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It was a problem that I encountered earlier in a harmonic analysis course.

One student asked some question, which leads to an interesting sub-question that: by definition, $sin(x)=\sum_{k} \frac{(-1)^k}{(2k+1)!}x^{(2k+1)}$ none of the summand $\frac{(-1)^k}{(2k+1)!}x^{(2k+1)}$ is periodic, why their sum $sin(x)$ is periodic of $2\pi$?

An immediate response was given much like the answer below:

How to prove periodicity of $\sin(x)$ or $\cos(x)$ starting from the Taylor series expansion?

But actually that does not answer the essence of the question(sometimes the summands may not even be differentiable, like wavelets), that is, Why a (convergent) series of aperiodic functions sum up to a periodic function? Is there any deeper theory lying behind the answer of this question?

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Henry.L
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  • Because the partial sums, none of which is periodic, converge to a periodic function? There is nothing magic about that. After all, if $f=g+h$, and $f$ is periodic, there's no reason whatsoever to think that $g$ or $h$ is periodic. For example, $\sin x = (\frac12\sin x + e^x)+(\frac12\sin x - e^x) $. – MPW Mar 30 '17 at 16:21
  • @MPW Even if it is picked as the definition of sine? I know what you mean, but for this case, I think the essence of the problem is actually how the periodicity comes into play, via the infinite series? or via the properties of summands? Or something else? Our original problem is actually about some construction which follows a similar construction of the sine series and we want to prove its periodicity. – Henry.L Mar 30 '17 at 16:26

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