Random walks in the plane

Transcription

1 Tulane University, New Orleans August 2, 2010 Joint work with: Jon Borwein Dirk Nuyens James Wan U. of Newcastle, AU K.U.Leuven, BE U. of Newcastle, AU

2 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

3 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

4 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

5 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

6 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

7 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction.

8 We study random walks in the plane consisting of n steps. Each step is of length 1 and is taken in a randomly chosen direction. We are interested in the distance traveled in n steps. For instance, how large is this distance on average?

9 Long walks Asked by Karl Pearson in Nature in 1905 K. Pearson. The random walk. Nature, 72, 1905.

10 Long walks Asked by Karl Pearson in Nature in 1905 For long walks, the probability density is approximately 2x /n n e x2 For instance, for n = 200: K. Pearson. The random walk. Nature, 72, Lord Rayleigh. The problem of the random walk. Nature, 72, 1905.

11 Densities n = 2 n = 3 n = n = 5 n = 6 n =

12 Densities n = 2 n = 3 n = n = 5 n = 6 n =

13 Moments Fact from probability theory: the distribution of the distance is determined by its moments.

14 Moments Fact from probability theory: the distribution of the distance is determined by its moments. Represent the kth step by the complex number e 2πix k. The sth moment of the distance after n steps is: W n (s) := [0,1] n n k=1 s e 2πx ki dx In particular, W n (1) is the average distance after n steps.

15 Moments Fact from probability theory: the distribution of the distance is determined by its moments. Represent the kth step by the complex number e 2πix k. The sth moment of the distance after n steps is: W n (s) := [0,1] n n k=1 s e 2πx ki dx In particular, W n (1) is the average distance after n steps. This is hard to evaluate numerically to high precision. For instance, Monte-Carlo integration gives approximations with an asymptotic error of O(1/ N) where N is the number of sample points.

16 Moments The sth moment of the distance after n steps: W n (s) := [0,1] n n k=1 s e 2πx ki dx n s = 1 s = 2 s = 3 s = 4 s = 5 s = 6 s =

17 Moments The sth moment of the distance after n steps: W n (s) := [0,1] n n k=1 s e 2πx ki dx n s = 1 s = 2 s = 3 s = 4 s = 5 s = 6 s = W 2 (1) = 4 π

18 Moments The sth moment of the distance after n steps: W n (s) := [0,1] n n k=1 s e 2πx ki dx n s = 1 s = 2 s = 3 s = 4 s = 5 s = 6 s = W 2 (1) = 4 π W 3 (1) = =?

19 Moments The sth moment of the distance after n steps: W n (s) := [0,1] n n k=1 s e 2πx ki dx n s = 1 s = 2 s = 3 s = 4 s = 5 s = 6 s = W 2 (1) = 4 π W 3 (1) = =?

20 Even moments n s = 2 s = 4 s = 6 s = 8 s = 10 Sloane s A A A

21 Even moments n s = 2 s = 4 s = 6 s = 8 s = 10 Sloane s A A A Sloane s, etc.: W 2 (2k) = ( 2k k W 3 (2k) = k ) j=0( k j W 4 (2k) = k j=0( k j ) 2 ( 2j ) j ) 2 ( 2j j )( 2(k j) k j )

22 Even moments n s = 2 s = 4 s = 6 s = 8 s = 10 Sloane s A A A Sloane s, etc.: W 2 (2k) = ( 2k k W 3 (2k) = k W 4 (2k) = k ) j=0( k j j=0( k j W 5 (2k) = k j=0( k j ) 2 ( 2j ) ) 2 ( 2j j )( 2(k j) ) j k j ) j l=0 ) 2 ( 2(k j) k j ( j 2 ( 2l ) l) l

23 Combinatorics Theorem (Borwein-Nuyens-S-Wan) W n (2k) = a 1 + +a n=k ( ) k 2. a 1,..., a n

24 Combinatorics Theorem (Borwein-Nuyens-S-Wan) W n (2k) = a 1 + +a n=k ( k a 1,..., a n ) 2. f n (k) := W n (2k) counts the number of abelian squares: strings xy of length 2k from an alphabet with n letters such that y is a permutation of x.

