Fourier Series for Nonperiodic Functions and Half-Range Expansions

Figure 6.8: A function, $ f(x)$, of interest on [0,3].
\includegraphics[scale=1]{figures/4.10.ps}
Figure 6.9: Left: Plot of a function reflected about the y-axis, an even function,Right: Plot of a function reflected about the origin, an odd function.
\includegraphics[scale=1]{figures/4.11.ps} \includegraphics[scale=1]{figures/4.12.ps}
Examples:
  1. Find the Fourier cosine series expansion of $ f(x)$, given that

    \begin{displaymath}
f(x)=\left\{
\begin{array}{ll}
0, & 0<x<1 \\
1, & 1<x<2 \\
\end{array}\right\rbrace
\end{displaymath}

    Figure 6.10left shows the even extension of the function. Because we are dealing with an even function on $ (-2,2)$ we know that the Fourier series will have only cosine terms. We get the $ A$s with

    $\displaystyle A_0=\frac{2}{2}\int_1^2(1)dx=1
$

    \begin{displaymath}
A_n=\frac{2}{2}\int_1^2(1)cos\left( \frac{n\pi x}{2}\right) ...
...c{2(-1)^{(n+1)/2}}{n\pi}, & n odd\\
\end{array}\right\rbrace
\end{displaymath}

    Then the Fourier cosine series is

    $\displaystyle f(x)\approx\frac{1}{2}+\frac{2}{\pi}\sum_{n=1}^{\infty}(\frac{(-1)^n cos\left( (2n-1)\pi x/2\right) }{(2n-1)}
$

  2. Find the Fourier sine series expansion for the same function. Figure 6.10right shows the odd extension of the function. We know that all of the $ A$-coefficients will be zero, so we need to compute only the $ B$s;

    $\displaystyle B_n=\frac{2}{2}\int_1^2(1)sin(\frac{n\pi x}{2})dx=\frac{2}{n\pi}\left[-cos(n\pi)+cos(\frac{n\pi}{2})\right], n=1,2,3,\ldots
$

    $\displaystyle f(x)=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\left[cos(n\pi /2)-cos(n\pi)\right]}{n}sin(\frac{n\pi x}{2})
$

Cem Ozdogan 2011-12-27