# On the Higher-Orderq-Euler Numbers and Polynomials with Weight α

Authors: K.-W. Hwang; D. V. Dolgy; T. Kim; S. H. Lee
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

The main purpose of this paper is to present a systemic study of some
families of higher-orderq-Euler numbers and polynomials with weight α. In particular, by
using the fermionic p-adic q-integral on ℤp, we give a new concept of q-Euler numbers
and polynomials with weight α.


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## Body

## 1. Introduction

Letp be a fixed odd prime. Throughout this paper ℤp, ℚp, ℂ, and ℂp, will, respectively, denote the ring of p-adic rational integers, the field, of p-adic rational numbers, the complex number field and the completion of algebraic closure of ℚp. Let ℕ be the set of natural numbers and ℤ+=ℕ∪{0}. Let νp be the normalized exponential valuation of ℂp with |p|p=p-νp(p)=p-1 (see [1–14]). When one speaks of q-extension, q can be regarded as an indeterminate, complex number q∈ℂ, or p-adic number q∈ℂp; it is always clear from context. If q∈ℂ, we assume |q|<1. If q∈ℂp, then we assume |1-q|p<1 (see [1–14]).In this paper, we use the notation ofq-number as follows: (1.1)[x]q=1-qx1-q(see [1–14]). Note that limq→1[x]q=x for any x with |x|p≤1 in the p-adic case.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 (1.2)I-q(f)=∫Zpf(x)dμ-q(x)=lim⁡N→∞1[pN]-q∑x=0pN-1f(x)(-q)x,=lim⁡N→∞[2]q2∑x=0pN-1f(x)(-q)x
(see [4–7]).From (1.2), we note that (1.3)qI-q(f1)+I-q(f)=[2]qf(0),
where f1(x)=f(x+1).It is well known that the ordinary Euler polynomials are defined by(1.4)2et+1ext=eE(x)t=∑n=0∞En(x)tnn!,
with the usual convention of replacing En(x) by En(x).In the special case,x=0 and En(0)=En are called the nth Euler numbers (see [1–14]).By (1.5), we get the following recurrence relation as follows:(1.5)E0=1,(E+1)n+E={2,ifn=0,0,ifn>0.Recently,(h,q)-Euler numbers are defined by (1.6)E0,q(h)=21+qh,qh(qEq(h)+1)n+Eq(h)={2,ifn=0,0,ifn>0,
with the usual convention about replacing (Eq(h))n by En,q(h) (see [1–12]).Note thatlimq→1En,q(h)=En.Forα∈ℕ, the weight q-Euler numbers are also defined by (1.7)Ẽ0,q(α)=1,q(qαẼq(α)+1)n+Ẽn,q(α)={[2]q,ifn=0,0,ifn>0,
with the usual convention about replacing (Ẽq(α))n by Ẽn,q(α) (see [4]).The purpose of this paper is to present a systemic study of some families of higher-orderq-Euler numbers and polynomials with weight α. In particular, by using the fermionic p-adic q-integral on ℤp, we give a new concept of q-Euler numbers and polynomials with weight α.


