# A Note on the Modifiedq-Bernstein Polynomials

Authors: Taekyun Kim; Lee-Chae Jang; Heungsu Yi
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We propose the modifiedq-Bernstein polynomials of degree n which are different q-Bernstein polynomials of Phillips (1997). From these modified q-Bernstein polynomials of degree n, we derive some recurrence formulae for the modified q-Bernstein polynomials.


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## Body

## 1. Introduction

LetC[0,1] denote the set of continuous function on [0,1]. For f∈C[0,1], Bernstein introduced the following well-known linear positive operators in [1]: (1.1)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!. Here Bn(f:x) is called the Bernstein operator of ordernforf. For k,n∈ℤ+, the Bernstein polynomial of degreen is defined by (1.2)Bk,n(x)=(nk)xk(1-x)n-k,
where x∈[0,1]. For example, (1.3)B0,1(x)=1-x,B1,1(x)=x,B0,2(x)=(1-x)2,B1,2(x)=2x(1-x),B2,2(x)=x2,….
Also, Bk,n(x)=0, for k>n, because (nk)=0.Some people have studied the Bernstein polynomials in the area of approximation theory (see [2] through [3]). Note that for k∈ℤ+ and x∈[0,1], (1.4)tke(1-x)txkk!=xkk!(tk∑n=0∞(1-x)ntnn!)=xkk!∑n=0∞(1-x)n(n+1)⋯(n+k)(n+k)!tn+k=∑n=k∞((nk)xk(1-x)n-k)tnn!=∑n=k∞Bk,n(x)tnn!.
Because Bk,0(x)=Bk,1(x)=⋯=Bk,k-1(x)=0, we obtain the generating function for Bk,n(x) as follows:(1.5)F(k)(t,x):=tke(1-x)txkk!=∑n=0∞Bk,n(x)tnn!
(see [4, 5]), where k∈ℤ+ and x∈[0,1]. Notice that (1.6)Bk,n(x)={(nk)xk(1-x)n-kifn≥k,0ifn<k,
for n,k∈ℤ+ (see [2]).Let0<q<1. Define theq-number ofx by (1.7)[x]q:=1-qx1-q.
See [2] through [3] for details and related facts. Note that limq→1[x]q=x. In [6], Phillips proposed a generalization of the classical Bernstein polynomials based on q-integers. In the last decade some new generalizations of well-known positive linear operators, based on q-integers were introduced and studied by several authors (see [1–13]). Recently, Simsek and Acikgoz have also studied the q-extension of Bernstein-type polynomials [5]. Their q-Bernstein-type polynomials are given by (1.8)Yn(k;x:q)=(nk)(-1)kk!(1-q)n-k∑m,l=0∞‍∑j=0n-k(k+l-1l)(n-kk)×((-1)jql+j(1-x)S(m,k)(xlnq)mm!),
where S(m,k) are the second-kind stirling number. In [5], we can find some interesting formulae related to q-extension of Bernstein polynomials which are different q-Bernstein polynomials of Phillips. In the conference of Jangjeon Mathematical Society which was held in IRAN (on Feb.2010), Acikgoz and Arci has introduced several-type Bernstein polynomials (see [2]). The Acikgoz paper [2] announced in the conference is actually what motivated us to write this paper. In this paper, we considered the q-extension of Bernstein polynomials which were introduced by Acikgoz at the conference of Jangjeon Mathematical Society on Feb. 2010. First, we consider the q-extension of the generating function of Bernstein polynomials in (1.5). Indeed, this generating function is also treated by Simsek and Acikgoz in a previous paper (see [5]). From this q-extension of the generating function for the Bernstein polynomials, we propose the modified q-Bernstein polynomials of degree n which are different q-Bernstein polynomials of Phillips. By using the properties of the modified q-Bernstein polynomials, we obtain some recurrence formulae for the modified q-Bernstein polynomials of degree n.


