AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation84894310.1155/2009/848943848943Research ArticleSome Identities of Symmetry for the Generalized Bernoulli Numbers and PolynomialsKimTaekyun1RimSeog-Hoon2LeeByungje3LittlejohnLance1Division of General Education-MathematicsKwangwoon UniversitySeoul 139-701South Koreakw.ac.kr2Department of Mathematics EducationKyungpook National UniversityTaegu 702-701South Koreaknu.ac.kr3Department of Wireless Communications EngineeringKwangwoon UniversitySeoul 139-701South Koreakw.ac.kr200915062009200904042009110520092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

Let p be a fixed prime number. Throughout this paper, the symbols , p, p, and p will 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 p, respectively. Let be the set of natural numbers and +={0}. Let vp be the normalized exponential valuation of p with |p|p=p-vp(p)=1/p. Let UD(p) be the space of uniformly differentiable function on p. For fUD(p), the p-adic invariant integral on p is defined as I(f)=pf(x)dx=limN1pNx=0pN-1f(x), (see ). From the definition (1.1), we have I(f1)=I(f)+f(0),where  f(0)=df(x)dx|x=0,f1(x)=f(x+1). Let fn(x)=f(x+n), (n). Then we can derive the following equation from (1.2): I(fn)=I(f)+i=0n1f(i), (see ). It is well known that the ordinary Bernoulli polynomials Bn(x) are defined as tet-1ext=n=0Bn(x)tnn!, (see ), and the Bernoulli number Bn are defined as Bn=Bn(0).

Let d be a fixed positive integer. For n, we set X=Xd=limN(/dpN),X1=p;X*=0<a<dp,(a,p)=1(a+dpp);a+dpNp={xXxa(mod dpN)}, where a lies in 0a<dpN. It is easy to see that Xf(x)dx=pf(x)dx,for  fUD(p).

In , the Witt's formula for the Bernoulli numbers are given by pxndx=Bn,n+.   Let χ be the Dirichlet's character with conductor d. Then the generalized Bernoulli polynomials attached to χ are defined as a=1dχ(a)teatedt-1ext=n=0Bn,χ(x)tnn!, (see ), and the generalized Bernoulli numbers attached to χ, Bn,χ are 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 invariant integral on p. Finally, we will give relationship between the power sum polynomials and the generalized Bernoulli numbers attached to χ.

2. Symmetry of Power Sum and the Generalized Bernoulli Polynomials

Let χ be the Dirichlet character with conductor d. From (1.3), we note that Xχ(x)extdx=ti=0d-1χ(i)eitedt-1=n=0Bn,χtnn!, where Bn,χ(x) are the nth generalized Bernoulli numbers attached to χ. Now, we also see that the generalized Bernoulli polynomials attached to χ are given by Xχ(y)e(x+y)tdy=ti=0d-1χ(i)eitedt-1ext=n=0Bn,χ(x)tnn!. By (2.1) and (2.2), we easily see that Xχ(x)xndx=Bn,χ,Xχ(y)(x+y)ndy=Bn,χ(x). From (2.2), we have Bn,χ(x)==0n(n)B,χxn-. From (2.2), we can also derive Xχ(x)extdx=i=0d-1χ(i)tedt-1e(i/d)dt=n=0(dn1i=0d-1χ(i)Bn(id))tnn!. Therefore, we obtain the following lemma.

Lemma 2.1.

For n+, one has Xχ(x)xndx=Bn,χ=dn1i=0d-1χ(i)Bi(id).

We observe that 1t(Xχ(x)e(nd+x)tdx-Xextχ(x)dx)=ndXχ(x)extdxXendxtdx=endt-1edt-1(i=0d-1χ(i)eit). Thus, we have 1t(Xχ(x)e(nd+x)tdx-Xχ(x)extdx)=k=0(=0nd-1χ()k)tkk!. Let us define the p-adic functional Tk(χ,n) as follows: Tk(χ,n)==0nχ()k,for  k+. By (2.8) and (2.9), we see that 1t(Xχ(x)e(nd+x)tdx-Xχ(x)extdx)=n=0(Tk(χ,nd-1))tkk!. By using Taylor expansion in (2.10), we have Xχ(x)(dn+x)kdx-Xχ(x)xkdx=kTk-1(χ,nd-1),for  k,n,d  . That is, Bk,χ(nd)-Bk,χ=kTk-1(χ,nd-1). Let w1,w2,d. Then we consider the following integral equation: dXχ(x1)χ(x2)e(w1x1+w2x2)tdx1dx2Xedw1w2xtdx=t(edw1w2t-1)(ew1dt-1)(ew2dt-1)(a=0d-1χ(a)ew1at)(b=0d-1χ(b)ew2bt). From (2.7) and (2.10), we note that dw1Xχ(x)extdxXedw1xtdx=k=0(Tk(χ,dw1-1))tkk!. Let us consider the p-adic functional Tχ(w1,w2) as follows: Tχ(w1,w2)=dXχ(x1)χ(x2)e(w1x1+w2x2+w1w2x)tdx1dx2Xedw1w2x3tdx3. Then we see that Tχ(w1,w2) is symmetric in w1 and w2, and Tχ(w1,w2)=t(edw1w2t-1)ew1w2xt(ew1dt-1)(ew2dt-1)(a=0d-1χ(a)ew1at)(b=0d-1χ(b)ew2bt). By (2.15) and (2.16), we have

