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