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Some new identities of Frobenius-Euler numbers and polynomials

Abstract

In this paper, we give some new and interesting identities which are derived from the basis of Frobenius-Euler. Recently, several authors have studied some identities of Frobenius-Euler polynomials. From the methods of our paper, we can also derive many interesting identities of Frobenius-Euler numbers and polynomials.

1 Introduction

Let λ(1)C. As is well known, the Frobienius-Euler polynomials are defined by the generating function to be

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

with the usual convention about replacing H n (x|λ) by H n (x|λ) (see [16]).

In the special case, x=0, H n (0|λ)= H n (λ) are called the n th Frobenius-Euler numbers.

Thus, by (1), we get

( H ( λ ) + 1 ) n λ H n (λ)= H n (1|λ)λ H n (λ)=(1λ) δ 0 , n ,
(2)

where δ 0 , n is the Kronecker symbol.

From (1), we can derive the following equation:

H n (x|λ)= ( H ( λ ) + x ) n = 0 l n ( n l ) H n l (λ) x l (see [6–16]).
(3)

Thus, by (3), we easily see that the leading coefficient of H n (x|λ) is H 0 (λ)=1. So, H n (x|λ) are monic polynomials of degree n with coefficients in Q(λ).

From (1), we have

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

Thus, by (4), we get

H n (x+1|λ)λ H n (x|λ)=(1λ) x n ,for n Z + .
(5)

It is easy to show that

d d x H n (x|λ)= d d x ( H ( λ ) + x ) n =n H n 1 (x|λ)(nN).
(6)

From (6), we have

0 1 H n (x|λ)dx= 1 n + 1 ( H n + 1 ( 1 | λ ) H n + 1 ( λ ) ) = λ 1 n + 1 H n + 1 (λ).
(7)

Let P n (λ)={p(x)Q(λ)[x]degp(x)n} be a vector space over Q(λ). Then we note that { H 0 (x|λ), H 1 (x|λ),, H n (x|λ)} is a good basis for P n (λ).

In this paper, we develop some new methods to obtain some new identities and properties of Frobenius-Euler polynomials which are derived from the basis of Frobenius-Euler polynomials. Those methods are useful in studying the identities of Frobenius-Euler polynomials.

2 Some identities of Frobenius-Euler polynomials

Let us take p(x) P n (λ). Then p(x) can be expressed as a Q(λ)-linear combination of H 0 (x|λ),, H n (x|λ) as follows:

p(x)= b 0 H 0 (x|λ)+ b 1 H 1 (x|λ)++ b n H n (x|λ)= 0 k n b k H k (x|λ).
(8)

Let us define the operator λ by

g(x)= λ p(x)=p(x+1)λp(x).
(9)

From (9), we can derive the following equation (10):

g(x)= λ p(x)= 0 k n b k ( H k ( x + 1 | λ ) λ H k ( x | λ ) ) =(1λ) 0 k n b k x k .
(10)

For r Z + , let us take the r th derivative of g(x) in (10) as follows:

g ( r ) (x)=(1λ) r k n k(k1)(kr+1) b k x k r ,where  g ( r ) (x)= d r g ( x ) d x r .
(11)

Thus, by (11), we get

g r (0)= d r g ( x ) d x r | x = 0 =(1λ)r! b r .
(12)

From (12), we have

b r = g ( r ) ( 0 ) ( 1 λ ) r ! = 1 ( 1 λ ) r ! ( p ( r ) ( 1 ) λ p ( r ) ( 0 ) ) ,
(13)

where r Z + and p ( r ) (0)= d r p ( x ) d x r | x = 0 . Therefore, by (13), we obtain the following theorem.

Theorem 1 For λ(1)C, n Z + , let p(x) P n (λ) with p(x)= 0 k n b k H k (x|λ). Then we have

b k = 1 ( 1 λ ) k ! g ( k ) (0)= 1 ( 1 λ ) k ! ( p ( k ) ( 1 ) λ p ( k ) ( 0 ) ) .

Let us take p(x)= H n (x| λ 1 ). Then, by Theorem 1, we get

H n ( x | λ 1 ) = 0 k n b k H k (x|λ),
(14)

where

b k = 1 ( 1 λ ) k ! n ! ( n k ) ! { H n k ( 1 | λ 1 ) λ H n k ( λ 1 ) } = 1 1 λ ( n k ) { H n k ( 1 | λ 1 ) λ H n k ( λ 1 ) } = 1 1 λ ( n k ) { ( 1 λ 1 ) 0 n k + 1 λ H n k ( λ 1 ) λ H n k ( λ 1 ) } .
(15)

By (14) and (15), we get

(16)

Therefore, by (16), we obtain the following theorem.

Theorem 2 For n Z + , we have

λ H n ( x | λ 1 ) + H n (x|λ)=(1+λ) 0 k n ( n k ) H n k ( λ 1 ) H k (x|λ).

Let

p(x)= 0 k n H k (x|λ) H n k (x|λ) P n (λ).
(17)

From Theorem 2, we note that p(x) can be generated by { H 0 (x|λ), H 1 (x|λ),, H n (x|λ)} as follows:

p(x)= 0 k n H k (x|λ) H n k (x|λ)= 0 k n b k H k (x|λ).
(18)

By (17), we get

p ( k ) (x)= ( n + 1 ) ! ( n k + 1 ) ! k l n H l k (x|λ) H n k (x|λ),
(19)

and

(20)

From (18) and (20), we have

0 k n H k ( x | λ ) H n k ( x | λ ) = ( n + 1 ) 0 k n 1 ( n k ) n k + 1 k l n { ( λ ) H l k ( λ ) H n l ( λ ) + 2 λ H n k ( λ ) } H k ( x | λ ) + ( n + 1 ) H n ( x | λ ) .
(21)

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

Theorem 3 For n Z + , we have

Let us consider

p(x)= k = 0 n 1 k ! ( n k ) ! H k (x|λ) H n k (x|λ) P n (λ).
(22)

By Theorem 1, p(x) can be expressed by

p(x)= k = 0 n b k H k (x|λ).
(23)

From (22), we have

p ( r ) (x)= 2 r k = r n H k r ( x | λ ) H n k ( x | λ ) ( k r ) ! ( n k ) ! (r Z + ).
(24)

By Theorem 1, we get

b k = 1 2 k ! { p ( k ) ( 1 ) p ( k ) ( 0 ) } = 2 k 1 k ! l = k n 1 ( l k ) ! ( n l ) ! { H l k ( 1 | λ ) H n l ( 1 | λ ) λ H l k ( λ ) H n l ( λ ) } = 2 k 1 k ! l = k n 1 ( l k ) ! ( n l ) ! { ( λ H l k ( λ ) + ( 1 λ ) δ 0 , l k ) ( λ H n l ( λ ) + ( 1 λ ) δ 0 , n l ) λ H l k ( λ ) H n l ( λ ) } = 2 k 1 k ! { l = k n λ ( λ 1 ) H l k ( λ ) H n l ( λ ) ( l k ) ! ( n l ) ! + 2 λ ( 1 λ ) H n k ( λ ) ( n k ) ! + ( 1 λ ) 2 δ n , k } = { 2 k 1 k ! l = k n { λ ( λ 1 ) H l k ( λ ) H n l ( λ ) ( l k ) ! ( n l ) ! + 2 λ ( 1 λ ) H n k ( λ ) ( n k ) ! } , if  k n , 2 n 1 ( 1 λ ) n ! , if  k = n .
(25)

Therefore, by (25), we obtain the following theorem.

Theorem 4 For n Z + , we have

3 Higher-order Frobenius-Euler polynomials

For n Z + , the Frobenius-Euler polynomials of order r are defined by the generating function to be

( 1 λ e t λ ) r e x t = e H ( r ) ( x | λ ) t = n = 0 H n ( r ) ( x | λ ) t n n ! ,
(26)

with the usual convention about replacing ( H ( r ) ( x | λ ) ) n by H n ( r ) (x|λ) (see [110]). In the special case, x=0, H n ( r ) (0|λ)= H n ( r ) (λ) are called the n th Frobenius-Euler numbers of order r (see [8, 9]).

