Identities of symmetry for (h;q)-extension of higher-order Euler polynomials
aa r X i v : . [ m a t h . N T ] D ec IDENTITIES OF SYMMETRY FOR ( h, q ) − EXTENSION OFHIGHER-ORDER EULER POLYNOMIALS
DAE SAN KIM, TAEKYUN KIM, JONG JIN SEO
Abstract.
In this paper, we study some symmetric properties of the multiple q − Eulerzeta function. From these properties, we derive several identities of symmetry for the( h, q ) − extension of higher-order Euler polynomials, which is an answer to a part of openquestion in [7].
1. Introduction
Let C be the complex number field. We assume that q ∈ C with | q | < q − numberis defined by [ x ] q = − q x − q . Note that lim q → [ x ] q = x. As is well known, the higher-orderEuler polynomials E ( r ) n ( x ) are defined by the generating function to be F ( r ) ( x, t ) = (cid:18) e t + 1 (cid:19) r e xt = ∞ X n =0 E ( r ) n ( x ) t n n ! , (see [4], [16]), (1.1)where | t | < π .When x = 0 , E ( r ) n = E ( r ) n (0) are called the Euler numbers of order r. Recently, the secondauthor defined the ( h, q ) − extension of higher-order Euler polynomials, which is given by thegenerating function to be F ( h,r ) q ( x, t ) =[2] rq ∞ X m , ··· ,m r =0 q P rj =1 ( h − j +1) m j ( − P rj =1 m j e [ m + ··· + m r + x ] q t = ∞ X n =0 E ( h,r ) n,q ( x ) t n n ! , (see [6], [8]), (1.2)where h ∈ Z and r ∈ Z ≥ . Note that lim q → F ( h,r ) q ( x, t ) = ( e t +1 ) r e xt = P ∞ n =0 E ( r ) n ( x ) t n n ! . By (1 . F ( h,r ) q ( t, x ) = [2] rq ∞ X m =0 (cid:18) m + r − m (cid:19) q ( − q h − r +1 ) m e [ m + x ] q t , (see [6], [8]), (1.3) Key words and phrases. multiple q − Euler zeta function, ( h, q ) − extension of higher-order Eulerpolynomials. . 1 Dae San Kim, Taekyun Kim, Jong Jin Seo where (cid:0) xm (cid:1) q = [ x ] q [ x − q [ x − q ··· [ x − m +1] q [ m ] q ! . From (1 . E ( h,r ) n,q ( x ) = [2] rq (1 − q ) n n X l =0 (cid:18) nl (cid:19) ( − q x ) l ( − q h − r + l +1 : q ) r =[2] rq ∞ X m =0 (cid:18) m + r − m (cid:19) q ( − q h − r +1 ) m [ m + x ] nq , (see [6]), (1.4)where ( x : q ) n = (1 − x )(1 − xq ) · · · (1 − xq n − ) . In [6] and [8], the second author constructed the multiple q − Euler zeta function which in-terpolates the ( h, q ) − extension of higher-order Euler polynomials at negative integers asfollows : ζ ( h ) q,r ( s, x ) = 1Γ( s ) Z ∞ F ( h,r ) q ( x, t ) t s − dt =[2] rq ∞ X m , ··· ,m r =0 ( − m + ··· + m r q P rj =1 ( h − j +1) m j [ m + · · · + m r + x ] sq , (see [6]), (1.5)where h, s ∈ C , x ∈ R with x = 0 , − , − , · · · . From (1 . ζ ( h ) q,r ( s, x ) = [2] rq ∞ X m =0 (cid:18) m + r − m (cid:19) q ( − q h − j +1 ) m m + x ] sq . (1.6)Using the Cauchy residue theorem and Laureut series in (1 . Lemma 1.1.
