General Mathematics Vol. 14, No. 4 (2006), 15–28
Simultaneous Approximation By Two
Dimensional Hybrid Positive Linear
Operators 1
Naokant Deo
In memoriam of Associate Professor Ph. D. Luciana Lupaş
Abstract
In this paper, we obtain some simultaneous approximation properties and asymptotic formulas of two dimensional hybrid (SzászMirakian and Lupaş Durrmeyer) positive linear operators and their
partial derivatives.
2000 Mathematics Subject Classification: 41A28, 41A36.
Key Words and Phrases: Szász-Mirakian operator, Lupaş operator,
Simultaneous approximation.
1
Received 1 October, 2006
Accepted for publication (in revised form) 9 November, 2006
15
16
Naokant Deo
1
Introduction
To approximate Lebesgue integrable functions on interval [0, ∞), Gupta and
Srivastava [6] proposed a sequence of linear positive operators, by combining
the well known Szász-Mirakian operator with the weight function of Lupaş
operator defined as:
(1)
(Sn f ) (x) = (n − 1)
∞
X
bn,k (x)
Z
∞
vn,k (t)f (t)dt,
x ∈ R = [0, ∞) ,
0
k=0
where
k
−nx
bn,k (x) =
e
(nx)
,
k!
and n ∈ N = {1, 2, ...}.

vn,k (t) = 
n+k−1


k
tk
(1 + t)n+k
Now we consider two dimensional hybrid positive linear operators as:
2
Mn[i,j]
(f(2)
; x, y) = (n − 1)
∞ X
∞
X
k=0 l=0
(i)
(j)
bn,k (x)bn,l (y)
Z
0
∞
Z
∞
vn,k (s)vn,l (t)f (s, t)dsdt,
0
where
(x, y) ∈ [0, ∞) × [0, ∞) and f ∈ C([0, ∞) × [0, ∞)).
[i,0]
In particular, if f (0) (x) = f (x) then the meanings of bn (f ; x, y) and
[0,j]
bn (f ; x, y) are clear.
By H[0, ∞)2 , we denote the class of all measurable functions defined on
[0, ∞) satisfying
Z
0
∞
Z
0
∞
|f (s)| |f (t)|
dsdt <∞, for some positive integer n.
{(1 + s)(1 + t)}n+1
Simultaneous Approximation By Two Dimensional Hybrid Positive...
17
This class is bigger than the class of all Lebesgue integrable functions on
[0, ∞).
Very recently author [1] has studied simultaneous approximation for two
dimensional Lupaş-Durrmeyer operators and in [2], he studied simultaneous
approximation for one variable. Gupta and other researchers also studied
simultaneous approximation for one as well as two variables for similar type
operators (see e.g. [3], [4], [5], [8]).
The main object of this paper is to obtain the properties of simultaneous
approximation by two dimensional Szász-Mirakian-Lupaş-Durrmeyer operators and obtained several asymptotic formulae for the partial derivative of
these operators (2).
2
Auxiliary Results
In this section, we shall mention certain results which are necessary to prove
our main theorem.
Lema 2.1.[7] For m ∈ N ∪ {0}, if we define
¶m
µ
∞
X
k
−x
,
bn,k (x)
Vn,m (x) =
n
k=0
then
£
¤
nVn,m+1 (x) = x V ′n,m (x) + mVn,m−1 (x) .
Consequently, we have
(i) Vn,m (x) is a polynomial in x of degree ≤ m.
¡
¢
(ii) Vn,m (x) = O n−[(m+1)/2] , where [γ] denotes the integral part of γ.
18
Naokant Deo
Lema 2.2.[6, 7] There exists the polynomials ϕc,h,r (x) independent of n and
k such that
xr
(3)
X
dr −nx
k
[e
(nx)
]
=
nc (k − nx)h ϕc,h,r (x)[e−nx (nx)k ].
dxr
2c+h≤r
c,h≥0
Lema 2.3. If f (x, y) is differentiable r1 + r2 times on [0, ∞) , then we get
∞
Mn[r1 ,r2 ] (f ; x, y) =
·
Z
0
∞
∞
Z
∞
nr1 +r2 (n − r1 − 1)!(n − r2 − 1)! X X
bn,k (x)bn,l (y)·
{(n − 2)!}2
k=0 l=0
vn−r1 ,k+r1 (s)vn−r2 ,l+r2 (t)
0
∂ r1 +r2
f (s, t) dsdt.
