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Poly-Cauchy and Peters mixed-type polynomials

Abstract

The Peters polynomials are a generalization of Boole polynomials. In this paper, we consider Peters and poly-Cauchy mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities of those polynomials associated with special polynomials.

1 Introduction

The Peters polynomials are defined by the generating function to be

n = 0 S n (x;λ,μ) t n n ! = ( 1 + ( 1 + t ) λ ) μ ( 1 + t ) x (see [1]).
(1)

The first few of them are given by

S 0 (x;λ,μ)= 2 μ , S 1 (x;λ,μ)= 2 ( μ + 1 ) (2xλμ),.

If μ=1, then S n (x;λ)= S n (x:λ,1) are called Boole polynomials.

In particular, for μ=1, λ=1, S n (x;1,1)= Ch n (x) are Changhee polynomials which are defined by

n = 0 Ch n (x) t n n ! = 1 t + 2 ( 1 + t ) x (see [2]).

The generating functions for the poly-Cauchy polynomials of the first kind C n ( k ) (x) and of the second kind C ˆ n ( k ) (x) are, respectively, given by

Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 C n ( k ) (x) t n n !
(2)

and

Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 C ˆ n ( k ) (x) t n n ! ,
(3)

where Lif k (t)= n = 0 t n n ! ( n + 1 ) k (kZ) (see [3, 4]).

In this paper, we consider the poly-Cauchy of the first kind and Peters (respectively the poly-Cauchy of the second kind and Peters) mixed-type polynomials as follows:

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 C P n ( k ) (x;λ,μ) t n n !
(4)

and

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) x = n = 0 C ˆ P n ( k ) (x;λ,μ) t n n ! .
(5)

For α Z 0 , the Bernoulli polynomials of order α are defined by the generating function to be

( t e t 1 ) α e x t = n = 0 B n ( α ) (x) t n n ! (see [2, 5–11]).
(6)

As is well known, the Frobenius-Euler polynomials of order α are given by

( 1 λ e t λ ) α e x t = n = 0 H n ( α ) (xλ) t n n ! (see [2–14]),
(7)

where λC with λ1 and α Z 0 .

When x=0, C P n ( k ) (0;λ,μ) (or C ˆ P n ( k ) (0;λ,μ)) are called the poly-Cauchy of the first kind and Peters (or the poly-Cauchy of the second kind and Peters) mixed-type numbers.

The higher-order Cauchy polynomials of the first kind are defined by the generating function to be

( t log ( 1 + t ) ) α ( 1 + t ) x = n = 0 C n ( α ) (x) t n n ! (α Z 0 ),
(8)

and the higher-order Cauchy polynomials of the second kind are given by

( t ( 1 + t ) log ( 1 + t ) ) α ( 1 + t ) x = n = 0 C ˆ n ( α ) (x) t n n ! (α Z 0 ).
(9)

The Stirling number of the first kind is given by

( x ) n =x(x1)(xn+1)= l = 0 n S 1 (n,l) x l .
(10)

Thus, by (10), we get

( log ( 1 + t ) ) m =m! l = m S 1 (l,m) t l l ! ,m Z 0  (see [1]).
(11)

It is easy to show that

x ( n ) =x(x+1)(x+n1)= ( 1 ) n ( x ) n = l = 0 n S 1 (n,l) ( 1 ) n l x l .
(12)

Let be the complex number field and let be the algebra of all formal power series in the variable t over as follows:

F= { f ( t ) = k = 0 a k t k k ! | a k C } .
(13)

Let P=C[x] and let P be the vector space of all linear functionals on . Lp(x) denotes 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+Mp(x)=Lp(x)+Mp(x), cLp(x)=cLp(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 (n0).
(14)

Then, by (13) and (14), we get

t k x n =n! δ n , k (n,k0) (see [1, 15]),
(15)

where δ n , k is the Kronecker symbol.

Let f L (t)= k = 0 L x k k ! t k . Then, by (14), we see that f L (t) x n =L x n . 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) 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))=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) of polynomials such that g(t)f ( t ) k s n (x)=n! δ n , k (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 and p(x)P, we have

f ( t ) g ( t ) p ( x ) = f ( t ) g ( t ) p ( x ) = g ( t ) f ( t ) p ( x )

and

f(t)= k = 0 f ( t ) x k t k k ! ,p(x)= k = 0 t k p ( x ) x k k ! .
(16)

By (16), we get

p ( k ) (0)= t k p ( x ) = 1 p ( k ) ( x ) (k0),
(17)

where p ( k ) (0)= d k p ( x ) d x k | x = 0 .

