Complex Fourier series

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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

\[\sum_{k=-\infty}^{\infty}c_ke^{ikx}\]

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

\[c_k = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}\,dx.\tag{eq:ck}\]

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 \(m \neq n\), then

\[\begin{eqnarray*} \int_{-\pi}^{\pi}e^{imx}e^{-inx}dx &=& \int_{-\pi}^{\pi}e^{i(m-n)x}dx\\ &=& \frac{1}{i(m-n)}\{e^{i(m-n)\pi} - e^{-i(m-n)\pi}\} = 0. \end{eqnarray*}\]

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

\[f(x) = \sum_{k=-\infty}^{\infty}c_ke^{ikx} \tag{eq:fxexp}\]

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

\[f(x) \sim \sum_{k=-\infty}^{\infty}c_k e^{ikx}.\]

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

\[\begin{eqnarray*} u(x) &=& \Re f(x) = \cos\frac{x}{2},\\ v(x) &=& \Im f(x) = \sin\frac{x}{2}. \end{eqnarray*}\]

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.

\[\begin{eqnarray*} 2\pi c_k &=& \int_{-\pi}^{\pi}f(x)e^{-ikx}dx = \int_{-\pi}^{\pi}e^{i(1/2-k)x}dx\\ &=& \frac{1}{i\left(\frac{1}{2} - k\right)}\left\{e^{i(1/2-k)\pi} - e^{-i(1/2-k)\pi}\right\}. \end{eqnarray*}\]

Noting that

\[\begin{eqnarray*} e^{i\pi/2} &=& i,\\ e^{-i\pi/2} &=& -i,\\ e^{ik\pi} &=& e^{-ik\pi} = (-1)^k ~~~ (k \in \mathbb{Z}), \end{eqnarray*}\]

we have

\[\begin{eqnarray*} c_k &=& \frac{1}{2\pi}\cdot\frac{1}{i\left(\frac{1}{2} - k\right)}\{i(-1)^{k} + i(-1)^{k}\}\\ &=& \frac{2}{\pi}\cdot\frac{(-1)^k}{1-2k} ~~~ (k \in \mathbb{Z}). \end{eqnarray*}\]

Thus,

\[\begin{eqnarray*} e^{ix/2} &\sim& \frac{2}{\pi}\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{1-2k}e^{ikx}\\ &=&\frac{2}{\pi}\left(\cdots + \frac{1}{5}e^{-i2x} -\frac{1}{3}e^{-ix} + 1 + e^{ix} - \frac{1}{3}e^{i2x} + \cdots \right). \end{eqnarray*}\]

□

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 \(c_n\):

\[\begin{eqnarray} 2c_n &=& a_n - ib_n,\\ 2c_{-n} &=& a_n + ib_n \end{eqnarray}\]

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

\[\begin{eqnarray} a_n &=& c_n + c_{-n},\\ b_n &=& i(c_n - c_{-n}). \end{eqnarray}\]

Consider the complex Fourier series of \(f\):

\[f \sim \sum_{k = -\infty}^{\infty}c_ke^{ikx}.\]

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

\[S_n = \sum_{k = -n}^{n}c_ke^{ikx}.\]

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

\[\begin{eqnarray*} S_n & = & \sum_{k = -n}^{n}c_ke^{ikx}\\ &=& c_0 + \sum_{k=1}^{n}\{(c_k + c_{-k})\cos(kx) + i(c_k - c_{-k})\sin(kx)\}\\ &=& \frac{1}{2}a_0 + \sum_{k=1}^{n}\{a_k\cos(kx) + b_k\sin(kx)\}. \end{eqnarray*}\]

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

\[\gamma_k = \int_{-\pi}^{\pi}f(x)e^{-ikx} dx = 0 ~~~ (k \in \mathbb{Z}),\tag{eq:zero}\]

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 \(\sum_{n}|c_n| < +\infty\), 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 \(f\).

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