/compile/output.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 3, 2016, 322-332 ISSN 1307-5543 – www.ejpam.com The Exact Order of Approximation for Bivariate Complex Bernstein- Schurer Polynomials Nurhayat İspir, Şule Yüksel Güngör∗ Department of Mathematics, Faculty of Science, Gazi University, Ankara, Turkey Abstract. In this paper we study the approximation properties of the tensor product kind bivariate complex Bernstein-Schurer polynomials. We obtain the order of simultaneous approximation and Voronovskaja-type results with quantitative estimate for bivariate complex Bernstein-Schurer polyno- mials attached to analytic functions on compact polydisks. 2010 Mathematics Subject Classifications: 30E10, 41A25, 41A28. Key Words and Phrases: Bivariate complex polynomials, Berstein-Schurer polynomials, Rate of con- vergence, Voronovskaja’s theorem, Exact order of approximation. 1. Introduction The Bernstein polynomial attached to f : [0,1]→ R was introduced by Bernstein to give a proof of the Weierstrass Theorem. If f is continuous on [0,1] then limn Bn( f )(x) = f (x) where Bn( f )(x) = ∑n k=0 � n k � xk (1− x)n−k f � k n � , x ∈ [0,1]. If f : G→ C is an analytic function in the open set G ⊂ C, with D1 ⊂ G (where D1 = {z ∈ C : |z|< 1}), then S. N. Bernstein [6] proved that the complex Bernstein polynomials defined by Bn( f )(z) = n ∑ k=0 � n k � zk (1− z)n−k f � k n � uniformly converge to f in D1. But Bernstein obtained this convergence result without any quantitative estimate. Recently, Voronovskaja- type results with quantitative estimates for the complex Bernstein, complex q-Bernstein, complex Bernstein- Kantorovich, complex Kan- torovich - Stancu polynomials attached to analytic functions on compact disks and the exact order of simultaneous approximation by these complex operators were obtained by S. G. Gal [5]. ∗Corresponding author. Email addresses: nispir@gazi.edu.tr (N. İspir), sulegungor@gazi.edu.tr (Ş. Güngör) http://www.ejpam.com 322 c© 2016 EJPAM All rights reserved. N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 323 The complex Bernstein-Schurer polynomials (introduced and studied in the case of real variable in [7]) are defined for any fixed p = 0,1,2, . . . by Bn,p( f )(z) = n+p ∑ k=0 � n+ p k � zk(1− z)n+p−k f (k/n) , z ∈ C. The approximation properties of these polynomials are investigated and the exact order of approximation with quantitative estimates were gave in [1]. It is clear that for p = 0 these polynomials become the classical complex Bernstein polynomials studied in [5]. In real case the approximation properties of