let f(x) =e^(−2x) arctan((3/x^2 )) find f^((n)) (x) and f^((n)) (1) 2) if f(x) =Σ_(n=0) ^∞ a_n (x−1)^n determinate a


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 99463 by mathmax by abdo last updated on 21/Jun/20

$let f (x) = e^{- 2 x} \arctan (\frac{3}{x^{2}})$ $find f^{(n)} (x) and f^{(n)} (1)$ $2) if f (x) = \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n} determinate a_{n}$
	Answered by mathmax by abdo last updated on 23/Jun/20
	$leibniz give f^((n)) (x) =Σ_(k=0) ^n C_n ^k (arctan((3/x^2 )))^((k)) (e^(−2x) )^((n−k)) =arctan((3/x^2 ))(−2)^n e^(−2x) +Σ_(k=1) ^(n ) C_n ^k (−2)^(n−k) e^(−2x) (arctan((3/x^2 )))^((k)) we have (arctan((3/x^2 )))^((1)) =−((3(2x))/(x^4 (1+(9/x^4 )))) =((−6)/(x^3 (1+(9/x^4 )))) =((−6)/(x^3 +(9/x))) =((−6x)/(x^4 +9)) =((−6x)/((x^2 −3i)(x^2 +3i))) =((−6x)/((x−(√3)e^((iπ)/4) )(x+(√3)e^((iπ)/4) )(x−(√3)e^(−((iπ)/4)) )(x+(√3)e^(−((iπ)/4)) ))) =(a/(x−(√3)e^((iπ)/4) )) +(b/(x+(√3)e^((iπ)/4) )) +(c/(x−(√3)e^(−((iπ)/4)) )) +(d/(x+(√3)e^(−((iπ)/4)) )) a =((−6(√3)e^((iπ)/4) )/(2(√3)e^((iπ)/4) (6i))) =−(1/(2i)) ,b =((6(√3)e^((iπ)/4) )/(−2(√3)e^((iπ)/4) (6i))) =−(1/(2i)) c =((−6(√3)e^(−((iπ)/4)) )/(2(√3)e^(−((iπ)/4)) (−6i)))=(1/(2i)) , d =((6 e^(−((iπ)/4)) )/((−2(√3)e^(−((iπ)/4)) )(−6i))) =(1/(2i)) ⇒ (arctan((3/x^2 )))^((k)) =(1/(2i)){−((1/(x−(√3)e^((iπ)/4) )))^((k−1)) −((1/(x+(√3)e^((iπ)/4) )))^((k−1)) +((1/(x−(√3)e^(−((iπ)/4)) )))^((k−1)) +((1/(x+(√3)e^(−((iπ)/4)) )))^((k−1)) } =(((−1)^(k−1) (k−1)!)/(2i)){(1/((x−(√3)e^(−((iπ)/4)) )^k )) +(1/((x+(√3)e^(−((iπ)/4)) )^k ))−(1/((x−(√3)e^((iπ)/4) )^k ))−(1/((x+(√3)e^((iπ)/4) )^k ))} =(((−1)^(k−1) (k−1)!)/(2i)){2i Im(x+(√3)e^((iπ)/4) )^k +2i Im(x−(√3)e^(−((iπ)/4)) )} =(−1)^(k−1) (k−1)!{ Im(x+(√3) e^((iπ)/4) )^k +Im(x−(√3)e^(−((iπ)/4)) )^k } ⇒ f^((n)) (x) =(−2)^n e^(−2x) arctan((3/x^2 )) +Σ_(k=1) ^n C_n ^k (−2)^(n−k) e^(−2x) (−1)^(k−1) (k−1)!{Im(x+(√3)e^((iπ)/4) )^k +Im(x−(√3)e^(−((iπ)/4)) )^k }$
