A combinatorial representation of the Bernstein operator
中 村 宗 敬
Munetaka NAKAMURA
Abstract
The generalized Bernstein operator is considered and and a representation theorem is given through q -Stirling numbers of the second kinds for polynomials.
1. Introduction
The Bernstein operator Bn is defined by
for a continuous function f on the unit interval [0,1], where {C（n,k）}0<－k<－n stands for binomial coefficients;
C（n,k）＝ . It is well known that Bn f （x） converges to f （x） uniformly in x as n → ∞. As is immediately and easily seen, the function Bn f （x） is a polynomial,
so that this gives a constructive concrete proof of the approximation theorem of Weierstrass. As for its rate of convergence, Voronovskaya's is fundamental, which says
if f is twicely differentiable. In order to obtain more elaborate estimation, the following representation is useful;
for f （x）＝ x r, where c （s, r－j） and S （r, s） are the Stirling number of the first and the second kind respectively （[1]）.
On the other hand, this decade, a generalization of Bn has been proposed and considered based on q -integers;
（See 3 for the details）. Here in this paper, we consider Bq,n and try to make a representation of Bq,n using
the q -Stirling numbers of the second kinds;
Bnƒ（x）＝
∑
C（n,k）ƒ（k / n）x（1－k x ）n－k n k＝0 n! k!（n－k）! Bnƒ（x）－ƒ（x）～x（1－x ）ƒ （x） 2n " Bnƒ（x）－ƒ（x）＝∑ ∑
（－1）s－r＋jS（r,s）c（s,r－j）（xs－xr） j＝1 r－1 r－1 s＝r－j 1 nj Bq,nƒ（x）＝∑
C（q n,k）ƒ（［k］）/［n］）x（1－k x）qn－k n k＝0 Bnƒ（x）＝∑
S（q r,s）［n］（S）x s r s＝1for f （x）＝ x r. We then apply this theorem to get a certain result concerned with the convergence of B
q,n f for
a continuous function f on [0,1].
2. Results on ordinary Bernstein operators
In this section, we briefly review the results in [1] with some other ones. Throughout the paper, n and r stand for positive integers, unless otherwise stated in a few exceptions. Firstly, the Stirling number of the first kind s （n, k） is defined by the following equation.
where x is a mere formal variable.
A simple observation immediately says that c （n, k）＝（－1）n－k s （n, k） is positive, and it is easy to see that, for this new c （n, k）， the following equation holds;
This positive number c （n, k） possesses a purely combinatorial meaning. That is, c （n, k） equals the number of the permutations of order n of which the number of cycles is exactly k . In view of this combinatorial meaning, one can obtain the following recurrence relations.
with the initial conditions c （n, 1）＝1 for all n. One should notice that the recurrence relation （2.4）is also obtained by comparing the coefficients of x k in （2.1）.
The Stirling number of the second kind S （n, k） is defined by the following equation.
The number S （n, k） possesses a purely combinatorial meaning as well as c （n, k） . That is, S （n, k） equals the number of the partitions of {1,2,..., n}into exactly k nonempty subsets （blocks）. In view of this combinatorial meaning, the interpretation of the following recurrence relations is quite easy.
with the initial conditions S （n, 1）＝1 for all n . Of course, this is obtained by comparing the coefficients of xk in （2.3）as before.
Let Bn be the Bernstein operator in the previous section. We induce one of its representations through Stirling numbers of both kinds in a probabilistic view point. Suppose that the distribution of a random variable
X on some probability space is B （n, x） with 0<－ x <－ 1 . In other words,
If we use the notation such as （x）n ＝ x （x－1）（x－2）… （x－n＋1）， we can compute the expectation E （Xr） as follows. n k＝1 x（x＋1）（x＋2）…（x＋n－1）＝
∑
c（n,k）xk （2.1） c（n＋1,k）＝nc（n,k－1）＋c（n,k） n≥1, k≥1 （2.2） xn＝∑
S（n,k）x（x－1）…（x－k＋1） （2.3） n k＝1 S（n＋1,k）＝S（n,k－1）＋kS（n,k） n≥1, k≥1 � （2.4） E（Xr）＝∑
krb（n,k,x）＝∑ ∑
S（r,s）（k）sb（n,k,x） n k＝0 n k＝0 r s＝1 ＝∑
S（r,s）∑
（k）b（n,k,x）＝∑
S（r,s）E［（X）］＝（n）∑
S（r,s）xs . r n r r n k＝1 x（x＋1）（x＋2）…（x－n＋1）＝∑
s（n,k）xk b（n,k,x）:＝P（X＝k）＝C（n,k）x（1－k x）n－k, k ＝0,1,...,n.Note that in the above computation of we use the equality E［（X）s］＝（n）s xs . which is easily proved by using the combinatorial relation （k）s C （n, k） ＝ （n）s C （n－s, k－s）．
Now let us switch to the Bernstein operators. In view of the stated results, one has, for f （x）＝ x r ,
To summarize, we have
If one compute according to（2.5）, Bn f （x）, viewed as a（negative）power series of n , the coefficient of n－1 is given by
which implies
This further suggests Voronovskaya’s theorem or the asymptotics of higher orders.
