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.

wild-chespin

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:

golden-converge

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):

golden-error

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 […]

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