Thus, by (17), we have

t k p(x)= p ( k ) (x)= d k p ( x ) d x k (see [1–3]).
(18)

Let s n (x)(g(t),f(t)). Then the following equations are known in [1]:

1 g ( f ¯ ( t ) ) e x f ¯ ( t ) = n = 0 s n (x) t n n ! for all xC,
(19)

where f ¯ (t) is the compositional inverse for f(t) with f( f ¯ (t))=t,

s n (x)= j = 0 n 1 j ! ( f ¯ ( t ) ) j g ( f ¯ ( t ) ) | x n x j ,
(20)
s n (x+y)= j = 0 n ( n j ) s j (x) P n j (y),where  P n (x)=g(t) s n (x),
(21)

and

s n + 1 (x)= ( x g ( t ) g ( t ) ) 1 f ( t ) s n (x),f(t) s n (x)=n s n 1 (x)(n0),
(22)

and

d d x s n (x)= l = 0 n 1 ( n l ) f ¯ ( t ) x n l s l (x).
(23)

As is well known, the transfer formula for p n (x)(1,f(t)), q n (x)(1,g(t)) (n1) is given by

q n (x)=x ( f ( t ) g ( t ) ) n x 1 p n (x).
(24)

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

s n (x)= m = 0 C n , m r n (x),
(25)

where

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) ( l ( f ¯ ( t ) ) ) m | x n (see [1]).
(26)

It is known that

f ( t ) x p ( x ) = t f ( t ) p ( x ) , e y t p(x)=p(x+y),
(27)

where f(t)F and p(x)P (see [13]).

In this paper, we consider Peters and poly-Cauchy mixed-type polynomials with umbral calculus viewpoint and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give some interesting identities of those polynomials associated with special polynomials.

2 Poly-Cauchy and Peters mixed-type polynomials

From (2), (3), and (19), we note that

C P n ( k ) (x;λ,μ) ( ( 1 + e λ t ) μ 1 Lif k ( t ) , e t 1 )
(28)

and

C ˆ P n ( k ) (x;λ,μ) ( ( 1 + e λ t ) μ 1 Lif k ( t ) , e t 1 ) .
(29)

It is not difficult to show that

( 1 + e λ t ) μ = 2 μ ( 1 + 1 2 j = 1 ( λ t ) j j ! ) μ = i = 0 j = 0 j 1 + + j i = j 2 μ i ( μ i ) ( j + i j 1 + 1 , , j i + 1 ) ( λ t ) j + i ( j + i ) !
(30)

and

( 1 + ( 1 + t ) λ ) μ = 2 μ ( 1 + 1 2 j = 0 ( λ j + 1 ) t j + 1 ) μ = i = 0 j = 0 j 1 + + j i = j 2 ( μ + i ) ( μ i ) ( λ j 1 + 1 ) ( λ j i + 1 ) t j + i .
(31)

From (14), we have

C P n ( k ) ( y ; λ , μ ) = l = 0 C P l ( k ) ( y ; λ , μ ) t l l ! | x n = ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) y x n = ( 1 + ( 1 + t ) λ ) μ | l = 0 n ( n l ) C l ( k ) ( y ) x n l = l = 0 n ( n l ) C l ( k ) ( y ) m = 0 S m ( 0 ; λ , μ ) t m m ! | x n l = l = 0 n ( n l ) S n l ( 0 ; λ , μ ) C l ( k ) ( y ) .
(32)

Therefore, by (32), we obtain the following theorem.

Theorem 1 For n0, we have

C P n ( k ) (x;λ,μ)= l = 0 n ( n l ) S n l (0;λ,μ) C l ( k ) (x).

Alternatively,

C P n ( k ) ( y ; λ , μ ) = l = 0 C P l ( k ) ( y ; λ , μ ) t l l ! | x n = Lif k ( log ( 1 + t ) ) ( 1 + ( 1 + t ) λ ) μ ( 1 + t ) y x n = Lif k ( log ( 1 + t ) ) | l = 0 n ( n l ) S l ( y ; λ , μ ) x n l = l = 0 n ( n l ) S l ( y ; λ , μ ) Lif k ( log ( 1 + t ) ) x n l = l = 0 n ( n l ) S l ( y ; λ , μ ) C n l ( k ) ( 0 ) .
(33)

Therefore, by (33), we obtain the following theorem.

Theorem 2 For n0, let C n l ( k ) (0)= C n l ( k ) . Then we have

C P n ( k ) (x;λ,μ)= l = 0 n ( n l ) C n l ( k ) S l (x;λ,μ).

Remark By the same method, we get

C ˆ P n ( k ) (x;λ,μ)= l = 0 n ( n l ) S n l (0;λ,μ) C ˆ l ( k ) (x)
(34)

and

C ˆ P n ( k ) (x;λ,μ)= l = 0 n ( n l ) C ˆ n l ( k ) S l (x;λ,μ).
(35)

From (20) and (28), we have

C P n ( k ) ( x ; λ , μ ) = j = 0 n 1 j ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j x n x j .
(36)

From (31), we note that

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j x n = m = 0 n j ( 1 ) j m ! ( m + 1 ) k l = 0 n j m ( m + j ) ! ( l + m + j ) ! S 1 ( l + m + j , m + j ) × ( n ) l + m + j ( 1 + ( 1 + t ) λ ) μ x n l m j = m = 0 n j ( 1 ) j m ! ( m + 1 ) k l = 0 n j m ( m + j ) ! ( l + m + j ) ! × S 1 ( l + m + j , m + j ) ( n ) l + m + j i = 0 n j m l r = 0 r 1 + + r i = r 2 ( μ + i ) × ( μ i ) ( λ r 1 + 1 ) ( λ r i + 1 ) t r + i x n l m j = 2 μ n ! m = 0 n j l = 0 n j m i = 0 n j m l r 1 + + r i = n j m l i 2 i ( 1 ) j ( m + j ) ! m ! ( m + 1 ) k ( l + m + j ) ! × ( μ i ) ( λ r 1 + 1 ) ( λ r i + 1 ) S 1 ( l + m + j , m + j ) .
(37)

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

Theorem 3 For n0, we have

C P n ( k ) ( x ; λ , μ ) = 2 μ n ! j = 0 n ( 1 ) j j ! { m = 0 n j l = 0 n j m i = 0 n j m l r 1 + + r i = n j m l i 2 i m ! ( m + 1 ) k × ( m + j ) ! ( l + m + j ) ! ( μ i ) ( λ r 1 + 1 ) ( λ r i + 1 ) S 1 ( l + m + j , m + j ) } x j .

Remark By the same method as Theorem 3, we get

C ˆ P n ( k ) ( x ; λ , μ ) = 2 μ n ! j = 0 n 1 j ! { m = 0 n j l = 0 n j m i = 0 n j m l r 1 + + r i = n j m l i 2 i ( 1 ) m m ! ( m + 1 ) k × ( m + j ) ! ( l + m + j ) ! ( μ i ) × ( λ r 1 + 1 ) ( λ r i + 1 ) S 1 ( l + m + j , m + j ) } x j .
(38)

From (28), we note that

( 1 + e λ t ) μ 1 Lif k ( t ) C P n ( k ) (x;λ,μ) ( 1 , e t 1 )
(39)

and

x n (1,t).
(40)

By (24), (39), and (40), we get

( 1 + e λ t ) μ 1 Lif k ( t ) C P n ( k ) ( x ; λ , μ ) = x ( t e t 1 ) n x n 1 = ( 1 ) n x ( t e t 1 ) n x n 1 = ( 1 ) n l = 0 n 1 ( 1 ) l B l ( n ) ( n 1 l ) x n l .
(41)

