The Restricted Toda Chain, Exponential Riordan Arrays, and Hankel Transforms

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract

We re-interpret results on the classification of Toda chain solutions given by Sheffer class orthogonal polynomials in terms of exponential Riordan arrays. We also examine associated Hankel transforms.

1 Introduction

The restricted Toda chain equation [15, 23] is simply described by

u̇_{n} = u_{n}(b_{n} − b_{n−1}), n = 1, 2, … ḃ_{n} = u_{n+1} − u_{n}, n = 0, 1, …  (1)

with u_{0} = 0, where the dot indicates differentiation with respect to t. In this note, we shall show how solutions to this equation can be formulated in the context of exponential Riordan arrays. The Riordan arrays we shall consider may be considered as parameterised (or “time”-dependent) Riordan arrays. We have already considered parameterized Riordan arrays [1], exploring the links between these Riordan arrays and orthogonal polynomials.

The restricted Toda chain equation is closely related to orthogonal polynomials, since the functions u_{n} and b_{n} can be considered as the coefficients in the usual three-term recurrence [4, 10, 22] satisfied by orthogonal polynomials:

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x), n = 1, 2, …  (2)

with initial conditions P_{0}(x) = 1 and P_{1}(x) = x − b_{0}.

Nakamura and Zhedanov [15, 16] give solutions of the restricted Toda chain equation in terms of Sheffer polynomials. Given the close links between Sheffer class polynomials and Riordan arrays [6, 11], we find it instructive to re-interpret these results in terms of Riordan arrays. Thus in the cases discussed below, we find that the coefficient arrays of the orthogonal polynomials linked to solutions of the restricted Toda chain are exponential Riordan arrays. We believe that this result concerning an application of exponential Riordan arrays to the Toda chain equation is new. In particular, the parameters of the solution are obtained by calculating the terms of the corresponding production matrices, using the procedures of Deutsch, Ferrari, and Rinaldi [8, 9].

The reader is referred to [4, 10, 22] for basic information on orthogonal polynomials (see also [18]), to [2, 3, 7, 19] for details of Riordan arrays, to [8, 9] for information on production (Stieltjes) matrices (see also [17]), and to [5, 14, 24] for examples of Hankel transforms. Many interesting examples of Riordan arrays (both ordinary and exponential), and Hankel transforms, can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS) [20, 21]. Integer equences are frequently referred to by their OEIS number. For instance, the binomial matrix B (“Pascal’s triangle”) is \underline{A007318}.

2 Integer sequences, Hankel transforms, exponential Riordan arrays, orthogonal polynomials

In this section, we recall known results on integer sequences, Hankel transforms, exponential Riordan arrays and orthogonal polynomials that will be useful for the sequel.

For an integer sequence a_{n}, that is, an element of ℤ^{ℕ}, the power series f_{o}(x) = ∑_{k=0}^{∞} a_{k}x^{k} is called the ordinary generating function or g.f. of the sequence, while f_{e}(x) = ∑_{k=0}^{∞} \frac{a_{k}}{k!} x^{k} is called the exponential generating function or e.g.f. of the sequence. a_{n} is thus the coefficient of x^{n} in f_{o}(x). We denote this by a_{n} = [x^{n}]f_{o}(x). Similarly, a_{n} = n![x^{n}]f_{e}(x). For instance, F_{n} = [x^{n}]\frac{x}{1−x−x^{2}} is the n-th Fibonacci number \underline{A000045}, while n! = n![x^{n}]\frac{1}{1−x}, which says that \frac{1}{1−x} is the e.g.f. of n! \underline{A000142}. For a power series f(x) = ∑_{n=0}^{∞} a_{n}x^{n} with f(0) = 0 and f′(0) ≠ 0 we define the reversion or compositional inverse of f to be the power series f̄(x) = f^{[−1]}(x) such that f(f̄(x)) = x. We sometimes write f̄ = Revf.

