# An Application of Pascal Distribution Series on Certain Analytic Functions Associated with Stirling Numbers and Sălăgean Operator

Authors: Basem Aref Frasin
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the present paper, we will observe that the Sălăgean differential operator can be written in terms of Stirling numbers. Furthermore, we find a necessary and sufficient condition and inclusion relation for Pascal distribution series to be in the classℙkλ,α of analytic functions with negative coefficients defined by the Sălăgean differential operator. Also, we consider an integral operator related to Pascal distribution series. Several corollaries and consequences of the main results are also considered.


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## Body

## 1. Preliminaries

Special functions are used in many applications of physics, engineering, and applied mathematics and statistics. Special polynomials have a close connection with number theory, and one of the most important sets of special numbers is the class of Stirling numbers (of the first and second kind), introduced in 1730 by the Scottish mathematician James Stirling.In combinatorics, a Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set ofk objects into j nonempty subsets and is denoted by Sk,j or by bk,j as used in this paper. These numbers occur in the field of mathematics called combinatorics and the study of partitions. In this paper, we will observe that the Sălăgean differential operator Dk can be written in terms of Stirling numbers.LetA denote the class of analytic functions f in the open unit disk D=z∈ℂ:z<1 and normalized by the conditions f0=0=f′0−1 has the following representation:(1)fz=z+∑n=2∞anzn.Furthermore, letT be a subclass of A consisting of functions of the form(2)fz=z−∑n=2∞anzn,z∈D.For a function fz in A, we define(3)D0fz=fz,(4)D1fz=zf′z,and in general, we have(5)Dkfz=zDk−1fz′,k∈ℕ.The differential operatorDk was introduced by Sălăgean [1].We note that(6)Dkfz=zDk−1fz′=bk,1zf′z+bk,2z2f′′z+bk,3z3f′′′z+⋯+bk,kzkfkz=∑j=1kbk,jzjfjz,k∈ℕ,where(7)bk,j=jbk−1,j+bk−1,j−1,andbk,1=bk,k=1.For example,(i)
Ifk=2, we have(8)D2fz=zD1fz′=zf′z+z2f′′z=b21zf′z+b22z2f′′z,where(9)b21=b22=1.(ii)
Ifk=3, we have(10)D3fz=zD2fz′=zf′z+3z2f′′z+z3f′′′z=b3,1zf′z+b3,2z2f′′z+b3,3z3f′′′z,where(11)b3,2=2b2,2+b2,1=3,b3,1=b3,3=1.(iii)
Ifk=4, we have(12)D4fz=zD3fz′=zf′z+7z2f′′z+6z3f′′′z+z4f4z=b4,1zf′z+b4,2z2f′′z+b4,3z3f′′′z+b4,4z4f4z,where(13)b4,2=2b3,2+b3,1=7,b4,3=3b3,3+b3,2=6,b4,1=b4,4=1.(iv)
Ifk=5, we have(14)D5fz=zD4fz′=zf′z+15z2f′′z+25z3f′′′z+10z4f4z+z5f5z=b5,1zf′z+b5,2z2f′′z+b5,3z3f′′′z+b5,4z4f4z+b5,5z5f5z,where(15)b5,2=2b4,2+b4,1=15,b5,3=3b4,3+b4,2=25,b5,4=4b4,4+b4,3=10,b5,1=b5,5=1.Table1 represents the coefficients bkj of zkfkz.Table 1
Stirling numbers of the second kind.
zf′zz2f″zz3f‴zz4f4zz5f5zz6f6zz7f7z⋯zkfkzD1fz100000000D2fz110000000D3fz131000000D4fz176100000D5fz115251010000D6fz1319065151000D7fz16330135014021100⋮1⋮⋮⋮⋮⋮⋮⋮0Dkfz1bk,2bk,3bk,4bk,5bk,6bk,7⋯1Table1 (see [2]) shows the first few possibilities for Stirling numbers of the second kind. Also, from this table, we note that:(1)
bk,j=jbk−1,j+bk−1,j−1(2)
bk,k−1=kk−1/2=k2(3)
bk,2=2k−1−1(4)
bk,3=1/63k−3.2k+3(5)
bk,1=bk,k=1(6)
bk,j=0, when j>k.(7)
