1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
5
  • $\begingroup$ 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

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.