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Fourier Series

More details about Fourier series can be found in Kreyszig, Chapter 10. Here we simply list some of the important results.

Firstly, if a function $f(x)$, defined over an interval $-L < x < L$, is extended as a periodic function so that

\begin{displaymath}
f(x+2L) = f(x),
\end{displaymath}

then $f(x)$ may be expressed in terms of a Fourier series as
\begin{displaymath}
f(x) = a_{0} + \sum_{n=1}^{\infty}\left ( a_{n}\cos {n\pi \over L}x +
b_{n}\sin {n\pi \over L}x \right ),
\end{displaymath} (2.11)

with the Fourier coefficients of $f(x)$ given by the Euler fromulae
$\displaystyle a_{0}$ $\textstyle =$ $\displaystyle {1\over 2L}\int_{-L}^{L}f(x) dx,$ (2.12)
$\displaystyle a_{n}$ $\textstyle =$ $\displaystyle {1\over L}\int_{-L}^{L}f(x) \cos {n\pi\over L}x dx,$ (2.13)
$\displaystyle b_{n}$ $\textstyle =$ $\displaystyle {1\over L}\int_{-L}^{L}f(x) \sin {n\pi \over L}x dx,$ (2.14)

for positive integer values of $n$. The derivation of theses coefficients requires some knowledge about the orthogonality properties of the trigonometric functions. The main properties that we need are given by the following integrals,

\begin{eqnarray*}
\int_{-L}^{L}\sin {m\pi \over L}x dx, & = & 0 \\
\int_{-L}^...
...
\int_{-L}^{L}\cos^{2} {m\pi \over L}x dx \hbox{ }= \hbox{ }L.
\end{eqnarray*}



The proof of these integrals all follows the same pattern and requires the use of some trigonometric identities. For example, we may express

\begin{displaymath}
\int_{-L}^{L}\cos {m\pi \over L}x \cos {n\pi \over L}x dx =...
...x dx + {1\over 2}\int_{-L}^{L}\cos
{(m - n)\pi \over L}x dx.
\end{displaymath}

The integrand of the first term on the right hand side is always periodic over $-L < x < L$ and so the integral will be zero. The second term on the right hand side is also periodic and will be zero provided that $m \ne n$. The case $m = n$ gives the integral as

\begin{displaymath}
{1\over 2}\int_{-L}^{L}\cos {(m - n)\pi \over L}x dx = {1\over
2}\int_{-L}^{L} dx = L.
\end{displaymath}

The other integrals are done in a similar manner, using

\begin{eqnarray*}
2 \sin mx \sin nx & = & \cos (m-n)x - \cos (m+n)x, \\
2 \cos mx \sin nx & = & \sin (m+n)x - \sin (m-n)x.
\end{eqnarray*}




next up previous
Next: Half-Range Expansions Up: Separation of Variables Previous: Separation of Variables
Prof. Alan Hood
2000-10-30