# 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 through 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

Clearly,

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

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

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