Section 5.5 Change of Interval
The period of a function in Fourier series sometimes appears in time like \(T\) or in length like \(2c\text{,}\) then in order to use Fourier series expansion such period must be converted into in angle like \(2\pi\text{.}\) In that case the independent variable \(x\) of a function \(f(x)\) must also be changed proportionally. Suppose we want to represent the function \(f(x)\) defined in the closed interval \((-c, c)\) by a Fourier series, \(c\) being any positive real number. We consider this interval as a result of elongating (or compressing) the interval \([-\pi, \pi]\text{.}\) The interval \(-c \leq x \leq c\) is transformed into the interval \(-\pi \leq x \leq \pi\) by the transformation \(z=\frac{\pi x}{c}\text{.}\)
Suppose the function \(F(z)\) is expanded in the interval of \(2\pi\) as
\begin{equation}
F(z) = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nz + b_{n}\sin nz\right] \tag{5.5.1}
\end{equation}
If \(f(x)\) has an interval of \(2c\text{,}\) then we can write the \(2c \) is a period of variable \(x\text{.}\) or, \(1\) is the period of variable \(x/2c,\)
\begin{equation*}
\therefore 2\pi \quad \text{is the period of variable}
\end{equation*}
\begin{equation*}
\frac{x}{2c}.2\pi = \frac{\pi x}{c}
\end{equation*}
i.e., \(z=\frac{\pi x}{c}\text{,}\) (because both the variables have same period of \(2\pi\text{.}\))
Now,
\begin{equation*}
f(x) = f\left(\frac{cz}{\pi}\right) = F(z)\quad \text{(say)}
\end{equation*}
Hence,
\begin{equation*}
a_{0}= \frac{1}{\pi}\int\limits_{-\pi}^{\pi}F(z)\,dz
\end{equation*}
\begin{equation*}
=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} f\left(\frac{cz}{\pi}\right)\,dz = \frac{1}{\pi}\int\limits_{-c}^{c} f(x)\,d\left(\frac{\pi x}{c}\right) = \frac{1}{c}\int\limits_{-c}^{c}f(x)\,dx
\end{equation*}
\begin{equation*}
a_{n}= \frac{1}{\pi}\int\limits_{-\pi}^{\pi}F(z) \cos nz\,dz = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f\left(\frac{cz}{\pi}\right)\cos nz\,dz
\end{equation*}
\begin{equation*}
= \frac{1}{\pi}\int\limits_{-c}^{c}f(x)\cos \left(\frac{n \pi x}{c}\right)\,d \left(\frac{\pi x}{c}\right) = \frac{1}{c}\int\limits_{-c}^{c}f(x)\cos \frac{n \pi x}{c}\,dx
\end{equation*}
and
\begin{equation*}
b_{n}= \frac{1}{\pi}\int\limits_{-\pi}^{\pi}F(z) \sin nz\,dz = \frac{1}{c}\int\limits_{-c}^{c}f(x)\sin \frac{n \pi x}{c}\,dx
\end{equation*}
The function \(f(x)\) is now expressed as
\begin{equation}
f(x) = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos \left(\frac{n\pi x}{c}\right) + b_{n}\sin \left(\frac{n\pi x}{c}\right)\right] \tag{5.5.2}
\end{equation}
for the period \(2c\text{.}\)
