Arias de Reyna, J.

Lune, J. (Jan) van de

English

CWI's Institutional Repository

A PROBLEM RELATED TO THE APPROXIMATIONOF pi BY ARCHIMEDES/HUYGENSJUAN ARIAS DE REYNA AND JAN VAN DE LUNEDedicated to Herman J. J. te Riele on the occasion of his retirement from theCWI in January 20121. The problem and its origin.It is well known that Archimedes approximated 2pi ( := the lengthof the circumference of a circle having radius r = 1 ) by the lengths ofinscribed and circumscribed regular n-gons.Denoting the length of such an inscribed n-gon by `n and that of acircumscribed one by Ln we have(1) `n = 2n sin2pi2nand Ln = 2n tan2pi2n.Huygens considered the question: Which of `n and Ln is the bestapproximation of 2pi and to what extent ?It should be clear that `n < 2pi < Ln. So, for a suitable λ ∈ (0, 1)one should take 2pi = λ`n + (1− λ)Ln.From this it is easily seen that the best λ would be λ = 1(1+ 2pi−`nLn−2pi ).So, one should consider the ratio 2pi−`nLn−2pi , or as we actually didLn − 2pi2pi − `n =tan pin− pinpin− sin pin.Writing x := pinwe are thus led to consider the ( even ) functionQ(x) := tanx−xx−sinx for x close to 0.It was known to Huygens that Q(x) > 2, and using l’Hoˆpital’s ruleit is easily seen that limx→0Q(x) = 2.Consequently one should ( in this context ) approximate 2pi by23`n +13Ln. Also note that23`n +13Ln > 2pi.( A similar analysis holds for the areas an and An of the n-gons. )For us it was just a matter of curiosity to have a closer look at thecoefficients in the power series of the function tanx−xx−sinx for x close to 0.Invoking Mathematica we found ( for various values of nMax ) forexample:117ARCHIMEDES/HUYGENS APPROXIMATION TO piand ( observing that all coefficients turned out to be positive ) arrivedat the conjecture that all coefficients cn in the power series expansiontanx− xx− sinx =∞∑n=0cn x2nare strictly positive indeed.We thus ran into the problem: If true, how can this be proved ?2. A proof of the conjecture.Q(x) is a meromorphic function on the complex plane. Its poles arethose of tanx at the points x = (2n + 1)pi/2 with n an integer and atthe zeros of x− sinx, except x = 0, which is a removable singularity ofQ(x).We consider the square R = [−2pi, 2pi]2. Inside this square thereare only four poles of Q(x) : at the points ± pi2and ± 3pi2. To see this itsuffices to show that x−sinx has only one ( triple ) zero inside R. Thiscan be proved formally by computing the variation of the argument ofx − sinx when moving along the rim of the rectangle with vertices at± 2pi ± iT with T a big real number.We compute the residuesResx=pi/2Q(x) =22− pi , Resx=−pi/2Q(x) = −22− pi ,Resx=3pi/2Q(x) = − 22 + 3pi, Resx=−3pi/2Q(x) =22 + 3pi.Hence, we may write(2) Q(x) =8pi(pi − 2)11− 4x2/pi2 +83pi(2 + 3pi)11− 4x2/9pi2 + h(x).where h is analytic on R.118ARCHIMEDES/HUYGENS APPROXIMATION TO piWe thus find the following value for cn(3) cn =8pi(pi − 2)( 2pi)2n+83pi(2 + 3pi)( 23pi)2n+ dn,with dn =12pii∫∂Rh(z)z2n+1dz.For x ∈ ∂R we have |Q(x)− h(x)| ≤ 14. In fact for |x| > 2pi we have|Q(x)− h(x)| ≤ 8pi(pi − 2)116− 1 +83pi(2 + 3pi)116/9− 1 = 0.244234 . . .Also, for x ∈ ∂R we will show that |Q(x)| ≤ 2. Since Q is even, we onlyhave to bound Q(2pi + iy) and Q(x+ 2pii) for |y| < 2pi and |x| < 2pi.First for y real and |y| < 2pi we haveQ(2pi + iy) =−2pi + i(tanh y − y)2pi + i(y − sinh y) .Then |Q(2pi + iy)| ≤ 2 is equivalent to4pi2 + (tanh y − y)2 < 16pi2 + 4(sinh y − y)2ortanh2 y − 2y tanh y < 12pi2 + 3y2 + 4 sinh2 y − 8y sinh y.So |Q(2pi + iy)| ≤ 2 follows from the two elementary inequalities :tanh2 y < 1 and 8y sinh y < 2 + 3y2 + 4 sinh2 y.On the other side of the rectangle, for −2pi < x < 2pi we have|Q(x+ 2pii)| =∣∣∣tan(x+ 2pii)− x− 2piix+ 2pii− sin(x+ 2pii)∣∣∣ ≤ coth 2pi + |x+ 2pii|sinh 2pi − |x+ 2pii|≤ coth 2pi + 2√2pisinh 2pi − 2√2pi = 0.0381898 . . .It follows that on ∂R we have |h(x)| ≤ |Q(x) − h(x)| + |Q(x)| ≤ 3,so that|dn| ≤ 12pi∫∂R3(2pi)2n+1|dz| ≤ 24(2pi)−2n−1.Hence with |θ| ≤ 1(4) cn =8pi(pi − 2)( 2pi)2n+83pi(2 + 3pi)( 23pi)2n+ θ · 24(2pi)−2n−1.Since 8/(pi(pi − 2)) is about 2.23064 . . . we havecn > 2( 2pi)2n− 24( 12pi)2n+1> 0 for all n ≥ 1completing our proof.119ARCHIMEDES/HUYGENS APPROXIMATION TO pi3. Further observations.In the previous Section we proved that in the power series expansiontanx− xx− sinx =∞∑n=0cn x2nall cn are positive.Writing tanx =∑∞n=1 tn x2n−1 and sinx =∑∞n=1 sn x2n−1 we definedT :=N∑n=1tn x2n−1 and S :=N∑n=1sn x2n−1and observed ( using Mathematica ) the following :The coefficients qn in the power series expansiontanx− TS − sinx =∞∑n=0qn x2n(1) are all positive if N ≡ 1 (mod 2)(2) are all negative if N ≡ 0 (mod 2).We have no proof for this and leave a proof ( or refutation ) as achallenge to the interested reader. One may want to try things out bymeans of the following program.A similar analysis of the inner and outer areas an and An leads to“similar” observations.Facultad de Matema´ticas, Universidad de Sevilla,Apdo. 1160, 41080-Sevilla, SpainE-mail address: arias@us.esLangebuorren 49, 9074 CH Hallum, The Neterlands(Formerly at CWI, Amsterdam )E-mail address: j.vandelune@hccnet.nl120

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