f-x-e-x-arctan-3-x-1-find-f-n-3-2-give-taylor-developpement-for-f-at-x-0-3-3-find-0-f-x-dx-

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 (\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 (\dots . .)