Some identities of symmetry for the generalized Bernoulli numbers and polynomials

Abstract

In this paper, by the properties of p-adic invariant integral on Zp, we establish various identities concerning the generalized Bernoulli numbers and polynomials. From the symmetric properties of p-adic invariant integral on Zp, we give some interesting relationship between the power sums and the generalized Bernoulli polynomials.

Full text

Page 1

arXiv:0903.2955v1 [math.NT] 17 Mar 2009
Some identities of symmetry for the generalized
Bernoulli numbers and polynomials
By
Taekyun Kim
Abstract.
we establish various identities concerning the generalized Bernoulli numbers and
polynomials.From the symmetric properties of p-adic invariant integral on Zp,
we give some interesting relationship between the power sums and the generalized
Bernoulli polynomials.
In this paper, by the properties of p-adic invariant integral on Zp,
2000 Mathematics Subject Classification: 11B68, 11M38, 11S80.
Key Words and Phrases: p-adic invariant integral, Bernoulli numbers, Bernoulli
polynomials.
§1. Introduction
Let p be a fixed prime number. Throughout this paper, the symbols Z,Zp,Qp, and
Cpwill denote the ring of rational integers, the ring of p-adic integers, the field of p-adic
rational numbers, and the completion of algebraic closure of Qp, respectively. Let N be the
set of natural numbers and Z+= N∪{0}. Let vpbe the normalized exponential valuation
of Cpwith |p|p = p−vp(p)= 1/p. Let UD(Zp) be the space of uniformly differentiable
function on Zp. For f ∈ UD(Zp), the p-adic invariant integral on Zpis defined as
I(f) =
?
Zp
f(x)dx = lim
N→∞
1
pN
pN−1
?
x=0
f(x),(see [6]). (1)
From the definition (1), we have
I1(f1) = I1(f) + f′(0), where f′(0) =
df(x)
dx|x=0 and f1(x) = f(x + 1). (2)
Let fn(x) = f(x + n), (n ∈ N). Then we can derive the following equation (3) from (2).
I(fn) = I(f) +
n
?
i=0
f′(i),(see [6]). (3)

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Some identities of symmetry for the generalized Bernoulli numbers and polynomials
It is well known that the ordinary Bernoulli polynomials Bn(x) are defined as
t
et− 1ext=
∞
?
n=0
Bn(x)tn
n!, ( see [1-25] ),
and the Bernoulli number Bnare defined as Bn= Bn(0).
Let d a fixed positive integer. For n ∈ N, we set
X = Xd= lim
← −
N
?
Z/dpNZ
?
,X1= Zp;
X∗=
?
(a,p)=1
0<a<dp,
(a + dpZp);
a + dpNZp= { x ∈ X| x ≡ a(mod dpN) },
where a ∈ Z lies in 0 ≤ a < dpN. In [6], it is known that
?
X
f(x)dx =
?
Zp
f(x)dx,for f ∈ UD(Zp).
Let us take f(x) = etx. Then we have
?
Zp
etxdx =
t
et− 1=
∞
?
n=0
Bntn
n!.
Thus, we note that
?
Zp
xndx = Bn,n ∈ Z+,(see [1-25]).
Let χ be the Dirichlet’s character with conductor d ∈ N. Then the generalized Bernoulli
polynomials attached to χ are defined as
d
?
a=1
χ(a)teat
edt− 1ext=
∞
?
n=0
Bn,χ(x)tn
n!,( see [22] ), (4)
and the generalized Bernoulli numbers attached to χ, Bn,χare defined as Bn,χ= Bn,χ(0).
In this paper, we investigate the interesting identities of symmetry for the generalized
Bernoulli numbers and polynomials attached to χ by using the properties of p-adic invari-
ant integral on Zp. Finally, we will give relationship between the power sum polynomials
and the generalized Bernoulli numbers attached to χ.

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Taekyun Kim
3
§2. Symmetry of power sum and the generalized Bernoulli polynomials
Let χ be the Dirichlet character with conductor d ∈ N. From (3), we note that
?
X
χ(x)extdx =t?d−1
i=0χ(i)eit
edt− 1
=
∞
?
n=0
Bn,χtn
n!, (5)
where Bn,χ(x) are n-th generalized Bernoulli numbers attached to χ. Now, we also see
that the generalized Bernoulli polynomials attached to χ are given by
?
X
χ(y)e(x+y)tdy =
?d−1
i=0χ(i)eit
edt− 1
ext=
∞
?
n=0
Bn,χ(x)tn
n!. (6)
By (5) and (6), we easily see that
?
X
χ(x)xndx = Bn,χ, and
?
X
χ(y)(x + y)ndy = Bn,χ(x). (7)
From (6), we have
Bn,χ(x) =
n
?
ℓ=0
?n
ℓ
?
Bℓ,χxn−ℓ. (8)
From (6), we can also derive
?
X
χ(x)extdx =
d−1
?
i=0
χ(i)
t
edt− 1e(i
d)dt=
∞
?
n=0
?
dn
d−1
?
i=0
χ(i)Bn(i
d)
?tn
n!.
Therefore, we obtain the following lemma.
Lemma1. For n ∈ Z+, we have
?
X
χ(x)xndx = Bn,χ= dn
d−1
?
i=0
χ(i)Bi
?i
d
?
.
We observe that
1
t
??
X
χ(x)e(nd+x)tdx −
?
X
extχ(x)dx
?
=nd?
Xχ(x)extdx
Xendxtdx
?
=endt− 1
edt− 1
?d−1
i=0
?
χ(i)eit?
. (9)
Thus, we have
1
t
??
X
χ(x)e(nd+x)tdx −
?
X
extdx
?
=
∞
?
k=0
?nd−1
ℓ=0
?
χ(ℓ)ℓk?tk
k!. (10)