## 2. The Modifiedq-Bernstein Polynomials

For0<q<1, consider the q-extension of (1.5) as follows:(2.1)Fq(k)(t,x):=tke[1-x]qt[x]qkk!=[x]qkk!∑n=0∞[1-x]qnn!tn+k=∑n=k∞(nk)[x]qk[1-x]qn-ktnn!,
where k,n∈ℤ+ and x∈[0,1]. Note that limq→1Fq(k)(t,x)=F(k)(t,x). We define the modifiedq-Bernstein polynomials as follows:(2.2)Fq(k)(t,x)=tke[1-x]qt[x]qkk!=∑n=0∞Bk,n(x,q)tnn!,
where k,n∈ℤ+ and x∈[0,1].Remark 2.
This generating function is also introduced by Simsek and Acikgoz in a previous paper (see [5]).By comparing the coefficients of (2.1) and (2.2), we obtain the following theorem.Theorem 2.1.
Fork,n∈ℤ+ and x∈[0,1],
(2.3)Bk,n(x,q)={(nk)[x]qk[1-x]qn-k,ifn≥k0,ifn<k.For0≤k≤n, we have (2.4)[1-x]qBk,n-1(x,q)+[x]qBk-1,n-1(x,q)=[1-x]q(n-1k)[x]qk[1-x]qn-1-k+[x]q(n-1k-1)[x]qk-1[1-x]qn-k=(n-1k)[x]qk[1-x]qn-k+(n-1k-1)[x]qk[1-x]qn-k=(nk)[x]qk[1-x]qn-k,
and the derivatives of the modified q-Bernstein polynomials of degree n are also polynomials of degree n-1, that is, (2.5)ddxBk,n(x,q)=(nk)k[x]qk-1[1-x]qn-klnqq-1qx+(nk)[x]qk(n-k)[1-x]qn-k-1(-lnqq-1)q1-x=lnqq-1{(nk)k[x]qk-1[1-x]qn-kqx-(nk)[x]qk(n-k)[1-x]qn-k-1q1-x}=n(qxBk-1,n-1(x,q)-q1-xBk,n-1(x,q))lnqq-1.
Therefore, we obtain the following recurrence formulae.Theorem 2.2 (recurrence formulae forBk,n(x,q)).
Fork,n∈ℤ+ and for x∈[0,1],
(2.6)[1-x]qBk,n-1(x,q)+[x]qBk-1,n-1(x,q)=Bk,n(x,q),ddxBk,n(x,q)=n(qxBk-1,n-1(x,q)-q1-xBk,n-1(x,q))lnqq-1.Letf be a continuous function on [0,1]. Then the modifiedq-Bernstein operator of ordernforf is defined by(2.7)Bn,q(f:x):=∑k=0nf(kn)Bk,n(x,q),
where 0≤x≤1, n∈ℤ+. We get from Theorem 2.1 and (2.7) that for f(x)=x,(2.8)Bn,q(f:x)=∑k=0nf(kn)(nk)[x]qk[1-x]qn-k=[x]q(1-[1-x]q[x]q(q-1))n-1=f([x]q)(1+(1-q)[x]q[1-x]q)n-1.
We also see from Theorem 2.1 that (2.9)Bn,q(1:x)=∑k=0nBk,n(x,q)=∑k=0n(nk)[x]qk[1-x]qn-k=∑k=0n(nk)[x]qk(1-q1-x[x]q)n-k=(1+(1-q)[x]q[1-x]q)n.The modifiedq-Bernstein polynomials are symmetric polynomials in the following sense: (2.10)Bn-k,n(1-x,q)=(nn-k)[1-x]qn-k[x]qk=Bk,n(x,q).
Therefore, we get the following theorem.Theorem 2.3.
Fork,n∈ℤ+ and x∈[0,1],
(2.11)Bn-k,n(1-x,q)=Bk,n(x,q),Bn,q(1:x)=(1+(1-q)[x]q[1-x]q)n.Forζ∈ℂ, x∈[0,1] and for n∈ℤ+, consider(2.12)n!2πi∮C([x]qζ)kk!e([1-x]qζ)dζζn+1,
