Problem with Fourier series

Hey guys,

I am having some serious trouble with this one.

We have this function:

$\displaystyle h(x) = 8+4x, 0<x<pi$

$\displaystyle h(x) = -8+4x, -pi<x<0$

We also know that:

$\displaystyle h(x+2\pi)=h(x)$

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

Re: Problem with Fourier series

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

Re: Problem with Fourier series

Quote:

Originally Posted by

**BAdhi** 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 $\displaystyle 2\pi$. 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 $\displaystyle a_{n}=0$ since the function is odd. I'm trying to figure out $\displaystyle b_{n}$ and have gotten this far:

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

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

Am I getting somewhere here?

Re: Problem with Fourier series

sounds right to me....

$\displaystyle b_n=\frac{2}{\pi}\int^{\pi}_0(8+6x)\sin (nx)dx$

$\displaystyle =\frac{2}{\pi}\left(\int^{\pi}_08sin(nx)dx+\int^{ \pi }_06xsin(nx)dx\right)$

$\displaystyle =\frac{2}{\pi}\left(\left[8\frac{-\cos(nx)}{n}\right]^{\pi}_0+\left[6x\frac{-\cos(nx)}{n}\right]^{\pi}_0+6\int^{ \pi }_0\frac{\cos(nx)}{n}dx\right)$

$\displaystyle =\frac{2}{\pi}\left(8\frac{1-(-1)^n}{n}-6\pi \frac{(-1)^n}{n}+6(0)\right)$

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