The Hankel transform [14] of a given sequence A = (a_{n})_{n≥0} is the sequence of Hankel determinants {h_{0}, h_{1}, h_{2}, … } where h_{n} = |a_{i+j}|_{i,j=0}^{n}, i.e

A = (a_{n})_{n∈ℕ_{0}} → h = (h_{n})_{n∈ℕ_{0}} : h_{n} = | \begin{matrix}a_{0} & a_{1} & ⋯ & a_{n} \\ a_{1} & a_{2} & & a_{n+1} \\ ⋮ & & ... \\ a_{n} & a_{n+1} & & a_{2n} \end{matrix} | .  (3)

The Hankel transform of a sequence a_{n} and that of its binomial transform are equal.

In the case that a_{n} has g.f. g(x) expressible in the form

g(x) = \frac{a_{0}}{1 − α_{0}x − \frac{β_{1}x^{2}}{1 − α_{1}x − \frac{β_{2}x^{2}}{1 − α_{2}x − \frac{β_{3}x^{2}}{1 − α_{3}x − ⋯}}}},

(with β_{i} ≠ 0 for all i) then we have [12, 13, 25]

h_{n} = a_{0}^{n}β_{1}^{n−1}β_{2}^{n−2} ⋯ β_{n−1}^{2}β_{n} = a_{0}^{n} ∏_{k=1}^{n} β_{k}^{n−k+1}.  (4)

Note that this is independent of α_{n}. In general α_{n} and β_{n} are not integers. Such a continued fraction is associated to a monic family of orthogonal polynomials which obey the three term recurrence

p_{n+1}(x) = (x − α_{n})p_{n}(x) − β_{n}p_{n−1}(x),  p_{0}(x) = 1,  p_{1}(x) = x − α_{0}.

The terms appearing in the first column of the inverse of the coefficient array of these polynomials are the moments of family.

The exponential Riordan group [2, 7, 9], is a set of infinite lower-triangular integer matrices, where each matrix is defined by a pair of generating functions g(x) = g_{0}+g_{1}x+g_{2}x^{2}+… and f(x) = f_{1}x + f_{2}x^{2} + … where f_{1} ≠ 0. The associated matrix is the matrix whose i-th column has exponential generating function g(x)f(x)^{i}/i! (the first column being indexed by 0). The matrix corresponding to the pair f, g is denoted by [g, f]. It is monic if g_{0} = 1. If in addition f_{1} = 1, then it is proper. The group law is given by

[g, f] ∗ [h, l] = [g(h ∘ f), l ∘ f].

The identity for this law is I = [1, x] and the inverse of [g, f] is [g, f]^{−1} = [1/(g ∘ f̄), f̄] where f̄ is the compositional inverse of f. We use the notation eR to denote this group. If M is the matrix [g, f], and u = (u_{n})_{n≥0} is an integer sequence with exponential generating function U (x), then the sequence Mu^{T} has exponential generating function g(x)U(f(x)). Thus the row sums of the array [g, f] are given by g(x)e^{f(x)} since the sequence 1, 1, 1, … has exponential generating function e^{x}.

Example 1. The binomial matrix is the matrix with general term \binom{n}{k}. It is realized by Pascal’s triangle. As an exponential Riordan array, it is given by [e^{x}, x]. We further have

([e^{x}, x])^{m} = [e^{mx}, x].

An important concept for the sequel is that of production matrix. The concept of a production matrix [8, 9] is a general one, but for this note we find it convenient to review it in the context of Riordan arrays. Thus let P be an infinite matrix (most often it will have integer entries). Letting r_{0} be the row vector

r_{0} = (1, 0, 0, 0, …),

we define r_{i} = r_{i−1}P, i ≥ 1. Stacking these rows leads to another infinite matrix which we denote by A_{P}. Then P is said to be the production matrix for A_{P}. If we let

u^{T} = (1, 0, 0, 0, …, 0, …)

then we have

A_{P} = ( u^{T} ; u^{T} P ; u^{T} P^{2} ; ⋮ )

and

DA_{P} = A_{P} P

where D = (δ_{i,j+1})_{i,j≥0} (where δ is the usual Kronecker symbol). Note that [17] P is also called the Stieltjes matrix associated to A_{P}. The following result [9] concerning matrices that are production matrices for exponential Riordan arrays is important for the sequel.

