f(x)=e^(−x) arctan((3/x)) 1)find f^((n)) (3) 2)give taylor developpement for f at x_0 =3 3)find ∫


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Question Number 145515 by mathmax by abdo last updated on 05/Jul/21

$f (x) = e^{- x} \arctan (\frac{3}{x})$ $1) find f^{(n)} (3)$ $2) give taylor developpement for f at x_{0} = 3$ $3) find \int_{0}^{\infty} f (x) dx$
	Answered by mathmax by abdo last updated on 06/Jul/21
	$1)f(x)=e^(−x) arctan((3/x)) ⇒f^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan((3/x)))^((k)) (e^(−x) )^((n−k)) =arctan((3/x))(−1)^n e^(−x) +Σ_(k=1) ^n (−1)^(n−k) C_n ^k (arctan((3/x)))^((k)) we have (arctan((3/x)))^((1)) =((−3)/(x^2 (1+(9/x^2 ))))=((−3)/(x^2 +9)) ⇒ (arctan((3/x)))^((k)) =−3((1/(x^2 +9)))^((k−1)) =−3((1/((x−3i)(x+3i))))^((k−1)) =−(1/(2i))((1/(x−3i))−(1/(x+3i)))^((k−1)) =−(1/(2i))((((−1)^(k−1) (k−1)!)/((x−3i)^k ))−(((−1)^(k−1) (k−1)!)/((x+3i)^k ))) =(((−1)^k (k−1)!)/(2i)){(((x+3i)^k −(x−3i)^k )/((x^2 +9)^k ))} ⇒ f^((n)) (x)=(−1)^n e^(−x) arctan((3/x)) +Σ_(k=1) ^n (−1)^(n−k) C_n ^k (((−1)^k (k−1)!)/(2i(x^2 +9)^k )){(x+3i)^k −(x−3i)^k } ⇒ f^((n)) (3)=(π/4)(−1)^n e^(−x) +(1/(2i))Σ_(k=1) ^(n ) (−1)^n C_n ^k (((k−1)!)/(18^k )){3^k ((1+i)^k −(1−i)^k } =(π/4)(−1)^n e^(−x) +Σ_(k=1) ^n (((−1)^n C_n ^k (k−1)!)/6^k )((√2))^k sin(((kπ)/4))$
	$1) f (x) = e^{- x} \arctan (\frac{3}{x}) \Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\arctan (\frac{3}{x}))}^{(k)} {(e^{- x})}^{(n - k)}$ $= \arctan (\frac{3}{x}) {(- 1)}^{n} e^{- x} + \sum_{k = 1}^{n} {(- 1)}^{n - k} C_{n}^{k} {(\arctan (\frac{3}{x}))}^{(k)}$ $we have {(\arctan (\frac{3}{x}))}^{(1)} = \frac{- 3}{x^{2} (1 + \frac{9}{x^{2}})} = \frac{- 3}{x^{2} + 9} \Rightarrow$ ${(\arctan (\frac{3}{x}))}^{(k)} = - 3 {(\frac{1}{x^{2} + 9})}^{(k - 1)}$ $= - 3 {(\frac{1}{(x - 3 i) (x + 3 i)})}^{(k - 1)} = - \frac{1}{2 i} {(\frac{1}{x - 3 i} - \frac{1}{x + 3 i})}^{(k - 1)}$ $= - \frac{1}{2 i} (\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - 3 i)}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 3 i)}^{k}})$ $= \frac{{(- 1)}^{k} (k - 1)!}{2 i} {\frac{{(x + 3 i)}^{k} - {(x - 3 i)}^{k}}{{(x^{2} + 9)}^{k}}} \Rightarrow$ $f^{(n)} (x) = {(- 1)}^{n} e^{- x} \arctan (\frac{3}{x})$ $+ \sum_{k = 1}^{n} {(- 1)}^{n - k} C_{n}^{k} \frac{{(- 1)}^{k} (k - 1)!}{2 i {(x^{2} + 9)}^{k}} {{(x + 3 i)}^{k} - {(x - 3 i)}^{k}} \Rightarrow$ $f^{(n)} (3) = \frac{π}{4} {(- 1)}^{n} e^{- x}$ $+ \frac{1}{2 i} \sum_{k = 1}^{n} {(- 1)}^{n} C_{n}^{k} \frac{(k - 1)!}{18^{k}} {3^{k} ({(1 + i)}^{k} - {(1 - i)}^{k}}$ $= \frac{π}{4} {(- 1)}^{n} e^{- x} + \sum_{k = 1}^{n} \frac{{(- 1)}^{n} C_{n}^{k} (k - 1)!}{6^{k}} {(\sqrt{2})}^{k} \sin (\frac{k π}{4})$
	Commented by mathmax by abdo last updated on 06/Jul/21

	$2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (3)}{n!} {(x - 3)}^{n}$ $f^{(n)} (3) is known$
	Commented by mathmax by abdo last updated on 06/Jul/21

	$f^{(n)} (3) = \frac{π}{4} {(- 1)}^{n} e^{- 3} + Σ (. . . . .)$
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