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Question Number 90018 by Rio Michael last updated on 20/Apr/20
expand , ln(1 + sin x) right up to the term in x^3
$$\mathrm{expand}\:,\:\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{sin}\:{x}\right)\:\mathrm{right}\:\mathrm{up}\:\mathrm{to}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:{x}^{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 21/Apr/20
sinx =x−(x^3 /6) +o(x^3 ) ⇒ln(1+sinx)=ln(1+x−(x^3 /6) +o(x^3 )) ln^′ (1+u) =(1/(1+u)) =1−u +u^2 −u^3 +o(u^3 ) ⇒ ln(1+u)=u−(u^2 /2) +(u^3 /3) +o(u^3 ) ⇒ ln(1+x−(x^3 /6))=x−(x^3 /6)−(1/2)(x−(x^3 /6))^2 +(((x−(x^3 /6))^3 )/3) +o(x^3 ) =x−(x^3 /6)−(1/2)(x^2 −(1/3)x^4 +(x^6 /(36))) +(1/3)x^3 (1−(x^2 /6))^3 +o(x^3 ) =x−(x^3 /6)−(1/2)x^2 +(1/6)x^4 −(x^6 /(72)) +(1/3)x^3 (1−3(x^2 /6)+...) +o(x^3 ) =x−(x^3 /6)−(1/2)x^2 +(1/3)x^3 +o(x^3 ) =x−(1/2)x^2 +(1/6)x^3 +o(x^3 )
$${sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow{ln}\left(\mathrm{1}+{sinx}\right)={ln}\left(\mathrm{1}+{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{3}} \right)\right) \\ $$$${ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{u}^{\mathrm{2}} −{u}^{\mathrm{3}} \:+{o}\left({u}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\:+{o}\left({u}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{2}} +\frac{\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{3}} }{\mathrm{3}}\:\:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{4}} \:+\frac{{x}^{\mathrm{6}} }{\mathrm{36}}\right)\:+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{4}} −\frac{{x}^{\mathrm{6}} }{\mathrm{72}}\:+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{3}\frac{{x}^{\mathrm{2}} }{\mathrm{6}}+…\right)\:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{3}} \right) \\ $$
Commented by Rio Michael last updated on 21/Apr/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 21/Apr/20
y=ln (1+sin x) ⇒y(0)=0 y^((1)) =((cos x)/(1+sin x)) ⇒y^((1)) (0)=1 y^((2)) =−((sin x)/(1+sin x))−((cos^2 x)/((1+sin x)^2 )) ⇒y^((2)) (0)=−1 y^((3)) =−((cos x)/(1+sin x))+((3 sin x cos x)/((1+sin x)^2 ))+((2cos^3 x)/((1+sin x)^3 ))⇒y^((3)) (0)=1 ⇒ln (1+sin x)=x−(x^2 /2)+(x^3 /6)+o(x^3 )
$${y}=\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\:\Rightarrow{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${y}^{\left(\mathrm{1}\right)} =\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:\Rightarrow{y}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$${y}^{\left(\mathrm{2}\right)} =−\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}−\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:\Rightarrow{y}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$$${y}^{\left(\mathrm{3}\right)} =−\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}+\frac{\mathrm{3}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2cos}^{\mathrm{3}} \:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{3}} }\Rightarrow{y}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{3}} \right) \\ $$
Commented by Rio Michael last updated on 21/Apr/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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