# Fourier series of x/3

1. Jan 22, 2012

### Kuma

1. The problem statement, all variables and given/known data

I'm trying to find the fourier series of x/3.

2. Relevant equations

3. The attempt at a solution

So i believe that it is an odd function so a0, ak = 0. To find bk:

bk = 1/3pi int (x sin kx dx) from -pi to pi. From integration by parts
= 1/3pi ([x sin kx from -pi to pi]+ 1/k int (cos kx dx) from -pi to pi)
= 1/3pi [(pi sin k pi + pi sin -k pi) +1/k^2 (sin k pi - sin -k pi)]
I think that works out to be 0? Is that right?

2. Jan 22, 2012

### A. Bahat

Well, by inspection $b_k$ cannot all be zero, because then the whole function would be zero.

3. Jan 22, 2012

### Kuma

Sorry I made a big mistake. Forgot to mention the period which is from 0 to 2pi. Also my integration by parts was wrong which resulted in bk being 0 when it shouldn't be.

First question, how do i do it with a period from 0 to 2pi. Is my integrand just from 0 to 2pi?
The formula for bk is 1/L*integral from L to -L f(x) sin k pi x/L dx

What is L in this case? 2pi??