# Pi and its Day

I thought I contribute to Pi day with a bit of mathematics and its foundations. The question is: What is Pi?

You will immediately answer that it is the quotient of the length of the half circle by the diameter. But that is a geometrical concept. Didn’t Bourbaki tell us that we should put all mathematics on a solid foundation? In fact, in my primary course about Analysis Pi was defined as the first positive route of

$$\cos(x) = 1 – \dfrac{x^2}{2} + \dfrac{x^4}{4!} -+ \ldots$$

If you start like this you have a hard time to derive a parametric representation of the circle. You have to prove that

$$\{ (\cos(t),\sin(t)) : 0 < t \le 2\pi \} = \{ (x,y) : x^2+y^2=1 \}.$$

It is not impossible and it sure is instructive, but for most students it is no fun.

But let us accept the geometrical description for a moment. If we want to go the constructive route we need to derive a way to actually compute Pi. Plus, we wish to have a proof for the existence of the limit, or even better an error estimate.

There is a lot of literature about this. One important figure of the ancient past was Archimedes. He used n-gons and derived that pi is between 3+10/71 and 3+10/70 (see Wikipedia). But it seems that other estimates were known in China. After all it is not difficult to get estimates with measurements on big circles, and drawing big circles is not difficult. Archimedes, however, may be the first to get an estimate by reasoning alone.

The first obvious thing is to compute the length of inscribed n-gons. With a little bit of Pythagoras we get a formula for the length of the side of the 2n-gon from the n-gon. We have

$$x^2 + \dfrac{r_n^2}{4} = r_{n+1}^2, \quad \dfrac{r_n^2}{4} + (1-x)^2 = 1.$$

It is easy to derive a recursive formula for this. Let us make an EMT program.

>function f(r) ...
$h=r^2/4; x=1-sqrt(1-h);$  return sqrt(x^2+h);
\$  endfunction
>longest (3*2^(0:10)*niterate("f",1,10))'
3
3.105828541230249
3.132628613281238
3.139350203046867
3.14103195089051
3.141452472285462
3.141557607911858
3.141583892148319
3.141590463228051
3.141592105999272
3.141592516692158
>longest pi
3.141592653589793

This is not quite in the Greek spirit because it involves square roots and non-rational numbers. It is also not easy to compute by hand and converges slowly. But we can speed up the convergence with a method due to Aitkens (see Wikipedia or my script about numerical mathematics).

>x=3*2^(0:10)*niterate("f",1,10);
>longest aitkens(x)'
3.141717032548338
3.141600361616389
3.141593134327089
3.14159268362007
3.141592655466439
3.14159265370708
3.141592653597124
3.141592653590251
3.141592653589822

But the Greeks were much more clever than this. They wanted some proof that the limit actually makes sense. To achieve this they ingeniously found lower and upper limits for Pi.

The modern definition of the length of a curve is the supremum of all lengths of chains of segments connecting points on the curve. For a curve gamma(t) defined on the interval [a,b] this is the sup of all

$$\sum\limits_{i=1}^n \|\gamma(x_i)- \gamma(x_{i-1})\|, \quad a =x_0 < \ldots < x_n = b.$$

A curve with a finite supremum has a length, and the supremum is defined to be the length. This is again an exact definition, which we use for our geometry problem in a somewhat inexact way. But with the exact definition, we can derive a lot of things among them the integral formula for the length.

$$L = \int\limits_a^b \|\gamma'(t)\| \, dt.$$

This leaves us with the problem of showing that the circle has a length.

From the exact definition given above it can be deduced that it suffices to look at any sequence of partitions of [a,b] such that the maximal distance of two points of the partition (the fineness of the partition) tends to 0. Now, we do not really have a parametric curve really. We only have our n-gons. With some effort, it can be shown that the lengths of the n-gons do indeed have a limit.

Should we have a problem with the lack of a parameterization before we compute the length? The Greeks certainly did fine with that. The concept of a transcendental function did not make sense to them anyway.

By the way, we can easily get a parameterization of the upper half circle with

$$\gamma(t) = (t, \sqrt{1-t^2}), \quad -1\le t \le 1.$$

This allows us to compute the length and prove that the circle has a length with ease.

What have we learned? I believe that this example shows how mathematics is inspired by nature, in our case by geometrical facts found in nature. Mathematics is not some heavenly subject of pure thinking. If we want to solve even the most simple things in nature we soon face considerable problems when we try to connect our exact definitions with the real objects.

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