## 2. Higher-Orderq-Euler Numbers and Polynomials with Weight α

Forh∈ℤ,α,k∈ℕ, and n∈ℤ+, let us consider the expansion of higher-order q-Euler polynomials with weight α as follows:(2.1)Ẽn,q(α)(h,k∣x)=∫Zp⋯∫Zp︸k-times[x1+⋯+xk+x]qαnqx1(h-1)+⋯+xk(h-k)dμ-q(x1)⋯dμ-q(xk).
From (1.2) and (2.1), we note that:(2.2)Ẽn,q(α)(h,k∣x)=[2]qk(1-qα)n∑l=0n(nl)(-1)lqαlx(1+qαl+h)⋯(1+qαl+h-k+1).
In the special case, x=0, Ẽn,q(α)(h,k∣0)=Ẽn,q(α)(h,k) are called the higher-order q-Euler numbers with weight α.By (2.1), we get (2.3)Ẽn,q(α)(h,k)=(qα-1)Ẽn+1,q(α)(h-α,k)+Ẽn,q(α)(h-α,k).From (2.1) and (2.2), we have(2.4)Ẽ0,q(α)(mα,k+1)=∫Zp⋯∫Zpq∑j=1k+1(mα-j)xjdμ-q(x1)⋯dμ-q(xk+1)=∑l=0m(ml)(qα-1)l∫Zp⋯∫Zp[x1+⋯+xk+1]qαlq-∑j=1k+1jxjdμ-q(x1)⋯dμ-q(xk+1)=∑l=0m(ml)(qα-1)lẼl,q(α)(0,k+1)=[2]qk+1(1+qαm)(1+qαm-1)⋯(1+qαm-k).From (2.1), we can derive the following equation:(2.5)∑j=0i(ij)(qα-1)j∫Zp⋯∫Zp[x1+⋯+xk]qαn-i+jq(h-α-1)x1+⋯+(h-α-k)xkdμ-q(x1)⋯dμ-q(xk)=∫Zp⋯∫Zp[x1+⋯+xk]qαn-iq(h-1)x1+⋯+(h-k)xkqα(x1+⋯+xk)(i-1)dμ-q(x1)⋯dμ-q(xk)=∑j=0i-1(qα-1)j(i-1j)Ẽn-i+j,q(α)(h,k).By (2.1), (2.2), (2.3), and (2.4), we see that(2.6)∑j=0i(qα-1)j(ij)Ẽn-i+j,q(α)(h-α,k)=∑j=0i-1(qα-1)j(i-1j)Ẽn-i+j,q(α)(h,k).Therefore, we obtain the following theorem.Theorem 2.1.
Forα,k∈ℕ and n,i∈ℤ+, one has
(2.7)∑j=0i(ij)(qα-1)jẼn-i+j,q(α)(h-α,k)=∑j=0i-1(qα-1)j(i-1j)Ẽn-i+j,q(α)(h,k).By simple calculation, we easily see that(2.8)∑j=0m(mj)(qα-1)jẼj,q(α)(0,k)=[2]qk(1+qαm)(1+qαm-1)⋯(1+qαm-k+1).


## 3. PolynomialsẼn,q(α)(0,k∣x)

We now consider the polynomialsẼn,q(α)(0,k∣x) (in qx) by(3.1)Ẽn,q(α)(0,k∣x)=∫Zp⋯∫Zp︸k-times[x+x1+⋯+xk]qαnq-∑j=1kjxjdμ-q(x1)⋯dμ-q(xk).By (3.1), we get(3.2)(qα-1)nẼn,q(α)(0,k∣x)=[2]qk∑l=0n(nl)qαlx(-1)n-l1(1+qαl)⋯(1+qαl-k+1).From (3.1) and (3.2), we can derive the following equation:(3.3)∫Zp⋯∫Zpq∑j=1k(αn-j)xj+αnxdμ-q(x1)⋯dμ-q(xk)=∑j=0n(nj)[α]qj(q-1)jẼj,q(α)(0,k∣x),∫Zp⋯∫Zpq∑j=1k(αn-j)xj+αnxdμ-q(x1)⋯dμ-q(xk)=[2]qkqαnx(1+qαn)⋯(1+qαn-k+1).Therefore, by (3.2) and (3.3), we obtain the following theorem.Theorem 3.1.
Forα∈ℕ and n,k∈ℤ+, one has
(3.4)Ẽn,q(α)(0,k∣x)=[2]qk[α]qn(1-q)n∑l=0n(nl)(-1)lqαlx1(-qαl-k+1:q)k,∑l=0n(nl)[α]ql(q-1)lẼl,q(α)(0,k∣x)=qαnx[2]qk(-qαn-k+1:q)k,
where (a:q)0=1 and (a:q)k=(1-a)(1-aq)⋯(1-aqk-1).Letd∈ℕ with d≡1(mod2). Then we have
(3.5)∫Zp⋯∫Zp[x+∑j=1kxj]qαnq-∑j=1kjxjdμ-q(x1)⋯dμ-q(xk)=[d]qαn[d]-qk∑a1,…,ak=0d-1q-∑j=2k(j-1)aj(-1)∑j=1kaj×∫Zp⋯∫Zp[x+∑j=1kajd+∑j=1kxj]qαdnq-d∑j=1kjxjdμ-qd(x1)⋯dμ-qd(xk).Thus, by (3.5), we obtain the following theorem.Theorem 3.2.
Ford∈ℕ with d≡1(mod2), one has
(3.6)Ẽn,q(α)(0,k∣x)=[d]qαn[d]-qk∑a1,…,ak=0d-1q-∑j=2k(j-1)aj(-1)∑j=1kajẼn,qd(α)(0,k∣x+a1+⋯+akd).
Moreover,
(3.7)Ẽn,q(α)(0,k∣dx)=[d]qαn[d]-qk∑a1,…,ak=0d-1q-∑j=2k(j-1)aj(-1)∑j=1kajẼn,qd(α)(0,k∣x+a1+⋯+akd).By (3.1), we get
(3.8)Ẽn,q(α)(0,k∣x)=∑l=0n(nl)[x]qαn-lqαlxẼl,q(α)(0,k),
where Ẽn,q(α)(0,k∣0)=Ẽn,q(α)(0,k).Thus, we note that(3.9)Ẽn,q(α)(0,k∣x+y)=∑l=0n(nl)[y]qαn-lqαlyẼl,q(α)(0,k∣x).