Proposition 2. Let A = (a_{n,k})_{n,k≥0} = [g(x), f(x)] be an exponential Riordan array and let

c(y) = c_{0} + c_{1}y + c_{2}y^{2} + …,  r(y) = r_{0} + r_{1}y + r_{2}y^{2} + …  (5)

be two formal power series that that

r(f(x)) = f′(x)  (6)
c(f(x)) = \frac{g′(x)}{g(x)}.  (7)

Then

(i)  a_{n+1,0} = ∑_{i} i!c_{i}a_{n,i}  (8)
(ii)  a_{n+1,k} = r_{0}a_{n,k−1} + \frac{1}{k!} ∑_{i≥k} i!(c_{i−k} + kr_{i−k+1})a_{n,i}  (9)

or, defining c_{−1} = 0,

a_{n+1,k} = \frac{1}{k!} ∑_{i≥k−1} i!(c_{i−k} + kr_{i−k+1})a_{n,i}.  (10)

Conversely, starting from the sequences defined by (5), the infinite array (a_{n,k})_{n,k≥0} defined by (10) is an exponential Riordan array.

A consequence of this proposition is that P = (p_{i,j})_{i,j≥0} where

p_{i,j} = \frac{i!}{j!}(c_{i−j} + jr_{r−j+1}) (c_{−1} = 0).

Furthermore, the bivariate exponential generating function

ϕ_{P} (t, z) = ∑_{n,k} p_{n,k}t^{k} \frac{z^{n}}{n!}

of the matrix P is given by

ϕ_{P} (t, z) = e^{tz}(c(z) + tr(z)).

Note in particular that we have

r(x) = f′(f̄(x))

and

c(x) = \frac{g′(f̄(x))}{g(f̄(x))}.

Example 3. We consider the exponential Riordan array [\frac{1}{1−x}, x], \underline{A094587}. This array [3] has elements

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 2 & 2 & 1 & 0 & 0 & 0 & … \\ 6 & 6 & 3 & 1 & 0 & 0 & … \\ 24 & 24 & 12 & 4 & 1 & 0 & … \\ 120 & 120 & 60 & 20 & 5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

and general term [k ≤ n]\frac{n!}{k!} with inverse

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ −1 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & −2 & 1 & 0 & 0 & 0 & … \\ 0 & 0 & −3 & 1 & 0 & 0 & … \\ 0 & 0 & 0 & −4 & 1 & 0 & … \\ 0 & 0 & 0 & 0 & −5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which is the array [1 − x, x]. In particular, we note that the row sums of the inverse, which begin 1, 0, −1, −2, −3, … (that is, 1 − n), have e.g.f. (1 − x) exp(x). This sequence is thus the binomial transform of the sequence with e.g.f. (1 − x) (which is the sequence starting 1, −1, 0, 0, 0, …). In order to calculate the production matrix P of [\frac{1}{1−x}, x] we note that f(x) = x, and hence we have f′(x) = 1 so f′(f̄(x)) = 1. Also g(x) = \frac{1}{1−x} leads to g′(x) = \frac{1}{(1−x)^{2}} , and so, since f̄(x) = x, we get

\frac{g′(f̄(x))}{g(f̄(x))} = \frac{1}{1 − x}.

Thus the generating function for P is

e^{tz} (\frac{1}{1 − z} + t) .

Thus P is the matrix [\frac{1}{1−x}, x] with its first row removed.

Example 4. We consider the exponential Riordan array [1, \frac{x}{1−x}]. The general term of this matrix [3] may be calculated as follows:

T_{n,k} = \frac{n!}{k!}[x^{n}]\frac{x^{k}}{(1 − x)^{k}}
= \frac{n!}{k!}[x^{n−k}](1 − x)^{−k}
= \frac{n!}{k!}[x^{n−k}] ∑_{j=0}^{∞} \binom{−k}{j}(−1)^{j}x^{j}

= \frac{n!}{k!}[x^{n−k}] ∑_{j=0}^{∞} \binom{k + j − 1}{j}x^{j}

= \frac{n!}{k!}\binom{k + n − k − 1}{n − k}
= \frac{n!}{k!}\binom{n − 1}{n − k}.