For n ∈ Z ≥ and h ∈ Z , we have ζ ( h ) q,r ( − n, x ) = E ( h,r ) n,q ( x ) , (see [6], [8]) . In [7], the second author introduced many identities of symmetry for Euler and Bernoullipolynomials which are derived from the p -adic integral expression of the generating functionand suggested an open problem about finding identities of symmetry for the Carlitz’s type q -Euler numbers and polynomials.When x = 0, E ( h,r ) n,q = E ( h,r ) n,q (0) are called the ( h, q )-Euler numbers of order r .From (1 .
3) and (1 . E ( h,r ) n,q ( x ) = ( q x E ( h,r ) q + [ x ] q ) n = n X l =0 (cid:18) nl (cid:19) q lx E ( h,r ) l,q [ x ] n − lq , (1.7)with the usual convention about replacing ( E ( h,r ) q ) n by E ( h,r ) n,q . Recently, Y. Simsek introduced recurrence symmetric identities for ( h, q )-Euler polynomialsand alternating sums of powers of consecutive ( h, q )- integers (see [16]).In this paper, we investigate some symmetric properties of the multiple q -Euler zeta function.From our investigation, we give some new identities of symmetry for the ( h, q )-extension ofhigher-order Euler polynomials, which is an answer to a part of open question in [7]. dentities of symmetry for ( h, q ) − extension of higher-order Euler polynomials 3
2. Identities for ( h, q ) − extension of higher-order Euler Polynomials In this section, we assume that h ∈ Z and a, b ∈ N with a ≡ mod
2) and b ≡ mod . Now, we observe that1[2] rq a ζ ( h ) q a ,r (cid:18) s, bx + b ( j + · · · + j r ) a (cid:19) = ∞ X m , ··· ,m r =0 ( − m + ··· + m r q a P rj =1 ( h − j +1) m j [ m + · · · + m r + bx + b ( j + ··· + j r ) a ] sq a = [ a ] sq ∞ X m , ··· ,m r =0 q a P rj =1 ( h − j +1) m j ( − m + ··· + m r [ b P rl =1 j l + abx + a P rl =1 m l ] sq = [ a ] sq ∞ X m , ··· ,m r =0 b − X i , ··· ,i r =0 ( − P rl =1 ( i l + bm l ) q a P rj =1 ( h − j +1)( i j + m j b ) [ ab ( x + P rl =1 m l ) + b P rl =1 j l + a P rl =1 i l ] sq . (2.1)Thus, by (2 . b ] sq [2] rq a a − X j , ··· ,j r =0 ( − P rl =1 j l q b P rl =1 ( h − l +1) j l ζ ( h ) q a ,r (cid:18) s, bx + b ( j + · · · + j r ) a (cid:19) =[ a ] sq [ b ] sq a − X j , ··· ,j r =0 b − X i , ··· ,i r =0 ∞ X m , ··· ,m r =0 ( − P rl =1 ( i l + j l + m l ) q a P rl =1 ( h − l +1) i l + b P rl =1 ( h − l +1) j l [ ab ( x + P rl =1 m l ) + P rl =1 ( bj l + ai l )] sq × q ab P rl =1 m l . (2.2)By the same method as (2 . a ] sq [2] rq b b − X j , ··· ,j r =0 ( − P rl =1 j l q a P rl =1 ( h − l +1) j l ζ ( h ) q b ,r (cid:18) s, ax + a ( j + · · · + j r ) b (cid:19) =[ b ] sq [ a ] sq b − X j , ··· ,j r =0 a − X i , ··· ,i r =0 ∞ X m , ··· ,m r =0 ( − P rl =1 ( i l + j l + m l ) q a P rl =1 ( h − l +1) j l + b P rl =1 ( h − l +1) i l [ ab ( x + P rl =1 m l ) + P rl =1 ( bi l + aj l )] sq × q ab P rl =1 m l . (2.3)Therefore, by(2 .
2) and (2 . Theorem 2.1.
For a, b ∈ N , with a ≡ mod and b ≡ mod , we have [2] rq b [ b ] sq a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l ζ ( h ) q a ,r (cid:18) s, bx + b ( j + · · · + j r ) a (cid:19) = [2] rq a [ a ] sq b − X j , ··· ,j r =0 ( − j + ··· + j r q a P rl =1 ( h − l +1) j l ζ ( h ) q b ,r (cid:18) s, ax + a ( j + · · · + j r ) b (cid:19) . From Lemma 1 . .
