# Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials

## Abstract

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

MSC:05A15, 05A40, 11B68, 11B75, 65Q05.

## 1 Introduction

In this paper, we consider the polynomials $D n ( k ) (x| a 1 ,…, a r )$ called the Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials, whose generating function is given by

$∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) x = ∑ n = 0 ∞ D n ( k ) (x| a 1 ,…, a r ) t n n ! ,$
(1)

where $a 1 ,…, a r ≠0$. Here, $Lif k (x)$ ($k∈Z$) is the polyfactorial function  defined by

$Lif k (x)= ∑ m = 0 ∞ x m m ! ( m + 1 ) k .$

When $x=0$, $D n ( k ) ( a 1 ,…, a r )= D n ( k ) (0| a 1 ,…, a r )$ is called Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type number.

Recall that the Barnes-type Daehee polynomials of the first kind, denoted by $D n (x| a 1 ,…, a r )$, are given by the generating function

$∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) x = ∑ n = 0 ∞ D n (x| a 1 ,…, a r ) t n n ! .$

If $a 1 =⋯= a r =1$, then $D n ( r ) (x)= D n (x| 1 , … , 1 ⏟ r )$ are the Daehee polynomials of the first kind of order r. Dahee polynomials were defined by the second author  and have been investigated in [3, 4].

The poly-Cauchy polynomials of the first kind, denoted by $c n ( k ) (x)$ [5, 6], are given by the generating function as

$Lif k ( ln ( 1 + t ) ) ( 1 + t ) x = ∑ n = 0 ∞ c n ( k ) (−x) t n n ! .$

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

## 2 Umbral calculus

Let be the complex number field and let be the set of all formal power series in the variable t:

$F= { f ( t ) = ∑ k = 0 ∞ a k k ! t k | a k ∈ C } .$
(2)

Let $P=C[x]$ and let $P ∗$ be the vector space of all linear functionals on . $〈L|p(x)〉$ is the action of the linear functional L on the polynomial $p(x)$, and we recall that the vector space operations on $P ∗$ are defined by $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant in . For $f(t)∈F$, let us define the linear functional on by setting

$〈 f ( t ) | x n 〉 = a n (n≥0).$
(3)

In particular,

$〈 t k | x n 〉 =n! δ n , k (n,k≥0),$
(4)

where $δ n , k$ is the Kronecker symbol.

For $f L (t)= ∑ k = 0 ∞ 〈 L | x k 〉 k ! t k$, we have $〈 f L (t)| x n 〉=〈L| x n 〉$. That is, $L= f L (t)$. The map $L↦ f L (t)$ is a vector space isomorphism from $P ∗$ onto . Henceforth, denotes both the algebra of formal power series in t and the vector space of all linear functionals on , and so an element $f(t)$ of will be thought of as both a formal power series and a linear functional. We call the umbral algebra and the umbral calculus is the study of the umbral algebra. The order $O(f(t))$ of a power series $f(t)$ (≠0) is the smallest integer k for which the coefficient of $t k$ does not vanish. If $O(f(t))=1$, then $f(t)$ is called a delta series; if $O(f(t))=0$, then $f(t)$ is called an invertible series. For $f(t),g(t)∈F$ with $O(f(t))=1$ and $O(g(t))=0$, there exists a unique sequence $s n (x)$ ($deg s n (x)=n$) such that $〈g(t)f ( t ) k | s n (x)〉=n! δ n , k$, for $n,k≥0$. Such a sequence $s n (x)$ is called the Sheffer sequence for $(g(t),f(t))$, which is denoted by $s n (x)∼(g(t),f(t))$.

For $f(t),g(t)∈F$ and $p(x)∈P$, we have

$〈 f ( t ) g ( t ) | p ( x ) 〉 = 〈 f ( t ) | g ( t ) p ( x ) 〉 = 〈 g ( t ) | f ( t ) p ( x ) 〉$
(5)

and

$f(t)= ∑ k = 0 ∞ 〈 f ( t ) | x k 〉 t k k ! ,p(x)= ∑ k = 0 ∞ 〈 t k | p ( x ) 〉 x k k !$
(6)

, Theorem 2.2.5]. Thus, by (6), we get

$t k p(x)= p ( k ) (x)= d k p ( x ) d x k and e y t p(x)=p(x+y).$
(7)

Sheffer sequences are characterized by the generating function , Theorem 2.3.4].

