# Some Identities on Bernstein Polynomials Associated withq-Euler Polynomials

Authors: A. Bayad; T. Kim; B. Lee; S.-H. Rim
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We investigate some interesting properties of theq-Euler polynomials. The purpose of this paper is to give some relationships between
Bernstein and q-Euler polynomials, which are derived by the p-adic integral representation of the Bernstein polynomials associated with q-Euler polynomials.


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## Body

## 1. Introduction

Letp be a fixed odd prime number. Throughout this paper ℤp, ℚp, and ℂp denote the ring of p-adic integers, the field of p-adic numbers, and the field of p-adic completion of the algebraic closure of ℚp, respectively (see [1–15]). Let ℕ be the set of natural numbers and ℤ+=ℕ∪{0}. The normalized p-adic absolute value is defined by |p|p=1/p. As an indeterminate, we assume that q∈ℂp with |1-q|p<1. Let UD(ℤp) be the space of uniformly differentiable function on ℤp. For f∈UD(ℤp), the p-adic invariant integral on ℤp is defined by
(1.1)I-1(f)=∫Zpf(x)dμ-1(x)=lim⁡N→∞∑x=0pN-1f(x)μ-1(x+pNZp)=lim⁡N→∞∑x=0pN-1f(x)(-1)x,
(see [7–10]). For n∈ℕ, we can derive the following integral equation from (1.1):(1.2)I-1(fn)=(-1)n∫Zpf(x)dμ-1(x)+2∑l=0n-1(-1)n-1-lf(l),
where fn(x)=f(x+n) (see [7–11]). As well-known definition, the Euler polynomials are given by the generating function as follows:(1.3)2et+1ext=eE(x)t=∑n=0∞En(x)tn!,
(see [3, 5, 7–15]), with usual convention about replacing En(x) by En(x). In the special case x=0, En(0)=En are called the nth Euler numbers. From (1.3), we can derive the following recurrence formula for Euler numbers: (1.4)E0=1,(E+1)n+En={2ifn=0,0ifn>0,
(see [12]), with usual convention about replacing En by En. By the definitions of Euler numbers and polynomials, we get
(1.5)En(x)=(E+x)n=∑l=0n(nl)xn-lEl,
(see [3, 5, 7–15]). Let C[0,1] denote the set of continuous functions on [0,1]. For f∈C[0,1], Bernstein introduced the following well-known linear positive operator in the field of real numbers ℝ: (1.6)Bn(f∣x)=∑k=0nf(kn)(nk)xk(1-x)n-k=∑k=0nf(kn)Bk,n(x),
where (nk)=n(n-1)⋯(n-k+1)/k!=n!/k!(n-k)! (see [1, 2, 7, 11, 12, 14]). Here, 𝔹n(f∣x) is called the Bernstein operator of order n for f. For k,n∈ℤ+, the Bernstein polynomials of degree n are defined by(1.7)Bk,n(x)=(nk)xk(1-x)n-k,forx∈[0,1].
In this paper, we study the properties of q-Euler numbers and polynomials. From these properties, we investigate some identities on the q-Euler numbers and polynomials. Finally, we give some relationships between Bernstein and q-Euler polynomials, which are derived by the p-adic integral representation of the Bernstein polynomials associated with q-Euler polynomials.


