# Some New Identities on the Bernoulli and Euler Numbers

Authors: Dae San Kim; Taekyun Kim; Sang-Hun Lee; D. V. Dolgy; Seog-Hoon Rim
Journal: Discrete Dynamics in Nature and Society
 (2011)

---

## Abstract

We give some new identities on the Bernoulli and Euler numbers by using the bosonicp-adic integral on ℤp and reflection symmetric properties of Bernoulli and Euler polynomials.


---

## Body

## 1. Introduction

Letp be a fixed prime number. Throughout this paper ℤp,ℚp, and ℂp will denote the ring of p-adic rational integers, the field of p-adic rational numbers, and the completion of algebraic closure of ℚp. Let UD(ℤp) be the space of uniformly differentiable functions on ℤp. For f∈UD(ℤp), the bosonic p-adic integral on ℤp is defined by
(1.1)I(f)=∫Zpf(x)dμ(x)=lim⁡N→∞⁡∑x=0pN-1f(x)μ(x+pNZp)=lim⁡N→∞1pN∑x=0pN-1f(x).
From (1.1), we note that(1.2)I(f1)=I(f)+f′(0),wheref1(x)=f(x+1),
see [1]. As is well known, the ordinary Bernoulli polynomials are defined by the generating function as follows:(1.3)F(t,x)=tet-1ext=eB(x)t=∑n=0∞Bn(x)tnn!,
see [1–19], where we use the technical notation by replacing Bn(x) by Bn(x)(n≥0), symbolically. In the special case, x=0, Bn(0)=Bn are called the n-th ordinary Bernoulli numbers. That is, the generating function of ordinary Bernoulli numbers is given by
(1.4)F(t)=F(t,0)=tet-1=∑n=0∞Bntnn!,
see [1–19]. From (1.4), we can derive the following relation:(1.5)B0=1,(B+1)n-Bn=δ1,n,
see [1, 10], where δ1,n is the Kronecker symbol.By (1.3) and (1.4), we easily get(1.6)Bn(x)=∑l=0n(nl)Blxn-l=∑l=0n(nl)Bn-lxl.
By (1.2) and (1.3), we easily get(1.7)∫Zpe(x+y)tdμ(y)=tet-1ext=∑n=0∞Bn(x)tnn!,
see [1, 10]. From (1.7), we can derive Witt’s formula for the n-th Bernoulli polynomials as follows:(1.8)∫Zp(x+y)ndμ(y)=Bn(x),wheren∈Z+,
see [11]. By (1.1) and (1.8), we easily see that(1.9)∫Zp(y+1-x)ndμ(y)=(-1)n∫Zp(y+x)ndμ(y).
Thus, by (1.8) and (1.9), we get reflection symmetric relation for the Bernoulli polynomials as follows:(1.10)Bn(1-x)=(-1)nBn(x)wheren∈Z+.
The ordinary Euler polynomials are defined by the generating function as follows:(1.11)Fe(t,x)=2et+1ext=∑n=0∞En(x)tnn!.
with the usual convention about replacing En(x) by En(x) (see [8, 9]). In the special case, x=0, En(0)=En are called the n-th Euler numbers (see [8, 9]).From (1.11), we note that(1.12)2et+1ext=21+e-te-(1-x)t=∑n=0∞(-1)nEn(1-x)(t)nn!,
By comparing the coefficients on both sides of (1.11) and (1.12), we obtain the following reflection symmetric relation for Euler polynomials as follows:(1.13)En(x)=(-1)nEn(1-x),wheren∈Z+.
The equations (1.10) and (1.13) are useful in deriving our main results in this paper.Forn,k∈ℤ+, the Bernstein polynomials are defined by(1.14)Bk,n(x)=(nk)xk(1-x)n-k,
see [13]. By (1.14), we easily get Bk,n(x)=Bn-k,n(1-x).In this paper we consider thep-adic integrals for the Bernoulli and Euler polynomials. From those p-adic integrals, we derive some new identities on the Bernoulli and Euler numbers.


