1-give-D-n-1-o-for-f-x-1-x-2-2-drcompose-inside-R-x-the-fraction-F-x-1-x-n-x-2-

Question Number 33335 by prof Abdo imad last updated on 14/Apr/18

1) g i v e ? D_{n - 1} (o) f o r f (x) = \frac{1}{x + 2}

2) d r 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^{n} (x + 2)}

Commented by prof Abdo imad last updated on 25/Apr/18

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

ξ {(x)}_{x \to 0} \to 0 b u t w e h a v e f^{(k)} (x) = \frac{{(- 1)}^{k} k!}{{(x + 2)}^{k + 1}} \Rightarrow

f^{(k)} (0) = \frac{{(- 1)}^{k} k!}{2^{k + 1}} \Rightarrow

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

2) F (x) = \sum_{k = 1}^{n} \frac{λ_{k}}{x^{k}} + \frac{a}{x + 2}

a = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{1}{{(- 2)}^{n}} = \frac{{(- 1)}^{n}}{2^{n} (x + 2)} \Rightarrow

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

λ_{n} = {l i m}_{x \to 0} x^{n} F (x) = \frac{1}{2} f r o m a n o t h e r s i d e

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

= \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{2^{k} x^{n - k}} + \frac{1}{n!} ξ (x) c h . o f i n d i c e n - k = p

F (x) = \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p}}{2^{n - p} x^{p}} + \frac{1}{n!} ξ (x)

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

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