∂sr1 ∂tr2
Proof. From (2), we have
Mn[r1 ,r2 ] (f ; x, y) =
Z ∞Z ∞
∞
∞ X
X
(r1 )
(r2 )
2
bn,k (x)bn,l (y)
vn,k (s)vn,l (t)f (s, t)dsdt
= (n − 1)
0
k=0 l=0
0
Using Leibnitz theorem, we get
Mn[r1 ,r2 ]
2
(f ; x, y) = (n−1)
r2 X
r2 X
∞
∞ X
X
n
r1 +r2
r1 −i
(−1)
r2 −j
(−1)
i=0 j=0 k=i l=j
e−nx (nx)k−i
·
(k − i)!
2 r1 +r2
= (n − 1) n
e−ny (ny)l−j
.
(l − j)!
r2 X
r2 X
∞
∞ X
X
Z
∞
0
Z
∞
vn,k (s)vn,l (t)f (s, t) dsdt =
0
r1 −i
(−1)
r2 −j
(−1)
i=0 j=0 k=0 l=0
·
Z
·
∞
2 r1 +r2
vn,k+i (s)vn,l+j (t)f (s, t) dsdt = (n−1) n
0
0
Z
0
∞
µ ¶µ ¶
r1
r2
bn,k (x)bn,l (y)·
i
j
∞
Z
µ ¶µ ¶
r1
r2
·
i
j
∞
∞ X
X
bn,k (x)bn,l (y)·
k=0 l=0
Z
0
r2 X
r2
∞X
i=0 j=0
r1 −i
(−1)
r2 −j
(−1)
µ ¶µ ¶
r1
r2
vn,k+i (s)vn,l+j (t)f (s, t) dsdt.
i
j
Simultaneous Approximation By Two Dimensional Hybrid Positive...
19
Once again applying Leibnitz theorem, we obtain


q
X
q
(n − 1)!
(q)
 vn,u+w (z),
vn−q,u+q (z) =
(−1)w 
(n − q − 1)! w=0
w
where q = r1 , r2 ; w = i, j; u = k, l. So we have
∞
Mn[r1 ,r2 ]
∞
nr1 +r2 (n − r1 − 1)!(n − r2 − 1)! X X
(f ; x, y) =
bn,k (x)bn,l (y)·
{(n − 2)!}2
k=0 l=0
·
Z
0
∞
Z
∞
0
(r )
(r )
1
2
(−1)r1 +r2 vn−r
(s)vn−r
(t)f (s, t) dsdt.
1 ,k+r1
2 ,l+r2
Further integrating by parts r1 + r2 times, we get the required result.
Lema 2.4. Let the i − th order moment be defined by
Z ∞
∞
X
bn,k (x)
vn−r,k+r (s)(t − x)i ds
µn,r,i (x) = (n − r − 1)
k=0
0
then, we have the following recurrence relation:
(n − i − r − 2)µn,r,i+1 (x) = xµ′n,r,i (x) + [(i + 1)(1 + 2x) + r(1 + x)] µn,r,i (x)
(4)
+ ix(x + 2)µn,r,i−1 (x),
where n > i + r + 2. Consequently,
(i) we have
µn,r,0 (x) = 1, µn,r,1 (x) =
r(x + 1) + (1 + 2x)
(n − r − 2)
and
µn,r,2 (x) =
x2 (n + r2 + 5r + 6) + 2x(n + r2 + 4r + 3) + (r2 + 3r + 2)
,
(n − r − 2)(n − r − 3)
¡
¢
(ii) for all x ∈ [0, ∞), we get µn,r,i (x) = O n−[(i+1)/2] .
20
Naokant Deo
Proof. First we prove (4), by using
xb′n,k (x) = (k − nx)bn,k (x) and t(1 + t)v ′n,k (t) = (k − nt)vn,k (t).
We have from the definition of µn,r,i (x)
xµ′n,r,i (x)
= (n − r − 1)
∞
X
xb′n,k (x)
k=0
Z
∞
vn−r,k+r (t)(t − x)i dt − ixµn,r,i−1 (x).
