# 9.2: Complex Exponential Fourier Series

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Before deriving the Fourier transform, we will need to rewrite the trigonometric Fourier series representation as a complex exponential Fourier series. We first recall from Chapter ?? the trigonometric Fourier series representation of a function defined on $$[-\pi, \pi]$$ with period $$2 \pi$$. The Fourier series is given by

$f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n x+b_{n} \sin n x\right),\label{eq:1}$

where the Fourier coefficients were found as

\begin{align} &a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x, \quad n=0,1, \ldots,\nonumber \\ &b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x, \quad n=1,2, \ldots\label{eq:2} \end{align}

In order to derive the exponential Fourier series, we replace the trigonometric functions with exponential functions and collect like exponential terms. This gives

\begin{align} f(x) & \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\left(\frac{e^{i n x}+e^{-i n x}}{2}\right)+b_{n}\left(\frac{e^{i n x}-e^{-i n x}}{2 i}\right)\right]\nonumber \\ &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(\frac{a_{n}-i b_{n}}{2}\right) e^{i n x}+\sum_{n=1}^{\infty}\left(\frac{a_{n}+i b_{n}}{2}\right) e^{-i n x} .\label{eq:3} \end{align}

The coefficients of the complex exponentials can be rewritten by defining

$c_{n}=\frac{1}{2}\left(a_{n}+i b_{n}\right), \quad n=1,2, \ldots .\label{eq:4}$

This implies that

$\bar{c}_{n}=\frac{1}{2}\left(a_{n}-i b_{n}\right), \quad n=1,2, \ldots\label{eq:5}$

So far the representation is rewritten as

$f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty} \bar{c}_{n} e^{i n x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x} .\nonumber$

Re-indexing the first sum, by introducing $$k=-n$$, we can write

$f(x) \sim \frac{a_{0}}{2}+\sum_{k=-1}^{-\infty} \bar{c}_{-k} e^{-i k x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x}\nonumber$

Since $$k$$ is a dummy index, we replace it with a new $$n$$ as

$f(x) \sim \frac{a_{0}}{2}+\sum_{n=-1}^{-\infty} \bar{c}_{-n} e^{-i n x}+\sum_{n=1}^{\infty} c_{n} e^{-i n x} .\nonumber$

We can now combine all of the terms into a simple sum. We first define $$c_{n}$$ for negative $$n^{\prime}$$s by

$c_{n}=\bar{c}_{-n}, \quad n=-1,-2, \ldots\nonumber$

Letting $$c_{0}=\frac{a_{0}}{2}$$, we can write the complex exponential Fourier series representation as

$f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n x},\label{eq:6}$

where

\begin{align} &c_{n}=\frac{1}{2}\left(a_{n}+i b_{n}\right), \quad n=1,2, \ldots\nonumber \\ &c_{n}=\frac{1}{2}\left(a_{-n}-i b_{-n}\right), \quad n=-1,-2, \ldots\nonumber \\ &c_{0}=\frac{a_{0}}{2}\label{eq:7} \end{align}

Given such a representation, we would like to write out the integral forms of the coefficients, $$c_{n}$$. So, we replace the $$a_{n}$$ ’s and $$b_{n}$$ ’s with their integral representations and replace the trigonometric functions with complex exponential functions. Doing this, we have for $$n=1,2, \ldots$$.

\begin{align} c_{n} &=\frac{1}{2}\left(a_{n}+i b_{n}\right)\nonumber \\ &=\frac{1}{2}\left[\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x+\frac{i}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x\right]\nonumber \\ &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x)\left(\frac{e^{i n x}+e^{-i n x}}{2}\right) d x+\frac{i}{2 \pi} \int_{-\pi}^{\pi} f(x)\left(\frac{e^{i n x}-e^{-i n x}}{2 i}\right) d x\nonumber \\ &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\label{eq:8} \end{align}

It is a simple matter to determine the $$c_{n}$$ ’s for other values of $$n$$. For $$n=0$$, we have that

$c_{0}=\frac{a_{0}}{2}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) d x .\nonumber$

For $$n=-1,-2, \ldots$$, we find that

$c_{n}=\bar{c}_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) \overline{e^{-i n x}} d x=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\nonumber$

Therefore, we have obtained the complex exponential Fourier series coefficients for all $$n$$. Now we can define the complex exponential Fourier series for the function $$f(x)$$ defined on $$[-\pi, \pi]$$ as shown below.

## Complex Exponential Series for $$f(x)$$ defined on $$[-\pi ,\pi ]$$

$f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n x},\label{eq:9}$

$c_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{i n x} d x .\label{eq:10}$

We can easily extend the above analysis to other intervals. For example, for $$x \in[-L, L]$$ the Fourier trigonometric series is

$f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos \frac{n \pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right)\nonumber$

with Fourier coefficients

\begin{aligned} &a_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L} d x, \quad n=0,1, \ldots, \\ &b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n \pi x}{L} d x, \quad n=1,2, \ldots . \end{aligned} \nonumber

This can be rewritten as an exponential Fourier series of the form

## Complex Exponential Series for $$f(x)$$ defined on $$[-L,L]$$.

$f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{-i n \pi x / L},\label{eq:11}$

$c_{n}=\frac{1}{2 L} \int_{-L}^{L} f(x) e^{i n \pi x / L} d x .\label{eq:12}$

We can now use this complex exponential Fourier series for function defined on $$[-L, L]$$ to derive the Fourier transform by letting $$L$$ get large. This will lead to a sum over a continuous set of frequencies, as opposed to the sum over discrete frequencies, which Fourier series represent.

This page titled 9.2: Complex Exponential Fourier Series is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform.