Some identities of Frobenius-Euler polynomials arising from Frobenius-Euler basis
aa r X i v : . [ m a t h . N T ] N ov Some identities of Frobenius-Euler polynomialsarising from Frobenius-Euler basis byDae San Kim and Taekyun Kim
Abstract
In this paper, we give some new and interesting identities which are de-rived from the basis of Frobenius-Euler. Recently, Simsek et als(see [13])have given some identities of q -analogue of Frobenius-Euler polynomialsrelated to q -Bernstein polynomials. From the methods of our paper, wecan also derive the results and identities of Simsek et als (cf.[13] ). Let λ ( = 1) ∈ C . As is well known, the Frobienius-Euler polynomials aredefined by the generating function to be1 − λe t − λ e xt = e H ( x | λ ) t = ∞ X 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 [1 − x = 0, H n (0 | λ ) = H n ( λ ) are called the n -th Frobenius-Eulernumbers.Thus, by(1), we get( H ( λ ) + 1) n − λH n ( λ ) = H n (1 | λ ) − λH n ( λ ) = (1 − λ ) δ ,n , (2)where δ ,n is the Kronecker symbol.From (1), we can derive the following equation : H n ( x | λ ) = ( H ( λ ) + x ) n = n X l =0 (cid:18) nl (cid:19) H n − l ( λ ) x l , (see [6 − . (3)1hus, by (3), we easily see that the leading coefficient of H n ( x | λ ) is H ( λ ) = 1 .So, H n ( x | λ ) is a monic polynomials of degree n with coefficients in Q ( λ ).From (1), we have ∞ X n =0 ( H n ( x + 1 | λ ) − λH n ( x | λ )) t n n ! = (1 − λ ) e ( x +1) t e t − λ − λ − λe t − λ e xt . (4)Thus, by (4), we get H n ( x + 1 | λ ) − λH n ( x | λ ) = (1 − λ ) x n , for n ∈ Z + . (5)It is easy to show that ddx H n ( x | λ ) = ddx ( H ( λ ) + x ) n = nH n − ( x | λ ) , ( n ∈ N ) . (6)From (6), we have Z H n ( x | λ ) dx = 1 n + 1 ( H n +1 (1 | λ ) − H n +1 ( λ )) = λ − n + 1 H n +1 ( λ ) . (7)Let P n ( λ ) = { p ( x ) ∈ Q ( λ )[ x ] | deg p ( x ) ≤ n } be a vector space over Q ( λ ).Then we note that { H ( x | λ ) , H ( 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 andproperties of Frobenius-Euler polynomials which are derived from the basis ofthe Frobenius-Euler polynomials. Those methods are useful in studying theidentities of Frobenius-Euler polynomials. Let us take p ( x ) ∈ P n ( λ ). Then p ( x ) can be expressed as a Q ( λ )-linear com-bination of H ( x | λ ) , · · · , H n ( x | λ ) as follows : p ( x ) = b H ( x | λ ) + b H ( x | λ ) + · · · + b n H n ( x | λ ) = n X k =0 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)2rom (9), we can derive the following equation (10) : g ( x ) = △ λ p ( x ) = n X k =0 b k ( H k ( x + 1 | λ ) − λH k ( x | λ )) = (1 − λ ) n X k =0 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 ) = (1 − λ ) n X k = r k ( k − · · · ( k − r +1) b k x k − r , where g ( r ) ( x ) = d r g ( x ) dx r . (11)Thus, by (11), we get g r (0) = d r g ( x ) dx r | x =0 = (1 − λ ) r ! b r . (12)From (12), we have b r = g ( r ) (0)(1 − λ ) r ! = 1(1 − λ ) r ! ( p ( r ) (1) − λp ( r ) (0)) , (13)where r ∈ Z + , and p ( r ) (0) = d r p ( x ) dx r | x =0 . Therefore, by (13), we obtain thefollowing theorem. Theorem 1.
