To recap, the Fibonacci sequence is the sequence you get if you start off with $F_1 =1$ and $F_2 = 1$ and then iterate using $F_{n+2} = F_{n+1}+F_n$. So:
$$
(F_n) = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...
$$
This all seems highly contrived, but somehow turns up in seed and leaf patterns. How?
The answer has to do with Phyllotaxis. This is the manner in which new leaves or seeds are formed. Imagine a long stem plant growing upwards and generating new leaves as it grows. The tip of the plant is known as the bud and embryonic leaves known as primordia appear on it and migrate to one side as the stem grows. This migration is what is known as phyllotaxis. It seems that - to a good approximation - the primordia choose an angle (around the circumference of the stem) early on and then stick with that.
The picture above shows what this looks like from above. The oldest leaves are the largest and appear furthest from the stem; the newest leaves are smaller and appear nearer the stem. Looking at the leaves on this long stemmed plant from above, it becomes obvious that the pattern of seeds on sunflowers is generated in exactly the same way, it's just that the $z$ dimension is flattened. So on a sunflower we can think of each seed as being created in the centre and choosing an angle, after which it migrates outwards radially.
Another feature is apparent from the picture above: the smaller leaves that are higher up can potentially shade the larger ones below. So the question naturally arises - what's a good way to choose the angle for the next primordia such that the amount of shading is minimized? (And remember that the mechanism must be encoded into the DNA and so has to be fairly simple.)
Some plants adopt a simple approach of producing 2 primordia at a time 180 degrees apart from each other and 90 degrees from the previous pair. This behaviour is called decussate and is obviously pretty trivial to encode in DNA. Also if there is enough stem separating each leaf pair from the last-but-one then shading of leaves directly below is unlikely to be a problem. However, there is another behaviour which is even easier to encode and works for shorter stems. This is to choose a fixed angle $\theta$ to separate each primordia from the last and to try to optimise the angle to produce minimal shading.
The remainder of this post will attempt to show two things: Firstly that the best angle to choose is $\frac{2\pi}{\phi^2}$ where $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio; and secondly that this results in numbers of clockwise and anti-clockwise spiral arms which are consecutive pairs from the Fibonacci sequence.
Let's call the angle $\theta$ and stick with that. It is clear that if $\frac{\theta}{2\pi} \in \mathbb{Q}$ - i.e. if $\theta$ is a rational fraction of the circumference - then eventually a leaf will completely overlap a leaf below it, and thereafter each new leaf will overlap another. So we're looking for irrational $\theta$. Let's expand $\frac{\theta}{2\pi}$ as a continued fraction (which we know must go on forever since $\theta$ is irrational):
$$
\frac{\theta}{2\pi} = \frac{1}{n_1+\frac{1}{n_2+\frac{1}{n_3+\frac{1}{n_4+\frac{1}{...}}}}}
$$
... where $n_i \in \mathbb{N}^+$. Now it is clear that if any of the $n_i$ in a continued fraction are "large" then we can approximate very well by a rational number - just set the $n_j = \infty, \forall j \ge i$. This means that in a sense the least approximable irrational number is
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{...}}}}}
$$
This is just $\phi - 1$. How do we know this? Well, because if you add one to it you get the postive solution of
$$
\begin{align}
\frac{1}{x} &= x - 1 \\
\iff x^2 &= x + 1
\end{align}
$$
and that's the definition of $\phi$.
So the best choice of $\theta$ is $2\pi (\phi - 1)$ which equals $\frac{2\pi}{\phi}$. Note that $\frac{2\pi}{\phi}$ clockwise is the same as $\frac{2\pi}{\phi^2}$ anti-clockwise since $\frac{1}{\phi}+\frac{1}{\phi^2} = 1$.
The picture below shows the result of choosing $\theta = \frac{2\pi}{(b/a)^2}$ where a, b are consecutive pairs from further and further along the Fibonacci sequence. Note that the ratio $b/a$ converges and since its limit $x$ must satisfy $x^2 = x + 1$ this ratio converges to $\phi$.
So, the next question is: Why should we expect the count of clockwise and anti-clockwise spiral arms to be consecutive pairs from the Fibonacci sequence? To see why it helps to plot the leaves in a slightly different way. Below is a plot of $r$ vs $\theta$, where the leaves are each placed an angle of $2\pi / \phi^2$ apart (and obviously we wrap every time $\theta$ exceeds $2\pi$).
What can we say about these? Well, suppose $n$ is the step between priomordia which take part in a spiral. Then the primordia involved in the first spiral arm are $1, 1 + n, 1 + 2n, 1+3n, ...$, the ones in the second are $2, 2 + n, 2 + 2n, 2+3n, ...$ and there are $n$ spiral arms. So there is a set of $n$ spiral arms for each choice of $n$, but which values of $n$ stand out? Well, note that if $n=F_m$ for some m then $n\phi \approx F_{m+1}$, which says that if we rotate $\frac{2\pi}{\phi}$ a total of $F_{m+1}$ times then we will almost get back to where we were. This tells us that the spiral defined by this step size is very nearly a straight radial line and therefore stands out clearly, and that the larger the value of $m$ in $n=F_m$ the more clearly the spiral stands out. But the step size is also the number of arms and so what we have discovered is that the number of arms is a number from the Fibonacci series. The same is true for the 2nd most obvious spiral.
Note that there is a little bit of ambiguity here since we could have chosen any $F_m$. However, there are two competing forces which work together to select the spirals which most stand out. These are i) $m$ should be as large as possible for the slope in the $r$ vs $\theta$ graph to be as near vertical as possible, and ii) there should be enough primordia involved in the spiral for it to be identifiable. This means that value of $m$ corresponding to the spiral that stands out the most actually changes as the number of primordia visible increases. This is why when we count the clockwise and anti-clockwise spiral arms we sometimes get 3,5, sometimes 5,8, sometimes 8,13, and so on.
Finally, observe that the sequence
$$
\frac{F_2}{F_1}, \frac{F_3}{F_2}, \frac{F_4}{F_3}, \frac{F_5}{F_4}, ...
$$
alternates between values greater than and less than $\phi$. This explains why the two spirals that stand out most in any given plant, corresponding to $n = F_m$ and $n=F_{m+1}$ (for some $m$) are a clockwise/anti-clockwise pair.
References
I found lots of references on the web like http://gofiguremath.org/natures-favorite-math/the-golden-ratio/the-golden-angle/ which simply state the facts about the golden angle (i.e. $2\pi/\phi^2$) appearing in some plants, and the numbers of spiral arms being pairs from the Fibonacci sequence. However, I found it difficult to find any web pages that explain why. There do seem to be papers in journals relating to this question but I didn't access any of them for the following reasons: i) I'm lazy ii) I'm mean iii) I wanted to figure it out for myself.In trying to figure it out for myself I got stuck and had to look for some extra clues: In https://goldenratiomyth.weebly.com/phyllotaxis-the-fibonacci-sequence-in-nature.html it calls $\phi$ the most irrational number and this provided a clue because clearly we do want $\theta/2\pi$ to be irrational. However I had no idea what most irrational number actually meant until I found this answer on math overflow.
So, although I figured this out for myself, it is unlikely to be all that original. But I did have fun doing it.
The illustrations were all created using python and matplotlib.
Postscript
For the skeptical I've taken a picture of a real sunflower from a friend's garden and annotated it. 34 clockwise spirals and 55 anti-clockwise ones!
Comments
Post a Comment