25 Combinatorics Theorem (Borwein-Nuyens-S-Wan) W n (2k) = a 1 + +a n=k ( k a 1,..., a n ) 2. f n (k) := W n (2k) counts the number of abelian squares: strings xy of length 2k from an alphabet with n letters such that y is a permutation of x. Introduced by Erdős and studied by others. L. B. Richmond and J. Shallit. Counting abelian squares. The Electronic Journal of Combinatorics, 16, 2009.

26 Combinatorics Theorem (Borwein-Nuyens-S-Wan) W n (2k) = a 1 + +a n=k ( k a 1,..., a n ) 2. f n (k) := W n (2k) counts the number of abelian squares: strings xy of length 2k from an alphabet with n letters such that y is a permutation of x. Introduced by Erdős and studied by others. f n (k) satisfies recurrences and convolutions. L. B. Richmond and J. Shallit. Counting abelian squares. The Electronic Journal of Combinatorics, 16, P. Barrucand. Sur la somme des puissances des coefficients multinomiaux et les puissances successives d une fonction de Bessel. C. R. Acad. Sci. Paris, 258, , 1964.

27 Functional Equations for W n (s) For integers k 0, (k + 2) 2 W 3 (2k + 4) (10k k + 23)W 3 (2k + 2) + 9(k + 1) 2 W 3 (2k) = 0.

28 Functional Equations for W n (s) For integers k 0, (k + 2) 2 W 3 (2k + 4) (10k k + 23)W 3 (2k + 2) + 9(k + 1) 2 W 3 (2k) = 0. Theorem (Carlson) If f(z) is analytic for Re (z) 0, nice, and f(0) = 0, f(1) = 0, f(2) = 0,..., then f(z) = 0 identically.

29 Functional Equations for W n (s) For integers k 0, (k + 2) 2 W 3 (2k + 4) (10k k + 23)W 3 (2k + 2) + 9(k + 1) 2 W 3 (2k) = 0. Theorem (Carlson) If f(z) is analytic for Re (z) 0, nice, and f(0) = 0, f(1) = 0, f(2) = 0,..., then f(z) = 0 identically. f(z) Ae α z, and f(iy) Be β y for β < π

30 Functional Equations for W n (s) For integers k 0, (k + 2) 2 W 3 (2k + 4) (10k k + 23)W 3 (2k + 2) + 9(k + 1) 2 W 3 (2k) = 0. Theorem (Carlson) If f(z) is analytic for Re (z) 0, nice, and f(0) = 0, f(1) = 0, f(2) = 0,..., then f(z) = 0 identically. W n (s) is nice! f(z) Ae α z, and f(iy) Be β y for β < π

31 Functional Equations for W n (s) So we get complex functional equations like (s+4) 2 W 3 (s+4) 2(5s 2 +30s+46)W 3 (s+2)+9(s+2) 2 W 3 (s) = 0.

32 Functional Equations for W n (s) So we get complex functional equations like (s+4) 2 W 3 (s+4) 2(5s 2 +30s+46)W 3 (s+2)+9(s+2) 2 W 3 (s) = 0. This gives analytic continuations of W n (s) to the complex plane, with poles at certain negative integers W 3 (s) W 4 (s)

33 W 3 (1) = =? Easy: W 2 (2k) = ( ) ( ) 2k s. In fact, W 2 (s) =. k s/2

34 W 3 (1) = =? ( 2k Easy: W 2 (2k) = k Again, from combinatorics: W 3 (2k) = ). In fact, W 2 (s) = j=0 ( ) s. s/2 k ( ) k 2 ( ) ( 2j 1 ) = 3 F 2, k, k 2 j j 1, 1 4 }{{} =:V 3 (2k)

35 W 3 (1) = =? ( 2k Easy: W 2 (2k) = k Again, from combinatorics: W 3 (2k) = ). In fact, W 2 (s) = j=0 ( ) s. s/2 k ( ) k 2 ( ) ( 2j 1 ) = 3 F 2, k, k 2 j j 1, 1 4 }{{} =:V 3 (2k) We discovered numerically that V 3 (1) i.