For λ ( = 1) ∈ C , n ∈ Z + ,let p ( x ) ∈ P n ( λ ) with p ( x ) = P nk =0 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) − λp ( k ) (0)) . Let us take p ( x ) = H n ( x | λ − ). Then, by Theorem 1, we get H n ( x | λ − ) = n X k =0 b k H k ( x | λ ) , (14)where b k = 1(1 − λ ) k ! n !( n − k )! { H n − k (1 | λ − ) − λH n − k ( λ − ) } (15)= 11 − λ (cid:18) nk (cid:19) { H n − k (1 | λ − ) − λH n − k ( λ − ) } = 11 − λ (cid:18) nk (cid:19) { (1 − λ − )0 n − k + 1 λ H n − k ( λ − ) − λH n − k ( λ − ) } .
3y (14) and (15), we get H n ( x | λ − ) (16)= − λ H n ( x | λ ) + n X k =0 { (cid:18) nk (cid:19) λ (1 − λ ) H n − k ( λ − ) − λ (cid:18) nk (cid:19) − λ H n − k ( λ − ) H k } ( x | λ )= − λ H n ( x | λ ) + n X k =0 (cid:18) nk (cid:19) λλ H n − k ( λ − ) H k ( x | λ ) . Therefore, by (16), we obtain the following theorem.
Theorem 2.
For n ∈ Z + , we have λH n ( x | λ − ) + H n ( x | λ ) = (1 + λ ) n X k =0 (cid:18) nk (cid:19) H n − k ( λ − ) H k ( x | λ ) . Let p ( x ) = n X k =0 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 ( x | λ ) , H ( x | λ ) , · · · , H n ( x | λ ) } as follows: p ( x ) = n X k =0 H k ( x | λ ) H n − k ( x | λ ) = n X k − b k H k ( x | λ ) . (18)By (17), we get p ( k ) ( x ) = ( n + 1)!( n − k + 1)! n X l = k H l − k ( x | λ ) H n − k ( x | λ ) , (19)4nd b k = 1(1 − λ ) k ! { p ( k ) (1) − λp ( k ) (0) } (20)= ( n + 1)!(1 − λ ) k !( n − k + 1)! n X l = k { H l − k (1 | λ ) H n − l (1 | λ ) − λH l − k ( λ ) H n − l ( λ ) } = n + 1(1 − λ )( n − k + 1) (cid:18) nk (cid:19) n X l = k { ( λH l − k ( λ ) + (1 − λ ) δ ,l − k )( λH n − l +(1 − λ ) δ ,n − l ) − λH l − k ( λ ) H n − l ( λ ) } = n + 1(1 − λ )( n − k + 1) (cid:18) nk (cid:19) n X l = k { λ (1 − λ ) δ ,l − k H n − l ( λ ) + λ (1 − λ ) × H l − k ( λ ) δ ,n − l + (1 − λ ) δ ,l − k δ ,n − l + λ ( λ − H l − k ( λ ) H n − l ( λ ) } = n + 1(1 − λ )( n − k + 1) (cid:18) nk (cid:19) n X l = k { λ ( λ − H l − k ( λ ) H n − l ( λ ) + λ (1 − λ ) × H n − k ( λ ) + λ (1 − λ ) H n − k ( λ ) + (1 − λ ) δ n,k } = n + 1 n − k + 1 (cid:18) nk (cid:19) n X l = k {− λH l − k ( λ ) H n − l ( λ ) + 2 λH n − k ( λ ) + (1 − λ ) δ n,k } . From (18) and (20), we have n X k =0 H k ( x | λ ) H n − k ( x | λ ) = ( n + 1) n − X k =0 (cid:18) nk (cid:19) n − k + 1 n X l = k { ( − λ ) H l − k ( λ ) H n − l ( λ )(21)+ 2 λH n − k ( λ ) } H k ( x | λ ) + ( n + 1) H n ( x | λ ) . Therefore, by (21), we obtain the following theorem
Theorem 3.
