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If one simplifies the infinite nested radical $$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$

Does one get $$\sqrt x\frac{1+\sqrt5}2$$ Any help would be appreciated!

S.C.B.
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mac07
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    Write "\sqrt{x+\sqrt{y+2}}" between dollars to get $\sqrt{x+\sqrt{y+2}}$. For bigger, centered formulas, write then between two pairs of dollars. – ajotatxe Aug 21 '16 at 14:14
  • @S.C.B Simplified answer is a product of square root of x and(1+sqrt(5))/2 . Not addition of square root of x and (1+sqrt(5))/2. Thanks S.C.B, ajotatxe and aakash for formatting/helping. – mac07 Aug 21 '16 at 14:31

2 Answers2

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$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}=\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ $$\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ now let

$$y=\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ so $$y^2-1=y$$ $$y=\frac{1\pm\sqrt{5}}{2}$$ select the $$y=\frac{1+\sqrt{5}}{2}$$

E.H.E
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You have

$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$ $$ = \sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ $$ = \sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$

The continued square root $\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$ is a well known form of $\phi = \frac{\sqrt{5}+1}{2}$ (see here) and so you end up with:

$$ \frac{\sqrt{x}(\sqrt{5}+1)}{2}$$ as required.

gowrath
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