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Delannoy Numbers and a Combinatorial Proof of the Orthogonality of the Jacobi Polynomials with Natural Number Parameters

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Delannoy Numbers and a Combinatorial Proof of the Orthogonality of the Jacobi Polynomials with Natural Number Parameters Delannoy numbers and a combinatorial proof of the orthogonality of the Jacobi polynomials with natural number parameters G´abor Hetyei Department of Mathematics and Statistics UNC Charlotte http://www.math.uncc.edu/∼ghetyei Delannoy numbers di,j = di−1,j + di,j−1 + di−1,j−1 H H j HH 012 3 4 i HH 0 111 1 1 di,j := 1 135 7 9 2 1 5 13 25 41 3 1 7 25 63 129 4 1 9 41 129 321 They count the number of lattice paths from (0, 0) to (m, n) using only steps (1, 0), (0, 1), and (1, 1). n n n+j ⇒ dn,n = j=0 j j . (Defined byP Henri Delannoy (1895), Sulanke has ≥ 29 interpretations.) A mysterious relation with the Legendre polynomials Good (1958), Lawden (1952), Moser and Zayachkowski (1963) observed that dn,n = Pn(3), where Pn(x) is the n-th Legendre polynomial. There has been a consensus that this link is not very relevant. Banderier and Schwer (2004): “there is no “natural” correspondence between Legendre polynomials and these lattice paths.” Sulanke (2003): “the definition of Legendre polynomials does not appear to foster any combinatorial interpretation leading to enumeration”. Jacobi and Legendre polynomials (α,β) Usual definition of the Jacobi polynomial Pn (x): (α,β) −n −1 −α −β Pn (x) = (−2) (n!) (1 − x) (1 + x) dn (1 − x)n+α(1 + x)n+β . dxn α,β > −1 “for integrability purposes”, α = β = 0 gives Legendre. The formula below extends to all α, β ∈ C (see Szeg˝o (4.21.2)): n + α + β + j n + α x − 1 j P (α,β)(x)= . n j n − j 2 Xj Substitute α = β = 0: n + j n x − 1 j P (0,0)(x)= n j j 2 Xj is the n-th Legendre polynomial. Properties of Jacobi polynomials (α,β) For α,β > −1 the Jacobi polynomials Pn (x) form an orthogonal basis with respect to the inner product 1 hf, gi := f(x) · g(x) · (1 − x)α(1 + x)β dx. Z−1 “Swapping rule:” n (α,β) (β,α) (−1) Pn (−x)= Pn (x), Asymmetric Delannoy numbers H H n HH 01 2 3 4 m HH 0 1 2 4 8 16 dm,n := 1 1 3 8 20 48 e 2 1 4 13 38 104 3 1 5 19 63 192 4 1 6 26 96 321 dm,n is the number of lattice paths from (0, 0) to N P (em, n +1) having steps (x, y) ∈ × . (Variant of A049600 in the On-Line Encyclopedia of Integer Sequences.) Lemma 1 The asymmetric Delannoy numbers satisfy n n m + j d = . m,n j j Xj=0 e Proof: We are enumerating sequences (0, 0) = (x0, y0), (x1, y1),..., (xj , yj), (xj+1, yj+1)=(m, n + 1), where 0 ≤ j ≤ n,0= x0 ≤ x1 ≤···≤ xj ≤ xj+1 = m, and 0 = y0 < y1 < ··· < yj < yj+1 = n + 1. For a given j m+j there are j ways to choose n 0= x0 ≤ x1 ≤···≤ xj ≤ xj+1 = m and j ways to choose 0 = y0 < y1 < ··· < yj < yj+1 = n+ 1. 3 Since n + β + j n x − 1 j P (0,β)(x)= , n j j 2 Xj we get (0,β) dn+β,n = Pn (3) for m ≥ n 3 − 1 because e= 1. 2 Shifted Jacobi and Legendre polynomials Shifted Legendre polynomials appear even in Abramowitz-Stegun: Pn(x) := Pn(2x − 1). e Shifted Jacobi polynomials seem to be less widely used: (α,β) (α,β) Pn (x) := Pn (2x − 1). e Well-known: n n n + k P (x)= (−1)n−k xk n k n kX=0 e n n + k 2k = (−1)n−k xk. n − k k kX=0 Generalization for shifted Jacobi polynomials (α ∈ N, β ∈ C): n+α n + α n + β + k (x−1)αP (α,β)(x)= (−1)n+α−kxk . n k n kX=0 e n n n + β + k ⇒ P (0,β)(x)= (−1)n−kxk . n k n kX=0 e Weighted Delannoy numbers Let u, v, w be commuting variables. We define the u,v,w weighted Delannoy numbers dm,n as the total weight of all Delannoy paths from (0, 0) to (m, n), where each step (0, 1) has weight u, each step (1, 0) has weight v, and each step (1, 1) has weight w. The weight of a