# Some Identities on Bernoulli and Euler Numbers

Authors: D. S. Kim; T. Kim; J. Choi; Y. H. Kim
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

Recently, Kim introduced the fermionicp-adic integral on ℤp. By
using the equations of the fermionic and bosonic p-adic integral on ℤp, we give some
interesting identities on Bernoulli and Euler numbers.


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## Body

## 1. Introduction/Preliminaries

Letp be a fixed 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 algebraic closure of ℚp, respectively. Let ℕ be the set of natural numbers and ℤ+=ℕ∪{0}. The p-adic absolute value |·|p is normally defined by |p|p=1/p.LetUD(ℤp) be the space of uniformly differentiable functions on ℤp and C(ℤp) the space of continuous function on ℤp. For f∈C(ℤp), the fermionic p-adic integral on ℤp is defined by Kim as follows:(1.1)I-1(f)=∫Zpf(x)dμ-1(x)=lim⁡N→∞∑x=0pN-1f(x)(-1)x,(see[1]).
The following fermionic p-adic integral equation on ℤp is well known (see [1–3]):(1.2)I-1(f1)+I-1(f)=2f(0),
where f1(x)=f(x+1).From (1.1) and (1.2), we can derive the generating function of Euler polynomials as follows:(1.3)∫Zpe(x+y)tdμ-1(y)=2et+1ext=∑n=0∞En(x)tnn!,
where En(x) is the nth ordinary Euler polynomial (see [1–4]). In the special case, x=0, En(0)=En is called the nth ordinary Euler number.By (1.3), we get Witt’s formula for the nth Euler polynomial as follows:(1.4)∫Zp(x+y)ndμ-1(y)=En(x),forn∈Z+.
Thus, by (1.4), we have(1.5)En(x)=(E+x)n=∑l=0n(nl)xn-lEl,
with the usual convention about replacing En by En (see [5, 6]). From (1.3), we note that(1.6)(E+1)n+En=2δ0,n,
where δk,n is the Kronecker symbol (see [3]). By (1.2) and (1.4), we get(1.7)∫Zp(x+y+1)ndμ-1(y)+∫Zp(x+y)ndμ-1(y)=2xn.
Thus, by (1.4) and (1.7), we have(1.8)En(x+1)+En(x)=2xn,forn∈Z+.
Equation (1.8) is equivalent to(1.9)xn=En(x)+12∑l=0n-1(nl)El(x).
From (1.6), we can derive the following equation:(1.10)En(2)=2-En(1)=2+En-2δ0,n,forn∈Z+.Forf∈UD(ℤp), the bosonic p-adic integral on ℤp is defined by(1.11)I1(f)=∫Zpf(x)dμ1(x)=lim⁡N→∞1pN∑x=0pN-1f(x),(see[4]).
From (1.11), we can easily derive the following I1-integral equation:(1.12)I1(f1)=I(f)+f′(0),(see[4,7,8]),
where f1(x)=f(x+1) and f′(0)=df(x)/dx|x=0.It is well known that the Bernoulli polynomial can be represented by the bosonicp-adic integral on ℤp as follows:(1.13)∫Zpe(x+y)tdμ1(y)=tet-1ext=∑n=0∞Bn(x)tnn!,
where Bn(x) is called the nth Bernoulli polynomial (see [4, 7–13]). In the special case, x=0, Bn(0)=Bn is called the nth Bernoulli number. By the definition of Bernoulli numbers and polynomials, we get(1.14)Bn(x)=∫Zp(x+y)ndμ1(y)=∑l=0n(nl)xn-lBl.
Thus, by (1.13) and (1.14), we see that(1.15)B0=1,(B+1)n-Bn=δ1,n,
with the usual convention about replacing Bn by Bn (see [1–22]).By (1.11), we easily get(1.16)∫Zp(1-x+y)ndμ1(y)=(-1)n∫Zp(x+y)ndμ1(y).
From (1.13), (1.14), and (1.16), we have(1.17)Bn(1-x)=(-1)nBn(x)forn∈Z+.
By (1.15), we get(1.18)Bn(2)=n+Bn(1)=n+Bn+δ1,n.
Thus, by (1.17) and (1.18), we have(1.19)(-1)nBn(-1)=Bn(2)=n+Bn+δ1,n,(see[4]).From (1.12) and (1.13), we get(1.20)∫Zp(x+1+y)n+1dμ1(y)-∫Zp(x+y)n+1dμ1(y)=(n+1)xn.
Thus, by (1.13) and (1.20), we have(1.21)Bn+1(x+1)-Bn+1(x)=(n+1)xnforn∈Z+.
Equation (1.21) is equivalent to the following equation:(1.22)xn=1n+1∑l=0n(n+1l)Bl(x)forn∈Z+.In this paper we derive some interesting and new identities for the Bernoulli and Euler numbers from thep-adic integral equations on ℤp.