Thus its row sums, which have e.g.f. exp (\frac{x}{1−x}), have general term ∑_{k=0}^{n} \frac{n!}{k!} \binom{n−1}{n−k}. This is \underline{A000262}, the ‘number of “sets of lists”: the number of partitions of \{1, .., n\} into any number of lists, where a list means an ordered subset’. Its general term is equal to (n−1)!L_{n−1}(1, −1). The inverse of [1, \frac{x}{1−x}] is the exponential Riordan array [1, \frac{x}{1+x}], \underline{A111596}. The row sums of this sequence have e.g.f. exp (\frac{x}{1+x}), and start 1, 1, −1, 1, 1, −19, 151, . . .. This is \underline{A111884}. To calculate the production matrix of [1, \frac{x}{1+x}] we note that g′(x) = 0, while f̄(x) = \frac{x}{1+x} with f′(x) = \frac{1}{(1+x)^{2}} . Thus

f′(f̄(x)) = (1 + x)^{2},

and so the generating function of the production matrix is given by

e^{tz}t(1 + z)^{2}.

The production matrix of the inverse begins

( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 2 & 4 & 1 & 0 & 0 & … \\ 0 & 0 & 6 & 6 & 1 & 0 & … \\ 0 & 0 & 0 & 12 & 8 & 1 & … \\ 0 & 0 & 0 & 0 & 20 & 10 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 5. The exponential Riordan array A = [\frac{1}{1−x}, \frac{x}{1−x}], or

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 2 & 4 & 1 & 0 & 0 & 0 & … \\ 6 & 18 & 9 & 1 & 0 & 0 & … \\ 24 & 96 & 72 & 16 & 1 & 0 & … \\ 120 & 600 & 600 & 200 & 25 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

has general term

T_{n,k} = \frac{n!}{k!}\binom{n}{k}.

Its inverse is [\frac{1}{1+x}, \frac{x}{1+x}] with general term (−1)^{n−k} \frac{n!}{k!} \binom{n}{k}. This is \underline{A021009}, the triangle of coefficients of the Laguerre polynomials L_{n}(x). The production matrix of [\frac{1}{1−x}, \frac{x}{1−x}] is given by

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 0 & 4 & 5 & 1 & 0 & 0 & … \\ 0 & 0 & 9 & 7 & 1 & 0 & … \\ 0 & 0 & 0 & 16 & 9 & 1 & … \\ 0 & 0 & 0 & 0 & 25 & 11 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 6. The exponential Riordan array [e^{x}, ln (\frac{1}{1−x})], or

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 1 & 8 & 6 & 1 & 0 & 0 & … \\ 1 & 24 & 29 & 10 & 1 & 0 & … \\ 1 & 89 & 145 & 75 & 15 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

is the coefficient array for the polynomials

2F_{0}(−n, x; −1)

which are an unsigned version of the Charlier polynomials (of order 0) [10, 18, 22]. This is \underline{A094816}. It is equal to

[e^{x}, x] [1, ln (\frac{1}{1 − x})] ,

or the product of the binomial array B and the array of (unsigned) Stirling numbers of the first kind. The production matrix of the inverse of this matrix is given by

( \begin{matrix}−1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & −2 & 1 & 0 & 0 & 0 & … \\ 0 & 2 & −3 & 1 & 0 & 0 & … \\ 0 & 0 & 3 & −4 & 1 & 0 & … \\ 0 & 0 & 0 & 4 & −5 & 1 & … \\ 0 & 0 & 0 & 0 & 5 & −6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

which indicates the orthogonal nature of these polynomials. We can prove this as follows. We have

[e^{x}, ln (\frac{1}{1 − x})]^{−1} = [e^{−(1−e^{−x})}, 1 − e^{−x}] .

Hence g(x) = e^{−(1−e^{−x})} and f(x) = 1 − e^{−x}. We are thus led to the equations

r(1 − e^{−x}) = e^{−x},
c(1 − e^{−x}) = −e^{−x},

with solutions r(x) = 1 − x, c(x) = x − 1. Thus the bivariate generating function for the production matrix of the inverse array is

e^{tz}(z − 1 + t(1 − z)),

which is what is required.

