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Some symmetry identities for the Apostol-type polynomials related to multiple alternating sums

Advances in Difference Equations20132013:32

https://doi.org/10.1186/1687-1847-2013-32

Received: 9 November 2012

Accepted: 22 January 2013

Published: 11 February 2013

Abstract

In recent years, symmetry properties of the Bernoulli polynomials and the Euler polynomials have been studied by a large group of mathematicians (He and Wang in Discrete Dyn. Nat. Soc. 2012:927953, 2012, Kim et al. in J. Differ. Equ. Appl. 14:1267-1277, 2008; Abstr. Appl. Anal. 2008, doi:11.1155/2008/914347, Yang et al. in Discrete Math. 308:550-554, 2008; J. Math. Res. Expo. 30:457-464, 2010). Luo (Integral Transforms Spec. Funct. 20:377-391, 2009), introduced the lambda-multiple power sum and proved the multiplication formulas for the Apostol-Bernoulli and Apostol-Euler polynomials of higher order. Ozarslan (Comput. Math. Appl. 2011:2452-2462, 2011), Lu and Srivastava (Comput. Math. Appl. 2011, doi:10.1016/j.2011.09.010.2011) gave some symmetry identities relations for the Apostol-Bernoulli and Apostol-Euler polynomials.

In this work, we prove some symmetry identities for the Apostol-Bernoulli and Apostol-Euler polynomials related to multiple alternating sums.

AMS Subject Classification: 11F20, 11B68, 11M35, 11M41.

Keywords

Bernoulli polynomialsEuler polynomialsApostol-Bernoulli polynomialsApostol-Euler polynomialssymmetry relationpower sumsalternating sums

1 Introduction, definitions and notations

The generalized Bernoulli polynomials B n ( α ) ( x ) of order α N 0 and the generalized Euler polynomials E n ( α ) ( x ) of order α N 0 , each of degree n as well as in α, are defined respectively by the following generating functions [13]:
(1)
(2)
The generalized Apostol-Bernoulli polynomials B n ( α ) ( x ; λ ) of order α N 0 and the generalized Apostol-Euler polynomials E n ( α ) ( x ; λ ) of order α N 0 are defined respectively by the following generating functions [3]:
(3)
(4)
Recently, Garg et al. in [4] introduced the following generalization of the Hurwitz-Lerch zeta function Φ ( z , s , a ) :
Φ μ , ν ( ρ , σ ) ( z , s , a ) : = n = 0 ( μ ) ρ n ( ν ) σ n z n ( n + a ) s
( μ C , a, υ C Z 0 , ρ , σ R + , ρ < σ when s , z C ( | z | < 1 ) ; ρ = σ and ( s μ + ν ) > 0 when | z | = 1 ). It is obvious that
Φ μ , 1 ( 1 , 1 ) ( z , s , a ) = Φ μ ( z , s , a ) = n = 0 ( μ ) n n ! z n ( n + a ) s
(5)

(for details on this subject, see [35]).

The multiple power sums and the λ-multiple alternating sums are defined by Luo [6] as follows:
(6)
(7)
From (6) and (7), we have
( 1 λ m e m t 1 λ e t ) l = λ ( l ) n = 0 { p = 0 n ( n p ) ( l ) n p S ( l ) ( m ; λ ) } t n n !
(8)
and
( 1 + ( 1 ) m + 1 ( λ e t ) m 1 + λ e t ) l = λ ( l ) n = 0 { p = 0 n ( n p ) ( l ) n p T ( l ) ( m ; λ ) } t n n !
(9)

(see [6]).

From (8) and (9), for l = 1 , we have respectively
(10)
(11)

Symmetry property and some recurrence relations of the Bernoulli polynomials, Euler polynomials, Apostol-Bernoulli polynomials and Apostol-Euler polynomials have been investigated by a lot of mathematicians [124]. Firstly, Yang [22] proved symmetry relation for Bernoulli polynomials. Wang et al. in [1, 20, 21] gave some symmetry relations for the Apostol-Bernoulli polynomials. Kim in [8, 10, 11, 14, 15] proved symmetric identities for the Bernoulli polynomials and Euler polynomials. Luo in [6, 17] gave multiplication formulas for the Apostol-Bernoulli and Apostol-Euler polynomials. Also, he defined λ-power sums. Srivastava et al. [2, 3, 5] proved some theorems and relations for these polynomials. They proved some symmetry identities for these polynomials.

In this work, we give some symmetry identities for the Apostol-type polynomials related to multiple alternating sums.

2 Symmetry identities for the Apostol-Bernoulli polynomials

We will prove the following theorem for the Apostol-Euler polynomials, which are symmetric in a and b.

