Automatisches Beweisen von Identitäten

Transcription

1 Automatisches Beweisen von Identitäten Stefan Gerhold 30. Oktober 2006

2 The General Scheme Parametric sums summand(n, k) = answer(n) Parametric integrals k integrand(x, y)dy = answer(x)

3 Zeilberger s Summation Algorithm ( 1990) Goal: derive a recurrence for the sum s(n) = k f (n, k). Summand must be hypergeometric in n and k: The quotients f (n+1,k) f (n,k) and f (n,k+1) f (n,k) must be rational functions of n and k. E.g., f (n, k) a product of binomal coefficients Provides algorithmic proofs of many combinatorial identities

4 A Simple Application of Zeilberger s Algorithm Input: f (n, k) := ( ) n k x k Output ( k is the forward difference operator): ) k (x + 1)f (n, k) + f (n + 1, k) = k ( f (n, k). n k + 1 Sum both sides for k = 0,..., n + 1: (x + 1) n n+1 f (n, k) + f (n + 1, k) = 0, k=0 k=0 hence n f (n, k) = (x + 1) n. k=0

5 Example Gallery n ( )( ) x k + 1 x 2k ( 1) k = k n k n ( ) n 2 ( ) 2n = k n k=0 k=0 2n ( ) 2n 3 ( 1) k = ( 1) n (3n)! k n! 3 n ( ) ( n x x + n ( 1) k k k + x = n k=0 k=0 ) 1 { 1 n even 0 n odd

6 What Zeilberger s Algorithm Does We want to do the sum k f (n, k). Define the forward shift S n f (n) := f (n + 1). Zb finds an identity P(S n, n)f (n, k) = (S k 1)Q(S n, n, S k )f (n, k) with polynomials P and Q. Upon summation, the right-hand side vanishes: P(S n, n) k f (n, k) = 0. P and Q are constructed by solving a system of linear equations with rational function coefficients.

7 Rogers-Ramanujan Identities Rogers (1894), Ramanujan, Schur (1917) k=1 k=1 q k2 (1 q)(1 q 2 )... (1 q k ) = 1 (1 q 5j+1 )(1 q 5j+4 ) j=0 q k(k+1) (1 q)(1 q 2 )... (1 q k ) = 1 (1 q 5j+2 )(1 q 5j+3 ) j=0 The formulas are the limits n of finite versions (Andrews 1974). Zeilberger (1990) gave a computer proof, greatly simplified by Paule (1994).

8 The Bieberbach Conjecture An injective holomorphic function f : { z < 1} C is called schlicht, if f (z) = z + n 2 a nz n. Conjecture (Bieberbach 1916): a n n for n 2. Bieberbach (1916): a 2 2. Nevanlinna (1920): Conjecture holds if image is star-shaped. Loewner (1923): a 3 3. Littlewood (1925): a n e n.... De Branges (1985) finally settled the conjecture.

9 The Bieberbach Conjecture De Branges proof uses that a certain 3 F 2 is non-negative. Askey, Gasper (1976) (α + 2) n n! ( n, n α, α+1 2 3F 2 α + 1, α+3 2 n k 2 j=0 ) x = ( 1 2 ) j( α 2 + 1) n j( α+3 2 ) n 2j(α + 1) n 2j j!( α+3 2 ) n+j( α+1 2 ) n 2j(n 2j)! ( 2j n, n 2j + α 1, α+1 2 3F 2 α + 1, α+2 2 ) x. The identity can be proven with Zeilberger s algorithm and implies the inequality.

10 The Irrationality of ζ(3) ζ(2) = n 1 n 2 = 1 6 π2 is irrational, and similarly for ζ(2n). Apéry (1978): ζ(3) = n 1 n 3 is irrational. The proof uses that the sequence b n := satisfies the recurrence n k=0 ( ) n 2 ( n + k k k n 3 b n +(n 1) 3 b n 2 = (34n 3 51n 2 +27n 5)b n 1, n 2. ) 2

11 Generalizations of Zeilberger s Algorithm (Chyzak 1998, Schneider 2001) General idea: summand satisfies recurrences, not necessarily of first order Compute recurrence for the sum Works also for integrands that depend on a continuous parameter Compute differential equation for the integral

