# How to use trigonometric Fourier series to verify this result

I'm studying signal processing. I've found the associated Fourier Series for a message $m(t)$ = $t^2$ over the interval $[-1, 1]$ with period $T = 2$.

However, I'm then asked to verify that

$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}.$$

The particular Fourier series not withstanding, this seems to be a common question framing in my text, to follow finding a Fourier Series with this verification, and I'm not quite sure what it is asking, or if it relates to my previous result at all. I thought I might be able to find it by equating the result to the energy integral, i.e.

$E_g = \frac{1}{2\pi} \int m(t)^2 dt$

and therefore not need to use the Fourier Series I've found at all, but the result is just similar.

The Fourier series here is $$x^2={1\over 3}+\sum_{n=1}^\infty {4(-1)^n\over n^2 \pi^2}\cos(n\pi x), \quad -1<x<1.$$
By Parseval's Identity, if $$f(x)={1\over 2}a_0+\sum_{n=1}^\infty [a_n\cos(n\pi x/\ell)+b_n\sin(n\pi x/\ell)], \quad -\ell<x<\ell,$$ then $$\|f\|^2=\ell\left({1\over 2}a_0^2+\sum_{n=1}^\infty (a_n^2+b_n^2)\right),$$ where the norm here is the $L^2$ norm on $-1<x<1$.
• Scroll to the very bottom of this definition of $L^2$ norm (not to be confused with $\ell^2$ norm). I'll edit to include that detail. – JohnD Dec 5 '14 at 2:22
• Basically both Parseval's and signal power method say that the square of the $L^2$ norm of the signal (also called the power of a signal) is equal to the interval length times the sum of the squares of the Fourier coefficients. Well, it almost says that... you have to put the $1/2$ on $a_0^2$, but otherwise that is the statement. Hope that helps explain the idea. – JohnD Dec 5 '14 at 2:26