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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) = \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})