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# On a class of generalized q-Bernoulli and q-Euler polynomials

## Abstract

The main purpose of this paper is to introduce and investigate a new class of generalized q-Bernoulli and q-Euler polynomials. The q-analogues of well-known formulas are derived. A generalization of the Srivastava-Pintér addition theorem is obtained.

## 1 Introduction

Throughout this paper, we always make use of the following notation: denotes the set of natural numbers, $N 0$ denotes the set of nonnegative integers, denotes the set of real numbers, denotes the set of complex numbers.

The q-numbers and q-factorial are defined by

$[ a ] q = 1 − q a 1 − q (q≠1); [ 0 ] q !=1; [ n ] q != [ 1 ] q [ 2 ] q ⋯ [ n ] q ,n∈N,a∈C,$

respectively. The q-polynomial coefficient is defined by

$[ n k ] q = ( q ; q ) n ( q ; q ) n − k ( q ; q ) k .$

The q-analogue of the function $( x + y ) n$ is defined by

$( x + y ) q n := ∑ k = 0 n [ n k ] q q 1 2 k ( k − 1 ) x n − k y k ,n∈ N 0 .$

The q-binomial formula is known as

$( 1 − a ) q n = ∏ j = 0 n − 1 ( 1 − q j a ) = ∑ k = 0 n [ n k ] q q 1 2 k ( k − 1 ) ( − 1 ) k a k .$

In the standard approach to the q-calculus, two exponential functions are used:

$e q ( z ) = ∑ n = 0 ∞ z n [ n ] q ! = ∏ k = 0 ∞ 1 ( 1 − ( 1 − q ) q k z ) , 0 < | q | < 1 , | z | < 1 | 1 − q | , E q ( z ) = ∑ n = 0 ∞ q 1 2 n ( n − 1 ) z n [ n ] q ! = ∏ k = 0 ∞ ( 1 + ( 1 − q ) q k z ) , 0 < | q | < 1 , z ∈ C .$

From this form, we easily see that $e q (z) E q (−z)=1$. Moreover,

$D q e q (z)= e q (z), D q E q (z)= E q (qz),$

where $D q$ is defined by

$D q f(z):= f ( q z ) − f ( z ) q z − z ,0<|q|<1,0≠z∈C.$

The above q-standard notation can be found in [1].

Carlitz firstly extended the classical Bernoulli and Euler numbers and polynomials, introducing them as q-Bernoulli and q-Euler numbers and polynomials [24]. There are numerous recent investigations on this subject by, among many other authors, Cenki et al. [57], Choi et al. [8] and [9], Kim et al. [1013], Ozden and Simsek [14], Ryoo et al. [15], Simsek [16, 17] and [18], and Luo and Srivastava [19], Srivastava et al. [20], Mahmudov [21, 22].

Recently, Natalini and Bernardini [23], Bretti et al. [24], Kurt [25, 26], Tremblay et al. [27, 28] studied the properties of the following generalized Bernoulli and Euler polynomials:

$( t m e t − ∑ k = 0 m − 1 t k k ! ) α e t x = ∑ n = 0 ∞ B n [ m − 1 , α ] ( x ) t n n ! , ( t m e t + ∑ k = 0 m − 1 t k k ! ) α e t x = ∑ n = 0 ∞ E n [ m − 1 , α ] ( x ) t n n ! , α ∈ C , 1 α : = 1 .$
(1)

Motivated by the generalizations in (1) of the classical Bernoulli and Euler polynomials, we introduce and investigate here the so-called generalized two-dimensional q-Bernoulli and q-Euler polynomials, which are defined as follows.

Definition 1 Let $q,α∈C$, $m∈N$, $0<|q|<1$. The generalized two-dimensional q-Bernoulli polynomials $B n , q [ m − 1 , α ] (x,y)$ are defined, in a suitable neighborhood of $t=0$, by means of the generating function

$( t m e q ( t ) − T m − 1 , q ( t ) ) α e q (tx) E q (ty)= ∑ n = 0 ∞ B n , q [ m − 1 , α ] (x,y) t n [ n ] q ! ,$

where $T m − 1 , q (t)= ∑ k = 0 m − 1 t k [ k ] q !$.

