Symmetric identities for degenerate generalized Bernoulli polynomials Taekyun Kim , Dmitry V. Dolgy

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J. Nonlinear Sci. Appl. 9 (2016), 677–683
Research Article
Symmetric identities for degenerate generalized
Bernoulli polynomials
Taekyun Kima,b , Dmitry V. Dolgyc , Dae San Kimd,∗
a
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China.
b
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.
c
Institute of Natural Sciences, Far Eastern Federal University, 690950 Vladivostok, Russaia.
d
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea.
Communicated by Seog-Hoon Rim
Abstract
In this paper, we give some interesting identities of symmetry for degenerate generalized Bernoulli polynomials attached to χ which are derived from the properties of symmetry of certain p-adic invariant integrals
c
on Zp . 2016
All rights reserved.
Keywords: Symmetry, identity, degenerate generalized Bernoulli polynomial.
2010 MSC: 11B68, 11B83, 11C08, 65D20, 65Q30, 65R20
1. Introduction
Let p be a fixed 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 . The p-adic
norm |·|p is normalized as |p|p = p1 . Let f (x) be a uniformly differentiable function on Zp . Then, the p-adic
invariant integral on Zp is defined as
N
p −1
1 X
f (x) dµ0 (x) = lim N
f (x) ,
N →∞ p
Zp
Z
(see [9, 11]) .
(1.1)
x=0
∗
Corresponding author
Email addresses: [email protected] (Taekyun Kim), [email protected] (Dmitry V. Dolgy), [email protected] (Dae San
Kim)
Received 2015-09-21
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
678
Thus, by (1.1), we get
Z
Z
f (x + n) dµ0 (x) −
f (x) dµ0 (x) =
Zp
Zp
n−1
X
f 0 (l) ,
(n ∈ N) .
(1.2)
l=0
In particular, for n = 1, we have
Z
Z
f (x + 1) dµ0 (x) −
f (x) dµ0 (x) = f 0 (0) ,
(see [11]) .
(1.3)
Zp
Zp
As is well known, the Bernoulli polynomials are given by the generating function
Z
∞
X
t
tn
e(x+y)t dµ0 (y) = t
ext =
Bn (x) .
e −1
n!
Zp
(1.4)
n=0
When x = 0, Bn = Bn (0) are called Bernoulli numbers (see [1]–[19]).
Carlitz introduced degenerate Bernoulli polynomials defined by the generating function
t
x
λ
(1 + λt) =
1
λ
(1 + λt) − 1
∞
X
βn (x | λ)
n=0
tn
,
n!
(see [4]) .
(1.5)
When x = 0, βn (λ) = βn (0 | λ) are called degenerate Bernoulli numbers.
Recently, T. Kim introduced degenerate Bernoulli polynomials which are different from Carlitz degenerate Bernoulli polynomials as follows:
1
log (1 + λt) λ
1
x
(1 + λt) λ =
(1 + λt) λ − 1
−
∞
X
n=0
βn,λ (x)
tn
,
n!
(1.6)
1
where λ, t ∈ Cp with |λt|p < p p−1 (see [11]).
Let d ∈ N with (d, p) = 1. Then, we set
X = lim
←−
N
Z
,
dpN Z
X∗ =
[
(a + dpZp ) ,
0<a<dp
p-a
and
a + dpN Zp = x | x ≡ a (mod dpN ) ,
where a ∈ Z with 0 ≤ a ≤ dpN − 1.
Let χ be a Dirichlet character with conductor d ∈ N. Then the generalized Bernoulli polynomials are
defined by the generating function
!
d−1
∞
X
X
t
tn
at
xt
χ
(a)
e
e
=
B
(x)
(see [1]–[19]) .
(1.7)
n,χ
n!
edt − 1
a=0
n=0
We define the degenerate generalized Bernoulli polynomials attached to χ by the generating function
!
1 d−1
∞
X
a
x
log (1 + λt) λ X
tn
λ
λ
χ (a) (1 + λt)
βn,λ,χ (x) .
(1 + λt) =
(1.8)
d
n!
(1 + λt) λ − 1 a=0
n=0
When x = 0, βn,λ,χ = βn,λ,χ (0) are called degenerate generalized Bernoulli numbers attached to χ.
Note that lim βn,λ,χ (x) = Bn,χ (x), (n ≥ 0).
λ→0
The purpose of this paper is to give some identities of symmetry for degenerate generalized Bernoulli
