Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find minimal distant from one point to a parabola? Find the minimal distance from the point (2,0) to the parabola $y^2=2x$.
| $$(2,0)to (x,y)\\dis=\sqrt{(x1-x2)^2+(y1-y2)^2}\\dis=\sqrt{(x-2)^2+(y-0)^2}\\dis=\sqrt{(x-2)^2+(y-0)^2}=\\dis=\sqrt{(x-2)^2+2x} \\dis=\sqrt{x^2-2x+4}=\\dis=\sqrt{x^2-2x+4}=\\dis=\sqrt{(x-1)^2+3}=\\min((x-1)^2+3)=0+3\\so\\min(dis=\sqrt{(x-1)^2+3} )=\sqrt{3}$$when put x=1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify a LHS of the trigonometric equation to obtain RHS Is this equality correct? If so, how to simplify the following expression on the LHS to get RHS:
$$\frac{\sin(x+\frac{nh}{2}) \sin(\frac{(n+1)h}{2})}{\sin\frac{h}{2}} +\sin(x+(n+1)h)=\frac{\sin(x+\frac{ (n+1)h}{2}) \sin(\frac{(n+2)h}{2})}{\sin \frac{h}{2}}$$
Anybody?
| The formula is equivalent to
$$
\sin(x+\frac{nh}2) \sin((n+1)\frac h2) +
\sin(x+(n+1)h)\sin \frac h2 = \sin(x + (n+1)\frac h2) \sin((n+2)\frac h2)
$$
Now use the formula
$$
\cos(A-B) - \cos(A+B) = 2\sin A\sin B
$$
This is equivalent to
$$
\cos(x-\frac h2) - \cos(x + nh + \frac h2) +
\cos(x+nh+\frac h2) - \cos(x+nh+3\frac h2)
\\=
\cos(x - \frac h2) - \cos(x + (n+1)h + \frac h2)
$$which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rewrite the expression in the form $A \sin(x+C)$ Rewrite the following expression in the form $A \sin(x+C)$
$$4 \sin x + 4\sqrt{3} \cos x$$
This is what I have so far, and I'm not even sure it's the right approach. I just dont understand this concept as a whole:
$$A \cos(c)\sin(x) + A\sin(c)\cos(x)$$
$$A\cos (c)=4$$
$$A\sin (c)=4\sqrt{3}$$
$$\sin^2(x) + \cos^2(x) = 1$$ ??
| Hint: As you wrote above, $\sin(x+c) = \cos(c) \sin(x) + \sin(c) \cos(x)$. Imagine that you could find some $c$ so that $\cos(c) = \frac{1}{2}$ and $\sin(c) = \frac{\sqrt{3}}{2}$, you would then have $$\sin(x + c) = \frac{1}{2} \sin(x) + \frac{\sqrt{3}}{2} \cos(x) = \frac{1}{2} \big[\sin(x) + \sqrt{3} \cos(x)\big].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999928",
"timestamp": "2023-03-29T00:00:00",
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$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$ - is there any other way to solve this? Here's a problem that I think I've managed to solve:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$$
Here's how I did it:
$(\sqrt{2^n+n^2}-\sqrt{2^n+1})\frac{\sqrt{2^n+n^2}+\sqrt{2^n+1}}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}=\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}$
And now you can see that (for sufficiently large $n$):
$0\le\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}\le\frac{n^2}{\sqrt{2^n}}$
Limit of $0$ clearly is $0$, and this is also true for $\frac{n^2}{\sqrt{2^n}}$, so that means that:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}=0$$
The thing is I got kind of lucky I've solved it and I'm interested if there are any other ways to solve this limit (not involving any derivatives nor integrals). Does anybody see any clever solution other than mine? And also - is my solution correct?
| $$\sqrt{2^n+n^2}-\sqrt{2^n+1}=2^{n/2} (\sqrt{1+n^22^{-n}}-\sqrt{1+2^{-n}})=2^{n/2}\left(1+\frac12n^22^{-n}+o(n^22^{-n})-1-\frac122^{-n}-o(2^{-n})\right)=\frac{n^2-1}22^{-n/2}+o(n^22^{-n/2}).$$
The expression converges to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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completing the square to solve equation Is it possible to use the method of completing the square to solve the equation $2x^2+18x+21=0$ ?
I have problem with how to remove the negative sign on the right side.
| First make the leading coefficient $1$:
$x^2 + 9x + \frac{21}{2} = 0$.
Then add and subtract $(\frac{9}{2})^2$:
$x^2 + 9x + \frac{21}{2} + (\frac{9}{2})^2 - (\frac{9}{2})^2 = 0$
Rearrange,
$x^2 + 9x + (\frac{9}{2})^2 = (\frac{9}{2})^2 - \frac{21}{2}$
Observe the LHS is a square:
$(x + \frac 92)^2 = \frac{81}{4} - \frac{42}{4} = \frac{39}{4}$
Take square root of both sides,
$x + \frac 92 = \pm \frac 12 \sqrt {39}$
Final solution:
$x = - \frac 92 \pm \frac 12 \sqrt {39}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003774",
"timestamp": "2023-03-29T00:00:00",
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How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| Both numbers are positive, so the equality is true if, and only if its squares are equal. By squaring both sides we get
$$14+4\sqrt{10}-2\sqrt{(14+4\sqrt10)(14-4\sqrt{10})}+14-4\sqrt{10} = 4^2$$
$$14-2\sqrt{(14^2-16 \cdot 10)}+14 = 16$$
$$28-12 = 16$$
$$16 = 16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does there exist regular planar pentagon? How to prove or disprove that the boundary of any convex body in $\mathbb{R}^3$ (treated as a surface) includes 5 points which form a regular planar pentagon?
| This is about a special case raised by Han de Bruijn: yes, you can inscribe a regular planar pentagon in a regular tetrahedron.
And you can do that in many ways, probably. Here is one of them.
Take a regular tetrahedron $ABCD$ with the edge length equal $1$.
Let $E$ be a point on the $BC$ edge at distance $x$ from $B$. Similary $F$ on $DC$, $x$ apart from $D$.
The triangle $\triangle EFA$ is isosceles, degenerating continuously from the equilateral $\triangle BDA$ to the segment $CA$ as $x$ changes from $0$ to $1$.
That means $\epsilon = \angle EAF$ changes from $60^\circ$ to $0$. 'Somewhere' during the change it is $\epsilon = 36^\circ$. Where?
Let $y=AE=AF$ and $z=EF$. Due to the law of cosines in $\triangle ABE$ $$y^2 = 1^2 + x^2 - 2\cdot 1\cdot x\,\cos(\angle ABE) = x^2 + x + 1$$ and by the similarity $\triangle EFC \sim \triangle BDC$ $$EF = EC\times BD:BC$$ $$z = 1-x$$
In the isosceles $\triangle EFA$
$$\sin\frac\epsilon 2 = \frac z2:y$$
We want $\epsilon=36^\circ$:
$$\left .\sin 18^\circ = \frac{1-x}{\sqrt{x^2+x+1}} \ \ \right\vert \big(\big)^2$$
$$\sin^2 18^\circ = \frac{x^2-2x+1}{x^2+x+1} = 1 - \frac{3x}{x^2+x+1}$$
$$\cos^2 18^\circ = \frac{3x}{x^2+x+1}$$
$$x^2+x+1 = \frac{3x}{\cos^2 18^\circ}$$
$$x^2+\left(1-\frac{3}{\cos^2 18^\circ}\right)x+1 = 0$$
The parenthesized expression is approx. $-2.317$ and one of the roots falls between $0$ and $1$:
$$x\approx 0.573$$
Once we have $36^\circ$ at $A$ we can fit a regular pentagon $JKLMN$ in the $\triangle EFA$ so that $JK \subset FA$, $LM \subset AE$, $KL \parallel EF$. If it is too small or too big, scale it so that $N$ becomes the midpoint of $EF$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Min/Max Problem Find the minimum of $$\frac{x^2}{x-1}$$ for $x > 1$.
I tried using AM-GM and Cauchy-Schwarz, but after using them I ended up with $x>1$ which was what was given. How should I start?
| The shape of $f(x)$ for $x>1$ is a steep drop from infinite heights at $x=1^+$ going dwon but womehow going up looking like $f(x)\approx x$ for very large $x$. So there will be a minimum.
A cute way to find the minimum is to simplify the expression, relying on the fact that $x \neq 1$:
$$
\frac{x^2}{x-1} = \frac{x^2-x}{x-1} + \frac{x}{x-1} = \frac{x^2-x}{x-1} + \frac{x-1}{x-1}
+ \frac{1}{x-1} = x + 1 + \frac{1}{x-1}$$
Then if we had calculus, we would find the minimum by taking the derivative of $x + 1 + \frac{1}{x-1}$ and setting it to zero, getting $x=2, f(x) = 4$, but in pre-calc that is, well, post-course.
We can find the minimum without calculus by rewriting
$$
f(x) = 2 + (x-1) + \frac{1}{x-1} = 2 + v + \frac{1}{v}$$
with $v = x-1$. Then you know that $v + \frac{1}{v}$ has a minimum at $v=1$.
Which means $f(x)$ is its minimum at $v=1 \Rightarrow x = 2$ as before.
By the way, we can see that $v+\frac{1}{v}$ is always at least $2$ by:
$$
v + \frac{1}{v} -2 = \left( \sqrt{v} - \sqrt{\frac{1}{v}} \right)^2 \geq 0
$$
Then where is $v + \frac{1}{v} = 2$? At $v=1$ of course.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equivalent condition for the divisibility by $2^{n-1}$. I guess that the equivalent condition that for any positive integer $n,m$
$$ \sum_{k \ge 0} \binom {n}{2k} m^k $$ is divisible by $2^{n-1}$ is that
$$m \equiv 1(mod 4).$$ Would you explain the reason why?
| $$ \sum_{k \ge 0} \binom {n}{2k} m^k =\sum_{k \ge 0} \binom {n}{2k} m^k 1^{n-2k}$$
$$=\frac{1}{2} [ \sum_{k \ge 0} \binom {n}{k} ( \sqrt{m})^k 1^{n-k} + \sum_{k \ge 0} \binom {n}{k} ( - \sqrt{m})^k 1^{n-k} ]
\\=\frac{1}{2} [ (1+ \sqrt m)^n + (1- \sqrt m )^n ]. $$
Note that if we let $r:=x+y, s:=xy$ and let $f_n (r,s):= \frac{x^n +y^n}{2}$
then $$ f_{n+1} (r,s)=rf_n(r,s)+sf_{n-1}(r,s) (n \ge 0).$$
$f_0 (r,s)= 1, f_1 (r,s)=r/2.$ Note also that $f_2(r,s)=(x^2+y^2)/2 =[(x+y)^2 -2xy]/2=(r^2-2s)/2.$
Now $x=1+ \sqrt{m}, y=1- \sqrt{m}$ then $r=2, s=1-m$ which is a multiple of 4 since $ m \equiv 1(mod 4)$. By induction on $n$ the claim follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| Hint: $x-y=(\sqrt x +\sqrt y)(\sqrt x - \sqrt y)$
| {
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"timestamp": "2023-03-29T00:00:00",
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An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? While Solving some integral problem, I encountered the following infinite series:
$$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$
I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.
| We have $\begin{align} \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2} & = \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{(n+k)^3 - n^3-k^3}{n^3k^3(n+k)^3} \\ &= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{n^3(n+k)^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{k^3(n+k)^3} \tag{*}\\ &=\frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n=1}^\infty\sum\limits_{k = n+1}^\infty \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{k=1}^\infty\sum\limits_{n=k+1}^\infty \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le n < k <\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le k < n < \infty} \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n=k=1}^\infty \frac{1}{n^3k^3} = \frac{1}{3}\zeta(6) \end{align}$
where, in line $(*)$ we reindexed the summation with the change of variables $n+k \mapsto k$ in the second sum and $n+k \mapsto n$ in the third sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to I solve this limit to -3 I need help solving this limit $\lim \limits_{x \to -3} \frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$\lim \limits_{x \to -3} \frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$.
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!
| $$
\begin{split}
\mathop {\lim }\limits_{x \to - 3} \frac{{4x + 12}}
{{3{x^3} - 27x}} &= \mathop {\lim }\limits_{x \to - 3} \frac{{4\left( {x + 3} \right)}}
{{3x\left( {{x^2} - 9} \right)}}\\
&= \mathop {\lim }\limits_{x \to - 3} \frac{{4\left( {x + 3} \right)}}
{{3x\left( {x - 3} \right)\left( {x + 3} \right)}} \\
&= \mathop {\lim }\limits_{x \to - 3} \frac{4}
{{3x\left( {x - 3} \right)}} \\
&= \frac{4}
{{3\cdot\left( { - 3} \right)\cdot\left( { - 3 - 3} \right)}} = \frac{2}
{{27}}
\end{split}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$ \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$$ = 2\sqrt{0}-2\sqrt{3} = -2\sqrt{3}$$
hi sorry i meant 3-x
| $u=3-x$, $dx=-du$, and as $x$ goes from $0$ to $3$, then $u$ goes from $3$ to $0$, so
$$
\int_0^3 \frac{dx}{\sqrt{3-x}} = \int_3^0\frac{-du}{\sqrt{u}} = \int_0^3\frac{du}{\sqrt{u}} = \left. 2\sqrt{u} \vphantom{\frac 1 1}\ \right|_0^3 = 2\sqrt{3}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Closed form for a zeta series :$\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$ It is not that diffcult to derive
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{k2^k}=&-\frac{\gamma}{2}+\ln\left(\frac{2}{\sqrt{\pi}}\right)\tag{1}\\
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+1)2^{k+1}}=&-\frac{4+\gamma}{8}+\ln\left(A^{3/2}2^{5/24}\right)\tag{2}
\end{align}
Hence, I would like to know if there exists a closed form in terms of known mathematical constants for the following series
$$\mathscr{S}=\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$$
As $(1)$ and $(2)$ follow immediately from the definitions of $\Gamma(z)$ and $G(z+1)$ respectively, my guess is that the evaluation of $\mathscr{S}$ involves the function $\Gamma_3(z)$. Unfortunately, I know almost nothing about higher order multiple gamma functions, and I would really appreciate it if someone can enlighten me on this matter and provide a viable solution to the series above. Thank you.
This is what I have managed to get so far.
Begin with the sum
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{k+1}z^{k+1}
=&\sum^\infty_{k=3}\sum^\infty_{m=1}\frac{(-1)^k}{k}\frac{z^{k}}{m^{k-1}}\\
=&\sum^\infty_{m=1}\left\{-m\ln\left(1+\frac{z}{m}\right)-\frac{z^2}{2m}+z\right\}\\
\end{align}
Compare this with $\ln{G(z+1)}$.
$$\ln{G(z+1)}=-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{m=1}\left\{m\ln\left(1+\frac{z}{m}\right)+\frac{z^2}{2m}-z\right\}$$
It follows that
$$\sum^\infty_{k=1}\frac{(-1)^{k-1}\zeta(k)}{k+1}z^{k+1}=-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}-\ln{G(z+1)}$$
Integrate from $0$ to $z$ to get
\begin{align}
&-\frac{z^2}{2}+\frac{z^2}{2}\ln(2\pi)-\frac{z^3}{2}-\frac{\gamma z^3}{2}-z\ln{G(z+1)}-\sum^\infty_{k=1}\frac{(-1)^{k-1}\zeta(k)}{(k+2)}z^{k+2}\\
=&\sum^\infty_{k=1}\left\{-k(k+z)\ln\left(\frac{k+z}{k}\right)+kz+\frac{z^2}{2}-\frac{z^3}{6k}\right\}
\end{align}
After letting $z=\frac12$, I have no idea how to proceed further as when I take the exponential of the partial sum, the portion with the $\ln$ term doesn't seem to telescope.
| As mentioned by Claude Leibovici, you have
$$
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}=-\frac{1}{8}-\frac{\gamma}{24}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2}. \tag1
$$
Here is a hint.
From the classic identity verified by the digamma function $\displaystyle \psi:=\Gamma'/\Gamma$, wich may be obtained from the Euler product giving $\Gamma(x+1)$, you have
$$ \psi(x+1) = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k(k+x)}\quad x\neq 0,-1,-2,-3,\dots$$
then you easily obtain, for $|x|<1$,
$$ \begin{align}
\psi(x+1) & = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k^2}\frac{1}{1+\dfrac{x}{k}} \\
&= -\gamma + \sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=0}^{\infty}\frac{(-1)^n}{k^n}x^{n+1} \\
&= -\gamma - \sum_{n=0}^{\infty}(-1)^{n-1} \zeta(n+2){x^{n+1}} \\
&= -\gamma - \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k-1}} \\
\end{align}
$$
and
$$-\gamma x^2 - x^2\psi(x+1) = \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}. \tag2$$
Using $(2)$ gives
$$
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}&=\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k)\int_0^{1/2}\!\!x^{k+1}dx\\
&= \int_0^{1/2}\!\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}\:dx \\
&= -\gamma \int_0^{1/2}\!x^2 dx - \int_0^{1/2}\! x^2\psi(x+1)\:dx \\
&=-\frac{\gamma}{24} - \int_0^{1/2}\! x^2\psi(x+1)\:dx,
\end{align}
$$ then integrating by parts twice leads to
$$ \begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}} & = -\frac{\gamma}{24} - \frac14\log \Gamma\left(\frac32\right)+2\int_0^{1/2}\! x\log \Gamma(x+1)\:dx\\
&= -\frac{\gamma}{24} - \frac18\log \pi- \frac14\ln2+\zeta'\left(-1,\frac32\right)+2\int_0^{1/2}\! \zeta'(-1,x+1)\:dx \\
&= -\frac{\gamma}{24}-\frac{1}{8}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2},
\end{align}
$$
where we have used the identity (25.11.34) and special values of $\zeta'(s,a)$.
| {
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Show that $\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$ I have to show that the left equation simplifies to $\tan\theta$:
Show that:
$$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$
I do have prior knowledge that:
$$\tan \theta = \frac{\sin\theta}{\cos \theta}$$
But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.
| Here is a detailed answer.Let's go!
$$
\require{cancel}
\begin{align}
\frac{1-\cos2\theta}{\sin2\theta}&=\frac{1-\left(\cos^2\theta-\sin^2\theta\right)}{2\sin\theta\cos\theta}\\
&=\frac{1-\cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\sin^2\theta\cancel{+\cos^2\theta}\cancel{-\cos^2\theta}+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\bcancel2\cancelto{\sin\theta}{\sin^2\theta}}{\bcancel2\cancel{\sin\theta}\cos\theta}\\
&=\frac{\sin\theta}{\cos\theta}\\
&=\tan\theta
\end{align}
$$
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$ $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$
For what i've found the part in brackets is an indeterminate form. I've tried to multiply the bracket part by $\frac{\sqrt{x - 1} + \sqrt{x+5}}{\sqrt{x - 1} + \sqrt{x+5}}$ , then multiply the numerator by $2x$. I don't know what to do next.
| You follow the right approach.
\begin{align}
f(x) &= 2x (\sqrt{x-1}-\sqrt{x+5}) = 2x
\frac{(\sqrt{x-1}-\sqrt{x+5})\cdot (\sqrt{x-1}+\sqrt{x+5})}{(\sqrt{x-1}+\sqrt{x+5})}
\\
&=2x\frac{x-1-x-5}{(\sqrt{x-1}+\sqrt{x+5})} = 2x\frac{-6}{(\sqrt{x-1}+\sqrt{x+5})}
= -12\frac{x}{(\sqrt{x-1}+\sqrt{x+5})}
\end{align}
Now you can try to bound $f$ from below and above by an expression of the form $$c\frac{x}{\sqrt{x}}$$
What do you know about an expression of that form?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Pluggin in numbers If $(a + c)(a − c) = 0$, which of the following must be true?
*
*$a = 0$,
*$c = 0$,
*$a = −c$,
*$a = c$,
*$a2 = c2$
The answer states: Try plugging in $a = 2$ and $c = 2$. This eliminates (A), (B), and (C). Now try $a = 2$ and $c = −2$. This eliminates (D). Only (E) remains.
