.
| Let $y=\sin x$. Then $$16(y^3-1)=\sqrt[3]{y+7}-2$$
$$16(y^3-1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=y-1$$
Then $y=1$ or $$16(y^2+y+1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$
$$4(4y^2+4y+4)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$
$$4\left((2y+1)^2+3\right)\left(\left(\sqrt[3]{y+7}+1\right)^2+3\right)=1$$
But $$4\left((2y+1)^2+3\right)\left(\left(\sqrt[3]{y+7}+1\right)^2+3\right) \ge 4 \cdot 3 \cdot 3= 36 >1$$.
Only solution $y=1$
| {
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Asymptotic behaviour of $\frac{n^n}{n^n - (n - 1)^n + 1}$ for large $n$ I am interested in the behaviour when $n$ is large of the following function:
$$f(n) := \frac{n^n}{n^n - (n - 1)^n + 1}.$$
The limit of this function as $n$ approaches infinity is
$$\lim_{n \to \infty} f(n) = \frac{e}{e - 1},$$
where $e$ is Napier's Constant.
However, I would like to have more information about this function, such as a series expansion at $n = \infty$, where the dominant term is $e/(e - 1)$, and there is an explicit error term that is $o(1)$ as $n \to \infty$.
Wolfram Alpha doesn't want to give any expansion. Mathematica gives some expansion, but it is not immediate from the expression even that the limit is $e/(e - 1)$.
| Here are the details of Greg Martin's answer.
First,
$$f(n) = \frac{1}{1 - (1 - \frac{1}{n})^n + n^{-n}}.$$
Since
$$\log(1 - \frac{1}{n})^n = n\log(1 - \frac{1}{n}) = n(-\frac{1}{n} - \frac{1}{2} \frac{1}{n^2} + O(\frac{1}{n^3})) = - 1 - \frac{1}{2}\frac{1}{n} + O(\frac{1}{n^2}),$$
exponentiating gives
$$(1 - \frac{1}{n})^n = e^{-1 - \frac{1}{2}\frac{1}{n} + O(\frac{1}{n^2})} = \frac{1}{e} -\frac{1}{2e} \frac{1}{n} + O(\frac{1}{n^2}).$$
Hence, as $n^{-n} = O(1/n^2)$,
$$(1 - \frac{1}{n})^n - n^{-n} = \frac{1}{e} -\frac{1}{2e} \frac{1}{n} + O(\frac{1}{n^2}),$$
also.
Therefore
\begin{align}
f(n) &= \frac{1}{1 - ((1 - \frac{1}{n})^n - n^{-n})} = \frac{1}{1 - (\frac{1}{e} -\frac{1}{2e} \frac{1}{n} + O(\frac{1}{n^2}))} = \frac{e}{e - 1 - \frac{1}{2}\frac{1}{n} + O(\frac{1}{n^2})} \\
&= \frac{e}{e - 1}\left(\frac{1}{1 - \frac{1}{2(e - 1)}\frac{1}{n} + O(\frac{1}{n^2})}\right) = \frac{e}{e - 1}(1 - \frac{1}{2(e - 1)} \frac{1}{n} + O(\frac{1}{n^2})).
\end{align}
| {
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
~
EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
| $ 5x + 3y = 100 $
Has the solution $ x = 20 - 3 t $ and $ y = 5 t $ where $ t= 1, 2, \dots $
Maximum $t $ is $\text{int}(\frac{20}{3}) = 6 $
$xy = 5 t (20 - 3 t) $
Its peak is at $\frac{1}{2} ( 0 + \frac{20}{3} ) = \frac{10}{3} $
Since this not an integer, the maximum is attained at $t = 3$ (because it closer to $\frac{10}{3}$ than $4$). Hence, the maximum is
$ \max(xy) = 15(11) = 165 $
| {
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Possible growth rates of a matrix entry with respect to exponentiation Let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, so $A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$. Thus, $(A^n)_{1,1} = 1 = \Theta(1)$, and $(A^n)_{1,2} = n = \Theta(n)$.
Given a constant $c$, if $B = \begin{pmatrix} c/2 & c/2 \\ c/2 & c/2\end{pmatrix}$ then $B^n = \begin{pmatrix} c^n/2 & c^n/2 \\ c^n/2 & c^n/2\end{pmatrix}$. Thus, $(B^n)_{1,1} = c^n/2 = \Theta(c^n)$.
We have observed three possible growth rates of a particular entry with respect to the exponent of a matrix, constant, linear, and exponential. What other growth rates are possible? For example, does there exist an $m\times m$ matrix $C$ and indices $i,j$ such that $(C^n)_{i,j} = \Theta(n^2)$?
| Due to the Cayley-Hamilton theorem, matrix elements with respect to exponentiation adhere to the linear recurrence defined by the characteristic polynomial of a matrix. And a linear recurrence $A_0, A_1, \dots$ with the characteristic polynomial
$$
P(x) = \prod\limits_{i=1}^k (x-x_i)^{d_i},
$$
has a generic solution
$$
A_n = \sum\limits_{i=1}^k C_i(n) x_i^n,
$$
where $x_1 < x_2 < \dots < x_k$ are distinct roots of $P(x)$ and $C_i(x)$ are polynomials of degree at most $d_i-1$.
With such characteristic polynomial, the asymptotics of $A_n$ is estimated as $A_n \in O(n^{d_i-1} x_k^n)$.
So, all possible growth rates have a form of $\Theta(n^d c^n)$, the simplest possible example to such growth rate is the Jordan cell that has a size $d+1$ and a characteristic polynomial $(x-c)^{d+1}$.
$$
J_{c,{d+1}} = \begin{pmatrix}
c & 1 & 0 & \cdots & 0 \\
0 & c & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & c & 1 \\
0 & 0 & 0 & 0 & c
\end{pmatrix}.
$$
Another common example is the companion matrix of the polynomial $(x-c)^{d+1} = x^{d+1}+\sum\limits_{i=0}^{d} a_i x^i$:
$$
C((x-c)^{d+1})=\begin{pmatrix}
0 & 0 & \dots & 0 & -a_0 \\
1 & 0 & \dots & 0 & -a_1 \\
0 & 1 & \dots & 0 & -a_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & -a_{d}
\end{pmatrix}.
$$
| {
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How can I calculate the area of a circle centered at (2,2) with radius 2 using double integrals
This is what the graph loosk like.
Obviously I know $\pi r^2$ but if I specifically wanted to find the area using double integrals in polar coordinates, how would I go about it?
My guess is that $\theta$ goes from $0$ to $\pi/2$
and for $r$ I need to do:
$(x-2)^2+(y-2)^2=4$
$x^2-4y+4+y^2-4y+4=4$
$(x^2+y^2)-4(x+y)=-4$
$r^2-4(r\cos\theta+r\sin\theta)=-4$
$r^2=4(r\cos\theta+r\sin\theta)-4$
Not exactly sure what to do from here.
| After getting $r^2=4(r\cos\theta+r\sin\theta)-4$ you can solve for $r$ to get $r_{1,2}(\theta) = 2\left(\sin\theta+\cos\theta\pm\sqrt{2\sin\theta\cos\theta}\right)$
The two solutions are the two intersections of the ray from the origin with the circle:
When using polar coordinates, the area can be obtained as $\frac{1}{2}\int_a^b r^2(\theta)\;d\theta$
So in this case you'll need to calculate
\begin{align}
&\frac{1}{2}\int_0^{\pi/2}r_1^2(\theta)\;d\theta - \frac{1}{2}\int_0^{\pi/2}r_2^2(\theta)\;d\theta \\
& = \frac{1}{2}\int_0^{\pi/2}(r_1^2(\theta)-r_2^2(\theta))\;d\theta\\
& = 8 \int_0^{\pi/2}(\sin\theta + \cos\theta)\sqrt{2\sin\theta\cos\theta}\;d\theta
\end{align}
which, after some calculations, gives $4\pi$.
| {
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Find number of points of discontinuity of $f(x)=\frac x5-\left\lfloor \frac x5\right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$ for $x\in[0,100]$ Find number of points of discontinuity of $$f(x)=\frac{x}{5}-\left\lfloor \frac{x}{5} \right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$$ for $x\in[0,100]$ .
My Attempt
The floor function is discontinuous for all integers .Now $\frac{x}{5}$ becomes integers at $21$ values of $x$ for $x\in[0,100]$ and $\frac{x}{2}$ becomes integer at $51$ values of $x$ for $x\in[0,100]$. So $f(x)$ should be discontinuous at $72$ values of $x$. But here the values of $x$ which are multiples of $10$ get counted twice so actually the function becomes discontinuous at $61$ values of $x$.
Given Answer
But in the answer given it was said that $f(x)$ is continuous at such values of $x$ which are multiples of $10$.I checked and found it to be true. So the function can be said to be discontinuous at $61-11=50$ points.
My Issue
How is one supposed to know for sure that $f(x)$ will be continuous at such $x$ which are multiples of $10$
| $\left\lfloor \frac{x}{5}\right\rfloor$ is continuous in the multiples of $5$.
$\left\lfloor \frac{x}{2}\right\rfloor$ is continuous in the multiples of $2$. So $f$ is continuous in the integers that are simultaneously multiples of $5$ and $2$, since $\gcd(2,5)=1$, those are the multiples of $\operatorname{lcm}(2,5)=10$.
| {
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Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$.
Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$.
Firstly if $k$ is even, then as $k=2t$ for $t \in \Bbb Z$ and so $x^k=(x^t)^2$ and $y^k=(y^t)^2$. These are both perfect squares and since $3^n \mid x^k +y^k$ we have that $3 \mid x^k +y^k \implies 3 \mid(x^t)^2+(y^t)^2$ but if $3 \mid a^2 +b^2$, then $3 \mid a$ and $3 \mid b$ so $3 \mid x$ and $3 \mid y$ implying that the $\gcd(x,y) \ne 1$ which is a contradiction.
Thus we can conclude that $k$ is odd. Now for odd $k$ and prime $p$ dividing $x+y$ we have that $$v_p(3^n)=v_p(x^k+y^k)=v_p(x+y)+v_p(k)$$ however as $p \mid x+y$ we have that $1=v_p(p) \le v_p(x+y)$ thus $$v_p(3^n) \ge1>0$$ so $p \mid 3 \implies p=3$.
We then have that $$v_3(3^n)=n = v_3(x+y)+v_3(k)$$ which implies that $$3^n=3^{v_3(x+y)}\cdot 3^{v_3(k)} = x^k+y^k.$$
I couldn't proceed further from here, but the solution I read stated that $$x^k+y^k = 3^{v_3(x+y)}\cdot 3^{v_3(k)} = \color{red}{(x+y)k}$$ and I have no idea where the rhs of the equality comes from?
Edit I think one has that from $n = v_3(3^n)=v_3(x+y)+v_3(k)$ we get $$3^n=3^{v_3(x+y)}\cdot 3^{v_3(k)} = x^k+y^k=(x+y)(x^{k-1}-x^{k-2}y+ \dots + y^{k-1})$$ so either $$3^{v_3(x+y)}=x+y, 3^{v_3(k)}=x^{k-1}-x^{k-2}y+ \dots + y^{k-1}$$ or vice versa.
| EDIT: It appears I read too fast and OP wasn't asking for a solution, but rather an explanation of an already given solution. Nevertheless, I think my solution is pretty neat and clean, so I'll leave it up.
As you've shown, $k$ must be odd. Let $p\mid k$ be prime, then
$$\frac{x^k+y^k}{x^p+y^p}=\frac{x^k-(-y)^k}{x^p-(-y)^p}\in\mathbb{Z}.$$
Hence, $x^p+y^p$ must be a power of three as well. Write
$$3^m=x^p+y^p$$
Now, note that $y$ is invertible modulo $3^m$, so
$$(x/y)^p\equiv -1\pmod {3^m}.$$
Now, $(\mathbb{Z}/3^m\mathbb{Z})^\times$ is cyclic of order $2\cdot 3^{m-1}$ and $p$ is odd. So if $p\neq 3$, we must have $x/y\equiv -1\pmod {3^m}$ and
$$x^p+y^p=3^m\mid x+y,$$
whence $x+y\ge x^p+y^p$. It follows that $x=y=1$, but this doesn't give a solution. We conclude that $p=3$ and
$$3^m=x^3+y^3.$$
Note that $x+y\mid x^3+y^3$, so there exists some positive integer $\ell$ satisfying
$$3^\ell=x+y\quad\text{and}\quad 3^{m-\ell}=\frac{x^3+y^3}{x+y}=x^2-xy+y^2.$$
This implies that
$$3^{m-\ell}=(x+y)^2-3xy=3^{2\ell}-3xy\equiv 3\pmod 9,$$
so $m-\ell=1$ and $3=x^2-xy+y^2\ge 2xy-xy=xy$. Since $3\nmid xy$, this gives only the options
$$(x,y)=(1,1),\quad (x,y)=(2,1),\quad (x,y)=(1,2).$$
Indeed we find that those last two are solutions.
| {
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Usage of Zsigmondy's theorem
Find the solutions for the equation $$3^x-5^y=z^2$$ where $x,y,z$ are positive integers.
We find one solution $(x,y,z)=(2,1,2)$ by checking small cases. Initially looking at this modulo $4$ one gets that $$(-1)^x-1^y \equiv 0,1 \pmod 4$$ as squares leave residues $0$ and $1$ modulo $4$. This in turn implies that $x$ must be even that is $x=2k$ for integer $k$.
We then have that $$\begin{align}3^{2k}-z^2&=5^y \\ (3^k-z)(3^k+z) &= 5^y \end{align}$$ thus $$\begin{align} 3^k-z &= 5^a \\ 3^k+z &= 5^b \end{align}$$ for such that $b>a\ge1$ and $a+b=y$.
Addin these two equations we eliminate $z$ and get that $$2\cdot 3^k=5^a+5^b =5^a(5^{b-a}+1)$$ the lhs is a multiple of $2$ and the rhs is a multiple of $5$ this makes $5^a=1$ and $a=0$.
Now apparently this is the place where I could apply Zsigmondy's theorem, but I don't know how. Could someone englighten me how to use this theorem? It states that
Let $a>b\geq1$ be coprime integers, and let $n\geq 2$. Then there exists a prime divisor of $a^n-b^n$ that does not divide $a^k-b^k$ for all $1\leq k< n$, except when: $n=2$, and $a+b$ is a power of $2$; or $(a,b,n)=(2,1,6).$
but I don't see how to put the pieces together.
Edit: I think that since $$2\cdot 3^k=5^a+5^b =5^a(5^{b-a}+1)$$ and the fact that $a$ must be $0$ we get the result $b=y$ from $a+b=y$ and so $$2 \cdot 3^k = 5^y +1$$ now due to $\gcd(5,1) =1$ and $5>1$ we can use Zsigmondy's for $y \ge 2$ and we have the existence of prime $p$ for which $p \mid 5^y +1^y$, but $p \nmid 5+1$ i.e $p \nmid 6$. I still don't see how this implies the result?
| There's 2 forms of Zsigmondy's theorem, one for forms $a^n-b^n$ and one for forms $a^n+b^n$. Your RHS is in the second form.
| {
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Find if the integral $ \int_{0}^{\infty}\frac{\tan^{-1}(x)}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$ converges without solving it I'm trying to solve this integral, but without any success.