Definition 2 Let $q,α∈C$, $0<|q|<1$, $m∈N$. The generalized two-dimensional q-Euler polynomials $E n , q [ m − 1 , α ] (x,y)$ are defined, in a suitable neighborhood of $t=0$, by means of the generating functions

$( 2 m e q ( t ) + T m − 1 , q ( t ) ) α e q (tx) E q (ty)= ∑ n = 0 ∞ E n , q [ m − 1 , α ] (x,y) t n [ n ] q ! .$

It is obvious that

$lim q → 1 − B n , q [ m − 1 , α ] ( x , y ) = B n [ m − 1 , α ] ( x + y ) , B n , q [ m − 1 , α ] = B n , q [ m − 1 , α ] ( 0 , 0 ) , lim q → 1 − B n , q [ m − 1 , α ] = B n [ m − 1 , α ] , lim q → 1 − E n , q [ m − 1 , α ] ( x , y ) = E n [ m − 1 , α ] ( x + y ) , E n , q [ m − 1 , α ] = E n , q [ m − 1 , α ] ( 0 , 0 ) , lim q → 1 − E n , q [ m − 1 , α ] = E n [ m − 1 , α ] , lim q → 1 − B n , q [ m − 1 , α ] ( x , 0 ) = B n [ m − 1 , α ] ( x ) , lim q → 1 − B n , q [ m − 1 , α ] ( 0 , y ) = B n [ m − 1 , α ] ( y ) , lim q → 1 − E n , q [ m − 1 , α ] ( x , 0 ) = E n [ m − 1 , α ] ( x ) , lim q → 1 − E n , q [ m − 1 , α ] ( 0 , y ) = E n [ m − 1 , α ] ( y ) .$

Here $B n [ m − 1 , α ] (x)$ and $E n [ m − 1 , α ] (x)$ denote the generalized Bernoulli and Euler polynomials defined in (1). Notice that $B n [ m − 1 , α ] (x)$ was introduced by Natalini [23], and $E n [ m − 1 , α ] (x)$ was introduced by Kurt [25].

In fact Definitions 1 and 2 define two different types $B n , q [ m − 1 , α ] (x,0)$ and $B n , q [ m − 1 , α ] (0,y)$ of the generalized q-Bernoulli polynomials and two different types $E n , q [ m − 1 , α ] (x,0)$ and $E n , q [ m − 1 , α ] (0,y)$ of the generalized q-Euler polynomials. Both polynomials $B n , q [ m − 1 , α ] (x,0)$ and $B n , q [ m − 1 , α ] (0,y)$ ($E n , q [ m − 1 , α ] (x,0)$ and $E n , q [ m − 1 , α ] (0,y)$) coincide with the classical higher-order generalized Bernoulli polynomials (Euler polynomials) in the limiting case $q→ 1 −$.

## 2 Preliminaries and lemmas

In this section we provide some basic formulas for the generalized q-Bernoulli and q-Euler polynomials to obtain the main results of this paper in the next section. The following result is a q-analogue of the addition theorem for the classical Bernoulli and Euler polynomials.

Lemma 3 For all $x,y∈C$ we have

$B n , q [ m − 1 , α ] ( x , y ) = ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] ( x + y ) q n − k , E n , q [ m − 1 , α ] ( x , y ) = ∑ k = 0 n [ n k ] q E k , q [ m − 1 , α ] ( x + y ) q n − k ,$
(2)
$B n , q [ m − 1 , α ] ( x , y ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 B k , q [ m − 1 , α ] ( x , 0 ) y n − k = ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] ( 0 , y ) x n − k ,$
(3)
$E n , q [ m − 1 , α ] ( x , y ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 E k , q [ m − 1 , α ] ( x , 0 ) y n − k = ∑ k = 0 n [ n k ] q E k , q [ m − 1 , α ] ( 0 , y ) x n − k .$
(4)

In particular, setting $x=0$ and $y=0$ in (3) and (4), we get the following formulae for the generalized q-Bernoulli and q-Euler polynomials, respectively,

$B n , q [ m − 1 , α ] ( x , 0 ) = ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] x n − k , B n , q [ m − 1 , α ] ( 0 , y ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 B k , q [ m − 1 , α ] y n − k , E n , q [ m − 1 , α ] ( x , 0 ) = ∑ k = 0 n [ n k ] q E k , q [ m − 1 , α ] x n − k , E n , q [ m − 1 , α ] ( 0 , y ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 E k , q [ m − 1 , α ] y n − k .$