Thus, by (41), we see that

C P n ( k ) ( x ; λ , μ ) = ( 1 ) n l = 0 n 1 ( 1 ) l ( n 1 l ) B l ( n ) ( 1 + e λ t ) μ Lif k ( t ) x n l = ( 1 ) n l = 0 n 1 ( 1 ) l ( n 1 l ) B l ( n ) m = 0 n l ( 1 ) m ( n l m ) ( m + 1 ) k ( 1 + e λ t ) μ x n l m = ( 1 ) n l = 0 n 1 ( 1 ) l ( n 1 l ) B l ( n ) m = 0 n l ( 1 ) m ( n l m ) ( m + 1 ) k i = 0 j = 0 j 1 + + j n = j 2 μ i ( μ i ) × ( j + i j 1 + 1 , , j i + 1 ) ( λ t ) j + i ( j + i ) ! x n l m = ( 1 ) n l = 0 n m = 0 n l i = 0 n l m j = 0 n l m i j 1 + + j i = n l m i r ( 1 ) n r 2 μ i λ n l m r ( m + 1 ) k ( n 1 l ) × ( n l m ) ( μ i ) ( n l m r j 1 + 1 , , j i + 1 ) ( n l m r ) B l ( n ) x r .
(42)

Therefore, by (42), we obtain the following theorem.

Theorem 4 For n0, we have

C P n ( k ) ( x ; λ , μ ) = λ n 2 μ r = 0 n ( λ 1 ) r { l = 0 n r m = 0 n r l i = 0 n r l m j 1 + + j i = n r l m i 2 i λ l m ( m + 1 ) k ( n 1 l ) × ( n l m ) ( μ i ) ( n r l m j 1 + 1 , , j i + 1 ) ( n l m r ) B l ( n ) } x r .

Remark By the same method as Theorem 4, we get

C ˆ P n ( k ) ( x ; λ , μ ) = λ n 2 μ r = 0 n λ r { l = 0 n r m = 0 n l r i = 0 n l m r j 1 + + j i = n r l m i ( 1 ) m 2 i λ l m ( m + 1 ) k ( n 1 l ) × ( n l m ) ( μ i ) ( n r l m j 1 + 1 , , j i + 1 ) ( n l m r ) B l ( m ) } x r .
(43)

From (12), we note that

x ( n ) =x(x+1)(x+n1) ( 1 , 1 e t ) .
(44)

Thus, by (44), we see that

( 1 ) n x ( n ) = ( x ) n = m = 0 n S 1 (n,m) ( x ) m ( 1 , e t 1 )
(45)

and

( 1 + e λ t ) μ 1 Lif k ( t ) C P n ( k ) (x;λ,μ) ( 1 , e t 1 ) .
(46)

From (45) and (46), we have

( 1 + e λ t ) μ 1 Lif k ( t ) C P n ( k ) ( x ; λ , μ ) = ( 1 ) n x ( n ) = l = 0 n S 1 ( n , l ) ( x ) l .
(47)

Thus, by (47), we get

C P n ( k ) ( x ; λ , μ ) = l = 0 n S 1 ( n , l ) ( 1 ) l ( 1 + e λ t ) μ Lif k ( t ) x l = l = 0 n S 1 ( n , l ) ( 1 ) l m = 0 l ( 1 ) m ( l m ) ( m + 1 ) k ( 1 + e λ t ) μ x l m = l = 0 n S 1 ( n , l ) ( 1 ) l m = 0 l ( 1 ) m ( l m ) ( m + 1 ) k i = 0 j = 0 j 1 + + j i = j 2 μ i × ( μ i ) ( j + i j 1 + 1 , , j i + 1 ) ( λ ) j + i ( j + i ) ! t j + i x l m = l = 0 n m = 0 l i = 0 l m r = 0 l m i j 1 + + j i = l m i r ( 1 ) r 2 μ i λ l m r ( m + 1 ) k × ( l m ) ( μ i ) ( l m r j 1 + 1 , , j i + 1 ) ( l m r ) S 1 ( n , l ) x r = 2 μ r = 0 n ( λ 1 ) r { l = r n m = 0 l r i = 0 l r m j 1 + + j i = l r m i 2 i λ l m ( m + 1 ) k × ( l m ) ( μ i ) ( l m r j 1 + 1 , , j i + 1 ) ( l m r ) S 1 ( n , l ) } x r .
(48)

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

Theorem 5 For n0, we have

C P n ( k ) ( x ; λ , μ ) = 2 μ r = 0 n ( λ 1 ) r { l = r n m = 0 l r i = 0 l r m j 1 + + j i = l r m i 2 i λ l m ( m + 1 ) k × ( l m ) ( μ i ) ( l m r j 1 + 1 , , j i + 1 ) ( l m r ) S 1 ( n , l ) } x r .

