The function $f$ is periodic with $T=2\pi$ and is given as $f(t)=t$ where $-\pi \leq t \leq \pi$ Is $f$ odd, even or periodic or neither? Also, where is $f$ discontinuous?
Normally if a function is even or odd we would plug in f(-t) and see if we got back the exact same function (even) etc, but here it is even with the period $2 \pi$. I wasn't too sure what to do so I instead found the fourier series corresponding to the restrictions above which gave me: $\sum^{\infty}_{n=1}\dfrac{2}{n}(-1)^{n+1}\sin{n t}$ Usually, sine is an odd function but I don't think this is the way to determine if $f(t)=t$ is odd or even or neither. Also, where does the discontinuity come into play? My guess is it is discontinuous in relation with its period, not sure if that makes sense but that's my only guess.