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 42482 by maxmathsup by imad last updated on 26/Aug/18

let f(x)=e^(−2x) arctan(x) 1) calculate f^((n)) (x) 2) calculate f^((n)) (0) 3) developp f at integr serie .

$${let}\:{f}\left({x}\right)={e}^{−\mathrm{2}{x}} \:{arctan}\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$

Commented by maxmathsup by imad last updated on 29/Aug/18

1) we have by leibniz formula f^((n)) (x) = Σ_(k=0) ^n C_n ^k arctan^((k)) x(x)(e^(−2x) )^((n−k)) =(−2)^n e^(−2x) arctan(x) +Σ_(k=1) ^n C_n ^k arctan^((k)) (x) (e^(−2x) )^((n−k)) but arctan^((1)) (x) =(1/(1+x^2 )) ⇒arctan^((k)) (x)=((1/(1+x^2 )))^((k−1)) =(1/(2i)){ (1/(x−i)) −(1/(x+i))}^((k−1)) =(1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−i)^k )) −(((−1)^(k−1) (k−1)!)/((x+i)^k ))} =(((−1)^(k−1) (k−1)!)/(2i)){ (1/((x−i)^k )) −(1/((x+i)^k ))} also we have (e^(−2x) )^((n−k)) =(−2)^(n−k) e^(−2x) ⇒ f^((n)) (x) =(−2)^n e^(−2x) arctanx +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){(1/((x−i)^k )) −(1/((x+i)^k ))}(−2)^(n−k) e^(−2x)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{by}\:{leibniz}\:{formula}\:{f}^{\left({n}\right)} \left({x}\right)\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{arctan}^{\left({k}\right)} {x}\left({x}\right)\left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\left(−\mathrm{2}\right)^{{n}} \:{e}^{−\mathrm{2}{x}} \:{arctan}\left({x}\right)\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{arctan}^{\left({k}\right)} \left({x}\right)\:\left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \:{but} \\ $$$${arctan}^{\left(\mathrm{1}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow{arctan}^{\left({k}\right)} \left({x}\right)=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{{x}−{i}}\:−\frac{\mathrm{1}}{{x}+{i}}\right\}^{\left({k}−\mathrm{1}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−{i}\right)^{{k}} }\:−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+{i}\right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{\left({x}−{i}\right)^{{k}} }\:−\frac{\mathrm{1}}{\left({x}+{i}\right)^{{k}} }\right\}\:\:{also}\:{we}\:{have}\:\left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} =\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \\ $$$$\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\left(−\mathrm{2}\right)^{{n}} \:{e}^{−\mathrm{2}{x}} {arctanx}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{\left({x}−{i}\right)^{{k}} }\:−\frac{\mathrm{1}}{\left({x}+{i}\right)^{{k}} }\right\}\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 29/Aug/18

2) x=0 ⇒f^((n)) (0) = Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){ (1/((−i)^k )) −(1/i^k )}(−2)^(n−k) f^((n)) (0) = (1/(2i))Σ_(k=1) ^n (−1)^(k−1) (k−1)! ((n!)/(k!(n−k)!)){ e^((ikπ)/2) − e^(−((ikπ)/2)) } = Σ_(k=1) ^n (((−1)^(k−1) n!)/(k(n−k)!)) sin(((kπ)/2))

$$\left.\mathrm{2}\right)\:{x}=\mathrm{0}\:\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{k}}−\mathrm{1}} \left(\boldsymbol{{k}}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{\left(−{i}\right)^{{k}} }\:−\frac{\mathrm{1}}{{i}^{{k}} }\right\}\left(−\mathrm{2}\right)^{{n}−{k}} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\:\:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\left\{\:{e}^{\frac{{ik}\pi}{\mathrm{2}}} \:−\:{e}^{−\frac{{ik}\pi}{\mathrm{2}}} \right\} \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {n}!}{{k}\left({n}−{k}\right)!}\:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right) \\ $$

Commented by maxmathsup by imad last updated on 29/Aug/18

3) we have f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = f(0) +Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n (f(o)=0) =Σ_(n=1) ^∞ ( Σ_(k=1) ^n (((−1)^(k−1) )/(k(n−k)!)) sin(((kπ)/2))) x^n .

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:=\:{f}\left(\mathrm{0}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:\:\:\:\left({f}\left({o}\right)=\mathrm{0}\right) \\ $$$$\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}\left({n}−{k}\right)!}\:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\right)\:{x}^{{n}} \:\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com