# Derivation of Identities Involving Bernoulli and Euler Numbers

Authors: Imju Lee; Dae San Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We derive some new and interesting identities involving Bernoulli and Euler numbers by using some polynomial identities andp-adic integrals on ℤp.


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## Body

## 1. Introduction and Preliminaries

Letp be a fixed odd prime. Throughout this paper, ℤp,ℚp,ℂp will, respectively, denote the ring of p-adic integers, the field of p-adic rational numbers, and the completion of algebraic closure of ℚp. The p-adic absolute value ||p on ℂp is normalized so that |p|p=1/p. Let ℤ>0 be the set of natural numbers and ℤ≥0=ℤ>0∪{0}.As is well known, the Bernoulli polynomialsBn(x) are defined by the generating function as follows:
(1.1)F(t,x)=tet-1ext=eB(x)t=∑n=0∞‍Bn(x)tnn!,
with the usual convention of replacing B(x)n by Bn(x).In the special case,x=0, Bn(0)=Bn is referred to as the nth Bernoulli number. That is, the generating function of Bernoulli numbers is given by
(1.2)F(t)=F(t,0)=tet-1=∑n=0∞‍Bntnn!=eBt,
with the usual convention of replacing Bn by Bn, (cf. [1–23]).From (1.2), we see that the recurrence formula for the Bernoulli numbers is
(1.3)(B+1)n-Bn=δ1,n,forn∈Z≥0,
where δk,n is the Kronecker symbol.By (1.1) and (1.2), we easily get the following:
(1.4)Bn(x)=(B+x)n=∑l=0n‍(nl)Blxn-l=∑l=0n‍(nl)Bn-lxl,forn∈Z≥0.
Let UD(ℤp) be the space of uniformly differentiable ℂp-valued functions on ℤp. For f∈UD(ℤp), the bosonic p-adic integral on ℤp is defined by
(1.5)I(f)=∫Zp‍f(x)dμ(x)=lim⁡N→∞1pN∑x=0pN-1‍f(x),(cf. [12]). Then it is easy to see that
(1.6)I(f1)=I(f)+f′(0),
where f1(x)=f(x+1) and f′(0)=df(x)/dx|x=0.By (1.6), we have the following:
(1.7)∫Zp‍e(x+y)tdμ(y)=tet-1ext=∑n=0∞‍Bn(x)tnn!,(cf. [12–14]). From (1.7), we can derive the Witt's formula for the nth Bernoulli polynomial as follows:
(1.8)∫Zp‍(x+y)ndμ(y)=Bn(x),forn∈Z≥0.By (1.1), we have the following:
(1.9)Bn(1-x)=(-1)nBn(x),forn∈Z≥0.
Thus, from (1.3), (1.4), and (1.9), we have the following:
(1.10)Bn(1)=Bn+δ1,n=(-1)nBn,forn∈Z≥0.
By (1.4), we have the following:
(1.11)Bn(x+y)=∑k=0n‍(nk)Bk(x)yn-k,forn∈Z≥0.
Especially, for x=1 and y=1,
(1.12)Bn(2)=∑k=0n‍(nk)Bk(1)=∑k=0n‍(nk)(Bk+δ1,k),forn∈Z≥0.
Therefore, from (1.9), (1.10), and (1.12), we can derive the following relation. For n∈ℤ≥0,
(1.13)(-1)nBn(-1)=Bn(2)=n+Bn(1)=n+Bn+δ1,n=n+(-1)nBn.
Let f(y)=(x+y)n+1. By (1.6), we have the following:
(1.14)∫Zp‍(x+y+1)n+1dμ(y)-∫Zp‍(x+y)n+1dμ(y)=(n+1)xn,forn∈Z≥0.
By (1.8) and (1.14), we have the following:
(1.15)Bn+1(x+1)-Bn+1(x)=(n+1)xn,forn∈Z≥0.
Thus, by (1.11) and (1.15), we have the following identity.
(1.16)xn=1n+1∑l=0n‍(n+1l)Bl(x),forn∈Z≥0.As is well known, the Euler polynomialsEn(x) are defined by the generating function as follows:
(1.17)G(t,x)=2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!,
with the usual convention of replacing E(x)n by En(x).In the special case,x=0, En(0)=En is referred to as the nth Euler number. That is, the generating function of Euler numbers is given by
(1.18)G(t)=G(t,0)=2et+1=∑n=0∞‍Entnn!=eEt,
with the usual convention of replacing En by En, (cf. [1–23]).From (1.18), we see that the recurrence formula for the Euler numbers is
(1.19)(E+1)n+En=2δ0,n,forn∈Z≥0.
By (1.17) and (1.18), we easily get the following:
(1.20)En(x)=(E+x)n=∑l=0n‍(nl)Elxn-l=∑l=0n‍(nl)En-lxl,forn∈Z≥0.
Let C(ℤp) be the space of continuous ℂp-valued functions on ℤp. For f∈C(ℤp), the fermionic p-adic integral on ℤp is defined by Kim as follows:
(1.21)I-1(f)=∫Zp‍f(x)dμ-1(x)=lim⁡N→∞∑x=0pN-1‍f(x)(-1)x,(cf. [9]).  Then it is easy to see that
(1.22)I-1(f1)+I-1(f)=2f(0),
where f1(x)=f(x+1).By (1.22), we have the following:
(1.23)∫Zp‍e(x+y)tdμ-1(y)=2et+1ext=∑n=0∞‍En(x)tnn!.
From (1.23), we can derive the Witt's formula for the n-th Euler polynomial as follows:
(1.24)∫Zp‍(x+y)ndμ-1(y)=En(x),forn∈Z≥0.
By (1.17), we have the following:
(1.25)En(1-x)=(-1)nEn(x),forn∈Z≥0.
Thus, from (1.19), (1.20), and (1.25), we have the following:
(1.26)En(1)=-En+2δ0,n=(-1)nEn,forn∈Z≥0.
By (1.20), we have the following:
(1.27)En(x+y)=∑k=0n‍(nk)Ek(x)yn-k,forn∈Z≥0.
Especially, for x=1 and y=1,
(1.28)En(2)=∑k=0n‍(nk)Ek(1)=∑k=0n‍(nk)(-En+2δ0,k),forn∈Z≥0.
Therefore, from (1.25), (1.26), and (1.28), we can derive the following relations. For n∈ℤ≥0,
(1.29)(-1)nEn(-1)=En(2)=2-En(1)=2+En-2δ0,n=2-(-1)nEn.
Let f(y)=(x+y)n. By (1.22), we have the following:
(1.30)∫Zp‍(x+y+1)ndμ-1(y)+∫Zp‍(x+y)ndμ-1(y)=2xn,forn∈Z≥0.
By (1.24) and (1.30), we have the following:
(1.31)En(x+1)+En(x)=2xn,forn∈Z≥0.
Thus, by (1.27) and (1.31), we get the following identity.
(1.32)xn=12∑l=0n-1‍(nl)El(x)+En(x),forn∈Z≥0.The Bernstein polynomials are defined by(1.33)Bk,n(x)=(nk)xk(1-x)n-k,fork,n∈Z≥0,
with 0≤k≤n (cf. [14]).By the definition ofBk,n(x), we note that
(1.34)Bk,n(x)=Bn-k,n(1-x).In this paper, we derive some new and interesting identities involving Bernoulli and Euler numbers from well-known polynomial identities. Here, we note that our results are “complementary” to those in [6], in the sense that we take a fermionic p-adic integral where a bosonic p-adic integral is taken and vice versa, and we use the identity involving Euler polynomials in (1.32) where that involving Bernoulli polynomials in (1.16) is used and vice versa. Finally, we report that there have been a lot of research activities on this direction of research, namely, on derivation of identities involving Bernoulli and Euler numbers and polynomials by exploiting bosonic and fermionic p-adic integrals (cf. [6–8]).