I don't understand how the plug in numbers were chosen. Could anyone help explain this?
| By choosing $a = 2$, $c = 2$, the author demonstrates that the equation can be satisfied when $a \neq 0$ since
$$(a + c)(a - c) = (2 + 2)(2 - 2) = 4 \cdot 0 = 0$$
Hence, $a = 2$, $c = 2$, is a counterexample to the claim that $a = 0$ must be true. There are others. For instance, $a = 1$, $c = 1$ or $a = 1$, $c= -1$. However, only one counterexample is required to demonstrate that a statement is false.
What the author is stating is that $a = 2$, $c = 2$ is also a counterexample to the claim that $b = 0$ must be true or that $a = -c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied.
Similarly, if $a = 2$ and $c = -2$, then
$$(a + c)(a - c) = [2 + (-2)][2 - (-2)] = 0 \cdot 4 = 0$$
so $a = 2$, $c = -2$ is a counterexample to the claim that $a = c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied. Again, there are other counterexamples such as $a = 1$, $c = -1$.
As for the remaining choice, since $(a + c)(a - c) = a^2 - c^2$, $$(a + c)(a - c) = 0 \Rightarrow a^2 - c^2 = 0 \Rightarrow a^2 = c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$ How to calculate $$\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$$
where $a>0, b>0?$
I could not calculate the limit with ordinary tool(ratio test, squeeze ...).
Does anyone have an idea of efficiency to find this limit?
Thank you!
| An approach that doesn't use asymptotics:
We consider the case $a>r$, that is, $a = b + r$ with $r>0$. Our sequence satisfies the recurrence
$$
f(n) = f(n-1)\frac{b+r+n}{b+n} = f(n-1)\left(1 + \frac{r}{b+n} \right)
$$
We note that our sequence is positive and monotonically increasing. It follows that
$$
f(n) - f(n-1) = f(n-1)\left(1 + \frac{r}{b+n} \right) - f(n-1) =\\
\frac{r}{b + n}f(n-1) >
\frac{r}{b + n}f(0) = \frac{a}{b}\frac{r}{b+n}
$$
We then note that
$$
\lim_{n \to \infty} f(n) = f(0) +
\lim_{n \to \infty} \sum_{k=1}^n (f(k) - f(k-1)) >
f(0) + f(0)\lim_{n \to \infty} \sum_{k=1}^n \frac{r}{b + k}
$$
By the divergence of the sum, we conclude that the limit is $\infty$.
In the case $a = b$, note that $f(n) = 1$, so our limit is $1$.
In the case $a < b$, our limit can be rewritten as
$$
\lim_{n\rightarrow\infty}\left(\frac{b(b+1)(b+2)\cdots(b+n)}{a(a+1)(a+2)\cdots(a+n)}\right)^{-1}
$$
so that from our previous analysis, the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$ I know the above expression equals to $\frac{n(2n−1)(2n+1)}{3}$, but how exactly can i come up with something from scratch?
| $$(2k-1)^2=8\binom{k}{2}+1$$
but since
$$\sum_{k=1}^{N}\binom{k}{2}=\binom{N+1}{3}$$
it follows that:
$$\sum_{k=1}^{N}(2k-1)^2 = N+8\sum_{k=1}^{N}\binom{k}{2}=N+8\binom{N+1}{3}=\frac{1}{3}N(2N+1)(2N-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why is the antiderivative of $\frac{1}{1+x^2}=\tan^{-1}(x)$? My textbook says the antiderivative of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$.
To confirm this to myself I took the derivative of $\tan^{-1}(x)$ expecting to get $\frac{1}{1+x^2}$ , but instead I ended up with $-\frac{1}{\sin^2(x)}$. So why is $\tan^{-1}(x)$ the antiderivative of $\frac{1}{1+x^2}$ if the derivative of $\tan^{-1}(x)$ is not $\frac{1}{1+x^2}$? Shouldn't the derivative of the antiderivative of a function give you the original function?
| Using the formula for geometric series,
$${1\over 1+x^2}=1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots
$$
Integrating,
$$
\begin{align}
\int 1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots dx&=x-{x^3\over3}+{x^5\over5}-{x^7\over7}+\cdots\\
&=\arctan x\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Calculation of $ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ Calculation of $\displaystyle \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ and $\displaystyle \cos\left(\frac{2\pi}{7}\right)\times \cos\left(\frac{4\pi}{7}\right) \times \cos\left(\frac{6\pi}{7}\right)$ is
$\bf{My\; Try::}$ Let $\displaystyle \frac{2\pi}{7}=\phi\;,$ Then $3\phi = 2\pi-4\phi\Rightarrow \cos \left(3\phi\right) = \cos \left(2\pi-4\phi\right)=\cos \left(4\phi\right)$
So $4\cos^3 \phi -3\cos \phi = 2\left[2\cos^2 \phi - 1\right]^2-1=2\left[4\cos^4 \phi+1-4\cos^2 \phi\right]-1 = 8\cos^4 \phi-8\cos^2 \phi+1$
So $8\cos^4\phi-4\cos^3 \phi-8\cos^2 \phi+3\cos \phi+1=0$
Now I did not understand how can i solve after that
Help me
Thanks
| One can use complex numbers to find the value of $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}.$$
Let $\omega=e^{\frac{2\pi i}{7}}=\cos\frac{2\pi}{7}+ i\sin\frac{2\pi}{7}.$ $\omega$ is a zero of $x^7-1=(x-1)(x^6+x^5+\dots +x+1)$. Since, $\omega\ne 1$, it is also a zero of the second factor. So,
$$\omega^6+\omega^5+\dots +\omega+ 1=0.$$
Using DeMoivre's Theorem and taking the real part of both sides, gives
$$\cos\frac{12\pi}{7}+\cos\frac{10\pi}{7}+\dots +\cos\frac{2\pi}{7}+1=0.$$
Let $S=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$. Now, $$\cos\frac{12\pi}{7}+\cos\frac{10\pi}{7}+\cos\frac{8\pi}{7}=\cos(2\pi-\frac{2\pi}{7})+\cos(2\pi-\frac{4\pi}{7})+\cos(2\pi-\frac{6\pi}{7})=S,$$ by the identity $\cos(2\pi-\theta)=\cos\theta$.
Therefore,
$$\cos\frac{12\pi}{7}+\cos\frac{10\pi}{7}+\dots +\cos\frac{2\pi}{7}+1=2S+1,$$
and $$2S+1=0.$$
So, $S=-\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Is there a general form for the determinant of this matrix? This came up in trying to deal with small oscillations of an $N$-pendulum. I obviously want to calculate the characteristic polynomial in $\omega^2$ to see if I can deal with the equation even in special cases. Is it the first time you see this kind of matrix? Does the characteristic polynomial have a reasonably nice form? Oh, $N$ is the dimension of the matrix.
You can suppose $m,g,l>0.$
$$A=\begin{pmatrix}
N(mgl-\omega^2ml^2) & -\omega^2ml^2(N-1)& -\omega^2ml^2(N-2)&...& -\omega^2ml^2 \\
-\omega^2ml^2(N-1)& (N-1)(mgl-\omega^2ml^2) &-\omega^2ml^2(N-2) &... &-\omega^2ml^2 \\
-\omega^2ml^2(N-2) &-\omega^2ml^2(N-2) &(N-2)(mgl-\omega^2ml^2) & ...& -\omega^2ml^2 \\
\vdots&\vdots&\vdots&\vdots&\vdots
\end{pmatrix}$$
Thank you in advance.
EDIT: As pointed out by Omnomnomnom (I had to copy paste that name) the problem reduces to finding the determinant of:
$$A = -\omega^2ml^2
\pmatrix{
N&N-1&N-2&\cdots&1\\
N-1&N-1&N-2&\cdots&1\\
N-2&N-2&N-2&\cdots&1\\
\vdots&\vdots&\ddots
}
+ mgl
\pmatrix{
N\\&N-1\\&&N-2\\&&&\ddots
}=-\omega^2ml^2T+mglV$$
Dividing by $mgl$ and calling $$\tilde\omega=\frac{\omega^2l}{g}$$
We have
$$\operatorname{Det}\bigg(\frac{A}{mgl}\bigg)=\operatorname{Det}B=\operatorname{Det}(V-\tilde\omega T)$$
I computed the first few polynomials in $\tilde \omega$.
For $N=2$
$$P_2(\tilde\omega)= \tilde\omega^2-4\tilde\omega+2$$
For $N=3$:
$$P_3(\tilde\omega)=-\tilde\omega^3+9\tilde\omega^2-18\tilde\omega+6$$
For $N=4$
$$P_4(\tilde\omega)=\tilde\omega^4-16\tilde\omega^3+72\tilde\omega^2-96\tilde\omega+24$$
For $N=5$
$$P_5(\tilde\omega)= -\tilde\omega^5+25\tilde\omega^4-200\tilde\omega^3+600\tilde\omega^2-600\tilde\omega+120$$
Do you see a pattern? The first coefficient might be $(-1)^N$, the second $N^2(-1)^{N+1}$, the last one is probably $N!$, the second to last one might be $-N!N$.
| Consider the matrix
$$
P^{-1}Q =
\pmatrix{
N\\&N-1\\&&N-2\\&&&\ddots
}^{-1}
\pmatrix{
N&N-1&N-2&\cdots&1\\
N-1&N-1&N-2&\cdots&1\\
N-2&N-2&N-2&\cdots&1\\
\vdots&\vdots&\ddots
} = \\
\pmatrix{
1&1-1/N&1-2/N&\cdots&1/N\\
1&1&1 - 1/(N-1)&\cdots&1/(N-1)\\
1&1&1&\cdots&1/(N-2)\\
\vdots&\vdots&\ddots
}
$$
Let $q_N(x)$ denote the characteristic polynomial $\det(P^{-1}Q - xI)$.
The answer to the determinant we seek will be
$$
\det(A) = N! \cdot a^N q_N(-b) =
N! (-\omega^2 m \ell^2)^N q_N(- mg \ell)
$$
Hopefully you find this expression useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the exact value of $\frac{1-2\sin\left(α\right)\cos\left(α\right)}{1-2\sin^2\left(α\right)}$. The task is: Given that
$0\leq \alpha \leq \pi$ is an angle with $\tan\left(\,\alpha\,\right)=3/4$, find the exact value of
$$
{1 - 2\sin\left(\,\alpha\,\right)\cos\left(\,\alpha\,\right)\over
1 - 2\sin^{2}\left(\,\alpha\,\right)}.
$$
I have stuck with this task. Can anybody help me with this ?
( I need detailed answer ).
| Although the OP has already accepted an answer, I'll provide a different approach, even because, together with the other, it did come to my mind earlier, but I had to leave the computer.
$$\displaystyle \frac{1-2\sin\alpha \cos\alpha}{1-2\sin^2 \alpha}=\frac{1-2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha},$$ which, recalling the double-angle formulae, becomes $$\displaystyle \frac{1-\sin2\alpha}{\cos2\alpha}.$$
Finally use $\displaystyle \sin\alpha=\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$ and $\displaystyle \cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$ to obtain $$\displaystyle \frac{1-\frac{2\tan\alpha}{1+\tan^2\alpha}}{\frac{1-\tan^2\alpha}{1+\tan^2\alpha}}=\frac{1-\frac{3}{2}\cdot\frac{16}{25}}{\frac{7}{16}\cdot \frac{16}{25}}=\frac{1}{25}\cdot\frac{25}{7}=\frac{1}{7}.$$
Edit: I notice Dr. Sonnhard had written something similar, but I guess this is more expanded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the sum of the cubes of the roots in a cubic polynomial? I have an equation, $x^3-x^2+x-2$, with three distinct roots, $p$, $q$ and $r$. What is the value of $p^3+q^3+r^3$?
I'm not sure how to do this. Using Vieta's formula, we know that:
$pqr= 2$
$pq+pr+qr= 1$
$p+q+r= 1$
After this, what should I do?
| Since p, q and r are solutions to the polynomial
$$x^3 - x^2 + x - 2 = 0,$$
i.e. $$ x^3 = x^2 - x + 2,$$
we have the following equations:
$$p^3 = p^2 - p + 2$$
$$q^3 = q^2 - q + 2$$
$$r^3 = r^2 - r + 2 .$$
Summing these equations yields
$$p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.$$
Next we make use of the following identity
$$p^2 + q^2 + r^2 = (p + q + r)^2 - 2(pq+pr+qr).$$
Substiting the above produces
$$p^3 + q^3 + r^3 = (p + q + r)^2 - 2(pq + pr + qr) - (p + q + r) + 6,$$
that is
$$p^3 + q^3 + r^3 = 1^2 - 2(1) - 1 + 6 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$ Give $a,b,c>0$ and $ab^2+bc^2+ca^2=1$
Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$
| For $a=b=c=\frac{1}{\sqrt[3]3}$ we get a value $\frac{1}{3}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a^4}{a+2b}\geq\frac{ab^2+bc^2+ca^2}{3}$$ or
$$\sum_{cyc}\left(\frac{a^4}{a+2b}-\frac{ab^2}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a-b)(3a^2+3ab+2b^2)}{a+2b}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a-b)(3a^2+3ab+2b^2)}{a+2b}-\frac{8}{9}(a^3-b^3)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(19a^2+22ab+16b^2)}{a+2b}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1\int_y^\sqrt{y}\frac{y}{\sqrt{x^2+y^2}}\,dx\,dy$ Not sure how to simplify this and start working it out. Any help and step through would be much appreciated thanks!
Ive gotten to $$\left[\frac{1}{2}y(x^2+y^2)^\frac{1}{2}+\frac{1}{3}x^3\right]^\sqrt{y}_y$$
Is this step right? And how do I carry on from here?
Thanks
| We have $$\int_0^1\int_y^\sqrt{y}\frac{y}{\sqrt{x^2+y^2}}dxdy=\int_0^1\int_{x^2}^{x}\frac{y}{\sqrt{x^2+y^2}}dydx=\int_0^1\left(\sqrt{x^2+y^2}\right)_{x^2}^{x}dx$$ also $$\int_0^1\left(\sqrt{x^2+y^2}\right)_{x^2}^{x}dx=\frac{\sqrt{2}}{2}-\int_0^1x\sqrt{x^2+1}dx$$ now let $t=x^2+1$. From here you can do
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
| Let us first substitute $y = x^{2} + 1$. Then
\begin{align*}
I
&:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\
&= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1}
\end{align*}
Using the substitution $y \mapsto 3-y$, it follows that
$$ I = \frac{1}{2} \int_{1}^{2} \frac{2-y}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy. \tag{2} $$
Thus adding (1) and (2) dividing by 2, we obtain
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}
= \frac{1}{12} \left[ \frac{3 - 2\sqrt{(3-y)y}}{2y-3} \right]_{1}^{2}
= \frac{3-2\sqrt{2}}{6}.
\end{align*}
Of course this may not be a comprehensive answer. My original approach utilized a chain of substitutions:
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}} \\
&= \frac{1}{2} \int_{3/2}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}, \qquad (\because \text{ by symmetry}) \\
&= \frac{1}{2} \int_{0}^{1} \frac{dy}{(9-s^{2}) + 3\sqrt{9-s^{2}}}, \qquad (s = 2x-3) \\
&= \frac{1}{6} \int_{0}^{\arcsin(1/3)} \frac{d\theta}{1+\cos\theta}, \qquad (s = 3\sin\theta) \\
&= \frac{1}{6} \int_{0}^{3-2\sqrt{2}} dt, \qquad (t = \tan(\theta/2)) \\
&= \frac{3-2\sqrt{2}}{6}.
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/1026268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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If $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$, find $x+y$ I was given this factorization problem and I tried many things, but couldn't solve it. Can someone, please, give me a hint?
If $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$, find $x+y$. Here $x, y$ and $p$ are real numbers.
Thank you
| This is a bad question, since:
The identity $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$ does not determine $x+y$.
Example: Let $p=4$, then $(x,y)=\left(0,\frac{15}8\right)$ and $(x,y)=\left(\frac34,\frac34\right)$ are both solutions, for $x+y=\frac{15}8$ and $x+y=\frac32$ respectively.
If one wishes to add the constraint that $$x=y$$ as the OP seems to be willing to do, then one should ask at the onset to solve $$x+\sqrt{x^2 + 1}=\sqrt{p}$$ Then $$(x+\sqrt{x^2+1})\cdot(x-\sqrt{x^2+1})=-1$$ hence every solution $x$ is such that $$2x=(x+\sqrt{x^2+1})+(x-\sqrt{x^2+1})=\sqrt{p}-\frac1{\sqrt{p}}$$ that is, since once again one assumed $x=y$, $$x+y=2x=\frac{p-1}{\sqrt{p}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Proving the inequality $\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$ If a,b,c are non-negative real numbers for which holds that $a+b+c=3$ then prove the following inequality: $$\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$$
| So, we have Schur's inequality which says that for non-negative real numbers a,b,c the following inequality holds: $a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)$.
Now using the an identity which says: $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ and the fact that $ab+c+=3$ we have:
$3(a^2+b^2+c^2-ab-bc-ca)+3abc+3abc\ge ab(3-c)+bc(3-a)+ca(3-b)$
$3(a^2+b^2+c^2)-3(ab+bc+ca)+6abc\ge 3(ab+bc+ca)-3abc$
$3(a^2+b^2+c^2)+9abc\ge 6(ab+bc+ca)$. Dividing both sides with 3 gives us:
$a^2+b^2+c^2+3abc\ge2(ab+bc+ca)$. Now adding to both sides $2(ab+bc+ca)$:
$(a^2+b^2+c^2+2(ab+bc+ca))+3abc\ge4(ab+bc+ca)$
$(a+b+c)^2+3abc\ge4(ab+bc+ca)$. Using now that $a+b+c=3$ and dividng both sides with 4 we get:
$\frac{9}{4}\ + \frac{3abc}4\ge ab+bc+ca$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\Rightarrow 6\sin t\cdot x^2-4\sin t \cdot x-2\sin t = 1-2x+5x^2$
$\displaystyle \Rightarrow (6\sin t-5)x^2+2(1-2\sin t)x-(2\sin t+1) = 0$ , Now for calculation of value of $2\sin t\;,$
equation must have real roots. So $\bf{D\geq 0}$.
So $4(1-2\sin t)^2+4(2\sin t+1)\cdot (6\sin t-5)\geq 0$
$\displaystyle (1-2\sin t)^2+(2\sin t+1)\cdot(6\sin t-5)\geq 0$
So we get $16\sin^2 t-8\sin t-4\geq 0\Rightarrow 4\sin^2 t-2\sin t-1\geq 0$
Now HGow can I solve after that, Help me
Thanks
| i think the answer is $ t \in [\frac{3\pi}{10}, \frac{6\pi}{10}] \cup [\frac{11\pi}{10}, \frac{19\pi}{10}]$ if i have not made any silly arithmetic errors.
here are steps:
(a) establish that the range of the rational function on the right hand side of your equation is $(-\infty, \frac{1-\sqrt 5}{2}] \cup [\frac{1+\sqrt 5}{2}, \infty).$
(b) now your equation reads as $\sin t$ is in $[-1, \frac{1-\sqrt 5}{4}] \cup [\frac{1+\sqrt 5}{4}, 1].$ the solution to this equation is what i claimed at the beginning of this answer.
you can establish (a) by turning $k = {{1-2x + 5x^2} \over {3x2 -2x -1}}$ into a quadratic equation $(3k-5)x^2 -2(k-1)x -(k+1) = 0$ whose discriminant is $4(k^2 - k -1)$ which is positive for $k$ in the range $(-\infty, \frac{1-\sqrt 5}{2}] \cup [\frac{1+\sqrt 5}{2}, \infty).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve $|z^2-1|<|z|^2$ where $z$ is a complex number? How to solve $|z^2-1|<|z|^2$ where $z$ is a complex number? I have tried it both with cartesian and polar coordinates but did not get a solution.
