J. Nonlinear Sci. Appl. 9 (2016), 443–451 Research Article
Symmetric identities of higher-order degenerate q−Euler polynomials
Dae San Kima, Taekyun Kimb,∗
aDepartment of Mathematics, Sogang University, Seoul 121-742, Republic of Korea.
bDepartment of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.
Communicated by Seog-Hoon Rim
Abstract
In this paper, we study the higher-order degenerateq-Euler polynomials and give some identities of symmetry on these polynomials derived from symmetric properties for certain multivariate fermionicp-adicq-integrals on Zp. c2016 All rights reserved.
Keywords: Symmetry, identity, higher-order degenerateq-Euler polynomial.
2010 MSC: 11B75, 11B83, 11S80.
1. Introduction
Let p be an odd prime number. Throughout this paper, Zp, Qp and Cp will denote the ring of p-adic integers, the field of p-adic rational numbers and the completion of the algebraic closure of Qp. Thep-adic norm is normalized so that |p|p = 1p. Let q be an indeterminate in Cp such that |1−q|p < p−p−11 . The q-analogue of the numberx is defined as [x]q = 1−q1−qx. Note that limq→1[x]q=x. Let f(x) be a continuous functional Zp. Then, the fermionicp-adicq-integral on Zp is defined by Kim as
Z
Zp
f(x)dµ−q(x) = lim
N→∞
1 [pN]−q
pN−1
X
x=0
f(x) (−q)x (1.1)
= [2]q 2 lim
N→∞
PN−1
X
x=0
f(x)qx(−1)x, (see [12, 14]),
∗Corresponding author
Email addresses: [email protected](Dae San Kim),[email protected](Taekyun Kim) Received 2015-09-10
where [x]−q = 1−(−q)1+q x. Note that
q→1lim Z
Zp
f(x)dµ−q(x) = lim
N→∞
pN−1
X
x=0
f(x) (−1)x (1.2)
= Z
Zp
f(x)dµ−1(x) is the ordinary fermonicp-adic integral onZp.
From (1.1), we can easily derive the following equation:
qn Z
Zp
f(x+n)dµ−q(x) + (−1)n−1 Z
Zp
f(x)dµ−q(x) = [2]q
n−1
X
x=0
f(x) (−1)n−1−x, (1.3) and
q Z
Zp
f(x+ 1)dµ−q(x) + Z
Zp
f(x)dµ−q(x) = [2]qf(0), (see [14]). (1.4) As is well known, the higher-order Euler polynomials are defined by the generating function
2 et+ 1
r
ext=
∞
X
n=0
En(r)(x)tn
n!. (1.5)
When x= 0,En(r) =En(r)(0) are called the higher-order Euler numbers (see [1]–[23]).
From (1.2), we note that Z
Zp
· · · Z
Zp
e(x1+···+xr+x)tdµ−1(x1)· · ·dµ−1(xr) = 2
et+ 1 r
ext
=
∞
X
n=0
En(r)(x)tn n!.
Carlitz considered q-Bernoulli numbers defined by the recurrence relation β0,q = 1, q(qβq+ 1)n−βn,q =
(1, ifn= 1,
0, ifn >1, (1.6)
with the usual convention about replacingβqn byβn,q (see [4]).
In [12, 14], Kim defined Carlitz’s type q-Euler numbers given by
E0,q = 1, q(qEq+ 1)n− En,q= [2]qδ0,n, (1.7) whereδn,k is the Kronecker’s symbol.
Recently, the higher-orderq-Euler polynomials are defined by the multivariate fermionicp-adicq-integral on Zp
Z
Zp
· · · Z
Zp
e[x1+···+xr+x]qtdµ−q(x1)· · ·dµ−q(xr) =
∞
X
n=0
En,q(r)(x)tn
n!, (see [14]). (1.8) When x = 0, En,q(r) = En,q(r)(0) are called the higher-order q-Euler numbers. In particular, r = 1, then En,q(1)(x) =En,q(x).
From (1.8), we have Z
Zp
· · · Z
Zp
[x1+· · ·+xr+x]nqdµ−q(x1)· · ·dµ−q(xr) =En,q(r)(x) (1.9)
= 1 (1−q)n
n
X
l=0
n l
(−1)l
[2]q 1 +ql+1
r qlx
= [2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mr[m1+· · ·+mr+x]nq , wherer∈Nand n≥0.
By (1.9), we get the generating function of the higher-order q-Euler polynomials as follows:
[2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mre[m1+···+mr]qt (1.10)
=
∞
X
n=0
En,q(r)(x)tn
n!, (see [14, 15]).