Lemma 1 The sequence $s n (x)$ is Sheffer for $(g(t),f(t))$ if and only if

$1 g ( f ¯ ( t ) ) e y f ¯ ( t ) = ∑ k = 0 ∞ s k ( y ) k ! t k (y∈C),$

where $f ¯ (t)$ is the compositional inverse of $f(t)$.

For $s n (x)∼(g(t),f(t))$, we have the following equations , Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$f(t) s n (x)=n s n − 1 (x)(n≥0),$
(8)
$s n (x)= ∑ j = 0 n 1 j ! 〈 g ( f ¯ ( t ) ) − 1 f ¯ ( t ) j | x n 〉 x j ,$
(9)
$s n (x+y)= ∑ j = 0 n ( n j ) s j (x) p n − j (y),$
(10)

where $p n (x)=g(t) s n (x)$.

Assume that $p n (x)∼(1,f(t))$ and $q n (x)∼(1,g(t))$. Then the transfer formula , Corollary 3.8.2] is given by

$q n (x)=x ( f ( t ) g ( t ) ) n x − 1 p n (x)(n≥1).$

For $s n (x)∼(g(t),f(t))$ and $r n (x)∼(h(t),l(t))$, assume that

$s n (x)= ∑ m = 0 n C n , m r m (x)(n≥0).$

Then we have , p.132]

$C n , m = 1 m ! 〈 h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n 〉 .$
(11)

## 3 Main results

From the definition (1), $D n ( k ) (x| a 1 ,…, a r )$ is the Sheffer sequence for the pair

$g(t)= ∏ j = 1 r ( e a j t − 1 t ) 1 Lif k ( t ) andf(t)= e t −1.$

So,

$D n ( k ) (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( e a j t − 1 t ) 1 Lif k ( t ) , e t − 1 ) .$
(12)

### 3.1 Explicit expressions

Recall that Barnes’ multiple Bernoulli polynomials $B n (x| a 1 ,…, a r )$ are defined by the generating function

$t r ∏ j = 1 r ( e a j t − 1 ) e x t = ∑ n = 0 ∞ B n (x| a 1 ,…, a r ) t n n ! ,$
(13)

where $a 1 ,…, a r ≠0$ [8, 9]. Let $( n ) j =n(n−1)⋯(n−j+1)$ ($j≥1$) with $( n ) 0 =1$. The (signed) Stirling numbers of the first kind $S 1 (n,m)$ are defined by

$( x ) n = ∑ m = 0 n S 1 (n,m) x m .$

Theorem 1

$D n ( k ) ( x | a 1 , … , a r ) = ∑ m = 0 n ∑ l = 0 m S 1 ( n , m ) ( m l ) ( m − l + 1 ) k B l ( x | a 1 , … , a r )$
(14)
$= ∑ j = 0 n ∑ l = 0 n − j ∑ i = 0 l ( n l ) ( l i ) S 1 (n−l,j) c i ( k ) D l − i ( a 1 ,…, a r ) x j$
(15)
$= ∑ l = 0 n ( n l ) D n − l ( a 1 ,…, a r ) c l ( k ) (−x)$
(16)
$= ∑ l = 0 n ( n l ) c n − l ( k ) D l (x| a 1 ,…, a r ).$
(17)

