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Question Number 135384 by Bird last updated on 12/Mar/21
let f(x)=ln(2+x^3 ) if f(x)=Σa_n x^n find a_n
$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{2}+{x}^{\mathrm{3}} \right) \\ $$$${if}\:{f}\left({x}\right)=\Sigma{a}_{{n}} {x}^{{n}} \\ $$$${find}\:{a}_{{n}} \\ $$
Answered by Dwaipayan Shikari last updated on 12/Mar/21
f(x)=Σ_(n=0) ^∞ ((f^n (0))/(n!))x^n a_n =((f^n (0))/(n!)) f(x)=log(α^3 +x^3 )=log(α+x)+log(α^2 −αx+x^2 ) =log(α+x)+log(x−β)+log(x−γ) β=αe^((2πi)/3) γ=αe^((−2πi)/3) f′(x)=(1/((α+x)))+(1/(x−β))+(1/(x−γ)) f′′(x)=((−1)/((α+x)^2 ))+((−1)/((x−β)^2 ))−(1/((x−γ)^2 )) f^n (x)=(−1)^(n+1) (n−1)!((1/((α+x)^n ))+(1/((x−β)^n ))+(1/((x−γ)^n ))) =(−1)^(n+1) (n−1)!((1/((α+x)^n ))+(((x−β)^n +(x−γ)^n )/((α^2 −αx+x^2 )^n ))) a_n =(((−1)^(n+1) )/n)((1/α^n )+(((−1)^n (e^((2πin)/3) +e^(−((2πin)/3)) ))/((α^2 )^n ))) α=(2)^(1/3) a_n =(((−1)^(n+1) )/n)((1/2^(n/3) )+((2(−1)^n cos(((2πn)/3)))/(((4)^(1/3) )^n )))
$${f}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} \\ $$$${a}_{{n}} =\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!} \\ $$$${f}\left({x}\right)={log}\left(\alpha^{\mathrm{3}} +{x}^{\mathrm{3}} \right)={log}\left(\alpha+{x}\right)+{log}\left(\alpha^{\mathrm{2}} −\alpha{x}+{x}^{\mathrm{2}} \right) \\ $$$$={log}\left(\alpha+{x}\right)+{log}\left({x}−\beta\right)+{log}\left({x}−\gamma\right)\:\:\:\:\:\:\:\:\beta=\alpha{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \:\gamma=\alpha{e}^{\frac{−\mathrm{2}\pi{i}}{\mathrm{3}}} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\left(\alpha+{x}\right)}+\frac{\mathrm{1}}{{x}−\beta}+\frac{\mathrm{1}}{{x}−\gamma} \\ $$$${f}''\left({x}\right)=\frac{−\mathrm{1}}{\left(\alpha+{x}\right)^{\mathrm{2}} }+\frac{−\mathrm{1}}{\left({x}−\beta\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}−\gamma\right)^{\mathrm{2}} } \\ $$$${f}^{{n}} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({n}−\mathrm{1}\right)!\left(\frac{\mathrm{1}}{\left(\alpha+{x}\right)^{{n}} }+\frac{\mathrm{1}}{\left({x}−\beta\right)^{{n}} }+\frac{\mathrm{1}}{\left({x}−\gamma\right)^{{n}} }\right) \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({n}−\mathrm{1}\right)!\left(\frac{\mathrm{1}}{\left(\alpha+{x}\right)^{{n}} }+\frac{\left({x}−\beta\right)^{{n}} +\left({x}−\gamma\right)^{{n}} }{\left(\alpha^{\mathrm{2}} −\alpha{x}+{x}^{\mathrm{2}} \right)^{{n}} }\right) \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\left(\frac{\mathrm{1}}{\alpha^{{n}} }+\frac{\left(−\mathrm{1}\right)^{{n}} \left({e}^{\frac{\mathrm{2}\pi{in}}{\mathrm{3}}} +{e}^{−\frac{\mathrm{2}\pi{in}}{\mathrm{3}}} \right)}{\left(\alpha^{\mathrm{2}} \right)^{{n}} }\right)\:\:\:\:\alpha=\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\left(\frac{\mathrm{1}}{\mathrm{2}^{\frac{{n}}{\mathrm{3}}} }+\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {cos}\left(\frac{\mathrm{2}\pi{n}}{\mathrm{3}}\right)}{\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{{n}} }\right) \\ $$
Commented by mathmax by abdo last updated on 12/Mar/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Olaf last updated on 12/Mar/21
f(x) = ln(2+x^3 ) f′(x) = ((3x^2 )/(2+x^3 )) = (3/2).(x^2 /(1+((x/2^(1/3) ))^3 )) f′(x) = (3/2)x^2 Σ_(n=0) ^∞ (−1)^n ((x/2^(1/3) ))^n f′(x) = (3/2)Σ_(n=0) ^∞ (((−1)^n )/2^(n/3) )x^(n+2) f(x) = (3/2)Σ_(n=0) ^∞ (((−1)^n )/((n+3)2^(n/3) ))x^(n+3) +C f(0) = ln(2+0) = ln2 ⇒ C = ln2 f(x) = 3Σ_(n=3) ^∞ (((−1)^n )/(n2^n ))x^n +ln2 a_0 = ln2, a_1 = a_2 = 0 and a_n = ((3(−1)^n )/(n2^n )), n≥3