## 2. Identities Involving Bernoulli Numbers

Taking the bosonicp-adic integral on both sides of (1.16), we have the following:
(2.1)∫Zp‍xmdμ(x)=∫Zp‍1m+1∑k=0m‍(m+1k)Bk(x)dμ(x)=1m+1∑k=0m‍(m+1k)∫Zp‍Bk(x)dμ(x)=1m+1∑k=0m‍(m+1k)∑j=0k‍(kj)Bk-j∫Zp‍xjdμ(x)=1m+1∑k=0m‍(m+1k)∑j=0k‍(kj)Bk-jBj.
Therefore, we obtain the following theorem.Theorem 2.1.
Letm∈ℤ≥0. Then on has the following:
(2.2)Bm=1m+1∑k=0m‍∑j=0k‍(m+1k)(kj)Bk-jBj.Let us apply (1.9) to the bosonic p-adic integral of (1.16). 						(2.3)∫Zp‍xmdμ(x)=1m+1∑k=0m‍(m+1k)∫Zp‍Bk(x)dμ(x)=1m+1∑k=0m‍(m+1k)(-1)k∫Zp‍Bk(1-x)dμ(x)=1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-j∫Zp‍(1-x)jdμ(x)=1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-j(-1)jBj(-1).
Then, we can express (2.3) in three different ways.By (1.13), (2.3) can be written as
(2.4)∫Zp‍xmdμ(x)=1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-j(j+Bj+δ1,j)=1m+1∑k=0m‍(m+1k)(-1)k(kBk-1(1)+kBk-1+∑j=0k‍(kj)Bk-jBj)=-∑k=0m-1‍(mk)(Bk+(-1)kBk)+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-jBj=-∑k=0m-1‍(mk)(Bk+Bk+δ1,k)+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-jBj=-2(Bm(1)-Bm)-(m-δ1,m)+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-jBj=-δ1,m-m+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-jBj.
Thus, we have the following theorem.Theorem 2.2.
Letm∈ℤ≥0. Then one has the following:
(2.5)Bm=-δ1,m-m+1m+1∑k=0m‍∑j=0k‍(m+1k)(kj)(-1)kBk-jBj.Corollary 2.3.
Letm be an integer ≥2. Then one has the following:
(2.6)Bm+m=1m+1∑k=0m‍∑j=0k‍(m+1k)(kj)(-1)kBk-jBj.Especially, for an odd integerm with m≥3, we obtain the following corollary.Corollary 2.4.
Letm be an odd integer with m≥3. Then one has the following:
(2.7)m(m+1)=∑k=0m‍∑j=0k‍(m+1k)(kj)(-1)kBk-jBj.By (1.13), (2.3) can be written as
(2.8)∫Zpxmdμ(x)‍=1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)Bk-j(j+(-1)jBj)=1m+1∑k=0m‍(m+1k)(-1)k(kBk-1(1)+∑j=0k‍(kj)(-1)jBk-jBj)=-∑k=0m-1‍(mk)Bk+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)(-1)jBk-jBj=-Bm(1)+Bm+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)(-1)jBk-jBj.
By (1.10), (2.8) can be written as
(2.9)∫Zp‍xmdμ(x)=(-1)m+1Bm+Bm+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)(-1)jBk-jBj.
So, we get the following theorem.Theorem 2.5.
Letm∈ℤ≥0. Then one has the following:
(2.10)Bm=1m+1∑k=0m‍∑j=0k‍(m+1k)(kj)(-1)m+k+jBk-jBj.By (1.10), (2.8) can also be written as
(2.11)∫Zp‍xmdμ(x)=-δ1,m+1m+1∑k=0m‍(m+1k)(-1)k∑j=0k‍(kj)(-1)jBk-jBj.
Thus, we have the following theorem.Theorem 2.6.
Letm∈ℤ≥0. Then one has the following:
(2.12)Bm=-δ1,m+1m+1∑k=0m‍∑j=0k‍(m+1k)(kj)(-1)k+jBk-jBj.


## 3. Identities Involving Euler Numbers

Taking the fermionicp-adic integral on both sides of (1.32), we have the following:
(3.1)∫Zpxmdμ-1(x)‍=∫Zp‍(Em(x)+12∑k=0m-1‍(mk)Ek(x))dμ-1(x)=∑l=0m‍(ml)Em-l∫Zp‍xldμ-1(x)+12∑k=0m-1‍(mk)∑j=0k‍(kj)Ek-j∫Zp‍xjdμ-1(x)=∑l=0m‍(ml)Em-lEl+12∑k=0m-1‍(mk)∑j=0k‍(kj)Ek-jEj.
So, we obtain the following theorem.Theorem 3.1.
Letm∈ℤ≥0. Then one has the following:
(3.2)Em=∑l=0m‍(ml)Em-lEl+12∑k=0m-1‍∑j=0k‍(mk)(kj)Ek-jEj.Let us apply (1.25) to the fermionic p-adic integral of (1.32).
(3.3)∫Zp‍xmdμ-1(x)=(-1)m∫Zp‍Em(1-x)dμ-1(x)+12∑k=0m-1‍(mk)(-1)k∫Zp‍Ek(1-x)dμ-1(x)=(-1)m∑k=0m‍(mk)Em-k(-1)kEk(-1)=+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-j(-1)jEj(-1).
Then, we can express (3.3) in two different ways.By (1.29), (3.3) can be written as
(3.4)∫Zpxmdμ-1(x)‍=(-1)m∑k=0m‍(mk)Em-k(2+Ek-2δ0,k)=+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-j(2+Ej-2δ0,j)=2Em+(-1)m∑k=0m‍(mk)Em-kEk+2(-1)m+1Em+∑k=0m-1‍(mk)Ek=+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-jEj+∑k=0m-1‍(mk)(-1)k+1Ek=2Em+(-1)m∑k=0m‍(mk)Em-kEk+2(-1)m+1Em+Em(1)-Em=+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-jEj+(-1)m+1(Em(-1)-Em)=-2+2δ0,m+(-1)m∑k=0m‍(mk)Em-kEk+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-jEj.
Thus, we get the following theorem.Theorem 3.2.
Letm∈ℤ≥0. Then one has the following:
(3.5)Em=-2+2δ0,m+(-1)m∑k=0m‍(mk)Em-kEk+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-jEj.Corollary 3.3.
Letm∈ℤ>0. Then one has the following:
(3.6)Em+2=(-1)m∑k=0m‍(mk)Em-kEk+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-jEj.By (1.29), (3.3) can be written as
(3.7)∫Zpxmdμ-1(x)‍=(-1)m∑k=0m‍(mk)Em-k(2-(-1)kEk)=+12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)kEk-j(2-(-1)jEj)=2Em+(-1)m+1∑k=0m‍(mk)(-1)kEm-kEk=+∑k=0m-1‍(mk)Ek-12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)k(-1)jEk-jEj=2Em+(-1)m+1∑k=0m‍(mk)(-1)kEm-kEk=+Em(1)-Em-12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)k(-1)jEk-jEj=2δ0,m+(-1)m+1∑k=0m‍(mk)(-1)kEm-kEk=-12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)k(-1)jEk-jEj.
So, we have the following theorem.Theorem 3.4.
Letm∈ℤ≥0. Then one has the following:
(3.8)Em=2δ0,m+(-1)m+1∑k=0m‍(mk)(-1)kEm-kEk-12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)k+jEk-jEj.Corollary 3.5.
Letm∈ℤ>1. Then one has the following:
(3.9)Em=(-1)m+1∑k=0m‍(mk)(-1)kEm-kEk-12∑k=0m-1‍∑j=0k‍(mk)(kj)(-1)k+jEk-jEj.