It is easy to see that

( 1 + e λ t ) μ 1 Lif k ( t ) C ˆ P n ( k ) (x;λ,μ) ( 1 , e t 1 )
(49)

and

( x ) n =x(x1)(xn+1)= l = 0 n S 1 (n,l) x l ( 1 , e t 1 ) .
(50)

By the same method as Theorem 5, we get

C ˆ P n ( k ) ( x ; λ , μ ) = 2 μ r = 0 n λ r { l = r n m = 0 l r i = 0 l r m j 1 + + j i = l r m i ( 1 ) m 2 i λ l m ( m + 1 ) k × ( l m ) ( μ i ) ( l m r j 1 + 1 , , j i + 1 ) ( l m r ) S 1 ( n , l ) } x r .
(51)

From (20) and (28), we have

C P n ( k ) ( x ; λ , μ ) = j = 0 n 1 j ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j x n x j .
(52)

Now, we observe that

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) j x n = ( 1 ) j log ( 1 + t ) j | m = 0 C P m ( k ) ( 0 ; λ , μ ) t m m ! x n = ( 1 ) j m = 0 n ( n m ) C P m ( k ) ( 0 ; λ , μ ) ( log ( 1 + t ) ) j x n m = ( 1 ) j m = 0 n ( n m ) C P m ( k ) ( 0 ; λ , μ ) j ! S 1 ( n m , j ) .
(53)

Therefore, by (52) and (53), we obtain the following theorem.

Theorem 6 For n0, we have

C P n ( k ) (x;λ,μ)= j = 0 n ( 1 ) j { m = 0 n ( n m ) S 1 ( n m , j ) C P m ( k ) ( 0 ; λ , μ ) } x j .

Remark By the same method as Theorem 6, we get

C ˆ P n ( k ) (x;λ,μ)= j = 0 n { m = 0 n ( n m ) S 1 ( n m , j ) C ˆ P n ( k ) ( 0 ; λ , μ ) } x j .
(54)

From (21), we have

C P n ( k ) (x+y;λ,μ)= j = 0 n ( 1 ) j ( n j ) C P n j ( k ) (x;λ,μ) y ( j )
(55)

and

C ˆ P n ( k ) (x+y;λ,μ)= j = 0 n ( n j ) C ˆ P n j ( k ) (x;λ,μ) ( y ) j .
(56)

By (22) and (28), we get

( e t 1 ) C P n ( k ) (x;λ,μ)=nC P n 1 ( k ) (x;λ,μ)
(57)

and

( e t 1 ) C P n ( k ) (x;λ,μ)=C P n ( k ) (x1;λ,μ)C P n ( k ) (x;λ,μ).
(58)

Therefore, by (57) and (58), we obtain the following theorem.

Theorem 7 For n0, we have

C P n ( k ) (x1;λ,μ)C P n ( k ) (x;λ,μ)=nC P n 1 ( k ) (x;λ,μ).

Remark By the same method as Theorem 7, we get

C ˆ P n ( k ) (x+1;λ,μ) C ˆ P n ( k ) (x;λ,μ)=n C ˆ P n 1 ( k ) (x;λ,μ).
(59)

From (22), (28), and (29), we have

C P n + 1 ( k ) ( x ; 1 , μ ) = x C P n ( k ) ( x + 1 ; 1 , μ ) + μ m = 0 n ( 1 2 ) m + 1 ( n ) m C P n m ( k ) ( x ; 1 , μ ) + 2 μ r = 0 n ( 1 ) r { m = r n l = r m i = 0 l r j 1 + + j i = l i r 2 i ( m l + 2 ) k ( m l ) × ( μ i ) ( l r j 1 + 1 , , j i + 1 ) ( l r ) S 1 ( n , m ) } ( x + 1 ) r
(60)

and

C ˆ P n + 1 ( k ) ( x ; 1 , μ ) = x C ˆ P n ( k ) ( x 1 ; 1 , μ ) + μ m = 0 n ( 1 2 ) m + 1 ( n ) m C ˆ P n m ( k ) ( x ; 1 , μ ) 2 μ r = 0 n { m = r n l = r m i = 0 l r j 1 + + j i = l i r ( 1 ) m l 2 i ( m l + 2 ) k ( m l ) ( μ i ) × ( l r j 1 + 1 , , j i + 1 ) ( l r ) S 1 ( n , m ) } ( x 1 ) r .
(61)

