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I'm currently working on this question:

Suppose a function is defined only on $[0,L]$. Show that we can write

$f(x) = \displaystyle\sum_{n=0}^{\infty} c_n \sin \left(\frac{(2n+1)\pi x}{2L} \right)$

For some constant $c_n$, which should be expressed as integrals of $f$.

Now, there were two earlier parts to this question, which were just the simple odd/even extensions of $f(x)$, however, I'm unsure with this one. I can see that the terms of the sum are just the odd terms of the odd extension, but I'm unsure exactly what effect this has on the coefficient $c_n$ compared to the normal coefficient of an odd extension, which would be

$c_n = \dfrac{1}{L} \displaystyle\int_{-L}^{L} f(x) \sin \left(\frac{n\pi x}{L} \right) \ \ dx = \dfrac{2}{L} \displaystyle\int_{0}^{L} f(x) \sin \left(\frac{n\pi x}{L} \right) \ \ dx $

Thanks! Sorry if I haven't put the best tag on this, I haven't posted an 'applied' question before so am somewhat unsure.

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up vote 1 down vote accepted

If you define $f(x)$ as odd and compute both integrals, they will be equal.
By definition, odd function satisfies $f(-x)=-f(x)$. This applies to $\sin()$ and $f()$ in this case. So, your product $f(x)\sin\left(\frac{n\pi x}{L}\right)$ is even. Then you can just do integration in the half of original domain, i.e. $[0,L]$ instead of $[-L,L]$ provided you multiply result by 2.

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