I got that far: $z=x+yi$ and then I got: $$\pm x >(\frac{y^2+0.5}{1+4y^2})^{0.5}$$ but I don't know how to visualise that in the coordinate system.
| With $\;z=x+iy\;$ :
$$|z^2-1|<|z|^2\iff |(x^2-y^2-1)+2xyi|<|x+iy|^2\iff$$
$$\sqrt{(x^2-y^2-1)^2+4x^2y^2}<x^2+y^2\iff(x^2-y^2-1)^2+4x^2y^2<(x^2+y^2)^2\iff$$
$$\color{red}{x^4}-\color{green}{2x^2y^2}+\color{blue}{y^4}-2x^2+2y^2+1+\color{green}{4x^2y^2}<\color{red}{x^4}+\color{green}{2x^2y^2}+\color{blue}{y^4}\iff$$
$$-2x^2+2y^2+1<0\iff x^2-y^2>\frac12$$
This already looks as the exterior of a rather simple hyperbola...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Ways to prove $ \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$ In how many ways can we prove the following theorem?
$$I(n):= \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$$
where $n$ is a nonnegative integer.
The proof I found is by considering $I(n+1)-I(n)$, which can be reduced to
$$
g(n):= \int_0^\pi \frac{\sin(2 n x) \cos x}{\sin x}dx
$$
I then showed that $g(n)=g(n+1)$, with $g(n) = ng(1) = n\pi$. This completes the proof.
I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions.
| I can post my solution , from my document. It can be found on page 83. If one truly want to shoot this problem with a canon one can use the following generalization
$$
\int _{0}^{\pi }\! \left( {\frac {\sin \left( nx \right) }{\sin
\left( x \right) }} \right)^{m}{dx}=\pi \sum _{l=0}^{\large\left\lfloor {\frac{m\left(n-1\right)}{2n}} \right\rfloor }\left( -1 \right) ^{l}{m\choose l}{\dfrac{m}{2}\left( n+1\right) -ln-1\choose m-1}\tag{10}
$$
Which is proved by Graham Hesketh, see equation 10 here. Set $m=1$ use the properties of the floor function and one is is done.
Another way to prove $(n)$ can be found in the same answer
Proof
Lemma: Let $k \in \mathbb{Z}$ then
$$\begin{align*}
\int_0^{\pi}\frac{\sin 2kx}{\sin x}\mathrm{d}x & = 0 \tag{1} \\ \int_0^{\pi}\frac{\sin (2k-1)x}{\sin x}\mathrm{d}x & = \pi \tag{2} \end{align*}$$
Proof: We first define the following function $\displaystyle I_n = \int_0^\pi \frac{\sin 2n x}{\sin x} \mathrm{d}x$. Note that we now have $I_0 = 0$ and $I_1=\pi$. since $\sin 0 = 0$ and $\sin 2x = 2\cos x \sin x$. We have the following relation for all $n$
$$
I_n-I_{n-2}
=\int_0^{\pi}\frac{\sin{nx}-\sin{(n-2)x}}{\sin x}\mathrm{d}x
=2\int_0^{\pi}\cos(n-1)x\mathrm{d}x
=2\left[\frac{\sin{(n-1)x}}{n-1}\right]_0^{\pi}
=0
$$
For $|n|\geq 3$. This means $I_{2k}=I_{2k-2}=\cdots=I_{2}=\pi$ and simmilarly $I_{2k+1} = I_{2k-1}=\cdots I_1=0$. Which is what we wanted to show.
Proposition: Let $k \in \mathbb{Z}$ then
$$
\ell_2(k)
= \int_0^\pi \left( \frac{\sin kx}{\sin x} \right)^2\mathrm{d}x
= \int_0^\pi \frac{1 - \cos kx}{1 - \cos x}\,\mathrm{d}x
= |k|\pi \tag{3}
$$
Proof 1: Here we will follow in the footsteps of the other answer and use the lemma above. We define $\displaystyle J_n=\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x$. We can note that $J_n - J_{n-1}$ is constant eg
$$
\begin{align*}
J_k - J_{k-1}
& = -\frac{1}{2}\int_0^{\pi}\frac{\cos{2kx}-\cos{(2n-2)x}}{(\sin x)^2}
= -\frac{1}{2}\int_0^{\pi}\frac{-2\sin\left(\frac{4k-2}{2}x\right)
\sin\left(\frac{2x}{2}\right)}{(\sin x)^2} \\
& = \int_0^{\pi}\frac{\sin{(2k-1)x}}{\sin x}
= I_{2k-1}=\pi.
\end{align*}
$$
Then we can write
$$J_n=J_{n-1}+\pi=J_{n-2}+2\pi=\cdots=J_1+(n-1)\pi=n\pi.$$
This completes the proof.
Proof 2: Here is another way to solve this problem. I got the idea from chat, but the tecnique is much older. We want to prove that $(I_{n+1} + I_{n-2}) / 2 = I_n$. Eg that $I_n$ is the average of the next and previous term
Some calculations show that
\begin{align*}
\frac{I_{n+1}+I_{n-1}}{2}
& = \frac{1}{2}\int_0^\pi \frac{1 - \cos(n+1)x }{1 - \cos x} + \frac{1-\cos(n-1)x}{1-\cos x} \,\mathrm{d}x \\
& = \frac{1}{2}\int_0^\pi \frac{2 -\bigl[ \cos(n+1)x + \cos(n-1)x\bigr]}{1-\cos x} \,\mathrm{d}x \\
& = \int_0^\pi \frac{1 - \cos nx \cos x}{1-\cos x} \,\mathrm{d}x \\
& = \int_0^\pi \frac{(1-\cos nx) + (1-\cos x) \cos nx}{1-\cos x} \,\mathrm{d}x \\
& = \int_0^\pi \frac{1 - \cos nx}{1-\cos x} \,\mathrm{d}x
= I_n
\end{align*}
Several things was used here. Like $\cos (n+1)x + \cos(n-1)x = 2 \cos nx \cos x$ and $ \int_0^\pi \cos n x\,\mathrm{d}x = 0 \ \forall \ n \in\mathbb{Z} \backslash \{0\}$. Now we have shown that
$$
I_n = \frac{I_{n+1}+I_{n-1}}{2}
\ \Rightarrow \
2I_n = I_{n+1}+I_{n-1}
\ \Rightarrow \
I_n - I_{n-1} = I_{n+1} - I_n
$$
Which is just an arithmetric sequence because the difference between two terms is constant (same as the previous answer). Hence
$I_n = I_0 + (n-0)d = nd$, where $d$ is the difference between two terms. We have $d = I_1 - I_0 = \pi - 0 = \pi$. This completes the second proof. $\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 1
} |
Evaluate $\sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}$
Evaluate $$\sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}$$
So far, I was able to get to
$$\frac{1}{64}\sum_{k=400}^{2000} \frac {1} {8^{2k}\cdot2^{4k}}$$
And then I'm completely stuck.
| $$\begin{align}
\large \sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}
&=\large\sum_{k=400}^{2000}2^{3-4k-3(2k+3)}\\
&=\large\sum_{k=400}^{2000}2^{-10k-6}\\
&=\large\frac 1{2^6}\sum_{k=400}^{2000}\frac 1{2^{10k}}\\
&=\large\frac 1{2^6}\sum_{k=400}^{2000}{u^{k}}\quad \text{where $u=\large \frac 1{2^{10}}$}\\
&=\large\frac 1{2^6}\frac {u^{400}(1-u^{1601})}{1-u}\\
&=\large\large\frac 1{2^6}\frac {\frac 1{2^{4000}}{(1-\frac 1{2^{16010}})}}{1-\frac 1{2^{10}}}\\
&=\large\frac {\frac 1{2^{3996}}{(1-\frac 1{2^{16010}})}}{2^{10}-1}\\
\end{align}$$
...it does look rather messy...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $\sqrt[n]{(x+1)...(x+n)} - x$ as $x \to +\infty$ Let $n \in \mathbb{N}^{\ast}$. I want to determine the following limit :
$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x.$$
Let $x = \frac{1}{t}$ with $t \to 0$. It is equivalent to the following limit :
$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$
$$
\begin{align*}
\sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm]
&= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm]
\end{align*}
$$
where $u(0)=0$ and $\displaystyle \lim \limits_{t \to 0} u(t) = 0$. Since $\displaystyle \sqrt[n]{1+t} = 1 + \frac{t}{n} + o(t)$ as $t \to 0$, $$\displaystyle \sqrt[n]{n! + u(t)} = \sqrt[n]{n!} \; \sqrt[n]{\displaystyle 1 + \frac{u(t)}{n!} } = \sqrt[n]{n!} \bigg( 1 + \frac{u'(0) t}{n(n!)} + o(t) \bigg).$$
I do not see how to go on from there !
| $\bf{My\; Solution::}$ Given $$\displaystyle \lim_{x\rightarrow \infty}\sqrt[n]{(x+1)(x+2)(x+3).....(x+n)}-x$$
Using $\bf{A.M\geq G.M},$ Here $$(x+1),(x+2),(x+3),............(x+n)>0\;,$$ when $x\rightarrow \infty$
So $$\displaystyle \frac{(x+1)+(x+2)+..............+(x+n)}{n}\geq \sqrt[n]{(x+1)(x+2)............(x+n)}$$
and equality hold when $$(x+1)=(x+2)=.............=(x+n)\;,$$ bcz here $x\rightarrow \infty$
So all are equal,
So $$\displaystyle \frac{nx+\frac{n(n+1)}{2}}{n} = \sqrt[n]{(x+1)(x+2)...........(x+n)}$$
So $$\displaystyle \lim_{n\rightarrow \infty}\frac{nx+\frac{n(n+1)}{2}}{n}-x = \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x$$
So $$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x = \frac{n+1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
Trig differentiation Prove that there is a
constant C such that
$$ \arcsin{\frac{1-x}{1+x}} + 2\arctan (\sqrt{x}) = C $$
for all $x$ in a certain domain. What is the largest domain on which this identity is
true? What is the value of the constant $C$?
=====================================================
Well, we can show that they differ by a constant if we show that $\arcsin\frac{1-x}{1+x}$ and $-2\arctan (x^{1/2}) $ have the same derivatives
I am trying to do that and i get $$ (1+x)/2(x^{1/2})$$ for arcsin and
$$ -2/(2(x^{1/2}) + 2x^{3/2}) $$ for arctan
that are not equal... what have I done wrong
| \begin{align}
& \frac d{dx} \arcsin\frac{1-x}{1+x} = \frac 1 {\sqrt{1-\left( \frac{1-x}{1+x} \right)^2}} \cdot \frac d {dx} \frac{1-x}{1+x} \\[10pt]
= {} & \frac 1 {\sqrt{1-\left( \frac{1-x}{1+x} \right)^2}} \cdot \frac{-2}{(1+x)^2} = \frac{1+x}{\sqrt{(1+x)^2- (1-x)^2}} \cdot\frac{-2}{(1+x)^2} \\[10pt]
= {} & \frac{-1}{(1+x)\sqrt{x}} \\[10pt]
& \text{and} \\[10pt]
& \frac d{dx} \arctan\sqrt x = \frac{1}{1+x}\cdot\frac d {dx} \sqrt x = \frac 1 {1+x} \cdot\frac 1 {2\sqrt x}
\end{align}
This is valid on an interval where $1+x$ is positive (otherwise where we had $\sqrt{(1+x)^2}$, it would have been $|1+x|$, etc.). And we must also have $x\ge 0$.
One should also figure out when $\dfrac{1-x}{1+x}$ is in the domain of the arcsine function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to prove $n < \left(1+\frac{1}{\sqrt{n}}\right)^n$ I want to know how to prove the following inequality.
For $n = 1, 2, 3, \ldots $
$$ n < \left(1+\frac{1}{\sqrt{n}} \right)^n $$
I tried with math induction but I failed.
| We have that
$$n < \left(1+\frac{1}{\sqrt{n}} \right)^n\iff \ln n< n\ln \left(1+\frac{1}{\sqrt{n}} \right)\iff \frac{\ln n}{n}<\ln \left(1+\frac{1}{\sqrt{n}} \right).$$
Now, we use the inequalities (see $(3)$ in http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf)
$$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.\tag{1}\end{equation}$$
In our case, we get from $(1),$
$$\ln \left(1+\frac{1}{\sqrt{n}} \right)\ge \frac{2}{\sqrt{n}+2}.$$
Thus we only have to show
$$\frac{2}{\sqrt{n}+2}>\frac{\ln n}{n}.$$ That is,
$$\ln n<\frac{2n}{\sqrt{n}+2}.$$ Since $\ln n=2\ln \sqrt{n}$ previous inequality is equivalent to
$$\ln \sqrt{n}<\frac{n}{\sqrt{n}+2}.$$ Now, using again $(1)$
$$\ln \sqrt{n}=\ln (1+\sqrt{n}-1)\le \frac{\sqrt{n}-1}{2}\frac{\sqrt{n}+1}{\sqrt{n}}=\frac{n-1}{2\sqrt{n}}.$$ Thus we have to show
$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}.$$ But
$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}\iff 2(n-1)<(n+1)\sqrt{n},$$ which holds for any $n\in\mathbb{N}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 2
} |
Find the determinants of the given matrices Consider scalars $a,b,c,d,e,f$ such that
$\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 1 & e\\
c & 1 & f\\ \end{array} \right) = 7$ and $\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 2 & e\\
c & 3 & f\\ \end{array} \right) = 11$
What is $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 3 & e\\
c & 3 & f\\ \end{array} \right)$ and $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 4 & e\\
c & 5 & f\\ \end{array} \right)$?
For the first one, I was thinking that the determinant would be $3\cdot 7 = 21$, and I think the determinant of the second one would still just be $11$ (since it's just a multiple of the second determinant matrix). Does this seem right? How can I solve these kinds of problems?
| Givan an $n \times n$ matrix $A = (a_1, a_2, \ldots, a_{n})$, where $a_{i}$ are column vectors. There are identities
$$det(a_1, \ldots, ca_i, \ldots, a_n) = cdet(A)$$ and
$$det(a_1, \ldots, a_{i-1}, a_i + v, a_{i+1} \ldots, a_n) = det(A) + det(a_1, \ldots, a_{i-1}, v, a_{i+1}, \ldots, a_n)$$, where $v$ is a column vector.
Therefore, returning to your questions,
$$det\begin{pmatrix}
a & 3 & d\\
b & 3 & e\\
c & 3 & f
\end{pmatrix} = 3 \cdot 7 = 21$$
and
$$det\begin{pmatrix}
a & 3 & d\\
b & 4 & e\\
c & 5 & f
\end{pmatrix} = 2 \cdot 7 + 11 = 25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\forall\ x,y,z\in \mathbb{R}$ Show that: $|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$
$\forall\ x,y,z\in \mathbb{R}$ Show that: $$|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$$
i tired,
i notice that $x,y,z$ plays a symmetrical role in the inequality
notice also that
\begin{align*}
|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z| & \Longleftrightarrow \\
(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x+y+z|
\end{align*}
note that $\forall a,b\in \mathbb{R}\quad |a|-|b|\leq |a+b| $
then $$(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x|+|y|+|z|$$
i'm stuck here
any help would be appreciated!
| We need to prove that
$$\left(|x+y+z|+|x|+|y|+|z|\right)^2\geq\left(|x+y|+|x+z|+|y+z|\right)^2$$ or
$$\sum_{cyc}\left(|x(x+y+z)|+|yz|\right)\geq\sum_{cyc}|(x+y)(x+z)|,$$
which is just a triangle inequality:
$$|x(x+y+z)|+|yz|\geq|x(x+y+z)+yz|=|(x+y)(x+z)|.$$
Also we can use Popoviciu.
For all convex function $f$ we have
$$f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\geq2\left(f\left(\frac{x+y}{2}\right)+f\left(\frac{x+z}{2}\right)+f\left(\frac{y+z}{2}\right)\right).$$
Since $f(x)=|x|$ is a convex function, we obtain
$$|x|+|y|+|z|+3\left|\frac{x+y+z}{3}\right|\geq2\left(\left|\frac{x+y}{2}\right|+\left|\frac{x+z}{2}\right|+\left|\frac{y+z}{2}\right|\right)$$ or
$$|x+y+z|+|x|+|y|+|z|\geq|x+y|+|x+z|+|y+z|$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Finding $\lim_{x\rightarrow 0}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}$ We know that $$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan({x}/{2})$$ so we can change the above function to another form as follow $$\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=\frac{(x/2)}{\tan(x/2)}$$ If we use the Taylor series of $\tan(x/2)$ we will get
$$\frac{(x/2)}{\tan(x/2)}=\frac{(x/2)}{\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+....}=\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}$$ Now we will find the limit for the two same functions
$$\lim_{x\rightarrow 0^{+}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^{-}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^+}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=1$$
$$\lim_{x\rightarrow 0^-}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=-1$$
My question is "Which solution is right?"
| Any factors in the limit that don't approach zero or infinity can simply be evaluated:
$$\begin{align}
\lim_{x\to0}\frac{x}{2}\sqrt{\frac{1+\cos x}{1-\cos x}}
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos x}}\\
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos x}}\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}\\
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x\sqrt{1+\cos x}}{\sqrt{1-\cos^2 x}}\\
&=\frac{\sqrt2}{2}\sqrt{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos^2 x}}\\
&=\frac{2}{2}\lim_{x\to0}\frac{x}{\sqrt{\sin^2 x}}\\
&=\lim_{x\to0}\frac{x}{\left|\sin x\right|}\\
\end{align}$$
I think we can see now that the limit does not exist (that it's $1$ from the right and $-1$ from the left).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| If you don't want to depend on the "trick" of raising to the sixth power, you can compare the logs: $\frac 13 \log \frac 12=\frac {- \log 2}3$ and $\frac 12 \log \frac 13=\frac {-\log 3}2$ Now $\frac 12 \gt \frac 13$ and $\log 3 \gt \log 2$, so $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} \lt \left(\frac{1}{2}\right)^{\frac{1}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 2
} |
Is $x^8+1$ irreducible?
How to decide if $f(x)=x^8+1$ is irreducible or not over the following fields:
*
*$F=\mathbb R$
*$F=\mathbb Q$.
I can't use Eisenstein's criterion. So the only possibility is computing the complex 8th-roots of -1?
| We can start with something similar.
Factor $x^4 + 4$:
$x^4 + 4x^2 + 4 - 4x^2$
$(x^2+2)^2 - (2x)^2$
$(x^2+2-2x)(x^2+2+2x)$
$\boxed{x^4 + 4 = (x^2-2x+2)(x^2+2x+2)}$
We could use the same method to factor $x^4+1$:
$x^4 + 2x^2 + 1 - 2x^2$
$(x^2+1)^2 - (\sqrt{2}x)^2$
$(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$
$\boxed{x^4 + 1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$
Here's what I discovered:
$x^4 + y^4 = (x^2+y^2-\sqrt{2}xy)(x^2+y^2+\sqrt{2}xy)$
This shows that every sum of fourth powers can be factored. Now let's factor $x^8+1$:
$(x^2)^4 + 1^4$
$((x^2)^2+1-\sqrt{2}x^2)((x^2)^2+1+\sqrt{2}x^2)$
$\boxed{(x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)}$
Hope it helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Investigate convergence of the following series Investigate the corvergence of the following series?
$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13}+\frac{1}{15}-\frac{1}{8}+ \ldots$
$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}+ \ldots$
I need a hit about which cirterion to use in each of those cases and i do not know how to find a general term of those twos sequence?
| For the first: rearrange the terms to get $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+ \ldots$$ Note that the first part of this (until $+\frac{1}{9}$) is exactly the alternating harmonic series, which is known to converge. However, there remain missing three even terms ($-\frac{1}{10}, -\frac{1}{12},-\frac{1}{14}$). In fact, there will always be even terms missing since the series you gave has two odd terms for every even term. Furthermore, as the series goes on the number of stray odd terms increases to infinity. So we know the first "half" of the series converges, since it is the alternating harmonic series, the question then becomes: does the harmonic series of only odd numbers converge or diverge?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Computing area of triangle via equations of medians For a triangle $ABC$,
$B=90^\circ , AC=6$, equations of medians through $A$ is $y=2x+4$ and through $C$ is $y=x+3$. What is the area of triangle $ABC$?
