let f(x)= e^(−2x) actan (3x+1) 1)calculste f^((n)) (x) and f^((n)) (0) 2) if f(x)=Σ a_n x^n determine the sequence a_n 3) calculate ∫


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Question Number 126179 by mathmax by abdo last updated on 17/Dec/20

$let f (x) = e^{- 2 x} actan (3 x + 1)$ $1) calculste f^{(n)} (x) and f^{(n)} (0)$ $2) if f (x) = Σ a_{n} x^{n} determine the sequence a_{n}$ $3) calculate \int_{0}^{\infty} f (x) dx$
	Answered by mathmax by abdo last updated on 19/Dec/20
	$1) by lebniz formulae f^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(3x+1))^((k)) (e^(−2x) )^()n−k)) =C_n ^o arctan(3x+1)(−2)^n e^(−2x) +Σ_(k=1) ^n C_n ^k (arctan(3x+1))^((k)) (−2)^(n−k) e^(−2x) let calculate (arctan(3x+1))^k we have (dx/dx)(arctan(3x+1))=(3/(1+(3x+1)^2 )) ⇒(arctan(3x+1))^((k)) =3((1/((3x+1)^2 +1)))^((k−1)) =3((1/((3x+1+i)(3x+1−i))))^((k−1)) =(1/3)((1/(x+((1+i)/3))(x+((1−i)/3)))))^((k−1)) =(1/(3(((1−i)/3)−((1+i)/3))))((1/(x+((1+i)/3)))−(1/(x+((1−i)/3))))^((k−1)) =−(1/(2i))((1/(x+((√2)/3)e^((iπ)/4) ))−(1/(x+((√2)/3)e^(−((iπ)/4)) )))^((k−1)) =(i/2){ (((−1)^(k−1) (k−1)!)/((x+((√2)/3)e^((iπ)/4) )^k ))−(((−1)^(k−1) (k−1)!)/((x+((√2)/3)e^(−((iπ)/4)) )^k ))} =(i/2)(−1)^(k−1) (k−1)!{ ((−2i Im(x+((√2)/3)e^((iπ)/4) )^k )/((x+((√2)/3)e^((iπ)/4) )^k (x+((√2)/3)e^(−((iπ)/4)) )^k ))} ⇒ f^((n)) (x)=(−2)^n e^(−2x) arctan(3x+1)+Σ_(k=1) ^n C_n ^k (−2)^(n−k) e^(−2x) (−1)^k (k−1)!×((Im(x+((√2)/3)e^((iπ)/4) )^k )/((x+((√2)/3)e^((iπ)/4) )^k (x+((√2)/3)e^(−((iπ)/4)) )^k ))$
	$1) by lebniz formulae f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\arctan (3 x + 1))}^{(k)} {(e^{- 2 x})}^{) n - k)}$ $= C_{n}^{o} \arctan (3 x + 1) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} {(\arctan (3 x + 1))}^{(k)} {(- 2)}^{n - k} e^{- 2 x}$ $let calculate {(\arctan (3 x + 1))}^{k} we have$ $\frac{dx}{dx} (\arctan (3 x + 1)) = \frac{3}{1 + {(3 x + 1)}^{2}} \Rightarrow {(\arctan (3 x + 1))}^{(k)} = 3 {(\frac{1}{{(3 x + 1)}^{2} + 1})}^{(k - 1)}$ $= 3 {(\frac{1}{(3 x + 1 + i) (3 x + 1 - i)})}^{(k - 1)} = \frac{1}{3} {(\frac{1}{x + \frac{1 + i}{3}) (x + \frac{1 - i}{3})})}^{(k - 1)}$ $= \frac{1}{3 (\frac{1 - i}{3} - \frac{1 + i}{3})} {(\frac{1}{x + \frac{1 + i}{3}} - \frac{1}{x + \frac{1 - i}{3}})}^{(k - 1)}$ $= - \frac{1}{2 i} {(\frac{1}{x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}}} - \frac{1}{x + \frac{\sqrt{2}}{3} e^{- \frac{i π}{4}}})}^{(k - 1)}$ $= \frac{i}{2} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}})}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{\sqrt{2}}{3} e^{- \frac{i π}{4}})}^{k}}}$ $= \frac{i}{2} {(- 1)}^{k - 1} (k - 1)! {\frac{- 2 i Im {(x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}})}^{k}}{{(x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}})}^{k} {(x + \frac{\sqrt{2}}{3} e^{- \frac{i π}{4}})}^{k}}} \Rightarrow$ $f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} \arctan (3 x + 1) + \sum_{k = 1}^{n} C_{n}^{k} {(- 2)}^{n - k} e^{- 2 x} {(- 1)}^{k} (k - 1)! \times \frac{Im {(x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}})}^{k}}{{(x + \frac{\sqrt{2}}{3} e^{\frac{i π}{4}})}^{k} {(x + \frac{\sqrt{2}}{3} e^{- \frac{i π}{4}})}^{k}}$
	Commented by mathmax by abdo last updated on 19/Dec/20

	$f^{(n)} (0) = {(- 2)}^{n} \frac{π}{4} + \sum_{k = 1}^{n} {(- 2)}^{n - k} C_{n}^{k} {(- 1)}^{k} (k - 1)! \times \frac{{(\frac{\sqrt{2}}{3})}^{k} \sin (\frac{k π}{4})}{{(\frac{\sqrt{2}}{3})}^{k}}$ $f^{(n)} (0) = \frac{π}{4} {(- 2)}^{n} + \sum_{k = 1}^{n} {(- 1)}^{k} {(- 2)}^{n - k} C_{n}^{k} (k - 1)! \sin (\frac{k π}{4})$
	Commented by mathmax by abdo last updated on 19/Dec/20

	$2) f (x) = Σ a_{n} x^{n} \Rightarrow a_{n} = \frac{f^{(n)} (0)}{n!} and f^{(n)} (0) is known$
	Commented by mathmax by abdo last updated on 19/Dec/20

	$sorry f^{(n)} (0) = \frac{π}{4} {(- 2)}^{n} + \sum_{k = 1}^{n} {(- 2)}^{n - k} C_{n}^{k} {(- 1)}^{k} (k - 1)! {(\frac{3}{\sqrt{2}})}^{k} \sin (\frac{k π}{4})$
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