Golden ratio and phyllotaxis, a clear mathematical link

Abstract

Exploiting Markoff’s theory for rational approximations of real numbers, we explicitly link how hard it is to approximate a given number to an idealized notion of growth capacity for plants which we express as a modular invariant function depending on this number. Assuming that our growth capacity is biologically relevant, this allows us to explain in a satisfying mathematical way why the golden ratio occurs in nature.

Appendix

We calculate here the limit of the averaging integral (7) in the case of the golden ratio $x=\varphi$, whose Hermite’s convergents are the quotients $F_{n+1}/F_{n}$. The required calculation is greatly simplified using the simplified expression

$$\begin{aligned} f_\varphi ^{(n)}(t)=\varphi ^{-2\,(n+1)}t+F_n^2\,t. \end{aligned}$$

(8)

for the function $f_\varphi ^{(n)}$, which follows from the fact that $(F_{n}-F_{n+1}\,\varphi )^2=\varphi ^{-2\,(n+1)}$. The corresponding minimum occurs at $F_n\,\varphi ^{n+1}$, and takes the value

$$\begin{aligned} 2F_n/\varphi ^{n+1}\approx 2/\sqrt{5}. \end{aligned}$$

These assertions follows from the well known Binet formula

$$\begin{aligned} F_n=\frac{\varphi ^{n+1}-(-1/\varphi )^{n+1}}{\sqrt{5}}. \end{aligned}$$

Exploiting that, $0\le \left| {(-1/\varphi )^{n+1}}\right| \ll 1$, we also deduce from it the very good approximation $F_n\approx {\varphi ^{n+1}}/{\sqrt{5}}$. Thus $f_\varphi ^{(n)}(t)$ is very well approximated by $\varphi ^{-2\,(n+1)}\,t+{\varphi ^{2\,(n+1)}}/{(5\,t)}$ when n is large enough. We may also calculate that

$$\begin{aligned} t_n(\varphi )= & {} \sqrt{\frac{F_n^2-F_{n-1}^2}{\varphi ^{-2\,n}-\varphi ^{-2\,(n+1)}}} \\= & {} \varphi ^{n}\,F_{n-1}\,\sqrt{((F_n/F_{n-1})^2-1)\,\varphi }\\\approx & {} \varphi ^{n+1}F_{n-1}\approx \varphi ^{2\,(n+1)}/\sqrt{5}, \end{aligned}$$

from which we get

$$\begin{aligned} \frac{t_{n+1}^2-t_n^2}{t_{n+1}-t_{n}}=t_{n+1}+t_{n}\approx \varphi ^{n+1}(\varphi \,F_{n}+F_{n-1})= \varphi ^{2\,(n+1)}, \end{aligned}$$

as well as

$$\begin{aligned} \frac{\log (t_{n+1}/t_n)}{t_{n+1}-t_{n}}\approx \sqrt{5}\,\frac{\log (\varphi ^{2\,n+3}/\varphi ^{2\,n+1})}{\varphi ^{2\,n+3}-\varphi ^{2\,n+1}}=2\,\sqrt{5}\,\frac{ \log (\varphi )}{\varphi ^{2\,(n+1)}} \end{aligned}$$

Hence, applying formula (7), we find that

$$\begin{aligned} g_\varphi= & {} \limsup _{n\rightarrow \infty }\frac{1}{t_{n+1}-t_{n}}\int _{t_{n}}^{t_{n+1}} f_\varphi ^{(n)}(t)\, dt \\= & {} \limsup _{n\rightarrow \infty } \frac{1}{t_{n+1}-t_{n}}\left[ \varphi ^{-2\,(n+1)}\,\frac{t^2}{2} + \frac{\varphi ^{2\,(n+1)}}{5}\log (t) \right] _{t=t_{n}}^{t=t_{n+1}} \\= & {} \limsup _{n\rightarrow \infty } \frac{\varphi ^{-2\,(n+1)}}{2}\,\left( \frac{t_{n+1}^2-t_n^2}{t_{n+1}-t_{n}}\right) + \frac{\varphi ^{2\,(n+1)}}{5}\left( \frac{\log (t_{n+1}/t_n)}{t_{n+1}-t_{n}} \right) \\= & {} \frac{1}{2} +\frac{2}{\sqrt{5}}\log (\varphi ). \end{aligned}$$

In the case of $\psi :=1+\sqrt{2}$, one replaces Fibonacci numbers by Pell numbers, $P_n$, and uses

$$\begin{aligned} f_\psi ^{(n)}(t)= {P_n}/{t} +\psi ^{-2\,n}\,t,\qquad P_n\approx {\varphi ^n}/{\sqrt{8}},\qquad \mathrm{and}\qquad t_n\approx {\psi ^{2\,n+1}}/{\sqrt{8}}, \end{aligned}$$

to show that $g_\psi ={1}/{2} +\log (\psi )/{\sqrt{8}}$ with a very similar calculation.