I'm really bad at geometry, and to make matters worse, the equations of the medians are given. I have no idea where to start. A hint would be appreciated.
|
Let denote the sides $|BC|=a$, $|AC|=b$, $|AB|=c$
and medians $|AA_m|=m_a$, $|BB_m|=m_b$, $|CC_m|=m_c$
of $\triangle ABC$.
Also, let
$x=|GA_m|=\tfrac13|AA_m|$,
$y=|GC_m|=\tfrac13|CC_m|$.
Since $\angle ABC=90^\circ$,
\begin{align}
|BB_m|=m_b=\tfrac12|AC|=\tfrac12b=3
.
\end{align}
It is known that for such right triangle
\begin{align}
a^2+c^2&=b^2
\tag{1}\label{1}
,\\
m_a^2+m_c^2&=5m_b^2
,\\
x^2+y^2&=\tfrac59\,m_b^2
=\tfrac5{36}\,b^2
\tag{2}\label{2}
.
\end{align}
\begin{align}
\triangle GCA_m:\quad
|CA_m|^2&=
|GA_m|^2+|GC|^2
-2|GA_m|\cdot|GC|\cos\theta
,\\
\tfrac14a^2&=
x^2+4y^2-4xy\cos\theta
,\\
\tfrac14a^2&=
\tfrac59b^2-3x^2-4xy\cos\theta
\tag{3}\label{3}
.
\end{align}
Similarly,
\begin{align}
\triangle GC_mA:\quad
|AC_m|^2&=
|GC_m|^2+|GA|^2
-2|GC_m|\cdot|GA|\cos\theta
,\\
\tfrac14c^2&=
4x^2+y^2-4xy\cos\theta
,\\
\tfrac14c^2&=
\tfrac59b^2-3y^2-4xy\cos\theta
\tag{4}\label{4}
.
\end{align}
Combination of \eqref{3}+\eqref{4}
with \eqref{1},\eqref{2}
results in
\begin{align}
y&=\frac{b^2}{18\,x\cos\theta}
\tag{5}\label{5}
\end{align}
Substitution of \eqref{5} into \eqref{2}
provides a quadratic equation in $x^2$
with the pair of roots,
which represents two combinations of $x^2,y^2$.
\begin{align}
x^2&=\tfrac5{72}\,b^2\left(1 \pm \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right)
,\\
y^2&=\tfrac5{72}\,b^2\left(1 \mp \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right)
.
\end{align}
The area of a general $\triangle ABC$
in terms of its medians is
\begin{align}
S&=\frac13\sqrt{4m_a^2m_c^2-(m_a^2+m_c^2-m_b^2)^2}
,
\end{align}
and in case of this right triangle we have
\begin{align}
S&=\frac13\sqrt{4m_a^2m_c^2-(4m_b^2)^2}
.
\end{align}
Using $m_a^2=9x^2$, $m_c^2=9y^2$,
we arrive at a nice
expression for the area of the right triangle
in terms of its hypotenuse $b$
and the angle $\theta$
between its medians $AA_m$ and $CC_m$
\begin{align}
S&=\tfrac13\,b^2\tan\theta
.
\end{align}
The length of hypotenuse is given,
and the angle $\theta=\angle A_mGC=\angle AGC_m$
and its tangent and cosine
can be derived
from the equations $y=2x+4$, $y=x+3$
of the lines through medians,
\begin{align}
\tan\theta&=\tan(\arctan(2)-\arctan(1))
=\frac{\tan(\arctan(2))-\tan(\arctan(1))}{1+\tan(\arctan(2))\tan(\arctan(1))}
=\frac13
,\\
\cos\theta&=\frac{3\sqrt{10}}{10}
.
\end{align}
Thus the sought area is
\begin{align}
S&=\tfrac13 \cdot 6^2 \cdot \tfrac13=4
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why are $x$, $x^3+1$ and $x^2+x+1$ always mutually co-prime, for any natural number $x$? I just read that for a natural number $x$, the three numbers $x$, $x^3 + 1$ and $x^2 + x + 1$ are all mutually co-prime.
I couldn't find a reason why this is true. OK, any of them does not divide either of the other two, but is it enough to conclude that their GCD is 1?
Couldn't they have any common factors?
Also on what basis, are two consecutive integers co-prime? I know they cannot have a common factor but don't know why.
| If integer $d$ divides both $x^2+x+1, x^3+1;$
$d$ must divide $x(x^2+x+1)-(x^3+1)=x^2+x-1$
$\implies d$ must divide $x^2+x+1-(x^2+x-1)=2$
But $2\mid x(x+1)\implies x^2+x\pm1=x(x+1)\pm1$ are odd $\implies d=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find coordinate vector for change of bases when matrix is non-square Find $x_{b2}$ if $x_{b1} = \begin{pmatrix}2 \\ 5 \end{pmatrix}_{b1}$
Let $$ B_{1} = \begin{pmatrix}
3 & -1 \\
-5 & 4 \\
5 & -3
\end{pmatrix}\,
\ \;\;\;\;\;\;\;\;\;\; B_{2}=
\begin{pmatrix}
2 & 1 \\
-1 & 3 \\
2 & -1
\end {pmatrix}
$$
I know I have to use the equation $x_{b2} = V^{-1}Ux_{b1}$, where $V =B_{2}$ by V is not invertible since it is not a square matrix. The textbook provides a solution in the back of the book, but no examples on how to deal with non-square matrices. How do I solve these type of questions.
| The coordinates of your vector, $x$, relative to the basis $B_1$ are $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$. This means that
$$ x = B_1 \begin{bmatrix} 2 \\ 5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -5 \\ 5 \end{bmatrix} + 5\begin{bmatrix} -1 \\ 4 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$
To find $x$'s coordinates relative to the other basis, you need to solve the system:
$$ B_2 \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 3 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $A^2$ is a symmetric matrix? Conjecture 1 :
Let $A$ be a real matrix such that $A^5=A A^T A A^T A$. Then $A^2$ is a symmetric matrix.
(here $A^T$ denotes the transpose of a matrix A).
I guess that the following is also true :
Conjecture 2 :
If $A^{2n+1}=AA^TAA^T\cdots AA^TA$ then
$A^n $ is symmetric.
PS: This second conjecture has been shown to be false when $A$ is invertible, see Robert Israel's answer below. But I still think that it's true when $A$ is not invertible.
The post if matrix such $AA^T=A^2$ then $A$ is symmetric? solves the $n=1$ case.
| Assume that $A\in M_3(\mathbb{R})$. Since the cases when $rank(A)=1,3$ are solved, assume that $rank(A)=2$.
Prop. Under the hypothesis above, $A^5=AA^TAA^TA$ implies that $A^2$ is symmetric.
Proof. Let $A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&k\end{pmatrix}$. We may assume that $AA^T=diag(u,v,0)$ where $u,v>0$. Clearly $g=h=k=0$ and $A^2=\begin{pmatrix}a^2+bd&b(a+e)&ac+bf\\d(a+e)&bd+e^2&cd+ef\\0&0&0\end{pmatrix}$.
Since $A$ is defined up to a real factor, we may assume $u=1$ and finally we study the system $AA^T=diag(1,v,0)$, $A^5=\begin{pmatrix}a&b&c\\v^2d&v^2e&v^2f\\0&0&0\end{pmatrix}$. We use the Grobner basis software of Maple ; here the difficulty is that there are an infinity of solutions over $\mathbb{C}$ and over $\mathbb{R}$ and consequently, we must work also with hand !
We obtain $ad+be+cf=0$ and $d(v^2-1)=b(v^2-1)=cf(v^2-1)=0$.
Case 1. $v=1$. Then $a^2+b^2+c^2=d^2+e^2+f^2=1$ and $(c^2+f^2)(c^2+f^2-2)=0$.
Case 1.1. $c^2+f^2=2$. Then $c=\pm 1,f=\pm 1$, $a=b=d=e=0$ that is contradictory because $ad+be+cf\not=0$.
Case 1.2. $c^2+f^2=0$. Then $A^2$ is symmetric.
Case 2. $v\not=1$ and $d=b=cf=0$. We obtain $c^2(c^2-2)=0,a^2+c^2=1$ that implies $c=0,a^2=1$.
Case 2.1. $f=0$. then $A^2$ is symmetric.
Case 2.2. $f\not=0$. Then $2e^2+f^2=0$, that is contradictory.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d\theta$
$v = \sec \theta$
$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$
$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$
I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?
| First integrate by parts, then substitute $x=\sec{\theta}$ $\implies \mathrm{dx}=\tan{\theta}\sec{\theta}\,\mathrm{d}\theta$:
$$\begin{align}
\int\frac{x\,\ln{\left(x\right)}}{\sqrt{x^2-1}}\,\mathrm{d}x
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{x^2-1}}{x}\,\mathrm{d}x\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{\sec^2{\theta}-1}}{\sec{\theta}}\,\tan{\theta}\sec{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\tan^2{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\left(\sec^2{\theta}-1\right)\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\tan{\theta}+\theta+\color{grey}{constant}\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\sqrt{x^2-1}+\sec^{-1}{x}+\color{grey}{constant}\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Why does this sequence converge? I have to deal with the following sequence :
$\lim \limits_{x \to \infty}\sqrt{x+\sqrt{x}} - \sqrt{x}$
If I factorize it to $\sqrt{x}(\sqrt{\sqrt{x}+1}-1)$, I would say it diverges since both factors diverge: $\lim \limits_{x \to \infty}\sqrt{x}= \infty $ and $\lim \limits_{x \to \infty} \sqrt{\sqrt{x}+1} = \infty$
But if I type it in WolframAlpha, I get $\frac12$ as limit. Can you help me out?
| Multiply with the conjugate $\sqrt{x+\sqrt{x}}+\sqrt{x}$ to obtain that $$\begin{align*}\sqrt{x+\sqrt{x}}-\sqrt{x}&=\frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\overset{(1)}=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\\[0.2cm]&=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\frac{\sqrt{\not x}}{\sqrt{\not x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\\[0.2cm]&=\frac{1}{\sqrt{\frac{1}{\sqrt{x}}+1}+1} \to \frac{1}{\sqrt{0+1}+1}=\frac12\end{align*}$$ as $x \to \infty$. In (1) we used the identity $$(a-b)(a+b)=a^2-b^2$$ with $a=\sqrt{x+\sqrt{x}}$ and $b=\sqrt{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Solving $3t^2-\frac{12}{3}t+\frac{4}{3}=0$ I need to to solve:
$$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$
The solution manual factorizes this to $\dfrac{1}{3}(3t-2)^2$. How can you do this easily?
| A problem like this is easy if you make a substitution, for example.
$$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$
$$\frac{9}{3}t^2-\frac{12}{3}t+\frac{4}{3}=0$$
$$\frac{1}{3}(9t^2-12t+4)=0$$
$$\frac{1}{3}((3t)^2-4(3t)+4)=0$$
Now, let $s=3t$
$$\frac{1}{3}(s^2-4s+4)=0$$
$$\frac{1}{3}(s-2)^2=0$$
Substituting back we have
$$\frac{1}{3}(3t-2)^2=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| The formal statement is
$$\sqrt{1 + 10^k + 5 \sum_{j=0}^{k-1} 10^j} = 2\sqrt{3\cdot 10^{k-1} + 8 \sum_{j=1}^{k-2} 10^j + 9} \quad\forall k\ge 2$$
The base case can be shown by verifying $\sqrt{156} = \sqrt{4\cdot 39} = 2\sqrt{39}$. Now try induction (first you should eliminate the square root). We get
$$\begin{align*}
1 + 10^{k+1} + 5 \sum_{j=0}^k 10^j &= 10^{k+1} + 1 + 5\cdot 10^k + 5\sum_{j=0}^{k-1} 10^j \\
&\stackrel{\text{i.h.}}= 14 \cdot 10^k +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\
&= 140 \cdot 10^{k-1} +4\cdot\left(3\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9\right) \\
&= 4 \cdot \left( 38 \cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\
&= 4 \cdot \left( 3 \cdot 10^k + 8\cdot 10^{k-1} + 8\sum_{j=1}^{k-2} 10^j + 9 \right) \\
&= 4 \cdot \left( 3 \cdot 10^k + 8\sum_{j=1}^{k-1} 10^j + 9 \right)
\end{align*}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How to find the coefficient of a power in a power series?? How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$
Is there a general method to this?
| We have
$(1+x+x^2+\cdots)(x^2+x^4+\cdots)(1+x^3+x^5)$
Then we want The coeficient of $x^{80}$, note that $x^nx^m=x^{n+m}$ then searching The ways of write $80$ using the sets $\{0,1,\cdots\},\{2,4,\cdots\},\{0,3,5\}$.
$$\begin{align}
80&=0+80+0\\
&=2+78+0\\
&=4+76+0\\
&\vdots\\
&=78+2+0\\
&=1+76+3\\
&=3+74+3\\
&=5+72+3\\
&\vdots\\
&=75+2+3\\
&=1+74+5\\
&=3+72+5\\
&\vdots\\
&=73+2+5
\end{align}$$
Then The coeficiente will be
$$\begin{align}
C_{80}&=\stackrel{2n}{(40-1+1)}+\stackrel{2n+1}{(37-0+1)}+\stackrel{2n+1}{(36-0+1)}\\
&=40+38+37\\
&=115
\end{align}$$
Then if nothing as wrong The answer must be $115$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $b_n=a_n+a_{n+1}$ is converge to $L$. How can I prove that $\lim_{n\rightarrow\infty}\frac{a_n}{n}=0$ If $b_n=a_n+a_{n+1}$ is converge to $L\in\mathbb{R}$. How can I prove that $$\lim_{n\rightarrow\infty}\frac{a_n}{n}=0$$
| Based on Arthur's answer, Here is an answer to my question.
Since $b_n$ converges to $L$, for any $\epsilon>0$ there is a $N\in\mathbb{N}$ such that
$n\geq N$ implies $|b_n-L|=|a_n+a_{n+1}-L|<\epsilon/2$. Then, for $n\geq N$, we have
\begin{align*}
|a_{n+2}-a_n|&\leq|a_{n+2}+a_{n+1}-L+L-a_{n+1}-a_{n}|\\
&\leq
|a_{n+2}+a_{n+1}-L|+|a_{n+1}+a_n-L|\\
&=\epsilon\,.\end{align*}
For a large enough $n>>N$, we get
\begin{align*}
|a_{n+2}-a_N|&=|a_{n+2}-a_n+a_n-a_{n-2}+a_{n-2}+a_{n-1}+\cdots+a_{N+2}-a_N|\\
&\leq |a_{n+2}-a_n|+|a_n-a_{n-2}|+\cdots+|a_{N+4}-a_{N+2}|+|a_{N+2}-a_N|\\
&\leq |a_{N+n+2-N}-a_{N+n-N}|+|a_{N+n-N}-a_{N+n-2-N}|+\cdots+|a_{N+4}-a_{N+2}|+|a_{N+2}-a_N|\\
&< \epsilon\left(\frac{n+2-N}{2}\right)
\end{align*}
Rewriting the last inequality:
$$-\left(a_N+\epsilon\left(\frac{n+2-N}{2}\right)\right)\leq a_{n+2}\leq a_N+\epsilon\left(\frac{n+2-N}{2}\right)\,,$$
and
$$-\left(\frac{|a_N|}{n+2}+\frac{\epsilon N}{n+2}+\frac{\epsilon}{2}\right)\leq \frac{ a_{n+2}}{n+2}\leq \frac{|a_N|}{n+2}+\frac{\epsilon N}{n+2}+\frac{\epsilon}{2}\,.$$
By taking a limit of $n\rightarrow\infty$, since $a_N$ and $N$ are fixed number, we get
$$-\frac{\epsilon}{2}<\lim_{n\rightarrow\infty}\frac{a_{n+2}}{n+2}<\frac{\epsilon}{2}\,.$$
Since the above inequality holds for any choice of $\epsilon,$ we get $$\lim_{n\rightarrow \infty}\frac{a_n}{n}=\lim_{n\rightarrow \infty}\frac{a_{n+2}}{n+2}=0\,.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of
$$x^4-2(a+b)x^2+(a-b)^2=0$$
are......
$$x=\sqrt{a}+\sqrt{b}$$
$$x=\sqrt{a}-\sqrt{b}$$
$$x=-\sqrt{a}+\sqrt{b}$$
$$x=-\sqrt{a}-\sqrt{b}$$
When $a$ and $b$ are real numbers(negative or positive)
I proved this by substituting $x=\sqrt{a}+\sqrt{b}$ in the main equation , but I think this way not effective because It needs to be repeated four times to complete the proving.
Is there another effective proving.
| since
$$(x^2-a-b)^2=4ab\Longrightarrow x^2=a+b\pm 2\sqrt{ab}=(\pm\sqrt{a}\pm \sqrt{b})^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Alternate way to Prove or disprove $6\mid n(n+1)(n+2)$ This is my proof, I'm wondering if I'm correct, and how to do without induction.
My Work
Basis Step
$$\frac{(1)(2)(3)}{6} = 1$$
Inductive Hypothesis
Assume that $\dfrac{k(k+1)(k+2)}{6} = d$ where $d \in \mathbb Z$
Induction Step
We must show $\frac{(k+1)(k+2)(k+3)}{6}$ is an integer
$$\frac{(k+1)(k+2)(k+3)}{6} = \frac{(k+1)(k+2)}{6} \cdot (k+3)$$
We know that:
$\dfrac{(k+1)(k+2)}{6} = \dfrac{1}{k}\cdot\dfrac{k(k+1)(k+2)}{6}$ by our inductive hypothesis we know this to be an integer
Therefore, by the closure property of integers under multiplication and addition, $\dfrac{(k+1)(k+2)}{6} \cdot (k+3)$ must be integer.
| If you insist on using induction:
Let $6|n(n+1)(n+2)$.
It is clear that one of the factors $n+1$, $n+2$ is even, so that $2|(n+1)(n+2)$.
If also $3|(n+1)(n+2)$ then $6|(n+1)(n+2)$ and consequently $6|(n+1)(n+2)(n+3)$ and we are ready.
If $3\nmid\left(n+1\right)\left(n+2\right)$ then we can conclude that
$3\mid n$ . Then also $3\mid n+3$ resulting again in $6\mid\left(n+1\right)\left(n+2\right)\left(n+3\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for a variable in the power when the base are two different values I would like to solve for $C$
$$7^C \times 2^{n-C-1} \le \frac{2^n}{100}$$
Real questions. The different base is really throwing me off. I got up to
$$7^C 2^{-C} \le \frac{1}{50} $$
| $$\begin{align*}7^C\cdot 100 \le 2^{n-n+C+1} &\iff 7^C\cdot 100 \le 2^{C+1}\\ &\iff 7^C\cdot100\le 2^C \cdot 2 \\&\iff \left(\frac{7}{2}\right)^C\le \frac{2}{100} \\&\iff \ln\left(\frac{7}{2}\right)^C \le \ln \frac{2}{100} \iff C \ln \frac{7}{2}\le \ln \frac{2}{100}\end{align*}$$
After editing your question, that you reached $7^C2^{-C}\le \frac{1}{50}$ then you only need to write the LHS as a fraction (optional actually) and to apply the logarithm on both sides in order to obtain $$\ln \left(\frac{7}{2}\right)^C\le \ln \frac{1}{50} \iff C \ln \frac{7}{2}\le \ln \frac{1}{50} \overset{\ln \frac72>0}\iff C\le \frac{-\ln 50}{\phantom{+}\ln \frac72}=-3.122$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what values of a will the system have a unique solution, and for which pair of values (a,b) will the system have more than one solution consider the following linear system
$$x+2y+2z=1\tag{1}$$
$$x+ay+3z=3\tag{2}$$
$$x+11y+az=b\tag{3}$$
in matrix form
$$\pmatrix{1&2&2&1\cr1&a&3&3\cr1&11&a&b\cr}$$
For what values of a will the system have a unique solution, and for which pair of values (a,b) will the system have more than one solution
My attempt at solving it:
-1*line 1 + line 2, and, -1*line 1 + line 3
$$\pmatrix{1&2&2&1\cr0&a-2&1&2\cr0&9
&a-2&b-1\cr}$$
1/(a-2) * line
$$\pmatrix{1&2&2&1\cr0&a-2&1&2\cr0&9/(a-2)&1&(b-1)/(b-2)\cr}$$
-1*line 3 + line 2
$$\pmatrix{1&2&2&1\cr0&((a-5)(a+1))/(a-2)&0&(-b+2a-3)/(a-2)\cr0&9/(a-2)&1&(b-1)/(b-2)\cr}$$
I don't know how to proceed beyond this, help guide me stack exchange!
