Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation

Linear Algebra and its Applications 436 (2012) 2105–2116

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Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation Wathek Chammam b , Francisco Marcellán a,∗ , Ridha Sfaxi b a b

Departamento de Matemáticas, Universidad Carlos III de Madrid, Avenida de la Universidad 30, 28911 Leganés, Spain Faculté des Sciences de Gabès, Département de Mathématiques, Cité Erriadh, 6072 Gabès, Tunisia

ARTICLE INFO

ABSTRACT

Article history: Received 11 January 2011 Accepted 10 October 2011 Available online 9 November 2011

In this paper we deal with Hankel determinants of the form det[ai+j+r (x)]ni,j=0 , where r is a non-negative integer, an+r (x) =

Submitted by R.A. Brualdi AMS classification: 05A10 33C45 11C20 11B65

an+r + an+r −1 x + · · · + a0 xn+r and (an )n0 is a sequence complex numbers. When a0  = 0 and the Hankel determinants associated with the sequence (an+r +1 )n0 are not identically zero, we show that (det[ai+j+r (x)]ni,j=0 )n0 is a sequence of polynomials satisfying a three-term recurrence relation. We illustrate our result by evaluating the Hankel determinant  associated with the sequence  l + k + r 2(l+k+r −ν) ν n 1 det ν=0 l+k+r +1−ν l+k+r −ν x l,k=0 , for r = 0 and r = 1. © 2011 Elsevier Inc. All rights reserved.

Keywords: Catalan numbers Hankel transform Orthogonal polynomials

1. Introduction

:= (an )n0 be a sequence of complex numbers. For any integer r  0, we set A(r ) := (an+r )n0 . By convention, A(0) = A. The Hankel transform of the sequence A(r ) is the sequence of the so-called shifted Hankel determinants H0 (A(r ) ), H1 (A(r ) ), . . . , (see [7]) given by Let A

Hn (A(r ) )

= det[al+k+r ]nl,k=0 .

(1.1)

Note that these are determinants of (n + 1) × (n + 1) matrices. Since the 19th century, the evaluation of Hankel determinants has attracted attention as one of the most interesting topics in the framework of the moment theory and orthogonal polynomials. In the ∗ Corresponding author. E-mail addresses: [email protected] (W. Chammam), [email protected] (F. Marcellán), [email protected] (R. Sfaxi). 0024-3795/$ - see front matter © 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2011.10.010

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literature (see [5,6]), several techniques for the evaluation of many classes of Hankel determinants are described. One also can find there an extensive bibliography on the subject. The aim of our contribution is focused on the evaluation of the Hankel determinants with a certain class of monic polynomials as entries. The basic idea of this work comes from [11], where the author evaluates the Hankel determinant associated with the Catalan numbers by applying the technique of γ -operator on the Hankel determinant associated with a particular sequence of monic polynomials satisfying a two-term recurrence relation. For more information on the technique of γ -operator, see [11,12]. We derive an explicit formula for this class of Hankel determinants using another approach based on some elementary properties of determinants and the theory of orthogonal polynomials.

2. Preliminaries Let P be the linear space of polynomials in one variable with complex coefficients and P its dual space. We denote by U , p the action of U ∈ P on p ∈ P and by (U )n := U , xn , n  0, the sequence of moments of U with respect to the polynomial sequence {xn }n0 . Let us define the following operations in P . For linear functionals U and V , any polynomial q, and any (a, b, c ) ∈ C∗ × C2 , let DU = U  , qU , (x − c )−1 U , τ−b U , ha U , and U V be the linear functionals defined by duality (see [3,8,9])

U  , p := −U , p , qU , p := U , qp, −1

(x − c )

U , p



:= U , θc p = U ,

p(x) − p(c ) x−c



,

τ−b U , p := U , τb p = U , p(x − b), ha U , p := U , ha p = U , f (ax), U V , p := U , V p, p ∈ P, where the right-multiplication of V by a polynomial p is a polynomial given by   xp(x) − yp(y) (V p)(x) := Vy , , p ∈ P. x−y Notice that deg(V p) = deg p if and only if (V )0  = 0. We can associate with the sequence A = (an )n0 a unique linear functional U

(U )n = an , n  0.

∈ P by setting (2.1)

The linear functional U is said to be quasi-definite if Hn (A)  = 0 for every integer n  0, [3]. In this case, there exists a unique sequence of monic polynomials (SMP) (Pn )n0 , i.e., Pn (x) = xn + lower degree terms, such that (i) U , xν Pn  = 0, ν (ii) U , xn Pn   = 0.

= 0, 1, . . . , n − 1.

