let-f-x-e-2x-arctan-3-x-2-find-f-n-x-and-f-n-1-2-if-f-x-n-0-a-n-x-1-n-determinate-a-n-

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) = \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}}