# A de Casteljau Algorithm forq-Bernstein-Stancu Polynomials

Authors: Grzegorz Nowak
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

This paper is concerned with a generalization of theq-Bernstein polynomials and Stancu operators, where the function is evaluated at intervals which are in geometric progression. It is shown that these polynomials can be generated by a de Casteljau algorithm, which is a generalization of that relating to the classical case and q-Bernstein case.


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## Body

## 1. Introduction

Letq>0. For any fixed real number q>0 and for n∈Z={0,±1,±2,…}, the q-integers of the number [n] are defined by(1.1)[n]=(1-qn)(1-q),forq≠1,[n]=n,forq=1.
The q-factorial  [n]!, for n∈N0={0,1,2,…}, is defined by(1.2)[n]!=[1][2]⋯[n](n=1,2,…),[0]!=1.
For the integers n, k, (n≥k≥0), the q-binomial or the Gaussian coefficients are defined by (see [1, page 12])(1.3)[nk]=[n]![k]![n-k]!.
For f∈C[0;1], q>0, α≥0 and each positive integer n, we introduce (see [2]) the following generalized q-Bernstein operators:(1.4)Bnq,α(f;x)=∑k=0npn,kq,α(x)f([k][n]),
where(1.5)pn,kq,α(x)=[nk]∏i=0k-1(x+α[i])∏s=0n-1-k(1-qsx+α[s])∏i=0n-1(1+α[i]).
Note, that an empty product in (1.5) denotes 1. In the case where α=0, Bnq,α(f;x) reduces to the well-known q-Bernstein polynomials introduced by Phillips [3, 4] in 1997(1.6)Bn,q(f;x)=∑k=0n[nk]xk∏i=0n-k-1(1-qix)f([k][n]).
In the case where q=1, Bnq,α(f;x) reduces to Bernstein-Stancu polynomials, introduced by Stancu [5] in 1968(1.7)Sn(f;x)=∑k=0n(nk)∏i=0k-1(x+αi)∏s=0n-k-1(1-x+sα)∏i=0n-1(1+iα)f(kn).
When q=1 and α=0, we obtain the classical Bernstein polynomial defined by(1.8)Bn(f;x)=∑k=0n(nk)xk(1-x)n-kf(kn).
Basic facts on Bernstein polynomials, their generalizations, and applications can be found for example in [6–8]. In recent years, the q-Bernstein polynomials have attracted much interest, and a great number of interesting results related to the Bn,q(f) polynomials have been obtained (see [3, 4, 9–12]). Some approximation properties of the Stancu operators are presented in [5, 13–15].LetΔq0fj=fj, for j=0,1,…,n, and recursively,(1.9)Δqk+1fj=Δqkfj+1-qkΔqkfj,
for k=0,1,…,n-j-1 and fj=f([j]/[n]). It is easily established by induction that q-differences satisfy the relation(1.10)Δqkfj=∑i=0k(-1)kqi(i-1)/2[ki]fj+k-i.
In [2], we prove that the operators Bnq,α(f;x) defined by (1.4) can be expressed in terms of q-differences(1.11)Bnq,αf(x)=∑k=0n[nk]Δqkf0∏i=0k-1x+α[i]1+α[i],
which generalized the well-known result [3, 4] for the q-Bernstein polynomial. In this paper, we show that polynomials defined by (1.4) can be generated by a de Castljau algorithm, which is a generalization of that relating to the classical case [16] and q-Bernstein case [4, 11].


