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Research ArticleOn a Class of 𝑞-Bernoulli, 𝑞-Euler, and 𝑞-Genocchi Polynomials

N. I. Mahmudov and M. Momenzadeh

Eastern Mediterranean University, Gazimagusa, Northern Cyprus, Turkey

Correspondence should be addressed to M. Momenzadeh; [email protected]

Received 20 January 2014; Revised 3 April 2014; Accepted 3 April 2014; Published 28 April 2014

Academic Editor: Sofiya Ostrovska

Copyright © 2014 N. I. Mahmudov and M. Momenzadeh. This is an open access article distributed under the Creative CommonsAttribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work isproperly cited.

The main purpose of this paper is to introduce and investigate a class of 𝑞-Bernoulli, 𝑞-Euler, and 𝑞-Genocchi polynomials. The𝑞-analogues of well-known formulas are derived. In addition, the 𝑞-analogue of the Srivastava-Pintér theorem is obtained. Somenew identities, involving 𝑞-polynomials, are proved.

1. Introduction

Throughout this paper, we always make use of the classicaldefinition of quantum concepts as follows.

The 𝑞-shifted factorial is defined by

(𝑎; 𝑞)0= 1, (𝑎; 𝑞)

𝑛=

𝑛−1

∏

𝑗=0

(1 − 𝑞𝑗𝑎) , 𝑛 ∈ N,

(𝑎; 𝑞)∞=

∞

∏

𝑗=0

(1 − 𝑞𝑗𝑎) ,

𝑞 < 1, 𝑎 ∈ C.

(1)

It is known that

(𝑎; 𝑞)𝑛=

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

𝑞(1/2)𝑘(𝑘−1)

(−1)𝑘𝑎𝑘. (2)

The 𝑞-numbers and 𝑞-factorial are defined by

[𝑎]𝑞 =1 − 𝑞𝑎

1 − 𝑞, (𝑞 ̸= 1, 𝑎 ∈ C) ;

[0]𝑞! = 1, [𝑛]𝑞! = [𝑛]𝑞[𝑛 − 1]𝑞!.

(3)

The 𝑞-polynomial coefficient is defined by

[𝑛

𝑘]

𝑞

=(𝑞; 𝑞)𝑛

(𝑞; 𝑞)𝑛−𝑘(𝑞; 𝑞)𝑘

, (𝑘 ⩽ 𝑛, 𝑘, 𝑛 ∈ N) . (4)

In the standard approach to the 𝑞-calculus two exponen-tial functions are used, these 𝑞-exponential functions andimproved type 𝑞-exponential function (see [1]) are defined asfollows:

𝑒𝑞(𝑧) =

∞

∑

𝑛=0

𝑧𝑛

[𝑛]𝑞!=

∞

∏

𝑘=0

1

(1 − (1 − 𝑞) 𝑞𝑘𝑧),

0 <𝑞 < 1, |𝑧| <

11 − 𝑞

,

𝐸𝑞 (𝑧) = 𝑒1/𝑞 (𝑧) =

∞

∑

𝑛=0

𝑞(1/2)𝑛(𝑛−1)

𝑧𝑛

[𝑛]𝑞!

=

∞

∏

𝑘=0

(1 + (1 − 𝑞) 𝑞𝑘𝑧) , 0 <

𝑞 < 1, 𝑧 ∈ C,

E𝑞(𝑧) = 𝑒

𝑞(𝑧

2)𝐸𝑞(𝑧

2) =

∞

∑

𝑛=0

(−1, 𝑞)𝑛

2𝑛

𝑧𝑛

[𝑛]𝑞!

=

∞

∏

𝑘=0

(1 + (1 − 𝑞) 𝑞𝑘(𝑧/2))

(1 − (1 − 𝑞) 𝑞𝑘 (𝑧/2)), 0 < 𝑞 < 1, |𝑧|<

2

1 − 𝑞.

(5)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014, Article ID 696454, 10 pageshttp://dx.doi.org/10.1155/2014/696454

2 Abstract and Applied Analysis

The form of improved type of 𝑞-exponential function E𝑞(𝑧)

motivated us to define a new 𝑞-addition and 𝑞-subtraction as

(𝑥 ⊕𝑞𝑦)𝑛

:=

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑘(−1, 𝑞)

𝑛−𝑘

2𝑛𝑥𝑘𝑦𝑛−𝑘,

𝑛 = 0, 1, 2, . . . ,

(𝑥⊖𝑞𝑦)𝑛

:=

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑘(−1, 𝑞)

𝑛−𝑘

2𝑛𝑥𝑘(−𝑦)𝑛−𝑘,

𝑛 = 0, 1, 2, . . . .

(6)

It follows that

E𝑞 (𝑡𝑥)E𝑞 (𝑡𝑦) =

∞

∑

𝑛=0

(𝑥 ⊕𝑞𝑦)𝑛 𝑡𝑛

[𝑛]𝑞!. (7)

The Bernoulli numbers {𝐵𝑚}𝑚≥0

are rational numbers in asequence defined by the binomial recursion formula:

𝑚

∑

𝑘=0

(𝑚

𝑘)𝐵𝑘− 𝐵𝑚= {

1, 𝑚 = 1,

0, 𝑚 > 1,(8)

or equivalently, the generating function

∞

∑

𝑘=0

𝐵𝑘

𝑡𝑘

𝑘!=

𝑡

𝑒𝑡 − 1. (9)

𝑞-Analogues of the Bernoulli numbers were first studiedby Carlitz [2] in the middle of the last century when heintroduced a new sequence {𝛽

𝑚}𝑚⩾0

:

𝑚

∑

𝑘=0

(𝑚

𝑘)𝛽𝑘𝑞𝑘+1

− 𝛽𝑚= {

1, 𝑚 = 1,

0, 𝑚 > 1.(10)

Here and in the remainder of the paper, for the parameter 𝑞we make the assumption that |𝑞| < 1. Clearly we recover (8)if we let 𝑞 → 1 in (10). The 𝑞-binomial formula is known as

(1 − 𝑎)𝑛

𝑞= (𝑎; 𝑞)

𝑛=

𝑛−1

∏

𝑗=0

(1 − 𝑞𝑗𝑎)

=

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

𝑞(1/2)𝑘(𝑘−1)

(−1)𝑘𝑎𝑘.

