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


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Question Number 94336 by mathmax by abdo last updated on 19/May/20

$let f (x) = \arctan (2 x) e^{- 3 x}$ $1) determine f^{(n)} (x) and f^{(n)} (0)$ $2) developp f at integr serie$
	Answered by prakash jain last updated on 18/May/20

	$e^{x} = \underset{k = 0}{\sum^{\infty}} \frac{x^{k}}{k!}$ $We need \frac{x^{4 n}}{(4 n)!} which is every 4^{th}$ $power .$ $Noting that i^{4} = 1 we start with$ $e^{i x} = 1 + i x - \frac{x^{2}}{2!} - i \frac{x^{3}}{3!} + \frac{x^{4}}{4} + i \frac{x^{5}}{5!} + . .$ $add e^{- i x} to remove all odd power$ $e^{i x} + e^{- i x} = 2 (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} . .)$ $To get required terms we need to$ $add all even power of x$ $e^{x} + e^{- x} = 2 (1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + - . . .)$ $sum$ $e^{i x} + e^{- i x} + e^{x} + e^{- x} = 4 (1 + \frac{x^{4}}{4!} + \frac{x^{8}}{8!} . . .)$ $\underset{k = 0}{\sum^{\infty}} \frac{x^{4 n}}{(4 n)!} = \frac{1}{2} (\cos x + \cosh x)$
	Commented by abdomathmax last updated on 18/May/20

	$thank you sir prakach$
	Commented by mathmax by abdo last updated on 20/May/20

	$the question here is calculate \sum_{n = 0}^{\infty} \frac{x^{4 n}}{(4 n)!}$
	Commented by prakash jain last updated on 20/May/20

	$May be I answered agaimst a wrong$ $question . I was answering$ $\underset{n = 0}{\sum^{\infty}} \frac{x^{4 n}}{(4 m)!}$
	Answered by mathmax by abdo last updated on 20/May/20
	$1) we have f(x) =e^(−3x) arctan(2x) ⇒ f^((n)) (x) =Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) (e^(−3x) )^((n−k)) =(−3)^n e^(−3x) arctan(2x) +Σ_(k=1) ^n C_n ^k (arctan(2x))^((k)) (−3)^(n−k) e^(−3x) we have (arctan(2x))^((1)) =(2/(1+4x^2 )) =(2/(4((1/4)+x^2 ))) =(1/(2(x−(i/2))(x+(i/2)))) =(1/(2i))((1/(x−(i/2)))−(1/(x+(i/2)))) ⇒(arctan(2x))^((k)) =(1/(2i)){((1/(x−(i/2))))^((k−1)) −((1/(x+(i/2))))^((k−1)) } =(1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−(i/2))^k ))−(((−1)^(k−1) (k−1)!)/((x+(i/2))^k ))} =(((−1)^(k−1) (k−1)!)/(2i))( (((x+(i/2))^k −(x−(i/2))^k )/((x^2 +(1/4))^k ))) =(−1)^(k−1) (k−1)!×((Im( (x+(i/2))^k ))/((x^2 +(1/4))^k )) ⇒ f^((n)) (x) =(−3)^n e^(−3x) arctan(2x) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)! .Im((x+(i/2))^k ))/((x^2 +(1/4))^k ))×(−3)^(n−k) e^(−3x) ⇒ f^((n)) (0) =Σ_(k=1) ^n C_n ^k 4^k (−1)^(k−1) (k−1)! ×(1/2^k )sin(((kπ)/2)) (−3)^(n−k) ⇒f^((n)) (0) =Σ_(k=1) ^n 2^k (−1)^(k−1) (k−1)! ((n!)/(k!(n−k)!)) sin(((kπ)/2))(−3)^(n−k) =Σ_(k=1) ^n (−1)^(k−1) 2^k ((n!)/(k(n−k)!))(−3)^(n−k) 2) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n ⇒ f(x) =Σ_(n=1) ^∞ (Σ_(k=1) ^n (−1)^(k−1) 2^k ×(1/(k(n−k)!))(−3)^(n−k) )x^n =Σ_(n=1) ^∞ ( Σ_(k=1) ^n (((−1)^(k−1) 2^k (−3)^(n−k) )/(k .(n−k)!))) x^n$
	$1) we have f (x) = e^{- 3 x} \arctan (2 x) \Rightarrow$ $f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\arctan (2 x))}^{(k)} {(e^{- 3 x})}^{(n - k)}$ $= {(- 3)}^{n} e^{- 3 x} \arctan (2 x) + \sum_{k = 1}^{n} C_{n}^{k} {(\arctan (2 x))}^{(k)} {(- 3)}^{n - k} e^{- 3 x}$ $we have {(\arctan (2 x))}^{(1)} = \frac{2}{1 + {4 x}^{2}} = \frac{2}{4 (\frac{1}{4} + x^{2})} = \frac{1}{2 (x - \frac{i}{2}) (x + \frac{i}{2})}$ $= \frac{1}{2 i} (\frac{1}{x - \frac{i}{2}} - \frac{1}{x + \frac{i}{2}}) \Rightarrow {(\arctan (2 x))}^{(k)} = \frac{1}{2 i} {{(\frac{1}{x - \frac{i}{2}})}^{(k - 1)} - {(\frac{1}{x + \frac{i}{2}})}^{(k - 1)}}$ $= \frac{1}{2 i} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - \frac{i}{2})}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{i}{2})}^{k}}}$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} (\frac{{(x + \frac{i}{2})}^{k} - {(x - \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}})$ $= {(- 1)}^{k - 1} (k - 1)! \times \frac{Im ({(x + \frac{i}{2})}^{k})}{{(x^{2} + \frac{1}{4})}^{k}} \Rightarrow$ $f^{(n)} (x) = {(- 3)}^{n} e^{- 3 x} \arctan (2 x)$ $+ \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)! . Im ({(x + \frac{i}{2})}^{k})}{{(x^{2} + \frac{1}{4})}^{k}} \times {(- 3)}^{n - k} e^{- 3 x}$ $\Rightarrow f^{(n)} (0) = \sum_{k = 1}^{n} C_{n}^{k} 4^{k} {(- 1)}^{k - 1} (k - 1)! \times \frac{1}{2^{k}} \sin (\frac{k π}{2}) {(- 3)}^{n - k}$ $\Rightarrow f^{(n)} (0) = \sum_{k = 1}^{n} 2^{k} {(- 1)}^{k - 1} (k - 1)! \frac{n!}{k! (n - k)!} \sin (\frac{k π}{2}) {(- 3)}^{n - k}$ $= \sum_{k = 1}^{n} {(- 1)}^{k - 1} 2^{k} \frac{n!}{k (n - k)!} {(- 3)}^{n - k}$ $2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \Rightarrow$ $f (x) = \sum_{n = 1}^{\infty} (\sum_{k = 1}^{n} {(- 1)}^{k - 1} 2^{k} \times \frac{1}{k (n - k)!} {(- 3)}^{n - k}) x^{n}$ $= \sum_{n = 1}^{\infty} (\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} 2^{k} {(- 3)}^{n - k}}{k . (n - k)!}) x^{n}$
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