(Pn )n0 is said to be the sequence of monic orthogonal polynomials (SMOP) with respect to U . It is very well known that Pn (x) can be represented as a determinant (see [3]) x . . . xn 1 a0 a1 . . . an n (−1) , n  1. a a . . . a P0 (x) = 1, Pn (x) = (2.2) 1 2 n + 1 Hn−1 (A) . .. .. . . . . an−1 an . . . a2n−1

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The orthogonality of (Pn )n0 can be characterized by a three-term recurrence relation (TTRR), according to the Favard’s theorem (see [3, p. 21]),

P−1 (x) = 0, P0 (x) = 1, (2.3) Pn+1 (x) = (x − βn )Pn (x) − γn Pn−1 (x), n  0, where βn ∈ C, γn ∈ C∗ for every integer n  0, and γ0 = (U )0 . Furthermore, from the general theory of orthogonal polynomials (see [9])

U , Pn2+1  U , xPn2  , γ , n  0, = n + 1 U , Pn2  U , Pn2  n  Hn (A) , n  0, U , Pn2  = γν = H−1 (A) = 1. H n−1 (A) ν=0 βn =

We can associate to the sequence A an (x)

:=

n

ν=0

an−ν xν ,

n

(2.4) (2.5)

= (an )n0 a sequence of polynomials (an (x))n0 given by [11]

 0.

(2.6)

Notice that the polynomials an (x) satisfy a−1 (x)

= 0, an (x) = xan−1 (x) + an , n  0.

(2.7)

Thus, deg an (x) = n if and only if a0  = 0. In this paper we investigate the Hankel transform of the sequence of polynomials (an+r (x))n0 where r is a fixed non-negative integer. Indeed, we define the sequence of determinants (Hn+r (Ar | x))n0 , where n  Hn+r (Ar | x) := det al+k+r (x) l,k=0 , n  0. (2.8)

= 0 we write Hn (A | x) = Hn (A0 | x). A (r ) From an+r (0) = an+r we  get Hn+r (r | 0) = Hn (A ). 2n For A = C , let Cn = n /(n + 1) denote the nth Catalan number. In [11] the γ -operator technique is introduced to obtain a differential equation satisfied by Hn (C | x) and, as a consequence, the following As a convention, for r

evaluation is proved Hn (C

| x) =

n

ν=0

ν n+ν

(−1)

n−ν

 xν .

In [11], it is also shown that Hn (C | x) has n real simple zeros on the interval (0, 4). Furthermore, (Hn (C | x))n0 is a Sturm sequence whose zeros satisfy the interlacing property. Here, we will show that (Hn (C | x))n0 is a SMOP with respect to a positive-definite linear functional. When a0  = 0 and the linear functional xr +1 U is quasi-definite, i.e., Hn (A(r +1) )  = 0, for all n  0, in Corollary 1 we will prove that Hn+r (Ar | x) is a polynomial with deg Hn+r (Ar | x) = n + r . Furthermore, the sequence of monic polynomials defined by ((−1)n (a0 Hn−1 (A(r +1) ))−1 Hn+r (Ar | x))n1 satisfies a

three-term recurrence relation. In particular, we prove that ((−1)n (a0 Hn−1 (A(r +1) ))−1 Hn+r (Ar | x))n1 is a subsequence of a sequence of monic orthogonal polynomials if and only if either r = 0 or r = 1. ˆ n (A | x))n0 defined by Hˆ 0 (A | x) = 1 and Hˆ n (A | x) = Indeed, for r = 0 the SMP (H (−1)n (a0 Hn−1 (A(1) ))−1 Hn (A | x), n  1, is a SMOP. More precisely, it is a co-recursive SMOP ˆ n (A1 | x))n0 where with respect to the linear functional xU . On the other hand, for r = 1 the SMP (H

ˆ 0 (A1 | x) = 1, Hˆ 1 (A1 | x) H also a SMOP.

1 n (2) ))−1 H A ˆ A = x + a− n+1 (1 | x), n  1, is 0 a1 and Hn (1 | x) = (−1) (a0 Hn−1 (A

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The paper is organized as follows. In Section 2, the proofs of our main results are given. We obtain

(see Theorem 3) a first evaluation of the Hankel determinants Hn+r (Ar | x) in the general setting. From there, the evaluation of these Hankel determinants (see Corollary 1) becomes more useful and

feasible, assuming the linear functional is quasi-definite. In Section 3, we focus our attention on the orthogonality of the polynomial sequence ((−1)n (a0 Hn−1 (A(r +1) ))−1 Hn+r (Ar | x))n1 , where r = 0 or r = 1. Finally, some illustrative examples based on the Catalan numbers are presented. 3. Main results Using (1.8), elementary row operations yield ar (x) ar +1 (x) . . . an+r (x) ar +1 ar +2 . . . an+r +1 Hn+r (Ar | x) = ar +2 ar +3 . . . an+r +2 , . .. .. . . . . an+r an+r +1 . . . a2n+r

n

 1.

