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Barnes-type Peters polynomial with umbral calculus viewpoint

Abstract

In this paper, we consider the Barnes-type Peters polynomials. We present several explicit formulas and recurrence relations for these polynomials. Also, we establish a connection between our polynomials and several known families of polynomials.

MSC:05A40, 11B83.

1 Introduction

The aim of this paper is to use umbral calculus to obtain several new and interesting identities of Barnes-type Peters polynomials. Umbral calculus has been used in numerous problems of mathematics (for example, see [110]). Umbral techniques have been used in different areas of physics; for example, it was used in group theory and quantum mechanics by Biedenharn et al. [11, 12] (for other examples, see [3, 10, 1318]).

Let r Z > 0 . Here we will consider the polynomials S n (x)= S n (x| λ 1 ,, λ r ; μ 1 ,, μ r ) and S ˆ n (x)= S ˆ n (x| λ 1 ,, λ r ; μ 1 ,, μ r ), which are called Barnes-type Peters polynomials of the first kind and of the second kind, respectively, and are given by

j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) x = n 0 S n (x) t n n ! ,
(1.1)
j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) x = n 0 S ˆ n (x) t n n ! ,
(1.2)

where λ 1 ,, λ r , μ 1 ,, μ r C with λ 1 ,, λ r 0. If r=1, then these polynomials are generalizations of Boole polynomials, see [19]. If μ 1 == μ r =1, then S n (x|λ)= S n (x| λ 1 ,, λ r )= S n (x| λ 1 ,, λ r ;1,,1) and S ˆ n (x|λ)= S ˆ n (x| λ 1 ,, λ r )= S ˆ n (x| λ 1 ,, λ r ;1,,1) are called Barnes-type Boole polynomials of the first kind and of the second kind. So,

j = 1 r ( 1 + ( 1 + t ) λ j ) 1 ( 1 + t ) x = n 0 S n ( x | λ ) t n n ! , j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) ( 1 + t ) x = n 0 S ˆ n ( x | λ ) t n n ! .

We introduce the polynomials E n (x|λ;μ)= E n (x| λ 1 ,, λ r ; μ 1 ,, μ r ) with the generating function

j = 1 r ( 2 1 + e λ j t ) μ j e x t = n 0 E n (x|λ;μ) t n n ! .

These polynomials may be called generalized Barnes-type Euler polynomials. When μ 1 == μ r =1, E n (x|λ)= E n (x| λ 1 ,, λ r )= E n (x| λ 1 ,, λ r ;1,,1) are called the Barnes-type Euler polynomials. If further λ 1 == λ r =1, E n ( r ) (x)= E n (x|1,,1;1,,1) are called the Euler polynomials of order r. When x=0, S n = S n (λ;μ)= S n (0|λ;μ) and S ˆ n = S ˆ n (λ;μ)= S ˆ n (0|λ;μ) are called Barnes-type Peters numbers of the first kind and of the second kind, respectively.

Let Π be the algebra of polynomials in a single variable x over , and let Π be the vector space of all linear functionals on Π. We denote the action of a linear functional L on a polynomial p(x) by L|p(x), and we define the vector space structure on Π by

c L + c L | p ( x ) =c L | p ( x ) + c L | p ( x ) ,

where c, c C (see [1922]). We define the algebra of formal power series in a single variable t to be

H= { f ( t ) = k 0 a k t k k ! | a k C } .
(1.3)

The formal power series in the variable t defines a linear functional on Π by setting

f ( t ) | x n = a n for all n0(see [19–22]).
(1.4)

By (1.3) and (1.4), we have

t k | x n =n! δ n , k for all n,k0(see [19–22]),
(1.5)

where δ n , k is the Kronecker symbol.

Let f L (t)= n 0 L| x n t n n ! . From (1.5), we have f L (t)| x n =L| x n . Thus, the map L f L (t) is a vector space isomorphism from Π onto . Therefore, is thought of as a set of both formal power series and linear functionals. We call umbral algebra. Umbral calculus is the study of umbral algebra.

