Babylonian Method Limit Question The Babylonian Method for finding square roots is a method that takes a guess, say $x$, and averages $x$ and $\frac{a}{x}$, where $a$ is the number you want to find the square root of. It then uses the average as a guess, and does the algorithm again. The value outputted will converge towards $\sqrt{a}$.
Translating this into algebraic terms, we get that
$$\lim_{n\to\infty}f^n(2)=\sqrt{2},$$
where $f(x)=\frac{x+\frac{2}{x}}{2}$.
I wondered what would happen if we changed the initial input value. So, this is my question. Let $f(x)=\frac{x+\frac{2}{x}}{2}$. What is
$$\lim_{n\to\infty} f^n(2^n)?$$
Plugging this into my calculator, I got an answer that was about $1.591891656$. I have no idea what is special about this number. If anyone could figure this out, I would appreciate it.
 A: In section 2 on the Babylonian method in the Wikipedia article
Methods of computing square roots
the convergence is shown to be quadratic. We give a precise
version of this convergence. Define the function
$$ f_a(x) := \frac12\left(x + \frac{a}{x}\right)
  \;\;\text{ where }\;\; a\in\mathbb{C}. \tag1 $$
It has three fixed points in the extended complex
plane: $\,\{\sqrt{a},-\sqrt{a}, \infty\}.\,$ The function
$\, z \mapsto z^2 \,$ also has three fixed points
$\,\{0,\infty,1\}\,$ and is conjugate to $\,f_a.\,$
More precisely, define the function
$$ g_a(x) := \sqrt{a}\frac{1+x}{1-x} \tag2 $$
whose inverse is given by the function
$$ h_a(x) := g_a^{(-1)}(x)  = \frac{x-\sqrt{a}}{x+\sqrt{a}}. \tag3 $$
These functions satisfy the equation
$$ f_a(x) = g_a(h_a(x)^2). \tag4 $$
This implies, by induction, that
$$ f^{(n)}_a(x) = g_a(h_a(x)^{2^n}). \tag5 $$
Some analysis shows that
$\, g_a(h_a(x)^x) \to L(a)\,$ as $\,x \to \infty\,$ where
$$ L(a) := \sqrt{a}\coth(\sqrt{a}). \tag6 $$
Apply this to the case $\,x = 2^n\,$ and we have
$$ \lim_{n\to\infty} f^{(n)}_a(2^n) = L(a). \tag7 $$
Now note that $\, L(2) \approx 1.59189165552048736452102946.$
