Complex Form of Fourier Series

I will talk about Fourier transforms in the next entry. To do so, I will first introduce a way to convert our Fourier series currently defined in terms of sines and cosines into Fourier series defined in terms of complex exponentials.

First of all, recall that $$\begin{array}{}\text{(1)}& f(x)\sim a_0+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos \frac{n\pi x}{L}+b_n\sin \frac{n\pi x}{L}),\end{array}$$ where, by orthogonality principles, $$\begin{array}{rl}{a}_0& =\frac{1}{2L}{\int }_{-L}^{L}f(x)dx,\\ a_n& =\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{n\pi x}{L}dx,\\ b_n& =\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{n\pi x}{L}dx.\end{array}$$ Now, recall Euler's formulas: $$\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2},\text{and}\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}.$$ It follows that we can rewrite equation $(1)$ as follows: $$f(x)\sim a_0+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}(a_n-i{b}_n)e^{\frac{n\pi ix}{L}}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}(a_n+i{b}_n)e^{-\frac{n\pi ix}{L}}.$$ Now, let's change the index of summation in the first term from $n$ to $-n$: $$f(x)\sim a_0+\frac{1}{2}\sum _{n=-1}^{-\mathrm{\infty }}(a_{-n}-i{b}_{-n})e^{-\frac{n\pi ix}{L}}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}(a_n+i{b}_n)e^{-\frac{n\pi ix}{L}}.$$ It follows from the definition of $a_n$ and $b_n$ that $a_{-n}=a_n$ and $b_{-n}=-b_n$. Therefore, if we let $$\begin{array}{rl}{c}_0& =a_0,\\ c_n& =\frac{a_n+i{b}_n}{2},\end{array}$$ we will have the following Fourier series: $$f(x)\sim \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{-\frac{in\pi x}{L}},$$ where the coefficients are $$c_n=\frac{1}{2L}{\int }_{-L}^{L}f(x)e^{\frac{in\pi x}{L}}dx.$$