The order O(f(t)) of the non-zero power series f(t) is the smallest integer k for which the coefficient of t k does not vanish (see [1922]). If O(f(t))=1 (respectively, O(f(t))=0), then f(t) is called a delta (respectively, an invertible) series. Suppose that O(f(t))=1 and O(g(t))=0, then there exists a unique sequence s n (x) of polynomials such that g(t) ( f ( t ) ) k | s n (x)=n! δ n , k , where n,k0 [[19], Theorem 2.3.1]. 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 [1922]). For f(t)H and p(x)Π, we have e y t |p(x)=p(y), f(t)g(t)|p(x)=g(t)|f(t)p(x) and

f(t)= n 0 f ( t ) | x n t n n ! ,p(x)= n 0 t n | p ( x ) x n n !
(1.6)

(see [1922]). From (1.6), we obtain

t k | p ( x ) = p ( k ) (0), 1 | p ( k ) ( x ) = p ( k ) (0),
(1.7)

where p ( k ) (0) denotes the k th derivative of p(x) with respect to x at x=0. So, by (1.7), we get that t k p(x)= p ( k ) (x)= d k d x k p(x) for all k0 (see [1922]).

Let s n (x)(g(t),f(t)). Then we have

1 g ( f ¯ ( t ) ) e y f ¯ ( t ) = n 0 s n (y) t n n ! ,
(1.8)

for all yC, where f ¯ (t) is the compositional inverse of f(t) (see [1922]). For s n (x)(g(t),f(t)) and r n (x)(h(t),(t)), let

s n (x)= k = 0 n c n , k r k (x),
(1.9)

then we have

c n , k = 1 k ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) ( ( f ¯ ( t ) ) ) k | x n
(1.10)

(see [1922]).

It is immediate from (1.1)-(1.2), we see that S n (x) and S ˆ n (x) are the Sheffer sequences for the pairs

S n (x) ( j = 1 r ( 1 + e λ j t ) μ j , e t 1 ) ,
(1.11)
S ˆ n (x) ( j = 1 r ( 1 + e λ j t e λ j t ) μ j , e t 1 ) .
(1.12)

The aim of the present paper is to present several new identities for the Peters polynomials by the use of umbral calculus.

2 Explicit expressions

It is well known that

( x ) n = m = 0 n S 1 (n,m) x m ( 1 , e t 1 ) ,
(2.1)

where S 1 (n,m) is the Stirling number of the first kind. By (1.11) and (1.12) we have

j = 1 r ( 1 + e λ j t ) μ j S n (x) ( 1 , e t 1 ) and j = 1 r ( 1 + e λ j t e λ j t ) μ j S ˆ n (x) ( 1 , e t 1 ) .
(2.2)

So

S n ( x ) = j = 1 r ( 1 + e λ j t ) μ j ( x ) n = m = 0 n S 1 ( n , m ) j = 1 r ( 1 + e λ j t ) μ j x m = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) j = 1 r ( 2 1 + e λ j t ) μ j x m = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) E m ( x | λ ; μ ) ,
(2.3)

which implies

S ˆ n ( x ) = j = 1 r ( e λ j t 1 + e λ j t ) μ j ( x ) n = e j = 1 r λ j μ j t j = 1 r ( 1 + e λ j t ) μ j ( x ) n = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) e j = 1 r λ j μ j t j = 1 r ( 2 1 + e λ j t ) μ j x m = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) e j = 1 r λ j μ j t E m ( x | λ ; μ ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) E m ( x + j = 1 r λ j μ j | λ ; μ ) .
(2.4)

Thus, we have the following result.

Theorem 1 For all n0,

S n ( x ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) E m ( x | λ ; μ ) , S ˆ n ( x ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) E m ( x + j = 1 r λ j μ j | λ ; μ ) .

By (1.6), (1.8), (1.11) and (1.12), we have

S n ( x ) = j = 0 n 1 j ! j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) j | x n x j , S ˆ n ( x ) = j = 0 n 1 j ! j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) j | x n x j ,

where

j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) j | x n = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | ( log ( 1 + t ) ) j x n = j ! = j n ( n ) S 1 ( , j ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | x n = j ! = j n ( n ) S 1 ( , j ) S n

and

j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) j | x n = j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | ( log ( 1 + t ) ) j x n = j ! = j n ( n ) S 1 ( , j ) j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | x n = j ! = j n ( n ) S 1 ( , j ) S ˆ n .

Hence, we can state the following formulas.

Theorem 2 For all n0,

S n (x)= j = 0 n ( = j n ( n ) S 1 ( , j ) S n ) x j and S ˆ n (x)= j = 0 n ( = j n ( n ) S 1 ( , j ) S ˆ n ) x j .

