# Some Identities on theq-Integral Representation of the Product of Several q-Bernstein-Type Polynomials

Authors: Taekyun Kim
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The purpose of this paper is to give some properties of severalq-Bernstein-type polynomials to express the q-integral on [0, 1] in terms of q-beta and q-gamma functions. Finally, we derive some identities on the q-integral of the product of several q-Bernstein-type polynomials.


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## Body

## 1. Introduction

Letq∈ℝ with 0≤q<1. We assume that q-number is defined by [x]q=(1-qx)/(1-q) and [0]q=0. Note that limq→1[x]q=x. The q-derivative of a map f:ℝ→ℝ at x∈ℝ∖{0} is given by(1.1)Dq(f)=dqf(x)dqx=f(qx)-f(x)(q-1)x
(see [1–6]). For n∈ℕ, by (1.1), we get Dqn(xn)=[n]q[n-1]q⋯[2]q[1]q=[n]1!. The q-binomial formula is given by(1.2)(a+b)qn=∏i=0n-1(a+bqi)=∑l=0n(nl)qq(l2)an-lbl
(see [2, 5, 7–11]), where (nk)q=[n]q!/[k]q![n-k]q!=[n]q[n-1]q⋯[n-k+1]q/[k]q!.Fora,b∈ℝ, the Jackson q-integral of f:ℝ→ℝ is defined by(1.3)∫abf(x)dqx=(1-q)∑n=0∞qn(bf(bqn)-af(aqn))
(see [1, 2, 5, 6, 9, 12, 13]). From (1.2), we note that(1.4)(n+1k)q=(nk-1)q+qk(nk)q=qn-k(nk-1)q+(nk)q.
By (1.2) and (1.4), we get(1.5)(1-b)qn=(b:q)n=∏i=0n-1(1-qib)=∑i=0n(ni)qq(i2)(-b)i,1(1-b)qn=1(b:q)n=1∏i=0n-1(1-qib)=∑i=0∞(n+i-1i)qbi.
Let C[0,1] denote the set of continuous function on [0,1]. For f∈C[0,1], Bernstein introduced the following well-known linear operators (see [1, 4, 9, 11, 14]):(1.6)Bn(f∣x)=∑k=0nf(kn)(nk)xk(1-x)n-k=∑k=0nf(kn)Bk,n(x).
Here 𝔹n(f∣x) is called Bernstein operator of order n for f. For k,n∈ℤ+(=ℕ∪{0}), the Bernstein polynomials of degree n are defined by(1.7)Bk,n(x)=(nk)xk(1-x)n-k
(see [1, 3, 4, 11–14]). By the definition of Bernstein polynomials (see (1.6) and (1.7)), we can see that Bernstein basis is the probability mass function of binomial distribution. A Bernoulli trial involves performing an experiment once and noting whether a particular event A occurs. The outcome of Bernoulli trial is said to be “success” if A occurs and a “failure” otherwise. Let k be the number of successes in n independent Bernoulli trials, the probabilities of k are given by the binomial probability law:(1.8)pn(k)=(nk)pk(1-p)n-k,fork=0,1,…,n,
where pn(k) is the probability of k successes in n trials. For example, a communication system transmits binary information over channel that introduces random bit errors with probability ξ=10-3. The transmitter transmits each information bit three times, an a decoder takes a majority vote of the received bits to decide on what the transmitted bit was. The receiver can correct a single error, but it will make the wrong decision if the channel introduces two or more errors. If we view each transmission as a Bernoulli trial in which a “success” corresponds to the introduction of an error, then the probability of two or more errors in three Bernoulli trials is(1.9)p(k≥2)=(32)(0.001)2(0.999)+(33)(0.001)3≈3(10-6),
see [9]. Based on the q-integers Phillips introduced the q-analogue of well-known Bernstein polynomials (see [4, 5, 9, 11, 15]). For f∈C([0,1]), Phillips introduced the q-extension of (1.6) as follows:(1.10)Bn,q(f∣x)=∑n=0nf([k]q[n]q)(nk)q(1-x)qn-k=∑n=0nf([k]q[n]q)Bk,n(x,q),fork,n∈Z+
(see [4, 5, 9, 11, 15]). Here 𝔹n,q(f∣x) is called the q-Bernstein operator of order n for f. For k,n∈ℤ+, the q-Bernstein polynomial of degree n is defined by(1.11)Bk,n(x,q)=(nk)qxk(1-x)qn-k,wherex∈[0,1].
Note that (1.11) is the q-extension of (1.7). That is, limq→1Bk,n(x,q)=Bk,n(x). For example, B0,1(x,q)=1-x, B1,1(x,q)=x, and B0,2(x,q)=1-[2]qx+qx2,…. Also Bk,n(x,q)=0 for k>n, because (nk)q=0. For n,k∈ℤ+, its probabilities are given by(1.12)p(x=k)=(nk)qxk(1-x)qn-k,wherex∈[0,1].
This distributions are studied by several authors and they have applications in physics as well as in approximation theory due to the q-Bernstein polynomials and the q-Bernstein operators (see [1–16]). By the definition of the q-Bernstein polynomials, we easily see that the q-Bernstein basis is the probability mass function of q-binomial distribution. In this paper we use the two q-analogues of exponential function as follows:(1.13)Eq(x)=((1-q)x:q)∞=(1+(1-q)x)q∞=∑n=0∞q(n2)xn[n]q!,(1.14)eq(x)=1((1-q)x:q)∞=1(1+(1-q)x)q∞=∑n=0∞xn[n]q!,
(see [2–4, 6, 10]). From (1.3), the improper q-integral is given by(1.15)∫0∞/Af(x)dqx=(1-q)∑n∈ZqnAf(qnA)
(see [6]), where the improper q-integral depends on A. The purpose of this paper is to give some properties of several q-Bernstein type polynomials to express the q-integral on [0,1] in terms of q-beta and q-gamma functions. Finally, we derive some identities on the q-integral of the product of several q-Bernstein type polynomials.