$${f}\left({x}\right)\:=\:\mathrm{ln}\left(\mathrm{2}+{x}^{\mathrm{3}} \right) \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{3}}{\mathrm{2}}.\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{x}}{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }\right)^{\mathrm{3}} } \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{{x}}{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }\right)^{{n}} \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}/\mathrm{3}} }{x}^{{n}+\mathrm{2}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{3}\right)\mathrm{2}^{{n}/\mathrm{3}} }{x}^{{n}+\mathrm{3}} +\mathrm{C} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{ln}\left(\mathrm{2}+\mathrm{0}\right)\:=\:\mathrm{ln2}\:\Rightarrow\:\mathrm{C}\:=\:\mathrm{ln2} \\ $$$${f}\left({x}\right)\:=\:\mathrm{3}\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\mathrm{2}^{{n}} }{x}^{{n}} +\mathrm{ln2} \\ $$$${a}_{\mathrm{0}} \:=\:\mathrm{ln2},\:{a}_{\mathrm{1}} \:=\:{a}_{\mathrm{2}} \:=\:\mathrm{0}\:\mathrm{and}\:{a}_{{n}} \:=\:\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{n}} }{{n}\mathrm{2}^{{n}} },\:{n}\geqslant\mathrm{3} \\ $$
Commented by mathmax by abdo last updated on 12/Mar/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\: \\ $$
Answered by mathmax by abdo last updated on 14/Mar/21
a_n =((f^()n)) (0))/(n!)) let determine f^()n)) (0) we have f^′ (x)=((3x^2 )/(x^3 +2)) let α=^3 (√2) ⇒f^′ (x)=((3x^2 )/(x^3 +α^3 ))=((3x^2 )/((x+α)(x^2 −αx +α^2 ))) =(a/(x+α))+((bx +c)/(x^2 −αx +α^2 )) we have a=((3α^2 )/(3α^2 ))=(3/2) lim_(x→+∞) xf^′ (x)=3 =a+b ⇒b=3−(3/2)=(3/2) f^′ (0)=0 =(a/α)+(c/α^2 ) ⇒o=αa +c ⇒c=−α.(3/2) ⇒ f^′ (x)=(3/(2(x+α)))+(((3/2)x−((3α)/2))/(x^2 −αx +α^2 )) =(3/(2(x+α)))+(3/2)((x−α)/(x^2 −αx +α^2 )) let decompose inside C(x) the fraction F(x)=((x−α)/(x^2 −αx +α^2 )) Δ=α^2 −4α^2 =−3α^2 ⇒z_1 =((α+iα(√3))/2) =α((1/2)+i((√3)/2))=α e^((iπ)/3) z_2 =αe^(−((iπ)/3)) ⇒F(x)=(a/(x−z_1 ))+(b/(x−z_2 ))(=((x−α)/((x−z_1 )(x−z_2 )))) a =((z_1 −α)/(z_1 −z_2 ))=((z_1 −α)/(2iα((√3)/2))) =(1/(iα(√3)))(z_1 −α) b =((z_2 −α)/(z_2 −z_1 )) =−(1/(iα(√3)))(z_2 −α) ⇒ f^′ (x)=(3/(2(x+α)))+((3(z_1 −α))/(2iα(√3)(x−z_1 )))−((3(z_2 −α))/(2iα(√3)(x−z_2 ))) ⇒ f^((n)) (x)=(3/2){(1/(x+α))}^((n−1)) +((3(z_1 −α))/(2i(√3))){(1/(x−z_1 ))}^((n−1)) −((3(z_2 −α))/(2i(√3))){(1/(x−z_2 ))}^((n−1)) =(3/2)(((−1)^(n−1) (n−1)!)/((x+α)^n ))+((3(z_1 −α))/(2i(√3)))×(((−1)^(n−1) (n−1)!)/((x−z_1 )^n )) −((3(z_2 −α))/(2i(√3)))×(((−1)^n (n−1)!)/((x−z_2 )^n )) ⇒ a_n =(1/(n!))f^((n)) (0) =(1/(n!)){(3/2)α^(−n) (−1)^(n−1) (n−1)!+((3(z_1 −α))/(2i(√3)))(−z_1 )^(−n) (−1)^(n−1) (n−1)! −((3(z_2 −α))/(2i(√3)))(−z_2 )^(−n) (−1)^n (n−1)!}
$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{f}^{\left.\right)\left.\mathrm{n}\right)} \left(\mathrm{0}\right)}{\mathrm{n}!}\:\:\mathrm{let}\:\mathrm{determine}\:\mathrm{f}^{\left.\right)\left.\mathrm{n}\right)} \left(\mathrm{0}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} \:+\mathrm{2}} \\ $$$$\mathrm{let}\:\alpha=^{\mathrm{3}} \sqrt{\mathrm{2}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} \:+\alpha^{\mathrm{3}} }=\frac{\mathrm{3x}^{\mathrm{2}} }{\left(\mathrm{x}+\alpha\right)\left(\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}\:+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{x}+\alpha}+\frac{\mathrm{bx}\:+\mathrm{c}}{\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}\:+\alpha^{\mathrm{2}} }\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}=\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{3}\alpha^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xf}^{'} \left(\mathrm{x}\right)=\mathrm{3}\:=\mathrm{a}+\mathrm{b}\:\Rightarrow\mathrm{b}=\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{f}^{'} \left(\mathrm{0}\right)=\mathrm{0}\:=\frac{\mathrm{a}}{\alpha}+\frac{\mathrm{c}}{\alpha^{\mathrm{2}} }\:\Rightarrow\mathrm{o}=\alpha\mathrm{a}\:+\mathrm{c}\:\Rightarrow\mathrm{c}=−\alpha.