Also, by the definitions, (2.1), (1.11) and (1.12), we have

S n ( y ) = i 0 S i ( y ) t i i ! | x n = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | ( 1 + t ) y x n = m = 0 n ( y ) m ( n m ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | x n m = m = 0 n ( y ) m ( n m ) S n m

and

S ˆ n ( y ) = i 0 S ˆ i ( y ) t i i ! | x n = j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n = j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | ( 1 + t ) y x n = m = 0 n ( y ) m ( n m ) j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | x n m = m = 0 n ( y ) m ( n m ) S ˆ n m ,

which implies the following formulas.

Theorem 3 For all n0,

S n (x)= j = 0 n S n j ( n j ) ( x ) j and S ˆ n (x)= j = 0 n S ˆ n j ( n j ) ( x ) j .

More generally, by (2.1) and (2.2) with p n (x)= j = 1 r ( 1 + e λ j t ) μ j S n (x)= ( x ) n (1, e t 1), we obtain that S n (x+y)= j = 0 b S j (x) ( y ) n j ( n j ) , and with p n (x)= j = 1 r ( 1 + e λ j t e λ j t ) μ j S ˆ n (x)= ( x ) n (1, e t 1), we obtain that S ˆ n (x+y)= j = 0 b S ˆ j (x) ( y ) n j ( n j ) , which gives the following corollary.

Corollary 1 For all n0,

S n (x+y)= j = 0 b S j (x) ( y ) n j ( n j ) and S ˆ n (x+y)= j = 0 b S ˆ j (x) ( y ) n j ( n j ) .

3 Recurrence relations

Note that if a n (x)(g(t),f(t)), then f(t) a n (x)=n a n 1 (x), Thus, by (1.11) and (1.12), we have that S n (x+1) S n (x)=( e t 1) S n (x)=n S n 1 (x) and S ˆ n (x+1) S ˆ n (x)=( e t 1) S ˆ n (x)=n S ˆ n 1 (x), which give the following recurrences.

Proposition 1 For all n1,

S n (x+1) S n (x)=n S n 1 (x)and S ˆ n (x+1) S ˆ n (x)=n S ˆ n 1 (x).

Note that for a n (x)(g(t),f(t)), we have that a n + 1 (x)=(x g (t)/g(t)) 1 f ( t ) a n (x). In the case (1.11), we obtain S n + 1 (x)=x S n (x1) e t g ( t ) g ( t ) S n (x) with g(t)= i = 1 r ( 1 + e λ j t ) μ j . Since g ( t ) g ( t ) = i = 1 r λ i μ i e λ i t 1 + e λ i t and by (2.3), we get

g ( t ) g ( t ) S n ( x ) = i = 1 r λ i μ i e λ i t 1 + e λ i t S n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t S n ( x ) = i = 1 r ( λ i μ i e λ i t 2 2 1 + e λ i t 2 j = 1 r μ j m = 0 n S 1 ( n , m ) j = 1 r ( 2 1 + e λ j t ) μ j x m ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r ( λ i μ i e λ i t 2 2 1 + e λ i t j = 1 r ( 2 1 + e λ j t ) μ j x m ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r λ i μ i 2 E m ( x + λ i | λ ; μ + e i ) ,

where e i =(0,,0,1,0,,0) is a vector with 1 in the i th coordinate. Thus,

S n + 1 (x)=x S n 1 (x) 2 1 i = 1 r μ j m = 0 n i = 1 r S 1 (n,m) λ i μ i E m (x+ λ i 1|λ;μ+ e i ).
(3.1)

On the other hand, by Theorem 2, we have

g ( t ) g ( t ) S n ( x ) = i = 1 r λ i μ i e λ i t 1 + e λ i t S n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t S n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t j = 0 n ( = j n ( n ) S 1 ( , j ) S n ) x j = j = 0 n ( = j n ( n ) S 1 ( , j ) S n i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t x j ) = j = 0 n ( = j n ( n ) S 1 ( , j ) S n i = 1 r λ i μ i e λ i t 2 λ i j E j ( x / λ i ) ) = j = 0 n ( = j n ( n ) S 1 ( , j ) S n i = 1 r λ i j + 1 μ i 2 E j ( 1 + x / λ i ) )

(note that E n (x)= 2 1 + e t x n = ( E + x ) n = j = 0 n ( n j ) E j x n j and 2 1 + e λ j t x j = λ i j E i (x/ λ i )), which implies

S n + 1 (x)=x S n (x1) j = 0 n = j n i = 1 r λ i j + 1 μ i 2 ( n ) S 1 (,j) S n E j ( 1 + ( x 1 ) / λ i ) .