(11)

The above 𝑞-standard notation can be found in [3].Carlitz has introduced the 𝑞-Bernoulli numbers and poly-

nomials in [2]. Srivastava and Pintér proved some relationsand theorems between the Bernoulli polynomials and Eulerpolynomials in [4]. They also gave some generalizationsof these polynomials. In [4–16], the authors investigatedsome properties of the 𝑞-Euler polynomials and 𝑞-Genocchipolynomials. They gave some recurrence relations. In [17],Cenkci et al. gave the 𝑞-extension of Genocchi numbers ina different manner. In [18], Kim gave a new concept for the𝑞-Genocchi numbers and polynomials. In [19], Simsek et al.investigated the 𝑞-Genocchi zeta function and 𝑙-function by

using generating functions andMellin transformation.Thereare numerous recent studies on this subject by, among manyother authors, Cigler [20], Cenkci et al. [17, 21], Choi et al.[22], Cheon [23], Luo and Srivastava [8–10], Srivastava et al.[4, 24], Nalci and Pashaev [25] Gaboury and Kurt, [26], Kimet al. [27], and Kurt [28].

We first give the definitions of the 𝑞-numbers and 𝑞-polynomials. It should be mentioned that the definition of 𝑞-Bernoulli numbers in Definition 1 can be found in [25].

Definition 1. Let 𝑞 ∈ C, 0 < |𝑞| < 1. The 𝑞-Bernoulli numbersb𝑛,𝑞

and polynomials B𝑛,𝑞(𝑥, 𝑦) are defined by means of the

generating functions:

B̂ (𝑡) :=𝑡𝑒𝑞(−𝑡/2)

𝑒𝑞 (𝑡/2) − 𝑒𝑞 (−𝑡/2)

=𝑡

E𝑞 (𝑡) − 1

=

∞

∑

𝑛=0

b𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 2𝜋,

𝑡

E𝑞 (𝑡) − 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦)

=

∞

∑

𝑛=0

B𝑛,𝑞(𝑥, 𝑦)

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 2𝜋.

(12)

Definition 2. Let 𝑞 ∈ C, 0 < |𝑞| < 1. The 𝑞-Euler numberse𝑛,𝑞

and polynomials E𝑛,𝑞(𝑥, 𝑦) are defined by means of the


Ê (𝑡) :=2𝑒𝑞(−𝑡/2)

𝑒𝑞 (𝑡/2) + 𝑒𝑞 (−𝑡/2)

=2

E𝑞 (𝑡) + 1

=

∞

∑

𝑛=0

e𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 𝜋,

2

E𝑞(𝑡) + 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦)

=

∞

∑

𝑛=0

E𝑛,𝑞(𝑥, 𝑦)

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 𝜋.

(13)

Definition 3. Let 𝑞 ∈ C, 0 < |𝑞| < 1.The 𝑞-Genocchi numbersg𝑛,𝑞

and polynomials G𝑛,𝑞(𝑥, 𝑦) are defined by means of the


Ĝ (𝑡) :=2𝑡𝑒𝑞(−𝑡/2)

𝑒𝑞(𝑡/2) + 𝑒

𝑞(−𝑡/2)

=2𝑡

E𝑞(𝑡) + 1

=

∞

∑

𝑛=0

g𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 𝜋,

2𝑡

E𝑞(𝑡) + 1

E𝑞 (𝑡𝑥)E𝑞 (𝑡𝑦)

=

∞

∑

𝑛=0

G𝑛,𝑞(𝑥, 𝑦)

𝑡𝑛

[𝑛]𝑞!, |𝑡| < 𝜋.

(14)

Abstract and Applied Analysis 3

Note that Cigler [20] defined 𝑞-Genocchi numbers as

𝑡𝑒𝑞 (𝑡) + 𝑒𝑞 (−𝑡)

𝑒𝑞(𝑡) + 𝑒

𝑞(−𝑡)

=

∞

∑

𝑛=0

𝑔2𝑛,𝑞

(−1)𝑛−1(−𝑞; 𝑞)

2𝑛−1𝑡2𝑛

[2𝑛]𝑞!. (15)

Then comparing g𝑛,𝑞

with 𝑔𝑛,𝑞, we see that

(−1)𝑛−122𝑛+1

g2𝑛+2,𝑞

= (−𝑞; 𝑞)2𝑛+1

𝑔2𝑛+2,𝑞

. (16)

Definition 4. Let 𝑞 ∈ C, 0 < |𝑞| < 1. The 𝑞-tangent numbersT𝑛,𝑞

are defined by means of the generating functions:

tanh𝑞𝑡 = −𝑖 tan

𝑞 (𝑖𝑡) =𝑒𝑞 (𝑡) − 𝑒𝑞 (−𝑡)


𝑞(−𝑡)

=E𝑞 (2𝑡) − 1

E𝑞(2𝑡) + 1

=

∞

∑

𝑛=1

T2𝑛+1,𝑞

(−1)𝑘𝑡2𝑛+1

[2𝑛 + 1]𝑞!.

