2018
DOI: 10.1186/s13660-018-1939-9
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Approximation by modified Kantorovich–Stancu operators

Abstract: In the present paper, we study a new kind of Kantorovich–Stancu type operators. For this modified form, we discuss a uniform convergence estimate. Some Voronovskaja-type theorems are given.

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Cited by 4 publications
(4 citation statements)
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“…, replaces (1 − x) in the original formula. In the papers dealing with this modification (see [2], [4], [18]) the superscript "M, 1" refers to this first disorder in the recursion.…”
Section: On the Recursion For The Fundamental Functions Of Bernstein ...mentioning
confidence: 99%
“…, replaces (1 − x) in the original formula. In the papers dealing with this modification (see [2], [4], [18]) the superscript "M, 1" refers to this first disorder in the recursion.…”
Section: On the Recursion For The Fundamental Functions Of Bernstein ...mentioning
confidence: 99%
“…Since they are functional in studying many problems and convenient in computer-aided studies, they have important generalizations and applications. We refer to [1,2,23,28,29,32].…”
Section: Introductionmentioning
confidence: 99%
“…The fact that they are functional in studying many problems, convenient in computer-aided studies and also they have a simple representation, important generalizations and applications has motivated a great number of authors to study intensively up to now. We refer the readers [1][2][3]11,22,27,28] for some studies about these operators.…”
Section: Introductionmentioning
confidence: 99%
“…x) = 15(β 2 + 6β + 7)x4 (1 − x)3 (1 + x) −[120x4 (1 − x)3 (1 + x) + 15(6α + 5)x 4 (1 − x) 2 (1 + x)](β + 3) +45x 5 (1 − x) 2 (1 + x)(β + 3) 2 + 130x 3 (1 − x) 4 (1 + x) +10(12α + 5)x 3 (1 − x) 3 (1 + x) + 5(9α 2 + 15α + 8)x 3 (1 − x) 2 (1 + x) +45x 4 (1 − x) 2 + 15(β + 5)x 3 (1 − x) 3 4 and A 2 (α, β, x) = 15(β 2 + 6β + 7)x 4 (1 − x) 3 (1 + x) −[120x 4 (1 − x) 3 (1 + x) + 15(6α + 5)x 4 (1 − x) 2 (1 + x)](β + 3) +45x 5 (1 − x) 2 (1 + x)(β + 3) 2 + 130x 3 (1 − x) 4 (1 + x) +10(12α + 5)x 3 (1 − x) 3 (1 + x) + 5(9α 2 + 15α + 8)x 3 (1 − x) 2 (1 + x) +45x 3 (1 − x) 2 − 15(β + 1)x 3 (1 − x) 3Because ρ is a continuous function, there exists an M > 0 such that |ρ(t; x)| < M , ∀ x, t ∈ [0, 1]. Using the above results for K(α,β) n,i ((t − x) 6 ; x), we obtain | K (α,β) n,i (ρ(t; x)(t − x) 6 ; x)| ≤ M 15x 4 (1 − x) 3 (1 + x)n 4 (n + β + 1) 6 + , i = 1, 2.So, (3.4) is proved.…”
mentioning
confidence: 99%