Carlitz introduced the higher-order degenerate Euler polynomials given by the generating function 2
(1 +λt)1λ+ 1
!r
(1 +λt)λx =
∞
X
n=0
En(r)(x|λ)tn
n!. (1.11)
When x = 0, En(r)(x) = En(r)(0|λ) are called the higher-order degenerate Euler numbers (see [5]). In particular,r= 1, En(1)(x|λ) =En(x|λ) are called degenerate Euler polynomials.
Note that limλ→0En(r)(x|λ) =E(r)n (x), (n≥0).
In this paper, we study the higher-order degenerate q-Euler polynomials and give some identities of symmetry on these polynomials derived from symmetric properties for certain multivariate fermionicp-adic q-integrals onZp.
2. Symmetric identities of higher-order degenerate q-Euler polynomials
Let λ, t∈Cp be such that|λt|p < p−p−11 . From (1.2) and (1.3), we note that Z
Zp
· · · Z
Zp
(1 +λt)
x1+···+xr+x
λ dµ−1(x1)· · ·dµ−1(xr) = 2 (1 +λt)λ1 + 1
!r
(1 +λt)xλ (2.1)
=
∞
X
n=0
En(r)(x|λ)tn n!.
In view of (1.8), we define the higher-order degenerate q-Euler polynomials by the generating function as
Z
Zp
· · · Z
Zp
(1 +λt)λ1[x1+···+xr+x]qdµ−q(x1)· · ·dµ−q(xr) =
∞
X
n=0
En,q,λ(r) (x)tn
n!. (2.2)
Thus, by (2.2), we get
λ→0limEn,q,λ(r) (x) =En,q(r)(x), (n≥0). From (2.2), we can derive
Z
Zp
· · · Z
Zp
[x1+· · ·+xr+x]q
n,λdµ−q(x1)· · ·dµ−q(xr) =En,q,λ(r) (x), (n≥0), (2.3) where
[x]q
n,λ= [x]q
[x]q−λ [x]q−2λ
· · ·
[x]q−(n−1)λ
, (n≥1)
and
[x]q
0,λ= 1.
By (2.3), we get En,q,λ(r) (x) =
Z
Zp
· · · Z
Zp
[x1+· · ·+xr]q
n,λdµ−q(x1)· · ·dµ−q(xr) (2.4)
=
n
X
l=0
S1(n, l)λn−l Z
Zp
· · · Z
Zp
[x1+· · ·+xr+x]lqdµ−q(x1)· · ·dµ−q(xr)
=
n
X
l=0
S1(n, l)λn−lEl,q(r)(x), whereS1(n, l) is the Stirling number of the first kind.
From (1.9) and (2.4), we have En,q,λ(r) (x) = [2]rq
n
X
l=0
∞
X
m1,...,mr=0
(−q)m1+···+mr[m1+· · ·+mr+x]lqS1(n, l)λn−l. (2.5)
Therefore, by (2.4), we obtain the following theorem.
Theorem 2.1. For n≥0, we have En,q,λ(r) (x) =
n
X
l=0
S1(n, l)λn−lEl,q(r)(x)
= [2]rq
n
X
l=0
∞
X
m1,...,mr=0
(−q)m1+···+mr[m1+· · ·+mr+x]lqS1(n, l)λn−l.
Now, we observe that [2]rq
n
X
l=0
∞
X
m1,...,mr=0
(−q)m1+···+mr[m1+· · ·+mr+x]lqS1(n, l)λn−l (2.6)
= [2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mr
[m1+· · ·+mr+x]q
n,λ.
Thus, by (2.6), we get
∞
X
n=0
En,q,λ(r) (x)tn n!= [2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mr
∞
X
n=0
[m1+· · ·+mr+x]q
n,λ
n! tn (2.7)
= [2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mr(1 +λt)
[m1+···+mr+x]q
λ .
Therefore, by (2.7), we obtain the following theorem.
Theorem 2.2. For r∈N, we have
∞
X
n=0
En,q,λ(r) (x)tn n! = [2]rq
∞
X
m1,...,mr=0
(−q)m1+···+mr(1 +λt)
[m1+···+mr+x]q
λ .
By replacing tby 1λ eλt−1
in (2.2), we get Z
Zp
e[x1+···+xr+x]qtdµ−q(x1)· · ·dµ−q(xr) =
∞
X
m=0
Em,q,λ(r) (x)λ−m eλt−1m
m! (2.8)
=
∞
X
n=0 n
X
m=0
λn−mEm,q,λ(r) (x)S2(n, m)
!tn n!, whereS2(n, m) is the Stirling number of the second kind.
Therefore, by (1.8) and (2.8), we obtain the following theorem.
