# On Generalized Bell Polynomials

Authors: Roberto B. Corcino; Cristina B. Corcino
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

It is shown that the sequence of the generalized Bell polynomialsSn(x) is convex under some restrictions of the parameters involved. A kind of recurrence relation for Sn(x) is established, and some numbers related to the generalized Bell numbers and their properties are investigated.


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## Body

## 1. Introduction

Hsu and Shiue [1] defined a kind of generalized Stirling number pair with three free parameters which is introduced via a pair of linear transformations between generalized factorials, viz, (1.1)(t∣α)n=∑k=0nS(n,k;α,β,γ)(t-γ∣β)k,(t∣β)n=∑k=0nS(n,k;β,α,-γ)(t+γ∣α)k,
where n∈N (set of nonnegative integers), α,β,andγ may be real or complex numbers with (α,β,γ) ≠ (0, 0, 0), and (t∣α)n denotes the generalized factorial of the form (1.2)(t∣α)n=∏j=0n-1(t-jα),n≥1,(t∣α)0=1.
In particular, (t∣1)n=(t)n with (t)0=1. Various well-known generalizations were obtained by special choices of the parameters α,β, and γ (cf. [1]), and the generalization of some properties of the classical Stirling numbers such as the recurrence relations(1.3)S(n+1,k;α,β,γ)=S(n,k-1;α,β,γ)+(kβ-nα+γ)S(n,k;α,β,γ),
the exponential generating function(1.4)(1+αt)γ/α[(1+αt)β/α-1β]k=k!∑n≥0S(n,k;α,β,γ)tnn!,
the explicit formula(1.5)S(n,k;α,β,γ)=1βkk!∑j=0k(-1)k-j(kj)(βj+γ∣α)n,
the congruence relation, and a kind of asymptotic expansion was established. As a follow-up study of these numbers, more properties were obtained in [2]. Furthermore, some combinatorial interpretations of S(n,k;α,β,γ) were given in [3] in terms of occupancy distribution and drawing of balls from an urn.Hsu and Shiue [1] also defined a kind of generalized exponential polynomials Sn(x)≡Sn(x;α,β,γ) in terms of generalized Stirling numbers S(n,k;α,β,γ) with α,β,andγ real or complex numbers as follows: (1.6)Sn(x)=∑k=0nS(n,k;α,β,γ)xk.
We may call these polynomials generalized Bell polynomials. Note that when x=1, we get (1.7)Wn=Sn(1)=∑k=0nS(n,k;α,β,γ),
the generalized Bell numbers. A kind of generating function of the sequence {Sn(x)} for the generalized exponential polynomials has been established by Hsu and Shiue, viz,(1.8)∑n≥0Sn(x)tnn!=(1+αt)γ/αexp⁡⁡[((1+αt)β/α-1)xβ],
where α,β≠0. In particular, (1.8) gives the generating function for the generalized Bell numbers:(1.9)∑n≥0Wntnn!=(1+αt)γ/αexp⁡⁡[((1+αt)β/α-1)β].
Note that, when α→0, (1+αt)γ/α→exp(γt). Hence,(1.10)(1+αt)γ/αexp⁡⁡[((1+αt)β/α-1)xβ]⟶eγtexp⁡⁡[(eβt-1)xβ].
If we define the polynomial Gn,β,r(x) as (1.11)Gn,β,r(x)=lim⁡α→0Sn(x;α,β,r),
then its exponential generating function is given by(1.12)∑n≥0Gn,β,r(x)tnn!=exp⁡⁡[rt+(eβt-1)xβ].
We may call Gn,β,r(x) the  (r,β)-Bell polynomial. Hence, with x=1, this yields the exponential generating function for the (r,β)-Bell numbers. Now, if we use S(n,k;β,γ) to denote the following limit: (1.13)S(n,k;β,γ)=lim⁡α→0S(n,k;α,β,γ),
then, by (1.5),(1.14)S(n,k;β,γ)=1βkk!∑j=0k(-1)k-j(kj)(βj+γ)n,(1.15)Gn,β,r(x)=∑k=0nS(n,k;β,γ)xk.Also obtained by Hsu and Shiue is an explicit formula forSn(x) of the form(1.16)Sn(x)=(1e)x/β∑k=0∞(x/β)kk!(kβ+γ∣α)n.
