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I was asked to compute the fourier series of $f(x)=x$ in two different intervals: $x \in [0,2\pi]$ and $x\in [-\pi,\pi]$. What is the real difference between the two series, cause we know that for periodic functions $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$, so I don't really get it.

Thank you so much

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    Thank you for the comments. so what should I do different from the way of computing what is needed in the interval $[-\pi,\pi]$? I mean where does it show?2011-12-27

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Let

$f(x):=\cases{x & (0 < x < 2\pi) \cr f(x+2\pi) & ({\rm all\ } x)\cr}\ ,\qquad g(x):=\cases{x & (-\pi < x < \pi) \cr g(x+2\pi) & ({\rm all\ } x)\cr}\ .$

Then $f(x)=\pi + g(x-\pi)\qquad(x\in{\mathbb R})\ .$

The function $g$ is odd; whence

$g(x)=\sum_{k=1}^\infty b_k\ \sin(k x)\ ,\qquad b_k={2\over\pi}\int_0^\pi x\ \sin(k x)\ dx={2\over k}(-1)^{k-1}\ .$ Therefore $g(x)=2\sum_{k=1}^\infty {(-1)^{k-1}\over k}\ \sin(k x)\ .$ From this we get the following expansion for $f$: $f(x)=\pi+g(x-\pi)=\pi + \sum_{k=1}^\infty b_k\ (-1)^k\sin(k x)=\pi - 2\sum_{k=1}^\infty{1\over k}\ \sin(k x)\ .$

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    @Jozef: By looking at the graphs over some interval containing several periods.2011-12-28
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Well the f functions are different in both cases (f is supposed periodic of period $2 \pi$):

$f(x)=x$ in $\left[0, 2 \pi \right[$ means that you'll have $f(x)= x + 2\pi$ in $\left[-2 \pi, 0 \right[$ and so on...

$f(x)=x$ in $\left[- \pi, \pi \right[$ is another function!

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The function g is odd; whence

g(x)=∑k=1∞bk sin(kx) ,bk=2π∫π0x sin(kx) dx=2k(−1)^k-1 . How do you get this

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    I'm not sure what you're trying to say with bk=2π∫π0x sin(kx) dx, so please view this post for a how-to on typesetting math: http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto2012-10-16