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If a Fourier series converges, will it converge to a periodic function? It seems logical since it is a trigonometric series. But often we are told to derive the Fourier series of functions like $x^2$, which are not periodic.

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    $\begingroup$ When you derive the Fourier series of $x^2$, what you are actually doing is deriving the Fourier series of the periodic function whose restriction to $[-\pi, \pi]$ equals $x^2$. If you want to do Fourier analysis with non-periodic functions defined on $\mathbb R$, you need the Fourier transform. $\endgroup$ – Giuseppe Negro Dec 1 '17 at 13:22
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The sum of a Fourier series is periodic. As said in a comment, when people talk about the Fourier series of $x^2$, they actually mean the Fourier series of this function:

periodic

which is the result of restricting $x^2$ to the interval $[-\pi, \pi]$ (or whatever interval you use for Fourier series), and then extending that periodically.

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  • $\begingroup$ "or whatever interval you use for the Fourier series" - how does one specify that when writing a series? $\endgroup$ – Hrit Roy Dec 1 '17 at 19:58
  • $\begingroup$ See Fourier series on general interval $[a,b]$ $\endgroup$ – user357151 Dec 1 '17 at 20:22
  • $\begingroup$ In the very beginning he says $f:[-π,π] →ℝ$. But isn't $f$ defined for all $x$? Like we define $f(x)=x^2$ in $(-π,π]$ and $f(x+2π)=f(x)$ for all $x$. $\endgroup$ – Hrit Roy Dec 1 '17 at 20:28
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    $\begingroup$ @HritRoy This is why the word "restricting" is there. Initially, the function defined for all $x$. We restrict it to $x\in [-\pi, \pi)$. Then we extend it periodically. $\endgroup$ – user357151 Dec 1 '17 at 22:11
  • $\begingroup$ Oh I missed the last part. Restrict, then extend. Makes sense. $\endgroup$ – Hrit Roy Dec 1 '17 at 22:12

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