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Poly-Cauchy numbers and polynomials of the second kind with umbral calculus viewpoint

Abstract

In this paper, we consider the poly-Cauchy polynomials and numbers of the second kind which were studied by Komatsu. We note that the poly-Cauchy polynomials of the second kind are the special generalized Bernoulli polynomials of the second kind. The purpose of this paper is to give various identities of the poly-Cauchy polynomials of the second kind which are derived from umbral calculus.

1 Introduction

As is well known, the Bernoulli polynomials of the second kind are defined by the generating function to be

t log ( 1 + t ) ( 1 + t ) x = n = 0 b n (x) t n n ! ( see [1, p.130] ).
(1)

When x=0, b n = b n (0) are called the Bernoulli numbers of the second kind (see [[1], p.131]).

Let Lif k (x) be the polylogarithm factorial function, which is defined by

Lif k (x)= n = 0 x m m ! ( m + 1 ) k (see [2–7]).
(2)

The poly-Cauchy polynomials of the second kind c ˆ n ( k ) (x) (kZ, n Z 0 ) are defined by the generating function to be

Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 c ˆ n ( k ) (x) t n n ! (see [2, 3]).
(3)

When x=0, c ˆ n ( k ) = c ˆ n ( k ) (0) are called the poly-Cauchy numbers of the second kind, defined by

n = 0 c ˆ n ( k ) t n n ! = Lif k ( log ( 1 + t ) ) .
(4)

In particular, if we take k=1, then we have

Lif 1 ( log ( 1 + t ) ) ( 1 + t ) x = t ( 1 + t ) log ( 1 + t ) ( 1 + t ) x = t ( 1 + t ) x 1 log ( 1 + t ) .
(5)

Thus, we note that

c ˆ n ( 1 ) (x)= b n (x1)= B n ( n ) (x),
(6)

where B n ( α ) (x) are the Bernoulli polynomials of order α (see [8]) as their numbers [[9], p.257 and p.259].

When x=0, c ˆ n ( 1 ) = c ˆ n ( 1 ) (0)= b n (1)= B n ( n ) , where B n ( α ) are the Bernoulli numbers of order α.

The falling factorial is defined by

( x ) n =x(x1)(xn+1)= l = 0 n S 1 (n,l) x l ,
(7)

where S 1 (n,l) is the signed Stirling number of the first kind.

For m Z 0 , it is well known that

( log ( 1 + t ) ) m = m ! l = m S 1 ( l , m ) t l l ! = l = 0 S 1 ( l + m , m ) m ! ( l + m ) ! t l + m ( see [10, p.62] ) .
(8)

For λC with λ1, the Frobenius-Euler polynomials of order r are defined by the generating function to be

( 1 λ e t λ ) r e x t = n = 0 H n ( r ) (x|λ) t n n ! (see [11–13]).

In this paper, we investigate the properties of the poly-Cauchy numbers and polynomials of the second kind with umbral calculus viewpoint. The purpose of this paper is to give various identities of the poly-Cauchy polynomials of the second kind which are derived from umbral calculus.

2 Umbral calculus

Let C 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 } .
(9)

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)=cL|p(x), where c is a complex constant in C. For f(t)F, let us define the linear functional on by setting

f ( t ) | x n = a n (n0).
(10)

Then, by (9) and (10), we get

t k | x n =n! δ n , k (n,k0),
(11)

where δ n , k is Kronecker’s 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 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 (see [10, 14, 15]). 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,k0. The 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)) (see [10, 15]).

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 ) ,
(12)

and

f(t)= k = 0 f ( t ) | x k t k k ! ,p(x)= k = 0 t k | p ( x ) x k k ! .
(13)

Thus, by (13), 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).
(14)

Let us assume that s n (x)(g(t),f(t)). Then the generating function of s n (x) is given by

1 g ( f ¯ ( t ) ) e x f ¯ ( t ) = n = 0 s n (x) t n n ! ,for all xC,
(15)

where f ¯ (t) is the compositional inverse of f(t) with f ¯ (f(t))=t (see [10, 15]).

