On the transition of Charlier polynomials to the Hermite function

It has been known for over 70 years that there is an asymptotic transition of Charlier polynomials to Hermite polynomials. This transition, which is still presented in its classical form in modern reference works, is valid if and only if a certain parameter is integer. In this light, it is surprising that a much more powerful transition exists from Charlier polynomials to the Hermite function, valid for any real value of the parameter. This greatly strengthens the asymptotic connections between Charlier polynomials and special functions, with applications for instance in queueing theory. It is shown in this paper that the convergence is uniform in bounded intervals, and a sharp rate bound is proved. It is also shown that there is a transition of derivatives of Charlier polynomials to the derivative of the Hermite function, again with a sharp rate bound. Finally, it is proved that zeros of Charlier polynomials converge to zeros of the Hermite function. While rigorous, the proofs use only elementary techniques.


Introduction
A unique feature of Charlier polynomials [3,8,19,18,14] is their affinity with the Poisson distribution. This has many important applications. Charlier polynomials concisely express the behaviour of Erlang loss systems, a fundamental concept in queueing theory [10,11,12]. Another example is the generalization of stochastic integrals over Poisson distributions to multiple stochastic integrals, which can be effectively computed using Charlier polynomials [7,20], while a third example is that of random matrices over Poisson distributions [13], which can be characterized by Charlier polynomial zeros.
High-dimensional or asymptotic problems typically engage Charlier polynomials of high degree and order (index). For instance, the asymptotic behavior in the number of servers of Erlang loss systems is described by Charlier polynomials whose degree and order tend to infinity simultaneously according to the Halfin-Whitt regime [9]. At a first glance, the classical formula ( [14,Eq. 9.14.12], [17, p. 532], [18,Eq. 18.21.9], [19,Eq. 2 appears useful for reducing Charlier polynomials in this limit, but unfortunately, this formula holds only for non-negative integer n. In light of the long standing of this formula, it can somewhat surprisingly be shown that for any real x and ν, a much stronger statement ( fig.1). Here, the ceiling function ⌈x⌉ denotes the smallest integer not smaller than x, and H ν (x) denotes the Hermite function [16,Ch. 10]. A proof of (1) has been given via Krawtchouk polynomials [19, pp. 36-37]. When ν is non-negative real and a − x √ 2a is integer, pointwise convergence of (2) (without rate bound) follows implicitly from [4].
In section 2, it is proved rigorously that convergence to the Hermite function holds for any real ν, and that convergence is uniform for ν and x in any bounded interval. A sharp rate bound is established. The same technique is then employed in section 3 to prove that there is a similar transition of the derivative with respect to ν. Also here, a sharp rate bound is provided. These results are used in section 4 for proving that zeros of Charlier polynomials converge to zeros of the Hermite function.
Below is first a recollection of some well-known definitions and recurrence relations from [8,14,16,18] in order to make the paper self-contained. This is followed by three sections, each proving an aspect of the transition of Charlier polynomials to Hermite functions. The notation "A B" is used for "A is defined B", in order to make the introduction of new symbols more explicit. The expression "bounded ν ≤ −3" is shorthand for "ν in any bounded interval [ν 0 , −3]". We will abbreviate Charlier polynomials C n (x, a) as c a n (x), c n (x), c a n , or even c n , unless there is a risk for misunderstanding. They can be defined for positive a and non-negative integer n by [8,Eq. 10.25.4], [18,Eq. 18.20.8], − xc a n (x) = ac a n+1 (x) − (n + a) c a n (x) + nc a n−1 (x) − nc a n (x) = ac a n (x + 1) − (x + a) c a n (x) + xc a n (x − 1) as well as the backward recurrence relation [14, Eq. 9.14.8], x a c a n−1 (x − 1) = c a n−1 (x) − c a n (x) The Hermite function H ν (x) is a solution of the differential equation [16,Eq. 10.2.3] y ′′ = 2xy ′ − 2νy (7) and satisfies the three-term recurrence [16,Eq. 10.4.7] where M is the confluent hypergeometric function of the first kind. When the expression involves a gamma function of a non-positive integer argument, the expression should be interpreted by its limiting value.