## 2. Identities on the Bernoulli and Euler Numbers

First, we consider thep-adic integral on ℤp for the nth ordinary Bernoulli polynomials as follows:
(2.1)I1=∫ZpBn(x)dμ(x)=∑l=0n(nl)Bn-l∫Zpxldμ(x)=∑l=0n(nl)Bn-lBl,wheren∈Z+.
On the other hand, by (1.3) and (1.10), one gets(2.2)I1=(-1)n∫ZpBn(1-x)dμ(x).
From (1.5), (1.6), (1.8), and (2.2), one notes that(2.3)I1=(-1)n∑l=0n(nl)Bn-l∫Zp(1-x)ldμ(x)=(-1)n∑l=0n(nl)Bn-l(l+Bl+δ1,l)=(-1)nnBn-l(1)+(-1)n∑l=0n(nl)Bn-lBl+(-1)nnBn-l.
Equating (2.1) and (2.3), one gets(2.4)(1+(-1)n+1)∑l=0n(nl)Bn-lBl=(-1)nn(δ1,n-l+Bn-1)+(-1)nnBn-1=2(-1)nnBn-l+(-1)nnδ1,n-1.
Let n∈ℕ with n≡1(mod2). Then, by (2.4), one has(2.5)∑l=02n-1(2n-1l)B2n-1-lBl=-(2n-1)B2n-2.
Therefore, by (2.4) and (2.5), we obtain the following theorem.Theorem 2.1.
Forn∈ℕ, one has
(2.6)(1+(-1)n+1)∑l=0n(nl)Bn-lBl=2(-1)nnBn-1+(-1)nnδ1,n-1.
In particular,
(2.7)∑l=02n-1(2n-1l)B2n-1-lBl=-(2n-1)B2n-2.By the same motivation, let us also consider thep-adic integral on ℤp for Euler polynomials as follows:(2.8)I2=∫ZpEn(x)dμ(x)=∑l=0n(nl)En-l∫Zpxldμ(x)=∑l=0n(nl)En-lBl,wheren∈Z+.
On the other hand, by (1.12) and (1.13), one gets(2.9)I2=(-1)n∫ZpEn(1-x)dμ(x)=(-1)n∑l=0n(nl)En-l∫Zp(1-x)ldμ(x)=(-1)n∑l=0n(nl)En-l(l+Bl+δ1,l)=n(-1)nEn-l(1)+(-1)n∑l=0n(nl)En-lBl+(-1)nnEn-l.
From (1.12) and the definition of Euler numbers, one has(2.10)En(x)=∑l=0n(nl)Elxn-l=∑l=0n(nl)En-lxl=(E+x)n,(2.11)E0=1,(E+1)n+En=2δ0,n,
see [8, 9] with the usual convention of replacing En by En. By (2.9), (2.10), and (2.11), one gets(2.12)I2=n(-1)n(2δ0,n-1-En-1)+(-1)nnEn-1+(-1)n∑l=0n(nl)En-lBl.
Equating (2.8) and (2.12), one has(2.13)(1+(-1)n-1)∑l=0n(nl)En-lBl=2n(-1)nδ0,n-1.
Therefore, by (2.13), we obtain the following theorem.Theorem 2.2.
Forn∈ℕ∪{0}, one has
(2.14)(1+(-1)n-1)∑l=0n(nl)En-lBl=2(-1)nnδ0,n-1.
In particular,
(2.15)∑l=02n+1(2n+1l)E2n+1-lBl=0,forn∈N.Let us consider the followingp-adic integral on ℤp for the product of Bernoulli and Euler polynomials as follows:(2.16)I3=∫ZpBm(x)En(x)dμ(x)=∑k=0m∑l=0n(mk)(nl)Bm-kEn-l∫Zpxk+l(x)dμ(x)=∑k=0m∑l=0n(mk)(nl)Bm-kEn-lBk+l.
