PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 265–279 DOI: 10.2298/PIM1613265G

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PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE
Nouvelle série, tome 99(113) (2016), 265–279
DOI: 10.2298/PIM1613265G
DIRECT ESTIMATIONS OF NEW
GENERALIZED BASKAKOV–SZÁSZ OPERATORS
Vijay Gupta and Neha Malik
Abstract. Several modifications of the discrete operators are available in
the literature. In the recent years, certain modifications of the well-known
Baskakov and Szász–Mirakyan operators have been discussed based on certain parameters. We propose mixed summation-integral type operators and
estimate the quantitative asymptotic formula and a global direct result for
the special case. For general case, we establish moments and some direct
convergence results in ordinary approximation, which includes pointwise approximation, asymptotic formula and a direct result in terms of modulus of
continuity.
1. Introduction
In order to generalize the Baskakov operators, Mihesan [15] proposed the following operators based on a non-negative constant a, independent of n as
Bna (f ; x) =
(1.1)
∞
X
ban,k (x)f
k=0
k
n
,
where
ban,k (x)
=e
ax
− 1+x
Pk
i=0
k
i
(n)i ak−i
xk
,
k!
(1 + x)n+k
and the rising factorial is given by (n)i = n(n + 1) · · · (n + i − 1), (n)0 = 1. Also,
in order to generalize the Szász–Mirakyan operators, Jain [13] introduced the following operators
Snβ (f, x) =
∞
X
(β)
Ln,k (x)f
k=0
k
,
n
x ∈ [0, ∞)
2010 Mathematics Subject Classification: 41A25; 41A30.
Key words and phrases: Baskakov operators, Szász-Mirakyan operators, direct results, modulus of continuity.
Communicated by Gradimir Milovanović.
265
266
GUPTA AND MALIK
where 0 6 β < 1 and the basis function is defined as
nx(nx + kβ)k−1 −(nx+kβ)
e
.
k!
P∞
P∞
(β)
It was seen in [15] and [13] that k=0 ban,k (x) = k=0 Ln,k (x) = 1. The integral
modification of the classical Bernstein polynomial was introduced by Durrmeyer [6].
Gupta et al. [10] proposed the hybrid Durrmeyer type operators by taking the gen(β)
eral basis function Ln,k (x) under summation sign, the actual Durrmeyer operators
for this basis function were recently considered by Gupta and Greubel [16]. In
order to approximate Lebesgue integrable functions on the interval [0, ∞), for a
non-negative parameter a and 0 6 β < 1, we propose the mixed hybrid Durrmeyer
type operators as follows:
(β)
Ln,k (x) =
Dna,β (f, x) =
(1.2)
(β)
∞
X
hLn,k (t), f (t)i
k=0
R∞
ban,k (x)
(β)
hLn,k (t), 1i
where hf, gi = 0 f (t) g(t) dt. Some approximation properties for the particular
case β = 0 were recently discussed by Agrawal et al. [1]. For the special case of
a = β = 0, these operators reduce to the Baskakov–Szász operators introduced
about twenty years ago by Gupta and Srivastava [12], which are defined as
Dn (f, x) := Dn0,0 (f, x) =
(0)
∞
X
hLn,k (t), f (t)i
k=0
∞
X
=n
(0)
hLn,k (t), 1i
b0n,k (x)
k=0
where
b0n,k (x)
n+k−1
xk
=
,
k
(1 + x)n+k
Z
∞
0
(0)
b0n,k (x)
(0)
Ln,k (t)f (t)dt,
Ln,k (t) = e−nt
(nt)k
.
k!
(β)
We may point out here that hLn,k (t), 1i is dependent of k, while for β = 0 this
term is just 1/n. Dragomir [3] established interesting approximation results of
continuous linear functionals in real normed spaces. Very recently, Gupta and
Agarwal [9] presented convergence estimates on several linear positive operators.
We also mention for the readers some of the related work as [2, 7, 8, 11, 17, 18] etc.
Here, we first estimate the quantitative asymptotic formula for the special
case and a global direct result. For general case, we establish moments and some
direct results in ordinary approximation, which includes pointwise approximation,
asymptotic formula, error-estimation in terms of first and second order modulus of
continuity.
2. Approximation for the case a = β = 0
Very recently Luo–Milovanovic–Agarwal [14] established some results on the
extended beta and extended hypergeometric functions. We observe that in the
special case, if we take a = β = 0, then the moments can be obtained in terms of
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 267
hypergeometric series as follows: Using the identities: k! = (1)k and Γ(k + m + 1) =
Γ(m + 1)(m + 1)k , we have
Z ∞
∞
X
Dn (tm , x) = n
b0n,k (x)
L0n,k (x)tm dt
=n
=
k=0
∞
X
0
(n)k xk
k!(1 + x)n+k
k=0
∞
X
1
nm
k=0
Z
0
∞
e−nt (nt)k m
t dt
k!
k
(n)k x
Γ(k + m + 1)
k!(1 + x)n+k
(1)k
∞
(1 + x)−n X (n)k (m + 1)k Γ(m + 1)xk
=
nm
k!(1 + x)k (1)k
k=0
m!(1 + x)−n
x =
F
n,
m
+
1;
1;
.
