decompose inside C(x) the fraction f(x)=(1/((x^2 +1)^n ))


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Question Number 74499 by mathmax by abdo last updated on 25/Nov/19
$decompose inside C(x) the fraction f(x)=(1/((x^2 +1)^n ))$
$d e c o m p o s e i n s i d e C (x) t h e f r a c t i o n$ $f (x) = \frac{1}{{(x^{2} + 1)}^{n}}$
Commented by mathmax by abdo last updated on 28/Nov/19
$f(x)=(1/((x^2 +1)^n )) =(1/((x−i)^n (x+i)^n )) =Σ_(k=1) ^n (a_k /((x−i)^k )) +Σ_(k=1) ^n (b_k /((x+i)^k )) ch. x−i =t give f(x)=g(t)=(1/(t^n (t+2i)^n )) let find D_(n−1) (o) for h(t)=(1/((t+2i)^n )) =(t+2i)^(−n) we have h(t)=Σ_(k=0) ^(n−1) ((h^((k)) (0))/(k!)) t^k +(t^n /(n!))ξ(t) we have {(t+2i)^(−n) }^((1)) =(−n) (t+2i)^(−n−1) {(t+2i)^(−n) }^((2)) =(−n)(−n−1)(t+2i)^(−n−2) =(−1)^2 n(n+1) (t+2i)^(−n−2) { (t+2i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(t+2i)^(−n−k) ⇒ h^((k)) (0) =(−1)^k n(n+1)....(n+k−1)×(2i)^(−n−k) ⇒ h(t)=Σ_(k=0) ^(n−1) (((−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) )/(k!)) t^k +(t^n /(n!))ξ(t) ⇒ g(t)=((h(t))/t^n ) =Σ_(k=0) ^(n−1) (((−1)^k n(n+1)....(n+k−1)(2i)^(−n−k) )/(k! t^(n−k) )) +(1/(n!))ξ(t) =_(n−k =j) Σ_(j=1) ^n (((−1)^(n+j) n(n+1)....(2n−j−1)(2i)^(−2n+j) )/((n−j)! t^j )) +(1/(n!))ξ(t) t=x−i ⇒ a_j =(((−1)^(n+j) n(n+1)....(2n−j−1)(2i)^(−2n+j) )/((n−j)!)) we have f^− =f ⇒Σ_(k=1) ^n (a_k ^− /((x+i)^k )) +Σ_(k=1) ^n (b_k ^− /((x−i)^k )) =f ⇒ b_k =a_k ^− ⇒ b_j =(((−1)^(n+j) n(n+1)....(2n−j−1)(−2i)^(−2n+j) )/((n−j)!))$
$f (x) = \frac{1}{{(x^{2} + 1)}^{n}} = \frac{1}{{(x - i)}^{n} {(x + i)}^{n}} = \sum_{k = 1}^{n} \frac{a_{k}}{{(x - i)}^{k}} + \sum_{k = 1}^{n} \frac{b_{k}}{{(x + i)}^{k}}$ $c h . x - i = t g i v e f (x) = g (t) = \frac{1}{t^{n} {(t + 2 i)}^{n}} l e t f i n d D_{n - 1} (o) f o r$ $h (t) = \frac{1}{{(t + 2 i)}^{n}} = {(t + 2 i)}^{- n} w e h a v e$ $h (t) = \sum_{k = 0}^{n - 1} \frac{h^{(k)} (0)}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t)$ $w e h a v e {{(t + 2 i)}^{- n}}^{(1)} = (- n) {(t + 2 i)}^{- n - 1}$ ${{(t + 2 i)}^{- n}}^{(2)} = (- n) (- n - 1) {(t + 2 i)}^{- n - 2} = {(- 1)}^{2} n (n + 1) {(t + 2 i)}^{- n - 2}$ ${{(t + 2 i)}^{- n}}^{(k)} = {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(t + 2 i)}^{- n - k} \Rightarrow$ $h^{(k)} (0) = {(- 1)}^{k} n (n + 1) . . . . (n + k - 1) \times {(2 i)}^{- n - k} \Rightarrow$ $h (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(2 i)}^{- n - k}}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t) \Rightarrow$ $g (t) = \frac{h (t)}{t^{n}} = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) . . . . (n + k - 1) {(2 i)}^{- n - k}}{k! t^{n - k}} + \frac{1}{n!} ξ (t)$ $=_{n - k = j} \sum_{j = 1}^{n} \frac{{(- 1)}^{n + j} n (n + 1) . . . . (2 n - j - 1) {(2 i)}^{- 2 n + j}}{(n - j)! t^{j}} + \frac{1}{n!} ξ (t)$ $t = x - i \Rightarrow a_{j} = \frac{{(- 1)}^{n + j} n (n + 1) . . . . (2 n - j - 1) {(2 i)}^{- 2 n + j}}{(n - j)!}$ $w e h a v e \bar{f} = f \Rightarrow \sum_{k = 1}^{n} \frac{{\bar{a}}_{k}}{{(x + i)}^{k}} + \sum_{k = 1}^{n} \frac{{\bar{b}}_{k}}{{(x - i)}^{k}} = f \Rightarrow$ $b_{k} = {\bar{a}}_{k} \Rightarrow$ $b_{j} = \frac{{(- 1)}^{n + j} n (n + 1) . . . . (2 n - j - 1) {(- 2 i)}^{- 2 n + j}}{(n - j)!}$
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