Transition of Charlier polynomials
Theorem 1. For real x, ν, and positive a, where c a n (ν) are Charlier polynomials and H ν (x) is the Hermite function. The is uniform for ν and x in any bounded interval, and is sharp in the sense that there are ν and x such that the error is proportional to 1/ √ a for arbitrarily large a.
Proving asymptotic properties of Charlier polynomials is difficult, since these do not satisfy a second-order linear ordinary differential equation with respect to the independent variable [5]. However, the three-term recurrence relation (4) is a discretization of such a differential equation (7). This can be used in order to prove the theorem in the following way: It is first proven for the special case x = 0 and ν ≤ −4 (Lemma 1-5), and then generalized to arbitrary real ν (Lemma 6). After that, the scaled polynomials are shown to approximate a Cauchy polygon converging to the H ν (x) solution of the Hermite differential equation initial value problem (Lemma 7).

Convergence for x = 0 and ν ≤ −4
For notational convenience, define A ⌈a⌉ and The superscript will be left out in y a ν and c a n unless there is a risk for misunderstanding. Consider the case x = 0 and ν ≤ −4. By the definition of Charlier polynomials (3), In order to prove that lim a→∞ y ν (0) = H ν (0) = 2 ν √ π Γ 1−ν 2 using the definition of H ν (x) in (10), y ν (0) can be expressed as a sum of terms When ν is negative, these are all positive. The series is difficult to sum due to multiple levels of numerical cancellation, but can be estimated by separating the factors. Another difficulty is the changing behaviour of T k with increasing a. This problem can be remedied by defining a border between "head" and "tail" sections that increases with a properly tuned power of a.
Proof. Define the "nuisance factor" due to truncation by the ceiling function by where 0 ≤ θ < 1. For 1 ≤ k ≤ A, taking the logarithm of β p(k) and Taylor expanding, where R p ≥ 0. By re-exponentiation, On the other hand, for 1 ≤ k ≤ A/2, by comparison with a geometric series, (14), The following lemma is similar to Gautschi's inequality [15], but while the inequality is restricted to −1 ≤ ν ≤ 0, the lemma here needs to hold for arbitrarily large negative ν.
Now it is time to take on the sum (12), split in a head and tail part at index Define ∆t 1/ √ A and the function Clearly, the functions f ν (t) ( fig. 2) and are continuous and bounded for bounded ν ≤ −3 and t ≥ 0.

Lemma 3. The following relations hold for ν ≤ −3:
and Proof. According to the well-known trapezoidal rule, since f ν (t) and f ′′ ν (t) are bounded for ν ≤ −3 and t ≥ 0, and for some τ ∈ [M∆t, A∆t], implying that when z increases, Γ(s, z) approaches zero faster than any negative power of z, including 1/ √ a, i.e., This proves the first relation. For the second relation, by (18), Proof. By Lemma 1 and 2, The term R head in (15) can be computed in a similar way.
This time k < M, and by Lemma 1 and 2, Using the identity f ν (t)t n = f ν−n (t), the error term ∆S is (20), again using Lemma 3, By (12), and combining Lemma 5 and 4,

Convergence for arbitrary x and arbitrary ν
In order to prove that y ν (x) in (11) converges to the solution of the Hermite differential equation (7) having initial conditions y (0) = H ν (0) and y ′ (0) = H ′ ν (0), it can be rewritten in normal form as where y y y(x) (y (x) , y ′ (x)) T and Let r √ 2a, ∆x 1/r, and x k k∆x. Define a Cauchy polygon u u u(x) for the differential equation (25) by linear interpolation between points (x k , u u u k ), where u u u 0 = y y y(0) and Given arbitrary ξ , ψ > 0 and L 1 + 4ψ 2 + 4ξ 2 , for x in [0, ξ ] and ν in [−ψ, ψ], by the definition of the Euclidean norm,

Bounds for |A A A (x) u u u(x)| and |∂ (A
and By Theorem 2, Theorem 3, and Lemma 6, for x ∈ [0, ξ ] and ν ∈ [−ψ, ψ], which is independent of x and ν, so the Cauchy polygon (26) converges uniformly to the Hermite function when a → ∞.
Define z z z 0 u u u 0 and Let m ⌈a − x k r⌉ = a − x k r + (⌈a⌉ − a) = a − x k r + θ , where 0 ≤ θ < 1. For simplicity of notation, the argument of c m is dropped when it is ν. Consequently, Multiplying the three-term recurrence relation (4) by two, and substituting x = ν and m = n gives the identity −2νc m = 2ac m+1 − (2m + 2a) c m + 2mc m−1 Rearranging, and using the facts that 2a = r 2 and m = a − x k r + θ , This is nearly the same expression as for the Cauchy polygon (26), with only the θ -term differing. Understanding the product sign below to multiply matrices in the proper order, and I I I to denote the identity matrix, Bounding the factor I I I + ∆x A A A x j ≤ exp(Lξ ) in the same way as in (27), demonstrates that z z z converges uniformly to u u u for x ∈ [0, ξ ] and ν ∈ [−ψ, ψ]. The proof for the descending direction from x = 0 is omitted, since it is exactly analogous. By (28) and Lemma 7, where the right hand side is independent of x and ν for these parameters in any bounded interval. In order to demonstrate the sharpness of the bound, choose ν = 2, any real x, and arbitrarily large a such that n = a − x √ 2a is integer. Since c a 2 (n) = 1 − (1 + 2a) n/a 2 + n 2 /a 2 and H 2 (x) = 4x 2 − 2, This completes the proof of Theorem 1.

