Unique solvability of the strong Hamburger moment problem

Njåstad, Olav; Thron, W. J.

doi:10.1017/s144678870002646x

Cited by 25 publications

(29 citation statements)

References 14 publications

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“…Set, as in Proposition 3.31 of [57], 11 (N e (z)) 12 ( e m ∞ (z)) 12 (N e (z)) 21 ( e m ∞ (z)) 21 (N e (z)) 22 = {z ′ ∈ C; (γ e (z)±(γ e (z)) −1 )| z=z ′ = 0}, one notes that the (simple) poles of ( e m ∞ (z)) 11 and ( e m ∞ (z)) 22 (resp., ( e m ∞ (z)) 12 and (…”

Section: Following Lemma 321 Of [57] Setmentioning

confidence: 99%

“…Unlike the moment theory for the SMP and the HMP, wherein the theory of orthogonal polynomials, and the analytic theory of continued fractions, enjoyed a prominent rôle, the extension of the moment theory to the SSMP and the SHMP introduced a 'rational generalisation' of the orthogonal polynomials, namely, the orthogonal Laurent (or L-) polynomials (as well as the introduction of special kinds of continued fractions commonly referred to as positive-T fractions), which are discussed below [10][11][12][13][14][15][16][17][18][19][20][21]. (The SHMP can also be solved using the spectral theory of unbounded self-adjoint operators in Hilbert space [22]; see, also, [23].…”

Section: Introductionmentioning

confidence: 99%

“…It is a fact [10,11,15,17] (on R) with the given (real) moments. For the case of the SSMP, there are four (Hankel determinant) inequalities (in this latter case, the c k , k ∈ Z, which appear in Equations (1.1) should be replaced by c SSMP k , k ∈ Z) which guarantee the existence of a non-negative measure µ SS MP (on [0, +∞)) with the given moments, namely [8] (see, also, [10,11] The φ n 's are normalised so that they all have real coefficients; in particular, the leading coefficients, LC(φ 2n ) := ξ [12,17]; see, also, [25]): φ 2m (ζ)(zφ 2m−1 (z)−ζφ 2m−1 (ζ))−ζφ 2m−1 (ζ)(φ 2m (z)−φ 2m (ζ)) = (z−ζ) ξ moreover, it can be shown that (see, for example, [15,17]), for n ∈ Z For each m ∈ Z + 0 , let µ 2m := card{z; π π π 2m (z) = 0} and µ 2m+1 := card{z; π π π 2m+1 (z) = 0}.…”

Section: Introductionmentioning

confidence: 99%

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Asymptotics of Laurent polynomials of even degree orthogonal with respect to varying exponential weights

McLaughlin

Vartanian

Zhou

2010

International Mathematics Research Papers

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Let Λ R denote the linear space over R spanned by z k , k ∈ Z. Define the real inner product (with varying exponential weights). . } with respect to · · ·, · · · L yields the even degree and odd degree orthonormal Laurent polynomialsn > 0, and φ 2n+1 (z) = ξ n , φ 2n (z) (in the entire complex plane), and Hankel determinant ratios associated with the real-valued, bi-infinite, strong moment sequenceare obtained by formulating the even degree monic orthogonal Laurent polynomial problem as a matrix Riemann-Hilbert problem on R, and then extracting the largen behaviour by applying the non-linear steepest-descent method introduced in [1] and further developed in [2,3].

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Section: Following Lemma 321 Of [57] Setmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

See 1 more Smart Citation

Asymptotics of Laurent polynomials of even degree orthogonal with respect to varying exponential weights

McLaughlin

Vartanian

Zhou

2010

International Mathematics Research Papers

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“…This problem is equivalent to the strong Hamburger moment problem. For definition and general treatment of this problem, see, e.g., [10,11,25]. The connection between the strong moment problem and the asymptotic expansions problem was treated in [10] for the case that the moment problem is nonsingular, and in [20][21][22][23] for the general case.…”

Section: Introductionmentioning

confidence: 99%

A Modified Schur Algorithm and an Extended Hamburger Moment Problem

Njåstad¹

1991

Transactions of the American Mathematical Society

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Abstract. An algorithm for a Pick-Nevanlinna problem where the interpolation points coalesce into a finite set of points on the real line is introduced, its connection with certain multipoint Padé approximation problems is discussed, and the results are used to obtain the solutions of an extended Hamburger moment problem. IntroductionThe Pick-Nevanlinna problem is the following: Let {za : a £ A} be a set of distinct points in the open upper half-plane H+ = {z : Imz > 0}, and let {wa : a £ A} be a set of points in C. Find a function F(z) which is analytic in //° such that ImF(z) > 0 for z e //J and F(z) = wa for all a £ A.(A function F(z) which is analytic for z £ HQ+ with ImF(z) > 0 is called a Nevanlinna function.) The problem was solved by Pick [27,28] in the case that A is finite, by Nevanlinna [15,16] in the case that A is countable, and by Krein and Rekhtman [14] in the general case.A variant of the problem for finite or countable A arises when all points za coalesce to a single point a , and given values wa at the points are replaced by the Taylor coefficients at a. A problem closely related to this is Carathéodory's coefficient problem (see [2,3,33]): Given a finite sequence {y0, ... , ym} or an infinite sequence {yn : n £ N}, find a function F(z) which is analytic in the open unit disc D° = {z : \z\ < 1} such that Re7(z) > 0 for z £ D° and F(z) = EZU?*2* + EZm+\ôkm)zk. or F(z) = Er=o^z¿-(For historical remarks on this problem, see [12].)By introducing the linear fractional transformation 7 -► j=£ we can reformulate the requirement Re7(z) > 0 to read |7(z)| < 1. (A function F(z) which is analytic for z £ D is called a Carathéodory function if Re F(z) > 0 for z € D°, or a Schur function if |7(z)| < 1 for z E 7J°.) Schur [31] invented an algorithm called the Schur algorithm to deal with this problem. The technique was adapted by Nevanlinna to deal with the Pick-Nevanlinna problem.

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“…Sri Ranga [2] For any strong distribution, di//(t), the relations (i) Q n {z) is a monic polynomial of degree n, for n > 0, In the three examples of strong symmetric distributions given in [4], namely, the strong Tchebycheff, the strong Legendre and the strong Hermite distributions, we note that the coefficients a n and P n of the associated recurrence relations satisfy, in addition to the expected condition fi n = 0, n > 1, the condition that a 2n are constant for all n > 1. We show here that this is due to these distributions being members of a "special" class of strong symmetric distributions.…”

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confidence: 99%