Bulletin T.CXXXI de l’Académie serbe des sciences et des arts − 2005 Classe des Sciences mathématiques et naturelles Sciences mathématiques, N\underline{o} 30

ORTHOGONAL POLYNOMIALS RELATED TO THE OSCILLATORY-CHEBYSHEV WEIGHT FUNCTION^{1}

G. V. MILOVANOVIĆ, A. S. CVETKOVIĆ

(Presented at the 8th Meeting, held on November 26, 2004)

A b s t r a c t. In this paper we discuss the existence question for polynomials orthogonal with respect to the moment functional

L(p) = ∫_{−1}^{1} p(x)x(1 − x^{2})^{−1/2}e^{iζx}dx, ζ ∈ ℝ.

Since the weight function alternates in sign in the interval of orthogonality, the existence of orthogonal polynomials is not assured. A nonconstructive proof of the existence is given. The three-term recurrence relation for such polynomials is investigated and the asymptotic formulae for recursion coefficients are derived.

AMS Mathematics Subject Classification (2000): Primary 30C10, 33C47 Key Words: Orthogonal polynomials; Moments; Moment functional;

Three-term recurrence relation; Oscillatory Chebyshev weight; Asymptotic formulae; Bessel functions

1The authors were supported in part by the Serbian Ministry of Science, Technology and Development (Project #2002: Applied Orthogonal Systems, Constructive Approximation and Numerical Methods).

1. Introduction

Let P be the space of all algebraic polynomials and P_{n} be the linear space of all algebraic polynomials of degree at most n.

Let a linear functional L be given on the linear space of all algebraic polynomials, i.e., let the functional L satisfy following equality, for each P, Q ∈ P,

L(αP + βQ) = αL(P) + βL(Q), α, β ∈ ℂ.

The value of the linear functional L at every polynomial is known if the values of L are known at the set of all monomials, due to linearity. The corresponding values of the linear functional L at the set of monomials are called the moments and we denote them by μ_{k}, k ∈ ℕ_{0},

L(x^{k}) = μ_{k}, k ∈ ℕ_{0}.

In [3, p. 7], the following definition can be found.

Definition 1 A sequence of polynomials {P_{n}(x)}_{n=0}^{+∞} is called the polynomial sequence orthogonal with respect to the moment functional L, provided for all nonnegative integers m and n,

• P_{n}(x) is polynomial of degree n,

• L(P_{n}(x)P_{m}(x)) = 0, if m ≠ n,

• L(P_{n}^{2}(x)) ≠ 0.

If the sequence of orthogonal polynomials exists for a given linear functional L, then L is called quasi-definite or regular linear functional. Under the condition L(P_{n}^{2}(x)) > 0, the functional L is called positive definite (see [3]).

Using only linear algebraic tools the following theorem can be stated (see [3, p. 11]).

Theorem 1. The necessary and sufficient conditions for the existence of a sequence of orthogonal polynomials with respect to the linear functional L are that for each n ∈ ℕ the Hankel determinants

Δ_{n} = | \begin{matrix}μ_{0} & μ_{1} & μ_{2} & … & μ_{n−1} \\ μ_{1} & μ_{2} & μ_{3} & … & μ_{n} \\ μ_{2} & μ_{3} & μ_{4} & … & μ_{n+1} \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ μ_{n−1} & μ_{n} & μ_{n+1} & … & μ_{2n−2} \end{matrix} | ≠ 0.  (1)

In this paper we consider the linear functional L given by

L(p) = ∫_{−1}^{1} p(x)x(1 − x^{2})^{−1/2}e^{iζx}dx,  (2)

where ζ ∈ ℝ. In order to prove the existence of the corresponding orthogonal polynomials, we need to compute the sequence of moments μ_{k}, k ∈ ℕ_{0}, and then to prove that all Hankel determinants are different from zero. The case of orthogonality with respect to the linear functional L(p) = ∫_{−1}^{1} p(x)xe^{imπx}dx, m ∈ ℕ, was investigated in [8].

Denote the sequence of moments by μ_{k}(ζ), k ∈ ℕ_{0}. Then, for each k ∈ ℕ_{0}, we can easily verify that

μ_{k}(ζ) = ∫_{−1}^{1} x^{k+1}(1−x^{2})^{−1/2}e^{iζx}dx = \overline{∫_{−1}^{1} x^{k+1}(1 − x^{2})^{−1/2}e^{−iζx}dx} = \overline{μ_{k}(−ζ)}.

This means that we need only to discuss the case ζ > 0, since the corresponding results for ζ < 0 can be obtained by a simple conjugation. We exclude the case ζ = 0 since for this case μ_{0} = Δ_{0} = 0, so that the linear functional L is not regular. In what follows we assume ζ > 0. Let J_{ν} be the Bessel function of the order ν defined by (cf. [14, p. 40])

J_{ν}(z) = ∑_{m=0}^{+∞} \frac{(−1)^{m}(z/2)^{ν+2m}}{m!Γ(ν + m + 1)} .  (3)

Theorem 2. The sequence of moments satisfy the following recurrence relation

μ_{k+2} = −\frac{k + 2}{iζ}μ_{k+1} + μ_{k} + \frac{k + 1}{iζ}μ_{k−1}, k ∈ ℕ,  (4)

with the initial conditions

μ_{0} = iπJ_{1}(ζ), μ_{1} = \frac{π}{ζ} (ζJ_{0}(ζ)−J_{1}(ζ)), μ_{2} = \frac{iπ}{ζ^{2}} (ζJ_{0}(ζ)+(ζ^{2}−2)J_{1}(ζ)). (5)

P r o o f: We start with the following simple equality

∫_{−1}^{1} x^{k+1}(1 − x^{2})(1 − x^{2})^{−1/2}e^{iζx}dx = μ_{k} − μ_{k+2}, k ∈ ℕ_{0}.

If we apply integration by parts to the integral which appears in the previous equality, we get

∫_{−1}^{1} x^{k+1}\sqrt{1 − x^{2}}e^{iζx}dx = \frac{x^{k+1}}{iζ} \sqrt{1 − x^{2}}e^{iζx}|_{−1}^{1} + \frac{1}{iζ} ∫_{−1}^{1} \frac{x^{k+2}}{\sqrt{1 − x^{2}}} e^{iζx}dx

−\frac{k + 1}{iζ} ∫_{−1}^{1} \frac{x^{k} − x^{k+2}}{\sqrt{1 − x^{2}}} e^{iζx}dx

= \frac{1}{iζ} μ_{k+1} − \frac{k + 1}{iζ}(μ_{k−1} − μ_{k+1}), k ∈ ℕ,

so that we have

μ_{k+2} = −\frac{k + 2}{iζ}μ_{k+1} + μ_{k} + \frac{k + 1}{iζ}μ_{k−1}, k ∈ ℕ.

In order to start the recursion we need to compute the moments μ_{0}, μ_{1} and μ_{2}. Because of the symmetry argument, we see that

μ_{0} = i ∫_{−1}^{1} x(1 − x^{2})^{−1/2} sin ζx dx,
μ_{1} = ∫_{−1}^{1} x^{2}(1 − x^{2})^{−1/2} cos ζx dx,
μ_{2} = i ∫_{−1}^{1} x^{3}(1 − x^{2})^{−1/2} sin ζx dx,

The series expansion for the function e^{iζx} is valid for |ζx| < +∞, and it converges uniformly everywhere in the complex plane (see [6]). This gives an opportunity to integrate term by term (see [13]), so that we have

μ_{0} = ∑_{k=0}^{+∞} \frac{(iζ)^{k}}{k!} ∫_{−1}^{1} \frac{x^{k+1}}{\sqrt{1 − x^{2}}} dx = \sqrt{π} ∑_{k=0}^{+∞} \frac{(iζ)^{k}}{k!} \frac{1 + (−1)^{k+1}}{2} \frac{Γ((k + 2)/2)}{Γ((k + 3)/2)}

= iπ ∑_{k=0}^{+∞}(−1)^{k} \frac{ζ^{2k+1}}{(2k + 1)!} \frac{(2k + 1)!!}{(2k + 2)!!} = π ∑_{k=0}^{+∞}(−1)^{k} \frac{(ζ/2)^{2k+1}}{k!(k + 1)!} = iπJ_{1}(ζ),

where J_{1}(ζ) is the Bessel function of the order one. Here, we used the known expressions for the moments of the Chebyshev weight of the first kind given in [7] and the series expansion for the Bessel function given in (3). Also, we used the simple fact that

\sqrt{π} \frac{1 + (−1)^{k+1}}{2} \frac{Γ((k + 2)/2)}{Γ((k + 3)/2)} = π \frac{(2n + 1)!!}{(2n + 2)!!}, k = 2n + 1,

which can be verified by the induction argument. Using the same method we derive the corresponding expression for the moments μ_{1} i μ_{2}. □

To explore further the moment sequence we adopt the following notation

μ_{k} = \frac{iπ}{(iζ)^{k}} (P_{k}J_{1} + ζQ_{k}J_{0}), k ∈ ℕ_{0}.  (6)

We have the following statement.

Theorem 3. The polynomials P_{k} and Q_{k} in ζ^{2} with integer coefficients of degrees 2[k/2] and 2[(k − 1)/2], respectively, satisfy the following recurrence relation

y_{k+2} = −(k + 2)y_{k+1} − ζ^{2}y_{k} − (k + 1)ζ^{2}y_{k−1},

with initial conditions

P_{0} = 1, P_{1} = −1, P_{2} = 2 − ζ^{2},
Q_{0} = 0, Q_{1} = 1, Q_{2} = −1.

The term with ζ^{0} in P_{k} is equal to (−1)^{k}k!, k ∈ ℕ_{0}.

P r o o f: Putting (6) into the recurrence relation for the moments (4), we get

\frac{iπ}{(iζ)^{k+2}} (P_{k+2}J_{1} + ζQ_{k+2}J_{0}) = −\frac{iπ(k + 2)}{(iζ)^{k+2}}(P_{k+1}J_{1} + ζQ_{k+1}J_{0})

−\frac{iπζ^{2}}{(iζ)^{k+2}} (P_{k}J_{1} + ζQ_{k}J_{0}) − \frac{(k + 1)iπζ^{2}}{(iζ)^{k+2}} (P_{k−1}J_{1} + ζQ_{k−1}J_{0}).

Since the equation is valid for all real ζ ≠ 0 and the functions J_{0} and J_{1} are linearly independent, it is just enough to read term with the functions J_{0} and J_{1} to obtain the recurrence relation stated.

The initial conditions for the moments give P_{0} = 1, P_{1} = −1, P_{2} = 2−ζ^{2} and Q_{0} = 0, Q_{1} = 1, Q_{2} = −1, for polynomials P_{k} and Q_{k}, respectively.

Now, obviously P_{0}, P_{1} and P_{2} are real polynomials with integer coefficients. According to the recurrence, P_{3} is also real polynomial in ζ and has degree 2, i.e.,

P_{3} = −(1 + 2)P_{2} − ζ^{2}P_{1} − ζ^{2}P_{0} = −6 + 2ζ^{2}.

Suppose that P_{k−1}, P_{k} and P_{k+1} are real polynomials in ζ, with degrees 2[(k−1)/2], 2[k/2] and 2[(k+1)/2], respectively. Then, using the recurrence relation

P_{k+2} = −(k + 2)P_{k+1} − ζ^{2}P_{k} − (k + 1)ζ^{2}P_{k−1},

we deduce the following recurrence

A_{2n+2} = −A_{2n}, A_{2n+3} = −(2n + 3)A_{2n+2} − A_{2n+1} − (2n + 2)A_{2n}, n ∈ ℕ_{0},

for the leading coefficient in P_{k}, k ∈ ℕ_{0}, with initial conditions A_{0} = 1, A_{1} = −1, A_{2} = −1. It can be checked by a direct calculation that we have the solution A_{2n} = (−1)^{n}, A_{2n+1} = (−1)^{n+1}(n + 1), n ∈ ℕ_{0}.

For the Q-sequence we have

Q_{0} = 0, Q_{1} = 1, Q_{2} = −1, Q_{3} = 3 − ζ^{2}.

We can easily verify the degrees of these initial polynomials. Suppose that Q_{k−1}, Q_{k} and Q_{k+1} are polynomials with degrees 2[(k − 2)/2], 2[(k − 1)/2] and 2[k/2], respectively. Using the recurrence for these polynomials, we can obtain the corresponding recurrence for their leading coefficients. Namely, we have

A_{2n+3} = −A_{2n+1}, A_{2n+2} = −(2n+2)A_{2n+1}−A_{2n}−(2n+1)A_{2n−1}, n ∈ ℕ_{0},

with initial conditions A_{0} = 0, A_{1} = 1, A_{2} = −1. It can be checked directly that the solution is A_{2n+1} = (−1)^{n} and A_{2n} = (−1)^{n+1}n, n ∈ ℕ_{0}.

To prove the statement on the coefficient A_{k} with ζ^{0} in P_{k}, k ∈ ℕ_{0}, we use the initial conditions A_{0} = 1, A_{1} = −1, and A_{2} = 2. Then, by the recurrence relation, we obtain

A_{k+2} = −(k + 2)A_{k+1}, k ∈ ℕ,

so that we conclude easily A_{k} = (−1)^{k}k!, k ∈ ℕ_{0}. □

Can we say anything about the existence of orthogonal polynomials? To illustrate the problem, we can calculate the Hankel determinant Δ_{2} in the form

Δ_{2} = \frac{π^{2}J_{1}^{2}(ζ)}{ζ^{2}} (−ζ^{2} \frac{J_{0}^{2}(ζ)}{J_{1}^{2}(ζ)} + ζ \frac{J_{0}(ζ)}{J_{1}(ζ)} + 1 − ζ^{2}) .

It is easy to conclude that Δ_{2} = 0, provided

\frac{J_{0}(ζ)}{J_{1}(ζ)} = \frac{1 ± \sqrt{5 − 4ζ^{2}}}{2ζ} ,

but any solution must be real so it must be |ζ| < \sqrt{5}/2. A careful numerical inspection shows that we cannot find solution for this equation, and it seems that it does not exist in the set of real numbers.

However, this is not the case with Δ_{3}. Using some computer algebra, it can be checked easily that

Δ_{3} = \frac{iπ^{3}J_{1}^{3}}{ζ^{6}} (7ζ^{3} \frac{J_{0}^{3}}{J_{1}^{3}} + (2ζ^{2} − 21)ζ^{2} \frac{J_{0}^{2}}{J_{1}^{2}} + ζ(5ζ^{2} + 12)\frac{J_{0}}{J_{1}} + 2ζ^{4} − 15ζ^{2} + 4) .

The smallest positive solution for the equation Δ_{3} = 0 is given by

ζ = 6.459008151994783455531721397032502543805710669120882 …,

so that for this particular number, the sequence of orthogonal polynomials does not exist.

However, there is the way to ensure the existence of orthogonal polynomials. Choose ζ to be any positive zero of the Bessel function J_{0}(ζ). Then our sequence of moments becomes

μ_{k} = \frac{iπ}{(iζ)^{k}} P_{k}J_{1}(ζ).

Because of the interlacing property of the positive zeros of the Bessel functions (see [14, p. 479]), we know that J_{1}(ζ) ≠ 0.

Theorem 4. Suppose ζ is a positive zero of the Bessel function J_{0}. Then the sequence of polynomials orthogonal with respect to the functional L, given by (2), exists.

P r o o f: We give the proof of this statement using the fact that all zeros of the Bessel function J_{0} are transcendental numbers [10] (see also [11], [4], [12]). Our sequence of moments is given by

μ_{k} = \frac{iπ}{(iζ)^{k}} P_{k}J_{1}(ζ), k ∈ ℕ_{0},

where we know the basic properties of the polynomials P_{k}, k ∈ ℕ_{0}, stated in Theorem 3.

Consider the Hankel determinants Δ_{k}, k ∈ ℕ, given in (1). We have to prove that the determinants Δ_{k} ≠ 0, k ∈ ℕ_{0}. Then, according to Theorem 1, the sequence of orthogonal polynomials exists. Consider determinant Δ_{k} and extract from every of its rows the factor iπJ_{1}(ζ). Denoting the obtained determinant with Δ′_{k}, we have

Δ_{k} = (iπJ_{1}(ζ))^{k+1}Δ′_{k}.

Now, we can consider the determinant Δ′_{k} as a Hankel determinant for the sequence of moments μ′_{ν} = (iζ)^{−ν}P_{ν}, ν ∈ ℕ_{0}. If we now extract from every ν-th row the factor 1/(iζ)^{ν−1} and after that from j-th column the factor 1/(iζ)^{j−1}, we obtain new determinant Δ′′_{k} and equality

Δ_{k} = (iπJ_{1}(ζ))^{k+1}Δ′_{k} = \frac{(iπJ_{1}(ζ))^{k+1}}{(iζ)^{k(k+1)}} Δ′′_{k}.

The determinant Δ′′_{k} is the Hankel determinant for the sequence of moments μ′′_{ν} = P_{ν}, ν ∈ ℕ_{0}. Hence, the value of Δ′′_{k} is certain polynomial in ζ^{2}, since all its elements are polynomials in ζ^{2}. Since ζ is transcendental number, the polynomial with integer coefficients Δ′′_{k} cannot be zero at ζ, because all its zeros must be algebraic numbers. There is only one possibility for Δ′′_{k} to have ζ as its zero, if the polynomial Δ′′_{k} is identically zero.

Thus, we have to prove that Δ′′_{k} is not identically zero. Since Δ′′_{k} is a polynomial in ζ and all its coefficients are polynomials in ζ, the term with ζ^{0} of the polynomial Δ′′_{k} equals the Hankel determinant which elements are terms with ζ^{0} in the polynomials P_{ν}, ν ∈ ℕ_{0}. According to Theorem 3, we know that P_{ν}(0) = (−1)^{ν}ν!, ν ∈ ℕ_{0}, so that the term with ζ^{0} in the polynomial Δ′′_{k} equals the Hankel determinant Δ̂_{k} for the sequence of moments μ̂_{ν} = (−1)^{ν}ν!, ν ∈ ℕ_{0}.

If we extract −1 from the rows 2ν + 1, ν = 0, 1, …, 2[k/2], and from the columns 2j + 1, j = 0, 1, …, 2[k/2], we get the Hankel determinant Δ̃_{k} for the sequence of moments μ̃_{ν} = ν!, ν ∈ ℕ_{0}, where the following equation holds

Δ̂_{k} = Δ̃_{k}.

Now, it is easy to recognize the sequence of moments μ̃_{ν} = ν!, as the sequence of moments for the Laguerre measure (see [7]). But, then it is easy to compute Δ̃_{k},

Δ̃_{k} = Δ̂_{k} = ∏_{ν=0}^{k}(ν!)^{2}.

This means that Δ′′_{k} is not a polynomial which is identically equal to zero. Hence Δ′′_{k} ≠ 0, which implies that Δ_{k} ≠ 0. The previous discussion is valid for any k ∈ ℕ, which means that Δ_{k} ≠ 0, k ∈ ℕ. □

In the rest of this paper we denote by p_{n}, n ∈ ℕ_{0}, the orthonormal sequence of polynomials with respect to the linear functional L, and by π_{n}, n ∈ ℕ_{0}, we denote the monic version of this polynomial sequence.

2. Asymptotic formulae

First we prove one auxiliary result, explaining the asymptotic properties of polynomials Q_{n} orthogonal with respect to the weight function w, defined by w(x) = χ_{[−1,1]}(x)(1 − x^{2})^{−1/2}e^{iζx}.

Since this weight function is in Magnus class of the complex weight functions (see [5]), the polynomials Q_{n}, n ∈ ℕ_{0}, exist asymptotically. Also, their three-term recurrence coefficients have the asymptotic behavior of the class M(0, 1) introduced and studied in [9]. Denote the three-term recurrence coefficients for the corresponding orthonormal polynomials by α_{n}^{Q} and β_{n}^{Q}. Then we have

lim_{n→+∞} α_{n}^{Q} = 0, lim_{n→+∞} β_{n}^{Q} = \frac{1}{2}.

Actually, using some recent results, we know even more.

Lemma 1. For the monic polynomial Q_{n}, n ∈ ℕ_{0}, orthogonal with respect to the weight function ω(x) = χ_{[−1,1]}(x)(1 − x^{2})^{−1/2} exp(iζx), we have the following asymptotic formula

\frac{Q_{n+1}(0)}{Q_{n}(0)} = \frac{1}{2} cosh \frac{ζ − i(n + 1)π}{2} ; cosh \frac{ζ − inπ}{2} + O(q^{n}),

where 0 < q < 1.

P r o o f. We use the following result proved in [1]. Suppose h is a complex function being analytic in some neighborhood of the interval [−1, 1], which is different from zero on the interval [−1, 1]. Then, the monic orthogonal polynomials Q_{n} with respect to h(x)χ_{[−1,1]}(x)(1 − x^{2})^{−1/2} exist asymptotically, and

γ_{n}Q_{n}(x) = φ_{+}(x) + φ_{−}(x) + O(q^{n}), x ∈ [−1, 1], 0 < q < 1,

where

γ_{n}^{−1} = 2^{−n} exp (\frac{1}{2π} ∫_{−1}^{1} \frac{log h(x)}{(1 − x^{2})^{−1/2}} dx)

and

φ(z) = (z + (z^{2} − 1)^{1/2})^{n} exp (−\frac{1}{2π} (z^{2} − 1)^{1/2} ∫_{−1}^{1} \frac{log h(x)}{z − x} \frac{dx}{(1 − x^{2})^{1/2}} ) .

Here, φ_{+}(x) is the limit of φ(z) as z approaches x ∈ [−1, 1] over the upper half-plane of the complex z plane and φ_{−}(x) is the limit of φ(z) as z approaches x ∈ [−1, 1] from the lower half-plane of the complex z plane. The square root is chosen such that it has cut along the interval [−1, 1] and it behaves as z as z approaches ∞.

We use x = cos θ ∈ [−1, 1], θ ∈ [0, π]. First, we calculate the integral which appears in the φ function for Im(z) > 0. So we have

∫_{−1}^{1} \frac{iζx}{z − x} \frac{dx}{(1 − x^{2})^{1/2}} = −iπζ + iζz ∫_{−1}^{1} \frac{1}{z − x} \frac{dx}{(1 − x^{2})^{1/2}}
= −iπζ + iζ ∫_{−1}^{1} \frac{dx}{(1 − x^{2})^{1/2}} ∑_{k=0}^{+∞} \frac{x^{k}}{z^{k}}

= −iπζ + iζ ∑_{k=0}^{+∞} \frac{m_{k}^{C}}{z^{k}}
= −iπζ (1 − \frac{1}{(1 − z^{−2})^{1/2}} ) ,

where m_{k}^{C} are the moments for Chebyshev weight of the first kind. This gives

φ_{+}(x) = e^{inθ} exp (−\frac{ζ}{2}(sin θ + i cos θ))

and

φ_{−}(x) = e^{−inθ} exp (\frac{ζ}{2}(sin θ − i cos θ)) .

Using the mentioned result, we can calculate directly

γ_{n}Q_{n}(x) = e^{inθ} exp(−\frac{ζ}{2}(sin θ + ix)) + e^{−inθ} exp(\frac{ζ}{2}(sin θ − ix)) + O(q^{n})

and also, for θ = π/2, we have

\frac{Q_{n+1}(0)}{Q_{n}(0)} = \frac{1}{2} cosh \frac{ζ − i(n + 1)π}{2} ; cosh \frac{ζ − inπ}{2} + O(q^{n}). □

Using the polynomials Q_{n}, n ∈ ℕ_{0}, we can express the polynomials π_{n}, n ∈ ℕ_{0}, and the corresponding three-term recurrence coefficients. Thus, we have the following statement:

Theorem 5. Suppose the sequence of orthogonal polynomials Q_{n} exists for n > N_{Q}. Then, for polynomials orthogonal with respect to the functional L given by (2), we have

π_{n}(x) = \frac{1}{x} (Q_{n+1}(x) − γ_{n}Q_{n}(x)) , n > N_{Q},  (7)

where

γ_{n} = \frac{Q_{n+1}(0)}{Q_{n}(0)}, n > N_{Q}.

The recursion coefficients for the sequence π_{n}, n ∈ ℕ_{0}, in the recurrence relation can be expressed in the following form

α_{n} = −γ_{n} − \frac{(β_{n+1}^{Q})^{2}}{γ_{n}} , β_{n+1}^{2} = \frac{γ_{n+1}}{γ_{n}} (β_{n+1}^{Q})^{2}, n > N_{Q}.

P r o o f. For the (monic) orthogonal polynomials Q_{k}, z ∈ ℂ, Q_{k}(z) ≠ 0, k ∈ ℕ, the polynomials

π_{n}(x; z) = \frac{1}{x − z} (Q_{n+1}(x) − \frac{Q_{n+1}(z)}{Q_{n}(z)} Q_{n}(x))

are known as the kernel polynomials (cf. [3]). Several results are known in the case when the point z is not in the supporting set of the measure of orthogonality and provided the sequence Q_{n} exists. In our case z = 0 and, therefore, we give the proof here.

Thus, supposing that the sequence of polynomials is given by (7), we have

∫_{−1}^{1} x^{ν}π_{n}(x)\frac{xe^{iζx}}{\sqrt{1 − x^{2}}} dx = ∫_{−1}^{1} x^{ν}(Q_{n+1}(x) − γ_{n}Q_{n}(x))\frac{e^{iζx}}{\sqrt{1 − x^{2}}} dx = 0,

provided n > ν. According to the uniqueness property, up to a multiplicative constant, the polynomials (7) are orthogonal with respect to the weight function x(1 − x^{2})^{−1/2} exp(iζx)χ_{[−1,1]}. We note that γ_{n} = Q_{n+1}(0)/Q_{n}(0).

If we assume that polynomials Q_{n}, n > N_{Q}, satisfy the following threeterm recurrence relation

Q_{n+1}(x) = (x − α_{n}^{Q})Q_{n}(x) − (β_{n}^{Q})^{2}Q_{n−1}(x), n > N_{Q} + 1,

it is clear that

γ_{n+1} = −α_{n+1}^{Q} − \frac{(β_{n+1}^{Q})^{2}}{γ_{n}} , n > N_{Q}.

Also, for polynomials π_{n}, n > N_{Q}, we have

xπ_{n+1} = Q_{n+2} − γ_{n+1}Q_{n+1}
= (x − α_{n+1}^{Q})Q_{n+1} − (β_{n+1}^{Q})^{2}Q_{n} − (−α_{n+1}^{Q} − \frac{(β_{n+1}^{Q})^{2}}{γ_{n}} )Q_{n+1}
= (x + γ_{n} + \frac{(β_{n+1}^{Q})^{2}}{γ_{n}} )Q_{n+1} − γ_{n}Q_{n+1} − (β_{n+1}^{Q})^{2}Q_{n}
= (x − α_{n})xπ_{n} + γ_{n}(β_{n}^{Q})^{2}Q_{n−1}

+(−γ_{n}(x − α_{n}^{Q}) − (β_{n+1}^{Q})^{2} + γ_{n}(x + γ_{n} + \frac{(β_{n+1}^{Q})^{2}}{γ_{n}} ))Q_{n}

= (x − α_{n})xπ_{n} + γ_{n}(β_{n}^{Q})^{2}Q_{n−1} + γ_{n} (α_{n}^{Q} + γ_{n}) Q_{n}
= (x − α_{n})xπ_{n} + γ_{n}(β_{n}^{Q})^{2}Q_{n−1} − \frac{γ_{n}}{γ_{n−1}} (β_{n}^{Q})^{2}Q_{n}
= (x − α_{n})xπ_{n} − \frac{γ_{n}}{γ_{n−1}} (β_{n}^{Q})^{2}xπ_{n−1},

wherefrom we read directly the corresponding expressions for the three-term recurrence coefficients. □

Theorem 6. For the three-term recurrence coefficients of the polynomial sequence p_{n}, n ∈ ℕ_{0}, we have the following asymptotic formulae

α_{2n+k} → \frac{i(−1)^{k}}{2} (tanh \frac{ζ}{2} − coth \frac{ζ}{2}) , n → +∞, k = 0, 1,

and

β_{2n+k}^{2} → \frac{1}{4} (tanh^{2} \frac{ζ}{2})^{(−1)^{k}} , n → +∞, k = 0, 1.

P r o o f. This theorem is a direct consequence of Lemma 1 and Theorem 5. □

Theorem 7. Let J be the associated Jacobi operator created using threeterm recurrence coefficients α_{k} and β_{k}, k ∈ ℕ_{0}. Then, we have

σ_{ess}(J) = [−1, 1].

P r o o f. According to Theorem presented in [2], we know that the essential spectrum of the Jacobi matrix, with periodic three-term recurrence coefficients of the basic period m, can be obtained as an inverse image of the interval [−2, 2] of the mapping

h(x) = \frac{p_{2m−1}}{p_{m−1}} .

In our case we have m = 2. Suppose that we give the sequences of threeterm recurrence coefficients α_{2n+k} = a_{k}, β_{2n+k} = b_{k}, k = 0, 1. Then, we can express h(x) as

b_{0}b_{1}h(x) = x^{2} − (a_{0} + a_{1})x + a_{0}a_{1} − b_{0}^{2} − b_{1}^{2}.

For our case

a_{0} = \frac{i}{2} (tanh \frac{ζ}{2} − coth \frac{ζ}{2}) = −a_{1} and b_{0} = \frac{1}{2} tanh \frac{ζ}{2} = \frac{1}{4b_{1}} .

We get easily h(x) = 4(x^{2} − 1/2). The inverse mappings of h are the mappings

h_{1}^{−1} = \frac{(h + 2)^{1/2}}{2} , h_{2}^{−1} = −\frac{(h + 2)^{1/2}}{2} ,

respectively.

If we let h change in [−2, 2] we get as a result exactly [−1, 1]. □

Acknowledgments. The authors would like to thank Professor Richard Askey, University of Wisconsin–Madison, WI, and Professor Michel Waldschmidt, Université Pierre et Marie Curie (Paris VI), for references on transcendentality of zeros of Bessel functions of a rational order.

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