where C is a circle around the origin and integration is in the positive direction. We see from the definition of the modified q-Bernstein polynomials and the basic theory of complex analysis including Laurent series that (2.13)∮C([x]qζ)kk!e[1-x]qζdζζn+1=∑m=0∞∮CBk,m(x,q)ζmm!dζζn+1=2πi(Bk,n(x,q)n!).
We get from (2.12) and (2.13) that(2.14)n!2πi∮C([x]qζ)kk!e[1-x]qζdζζn+1=Bk,n(x,q),(2.15)∮C([x]qζ)kk!e[1-x]qζdζζn+1=[x]qkk!∑m=0∞([1-x]qmm!∮Cζm-n-1+kdζ)=2πi([x]qk[1-x]qn-kk!(n-k)!).
We also get from (2.12) and (2.15) that(2.16)n!2πi∮C([x]qζ)kk!e([1-x]qζ)dζζn+1=(nk)[x]qk[1-x]qn-k.
Therefore, we see from (2.14) and (2.16) that(2.17)Bk,n(x,q)=(nk)[x]qk[1-x]qn-k.
Note that (2.18)(n-kn)Bk,n(x,q)+(k+1n)Bk+1,n(x,q)=(n-1)!k!(n-k-1)![x]qk[1-x]qn-k+(n-1)!k!(n-k-1)![x]qk+1[1-x]qn-k-1=([1-x]q+[x]q)Bk,n-1(x,q)=(1+[x]q(1-q1-x))Bk,n-1(x,q)=(1+(1-q)[x]q[1-x]q)Bk,n-1(x,q).
Therefore, we can write the modified q-Bernstein polynomials as a linear combination of polynomials of higher order as follows.Theorem 2.4.
Fork,n∈ℤ+ and x∈[0,1],
(2.19)(n+1-kn+1)Bk,n+1(x,q)+(k+1n+1)Bk+1,n+1(x,q)=(1+(1-q)[x]q[1-x]q)Bk,n(x,q).We easily see from (2.17) that for n,k∈ℕ, (2.20)(n-k+1k)([x]q[1-x]q)Bk-1,n(x,q)=(n-k+1k)([x]q[1-x]q)(nk-1)[x]qk-1[1-x]qn-k+1=n!k!(n-k)![x]qk[1-x]qn-k=Bk,n(x,q).
Thus, the following corollary holds.Corollary 2.5.
Forn,k∈ℕ and x∈[0,1],
(2.21)(n-k+1k)([x]q[1-x]q)Bk-1,n(x,q)=Bk,n(x,q).Note from the definition of the modifiedq-Bernstein polynomials and the binomial theorem that for k,n∈ℤ+, (2.22)Bk,n(x,q)=(nk)[x]qk[1-x]qn-k=(nk)[x]qk(1-q1-x[x]q)n-k=(nk)[x]qk∑l=0n-k(n-kl)(-1)lql(1-x)[x]ql=∑l=0n-k(k+lk)(nk+l)(-1)lql(1-x)[x]ql+k=∑j=kn(nk)(nj)(-1)j-kq(1-x)(j-k)[x]qj.
Therefore, we showed that the following theorem holds.Theorem 2.6.
Fork,n∈ℤ+ and x∈[0,1],
(2.23)Bk,n(x,q)=∑j=kn(jk)(nj)(-1)j-kq(1-x)(j-k)[x]qj.It is possible to write[x]qk as a linear combination of the modified q-Bernstein polynomials by using the degree evaluation formulae and mathematical induction. We easily see from the property of the modified q-Bernstein polynomials that (2.24)∑k=1n(kn)Bk,n(x,q)=∑k=1n(n-1k-1)[x]qk[1-x]qn-k=∑k=0n-1(n-1k)[x]qk+1[1-x]qn-1-k=[x]q([x]q+[1-x]q)n-1,
and that (2.25)∑k=2n(k2)(n2)Bk,n(x,q)=∑k=2n(n-2k-2)[x]qk[1-x]qn-k=∑k=0n-2(n-2k)[x]qk+2[1-x]qn-2-k=[x]q2([x]q+[1-x]q)n-2.
Continuing this process, we obtain (2.26)∑k=jn(kj)(nj)Bk,n(x,q)=[x]qj([x]q+[1-x]q)n-j,
for j∈ℕ. Therefore, we obtain the following theorem.Theorem 2.7.
Forn,j∈ℤ+ and x∈[0,1],
