MSC 2010: 33C45, 40G05Series in Hermite polynomials with poles on the boundaries of their regions of convergence are considered

Boychev, Georgi S.

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M a t h e m a t i c aB a l k a n i c a—————————NewSeries Vol. 26, 2012, Fasc. 1-2Hermite Series with Polar SingularitiesGeorgi S. BoychevPresented at 6th International Conference “TMSF’ 2011”Series in Hermite polynomials with poles on the boundaries of their regions of conver-gence are considered.MSC 2010: 33C45, 40G05Key Words: Hermite polynomials, Hermite series, poles1. IntroductionDefinition 1. The polynomials {Hn(z)}+∞n=0 defined by equalitiesHn(z) = (−1)n exp(z2) dndzn{exp(−z2)}, n = 0, 1, 2, . . . ; z ∈ C,where C is the complex plane, are called Hermite polynomials.Let λ(z) =√2 exp(z2/2) and cn(z) = (2n/e)n/2 cos[(2n+1)1/2z−npi/2].Then the asymptotic formula for the Hermite polynomials (n ≥ 1) [1, ChapterIII, (2.2)] yield thatHn(z) = λ(z)cn(z){1 + hn(z)}, (1.1)where {hn(z)}+∞n=1 are holomorphic functions in the open set G = C\(−∞, +∞)andhn(z) = O(n−1/2) (n→ +∞)uniformly on every compact subset of G.Definition 2. The series of the kind+∞∑n=0anHn(z), (1.2)we shall call Hermite series.Let 0 < τ < +∞ and define S(τ) = {z ∈ C : |=z| < τ} and S∗(τ) =C\S(τ) . We assume S(0) = ∅, S(∞) = C, S∗(0) = C\R and S∗(∞) = ∅. Then:50 G. BoychevTheorem 1. Ifτ = max{0,− limn→+∞ sup (2n+ 1)−1/2 log | (2n/e)n/2an|} (1.3)then the series (1.2) is absolutely uniformly convergent on every compact subsetof the set S(τ) and diverges in S∗(τ) ([1], Theorem IV.3.1, b)).Remark 1. The equality (1.3) can be regarded as a formula of Cauchy-Hadamard type for the series of kind (1.2).Remark 2. In the proof of Theorem 1 it is used the asymptotic formula(1.1).In [2] we prove the followingTheorem 2. Let z0 ∈ G and anHn(z0) = O(np)(n → +∞), wherep ≥ −1. Then the series (1.2) is absolutely convergent in the strip S(|=z0|).Remark 3. If the conditions of Theorem 2 are satisfied, then theseries (1.2) is absolutely uniformly convergent on every compact subset of thestrip S(|=z0|) and the sum of this series is a complex function holomorphic inS(|=z0|).2. The main resultThe basic result is given by the followingTheorem 3. Let z0 ∈ G and anHn(z0) = o(np)(n → +∞), where p isa nonnegative integer. If f(z) is the sum of the series (1.2) in the strip S(τ0)with τ0 = |=z0| and f(z) has a pole of order m on ∂S(τ0), then m ≤ 2p+ 1.P r o o f. Suppose that there is a point ζ ∈ ∂S(τ0) such that the functionf(z) has a pole at ζ of order m and m > 2p+ 1. Thenm ≥ 2p+ 2. (2.1)Since the Hermite polynomials have no zeros outside real line, we canwrite thatg(z) = (z − ζ)mf(z) = (z − ζ)m+∞∑n=0anHn(z0)Hn(z)Hn(z0),where z ∈ S(τ0)\(−∞, +∞). Hence,|g(z)|≤|(z−ζ)mN∑n=0anHn(z)|+|(z−ζ)m+∞∑n=N+1anHn(ζ)Hn(z)Hn(ζ)|=gN,1(z)+gN,2(z)whatever the non-negative integer N can be.Let ε > 0. Then it follows that exists a positive N0 such thatHermite Series with . . . 51|anHn(z0)| < εnpfor n > N0. Then for N > N0 we obtain thatgN,2(z) ≤ ε|z − ζ|m+∞∑n=N+1np| Hn(z)Hn(z0)|. (2.2)Let max(0, τ0 − 1) < δ < τ0 and D(ζ; δ) = {z ∈ C : <z = <ζ} ∩ {z ∈S(τ0) − (−∞, +∞) : |=z| ≥ δ} . Since D(ζ; δ) is a compact subset of G, theasymptotic formula (1.1) yields thatHn(z)Hn(z0)= O{exp(−√2n+ 1(τ0 − |=z|)}, n→ +∞,uniformly on D(ζ; δ). Then,+∞∑n=N+1np| Hn(z)Hn(z0)| = O{+∞∑n=N+1np exp[−√2n+ 1(τ0 − |=z|)]}= O(∫ +∞1tp exp[−(τ0 − |=z|)√2t+ 1]dt).It is not difficult to prove that∫ +∞1tp exp[−(τ0 − |=z|)√2t+ 1]dt ≤M(τo − |=z|)−2p−2,where M is a constant not depending of N. Hence,+∞∑n=N+1np| Hn(z)Hn(z0)| ≤ K(τo − |=z|)−2p−2, z ∈ D(ζ; δ),where K is a constant not depending of N. Then from (2.2) it follows thatgN,2(z) ≤ εK|z − ζ|m(τ0 − |=z|)−2p−2.Obviously, |z − ζ|m = (τ0 − |=z|)m for z ∈ D(ζ; δ). Using (2.1) and theinequality τ0 − |=z| < 1, we obtain thatgN,2(z) ≤ εK (2.3)for each N > N0 and z ∈ D(ζ; δ). Let such N be fixed, then there exists apositive constant L such thatgN,1(z) ≤ L(τ0 − |=z|)m,for z ∈ D(ζ; δ). Moreover let (τ0 − |=z|)m < ε. ThengN,1(z) ≤ εL. (2.4)From the inequalities (2.3) and (2.4) it follows that52 G. Boychev|g(z)| ≤ ε(K + L)for z ∈ D(ζ; δ) and sufficiently closed to ζ. This means thatlimz→ζ(z − ζ)mf(z) = 0, z ∈ D(ζ; δ).However, this contradicts the assumption that the function f(z) has apole of order m at the point ζ. This completes the proof of Theorem 3.Corollary. Let z0 ∈ G, limn→+∞ anHn(z0) = 0 and f(z) is the sum of(1.2) in the strip S(τ0), where τ0 = |=z0|. If the function f(z) has a pole on∂S(τ0), then it is a simple pole.Finally we shall note that the following assertion holds:Theorem 4. Let z0 ∈ G and anHn(z0) = O(np)(n→ +∞), where p isa nonnegative integer. If f(z) is the sum of the series (1.2) in the strip S(τ0)with τ0 = |=z0| and f(z) has a pole of order m on ∂S(τ0), then m ≤ 2p+ 2.Acknowledgements. The author wishes to thank Professor Peter Ru-sev for his most helpful comments on the preparation of this paper.References[1] P. Rusev, Classical Orthogonal Polynomials and Their Associated Functionsin Complex Plane. Marin Drinov Acad. Publ. House, Sofia (2005).[2] G. Boychev, On Hermite series in complex plane. Compt. rend. Acad. bulg.Sci. 61, No 6 (2008), 689-694.Trakia UniversityStara Zagora – 6000, BULGARIAe-mail: GBoychev@hotmail.com Received: October 21, 2011

Hermite Series with Polar Singularities

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Hermite Series with Polar Singularities

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