# Problem with Fourier series

• May 22nd 2012, 08:46 AM
hampos
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!
• May 22nd 2012, 08:55 PM
Re: Problem with Fourier series
can you tell us where you are stuck? because we cannot just give out the answer.
• May 22nd 2012, 11:29 PM
hampos
Re: Problem with Fourier series
Quote:

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?
• May 23rd 2012, 10:25 AM
$b_n=\frac{2}{\pi}\int^{\pi}_0(8+6x)\sin (nx)dx$
$=\frac{2}{\pi}\left(\int^{\pi}_08sin(nx)dx+\int^{ \pi }_06xsin(nx)dx\right)$
$=\frac{2}{\pi}\left(\left[8\frac{-\cos(nx)}{n}\right]^{\pi}_0+\left[6x\frac{-\cos(nx)}{n}\right]^{\pi}_0+6\int^{ \pi }_0\frac{\cos(nx)}{n}dx\right)$
$=\frac{2}{\pi}\left(8\frac{1-(-1)^n}{n}-6\pi \frac{(-1)^n}{n}+6(0)\right)$
$b_n=\frac{2}{\pi}\left[\frac{8-(8+6\pi)(-1)^n}{n}\right]$