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So i have a question which asks to find the fourier series of $\left\vert\,\sin\left(x\right)\,\right\vert\,$. I have worked out the solution as $$ {2 \over\pi} - {4 \over \pi}\sum_{k = 1}^{\infty}{\cos\left(2kx\right) \over 4k^{2} - 1} $$

Which i am pretty sure is correct as i have the solution in my book.

The second part of the question asks to work out the sums of $$ \sum_{k = 1}^{\infty}{1 \over 4k^{2} - 1}\qquad\mbox{and}\qquad \sum_{k = 1}^{\infty}{\left(-1\right)^{k} \over 4k^{2} - 1} $$

Im sure this is probably very simple but i have no solution for this and I am struggling to search for an explanation of how to do this on google. Could someone please tell me know it is done ?.

Many thanks.

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Solve for the infinite sum:

$$ \sum_{k=1}^\infty \frac{\cos(2nx)}{4n^2-1} = \frac{1}{2} - \frac{\pi}{4} \left| \sin x \right| $$

What happens if you let $x = 1$?

What happens if you let $x = \frac{\pi}{2}$?

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  • $\begingroup$ Sorry yeah i deleted my comment straight away as i soon realised what you had written!for x = 0 i got 1/2, for x=pi/2 i have $1/2-\pi/4$ $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:51
  • $\begingroup$ Good observation -- I should have written $x = \pi/2$ (I will edit that now). Do you see how $x = 0$ gives an answer to your first question? $\endgroup$ – Kyle Dec 12 '13 at 0:53
  • $\begingroup$ yeah i do! amazing thank you very much, both combined answers have led me to a solution! :) $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:54
  • $\begingroup$ Glad to help! If you wouldn't mind, please consider accepting one of our answers -- helps to keep the site organized! (I would do Igor's since we were both helpful but he answered first!) $\endgroup$ – Kyle Dec 12 '13 at 1:07
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Well, if the fourier series converges to the function, plug in zero as the argument to $|\sin x|$ and to the fourier series. See what happens. In the second instance, plug in $\pi/2$ as argument.

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  • $\begingroup$ but what do i plug 0 into? $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:44
  • $\begingroup$ would i plug 0 into $$ \sum_{k=1}^\infty \frac{\cos(2nx)}{4n^2-1} = \frac{1}{2} - \frac{\pi}{4} \left| \sin x \right| $$ giving $1/2$ $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:48
  • $\begingroup$ and plugging in $\pi/2$ gives $1/2 - \pi/4$ $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:49
  • $\begingroup$ also could you please tell me why 0 and $\pi/2$? $\endgroup$ – Bernard.Mathews Dec 12 '13 at 0:50
  • $\begingroup$ because you want $\cos(2n x) \equiv 1$ for the first question and $\cos (2n x) = (-1)^n$ for the second. $\endgroup$ – Igor Rivin Dec 12 '13 at 0:51
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what a beautiful result! and if one plugs in $\frac{\pi}4$ it appears one gets (something like!) $$\frac{\sqrt{2}}2=\frac2{\pi} \left(1+\sum_{n=1} \frac{(-1)^{n+1}}{(4n+1)(4n-1)} \right)$$

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