extracting every $n^{\mathrm{th}}$ term of a series

Roots of unity can be used to extract every $n^{\mathrm{th}}$ term of a series. This method is due to Simpson [1759].

Theorem. Let $\omega =e^{2\pi i/k}$ be a primitive $k^{\mathrm{th}}$ root of unity. If $f(x)={\sum }_{j=0}^{\mathrm{\infty }}{a}_j{x}^j$ and $n\not\equiv 0(modk)$, then

$$\sum _{j=0}^{\mathrm{\infty }}{a}_{kj+n}{x}^{kj+n}=\frac{1}{k}\sum _{j=0}^{k-1}{\omega }^{-jn}f({\omega }^{j}x)$$

Proof.This is a consequence of the fact that ${\sum }_{j=0}^{k-1}{\omega }^{jm}=0$ for $m\not\equiv 0(modk)$.

Consider the term involving $x^r$ on the right-hand side. It is

$$\frac{1}{k}\sum _{j=0}^{k-1}{\omega }^{-jn}{a}_r{\omega }^{jr}{x}^r=\frac{1}{k}{a}_r{x}^r\sum _{j=0}^{k-1}{\omega }^{j(r-n)}$$

If $r\not\equiv n(modk)$, the sum is zero. So the term involving $x^r$ is zero unless $r\equiv n(modk)$, in which case it is $a_r{x}^r$ since each element of the sum is $1$.

Note that this method is a generalization of the commonly known trick for extracting alternate terms of a series:

$$\frac{1}{2}(f(x)-f(-x))$$

produces the odd terms of $f$.