let-U-n-z-C-z-n-1-simplify-p-0-2n-1-w-p-with-w-U-n-and-p-0-2n-1-2w-1-p-

Question Number 148501 by mathmax by abdo last updated on 28/Jul/21

let U_{n} = {z \in C / z^{n} = 1} simplify

\sum_{p = 0}^{2 n - 1} w^{p} with w \in U_{n}

and \sum_{p = 0}^{2 n - 1} {(2 w + 1)}^{p}

Answered by Olaf_Thorendsen last updated on 28/Jul/21

∙ w = 1

S_{n} (1) = \underset{p = 0}{\sum^{2 n - 1}} 1^{p} = 2 n (trivial)

∙ w^{n} = 1 and w \neq 1

S_{n} (w) = \underset{p = 0}{\sum^{2 n - 1}} w^{p} = \frac{1 - w^{2 n}}{1 - w} = \frac{1 - 1}{1 - w} = 0

T_{n} (w) = \underset{p = 0}{\sum^{2 n - 1}} {(2 w + 1)}^{p}

T_{n} (w) = \frac{1 - {(2 w + 1)}^{2 n}}{1 - (2 w + 1)} = \frac{{(2 w + 1)}^{2 n} - 1}{2 w}

Answered by mathmax by abdo last updated on 28/Jul/21

if w = 1 A_{n} = \sum_{p = 0}^{2 n - 1} w^{p} = 2 n

if w \neq 1 \sum_{p = 0}^{2 n - 1} w^{p} = \frac{1 - w^{2 n}}{1 - w} = \frac{1 - {(w^{n})}^{2}}{1 - w} = \frac{1 - 1}{1 - w} = 0

{alsoB}_{n} = \sum_{p = 0}^{2 n - 1} {(2 w + 1)}^{p} the roots of z^{n} = 1 are z_{k} = e^{i \frac{2 k π}{n}}

and 0 ⩽ k ⩽ n - 1 \Rightarrow w \neq - \frac{1}{2} \Rightarrow B_{n} = \frac{1 - {(2 w + 1)}^{2 n}}{1 - (2 w + 1)}

= \frac{1}{2 w} {{(2 w + 1)}^{2 n} - 1}

= \frac{1}{2 w} {\sum_{k = 0}^{2 n} C_{2 n}^{k} {(2 w)}^{k} - 1}

= \frac{1}{2 w} \sum_{k = 0}^{2 n} C_{2 n}^{k} 2^{k} w^{k}