(3.1)

Expanding the above determinant by the first row, we get Hn+r (Ar | x)

=

n j=0

(−1)j Dj+1,r aj+r (x),

(3.2)

where Dj,r denotes the determinant of the n×n matrix obtained by deleting the jth column of the n×(n + 1) matrix ⎤ ⎡ a ar +2 . . . an+r +1 ⎥ ⎢ r +1 ⎥ ⎢ ⎢ ar +2 ar +3 . . . an+r +2 ⎥ ⎥ ⎢ ⎢ . , n  1. (3.3) .. ⎥ .. ⎥ ⎢ . ⎥ ⎢ . . . ⎦ ⎣ an+r an+r +1 . . . a2n+r Notice that Dn+1,r = Hn−1 (A(r +1) ). From (1.7) and (2.2), we get the following result. Lemma 1. Let A = (an )n0 be a sequence of complex numbers and let U n  0. The following statements are equivalent.

∈ P be such that (U )n = an ,

(a) deg Hn+r (Ar | x) (b) a0 (c) a0

= n + r , n  0. = 0 and Hn−1 (A(r +1) ) = 0, n  1. = 0 and xr +1 U is quasi-definite.

The next lemma will play an important role in the sequel. Lemma 2. Let A = (an )n0 be a sequence of complex numbers and let U n  0. Suppose (an (x))n0 is defined as in (1.7). Then an (x)

∈ P be such that (U )n = an ,

= (U xn )(x), n  0.

Proof. From (1.2) and (1.7), we obtain   n xn+1 − yn+1 (U xn )(x) = Uy , = (U )n−ν xν x−y ν=0 Thus, our statement holds. 

(3.4)

=

n

ν=0

an−ν xν

= an (x), n  0.

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As a straightforward consequence of Lemma 2 and (2.1) the following expression of the Hankel determinants Hn (Ar | x) holds. Theorem 3. Let A = (an )n0 be a sequence of complex numbers and let U n  0. For any integer r  0, we have Hn+r (Ar | x)

= U xr Qn (x; A(r +1) ), n  0,

where Q0 (x; A(r +1) )

= 1,

Qn (x; A(r +1) )

According to (1.3), if Hn−1 (A(r +1) ) −1

by ((−1)n Hn−1 (A(r +1) ) Furthermore, Qˆ n (x; A(r +1) )

:=

=

(3.5)

x 1 ar +1 ar +2 ar +2 ar +3 . .. . . . an+r an+r +1

... ... ... ...

xn an+r +1 an+r +2 , n .. . a2n+r

 1.

= 0, n  1, then the sequence of monic polynomials defined

Qn (x; A(r +1) ))n0 is orthogonal with respect to the linear functional xr +1 U .

(−1)n Qn (x; A(r +1) ), n  0, Hn−1 (A(r +1) )

then, the SMOP (Qˆ n (x; A(r +1) ))n0 satisfies the TTRR, ⎧ (r +1) ) = 0, Q ˆ 0 (x; A(r +1) ) = 1, ˆ ⎪ ⎪ ⎨Q−1 (x; A (r +1) ˆ ( r + 1 ) Qˆ n+1 (x; A ) = (x − αn )Qn (x; A(r +1) )− ⎪ ⎪ (r +1) ˆ ⎩ Q (x; A(r +1) ), λ n−1

n

(r +1)

∈ P be such that (U )n = an ,

(r +1)

(3.6)

(3.7)

 0,

n

(r +1)

∈ C, λn ∈ C∗ , and λ0 = ar +1 . ˆ n+r (Ar | x), n  1, given by Assuming a0  = 0 and Hn−1 (A(r +1) )  = 0, n  1, the polynomials H

where αn

ˆ n+r (Ar | x) H

:=

(−1)n −1 Hn+r (Ar | x) = a0 U xr Qˆ n (x; A(r +1) ), a0 Hn−1 (A(r +1) )

(3.8)

are monic and deg Hn+r (Ar | x) = n + r . Under the above assumptions, we get the following:

= (an )n0 be a sequence of complex numbers and let U ∈ P be such that (U )n = an , ˆ n+r (Ar | n  0. Suppose that a0  = 0 and the linear functional xr +1 U is quasi-definite. Then the SMP (H x))n1 given by

Corollary 1. Let A

ˆ n+r (Ar | H

x)

=

(−1)n

⎡

n−1

a0 det[al+k+r +1 ]l,k=0

det ⎣

l+ k+r

ν=0

⎤n

al+k+r −ν x

ν⎦

, n  1,

l,k=0

satisfies the following three term recurrence relation (TTRR) ⎧ 1 r ˆ r (Ar | x) = a− ⎪ ⎪H 0 Ux , ⎨ (r +1)

−1

ˆ r +1 (Ar | x) = (x − α0 H )Hˆ r (Ar | x) + a0 ar +1 , ⎪ ⎪ ( r + 1) (r +1) ˆ ⎩H ˆ (A | x) = (x − α )Hˆ (A | x) − λ H n+r +1 r

(r +1)

where αn

n

(r +1)

and λn

n+r r

n

(3.9) A n+r −1 (r | x), n  1,

are the coefficients in the TTRR for the SMOP with respect to xr +1 U .

