Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series

Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira M. Gessel with Jiang Zeng and Guoce Xin LaBRI June 8, 2007

Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials.

Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things?

Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things? Continued fractions count certain weighted Motzkin paths which encode objects of enumerative interest such as partitions and permutations. Flajolet)

Orthogonal polynomials A sequence of polynomials p n x) ) n 0, where p nx) has degree n, is orthogonal if there is a linear functional L on polynomials such that L p m x)p n x) ) = 0 for m n, but L p m x) 2) 0. We will assume that L1) = 1. The moments of the sequence p n x) ) are µ n = Lx n ). So µ 0 = 1. Theorem. The sequence p n x) ) of monic polynomials is orthogonal if and only if there exist numbers a n )n 0 and bn ) n 1, with b n 0 for all n 1 such that xp n x) = p n+1 x) + a n p n x) + b n p n 1 x), n 1 with p 0 x) = 1 and p 1 x) = x a 0.

The connection We then have the continued fraction µ k x k = k=0 1 b 1 x 2 1 a 0 x b 2 x 2 1 a 1 x 1 a 2 x and the Hankel determinant detµ i+j ) 0 i,j<n is equal to b n 1 1 b n 2 2 bn 2 2 b 1.

If we re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials?

If we re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the classical orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them... except the moments. What are the moments of the classical orthogonal polynomials?

How do we find the moments of a sequence of orthogonal polynomials? The key is to find numbers α 0, α 1, α 2,... such that we can evaluate L x + α 0 )x + α 1 ) x + α k ) ) for each k. Usually this isn t too hard to do.) We then apply the following Lemma: t n x+α 0 )x+α 1 ) x+α n 1 ) 1 + α 0 t) 1 + α n t) = 1 1 xt. Proof. We have the indefinite sum m t n x + α 0 ) x + α n 1 ) 1 + α 0 t) 1 + α n t) = 1 [ t m+1 ] 1 x + α 0 ) x + α n ), 1 xt 1 + α 0 t) 1 + α n t) which is easily proved by induction. The lemma follows by taking m.

Now suppose that L x + α 0 )x + α 1 ) x + α n 1 ) ) = M n. We apply L to t n x+α 0 )x+α 1 ) x+α n 1 ) 1 + α 0 t) 1 + α n t) = 1 1 xt. to get M n t n 1 + α 0 t) 1 + α n t) = L = 1 ) 1 xt Lx k )t k = k=0 µ k t k. k=0

Let s look at an example. First, the standard notation: α) n = αα + 1) α + n 1) ) a1,..., a p pf q b 1,..., b p z a 1 ) n a p ) n = z n n! b 1 ) n b q ) n The Hahn polynomials are defined by Q n x; α, β, N) = 3 F 2 n, n + α + β + 1, x α + 1, N ) 1 They are orthogonal with respect to the linear functional L given by L px) ) N ) ) α + x β + N x = px), x N x x=0 By applying Vandermonde s theorem we find that L x + α + 1) k ) = α + β + 2) N N! α + 1) k α + β + 2 + N) k α + β + 2) k.

Since we want L1) to be 1, we normalize this by dividing L by α + β + 2) N /N!. This is independent of k.) Then with our new normalization, L x + α + 1) k ) = α + 1) k α + β + 2 + N) k α + β + 2) k. Since the right side is a rational function of α, β, and N, we don t need N to be a nonnegative integer and we can therefore make a change of variables, α = A 1, β = C A 1, N = B C. Then L x +A) n ) = L x +A)x +A+n) x +A+n 1) ) = A) n B) n C) n. Now we can apply our lemma to get the ordinary generating function for the moments µ n of the Hahn polynomials: µ n t n = A) n B) n C) n t n n j=0 1 + A + j)t).

This can be written as a rather strange-looking hypergeometric series 1 A) n B n ) 1 + At C) n 1 + A + t 1 = 1 ) A, B, 1 ) n 1 + At 3 F 2 C, 1 + A + t 1 1.

This can be written as a rather strange-looking hypergeometric series 1 A) n B n ) 1 + At C) n 1 + A + t 1 = 1 ) A, B, 1 ) n 1 + At 3 F 2 C, 1 + A + t 1 1. How do we get an exponential generating function for the moments? We define a linear operator ε on formal power series by ) ε u n t n = u n t n n!.

Lemma. t n ) ε = e At 1 e t ) n. 1 + At)1 + A + 1)t) 1 + A + n)t) n! Proof #1. Expand the left side by partial fractions and the right side by the binomial theorem. Proof #2. Without explicitly computing the partial fraction on the left, we can see that it will be a linear combination of e At, e A+1)t,..., e A+n)t and the first nonzero term is t n /n!. The right side is the only possibility.

