let-f-x-arctan-2x-x-3-1-calculate-f-n-x-snd-f-n-0-2-developp-f-at-integr-serie-

Question Number 98188 by abdomathmax last updated on 12/Jun/20

let f (x) = \frac{\arctan (2 x)}{x + 3}

1) calculate f^{(n)} (x) snd f^{(n)} (0)

2) developp f at integr serie

Answered by mathmax by abdo last updated on 12/Jun/20

1) we have f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\arctan (2 x))}^{(k)} \times {(\frac{1}{x + 3})}^{(n - k)}

= \arctan (2 x) \times \frac{{(- 1)}^{n} n!}{{(x + 3)}^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} {(\arctan (2 x))}^{(k)} \times \frac{{(- 1)}^{n - k} (n - k)!}{{(x + 3)}^{n - k + 1}}

let determine {(\arctan (2 x))}^{(k)} we have

{(\arctan (2 x))}^{(1)} = \frac{2}{1 + {4 x}^{2}} = \frac{2}{4 (x^{2} + \frac{1}{4})} = \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{2 i Im {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}}} = \frac{{(- 1)}^{k - 1} (k - 1)! Im {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x + 3)}^{n + 1}} \arctan (2 x)

+ \sum_{k = 1}^{n} {(- 1)}^{n - k} (n - k)! C_{n}^{k} \times \frac{{(- 1)}^{k - 1} (k - 1)! Im {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k} {(x + 3)}^{n - k + 1}}

= \frac{{(- 1)}^{n} n!}{{(x + 3)}^{n + 1}} \arctan (2 x) + \sum_{k = 1}^{n} {(- 1)}^{n - 1} (n - k)! \frac{n!}{k! (n - k)!} (k - 1)! \times \frac{Im {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k} {(x + 3)}^{n - k + 1}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x + 3)}^{n + 1}} \arctan (2 x) + \sum_{k = 1}^{n} {(- 1)}^{n - 1} \frac{n!}{k} \times \frac{Im {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k} {(x + 3)}^{n - k + 1}}

Commented by mathmax by abdo last updated on 13/Jun/20

f^{(n)} (0) = \sum_{k = 1}^{n} \frac{{(- 1)}^{n - 1} n!}{k} \times \frac{Im {(\frac{i}{2})}^{k}}{{(\frac{1}{4})}^{k} . 3^{n - k + 1}}

= \sum_{k = 1}^{n} \frac{{(- 1)}^{n - 1} n!}{k} . \frac{4^{k}}{3^{n - k + 1}} \times \frac{1}{2^{k}} \sin (\frac{k π}{2})

= \sum_{k = 1}^{n} \frac{{(- 1)}^{n - 1} n!}{k . 3^{n - k + 1}} . 2^{k} \sin (\frac{k π}{2})

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\sum_{k = 1}^{n} \frac{2^{k}}{k . 3^{n - k + 1}} \sin (\frac{k π}{2})) x^{n}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3^{n}} (\sum_{k = 1}^{n} \frac{2^{k} . 3^{k - 1}}{k} \sin (\frac{k π}{2})) x^{n}