Coefficients of a Fourier cosine series

Question

I know that, for instance, if I have the following cosine series: $$ f(x)=\sum_{n=1}^{\infty}A_n\cos\left(\frac{n\pi x}{L}\right) $$ The coefficients $A_n$ can be calculated by: $$ A_n=\frac{2}{L}\int_{0}^{L}f(x)\cos\left(\frac{n\pi x}{L}\right)dx $$ But suppose now that I change the argument of my cosine series to: $$ f(x)=\sum_{n=1}^{\infty}A_n\cos\left(\frac{\pi(2n+1)x}{2L}\right) $$ Does it change the way that I applied the orthogonality to evaluate the coefficients? I believe so, since my results do not correspond to what I was looking for. Any help would be truly apreciated. Thank you in advance!

Answer 1

As $x$ ranges over $[0,L]$, the argument of $\cos$ in the following ranges over $[0,(n+1/2)\pi]$: $$ c_n(x)=\cos\left(\frac{\pi(2n+1)x}{2L}\right) $$ That means $c_n'(0)=0$ and $c_n(L)=0$. So these functions are solutions of the following Sturm-Liouville problem on $[0,L]$: $$ y''=\lambda y,\;\; y'(0)=0,\; y(L)=0. $$ That makes the solutions mutually orthogonal in $L^2[0,L]$. In fact, these are all such solutions, which makes them a complete orthogonal basis of $L^2[0,L]$. Therefore, the coefficients $A_n$ in $$ f(x)=\sum_{n=0}^{\infty}A_n\cos\left(\frac{\pi(2n+1)x}{2L}\right) $$ are determined for $n=0,1,2,3\cdots$ by the usual Fourier trick of multiplying both sides by an eigenfunction, integrating, and using integral "orthogonality" of the different eigenfunctions: $$ \int_0^Lf(x)\cos\left(\frac{\pi(2n+1)x}{2L}\right)dx = A_n\int_0^L\cos^2\left(\frac{\pi(2n+1)x}{2L}\right)dx \\ \implies A_n = \frac{\int_0^Lf(x)\cos\left(\frac{\pi(2n+1)x}{2L}\right)dx}{\int_0^L\cos^2\left(\frac{\pi(2n+1)x}{2L}\right)dx}. $$