Is there a way to tell if the integral is convergent or divergent without actually solving it? Am I missing something?
$$\int_{0}^{\infty} \dfrac{\arctan x}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$$
| We want to show $$\int_{0}^{\infty} \dfrac{\arctan x}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$$ converges. We'll consider an initial and final segment separately and bound them both.
Let's look at the factors. For large $x$ we have
$$\arctan(x) \le \pi/2 \qquad \dfrac{1}{(2x+1) } \simeq \frac{1}{2x} \qquad \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}} \simeq \frac{1}{x^{4/3}}$$
. . . since in the last factor the $x^4$ dominates.
Make the above formal and use it to compare some final segment of the integral to
$$\int_{x}^{\infty} \dfrac{\pi/2}{(2x )x^{4/3}}\,dx = \frac{\pi}{2}\int_{x}^{\infty} \dfrac{dx}{ x^{7/3}} $$
For small $x$ we have
$$\arctan(x) \simeq x -\frac{x^3}{3}+\frac{x^5}{5} - \ldots \qquad \dfrac{1}{(2x+1) } \simeq \frac{1}{2} \qquad \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}} \simeq \dfrac{1}{ ( 9x^2)^{1/3}}$$
. . . since in the last factor the $x^2$ term dominates.
Use the above similar to before to bound some integral over an initial segment.
Now consider the integral near $3$ where it blows up:
This time write $ x^4-6x^3+9x^2= x^2 (x-3)^2 $ and so
$$ \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}} = \frac{1}{x^{2/3} (x-3)^{2/3}}$$
Since everything else is bounded you only need to show the integral of $\frac{1}{ (x-3)^{2/3}}$ around $3$ converges.
We conclude the integrals at $0,3$ and $\infty$ all converge. Since the function is continuous everywhere else it is bounded over the remaining domain hence has finite integral.
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"url": "https://math.stackexchange.com/questions/4476374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x$? By my post, I had found that
$$
\begin{aligned}
\frac{\pi^{3}}{8}&= \int_{0}^{\infty} \frac{\ln ^{2} x}{x^{2}+1} d x \\&\stackrel{x\mapsto\tan x}{=} \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x \\
&= \int_{0}^{\frac{\pi}{2}}[\ln (\sin x)-\ln (\cos x)]^{2} d x \\
&= \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin x) d x-2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x +\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) d x \\
&=2\left[\frac{\pi^{2}}{24}\left(\pi^{2}+3 \ln ^{2} 4\right)\right]-2 I,
\end{aligned}
$$
the last line comes from the post.
Rearranging yields the result
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x \\
&=\frac{1}{2}\left(\frac{\pi^{3}}{12}+\pi \ln^{2} 2-\frac{\pi^{3}}{8}\right) \\
&=-\frac{\pi^{3}}{48}+\frac{\pi}{2} \ln ^{2} 2
\end{aligned}
$$
Is there any other solution? Comments and solutions are highly appreciated.
| If you are familiar with beta function and feymman's trick then
$$ B(x,y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt $$
$$ \frac{\partial }{\partial x} B(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$
$$ \frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} B(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt $$
and we know that $$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$
so differentiate and put $ x =\frac{1}{2} , y = \frac{1}{2} $
$$\frac{\partial^2 B}{\partial x\partial y}=(\psi(x)-\psi(x+y))(\psi(y)-\psi(x+y))B(x,y)+B(x,y)(-\psi'(x+y)))$$
$$ \psi \left(\frac{1}{2} \right) = -2\ln 2 - \gamma $$
$$ \psi(1) = -\gamma $$
$$ \psi^{(1)}(1) = \frac{\pi^2}{6} $$
$$ B\left( \frac{1}{2} , \frac{1}{2} \right) = \frac{\Gamma \left( \frac{1}{2} \right)^2}{\Gamma(1)} = \pi $$
thus you'll have the integral $ = \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!},
\end{align*}
I managed to get
$\displaystyle \lim _{x\to 0}\frac{\cos x-1+\dfrac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}=\lim _{x\to 0}\frac{\left(\dfrac{\left(-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}\right)}{x^{4}}+\dfrac{x^{2}-\frac{x^{4}}{3!}}{2x^{4}}\right)}{\left(\dfrac{\ln\left(1+x\right)-x}{x}+1\right)^{4}}$
but I can't see how to use that $\displaystyle \lim _{x\to 0}\frac{\ln\left(1+x\right)-x}{x}=0$ in here.
Am I missing something? I didn't learn the little-$\mathcal{O}$ notation yet.
Thanks
| You can use
$$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1 $$
to handle the limit. In fact
\begin{eqnarray}
&&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}\\
&=&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{x^4}\bigg[\frac{x}{\ln \left(x+1\right)}\bigg]^4\\
&=&\lim _{x\to 0}\frac{-\sin x+\frac{1}{2} \sin x+\frac{x}{2}\cos x}{4x^3}\\
&=&\lim _{x\to 0}\frac{-\frac{1}{2} \cos x+\frac{1}{2}\cos x-\frac{x}{2}\sin x}{12x^2}\\
&=&\lim _{x\to 0}\frac{-\sin x}{24x}\\
&=&-\frac{1}{24}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$ On $[a,b]$ function $f$ is differentiable for arbitrary order, and $f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$.
I first try Taylor expansion at $x_0=\frac{a+b}{2}$, and drop higher order term ($(x-x_0)^3$ terms). But this works only locally at the neighborhood of $x_0$. How to prove this inequality on a finite interval? Thank you.
| Let $L(x) = f'\left(\frac{a+b}{2}\right)\left(x - \frac{a+b}{2}\right) + f\left(\frac{a+b}{2}\right)$. Then, $L$ describes the line tangent to $f$ at $x=\frac{a+b}{2}$. Note that:
$$\int_{a}^{b}L(x)\ dx = (b-a)f\left(\frac{a+b}{2}\right)$$
Now, consider $g(x) = f(x)-L(x)$. Note that $g''(x) > 0$ because $L''(x)=0$ and $f''(x)>0$. Additionally, note that $ g\left(\frac{a+b}{2}\right) = g'\left(\frac{a+b}{2}\right) = 0$.
Note that $g'(x)$ is an increasing function because $g''(x)>0$. Then, since $g'\left(\frac{a+b}{2}\right) = 0$, we must have $g'(x) > 0$ for $x>\frac{a+b}{2}$. Repeating this argument for $g(x)$, we can see that $g(x)>0$ for $x>\frac{a+b}{2}$.
Similarly, since $g'(x)$ is increasing and $g'\left(\frac{a+b}{2}\right) = 0$, we must have $g'(x)<0$ for $x<\frac{a+b}{2}$. This shows that $g(x)$ is decreasing for $x<\frac{a+b}{2}$. Combining this with $g\left(\frac{a+b}{2}\right) = 0$, we see that $g(x) > 0$ for $x<\frac{a+b}{2}$.
We can conclude that $g(x) > 0$ for all $x\neq \frac{a+b}{2}$. Thus:
$$\int_{a}^{b}g(x)\ dx = \int_{a}^{b}f(x) - L(x)\ dx > 0$$
$$\int_{a}^{b}f(x)\ dx > \int_{a}^{b}L(x)\ dx$$
$$\boxed{\int_{a}^{b}f(x)\ dx > (b-a)f\left(\frac{a+b}{2}\right)}$$
An intuitive but less rigorous way to understand this is that $f(x)$ is always above $L(x)$ because convex functions (defined by $f''(x)>0$) "curve" upwards from their tangent lines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding the roots of $x^5+x^2-9x+3$ I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$
I can take it from here employing the quadratic and cubic formula, but how can I get the same answer w/o using any software?
| Here's one way to think about it:
$$\begin{align}x^5+x^2-9x + 3 &= x^5-9x+x^2+3\\ &= x(x^4-9)+x^2+3 \\ &= x(x^2-3)(x^2+3)+x^2+3 \\ &= (x^2+3)(x(x^2-3)+1) \\ &= (x^2+3)(x^3-3x+1). \end{align}$$
In words, the trick is to recognize that $x^5-9x$ is "almost" a difference of squares (i.e., up to the factor of $x$ multiplying it). Then everything falls into place.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.0951875 ... $,
and its possible closed form is $ - \frac18\ln2\ln3$. But I don't know how to do it.
Thanks in advance.
| Firstly, recognise the "symmetry" in the integral
$$I = \int_{0}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln \left(\frac{\ln(1/2+x)}{\ln(1/2-x)}\right)\, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln\left(-\ln\left(\frac{1}{2}+x\right)\right)\,dx$$
Now enforce the substitution $-\ln(1/2+x)=u$:
$$\implies I = \frac{1}{8} \int_{0}^{\infty} \frac{\ln(u) (2-e^u)}{e^{2u}-e^u+1}\,du$$
Now consider firstly the Mellin transform
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-z}\,dx = \frac{1}{z} \Gamma (s) \operatorname{Li}_s (z)$$
which can be found here. Now, through partial fractions, we can determine that
\begin{align}\int_{0}^{\infty} \frac{x^{s-1}}{(e^x+p-q)(e^x+p+q)}\,dx &= \int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}+2pe^x+p^2-q^2}\,dx\\&=\frac{\Gamma(s)}{2q(p+q)} \operatorname{Li}_s (-p-q) - \frac{\Gamma(s)}{2q(p-q)} \operatorname{Li}_s (-p+q)\end{align}
Thus setting $p=-\frac{1}{2}$ and $q=\frac{i \sqrt{3}}{2}$:
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}-e^x+1}\,dx=-\frac{\Gamma(s)}{2\sqrt{3}}\left(\left(\sqrt{3}-i\right)\operatorname{Li}_s\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)+\left(\sqrt{3}+i\right)\operatorname{Li}_s\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\right)$$
Through a similar process, we can determine $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{(e^x+p-q)(e^x+p+q)}\,dx = \frac{\Gamma(s)}{2q} \left(\operatorname{Li}_s (-p+q)-\operatorname{Li}_s (-p-q)\right)$$
In general:
$$\int_{0}^{\infty} \frac{x^{s-1}}{a \cosh x+b\sinh x+c}\,dx=\frac{\Gamma(s)}{\sqrt{b^2+c^2-a^2}} \left(\operatorname{Li}_s \left(\frac{\sqrt{b^2+c^2-a^2}-c}{a+b}\right)-\operatorname{Li}_s \left(-\frac{\sqrt{b^2+c^2-a^2}+c}{a+b}\right)\right)$$
which I utilised here.
This gives $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{e^{2x}-e^x+1} \,dx=-\frac{i\cdot\Gamma(s)}{\sqrt{3}}\left(\operatorname{Li}_s\left(\frac{1+i\sqrt{3}}{2}\right)-\operatorname{Li}_s\left(\frac{1-i\sqrt{3}}{2}\right)\right)$$
Thus $$\int_{0}^{\infty} \frac{x^{s-1}(2-e^x)}{e^{2x}-e^x+1}\,dx=-\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)$$
Giving finally
$$I = -\frac{1}{8}\cdot\frac{d}{ds}\left(\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)\right)\Bigg|_{s=1} = -\frac{1}{8} \ln(2) \ln(3)$$ as desired. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Assistance with solving the integral Can you give me an idea how to handle this integral?
$\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\,dx$
| Using the tangent half-angle substitution
$$I=\int_{0}^{2\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx=2\int_{0}^{\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx$$
$$I=8 \sqrt{2} \int_0^\infty \sqrt{\frac{t^2+2}{\left(t^2+1\right)^5}}\,dt=4 \sqrt{2}\int_1^\infty \frac 1 {x^2}\sqrt{\frac{x+1}{(x-1) x}}$$ The antiderivative is nasty but its evaluation at $\infty$ is simply
$$I=\frac{2 }{3 \sqrt{\pi }}\left(\Gamma \left(\frac{1}{4}\right)^2+12\, \Gamma
\left(\frac{3}{4}\right)^2\right)$$ which is the same as
$$I=16 \left(E\left(\frac{1}{2}\right)-\frac{1}{3}K\left(\frac{1}{2}\right)\right)$$ and
$$I=8 \sqrt{2} \left(E(-1)-\frac{2 }{3}K(-1)\right)$$ already given by @Robert Lee.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
| Write $P\equiv \sqrt{x^2-1}$ and $Q\equiv x^3-3x+1$ then the first equation can be written as
$$
Q(x+P)+1=0 \tag{1}
$$
It's clear that $|x|\geq 1$ otherwise there is no solution for $P$. By inspection, $x=1$ is one of the solutions.
Assuming $|x|>1$, multiply $(1)$ by $(P-x)$:
$$
Q(P^2-x^2)+P-x=0
$$
Since $P^2-x^2=-1$,
$$
P=Q+x=x^3-2x+1
$$
Define $y=x^2-2$ then this becomes
$$
P=yx+1.
$$
Squaring,
$$
P^2\equiv y+1=(yx+1)^2
$$
and so
$$
y+1=(yx+1)^2
$$
If $x>1$, you can easily convince yourself that the only way for this to be true is if $y=0$, hence $x=\pm \sqrt2$.
However, there is a flaw that I have not yet convinced myself that if $x<-1$ there is no other obvious solution than $x=-\sqrt2$. Sorry to leave a slightly incomplete answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ and $-x+1$ and $x-1$ and $-x-1$. All of which have real roots. So the answer is $A$.
For quadratic we have $8$ polynomials which are $x^2+x+1$ and $x^2+x-1$ and $x^2-x+1$ and $x^2-x-1$ and $-x^2+x+1$ and $-x^2-x+1$ and $-x^2+x-1$ and $-x^2-x-1$ Here four of the polynomials have real roots which are $x^2+x-1$ and $-x^2-x+1$ and $-x^2+x+1$ and $x^2-x-1$. So we can say that $f(x)$ can be of degree two. So $B$ is also the answer.
For cubic and quartic I don't know how to proceed. Any hints or solutions are appreciated.
| By the Rational Root Theorem, if such a polynomial has a rational root, it must be $x = \pm 1$. It turns out that 8 of the 16 possible cubics (the ones with even numbers of + and - coefficients) have such a root.
For a quartic, we get $f(x) = \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1 = \pm 1 \pm 1 \pm 1 \pm 1 \pm 1$, which always works out to an odd number, never zero. So $f$ has no rational roots. Proving that there are no real solutions can be done using Descartes' Rule of Signs, as in the other answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Factors of central binomial coefficient The central binomial coefficient $\binom{2n}{n}$ is divisible by $n+1$, as seen from the
identity
$$\binom{2n}{n} = (n+1)\binom{2n}{n} -(n+1)\binom{2n}{n+1}.$$
In fact the Catalan numbers
$$C_n=\frac{1}{n+1}\binom{2n}{n}$$
have a very large (over 200 according to R. Stanley's book) combinatorial interpretations. The identity
$$ \binom{2n}{n} =2(2n-1)C_{n-1}$$
shows that $\binom{2n}{n}$ is also divisible by $2n-1$. So it is natural to ask when
$\binom{2n}{n}$ is divisible by the product $(n+1)(2n-1)$. If $n\equiv 0$ or $1$ $\pmod{3}$, then $\gcd(n+1,2n-1)=1$, and so $(n+1)(2n-1)$ will be a factor of $\binom{2n}{n}$ in this case. But if $n\equiv 2 \pmod{3}$, then $\gcd(n+1,2n-1)=3$ so $(n+1)(2n-1)$ will always be a factor of $3\binom{2n}{n}$, but it may or may not be a factor of $\binom{2n}{n}$. According to (limited) computer calculations it seems that it is more likely that it is a factor. So I would be interested in understanding the set
$$\left\{n: (n+1)(2n-1) \mbox{ is not a factor of } \binom{2n}{n}\right\}.$$
According to computer calculations the first few elements of the set are
$$2,5,8,11,14,26,29,32,35,38,41,80,83,86,89,92,95,107,110,113,116,119,122,242,245,248,251,254,257,269,272,275,278,281,284.$$
So looking at the first 300 integers, about 35% of those that are congruent to $2\pmod{3}$ will be such that $(n+1)(2n-1)$ is not a factor of $\binom{2n}{n}$.
| Here is a different proof of the nice result found by Dark Malthorp.