lattice path is the product of the weights of its steps. Easy to show: n 2n − k 2n − 2k du,v,w = un−kvn−kwk. n,n k n − k Xk=0 Since u,v,w n u,−v/w,−1 n 1,−uv/w,−1 dn,n =(−w) dn,n =(−w) dn,n we have n 2n − k 2n − 2k uv n−k d1,−uv/w,−1 = − (−1)k. n,n k n − k w kX=0 u,v,w n uv dn,n =(−w) Pn − . w Now e 1,1,1 n n dn,n = dn,n =(−1) Pn(−1)=(−1) Pn(−3) = Pn(3) n since (−1) Pn(−x)= Pen(x). Generalization to shifted Jacobi polynomials n m + n − k m + n − 2k du,v,w = um−kvn−kwk. m,n k n − k Xk=0 u,v,w β n (0,β) uv dn+β,n = u (−w) Pn − . w Here β ∈ Z is any integer satisfyinge β ≥−n. n (0,β) n (0,β) dn+β,n =(−1) Pn (−1)=(−1) Pn (−3). Using the “swapping rule”e n (α,β) (β,α) (−1) Pn (−x)= Pn (x), we get (β,0) dn+β,n = Pn (3). “Swapped” variant of the formula for weighted Delannoy numbers: u,v,w β n (β,0) uv dn+β,n = u w Pn + 1 . w e Many arrays, same diagonal r,2/r,−1 R dn,n = dn,n for all r ∈ \ {0}, and r,1/r,1 R dn,n = dn,n for all r ∈ \ {0}. Lattice path model for the shifted Legendre and Jacobi polynomials 1,x−1,1 1,x,−1 Pn(x)= dn,n = dn,n e (0,β) 1,x,−1 Pn (x)= dn+β,n e (α,β) Fact: The Jacobi polynomials Pn (x) form an orthogonal basis with respect to the inner product 1 hf, gi := f(x) · g(x) · (1 − x)α(1 + x)β dx. Z−1 Goal: to provide a combinatorial, non-inductive proof of this fact for all α, β ∈ N A linear substitution gives the following equivalent form. (α,β) The shifted Jacobi polynomials Pn (x) form an orthogonal basis with respect toe the inner product 1 hf, gi := f(x) · g(x) · (1 − x)αxβ dx. Z0 The case α = 0 Assume m < n. 1 m+β (0,β) (n + m + β + 1)! x Pn (x) dx Z0 n n n + β + ke (n + m + β + 1)! = (−1)n−k k n m + β + k + 1 kX=0 total weight of all pairs (L, σ) where L is a Delannoy path from (0, 0) to (n + β, n) and σ is a bijection {r, a1,...,an+β,b1,...,bm} → {1,...,m + n + β + 1}, subject to: (i) σ(r) < σ(ai) holds for all i such that there is an east step in L from (i − 1, y) to (i, y) for some y; (ii) σ(r) < σ(bj ) holds for j = 1, 2,...,m. Diagonal steps contribute a factor of (−1), all others contribute 1. Cancelling terms Cancel the diagonal steps with the ((1, 0), (0, 1)) sequences, when possible. You will be left with pairs of lattice paths and permutations such that (a) ((1, 0), (0, 1)) is forbidden; (b) σ(r) > σ(ai) holds for all i such that there is a northeast east step in L from (i − 1, y) to (i, y + 1) for some y. (b) makes σ(r) unique, (a) makes the lattice path depend on the position of the diagonal steps only (∼ “rook placements”). Example α = 0, n = 6, m = 2. (6, 6) (0, 0) 61 8 2 5 9 3 4 7 Connection to the orthogonality of Laguerre polynomials We obtained 1 m+β (0,β) (n + m + β + 1)! · x · Pn (x) dx Z0 n n + β n e = (−1)k · k!(n + m + β − k)! k k kX=0 The right hand side is ∞ m (β) β −x N x ln (x)x e dx for all m, n ∈ . Z0 Here n n + β n l(β)(x) := (−1)k k!xn−k n k k kX=0 is the n-th generalized Laguerre polynomial associated to the rectangular board [n + β] × [n]. Rook polynomials Board: B ⊆ [n] × [n]. S ⊆ B compatible if no two elements of S agree in either coordinate. The rook polynomial of B is n k n−k rB(x) := (−1) rkx kX=0 where rk is the number of compatible k-subsets of B. Let L be the linear functional defined by L(xn) := n!. Then ∞ L(p(x)) = e−xp(x) dx Z0 and the number of permutations π of [n] × [n] such that no (i, π(i)) belongs to B is L(rB(x)). The rook polynomial of [n] × [n] is the Laguerre polynomial n n 2 l (x) := (−1)k k!xn−k (1) n k kX=0 n ln(x)=(−1) n!Ln(x). Laguerre polynomials form an orthogonal basis: L(lm(x)ln(x)) = δm,nn! Just for completeness sake The right hand side is (m + β)! times n n p(m) := (−1)k (n + β) (n + m + β − k) k k n−k kX=0 n n =(−1)n (n + β) (−m − β − 1) .
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