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Some properties of higher-order Daehee polynomials of the second kind arising from umbral calculus

Abstract

In this paper, we study the higher-order Daehee polynomials of the second kind from the umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

1 Introduction

Let k Z 0 . The Daehee polynomials of the second kind of order k are defined by the generating function to be

( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) x = n = 0 D ˆ n ( k ) (x) t n n !
(1)

(see [1]).

When x=0, D ˆ n ( k ) = D ˆ n ( k ) (0) are called the Daehee numbers of the second kind of order k.

The Stirling number of the first kind is defined by the falling factorial to be

( x ) n =x(x1)(xn+1)= l = 0 n S 1 (n,l) x l .
(2)

Thus, by (2), we get

( log ( 1 + t ) ) m =m! l = m S 1 (l,m) t l l !
(3)

(see [24]), where m Z 0 .

For λC with λ1, the Frobenius-Euler polynomials of order s (N) are given by

( 1 λ e t λ ) s e x t = n = 0 H n ( s ) (x|λ) t n n !
(4)

(see [118]).

When x=0, H n ( s ) (λ)= H n ( s ) (λ|0) are called the Frobenius-Euler numbers of order s.

As is well known, the Bernoulli polynomials of order k (N) are defined by the generating function to be

( t e t 1 ) k e x t = n = 0 B n ( k ) (x) t n n !
(5)

(see [118]).

When x=0, B n ( k ) = B n ( k ) (0) are called the Bernoulli numbers of order k.

In this paper, we study the higher-order Daehee polynomials of the second kind with umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

2 Umbral calculus

Let be the complex number field and let be the set of all formal power series

F= { f ( t ) = k = 0 a k t k k ! | a k C } .

Let P=C[x], and let P be the vector space of all linear functionals on . L|p(x) indicates the action of the linear functional L on the polynomial p(x). Then the vector space operations on P are given by L+M|p(x)=L|p(x)+M|p(x), and cL|p(x)=cL|p(x), where c is a complex constant in . For f(t)F, the linear functional on is defined by f(t)| x n = a n . Then, in particular, we have

t k | x n =n! δ n , k (n,k0)
(6)

(see [3, 18]), where δ n , k is the Kronecker symbol.

Let f L (t)= k = 0 L | x k k ! t k . By (6), we get 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 the 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 the power series f(t) (≠0) is the smallest integer for which the coefficient of t k does not vanish. If o(f(t))=0, then f(t) is called an invertible series; if o(f(t))=1, then f(t) is called a delta series.

Let f(t),g(t)F with o(f(t))=1 and o(g(t))=0. Then 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)). For f(t),g(t)F, we have

f ( t ) g ( t ) | p ( x ) = f ( t ) | g ( t ) p ( x ) = g ( t ) | f ( t ) p ( x ) .
(7)

From (6), we note that

f(t)= k = 0 f ( t ) | x k t k k ! ,p(x)= k = 0 t k | p ( x ) x k k !
(8)

and, by (8), 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)
(9)

(see [3, 18]).

For s n (x)(g(t),f(t)), we have

d s n ( x ) d x = l = 0 n 1 ( n l ) f ¯ ( t ) | x n l s l (x),
(10)

where f ¯ (t) is the compositional inverse of f(t) with f ¯ (f(t))=t. We have

1 g ( f ¯ ( t ) ) e x f ¯ ( t ) = n = 0 s n (x) t n n ! ,for all xC,
(11)
f(t) s n (x)=n s n 1 (x)(n1), s n (x)= j = 0 n 1 j ! g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n x j ,
(12)
s n (x+y)= j = 0 n ( n j ) s j (x) p n j (y),
(13)

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

f ( t ) | x p ( x ) = t f ( t ) | p ( x ) ,
(14)

with t f(t)= d f ( t ) d t , and

s n + 1 (x)= ( x g ( t ) g ( t ) ) 1 f ( t ) s n (x)(n0)
(15)

(see [3, 18]).

Let us assume that s n (x)(g(t),f(t)) and r n (x)(h(t),l(t)). Then we see that

s n (x)= m = 0 n C n , m r m (x)(n0),
(16)

where

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n
(17)

(see [3, 18]).