Let
$$K_n=\frac{3}{(2n-1)(n+1)}\binom{2n}{n}$$
We will use the easily verified identities:
$$ K_n=\frac{6}{n(n+1)}\binom{2(n-1)}{n-1}\hspace{10ex} (1) $$
and
$$K_n=\frac{3}{2(2n-1)(2n+1)}\binom{2(n+1)}{n+1} \hspace{10ex} (2)$$
We will also make use of Legendre's formula for the $p$-adic valuation of the factorial. If
$p$ is a prime number, let $\nu_p(n)=\max\{k\in{\mathbb Z}^+: p^k|n\}$ (so in the notation of the previous answer,
$\ell_m=\nu_3(K_{3m+2})$). We also denote by $s_p(n)$ the sum of the digits of $n$ in its base $p$ expansion. Legendre's formula is
$$
\nu_p(n!)=\frac{n-s_p(n)}{p-1}
$$
The following lemma follows immediately from the observation that the presence of a $2$ in the base $3$ expansion of
$n$ would necessarily result in a carry when $n$ is multiplied by $2$.
Lemma
For each positive integer $n$, we have
$$
s_3(2n)\leq 2s_3(n)
$$
and equality occurs if and only if there are no $2$'s in the base $3$ expansion of $n$.
Using Legendre's formula on (1) and (2), we find
$$\nu_3(K_n)=s_3(n-1)-\frac{1}{2}s_3(2(n-1))-\nu_3(n) -\nu_3(n+1)+1 \hspace{10ex} (3)$$
$$\nu_3(K_n)=s_3(n+1)-\frac{1}{2}s_3(2(n+1))-\nu_3(2n-1)-\nu_3(2n+1) +1
\hspace{10ex} (4)$$
Suppose now $n\equiv 2 \pmod{3}$. Then $\nu_3(n)=\nu_3(2n+1)=0$.
If $n-1$ has no $2$'s in its base $3$ expansion, since $\nu_3(n-1)=0$,
the base $3$ expansion of $n-1$ ends in $d1$, where $d\in\{0,1\}$. This means that $n+1$ ends in
$(d+1)0$, and since $d+1\leq 2$, $\nu_3(n+1)=1$.
It then follows from (3) and the Lemma that $\nu_3(K_n)=0$.
If $n+1$ has no $2$'s in its base $3$ expansion, then since $\nu_3(n+1) >0$,
$n+1$ (in base $3$) ends in $0$, so $n$ ends in $2$, $2n$ ends in $21$, and $2n-1$ ends in $20$.
So $\nu_3(2n-1)=1$, and it follows from (4) and the Lemma that $\nu_3(K_n)=0$. This proves that if
$n\equiv 2 \pmod{3}$ and either $n-1$ or $n+1$ have no $2$'s in their base $3$ expansion, then $K_n$ is an integer.
For the converse, suppose that $\nu_3(K_n)=0$. Then (3) and (4) give us
$$s_3(n-1)=\frac{1}{2}s_3(2(n-1))+\nu_3(n+1)-1 \hspace{10ex} (5)$$
$$s_3(n+1)=\frac{1}{2}s_3(2(n+1))+\nu_3(2n-1)-1 \hspace{10ex} (6)$$
If $n-1$ has some $2$'s in its base $3$ expansion, then by the Lemma and (5) $\nu_3(n+1)\geq 2$.
So $n+1$ (in base 3) ends in $00$, $n$ ends in $22$, $2n$ ends in $21$, and $2n-1$ ends in $20$. Hence
$\nu_3(2n-1)=1$ and from (6) and the Lemma, $n+1$ has no $2$'s in its base $3$ expansion.
| {
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"url": "https://math.stackexchange.com/questions/4494256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Painting the board by selecting $2\times 2$ type squares from the $m\times n$ type board Problem: Initially we have an $m\times n$ board with all unit squares white. As a result, we want to paint any two squares with a common edge, one black and one white. (That is, we want to get it like a chessboard pattern after the final step.)
In each step of the painting process, a $2\times 2$ square is selected on the board and its white unit squares are painted black and its black unit squares painted white. For which $(m, n)$ pairs can the board be painted as desired?
Note: I solved some parts of the problem. I'm adding them to the post. Remaining part of the problem is at the bottom of the page. Thanks.
My attempts and some parts about the solution:
*
*For $4\mid m$ and $4 \mid n$, it is possible. Firstly, let's take $m=4, n=4$.
$
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \text{ } & \text{ } & \text{ } \\ \hline
\text{ } & \text{ } & \text{ } & \text{ } \\ \hline
\text{ } & \text{ } & \text{ } & \text{ } \\ \hline
\text{ } & \text{ } & \text{ } & \text{ } \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \blacksquare & \text{ } & \text{ } \\ \hline
\blacksquare & \blacksquare & \text{ } & \text{ } \\ \hline
\text{ } & \text{ } & \blacksquare & \blacksquare \\ \hline
\text{ } & \text{ } & \blacksquare & \blacksquare \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \text{ } & \blacksquare & \text{ } \\ \hline
\blacksquare& \text{ } & \blacksquare & \text{ } \\ \hline
\text{ } & \blacksquare & \text{ } & \blacksquare \\ \hline
\text{ } & \blacksquare & \text{ } & \blacksquare \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \text{ } & \blacksquare & \text{ } \\ \hline
\text{ } & \blacksquare & \text{ } & \blacksquare \\ \hline
\blacksquare & \text{ } & \blacksquare & \text{ } \\ \hline
\text{ } & \blacksquare & \text{ } & \blacksquare \\ \hline
\end{array}
$
That is, a $4 \times 4$ board can be paint as desired. Easily, we expand that, if $4\mid m$ and $4\mid n$ then a $m \times n$ board can be paint as desired.
*
*If $m=2$ or $n=2$, $m \times n$ board can not be paint as desired. Let $m=2$.
\begin{array}{|c|c|c|c|c|}
\hline
\text{x} & \text{ } & \text{ } & \text{ }& \text{ } & \text{ } \\ \hline
\text{x} & \text{ } & \text{ } & \text{ }& \text{ } & \text{ } \\ \hline
\end{array}
Let's examine the first column of the $2\times n$ table (Cells with x in them). Since the colors of these cells will always be the same, a desired format cannot be done.
*
*If $m$ or $n$ is an odd number, a painting in the desired format can not be done. Suppose that $m \geq 3$ odd number. Let's examine difference of number of white squares and number of black squares in the first two columns. Let $w, b$ be number of white squares and number of black squares in the first two columns. Difference of they is $D = w- b$.
$
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\end{array}
$
Initially $D = 2m - 0 = 2m$.
$
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \blacksquare \\ \hline
\blacksquare & \text{ } \\ \hline
\text{ } & \blacksquare \\ \hline
\blacksquare & \text{ } \\ \hline
\text{ } & \blacksquare \\ \hline
\end{array}
$
After the final step $D= m - m = 0$.
