decompose the fraction F(x) =(x^n /(x^(2n) −1)) inside C(x) and R(x)


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Question Number 57923 by maxmathsup by imad last updated on 14/Apr/19
$decompose the fraction F(x) =(x^n /(x^(2n) −1)) inside C(x) and R(x)$
$d e c o m p o s e t h e f r a c t i o n F (x) = \frac{x^{n}}{x^{2 n} - 1} i n s i d e C (x) a n d R (x)$
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
$we have Z_k =e^((ikπ)/n) with k∈[[0,2n−1]] Z_0 =1 , Z_1 =e^((iπ)/n) , Z_(2 ) = e^((i2π)/n) , Z_(n−1) =e^(i(((n−1)π)/n)) ,Z_n =−1 ,Z_(n+1) =e^(i(((n+1)π)/n)) , ....Z_(2n−1) =e^(i(((2n−1)π)/n)) ⇒Z_(2n−1) =Z_1 ^− ,Z_(2n−2) =Z_2 ^− .... ⇒ F(x) =(1/(2n)) { (1/(Z−1)) +(((−1)^(n+1) )/(x+1)) +Σ_(k=1) ^(n−1) ( (Z_k ^(n+1) /(x−Z_k )) +(((Z_k ^− )^(n+1) )/(x−Z_k ^− )))} ⇒ F(x)= (1/(2n(x−1))) +(((−1)^(n+1) )/(2n(x+1))) +(1/(2n)) +Σ_(k=0) ^(n−1) ((Z_k ^(n+1) (x−Z_k ^− )+(x−Z_k )(Z_k ^− )^(n+1) )/(x^2 −2xRe(Z_k )+1)) F(x) =(1/(2n(x−1))) +(((−1)^(n+1) )/(2n(x+1))) +Σ_(k=0) ^(n−1) (((Z_k ^(n+1) +(Z_k ^− )^(n+1) )x −Z_k ^(n+1) Z_k ^− −Z_k (Z_k ^− )^(n+1) )/(x^2 −2cos(((kπ)/n))x +1)) is the decomposition inside R(x) with Z_k ^(n+1) +(Z_k ^− )^(n+1) = 2 Re(Z_k ^(n+1) ) =2cos(k(((n+1)π)/n)) −Z_k ^(n+1) Z_k ^− −Z_k (Z_k ^− )^(n+1) =−( Z_k ^n +(Z_k ^− )^n ) =−2cos(kπ) =−2(−1)^k .$
$w e h a v e Z_{k} = e^{\frac{i k π}{n}} w i t h k \in [[0, 2 n - 1]]$ $Z_{0} = 1, Z_{1} = e^{\frac{i π}{n}}, Z_{2} = e^{\frac{i 2 π}{n}}, Z_{n - 1} = e^{i \frac{(n - 1) π}{n}}, Z_{n} = - 1, Z_{n + 1} = e^{i \frac{(n + 1) π}{n}},$ $. . . . Z_{2 n - 1} = e^{i \frac{(2 n - 1) π}{n}} \Rightarrow Z_{2 n - 1} = {\bar{Z}}_{1}, Z_{2 n - 2} = {\bar{Z}}_{2} . . . . \Rightarrow$ $F (x) = \frac{1}{2 n} {\frac{1}{Z - 1} + \frac{{(- 1)}^{n + 1}}{x + 1} + \sum_{k = 1}^{n - 1} (\frac{Z_{k}^{n + 1}}{x - Z_{k}} + \frac{{({\bar{Z}}_{k})}^{n + 1}}{x - {\bar{Z}}_{k}})} \Rightarrow$ $F (x) = \frac{1}{2 n (x - 1)} + \frac{{(- 1)}^{n + 1}}{2 n (x + 1)} + \frac{1}{2 n} + \sum_{k = 0}^{n - 1} \frac{Z_{k}^{n + 1} (x - {\bar{Z}}_{k}) + (x - Z_{k}) {({\bar{Z}}_{k})}^{n + 1}}{x^{2} - 2 x R e (Z_{k}) + 1}$ $F (x) = \frac{1}{2 n (x - 1)} + \frac{{(- 1)}^{n + 1}}{2 n (x + 1)} + \sum_{k = 0}^{n - 1} \frac{(Z_{k}^{n + 1} + {({\bar{Z}}_{k})}^{n + 1}) x - Z_{k}^{n + 1} {\bar{Z}}_{k} - Z_{k} {({\bar{Z}}_{k})}^{n + 1}}{x^{2} - 2 c o s (\frac{k π}{n}) x + 1} i s t h e d e c o m p o s i t i o n i n s i d e R (x)$ $w i t h Z_{k}^{n + 1} + {({\bar{Z}}_{k})}^{n + 1} = 2 R e (Z_{k}^{n + 1}) = 2 c o s (k \frac{(n + 1) π}{n})$ $- Z_{k}^{n + 1} {\bar{Z}}_{k} - Z_{k} {({\bar{Z}}_{k})}^{n + 1} = - (Z_{k}^{n} + {({\bar{Z}}_{k})}^{n}) = - 2 c o s (k π) = - 2 {(- 1)}^{k} .$
Commented by maxmathsup by imad last updated on 29/Apr/19

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