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eISSN 0454-8124
pISSN 1225-6951

### Article

KYUNGPOOK Math. J. 2019; 59(2): 301-314

Published online June 23, 2019

Copyright © Kyungpook Mathematical Journal.

### Pascal Distribution Series Connected with Certain Subclasses of Univalent Functions

Sheeza M. El-Deeb*,Teodor Bulboacă, Jacek Dziok

Department of Mathematics, Faculty of Science, Damietta University, New Damietta 34517, Egypt
e-mail : shezaeldeeb@yahoo.com

Faculty of Mathematics and Computer Science, Babe¸s-Bolyai University, 400084 Cluj-Napoca, Romania
e-mail : bulboaca@math.ubbcluj.ro

Faculty of Mathematics and Natural Sciences, University of Rzeszow, 35-310 Rzeszow, Poland
e-mail : jdziok@ur.edu.pl

Received: November 5, 2018; Revised: March 13, 2019; Accepted: March 18, 2019

### Abstract

The aim of this article is to make a connection between the Pascal distribution series and some subclasses of normalized analytic functions whose coefficients are probabilities of the Pascal distribution. For these functions, for linear combinations of these functions and their derivatives, for operators defined by convolution products, and for the Alexander-type integral operator, we find simple sufficient conditions such that these mapping belong to a general class of functions defined and studied by Goodman, Rønning, and Bharati et al.

### 1. Introduction

Let denote the class of analytic functions of the form

$f(z)=z+∑n=2∞anzn, z∈U:={z∈ℂ:∣z∣ <1},$

and let be the subclass of consisting of functions that are univalent in .

Furthermore, for given by

$g(z)=z+∑n=2∞bnzn, z∈U$

the Hadamard (or convolution) product of the functions f and g is given by (see [5])

$(f*g)(z):=z+∑n=2∞anbnzn, z∈U.$

If f and g are analytic functions in , we say that f is subordinate to g, written f(z) ≺ g(z), if there exists a Schwarz function w, which is analytic in with w(0) = 0 and |w(z)| < 1 for all , such that f(z) = g(w(z)), . Furthermore, if the function g is univalent in , then we have the following equivalence (see [3] and [13]):

$f(z)≺g(z)⇔f(0)=g(0) and f(U)⊂g(U).$

Goodman in [6, 7], Rønning in [15, 16], and Bharati et al. in [2] introduced and studied the following subclasses:

• A function is said to be in the class of k-uniformly starlike functions of order β if it satisfies the condition $Re (zf′(z)f(z)-β)>k|zf′(z)f(z)-1|, z∈U,$

where 0 ≤ β < 1 and k ≥ 0.

• A function is said to be in the class of k-uniformly convex functions of order β if it satisfies the condition $Re (1+zf″(z)f′(z)-β)>k|zf″(z)f′(z)|, z∈U,$

where 0 ≤ β < 1 and k ≥ 0.

It follows from (1.2) and (1.3) that for a function we have the equivalence

$f∈k-UCV(β)⇔zf′(z)∈k-Sp(β).$

For β = 0 the classes and reduce to the classes and respectively, that have been introduced and studied by Kanas and Wisniowska [10, 11].

### Definition 1.1.([18])

For −1 ≤ A < B ≤ 1 and $∣θ∣ <π2$, a function is said to be in the class ℒ(A,B, θ) if it satisfies the subordination condition

$eiθf′(z)≺1+Az1+Bzcos θ+i sin θ.$

Using the definition of the subordination, a function belongs to the class ℒ(A,B, θ) if and only if there exists an analytic function w, satisfying w(0) = 0 and |w(z)| < 1 for all , such that

$eiθf′(z)=1+Aw(z)1+Bw(z)cos θ+i sin θ, z∈U,$

or equivalently

$|eiθ(f′(z)-1)Beiθf′(z)-[Beiθ+(A-B)cosθ]|<1, z∈U.$

We note that for suitable choices of A, B and θ we obtain the following subclasses defined and studied by different authors:

• $L (2α-1,1,θ)=:L (0;α) (0≤α<1,∣θ∣<π2)$ (see Kanas and Srivastava [9]);

• ℒ(β(2α − 1), β, 0) =: ℛ (β; α) (0 < β ≤ 1, 0 ≤ α < 1) (see Juneja and Mogra [8]);

• (see Caplinger and Causey [4]);

• $L (A+(B-A)α,B,θ)=:L (A,B,θ;α) (-1≤A (see Aouf [1]).