We will prove that $D$ is an invariant in $\mod 4$. Because,
$
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \blacksquare \\ \hline
\blacksquare & \blacksquare \\ \hline
\end{array}
$
After the step, $D$ will decrease $8$.
$
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \text{ } \\ \hline
\text{ } & \text{ } \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \blacksquare \\ \hline
\blacksquare & \blacksquare \\ \hline
\end{array}
$
After the step, $D$ will decrease $4$.
$
\begin{array}{|c|c|c|c|c|}
\hline
\blacksquare & \blacksquare \\ \hline
\text{ } & \text{ } \\ \hline
\end{array}
\to
\begin{array}{|c|c|c|c|c|}
\hline
\text{ } & \text{ } \\ \hline
\blacksquare & \blacksquare \\ \hline
\end{array}
$
After the step, $D$ will same...etc.
Thus, $D$ is an invariant in $\mod 4$.
When $m\geq 3$ odd number, initially $D = 2m \equiv 2 \pmod{4}$ and after the final step $D\equiv 0 \pmod{4}$. Since $2\not\equiv 0 \pmod{4}$, there is a contradiction. That is, when $m$ or $n$ is an odd number, a painting in the desired format can not be done.
Remaining Problem: Let $m,n>2$ even numbers.
(a) If $m\equiv 2 \pmod{4}$ and $n\equiv 2 \pmod{4}$, can we paint in the desired format?
(b) If $m\equiv 2 \pmod{4}$ and $n\equiv 0 \pmod{4}$, can we paint in the desired format?
| Hint As you noticed, you need some sort of invariant to prove you cannot succeed. You used the difference $b-w$ for the squares in two columns. Instead, find a modulo invariant for the number of black squares alone in a single column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.
| (The following is about the other roots in ConMan's answer, but it was too long for a comment.)
Courtesy WA, the original equation can be brought to polynomial form with groebnerBasis[ {u - v - w, x - 2 + (u^2 - 2)^2, v^2 x - x^2 + 1, w^2 x - x + 1}, {x}, {u, v, w} ] which returns and factors:
$$
x^6 - 10 x^5 + 39 x^4 - 22 x^3 - 63 x^2 + 32 x + 32 = (x^2 - x - 1) (x^4 - 9 x^3 + 31 x^2 - 32)
$$
*
*The positive root $\,\varphi\,$ of the quadratic and the negative real root $\,\simeq -0.8959\,$ of the quartic satisfy the original equation $\,\sqrt{2 + \sqrt{2-x}} = x = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
*The negative root $\,-\dfrac{1}{\varphi}\,$ of the quadratic satisfies $\,\sqrt{2 - \sqrt{2-x}} = -x = - \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
*The positive real root $\,\simeq 1.2197\,$ of the quartic satisfies $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
*The two complex roots $\,\simeq 4.3381 \pm 3.2349 i\,$ of the quartic satisfy $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain the 5,6,7 equation line? Thank you
| You can use this transformation:
$$\lim_{x \to 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}
=24\lim _{x \rightarrow 0} \frac{\sin^2{\frac{x}{2}}-\frac{x^{2}}{4}}{x^{4}}
=\frac{24}{16}\lim _{y \rightarrow 0} \frac{\sin^2{y}-y^2}{y^{4}}
=\frac{24}{16}\lim _{y \rightarrow 0} \left(\frac{\sin{y}-y}{y^{3}}.\frac{\sin{y}+y}{y}\right)$$
The first factor can be evaluated without using L'H rule by this link.
The second factor equals to 2.
Then your limit equals $$\frac{24}{16}.2.\frac{-1}{6}=\frac{-1}{2}$$.
Thanks for reading.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
| Premise
$0 < |x| < \delta \implies -\delta < x < \delta.$
To Do
Establish a relationship between $0 < \epsilon$ and $\delta$ such that the premise will imply that
$\displaystyle \left|\frac{x}{x+1}\right| < \epsilon \iff -\epsilon < \frac{x}{x+1} < \epsilon.$
In addition to establishing a relationship between $\delta$ and $\epsilon$, I will impose the artificial constraint that
$\displaystyle \delta \leq (1/2) \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \implies \frac{1}{2} \leq (x + 1) \leq \frac{3}{2}.$
Note that the constraint $~\displaystyle \delta \leq \frac{1}{2}~$ forces $(x+1)$ to always be positive, which simplifies the analysis.
To find the minimum value of $~\displaystyle \frac{x}{x+1}$, since the numerator will be negative, and the denominator will be positive, you will want to minimize both the numerator and denominator.
Therefore $~\displaystyle -\frac{\delta}{(1/2)} < \frac{x}{x+1}.~$
To find the maximum value of $~\displaystyle \frac{x}{x+1}$, since the numerator will be positive, and the denominator will be positive, you will want to maximize the numerator and minimize the denominator.
Therefore $~\displaystyle \frac{\delta}{(1/2)} > \frac{x}{x+1}.~$
From the previous discussion, you have that
$$- 2\delta < \frac{x}{x+1} < 2\delta. \tag1 $$
Therefore, setting $\displaystyle \delta = \min\left[\frac{1}{2}, ~\frac{\epsilon}{3}\right] \implies $
$$-\epsilon < \frac{x}{x+1} < \epsilon,$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \sin x \cos x} \\
=& \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^{2}}-\int \frac{d(\sin x+\cos x)}{3-(\sin x+\cos x)^{2}} \\
=& \tan ^{-1}(\sin x-\cos x)+\frac{1}{2 \sqrt{3}} \ln \left| \frac{\sin x+\cos x-\sqrt{3}}{\sin x+\cos x+\sqrt{3}} \right|+C
\end{aligned}
$$
Is there any other method?
| Using the tangent half-angle substitution
$$\int \frac{\sin (x)}{1-\sin (x) \cos (x)} \,d x=4\int \frac{t}{t^4+2 t^3+2 t^2-2 t+1}\,dt$$
The denominator has four roots
$$r_1=\frac{1}{2} \left(-1+\sqrt{3}-i \sqrt{2 \left(2-\sqrt{3}\right)}\right)\qquad r_2=\frac{1}{2} \left(-1+\sqrt{3}+i \sqrt{2 \left(2-\sqrt{3}\right)}\right)$$
$$r_3=\frac{1}{2} \left(-1-\sqrt{3}-i \sqrt{2 \left(2+\sqrt{3}\right)}\right)\qquad r_4=\frac{1}{2} \left(-1-\sqrt{3}+i \sqrt{2 \left(2+\sqrt{3}\right)}\right)$$ So, using partial fraction decomposition, four logarithms to be recombined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ and
$\frac{1}{x}=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$
How can I find $x^{2022}+\frac{1}{x^{2022}}$?
| The trick with this question is to try finding $x^2+\frac{1}{x^2}$, via squaring $x+\frac{1}{x}$:
$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$
$2=x^2+\frac{1}{x^2}+2$
$0=x^2+\frac{1}{x^2}$
The fact that this equals 0 is key to solving the problem, since subtracting $\frac{1}{x^2}$ from both sides and then putting both sides to the power of 1011 solves the problem:
$x^2=-\frac{1}{x^2}$
$(x^2)^{1011}=(-\frac{1}{x^2})^{1011}$
$x^{2022}=-\frac{1}{x^{2022}}$
$x^{2022}+\frac{1}{x^{2022}}=0$
So the answer is 0 and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) \ \ \text{ none of } (\text{a})-(\text{c}).\end{equation*}
$$$$
I have done the following :
(a) It cannot be true since $2\cdot \left (2^{2022}-1\right )$ exceeds the sum of the series, which is a subseries of the infinite geometric series $\left \{\frac{1}{2^n}\right \}$ which converges to
$\frac{1}{1-\frac{1}{2}}=2$.
So we have that $2\cdot \left (2^{2022}-1\right )\gg2>S$.