For s n (x)(g(t),f(t)), we have the following equation:

f(t) s n (x)=n s n 1 (x)(n0),
(16)
s n (x)= j = 0 n 1 j ! g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n x j ,
(17)

and

s n (x+y)= j = 0 n ( n j ) s j (x) p n j (y),
(18)

where p n (x)=g(t) s n (x) (see [[10], p.21]).

Let us assume that p n (x)(1,f(t)), q n (x)(1,g(t)). Then the transfer formula is given by

q n (x)=x ( f ( t ) g ( t ) ) n x 1 p n (x)(n0)( see [10, p.51] ).

For s n (x)(g(t),f(t)), r n (x)(h(t),l(t)), let us assume that

s n (x)= m = 0 n C n , m r n (x)(n0).
(19)

Then we have

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n ( see [10, p.132] ).
(20)

3 Poly-Cauchy numbers and polynomials of the second kind

From (3), we note that c ˆ n ( k ) (x) is the Sheffer sequence for the pair

( g ( t ) = 1 Lif k ( t ) , f ( t ) = e t 1 ) ,

that is,

c ˆ n ( k ) (x) ( 1 Lif k ( t ) , e t 1 ) .
(21)

Because for f ¯ (t)=log(1+t), using the formula (15), we get

Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 s n (x) t n n !

which is the generating function of c ˆ n ( k ) (x) in (3).

From (21), we have

1 Lif k ( t ) c ˆ n ( k ) (x) ( 1 , e t 1 ) ,
(22)

and

( x ) n = l = 0 n S 1 (n,l) x l ( 1 , e t 1 ) .
(23)

By (22) and (23), we get

c ˆ n ( k ) ( x ) = Lif k ( t ) ( x ) n = m = 0 n S 1 ( n , m ) Lif k ( t ) x m = m = 0 n S 1 ( n , m ) a = 0 m ( 1 ) a a ! ( a + 1 ) k t a x m = m = 0 n a = 0 m S 1 ( n , m ) ( 1 ) a ( m a ) ( a + 1 ) k x m a = m = 0 n j = 0 m S 1 ( n , m ) ( 1 ) m j ( m j ) ( m j + 1 ) k x j = j = 0 n { m = j n S 1 ( n , m ) ( 1 ) m j ( m j ) ( m j + 1 ) k } x j .
(24)

By (17) and (21), we get

c ˆ n ( k ) (x)= j = 0 n 1 j ! Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j | x n x j .
(25)

Now, we observe that

Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j | x n = m = 0 ( 1 ) m m ! ( m + 1 ) k ( log ( 1 + t ) ) m + j | x n = m = 0 n j ( 1 ) m m ! ( m + 1 ) k l = 0 n j m S 1 ( l + m + j , m + j ) ( l + m + j ) ! ( m + j ) ! t m + j + l | x n = m = 0 n j ( 1 ) m m ! ( m + 1 ) k l = 0 n m j S 1 ( l + m + j , m + j ) ( l + m + j ) ! ( m + j ) ! n ! δ n , l + m + j = m = 0 n j ( 1 ) m ( m + j ) ! m ! ( m + 1 ) k S 1 ( n , m + j ) .
(26)

From (25) and (26), we have

c ˆ n ( k ) ( x ) = j = 0 n 1 j ! m = 0 n j ( 1 ) m ( m + j ) ! m ! ( m + 1 ) k S 1 ( n , m + j ) x j = j = 0 n { m = 0 n j ( 1 ) m ( m + j m ) ( m + 1 ) k S 1 ( n , m + j ) } x j = j = 0 n { m = j n ( 1 ) m j ( m j ) ( m j + 1 ) k S 1 ( n , m ) } x j ,
(27)

which is the same as the expression in (24). From (1), we note that

1 Lif k ( t ) c ˆ n ( k ) (x) ( 1 , e t 1 ) , x n (1,t).
(28)

For n1, by (19) and (28), we get

1 Lif k ( t ) c ˆ n ( k ) ( x ) = x ( t e t 1 ) n x 1 x n = x ( t e t 1 ) n x n 1 = x B n 1 ( n ) ( x ) = l = 0 n 1 ( n 1 l ) B n 1 l ( n ) x l + 1 .
(29)

