Fourier Series *

Fourier series allow us to expand any periodic function on the range \bgroup\color{black}$(-L,L)$\egroup, in terms of sines and cosines also periodic on that interval.

\begin{displaymath}\bgroup\color{black}f(x)=\sum\limits_{n=0}^\infty A_n\cos\lef...
...\limits_{n=1}^\infty B_n\sin\left({n\pi x\over L}\right)\egroup\end{displaymath}

Since the sines and cosines can be made from the complex exponentials, we can equally well use them for our basis for expansion. This has the nice simplification of having only one term in the sum, using negative \bgroup\color{black}$n$\egroup to get the other term.

\begin{displaymath}\bgroup\color{black}f(x)=\sum\limits_{n=-\infty}^\infty a_ne^{in\pi x\over L}\egroup\end{displaymath}

The exponentials are orthogonal and normalized over the interval (as were the sines and cosines)

\begin{displaymath}\bgroup\color{black}{1\over 2L}\int\limits_{-L}^L\; e^{in\pi x\over L}\; e^{-im\pi x\over L} dx = \delta_{nm}\egroup\end{displaymath}

so that we can easily compute the coefficients.

\begin{displaymath}\bgroup\color{black}a_n={1\over 2L} \int\limits_{-L}^L f(x)e^{-in\pi x\over L}dx \egroup\end{displaymath}

In summary, the Fourier series equations we will use are

\begin{displaymath}\bgroup\color{black}f(x)=\sum\limits_{n=-\infty}^\infty a_ne^{in\pi x\over L}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}a_n={1\over 2L} \int\limits_{-L}^L f(x)e^{-in\pi x\over L}dx .\egroup\end{displaymath}

We will expand the interval to infinity.

Jim Branson 2013-04-22