(b) $\frac{2^{2022}-1}{2}=2^{2021}-\frac{1}{2}\gg2>S$
Therefore it cannot be true.
(c) We want to check if $S=\frac{2^{2022}-1}{2^{2021}}=2-\frac{1}{2^{2021}}$.
Let $\hat{S}$ be the sum of the infinite series : \begin{equation*}\hat{S}=1+\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2^n}=2 \text{ when } n\to \infty\end{equation*}
We therefore want to check if $S=\hat{S}-\frac{1}{2^{2021}}$ or $\hat{S}-S=\frac{1}{2^{2021}}$ ?
We know that $\hat{S}-S=\frac{1}{2^{2023}}+\frac{1}{2^{2024}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty \ \ \ (\star)$
We can rewrite $\frac{1}{2^{2021}}=\frac{2}{2^{2022}}=\frac{1}{2^{2022}}+\frac{1}{2^{2022}}=\frac{1}{2^{2022}}+\frac{2}{2^{2023}} \Rightarrow$ etc $\Rightarrow \frac{1}{2^{2021}}=\frac{1}{2^{2022}}+\frac{1}{2^{2023}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty$
Comparison with (⋆) gives that $\frac{1}{2^{2021}}>\hat{S}-S \Rightarrow $ c) cannot be true.
So (d) must be the correct answer.
$$$$
Is everything correct ?
| $\sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}$ $($ see here$) $
$\begin{align}S &=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} \\\end{align}$
Put $r=\frac{1}{2}, n=2022$.
Hence $S=2 (1-\frac{1}{2^{2023}})=\frac{2^{2023}-1}{2^{2022}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that If $n,k\in\mathbb{N}$ and $\binom{n}{k}$ is a prime number, then $k=1$ or $k=n-1$. My attempt was the following:
Suppose $p=\binom{n}{k}$ is prime. By definition,
\begin{align}
p&=\binom{n}{k}\nonumber\\
p&=\frac{n}{k}\binom{n-1}{k-1}\nonumber\\
\end{align}
Since $p,\binom{n-1}{k-1}\in\mathbb{Z}$, it follows that $k\mid n$ or $k\mid\binom{n-1}{k-1}$. We will look into both cases separately.
Case 1. Suppose $k\mid n$. Thus, $n=kx$ for some integer $x$. Hence, $p=\frac{kx}{k}\binom{n-1}{k-1}=x\binom{n-1}{k-1}$. Since $p$ is prime, it follows that $x=1$ and $\binom{n-1}{k-1}=p$ or $x=p$ and $\binom{n-1}{k-1}=1$. If $\binom{n-1}{k-1}=p=\binom{n}{k}$, it follows that $n=k$ and $p=1$ which is not prime, so it must be the case that $x=p$ and $\binom{n-1}{k-1}=1$, which implies $k=1$ or $k=n-1$.
Case 2. Suppose $k\mid\binom{n-1}{k-1}$. Thus, $\binom{n-1}{k-1}=ky$, for some integer $y$. Hence, $p=\frac{n}{k}ky=ny$. Thus, $n=1$ and $y=p$ or $n=p$ and $y=1$. If $n=1$, then $p=\binom{n}{k}=1$ or $p=\binom{n}{k}=0$, independent of the value of $k$. Either way, $p$ would not be prime, and so, it must be the case that $n=p$ and $y=1$. Since $n=p$, it follows that $\binom{n-1}{k-1}=ky=k$. Hence, $\frac{(n-1)!}{(n-k)!}=k!$. We can see clearly that $k=1$ satisfies the equation. If $k\neq 1$, we have that $(n-1)(n-2)\ldots(n-k+1)=k!$. Note that the left-hand side of the equation is the product of $k-1$ consecutive integers, for which the result is $k!$. This is only possible if $(n-1)(n-2)\ldots(n-k+1)=(n-1)!$, which implies $k=n-1$.
Either way, $k=1$ or $k=n-1$. This completes the proof. $\blacksquare$
My question is on the assertion "If $k\neq 1$, we have that $(n-1)(n-2)\ldots(n-k+1)=k!$. Note that the left-hand side of the equation is the product of $k-1$ consecutive integers, for which the result is $k!$. This is only possible if $(n-1)(n-2)\ldots(n-k+1)=(n-1)!$, which implies $k=n-1$."
Is this ok to assume true? if a product of $n-1$ consecutive integers equals $n!$, then those integers should be $n,n-1,n-2,\ldots,2$ right?
| Notice that if $k = n$, then $p = \binom{n}{k} = 1$ is not prime. So let's assume $1 < k < n-1$. Then $p > n$ that implies $\gcd(p, n) = 1$ and by the equation $kp = n\binom{n-1}{k-1}$ we have $n \mid kp$. But this implies $n \mid k$ that is impossible because $1 < k < n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $e^{At}$ with a $2\times2$ parametric matrix. I would like to calculate the exponential $e^{At}$ where $A$ is a $2\times2$ matrix:
$$
A=\left[\begin{array}{cc}
0 & 1 \\
a & b
\end{array}\right]
$$
and $a,b\in\mathbb R$. The calculus is based on the powers of this matrix, i.e.
$$e^{At}=I+tA+\frac{1}{2}t^2A^2+\frac{1}{3!}t^3A^3+ \cdots$$
But the power of $A$ seems to be very hard to unravel. For example:
$$A^2=\left[\begin{array}{cc}
a & b \\
ab & a+b^2
\end{array}\right], \quad A^3=\left[\begin{array}{cc}
ab & a+b^2 \\
a(a+b^2) & ab+b(a+b^2)
\end{array}\right], \quad \ldots
$$
Could you do better? Also, I wonder if anyone has already posed this problem by publishing (in literature) a few notes about it.
| For general $A,$ the trick would be to diagonalize $A.$
This particular matrix is easier if you know the general formula for linear recurrences. $A$ has the property that $A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\ax+by\end{pmatrix},$ so we can use it to define sequences $p_n,q_n$ as $$A^n\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}p_{n}\\p_{n+1}\end{pmatrix}\\A^n\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}q_{n}\\q_{n+1}\end{pmatrix}$$
where $p_0=1,p_1=0,q_0=0,q_1=1$ and $$p_{n+2}=bp_{n+1}+ap_n\\ q_{n+2}=bq_{n+1}+aq_n$$
The general closed formula for $p_n,q_n$, when $b^2+4a$ is not zero, so we don't have repeated roots, is in terms of $x_{+},x_-=\frac{b\pm \sqrt{b^2+4a}}2:$
$$q_n = \frac{1}{\sqrt{b^2+4a}}\left(x_+^n-x_-^n\right)\\
p_n = \frac{1}{\sqrt{b^2+4a}}\left(x_+x_-^n-x_-x_+^n\right).$$
So: $$\sum_{k=0}^{\infty}\frac1{k!}(At)^k\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}\sum_{k=0}^\infty\frac{q_{k}t^k}{k!}\\\sum_{k=0}^\infty\frac{q_{k+1}t^k}{k!}\end{pmatrix}=\frac{1}{\sqrt{b^2+4a}}\begin{pmatrix}e^{x_+t}-e^{x_-t}\\x_+e^{x_+t}-x_-e^{x_-t}\end{pmatrix}
$$
That should be the right column of $e^{At}.$ The left column is similarly calculated as:
$$\frac{1}{\sqrt{b^2+4a}}\begin{pmatrix}x_{+}e^{x_-t}-x_{-}e^{x_+t}\\a\left(e^{x_+t}-e^{x_-t}\right)\end{pmatrix}$$
This is all essentially equivalent to diagonalizing.
A bit more work will solve the case when $b^2+4a=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If a and b are the distinct roots of the equation $x^2+3^{1/4}x + 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$.
