# Finite Fourier Series

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## 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.