Consequently, with x=1, we have (1.17)Wn=(1e)1/β∑k=0∞(kβ+γ∣α)nβkk!.
Note that, by taking α=0, (1.16) gives(1.18)Gn,β,r(x)=(1e)x/β∑k=0∞(x/β)kk!(kβ+γ)n,
the explicit formula for (r,β)-Bell polynomial. When x=1, this gives (1.19)Gn,β,r=(1e)1/β∑k=0∞(1/β)kk!(kβ+γ)n,
a kind of the Dobinski formula for (r,β)-Bell numbers. This reduces further to the Dobinski formula for r-Bell numbers [4] when β=1. Moreover, with γ=0, we get (1.20)Bn=1e∑k=0∞kk!,
which is the Dobinski formula for the ordinary Bell numbers [5].In this paper, a recurrence relation and convexity of the generalized Bell numbers will be established and some numbers related toWn will be investigated. Some theorems on (r,β)-Bell polynomials will be established including the asymptotic approximation of the (r,β)-Bell numbers.


## 2. More Properties ofSn(x)

Recurrence relation is one of the useful tools in constructing tables of values. The recurrence relation for the ordinary Bell numbers [6] is given by(2.1)Bn+1=∑k=0n(nk)Bn-k,
with initial condition B0=1. Carlitz’s Bell numbers [7] also satisfy the recurrence relation:(2.2)An+1(λ)=-λnAn(λ)+∑k=0nk!(nk)(μk)λkAn-k(λ),μ=1λ,
with A0(λ)=1. Note that for λ=1,An(1)=Bn and (2.2) will reduce to (2.1). Moreover, Mező [4] obtained certain recurrence relations for the r-Bell polynomials, respectively, as(2.3)Bn,r(x)=rBn-1,r(x)+x∑k=0n-1(n-1k)Bk,r(x).
The following theorem will generalize all of these recurrence relations.Theorem 2.1.
The generalized exponential polynomials satisfy the following recurrence relation:(2.4)Sn+1(x)=(γ-αn)Sn(x)+∑k=0nx(nk)(β∣α)kSn-k(x)
with S0(x)=1.  Moreover, the generalized Bell numbers Wn=Sn(1) satisfy
(2.5)Wn+1=(γ-αn)Wn+∑k=0n(nk)(β∣α)kWn-k.Proof.
Differentiating both sides of (1.8) with respect to t will give
(2.6)∑n≥0Sn(x)tn-1(n-1)!=(1+αt)γ/αexp⁡⁡[((1+αt)β/α-1)xβ]((1+αt)β/αx+γ1+αt).
Applying binomial theorem and Cauchy’s rule for product of two power series will yield
(2.7)(1+αt)∑n≥0Sn(x)tn-1(n-1)!=(∑n≥0Sn(x)tnn!)(∑n≥0(βαn)xαntn+γ),∑n≥0Sn(x)tn-1(n-1)!+∑n≥0nαSn(x)tnn!=∑n≥0(∑k=0nxk!(nk)(βαk)αkSn-k(x))tnn!.
Comparing the coefficients of tn/n!, we obtain
(2.8)Sn+1(x)+αnSn(x)=γSn(x)+∑k=0nxk!(nk)(βαk)αkSn-k(x),
which is precisely equivalent to (1.10).By takingα=0, Theorem 2.1 yields the recurrence relations for the (r,β)-Bell polynomials. More precisely, (2.9)Gn+1,β,r(x)=rGn,β,r(x)+∑k=0nx(nk)βkGn-k,β,r(x).
