Evaluating an infinite sum using a fourier series Watch

wanderlust.xx
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#1
Report Thread starter 7 years ago
#1
I've got

f(x) = x(\pi-x)

and
f(x) = -\dfrac{\pi^2}{3} - 2 \displaystyle\sum_{n=1}^{\infty} (-1)^n \left(\dfrac{2}{n^2} cos(nx) + \dfrac{\pi}{n} sin(nx)\right)

I want to find \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}

I know I need to look at \pm\pi ideally, since we need our sin term to disappear and our cos to turn into (-1)^n 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 \frac{\pi^2}{6}.
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DFranklin
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#2
Report 7 years ago
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It's simpler than that. Set x = 0.
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wanderlust.xx
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#3
Report Thread starter 7 years ago
#3
What happens to the (-1)^n?


f(0) = 0 = -\dfrac{\pi^2}{3} - 4 \displaystyle\sum_{n=1}^{\infty} \dfrac{ (-1)^n}{n^2}

\implies -\dfrac{\pi^2}{12}  = \displaystyle\sum_{n=1}^{\infty} \dfrac{ (-1)^n}{n^2}
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Lord_Tachanka
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#4
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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)
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Lord_Tachanka
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#5
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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.
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DFranklin
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#6
Report 8 months ago
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(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.
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