Problem with Fourier series

**hampos** · May 22nd 2012, 08:46 AM

Hey guys,

I am having some serious trouble with this one.

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

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!

**BAdhi** · May 22nd 2012, 08:55 PM

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

**hampos** · May 22nd 2012, 11:29 PM

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 $2\pi$ . 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?

**BAdhi** · May 23rd 2012, 10:25 AM

sounds right to me....

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

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

Thread: Problem with Fourier series

LinkBack

Thread Tools

Display

Problem with Fourier series

Re: Problem with Fourier series

Re: Problem with Fourier series

Re: Problem with Fourier series

Similar Math Help Forum Discussions

Fourier Series Problem.

fourier series problem

Fourier series problem

Fourier Series problem

a problem about fourier series

Search Tags