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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.

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  • $\begingroup$ You mean the Fourier series of $f(x) = x \sin(x) 1_{x \in [0,4\pi]}$ periodized, not the Fourier transform. And $f(x)$ integrable means its Fourier coefficients are well-defined, but not that its Fourier series converges. Here its Fourier series converges, because except at $x=0$, it is $C^1$, and its discontinuity at $x=0$ is of the same kind as the sawtooth $s(x) = x 1_{x \in [0,4\pi]}$ whose Fourier series converges (you have to prove it). Note that from the Fourier series of $s(x)$ you can deduce the Fourier series of $(x)$ directly $\endgroup$
    – reuns
    Commented Sep 14, 2016 at 9:57
  • $\begingroup$ The Fourier transform of a periodic function is well-defined only as a distribution, and is nothing more than a different representation of the Fourier series. $\endgroup$
    – reuns
    Commented Sep 14, 2016 at 9:59

1 Answer 1

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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 $1/n^2$.

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