let f(x) =(x+1)^n arctan(nx) calculate f^((n)) (0).


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Question Number 62210 by maxmathsup by imad last updated on 17/Jun/19

$l e t f (x) = {(x + 1)}^{n} a r c t a n (n x)$ $c a l c u l a t e f^{(n)} (0) .$
Commented by mathmax by abdo last updated on 24/Jun/19
$let determine f^((p)) (x) leibniz formulse give f^((p)) (x) =Σ_(k=0) ^p C_p ^k {(x+1)^n }^((k)) (arctan(nx))^((p−k)) we have (X^n )^((k)) =0 si k>n (X^n )^((1)) =nX^(n−1) ⇒(X^n )^((k)) =n(n−1)..(n−k+1)X^(n−k) =((n!)/((n−k)!))X^(n−k) if k≤n and (X^n )^((n)) =n! so for p≥n we get f^((p)) (x) =Σ_(k=0) ^n C_p ^k {(x+1)^n }^((k)) {arctan(nx)}^((p−k)) +Σ_(k=n+1) ^p C_p ^k {(x+1)^n }^((k)) {arctan(nx)}^((p−k)) =Σ_(k=0) ^n C_p ^k ((n!)/((n−k)!))(x+1)^(n−k) {arctan(nx)}^((p−k)) ⇒ f^((n)) (x) =Σ_(k=0) ^n C_n ^k ((n!)/((n−k)!)){ arctan(nx)}^((n−k)) =Σ_(k=0) ^n C_n ^k k! ((n!)/(k!(n−k)!)) {arctan(nx)}^((n−k)) =Σ_(k=0) ^n k! (C_n ^k )^2 { arctan(nx)}^((n−k)) we have (arctan(nx))^((1)) =(n/(1+n^2 x^2 )) =(n/(n^2 (x^2 +(1/n^2 )))) =(1/(n(x−(i/n))(x+(i/n)))) =(1/(n(2(i/n)))){ (1/(x−(i/n))) −(1/(x+(i/n)))} =(1/(2i)){ (1/(x−(i/n))) −(1/((x+(i/n))))} ⇒ {arctan(nx)}^((p)) =(1/(2i)){ ((1/(x−(i/n))))^((p−1)) −((1/(x+(i/n))))^((p−1)) } =(1/(2i)){ (((−1)^(p−1) (p−1)!)/((x−(i/n))^p )) −(((−1)^(p−1) (p−1)!)/((x+(i/n))^p ))} ⇒ {arctan(nx)}^((n−k)) =(1/(2i))(−1)^(n−k−1) (n−k−1)!{(1/((x−(i/n))^(n−k) )) −(1/((x+(i/n))^(n−k) ))} ⇒ f^((n)) (x) =(1/(2i))Σ_(k=0) ^n k!(C_n ^k )^2 (−1)^(n−k−1) (n−k−1)!{(1/((x−(i/n))^(n−k) )) −(1/((x+(i/n))^(n−k) ))}$
$l e t d e t e r m i n e f^{(p)} (x) l e i b n i z f o r m u l s e g i v e$ $f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {(a r c t a n (n x))}^{(p - k)}$ $w e h a v e {(X^{n})}^{(k)} = 0 s i k > n$ ${(X^{n})}^{(1)} = {n X}^{n - 1} \Rightarrow {(X^{n})}^{(k)} = n (n - 1) . . (n - k + 1) X^{n - k} = \frac{n!}{(n - k)!} X^{n - k} i f k ⩽ n a n d$ ${(X^{n})}^{(n)} = n! s o f o r p ⩾ n w e g e t$ $f^{(p)} (x) = \sum_{k = 0}^{n} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {a r c t a n (n x)}^{(p - k)} + \sum_{k = n + 1}^{p} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {a r c t a n (n x)}^{(p - k)}$ $= \sum_{k = 0}^{n} C_{p}^{k} \frac{n!}{(n - k)!} {(x + 1)}^{n - k} {a r c t a n (n x)}^{(p - k)} \Rightarrow$ $f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} \frac{n!}{(n - k)!} {a r c t a n (n x)}^{(n - k)}$ $= \sum_{k = 0}^{n} C_{n}^{k} k! \frac{n!}{k! (n - k)!} {a r c t a n (n x)}^{(n - k)} = \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {a r c t a n (n x)}^{(n - k)}$ $w e h a v e {(a r c t a n (n x))}^{(1)} = \frac{n}{1 + n^{2} x^{2}} = \frac{n}{n^{2} (x^{2} + \frac{1}{n^{2}})} = \frac{1}{n (x - \frac{i}{n}) (x + \frac{i}{n})}$ $= \frac{1}{n (2 \frac{i}{n})} {\frac{1}{x - \frac{i}{n}} - \frac{1}{x + \frac{i}{n}}} = \frac{1}{2 i} {\frac{1}{x - \frac{i}{n}} - \frac{1}{(x + \frac{i}{n})}} \Rightarrow$ ${a r c t a n (n x)}^{(p)} = \frac{1}{2 i} {{(\frac{1}{x - \frac{i}{n}})}^{(p - 1)} - {(\frac{1}{x + \frac{i}{n}})}^{(p - 1)}}$ $= \frac{1}{2 i} {\frac{{(- 1)}^{p - 1} (p - 1)!}{{(x - \frac{i}{n})}^{p}} - \frac{{(- 1)}^{p - 1} (p - 1)!}{{(x + \frac{i}{n})}^{p}}} \Rightarrow$ ${a r c t a n (n x)}^{(n - k)} = \frac{1}{2 i} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(x - \frac{i}{n})}^{n - k}} - \frac{1}{{(x + \frac{i}{n})}^{n - k}}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{2 i} \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(x - \frac{i}{n})}^{n - k}} - \frac{1}{{(x + \frac{i}{n})}^{n - k}}}$
Commented by mathmax by abdo last updated on 24/Jun/19
$f^((n)) (0) =(1/(2i))Σ_(k=0) ^n k!(C_n ^k )^2 (−1)^(n−k−1) (n−k−1)!{(1/((−(i/n))^(n−k) ))−(1/(((i/n))^(n−k) ))} .$
$f^{(n)} (0) = \frac{1}{2 i} \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(- \frac{i}{n})}^{n - k}} - \frac{1}{{(\frac{i}{n})}^{n - k}}} .$
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