3 Hermite polynomials and the Toda chain

We recall that the Hermite polynomials may be defined as

H_{n}(x) = n! ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \frac{(−1)^{k}(2x)^{n−2k}}{k!(n − 2k)!} .

The generating function for H_{n}(x) is given by

e^{2xt−t^{2}} = ∑_{n=0}^{∞} H_{n}(x)\frac{t^{n}}{n!}.

The following result was proved [1]:

Proposition 7. The proper exponential Riordan array

L = [e^{2rx−x^{2}}, x]

has as first column the Hermite polynomials H_{n}(r). This array has a tri-diagonal production array.

Proof. The first column of L has generating function e^{2rx−x^{2}}, from which the first assertion follows. Standard Riordan array techniques show us that the production array P of L is indeed tri-diagonal, beginning with

( \begin{matrix}2r & 1 & 0 & 0 & 0 & 0 & … \\ −2 & 2r & 1 & 0 & 0 & 0 & … \\ 0 & −4 & 2r & 1 & 0 & 0 & … \\ 0 & 0 & −6 & 2r & 1 & 0 & … \\ 0 & 0 & 0 & −8 & 2r & 1 & … \\ 0 & 0 & 0 & 0 & −10 & 2r & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This is so since f(x) = x gives us f̄(x) = x and f′(x) = 1, and thus

r(x) = f′(f̄(x)) = f′(x) = 1.

Similarly,

c(x) = \frac{g′(f̄(x))}{g(f̄(x))} = 2(r − x),

and thus the bivariate generating function ϕ_{P} (y, w) of P is given by

ϕ_{P} (y, z) = e^{yz}(2(r − z) + y),

as required.

We note that L starts with

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 2r & 1 & 0 & 0 & 0 & 0 & … \\ 2(2r^{2} − 1) & 4r & 1 & 0 & 0 & 0 & … \\ 4r(2r^{2} − 3) & 6(2r^{2} − 1) & 6r & 1 & 0 & 0 & … \\ 4(4r^{3} − 12r^{2} + 3) & 16r(2r^{2} − 3) & 12(2r^{2} − 1) & 8r & 1 & 0 & … \\ 8r(4r^{4} − 20r^{2} + 15) & 20(4r^{4} − 12r^{2} + 3) & 40r(2r^{2} − 3) & 20(2r^{2} − 1) & 10r & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Thus

L^{−1} = [e^{−2rx+x^{2}}, x]

is the coefficient array of a set of orthogonal polynomials which have as moments the Hermite polynomials. These new orthogonal polynomials satisfy the three-term recurrence

ℌ_{n+1}(x) = (x − 2r)ℌ_{n}(x) + 2nℌ_{n−1}(x),

with ℌ_{0} = 1, ℌ_{1} = x − 2r.

We can now modify this result to give us our first Toda chain result.

Proposition 8. The exponential Riordan array

[e^{−2(z−t)x+x^{2}}, x]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where (u_{n}, b_{n}) is a solution to the restricted Toda chain. Proof. We easily determine that the inverse matrix

[e^{2(z−t)x−x^{2}}, x]

has the production matrix P given by

( \begin{matrix}2(z − t) & 1 & 0 & 0 & 0 & 0 & … \\ −2 & 2(z − t) & 1 & 0 & 0 & 0 & … \\ 0 & −4 & 2(z − t) & 1 & 0 & 0 & … \\ 0 & 0 & −6 & 2(z − t) & 1 & 0 & … \\ 0 & 0 & 0 & −8 & 2(z − t) & 1 & … \\ 0 & 0 & 0 & 0 & −10 & 2(z − t) & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This is so since f(x) = x gives us f̄(x) = x and f′(x) = 1, and thus

r(x) = f′(f̄(x)) = f′(x) = 1.

Similarly,

c(x) = \frac{g′(f̄(x))}{g(f̄(x))} = −2(x − z + t),

and thus the bivariate generating function ϕ_{P} (y, w) of P is given by

ϕ_{P} (y, w) = e^{yw}(−2(w − z + t) + y),

as required.

This verifies that P_{n}(x) is indeed a family of orthogonal polynomials, for which

u_{n}(t) = −2n, b_{n}(t) = 2(z − t).