Proof Since

$∏ j = 1 r ( e a j t − 1 t ) 1 Lif k ( t ) D n ( k ) (x| a 1 ,…, a r )∼ ( 1 , e t − 1 )$
(18)

and

$( x ) n ∼ ( 1 , e t − 1 ) ,$
(19)

we have

$D n ( k ) ( x | a 1 , … , a r ) = ∏ j = 1 r ( t e a j t − 1 ) Lif k ( t ) ( x ) n = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( t e a j t − 1 ) Lif k ( t ) x m = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( t e a j t − 1 ) ∑ l = 0 m t l l ! ( l + 1 ) k x m = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( t e a j t − 1 ) ∑ l = 0 m ( m ) l l ! ( l + 1 ) k x m − l = ∑ m = 0 n S 1 ( n , m ) ∑ l = 0 m ( m ) l l ! ( l + 1 ) k ∏ j = 1 r ( t e a j t − 1 ) x m − l = ∑ m = 0 n ∑ l = 0 m S 1 ( n , m ) ( m l ) ( l + 1 ) k B m − l ( x | a 1 , … , a r ) = ∑ m = 0 n ∑ l = 0 m S 1 ( n , m ) ( m l ) ( m − l + 1 ) k B l ( x | a 1 , … , a r ) .$

Thus, we get (14).

By (9) with (12), we get

$〈 g ( f ¯ ( t ) ) − 1 f ¯ ( t ) j | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) j | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) | ∑ l = 0 ∞ j ! ( l + j ) ! S 1 ( l + j , j ) t l + j x n 〉 = ∑ l = 0 n − j j ! ( l + j ) ! S 1 ( l + j , j ) ( n ) l + j 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) | x n − l − j 〉 = ∑ l = 0 n − j j ! ( n l + j ) S 1 ( l + j , j ) 〈 ∑ i = 0 ∞ D i ( k ) ( a 1 , … , a r ) t i i ! | x n − l − j 〉 = ∑ l = 0 n − j j ! ( n l + j ) S 1 ( l + j , j ) D n − l − j ( k ) ( a 1 , … , a r ) = ∑ l = 0 n − j j ! ( n l ) S 1 ( n − l , j ) D l ( k ) ( a 1 , … , a r ) .$

On the other hand,

$〈 g ( f ¯ ( t ) ) − 1 f ¯ ( t ) j | x n 〉 = ∑ l = 0 n − j j ! ( l + j ) ! S 1 ( l + j , j ) ( n ) l + j 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | Lif k ( ln ( 1 + t ) ) x n − l − j 〉 = ∑ l = 0 n − j j ! ( n l + j ) S 1 ( l + j , j ) 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ i = 0 n − l − j c i ( k ) t i i ! x n − l − j 〉 = ∑ l = 0 n − j j ! ( n l + j ) S 1 ( l + j , j ) ∑ i = 0 n − l − j c i ( k ) ( n − l − j ) i i ! 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l − j − i 〉 = ∑ l = 0 n − j j ! ( n l + j ) S 1 ( l + j , j ) ∑ i = 0 n − l − j c i ( k ) ( n − l − j ) i i ! 〈 ∑ m = 0 ∞ D m ( a 1 , … , a r ) t m m ! | x n − l − j − i 〉 = ∑ l = 0 n − j ∑ i = 0 n − l − j j ! ( n l + j ) ( n − l − j i ) S 1 ( l + j , j ) c i ( k ) D n − l − j − i ( a 1 , … , a r ) = ∑ l = 0 n − j ∑ i = 0 l j ! ( n l ) ( l i ) S 1 ( n − l , j ) c i ( k ) D l − i ( a 1 , … , a r ) .$

Thus, we obtain

$D n ( k ) ( x | a 1 , … , a r ) = ∑ j = 0 n ∑ l = 0 n − j ( n l ) S 1 ( n − l , j ) D l ( k ) ( a 1 , … , a r ) x j = ∑ j = 0 n ∑ l = 0 n − j ∑ i = 0 l ( n l ) ( l i ) S 1 ( n − l , j ) c i ( k ) D l − i ( a 1 , … , a r ) x j ,$

which is the identity (15).

Next,

$D n ( k ) ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ D i ( k ) ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | Lif k ( ln ( 1 + t ) ) ( 1 + t ) y x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ l = 0 n c l ( k ) ( − y ) t l l ! x n 〉 = ∑ l = 0 n ( n l ) c l ( k ) ( − y ) 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l 〉 = ∑ l = 0 n ( n l ) c l ( k ) ( − y ) 〈 ∑ i = 0 ∞ D i ( a 1 , … , a r ) t i i ! | x n − l 〉 = ∑ l = 0 n ( n l ) c l ( k ) ( − y ) D n − l ( a 1 , … , a r ) .$

Thus, we obtain (16).