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W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

Proof. First, we need the following formula (see [9])

(U xp)(x) = x(U p)(x) + yUy , p(y), p ∈ P. From (2.7), (2.8), and (2.10) with p(x)

ˆ n+r +1 (Ar | x) H

=

(3.10)

x Qˆ n (x, A(r +1) ), we get r

= (x − αn(r +1) )Hˆ n+r (Ar | x) − λ(nr +1) Hˆ n+r −1 (Ar | x)+ −1 a0 yr +1 Uy , Qˆ n (y, A(r +1) ), n  2.

Since (Qˆ n (x, A(r +1) )n0 is a SMOP with respect to xr +1 U , then yr +1 Uy , Qˆ n (y, A(r +1) ) n  0. Therefore,

ˆ n+r +1 (Ar | x) H ˆ r (Ar | x) By adding H −1

x) + a0 ar +1 . 

= ar +1 δn,0 ,

= (x − αn(r +1) )Hˆ n+r (Ar | x) − λ(nr +1) Hˆ n+r −1 (Ar | x), n  2, (r +1)

1 r A ˆ = a− 0 U x and, proceeding as above, we get Hr +1 (r | x) = (x − α0

)Hˆ r (Ar |

4. Special cases and illustrative examples

 2 the SMP (Hˆ n (Ar | x))nr is not a ˆ n (Ar | x)}nr is a subsequence of a subsequence of a SMOP. Whereas for r = 0, 1 we will show that {H A ˆ n (r | x)}n0 . SMOP {H From Favard’s Theorem and (2.9) , we can deduce that for r

l+k 4.1. Evaluation of det[ ν=0 al+k−ν xν ]nl,k=0

ˆ 0 (A | x) = 1. Then, the SMP (Hˆ n (A Let us take r = 0 in Corollary 1 and assume H the following TTRR, ⎧ (1) 1 ⎨H ˆ 0 (A | x) = 1, Hˆ 1 (A | x) = x − α0 + a− 0 a1 , ( 1 ) ( 1 ) ⎩H ˆ (A | x) = (x − αn )Hˆ (A | x) − λn Hˆ (A | x), n  1. n+1

| x))n0 satisfies

(4.1)

n−1

n

| x))n0 is the co-recursive of the SMOP (Qˆ n (x, A(1) ))n0 , since it is generated by (1) (1) −1 the TTRR (2.7) with r = 0, where α0 is replaced by α0 − a1 a0 . For more information, see [3,9]. ( 1) ( 1 ) Denoting by (Qˆ n (x, A ))n0 the first kind associated SMOP of the sequence (Qˆ n (x, A(1) ))n0 that ˆ n (A So, the SMP (H

is defined by the following TTRR (see [3]) ⎧ ⎨Qˆ (1) (x, A(1) ) = 0, Qˆ (1) (x, A(1) ) = 1, −1 0 ⎩Qˆ (1) (x, A(1) ) = (x − α (1) )Qˆ n(1) (x, A(1) ) − λ(1) Qˆ (1) (x, A(1) ), n+1 n+1 n+1 n−1

n

 0,

(4.2)

we get

ˆ n (A H

(1)

1ˆ (1) | x) = Qˆ n (x, A(1) ) + a1 a− ), n  0. 0 Qn−1 (x, A

From (3.3) and Corollary 1 with r l+k det[ ν=0 al+k−ν xν ]nl,k=0 holds.

(4.3)

= 0, the following evaluation of the Hankel determinant

= (an )n0 be a sequence of complex numbers and let U ∈ P be such that (U )n = an , n  0. Assuming that a0  = 0 and the linear functional xU is quasi-definite, if (Qˆ n (x, A(1) ))n0 is the Corollary 2. Let A

corresponding SMOP, then ⎤n ⎡ l+k ν al+k−ν x ⎦ det ⎣ ν=0

l,k=0

1 ˆ∗ (1) = a0 (−1)n det[al+k+1 ]nl,− ), n  1, k=0 Qn (x, A

W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

2111

where (Qˆ n∗ (x, A(1) ))n0 is the co-recursive SMOP of the SMOP (Qˆ n (x, A(1) ))n0 , given by Qˆ n∗ (x, A(1) )

1 (1) ˆ (1) = Qˆ n (x, A(1) ) + a− ), n  0. 0 a1 Qn−1 (x, A

As an application of the previous result, we will recover (see [11]) the evaluation of Hn (C | x) =    n l+k 2(l+k)−2ν) ν 1 det . x ν=0 l+k+1−ν l+k−ν l,k=0   2n First, let us take in Corollary 2, A = C where Cn = n /(n + 1), the nth Catalan number. Notice that (n + 2)Cn+1 = (4n + 2)Cn , n  0. So, the related linear functional U with (U )n is a solution of the functional equation   x(x − 4)U + 2(−x + 1)U = 0.

