Question Number 25114 by Mr easy last updated on 04/Dec/17
Commented by moxhix last updated on 04/Dec/17
$${x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}}>\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{0}}} \\ $$$$\therefore{x}>\sqrt{\mathrm{2}} \\ $$$${put}\:\phi:\:\:\phi=\sqrt{\mathrm{1}+\phi}\:\:\left(\phi>\mathrm{1}\right) \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0} \\ $$$$\therefore\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\left({golden}\:{ratio}\right) \\ $$$$ \\ $$$$\phi=\sqrt{\mathrm{1}+\phi}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\phi}}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\phi}}} \\ $$$${so} \\ $$$${x}=\phi \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$${golden}\:{ratio}: \\ $$$$\phi=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+....}}}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{1}.\mathrm{618}... \\ $$