Transition of the derivative Theorem 4.
For real x, ν, and positive a, where c a n (ν) are Charlier polynomials and H ν (x) is the Hermite function. The

uniform for ν and x in any bounded interval, and is sharp.
The proof of this theorem uses same technique as the proof of Theorem 1, so the procedure can be abbreviated. First, the theorem is proved for the special case x = 0 and ν ≤ 5, then generalized to arbitrary ν, and finally shown to converge to the solution of a differential equation uniquely solved by the derivative of the Hermite function.

Convergence for x = 0 and ν ≤ −5
Differentiating (12) with respect to ν, The first sum ∑ T k is given by Lemmas 4 and 5. Consider the second sum Here and ψ(z) = Γ ′ (z)/Γ(z Since t lnt → 0 when t → 0 + , taking zero as the value at t = 0, the functions g ν (t) and are continuous and bounded for bounded ν ≤ −4 and t ≥ 0.
Let y y y(x) (y (x) , y ′ (x), w(x), w ′ (x)) T . Equation (39) can be rewritten in the normal form (25) where As in the proof of Lemma 7, this means that for bounded x and ν, u u u(x) converges uniformly to the solution y y y with an error bound Now, extending the definition of z z z k (29) to four components, z z z 0 u u u 0 and and writing d m ∂ c m /∂ ν, Differentiating (30) with respect to ν, By a procedure similar to the application of equations (31)-(33) in section 2, where the right hand side is independent of x and ν for these parameters in any bounded interval. The sharpness of the bound can be proved by contradiction: Suppose that ∂ y a ν (x)/∂ ν − ∂ H ν (x)/∂ ν = O (b(a)) where O(b(a)) is tighter than O (1/ √ a). Integrating this difference, Choosing ν 1 = 0 and ν 2 = 2, arbitrary x, arbitrarily large a such that a − x √ 2a is integer, and using (34), which is a contradiction. This completes the proof of Theorem 4.

Convergence of zeros
Theorem 5. For fixed real x and positive a → ∞, let n ⌈a − x √ 2a⌉. For a convergent sequence of zeros ν n → ν such that c a n (ν n ) = 0, the limit ν is a zero of the Hermite function, H ν (x) = 0, satisfying ν = ν n + O (1/ √ a ). Conversely, for a positive real zero ν of the Hermite function, there is a convergent sequence ν n → ν of zeros of c a n satisfying ν = ν n + O (1/ √ a ). Similarly, z = ν n − ε implies that H z (x) < 0. Since H z (x) is an entire function and changes sign for z in [ν n − ε, ν n + ε], it must have a zero there. By letting a → ∞, the theorem is proved in one direction. For the reverse direction, switch the roles of w and H. Assume that H ν (x) = 0. Since H 0 (x) ≡ 1, ν cannot be zero. Expand H z (x) around z = ν, writing ∂ H ν (x)/∂ ν for ∂ H z (x) /∂ z at z = ν, Let x(ν) be defined as the pth zero in x of H ν (x) = 0. It is known that x(ν) is a strictly monotonic function of ν for ν ≥ 0, so dx/dν = 0 [6]. Differentiating the equation by ν, so obviously, ∂ H ν (x)/∂ ν = 0 if and only if ∂ H ν (x)/∂ x = 0. But if the latter derivative is zero, then H ν−1 (x) = 0 by the derivative rule (9), and according to the three-term recurrence for Hermite functions (8), all derivatives of H z (x) would be zero at z = ν, entailing that H, being analytic, would be identically zero. In other words, all positive real zeros ν of H ν (x) are simple. Consequently, ∂ H ν (x)/∂ ν = 0, and similarly to the first half of the proof, Z (ν, ε) must be non-zero for ε in some sufficiently small interval. It follows that w n (z) must be zero for some z ∈ [ν − ε, ν + ε], where ε = O (1/ √ a).