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Exact value of infinity sum of Fourier series coefficients
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[QUOTE="Inertigratus, post: 3538448, member: 303466"] [h2]Homework Statement [/h2] 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. [h2]Homework Equations[/h2] [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]) [h2]The Attempt at a Solution[/h2] 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. [/QUOTE]
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Exact value of infinity sum of Fourier series coefficients
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