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

$$studytheconvergenceof{u}_{n+1}=2\sqrt{1+u_n^2}-u_n-1with{u}_0=0$$

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

$$wehave{u}_{n+1}=f\left(u_n\right)withf\left(x\right)=2\sqrt{1+x^2}-x-1$$$$letprovethat{u}_n>0wehave{u}_1=2-1=1>0letsuppose{u}_n>0$$$$2\sqrt{1+u_n^2}-u_n-1=2\sqrt{1+u_n^2}-(u_n+1)and4(1+u_n^2)-{(u_n+1)}^2$$$$=4u_n^2+4-u_n^2-2u_n-1=2u_n^2-2u_n+3\to \mathrm{\Delta }=4-4\left(2\right).3<0(a=2>0)\Rightarrow$$$$2u_n^2-u_n+3forthatletstudythevariationoffon[0,+\mathrm{\infty }[$$$$f^{{}^{\prime }}\left(x\right)=2\frac{2x}{2\sqrt{1+x^2}}-1=\frac{2x}{\sqrt{1+x^2}}-1=\frac{2x-\sqrt{1+x^2}}{\sqrt{1+x^2}}=\frac{4x^2-1-x^2}{\sqrt{1+x^2}(2x+\sqrt{1+x^2}})$$$$=\frac{3x^2-1}{\sqrt{1+x^2(2x+\sqrt{1+x^2)}}}so{f}^{{}^{\prime }}\left(x\right)=0\iff x=\frac{1}{\sqrt{3}}$$$$f\left(\frac{1}{\sqrt{3}}\right)=2\sqrt{1+\frac{1}{3}}-\frac{1}{\sqrt{3}}-1=\frac{4}{\sqrt{3}}-\frac{1}{\sqrt{3}}-1=\sqrt{3}-1$$$$x0\frac{1}{\sqrt{3}}+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)-+$$$$f\left(x\right)1decr\sqrt{3}-1incr+\mathrm{\infty }$$$$letdeterminethefixedpoint$$$$f\left(x\right)=x\Rightarrow 2\sqrt{1+x^2}-x-1=x\Rightarrow 2\sqrt{1+x^2}=2x+1\Rightarrow 4(1+x^2)=4x^2+4x+1\Rightarrow$$$$4=4x+1\Rightarrow 4x=3\Rightarrow x=\frac{3}{4}fiscontinueon[0,+\mathrm{\infty }[so\left(u_n\right)convergesand$$$${lim}_{n\to +\mathrm{\infty }}{u}_n=\frac{3}{4}.$$

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

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

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