n-integr-decompose-imsidr-R-x-the-fraction-F-x-1-x-2-1-n-

Question Number 34019 by prof Abdo imad last updated on 29/Apr/18

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

F (x) = \frac{1}{{(x^{2} - 1)}^{n}}

Commented by abdo mathsup 649 cc last updated on 06/May/18

F (x) = \frac{1}{{(x - 1)}^{n} {(x + 1)}^{n}} c h a 7 g e m e n t x - 1 = t g i v e

F (x) = g (t) = \frac{1}{t^{n} {(t + 2)}^{n}} l e t f i n d D_{n - 1} (0) f o r \frac{1}{{(t + 2)}^{n}}

h (x) = \frac{1}{{(t + 2)}^{n}} = \sum_{k = 0}^{n - 1} \frac{h^{(k)} (0)}{k!} x^{k} + \frac{x^{n}}{n!} ξ (x)

h^{(k)} (x) = {{(t + 2)}^{- n}}^{(k)}

h^{^{'}} (x) = - n {(t + 2)}^{- (n + 1)}

h^{(2)} (x) = {(- 1)}^{2} n (n + 1) {(t + 2)}^{- (n + 2)}

h^{(k)} (x) = {(- 1)}^{k} n (n + 1) \dots . . (n + k - 1) {(t + 2)}^{- (n + k)}

h^{(k)} (0) = {(- 1)}^{k} n (n + 1) \dots . . (n + k - 1) 2^{- (n + k)}

\Rightarrow h (x) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots . (n + k - 1)}{k! 2^{n + k}} x^{k}

+ \frac{x^{n}}{n!} ξ (x)

Commented by abdo mathsup 649 cc last updated on 07/May/18

h (t) = \frac{1}{{(t + 2)}^{n}} \Rightarrow

g (t) = \frac{1}{t^{n}} \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots (n + k - 1)}{k! 2^{n + k}} t^{k}

+ \frac{1}{n!} ξ (t)

= \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots . . (n + k - 1)}{k! 2^{n + k} t^{n - k}}

c h a n g e m e n t o f i n d i c e n - k = p g i v e

g (t) = \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) \dots . (n + n - p - 1)}{(n - p! 2^{n + n - p} t^{p}}

= \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) \dots . (2 n - p - 1)}{(n - p)! 2^{2 n - p} t^{p}}

f r o m a n o t h e r s i d e

g (t) = \sum_{p = 1}^{n} \frac{λ_{p}}{t^{p}} + \sum_{k = 1}^{n} \frac{a_{k}}{{(t + 2)}^{k}} \Rightarrow

λ_{p} = \frac{{(- 1)}^{n - p} n (n + 1) \dots . ((n - p - 1)}{(n - p)! 2^{2 n - p}}

b e c o n t i n u e d \dots

Commented by abdo mathsup 649 cc last updated on 07/May/18

λ_{p} = \frac{{(- 1)}^{n - p} n (n + 1) \dots . . (2 n - p - 1)}{(n - p)! 2^{2 n - p}}

Commented by prof Abdo imad last updated on 07/May/18

w e h a v e F (x) = \frac{1}{{(x - 1)}^{n} {(x + 1)}^{n}} n o w w e u s e

t h e c h a n g e m e n t x + 1 = t \Rightarrow

F (x) = g (t) = \frac{1}{{(t - 2)}^{n} t^{n}} l e t p u t

h (t) = \frac{1}{{(t - 2)}^{n}} = \sum_{k = 0}^{n - 1} \frac{h^{(k)} (0)}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t)

h^{(k)} (x) = {(- 1)}^{k} n (n + 1) \dots . (n + k - 1) {(t - 2)}^{- (n + k)}

h^{(k)} (0) = {(- 1)}^{k} n (n + 1) \dots . . (n + k - 1) {(- 2)}^{- (n + k)}

\Rightarrow h (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots . (n + k - 1)}{k! {(- 2)}^{n + k}} t^{k}

+ \frac{t^{n}}{n!} ξ (t) \Rightarrow g (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots (n + k - 1)}{k! {(- 2)}^{n + k} t^{n - k}}

a n d w e f o l o w t b e s a m e r o a d t o f i n d a_{k} \dots .