The Asymptotics of the Geometric Polynomials

1. Introduction

Geometric polynomials ω_n(x) are defined by the exponential generating function

Alternatively, the geometric polynomial of degree n is given by

where S(n, k) stand for the Stirling numbers of the second kind. An overview of properties and applications of the geometric polynomials can be found in [8,11,19]. Here are the first five geometric polynomials,

The geometric polynomials, as well as related exponential polynomials, are important instruments in obtaining limit theorems for combinatorial numbers satisfying a class of triangular arrays. In a series of works [4–7], we have received such central and local limit theorems for the combinatorial numbers associated with the Riemann zeta function.

The asymptotic expansion for the exponential polynomials recently has been established by Paris [21]. Similar asymptotic formulas for different polynomials also have attracted the attention of many researchers; see, for instance, the results of Alfaro et al. [1] for the generalized Freud polynomials, Barbero et al. [2] for the Appell polynomials, Corcino and Corcino [13] for the Genocchi polynomials, Lee and Wong [16] for the Tricomi–Carlitz polynomials, Li et al. [17] for the Wilson polynomials, Paris [20] for the generalized Hermite–Bell polynomials, Wang et al. [22, 23] for the Racah polynomials. This paper aims to extend these investigations and receive the geometric polynomials’ asymptotics. The asymptotics will be applied to derive limit theorems for combinatorial numbers. Note that the asymptotic approximation for the special case of x = 1 has been addressed in [14, 15, 18].

2. The asymptotics of ω_n(x) for large n

The exponential polynomials [3, 9, 10],

and the geometric polynomials are connected by the relation [8],

The exponential polynomials, for their part, have the integral representation (cf. formula (2.3) of [12: Section 2]),

Let c₀ = min(ln(1 + x⁻¹), π/2). Note that for x > 0, θ ∈ (0, π) and τ ∈ (0, c₀), we have u(x, τ, θ) > 0. Thus,

Next, we notice that the integral expression (2.6) does not yield to the techniques of elementary calculus. We will evaluate it using contour integration. Let us denote the denominator in (2.6) by D(x, τ, θ),

Zeros of the function are

where n, k∈ℤ.

Suppose x₀ = 1/(e – 1). Let us consider two cases.

Case 1. First we consider the integral (2.6) for x ∈ (0, x₀). Noticing that u(x, τ, θ) is even in θ and v(x, τ, θ) is odd in θ, we get

To evaluate the integral (2.9), we apply the contour integral of the corresponding complex-valued function over the rectangular contour,

Let us denote the vertices of the rectangular contour γ as A(−π, 0), B(π, 0), C(π, R) and D(−π, R). By the residue theorem, we have that, while R goes to infinity,

here

and f(z) stands for an integrand. Note that

Next we evaluate integrals over the sides of the rectangular contour γ (cf. (2.11)).

Side BC. We parametrize the side BC by z = π + it, dz = idt and evaluate the integral using Watson’s lemma. We have

Side DA. Similarly, we parametrize the side DA by z = −π + it, dz = idt and evaluate the integral using Watson’s lemma. We receive

Side CD. We parametrize the side CD by z = t + iR, dz = dt. We get

Next, by applying the residue theorem to the integral (2.10), we obtain

Note that (cf. (2.12))

hence, by (2.13), we have

and, by (2.13), (2.18) and (2.19), we calculate the residues

Combining (2.11), (2.14), (2.15) and (2.16), we obtain that

Next, combining (2.9) and (2.21), we get

where ηk=arctan(2πkln(1+x−1)). Note that ln (1 + x⁻¹) > 1. Thus, we receive the statement of the theorem

for the first interval.

Case 2. Now let us consider the integral (2.6) for x ∈ (x₀, +∞). We compute ω_n(x) (cf. (2.9)),

integrating

over the rectangular contour δ with vertices A(−π, 0), B(π, 0), C′(π, −R) and D′(−π, −R). By the residue theorem, we have that, while R goes to infinity,

here

and g(z) stands for an integrand. Note that

Next we evaluate integrals over the sides of the rectangular contour δ (cf. (2.26)).

Side BC′. We parametrize the side BC′ by z = π – it, dz = −idt and evaluate the integral using Watson’s lemma. We get

Side D′A. Similarly, we parametrize the side D′A by z = −π – it, dz = −idt and evaluate the integral using Watson’s lemma. We have

Side C′D′. We parametrize the side C′D′ by z = t – iR, dz = dt. We get

Next, by applying the residue theorem to the integral (2.25), we receive

Note that (cf. (2.27) and (2.28))

hence,

and, by (2.28), (2.34) and (2.35), we obtain the residues

Combining (2.26), (2.29), (2.30) and (2.31), we receive that

Next, combining (2.24) and (2.36), we get

where ηk=arctan(2πkln(1+x−1)). Note that ln (1 + x⁻¹) ∈ (0, 1). Thus, we receive the statement of the theorem

for the second interval.

Finally, considering two one-sided limits, limx↗x0ωn(x) (see (2.23), Case 1) and limx↘x0ωn(x) (see (2.38), Case 2), we receive

thus concluding the proof of the theorem. □