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Math Help - Fourier Series of odd function

  1. #1
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    Fourier Series of odd function

    I am having trouble with this fourier series:

    f(x)=\left\{\begin{array}{cc}\cos(x),&\mbox{ if }<br />
 0\leq x< \pi\\-\cos(x), & \mbox{ if }-\pi\leq x< 0 \end{array}\right.

    I understand that it is an odd function and as such there are no \cos terms, only \sin but i am told that the function should equal

    \frac{8}{\pi} \sum_{m\ =\ 1}^\infty \frac{m}{4m^2-1}\sin(2mx)

    and i am finding it impossible to do so.

    Could someone please show me how this is done?
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  2. #2
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    Hi

    \int_{-\pi}^{\pi} f(x) \sin(nx) dx = \int_{-\pi}^{0} -\cos x \sin(nx) dx + \int_{0}^{\pi} \cos x \sin(nx) dx

    Now using \sin a \cos b = \frac12\:\left(\sin(a+b) + \sin(a-b)\right)

    \int_{-\pi}^{\pi} f(x) \sin(nx) dx = \int_{-\pi}^{0} -\frac12\:\left(\sin(n+1)x + \sin(n-1)x\right) dx + \int_{0}^{\pi} \frac12\:\left(\sin(n+1)x + \sin(n-1)x \right) dx

    Can you get it from here ?
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  3. #3
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    Each time i try to find the series it seems to cancel out.

    For example i managed to reach

    \frac{\cos(n-1)\pi}{n-1} - \frac{\cos(n+1)\pi}{n+1}

    but either this is wrong or i simply can't progress anywhere from it without it seeming to cancel to 0.
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  4. #4
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    Remember that \cos(n+1)\pi = (-1)^{n+1}

    Therefore you need to separate the cases : n odd or n even

    If n is odd (n = 2m+1) then \frac{1}{n+1} - \frac{(-1)^{n+1}}{n+1} = 0
    The same for the other terms and finally the sum is equal to 0

    If n is even (n = 2m) then \frac{1}{n+1} - \frac{(-1)^{n+1}}{n+1} = \frac{2}{n+1}
    The other terms also do not cancel but double and at the end you can find \frac{4}{2m+1}+\frac{4}{2m-1} = \frac{16m}{4m^2-1}
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  5. #5
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    Thanks, i guess i need to remember those identities a lot better .
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