Finite Fourier Series

Theorem

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

Let $\xi = e^{2\pi i/b}$ is the first $b$th root of unity, then:

$\displaystyle a \left({n}\right) = \sum_{k=0}^{b-1} a_* \left({k}\right) \xi^ {nk}$

where

$\displaystyle a_* \left({n}\right) = \frac{1}{b} \sum_{k=0}^{b-1} a \left({k}\right) \xi^{-nk}$

Proof

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

$a \left({n + b}\right) = a \left({n}\right)$

So if we define:

$\displaystyle F \left({z}\right) = \sum_{n \ge 0} a \left({n}\right) z^n$

we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle F \left({z}\right)\)	\(=\)	\(\displaystyle \left({\sum_{k=0}^{b-1} a \left({k}\right) z^k}\right) + z^b \left({\sum_{k=0}^{b-1} a \left({k}\right) z^k }\right) + z^{2b} \left({\sum_{k=0}^{b-1} a \left({k}\right)z^k }\right) + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{1} {1 - z^b} \left({\sum_{k=0}^{b-1} a \left({k}\right)z^k}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{P \left({z}\right)} {1 - z^b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.