Symmetric Properties for Carlitz s Type (h, q)-twisted Tangent Polynomials Using Twisted (h, q)-tangent Zeta Function

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1 International Journal of Algebra, Vol 11, 2017, no 6, HIKARI Ltd, wwwm-hiaricom Symmetric Properties for Carlitz s Type h, -Twisted Tangent Polynomials Using Twisted h, -Tangent Zeta Function Cheon Seoung Ryoo Department of Mathematics, Hannam University, Daejeon 34430, Korea Copyright c 2017 Cheon Seoung Ryoo This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original wor is properly cited Abstract Our aim in this paper is to obtain special symmetric properties for Carlitz s type twisted h, -tangent polynomials We are going to find a symmetric identity for Carlitz s type twisted h, -tangent zeta function From property of the Carlitz s type twisted h, -tangent zeta function, we derive some symmetric properties of Carlitz s type twisted h, -tangent polynomials Mathematics Subject Classification: 11B68, 11S40, 11S80 Keywords: Carlitz s type twisted h, -tangent polynomials, Carlitz s type twisted h, -tangent zeta function, symmetric identities of Carlitz s type twisted h, -tangent polynomials, symmetric properties of Carlitz s type twisted h, -tangent zeta function 1 Introduction The area of the Euler, Bernoulli and Genocchi, tangent polynomials have been studied by many authors Those polynomials possess many interesting properties and are of great importance in pure mathematics, for example, number theory, mathematical analysis and in the calculus of finite differences Those polynomials also have various applications in other branches of sciencesee [1-8] Throughout this paper, we always mae use of the following notations:

2 256 C S Ryoo N = {1, 2, 3, } denotes the set of natural numbers, Z denotes the set of integers, R denotes the set of real numbers, C denotes the set of complex numbers Let be a complex number with < 1 and h Z Then we use the notation: [x] = 1 x 1, [x] = 1 x 1 + cf [1-4] Note that lim 1 [x] = x for any x The tangent polynomials are defined by F t, x = 2 e xt = e 2t + 1 T n x tn, 2t < π The tangent numbers, T n = T n 0, implies that they are rational numberssee [3] Definition 11 [4] Let r be a positive integer, and let ω be rth root of unity The twisted h, -tangent polynomials are defined as T h n,,ωx tn = 2 m ω m hm e [2m+x]t, where we use the technical method s notation by replacing symbolically T h,ω n x by T h n,,ωx, In the special case x = 0, T n,,ω0 h = T n,,ω h are called the nth twisted h, - tangent numbers The following elementary properties of the h, - tangent numbers T n,,ω h and polynomials T n,,ωx h are readily derived form Definition 11 see, for details, [4] We, therefore, choose to omit details involved Theorem 12 Let C with < 1 and ω be the rth root of unity Then we have T n,,ωx h = m ω m hm [x + 2m] n Theorem 13 Let C with < 1 and ω be the rth root of unity Then we have n n T n,,ωx h = [x] n l xl T h l,,ω l l=0 = x T h,ω + [x] n Theorem 14 Let x, y C Then we have n n T n,,ωx h + y = [x] n l xl T h l,,ω l y l=0

3 Symmetric properties for Carlitz s type twisted h, -tangent polynomials 257 Theorem 15 Property of complement Let n be a positive integer Then one has T h n,,ω 2 x = n ω n+h T n,,ωx h Definition 16 [4] For s C and Res > 0, the Hurwitz-type twisted h, -tangent zeta function is defined by ζ h,ωs, x = 2 n hn ω n [2n + x] s Note that ζ,ωs, h x is a meromorphic function on C Obverse that, if 1 and ω = 1, then ζ,ωs, h x = ζ T s, x which is the Hurwitz tangent zeta functionssee [3] Relation between ζ,ωs, h x and T h,,ω x is given by the following theorem Theorem 17 For N, we have ζ h,ω, x = T h,,ω x Observe that ζ,ω, h x function interpolates T h,,ω x numbers at non-negative integers 2 Symmetric properties about Hurwitz-type twisted h, -tangent zeta function In this section, we establish some interesting symmetric identities for Carlitz s type twisted h, -tangent polynomials by using Carlitz s type twisted h, - tangent zeta function Let, be any positive odd integers Our main result of symmetry of Carlitz s type twisted h, -tangent zeta function is given the following theorem, which is symmetric in and Theorem 21 Let s C with Res > 0 Then we have [ ] s = [ ] s i hi ω i ζ h,ω j hj ω j ζ h,ω s, x + 2i s, x + 2j

4 258 C S Ryoo Proof Observe that [xy] = [x] y[y] for any x, y C By substitute x+ 2i for x in Definition 16, replace by and replace ω by ω, respectively, we derive ζ h,ω s, x + 2w 1i n hw2n ω n = 2 [w 1 x + 2i + 2n] s = 2[ ] s n hw2n ω n [ x + 2 i + 2 n] s Since for any non-negative integer m and odd positive integer, there exist uniue non-negative integer r such that m = r + j with 0 j 1 Hence, this can be written as ζ h,ω s, x + 2w 1i = 2[ ] s = 2[ ] s r+ 0 j r=0 r+j h r+j ω r+j [2 r + j + x + 2 i] s w1r+j h r+j ω r+j [ 2r + x + 2 i + 2 j] s It follows from the above euation that [ ] s = 2[ ] s [ ] s i hi ω i ζ h,ω r=0 s, x + 2w 1i r+i+j h r+ i+ j ω r+ i+ j [ 2r + x + 2 i + 2 j] s 21 From the similar method, we can have that ζ h,ω s, x + 2w 2j n hw1n ω n = 2 [w 2 x + 2j + 2n] s = 2[ ] s n hw1n ω n [ x + 2 j + 2 n] s After some calculations in the above, we have [ ] s = 2[ ] s [ ] s j hj ω j ζ h,ω r=0 s, x + 2w 2j r+i+j h r+ i+ j ω r+ i+ j [ 2r + x + 2 i + 2 j] s 22