Thus, by (29), we see that

c ˆ n ( k ) ( x ) = l = 0 n 1 ( n 1 l ) B n 1 l ( n ) Lif k ( t ) x l + 1 = l = 0 n 1 m = 0 l + 1 ( 1 ) m ( n 1 l ) ( l + 1 m ) B n 1 l ( n ) ( m + 1 ) k x l + 1 m = l = 0 n 1 j = 0 l + 1 ( 1 ) l + 1 j ( n 1 l ) ( l + 1 j ) B n 1 l ( n ) ( l + 2 j ) k x j = l = 0 n 1 ( 1 ) l + 1 ( n 1 l ) B n 1 l ( n ) ( l + 2 ) k + j = 1 n { l = j 1 n 1 ( 1 ) l + 1 j ( n 1 l ) ( l + 1 j ) B n 1 l ( n ) ( l + 2 j ) k } x j .
(30)

Therefore, by (27) and (30), we obtain the following theorem.

Theorem 1 For n1, 1jn, we have

m = j n ( 1 ) m j ( m j ) ( m j + 1 ) k S 1 (n,m)= l = j 1 n 1 ( 1 ) l + 1 j ( n 1 l ) ( l + 1 j ) B n 1 l ( n ) ( l + 2 j ) k .

In addition, for n1, we have

c ˆ n ( k ) = m = 0 n S 1 (n,m) ( 1 ) m ( m + 1 ) k = l = 0 n 1 ( 1 ) l + 1 ( n 1 l ) B n 1 l ( n ) ( l + 2 ) k .

From (18), we note that

c ˆ n ( k ) (x+y)= j = 0 n ( n j ) c ˆ j ( k ) (x) p n j (y),
(31)

where p n (y)= 1 Lif k ( t ) c ˆ n ( k ) (y)(1, e t 1).

By (22) and (23), we get

( y ) n = p n (y) ( 1 , e t 1 ) .
(32)

Thus, from (31) and (32), we have

c ˆ n ( k ) (x+y)= j = 0 n ( n j ) c ˆ j ( k ) (x) ( y ) n j .
(33)

By (14), (16), and (21), we get

c ˆ n ( k ) (x+1) c ˆ n ( k ) (x)= ( e t 1 ) c ˆ n ( k ) (x)=n c ˆ n 1 ( k ) (x).

For s n (x)(g(t),f(t)), the recurrence formula for s n (x) is given by

s n + 1 (x)= ( x g ( t ) g ( t ) ) 1 f ( t ) s n (x)(see [10]).
(34)

By (21) and (34), we get

c ˆ n + 1 ( k ) ( x ) = ( x Lif k ( t ) Lif k ( t ) ) e t c ˆ n ( k ) ( x ) = x c ˆ n ( k ) ( x 1 ) e t Lif k ( t ) Lif k ( t ) c ˆ n ( k ) ( x ) .
(35)

We observe that

Lif k ( t ) Lif k ( t ) Lif k ( t ) c ˆ n ( k ) ( x ) = Lif k ( t ) 1 Lif k ( t ) c ˆ n ( k ) ( x ) = Lif k ( t ) ( x ) n = l = 0 n S 1 ( n , l ) Lif k ( t ) x l = l = 0 n S 1 ( n , l ) m = 0 l ( 1 ) m ( l m ) ( m + 2 ) k x l m = j = 0 n { l = j n ( 1 ) l j ( l j ) ( l j + 2 ) k S 1 ( n , l ) } x j .
(36)

Therefore, by (35) and (36), we obtain the following theorem.

Theorem 2 For n0, we have

c ˆ n + 1 ( k ) (x)=x c ˆ n ( k ) (x1) j = 0 n { l = j n S 1 ( n , l ) ( 1 ) l j ( l j + 2 ) k ( l j ) } ( x 1 ) j .

