# Some Properties and Identities of Bernoulli and Euler Polynomials Associated withp-adic Integral on ℤp

Authors: D. S. Kim; T. Kim; D. V. Dolgy; S. H. Lee; S.-H. Rim
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We investigate some properties and identities of Bernoulli and Euler polynomials. Further, we give some formulae on Bernoulli and Euler polynomials by usingp-adic integral on ℤp.


---

## Body

## 1. Introduction

Letp be a fixed odd 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 the algebraic closure of ℚp. Let vp be the normalized exponential valuation of ℂp with |p|p=p-vp(p)=1/p.Forf∈UD(ℤp), the p-adic invariant integral on ℤp in the bosonic sense is defined by
(1.1)I(f)=∫Zpf(x)dμ(x)=lim⁡N→∞1pN∑x=0pN-1f(x)dμ(x)
(see [1, 2]). The fermionic p-adic integral on ℤp is defined by Kim as follows:
(1.2)I-1(f)=∫Zpf(x)dμ-1(x)=lim⁡N→∞∑x=0pN-1f(x)(-1)x
(see [3]). As is well known, Bernoulli polynomials are defined by
(1.3)tet-1ext=eB(x)t=∑n=0∞Bn(x)tnn!,
with the usual convention about replacing Bn(x) by Bn(x), symbolically (see [1–19]). In the special case x=0, Bn(0)=Bn is called the nth Bernoulli number.The Euler polynomials are also defined by the generating function as follows:(1.4)2et+1ext=eE(x)t=∑n=0∞En(x)tnn!,
with the usual convention about replacing En(x) by En(x), symbolically (see [1–19]). In the special case x=0, En(0)=En is called the n-th Euler number.By (1.3) and (1.4), we easily see that
(1.5)Bn(x)=(B+x)n=∑l=0n(nl)xn-lBl,En(x)=(E+x)n=∑l=0n(nl)xn-lEl,
where (nl)=n!/(n-l)!l!=n(n-1)(n-2)⋯(n-l+1)/l!  (see [14, 16, 19]).The following properties of Bernoulli numbers and polynomials are well known (see [10, 11]).Forn∈ℤ+=ℕ∪{0},
(1.6)∑j=0n(nj)yn-jBj+1(x)j+1=Bn+1(x+y)-yn+1n+1,(1.7)∑j=0[n/2](n2j)yn-2jB2j+1(x)2j+1=Bn+1(x+y)+(-1)nBn+1(x-y)2n+2,(1.8)∑j=1[(n+1)/2](n2j-1)yn+1-2jB2j(x)2j=Bn+1(x+y)+(-1)n-1Bn+1(x-y)-2yn+12n+2,
where [·] is Gauss’ symbol.First, we investigate some identities of Euler polynomials corresponding to (1.6), (1.7) and (1.8). From those identities, we derive some interesting identities and properties by using p-adic integral on ℤp.


