Is The Infinite Nested Radical $\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{x^3+...}}}}$ Analytical?

Question

I first started thinking about this problem a while back and being reminded of the famous representation of the golden ratio which includes an infinite nested radical like this one except $x$ is simply 1. I then wondered if any other well defined results appear, to which the answer is a pseudo-yes. An $x$ value of 4 approaches exactly 2, for reasons I have yet to rigorously prove. And based on what you consider $0^0$ to be, an $x$ value of 0 can approach either 1 or 0, although for my purposes I've made the first term $x^0$ 1 for all cases (hence why it's just 1 and not $x^0$ in the title). I've found several forums relating to this topic, but only over the specific form of $$\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+...}}}}$$ in which a value (1.783165809...) was found that the series converges to. However, no one was able to find a analytical approach that could produce said value. So of course the next logical thing to ask was if there are well defined values for some $x$, in which $x$ is also well defined, like 1 or 4, is there an analytical approach to take for other well defined $x$ such as 2 (in other words, is 1.78316... a constant or a result of some function $f(x)$)? And if it is analytical, what is the function itself?

Answer 1

Here is perhaps a related problem with an upper bound.

Call your number $F(2)$.

As it was marked in the comments, the sequence converges $\forall x \ge 0$. Note first that if $$ a = \sqrt{1+\sqrt{1+\sqrt{1+\ldots}}} $$ then $a \in \mathbb{R}$ since $a$ is less than any of our cases for $x>1$, and we clearly have $$a^2=1+a \quad \text{and} \quad a>0 \implies a = \frac{1+\sqrt5}{2}.$$

Now note that $$ \begin{split} G(x) &= \sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\ldots}}}}}\\ &= \sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}\\ &= \sqrt{1+a\sqrt{x}} \end{split} $$ and in particular $G(2) = \sqrt{1+a\sqrt{2}} \approx 1.8133$, which would be an upper bound for your result, since finite sequences of roots leading to $G(2)$ is strictly larger than the sequence leading to $F(2)$.