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Math Help - Problem with Fourier series

  1. #1
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    Problem with Fourier series

    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!
    Last edited by hampos; May 22nd 2012 at 09:49 AM.
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Problem with Fourier series

    can you tell us where you are stuck? because we cannot just give out the answer.
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  3. #3
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    Re: Problem with Fourier series

    Quote Originally Posted by BAdhi View Post
    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?
    Last edited by hampos; May 23rd 2012 at 01:51 AM.
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  4. #4
    Senior Member BAdhi's Avatar
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    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]
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