## 4. Identities Involving Bernoulli and Euler Numbers

By (1.16) and (1.32), we have the following:
(4.1)∫Zpxm+ndμ(x)‍=∫Zp‍(1m+1∑k=0m‍(m+1k)Bk(x))(En(x)+12∑l=0n-1‍(nl)El(x))dμ(x)=1m+1∑k=0m‍(m+1k)∫Zp‍Bk(x)En(x)dμ(x)=+12(m+1)∑k=0m‍(m+1k)∑l=0n-1‍(nl)∫Zp‍Bk(x)El(x)dμ(x)=1m+1∑k=0m‍∑j=0k‍∑l=0n‍(m+1k)(kj)(nl)Bk-jEn-lBj+l=+12(m+1)∑k=0m‍∑l=0n-1‍∑j=0k‍∑i=0l‍(m+1k)(nl)(kj)(li)Bk-jEl-iBj+i.
Therefore, we get the following theorem.Theorem 4.1.
Letm,n∈ℤ≥0. Then one has the following:
(4.2)Bm+n=1m+1∑k=0m‍∑j=0k‍∑l=0n‍(m+1k)(kj)(nl)Bk-jEn-lBj+l+12(m+1)∑k=0m‍∑l=0n-1‍∑j=0k‍∑i=0l‍(m+1k)(nl)(kj)(li)Bk-jEl-iBj+i.By (1.16) and (1.33), we have the following:
(4.3)∫ZpxmBk,n(x)dμ(x)‍=∫Zp‍1m+1∑l=0m‍(m+1l)Bl(x)Bk,n(x)dμ(x)=1m+1(nk)∑l=0m‍∑i=0l‍(m+1l)(li)Bl-i∫Zp‍xi+k(1-x)n-kdμ(x)=1m+1(nk)∑l=0m‍∑i=0l‍∑j=0n-k‍(m+1l)(li)(n-kj)(-1)jBl-i∫Zp‍xi+k+jdμ(x)=1m+1(nk)∑l=0m‍∑i=0l‍∑j=0n-k‍(m+1l)(li)(n-kj)(-1)jBl-iBi+k+j.
By (1.33), we have the following:
(4.4)∫Zp‍xmBk,n(x)dμ(x)=(nk)∫Zp‍xm+k(1-x)n-kdμ(x)=(nk)∑j=0n-k‍(n-kj)(-1)j∫Zp‍xm+k+jdμ(x)=(nk)∑j=0n-k‍(n-kj)(-1)jBm+k+j.
By (4.3) and (4.4), we obtain the following theorem.Theorem 4.2.
Letm,n,k∈ℤ≥0. Then one has the following:
(4.5)∑j=0n-k‍(n-kj)(-1)jBm+k+j=1m+1∑l=0m‍∑i=0l‍∑j=0n-k‍(m+1l)(li)(n-kj)(-1)jBl-iBi+k+j.
Especially, one has the following:
(4.6)(m+1)Bm+n=∑l=0m‍∑i=0l‍(m+1l)(li)Bl-iBi+n.By (4.2) and (4.6), we have the following theorem. Note that (4.8) in the following was obtained in [6].Theorem 4.3.
Letm,n∈ℤ≥0. Then one has the following:
(4.7)Bm+n=∑l=0n‍(nl)En-lBm+l+12∑l=0n-1‍∑i=0l‍(nl)(li)El-iBm+i.In particular, we have the following:(4.8)Bn=∑l=0n‍(nl)En-1Bl+12∑l=0n-1‍∑i=0l‍(nl)(li)El-iBi.



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