Value of the infinite sum of fourier coefficients?

  • #1
Inertigratus
128
0

Homework Statement


Calculate the exact value of the sum from minus infinity to infinity of Ck.


Homework 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])


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:

Answers and Replies

  • #2
Eynstone
336
0
I don't see why the series should converge, let alone the exact sum.
 
  • #3
Inertigratus
128
0
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?
 
  • #4
Inertigratus
128
0
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.
 
  • #5
Inertigratus
128
0
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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