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I want the value of $n$ for which this relation must satisfy: $$ \frac{x}{\pi}+\sum_{n \geq 1}\frac{2}{n\pi}\cos(n\pi)\sin(\pi x)=0 $$ How to solve this?

amWhy
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user3563283
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  • please use back slashes for pi, cos, sin, etc. \sin(\pi x) = $\sin(\pi x)$. Compare to $sin(pix)$ without the backslashes. – amWhy Sep 12 '14 at 12:17
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    As it stands it makes little sense to ask "for which $n$". Do you mean to ask where to start and stop the summation? – Arthur Sep 12 '14 at 12:18
  • @amWhy: I believe the summation is with respect to $n$. – hardmath Sep 12 '14 at 12:18
  • I think the Question must be for what value of $x$ the series expansion holds. – hardmath Sep 12 '14 at 12:22
  • Yes, I agree. The op had clearly indicated using underscore, $n\geq 1$. I suspect the OP is trying to find the range of summation for $n$, perhaps? – amWhy Sep 12 '14 at 12:23
  • Actually I am new to this site. I am learning how to post a question. Thanks for support. – user3563283 Sep 12 '14 at 12:23
  • The original post (title) had $\frac x{\pi} + $\underscore{n\geq 1}$\sum\frac{2}{n\pi}\cos(n\pi)\sin(\pi x)=0$ – amWhy Sep 12 '14 at 12:25
  • @user3563283 After all these edition and roll back, the question still didn't make any sense. Please restate what you really want to ask. If you don't know how to type the correct formula, just state it in words, we can fix the formula for you. – achille hui Sep 12 '14 at 12:27
  • I framed the equation asif it is in LaTex. I want the range of $n$. – user3563283 Sep 12 '14 at 12:28
  • @user3563283: That doesn't make any sense! Your expression depends only on $x$, and $n$ is just a dummy summation variable. – TonyK Sep 12 '14 at 13:23

1 Answers1

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You have $$ \sum_{n \geq 1}\frac{2}{n\pi}\cos(n\pi)=\frac{2}{\pi}\sum_{n \geq 1}\frac{(-1)^n}{n}=-\frac{2}{\pi}\ln 2 $$ Hence, your equation is $$ \frac{x}{\pi}-\frac{2}{\pi}\sin(\pi x)\ln 2=0 $$ which is equivalent to $$ x-2\ln 2\:\sin(\pi x)=0 $$ Setting $u=\pi x$, you get $$ \sin u=\alpha \:u $$ which admits a transcendental solution $u:=u ( \alpha )$ with $ \alpha=\dfrac{1}{2\pi\ln 2}$.

Fourier
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