By (14) and (27), we get

C P n ( k ) ( y ; λ , μ ) = l = 0 C P l ( k ) ( y ; λ , μ ) t l l ! | x n = ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) y x x n 1 = t ( ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) y ) x n 1 = ( t ( 1 + ( 1 + t ) λ ) μ ) Lif k ( log ( 1 + t ) ) ( 1 + t ) y x n 1 + ( 1 + ( 1 + t ) λ ) μ ( t Lif k ( log ( 1 + t ) ) ) ( 1 + t ) y x n 1 + ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( t ( 1 + t ) y ) x n 1 = μ λ ( 1 + ( 1 + t ) λ ) μ 1 Lif k ( log ( 1 + t ) ) ( 1 + t ) ( y λ + 1 ) x n 1 y ( 1 + ( 1 + t ) λ ) μ ( Lif k ( log ( 1 + t ) ) ) ( 1 + t ) y 1 x n 1 + ( 1 + ( 1 + t ) λ ) μ ( t Lif k ( log ( 1 + t ) ) ) ( 1 + t ) y x n 1 = μ λ C P n 1 ( k ) ( y λ + 1 ; λ , μ + 1 ) y C P n 1 ( k ) ( y + 1 ; λ , μ ) + ( 1 + ( 1 + t ) λ ) μ Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) ( 1 + t ) log ( 1 + t ) ( 1 + t ) y x n 1 .
(62)

Now, we observe that

( 1 + ( 1 + t ) λ ) μ Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) ( 1 + t ) log ( 1 + t ) ( 1 + t ) y | x n 1 = ( 1 + ( 1 + t ) λ ) μ Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) y | t ( 1 + t ) log ( 1 + t ) x n 1 = l = 0 n 1 ( n 1 l ) C ˆ n 1 l ( 1 ) ( 0 ) × ( 1 + ( 1 + t ) λ ) μ Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) y | x l = l = 0 n 1 ( n 1 l ) C ˆ n 1 l ( 1 ) ( 0 ) × ( 1 + ( 1 + t ) λ ) μ Lif k 1 ( log ( 1 + t ) ) Lif k ( log ( 1 + t ) ) t ( 1 + t ) y | t x l + 1 l + 1 = 1 n l = 0 n 1 ( n l + 1 ) C ˆ n 1 l ( 1 ) ( 0 ) { C P l + 1 ( k 1 ) ( y ; λ , μ ) C P l + 1 ( k ) ( y ; λ , μ ) } .
(63)

Therefore, by (62) and (63), we obtain the following theorem.

Theorem 8 For n0, we have

C P n ( k ) ( x ; λ , μ ) = μ λ C P n 1 ( k ) ( x λ + 1 ; λ , μ + 1 ) x C P n 1 ( k ) ( x + 1 ; λ , μ ) + 1 n l = 0 n 1 ( n l + 1 ) C ˆ n 1 l { C P l + 1 ( k 1 ) ( x ; λ , μ ) C P l + 1 ( k ) ( x ; λ , μ ) } ,

where C ˆ n 1 l = C ˆ n 1 l ( 1 ) (0).

Remark By the same method as Theorem 8, we get

C ˆ P n ( k ) ( x ; λ , μ ) = μ λ C ˆ P n 1 ( k ) ( x + λ 1 ; λ , μ + 1 ) + x C ˆ P n 1 ( k ) ( x 1 ; λ , μ ) + 1 n l = 0 n 1 ( n l + 1 ) C ˆ n 1 l ( C ˆ P l + 1 ( k 1 ) ( x ; λ , μ ) C ˆ P l + 1 ( k ) ( x ; λ , μ ) ) .
(64)

