In his Measurement of a Circle (here in a translation by Heath), Archimedes proves that

the ratio of the circumference of any circle to its diameter is less than \(3\!\tfrac{1}{7}\) but greater than \(3\!\tfrac{10}{71}\)

by repeatedly applying Euclid's third proposition from book six.
The ratio of the circumference of any circle to its diameter

is now called π.
Archimedes first overestimates then underestimates the circumference of a circle by constructing 96-gons; dividing their perimeters by the circle's diameter yields bounds for π.
I spent a good dozen hours yesterday poring over the proofs,
trying to abstract away an algorithm that would go beyond Archimedes' four bisections.
I unearthed a symmetric surprise:
the four-fold bookkeeping in the proofs conceals two *identical*
mutually recursive sequences converging to π—one from above, the other from below.
In the exposition below I shall separate construction from measurement.
Throughout this document, *n* ≥ 0.

Archimedes only uses this once (on p.96, bottom, in passing from the 3^{rd} line to the 4^{th}) but he could have used it previously in an analogous claim (on p.94, 3^{rd} line after

), as I shall do:
if \[\frac{a}{b} = \frac{c}{d}\]
then \[\frac{a}{b} = \frac{a+c}{b+d} = \frac{c}{d}\]
(when *First**b* + *d* ≠ 0, of course (but everything is positive in geometry)).

LetABbe the diameter of any circle,Oits centre,ACthe tangent atA; and let the angleAOCbe one-third of a right angle. [...]

First, drawODbisecting the angleAOCand meetingACinD. [...]

Secondly, letOEbisect the angleAOD, meetingADinE. [...]

Thirdly, letOFbisect the angleAOEand meetAEinF. [...]

Fourthly, letOGbisect the angleAOF, meetingAFinG.

These repeating instructions prompt us to define a sequence of points on the line *AC*:
let *P*_{0} = *C*,
and let *P*_{n+1} be such that
*OP*_{n+1} bisects ∠*AOP _{n}*.
Have a look at my Paint-drawn diagram (I've exaggerated the central angle for clarity):

Now we follow Archimedes in measurement.
By Euclid's proposition VI.3,
*OA* : *AP*_{n+1} =
*OP _{n}* :

Starting with *x*_{0} = √3 and *y*_{0} = 2, Archimedes repeats four times the steps captured in the recurrence relations for *x _{n}* (which, oops, I never stated outright—here it is:

a regular polygon of 96 sides circumscribed to the given circle(p.95). 96 being 2

Next letABbe the diameter of a circle, and letAC, meeting the circle inC, make the angleCABequal to one-third of a right angle. JoinBC. [...]

First, letADbisect the angleBACand meetBCindand the circle inD. JoinBD. [...]

Secondly, letAEbisect the angleBAD, meeting the circle inE; and letBEbe joined. [...]

Thirdly, letAFbisect the angleBAE, meeting the circle inF. [...]

Fourthly, let the angleBAFbe bisected byAGmeeting the circle inG.

*These* repeating instructions also prompt us to define a sequence of points,
this time on the given circle: let *P*_{0} = *C* and let *P*_{n+1}
be such that *AP*_{n+1} bisects ∠*BAP _{n}*.
Draw

The measurement is a little trickier this time.
By construction, the angles *P _{n}AQ*

Geometrically, the ratio *AP _{n}* :

the other summand) and (surprise!) we get

Starting (again) with *x*_{0} = √3 and *y*_{0} = 2
(but this time estimating √3 from above),
Archimedes works up to *n* = 4,
at which point

(p.98).
For us, *BG* is a side of a regular inscribed polygon of 96 sides*BP _{n}* is the side-length of a regular 2

n |
error below | lower bound | x_{n} |
upper bound | error above |
---|---|---|---|---|---|

0 | 0.000 024 663… | \( \frac{265}{153} \) | \( \sqrt{3} \) | \( \frac{1351}{780} \) | 0.000 000 474… |

1 | ″ | \( \frac{571}{153} \) | \( 2 + \sqrt{3} \) | \( \frac{2911}{780} \) | ″ |