0
Therefore, we have
£
¤
x µ′n,r,i (x) + iµn,r,i−1 (x) =
= (n − r − 1)
∞
X
(k − nx)bn,k (x)
= (n − r − 1)
bn,k (x)
k=0
∞
vn−r,k+r (t)(t − x)i dt
0
k=0
∞
X
Z
Z
∞
[(k + r) − (n − r)t] vn−r,k+r (t)(t − x)i dt+
0
+(n − r)µn,r,i+1 (x) + (n − r)xµn,r,i (x) − (nx + r)µn,r,i (x) =
= (n−r−1)
∞
X
bn,k (x)
k=0
Z
∞
0
t(1 + t)v ′n−r,k+r (t)(t − x)i dt + (n − r)µn,r,i+1 (x)−
−(1 + x)rµn,r,i (x) =
= (n − r − 1)
∞
X
k=0
bn,k (x)
Z
∞
0
£
¤
(1 + 2x)(t − x) + (t − x)2 + x(1 + x) ·
·b′n−r,k+r (t)(t − x)i dt + (n − r)µn,r,i+1 (x) − (1 + x)rµn,r,i (x) =
= −(i + 1)(1 + 2x)µn,r,i (x) − (i + 2)µn,r,i+1 (x) − x(1 + x)iµn,r,i−1 (x)+
+(n − r)µn,r,i+1 (x) − (1 + x)rµn,r,i (x).
This leads to the proof of (4).
Simultaneous Approximation By Two Dimensional Hybrid Positive...
21
Lema 2.5. Suppose that
Bn,r1 ,r2 ,i,j (x, y) = (n − r1 − 1)(n − r2 − 1)
∞
∞ X
X
bn,k (x)bn,l (y)·
k=0 l=0
·
Z
0
(5)
∞
Z
∞
vn−r1 ,k+r1 (s)vn−r2 ,l+r2 (t) (s − x)i (t − y)j dsdt =
0
= µn,r1 ,i (x).µn,r2 ,j (y),
then we obtain the following results by Lemma 2.4
³
´
j+1
i+1
Bn,r1 ,r2 ,0,0 (x, y) = 1, Bn,r1 ,r2 ,i,j (x, y) = O n−([ 2 ]+[ 2 ]) ,
Bn,r1 ,r2 ,1,0 (x, y) =
r1 (1 + x) + (1 + 2x)
,
(n − r1 − 2)
Bn,r1 ,r2 ,0,1 (x, y) =
r2 (1 + y) + (1 + 2y)
,
(n − r2 − 2)
Bn,r1 ,r2 ,1,1 (x, y) =
{r1 (1 + x) + (1 + 2x)}{r2 (1 + y) + (1 + 2y)}
,
(n − r1 − 2)(n − r2 − 2)
Bn,r1 ,r2 ,2,0 (x, y) =
x2 (n + r12 + 5r1 + 6) + 2x(n + r12 + 4r1 + 3) + (r12 + 3r1 + 2)
,
(n − r1 − 2)(n − r1 − 3)
Bn,r1 ,r2 ,0,2 (x, y) =
y 2 (n + r22 + 5r2 + 6) + 2y(n + r22 + 4r2 + 3) + (r22 + 3r2 + 2)
.
(n − r2 − 2)(n − r2 − 3)
3
Main Results
Now we consider slightly modified operators Mn∗ , for our convenience,
Mn∗[r1 ,r2 ] f =
1
M [r1 ,r2 ] f,
C1 (n, r1 , r2 ) n
where Mn preserve constants and
C1 (n, r1 , r2 ) =
nr1 +r2 (n − r1 − 2)!(n − r2 − 2)!
.
{(n − 2)!}2
22
Naokant Deo
Theorem 3.1. Suppose f is bounded on every finite subinterval of [0, ∞)
and
f ∈ H[0, ∞)¯2 . If f (r+2) exists at a fixed point x ∈ [0, ∞) and
¯
¯
¯
∂ r+2
¯ ≤ µxα y β , (x → ∞, y → ∞); j = 1, ..., r + 2 for
¯
f
(x,
y)
¯
¯ ∂xj ∂y r+2−j
some α, β ≥ 0, then we get
¸
·
∂r
∗[r,0]
lim Mn (f ; x, y) − r f (x, y) =
n→∞
∂x
(6)
= (1 + 2y)
∂ r+1
∂ r+1
f
(x,
y)
+
{r(1
+
x)
+
(1
+
2x)}
f (x, y)+
∂xr ∂y
∂xr+1
x(2 + x) ∂ r+2
∂ r+2
f (x, y).
