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Math Help - fourier series

  1. #1
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    fourier series

    Hey everyone, I need a little bit of help with this question so any help would be much appreciated.

    Find the Fourier series of the function and discuss the divergence of it.

    f(x)=\begin{cases}  1 \mbox{ for } |x| \le \frac{\pi}{2} \\  0\mbox{ for } \frac{\pi}{2} < |x| < \pi \end{cases}

    Thanks for any help everyone.
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  2. #2
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    Re: fourier series

    Actually obtaining the series shouldn't be that bad. For n\geq 1

    a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)cos(nx)dx

    a_n = \frac{1}{\pi}\Big (\int_{-\pi}^{-\frac{\pi}{2}} 0\cdot cos(nx)dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\cdot cos(nx)dx + \int_{\frac{\pi}{2}}^\pi 0\cdot cos(nx)dx \Big )

    a_n = \frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos(nx)dx

    And you'll get something very similar for b_n

    Can you take it from here?
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  3. #3
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    Re: fourier series

    oh yes, stupid me.

    I get b_n = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin(nx)dx

    But what about the divergence of the Fourier series?
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  4. #4
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    Re: fourier series

    hi,
    first the function must be periodic to be developed in Fourier series.
    f\left ( x \right )= \frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left ( a_{n} cos\left ( nx \right )+b_{n}sin\left ( nx \right )\right )
    I consider that the function has a period of 2pi look the attachement. And it's clear that the function is even thus the (bn) coefficients are null.
    a_{0}=\frac{1}{\pi }\int_{-\pi }^{\pi }f\left ( x \right )dx
    a_{0}=1
    a_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f\left ( x \right )cos\left ( nx \right )dx
    a_{n}= sinc\left ( \frac{n}{2} \right )

    and about the convergence: The serie converge to f(x) where f(x) is continous
    and diverge where f(x) is discontinous
    this is from the lejeune-dirichlet theorem:
    Attached Thumbnails Attached Thumbnails fourier series-function.jpg  
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  5. #5
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    Re: fourier series

    Thanks salim, get it now.
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