Where does the following formula come from?

Question

Where does the following formula come from?

For a Laurent polynomial $f(z)=\sum a_j z^j$ and a positive integer $n$ we have $$\sum_{k\equiv \alpha\pmod n} a_k=\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega).$$

I hope someone can answer this question or give some refferences about it ! Thanks a lot!

Answer 1

first we shall see that:

$\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega) = \frac1n\sum_{\omega:\omega^n=1}\omega^{-\alpha }\sum_{j}a_j\omega^j = \frac1n\sum_{\omega:\omega^n=1}\sum_{j}\omega^{j-\alpha }a_j =$

$ = \frac1n\sum_{j}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j$

I'm leaving for you to think why we can preform the last move (change the order). also put notice that $\omega^{j-\alpha} = 1 \iff j \equiv \alpha \ (mod \ n)$ and also because there are n roots of unity in $\mathbb{C}$ we get:

$ \frac1n\sum_{j}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j = \frac1n\sum_{j \equiv \alpha (mod \ n)}\sum_{\omega:\omega^n=1}a_j + \frac1n\sum_{j \neq \alpha (mod \ n)}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j = \sum_{j \equiv \alpha (mod \ n)} \frac{n}{n} a_j + \frac1n\sum_{j \neq \alpha (mod \ n)}a_j\sum_{\omega:\omega^n=1}\omega^{j-\alpha }$

but we know that $\forall \ 0<m< \ n: \sum_{\omega:\omega^n=1}\omega^{m } = 0 $ and therefor we get: $\frac1n\sum_{j \neq \alpha (mod \ n)}a_j\sum_{\omega:\omega^n=1}\omega^{j-\alpha } = 0 $ (as $0<j-\alpha < n$) and were done (-:

Answer 2

This formula comes from Thomas Simpson's Series Multisection Theory. Speak in the concrete， a multisection of the series of an analytic function

$$f(z) = \sum_{n=0}^\infty a_n\cdot z^n.$$

has a closed-form expression in terms of the function $f(x)$:

$$\sum_{m=0}^\infty a_{qm+p}\cdot z^{qm+p} = \frac{1}{q}\cdot \sum_{k=0}^{q-1} \omega^{-kp}\cdot f(\omega^k\cdot z),$$

where $\omega = e^{\frac{2\pi i}{q}}$ is a Primitive nth root of unity primitive ''q''-th root of unity . This solution was first discovered by Thomas Simpson.