Tχ(w1,w2)=(1w1Xχ(x1)ew1(x1+w2x)tdx1)(dw1Xχ(x2)ew2x2tdx2Xedw1w2xtdx)=(1w1i=0Bi,χ(w2x)w1itii!)(k=0Tk(χ,dw1-1)w2ktkk!)=1w1(=0(i=0Bi,χ(w2x)T-i(χ,dw1-1)w1iw2-i!i!(-i)!)t!)==0(i=0(i)Bi,χ(w2x)T-i(χ,dw1-1)w1i-1w2-i)t!. From the symmetric property of Tχ(w1,w2) in w1 and w2, we note that Tχ(w1,w2)=(1w2Xχ(x2)ew2(x2+w1x)tdx2)(dw2Xχ(x1)ew1x1tdx1Xedw1w2xtdx)=(1w2i=0Bi,χ(w1x)w2itii!)(k=0Tk(χ,dw2-1)w1ktkk!)=1w2(=0(i=0Bi,χ(w1x)w2iT-i(χ,dw2-1)w1-i!i!(-i)!)t!)==0(i=0(i)w2i-1w1-iBi,χ(w1x)T-i(χ,dw2-1))t!. By comparing the coefficients on the both sides of (2.17) and (2.18), we obtain the following theorem.

Theorem 2.2.

For w1,w2,d, one has i=0(i)Bi,χ(w2x)T-i(χ,dw1-1)w1i-1w2-i=i=0(i)Bi,χ(w1x)T-i(χ,dw2-1)w2i-1w1-i.

Let x=0 in Theorem 2.2. Then we have i=0(i)Bi,χT-i(χ,dw1-1)w1i-1w2-i=i=0(i)Bi,χT-i(χ,dw2-1)w2i-1w1-i. By (2.14) and (2.16), we also see that Tχ(w1,w2)=(ew1w2xtw1Xχ(x1)ew1x1tdx1)(dw1Xχ(x2)ew2x2tdx2Xedw1w2xtdx)=(ew1w2xtw1Xχ(x1)ew1x1tdx1)(edw1w2t-1ew2dt-1)(i=0d-1χ(i)ew2it)=(ew1w2xtw1Xχ(x1)ew1x1tdx1)(=0w1-1i=0d-1ew2(i+d)tχ(i+d))=(ew1w2xtw1Xχ(x1)ew1x1tdx1)(i=0dw1-1ew2itχ(i))=1w1i=0dw1-1χ(i)Xχ(x1)ew1(x1+w2x+(w2/w1)i)tdx1=1w1i=0dw1-1χ(i)k=0Bk,χ(w2x+w2w1i)w1ktkk!=k=0(i=0dw1-1χ(i)Bk,χ(w2x+w2w1i)w1k-1)tkk!. From the symmetric property of Tχ(w1,w2) in w1 and w2, we can also derive the following equation: Tχ(w1,w2)=(ew1w2xtw2Xχ(x2)ew2x2tdx2)(dw2Xχ(x1)ew1x1tdx1Xedw1w2xtdx)=(ew1w2xtw2Xχ(x2)ew2x2tdx2)(edw1w2t-1ew1dt-1)(i=0d-1χ(i)ew1it)=(ew1w2xtw2Xχ(x2)ew2x2tdx2)(=0w2-1ew1dt)(i=0d-1χ(i)ew1it)=1w2i=0dw2-1χ(i)Xχ(x2)ew2(x2+w1x+(w1/w2)i)tdx2=1w2i=0dw2-1χ(i)k=0Bk,χ(w1x+w1w2i)w2ktkk!=k=0{i=0dw2-1χ(i)Bk,χ(w1x+w1w2i)w2k-1}tkk!. By comparing the coefficients on the both sides of (2.21) and (2.22), we obtain the following theorem.

Theorem 2.3.

For w1,w2,d, one has i=0dw1-1χ(i)Bk,χ(w2x+w2w1i)w1k-1=i=0dw2-1χ(i)Bk,χ(w1x+w1w2i)w2k-1.

Remark 2.4.

Let x=0 in Theorem 2.3. Then we see that i=0dw1-1χ(i)Bk,χ(w2w1i)w1k-1=i=0dw2-1χ(i)Bk,χ(w1w2i)w2k-1.

If we take w2=1, then we have i=0dw1-1χ(i)Bk,χ(iw1)w1k-1=i=0d-1χ(i)Bk,χ(w1i).

Remark 2.5.

Let χ be trivial character. Then we can easily derive the “multiplication theorem for Bernoulli polynomials’’ from Theorems 2.2 and 2.3 (see ).

Acknowledgment

The present research has been conducted by the research grant of the Kwangwoon University in 2009.