polynomials attached to χ which are derived from certain symmetric identities of p-adic invariant integrals
on Zp .
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
679
2. Symmetric identities of degenerate generalized Bernoulli polynomials
−
1
Let us assume that λ, t ∈ Cp with |λt|p < p p−1 . It is known that
Z
Z
f (x) dµ0 (x) , (see [9]) ,
f (x) dµ0 (x) =
X
(2.1)
Zp
where f (x) is uniformly differentiable function on Zp . From (1.2), we note that
Z
x
λ
χ (x) (1 + λt) dµ0 (x) =
X
1 d−1
log (1 + λt) λ X
d
a
χ (a) (1 + λt) λ =
(1 + λt) λ − 1 a=0
∞
X
βn,λ,χ
n=0
tn
.
n!
(2.2)
Thus, by (2.2), we get
Z
χ (y) (1 + λt)
x+y
λ
dµ0 (y) =
X
1 d−1
log (1 + λt) λ X
d
χ (a) (1 + λt)
a+x
λ
=
(1 + λt) λ − 1 a=0
∞
X
βn,λ,χ (x)
n=0
tn
,
n!
(2.3)
where χ is a Dirichlet character with conductor d ∈ N. Now, we observe that
Z
χ (x) (1 + λt)
x+nd
λ
Z
dµ0 (x) −
x
λ
χ (x) (1 + λt) dµ0 (x) = log (1 + λt)
X
X
1
λ
nd−1
X
a
χ (a) (1 + λt) λ .
(2.4)
a=0
By (2.4), we get
!
t
1
log (1 + λt) λ
∞
∞ nd−1
X
X
1X
tm
tm
(βm,λ,χ (nd) − βm,λ,χ )
=
χ (a) (a | λ)m
,
t
m!
m!
m=0
(2.5)
m=0 a=0
where (a | λ)m = a (a − λ) · · · (a − λ (m − 1)) and (a | λ)0 = 1. It is not difficult to show that
∞
1X
tm
(β
(nd)
−
β
)
m,λ,χ
m,λ,χ
1
m!
log (1 + λt) λ t m=0
!
!
∞
∞
X
X
βk+1,λ,χ (nd) − βk+1,λ,χ tk
λl tl
=
bl
l!
k+1
k!
l=0
k=0
!
∞
m X
X
m βk+1,λ,χ (nd) − βk+1,λ,χ
tm
bm−k λm−k
.
=
k
k+1
m!
t
m=0
(2.6)
k=0
From (2.5) and (2.6), we have
m X
βk+1,λ,χ (nd) − βk+1,λ,χ
m
k=0
k
k+1
bm−k λm−k =
nd−1
X
χ (a) (a | λ)m ,
(2.7)
a=0
where bm is the m-th Bernoulli number of the second kind. Let us define Sk (n, λ | χ) as follows:
n
X
χ (l) (l | λ)k = Sk (n, λ | χ) ,
(n ∈ N) .
(2.8)
l=0
We observe that
Z
1
χ (x) (1 + λt)
1
log (1 + λt) λ
X
nd+x
λ
Z
dµ0 (x) −
x
λ
χ (x) (1 + λt) dµ0 (x)
X
(2.9)
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
nd
=
x
R
nd
λ
(1 + λt) λ − 1
X χ (x) (1 + λt) dµ0 (x)
=
R
d
ndx
λ dµ (x)
(1 + λt) λ − 1
0
X (1 + λt)
d−1
X
680
χ (i) (1 + λt)
i
λ
!
.
i=0
From (2.9), we have
Z
1
χ (x) (1 + λt)
1
log (1 + λt) λ
Z
x
λ
dµ0 (x) −
X
nd−1
X
=
nd+x
λ
χ (x) (1 + λt) dµ0 (x)
(2.10)
X
a
λ
χ (a) (1 + λt) =
a=0
Thus, by (2.8) and (2.10), we get
Z
1
χ (x) (1 + λt)
1
X
log (1 + λt) λ
∞
X
tk
=
(Sk (nd − 1, λ | χ)) .
k!
∞
X
nd−1
X
k=0
a=0
nd+x
λ
!
χ (a) (a | λ)k
Z
dµ0 (x) −
tk
.
k!
χ (x) (1 + λt) dµ0 (x)
x
λ
X
(2.11)
k=0
From (2.7) and (2.11), we have
m X
βk+1,λ,χ (nd) − βk+1,λ,χ
m
bm−k λm−k = Sm (nd − 1, λ | χ) ,
k
k+1
(2.12)
k=0
where m ≥ 0. Let w1 , w2 , d ∈ N. Then, we consider the following integral equation:
d
w1 x1 +w2 x2
R R
λ
χ (x1 ) χ (x2 ) (1 + λt)
dµ0 (x1 ) dµ0 (x2 )
dw1 w2 x
R
λ
dµ0 (x)
X (1 + λt)
dw1 w2
1
!
d−1
log (1 + λt) λ (1 + λt) λ − 1
X
w1 a
χ (a) (1 + λt) λ
=
w2 d
w1 d
(1 + λt) λ − 1 (1 + λt) λ − 1
a=0
!
d−1
X
w2 b
×
χ (b) (1 + λt) λ .
X
X
(2.13)
b=0
By (2.9), we get
R
x
∞
dw1 X χ (x) (1 + λt) λ dµ0 (x) X
tk
=
S
(dw
−
1,
λ
|
χ)
.
1
k
dw1 x
R
k!
λ
(1
+
λt)
dµ
(x)
k=0
0
X
Let
Tχ (w1 , w2 | λ) =
d
w1 x1 +w2 x2 +w1 w2 x
R R
X
(2.14)
X
λ
χ (x1 ) χ (x2 ) (1 + λt)
dµ0 (x1 ) dµ0 (x2 )
.
dw1 w2 x
R
λ
(1
+
λt)
dµ
(x)
0
X
(2.15)
From (2.15), we note that Tχ (w1 , w2 | λ) is symmetric in w1 and w2 , and
Tχ (w1 , w2 | λ)
dw1 w2
w1 w2 x
1
log (1 + λt) λ (1 + λt) λ − 1 (1 + λt) λ
=
w1 d
w2 d
(1 + λt) λ − 1 (1 + λt) λ − 1
! d−1
!
d−1
X
X
w1 a
w2 b
×
χ (a) (1 + λt) λ
χ (b) (1 + λt) λ .
a=0
b=0
(2.16)
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
Thus, by (2.15) and (2.16), we get
Z
w1 (x1 +w2 x)
1
λ
χ (x1 ) (1 + λt)
Tχ (w1 , w2 | λ) =
dµ0 (x1 )
w1 X