These further give (2.3) when β=1. Surely, (2.2) can be deduced from (2.5) by letting (α,β,γ)=(λ,1,0). Furthermore, for (α,β,γ)=(0,0,1), (2.4) gives
(2.10)B¯n+1(x)=2B¯n(x),
where B¯n(x)=∑k=0n(nk)xk. If we let B¯n=B¯n(1), we get (2.11)B¯n+1=2B¯n,
which implies (2.12)∑k=0n(n+1k)=2n+1-1,
the number of distinct partitions of an (n+2)-set into 2 nonempty subsets, or simply S(n+2,2), the classical Stirling number of the second kind.Mathematicians have been aware for quite a while that the global behaviour of combinatorial sequences can be used in asymptotic estimates. One of these interesting behaviours is convexity [5]. A real sequence vk,k=0,1,2,… is called convex on an interval [a,b] (containing at least 3 consecutive integers) when (2.13)vk≤12(vk-1+vk+1),k∈[a+1,b-1].
For instance, the sequence of binomial coefficients (nk) satisfies the convexity property since (2.14)(n+2k)-2(n+1k)+(nk)=(nk-2)>0,fork≥2.
This implies that(2.15)B¯n+1≤12(B¯n+B¯n+2),
that is, B¯n is convex.The next theorem asserts that the sequence of generalized exponential polynomials as well as the generalized Bell numbers is convex under some restrictions.Theorem 2.2.
The sequence of generalized exponential polynomialsSn(x) with x>0,α≤0, and β,γ≥0 possesses the convexity property, viz,
(2.16)Sn+1(x)≤12(Sn(x)+Sn+2(x)),n=1,2,….Proof.
Sinceα≤0 and (kβ+γ-nα)≥0, we have
(2.17)0≤[1-(kβ+γ-nα)]2-α(kβ+γ-nα),0≤1-2(kβ+γ-nα)+(kβ+γ-nα)2-α(kβ+γ-nα),2(kβ+γ-nα)≤1+(kβ+γ-nα)(kβ+γ-nα-α).
Multiplying both sides by (kβ+γ∣α)n, we get
(2.18)2(kβ+γ∣α)n+1≤(kβ+γ∣α)n+(kβ+γ∣α)n+2.
Thus, making use of (1.16), we obtain (2.16).Note that, for(α,β,γ,x)=(0,β,r,1), (2.16) asserts the convexity of (r,β)-Bell polynomials which further imply the convexity of r-Bell polynomials when β=1. Moreover, letting (α,β,γ,x)=(0,1,0,1), (2.16) yields (2.15) and implies the convexity of B¯n.


## 3. A Variation of Generalized Bell Numbers

Let us denoteA¯(n,k;α,β,γ)=k!βkS(n,k;α,β,γ) and define (3.1)Bn(α,β,γ)=∑k=1nA¯(n,k;α,β,γ).
The numbers A¯(n,k;α,β,γ) were given combinatorial interpretation in [2], for nonnegative integers α,β,andγ, as the number of ways to distribute n distinct balls, one ball at a time, into k+1 distinct cells, first k of which has β distinct compartments and the last cell with γ distinct compartments such that (i)
the compartments in each cell are given cyclic ordered numbering,(ii)
the capacity of each compartment is limited to one ball,(iii)
each successiveα available compartments in a cell can only have the leading compartment getting the ball,(iv)
the firstk cells are nonempty.Illustration of (iii)
Suppose the first ball lands in compartment 3 of cell 2. The compartment numbered 4, 5, 6,…,α,α+1,α+2 will be closed. And suppose the second ball lands in compartment β-2 also of cell 2. Then compartments numbered β-1,β,1,2,α+3,α+4,α+5,…,2α-3 of cell 2 will be closed.Ifk+1 cells will be changed to any number of cells with the last cell containing γ distinct compartments and the rest of the cells each has β distinct compartments such that only the last cell could be empty, then this gives the combinatorial interpretation of Bn(α,β,γ).The following theorem contains a kind of exponential generating function forBn(α,β,γ).Theorem 3.1.
The numbersBn(α,β,γ) have the following exponential generating function:
(3.2)∑n≥0Bn(α,β,γ)tnn!=(1+αt)γ/α2-(1+αt)β/α.Proof.
Using the exponential generating function in (1.4), we get
(3.3)∑n≥0Bn(α,β,γ)tnn!=∑n≥0∑k≥0βkk!S(n,k;α,β,γ)tnn!=(1+αt)γ/α∑k≥0[(1+αt)β/α-1]k=(1+αt)γ/α11-[(1+αt)β/α-1].