bivariate Berstein-Schurer polynomials is studied by D. Barbasu [2–4]. In this note we would like to extend the approximation results from the univariate case, obtained for the complex Bernstein-Schurer polynomials, to the bivariate case. First we present a few concepts in the bivariate case which are natural extensions of the usual concepts in the univariate case. Let DR j := � z j ∈ C : � �z j � �< R j , j = 1,2 and P(0; R) = DR1 × DR2 denotes an open polydisk (of center 0 and radius R) where R = (R1,R2) and � �z1 � �≤ r1, � �z2 � �≤ r2, r1 < R1 with r2 < R2. Let also PR := P (0,R) = �� z1, z2 � ∈ C2 : � �z j � �≤ R j , j = 1,2 denotes the closed polydisk. For f (z1, z2) is an analytic function of two complex variables (z1, z2) in the polydisk P(0; R) we can define the tensor product kind Bernstein-Schurer poly- nomials as follows Bn,m,p,q( f ) � z1, z2 � = n+p ∑ k=0 m+q ∑ j=0 bn,k � z1 � bm, j � z2 � f � k/n, j/m � (1) where bn,k � z1 � = � n+p k � zk 1(1− z1) n+p−k, bm, j � z2 � = � m+q j � z j 2 (1− z2) m+q− j , n, m ∈ N and p,q ∈ N∪ {0}. The goal of this paper is to obtain the exact order of approximation for the polynomi- als given by (1) on compact polydisks. First we give the order of approximation and the Voronovskaja- type theorems with quantitative estimate for the polynomials Bn,m � f � (z) de- fined by (1). These results allow us to obtain the exact order in approximation by the polyno- mials Bn,m � f � (z). 2. The Convergence Results with Quantitative Estimates Theorem 1. For fixed p,q ∈ N∪{0} and R1 > p+1, R2 > q+1 suppose that f : P(0; R)→ C is analytic in P(0; R) = DR1 × DR2 , that is f (z1, z2) = ∑∞ k=0 ∑∞ j=0 ck, jz k 1z j 2 for all (z1,z2) ∈ P(0; R), R= (R1; R2). Then we have (i) For all � �z1 � �≤ r1, � �z2 � �≤ r2 with 1< r1, (p+1)r1 < R1, 1< r2, (q+1)r2 < R2 and n, m ∈ N � �Bn,m,p,q( f ) � z1, z2 � − f (z1; z2) � �≤ M p,q r1,r2,n,m ( f ) N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 324 where M p,q r1,r2,n,m ( f ) = ∞ ∑ k=1 ∞ ∑ j=1 � �ck, j � � r j 2 � 3k (k− 1) � n+ p � �� p+ 1 � r1 �k + 1 n �� p+ 1 � r1 �k − rk 1 n � + ∞ ∑ k=1 ∞ ∑ j=1 � �ck, j � � rk 1   3 j � j − 1 � � m+ q � �� q+ 1 � r2 � j + 1 m �� q+ 1 � r2 � j − r j 2 m   and M p,q r1,r2,n,m ( f )<∞. (ii) Let k1, k2 ∈ N be with k1 + k2 ≥ 1, 1≤ r1 < r∗1 ≤ (1+ p)r1 < R1, 1≤ r2 < r∗2 ≤ (1+q)r2 < R2. Then for all � �z1 � �≤ r1, � �z2 � �≤ r2 and n, m ∈ N, p,q ∈ N∪{0} we have � � � � � ∂ k1+k2 Bn,m,p,q( f ) ∂ k1z1∂ k2z2 � z1, z2 � − ∂ k1+k2 f ∂ k1z1∂ k2z2 � z1, z2 � � � � � � ≤ M p,q r∗ 1, r∗ 2,n,m ( f ). k1! � r∗ 1 − r1 �k1+1 k2! � r∗ 2 − r2 �k2+1 where M p,q r∗ 1, r∗ 2,n,m ( f ) is given as at the above point (i). Proof. (i) Denote ek, j(z1, z2) = ek(z1)e j(z2) where ek(u) = uk. Since f (z1, z2) = ∑∞ k=0 ∑∞ j=0 ck, jek, j(z1, z2) and by the definition of the operator (1) we get � �Bn,m,p,q( f ) � z1, z2 � − f (z1, z2) � �≤ ∞ ∑ k=0 ∞ ∑ j=0 � �ck, j � � � �Bn,m,p,q(ek, j) � z1, z2 � − ek, j(z1, z2) � � . By the simple calculation we can write � �Bn,m,p,q(ek, j) � z1, z2 � − ek, j(z1, z2) � �= � � �Bn,p(ek)(z1).Bm,q(e j)(z2)− zk 1z j 2 � � � ≤ � � �z j 2 � � � � �Bn,p(ek)(z1)− zk 1 � �+ � �Bn,p(ek)(z1) � � � � �Bm,q(e j)(z2)− z j 2 � � � ≤r j 2 A+ � �Bn,p(ek)(z1) � �B, (2) say. By the Stirling numbers of second kind S � k, j � , we can write Bn,p(ek)(z1) = k ∑ j=1 S � k, j � � n+ p � . . . � n+ p− � j − 1 �� � n+ p � j z j (for complex Bernstein polynomials in one variable see [5, page 27]). Since S � k, j � ¾ 0, for all n, k ∈ N, p ∈ N∪ {0} and ∑k j=1 S � k, j � � n+ p � . . . � n+ p− � j − 1 �� = � n+ p �k it follows � �Bn,p(ek)(z1) � �¶ k ∑ j=1 S � k, j � � n+ p � . . . � n+ p− � j − 1 �� � n+ p �k r j ¶ rk N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 325 for all � �z1 � �≤ r1, with 1< r1, (p+1)r1 < R1. To estimate A and B for fixed n, m ∈ N, we should consider two possible cases: (1) 0≤ k ≤ n+ p, 0≤ j ≤ m+ q and (2) k > n+ p, j > m+ q. We start with case (1). If k = 0, j = 0 then obviously we get � �Bn,p(ek)(z1)− (ek)(z1) � �= 0 and � �Bm,q(e j)(z2)− (e j)(z2) � �= 0. Therefore, let us suppose that 1≤ k ≤ n+ p, 1 ≤ j ≤ m+ q. Denoting by ∆k the finite difference of order k, as in the case of the classical Bernstein polynomials we easily can write the representation formulas Bn,p( f )(z1) = n+p ∑ v=0 ∆v 1/n f (0)ev(z1), Bm,q( f )(z2) = m+q ∑ w=0 ∆w 1/m f (0)ew(z2). For simplicity, we use the following notations C p n,v,k = � n+ p v � ∆v 1/n ek(0) = � n+ p v �� 0, 1 n . . . , j n ; ek � v!/nv , C q n,w, j = � m+ q w � ∆w 1/m e j(0) = � m+ q w �� 0, 1 m , . . . , j m ; e j � w!/mw. Since ek, e j are convex of any order, it follows that all C p n,v,k ≥ 0, C q n,w, j ≥ 0 and taking into account that Bn,p( f )(1) = f � (n+ p)/p � , Bm,q( f )(1) = f �� m+ q � /m � we get n+p ∑ v=0 C p n,v,k = Bn,p(e1)(1) = � n+ p n �k , m+q ∑ w=0 C q n,w, j = Bn,p(e2)(1) = � m+ q m � j . (3) Using the result in the proof of Theorem 2.1 in [1] for any � �z1 � �≤ r1, � �z2 � �≤ r2 with 1≤ r1 ≤ (p+ 1)r1 < R1, 1≤ r2 ≤ (q+ 1)r2 < R2, directly we can write A= � � �Bm,q(e j)(z2)− z j 2 � � �≤ j � j − 1 � � m+ q � �� q+ 1 � r2 � j + 1 m �� q+ 1 � r2 � j − r j 2 m and B = � �Bn,p(ek)(z1)− ek(z1) � �≤ k (k− 1) � n+ p � �� p+ 1 � r1 �k + 1 n �� p+ 1 � r1 �k − rk 1 n . We now consider second case. For k > n+ p, � �z1 � � ≤ r1 with 1 ≤ r1 ≤ (p + 1)r1 < R1 and for j > m+ q, � �z2 � �≤ r2 with 1≤ r2 ≤ (q+ 1)r2 < R2, and considering (3) we get B = � �Bn,p(ek)(z1)− ek(z1) � �≤ � �Bn,p(ek)(z1) � �+ rk 1 ≤ � n+ p n �k r n+p 1 + rk 1 ≤ 2 �� p+ 1 � r1 �k ≤ 2 (k− 1) n+ p �� p+ 1 � r1 �k , and in similar way, A= � �Bm,q(e j)(z2)− (e j)(z2) � �≤ 2( j−1) m+q �� q+ 1 � r2 � j . N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 326 Combining all of the results obtained for A and B, we have the desired inequality. (ii) Now we give the rate of convergence in simultaneous approximation. Let 1 ≤ r1 < r∗1 < R1 2 , 1 ≤ r2 < r∗2 < R2 2 and γ1 = � �u1 − z1 � � = r∗1 , γ2 = � �u2 − z2 � � = r∗2 . By the Cauchy’s formula ∂ k1+k2 Bn,m,p,q( f ) ∂ k1z1∂ k2z2 � z1, z2 � − ∂ k1+k2 f ∂ k1z1∂ k2z2 � z1, z2 � = k1!k2! (2πi)2 ∫ γ2 ∫ γ1 � Bn,m,p,q(u1,u2)− f (u1,u2) � du1du2 � u1 − z1 �k1+1 � u2 − z2 �k2+1 passing to absolute value with � �z1 � �≤ r1, � �z2 � �≤ r2 and taking into account that � �u1 − z1 � �≥ r∗1 − r1, � �u2 − z2 � �≥ r∗2 − r2, by applying the estimate in (i) we easily obtain � � � � � ∂ k1+k2 Bn,m,p,q( f ) ∂ k1z1∂ k2z2 � z1, z2 � − ∂ k1+k2 f ∂ k1z1∂ k2z2 � z1, z2 � � � � � � ≤ M p,q r∗ 1, r∗ 2,n,m ( f ). k1! � r∗ 1 − r1 �k1+1 k2! � r∗ 2 − r2 �k2+1 which proves the theorem. In what follows a Voronovskaja’s result for Bn,m,p,q( f ) is presented. It will be the product of the parametric extensions generated by the Voronovskaja’s formula in univariate case in The- orem 2.2 in [1]. Indeed, for f (z1, z2) defining the parametric extensions of the Voronovskaja’s formula by z1 Ln( f )(z1, z2) :=Bn,p � f (., z2) � (z1)− f (z1, z2)− p n z1 ∂ f ∂ z1 � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, z2), z2 Lm( f )(z1, z2) :=Bm,q � f (z1, .) � (z2)− f (z1, z2)− q m z2 ∂ f ∂ z2 � z1, z2 � − z2(1− z2) 2m ∂ 2 f ∂ z2 2 (z1, z2), their product (composition) gives z2 Lm( f )(z1, z2)oz1 Ln( f )(z1, z2) =Bm,q � Bn,p � f (., .) � (z1)− f (z1, .)− p n z1 ∂ f ∂ z1 � z1, . � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, .) � � z2 � − � Bn,p � f (., z2) � (z1)− f (z1, z2)− p n z1 ∂ f ∂ z1 � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, z2) � − q m z2 � Bn,p � ∂ f ∂ z2 (., z2) � (z1)− ∂ f ∂ z2 (z1, z2) − p n z1 ∂ f ∂ z1 � ∂ f ∂ z2 � � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 � ∂ f ∂ z2 � � z1, z2 � � − z2(1− z2) 2m . � Bn,p � ∂ 2 f ∂ z2 2 (., z2) � (z1)− ∂ 2 f ∂ z2 2 (z1, z2)− − p n z1 ∂ f ∂ z1 � ∂ 2 f ∂ z2 2 � � z1, z2 � − z1(1− z1) 2n ∂ 2 ∂ z2 1 � ∂ 2 f ∂ z2 2 � (z1, z2) � N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 