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Question Number 145515 by mathmax by abdo last updated on 05/Jul/21

f(x)=e^(−x) arctan((3/x)) 1)find f^((n)) (3) 2)give taylor developpement for f at x_0 =3 3)find ∫_0 ^∞ f(x)dx

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{−\mathrm{x}} \mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{1}\right)\mathrm{find}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{3}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{give}\:\mathrm{taylor}\:\mathrm{developpement}\:\mathrm{for}\:\mathrm{f}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{3} \\ $$$$\left.\mathrm{3}\right)\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Answered by mathmax by abdo last updated on 06/Jul/21

1)f(x)=e^(−x) arctan((3/x)) ⇒f^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan((3/x)))^((k)) (e^(−x) )^((n−k)) =arctan((3/x))(−1)^n e^(−x) +Σ_(k=1) ^n (−1)^(n−k) C_n ^k (arctan((3/x)))^((k)) we have (arctan((3/x)))^((1)) =((−3)/(x^2 (1+(9/x^2 ))))=((−3)/(x^2 +9)) ⇒ (arctan((3/x)))^((k)) =−3((1/(x^2 +9)))^((k−1)) =−3((1/((x−3i)(x+3i))))^((k−1)) =−(1/(2i))((1/(x−3i))−(1/(x+3i)))^((k−1)) =−(1/(2i))((((−1)^(k−1) (k−1)!)/((x−3i)^k ))−(((−1)^(k−1) (k−1)!)/((x+3i)^k ))) =(((−1)^k (k−1)!)/(2i)){(((x+3i)^k −(x−3i)^k )/((x^2 +9)^k ))} ⇒ f^((n)) (x)=(−1)^n e^(−x) arctan((3/x)) +Σ_(k=1) ^n (−1)^(n−k) C_n ^k (((−1)^k (k−1)!)/(2i(x^2 +9)^k )){(x+3i)^k −(x−3i)^k } ⇒ f^((n)) (3)=(π/4)(−1)^n e^(−x) +(1/(2i))Σ_(k=1) ^(n ) (−1)^n C_n ^k (((k−1)!)/(18^k )){3^k ((1+i)^k −(1−i)^k } =(π/4)(−1)^n e^(−x) +Σ_(k=1) ^n (((−1)^n C_n ^k (k−1)!)/6^k )((√2))^k sin(((kπ)/4))

$$\left.\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{−\mathrm{x}} \:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\right)^{\left(\mathrm{k}\right)} \:\left(\mathrm{e}^{−\mathrm{x}} \right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{x}} \:+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\right)^{\left(\mathrm{k}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\right)^{\left(\mathrm{1}\right)} \:=\frac{−\mathrm{3}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }\right)}=\frac{−\mathrm{3}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{9}}\:\Rightarrow \\ $$$$\left(\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\right)^{\left(\mathrm{k}\right)} \:=−\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{9}}\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \\ $$$$=−\mathrm{3}\left(\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{3i}\right)\left(\mathrm{x}+\mathrm{3i}\right)}\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \:=−\frac{\mathrm{1}}{\mathrm{2i}}\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{3i}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{3i}}\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2i}}\left(\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\left(\mathrm{x}−\mathrm{3i}\right)^{\mathrm{k}} }−\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\left(\mathrm{x}+\mathrm{3i}\right)^{\mathrm{k}} }\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\frac{\left(\mathrm{x}+\mathrm{3i}\right)^{\mathrm{k}} −\left(\mathrm{x}−\mathrm{3i}\right)^{\mathrm{k}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{k}} }\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right) \\ $$$$+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{2i}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{k}} }\left\{\left(\mathrm{x}+\mathrm{3i}\right)^{\mathrm{k}} −\left(\mathrm{x}−\mathrm{3i}\right)^{\mathrm{k}} \right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{3}\right)=\frac{\pi}{\mathrm{4}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{x}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2i}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\:\:\frac{\left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{18}^{\mathrm{k}} }\left\{\mathrm{3}^{\mathrm{k}} \left(\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{k}} −\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{k}} \right\}\right. \\ $$$$=\frac{\pi}{\mathrm{4}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{x}} \:+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{6}^{\mathrm{k}} }\left(\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{4}}\right) \\ $$

Commented by mathmax by abdo last updated on 06/Jul/21

2)f(x)=Σ_(n=0) ^∞ ((f^((n)) (3))/(n!))(x−3)^n f^((n)) (3)is known

$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{3}\right)}{\mathrm{n}!}\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{n}} \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{3}\right)\mathrm{is}\:\mathrm{known} \\ $$

Commented by mathmax by abdo last updated on 06/Jul/21

f^((n)) (3)=(π/4)(−1)^n e^(−3) +Σ(.....)

$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{3}\right)=\frac{\pi}{\mathrm{4}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{3}} \:+\Sigma\left(.....\right) \\ $$

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