Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 55270 by maxmathsup by imad last updated on 20/Feb/19

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

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

Commented by maxmathsup by imad last updated on 02/Mar/19

1) leibniz formula give f^((n)) (x)=Σ_(k=0) ^n C_n ^k (x^n )^((k)) (arctan(x^2 ))^((n−k)) but (x^n )^((k)) =n(n−1)...(n−k+1)x^(n−k) =((n!)/((n−k)!)) x^(n−k) if k≤n ⇒ f^((n)) (x) =Σ_(k=0) ^n ((n!)/(k!(n−k)!)) ((n!)/((n−k)!)) x^(n−k) {arctan(x^2 )}^()n−k)) =Σ_(k=0) ^n (((n!)^2 )/(k!{(n−k)!}^2 )) { arctan(x^2 )}^((n−k)) let find (arctan(x^2 ))^((m)) we have {arctan(x^2 )}^((1)) =((2x)/(1+x^4 )) =((2x)/((x^2 −i)(x^2 +i))) =((2x)/((x−(√i))(x+(√i))(x−(√(−i)))(x+(√(−i))))) =((2x)/((x−e^((iπ)/4) )(x+e^((iπ)/4) )(x−e^(−((iπ)/4)) )(x+e^(−((iπ)/4)) ))) =F(x)=(a/(x−e^((iπ)/4) )) +(b/(x+e^((iπ)/4) )) +(c/(x−e^(−((iπ)/4)) )) +(d/(x+e^(−((iπ)/4)) )) a =((2 e^((iπ)/4) )/(2e^((iπ)/4) (2i sin((π/4)))(2cos((π/4))))) = (1/(4i(1/2))) =(1/(2i)) b =((−2 e^((iπ)/4) )/((−2e^((iπ)/4) )(−2cos((π/4)))(−2i sin((π/4))))) =(1/(2i)) c =((2e^(−((iπ)/4)) )/((−2isin((π/4)))(2cos((π/4)))(2e^(−((iπ)/4)) ))) =−(1/(2i)) d =((−2e^(−((iπ)/4)) )/((−2cos((π/4)))(2i sin((π/4)))(−2e^(−((iπ)/4)) ))) =−(1/(2i)) ⇒ (arctan(x^2 ))^((1)) =(1/(2i)){(1/(x−e^((iπ)/4) )) +(1/(x+e^((iπ)/4) )) −(1/(x−e^(−((iπ)/4)) )) −(1/(x+e^(−((iπ)/4)) ))} ⇒ (arctan(x^2 ))^((m)) =(1/(2i))(−1)^(m−1) (m−1)!{ (1/((x−e^((iπ)/4) )^m )) +(1/((x+e^((iπ)/4) )^m )) −(1/((x−e^(−((iπ)/4)) )^m ))−(1/((x+e^(−((iπ)/4)) )))} ⇒ f^((n)) (x) =(1/(2i))Σ_(k=0) ^n (((n!)^2 )/(k!{(n−k)!}^2 ))(−1)^(n−k−1) (n−k−1)!{(1/((x−e^((iπ)/4) )^(n−k) ))+(1/((x+e^((iπ)/4) )^(n−k) ))−(1/((x−e^(−((iπ)/4)) )^(n−k) )) −(1/((x+e^(−((iπ)/4)) )^(n−k) ))}

$$\left.\mathrm{1}\right)\:{leibniz}\:{formula}\:{give}\:\:{f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({x}^{{n}} \right)^{\left({k}\right)} \:\:\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({n}−{k}\right)} \\ $$$${but}\:\left({x}^{{n}} \right)^{\left({k}\right)} ={n}\left({n}−\mathrm{1}\right)...\left({n}−{k}+\mathrm{1}\right){x}^{{n}−{k}} \:=\frac{{n}!}{\left({n}−{k}\right)!}\:{x}^{{n}−{k}} \:\:\:{if}\:{k}\leqslant{n}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\frac{{n}!}{\left({n}−{k}\right)!}\:{x}^{{n}−{k}} \:\left\{{arctan}\left({x}^{\mathrm{2}} \right)\right\}^{\left.\right)\left.{n}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left({n}!\right)^{\mathrm{2}} }{{k}!\left\{\left({n}−{k}\right)!\right\}^{\mathrm{2}} }\:\left\{\:{arctan}\left({x}^{\mathrm{2}} \right)\right\}^{\left({n}−{k}\right)} \:\:\:{let}\:{find}\:\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({m}\right)} \\ $$$${we}\:{have}\:\:\left\{{arctan}\left({x}^{\mathrm{2}} \right)\right\}^{\left(\mathrm{1}\right)} =\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} \:+{i}\right)}\:=\frac{\mathrm{2}{x}}{\left({x}−\sqrt{{i}}\right)\left({x}+\sqrt{{i}}\right)\left({x}−\sqrt{−{i}}\right)\left({x}+\sqrt{−{i}}\right)} \\ $$$$=\frac{\mathrm{2}{x}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:={F}\left({x}\right)=\frac{{a}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:+\frac{{b}}{{x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:+\frac{{c}}{{x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:+\frac{{d}}{{x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \\ $$$${a}\:=\frac{\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}} \\ $$$${b}\:=\frac{−\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\left(−\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}} \\ $$$${c}\:=\frac{\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\left(−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{2}{i}} \\ $$$${d}\:=\frac{−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\left(−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left(\mathrm{1}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:+\frac{\mathrm{1}}{{x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$$\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({m}\right)} =\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\mathrm{1}\right)^{{m}−\mathrm{1}} \left({m}−\mathrm{1}\right)!\left\{\:\frac{\mathrm{1}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{m}} }\:+\frac{\mathrm{1}}{\left({x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{m}} }\:−\frac{\mathrm{1}}{\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{m}} }−\frac{\mathrm{1}}{\left({x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left({n}!\right)^{\mathrm{2}} }{{k}!\left\{\left({n}−{k}\right)!\right\}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}−{k}} }+\frac{\mathrm{1}}{\left({x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}−{k}} }−\frac{\mathrm{1}}{\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}−{k}} }\:−\frac{\mathrm{1}}{\left({x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}−{k}} }\right\} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com