I have a question about recursive functions and their convergence as sequences.
We know that the golden ratio can be represented by: $$ \phi= 1+\frac{1}{1+\frac{1}{1+\frac{1}{...}}} $$ Rewriting the above infinite expression as a recursive function gives us: $$ \phi(x)=1+\frac{1}{x} $$ $\lim_{p \to \infty} \phi^p(x)$ converges to the golden ratio for $x \in \mathbb{R}, x\neq0,\Phi$ (the golden ratio conjugate). The solution for this convergence can be found when substituting $\phi(x)$ in for $x$ if you assume $\lim_{p \to \infty} \phi^p(x)=\phi^{p-1}(x)$. My question is: for which recursive functions (as a sequence) can you assume that for in order to solve for convergence? Also, how will you know how many solutions there are for any of these functions? Here's a few examples I've found/wondered about: $$ \phi(x)=1+\frac{1}{x}, \lim_{p \to \infty} \phi^p(x)=1.618... \text{ (unless $x=0,\Phi$), two convergences ($\phi$ and $\Phi$), for } x \in \mathbb{R} $$ $$ f(x)=\frac{x}{2}, \lim_{p \to \infty} f^p(x)=0, \text{one convergence, for } x \in \mathbb{R} $$ $$ d(x)=x-1+0^{x-1}, \lim_{p \to \infty} d^p(x)=1, \text{one convergence, for } x \in \mathbb{N} \text{ (a strange function, but I'm still curious)} $$
You were able to see that your recursion was of the form $x_{n+1} = \phi(x_n)$. Therefore, if you assume that $(x_n)$ is a convergent sequence, it must converge to a fixed point of $\phi$, .i.e. a solution of the equation $x = \phi(x)$. In your example, $x = \phi(x) \Leftrightarrow x = \frac 12 (1\pm \sqrt{5})$, where one of the solutions can be discarded based on the signal of the terms.
This "method" can be used whenever you can assure that $(x_n)$ is a convergent sequence. The fixed point theorem gives nice sufficient conditions for convergence, just based on the behavior of $\phi(x)$ and $x_0$.