Thus, by (3.1), we can state the following result.

Theorem 4 For all n0,

S n + 1 ( x ) = x S n ( x 1 ) 2 1 i = 1 r μ j m = 0 n i = 1 r S 1 ( n , m ) λ i μ i E m ( x + λ i 1 | λ ; μ + e i ) , S n + 1 ( x ) = x S n ( x 1 ) j = 0 n = j n i = 1 r λ i j + 1 μ i 2 ( n ) S 1 ( , j ) S n E j ( 1 + ( x 1 ) / λ i ) .

As a corollary, we get the following identity.

Corollary 2 For all n0,

2 1 i = 1 r μ j m = 0 n i = 1 r S 1 ( n , m ) λ i μ i E m ( x + λ i 1 | λ ; μ + e i ) , = j = 0 n = j n i = 1 r λ i j + 1 μ i 2 ( n ) S 1 ( , j ) S n E j ( 1 + ( x 1 ) / λ i ) .

In the case (1.12), we obtain S ˆ n + 1 (x)=x S ˆ n (x1) e t g ( t ) g ( t ) S ˆ n (x) with g(t)= i = 1 r ( 1 + e λ j t e λ i t ) μ j . Since g ( t ) g ( t ) = i = 1 r λ i μ i e λ i t 1 + e λ i t i = 1 r λ i μ i and by (2.4), we get

g ( t ) g ( t ) S n (x)= i = 1 r λ i μ i e λ i t 1 + e λ i t S ˆ n (x)λμ S ˆ n (x),

where λμ= j = 1 r λ j μ j and

i = 1 r λ i μ i e λ i t 1 + e λ i t S ˆ n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t S ˆ n ( x ) = i = 1 r ( λ i μ i e λ i t 2 2 1 + e λ i t 2 j = 1 r μ j m = 0 n S 1 ( n , m ) e j = 1 r λ j μ j t j = 1 r ( 2 1 + e λ j t ) μ j x m ) = 2 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r ( λ i μ i 2 e λ i t + j = 1 r λ j μ j t 2 1 + e λ i t j = 1 r ( 2 1 + e λ j t ) μ j x m ) = 2 1 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r λ i μ i E n ( x + λ ( μ + e i ) | λ ; μ + e i ) .

So

S ˆ n + 1 ( x ) = ( x + λ μ ) S ˆ n ( x 1 ) 2 1 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r λ i μ i E n ( x + λ ( μ + e i ) 1 | λ ; μ + e i ) .
(3.2)

On the other hand, by Theorem 2, we have

g ( t ) g ( t ) S ˆ n ( x ) = i = 1 r λ i μ i e λ i t 1 + e λ i t S ˆ n ( x ) λ μ S ˆ n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t S ˆ n ( x ) λ μ S ˆ n ( x ) = i = 1 r λ i μ i e λ i t 2 2 1 + e λ i t j = 0 n ( = j n ( n ) S 1 ( , j ) S ˆ n ) x j λ μ S ˆ n ( x ) = j = 0 n ( = j n ( n ) S 1 ( , j ) S ˆ n i = 1 r λ i j + 1 μ i 2 E j ( 1 + x / λ i ) ) λ μ S ˆ n ( x ) .

Therefore, by (3.2), we have the following result.

Theorem 5 For all n0,

S ˆ n + 1 ( x ) = ( x + λ μ ) S ˆ n ( x 1 ) 2 1 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r λ i μ i E n ( x + λ ( μ + e i ) 1 | λ ; μ + e i ) , S ˆ n + 1 ( x ) = ( x + λ μ ) S ˆ n ( x 1 ) j = 0 n = j n i = 1 r λ i j + 1 μ i 2 ( n ) S 1 ( , j ) S ˆ n E j ( 1 + ( x 1 ) / λ i ) .

As a corollary, we get the following identity.