327 :=E1 − E2 − E3 − E4. After simple calculation, we can write z2 Lm( f )(z1, z2)oz1 Ln( f )(z1, z2) = Bn,m,p,q � f � � z1, z2 � − Bm,q � f � z1, . �� � z2 � − p n z1.Bm,q � ∂ f ∂ z1 � z1, . � � � z2 � − z1(1− z1) 2n .Bm,q � ∂ 2 f ∂ z2 1 (z1, .) � (z2) − Bn,p � f (., z2) � (z1)− f (z1, z2)− p n z1 ∂ f ∂ z1 � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, z2) − q m z2 Bn,p � ∂ f ∂ z2 (., z2) � (z1) + q m z2 ∂ f ∂ z2 (z1, z2) + p n z1 q m z2 ∂ 2 f ∂ z1∂ z2 � z1, z2 � + q m z2 z1(1− z1) 2n ∂ 2 f ∂ z2 1 � ∂ f ∂ z2 � � z1, z2 � − z2(1− z2) 2m .Bn,p � ∂ 2 f ∂ z2 2 (., z2) � (z1) + z2(1− z2) 2m ∂ 2 f ∂ z2 2 (z1, z2) + z2(1− z2) 2m p n z1 ∂ f ∂ z1 � ∂ 2 f ∂ z2 2 � � z1, z2 � + z1(1− z1) 2n z2(1− z2) 2m ∂ 4 f ∂ z2 1 ∂ z2 2 (z1, z2) from which immediately can be derived the commutativity property z2 Lm( f )(z1, z2)oz1 Ln( f )(z1, z2) =z1 Ln( f )(z1, z2)oz2 Lm( f )(z1, z2). Now we can give the Voronovskaja- type theorem. Theorem 2. For fixed p,q ∈ N∪ {0} and R1 > p+ 1, R2 > q+ 1 suppose that f : P(0; R)→ C is analytic in P(0; R) = DR1 × DR2 , that is f (z1; z2) = ∑∞ k=0 ∑∞ j=0 ck, jz k 1z j 2 for all � z1,z2 � ∈ P(0; R), R = (R1,R2). For all � �z1 � � ≤ r1, � �z2 � � ≤ r2 with 1 < r1, (p + 1)r1 < R1, 1 < r2, (q+ 1)r2 < R2 and n, m ∈ N we have � � z2 Lm( f )(z1, z2)oz1 Ln( f )(z1, z2) � �≤ M p,q r1,r2 ( f ) � 1 n2 + 1 m2 � , where M p,q r1,r2 ( f ) =max ( ∞ ∑ k=2 ∞ ∑ j=0 ck, j �� q+ 1 � r2 � j Dk,p,r1 , ∞ ∑ k=1 ∞ ∑ j=1 � �ck, j � � j �� q+ 1 � r2 � j−1 Dk,p,r1 , ∞ ∑ k=2 ∞ ∑ j=2 ck, j j( j − 1) �� q+ 1 � r2 � j−2 Dk,p,r1 ) , Dk,p,r1 = (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 ,Ak = (k− 1) [4 (k− 1) (k− 2) + 2] and Ck,p = (k− 1) � p (5k− 4) + p2 + k (4k− 7) � . N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 328 Proof. Since f (z1; z2) = ∑∞ k=0 ∑∞ j=0 ck, jz k 1z j 2 for all � z1,z2 � ∈ P(0; R), we can write f (z1; z2) = ∑∞ k=0 fk(z2)z k 1 with fk(z2) = ∑∞ j=0 ck, jz j 2 . It follows ∂ f ∂ z1 (z1, z2) = ∑∞ k=1 fk � z2 � kzk−1 1 , ∂ 2 f ∂ z2 1 (z1, z2) = ∑∞ k=2 fk � z2 � k (k− 1) zk−2 1 and ∂ 2 f ∂ z2 2 (z1, z2) = ∑∞ k=0 ∂ 2 ∂ z2 2 fk � z2 � zk 1 where ∂ 2 ∂ z2 2 fk � z2 � = ∑∞ j=2 ck, j j( j − 1)z j−2 2 . Hence Bn,p � f (., z2) � (z1) = ∑∞ k=0 fk � z2 � Bn,p � ek 1 � � z1 � and Bn,p � f (., z2) � (z1)− f (z1, z2)− p n z1 ∂ f ∂ z1 � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, z2) = ∞ ∑ k=2 fk � z2 � � Bn,p � ek 1 � � z1 � − � ek 1 � � z1 � − p n kzk 1 − zk−1 1 (1− z1)k (k− 1) 2n � . Applying Bm,q to the last expression with respect to z2, we obtain E1 = ∞ ∑ k=2 Bm,q � fk � � z2 � � Bn,p � ek 1 � � z1 � − � ek 1 � � z1 � − p n kzk 1 − zk−1 1 (1− z1)k (k− 1) 2n � = ∞ ∑ k=2 ∞ ∑ j=0 ck, jBm,q � e j 1 � � z2 � ! � Bn,p � ek 1 � � z1 � − � ek 1 � � z1 � − p n kzk 1 − zk−1 1 (1− z1)k (k− 1) 2n � . Passing now to absolute value with � �z1 � � ≤ r1, � �z2 � � ≤ r2 and considering the estimates in the proof of Theorem 2.1 and Theorem 2.2 in [1], we can write � �E1 � �≤ ∞ ∑ k=2 ∞ ∑ j=0 � �ck, j � � �� 1+ q � r2 � j   (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 n2   where Ak = (k− 1) [4 (k− 1) (k− 2) + 2] and Ck,p = (k− 1) � p (5k− 4) + p2 + k (4k− 7) � . Similarly, � �E2 � �≤ ∞ ∑ k=2 � � fk � z2 �� �   (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 n2   ≤ ∞ ∑ k=2 ∞ ∑ j=0 � �ck, j � � �� 1+ q � r2 � j   (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 n2   . Then Bn,p � ∂ f ∂ z2 (., z2) � (z1) = ∞ ∑ k=0 ∂ fk ∂ z2 � z2 � Bn,p � ek 1 � � z1 � = ∞ ∑ k=0 ∞ ∑ j=1 ck, j jz j−1 2 Bn,p � ek 1 � � z1 � N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 329 and � Bn,p � ∂ f ∂ z2 (., z2) � (z1)− ∂ f ∂ z2 (z1, z2)− p n z1 ∂ f ∂ z1 � ∂ f ∂ z2 � � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 � ∂ f ∂ z2 � � z1, z2 � � = ∞ ∑ k=1 ∞ ∑ j=1 ck, j jz j−1 2 � Bn,p � ek 1 � � z1 � − � ek 1 � � z1 � − p n kzk 1 − zk−1 1 (1− z1)k (k− 1) 2n � , in the same way, we get � �E3 � �≤ ∞ ∑ k=1 ∞ ∑ j=1 � �ck, j � � j �� q+ 1 � r2 � j−1   (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 n2   . Similarly Bn,p � ∂ 2 f ∂ z2 2 (., z2) � (z1) = ∞ ∑ k=0 ∂ 2 fk ∂ z2 2 � z2 � Bn,p � ek 1 � � z1 � = ∞ ∑ k=0 ∞ ∑ j=2 ck, j j � j − 1 � z j−2 2 Bn,p � ek 1 � � z1 � and Bn,p � ∂ 2 f ∂ z2 2 (., z2) � (z1)− ∂ 2 f ∂ z2 2 (z1, z2)− p n z1 ∂ f ∂ z1 � ∂ 2 f ∂ z2 2 � � z1, z2 � − z1(1− z1) 2n ∂ 2 ∂ z2 1 � ∂ 2 f ∂ z2 2 � (z1, z2) = ∞ ∑ k=2 ∞ ∑ j=2 ck, j j � j − 1 � z j−2 2 � Bn,p � ek 1 � � z1 � − � ek 1 � � z1 � − p n kzk 1 − zk−1 1 (1− z1)k (k− 1) 2n � hence by similar opinion we have � �E4 � �≤ ∞ ∑ k=2 ∞ ∑ j=2 � �ck, j � � j � j − 1 � �� q+ 1 � r2 � j−2   (k− 1)Ak �� p+ 1 � r1 �k−1 + Ck,p �� p+ 1 � r1 �k−2 n2   . Interchanging above the places of n and m we obtain a similar order of approximation for � � z1 Lm( f )(z1, z2)oz2 Ln( f )(z1, z2) � � therefore � � z2 Lm( f )(z1, z2)oz1 Ln( f )(z1, z2) � �≤ � �E1 � �+ � �E2 � �+ � �E3 � �+ � �E4 � � ≤M p,q r1,r2 ( f ) � 1 n2 + 1 m2 � with M p,q r1,r2 ( f ) given by the statement. The Voronovskaja- type theorem will be used to find the exact order in approximation by Bn,n,p,p � f � . We present the following Theorem. N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 330 Theorem 3. For fixed p,q ∈ N∪ {0} and R1 > p+ 1, R2 > q+ 1 suppose that