We give some new identities on the Bernoulli and Euler numbers by using the bosonic p-adic integral on ℤp and reflection symmetric properties of Bernoulli and Euler polynomials.
1. Introduction
Let p 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
I(f)=∫Zpf(x)dμ(x)=limN→∞∑x=0pN-1f(x)μ(x+pNZp)=limN→∞1pN∑x=0pN-1f(x).
From (1.1), we note thatI(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: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
F(t)=F(t,0)=tet-1=∑n=0∞Bntnn!,
see [1–19]. From (1.4), we can derive the following relation: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 getBn(x)=∑l=0n(nl)Blxn-l=∑l=0n(nl)Bn-lxl.
By (1.2) and (1.3), we easily get∫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:∫Zp(x+y)ndμ(y)=Bn(x),wheren∈Z+,
see [11]. By (1.1) and (1.8), we easily see that∫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:Bn(1-x)=(-1)nBn(x)wheren∈Z+.
The ordinary Euler polynomials are defined by the generating function as follows: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 that2et+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: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.
For n,k∈ℤ+, the Bernstein polynomials are defined byBk,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 the p-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 the p-adic integral on ℤp for the nth ordinary Bernoulli polynomials as follows:
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 getsI1=(-1)n∫ZpBn(1-x)dμ(x).
From (1.5), (1.6), (1.8), and (2.2), one notes thatI1=(-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(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∑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
(1+(-1)n+1)∑l=0n(nl)Bn-lBl=2(-1)nnBn-1+(-1)nnδ1,n-1.
In particular,
∑l=02n-1(2n-1l)B2n-1-lBl=-(2n-1)B2n-2.
By the same motivation, let us also consider the p-adic integral on ℤp for Euler polynomials as follows: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 getsI2=(-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 hasEn(x)=∑l=0n(nl)Elxn-l=∑l=0n(nl)En-lxl=(E+x)n,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 getsI2=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(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.
For n∈ℕ∪{0}, one has
(1+(-1)n-1)∑l=0n(nl)En-lBl=2(-1)nnδ0,n-1.
In particular,
∑l=02n+1(2n+1l)E2n+1-lBl=0,forn∈N.
Let us consider the following p-adic integral on ℤp for the product of Bernoulli and Euler polynomials as follows: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 getsI3=(-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((-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((-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.
For n∈ℕ, one has
((-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
∑k=02m+1∑l=02n(2m+1k)(2nl)B2m+1-kE2n-lBk+l=0.
By the same motivation, we consider the p-adic integral on ℤp for the product of Bernoulli and Bernstein polynomials as follows:I4=∫ZpBm(x)Bk,n(x)dμ(x)wherem,n,k∈N∪{0}.
From (1.6) and (1.14), one getsI4=∑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,
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(-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.
For n,m∈ℕ, one has
∑l=02m∑j=02n(-1)j(2ml)(2nj)B2m-lBl+j=2nB2m+∑l=02m(2ml)B2m-lB2n+l.
Now, we consider the p-adic integral on ℤp for the product of Euler and Bernstein polynomials as follows: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
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(-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.
For n,m∈ℕ, one has
∑l=02m∑j=02n(-1)j(2ml)(2nj)E2m-lBl+j=-2mE2m-1+B2m+2n.
Finally, we consider the p-adic integral on ℤp for the product of Euler, Bernoulli, and Bernstein polynomials as follows: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 getsI6=(-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(-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.
For r,n,s∈ℕ, one has
∑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.
Acknowledgments
The authors express their sincere gratitude to the referees for their valuable suggestions and comments. This paper is supported in part by the Research Grant of Kwangwoon University in 2011.
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