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Question Number 35235 by abdo.msup.com last updated on 17/May/18

l e t f (x) = e^{- 2 x} a r c t a n x

1) c a l c u l a t e f^{(n)} (x)

2) f i n d f^{(n)} (0)

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

Commented by abdo mathsup 649 cc last updated on 18/May/18

l e i b n i z f o r m u l a g i v e f^{) n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n x)}^{(k)} {(e^{- 2 x})}^{(n - k)}

w e h a v e {(a r c t a n x)}^{(1)} = \frac{1}{1 + x^{2}} \Rightarrow

{(a r c t a n x)}^{(k)} = {(\frac{1}{1 + x^{2}})}^{(k - 1)} = {(\frac{1}{(x - i) (x + i)})}^{(k - 1)}

= \frac{1}{2 i} {\frac{1}{x - i} - \frac{1}{x + i}}^{(k - 1)}

= \frac{1}{2 i} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - i)}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{k}}}

= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(x + i)}^{k} - {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}}

l e t f i n d {(e^{- 2 x})}^{(p)} w e h a v e

{(e^{- 2 x})}^{(1)} = - 2 e^{- 2 x}, {(e^{- 2 x})}^{(2)} = {(- 2)}^{2} e^{- 2 x}

\dots . {(e^{- 2 x})}^{(p)} = {(- 2)}^{p} e^{- 2 p} \Rightarrow

f^{(n)} (x) = a r c t a n x {(- 2)}^{n} e^{- 2 x}

+ \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(x + i)}^{k} - {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}

2) f^{(n)} (0) =

\sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{i^{k} - {(- i)}^{k}}{1}} {(- 2)}^{n - k}

b u t i^{k} - {(- i)}^{k} = e^{i \frac{k π}{2}} - e^{- i \frac{k π}{2}} = 2 i s i n (\frac{k π}{2}) \Rightarrow

f^{(n)} (0) = \sum_{k = 1}^{n} C_{n}^{k} (k - 1)! {(- 1)}^{k - 1} {(- 2)}^{n - k} s i n (\frac{k π}{2})

Commented by abdo mathsup 649 cc last updated on 18/May/18

3) w e h a v e C_{n}^{k} (k - 1)! = \frac{n!}{k! (n - k)!} (k - 1)!

= \frac{n!}{k (n - k)!} t h e d e v e l o p p e n t o f f i s

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

= \sum_{n = 0}^{\infty} {\sum_{k = 1}^{n} \frac{1}{k (n - k)!} {(- 1)}^{k - 1} {(- 2)}^{n - k} s i n (\frac{k π}{2})} x^{n}