# Why does this process, when iterated, tend towards a certain number? (the golden ratio?)

1. Take any number $x$ (edit: x should be positive, heh)
2. Add 1 to it $x+1$
3. Find its reciprocal $1/(x+1)$
4. Repeat from 2

So, taking $x = 1$ to start:

• 1
• 2 (the + 1)
• 0.5 (the reciprocal)
• 1.5 (the + 1)
• 0.666... (the reciprocal)
• 1.666... (the + 1)
• 0.6 (the reciprocal)
• 1.6 (the + 1)
• 0.625
• 1.625
• 0.61584...
• 1.61584...
• 0.619047...
• 1.619047...
• 0.617647058823..

etc.

If we look at just the "step 3"'s (the reciprocals), we get:

• 1
• 0.5
• 0.666...
• 0.6
• 0.625
• 0.61584...
• 0.619047...
• 0.617647058823..

This appears to always converge to 0.61803399... no matter where you start from. I looked up this number and it is often called "The golden ratio" - 1, or $\frac{1+\sqrt{5}}{2}-1$.

1. Is there any "mathematical" way to represent the above procedure (or the terms of the second series, of "only reciprocals") as a limit or series?
2. Why does this converge to what it does for every starting point $x$?

edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1).

However, I think I could easily make it the golden ratio by taking the +1 "steps" of the original series, instead of the reciprocation steps of the original series:

• 2
• 1.5
• 1.666...
• 1.6
• 1.625
• 1.61584...
• 1.619047...
• 1.617647058823..

which does converge to $\frac{1+\sqrt{5}}{2}-1$

Explaining either of these series is adequate as I believe that explaining one also explains the other.

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1
– Hans Lundmark Oct 20 '10 at 8:07

## 5 Answers

We want to show that the function $f(x) = \frac{1}{1+x}$ has a unique fixed point to which it converges when iterated on positive $x$. If $x$ is positive, then $f(x)$ is between $1$ and $0$, so $f(f(x))$ is between $1$ and $\frac{1}{2}$. It is not hard to see that in fact $f$ fixes the interval $\left[ \frac{1}{2}, 1 \right]$. Now, for $x, y$ in this interval,

$$\left| \frac{1}{x+1} - \frac{1}{y+1} \right| = \left| \frac{y-x}{(1+x)(1+y)} \right| \le \frac{4}{9} |x - y|$$

so on this interval $f$ satisfies the conditions of the Banach fixed point theorem. The unique fixed point to which everything always converges is the unique solution to $f(x) = x$, which you have already found.

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1
Note that this argument is extremely general; in the right intervals it applies to many other functions one might idly be tempted to iterate on a calculator... – Qiaochu Yuan Oct 20 '10 at 15:09

1
– J. M. Oct 20 '10 at 15:17
+1: For talking about Banach's fixed point theorem. – Aryabhata Oct 20 '10 at 16:32

1
The generality of this answer is pretty striking. It even showed why iterating $f(x) = \sqrt{x+1}$ also tends towards the golden ratio. and...basically every other function I've been trying out on my shiny new hp 35s rpn calculator. – Justin L. Oct 20 '10 at 16:49

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Maybe it's also worth mentioning what happens for negative x. If x equals -F_{n+1}/F_n for any index n (where F_n is the Fibonacci sequence) then some iterate of x is equal to -1 and the sequence blows up. On the other hand, for x less than -2, f(x) is in (-1, 0] so f(f(x)) is positive. And for x greater than -1.5, f(x) is less than -2. The danger zone is [-2, -1.5] and there is some repelling behavior here away from -phi, in addition to the blowing up; note that on this interval f is expansive rather than contractive. – Qiaochu Yuan Oct 20 '10 at 21:42

You are iterating the operation $$x\mapsto\frac1{x+1}.$$ We can represent this in matrix terms. Set $$A=\begin{pmatrix} 0&1\\ 1&1\end{pmatrix}.$$ If $$A \begin{pmatrix} x\\ 1\end{pmatrix} =\begin{pmatrix} y_1\\ z_1\end{pmatrix}$$ then $1/(x+1)=y_1/z_1$. After $n$ iterations one gets $$A^n \begin{pmatrix} x\\ 1\end{pmatrix} =\begin{pmatrix} y_n\\ z_n\end{pmatrix}$$ and $x_n=y_n/z_n$ is got by applying the map $x\mapsto 1/(x+1)$ $n$ times to $x$.