Finally, we obtain

$D n ( k ) ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ D i ( k ) ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n 〉 = 〈 Lif k ( ln ( 1 + t ) ) | ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y x n 〉 = 〈 Lif k ( ln ( 1 + t ) ) | ∑ l = 0 n D l ( y | a 1 , … , a r ) t l l ! x n 〉 = ∑ l = 0 n D l ( y | a 1 , … , a r ) ( n l ) 〈 Lif k ( ln ( 1 + t ) ) | x n − l 〉 = ∑ l = 0 n D l ( y | a 1 , … , a r ) ( n l ) 〈 ∑ i = 0 ∞ c i ( k ) t i i ! | x n − l 〉 = ∑ l = 0 n ( n l ) D l ( y | a 1 , … , a r ) c n − l ( k ) .$

Thus, we get the identity (17). □

### 3.2 Sheffer identity

Theorem 2

$D n ( k ) (x+y| a 1 ,…, a r )= ∑ j = 0 n ( n j ) D j ( k ) (x| a 1 ,…, a r ) ( y ) n − j .$
(20)

Proof By (12) with

$p n ( x ) = ∏ j = 1 r ( e a j t − 1 t ) 1 Lif k ( t ) D n ( x | a 1 , … , a r ) = ( x ) n ∼ ( 1 , e t − 1 ) ,$

using (10), we have (20). □

### 3.3 Difference relations

Theorem 3

$D n ( k ) (x+1| a 1 ,…, a r )− D n ( k ) (x| a 1 ,…, a r )=n D n − 1 ( k ) (x| a 1 ,…, a r ).$
(21)

Proof By (8) with (12), we get

$( e t − 1 ) D n ( k ) (x| a 1 ,…, a r )=n D n − 1 ( k ) (x| a 1 ,…, a r ).$

By (7), we have (21). □

### 3.4 Recurrence

Theorem 4

$D n + 1 ( k ) ( x | a 1 , … , a r ) = x D n ( k ) ( x − 1 | a 1 , … , a r ) − ∑ m = 0 n ∑ j = 1 r ∑ l = 1 m + 1 ∑ i = 0 l ( m + 1 l ) ( l i ) ( m + 1 ) ( l − i + 1 ) k S 1 ( n , m ) × ( − a j ) m + 1 − l B m + 1 − l B i ( x − 1 | a 1 , … , a r ) + ∑ m = 0 n ∑ l = 0 m ( m l ) ( m + 2 − l ) k S 1 ( n , m ) B l ( x − 1 | a 1 , … , a r ) ,$
(22)

where $B n$ is the nth ordinary Bernoulli number.

Proof By applying

$s n + 1 (x)= ( x − g ′ ( t ) g ( t ) ) 1 f ′ ( t ) s n (x)$
(23)

, Corollary 3.7.2] with (12), we get

$D n + 1 ( k ) (x| a 1 ,…, a r )=x D n ( k ) (x−1| a 1 ,…, a r )− e − t g ′ ( t ) g ( t ) D n ( k ) (x| a 1 ,…, a r ).$

Now,

$g ′ ( t ) g ( t ) = ( ln g ( t ) ) ′ = ( ∑ j = 1 r ln ( e a j t − 1 ) − r ln t − ln Lif k ( t ) ) ′ = ∑ j = 1 r a j e a j t e a j t − 1 − r t − Lif k ′ ( t ) Lif k ( t ) = ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) t ∏ j = 1 r ( e a j t − 1 ) − Lif k ′ ( t ) Lif k ( t ) .$

