Inspired by this question that I recently saw, I was wondering if there is a closed form for $$y = \sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+...}}}}}$$

as a function of $x$.

Usually, in problems like these, I would try to plug the value back into the equation. When I took $$y^2-1 = \sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+...}}}}$$ and tried dividing by $\sqrt{x}$, I got $$\frac{y^2-1}{\sqrt{x}} = \sqrt{1+\sqrt{1+\sqrt{x^{-1}+\sqrt{x^{-4}+...}}}}$$ which definitely does not equal $y$.

Unfortunately, the linked question does not address a closed form, instead proving that when $y = 2$, $x = 4$. Are there any ideas about getting a closed form?

  • $\begingroup$ You may find MSE question 2255364 "Crazy Iterated Square Roots" of interest. It is very unlikely that there is a closed form. $\endgroup$ – Somos Oct 17 '19 at 1:25

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