The Computation of π by Archimedes Bill McKeeman Dartmouth College 2012.02.15
Abstract It is famously known that Archimedes approximated π by computing the perimeters of manysided regular polygons, one polygon inside the circle and one outside. This presentation recapitulates Archimedes' computation, using MATLAB instead of hand, ink and papyrus. The surprise to me was how many "tweaks" Archimedes applied at various stages of an otherwise systematic approach.
follow along http://www.mathworks.com/matlabcentral/fileexchange/29504- the-computation-of-pi-byarchimedes/content/html/computationofpibyarchimedes.html MATLAB publication is a sort of self-checking paper. This URL takes you to the MATLAB version of the talk. Grey background is the MATLAB code; typeset material is MATLAB comments, everything else is MATLAB output.
Earlier that 200BCE Archimedes lived in Sicily, Euclid in Egypt, Pythagoras in Turkey and Italy
Archimedes did other things Syracuse was often under attack. Archimedes invented weapons to defend his city. This iron ship-dumper may be a myth. Other weapons such as fireballs were not. When a Roman sneak attack finally conquered Syracuse, Archimedes was considered by the commanding general as the most important catch. Unfortunately a Roman soldier killed old Archimedes for not being instantly obedient. Too bad for the Roman soldier.
A polygon inside the circle and a polygon outside the circle. The Plan The perimeters of the polygons bound the circumference of the circle which (by definition) is 2πR. The inner hexagon has perimeter 6R which implies 6R<2πR, or 3<π. Increasing the number of edges tightens the bounds. Thus thought Archimedes.
Greek Numbers (1) Greeks used uppercase letters to represent decimal digits (lower case is used here for readability when mixed with English). For example ˏαυκδ means 1424. The prefix ˏ turns α into 1000. The postfix ends the number (identifying it amongst Greek text).
Greek Numbers (2) In case you noticed it, 27 letters are needed for this scheme, so 3 ancient letters were used in addition to the 24 we know. Greeks had no digit 0. Greeks had no decimal point. Greeks did have rational numbers. For example: γ ζ means 3⅟₇ Greek numbers evolved over the centuries.
Greek Math Finally, algebra was unavailable, so words were used to describe computations and algorithms. The flavor of ancient presentation is provided by Archimedes' statement "The surface of any sphere is four times its greatest circle." We would write A = 4πR².
The Computation of π The result presented by Archimedes translates to 3¹⁰/₇₁ < π < 3⅟₇ This is what we want to compute in MATLAB using only the tools available to Archimedes. (I will compromise by using decimal integers, fractions and algebraic notation. You will thank me.)
The Outer N-gon The first equation is a consequence of similar triangles. The second and third equations are algebraic manipulations of the first. The fourth equation is an application of the Pythagorean Theorem.
The Iteration The last two formulas on the previous page are combined into an iteration. So, if a n is known, so is e n, and therefore the halfperimeter ne n, and, finally, an upper bound on π.
The Computation Archimedes started with the hexagon. That is: R = 1; e₆ = R (4/3); a₆ = R (1/3) It s easy for MATLAB to check the iteration formula is converging to 3.141592653589793 : % a π a₆ = R/e₆ % 1.73205 3.4641 a₁₂ = a₆ + (a₆² + 1) % 3.73205 3.31539 a₂₄ = a₁₂ + (a₁₂² + 1) % 7.59575 3.15966 a₄₈ = a₂₄ + (a₂₄² + 1) % 15.2571 3.14609 a₉₆ = a₄₈ + (a₄₈² + 1) % 30.5468 3.14271 e₉₆ = R/a₉₆
Hand Calculation Unfortunately for Archimedes, he did not have double precision floating point so he, and therefore we, still have some work to do. It is simpler for hand calculation if intermediate results are improper fractions with a large integer part. To make this happen, the iteration formula is multiplied by an arbitrary integer constant c (to be chosen later).
MATLAB Fractions Switch to Greek Mode (rational arithmetic) Need to take square roots. Archimedes picked approximations that led to an elegant result. Nobody knows how Archimedes did square roots. So, I ll skip my speculations here. Archimedes stated without explanation: Its easy to check in MATLAB assert((265/153)^2 < 3 && 3 < (1351/780)^2)
The Upper Bound (1) For the upper bound computation, Archimedes chose the following starting values: c = 153 a₆ = 256 (a little less than 265.0038) b₆ = 306 Therefore a₉₆ will be less than the fully accurate irrational value and e₉₆ will be greater than the fully accurate half-side of the 96-gon, and finally 96*e₉₆ is greater than the half-perimeter and therefore greater than π.
The Upper Bound (2) c = 153 a₆ = 265 b₆ = 306 a₁₂ = 571 Archimedes (very carefully ) chose b₁₂ Note that the difference between the true irrational value 591.143 and Archimedes choice 591.125 is very small
The Upper Bound (3) c = 153 a₆ = 265 b₆ = 306 a₁₂ = 571 b₁₂ = 591 a₂₄ = 1162⅟₈ b₂₄ = 1172⅟₈ a₄₈ = 2334⅟₄ b₄₈ = 2339⅟₄ a₉₆ = 4673⅟₂ continuing, Archimedes chose again and again
The Upper Bound (4) c = 153 a₆ = 265 b₆ = 306 a₁₂ = 571 b₁₂ = 591 a₂₄ = 1162⅟₈ b₂₄ = 1172⅟₈ a₄₈ = 2334⅟₄ b₄₈ = 2339⅟₄ a₉₆ = 4673⅟₂ Using e₉₆ = c/a₉₆ hi = 96c/a₉₆ = 96*153/4673⅟₂ = 3¹³³⁵/₉₃₄₇ Then Archimedes made another choice, picking the elegant 3¹/₇ because π < 3¹³³⁵/₉₃₄₇ < 3¹/₇ This establishes the upper bound on π.
The Lower Bound The lower bound computation is similar. The starting values and numbers are different. The final result is a₆ = 1351 b₆ = 1560 c = 780 a₁₂ = 2911 b₁₂ = 3013³/₄ a₂₄ = 5924³/₄ b₂₄ = 1838⁹/₁₁ c = 240 a₄₈ = 3661⁹/₁₁ b₄₈ = 1009⅟₆ a₉₆ = 2016⅟₆ b₉₆ = 2017⅟₄ c = 66 Using e₉₆ = c/a₉₆ lo = 96c/a₉₆ = 96*66/2017⅟₄ = 3¹¹³⁷/₈₀₆₉ Then Archimedes made his final choice, picking 3¹⁰/₇₁
The Result The values Archimedes provided: 3¹⁰/₇₁ < 3¹¹³⁷/₈₀₆₉ < π < 3¹³³⁵/₉₃₄₇ < 3¹/₇ established the bounds on π. Archimedes wrote ΟΕΔ meaning ὅπερ ἔδει δεῖξαι These days we say what was required to be proved or QED.
Summary Calculation