# Finite Fourier Series

## Theorem

Let $\map a n$ be any finite periodic function on $\Z$ with period $b$.

Let $\xi = e^{2 \pi i/ b}$ be the first $b$th root of unity.

Then:

$\displaystyle \map a n = \sum_{k \mathop = 0}^{b - 1} \map {a_*} k \xi^{n k}$

where:

$\displaystyle \map {a_*} n = \frac 1 b \sum_{k \mathop = 0}^{b - 1} \map a k \xi^{-n k}$

## Proof

Since $a$ has period $b$, we have:

$\map a {n + b} = \map a n$

So if we define:

$\displaystyle \map F z = \sum_{n \mathop \ge 0} \map a n z^n$

we have:

 $\displaystyle \map F z)$ $=$ $\displaystyle \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^b \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^{2 b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + \cdots$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1 - z^b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k}$ $\displaystyle$ $=$ $\displaystyle \frac {\map P z} {1 - z^b}$

where the last step defines the polynomial $P$.

If we expand $F$ now using partial fractions, we get

## Source of Name

This entry was named for Joseph Fourier.