# On theq-Euler Numbers and Polynomials with Weight 0

Authors: T. Kim; J. Choi
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The purpose of this paper is to investigate some properties ofq-Euler numbers and polynomials with weight 0. From those q-Euler numbers with weight 0, we derive some identities on the q-Euler numbers and polynomials with weight 0.


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## 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 algebraic closure of ℚp. The p-adic absolute value is defined by |x|p=1/pr where x=prs/t for s,t∈ℤ with (p,t)=(p,s)=1 and r∈ℚ. In this paper, we assume that α∈ℚ and q∈ℂp with |1-q|p<1. As well-known definition, the Euler polynomials are defined by(1.1)2et+1ext=eE(x)t=∑n=0∞En(x)tnn!,
with the usual convention about replacing En(x) by En(x) (see [1–15]).In this special case,x=0, En(0)=En are called the nth Euler numbers (see [1]). Recently, the q-Euler numbers with weight α are defined by(1.2)Ẽ0,q(α)=1,q(qαẼq(α)+1)n+Ẽn,q(α)=0ifn>0,
with the usual convention about replacing (Ẽq(α))n by Ẽn,q(α) (see [3, 12]). The q-number of x is defined by [x]q=(1-qx)/(1-q) (see [1–15]). Note that limq→1[x]q=x. Let us define the notation of q-Euler numbers with weight 0 as Ẽn,q(0)=Ẽn,q. The purpose of this paper is to investigate some interesting identities on the q-Euler numbers with weight 0.