## 2.q-Euler Numbers and Polynomials

In this section, we assume thatq∈ℂp with |1-q|p<1. Let f(x)=qxext. From (1.1) and (1.2), we have(2.1)∫Zpf(x)dμ-1(x)=2qet+1.
Now, we define the q-Euler numbers as follows:(2.2)2qet+1=eEqt=∑n=0∞En,qtnn!,
with the usual convention about replacing Eqn by En,q.By (2.2), we easily get(2.3)E0,q=2q+1,q(Eq+1)n+En,q={2ifn=0,0ifn>0,
with usual convention about replacing Eqn by En,q.We note that(2.4)2qet+1=2et+q-1⋅21+q=21+q∑n=0∞Hn(-q-1)tnn!,
where Hn(-q-1) is the nth Frobenius-Euler numbers.From (2.1), (2.2), and (2.4), we have(2.5)∫Zpqxextdμ-1(x)=En,q=21+qHn(-q-1),forn∈Z+.
Now, we consider the q-Euler polynomials as follows:(2.6)2qet+1ext=eEq(x)t=∑n=0∞En,q(x)tnn!,
with the usual convention Eqn(x) by En,q(x).From (1.2), (2.1), and (2.6), we get(2.7)∫Zpqxe(x+y)tdμ-1(y)=2qet+1ext=∑n=0∞En,q(x)tnn!.
By comparing the coefficients on the both sides of (2.6) and (2.7), we get the following Witt's formula for the q-Euler polynomials as follows:(2.8)∫Zpqy(x+y)ndμ-1(y)=En,q(x)=∑l=0n(nl)xn-lEl,q.
From (2.6) and (2.8), we can derive the following equation:(2.9)2qqet+1e(1-x)t=21+q-1e-te-xt=∑n=0∞En,q-1(x)(-1)ntn!.
By (2.6) and (2.9), we obtain the following reflection symmetric property for the q-Euler polynomials.Theorem 2.1.
Forn∈ℤ+, one has
(2.10)(-1)nEn,q-1(x)=qEn,q(1-x).From (2.5), (2.6), (2.7), and (2.8), we can derive the following equation: for n∈ℕ, (2.11)En,q(2)=(Eq+1+1)n=∑l=0n(nl)El,q(1)=E0,q+1q∑l=1n(nl)qEl,q(1)=21+q-1q∑l=1n(nl)El,q=21+q+2q(1+q)-1q∑l=0n(nl)El,q=2q-1q2qEn,q(1)=2q+1q2En,q,
by using recurrence formula (2.3). Therefore, we obtain the following theorem.Theorem 2.2.
Forn∈ℕ, one has
(2.12)qEn,q(2)=2+1qEn,q.By using (2.5) and (2.8), we get (2.13)∫Zpq-x(1-x)ndμ-1(x)=(-1)n∫Zpq-x(x-1)ndμ-1(x)=(-1)nEn,q-1(-1)=q∫Zp(x+2)ndμ-1(x)=q(2q+1q2En,q)=2+1qEn,q=2+1q∫Zpxnqxdμ-1(x),forn>0.
Therefore, we obtain the following theorem.Theorem 2.3.
Forn∈ℕ, one has
(2.14)∫Zpq-x(1-x)ndμ-1(x)=2+1q∫Zpxnqxdμ-1(x).By using Theorem2.3, we will study for the p-adic integral representation on ℤp of the Bernstein polynomials associated with q-Euler polynomials in Section 3.