+y(2 + y) r 2 f (x, y) +
∂x ∂y
2
∂xr+2
and
¸
·
∂r
∗[0,r]
lim Mn (f ; x, y) − r f (x, y) =
n→∞
∂y
(7)
= (1 + 2x)
∂ r+1
∂ r+1
f
(x,
y)
+
{r(1
+
y)
+
(1
+
2y)}
f (x, y)+
∂y r ∂x
∂y r+1
+x(2 + x)
Proof.
y(2 + y) ∂ r+2
∂ r+2
f
(x,
y)
+
f (x, y).
∂y r ∂x2
2
∂y r+2
Since the proof of (7) is identical therefore we shall give the
prove (6) only. By Taylorş expansion of f (s, t), we have
¶
µ
r+2 X
X
1
∂d
f (s, t) =
f (x, y) (s − x)i (t − y)j +
i ∂y j
i!j!
∂x
d=0 i+j=d
X
+
ε(s, t, x, y)(s − x)i (t − y)j .
i+j=r+2
where ε(s, t, x, y) → 0 as s → x, t → y and ε(s, t, x, y) ≤ µ(s − x)α (t − y)β
as s → ∞ , x → ∞ for some α, β > 0 then
¸
·
∂r
∗[r,0]
n Mn (f ; x, y) − r f (x, y) =
∂x
Simultaneous Approximation By Two Dimensional Hybrid Positive...
23
µ
¶
r+2 X
X
¡
¢
∂d
1
∗[r,0]
i
j
f
(x,
y)
M
(s
−
x)
(t
−
y)
;
x,
y
+
=n
n
i!j! ∂xi ∂y j
d=0 i+j=d
+n
¡
¢
∂r
Mn∗[r,0] ε(s, t, x, y)(s − x)i (t − y)j ; x, y − n r f (x, y) =
∂x
i+j=r+2
X
= Q1 + Q2 − n
∂r
f (x, y).
∂xr
From Lemma 2.3, we get
Q1 = n
∞ X
r+2 X
∞
X
X
∂d
1
f
(x,
y)(n
−
1)(n
−
r
−
1)
bn,k (x)bn,l (y)·
i ∂y j
i!j!
∂x
k=0 l=0
d=0 i+j=d
·
Z
0
∞
Z
∞
vn−r,k+r (s)vn,l (t)
0
¢
∂r ¡
(s − x)i (t − y)j dsdt =
r
∂s
∞
∞ X
X
n ∂r
bn,k (x)bn,l (y)·
f (x, y)(n − 1)(n − r − 1)
=
r! ∂xr
k=0 l=0
Z ∞Z ∞
·
vn−r,k+r (s)vn,l (t)r!dsdt+
0
0
∞ X
∞
X
n ∂ r+1
+
f (x, y)(n − 1)(n − r − 1)
bn,k (x)bn,l (y)·
r!1! ∂xr ∂y
k=0 l=0
Z ∞Z ∞
vn−r,k+r (s)vn,l (t)r!(t − y)dsdt+
·
0
0
∞
∞ X
X
∂ r+1
n
bn,k (x)bn,l (y)·
f (x, y)(n − 1)(n − r − 1)
+
(r + 1)! ∂xr+1
k=0 l=0
Z ∞Z ∞
∂r
·
vn−r,k+r (s)vn,l (t) r (s − x)r+1 dsdt+
∂s
0
0
∞ X
∞
X
n ∂ r+2
+
f (x, y)(n − 1)(n − r − 1)
bn,k (x)bn,l (y)·
r!2! ∂xr ∂y 2
k=0 l=0
Z ∞Z ∞
·
vn−r,k+r (s)vn,l (t)r!(t − y)2 dsdt+
0
0
24
Naokant Deo
∞
∞ X
X
∂ r+2
n
bn,k (x)bn,l (y)·
f (x, y)(n − 1)(n − r − 1)
+
(r + 1)! ∂xr+1 ∂y
k=0 l=0
Z ∞Z ∞
r ©
ª
∂
·
vn−r,k+r (s)vn,l (t) r (s − x)r+1 (t − y) dsdt+
∂s
0
0
+
∞
∞ X
X
∂ r+2
n
bn,k (x)bn,l (y)·
f
(x,
y)(n
−
1)(n
−
r
−
1)
(r + 2)! ∂xr+2
k=0 l=0
Z ∞Z ∞
r
