## A random Golden Ratio Appears!

Well, we all make errors once in a while, but sometimes they’re more interesting than others. So I inadvertently forgot a square in a calculation and the series suddenly started converging, to the most beautiful of convergences of all: $\phi$. So, let’s see how I got the golden ratio to emerge from a broken algorithm.

First, I wrote the recurrence as (remember, I forgot a square out somewhere): $r(n)=\begin{cases}1&\text{if }n\leqslant{}1\\\sqrt{1+r(n-1)}&\text{otherwise}\end{cases}$

The first few terms of the series: $1=\sqrt{1}$, $\sqrt{1+\sqrt{1}}$, $\sqrt{1+\sqrt{1+\sqrt{1}}}$, $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}}}}$, $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}}}}}$, $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}}}}}}$, $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}}}}}}}$,…

and if you plot them, you see that they do converge rapidly: And the value 1.61-ish should be familiar. Could it converge to the golden ratio, $\displaystyle\phi=\frac{1+\sqrt{5}}{2}$ ?

Well, numerically it does. Very quickly. The following log-plot shows the error $\phi-r(n)$: Empirically, we are now fairly convinced that the limit of $r(n)$ as $n$ goes to infinity is $\phi$. Can we show it algebraically? Yes! And, no, it wont be überschmerzlich1.

Let $x=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

Let’s square it, and see what happens: $x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

So it’s… $x^2=1+x$

and we find that the unique $x$ that satisfies the above equation is $\displaystyle x=\phi=\frac{1+\sqrt{5}}{2}$

and therefore, $\lim_{n\to\infty} r(n)=\phi$.

1 Superpainful.

### One Response to A random Golden Ratio Appears!

1. El corte de Leones | Harder, Better, Faster, Stronger says:

[…] with an infinite number of layer, for that, we would need to use another process, maybe like the wild golden ratio of last […]