A variable x is said to have the Pascal distribution if it takes the values 0, 1, 2, 3, … with the probabilities (1−q)r, $qr(1-q)r1!, q2r(r+1)(1-q)r2!, q3r(r+1)(r+2)(1-q)r3!$, …, respectively, where q, r are called the parameters, and thus

$P(X=k)=(k+r-1r-1)qk(1-q)r,k∈{0,1,2,3,…}.$

Now we introduced a power series whose coefficients are probabilities of the Pascal distribution, that is

$Qqr(z):=z+∑n=2∞(n+r-2r-1)qn-1(1-q)rzn, z∈U (r≥1, 0≤q≤1).$

Note that, by using ratio test we deduce that the radius of convergence of the above power series is infinity.

We will define the functions

$Nq,λr(z):=(1-λ)Qqr(z)+λz (Qqr(z))′=z+∑n=2∞(n+r-2r-1) [1+λ(n-1)] qn-1(1-q)rzn, z∈U, (r≥1, 0≤q≤1, λ≥0)$

and

$Mq,λ,γr(z):=(1-λ+γ)Qqr(z)+(λ-γ)z (Qqr(z))′+λγz2 (Qqr(z))″=z+∑n=2∞(n+r-2r-1)[1+(n-1)(λ-γ+nλγ)] qn-1(1-q)rzn, z∈U, (r≥1, 0≤q≤1, λ,γ≥0),$

and we introduce the linear operator $Pqr:A→A$ defined by

$Pqr(f)(z):=Qqr(z)*f(z)=z+∑n=2∞(n+r-2r-1)qn-1(1-q)ranzn, z∈U,(r≥1, 0≤q≤1).$

In the present paper we established some sufficient conditions for the Pascal distribution series and other related series to be in some subclasses of analytic functions. Also, we studied similar properties for an integral transform related to this series.

### 2. Preliminaries

Unless otherwise mentioned, we assume that −1 ≤ A < B ≤ 1, $∣θ∣ <π2$, r ≥ 1, 0 ≤ q ≤ 1, λ, γ ≥ 0 and λγ. To prove our results we need the following lemmas.

### Lemma 2.1.([1, Theorem 4 for p = 1 and α = 0])

If the functionis of the form (1.1) and

$∑n=2∞n(1+∣B∣)∣an∣ ≤(B-A) cos θ,$

then f ∈ ℒ (A,B, θ).

### Lemma 2.2.([1, Theorem 1 for p = 1 and α = 0])

If the function f ∈ ℒ (A,B, θ) is of the form (1.1), then

$∣an∣ ≤(B-A) cos θn$

for every n ≥ 2. The estimate is sharp.

### Lemma 2.3.[10, Theorem 3.3]

If, and for some k (0 ≤ k < ∞) the following inequality holds

$∑n=2∞n(n-1)∣an∣≤1k+2,$

then. The number$1k+2$cannot be increased.

### Lemma 2.4.([17, Theorem 2.1])

If the functionis of the form (1.1) and

$∑n=2∞[n(1+k)-(k+β)]∣an∣≤1-β,$

then

### Theorem 3.1

A sufficient condition for the function$Qqr$given by (1.4) to be in the class ℒ(A,B, θ) is

$qr1-q-(1-q)r+1≤(B-A) cos θ1+∣B∣.$
Proof

To prove that $Qqr∈L(A,B,θ)$, according to Lemma 2.1 it is sufficient to show that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)∣q∣n-1∣1-q∣r≤(B-A) cos θ.$

In the proofs of all our results we will use the relation

$∑n=0∞(n+r-1r-1)qn=1(1-q)r, 0≤q≤1,$

and the corresponding ones obtained by replacing the value of r with r − 1, r + 1, and r + 2.

Thus, from the definition relation (1.4) combined with the assumption (3.1) we get

$∑n=2∞(n+r-2r-1)n(1+∣B∣)qn-1(1-q)r=(1+∣B∣)(1-q)r [∑n=2∞(n+r-2r-1)(n-1)qn-1+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [∑n=2∞qr(n+r-2r) qn-2+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [qr∑n=0∞(n+rr) qn+∑n=1∞(n+r-1r-1) qn]=(1+∣B∣)(1-q)r [qr∑n=0∞(n+rr) qn+∑n=0∞(n+r-1r-1) qn-1]=(1+∣B∣) [qr1-q+1-(1-q)r]≤(B-A) cos θ,$

and from Lemma 2.1 it follows that $Qqr∈L(A,B,θ)$.