1, we can derive the following theorem.
Dae San Kim, Taekyun Kim, Jong Jin Seo
Theorem 2.2.
For n ∈ Z ≥ and a, b ∈ N , with a ≡ mod and b ≡ mod , we have [2] rq b [ a ] nq a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l E ( h,r ) n,q a (cid:18) bx + b ( j + · · · + j r ) a (cid:19) = [2] rq a [ b ] nq b − X j , ··· ,j r =0 ( − j + ··· + j r q a P rl =1 ( h − l +1) j l E ( h,r ) n,q b (cid:18) ax + a ( j + · · · + j r ) b (cid:19) . By (1 . E ( h,k ) n,q ( x + y ) = ( q x + y E ( h,k ) q + [ x + y ] q ) n = ( q x + y E ( h,k ) q + q x [ y ] q + [ x ] q ) n = (cid:16) q x ( q y E ( h,k ) q + [ y ] q ) + [ x ] q (cid:17) n = n X i =0 (cid:18) ni (cid:19) q ix E ( h,k ) i,q ( y )[ x ] n − iq . (2.4)Therefore, by (2 . Proposition 2.3.
For n ≥ , we have E ( h,k ) n,q ( x + y ) = n X i =0 (cid:18) ni (cid:19) q ix E ( h,k ) i,q ( y )[ x ] n − iq = n X i =0 (cid:18) ni (cid:19) q ( n − i ) x E ( h,k ) n − i,q ( y )[ x ] iq . From Proposition2 .
3, we note that a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l E ( h,r ) n,q a (cid:18) bx + b ( j + · · · + j r ) a (cid:19) = a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l n X i =0 (cid:18) ni (cid:19) q ia (cid:16) b ( j ··· + jr ) a (cid:17) E ( h,r ) i,q a ( bx ) × (cid:20) b ( j + · · · + j r ) a (cid:21) n − iq a = a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l n X i =0 (cid:18) ni (cid:19) q ( n − i ) b ( j + ··· + j r ) E ( h,r ) n − i,q a ( bx ) × (cid:20) b ( j + · · · + j r ) a (cid:21) iq a = n X i =0 (cid:18) ni (cid:19) (cid:18) [ b ] q [ a ] q (cid:19) i E ( h,r ) n − i,q a ( bx ) a − X j , ··· j r =0 ( − j + ··· + j r q b P rl =1 ( h + n − l − i +1) j l [ j + · · · + j r ] iq b = n X i =0 (cid:18) ni (cid:19) (cid:18) [ b ] q [ a ] q (cid:19) i E ( h,r ) n − i,q a ( bx ) S ( h,r ) n,i,q b ( a ) , (2.5) where S ( h,r ) n,i,q ( a ) = a − X j , ··· j r =0 ( − j + ··· + j r q P rl =1 ( h + n − l − i +1) j l [ j + · · · + j r ] iq . (2.6) dentities of symmetry for ( h, q ) − extension of higher-order Euler polynomials 5 By (2 . rq b [ a ] nq a − X j , ··· ,j r =0 ( − j + ··· + j r q b P rl =1 ( h − l +1) j l E ( h.r ) n,q a (cid:18) bx + b ( j + · · · + j r ) a (cid:19) = [2] rq b n X i =0 (cid:18) ni (cid:19) [ a ] n − iq [ b ] iq E ( h,r ) n − i,q a ( bx ) S ( h,r ) n,i,q b ( a ) . (2.7)By the same method as (2 . rq a [ b ] nq b − X j , ··· j r =0 ( − j + ··· + j l q a P rl =1 ( h − l +1) j l E ( h,r ) n,q b (cid:18) ax + a ( j + · · · + j r ) b (cid:19) = [2] rq a n X i =0 (cid:18) ni (cid:19) [ b ] n − iq [ a ] iq E ( h,r ) n − i,q b ( ax ) S ( h,r ) n,i,q a ( b ) . (2.8)Therefore, by (2 .