Theorem 2.1 There is the following relation between Apostol-Bernoulli polynomials and the Hurwitz-Lerch zeta function Φ ( z , s , a ) :
(12)
Proof Let f ( t ) = t α 1 e a b x t ( 1 λ a b e a b t ) e a b y t ( 1 λ a e a t ) α ( 1 λ b e b t ) α . Then
f ( t ) = 1 b α 1 e a b x t ( 1 λ a e a t ) α ( 1 λ a b e a b t 1 λ b e b t ) ( b t 1 λ b e b t ) α 1 e a b y t .
From (3) and (10), we write
f ( t ) = ( 1 ) α 1 b α 1 β = 0 ( β + α 1 β ) λ α β e a t ( β + b x ) 1 λ b r = 0 p = 0 r ( r p ) ( 1 ) r p S p ( a ; λ b ) b r t r r ! × k = 0 B k ( α 1 ) ( a y ; λ b ) b k t k k ! ,
where | log λ + t | < min ( 2 π a , 2 π b ) . After the Cauchy product, we have
= n = 0 { s = 0 n ( n s ) b s + 1 α a n s ( 1 ) α 1 k = 0 s ( s k ) p = 0 r ( r p ) ( 1 ) r p λ b × S p ( a ; λ b ) B k ( α 1 ) ( a y ; λ b ) β = 0 ( β + α 1 β ) λ α β ( β + b x ) s n } t n n ! .
In a similar manner,
f ( t ) = t α 1 e a b y t ( 1 λ a b e a b t ) e a b x t ( 1 λ b e b t ) α ( 1 λ a e a t ) α = 1 a α 1 e a b x t ( 1 λ b e b t ) α ( 1 λ a b e a b t 1 λ a e a t ) ( a t 1 λ a e a t ) α 1 e a b y t .
From (3) and (10), we write
f ( t ) = ( 1 ) α 1 a α 1 β = 0 ( β + α 1 β ) λ α β e b t ( β + a x ) 1 λ a r = 0 p = 0 r ( r p ) ( 1 ) r p S p ( b ; λ a ) a r t r r ! × k = 0 B k ( α 1 ) ( b y ; λ a ) a k t k k ! .
Since | log λ + t | < min ( 2 π a , 2 π b ) , after the Cauchy product, we have
= n = 0 { s = 0 n ( n s ) a s + 1 α b n s ( 1 ) α 1 k = 0 s ( s k ) p = 0 r ( r p ) ( 1 ) r p λ a × S p ( b ; λ a ) B k ( α 1 ) ( b y ; λ a ) β = 0 ( β + α 1 β ) λ b β ( β + a y ) s n } t n n ! .

Compressing to coefficients t n n ! and by using (5), we prove the theorem. □

Theorem 2.2 For all a , b , m N , n N 0 , we have the following symmetry identity:
(13)
Proof Let h ( t ) = t 2 m + 1 e a b x t ( 1 λ a b e a b t ) m e a b y t ( 1 λ a e a t ) m + 1 ( 1 λ b e b t ) m + 1 . Then
h ( t ) = 1 a m + 1 b ( a t 1 λ a e a t ) m + 1 e a b x t ( 1 λ a b e a b t 1 λ b e b t ) m ( b t 1 λ b e b t ) e a b y t .
From (3) and (8), we have
h ( t ) = ( 1 ) m a m + 1 b n = 0 { r = 0 n ( n r ) B n r ( m + 1 ) ( b x ; λ a ) a n r λ m b k = 0 r ( r k ) × p = 0 k ( k p ) ( l ) k p S p ( m ) ( a ; λ b ) B r k ( a y ; λ b ) b r } t n n ! .
In a similar manner,
h ( t ) = ( 1 ) m b m + 1 a n = 0 { r = 0 n ( n r ) B n r ( m + 1 ) ( a y ; λ b ) b n r λ m a k = 0 r ( r k ) × p = 0 k ( k p ) ( l ) k p S p ( m ) ( b ; λ a ) B r k ( a x ; λ a ) a r } t n n ! .

Comparing the coefficients of t n n ! , we proved the theorem. □

Corollary 2.3 We put a = b = λ = 1 in (13). We have
k = 0 n ( n k ) B k ( m + 1 ) ( y ) B n k ( x ) = k = 0 n ( n k ) B k ( m + 1 ) ( x ) B n k ( y ) .