2
1
nm
1+x
Applying the well known transformation
2 F1 (a, b; c; x)
we obtain at once
= (1 − x)−a 2 F1 a, c − b; c;
Dn (tm , x) =
x ,
x−1
m!
2 F1 (n, −m; 1; −x).
nm
Furthermore, we obtain
Dn (t, x) = x +
1
,
n
Dn (t2 , x) =
2 + 4nx + n(n + 1)x2
n2
and
2 + 2nx + nx2
1
, Dn ((t − x)2 , x) =
.
n
n2
In general, using the similar approach, one can show that:
(2.2)
Dn ((t − x)r , x) = O n−[(r+1)/2] ,
(2.1)
Dn (t − x, x) =
where [α] denotes the integral part of α.
Let Bx2 [0, ∞) be the set of all functions f defined on R+ satisfying the condition
|f (x)| 6 Mf (1+x2 ) with some constant Mf , depending only on f , but independent
of x. Bx2 [0, ∞) is called weighted space and it is a Banach space endowed with the
norm
f (x)
kf kx2 = sup
.
+
1
+ x2
x∈R
Let Cx2 [0, ∞) = C[0, ∞) ∩ Bx2 [0, ∞) and by Cxk2 [0, ∞), we denote subspace of
f (x)
all continuous functions f ∈ Bx2 [0, ∞) for which limx→∞ 1+x
2 is finite.
We know that usual first modulus of continuity ω(δ) does not tend to zero, as
δ → 0 on infinite interval. Thus, we use weighted modulus of continuity Ω(f, δ)
268
GUPTA AND MALIK
defined on infinite interval R+ (see [17]). Let
Ω(f, δ) =
sup
|h|<δ, x∈R+
|f (x + h) − f (x)|
(1 + h2 )(1 + x2 )
for each f ∈ Cx2 [0, ∞).
Now, some elementary properties of Ω(f, δ) are collected in the following lemma.
Lemma 2.1. Let f ∈ Cxk2 [0, ∞). Then,
i) Ω(f, δ) is a monotonically increasing function of δ, δ > 0.
ii) For every f ∈ Cxk2 [0, ∞), limδ→0 Ω(f, δ) = 0.
iii) For each λ > 0,
(2.3)
Ω(f, λδ) 6 2(1 + λ)(1 + δ 2 )Ω(f, δ).
From the inequality (2.3) and definition of Ω(f, δ), we get
|t − x| (1 + δ 2 )Ω(f, δ)
(2.4)
|f (t) − f (x)| 6 2(1 + x2 ) 1 + (t − x)2 1 +
δ
for every f ∈ Cx2 [0, ∞) and x, t ∈ R+ . The following estimate is quantitative
Voronovskaya type asymptotic formula:
Theorem 2.1. Let f ′′ ∈ Cxk2 [0, ∞), a = β = 0 and x > 0. Then we have
1
x(x + 2) ′′ f ′′ (x)
1 f (x) 6
+8(1+x2 )O(n−1 )Ω f ′′ , √ .
Dn (f, x)−f (x)− f ′ (x)−
2
n
2n
n
n
Proof. By Taylor’s formula, there exist η lying between x and y such that
f (y) = f (x) + f ′ (x)(y − x) +
where
f ′′ (x)
(y − x)2 + h(y, x)(y − x)2 ,
2
f ′′ (η) − f ′′ (x)
2
and h is a continuous function which vanishes at 0. Applying the operator Dn to
above equality, we get
f ′ (x) f ′′ (x) h 2 + 2nx + nx2 i
Dn (f, x) − f (x) =
+
+ Dn h(y, x)(y − x)2 , x .
2
n
2
n
Also, we can write that
f ′ (x) f ′′ (x) 2x + x2 f ′′ (x)
−
+ Dn |h(y, x)|(y − x)2 , x
Dn (f, x) − f (x) −
6
2
n
2
n
n
To estimate last inequality using (2.4) and the inequality |η − x| 6 |y − x|, we can
write
|y − x| |h(y, x)| 6 (1 + (y − x)2 )(1 + x2 ) 1 +
(1 + δ 2 )Ω(f ′′ , δ).
δ
Also,
(
2(1 + x2 )(1 + δ 2 )2 Ω(f ′′ , δ),
|y − x| < δ
|h(y, x)| 6
|y−x| 2
2
2
′′
(1 + (y − x) )(1 + x ) 1 + δ (1 + δ )Ω(f , δ), |y − x| > δ
h(y, x) :=
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 269
Now choosing δ < 1, we have
(y − x)4 (1 + δ 2 )2 Ω(f ′′ , δ)
|h(y, x)| 6 2(1 + x2 ) 1 +
δ4
(y − x)4 6 8(1 + x2 ) 1 +
Ω(f ′′ , δ).