## 4. PolynomialsẼn,q(α)(h,1∣x)

Let us define polynomialsẼn,q(α)(h,1∣x) as follows: (4.1)Ẽn,q(α)(h,1∣x)=∫Zp[x+x1]qαnqx1(h-1)dμ-q(x1).From (4.1), we have (4.2)Ẽn,q(α)(h,1∣x)=[2]q(1-qα)n∑l=0n(nl)(-1)lqαlx1(1+qαl+h).By the calculation of the fermionicp-adic q-integral on ℤp, we see that(4.3)qαx∫Zp[x+x1]qαnqx1(h-1)dμ-q(x1)=(qα-1)∫Zp[x+x1]qαn+1qx1(h-α-1)dμ-q(x1)+∫Zp[x+x1]qαnqx1(h-α-1)dμ-q(x1).Thus, by (4.3), we obtain the following theorem.Theorem 4.1.
Forα∈ℕ and h∈ℤ, one has
(4.4)qαxẼn,q(α)(h,1∣x)=(qα-1)Ẽn+1,q(α)(h-α-1,1∣x)+Ẽn,q(α)(h-α-1,1∣x).It is easy to show that(4.5)Ẽn,q(α)(h,1∣x)=∫Zp[x+x1]qαnqx1(h-1)dμ-q(x1)=∑l=0n(nl)[x]qαn-lqαlx∫Zp[x1]qαlqx1(h-1)dμ-q(x1)=∑l=0n(nl)[x]qαn-lqαlxẼl,q(α)(h,1)=(qαxẼq(α)(h,1)+[x]qα)n,forn≥1,
with the usual convention about replacing (Ẽq(α)(h,1))n by Ẽn,q(α)(h,1).FromqI-q(f1)+I-q(f)=[2]qf(0), we have(4.6)qh∫Zp[x+x1+1]qαnqx1(h-1)dμ-q(x1)+∫Zp[x+x1]qαnqx1(h-1)dμ-q(x1)=[2]q[x]qαn.By (4.3) and (4.6), we get (4.7)qhẼn,q(α)(h,1∣x+1)+Ẽn,q(α)(h,1∣x)=[2]q[x]qαn.Forx=0 in (4.7), we have(4.8)qhẼn,q(α)(h,1∣1)+Ẽn,q(α)(h,1)={[2]q,ifn=0,0,ifn>0.Therefore, by (4.8), we obtain the following theorem.Theorem 4.2.
Forh∈ℤ and n∈ℤ+, one has
(4.9)qh(qαẼq(α)(h,1)+1)n+Ẽn,q(α)(h,1)={[2]q,ifn=0,0,ifn>0,
with the usual convention about replacing (Ẽq(α)(h,1))n by Ẽn,q(α)(h,1).From the fermionicp-adic q-integral on ℤp, we easily get
(4.10)Ẽ0,q(α)(h,1)=∫Zpqx1(h-1)dμ-q(x1)=[2]q[2]qh.By (4.1), we see that
(4.11)Ẽn,q-1(α)(h,1∣1-x)=∫Zp[1-x+x1]q-αnq-x1(h-1)dμ-q-1(x1)=(-1)nqαn+h-1[2]q(1-qα)n∑l=0n(nl)(-1)lqαlx11+qαl+h=(-1)nqαn+h-1Ẽn,q-1(α)(h,1∣x).Therefore, by (4.11), we obtain the following theorem.Theorem 4.3.
Forα∈ℕ, h∈ℤ, and n∈ℤ+, one has
(4.12)Ẽn,q-1(α)(h,1∣1-x)=(-1)nqαn+h-1Ẽn,q(α)(h,1∣x).
In particular, for x=1, one gets
(4.13)Ẽn,q(α)(h,1)=(-1)nqαn+h-1Ẽn,q(α)(h,1∣1)=(-1)n-1qαn-1Ẽn,q(α)(h,1)ifn≥1.Letd∈ℕ with d≡1(mod2). Then one has
(4.14)∫Zpqx1(h-1)[x+x1]qαndμ-q(x1)=[d]qαn[d]-q∑a=0d-1qha(-1)a∫Zp[x+ad+x1]qαdnqx1(h-1)ddμ-qd(x1).Therefore, by (4.14), we obtain the following theorem.Theorem 4.4 (Multiplication formula).
Ford∈ℕ with d≡1(mod2), we have
(4.15)Ẽn,q(α)(h,1∣x)=[d]qαn[d]-q∑a=0d-1qha(-1)aẼn,qd(α)(h,1∣x+ad).


