Computing π with Bouncing Balls

Transcription

1 by Let the mass of two balls be M and m, where M = (6 n )m for n N. The larger ball s rolled towards the lghter ball, whch ear a wall and ntally at rest. Fnd the number of collsons between the two balls untl the larger ball begns to move n the drecton opposte to ts ntal velocty. Assume all collsons are perfectly elastc.

2 Our Soluton Ths problem, unsurprsngly, begns wth conservaton of momentum and, due to the elastc nature of the collsons, conservaton of energy. In defnng we have x m M, () u n x = u n+ + x+ () u n + xs n = u n+ + xs n+, (3) where u n and are the velocty of the large and small ball, respectvely, after the nth collson. Do due the lnearty of momentum, we guess a lnear soluton of the form: u n+ = αu n + β + = γu n + δ, = un+ + α β un =, (4) γ δ where all α, β, γ, δ R are are constant. Substtutng (4) nto the momentum and energy expressons () and (3), respectvely, we have u n x = u n (α + xγ) + (β + xδ) u n + xs n = u n(α + xγ ) + u n (αβ + γδx) + s n(β + xδ ). By matchng coeffcents wth the constants wthn each expresson, we establsh the system: α + xγ = α + xγ = αβ + γδx = β + xδ = x β + xδ = x From here, we wsh to derve non-trval solutons for the constants α, β, γ, and δ, whch we clam are nonzero and n terms of x. Applyng ths restrcton and trudgng through the algebra, one obtans the unque soluton α = x + x, β = x + x, γ = + x, and δ = x + x..

3 These mply the transformaton matrx Λ = x +x +x x +x x +x, makng the recurrence n (4) become or more succnctly where un+ + = x +x +x x +x x +x v n+ = Λ v n, v n = un. un, (5) We now acknowledge that (5), though correct n applcaton, ot helpful n ts present form. We are certan that wth n = (essentally at t = ), v = Λ v as ths s the frst collson and so only one transformaton of Λ ecessary. Though for the next collson, Λ s surrepttously appled twce: v = Λ v = v = Λ(Λ v ) = Λ v. And ths s true n general: v n+ = Λ n+ v, promptng a more useful expresson of (5) to be un+ + = x +x +x x +x u n+ x s +x. (6) Ths expresson s ndcatve that the matrx Λ s lkely dagonalzable, meanng there may exst matrces D and P such that Λ = P DP λ, where D = and P = z λ z λ, λ where each z λ s a column and egenvector for the egenvalue λ of Λ. Notce, then, that f Λ s dagonalzable, Λ n+ = (P DP )(P DP ) (P DP ) = P D n+ P, (7) }{{} n+ factors 3

4 whch vastly smplfes the computaton. To proceed, we frst fnd the egenvalues and egenvectors of Λ. Settng cos θ = x + x = x = cos θ + cos θ, t follows from both the functonal expresson det(λ λi ) =, where I s the dentty matrx, and Euler s equaton e θ = cos θ + sn θ that λ = e ±θ = λ = e θ and λ = e θ. To compute the egenvectors z λ and z λ, we revert back to Λ n terms of x. Ths prompts the egenvectors x x z λ = and z λ =. Because the egenvalues of Λ are dstnct and the above egenvectors are lnearly ndependent, Λ s dagonalzable, thereby establshng the matrces P = x and P = x n the dagonalzaton of Λ. Furthermore, we have e θ e D = = D n+ (n+)θ =, e θ e (n+)θ makng Λ n+ = x e (n+)θ e (n+)θ by the logc n (7). Though t may seem abrupt, t s convenent for computatonal sake to now defne p as the number of collsons that occur untl the velocty vector of the larger ball egatve. Now, multplyng the above matrx expresson wth p a + and through rearrangements of Euler s equaton, we obtan Λ p = cos(θp) sn(θp) x sn(θp). cos(θp) For sake of brevty (and to not underwhelm the result of ths problem), we have excluded a majorty of the ntermedate algebra. 4

5 By (6), we have up s p = cos(θp) sn(θp) x sn(θp) u cos(θp) s As specfed n the problem statement, s = ; hence the matrx-vector product above becomes up s p = cos(θp), sn(θp) whch mples u p = cos(θp). By the way n whch we ve defned p (the number of collsons before the drecton of the velocty vector of the larger ball s reversed), t follows that cos(θp) and cos(θ(p+)) <, whch formulates the compound nequaltes cos(θp) > cos(θ(p + )) = θp π < θ(p + ), (8) because the cosne functon changes sgn at π. We are now nearly there. Recallng the defnton cos θ = x + x, t follows from the Taylor polynomals cos θ = n= ( ) n θ n (n)! that, to a second-order approxmaton n θ, and x + x = + ( ) n x n n= θ x = θ. By the defnton for x n () and the value of x n the problem statement, we have that θ n. Hence, substtutng nto (8), we obtan the nequalty p n π < p +. A compound nequalty of ths form s the defnton of the floor functon beng operated on p. Consequently, t follows that p = n π, suggestng that the number of collsons p that occur untl the larger ball commences to retreat s the nteger formed by the frst n + dgts of π. Qute spectacular! 5