\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{x}+\alpha\right)}+\frac{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{3}\alpha}{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}\:+\alpha^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{x}+\alpha\right)}+\frac{\mathrm{3}}{\mathrm{2}}\frac{\mathrm{x}−\alpha}{\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}\:+\alpha^{\mathrm{2}} } \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{inside}\:\mathrm{C}\left(\mathrm{x}\right)\:\mathrm{the}\:\mathrm{fraction}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}−\alpha}{\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}\:+\alpha^{\mathrm{2}} } \\ $$$$\Delta=\alpha^{\mathrm{2}} −\mathrm{4}\alpha^{\mathrm{2}} \:=−\mathrm{3}\alpha^{\mathrm{2}} \:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{\alpha+\mathrm{i}\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\alpha\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\alpha\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\alpha\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}−\mathrm{z}_{\mathrm{1}} }+\frac{\mathrm{b}}{\mathrm{x}−\mathrm{z}_{\mathrm{2}} }\left(=\frac{\mathrm{x}−\alpha}{\left(\mathrm{x}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{x}−\mathrm{z}_{\mathrm{2}} \right)}\right) \\ $$$$\mathrm{a}\:=\frac{\mathrm{z}_{\mathrm{1}} −\alpha}{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} }=\frac{\mathrm{z}_{\mathrm{1}} −\alpha}{\mathrm{2i}\alpha\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{i}\alpha\sqrt{\mathrm{3}}}\left(\mathrm{z}_{\mathrm{1}} −\alpha\right) \\ $$$$\mathrm{b}\:=\frac{\mathrm{z}_{\mathrm{2}} −\alpha}{\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} }\:=−\frac{\mathrm{1}}{\mathrm{i}\alpha\sqrt{\mathrm{3}}}\left(\mathrm{z}_{\mathrm{2}} −\alpha\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{x}+\alpha\right)}+\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{1}} −\alpha\right)}{\mathrm{2i}\alpha\sqrt{\mathrm{3}}\left(\mathrm{x}−\mathrm{z}_{\mathrm{1}} \right)}−\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{2}} −\alpha\right)}{\mathrm{2i}\alpha\sqrt{\mathrm{3}}\left(\mathrm{x}−\mathrm{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{x}+\alpha}\right\}^{\left(\mathrm{n}−\mathrm{1}\right)} \:+\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{1}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{z}_{\mathrm{1}} }\right\}^{\left(\mathrm{n}−\mathrm{1}\right)} −\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{2}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{z}_{\mathrm{2}} }\right\}^{\left(\mathrm{n}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}+\alpha\right)^{\mathrm{n}} }+\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{1}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}×\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{n}} } \\ $$$$−\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{2}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}×\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{n}} }\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}!}\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{n}!}\left\{\frac{\mathrm{3}}{\mathrm{2}}\alpha^{−\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!+\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{1}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}\left(−\mathrm{z}_{\mathrm{1}} \right)^{−\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!\right. \\ $$$$\left.−\frac{\mathrm{3}\left(\mathrm{z}_{\mathrm{2}} −\alpha\right)}{\mathrm{2i}\sqrt{\mathrm{3}}}\left(−\mathrm{z}_{\mathrm{2}} \right)^{−\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{1}\right)!\right\} \\ $$

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