## 2. Some Identities of Bernoulli and Euler Polynomials

By (1.4), we get
(2.1)Ek(x+y)=∑j=0k(kj)yk-jEj(x),for∈Z+.
From (2.1), we note that
(2.2)Ek(x+y)=∑j=0k(kj)yk-jEj(x)=yk+∑j=1kkj(k-1j-1)yk-jEj(x).
Thus, we have
(2.3)∑j=0k-1(k-1j)yk-1-jEj+1(x)j+1=Ek(x+y)-ykk.
Replacing k by k+1 in (2.3), we obtain the following proposition.Proposition 2.1.
Fork∈ℤ+, one has
(2.4)∑j=0k(kj)yk-jEj+1(x)j+1=Ek+1(x+y)-yk+1k+1.Let us replacey by -y in Proposition 2.1. Then we have
(2.5)∑j=0k(kj)(-1)k-jyk-jEj+1(x)j+1=Ek+1(x-y)-(-1)k+1yk+1k+1.
Thus, we see that
(2.6)∑j=0k(kj)(-1)jyk-jEj+1(x)j+1=(-1)kEk+1(x-y)+yk+1k+1.
Therefore, adding (2.4) and (2.6), we obtain the following proposition.Proposition 2.2.
Fork∈ℤ+, one has
(2.7)∑j=0[k/2](k2j)yk-2jE2j+1(x)2j+1=Ek+1(x+y)+(-1)kEk+1(x-y)2k+2.From (2.2), we note that
(2.8)∑j=1k1j(k-1j-1)yk-j(-1)jEj(x)=(-1)kEk(x-y)-ykk.
By (2.3) and (2.8), we get
(2.9)∑j=1[k/2]12j(k-12j-1)yk-2jE2j(x)=Ek(x+y)+(-1)kEk(x-y)-2yk2k.
Therefore, replacing k by k+1, we obtain the following proposition.Proposition 2.3.
Fork∈ℕ, one has
(2.10)∑j=1[(k+1)/2](k2j-1)yk+1-2jE2j(x)2j=Ek+1(x+y)+(-1)k+1Ek+1(x-y)-2yk+12k+2.Lettingy=1 in Proposition 2.1, we have
(2.11)∑j=0k(kj)Ej+1(x)j+1=Ek+1(x+1)-1k+1,(2.12)Ek+1(x+1)=∑l=0k+1(k+1l)(E+1)lxk+1-l=(2-E0)xk+1-∑l=1k+1(k+1l)Elxk+1-l=2xk+1-∑l=0k+1(k+1l)Elxk+1-l=2xk+1-Ek+1(x).
Therefore, by (2.11) and (2.12), we obtain the following corollary.Corollary 2.4.
Fork∈ℤ+, one has
(2.13)∑j=0k(kj)Ej+1(x)j+1=-Ek+1(x)k+1+2xk+1-1k+1.Replacingy by 1 and k by 2k in Proposition 2.2, we have
(2.14)∑j=0k(2k2j)E2j+1(x)2j+1=E2k+1(x+1)+E2k+1(x-1)4k+2=E2k+1(x+1)+E2k+1(x)+E2k+1(x)+E2k+1(x-1)4k+2-2E2k+1(x)4k+2=2x2k+1+2(x-1)2k+14k+2-E2k+1(x)2k+1.
Therefore, by (2.14), we obtain the following corollary.Corollary 2.5.
Fork∈ℤ+, one has
(2.15)∑j=0k(2k2j)E2j+1(x)2j+1=-E2k+1(x)2k+1+x2k+1+(x-1)2k+12k+1.Replacingy by 1 and k by 2k in Proposition 2.3, we have
(2.16)∑j=1k(2k2j-1)E2j(x)2j=E2k+1(x+1)-E2k+1(x-1)-24k+2=(E2k+1(x+1)+E2k+1(x))-(E2k+1(x)+E2k+1(x-1))4k+2-12k+1=2x2k+1-2(x-1)2k+14k+2-12k+1=x2k+1-(x-1)2k+12k+1-12k+1.
Therefore, by (2.16), we obtain the following corollary.Corollary 2.6.
Fork∈ℕ, one has
(2.17)∑j=1k(2k2j-1)E2j(x)2j=x2k+1-(x-1)2k+12k+1-12k+1.Replacingy by 1/2 and k by 2k in Proposition 2.3, we get
(2.18)∑j=1k(2k2j-1)(12)2k+1-2jE2j(x)2j=E2k+1(x+1/2)-E2k+1(x-1/2)-2-2k4k+2.
Thus, we have