## 2. Auxiliary Results

We note thatBnq,α(f;x) defined by (1.4), is a monotone linear operator for any 0<q≤1 and α≥0. These operators reproduces linear functions [2], that is,(2.1)Bnq,α(ax+b;x)=ax+b,a,b∈R.
They also satisfy the end point interpolation conditions Bnq,α(f;0)=f(0) and Bnq,α(f;1)=f(1). These properties are significant in designing curves and surfaces.Moreover, the following holds.Lemma 2.1.
Let0<q≤1, α≥0. Then,
(2.2)∏u=0m-1(qr-qux+α([u]-[r]))=∑s=0m(-1)sqs(s-1)/2+(m-s)r[ms]∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j]),
for all m∈N, r∈N0=N∪{0} and x∈[0;1].Proof.
We use induction onm. First, we see from equality [-r]=-q-r[r], (r∈N), that (2.2) is evident for m=1. Let us assume that (2.2) holds for a given m∈N. Then, using (2.2), we obtain
(2.3)∏u=0m(qr-qux+α([u]-[r]))=(qr-qmx+α([m]-[r]))∑s=0m(-1)sqs(s-1)/2+(m-s)r[ms]·∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j])=∑s=0m(-1)sqs(s-1)/2+(m-s)r(qr+α[m]-α[r]+αqm[s])[ms]·∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j])+∑s=1m+1(-1)sq(s-1)(s-2)/2+(m-s+1)r+m(1+α[s-r-1])[ms-1]·∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j])=qmr(qr+α[m]-α[r])∏j=-rm-r-1(1+α[j])+(-1)m+1qm(m-1)/2+m(1+α[m-r])∏i=0m(x+α[i])+∑s=1m(-1)sqs(s-1)/2+(m+1-s)rUs∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j]),
where
(2.4)Us=[ms](qr+α[m]-α[r]+αqm[s])q-r+qm-s+1[ms-1](1+α[s-r-1]).
Using the obvious equalities
(2.5)(qr+α[m]-α[r])q-r=1+α[m-r],[ms][s]=[ms-1][m-s+1],
we have
(2.6)Us=[ms](1+α[m-r])+[ms-1]qm-s+1(1+α([m-s+1]qs-r-1+[s-r-1])).
It is easy to see that
(2.7)[m-s+1]qs-r-1+[s-r-1]=[m-r],[ms]+[ms-1]qm-s+1=[m+1s].
Therefore,
(2.8)Us=(1+α[m-r])[m+1s].
From last equality and (2.3), we obtain
(2.9)∏u=0m(qr-qux+α([u]-[r]))=qmr(qr+α[m]-α[r])∏j=-rm-r-1(1+α[j])+(-1)m+1qm(m-1)/2+m(1+α[m-r])∏i=0m(x+α[i])+∑s=1m(-1)sqs(s-1)/2+(m+1-s)r[m+1s](1+α[m-r])∏i=0s-1(x+α[i])∏j=s-rm-r-1(1+α[j])=∑s=0m+1(-1)sqs(s-1)/2+(m+1-s)r[m+1s]∏i=0s-1(x+α[i])∏j=s-rm-r(1+α[j]).
This completes the proof of the lemma.


