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let-f-x-e-2x-arctanx-1-calculate-f-n-x-2-find-f-n-0-3-developp-f-at-integr-serie-

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Question Number 35235 by abdo.msup.com last updated on 17/May/18
let f(x)= e^(−2x) arctanx 1) calculate f^((n)) (x) 2) find f^((n)) (0) 3) developp f at integr serie
$${let}\:{f}\left({x}\right)=\:{e}^{−\mathrm{2}{x}} \:{arctanx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by abdo mathsup 649 cc last updated on 18/May/18
leibniz formula give f^()n)) (x)= Σ_(k=0) ^n C_n ^k (arctanx)^((k)) (e^(−2x) )^((n−k)) we have (arctanx)^((1)) = (1/(1+x^2 )) ⇒ (arctanx)^((k)) = ((1/(1+x^2 )))^((k−1)) =((1/((x−i)(x+i))))^((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)) {(((x+i)^k −(x−i)^k )/((x^2 +1)^k ))} let find (e^(−2x) )^((p)) we have (e^(−2x) )^((1)) =−2 e^(−2x) , (e^(−2x) )^((2)) =(−2)^2 e^(−2x) ....(e^(−2x) )^((p)) = (−2)^p e^(−2p) ⇒ f^((n)) (x) =arctanx (−2)^n e^(−2x) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){ (((x+i)^k −(x−i)^k )/((x^2 +1)^k ))}(−2)^(n−k) e^(−2x) 2) f^((n)) (0)= Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){((i^k −(−i)^k )/1)}(−2)^(n−k) but i^k −(−i)^k =e^(i((kπ)/2)) −e^(−i((kπ)/2)) =2i sin(((kπ)/2)) ⇒ f^((n)) (0)=Σ_(k=1) ^n C_n ^k (k−1)!(−1)^(k−1) (−2)^(n−k) sin(((kπ)/2))
$$\:{leibniz}\:{formula}\:{give}\:{f}^{\left.\right)\left.{n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({arctanx}\right)^{\left({k}\right)} \:\left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \\ $$$${we}\:{have}\:\left({arctanx}\right)^{\left(\mathrm{1}\right)} =\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left({arctanx}\right)^{\left({k}\right)} =\:\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\left({k}−\mathrm{1}\right)} =\left(\frac{\mathrm{1}}{\left({x}−{i}\right)\left({x}+{i}\right)}\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\mathrm{1}}{{x}−{i}}\:−\frac{\mathrm{1}}{{x}+{i}}\right\}^{\left({k}−\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:\left({k}−\mathrm{1}\right)!}{\left({x}−{i}\right)^{{k}} }\:\:−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+{i}\right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\left\{\frac{\left({x}+{i}\right)^{{k}} \:−\left({x}−{i}\right)^{{k}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{k}} }\right\} \\ $$$${let}\:{find}\:\left({e}^{−\mathrm{2}{x}} \right)^{\left({p}\right)} \:\:{we}\:{have} \\ $$$$\left({e}^{−\mathrm{2}{x}} \right)^{\left(\mathrm{1}\right)} =−\mathrm{2}\:{e}^{−\mathrm{2}{x}} \:\:\:,\:\:\left({e}^{−\mathrm{2}{x}} \right)^{\left(\mathrm{2}\right)} =\left(−\mathrm{2}\right)^{\mathrm{2}} \:\:\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} \\ $$$$….\left(\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} \right)^{\left(\boldsymbol{{p}}\right)} \:\:=\:\left(−\mathrm{2}\right)^{\boldsymbol{{p}}} \:\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{p}}} \Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={arctanx}\:\left(−\mathrm{2}\right)^{{n}} \:{e}^{−\mathrm{2}{x}} \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\frac{\left({x}+{i}\right)^{{k}} \:−\left({x}−{i}\right)^{{k}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{k}} }\right\}\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \\ $$$$\left.\mathrm{2}\right)\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)=\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:{C}_{{n}} ^{{k}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{{i}^{{k}} \:−\left(−{i}\right)^{{k}} }{\mathrm{1}}\right\}\left(−\mathrm{2}\right)^{{n}−{k}} \\ $$$${but}\:{i}^{{k}} \:−\left(−{i}\right)^{{k}} \:={e}^{{i}\frac{{k}\pi}{\mathrm{2}}} \:−{e}^{−{i}\frac{{k}\pi}{\mathrm{2}}} \:=\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {C}_{{n}} ^{{k}} \:\left({k}−\mathrm{1}\right)!\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left(−\mathrm{2}\right)^{{n}−{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 18/May/18
3) we have C_n ^k (k−1)! =((n!)/(k!(n−k)!))(k−1)! = ((n!)/(k(n−k)!)) the developpent of f is f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ { Σ_(k=1) ^n (1/(k(n−k)!))(−1)^(k−1) (−2)^(n−k) sin(((kπ)/2))}x^n
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{C}_{{n}} ^{{k}} \:\:\left({k}−\mathrm{1}\right)!\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}\left({k}−\mathrm{1}\right)! \\ $$$$=\:\frac{{n}!}{{k}\left({n}−{k}\right)!}\:\:{the}\:{developpent}\:{of}\:{f}\:{is} \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \: \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\left\{\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\left({n}−{k}\right)!}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left(−\mathrm{2}\right)^{{n}−{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\right\}{x}^{{n}} \\ $$

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