3. Generalized counterparts
In this section, we introduce q -analogue of binomial coefficients and Stirling numbers. Let 0＜ q ＜－ 1 . The q -integer [n] and the q -factorial [n]! are defined by
Formally, we introduce [0]!＝1 as in the ordinary case.
The q -analogue of C （n, k）， which is denoted by C q （n, k） is naturally defined by
We also put C（q n, k）＝ 0 if k ＜ 0 or k ＞ n for convenience. For some interesting combinatorial interpretations of Cq （n, k）， see [2] for example.
Bnƒ（x）－ƒ（x）＝
∑
krb（n,k,x）－xr＝∑�
S（r,s）（n）sxs－xr r－1 j＝1 r s＝1 1 nrt 1 nr ＝∑
S（r,s）（n）s（xs－xr）＝∑�
S（r,s）∑
（－1）s－kc（s,k）n（k xs－xr） r s＝1 r s＝1 s k＝1 1 nr 1 nr ＝∑
∑
（－1）s－kS（r,s）c（s,k）（xs－xr） r－1 k＝1 r－1 s＝k 1 nr－k ＝∑
∑
（－1）s－r＋j S（r,s）c（s,r－j）（xs－xr）.� r－1 j＝1 r－1 s＝r－j 1 nj Bnƒ（x）－ƒ（x）＝∑
∑�
（－1）s－r＋j S（r,s）c（s,r－j）（xs－xr） （2.5） r－1 j＝1 r－1 s＝r－j 1 nj （r）2 2 S（r,r－1）c（r－1,r－1）＝S（r,r－1）＝C（r,2）＝ , Bnƒ（x）＝ƒ（x）＋ （x（r）2 r－1－xr）＋O（n－2） 2n ＝ƒ（x）＋ ＋O（n（r）2�x －2）＝ƒ（x）＋ ＋O（n－2）. r－2×x（1－x ） 2n ƒ "（x ）x（1－x ） 2n ［n］＝1＋q＋q2＋…＋qn－1＝ ,1－q n 1－q ［n］!＝［n］×［n－1］×…×［2］×［1］. Cq（n,k）＝ ��, k＝0,1,...,n.［n］! ［k］!［n－k］!With definition of the power of the sum , we can prove the following Gauss’s binomial formula.
Concerned with C q （n, k）, there are two q -analogue of the recurrence relations.
Based on C q （n, k）, we can define q -analogue of Bernstein operator by
for a continuous function f on [0,1].
It may be appropriate to define the q -Stirling number of the first kind c q （n, k） and the q -Stirling number of the second kind S q （n, k） in the following recurrent way respectively.
with the conditions c q （1,1）＝ S q （1,1）＝1 and c q （n, k）＝ S q （n, k）＝0 if k ＜ 0 or k ＞ n. 4. Representation of the Bernstein operator
Firstly let us consider the generating functions of the q -Stirling numbers as in the ordinary cases. Though in order to prove the main theorem （Theorem 4）， we do not use the following Theorem 1 for the q -Stirling number of the first kind, we present here for the comparison to （2.1）.
Theorem 1. The following equation holds for any positive integer x .
where
Proof. We prove （4.1）by the induction on n . It is clear that the equality is true if n＝1 . Hence we assume （4.1）is true for n－1. Then, we have
（by the recurrence relation （3.3）） （x＋y）qn＝
∏
（x＋qk－1y） n k＝1 （x＋y）qn＝∑
C（q n,k）qk（k－1）/2xkyn－k. （3.1） C（q n,k）＝C（q n－1,k－1）＋qkC（q n－1,k）, C（q n,k）＝qn－kC（q n－1,k－1）＋C（q n－1,k）. Bq,nƒ（x）＝∑
C（q n,k）ƒ（［k］/［n］）x（1－k x ）qn－k n k＝0 C（q n,k）＝qn－1c（q n－1,k－1）＋［n－1］c（q n－1,k）,��n≥2,��1≤k≤n, （3.2） S（q n,k）＝qk－1S（q n－1,k－1）＋［k］S（q n－1,k）,��n≥2,��1≤k≤n （3.3） ［x］（n）＝∑
c（q n,k）［x］k � （4.1） n k＝1 ［x］（n）＝∏
［x＋j］. n－1 j＝0∑
c（q n,k）［x］k＝qn－1�∑
c（q n－1,k－1）［x］k＋［n－1］�∑
c（q n－1,k）［x］k n k＝1 n k＝1 n k＝1 ＝�∑
c（q n－1,k）（qn－1［x］k＋1＋［n－1］［x］k）＝∑�
c（q n－1,k）［x］（k qn－1［x］＋［n－1］） n－1 k＝1 n－1 k＝1This completes the proof.
Theorem 2. The following equation holds for any positive integer x .
where
Proof. The proof is carried out by the induction on n as in the previous theorem. It is clear that the equality is true if n＝1 . Hence we assume （4.2）is true for n－1 . Then, we have
（by the recurrence relation （3.4））
Note that [ x ] （ k ）＝ 0 if x ＜ k and all of the above equalities are valid. This completes the proof.