## 2. Some Identities on Bernoulli and Euler Numbers

From (1.1), we note that(2.1)∫Zp(1-x+y)ndμ-1(y)=(-1)n∫Zp(x+y)ndμ-1(y).
By (1.14) and (2.1), we get(2.2)En(1-x)=(-1)nEn(x),wheren∈Z+.
In the special case, x=-1, we have(2.3)En(2)=(-1)nEn(-1)=2+En-2δ0,n.Let us consider the following fermionicp-adic integral on ℤp as follows:(2.4)∫Zpxndμ-1(x)=1n+1∑l=0n(n+1l)∫ZpBl(x)dμ-1(x)=1n+1∑l=0n(n+1l)∑k=0l(lk)Bl-k∫Zpxkdμ-1(x)=1n+1∑l=0n(n+1l)∑k=0l(lk)Bl-kEk.
Therefore, by (1.4) and (2.4), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.5)En=1n+1∑l=0n(n+1l)∑k=0l(lk)Bl-kEk.It is known thatBn(x)=(-1)nBn(1-x). If we take the fermionic p-adic integral on both sides of (1.22), then we have(2.6)∫Zpxndμ-1(x)=1n+1∑l=0n(n+1l)∫ZpBl(x)dμ-1(x)=1n+1∑l=0n(n+1l)(-1)l∫ZpBl(1-x)dμ-1(x)=1n+1∑l=0n(n+1l)(-1)l∑k=0l(lk)Bl-k∫Zp(1-x)kdμ-1(x)=1n+1∑l=0n(n+1l)(-1)l∑k=0l(lk)Bl-k(-1)kEk(-1).
From (2.2) and (2.6), we note that(2.7)∫Zpxndμ-1(x)=1n+1∑l=0n(n+1l)(-1)l∑k=0l(lk)Bl-kEk(2)=1n+1∑l=0n(n+1l)(-1)l∑k=0l(lk)Bl-k(2+Ek-2δ0,k)=1n+1∑l=0n(n+1l)(-1)l(2Bl(1)+∑k=0l(lk)Bl-kEk-2Bl)=1n+1∑l=0n(n+1l)(-1)l(∑k=0l(lk)Bl-kEk+2δ1,l).
Therefore, by (1.4) and (2.7), we obtain the following theorem.Theorem 2.2.
Forn∈ℤ+, one has
(2.8)En=1n+1∑l=0n(n+1l)(-1)l(∑k=0l(lk)Bl-kEk+2δ1,l).Corollary 2.3.
Forn∈ℕ, one has
(2.9)2+En=1n+1∑l=0n(n+1l)(-1)l(∑k=0l(lk)Bl-kEk).Let us take the bosonicp-adic integral on both sides of (1.9) as follows:(2.10)∫Zpxndμ1(x)=∫Zp(En(x)+12∑l=0n-1(nl)El(x))dμ1(x)=∑l=0n(nl)En-l∫Zpxldμ1(x)+12∑l=0n-1(nl)∑k=0l(lk)El-k∫Zpxkdμ1(x)=∑l=0n(nl)En-lBl+12∑l=0n-1(nl)∑k=0l(lk)El-kBk.
Thus, by (1.14) and (2.10), we obtain the following theorem.Theorem 2.4.
Forn∈ℤ+, one has
(2.11)Bn=∑l=0n(nl)En-lBl+12∑l=0n-1(nl)∑k=0l(lk)El-kBk.On the other hand, by (2.2) and (2.10), we