Theorem 2.3. For n≥0, we have
En,q(r)(x) =
n
X
m=0
λn−mEm,q,λ(r) (x)S2(n, m). Let w1, w2 ∈N be such thatw1 ≡1, w2 ≡1 (mod 2). Then, by (2.2), we get
1 [w1]r−q
Z
Zp
· · · Z
Zp
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr l=1yl]q
λ dµ−qw1(y1)· · ·dµ−qw1 (yr) (2.9)
= 1
[w1]r−q lim
N→∞
pN−1
X
y1,...,yr=0
1 [pN]r−qw1
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr l=1yl]q
λ
×(−qw1)y1+···+yr
= 1
[w1]r−q lim
N→∞
1 [w2pN]r−qw1
w2pN−1
X
y1,...,yr=0
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr l=1yl]q
λ
×(−q)w1y1+···+w1yr
= [2]rq 2r lim
N→∞
w2−1
X
i1,i2,...,ir=0 pN−1
X
y1,...,yr=0
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr
l=1(il+w2yl)]q
λ
×(−1)y1+···+yrqw1(i1+w2y1)+w1(i2+w2y2)+···+w1(ir+w2yr)×(−1)i1+···+ir
= [2]rq 2r
w2−1
X
i1,...,ir=0
(−1)Prl=1ilqw1Prl=1il
× lim
N→∞
pN−1
X
y1,...,yr=0
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr
l=1(il+w2yl)]q
λ
×(−1)y1+···+yrqw1w2y1+w1w2y2+···+w1w2yr. From (2.9), we note that
1 [w1]r−q
w1−1
X
j1,...,jr=0
qw2Prl=1jl(−1)Prl=1jl (2.10)
× Z
Zp
· · · Z
Zp
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr l=1yl]q
λ dµ−qw1 (y1)· · ·dµ−qw1 (yr)
= [2]rq 2r lim
N→∞
w2−1
X
i1,...,ir=0 w1−1
X
j1,...,jr=0 pN−1
X
y1,...,yr=0
(−1)Prl=1(jl+il+yl)
×qw1Prl=1il+w2Prl=1jl+w1w2Prl=1yl
×(1 +λt)
1
λ[w1w2x+w2Prl=1jl+w1Pr
l=1il+w1w2Pr l=1yl]q. On the other hand,
1 [w2]r−q
w2−1
X
j1,...,jr=0
qw1Prl=1jl(−1)Prl=1jl (2.11)
× Z
Zp
· · · Z
Zp
(1 +λt)
[w1w2x+w1Pr
l=1jl+w2Pr l=1yl]q
λ dµ−qw2 (y1)· · ·dµ−qw2 (yr)
= [2]rq 2r lim
N→∞
w1−1
X
i1,...,ir=0 w2−1
X
j1,...,jr=0 pN−1
X
y1,...,yr=0
(−1)Prl=1(il+jl+yl)
×qw1Prl=1jl+w2Prl=1il+w1w2Prl=1yl
×(1 +λt)
1
λ[w1w2x+w1Pr
l=1jl+w2Pr
l=1il+w1w2Pr l=1yl]q. Therefore, by (2.10) and (2.11), we obtain the following theorem.
Theorem 2.4. Let w1, w2 ∈Nsuch that w1≡1 (mod 2) andw2 ≡1 (mod 2). Then, we have 1
[w1]r−q
w1−1
X
j1,...,jr=0
qw2Prl=1jl(−1)Prl=1jl
× Z
Zp
· · · Z
Zp
(1 +λt)
[w1w2x+w2Pr
l=1jl+w1Pr l=1yl]q
λ dµ−qw1(y1)· · ·dµ−qw1(yr)
= 1
[w2]r−q
w2−1
X
j1,...,jr=0
qw1Prl=1jl(−1)Prl=1jl
× Z
Zp
· · · Z
Zp
(1 +λt)
[w1w2x+w1Pr
l=1jl+w2Pr l=1yl]q
λ dµ−qw2(y1)· · ·dµ−qw2(yr). We observe that
"
w1w2x+
r
X
l=1
jlw2+
r
X
l=1
ylw1
#
q
= [w1]q
"
w2x+ w2 w1
r
X
l=1
jl+
r
X
l=1
yl
#
qw1
. (2.12)
From (2.12), we have Z
Zp
· · · Z
Zp
(1 +λt)
1
λ[w1w2x+Pr
l=1jlw2+Pr l=1ylw1]q
dµ−qw1 (y1)· · ·dµ−qw1(yr) (2.13)
= Z
Zp
· · · Z
Zp
(1 +λt)
[w1]q λ
h w2x+ww2
1
Pr
l=1jl+Pr l=1yli
qw1 dµ−qw1 (y1)· · ·dµ−qw1(yr)
=
∞
X
n=0
E(r)
n,qw1,[wλ
1]q
w2x+w2
w1
(j1+· · ·+jr)
[w1]nq tn n!.
