So I have this question here (I write out the whole question):
If $f(t)=(t+1)\cos (t)$ for $-\pi < t < \pi$, what is the sum for the Fourier series for $f(t)$ at $t=3 \pi$?
My solution is this:
IF the function is $2\pi$-periodic I can just calculate the mean value of $f(-\pi)$ and $f(\pi)$ since the Fourier series converge to the mean value at discontinuities, right? That is also the answer stated in the Solution manual. But how do I know that function is in fact $2\pi$-periodic? How do I know that the function doesn't have any other properties at the intervalls outside this one? I am just asking this to be sure that I'm not thinking to hard about things like this..
Expanding on Dr. MV's comment, consider a Fourier series for $f$ on $-L<x<L$,
$$f(x) = \sum_\limits{n=-\infty}^\infty c_n e^{i n \pi x/L}$$
What's the period of the extended series? Now what if $L=\pi$ as in your case - what's the period?
When you restrict the sum's domain to $[-L,L]$ you get $f$ on that interval, which recreates the original function. Note that the length of this interval is $2L$.
When you loosen the restriction, such as to define the sum for any $x \in \mathbb{R}$ , the RHS's periodicity comes into play. It will periodically map out $f$ in "chunks" of length $2L$.