find-the-decomposition-in-C-x-then-R-x-for-the-rationsl-fraction-F-x-1-x-2n-1-with-n-integer-not-0-

Question Number 26583 by abdo imad last updated on 27/Dec/17

f i n d t h e d e c o m p o s i t i o n i n C [x] t h e n R [x]

f o r t h e r a t i o n s l f r a c t i o n

F (x) = \frac{1}{x^{2 n} - 1} . w i t h n i n t e g e r n o t 0

Commented by abdo imad last updated on 28/Dec/17

l e t f i n d t h e p o l e s o f F l e t p u t z = r c o s θ

z^{2 n} = 1 \Leftrightarrow 2 n θ = 2 k π a n d r = 1 s o t h e p o l e s o f a r e

z_{k} = e^{\frac{i k π}{n}} w i t h k f r o m [[0, 2 n - 1]]

F (x) = \sum_{k = 0}^{2 n - 1} \frac{α_{k}}{x - z_{k}} a n d α_{k} = \frac{1}{2 n z_{k}^{2 n - 1}} = \frac{1}{2 n} z_{k}

\Rightarrow F (x) = \frac{1}{2 n} \sum_{k = 0}^{2 n - 1} \frac{z_{k}}{x - z_{k}} i s t h e d e c o m p o s i t i o n o f F (x)

i n C [x] .

Commented by abdo imad last updated on 28/Dec/17

w e h a v e F (x) = \frac{1}{2 n} \sum_{k = 0}^{2 n - 1} \frac{z k}{x - x_{k}} b u t

z_{0} = 1, z_{1} = e^{i \frac{π}{n}}, z_{2} = e^{i \frac{2 π}{n}}, z_{n - 1} = e^{i \frac{n - 1}{n} π} z_{n} = - 1

z_{n + 1} = e^{i \frac{(n + 1) π}{n}} = z_{1}^{-}, z_{n + 2} = e^{i \frac{(n + 2) π}{n}} = z_{2}^{-}, z_{n - 1} = z_{1}^{-}

\Rightarrow F (x) = \frac{1}{2 n} (\frac{z_{0}}{x - z_{0}} + \frac{z_{n}}{x - z_{n}} + \sum_{k = 1}^{k = n - 1} (\frac{z_{k}}{x - z_{k}} + \frac{z_{k}^{-}}{x - z_{k}^{-}}))

= \frac{1}{2 n} (\frac{1}{x - 1} - \frac{1}{x + 1}) + \frac{1}{2 n} \sum_{k = 1}^{n - 1} \frac{(z_{k} + z_{k}^{-}) x - 2}{x^{2} - 2 c o s (\frac{k π}{n}) x + 1}

F (x) = \frac{1}{2 n} (\frac{1}{x - 1} - \frac{1}{x + 1}) + \frac{1}{n} \sum_{k = 1}^{k = n - 1} \frac{c o s (\frac{k π}{n}) x - 1}{x^{2} - 2 c o s (\frac{k π}{n}) x + 1}

i s t h e d e c o m p o s i t i o n o f F (x) i n s i d e R [x] .