# Peeter Joot's (OLD) Blog.

• ## Archives

 papasu on PHY450H1S. Relativistic Electr… papasu on Energy term of the Lorentz for… lidiodu on PHY450H1S. Relativistic Electr… lidiodu on PHY450H1S. Relativistic Electr… lidiodu on bivector form of Stokes t…

• 307,992

## 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.