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

(1.1)  B_{n}^{a}(f; x) = ∑_{k=0}^{∞} b_{n,k}^{a} (x)f( \frac{k}{n}) ,

where

b_{n,k}^{a} (x) = e^{− \frac{ax}{1+x}} \frac{∑ _{i=0}^{k} \binom{k}{i}(n)_{i}a^{k−i}}{k!} \frac{x^{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

S_{n}^{β}(f, x) = ∑_{k=0}^{∞} L_{n,k}^{(β)} (x)f( \frac{k}{n}) , x ∈ [0, ∞)

2010 Mathematics Subject Classification: 41A25; 41A30. Key words and phrases: Baskakov operators, Szász-Mirakyan operators, direct results, mod-

ulus of continuity.

Communicated by Gradimir Milovanović.

where 0 ⩽ β < 1 and the basis function is defined as

L_{n,k}^{(β)} (x) = \frac{nx(nx + kβ)^{k−1}}{k!} e^{−(nx+kβ)} .

It was seen in [15] and [13] that ∑ _{k=0}^{∞} b_{n,k}^{a} (x) = ∑ _{k=0}^{∞} L_{n,k}^{(β)} (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 general basis function L_{n,k}^{(β)} (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 ⩽ β < 1, we propose the mixed hybrid Durrmeyer type operators as follows:

(1.2)  D_{n}^{a,β} (f, x) = ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(β)} (t), f(t)⟩}{⟨L_{n,k}^{(β)} (t), 1⟩} b_{n,k}^{a} (x)

where ⟨f, g⟩ = ∫_{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

D_{n}(f, x) := D_{n}^{0,0}(f, x) = ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(0)} (t), f(t)⟩}{⟨L_{n,k}^{(0)} (t), 1⟩} b_{n,k}^{0} (x)

= n ∑_{k=0}^{∞} b_{n,k}^{0} (x) ∫_{0}^{∞} L_{n,k}^{(0)} (t)f(t)dt,

where

b_{n,k}^{0} (x) = \binom{n + k − 1}{k} \frac{x^{k}}{(1 + x)^{n+k}} , L_{n,k}^{(0)} (t) = e^{−nt} \frac{(nt)^{k}}{k!} .

We may point out here that ⟨L_{n,k}^{(β)} (t), 1⟩ 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

DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 267

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 hypergeometric series as follows: Using the identities: k! = (1)_{k} and Γ(k +m+1) = Γ(m + 1)(m + 1)_{k}, we have

D_{n}(t^{m}, x) = n ∑_{k=0}^{∞} b_{n,k}^{0} (x) ∫_{0}^{∞} L_{n,k}^{0} (x)t^{m}dt

= n ∑_{k=0}^{∞} \frac{(n)_{k}x^{k}}{k!(1 + x)^{n+k}} ∫_{0}^{∞} \frac{e^{−nt} (nt)^{k}}{k!} t^{m}dt

= \frac{1}{n^{m}} ∑_{k=0}^{∞} \frac{(n)_{k}x^{k}}{k!(1 + x)^{n+k}} \frac{Γ(k + m + 1)}{(1)_{k}}

= \frac{(1 + x)^{−n}}{n^{m}} ∑_{k=0}^{∞} \frac{(n)_{k}(m + 1)_{k} Γ(m + 1)x^{k}}{k!(1 + x)^{k}(1)_{k}}

= \frac{m!(1 + x)^{−n}}{n^{m}} _{2}F_{1}( n, m + 1; 1; \frac{x}{1 + x} ) .

Applying the well known transformation

2F_{1}(a, b; c; x) = (1 − x)^{−a} _{2}F_{1}( a, c − b; c; \frac{x}{x − 1} ) ,

we obtain at once

D_{n}(t^{m}, x) = \frac{m!}{n^{m}} _{2}F_{1}(n, −m; 1; −x).

Furthermore, we obtain

D_{n}(t, x) = x + \frac{1}{n}, D_{n}(t^{2}, x) = \frac{2 + 4nx + n(n + 1)x^{2}}{n^{2}}

and

(2.1)  D_{n}(t − x, x) = \frac{1}{n}, D_{n}((t − x)^{2}, x) = \frac{2 + 2nx + nx^{2}}{n^{2}} .

In general, using the similar approach, one can show that:

(2.2)  D_{n}((t − x)^{r}, x) = O(n^{−[(r+1)/2]} ),

where [α] denotes the integral part of α.

Let B_{x^{2}} [0, ∞) be the set of all functions f defined on ℝ^{+} satisfying the condition |f(x)| ⩽ M_{f} (1+x^{2}) with some constant M_{f} , depending only on f, but independent of x. B_{x^{2}} [0, ∞) is called weighted space and it is a Banach space endowed with the norm

∥f∥_{x^{2}} = sup_{x∈ℝ^{+}} \frac{f(x)}{1 + x^{2}} .