Observe that

$∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) ∏ j = 1 r ( e a j t − 1 ) = 1 2 ( ∑ j = 1 r a 1 ⋯ a j − 1 a j 2 a j + 1 ⋯ a r ) t r + 1 + ⋯ ( a 1 ⋯ a r ) t r + ⋯ = 1 2 ( ∑ j = 1 r a j ) t + ⋯$

is a series with order ≥1. Since

$D n ( k ) (x| a 1 ,…, a r )= ∏ j = 1 r ( t e a j t − 1 ) Lif k (t) ( x ) n = ∑ m = 0 n S 1 (n,m) ∏ j = 1 r ( t e a j t − 1 ) Lif k (t) x m ,$

we have

$g ′ ( t ) g ( t ) D n ( k ) ( x | a 1 , … , a r ) = ∑ m = 0 n S 1 ( n , m ) g ′ ( t ) g ( t ) ( ∏ j = 1 r t e a j t − 1 ) Lif k ( t ) x m = ∑ m = 0 n S 1 ( n , m ) Lif k ( t ) ( ∏ j = 1 r t e a j t − 1 ) × ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) t ∏ j = 1 r ( e a j t − 1 ) x m − ∑ m = 0 n S 1 ( n , m ) Lif k ′ ( t ) ( ∏ j = 1 r t e a j t − 1 ) x m .$
(24)

Since

$∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) t ∏ j = 1 r ( e a j t − 1 ) x m = ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) ∏ j = 1 r ( e a j t − 1 ) x m + 1 m + 1 = 1 m + 1 ∑ j = 1 r ( a j t e a j t e a j t − 1 − 1 ) x m + 1 = 1 m + 1 ∑ j = 1 r ( ∑ l = 0 ∞ ( − 1 ) l B l a j l l ! t l − 1 ) x m + 1 = 1 m + 1 ∑ j = 1 r ( ∑ l = 0 m + 1 ( m + 1 l ) ( − a j ) l B l x m + 1 − l − x m + 1 ) = 1 m + 1 ∑ j = 1 r ∑ l = 1 m + 1 ( m + 1 l ) ( − a j ) l B l x m + 1 − l = 1 m + 1 ∑ j = 1 r ∑ l = 1 m + 1 ( m + 1 l ) ( − a j ) m + 1 − l B m + 1 − l x l ,$

the first term in (24) is

$∑ m = 0 n S 1 ( n , m ) m + 1 ∑ j = 1 r ∑ l = 1 m + 1 ( m + 1 l ) ( − a j ) m + 1 − l B m + 1 − l Lif k ( t ) ( ∏ j = 1 r t e a j t − 1 ) x l = ∑ m = 0 n S 1 ( n , m ) m + 1 ∑ j = 1 r ∑ l = 1 m + 1 ( m + 1 l ) ( − a j ) m + 1 − l B m + 1 − l ∑ i = 0 l t i i ! ( i + 1 ) k B l ( x | a 1 , … , a r ) = ∑ m = 0 n S 1 ( n , m ) m + 1 ∑ j = 1 r ∑ l = 1 m + 1 ( m + 1 l ) ( − a j ) m + 1 − l B m + 1 − l ∑ i = 0 l ( l i ) ( i + 1 ) k B l − i ( x | a 1 , … , a r ) = ∑ m = 0 n ∑ j = 1 r ∑ l = 1 m + 1 ∑ i = 0 l ( m + 1 l ) ( l i ) ( m + 1 ) ( l − i + 1 ) k S 1 ( n , m ) ( − a j ) m + 1 − l B m + 1 − l B i ( x | a 1 , … , a r ) .$

Since

$Lif k − 1 (t)− Lif k (t)= ( 1 2 k − 1 − 1 2 k ) t+⋯,$
(25)