## 3. Main Result

The generalizedq-Bernstein polynomials, defined by (1.4), may be evaluated by Algorithm 1.Algorithm 1:De Casteljau type algorithm.
input:q;f([0]/[n]),f([1]/[n]),…,f([n]/[n])forr=0 to nfr[0]:=f([r][n])nextrform=1 to nforr=0 to n-mfr[m]:={(qr-qm-1x+α([m-1]-[r]))fr[m-1]+(x+α[r])fr+1[m-1]}1+α[m-1]nextrnextmIn the case, whereα=0, this is the de Casteljau algorithm for evaluating the q-Bernstein polynomial [3, 4]. Note that with q=1 and α=0, we recover the original classical de Casteljau algorithm (see Hoschek and Lasser [16]). The algorithm is justifed by the following theorem.Theorem 3.1.
Each intermediate pointfr[m] of the algorithm can be expressed as
(3.1)fr[m]=(∏i=0m-1(1+α[i]))-1·∑t=0mfr+t[mt]∏s=0t-1(x+α[r+s])∏u=0m-t-1(qr-qux+α([u]-[r])),
and, in particular
(3.2)f0[n]=Bnq,α(f;x).Proof.
We use induction onm. From the initial conditions in the algorithm, fr[0]=f([r]/[n])=fr, 0≤r≤n, it is clear that (3.1) holds for m=0 and 0≤r≤n. Let us assume that (3.1) holds for some m such that 0≤m<n, and for all r such that 0≤r≤n-m. Then, for 0≤r≤n-m-1, it follows from the algorithm that
(3.3)fr[m+1]:={(qr-qmx+α([m]-[r]))fr[m]+(x+α[r])fr+1[m]}11+α[m],
and using (3.1), we obtain
(3.4)fr[m+1](∏i=0m(1+α[i])):=(qr-qmx+α([m]-[r]))·∑t=0mfr+t[mt]∏s=0t-1(x+α[r+s])·∏u=0m-t-1(qr-qux+α([u]-[r]))+(x+α[r])·∑t=0mfr+t+1[mt]∏s=0t-1(x+α[r+s+1])·∏u=0m-t-1(qr+1-qux+α([u]-[r+1]))=(qr-qmx+α([m]-[r]))fr∏u=0m-1(qr-qux+α([u]-[r]))+(qr-qmx+α([m]-[r]))·∑t=1mfr+t[mt]·∏s=0t-1(x+α[r+s])·∏u=0m-t-1(qr-qux+α([u]-[r]))+(x+α[r])·∑t=1mfr+t[mt-1]∏s=0t-2(x+α[r+s+1])·∏u=0m-t(qr+1-qux+α([u]-[r+1]))+(x+α[r])fr+m+1∏s=0m-1(x+α[r+s+1])=fr∏u=0m(qr-qux+α([u]-[r]))+∑t=1m{∏u=0m-t(qr+1-qux+α([u]-[r+1]))(qr-qmx+α([m]-[r]))[mt]·∏s=0t-1(x+α[r+s])∏u=0m-t-1(qr-qux+α([u]-[r]))+(x+α[r])[mt-1]∏s=0t-2(x+α[r+s+1])∏u=0m-t(qr+1-qux+α([u]-[r+1]))}fr+t+fr+m+1∏s=0m(x+α[r+s]).
We see that
(3.5)∏u=0m-t(qr+1-qux+α([u]-[r+1]))=(qr+1-x-α[r+1])∏u=0m-t-1(qr+1-qu+1x+α([u+1]-[r+1]))=(qr+1-x-α[r+1])∏u=0m-t-1(qr+1-qu+1x+αq([u]-[r]))=(qr+1-x-α[r+1])qm-t∏u=0m-t-1(qr-qux+α([u]-[r])),
and hence,
(3.6)fr[m+1](∏i=0m(1+α[i])):=fr∏u=0m(qr-qux+α([u]-[r]))+∑t=1m{[mt](qr-qmx+α([m]-[r]))+[mt-1](qr+1-x-α[r+1])qm-t}·fr+t∏s=0t-1(x+α[r+s])∏u=0m-t-1(qr-qux+α([u]-[r]))+fr+m+1∏s=0m(x+α[r+s]).
It is easy to verify that
(3.7)[mt]+qm-t+1[mt-1]=[m+1t],[mt-1]+qt[mt]=[m+1t].
Therefore,
(3.8)[mt](qr-qmx+α([m]-[r]))+[mt-1](qr+1-x-α[r+1])qm-t=qr([mt]+qm-t+1[mt-1])-xqm-t([mt-1]+qt[mt])+α1-q{qr([mt]+qm-t+1[mt-1])-qm-t([mt-1]+qt[mt])}=[m+1t]{(qr-xqm-t)+α([m-t]-[r])}.
Consequently,