Corollary 3 For all n0,

2 1 j = 1 r μ j m = 0 n S 1 ( n , m ) i = 1 r λ i μ i E n ( x + λ ( μ + e i ) 1 | λ ; μ + e i ) , = j = 0 n = j n i = 1 r λ i j + 1 μ i 2 ( n ) S 1 ( , j ) S ˆ n E j ( 1 + ( x 1 ) / λ i ) .

Recall that for a n (x)(g(t),f(t)), we have d d x a n (x)= = 0 n 1 ( n ) f ¯ (t)| x n a (x). Hence, in the case (1.11), namely a n (x)= S n (x), we have

f ¯ ( t ) | x n = log ( 1 + t ) | x n = m 1 ( 1 ) m 1 x m m | x n = ( 1 ) n 1 ( n 1 ) ! ,

which implies d/dx S n (x)=n! = 0 n 1 ( 1 ) n 1 ! ( n ) S (x). In the same way, we obtain the case S ˆ n (x), which leads to the following result.

Theorem 6 For all n1,

d d x S n (x)=n! = 0 n 1 ( 1 ) n 1 ! ( n ) S (x)and d d x S ˆ n (x)=n! = 0 n 1 ( 1 ) n 1 ! ( n ) S ˆ (x).

Now we find another recurrence relation by using the derivative operator. For n1, by (1.11) we have

S n ( y ) = i 0 S i ( y ) t i i ! | x n = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n = d d t ( j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y ) | x n 1 = d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 + j = 1 r ( 1 + ( 1 + t ) λ j ) μ j d d t ( 1 + t ) y | x n 1 = d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 + y S n 1 ( y 1 ) .

Observe that d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j i = 1 r λ i μ i ( 1 + t ) λ i 1 1 + ( 1 + t ) λ i . Thus,

d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 = i = 1 r λ i μ i ( 1 + ( 1 + t ) λ i ) 1 j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y + λ i 1 | x n 1 = i = 1 r λ i μ i S n 1 ( y + λ i 1 | λ ; μ + e i ) .

Hence,

S n (x)=x S n 1 (x1) i = 1 r λ i μ i S n 1 (x+ λ i 1|λ;μ+ e i ).
(3.3)

Also, for n1, by (1.12) we have

S ˆ n ( y ) = i 0 S ˆ i ( y ) t i i ! | x n = j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n = d d t [ j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y ] | x n 1 = d d t j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 + j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j d d t ( 1 + t ) y | x n 1 = d d t j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 + y S ˆ n 1 ( y 1 ) .

Observe that d d t j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j = j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j i = 1 r λ i μ i ( 1 + t ) λ i 1 ( 1 + t ) λ i 1 + ( 1 + t ) λ i . So

d d t j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y | x n 1 = i = 1 r λ i μ i ( 1 + t ) λ i 1 + ( 1 + t ) λ i j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) y λ i 1 | x n 1 = i = 1 r λ i μ i S ˆ n 1 ( y λ i 1 | λ ; μ + e i ) .

Thus,

S ˆ n (x)=x S ˆ n 1 (x1)+ i = 1 r λ i μ i S ˆ n 1 (x λ i 1|λ;μ+ e i ).
(3.4)

Hence, by (3.3) and (3.4), we obtain the following result.

Theorem 7 For n1,

S n ( x ) = x S n 1 ( x 1 ) i = 1 r λ i μ i S n 1 ( x + λ i 1 | λ ; μ + e i ) , S ˆ n ( x ) = x S ˆ n 1 ( x 1 ) + i = 1 r λ i μ i S ˆ n 1 ( x λ i 1 | λ ; μ + e i ) .

Another result that can be obtained is the following.

Theorem 8 For n1m1,

= 0 n m ( n ) S 1 ( n , m ) S = = 0 n m ( n 1 ) S 1 ( n 1 , m 1 ) S ( 1 ) = 0 n 1 m ( n 1 ) S 1 ( n 1 , m ) i = 1 r λ i μ i S ( λ i 1 | λ ; μ + e i ) , = 0 n m ( n ) S 1 ( n , m ) S ˆ = = 0 n m ( n 1 ) S 1 ( n 1 , m 1 ) S ˆ ( 1 ) + = 0 n 1 m ( n 1 ) S 1 ( n 1 , m ) i = 1 r λ i μ i S ˆ ( λ i 1 | λ ; μ + e i ) .

