The Mathematics of

Growth, Form and Size

Part 2

Fibonacci numbers & the Golden Flower

A page from the Liber Abbaci of Leonardo of Pisa (Fibonacci)

Fibonacci's Rabbit Problem: How many pairs of rabbits can be bred in one year from one pair?

A certain person places one pair of rabbits in a certain place surrounded on all sides by a wall. We want to know how many pairs can be bred from that pair in one year, assuming that it is in their nature that each month they give birth to another pair, and in the second month after birth each new pair can also breed.

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...

In the reproductive cycle of the honey bee:

(1) Each male is born asexually from a single female; males are called drones.

(2) Each female is born from the mating of a male and a female.

Look at the family tree of a male bee …

• F • M • F 3 g'g'parents

• F 2 g'parents • F 1 parent

• M

• M

Steps Problem Taking either one or two steps at a time, how many different ways are there to climb n steps?

Sn = Sn -1 + Sn - 2

Population growth-rate

Theorem: Rn = fn+1 /fn → φ 1/1 , 2/1 , 3/2 , 5/3 , 8/5, 13/8 … → 1.618... = 1/2(1 +√5 )1 2 1.5 1.66.. 1.6 1.625

Proof:

Remember: φ = 1 + 1/φ = 1 + 0.618... = 1.618...

GRAPHING THE GROWTH-RATE

Fibonacci numbers converge from both sides on the

Golden Ratio

Counting petals, leaves, spirals

Counting petals, leaves, spirals

Why Fibonacci numbers? Why spirals?

Principles of growth:

Leaves appear on the plant stem as it grows longer, spiralling around for optimal light-sharing;

Seed primordia appear in the growing flower-head or fircone, moving outward to make room for the next ones and spiralling around for optimal packing strength and efficiency.

Count t circuits around stem before a leaf appears (almost) directly above the first leaf.

Count n leaves along this spiral.

Then t/n is the divergency constant for this plant species. It's often 2/3 , or 3/5, or 5/8.

Why? If the angle turned through between leaves is (say) ¼ of a turn, then every 4th leaf will be directly above the first – there will be four vertical columns of leaves, giving very poor sunlight sharing. All fractions have the same problem. Nature prefers irrational angles! The best is the Golden Ratio φ. Or equivalently 1/φ = φ − 1≈ 0.618.

Suppose that we count t, n, and the divergency constant is 3/5. After 5 leaves how many times have we circled the stem?

5(1/φ) ≈ 3 times, a whole number. Or, if you want, 5φ ≈ 8 times.

Each new baby primordia appears at an angle φ = 1.618... around from the previous, or, equivalently, an angle 1/φ ≈ 0.618 ≈ 222.5o , which is equivalent to 137.5o in the other direction.

Plotting flower-heads

Each new baby primordia appears at an angle φ = 1.618... around from the previous, or, equivalently, an angle 1/φ ≈ 0.618 ≈ 222.5o , which is equivalent to 137.5o in the other direction.

Each new baby primordia appears at an angle φ = 1.618... around from the previous, or, equivalently, an angle 1/φ ≈ 0.618 ≈ 222.5o , which is equivalent to 137.5o in the other direction.

And meanwhile, the primordia have been growing as they move outwards ...

The Golden Flower

Fibonacci numbers:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Successive ratios:

1/1 , 2/1 , 3/2 , 5/3 , 8/5 , … → φ = 1.618... = ½ (1 +√5)1/1 , 1/2 , 2/3 , 3/5 , 5/8 , ...→ p = 1/φ = 0.618...= ½ (√5 − 1)

1/2 < 3/5 < 8/13 < 21/34 < … < p < … < 13/21 < 5/8 < 2/3180o 216o 221.5o 222.4o 222.9o 225o 240o

What if the primordia used another angle?

Any fraction will give spokes, eventually.

Try 2/3, 3/5, 5/8, 8/13No. of spokes: 3 5 8 13

What if the primordia used another angle?

Any fraction will give spokes, eventually.

Try 2/3, 3/5, 5/8, 8/13No. of spokes: 3 5 8 13

What if the primordia used an angle between two Fibonacci ratios 8/13 and 5/8?