	$leibniz give f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\arctan (\frac{3}{x^{2}}))}^{(k)} {(e^{- 2 x})}^{(n - k)}$ $= \arctan (\frac{3}{x^{2}}) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} {(- 2)}^{n - k} e^{- 2 x} {(\arctan (\frac{3}{x^{2}}))}^{(k)}$ $we have {(\arctan (\frac{3}{x^{2}}))}^{(1)} = - \frac{3 (2 x)}{x^{4} (1 + \frac{9}{x^{4}})} = \frac{- 6}{x^{3} (1 + \frac{9}{x^{4}})} = \frac{- 6}{x^{3} + \frac{9}{x}}$ $= \frac{- 6 x}{x^{4} + 9} = \frac{- 6 x}{(x^{2} - 3 i) (x^{2} + 3 i)} = \frac{- 6 x}{(x - \sqrt{3} e^{\frac{i π}{4}}) (x + \sqrt{3} e^{\frac{i π}{4}}) (x - \sqrt{3} e^{- \frac{i π}{4}}) (x + \sqrt{3} e^{- \frac{i π}{4}})}$ $= \frac{a}{x - \sqrt{3} e^{\frac{i π}{4}}} + \frac{b}{x + \sqrt{3} e^{\frac{i π}{4}}} + \frac{c}{x - \sqrt{3} e^{- \frac{i π}{4}}} + \frac{d}{x + \sqrt{3} e^{- \frac{i π}{4}}}$ $a = \frac{- 6 \sqrt{3} e^{\frac{i π}{4}}}{2 \sqrt{3} e^{\frac{i π}{4}} (6 i)} = - \frac{1}{2 i}, b = \frac{6 \sqrt{3} e^{\frac{i π}{4}}}{- 2 \sqrt{3} e^{\frac{i π}{4}} (6 i)} = - \frac{1}{2 i}$ $c = \frac{- 6 \sqrt{3} e^{- \frac{i π}{4}}}{2 \sqrt{3} e^{- \frac{i π}{4}} (- 6 i)} = \frac{1}{2 i}, d = \frac{6 e^{- \frac{i π}{4}}}{(- 2 \sqrt{3} e^{- \frac{i π}{4}}) (- 6 i)} = \frac{1}{2 i} \Rightarrow$ ${(\arctan (\frac{3}{x^{2}}))}^{(k)} = \frac{1}{2 i} {- {(\frac{1}{x - \sqrt{3} e^{\frac{i π}{4}}})}^{(k - 1)} - {(\frac{1}{x + \sqrt{3} e^{\frac{i π}{4}}})}^{(k - 1)} + {(\frac{1}{x - \sqrt{3} e^{- \frac{i π}{4}}})}^{(k - 1)}$ $+ {(\frac{1}{x + \sqrt{3} e^{- \frac{i π}{4}}})}^{(k - 1)}}$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{1}{{(x - \sqrt{3} e^{- \frac{i π}{4}})}^{k}} + \frac{1}{{(x + \sqrt{3} e^{- \frac{i π}{4}})}^{k}} - \frac{1}{{(x - \sqrt{3} e^{\frac{i π}{4}})}^{k}} - \frac{1}{{(x + \sqrt{3} e^{\frac{i π}{4}})}^{k}}}$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {2 i Im {(x + \sqrt{3} e^{\frac{i π}{4}})}^{k} + 2 i Im (x - \sqrt{3} e^{- \frac{i π}{4}})}$ $= {(- 1)}^{k - 1} (k - 1)! {Im {(x + \sqrt{3} e^{\frac{i π}{4}})}^{k} + Im {(x - \sqrt{3} e^{- \frac{i π}{4}})}^{k}} \Rightarrow$ $f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} \arctan (\frac{3}{x^{2}})$ $+ \sum_{k = 1}^{n} C_{n}^{k} {(- 2)}^{n - k} e^{- 2 x} {(- 1)}^{k - 1} (k - 1)! {Im {(x + \sqrt{3} e^{\frac{i π}{4}})}^{k} + Im {(x - \sqrt{3} e^{- \frac{i π}{4}})}^{k}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum