UVAROV’S FORMULA FOR RATIONAL DEFORMATIONS OF ORTHOGONAL POLYNOMIALS AND ITS ASSOCIATED IDENTITY

— A THOROUGH DISCUSSION

JINHO BAIK, PERCY DEIFT, CHRISTIAN KRATTENTHALER^{†}, AND EVGENY STRAHOV

Abstract. Let p_{n}(x), n = 0, 1, …, be the orthogonal polynomials with respect to a given density dμ(x). Furthermore, let dν(x) be a density which arises from dμ(x) by multiplication by a rational function in x. We prove a formula that expresses the Hankel determinants of moments of dν(x) in terms of a determinant involving the orthogonal polynomials p_{n}(x) and associated functions q_{n}(x) = ∫ p_{n}(u) dμ(u)/(x−u). Uvarov’s formula for the orthogonal polynomials with respect to dν(x) is a corollary of our theorem. Our result generalises a Hankel determinant formula for the case where the rational function is a polynomial that existed somehow hidden in the folklore of the theory of orthogonal polynomials but has been stated explicitly only relatively recently (see [arχiv:2101.04225]). Our theorem can be interpreted in a two-fold way: analytically or in the sense of formal series. We apply our theorem to derive several curious Hankel determinant evaluations, and to show linear recursiveness for a certain class of such Hankel determinants.

1. Introduction

Recently, in [9] this author discovered a formula that expresses the Hankel determinant of linear combinations of moments of orthogonal polynomials in terms of a determinant involving these orthogonal polynomials. A literature search revealed that this formula existed in a hidden form behind a theorem (cf. [11, Theorem 2.5] or [7, Theorem 2.7.1]) that is commonly attributed to Christoffel [3] (although he had only proved it in a very special case); only recently it had been stated explicitly, by Lascoux in [10, Prop. 8.4.1] (although incorrectly), by Elouafi [5, Theorem 1] (however with an incomplete proof), and by Brézin and Hikami [2]. Four fundamentally different proofs are given in [9]: one due to this author, one following Lascoux’s arguments, one completing Elouafi’s arguments, and one presenting the random matrix-inspired proof of Brézin and Hikami.

The purpose of this article is to present and prove a generalisation of the aforementioned formula that is inspired by Uvarov’s formula [12, 13] for the orthogonal polynomials with respect to a density that is related to another given density by the multiplication by a rational function, see Theorem 1 below.

2020 Mathematics Subject Classification. Primary 33C45; Secondary 05A10 05A19 11C20 15A15 42C05.

Key words and phrases. Hankel determinants, moments of orthogonal polynomials, Dodgson condensation, Chebyshev polynomials, Catalan numbers.

†Research partially supported by the Austrian Science Foundation FWF (grant S50-N15) in the framework of the Special Research Program “Algorithmic and Enumerative Combinatorics”.

Let (p_{n}(x))_{n≥0} be a sequence of monic polynomials over a field K of characteristic zero^{1} with deg p_{n}(x) = n, and assume that they are orthogonal with respect to the linear functional L, i.e., they satisfy L(p_{m}(x)p_{n}(x)) = ω_{n}δ_{m,n} with ω_{n} ≠ 0 for all n, where δ_{m,n} is the Kronecker delta. Furthermore, we write μ_{n} for the n-th moment L(x^{n}) of the functional L. For convenience (and in order to keep the usual analytic meaning in mind), we shall write ∫ f(u) dμ(u) instead of L(f(x)). This can be either read in a purely formal way, or the analyst may think of it as a concrete integral with respect to the measure given by the density dμ(u).

For the statement of our theorem, we need the “functions”^{2}

q_{n}(y) = ∫ \frac{p_{n}(u)}{y − u} dμ(u).  (1.1)

These can be understood in the “ordinary” analytic sense if μ(u) is a concrete measure, or, alternatively, these can be understood in the sense of formal power series in 1/y, see Lemma 6, Equation (2.7).

Here is the main result of this article.

Theorem 1. Let k, m and n be non-negative integers and x_{1}, x_{2}, …, x_{m} and y_{1}, y_{2}, …, y_{k} be variables. Then, with the above notations, for n ≥ k we have

\begin{matrix}det_{0≤i,j≤n−1} & (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) \\ det_{0≤i,j≤n−k−1} & (μ_{i+j}) \end{matrix}

= (−1)^{n(m−k)+km} \frac{det M_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k})}{( ∏_{1≤i<j≤m}(x_{j} − x_{i}))( ∏_{1≤i<j≤k}(y_{i} − y_{j}))}, (1.2)

where

M_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k}) = (\begin{matrix}p_{n−k}(x_{1}) & & & p_{n−k+1}(x_{1}) & & … & & & p_{n+m−1}(x_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ p_{n−k}(x_{m}) & & & p_{n−k+1}(x_{m}) & & … & & & p_{n+m−1}(x_{m}) \\ q_{n−k}(y_{1}) & & & q_{n−k+1}(y_{1}) & & … & & & q_{n+m−1}(y_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ q_{n−k}(y_{k}) & & & q_{n−k+1}(y_{k}) & & … & & & q_{n+m−1}(y_{k}) \end{matrix}) .

the second kind”.

1For the analyst, (usually) this field is the field of real numbers, and a further restriction is that the linear functional L is defined by a measure with non-negative density. However, the formulae in this paper do not need these restrictions and are valid in this wider context of “formal orthogonality”.

2We put “functions” in quotation marks since they are functions in the literal sense only if we “put on the analytic glasses” and regard dμ(u) as the density of a concrete measure, but they are not if we read this in the formal sense. In [7, Sec. 3.6], up to normalisation, the q_{n}(y)’s are called “functions of

If n < k, then

det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u))

= (−1)^{n(m−k)+km} \frac{det N_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k})}{( ∏_{1≤i<j≤m}(x_{j} − x_{i}))( ∏_{1≤i<j≤k}(y_{i} − y_{j}))}, (1.3)

where

N_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k})

= (\begin{matrix}0 & & … & & & 0 & & & 0 & & & 0 & & & p_{0}(x_{1}) & & p_{1}(x_{1}) & & … & & p_{n+m−1}(x_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 0 & & … & & & 0 & & & 0 & & & 0 & & & p_{0}(x_{m}) & & p_{1}(x_{m}) & & … & & p_{n+m−1}(x_{m}) \\ y_{1}^{k−n−1} & & … & & & y_{1}^{2} & & & y_{1} & & & 1 & & & q_{0}(y_{1}) & & q_{1}(y_{1}) & & … & & q_{n+m−1}(y_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ y_{k}^{k−n−1} & & … & & & y_{k}^{2} & & & y_{k} & & & 1 & & & q_{0}(y_{k}) & & q_{1}(y_{k}) & & … & & q_{n+m−1}(y_{k}) \end{matrix}) .

Here, determinants of empty matrices and empty products are understood to equal 1.

Remarks. (1) The numerator determinant on the left-hand sides of (1.2) and (1.3) is the Hankel determinant det_{0≤i,j≤n−1}(ρ_{i+j}), where the ρ_{s}’s are the moments of the linear functional

p(x) ↦ ∫ p(u)\frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u).

(2) The theory of orthogonal polynomials guarantees that in our setting (namely due to the condition ω_{n} ≠ 0 in the orthogonality) the Hankel determinant of moments in the denominator on the left-hand side of (1.2) is non-zero.

(3) The main theorem in [9] is equivalent to the special case of Theorem 1 where k = 0. It is worth stating this special case separately.

Corollary 2. Let m and n be non-negative integers and x_{1}, x_{2}, …, x_{m} be variables. Then we have

\begin{matrix}det_{0≤i,j≤n−1} & (∫ u^{i+j}∏_{ℓ=1}^{m}(u − x_{ℓ}) dμ(u)) \\ det_{0≤i,j≤n−1} & (μ_{i+j}) \end{matrix} = (−1)^{nm} \frac{det_{1≤i,j≤m} (p_{n+j−1}(x_{i}))}{∏_{1≤i<j≤m}(x_{j} − x_{i})}.  (1.4)

(4) Similarly, it is worth stating the special case of Theorem 1 where m = 0 separately.

Corollary 3. Let k and n be non-negative integers and y_{1}, y_{2}, …, y_{k} be variables. Then, for n ≥ k we have

\begin{matrix}det_{0≤i,j≤n−1} & (∫ \frac{u^{i+j} dμ(u)}{∏_{ℓ=1}^{k}(u − y_{ℓ})}) \\ det_{0≤i,j≤n−k−1} & (μ_{i+j}) \end{matrix} = (−1)^{nk} \frac{det_{1≤i,j≤k} (q_{n−k+j−1}(y_{i}))}{∏_{1≤i<j≤k}(y_{i} − y_{j})}.  (1.5)

If n < k, then

det_{0≤i,j≤n−1} (∫ \frac{u^{i+j} dμ(u)}{∏_{ℓ=1}^{k}(u − y_{ℓ})}) = (−1)^{nk} \frac{det N_{k,n}(y_{1}, …, y_{k})}{∏_{1≤i<j≤k}(y_{i} − y_{j})},  (1.6)

where

N_{k,n}(y_{1}, …, y_{k}) = (\begin{matrix}y_{1}^{k−n−1} & & … & & & y_{1}^{2} & & y_{1} & & & 1 & & q_{0}(y_{1}) & & q_{1}(y_{1}) & & … & & & q_{n−1}(y_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ y_{k}^{k−n−1} & & … & & & y_{k}^{2} & & y_{k} & & & 1 & & q_{0}(y_{k}) & & q_{1}(y_{k}) & & … & & & q_{n−1}(y_{k}) \end{matrix}) .

(5) Let dμ(u) be a given density. We will explain in Section 6 how Theorem 1 is related to Uvarov’s formula [12, 13] for the orthogonal polynomials with respect to the linear functional defined by the density

\frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u).

(6) In [9], four proofs of the special case of Theorem 1 where k = 0 — that is, of Corollary 2 — are given, one using the method of condensation, one using classical results from the theory of orthogonal polynomials, one using a vanishing argument, and one using a random matrix-inspired argument. It is interesting to note that neither the second nor the third proof seem to extend to a proof of Theorem 1, only the condensation argument does. This is indeed the argument that we apply here in Section 4.

The rest of this article is organised as follows. In the next section, we review some classical facts from the theory of orthogonal polynomials that will be relevant for the proof of Theorem 1. Furthermore, in Lemma 6, we provide some information on the “functions” q_{n}(y) that are so central in Theorem 1. Our proof of Theorem 1 requires several determinant identities. These are presented in Section 3, among which Jacobi’s condensation formula (see Lemma 8). The actual proof of Theorem 1 is then the subject of Section 4. In Proposition 13 in Section 5 it is explained how to “read” Theorem 1 in cases where two (or more) of the x_{i}’s or the y_{i}’s are equal. We come back to the title of this article in Section 6 by outlining how Theorem 1 relates to Uvarov’s formula [12, 13]. By considering the special case of Theorem 1 where dμ(u) = \sqrt{1 − u^{2}} du — corresponding to Chebyshev polynomials of the second kind —, we derive evaluations of several curious Hankel determinants in Section 7. Finally, in Section 8, we prove that, if the coefficients of the three-term recurrence (2.1) satisfied by the orthogonal polynomials p_{n}(x) are constants, the (scaled) Hankel determinants of moments that appear on the left-hand side of (1.2) satisfy a linear recurrence with constant coefficients.

2. Preliminaries on orthogonal polynomials

In this section we survey classical facts about orthogonal polynomials that we shall need in the sequel. We also prove some properties of the “functions” q_{n}(y).

As in the introduction, let (p_{n}(x))_{n≥0} be a sequence of monic polynomials over a field K of characteristic zero that is orthogonal with respect to the linear function L given by

L : p(x) ↦ ∫ p(u) dμ(u),

where this may be read formally, or — when we want to interpret this analytically — where dμ(u) is some given density. Such a sequence of orthogonal polynomials exists if and only if all Hankel determinants det_{0≤i,j≤n−1}(μ_{i+j}) of moments μ_{s} = ∫ u^{s} dμ(u) do not vanish. For explicit formulae for the orthogonal polynomials p_{n}(x) in terms of the moments see Lemmas 4 and 5 below.

By Favard’s theorem (see e.g. [8, Theorems 11–13]), the sequence (p_{n}(x))_{n≥0} is orthogonal if and only if it satisfies a three-term recurrence

p_{n}(x) = (x − s_{n−1})p_{n−1}(x) − t_{n−2}p_{n−2}(x), for n ≥ 1,  (2.1)

with initial values p_{−1}(x) = 0 and p_{0}(x) = 1, for some sequences (s_{n})_{n≥0} and (t_{n})_{n≥0} of elements of K with t_{n} ≠ 0 for all n.

The t_{n}’s are connected with the Hankel determinants of moments by the formula (see e.g. [14, Ch. IV, Cor. 6])

det_{0≤i,j≤n−1} (μ_{i+j}) = ∏_{i=0}^{n−1} t_{i}^{n−i−1}.  (2.2)

Conversely, we may use this formula to express t_{n−1} in terms of the Hankel determinants of moments. For convenience, we introduce the abbreviation

H(n) := det_{0≤i,j≤n−1} (μ_{i+j}) .  (2.3)

Then, from (2.2) we infer

t_{n−1} = \frac{H(n + 1)/H(n)}{H(n)/H(n − 1)}.  (2.4)

Next we quote two formulae that express orthogonal polynomials in terms of their associated moments. The first can be found in [11, Eq. (2.2.6)] or [7, Eq. (2.1.11)], and the second in [11, Eq. (2.2.9)]) or [7, Eq. (2.1.10)].

Lemma 4. Let M be a linear functional on polynomials in x with moments ν_{n}, n = 0, 1, …, such that all Hankel determinants det_{0≤i,j≤n}(ν_{i+j}), n = 0, 1, …, are non-zero. Then the determinants

\frac{1}{det_{0≤i,j≤n−1} (ν_{i+j})} det (\begin{matrix}ν_{0} & & ν_{1} & & & ν_{2} & & & … & & & ν_{n} \\ ν_{1} & & ν_{2} & & & ν_{3} & & & … & & & ν_{n+1} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ ν_{n−1} & & ν_{n} & & & ν_{n+1} & & & … & & & ν_{2n−1} \\ 1 & & x & & & x^{2} & & & … & & & x^{n} \end{matrix})

are a sequence of monic orthogonal polynomials with respect to M.

Lemma 5. Let M be a linear functional on polynomials in x with moments ν_{n}, n = 0, 1, …, such that all Hankel determinants det_{0≤i,j≤n}(ν_{i+j}), n = 0, 1, …, are non-zero. Then the determinants

det_{0≤i,j≤n−1} (ν_{i+j+1} − ν_{i+j}x)  (2.5)

are a sequence of orthogonal polynomials with respect to M.

We now turn to the “functions” q_{n}(y) defined in (1.1). They satisfy the same threeterm recurrence relation as the original orthogonal polynomials p_{n}(x). Moreover, it follows from Lemma 4 that q_{n}(y) can be seen as (formal) power series in 1/y of degree −n − 1.^{3}

Lemma 6. We have

q_{n}(y) = (y − s_{n−1})q_{n−1}(y) − t_{n−2}q_{n−2}(y), for n ≥ 2,  (2.6)

with initial values q_{0}(y) = ∫ \frac{dμ(u)}{y−u} and q_{1}(y) = (y − s_{0}) ∫ \frac{dμ(u)}{y−u} − μ_{0}, with the sequences (s_{n})_{n≥0} and (t_{n})_{n≥0} of elements of K that feature in the three-term recurrence (2.1) for the underlying orthogonal polynomials p_{n}(x).

Moreover, for all non-negative integers n, as a formal power series in 1/y the “function” q_{n}(y) starts as

q_{n}(y) = \frac{H(n + 1)}{H(n)}y^{−n−1} + O (y^{−n−2}) ,  (2.7)

where H(n) and H(n + 1) is the short notation introduced in (2.3).

Proof. Let n ≥ 2. From (2.1), we get

∫ \frac{p_{n}(u)}{y − u} dμ(u) = ∫ \frac{(u − s_{n−1})p_{n−1}(u)}{y − u} dμ(u) − t_{n−2} ∫ \frac{p_{n−2}(u)}{y − u} dμ(u).  (2.8)

The first term on the right-hand side can be simplified as follows:

∫ \frac{(u − s_{n−1})p_{n−1}(u)}{y − u} dμ(u) = ∫ \frac{(y − s_{n−1})p_{n−1}(u)}{y − u} dμ(u) − ∫ \frac{(y − u)p_{n−1}(u)}{y − u} dμ(u)
= (y − s_{n−1})q_{n−1}(y).

If this is used in (2.8) together with the definition (1.1) of q_{n}(y), the recurrence (2.6) results immediately. The initial values for q_{0}(y) and q_{1}(y) are straightforward to derive from p_{0}(x) = 1 and p_{1}(x) = x − s_{0}.

In order to show the second assertion, we note that, by definition of q_{n}(y), we have

q_{n}(y) = ∫ \frac{p_{n}(u)}{y − u} dμ(u) = ∑_{i=0}^{∞} ∫ p_{n}(u)u^{i}y^{−i−1} dμ(u).

Because of the orthogonality of p_{n}(u) with respect to the density dμ(u), all terms of the above sum with i < n vanish. Thus,

q_{n}(y) = ∑_{i=n}^{∞} ∫ p_{n}(u)u^{i}y^{−i−1} dμ(u) = y^{−n−1} ∫ p_{n}(u)u^{n} dμ(u) + O (y^{−n−2}) . (2.9)

3By “degree” of a formal power series f(y) in 1/y we mean the maximal exponent e such that y^{e} appears in f(y) with non-zero coefficient. We warn the reader that the recurrence (2.6) is deceiving in regard of the degree of q_{n}(y): it seems to suggest that — similar to the situation for the underlying orthogonal polynomials p_{n}(x) — the degree would rise by 1 when going from q_{n−1}(y) to q_{n}(y). However, as (2.7) shows, on the contrary the degree drops by 1, caused by a cancellation of leading terms in (2.6).

Now we use the formula of Lemma 4 with ν_{s} = μ_{s} for all s, to obtain

∫ p_{n}(u)u^{n} dμ(u) = \frac{1}{H(n)} ∫ det (\begin{matrix}μ_{0} & & μ_{1} & & μ_{2} & & & … & & μ_{n} \\ μ_{1} & & μ_{2} & & μ_{3} & & & … & & μ_{n+1} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ μ_{n−1} & & μ_{n} & & μ_{n+1} & & & … & & μ_{2n−1} \\ 1 & & u & & u^{2} & & & … & & u^{n} \end{matrix}) u^{n} dμ(u)

= \frac{1}{H(n)} (\begin{matrix}μ_{0} & & μ_{1} & & μ_{2} & & & … & & μ_{n} \\ μ_{1} & & μ_{2} & & μ_{3} & & & … & & μ_{n+1} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ μ_{n−1} & & μ_{n} & & μ_{n+1} & & & … & & μ_{2n−1} \\ μ_{n} & & μ_{n+1} & & μ_{n+2} & & & … & & μ_{2n} \end{matrix}) = \frac{H(n + 1)}{H(n)}.

If this is substituted back in (2.9), the assertion of the lemma follows immediately. □

3. Auxiliary determinant identities

The purpose of this section is to collect three determinant formulae that will turn out to be crucial in our proof of Theorem 1.

The proof method of our proof in Section 4 is the method of condensation (frequently referred to as “Dodgson condensation”; see [8, Sec. 2.3]). This method provides inductive proofs that are based on a determinant identity due to Jacobi, which we recall in the following proposition.

Proposition 7. Let A be an N × N matrix. Denote the submatrix of A in which rows i_{1}, i_{2}, …, i_{k} and columns j_{1}, j_{2}, …, j_{k} are omitted by A_{i_{1},i_{2},...,i_{k}}^{j_{1},j_{2},...,j_{k}}. Then we have

det A · det A_{i_{1},i_{2}}^{j_{1},j_{2}} = det A_{i_{1}}^{j_{1}} · det A_{i_{2}}^{j_{2}} − det A_{i_{1}}^{j_{2}} · det A_{i_{2}}^{j_{1}}  (3.1)

for all integers i_{1}, i_{2}, j_{1}, j_{2} with 1 ≤ i_{1} < i_{2} ≤ N and 1 ≤ j_{1} < j_{2} ≤ N.

We need two further determinantal formulae, which involve Hankel determinants of linear combinations of sequence elements.

The following is [9, Lemma 3 with n replaced by n − 1].

Lemma 8. Let (c_{n})_{n≥0} be a given sequence, and α and β be variables. Then, for all positive integers n, we have

(β − α) det_{0≤i,j≤n−2} (αβc_{i+j} + (α + β)c_{i+j+1} + c_{i+j+2}) det_{0≤i,j≤n−1} (c_{i+j})

= det_{0≤i,j≤n−2} (αc_{i+j} + c_{i+j+1}) det_{0≤i,j≤n−1} (βc_{i+j} + c_{i+j+1})

− det_{0≤i,j≤n−2} (βc_{i+j} + c_{i+j+1}) det_{0≤i,j≤n−1} (αc_{i+j} + c_{i+j+1}). (3.2)

We require another, similarly looking determinant identity.

Lemma 9. Let (c_{n})_{n≥0} be a given sequence, and α and β be variables. Then, for all positive integers n, we have

det_{0≤i,j≤n−1} (αc_{i+j} + c_{i+j+1}) det_{0≤i,j≤n−1} (βc_{i+j} + c_{i+j+1})

= − det_{0≤i,j≤n} (c_{i+j}) det_{0≤i,j≤n−2} (αβc_{i+j} + (α + β)c_{i+j+1} + c_{i+j+2})

+ det_{0≤i,j≤n−1} (c_{i+j}) det_{0≤i,j≤n−1} (αβc_{i+j} + (α + β)c_{i+j+1} + c_{i+j+2}). (3.3)

Proof. In principle, it should be possible to prove this directly in the style of the proof of Lemma 8 given in [9], that is, by extracting the coefficient of α^{s}β^{t} on both sides of (3.3), and by then reducing everything to some known determinant identity. Embarrassingly, I failed to carry this through. Therefore, instead I take recourse to a dirty trick that is based on the fact that Theorem 1 had been proved earlier for the special case in which k = 0 (cf. [9, Theorem 1] and Proposition 11 below). Namely, we interpret the sequence (c_{n})_{n≥0} as the sequence of moments of some linear functional, and we let (r_{n}(x))_{n≥0} be the corresponding sequence of monic orthogonal polynomials with respect to that functional. As was said earlier, for the orthogonal polynomials to exist we must assume that all Hankel determinants det_{0≤i,j≤n−1}(c_{i+j}) are non-zero, which we do for the moment. Since both sides of (3.3) are polynomials in the c_{i}’s, α and β, this restriction can be removed in the end.

Now, in the above setting, by Theorem 1 with k = 0, m = 1, and x_{1} = −α, we have

det_{0≤i,j≤n−1} (αc_{i+j} + c_{i+j+1}) = H(n)r_{n}(−α),

and, by Theorem 1 with k = 0, m = 2, x_{1} = −α and x_{2} = −β, we have

det_{0≤i,j≤n−1} (αβc_{i+j} + (α + β)c_{i+j+1} + c_{i+j+2}) = H(n)\frac{r_{n}(−α)r_{n+1}(−β) − r_{n+1}(−α)r_{n}(−β)}{α − β}.

Consequently, Equation (3.3) turns out to be equivalent with

(H̃(n))^{2}r_{n}(−α)r_{n}(−β) = H̃(n + 1)H̃(n − 1)\frac{r_{n−1}(−α)r_{n}(−β) − r_{n}(−α)r_{n−1}(−β)}{α − β}

+ (H̃(n))^{2} \frac{r_{n}(−α)r_{n+1}(−β) − r_{n+1}(−α)r_{n}(−β)}{α − β}.

Here we used the short notation H̃(n) = det_{0≤i,j≤n−1}(c_{i+j}). After cancellation of common factors using (2.2) (with H̃(n) in place of H(n) and the appropriate sequence (t̃_{n})_{n≥0} instead of (t_{n})_{n≥0}), this becomes

(α − β)r_{n}(−α)r_{n}(−β) = t̃_{n−1}(r_{n−1}(−α)r_{n}(−β) − r_{n}(−α)r_{n−1}(−β))

+ (r_{n}(−α)r_{n+1}(−β) − r_{n+1}(−α)r_{n}(−β)).

Indeed, this is trivially true because of (2.1) with n replaced by n + 1, p_{n}(x) replaced by r_{n}(x), s_{n} replaced by s̃_{n}, t_{n} replaced by t̃_{n}, and x = −β, respectively x = −α. □

4. Proof of Theorem 1

Here we prove Theorem 1. As announced, we shall apply the method of condensation (see Proposition 7) in order to set up an inductive proof. Propositions 11 and 12 will provide the start of the induction. In order to “keep the induction running”, we need a non-vanishing result that we present first.

Lemma 10. Under the assumption that the Hankel determinants H(n) in (2.3) do not vanish, the numerator determinant on the left-hand sides of (1.2) and (1.3),

det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) ,  (4.1)

does not vanish identically.

Proof. Adopting the viewpoint of formal series in the variables x_{1}, x_{2}, …, x_{m} and 1/y_{1}, 1/y_{2}, …, 1/y_{k}, the highest degree term in (4.1) is

det_{0≤i,j≤n−1} ((−1)^{m−k}μ_{i+j} \frac{∏_{ℓ=1}^{m} x_{ℓ}}{∏_{ℓ=1}^{k} y_{ℓ}} ) = (−1)^{n(m−k)} \frac{∏_{j=1}^{m} x_{j}^{n}}{∏_{j=1}^{k} y_{j}^{n}} det_{0≤i,j≤n−1} (μ_{i+j})

= (−1)^{n(m−k)}H(n)\frac{∏_{j=1}^{m} x_{j}^{n}}{∏_{j=1}^{k} y_{j}^{n}} ,

which is non-zero by our assumption of non-vanishing of H(n).

If we adopt the analytic point of view, then one would multiply the determinant (4.1) by ∏_{j=1}^{m} x_{j}^{−n}/∏_{j=1}^{k} y_{j}^{−n} and then compute the limit^{4} as x_{i} → ∞ and y_{j} → ∞ for all i and j. The result would be the determinant

det_{0≤i,j≤n−1} ((−1)^{m−k}μ_{i+j}) = (−1)^{n(m−k)}H(n),

with the same conclusion. □

Proposition 11. Theorem 1 holds for k = 0.

This is the main theorem (namely Theorem 1) in [9], for which four fundamentally different proofs are provided there.

Proposition 12. Theorem 1 holds for m = 0.

Proof. We prove the claim by induction on k.

For the start of the induction we need the validity of (1.2) for k = m = 0 — this is obvious since both left-hand and right-hand side equal 1 in that case —, of (1.3) for m = 0 and k = 1 — also this is easy to see since the only case to consider is n = 0, which makes the left-hand side reduce to 1, and also the right-hand side, due to the evaluation of the Vandermonde determinant —, and of (1.2) for m = 0 and k = 1. The latter needs an argument, which we provide next.

For m = 0 and k = 1, the numerator of the left-hand side of (1.2) reads

det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{dμ(u)}{u − y_{1}} ) .  (4.2)

We have

∫ u^{i+j} \frac{dμ(u)}{u − y_{1}} = ∫ u^{i+j−1} \frac{u − y_{1} + y_{1}}{u − y_{1}} dμ(u) = μ_{i+j−1} + y_{1} ∫ u^{i+j−1} \frac{dμ(u)}{u − y_{1}} .

4This limit is unproblematic if dμ(u) is a measure with finite support.

We use this relation in the last row, that is, for i = n − 1. Subsequently, we subtract the (n − 2)-nd row multiplied by y_{1} from the last row (the (n − 1)-st row). Thereby the entry in the j-th column of the last row becomes μ_{n+j−2}. We repeat this operation with the (n − 2)-nd and the (n − 3)-rd row, with the (n − 3)-rd and the (n − 4)-th row, …, and with the first row and the 0-th row. As a result, we have converted the determinant in (4.2) into the determinant

det (\begin{matrix}∫ u^{j} \frac{dμ(u)}{u − y_{1}} & for i = 0, 0 ≤ j ≤ n − 1 \\ μ_{i+j−1} & for 1 ≤ i ≤ n − 1, 0 ≤ j ≤ n − 1 \end{matrix})

= ∫ det (\begin{matrix}1 & & u & & u^{2} & & & … & & u^{n−1} \\ μ_{0} & & μ_{1} & & μ_{2} & & & … & & μ_{n−1} \\ μ_{1} & & μ_{2} & & μ_{3} & & & … & & μ_{n} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ μ_{n−2} & & μ_{n−1} & & μ_{n+1} & & & … & & μ_{2n−3} \end{matrix}) \frac{dμ(u)}{u − y_{1}}

= (−1)^{n−1} det_{0≤i,j≤n−2} (μ_{i+j}) ∫ \frac{p_{n−1}(u)}{u − y_{1}} dμ(u)

= (−1)^{n} det_{0≤i,j≤n−2} (μ_{i+j}) q_{n−1}(y_{1}),

where the next-to-last line is due to Lemma 4. This confirms (1.2) for m = 0 and k = 1.

We now turn to the induction step. We have to distinguish between two cases, depending on whether n ≥ k or not. For convenience, we rewrite (1.2) (with m = 0) in the form

(−1)^{nk} \frac{∏_{1≤i<j≤k}(y_{i} − y_{j})}{det_{0≤i,j≤n−k−1} (μ_{i+j})} det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=1}^{k}(u − y_{ℓ})}) = det M_{k,0,n}(y_{1}, …, y_{k}),

(4.3)

and we apply a similar rewriting to (1.3), leading to (4.3) with denominator omitted on the left-hand side and M replaced by N on the right-hand side.

Let k ≥ 2 and assume that (4.3), and also the analogue corresponding to (1.3), is true for “smaller k”.

We are going to use the condensation formula of Proposition 7. For n ≥ k, the identity (3.1) with N = k, A = M_{k,0,n}(y_{1}, …, y_{k}), i_{1} = j_{1} = 1 and i_{k} = j_{k} = k gives

det M_{k,0,n}(y_{1}, …, y_{k}) · det M_{k−2,0,n−1}(y_{2}, …, y_{k−1})

= det M_{k−1,0,n}(y_{2}, …, y_{k}) · det M_{k−1,0,n−1}(y_{1}, …, y_{k−1}) − det M_{k−1,0,n−1}(y_{2}, …, y_{k}) · det M_{k−1,0,n}(y_{1}, …, y_{k−1}). (4.4)

For n < k, with the same choices of N, i_{1}, i_{2}, j_{1}, j_{2}, but with A = N_{k,0,n}(y_{1}, …, y_{k}), we obtain the same identity, but with N instead of M. A little detail is that, if n = k −1, we encounter the terms N_{k−2,0,n−1}(…) = N_{k−2,0,k−2}(…) and N_{k−1,0,n}(…) = N_{k−1,0,k−1}(…) in (4.4) (with M replaced by N). It can be seen by inspection that N_{k−2,0,k−2}(…) = M_{k−2,0,k−2}(…) and N_{k−1,0,k−1}(…) = M_{k−1,0,k−1}(…). This is the interpretation that we give these terms in that special case.

The identity (4.4) can be seen as a recurrence formula for det M_{k,0,n}(y_{1}, …, y_{k}), as one can use it to express det M_{k,0,n}(y_{1}, …, y_{k}) in terms of expressions of the form det M_{l,0,s}(y_{a}, …, y_{b}) with l smaller than k (and similarly with M replaced by N), provided the determinants det M_{k−2,0,n−1}(y_{2}, …, y_{k−1}) and det N_{k−2,0,n−1}(y_{2}, …, y_{k−1}) are all non-zero. (We shall address the latter point later.) Hence, for carrying out the induction step it suffices to verify that the left-hand side of (4.3) satisfies the same recurrence. Consequently, we substitute this left-hand side in (4.4). After cancellation of factors that are common to both sides, we arrive at

(y_{k} − y_{1}) det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=1}^{k}(u − y_{ℓ})}) · det_{0≤i,j≤n−2} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=2}^{k−1}(u − y_{ℓ})})

= det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=2}^{k}(u − y_{ℓ})}) · det_{0≤i,j≤n−2} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=1}^{k−1}(u − y_{ℓ})})

− det_{0≤i,j≤n−2} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=2}^{k}(u − y_{ℓ})}) · det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=1}^{k−1}(u − y_{ℓ})}) .

This is the special case of Lemma 8 where

c_{n} = ∫ u^{i+j} \frac{dμ(u)}{∏_{ℓ=1}^{k}(u − y_{ℓ})},

α = −y_{k} and β = −y_{1}. Since Lemma 10 with m = 0 says that all these determinants do not vanish identically, this establishes the induction step and proves (4.3) and thus the proposition. □

We are now ready to prove Theorem 1.

Proof of Theorem 1. We apply induction with respect to k+m. As start of the induction we use Propositions 11 and 12. In other words, we know that Theorem 1 holds for k = 0 and for m = 0.

In preparation of the induction step, we again rewrite (1.2),

(−1)^{n(m−k)+km} \frac{( ∏_{1≤i<j≤m}(x_{j} − x_{i}))( ∏_{1≤i<j≤k}(y_{i} − y_{j}))}{det_{0≤i,j≤n−k−1} (μ_{i+j})}

× det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u))

= det M_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k}), (4.5)

and similarly (1.3), leading to (4.5) with denominator omitted on the left-hand side and M replaced by N on the right-hand side.

Let now k and m be positive integers, and assume that (4.5), and also its analogue corresponding to (1.3), hold for “smaller k + m”.

Here again, we are going to use the condensation formula of Proposition 7. For n ≥ k, the identity (3.1) with N = k, A = M_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k}), i_{1} = j_{1} = 1 and i_{k} = j_{k} = k + m gives

det M_{k,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k}) · det M_{k−1,m−1,n}(x_{2}, …, x_{m}, y_{1}, …, y_{k−1})

= det M_{k,m−1,n+1}(x_{2}, …, x_{m}, y_{1}, …, y_{k}) · det M_{k−1,m,n−1}(x_{1}, …, x_{m}, y_{1}, …, y_{k−1}) − det M_{k,m−1,n}(x_{2}, …, x_{m}, y_{1}, …, y_{k}) · det M_{k−1,m,n}(x_{1}, …, x_{m}, y_{1}, …, y_{k−1}). (4.6)

For n < k, with the same choices of N, i_{1}, i_{2}, j_{1}, j_{2}, but with A = N_{k,m,n}(y_{1}, …, y_{k}), we obtain the same identity, but with N instead of M. Again we have to take notice of the little detail that, if n = k − 1, we encounter terms such as N_{k,m−1,n+1}(…) = N_{k,m−1,k}(…) and N_{k−1,m,n}(…) = N_{k−1,m,k−1}(…) in (4.6) (with M replaced by N). Here also they have to be interpreted as the corresponding “M-terms”.

In order to accomplish the induction step, we have to prove that the left-hand side of (4.5) satisfies the same relation. Consequently, we substitute this left-hand side in (4.6). After cancellation of factors that are common to both sides, we arrive at

det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) · det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k−1}(u − y_{ℓ})} dμ(u))

= − det_{0≤i,j≤n} (∫ u^{i+j} \frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) · det_{0≤i,j≤n−2} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k−1}(u − y_{ℓ})} dμ(u))

+ det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) · det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k−1}(u − y_{ℓ})} dμ(u)) .

This is the special case of Lemma 9 where

c_{n} = ∫ u^{i+j} \frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u),

α = −x_{1} and β = −y_{k}. Since Lemma 10 says that all these determinants do not vanish identically, this establishes the induction step and proves (4.5), and thus Theorem 1. □

5. The case of equal parameters

Let

R(x_{1}, x_{2}, …, x_{m}, y_{1}, y_{2}, …, y_{k})

denote the right-hand side of (1.2) if n ≥ k, and the right-hand side of (1.3) if n < k. Since the numerator (regardless whether we are considering (1.2) or (1.3)) is skewsymmetric in the x_{i}’s and skew-symmetric in the y_{i}’s, it is divisible by the Vandermonde products

( ∏_{1≤i<j≤m}(x_{j} − x_{i}))( ∏_{1≤i<j≤k}(y_{i} − y_{j}))

in the denominator. Thus, while in its definition it seems problematic to substitute the same value for two different x_{i}’s or for two different y_{i}’s in R(x_{1}, x_{2}, …, x_{m}, y_{1}, y_{2}, …, y_{k}), this is actually not the case. The proposition below provides an explicit expression for such substitutions of equal parameters.

Proposition 13. Let r, s, k_{1}, k_{2}, …, k_{s}, m_{1}, m_{2}, …, m_{r}, and n be non-negative integers and ξ_{1}, ξ_{2}, …, ξ_{r} and ω_{1}, ω_{2}, …, ω_{s} be variables. We write k for the sum ∑_{i=1}^{s} k_{i}, and m for the sum ∑_{i=1}^{r} m_{i}.

Then

R(ξ_{1}, …, ξ_{1}, ξ_{2}, …, ξ_{2}, …, ξ_{r}, …, ξ_{r}, ω_{1}, …, ω_{1}, ω_{2}, …, ω_{2}, …, ω_{s}, …, ω_{s})

= (−1)^{n(m−k)+km} \frac{det M_{k_{1},...,k_{s},m_{1},...,m_{r},n}(ξ_{1}, …, ξ_{r}, ω_{1}, …, ω_{s})}{( ∏_{1≤i<j≤r}(x_{j} − x_{i})^{m_{i}m_{j}})( ∏_{1≤i<j≤s}(y_{i} − y_{j})^{k_{i}k_{j}})},

where ξ_{i} is repeated m_{i} times and ω_{i} is repeated k_{i} times in the argument of R on the left-hand side. The matrix in the numerator on the right-hand side is defined by

M_{k_{1},...,k_{s},m_{1},...,m_{r},n}(ξ_{1}, …, ξ_{r}, ω_{1}, …, ω_{s}) = (P_{m_{1},k,n}(ξ_{1}) ; ⋮ ; P_{m_{r},k,n}(ξ_{r}) ; Q_{k_{1},k,n}(ω_{1}) ; ⋮ ; Q_{k_{s},k,n}(ω_{s})) ,

with

P_{a,k.n}(ξ) = (\frac{p_{n−k+j−1}^{(i−1)}(ξ)}{(i − 1)!})_{1≤i≤a, 1≤j≤k+m}

and

Q_{a,k,n}(ω) = (\frac{q_{n−k+j−1}^{(i−1)}(ω)}{(i − 1)!})_{1≤i≤a, 1≤j≤k+m} .

If b < 0 the polynomial p_{b}(ξ) has to be interpreted as 0, while the function q_{b}(ω) has to be interpreted as ω^{−b−1},

Proof. This can be proved in the same way as [9, Prop. 5]. We leave the details to the reader. □

Remark. Obviously, we have

q_{n}^{(i−1)}(y) = (−1)^{i−1}(i − 1)! ∫ \frac{p_{n}^{(i−1)}(u)}{(y − u)^{i}} dμ(u).

6. Uvarov’s formula

In this section, we show that our identity in Theorem 1 implies Uvarov’s formula for the orthogonal polynomials of a density that arises from a given density by the multiplication of a rational function. The connection between Theorem 1 and Uvarov’s formula is set up by Lemma 5.

Let dμ(u) be the density of a positive measure with infinite support all of whose moments exist. We let dν(u) be the modified density

dν(u) = \frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u),  (6.1)

where x_{2}, x_{3}, …, x_{m} and y_{1}, y_{2}, …, y_{k} are real numbers chosen so that the modification factor

\frac{∏_{ℓ=2}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})}

is positive for all u in the support of the measure μ. In [12, 13] (cf. [7, Theorem 2.7.3]), Uvarov provides a formula for the orthogonal polynomials with respect to the density dν(u) in (6.1), expressed in terms of the orthogonal polynomials with respect to the original density dμ(u).

Theorem 14. Under the above assumptions, for n ≥ k, the orthogonal polynomials with respect to the density dν(u) given in (6.1), as polynomials in x_{1}, can be expressed as

\frac{1}{∏_{ℓ=2}^{m}(x_{1} − x_{ℓ})} det (\begin{matrix}p_{n−k}(x_{1}) & & & p_{n−k+1}(x_{1}) & & … & & & p_{n+m−1}(x_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ p_{n−k}(x_{m}) & & & p_{n−k+1}(x_{m}) & & … & & & p_{n+m−1}(x_{m}) \\ q_{n−k}(y_{1}) & & & q_{n−k+1}(y_{1}) & & … & & & q_{n+m−1}(y_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ q_{n−k}(y_{k}) & & & q_{n−k+1}(y_{k}) & & … & & & q_{n+m−1}(y_{k}) \end{matrix}) ,  (6.2)

while, for n < k, they can be expressed as

\frac{1}{∏_{ℓ=2}^{m}(x_{1} − x_{ℓ})} det (\begin{matrix}0 & & … & & & 0 & & & 0 & & & 0 & & & p_{0}(x_{1}) & & p_{1}(x_{1}) & & … & & p_{n+m−1}(x_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 0 & & … & & & 0 & & & 0 & & & 0 & & & p_{0}(x_{m}) & & p_{1}(x_{m}) & & … & & p_{n+m−1}(x_{m}) \\ y_{1}^{k−n−1} & & … & & & y_{1}^{2} & & & y_{1} & & & 1 & & & q_{0}(y_{1}) & & q_{1}(y_{1}) & & … & & q_{n+m−1}(y_{1}) \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ y_{k}^{k−n−1} & & … & & & y_{k}^{2} & & & y_{k} & & & 1 & & & q_{0}(y_{k}) & & q_{1}(y_{k}) & & … & & q_{n+m−1}(y_{k}) \end{matrix}) ,

(6.3)

with q_{n}(y) the “functions” given in (1.1).

Proof. In Lemma 5 we choose for the moments ν_{i} the moments corresponding to the density in (6.1). Then the determinant (2.5) turns out to be exactly the determinant on the left-hand sides of (1.2) and (1.3). Hence, in view of Lemma 5, it is obvious that the left-hand sides of (1.2) and (1.3) are orthogonal polynomials with respect to the density in (6.1), as polynomials in x_{1}. These right-hand sides are, up to (non-zero) factors that are independent of x_{1}, proportional to the expressions in (6.2) and (6.3), respectively. There is one “catch” though: Lemma 5 says that this is only the case if all Hankel determinants of moments of (6.1) are non-zero. Now, the choice of the x_{i}’s and y_{i}’s is such that (6.1) is the density of a positive measure. It then follows from Heine’s formula (cf. [11, Eq. (2.2.11)] or [7, Cor. 2.1.3]) that all Hankel determinants of moments are non-zero (in fact, positive). □

More generally, our Theorem 1 implies that, whenever the right-hand sides of (1.2) and (1.3) are polynomials in x_{1} of degree n, then they are orthogonal polynomials in x_{1} for the density in (6.1).

7. Applications: some Hankel determinant evaluations

In this final section, we apply Theorem 1 to derive some Hankel determinant evaluations featuring Catalan numbers and central binomial coefficients.

We restrict ourselves to the special case of Theorem 1 where

dμ(u) = \sqrt{1 − \frac{u^{2}}{4}} du,  −2 ≤ u ≤ 2.  (7.1)

The polynomials which are orthogonal with respect to this density are, essentially, the Chebyshev polynomials of the second kind U_{n}(x), given by the generating function

∑_{n≥0} U_{n}(x) z^{n} = \frac{1}{1 − 2xz + z^{2}} .

More precisely, we have

\frac{1}{π} ∫_{−2}^{2} U_{m}(u/2)U_{n}(u/2)\sqrt{1 − \frac{u^{2}}{4}} du = δ_{n,m},

with, again, δ_{n,m} denoting the Kronecker delta. The polynomials U_{n}(x/2) are monic, as can be seen from

∑_{n=0}^{∞} U_{n}(x/2)z^{n} = \frac{1}{1 − xz + z^{2}} .  (7.2)

The non-vanishing moments of the density (7.1) are given by the Catalan numbers C_{n} = \frac{1}{n+1} \binom{2n}{n}. To be precise, we have

\frac{1}{π} ∫_{−2}^{2} u^{n}\sqrt{1 − \frac{u^{2}}{4}} du = \begin{cases}C_{n/2}, & if n is even, \\ 0, & if n is odd, \end{cases} n ≥ 0.

We shall apply Theorem 1 with k = 1 and m = 0, and with k = m = 1. For carrying out the corresponding calculations, we need the well-known evaluation of the Hankel determinant of moments of the density (7.1) (see e.g. [4, Eq. (2.6) in combination with Case (xvi) in Sec. 4])

det_{0≤i,j≤n−1}(C_{(i+j)/2}) = 1  (7.3)

(where C_{s} is defined to be zero if s is not an integer), and the following fundamental integral evaluation:^{5}

\frac{1}{π} ∫_{−1}^{1} \frac{\sqrt{1 − u^{2}}}{u + a} du = \begin{cases}a − \sqrt{a^{2} − 1}, & for a ≥ 1, \\ & for ℜa > 0 and ℑa ≠ 0, \\ & for ℜa = 0 and ℑa > 0, \\ a, & for − 1 ≤ a ≤ 1, \\ a + \sqrt{a^{2} − 1}, & for a ≤ −1, \\ & for ℜa < 0 and ℑa ≠ 0, \\ & for ℜa = 0 and ℑa < 0. \end{cases}  (7.4)

5The substitution u = 2t/(1 + t^{2}) transforms the integral into an integral of a rational function.

Subsequently, a thorough case analysis has to be performed.

For a “test”, we may use this to confirm the orthogonality of the U_{n}(x/2)’s. We have

∑_{n≥0} z^{n} \frac{1}{π} ∫_{−2}^{2} U_{n}(u/2)\sqrt{1 − \frac{u^{2}}{4}} du = \frac{1}{π} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{1 − uz + z^{2}} du

= \frac{2}{π} ∫_{−1}^{1} \frac{\sqrt{1 − u^{2}}}{1 − 2uz + z^{2}} du

= −\frac{1}{zπ} ∫_{−1}^{1} \frac{\sqrt{1 − u^{2}}}{u − \frac{1+z^{2}}{2z}} du.  (7.5)

This calculation is meant as a calculation for (formal) power series in z. Thus, we must think of z as “small”, which implies that \frac{1+z^{2}}{2z} is “large”. Hence, when we apply (7.4) with a = −\frac{1+z^{2}}{2z}, we find ourselves in the third case of the case distinction. This gives

∑_{n≥0} z^{n} \frac{1}{π} ∫_{−2}^{2} U_{n}(u/2)\sqrt{1 − \frac{u^{2}}{4}} du = −\frac{1}{z} (−\frac{1 + z^{2}}{2z} + \sqrt{\frac{(1 + z^{2})^{2}}{4z^{2}} − 1}) = 1, (7.6)

as expected. Another “test” that we may perform is the calculation of moments (of the measure) of the Chebyshev polynomials. We have

∑_{n≥0} z^{n} \frac{1}{π} ∫_{−2}^{2} u^{n}\sqrt{1 − \frac{u^{2}}{4}} du = \frac{1}{π} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{1 − uz} du = \frac{2}{π} ∫_{−1}^{1} \frac{\sqrt{1 − u^{2}}}{1 − 2uz} du

= −\frac{1}{zπ} ∫_{−1}^{1} \frac{\sqrt{1 − u^{2}}}{u − \frac{1}{2z}} du = \frac{1}{2z^{2}} (1 − \sqrt{1 − 4z^{2}})

= ∑_{n≥0} C_{n}z^{2n},  (7.7)

again as expected.

We intend to apply Theorem 1 with k = 1 and y_{1} = −2a. For convenience, let us write ω(a) for the quantity in (7.4). Then, by (7.4) and (7.7), for the moments of the

density \frac{1}{u−y_{1}} \sqrt{1 − \frac{u^{2}}{4}} du we have

∑_{n≥0} z^{n} \frac{1}{π} ∫_{−2}^{2} \frac{u^{n}\sqrt{1 − \frac{u^{2}}{4}}}{u + 2a} du = \frac{1}{π} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{(u + 2a)(1 − uz)} du

= \frac{1}{π(1 + 2az)} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{u + 2a} du + \frac{z}{π(1 + 2az)} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{1 − uz} du

= \frac{ω(a)}{1 + 2az} + \frac{z}{1 + 2az} ∑_{n≥0} C_{n}z^{2n}

= ∑_{n≥0} ω(a)(−2a)^{n}z^{n} + ∑_{n≥0} z^{n} ∑_{k=0}^{⌊(n−1)/2⌋}(−2a)^{n−2k−1}C_{k}.

Hence,

\frac{1}{π} ∫_{−2}^{2} \frac{u^{n}\sqrt{1 − \frac{u^{2}}{4}}}{u + 2a} du = ω(a)(−2a)^{n} + ∑_{k=0}^{⌊(n−1)/2⌋}(−2a)^{n−2k−1}C_{k}.  (7.8)

We next compute the functions q_{n}(−2a) associated with the Chebyshev polynomials U_{n}(x/2) (cf. (1.1)),

∑_{n≥0} z^{n}q_{n}(−2a) = ∑_{n≥0} z^{n} \frac{1}{π} ∫_{−2}^{2} \frac{U_{n}(u/2)\sqrt{1 − \frac{u^{2}}{4}}}{−2a − u} du = −\frac{1}{π} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{(u + 2a)(1 − uz + z^{2})} du

= −\frac{1}{π(1 + 2az + z^{2})} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{u + 2a} du

− \frac{z}{π(1 + 2az + z^{2})} ∫_{−2}^{2} \frac{\sqrt{1 − \frac{u^{2}}{4}}}{1 − uz + z^{2}} du

= −\frac{ω(a) + z}{1 + 2az + z^{2}} = − ∑_{n≥0} (ω(a)U_{n}(−a) + U_{n−1}(−a))z^{n},

where we used (7.4), (7.5)/(7.6), and finally (7.2) to obtain the last line. Hence,

q_{n}(−2a) = −(ω(a)U_{n}(−a) + U_{n−1}(−a)).  (7.9)

If we put these findings together, we get the following result from Theorem 1 with k = 1 and m = 0.

Theorem 15. Let X and a be variables. For all positive integers n, we have

det_{0≤i,j≤n−1} (X(−2a)^{i+j} + ∑_{k=0}^{⌊(i+j−1)/2⌋}(−2a)^{i+j−2k−1}C_{k})

= (−1)^{n−1}(XU_{n−1}(−a) + U_{n−2}(−a)). (7.10)

Proof. Using (7.3), (7.8) and (7.9), we see that Theorem 1 with k = 1, m = 0, y_{1} = −2a,

and dμ(u) = \sqrt{1 − \frac{u^{2}}{4}} du reads

det_{0≤i,j≤n−1} (ω(a)(−2a)^{i+j} + ∑_{k=0}^{⌊(i+j−1)/2⌋}(−2a)^{i+j−2k−1}C_{k})
= (−1)^{n−1}(ω(a)U_{n−1}(−a) + U_{n−2}(−a)).

This holds for all complex numbers a. Now, if we regard the above determinant (formally) as a polynomial in ω(a), then it is not difficult to see that the degree in ω(a) is at most 1. On the other hand, the right-hand side of the above equation is visibly a polynomial of degree 1 in ω(a). If we now choose a = −2 (say), so that a+\sqrt{a^{2} − 1} = −2+\sqrt{3} and 1 are linearly independent over the rational numbers, then we may conclude that ω(a) can be replaced by a variable, X say. This establishes the assertion of the theorem. □

If we choose a = −1 in the above theorem, the sum in the determinant can be simplified. Namely, we have

(−1)2^{n} + ∑_{k=0}^{⌊(n−1)/2⌋} 2^{n−2k−1}C_{k} = \begin{cases}−\binom{n}{n/2}, & if n is even, \\ −\frac{1}{2} \binom{n+1}{(n+1)/2}, & if n is odd, \end{cases}  (7.11)

as is straightforward to verify by an induction on n. Since U_{n}(1) = n + 1 for all n, the choice of X = −1 in Theorem 15 leads to the Hankel determinant evaluation

det_{0≤i,j≤n−1} (2^{−2⌈(i+j)/2⌉}\binom{2 ⌈(i + j)/2⌉}{⌈(i + j)/2⌉}) = 2^{−n(n−1)}.  (7.12)

More generally, if we choose X = −Y − 1, then we get

det_{0≤i,j≤n−1} (Y + 2^{−2⌈(i+j)/2⌉}\binom{2 ⌈(i + j)/2⌉}{⌈(i + j)/2⌉}) = 2^{−n(n−1)}(Y n + 1).  (7.13)

In a similar fashion, Theorem 1 with k = m = 1, x_{1} = b, y_{1} = −2a, and dμ(u) =

\sqrt{1 − \frac{u^{2}}{4}} du implies the following result.