We have µ n t n n! = A) n B) n e At 1 e t ) n C) n n! ) 1 e t. = e At 2F 1 A, B C Let s look at a few examples. First let s see how this generalizes the Meixner polynomials, which are the most general Sheffer orthogonal polynomials. To get one form of) the Meixner polynomials, we take B = uc and then take the limit as C, getting as the exponential generating function for the moments 1 u)e t + u ) A

We have µ n t n n! = A) n B) n e At 1 e t ) n C) n n! ) 1 e t. = e At 2F 1 A, B C As another example, if we take A = B = 1, C = 2, we get the Bernoulli numbers: so µ n = B n. µ n t n n! = t e t 1,

We have µ n t n n! = A) n B) n e At 1 e t ) n C) n n! ) 1 e t. = e At 2F 1 A, B C Similarly, if we take A = 1, B = 2, C = 3, we get the Bernoulli numbers shifted by 1: so µ n = 2B n+1. t n µ n n! = 2d t dt e t 1,

We can use the same approach on the Wilson polynomials, which are the most general classical orthogonal polynomials with q = 1). They are defined by W n x 2 ) = 4 F 3 n, n + a + c + d 1, a + ix, a ix a + b, a + c, a + d ) 1, in this form they are not monic). The corresponding linear functional is L px) ) = 1 Γa + ix)γb + ix)γc + ix)γd + ix) 2 2π Γ2ix) px 2 ) dx 0

It follows easily from known facts that L x + a 2 )x + a + 1) 2 ) x + a + n 1) 2 ) ) = a + b) na + c) n a + d) n a + b + c + d) n so from our first lemma, µ n t n a + b) n a + c) n a + d) n t n = a + b + c + d) n n j=0 1 + a + j) 2 t )

It follows easily from known facts that L x + a 2 )x + a + 1) 2 ) x + a + n 1) 2 ) ) = a + b) na + c) n a + d) n a + b + c + d) n so from our first lemma, µ n t n a + b) n a + c) n a + d) n t n = a + b + c + d) n n j=0 1 + a + j) 2 t ) What about an exponential generating function? If we replace t with t 2, then the denominator factors into linear factors: n 1 a + j) 2 t 2) = ) ) ) 1 a + n)t 1 a + 1)t 1 at j=0 1 + at ) 1 + a + 1 ) t 1 + a + n)t ).

We want to apply our lemma t m ) ε = e At 1 e t ) m. 1 + At)1 + A + 1)t) 1 + A + m)t) m! to t 2n 1 a + n)t ) 1 a + 1)t ) 1 at ) 1 + at ) 1 + a + 1 ) t 1 + a + n)t ).

We want to apply our lemma t m ) ε = e At 1 e t ) m. 1 + At)1 + A + 1)t) 1 + A + m)t) m! to t 2n 1 a + n)t ) 1 a + 1)t ) 1 at ) 1 + at ) 1 + a + 1 ) t 1 + a + n)t ). We can do this if a = 0 or if a = 1/2. If a = 1/2, we must multiply by t.) Our conclusion is

For a = 0, and for a = 1/2, t 2n µ n 2n)! = 3F 2 µ n t 2n+1 2n + 1)! = 2 sin t 2 3 F 2 b, c, d t ) b + c + d, 1 sin2 2 2 b + 1 2, c + 1 2, d + 1 2 b + c + d + 1 2, 3 2 ) t sin2 2

If we take the limit as d in the Wilson polynomials ) n, n + a + c + d 1, a + ix, a ix W n x 2 ) = 4 F 3 a + b, a + c, a + d 1, we get the continuous dual Hahn polynomials ) n, a + ix, a ix p n x 2 ) = 3 F 2 a + b, a + c 1. The generating function for the moments of the continuous dual Hahn polynomials is µ n a, b, c)t n = a + b) n a + c) n t n n j=0 1 + a + j) 2 t ) Then it is clear that the moments µ n a, b, c) are polynomials in a, b, and c, and they have nonnegative coefficients, as is clear from the corresponding continued fraction.

In fact µ n 1, 1, 1) = G 2n+4 and µ n 0, 1, 1) = G 2n+2, where the Genocchi numbers are given by x tan 1 2 x = n=2 G nx n /n!, and the polynomials µ n a, b, c) are polynomials studied by Dumont and Foata as refinements of the Genocchi numbers. For a = 0 we have the generating function µ n 0, b, c) t2n 2n)! = 2F 1 and it s not hard to check that indeed, 2F 1 1, 1 1 2 t ) sin2 = 2 b, c 1 2 t sin2 2 ), G n+2 x n n! = d 2 dx 2 x tan x 2 using 2F 1 1, 1 1 2 ) y 2 = 1 y 2 + y sin 1 y 1 y 2 ) 3/2.

