I have the following fuction with T=2 and x definde for [-1,1]
$f(x) = \begin{cases} 2(x+1) & -1 \leq x\leq -0.5 \\ 1 & -0.5 \leq x\leq 0.5 \\ 2(1-x) & 0.5 \leq x\leq 1 \\ \end{cases} $
Now I have this question:
What's the value of $\sum_{n} |c_n|^2 $?
Options are 4/3, 2/3, 1/3 or 1.
We have T = 2. We use Parseval (see the Wikipedia link someone commented). In that formula we have to use T instead of 2$\pi$.
$ \int_{-1}^{-0.5} (2x+2)^2 dx + \int_{-0.5}^{0.5} (1)^2 dx + \int_{0.5}^{1} (2-2x)^2 dx = 4/3$
Now we just have to divide by T.
$4/3$ divided by 2 equals $2/3$.
So our answer is 2/3.