On the other hand, by (1.10) and (1.13), one gets(2.17)I3=(-1)m+n∫ZpBm(1-x)En(1-x)dμ(x)=(-1)m+n∑k=0m∑l=0n(mk)(nl)Bm-kEn-l∫Zp(1-x)k+ldμ(x)=(-1)m+n∑k=0m∑l=0n(mk)(nl)Bm-kEn-l(k+l+Bk+l+δ1,k+l)=(-1)m+nmBm-1(1)En(1)+(-1)m+nnBm(1)En-1(1)+(-1)m+n∑k=0m∑l=0n(mk)(nl)Bm-kEn-lBk+l+(-1)m+n(mBm-1En+nBmEn-1).
Equating (2.16) and (2.17), one gets(2.18)((-1)m+n+1+1)∑k=0m∑l=0n(mk)(nl)Bm-kEn-lBk+l=(-1)m+nm(Bm-1+δ1,m-1)(2δ0,n-En)+(-1)m+nn(Bm+δ1,m)(2δ0,n-1-En-1)+(-1)m+n(nBmEn-1+mBm-1En).
For n∈ℕ, by (2.18), one gets(2.19)((-1)m+1+1)∑k=0m∑l=02n(mk)(2nl)Bm-kE2n-lBk+l=(-1)m+12n(Bm+δ1,m)E2n-1+(-1)m(2nBmE2n-1)=(-1)m+12nδ1,mE2n-1.
Therefore, by (2.19), one obtains the following theorem.Theorem 2.3.
Forn∈ℕ, one has
(2.20)((-1)m+1+1)∑k=0m∑l=02n(mk)(2nl)Bm-kE2n-lBk+l=(-1)m+12nδ1,mE2n-1.
In particular, for m∈ℕ, one has
(2.21)∑k=02m+1∑l=02n(2m+1k)(2nl)B2m+1-kE2n-lBk+l=0.By the same motivation, we consider thep-adic integral on ℤp for the product of Bernoulli and Bernstein polynomials as follows:(2.22)I4=∫ZpBm(x)Bk,n(x)dμ(x)wherem,n,k∈N∪{0}.
From (1.6) and (1.14), one gets(2.23)I4=∑l=0m(ml)Bm-l∫ZpxlBk,n(x)dμ(x)=(nk)∑l=0m(ml)Bm-l∫Zpxk+l(1-x)n-kdμ(x)=(nk)∑l=0m∑j=0n-k(-1)j(ml)(n-kj)Bm-lBk+l+j.
On the other hand,
(2.24)I4=(-1)m∫ZpBm(1-x)Bn-k,n(1-x)dμ(x)=(-1)m(nk)∑l=0m∑j=0k(-1)j(ml)(kj)Bm-l(n-k+j+l+Bn-k+l+j+δ1,n-k+l+j)=(-1)m(nk)(n-k)Bm(1)δ0,k+(-1)m(nk)mBm-1(1)δ0,k-(-1)m(nk)mBm(1)kδ0,k-1+(-1)m(nk)∑l=0m∑j=0k(-1)j(ml)(kj)Bm-lBn-k+l+j+(-1)m(nk)(mBm-1-kBm)δn,k+(-1)m(nk)Bmδn,k+1.
Equating (2.23) and (2.24), one gets(2.25)(-1)m∑l=0m∑j=0n-k(-1)j(ml)(n-kj)Bm-lBk+l+j=((n-k)Bm(1)+mBm-1(1))δ0,k-kBm(1)δ0,k-1+(mBm-1-kBm)δn,k+Bmδn,k+1+∑l=0m∑j=0k(-1)j(ml)(kj)Bm-lBn-k+l+j.
By (2.25), we obtain the following theorem.Theorem 2.4.
Forn,m∈ℕ, one has
