Archimedes' Recurrence Formula

Let $a_n$ and $b_n$ be the Perimeters of the Circumscribed and Inscribed $n$-gon and $a_{2n}$ and $b_{2n}$ the Perimeters of the Circumscribed and Inscribed $2n$-gon. Then

$$a_{2n} = {2a_nb_n\over a_n+b_n}$$ (1)

$$b_{2n} = \sqrt{a_{2n}b_n}\,.$$ (2)

The first follows from the fact that side lengths of the Polygons on a Circle of Radius $r=1$ are

$$s_R = 2\tan\left({\pi\over n}\right)$$ (3)

$$s_r = 2\sin\left({\pi\over n}\right),$$ (4)

so

$$a_n = 2n\tan\left({\pi\over n}\right)$$ (5)

$$b_n = 2n\sin\left({\pi\over n}\right).$$ (6)

But

$${2a_nb_n\over a_n+b_n} = {2\cdot 2n\tan\left({\pi\over n}\right)\cdot 2n\sin\left({\pi\over n}\right)\over 2n\tan\left({\pi\over n}\right)+2n\sin\left({\pi\over n}\right)}$$

$$= 4n {\tan\left({\pi\over n}\right)\sin\left({\pi\over n}\right)\over\tan\left({\pi\over n}\right)+\sin\left({\pi\over n}\right)}.$$ (7)

Using the identity

$$\tan({\textstyle{1\over 2}}x)={\tan x\sin x\over\tan x+\sin x}$$ (8)

then gives

$${2a_nb_n\over a_n+b_n}=4n\tan\left({\pi\over 2n}\right)=a_{2n}.$$ (9)

The second follows from

$$\sqrt{a_{2n}b_n}=\sqrt{4n\tan\left({\pi\over 2n}\right)\cdot 2n\sin\left({\pi\over n}\right)}$$ (10)

Using the identity

$$\sin x=2\sin({\textstyle{1\over 2}}x)\cos({\textstyle{1\over 2}}x)$$ (11)

gives

$\sqrt{a_{2n}b_n} = 2n\sqrt{2\tan\left({\pi\over 2n}\right)\cdot 2\sin\left({\pi\over 2n}\right)\cos\left({\pi\over 2n}\right)}$
$= 4n\sqrt{\sin^2\left({\pi\over 2n}\right)} = 4n\sin\left({\pi\over 2n}\right)=b_{2n}.\quad$	(12)

Successive application gives the Archimedes Algorithm, which can be used to provide successive approximations to Pi ($\pi$).

See also Archimedes Algorithm, Pi

References

Dörrie, H. 100 Great Problems of Elementary Mathematics: Their History and Solutions. New York: Dover, p. 186, 1965.