Why does the Fibonacci sum of ANY 2 real numbers converge to the golden ratio?

Question

You're probably wondering what I mean by "Fibonacci sum". I mean that if you start a sequence with any 2 real numbers, not necessarily positive ones, and add them up to generate the next number in the sequence, and then add that number with the previous number in the sequence, on and on, will always approach the golden ratio, approximately $1.6180339887$. An example might clarify things further.

Start with $5$ and $6$. Their ratio is $\frac{6}{5}$ which is $1.2$, and their sum is $11$. Now you have $6$ and $11$. Their ratio is $\frac{11}{6}$, which is approximately $1.83333$, and their sum is $17$. Now you have $11$ and $17$. Their ratio is $\frac{17}{11}$, which is around $1.54545$, and their sum is $28$. Now you have $17$ and $28$. Their ratio is $\frac{28}{17}$, which is around $1.64706$, and their sum is $45$. Now you have $28$ and $45$. Their ratio is $\frac{45}{28}$, which is around $1.60714$, and their sum is $73$. Now you have $45$ and $73$. Their ratio is $\frac{73}{45}$, which is around $1.62222$, and their sum is $118$. Now you have $73$ and $118$. Their ratio is $\frac{118}{73}$, which is around $1.61644$, and their sum is $191$. Now you have $191$ and $118$. Their ratio is $\frac{191}{118}$, which is about $1.61864$. And you continue on like this.

And it doesn't have to be positive numbers or whole numbers. Regardless of which 2 numbers you start with, it will always converge to the golden ratio (with the exception of $0$ and $0$, of course). Try it yourself.

I want to know why this fact is true and why it is connected to various other ways you might define the golden ratio.

Answer 1

Your observation applies not only to the Fibonacci sequence, but to linear second-order sequences in general. I have written frequently on these pages of the general solution to the sequence $f_n=af_{n-1}+bf_{n-2}$, with arbitrary initial conditions, $f_{0,1}$. See, for example, here.

The general solution can be expressed as

$$ f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n) $$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.

This can also be expressed as

$$ f_n=\frac{f_0-f_1\beta}{\alpha-\beta}\alpha^n-\frac{f_0-f_1\alpha}{\alpha-\beta}\beta^n $$

Since $\alpha$ is typically greater than $\beta$, the limiting ratio will be

$$ \lim_{n\to \infty} \frac{f_{n+1}}{f_n}=\alpha $$

for $f_{0,1}$ positive, negative, or complex. That is, unless $f_0-f_1\beta=0$, and the limit is equal to $\beta$.

Answer 2

Here is a proof for rational initial conditions and for the quotient $f_{2k+1}/f_{2k}$. Perhaps you can extend it to your case.