w2 x2
R
λ
dµ0 (x2 )
dw1 X χ (x2 ) (1 + λt)

×
dw1 w2 x
R
λ
dµ0 (x)
X (1 + λt)
!
k k!
∞
∞
X
w1i ti
w2 t
λ 1 X
χ
βi, λ ,χ (w2 x)
Sk dw1 − 1,
=
w1
w1
i!
w2
k!
i=0
k=0


λ
i l−i
∞
l
1 X X βi, wλ1 ,χ (w2 x) Sl−i dw1 − 1, w2 | χ w1 w2 l!  tl
=


w1
i!(l − i)!
l!
681
(2.17)
i=0
l=0
!
∞
l X
X
λ tl
l
i−1 l−i
χ
w
w
=
βi, λ ,χ (w2 x) Sl−i dw1 − 1,
.
1
2
w1
w2 l!
i
l=0
i=0
On the other hand,
!
∞
l X
X
tl
λ l
i−1 l−i
w
.
Tχ (w1 , w2 | λ) =
βi, λ ,χ (w1 x) Sl−i dw2 − 1,
χ
w
1
2
w2
w1 l!
i
l=0
(2.18)
i=0
By comparing the coefficients on both sides of (2.17) and (2.18), we obtain the following theorem.
Theorem 2.1. For w1 , w2 , d ∈ N, let χ be a Dirichlet character with conductor d. Then we have
l X
l
i=0
λ β λ (w2 x) Sl−i dw1 − 1,
χ w1i−1 w2l−i
w2 i i, w1 ,χ
l X
l
λ =
β λ (w1 x) Sl−i dw2 − 1,
χ w2i−1 w1l−i .
i i, w2 ,χ
w1 i=0
In particular, taking x = 0, we have
l
X
i=0
βi,
λ
w1
,χ Sl−i
dw1 − 1,
X
l
λ λ i−1 l−i l
i−1 l−i l
χ
w
w
=
β
S
dw
−
1,
χ
w
w
.
2
2
1
1
2
i, wλ ,χ l−i
w2 i
w1 i
2
i=0
From (2.9), (2.15) and (2.17), we have
!
w1 w2
Z
w1 x1
(1 + λt) λ x
Tχ (w1 , w2 | λ) =
χ (x1 ) (1 + λt) λ dµ0 (x1 )
w1
X


w2 x2
R
λ
dw1 X χ (x2 ) (1 + λt)
dµ0 (x2 )

