Fourier Series for Nonperiodic Functions and Half-Range Expansions

Figure 6.8: A function, $f(x)$, of interest on [0,3].

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.

• The development until now has been for a periodic function. What if $f(x)$ is not periodic? Can we approximate it by a trigonometric series? We assume that we are interested in approximating the function only over a limited interval and we do not care whether the approximation holds outside of that interval.
• Suppose we have a function defined for all $x$-values, but we are only interested in representing it over (0, L). Figure 6.8 is typical.
• Because we will ignore the behavior of the function outside of (0, L), we can redefine the behavior outside that interval as we wish Figs. 6.9left and -right show two possible redefinitions.
  • In the first redefinition, we have reflected the portion of $f(x)$ about the $y$-axis and have extended it as a periodic function of period $2L$. This creates an even periodic function.
    $$f(x) is even if f(-x)=f(x)$$

  • If we reflect it about the origin and extend it periodically, we create an odd periodic function of period $2L$.
    $$f(x) is odd if f(-x)=-f(x)$$

  It is easy to see that $cos(Cx)$ is an even function and that $sin(Cx)$ is an odd function for any real value of $C$.
• There are two important relationships for integrals of even and odd functions.
  $$\begin{array}{ll} if f(x) is even,& \int_{-L}^{L}f(x)dx=2\in... ...x)dx \\ if f(x) is odd, &\int_{-L}^{L}f(x)dx=0 \\ \end{array}$$
  • the product of two even functions is even;
    if $f(x)$ is even, $f(x)cos(nx)$ is even
  • the product of two odd functions is even;
    if $f(x)$ is odd, $f(x)sin(nx)$ is even
  • the product of an even and an odd function is odd;
    if $f(x)$ is even, $f(x)sin(nx)$ is odd
    if $f(x)$ is odd, $f(x)cos(nx)$ is odd
  • The Fourier series expansion of an even function will contain only cosine terms (all the $B$-coefficients are zero).
  • The Fourier series expansion of an odd function will contain only sine terms (all the $A$-coefficients are zero).
• If we want to represent $f(x)$ between 0 and $L$ as a Fourier series and are interested only in approximating it on the interval $(0,L)$, we can redefine $f$ within the interval $(-L, L)$ in two importantly different ways;
  • We can redefine the portion from $-L$ to 0 by reflecting about the $y$-axis. We then generate an even function.
  • We can reflect the portion between 0 and $L$ about the origin to generate an odd function.

  Figure 6.10: Left: Plot of the function reflected about the y-axis, Right: Plot of the function reflected about the origin.
  Figure 6.10 shows these two possibilities.
• Thus two different Fourier series expansions of $f(x)$ on $(0,L)$ are possible, one that has only cosine terms or one that has only sine terms. We get the $A$s for the even extension of $f(x)$ on $(0,L)$ from
  $$A_n=\frac{2}{L}\int_0^Lf(x)cos\left( \frac{n\pi x}{L}\right) dx, n=0,1,2,\ldots$$
  We get the $B$s for the odd extension of $f(x)$ on $(0,L)$ from
  $$B_n=\frac{2}{L}\int_0^Lf(x)sin\left( \frac{n\pi x}{L}\right) dx, n=1,2,3,\ldots$$

Examples:

1. Find the Fourier cosine series expansion of $f(x)$, given that
   $$f(x)=\left\{ \begin{array}{ll} 0, & 0<x<1 \\ 1, & 1<x<2 \\ \end{array}\right\rbrace$$
   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
   $$A_0=\frac{2}{2}\int_1^2(1)dx=1$$
   $$A_n=\frac{2}{2}\int_1^2(1)cos\left( \frac{n\pi x}{2}\right) ... ...c{2(-1)^{(n+1)/2}}{n\pi}, & n odd\\ \end{array}\right\rbrace$$
   Then the Fourier cosine series is
   $$f(x)\approx\frac{1}{2}+\frac{2}{\pi}\sum_{n=1}^{\infty}(\frac{(-1)^n cos\left( (2n-1)\pi x/2\right) }{(2n-1)}$$

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