Question Number 147071 by alcohol last updated on 17/Jul/21 | ||
$${prove}\:{that} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\ddots}}} \\ $$ | ||
Answered by Olaf_Thorendsen last updated on 17/Jul/21 | ||
$$\varphi^{\mathrm{2}} \:=\:\varphi+\mathrm{1}\:\left({golden}\:{ratio}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\varphi} \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\varphi}} \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\varphi}}} \\ $$$$...{etc}... \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{...}}} \\ $$$$\mathrm{Let}\:{x}\:=\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{1}+{x} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\:\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{But}\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\::\:\mathrm{impossible}\:\mathrm{because}\:{x}>\mathrm{1} \\ $$$$\mathrm{Finally}\:{x}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$ | ||