δ4
Using (2.2), we deduce that
Dn (|h(y, x)|(y − x)2 , x) = 8(1 + x2 )Ω(f ′′ , δ) Dn ((t − x)2 , x) + δ −4 Dn ((t − x)6 , x)
= 8(1 + x2 )Ω(f ′′ , δ) O(n−1 ) + δ −4 O(n−3 ) .
√
Choosing δ = 1/ n we have
√ Dn (|h(y, x)|(y − x)2 , x) 6 8(1 + x2 )O(n−1 )Ω f ′′ , 1/ n .
By CB [0, ∞), we mean the class of all real valued continuous and bounded
functions f on [0, ∞). The second order Ditzian–Totik modulus of smoothness is
defined by
ωϕ2 (f, δ) = sup
sup
06h6δ x±hϕ(x)∈[0,∞)
ϕ(x) =
|f (x + hϕ(x)) − 2f (x) + f (x − hϕ(x))|,
p
(x + 1)(x + 2), x > 0. The corresponding K-functional is
K2,ϕ (f, δ 2 ) =
inf
{kf − hk + δ 2 kϕ2 h′′ k},
2 (ϕ)
h∈W∞
2
where W∞
(ϕ) = {h ∈ CB [0, ∞) : h′ ∈ ACloc [0, ∞) : ϕ2 h′′ ∈ CB [0, ∞)}. By [5,
Thm. 2.1.1], it follows that
C −1 ωϕ2 (f, δ) 6 K2,ϕ (f, δ 2 ) 6 Cωϕ2 (f, δ)
for some absolute constant C > 0. Also, the Ditzian–Totik modulus of the first
order is given by
−
→
ω ϕ (f, δ) = sup
sup
|f (x + hϕ(x)) − f (x)|,
06h6δ x±hϕ(x)∈[0,∞)
where ϕ is an admissible step-weight function on [0, ∞).
Theorem 2.2. If f ∈ CB [0, ∞) and n ∈ N, then we have the inequality
√ kDn (f, x) − f (x)k 6 4ωϕ2 f, 1/ n + ω~ϕ f, 1/n .
p
Proof. We set ϕ(x) = (x + 1)(x + 2),
Wϕ2 [0, ∞) = {g ∈ ACloc [0, ∞) : ϕ2 g ′′ ∈ CB [0, ∞)}
D̃n (f, x) = Dn (f, x) − f (x + 1/n) + f (x).
|t−x|
˜
Then ϕ|t−u|
2 (u) 6 ϕ2 (x) for u between x and t, Dn (t − x, x)
2
g ∈ Wϕ [0, ∞), by Taylor’s formula, we have
g(t) = g(x) + g ′ (x)(t − x) +
Z
= 0 by using (2.1), and for
t
x
g ′′ (u)(t − u) du.
Applying the operator Dn to above equality and then taking modulus, we get
270
GUPTA AND MALIK
|D̃n (g, x) − g(x)|
Z
Z t
′′
6 Dn (t − u) g (u) du, x + 1
x+ n
1
′′
x + − u g (u)du
n
x
x
1
Z
x+
n
|x + n1 − u|
Dn ((t − x)2 , x)
+ kϕ2 g ′′ k
du
6 kϕ2 g ′′ k
(x + 1)(x + 2)
(x + 1)(x + 2)
x
n 1 1 2 o
2
6 kϕ2 g ′′ k
+
6 kϕ2 g ′′ k.
n
n
n
Now for f ∈ CB [0, ∞), we have
|Dn (f, x) − f (x)|
1
= |D̃n (f − g, x) − (f − g)(x)| + |D̃n (g, x) − g(x)| + f x +
− f (x)
n
p
1
2
6 4kf − gk + kϕ2 g ′′ k + f x + (x + 1)(x + 2) p
− f (x)
n
n (x + 1)(x + 2)
n
o
1
1
6 4 kf − gk + kϕ2 g ′′ k + sup f t + ϕ(t) p
− f (t)
n
n (x + 1)(x + 2)
t>0
n
o
1
1
6 4 kf − gk + kϕ2 g ′′ k + ~ωϕ f, p
n
n (x + 1)(x + 2)
n
o
1
1
6 4 kf − gk + kϕ2 g ′′ k + ~ωϕ f,
.
