Exact value of infinity sum of Fourier series coefficients

  • #1
Inertigratus
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Homework Statement


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.

Homework 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])

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:

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