I was doing Fourier series problem sets but encountered a rather surprising "problem".
The first problem stated:
Find the Fourier series for $f(x)=|\sin x|$ for $-\pi < x < \pi$.
Thus, the implied period of the function that I used for calculations later on was $2\pi$. The second didn't specify an interval, so I used $\pi$, but the actual period of $|\sin(x)|$ is just $\pi$.
Is it because in the problem it said "$-\pi < x < \pi$" the reason one must use $2\pi $ as a period? This changes the angular frequency ($1$ for period $\pi$ and $2$ for period $2\pi$). Or is there another reason?
Thanks!
If I understood your question, the reason the period of the function is $\;\pi\;$ is because (using trigonometric identity)
$$\sin(x+\pi)=\sin x\cos \pi+\sin \pi\cos x=-\sin x\implies$$
$$|\sin(x+\pi)|=|-\sin x|=|\sin x|$$