# Approximating pi

Mathematical calculus is based on the concept of limits. Some of the most famous examples using limits, are the attempts throughout history to find an approximation for $$\pi$$. By trying to find the limit of the area/perimeter of a regular polygon as the number of vertices tends to infinity, you get an approximation of the area/circumference of the circle. The area/circumference of a circle can then be used to find an approximation of $$\pi$$.

$$\pi$$ is defined to be the ratio of the circumference of a circle to its diameter.

$\pi=\frac{C}{d}=\frac{C}{2r}$

You can use either the definition (the circumference), or you can use the ratio between the area of a circle and the square of the radius, to approximate $$\pi$$.

$\pi=\frac{A}{r^2}$

Throughout history, both these approaches have been used.

## Archimedes' pi (≈250 BC)

Archimedes used the fact that the circumference of a circle is bounded by the perimeter of an inscribed polygon and the perimeter of a circumscribed polygon.This fact was used to approximate $$\pi$$.

Archimedes used a 96-sided polygon to find following approximation:

$\frac{223}{71} \lt \pi \lt \frac{22}{7}$

It is easy to use this method by using trigonometry, Archimedes however only used geometry and Greek numerals!

## Liu Hui's pi (≈250 AD)

Liu Hui used areas and realized that:

$\lim_{n\rightarrow \infty } \text{area of an equilateral} n \text{-gon}=\text{area of circle}$

He used the fact that if you know the side in a regular n-gon, you can find the side in a regular 2n-gon. (Just as Archimedes he didn’t use trigonometry and he didn’t have decimal numbers)

### Liu Hui's method

This isn’t the entire method, it is just a few details. (Google on it for more information)

Start with a circle with radius $$r$$ and a $$n$$-gon with a known side $$s_0$$.

Find the side of the $$2n$$-gon, $$s_1$$, in terms of $$r$$ and $$s_0$$. Do not use trigonometry.

## Exercise

### Liu Hui's method using a spreadsheet

1. Find the side of the hexagon in terms of the radius r. Why is a hexagon a good choice for the first polygon?
2. Pick a radius of your own choice. Let $$s_0$$ be the length of a side in a regular hexagon, $$s_1$$ be the length of a side in a regular dodecahedron, $$s_2$$ the length of a side in a regular 24-gon, and so on. Find a recursive formula for sn.
3. Use the recursive formula obtained and a spreadsheet to approximate π. It’s easier to use the circumference than the area (since you, unlike Liu Hui, can use clever numerals and a computer).
4. Use relative copies to obtain better approximations of $$\pi$$. Show 15 decimals by choosing Option->Rounding->15 Decimal Places. After a number of steps, GeoGebra recognizes that the number in question is $$\pi$$. What happens after these steps and why?

## And then what?

The reason Liu Hui used areas instead of circumferences was that he found a clever way of approximating the area of a polygon with a rational number, thus avoiding having to taking successive square roots.

There is no known record of any approximation of π prior to Archimedes. Those claiming there is, do not refer to any known references where any significant property of $$\pi$$ is mentioned.

Archimedes used rational numbers as approximation of square roots. He approximated $$\sqrt{3}$$ correct to 5 significant figures. There is no record of how he did it.

Archimedes (≈250 BC) used a 96-gon.

Liu Hui (≈250 AD) used a 3072-gon.

Zu Chongzhi (≈500 AD) used a 12 288-gon to obtain π ≈3.1415962, correct to 8 significant figures.

↓ 900 years later

Madhava of Sanganagrama (≈1400 AD) obtained 10 correct significant figures by using a series later called the Leibniz' series.

Leibniz discovered this series ≈1700 AD.

↑ 300 years later

## The Formula, solution

Start with a circle with radius r and a n-gon with a known side $$s_0$$.

In the picture to the right, $$OB=OC=r$$.

Using Pythagoras' theorem on the triangle $$\Delta OBD$$ we get

$OD=\sqrt{r^2-\frac{s_0^2}{4}}$

The length

$CD=r-OD=r-\sqrt{r^2-\frac{s_0^2}{4}}$

Using Pythagoras' theorem on the triangle $$\Delta BCD$$ we get

$BC=\sqrt{BD^2+CD^2}=\sqrt{\frac{s_0^2}{4}+\left(r-\sqrt{r^2-\frac{s_0^2}{4}}\right)^2}$

This can be simplified as below.

$BC=\sqrt{\frac{s_0^2}{4}+r^2-2r\sqrt{r^2-\frac{s_0^2}{4}}+\left(r^2-\frac{s_0^2}{4} \right)} = \sqrt{2r^2-2r\sqrt{r^2-\frac{s_0^2}{4}}}$

The last simplification is not needed in order to enter it into a spreadsheet.

### A hexagon

A regular hexagon can be divided into six equilateral triangles with the interior angles 60°. If the hexagon is inscribed in a circle with radius 1, the side length of the hexagon is 1. We get the recursive formula for the side lengths of the successive polygons

$\left\{ \begin{eqnarray} s_0 &=& 1 \\ s_{n+1} &=& \sqrt{2-2\sqrt{1-\frac{s_n^2}{4}}} \end{eqnarray} \right.$

where $$s_0$$ is the side of the hexagon and $$s_n$$ is the side of a 6⋅2n-gon when n>1.