decompose-tbe-fraction-F-x-1-x-n-x-1-with-n-integr-natural-

Question Number 57232 by maxmathsup by imad last updated on 31/Mar/19

d e c o m p o s e t b e f r a c t i o n F (x) = \frac{1}{x^{n} (x + 1)} w i t h n i n t e g r n a t u r a l .

Commented by maxmathsup by imad last updated on 14/Apr/19

t h e d c o m p o s i t i o n o f F i s F (x) = \frac{a}{x + 1} + \sum_{i = 1}^{n} \frac{λ_{i}}{x^{i}} l e t w (x) = \frac{1}{x + 1}

l e t f i n d D_{n - 1} (o) (w) \Rightarrow w (x) = \sum_{k = 0}^{n - 1} \frac{x^{k}}{k!} w^{(k)} (0) + \frac{x^{n}}{n!} ξ (x) (ξ (x) \to 0 f x \to 0)

b u t w^{(k)} (x) = \frac{{(- 1)}^{k} k!}{{(x + 1)}^{k + 1}} \Rightarrow w^{(k)} (0) = {(- 1)}^{k} k! \Rightarrow w (x) = \sum_{k = 0}^{n - 1} {(- 1)}^{k} x^{k} + \frac{x^{n}}{n!} ξ (x)

\Rightarrow F (x) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{x^{n - k}} + \frac{1}{n!} ξ (x) c h a n g e m e n t o f i n d i c e n - k = i g i v e

F (x) = \sum_{i = 1}^{n} \frac{{(- 1)}^{n - i}}{x^{i}} + \frac{1}{n!} ξ (x) b u t F (x) = \frac{a}{x + 1} + \sum_{i = 1}^{n} \frac{λ_{i}}{x^{i}} \Rightarrow λ_{i} = {(- 1)}^{n - i}

i \in [[1, n]] w e h a v e a = {l i m}_{x \to - 1} (x + 1) F (x) = {(- 1)}^{n} \Rightarrow

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