## 5. PolynomialsẼn,q(α)(h,k∣x) and k=h

In (2.1), we know that(5.1)Ẽn,q(α)(h,k∣x)=∫Zp⋯∫Zp[x1+⋯+xk+x]qαnq(h-1)x1+⋯+(h-k)xkdμ-q(x1)⋯dμ-q(xk).
Thus, we get(5.2)(qα-1)nẼn,q(α)(h,k∣x)=[2]qk∑l=0n(nl)(-1)n-lqαlx(1+qαl+h)⋯(1+qαl+h-k+1),qh∫Zp⋯∫Zp[x+1+x1+⋯+xk]qαnq(h-1)x1+⋯+(h-k)xkdμ-q(x1)⋯dμ-q(xk)=-∫Zp⋯∫Zp[x+x1+⋯+xk]qαnq(h-1)x1+⋯+(h-k)xkdμ-q(x1)⋯dμ-q(xk)+[2]q∫Zp⋯∫Zp[x+x2+⋯+xk]qαnq(h-2)x2+⋯+(h-k)xkdμ-q(x2)⋯dμ-q(xk).Therefore, by (2.1) and (5.2), we obtain the following theorem.Theorem 5.1.
Forh∈ℤ,α∈ℕ, and n∈ℤ+, one has
(5.3)qhẼn,q(α)(h,k∣x+1)+Ẽn,q(α)(h,k∣x)=[2]qẼn,q(α)(h-1,k-1∣x).Note that(5.4)qαx∫Zp⋯∫Zp[x+x1+⋯+xk]qαnqhx1+(h-1)x2+⋯+(h+1-k)xkdμ-q(x1)⋯dμ-q(xk)=(qα-1)∫Zp⋯∫Zp[x+x1+⋯+xk]qαn+1q(h-α)x1+⋯+(h+1-α-k)xkdμ-q(x1)⋯dμ-q(xk)+∫Zp⋯∫Zp[x+x1+⋯+xk]qαnq(h-α)x1+⋯+(h+1-α-k)xkdμ-q(x1)⋯dμ-q(xk)=(qα-1)Ẽn+1,q(α)(h+1-α,k∣x)+Ẽn,q(α)(h+1-α,k∣x).Therefore, by (5.4), we obtain the following theorem.Theorem 5.2.
Forn∈ℤ+, one has
(5.5)qαxẼn,q(α)(h+1,k∣x)=(qα-1)Ẽn+1,q(α)(h+1-α,k∣x)+Ẽn,q(α)(h+1-α,k∣x).Letd∈ℕ with d≡1(mod2). Then we get
(5.6)∫Zp⋯∫Zp[x+∑j=1kxj]qαnq∑j=1k(h-j)xjdμ-q(x1)⋯dμ-q(xk)=[d]qαn[d]-qk∑a1,…,ak=0d-1qh∑j=1kaj-∑j=2k(j-1)aj(-1)∑j=1kaj×∫Zp⋯∫Zp[x+∑j=1kajd+∑j=1kxj]qαdnqd∑j=1k(h-j)xjdμ-qd(x1)⋯dμ-qd(xk).Therefore, by (5.6), we obtain the following theorem.Theorem 5.3.
Ford∈ℕ with d≡1(mod2), one has
(5.7)Ẽn,q(α)(h,k∣dx)=[d]qαn[d]-qk∑a1,…,ak=0d-1qh∑j=1kaj-∑j=2k(j-1)aj(-1)∑j=1kajẼn,qd(α)(h,k∣x+a1+⋯+akd).LetẼn,q(α)(k,k∣x)=Ẽn,q(α)(k∣x). Then we get
