BINOMIAL DISTRIBUTION SERIES CONNECTED WITH CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS

ISSN: 2005-4238 IJAST 70

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BINOMIAL DISTRIBUTION SERIES CONNECTED WITH CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS

S.Santhiya¹ , K.Thilagavathi ^*2 School of Advanced Sciences, Vellore Institute of technology,

Vellore-632014, India.

Abstract

The purpose of present paper is to obtain the necessary and sufficient conditions and inclusion relations for Binomial distribution series 𝐾(𝑚, 𝑝, 𝑧) = 𝑧 + ∑^∞_𝑛=2(𝑚−𝑛)!(𝑛−1)!^(𝑚−1)! 𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛 belonging to the subclasses 𝑇(𝛿, 𝜆) and 𝐶(𝛿, 𝜆) of analytic functions with negative coefficients.

Further the integral operator 𝐻(𝑚, 𝑝, 𝑧) = ∫₀^{𝑧𝐹(𝑚,𝑝,𝑡)}_𝑡 𝑑𝑡 is related to this series is also considered.

2010 Mathematics Subject Classification: 30C45, 30C55.

Keywords: Analytic functions, Starlike functions, Convex functions, Binomial distribution series.

1. Introduction

Let 𝐴 denote the class of functions 𝑓 of the form 𝑓(𝑧) = 𝑧 + ∑ 𝑎_𝑛𝑧^𝑛

∞ 𝑛=2

. (1) Which are analytic in the punctured unit disk 𝑈 = {𝑧 ∈ 𝐶: 0 < |𝑧| < 1}.

Also 𝑇 be the subclass of 𝐴 containing the functions of the form 𝑓(𝑧) = 𝑧 − ∑|𝑎_𝑛|𝑧^𝑛

∞ 𝑛=2

. (2) Let 𝑇(𝛿, 𝜆) be a subclass of 𝑇 consisting the functions which satisfy the given condition

|

𝑧𝑓^′(𝑧)

(1 − 𝜆)𝑓(𝑧) + 𝜆𝑧𝑓^′(𝑧) − 1 𝑧𝑓^′(𝑧)

(1 − 𝜆)𝑓(𝑧) + 𝜆𝑧𝑓^′(𝑧) + 1

| < 𝛿, (0 < 𝛿 ≤ 1, 0 ≤ 𝜆 < 1). (3) and 𝑓 ∈ 𝐶(𝛿, 𝜆) if and only if 𝑧𝑓^′ ∈ 𝑇(𝛿, 𝜆) . 𝑇he class 𝑇(𝛿, 𝜆) and 𝐶(𝛿, 𝜆) is introduce by Frasin. et al [4].

A function 𝑓 ∈ 𝐴 is said to be in the class 𝑅^𝜏(𝐴, 𝐵), 𝜏 ∈ ℂ\{0}, −1 ≤ 𝐴 < 𝐵 ≤ 1, if it satisfies the inequality

|_{(𝐴−𝐵)𝜏−𝐵[𝑓}^{𝑓′(𝑧)−1}_{′(𝑧)−1]}| < 1. (4)

This class was introduced by Dixit and Pal [3]. It is interest to note that if 𝜏 = 1, 𝐴 = 𝛽 , 𝐵 = −𝛽(0 <

𝛽 ≤ 1) . We obtain the class of functions 𝑓 ∈ 𝐴 satisfying the inequality |^𝑓_𝑓^′_′^(𝑧)−1_(𝑧)+1| < 𝛽 (z ∈U, 0 < 𝛽 ≤ 1 )

which was studied by (among others) Padmanabhan [7] and Caplinger and Causey [1].

Recently Porwal[8] introduce a power series whose coefficients are probabilities of Poisson distribution. Motivated by results on connection between various subclasses of analytic functions and poisson distribution given in Murugusundaramoorthy[5]

Let 𝑓(𝑚, 𝑝) be a binomial distribution defined by

𝑓(𝑚, 𝑝) = 𝑝𝑟(𝑋 = 𝑛) =_{(𝑚−𝑛)!𝑛!}^𝑚! 𝑝^𝑛(1 − 𝑝)^𝑚−𝑛, 𝑛 = 0,1,2, … … … … . . 𝑚.

when 𝑛 > 𝑚, then 𝑓(𝑚, 𝑝) = 0.