Also if someone could touch up my matrices formatting that would be soooo cool.
| The original system is equivalent to the following under-determined system:
$$x+2y+2z+w=0\tag{1}$$
$$x+ay+3z+3w=0\tag{2}$$
$$x+11y+az+bw=0\tag{3}$$
Solve $z,w$ from (1) and (2) and substitution of it into (3) leads to:
$$(3-2a+b)x+(33+a^2+6b-2a(3+b))y=f(a,b)x+g(a,b)y=0\tag{4}$$
(1)If $a=-1,b=-5$ or $a=5,b=7$, then $f(a,b)=g(a,b)=0$. Thus $x,y$ can take any values.
(2) If $a\not=-1,5$ and $b=2a-3$, then $f(a,b)=0$ but $g(a,b)\not=0$, so $y=0$ and $x$ can take any value.
(3) If $a\not=-1,5$ and $b=\frac{33-6a+a^2}{2(a-3)}$, then $g(a,b)=0$ but $f(a,b)\not=0$, so $x=0$ and $y$ can take any value.
(4) If $a\not=-1,5$, $b\not=2a-3,\frac{33-6a+a^2}{2(a-3)}$, then $f(a,b)\not =0,g(a,b)\not=0$, so $x=-\frac{g(a,b)}{f(a,b)}y$ and $y$ can take any value.
| {
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"url": "https://math.stackexchange.com/questions/1064530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Determining the last two digits of $229^{10} +37^{10}$ Determine the last two digits of: $229^{10}+37^{10}.$
I do not want to use the Euler-totient function or the carmichael function please!
Thanks
| $37^{10}\equiv 7^{10}\equiv 49\pmod{100}$
$229^{10}\equiv 9^{10}\pmod {100}$
$\begin{align}9^3\equiv 29\pmod {100}, 9^2\equiv-19\pmod{100} & \implies 9^5\equiv -(29\cdot19)\pmod{100}\\&\implies 9^5\equiv -551\pmod{100}\\&\implies 9^5\equiv -51\pmod{100}\\&\implies 9^{10}\equiv 1\pmod{100}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the integral solutions to $ x^2+y^2+z^2=x^2y^2$ I am unfamiliar with this type of problem. How does one solve this and under what category of math does this fall under.
Find the integral solutions for $x^2+y^2+z^2=x^2y^2$
| $$ z^2 + 1 = (x^2 - 1)(y^2 - 1). $$
The main thing you need to know from quadratic forms is that $z^2 + 1$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4,$ indeed not divisible by any (positive) number $n \equiv 3 \pmod 4.$
If either $x$ or $y$ is even and nonzero, let's say $x,$ then $x^2 - 1 \equiv 3 \pmod 4,$ not allowed.
When both $x,y$ are odd, both $x^2 -1$ and $y^2-1$ are even and $z^2+1$ is divisible by $4,$ again not allowed.
Therefore $(0,0,0)$ is it.
| {
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"timestamp": "2023-03-29T00:00:00",
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$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.
| For me,
$4x + 3y + 2z = \langle\begin{pmatrix} \sqrt{2} x \\ \sqrt{3}y \\ 2z\end{pmatrix}, \begin{pmatrix} 2\sqrt{2} \\\sqrt{3} \\\ 1\end{pmatrix}\rangle \leq 1\times 2\sqrt{3}=2\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proving limit through definition Prove
$$\lim_{x\to 2}\frac{x^2+4}{x+2}=2$$
through definition.
My solution:
Fix $\epsilon >0$ and find $\delta$
\begin{align}
0<|x-2|<\delta &\Rightarrow \left| \frac{x^2+4}{x+2}-2 \right| < \epsilon\\
&\Rightarrow\left|\frac{x(x-2)}{x+2}\right| < \epsilon
\end{align}
Let $\delta <1$, then $0<|x-2|<1$ then $x\in (1,3)$ and $x>0$ and $x+2>0$
$$0<|x-2|<1 \Rightarrow \frac{x}{x+2}|x-2|<\epsilon$$
In conclusion $\delta :=\min \left\{ 1,\frac{5\epsilon}3\right\}$
Unfortunately the answer is not correct (according to my book). It says $\delta :=\min \left\{ 1,\epsilon \right\}$. Where did I make a mistake?
| I believe you got it wrong in the end. Remember that $1 < |x| < 3$ if we suppose $\delta < 1$. So: $$\begin{align}
0<|x-2|<\delta &\Rightarrow \left| \frac{x^2+4}{x+2}-2 \right| < \epsilon\\
&\Rightarrow\left|\frac{x(x-2)}{x+2}\right| < \frac{3\delta}{|3| - 2} = 3 \delta < \epsilon
\end{align}$$
So take $\delta = \min\{1, \epsilon/3\}$. But $\epsilon/3 < \epsilon$, so we're good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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More rigorous method for this elementary problem? The problem is:
Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$
One solution I received was as follows:
$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to solve $$(\sqrt{3}+\sqrt{2})^{2x}+(\sqrt{3}-\sqrt{2})^{2x}=2\sqrt{3}$$
for $x$. Note that $$(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})=2\sqrt{3}$$
so $x=\dfrac{1}{2}$ is a solution.
The problem with the solution is that it doesn't prove that $\dfrac{1}{2}$ is the only solution, not to mention its absolute inelegancy. Is there a more mathematically rigorous way of doing this?
I don't mind if higher algebra is used. I'm currently attempting to self-teach myself group theory, so including some of that in a solution would do me good.
| $\left(\sqrt{3}+\sqrt{2}\right)^{2x}+\left(\sqrt{3}-\sqrt{2}\right)^{2x}=\left(\left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}\right)^2-2$
$\therefore \left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2+2\sqrt{3}}\tag{1}$
Similarly,
$\left(\sqrt{3}+\sqrt{2}\right)^{x}-\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2\sqrt{3}-2}\tag{2}$
$\begin{align}\therefore (1)+(2) \implies \left(\sqrt{3}+\sqrt{2}\right)^{x} & =\dfrac{1}{2}\left(\sqrt{2\sqrt{3}+2}+\sqrt{2\sqrt{3}-2}\right)\\&= \dfrac{1}{\sqrt{2}}\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)\end{align}$
$$\color{blue}{\boxed{\begin{align}\therefore \left(\sqrt{3}+\sqrt{2}\right)^{2x} & = \dfrac{1}{{2}}\left(2\sqrt{3}+2\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)\end{align}}}$$
| {
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"url": "https://math.stackexchange.com/questions/1071625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| $x^4+y^4+z^4=\left(x^2+y^2+z^2\right)^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\tag{1}$
$\left(x^2y^2+y^2z^2+z^2x^2\right) =(xy+yz+zx)^2-2xyz(x+y+z)\tag{2}$
Now note that
$$\color{red}{\boxed{(xy+yz+zx)=\dfrac{1}{2}\left((x+y+z)^2-\left(x^2+y^2+z^2\right)\right)}}$$
from which you get the value of $xy+yz+zx$. Also, note that,
$$\color{blue}{\boxed{x^3+y^3+z^3-3xyz=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)}}$$
from which you get the value of $xyz$.
Can you take it from here?
| {
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Find $ \int \frac {1-x^2}{1+3x^2+x^4} \, \mathrm{d}x $ Today, the CalcBee sample problems got released. The following problem was my creation and I wanted to see how many solutions people can come up with. The result is very beautiful and I thought it would be instructive to see multiple solutions to this if they exist. Note that, on the real contest, none of the problems will be this hard. This was meant as a challenge since partial fractions do not exactly work right away.
Find $ \displaystyle\int \frac {1-x^2}{1+3x^2+x^4} \, \mathrm{d}x $.
| For the record, for ease of access, here is my solution.
Notice that the degree of the denominator is double the degree of the numerator, so there is a possibility that the answer is of the form $ \arctan \left( f(x) \right) $, where $f(x)$ is a rational function. So we try it out. If this is the case, we have $$ \frac {f'(x)}{1 + \left[ f(x) \right]^2} = \frac {1-x^2}{1+3x^2+x^4}. $$Then, let $ f(x) = \frac {p(x)}{q(x)} $, where $p(x)$ and $q(x)$ are polynomials with degree $a$ and $b$, respectively. Then, $$ f'(x) = \frac {q(x) p'(x) - p(x) q'(x)}{\left[ q(x) \right]^2}, $$where the numerator has degree less than or equal to $a+b-1$ and the denominator has degree $2b$. Therefore, $b=2$ and $a=1$. Then, $$ f(x) = \frac {rx + s}{tx^2 + ux + v} \implies f'(x) = \frac {(rv-su) - 2stx - rtx^2}{t^2 x^4 + 2tux^3 + (2tv+u^2) x^2 + 2uvx + v^2}. $$Then, the derivative of $ \arctan \left( f(x) \right) $ is $$ \frac {f'(x)}{1 + \left[ f(x) \right]^2} = \frac {\frac {(rv-su) - 2stx - rtx^2}{t^2 x^4 + 2tux^3 + (2tv+u^2) x^2 + 2uvx + v^2}}{1 + \left( \frac {rx+s}{tx^2 + ux + v} \right)^2}. $$We find that $p(x)=x$ and $q(x)=1+x^2$, so our answer is $$ \boxed {\arctan \left( \frac {x}{1+x^2} \right)}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions of $z^6 + 1 = 0$ Solve:
$$z^6 + 1 = 0$$
That lie in the top region of the plane.
We know that:
$$(z^2 + 1)(z^4 - z^2 + 1) = 0$$
$$z = -i, i$$
We need to solve:
$$((z^2)^2 - (z)^2 + 1) = 0$$
$$z = \frac{1 \pm \sqrt{-3}}{2}$$
But this is incorrect. How to do this then?
| In order to solve $\sqrt[2n]{-1}$:
*
*Draw the unit circle
*Draw the first solution, which is obviously $0+1i=\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})i$
*Repeat $2n-1$ times: find the next solution by rotating the previous solution $\frac{\pi}{n}$ radians
For example, $\sqrt[6]{-1}$:
*
*$\cos(\frac{ 3\pi}{6})+\sin(\frac{ 3\pi}{6})i$
*$\cos(\frac{ 5\pi}{6})+\sin(\frac{ 5\pi}{6})i$
*$\cos(\frac{ 7\pi}{6})+\sin(\frac{ 7\pi}{6})i$
*$\cos(\frac{ 9\pi}{6})+\sin(\frac{ 9\pi}{6})i$
*$\cos(\frac{11\pi}{6})+\sin(\frac{11\pi}{6})i$
*$\cos(\frac{13\pi}{6})+\sin(\frac{13\pi}{6})i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077510",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Why does the square root of a square involve the plus-minus sign? If $\sqrt{x^2}$ can be simplified as follows:
$\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{\frac{2}{1}\times\frac{1}{2}} =x^\frac{2}{2} = x^1 = x$
Then why would $\sqrt{x^2} = \pm x$?
| Because $\sqrt{x^2}$ can't be simplified "as follows." To be more precise: the notation $\sqrt{x^2}$ naturally designates a number $y$ such that $y^2=x^2$. Among real numbers there are two such $y$, namely, $y=-x$ and $y=x$. Your mistake is in extending the rule $(x^a)^b=x^{ab}$, valid for $a$ and $b$ integers, to allow $a$ and $b$ to be fractions.
| {
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Integral along a contour is $0$, how? I recently had an extremely failed attempt at asking the same question, so I am posting the same question more or less to hope that someone can give me feedback.
Consider the integral:
$$\int_{0}^{\infty} \frac{\log^2(x)}{x^2 + 1} dx$$
$\hskip1in$
Image taken and modified from: Complex Analysis Solution (Please Read for background information).
$R$ is the big radius, $\delta$ is the small radius.
Actually, lets consider $u$ the small radius. Let $\delta = u$
Ultimately the goal is to let $u \to 0$
We can parametrize,
$$z = ue^{i\theta}$$
$$\int_{\delta} f(z)dz = (-)\cdot\int_{0}^{\pi} \frac{(i\theta + \log(u))^2\cdot (uie^{i\theta})}{(ue^{i\theta})^2 + 1} d\theta$$
$$\left | \int_{0}^{\pi} \frac{(i\theta + \log(u))^2\cdot (uie^{i\theta})}{(ue^{i\theta})^2 + 1} d\theta \right | \le \int_{0}^{\pi} \frac{|(i\theta + \log(u))|^2\cdot(u)}{|(ue^{i\theta})^2 + 1 |} d\theta$$
$$|(ue^{i\theta})^2 + 1 | < u^2 + 1 $$
$$\frac{1}{u^2 + 1} < \frac{1}{|(ue^{i\theta})^2 + 1 |}$$
Since the maximum value of $\theta$ is $\theta = \pi$
$$|(i\theta + \log(u))| = \sqrt{\log^2(u) - \theta^2} \le \sqrt{\log^2(u) + \pi^2}$$
So:
$$|(i\theta + \log(u))|^2 \le \log^2(u) + \pi^2$$
Then:
$$|(i\theta + \log(u))|^2 \le \log^2(u) + \pi^2$$
For values $u$ near $0$.
$$(u)|(i\theta + \log(u))|^2 \le (\log^2(u) + \pi^2)u \le (\pi^2)u + 5\pi^2$$
Therefore,
$$\frac{|\log(z)|}{|z^2 + 1|} \le \frac{(\pi^2)u + 5\pi^2}{u^2 + 1}$$
Then we take the limit as $u \to 0$ which makes the RHS of the inequality 0.
hence the LHS upperbound is $0$.
So is the contour integral around the small semi circle $\delta = 0$?
How do I do this?
Thanks
| We want to show that $$\int_{0}^{\infty}\frac{\ln^2 x}{x^2+1} = \frac{\pi^3}{8} \tag{1}$$
Breaking through, let's take $f(z) = \frac{\ln z}{z^2+1}$ with branch $\Big(|z|> 0 -\frac{\pi}{2} < \arg z < \frac{3\pi}{2}\Big)$ of the multiple-valued function $\ln z / (z^2+1)$. As long as we are isolating $z = i$ we're going to take $\delta < 1 < R$. According to Cauchy's Residue Theorem,
$$\int_{L_1} f(z)dz + \int_{C_R} f(z)dz + \int_{L_2} f(z)dz + \int_{C_\delta} f(z)dz= 2\pi i Res_{z=i}f(z)$$
That is,
$$\int_{L_1} f(z)dz + \int_{L_2} f(z)dz = 2\pi i Res_{z=i}f(z) - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \tag{2}$$
Since $$f(z) = \frac{(\ln r + i\theta)^2}{r^2e^{i0} + 1} \ \ \ \ \ \ (z=re^{i\theta})$$
the parametric representations
$$z = r e^{i0} = r \ \ (\delta\leq r\leq R) \ \ \text{and}\ \ z = re^{i\pi} = -r \ \ (\delta\leq r \leq R)$$
for the legs $L_1$ and $-L_2$ can be used to write the LHS pf equation $(2)$ as
$$\int_{L_1} f(z)dz - \int_{-L_2} f(z)dz = \int_{\delta}^{R} \frac{\ln^2 r}{r^2 + 1}dr + \int_{\delta}^{R} \frac{(\ln r + i\pi)^2}{r^2 + 1}dr$$
Also, since
$$Res_{z=i}f(z) = \frac{p(z)}{\phi'(z)}\ \ \text{where}\ \ p(z) = \ln^2 z \ \ \text{and}\ \ \phi(z) = z^2 + 1$$
then
$$Res_{z=i}f(z) = \frac{\Big(\ln (1) + i\frac{\pi}{2}\Big)^2}{2i}$$
Thus equation $(2)$ becomes
$$\begin{align}&2\int_{\delta}^{R} \frac{\ln^2 r}{r^2 + 1}dr + 2\pi i\int_{\delta}^{R} \frac{\ln r }{r^2 + 1}dr - \pi^2\int_{\delta}^{R} \frac{1}{r^2 + 1}dr\\ & = 2\pi i \frac{\Big(\ln (1) + i\frac{\pi}{2}\Big)^2}{2i} - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \\ &= -\frac{\pi^3}{4} - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \end{align}$$
Evaluating integrals
1- $\lim_{\delta \to 0}_{R\to \infty}\int_{\delta}^{R} \frac{\ln r }{r^2 + 1}dr = 0$
2- $\lim_{\delta \to 0}_{R \to \infty}\int_{\delta}^{R} \frac{1}{r^2 + 1}dr = \frac{\pi}{2}$
3- $\lim_{R\to\infty}\int_{C_R} f(z)dz = 0 $
4- $\lim_{\delta \to 0}\int_{C_\delta} f(z)dz = 0 $
Showing $4$.
Take $z = \delta e^{i\theta}$. Notice that if $\delta < 1$ and $z = \delta e^{i\theta}$, $$\begin{align}|\log^2 z| &=|(\ln \delta + i\theta )^2| = |\ln^2\delta + 2i\theta\ln\delta - \theta^2|\\&\leq |\ln^2\delta| + 2|i\theta\ln\delta|+\theta^2 \leq \ln^2\delta -2\pi\ln \delta + \pi^2\end{align}$$ and $$|z^2+1| \geq ||z^2| - 1| = 1 - \delta^2$$ then
$$\Bigg|\int_{C_\delta} f(z)dz\Bigg| \leq \int_{C_\delta} |f(z)| |dz| \leq \frac{\ln^2\delta -2\pi\ln \delta + \pi^2}{1 - \delta^2} \pi\delta$$
the RHS of the inequality goes to $0$ as $\delta \to 0$.
There fore we get
$$2\int_{0}^{\infty} \frac{\ln^2 r}{r^2 + 1}dr = \frac{\pi^3}{4} \Rightarrow \int_{0}^{\infty} \frac{\ln^2 r}{r^2 + 1}dr = \frac{\pi^3}{8}$$
| {
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"url": "https://math.stackexchange.com/questions/1080172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?
Let $p>2$ be an odd prime. Let $\mathbb F_{p^n}$ be the field with $p^n$ elements. How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?
My work:
Char $F=p$.
$x^2-y^2=(x+y)(x-y)=1$. Since we are looking for $x,y\in F$, $x+y,x-y\in F$ and if $x+y=\alpha$, $x-y=\alpha^{-1}$. Hence, $\displaystyle x=\frac{\alpha+\alpha^{-1}}{2}, y=\frac{\alpha-\alpha^{-1}}{2}$. I was stuck afterwards since I cannot find the distinct sets of $x,y$ from these relations. Can anyone please give me a hint?
| Here is a partial answer for $n = 1$. Is it possible to extend this proof generally? I have having trouble doing this since we can no longer use the legendre symbol.
Anyway, denote $N(x^2-y^2 = 1)$ to be the number of solutions. Then
$$N(x^2-y^2 = 1) = \sum_{a+b = 1} N(x^2 = a)N(y^2 = -b).$$
Now $N(x^2 = a) = 1+ \left( \frac{a}p \right)$ so
$$N(x^2-y^2 = 1) = \sum_{a+b= 1}1 + \sum_a \left( \frac{a}p \right)+\sum_{-b} \left( \frac{-b}p \right)+\sum_{a+b= 1}\left( \frac{a}p \right) \left( \frac{-b}p \right).$$
The first sum is obviously $p$ and it is a well known result that the next two sums are $0$. Now using Jacobi sums (see $\star$) we can arrive at
$$ \sum_{a+b= 1}\left( \frac{a}p \right) \left( \frac{b}p \right) = - (-1)^{\frac{p-1}2}.$$
So in the sum that we are interested in, we can factor out $\left( \frac{-1}p \right)$ and depending on if $p \equiv \pm 1 \pmod 4$, we have $\left( \frac{-1}p \right) = \pm 1$ and we can find the value of our sum using the result from $\star$. In anycase, we have $$N(x^2-y^2 = 1) = p-1.$$
$\star$ A Classical Introduction to Modern Number Theory by Ireland and Rosen.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Is the digital root of twin primes product larger than (3,5) always 8
Is the digital root of twin primes product larger than $(3,5)$ always $8$?