From (26), we have

H n ( r ) (x|λ)= ( H ( r ) ( λ ) + x ) n = l = 0 n ( n l ) H n l ( r ) (λ) x l ,
(27)

with the usual convention about replacing ( H ( r ) ( λ ) ) n by H n ( r ) (λ).

By (26), we get

H n ( r ) (λ)= n 1 + + n r = n ( n n 1 , n 2 , , n r ) H n 1 (λ) H n r (λ),
(28)

where ( n n 1 , n 2 , , n r ) = n ! n 1 ! n 2 ! n r ! . From (27) and (28), we note that the leading coefficient of H n ( r ) (x|λ) is given by

H 0 ( r ) ( λ ) = n 1 + + n r = 0 ( n n 1 , n 2 , , n r ) H n 1 ( λ ) H n r ( λ ) = H 0 ( λ ) H 0 ( λ ) = 1 .
(29)

Thus, by (29), we see that H n ( r ) is a monic polynomial of degree n with coefficients in Q(λ). From (26), we have

H n ( 0 ) (x|λ)= x n ,for n Z + ,
(30)

and

x H n ( r ) (x|λ)= x ( H ( r ) ( λ ) + x ) n =n H n 1 ( r ) (x|λ)(r0).
(31)

It is not difficult to show that

H n ( r ) (x+1|λ)λ H n ( r ) (x|λ)=(1λ) H n ( r 1 ) (x|λ).
(32)

Now, we note that { H 0 ( r ) (x|λ), H 1 ( r ) (x|λ),, H n ( r ) (x|λ)} is also a good basis for P n (λ).

Let us define the operator D as Df(x)= d f ( x ) d x and let p(x) P n (λ). Then p(x) can be written as

p(x)= k = 0 n C k H k ( r ) (x|λ).
(33)

From (9) and (32), we have

λ H n ( r ) (x|λ)= H n ( r ) (x+1|λ)λ H n ( r ) (x|λ)=(1λ) H n ( r 1 ) (x|λ).
(34)

Thus, by (33) and (34), we get

λ r p(x)= ( 1 λ ) r k = 0 n C k H k ( 0 ) (x|λ)= ( 1 λ ) r k = 0 n C k x k .
(35)

Let us take the k th derivative of λ r p(x) in (35).

Then we have

D k ( λ r p ( x ) ) = ( 1 λ ) r l = k n l ! ( l k ) ! C l x l k .
(36)

Thus, from (36), we have

D k ( λ r p ( 0 ) ) = ( 1 λ ) r l = k n l ! C l ( l k ) ! 0 l k = ( 1 λ ) r k! C k .
(37)

Thus, by (37), we get

C k = D k ( λ r p ( 0 ) ) ( 1 λ ) r k ! = λ r ( D k p ( 0 ) ) ( 1 λ ) r k ! = 1 ( 1 λ ) r k ! j = 0 r ( r j ) ( λ ) ( r j ) D k p ( j ) .
(38)

Therefore, by (33) and (38), we obtain the following theorem.

Theorem 5 For r Z + , let p(x) P n (λ) with

p(x)= 1 ( 1 λ ) r 0 k n C k H k ( r ) (x|λ) ( C k Q ( λ ) ) .

Then we have

C k = 1 ( 1 λ ) r k ! 0 j r ( r j ) ( λ ) r j D k p(j).

That is,

p(x)= 1 ( 1 λ ) r 0 k n ( 0 j r 1 k ! ( r j ) ( λ ) r j D k p ( j ) ) H k ( r ) (x|λ).

Let us take p(x)= H n (x|λ) P n (λ). Then, by Theorem 5, p(x)= H n (x|λ) can be generated by { H 0 ( r ) (x|λ), H 1 ( r ) (λ),, H n ( r ) (x|λ)} as follows:

H n (x|λ)= 0 k n C k H k ( r ) (x|λ),
(39)

where

C k = 1 ( 1 λ ) r 1 k ! 0 j r ( r j ) ( λ ) r j D k p(j),
(40)

and

p ( k ) (x)= D k p(x)=n(n1)(nk+1) H n k (x|λ)= n ! ( n k ) ! H n k (x|λ).
(41)

By (40) and (41), we get

C k = 1 ( 1 λ ) r ( n k ) 0 j r ( r j ) ( λ ) r j H n k (j|λ).
(42)

Therefore, by (39) and (42), we obtain the following theorem.