## 2. On the Extendedq-Euler Numbers of Higher-Order with Weight 0

LetC(ℤp) be the space of continuous functions on ℤp. For f∈C(ℤp), the fermionic p-adic q-integral on ℤp is defined by Kim as follows:(2.1)Iq(f)=∫Zpf(x)dμ-q(x)=lim⁡N→∞[2]q1+qpN∑x=0pN-1f(x)(-q)x,
(see [1–12]). By (2.1), we get(2.2)qnIq(fn)+(-1)n-1Iq(f)=[2]q∑l=0n-1(-1)n-1-lf(l)ql,
where fn(x)=f(x+n) and n∈ℕ (see [4, 5]).By (1.2), (2.1), and (2.2), we see that(2.3)∫Zp[x]qαndμ-q(x)=Ẽn,q(α)=[2]q(1-q)n[α]qn∑l=0n(nl)(-1)l11+qαl+1.In the special case,n=1, we get(2.4)∫Zpextdμ-q(x)=[2]qqet+1=1+q-1et+q-1=∑n=0∞Hn(-q-1)tnn!,
where Hn(-q-1) are the nth Frobenius-Euler numbers. From (2.4), we note that the q-Euler numbers with weight 0 are given by(2.5)Ẽn,q=∫Zpxndμ-q(x)=Hn(-q-1),forn∈Z+.Therefore, by (2.5), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.6)Ẽn,q=Hn(-q-1),
where Hn(-q-1) are called the nth Frobenius-Euler numbers.Let us define the generating function of theq-Euler numbers with weight 0 as follows:(2.7)F̃q(t)=∑n=0∞Ẽn,qtnn!.Then, by (2.3) and (2.7), we get(2.8)F̃q(t)=[2]q∑m=0∞(-1)mqmemt=1+qqet+1.Now we define theq-Euler polynomials with weight 0 as follows:(2.9)∑n=0∞Ẽn,q(x)tnn!=1+qqet+1ext.Thus, (2.4) and (2.9), we get(2.10)∫Zpe(x+y)tdμ-q(y)=1+qqet+1ext=∑n=0∞Ẽn,q(x)tnn!.From (2.10), we have(2.11)∑n=0∞Ẽn,q(x)tnn!=(1+q-1et+q-1)ext=∑n=0∞Hn(-q-1,x)tnn!,
where Hn(-q-1,x) are called the nth Frobenius-Euler polynomials (see [9]).Therefore, by (2.11), we obtain the following theorem.Theorem 2.2.
Forn∈ℤ+, one has
(2.12)Ẽn,q(x)=∫Zp(x+y)ndμ-q(x)=Hn(-q-1,x),
where Hn(-q-1,x) are called the nth Frobenius-Euler polynomials.From (2.2) and Theorem 2.2, we note that(2.13)qnHm(-q-1,n)+Hm(-q-1)=[2]q∑l=0n-1(-1)llmql,
where n∈ℕ with n≡1 (mod 2).Therefore, by (2.13), we obtain the following corollary.Corollary 2.3.
Forn∈ℕ, with n≡1 (mod 2) and m∈ℤ+, one has
(2.14)qnHm(-q-1,n)+Hm(-q-1)=[2]q∑l=0n-1(-1)llmql.In particular,q=1, we get Em(n)+Em=2∑l=0n-1(-1)llm, where Em and Em(n) are called the mth Euler numbers and polynomials which are defined by(2.15)2et+1=∑m=0∞Emtmm!,2et+1ext=∑m=0∞Em(x)tmm!.By (2.2), we easily see that(2.16)q∫Zpf(x+1)dμ-q(x)+∫Zpf(x)dμ-q(x)=[2]qf(0).Thus, by (2.16), we get(2.17)[2]q=q∫Zpe(x+1)tdμ-q(x)+∫Zpextdμ-q(x)=∑n=0∞(q∫Zp(x+1)ndμ-q(x)+∫Zpxndμ-q(x))tnn!=∑n=0∞(qHn(-q-1,1)+Hn(-q-1))tnn!.Therefore, by (2.16), we obtain the following theorem.Theorem 2.4.
Forn∈ℤ+, one has
(2.18)qHn(-q-1,1)+Hn(-q-1)={1+q,ifn=0,0,ifn>0,whereHn(-q-1,x) are called the nth Frobenius-Euler polynomials and Hn(-q-1) are called the nth Frobenius-Euler numbers. In particular, q=1, we have(2.19)En(1)+En={2,ifn=0,0,ifn>0,
where En are called the nth Euler numbers.From (2.5) and Theorem 2.2, we note that(2.20)Ẽn,q(x)=∫Zp(x+y)ndμ-q(y)=∑l=0n(nl)∫Zpyldμ-q(y)xn-l=∑l=0n(nl)Ẽn,qxn-l=(x+Ẽq)n,
where the usual convention about replacing (Ẽq)l by Ẽl,q. By Theorems 2.2 and 2.4, we get(2.21)qẼn,q(1)+Ẽn,q={[2]q,ifn=0,0,ifn>0.From (2.20) and (2.21), we have(2.22)q(Ẽq+1)n+Ẽn,q={[2]q,ifn=0,0,ifn>0.Forn∈ℕ, by (2.20) and (2.22), we have(2.23)q2Ẽn,q(2)=q2(Ẽq+1+1)n=q2∑l=1n(nl)(Ẽq+1)l+q(1+q-Ẽ0,q)=q+q2-q∑l=0n(nl)Ẽl,q=q+q2-q(Ẽq+1)n=q+q2+Ẽn,q-q[2]qδ0,n.Therefore, by (2.23), we obtain the following theorem.Theorem 2.5.
Forn∈ℕ, one has
(2.24)q2Ẽn,q(2)=q+q2+Ẽn,q.Forn∈ℤ+, we have(2.25)Ẽn,q-1(1-x)=∫Zp(1-x+x1)ndμ-q-1(x1)=(-1)n∫Zp(x1+x)ndμ-q(x1)=(-1)nẼn,q(x).Therefore, by (2.25), we obtain the following theorem.Theorem 2.6.
Forn∈ℤ+, one has
(2.26)Ẽn,q-1(1-x)=(-1)nẼn,q(x).From (2.20), we have(2.27)∫Zp(1-x)ndμ-q(x)=(-1)n∫Zp(x-1)ndμ-q(x)=(-1)nẼn,q(-1).By Theorem2.6 and (2.27), we get(2.28)∫Zp(1-x)ndμ-q(x)=Ẽn,q-1(2)=1+q+q2Ẽn,q-1ifn>0.Therefore, by (2.28), we obtain the following theorem.Theorem 2.7.
Forn∈ℕ, one has
(2.29)∫Zp(1-x)ndμ-q(x)=1+q+q2Ẽn,q-1.LetC(ℤp) be the space of continuous functions on ℤp. For f∈C(ℤp), p-adic analogue of Bernstein operator of order n for f is given by(2.30)Bn(f∣x)=∑k=0nBk,n(x)f(kn)=∑k=0nf(kn)(nk)xk(1-x)n-k,
where n,k∈ℤ+ (see [1, 6, 7]).Forn,k∈ℤ+, p-adic Bernstein polynomial of degree n is defined by(2.31)Bk,n(x)=(nk)xk(1-x)n-k,x∈Zp
(see [1, 6, 7]).Let us take the fermionicp-adic q-integral on ℤp for one Bernstein polynomials in (2.31) as follows:(2.32)∫ZpBk,n(x)dμ-q(x)=(nk)∫Zpxk(1-x)n-kdμ-q(x)=(nk)∑l=0n-k(n-kl)(-1)l∫Zpxk+ldμ-q(x)=(nk)∑l=0n-k(n-kl)(-1)lẼk+l,q.By simple calculation, we easily get(2.33)∫ZpBk,n(x)dμ-q(x)=∫ZpBn-k,n(1-x)dμ-q(x)=(nk)∑l=0k(kl)(-1)k+l∫Zp(1-x)n-ldμ-q(x)=(nk)∑l=0k(kl)(-1)k+l(1+q+q2Ẽn-l,q-1)=(nk)∑l=0k(kl)(-1)k+lq2Ẽn-l,q-1+[2]q(nk)(-1)kδ0,kifn>k.Therefore, by (2.32) and (2.33), we obtain the following theorem.Theorem 2.8.
Forn∈ℤ+ with n>k>0, one has
(2.34)∑l=0n-k(n-kl)(-1)lẼk+l,q=∑l=0k(kl)(-1)k+lq2Ẽn-l,q-1.In particular,k=0, we get(2.35)∑l=0n(nl)(-1)lẼl,q=q2Ẽn,q-1+[2]q.By Theorems2.1 and 2.2, we get(2.36)∑l=0n-k(n-kl)(-1)lHk+l(-q-1)=∑l=0k(kl)(-1)k+lq2Hn-l(-q),
where n,k∈ℤ+ with n>k>0.



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*Source: 795304-2012-01-15.xml*