Hi I'm having trouble with this question where it asks me to evaluate a fourier series at a particular value to derive an identity. The question is:
"Let $f(x)$ is given by $f(x) = x$, for $0\leq x \leq \pi$. The function is then defined on the interval $-3\pi\leq x \leq 3\pi$ by extending it as an even function. It is then defined on the whole real line by extending it as an $2π$-periodic function.
a)Sketch the graph of $f(x)$ between $−3π$ and $3π$
b)Calculate the Fourier series of $f(x)$.
c)By evaluating the Fourier series of $f(x)$ at $x = 0$ show that:
$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}"$$
So the first question was easy enough it looks similar to a triangular wave with its minimum at $x=0$.
For the second question I got used the fourier series even extensions to get the fouier series for $f(x)$. The fourier series I got was: $$\frac{\pi}{2}+ \sum_{m=1}^{\infty} \frac{2((-1)^m-1)}{m^2\pi}cos(mx) = x$$
For the third question I set $x=0$ and rearranged the FS above to get:
$$\sum_{m=1}^{\infty} \frac{(-1)^{m+1}+1}{m^2} = \frac{\pi^2}{4}$$ At this point I was unsure so I let $m=2k-1$. This gave me:
$$\sum_{k=1}^{\infty} \frac{(-1)^{2k}+1}{(2k-1)^2} = \frac{\pi^2}{4}$$
noting that $(-1)^{2k}$ is $1$:
$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$$. Which is the desired result.
The only part that i'm unsure about is the part where I let $m=2k-1$ because I am usually quite poor at changing the limits of a summation.
Any verification/ help would be appreciated.
Your answer is correct except for the part where you simply let $m=2k-1$. Instead you should write: $$ \sum_{m=1}^{\infty} \frac{(-1)^{m+1}+1}{m^2} =\sum_{k=1}^{\infty} \left[\frac{(-1)^{2k}+1}{(2k-1)^2}+\frac{(-1)^{2k+1}+1}{(2k)^2}\right], $$ effectively summing separately over odd and even $m$ (which together form the complete set of positive integers).
Do you see now how the result follows?