decompose-inside-C-x-the-fraction-f-x-1-x-2-1-n-

Question Number 74499 by mathmax by abdo last updated on 25/Nov/19

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) = \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) \dots (n + k - 1) {(t + 2 i)}^{- n - k} \Rightarrow

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

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