Analytic Continuation of q-Euler numbers and polynomials

0.09 MB·English

by  T. Kim

Analytic Continuation of q-Euler numbers and
polynomials
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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: [e-mail]

]
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
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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.
Deformation of the curve E (2,w) into the curve of E (3,w) via the real
q q
analytic continuation E (s,w),2≤s≤3,−0.5≤w ≤0.5.
q
ACKNOWLEDGEMENTS. This paper is supported by Jangjeon Mathe-
matical Society and Jangjeon Research Institute for Mathematical Science and
Physics( 2007-001-JRIMS1234567)
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