n
n
Hence, by definition of K2,ϕ (f, δ 2 ), we have the inequality
1
1
1
1 −
→
kDn (f, x) − f (x) 6 4K2,ϕ f,
+−
ω ϕ f,
6 4ωϕ2 f, √
+→
ω ϕ f,
. n
n
n
n
3. Moments for the general case
For the general case, we obtain moments as follows:
Lemma 3.1. For the operators defined by (1.1), if we denote
∞
k m
X
µan,m (x) =
ban,k (x)
,
n
k=0
then, we have the following recurrence relation:
x(1 + x) a
ax a
µan,m+1 (x) =
[µn,m (x)]′ + x +
µ (x).
n
n(1 + x) n,m
Proof. Using the identity
x(1 + x)2 [ban,k (x)]′ = [(k − nx)(1 + x) − ax]ban,k (x),
we have
x(1 + x)2 [µan,m (x)]′
=
∞
X
k=0
k m
[(k − nx)(1 + x) − ax]ban,k (x)
n
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 271
= n(1 + x)
∞
X
k=0
∞
k m
k m+1
X
− [nx(1 + x) + ax]
ban,k (x)
ban,k (x)
n
n
k=0
= n(1 + x)µan,m+1 (x) − [nx(1 + x) + ax]µan,m (x).
Remark 3.1. Using Lemma 3.1, the first few moments are as follows:
ax
µan,0 (x) = 1, µan,1 (x) = x +
,
n(1 + x)
µan,2 (x) = x2 +
µan,3 (x) = x3 +
+
+
µan,4 (x) = x4 +
+
+
+
1h
2ax2 i
1 h a2 x2
ax i
x + x2 +
+ 2
+
,
n
(1 + x)
n (1 + x)2
(1 + x)
1h 3
3ax3 i
3x + 3x2 +
n
(1 + x)
h
1
3ax3
6ax2
3a2 x3 i
3
2
2x
+
3x
+
x
+
+
+
n2
(1 + x) (1 + x) (1 + x)2
h
2 2
1 3a x
ax
a3 x3 i
+
+
n3 (1 + x)2
(1 + x) (1 + x)3
h
1
4ax4 i
6x4 + 6x3 +
n
(1 + x)
1h
12ax4
18ax3
6a2 x4 i
4
3
2
11x
+
18x
+
7x
+
+
+
n2
(1 + x) (1 + x) (1 + x)2
h
1
8ax4
18ax3
14ax2
4
3
2
6x
+
12x
+
7x
+
x
+
+
+
n3
(1 + x) (1 + x) (1 + x)
3a2 x3 (x2 + 4x + 3)
9a2 x3
4a3 x4
3a2 x4 i
+
+
+
+
(1 + x)3
(1 + x)2
(1 + x)3
(1 + x)2
h
2 2
3 3
4 4 i
1 7a x
ax
6a x
a x
+
+
+
.
4
2
3
n (1 + x)
(1 + x) (1 + x)
(1 + x)4
Lemma 3.2. [16, Lemma 2] For 0 6 β < 1, we have
(β)
hLn,k (t), tr i
(β)
hLn,k (t), 1i
= Pr (k; β),
R∞
where hf, gi = 0 f (t) g(t) dt and Pr (k; β) is a polynomial of order r in the variable k. In particular
P0 (k; β) = 1,
1h
1 i
(1 − β)k +
,
P1 (k; β) =
n
1−β
h
1
2! i
P2 (k; β) = 2 (1 − β)2 k 2 + 3k +
,
n
1−β
1h
(11 − 8β)k
3! i
P3 (k; β) = 3 (1 − β)3 k 3 + 6(1 − β)k 2 +
+
,
n
1−β
1−β
272
GUPTA AND MALIK
1h
10(5 − 3β)k
4! i
(1 − β)4 k 4 + 10(1 − β)2 k 3 + 5(7 − 4β)k 2 +
+
,
4
n
1−β
1−β
1h
P5 (k; β) = 5 (1 − β)5 k 5 + 15(1 − β)3 k 4 + 5(1 − β)(17 − 8β)k 3
n
15(15 − 20β + 6β 2 )k 2
(274 − 144β)k
5! i
+
+
+
.
1−β
1−β
1−β
P4 (k; β) =
Lemma 3.3. If the r-th order moment with monomials er (t) = tr , r = 0, 1, . . .
of the operators (1.2) be defined by
a,β
Tn,r
(x)
:=
Dna,β (er , x)
=
(β)
∞
X
hLn,k (t), tr i
(β)
k=0
The first few are
a,β
Tn,0
(x) = 1,
hLn,k (t), 1i
ban,k (x) =
a,β
Tn,1
(x) = x(1 − β) +
∞
X
Pr (k; β)ban,k (x).