bp,j≡0modp iff 1<j<p, where p is a prime number.Furthermore, fork=2,3,4,5, we observe that(16)D2fz=z2f′′z+zf′z,z+∑n=2∞n2anzn=z+∑n=2∞nn−1+nanzn,(17)D3fz=z3f′′′z+3z2f′′z+zf′z,z+∑n=2∞n3anzn=z+∑n=2∞nn−1n−2+3nn−1+nanzn,(18)D4fz=z4f4z+6z3f′′′z+7z2f′′z+zf′z,z+∑n=2∞n4anzn=z+∑n=2∞nn−1n−2n−3+6nn−1n−2+7nn−1+n.and(19)D5fz=z5f5z+15z4f4z+25z3f′′′z+10z2f′′z+zf′z,z+∑n=2∞n5anzn=z+∑n=2∞nn−1n−2n−3n−4+10nn−1n−2n−3+25nn−1n−2+15nn−1+nanzn.From (16)–(19), we conclude that(20)n=n−1+1=b2,2n−1+b2,1,n2=n−1n−2+3n−1+1=b3,3n−1n−2+b3,2n−1+b3,1,n3=n−1n−2n−3+6n−1n−2+7n−1+1=b4,4n−1n−2n−3+b4,3n−1n−2+b4,2n−1+b4,1,(21)n4=n−1n−2n−3n−4+10n−1n−2n−3+25n−1n−2+15n−1+1=b5,5n−1n−2n−3n−4+b5,4n−1n−2n−3+b5,3n−1n−2+b5,2n−1+b5,1.In general, we have(22)nk=bk+1,1+bk+1,2n−1+bk+1,3n−1n−2+⋯+bk+1,k+1n−1n−2n−3⋯n−k=bk+1,1+∑j=1kbk+1,j+1n−1n−2n−3⋯n−j,k=1,2,3,⋯.For functionsf∈A given by (1) and g∈A given by gz=z+∑n=2∞bnzn, we recall that the well-known Hadamard product of f and g is given by(23)f∗gz≔z+∑n=2∞anbnzn,z∈D.Forϵ∈ℂ/0 and −1≤D<ℭ≤1, we say that a function f∈A is in the class ℛϵℭ,D if it satisfies the inequality(24)f′z−1ℭ−Dε−Df′z−1<1,z∈D.The function classℛϵℭ,D was introduced by Dixit and Pal [3].With the help of the differential operatorDk, we say that a function fz belonging to A is said to be in the class Qλ,α if and only if(25)ℜ1−λzDkfz′+λzDk+1fz′1−λDkfz+λDk+1fz>α,k∈ℕ0=ℕ∪0,for some α0≤α<1, λ0≤λ≤1, and for all z in D.Furthermore, we define the classℙkλ,α by(26)ℙkλ,α=Qλ,α∩T.The classℙTλ,α was introduced and studied by Aouf and Srivatava [4].We note that, by specializing the parametersk and λ, we obtain the following subclasses:(i)
ℙ00,α=T∗α and ℙ01,α=Cα, where T∗α and Cα represent the classes of starlike functions of order α and convex functions of order α with negative coefficients, respectively, introduced and studied by Silverman [5](ii)
ℙk1,α=Ckα (see [4]), where Ckα represents the class of functions fz∈T satisfying the inequality(27)ℜzDk+1fz′Dk+1fz>α,k∈ℕ0.(iii)
ℙk0,α=Tk∗α (see [4]), where Tk∗α represents the class of functions fz∈T satisfying the inequality(28)ℜzDkfz′Dkfz>α,k∈ℕ0.In statistics and probability, distributions of random variables play a basic role and are used extensively to describe and model a lot of real-life phenomenon; they describe the distribution of the probabilities over the values of the random variable. In recent years, many researchers have examined some important features in the geometric function theory, such as coefficient estimates, inclusion relations, and conditions of being in some known classes, using different probability distributions such as the Poisson, Pascal, Borel, Mittag–Leffler-type Poisson distribution, etc.(see, for example, [6–10]).The probability density function of a discrete random variableX which follows the Pascal distribution is given by(29)ProbX=r=r+m−1m−1σr1−σm,r=0,1,2,3,….Very recently, El-Deeb et al. [11] introduced a power series whose coefficients are probabilities of the Pascal distribution(30)Λσmz=z+∑n=2∞n+m−2m−1σn−11−σmzn,z∈Dwhere m≥1and0≤σ≤1 and we note that, by a ratio test, the radius of convergence of above series is infinity. We also define the series(31)Υσmz=2z−Λσmz=z−∑n=2∞n+m−2m−1σn−11−σmzn,z∈D.Now, we considered the linear operator(32)ℐσmz:A⟶Adefined by the Hadamard product(33)ℐσmfz=Λσmz∗fz=z+∑n=2∞n+m−2m−1σn−11−σmanzn,z∈D.Motivated by several earlier results on connections between various subclasses of analytic and univalent functions, using hypergeometric functions, generalized Bessel functions, Struve functions, Poisson distribution series, and Pascal distribution series (see, for example, [12], [13–15], [7–9, 16–23], [24]), we determine a necessary and sufficient condition for Υσmz to be in our class ℙkλ,α. Furthermore, we give sufficient conditions for ℐσmℛϵℭ,D⊂ℙkλ,α. Finally, we give conditions for the integral operator Gσmfz=∫0zYσmt/tdt belonging to the class ℙkλ,α.The following results will be required in our investigation.Lemma 1 (see [4]).
Let the functionfzbe defined by (2). Then,fz∈ℙkλ,αif and only if(34)∑n=2∞nkn−α1+n−1λan≤1−α,z∈D.The result (34) is sharp.Lemma 2 (see [3]).
Iff∈ℛϵℭ,Dis of the form (1), then(35)an≤ℭ−Dϵn,n∈ℕ−1.The result is sharp for the function(36)fz=∫0z1+ℭ−Dϵtn−11+Dtn−1dt,z∈D;n∈ℕ−1.