3 Higher-order Daehee polynomials of the second kind

By (1), we see that

D ˆ n ( k ) (x) ( ( e t 1 t e t ) k , e t 1 ) .
(18)

From (18), we have

( e t 1 t e t ) k D ˆ n ( k ) (x) ( 1 , e t 1 ) and ( x ) n ( 1 , e t 1 ) .
(19)

By (19), we get

D ˆ n ( k ) ( x ) = ( t e t e t 1 ) k ( x ) n = m = 0 n S 1 ( n , m ) ( t e t e t 1 ) k x m = m = 0 n S 1 ( n , m ) e k t B n ( k ) ( x ) = m = 0 n S 1 ( n , m ) B m ( k ) ( x + k ) .
(20)

From (12) and (18), we have

D ˆ n ( k ) (x)= j = 0 n 1 j ! ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) j | x n x j ,
(21)

where

( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) j | x n = ( log ( 1 + t ) t ) k + j ( 1 + t ) k | t j x n = ( n ) j ( log ( 1 + t ) t ) k + j | m = 0 min { k , n j } ( k m ) t m x n j = ( n ) j m = 0 n j ( k m ) ( n j ) m l = 0 ( k + j ) ! ( l + k + j ) ! S 1 ( l + k + j , k + j ) t l | x n j m = ( n ) j m = 0 n j ( k m ) ( n j ) m ( k + j ) ! ( n + k m ) ! S 1 ( n + k m , k + j ) ( n j m ) ! = ( n ) j m = 0 n j ( k m ) ( n j ) m S 1 ( n + k m , k + j ) ( n + k m k + j ) .
(22)

Therefore, by (21) and (22), we obtain the following theorem.

Theorem 1 For n Z 0 and k1, we have

D ˆ n ( k ) (x)= j = 0 n { ( n j ) m = 0 n j ( k m ) ( n j ) m S 1 ( n + k m , k + j ) ( n + k m k + j ) } x j .

By (1) and (6), we get

D ˆ n ( k ) ( y ) = l = 0 D ˆ l ( k ) ( y ) t l l ! | x n = ( log ( 1 + t ) t ) k ( 1 + t ) y | ( 1 + t ) k x n = 0 r min { k , n } ( k r ) ( n ) r ( log ( 1 + t ) t ) k ( 1 + t ) y | x n r = 0 r min { k , n } ( k r ) ( n ) r 0 m n r ( y m ) ( n r ) m × 0 l n r m k ! S 1 ( l + k , k ) ( l + k ) ! t l | x n r m = 0 r n 0 m n r ( n ) r ( k r ) ( n r m ) ( n r m + k k ) S 1 ( n r m + k , k ) ( y ) m .
(23)

Therefore, by (23), we obtain the following theorem.

Theorem 2 For n0, we have

D ˆ n ( k ) ( x ) = 0 m n { 0 r n m ( n ) r ( k r ) ( n r m ) ( n r m + k k ) S 1 ( n r m + k , k ) } ( x ) m = 0 m n { 0 r n m ( n ) r ( k r ) ( n r n m ) ( m r + k k ) S 1 ( m r + k , k ) } ( x ) n m .

From (12) and (18), we have

( e t 1 ) D ˆ n ( k ) (x)=n D ˆ n 1 ( k ) (x)
(24)

and

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

Thus, by (24), we get

D ˆ n ( k ) (x+1) D ˆ n ( k ) (x)=n D ˆ n 1 ( k ) (x)(n1).
(25)

From (15) and (18), we derive the following equation:

D ˆ n + 1 ( k ) ( x ) = ( x + k e t 1 t t ( e t 1 ) ) e t D ˆ n ( k ) ( x ) = x D ˆ n ( k ) ( x 1 ) + k e t e t 1 t t ( e t 1 ) D ˆ n ( k ) ( x ) ,
(26)

where

e t e t 1 t t ( e t 1 ) D ˆ n ( k ) ( x ) = e t e t 1 t t ( e t 1 ) 0 j n { ( n j ) 0 m n j m ! ( k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) } x j = 0 j n ( n j ) 0 m n j m ! ( k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) e t e t 1 t t ( e t 1 ) x j = 0 j n ( n j ) 0 m n j m ! ( m + k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) e t ( e t 1 t e t 1 ) x j + 1 j + 1 = 0 j n ( n j ) 0 m n j m ! ( m + k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) j + 1 e t ( x j + 1 B j + 1 ( x ) ) = 0 j n ( n j ) 0 m n j m ! ( m + k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) j + 1 e t ( ( x 1 ) j + 1 B j + 1 ( x 1 ) ) .
(27)

Therefore, from (26) and (27), we obtain the following theorem.