×
dw1 w2 x
R
λ
dµ0 (x)
X (1 + λt)
!
w1 w2 x Z
w1 x1
(1 + λt) λ
=
χ (x1 ) (1 + λt) λ dµ0 (x1 )
w1
X
! d−1
!
dw1 w2
X
w2 i
(1 + λt) λ − 1
×
χ (i) (1 + λt) λ
w2 d
(1 + λt) λ − 1
i=0
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
(1 + λt)
w1
=
w1 w2 x
λ
χ (x1 ) (1 + λt)
w1 x1
λ
!
dµ0 (x1 )
X
wX
d−1
1 −1 X
×
Z
682
(1 + λt)
w2 (i+ld)
λ
!
χ (i + ld)
l=0 i=0
=
(1 + λt)
w1
w1 w2 x
λ
Z
χ (x1 ) (1 + λt)
w1 x1
λ
dw
1 −1
X
!
dµ0 (x1 )
X
(1 + λt)
w2 i
λ
!
χ (i)
(2.19)
i=0
Z
dw1 −1
w1
w
1 X
x +w x+ 2 i
χ (x1 ) (1 + λt) λ 1 2 w1 dµ0 (x1 )
χ (i)
w1
X
i=0
k k
dw1 −1
∞
X
w2
w1 t
1 X
χ (i)
i
=
βk, λ ,χ w2 x +
w1
w1
w1
k!
i=0
k=0
!
dw
∞
1 −1
X
X
tk
w2
=
χ (i) βk, λ ,χ w2 x +
i w1k−1
.
w1
w1
k!
=
i=0
k=0
On the other hand,
!
w1 w2 x Z
w2 x2
(1 + λt) λ
χ (x2 ) (1 + λt) λ dµ0 (x2 )
Tχ (w1 , w2 | λ) =
w2
X


w1 x1
R
dw2 X χ (x1 ) (1 + λt) λ dµ0 (x1 )

×
dw1 w2 x
R
λ
(1
+
λt)
dµ
(x)
0
X
!
w1 w2
Z
x
w2 x2
(1 + λt) λ
=
χ (x2 ) (1 + λt) λ dµ0 (x2 )
w2
X
! d−1
!
dw1 w2
X
w1 i
(1 + λt) λ − 1
χ (i) (1 + λt) λ
×
w1 d
(1 + λt) λ − 1
i=0
!
w1 w2 x Z
w2 x2
(1 + λt) λ
=
χ (x2 ) (1 + λt) λ dµ0 (x2 )
w2
X
!
wX
d−1
2 −1 X
w1 (i+ld)
×
(1 + λt) λ χ (i + ld)
(2.20)
l=0 i=0
=
(1 + λt)
w2
w1 w2 x
λ
Z
χ (x2 ) (1 + λt)
w2 x2
λ
!
dµ0 (x2 )
X
dw
2 −1
X
i=0
Z
dw2 −1
w2
w
1 X
x +w x+ 1 i
χ (i)
χ (x2 ) (1 + λt) λ 2 1 w2 dµ0 (x2 )
w2
X
i=0
k k
dw2 −1
∞
X
1 X
w1
w2 t
=
χ (i)
βk, λ ,χ w1 x +
i
w2
w2
w2
k!
i=0
k=0
!
dw
∞
2 −1
X
X
w1
tk
=
χ (i) βk, λ ,χ w1 x +
i w2k−1
.
w2
w2
k!
=
k=0
i=0
Therefore, by (2.19) and (2.20), we obtain the following theorem.
(1 + λt)
w1 i
λ
!
χ (i)
T. Kim, D. V. Dolgy, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 677–683
683
Theorem 2.2. For w1 , w2 , d ∈ N, let χ be a Dirichlet character with conductor d. Then we have
w1k−1
dw
1 −1
X
i=0
dw
2 −1
X
w2
w1
k−1
i = w2
χ (i) βk, λ ,χ w1 x +
i .
χ (i) βk, λ ,χ w2 x +
w1
w2
w1
w2
i=0
Remark 2.3. Let x = 0 in Theorem 2.2. Then we note that
w1k−1
dw
1 −1
X
χ (i) βk,
i=0
λ
,χ
w1
w2
i
w1
=
w2k−1
dw
2 −1
X
i=0
χ (i) βk,
λ
,χ
w2
w1
i .
w2
Further, if we take w2 = 1, then we get
w1k−1
dw
1 −1
X
i=0
χ (i) βk,
λ
,χ
w1
X
d−1
1
i =
χ (i) βk,λ,χ (w1 i) .
w1
i=0
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