(17)

It is obvious that, by letting 𝑞 tend to 1 from the left side,we lead to the classic definition of these polynomials:

b𝑛,𝑞:= B𝑛,𝑞 (0) , lim

𝑞→1−

B𝑛,𝑞 (𝑥) = 𝐵𝑛 (𝑥) ,

lim𝑞→1

−

B𝑛,𝑞(𝑥, 𝑦) = 𝐵

𝑛(𝑥 + 𝑦) , lim

𝑞→1−

b𝑛,𝑞= 𝐵𝑛,

e𝑛,𝑞:= E𝑛,𝑞 (0) , lim

𝑞→1−

E𝑛,𝑞 (𝑥) = 𝐸𝑛 (𝑥) ,

lim𝑞→1

−

E𝑛,𝑞(𝑥, 𝑦) = 𝐸

𝑛(𝑥 + 𝑦) , lim

𝑞→1−

e𝑛,𝑞= 𝐸𝑛,

g𝑛,𝑞:= G𝑛,𝑞(0) , lim

𝑞→1−

G𝑛,𝑞(𝑥) = 𝐺

𝑛(𝑥) ,

lim𝑞→1

−

G𝑛,𝑞(𝑥, 𝑦) = 𝐺

𝑛(𝑥 + 𝑦) lim

𝑞→1−

g𝑛,𝑞= 𝐺𝑛.

(18)

Here 𝐵𝑛(𝑥), 𝐸

𝑛(𝑥), and 𝐺

𝑛(𝑥) denote the classical Bernoulli,

Euler, and Genocchi polynomials, respectively, which aredefined by

𝑡

𝑒𝑡 − 1𝑒𝑡𝑥=

∞

∑

𝑛=0

𝐵𝑛(𝑥)

𝑡𝑛

𝑛!,

2

𝑒𝑡 + 1𝑒𝑡𝑥=

∞

∑

𝑛=0

𝐸𝑛(𝑥)

𝑡𝑛

𝑛!,

2𝑡

𝑒𝑡 + 1𝑒𝑡𝑥=

∞

∑

𝑛=0

𝐺𝑛 (𝑥)

𝑡𝑛

𝑛!.

(19)

The aim of the present paper is to obtain some resultsfor the above newly defined 𝑞-polynomials. It should bementioned that 𝑞-Bernoulli and 𝑞-Euler polynomials in ourdefinitions are polynomials of 𝑥 and 𝑦 and when 𝑦 = 0,they are polynomials of 𝑥. First advantage of this approach isthat for 𝑞 → 1−, B

𝑛,𝑞(𝑥, 𝑦) (E

𝑛,𝑞(𝑥, 𝑦), G

𝑛,𝑞(𝑥, 𝑦)) becomes

the classical BernoulliB𝑛(𝑥 + 𝑦) (EulerE

𝑛(𝑥 + 𝑦), Genocchi

G𝑛,𝑞(𝑥, 𝑦)) polynomial and wemay obtain the 𝑞-analogues of

well-known results, for example, Srivastava and Pintér [11],Cheon [23], and so forth. Second advantage is that, similarto the classical case, odd numbered terms of the Bernoullinumbers b

𝑘,𝑞and the Genocchi numbers g

𝑘,𝑞are zero, and

even numbered terms of the Euler numbers e𝑛,𝑞

are zero.

2. Preliminary Results

In this section we will provide some basic formulae for the𝑞-Bernoulli, 𝑞-Euler, and 𝑞-Genocchi numbers and polyno-mials in order to obtain the main results of this paper in thenext section.

Lemma 5. The 𝑞-Bernoulli numbers b𝑛,𝑞

satisfy the following𝑞-binomial recurrence:

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘b𝑘,𝑞− b𝑛,𝑞= {

1, 𝑛 = 1,

0, 𝑛 > 1.(20)

Proof. By a simple multiplication of (8) we see that

B̂ (𝑡)E𝑞(𝑡) = 𝑡 + B̂ (𝑡) . (21)

So∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘b𝑘,𝑞

𝑡𝑛

[𝑛]𝑞!= 𝑡 +

∞

∑

𝑛=0

b𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!. (22)

The statement follows by comparing 𝑡𝑚 coefficients.

We use this formula to calculate the first few b𝑘,𝑞:

b0,𝑞= 1,

b1,𝑞= −

1

2,

b2,𝑞=1

4

𝑞 (𝑞 + 1)

𝑞2 + 𝑞 + 1=𝑞[2]𝑞

4[3]𝑞

,

b3,𝑞= 0.

(23)

The similar property can be proved for 𝑞-Euler numbers

𝑚

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘e𝑘,𝑞+ e𝑚,𝑞

= {2, 𝑚 = 0,

0, 𝑚 > 0.(24)

and 𝑞-Genocchi numbers

𝑚

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘g𝑘,𝑞+ g𝑚,𝑞

= {2, 𝑚 = 1,

0, 𝑚 > 1.(25)

Using the above recurrence formulae we calculate the firstfew e𝑛,𝑞

and g𝑛,𝑞

terms as well:

e0,𝑞= 1, g

0,𝑞= 0,

e1,𝑞= −

1

2, g

1,𝑞= 1,

e2,𝑞= 0, g

2,𝑞= −

[2]𝑞

2= −

𝑞 + 1

2,

e3,𝑞=[3]𝑞[2]𝑞 − [4]𝑞

8=𝑞 (1 + 𝑞)

8, g

3,𝑞= 0.

(26)


Remark 6. The first advantage of the new 𝑞-numbersb𝑘,𝑞, e𝑘,𝑞, and g

𝑘,𝑞is that similar to classical case odd num-

bered terms of the Bernoulli numbers b𝑘,𝑞

and the Genocchinumbers g

𝑘,𝑞are zero, and even numbered terms of the Euler

numbers e𝑛,𝑞

are zero.

Next lemma gives the relationship between 𝑞-Genocchinumbers and 𝑞-Tangent numbers.

Lemma 7. For any 𝑛 ∈ N, we have

T2𝑛+1,𝑞

= g2𝑛+2,𝑞

(−1)𝑘−122𝑛+1

[2𝑛 + 2]𝑞

. (27)

Proof. First we recall the definition of 𝑞-trigonometric func-tions:

cos𝑞𝑡 =

𝑒𝑞(𝑖𝑡) + 𝑒

𝑞(−𝑖𝑡)

2, sin

𝑞𝑡 =

𝑒𝑞(𝑖𝑡) − 𝑒

𝑞(−𝑖𝑡)

2𝑖,

𝑖 tan𝑞𝑡 =

𝑒𝑞(𝑖𝑡) − 𝑒

𝑞(−𝑖𝑡)

𝑒𝑞 (𝑖𝑡) + 𝑒𝑞 (−𝑖𝑡)

, cot𝑞𝑡 = 𝑖

𝑒𝑞(𝑖𝑡) + 𝑒

𝑞(−𝑖𝑡)

𝑒𝑞 (𝑖𝑡) − 𝑒𝑞 (−𝑖𝑡)

.

(28)

Now by choosing 𝑧 = 2𝑖𝑡 in B̂(𝑧), we get

B̂ (2𝑖𝑡) =2𝑖𝑡

E𝑞(2𝑖𝑡) − 1

=𝑡𝑒𝑞(−𝑖𝑡)

sin𝑞𝑡

=

∞

∑

𝑛=0

b𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!. (29)

It follows that

B̂ (2𝑖𝑡) =𝑡𝑒𝑞(−𝑖𝑡)

sin𝑞𝑡

=𝑡

sin𝑞𝑡(cos𝑞𝑡 − 𝑖 sin

𝑞𝑡) = 𝑡 cot

𝑞𝑡 − 𝑖𝑡

= b0,𝑞+ 2𝑖𝑡b

1,𝑞+

∞

∑

𝑛=2

b𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!

= 1 − 𝑖𝑡 +

∞

∑

𝑛=2

b𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!.

(30)

Since the function 𝑡 cot𝑞𝑡 is even in the above sum odd

coefficients b2𝑘+1,𝑞

, 𝑘 = 1, 2, . . ., are zero, and we get

𝑡 cot𝑞𝑡 = 1 +

∞

∑

𝑛=2

b𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!= 1 +

∞

∑

𝑛=1

b𝑛,𝑞

(2𝑖𝑡)2𝑛

[2𝑛]𝑞!. (31)

By choosing 𝑧 = 2𝑖𝑡 in Ĝ(𝑧), we get

Ĝ (2𝑖𝑡) =4𝑖𝑡

E𝑞(2𝑖𝑡) + 1

=2𝑖𝑡𝑒𝑞 (−𝑖𝑡)

cos𝑞𝑡

=

∞

∑

𝑛=0

g𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!,

Ĝ (2𝑖𝑡) =4𝑖𝑡

E𝑞 (2𝑖𝑡) + 1

=2𝑖𝑡𝑒𝑞 (−𝑖𝑡)

cos𝑞𝑡

=2𝑖𝑡

cos𝑞𝑡(cos𝑞𝑡 − 𝑖 sin

𝑞𝑡)

= 2𝑖𝑡 + 2𝑡 tan𝑞𝑡 = g0,𝑞+ 2𝑖𝑡g

1,𝑞+

∞

∑

𝑛=2

g𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!

= 2𝑖𝑡 +

∞

∑

𝑛=2

g𝑛,𝑞

(2𝑖𝑡)𝑛

[𝑛]𝑞!.

(32)

It follows that

2𝑡 tan𝑞𝑡 =

∞

∑

𝑛=1

g2𝑛,𝑞

(2𝑖𝑡)2𝑛

[2𝑛]𝑞!,

tan𝑞𝑡 =

∞

∑

𝑛=1

g2𝑛,𝑞

(−1)𝑛(2𝑡)2𝑛−1

[2𝑛]𝑞!,

tanh𝑞𝑡 = −𝑖 tan

𝑞(𝑖𝑡) = −𝑖

∞

∑

𝑛=1

g2𝑛,𝑞

(−1)𝑛(2𝑖𝑡)2𝑛−1

[2𝑛]𝑞!

= −

∞

∑

𝑛=1

g2𝑛,𝑞

(2𝑡)2𝑛−1

[2𝑛]𝑞!= −

∞

∑

𝑛=1

g2𝑛+2,𝑞

(2𝑡)2𝑛+1

[2𝑛 + 2]𝑞!,

(33)

Thus

tanh𝑞𝑡 = −𝑖tan

𝑞 (𝑖𝑡) =𝑒𝑞(𝑡) − 𝑒

𝑞(−𝑡)


𝑞(−𝑡)

=E𝑞(2𝑡) − 1

E𝑞(2𝑡) + 1

=

∞

∑

𝑛=1

T2𝑛+1,𝑞

(−1)𝑘𝑡2𝑛+1

[2𝑛 + 1]𝑞!,

T2𝑛+1,𝑞

= g2𝑛+2,𝑞

(−1)𝑘−122𝑛+1

[2𝑛 + 2]𝑞

.

(34)

The following result is a 𝑞-analogue of the additiontheorem, for the classical Bernoulli, Euler, and Genocchipolynomials.

Lemma 8 (addition theorems). For all 𝑥, 𝑦 ∈ C we have

B𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

b𝑘,𝑞(𝑥 ⊕𝑞𝑦)𝑛−𝑘

,

B𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞 (𝑥) 𝑦

𝑛−𝑘,


E𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

e𝑘,𝑞(𝑥 ⊕𝑞𝑦)𝑛−𝑘

,

E𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘E𝑘,𝑞(𝑥) 𝑦𝑛−𝑘,

G𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

g𝑘,𝑞(𝑥 ⊕𝑞𝑦)𝑛−𝑘

,

G𝑛,𝑞(𝑥, 𝑦) =

𝑛

∑

𝑘=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘G𝑘,𝑞(𝑥) 𝑦𝑛−𝑘.

(35)

Proof. We prove only the first formula. It is a consequence ofthe following identity:

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!=

𝑡

E𝑞(𝑡) − 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦)

=

∞

∑

𝑛=0

b𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0

(𝑥 ⊕𝑞𝑦)𝑛 𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

b𝑘,𝑞(𝑥 ⊕𝑞𝑦)𝑛−𝑘 𝑡𝑛

[𝑛]𝑞!.

(36)

In particular, setting 𝑦 = 0 in (35), we get the followingformulae for 𝑞-Bernoulli, 𝑞-Euler and 𝑞-Genocchi polynomi-als, respectively:

B𝑛,𝑞(𝑥) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘b𝑘,𝑞𝑥𝑛−𝑘,

E𝑛,𝑞(𝑥) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘e𝑘,𝑞𝑥𝑛−𝑘,

(37)

G𝑛,𝑞 (𝑥) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘g𝑘,𝑞𝑥𝑛−𝑘. (38)

Setting 𝑦 = 1 in (35), we get

B𝑛,𝑞(𝑥, 1) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞(𝑥) ,

E𝑛,𝑞(𝑥, 1) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘E𝑘,𝑞(𝑥) ,

G𝑛,𝑞(𝑥, 1) =

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘G𝑘,𝑞(𝑥) .

(39)

Clearly (39) is 𝑞-analogues of

𝐵𝑛 (𝑥 + 1) =

𝑛

∑

𝑘=0

(𝑛

𝑘)𝐵𝑘 (𝑥) ,

𝐸𝑛(𝑥 + 1) =

𝑛

∑

𝑘=0

(𝑛

𝑘)𝐸𝑘(𝑥) ,

𝐺𝑛(𝑥 + 1) =

𝑛

∑

𝑘=0

(𝑛

𝑘)𝐺𝑘(𝑥) ,

(40)

respectively.

Lemma 9. The odd coefficients of the 𝑞-Bernoulli numbers,except the first one, are zero. That means b

𝑛,𝑞= 0 where

𝑛 = 2𝑟 + 1 (𝑟 ∈ N).

Proof. It follows from the fact that the function

𝑓 (𝑡) =

∞

∑

𝑛=0

b𝑛,𝑞

𝑡𝑛

[𝑛]𝑞!− b1,𝑞𝑡

=𝑡

E𝑞 (𝑡) − 1

+𝑡

2=𝑡

2(E𝑞 (𝑡) + 1

E𝑞 (𝑡) − 1

) ,

(41)

By using 𝑞-derivative we obtain the next lemma.

Lemma 10. One has

𝐷𝑞,𝑥B𝑛,𝑞 (𝑥) = [𝑛]𝑞

B𝑛−1,𝑞 (𝑥) +B𝑛−1,𝑞 (𝑞𝑥)

2,

𝐷𝑞,𝑥E𝑛,𝑞(𝑥) = [𝑛]𝑞

E𝑛−1,𝑞

(𝑥) + E𝑛−1,𝑞

(𝑞𝑥)

2,

𝐷𝑞,𝑥G𝑛,𝑞 (𝑥) = [𝑛]𝑞

G𝑛−1,𝑞

(𝑥) +G𝑛−1,𝑞

(𝑞𝑥)

2.

(42)

Lemma 11 (difference equations). One has

B𝑛,𝑞(𝑥, 1) −B

𝑛,𝑞(𝑥) =

(−1; 𝑞)𝑛−1

2𝑛−1[𝑛]𝑞𝑥𝑛−1, 𝑛 ≥ 1, (43)

E𝑛,𝑞 (𝑥, 1) + E𝑛,𝑞 (𝑥) = 2

(−1; 𝑞)𝑛

2𝑛𝑥𝑛, 𝑛 ≥ 0, (44)

G𝑛,𝑞(𝑥, 1) +G

𝑛,𝑞(𝑥) = 2

(−1; 𝑞)𝑛−1

2𝑛−1[𝑛]𝑞𝑥𝑛−1, 𝑛 ≥ 1. (45)

Proof. Weprove the identity for the 𝑞-Bernoulli polynomials.From the identity

𝑡E𝑞(𝑡)

E𝑞(𝑡) − 1

E𝑞 (𝑡𝑥) = 𝑡E𝑞 (𝑡𝑥) +

𝑡

E𝑞(𝑡) − 1

E𝑞 (𝑡𝑥) , (46)

it follows that∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞 (𝑥)

𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

(−1, 𝑞)𝑛

2𝑛𝑥𝑛 𝑡𝑛+1

[𝑛]𝑞!+

∞

∑

𝑛=0

B𝑛,𝑞(𝑥)

𝑡𝑛

[𝑛]𝑞!.

(47)


From (43) and (37), (44) and (38), we obtain the followingformulae.

Lemma 12. One has

𝑥𝑛=

2𝑛

(−1; 𝑞)𝑛[𝑛]𝑞

𝑛

∑

𝑘=0

[𝑛 + 1

𝑘]

𝑞

(−1; 𝑞)𝑛+1−𝑘

2𝑛+1−𝑘B𝑘,𝑞(𝑥) ,

𝑥𝑛=

2𝑛−1

(−1; 𝑞)𝑛

(

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1; 𝑞)𝑛−𝑘

2𝑛−𝑘E𝑘,𝑞 (𝑥) + E𝑛,𝑞 (𝑥)) ,

𝑥𝑛=

2𝑛−1

(−1; 𝑞)𝑛[𝑛 + 1]𝑞

× (

𝑛+1

∑

𝑘=0

[𝑛 + 1

𝑘]

𝑞

(−1; 𝑞)𝑛+1−𝑘

2𝑛+1−𝑘G𝑘,𝑞(𝑥) +G

𝑛+1,𝑞(𝑥)) .

(48)

The above formulae are 𝑞-analogues of the followingfamiliar expansions:

𝑥𝑛=

1

𝑛 + 1

𝑛

∑

𝑘=0

(𝑛 + 1

𝑘)𝐵𝑘(𝑥) ,

𝑥𝑛=1

2[

𝑛

∑

𝑘=0

(𝑛

𝑘)𝐸𝑘 (𝑥) + 𝐸𝑛 (𝑥)] ,

𝑥𝑛=

1

2 (𝑛 + 1)[

𝑛+1

∑

𝑘=0

(𝑛 + 1

𝑘)𝐸𝑘(𝑥) + 𝐸

𝑛+1(𝑥)] ,

(49)

respectively.