(2.26)∑l=02m∑j=02n(-1)j(2ml)(2nj)B2m-lBl+j=2nB2m+∑l=02m(2ml)B2m-lB2n+l.Now, we consider thep-adic integral on ℤp for the product of Euler and Bernstein polynomials as follows:(2.27)I5=∫ZpEm(x)Bk,n(x)dμ(x)=∑l=0m(ml)Em-l∫ZpxlBk,n(x)dμ(x)=(nk)∑l=0m∑j=0n-k(-1)j(ml)(n-kj)Em-lBk+l+j.
On the other hand, by (1.13) and (1.14), one gets
(2.28)I5=(-1)m∫ZpBn-k,n(1-x)Em(1-x)dμ(x)=(-1)m(nk)∑l=0m∑j=0k(-1)j(ml)(kj)Em-l∫Zp(1-x)n-k+l+jdμ(x)=(-1)m(nk)∑l=0m∑j=0k(-1)j(ml)(kj)(n-k+l+j+Bn-k+l+j+δ1,n-k+l+j)Em-l=(-1)m(n-k)(nk)Em(1)δ0,k+(-1)m(nk)mEm-1(1)δ0,k-(-1)m(nk)Em(1)kδ0,k-1+(-1)m(nk)∑l=0m∑j=0k(-1)j(ml)(kj)Em-lBn-k+l+j+(-1)m(nk)(δn,k+1Em+δn,k(mEm-1-kEm)).
Equating (2.27) and (2.28), one gets(2.29)(-1)m∑l=0m∑j=0n-k(-1)j(ml)(n-kj)Em-lBk+l+j=(n-k)Em(1)δ0,k+mδ0,kEm-1(1)-kEm(1)δ0,k-1+∑l=0m∑j=0k(-1)j(ml)(kj)Em-lBn-k+l+j+δn,k+1Em+(mEm-1-kEm)δn,k.
Therefore, by (2.11) and (2.29), we obtain the following theorem.Theorem 2.5.
Forn,m∈ℕ, one has
(2.30)∑l=02m∑j=02n(-1)j(2ml)(2nj)E2m-lBl+j=-2mE2m-1+B2m+2n.Finally, we consider thep-adic integral on ℤp for the product of Euler, Bernoulli, and Bernstein polynomials as follows:(2.31)I6=∫ZpBr(x)Es(x)Bk,n(x)dμ(x)=(nk)∑l=0r∑j=0s(rl)(sj)Br-lEs-j∫Zpxk+l+j(1-x)n-kdμ(x)=(nk)∑l=0r∑j=0s∑i=0n-k(-1)i(rl)(sj)(n-ki)Br-lEs-jBk+l+i+j.On the other hand, by (1.10), (1.13), and (1.14), one gets(2.32)I6=(-1)r+s∫ZpBr(1-x)Es(1-x)Bn-k,n(1-x)dμ(x)=(-1)r+s(nk)∑l=0r∑j=0s∑i=0k(-1)i(rl)(sj)(ki)Br-lEs-j∫Zp(1-x)n-k+l+i+jdμ(x).
Equating (2.31) and (2.32), we easily see that(2.33)(-1)r+s∑l=0r∑j=0s∑i=0n-k(-1)i(rl)(sj)(n-ki)Br-lEs-jBk+l+i+j=∑l=0r∑j=0s∑i=0k(-1)i(rl)(sj)(ki)(n-k+l+i+j+Bn-k+l+i+j+δ1,n-k+l+i+j)Br-lEs-j=(n-k)Br(1)Es(1)δ0,k+rBr-1(1)δ0,kEs(1)+sBr(1)Es-1(1)δ0,k-kBr(1)Es(1)δ0,k-1+∑l=0r∑j=0s∑i=0k(-1)i(rl)(sj)(ki)Br-lEs-jBn-k+l+i+j+δn,k+1BrEs+(rBr-1Es+sBrEs-1-kBrEs)δn,k.
Therefore, by (1.5) and (2.11), we obtain the following theorem.Theorem 2.6.
Forr,n,s∈ℕ, one has
(2.34)∑l=02r∑j=02s∑i=02n(-1)i(2rl)(2sj)(2ni)B2r-lE2s-jBl+i+j=-2sB2rE2s-1+∑l=0r(2r2l)B2r-2lB2n+2l+2s-r∑j=1s(2s2j-1)E2s-2j+1B2n+2r+2j-2.



---

*Source: 856132-2011-12-07.xml*