36 W 3 (1) = =? ( 2k Easy: W 2 (2k) = k Again, from combinatorics: W 3 (2k) = ). In fact, W 2 (s) = j=0 ( ) s. s/2 k ( ) k 2 ( ) ( 2j 1 ) = 3 F 2, k, k 2 j j 1, 1 4 }{{} =:V 3 (2k) We discovered numerically that V 3 (1) i. Theorem (Borwein-Nuyens-S-Wan) For integers k we have W 3 (k) = Re 3 F 2 ( 1 2, k 2, k 2 1, 1 ). 4

37 W 3 (1) = =? Corollary (Borwein-Nuyens-S-Wan) W 3 (1) = 3 2 1/3 16 π 4 Γ6 ( ) /3 π 4 Γ6 ( ) 2 3 Similar formulas for W 3 (3), W 3 (5),...

38 A generating function Recall: W n (2k) = a 1 + +a n=k ( k a 1,..., a n ) 2

39 A generating function Recall: Therefore: k=0 W n (2k) = W n (2k) ( x)k (k!) 2 a 1 + +a n=k = = ( k=0 a 1 + +a n=k ( a=0 ( x) a (a!) 2 k a 1,..., a n ) 2 ( x) k (a 1!) 2 (a n!) 2 ) n = J 0 (2 x) n

40 Ramanujan s Master Theorem Theorem (Ramanujan s Master Theorem) For nice analytic functions ϕ, ( ) x ν 1 ( 1) k ϕ(k)x k dx = Γ(ν)ϕ( ν). k! 0 k=0

41 Ramanujan s Master Theorem Theorem (Ramanujan s Master Theorem) For nice analytic functions ϕ, ( ) x ν 1 ( 1) k ϕ(k)x k dx = Γ(ν)ϕ( ν). k! 0 k=0 Begs to be applied to k=0 W n (2k) ( x)k (k!) 2 = J 0(2 x) n by setting ϕ(k) = W n(2k) k!

42 Ramanujan s Master Theorem We find: 1 s Γ(1 s/2) W n ( s) = 2 Γ(s/2) 0 x s 1 J n 0 (x) dx

43 Ramanujan s Master Theorem We find: 1 s Γ(1 s/2) W n ( s) = 2 Γ(s/2) A 1-dimensional representation! Useful for symbolical computations as well as for high-precision integration 0 x s 1 J n 0 (x) dx

44 Ramanujan s Master Theorem We find: 1 s Γ(1 s/2) W n ( s) = 2 Γ(s/2) A 1-dimensional representation! Useful for symbolical computations as well as for high-precision integration 0 x s 1 J n 0 (x) dx First and more inspiredly found by David Broadhurst building on work of J.C. Kluyver David Broadhurst. Bessel moments, random walks and Calabi-Yau equations. Preprint, Nov J.C. Kluyver. A local probability problem. Nederl. Acad. Wetensch. Proc., 8, , 1906.

45 A convolution formula Conjecture For even n, W n (s) =? ( ) s/2 2 W n 1(s 2j). j j=0

46 A convolution formula Conjecture For even n, W n (s)? = ( ) s/2 2 W n 1(s 2j). j j=0 Inspired by the combinatorial convolution for f n (k) = W n (2k): f n+m (k) = k j=0 ( ) k 2 f n (j) f m (k j) j

47 A convolution formula Conjecture For even n, W n (s)? = ( ) s/2 2 W n 1(s 2j). j j=0 Inspired by the combinatorial convolution for f n (k) = W n (2k): True for even s True for n = 2 f n+m (k) = k j=0 ( ) k 2 f n (j) f m (k j) j Now proven up to some technical growth conditions

48 You will have to look at the papers to find... a hyper-closed form for W 4 (1), Meijer-G and hypergeometric expressions for W 3 (s) and W 4 (s), evaluations of derivatives including W 3(0) = 1 π Cl ( π 3 ), W 4(0) = 7ζ(3) 2π 2, expressions for residues at the poles of W n (s),...

49 References J. Borwein, D. Nuyens, A. Straub, and J. Wan. Random Walk Integrals. Preprint, Oct J. Borwein, A. Straub, and J. Wan. Three-Step and Four-Step Random Walk Integrals. Preprint, May Both preprints as well as this talk are/will be available from: THANK YOU! Special thanks to: Tewodros Amdeberhan, David Bailey, David Broadhurst, Richard Crandall, Peter Donovan, Victor Moll, Michael Mossinghoff, Sinai Robins, Bruno Salvy, Wadim Zudilin