Pi Day!

Pi Day is coming up again (3/14 as a US date). The number π is, of course, 3.14159265… Here are some possible activities for children:

  • Search for your birthday (or any other number) in the digits of π
  • Follow in the footsteps of Archimedes, showing that π is between 22/7 = 3.1429 and 223/71 = 3.1408.
  • Calculate 333/106 = 3.1415 and 355/113 = 3.1415929, which are better approximations than 22/7.
  • Measure the circumference and diameter of a round plate and divide. Use a ruler to measure the diameter and a strip of paper (afterwards measured with a ruler) for the circumference. For children who cannot yet divide, try to find a plate with diameter 7, 106, or 113.
  • Calculate π by measuring the area of a circle (most simply, with radius 10 or 100), using A = πr2. An easy way is to draw an appropriate circle on a sheet of graph paper.

You can also try estimating π using Buffon’s needle. You will need some toothpicks (or similar) of length k and some parallel lines (such as floorboards) a distance d apart (greater than or equal to k). Then the fraction of dropped toothpicks that touch or cross a line will be 2 k / (π d), or 2 / π if k = d. There is an explanation and simulator here (see also the picture below). And, of course, you can bake a celebratory pie and listen to Kate Bush singing π, mostly correctly!

This picture by McZusatz has 11 of 17 matches touching a line, suggesting the value of 2×17/11 = 3.1 for π (since k = d).

Actually, of course, π = 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 … (digits in red are sung by Kate Bush, accurately, although some have said otherwise).

Fast Fibonacci numbers

There was some discussion on reddit recently of the Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1,597, 2,584, 4,181, 6,765, 10,946, 17,711, 28,657, 46,368, 75,025, 121,393, 196,418, 317,811, 514,229, 832,040, …) and efficient ways of calculating them.

One way of doing so is using numbers of the form a + b σ where  σ is the square root of 5. Multiplication of such numbers satisfies:

(a + b σ) × (c + d σ) = ac + 5bd + (ad + bc) σ.

We can define the golden ratio φ = (1 + σ) / 2 and also ψ = 1 − φ = (1 − σ) / 2, in which case the nth Fibonacci number Fn will be exactly (φn − ψn) / σ. This is known as Binet’s formula.

We can use this formula to calculate Fibonacci numbers using only integer arithmetic, without ever evaluating the σ. We will have:

(2 φ)n − (2 ψ)n = (1 + σ)n − (1 − σ)n = 0 + p σ

for some integer p, and division by a power of two will give Fn = p / 2n.

I am using the R language, with the gmp package, which provides support for large integer matrices, and this allows us to use the relationship:

If we call this matrix A and calculate An−1, the first number in the resultant matrix will be the nth Fibonacci number Fn. The following R code calculates F10 = 55 using a combination of multiplication and squaring:

n <- 10

A <- matrix.bigz(c(
	1, 1,
	1, 0), 2)

p <- function(n) {
	if (n == 1) A
	else if (n %% 2 == 1) A %*% p(n-1)
	else {
		b <- p(n/2)
		b %*% b


This same code will calculate, for example:

The time taken to calculate Fn is approximately proportional to n1.156, with the case of n = 1,000,000,000 (giving a number with 208,987,640 digits) taking about a minute.

Some knots

I have been reading up on knots recently, so here is a table of 8 knots with up to 6 crossings (plus 2 extras), along with their Conway polynomials. These are polynomials that can be associated with each knot. Different polynomials imply different knots, but sometimes different knots have the same polynomial. Some examples are shown in red below. I have also included pictures of the ten knots mentioned (not drawn by me).

Knot Conway polynomial
01 (Unknot) 1
31 (Trefoil knot) z2 + 1
41 (Figure-eight knot) 1 − z2
51 (Cinquefoil knot) z4 + 3 z2 + 1
52 (Three-twist knot) 2 z2 + 1
61 (Stevedore’s knot) 1 − 2 z2
62 (Miller Institute knot)  − z4 − z2 + 1
63 z4 + z2 + 1
946 1 − 2 z2
10132 z4 + 3 z2 + 1

The pancake theorem

A previous post mentioned the Borsuk–Ulam theorem. A corollary (for the case of a circle) is the pancake theorem: given two plane regions, there is a cut (line) which divides both in half by area.

The example shows a cut dividing both South America and Africa in half (the theorem doesn’t tell us how to find the cut; I used simulated annealing).

The corollary for the 2-sphere is the ham sandwich theorem: given three 3-dimensional objects, there is a cut (plane) which divides all three in half by volume.

The Borsuk–Ulam theorem

The mathematical tidbit for today is the Borsuk–Ulam theorem, which states that every continuous function f from the n-dimensional sphere to n-dimensional space must satisfy f(p) = f(−p) for some point p.

In particular, every continuous function f from a 2-dimensional sphere (say, the Earth’s surface) to the plane must satisfy f(p) = f(q) for some antipodal pair of points p and q.

Thus, if we can describe weather by a pair of numbers (say, temperature and rainfall), there must be an antipodal pair of points p and q with the same weather (because two numbers specify a point in the plane).

The maps above (for average maximum temperature) and below (for rainfall) show July weather at various places on Earth, and a pair of points with the same weather is highlighted.

It’s a miracle that it works in this case, of course, because the maps only define temperature and rainfall on the land; I would not have been able to recognise a suitable antipodal pair of points if one or both were at sea.

The Projective Plane

I have been thinking some more about the famous Möbius strip (see also my post on the Klein bottle). The so-called “Sudanese Möbius Band” in the video above is a Möbius strip stretched so as to make the boundary perfectly circular (it is not named after the country, but after the topologists Sue E. Goodman and Daniel Asimov, and you can purchase a plastic one here).

If we glue two of these Möbius strips together (not actually possible in 3 dimensions), we get a Klein bottle. If we glue one to a disc (also not possible in 3 dimensions), we get a projective plane.

Just for fun, the video below shows a Game of Life glider on the projective plane. The top and bottom of the square are considered to be joined, as are the left and right sides. In both cases, there is a reversal of orientation (a manoeuvre not really possible in 3 dimensions). The glider changes colour as it changes orientation.

Video produced using the R animation package.

The Klein Bottle

I have been thinking about the famous Klein bottle (above). To quote a limerick by Leo Moser:

A mathematician named Klein
Thought the Möbius band was divine.
Said he: “If you glue
The edges of two,
You’ll get a weird bottle like mine.”

Just for fun, here is a Game of Life glider on a Klein bottle. The top and bottom of the square are considered to be joined, so as to form a tube. The ends of the tube (vertical sides of the square) are also joined, but with a reversal of orientation (a manoeuvre not really possible in three dimensions). The glider changes colour as it changes orientation.

Video produced using the R animation package.

Some posters

For people with children, these high-resolution posters are intended for printing on A3 paper, and can be freely downloaded:

Mathematics posters: Flat shapes, Solid Shapes, Prime Factors (colour-coded)

Biology posters: Ladybirds of Australia, Plants and Fungi, Some Flowering Plants (monocots marked with a dot)

Astronomy/geography posters: Southern Cross (showing colours and magnitudes of stars), Orion (ditto), Geographical Features

Technology posters: Vehicles, Milestones in Materials

Australia Day Honours, 2021

It’s Australia Day again, which means the annual Australia Day honours. Some recipients of note this year in the field of the sciences include:

Congratulations to all four of these outstanding Australians, and to the many others on the list.

For another perspective: