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Question Number 145391 by puissant last updated on 04/Jul/21

Developpement limite^� a l′ordre 2 de g(x)=((√(1+x^2 ))/(1+x+(√(1+x^2 ))))

$$\mathrm{Developpement}\:\mathrm{limit}\acute {\mathrm{e}}\:\mathrm{a}\:\mathrm{l}'\mathrm{ordre}\:\mathrm{2}\:\mathrm{de}\: \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

g(x) = ((√(1+x^2 ))/(1+x+(√(1+x^2 )))) g(x) = ((√(1+x^2 ))/((1+x)^2 −(1+x^2 )))(1+x−(√(1+x^2 ))) g(x) = ((√(1+x^2 ))/(2x))(1+x−(√(1+x^2 ))) g(x) = ((1+(x^2 /2))/(2x))(1+x−1−(x^2 /2))+o(x^2 ) g(x) = ((1+(x^2 /2))/2)(1−(x/2))+o(x^2 ) g(x) = (1/2)(1+(x^2 /2)−(x/2))+o(x^2 ) g(x) = (1/2)−(x/4)+(x^2 /8)+o(x^2 )

$${g}\left({x}\right)\:=\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${g}\left({x}\right)\:=\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\left(\mathrm{1}+{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$${g}\left({x}\right)\:=\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}{x}}\left(\mathrm{1}+{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$${g}\left({x}\right)\:=\:\frac{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{2}{x}}\left(\mathrm{1}+{x}−\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${g}\left({x}\right)\:=\:\frac{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$

Commented by puissant last updated on 04/Jul/21

merci

$$\mathrm{merci} \\ $$

Answered by mathmax by abdo last updated on 04/Jul/21

g(x)=g(0)+xg^′ (0)+(x^2 /2)g^((2)) (0)+x^3 ξ(x) g(0)=(1/2) we have g^′ (x)=((((2x)/(2(√(1+x^2 ))))(1+x+(√(1+x^2 )))−(1+((2x)/(2(√(1+x^2 )))))(√(1+x^2 )))/((1+x+(√(1+x^2 )))^2 )) =(((x/( (√(1+x^2 ))))(1+x+(√(1+x^2 )))−(((x+(√(1+x^2 )))/( (√(1+x^2 )))))(√(1+x^2 )))/((1+x+(√(1+x^2 )))^2 )) =(((1+x+(√(1+x^2 )))−(√(1+x^2 ))(x+(√(1+x^2 ))))/( (√(1+x^2 ))(1+x+(√(1+x^2 )))^2 )) ⇒ g^′ (0)=((2−1)/2)=(1/2) g^((2)) (x)=(d/dx)(((u−v)/w))=(((u^′ −v^′ )w−(u−v)w^′ )/w^2 ) u=1+x+(√(1+x^2 )) ⇒u^′ =1+(x/( (√(1+x^2 )))) v=(√(1+x^2 ))(x+(√(1+x^2 ))) ⇒v^′ =(x/( (√(1+x^2 ))))(x+(√(1+x^2 )))+(√(1+x^2 ))(1+(x/( (√(1+x^2 ))))) w=(√(1+x^2 ))(1+x+(√(1+x^2 )))^2 ⇒ w^′ =(x/( (√(1+x^2 ))))(1+x+(√(1+x^2 )))^2 +2(√(1+x^2 ))(1+x+(√(1+x^2 )))(1+(x/( (√(1+x^2 ))))) ⇒ g^((2)) (0)=(((u^′ (0)−v^′ (0))w(0)−(u(0)−v(0))w^′ (0))/(w^2 (0))) =(((1−1)4−(2−1)4)/4^2 )=−(1/4) ⇒ g(x)=(1/2)+(x/2)−(x^2 /8) +x^3 ξ(x)

$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{0}\right)+\mathrm{xg}^{'} \left(\mathrm{0}\right)+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\mathrm{x}^{\mathrm{3}} \xi\left(\mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{g}^{'} \left(\mathrm{x}\right)=\frac{\frac{\mathrm{2x}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}+\frac{\mathrm{2x}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)−\left(\frac{\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{g}^{'} \left(\mathrm{0}\right)=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}−\mathrm{v}}{\mathrm{w}}\right)=\frac{\left(\mathrm{u}^{'} −\mathrm{v}^{'} \right)\mathrm{w}−\left(\mathrm{u}−\mathrm{v}\right)\mathrm{w}^{'} }{\mathrm{w}^{\mathrm{2}} } \\ $$$$\mathrm{u}=\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{u}^{'} \:=\mathrm{1}+\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{v}=\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\:\Rightarrow\mathrm{v}^{'} \:=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{w}=\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{w}^{'} \:=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\:\Rightarrow \\ $$$$\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\frac{\left(\mathrm{u}^{'} \left(\mathrm{0}\right)−\mathrm{v}^{'} \left(\mathrm{0}\right)\right)\mathrm{w}\left(\mathrm{0}\right)−\left(\mathrm{u}\left(\mathrm{0}\right)−\mathrm{v}\left(\mathrm{0}\right)\right)\mathrm{w}^{'} \left(\mathrm{0}\right)}{\mathrm{w}^{\mathrm{2}} \left(\mathrm{0}\right)} \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{1}\right)\mathrm{4}−\left(\mathrm{2}−\mathrm{1}\right)\mathrm{4}}{\mathrm{4}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{8}}\:+\mathrm{x}^{\mathrm{3}} \xi\left(\mathrm{x}\right) \\ $$

Commented by mathmax by abdo last updated on 04/Jul/21

sorry g^′ (0)=(1/4) ⇒g(x)=(1/2)+(x/4)−(x^2 /8) +x^3 ξ(x)

$$\mathrm{sorry}\:\mathrm{g}^{'} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{x}}{\mathrm{4}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{8}}\:+\mathrm{x}^{\mathrm{3}} \xi\left(\mathrm{x}\right) \\ $$

Commented by puissant last updated on 04/Jul/21

thanks

$$\mathrm{thanks} \\ $$

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