decompose-the-fraction-F-x-x-n-x-2n-1-inside-C-x-and-R-x-

Question Number 57923 by maxmathsup by imad last updated on 14/Apr/19

d e c o m p o s e t h e f r a c t i o n F (x) = \frac{x^{n}}{x^{2 n} - 1} i n s i d e C (x) a n d R (x)

Commented by maxmathsup by imad last updated on 28/Apr/19

r o o t s o f z^{2 n} - 1 = 0 l e t z = {r e}^{i θ} s o z^{2 n} = 1 \Rightarrow r^{n} = 1 a n d 2 n θ = 2 k π \Rightarrow

θ_{k} = \frac{k π}{n} \Rightarrow t h e r o o t s {a r e Z}_{k} = e^{\frac{i k}{n}} a n d k \in [[0, 2 n - 1]] \Rightarrow

F (x) = \frac{x^{n}}{\prod_{k = 0}^{2 n - 1} (x - Z_{k})} = \sum_{k = 0}^{2 n - 1} \frac{λ_{k}}{x - Z_{k}} a n d λ_{k} = \frac{Z_{k}^{n}}{2 n Z_{k}^{2 n - 1}} = \frac{1}{2 n} Z_{k}^{n + 1} \Rightarrow

F (x) = \frac{1}{2 n} \sum_{k = 0}^{2 n - 1} \frac{Z_{k}^{n + 1}}{x - Z_{k}} i s t h e d e c o m p o s i t i o n i n s i d e C (x) .

Commented by maxmathsup by imad last updated on 28/Apr/19

w e h a v e Z_{k} = e^{\frac{i k π}{n}} w i t h k \in [[0, 2 n - 1]]

Z_{0} = 1, Z_{1} = e^{\frac{i π}{n}}, Z_{2} = e^{\frac{i 2 π}{n}}, Z_{n - 1} = e^{i \frac{(n - 1) π}{n}}, Z_{n} = - 1, Z_{n + 1} = e^{i \frac{(n + 1) π}{n}},

\dots . Z_{2 n - 1} = e^{i \frac{(2 n - 1) π}{n}} \Rightarrow Z_{2 n - 1} = {\bar{Z}}_{1}, Z_{2 n - 2} = {\bar{Z}}_{2} \dots . \Rightarrow

F (x) = \frac{1}{2 n} {\frac{1}{Z - 1} + \frac{{(- 1)}^{n + 1}}{x + 1} + \sum_{k = 1}^{n - 1} (\frac{Z_{k}^{n + 1}}{x - Z_{k}} + \frac{{({\bar{Z}}_{k})}^{n + 1}}{x - {\bar{Z}}_{k}})} \Rightarrow

F (x) = \frac{1}{2 n (x - 1)} + \frac{{(- 1)}^{n + 1}}{2 n (x + 1)} + \frac{1}{2 n} + \sum_{k = 0}^{n - 1} \frac{Z_{k}^{n + 1} (x - {\bar{Z}}_{k}) + (x - Z_{k}) {({\bar{Z}}_{k})}^{n + 1}}{x^{2} - 2 x R e (Z_{k}) + 1}

F (x) = \frac{1}{2 n (x - 1)} + \frac{{(- 1)}^{n + 1}}{2 n (x + 1)} + \sum_{k = 0}^{n - 1} \frac{(Z_{k}^{n + 1} + {({\bar{Z}}_{k})}^{n + 1}) x - Z_{k}^{n + 1} {\bar{Z}}_{k} - Z_{k} {({\bar{Z}}_{k})}^{n + 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 i n s i d e R (x)

w i t h Z_{k}^{n + 1} + {({\bar{Z}}_{k})}^{n + 1} = 2 R e (Z_{k}^{n + 1}) = 2 c o s (k \frac{(n + 1) π}{n})

- Z_{k}^{n + 1} {\bar{Z}}_{k} - Z_{k} {({\bar{Z}}_{k})}^{n + 1} = - (Z_{k}^{n} + {({\bar{Z}}_{k})}^{n}) = - 2 c o s (k π) = - 2 {(- 1)}^{k} .

Commented by maxmathsup by imad last updated on 29/Apr/19

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