let f(x) = arctan(nx) with n integr natural 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie .


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Question Number 62809 by mathmax by abdo last updated on 25/Jun/19

$l e t f (x) = a r c t a n (n x) w i t h n i n t e g r n a t u r a l$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by mathmax by abdo last updated on 02/Jul/19
$1) we have f^′ (x) =(n/(1+n^2 x^2 )) =(n/(n^2 (x^2 +(1/n^2 )))) =(1/(n(x−(i/n))(x+(i/n)))) =(1/n)(n/(2i)){(1/(x−(i/n))) −(1/(x+(i/n)))} =(1/(2i)){ (1/(x−(i/n))) −(1/(x+(i/n)))} ⇒ f^((p)) (x) =(1/(2i)){ ((1/(x−(i/n))))^((p−1)) −((1/(x+(i/n))))^((p−1)) }=(1/(2i)){ (((−1)^(p−1) (p−1)!)/((x−(i/n))^p )) −(((−1)^(p−1) (p−1)!)/((x+(i/n))^p ))} =(((−1)^(p−1) (p−1)!)/(2i)){(((x+(i/n))^p −(x−(i/n))^p )/((x^2 +(1/n^2 ))^p ))} ⇒f^((p)) (0) =(((−1)^(p−1) (p−1)!)/(2i)) (n^(2p) /n^p )(((nx+i)^p −(nx−i)^p )/((x^2 n^2 +1)^p )) f^((n)) (x) =(((−1)^(n−1) (n−1)!)/(2i)){ (((x+(i/n))^n −(x−(i/n))^n )/((x^2 +(1/n^2 ))^n ))} =((n^(2n) (−1)^(n−1) (n−1)!)/(2i)) (((nx+i)^n −(nx−i)^n )/((n^2 x^2 +1)^n ))×(1/n^n ) ⇒ f^((n)) (x)=((n^n (−1)^(n−1) (n−1)!)/(2i))(((nx+i)^n −(nx−i)^n )/((n^2 x^2 +1)^n )) with n≥1 x=0 ⇒f^((n)) (0) =((n^n (−1)^(n−1) (n−1)!)/(2i)){ i^n −(−i)^n } =((n^n (−1)^(n−1) (n−1)!)/(2i)) ×2iIm(i^n ) = n^n (−1)^(n−1) (n−1)!sin(((nπ)/2)) 2) f(x) =Σ_(p=0) ^∞ ((f^((p)) (0))/(p!)) x^p =Σ_(p=0) ^∞ ((n^p (−1)^(p−1) )/(2i)) (((nx+i)^p −(nx−i)^p )/((n^2 x^2 +1)^p )) (x^p /p)$
$1) w e h a v e f^{^{'}} (x) = \frac{n}{1 + n^{2} x^{2}} = \frac{n}{n^{2} (x^{2} + \frac{1}{n^{2}})} = \frac{1}{n (x - \frac{i}{n}) (x + \frac{i}{n})}$ $= \frac{1}{n} \frac{n}{2 i} {\frac{1}{x - \frac{i}{n}} - \frac{1}{x + \frac{i}{n}}} = \frac{1}{2 i} {\frac{1}{x - \frac{i}{n}} - \frac{1}{x + \frac{i}{n}}} \Rightarrow$ $f^{(p)} (x) = \frac{1}{2 i} {{(\frac{1}{x - \frac{i}{n}})}^{(p - 1)} - {(\frac{1}{x + \frac{i}{n}})}^{(p - 1)}} = \frac{1}{2 i} {\frac{{(- 1)}^{p - 1} (p - 1)!}{{(x - \frac{i}{n})}^{p}} - \frac{{(- 1)}^{p - 1} (p - 1)!}{{(x + \frac{i}{n})}^{p}}}$ $= \frac{{(- 1)}^{p - 1} (p - 1)!}{2 i} {\frac{{(x + \frac{i}{n})}^{p} - {(x - \frac{i}{n})}^{p}}{{(x^{2} + \frac{1}{n^{2}})}^{p}}} \Rightarrow f^{(p)} (0) = \frac{{(- 1)}^{p - 1} (p - 1)!}{2 i} \frac{n^{2 p}}{n^{p}} \frac{{(n x + i)}^{p} - {(n x - i)}^{p}}{{(x^{2} n^{2} + 1)}^{p}}$ $f^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{{(x + \frac{i}{n})}^{n} - {(x - \frac{i}{n})}^{n}}{{(x^{2} + \frac{1}{n^{2}})}^{n}}}$ $= \frac{n^{2 n} {(- 1)}^{n - 1} (n - 1)!}{2 i} \frac{{(n x + i)}^{n} - {(n x - i)}^{n}}{{(n^{2} x^{2} + 1)}^{n}} \times \frac{1}{n^{n}} \Rightarrow$ $f^{(n)} (x) = \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} \frac{{(n x + i)}^{n} - {(n x - i)}^{n}}{{(n^{2} x^{2} + 1)}^{n}} w i t h n ⩾ 1$ $x = 0 \Rightarrow f^{(n)} (0) = \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} {i^{n} - {(- i)}^{n}}$ $= \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} \times 2 i I m (i^{n}) = n^{n} {(- 1)}^{n - 1} (n - 1)! s i n (\frac{n π}{2})$ $2) f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (0)}{p!} x^{p}$ $= \sum_{p = 0}^{\infty} \frac{n^{p} {(- 1)}^{p - 1}}{2 i} \frac{{(n x + i)}^{p} - {(n x - i)}^{p}}{{(n^{2} x^{2} + 1)}^{p}} \frac{x^{p}}{p}$
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