the second term in (24) is

$∑ m = 0 n S 1 ( n , m ) Lif k − 1 ( t ) − Lif k ( t ) t B m ( x | a 1 , … , a r ) = ∑ m = 0 n S 1 ( n , m ) ( Lif k − 1 ( t ) − Lif k ( t ) ) B m + 1 ( x | a 1 , … , a r ) m + 1 = ∑ m = 0 n S 1 ( n , m ) m + 1 ( Lif k − 1 ( t ) − Lif k ( t ) ) B m + 1 ( x | a 1 , … , a r ) = ∑ m = 0 n S 1 ( n , m ) m + 1 ( ∑ l = 0 m + 1 t l l ! ( l + 1 ) k − 1 B m + 1 ( x | a 1 , … , a r ) − ∑ l = 0 m + 1 t l l ! ( l + 1 ) k B m + 1 ( x | a 1 , … , a r ) ) = ∑ m = 0 n S 1 ( n , m ) m + 1 ( ∑ l = 0 m + 1 ( m + 1 l ) ( l + 1 ) k − 1 B m + 1 − l ( x | a 1 , … , a r ) − ∑ l = 0 m + 1 ( m + 1 l ) ( l + 1 ) k B m + 1 − l ( x | a 1 , … , a r ) ) = ∑ m = 0 n S 1 ( n , m ) m + 1 ∑ l = 1 m + 1 ( m + 1 l ) l ( l + 1 ) k B m + 1 − l ( x | a 1 , … , a r ) = ∑ m = 0 n ∑ l = 1 m + 1 ( m l − 1 ) S 1 ( n , m ) 1 ( l + 1 ) k B m + 1 − l ( x | a 1 , … , a r ) = ∑ m = 0 n ∑ l = 0 m ( m l ) ( m + 2 − l ) k S 1 ( n , m ) B l ( x | a 1 , … , a r ) .$

Thus, we have

$D n + 1 ( k ) ( x | a 1 , … , a r ) = x D n ( k ) ( x − 1 | a 1 , … , a r ) − ∑ m = 0 n ∑ j = 1 r ∑ l = 1 m + 1 ∑ i = 0 l ( m + 1 l ) ( l i ) ( m + 1 ) ( l − i + 1 ) k S 1 ( n , m ) × ( − a j ) m + 1 − l B m + 1 − l B i ( x − 1 | a 1 , … , a r ) + ∑ m = 0 n ∑ l = 0 m ( m l ) ( m + 2 − l ) k S 1 ( n , m ) B l ( x − 1 | a 1 , … , a r ) ,$

which is the identity (22). □

### 3.5 Differentiation

Theorem 5

$d d x D n ( k ) (x| a 1 ,…, a r )=n! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) D l ( k ) (x| a 1 ,…, a r ).$
(26)

Proof We shall use

$d d x s n (x)= ∑ l = 0 n − 1 ( n l ) 〈 f ¯ ( t ) | x n − l 〉 s l (x)$

(cf. , Theorem 2.3.12]). Since

$〈 f ¯ ( t ) | x n − l 〉 = 〈 ln ( 1 + t ) | x n − l 〉 = 〈 ∑ m = 1 ∞ ( − 1 ) m − 1 t m m | x n − l 〉 = ∑ m = 1 n − l ( − 1 ) m − 1 m 〈 t m | x n − l 〉 = ∑ m = 1 n − l ( − 1 ) m − 1 m ( n − l ) ! δ m , n − l = ( − 1 ) n − l − 1 ( n − l − 1 ) ! ,$

with (12), we have

$d d x D n ( k ) ( x | a 1 , … , a r ) = ∑ l = 0 n − 1 ( n l ) ( − 1 ) n − l − 1 ( n − l − 1 ) ! D l ( k ) ( x | a 1 , … , a r ) = n ! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) D l ( k ) ( x | a 1 , … , a r ) ,$

which is the identity (26). □

### 3.6 One more relation

The classical Cauchy numbers $c n$ are defined by

$t ln ( 1 + t ) = ∑ n = 0 ∞ c n t n n !$

(see e.g. [1, 10]).

Theorem 6

$D n ( k ) ( x | a 1 , … , a r ) = x D n − 1 ( k ) ( x − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n ( n l ) c l D n − l ( k − 1 ) ( x − 1 | a 1 , … , a r ) + r − 1 n ∑ l = 0 n ( n l ) c l D n − l ( k ) ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( k ) ( x + a j − 1 | a 1 , … , a r , a j ) .$
(27)

Proof For $n≥1$, we have

$D n ( k ) ( y | a 1 , … , a r ) = 〈 ∑ l = 0 ∞ D l ( k ) ( y | a 1 , … , a r ) t l l ! | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y ) | x n − 1 〉 = 〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n − 1 〉 + 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ∂ t Lif k ( ln ( 1 + t ) ) ) ( 1 + t ) y | x n − 1 〉 + 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ∂ t ( 1 + t ) y ) | x n − 1 〉 .$

