Daily Archives: April 15, 2021

Truth vs. Intellect

It was known long ago that \pi, the ratio of the circumference to the diameter of a circle, is a constant. Nearly all people of the ancient world used number 3 for \pi. As an approximation obtained though physical measurements with limited accuracy, it is sufficient for everyday needs.

An ancient Chinese text (周髀算经,100 BC) stated that for a circle with unit diameter, the ratio is 3.

In the Bible, we find the following description of a large vessel in the courtyard of King Solomon’s temple:

He made the Sea of cast metal, circular in shape, measuring ten cubits from rim to rim and five cubits high, It took a line of thirty cubits to measure around it. (1 Kings 7:23, New International Version)

This infers a value of \pi = \frac{30}{10} = 3.

It is fairly obvious that a regular polygon with many sides is approximately a circle. Its perimeter is approximately the circumference of the circle. The more sides the polygon has, the more accurate the approximation.

To find an accurate approximation for \pi, we inscribe regular polygons in a circle of diameter 1. Let L_{n}, L_{n+1} denotes the side’s length of regular polygon with 2^n and 2^{n+1} sides respectively, n=2, 4, ....

Fig. 1

From Fig. 1, we have

\begin{cases} L_{n+1}^2 = x^2 + (\frac{1}{2} L_n)^2\quad\quad\quad(1) \\ (\frac{1}{2})^2 = (\frac{1}{2}L_n)^2 + y^2\quad\quad\;\quad(2)\\ x+y = \frac{1}{2}\;\quad\quad\quad\quad\quad\quad(3) \end{cases}

It follows that

y\overset{(2)}{=}\sqrt{(\frac{1}{2})^2-(\frac{1}{2} L_n)^2} \overset{(3)}{ \implies} x=\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}.\quad\quad\quad(4)

Substituting (4) into (1) yields

L_{n+1}^2 = \left(\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}\right)^2+(\frac{1}{2}L_n)^2

That is,

L_{n+1}^2 = \frac{1}{4}\left(L_n^2 + \left(1-\sqrt{1-L_n^2}\right)^2\right).

Further simplification gives

L_{n+1}^2 = \frac{1}{2}\left(1-\sqrt{1-L_n^2}\right),\quad\quad\quad(5)

Starting with an inscribed square (L_2^2 =\frac{1}{2}), we compute L_{n+1}^2 from L_{n}^2 (see Fig. 2). The perimeter of the polygon with 2^{n+1} sides is 2^{n+1} \cdot L_{n+1}.

Fig. 2


\lim\limits_{n \rightarrow \infty} 2^n \cdot L_{n} = \pi.

Exercise-1 Explain, and then make the appropriate changes:

Hint: (5) is equivalent to L_{n+1}^2 = \frac{L_n^2}{2\left(1+\sqrt{1-L_{n}^2}\right)}.