let g(x)= e^(−2x) arctan(x+3) developp g at integr serie .


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Question Number 39892 by math khazana by abdo last updated on 13/Jul/18

$l e t g (x) = e^{- 2 x} a r c t a n (x + 3)$ $d e v e l o p p g a t i n t e g r s e r i e .$
Commented by math khazana by abdo last updated on 13/Jul/18
$we have g^((n)) (x) = Σ_(k=0) ^n (arctan(x+3))^((k)) (e^(−2x)) )^((n−k)) but arctan(x+3)^((1)) = (1/((1+(x+3)^2 )) ⇒ (arctan(x+3)}^((n)) = { (1/((x+3)^2 +1))}^((n−1)) let w(x)= (1/((x+3)^2 +1)) w(x)= (1/((x+3+i)(x+3−i))) = (a/(x+3+i)) +(b/(x+3−i)) a =lim_(x→−3−i) (x+3+i)w(z) = (1/(−3−i+3−i)) =−(1/(2i)) b=lim_(x→−3+i) (x+3−i)w(z)= (1/(−3+i+3+i)) =(1/(2i)) w(x)=(1/(2i)){ (1/(x+3−i)) −(1/(x+3+i))} ⇒(arctan(x+3))^((k)) =(1/(2i)){ ( (1/(x+3−i)))^((k−1))) −((1/(x+3+i)))^((k−1)) } = (1/(2i)) (((−1)^(k−1) (k−1)!)/((x+3−i)^k )) −(1/(2i))(((−1)^(k−1) (k−1)!)/((x+3+i)^k )) ⇒ g^((n)) (x) =Σ_(k=0) ^n (((−1)^(k−1) (k−1)!)/(2i)){ (1/((x+3−i)^k )) −(1/((x+3+i)^k ))}(−2)^(n−k) e^(−2x)$
$w e h a v e g^{(n)} (x) = \sum_{k = 0}^{n} {(a r c t a n (x + 3))}^{(k)} {(e^{- 2 x)})}^{(n - k)}$ $b u t a r c t a n {(x + 3)}^{(1)} = \frac{1}{(1 + {(x + 3)}^{2}}$ $\Rightarrow {(a r c t a n (x + 3)}}^{(n)} = {\frac{1}{{(x + 3)}^{2} + 1}}^{(n - 1)}$ $l e t w (x) = \frac{1}{{(x + 3)}^{2} + 1}$ $w (x) = \frac{1}{(x + 3 + i) (x + 3 - i)} = \frac{a}{x + 3 + i} + \frac{b}{x + 3 - i}$ $a = \lim_{x \to - 3 - i} (x + 3 + i) w (z) = \frac{1}{- 3 - i + 3 - i} = - \frac{1}{2 i}$ $b = {l i m}_{x \to - 3 + i} (x + 3 - i) w (z) = \frac{1}{- 3 + i + 3 + i} = \frac{1}{2 i}$ $w (x) = \frac{1}{2 i} {\frac{1}{x + 3 - i} - \frac{1}{x + 3 + i}}$ $\Rightarrow {(a r c t a n (x + 3))}^{(k)} = \frac{1}{2 i} {{(\frac{1}{x + 3 - i})}^{(k - 1))} - {(\frac{1}{x + 3 + i})}^{(k - 1)}}$ $= \frac{1}{2 i} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 3 - i)}^{k}} - \frac{1}{2 i} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 3 + i)}^{k}} \Rightarrow$ $g^{(n)} (x) = \sum_{k = 0}^{n} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x + 3 - i)}^{k}} - \frac{1}{{(x + 3 + i)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}$
Commented by math khazana by abdo last updated on 13/Jul/18
$g^((n)) (x) = arctan(x+3) +Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/(2i)){ (1/((x+3−i)^k )) −(1/((x+3+i)^k ))}(−2)^(n−k) e^(−2x)$
$g^{(n)} (x) = a r c t a n (x + 3) + \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x + 3 - i)}^{k}} - \frac{1}{{(x + 3 + i)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}$
Commented by math khazana by abdo last updated on 13/Jul/18
$g^((n)) (0) =arctan(3)+Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/(2i)){ (1/((3−i)^k )) −(1/((3+i)^k ))} =arctan(3) +Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/(2i)){(((3+i)^k −(3−i)^k )/(10^k ))} f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n and f^((n)) (0) is known .$
$g^{(n)} (0) = a r c t a n (3) + \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(3 - i)}^{k}} - \frac{1}{{(3 + i)}^{k}}}$ $= a r c t a n (3) + \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(3 + i)}^{k} - {(3 - i)}^{k}}{10^{k}}}$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} a n d f^{(n)} (0) i s k n o w n .$
Commented by maxmathsup by imad last updated on 13/Jul/18
$g^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(x+3))^((k)) (e^(−2x) )^((n−k)) and at the final line g^((n)) (x) =arctan(x+3) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){ (1/((x+3−i)^k )) −(1/((x+3+i)^k ))}(−2)^(n−k) e^(−2x)$
$g^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n (x + 3))}^{(k)} {(e^{- 2 x})}^{(n - k)} a n d a t t h e f i n a l l i n e$ $g^{(n)} (x) = a r c t a n (x + 3) + \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x + 3 - i)}^{k}} - \frac{1}{{(x + 3 + i)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}$
Commented by maxmathsup by imad last updated on 13/Jul/18
$g^((n)) (0)= arctan(3) +Σ_(k=1) ^n C_n ^k (−2)^(n−k) (((−1)^(k−1) (k−1)!)/(2i)){ (((3+i)^k −(3−i)^k )/(10^k ))}$
$g^{(n)} (0) = a r c t a n (3) + \sum_{k = 1}^{n} C_{n}^{k} {(- 2)}^{n - k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(3 + i)}^{k} - {(3 - i)}^{k}}{10^{k}}}$
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