Lemma 13. The following identities hold true:

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞(𝑥, 𝑦) −B

𝑛,𝑞(𝑥, 𝑦)

= [𝑛]𝑞(𝑥 ⊕𝑞 𝑦)𝑛−1

,

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘E𝑘,𝑞(𝑥, 𝑦) + E

𝑛,𝑞(𝑥, 𝑦) = 2(𝑥 ⊕

𝑞𝑦)𝑛

,

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘G𝑘,𝑞(𝑥, 𝑦) +G

𝑛,𝑞(𝑥, 𝑦)

= 2[𝑛]𝑞(𝑥 ⊕𝑞 𝑦)𝑛−1

.

(50)

Proof. We prove the identity for the 𝑞-Bernoulli polynomi-als. From the identity

𝑡E𝑞(𝑡)

E𝑞(𝑡) − 1


= 𝑡E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦) +

𝑡

E𝑞 (𝑡) − 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦) ,

(51)

it follows that∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞(𝑥, 𝑦)

𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑘(−1, 𝑞)

𝑛−𝑘

2𝑛𝑥𝑘𝑦𝑛−𝑘 𝑡𝑛+1

[𝑛]𝑞!

+

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!.

(52)

3. Some New Formulae

The classical Cayley transformation 𝑧 →Cay(𝑧, 𝑎) := (1 +𝑎𝑧)/(1−𝑎𝑧)motivated us to derive the formula forE

𝑞(𝑞𝑡). In

addition, by substituting Cay(𝑧, (𝑞 − 1)/2) in the generatingformula we have

B̂𝑞(𝑞𝑡) B̂

𝑞 (𝑡) = (B̂q (𝑞𝑡) − 𝑞B̂𝑞 (𝑡) (1 + (1 − 𝑞)𝑡

2))

×1

1 − 𝑞×

2

E𝑞 (𝑡) + 1

.

(53)

The right hand side can be presented by 𝑞-Euler numbers.Now equating coefficients of 𝑡𝑛 we get the following identity.In the case that 𝑛 = 0, we find the first improved 𝑞-Eulernumber which is exactly 1.

Proposition 14. For all 𝑛 ≥ 1,𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

B𝑘,𝑞B𝑛−𝑘,𝑞

𝑞𝑘

= −𝑞

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

2𝑛−𝑘B𝑘,𝑞E𝑛−𝑘,𝑞[𝑘 − 1]𝑞

−𝑞

2

𝑛−1

∑

𝑘=0

[𝑛 − 1

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘−1

2𝑛−𝑘−1B𝑘,𝑞E𝑛−𝑘−1,𝑞[𝑛]𝑞.

(54)

Let us take a 𝑞-derivative from the generating function,after simplifying the equation, by knowing the quotient rulefor quantum derivative, and also using

E𝑞(𝑞𝑡) =

1 − (1 − 𝑞) (𝑡/2)

1 + (1 − 𝑞) (𝑡/2)E𝑞(𝑡) ,

𝐷𝑞(Eq (𝑡)) =

E𝑞(𝑞𝑡) +E

𝑞 (𝑡)

2,

(55)

one has

B̂𝑞(𝑞𝑡) B̂

𝑞(𝑡) =

2 + (1 − 𝑞) 𝑡

2E𝑞 (𝑡) (𝑞 − 1)

(𝑞B̂𝑞(𝑡) − B̂

𝑞(𝑞𝑡)) . (56)

It is clear that E−1𝑞(𝑡) = E

𝑞(−𝑡). Now, by equating

coefficients of 𝑡𝑛 we obtain the following identity.


Proposition 15. For all 𝑛 ≥ 1,

2𝑛

∑

𝑘=0

[2𝑛

𝑘]

𝑞

B𝑘,𝑞B2𝑛−𝑘,𝑞

𝑞𝑘

= −𝑞

2𝑛

∑

𝑘=0

[2𝑛

𝑘]

𝑞

(−1, 𝑞)2𝑛−𝑘

22𝑛−𝑘B𝑘,𝑞[𝑘 − 1]𝑞(−1)

𝑘

+𝑞 (1 − 𝑞)

2

2𝑛−1

∑

𝑘=0

[2𝑛 − 1

𝑘]

𝑞

(−1, 𝑞)2𝑛−1−𝑘

22𝑛−1−𝑘B𝑘,𝑞

× [𝑘 − 1]𝑞(−1)𝑘,

2𝑛+1

∑

𝑘=0

[2𝑛 + 1

𝑘]

𝑞

B𝑘,𝑞B2𝑛−𝑘+1,𝑞

𝑞𝑘

= 𝑞

2𝑛+1

∑

𝑘=0

[2𝑛 + 1

𝑘]

𝑞

(−1, 𝑞)2𝑛+1−𝑘

22𝑛+1−𝑘B𝑘,𝑞[𝑘 − 1]𝑞(−1)

𝑘

−𝑞 (1 − 𝑞)

2

2𝑛

∑

𝑘=0

[2𝑛

𝑘]

𝑞

(−1, 𝑞)2𝑛−𝑘

22𝑛−𝑘B𝑘,𝑞[𝑘 − 1]𝑞(−1)

𝑘.

(57)

We may also derive a differential equation for B̂𝑞(𝑡). If

we differentiate both sides of the generating function withrespect to 𝑡, after a little calculation we find that

𝜕

𝜕𝑡B̂𝑞(𝑡)

= B̂𝑞 (𝑡) (

1

𝑡−(1 − 𝑞)E

𝑞(𝑡)

E𝑞(𝑡) − 1

(

∞

∑

𝑘=0

4𝑞𝑘

4 − (1 − 𝑞)2𝑞2𝑘

)) .