If $a$ and $b$ are the distinct roots of the equation $x^2+3^{1/4}x +
3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$
My attempt:
LHS = $(a^{108}+b^{108})-(a^{96}+b^{96})$
$a^{n} + b^{n} = -[3^{1/4}(a^{n-1}+b^{n-1})+3^{1/2}(a^{n-2}+b^{n-2})]$
$a^{108} + b^{108} = -[(3^{1/4}(a^{107}+b^{107})+3^{1/2}(a^{106}+b^{106})]$
I suppose i could repeat this process until every term in is terms of $a^{96}$ and $b^{96}$ but that would be very tedious
| $x^2+3^{\frac{1}{4}}x + 3^{\frac{1}{2}} =0$
$\implies \left(x^2+3^{\frac{1}{2}}\right)^2=\left(-3^{\frac{1}{4}}x\right)^2$
$\implies x^4+2x^2{3}^{\frac{1}{2}}+3=3^{\frac{1}{2}}x^2$
$\implies\left(x^4+3\right)^2=3x^4$
$\implies x^8+6x^4+9=3x^4$
$\implies x^8=-3x^4-9$
$ \begin{align}\implies x^{12}&=-3x^8-9x^4\\&=9x^4+27-9x^4\\&=27\end{align}$
Hence $a^{12}=27$ and $b^{12}=27$
$\begin{align}&a^{96}\left(a^{12}-1\right)+b^{96}\left(b^{12}-1\right)\\&=26\cdot 27^{8}+26\cdot27^{8}\\&=52\cdot 3^{24}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| Render the equation as
$x^2+7x+(4-2^y)=0$
and treat this as a quadratic equation for $x$. If this is to have natural number roots the discriminant must be a perfect square, which you can work out:
$\Delta=33+(4×2^y)=m^2$
where wlog $m$ may be rendered positive.
If $y$ is odd, then $2^y\equiv2\bmod3$, and substituting that into the discriminant above then gives $m^2\equiv2\bmod3$. That clearly fails, so $y$ must be even and we put in $y=2z$.
Then we have
$33+(4×2^y)=m^2$
$4×2^y=(2×2^z)^2=n^2, n=2×2^z$
where we also render $n$ positive.
Thus by difference
$m^2-n^2=(m+n)(m-n)=33,$
and evidently we must take one of the following:
$m+n=33,m-n=1\implies m=17,n=16$
$m+n=11,m-n=3\implies m=7,n=4$
Comparing the values of $n$ with $n=2×2^z$ then gives $z\in\{1,3\}$, from which $y=2z\in\{2,6\}$ and the complete solution set follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is there any method to compute $\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)$ other than complex numbers? In this couple of days, I need to know the high derivatives of $cos^kx$ whose power $k$ make the differentiation much harder. Then I attempt to
use the identity $$
\cos x=\frac{1}{2}\left(e^{x i}+e^{-x i}\right),
$$
to expand it with Binomial expansion
$$
\begin{aligned}
\cos ^{k} x &=\frac{1}{2 ^k} \sum_{j=0}^{k}\binom{k}{j}e^{x(k-j) i} e^{-x i j}
=\frac{1}{2^{k}} \sum_{j=0}^{k}\binom{k}{j} e^{x(k-2 j) i}
\end{aligned}
$$
Differentiating it by $n$ times yields
$$
\begin{aligned}
\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right) &=\frac{1}{2^k} \sum_{j=0}^{k}\binom{k}{j}[(k-2 j) i]^{n} e^{x(k-2 j) i} \\
&=\frac{i^{n}}{2 ^k} \sum_{j=0}^{k}\binom{k}{j}(k-2 j)^{n} e^{x(k-2 j) i} \\
&=\frac{i^{n}}{2^{k}} \sum_{j=0}^{k}\binom{k}{j}(k-2 j)^{n}[\cos ((k-2 j) x)+i \sin ((k-2 j) x)]
\end{aligned}
$$
If $x$ is real, then comparing the imaginary and real parts on both sides yields \begin{equation}
\displaystyle \frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)=\left\{\begin{array}{ll}
\displaystyle \frac{(-1)^{\frac{n}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n}\cos((k-2j)x) \quad \textrm{ if n is even.}\\\displaystyle
\frac{(-1)^{\frac{n+1}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n} \sin((k-2j)x) \quad \textrm{ if n is odd.}
\end{array}\right.
\end{equation}
For examples,
*
*When $n$ is even,
$$
\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)\right|_{x=0}=\frac{(-1)^{\frac{n}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n}
$$
In particular,
$$
\begin{aligned}
\left.\frac{d^{6}}{d x^{6}}\left(\cos ^{5} x\right)\right|_{x=0}&=\frac{(-1)^{3}}{2^{4}} \sum_{j=0}^{2}\binom{5}{j}(5-2 j)^{6} \\
&=\frac{1}{16}\left[5^{6}+5 \cdot 3^{6}+10 \cdot 1^{6}\right] \\
&=1205
\end{aligned}
$$
*When $n$ is odd,$$
\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)\right|_{x=0}=0;
$$
$$\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right) \right|_{x=\frac{\pi}{4} }=
\frac{(-1)^{\frac{n+1}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n} \sin \frac{(k-2 j) \pi}{4} $$
Can we find its closed form without using the complex numbers?
| You could use Faà di Bruno's formula, with $g(x)=\cos(x)$ and $f(x)=x^k$.
Just for an example, with $\frac{d^3}{dx^3}\cos^4(x)$ you need to sum over $(m_1,m_2,m_3)$ such that $m_1+2m_2+3m_3=3$, with each $m_i\geq0$. Such tuples are: $(3,0,0)$, $(1,1,0)$, and $(0,0,1)$. So
$$
\begin{align}
\frac{d^3}{dx^3}\cos^4(x)
&=\frac{3!}{3!}f^{(3)}(\cos(x))\left(\cos^{(1)}(x)\right)^3
+\frac{3!}{2!}f^{(2)}(\cos(x))\cos^{(1)}(x)\cos^{(2)}(x)
+\frac{3!}{3!}f^{(1)}(\cos(x))\cos^{(3)}(x)\\
&=24\cos(x)\left(-\sin^3(x)\right)+3(12)\cos^2(x)\left(-\sin(x)\right)\left(-\cos(x)\right)+4\cos^3(x)\sin(x)\\
&=-24\sin^3(x)\cos(x)+40\sin(x)\cos^3(x)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$ for any $\triangle ABC$ I found this problem on Mathematics, Textbook for Class XI by NCERT, ed. January 2021 that uses the sine and cosine formulas along with standard trigonometric identities for proving an identity.
By triangle ABC, the question assumes that the angles of the triangle are A, B and C, and the sides opposite the angles A, B and C, are a, b and c respectively.
For any triangle $ABC$, prove that:
$$(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$$
I've tried to approach this problem by simplifying the
$$\cos \frac{B+C}{2}$$
part into
$$\sin\frac{A}{2}$$
and assuming that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$$
keeping in mind the sine law. However, I haven't been able to solve it yet.
A guidance would be helpful even if the question is not solved in its entirety.
| First you need to know The sum to product identities:
$$\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2}) $$
Then the problem is solved:
$$
\begin{equation*}
\begin{aligned}
(b+c)\cos(\frac{B+C}{2})-a\cos(\frac{B-C}{2}) &= R[(\sin B+\sin C)\sin(\frac{A}{2})-\sin A\cos(\frac{B-C}{2})] \\
&= R\sin(\frac{A}{2})[(\sin B+\sin C)-2\cos (\frac{A}{2})\cos(\frac{B-C}{2})] \\
&= R\sin(\frac{A}{2})[(\sin B+\sin C)-2\sin (\frac{B+C}{2})\cos(\frac{B-C}{2})] \\
&= 0\\
\end{aligned}
\end{equation*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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