It is immediate that these satisfy Eq. (1).

We now note that the moments of this polynomial family (the first column of the inverse matrix) m_{n} satisfy the following relation:

m_{n} = n![x^{n}]e^{2(z−t)x−x^{2}}
= \frac{1}{e^{−t^{2}+2tz}} \frac{d^{n}}{dt^{n}} e^{−t^{2}+2tz}

= n! ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}(−1)^{k} \frac{(2(z − t))^{n−2k}}{(n − k)!}

= ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{k}\binom{n − k}{k}k!(2(z − t))^{n−2k}
= H_{n}(z − t).

From the form of the production matrix P, we infer that the generating function of m_{n} may be expressed as

\frac{1}{1 − 2(z − t)x + \frac{2x^{2}}{1 − 2(z − t)x + \frac{4x^{2}}{1 − 2(z − t)x + \frac{6x^{2}}{1 − 2(z − t)x + ⋯}}}}.

The Hankel transform of m_{n}, defined as the sequence h_{n} where

h_{n} = |m_{i+j}|_{0≤i,j≤n}

is then given by

h_{n} = (−2)^{\binom{n+1}{2}} ∏_{k=0}^{n} k!.

4 Charlier polynomials and the Toda chain

Proposition 9. The exponential Riordan array

[e^{−xe^{t}}, ln(1 + x)]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where (u_{n}, b_{n}) is a solution to the restricted Toda chain.

Proof. We determine that the inverse matrix

[e^{e^{t+x}−e^{t}}, e^{x} − 1]

has the production matrix

( \begin{matrix}e^{t} & 1 & 0 & 0 & 0 & 0 & … \\ e^{t} & e^{t} + 1 & 1 & 0 & 0 & 0 & … \\ 0 & 2e^{t} & e^{t} + 2 & 1 & 0 & 0 & … \\ 0 & 0 & 3e^{t} & e^{t} + 3 & 1 & 0 & … \\ 0 & 0 & 0 & 4e^{t} & e^{t} + 4 & 1 & … \\ 0 & 0 & 0 & 0 & 5e^{t} & e^{t} + 5 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

as follows. We have f̄(x) = ln(1 + x) and f′(x) = e^{x}, and thus

r(x) = f′(f̄(x)) = 1 + x.

Similarly,

c(x) = \frac{g′(f̄(x))}{g(f̄(x))} = \frac{e^{xe^{t}+t}(1 + x)}{e^{xe^{t}}} = e^{t}(1 + x).

Thus the bivariate generating function for the production matrix P is given by

ϕ_{P} (y, z) = e^{yz}(e^{t}(1 + z) + y(1 + z)),

as required. This verifies that P_{n}(x) is indeed a family of orthogonal polynomials, for which

u_{n}(t) = ne^{t}, b_{n}(t) = n + e^{t}.

It is easy now to verify that, with these values, (u_{n}, b_{n}) satisfies the Toda chain equations Eq. (1).

The moments m_{n} of this family of orthogonal polynomials may be expressed as:

m_{n} = n![x^{n}]e^{e^{t+x}−e^{t}}
= \frac{1}{e^{e^{t}−1}} \frac{d^{n}}{dt^{n}} e^{e^{t}−1}
= ∑_{k=0}^{n} \left\{{n \atop k}\right\}e^{kt},

where

\left\{{n \atop k}\right\} = \frac{1}{k!} ∑_{j=0}^{k}(−1)^{k−j}\binom{k}{j}j^{n}

are the Stirling numbers of the second kind. From the form of the production matrix P, we infer that the generating function of m_{n} may be expressed as

\frac{1}{1 − e^{t}x − \frac{e^{t}}{1 − (e^{t} + 1)x − \frac{2e^{t}}{1 − (e^{t} + 2)x − \frac{3e^{t}}{1 − (e^{t} + 3)x − ⋯}}}}.

The Hankel transform of m_{n} is then given by

h_{n} = (e^{t})^{\binom{n+1}{2}} ∏_{k=0}^{n} k!

In particular we have

m_{n}(0) = ∑_{k=0}^{n} \left\{{n \atop k}\right\} = Bell(n),

where Bell(n) denotes the n-th Bell number, with Hankel transform

h_{n} = ∏_{k=0}^{n} k!.

