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

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

$${let}\:{f}\left({x}\right)\:=\left({x}+\mathrm{1}\right)^{{n}} \:{arctan}\left({nx}\right) \\ $$$${calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right). \\ $$

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) ))}

$${let}\:{determine}\:{f}^{\left({p}\right)} \left({x}\right)\:\:{leibniz}\:{formulse}\:{give} \\ $$$${f}^{\left({p}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}} ^{{k}} \:\:\:\left\{\left({x}+\mathrm{1}\right)^{{n}} \right\}^{\left({k}\right)} \:\:\left({arctan}\left({nx}\right)\right)^{\left({p}−{k}\right)} \\ $$$$\:\:{we}\:{have}\:\:\:\left({X}^{{n}} \right)^{\left({k}\right)} \:=\mathrm{0}\:{si}\:{k}>{n} \\ $$$$\left({X}^{{n}} \right)^{\left(\mathrm{1}\right)} \:={nX}^{{n}−\mathrm{1}} \:\Rightarrow\left({X}^{{n}} \right)^{\left({k}\right)} \:={n}\left({n}−\mathrm{1}\right)..\left({n}−{k}+\mathrm{1}\right){X}^{{n}−{k}} =\frac{{n}!}{\left({n}−{k}\right)!}{X}^{{n}−{k}} \:\:{if}\:{k}\leqslant{n}\:\:{and}\: \\ $$$$\left({X}^{{n}} \right)^{\left({n}\right)} \:={n}!\:\:\:{so}\:{for}\:{p}\geqslant{n}\:{we}\:{get} \\ $$$${f}^{\left({p}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{p}} ^{{k}} \:\:\left\{\left({x}+\mathrm{1}\right)^{{n}} \right\}^{\left({k}\right)} \:\left\{{arctan}\left({nx}\right)\right\}^{\left({p}−{k}\right)} \:+\sum_{{k}={n}+\mathrm{1}} ^{{p}} \:{C}_{{p}} ^{{k}} \:\left\{\left({x}+\mathrm{1}\right)^{{n}} \right\}^{\left({k}\right)} \left\{{arctan}\left({nx}\right)\right\}^{\left({p}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{p}} ^{{k}} \:\:\:\frac{{n}!}{\left({n}−{k}\right)!}\left({x}+\mathrm{1}\right)^{{n}−{k}} \:\:\left\{{arctan}\left({nx}\right)\right\}^{\left({p}−{k}\right)} \:\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{{n}!}{\left({n}−{k}\right)!}\left\{\:{arctan}\left({nx}\right)\right\}^{\left({n}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{k}!\:\:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\left\{{arctan}\left({nx}\right)\right\}^{\left({n}−{k}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{k}!\:\:\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \:\left\{\:{arctan}\left({nx}\right)\right\}^{\left({n}−{k}\right)} \\ $$$${we}\:{have}\:\left({arctan}\left({nx}\right)\right)^{\left(\mathrm{1}\right)} \:=\frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} \:{x}^{\mathrm{2}} }\:=\frac{{n}}{{n}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}}{{n}\left({x}−\frac{{i}}{{n}}\right)\left({x}+\frac{{i}}{{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{{n}\left(\mathrm{2}\frac{{i}}{{n}}\right)}\left\{\:\:\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\:−\frac{\mathrm{1}}{{x}+\frac{{i}}{{n}}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\:−\frac{\mathrm{1}}{\left({x}+\frac{{i}}{{n}}\right)}\right\}\:\Rightarrow \\ $$$$\left\{{arctan}\left({nx}\right)\right\}^{\left({p}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\left(\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\right)^{\left({p}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+\frac{{i}}{{n}}}\right)^{\left({p}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left({x}−\frac{{i}}{{n}}\right)^{{p}} }\:−\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left({x}+\frac{{i}}{{n}}\right)^{{p}} }\right\}\:\Rightarrow \\ $$$$\left\{{arctan}\left({nx}\right)\right\}^{\left({n}−{k}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left({x}−\frac{{i}}{{n}}\right)^{{n}−{k}} }\:−\frac{\mathrm{1}}{\left({x}+\frac{{i}}{{n}}\right)^{{n}−{k}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}!\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \:\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left({x}−\frac{{i}}{{n}}\right)^{{n}−{k}} }\:−\frac{\mathrm{1}}{\left({x}+\frac{{i}}{{n}}\right)^{{n}−{k}} }\right\} \\ $$

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) ))} .

$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}!\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \:\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left(−\frac{{i}}{{n}}\right)^{{n}−{k}} }−\frac{\mathrm{1}}{\left(\frac{{i}}{{n}}\right)^{{n}−{k}} }\right\}\:. \\ $$

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