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Question Number 96593 by mathmax by abdo last updated on 03/Jun/20

let f (x) = \arctan (x^{n}) with n integr natural

1) calculate f^{^{'}} (x) and f^{(2)} (x)

2) calculate f^{(n)} (x) and f^{(n)} (0)

3) developp f at integr serie

Answered by mathmax by abdo last updated on 03/Jun/20

1) we have f (x) = \arctan (x^{n}) \Rightarrow f^{^{'}} (x) = \frac{{nx}^{n - 1}}{1 + x^{2 n}}

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

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

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

Answered by mathmax by abdo last updated on 03/Jun/20

2) we have f^{^{'}} (x) = \frac{{nx}^{n - 1}}{1 + x^{2 n}} poles of f^{^{'} ?}

x^{2 n} + 1 = 0 \Rightarrow x^{2 n} = e^{i (2 k + 1) π} \Rightarrow x_{k} = e^{i \frac{(2 k + 1) π}{2 n}} k \in [[0, 2 n - 1]]

\Rightarrow f^{^{'}} (x) = \frac{{nx}^{n - 1}}{\prod_{k = 0}^{2 n - 1} (x - x_{k})} = \sum_{k = 0}^{2 n - 1} \frac{a_{k}}{x - x_{k}}, a_{k} = \frac{{nx}_{k}^{n - 1}}{2 n x_{k}^{2 n - 1}} = \frac{1}{2} \times \frac{x_{k}^{n}}{(- 1)}

= - \frac{1}{2} e^{i \frac{(2 k + 1) π}{2}} \Rightarrow f^{^{'}} (x) = - \frac{1}{2} \sum_{k = 0}^{2 n - 1} \frac{e^{i (\frac{2 k + 1) π}{2})}}{x - x_{k}} \Rightarrow

f^{(p)} (x) = - \frac{1}{2} \sum_{k = 0}^{2 n - 1} e^{i (π k + \frac{π}{2})} {\frac{1}{x - x_{k}}}^{(p - 1)}

= - \frac{1}{2} \sum_{k = 0}^{2 n - 1} i {(- 1)}^{k} \times \frac{{(- 1)}^{p - 1} (p - 1)!}{{(x - x_{k})}^{p}}

= \frac{i {(- 1)}^{p} (p - 1)!}{2} \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(x - x_{k})}^{p}}

p = n \Rightarrow f^{(n)} (x) = \frac{i}{2} {(- 1)}^{n} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(x - x_{k})}^{n}}

f^{(n)} (0) = \frac{i}{2} {(- 1)}^{n} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(- 1)}^{n} x_{k}^{n}}

= \frac{i}{2} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{i {(- 1)}^{k}} = \frac{1}{2} \sum_{k = 0}^{2 n - 1} (1) = n \Rightarrow f^{(n)} (0) = n

Commented by mathmax by abdo last updated on 03/Jun/20

3) f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (0)}{p!} x^{p} = \sum_{p = 1}^{\infty} \frac{1}{p!} {\frac{i}{2} {(- 1)}^{p} (p - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(- 1)}^{p} x_{k}^{p}}} x^{p}

= \frac{i}{2} \sum_{p = 1}^{\infty} (\frac{1}{p} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} x_{k}^{- p}) x^{p} with x_{k} = e^{i (\frac{2 k + 1) π}{2 n})}