E.g. I just checked the following and it is true was wondering if it is more widely true
$5\times 7=35$, digital root $=3+5=8$
$11\times 13=143$, digital root $=1+4+3=8$
$17\times 19=323$, digital root $=3+2+3=8$
$101\times 103=10403$, digital root $=1+0+4+0+3=8$
| One of the basic properties of twin primes $k-1$, $k+1$ is that $k$ must be a multiple of $6$ (except $k=4$). Set $k=6t$, and we have $$(k-1)(k+1)=(6t+1)(6t-1)=36t-1$$
Hence the product of any pair of twin primes (except $3,5$) will be congruent to $-1$ modulo $9$. Hence the digital root of same will be $8\equiv -1\pmod{9}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of the following infinite series Find the sum of the following infinite series in which numerator and denominator contains term which are product of integers in arithmetic progression:
$$\frac15+ \frac{1\times4}{5\times10}+\frac{1\times4\times7}{5\times10\times15}+\dots$$
I found this problem in an Indian competitive exam. I tried some conventional methods, but could not find the sum. Please offer any help you think is useful. Thanks!
| Consider the series expansion for $f(x)=(1-x)^{-1/3}$:
$$\begin{align}f(x) &= 1+\frac13 x + \frac1{2!} \left ( -\frac13\right ) \left ( -\frac{4}{3}\right ) x^2 - \frac1{3!} \left ( -\frac13\right ) \left ( -\frac{4}{3}\right )\left ( -\frac{7}{3}\right ) x^3 +\cdots\\ &= 1+\frac13 x + \frac{1 \cdot 4}{2! 3^2} x^2 + \frac{1 \cdot 4 \cdot 7}{3! 3^3} x^2+\cdots\end{align}$$
Thus, the stated sum is simply
$$f \left ( \frac{3}{5} \right ) - 1 = \left ( \frac{5}{2} \right )^{1/3}-1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why is the ratio of the number of terms needed to achieve successive integer values in the harmonic series approximately $e$? Consider the harmonic series:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} + \cdots .$$
It takes $1$ term to achieve a partial sum of $1$, since $1$ is the first number.
It takes $4$ terms to achieve a partial sum of $2$:
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} = 2.08\bar{3}$ is the first partial sum of value at least $2$.
To get to $3$ it takes $11$ terms.
To get to $4$ it takes $31$ terms.
To get to $5$ it takes $83$ terms.
If you consider this equation: $x_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ and replace $x$ with any natural number then: ${(x+1)}_{n*e}$
I might not be wording this correctly, but I want to know whether and how it can be shown mathematically why $e$ appears to show up in the ratio of the numbers of terms required for partial sums that take on successive integer values.
More examples
The first partial sum in the harmonic series larger than $15$ is
$$1+\frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{1835421} > 15$$
The first partial sum larger than 16 is
$$1+\frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{4989191} > 16$$
Note that $\frac{4989191}{1835421}$ is an approximation of $e$.
I've calculated the first five of the quotients of the iterations needed to reach partial sums of integers:
$\frac{4}{1}$
$\frac{11}{4}$
$\frac{31}{11}$
$\frac{83}{31}$
$\frac{227}{83}$
| Let's $$1+\frac{1}{2}+\cdots+\frac{1}{x_{n}-1}\lt n$$ and $$1+\frac{1}{2}+\cdots+\frac{1}{x_n}\gt n.$$ If for some $N$, $$\frac{1}{x_n+1}+\cdots+\frac{1}{N}\ge 1,$$ then $x_{n+1}\le N$ and if for some $M$, $$\frac{1}{x_n+1}+\cdots+\frac{1}{M}\lt 1-\frac{1}{x_n},$$ then $x_{n+1}\gt M$. Now it is easy to see that if $$
\int_{x_n+1}^{N+1}\frac{1}{x}dx\ge1\Rightarrow N\ge-1+e(x_n+1)=ex_n+e-1,$$then $\frac{1}{x_n+1}+\cdots+\frac{1}{N}\ge 1$, so $$x_{n+1}\le ex_n+e,$$ and if $$\int_{x_n}^{M}\frac{1}{x}dx\lt1-\frac{1}{x_n}\Rightarrow M\lt e^{1-\frac{1}{x_n}}x_n,$$then $\frac{1}{x_n+1}+\cdots+\frac{1}{M}\lt 1-\frac{1}{x_n},$so $$x_{n+1}\ge e^{1-\frac{1}{x_n}}x_n -1.$$ These bounds show that, indeed, $$\frac{x_{n+1}}{x_n}\rightarrow e.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conditional inequality Let x,y,z be positive reals with $xy+yz+zx=1$. Prove the inequality $$\sum_{cyc(x,y,z)}\frac {2x(1-x^2)}{(1+x^2)^2} \le \sum_{cyc(x,y,z)} \frac x{1+x^2}.$$ I substituted $x=tan\frac{\theta}2, y=tan\frac {\phi}2,z=tan\frac {\epsilon}2$ where $0\le\theta,\phi,\epsilon\le \pi/2$. Therefore the equality gives $tan\frac {\theta}2*tan\frac {\phi}2+tan\frac {\phi}2*tan\frac{\epsilon}2+tan\frac {\epsilon}2*tan\frac {\theta}2=1=>\frac {\theta+\phi+\epsilon}2=\frac {\pi}2=>\theta+\phi+\epsilon=\pi$. Now using trigonometric formulae we can reduce the given inequality into $$\begin{align} &=> 2sin \theta* cos\theta+2sin \phi* cos\phi+2sin\epsilon* cos\epsilon \le sin\theta+sin\phi+\sin\epsilon \\&=>sin2\theta+sin2\phi+sin2\epsilon\le sin\theta+sin\phi+sin\epsilon \\&=> 4sin \theta*sin\phi*sin \epsilon \le 4cos \frac{\theta}2*cos \frac {\phi}2*cos \frac{\epsilon}2 \\&=>8sin \frac {\theta}2*sin \frac {\phi}2*sin \frac {\epsilon}2 \le 1\end{align}$$ How do i prove this?
| We need to prove that
$$\sum_{cyc}\frac{x}{x^2+xy+xz+yz}\geq2\sum_{cyc}\frac{x(xy+xz+yz-x^2)}{(x^2+xy+xz+yz)^2}$$ or
$$\sum_{cyc}\frac{x}{(x+y)(x+z)}\geq2\sum_{cyc}\frac{x(xy+xz+yz-x^2)}{(x+y)^2(x+z)^2}$$ or
$$\sum_{cyc}x(x+y)(x+z)(y+z)^2\geq2\sum_{cyc}x(xy+xz+yz-x^2)(y+z)^2$$ or
$$\sum_{cyc}x(3x^2-xy-xz-yz)(y+z)^2\geq0$$ or
$$\sum_{cyc}x((x-y)(3x+z)-(z-x)(2x+y))(y+z)^2\geq0$$ or
$$\sum_{cyc}(x-y)((3x^2+xz)(y+z)^2-(3y^2+yz)(x+z)^2)\geq0$$ or
$$\sum_{cyc}(x-y)^2z(5xy+3xz+3yz+z^2)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| I would start by using the substitution $u=\frac{1-x}{1+x}$. This is a handy substitution because it is its own inverse:
$$u=\frac{1-x}{1+x}\implies x=\frac{1-u}{1+u},~\mathrm{d}x=\frac{-2}{(1+u)^2}\,\mathrm{d}u.$$
Then,
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=\int\left(\frac{1+u}{1-u}\right)\sqrt{u}\frac{(-2)}{(1+u)^2}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{(1-u)(1+u)}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u.\\
\end{align}$$
Another substitution such as $u=t^2$ would transform this to an integral of a rational function:
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u\\
&=-2\int\frac{t}{1-t^4}\left(2t\,\mathrm{d}t\right)\\
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t.\\
\end{align}$$
From there, partial fraction decomposition will break the integral down into elementary integrals, and back-substitution will yield the desired anti-derivative. PFD gives us,
$$-\frac{4t^2}{1-t^4}=\frac{2}{1+t^2}-\frac{1}{1+t}-\frac{1}{1-t},$$
so,
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t\\
&=2\int\frac{\mathrm{d}t}{1+t^2}-\int\frac{\mathrm{d}t}{1+t}-\int\frac{\mathrm{d}t}{1-t}\\
&=2\tan^{-1}{\left(t\right)}-\ln{\left(1+t\right)}+\ln{\left(1-t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(t\right)}-2\tanh^{-1}{\left(t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{u}\right)}-2\tanh^{-1}{\left(\sqrt{u}\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}-2\tanh^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}+\color{grey}{constant}.~\blacksquare\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Probability of selecting the winning numbers in a lottery I've been studying combinatorics for a while. I've solved a problem but I'm not sure if I'm right. I'll just copy-paste the problem here.
In a lottery, six distinct numbers are selected at random from the set $1, \ldots, 50$ and designated as winning numbers. A player selects six distinct numbers in advance, hoping to include as many winners as possible. Find the probability that the player selects exactly $k$ winning numbers, for each $k$ from $0$ to $6$.
Here's my solution.
The probability of selecting one winning number is $\frac{\binom{6}{1}\cdot\binom{49}{5}}{\binom{50}{6}}$. And the probabilities of selecting 2, 3, 4, 5 and 6 winning numbers are $\frac{\binom{6}{2}\cdot\binom{48}{4}}{\binom{50}{6}}$, $\frac{\binom{6}{3}\cdot\binom{47}{3}}{\binom{50}{6}}$, $\frac{\binom{6}{4}\cdot\binom{46}{2}}{\binom{50}{6}}$, $\frac{\binom{6}{5}\cdot\binom{45}{1}}{\binom{50}{6}}$, and $\frac{1}{\binom{50}{6}}$ respectively. So the probability of selecting 0 winning number is
$1 - \left( \frac{\binom{6}{1}\cdot\binom{49}{5}}{\binom{50}{6}} + \frac{\binom{6}{2}\cdot\binom{48}{4}}{\binom{50}{6}} + \frac{\binom{6}{3}\cdot\binom{47}{3}}{\binom{50}{6}} + \frac{\binom{6}{4}\cdot\binom{46}{2}}{\binom{50}{6}} + \frac{\binom{6}{5}\cdot\binom{45}{1}}{\binom{50}{6}} + \frac{1}{\binom{50}{6}}\right)$
Am I wrong?
| Just so this one doesn't stay unanswered, as Kamster says the chance of one winning number is $\frac{\binom{6}{1}\cdot\binom{44}{5}}{\binom{50}{6}}$ because when you choose the non-winning numbers there are only $44$ of them to choose. Once you adjust your other answers with this in mind you will be correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
| You can use this way to solve:
\begin{eqnarray}
\underbrace{55\cdots5}_{100\ \text{times}} \ \text{mod} \ 7&\equiv&\sum_{i=0}^{9}\underbrace{55\cdots5}_{10\ \text{times}}\times 10^{10i}\ \text{mod} \ 7\\
&\equiv&4\sum_{i=0}^{9}3^{10i}\ \text{mod} \ 7\\
&\equiv&4\sum_{i=0}^{9} 59049^{i}\ \text{mod} \ 7\\
&\equiv&4\sum_{i=0}^{9} 4^{i}\ \text{mod} \ 7\\
&\equiv&4\times\frac{4^{10}-1}{4-1}\ \text{mod} \ 7\\
&\equiv&4 349525 \ \text{mod} \ 7\\
&\equiv&4 \ \text{mod} \ 7\\
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$$
Transforming it into $2 \sin \frac{3\alpha-180}4 \cos \frac{\beta - \gamma}4 + \sin \frac{3\alpha}2 =\frac{3}2$
and
$\cos \frac{\beta - \gamma}4 =\cos \frac{180- \alpha}4 + 2 \sin \frac{\beta}4 \sin \frac{\gamma}4$ is as far as I came.
| I took an approach which may be only superficially different from abel's because of the underlying trig identities and follows a lead SoulEater was taking, but along a slightly different path.
We can get rid of $ \ \alpha \ $ , to start off, by writing $ \ \alpha - \beta \ = \ \pi - 2 \beta - \gamma \ $ and, similarly, $ \ \alpha - \gamma \ = \ \pi - \beta - 2\gamma \ $ . From this, we have
$$ \sin \left( \frac{\alpha - \beta}{2} \right) \ = \ \sin \left( \frac{\pi}{2} - [ \ \beta + \frac{\gamma}{2} \ ] \right) \ = \ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ \ $$
and $ \ \sin \left( \frac{\alpha - \gamma}{2} \right) \ = \ \cos \left( \ \frac{\beta}{2} + \gamma \ \right) \ $ . [So far, this looks like what abel did, while deriving the variable transformations.] We will also write
$$ \sin \frac{3 \alpha}{2} \ = \ \sin \left( \frac{3}{2} [ \pi - \beta - \gamma ] \right) \ = \ \sin \left( \frac{3 \pi}{2} - \ \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ -\cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ \ . $$
The original equation is now
$$ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ + \ \cos \left( \ \frac{\beta}{2} + \gamma \ \right) \ - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ \frac{3}{2} \ \ . $$
Applying a "sum-to-product rule" on the first two terms here gives
$$ 2 \ \cos \left( \frac{3}{4} [\beta + \gamma ] \ \right) \cos \left( \frac{1}{4} [\beta - \gamma ] \right) - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ \frac{3}{2} \ \ . $$
This seemingly doesn't appear to have done much of anything useful, but it turns out we've actually finished the "hard part"...
We will label the angles $ \ \Theta \ = \ \frac{3}{4} [\beta + \gamma ] \ $ and $ \ \Phi \ = \ \frac{1}{4} [\beta - \gamma ] \ $ , allowing us to write
$$ 2 \ \cos \Theta \ \cos \Phi \ - \ \cos \ ( 2 \Theta ) \ = \ \frac{3}{2} \ \ \Rightarrow \ \ 2 \ \cos \Theta \ \cos \Phi \ - \ 2 \ \cos^2 \Theta \ + \ 1 \ = \ \frac{3}{2} $$
$$ \Rightarrow \ \ \cos \Theta \ \cos \Phi \ - \ \cos^2 \Theta \ = \ \frac{1}{4} \ \ . $$
Now, if $ \ \beta \ = \ \gamma \ $ , then $ \ \Phi \ = \ 0 \ $ and we just have the quadratic equation in $ \ \cos \Theta \ $ ,
$$ \cos^2 \Theta \ - \ \cos \Theta \ + \ \frac{1}{4} \ = \ \left( \cos \Theta \ - \ \frac{1}{2} \right)^2 \ = \ 0 \ \ . $$
Since we are limiting ourselves to angles of a triangle, we obtain
$$ \cos \left( \frac{3}{4} [\beta + \gamma ] \ \right) \ = \ \frac{1}{2} \ \ \Rightarrow \ \ \frac{3}{4} [\beta + \gamma ] \ = \ \frac{\pi}{3} \ \ \Rightarrow \ \ \beta \ + \ \gamma \ = \ \frac{4 \pi}{9} \ \ . $$
Thus, $ \ \beta \ = \ \gamma \ = \ \frac{2 \pi}{9} \ \ \Rightarrow \ \ \alpha \ = \ \frac{5 \pi}{9} \ $ .
Are other solutions permissible? If we allow $ \ \beta \ \neq \ \gamma \ $ and call $ \ \cos \Phi \ = \ \varphi \ $ , our quadratic equation becomes $ \ \cos^2 \Theta \ - \ \varphi \ \cos \Theta \ + \ \frac{1}{4} = \ 0 \ $ , which has potential solutions
$$ \cos \Theta \ = \ \frac{\varphi \ \pm \ \sqrt{ \ \varphi^2 \ - \ 1 } } {2} \ \ . $$
But real solutions for $ \ \cos \Theta \ $ would only be possible here for $ \ \varphi \ = \ \cos \Phi \ \ge \ 1 \ $ . So the only solution for our original equation is the one we've already found with $ \ \varphi \ = \ 1 \ $ or $ \ \beta \ = \ \gamma \ $ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Integral question showing the primitive functions differ only by a constant? $$\int \frac{dx}{\sqrt{x^2-6x+13}}$$
$$\int \frac{dx}{\sqrt{\left(x-3\right)^2+4}}$$
It can be solved by
Method 1
Let
$$x-3=2\tan u$$
$$dx=2\sec^2 u\,du$$
Therefore, using the trigonometric identity $\tan^2 u+1=\sec^2 u$
$$\int \frac{2\sec^2 u\,du}{\sqrt{4\tan^2 u +4}}$$
$$\int \frac{2\sec^2 u\,du}{|2\sec u |}$$
$$\int |\sec u|\,du$$
To focus on the problem , I skipped the steps in evaluating the sec integral by using this
$$\int \left|\sec u\right|\,du=\begin{cases}\int -\sec u\,du, & \sec u<0\\ \int \sec u\,du, & \sec u>0 \end{cases} = \begin{cases}-\ln |\sec u+\tan u |+C, & \sec u<0\\ \ln |\sec u+\tan u |+C, & \sec u>0 \end{cases}$$
$$=|\ln |\sec u+\tan u ||+C$$
Reverting to $x$ variables
$$\left|\ln\left|\sec \left(\arctan \frac{x-3}{2} \right) +\tan \left(\arctan \frac{x-3}{2} \right) \right|\right|+C$$
Method 2
Let
$$x-3=2\sinh u$$
$$dx=2\cosh u\,du$$
Therefore, using the trigonometirc identity $\sinh^2 u+1=\cosh^2 u$
$$\int \frac{2\cosh u\,du}{\sqrt{4\cosh^2 u +4}}$$
$$\int \frac{2\cosh u\,du}{|2\cosh u|}=\left\{\begin{matrix}\int -du,\cosh u<0\\ \int du,\cosh u>0 \end{matrix}\right.=\left\{\begin{matrix}-u+C,\cosh u<0\\ u+C,\cosh u>0 \end{matrix}\right.$$
$$=|u|+C$$
Reverting to $x$ variables
$$\left|\sinh^{-1} \left(\frac{x-3}{2}\right) \right|+C=\left|\ln \left|\left(\frac{x-3}{2}\right)+\sqrt{\left(\frac{x-3}{2}\right)^2+1}\right|\right|$$
I understand that in general, by using the results here, we can prove in general that all primitive of a function f must differ only by a constant
But
$$\left|\ln\left|\sec \left(\arctan \frac{x-3}{2} \right) +\tan \left(\arctan \frac{x-3}{2} \right) \right|\right|+C$$
and
$$|\sinh^{-1} \left(\frac{x-3}{2}\right)|+C$$
look really different. how to show that they are really the same thing save for a constant?
| $$
\tan \left(\arctan \left(\frac{x-3}{2}\right)\right) = \frac{x-3}{2}\\
\sec \left(\arctan \left(\frac{x-3}{2}\right)\right) = \pm\sqrt{1+\tan^2\left(\arctan \left(\frac{x-3}{2}\right)\right)} \\= \pm\sqrt{1+\left(\tan \left(\arctan \left(\frac{x-3}{2}\right)\right)\right)^2}=\pm \sqrt{1+\left(\frac{x-3}{2}\right)^2}
$$
I utilize the fact that
$$
\tan^2 y + 1 = \sec^2 y
$$
thus
$$
\ln\left|\sec \left(\arctan \left(\frac{x-3}{2}\right)\right) +\tan \left(\arctan \left(\frac{x-3}{2}\right)\right) \right| = \ln\left|\pm \sqrt{1+\left(\frac{x-3}{2}\right)^2} + \frac{x-3}{2}\right|
$$
or
$$
\ln\left|\frac{x-3}{2} +\sqrt{1+\left(\frac{x-3}{2}\right)^2}\right|
$$
now you can compare with log formulation of $\mathrm{sinh}^{-1} z$
| {
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"timestamp": "2023-03-29T00:00:00",
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if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
| $$x=\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}$$
$$x-\sqrt{x+1}=\sqrt{x+2}+\sqrt{x+3}$$
$$x^2+x+1-2x\sqrt{x+1}=2x+5+2\sqrt{x^2+5x+6}$$
$$x^2-x-4=2(\sqrt{x^3+x^2}+\sqrt{x^2+5x+6})$$
$$x^4-2x^3-7x^2+8x+16=4(x^3+2x^2+5x+6+2\sqrt{x^5+6x^4+11x^3+6x^2})$$
$$x^4-6x^3-15x^2-12x-8=8\sqrt{x^5+6x^4+11x^3+6x^2}$$
$$x^8-12x^7+6x^6+156x^5+353x^4+456x^3+384x^2+192x+64=64x^5+384x^4+704x^3+384x^2$$
$$x^8-12x^7+6x^6+92x^5-31x^4-248x^3+192x+64=0$$
This equation has $6$ real roots, but only the one I mentioned in my comment fulfills the original equation. Therefore, it is the only real solution.