7) and (2 . Theorem 2.4.
For a, b ∈ N with a ≡ mod and b ≡ mod , n ∈ Z ≥ ,let S ( h,r ) n,i,q ( a ) = a − X j , ··· ,j r =0 ( − j + ··· + j r q P rl =1 ( h + n − l − i +1) j l [ j + · · · + j r ] iq . Then we have [2] rq b n X i =0 (cid:18) ni (cid:19) [ a ] n − iq [ b ] iq E ( h,r ) n − i,q a ( bx ) S ( h,r ) n,i,q b ( a ) = [2] rq a n X i =0 (cid:18) ni (cid:19) [ b ] n − iq [ a ] iq E ( h,r ) n − i,q b ( ax ) S ( h,r ) n,i,q a ( b ) . It is not difficult to show that[ x + y + m ] q ( u + v ) − [ x ] q v = [ x ] q u + q x [ y + m ] q ( u + v ) . (2.9)From (2 . e [ x ] q u ∞ X m , ··· ,m r =0 q P rj =1 ( h − j +1) m j ( − P rj =1 m j e [ m + ··· + m r + y ] q q x ( u + v ) = e − [ x ] q v ∞ X m , ··· ,m r =0 q P rj =1 ( h − j +1) m j ( − P rj =1 m j e [ x + y + m + ··· + m r ] q ( u + v ) . (2.10) Dae San Kim, Taekyun Kim, Jong Jin Seo
The left hand side of (2 .
10) multiplied by [2] rq is given by[2] rq e [ x ] q u ∞ X m , ··· ,m r =0 q P rj =1 ( h − j +1) m j ( − P rj =1 m j e [ m + ··· + m r + y ] q q x ( u + v ) = e [ x ] q u ∞ X n =0 q nx E ( h,r ) n,q ( y ) 1 n ! ( u + v ) n = ∞ X l =0 [ x ] lq u l l ! ! ∞ X n =0 q nx E ( h,r ) n,q ( y ) n X k =0 u k k !( n − k )! v n − k ! = ∞ X l =0 [ x ] lq u l l ! ! ∞ X k =0 ∞ X n =0 q ( n + k ) x E ( h,r ) n + k,q ( y ) u k k ! v n n ! ! = ∞ X m =0 ∞ X n =0 m X k =0 (cid:18) mk (cid:19) q ( n + k ) x E ( h,r ) n + k,q ( y )[ x ] m − kq ! u m m ! v n n ! (2.11)The right hand side of (2 .
10) multiplied by [2] rq is given by[2] rq e − [ x ] q v ∞ X m , ··· ,m r =0 q P rj =1 ( h − j +1) m j ( − P rj =1 m j e [ x + y + m + ··· + m r ] q ( u + v ) = e − [ x ] q v ∞ X n =0 E ( h,r ) n,q ( x + y ) 1 n ! ( u + v ) n = ∞ X l =0 ( − [ x ] q ) l l ! v l ! ∞ X m =0 ∞ X k =0 E ( h,r ) m + k,q ( x + y ) u m m ! v k k ! ! = ∞ X n =0 ∞ X m =0 n X k =0 (cid:18) nk (cid:19) E ( h,r ) m + k,q ( x + y )( − [ x ] q ) n − k ! u m m ! v n n != ∞ X n =0 ∞ X m =0 n X k =0 (cid:18) nk (cid:19) E ( h,r ) m + k,q ( x + y ) q ( n − k ) x [ − x ] n − kq ! u m m ! v n n ! . (2.12)Therefore, by (2 . .
11) and (2 . Theorem 2.5.