## 2. Necessary and Sufficient Conditions

By simple calculations, we derive the following relations:(37)∑n=2∞n+m−2m−1σn−1=11−σm−1,∑n=5∞n−1n−2n−3n−4n+m−2m−1σn−1=24σ4m+3m−11−σm+4,∑n=4∞n−1n−2n−3n+m−2m−1σn−1=6σ3m+2m−11−σm+3,∑n=3∞n−1n−2n+m−2m−1σn−1=2σ2m+1m−11−σm+2,∑n=2∞n−1n+m−2m−1σn−1=σmm−11−σm+1,and, in general, for s=1,2,3,…, we have(38)∑n=s+1∞n−1n−2n−3⋯n−sn+m−2m−1σn−1=s!σsm+s−1m−11−σm+s.Unless otherwise mentioned, we shall assume in this paper that0≤α<1,0≤λ≤1m≥1, and 0≤σ<1.First of all, with the help of Lemma1, we obtain the following necessary and sufficient condition for Υσmz to be in ℙkλ,α.Theorem 1.
Letk≥1.Then,Υσmz∈ℙkλ,αif and only if(39)∑j=1kλbk+3,j+1+1−λ−αλbk+2,j+1+αλ−1bk+1,j+1j!σjm+j−1m−11−σm+j+λbk+3,k+2+1−λ−αλk+1!σk+1m+km−11−σm+k+1+λk+2!σk+2m+k+1m−11−σm+k+2≤1−α.Proof.
In view of Lemma1, we only need to show that Q≤1−α, where(40)Q=∑n=2∞nkn−α1+n−1λn+m−2m−1σn−11−σm.
Using (22) and (38), we have(41)Q=∑n=2∞λnk+2+1−λ−αλnk+1−α1−λnkn+m−2m−1σn−11−σm=∑n=2∞λbk+3,1+∑j=1k+2bk+3,j+1n−1n−2n−3⋯n−j+1−λ−αλbk+2,1+∑j=1k+1bk+2,j+1n−1n−2n−3⋯n−j+αλ−1bk+1,1+∑j=1kbk+1,j+1n−1n−2n−3⋯n−j×n+m−2m−1σn−11−σm=∑n=2∞λbk+3,1n+m−2m−1σn−11−σm+∑n=2∞λbk+3,k+3n−1n−2n−3⋯n−k+2n+m−2m−1σn−11−σm+∑n=2∞λbk+3,k+2n−1n−2n−3⋯n−k+1n+m−2m−1σn−11−σm+∑n=2∞∑j=1kλbk+3,j+1n−1n−2n−3⋯n−jn+m−2m−1σn−11−σm+∑n=2∞1−λ−αλbk+2,1n+m−2m−1σn−11−σm+∑n=2∞1−λ−αλbk+2,k+2n−1n−2n−3⋯n−k+1n+m−2m−1σn−11−σm+∑n=2∞∑j=1k1−λ−αλbk+2,j+1n−1n−2n−3⋯n−jn+m−2m−1σn−11−σm+∑n=2∞αλ−1bk+1,1n+m−2m−1σn−11−σm+∑n=2∞∑j=1kαλ−1bk+1,j+1n−1n−2n−3⋯n−jn+m−2m−1σn−11−σm=λbk+3,1+1−λ−αλbk+2,1+αλ−1bk+1,1∑n=2∞n+m−2m−1σn−11−σm+λbk+3,k+3∑n=k+2∞n−1n−2n−3⋯n−k+2n+m−2m−1σn−11−σm+λbk+3,k+2+1−λ−αλbk+2,k+2∑n=k+1∞n−1n−2n−3⋯n−k+1×n+m−2m−1σn−11−σm+∑j=1kλbk+3,j+1+1−λ−αλbk+2,j+1+αλ−1bk+1,j+1×∑n=j+1∞n−1n−2n−3⋯n−jn+m−2m−1σn−11−σm=1−α1−1−σm+λk+2!σk+2m+k+1m−11−σk+2+λbk+3,k+2+1−λ−αλk+1!σk+1m+km−11−σk+1+∑j=1kλbk+3,j+1+1−λ−αλbk+2,j+1+αλ−1bk+1,j+1j!σjm+j−1m−11−σj.
Therefore, we see that the last expression is bounded above by1−α if (39) is satisfied.