## 2.q-Integral Representation of q-Bernstein Polynomials

The gamma and beta functions are defined as the following definite integrals (α>0,β>0):(2.1)Γ(α)=∫0∞e-ttα-1dt,
(see [1–11, 14–16])(2.2)B(α,β)=∫01tα-1(1-t)β-1dt=∫0∞tα-1(1+t)α+βdt.
From (2.1) and (2.2), we can derive the following equations:(2.3)Γ(α+1)=αΓ(α),B(α,β)=Γ(α)Γ(β)Γ(α+β).
As the q-extensions of (2.1) and (2.2), the q-gamma and q-beta functions are defined as the following q-integrals (α>0,β>0):(2.4)Γq(α)=∫01/(1-q)xα-1Eq(-qx)dqx
(see [2–6, 10]),(2.5)Bq(α,β)=∫01xα-1(1-qx)qβ-1dqx
(see [2, 4, 6, 10]).By (2.4) and (2.5), we obtain the following lemma.Lemma 2.1 (see [2, 6]).
(a)Γq can be equivalently expressed as
(2.6)Γq(α)=(1-q)qα-1(1-q)α-1,whereα>0.
In particular, one has
(2.7)Γq(α+1)=[α]qΓq(α),forα>0,Γq(1)=1.(b) The q-gamma and q-beta functions are related to each other by the following two equations:
(2.8)Γq(α)=Bq(α,∞)(1-q)α,Bq(α,β)=Γq(α)Γq(β)Γq(α+β),whereα>0,β>0.Now one takes theq-integral for one q-Bernstein polynomial as follows: for n,k∈ℤ+,
(2.9)q-k∫01Bk,n(qx,q)dqx=(nk)q∫01xk(1-qx)qn-kdqx=(nk)q∑l=0n-k(n-kl)q(-1)lq(l+12)∫01xl+kdqx=(nk)q∑l=0n-k(n-kl)q(-1)n-k-lq(n-k-l+12)1[n-l+1]q.
Therefore, by (2.9), one obtains the following proposition.Proposition 2.2.
Forn,k∈ℤ+, one has
(2.10)∫01Bk,n(qx,q)dqx=qk(nk)q∑l=0n-k(n-kl)q(-1)n-k-lq(n-k-l+12)1[n-l+1]q.The Proposition2.2 is closely related to the q-beta function which is given by(2.11)Bq(n,m)=∫01xn-1(1-qx)qm-1dqx,(2.12)Γq(m)=∫01/(1-q)xn-1Eq(-qx)dqx,
(see (2.5)). From Lemma 2.1, one has(2.13)Bq(n,m)=Γq(m)Γq(n)Γq(n+m),wherem,n∈N.
By (2.9) and (2.13), one gets(2.14)q-k∫01Bk,n(qx,q)dqx=(nk)qBq(k+1,n-k+1)=(nk)qΓq(k+1)Γq(n-k+1)Γq(n+2),wherek>-1,n>k-1.
Therefore, by (2.14), one obtains the following theorem.Theorem 2.3.
Forn,k∈ℤ+ with k>-1 and n>k-1, one has
(2.15)∫01Bk,n(qx,q)dqx=(nk)q[k]q[n-k]q((q-1)[k]q+1)Γq(k)Γq(n-k)Γq(n+2).By comparing the coefficients on the both sides of Proposition2.2 and Theorem 2.3, one obtains the following corollary.Corollary 2.4.
Forn,k∈ℤ+ with k>-1 and n>k-1, one has