Theorem 16. Let X, a and b be variables. For all positive integers n, we have

det_{0≤i,j≤n−1} (X(−2a)^{i+j+1} + ∑_{k=0}^{⌊(i+j)/2⌋}(−2a)^{i+j−2k}C_{k}

−b(X(−2a)^{i+j} + ∑_{k=0}^{⌊(i+j−1)/2⌋}(−2a)^{i+j−2k−1}C_{k}))

= U_{n−1}(b/2)(XU_{n}(−a) + U_{n−1}(−a)) − U_{n}(b/2)(XU_{n−1}(−a) + U_{n−2}(−a)). (7.14)

Again, it is worth stating the identities that one obtains in the special case where a = −1 explicitly. Namely, in view of the simplification (7.11) and of U_{n}(1) = n + 1, in that case Equation (7.14) reduces to

det_{0≤i,j≤n−1} (Y + 2^{−2⌈(i+j+1)/2⌉}\binom{2 ⌈(i + j + 1)/2⌉}{⌈(i + j + 1)/2⌉}

−b(\frac{1}{2}Y + 2^{−2⌈(i+j)/2⌉−1}\binom{2 ⌈(i + j)/2⌉}{⌈(i + j)/2⌉}))

= (−1)^{n−1}2^{−n^{2}}(U_{n−1}(b/2)(Y (n + 1) + 1) − U_{n}(b/2)(Y n + 1)). (7.15)

Since U_{n}(0) = (−1)^{n/2} if n is even and U_{n}(0) = 0 otherwise, for b = 0 Equation (7.15) becomes

det_{0≤i,j≤n−1} (Y + 2^{−2⌈(i+j+1)/2⌉}\binom{2 ⌈(i + j + 1)/2⌉}{⌈(i + j + 1)/2⌉})) = (−1)^{\binom{n}{2}}2^{−n^{2}}(2 ⌈n/2⌉ Y + 1).

(7.16)

Finally, since

U_{n}(1/2) = \begin{cases}1, & if n ≡ 0, 1 mod 6, \\ 0, & if n ≡ 2 mod 3, \\ −1, & if n ≡ 3, 4 mod 6, \end{cases}

for b = 1 Equation (7.15) becomes

det_{0≤i,j≤n−1} (Y + 2^{−2⌈(i+j)/2⌉}\binom{2 ⌈(i + j)/2⌉}{⌈(i + j + 1)/2⌉}))

= \begin{cases}2^{−n(n−1)}Y, & if n ≡ 0 (mod 3), \\ −2^{−n(n−1)}(Y (n + 1) + 1), & if n ≡ 1 (mod 3), \\ 2^{−n(n−1)}(Y n + 1), & if n ≡ 2 (mod 3). \end{cases} (7.17)

When testing (7.13) and (7.16) with the help of Mathematica, initially I mistyped the exponent of 2. This led to the discovery of two more — conjectural — Hankel determinant evaluations.

Conjecture 17. Let Y be a variable. For all positive integers n, we have

det_{0≤i,j≤n−1} (Y + 2^{−i−j}\binom{2 ⌈(i + j)/2⌉}{⌈(i + j)/2⌉})

= 2^{−(n−1)^{2}} × \begin{cases}2(−1)^{s}(2sY + 1), & if n = 3s, \\ (−1)^{s}((2s + 1)Y + 1), & if n = 3s + 1, \\ (−1)^{s+1}((2s + 1)Y + 1), & if n = 3s + 2, \end{cases} (7.18)

and

det_{0≤i,j≤n−1} (Y + 2^{−i−j−1}\binom{2 ⌈(i + j + 1)/2⌉}{⌈(i + j + 1)/2⌉}))

= \begin{cases}(−1)^{⌊n/6⌋}2^{−n(n−1)} (\frac{1}{2} ⌊\frac{4n+2}{3}⌋ Y + 1) , & if n is even, \\ (−1)^{⌊n/6⌋}2^{−n(n−1)} (\frac{1}{2} ⌊\frac{4n+4}{3}⌋ Y + 1) , & if n is odd. \end{cases} (7.19)

8. Recursiveness of Hankel determinants of moments

In [9, Cor. 8], it was shown that, if the coefficients of the three-term recurrence (2.1) satisfied by the orthogonal polynomials p_{n}(x) are constants, the scaled Hankel determinants of moments on the left-hand side of (1.4) (i.e., Equation (1.2) with k = 0) satisfy a linear recurrence with constant coefficients. Since, by (2.6), the “functions” q_{n}(y) satisfy the same type of three-term recurrence, this result — and its proof — extend straightforwardly to the scaled Hankel determinants of moments on the left-hand side of (1.2). It should be noted that the form of the scaling in (8.1) below is based on (2.2).

Corollary 18. Within the setup in Sections 1 and 2, let s_{i} ≡ s and t_{i} ≡ t for i ≥ 1. Furthermore, let

H_{n} = t_{0}^{−(n−k−1)}t^{−\binom{n−k−1}{2}} det_{0≤i,j≤n−1} (∫ u^{i+j} \frac{∏_{ℓ=1}^{m}(u − x_{ℓ})}{∏_{ℓ=1}^{k}(u − y_{ℓ})} dμ(u)) .  (8.1)

Then the sequence (H_{n})_{n≥0} of (scaled) Hankel determinants of moments satisfies a linear recurrence of the form

∑_{i=0}^{2^{k+m}} c_{i}H_{n−i} = 0, for n > 2^{k+m},  (8.2)

for some constants c_{i} ∈ K, normalised by c_{0} = 1. Explicitly, these constants can be computed as the coefficients of the characteristic polynomial (in x) ∑_{i=0}^{2^{k+m}} c_{i}x^{2^{k+m}−i} of the tensor product of 2 × 2 matrices

(\begin{matrix}s − x_{1} & t \\ −1 & 0 \end{matrix}) ⊗ ⋯ ⊗ (\begin{matrix}s − x_{m} & t \\ −1 & 0 \end{matrix}) ⊗ (\begin{matrix}s − y_{1} & t \\ −1 & 0 \end{matrix}) ⊗ ⋯ ⊗ (\begin{matrix}s − y_{k} & t \\ −1 & 0 \end{matrix}) .  (8.3)

In particular, we have

c_{1} = −( ∏_{ℓ=1}^{m}(s − x_{ℓ}))( ∏_{ℓ=1}^{k}(s − y_{ℓ}))  (8.4)

and

c_{2^{k+m}−i} = t^{(k+m)(2^{k+m−1}−i)}c_{i}, for i = 0, 1, …, 2^{k+m}.  (8.5)

Acknowledgement

I thank Mourad Ismail for insisting that, after having written [9], I throw a “combinatorial eye” on Uvarov’s formula. Moreover, I am indebted to Arno Kuijlaars for bringing [1] to my attention and for helpful discussion. Finally, I thank Alexander Dyachenko for pointing out [15].

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Department of Mathematics, University of Michigan, Ann Arbor, MI 48109, USA.

WWW: http://www.math.lsa.umich.edu/~baik/Welcome.html.

Courant Institute of Mathematical Sciences, New York University, New York NY

10012, USA. WWW: https://math.nyu.edu/~deift/.

Fakultät für Mathematik, Universität Wien, Oskar-Morgenstern-Platz 1, A-1090

Vienna, Austria. WWW: https://www.mat.univie.ac.at/~kratt.

Department of Mathematics, The Hebrew University of Jerusalem, 91904 Givat Ram,

Jerusalem, Israel. WWW: https://mathematics.huji.ac.il/people/evgeny-strahov.