= Cn , n  0, (4.4)

h−2 ◦ τ−1 J ( 12 , − 12 ), where J ( 12 , − 12 ) is the Jacobi monic linear

According to [8,10], U = functional with parameters α = −β = 1/2, i.e., Chebyshev linear functional of the third kind. In the same way, U1 := xU is monic (i.e., (U1 )0 = 1) and positive-definite. Indeed, according to (3.4) U1 satisfies   x(x − 4)U1 + 3(−x + 2)U1 = 0. (4.5) From [10], U1 = h−2 ◦ τ−1 J ( 12 , 12 ), where J ( 12 , 12 ) is the Chebyshev linear functional of the second kind. Here, the SMOP (Qˆ n (x, C (1) ))n0 with respect to U1 is given by   2−x ˆn Qˆ n (x, C (1) ) = (−2)n U , n  0, (4.6) 2

ˆ n (x))n0 is the Chebyshev SMOP of second kind. where (U Using the TTRR satisfied by

(Qˆ n (x, C (1) ))n0 ,

(Uˆ n (x))n0 (see [10]) we get the following TTRR satisfied by

= 0, Qˆ 0 (x, C (1) ) = 1, ( 1) ˆ Qn+1 (x, C ) = (x − 2)Qˆ n (x, C (1) ) − Qˆ n−1 (x, C (1) ), n  0. Qˆ −1 (x, C (1) )

(4.7)

From (3.7) and (1.6), we recover the well-known Hankel determinant evaluation [1,4], det[Cl+k+1 ]nl,k=0

= 1, n  0.

(4.8)

Using again (3.7) we obtain the following relation: Qˆ n(1) (x, C (1) )

= Qˆ n (x, C (1) ), n  0.

(4.9)

From Corollary 2 with A = C and by taking into account (3.6), (3.8), and (3.9) we obtain ⎤n ⎡     l+k 2−x 2−x ν Cl+k−ν x ⎦ = 2n Uˆ n − 2n−1 Uˆ n−1 , n  1. det ⎣ 2 2 ν=0

(4.10)

l,k=0

ˆ n (C By (3.3) with A = C and (3.9), the SMOP (H connection relation: ˆ n (C H

| x))n0 and (Qˆ n (x, C (1) ))n0 verify the following

| x) = Qˆ n (x, C (1) ) + Qˆ n−1 (x, C (1) ), n  0.

ˆ n (C From (3.1) with A = C , we get the TTRR satisfied by (H

ˆ 0 (C | x) = 1, Hˆ 1 (C | x) = x − 1, H ˆ n+1 (C H

| x))n0 ,

| x) = (x − 2)Hˆ n (C | x) − Hˆ n−1 (C | x), n  1.

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ˆ n (C From [10], we get H

| x) = 2n ˆJn



x−2 ; − 21 , 12 2



 ,n



 0, where ˆJn (x; − 12 , 12 )

SMOP of the fourth kind. Hence, the following evaluation holds ⎡ ⎤n   l+k 1 1 x−2 ν⎦ ⎣ det Cl+k−ν x = (−1)n ˆJn ;− , , n 2 2 2 ν=0

n0

is the Chebyshev

 0.

(4.11)

l,k=0

ˆ n (C | x))n0 is a SMOP with respect to a positive definite linear functional. Thus their Notice that (H zeros satisfy the interlacing property (see [3]). ˆ n (x))n0 , we can prove the Based on some properties of the Chebyshev SMOP of the second kind (U following result. Lemma 4

⎡

det ⎣

l+k

ν=0

where x



⎤n Cl+k−ν xν ⎦ l,k=0





=

cos

  n + 12 z   , cos 2z

(4.12)

= 2 1 − cos(z) . On the other hand, ⎡

det ⎣

l+k

ν=0

⎤n

Cl+k−ν x

ν⎦

= (−1)

n

−1 n

ν=0

l,k=0



x − 4 sin

2

(2ν + 1) π 2n + 1 2



, n  1.

