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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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    Why would just the sines be a basis for $L^2[-\pi,\pi]$? How would you get the even functions (many of which are square-integrable) with just sines?2012-05-15
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    No, I wanted to say they are a basis of $L^2[0,\pi]$, since we are interested in that interval in the equation2012-05-15
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    The sines and cosines are a basis for $L^2[0,2 \pi]$. The sines alone will not do (the representation of cosine would be $0$, if confined to sines).2012-05-15
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    Ok. What about the expansion of $g$ in series of cosines? Is it correct my first 'proof', extending it to an even function?2012-05-15
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    Well, a function on $[0,\pi]$ can be extended in the obvious way to be an even function on $[-\pi, \pi]$. This has a representation in cosines, which can then be restricted back to $[0,\pi]$. The same is true with odd functions (extended in the analogous way), but unless it is $0$ at $0$, there will be some discontinuity at $0$ with the subsequent representation in sines.2012-05-16
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    So in this way it seems that the cosines are a basis of $L^2[0,\pi]$. (And so do sines), but the representation of cosine in terms of sines should be 0 according to what you said...?2012-05-16
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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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