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Section 5.2 Examples A

Example 5.2.1.

Find the Fourier series expansion of \(f(x) =x\text{,}\) where \(0\lt x \lt 2\pi\text{,}\) and sketch its graph from \(x=-4\pi\) to \(4\pi\text{.}\)
Solution.
Let
\begin{equation*} f(x) = x = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nx + b_{n}\sin nx\right] \end{equation*}
where
\begin{equation*} a_{o} = \frac{1}{\pi}\int\limits_{0}^{2\pi}f(x)\,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi}x\,dx = \frac{1}{\pi}\left[\frac{x^{2}}{2}\right]^{2\pi}_{0} = 2\pi. \end{equation*}
\begin{equation*} a_{n}= \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\cos nx \,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi} x\cos nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[x.\frac{\sin nx}{n}-1.\frac{\cos nx}{-n^{2}}\right]_{0}^{2\pi} \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[0+\frac{1}{n^{2}}(\cos 2n\pi-\cos 0)\right]= \frac{1}{\pi}\left[\frac{1}{n^{2}}(1-1)\right] = 0. \end{equation*}
and
\begin{equation*} b_{n}= \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\sin nx \,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi} x\sin nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[x.\frac{\cos nx}{-n}-1.\frac{\sin nx}{-n^{2}}\right]_{0}^{2\pi} \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[\frac{2\pi \cos 2n \pi}{-n}+\frac{1}{n^{2}}\left(\sin nx\right)_{0}^{2\pi}\right]= -\frac{2}{n}. \end{equation*}
Substituting these values in expression \(f(x)\text{,}\) we get -
\begin{equation*} x=\pi - 2\left[\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots\right] \end{equation*}
The graphical representation of the given function is shown in figure below.

Example 5.2.2.

Obtain the Fourier series for \(f(x) = e^{-x}\) in the interval \(0 \lt x\lt 2\pi\text{.}\)
Solution.
Let
\begin{equation*} f(x) =e^{-x} = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nx + b_{n}\sin nx\right] \end{equation*}
where
\begin{equation*} a_{o} = \frac{1}{\pi}\int\limits_{0}^{2\pi}f(x)\,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi}e^{-x}\,dx = \frac{1}{\pi}\left[\frac{e^{-x}}{-1}\right]^{2\pi}_{0} \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[-e^{-2\pi}+1\right] = \frac{1-e^{-2\pi}}{\pi}. \end{equation*}
\begin{equation*} a_{n} = \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\cos nx \,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi} e^{-x}\cos nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[\frac{e^{-x}}{1+n^{2}}(-\cos nx +n \sin nx)\right]_{0}^{2\pi} \end{equation*}
\begin{equation*} =\frac{1}{\pi(1+n^{2})}\left[e^{-x}(-\cos 2n\pi+n\sin 2n\pi) - e^{0}(-\cos 0+n\sin 0)\right] \end{equation*}
\begin{equation*} =\frac{1}{\pi(1+n^{2})}\left[1-e^{-2\pi}\right]. \end{equation*}
\begin{equation*} b_{n} = \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\sin nx \,dx = \frac{1}{\pi}\int\limits_{0}^{2\pi} e^{-x}\sin nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[\frac{e^{-x}}{1+n^{2}}(-\sin nx +n \cos nx)\right]_{0}^{2\pi} \end{equation*}
\begin{equation*} =\frac{1}{\pi (1+n^{2})}\left[e^{-2\pi}(-\sin 2n\pi - n \cos 2n \pi) - e^{0}(-\sin 0 - n\cos 0)\right] \end{equation*}
\begin{equation*} =\frac{n}{\pi (1+n^{2})}\left[e^{-2\pi}(-n) - e^{0}(-n)\right]. \end{equation*}
From expression \(f(x)\text{,}\) we get -
\begin{equation*} e^{-x} = \frac{1-e^{-2\pi}}{\pi} + \sum\limits_{n=1}^{\infty}\left[\frac{1-e^{-2\pi}}{\pi (1+n^{2})} \cos nx + \frac{1-e^{-2\pi}}{\pi (1+n^{2})}n \sin nx\right] \end{equation*}
\begin{equation*} =\frac{1-e^{-2\pi}}{\pi}\left[\frac{1}{2}+\left\{\frac{\cos x}{2}+\frac{\cos 2x}{5}+\cdots + \right\} \right. \end{equation*}
\begin{equation*} \left. + \left\{\frac{\sin x}{2}+\frac{2\sin 2x}{5}+\cdots + \right\}\right]. \end{equation*}
which is the required Fourier series.

