EXACT ORDERS IN SIMULTANEOUS APPROXIMATION BY COMPLEX BERNSTEIN-STANCU POLYNOMIALS

. In this paper the exact orders in approximation by the complex Bernstein-Stancu polynomials (depending on two parameters) and their derivatives on compact disks are obtained.


INTRODUCTION
In the recent paper [1] the following upper estimates and Voronovskaja's theorem in approximation by complex Bernstein-Stancu polynomials depending on two parameters were proved.
(ii) If 1 ≤ r < r 1 < R, then for all n, p ∈ N, we have where e 1 (z) = z and 0 < M (α,β) 1 (f ) < ∞ depends only on α, β and f .Remark 1.2.Following exactly the lines in the proof of Theorem 1.1, (iii) in [1], it is immediate that in fact for any 1 ≤ r < R we have an upper estimate of the form where the constant M (α,β) r (f ) > 0 is independent of n and depends on f , r, α and β.This estimate will be useful in Section 3.
The goal of this paper is to show that in Theorem 1.1, (i) and (ii), also lower estimates hold.Thus, in Section 2 we prove that if the analytic function f is not a polynomial of degree ≤ 0 and 1 ≤ r < R, then we have In Section 3 we prove that for any p ∈ N and 1 r ∼ 1 n , where the constants in the equivalence depend only on f , α, β, r and p.
Since the case α = β = 0 (i.e. the case of classical Bernstein polynomials) was already considered in [2], in the rest of the paper we will exclude it.

EXACT ORDER OF APPROXIMATION FOR COMPLEX BERNSTEIN-STANCU POLYNOMIALS
The main result of this section is the following.
Proof.For all z ∈ D R and n ∈ N we have Note that in the case α = β = 0 in [2], necessarily f was supposed to be not a polynomial of degree ≤ 1.
In what follows we will apply to the above identity the following obvious property: It follows Since by Remark 1.2 we have and denoting 2 f (z), if we prove that H r > 0, then it is clear that there exists an index n 0 depending only on f , α and β, such that Therefore it remains to show that H r > 0. Indeed, suppose that H r = 0. We have two possibilities: 1) 0 = α < β or 2) 0 < α ≤ β .
Case 1).We obtain Combining now Theorem 2.1 with Theorem 1.1, (i), we immediately get the following.
where the constants in the equivalence depend on f , r, α and β.

EXACT ORDERS OF APPROXIMATION FOR DERIVATIVES OF COMPLEX BERNSTEIN-STANCU POLYNOMIALS
The main result of this section is the following.
where the constants in the equivalence depend on f , α, β, r, r 1 and p.
Proof.Taking into account Theorem 1.1, (ii), it remains only to prove the lower estimate for [S r .Denoting by Γ the circle of radius r 1 > r (with r ≥ 1) and center 0, by the Cauchy's formulas it follows that for all |z| ≤ r and n ∈ N we have dv, where we have the inequality |v − z| ≥ r 1 − r valid for all |z| ≤ r and v ∈ Γ.
As in the proof of Theorem 2.1 (keeping the notation for H), for all v ∈ Γ and n ∈ N we have which replaced in the above Cauchy's formula implies Passing now to absolute value, for all |z| ≤ r and n ∈ N it follows where by using the Remark 1.2, for all |z| ≤ r and n ∈ N we get .
Denoting now F p (z) = H (p) (z), we prove that F p r > 0. Indeed, if we suppose that F p r = 0 then it follows that f satisfies the differential equation

Theorem 1 . 1 .
Let D R = {z ∈ C; |z| < R} be with R > 1 and let us suppose that f : D R → C is analytic in D R , i.e. f (z) = ∞ k=0 c k z k , for all z ∈ D R .Also, for 0 ≤ α ≤ β (independent of n) let us define the complex Bernstein-Stancu polynomials by Here f r = sup{|f (z)|; |z| ≤ r}.† This work has been supported by the Romanian Ministry of Education and Research, under CEEX grant: 2-CEx 06-11-96.

2 f 2 y 2 y
(z) = 0, for all |z| ≤ r and denoting y(z) = f (z), it follows that y(z) is an analytic function in D R , solution of the differential equation −βzy(z) + z(1−z) (z) = 0, |z| ≤ r, which after simplification with z = 0 becomes −βy(z) + (1−z) 2 y (z) = 0, |z| ≤ r.Now, seeking y(z) in the form y(z) = ∞ k=0 b k z k and replacing it in the differential equation, by the identification of the coefficients we easily obtain b k = 0 for all k = 0, 1, . ... Therefore y(z) = 0 for all |z| ≤ r, which by the identity theorem on analytic (holomorphic) functions implies y(z) = 0 for all z ∈ D R and the contradiction that f is a polynomial of degree ≤ 0. Case 2).Denoting y(z) = f (z) by hypothesis it follows that y(z) is an analytic function in D R solution of the differential equation (−βz + α)y(z) + z(1−z) (z) = 0, |z| ≤ r.Taking z = 0 it follows αy(0) = 0, which means y(0) = 0. Seeking y(z) in the form y(z) = ∞ k=1 b k z k and replacing it in the differential equation, by the identification of the coefficients we easily obtain b k = 0 for all k = 1, 2, . .., which finally leads to the contradiction that f is a constant.

1 n • C 0 2 .
r, where Q p−1 (z) is a polynomial of degree ≤ p − 1. Simplifying with z, making the substitution y(z) = f (z), searching y(z) in the form y(z) = ∞ k=0 b k z k and then replacing in the differential equation, by simple calculations we easily obtain that bk = 0 for all k ≥ p − 1, that is y(z) is a polynomial of degree ≤ p−2.This implies the contradiction that f is a polynomial of degree ≤ p−1.Continuing exactly as in the proof of Theorem 2.1 (withS (α,β) n (f ) − f r replaced by [S (α,β) n (f )] (p) − f (p)r ), finally there exists an index n 0 ∈ N depending on f , r, r 1 and p, such that for all n ≥ n 0 we have[S (α,β) n (f )] (p) − f (p)r ≥ Also, the cases when n ∈ {1, 2, . . ., n 0 − 1} are similar with those in the proof of Theorem 2.1.