## 3. Bernstein Polynomials Associated withq-Euler Numbers and Polynomials

Now, we take thep-adic integral on ℤp for the Bernstein polynomials in (1.7) as follows:(3.1)∫ZpBk,n(x)qxdμ-1(x)=∫Zp(nk)xk(1-x)n-kqxdμ-1(x)=(nk)∑j=0n-k(n-kj)(-1)n-k-j∫Zpxn-jqxdμ-1(x)=(nk)∑j=0n-k(n-kj)(-1)n-k-jEn-j,q=(nk)∑j=0n-k(n-kj)(-1)jEk+j,q,wheren,k∈Z+.
By the definition of Bernstein polynomials, we see that(3.2)Bk,n(x)=Bn-k,n(1-x),wheren,k∈Z+.
Let n,k∈ℤ+ with n>k. Then, by (3.2), we get (3.3)∫ZpqxBk,n(x)dμ-1(x)=∫ZpqxBn-k,n(1-x)dμ-1(x)=(nn-k)∑j=0k(kj)(-1)k-j∫Zp(1-x)n-jqxdμ-1(x)=(nk)∑j=0k(kj)(-1)k-j(2+q∫Zpxn-jqxdμ-1(x))=(nk)∑j=0k(kj)(-1)k-j(2+qEn-j,q-1)={2+qEn,q-1ifk=0,(nk)q∑j=0k(kj)(-1)k-jEn-j,q-1ifk>0.
Thus, we obtain the following theorem.Theorem 3.1.
Forn,k∈ℤ+ with n>k, one has
(3.4)∫Zpq1-xBk,n(x)dμ-1(x)={2q+En,qifk=0,(nk)∑j=0k(kj)(-1)k-jEn-j,qifk>0.By (3.1) and Theorem 3.1, we get the following corollary.Corollary 3.2.
Forn,k∈ℤ+ with n>k, one has
(3.5)∑j=0n-k(n-kj)(-1)jEk+j,q-1={2+1qEn,qifk=0,∑j=0k(kj)(-1)k-j1qEn-j,qifk>0.Form,n,k∈ℤ+ with m+n>2k. Then, we get (3.6)∫ZpBk,n(x)Bk,m(x)q-xdμ-1(x)=(nk)(mk)∑j=02k(2kj)(-1)j+2k∫Zpq-x(1-x)n+m-jdμ-1(x)=(nk)(mk)∑j=02k(2kj)(-1)j+2kq∫Zp(x+2)n+m-jqxdμ-1(x)=(nk)(mk)∑j=02k(2kj)(-1)j+2kq(2q+1q2En+m-j,q)={2+1qEn+m,qifk=0,(nk)(mk)∑j=02k(2kj)(-1)j+2k1qEn+m-j,qifk>0.
Therefore, we obtain the following theorem.Theorem 3.3.
Form,n,k∈ℤ+ with m+n>2k, one has
(3.7)∫ZpBk,n(x)Bk,m(x)q1-xdμ-1(x)={2q+En+m,qifk=0,(nk)(mk)∑j=02k(2kj)(-1)j+2kEn+m-j,qifk>0.By using binomial theorem, form,n,k∈ℤ+, we get(3.8)∫ZpBk,n(x)Bk,m(x)q1-xdμ-1(x)(3.9)=(nk)(mk)∑j=0n+m-2k(n+m-2kj)(-1)j∫Zpxj+2kq1-xdμ-1(x)=q(nk)(mk)∑j=0n+m-2k(n+m-2kj)(-1)jEj+2k,q-1.
By comparing the coefficients on the both sides of (3.8) and Theorem 3.3, we obtain the following corollary.Corollary 3.4.
Letm,n,k∈ℤ+ with m+n>2k. Then, we get
(3.10)∑j=0n+m-2k(n+m-2kj)(-1)jEj+2k,q-1={2+1qEn+m,qifk=0,1q∑j=02k(2kj)(-1)j+2kEn+m-j,qifk>0.Fors∈ℕ, let n1,n2,…,ns, k∈ℤ+ with n1+n2+⋯+ns>sk. By induction, we get
(3.11)∫ZpBk,n1(x)⋯Bk,ns(x)q-xdμ-1(x)=(∏i=1s(nik))∫Zpxsk(1-x)n1+⋯+ns-skq-xdμ-1(x)=(∏i=1s(nik))∑j=0sk(skj)(-1)sk+j∫Zp(1-x)n1+⋯+ns-jq-xdμ-1(x)=(∏i=1s(nik))∑j=0sk(skj)(-1)sk+jq∫Zp(x+2)n1+⋯+ns-jqxdμ-1(x)=(∏i=1s(nik))∑j=0sk(skj)(-1)sk+jq(2q+1q2En1+⋯+ns-j,q)={2+1qEn1+⋯+ns,qifk=0,(∏i=1s(nik))1q∑j=0sk(skj)(-1)sk+jEn1+⋯+ns-j,qifk>0.
Therefore, we obtain the following theorem.Theorem 3.5.
Lets∈ℕ. For n1,n2,…,ns,k∈ℤ+ with n1+n2+⋯+ns>sk, one has
(3.12)∫Zp(∏i=1sBk,ni(x))q1-xdμ-1(x)={2q+En1+n2+⋯+ns,qifk=0,(∏i=1s(nik))∑j=0sk(skj)(-1)sk+jEn1+n2+⋯+ns-j,qifk>0.Forn1,n2,…,ns,k∈ℤ+ by binomial theorem, we get(3.13)∫Zp(∏i=1sBk,ni(x))q-xdμ-1(x)=(n1k)⋯(nsk)∑j=0n1+⋯+ns-sk(n1+⋯+ns-skj)(-1)j∫Zpxj+skq-xdμ-1(x)=(n1k)⋯(nsk)∑j=0n1+⋯+ns-sk(n1+⋯+ns-skj)(-1)jEj+sk,q-1.By using (3.13) and Theorem 3.5, we obtain the following corollary.Corollary 3.6.
Lets∈ℕ. For n1,n2,…,ns,k∈ℤ+ with n1+n2+⋯+ns>sk, one has
(3.14)∑j=0n1+⋯+ns-sk(n1+⋯+ns-skj)(-1)jEj+sk,q-1={2+1qEn1+n2+⋯+ns,qifk=0,1q∑j=0sk(skj)(-1)sk+jEn1+n2+⋯+ns-j,qifk>0.



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*Source: 294715-2011-04-05.xml*