## 3. Inclusion Properties

Making use of Lemma2, we will study the action of the Pascal distribution series on the class ℙkλ,α.Theorem 2.
Letk≥2andf∈ℛϵℭ,D. Then,ℐσmfz∈ℙkλ,αif(42)ℭ−Dϵ∑j=1k−1λbk+2,j+1+1−λ−αλbk+1,j+1+αλ−1bk,j+1j!σjm+j−1m−11−σj+λk+1!σk+1m+km−11−σk+1+λbk+2,k+1+1−λ−αλk!σkm+k−1m−11−σk+1−α1−1−σm∑j=1k−1λbk+2,j+1+1−λ−αλbk+1,j+1+αλ−1bk,j+1j!σjm+j−1m−11−σj+λk+1!σk+1m+km−11−σk+1+λbk+2,k+1+1−λ−αλk!σkm+k−1m−11−σk+1−α1−1−σm≤1−α.Proof.
In view of Lemma1, it suffices to show that L≤1−α, where(43)L=∑n=2∞nkn−α1+n−1λn+m−2m−1σn−11−σman.
Applying Lemma2, we find from equations (22) and (38) that(44)L≤ℭ−Dϵ∑n=2∞λnk+1+1−λ−αλnk−α1−λnk−1n+m−2m−1σn−11−σm=ℭ−Dϵ1−σm∑n=2∞λbk+2,1+∑j=1k+1bk+2,j+1n−1n−2n−3⋯n−j+1−λ−αλbk+1,1+∑j=1kbk+1,j+1n−1n−2n−3⋯n−j+αλ−1bk,1+∑j=1k−1bk,j+1n−1n−2n−3⋯n−jn+m−2m−1σn−1=ℭ−Dϵ1−σm∑n=2∞λbk+2,1+1−λ−αλbk+1,1+αλ−1bk,1n+m−2m−1σn−1+λbk+2,k+2∑n=2∞n−1n−2n−3⋯n−k+1n+m−2m−1σn−1+λbk+2,k+1∑n=2∞n−1n−2n−3⋯n−kn+m−2m−1σn−1+1−λ−αλbk+1,k+1∑n=2∞n−1n−2n−3⋯n−kn+m−2m−1σn−1+λ∑j=1k−1bk+2,j+1+1−λ−αλ∑j=1k−1bk+1,j+1+αλ−1∑j=1k−1bk,j+1×∑n=2∞n−1n−2n−3⋯n−jn+m−2m−1σn−1=ℭ−Dϵλbk+2,1+1−λ−αλbk+1,1+αλ−1bk,11−1−σm+λbk+2,k+2k+1!σk+1m+km−11−σk+1+λbk+2,k+1k!σkm+k−1m−11−σk+λ∑j=1k−1bk+2,j+1j!σjm+j−1m−11−σj+1−λ−αλbk+1,k+1k!σkm+k−1m−11−σk+1−λ−αλ∑j=1k−1bk+1,j+1j!σjm+j−1m−11−σj+αλ−1∑j=1k−1bk,j+1j!σjm+j−1m−11−σj=ℭ−Dϵ∑j=1k−1λbk+2,j+1+1−λ−αλbk+1,j+1+αλ−1bk,j+1j!σjm+j−1m−11−σj+λk+1!σk+1m+km−11−σk+1+λbk+2,k+1+1−λ−αλk!σkm+k−1m−11−σk+1−α1−1−σm.
However, this last expression is bounded by1−α, if (42) holds. This completes the proof of Theorem 2.


