A HIGH-SCHOOL ALGEBRA1 , WALLET-SIZED PROOF, OF THE BIEBERBACH
그는 루이스 드 브랑스(Louis de Branges)의 증명을 단순화하고, 고급 수학적 도구를 사용하지 않고도 증명할 수 있음을 보여준다.
논문에서 제시하는 방법으로, 비어버치 추측을 증명하기 위해 두 가지 사실(Fact 1과 Fact 2)을 사용한다.
Fact 1은 formal Laurent series의 특성을 이용하여 formally identity를 증명하고,
Fact 2는 polynomials Ak,n(c)가 non-negative인지를 증명한다.
사실이 proved되면 Bieberbach conjecture 또한 proved된다.
논문은 비어버치 추측에 대한 단순한 계산을 통한 증명을 제공하며, 이 방법은 고급 수학적 도구를 사용하지 않고도 가능하다.
이 논문의 중요한 기여는 비어버치 추측의 증명에서 formal Laurent series와 polynomials의 특성을 사용하는 것이며,
사실의 proof가 매우 단순하여 쉽게 이해할 수 있다.
A HIGH-SCHOOL ALGEBRA1 , WALLET-SIZED PROOF, OF THE BIEBERBACH
arXiv:math/9307209v1 [math.CA] 9 Jul 1993ZeilbergerA HIGH-SCHOOL ALGEBRA1 , WALLET-SIZED PROOF, OF THE BIEBERBACHCONJECTURE [After L. Weinstein]Shalosh B. Ekhad2 and Doron Zeilberger 2Dedicated to Leonard Carlitz 3 , master of formal mathematicsWeinstein’s[2] brilliant short proof of de Branges’[1] theorem can be made yet much shorter(moduloroutine calculations), completely elementary (modulo L¨owner theory), self contained(no need forthe esoteric Legendre polynomials’ addition theorem), and motivated(ditto), as follows. Replacethe text between p. 62, line 7 and p. 63, line 7, by Fact 1 below, and the text between the last lineof p.63 and p.64, line 7, by Fact 2 below.FACT 1: Let ft(z) = etz exp(P∞k=0 ck(t)zk) where ck(t) are formal functions of t. Let z and wbe related by z/(1 −z)2 = etw/(1 −w)2.
The following formal identity holds. (For any formalLaurent series f(z), CTzf(z) denotes the Constant Term of f(z).
)(1 + w) ddt{∞Xk=1(4/k −kck(t)ck(t))wk} =(1−w)∞Xk=1Re CTz( ∂ft(z)∂tz∂ft(z)∂z· (2(1 + . .
. + kck(t)zk) −kck(t)zk) · (2(1 + .
. .
+ kck(t)z−k) −kck(t)z−k))wkProof: Routine. (Obviously computer-implementable.
)FACT 2: The polynomials Ak,n(c), defined in terms of the formal power series (Laurent in w)expansion (1−z(2c+(1−c)(w+1/w))+z2)−1 = P∞n=0Pnk=0 Ak,n(c)(wk+w−k)zn are non-negative.Proof: This follows immediately from the stronger fact that the polynomialsBk,n(c), defined bythe expansion (1 −z(2c + (1 −c)(w + 1/w)) + z2)−1/2 = P∞n=0Pnk=0 Bk,n(c)(wk + w−k)zn areperfect squares. To prove thatBk,n(c) := CTz,wF(z, w, c, k, n) := CTz,w[(1 −z(2c + (1 −c)(w + 1/w)) + z2)−1/2znwk]are indeed perfect squares, the reader4 can easily find polynomials in (n, k, c), p0, p1, p2, p3, andpolynomials in (n, k, c, z, w), G1, and G2, both of degree 2 in both z and w, such that1 and high-school (purely formal) calculus.2 DepartmentofMathematics,TempleUniversity,Philadelphia,PA19122,USA.ekhad@euclid.math.temple.edu , zeilberg@euclid.math.temple.edu .
Supported in part by the NSF. We wouldlike to thank Richard Askey and JeffLagarias for comments, that improved readability.3 L. Carlitz was, for many years, editor of the Duke Journal, until he was relieved from his duties by the proponents ofso-called “modern math”, who proceeded to reject anything that smacked, even faintly, of Carlitz-style mathematics.But those who have drowned Carlitz will soon be drowned themselves, as post-modern, computer-assisted andcomputer-generated mathematics, that by its very nature is purely formal, will soon take over and make so-called“modern” mathematics a relic of the past.4 Human readers: Find a computer friend to help you with this.
All you have to do is express G1, G2, qua polynomialsin z, w (of degree 2 in each), generically, with ”indeterminate coefficients”, then divide (WZ) by F , simplify cleardenominators, and equate all the coefficients of the monomials ziwj in the resulting identity to 0, getting a linearsystem of equations, with 2[(2+1)·(2+1)]+4 unknowns, that your computer friend can easily solve. This method iscalled the WZ method, and the fact that such a recurrence always exists follows from the WZ theory (Wilf andZeilberger, Invent.
Math.108(1992), 575-633. ), but at any particular instance, like in this case, no explicit referenceto WZ theory need be made.1
p0F(z, w, c, k, n) + p1F(z, w, c, k, n + 1) + p2F(z, w, c, k, n + 2) + p3F(z, w, c, k, n + 3) =z ddz (G1Fz3w ) + w ddw (G2Fzw3 ). (WZ)Applying CT to both sides of (WZ), remembering the obvious fact that for any formal Laurentseries f(z), CT(z(d/dz)f(z)) = 0, we get that the Bk,n satisfy the linear recurrence, in n:p0Bk,n(c) + p1Bk,n+1(c) + p2Bk,n+2(c) + p3Bk,n+3(c) = 0.
(Rec2)The recurrence (Rec2) can be used to generate many Bk,n(c), and it turns out, empirically fornow, that they are all perfect squares Bk,n(c) = Lk,n(c)2, for some double-sequence of polynomi-als Lk,n(c). These empirically-generated polynomials can be used to find a (conjectured) linearrecurrenceq0Lk,n(c) + q1Lk,n+1(c) + q2Lk,n+2(c) = 0,(Rec)where q0, q1, q2 are polynomials of (n, k, c).
Let’s now define L′k,n(c) to be the solution of (Rec)under the appropriate initial conditions Lk,0, Lk,1, and define B′k,n(c) := L′k,n(c)2. Using high-school linear algebra (which is implemented in the gfun Maple package developed by Salvy andZimmerman) one can easily find a (third order) recurrence satisfied by B′k,n(c), that turns out tobe identical with (Rec2).
Matching the three initial values n = 0, 1, 2 completes the proof.References1. L. de Branges, A proof of the Bieberbach conjecture, Acta Math.
154(1985), 137-152.2. L. Weinstein, The Bieberbach Conjecture, Duke Math.
J. 59(1991) 61-64.2
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