Very recently,Waqas Nazeer. et al [12] introduce a power series whose coefficients are probabilities of Binomial distribution

Consider a power series defined as

ISSN: 2005-4238 IJAST 71

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𝐾(𝑚, 𝑝, 𝑧) = 𝑧 + ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛.

∞

𝑛=2

Now we obtain the series

𝐹(𝑚, 𝑝, 𝑧) = 2𝑧 − 𝐾(𝑚, 𝑝, 𝑧) = 𝑧 − ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛.

∞

Very recently Porwal and Kumar[9] introduced a linear operator 𝐼(𝑚, 𝑝, 𝑧): 𝐴 → 𝐴 𝑛=2

𝐼(𝑚, 𝑝, 𝑧)𝑓 = 𝐾(𝑚, 𝑝, 𝑧) ∗ 𝑓(𝑧)

= 𝑧 + ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛 𝑎_𝑛 𝑧^𝑛.

∞

Where ∗ denote the convolution or Hadamard product of two series. 𝑛=2

Most of literatures (see[2],[6],[10],[11]) used hypergeometric function to relate subclasses of analytic and univalent functions. Hence it motivate the authors to determine the necessary and sufficient conditions for 𝐹(𝑚, 𝑝, 𝑧) to be in the classes 𝑇(𝛿, 𝜆) 𝑎𝑛𝑑 𝐶(𝛿, 𝜆) and connections of these subclasses with 𝑅^𝜏(𝐴, 𝐵).Finally the condition for integral operator 𝐻(𝑚, 𝑝, 𝑧) = ∫₀^{𝑧𝐹(𝑚,𝑝,𝑡)}_𝑡 𝑑𝑡 belongs to the classes 𝑇(𝛿, 𝜆) 𝑎𝑛𝑑 𝐶(𝛿, 𝜆).

In our investigation, we shall require the following lemmas.

LEMMA 1.1.[4] A function f of the form (2) is in 𝑇(𝛿, 𝜆) if and only if satisfies

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) − (1 − 𝜆)(1 − 𝛿)]|𝑎_𝑛| ≤ 2𝛿 (5)

∞

Where 0 < 𝛿 ≤ 1,0 ≤ 𝜆 < 1. 𝑛=2

LEMMA 1.2.[4] A function f of the form (2) is in 𝐶(𝛿, 𝜆) if and only if satisfies

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) − (1 − 𝜆)(1 − 𝛿)]|𝑎𝑛| ≤ 2𝛿 (6)

∞

Where 0 < 𝛿 ≤ 1,0 ≤ 𝜆 < 1. 𝑛=2

2. Main results

THEOREM 2.1. The function 𝐹(𝑚. 𝑝. 𝑧) belong to the class 𝑇(𝛿, 𝜆) if and only if 𝑃((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1) + 2𝛿𝐸 ≤ 2𝛿

Where

𝐸 = ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1

Proof. Since

𝐹(𝑚. 𝑝. 𝑧) = 𝑧 − ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛.

∞

𝑛=2

According to lemma 1.1 we must show that

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛≤ 2𝛿 Now

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛 Writing 𝑛 = (𝑛 − 1) + 1, 𝑤𝑒 ℎ𝑎𝑣𝑒

∑[(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆)) + 2𝛿]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

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= ∑(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆))

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛+ 2𝛿 × ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 2)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛+ 2𝛿

∞

𝑛=2

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 2)! 𝑛!𝑝^𝑛+1(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1

= ((1 − 𝜆) + 𝛿(1 + 𝜆))𝑝 ∑(𝑚 − 1)(𝑚 − 2)!

(𝑚 − 𝑛 − 2)𝑛! 𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1

= 𝑝((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1) ∑ (𝑚 − 2)!

(𝑚 − 𝑛 − 2)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1= 𝑝((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1) + 2𝛿𝐸

≤ 2𝛿 . This completes the proof.