(5.8)(qα-1)nẼn,q(α)(k∣x),=∑l=0n(nl)(-1)n-lqαlx[2]qk(1+qαl+k)⋯(1+qαl+1)∫Zp⋯∫Zp[k-x+x1+⋯+xk]q-αnq-(k-1)x1-⋯-(k-k)xkdμ-q-1(x1)⋯dμ-q-1(xk)=qα(k+12)-k(1-q-α)n[2]qk∑l=0n(nl)(-1)lqαlx1(1+qαl+1)⋯(1+qαl+k)=(-1)nqnαqα(k+12)-k[2]qk(1-qα)n∑l=0n(nl)(-1)lqαlx(1+qαl+1)⋯(1+qαl+k)=(-1)nqα(n+(k+12))-kẼn,q(α)(k∣x).Therefore, by (5.8), we obtain the following theorem.Theorem 5.4.
Forn∈ℤ+, one has
(5.9)Ẽn,q-1(α)(k∣k-x)=(-1)nqα(n+(k+12))-kẼn,q(α)(k∣x).Letx=k in Theorem 5.4. Then we see that(5.10)Ẽn,q-1(α)(k∣0)=(-1)nqα(n+(k+12))-kẼn,q(α)(k∣k).From (4.6) and Theorem 5.1, we note that(5.11)qkẼn,q(α)(k∣x+1)+Ẽn,q(α)(k∣x)=[2]qẼn,q(α)(k-1∣x).It is easy to show that(5.12)(qα-1)nẼn,q(α)(k∣0)=∑l=0n(nl)(-1)l+n[2]qk(1+qαl+1)⋯(1+qαl+k).By simple calculation, we get(5.13)∑l=0n(nl)(qα-1)l∫Zp⋯∫Zp[x1+⋯+xk]qαlq∑l=1k(k-l)xldμ-q(x1)⋯dμ-q(xk)=[2]qk(1+qαn+k)(1+qαn+k-1)⋯(1+qαn+1).From (5.13), we note that(5.14)∑l=0n(nl)(qα-1)lẼl,q(α)(k∣0)=[2]qk(1+qαn+k)(1+qαn+k-1)⋯(1+qαn+1),Ẽn,q(α)(k∣x)=∫Zp⋯∫Zp[x+x1+⋯+xk]qαnq(k-1)x1+⋯+(k-k)xkdμ-q(x1)⋯dμ-q(xk)=∑l=0n(nl)qαlxẼl,q(α)(k∣0)[x]qαn-l=(qxαẼq(α)(k∣0)+[x]qα)nforn∈Z+,
with the usual convention about replacing (Ẽq(α)(k∣0))n by Ẽn,q(α)(k∣0).Putx=0 in (5.11); we get (5.15)qkẼn,q(α)(k∣1)+Ẽn,q(α)(k∣0)=[2]qẼn,q(α)(k-1∣0).Thus, we have(5.16)qk(qαẼq(α)(k∣0)+1)n+Ẽn,q(α)(k∣0)=[2]qẼn,q(α)(k-1∣0),
with the usual convention about replacing (Ẽq(α)(k∣0))n by Ẽn,q(α)(k∣0).



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