Subsection 5.1.1 Determination of Fourier Coefficients
The Fourier series representation is given by
\begin{equation}
f(x) = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nx + b_{n}\sin nx\right]\tag{5.1.1}
\end{equation}
- \begin{equation*} \int\limits_{0}^{2\pi}f(x)\,dx = \frac{a_{o}}{2}\int\limits_{0}^{2\pi}\,dx + a_{1}\int\limits_{0}^{2\pi} \cos x\,dx + \cdots + a_{n}\int\limits_{0}^{2\pi} \cos nx\,dx \end{equation*}\begin{equation*} + \cdots+b_{1}\int\limits_{0}^{2\pi} \sin x\,dx + \cdots + b_{n}\int\limits_{0}^{2\pi} \sin nx\,dx + \cdots \end{equation*}\begin{equation*} =\frac{a_{o}}{2}\left[x\right]^{2\pi}_{0}+0+0 +\cdots.+ 0+0+ \cdots =\frac{a_{o}}{2}.2\pi = a_{o}\pi \end{equation*}\begin{equation} \therefore a_{o} = \frac{1}{\pi}\int\limits_{0}^{2\pi}f(x)\,dx.\tag{5.1.2} \end{equation}
- To find \(a_{n}\text{:}\) Multiply equation, (5.1.1) by \(\cos nx\) and integrate from \(x=0\) to \(2\pi\text{,}\) we get -\begin{equation*} \int\limits_{0}^{2\pi} f(x)\cos nx \,dx = \frac{a_{o}}{2}\int\limits_{0}^{2\pi} \cos nx \,dx + \cdots \end{equation*}\begin{equation*} + a_{n}\int\limits_{0}^{2\pi} \cos^{2} nx \,dx + \cdots + b_{n}\int\limits_{0}^{2\pi} \sin nx .\cos nx \,dx + \cdots \end{equation*}\begin{equation*} =0+\cdots + a_{n}\int\limits_{0}^{2\pi} \cos^{2} nx \,dx + 0+ \cdots = a_{n}\pi \end{equation*}\begin{equation} \therefore \quad a_{n}= \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\cos nx \,dx.\tag{5.1.3} \end{equation}
- To find \(b_{n}\text{:}\) Multiply equation (5.1.1) by \(\sin nx\) and integrate it from \(x=0\) to \(2\pi\text{,}\) we get -\begin{equation*} \int\limits_{0}^{2\pi} f(x)\sin nx \,dx = \frac{a_{o}}{2}\int\limits_{0}^{2\pi} \sin nx \,dx + \cdots + a_{n}\int\limits_{0}^{2\pi} \cos nx.\sin nx \,dx + \cdots \end{equation*}\begin{equation*} + b_{n}\int\limits_{0}^{2\pi} \sin^{2} nx \,dx + \cdots =b_{n}\int\limits_{0}^{2\pi} \sin^{2} nx \,dx = b_{n}\pi \end{equation*}\begin{equation} \therefore \quad b_{n}= \frac{1}{\pi}\int\limits_{0}^{2\pi} f(x)\sin nx \,dx\tag{5.1.4} \end{equation}
