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Question Number 120037 by floor(10²Eta[1]) last updated on 28/Oct/20

$$\mathrm{Suppose}\mathrm{that}\mathrm{R}>0,{\mathrm{x}}_0>0,\mathrm{and}$$ $${\mathrm{x}}_{\mathrm{n}+1}=\frac{1}{2}(\frac{\mathrm{R}}{{\mathrm{x}}_{\mathrm{n}}}+{\mathrm{x}}_{\mathrm{n}}),\mathrm{n}⩾0$$ $$\mathrm{Prove}:\mathrm{For}\mathrm{n}⩾1,{\mathrm{x}}_{\mathrm{n}}>{\mathrm{x}}_{\mathrm{n}+1}>\sqrt{\mathrm{R}}\mathrm{and}$$ $${\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}⩽\frac{1}{2^{\mathrm{n}}}\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{\mathrm{x}}_0}$$

Answered by floor(10²Eta[1]) last updated on 29/Oct/20

$${2\mathrm{x}}_{\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}=\mathrm{R}+{\mathrm{x}}_{\mathrm{n}}^2$$ $${2\mathrm{x}}_{\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}-2\sqrt{\mathrm{R}}{\mathrm{x}}_{\mathrm{n}}={({\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}})}^2⩾0$$ $${2\mathrm{x}}_{\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}-2\sqrt{\mathrm{R}}{\mathrm{x}}_{\mathrm{n}}⩾0$$ $${\mathrm{x}}_{\mathrm{n}+1}⩾\sqrt{\mathrm{R}}$$ $$\Rightarrow {\mathrm{x}}_{\mathrm{n}+1}>\sqrt{\mathrm{R}}$$ $${\mathrm{x}}_{\mathrm{n}}^2-{2\mathrm{x}}_{\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}+\mathrm{R}=0$$ $${\mathrm{x}}_{\mathrm{n}}={\mathrm{x}}_{\mathrm{n}+1}\pm \sqrt{{\mathrm{x}}_{\mathrm{n}+1}^2-\mathrm{R}}>\sqrt{\mathrm{R}}$$ $$\mathrm{suppose}\mathrm{that}{\mathrm{x}}_{\mathrm{n}}<{\mathrm{x}}_{\mathrm{n}+1}$$ $${\mathrm{x}}_{\mathrm{n}}<\frac{1}{2}(\frac{\mathrm{R}}{{\mathrm{x}}_{\mathrm{n}}}+{\mathrm{x}}_{\mathrm{n}})$$ $${2\mathrm{x}}_{\mathrm{n}}^2<\mathrm{R}+{\mathrm{x}}_{\mathrm{n}}^2\Rightarrow {\mathrm{x}}_{\mathrm{n}}<\sqrt{\mathrm{R}}(\mathrm{imp}.)$$ $$\Rightarrow {\mathrm{x}}_{\mathrm{n}}>{\mathrm{x}}_{\mathrm{n}+1}>\sqrt{\mathrm{R}}$$ $${2\mathrm{x}}_{\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}-2\sqrt{\mathrm{R}}{\mathrm{x}}_{\mathrm{n}}={({\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}})}^2$$ $$2({\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}})>2({\mathrm{x}}_{\mathrm{n}+1}-\sqrt{\mathrm{R}})=\frac{{({\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}})}^2}{{\mathrm{x}}_{\mathrm{n}}}$$ $${\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}>\frac{{({\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_{\mathrm{n}}}<\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}$$ $$\left(\mathrm{i}\right):{\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}⩾\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}$$ $${\mathrm{x}}_0-\sqrt{\mathrm{R}}>{\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}⩾\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}$$ $${2\mathrm{x}}_0^2-2\sqrt{\mathrm{R}}{\mathrm{x}}_0>{\mathrm{x}}_0^2-2\sqrt{\mathrm{R}}{\mathrm{x}}_0+\mathrm{R}$$ $${\mathrm{x}}_0^2>\mathrm{R}\Rightarrow {\mathrm{x}}_0>\sqrt{\mathrm{R}}$$ $$\left(\mathrm{ii}\right):$$ $${\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}⩽\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}$$ $${\mathrm{x}}_0-\sqrt{\mathrm{R}}<\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}$$ $${\mathrm{x}}_0<-\sqrt{\mathrm{R}}$$ $$\Rightarrow {\mathrm{x}}_{\mathrm{n}}-\sqrt{\mathrm{R}}⩾\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{{2\mathrm{x}}_0}⩾\frac{{({\mathrm{x}}_0-\sqrt{\mathrm{R}})}^2}{2^{\mathrm{n}}{\mathrm{x}}_0}$$

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