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Some Approximation Results by $(p,q)$-analogue of Bernstein-Stancu Operators (Revised) PDF

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Some Approximation Results by (p,q)-analogue of Bernstein-Stancu Operators (Revised) M. Mursaleen, Khursheed J. Ansari and Asif Khan Department of Mathematics, Aligarh Muslim University, Aligarh–202002, India [email protected]; [email protected]; [email protected] 6 1 0 Abstract 2 In this paper, We have given a corrigendum to our paper “Some Approximation Results by t c (p,q)-analogue of Bernstein-Stancu Operators” published in Applied Mathematics and Computation O 264 (2015) 392-402. We introduce a new analogue of Bernstein-Stancu operators and we call it as 1 (p,q)-Bernstein-Stancu operators. We study approximation properties based on Korovkins type 1 approximationtheoremof (p,q)-Bernstein-Stancuoperators. We also establishsome direct theorems. ] A C Keywords and phrases: (p,q)-integers; (p,q)-Bernstein operator; modulus of continuity; positive . linear operator; Korovkintype approximationtheorem. h t AMS Subject Classifications (2010): 41A10, 41A25, 41A36, 40A30 a m [ 1 Construction of Revised Operators 2 v Mursaleen et. al [3] introduced (p,q)-analogue of Bernstein-Stancu operators as 8 8 n n−k−1 2 S (f;x) = n xk (ps −qsx) f [k]p,q +α , x ∈ [0,1]. (1) 6 n,p,q k [n] +β 0 k=0(cid:20) (cid:21)p,q s=0 (cid:18) p,q (cid:19) . X Y 2 But S (1;x) 6= 1 for all x ∈ [0,1) and S (1;x) = 1 for x = 1 only. Hence, we are n,p,q n,p,q 0 re-introducing our operators as follows: 6 1 Xiv: Sn,p,q(f;x) = pn(1n2−1) k=n0(cid:20) nk (cid:21)p,qpk(k2−1)xkn−s=k−01(ps−qsx) f (cid:18)pn−[nk][pk,q]p+,q +β α(cid:19), x ∈ [0,1]. X Y r (2) a Note that for α = β = 0, (p,q)-Bernstein-Stancu operators given by (2) reduces into (p,q)-Bernstein operators as given in [1, 2]. Also for p = 1, (p,q)-Bernstein-Stancu operators given by (2) turn out to be q-Bernstein-Stancu operators. For any p > 0 and q > 0, the (p,q) integers [n] are defined by p,q pn−qn, when p 6= q 6= 1 p−q  [n] = pn−1+pn−2q+pn−3q2+...+pqn−2+qn−1 =  n pn−1, when p = q 6= 1 p,q      [n] , when p = 1 q n, when p = q = 1        1 where [n] denotes the q-integers and n = 0,1,2,···. q Obviously, it may be seen that [n]p,q = pn−1[n]q. p We have the following basic result: Lemma 1. For x ∈ [0,1], 0 < q < p ≤ 1, we have (i) S (1;x) = 1, n,p,q (ii) S (t;x) = [n]p,q x+ α , n,p,q [n]p,q+β [n]p,q+β (iii) S (t2;x) = q[n]p,q[n−1]p,qx2 + [n]p,q(2α+pn−1)x+ α2 . n,p,q ([n]p,q+β)2 ([n]p,q+β)2 ([n]p,q+β)2 Proof. (i) n n−k−1 Sn,p,q(1;x) = n(1n−1) nk pk(k2−1)xk (ps −qsx) = 1. p 2 k=0(cid:20) (cid:21)p,q s=0 X Y (ii) Sn,p,q(t;x) = pn(1n2−1) k=n0(cid:20) nk (cid:21)p,qpk(k2−1)xkn−s=k−01(ps −qsx) pn−[nk][pk,q]p+,q +β α X Y = 1 n n pk(k2−1)xkn−k−1(ps −qsx) pn−k[k]p,q pn(n2−1) k=0(cid:20) k (cid:21)p,q s=0 [n]p,q +β X Y n n−k−1 + α 1 n pk(k2−1)xk (ps −qsx) [n]p,q +β pn(n2−1) k=0(cid:20) k (cid:21)p,q s=0 X Y n−1 n−k−2 = [n]p,q n−1 p(k+12)(k)xk+1 (ps −qsx) 1 + α pn(n2−3) k=0(cid:20) k (cid:21)p,q s=0 pk+1([n]p,q +β) [n]p,q +β X Y n−1 n−k−2 = [n]p,q x 1 n−1 pk(k2−1)xk (ps −qsx)+ α [n]p,q +β p(n−1)2(n−2) k=0(cid:20) k (cid:21)p,q s=0 [n]p,q +β X Y [n] α p,q = x+ . [n] +β [n] +β p,q p,q 2 (iii) Sn,p,q(t2;x) = pn(1n2−1) k=n0(cid:20) nk (cid:21)p,qpk(k2−1)xkn−s=k−01(ps −qsx) (cid:18)pn−[nk][pk,q]p+,q +β α(cid:19)2 X Y = 1 1 p2n n n pk(k2−1)xkn−k−1(ps −qsx) [k]2p,q ([n]p,q +β)2 pn(n2−1) " k=0(cid:20) k (cid:21)p,q s=0 p2k X Y n n−k−1 +2α pn n pk(k2−1)xk (ps −qsx) [k]p,q k pk k=0(cid:20) (cid:21)p,q s=0 X Y n n−k−1 +α2 n pk(k2−1)xk (ps −qsx) . k k=0(cid:20) (cid:21)p,q s=0 # X Y On shifting the limits and using [k +1] = pk +q[k] , we get our desired result. p,q p,q Lemma 2.2. For x ∈ [0,1], 0 < q < p ≤ 1 (i) S (t−x);x = α−βx , n,p,q [n]p,q+β (ii) Sn,p,q(cid:0)(t−x)2;x(cid:1) = q[n]p,q[n([−n1]p],pq,+q−β[)n2]2p,q+β2 x2 + pn(−[n1[]np,]qp+,qβ−)22αβ x+ ([n]pα,q2+β)2. (cid:0) (cid:1) (cid:8) (cid:9) (cid:8) (cid:9) Proof. (ii) By linearity of the operator we have S (t−x)2;x = S (t2;x)−2xS (t;x)+x2S (1;x) n,p,q n,p,q n,p,q n,p,q q[n] [n−1] [n] (2α+pn−1) (cid:0) (cid:1) = p,q p,qx2 + p,q x ([n] +β)2 ([n] +β)2 p,q p,q α2 [n] α + −2x p,q −2x +x2 ([n] +β)2 [n] +β [n] +β p,q p,q p,q q[n] [n−1] −[n]2 +β2 pn−1[n] −2αβ α2 = p,q p,q p,q x2 + p,q x+ . ([n] +β)2 ([n] +β)2 ([n] +β)2 (cid:26) p,q (cid:27) (cid:26) p,q (cid:27) p,q 2 Main Results: 2.1 Korovkin type approximation theorem Remark 2.1 For q ∈ (0,1) and p ∈ (q,1], it is obvious that lim[n] = 0 or 1 . n→∞ p,q p−q In order to reach to convergence results of the operator Ln (f;x), we take a sequence p,q q ∈ (0,1) and p ∈ (q ,1] such that lim p = 1, lim q = 1 and lim pn = 1, n n n n→∞ n n→∞ n n→∞ n lim qn = 1. So we get lim[n] = ∞. n→∞ n n→∞ pn,qn Theorem 3.1.1. Let sequence q and p be such that it satisfies Remarks 2.1, then n n S (f,x) converges uniformly to f on [0,1]. n,p,q 3 Proof. By the Bohman-Korovkin Theorem it is sufficient to show that lim kS (tm;x)−xmk = 0, m = 0,1,2. n→∞ n,pn,qn C[0,1] By Lemma 2.1 (i)-(ii), it is clear that lim kS (1;x)−1k = 0; n→∞ n,pn,qn C[0,1] lim kS (t;x)−xk = 0. n→∞ n,pn,qn C[0,1] Using q [n−1] = [n] −pn−1 and by Lemma 2.1(iii), we have n pn,qn pn,qn n q[n] [n−1] [n] (2α+pn−1) α2 |S (t2;x)−x2| = p,q p,qx2+ p,q x+ −x2 . n,pn,qn C[0,1] ([n] +β)2 ([n] +β)2 ([n] +β)2 (cid:12) p,q p,q p,q (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Taking maximum of both sides of the above inequality, we get (cid:12) (cid:12) q[n] [n−1] [n] (2α+pn−1) α2 kS (t2;x)−x2k ≤ p,q p,q −1+ p,q + n,pn,qn C[0,1] ([n] +β)2 ([n] +β)2 ([n] +β)2 p,q p,q p,q which yields lim kS (t2;x)−x2k = 0. n→∞ n,pn,qn C[0,1] Thus the proof is completed and all other results follow similarly. References [1] M. Mursaleen, Khursheed J. Ansari, Asif Khan, On (p,q)-analogue of Bernstein operators, Applied Mathematics and Computation, 266 (2015) 874-882. [2] M. Mursaleen, Khursheed J. Ansari, Asif Khan, On (p,q)-analogue of Bernstein operators (revised), arxiv:1503.07404v2[math CA]. [3] M.Mursaleen, K.J.AnsariandAsifKhan,, SomeApproximationResultsby(p,q)- analogue of Bernstein-Stancu Operators, Applied Mathematics and Computation 264,(2015), 392-402. 4

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