(2.16)∑l=0n-k(n-kl)q(-1)n-k-lq(n-k-l+12)[n-l+1]q=Γq(k+1)Γq(n-k+1)Γq(n+2).According to this result one can say that theq-integral of q-Bernstein polynomials from 0 to 1 is symmetric. Now one considers the q-integral for the multiplication of two q-Bernstein polynomials which is given by the following relation:(2.17)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)dqxqnk-k2+2k=(nk)q(mk)q∫01x2k(1-qx)qn+m-2kdqx=(nk)q(mk)q∫01un+m-2k(1-qu)q2kdqu.
For n,k,m∈ℤ+, one can derive the following equation (2.20) from (2.17):(2.18)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)dqxqnk-k2+2k=(nk)q(mk)q∑l=02k(2kl)q(-1)lq(l+12)[n+m+l-2k+1]q=(nk)q(mk)q∑l=02k(2kl)q(-1)2k-lq(2k-l+12)[n+m-l+1]q.
Therefore, one obtains the following theorem.Theorem 2.5.
Form,n,k∈ℤ+, one has
(2.19)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)dqx=qnk-k2+2k(nk)q(mk)q∑l=02k(2kl)q(-1)2k-lq(2k-l+12)[n+m-l+1]q.Form,n,k∈ℤ+, by (2.5) and (2.9), one gets(2.20)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)dqxqnk-k2+2k=(nk)q(mk)qBq(n+m-2k+1,2k+1).
Therefore, by Theorem 2.5 and (2.20), one obtains the following corollary.Corollary 2.6.
Fork>-1 and n+m-2k>-1, one has
(2.21)∑l=02k(2kl)q(-1)2k-lq(2k-l+12)[n+m-l+1]q=Γq(n+m-2k+1)Γq(2k+1)Γq(n+m+2).By the same method, the multiplication of threeq-Bernstein polynomials is given by the following relation: for k,n,m,s∈ℤ+,(2.22)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)Bk,s(qn+m-2k+1x,q)dqxq3k+2nk-3k2+mk=(nk)q(mk)q(sk)q∫01x3k(1-qx)qn+m+s-3kdqx=(nk)q(mk)q(sk)q∫01un+m+s-3k(1-qu)q3kdqu=(nk)q(mk)q(sk)q∑l=03k(3kl)qq(l+12)(-1)l∫01un+m+s-3k+ldqu=(nk)q(mk)q(sk)q∑l=03k(3kl)qq(3k-l+12)(-1)l+3k1[n+m+s-l+1]q.
Therefore, by (2.22), one obtains the following theorem.Theorem 2.7.
Forn,m,s,k∈ℤ+, one has
(2.23)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)Bk,s(qn+m-2k+1x,q)dqx=q3k+2nk-3k2+mk(nk)q(mk)q(sk)q∑l=03k(3kl)qq(3k-l+12)(-1)l+3k[n+m+s-l+1]q.From (2.5) and (2.22), one has(2.24)∫01Bk,n(qx,q)Bk,m(qn-k+1x,q)Bk,s(qn+m-2k+1x,q)dqxq3k+2nk-3k2+mk=(nk)q(mk)q(sk)qBq(n+m+s-3k+1,3k+1).
Therefore, by Theorem 2.7 and (2.24), one obtains the following corollary.Corollary 2.8.
Fork>-1/3 and n+m+s-3k>-1, one has