### Theorem 3.2

A sufficient condition for the function$Nq,λr$defined by (1.5) to be in the class ℒ(A,B, θ) is

$(1+2λ)qr1-q+1-(1-q)r+λq2r(r+1)(1-q)2≤(B-A) cos θ1+∣B∣.$
Proof

From Lemma 2.1, to prove that $Nq,λr∈L(A,B,θ)$ it is sufficient to show that

$∑n=2∞(n+r-2r-1) [n(1+∣B∣)] [1+λ(n-1)] ∣q∣n-1∣1-q∣r≤(B-A) cos θ.$

Thus,

$∑n=2∞(n+r-2r-1) [n (1+∣B∣)] [1+λ(n-1)] qn-1(1-q)r=(1+∣B∣)(1-q)r [∑n=2∞(n+r-2r-1) (1+2λ)(n-1)qn-1+∑n=2∞(n+r-2r-1) qn-1+∑n=2∞(n+r-2r-1)λ(n-1)(n-2)qn-1]=(1+∣B∣)(1-q)r[∑n=2∞qr(1+2λ)(n+r-2r) qn-2+∑n=2∞(n+r-2r-1) qn-1+λq2∑n=3∞(n+r-2r-1)(n-1)(n-2)qn-3]=(1+∣B∣)(1-q)r [(1+2λ)qr∑n=0∞(n+rr) qn+∑n=0∞(n+r-1r-1) qn-1+λq2r(r+1)∑n=0∞(n+r+1r+1) qn]=(1+∣B∣)(1-q)r [(1+2λ)qr1(1-q)r+1+1(1-q)r-1+λq2r(r+1)1(1-q)r+2]=(1+∣B∣) [(1+2λ)qr1-q+1-(1-q)r+λq2r(r+1)(1-q)2].$

According to (3.2), from the above identity we conclude it follows that the inequality (3.3) holds, and therefore $Nq,λr∈L(A,B,θ)$.

### Theorem 3.3

A sufficient condition for the function$Mq,λ,γr$given by (1.6) to be in the class ℒ(A,B, θ) is

$(1+2λ-2γ+4λγ) qr1-q+1-(1-q)r+(λ-γ+5λγ) q2r(r+1)(1-q)2+λγq3r(r+1)(r+2)(1-q)3≤(B-A) cos θ1+∣B∣.$
Proof

To prove that $Mq,λ,γr∈L(A,B,θ)$, from Lemma 2.1 it is sufficient to show that

$∑n=2∞(n+r-2r-1)n(1+∣B∣) [1+(n-1)(λ-γ+nλγ)] ∣q∣n-1∣1-q∣r≤(B-A) cos θ.$

From (3.4) it follows that

$∑n=2∞(n+r-2r-1)n(1+∣B∣) [1+(n-1) (λ-γ+nλγ)] qn-1(1-q)r=(1+∣B∣)(1-q)r [∑n=2∞(n+r-2r-1) (1+2λ-2γ+4λγ)(n-1)qn-1+∑n=2∞(n+r-2r-1) qn-1+∑n=2∞(n+r-2r-1) (λ-γ+5λγ)(n-1)(n-2)qn-1+∑n=2∞(n+r-2r-1)λγ(n-1)(n-2)(n-3)qn-1]=(1+∣B∣)(1-q)r [∑n=2∞qr(1+2λ-2γ+4λγ)(n+r-2r) qn-2+∑n=2∞(n+r-2r-1) qn-1+∑n=3∞(λ-γ+5λγ) q2(n+r-2r-1)(n-1)(n-2)qn-3+∑n=4∞λγq3(n+r-2r-1) qn-4(n-1)(n-2)(n-3)]=(1+∣B∣)(1-q)r [(1+2λ-2γ+4λγ)qr∑n=0∞(n+rr) qn+∑n=0∞(n+r-1r-1) qn-1+(λ-γ+5λγ)q2r(r+1)∑n=0∞(n+r+1r+1) qn+λγq3r(r+1)(r+2)∑n=0∞(n+r+2r+2) qn]=(1+∣B∣)(1-q)r [(1+2λ-2γ+4λγ) qr1(1-q)r+1+1(1-q)r-1+(λ-γ+5λγ) q2r(r+1)1(1-q)r+2+λγq3r(r+1)(r+2)1(1-q)r+3]=(1+∣B∣) [(1+2λ-2γ+4λγ) qr1-q+1-(1-q)r+(λ-γ+5λγ) q2r(r+1)(1-q)2+λγq3r(r+1)(r+2)(1-q)3]≤(B-A) cos θ,$

that is (3.5) holds, hence $Mq,λ,γr∈L(A,B,θ)$.

