# Some identities of special q-polynomials

## Abstract

In this paper, we investigate some identities of q-extensions of special polynomials which are derived from the fermonic q-integral on $Z p$ and the bosonic q-integral on $Z p$.

## 1 Introduction

Let p be a fixed odd prime number. Throughout this paper, $Z p$, $Q p$, and $C p$ will denote the ring of p-adic integers, the field of p-adic rational numbers and the completion of algebraic closure of $Q p$, respectively. Let q be an indeterminate in $C p$ with $|1−q | p < p − 1 p − 1$ and $UD( Z p )$ be the space of all uniformly differentiable functions on $Z p$. The q-analog of x is defined as $[ x ] q = 1 − q x 1 − q$. Note that $lim q → 1 [ x ] q =x$. For $f∈UD( Z p )$, the bosonic p-adic q-integral on $Z p$ is defined by Kim to be

$I q (f)= ∫ Z p f(x)d μ q (x)= lim N → ∞ 1 [ p N ] q ∑ x = 0 p N − 1 f(x) q x (see [1, 2])$
(1.1)

and the fermionic p-adic q-integral on $Z p$ is also defined by Kim to be

$I − q (f)= ∫ Z p f(x)d μ − q (x)= lim N → ∞ 1 [ p N ] − q ∑ x = 0 p N − 1 f(x) ( − q ) x (see [1–3]).$
(1.2)

From (1.1) and (1.2), we have

$q I q ( f 1 )− I q (f)=(q−1)f(0)+ q − 1 log q f ′ (0)$
(1.3)

and

$q I − q ( f 1 )+ I − q (f)= [ 2 ] q f(0)(see [1–3]).$
(1.4)

As is well known, the q-analog of the Bernoulli polynomials is given by the generating function to be

$q − 1 + q − 1 log q t q e t − 1 e x t = ∑ n = 0 ∞ B n , q (x) t n n ! (see [1, 2, 4–20]),$
(1.5)

and the q-analog of the Euler polynomials is given by

$[ 2 ] q q e t + 1 e x t = ∑ n = 0 ∞ E n , q (x) t n n ! (see [1, 2, 4–21]).$
(1.6)

The higher-order q-Daehee polynomials are given by

$q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ( 1 + t ) x = ∑ n = 0 ∞ D n , q (x) t n n ! ,$
(1.7)

where $t∈ C p$ with $|t | p < p − 1 p − 1$.

Now, we define the q-analog of the Changhee polynomials, which are given by the generating function to be

$( [ 2 ] q q t + [ 2 ] q ) ( 1 + t ) x = ∑ n = 0 ∞ Ch n , q (x) t n n ! .$
(1.8)

In this paper, we investigate some properties for the q-analog of several special polynomials which are derived from the bosonic or fermionic p-adic q-integral on $Z p$.

## 2 Some special q-polynomials

In this section, we assume that $t∈ C p$ with $|t | p < p − 1 p − 1$. Now, we define the higher-order q-Bernoulli numbers,

$( q − 1 + q − 1 log q t q e t − 1 ) r e x t = ∑ n = 0 ∞ B n , q ( r ) (x) t n n ! .$
(2.1)

When $x=0$, $B n , q ( r ) = B n , q ( r ) (0)$ are called the higher-order q-Bernoulli numbers.

We also consider the higher-order q-Daehee polynomials as follows:

$( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ( 1 + t ) x = ∑ n = 0 ∞ D n , q ( r ) (x) t n n ! .$
(2.2)

When $x=0$, $D n , q ( r ) = D n , q ( r ) (0)$ are called the higher-order q-Daehee numbers.