From (11), we note that

c ˆ n ( k ) ( y ) = l = 0 c ˆ l ( k ) ( y ) t l l ! | x n = Lif k ( log ( 1 + t ) ) ( 1 + t ) y | x n = Lif k ( log ( 1 + t ) ) ( 1 + t ) y | x x n 1 = t ( Lif k ( log ( 1 + t ) ) ( 1 + t ) y ) | x n 1 = t ( Lif k ( log ( 1 + t ) ) ) ( 1 + t ) y | x n 1 + Lif k ( log ( 1 + t ) ) t ( 1 + t ) y | x n 1 = t ( Lif k ( log ( 1 + t ) ) ) ( 1 + t ) y | x n 1 + y c ˆ n 1 ( k ) ( y 1 ) ,
(37)

where t f(t)= d f ( t ) d t .

Since t Lif k (t)= Lif k 1 (t) Lif k (t), we get

Lif k (t)= Lif k 1 ( t ) Lif k ( t ) t .
(38)

By (37) and (38), we see that

c ˆ n ( k ) ( y ) = y c ˆ n 1 ( k ) ( y 1 ) + Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) ( 1 + t ) log ( 1 + t ) ( 1 + t ) y | x n 1 = y c ˆ n 1 ( k ) ( y 1 ) + Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) ( 1 + t ) y | t log ( 1 + t ) x n 1 .
(39)

From (1), (6), and (38), we note that

c ˆ n ( k ) ( y ) = y c ˆ n 1 ( k ) ( y 1 ) + l = 0 n 1 B l ( l ) ( 1 ) l ! ( n 1 ) l × Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) y 1 | x n l 1 = y c ˆ n 1 ( k ) ( y 1 ) + l = 0 n 1 B l ( l ) ( 1 ) l ! ( n 1 ) l × Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) y 1 | t x n l n l = y c ˆ n 1 ( k ) ( y 1 ) + l = 0 n 1 ( n 1 l ) B l ( l ) ( 1 ) n l { c ˆ n l ( k 1 ) ( y 1 ) c ˆ n l ( k ) ( y 1 ) } = y c ˆ n 1 ( k ) ( y 1 ) + 1 n l = 0 n 1 ( n l ) B l ( l ) ( 1 ) { c ˆ n l ( k 1 ) ( y 1 ) c ˆ n l ( k ) ( y 1 ) } .
(40)

It is not difficult to show that c ˆ 0 ( k ) (y1)= c ˆ 0 ( k 1 ) (y1). Since c ˆ 0 ( k ) (y1)= c ˆ 0 ( k 1 ) (y1), by (40), we obtain the following theorem.

Theorem 3 For n1, we have

c ˆ n ( k ) (x)=x c ˆ n 1 ( k ) (x1)+ 1 n l = 0 n ( n l ) B l ( l ) (1) { c ˆ n l ( k 1 ) ( x 1 ) c ˆ n l ( k ) ( x 1 ) } .

For nm1, we compute

( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) | x n

in two different ways.

On the one hand,

( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) | x n = Lif k ( log ( 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 Lif k ( log ( 1 + t ) ) | x n l m = l = 0 n m m ! ( n l + m ) S 1 ( l + m , m ) c ˆ n l m ( k ) .
(41)

On the other hand, we get

( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) | x n = ( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) | x x n 1 = t ( ( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) ) | x n 1 .
(42)

Now, we observe that

t ( ( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) ) = m ( log ( 1 + t ) ) m 1 1 1 + t Lif k ( log ( 1 + t ) ) + ( log ( 1 + t ) ) m Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) ( 1 + t ) log ( 1 + t ) = ( log ( 1 + t ) ) m 1 ( 1 + t ) 1 { m Lif k ( log ( 1 + t ) ) + Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) } .
(43)