get(2.12)∫Zpxndμ1(x)=(-1)n∫ZpEn(1-x)dμ1(x)+12∑l=0n-1(nl)(-1)l∫ZpEl(1-x)dμ1(x)=(-1)n∑l=0n(nl)En-l∫Zp(1-x)ldμ1(x)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-k∫Zp(1-x)kdμ1(x)=(-1)n∑l=0n(nl)En-l(-1)lBl(-1)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-k(-1)kBk(-1)=(-1)n∑l=0n(nl)En-lBl(2)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-kBk(2)=(-1)n∑l=0n(nl)En-l(l+Bl+δ1,l)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-k(k+Bk+δ1,k)=(-1)nnEn-1(1)+(-1)n∑l=0n(nl)En-lBl+(-1)nnEn-1+12∑l=0n-1(nl)(-1)llEl-1(1)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-kBk+12∑l=1n-1(nl)(-1)llEl-1=(-1)nn(2+En-1-2δ0,n-1)+(-1)n∑l=0n(nl)En-lBl+(-1)nnEn-1+12∑l=0n-1(nl)(-1)ll(2+El-1-δ0,l-1)+12∑l=0n-1(nl)(-1)l∑k=0l(lk)El-kBk+12∑l=1n-1(nl)(-1)llEl-1,
where n∈ℕ with n≥2. Therefore, by (2.12), we obtain the following theorem.Theorem 2.5.
Forn∈ℕ with n≥2, one has
(2.13)B2n-1=-2n-12-(2n-1)E2n-2(-1)-∑l=02n-1(2n-1l)E2n-1-lBl+12∑l=02n-2(2n-1l)(-1)l∑k=0l(lk)El-kBk.By (1.9) and (1.22), we get(2.14)∬Zpxmyndμ-1(x)dμ1(y)=∬Zp(1m+1∑k=0m(m+1k)Bk(x))(En(y)+12∑l=0n-1(nl)El(y))dμ-1(x)dμ1(y)=1m+1∑k=0m(m+1k)∬ZpBk(x)En(y)dμ-1(x)dμ1(y)+12(m+1)∑k=0m∑l=0n-1(m+1k)(nl)∬ZpBk(x)El(y)dμ-1(x)dμ1(y)=1m+1∑k=0m∑l=0k∑p=0n(m+1k)(kl)(np)Bk-lEn-pBpEl+12(m+1)∑k=0m∑l=0n-1∑s=0k∑p=0l(m+1k)(nl)(ks)(lp)Bk-sEl-pEsBp.
Therefore, by (1.4), (1.14), and (2.14), we obtain the following theorem.Theorem 2.6.
Form∈ℤ+ and n∈ℕ, one has
(2.15)EmBn=1m+1∑k=0m∑l=0k∑p=0n(m+1k)(kl)(np)Bk-lEn-pBpEl+12(m+1)∑k=0m∑l=0n-1∑s=0k∑p=0l(m+1k)(nl)(ks)(lp)Bk-sEl-pEsBp.It is easy to show that(2.16)∫Zpxm+ndμ-1(x)=∫Zp(1m+1∑k=0m(m+1k)Bk(x))(En(x)+12∑l=0n-1(nl)El(x))dμ-1(x)=1m+1∑k=0m∑i=0k∑j=0n(m+1k)(ki)(nj)Bk-iEn-j∫Zpxi+jdμ-1(x)+12(m+1)∑k=0m∑l=0n-1∑i=0k∑j=0l(m+1k)(nl)(ki)(lj)Bk-iEl-j∫Zpxi+jdμ-1(x)=1m+1∑k=0m∑i=0k∑j=0n(m+1k)(ki)(nj)Bk-iEn-jEi+j+12(m+1)∑k=0m∑l=0n-1∑i=0k∑j=0l(m+1k)(nl)(ki)(lj)Bk-iEl-jEi+j.
Therefore, by (2.16), we obtain the following corollay.Corollary 2.7.
Form∈ℤ+ and n∈ℕ, one has
(2.17)Em+n=1m+1∑k=0m∑i=0k∑j=0n(m+1k)(ki)(nj)Bk-iEn-jEi+j+12(m+1)∑k=0m∑l=0n-1∑i=0k∑j=0l(m+1k)(nl)(ki)(lj)Bk-iEl-jEi+j.Forf∈C(ℤp),  p-adic analogue of Bernstein operator of order n for f is given by(2.18)Bn(f∣x)=∑k=0nf(kn)(nk)xk(1-x)n-k=∑k=0nf(kn)Bk,n(x),