| {
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"url": "https://math.stackexchange.com/questions/1089589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum of $f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|$
For $z\in\mathbb{C}$, calculate the minimum value of
$$
f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|
$$
My Attempt:
Let $z= x+iy$. Then
$$
\begin{align}
z^2+z+1 &= (x+iy)^2+(x+iy)+1\\
&= (x^2-y^2+x+1)+i(2xy+y)
\end{align}
$$
and
$$
\begin{align}
z^2-z+1 &= (x+iy)^2-(x+iy)+1\\
&= (x^2-y^2-x+1)+i(2xy-y)
\end{align}
$$
Define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by
$$
f(x,y) = \sqrt{\big(x^2-y^2+x+1\big)^2+(2xy+y)^2}+\sqrt{\big(x^2-y^2-x+1\big)^2+(y-2xy)^2}
$$
Using the Triangle Inequality, we know that
$$
f(x,y) \geq 2\sqrt{\left(x^2-y^2+1\right)^2+y^2}
$$
How can the problem be solved from this point?
| We can write $f$ in the form
$$f(z)=2|z|\>\left(\left|J(z)-{1\over2}\right|+\left|J(z)+{1\over2}\right|\right)\ ,$$
where $J$ denotes the so-called Joukowski function:
$$J(z):={1\over2}\left(z+{1\over z}\right)\ .$$
It is well known that $J$ maps circles $|z|=r$ onto ellipses centered at $0$ with foci $\pm1$ and semiaxes $$a={1\over2}\left(r+{1\over r}\right),\quad b={1\over2}\left|r-{1\over r}\right|\ .$$ Looking at such an ellipse we see that the points minimizing $f$ are the points $J(\pm i r)$ on the imaginary axis. It follows that
$$g(r):=\min_{|z|=r} f(z)=f(ir)=2\sqrt{(r^2-1)^2+r^2}\ .$$
Now we have to minimize the right side over $r\geq0$. This leads to the equation
$4r^3-2r=0$ with the solutions $r_1=0$, $r_2={1\over\sqrt{2}}$. Since $g(0)=2$ and $g\bigl({1\over\sqrt{2}}\bigr)=\sqrt{3}$ we conclude that the global minimum of $f$ on the complex plane is $\sqrt{3}$, and this minimum is taken at the points $\pm{i\over\sqrt{2}}$.
| {
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"url": "https://math.stackexchange.com/questions/1090447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively
I want the multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively.
Now I have had a search of similar questions, but I may have missed it since I am on a phone.
This question should be trivial to me, but I am having trouble, so that is depressing enough, please just give me a hint.
It would appear that I am dealing with:
$$\frac{7x}2\equiv 1\pmod 7$$
$$\frac{7x}{3} \equiv 2\pmod 7 $$
$$\frac{7x}{4} \equiv 3 \pmod 7$$
and then I find inverses for $2,3,4$ and multiply both sides by these, then apply Chinese remainder theorem. But having the different remainders under the same modulus makes no sense, so this leads me to believe I have formulated the question wrong, and chosen the wrong modulus. Thanks
| Solution:
First recognise that $x \equiv 3 \pmod 4 \implies x \equiv 1 \pmod 2$. The latter criterion is redundant.
So you can simply solve this system:
$$x \equiv 3 \pmod 4$$
$$x \equiv 2 \pmod 3$$
From the second equation, $x = 3s+2$.
Subbing into second equation, $3s+2 \equiv 3 \pmod 4$, giving $3s \equiv 1 \implies s \equiv 3^{-1} = 3 \pmod 4$. Hence $s = 4t + 3$.
So $x = 3(4t+3) + 2 = 12t + 11$ or $x \equiv -1 \pmod{12}$, which is the solution of all three congruences.
Finally, you have to incorporate the divisibility by seven criterion:
$x = 7y \implies 7y \equiv -1 \pmod{12} \implies y \equiv -1.7^{-1} = -7 \equiv 5 \pmod{12}$.
Hence the required solution is $x = 7(12v + 5)$, also expressible as $x \equiv 35 \pmod{84}$. Some possible particular solutions are: $35, 119, 203,..$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality $\sum\limits_{\text{cyc}} \frac{x}{y+z+\sqrt{x}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}$ How do I show the following?
$$
\frac{x}{y+z+\sqrt{x}}+\frac{y}{z+x+\sqrt{y}}+\frac{z}{x+y+\sqrt{z}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}
$$
Any hint would do; I don't require a complete solution.
| use that e.g. for $x>0$ we have $\frac{x}{y+z+\sqrt{x}}=\frac{x^2}{xy+xz+x\sqrt{x}}$ and use Titu's Lemma$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help solving this differential equation $$p^3 - 2xyp + 4y^2 = 0$$
where $p = \mathrm dy / \mathrm dx$.
I don't know which type of equation it is or how to simplify it.
Though there is an observation that
$$(xy^2)′=y^2+2xyy′$$
| The given equation is:
$$\tag 1 p^3 - 2xyp + 4y^2 = 0$$
Solving for $x$ yields:
$$x = \dfrac{2y}{p} + \dfrac{p^2}{2y}$$
Differentiating w.r.t. $y$ yields:
$$\dfrac{dx}{dy} = \dfrac 1p = \dfrac 2p - \dfrac{2y}{p^2}\dfrac{dp}{dy}-\dfrac {p^2}{2y^2} + \dfrac{p}{y} \dfrac{dp}{dy}$$
Simplifying yields:
$$p = 2p - 2y \dfrac{dp}{dy} - \dfrac{p^4}{2y^2} + \dfrac{p^3}{y} \dfrac{dp}{dy}$$
$$p - 2y \dfrac{dp}{dy} - \dfrac{p^4}{2y^2} + \dfrac{p^3}{y} \dfrac{dp}{dy} = 0$$
$$p \left( 1 - \dfrac{p^3}{2y^2} \right) - 2y \dfrac{dp}{dy} \left( 1 - \dfrac{p^3}{2y^2} \right) = 0$$
$$\left( 1 - \dfrac{p^3}{2y^2} \right) \left( p - 2y \dfrac{dp}{dy} \right) = 0$$
Ignoring the first term since it does not include $\dfrac{dp}{dy}$, we have:
$$p - 2y \dfrac{dp}{dy} = 0 \implies 2 \int \dfrac{dp}{p}- \int \dfrac{dy}{y} = 0 \implies 2 \ln p - \ln y = \ln c$$
This yields:
$$\tag 2 p^2 = cy$$
Now, we have to eliminate $p$ from $(1)$, yielding:
$$2xyp = p^3 + 4y^2 \implies 2xyp-p^3 = 4y^2 \implies p(2xy-p^2) = 4y^2$$
Squaring yields:
$$p^2(2xy-p^2)^2 = 16y^4$$
Substituting $(2)$ yields:
$$cy(2xy - cy)^2 = 16 y^4$$
Maybe it is possible to simplify, but I am sure you can proceed with that.
| {
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"url": "https://math.stackexchange.com/questions/1096242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Explanation for applying Cauchy Integral Formula
I do not understand the last part.
How do you get:
$$\oint_{C_N} f(z) dz = \frac{-7\pi^3}{45} + 4\sum_{n=1}^{N} \frac{\coth(n\pi)}{n^3}$$
How do you derive this, and what part of cauchy's formula is this?
Thanks!
| You get the result from the residue theorem. As has been pointed out, the integrand
$$f(z) = \frac{\pi \cot{\pi z} \coth{\pi z}}{z^3} $$
has poles at $z= \pm n$ and $z=\pm i n$, for $n \in \mathbb{Z}$. The poles for $n \ne 0$ are simple, while the pole for $n=0$ has order $5$.
Note that $\lim_{x\to 0} x \cot{x} = 1$. Thus, the residue for $n \ne 0$ at $z=\pm n$ is
$$\operatorname*{Res}_{z=\pm n} \frac{\pi \cot{\pi z} \coth{\pi z}}{z^3} = \frac{\coth{\pi n}}{n^3} $$
Note also that $\lim_{x\to 0} x \coth{x} = 1$ and $i \cot{i \pi n} = \coth{\pi n}$. Thus,
$$\operatorname*{Res}_{z= \pm i n} \frac{\pi \cot{\pi z} \coth{\pi z}}{z^3} = \frac{\coth{\pi n}}{n^3} $$
The above explains the sum on the RHS.
The other piece is the residue at $z=0$. Given that the pole is of order 5, the residue is
$$\operatorname*{Res}_{z= 0} \frac{\pi \cot{\pi z} \coth{\pi z}}{z^3} = \frac{\pi}{4!} \left [\frac{d^4}{dz^4}(z^2\cot{\pi z} \coth{\pi z}) \right ]_{z=0} $$
That is a pretty horrifying calculation. An easier way is to return to the definition of the residue, which is simply the coefficient of $1/z$ in the Laurent series of $f(z)$. Now, use the fact that
$$\cot{\pi z} = \frac1{\pi z} \left (1-\frac13 \pi^2 z^2 - \frac1{45} \pi^4 z^4 + O(z^6) \right ) $$
$$\coth{\pi z} = \frac1{\pi z} \left (1+\frac13 \pi^2 z^2 - \frac1{45} \pi^4 z^4 + O(z^6) \right ) $$
Then
$$\frac{\pi \cot{\pi z} \coth{\pi z}}{z^3} = \frac{1}{\pi z^5} \left (1 - \frac{2}{45} \pi^4 z^4 - \frac1{9} \pi^4 z^4 + O(z^6) \right ) $$
The residue is then
$$-\frac{2}{45} \pi^3 - \frac19 \pi^3 = -\frac{7 \pi^3}{45}$$
as shown above.
| {
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"url": "https://math.stackexchange.com/questions/1098056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$? I found the integral
$$\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$$
in Gradshteyn and Ryzhik book and I try to solve it, but can't.
Taking the derivative, you get
$$\int_0^1 \frac{1}{(1 + a^2 x^2) \sqrt{1 - x^2}} dx = \frac{\pi}{2} \frac{1}{\sqrt{1 + a^2}}$$,
but this integral doesn't seem any easier. Have hints? Highschool methods would be best for me.
| $$\large I(\alpha)=\int_0^1 \frac{\text{arctan }(\alpha x)}{x\sqrt{1-x^2}}$$
$$\large I'(\alpha)=\frac{\pi}{2}\frac{1}{\sqrt{1+\alpha^2}}$$
as per your question
$$ I(\alpha)=\frac{\pi}{2}\int\frac{1}{\sqrt{1+\alpha^2}}$$
$$ I(\alpha)=\frac{\pi}{2}\int\frac{\cosh\theta}{\sqrt{1+\sinh\theta^2}}$$
$$ I(\alpha)=\frac{\pi}{2}\int\frac{1}{1}$$
$$ I(\alpha)=\frac{\pi}{2}\theta+C$$
$$ I(\alpha)=\frac{\pi}{2}\text{arcsinh } \alpha+C$$
Now we find $C$.
$$ I(0)=\int_0^1 \frac{0}{x\sqrt{1-x^2}}=0=\frac{\pi}2 \times 0 +C$$
Therefore $C=0,$ and therefore
$$ I(\alpha)=\frac\pi 2 \text{arcsinh } \alpha$$
$$ \large I(\alpha)=\frac\pi 2 (\log[\alpha+\sqrt{\alpha^2+1}])$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{x \to 4}\frac{(2x)^{1/3} - 2}{\sqrt{x} - 2}$ without L'Hospital rule I know how to solve this limit with the L'Hospital rule:
$$\lim_{x \rightarrow 4} \frac{(2x)^{1/3} - 2}{\sqrt{x} -2}$$
The answer is $\frac{2}{3}$. I am trying to solve it without the L'Hospital rule.
| $$\frac{(2x)^{1/3}-2}{\sqrt x-2}=\sqrt[3]2\frac{\sqrt[3]x-\sqrt[3]{2^2}}{\sqrt x-2}=\sqrt[3]2\frac{\overbrace{x-4}^{=(\sqrt x-2)(\sqrt x+2)}}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\frac1{\sqrt x-2}=$$
$$=\frac{\sqrt[3]2(\sqrt x+2)}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\xrightarrow[x\to 4]{}\frac{\sqrt[3]2\cdot4}{3\cdot2\sqrt[3]2}=\frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098383",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How can i prove that this function is in the given bounds? So, I have been struggling to show the following:
Let $a \in [0,1)$, $b \in \left[0,\frac{1-a^2}{1+a^2}\right)$, and $f(a,b)=\sqrt{a^2+b^2-a^2b^2 + 2ab\sqrt{1-b^2}}$.
Prove that $a \leq f(a,b) < 1$
The $a \leq f()$ is pretty trivial, but the other inequality escapes me.
I have verified that $f(a,0)=a$ (obvious) and $f\left(a,\frac{1-a^2}{1+a^2}\right)=1$, and $f$ appears to be monotone increasing in $b$ for each $a$, but this just led to another bound that was difficult to show.
| Notice that:
$$ \left(a\sqrt{1-b^2} + b\right)^2 = a^2(1-b^2) + b^2 + 2ab\sqrt{1-b^2} = a^2 + b^2 - a^2b^2 + 2ab\sqrt{1-b^2}$$
Thus your expression is $a\sqrt{1-b^2} + b$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate via substitution and derivation rule I have to solve
this integral
$$\int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x$$
with substitution and then the trick that $\dfrac 1 {\sqrt{1-x^2}} = \dfrac{\mathsf d\;\arcsin(x)}{\mathsf d\;x\qquad\quad\;\,} $
can someone give me a tip on what I should substitute in order to continue?
| I would say
$\displaystyle \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x=4\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}} \sqrt{\frac{r^2}{r^2-x^2-y^2}} \operatorname d y \operatorname d x=$
$\displaystyle=4\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}} \frac{r}{\sqrt{r^2-x^2}}\frac{1}{\sqrt{1-\left(\frac{y}{\sqrt{r^2-x^2}}\right)^2}} \operatorname d y \operatorname d x=$
$\displaystyle =4r\int_{0}^{r}\left[\arcsin \frac{y}{\sqrt{r^2-x^2}}\right]_{0}^{\sqrt{r^2-x^2}}\operatorname dx=\cdots $
| {
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"url": "https://math.stackexchange.com/questions/1098755",
"timestamp": "2023-03-29T00:00:00",
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Are there solutions to FLT which are linearly independent over $\mathbb{Z}$ Specifically, I would like to know if there is some $R$, where
*
*$R$ is a ring with unity
*$\mathbb{Z} \subseteq R$
*there are $x,y,z \in R$ and a prime $p \in \mathbb{Z}$ such that $x^p + y^p + z^p = 0$
*$ax+by+cz=0 \implies a=b=c=0$ for $a,b,c \in \mathbb{Z}\quad$ (linear independence over $\mathbb{Z}$)
If so, I'd be happy to see an example.
This non-example in $M_2(\mathbb{Z})$ lead me to me ask this question:
Let
$$
x=
\begin{pmatrix}
1 & 25 \\
0 & 1 \\
\end{pmatrix}
\!,\;y=
\begin{pmatrix}
-1 & 0 \\
5 & -1 \\
\end{pmatrix}
\,\text{and}\;z=
\begin{pmatrix}
0 & -5 \\
-1 & 0 \\
\end{pmatrix}
\!\begin{matrix}\\.\end{matrix}
$$
Then $x^5+y^5+z^5=0$, but also $x+y+5z=0$.
I realize that this phenomenon (of solutions to an equation being linearly dependent over $\mathbb{Z}$) is probably not FLT specific, so if someone wants to share something on the general phenomenon I'd be happy, too.
| We can take
$$
x=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\;
y=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\;
z=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{pmatrix},\;
$$
then we have $x^2=y^2=z^2=0$, so that $x^2+y^2=z^2$ for $p=2$, and $x,y,z$ are linearly independent. Also $x^p=y^p+z^p$ for all $p\ge 3$.
| {
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"url": "https://math.stackexchange.com/questions/1098979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$ I would like this integral:
$$\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$$
I used substitution $x+ \frac{a}{2k} = \frac{\sqrt{4kb-a^{2}}}{2k} \tan \theta$.
This will reduce the denominator to
$$
\left(\left(\frac{\sqrt{4kb-a^{2}}}{2k}\tan\theta \right)^{2} + \frac{4kb-a^{2}}{4k^{2}}\right)^{2} = \left(\frac{4kb-a^{2}}{4k^{2}}\right)^{2} \sec^{4}\theta ,$$
and integral is reduced to
$$\frac{1}{4k^{2}}\frac{16k^{4}}{(4kb-a^{2})^{\frac{3}{2}}2k} \int \cos^{2} \theta d\theta = \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}}\left(\frac{\theta}{2}+\frac{1}{4}\sin(2\theta)+c\right)$$
$$= \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}} \left(\frac{\arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{2}+\frac{\sin(2 \arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{4}\right).$$
Is there a simpler solution?
| The method is correct. You just made some computational mistakes.
You got the substitution and denominator correctly.
The integral is reduced to $\frac{1}{2\sqrt{4bk-a^2}}\int \cos^2{\theta} d\theta$. You also did the integral correctly.
| {
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"url": "https://math.stackexchange.com/questions/1101714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$? How to prove the following result?
$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$
For my part no idea?
| $\bf{Modified:}$
Based on an idea from the solution of @user111187:
Consider a general integral of the form
$$I_b =\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx$$
with derivative
$$J_b \colon =\frac{dI_b}{db} = \int_0^1 \frac{x\, R(x)}{(1+b^2 x^2)\sqrt{1-x^2}} dx$$
and $I_0=0$.
The integral $J_b$ is doable if $R$ is a rational function. We get
$$I =I_1 = \int_{0}^1 J_b\, db$$
This last integral is doable in some cases.
