How to simplify this expression of square roots?

Question

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!

Answer 1

$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}=\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ $$\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$ 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 $$y=\frac{1+\sqrt{5}}{2}$$

Answer 2

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.