Setting $y=1$ and $x=1$ in (3) and (4), we get, respectively,

$B n , q [ m − 1 , α ] ( x , 1 ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 B k , q [ m − 1 , α ] ( x , 0 ) , B n , q [ m − 1 , α ] ( 1 , y ) = ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] ( 0 , y ) ,$
(5)
$E n , q [ m − 1 , α ] ( x , 1 ) = ∑ k = 0 n [ n k ] q q ( n − k ) ( n − k − 1 ) / 2 E k , q ( α ) ( x , 0 ) , E n , q [ m − 1 , α ] ( 1 , y ) = ∑ k = 0 n [ n k ] q E k , q [ m − 1 , α ] ( 0 , y ) .$
(6)

Clearly, (5) and (6) are the generalization of q-analogues of

$B n (x+1)= ∑ k = 0 n ( n k ) B k (x), E n (x+1)= ∑ k = 0 n ( n k ) E k (x),$

respectively.

Lemma 4 The generalized q-Bernoulli and q-Euler polynomials satisfy the following relations:

$B n , q [ m − 1 , α + β ] ( x , y ) = ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] ( x , 0 ) B k , q [ m − 1 , β ] ( 0 , y ) , E n , q [ m − 1 , α + β ] ( x , y ) = ∑ k = 0 n [ n k ] q E k , q [ m − 1 , α ] ( x , 0 ) E k , q [ m − 1 , β ] ( 0 , y ) .$

Lemma 5 We have

$D q , x B n , q [ m − 1 , α ] ( x , y ) = [ n ] q B n − 1 , q [ m − 1 , α ] ( x , y ) , D q , y B n , q [ m − 1 , α ] ( x , y ) = [ n ] q B n − 1 , q [ m − 1 , α ] ( x , q y ) , D q , x E n , q [ m − 1 , α ] ( x , y ) = [ n ] q E n − 1 , q [ m − 1 , α ] ( x , y ) , D q , y E n , q [ m − 1 , α ] ( x , y ) = [ n ] q E n − 1 , q [ m − 1 , α ] ( x , q y ) .$

Lemma 6 The generalized q-Bernoulli and q-Euler polynomials satisfy the following relations:

$B n , q [ m − 1 , α ] ( 1 , y ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) = [ n ] q ! [ n − m ] q ! B n − m , q [ m − 1 , α − 1 ] ( 0 , y ) , n ≥ m , E n , q [ m − 1 , α ] ( 1 , y ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q E n , q [ m − 1 , α ] ( 0 , y ) = 2 m E n , q [ m − 1 , α − 1 ] ( 0 , y ) , B n , q [ m − 1 , α ] ( x , 0 ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n , q [ m − 1 , α ] ( x , − 1 ) = [ n ] q ! [ n − m ] q ! B n − m , q [ m − 1 , α − 1 ] ( x , − 1 ) , n ≥ m , E n , q [ m − 1 , α ] ( x , 0 ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q E n , q [ m − 1 , α ] ( x , − 1 ) = 2 m E n , q [ m − 1 , α − 1 ] ( x , − 1 ) .$
(7)

Proof We prove only (7). The proof is based on the following equality:

$∑ n = 0 ∞ ( B n , q [ m − 1 , α ] ( 1 , y ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) ) t n [ n ] q ! = ( t m e q ( t ) − T m − 1 , q ( t ) ) α e q ( t ) E q ( t y ) − T m − 1 , q ( t ) ( t m e q ( t ) − T m − 1 , q ( t ) ) α E q ( t y ) = ( t m e q ( t ) − T m − 1 , q ( t ) ) α E q ( t y ) ( e q ( t ) − T m − 1 , q ( t ) ) = t m ( t m e q ( t ) − T m − 1 , q ( t ) ) α − 1 E q ( t y ) = ∑ n = 0 ∞ [ n + m ] q ! [ n ] q ! B n , q [ m − 1 , α − 1 ] ( 0 , y ) t n + m [ n + m ] q ! .$

Here we used the following relation:

