let f(x)=e^(−2x) arctan(x) 1) calculate f^((n)) (x) 2) calculate f^((n)) (0) 3) developp f at integr serie .


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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
$1) we have by leibniz formula f^((n)) (x) = Σ_(k=0) ^n C_n ^k arctan^((k)) x(x)(e^(−2x) )^((n−k)) =(−2)^n e^(−2x) arctan(x) +Σ_(k=1) ^n C_n ^k arctan^((k)) (x) (e^(−2x) )^((n−k)) but arctan^((1)) (x) =(1/(1+x^2 )) ⇒arctan^((k)) (x)=((1/(1+x^2 )))^((k−1)) =(1/(2i)){ (1/(x−i)) −(1/(x+i))}^((k−1)) =(1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−i)^k )) −(((−1)^(k−1) (k−1)!)/((x+i)^k ))} =(((−1)^(k−1) (k−1)!)/(2i)){ (1/((x−i)^k )) −(1/((x+i)^k ))} also we have (e^(−2x) )^((n−k)) =(−2)^(n−k) e^(−2x) ⇒ f^((n)) (x) =(−2)^n e^(−2x) arctanx +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){(1/((x−i)^k )) −(1/((x+i)^k ))}(−2)^(n−k) e^(−2x)$
$1) w e h a v e b y l e i b n i z f o r m u l a f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {a r c t a n}^{(k)} x (x) {(e^{- 2 x})}^{(n - k)}$ $= {(- 2)}^{n} e^{- 2 x} a r c t a n (x) + \sum_{k = 1}^{n} C_{n}^{k} {a r c t a n}^{(k)} (x) {(e^{- 2 x})}^{(n - k)} b u t$ ${a r c t a n}^{(1)} (x) = \frac{1}{1 + x^{2}} \Rightarrow {a r c t a n}^{(k)} (x) = {(\frac{1}{1 + x^{2}})}^{(k - 1)}$ $= \frac{1}{2 i} {\frac{1}{x - i} - \frac{1}{x + i}}^{(k - 1)} = \frac{1}{2 i} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - i)}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{k}}}$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x - i)}^{k}} - \frac{1}{{(x + i)}^{k}}} a l s o w e h a v e {(e^{- 2 x})}^{(n - k)} = {(- 2)}^{n - k} e^{- 2 x}$ $\Rightarrow$ $f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} a r c t a n x + \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x - i)}^{k}} - \frac{1}{{(x + i)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}$
Commented by maxmathsup by imad last updated on 29/Aug/18
$2) x=0 ⇒f^((n)) (0) = Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){ (1/((−i)^k )) −(1/i^k )}(−2)^(n−k) f^((n)) (0) = (1/(2i))Σ_(k=1) ^n (−1)^(k−1) (k−1)! ((n!)/(k!(n−k)!)){ e^((ikπ)/2) − e^(−((ikπ)/2)) } = Σ_(k=1) ^n (((−1)^(k−1) n!)/(k(n−k)!)) sin(((kπ)/2))$
$2) x = 0 \Rightarrow f^{(n)} (0) = \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(- i)}^{k}} - \frac{1}{i^{k}}} {(- 2)}^{n - k}$ $f^{(n)} (0) = \frac{1}{2 i} \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! \frac{n!}{k! (n - k)!} {e^{\frac{i k π}{2}} - e^{- \frac{i k π}{2}}}$ $= \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} n!}{k (n - k)!} s i n (\frac{k π}{2})$
Commented by maxmathsup by imad last updated on 29/Aug/18

$3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} (f (o) = 0)$ $= \sum_{n = 1}^{\infty} (\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k (n - k)!} s i n (\frac{k π}{2})) x^{n} .$
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