The third term is

$y 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | x n − 1 〉 = y D n − 1 ( k ) ( y − 1 | a 1 , … , a r ) .$

By (25), the second term is

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) ( 1 + t ) ln ( 1 + t ) ( 1 + t ) y | x n − 1 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) t ( 1 + t ) y − 1 | t ln ( 1 + t ) x n − 1 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) t ( 1 + t ) y − 1 | ∑ l = 0 ∞ c l t l l ! x n − 1 〉 = ∑ l = 0 n − 1 ( n − 1 l ) c l × 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y − 1 | Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) t x n − 1 − l 〉 = ∑ l = 0 n − 1 ( n − 1 l ) c l × 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y − 1 | ( Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) ) x n − l n − l 〉 = 1 n ∑ l = 0 n − 1 ( n l ) c l ( 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k − 1 ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | x n − l 〉 − 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | x n − l 〉 ) = 1 n ∑ l = 0 n − 1 ( n l ) c l ( D n − l ( k − 1 ) ( y − 1 | a 1 , … , a r ) − D n − l ( k ) ( y − 1 | a 1 , … , a r ) ) .$

Since

$∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) = 1 1 + t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ∑ i = 1 r ( t ln ( 1 + t ) − a i t ( 1 + t ) a i ( 1 + t ) a i − 1 ) t ,$

with

$∑ i = 1 r ( t ln ( 1 + t ) − a i t ( 1 + t ) a i ( 1 + t ) a i − 1 ) =− 1 2 ( ∑ i = 1 r a i ) t+⋯$

a series with order (≥1), the first term is

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | ∑ i = 1 r ( t ln ( 1 + t ) − a i t ( 1 + t ) a i ( 1 + t ) a i − 1 ) t x n − 1 〉 = 1 n 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | ∑ i = 1 r ( t ln ( 1 + t ) − a i t ( 1 + t ) a i ( 1 + t ) a i − 1 ) x n 〉 = r n 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | t ln ( 1 + t ) x n 〉 − 1 n ∑ i = 1 r a i 〈 ln ( 1 + t ) ( 1 + t ) a i − 1 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y + a i − 1 | t ln ( 1 + t ) x n 〉 = r n 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 − 1 n ∑ i = 1 r a i 〈 ln ( 1 + t ) ( 1 + t ) a i − 1 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y + a i − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 = r n ∑ l = 0 n ( n l ) c l 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y − 1 | x n − l 〉 − 1 n ∑ i = 1 r a i ∑ l = 0 n ( n l ) c l × 〈 ln ( 1 + t ) ( 1 + t ) a i − 1 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y + a i − 1 | x n − l 〉 = r n ∑ l = 0 n ( n l ) c l D n − l ( k ) ( y − 1 | a 1 , … , a r ) − 1 n ∑ i = 1 r a i ∑ l = 0 n ( n l ) c l D n − l ( k ) ( y + a i − 1 | a 1 , … , a r , a i ) .$

Therefore, we obtain

$D n ( k ) ( x | a 1 , … , a r ) = x D n − 1 ( k ) ( x − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n − 1 ( n l ) c l ( D n − l ( k − 1 ) ( x − 1 | a 1 , … , a r ) − D n − l ( k ) ( x − 1 | a 1 , … , a r ) ) + r n ∑ l = 0 n ( n l ) c l D n − l ( k ) ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( k ) ( x + a j − 1 | a 1 , … , a r , a j ) = x D n − 1 ( k ) ( x − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n − 1 ( n l ) c l D n − l ( k − 1 ) ( x − 1 | a 1 , … , a r ) + r − 1 n ∑ l = 0 n ( n l ) c l D n − l ( k ) ( x − 1 | a 1 , … , a r ) + 1 n c n − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( k ) ( x + a j − 1 | a 1 , … , a r , a j ) = x D n − 1 ( k ) ( x − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n ( n l ) c l D n − l ( k − 1 ) ( x − 1 | a 1 , … , a r ) + r − 1 n ∑ l = 0 n ( n l ) c l D n − l ( k ) ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( k ) ( x + a j − 1 | a 1 , … , a r , a j ) ,$

