show a derivation of this recurrence relation One of the approximations for $\sqrt3$ is $x_{n+1} = \frac{x_{n} + 3}{x_{n} + 1}$. I can see clearly why this is true, since if the sequence converges, $x_{n}$ approaches $x_{n+1}$ (then multiply out). But its really bugging me because I can't find a formal way in which this is derived. I mean does this come from Newton's procedure, the $x_{n+1} = f(x_{n})$ method.. I'm pretty sure this is a stupid question but an answer would be appreciated anyway
 A: By solving a separable ODE we may see that the iteration $x\mapsto\frac{x+3}{x+1}$ is just Newton's iteration for the function:
$$ f(x) = (\sqrt{3}-x)^{\frac{3+\sqrt{3}}{6}}(\sqrt{3}+x)^{\frac{3-\sqrt{3}}{6}}.$$
A: There might be no derivation of the method. One could just take a linear fractional mapping
$$
x_{n+1} = g(x_n) \equiv \frac{\alpha x_n + \beta}{x_n + \gamma}
$$
with random numbers $\alpha, \beta, \gamma$, eventually it either diverges or converges to a solution of the quadratics
$$
x^2 + (\gamma - \alpha) x - \beta = 0\\
x_{1,2} = \frac{\alpha - \gamma \pm \sqrt{(\alpha - \gamma)^2 + 4 \beta}}{2}.
$$
Synthetically, one might derive the method in the following steps:
$$
x^2 = 3\\
x^2 - 1 = 2\\
(x+1)(x-1) = 2
$$
Let the first $x$ be the $x_n$ and the second be the $x_{n+1}$:
$$
x_{n+1} = 1 + \frac{2}{x_n + 1} = \frac{x_n + 3}{x_n + 1}
$$
It is easy to verify that this method converges linearly with speed $q \approx 0.267$, so it does not behave like a Newton's method applied to a smooth function with a simple root.
