# Fourier Series

1. May 12, 2016

### Poirot

1. The problem statement, all variables and given/known data
The following function is periodic between -π and π:

f(x) = |x|
Find the Coefficients of the Fourier series and, by examining the Fourier series at x=π or otherwise, determine:
1 + 1/32 + 1/52 + 1/72 ... = Σj=1 1/(2j - 1)2

2. Relevant equations
f(x) = a0/2 + ∑n=1 ancos(nx) + bn sin(nx)

a0 = 1/π ∫π f(x) dx
an = 1/π ∫π f(x) cos(nx) dx
bn = 1/π ∫π f(x) sin(nx) dx
3. The attempt at a solution
So I've found the coefficients:
a0 = π
an = -2(1 - (-1)n)/πn2
bn = 0 (as even function)

and so f(x) = π/2 + ∑-2(1 - (-1)n)/πn2 cos(nx)

I don't know how to do the last bit however, I don't really understand if I'm meant to come out with a number or something...
Any help would be greatly appreciated thank you.

2. May 12, 2016

### stevendaryl

Staff Emeritus
Yes. Just plug $x=\pi$ into your expression. (You can see that when $n$ is even, you get 0, so you only have to consider the case for $n$ is odd.)

3. May 12, 2016

### Samy_A

Yes, they expect a value for $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)²}$.

You have computed the Fourier series, and found $\displaystyle |x|=\frac{\pi}{2}- \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(1-(-1)^n)}{n²}\cos(nx)$.

What you have to do next is actually stated in the question: "examining the Fourier series at x=π". In other words, set x=π in your Fourier series.

Last edited: May 12, 2016
4. May 12, 2016

### Poirot

Ok, Yeah I've done that and get

π/2 +4/π∑1/n^2 (From n=1 to ∞ for n odd only)

And can't see how to get rid of the π/2 or 4/π

Thanks

5. May 12, 2016

Take another look at it: You have $\pi=\pi/2+(4/\pi)(...)$ The rest is algebra.