∂
·
vn−r,k+r (s)vn,l (t) r (s − x)r+2 dsdt =
∂s
0
0
=n
+n
+n
=n
∂r
∂ r+1
f
(x,
y)B
(x,
y)
+
n
f (x, y)Bn,r,0,0,1 (x, y)+
n,r,0,0,0
∂xr
∂xr ∂y
∂ r+1
n ∂ r+2
f
(x,
y)B
(x,
y)
+
f (x, y)Bn,r,0,0,2 (x, y)+
n,r,0,1,0
∂xr+1
2 ∂xr ∂y 2
∂ r+2
n ∂ r+2
f
(x,
y)B
(x,
y)
+
f (x, y)Bn,r,0,2,0 (x, y) =
n,r,0,1,1
∂xr+1 ∂y
2 ∂xr+2
∂r
n(1 + 2y) ∂ r+1
n{r(1 + x) + (1 + 2x)} ∂ r+1
f
(x,
y)+
f
(x,
y)+
f (x, y)+
∂xr
(n − 2) ∂xr ∂y
(n − r − 2)
∂xr+1
+
+
+
n{y 2 (n + 6) + 2y(n + 3) + 2} ∂ r+2
f (x, y)+
(n − 2)(n − 3)
∂xr ∂y 2
n(1 + 2y){r(1 + x) + (1 + 2x)} ∂ r+2
f (x, y)+
(n − 2)(n − r − 2)
∂xr+1 ∂y
n{x2 (n + r2 + 5r + 6) + 2x(n + r2 + 4r + 3) + (r2 + 3r + 2)} ∂ r+2
f (x, y),
2(n − r − 2)(n − r − 3)
∂xr+2
by Lemma 2.4, we obtain the above results. In order to prove the theorem,
it is sufficient to show that
En ∼
= xr Q2 = n(n − 1)2
∞
∞ X
X X
(r)
xr bn,k (x)bn,l (y)·
i+j=r+2 k=0 l=0
·
Z
∞
0
Z
0
∞
vn,k (s)vn,l (t)ε(s, t, x, y)(s − x)i (t − y)j dsdt → 0 as (n → ∞).
Simultaneous Approximation By Two Dimensional Hybrid Positive...
25
Using Lemma 2.2, we get
∞
∞ X
X
X X
2
|En | ≤ n(n − 1)
nc |k − nx|h |ϕc,h,r (x)|bn,k (x)bn,l (y)·
i+j=r+2 k=0 l=0 2c+h≤r
·
Z
∞
Z
∞
vn,k (s)vn,l (t)|ε(s, t, x, y)||s − x|i |t − y|j dsdt ≤
0
0
X
≤ nχ(x)
X
nc
i+j=r+2 2c+h≤r
1
· (bn,k (x)bn,l (y)) 2 (n−1)2
Z
0
X
≤ nχ(x)
k=0 l=0
∞Z
nc
i+j=r+2 2c+h≤r
½
Z
2
· (n − 1)
∞
0
Z
∞
0
∞
vn,k (s)vn,l (t)|ε(s, t, x, y)||s − x|i |t − y|j dsdt ≤
0
X
·
∞ ³
∞ X
´
X
1
(bn,k (x)bn,l (y)) 2 |k − nx|h ·
"
"
∞
∞ X
X
bn,k (x)bn,l (y) (k − nx)
k=0 l=0
∞
∞ X
X
2h
# 21
·
bn,k (x)bn,l (y)·
k=0 l=0
¾2 # 21
vn,k (s)vn,l (t)|ε(s, t, x, y)||s − x|i |t − y|j dsdt
,
where
χ(x) =
X
sup |ϕc,h,r (x)| .
2c+h≤r
c,h≥0
From Lemma 2.1, we have
∞ X
∞
X
2h
bn,k (x)bn,l (y) (k − nx)
k=0 l=0
=n
2h
∞
X
¶2h
µ
k
=
−x
bn,k (x)
n
k=0
³
´
¡
¢
−[ 2h+1
]
2
=n O n
= n2h O n−h = nh O(1)
2h
and let
ρn =
∞ X
∞
X
k=0 l=0
bn,k (x)bn,l (y)·
26
Naokant Deo
∞
·
Z
2
· (n − 1)
0
∞
Z
¸2
vn,k (s)vn,l (t)|ε(s, t, x, y)(s − x) (t − y) |dsdt .