THEOREM 2.2. The function 𝐹(𝑚. 𝑝. 𝑧) belong to the class 𝐶(𝛿, 𝜆) if and only if

𝑝²((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1)(𝑚 − 2) + 2𝑝(1 + 2𝛿 + 𝛿𝜆 − 𝜆)(𝑚 − 1) + 2𝛿𝑅 ≤ 2𝛿 Where

𝑅 = ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞ 𝑛=1

. Proof. As

𝐹(𝑚. 𝑝. 𝑧) = 𝑧 − ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛,

∞

𝑛=2

According to lemma 1.2 we must show that

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛≤ 2𝛿 Now

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)! 𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛 Writing 𝑛 = (𝑛 − 1) + 1 𝑎𝑛𝑑 𝑛²= (𝑛 − 1)(𝑛 − 2) + 3(𝑛 − 1) + 1, 𝑤𝑒 ℎ𝑎𝑣𝑒

= ∑(𝑛 − 1)(𝑛 − 2)((1 − 𝜆) + 𝛿(1 + 𝜆))

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

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+ ∑(𝑛 − 1)[3((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!

× 𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛+ ∑ 2𝛿

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 3)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛+ 2(1 + 2𝛿 + 𝛿𝜆 − 𝜆)

∞

𝑛=3

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 2)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛+ 2𝛿 ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛

∞ 𝑛=2

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 3)! 𝑛!𝑝^𝑛+2(1 − 𝑝)^{𝑚−𝑛−3}+ 2(1 + 2𝛿 + 𝛿𝜆 − 𝜆)

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 2)! 𝑛!𝑝^𝑛+1(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿 ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞ 𝑛=1

∞

𝑛=0

= 𝑝²((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 3)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−3}+ 2𝑝(1 + 2𝛿 + 𝛿𝜆 − 𝜆)

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 2)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿 ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞ 𝑛=1

∞

𝑛=0

= 𝑝²((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1)(𝑚 − 2) ∑ (𝑚 − 3)!

(𝑚 − 𝑛 − 3)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−3}

∞

𝑛=0

+2𝑝(1 + 2𝛿 + 𝛿𝜆 − 𝜆)(𝑚 − 1) ∑ (𝑚 − 2)!

(𝑚 − 𝑛 − 2)! 𝑛!

∞ 𝑛=0

𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

= 𝑝𝑛=1²((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1)(𝑚 − 2) + 2𝑝(1 + 2𝛿 + 𝛿𝜆 − 𝜆)(𝑚 − 1) + 2𝛿𝑅

≤ 2𝛿.

This completes the proof.

In the following theorem, we obtain the analogous results in connection with the particular integral operator 𝐻(𝑚, 𝑝, 𝑧) as follows

𝐻(𝑚, 𝑝, 𝑧) = ∫𝐹(𝑚, 𝑝, 𝑡)

𝑡 𝑑𝑡 (7)

𝑧

THEOREM 2.3. The operator 𝐻(𝑚. 𝑝. 𝑧) characterized by (7) to the class 𝐶(𝛿, 𝜆) if and only if 0

𝑝((1 − 𝜆) + 𝛿(1 + 𝜆)(𝑚 − 1) + 2𝛿𝑉 ≤ 2𝛿 Where

𝑉 = ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞ 𝑛=1

. Proof. Since

𝐻(𝑚. 𝑝. 𝑧) = 𝑧 − ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛,

∞

𝑛=2

According to lemma 1.2 we must show that

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛≤ 2𝛿

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Or, equivalently

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛≤ 2𝛿 Now

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

= ∑[(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆)) + 2𝛿]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

= ∑(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆))

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛+ 2𝛿 × ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 2)! 𝑛!𝑝^𝑛+1(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1

= ((1 − 𝜆) + 𝛿(1 + 𝜆))𝑝 ∑(𝑚 − 1)(𝑚 − 2)!

(𝑚 − 𝑛 − 2)! 𝑛! 𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

∞

𝑛=0

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

= 𝑝((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1) + 2𝛿𝑉 𝑛=1

≤ 2𝛿.

This completes the proof.

THEOREM 2.4. The operator 𝐻(𝑚. 𝑝. 𝑧) characterized by (7) to the class 𝑇(𝛿, 𝜆) if and only if 𝑝((1 − 𝜆) + 𝛿(1 + 𝜆)𝐷 + 2𝛿𝐽 ≤ 2𝛿

Where

𝐷 = ∑ (𝑚 − 1)(𝑚 − 2)!