From (1.3), we can derive the following equation:

$∫ Z p ⋯ ∫ Z p ( 1 + t ) x 1 + ⋯ + x r + x d μ q ( x 1 ) ⋯ d μ q ( x r ) = ( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ( 1 + t ) x = ∑ n = 0 ∞ D n , q ( r ) ( x ) t n n ! .$
(2.3)

Thus, by (2.3), we get

$∫ Z p ⋯ ∫ Z p ( x 1 + ⋯ + x r + x n ) d μ q ( x 1 )⋯d μ q ( x r )= D n , q ( r ) ( x ) n ! (n≥0).$
(2.4)

By replacing t by $e t −1$ in (2.2), we get

$∑ n = 0 ∞ D n , q ( r ) (x) ( e t − 1 ) n n ! = ( q − 1 + q − 1 log q t q e t − 1 ) r e x t = ∑ n = 0 ∞ B n , q ( r ) (x) t n n !$
(2.5)

and

$∑ n = 0 ∞ D n , q ( r ) ( x ) 1 n ! ( e t − 1 ) n = ∑ n = 0 ∞ D n , q ( r ) ( x ) 1 n ! n ! ∑ m = n ∞ S 2 ( m , n ) t m m ! = ∑ m = 0 ∞ ( ∑ n = 0 m D n , q ( r ) ( x ) S 2 ( m , n ) ) t m m ! .$
(2.6)

Thus, by (2.5) and (2.6), we get

$B n , q ( r ) (x)= ∑ m = 0 n D m , q ( r ) (x) S 2 (n,m).$
(2.7)

Therefore, by (2.4) and (2.7), we obtain the following theorem.

Theorem 1 For $n≥0$, we have

$B n , q ( r ) (x)= ∑ m = 0 n D m , q ( r ) (x) S 2 (n,m)$

and

$∫ Z p ⋯ ∫ Z p ( x 1 + ⋯ + x r + x n ) d μ q ( x 1 ) ⋯ d μ q ( x r ) = D n , q ( r ) ( x ) n ! ,$

where $S 2 (n,m)$ is the Stirling number of the second kind.

From (2.1), by replacing t by $log(1+t)$, we obtain

$( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ( 1 + t ) x = ∑ n = 0 ∞ B n , q ( r ) ( x ) 1 n ! ( log ( 1 + t ) ) n = ∑ n = 0 ∞ B n , q ( r ) ( x ) 1 n ! n ! ∑ m = n ∞ S 1 ( m , n ) t m m ! = ∑ m = 0 ∞ ( ∑ n = 0 m S 1 ( m , n ) B n , q ( r ) ( x ) ) t m m ! ,$
(2.8)

where $S 1 (n,m)$ is the Stirling number of the first kind.

Therefore, by (2.2) and (2.8), we obtain the following theorem.

Theorem 2 For $n≥0$, we have

$D n , q ( r ) (x)= ∑ m = 0 n S 1 (n,m) B m , q ( r ) (x).$

Now, we define the higher-order q-Changhee polynomials as follows:

$( [ 2 ] q q t + [ 2 ] q ) r ( 1 + t ) x = ∑ n = 0 ∞ Ch n , q ( r ) (x) t n n ! .$
(2.9)

When $x=0$, $Ch n , q ( r ) = Ch n , q ( r ) (0)$ are called the higher-order q-Changhee numbers.

From (1.4), we note that

$∫ Z p ⋯ ∫ Z p ( 1 + t ) x 1 + ⋯ + x r + x d μ − q ( x 1 )⋯d μ − q ( x r )= ( [ 2 ] q q t + [ 2 ] q ) r ( 1 + t ) x .$
(2.10)

Thus, by (2.10), we get

$∫ Z p ⋯ ∫ Z p ( x 1 + ⋯ + x r + x n ) d μ − q ( x 1 )⋯d μ − q ( x r )= Ch n , q ( r ) ( x ) n ! .$
(2.11)

In view of (1.6), we define the higher-order q-Euler polynomials which are given by the generating function to be

$( [ 2 ] q q e t + 1 ) r e x t = ∑ n = 0 ∞ E n , q ( r ) (x) t n n ! .$
(2.12)

From (2.10), we note that

$∫ Z p ⋯ ∫ Z p ( 1 + t ) x 1 + ⋯ + x r + x d μ − q ( x 1 ) ⋯ d μ − q ( x r ) = ( [ 2 ] q q e log ( 1 + t ) + 1 ) r e x log ( 1 + t ) = ∑ n = 0 ∞ E n , q ( r ) ( x ) 1 n ! ( log ( 1 + t ) ) n = ∑ n = 0 ∞ E n , q ( r ) ( x ) ∑ m = n ∞ S 1 ( m , n ) t m m ! = ∑ m = 0 ∞ ( ∑ n = 0 m E n , q ( r ) ( x ) S 1 ( m , n ) ) t m m ! .$
(2.13)

Therefore, by (2.11) and (2.13), we obtain the following theorem.

