Analytic Continuation of q-Euler numbers and polynomials 8 0 0 Taekyun Kim1 2 n a 1 School of Electrical Engineering and Computer Science , J Kyungpook National University, Taegu 702-701, S. Korea 3 e-mail: [email protected] ] T N Abstract Inthispaperwestudythattheq-Eulernumbersandpolynomialsareanalyt- h. icallycontinued toEq(s). A new formulaforthe Euler’sq-Zeta function ζE,q(s)interms of t nested seriesof ζE,q(n)is derived. Finallyweintroduce the new concept of the dynamics of a m analyticallycontinued q-Eulernumbersandpolynomials. 2000 Mathematics Subject Classification-11B68,11S40 [ Key words-q-Bernoullipolynomial,q-RiemannZetafunction 1 v 1 Introduction 0 8 4 Throughout this paper, Z,R and C will denote the ring of integers, the field of 0 real numbers and the complex numbers, respectively. . 1 Whenonetalksofq-extension,qisvariouslyconsideredasanindeterminate, 0 a complex numbers or p-adic numbers. Throughout this paper, we will assume 8 that q ∈C with |q|<1. The q-symbol [x] denotes [x] = 1−qx, (see [1-16]). 0 q q 1−q In this paper we study that the q-Euler numbers and polynomials are an- : v alytically continued to E (s). A new formula for the Euler’s q-Zeta function q i X ζ (s) in terms of nested series of ζ (n) is derived. Finally we introduce the E,q E,q r new concept of the dynamics of analytically continued q-Euler numbers and a polynomials. 2 Generating q-Euler polynomials and numbers For h∈Z, the q-Euler polynomials were defined as ∞ ∞ E (x,h|q) n tn =[2] (−1)nqhne[n+x]qt, (2.1) q n! nX=0 nX=0 1 for x,q ∈C, cf. [1,7]. In the special case x=0, E (0,h|q)=E (h|q) are called n n q-Euler numbers, cf. [1,2,3,4]. By (2.1), we easily see that n [2] n 1 E (x,h|q)= q (−1)l qlx, cf.[7,8], (2.2) n (1−q)n (cid:18)l(cid:19) 1+ql+h Xl=0 where n is binomial coefficient. From (2.1), we derive j (cid:0) (cid:1) E (x,h|q)=(qxE(h|q)+[x] )n n,q q with the usual convention of replacing En(h|q) by E (h|q). In the case h = 0, n E (x,0|q) will be symbolically written as E (x). Let G (x,t) be generating n n,q q function of q-Euler polynomials as follows: ∞ tn G (x,t)= E (x) . (2.3) q n,q n! nX=0 Then we easily see that ∞ G (x,t)=[2] (−1)ke[k+x]qt. (2.4) q q kX=0 For x=0,E =E (0) will be called q-Euler numbers. n,q n,q From(2.3),(2.4),we easilyderivethe following: Fork(= even)andn∈Z , + we have k−1 E (k)−E =[2] (−1)l[l]n. (2.5) n,q n,q q q Xl=0 For k(= odd) and n∈Z , we have + k−1 E (k)+E =[2] (−1)l[l]n. (2.6) n,q n,q q q Xl=0 By (2.4), we easily see that m m E (x)= qxlE [x]m−l. (2.7) m,q (cid:18)l(cid:19) l,q q Xl=0 From (2.5), (2.6), and (2.7), we derive k−1 k−1 n [2] (−1)l−1[l]n =(qkn−1)E + qklE [k]n−l, (2.8) q q n,q (cid:18)l(cid:19) l,q q Xl=0 Xl=0 where k(= even) ∈N. For k(= odd) and n∈Z , we have + k−1 k−1 n [2] (−1)l[l]n =(qkn+1)E + qklE [k]n−l. (2.9) q q n,q (cid:18)l(cid:19) l,q q Xl=0 Xl=0 2 3 q-Euler zeta function It was known that the Euler polynomials are defined as ∞ 2 E (x) ext = n tn, |t|<π, cf. [1-16]. (3.1) et+1 n! nX=0 For s∈C,x∈R with 0≤x<1, define ∞ ∞ (−1)n (−1)n ζ (s,x)=2 , and ζ (s)=2 . (3.2) E (n+x)s E ns nX=0 nX=1 By (3.1) and (3.2) we see that Euler numbers are related to the Euler zeta function as ζ (−n)=E , ζ (−n,x)=E (x). E n E n For s,q,h∈C with |q|<1, we define q-Euler zeta function as follows: ∞ ∞ (−1)nqnh (−1)nqnh ζ (s,x|h)=[2] , and ζ (s|h)=[2] . (3.3) E,q q [n+x]s E,q q [n]s nX=0 q nX=1 q For k ∈N,h∈Z, we have ζ (−n|h)=E (h|q). E,q n In the special case h = 0, ζ (s|0) will be written as ζ (s). For s ∈ C, we E,q E,q note that ∞ (−1)n ζ (s)=[2] . E,q q [n]s nX=1 q We now consider the function E (s) as the analytic continuation of Euler q numbers. All the q-Euler numbers E agree with E (n), the analytic contin- n,q q uation of Euler numbers evaluated at n, E (n)=E for n≥0. q n,q Ordinary Euler numbers are defined by ∞ 2 tn = E , |t|<π. (3.4) et+1 nn! nX=0 By (3.4), it is easy to see that n−1 1 n E =1, and E =− E , n=0,1,2,··· . 0 n 2 (cid:18)l(cid:19) l Xl=0 From (2.9) and (3.3), we can consider the q-extension of Euler numbers E as n follows: [2] 1 n−1 n E = q, and E =− qlE ,n=1,2,3,··· , (3.5) 0,q 2 n,q [2]qn Xl=0(cid:18)l(cid:19) l,q 3 In fact, we can express E′(s) in terms of ζ′ (s), the derivative of ζ (s). q E,q E,q ′ ′ ′ ′ E (s)=ζ (−s),E (s)=ζ (−s),E (2n+1)=ζ (−2n−1), (3.6) q E,q q E,q q E,q for n ∈ N∪{0}. This is just the differential of the functional equation and so verifies the consistency of E (s) and E′(s) with E and ζ(s). q q n,q From the above analytic continuation of q-Euler numbers, we derive E (s)=ζ (−s),E (−s)=ζ (s) q E,q q E,q (3.7) ⇒E−n,q =Eq(−n)=ζE,q(n),n∈Z+. The curve Eq(s) runs through the points E−n,q and grows ∼ n asymptoti- cally as (−n) → −∞. The curve E (s) runs through the point E (−n) and q q limn→∞Eq(−n) = limn→∞ζE,q(n) = −2. From (3.5), (3.6) and (3.7), we note that ζ (−n)=E (n)7→ζ (−s)=E (s). E,q q E,q q 4 Analytic continuation of q-Euler polynomials ForconsistencywiththeredefinitionofE =E (n)in(4.5)and(4.6),wehave n,q q n n E (x)= E qkx[x]n−k. n,q (cid:18)k(cid:19) k,q q Xk=0 Let Γ(s) be the gamma function. Then the analytic continuation can be ob- tained as n7→s∈R,x7→w∈C, E 7→E (k+s−[s])=ζ (−(k+(s−[s]))), k,q q E,q n Γ(1+s) 7→ (cid:18)k(cid:19) Γ(1+k+(s−[s]))Γ(1+[s]−k) [s] Γ(1+s)E (k+s−[s])q(k+s−[s])w[w][s]−k ⇒E (s)7→E (s,w)= q q n,q q Γ(1+k+(s−[s]))Γ(1+[s]−k) kX=−1 [s]+1Γ(1+s)E ((k−1)+s−[s])q((k−1)+s−[s])w[w][s]+1−k q q = , Γ(k+(s−[s]))Γ(2+[s]−k) Xk=0 where [s] gives the integer part of s, and so s−[s] gives the fractional part. 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