Finite Fourier Series

Theorem

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

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

Then:

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

where:

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

Proof

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

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

So if we define:

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

we have:

\(\ds \map F z)\)

\(=\)

\(\ds \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^b \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^{2 b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + \cdots\)

\(\ds \)

\(=\)

\(\ds \frac 1 {1 - z^b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k}\)

\(\ds \)

\(=\)

\(\ds \frac {\map P z} {1 - z^b}\)

where the last step defines the polynomial $P$.

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

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Source of Name

This entry was named for Joseph Fourier.