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In solving the following problem with the method of separation of variables $ u_{tt}=u_{xx} \quad 00 \\ u(x,0)=g \quad 0 \leq x \leq \pi \\ u_x(t,0)=u_x(t,\pi)=0 \quad t>0 $

i came across the problem of expanding $g$ in series of cosines. $g$ is of cass $C^1$, so if I extend it to an even function $G$ on $[-\pi,\pi]$ I should be able to expand it in series of cosine since $G$ is piecewise $C^1$. But in class we said for an analogous problem (with the data $u(t,0)=u(t,\pi)=0$) that since by the separation of variables we obtain the eigenvectors of the laplacian with 0-boundary data (in that case they are sines, which are a basis of $L^2[0,\pi]$) we can write g as a series of sines (in the sense of L^2, so a.e.). Can I say something analogous here, without extending $g$ to an even function. In this case, how can I say that cosines are a basis of $L^2[0,\pi]$?

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    For example extending $sin(x)$ to an even function on $[-\pi,\pi]$ if the calculations are correct i found $sin(x)=\sum\frac{1+(-1)^k}{1-k^2}cos(kx)$. Is this possible?2012-05-16

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