Plane Geometry - Bruce E. Shapiro

SECTION 44. ESTIMATING π 243As the number of sides of the corresponding polygons gets larger and larger,these two numbers approach one another.If we let Γ = C(O, r) be any circle with radius 1/2, then C = 2πr = π.HenceI n < π < C nArchimedes obtained the following recurrence relations: 1C 2n =2C nI nC n + I n(circumscribed polygon)I 2n = √ C 2n I n (inscribed polygon)where C 2n and I 2n are the circumferences of the 2n-sided regular circumbscribedand inscribed polygons.Archimedes obtained the bounds22371 < π < 227by setting n = 6·2 k for k = 0, 1, .., 4 (i.e., triangle, hexagon, 12-gon, 24-gon,48-gon, 96-gon). His limits were rational numbers rather than real numbersbecause he approximated all of his square roots with real numbers by somemethod that he did not describe. 2Taylor SeriesUsing the Taylor Series expansion for the arctangent,tan −1 x = x − 1 3 x3 + 1 5 x5 − 1 7 x7 + ...gives (using tan π/4 = 1),π = 4Å1 − 1 3 + 1 5 − 1 ã7 + ...(44.1)Unfortunately this converges very slowly - here are the results after up toa million iterations:1 If you are interested, you can find a derivation of these formulas using trigonometry athttp://mathworld.wolfram.com/ArchimedesRecurrenceFormula.html.2 A likely method that was known by the ancients to estimate square roots was therecursion relation x n+1 = (x n + a/x n)/2 whic converges to √ a as n → ∞. A perfectlygood starting guess is x 0 = a.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.