Another application of the continuous dual Hahn polynomials We have µ n 1 2, 1 2 γ, 1 t2n 2γ) 2n)! = 2F 1 = sin 1 2 γt γ sin 1 2 t 1 2 1 γ), ) 1 2 1 + γ) t sin2 2 In the case γ = 1/3, the corresponding Hankel determinant counts alternating sign matrices Colomo and Pronko). 3 2

Hankel Determinants and Ternary Tree Numbers Let a n = ) 1 3n, 2n + 1 n so that a n is the number of trees with n vertices. Michael Somos conjectured that deta i+j ) 0 i,j n 1 is the number of cyclically symmetric transpose complement plane partitions whose Ferrers diagrams fit in an n n n box. This number is known to be n 3i + 1) 6i)! 2i)!. 4i + 1)! 4i)! i=1 Somos had similar conjectures for the Hankel determinants deta i+j ) 0 i,j n 1 and deta i+j+1)/2 ) 0 i,j n 1, relating them to alternating sign matrices invariant under vertical reflection and alternating sign matrices invariant under both vertical and horizontal reflection.

Let gx) = Then gx) satisfies ) 1 3n x n = 2n + 1 n gx) = 1 + xgx) 3. a n x n. If we compute the continued fraction for gx), we find empirically that it has a simple formula, which implies the determinant evaluation. I found a complicated) proof of this continued fraction. Then Guoce Xin found a much nicer proof, using Gauss s continued fraction. However, the same method had already been used by Ulrich Tamm to evaluate these determinants, before Somos stated his conjectures. But there is no known combinatorial connection between these Hankel determinants and plane partitions or alternating sign matrices.)

The key fact is that the continued fraction for gx) is a special case of Gauss s continued fraction: where 2F 1 a, b + 1 c + 1 ) / x 2F 1 a, b c ) x = Sx; λ 1, λ 2,...), 1) a + n 1)c b + n 1) λ 2n 1 =, c + 2n 2)c + 2n 1) n = 1, 2,..., b + n)c a + n) λ 2n =, c + 2n 1)c + 2n) n = 1, 2,..., and Sx; λ 1, λ 2, λ 3,...) denotes the continued fraction 1 Sx; λ 1, λ 2, λ 3,...) = λ 1 x 1 1 λ 2x 1 λ 3x... 2)

In fact we have g = 2 F 1 2 3, 4 3 ; 3 2 27 4 x ) /2F 1 2 3, 1 3 ; 1 2 27 4 x ). and this isn t too hard to prove; if we set f = g 1 by Lagrange inversion one can show that the numerator is 1 + f ) 2 /1 2f ) and the denominator is 1 + f )/1 2f ). Xin and I tried to find all related cases where an instance of Gauss s continued fraction is a polynomial in f and thus the coefficients have an explicit formula).

Five of them are 1 + f = 2 F 1 2 3, 4 3 ; 3 2 1 + f ) 2 = 2 F 1 4 3, 5 3 ; 5 2 1 + f )1 + 1 2 f ) = 2F 1 5 3, 7 3 ; 7 2 1 + f )1 1 2 f ) = 2F 1 5 3, 7 3 ; 5 2 1 + f )1 + 2 5 f ) = 2F 1 2 3, 4 3 ; 5 2 Why do they factor? ) ) 27 4 /2F x 2 1 3, 1 3 ; 1 2 27 4 x ) ) 27 4 /2F x 4 1 3, 2 3 ; 3 2 27 4 x ) ) 27 4 /2F x 5 1 3, 4 3 ; 5 2 27 4 x ) ) 27 4 /2F x 5 1 3, 4 3 ; 3 2 27 4 x ) ) 27 4 /2F x 2 1 3, 1 3 ; 3 2 27 4 x

Five of them are 1 + f = 2 F 1 2 3, 4 3 ; 3 2 1 + f ) 2 = 2 F 1 4 3, 5 3 ; 5 2 1 + f )1 + 1 2 f ) = 2F 1 5 3, 7 3 ; 7 2 1 + f )1 1 2 f ) = 2F 1 5 3, 7 3 ; 5 2 1 + f )1 + 2 5 f ) = 2F 1 2 3, 4 3 ; 5 2 Why do they factor? We don t know. ) ) 27 4 /2F x 2 1 3, 1 3 ; 1 2 27 4 x ) ) 27 4 /2F x 4 1 3, 2 3 ; 3 2 27 4 x ) ) 27 4 /2F x 5 1 3, 4 3 ; 5 2 27 4 x ) ) 27 4 /2F x 5 1 3, 4 3 ; 3 2 27 4 x ) ) 27 4 /2F x 2 1 3, 1 3 ; 3 2 27 4 x