Proof Because of the similarity in the two cases S n (x) and S ˆ n (x), we only give the proof of the first identity. In order to prove the first identity, we compute the following in two different ways:

A= j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n .

On the one hand, it is equal to

A = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | ( log ( 1 + t ) ) m x n = j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | m ! m S 1 ( , m ) t ! x n = m ! = m n S 1 ( , m ) ( n ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | x n = m ! = m n S 1 ( , m ) ( n ) S n = m ! = 0 n m S 1 ( n , m ) ( n ) S .
(3.5)

On the other hand,

A = d d t [ j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m ] | x n 1 = d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n 1 + j = 1 r ( 1 + ( 1 + t ) λ j ) μ j d d t ( log ( 1 + t ) ) m | x n 1 .

Here,

j = 1 r ( 1 + ( 1 + t ) λ j ) μ j d d t ( log ( 1 + t ) ) m | x n 1 = m j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) 1 | ( log ( 1 + t ) ) m 1 x n 1 = m ! = m 1 n 1 S 1 ( , m 1 ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) 1 | t ! x n 1 = m ! = 0 n m ( n 1 ) S 1 ( n 1 , m 1 ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) 1 | x = m ! = 0 n m ( n 1 ) S 1 ( n 1 , m 1 ) S ( 1 )

and

d d t j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n 1 = i = 1 r λ i μ i ( 1 + ( 1 + t ) λ i ) 1 j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) λ i 1 | ( log ( 1 + t ) ) m x n 1 = i = 1 r λ i μ i ( 1 + ( 1 + t ) λ i ) 1 j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) λ i 1 | m ! m S 1 ( , m ) t ! x n 1 = m ! i = 1 r = m n 1 λ i μ i ( n 1 ) S 1 ( , m ) × ( 1 + ( 1 + t ) λ i ) 1 j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) λ i 1 | x n 1 = m ! i = 1 r = 0 n 1 m λ i μ i ( n 1 ) S 1 ( n 1 , m ) × ( 1 + ( 1 + t ) λ i ) 1 j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( 1 + t ) λ i 1 | x = m ! i = 1 r = 0 n 1 m λ i μ i ( n 1 ) S 1 ( n 1 , m ) S ( λ i 1 | λ ; μ + e i ) .

Altogether, we have, for n1m1,

m ! = 0 n m ( n ) S 1 ( n , m ) S = m ! = 0 n m ( n 1 ) S 1 ( n 1 , m 1 ) S ( 1 ) m ! i = 1 r = 0 n 1 m λ i μ i ( n 1 ) S 1 ( n 1 , m ) S ( λ i 1 | λ ; μ + e i ) .

By dividing by m!, we complete the proof. □

4 Identities

Let S n (x)= m = 0 n c n , m ( x ) m and S ˆ n (x)= m = 0 n c ˆ n , m ( x ) m . By (1.9), (1.10) and (1.11), we obtain

c n , m = 1 m ! j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | t m x n = ( n m ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j | x n m = ( n m ) S n m ,

and by (1.9), (1.10) and (1.12), we obtain

c ˆ n , m = 1 m ! j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | t m x n = ( n m ) j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j | x n m = ( n m ) S ˆ n m .

Hence, we have the following identities.

Theorem 9 For all n0,

S n (x)= m = 0 n S n m ( n m ) ( x ) m and S ˆ n (x)= m = 0 n S ˆ n m ( n m ) ( x ) m .

Now, let S n (x)= m = 0 n c n , m H m ( s ) (x|α) and S ˆ n (x)= m = 0 n c ˆ n , m H m ( s ) (x|α), where H n ( s ) (x|α)( ( e t α 1 α ) s ,t), with α1. Then, by (1.9), (1.10) and (1.11), we obtain

c n , m = 1 m ! ( e log ( 1 + t ) α 1 α ) s j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n = 1 m ! ( 1 α ) s j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m ( 1 α + t ) s | x n = 1 m ! ( 1 α ) s j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | j = 0 min { s , n } ( s j ) ( 1 α ) t j x n = 1 m ! ( 1 α ) s j = 0 n m ( s j ) ( 1 α ) s j ( n ) j j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n j ,

and by Theorem 8, we have

c n , m = 1 m ! ( 1 α ) s j = 0 n m ( s j ) ( 1 α ) s j ( n ) j ( m ! = 0 n j m ( n j ) S 1 ( n j , m ) S ) = j = 0 n m = 0 n m j ( s j ) ( n j ) ( 1 α ) j ( n ) j S 1 ( n j , m ) S .

