Complex Fourier series

The Fourier series we have studied so far is a special case of the complex Fourier series. In many cases, the complex Fourier series is more convenient than the (real) Fourier series.

The complex Fourier series is defined as a series of the form

$$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikx}$$

where $i=\sqrt{-1}$ is the imaginary unit and $c_k\in \mathbb{C}$. Note that the variable $x$ is still real, $x\in \mathbb{R}$. 

In the following, the collections ${\mathcal{R}}_{2\pi }^1$ and ${\mathcal{R}}_{2\pi }^2$ also include complex-valued functions.

If $f\in {\mathcal{R}}_{2\pi }^1$, the complex Fourier coefficients $c_k$ of $f$ are defined by

$$\begin{array}{}\text{(eq:ck)}& c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-ikx}dx.\end{array}$$

Here, the right-hand side is the definite integral of a complex-valued function.

See also: Calculus of complex-valued functions

For $m,n\in \mathbb{Z}$, if $m\ne n$, then

$$\begin{array}{rcl}{\int }_{-\pi }^{\pi }{e}^{imx}{e}^{-inx}dx& =& {\int }_{-\pi }^{\pi }{e}^{i(m-n)x}dx\\ & =& \frac{1}{i(m-n)}\{e^{i(m-n)\pi }-e^{-i(m-n)\pi }\}=0.\end{array}$$

On the other hand,

$${\int }_{-\pi }^{\pi }{e}^{imx}{e}^{-imx}dx={\int }_{-\pi }^{\pi }1dx=2\pi .$$

Using Kronecker's delta, these results can be summarized as

$${\int }_{-\pi }^{\pi }{e}^{imx}{e}^{-inx}dx=2\pi {\delta }_{m,n}(m,n\in \mathbb{Z}).$$

This is the orthogonality of $\{e^{ik\pi }\}$. That is, the set of functions $\{e^{ik\pi }\mid k\in \mathbb{Z}\}$ comprises an orthogonal basis (of some vector space).

Suppose that $f\in {\mathcal{R}}_{2\pi }^1$ can be represented as

$$\begin{array}{}\text{(eq:fxexp)}& f(x)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikx}\end{array}$$

and that the term-wise integration of $f(x)e^{inx}$ is allowed. Then,

$${\int }_{-\pi }^{\pi }f(x)e^{-inx}dx=2\pi c_n(n\in \mathbb{Z}).$$

Therefore, the coefficients in the expansion (eq:fxexp) are the same as the complex Fourier coefficients defined by (eq:ck). Based on this fact, for any $f\in {\mathcal{R}}_{2\pi }^1$, we define the complex Fourier series of $f$ by (eq:fxexp) with the complex Fourier coefficients $\{c_k\}$ of $f$, defined by (eq:ck). The complex Fourier series of $f$ is denoted as $S_c[f]$ or simply $S[f]$, and we write

$$f(x)\sim \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikx}.$$

Example. Consider the function $f(x)=e^{ix/2}\in {\mathcal{P}}_{2\pi }$ $(-\pi \le x<\pi )$. Let us define

$$\begin{array}{rcl}u(x)& =& \mathrm{\Re }f(x)=\cos \frac{x}{2},\\ v(x)& =& \mathrm{\Im }f(x)=\sin \frac{x}{2}.\end{array}$$

Since $u(-\pi )=u(\pi -0)=\cos (\pi /2)=0$, $u$ is continuous on $\mathbb{R}$. On the other hand, $v(-\pi )=\sin (-\pi /2)=-1$, $v(\pi -0)=\sin (\pi /2)=1$ so that $v$ is discontinuous at $x=(2k-1)\times \pi ,k\in \mathbb{Z}$. Let us calculate the complex Fourier coefficients.

$$\begin{array}{rcl}2\pi c_k& =& {\int }_{-\pi }^{\pi }f(x)e^{-ikx}dx={\int }_{-\pi }^{\pi }{e}^{i(1/2-k)x}dx\\ & =& \frac{1}{i(\frac{1}{2}-k)}\{e^{i(1/2-k)\pi }-e^{-i(1/2-k)\pi }\}.\end{array}$$