For n ∈ Z + , we have n + 1 n X k =0 H k ( x | λ ) H n − k ( x | λ )= n − X k =0 (cid:18) nk (cid:19) n − k + 1 n X l = k { ( − λ ) H l − k ( λ ) H n − l ( λ ) + 2 λH n − k ( λ ) } H k ( x | λ ) + H n ( x | λ ) . p ( x ) = n X k =0 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 ) = n X k =0 b k H k ( x | λ ) . (23)From (22), we have p ( r ) ( x ) = 2 r n X k = r H k − r ( x | λ ) H n − k ( x | λ )( k − r )!( n − k )! , ( r ∈ Z + ) . (24)By Theorem 1, we get b k = 12 k ! { p ( k ) (1) − p ( k ) (0) } (25)= 2 k − k ! n X l = k l − k )!( n − l )! { H l − k (1 | λ ) H n − l (1 | λ ) − λH l − k ( λ ) H n − l ( λ ) } = 2 k − k ! n X l = k l − k )!( n − l )! { ( λH l − k ( λ ) + (1 − λ ) δ ,l − k )( λH n − l ( λ )+ (1 − λ ) δ ,n − l ) − λH l − k ( λ ) H n − l ( λ ) } = 2 k − k ! { n X l = k λ ( λ − H l − k ( λ ) H n − l ( λ )( l − k )!( n − l )! + 2 λ (1 − λ ) H n − k ( λ )( n − k )! + (1 − λ ) δ n,k } = ( k − k ! P nl = k { λ ( λ − H l − k ( λ ) H n − l ( λ )( l − k )!( n − l )! + λ (1 − λ ) H n − k ( λ )( n − k )! } , if k = n n − (1 − λ ) n ! , if k = n Therefore, by (25), we obtain the following theorem .
Theorem 4.
For n ∈ Z + , we have n X k =0 k !( n − k )! H k ( x | λ ) H n − k ( x | λ )= n − X k =0 k − k ! n X l = k { λ ( λ − H l − k ( λ ) H n − l ( λ )( l − k )!( n − l )! + 2 λ (1 − λ ) H n − k ( λ )( n − k )! } H k ( x | λ )+ 2 n − (1 − λ ) n ! H n ( x | λ ) . Higher-order Frobenius-Euler polynomials
For n ∈ Z + , the Frobenius-Euler polynomials of order r are defined by thegenerating function to be( 1 − λe t − λ ) r e xt = e H ( r ) ( x | λ ) t (26)= ∞ X n =0 H ( r ) n ( x | λ ) t n n ! , with the usual convention about replacing ( H ( r ) ( x | λ )) n by H ( r ) n ( x | λ ), (see[1 − x = 0, H ( r ) n (0 | λ ) = H ( r ) n ( λ ) are called the n -thFrobenius-Euler numbers of order r , (see [8 − H ( r ) n ( x | λ ) = H ( r ) ( λ ) + x ) n = n X l =0 (cid:18) nl (cid:19) H ( r ) n − l ( λ ) x l , (27)with the usual convention about replacing ( H ( r ) ( λ )) n by H ( r ) n ( λ ) .By (26), we get H ( r ) n ( λ ) = X n + ··· + n r = n (cid:18) nn , n , · · · , n r (cid:19) H n ( λ ) · · · H n r ( λ ) , (28)where (cid:18) nn , n , · · · , n r (cid:19) = n ! n ! n ! ··· n r ! . From (27) and (28), we note that theleading coefficient of H ( r ) n ( x | λ ) is given by H ( r )0 ( λ ) = X n + ··· + n r =0 (cid:18) nn , n , · · · , n r (cid:19) H n ( λ ) · · · H n r ( λ )= H ( λ ) · · · H ( λ ) = 1 . (29)Thus, by (29), we see that H ( r ) n is a monic polynomial of degree n with coeffi-cients in Q ( λ ) . From (26), we have H (0) n ( x | λ ) = x n , for n ∈ Z + , (30)and ∂∂x H ( r ) n ( x | λ ) = ∂∂x ( H ( r ) ( λ ) + x ) n = nH ( r ) n − ( x | λ ) , ( r ≥ . (31)7t is not difficult to show that H ( r ) n ( x + 1 | λ ) − λH ( r ) n ( x | λ ) = (1 − λ ) H ( r − n ( x | λ ) . (32)Now, we note that { H ( r )0 ( x | λ ) , H ( r )1 ( x | λ ) , · · · , H ( r ) n ( x | λ ) } is also a good basisfor P n ( λ ) .Let us define the operator D as Df ( x ) = df ( x ) dx and let p ( x ) ∈ P n ( λ ) . Then p ( x ) can be written as p ( x ) = n X k =0 C k H ( r ) k ( x | λ ) . (33)From (9) and (32), we have △ λ H ( r ) n ( x | λ ) = H ( r ) n ( x + 1 | λ ) − λH ( r ) n ( x | λ ) = (1 − λ ) H ( r − n ( x | λ ) . (34)Thus, by (33) and (34), we get △ rλ p ( x ) = (1 − λ ) r n X k =0 C k H (0) k ( x | λ ) = (1 − λ ) r n X k =0 C k x k . (35)Let us take the k -th derivative of △ rλ p ( x ) in (35) .Then we have D k ( △ rλ p ( x )) = (1 − λ ) r n X l = k l !( l − k )! C l x l − k . (36)Thus, from (36), we have D k ( △ rλ p (0)) = (1 − λ ) r n X l = k l ! C l ( l − k )! 0 l − k = (1 − λ ) r k ! C k . (37)Thus, by(37), we get C k = D k ( △ rλ p (0))(1 − λ ) r k ! (38)= △ rλ ( D k p (0))(1 − λ ) r k ! = 1(1 − λ ) r k ! r X j =0 (cid:18) rj (cid:19) ( − λ ) ( r − j ) D k p ( j ) . Therefore, by (33) and (38), we obtain the following theorem.8 heorem 5.
For r ∈ Z + , let p ( x ) ∈ P n ( λ ) with p ( x ) = 1(1 − λ ) r n X k =0 C k H ( r ) k ( x | λ ) , ( C k ∈ Q ( λ )) . Then we have C k = 1(1 − λ ) r k ! r X j =0 (cid:18) rj (cid:19) ( − λ ) r − j D k p ( j ) . That is, p ( x ) = 1(1 − λ ) r n X k =0 ( r X j =0 k ! (cid:18) rj (cid:19) ( − λ ) r − j D k p ( j )) H ( r ) k ( 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 ( r )0 ( x | λ ) , H ( r )1 ( λ ) , · · · , H ( r ) n ( x | λ ) } as follows : H n ( x | λ ) = n X k =0 C k H ( r ) k ( x | λ ) , (39)where C k = 1(1 − λ ) r k ! r X j =0 (cid:18) rj (cid:19) ( − λ ) r − j D k p ( j ) , (40)and p ( k ) ( x ) = D k p ( x ) = n ( n − · · · ( n − k + 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 (cid:18) nk (cid:19) r X j =0 (cid:18) rj (cid:19) ( − 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 n X k =0 (cid:18) nk (cid:19) ( r X j =0 (cid:18) rj (cid:19) ( − λ ) r − j H n − k ( j | λ )) H ( r ) k ( x | λ ) . p ( x ) = H ( r ) n ( x | λ ) .Then we have p k ( x ) = n ( n − · · · ( n − k + 1) H ( r ) n − k ( x | λ ) (43)= n !( n − k )! H ( r ) n − k ( x | λ ) . From Theorem 1, we note that p ( x ) = H ( r ) n ( x | λ ) can be expressed as a linearcombination of H ( x | λ ) , H ( x | λ ) , · · · , H n ( x | λ ) : H ( r ) n ( x | λ ) = n X k =0 b k H k ( x | λ ) , (44)where b k = 1(1 − λ ) k ! { p k (1) − λp ( k ) (0) } (45)= n !(1 − λ ) k !( n − k )! { H ( r ) n − k (1 | λ ) − λH ( r ) n − k ( λ ) } . By (34) and (45), we get b k = 1(1 − λ ) (cid:18) nk (cid:19) H ( r − n − k ( λ ) . (46)Therefore, by (44) and (46), we obtain the following theorem. Theorem 7.
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