Question-25114

Question Number 25114 by Mr easy last updated on 04/Dec/17

Commented by moxhix last updated on 04/Dec/17

x = \sqrt{1 + \sqrt{1 + \sqrt{1 + x}}} > \sqrt{1 + \sqrt{1 + 0}}

∴ x > \sqrt{2}

p u t ϕ : ϕ = \sqrt{1 + ϕ} (ϕ > 1)

ϕ^{2} - ϕ - 1 = 0

∴ ϕ = \frac{1 + \sqrt{5}}{2} (g o l d e n r a t i o)

ϕ = \sqrt{1 + ϕ} = \sqrt{1 + \sqrt{1 + ϕ}} = \sqrt{1 + \sqrt{1 + \sqrt{1 + ϕ}}}

s o

x = ϕ

- - - - - - - - - - - - - - - - - -

g o l d e n r a t i o :

ϕ = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots .}}}}} = \frac{1 + \sqrt{5}}{2} = 1.618 \dots

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