(58)

If we differentiate B̂𝑞(𝑡) with respect to 𝑞, we obtain,

instead,

𝜕

𝜕𝑞B̂𝑞(𝑡) = −B̂

2

𝑞(𝑡)E𝑞(𝑡)

∞

∑

𝑘=0

4𝑡 (𝑘𝑞𝑘−1

− (𝑘 + 1) 𝑞𝑘)

4 − (1 − 𝑞)2𝑞2𝑘

. (59)

Again, using the generating function and combining thiswith the derivative we get the partial differential equation.

Proposition 16. Consider the following:

𝜕

𝜕𝑡B̂𝑞(𝑡) −

𝜕

𝜕𝑞B̂𝑞(𝑡)

=B̂𝑞(𝑡)

𝑡+

B̂2𝑞(𝑡)E𝑞(𝑡)

𝑡

×

∞

∑

𝑘=0

4𝑡 (𝑘𝑞𝑘−1

− (𝑘 + 1) 𝑞𝑘) − 𝑞𝑘(1 − 𝑞)

4 − (1 − 𝑞)2𝑞2𝑘

.

(60)

4. Explicit Relationship between the𝑞-Bernoulli and 𝑞-Euler Polynomials

In this section, we give some explicit relationship betweenthe 𝑞-Bernoulli and 𝑞-Euler polynomials.We also obtain newformulae and some special cases for them.These formulae areextensions of the formulae of Srivastava and Pintér, Cheon,and others.

We present natural 𝑞-extensions of themain results in thepapers [9, 11]; see Theorems 17 and 19.

Theorem 17. For 𝑛 ∈ N0, the following relationships hold true:


=1

2

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

𝑚𝑘−𝑛 [

[

B𝑘,𝑞 (𝑥) +

𝑘

∑

𝑗=0

[𝑛

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

B𝑗,𝑞 (𝑥)

2𝑘−𝑗𝑚𝑘−𝑗]

]

× E𝑛−𝑘,𝑞

(𝑚𝑦)

=1

2

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

𝑚𝑘−𝑛

[B𝑘,𝑞(𝑥) +B

𝑘,𝑞(𝑥,

1

𝑚)]E𝑛−𝑘,𝑞

(𝑚𝑦) .

(61)

Proof. Using the following identity

𝑡

E𝑞(𝑡) − 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦)

=𝑡

E𝑞(𝑡) − 1

E𝑞 (𝑡𝑥)

⋅E𝑞 (𝑡/𝑚) + 1

2⋅

2

E𝑞(𝑡/𝑚) + 1

E𝑞(𝑡

𝑚𝑚𝑦)

(62)

we have

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

=1

2

∞

∑

𝑛=0

E𝑛,𝑞(𝑚𝑦)

𝑡𝑛

𝑚𝑛[𝑛]𝑞!

∞

∑

𝑛=0

(−1, 𝑞)𝑛

𝑚𝑛2𝑛

𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0

B𝑛,𝑞(𝑥)

𝑡𝑛

[𝑛]𝑞!

+1

2

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

∞

∑

𝑛=0

B𝑛,𝑞 (𝑥)

𝑡𝑛

[𝑛]𝑞!

=: 𝐼1+ 𝐼2.

(63)

It is clear that

𝐼2=1

2

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

∞

∑

𝑛=0

B𝑛,𝑞(𝑥)

𝑡𝑛

[𝑛]𝑞!

=1

2

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑗]

𝑞

𝑚𝑘−𝑛

B𝑘,𝑞 (𝑥)E𝑛−𝑘,𝑞 (𝑚𝑦)

𝑡𝑛

[𝑛]𝑞!.

(64)


On the other hand

𝐼1=1

2

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

×

∞

∑

𝑛=0

𝑛

∑

𝑗=0

[𝑛

𝑗]

𝑞

B𝑗,𝑞 (𝑥)

(−1, 𝑞)𝑛−𝑗

𝑚𝑛−𝑗2𝑛−𝑗

𝑡𝑛

[𝑛]𝑞!

=1

2

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

E𝑛−𝑘,𝑞

(𝑚𝑦)

×

𝑘

∑

𝑗=0

[𝑛

𝑗]

𝑞

B𝑗,𝑞(𝑥) (−1, 𝑞)

𝑘−𝑗

𝑚𝑛−𝑘𝑚𝑘−𝑗2𝑘−𝑗

𝑡𝑛

[𝑛]𝑞!.

(65)

Therefore∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

=1

2

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

𝑚𝑘−𝑛

× [

[

B𝑘,𝑞 (𝑥) +

𝑘

∑

𝑗=0

[𝑛

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

B𝑗,𝑞(𝑥)

2𝑘−𝑗𝑚𝑘−𝑗]

]

× E𝑛−𝑘,𝑞

(𝑚𝑦)𝑡𝑛

[𝑛]𝑞!.

(66)

It remains to equate the coefficients of 𝑡𝑛.

Next we discuss some special cases of Theorem 17.

Corollary 18. For 𝑛 ∈ N0the following relationship holds true:


=

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(B𝑘,𝑞 (𝑥) +

(−1; 𝑞)𝑘−1

2𝑘[𝑘]𝑞𝑥𝑘−1)E𝑛−𝑘,𝑞

(𝑦) .

(67)

The formula (67) is a 𝑞-extension of the Cheon’s mainresult [23].

Theorem 19. For 𝑛 ∈ N0, the following relationships


=1

[𝑛 + 1]𝑞

×

𝑛+1

∑

𝑘=0

1

𝑚𝑛+1−𝑘[𝑛 + 1

𝑘]

𝑞

× (

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

𝑚𝑘−𝑗2𝑘−𝑗E𝑗,𝑞(𝑦) − E

𝑘,𝑞(𝑦))

×B𝑛+1−𝑘,𝑞 (𝑚𝑥)

(68)

hold true between the 𝑞-Bernoulli polynomials and 𝑞-Eulerpolynomials.