k=0
ax(1 − β)
1
+
,
n(1 + x)
n(1 − β)
i
1h 2
2ax2 (1 − β)2
x (1 − β)2 + x(1 − β)2 +
+ 3x
n
1+x
1 h a2 x2 (1 − β)2
ax(1 − β)2
3ax
2 i
+ 2
+
+
+
,
n
(1 + x)2
1+x
1+x 1−β
a,β
Tn,2
(x) = x2 (1 − β)2 +
a,β
Tn,3
(x) = x3 (1 − β)3
i
1h 3
3ax3 (1 − β)3
+
3x (1 − β)3 + 3x2 (1 − β)3 +
+ 6x2 (1 − β)
n
(1 + x)
1h 3
3ax3 (1 − β)3
+ 2 2x (1 − β)3 + 3x2 (1 − β)3 + x(1 − β)3 +
n
(1 + x)
6ax2 (1 − β)3
3a2 x3 (1 − β)3
+
+
+ 6x2 (1 − β)
(1 + x)
(1 + x)2
12ax2 (1 − β) (11 − 8β)x i
+ 6x(1 − β) +
+
(1 + x)
(1 − β)
h
2 2
3
3
3 3
3
1 3a x (1 − β)
ax(1 − β)
a x (1 − β)
+ 3
+
+
n
(1 + x)2
(1 + x)
(1 + x)3
6a2 x2 (1 − β) 6ax(1 − β)
(11 − 8β)ax
6 i
+
+
+
+
(1 + x)2
(1 + x)
(1 − β)(1 + x) (1 − β)
a,β
Tn,4
(x) = x4 (1 − β)4
i
1h 4
4ax4 (1 − β)4
+
6x (1 − β)4 + 6x3 (1 − β)4 +
+ 10x3 (1 − β)2
n
(1 + x)
h
1
12ax4 (1 − β)4
+ 2 11x4 (1 − β)4 + 18x3 (1 − β)4 + 7x2 (1 − β)4 +
n
(1 + x)
3
4
2 4
4
18ax (1 − β)
6a x (1 − β)
+
+
+ 30x3 (1 − β)2
(1 + x)
(1 + x)2
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 273
+ 30x2 (1 − β)2 +
i
30ax3 (1 − β)2
+ (35 − 20β)x2
(1 + x)
1h 4
6x (1 − β)4 + 12x3 (1 − β)4 + 7x2 (1 − β)4 + x(1 − β)4
n3
8ax4 (1 − β)4
18ax3 (1 − β)4
14ax2 (1 − β)4
+
+
+
(1 + x)
(1 + x)
(1 + x)
2 3 2
4
2 3
3a x (x + 4x + 3)(1 − β)
9a x (1 − β)4
4a3 x4 (1 − β)4
+
+
+
(1 + x)3
(1 + x)2
(1 + x)3
2 4
4
3a x (1 − β)
+
+ 20x3 (1 − β)2 + 30x2 (1 − β)2 + 10x(1 − β)2
(1 + x)2
30ax3 (1 − β)2
60ax2 (1 − β)2
30a2 x3 (1 − β)2
+
+
+
(1 + x)
(1 + x)
(1 + x)2
(70 − 40β)ax2
(50 − 30β)x i
+ (35 − 20β)x2 + (35 − 20β)x +
+
(1 + x)
(1 − β)
h
2 2
4
4
3 3
4
4 4
1 7a x (1 − β)
ax(1 − β)
6a x (1 − β)
a x (1 − β)4
+
+
+
+ 4
n
(1 + x)2
(1 + x)
(1 + x)3
(1 + x)4
2
3 3
2
2 2
2
30a x (1 − β)
10ax(1 − β)
10a x (1 − β)
+
+
+
(1 + x)2
(1 + x)
(1 + x)3
2 2
(35 − 20β)a x
(35 − 20β)ax
(50 − 30β)ax
24 i
+
+
+
+
.
(1 + x)2
(1 + x)
(1 − β)(1 + x) (1 − β)
+
a,β
Proof. Obviously by (1.2), we have Tn,0
(x) = 1. Next, by definition of
a,β
Tn,r
(x), we have
a,β
Tn,r
(x) =
(β)
∞
X
hLn,k (t), tr i
(β)
k=0 hLn,k (t), 1i
ban,k (x) =
∞
X
Pr (k; β) ban,k (x).
k=0
Thus, using Lemma 3.1 and Lemma 3.2, we have
∞
∞
X
X
1h
1 i
a,β
Tn,1
(x) =
ban,k (x)P1 (k; β) =
ban,k (x) (1 − β)k +
n
1−β
k=0
k=0
1
ax(1 − β)
1
µa (x) = x(1 − β) +
+
n(1 − β) n,0
n(1 + x)
n(1 − β)
∞
∞
X
X
1 h
2 i
a,β
Tn,2
(x) =
ban,k (x)P2 (k; β) =
ban,k (x) 2 (1 − β)2 k 2 + 3k +
n
1−β
= (1 − β)µan,1 (x) +
k=0
k=0
3
2
µa (x)
= (1 −
+ µan,1 (x) + 2
n
n (1 − β) n,0
h
i
x2
x
a2 x2
2ax2
ax
= (1 − β)2 x2 +
+ + 2
+
+
n
n n (1 + x)2
n(1 + x) n2 (1 + x)
h
i
3
ax
2
+
x+
+ 2
n
n(1 + x)
n (1 − β)
β)2 µan,2 (x)
274
GUPTA AND MALIK
i
2ax2 (1 − β)2
1h 2
x (1 − β)2 + x(1 − β)2 +
+ 3x
n
1+x
1 h a2 x2 (1 − β)2
ax(1 − β)2
3ax
2 i
+ 2
+
+
+
.