(2.19)∑j=1k(2k2j-1)22jE2j(x)2j=22k(E2k+1(x+1/2)-E2k+1(x-1/2))-12k+1,(2.20)E2k+1(x+12)=E2k+1(x-12+1)=∑l=02k+1(2k+1l)(x-12)2k+1-l(E+1)l=2(x-12)2k+1-∑l=02k+1(2k+1l)(x-12)2k+1-lEl=2(x-12)2k+1-E2k+1(x-12).
Therefore, by (2.19) and (2.20), we obtain the following corollary.Corollary 2.7.
Fork∈ℕ, we have
(2.21)∑j=1k(2k2j-1)22jE2j(x)2j=-22k+1E2k+1(x-1/2)2k+1+22k+1(x-1/2)2k+12k+1-12k+1.Replacingy by 1 and k by 2k+1 in Proposition 2.2, we get
(2.22)∑j=0k(2k+12j)E2j+1(x)2j+1=E2k+2(x+1)-E2k+2(x-1)4k+4=(E2k+2(x+1)+E2k+2(x))-(E2k+2(x)+E2k+2(x-1))4k+4=2x2k+2-2(x-1)2k+24k+4=x2k+2-(x-1)2k+22k+2.
Therefore, by (2.22), we obtain the following corollary.Corollary 2.8.
Fork∈ℤ+, one has
(2.23)∑j=0k(2k+12j)E2j+1(x)2j+1=x2k+2-(x-1)2k+22k+2.Replacingk by 2k+1 and y by 1 in Proposition 2.3, we get(2.24)∑j=1k+1(2k+12j-1)E2j(x)2j=E2k+2(x+1)+E2k+2(x-1)-24k+4=(E2k+2(x+1)+E2k+2(x))+(E2k+2(x)+E2k+2(x-1))4k+4-E2k+2(x)+12k+2=x2k+2+(x-1)2k+22k+2-E2k+2(x)+12k+2.
Therefore, by (2.24), we obtain the following corollary.Corollary 2.9.
Fork∈ℤ+, we have
(2.25)∑j=1k+1(2k+12j-1)E2j(x)2j=x2k+2+(x-1)2k+22k+2-E2k+2(x)+12k+2.Replacingk by 2k+1 and y by 1/2 in Proposition 2.2, we have
(2.26)∑j=0k(2k+12j)(12)2k+1-2jE2j+1(x)2j+1=E2k+2(x+1/2)-E2k+2(x-1/2)4k+4.
Thus, by multipling 22k+1 on both sides, we get
(2.27)∑j=0k(2k+12j)22jE2j+1(x)2j+1=22k+1{E2k+2(x+1/2)-E2k+2(x-1/2)}4k+4.
By (2.20) and (2.27), we see that
(2.28)∑j=0k(2k+12j)22jE2j+1(x)2j+1=22k(2(x-1/2)2k+2-2E2k+2(x-1/2))2k+2=22k(x-1/2)2k+2-22kE2k+2(x-1/2)k+1.
Therefore, by (2.28), we obtain the following corollary.Corollary 2.10.
Fork∈ℤ+, we have
(2.29)∑j=0k(2k+12j)22jE2j+1(x)2j+1=-22kE2k+2(x-1/2)k+1+22k(x-1/2)2k+2k+1.From (1.6), we can derive the following equation:
(2.30)∑j=0k-1(kj)Bj+1(x)j+1=xk-1k+1,fork∈N.
Let us take the p-adic integral on both sides in (2.30) as follows: for k∈ℕ,
(2.31)I1=∑j=0k-1(kj)∫ZpBj+1(x)j+1dμ(x)=∑j=0k-1(kj)1j+1∑l=0j+1(j+1l)Bj+1-l∫Zpxldμ(x)=∑j=0k-1∑l=0j+11j+1(kj)(j+1l)Bj+1-lBl.
On the other hand,
(2.32)I1=∫Zpxkdμ(x)-1k+1∫Zpdμ(x)=Bk-1k+1.
Therefore, by (2.31) and (2.32), we obtain the following theorem.Theorem 2.11.
Fork∈ℕ, one has
(2.33)∑j=0k-1∑l=0j+11j+1(kj)(j+1l)Bj+1-lBl=Bk-1k+1.In (2.30), let us take the fermionic p-adic integral on both sides as follows:
(2.34)I2=∑j=0k-1(kj)1j+1∫ZpBj+1(x)dμ-1(x)=∑j=0k-1(kj)j+1∑l=0j+1(j+1l)Bj+1-l∫Zpxldμ-1(x)=∑j=0k-1∑l=0j+11j+1(kj)(j+1l)Bj+1-lEl.
On the other hand