By (42) and (43), we get

( log ( 1 + t ) ) m Lif k ( log ( 1 + t ) ) | x n = l = 0 n m ( m 1 ) ! ( l + m 1 ) ! S 1 ( l + m 1 , m 1 ) × { ( m 1 ) Lif k ( log ( 1 + t ) ) ( 1 + t ) 1 | t l + m 1 x n 1 + Lif k 1 ( log ( 1 + t ) ) ( 1 + t ) 1 | t l + m 1 x n 1 } = ( m 1 ) l = 0 n m ( m 1 ) ! ( l + m 1 ) ! S 1 ( l + m 1 , m 1 ) ( n 1 ) l + m 1 × Lif k ( log ( 1 + t ) ) ( 1 + t ) 1 | x n m l + l = 0 n m ( m 1 ) ! ( l + m 1 ) ! S 1 ( l + m 1 , m 1 ) ( n 1 ) l + m 1 × Lif k 1 ( log ( 1 + t ) ) ( 1 + t ) 1 | x n m l = l = 0 n m ( m 1 ) ! ( n 1 l + m 1 ) S 1 ( l + m 1 , m 1 ) × { ( m 1 ) c ˆ n l m ( k ) ( 1 ) + c ˆ n l m ( k 1 ) ( 1 ) } .
(44)

Therefore, by (41) and (44), we obtain the following theorem.

Theorem 4 For nm1, we have

l = 0 n m m ! ( n l + m ) S 1 ( l + m , m ) c ˆ n l m ( k ) = l = 0 n m ( m 1 ) ! ( n 1 l + m 1 ) S 1 ( l + m 1 , m 1 ) × { ( m 1 ) c ˆ n l m ( k ) ( 1 ) + c ˆ n l m ( k 1 ) ( 1 ) } .

In particular, if we take m=1, then we get

c ˆ n ( k 1 ) (1)= l = 0 n 1 ( 1 ) l l! ( n l + 1 ) c ˆ n l 1 ( k ) .

Remark For s n (x)(g(t),f(t)), it is known that

d d x s n (x)= l = 0 n 1 ( n l ) f ¯ ( t ) | x n l s l (x)( see [10, p.108] ).
(45)

By (21) and (45), we easily show that

d d x c ˆ n ( k ) (x)= ( 1 ) n n! l = 0 n 1 ( 1 ) l 1 ( n l ) l ! c ˆ l ( k ) (x),

which is a special case of Proposition 2 in [4].

Let us consider the following two Sheffer sequences:

c ˆ n ( k ) (x) ( 1 Lif k ( t ) , e t 1 ) ,
(46)

and

B n ( r ) (x) ( ( e t 1 t ) r , t ) .

Suppose that

c ˆ n ( k ) (x)= m = 0 n C n , m B m ( r ) (x).
(47)

By (20), we see that

C n , m = 1 m ! ( t log ( 1 + t ) ) r 1 Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n = 1 m ! Lif k ( log ( 1 + t ) ) ( t log ( 1 + t ) ) r ( log ( 1 + t ) ) m | x n = 1 m ! l = 0 n m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m × Lif k ( log ( 1 + t ) ) ( t log ( 1 + t ) ) r | x n l m = 1 m ! l = 0 n m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m a = 0 n l m B a ( a r + 1 ) 1 a ! × Lif k ( log ( 1 + t ) ) | t a x n l m = l = 0 n m ( n l + m ) S 1 ( l + m , m ) a = 0 n l m B a ( a r + 1 ) ( n l m ) a a ! × Lif k ( log ( 1 + t ) ) | x n l m a = l = 0 n m a = 0 n l m ( n l + m ) ( n m l a ) S 1 ( l + m , m ) B a ( a r + 1 ) ( 1 ) c ˆ n l m a ( k ) .
(48)

Therefore, by (47) and (48), we obtain the following theorem.

Theorem 5 For n0, we have

c ˆ n ( k ) (x)= m = 0 n { l = 0 n m a = 0 n m l ( n l + m ) ( n m l a ) S 1 ( l + m , m ) B a ( a r + 1 ) ( 1 ) c ˆ n m l a ( k ) } B m ( r ) (x).

Remark The Narumi polynomials of order a are defined by the generating function to be

k = 0 N k ( a ) ( x ) k ! t k = ( t log ( 1 + t ) ) a ( 1 + t ) x ( see [10, p.127]) .
(49)

Indeed, N a ( k ) (x)= B k ( k + a + 1 ) (x+1), N k ( a ) (x)( ( e t 1 t ) a , e t 1).