5 Laguerre polynomials and the Toda chain

We have seen that the exponential Riordan array

[\frac{1}{1 − x}, \frac{x}{1 − x}]

is closely related to the Laguerre polynomials. We introduce a “time” parameter t as follows:

[\frac{1}{1 − \frac{x}{1+t}} , \frac{x}{1 − \frac{x}{1+t}} ] = [(1 − \frac{x}{1 + t})^{−1} , \frac{x}{1 − \frac{x}{1+t}} ] ,

and further generalize this array by introducing a general power factor α, to get the array

[(1 − \frac{x}{1 + t})^{α} , \frac{x}{1 − \frac{x}{1+t}} ] .

We then have the following proposition.

Proposition 10. The exponential Riordan array

[(1 − \frac{x}{1 + t})^{α} , \frac{x}{1 − \frac{x}{1+t}} ]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where (u_{n}, b_{n}) is a solution to the restricted Toda chain.

Proof. The inverse matrix

[(\frac{1 + t + x}{1 + t})^{α} , \frac{(1 + t)x}{1 + t + x}]

has the production matrix

( \begin{matrix}\frac{α}{1+t} & 1 & 0 & 0 & 0 & 0 & … \\ \frac{−α}{(1+t)^{2}} & \frac{α−2}{1+t} & 1 & 0 & 0 & 0 & … \\ 0 & \frac{2(1−α)}{(1+t)^{2}} & \frac{α−4}{1+t} & 1 & 0 & 0 & … \\ 0 & 0 & \frac{3(2−α)}{(1+t)^{2}} & \frac{α−6}{1+t} & 1 & 0 & … \\ 0 & 0 & 0 & \frac{4(3−α)}{(1+t)^{2}} & \frac{α−8}{1+t} & 1 & … \\ 0 & 0 & 0 & 0 & \frac{5(4−α)}{(1+t)^{2}} & \frac{α−10}{1+t} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This verifies that P_{n}(x) is indeed a family of orthogonal polynomials, for which

u_{n}(t) = \frac{n(n − α − 1)}{(1 + t)^{2}}, b_{n}(t) = \frac{α − 2n}{1 + t}.

It is easy now to verify that, with these values, (u_{n}, b_{n}) satisfies the Toda chain equations Eq. (1).

For this family of orthogonal polynomials, the moments m_{n} may be expressed as:

m_{n} = n![x^{n}] (1 + \frac{x}{1 + t})^{α} = \frac{1}{(1 + t)^{α}} \frac{d^{n}}{dt^{n}} (1 + t)^{α} = \frac{(α)_{n}}{(1 + t)^{n}} ,  (11)

where

(α)_{n} = α(α − 1) ⋯ (α − n + 1).

From the form of the production matrix, we infer that the generating function of m_{n} may be expressed as

\frac{1}{1 − \frac{α}{1+t}x − \frac{\frac{0−α}{(1+t)^{2}} x^{2}}{1 − \frac{α−2}{1+t}x − \frac{\frac{2(1−α)}{(1+t)^{2}}x^{2}}{1 − \frac{α−4}{1+t}x − \frac{\frac{3(2−α)}{(1+t)^{2}}x^{2}}{1 − \frac{α−6}{1+t}x − ⋯}}}}.

The Hankel transform of m_{n} is then given by

h_{n} = \frac{(−1)^{\binom{n+1}{2}}}{(1 + t)^{n(n+1)}} ∏_{k=0}^{n} k!(α − k)^{n−k}.

Noticing that

lim_{c→1} \frac{c(1 + t)}{c − 1} ln (\frac{1 − \frac{x}{c(1+t)}}{1 − \frac{x}{1+t}} ) = \frac{x}{1 − \frac{x}{1+t}} ,

we can ’embed’ this previous result in the following: Proposition 11. The exponential Riordan array

[(1 − \frac{x}{1 + t})^{α} , \frac{c(1 + t)}{c − 1} ln (\frac{1 − \frac{x}{c(1+t)}}{1 − \frac{x}{1+t}} )]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where for c = 1, (u_{n}, b_{n}) is a solution to the restricted Toda chain. Proof. We can show that the inverse array has a tri-diagonal production matrix with

b(n) = \frac{αc − n(c + 1)}{c(1 + t)},

and

u(n) = \frac{n(n − 1 − α)}{c(1 + t)^{2}}.