(4.13)

Proof. The Chebyshev polynomials of second kind are defined by (see [2]),   sin (n + 1)z ˆUn (cos(z)) = , n  0. 2n sin(z ) Letting x

= 2(1 − cos(z)) into the right-hand side of (3.10), we get ⎡

det ⎣

l+k

ν=0

⎤n Cl+k−ν x

ν⎦

=

sin

l,k=0





(n + 1)z − sin(nz) sin(z )



=

cos

  n + 12 z   . cos 2z

Hence (3.12) holds.    (2ν+1)π Since zν = 2n+1 , ν = 0, 1, . . . , n − 1, are the zeros of cos (n + 12 )z then xν = 2 1 −  (2ν+1)  (2ν+1)π  ˆ n ( 2−x ) − 2n−1 Uˆ n−1 ( 2−x ). cos( 2n+1 ) = 4 sin2 2n+1 π2 are the zeros of the polynomial 2n U 2 2 Taking into account the degree of this polynomial is n, then       −1 n 2−x 2−x nˆ n−1 ˆ n 2 (2ν + 1) π x − 4 sin − 2 Un−1 = (−1) , n  1. 2 Un 2 2 2n + 1 2 ν=0 Hence (3.13) holds.  Furthermore, we need the following identities which can be derived in a straightforward way from

(3.11), (3.12), and (3.13). ˆJn



x−2 2

1 1

;− ,

for any integer n

2 2





=

cos

n+

(−1)n cos

  z  

1 2

z 2

=

 1, where x = 2(1 − cos(z)).

n −1

ν=0



x − 4 sin

2

(2ν + 1) π 2n + 1 2



,

(4.14)

W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

2113

 l+k ν n Next, we evaluate det ν=0 Cl+k−ν x l,k=0 for some special values of x. • For x = 0 and z = 0 in (3.12) and (3.13), we find  −1 n n  (2ν + 1) π n det Cl+k l,k=0 = 2 sin = 1. 2n + 1 2 ν=0 π

• For x = 1 and z = ⎡ det ⎣

l+k

ν=0

in (3.12) and (3.13),

3

⎤n Cl+k−ν ⎦

  n −1 (2ν + 1)π 1 − 2 cos 2n + 1 ν=0

=

l,k=0

⎧ ⎪ 1, ⎪ ⎪ ⎨

if n

= ⎪ 0,

if ⎪ ⎪ ⎩ −1, if

≡ 0, 5 (mod 6)

n

≡ 1, 4 (mod 6)

n

≡ 2, 3 (mod 6).

π

• For x = 2 and z =

in (3.12) and (3.13), ⎤n  l+k −1 n (2ν + 1)π ν n det ⎣ Cl+k−ν 2 ⎦ = (−2) cos 2n + 1 ν=0 ν=0 2

⎡

= (−1)

n(n+1) 2

.

l,k=0

2π 3

• For x = 3 and z = ⎡

det ⎣

l+k

ν=0

⎤n

in (3.12) and (3.13),

Cl+k−ν 3ν ⎦

= (−1)n l,k=0

=

⎧ ⎨ 1,

−1 n



1 + 2 cos

ν=0

(2ν + 1)π 2n + 1



≡ 0, 2 (mod 3)

if n ⎩ −2, if n

≡ 1 (mod 3).

• For x = 4 and z = π in (3.12) and (3.13), ⎡

det ⎣

l+k

ν=0 n −1

⎤n

Cl+k−ν 4

ν⎦

= (−1)n (2n + 1) l,k=0



√ = 2−n 2n + 1.

(2ν + 1) π cos 2n + 1 2 ν=0 Moreover, since (see [11])  det 22(l+k)+1

• For x = 2 −

√

−

n

ν=0 Cn−ν 4

ν



2(l + k) + 1 l+k+1 π





2n+1 n+1

, n  0, we obtain

n l,k=0

= (−1)n (2n + 1).

in (3.12) and (3.13), ⎤n   l+k −1 n √ ν √ (2ν + 1)π 2 cos det ⎣ Cl+k−ν (2 − 2) ⎦ = (−1)n − 2 2n + 1 ν=0 ν=0 l,k=0 ⎧ ⎪ if n ≡ 0, 7 (mod 8) 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ −1, if n ≡ 3, 4 (mod 8) = √ ⎪ ⎪ 2 − 1, if n ≡ 1, 6 (mod 8) ⎪ ⎪ ⎪ √ ⎪ ⎩ 1 − 2, if n ≡ 2, 5 (mod 8). ⎡

2 and z

=

= 22n+1 −

4

2114

W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

l+k+1 4.2. Evaluation of det[ ν=0 al+k+1−ν xν ]nl,k=0 Let us take r SMP

(Hˆ n (A1 |

1 = 1 in Corollary 1 and let us put Hˆ 0 (A1 | x) = 1 and Hˆ 1 (A1 | x) = x + a− 0 a1 . Then the

x))n0 satisfies the following TTRR, ⎧ −1 ˆ A ˆ A ⎪ ⎪ ⎨H0 (1 | x) = 1, H1 (1 | x) = x + a0 a1 , 1 ˆ A ˆ 2 (A1 | x) = (x − α0(2) )Hˆ 1 (A1 | x) + a− H 0 a2 H0 (1 | x), ⎪ ⎪ ( 2 ) ( ⎩H ˆ (A | x) = (x − α )Hˆ (A | x) − λ 2) Hˆ (A | x), n+2 1

n

n+1 1

n

n 1

(4.15) n

 1,

ˆ n (A1 | x))n0 is a SMOP. If the linear functional x U is quasi-definite then a2 = (x2 U )0  = 0. Thus, (H ν On the other hand, if we assume that the linear functionals x U , ν = 1, 2 , are quasi-definite, ˆ n (A1 | x) in terms of the polynomials Hˆ n (A | x) can be another expression of the Hankel determinant H evaluated. In this case, the corresponding SMOP (Qˆ n (x, A(ν) ))n0 , ν = 1, 2, satisfies the following connection relation (see [9]) 2

xQˆ n (x, A(2) ) From (2.8) with r

ˆ n+1 (A1 | H

x)