### Theorem 3.4

• If the condition$q2r(r+1)(1-q)2+3qr1-q+1-(1-q)r≤(B-A) cos θ1+∣B∣$

holds, then the operator$Pqr$defined by (1.7) maps the classto the class ℒ(A,B, θ), that is$Pqr(S*)⊂L (A,B,θ)$.

• If the condition (3.1) is satisfied, then the operator$Pqr$maps the classto the class ℒ(A,B, θ), that is$Pqr(K)⊂L(A,B,θ)$.

Proof

According to Lemma 2.1, to prove that $Pqr(f)∈L(A,B,θ)$ it is sufficient to show that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)∣q∣n-1∣1-q∣r∣an∣ ≤(B-A) cos θ.$

(i) If has the form (1.1), then the well-known inequality |an| ≤ n holds for all n ≥ 2 (see [12] and [14]), and using (3.6) we obtain that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)qn-1(1-q)r∣an∣≤∑n=2∞(n+r-2r-1) n2(1+∣B∣)qn-1(1-q)r=(1+∣B∣)(1-q)r [∑n=2∞(n+r-2r-1)(n-1)(n-2)qn-1+3∑n=2∞(n+r-2r-1)(n-1)qn-1+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [q2∑n=3∞(n+r-2r-1)(n-1)(n-2)qn-3+3q∑n=2∞(n+r-2r-1)(n-1)qn-2+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [q2r(r+1)∑n=0∞(n+r+1r+1) qn+3qr∑n=0∞(n+rr) qn+∑n=1∞(n+r-1r-1) qn]=(1+∣B∣)(1-q)r[q2r(r+1)(1-q)r+2+3qr(1-q)r+1+1(1-q)r-1]=(1+∣B∣) [q2r(r+1)(1-q)2+3qr1-q+1-(1-q)r]≤(B-A) cos θ,$

that is (3.7) holds, hence $Pqr(f)∈L(A,B,θ)$.

(ii) If is of the form (1.1), then the well-known inequality |an| ≤ 1 holds for all n ≥ 2 (see [12]), and according to (3.1) we obtain that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)qn-1(1-q)r∣an∣≤∑n=2∞(n+r-2r-1)n(1+∣B∣)qn-1(1-q)r=(1+∣B∣)(1-q)r [∑n=2∞(n+r-2r-1)(n-1)qn-1+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [∑n=2∞q(n+r-2r-1)(n-1)qn-2+∑n=2∞(n+r-2r-1) qn-1]=(1+∣B∣)(1-q)r [qr∑n=0∞(n+rr) qn+∑n=1∞(n+r-1r-1) qn]=(1+∣B∣)(1-q)r [qr∑n=0∞(n+rr) qn+∑n=0∞(n+r-1r-1) qn-1]=(1+∣B∣) [qr1-q+1-(1-q)r]≤(B-A) cos θ,$

hence (3.7) holds, and therefore $Pqr(f)∈L(A,B,θ)$.

### Theorem 3.5

If the condition

$(B-A)qr cos θ≤1-qk+2$

holds, then the operator$Pqr$maps the class ℒ(A,B, θ) to the class, that is$Pqr(L(A,B,θ))⊂k-UCV$.

Proof

If f ∈ ℒ(A,B, θ) has the form (1.1), since $Pqr$ is given by (1.7), using Lemma 2.3 we need to prove that

$∑n=2∞(n+r-2r-1)n(n-1)∣q∣n-1∣1-q∣r∣an∣≤1k+2.$

From Lemma 2.2 and the assumption (3.8) we obtain that

$∑n=2∞(n+r-2r-1)n(n-1)qn-1(1-q)r∣an∣≤∑n=2∞(n+r-2r-1)(n-1)(B-A)qn-1(1-q)r cos θ=(B-A)(1-q)r cos θ∑n=2∞(n-1)(n+r-2r-1) qn-1=(B-A)(1-q)rqr cos θ∑n=0∞(n+rr) qn=(B-A)qr cos θ1-q≤1k+2,$

hence (3.9) holds, and consequently $Pqr(f)∈k-UCV$.

### Theorem 3.6

If the condition

$(B-A) cos θ [k+1-(1-q)r-k(1-q)q(r-1)(1-(1-q)r-1)]≤1$

holds, then$Pqr$maps the class ℒ(A,B, θ) to the class, that is$Pqr(L(A,B,θ))⊂k-Sp$.

Proof

If f ∈ ℒ(A,B, θ) has the power expansion series (1.1), and $Pqr$ is given by (1.7), according to Lemma 2.4 for β = 0 we need to prove that

$∑n=2∞(n+r-2r-1)[n(k+1)-k] ∣q∣n-1∣1-q∣r∣an∣≤1.$

From Lemma 2.2 and the assumption (3.10) we may easily show that

$∑n=2∞(n+r-2r-1) [n(k+1)-k] ∣q∣n-1∣1-q∣r∣an∣≤∑n=2∞(n+r-2r-1) [n(k+1)-k] qn-1(1-q)r(B-A) cos θn=(B-A)(1-q)r cos θ [∑n=2∞(k+1)(n+r-2r-1) qn-1-∑n=2∞kn(n+r-2r-1) qn-1]=(B-A)(1-q)r cos θ [(k+1)(∑n=0∞(n+r-1r-1) qn-1)-kq(r-1)(∑n=0∞(n+r-2r-2) qn-1-(r-1)q)]=(B-A) cos θ[(k+1) [1-(1-q)r]-kq(r-1)[(1-q)-(1-q)r-q(r-1)(1-q)r]]=(B-A) cos θ [k+1-(1-q)r-k(1-q)q(r-1)(1-(1-q)r-1)]≤1,$

that is (3.11) holds, and thus $Pqr(f)∈k-Sp$.

In the following two theorems we will obtain analogues results in connection with the function $Iqr$ defined by

$Iqr(z):=∫0zQqr(t)tdt, z∈U.$

### Theorem 3.7

A sufficient condition for the function$Iqr$to be in the class ℒ(A,B, θ) is

$(1+∣B∣) [1-(1-q)r]≤(B-A) cos θ.$
Proof

Since

$Iqr(z)=z+∑n=2∞(n+r-2r-1) qn-1(1-q)rznn, z∈U,$

to prove that $Iqr∈L (A,B,θ)$, according to Lemma 2.1 it is sufficient to show that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)n∣q∣n-1∣1-q∣r≤(B-A) cos θ.$

Using the assumption (3.12), a simple computation shows that

$∑n=2∞(n+r-2r-1)n(1+∣B∣)nqn-1(1-q)r=(1+∣B∣)(1-q)r∑n=2∞(n+r-2r-1) qn-1=(1+∣B∣)(1-q)r∑n=1∞(n+r-1r-1) qn=(1+∣B∣)(1-q)r [1(1-q)r-1]=(1+∣B∣) [1-(1-q)r]≤(B-A) cos θ.$

and from (3.14) if follows that $Iqr∈L (A,B,θ)$.

### Theorem 3.8

A sufficient condition for the function$Iqr$to be in the classis

$k+1-(1-β)(1-q)r-(k+β)(1-q)q(r-1)(1-(1-q)r-1)≤1-β.$
Proof

Since $Iqr$ has the form (3.13), to prove that $Iqr∈k-Sp(β)$, according to Lemma 2.4 it is sufficient to prove that

$∑n=2∞(n+r-2r-1)n(k+1)-(k+β)n∣q∣n-1∣1-q∣r≤1-β.$

Thus, from the assumption (3.15) we get

$∑n=2∞(n+r-2r-1)[n(k+1)-(k+β)]nqn-1(1-q)r=(1-q)r [(k+1)∑n=2∞(n+r-2r-1) qn-1-∑n=2∞k+βn(n+r-2r-1) qn-1]=(1-q)r [(k+1)∑n=1∞(n+r-1r-1) qn-k+βq(r-1)∑n=2∞(n+r-2r-2) qn]=(1-q)r [(k+1) (∑n=0∞(n+r-1r-1) qn-1)-k+βq(r-1)(∑n=0∞(n+r-2r-2) qn-1-q(r-1))]=(k+1) [1-(1-q)r]-k+βq(r-1)[1-q-[1+q(r-1)] (1-q)r]=k+1-(1-β)(1-q)r-(k+β)(1-q)q(r-1)(1-(1-q)r-1)≤1-β.$

and from (3.16) we conclude that $Iqr∈k-Sp(β)$.

### Acknowledgements

The authors are grateful to the reviewers of this article, that gave valuable remarks, comments, and advices, in order to revise and improve the results of the paper.

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