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Archimedes of Syrakus prooved the identity of the two numbers a and b, where a is the quotient of the perimeter and the diameter of a circle and b is the quotient of the area and the square of the radius of the same circle. He did not yet use the name pi for this number, but he described how this number could be calculated using inscribed and circumscribed polygons. This is presumably the oldest numerical approximation in mathematical history. Archimedes himself calculated pi using polygons with up to 96 vertices and got his famous result:

${\displaystyle 3{,}1408450\dots =3+{\frac {10}{71}}<\pi <3+{\frac {10}{70}}=3{,}1428571\dots }$

In modern mathematical language the approximation uses an unit circle (r = 1) and starts with simple polygons with small vertice numbers like n = 2, 3, 4 or 6 and the respective edge lengths ${\displaystyle s_{2}=2}$, ${\displaystyle s_{3}={\sqrt {3}}}$, ${\displaystyle s_{4}={\sqrt {2}}}$ or ${\displaystyle s_{6}=1}$.

Then the vertice number is doubled using the line ${\displaystyle \rho _{n}=MS}$ and in two rectangled triangles the theorem of Pythagoras (${\displaystyle AM^{2}=MS^{2}+AS^{2},1=\rho _{n}^{2}+s_{n}^{2}/4,\rho _{n}={\sqrt {1-s_{n}^{2}/4}}}$ and ${\displaystyle AC^{2}=AS^{2}+SC^{2},s_{2n}^{2}=s_{n}^{2}/4+(1-\rho _{n})^{2}=s_{n}^{2}/4+1-2\rho _{n}+\rho _{n}^{2}}$). The result is the edge length ${\displaystyle s_{2n}=AC}$:

${\displaystyle s_{2n}={\sqrt {2-2{\sqrt {1-{\frac {s_{n}^{2}}{4}}}}}}}$

With this formula an approximation of pi can be calculated using elementary arithmetic (addition, subtraction, multiplication, division and square root). The following table contains the begin of this approximation starting with a "two-angle". It gives the numbers n (vertice number), the distance ${\displaystyle 1-\rho _{n}}$ between line mid S of AB and perimeter of the circle, the edge lengths ${\displaystyle s_{n}=AB}$ and ${\displaystyle S_{n}=A'B'=s_{n}\cdot 1/\rho _{n}}$ of the inscribed and circumscribed polygon and the respective areas ${\displaystyle a_{n}=ns_{n}\rho _{n}/2}$ and ${\displaystyle A_{n}=nS_{n}/2}$. These areas should enclose pi and they should converge to pi.

The calculation is numerically unstable because the subtraction produces a cancellation of digits and can be used to demonstrate cancellation.

${\displaystyle n}$             ${\displaystyle 1-\rho _{n}}$        ${\displaystyle s_{n}}$         ${\displaystyle S_{n}}$            ${\displaystyle a_{n}}$                 ${\displaystyle A_{n}}$
2          1.000e+00  2.00e+00       Inf  0.00000000000000               Inf
4          2.929e-01  1.41e+00  2.00e+00  2.00000000000000  4.00000000000000
8          7.612e-02  7.65e-01  8.28e-01  2.82842712474619  3.31370849898476