$T m − 1 , q ( t ) ( t m e q ( t ) − T m − 1 , q ( t ) ) α E q ( t y ) = ∑ n = 0 m − 1 t n [ n ] q ! ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! = ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) ( t n [ n ] q ! + t n + 1 [ n ] q ! + t n + 2 [ n ] q ! [ 2 ] q ! + ⋯ + t n + m − 1 [ n ] q ! [ m − 1 ] q ! ) = ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! + ∑ n = 0 ∞ [ n ] q B n − 1 , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! + ∑ n = 0 ∞ [ n ] q [ n − 1 ] q [ 2 ] q ! B n − 2 , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! + ⋯ + ∑ n = 0 ∞ [ n ] q ⋯ [ n − m + 2 ] q [ m − 1 ] q ! B n − m + 1 , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! = ∑ n = 0 ∞ ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! .$

□

Corollary 7 Taking $q→ 1 −$, we have

$B n [ m − 1 , α ] ( y + 1 ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k [ m − 1 , α ] ( y ) = [ n ] q ! [ n − m ] q ! B n − m [ m − 1 , α − 1 ] ( y ) , n ≥ m , E n [ m − 1 , α ] ( y + 1 ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q E n [ m − 1 , α ] ( y ) = 2 m E n [ m − 1 , α − 1 ] ( y ) .$

Lemma 8 The generalized q-Bernoulli polynomials satisfy the following relations:

$B n , q [ m − 1 , α ] ( 1 , y ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) = [ n ] q ∑ k = 0 n − 1 [ n − 1 k ] q B k , q [ m − 1 , α ] ( 0 , y ) B n − 1 − k , q [ 0 , − 1 ] .$
(8)

Proof

Indeed,

$∑ n = 0 ∞ ( B n , q [ m − 1 , α ] ( 1 , y ) − ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) ) t n [ n ] q ! = ( t m e q ( t ) − T m − 1 , q ( t ) ) α e q ( t ) E q ( t y ) − T m − 1 , q ( t ) ( t m e q ( t ) − T m − 1 , q ( t ) ) α E q ( t y ) = ( t m e q ( t ) − T m − 1 , q ( t ) ) α E q ( t y ) e q ( t ) − T m − 1 , q ( t ) t t = ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! ∑ n = 0 ∞ B n , q [ 0 , − 1 ] t n + 1 [ n ] q ! = ∑ n = 1 ∞ [ n ] q ∑ k = 0 n − 1 [ n − 1 k ] q B k , q [ m − 1 , α ] ( 0 , y ) B n − 1 − k , q [ 0 , − 1 ] t n [ n ] q ! .$

□

Remark 9 Notice taking limit in (8) as $q→ 1 −$, we get

$B n [ m − 1 , α ] (y+1)− ∑ k = 0 min ( n , m − 1 ) ( n k ) B n − k [ m − 1 , α ] (y)=n ∑ k = 0 n − 1 ( n − 1 k ) B k [ m − 1 , α ] (y) B n − 1 − k [ 0 , − 1 ] .$

It is a correct form of formula (2.7) from [27] for $λ=1$.

Lemma 10 We have

$x n = ∑ k = 0 n [ n k ] q [ k ] q ! [ k + m ] q ! B n − k , q [ m − 1 , 1 ] ( x , 0 ) , y n = 1 q n ( n − 1 ) 2 ∑ k = 0 n [ n k ] q [ k ] q ! [ k + m ] q ! B n − k , q [ m − 1 , 1 ] ( 0 , y ) , x n = 1 2 m ( ∑ k = 0 n [ n k ] q E k , q [ m − 1 , 1 ] ( x , 0 ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q E k , q [ m − 1 , 1 ] ( x , 0 ) ) , y n = 1 2 m q n ( n − 1 ) 2 ( ∑ k = 0 n [ n k ] q E k , q [ m − 1 , 1 ] ( 0 , y ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q E n , q [ m − 1 , 1 ] ( 0 , y ) ) .$

From Lemma 10 we obtain the list of generalized q-Bernoulli polynomials as follows