Proof. The proof is based on the following identity:2

E𝑞(𝑡) + 1


=2

E𝑞 (𝑡) + 1

E𝑞(𝑡𝑦)

⋅E𝑞(𝑡/𝑚) − 1

𝑡⋅

𝑡

E𝑞(𝑡/𝑚) − 1

E𝑞(𝑡

𝑚𝑚𝑥) .

(69)

Indeed∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

E𝑛,𝑞(𝑦)

𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0

(−1, 𝑞)𝑛

𝑚𝑛2𝑛

𝑡𝑛−1

[𝑛]𝑞!

×

∞

∑

𝑛=0

B𝑛,𝑞(𝑚𝑥)

𝑡𝑛

𝑚𝑛[𝑛]𝑞!

−

∞

∑

𝑛=0

E𝑛,𝑞(𝑦)

𝑡𝑛−1

[𝑛]𝑞!

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

=: 𝐼1− 𝐼2.

(70)

It follows that

𝐼2=1

𝑡

∞

∑

𝑛=0

E𝑛,𝑞(𝑦)

𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

=1

𝑡

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

1

𝑚𝑛−𝑘E𝑘,𝑞(𝑦)B

𝑛−𝑘,𝑞 (𝑚𝑥)𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

1

[𝑛 + 1]𝑞

×

𝑛+1

∑

𝑘=0

[𝑛 + 1

𝑘]

𝑞

1

𝑚𝑛+1−𝑘E𝑘,𝑞(𝑦)B

𝑛+1−𝑘,𝑞 (𝑚𝑥)𝑡𝑛

[𝑛]𝑞!,

𝐼1=1

𝑡

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

×

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

𝑚𝑛−𝑘2𝑛−𝑘E𝑘,𝑞(𝑦)

𝑡𝑛

[𝑛]𝑞!

=1

𝑡

∞

∑

𝑛=0

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

1

𝑚𝑛−𝑘B𝑛−𝑘,𝑞

(𝑚𝑥)

×

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

𝑚𝑘−𝑗2𝑘−𝑗E𝑗,𝑞(𝑦)

𝑡𝑛−1

[𝑛]𝑞!.

(71)

Next we give an interesting relationship between the 𝑞-Genocchi polynomials and the 𝑞-Bernoulli polynomials.


Theorem 20. For 𝑛 ∈ N0, the following relationship


=1

[𝑛 + 1]𝑞

×

𝑛+1

∑

𝑘=0

1

𝑚𝑛−𝑘[𝑛 + 1

𝑘]

𝑞

× (

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

𝑚𝑘−𝑗2𝑘−𝑗G𝑗,𝑞(𝑥) −G

𝑘,𝑞(𝑥))

×B𝑛+1−𝑘,𝑞

(𝑚𝑦) ,


=1

2[𝑛 + 1]𝑞

×

𝑛+1

∑

𝑘=0

1


𝑘]

𝑞

× (

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗

𝑚𝑘−𝑗2𝑘−𝑗B𝑗,𝑞(𝑥) +B

𝑘,𝑞(𝑥))

×G𝑛+1−𝑘,𝑞

(𝑚𝑦)

(72)

holds true between the 𝑞-Genocchi and the 𝑞-Bernoulli polyno-mials.

Proof. Using the following identity

2𝑡

E𝑞(𝑡) + 1

E𝑞(𝑡𝑥)E

𝑞(𝑡𝑦)

=2𝑡

E𝑞 (𝑡) + 1

E𝑞(𝑡𝑥) ⋅ (E

𝑞(𝑡

𝑚) − 1)

𝑚

𝑡

⋅𝑡/𝑚

E𝑞(𝑡/𝑚) − 1

⋅E𝑞(𝑡

𝑚𝑚𝑦)

(73)

we have

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

=𝑚

𝑡

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

×

∞

∑

𝑛=0

(−1, 𝑞)𝑛

𝑚𝑛2𝑛

𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0

B𝑛,𝑞(𝑚𝑦)

𝑡𝑛

𝑚𝑛[𝑛]𝑞!

−𝑚

𝑡

∞

∑

𝑛=0


𝑡𝑛

[𝑛]𝑞!

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

=𝑚

𝑡

∞

∑

𝑛=0

(

𝑛

∑

𝑘=0

[𝑛

𝑘]

𝑞

(−1, 𝑞)𝑛−𝑘

𝑚𝑛−𝑘2𝑛−𝑘G𝑘,𝑞 (𝑥) −G𝑛,𝑞 (𝑥))

𝑡𝑛

[𝑛]𝑞!

×

∞

∑

𝑛=0


𝑡𝑛

𝑚𝑛[𝑛]𝑞!

=𝑚

𝑡

∞

∑

𝑛=0

𝑛

∑

𝑘=0

1

𝑚𝑛−𝑘[𝑛

𝑘]

𝑞

× (

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗


𝑘,𝑞(𝑥))

×B𝑛−𝑘,𝑞

(𝑚𝑦)𝑡𝑛

[𝑛]𝑞!

=

∞

∑

𝑛=0

1

[𝑛 + 1]𝑞

𝑛+1

∑

𝑘=0

1


𝑘]

𝑞

× (

𝑘

∑

𝑗=0

[𝑘

𝑗]

𝑞

(−1, 𝑞)𝑘−𝑗


𝑘,𝑞(𝑥))

×B𝑛+1−𝑘,𝑞

(𝑚𝑦)𝑡𝑛

[𝑛]𝑞!.

(74)

The second identity can be proved in a like manner.

Conflict of Interests

The authors declare that there is no conflict of interests withany commercial identities regarding the publication of thispaper.

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International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation http://www.hindawi.com Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttp://www.hindawi.com

Volume 2014 Hindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

Stochastic AnalysisInternational Journal of

Documents

Research Article -Bernoulli, -Euler, and -Genocchi Polynomialsdownloads.hindawi.com/journals/aaa/2014/696454.pdf-Bernoulli, -Euler, and -Genocchi numbers and polyno-mials in order