(2.35)I2=∫Zpxkdμ-1(x)-1k+1∫Zpdμ-1(x)=Ek-1k+1.
Therefore, by (2.34) and (2.35), we obtain the following theorem.Theorem 2.12.
Fork∈ℕ, one has
(2.36)∑j=0k-1∑l=0j+11j+1(kj)(j+1l)Bj+1-lEl=Ek-1k+1.From (1.7), we can easily derive the following equation:
(2.37)∑j=0k-1(2k2j)B2j+1(x)2j+1=x2k-(x-1)2k2.
Let us take ∫ℤpdμ(x) on both sides in (2.37). Then we have
(2.38)I3=∑j=0k-1(2k2j)12j+1∫ZpB2j+1(x)dμ(x)=∑j=0k-1(2k2j)12j+1∑l=02j+1(2j+1l)B2j+1-l∫Zpxldμ(x)=∑j=0k-1∑l=02j+112j+1(2k2j)(2j+1l)B2j+1-lBl.
On the other hand,
(2.39)I3=12(∫Zpx2kdμ(x)-∫Zp(x-1)2kdμ(x))=12(B2k-B2k(-1))=12(B2k-B2k(2))=12(B2k-(2k+δ1,2k+B2k)),
where δn,k is a Kronecker symbol.Therefore, by (2.38) and (2.39), we obtain the following theorem.Theorem 2.13.
Fork∈ℕ, one has
(2.40)∑j=0k-1∑l=02j+112j+1(2k2j)(2j+1l)B2j+1-lBl=-k.Taking∫ℤpdμ-1(x) on both sides in (2.37), we get
(2.41)I4=∑j=0k-1(2k2j)12j+1∑l=02j+1(2j+1l)B2j+1-l∫Zpxldμ-1(x)=∑j=0k-1∑l=02j+112j+1(2k2j)(2j+1l)B2j+1-lEl.
On the other hand(2.42)I4=12(∫Zpxkdμ-1(x)-∫Zp(x-1)2kdμ-1(x))=12(∫Zpx2kdμ-1(x)-∫Zp(x+2)2kdμ-1(x))=12(E2k-E2k(2))=12{E2k-(2+E2k-2δ0,2k)}=-1+δ0,k.
Therefore, by (2.41) and (2.42), we obtain the following theorem.Theorem 2.14.
Fork∈ℕ, one has
(2.43)∑j=0k-1∑l=02j+112j+1(2k2j)(2j+1l)B2j+1-lEl=-1.From (1.8), we can also derive the following equation:
(2.44)∑j=1k(2k2j-1)B2j(x)2j=x2k+(x-1)2k2-12k+1.
Let us take the bosonic p-adic integral on both sides in (2.44). Then we get(2.45)I5=∑j=1k(2k2j-1)12j∫ZpB2j(x)dμ(x)=∑j=1k(2k2j-1)12j∑l=02j(2jl)B2j-l∫Zpxldμ(x)=∑j=1k∑l=02j12j(2k2j-1)(2jl)B2j-lBl.
On the other hand,
(2.46)I5=12∫Zp(x2k+(-1+x)2k)dμ(x)-12k+1∫Zpdμ(x)=12∫Zp(x2k+(x+2)2k)dμ(x)-12k+1=12(B2k+B2k(2))-12k+1=12(B2k+2k+B2k+δ1,2k)-12k+1.
Therefore, by (2.45) and (2.46), we obtain the following theorem.Theorem 2.15.
Fork∈ℕ, one has
(2.47)∑j=1k∑l=02j12j(2k2j-1)(2jl)B2j-lBl=B2k+k-12k+1.Now, let us consider the fermionicp-adic integral on both sides in (2.44):
(2.48)I6=∑j=1k(2k2j-1)12j∑l=02j(2jl)B2j-l∫Zpxldμ-1(x)=∑j=1k∑l=02j12j(2k2j-1)(2jl)B2j-lEl.
On the other hand,
(2.49)I6=12∫Zp(x2k+(x-1)2k)dμ-1(x)-12k+1∫Zpdμ-1(x)=12∫Zp(x2k+(x+2)2k)dμ-1(x)-12k+1=12(E2k+E2k(2))-12k+1=12(E2k+(2+E2k-2δ0,2k))-12k+1=E2k+1-δ0,2k-12k+1=2k2k+1.
Therefore, by (2.48) and (2.49), we obtain the following theorem.Theorem 2.16.
Fork∈ℕ, one has
(2.50)∑j=1k∑l=02j12j(2k2j-1)(2jl)B2j-lEl=2k2k+1.



---

*Source: 847901-2012-04-09.xml*