This is exactly the desired generating function.Differentiating both sides of (1.9) with respect to t, we yield (3.4)A¯(n,k;α,β,γ)=dndtn[(1+αt)γ/α((1+αt)β/α-1)k]t=0.
Since A¯(n,k;α,β,γ) vanishes when k=0 and k>n, we have (3.5)Bn(α,β,γ)=∑k=0∞dndtn[(1+αt)γ/α((1+αt)β/α-1)k]t=0=dndtn[(1+αt)γ/α(2-(1+αt)β/α)-1]t=0=12dndtn[(1+αt)γ/α∑ν=0∞(12(1+αt)β/α)ν]t=0=12∑ν=0∞dndtn[(1+αt)(γ+βν)/α]t=012ν.
This result is embodied in the following theorem.Theorem 3.2.
The numberBn(α,β,γ) is equal to
(3.6)Bn(α,β,γ)=12∑ν=0∞(γ+βν∣α)n2-ν,n≥1.The next theorem provides a recurrence relation for the numberBn(α,β,γ) which can be used as a quick tool in computing its first values.Theorem 3.3.
The following recurrence relation holds:(3.7)Bn(α,β,γ)=(γ∣α)n+(β∣α)n+∑j=1n-1(nj)(β∣α)jBn-j(α,β,γ),
where n≥1.Proof.
Making use of (3.6), we have
(3.8)(βαj)Bn-j(α,β,γ)αn-j(n-j)!=12∑ν=0∞(βαj)(βν+γαn-j)2-ν.
Summing up both sides from j=0 to n-1 and using Vandermonde’s formula, we get
(3.9)∑j=0n-1(nj)(β∣α)jBn-j(α,β,γ)αnn!=12∑ν=0∞(∑j=0n-1(βαj)(βν+γαn-j))2-ν=12∑ν=0∞(β+νβ+γαn)2-ν-12∑ν=0∞(βαn)2-ν.
Hence, we have
(3.10)∑j=0n-1(nj)(β∣α)jBn-j(α,β,γ)=12∑ν=0∞((ν+1)β+γ∣α)n2-ν-(β∣α)n12∑ν=0∞2-ν.
Now, by (3.6),
(3.11)12∑ν=0∞(β(ν+1)+γ∣α)n2-ν=∑ν=0∞(β(ν+1)+γ∣α)n2-(ν+1)=∑ν=0∞(βν+γ∣α)n2-ν-(γ∣α)n=2Bn(α,β,γ)-(γ∣α)n
and (1/2)∑ν=0∞2-ν=1. Thus,
(3.12)∑j=1n-1(nj)(β∣α)jBn-j(α,β,γ)=Bn(α,β,γ)-(γ∣α)n-(β∣α)n
which is precisely equivalent to (3.7).Note that whenn=1, (3.7) gives (3.13)B1(α,β,γ)=γ+β,
while (3.6) gives (3.14)B1(α,β,γ)=γ+β(∑ν=1∞ν2ν+1).
This implies that (3.15)∑ν=1∞ν2ν+1=1.The following theorem gives a kind of congruence relation forBn(α,β,γ) with the restriction that α→0. We use Ĝn,β,r to denote the following limit: (3.16)Ĝn,β,r=lim⁡α→0Bn(α,β,γ).Theorem 3.4.
Letr and β be integers. Then for any odd prime p and n≥1, one has the following congruence relation:
(3.17)Ĝn+p-1,β,r-Ĝn,β,r≡0(mod⁡2p).Proof.
Note that the explicit formula in (1.14) can be expressed in terms of a kth difference operator. That is,
(3.18)[Δk(βt+r)n]t=0=∑j=0k(-1)k-j(kj)(βj+r)n,
where Δk denotes the kth difference operator. Hence,
(3.19)Ĝn+p-1,β,r=∑k=0∞[Δk(βt+r)n+p-1]t=0.
Thus,
(3.20)Ĝn+p-1,β,r-Ĝn,β,r=∑k=0∞Δk{(βt+r)n-1[(βt+r)p-(βt+r)]}t=0.
