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Complex exponential and sine-cosine Fourier series

  1. Apr 15, 2013 #1
    The sine-cosine (SC) Fourier series: $$f(x) = \frac{A_0}{2} + \sum_{j=1}^{+\infty} A_j cos(jx) + \sum_{j=1}^{+\infty} B_jsin(jx) $$

    This form can also be expanded into a complex exponential (CE) Fourier series of the form: $$ f(x) = \sum_{n=-\infty}^{+\infty} C_n e^{inx} $$

    and vice versa. Here we define, for convenience: $$ B_n := 0 $$

    (a) Prove that the SC Fourier coefficients ##A_j## and ##B_j## for non-negative integer ## j = 0,1,2,3...## are related to the CE Fourier coefficients ##C_n## by: $$ A_j = C_j + C_{-j} ; \\ B_j = i(C_j - C_{-j}) ; \\ for: j = 0,1,2,3... $$

    Hint: All you need here is Euler: ## e^{i\gamma} = cos(\gamma) + isin(\gamma) ## or ## cos(\gamma) = (e^{i\gamma} + e^{-i\gamma})/2 ## or ## sin(\gamma) = (e^{i\gamma} - e^{-i\gamma})/(2i) ##


    I plugged in the Euler formula to the CE Fourier series and evaluated the definite integral from -pi to pi, which equaled zero, soooo i'm not too sure what to do...
     
  2. jcsd
  3. Apr 15, 2013 #2

    jambaugh

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    Gold Member

    You don't need to integrate. Just expand the (CE) series using Euler's formula and group the sines and cosines into their own sums. Then judiciously apply the evenness and oddness of cosine and sine respectively to get your coefficient identities.
     
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