3 Some symmetry identities for the Apostol-Euler polynomials

Theorem 3.1 Let a and b be positive integers with the same parity. Then
(14)
Proof Let h ( t ) = 2 e a b x t λ a e a t + 1 1 + ( 1 ) a + 1 ( λ b e b t ) a λ b e b t + 1 . From (4) and (9) for l = 1 , we have
h ( t ) = k = 0 E k ( b x ; λ a ) a k t k k ! 1 λ a l = 0 p = 0 l ( l p ) ( 1 ) l p T p ( a ; λ b ) b l t l l ! = n = 0 ( k = 0 n ( n k ) E k ( b x ; λ a ) λ b a k b n k p = 0 n k ( n k p ) ( 1 ) n k p T n k ( a ; λ b ) ) t n n ! .
Since ( 1 ) a + 1 = ( 1 ) b + 1 , the expression for h ( t ) = 2 e a b x t λ b e b t + 1 1 + ( 1 ) b + 1 ( λ a e a t ) b λ a e a t + 1 is symmetric in a and b. Therefore, we obtain the following power series for h ( t ) by symmetry:
h ( t ) = n = 0 ( k = 0 n ( n k ) E k ( a x ; λ b ) λ a b k a n k p = 0 n k ( n k p ) ( 1 ) n k p T n k ( b ; λ a ) ) t n n ! .

Equating the coefficient of t n n ! in the two expressions for h ( t ) gives us the desired result. □

Theorem 3.2 Let a and b be positive integers with the same parity. Then
(15)
Proof Let k ( t ) = 2 α + 2 e a b x t ( 1 + ( 1 ) a + 1 ( λ a b e a b t ) ) α ( λ a e a t + 1 ) α + 1 ( λ b e b t + 1 ) α + 1 e a b y t . From (4) and (9), we write
k ( t ) = ( 2 λ a e a t + 1 ) ( α + 1 ) e a b x t ( 1 + ( 1 ) a + 1 ( λ a b e a b t ) λ b e b t + 1 ) α ( 2 λ b e b t + 1 ) e a b y t = n = 0 E n ( α + 1 ) ( b x ; λ a ) a n t n n ! 1 λ b α n = 0 n = 0 p ( p n ) ( α ) n p T p ( α ) ( a ; λ b ) b n t n n ! × n = 0 E n ( a y ; λ b ) b n t n n ! = n = 0 { s = 0 n ( n s ) E n s ( α + 1 ) ( b x ; λ a ) λ ( b α ) a n s × k = 0 s ( s k ) p = 0 k ( k p ) ( α ) k p T k ( α ) ( a ; λ b ) E s k ( a y ; λ b ) b s } t n n ! .

Since ( 1 ) a + 1 = ( 1 ) b + 1 , the expression for h ( t ) is symmetric in a and b.

In a similar manner, we have
k ( t ) = n = 0 { s = 0 n ( n s ) E n s ( α + 1 ) ( a y ; λ b ) λ ( a α ) b n s × k = 0 s ( s k ) p = 0 k ( k p ) ( α ) k p T k ( α ) ( b ; λ a ) E s k ( b x ; λ b ) a s } t n n ! .

Equating the coefficient of t n n ! in the two expressions for k ( t ) gives us the desired result. □

Theorem 3.3 Let p, l, a, b and n be positive integers and a, b be of the same parity. Then
(16)
Proof Let k ( t ) = a t e a n t λ a e a t 1 1 + ( 1 ) b ( λ b e b t ) a λ a e a t + 1 . From (3) and (10), we have
k ( t ) = a t e a n t λ a e a t 1 1 + ( 1 ) b + 1 ( λ b e b t ) a λ a e a t + 1 = n = 0 B n ( n ; λ a ) a n t n n ! λ α n = 0 n = 0 p ( p n ) ( 1 ) n p T p ( b ; λ a ) a n t n n ! = n = 0 ( l = 0 n ( n l ) B n l ( n ; λ a ) a n l λ a p = 0 l ( l p ) ( 1 ) l p T p ( b ; λ a ) a l ) t n n ! .
On the other hand, we write the function k ( t ) as
k ( t ) = 1 2 2 a e n 2 ( 2 a t ) λ 2 a e 2 a t 1 + ( 1 ) b + 1 λ a b 2 a t e 2 a t ( n + b 2 ) 2 ( λ 2 a e 2 a t 1 ) = 1 2 n = 0 B n ( n 2 ; λ 2 a ) 2 n a n t n n ! + ( 1 ) b + 1 λ a b 2 n = 0 B n ( b + n 2 ; λ 2 a ) 2 n a n t n n ! = n = 0 ( 2 n l a n ( B n ( n 2 ; λ 2 a ) + ( 1 ) b + 1 λ a b 2 B n ( b + n 2 ; λ n ) ) ) t n n ! .

Equating the coefficient of t n n ! , we obtain (16). □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

This paper was supported by the Scientific Research Project Administration of Akdeniz University.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Akdeniz University

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