f : P(0; R)→ C is analytic in P(0; R) = DR1 × DR2 , that is f (z1, z2) = ∑∞ k=0 ∑∞ j=0 ck, jz k 1z j 2 for all (z1, z2) ∈ P(0; R), R = (R1; R2). Denoting f r1,r2 = sup �� � f (z1; z2) � � : � �z1 � �≤ r1, � �z2 � �≤ r2 , if f is not a solution of the complex partial differential equation pz1 ∂ f ∂ z1 + z1(1− z1) 2 ∂ 2 f ∂ z2 1 (z1, z2) + qz2 ∂ f ∂ z2 + z2(1− z2) 2 ∂ 2 f ∂ z2 2 (z1, z2) = 0, (4) for any (z1, z2) ∈ P(0; R), then we have Bn,n,p,p − f r1,r2 ≥ Kr1,r2, f n , for all n ∈ N (5) where Kr1,r2, f is independent on n. Proof. We can write Bn,n,p,p � f � � z1, z2 � − f � z1, z2 � = 2 n ¨ p 2 z1 ∂ f ∂ z1 + z1(1− z1) 4 ∂ 2 f ∂ z2 1 (z1, z2) + p 2 z2 ∂ f ∂ z2 + z2(1− z2) 4 ∂ 2 f ∂ z2 2 (z1, z2) + 2 n � n2 4 � z2 Ln( f )oz1 Ln( f ) � (z1, z2) � + Rn � f � (z1, z2) � where Rn � f � (z1, z2) = n 2 � Bn,p � f (z1, .) � (z2)− f (z1, z2)− p n z2 ∂ f ∂ z2 � z1, z2 � − z2(1− z2) 2n ∂ 2 f ∂ z2 2 (z1, z2) � + n 2 � Bn,p � f (., z2) � (z1)− f (z1, z2)− p n z1 ∂ f ∂ z1 � z1, z2 � − z1(1− z1) 2n ∂ 2 f ∂ z2 1 (z1, z2) � + p 2n z1 � Bn,p � ∂ f ∂ z1 � z1, . � � � z2 � − ∂ f ∂ z1 � z1, z2 � � +z2 � Bn,p � ∂ f ∂ z2 (., z2) � (z1)− ∂ f ∂ z2 � z1, z2 � � ! + z2(1− z2) 4 � Bn,p � ∂ 2 f ∂ z2 2 (., z2) � (z1)− ∂ 2 f ∂ z2 2 (z1, z2)− p n z1 ∂ f ∂ z1 � ∂ 2 f ∂ z2 2 � � z1, z2 � � + z1(1− z1) 4 � Bn,p � ∂ 2 f ∂ z2 1 (z1, .) � (z2)− ∂ 2 f ∂ z2 1 (z1, z2)− p n z2 ∂ 2 f ∂ z2 1 � ∂ f ∂ z2 � � z1, z2 � � + z1(1− z1)z2(1− z2) 8n ∂ 4 f ∂ z2 1 ∂ z2 2 (z1, z2) From Theorem 2.2 in [1] and by the reasonings in the above Theorem 2, it is immediate that Rn � f � r1,r2 → 0 as n→∞. Also, by Theorem 2 we obtain n2 4 z2 Lm( f )oz1 Ln( f ) r1,r2 ≤ M p,q r1,r2 ( f ) 2 N. İspir, Ş. Güngör / Eur. J. Pure Appl. Math, 9 (2016), 322-332 331 which implies 2 n � n2 4 � z2 Lm( f )oz1 Ln( f ) � � + Rn � f � r1,r2 → 0, as n→∞. Denoting H � z1, z2 � = p 2 z1 ∂ f ∂ z1 + z1(1− z1) 4 ∂ 2 f ∂ z2 1 (z1, z2) + q 2 z2 ∂ f ∂ z2 + z2(1− z2) 4 ∂ 2 f ∂ z2 2 (z1, z2) and from the inequalities ‖F + G‖r1,r2 ≥ � �‖F‖r1,r2 − ‖G‖r1,r2 � �≥ ‖F‖r1,r2 − ‖G‖r1,r2 it follows Bn,n,p,p − f r1,r2 ≥ 2 n ¨ ‖H‖r1,r2 − 2 n � n2 4 � z2 Lm( f )oz1 Ln( f ) � � + Rn � f � r1,r2 « ≥ 2 n 1 2 ‖H‖r1,r2 = 1 n ‖H‖r1,r2 , for all n ≥ n0, with n0 depending only on f , r1 and r2. We used here that by hypothesis we have ‖H‖r1,r2 > 0. For n ∈ � 1,2, . . . n0 − 1 it is easily seen that Bn,n,p,p − f r1,r2 ≥ Ar1,r2,n,p( f ) n with Ar1,r2,n,p � f � = n Bn,n,p,p − f r1,r2 which finally implies (5) where Kr1,r2, f =max § Ar1,r2,1,p � f � , . . . ,Ar1,r2,n0−1,p � f � , 1 2 ‖H‖r1,r2 ª . This completes the proof. 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