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Some identities of symmetry for the generalized Bernoulli numbers and polynomials
Let us define the p-adic functional Tk(χ,n) as follows:
Tk(χ,n) =
n
?
ℓ=0
χ(ℓ)ℓk,for k ∈ Z+. (11)
By (10) and (11), we see that
1
t
??
X
χ(x)e(nd+x)tdx −
?
X
extdx
?
=
∞
?
n=0
?
Tk(χ,nd − 1)
?tk
k!. (12)
By using Taylor expansion in (12), we have
?
X
χ(x)(dn + x)kdx −
?
X
χ(x)xkdx = kTk−1(χ,nd − 1), for k,n,d ∈ N. (13)
That is,
Bk,χ(nd) − Bk,χ= kTk−1(χ,nd − 1).
Let w1,w2,d ∈ N. Then we consider the following integral equation
d?
X
?
Xχ(x1)χ(x2)e(w1x1+w2x2)tdx1dx2
?
t(edw1w2t− 1)
(ew1dt− 1)(ew2dt− 1)
Xedw1w2xtdx
=
?d−1
a=0
?
χ(a)ew1at??d−1
?
b=0
χ(b)ew2bt?
.
(14)
From (9) and (12), we note that
dw1
?
Xedw1xtdx
Xχ(x)extdx
?
=
∞
?
k=0
?
Tk(χ,dw1− 1)
?tk
k!. (15)
Let us consider the p-adic functional Tχ(w1,w2) as follows:
Tχ(w1,w2) =d?
X
?
Xχ(x1)χ(x2)e(w1x1+w2x2+w1w2x)tdx1dx2
?
Xedw1w2x3tdx3
.(16)
Then we see that Tχ(w1,w2) is symmetric in w1and w2, and
Tχ(w1,w2) =t(edw1w2t− 1)ew1w2xt
(ew1dt− 1)(ew2dt− 1)
?d−1
a=0
?
χ(a)ew1at??d−1
?
b=0
χ(b)ew2bt?
.(17)

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Taekyun Kim
5
By (16) and (17), we have
Tχ(w1,w2) =
?1
w1
?1
w1
?
X
χ(x1)ew1(x1+w2x)tdx1
??dw1
Tk(χ,dw1− 1)wk
?
?
Xχ(x2)ew2x2tdx2
Xedw1w2xtdx
?
=
∞
?
i=0
Bi,χ(w2x)wi
1ti
i!
?? ∞
?
k=0
2tk
k!
?
=
1
w1
? ∞
ℓ=0
?
(
ℓ
?
i=0
Bi,χ(w2x)Tℓ−i(χ,dw1− 1)wi
i!(ℓ − i)!
1wℓ−i
2 ℓ!
)tℓ
ℓ!
?
=
∞
?
ℓ=0
?
ℓ
?
i=0
?ℓ
i
?
Bi,χ(w2x)Tℓ−i(χ,dw1− 1)wi−1
1 wℓ−i
2
?tℓ
ℓ!.
(18)
From the symmetric property of Tχ(w1,w2) in w1and w2, we note that
Tχ(w1,w2) =
?1
w2
?1
w2
?
X
χ(x2)ew2(x2+w1x)tdx2
??dw2
Tk(χ,dw2− 1)wk
?
?
Xχ(x1)ew1x1tdx1
Xedw1w2xtdx
?
=
∞
?
i=0
Bi,χ(w1x)wi
2ti
i!
?? ∞
?
k=0
1tk
k!
?
=
1
w2
? ∞
ℓ=0
?
(
ℓ
?
i=0
Bi,χ(w1x)wi
2Tℓ−i(χ,dw2− 1)wℓ−i
i!(ℓ − i)!
1 ℓ!
)tℓ
ℓ!
?
=
∞
?
ℓ=0
?
ℓ
?
i=0
?ℓ
i
?
wi−1
2 wℓ−i
1 Bi,χ(w1x)Tℓ−i(χ,dw2− 1)
?tℓ
ℓ!.
(19)
By comparing the coefficients on the both sides of (18) and (19), we obtain the following
theorem.
Theorem 2. For w1,w2,d ∈ N, we have
ℓ
?
i=0
?ℓ
i
?
Bi,χ(w2x)Tℓ−i(χ,dw1− 1)wi−1
1 wℓ−i
2
=
ℓ
?
i=0
?ℓ
i
?
Bi,χ(w1x)Tℓ−i(χ,dw2− 1)wi−1
2 wℓ−i
1 .
Let x = 0 in Theorem 2. Then we have
ℓ
?
i=0
?ℓ
i
?
Bi,χTℓ−i(χ,dw1− 1)wi−1
1 wℓ−i
2
=
ℓ
?
i=0
?ℓ
i
?
Bi,χTℓ−i(χ,dw2− 1)wi−1
2 wℓ−i
1 .