By (48) and (49), we get

C n , m = l = 0 n m a = 0 n m l ( n l + m ) ( n l m a ) S 1 (l+m,m) N a ( r ) c ˆ n l m a ( k ) .
(50)

From (47) and (50), we have

c ˆ n ( k ) ( x ) = m = 0 n { l = 0 n m a = 0 n m l ( n l + m ) ( n l m a ) × S 1 ( l + m , m ) N a ( r ) c ˆ n l m a ( k ) } B m ( r ) ( x ) .
(51)

By (1), we easily show that

C n , m = l = 0 n m a = 0 n m l a 1 + + a r = a ( n l + m ) ( n l m a ) ( a a 1 , , a r ) × S 1 ( l + m , m ) b a 1 b a r c ˆ n m l a ( k ) .
(52)

From (47) and (52), we can derive the following equation:

c ˆ n ( k ) ( x ) = m = 0 n { l = 0 n m a = 0 n m l a 1 + + a r = a ( n l + m ) ( n l m a ) ( a a 1 , , a r ) × S 1 ( l + m , m ) ( i = 1 r b a i ) c ˆ n m l a ( k ) } B m ( r ) ( x ) .
(53)

For (20) and (24), let

c ˆ n ( k ) (x)= m = 0 n C n , m H m ( r ) (x|λ),
(54)

where, by (20), we get

C n , m = 1 m ! ( 1 λ ) r Lif k ( log ( 1 + t ) ) ( 1 + t λ ) r | ( log ( 1 + t ) ) m x n = 1 m ! ( 1 λ ) r l = 0 n m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m × Lif k ( log ( 1 + t ) ) ( 1 + t λ ) r | x n l m .
(55)

We observe that

Lif k ( log ( 1 + t ) ) ( 1 + t λ ) r | x n l m = a = 0 r ( r a ) ( 1 λ ) r a Lif k ( log ( 1 + t ) ) | t a x n l m = a = 0 r ( r a ) ( 1 λ ) r a ( n m l ) a Lif k ( log ( 1 + t ) ) | x n l m a = a = 0 r ( r a ) ( 1 λ ) r a ( n m l ) a c ˆ n l m a ( k ) .
(56)

Thus, by (55) and (56), we get

C n , m = l = 0 n m a = 0 r ( n l + m ) ( r a ) ( n m l ) a ( 1 λ ) a S 1 (l+m,m) c ˆ n m l a ( k ) .
(57)

Therefore, by (54) and (57), we obtain the following theorem.

Theorem 6 For n0, we have

c ˆ n ( k ) ( x ) = m = 0 n { l = 0 n m a = 0 r ( n l + m ) ( r a ) ( n m l ) a ( 1 λ ) a S 1 ( l + m , m ) × c ˆ n m l a ( k ) } H m ( r ) ( x | λ ) .

For c ˆ n ( k ) (x)( 1 Lif k ( t ) , e t 1), and ( x ) n (1, e t 1), let us assume that

c ˆ n ( k ) (x)= m = 0 n C n , m ( x ) m .
(58)

From (20), we note that

C n , m = 1 m ! Lif k ( log ( 1 + t ) ) t m | x n = 1 m ! Lif k ( log ( 1 + t ) ) | t m x n = ( n m ) Lif k ( log ( 1 + t ) ) | x n m = ( n m ) c ˆ n m ( k ) .
(59)

Therefore, by (58) and (59), we obtain the following theorem.

Theorem 7 For n0, we have

c ˆ n ( k ) (x)= m = 0 n ( n m ) c ˆ n m ( k ) ( x ) m .

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Acknowledgements

The authors express their sincere gratitude to the referees for their valuable suggestions and comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786).

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Correspondence to Taekyun Kim.

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Kim, D.S., Kim, T. Poly-Cauchy numbers and polynomials of the second kind with umbral calculus viewpoint. Adv Differ Equ 2014, 36 (2014). https://doi.org/10.1186/1687-1847-2014-36

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Keywords

  • Vector Space
  • Formal Power Series
  • Linear Functional
  • Bernoulli Number
  • Bernoulli Polynomial