By (23), we get

d d x C P n ( k ) ( x ; λ , μ ) = l = 0 n 1 ( n l ) log ( 1 + t ) x n l C P l ( k ) ( x ; λ , μ ) = l = 0 n 1 ( n l ) m = 1 ( 1 ) m m t m | x n l C P l ( k ) ( x ; λ , μ ) = l = 0 n 1 ( n l ) m = 1 ( 1 ) m m t m x n l C P l ( k ) ( x ; λ , μ ) = l = 0 n 1 ( n l ) ( 1 ) n l C P l ( k ) ( x ; λ , μ ) ( n l 1 ) ! = n ! l = 0 n 1 ( 1 ) n l ( n l ) l ! C P l ( k ) ( x ; λ , μ ) .
(65)

By the same method as (65), we get

d d x C ˆ P n ( k ) ( x ; λ , μ ) = n ! l = 0 n 1 ( 1 ) n l 1 ( n l ) l ! C ˆ P l ( k ) ( x ; λ , μ ) .
(66)

Now, we compute the following equation in two different ways:

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m x n .

On the one hand,

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m x n = ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m x n = l = 0 n m m ! ( n l + m ) S 1 ( l + m , m ) ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) x n l m = l = 0 n m m ! ( n l ) S 1 ( n l , m ) C ˆ P l ( k ) ( 0 ; λ , μ ) .
(67)

On the other hand,

( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m x n = t ( ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m ) x n 1 = ( t ( 1 + ( 1 + t ) λ ) μ ) Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m x n 1 + ( 1 + ( 1 + t ) λ ) μ ( t Lif k ( log ( 1 + t ) ) ) ( log ( 1 + t ) ) m x n 1 + ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( t ( log ( 1 + t ) ) m ) x n 1 .
(68)

Therefore, by (67) and (68), we obtain the following theorem.

Theorem 9 For nN with n2, let n1m1. Then we have

m l = 0 n m ( n l ) S 1 ( n l , m ) C ˆ P l ( k ) ( 0 ; λ , μ ) = μ λ m l = 0 n 1 m ( n 1 l ) S 1 ( n 1 l , m ) C ˆ P l ( k ) ( λ 1 ; λ , μ + 1 ) + l = 0 n m ( n 1 l ) S 1 ( n 1 l , m 1 ) C ˆ P l ( k 1 ) ( 1 ; λ , μ ) + ( m 1 ) l = 0 n m ( n 1 l ) S 1 ( n 1 l , m 1 ) C ˆ P l ( k ) ( 1 ; λ , μ ) .

Remark By the same method as Theorem 9, we get

m l = 0 n m ( n l ) S 1 ( n l , m ) C P l ( k ) ( 0 ; λ , μ ) = μ λ m l = 0 n 1 m ( n 1 l ) S 1 ( n 1 l , m ) C P l ( k ) ( 1 λ ; λ , μ + 1 ) + l = 0 n m ( n 1 l ) S 1 ( n 1 l , m 1 ) C P l ( k 1 ) ( 1 ; λ , μ ) + ( m 1 ) l = 0 n m ( n 1 l ) S 1 ( n 1 l , m 1 ) C P l ( k ) ( 1 ; λ , μ ) ,

where n1m1.

Let us consider the following two Sheffer sequences:

C P n ( k ) (x;λ,μ) ( ( 1 + e λ t ) μ 1 Lif k ( t ) , e t 1 )
(69)

and

B n ( s ) (x) ( ( e t 1 t ) s , t ) (s Z 0 ).
(70)

Let

C P n ( k ) (x;λ,μ)= m = 0 n C n , m B m ( s ) (x).
(71)

Then, by (26), we get

C n , m = 1 m ! ( e log ( 1 + t ) 1 log ( 1 + t ) ) s ( 1 + e λ log ( 1 + t ) ) μ Lif k ( log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n = ( 1 ) m m ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) × ( 1 + t ) s ( t log ( 1 + t ) ) s | ( log ( 1 + t ) ) m x n = ( 1 ) m m ! l = 0 n m m ! ( n l + m ) S 1 ( l + m , m ) i = 0 n l m ( n l m i ) C i ( s ) × ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) s x n l m i = ( 1 ) m l = 0 n m ( n l ) S 1 ( n l , m ) i = 0 l ( l i ) C i ( s ) C P l i ( k ) ( s ; λ , μ ) .
(72)

Therefore, by (71) and (72), we obtain the following theorem.

