Fourier Series problem - help craved

Eliot Rosewater

Hi guys. Just doing a bit of Fourier Series there, and having problems getting correct answers. If anyone could help, I'd seriously love them.

I'm considering the piecewise function
[LATEX]\displaystyle f(x) = -\pi, 0<x</pi[/LATEX]
[LATEX]\displaystyle f(x) = x-\pi, \pi<x<2\pi[/LATEX]

We want to evaluate the Fourier series at [LATEX]\displaystyle f(\pi) = 0[/LATEX] in order to derive the identity
[LATEX]\displaystyle \sum_{m=0}^{\infty} \frac{1}{(2m+1)^2}= \frac{\pi^2}{8}[/LATEX]

The Fourier Series is given by
[LATEX]\displaystyle S(x) = \frac{1}{2}a_{0}+\sum_{n=1}^{\infty}\left(
a_{n}\cos(n x) + b_{n} sin(n x)
\right)[/LATEX]

The Sin part of the Fourier series is not important, as we need to evaluate the series at x=pi, and all the Sin terms will disappear as Sin(n pi)=0 for all n in the natural numbers.

I got the co-efficients by hand, and verified with Mathematica:

[LATEX]\displaystyle a_{0} = \int_{0}^{2\pi} f(x) dx = \frac{-\pi^2}{2} [/LATEX]

[LATEX]\displaystyle a_{n} = \int_{0}^{2\pi} f(x) \cos(n x) dx[/LATEX]

[LATEX]\displaystyle a_{n} = \int_{0}^{\pi} -\pi \cos(nx)dx +
\int_{\pi}^{2\pi} x \cos(nx)dx +
\int_{\pi}^{2\pi} -\pi \cos(nx)dx
[/LATEX]

The first and third intergrals are 0, so

[LATEX]\displaystyle a_{n} = \int_{\pi}^{2\pi} x \cos(nx)dx[/LATEX]

[LATEX]\displaystyle a_{n} = \left[ \frac{x \sin(nx)}{n} +\frac{\cos(nx)}{n^2} \right]_{\pi}^{2\pi}[/LATEX]

[LATEX]\displaystyle a_{n} = \frac{\cos{2\pi n}-cos{\pi n}}{n^2}[/LATEX]

[LATEX]\displaystyle a_{n} = \frac{1-(-1)^n}{n^2}[/LATEX]

as n is a natural number.

The Cosine Fourier Series is thus

[LATEX]\displaystyle S(x) = \frac{1}{2}\frac{-\pi^2}{2}+\sum_{n=1}^{\infty}\frac{1-(-1)^n}{n^2}\cos(n x) [/LATEX]

For n even, the terms disapear, so

[LATEX]\displaystyle S(x) = \frac{1}{2}\frac{-\pi^2}{2}+\sum_{m=0}^{\infty}\frac{1--1}{(2m+1)^2}\cos((2m+1) x) [/LATEX]`

[LATEX]\displaystyle S(x) = \frac{-\pi^2}{4}+\sum_{m=0}^{\infty}\frac{2}{(2m+1)^2}\cos((2m+1) x) [/LATEX]

Let x=pi. Then Cos((2m+1)pi) = -1, so

[LATEX]\displaystyle S(\pi) = \frac{-\pi^2}{4}+\sum_{m=0}^{\infty}\frac{2}{(2m+1)^2}(-1) =0[/LATEX]

[LATEX]\displaystyle \sum_{m=0}^{\infty}\frac{-2}{(2m+1)^2} = \frac{\pi^2}{4}[/LATEX]

[LATEX]\displaystyle \sum_{m=0}^{\infty}\frac{1}{(2m+1)^2} = \frac{-\pi^2}{8}[/LATEX]

Which is obviously a silly answer: the left is positive; the right is negative. Can anyone tell me where I'm going wrong?

Eliot Rosewater

Okay, I'm an idiot. (This isn't the first time I've replied to one of my own threads here with that!)

I made two mistakes, the second partly stemming from the first.

For my Fourier co-efficients I wasn't multiplying by 1/pi. So my series should have been:

[LATEX]\displaystyle S(x) = \frac{1}{\pi}\left( \frac{-\pi^2}{4}+\sum_{m=0}^{\infty}\frac{2}{(2m+1)^2}\cos((2m+1) x) \right) [/LATEX]

Because I was getting a pi squared for the first term (whereas, by the correction above, I should have gotten just pi) I took f(x) to be 0 at x=pi, so that the term coming over would be pi squared. Silly assumption. The actual convergance forumula is

[LATEX]\displaystyle S(\pi)=\frac{1}{2}\left(\lim_{x \to \pi^{+}}f(x) + \lim_{x \to \pi^{-}}f(x) \right)[/LATEX]

In this particular case,
[LATEX]\displaystyle S(\pi)=\frac{1}{2}\left( (-\pi) + (x -\pi) \right)[/LATEX]

[LATEX]\displaystyle S(\pi)=\frac{1}{2}\left( -\pi \right)[/LATEX]

And thus:

[LATEX]\displaystyle \frac{1}{\pi}\left( \frac{-\pi^2}{4}+\sum_{m=0}^{\infty}\frac{2}{(2m+1)^2}\cos((2m+1) \pi) \right) = \frac{1}{2}(-\pi)[/LATEX]

[LATEX]\displaystyle \sum_{m=0}^{\infty}\frac{2}{(2m+1)^2}(-1) = \frac{-\pi^2}{2}+\frac{\pi^2}{4}[/LATEX]

[LATEX]\displaystyle \sum_{m=0}^{\infty}\frac{-2}{(2m+1)^2}= \frac{-\pi^2}{4}[/LATEX]

[LATEX]\displaystyle \sum_{m=0}^{\infty}\frac{1}{(2m+1)^2}= \frac{\pi^2}{8}[/LATEX]

As required. Phew.

Michael Collins

Thanks for the resolution Eliot.

For anyone else watching this is a perfect example of how people should ask a question in my opinion.

And if you answer it yourself, all the better...:D