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Question Number 99463 by mathmax by abdo last updated on 21/Jun/20

let f(x) =e^(−2x) arctan((3/x^2 )) find f^((n)) (x) and f^((n)) (1) 2) if f(x) =Σ_(n=0) ^∞ a_n (x−1)^n determinate a_n

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\mathrm{2x}} \:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\mathrm{find}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} \:\:\:\:\mathrm{determinate}\:\mathrm{a}_{\mathrm{n}} \\ $$

Answered by mathmax by abdo last updated on 23/Jun/20

leibniz give f^((n)) (x) =Σ_(k=0) ^n C_n ^k (arctan((3/x^2 )))^((k)) (e^(−2x) )^((n−k)) =arctan((3/x^2 ))(−2)^n e^(−2x) +Σ_(k=1) ^(n ) C_n ^k (−2)^(n−k) e^(−2x) (arctan((3/x^2 )))^((k)) we have (arctan((3/x^2 )))^((1)) =−((3(2x))/(x^4 (1+(9/x^4 )))) =((−6)/(x^3 (1+(9/x^4 )))) =((−6)/(x^3 +(9/x))) =((−6x)/(x^4 +9)) =((−6x)/((x^2 −3i)(x^2 +3i))) =((−6x)/((x−(√3)e^((iπ)/4) )(x+(√3)e^((iπ)/4) )(x−(√3)e^(−((iπ)/4)) )(x+(√3)e^(−((iπ)/4)) ))) =(a/(x−(√3)e^((iπ)/4) )) +(b/(x+(√3)e^((iπ)/4) )) +(c/(x−(√3)e^(−((iπ)/4)) )) +(d/(x+(√3)e^(−((iπ)/4)) )) a =((−6(√3)e^((iπ)/4) )/(2(√3)e^((iπ)/4) (6i))) =−(1/(2i)) ,b =((6(√3)e^((iπ)/4) )/(−2(√3)e^((iπ)/4) (6i))) =−(1/(2i)) c =((−6(√3)e^(−((iπ)/4)) )/(2(√3)e^(−((iπ)/4)) (−6i)))=(1/(2i)) , d =((6 e^(−((iπ)/4)) )/((−2(√3)e^(−((iπ)/4)) )(−6i))) =(1/(2i)) ⇒ (arctan((3/x^2 )))^((k)) =(1/(2i)){−((1/(x−(√3)e^((iπ)/4) )))^((k−1)) −((1/(x+(√3)e^((iπ)/4) )))^((k−1)) +((1/(x−(√3)e^(−((iπ)/4)) )))^((k−1)) +((1/(x+(√3)e^(−((iπ)/4)) )))^((k−1)) } =(((−1)^(k−1) (k−1)!)/(2i)){(1/((x−(√3)e^(−((iπ)/4)) )^k )) +(1/((x+(√3)e^(−((iπ)/4)) )^k ))−(1/((x−(√3)e^((iπ)/4) )^k ))−(1/((x+(√3)e^((iπ)/4) )^k ))} =(((−1)^(k−1) (k−1)!)/(2i)){2i Im(x+(√3)e^((iπ)/4) )^k +2i Im(x−(√3)e^(−((iπ)/4)) )} =(−1)^(k−1) (k−1)!{ Im(x+(√3) e^((iπ)/4) )^k +Im(x−(√3)e^(−((iπ)/4)) )^k } ⇒ f^((n)) (x) =(−2)^n e^(−2x) arctan((3/x^2 )) +Σ_(k=1) ^n C_n ^k (−2)^(n−k) e^(−2x) (−1)^(k−1) (k−1)!{Im(x+(√3)e^((iπ)/4) )^k +Im(x−(√3)e^(−((iπ)/4)) )^k }

$$\mathrm{leibniz}\:\mathrm{give}\:\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)\right)^{\left(\mathrm{k}\right)} \:\left(\mathrm{e}^{−\mathrm{2x}} \right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)\left(−\mathrm{2}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2x}} \:+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:\:} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(−\mathrm{2}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{e}^{−\mathrm{2x}} \:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)\right)^{\left(\mathrm{k}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)\right)^{\left(\mathrm{1}\right)} \:=−\frac{\mathrm{3}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{4}} }\right)}\:=\frac{−\mathrm{6}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{4}} }\right)}\:=\frac{−\mathrm{6}}{\mathrm{x}^{\mathrm{3}} \:+\frac{\mathrm{9}}{\mathrm{x}}} \\ $$$$=\frac{−\mathrm{6x}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{9}}\:=\frac{−\mathrm{6x}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3i}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3i}\right)}\:=\frac{−\mathrm{6x}}{\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\:+\frac{\mathrm{b}}{\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\:+\frac{\mathrm{c}}{\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\:+\frac{\mathrm{d}}{\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } \\ $$$$\mathrm{a}\:=\frac{−\mathrm{6}\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{6i}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2i}}\:\:,\mathrm{b}\:=\frac{\mathrm{6}\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{6i}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2i}} \\ $$$$\mathrm{c}\:=\frac{−\mathrm{6}\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{6i}\right)}=\frac{\mathrm{1}}{\mathrm{2i}}\:,\:\:\mathrm{d}\:=\frac{\mathrm{6}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\left(−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{6i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2i}}\:\Rightarrow \\ $$$$\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right)\right)^{\left(\mathrm{k}\right)} \:=\frac{\mathrm{1}}{\mathrm{2i}}\left\{−\left(\frac{\mathrm{1}}{\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right)^{\left(\mathrm{k}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \:+\left(\frac{\mathrm{1}}{\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \right. \\ $$$$\left.+\left(\frac{\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\frac{\mathrm{1}}{\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} }\:+\frac{\mathrm{1}}{\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} }−\frac{\mathrm{1}}{\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} }−\frac{\mathrm{1}}{\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\mathrm{2i}\:\mathrm{Im}\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} +\mathrm{2i}\:\mathrm{Im}\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!\left\{\:\mathrm{Im}\left(\mathrm{x}+\sqrt{\mathrm{3}}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} \:+\mathrm{Im}\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} \right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\left(−\mathrm{2}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2x}} \:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(−\mathrm{2}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{e}^{−\mathrm{2x}} \:\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!\left\{\mathrm{Im}\left(\mathrm{x}+\sqrt{\mathrm{3}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} \:+\mathrm{Im}\left(\mathrm{x}−\sqrt{\mathrm{3}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{k}} \right\} \\ $$

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