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 $-1<x<1$.

In this context, this leads to \begin{align} \|x^2\|^2&={1\over 2}\left({2\over 3}\right)^2+\sum_{n=1}^\infty \left({4(-1)^n\over n^2 \pi^2}\right)^2\\ \int_{-1}^1 (x^2)^2\,dx&={2\over 9}+\sum_{n=1}^\infty {16\over n^4 \pi^4}\\ {2\over 5}&={2\over 9}+\sum_{n=1}^\infty {16\over n^4 \pi^4}\\ {8\over 45}\cdot{\pi^4\over 16}&=\sum_{n=1}^\infty {1\over n^4}\\ {\pi^4\over 90}&=\sum_{n=1}^\infty {1\over n^4}. \end{align}

  • $\begingroup$ Indepth and informative answer! However, I am still wondering two things about your answer: 1) How did you get the 2/5 term? 2) Should not the 2/3 which is being squared be 1/3, since that was found to be ao? $\endgroup$ – Rome_Leader Dec 5 '14 at 2:18
  • $\begingroup$ Just found an example that mentions use of a "signal power method". Might that be a simpler way to achieve that result? Not sure what it is referring to, however. $\endgroup$ – Rome_Leader Dec 5 '14 at 2:21
  • $\begingroup$ 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. $\endgroup$ – JohnD Dec 5 '14 at 2:22
  • 1
    $\begingroup$ "signal power method" is just another name for Parseval's Identity. $\endgroup$ – JohnD Dec 5 '14 at 2:22
  • $\begingroup$ 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. $\endgroup$ – JohnD Dec 5 '14 at 2:26

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