Theorem 6 For n Z + , we have

H n (x|λ)= 1 ( 1 λ ) r 0 k n ( n k ) ( 0 j r ( r j ) ( λ ) r j H n k ( j | λ ) ) H k ( r ) (x|λ).

Let us assume that p(x)= H n ( r ) (x|λ).

Then we have

p k ( x ) = n ( n 1 ) ( n k + 1 ) H n k ( r ) ( x | λ ) = n ! ( n k ) ! H n k ( r ) ( x | λ ) .
(43)

From Theorem 1, we note that p(x)= H n ( r ) (x|λ) can be expressed as a linear combination of H 0 (x|λ), H 1 (x|λ),, H n (x|λ)

H n ( r ) (x|λ)= 0 k n b k H k (x|λ),
(44)

where

b k = 1 ( 1 λ ) k ! { p k ( 1 ) λ p ( k ) ( 0 ) } = n ! ( 1 λ ) k ! ( n k ) ! { H n k ( r ) ( 1 | λ ) λ H n k ( r ) ( λ ) } .
(45)

By (34) and (45), we get

b k = ( n k ) H n k ( r 1 ) (λ).
(46)

Therefore, by (44) and (46), we obtain the following theorem.

Theorem 7 For n Z + , we have

H n ( r ) (x|λ)= 0 k n ( n k ) H n k ( r 1 ) (λ) H k (x|λ).

Remark From (2) and (37), we note that

d d λ ( 1 λ e t λ ) = 1 e t ( e t λ ) 2 = 1 ( 1 λ ) 2 ( ( 1 λ ) 2 ( e t λ ) 2 ( 1 λ ) 2 ( e t λ ) 2 e t ) = 1 ( 1 λ ) 2 ( ( 1 λ ) 2 ( e t λ ) 2 ( 1 λ ) 2 ( e t λ ) 2 ( e t λ + λ ) ) = 1 1 λ ( ( 1 λ ) 2 ( e t λ ) 2 1 λ e t λ ) = 1 1 λ n = 0 ( H n ( 2 ) ( λ ) H n ( λ ) ) t n n ! ,
(47)

and

d 2 d λ 2 ( 1 λ e t λ ) = 2 ! 1 e t ( e t λ ) 3 = 2 ! ( 1 λ ) 3 ( ( 1 λ ) 3 ( e t λ ) 3 ( 1 λ ) 3 ( e t λ ) 3 e t ) = 2 ! ( 1 λ ) 3 ( ( 1 λ ) 3 ( e t λ ) 3 ( 1 λ ) 3 ( e t λ ) 3 ( e t λ + λ ) ) = 2 ! ( 1 λ ) 2 ( ( 1 λ ) 3 ( e t λ ) 3 ( 1 λ ) 2 ( e t λ ) 2 ) = 2 ! ( 1 λ ) 2 n = 0 ( H n ( 3 ) ( λ ) H n ( 2 ) ( λ ) ) t n n ! .
(48)

Continuing this process, we obtain the following equation:

d k d λ k ( 1 λ e t λ ) = k ! ( 1 λ ) k ( ( 1 λ ) k + 1 ( e t λ ) k + 1 ( 1 λ ) k ( e t λ ) k ) = k ! ( 1 λ ) k n = 0 ( H n ( k + 1 ) ( λ ) H n ( k ) ( λ ) ) t n n ! (see [8]) .
(49)

By (1), (2) and (49), we get

d k d λ k H n (λ)= k ! ( 1 λ ) k ( H n ( k + 1 ) ( λ ) H n ( k ) ( λ ) ) ,

where k is a positive integer (see [7, 8]).

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Acknowledgements

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology 2012R1A1A2003786.

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Kim, D.S., Kim, T. Some new identities of Frobenius-Euler numbers and polynomials. J Inequal Appl 2012, 307 (2012). https://doi.org/10.1186/1029-242X-2012-307

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Keywords

  • Positive Integer
  • Linear Combination
  • Vector Space
  • Good Basis
  • Monic Polynomial