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Question Number 42482 by maxmathsup by imad last updated on 26/Aug/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) c a l c u l a t e 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 maxmathsup by imad last updated on 29/Aug/18

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

= {(- 2)}^{n} e^{- 2 x} a r c t a n (x) + \sum_{k = 1}^{n} C_{n}^{k} {a r c t a n}^{(k)} (x) {(e^{- 2 x})}^{(n - k)} b u t

{a r c t a n}^{(1)} (x) = \frac{1}{1 + x^{2}} \Rightarrow {a r c t a n}^{(k)} (x) = {(\frac{1}{1 + x^{2}})}^{(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{1}{{(x - i)}^{k}} - \frac{1}{{(x + i)}^{k}}} a l s o w e h a v e {(e^{- 2 x})}^{(n - k)} = {(- 2)}^{n - k} e^{- 2 x}

\Rightarrow

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

Commented by maxmathsup by imad last updated on 29/Aug/18

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

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

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

Commented by maxmathsup by imad last updated on 29/Aug/18

3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} (f (o) = 0)

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