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Question Number 147071 by alcohol last updated on 17/Jul/21

prove that (√(1+(√(1+(√(1+(√(1+...)))))))) = 1+(1/(1+(1/(1+(1/(1+⋱))))))

$${prove}\:{that} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\ddots}}} \\ $$

Answered by Olaf_Thorendsen last updated on 17/Jul/21

ϕ^2 = ϕ+1 (golden ratio ((1+(√5))/2)) ⇒ ϕ = 1+(1/ϕ) ϕ = 1+(1/(1+(1/ϕ))) ϕ = 1+(1/(1+(1/(1+(1/ϕ))))) ...etc... ϕ = 1+(1/(1+(1/(1+(1/(...)))))) Let x = (√(1+(√(1+(√(1+(√(1+...)))))))) x^2 = 1+(√(1+(√(1+(√(1+...)))))) x^2 = 1+x x^2 −x−1 = 0 (x−((1−(√5))/2))(x−((1+(√5))/( 2))) = 0 x = ((1±(√5))/2) But ((1−(√5))/2)<0 : impossible because x>1 Finally x = ((1+(√5))/2) = ϕ

$$\varphi^{\mathrm{2}} \:=\:\varphi+\mathrm{1}\:\left({golden}\:{ratio}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\varphi} \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\varphi}} \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\varphi}}} \\ $$$$...{etc}... \\ $$$$\varphi\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{...}}} \\ $$$$\mathrm{Let}\:{x}\:=\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+...}}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{1}+{x} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\:\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{But}\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\::\:\mathrm{impossible}\:\mathrm{because}\:{x}>\mathrm{1} \\ $$$$\mathrm{Finally}\:{x}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$

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