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


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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
	$2)we have f^′ (x) =((nx^(n−1) )/(1+x^(2n) )) poles of f^(′?) x^(2n) +1 =0 ⇒x^(2n) =e^(i(2k+1)π) ⇒x_k =e^(i(((2k+1)π)/(2n))) k∈[[0,2n−1]] ⇒f^′ (x) =((nx^(n−1) )/(Π_(k=0) ^(2n−1) (x−x_k ))) =Σ_(k=0) ^(2n−1) (a_k /(x−x_k )) , a_k =((nx_k ^(n−1) )/(2n x_k ^(2n−1) )) =(1/2)×(x_k ^n /((−1))) =−(1/2) e^(i(((2k+1)π)/2)) ⇒f^′ (x) =−(1/2) Σ_(k=0) ^(2n−1) (e^(i(((2k+1)π)/2))) /(x−x_k )) ⇒ f^((p)) (x) =−(1/2) Σ_(k=0) ^(2n−1) e^(i(πk +(π/2))) {(1/(x−x_k ))}^((p−1)) =−(1/2)Σ_(k=0) ^(2n−1) i(−1)^k ×(((−1)^(p−1) (p−1)!)/((x−x_k )^p )) =((i(−1)^p (p−1)!)/2) Σ_(k=0) ^(2n−1) (((−1)^k )/((x−x_k )^p )) p=n ⇒f^((n)) (x) =(i/2)(−1)^n (n−1)!Σ_(k=0) ^(2n−1) (((−1)^k )/((x−x_k )^n )) f^((n)) (0) =(i/2)(−1)^n (n−1)! Σ_(k=0) ^(2n−1) (((−1)^k )/((−1)^n x_k ^n )) =(i/2)(n−1)! Σ_(k=0) ^(2n−1) (((−1)^k )/(i(−1)^k )) =(1/2) Σ_(k=0) ^(2n−1) (1) =n ⇒f^((n)) (0) =n$
	$2) we have f^{^{'}} (x) = \frac{{nx}^{n - 1}}{1 + x^{2 n}} poles of f^{^{'} ?}$ $x^{2 n} + 1 = 0 \Rightarrow x^{2 n} = e^{i (2 k + 1) π} \Rightarrow x_{k} = e^{i \frac{(2 k + 1) π}{2 n}} k \in [[0, 2 n - 1]]$ $\Rightarrow f^{^{'}} (x) = \frac{{nx}^{n - 1}}{\prod_{k = 0}^{2 n - 1} (x - x_{k})} = \sum_{k = 0}^{2 n - 1} \frac{a_{k}}{x - x_{k}}, a_{k} = \frac{{nx}_{k}^{n - 1}}{2 n x_{k}^{2 n - 1}} = \frac{1}{2} \times \frac{x_{k}^{n}}{(- 1)}$ $= - \frac{1}{2} e^{i \frac{(2 k + 1) π}{2}} \Rightarrow f^{^{'}} (x) = - \frac{1}{2} \sum_{k = 0}^{2 n - 1} \frac{e^{i (\frac{2 k + 1) π}{2})}}{x - x_{k}} \Rightarrow$ $f^{(p)} (x) = - \frac{1}{2} \sum_{k = 0}^{2 n - 1} e^{i (π k + \frac{π}{2})} {\frac{1}{x - x_{k}}}^{(p - 1)}$ $= - \frac{1}{2} \sum_{k = 0}^{2 n - 1} i {(- 1)}^{k} \times \frac{{(- 1)}^{p - 1} (p - 1)!}{{(x - x_{k})}^{p}}$ $= \frac{i {(- 1)}^{p} (p - 1)!}{2} \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(x - x_{k})}^{p}}$ $p = n \Rightarrow f^{(n)} (x) = \frac{i}{2} {(- 1)}^{n} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(x - x_{k})}^{n}}$ $f^{(n)} (0) = \frac{i}{2} {(- 1)}^{n} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(- 1)}^{n} x_{k}^{n}}$ $= \frac{i}{2} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{i {(- 1)}^{k}} = \frac{1}{2} \sum_{k = 0}^{2 n - 1} (1) = n \Rightarrow f^{(n)} (0) = n$
	Commented by mathmax by abdo last updated on 03/Jun/20
	$3) f(x) =Σ_(p=0) ^∞ ((f^((p)) (0))/(p!)) x^p =Σ_(p=1) ^∞ (1/(p!)){(i/2)(−1)^p (p−1)! Σ_(k=0) ^(2n−1) (((−1)^k )/((−1)^p x_k ^p ))}x^p =(i/2) Σ_(p=1) ^∞ ( (1/p) Σ_(k=0) ^(2n−1) (−1)^k x_k ^(−p) )x^p with x_k =e^(i(((2k+1)π)/(2n))))$
	$3) f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (0)}{p!} x^{p} = \sum_{p = 1}^{\infty} \frac{1}{p!} {\frac{i}{2} {(- 1)}^{p} (p - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(- 1)}^{p} x_{k}^{p}}} x^{p}$ $= \frac{i}{2} \sum_{p = 1}^{\infty} (\frac{1}{p} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} x_{k}^{- p}) x^{p} with x_{k} = e^{i (\frac{2 k + 1) π}{2 n})}$
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