(𝑚 − 𝑛 − 2)! (𝑛 + 2)𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−2}

∞

𝑛=0

and

𝐽 = ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛

∞ 𝑛=2

. Proof. Since

𝐻(𝑚. 𝑝. 𝑧) = 𝑧 − ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛𝑧^𝑛,

∞

𝑛=2

According to lemma 1.1 we must show that

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛≤ 2𝛿 Now

∑[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

= ∑[(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆)) + 2𝛿]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

ISSN: 2005-4238 IJAST 75

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= ∑(𝑛 − 1)((1 − 𝜆) + 𝛿(1 + 𝜆)) (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛+ 2𝛿

∞

𝑛=2

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛(𝑛 − 2)!

∞ 𝑛=2

𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛+ 2𝛿

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

∞ 𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆)) ∑ (𝑚 − 1)(𝑚 − 2)!

(𝑚 − 𝑛 − 2)! (𝑛 + 2)𝑛!

∞ 𝑛=0

𝑝^𝑛+1 (1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

∞

𝑛=2

= ((1 − 𝜆) + 𝛿(1 + 𝜆))𝑝 ∑ (𝑚 − 1)(𝑚 − 2)!

(𝑚 − 𝑛 − 2)! (𝑛 + 2)𝑛!

∞ 𝑛=0

𝑝^𝑛 (1 − 𝑝)^{𝑚−𝑛−2}+ 2𝛿

× ∑ (𝑚 − 1)!

(𝑚 − 𝑛)! 𝑛!𝑝^𝑛−1 (1 − 𝑝)^𝑚−𝑛

∞

= 𝑝((1 − 𝜆) + 𝛿(1 + 𝜆))𝐷 + 2𝛿𝐽 𝑛=2

≤ 2𝛿.

This completes the proof.

3. Inclusion Properties

Making use of the following Lemma we will study the action of the Binomial distribution series on the class 𝐶(𝛿, 𝜆). .

Lemma 3.1. [2] If 𝑓 ∈ 𝑅^𝜏(𝐴, 𝐵) is of form (1), then |𝑎_𝑛|≤ (A−B)^|τ| _𝑛, n ∈N\{1}.

The result is sharp.

THEOREM 3.1. Let 𝑛 > 𝑚, 0 < 𝛿 ≤ 1 𝑎𝑛𝑑 0 ≤ 𝜆 < 1 𝑖𝑓 𝑓 ∈ 𝑅^𝜏(𝐴, 𝐵), 𝑡ℎ𝑒𝑛 𝐼(𝑚, 𝑝, 𝑧)𝑓 is in 𝐶(𝛿, 𝜆) if and only if

(𝐴 − 𝐵)|𝜏|[𝑝((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1)]

(1 − (𝐴 − 𝐵)|𝜏|𝐺) ≤ 2𝛿

Where

𝐺 = ∑ (𝑚 − 1)!

(𝑚 − 𝑛 − 1)! 𝑛!𝑝^𝑛(1 − 𝑝)^{𝑚−𝑛−1}

∞

𝑛=1

Proof. In view of lemma (1.1) it suffices to show that

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛|𝑎_𝑛| ≤ 2𝛿 Since 𝑓 ∈ 𝑅^𝜏(𝐴, 𝐵), then by lemma (3.1) we get

|𝑎_𝑛| ≤ (𝐴 − 𝐵)|𝜏|

𝑛 Thus we have

∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛|𝑎_𝑛|

ISSN: 2005-4238 IJAST 76

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≤ ∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞ 𝑛=2

(𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛 × (𝐴 − 𝐵)|𝜏|

𝑛

= (𝐴 − 𝐵)|𝜏| ∑ 𝑛[𝑛((1 − 𝜆) + 𝛿(1 + 𝜆)) + (1 − 𝜆)(𝛿 − 1)]

∞

𝑛=2

× (𝑚 − 1)!

(𝑚 − 𝑛)! (𝑛 − 1)!𝑝^𝑛−1(1 − 𝑝)^𝑚−𝑛 Proceeding as in Theorem 2.1 we get

= (𝐴 − 𝐵)|𝜏|[𝑃((1 − 𝜆) + 𝛿(1 + 𝜆))(𝑚 − 1) + 2𝛿𝐺]

≤ 2𝛿.

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