# Thread: Problem with Fourier series

1. ## Problem with Fourier series

Hey guys,

I am having some serious trouble with this one.

We have this function:
$h(x) = 8+4x, 0
$h(x) = -8+4x, -pi

We also know that:
$h(x+2\pi)=h(x)$

The task is to decide the Fourier series for the function h. Can anyone help me with this one? Would really appreciate it!

2. ## Re: Problem with Fourier series

can you tell us where you are stuck? because we cannot just give out the answer.

3. ## Re: Problem with Fourier series

can you tell us where you are stuck? because we cannot just give out the answer.
Well, I've plotted h(x) and realized it's an odd function with period $2\pi$. But when trying to get the Fourier series I keep ending up with weird integrals, which makes me feel I've done it all wrong.

Edit: So, I assume that $a_{n}=0$ since the function is odd. I'm trying to figure out $b_{n}$ and have gotten this far:

$b_{n}=2/\pi\int\limits_0^\pi h(x) sin(nx) dx$

...where $h(x)=8+6x$

Am I getting somewhere here?

4. ## Re: Problem with Fourier series

sounds right to me....

$b_n=\frac{2}{\pi}\int^{\pi}_0(8+6x)\sin (nx)dx$
$=\frac{2}{\pi}\left(\int^{\pi}_08sin(nx)dx+\int^{ \pi }_06xsin(nx)dx\right)$
$=\frac{2}{\pi}\left(\left[8\frac{-\cos(nx)}{n}\right]^{\pi}_0+\left[6x\frac{-\cos(nx)}{n}\right]^{\pi}_0+6\int^{ \pi }_0\frac{\cos(nx)}{n}dx\right)$
$=\frac{2}{\pi}\left(8\frac{1-(-1)^n}{n}-6\pi \frac{(-1)^n}{n}+6(0)\right)$

$b_n=\frac{2}{\pi}\left[\frac{8-(8+6\pi)(-1)^n}{n}\right]$