$B 0 , q [ m − 1 , 1 ] ( x , 0 ) = [ m ] q ! , B 0 , q [ m − 1 , 1 ] ( 0 , y ) = [ m ] q ! , B 1 , q [ m − 1 , 1 ] ( x , 0 ) = [ m ] q ! ( x − 1 [ m + 1 ] q ) , B 1 , q [ m − 1 , 1 ] ( 0 , y ) = [ m ] q ! ( y − 1 [ m + 1 ] q ) , B 2 , q [ m − 1 , 1 ] ( x , 0 ) = x 2 − [ 2 ] q [ m ] q ! [ m + 1 ] q x + [ 2 ] q q m + 1 [ m ] q ! [ m + 1 ] q 2 [ m + 2 ] q , B 2 , q [ m − 1 , 1 ] ( 0 , y ) = q y 2 − [ 2 ] q [ m ] q ! [ m + 1 ] q y + [ 2 ] q q m + 1 [ m ] q ! [ m + 1 ] q 2 [ m + 2 ] q .$

## 3 Explicit relationship between the q-Bernoulli and q-Euler polynomials

In this section, we give some generalizations of the Srivastava-Pintér addition theorem. We also obtain new formulae and their some special cases below.

We present natural q-extensions of the main results of the papers [29, 30].

Theorem 11 The relationships

$B n , q [ m − 1 , α ] ( x , y ) = 1 2 ∑ k = 0 n [ n k ] q [ 1 l n − k B k , q [ m − 1 , α ] ( x , 0 ) + ∑ j = 0 k [ k j ] q 1 l k − j B j , q [ m − 1 , α ] ( x , 0 ) ] E n − k , q ( 0 , l y ) ,$
(9)
$B n , q [ m − 1 , α ] (x,y)= 1 2 ∑ k = 0 n [ n k ] q 1 l n − k [ B k , q [ m − 1 , α ] ( 0 , y ) + B k , q [ m − 1 , α ] ( 1 l , y ) ] E n − k , q (lx,0)$
(10)

hold true between the generalized q-Bernoulli polynomials and q-Euler polynomials.

Proof First we prove (9). Using the identity

$( t m e q ( t ) − ∑ i = 0 m − 1 t i [ i ] q ! ) α e q ( t x ) E q ( t y ) = 2 e q ( t l ) + 1 ⋅ E q ( t l l y ) ⋅ e q ( t l ) + 1 2 ⋅ ( t m e q ( t ) − ∑ i = 0 m − 1 t i [ i ] q ! ) α e q ( t x ) ,$

we have

$∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , y ) t n [ n ] q ! = 1 2 ∑ n = 0 ∞ E n , q ( 0 , l y ) t n l n [ n ] q ! ∑ k = 0 ∞ t k l k [ k ] q ! ∑ j = 0 ∞ B j , q [ m − 1 , α ] ( x , 0 ) t j [ j ] q ! + 1 2 ∑ k = 0 ∞ E k , q ( 0 , l y ) t k l k [ k ] q ! ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , 0 ) t n [ n ] q ! = : I 1 + I 2 .$

It is clear that

$I 2 = 1 2 ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , 0 ) t n [ n ] q ! ∑ k = 0 ∞ E k , q ( 0 , l y ) t k l k [ k ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q l k − n B k , q [ m − 1 , α ] ( x , 0 ) E n − k , q ( 0 , l y ) t n [ n ] q ! .$

On the other hand,

$I 1 = 1 2 ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , 0 ) t n [ n ] q ! ∑ k = 0 ∞ E k , q ( 0 , l y ) t k l k [ k ] q ! ∑ j = 0 ∞ t j l j [ j ] q ! = 1 2 ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , 0 ) t n [ n ] q ! ∑ k = 0 ∞ ∑ j = 0 k [ k j ] q E j , q ( 0 , l y ) t k l k [ k ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q B k , q [ m − 1 , α ] ( x , 0 ) ∑ j = 0 n − k [ n − k j ] q 1 l n − k E j , q ( 0 , l y ) t n [ n ] q ! = 1 2 ∑ n = 0 ∞ ∑ j = 0 n [ n j ] q E j , q ( 0 , l y ) ∑ k = 0 n − j [ n − j k ] q 1 l n − k B k , q [ m − 1 , α ] ( x , 0 ) t n [ n ] q ! .$

