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The background of this question is Fourier series. I was suppose to find the Fourier series of $f(x)=x$, in the interval $-2<x<2$ with $f(x+4)=f(x)$. enter image description here It is mandatory to have a periodic function to find it's Fourier series.

How come $f(x)=x$ is periodic?

A function $f$ is said to be periodic if there exists a positive number $P$, such that $f (x + P ) = f (x)$

zhk
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    You look at the periodic extension (with period $4$) of the function $g\colon [-2,2) \to \mathbb{C}$ with $g(x) = x$. (Possibly, you change the value for $g(-2)$, setting it maybe to $0$. That doesn't have any effect on the Fourier series.) – Daniel Fischer Nov 02 '16 at 16:20
  • @DanielFischer Could you please make things a little more simpler in the light of the definition of a periodic function? – zhk Nov 02 '16 at 16:27
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    You define $f(x) = t$, where $x = 4k + t$ for some $k\in \mathbb{Z}$, and $t \in [-2,2)$. In a single formula: $$f(x) = x - 4\biggl\lfloor \frac{x+2}{4}\biggr\rfloor.$$ – Daniel Fischer Nov 02 '16 at 16:34
  • @DanielFischer It means we are dealing with a periodic extension of $f(x)=x$ with period $4$. To be honest, I am still confused. – zhk Nov 02 '16 at 17:20
  • @MMM it is not $f(x) = x$ for every $x\in \mathbb R$ ... – user251257 Nov 02 '16 at 17:30

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The source of confusion here seems to be that you are conflating two different functions. One is the function $f$ in the title of your question and in the next-to-last sentence of your question; it is defined by $f(x)=x$ for all $x$, and it is not periodic. The other is the function, which you also called $f$ but which I'll call $g$ to avoid confusion, whose graph you included in the question. It is periodic, with period 4, because $g(x+4)=g(x)$ for all $x$. The two functions are related by the fact that $f(x)=g(x)$ when $x$ is between $-2$ and $2$; for other values of $x$, $f(x)$ and $g(x)$ are quite different.

As $g$ is periodic, it makes sense to talk about its Fourier series. In fact, that series will converge to $g$ at all points except where $g$ is discontinuous (or undefined, it's hard to tell from the graph). Since $f$ and $g$ agree on the interval from $-2$ to $2$, the Fourier series of $g$ will converge to $f$ on that interval (but not outside that interval).

Andreas Blass
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  • So, we can take a periodic function out of the blue on some random finite interval without needing to show that mathematically $f(x+P)=f(x)$ holds? In the same breath I can take $f(x)=x$ to be a periodic function in the interval $0<x<4$ with period $2$? – zhk Nov 03 '16 at 01:36
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    @MMM In a formal proof (as opposed to a stackexchange answer), you'd have to prove that your second $f$, the one I called $g$, is periodic with period 4, rather than just taking it as obvious in light of the picture. $g$ is neither random nor out of the blue; it is the unique periodic function with period 4 that agrees with $f$ on the specified interval from $-2$ to $2$. And you cannot take $f(x)=4$ to be periodic on $0<x<4$ with period $2$ because $f(1)\neq f(1+2)$ and both $1$ and $1+2$ are in the range from $0$ to $4$. – Andreas Blass Nov 03 '16 at 02:29
  • How you know that $f(1)\neq f(1+2)$? My second question is, how we know that $4$ is a period of $f(x)=x$ in $−2<x<2$ or in $0<x<4$ without using a graph? – zhk Nov 03 '16 at 10:07
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    @MMM I know that $f(1)\neq f(1+2)$ because you defined $f(x)$ to be $x$, so $f(1)=1$ and $f(1+2)=1+2$; I also know $1\neq 1+2$ because I can count. As to your other question, "how do we know that $4$ is a period of $f(x)$ in $-2<x<2$ or in $0<x<4$?" that makes no sense. "Period 4" involves (as you may know since you quoted the definition) comparing the values of $f$ at two inputs $x$ and $x+4$ that differ by $4$; neither of the intervals you mention contains two such inputs. – Andreas Blass Nov 03 '16 at 16:25
  • I am sorry if I have offended you? Finally, my thick brain has start to process. Thanks – zhk Nov 04 '16 at 03:11