Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ I recently became fascinated by infinite nested radicals, first drawn attention to me from a question in my textbook about the value of $\sqrt{1+\sqrt{{1}+\sqrt{{1}+\sqrt{{1}...}}}}$ which turned out to be $\phi$ when I worked it out, a rather beautiful result.
I then tried to find a formula to evaluate the general case
$$\sqrt{x+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}...}}}}$$
which I succeeded in; it can be evaluated as $$\frac{1+\sqrt{1+4x}}{2}$$
Multiplying the nested radical which was equal to $\phi$ by $x$ produces the following nested radical:
$$\sqrt{{x^2}+\sqrt{{x^4}+\sqrt{{x^8}+\sqrt{{x^{16}}...}}}}$$
so this is equal to $x\left(\frac{1+\sqrt5}{2}\right)$.
However, I have tried and failed to find the value of the following infinite square root:
$$\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$$
 A: Not an answer, but something I find quite interesting is that when $x = 4$, this converges to $3$.
For the comment regarding my code, it's quite simple. Here it is:
import numpy as np

x = 2 # Or whatever you want
iterations = 10 # Or whatever you want

value = np.sqrt(x ** iterations)
for i in reversed(range(1, iterations)):
  value = np.sqrt(value + x ** i)

print (value)

Another interesting aspect is this:
Define a function as the expression in the title:
$f(x) = \sqrt{x + \sqrt{x^2 + \sqrt{x^3 + \sqrt{x^4 ...}}}}$
For any positive x, $f(f(f(f(...x)))) \approx 2.340649036282968$
This is the intersection between $y=f(x)$ and $y=x$.
Edit:
Just another neat fact: this function can be approximated very closely by the function $f(x) = \sqrt{2x} + 0.17555$ for most relatively small values ($x \in (0, 10^{10}]$)
Edit:
The case for x = 1 isn’t too difficult.
$a = \sqrt{1 + \sqrt{1^2 + \sqrt{1^3 + \sqrt{1^4 ...}}}}$
$a = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 ...}}}}$
$a^2 - 1 = a$
$a^2 - a - 1 = 0$
Now you have a quadratic that you can easily solve.
I’m still unsure how to do the other cases.