Therefore, by Theorem 2.4, (2.12) and (2.13), we obtain the following theorem.
Theorem 2.5. For n≥0, w1, w2 ∈Nwith w1≡1 (mod 2),w2 ≡1 (mod 2), we have [w1]nq
[w1]r−q
w1−1
X
j1,...,jr=0
(−1)j1+···+jrqw2(j1+···+jr)E(r)
n,qw1,[wλ
1]q
w2x+w2
w1 (j1+· · ·+jr)
= [w2]nq [w2]r−q
w2−1
X
j1,...,jr=0
(−1)j1+···+jrqw1(j1+···+jr)E(r)
n,qw2,[wλ
2]q
w1x+w1
w2 (j1+· · ·+jr)
.
From (2.3), we have Z
Zp
· · · Z
Zp
1 [w1]q
!n
[w1]q
"
w2x+ w2 w1
r
X
l=1
jl+
r
X
l=1
yl
#
qw1
n,λ
(2.14)
×dµ−qw1 (y1)· · ·dµ−qw1 (yr)
=
n
X
l=0
S1(n, l) λ [w1]q
!n−l
Z
Zp
· · · Z
Zp
"
w2x+w2 w1
r
X
l=1
jl+
r
X
l=1
yl
#l
qw1
×dµ−qw1 (y1)· · ·dµ−qw1 (yr)
=
n
X
l=0
S1(n, l) λ [w1]q
!n−l l
X
i=0
l i
[w2]q [w1]q
!i
[j1+· · ·+jr]iqw2qw2(l−i)Prk=1jk
× Z
Zp
· · · Z
Zp
[w2x+y1+· · ·+yr]l−iqw1dµ−qw1(y1)· · ·dµ−qw1(yr)
=
n
X
l=0 l
X
i=0
S1(n, l)λn−l[w1]l−n−iq [w2]iq[j1+· · ·+jr]iqw2
×qw2(l−i)Prk=1jk l
i
El−i,q(r) w1 (w2x). Thus, by (2.14), we get
[w1]nq [w1]r−q
w1−1
X
j1,...,jr=0
qw2Prl=1jl(−1)Prl=1jl 1 [w1]q
!n
(2.15)
× Z
Zp
· · · Z
Zp
[w1]q
"
w2x+w2 w1
r
X
l=1
jl+
r
X
l=1
yl
#
qw1
n,λ
dµ−qw1 (y1)· · ·dµ−qw1 (yr)
= 1
[w1]r−q
n
X
l=0 l
X
i=0
l i
S1(n, l)λn−l[w1]l−iq [w2]iqEl−i,q(r) w1 (w2x) ˜Tl+1,i(r) (w1 |qw2), where
T˜n,i(r)(w|q) =
w−1
X
j1,...,jr=0
(−1)j1+···+jr[j1+· · ·+jr]iqq(n−i)Prl=1jl. (2.16) On the other hand,
[w2]nq [w2]r−q
w2−1
X
j1,...,jr=0
qw1Prl=1jl(−1)Prl=1jl λ [w2]q
!n
(2.17)
× Z
Zp
· · · Z
Zp
[w2]q
λ
"
w1x+w1
w2
r
X
l=1
jl+
r
X
l=1
yl
#
qw2
n,λ
dµ−qw2 (y1)· · ·dµ−qw2 (yr)
= 1
[w2]r−q
n
X
l=0 l
X
i=0
l i
S1(n, l)λn−l[w2]l−iq [w1]iqEl−i,q(r) w2 (w1x) ˜Tl+1,i(r) (w2|qw1).
Therefore, by (2.15) and (2.17), we obtain the following theorem.
Theorem 2.6. For n≥0, w1, w2 ∈Nwith w1≡1 (mod 2),w2 ≡1 (mod 2), we have 1
[w1]r−q
n
X
l=0 l
X
i=0
l i
S1(n, l)λn−l[w1]l−iq [w2]iqEl−i,q(r) w1 (w2x) ˜Tl+1,i(r) (w1 |qw2)
= 1
[w2]r−q
n
X
l=0 l
X
i=0
l i
S1(n, l)λn−l[w2]l−iq [w1]iqEl−i,q(r) w2 (w1x) ˜Tl+1,i(r) (w2 |qw1). Remark 2.7. If we take λ→0, then we get
1 [w1]r−q
n
X
i=0
n i
[w1]n−iq [w2]iqEn−i,q(r) w1 (w2x) ˜Tn+1,i(r) (w1 |qw2)
= 1
[w2]r−q
n
X
i=0
n i
[w2]n−iq [w1]iqEn−i,q(r) w2 (w1x) ˜Tn+1,i(r) (w2 |qw1).
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