I'm having trouble visualizing the Fourier series of $x \sin(x)$, say if it's graphed over $[0,4 \pi]$ and then repeated.
This is an integrable function, so I should be able take the Fourier series right? I'm just having trouble with the intuition here.
I know the Fourier transform gives a picture of 'how much' of each frequency of sin functions is used in the construction of the function as a Fourier series. Each 'loop' could be thought of as a sum of sines with different frequencies, and so this would certainly not be a Dirac impulse like $sin(x)$.
I was initially worried about the lack of differentiability at each integer multiple of $4 \pi$ but realized that's nonsense, as a saw tooth function has a Fourier series.
Can someone clear up my confusion? Sorry if I'm casually throwing around incorrect terminology; I haven't had to think about Fourier series for over a year.
A smooth function has a Fourier series with fast decaying coefficients. The best example is the sinusoid (or more generally any trigonometric polynomial), with all coefficients zero but a finite number of them. Another is $e^{\cos(x)}\sin(\sin x)$, with coefficients $1/n!$
On the opposite, discontinuities correspond to small decays because smooth functions have trouble recreating them.
In the given case the function is continous,
$$0\sin(0)=4\pi\sin(4\pi)$$
but its derviative isn't,
$$\sin(0)+0\cos(0)\ne\sin(4\pi)+4\pi\cos(4\pi).$$
Zero-th order discontinuities give decays in $1/n$, while first order discontinuities (derivative) are in $1/n^2$.