n
(1 + x)2
1+x
1+x 1−β
= x2 (1 − β)2 +
a,β
A continuation of this process will provide Tn,r
(x) for cases of r > 3.
a,β
Remark 3.2. If we denote the central moment as Un,r
(x) = Dna,β ((t − x)r , x),
then
ax(1 − β)
1
a,β
Un,1
(x) = −xβ +
+
,
n(1 + x)
n(1 − β)
1h
2ax2 β(1 − β)
2x i
3x + x(1 + x)(1 − β)2 −
−
n
1+x
1−β
1 h x2 a2 (1 − β)2
ax(1 − β)2
3ax
2 i
+ 2
+
+
+
,
n
(1 + x)2
1+x
1+x 1−β
a,β
Un,2
(x) = x2 β 2 +
a,β
Un,3
(x) = x3 (1 − β)3 − x3 − 3x3 (1 − β)2 + 3x3 (1 − β)
1h 3
3ax3 (1 − β)3
+
3x (1 − β)3 + 3x2 (1 − β)3 +
n
(1 + x)
6ax3 (1 − β)2
+ 6x2 (1 − β) − 3x3 (1 − β)2 −
(1 + x)
3
3ax (1 − β)
3x2 i
− 3x2 (1 − β)2 − 9x2 +
+
(1 + x)
(1 − β)
1 h 3
3ax3 (1 − β)3
+ 2 2x (1 − β)3 + 3x2 (1 − β)3 + x(1 − β)3 +
n
(1 + x)
6ax2 (1 − β)3
3a2 x3 (1 − β)3
+
+
+ 6x2 (1 − β)
(1 + x)
(1 + x)2
12ax2 (1 − β) (11 − 8β)x
+ 6x(1 − β) +
+
(1 + x)
(1 − β)
3a2 x3 (1 − β)2
3ax2 (1 − β)2
9ax2
6x i
−
−
−
−
(1 + x)2
(1 + x)
(1 + x) (1 − β)
h
2 2
1 3a x (1 − β)3
ax(1 − β)3
a3 x3 (1 − β)3
+ 3
+
+
n
(1 + x)2
(1 + x)
(1 + x)3
2 2
6a x (1 − β) 6ax(1 − β)
(11 − 8β)ax
6 i
+
+
+
+
(1 + x)2
(1 + x)
(1 − β)(1 + x) (1 − β)
a,β
Un,4
(x) = x4 (1 − β)4 + x4 + 6x4 (1 − β)2 − 4x4 (1 − β) − 4x4 (1 − β)3
1h 4
4ax4 (1 − β)4
+
6x (1 − β)4 + 6x3 (1 − β)4 +
n
1+x
12ax4 (1 − β)2
+ 10x3 (1 − β)2 + 6x4 (1 − β)2 +
1+x
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 275
4x3
4ax4 (1 − β)
−
1+x
1−β
i
4
12ax (1 − β)3
− 12x4 (1 − β)3 − 12x3 (1 − β)3 −
− 24x3 (1 − β)
1+x
1 h
12ax4 (1 − β)4
+ 2 11x4 (1 − β)4 + 18x3 (1 − β)4 + 7x2 (1 − β)4 +
n
1+x
18ax3 (1−β)4
6a2 x4 (1−β)4
+
+
+ 30x3 (1−β)2 + 30x2 (1−β)2
1+x
(1 + x)2
30ax3 (1−β)2
6a2 x4 (1−β)2
6ax3 (1−β)2
+
+ (35 − 20β)x2 +
+
1+x
(1 + x)2
1+x
2
3
12x
18ax
+
− 8x4 (1 − β)3 − 12x3 (1 − β)3 − 4x2 (1 − β)3
+
1+x
1−β
12ax4 (1 − β)3
24ax3 (1 − β)3
12a2 x4 (1 − β)3
−
−
−
1+x
1+x
(1 + x)2
3
48ax (1−β) 4(11 − 8β)x2 i
− 24x3 (1−β) − 24x2 (1−β) −
−
1+x
1−β
h
8ax4 (1−β)4
1
+ 3 6x4 (1−β)4 + 12x3 (1−β)4 + 7x2 (1−β)4 + x(1−β)4 +
n
1+x
2
4
2 3 2
3
4
18ax (1 − β)
14ax (1 − β)
3a x (x + 4x + 3)(1 − β)4
+
+
+
1+x
1+x
(1 + x)3
9a2 x3 (1 − β)4
4a3 x4 (1 − β)4
3a2 x4 (1 − β)4
+
+
+
+ 20x3 (1 − β)2
2
3
(1 + x)
(1 + x)
(1 + x)2