(2.25)∑k=03k(3kl)q(-1)l+3kq(3k-l+12)[n+m+s-l+1]q=Γq(n+m+s-3k+1)Γq(3k+1)Γq(n+m+s+2).Fors∈ℕ, let n1,n2,…,ns,k∈ℤ+. Then one has(2.26)∫01Bk,n1(qx,q)(∏i=1s-1Bk,ni+1(q∑l=1inl-ik+1x,q)dqx)qsk+k∑i=1s-1ins-i-k2(s2)=(n1k)q(n2k)q⋯(nsk)q∫01xsk(1-qx)qn1+⋯+ns-skdqx=(n1k)q(n2k)q⋯(nsk)q∑l=0sk(skl)q(-1)lq(l+12)∫01xn1+⋯+ns-sk+ldqx=(n1k)q(n2k)q⋯(nsk)q∑l=0sk(skl)q(-1)l+skq(sk-l+12)[n1+⋯+ns-l+1]q.
Therefore, by (2.26), one obtains the following theorem.Theorem 2.9.
Fors∈ℕ, let n1,n2,…,ns,k∈ℤ+. Then one has
(2.27)∫01Bk,n1(qx,q)(∏i=1s-1Bk,ni+1(q∑l=1inl-ik+1x,q))dqx=qsk+k∑i=1s-1ins-i-k2(s2)(n1k)q⋯(nsk)q∑l=0sk(skl)q(-1)l+skq(sk-l+12)[n1+⋯+ns-l+1]q.By (2.5) and (2.26), we get(2.28)∫01Bk,n1(qx,q)(∏i=1s-1Bk,ni+1(q∑l=1inl-ik+1x,q)dqx)qsk+k∑i=1s-1ins-i-k2(s2)=(n1k)q(n2k)q⋯(nsk)qBq(sk+1,n1+⋯+ns-sk+1)=(n1k)q(n2k)q⋯(nsk)qΓq(sk+1)Γq(n1+⋯+ns-sk+1)Γq(n1+⋯+ns+2).
By comparing the coefficients on the both sides of Theorem 2.9 and (2.28), one obtains the following corollary.Corollary 2.10.
Fors∈ℕ, let k>-1/s and n1+⋯+ns-sk>-1. Then one has
(2.29)∑l=0sk(skl)q(-1)l+skq(sk-l+12)[n1+⋯+ns-l+1]q=Γq(sk+1)Γq(n1+⋯+ns-sk+1)Γq(n1+⋯+ns+2).Forn∈ℤ+, one gets(2.30)∫01B0,n(qx,q)(∏l=1nBl,n(qnl-(l2)+1x,q))dqxq∑l=1n(nl-(l2)+1)l=(∏i=0n(ni)q)∫01x(n+12)(1-qx)q(n+12)dqx=(∏i=0n(ni)q)Bq((n+12)+1,(n+12)+1)=((Γq(n+1))n+1(∏i=1nΓq(i+1))2)((Γq(n(n+1)/2+1))2Γq(n(n+1)+2)).
Therefore, by (2.30), one obtains the following theorem.Theorem 2.11.
Forn∈ℤ+, one has
(2.31)∫01B0,n(qx,q)(∏l=1nBl,n(qnl-(l2)+1x,q))dqx=q∑l=1n(nl-(l2)+1)l((Γq(n+1))n+1(∏i=1nΓq(i+1))2)((Γq(n(n+1)/2+1))2Γq(n(n+1)+2)).From (2.30), one can also derive the following equation:(2.32)∫01B0,n(qx,q)(∏l=1nBl,n(qnl-(l2)+1x,q))dqxq∑l=1n(nl-(l2)+1)l=(∏i=0n(ni)q)∑l=0(2n+1)((n+12)l)q(-1)lq(l+12)∫01x(n+12)+ldqx=(∏i=0n(ni)q)∑l=0(2n+1)((n+12)l)q(-1)lq(l+12)1[n(n+1)/2+l+1]q.
By comparing the coefficients on the both sides of Theorem 2.11 and (2.30), one can see that(2.33)∑l=0n(n+1)/2(n(n+1)/2l)q(-1)lq(l+12)[n(n+1)/2+l+1]q=Bq(n(n+1)2+1,n(n+1)2+1).
Therefore, by (2.33), one obtains the following corollary.Corollary 2.12.
Forn∈ℤ+, one has
(2.34)∑l=0n(n+1)/2(n(n+1)/2l)q(-1)lq(l+12)[n(n+1)/2+l+1]q=((Γq(n(n+1)/2+1))2Γq(n(n+1)+2)).



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