Theorem 3 For $n≥0$, we have

$∫ Z p ⋯ ∫ Z p ( x 1 + ⋯ + x r + x n ) d μ − q ( x 1 ) ⋯ d μ − q ( x r ) = Ch n , q ( r ) ( x ) n ! = 1 n ! ∑ m = 0 n E m , q ( r ) ( x ) S 1 ( n , m ) .$

By replacing t by $e t −1$ in (2.9), we get

$∑ n = 0 ∞ Ch n , q ( r ) (x) ( e t − 1 ) n n ! = ( [ 2 ] q q e t + 1 ) r e x t$
(2.14)

and

$∑ n = 0 ∞ Ch n , q ( r ) ( x ) 1 n ! ( e t − 1 ) n = ∑ n = 0 ∞ Ch n , q ( r ) ( x ) ∑ m = n ∞ S 2 ( m , n ) t m m ! = ∑ m = 0 ∞ ( ∑ n = 0 m Ch n , q ( r ) ( x ) S 2 ( m , n ) ) t m n ! .$
(2.15)

Therefore, by (2.12), (2.14), and (2.15), we obtain the following theorem.

Theorem 4 For $m≥0$, we have

$E m , q ( r ) (x)= ∑ n = 0 m Ch n , q ( r ) (x) S 2 (m,n).$

Now, we consider the q-analog of the higher-order Cauchy polynomials, which are defined by the generating function to be

$( q ( 1 + t ) − 1 ( q − 1 ) + q − 1 log q log ( 1 + t ) ) r ( 1 + t ) x = ∑ n = 0 ∞ C n , q ( r ) (x) t n n ! .$
(2.16)

When $x=0$, $C n , q ( r ) = C n , q ( r ) (0)$ are called the higher-order q-Cauchy numbers. Indeed,

$lim q → 1 ( q ( 1 + t ) − 1 q − 1 + q − 1 log q log ( 1 + t ) ) r ( 1 + t ) x = ( t log ( 1 + t ) ) r ( 1 + t ) x = ∑ n = 0 ∞ C n ( r ) ( x ) t n n ! ,$
(2.17)

where $C n ( r ) (x)$ are called the higher-order Cauchy polynomials.

We observe that

$( 1 + t ) x = ( q ( 1 + t ) − 1 q − 1 + q − 1 log q log ( 1 + t ) ) r ( 1 + t ) x ( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r = ( ∑ l = 0 ∞ C l , q ( r ) ( x ) t l l ! ) ( ∑ m = 0 ∞ D m , q ( r ) t m m ! ) = ∑ n = 0 ∞ ( ∑ l = 0 n ( n l ) C l , q ( r ) ( x ) D n − l , q ( r ) ) t n n !$
(2.18)

and

$( 1 + t ) x = ∑ n = 0 ∞ ( x ) n t n n ! .$
(2.19)

By (2.18) and (2.19), we get

$( x ) n = ∑ l = 0 n ( n l ) C l , q ( r ) (x) D n − l , q ( r ) .$
(2.20)

Therefore, by (2.20), we obtain the following theorem.

Theorem 5 For $n≥0$, we have

$( x n ) = 1 n ! ∑ l = 0 n ( n l ) C l , q ( r ) (x) D n − l , q ( r ) .$

For $n∈N∪{0}$, we define the q-analog of the Bernoulli-Euler mixed-type polynomials of order $(r,s)$ as follows:

$B E n , q ( r , s ) (x)= ∫ Z p ⋯ ∫ Z p E n , q ( s ) (x+ y 1 +⋯+ y r )d μ q ( y 1 )⋯d μ q ( y r ).$
(2.21)