Since, by Fermat’s little theorem, (βt+r)p-(βt+r) is divisible by p,
(3.21)(βt+r)n-1[(βt+r)p-(βt+r)]=px,
for some integer x. Also, since (βt+r)n and (βt+r)p-1-1 are of different parity, (βt+r)n[(βt+r)p-1-1] is divisible by 2. Hence,
(3.22)(βt+r)n[(βt+r)p-1-1]=2py,
for some integer y. Thus, we have
(3.23)(βt+r)n[(βt+r)p-1-1]≡0(mod⁡2p).
This completes the proof of the theorem.


## 4. Some Theorems on(r,β)-Bell Polynomials

The(r,β)-Bell polynomials Gn,β,r(x) have already possessed numerous properties. Some of them are obtained as special case of the properties of Sn(x). However, there are properties of the ordinary Bell numbers or r-Bell numbers which are difficult to establish in Sn(x) but can be done in Gn,β,r(x). For instance, using the rational generating function for S(n,k;β,r) in [2] which is given by(4.1)∑n≥kS(n,k;β,r)tn=tk∏j=0k[1-(βj+r)t],
we can have (4.2)∑n≥0S(n,k;β,r)tn=1βk+1t1∏i=0k((1-rt)/(βt)-i)=1βk+1t1((1-rt)/(βt))∏i=1k((1-rt)/(βt)-i)=-1βk(rt-1)(-1)k∏i=1k((rt-1)/(βt)+i).
It can easily be shown that (4.3)∏i=1k(rt-1βt-i)=((β+r)t-1βt)k.Thus,(4.4)∑k≥0(∑n≥0S(n,k;β,r)tn)xk=∑k≥0(-1βk(rt-1)(-1)k(((β+r)t-1)/βt)k)xk,∑n≥0(∑k=0nS(n,k;β,r)xk)tn=-1rt-1∑k≥0(1)k(((β+r)t-1)/βt)k(-x/β)kk!.This can be expressed further as(4.5)∑n≥0Gn,β,r(x)tn=-1rt-1⋅F11(1(β+r)t-1βt|-xβ),whereF11 is the hypergeometric function which is defined by (4.6)Fqp(a1,a2,…,apb1,b2,…,bq|t)=∑k=0∞(a1)k(a2)k⋯(ap)k(b1)k(b2)k⋯(bq)ktkk!,where(ai)j=ai(ai+1)(ai+2)⋯(ai+j-1). Applying Kummer’s formula [8], (4.7)ex1F1(ab|-x)=F11(b-ab|x),we obtain the following generating function.Theorem 4.1.
The(r,β)-Bell polynomials satisfy the following generating function:
(4.8)∑n≥0Gn,β,r(x)tn=-1rt-1⋅1ex/β⋅F11(rt-1βtβt+rt-1βt|xβ).It will be interesting if one can also obtain a generating function of this form forSn(x).Now, using the integral identity in [9],(4.9)Im⁡∫0πejeiθsin⁡(nθ)dθ=π2jnn!,
and the explicit formula in (1.14), we get (4.10)π2n!S(n,k;β,r)=1βkk!∑j=0k(-1)k-j(kj)Im⁡∫0πe(βj+r)eiθsin⁡(nθ)dθ=1βkk!Im⁡∫0π[∑j=0k(-1)k-j(kj)(eβeiθ)j]ereiθsin⁡(nθ)dθ=Im⁡∫0π[(eβeiθ-1)/β]kk!ereiθsin⁡(nθ)dθ.
Hence, (4.11)∑k=0∞S(n,k;β,r)xk=2n!πIm⁡∫0π{∑k=0∞[(eβeiθ-1)/β]kk!xk}ereiθsin⁡(nθ)dθ=2n!πIm⁡∫0πex(eβeiθ-1)/βereiθsin⁡(nθ)dθ.
Thus, (4.12)Gn,β,r(x)=2n!πex/βIm⁡∫0πexβ-1eβeiθereiθsin⁡(nθ)dθ,
where β≠0. By simple algebraic manipulation, this can further be expressed as follows.Theorem 4.2.
