2
$\begingroup$

I need to prove this: $$ \sum_{n = 1}^{\infty}{8 \over \left(\,2n - 1\,\right)^{2}\pi^{2}}\, \sin\left(\,\left[\,2n - 1\,\right]\,{\pi x \over 2}\,\right) \sin\left(\,\left[\,2n - 1\,\right]\,{\pi z \over 2}\,\right) = \min\left\{x, z\right\} $$

I got this: $$ \frac{8}{\pi ^ 2} \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} \cos((2n -1)\frac{\pi (x - z)}{2}) - \frac{8}{\pi ^ 2} \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} \cos((2n -1)\frac{\pi (x + z)}{2})$$

But now I'm completely stuck. Can anybody please help me?

Thanks a lot!!

$\endgroup$
  • $\begingroup$ I think the index of the sums should be $n$ instead of $i$? $\endgroup$ – Harto Saarinen Sep 19 '14 at 20:09
  • $\begingroup$ are $x,z\ge0$ ? $\endgroup$ – robjohn Sep 19 '14 at 20:58
  • $\begingroup$ Yes, they are. $x$ and $z$ are defined in [0, 1] $\endgroup$ – Maria Sep 19 '14 at 21:30
3
$\begingroup$

Big Hint:

Integrating the negative of $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(nx)}{n} &=\mathrm{Im}\left(\sum_{k=1}^\infty\frac{e^{inx}}{n}\right)\\ &=-\mathrm{Im}\left(\log(1-e^{ix})\right)\\ &=\frac{x}{2|x|}(\pi-|x|)\tag{1} \end{align} $$ we get $$ \sum_{n=1}^\infty\frac{\cos(nx)}{n^2}=\frac{2\pi^2-6\pi|x|+3x^2}{12}\tag{2} $$ and subtracting $\frac14$ of $(2)$ at $2x$, which is the even terms of $(2)$, $$ \sum_{n=1}^\infty\frac{\cos(2nx)}{4n^2}=\frac{\pi^2-6\pi|x|+6x^2}{24}\tag{3} $$ we get $$ \sum_{n=1}^\infty\frac{\cos((2n-1)x)}{(2n-1)^2}=\frac{\pi^2-2\pi|x|}8\tag{4} $$ for $x\in(-\pi,\pi)$.

Next, recall that $\min(x,y)=\dfrac{x+y-|x-y|}2$.

$\endgroup$
  • $\begingroup$ Thank you so much!! I am trying to figure out the expression for $\operatorname{Im}(\log(1-\exp(ix)))$ I have to solve $\arctan(\frac{-\sin(x)}{1 - \cos(x)})$, right? I obtain $\frac{\pi - x}{2}$. Could you please explain to me why you add the sign of x to the $\pi$ term? Thanks again, this is so helpful!! $\endgroup$ – Maria Sep 19 '14 at 21:25
  • $\begingroup$ Note that $\displaystyle\sum_{k=1}^\infty\frac{\sin(nx)}{n}$ is an odd function of $x$. The sign trickery is there simply to make the result an odd function. $\endgroup$ – robjohn Sep 19 '14 at 21:32
  • $\begingroup$ Ah, ok I got it! Thank you very much! $\endgroup$ – Maria Sep 19 '14 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.