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Question Number 56828 by maxmathsup by imad last updated on 24/Mar/19

study the convergence of u_(n+1) =2(√(1+u_n ^2 ))−u_n −1 with u_0 =0

$${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}+\mathrm{1}} =\mathrm{2}\sqrt{\mathrm{1}+{u}_{{n}} ^{\mathrm{2}} }−{u}_{{n}} −\mathrm{1}\:\:\:\:{with}\:{u}_{\mathrm{0}} =\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 16/Apr/19

we have u_(n+1) =f(u_n ) withf(x)=2(√(1+x^2 ))−x−1 let prove that u_n >0 we have u_1 =2−1=1>0 let suppose u_n >0 2(√(1+u_n ^2 ))−u_n −1 =2(√(1+u_n ^2 )) −(u_n +1) and 4(1+u_n ^2 )−(u_n +1)^2 =4u_n ^2 +4 −u_n ^2 −2u_n −1 =2u_n ^2 −2u_n +3 →Δ =4−4(2).3 <0 (a=2 >0)⇒ 2u_n ^2 −u_n +3 for that let study the variation of f on[0,+∞[ f^′ (x) =2 ((2x)/(2(√(1+x^2 )))) −1 =((2x)/(√(1+x^2 ))) −1 =((2x−(√(1+x^2 )))/(√(1+x^2 ))) =((4x^2 −1−x^2 )/((√(1+x^2 ))(2x+(√(1+x^2 ))))) =((3x^2 −1)/(√(1+x^2 (2x +(√(1+x^2 )))))) so f^′ (x)=0 ⇔ x =(1/(√3)) f((1/(√3))) =2(√(1+(1/3)))−(1/(√3)) −1 =(4/(√3)) −(1/(√3)) −1 =(√3)−1 x 0 (1/(√3)) +∞ f^′ (x) − + f(x) 1 decr (√3)−1 incr +∞ let determine the fixed point f(x) =x ⇒2(√(1+x^2 )) −x−1 =x ⇒2(√(1+x^2 ))=2x +1 ⇒4(1+x^2 ) =4x^2 +4x +1 ⇒ 4 =4x+1 ⇒4x =3 ⇒x =(3/4) f is continue on [0,+∞[ so (u_n )converges and lim_(n→+∞) u_n =(3/4) .

$${we}\:{have}\:{u}_{{n}+\mathrm{1}} ={f}\left({u}_{{n}} \right)\:{withf}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}−\mathrm{1} \\ $$$${let}\:{prove}\:{that}\:{u}_{{n}} >\mathrm{0}\:\:\:{we}\:{have}\:{u}_{\mathrm{1}} =\mathrm{2}−\mathrm{1}=\mathrm{1}>\mathrm{0}\:{let}\:{suppose}\:{u}_{{n}} >\mathrm{0} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}+{u}_{{n}} ^{\mathrm{2}} }−{u}_{{n}} −\mathrm{1}\:=\mathrm{2}\sqrt{\mathrm{1}+{u}_{{n}} ^{\mathrm{2}} }\:−\left({u}_{{n}} \:+\mathrm{1}\right)\:\:{and}\:\mathrm{4}\left(\mathrm{1}+{u}_{{n}} ^{\mathrm{2}} \right)−\left({u}_{{n}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}{u}_{{n}} ^{\mathrm{2}} \:+\mathrm{4}\:−{u}_{{n}} ^{\mathrm{2}} \:−\mathrm{2}{u}_{{n}} −\mathrm{1}\:=\mathrm{2}{u}_{{n}} ^{\mathrm{2}} \:−\mathrm{2}{u}_{{n}} \:+\mathrm{3}\:\rightarrow\Delta\:=\mathrm{4}−\mathrm{4}\left(\mathrm{2}\right).\mathrm{3}\:<\mathrm{0}\:\:\:\left({a}=\mathrm{2}\:>\mathrm{0}\right)\Rightarrow \\ $$$$\mathrm{2}{u}_{{n}} ^{\mathrm{2}} −{u}_{{n}} +\mathrm{3}\:\:\:{for}\:{that}\:{let}\:{study}\:{the}\:{variation}\:{of}\:{f}\:{on}\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$$\left.{f}^{'} \left({x}\right)\:=\mathrm{2}\:\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:−\mathrm{1}\:=\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:−\mathrm{1}\:=\frac{\mathrm{2}{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{2}{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right.}\right) \\ $$$$=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{2}{x}\:+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.}}\:{so}\:{f}^{'} \left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{x}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${f}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:=\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:−\mathrm{1}\:=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:−\mathrm{1}\:=\sqrt{\mathrm{3}}−\mathrm{1}\: \\ $$$${x}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\left({x}\right)\:\:\:\mathrm{1}\:\:\:\:{decr}\:\:\:\:\:\sqrt{\mathrm{3}}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:{incr}\:\:\:\:+\infty \\ $$$${let}\:{determine}\:\:{the}\:{fixed}\:\:{point}\:\: \\ $$$${f}\left({x}\right)\:={x}\:\:\Rightarrow\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−{x}−\mathrm{1}\:={x}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{2}{x}\:+\mathrm{1}\:\Rightarrow\mathrm{4}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:=\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{4}\:=\mathrm{4}{x}+\mathrm{1}\:\Rightarrow\mathrm{4}{x}\:=\mathrm{3}\:\Rightarrow{x}\:=\frac{\mathrm{3}}{\mathrm{4}}\:\:{f}\:{is}\:{continue}\:{on}\:\left[\mathrm{0},+\infty\left[\:{so}\:\left({u}_{{n}} \right){converges}\:{and}\right.\right. \\ $$$${lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\:. \\ $$

Answered by kaivan.ahmadi last updated on 25/Mar/19

l=2(√(1+l^2 ))−l−1⇒2l+1=2(√(1+l^2 ))⇒ 4l^2 +4l+1=4+4l^2 ⇒4l=3⇒l=(3/4)

$${l}=\mathrm{2}\sqrt{\mathrm{1}+{l}^{\mathrm{2}} }−{l}−\mathrm{1}\Rightarrow\mathrm{2}{l}+\mathrm{1}=\mathrm{2}\sqrt{\mathrm{1}+{l}^{\mathrm{2}} }\Rightarrow \\ $$$$\mathrm{4}{l}^{\mathrm{2}} +\mathrm{4}{l}+\mathrm{1}=\mathrm{4}+\mathrm{4}{l}^{\mathrm{2}} \Rightarrow\mathrm{4}{l}=\mathrm{3}\Rightarrow{l}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$

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