## Introduction

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.

## A trick

Archimedes only uses this once (on p.96, bottom, in passing from the 3rd line to the 4th) but he could have used it previously in an analogous claim (on p.94, 3rd line after First), as I shall do: if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b} = \frac{a+c}{b+d} = \frac{c}{d}$ (when b + d ≠ 0, of course (but everything is positive in geometry)).

## The apothematic sequence

Let AB be the diameter of any circle, O its centre, AC the tangent at A; and let the angle AOC be one-third of a right angle. [...]
First, draw OD bisecting the angle AOC and meeting AC in D. [...]
Secondly, let OE bisect the angle AOD, meeting AD in E. [...]
Thirdly, let OF bisect the angle AOE and meet AE in F. [...]
Fourthly, let OG bisect the angle AOF, meeting AF in G.

These repeating instructions prompt us to define a sequence of points on the line AC: let P0 = C, and let Pn+1 be such that OPn+1 bisects ∠AOPn. 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 : APn+1 = OPn : PnPn+1, so that, by the Trick, both equal OA + OPn : APn+1 + PnPn+1, which is equal to OA + OPn : APn. Hey! We've got OA : APn+1 in terms of OA : APn—let's call that latter quantity xn. How does Archimedes address the other summand, OPn : APn, which we'll call yn? He uses the Pythagorean theorem: yn+12OPn+12 : APn+12 = (OA2 + APn+12) : APn+12 = (OA : APn+1)2 + 1 = xn+12 + 1, whence yn+1 = xn+12 + 1. (I could have written this for yn, but Archimedes starts with D = P1.)

Starting with x0 = √3 and y0 = 2, Archimedes repeats four times the steps captured in the recurrence relations for xn (which, oops, I never stated outright—here it is: xn+1xn + yn) and yn. Stopping after P4G, he concludes that AG is half the side-length of a regular polygon of 96 sides circumscribed to the given circle (p.95). 96 being 24·6 should be enough to convince you that, in our construction, APn is half the side-length of a regular 2n·6-gon (I keep the 6 because of the initial hexagon) with apothem OA, whose perimeter, 2n·6·(2APn), exceeds the circle's circumference. Dividing through this inequality by the circle's diameter AB = 2OA yields π < 2n·6/xn.

Next let AB be the diameter of a circle, and let AC, meeting the circle in C, make the angle CAB equal to one-third of a right angle. Join BC. [...]
First, let AD bisect the angle BAC and meet BC in d and the circle in D. Join BD. [...]
Secondly, let AE bisect the angle BAD, meeting the circle in E; and let BE be joined. [...]
Thirdly, let AF bisect the angle BAE, meeting the circle in F. [...]
Fourthly, let the angle BAF be bisected by AG meeting the circle in G.

These repeating instructions also prompt us to define a sequence of points, this time on the given circle: let P0 = C and let Pn+1 be such that APn+1 bisects ∠BAPn. Draw BPn and let it cross APn+1 at Qn+1 (though it'd be nice were the triangle called APnQn, I picked Qn+1 and not Qn because it's on the line APn+1 and because Q0 doesn't exist). Here's another Paint-drawn diagram (again, distorted for clarity's sake):

The measurement is a little trickier this time. By construction, the angles PnAQn+1 and Pn+1AB are equal (each being half of ∠BAPn). By Thales' theorem, each angle APnB is right (this includes ∠APn+1B). Therefore, triangles APnQn+1 and APn+1B are similar. This is key. (Archimedes points out that ΔBPn+1Qn+1 is also similar to these, which is true, but not needed).

Geometrically, the ratio APn : BPn corresponds to the apothematic sequence's yn, but I'll call it xn (the answer to “Why?” is the punchline). Following Archimedes, we have, by similarity, that xn+1APn+1 : BPn+1APn : PnQn+1, which, by Euclid's proposition VI.3, equals AB : BQn+1. Here Archimedes uses the Trick with the last equality to get xn+1APn + AB : PnQn+1 + BQn+1APn + AB : BPn. Let ynAB : BPn (the other summand) and (surprise!) we get xn+1xn + yn. And, exactly as before, Archimedes uses the Pythagorean theorem to compute yn+1xn+12 + 1. Despite a totally different construction, the sequences of ratios satisfy the same recurrences!

Starting (again) with x0 = √3 and y0 = 2 (but this time estimating √3 from above), Archimedes works up to n = 4, at which point BG is a side of a regular inscribed polygon of 96 sides (p.98). For us, BPn is the side-length of a regular 2n·6-gon with circumradius ½AB; its perimeter, 2n·6·BPn, falls short of the circle's circumference. Dividing through this inequality by the circle's diameter AB yields π > 2n·6/yn.