12 Example Gallery i+j+k=n ( )( )( ) i + j j + k k + i = i j k P n (x)y n = n=0 n k=0 ( ) 2k k 1 1 2xy + y 2 e px T n (x) dx = 1 x 2 π( 1)n I n (p) xe px2 J ν (ax)j ν (bx)dx = 1 + b 2 2p exp( a2 ) I ν ( ab 4p 2p )

13 When is equal to 1? Balogh and Pemantle (2004) came across the sum S := j=1 k=1 H j (H k+1 1) jk(k + 1)(j + k) in a run time analysis of the simplex algorithm on a certain polytope. (H n := n k=1 1/k.) Easy bound: S Theorem (Schneider 2006) S = 4ζ(2) 2ζ(3) + 4ζ(2)ζ(3) + 2ζ(5) = Proof: truncate the sums, find a closed form, pass to the limit

14 Further Uses of Recurrences Recurrences are not only useful for proving identities Rapid computation of sums and integrals Example: Recurrences for basis functions in higher order finite element schemes (Paule, Schöberl et al. 2006) Asymptotics via recurrences and differential equations for generating functions

15 From Recurrences to Asymptotics Weiss, Glebsky (2005): The number of limit states in a certain Schelling population model is s(n) =4 2n k=1 ( n 1 k 1 )( n k 1 k 1 + three similar sums. What is the behaviour as n? Let a(n) := 4 denote the first sum. 2n k=1 ) + 2 2n k=1 ( )( ) n 1 n k 1 k 1 k 1 ( )( ) n 1 n k 1 k k 1

16 From Recurrences to Asymptotics Generating function A(z) := n 0 a(n)z n. Zeilberger s algorithm recurrence for a(n) ODE for A(z). Fuchs-Frobenius theory yields A(z) const (1 3z) 1/2, z 1 3. Singularity analysis (Flajolet, Odlyzko 1990) then gives a(n) const 3 n / n, n.

17 Automatic Proofs of Inequalities (M. Kauers, SG 2005) Inequality must depend on some discrete parameter n Quantities must satisfy polynomial recurrences Induction step is reduced to CAD (Cylindrical Algebraic Decomposition) Check finitely many initial values

18 Turán s Inequality for Legendre Polynomials Turán (1946): n (x) := P n (x) 2 P n 1 (x)p n+1 (x) 0, x [ 1, 1], n 1. Introduce real variables N, Y 1, Y 0, Y 1, Y 2 representing n, P n 1 (x), P n (x), P n+1 (x), P n+2 (x) Sufficient condition for induction step: N, X, Y 1,Y 0, Y 1, Y 2 R : ( N 1 1 X 1 (N + 2)Y 2 = (N + 1)Y 0 + (3X + 2NX )Y 1 (N + 1)Y 1 = NY 1 + (X + 2NX )Y 0 ) = ( Y 2 0 Y 1 Y 1 0 = Y 2 1 Y 0 Y 2 0 ).

19 Turán s Inequality: Refinements SG, Kauers (2005): x P n (x) 2 P n 1 (x)p n+1 (x) 0, x [ 1, 1], n 1, Alzer, SG, Kauers (2006): α n (1 x 2 ) P n (x) 2 P n 1 (x)p n+1 (x) β n (1 x 2 ) with the best possible factors α n = µ n/2 µ (n+1)/2 and β n = 1 2, µ n := 2 2n ( 2n n ).

20 Other Inequalities The following inequalities can be proven automatically: ( n ) 2 x k y k k=1 n n xk 2 yk 2 k=1 k=1 (Cauchy-Schwarz) (x + 1) n 1 + n x, n 0, x 1 (Bernoulli) n n (1 a k ) > 1 a k, 0 < a k < 1, a k < 1 (Weierstraß) k=1 k=1 n 34 a n 12 n + 1 4, a 1 = 1, a n+1 = 1 + n a n

21 Conclusion Symbolic methods are useful for working with special functions Classical Tables like Abramowitz-Stegun can be partially reproduced by computer algebra NIST s Digital Library of Mathematical Functions will have a chapter on these methods (Chyzak, Paule)