The(r,β)-Bell polynomials have the following integral representation:
(4.13)Gn,β,r(x)=2n!πex/β∫0πeJ1(θ)sin⁡(J2(θ))sin⁡(nθ)dθ,
where
(4.14)J1(θ)=rcos⁡θ+xeβcos⁡θcos⁡(βsin⁡θ)β,J2(θ)=rsin⁡θ+xeβcos⁡θsin⁡(βsin⁡θ)β.
It will also be compelling to establish such integral representation for Sn(x).The Bell polynomialsBn(λ) are known to be connected to the Poisson distribution. More precisely, Bn(λ) can be expressed in terms of the moment of the Poisson random variable Z with parameter λ>0 as (4.15)Bn(λ)=Eλ[Zn].
The exponential generating function for the (r,β)-Bell polynomials in (1.12) can be written as follows:
(4.16)e(r/β)βte(x/β)(eβt-1)=e(r/β)βtEx/β[e(βt)Z]=∑n≥0{∑k=0n(nk)rn-kβkEx/β[Zk]}tnn!.
Hence, we can also express the (r,β)-Bell polynomials in terms of the following moment:(4.17)Gn,β,r(x)=Ex/β[(βZ+r)n].
Now, (4.18)Gn,β,r(x)=∑k=0n(nk)rn-kβkEx/β[Zk]=∑k=0n(nk)rn-kβkBk(xβ)=∑k=0n(nk)rn-kβk∑j=0kS(k,j)(xβ)j.
Thus, we have the following theorem.Theorem 4.3.
The(r,β)-Bell polynomials equal
(4.19)Gn,β,r(x)=∑k=0n(nk)rn-k∑j=0kβk-jS(k,j)xj.An extension of the Bell polynomialsBn(y,λ), defined by Privault [10] as (4.20)∑n=0∞Bn(y,λ)tnn!=eyt-λ(et-t-1),
can be expressed in terms of the (r,β)-Bell polynomials as (4.21)Bn(y,λ)=Gn,1,λ+y(-λ).
Using Theorem 4.3, we obtain (4.22)Bn(y,-λ)=Gn,1,-λ+y(λ)=∑k=0n(nk)(y-λ)n-k∑j=0kS(k,j)λj.
This is exactly the identity obtained by Privault in [10].


## 5. An Asymptotic Approximation forGn,β,r

Using the exponential generating function forGn,r,β in (1.12) with x=1 and Cauchy’s theorem for integrals, we obtain the integral representation (5.1)Gn,r,β=n!2πi∫γexp⁡[rz+(eβz-1/β)]zn+1dz,
where γ is the circle z=Reiθ,-π≤θ≤π. Contour integration yields (5.2)Gn,r,β=n!2πiRn∫-ππexp⁡(β-1eβReiθ+rReiθ-inθ-β-1)dθ,
which can be written into the compact form(5.3)Gn,r,β=A∫-ππexp⁡(F(θ))dθ,
where (5.4)A=n!exp⁡(rR+β-1eβR-β-1)2πRn,F(θ)=β-1eβReiθ+rReiθ-inθ-(rR+β-1eβR).
Define ϵ=e-3R/8 and let (5.5)J1=∫-πϵexp⁡(F(θ))dθ,J2=∫ϵπexp⁡(F(θ))dθ.
Thus (5.3) can be written as (5.6)Gn,r,β=AJ1+A∫ϵϵexp⁡(F(θ))dθ+AJ2.Lemma 5.1.
There exists a constantk>0 such that
(5.7)|J2|<e-kβ-1eβR(π-ϵ).Proof.
It can be shown that(5.8)|exp⁡(F(θ))|=e-[(rR+β-1eβR)+β-1cos⁡(βRsin⁡θ)eβRcos⁡θ].
Since cosθ<1 for 0<ϵ<θ≤π, we have
(5.9)|exp⁡(F(θ))|=e-β-1eβR[1-cos⁡(βRsin⁡θ)].