5 Symmetric properties for Carlitz s type twisted h, -tangent polynomials 259 Thus, we complete the proof of the theorem from 21 and 22 In Theorem 21, we get the following formulas for the twisted h, -tangent zeta function Corollary 22 Let = 1 in Theorem 21 Then we get ζ,ωs, h x = [ ] s j hj ω j ζ h,ω s, x + 2j Corollary 23 Let = 2, = 1 in Theorem 21 Then we have ζ h 2,ω s, x h ωζ h 2 2 2,ω s, x + 1 = [2] s 2 2 ζ,ωs, h x Theorem 24 Let, be any odd positive integer Then for non-negative integers n, one has [ ] n = [ ] n i hi ω i T h n,,ω i hi ω i T h n,,ω x + 2i x + 2i Proof By substituting T n,,ωx h for ζ,ωs, h x in Theorem 21 and Theorem 17, we can derive that [ ] n = [ ] n i hi ω i ζ h,ω i hi ω i T h n,,ω n, x + 2i x + 2i, and [ ] n = [ ] n j hj ω j ζ h,ω j hj ω j T h n,,ω n, x + 2w 2j x + 2w 2j Thus, we can complete the proof of the theorem from Theorem 21

6 260 C S Ryoo 3 Some symmetric identities about twisted h, - tangent polynomials In this section, we derive the symmetric results by using definition and theorem of twisted h, -tangent polynomials We obtain another result by applying the addition theorem for the twisted h, -tangent polynomialstheorem 13 Theorem 31 Let, be any odd positive integer Then we have n n [ ] l l [ ] n l T h n l,,ω xt h n,l,,ω l=0 n n = [ ] l l [ ] n l T h n l,,ω xt h n,l,,ω, l=0 where T h n,l,,ω = i h+2n 2li ω i [2i] l is called as the sums of powers Theorem 32 Let n, m be any non-negative integers Then we obtain that n n m T h m m+,,ω x + yn x [ x] n = T h +n,,ω yn+x [x] m Proof Observe that [x] u + x [y + 2m] u + v = [x + y + 2m] u + v [x] v 31 From Definition 11, we easily see that 2 m hm ω m e [x+y+2m]u+v = 2e [x]u+v Since [x + y] = [x] + x [y], we can find out e [x]v 2 = e [x]u 2 m hm ω m e [x+y+2m]u+v m hm ω m e x [y+2m] u+v Note that m hm ω m e [x+y+2m]u+v = = m hm ω m e x [y+2m] u+v T n,,ωx h u + vn + y T n+m,,ωx h + y um v n m! 32

7 Symmetric properties for Carlitz s type twisted h, -tangent polynomials 261 From the above euation, the left-hand side of 32 can be expressed as e [x]v m hm ω m e [x+y+2m]u+v = [x] n v n n n = Observe that m hm ω m e x [y+2m] u+v = T h n+m,,ωx + y um m! n x T h +m,,ω x + y[ x]n = v n u m v n m! nx T n,,ωy h u + vn n+mx T n+m,,ωy h vn u m m! 33 From the above euation, the right-hand side of 32 can be expressed as follows: e [x]u m hm ω m e x [y+2m] u+v = [x] m u m m! m m = n+mx T n+m,,ωy h vn +nx T h n+,,ω y[x]m u m m! v n u m m! 34 By comparing the coefficients of vn u m in 33 and 34, we assert that the m! theorem is right As another special case, we discover that for any non-negative integers m, n, m m n m n+ T h n n+,,ω x = n 1 m h ω T h m+,,ω 1 x 35 Theorem 33 Let, n, m be non-negative integers Then we have m + 1 m + n + 1 +n T h +n,,ω x n+1 n n + m + 1 +m+h w T h +m,,ω 1 x = 0

8 262 C S Ryoo Proof Since = m + 1 m for any non-negative integers and m then m + 1 m + n + 1 +n T h +n,,ω x = m n + 1 It follows from 3,5 that + m m + 1 m + 1 +n T h +n,,ω x m 1 +n T h +n,,ω x m + 1 m n + 1 +n T h +n,,ω x n+1 n + 1 = n+1 1 m h ω T h +m,,ω 1 x, and m m m + 1 +n T h +n,,ω 1 x n+1 n + 1 = n+1 m m h ω T h +m,,ω 1 x, 38 Thus, putting 37 and 38 to the right hand side of 36 gives the desired result and this completes the proof References [1] J Y Kang, C S Ryoo, On symmetric property for -Genocchi polynomials and zeta function, Int Journal of Math Analysis, , [2] Y He, Symmetric identities for Carlitz s -Bernoulli numbers and polynomials, Advances in Difference Euations, , [3] C S Ryoo, A note on the tangent numbers and polynomials, Adv Studies Theor Phys, ,

9 Symmetric properties for Carlitz s type twisted h, -tangent polynomials 263 [4] C S Ryoo, Carlitz s type twisted h, -tangent numbers and polynomials, Applied Mathematical Sciences, , [5] CS Ryoo, A numerical investigation on the zeros of the tangent polynomials, J Appl Math & Informatics, , [6] CS Ryoo, Differential euations associated with tangent numbers, J Appl Math & Informatics, , [7] CS Ryoo, A Note on the Zeros of the -Bernoulli Polynomials, J Appl Math & Informatics, , [8] CS Ryoo, Reflection Symmetries of the -Genocchi Polynomials, J Appl Math & Informatics, , Received: July 17, 2017; Published: August 2, 2017