(3.9)fr[m+1](∏i=0m(1+α[i])):=fr∏u=0m(qr-qux+α([u]-[r]))+∑t=1m[m+1t]{(qr-xqm-t)+α([m-t]-[r])}·fr+t∏s=0t-1(x+α[r+s])∏u=0m-t-1(qr-qux+α([u]-[r]))+fr+m+1∏s=0m(x+α[r+s])=∑t=0m+1[m+1t]·fr+t∏s=0t-1(x+α[r+s])∏u=0m-t(qr-qux+α([u]-[r])).
Thus, one has the desired result.Theorem 3.2.
For0≤m≤n and 0≤r≤n-m, we have
(3.10)fr[m]=∑s=0mq(m-s)r[ms]Δqsfr∏i=rs+r-1(x+α[i])∏j=0s-1(1+α[j]),
for all x∈[0;1].Proof.
Using (2.2) and (3.1), we have
(3.11)fr[m]∏i=0m-1(1+α[i])=∑t=0m[mt]fr+tSt(m),
where
(3.12)St(m)=∑u=0m-t(-1)uqu(u-1)/2+(m-t-u)r[m-tu]×∏s=rt+r-1(x+α[s])∏i=0u-1(x+α[i])∏j=u-rm-t-r-1(1+α[j])(0≤t≤m).
First, we prove that
(3.13)St(m)=∑u=0m-t(-1)uqu(u-1)/2+(m-t-u)r[m-tu]·∏i=rt+u+r-1(x+α[i])∏j=u+tm-1(1+α[j])
for all m∈N0={0,1,2,…}, t∈N0, and x∈[0;1]. Note that an empty sum denotes 0.
We use the induction onm. First, we see that (3.13) holds for m=0 and all t∈N0. Let us assume that (3.13) holds for a given m, and for all t∈N0. Then, from (3.12) and (3.13), we obtain
(3.14)St(m+1)=(x+α[t+r-1])∑u=0m+1-t(-1)uqu(u-1)/2+(m-t+1-u)r[m-t+1u]·∏i=rt+u-2+r(x+α[i])∏j=u+t-1m-1(1+α[j])=∑u=0m+1-t(-1)uqu(u-1)/2+(m-t+1-u)r[m-t+1u]·∏i=rt+u+r-1(x+α[i])∏j=u+t-1m-1(1+α[j])+α∑u=1m-t+1(-1)uqu(u-1)/2+(m-t+1-u)r[m-t+1u]·([t+r+1]-[t+u+r+1])∏i=rt+u-2+r(x+α[i])∏j=u+t-1m-1(1+α[j]).
We see that
(3.15)[m-t+1u]([t+r-1]-[t+u+r-1])=-qt+r-1[m-t-u+2][m-t+1u-1],
and hence,
(3.16)St(m+1)=∑u=0m-t+1(-1)uqu(u-1)/2+(m-t+1-u)r[m-t+1u]·∏i=rt+u+r-1(x+α[i])∏j=u+t-1m-1(1+α[j])+α∑u=0m-t(-1)uqu(u-1)/2+(m-t+1-u)rqu+t-1[m-t+1u]·[m-t-u+1]∏i=rt+u+r-1(x+α[i])∏j=u+tm-1(1+α[j])=(-1)m-t+1q(m-t+1)(m-t)/2∏i=rm+r(x+α[i])+∑u=0m-t(-1)uqu(u-1)/2+(m-t+1-u)r·(1+α[u+t-1]+αqu+t-1[m-t-u+1])[m-t+1u]·∏i=rt+u+r-1(x+α[i])∏j=u+tm-1(1+α[j]).
Next, in view of the equality
(3.17)(1+α[u+t-1]+αqu+t-1[m-t-u+1])=1+α[m],
we obtain (3.13). Consequently, in view of (3.11) and (3.13), we get
(3.18)fr[m]∏i=0m-1(1+α[i])=∑t=0m[mt]fr+t∑u=0m-t(-1)uqu(u-1)/2+(m-t-u)r·[m-tu]∏i=rt+u+r-1(x+α[i])∏j=u+tm-1(1+α[j])=∑t=0m∑u=tm[mt]fr+t(-1)u-tq(u-t)(u-t-1)/2+(m-u)r·[m-tu-t]∏i=ru+r-1(x+α[i])∏j=um-1(1+α[j]).
Next, in view of the equality
(3.19)[mt][m-tu-t]=[mu][ut],
we obtain
(3.20)fr[m]∏i=0m-1(1+α[i])=∑u=0m∑t=0u[mu]fr+t(-1)u-tq(u-t)(u-t-1)/2+(m-u)r·[ut]∏i=ru+r-1(x+α[i])∏j=um-1(1+α[j])=∑u=0m[mu]q(m-u)r∏i=ru+r-1(x+α[i])∏j=um-1(1+α[j])·∑t=0u[ut](-1)u-tq(u-t)(u-t-1)/2fr+t.
The condition (1.10) completes the proof.Theorems3.1 and 3.2 are generalizations of Theorems 2.1 and 2.3 in [11].Note that whenm=n and r=0, (3.10) does indeed reduce to (1.11)



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