Since [1-cos(βRsinθ)]>0 for cosθ<1 for 0<ϵ<θ≤π, there exists a constant k>0 such that [1-cos(βRsinθ)]<k. Hence
(5.10)|J2|<e-kβ-1eβR(π-ϵ).It will be seen later thatR→∞ as n→∞. With the result in Lemma 5.1 we see that J1 and J2 will tend to zero as n→∞. Hence(5.11)Gn,r,β~A∫-ϵϵexp⁡(F(θ))dθ.
Observe that F(θ) is analytic at θ=0.Thus F(θ) has a Maclaurin series expansion about θ=0. This Maclaurin expansion can be written in the form(5.12)F(θ)=(ReβR+rR-n)iθ+12(βR2+ReβR+rR)i2θ+∑k=3∞[β-1ρk(eβR)+rR](iθ)k,
where we define ρ to be the operator ρ=R(dθ/dR). Choose R such that ReβR+rR-n=0; that is, R satisfies the equation xeβR+rx-n=0. This R is shown to exist in the following lemma.Lemma 5.2.
There exists a unique positive real solution to the equationxeβR+rx-n=0.Proof.
We can rewrite the given equation in the form(5.13)xn-rx=e-βx.
The desired solution is the x-coordinate of the intersection of the functions h(x)=x/(n-rx) and g(x)=e-βx.It can be seen from the preceding lemma thatR→∞ as n→∞. With this choice of R, we have (5.14)F(θ)=-12(βR2+ReβR+rR)θ+∑k=3∞[β-1ρk(eβR)+rR](iθ)k.
We now introduce the following notations:(5.15)ϕ=[(1/2)(βR2eβR+ReβR+rR)1/2]θ,ak=[β-1e-βRρk+2(eβR)+rRe-βR](iϕ)k+2(k+1)![1/2(βR2+R+rRe-βR)]k+2/2,z=e-βR/2,f(z)=∑k=1∞akzk.
Then F(θ)=-ϕ2+f(z) and(5.16)Gn,r,β~C∫-hhexp⁡[-ϕ2+f(z)]dz,
where h=(1/2)(βR2eβR+ReβR+rR)1/2e-3R/8 and C=A/[(1/2)(βR2eβR+rR)]1/2.We have definedz as a function of R. However, for the moment we consider z to be an independent variable and expand ef(z) into a convergent Maclaurin series expansion of the form(5.17)ef(z)=∑k=0∞bkzk,
where b0=ef(0)=1,b1=ef(0)f′(0)=a1,andb2=a2+(a12/2).Lemma 5.3.
There is a constantRo such that for all R>Ro,
(5.18)|ak|<|2ϕ|k+2.Proof.
We see that(5.19)|ak|=Rk+2[1+o(Rk+2)](2)(k+2)/2(k+2)!(βR2)(k+2)/2[1+o(R2)]|ϕ|k+2
which tends to
(5.20)2k+2/2(k+2)!<2k+2|ϕ|k+2
as R→∞. From this, it follows that there is a constant Ro satisfying (5.18).Now, it will follow from Lemma5.3 that the radius of convergence of (5.17) becomes large when θ is near zero. Thus, z=e-βR/2 is within the domain of convergence.Withz=e-βR/2,(5.21)Gn,r,β~C∑k=0s-1(∫-hhe-ϕ2bkdϕ)zk+Qs,
where (5.22)Qs=∫-hh(∑k=s∞e-ϕ2bkzk)dϕ.Note thatR→∞ as n→∞. Furthermore with (5.23)h=12(βR2eβR+ReβR+rR)1/2e-3R/8=12(βR2+R+rRe-βR)1/2e(R(4β-3))/8,h→∞ as R→∞. From these facts and the known asymptotic expansion of the function of the form (5.24)∫-hhe-ϕ2(polynomialin|ϕ|)dϕ,
the replacement of h  by ∞ in (5.16) is easily justified (see [11]). Hence (5.25)Gn,r,β~C∑k=0s-1(∫-∞∞e-ϕ2bkdϕ)zk+Qs.