Naturally you won't be surprised to find that the eigenvalues of $A$ are $\tau=\frac12(1+\sqrt5)$ and $\tau=\frac12(1-\sqrt5)$. The eigenvectors are $v=(1\ \tau)^t$ and $w=(1\ \tau')^t$. We can write $$\begin{pmatrix} x\\ 1\end{pmatrix}=av+bw$$ and so $$\begin{pmatrix} y_n\\ z_n\end{pmatrix}=a\tau^n v+b\tau'^nw.$$ Now $\tau>1>|\tau'|$ so that for large $n$, $y_n$ and $z_n$ are very close to $a\tau^n$and $a\tau^{n+1}$ so that $x_n\to\infty$ if $a=0$. The only exception is when $a=0$ which only arises for $x=-\frac12(1+\sqrt5)$. (I bet you didn't test that one!)

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2
I also bet he didn't test $x=\frac(1-\sqrt{5})$. ;-) – Hans Lundmark Oct 20 '10 at 7:55

1
It's hard to test that $x = -(1+\sqrt{5})/2$ is a fixed point numerically, because it's a repelling fixed point. – Michael Lugo Oct 20 '10 at 16:31

1
(Just to clarify: My comment refers to an earlier version where there was a typo. It's been corrected now.) – Hans Lundmark Oct 20 '10 at 17:29
as I remark in the comments to my answer there are a few other exceptions, namely cases where z_n = 0 for some finite n. Of course one can continue the recursion beyond this if one is willing to work in P^1(R) instead of R... – Qiaochu Yuan Oct 21 '10 at 10:24

This was supposed to be an addendum to Hans's comment, but it got too long.

The iteration Justin considered formally generates a continued fraction:

$$\cfrac{1}{1+\cfrac{1}{1+\dots}}$$

and as I mentioned in this answer, the numerators and denominators of the nth convergent of a continued fraction can be computed recursively.

If we apply the formula in that answer to this situation, we get

$$\begin{bmatrix}C_n\\D_n\end{bmatrix}=\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$$

with initial conditions

$$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix},\qquad \begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$

From here, the setup is now similar to Robin's answer, since the two recursions in fact generate the Fibonacci numbers: $C_n=F_n$ and $D_n=F_{n+1}$

Your continued fraction then is the limit

$$\lim_{n\to\infty}\frac{F_n}{F_{n+1}}$$

and the equivalence to Robin's answer is due to the Binet formula:

$$F_n=\frac{\varphi^n-\left(-\varphi^{-1}\right)^n}{\sqrt{5}}$$

(which in fact can be derived from Robin's answer).

Substituting that into the limit and evaluating gets you the answer.

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Here is another way of looking at it.

First consider the special case of starting with $1$.

Consider what happens when $\displaystyle x = \frac{f_n}{f_{n+1}}$ where $f_n$ is the $n^{th}$ fibonacci number.

You get $$\frac{1}{\frac{f_n}{f_{n+1}} + 1} = \frac{f_{n+1}}{f_n + f_{n+1}} = \frac{f_{n+1}}{f_{n+2}}$$

Since $\displaystyle 1 = \frac{f_1}{f_2}$

we see that after $n$ iterations, $\displaystyle x = \frac{f_{n+1}}{f_{n+2}}$

This can be generalized to any other starting value, by using a Fibonacci like sequence, which satisfies the recurrence $\displaystyle a_{n+2} = a_{n+1} + a_{n}$ and choosing appropriate $a_{2}$ and $a_{1}$ so that $\displaystyle \frac{a_1}{a_2}$ is the initial guess for $x$.

The $n^{th}$ value for $x$ will be given by $\displaystyle \frac{a_n}{a_{n+1}}$

The general formula for such sequences is given by $a_{n} = A\alpha^n + B\beta^n$ where $\alpha,\beta$ are roots of $\displaystyle z^2 = z + 1$ and thus the limit of $\displaystyle \frac{a_n}{a_{n+1}}$ can be easily found, which will be one of $1/\alpha$ or $1/\beta$ (which you can also see, by assuming there is a limit $1/L$ and setting $\displaystyle 1/L = \frac{1}{1+1/L}$).

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Define the function $f(x) = \frac1{1+x}$ so the process you defined is just going from x to f(x). let $x_0\in\mathbb{R}$ and $x_{i+1} = f(x_i)$ then your claim is that $\lim\; x_i = \frac {1+\sqrt{5}}{2}$. if this series converges then $x = \lim\; x_i = \lim\; f(x_{i-1}) = f( \lim\; x_{i-1} ) = f(x)$ so the limit is $x=\frac{1}{1+x}$ or in other words you have $x^2+x-1=0$. the solutions are $\frac{-1\pm \sqrt{5}}{2}$ this isn't exactly the golden ratio. if you take $\frac1{x-1}$ instead then you'll have it. you still need to find out for which $x_0$ this sequence converges - this could be done using numerical analysis (sorry, I'm a bit rusty on this, so I'll leave this part to someone else).

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