(2.27)1([1-x]q+[x]q)n-j∑k=jn(kj)(nj)Bk,n(x,q)=[x]qj.Fork∈ℕ, the Bernoulli polynomial of orderk is defined by (2.28)(tet-1)kext=(tet-1)×⋯×(tet-1)︸k-timesext=∑n=0∞Bn(k)(x)tnn!,
and Bn(k)=Bn(k)(0) are called the  nthBernoulli numbers of orderk. It is well known that the second kind stirling number is defined by(2.29)(et-1)kk!:=∑n=0∞S(n,k)tnn!,
for k∈ℕ. We note from (2.2) that(2.30)([x]qt)ke[1-x]qtk!=[x]qk(et-1)kk!(tet-1)ke[1-x]qt=[x]qk(∑m=0∞S(m,k)tmm!)(∑n=0∞Bn(k)([1-x]q)tnn!)=[x]qk∑l=0∞(∑n=0lBn(k)([1-x]q)S(l-n,k)l!n!(l-n)!)tll!.
We have from (2.2) and (2.30) that(2.31)Bk,l(x,q)=[x]qk∑n=0l(ln)Bn(k)([1-x]q)S(l-n,k),
and Bk,0(x,q)=Bk,1(x,q)=⋯=Bk,k-1(x,q)=0.Remark 2.
The Equations (2.30) and (2.31) are already known by Simsek and Acikgoz in a previous paper [5, page 7].LetΔ be the shift difference operator defined by Δf(x)=f(x+1)-f(x). We see from the iterative method that(2.32)Δnf(0)=∑k=0n(nk)(-1)n-kf(k),
for n∈ℕ. We get from (2.29) and (2.32) that (2.33)∑n=0∞S(n,k)tnn!=1k!∑l=0k(kl)(-1)k-lelt=∑n=0∞{1k!∑l=0k(kl)(-1)k-lln}tnn!=∑n=0∞Δk0nk!tnn!.
By comparing the coefficients on both sides above, we have(2.34)S(n,k)=Δk0nk!,
for n,k∈ℤ+. Thus, we get from (2.31) and (2.34) that (2.35)Bk,l(x,q)=[x]qk∑n=0l(ln)Bn(k)([1-x]q)Δk0l-nk!.
Let (Eh)(x)=h(x+1) be the shift operator. Then the  q-difference operator is defined by (2.36)Δqn=Πj=0n-1(E-qjI),
where I is an identity operator(see [7] through [11]). For f∈C[0,1] and n∈ℕ, we have (2.37)Δqnf(0)=∑k=0n(nk)q(-1)kq(n2)f(n-k),
where (nk)q is the Gaussian binomial coefficient defined by (2.38)(xk)q=[x]q[x-1]q⋯[x-k+1]q[k]q!.LetFq(t) be the generating function of the  q-extension of the second kind stirling number as follows:(2.39)Fq(t):=q-(k2)[k]q!∑j=0k(-1)k-j(kj)qq(k-j2)e[j]qt=∑n=0∞S(n,k:q)tnn!.
We have from (2.39) that (2.40)S(n,k:q)=q-(k2)[k]q!∑j=0k(-1)jq(j2)(kj)q[k-j]qn=q-(k2)[k]q!Δqk0n,
where [k]q!=[k]q[k-1]q⋯[2]q[1]q. It is not difficult to see that(2.41)[x]qn=∑k=0nq(k2)(xk)q[k]q!S(n,k:q).
See also [7] through [11] for details and related facts for above. Then, we get from (2.41) and Theorem 2.7 that (2.42)∑k=0jq(k2)(xk)q[k]q!S(j,k:q)=1([1-x]q+[x]q)n-j∑k=jn(kj)(nj)Bk,n(x,q).
Therefore, this completes the proof of the following theorem.Theorem 2.8.
Forn,j∈ℤ+ and x∈[0,1],
(2.43)1([1-x]q+[x]q)n-j∑k=jn(kj)(nj)Bk,n(x,q)=∑k=0jq(k2)(xk)q[k]q!S(j,k:q).



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