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

0.07 MB·

by  M. Mursaleen

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]
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0 Abstract
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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

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(p,q)-Bernstein-Stancu operators. We study approximation properties based on Korovkins type
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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.
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t AMS Subject Classifications (2010): 41A10, 41A25, 41A36, 40A30
a
m
[ 1 Construction of Revised Operators

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v
Mursaleen et. al [3] introduced (p,q)-analogue of Bernstein-Stancu operators as
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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:
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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.
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