30ax3 (1 − β)2
60ax2 (1 − β)2
+ 30x2 (1 − β)2 + 10x(1 − β)2 +
+
1+x
1+x
30a2 x3 (1 − β)2
(70
− 40β)ax2
2
+
+
(35
−
20β)x
+
(35
−
20β)x
+
(1 + x)2
1+x
(50 − 30β)x 12a2 x3 (1 − β)3
4ax2 (1 − β)3
4a3 x4 (1 − β)3
+
−
−
−
1−β
(1 + x)2
1+x
(1 + x)3
2 3
2
2
24a x (1 − β) 24ax (1 − β) 4(11 − 8β)ax
24x i
−
−
−
−
(1 + x)2
1+x
(1 − β)(1 + x) 1 − β
h
2 2
1 7a x (1 − β)4
ax(1 − β)4
6a3 x3 (1 − β)4
a4 x4 (1 − β)4
+ 4
+
+
+
n
(1 + x)2
1+x
(1 + x)3
(1 + x)4
2 2
2
2
3 3
2
30a x (1 − β)
10ax(1 − β)
10a x (1 − β)
+
+
+
2
(1 + x)
1+x
(1 + x)3
(35 − 20β)a2 x2
(35 − 20β)ax
(50 − 30β)ax
24 i
+
+
+
.
+
(1 + x)2
1+x
(1 − β)(1 + x) 1 − β
+ 6x3 (1 − β)2 + 18x3 −
4. Direct estimates for general case
In this section, we establish the following direct results:
276
GUPTA AND MALIK
Proposition 4.1. Let f be a continuous function on [0, ∞) and β = βn → 0
as n → ∞, the sequence {Dna,β (f, x)} converges uniformly to f (x) in [a, b] ⊂ [0, ∞).
Proof. Using Lemma 3.3, we have
Dna,β (e0 , x) = 1,
Dna,β (e1 , x) = x(1 − β) +
and
ax(1 − β)
1
+
n(1 + x)
n(1 − β)
i
1h 2
2ax2 (1 − β)2
x (1 − β)2 + x(1 − β)2 +
+ 3x
n
1+x
1 h a2 x2 (1 − β)2
ax(1 − β)2
3ax
2 i
+ 2
+
+
+
.
n
(1 + x)2
1+x
1+x 1−β
Dna,β (e2 , x) = x2 (1 − β)2 +
Obviously, Dna,β (e0 , x), Dna,β (e1 , x) and Dna,β (e2 , x) converges uniformly to 1, x and
x2 respectively on every compact subset of [0, ∞). Thus, the required result follows
from the well known Bohman–Korovkin theorem.
Theorem 4.1. Let f be a bounded integrable function on [0, ∞) and has the
second derivative at a point x ∈ [0, ∞), then with the condition β = βn → 0 as
n → ∞, it follows that
h x2 + 2x i
h
ax i ′
f (x) +
f ′′ (x).
lim n[Dna,βn (f, x) − f (x)] = 1 +
n→∞
1+x
2
Proof. By Taylor’s expansion of f , we have
1
(4.1)
f (t) = f (x) + f ′ (x)(t − x) + f ′′ (x)(t − x)2 + r(t, x)(t − x)2 ,
2
where r(t, x) is the remainder term and limt→x r(t, x) = 0. Operating Dna,β to the
equation (4.1), we obtain
Dna,β (f, x) − f (x) = Dna,β (t − x, x)f ′ (x) + Dna,β ((t − x)2 , x)
f ′′ (x)
2
+ Dna,β (r(t, x)(t − x)2 , x)
Using the Cauchy–Schwarz inequality, we have
q
q
(4.2)
Dna,β (r(t, x)(t − x)2 , x) 6 Dna,β (r2 (t, x), x) Dna,β ((t − x)4 , x).
