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Take Euler's famous example:

$$\dfrac{1}{2} x= \sin x-\dfrac{1}{2} \sin 2x+ \dfrac{1}{3} \sin 3x- \dfrac{1}{4} \sin 4x+\cdots $$

What is the reason this only works on $[-π,π]$?

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    $\begingroup$ Because the functions involved are periodic and their period is $2\pi$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 14 '14 at 17:31
  • $\begingroup$ Sine function is periodic while taking $x=(x+2\pi)$ would yield the same number $\endgroup$ – kingW3 Apr 14 '14 at 17:36
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    $\begingroup$ They don't work just for $2\pi$-periods. They work for any period. $\endgroup$ – Git Gud Apr 14 '14 at 17:37
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Because you are not approximating the function $f(x)=x/2$, you are approximating the function $$f(x)=\begin{cases}x/2&x\in[-\pi,\pi]\\f\left(x-\lfloor\frac{x+\pi}{2\pi}\rfloor(2\pi)\right)&x\notin[-\pi,\pi]\end{cases}$$ which is periodic. So technically it does work for all $x\in\mathbb{R}$, just not for the function you were thinking of.

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