which is the identity (27). □

### 3.7 A relation including the Stirling numbers of the first kind

Theorem 7 For $n≥m≥1$, we have

$m ∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) D l ( k ) ( a 1 , … , a r ) = m r n ∑ l = 0 n − m ∑ i = 0 l ( n l ) ( l i ) S 1 ( n − l , m ) c l − i D i ( k ) ( − 1 | a 1 , … , a r ) − m n ∑ l = 0 n − m ∑ i = 0 l ∑ j = 1 r ( n l ) ( l i ) S 1 ( n − l , m ) a j c l − i D i ( k ) ( a j − 1 | a 1 , … , a r , a j ) + ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( k − 1 ) ( − 1 | a 1 , … , a r ) + ( m − 1 ) ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( k ) ( − 1 | a 1 , … , a r ) .$
(28)

Proof We shall compute

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n 〉$

in two different ways. On the one hand,

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) | ( ln ( 1 + t ) ) m x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) | ∑ l = 0 ∞ m ! ( l + m ) ! S 1 ( l + m , m ) t l + m x n 〉 = ∑ l = 0 n − m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) | x n − l − m 〉 = ∑ l = 0 n − m m ! ( n l + m ) S 1 ( l + m , m ) D n − l − m ( k ) ( a 1 , … , a r ) = ∑ l = 0 n − m m ! ( n l ) S 1 ( n − l , m ) D l ( k ) ( a 1 , … , a r ) .$

On the other hand,

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m ) | x n − 1 〉 = 〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 + 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ∂ t Lif k ( ln ( 1 + t ) ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 + 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( ∂ t ( ln ( 1 + t ) ) m ) | x n − 1 〉 .$
(29)

The third term of (29) is equal to

$m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 | ( ln ( 1 + t ) ) m − 1 x n − 1 〉 = m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 | ∑ l = 0 n − m ( m − 1 ) ! ( l + m − 1 ) ! S 1 ( l + m − 1 , m − 1 ) t l + m − 1 x n − 1 〉 = m ∑ l = 0 n − m ( m − 1 ) ! ( l + m − 1 ) ! S 1 ( l + m − 1 , m − 1 ) ( n − 1 ) l + m − 1 × 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 | x n − l − m 〉 = m ! ∑ l = 0 n − m ( n − 1 l + m − 1 ) S 1 ( l + m − 1 , m − 1 ) D n − l − m ( k ) ( − 1 | a 1 , … , a r ) = m ! ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( k ) ( − 1 | a 1 , … , a r ) .$

The second term of (29) is equal to

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( Lif k − 1 ( ln ( 1 + t ) ) − Lif k ( ln ( 1 + t ) ) ( 1 + t ) ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k − 1 ( ln ( 1 + t ) ) ( 1 + t ) − 1 | ( ln ( 1 + t ) ) m − 1 x n − 1 〉 − 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 | ( ln ( 1 + t ) ) m − 1 x n − 1 〉 = ( m − 1 ) ! ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( k − 1 ) ( − 1 | a 1 , … , a r ) − ( m − 1 ) ! ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( k ) ( − 1 | a 1 , … , a r ) .$

The first term of (29) is equal to

$〈 1 1 + t ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 = 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 ( ln ( 1 + t ) ) m | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t x n − 1 〉 = 1 n 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 ( ln ( 1 + t ) ) m | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) x n 〉 = 1 n 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 × ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) | ( ln ( 1 + t ) ) m x n 〉 = 1 n 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 × ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) | ∑ l = 0 ∞ m ! ( l + m ) ! S 1 ( l + m , m ) t l + m x n 〉 = 1 n ∑ l = 0 n − m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 × ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) | x n − l − m 〉 = m ! n ∑ l = 0 n − m ( n l + m ) S 1 ( l + m , m ) × ( r 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) − 1 | t ln ( 1 + t ) x n − l − m 〉 − ∑ j = 1 r a j 〈 ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) a j − 1 | t$