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Question Number 141525 by hgrocks last updated on 20/May/21

$$\mathrm{Let}<{\mathrm{x}}_{\mathrm{n}}>\mathrm{be}\mathrm{a}\mathrm{sequence}\mathrm{defined}\mathrm{by}$$$${\mathrm{x}}_{\mathrm{n}+1}=\frac{1}{\mathrm{k}}({\mathrm{x}}_{\mathrm{n}}+\frac{\mathrm{k}}{{\mathrm{x}}_{\mathrm{n}}})\mathrm{\forall }\mathrm{n}\in \mathbb{N}$$$$\mathrm{Show}\mathrm{that}<{\mathrm{x}}_{\mathrm{n}}>\mathrm{converges}\mathrm{to}\sqrt{\frac{\mathrm{k}}{\mathrm{k}-1}}$$$${\mathrm{x}}_1>0,\mathrm{k}>1$$

Answered by mathmax by abdo last updated on 21/May/21

$${\mathrm{x}}_{\mathrm{n}+1}=\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}}\right)\mathrm{with}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{k}}(\mathrm{x}+\frac{\mathrm{k}}{\mathrm{x}})\mathrm{f}\mathrm{is}\mathrm{continue}\mathrm{on}{\mathrm{R}}^{★}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{\mathrm{k}}(1-\frac{\mathrm{k}}{{\mathrm{x}}^2})=\frac{1}{\mathrm{k}}\times \frac{{\mathrm{x}}^2-\mathrm{k}}{{\mathrm{x}}^2}=\frac{1}{{\mathrm{kx}}^2}(\mathrm{x}-\sqrt{\mathrm{k}})(\mathrm{x}+\sqrt{\mathrm{k}})$$$$\mathrm{we}\mathrm{can}\mathrm{limit}\mathrm{the}\mathrm{variation}\mathrm{on}]0,+\mathrm{\infty }[\mathrm{due}\mathrm{to}{\mathrm{u}}_{\mathrm{n}}>0\mathrm{for}\mathrm{all}\mathrm{n}$$$$\mathrm{x}0\sqrt{\mathrm{k}}+\mathrm{\infty }$$$${\mathrm{f}}^{{}^{\prime }}\mid \mid -0+$$$$\mathrm{f}\mid \mid \mathrm{dec}\mathrm{f}\left(\sqrt{\mathrm{k}}\right)\mathrm{incr}+\mathrm{\infty }$$$$\mathrm{due}\mathrm{to}\mathrm{continuity}\mathrm{of}\mathrm{f}\mathrm{the}\mathrm{limit}\mathrm{of}{\mathrm{u}}_{\mathrm{n}}\mathrm{is}\mathrm{the}\mathrm{fix}\mathrm{point}\mathrm{of}\mathrm{f}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow \frac{1}{\mathrm{k}}(\mathrm{x}+\frac{\mathrm{k}}{\mathrm{x}})=\mathrm{x}\Rightarrow {\mathrm{x}}^2+\mathrm{k}={\mathrm{kx}}^2\Rightarrow (\mathrm{k}-1){\mathrm{x}}^2=\mathrm{k}\Rightarrow$$$$\mathrm{but}\mathrm{x}>0\Rightarrow \mathrm{x}=\sqrt{\frac{\mathrm{k}}{\mathrm{k}-1}}(\mathrm{k}>1)$$

Commented by hgrocks last updated on 21/May/21

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}.★★$$$$$$

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