For m, n ≥ , we have m X k =0 (cid:18) mk (cid:19) q ( n + k ) x E ( h,r ) n + k,q ( y )[ x ] m − kq = n X k =0 (cid:18) nk (cid:19) E ( h,r ) m + k,q ( x + y ) q ( n − k ) x [ − x ] n − kq . Remark. Recently, several authors have studied ( h, q ) − extension of Bernoulli and Eulerpolynomials (see[1]-[5], [9]-[17]) . References
1. S. Araci, J. Seo, D. Erdal,
New construction weighted ( h, q ) − Genocchi numbers and polynomials relatedto zeta type functions , Discrete Dyn. Nat. Soc. (2011), Art. ID 487490, 7 pp.2. I. N. Cangul, H. Ozden, Y. Simsek,
Generating functions of the ( h, q ) − extension of twisted Eulerpolynomials and numbers . Acta Math. Hungar. 120 (2008), no. 3, 281-299.3. M. Cenkci, The p − adic generalized twisted ( h, q ) − Euler-l-function and its applications , Adv. Stud. Con-temp. Math. 15 (2007), no. 1, 37-47. dentities of symmetry for ( h, q ) − extension of higher-order Euler polynomials 7
4. D. V. Dolgy, D. J. Kang, T. Kim, B. Lee,
Some new identities on the twisted ( h, q ) − Euler numbers and q − Bernstein polynomials , J. Comput. Anal. Appl. 14(2012), no. 5, 974-984.5. D. S. Kim, N. Lee, J. Na, K. H. Park,
Identities of symmetry for higher-order Euler polynomials in threevariables (II) , J. Math. Anal. Appl. 379 (2011), no. 1, 388-400.6. T. Kim,
New approach to q − Euler polynomials of higher order , Russ. J. Math. Phys. 17 (2010), no. 2,218-225.7. T. Kim,
Symmetry p − adic invariant integral on Z p for Bernoulli and Euler polynomials , J. DifferenceEqu. Appl. 14 (2008), no. 12, 1267-1277.8. T. Kim, q-Euler numbers and polynomials associated with p − adic q − integrals , J. Nonlinear Math. Phys.14 (2007), no. 1, 15-27.9. T. Kim, A family of ( h, q ) − zeta function associated with ( h, q ) − Bernoulli numbers and polynomials , J.Comput. Anal. Appl. 14 (2012), no. 3, 402-409.10. T. Mansour, A.Sh. Shabani,
Generalization of some inequalities for the ( q , · · · , q s ) − gamma function ,Matematiche (Catania) 67 (2012), no. 2, 119-130.11. B. Kurt, Some formulas for the multiple twisted ( h, q ) − Euler polynomials and numbers , Appl. Math.Sci. (Ruse) 5 (2011), no. 25-28, 1263-1270.12. H. Ozden, Y. Simsek,
Interpolation function of the ( h, q ) − extension of twisted Euler numbers , Comput.Math. Appl. 56 (2008), no. 4, 898-908.13. H. Ozden, I. N. Cangul, Y. Simsek, Remarks on sum of procucts of ( h, q ) − twisted Euler polynomialsand numbers , J. Inequal. Appl.(2008). Art. ID 816129, 8 pp.14. K. H. Park, On interpolation functions of the generalized twisted ( h, q ) − Euler polynomials , J.Inequal.Appl. (2009), Art. ID 946569,17 pp.15. S.-H. Rim, S.-J. Lee,
Some identities on the twisted ( h.q ) − Genocchi numbers and polynomials associatedwith q − Bernstein polynomials , Int. J. Math. Math. Sci. (2011), Art. ID 482840, 8 pp.16. Y. Simsek,
Complete sum of products of ( h.q ) − extension of Euler polynomials and numbers , J. DifferenceEqu. Appl. 16 (2010), no. 11, 1331-1348.17. Y. Simsek, Twisted ( h.q ) − Bernoulli numbers and polynomials related to twisted ( h.q ) − zeta functionand L-function , J. Math. Anal. Appl. 324 (2006), no.2, 790-804.
1, Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
E-mail address : [email protected]
2, Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
E-mail address : [email protected]
3, Department of Applied Mathematics, Pukyong National University, Busan 608-737, Republicof Korea
E-mail address ::