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Value of the infinite sum of fourier coefficients?

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the exact value of the sum from minus infinity to infinity of Ck.


    2. Relevant equations
    Ck = [itex]\frac{672}{κ^{2}π^{2}}[/itex]([itex]e^{-iκ\frac{π}{2}}[/itex][itex]-[/itex][itex]e^{-iκπ}[/itex])[itex]-[/itex]2744([itex]\frac{e^{-iκ\frac{π}{14}}}{iκπ}[/itex][itex]+[/itex][itex]\frac{e^{iκ\frac{π}{14}}}{κ^{2}π^{2}}[/itex][itex]-[/itex][itex]\frac{14e^{-iκ\frac{π}{14}}}{iκ^{3}π^{3}}[/itex][itex]+[/itex][itex]\frac{14e^{ik\frac{π}{14}}}{iκ^{3}π^{3}}[/itex])[itex]+[/itex][itex]\frac{392}{κ^{2}π^{2}}[/itex]([itex]e^{iκ\frac{π}{14}}[/itex][itex]-[/itex][itex]e^{iκ\frac{π}{2}}[/itex][itex]-[/itex][itex]e^{-iκ\frac{π}{14}}[/itex][itex]+[/itex][itex]e^{-iκ\frac{π}{2}}[/itex])


    3. The attempt at a solution
    I'm not good with sums, I really have no idea how to get started.

    (it's somewhat urgent, I'm supposed to solve it until tomorrow)
     
    Last edited: Oct 3, 2011
  2. jcsd
  3. Oct 3, 2011 #2
    I don't see why the series should converge, let alone the exact sum.
     
  4. Oct 3, 2011 #3
    I did a problem before where I had to calculate the trigonometric coefficients (the ones I wrote here) and in this problem it says to sum the coefficients and find the exact value.
    However, one of my teachers mentioned something that it was supposed to be the sum of complex coefficients. If there wouldn't be a sine, and there would be the number "e" instead (powered to something). Would it then be possible to find an exact value?
     
  5. Oct 3, 2011 #4
    Took a little pause, now trying to calculate the complex coefficients... takes some time since the function is piecewise. It seems that the result will have various e to the power of plus minus i times k times some radian angle. The e's have some coefficients also.
    Is this something that seems like it could converge if summed?
    Will try to post the complex coefficients once I'm done calculating.
     
  6. Oct 3, 2011 #5
    I changed to complex coefficients now, any ideas?
    Hope I did it right, was even more terms before but they cancelled out.
    The numbers are divisible by 7 (more then once) just so you know.
     
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