Home Turing's sunflower

In a letter to the zoologist J.Z. Young, British mathematician Alan Turing announced that his theory of embriology could explain, among other things, the connection between leaf arrangements and Fibonacci numbers.

At present I am not working on the problem at all, but on my mathematical theory of embriology [...]. This is yielding to treatment, and it will so far as I can see, give satisfactory explanations of -

[...]

iii) Leaf arrangement, in particular the way the Fibonacci series (0, 1, 1 = 2, 3, 5, 8, 13,.......) comes to be involved.

A plausible explanation of this fact was given by German physicist Helmut Vogel in "A better Way to Construct the Sunflower Head", where he proposed that spiral branches of seeds in a sunflower head are added from the center at an angle of \(137.5^{\circ}\) from the preceding one.

The interactive visualization shows that even small deviations from this value determine different shapes, not always similar to how seeds are arranged in a sunflower. Why \(137.5^{\circ}\)? Vogel postulated that

each new branch fits into the largest still-existing gap between older branches, cutting a constant fraction off that gap.

Let \(\delta = 2z\pi\) be the angle between two branches. We see that \(z\) can't be a rational, otherwise after enough steps we'd have gaps that will never be filled. Let's instead construct a sequence of rationals \(z_n\) converging to the real \(z\).

At the \(n\)-th step we'd have \(z_n = \frac{p_n}{q_n}\), which means, counting from \(0\), that the \(q_n\)-th branch coincides with the \(0\)-th and that we've done \(p_n\) turns around the center.

At the \((n + 1)\)-th step the \(q_n\)-th branch cuts a fraction \(z_{n+1}\) off the gap between branch \(0\) and its neighbor, which was \(\frac{z_n}{p_n} = \frac{1}{q_n}\). Thus the \(q_n\)-th branch lies at \(p_n + \frac{z_{n+1}}{q_n} = q_{n}z_{n + 1}\)turns, therefore \[z_{n + 1} = \frac{p_n}{q_n} + \frac{z_{n+1}}{q_n^2} = \frac{p_n}{q_n}\left(1 - \frac{1}{q_n^2}\right)^{-1} = \frac{p_{n}q_{n}}{q_n^2 - 1}\text{.}\]

But we also have \(z_{n + 1} = \frac{p_{n + 1}}{q_{n + 1}}\), thus we have the equation \[z_{n + 1} = \frac{p_{n + 1}}{q_{n + 1}} = \frac{p_{n}q_{n}}{q_n^2 - 1}\text{.}\] By substituting \(p_n = F_{2m}\) and \(q_n = F_{2m + 2}\), consecutive Fibonacci numbers of even order, the previous equation becomes \[F_{2m + 2}^2 - 1 = F_{2m}F_{2m + 4}\text{,}\] which follows from Cassini's Identity.

Thus, when \(n\rightarrow\infty\), we have that \(z_n\rightarrow \varphi^{-2}\) where \(\varphi = \frac{1 + \sqrt{5}}{2}\), the golden ratio, because it is well known that the ratio of consecutive Fibonacci numbers tends to \(\varphi\). Therefore \[\delta = 2\varphi^{-2}\pi \approx 137.5^{\circ}\text{.}\]