i
0
Therefore, we obtain
X
|En | ≤ nχ(x)
X
i+j=r+2 2c+h≤r
Now
·
Z
2
(n − 1)
∞
0
∞
Z
¸2
vn,k (s)vn,l (t)|ε(s, t, x, y)(s − x) (t − y) |dsdt ≤
0
≤ (n − 1)
Z
∞
0
·(n − 1)
Z
∞
0
Z
∞
Z
j
∞
vn,k (s)vn,l (t)dsdt·
0
vn,k (s)vn,l (t)ε2 (s, t, x, y)(s − x)2i (t − y)2j dsdt =
0
∞
Z
= (n − 1)2
¡
¢1
1
nc nh O(1) 2 (ρn ) 2 .
i
2
2
j
0
∞
Z
vn,k (s)vn,l (t)ε2 (s, t, x, y)(s − x)2i (t − y)2j dsdt =
0
2
= (n − 1)
·Z
(s−x)2 +(t−y)2 ≤δ 2
Z
+
(s−x)2 +(t−y)2 >δ 2
¸
·
·vn,k (s)vn,l (t)ε2 (s, t, x, y)(s − x)2i (t − y)2j dsdt.
For a given η > 0, there exists a δ > 0 such that |ε(s, t, x, y)| < η whenever (s − x)2 + (t − y)2 ≤ δ 2 . For (s − x)2 + (t − y)2 > δ 2 , we obtain
|ε(s, t, x, y)| < K(s − x)α (t − y)β .
ρn = η
2
∞
∞ X
X
bn,k (x)bn,l (y)(n − 1)2 ·
k=0 l=0
·
Z
0
+K
∞ X
∞
X
k=0 l=0
∞Z
∞
vn,k (s)vn,l (t)(s − x)2i (t − y)2j dsdt+
0
2
bn,k (x)bn,l (y)(n − 1)
(s − x)2 + (t − y)2
·
δ2
(s−x)2 +(t−y)2 >δ 2
Z
·(s − x)2(i+α) (t − y)2(j+β) vn,k (s)vn,l (t)dsdt =
Simultaneous Approximation By Two Dimensional Hybrid Positive...
27
´ 1 ³
´
³
2i+2α+2+1
2i+1
2i+1
]+[ 2j+2β+1
]) +
2
2
= η 2 O n−([ 2 ]+[ 2 ]) + 2 O n−([
δ
´
³
2i+2α+1
1
+[ 2j+2β+2+1
−([
]
])
2
2
+ 2O n
=
δ
´
¡
¢ 1 ³
2β+1
2α+1
= η 2 O n−(i+j) + 2 O n−(i+j) n−([ 2 ]+1+[ 2 ]) +
δ
´
1 ³ −(i+j) −([ 2α+1
+1+[ 2β+1
2 ]
2 ])
n
=
+ 2O n
δ
µ
¶
·
¸
·
¸
¡ −(i+j) ¢ 2
2 −ζ
2α + 1
2β + 1
=O n
η + 2n
, where ζ =
+1+
> 0.
δ
2
2
Thus, we get
¶¸ 1
µ
·
X X
¡ −(i+j) ¢ 2
£ h
¤ 21
2 −ζ 2
c
≤
η + 2n
O n
|En | ≤ nχ(x)
n n O(1)
δ
i+j=r+2 2c+h≤r
≤ nχ(x)
X
2c+h≤r
c
¡
n n
h
¢ 21
¶¸ 1
µ
X · ¡
¢ 2
2 −ζ 2
−(i+j)
=
η + 2n
O(1)
O n
δ
i+j=r+2
¶1
µ
2 −ζ 2
2
= O(1)n n
=
η + 2n
δ
µ
¶ 21
2
→ 0, as n → ∞.
= O(1) η 2 + 2 n−ζ
δ
This completes the proof of (6).
r+2
2
− r+2
2
References
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180-188.
[2] Deo N., Simultaneous approximation by Lupaş modified operators with
weighted functions of Szász operators, J. Inequal. Pure and Appl. Math.,
5(4), Art.113, (2004), 1-5.
28
Naokant Deo
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School of Information Science and Engineering,
Graduate University of Chinese Academy of Sciences,
Zhongguancun Nan Yi Tiao NO.3, Haidian District,
Beijing 100080, P. R. CHINA.
E-mail address:dr naokant deo@yahoo.com