## Conclusion

From Archimedes' Measurement of a Circle we have gleaned a pair of sequences xn and yn such that 2n·6/xn ↘ π and 2n·6/yn ↗ π as n tends to infinity. The sequences start with x0 = √3 and y0 = 2; subsequent terms are given by xn+1xn + yn and yn+1xn+12 + 1 (these can be untwined; e.g. xn+1xn + xn2 + 1). Here are Archimedes' bounds for the xn:
n error below lower bound xn 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…
and here, for the yn:
n error below lower bound yn 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…
For completeness' sake, the final approximation is $3\!\tfrac{1}{7} > \frac{96\cdot 153}{4673\!\tfrac{1}{2}} > \frac{96}{x_4} = 3.142... > \pi > \frac{96}{y_4} = 3.141... > \frac{96\cdot 66}{2017\!\tfrac{1}{4}} > 3\!\tfrac{10}{71}.$

I finish with some questions, whose answers I know not: how sensitive are the limits of 6·2n/xn and 6·2n/yn to the initial values of x0 and y0? If given xn and yn without context, how would I prove 6·2n/xn and 6·2n/yn converge to π?

—Mateusz Olechnowicz, 18 August 2013.

## Update (28 October 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 x1 = cos θ, y1 = 1 and
xn+1 = ½(xn + yn), yn+1 = √(xn+1yn), n = 1, 2, …,
show that xn and yn tend to the common limit sin θ/θ.

This is exactly what we need! For if (in terms of our old sequences) we let Xn = xn−1/2n and Yn = yn−1/2n then X1 = √3/2 = cos(π/6) and Y1 = 1. Furthermore, $X_{n+1} = \frac{x_n}{2^{n+1}} = \frac{x_{n-1} + y_{n-1}}{2^{n+1}} = \frac{1}{2}\left(\frac{x_{n-1}}{2^n}+\frac{y_{n-1}}{2^n}\right)=\frac{1}{2}(X_n+Y_n).$ As for Yn+1, note that the radicand in the definition of yn+1 can be simplified: xn+12 + 1 = (xn + yn)2 + 1 = xn2 + 2xnyn + yn2 + 1 = 2yn2 + 2xnyn (because yn2 = xn2 + 1), and that is equal to 2(yn)(yn + xn), whence yn+1 = √2⋅xn+1yn. Therefore, $Y_{n+1} = \frac{y_n}{2^{n+1}} = \frac{\sqrt{2 x_n y_{n-1}}}{2^{n+1}} = \sqrt{\frac{x_n y_{n-1}}{2^{2n+1}}} = \sqrt{\frac{x_n}{2^{n+1}}\frac{y_{n-1}}{2^n}} = \sqrt{X_{n+1}Y_n}$ since 2n + 1 = (n + 1) + n. By Hyslop's exercise 12, Xn and Yn tend to sin(π/6)/(π/6) = (1/2)/(π/6) = 3/π, so the sequences 6·2n/xn and 6·2n/yn, which equal 3/Xn+1 and 3/Yn+1 respectively, converge to 3/(3/π) = π.

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

## Proof of Exercise 12

The angle addition formula for cosine implies that (cos φ + 1)/2 = cos2(φ/2) for all angles φ. If for n ≥ 0 we define γn+1 = n

k=1
cos(θ/2k) and χn+1 = γn+1⋅cos(θ/2n) then γ1 = 1 (empty product) and χ1 = cos θ. Furthermore, (χn + γn)/2 = γn⋅[cos(θ/2n−1) + 1]/2 = γn⋅cos2(θ/2n) = γn+1⋅cos(θ/2n) = χn+1, and √{χn+1γn} = √{γn+1⋅cos(θ/2n)⋅γn} = √{γn+12} = γn+1, so we see that χn and γn satisfy the same initial conditions and recurrence relations as xn and yn. Since the latter are well-defined, they equal the former.

Now, xn+1/yn+1 = cos(θ/2n), which tends to cos(0) = 1 as n goes to infinity, so xn and yn tend to the same limit. What is that limit?

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) = 2n⋅sin(θ/2n)⋅yn+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 yn is sin(θ)/θ and we have also proved $\cos \frac{x}{2} \cos\frac{x}{4} \cos\frac{x}{8} \dots = \frac{\sin x}{x}.$