decompose-inside-R-x-the-fraction-F-x-1-x-3-n-x-1-with-n-integr-

Question Number 33843 by prof Abdo imad last updated on 25/Apr/18

d e c o m p o s e i n s i d e R (x) t h e f r a c t i o n

F (x) = \frac{1}{{(x + 3)}^{n} (x + 1)} w i t h n i n t e g r .

Commented by abdo imad last updated on 28/Apr/18

l e t u s e x + 3 = t \Rightarrow F (x) = g (t) = \frac{1}{t^{n} (t - 2)} l e t f i n d D_{n - 1} (0) f o r

g (t) = \frac{1}{t - 2} w e h a v e g (t) = \sum_{k = 0}^{n - 1} \frac{g^{(k)} (0)}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t) w i t h

ξ {(t)}_{t \to 0} \to 0 b u t g^{(k)} (t) = \frac{{(- 1)}^{k} k!}{{(t - 2)}^{k + 1}} \Rightarrow g^{(k)} (0) = \frac{{(- 1)}^{k} k!}{{(- 1)}^{k + 1} 2^{k + 1}}

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

g (t) = \sum_{k = 1}^{n - 1} \frac{λ_{k}}{t^{k}} + \frac{λ_{n}}{t^{n}} + \frac{a}{t - 2}

a = {l i m}_{t \to 2} (g - 2) g (t) = \frac{1}{2^{n}}

λ_{n} = {l i m}_{t \to 0} t^{n} g (t) = - \frac{1}{2} \Rightarrow g (t) = \sum_{k = 1}^{n} \frac{λ_{k}}{t^{k}} - \frac{1}{2 t^{n}} + \frac{1}{2^{n} (t - 2)}

f r o m a n o t h e r s i d e g (t) = - \frac{1}{t^{n}} \sum_{k = 0}^{n - 1} \frac{t^{k}}{2^{k + 1}} + \frac{1}{n!} ξ (t)

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

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

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

F (x) = \sum_{k = 1}^{n - 1} \frac{- 1}{2^{n - k + 1} (x + 3)} - \frac{1}{2 {(x + 3)}^{n}} + \frac{1}{2^{n} (x + 1)} .

Commented by abdo imad last updated on 28/Apr/18

★ F (x) = \sum_{k = 1}^{n - 1} \frac{- 1}{2^{n - k + 1} {(x + 3)}^{k}} - \frac{1}{2 {(x + 3)}^{n}} + \frac{1}{2^{n} (x + 1)} . ★