While playing around with Python, I saw that if you take any non-zero integers p & q and perform the following operation, the values tend towards phi.
Flip the numerator and denominator and complete the process by adding the value of the denominator to the numerator.
$\dfrac{3}{2},\dfrac{5}{3}, ... \text{after 38 repetitions}... \dfrac{267914296}{165580141} $
Can someone explain how this works?
Kepler observed the fact you recite:
$$ \lim_{n \rightarrow \infty} \frac{F_{n+1}}{F_n} = \varphi \text{,} $$ where $F_n$ is the $n^\text{th}$ Fibonacci number and $\varphi$ is the golden ratio.
The usual way to show this is with Binet's formula. $$ F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi} = \frac{\varphi^n - \psi^n}{\sqrt{5}} \text{,} $$ where $\psi = -1/\varphi$ is the conjugate golden ratio -- the other root of a quadratic equation constructed from the recurrence for the Fibonacci numbers. Using Binet's formula, \begin{align*} \frac{F_{n+1}}{F_n} &= \frac{\varphi^{n+1} - \psi^{n+1}}{\varphi^{n} - \psi^{n}} \\ &= \frac{\varphi^{n+1}}{\varphi^n} \cdot \frac{1 - \psi^{n+1}/\varphi^{n+1}}{1 - \psi^{n}/\varphi^{n}} \\ &= \varphi \frac{1 - (-1)^{n+1}/\varphi^{2(n+1)}}{1 - (-1)^{n}/\varphi^{2n}} \end{align*}
As $n$ increases, since $\varphi > 1$, the fractions $(-1)^{n+1}/\varphi^{2(n+1)}$ and $(-1)^{n}/\varphi^{2n}$ approach $0$. So, in the limit as $n \rightarrow \infty$, this is $$ \varphi \frac{1 - 0}{1 - 0} = \varphi \text{.} $$
You're sending $\frac{a}{b}$ to $\frac{a + b}{a} = 1 + \frac{b}{a}$. In other words, you're sending $x$ to $1 + 1/x$. The fixed point of this operation is $\phi$.
