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

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

$${let}\:{f}\left({x}\right)\:=\:{arctan}\left({nx}\right)\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$

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)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }\:=\frac{{n}}{{n}^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}}{{n}\left({x}−\frac{{i}}{{n}}\right)\left({x}+\frac{{i}}{{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{{n}}\frac{{n}}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\:−\frac{\mathrm{1}}{{x}+\frac{{i}}{{n}}}\right\}\:\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\:−\frac{\mathrm{1}}{{x}+\frac{{i}}{{n}}}\right\}\:\Rightarrow \\ $$$${f}^{\left({p}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\left(\frac{\mathrm{1}}{{x}−\frac{{i}}{{n}}}\right)^{\left({p}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+\frac{{i}}{{n}}}\right)^{\left({p}−\mathrm{1}\right)} \right\}=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left({x}−\frac{{i}}{{n}}\right)^{{p}} }\:−\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left({x}+\frac{{i}}{{n}}\right)^{{p}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{\left({x}+\frac{{i}}{{n}}\right)^{{p}} \:−\left({x}−\frac{{i}}{{n}}\right)^{{p}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{p}} }\right\}\:\Rightarrow{f}^{\left({p}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\frac{{n}^{\mathrm{2}{p}} }{{n}^{{p}} }\frac{\left({nx}+{i}\right)^{{p}} −\left({nx}−{i}\right)^{{p}} }{\left({x}^{\mathrm{2}} {n}^{\mathrm{2}} +\mathrm{1}\right)^{{p}} } \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\frac{\left({x}+\frac{{i}}{{n}}\right)^{{n}} \:−\left({x}−\frac{{i}}{{n}}\right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}} }\right\} \\ $$$$=\frac{{n}^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\frac{\left({nx}+{i}\right)^{{n}} −\left({nx}−{i}\right)^{{n}} }{\left({n}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }×\frac{\mathrm{1}}{{n}^{{n}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{{n}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\frac{\left({nx}+{i}\right)^{{n}} −\left({nx}−{i}\right)^{{n}} }{\left({n}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}\geqslant\mathrm{1} \\ $$$${x}=\mathrm{0}\:\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{{n}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:{i}^{{n}} −\left(−{i}\right)^{{n}} \right\} \\ $$$$=\frac{{n}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:×\mathrm{2}{iIm}\left({i}^{{n}} \right)\:=\:{n}^{{n}} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)\:=\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({p}\right)} \left(\mathrm{0}\right)}{{p}!}\:{x}^{{p}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\:\frac{{n}^{{p}} \left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{2}{i}}\:\frac{\left({nx}+{i}\right)^{{p}} −\left({nx}−{i}\right)^{{p}} }{\left({n}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{p}} }\:\frac{{x}^{{p}} }{{p}} \\ $$

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