Theorem 10 For n0, we have

C P n ( k ) (x;λ,μ)= m = 0 n ( 1 ) m { l = 0 n m i = 0 l ( n l ) ( l i ) S 1 ( n l , m ) C i ( s ) C P l i ( k ) ( s ; λ , μ ) } B n ( s ) (x).

Remark By the same method as Theorem 10, we have

C ˆ P n ( k ) ( x ; λ , μ ) = m = 0 n { l = 0 n m i = 0 l ( n l ) ( l i ) S 1 ( n l , m ) C ˆ i ( s ) C ˆ P l i ( k ) ( s ; λ , μ ) } B m ( s ) ( x ) .
(73)

For C P n ( k ) (x;λ,μ)( ( 1 + e λ t ) μ 1 Lif k ( t ) , e t 1), H n ( s ) (xλ)( ( e t λ 1 λ ) s ,t), s Z 0 , λC with λ1, let us assume that

C P n ( k ) (x;λ,μ)= m = 0 n C n , m H m ( s ) (x;λ).
(74)

From (26), we have

C n , m = ( 1 ) m m ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) × ( 1 + t ) s ( 1 + λ λ 1 t ) s | ( log ( 1 + t ) ) m x n = ( 1 ) m m ! l = 0 n m m ! ( n l + m ) S 1 ( l + m , m ) i = 0 min { s , n l m } ( s i ) ( λ λ 1 ) i × ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( 1 + t ) s t i x n l m = ( 1 ) m l = 0 n m i = 0 min { s , n l m } ( n l + m ) ( s i ) × ( n l m ) i ( λ λ 1 ) i S 1 ( l + m , m ) C P n l m i ( k ) ( s ; λ , μ ) = ( 1 ) m l = 0 n m i = 0 min { s , l } ( n l ) ( s i ) ( l ) i ( λ λ 1 ) i S 1 ( n l , m ) C P l i ( k ) ( s ; λ , μ ) .
(75)

Therefore, by (75) and (76), we obtain the following theorem.

Theorem 11 For λC with λ1, n0, we have

C P n ( k ) ( x ; λ , μ ) = m = 0 n ( 1 ) m { l = 0 n m i = 0 min { s , l } ( n l ) ( s i ) ( l ) i ( λ λ 1 ) i S 1 ( n l , m ) C P l i ( k ) ( s ; λ , μ ) } H m ( s ) ( x ; λ ) .

Remark By the same method as Theorem 11, we get

C ˆ P n ( k ) ( x ; λ , μ ) = m = 0 n { l = 0 n m i = 0 min { s , l } ( n l ) ( s i ) ( l ) i ( 1 1 λ ) i S 1 ( n l , m ) C ˆ P l i ( k ) ( 0 ; λ , μ ) } H m ( s ) ( x ; λ ) .

For C P n ( k ) (x;λ,μ)( ( 1 + e λ t ) μ 1 Lif k ( t ) , e t 1) and x ( n ) (1,1 e t ), let us assume that

C P n ( k ) (x;λ,μ)= m = 0 C n , m x ( m ) .
(76)

By (26), we get

C n , m = 1 m ! 1 ( 1 + e λ log ( 1 + t ) ) μ Lif k ( log ( 1 + t ) ) ( 1 e log ( 1 + t ) ) m | x n = 1 m ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) ( t ) m x n = ( 1 ) m m ! ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) t m x n = ( 1 ) m ( n m ) ( 1 + ( 1 + t ) λ ) μ Lif k ( log ( 1 + t ) ) x n m = ( 1 ) m ( n m ) C P n m ( k ) ( 0 ; λ , μ ) .
(77)

Therefore, by (77) and (78), we obtain the following theorem.

Theorem 12 For n0, we have

C P n ( k ) (x;λ,μ)= m = 0 n ( 1 ) m ( n m ) C P n m ( k ) (0;λ,μ) x ( m ) .

Remark By the same method as Theorem 12, we get

C ˆ P n ( k ) (x;λ,μ)= m = 0 n ( n m ) C ˆ P n m ( k ) (0;λ,μ) ( x ) m .
(78)

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This work was supported by Kwangwoon University 2013.

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Kim, D.S., Kim, T. Poly-Cauchy and Peters mixed-type polynomials. Adv Differ Equ 2014, 4 (2014). https://doi.org/10.1186/1687-1847-2014-4

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Keywords

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