= Qˆ n+1 (x, A(1) ) −

Qˆ n+1 (0, A(1) ) Qˆ n (0, A(1) )

n

 0.

(4.16)

= 1 and (3.16), we get 1 (2) ˆ = a− ) 0 U xQn (x; A 1 (1) ˆ )− = a− 0 U Qn+1 (x; A

and, again, from (2.8) with r

ˆ n+1 (A1 | x) H

Qˆ n (x, A(1) ),

Qˆ n+1 (0, A(1) ) −1 a0 U Qˆ n (x; A(1) ), Qˆ n (0, A(1) )

n

 1,

= 0, we deduce

= Hˆ n+1 (A | x) −

Qˆ n+1 (0, A(1) ) Qˆ n (0, A(1) )

ˆ n (A H

| x), n  1.

(4.17)

As a consequence of Corollary 2, (2.8), and (3.17), we get Corollary 3. Let A = (an )n0 be a sequence of complex numbers and let U ∈ P be such that (U )n = an , n  0. Suppose that a0  = 0, and xr +1 U , r = 0, 1, are quasi-definite linear functionals. Then, we have ⎡ ⎤n l+ k+1   ν det ⎣ al+k+1−ν x ⎦ = ϑn Hˆ n+1 (A | x) + ξn Hˆ n (A | x) , n  1, ν=0

l,k=0

where 1 ϑn = (−1)n a0 det[al+k+2 ]nl,− k=0 , ξn = −

ˆ n (A H

(1)

Qˆ n+1 (0, A(1) ) Qˆ n (0, A(1) )

,

1ˆ (1) | x) = Qˆ n (x, A(1) ) + a1 a− ), 0 Qn−1 (x, A

and {Qˆ n (x, A(1) )}n0 is the SMOP with respect to xU .

det

From Corollary 3, with A = C , and using  l+k+1 2(l+k+1−ν) ν n 1 ν=0 l+k+2−ν l+k+1−ν x l,k=0 .

(3.11) and (3.14), we next evaluate

Indeed, from (3.7) with x = 0 and using induction, it is easy to show that Qˆ n (0, C (1) ) = (−1)n (n + n−1 2 1), n  0. As a consequence, ξn = nn+ +1 , n  0. On the other hand, let us remind that det[Cl+k+2 ]l,k=0

W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

2115

= n + 1, n  1, (see [4]). Therefore, taking into account (3.11) , ⎡

l+ k+1

det ⎣

ν=0

⎤n

Cl+k+1−ν x



ν⎦

= (−2)n (n + 1) 2ˆJn+1

l,k=0

+

n+2 n+1

ˆJn



x−2 2



x−2 2 1 1

;− ,

1 1

;− ,



2 2



2 2

, n  1.

(4.18)

Substituting (3.14) into (3.18), we get ⎡ ⎤n l+ k+1 det ⎣ Cl+k+1−ν xν ⎦ = ν=0

l,k=0

(n + 2) cos



n+

     3 n+ z − (n + 1) cos z 2 2   , z cos 2 1

n

 1,

where x = 2(1 − cos(z )).  l+k+1 ν n Next, we evaluate det ν=0 Cl+k+1−ν x l,k=0 at the same values of x as above. • For x = 0 and z = 0 in (3.19), we get  n det Cl+k+1 l,k=0 = 1.

• For x = 1 and z =

π 3

⎡

l+ k+1

det ⎣

ν=0

⎤n Cl+k+1−ν ⎦

⎡

l+ k+1

ν=0

π 2

ν=0

= (−1)

n(n+1) 2

if n

≡ 0 (mod 6)

if n

≡ 1 (mod 6)

if n

≡ 2 (mod 6)

if n

≡ 3 (mod 6)

if n

≡ 4 (mod 6)

if n

≡ 5 (mod 6).



 n + 2 + (−1)n (n + 1) .

l,k=0 2π 3

in (3.19), ⎤n

⎡

l+ k+1

⎧ ⎪ ⎪ ⎪ n + 2, ⎪ ⎪ ⎪ ⎪ n + 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ −1, ⎪ ⎪ −(n + 2), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −(n + 1), ⎪ ⎪ ⎪ ⎩ 1,

in (3.19), ⎤n

Cl+k+1−ν 2ν ⎦

• For x = 3 and z = det ⎣

= l,k=0

• For x = 2 and z = det ⎣

in (3.19),

ν⎦

=

Cl+k+1−ν 3

l,k=0

⎧ ⎪ 3n + 4, if n ⎪ ⎪ ⎨ −3n − 5, if n ⎪ ⎪ ⎪ ⎩ 1, if n

≡ 0 (mod 3) ≡ 1 (mod 3) ≡ 2 (mod 3).