Lemma 3. Set bq （n, k, x）＝ Cq （n, k）x k （1－x）qn－k . Then
[k] （s）bq （n, k, x）＝[n] （s）x s bq （ n－s, k－s, x ）． Proof. By the definition of Cq （n, k）， the equation is shown as follows.
＝
∑
��（� �－1,�）（�［� �］（�）＋［�］［�］（�））＝∑
��（� �－1,�）［�］（�）（�［� �－�］＋［�］） �－1 �＝1 �－1 �＝1 ＝［�］∑
��（� �－1,�）［�］（�） （�����［�］＝�［� �－�］＋［�－1］） �－1 �＝1 ＝［�］［�］�－1 （����������������������������������） ＝［�］�. ＝［�＋�－1］�∑�
�（� �－1,�）［�］� （�����［�＋�－1］＝��－1［�］＋［�－1］） �－1 �＝1 ＝［�］（�）. ＝［�＋�－1］�［�］（�－1） （����������������������������������） ［�］�� ＝∑�
�（� �,�）［�］（�） （4.2） � �＝1 ［�］（�）＝∏�
［�－�］. �－1 �＝0∑
��（� �,�）［�］（�）� ＝∑
���－1�（� �－1,�－1）［�］（�）＋［�］∑
��（� �－1,�）［�］（�） � �＝1 � �＝1 � �＝1 ［�］（�）�（� �,���）＝［�］（�）�（� �,�）� � （1－�）��－� ＝［�］（�）× ×�（1－� �）� �－� ［�］! ［�］!［�－�］! ＝��× ×［�］（�）［�－�］! ��－�（1－�）��－�－（�－�） ［�－�］!［（�－�）－（�－�）］! ＝［�］（�）����（�－�,�－�）� �－� （1－�）��－�－（�－�） ＝［�］（�）����（�－�,�－�,�）.Theorem 4. The following equality holds for f （x）＝ x r .
Proof. Using bq （n, k, x） and Theorem 2 as in the ordinary case, we can compute as follows.
On the other hand, by Lemma 3 and Gauss's binomial formula （3.1）， it follows that
We conclude the paper stating an application of the above representation formula. The following theorem is shown in [4] in a different manner. Our proof is more straightforward on the basis of （4.4）.
Theorem 5. If qn ∈ （0,1）and q n → 1 as n → ∞ , B q n . n f （x） converges f （x） uniformly in x . Proof. Putting q ＝ qn in （4.3）， we obtain
for f （x）＝ x r . It is obvious from the definition （3.3）that S q （r, s） is a polynomial, and is continuous in q,
which implies S q n （r, s） →S 1 （r, s） ＝ S（r, s） as n → ∞ for any fixed r and s . This further means that
the set ｛S qn （r, s）：0≤s≤r,n＝1,2,...｝ is bounded. Since [n] ＝1＋q n＋q n2＋…＋q nn－1 → ∞ as n→ ∞ , the
is a polynomial in [n] of degree s . By the way, again by the definition （3.3）we get the relation S q （r, r）
＝ qr－1 S
q （r－1, r－1）， which leads to S q（r, r）＝ qr （r－1）/2 in general. Therefore we have S qn （r, r）
theorem mean the desired result.
Bq,nƒ（x）＝
∑
S（q r,s）［n］（s）xs. （4.3） r s＝1 1 nr Bq,nƒ（x）＝ ��∑
［k］rb（q n,k,x） n k＝0 1 ［n ］r ＝ ��∑
b（q n,k,x）∑
S（q r,s）［k］（s） n k＝0 r s＝1 1 ［n ］r ＝ ��∑
S（q r,s）∑
［k］（s）b（q n,k,x）. r s＝1 n k＝0 1 ［n ］r Bq,nƒ（x）＝ ��∑
S（q r,s）［n］（s）�xs∑
b（q n－s,k－s,x） r s＝1 r－s k＝s 1 ［n ］r ＝ ��∑
S（q r,s）［n］（s）�xs∑
C（q n－s,k－s,x）x k－s （1－x）qn－s－（k－s） r s＝1 r－s k＝s 1 ［n ］r ＝ ��∑
S（q r,s）［n］（s）��xs （x＋1－x）q n－s r s＝1 1 ［n ］r ＝ ��∑
S（q r,s）［n］（s）. r s＝1 1 ［n ］r s＝1 Bqn,nƒ（x）＝∑
Sq n（r,s）［n］（s）�xs （4.4） r 1 nr Sq n（r,r）［n］（r） ［n ］rterms other than in （4.4）converge to 0 uniformly in x as n → ∞ . Recall that [n] （s）
Sq n（r,r）［n］（r）
［n ］r
References
[1] U.Abel, The complete asymptotic expansion for the Meyer-Konig and Zeller operators, J. Mathematical Analysis and applications 208,109-119, 1997.
[2] V.Kac and P.Cheung, Quantum calculus, Springer, 2002.
[3] H.Oruc and G.M. Philips, A generalization of the Bernstein polynomials, Proc. Edin. Math. Soc. 42, 403-412,1999. [4] G.M. Philips, Interpolation and approximation by polynomials, Canad. Math. Soc. 2003.
