Given the Fourier series of the $2\pi$-periodic function defined for $$-\pi\leqslant x \leqslant \pi$$ by $$f(x) = |x|$$ is $$ \frac{\pi}{2} -\frac{4}{\pi} \sum_{k\geq 1, k\ odd}^{\infty} \frac{cos(kx)}{k^2} $$ use this to evaluate the sum of the series $$\sum_{k\geq 1,\, k\ odd}^{\infty} \frac{1}{k^4} $$
I don't know why I'm struggling with this, the answer is $s = \frac {\pi^4} {96}$ but I can't seem to get that. My approach is to let $x = \pi$ and this sets the given equation to $\frac {-1}{k^2}$, then I equate that with pi and get $\frac{\pi^2} {8}$. I've tried a number of things including Parsevals formula but I keep getting the wrong answer.
Thank you.
I think I've solved it with the help of the comments 🙂
$\frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 = \frac{1}{4}(\pi)^2 + \frac{1}{2}(\frac{4}{\pi})^2 \sum_{k\geq 1, k odd}^{\infty } \frac{1}{k^4}$
As per Parsevals formula. My only remaining question is why can I just set $x = \pi$ like this, and get rid of the $cos(kx)$ term? The function is left and right differentiable at $x = pi$ but why does that help?
Anyway, solving the above equation yields $\frac{\pi^2}{6} = (\frac{4}{\pi})^2 \sum_{k\geq 1, k odd}^{\infty } \frac{1}{k^4}$.
This is equal to $\frac{\pi^4}{96}$ as required.
Thank you and I hope the solution saves somebody the trouble in the future.
Best Answer
Parseval says $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} |x|^2 dx = \frac{1}{4}|A_0|^2 + \frac{1}{2} \sum_{n=1}^{\infty} |A_n|^2 $$ (since all $B_n$ are zero). What's easy to forget is that the constant term, $\pi/2$ is not $A_0$ but $A_0/2$; hence $$ \frac{\pi^2}{3} = \frac{1}{4} \left( 2 \frac{\pi}{2} \right)^2 + \dots $$ Maybe that's why you're not getting the right answer?