• For x = 4 and z = π in (3.19), ⎤n

⎡

l+ k+1

det ⎣

ν=0

Cl+k+1−ν 4

ν⎦ l,k=0

= (−1)n (4n2 + 10n + 5).

(4.19)

2116

Since

W. Chammam et al. / Linear Algebra and its Applications 436 (2012) 2105–2116

n+1

ν=0 Cn+1−ν 4

ν

= 22n+3 −



det 22(l+k)+3

• For x = 2 −

⎡ det ⎣

l+ k+1

ν=0

√

−





2n+3 n+2



2(l + k) + 3 l+k+2

2 and z

=

π 4

Cl+k+1−ν (2 −

, n  0, then

n l,k=0

= (−1)n (4n2 + 10n + 5).

in (3.19),

√

⎤n 2)ν ⎦

= l,k=0

⎧ √ √ ⎪ ⎪ n(2 − 2) + 3 − 2, ⎪ ⎪ ⎪ ⎪ √ ⎪ ⎪ ( 2 − 1)(2n + 3), ⎪ ⎪ ⎪ √ √ ⎪ ⎪ ⎪ (2 − 2)n + 3 − 2 2, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ −1, √ √ ⎪ ⎪ −n 2 − 1 − 2, ⎪ ⎪ ⎪ √ ⎪ ⎪ ⎪ (1 − 2)(2n + 3), ⎪ ⎪ ⎪ √ √ ⎪ ⎪ ⎪ ⎪ ( 2 − 2)n + 2 2 − 3, ⎪ ⎪ ⎪ ⎪ ⎩ 1,

if n

≡ 0 (mod 8)

if n

≡ 1 (mod 8)

if n

≡ 2 (mod 8)

if n

≡ 3 (mod 8)

if n

≡ 4 (mod 8)

if n

≡ 5 (mod 8)

if n

≡ 6 (mod 8)

if n

≡ 7 (mod 8).

Acknowledgements We thank the comments and suggestions by the referee which have contributed to improve the presentation of the manuscript. The work of the second author (F.M.) was supported by Dirección General de Investigación, Ministerio de Ciencia e Innovación of Spain, under grant MTM 2009-12740C03-01. The work of the third author (R.S.) was supported by Faculté des Sciences de Gabès and Faculté des Sciences de Sfax, Tunisia. References [1] A.T. Benjamin, N.T. Cameron, J.J. Quinn, C.R. Yerger, Catalan determinants. A combinatorial approach, Congr. Numer. 200 (2010) 27–34. [2] M.M. Chawla, On the Chebyshev polynomials of the second kind, SIAM Rev. 9 (1967) 729–733. [3] T.S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach, New York, 1978. [4] J. Cigler, C.F. Krattenthaler, Some determinants of path generating functions, Adv. Appl. Math. 46 (2011) 144–174. [5] C.F. Krattenthaler, Advanced determinant calculus, Sém. Lothar. Combin. 42 (1999) 67 (The Andrews Festschrift, Maratea, 1998), Article B42q. [6] C.F. Krattenthaler, Advanced determinant calculus: a complement, Linear Algebra Appl. 411 (2005) 68–166. [7] J.W. Layman, The Hankel transform and some of its properties, J. Integer Seq. 4 (2001) 11, Article 01.1.5. [8] F. Marcellán, R. Sfaxi, A characterization of weakly regular linear functionals, Rev. Acad. Colombiana Cienc. Exact. Fis. Natur. 31 (119) (2007) 285–295. [9] P. Maroni, Une théorie algébrique des polynômes orthogonaux. Applications aux polynômes orthogonaux semi-classiques, in: C. Brezinski (Ed.), et al., Orthogonal Polynomials and their Applications, Annals Comput. Appl. Math., vol. 9, Baltzer, Basel, 1991, pp. 95–130. [10] P. Maroni, Fonctions eulériennes, Polynômes orthogonaux classiques, Tech. l’ingénieur A 154 (1994) 1–30. [11] Ö. Egecioglu, A Catalan–Hankel determinant, Congr. Numer. 195 (2009) 49–63. [12] Ö. Egecioglu, T. Redmond, C. Ryavec, A multilinear operator for almost product evaluation of Hankel determinants, J. Combin. Theory Ser. A 117 (2010) 77–103.