Recall that : 3/5 < 8/13 < 5/8 < 2/3No. of spokes: 5 13 8 3Now choose an angle slightly bigger than 8/13 : the 13 spokes will be pulled into anticlockwise spirals.

3/5 < 8/13 < 5/8 < 2/3No. of spokes: 5 13 8 3Choose an angle slightly smaller than 5/8: the 8 spokes will be pulled into clockwise spirals.

3/5 < 8/13 < p < 5/8 < 2/3

The golden angle p gives such perfect organization we can see both sets of spirals, the 8-spirals and the 13-spirals. And also the 21-spirals, 34-spirals, 55-spirals, etc.

Angle = 8/13 + 0.001

Angle = 5/8 − 0.001

Angle = p

As we move inwards towards the golden angle, the packing efficiency and strength improve.

The ultimate selection of the golden angle is a natural consequence of plant growth

dynamics – getting the best packing and strength, and maximising the number of seeds

in the available space.

Some computer modellingAssumptions: ignore seed growth, and plot the kth seed a

distance √k from the centre, to maintain constant density of seeds in concentric circles.

Area of circle / number of seeds = π(√k)2 / k = π.

Let the species seed growth angle be t, expressed as a proportion of the whole angle 2π. Take the coordinates of the kth seed to be

(x,y) = (√k sin 2πtk, √k cos 2πtk)

Let there be n seeds, of certain size, shape and colour. Plot the seed positions for k = 1,2,...,n

Script for golden flower and spiralsMATLAB R

t = 1/2*(1 + sqrt(5));

n = 2^10;

c = ones(n,1);

c(:) = 1:1024;

x = sqrt(c) .* cos(2*pi*t*c);

y = sqrt(c) .* sin(2*pi*t*c);

# To draw 34-spiral for Fibonacci

# number 34, starting from 1

plot(x,y, 'o')

v = 1:34:n;

plot(x,y, 'o')

hold on

plot(x(v), y(v))

t=1/2*(1+sqrt(5))

n=2^10

c=seq(1,n)

x=sqrt(c)*cos(2*pi*t*c)

y=sqrt(c)*sin(2*pi*t*c)

plot(x,y, pch=19, cex=1.5, col="red4")

# To draw 34-spiral for Fibonacci

# number 34, starting from 1

plot(x,y, pch=1, cex=1)

v=seq(1,n,34)

lines (x[v], y[v])

The Golden Flower drawn with R: n=2^10, cex=1

The Golden Flower drawn with R, n=2^11, cex=0.5, pch=19

34-spiral drawn with R

Golden flower with the first 8-spiral and the first 13-spiral

Golden flower with the eight 8-spirals

Golden flower with all five 5-spirals and all eight 8-spirals

Pine cone n = 2^7, with the 8-spirals and 13-spirals

Angle θ = √2

Exercises in the Golden Flower

(1) Draw the five 5-spirals, through 1, 6, 11, 16, 21, …, and through 2, 7, 12, 17, 22, ..., etc.

(2) Draw (in another colour) the eight 8-spirals with starting points 1,2,3,4,5,6,7,8.

(3) Choose any starting point N in the flower, choose any Fibonacci number F. Observe that the arithmetic series with first term N and common difference F forms a perfect spiral.

(4) Ring the Fibonacci numbers in the Golden Flower. What pattern do you see? Why is this?

Solution to Exercise 4 Ring the Fibonacci numbers in the Golden Flower. What

pattern do you see? Why is this?

Take 55, for example. The ratio 34/55 approximates very closely to the golden angle p from above, so 55p ≈ 34, slightly smaller.

Hence 34 clockwise revolutions, minus a bit!

Successive ratios of higher Fibonacci numbers get ever closer to p, alternating from above and below. Hence the 89th seed is placed at 55 revolutions plus a tiny bit. And so on.

Exercise: Show graphically that successive ratios of Fibonacci

numbers are the best rational approximations to φ

The logarithmic spiral or equi-angular spiral is the basis of the growth pattern of many biological organisms

Nautilus shell with logarithmic spiral