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Question Number 140638 by Mathspace last updated on 10/May/21

l e t f (x) = a r c t a n (\frac{2}{x})

d e v e l o p p f a t i n t e g r s e r i e

Answered by Dwaipayan Shikari last updated on 10/May/21

f (x) = f (0) + \frac{f^{'} (0)}{1!} x + \frac{f^{″} (0)}{2!} x^{2} + . .

f (x) = {t a n}^{- 1} (\frac{2}{x})

f^{'} (x) = - \frac{2}{x^{2}} . \frac{1}{1 + \frac{4}{x^{2}}} = \frac{- 2}{x^{2} + 4} = \frac{1}{2 i} (\frac{1}{(x + 2 i)} - \frac{1}{(x - 2 i)})

f^{″} (x) = - \frac{1}{2 i} . \frac{1}{{(x + 2 i)}^{2}} + \frac{1}{2 i} . \frac{1}{{(x - 2 i)}^{2}}

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

f^{n} (0) = \frac{1}{2 i} {(- 1)}^{n + 1} (n - 1)! \frac{1}{2^{n}} (e^{- \frac{π i n}{2}} - e^{\frac{π i n}{2}}) = \frac{{(- 1)}^{n} s i n (\frac{π n}{2})}{2^{n}} n (n - 1)!

{t a n}^{- 1} (\frac{2}{x}) = \frac{π}{2} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} s i n (\frac{π n}{2}) (n - 1)!}{2^{n} n!} x^{n}

= \frac{π}{2} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} s i n (\frac{π n}{2})}{2^{n} n} x^{n}

Answered by Dwaipayan Shikari last updated on 10/May/21

{t a n}^{- 1} (\frac{2}{x}) = \frac{π}{2} - {t a n}^{- 1} (\frac{x}{2}) = \frac{π}{2} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1)} {(\frac{x}{2})}^{2 n + 1}

Answered by mathmax by abdo last updated on 11/May/21

if x > 0 \Rightarrow f (x) = \frac{π}{2} - \arctan (\frac{x}{2}) we know

\arctan^{^{'}} u = \frac{1}{1 + u^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n} for ∣ u ∣< 1

\Rightarrow \arctan (u) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{u^{2 n + 1}}{2 n + 1} \Rightarrow

f (x) = \frac{π}{2} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{{(\frac{x}{2})}^{2 n + 1}}{2 n + 1} = \frac{π}{2} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1) 2^{2 n + 1}}

if x < 0 \Rightarrow f (x) = - \frac{π}{2} - \arctan (\frac{x}{2})

= - \frac{π}{2} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1) 2^{2 n + 1}} with ∣ x ∣< 2