Noting that

$$\begin{array}{rcl}{e}^{i\pi /2}& =& i,\\ e^{-i\pi /2}& =& -i,\\ e^{ik\pi }& =& e^{-ik\pi }=(-1{)}^k(k\in \mathbb{Z}),\end{array}$$

we have

$$\begin{array}{rcl}{c}_k& =& \frac{1}{2\pi }\cdot \frac{1}{i(\frac{1}{2}-k)}\{i(-1{)}^k+i(-1{)}^k\}\\ & =& \frac{2}{\pi }\cdot \frac{(-1{)}^k}{1-2k}(k\in \mathbb{Z}).\end{array}$$

Thus,

$$\begin{array}{rcl}{e}^{ix/2}& \sim & \frac{2}{\pi }\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(-1{)}^k}{1-2k}{e}^{ikx}\\ & =& \frac{2}{\pi }(\cdots +\frac{1}{5}{e}^{-i2x}-\frac{1}{3}{e}^{-ix}+1+e^{ix}-\frac{1}{3}{e}^{i2x}+\cdots ).\end{array}$$

□

Let $f\in {\mathcal{R}}_{2\pi }^1.$ We have the following relations between the (ordinary) Fourier coefficients $a_n,b_n$ and the complex Fourier coefficients $c_n$:

$$\begin{array}{rcl}2c_n& =& a_n-i{b}_n,\\ 2c_{-n}& =& a_n+i{b}_n\end{array}$$

for $n\in \mathbb{N}$; and $2c_0=a_0.$ Conversely,

$$\begin{array}{rcl}{a}_n& =& c_n+c_{-n},\\ b_n& =& i(c_n-c_{-n}).\end{array}$$

Consider the complex Fourier series of $f$:

$$f\sim \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikx}.$$

For an arbitrary $n\in \mathbb{N}$, let's define the partial sum $S_n$ as

$$S_n=\sum _{k=-n}^n{c}_k{e}^{ikx}.$$

Using Euler's formula $e^{ikx}=\cos (kx)+i\sin (kx)$, the partial sum is rearranged as

$$\begin{array}{rcl}{S}_n& =& \sum _{k=-n}^n{c}_k{e}^{ikx}\\ & =& c_0+\sum _{k=1}^n\{(c_k+c_{-k})\cos (kx)+i(c_k-c_{-k})\sin (kx)\}\\ & =& \frac{1}{2}{a}_0+\sum _{k=1}^n\{a_k\cos (kx)+b_k\sin (kx)\}.\end{array}$$

Thus, $S_n$ is also the partial sum of the ordinary Fourier series. Based on this observation, we have a complex version of the Lemma (Zero function):

Lemma (Zero function)

If $f\in {\mathcal{R}}_{2\pi }^1$ satisfies the condition

$$\begin{array}{}\text{(eq:zero)}& {\gamma }_k={\int }_{-\pi }^{\pi }f(x)e^{-ikx}dx=0(k\in \mathbb{Z}),\end{array}$$

then $f(x_0)=0$ where $x_0$ is a continuous point of $f$. If, in addition to the condition (eq:zero), $f$ is also a continuous function with period $2\pi$, then $f(x)=0$ (identically zero).

Proof. Exercise. ■

See also: Uniform convergence of Fourier series

The following theorems should also be ``trivial'':

Theorem (If $S[f]$ converges uniformly, then $f=S[f]$)

If the complex Fourier series $S[f]$ of $f\in {\mathcal{R}}_{2\pi }^1$ converges uniformly, then $S[f](x_0)$ converges to $f(x_0)$ where $x_0$ is any point at which $f$ is continuous. In particular, if $f$ is continuous with a period $2\pi$ and $S[f]$ converges uniformly, then $f=S[f]$.

Theorem

Let $f\in {\mathcal{R}}_{2\pi }^1$. If the complex Fourier coefficients $c_n$ of $f$ satisfy $\sum _n|c_n|<+\mathrm{\infty }$, then the complex Fourier series $S[f]$ of $f$ converges uniformly and $S[f](x)=f(x)$ at each continuous point $x$ of $f$.

That is, if the series of the Fourier coefficients of $f$ converges absolutely, then the Fourier series $S[f]$ converges uniformly to $f$ at each continuous point of $f$.