n integr decompose imsidr R[x] the fraction F(x) = (1/((x^2 −1)^n ))


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Question Number 34019 by prof Abdo imad last updated on 29/Apr/18
$n integr decompose imsidr R[x] the fraction F(x) = (1/((x^2 −1)^n ))$
$n i n t e g r d e c o m p o s e i m s i d r R [x] t h e f r a c t i o n$ $F (x) = \frac{1}{{(x^{2} - 1)}^{n}}$
Commented by abdo mathsup 649 cc last updated on 06/May/18
$F(x)= (1/((x−1)^n (x+1)^n )) cha7gement x−1 =t give F(x)=g(t)= (1/(t^n (t+2)^n )) let find D_(n−1) (0) for (1/((t+2)^n )) h(x)= (1/((t+2)^n )) = Σ_(k=0) ^(n−1) ((h^((k)) (0))/(k!)) x^k +(x^n /(n!)) ξ(x) h^((k)) (x)={(t+2)^(−n) }^((k)) h^′ (x)= −n(t+2)^(−(n+1)) h^((2)) (x) =(−1)^2 n(n+1) (t+2)^(−(n+2)) h^((k)) (x)= (−1)^k n(n+1).....(n+k−1)(t+2)^(−(n+k)) h^((k)) (0) =(−1)^k n(n+1).....(n+k−1) 2^(−(n+k)) ⇒ h(x) = Σ_(k=0) ^(n−1) (((−1)^k n(n+1)....(n+k−1))/(k! 2^(n+k) )) x^k + (x^n /(n!)) ξ(x)$
$F (x) = \frac{1}{{(x - 1)}^{n} {(x + 1)}^{n}} c h a 7 g e m e n t x - 1 = t g i v e$ $F (x) = g (t) = \frac{1}{t^{n} {(t + 2)}^{n}} l e t f i n d D_{n - 1} (0) f o r \frac{1}{{(t + 2)}^{n}}$ $h (x) = \frac{1}{{(t + 2)}^{n}} = \sum_{k = 0}^{n - 1} \frac{h^{(k)} (0)}{k!} x^{k} + \frac{x^{n}}{n!} ξ (x)$ $h^{(k)} (x) = {{(t + 2)}^{- n}}^{(k)}$ $h^{^{'}} (x) = - n {(t + 2)}^{- (n + 1)}$ $h^{(2)} (x) = {(- 1)}^{2} n (n + 1) {(t + 2)}^{- (n + 2)}$ $h^{(k)} (x) = {(- 1)}^{k} n (n + 1) . . . . . (n + k - 1) {(t + 2)}^{- (n + k)}$ $h^{(k)} (0) = {(- 1)}^{k} n (n + 1) . . . . . (n + k - 1) 2^{- (n + k)}$ $\Rightarrow h (x) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . . (n + k - 1)}{k! 2^{n + k}} x^{k}$ $+ \frac{x^{n}}{n!} ξ (x)$
Commented by abdo mathsup 649 cc last updated on 07/May/18

$h (t) = \frac{1}{{(t + 2)}^{n}} \Rightarrow$ $g (t) = \frac{1}{t^{n}} \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . (n + k - 1)}{k! 2^{n + k}} t^{k}$ $+ \frac{1}{n!} ξ (t)$ $= \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . . . (n + k - 1)}{k! 2^{n + k} t^{n - k}}$ $c h a n g e m e n t o f i n d i c e n - k = p g i v e$ $g (t) = \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) . . . . (n + n - p - 1)}{(n - p! 2^{n + n - p} t^{p}}$ $= \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) . . . . (2 n - p - 1)}{(n - p)! 2^{2 n - p} t^{p}}$ $f r o m a n o t h e r s i d e$ $g (t) = \sum_{p = 1}^{n} \frac{λ_{p}}{t^{p}} + \sum_{k = 1}^{n} \frac{a_{k}}{{(t + 2)}^{k}} \Rightarrow$ $λ_{p} = \frac{{(- 1)}^{n - p} n (n + 1) . . . . ((n - p - 1)}{(n - p)! 2^{2 n - p}}$ $b e c o n t i n u e d . . .$
Commented by abdo mathsup 649 cc last updated on 07/May/18

$λ_{p} = \frac{{(- 1)}^{n - p} n (n + 1) . . . . . (2 n - p - 1)}{(n - p)! 2^{2 n - p}}$
Commented by prof Abdo imad last updated on 07/May/18

$w e h a v e F (x) = \frac{1}{{(x - 1)}^{n} {(x + 1)}^{n}} n o w w e u s e$ $t h e c h a n g e m e n t x + 1 = t \Rightarrow$ $F (x) = g (t) = \frac{1}{{(t - 2)}^{n} t^{n}} l e t p u t$ $h (t) = \frac{1}{{(t - 2)}^{n}} = \sum_{k = 0}^{n - 1} \frac{h^{(k)} (0)}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t)$ $h^{(k)} (x) = {(- 1)}^{k} n (n + 1) . . . . (n + k - 1) {(t - 2)}^{- (n + k)}$ $h^{(k)} (0) = {(- 1)}^{k} n (n + 1) . . . . . (n + k - 1) {(- 2)}^{- (n + k)}$ $\Rightarrow h (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . . (n + k - 1)}{k! {(- 2)}^{n + k}} t^{k}$ $+ \frac{t^{n}}{n!} ξ (t) \Rightarrow g (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . (n + k - 1)}{k! {(- 2)}^{n + k} t^{n - k}}$ $a n d w e f o l o w t b e s a m e r o a d t o f i n d a_{k} . . . .$
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