It remains to show that Qs=o(|z|s) as R→∞, that is, z→0. From a lemma in [12], |bk|≤|2ϕ|k+2(1+|2ϕ|2)k-1. Thus, (5.26)|∑k=s∞bkzk|≤[|2ϕ|s+2(1+|2ϕ|2)s-1|z|s][1+μ+μ2+⋯],
where μ=|2ϕ|(1+|2ϕ|2)|z|.Now, forμ<1, we have(5.27)|∑k=s∞bkzk|≤|2ϕ|s+2(1+|2ϕ|2)s-1|z|s1-|z||2ϕ|(1+|2ϕ|2).LetM and Ps(|ϕ|)|z|s denote the denominator and the numerator, respectively, in (5.27). Since |ϕ|≤h and z=e-βR/2, we have (5.28)|ϕ3||z|≤18(βR2+R+rRe-βR)3/2e-3R/8⟶0asR⟶∞.
Hence for sufficiently large R, M≥1/2. Moreover,(5.29)∫-∞∞e-ϕ2Ps(|ϕ|)dϕ
exists and tends to zero as R→∞. Therefore, (5.30)|Qs||z|s≤∫-∞∞e-ϕ2Ps(|ϕ|)Mdϕ.
Thus, |Qs|=o(|z|s). Consequently,(5.31)Gn,r,β~C∑k=0∞(∫-∞∞e-ϕ2bkdϕ)e(-kβR)/2.
Since ∫-∞∞e-x2xn=0 for odd n, and b2k+1, as a polynomial in ϕ, contain only odd powers of ϕ, it follows that(5.32)Gn,r,β~C∑k=0∞(∫-∞∞e-ϕ2b2kdϕ)e-kβR.Calculation yields(5.33)a1=βR3+3R2+β-1R+rRe-βR3![(1/2)(βR2+R+rRe-βR)]3/2(iϕ)3,a2=βR4+6βR3+7R2+β-1R+rRe-βR4![(1/2)(βR2+R+rRe-βR)]2(iϕ)4.
Taking the first two terms of the asymptotic expansion of (5.32), we have (5.34)Gn,r,β~C∫-∞∞e-ϕ2bodϕ+Cz2∫-∞∞e-ϕ2b2dϕ.
Since b2=a2+a12/2 and bo=1,(5.35)Gn,r,β~C∫-∞∞e-ϕ2dϕ+Cz2∫-∞∞a2e-ϕ2dϕ+Cz22∫-∞∞e-ϕ2(a12)dϕ.
Let I1,I2,andI3 denote, respectively, the integrals in (5.35). Then evaluating the last two integrals by parts and since ∫-∞∞e-ϕ2dϕ=π, we obtain (5.36)I1=Cπ,I2=Ce-Rπ(βR3+6βR2+β-1+re-βR)8R(βR+1+re-βR)2,I3=-5Ce-Rπ(βR2+3β-1R2+re-βR)224R(βR+1+re-βR)3.
Substituting the results in (5.35) and simplifying, we obtain (5.37)Gn,r,β~Cπ(1+D+EF),
where (5.38)D=(3β2R3+8βR3+3βR+3-10β-1-2re-βR)re-βR,E=(3β3-5β2)R4+(21β2-30β)R3+(39β-55)R2+(24-30β-1)R+(3β-1-5β-2)F=24ReβR(βR+1+re-βR)3.,
Since ReβR=(n-rR)β-1 and Rn=nn(βeβR+r)-n,(5.39)C=n!exp⁡(rR+β-1eβR-β)π[nn(βeβR+r)-n][2(n-rR)β-1]1/2(βR+1+re-βR)1/2.
Using Stirling's approximation for n!, viz, (5.40)n!~(2π)e-nnn+(1/2)(1+112n),
we obtain (5.41)C~n1/2(1+(1/12n))exp⁡(rR+β-1eβR-β)(ββR+r)nπ1/2[(n-rR)β-1]1/2(βR+1+re-βR)1/2en.
Using (5.37), we obtain (5.42)Gn,r,β~n1/2(1+(1/12n))exp⁡(rR+β-1eβR-β-n)(ββR+r)n[(n-rR)β-1]1/2(βR+1+re-βR)1/2(1+D+EF).



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*Source: 623456-2011-10-11.xml*