As r2 (x, x) = 0 and r2 (t, x) ∈ C2∗ [0, ∞), we have
(4.3)
lim Dna,βn (r2 (t, x), x) = r2 (x, x) = 0
n→∞
uniformly with respect to x ∈ [0, A], for some A > 0. Now from (4.2), (4.3) and
from Remark 3.2, we get limn→∞ nDna,βn (r(t, x)(t − x)2 , x) = 0. Thus
lim n(Dna,βn (f, x) − f (x))
h
i
1
= lim n Dna,βn (t−x, x)f ′ (x)+ f ′′ (x)Dna,βn ((t−x)2 , x)+Dna,βn (r(t, x)(t−x)2 , x)
n→∞
2
h
h x2 + 2x i
ax i ′
= 1+
f (x) +
f ′′ (x). 1+x
2
n→∞
DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 277
We denote the norm kf k = supx∈[0,∞) |f (x)|. For f ∈ CB [0, ∞) and δ > 0, the
m-th order modulus of continuity is defined as
ωm (f, δ) = sup
sup |∆m
h f (x)|,
06h6δ x∈[0,∞)
where ∆ is the forward difference. In case m = 1, we mean the usual modulus of
continuity denoted by ω(f, δ). Peetre’s K-functional is defined as
K2 (f, δ) =
inf
2 [0,∞)
g∈CB
2
{kf − gk + δkg ′′ k : g ∈ CB
[0, ∞)},
where
2
CB
[0, ∞) = {g ∈ CB [0, ∞) : g ′ , g ′′ ∈ CB [0, ∞)}.
Theorem 4.2. Let f ∈ CB [0, ∞) and 0 < β < 1, then
ax(1 − β)
p 1
|Dna,β (f, x) − f (x)| 6 Cω2 f, δn + ω f,
+
− xβ
n(1 + x)
n(1 − β)
where C is a positive constant and δn is given by
1h
4ax2 β(1 − β)
2x
2xβ i
δn = 2x2 β 2 +
3x + x(x + 1)(1 − β)2 −
−
−
n
1+x
1−β
1−β
h
i
2 2
2
2
5ax
ax(1 − β)
2
1
1 2a x (1 − β)
+ 2
+
+
+
+
.
2
2
n
(1 + x)
1+x
1+x
1−β
(1 − β)
with β = βn → 0 as n → ∞.
Proof. We introduce the auxiliary operators D̄na,β : CB [0, ∞) → CB [0, ∞) by
ax(1 − β)
1
(4.4) D̄na,β (f, x) = Dna,β (f, x) − f x(1 − β) +
+
+ f (x).
n(1 + x)
n(1 − β)
These operators are linear and preserve the linear functions in view of Lemma 3.3.
2
Let g ∈ CB
[0, ∞) and x, t ∈ [0, ∞). By Taylor’s expansion, we have
Z t
g(t) = g(x) + (t − x) g ′ (x) +
(t − u) g ′′ (u) du,
x
Hence,
|D̄na,β (g, x) − g(x)|
Z t
Z t
a,β ′′
a,β ′′
6 D̄n (t − u)g (u)du, x 6 Dn (t − u)g (u)du, x
x
x
Z
+ x(1−β)+
x
ax(1−β)
1
+ n(1−β)
n(1+x)
ax(1 − β)
1
′′
x(1 − β) +
+
− u g (u)du
n(1 + x)
n(1 − β)
6 Dna,β ((t − x)2 , x)kg ′′ k
Z x(1−β)+ ax(1−β) + 1 n(1+x)
n(1−β)
ax(1 − β)
1
+
+
− xβ dukg ′′ k
n(1 + x)
n(1 − β)
x
278
GUPTA AND MALIK
Next, using Remark 3.2, we have
h
|D̄na,β (g, x) − g(x)| 6 Dna,β ((t − x)2 , x)
ax(1 − β)
2 i
(4.5)
1
+
+
− xβ
kg ′′ k = δn kg ′′ k.
n(1 + x)
n(1 − β)
Since
|Dna,β (f, x)| 6
(β)
∞
X
hLn,k (t), |f (t)|i
k=0
(β)
hLn,k (t), 1i
ban,k (x) 6 kf k
(β)
∞
X
hLn,k (t), 1i
ban,k (x)
(β)
k=0 hLn,k (t), 1i
6 kf k.
Now by (4.4), we have
(4.6)
|kD̄na,β (f, x)|| 6 k|Dna,β (f, x)k + 2kf k 6 3kf k,
f ∈ CB [0, ∞).
Using (4.4), (4.5) and (4.6), we have
|Dna,β (f, x) − f (x)| 6 |D̄na,β (f − g, x) − (f − g)(x)| + |D̄na,β (g, x) − g(x)|
ax(1 − β)
1
+
− f (x)
+ f x(1 − β) +
n(1 + x)
n(1 − β)
6 4kf − gk + δn kg ′′ k
1
ax(1 − β)
+ f x(1 − β) +
+
− f (x)
n(1 + x)
n(1 − β)
ax(1 − β)
1
6 C{kf − gk + δn kg ′′ k} + ω f,
+
− xβ .
n(1 + x)
n(1 − β)
2
Taking infimum over all g ∈ CB
[0, ∞), and using the inequality
√
K2 (f, δ) 6 Cω2 (f, δ), δ > 0
due to [4], we get the desired assertion.
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Department of Mathematics,
Netaji Subhas Institute of Technology
New Delhi
India
vijaygupta2001@hotmail.com
neha.malik_nm@yahoo.com
(Received 12 01 2015)
(Revised 04 07 2015)
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