Example 5.2.3.

Find the Fourier series of the function
\begin{equation*} f(x) = \begin{cases} -1, & \text{for } -\pi \lt x \lt -\frac{\pi}{2}\\ 0, & \text{for} -\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\\ +1, & \text{for} +\frac{\pi}{2} \lt x \lt \pi. \end{cases} \end{equation*}
Solution.
Let
\begin{equation*} f(x) = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nx + b_{n}\sin nx\right] \end{equation*}
where
\begin{equation*} a_{o} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\,dx = \frac{1}{\pi}\left[\int\limits_{-\pi}^{-\pi/2}(-1)\,dx + \int\limits_{-\pi/2}^{\pi/2} 0. \,dx + \int\limits_{\pi/2}^{\pi}(+1)\,dx\right] \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[-\left(x\right)_{-\pi}^{-\pi/2}+\left(x\right)_{\pi/2}^{\pi}\right] \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[-\left\{-\frac{\pi}{2}+\pi \right\} + \left\{\pi -\frac{\pi}{2} \right\}\right] = \frac{1}{\pi}\left[-\left\{-\frac{\pi}{2} +\frac{\pi}{2} \right\}\right] = 0. \end{equation*}
\begin{equation*} a_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi} f(x)\cos nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[\int\limits_{-\pi}^{-\pi/2} (-1)\cos nx \,dx + \int\limits_{-\pi/2}^{\pi/2} 0.\cos nx \,dx + \int\limits_{\pi/2}^{\pi} (+1)\cos nx \,dx\right] \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[-\left(\frac{\sin nx}{n}\right)_{-\pi}^{-\pi/2}+\left(\frac{\sin nx}{n}\right)_{\pi/2}^{\pi}\right] \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[-\left(-\frac{\sin n\pi/2}{n}+ \frac{\sin n\pi}{n}\right)+\left(\frac{\sin n\pi}{n} - \frac{\sin n\pi/2}{n}\right)\right] = 0. \end{equation*}
\begin{equation*} b_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi} f(x)\sin nx \,dx \end{equation*}
\begin{equation*} = \frac{1}{\pi}\left[\int\limits_{-\pi}^{-\pi/2} (-1)\sin nx \,dx + \int\limits_{-\pi/2}^{\pi/2} 0.\sin nx \,dx + \int\limits_{\pi/2}^{\pi} (+1)\sin nx \,dx\right] \end{equation*}
\begin{equation*} =\frac{1}{\pi}\left[+\left(\frac{\cos nx}{n}\right)_{-\pi}^{-\pi/2}+\left(\frac{\cos nx}{-n}\right)_{\pi/2}^{\pi}\right] \end{equation*}
now, from expression \(f(x)\text{,}\) we get -
\begin{equation*} f(x)=\sum\limits_{n=1}^{\infty} \left[+\frac{2}{n\pi}\left(\cos \frac{n\pi}{2}-\cos n\pi\right)\sin nx\right] \end{equation*}
\begin{equation*} = +\frac{2}{\pi}\left[\frac{\sin x}{1}-\frac{2\sin 2x}{2}+\frac{\sin 3x}{3}-\cdots\right] \end{equation*}
\begin{equation*} = +\frac{2}{\pi}\left[\sin x-\frac{2\sin 2x}{2}+\frac{\sin 3x}{3}-\cdots\right]. \end{equation*}
which is the required Fourier series expansion.