Let $a,b \in \mathbb{Q}$ with $(a,b) \neq (0,0)$. Define $$\begin{aligned}[b]f_0 &= a\\f_1 &= b\\f_{n+2} &= f_{n+1} + f_{n}\\\end{aligned}.$$ Then, $$\begin{aligned}[b]f_{n+2} &= f_{n+1} + f_{n}\\f_{n+3} &= f_{n+2} + f_{n+1} = 2f_{n+1} + f_{n}\\\end{aligned}.$$ That is, $$\left(\begin{array}{c}f_{n+2}\\f_{n+3}\end{array}\right) = \left(\begin{array}{cc}1&1\\1&2\end{array}\right)\left(\begin{array}{c}f_n\\f_{n+1}\end{array}\right).$$ Thus, $$\left(\begin{array}{c}f_{2k}\\f_{2k+1}\end{array}\right) = \left(\begin{array}{cc}1&1\\1&2\end{array}\right)^k\left(\begin{array}{c}f_0\\f_{1}\end{array}\right).$$ The matrix is diagonalizable: $$\left(\begin{array}{cc}1&1\\1&2\end{array}\right) = \underbrace{\left(\begin{matrix}\frac{-\sqrt{5}-1}{2} & \frac{\sqrt{5}-1}{2} \\1 & 1\end{matrix}\right)}_{P}\underbrace{\left(\begin{matrix}\frac{-\sqrt{5}+3}{2} & 0 \\0 & \frac{\sqrt{5}+3}{2}\end{matrix}\right)}_{D}\underbrace{\left(\begin{matrix}\frac{-\sqrt{5}}{5} & \frac{-\sqrt{5}+5}{10} \\\frac{\sqrt{5}}{5} & \frac{\sqrt{5}+5}{10}\end{matrix}\right)}_{P^{-1}}.$$ Then, $$\left(\begin{array}{c}f_{2k}\\f_{2k+1}\end{array}\right) = P\left(\begin{matrix}\left(\frac{-\sqrt{5}+3}{2}\right)^k & 0 \\0 & \left(\frac{\sqrt{5}+3}{2}\right)^k\end{matrix}\right)P^{-1}\left(\begin{array}{c}f_0\\f_{1}\end{array}\right).$$ Define $$\operatorname{Ra}(x) = \frac{x+\bar x}{2}, \qquad \operatorname{Co}(x) = \frac{x - \operatorname{Ra}(x)}{\sqrt{5}},$$where $\bar x$ is the $\sqrt{5}$-conjugate. i.e. $$x = c + d\sqrt{5},\quad \bar x = c - d\sqrt{5},$$for $c,d \in \mathbb{Q}$. Then, we can write $$x = \operatorname{Ra}(x) + \operatorname{Co}(x) \sqrt{5}.$$ Let $\Phi = \frac{\sqrt{5} - 1}{2}, \gamma = \frac{\sqrt{5} + 3}{2}, \alpha = \frac{\sqrt{5}}{5}$ and $\beta = \frac{\sqrt{5}+5}{10}$. Then, $$ \left(\begin{array}{c}f_{2k}\\f_{2k+1}\end{array}\right) = \left(\begin{matrix}\bar \Phi & \Phi \\1 & 1\end{matrix}\right)\left(\begin{matrix}\bar\gamma^k(\bar\alpha f_0 + \bar\beta f_1)\\\gamma^k(\alpha f_0 + \beta f_1)\end{matrix}\right). $$ Let $\Gamma_k = \gamma^k(\alpha f_0 + \beta f_1)$. Then, $$ \left(\begin{array}{c}f_{2k}\\f_{2k+1}\end{array}\right) = \left(\begin{matrix}\bar\Phi\bar\Gamma_k + \Phi\Gamma_k\\ \bar\Gamma_k + \Gamma_k\end{matrix}\right). $$ So, $$ \begin{aligned} \lim_{k \to \infty}\frac{f_{2k}}{f_{2k+1}} &= \lim_{k \to \infty}\frac{\operatorname{Ra}(\Phi \Gamma_k)}{\operatorname{Ra}(\Gamma_k)}\\ &= \operatorname{Ra}(\Phi) + 5\operatorname{Co}(\Phi) \lim_{k \to \infty}\frac{\operatorname{Co(\Gamma_k)}}{\operatorname{Ra}(\Gamma_k)}\\ &= \Phi. \end{aligned} $$ This follows because:

1. $\frac{\Gamma_k}{\operatorname{Ra}(\Gamma_k)} = 1 + \frac{\operatorname{Co(\Gamma_k)}}{\operatorname{Ra}(\Gamma_k)} \sqrt{5}$, from definition.
2. $\frac{\operatorname{Ra}(\Gamma_k)}{\Gamma_k}=\frac{\operatorname{Ra}(\gamma^k)}{\gamma^k} \frac{\operatorname{Ra}(\omega)}{\omega} + 5\frac{\operatorname{Co}(\gamma^k)}{\gamma^k}\frac{\operatorname{Co}(\omega)}{\omega}$, where $\omega = \alpha f_0 + \beta f_1$, from definition.
3. $\lim\limits_{k \to \infty}\frac{\operatorname{Ra}(\gamma^k)}{\gamma^k} = \lim\limits_{k \to \infty} \frac{(c+d\sqrt{5})^k + (c-d\sqrt{5})^k}{2(c+d\sqrt{5})^k} = \frac{1}{2}$, with $c = 3/2$ and $d = 1/2$.
4. $\frac{\operatorname{Ra}(\gamma^k)}{\gamma^k} + \frac{\operatorname{Co}(\gamma^k)}{\gamma^k}\sqrt{5} = 1$, from definition.
5. $\lim\limits_{k \to \infty}\frac{\operatorname{Ra}(\Gamma_k)}{\Gamma_k} = \frac{1}{2}$, substituting from (3) and (4) into (2).
6. $\lim\limits_{k \to \infty}\frac{\Gamma_k}{\operatorname{Ra}(\Gamma_k)} = 2$, from 5.
7. $\lim\limits_{k \to \infty}\frac{\operatorname{Co(\Gamma_k)}}{\operatorname{Ra}(\Gamma_k)} = \frac{\sqrt{5}}{5}$, substituting (6) into (1).
8. $\lim\limits_{k \to \infty}\frac{f_{2k}}{f_{2k+1}} =\Phi$, using (7).

Finally, $$\lim\limits_{k \to \infty}\frac{f_{2k+1}}{f_{2k}} = \frac{\sqrt{5}+1}{2} = \phi,$$ derived from (8).