Theorem 3 For n0, k1, we have

D ˆ n + 1 ( k ) ( x ) = x D ˆ n ( k ) ( x 1 ) + k 0 j n ( n j ) 0 m n j m ! ( m + k m ) ( n j m ) ( n + k m k + j ) × S 1 ( n + k m , k + j ) j + 1 { ( x 1 ) j + 1 B j + 1 ( x 1 ) } .

Now, we observe that

e t e t 1 t t ( e t 1 ) D ˆ n ( k ) ( x ) = j = 0 n ( n j ) ( n + k j + k ) S 1 ( n + k , j + k ) e t e t 1 t t ( e t 1 ) ( x + k ) j = j = 0 n ( n j ) ( n + k j + k ) S 1 ( n + k , j + k ) e ( k 1 ) t e t 1 t t ( e t 1 ) x j = j = 0 n ( n j ) ( n + k j + k ) S 1 ( n + k , j + k ) j + 1 e ( k 1 ) t ( x j + 1 B j + 1 ( x ) ) = j = 0 n ( n j ) ( n + k j + k ) S 1 ( n + k , j + k ) j + 1 ( ( x + k 1 ) j + 1 B j + 1 ( x + k 1 ) ) .
(28)

Thus, by (28), we get

D ˆ n + 1 ( k ) (x)=x D ˆ n ( k ) (x1)+k j = 0 n ( n j ) ( n + k j + k ) S 1 ( n + k , j + k ) j + 1 ( ( x + k 1 ) j + 1 B j + 1 ( x + k 1 ) ) .

From (10) and (18), we note that

d d x D ˆ n ( k ) (x)=n! l = 0 n 1 ( 1 ) n l 1 l ! ( n l ) D ˆ l ( k ) (x).
(29)

By (6) and (18), we see that

D ˆ n ( k ) ( y ) = l = 0 D ˆ l ( k ) ( y ) t l l ! | x n ( n 1 ) = ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) y | x n = t ( ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) y ) | x n 1 = ( t ( ( 1 + t ) log ( 1 + t ) t ) k ) ( 1 + t ) y | x n 1 + y ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) y 1 | x n 1 = y D ˆ n 1 ( k ) ( y 1 ) + k ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( 1 + t ) y | ( log ( 1 + t ) + 1 ( 1 + t ) log ( 1 + t ) t ) x n n = y D ˆ n 1 ( k ) ( y 1 ) + k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( 1 + t ) y | log ( 1 + t ) x n + k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( 1 + t ) y | x n k n ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) y | x n = y D ˆ n 1 ( k ) ( y 1 ) + k n D ˆ n ( k 1 ) ( y ) k n D ˆ n ( k ) ( y ) + k n 1 l n ( 1 ) l 1 ( n ) l l ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( 1 + t ) y | x n l .
(30)

Thus, by (30), we get

D ˆ n ( k ) ( x ) = n n + k x D ˆ n 1 ( k ) ( x 1 ) + k n + k D ˆ n ( k 1 ) ( x ) + k n + k 1 l n ( 1 ) l 1 ( n l ) ( l 1 ) ! D ˆ n l ( k 1 ) ( x ) .
(31)

Therefore, by (31), we obtain the following theorem.

Theorem 4 For n0, k1, we have

D ˆ n ( k ) ( x ) = n n + k x D ˆ n 1 ( k ) ( x 1 ) + k n + k D ˆ n ( k 1 ) ( x ) + k n + k 1 l n ( 1 ) l 1 ( n l ) ( l 1 ) ! D ˆ n l ( k 1 ) ( x ) .

