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Calculate the Fourier series expansion for the following function of period 2:

$f(t)=2+2t^2$ for $-1<t<1$

I just have a small question for this problem.

I've already gotten $A_0$ to be $\displaystyle \frac{4}{3}$ and worked through the integral of $A_n$ to get to $\displaystyle \frac{2t^2\sin(n\pi t)}{n\pi}-\frac{4t\cos(n\pi t)}{n^2\pi^2}+\frac{4\sin(n\pi t)}{n^3\pi^3}$

However, I'm confused as to how to get from this answer to the final $A_n$ of $\displaystyle \frac{8(-1)^n}{n^2\pi^2}$.

I just need to know how this part works - I understand the rest of the problem. Thanks!

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First it looks like you have not actually evaluated the definite integral, ie. putting in t = 1 and t = -1. Once you've done this, you should recall the fact that $ \sin(\pi n) = 0 $ for all n, and $ \cos(\pi n) = -1^n $ for all n.

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  • $\begingroup$ Ah, thank you for correcting my silly mistake. I took a break from Fourier Series for a few weeks and I just needed a refresher. Thank you for your help! $\endgroup$
    – Outlaw94
    May 7, 2013 at 2:50

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