2 | 0.000 165 877… | \( \frac{1162\tfrac{1}{8}}{153} \) | \( 2 + \sqrt{2} + \sqrt{3} + \sqrt{6} \) | \( \frac{1823}{240} \) | 0.000 079 220… |

3 | 0.000 515 740… | \( \frac{2234\tfrac{1}{4}}{153} \) | \( 2 + \sqrt{2} + \sqrt{3} + \sqrt{6} + \sqrt{16 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}} \) | \( \frac{1007}{66} \) | 0.000 524 069… |

4 | 0.001 088 346… | \( \frac{4673\tfrac{1}{2}}{153} \) | 30.546 839 986… | \( \frac{2016\tfrac{1}{6}}{66} \) | 0.001 139 817… |

n |
error below | lower bound | y_{n} |
upper bound | error above |
---|---|---|---|---|---|

0 | = | \( \frac{306}{153} \) | \( 2 \) | \( \frac{1560}{780} \) | = |

1 | 0.000 141 213… | \( \frac{591\tfrac{1}{8}}{153} \) | \( \sqrt{2} + \sqrt{6} \) | \( \frac{3013\tfrac{3}{4}}{780} \) | 0.000 078 746… |

2 | 0.000 349 863… | \( \frac{1172\tfrac{1}{8}}{153} \) | \( \sqrt{16+10\sqrt{2}+8\sqrt{3}+6\sqrt{6}} \) | \( \frac{1838\tfrac{9}{11}}{240} \) | 0.000 444 848… |

3 | 0.000 572 603… | \( \frac{2339\tfrac{1}{4}}{153} \) | 15.289 788 298… | \( \frac{1009\tfrac{1}{6}}{66} \) | 0.000 615 750… |

4 | n/a | — | 30.563 203 896… | \( \frac{2017\!\tfrac{1}{4}}{66} \) | 0.001 190 042… |

I finish with some questions, whose answers I know not:
how sensitive are the limits of 6·2^{n}/*x _{n}*
and 6·2

—Mateusz Olechnowicz, 18 August 2013.

Infinite Series by James M. Hyslop (5th edition, reprinted in 1965, published by Oliver and Boyd), a little brown book which I picked up from the library on a whim today, has the following exercise on page 35:

12. If

x_{1}= cosθ,y_{1}= 1 and

x_{n+1}= ½(x+_{n}y),_{n}y_{n+1}= √(x_{n+1}y),_{n}n= 1, 2, …,

show thatxand_{n}ytend to the common limit sin_{n}θ/θ.

This is exactly what we need!
For if (in terms of our old sequences) we let
*X _{n}* =

This kind of recurrence has been called an "Archimedean process". Note the off-by-one resemblance to the arithmetic-geometric mean.

The angle addition formula for cosine implies that
(cos φ + 1)/2 = cos^{2}(φ/2) for all angles φ.
If for *n* ≥ 0 we define
*γ*_{n+1} =
*n*

∏*k*=1
cos(*θ*/2^{k})
and *χ*_{n+1} = *γ*_{n+1}⋅cos(*θ*/2^{n})
then *γ*_{1} = 1 (empty product)
and *χ*_{1} = cos *θ*.
Furthermore,
(*χ _{n}* +

Now, *x*_{n+1}/*y*_{n+1}
= cos(*θ*/2^{n}),
which tends to cos(0) = 1 as *n* goes to infinity,
so *x _{n}* and

The angle addition formula for sine implies that
sin(*θ*) = 2⋅sin(*θ*/2)⋅cos(*θ*/2),
so by repeatedly replacing the sine term in the RHS
by the entire RHS appropriately scaled, we get the identity
sin(*θ*)
= 2^{n}⋅sin(*θ*/2^{n})⋅*y*_{n+1}.
By the sequential characterization of continuity,
\[\lim_{n\to\infty} 2^n\sin\left(\frac{\theta}{2^n}\right)
= \lim_{x\to\infty} x\sin\left(\frac{\theta}{x}\right)
= \theta\lim_{\theta/x \to 0}\frac{\sin(\theta/x)}{\theta/x}
= \theta\cdot 1 = \theta.\]
Therefore the limit of *y _{n}*
is sin(

See this article.