Now, we compute ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n in two different ways:

( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n = ( ( 1 + t ) log ( 1 + t ) t ) k | ( log ( 1 + t ) ) m x n = 0 l n m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m ( ( 1 + t ) log ( 1 + t ) t ) k | x n l m = 0 l n m m ! ( n l + m ) S 1 ( l + m , m ) D ˆ n l m ( k ) = 0 l n m m ! ( n l ) S 1 ( n l , m ) D ˆ l ( k ) .
(32)

On the other hand,

( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n = t ( ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m ) | x n 1 = k ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( log ( 1 + t ) + 1 ( 1 + t ) log ( 1 + t ) t t ) ( log ( 1 + t ) ) m | x n 1 + m ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) 1 ( log ( 1 + t ) ) m 1 | x n 1 = k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( log ( 1 + t ) ) m + 1 | x n + k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( log ( 1 + t ) ) m | x n k n ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n + m ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) 1 ( log ( 1 + t ) ) m 1 | x n 1 .
(33)

Thus, by (33), we get

n + k n ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n = k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( log ( 1 + t ) ) m + 1 | x n + k n ( ( 1 + t ) log ( 1 + t ) t ) k 1 ( log ( 1 + t ) ) m | x n + m ( ( 1 + t ) log ( 1 + t ) t ) k ( 1 + t ) 1 ( log ( 1 + t ) ) m 1 | x n 1 .
(34)

From (34), we derive the following equation:

n + k k 0 l n m m ! ( n l ) S 1 ( n l , m ) D ˆ l ( k ) = k n 0 l n m 1 ( m + 1 ) ! ( n l ) S 1 ( n l , m + 1 ) D ˆ l ( k 1 ) + k n 0 l n m m ! ( n l ) S 1 ( n l , m ) D ˆ l ( k 1 ) + m 0 l n m ( m 1 ) ! ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 ) .
(35)

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

Theorem 5 For n1m1, we have

l = 0 n m ( n l ) S 1 ( n l , m ) D ˆ l ( k ) = k ( m + 1 ) n + k 0 l n m 1 ( n l ) S 1 ( n l , m + 1 ) D ˆ l ( k 1 ) + k n + k 0 l n m ( n l ) S 1 ( n l , m ) D ˆ l ( k 1 ) + n n + k 0 l n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 ) .

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

D ˆ n ( k ) (x)= m = 0 n C n , m ( x ) m .
(36)

Then, by (16) and (17), we get

C n , m = 1 m ! ( ( 1 + t ) log ( 1 + t ) t ) k | t m x n = ( n m ) ( ( 1 + t ) log ( 1 + t ) t ) k | x n m = ( n m ) D ˆ n m ( k ) .
(37)

Therefore, by (36) and (37), we obtain the following theorem.

Theorem 6 For n0, we have

D ˆ n ( k ) ( x ) = 0 m n ( n m ) D ˆ n m ( k ) ( x ) m = 0 m n m ! ( n m ) D ˆ n m ( k ) ( x m ) .

Now, we consider the following two Sheffer sequences:

D ˆ n ( k ) (x) ( ( e t 1 t e t ) k , e t 1 )
(38)

and

H n ( s ) (x|λ) ( ( e t λ 1 λ ) s , t ) ,sN,λC with λ1.

Let

D ˆ n ( k ) (x)= m = 0 n C n , m H m ( s ) (x|λ).
(39)

Here

C n , m = 1 m ! ( 1 λ ) s ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m ( 1 λ + t ) s | x n = 1 m ! ( 1 λ ) s j = 0 n ( s j ) ( 1 λ ) s j ( n ) j × ( ( 1 + t ) log ( 1 + t ) t ) k ( log ( 1 + t ) ) m | x n j = j = 0 n m ( s j ) ( 1 λ ) j ( n ) j l = 0 n m j ( n j l + m ) S 1 ( l + m , m ) D ˆ n j l m ( k ) = j = 0 n m l = 0 n m j ( s j ) ( n j l ) ( n ) j ( 1 λ ) j S 1 ( n j l , m ) D ˆ l ( k ) .
(40)

Therefore, by (39) and (40), we obtain the following theorem.

