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Exact value of infinity sum of Fourier series coefficients

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Sorry for doing another thread but I can't edit the old one any longer and I found out I made some calculation error but I'm pretty sure it's right now.

    The problem is to find the exact value of the series.

    2. Relevant equations
    [itex]\sum a_{k}[/itex]
    The summation is to be done from minus infinity to infinity.

    [itex]a_{k} = [/itex]-[itex]\frac{12}{49}[/itex]cos(nπ) - [itex]\frac{16}{7}[/itex]cos(n[itex]\frac{π}{14}[/itex]) + [itex]\frac{26}{49}[/itex]cos(n[itex]\frac{π}{2}[/itex])

    3. The attempt at a solution
    I was googling some and read that this sum results from when t = 0, x(0).
    Then the Fourier series changes to [itex]\frac{a_{0}}{2}[/itex] + [itex]\sum a_{k}[/itex].
    However, isn't the sum in the Fourier series from n = 1 to infinity? While here it's from minus to plus infinity. Can I change the index?
    I was thinking that this sum is the same, as the sum from minus infinity to 1 plus the sum from 1 to infinity. And since the coefficients only have cosines which are even functions then the sum from minus infinity to 1 is equal to the sum from 1 to infinity. Therefor the whole sum is simply equal to 2 times the sum from 1 to infinity. Is this correct?
    If it is, then I think I can solve it.
     
    Last edited: Oct 4, 2011
  2. jcsd
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