# Evaluating an infinite sum using a fourier seriesWatch

#1
I've got

and

I want to find

I know I need to look at ideally, since we need our sin term to disappear and our cos to turn into so we can somehow get a 1 on the numerator.

However, both plus pi and minus pi yield answers that aren't consistent with what I think I should get, which I believe should be .
0
7 years ago
#2
It's simpler than that. Set x = 0.
0
#3
What happens to the (-1)^n?

0
8 months ago
#4
Set x to pi, the sin(npi)=0 and cos(npi)=(-1)^n which becomes 1 when multiplied to the (-1)^n you already have, then solve from there (and I know its 7 years later, but idk, math is math)
0
8 months ago
#5
and if you want a 1 on the numerator, divide by 2, a sum is a linear operator so you can factor it out and then divide both sides by 2.
0
8 months ago
#6
(Original post by Lord_Tachanka)
Set x to pi, the sin(npi)=0 and cos(npi)=(-1)^n which becomes 1 when multiplied to the (-1)^n you already have, then solve from there (and I know its 7 years later, but idk, math is math)
It's against the rules (and obviously pointless) to resurrect posts that are several years old.
0
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