where Bk,n(x)=(nk)xk(1-x)n-k for n,k∈ℤ+ is called the Bernstein polynomial of degree n (see [8]). From the definition of Bk,n(x), we note that Bn-k,n(1-x)=Bk,n(x).Let us take the fermionicp-adic integral on ℤp for the product of xm and Bk,n(x) as follows:(2.19)∫ZpxmBk,n(x)dμ-1(x)=1m+1∑l=0m(m+1l)∫ZpBl(x)Bk,n(x)dμ-1(x)=(nk)m+1∑l=0m∑j=0l(m+1l)(lj)Bl-j∫Zpxj+k(1-x)n-kdμ-1(x)=(nk)m+1∑l=0m∑j=0l∑i=0n-k(-1)iBl-j(m+1l)(lj)(n-ki)∫Zpxi+j+kdμ-1(x)=(nk)m+1∑l=0m∑j=0l∑i=0n-k(-1)i(m+1l)(lj)(n-ki)Bl-jEi+j+k.
From (2.18), we note that(2.20)∫ZpxmBk,n(x)dμ-1(x)=(nk)∫Zpxm+k(1-x)n-kdμ-1(x)=(nk)∑j=0n-k(n-kj)(-1)j∫Zpxm+k+jdμ-1(x)=(nk)∑j=0n-k(n-kj)(-1)jEm+k+j.
Therefore, by (2.19) and (2.20), we obtain the following theorem.Theorem 2.8.
Form,n,k∈ℤ+, one has
(2.21)∑j=0n-k(n-kj)(-1)jEm+k+j=1m+1∑l=0m∑j=0l∑i=0n-k(-1)i(m+1l)(lj)(n-ki)Bl-jEi+j+k.
In particular,
(2.22)(m+1)Em+n=∑l=0m∑j=0l(m+1l)(lj)Bl-jEj+n.By (1.17) and the symmetric property of Bk,n(x), we get(2.23)∫ZpxmBk,n(x)dμ-1(x)=∫ZpxmBn-k,n(1-x)dμ-1(x)=1m+1∑l=0m(-1)l(m+1l)∫ZpBl(1-x)Bn-k,n(1-x)dμ-1(x)=(nk)m+1∑l=0m∑j=0l∑i=0k(-1)i+l(m+1l)(lj)(ki)Bl-j∫Zp(1-x)i+j+n-kdμ-1(x).
From (1.4) and (2.2), we note that(2.24)∫Zp(1-x)ndμ-1(x)=(-1)nEn(-1)=En(2)=2+En-2δ0,n.
By (2.23) and (2.24), we see that(2.25)∫ZpxmBk,n(x)dμ-1(x)=(nk)m+1∑l=0m∑j=0l∑i=0k(-1)i+l(m+1l)(lj)(ki)Bl-j(2+Ei+j+n-k-2δ0,i+j+n-k).
From (2.20) and (2.25), we have(2.26)∑j=0n-k(n-kj)(-1)jEm+k+j=2δ0,km+1∑l=0m∑j=0l(-1)l(m+1l)(lj)Bl-j-2m+1∑l=0m(-1)l(m+1l)Blδk,n+1m+1∑l=0m∑j=0l∑i=0k(-1)i+l(m+1l)(lj)(ki)Bl-jEi+j+n-k=2δ0,km+1∑l=0m∑j=0l(-1)l(m+1l)(lj)Bl-j-2m+1(Bm+1(2)+(-1)mBm+1)δk,n+1m+1∑l=0m∑j=0l∑i=0k(-1)i+l(m+1l)(lj)(ki)Bl-jEi+j+n-k.
Therefore, by (1.19) and (2.26), we obtain the following theorem.Theorem 2.9.
Form,n,k∈ℕ with n≥k, one has
(2.27)∑j=0n-k(n-kj)(-1)jEm+k+j=1m+1∑l=0m∑j=0l∑i=0k(-1)i+l(m+1l)(lj)(ki)Bl-jEi+j+n-k-2m+1(Bm+1+m+1+(-1)mBm+1).
In particular,
(2.28)(2m+2)(E2m+n+1+2)=∑l=02m+1∑j=0l∑i=0n(-1)i+l(2m+2l)(lj)(ni)Bl-jEi+j.



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