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Question Number 37277 by abdo.msup.com last updated on 11/Jun/18

l e t f (x) = \frac{1}{1 + x^{n}} w i t h n i n t e g r

1) f i n d f^{^{'}} (x) a n d f^{^{″}} (x)

2) f i n d t h e p o l e s o f f

3) c a l c u l a t e f^{(n)} (0)

4) d e v e l o p p f a t i n t e g r s e r i e .

Commented by prof Abdo imad last updated on 16/Jun/18

1) f^{^{'}} (x) = - \frac{{n x}^{n - 1}}{{(1 + x^{n})}^{2}} a n d

f^{^{″}} (x) = - \frac{n (n - 1) x^{n - 2} {(1 + x^{n})}^{2} - 2 (1 + x^{n}) {n x}^{n - 1} {n x}^{n - 1}}{{(1 + x^{n})}^{4}}

= - \frac{n (n - 1) x^{n - 2} (1 + x^{n}) - 2 n^{2} x^{2 n - 2}}{{(1 + x^{n})}^{3}}

2) z^{n} + 1 = 0 \Leftrightarrow z^{n} = e^{i π} s o i f z = {r e}^{i θ} w e g e t

r = 1 a n d n θ = (2 k + 1) π \Rightarrow θ_{k} = \frac{(2 k + 1) π}{n}

s o t h e p o l e s o f f a r e z_{k} = e^{i \frac{(2 k + 1) π}{n}}

a n d k \in [[0, n - 1]] .

Commented by prof Abdo imad last updated on 16/Jun/18

3) f (x) = \sum_{k = 0}^{n - 1} \frac{λ_{k}}{x - z_{k}}

λ_{k} = \frac{1}{n z_{k}^{n - 1}} = - \frac{z_{k}}{n} \Rightarrow f (x) = - \frac{1}{n} \sum_{k = 0}^{n - 1} \frac{z_{k}}{x - z_{k}} \Rightarrow

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

= \frac{p! {(- 1)}^{p + 1}}{n} \sum_{k = 0}^{n - 1} \frac{z_{k}}{{(x - z_{k})}^{p + 1}} \Rightarrow

f^{(p)} (0) = \frac{p! {(- 1)}^{p + 1}}{n} \sum_{k = 0}^{n - 1} \frac{z_{k}}{{(- 1)}^{p + 1} z_{k}^{p + 1}}

= \frac{p!}{n} \sum_{k = 0}^{n - 1} z_{k}^{- p} .

Commented by prof Abdo imad last updated on 16/Jun/18

f (x) = \sum_{p = 0}^{\infty} \frac{x^{p}}{p!} f^{(p)} (0)

= \sum_{p = 0}^{\infty} {\frac{p!}{n} \sum_{k = 0}^{n - 1} z_{k}^{- p}} \frac{x^{p}}{p!}

= \frac{1}{n} \sum_{p = 0}^{\infty} \sum_{k = 0}^{n - 1} z_{k}^{- p} x^{p}

b u t f o r ∣ x ∣< 1 w e h a v e a l s o

f (x) = \frac{1}{1 + x^{n}} = \sum_{p = 0}^{\infty} {(- 1)}^{p} x^{n p} .