By (1.9), (1.10) and (1.12), we obtain

c ˆ n , m = 1 m ! ( e log ( 1 + t ) α 1 α ) s j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n = 1 m ! ( 1 α ) s j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | ( 1 α + t ) s x n = 1 m ! ( 1 α ) s j = 0 n m ( s j ) ( 1 α ) s j ( n ) j j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n j ,

and by Theorem 8, we have

c ˆ n , m = 1 m ! ( 1 α ) s j = 0 n m ( s j ) ( 1 α ) s j ( n ) j ( m ! = 0 n j m ( n j ) S 1 ( n j , m ) S ˆ ) = j = 0 n m = 0 n m j ( s j ) ( n j ) ( 1 α ) j ( n ) j S 1 ( n j , m ) S ˆ .

Therefore, we can state the following result.

Theorem 10 For all n0,

S n ( x ) = m = 0 n ( j = 0 n m = 0 n m j ( s j ) ( n j ) ( 1 α ) j ( n ) j S 1 ( n j , m ) S ) H m ( s ) ( x | α ) , S ˆ n ( x ) = m = 0 n ( j = 0 n m = 0 n m j ( s j ) ( n j ) ( 1 α ) j ( n ) j S 1 ( n j , m ) S ˆ ) H m ( s ) ( x | α ) .

Finally, we express our polynomials S n (x) and S ˆ n (x) in terms of Bernoulli polynomials of order s. Let S n (x)= m = 0 n c n , m B m ( s ) (x) and S ˆ n (x)= m = 0 n c ˆ n , m B m ( s ) (x), where B n ( s ) (x)( ( e t 1 t ) s ,t). Then, by (1.9), (1.10) and (1.11), we obtain

c n , m = 1 m ! ( e log ( 1 + t ) 1 log ( 1 + t ) ) s j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n = 1 m ! j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | ( t log ( 1 + t ) ) s x n ,

and by the fact that ( t log ( 1 + t ) ) s = n 0 C n ( s ) t n n ! , where C n ( s ) is the Cauchy number of the first kind of order s, we derive

c n , m = 1 m ! i = 0 n m ( n i ) C i ( s ) j = 1 r ( 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n i ,

and by Theorem 8, we obtain

c n , m = 1 m ! i = 0 n m ( n i ) C i ( s ) ( m ! = 0 n i m ( n i ) S 1 ( n i , m ) S ) = i = 0 n m = 0 n i m ( n i ) ( n i ) C i ( s ) S 1 ( n i , m ) S .

Also, by (1.9), (1.10) and (1.12), we obtain

c ˆ n , m = 1 m ! ( e log ( 1 + t ) 1 log ( 1 + t ) ) s j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n = 1 m ! j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | ( t log ( 1 + t ) ) s x n = 1 m ! i = 0 n m ( n i ) C i ( s ) j = 1 r ( ( 1 + t ) λ j 1 + ( 1 + t ) λ j ) μ j ( log ( 1 + t ) ) m | x n i ,

and by Theorem 8, we obtain

c n , m = 1 m ! i = 0 n m ( n i ) C i ( s ) ( m ! = 0 n i m ( n i ) S 1 ( n i , m ) S ˆ ) = i = 0 n m = 0 n i m ( n i ) ( n i ) C i ( s ) S 1 ( n i , m ) S ˆ .

Hence, we have the following identities.

Theorem 11 For all n0,

S n ( x ) = m = 0 n ( j = 0 n m = 0 n m j ( n j ) ( n j ) C j ( s ) S 1 ( n j , m ) S ) B m ( s ) ( x ) , S ˆ n ( x ) = m = 0 n ( j = 0 n m = 0 n m j ( n j ) ( n j ) C j ( s ) S 1 ( n j , m ) S ˆ ) 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 Korea government (MOE) (No. 2012R1A1A2003786) and was partially supported by Kwangwoon University in 2014.

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Kim, D.S., Kim, T., Kwon, H.I. et al. Barnes-type Peters polynomial with umbral calculus viewpoint. J Inequal Appl 2014, 324 (2014). https://doi.org/10.1186/1029-242X-2014-324

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Keywords

  • Barnes-type Peters polynomials
  • umbral calculus