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Math Help - Fourier series of sinx

  1. #1
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    Fourier series of sinx

    Find the full series of sinx on (-l,l) in its real and complex forms. Assume that l is not an integer multiple of \pi. Hint: First find the series for e^{ix}.

    I don't see how finding the fourier series of e^{ix} helps in finding the series for sinx.

    THanks
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  2. #2
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    The reason they said that is because:

    sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
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  3. #3
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    So I simplified the complex version for the coefficient of the fourier series to

    \frac{1}{-2i{\pi}n}(e^{-in\pi+il}-e^{in\pi-il})

    I'm not sure how to simplify it from here.

    THanks
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  4. #4
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    Simplifying…

    To simplify \frac{1}{-2 \imath \pi n}(e^{-\imath n \pi+\imath l}-e^{\imath n\pi-\imath l}) we can use the fact that e^{\imath x} = \cos x + \imath\sin x to get that \frac{e^{\imath x} - e^{-\imath x}}{2i} = sin x

    Thus \frac{1}{-2 \imath \pi n} (e^{-\imath n \pi+\imath l}-e^{\imath n\pi- \imath l})=\frac{1}{2\imath \pi n}(e^{\imath (n\pi -l)}-e^{-\imath( n\pi -l)})
     =\frac{1}{\pi n} \sin (n\pi -l)

    --Kevin C.
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