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Question Number 62440 by mathsolverby Abdo last updated on 21/Jun/19

l e t h (x) = a r c t a n (x + \frac{1}{x})

1) c a l c u l a t e h^{(n)} (x) a n d h^{(n)} (1)

2) d e v e l o p p f (x) a t i n t e g r s e r i e a t x_{0} = 1

Commented by mathmax by abdo last updated on 23/Jun/19

1) w e h a v e h^{^{'}} (x) = \frac{1 - \frac{1}{x^{2}}}{1 + {(x + \frac{1}{x})}^{2}} = \frac{x^{2} - 1}{x^{2} + {(x^{2} + 1)}^{2}} = \frac{x^{2} - 1}{x^{2} + x^{4} + 2 x^{2} + 1} = \frac{x^{2} - 1}{x^{4} + 3 x^{2} + 1}

x^{2} = t \Rightarrow h^{^{'}} (x) = g (t) = \frac{t - 1}{t^{2} + 3 t + 1}

Δ = 9 - 4 = 5 \Rightarrow t_{1} = \frac{- 3 + \sqrt{5}}{2} a n d t_{2} = \frac{- 3 - \sqrt{5}}{2} \Rightarrow g (t) = \frac{t - 1}{(t - t_{1}) (t - t_{2})}

= \frac{x^{2} - 1}{(x^{2} - \frac{- 3 + \sqrt{5}}{2}) (x^{2} + \frac{3 + \sqrt{5}}{2})} = \frac{x^{2} - 1}{(x^{2} + \frac{3 - \sqrt{5}}{2}) (x^{2} + \frac{3 + \sqrt{5}}{2})}

= (x^{2} - 1) \frac{1}{\sqrt{5}} {\frac{1}{x^{2} + \frac{3 - \sqrt{5}}{2}} - \frac{1}{x^{2} + \frac{3 + \sqrt{5}}{2}}} = \frac{1}{\sqrt{5}} {\frac{x^{2} - 1}{x^{2} + \frac{3 - \sqrt{5}}{2}} - \frac{x^{2} - 1}{x^{2} + \frac{3 + \sqrt{5}}{2}}}

= \frac{1}{\sqrt{5}} {1 - \frac{1 + \frac{3 - \sqrt{5}}{2}}{x^{2} + \frac{3 - \sqrt{5}}{2}} - (1 - \frac{1 + \frac{3 + \sqrt{5}}{2}}{x^{2} + \frac{3 + \sqrt{5}}{2}})}

= \frac{5 - \sqrt{5}}{2 \sqrt{5}} {\frac{1}{x^{2} + \frac{3 - \sqrt{5}}{2}}} + \frac{5 + \sqrt{5}}{2 \sqrt{5}} {\frac{1}{x^{2} + \frac{3 + \sqrt{5}}{2}}} l e t α = \sqrt{\frac{3 - \sqrt{5}}{2}} a n d β = \sqrt{\frac{3 + \sqrt{5}}{2}}

\Rightarrow h^{^{'}} (x) = \frac{\sqrt{5} - 1}{2} {\frac{1}{x^{2} + α^{2}} + \frac{1}{x^{2} + β^{2}}}

= \frac{\sqrt{5} - 1}{2} {\frac{1}{(x + i α) (x - i α)} + \frac{1}{(x - i β) (x + i β)}}

= \frac{\sqrt{5} - 1}{2} {\frac{1}{2 i α} (\frac{1}{x - i α} - \frac{1}{x + i α}) + \frac{1}{2 i β} (\frac{1}{x - i β} - \frac{1}{x + i β})} \Rightarrow

h^{(n)} (x) = \frac{\sqrt{5} - 1}{4 i α} {{(\frac{1}{x - i α})}^{(n - 1)} - {(\frac{1}{x + i α})}^{n - 1}} + \frac{\sqrt{5} - 1}{4 i β} {{(\frac{1}{x - i β})}^{(n - 1)} - {(\frac{1}{x + i β})}^{(n - 1)}}

= \frac{\sqrt{5} - 1}{4 i α} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - i α)}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + i α)}^{n}}}

+ \frac{\sqrt{5} - 1}{4 i β} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - i β)}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + i β)}^{n}}}

= \frac{(\sqrt{5} - 1) {(- 1)}^{n - 1} (n - 1)!}{4 i} {\frac{1}{α} (\frac{1}{{(x - i α)}^{n}} - \frac{1}{{(x + i α)}^{n}}) + \frac{1}{β} (\frac{1}{{(x - i β)}^{n}} - \frac{1}{{(x + i β)}^{n}})}

= \frac{(\sqrt{5} - 1) {(- 1)}^{n - 1} (n - 1)!}{4 i} {\frac{2 i}{α} \frac{I m {(x + i α)}^{n}}{{(x^{2} + α^{2})}^{n}} + \frac{2 i}{β} \frac{I m {(x + i β)}^{n}}{{(x^{2} + β^{2})}^{n}}}

h^{(n)} (x) = \frac{\sqrt{5} - 1}{2} {(- 1)}^{n - 1} (n - 1)! {\frac{I m {(x + i α)}^{n}}{α {(x^{2} + α^{2})}^{n}} + \frac{I m {(x + i β)}^{n}}{{(x^{2} + β^{2})}^{n}}}

h^{(n)} (1) = \frac{\sqrt{5} - 1}{2} {(- 1)}^{n - 1} (n - 1)! {\frac{I m {(1 + i α)}^{n}}{α {(1 + α^{2})}^{n}} + \frac{I m {(1 + i β)}^{n}}{{(1 + β^{2})}^{n}}} .

3) h (x) = \sum_{n = 0}^{\infty} \frac{h^{(n)} (1)}{n!} {(x - 1)}^{n} a n d h^{(n)} (1) i s k n o w n .

Commented by mathmax by abdo last updated on 23/Jun/19

e r r o r o f t y p o

h^{(n)} (x) = \frac{\sqrt{5} - 1}{2} {(- 1)}^{n - 1} (n - 1)! {\frac{I m {(x + i α)}^{n}}{α {(x^{2} + α^{2})}^{n}} + \frac{I m {(x + i β)}^{n}}{β {(x^{2} + β^{2})}^{n}}} .