## 4. An Integral Operator

In this section, we consider the integral operatorGσm defined by(45)Gσmfz=∫0zΥσmttdt.Theorem 3.
Letk≥2.Then, the integral operatorGσmfzdefined by (45) is in the classℙkλ,αif and only if(46)∑j=1k−1λbk+2,j+1+1−λ−αλbk+1,j+1+αλ−1bk,j+1j!σjm+j−1m−11−σm+j+λk+1!σk+1m+km−11−σm+k+1+λbk+2,k+1+1−λ−αλk!σkm+k−1m−11−σm+k≤1−α.Proof.
From definitions (31) and (45), it is easily verified that(47)Gσmfz=z−∑n=2∞n+m−2m−1σn−11−σmznn.
Then, by Lemma1, we need only to show that H≤1−α, where(48)H=∑n=2∞nkn−α1+n−1λ×1nn+m−2m−1σn−11−σm.or equivalently,(49)H=∑n=2∞λnk+1+1−λ−αλnk+αλ−1nk−1n+m−2m−1σn−11−σm.The remaining part of the proof of Theorem 3 is similar to that of Theorem 2, and so, we omit the details.


## 5. Corollaries and Consequences

By specializing the parameterλ=1 in Theorems 1–3, we obtain the following corollaries.(52)∑j=1k−1bk+2,j+1−αbk+1,j+1j!σjm+j−1m−11−σj+bk+2,k+1−αbk+1,k+1k!σkm+k−1m−11−σk+bk+2,k+2(k+1)!σk+1m+km−11−σk+1+bk+2,1−αbk+1,11−1−σm≤1−α.Corollary 1.
Letk≥1.Then,Υσmz∈Ckαif and only if(50)∑j=1kbk+3,j+1−αbk+2,j+1j!σjm+j−1m−11−σm+j+bk+3,k+2−αk+1!σk+1m+km−11−σm+k+1+k+2!σk+2m+k+1m−11−σm+k+2≤1−α.Corollary 2.
Letk≥2andf∈ℛϵℭ,D.Then,ℐσmfz∈Ckαif(51)ℭ−Dϵ∑j=1k−1bk+2,j+1−αbk+1,j+1j!σjm+j−1m−11−σj+bk+2,k+1−αk!σkm+k−1m−11−σk+k+1!σk+1m+km−11−σk+1+1−α1−1−σm≤1−α.Corollary 3.
Letk≥2.Then, the integral operatorGσmfzdefined by (45) is in the classCkαif and only ifBy specializing the parameterk=2 in Theorems 1–3, we obtain the following corollaries.Corollary 4.
The seriesΥσmz∈ℙ2λ,αif and only if(53)8λ−4αλ−3α+7σmm−11−σm+1+219λ−5αλ−α+6σ2m+1m−11−σm+2+69λ−αλ+1σ3m+2m−11−σm+3+24λσ4m+3m−11−σm+4≤1−α.Corollary 5.
Letf∈ℛϵℭ,D.Then,ℐσmfz∈ℙ2λ,αif(54)ℭ−Dϵ4λ−2αλ+3−ασmm−11−σ+25λ−αλ+1σ2m+1m−11−σ2+6λσ3m+2m−11−σ3+1−α1−1−σm≤1−α.Corollary 6.
The integral operatorGσmfzdefined by (45) is in the classℙ2λ,αif and only if(55)4λ−2αλ+3−ασmm−11−σ+25λ−αλ+1σ2m+1m−11−σ2+6λσ3m+2m−11−σ3≤1−α.Remark 1.
Using relation (22) and Lemma 1, we can obtain new necessary and sufficient conditions and inclusion relations for the Pascal distribution series to be in the class ℙkλ,α for k=3,4,….


## 6. Conclusions

The Sălăgean differential operator plays an important role in the geometric function theory. Several authors have used this operator to define and consider the properties of certain known and new classes of analytic univalent functions (see, for example, [25, 26]). In the present paper, and due to the earlier works (see, for example, [11, 16, 18]), we find a necessary and sufficient condition and inclusion relation for the Pascal distribution series to be in the class ℙkλ,α of analytic functions associated with the Stirling numbers and Sălăgean differential operator. Furthermore, we consider an integral operator related to the Pascal distribution series. Some interesting corollaries and applications of the results are also discussed. Making use of the relation (22) could inspire researchers to find new necessary and sufficient conditions and inclusion relations for the Pascal distribution series to be in different classes of analytic functions with negative coefficients defined by the Sălăgean differential operator.



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