# Peeter Joot's (OLD) Blog.

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## A Fourier series refresher.

Posted by peeterjoot on May 3, 2012

# Motivation.

I’d used the wrong scaling in a Fourier series over a $[0, 1]$ interval. Here’s a reminder to self what the right way to do this is.

# Guts

Suppose we have a function that is defined in terms of a trigonometric Fourier sum

\begin{aligned}\phi(x) = \sum c_k e^{i \omega k x},\end{aligned} \hspace{\stretch{1}}(2.1)

where the domain of interest is $x \in [a, b]$. Stating the problem this way avoids any issue of existence. We know $c_k$ exists, but just want to find what they are given some other representation of the function.

Multiplying and integrating over our domain we have

\begin{aligned}\begin{aligned}\int_a^b \phi(x) e^{-i \omega m x} dx &= \sum c_k \int_a^b e^{i \omega (k -m) x} dx \\ &= c_m (b - a) + \sum_{k \ne m} \frac{e^{i \omega(k-m) b} - e^{i \omega(k-m)a}}{i \omega (k -m)} .\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.2)

We want all the terms in the sum to be be zero, requiring equality of the exponentials, or

\begin{aligned}e^{i \omega (k -m) (b -a )} = 1,\end{aligned} \hspace{\stretch{1}}(2.3)

or

\begin{aligned}\omega = \frac{2 \pi}{b - a}.\end{aligned} \hspace{\stretch{1}}(2.4)

This fixes our Fourier coefficients

\begin{aligned}c_m = \frac{1}{{b - a}} \int_a^b \phi(x) e^{- 2 \pi i m x/(b - a)} dx.\end{aligned} \hspace{\stretch{1}}(2.5)

Given this, the correct (but unnormalized) Fourier basis for a $[0, 1]$ interval would be the functions $e^{2 \pi i x}$, or the sine and cosine equivalents.