Therefore

$∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , y ) t n [ n ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q [ 1 l n − k B k , q [ m − 1 , α ] ( x , 0 ) + ∑ j = 0 k [ k j ] q 1 l k − j B j , q [ m − 1 , α ] ( x , 0 ) ] × E n − k , q ( 0 , l y ) t n [ n ] q ! .$

Next we prove (10). Using the identity

$( t m e q ( t ) − ∑ i = 0 m − 1 t i [ i ] q ! ) α e q ( t x ) E q ( t y ) = 2 e q ( t l ) + 1 ⋅ e q ( t l l x ) ⋅ e q ( t l ) + 1 2 ⋅ ( t m e q ( t ) − ∑ i = 0 m − 1 t i [ i ] q ! ) α E q ( t y ) ,$

we have

$∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , y ) t n [ n ] q ! = 1 2 ∑ n = 0 ∞ E n , q ( l x , 0 ) t n l n [ n ] q ! ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 1 l , y ) t n [ n ] q ! + 1 2 ∑ k = 0 ∞ E k , q ( l x , 0 ) t k l k [ k ] q ! ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! = : I 1 + I 2 .$

It is clear that

$I 2 = 1 2 ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 0 , y ) t n [ n ] q ! ∑ k = 0 ∞ E k , q ( l x , 0 ) t k l k [ k ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q l k − n B k , q [ m − 1 , α ] ( 0 , y ) E n − k , q ( l x , 0 ) t n [ n ] q ! .$

On the other hand,

$I 1 = 1 2 ∑ n = 0 ∞ B n , q [ m − 1 , α ] ( 1 l , y ) t n [ n ] q ! ∑ k = 0 ∞ E k , q ( l x , 0 ) t k m k [ k ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q l k − n B k , q [ m − 1 , α ] ( 1 l , y ) E n − k , q ( l x , 0 ) t n [ n ] q ! .$

Therefore

$∑ n = 0 ∞ B n , q [ m − 1 , α ] ( x , y ) t n [ n ] q ! = 1 2 ∑ n = 0 ∞ ∑ k = 0 n [ n k ] q l k − n [ B k , q [ m − 1 , α ] ( 0 , y ) + B k , q [ m − 1 , α ] ( 1 l , y ) ] E n − k , q ( l x , 0 ) t n [ n ] q ! .$

□

Next we discuss some special cases of Theorem 11.

Theorem 12 The relationship

$B n , q [ m − 1 , α ] ( x , y ) = 1 2 ∑ k = 0 n [ n k ] q [ B k , q [ m − 1 , α ] ( 0 , y ) + ∑ k = 0 min ( n , m − 1 ) [ n k ] q B n − k , q [ m − 1 , α ] ( 0 , y ) + [ k ] q ∑ j = 0 k − 1 [ k − 1 j ] q B j , q [ m − 1 , α ] ( 0 , y ) B k − 1 − j , q [ 0 , − 1 ] ] E n − k , q ( x , 0 )$

holds true between the generalized q-Bernoulli polynomials and the q-Euler polynomials.

Remark 13 Taking $q→ 1 −$ in Theorem 12, we obtain the Srivastava-Pintér addition theorem for the generalized Bernoulli and Euler polynomials.

$B n [ m − 1 , α ] ( x + y ) = 1 2 ∑ k = 0 n ( n k ) [ B k [ m − 1 , α ] ( y ) + ∑ k = 0 min ( n , m − 1 ) ( n k ) B n − k [ m − 1 , α ] ( y ) + k ∑ j = 0 k − 1 ( k − 1 j ) B j [ m − 1 , α ] ( y ) B k − 1 − j [ 0 , − 1 ] ] E n − k ( x ) .$
(11)

Notice that the Srivastava-Pintér addition theorem for the generalized Apostol-Bernoulli polynomials and the Apostol-Euler polynomials was given in [27]. The formula (11) is a correct version of Theorem 3 [27] for $λ=1$.

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## Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to thank the reviewers for their valuable comments and helpful suggestions that improved the note’s quality.

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Correspondence to Nazim I Mahmudov.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Mahmudov, N.I., Keleshteri, M.E. On a class of generalized q-Bernoulli and q-Euler polynomials. Adv Differ Equ 2013, 115 (2013). https://doi.org/10.1186/1687-1847-2013-115