Theorem 7 For n0, k1 and λC with λ1, we have

D ˆ n ( k ) ( x ) = m = 0 n { j = 0 n m l = 0 n m j ( s j ) ( n j l ) ( n ) j × ( 1 λ ) j S 1 ( n j l , m ) D ˆ l ( k ) } H m ( s ) ( x | λ ) .

We consider the following two Sheffer sequences:

D ˆ n ( k ) (x) ( ( e t 1 t e t ) k , e t 1 ) , B n ( s ) (x) ( ( e t 1 t ) s , t ) .

Let

D ˆ n ( k ) (x)= m = 0 n C n , m B m ( s ) (x).
(41)

Here

C n , m = 1 m ! ( t log ( 1 + t ) ) s ( t ( 1 + t ) log ( 1 + t ) ) k ( log ( 1 + t ) ) m | x n = 1 m ! ( 1 + t ) s ( t ( 1 + t ) log ( 1 + t ) ) s ( t ( 1 + t ) log ( 1 + t ) ) k ( log ( 1 + t ) ) m | x n .
(42)

Case 1. For s>k, we have

C n , m = 1 m ! ( t ( 1 + t ) log ( 1 + t ) ) s k ( log ( 1 + t ) ) m | ( 1 + t ) s x n = 1 m ! 0 j n ( s j ) ( n ) j ( t ( 1 + t ) log ( 1 + t ) ) s k | ( log ( 1 + t ) ) m x n j = 0 j n m ( s j ) ( n ) j m l n j S 1 ( l , m ) × ( n j l ) ( t ( 1 + t ) log ( 1 + t ) ) s k | x n j l = 0 j n m m l n j ( s j ) ( n j l ) ( n ) j S 1 ( l , m ) C ˆ n j l ( s k ) ,
(43)

where C ˆ i ( s k ) is the i th Cauchy number of the second kind of order sk (see [14]).

Case 2. For s=k, we have

C n , m = 1 m ! ( log ( 1 + t ) ) m | ( 1 + t ) s x n = 1 m ! ( log ( 1 + t ) ) m | j = 0 s ( s j ) t j x n = 0 j n m ( s j ) ( n ) j m l < S 1 ( l , m ) l ! t l | x n j = 0 j n m ( s j ) ( n ) j S 1 ( n j , m ) .
(44)

Case 3. For s<k, we have

C n , m = 1 m ! ( ( 1 + t ) log ( 1 + t ) t ) k s ( log ( 1 + t ) ) m | ( 1 + t ) s x n = 0 j n m m l n j ( s j ) ( n j l ) ( n ) j S 1 ( l , m ) D ˆ n j l ( k s ) .
(45)

Therefore, by (41), (42), (43), (44), and (45), we obtain the following theorem.

Theorem 8 Let n0, we have:

  1. (I)

    For s>k, we have

    D ˆ n ( k ) ( x ) = 0 m n { 0 j n m m l n j ( s j ) ( n j l ) × ( n ) j S 1 ( l , m ) C ˆ n j l ( s k ) } B m ( s ) ( x ) .
  2. (II)

    For s=k, we have

    D ˆ n ( k ) (x)= 0 m n { 0 j n m ( s j ) ( n ) j S 1 ( n j , m ) } B m ( s ) (x).
  3. (III)

    For s<k, we have

    D ˆ n ( k ) ( x ) = 0 m n { 0 j n m m l n j ( s j ) ( n j l ) × ( n ) j S 1 ( l , m ) D ˆ n j l ( k s ) } B m ( s ) ( x ) .

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Acknowledgements

The authors would like to thank the referees for their valuable comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MOE) (No. 2012R1A1A2003786).

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Correspondence to Taekyun Kim.

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All authors contributed equally to this work. All authors read and approved the final manuscript.

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Kim, D.S., Kim, T. Some properties of higher-order Daehee polynomials of the second kind arising from umbral calculus. J Inequal Appl 2014, 195 (2014). https://doi.org/10.1186/1029-242X-2014-195

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  • DOI: https://doi.org/10.1186/1029-242X-2014-195

Keywords

  • Formal Power Series
  • Linear Functional
  • Bernoulli Number
  • Bernoulli Polynomial
  • Stirling Number