Letting c → 1 now gives us the result. Another direction of generalization is given be looking at the related exponential Riordan array

[e^{αx}, \frac{x}{1 + \frac{x}{1+t}} ] ,

which has inverse

[e^{\frac{αx}{1− \frac{x}{1+t}}} , \frac{x}{1 − \frac{x}{1+t}} ] .

For this, we have the following proposition. Proposition 12. The exponential Riordan array

[e^{\frac{αx}{1− \frac{x}{1+t}}} , \frac{x}{1 − \frac{x}{1+t}} ]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where (u_{n}, b_{n}) is a solution to the restricted Toda chain.

Proof. We find that

u_{n}(t) = \frac{n(n − 1)}{(1 + t)^{2}}

and

b_{n}(t) = −\frac{α(1 + t) + 2n}{1 + t}.

In this case, the moments are simply

m_{n} = (−α)^{n}.

6 Meixner polynomials and the Toda chain

Proposition 13. The exponential Riordan array

[\frac{1}{\sqrt{1 − 2x tanh(t) − x^{2}sech(t)^{2}}} , ln (\sqrt{\frac{1 + xe^{−t}sech(t)}{1 − xe^{t}sech(t)}})]

is the coefficient array of a family of orthogonal polynomials P_{n}(x) with

P_{n+1}(x) + b_{n}P_{n}(x) + u_{n}P_{n−1}(x) = xP_{n}(x),

where (u_{n}, b_{n}) is a solution to the restricted Toda chain. Proof. The inverse matrix

[\frac{sech(x + t)}{sech(t)}, sinh(x)\frac{sech(x + t)}{sech(t)}]

has the production matrix

( \begin{matrix}− tanh(t) & 1 & 0 & 0 & 0 & 0 & … \\ −sech^{2}(t) & −3 tanh(t) & 1 & 0 & 0 & 0 & … \\ 0 & −4 sech^{2}(t) & −5 tanh(t) & 1 & 0 & 0 & … \\ 0 & 0 & −9 sech^{2}(t) & −7 tanh(t) & 1 & 0 & … \\ 0 & 0 & 0 & −16 sech^{2}(t) & −9 tanh(t) & 1 & … \\ 0 & 0 & 0 & 0 & −25 sech^{2}(t) & −11 tanh(t) & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This verifies that P_{n}(x) is indeed a family of orthogonal polynomials, for which

u_{n}(t) = −n^{2}sech^{2}(t), b_{n}(t) = −(2n + 1) tanh(t).

It is easy now to verify that, with these values, (u_{n}, b_{n}) satisfies the Toda chain equations Eq. (1).

We may describe the moments m_{n} of this family of polynomials by

m_{n} = n![x^{n}]\frac{sech(x + t)}{sech(t)} = \frac{1}{sech(t)} \frac{d^{n}}{dt^{n}} sech(t).  (12)

The Hankel transform of m_{n} is then given by

h_{n} = (−1)^{\binom{n+1}{2}}sech(t)^{n(n+1)} ∏_{k=0}^{n}(k!)^{2}.

7 Acknowledgements

The author would like to express his gratitude for the careful reading and helpful remarks of an anonymous reviewer, which have resulted in what is hopefully a clearer paper.

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2010 Mathematics Subject Classification: Primary 42C05; Secondary 11B83, 11C20, 15B05, 15B36, 33C45.

Keywords: Toda chain, orthogonal polynomials, moments, Hermite polynomial, Charlier polynomial, Meixner polynomial, Riordan array, Hankel determinant, Hankel transform.

(Concerned with sequences \underline{A000045}, \underline{A000142}, \underline{A000262}, \underline{A007318}, \underline{A021009}, \underline{A094587}, \underline{A094816}, \underline{A111596}, and \underline{A111884}.)

Received May 27 2010; revised version received August 3 2010. Published in Journal of Integer Sequences, August 4 2010.

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