Then, by (2.21), we get

$∑ n = 0 ∞ B E n , q ( r , s ) ( x ) t n n ! = ∫ Z p ⋯ ∫ Z p ∑ n = 0 ∞ E n , q ( s ) ( x + y 1 + ⋯ + y r ) t n n ! d μ q ( y 1 ) ⋯ d μ q ( y r ) = ( [ 2 ] q q e t + 1 ) s ∫ Z p ⋯ ∫ Z p e ( x + y 1 + ⋯ + y r ) t d μ q ( y 1 ) ⋯ d μ q ( y r ) = ( [ 2 ] q q e t + 1 ) s ( q − 1 + q − 1 log q t q e t − 1 ) r e x t .$
(2.22)

It is easy to show that

$( [ 2 ] q q e t + 1 ) s ( q − 1 + q − 1 log q t q e t − 1 ) r e x t = ∑ n = 0 ∞ ( ∑ l = 0 n ( n l ) E l , q ( s ) B n − l , q ( r ) ( x ) ) t n n ! .$
(2.23)

Therefore, by (2.22) and (2.23), we obtain the following theorem.

Theorem 6 For $n≥0$, we have

$B E n , q ( r , s ) (x)= ∑ l = 0 n ( n l ) E l , q ( s ) B n − l , q ( r ) (x).$

By replacing t by $log(1+t)$ in (2.22), we get

$∑ n = 0 ∞ B E n , q ( r , s ) ( x ) ( log ( 1 + t ) ) n n ! = ( [ 2 ] q q t + [ 2 ] q ) s ( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ( 1 + t ) x = ∑ n = 0 ∞ { ∑ m = 0 n ( n m ) D m , q ( r ) ( x ) Ch n − m , q ( s ) } t n n !$
(2.24)

and

$∑ m = 0 ∞ B E m , q ( r , s ) ( x ) ( log ( 1 + t ) ) m m ! = ∑ n = 0 ∞ { ∑ m = 0 n B E m , q ( r , s ) ( x ) S 1 ( n , m ) } t n n ! .$
(2.25)

Therefore, by (2.24) and (2.25), we obtain the following theorem.

Theorem 7 For $n≥0$, we have

$∑ m = 0 n ( n m ) D m , q ( r ) (x) Ch n − m , q ( s ) = ∑ m = 0 n B E m , q ( r , s ) (x) S 1 (n,m).$

Let us consider the q-analog of the Daehee-Changhee mixed-type polynomials of order $(r,s)$ as follows: for $n≥0$,

$D C n , q ( r , s ) (x)= ∫ Z p ⋯ ∫ Z p D n , q ( r ) (x+ y 1 +⋯+ y s )d μ − q ( y 1 )⋯d μ − q ( y s ).$
(2.26)

Thus, by (2.26), we get

$∑ n = 0 ∞ D C n , q ( r , s ) ( x ) t n n ! = ∫ Z p ⋯ ∫ Z p ∑ n = 0 ∞ D n , q ( r ) ( x + y 1 + ⋯ + y s ) t n n ! d μ − q ( y 1 ) ⋯ d μ − q ( y s ) = ( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ∫ Z p ⋯ ∫ Z p ( 1 + t ) x + y 1 + ⋯ + y s d μ − q ( y 1 ) ⋯ d μ − q ( y s ) = ( q − 1 + q − 1 log q log ( 1 + t ) q ( 1 + t ) − 1 ) r ( [ 2 ] q q t + [ 2 ] q ) s ( 1 + t ) x = ( ∑ m = 0 ∞ D m , q ( r ) t m m ! ) ( ∑ l = 0 ∞ Ch l , q ( s ) ( x ) t l l ! ) = ∑ n = 0 ∞ { ∑ m = 0 n ( n m ) D m , q ( r ) Ch n − m , q ( s ) ( x ) } t n n !$
(2.27)

and

$∑ n = 0 ∞ D C n , q ( r , s ) ( x ) ( e t − 1 ) n n ! = ( q − 1 + q − 1 log q t q e t − 1 ) r ( [ 2 ] q q e t + 1 ) s e x t = ∑ n = 0 ∞ { ∑ m = 0 n ( n m ) B m , q ( r ) E n − m , q ( s ) ( x ) } t n n ! .$
(2.28)