In our case $R(x) = \frac{x^3}{(3-x^2)^2}$ so we get
$$J_b = \int_0^1 \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}} dx$$
We find first the indefinite integral
$$\int \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}} dx $$
One can check that it equals
$$
F_b(x):=\frac{3 x \sqrt{1-x^2}}{4 \left(1+3 b^2\right) \left(-3+x^2\right)}+\frac{3 \sqrt{\frac{3}{2}} \left(-1+b^2\right) \text{ArcTan}\left[\frac{\sqrt{\frac{2}{3}}
x}{\sqrt{1-x^2}}\right]}{4 \left(1+3 b^2\right)^2}+\frac{\text{ArcTan}\left[\frac{\sqrt{1+b^2} x}{\sqrt{1-x^2}}\right]}{\sqrt{1+b^2} \left(1+3 b^2\right)^2}$$
Indeed, by direct calculations
$$F_b'(x) = \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}}$$
We get
$$J_b = F_b(1_{-})- F_b(0) = \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2}$$
and therefore
$$I = I_1 = \int_0^1 \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2} db = \frac{\pi \sqrt{2}}{192}\cdot (\,18 - 6 \sqrt{3} - \pi) $$
This approach works for all integrals of form
$$\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx=\int_0^1 \frac{x \, Q(x^2)}{\sqrt{1-x^2}} \arctan(x) dx$$
where $R(x) = xQ(x^2)$ is an odd rational function.
| {
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"url": "https://math.stackexchange.com/questions/1101966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
| Hint: $\dfrac{\sqrt{1 + x+ x^2} -1}{x} = \dfrac{x + x^2}{x (\sqrt{1 + x+ x^2} + 1)} = \dfrac{1+x}{\sqrt{1 + x+ x^2} + 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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First order DE, need help I am trying to solve this equation by inspection: $$(xy-y)dx+(x^2-2x+y)dy=0$$ Hints would be very helpful.. Thanks
| first observe that $x-1=(xy - y)_y \neq (x^2-2x+y)_x = 2x - 2.$ so it is not a complete integral, but close.
here is one way to do this. write your system $$ \dfrac{dx}{dy} = \dfrac{x^2 - 2x + y}{(1-x)y}$$ as a system of two equations $$\dfrac{dx}{dt} = \dfrac{1}{1-x},
\dfrac{dt}{dy} = \dfrac{x^2 - 2x + y}{y} $$
then the integrals are $$2x - x^2 = 2(t + C)$$ and
substituting for $x$ we get a linear differential equation for $t$:
$$\dfrac{dt}{dy} = \dfrac{y - 2(t + C)}{y} = -\dfrac{2}{y}t +\left(1- \dfrac{2C}{y}\right) \tag 1$$ whose homogeneous solution is $t_H = \dfrac{B}{y^2}$ we will use variation of parameters to determine the solution for
$t = \dfrac{B}{y^2}, \dfrac{dt}{dy} = \dfrac{1}{y^2} \dfrac{dB}{dy} - \dfrac{2B}{y^3} $ putting this in (1), we find $$\dfrac{dB}{dy} = y^2 - 2Cy$$ whose solution is $B = \dfrac{1}{3}y^3- Cy^2 + D$ and the solution for $t$ is $$ t = \dfrac{1}{3}y- C + \dfrac{D}{y^2}$$
eliminating the intermediate variable $t$ between $x$ and $y$ we find, $$2x - x^2 = 2( \dfrac{1}{3}y + \dfrac{D}{y^2}) \text{ where $D$ is an arbitrary constant}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate rational function with multiple complex roots I want to integrate
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx.
$$
By WolframAlpha I know the solution is
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx = \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}.
$$
Do derive it on my own, I did a partial fraction decomposition and found
$$
\frac{x^2}{(1+n^2x^2)^2} = \frac{\frac{1}{2n^2}}{1+n^2x^2} + \frac{\frac{1}{4n^2}}{(x-i/n)^2} + \frac{\frac{1}{4n^2}}{(x+i/n)^2}.
$$
I checked this multiple times, but it must be wrong, because when I proceed with it I get
\begin{align*}
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx
& = \frac{1}{n^2} \left( \frac{1}{2} \int_{-1}^1 \frac{1}{1+n^2x^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x-i/n)^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x+i/n)^2} d x \right) \\
& = \frac{1}{n^2}\left( \frac{1}{2} \frac{\arctan(xn)}{n}\bigg\vert_{-1}^1 + \frac{1}{4} \left( \frac{-1}{x-i/n}\bigg\vert_{-1}^1 + \frac{-1}{x+i/n}\bigg\vert_{-1}^1 \right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} + \frac{1}{4} \left( \frac{-2}{1+1/n^2} + \frac{-2}{1+1/n^2}\right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} -\frac{1}{1+1/n^2} \right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} - \frac{n^2}{n^2+1} \right) \\
& = \frac{\arctan(n)}{n^3} - \frac{1}{n^2+1}
\end{align*}
This result is consistent which what I found with WolframAlpha, but surely
$$
\frac{\arctan(n)}{n^3} - \frac{1}{n^2+1} \ne \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}
$$
so what went wrong, I am sitting here since hours and do not see any fault...
| A quick and dirty way to check this. Let
$$f(m) = \int_{-1}^1 \frac{dx}{1+m x^2} = \frac{2}{m} \int_0^1 \frac{dx}{1/m+x^2} = \frac{2}{\sqrt{m}}\arctan \sqrt{m}$$
$$f'(m) = -\int_{-1}^1 dx \frac{x^2}{(1+m x^2)^2} = -m ^{-3/2} \arctan{\sqrt{m}} + \frac1{m (1+m)}$$
Your integral is thus
$$-f'(n^2) = \frac{\arctan{n}}{n^3} - \frac1{n^2 (1+n^2)} $$
| {
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"url": "https://math.stackexchange.com/questions/1103142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$4x^2+1$ factors only into $4y+1$ primes How can one prove that numbers of the form $4x^2+1$ can only be divided by primes of the form $4y+1$ (e.g. there is no $x$ for which $7$ divides $4x^2+1$)?
On a quick lookup, the statement is given as the first comment in https://oeis.org/A053755 and of course, manual testing concurs with this
| In general; note that your question is about the form $4x^2 + y^2$ with $\Delta = -16.$ If $q \equiv 3 \pmod 4$ is a prime that divides the form, then it divides both $x,y.$ However, in your case $y=1.$
Given a binary quadratic form
$$ f(x,y) = a x^2 + b xy+ c y^2 $$
with $a,b,c$ integers.
Given
$$ \Delta = b^2 - 4 a c, $$
where we require that $\Delta,$ if non-negative, is not a square (so also $\Delta \neq 0,1$).
Proposition: given an odd prime $q$ such that $q$ does not divide $\Delta$ and, in fact, $$ (\Delta| q) = -1, $$
whenever $$ f(x,y) \equiv 0 \pmod q, $$ then BOTH
$$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$
Proof: the integers taken $\pmod q$ form a field; every nonzero element has a multiplicative inverse. If either $a,c$ were divisible by $q,$ we would have $\Delta $ equivalent to $b^2$ mod $q,$ which would cause $(\Delta|q)$ to be $1$ rather than $-1.$ As a result, neither $a$ nor $c$ is divisible by $q.$
Next, $q$ is odd, so that $2$ and $4$ have multiplicative inverses mod $q,$ they are non zero in the field. Put togethaer, $$4a \neq 0 \pmod q$$
Now, complete the square:
$$ a x^2 + b xy+ c y^2 \equiv 0 \pmod q $$
if and only if
$$ 4a(a x^2 + b xy+ c y^2) \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + 4ac y^2 \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + (b^2 y^2 - b^2 y^2) + 4ac y^2 \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + b^2 y^2 - (b^2 y^2 - 4ac y^2) \equiv 0 \pmod q, $$
$$ (4a^2 x^2 + 4abxy + b^2 y^2) - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 - \Delta y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 \equiv \Delta y^2 \pmod q. $$
Now, ASSUME that $y \neq 0 \pmod q.$ Then $y$ has a multiplicative inverse which we are allowed to call $1/y,$ and we have
$$ \frac{(2ax + by)^2}{y^2} \equiv \Delta \pmod q, $$
$$ \left( \frac{2ax + by}{y} \right)^2 \equiv \Delta \pmod q. $$
However, the HYPOTHESIS that $ (\Delta| q) = -1 $ says that this is impossible, thus contradicting the assumption that $y \neq 0 \pmod q.$
So, in fact, $y \equiv 0 \pmod q.$ The original equation now reads
$$ a x^2 \equiv 0 \pmod q $$ with the knowledge that $a \neq 0 \pmod q,$ so we also get
$$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume that $n$ is a positive number $> 1$ and that $P(n)$ holds. Hence,
$$
\frac{1}{2}+\ldots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I now add ${\displaystyle \frac{n + 1}{2\cdot 2^n}}$ to both sides to get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2\cdot 2^n}=2-\frac{2+n}{2^n}+\frac{n+1}{2\cdot 2^n}.
$$
But after manipulating the right side I get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{3n + 5}{2^{n+1}}.
$$
This is definitely not right since I should be getting
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n + 3}{2^{n+1}}=2-\frac{2+(n+1)}{2^{n+1}}
$$
to prove that $P(n)$ implies $P(n+1)$.
What am I doing wrong?
| Since there is always the possibility that the original statement that you try to prove is false, I tried $n = 1, 2, 3$.
For $n = 3$, the sum is $\frac{11}{8}=2-\frac{5}{8}$ as the statement claims, but not $2-\frac{11}{8}$ as you calculated. So it seems that you are making a mistake in your calculation. Note that $a-b+c$ is not $a-(b+c)$ but $a-(b-c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Elliptic integral evaluation How to integrate ( $ r_o, r_b$ constants)
$$ \int \sqrt{\dfrac{r_o^2- r^2}{r^2-r_b^2}} \, dr, (r_o > r > r_b > 0)\, ? $$
With Mathematica got its coefficient imaginary, needing to take the condition properly.
EDIT1:
As a stand alone its evaluation is problematic. However, using NDSolve of Mathematica along with other evaluated variables it solves OK in a sort of self-regulatory mode, the puzzling aspect had me post it over here.
| We are given $0<r_{b}<r<r_{o}$, which implies $0<\frac{r_{b}}{r_{o}}<\frac{r}{r_{o}}<1$. Defining $b:=\frac{r_{b}}{r_{o}}$ and substituting $x=\frac{r}{r_{o}}$, the integral becomes:
$$\begin{align}
\int\sqrt{\frac{r_{o}^2-r^2}{r^2-r_{b}^2}}\,\mathrm{d}r
&=\int\sqrt{\frac{1-\left(\frac{r}{r_{o}}\right)^2}{\left(\frac{r}{r_{o}}\right)^2-\left(\frac{r_{b}}{r_{o}}\right)^2}}\,\mathrm{d}r
\\
&=r_{o}\int\sqrt{\frac{1-x^2}{x^2-b^2}}\,\mathrm{d}x;~~[0<b<x<1].
\\
\end{align}$$
To solve the rescaled indefinite integral $\int\sqrt{\frac{1-x^2}{x^2-b^2}}\,\mathrm{d}x$, start by substituting $y=\sqrt{1-x^2}$, and then $z=\frac{y}{\sqrt{1-b^2}}$:
$$\begin{align}
\int\sqrt{\frac{1-x^2}{x^2-b^2}}\,\mathrm{d}x
&=\int\frac{\sqrt{1-x^2}}{\sqrt{x^2-b^2}}\,\mathrm{d}x\\
&=\int\frac{\sqrt{1-x^2}}{\sqrt{(1-b^2)-(1-x^2)}}\,\mathrm{d}x\\
&=\int\frac{y}{\sqrt{(1-b^2)-y^2}}\cdot\frac{(-y)}{\sqrt{1-y^2}}\,\mathrm{d}y\\
&=-\int\frac{y^2}{\sqrt{(1-b^2)-y^2}\,\sqrt{1-y^2}}\,\mathrm{d}y\\
&=-\frac{1}{\sqrt{1-b^2}}\int\frac{(1-b^2)\,z^2}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\sqrt{1-b^2}\,\mathrm{d}z\\
&=\int\frac{-(1-b^2)z^2}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\,\mathrm{d}z\\
&=\int\frac{-1+1-(1-b^2)z^2}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\,\mathrm{d}z\\
&=\int\frac{1-(1-b^2)z^2}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\,\mathrm{d}z-\int\frac{\mathrm{d}z}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\\
&=\int\frac{\sqrt{1-(1-b^2)z^2}}{\sqrt{1-z^2}}\,\mathrm{d}z-\int\frac{\mathrm{d}z}{\sqrt{1-z^2}\,\sqrt{1-(1-b^2)z^2}}\\
&=\int\frac{\sqrt{1-b^{~\prime\,2}z^2}}{\sqrt{1-z^2}}\,\mathrm{d}z-\int\frac{\mathrm{d}z}{\sqrt{1-z^2}\,\sqrt{1-b^{~\prime\,2}z^2}}\\
&=\int\frac{\sqrt{1-b^{~\prime\,2}z^2}}{\sqrt{1-z^2}}\,\mathrm{d}z-\int\frac{\mathrm{d}z}{\sqrt{1-z^2}\,\sqrt{1-b^{~\prime\,2}z^2}}\\
&=E{\left(\arcsin{\left(z\right)}\,|\,b^{~\prime\,2}\right)}-F{\left(\arcsin{\left(z\right)}\,|\,b^{~\prime\,2}\right)}+\color{grey}{constant}.\\
\end{align}$$
The rest of the derivation is just straightforward back-substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
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probablity - recursion and random balls In each of $k$ urns numbered $1,\ldots,k$ are
$b$ white balls and $c$ black balls. Balls are randomly selected (always one ball from one urn) from the urn and throw to the next urn. From 1. to 2., 2. to 3....
What is probablity that from last urn ($k$) we select white ball ?
Look at my attempt and correct me when I am wrong.
$B_k$ - from $k^{th}$ urn we selected white ball.
$C_k$ - from $k^{th}$ urn we selected black ball.
$$P(B_k) = P(B_k|B_{k-1}) P(B_{k-1}) +
P(B_k|C_{k-1})P(C_{k-1}) =
\frac{b+1}{b+c+1}P(B_{k-1}) + \frac{b}{b+c+1} (1-P(B_{k-1})=
\frac{1}{b+c+1}P(B_{k-1}) + \frac{b}{b+c+1} $$
$$Pr(B_1) = \frac{b}{b+c}$$
So we have:
$$Pr(B_k) = \frac{\frac{b}{b+c}}{(b+c+1)^{k-1}} + \frac{(k-1)b}{b+c+1}$$
Is it ok ?
| All OK until you solve the recurrence relation. I think it should be:
\begin{eqnarray*}
P(B_k) &=& \dfrac{1}{b+c+1}P(B_{k-1}) + \dfrac{b}{b+c+1} \\
&& \\
&=& \dfrac{1}{b+c+1}\left(\dfrac{1}{b+c+1}P(B_{k-2}) + \dfrac{b}{b+c+1}\right) + \dfrac{b}{b+c+1} \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^2P(B_{k-2}) + \dfrac{b}{b+c+1}\left(1+\dfrac{1}{b+c+1}\right) \\
&& \\
&& \cdots \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}P(B_1) + \dfrac{b}{b+c+1}\left(1+\dfrac{1}{b+c+1}+\cdots +\left(\dfrac{1}{b+c+1}\right)^{k-1}\right) \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}\dfrac{b}{b+c} + \dfrac{b}{b+c+1}\dfrac{1-\left(\dfrac{1}{b+c+1}\right)^k}{1-\dfrac{1}{b+c+1}} \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}\dfrac{b}{b+c} + \dfrac{b-b\left(\dfrac{1}{b+c+1}\right)^k}{b+c}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^4-3x^2+1=0$ in terms of cosine. I put the equation in the form of a quadratic:
$(x^2)^2-3x^2+1=0$
Then using the quadratic formula,
$x^2=\frac{3\pm\sqrt{9-4}}{2}$
$x^2=\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$
$x=\pm\frac{1+\sqrt{5}}{2}$ and $\pm\frac{1-\sqrt{5}}{2}$
So there are four roots as expected given the equation is a quartic. But I really don't know how to put the answers in terms of cosine. Any hints?
| let $\tau = \dfrac{1 + \sqrt 5}{2}$ be the golden ratio which is the positive solution of the quadratic equation $x^2 - x - 1=0$
we also have all powers of $\tau$ as linear combination of $1, \tau$ for example, $ \dfrac{1}{\tau} = \dfrac{\sqrt 5 -1}{2}, \tau^2 = \tau + 1 =\dfrac{3+\sqrt 5}{2}, \dfrac{1}{\tau ^2} = 2 - \tau = \dfrac{3 - \sqrt 5}{2}, \cdots$
from the regular pentagon you get $\cos 36^\circ = \tau^2/2, \cos 72^\circ = \dfrac{1}{2\tau}$ the quadratic equation that has $\tau^2, 1/\tau^2$ for roots is $(x - \tau^2)(x - 1/\tau^2) = x^2 - 3x + 1$
so the roots of the quartic equation $x^4 - 3x^2 + 1 = 0$ are $\pm \tau, \pm \dfrac{1}{\tau}.$
we can express the two roots $\dfrac{1}{\tau} = 2 \cos 72^\circ, \tau^2 = 2\cos 36^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral of $\cos^4(2t)\,dt$ with bounds from $0$ to $\pi$ $$\int_0^\pi\cos^4(2t)\,dt=?$$ I have attempted this problem two different ways and got two different answers that are nowhere near the correct answer. Could you please show me detailed steps on how to work this problem out. The final answer is $\frac{3\pi}{8}$.
| $$I=\int_0^\pi\cos^4(2t)dt$$
Substitute: $u=2t\rightarrow du=2dt$,
$$I=\dfrac{1}{2}\int_{0\times 2}^{\pi\times 2}\cos^4(u)du=\dfrac{1}{2}\int_0^{2\pi}\cos^4(u)du$$
Apply the reduction formula: $\int\cos^n(x)dx=\dfrac{\sin(x)\cos^{n-1}(x)}{n}+\dfrac{n-1}{n}\int\cos^{n-2}(x)dx$, with $n=4$ :
$$I=\left.\dfrac{1}{8}\sin(u)\cos^3(u)\right\vert_0^{2\pi}+\dfrac{3}{8}\int_0^{2\pi}\cos^2(u)du$$
Use the trigonometric identity: $\cos^2(x)=\dfrac{1}{2}\cos(2x)+\dfrac{1}{2}$,
$$I=0+\dfrac{3}{8}\int_0^{2\pi}\left(\dfrac{1}{2}\cos(2u)+\dfrac{1}{2}\right)du$$
Substitute: $v=2u\rightarrow dv=2du$,
$$I=\dfrac{3}{32}\int_{0\times 2}^{2\pi\times 2}\cos(v)dv+\dfrac{3}{16}\int_0^{2\pi}du$$
$$I=\left.\dfrac{3}{32}\sin(v)\right\vert_0^{4\pi}+\left.\dfrac{3}{16}u\right\vert_0^{2\pi}$$
$$I=0+\dfrac{3}{16}\left(2\pi-0\right)$$
$$I=\dfrac{3\pi}{8}$$
Conclude:
$$\int_0^{\pi}\cos^4(2t)dt=\dfrac{3\pi}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction
My proof so far:
Let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$
Base Case
$P(1):$
LHS = $1^3 = 1$
RHS = $(1)^2 = 1$
Since LHS = RHS, therefore base case holds
Induction Hypothesis
Let $n \in \mathbb{N}$ be arbitrary
Assume $P(n)$ holds
Induction Step
Prove $P(n+1)$ holds:
$$
\begin{align}
& 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1)
\end{align}
$$
This is where I get stuck. I don't know how to prove that my last step is equivalent to:
$$(1 + 2 + \cdots + \;n + (n+1))^2$$
| Hint: by the binomial theorem,
$(1+2+\cdots +n +(n+1))^2=(1+2+\cdots +n)^2+(n+1)^2+2(n+1)(1+2+ \cdots n)$.
Now, you can show by induction (or by Gauss' trick) that $1+2+\cdots +n= \dfrac{n(n+1)}{2}$, and hus reach the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How is $x^2+1=(1/{x^2})[1-{1}/{x^2}+{1}/{x^4}-{1}/{x^6}+\cdots]$? The author of my book writes:
$$x^2+1=x^2\left(1+\frac{1}{x^2}\right)$$
$$=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]$$
I do not understand the last step. How did the author write the last step. Please help.
| Formula is not correct. Take the sum of a geometric series:
$$\sum_{i=0}^{\infty} ar^i=\frac{a}{1-r}$$
Lets work backward from your last equation:
$$\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]=\frac{1}{x^2}\left[\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i-x^2\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i+x^2\right]=$$
The two sums in the brackets are simply geometric sums with $r=\frac{1}{x^4}$, thus:
$$\left[\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i-x^2\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i+x^2\right]=\left[\frac{1}{1-\frac{1}{x^4}}-\frac{x^2}{1-\frac{1}{x^4}} +x^2\right]=\left[\frac{1-x^2+(1-\frac{1}{x^4})x^2}{1-\frac{1}{x^4}}\right]$$
The last equation simplifies to:
$$\left[\frac{1-\frac{1}{x^2}}{1-\frac{1}{x^4}}\right]=\frac{x^2}{1+x^2}\implies x^2+1=\frac{1}{x^2+1}???$$ This is not correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Subsets and Splits
Fractions in Questions and Answers
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