Now, we observe that

$∑ n = 0 ∞ D C n , q ( r , s ) ( x ) ( e t − 1 ) n n ! = ∑ n = 0 ∞ D C n , q ( r , s ) ( x ) 1 n ! n ! ∑ m = n ∞ S 2 ( m , n ) t m m ! = ∑ m = 0 ∞ { ∑ n = 0 m D C n , q ( r , s ) ( x ) S 2 ( m , n ) } t m m ! .$
(2.29)

Therefore, by (2.27), (2.28), and (2.29), we obtain the following theorem.

Theorem 8 For $n≥0$, we have

$D C n , q ( r , s ) (x)= ∑ m = 0 n ( n m ) D m , q ( r ) Ch n − m , q ( s ) (x)$

and

$∑ m = 0 n ( n m ) B m , q ( r ) E n − m , q ( s ) (x)= ∑ m = 0 n D C m , q ( r , s ) (x) S 2 (n,m).$

Now, we consider the q-extension of the Cauchy-Changhee mixed-type polynomials of order $(r,s)$ as follows: for $n≥0$,

$C C n , q ( r , s ) (x)= ∫ Z p ⋯ ∫ Z p C n , q ( r ) (x+ y 1 +⋯+ y s )d μ − q ( y 1 )⋯d μ − q ( y r ).$
(2.30)

Thus, by (2.30), we get

$∑ n = 0 ∞ C C n , q ( r , s ) ( x ) t n n ! = ∫ Z p ⋯ ∫ Z p ∑ n = 0 ∞ C n , q ( r ) ( x + y 1 + ⋯ + y s ) t n n d μ − q ( y 1 ) ⋯ d μ − q ( y s ) = ( q ( 1 + t ) − 1 q − 1 + q − 1 log q log ( 1 + t ) ) r ( [ 2 ] q q t + [ 2 ] q ) s ( 1 + t ) x = ∑ n = 0 ∞ { ∑ m = 0 n ( n m ) C m , q ( r ) Ch n − m , q ( s ) ( x ) } t n n ! ,$
(2.31)
$∑ n = 0 ∞ C C n , q ( r , s ) ( x ) ( e t − 1 ) n n ! = ( q e t − 1 q − 1 + q − 1 log q t ) r ( [ 2 ] q q e t + 1 ) s e t x = ( ∑ m = 0 ∞ B m , q ( − r ) t m m ! ) ( ∑ l = 0 ∞ E l , q ( s ) ( x ) t l l ! ) = ∑ n = 0 ∞ ( ∑ m = 0 n ( n m ) B m , q ( − r ) E n − m , q ( s ) ( x ) ) t n n ! .$
(2.32)

Note that

$∑ n = 0 ∞ C C n , q ( r , s ) (x) ( e t − 1 ) n n ! = ∑ n = 0 ∞ ( ∑ m = 0 n C C m , q ( r , s ) ( x ) S 2 ( n , m ) ) t n n ! .$
(2.33)

Therefore, by (2.31), (2.32), and (2.33), we obtain the following theorem.

Theorem 9 For $n≥0$, we have

$C C n , q ( r , s ) (x)= ∑ m = 0 n ( n m ) C m , q ( r ) Ch n − m , q ( s ) (x)$

and

$∑ m = 0 n ( n m ) B m , q ( − r ) E n − m , q ( s ) (x)= ∑ m = 0 n C C m , q ( r , s ) (x) S 2 (n,m).$

Finally, we define the q-extension of the Cauchy-Daehee mixed-type polynomials of order $(r,s)$ as follows:

$C D n , q ( r , s ) (x)= ∫ Z p ⋯ ∫ Z p C n , q ( r ) (x+ y 1 +⋯+ y r )d μ q ( x 1 )⋯d μ q ( x r ).$
(2.34)

Thus, by (2.34), we get

(2.35)

Therefore, by (2.35), we obtain the following equation:

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## Acknowledgements

This paper is supported by grant No. 14-11-00022 of Russian Scientific fund.

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Correspondence to Taekyun Kim.

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