Let C_{x^{2}} [0, ∞) = C[0, ∞) ∩ B_{x^{2}} [0, ∞) and by C_{x^{2}}^{k} [0, ∞), we denote subspace of all continuous functions f ∈ B_{x^{2}} [0, ∞) for which lim_{x→∞} \frac{f(x)}{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) defined on infinite interval ℝ^{+} (see [17]). Let

Ω(f, δ) = sup_{|h|<δ, x∈ℝ^{+}} \frac{|f(x + h) − f(x)|}{(1 + h^{2})(1 + x^{2})} for each f ∈ C_{x^{2}} [0, ∞).

Now, some elementary properties of Ω(f, δ) are collected in the following lemma.

Lemma 2.1. Let f ∈ C_{x^{2}}^{k} [0, ∞). Then, i) Ω(f, δ) is a monotonically increasing function of δ, δ ⩾ 0. ii) For every f ∈ C_{x^{2}}^{k} [0, ∞), lim_{δ→0} Ω(f, δ) = 0. iii) For each λ > 0,

(2.3)  Ω(f, λδ) ⩽ 2(1 + λ)(1 + δ^{2})Ω(f, δ).

From the inequality (2.3) and definition of Ω(f, δ), we get

(2.4)  |f(t) − f(x)| ⩽ 2(1 + x^{2})(1 + (t − x)^{2})( 1 + \frac{|t − x|}{δ} ) (1 + δ^{2})Ω(f, δ)

for every f ∈ C_{x^{2}} [0, ∞) and x, t ∈ ℝ^{+} . The following estimate is quantitative Voronovskaya type asymptotic formula:

Theorem 2.1. Let f′′∈ C_{x^{2}}^{k} [0, ∞), a = β = 0 and x > 0. Then we have

|D_{n}(f, x)−f(x)− \frac{1}{n}f′(x)− \frac{x(x + 2)}{2n} f′′(x)| ⩽ \frac{f′′(x)}{n^{2}} +8(1+x^{2})O(n^{−1} )Ω( f′′, \frac{1}{\sqrt{n}}) .

Proof. By Taylor’s formula, there exist η lying between x and y such that

f(y) = f(x) + f′(x)(y − x) + \frac{f′′(x)}{2} (y − x)^{2} + h(y, x)(y − x)^{2},

where

h(y, x) := \frac{f′′(η) − f′′(x)}{2}

and h is a continuous function which vanishes at 0. Applying the operator D_{n} to above equality, we get

D_{n}(f, x) − f(x) = \frac{f′(x)}{n} + \frac{f′′(x)}{2} [ \frac{2 + 2nx + nx^{2}}{n^{2}} ] + D_{n}(h(y, x)(y − x)^{2}, x).

Also, we can write that

|D_{n}(f, x) − f(x) − \frac{f′(x)}{n} − \frac{f′′(x)}{2} ( \frac{2x + x^{2}}{n} ) | ⩽ \frac{f′′(x)}{n^{2}} + D_{n}(|h(y, x)|(y − x)^{2}, x)

To estimate last inequality using (2.4) and the inequality |η − x| ⩽ |y − x|, we can write

|h(y, x)| ⩽ (1 + (y − x)^{2})(1 + x^{2})( 1 + \frac{|y − x|}{δ} ) (1 + δ^{2})Ω(f′′, δ).

Also,

|h(y, x)| ⩽ \{ \begin{matrix}2(1 + x^{2})(1 + δ^{2})^{2}Ω(f′′, δ), & |y − x| < δ \\ (1 + (y − x)^{2})(1 + x^{2})(1 + \frac{|y−x|}{δ} )(1 + δ^{2})Ω(f′′, δ), & |y − x| ⩾ δ \end{matrix}

Now choosing δ < 1, we have

|h(y, x)| ⩽ 2(1 + x^{2})( 1 + \frac{(y − x)^{4}}{δ^{4}} ) (1 + δ^{2})^{2}Ω(f′′, δ)

⩽ 8(1 + x^{2})( 1 + \frac{(y − x)^{4}}{δ^{4}} ) Ω(f′′, δ).

Using (2.2), we deduce that

D_{n}(|h(y, x)|(y − x)^{2}, x) = 8(1 + x^{2})Ω(f′′, δ){D_{n}((t − x)^{2}, x) + δ^{−4} D_{n}((t − x)^{6}, x)}
= 8(1 + x^{2})Ω(f′′, δ){O(n^{−1} ) + δ^{−4} O(n^{−3} )}.

Choosing δ = 1/\sqrt{n} we have

D_{n}(|h(y, x)|(y − x)^{2}, x) ⩽ 8(1 + x^{2})O(n^{−1} )Ω(f′′, 1/\sqrt{n} ).  □

By C_{B}[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

ω_{φ}^{2}(f, δ) = sup_{0⩽h⩽δ} sup_{x±hφ(x)∈[0,∞)} |f(x + hφ(x)) − 2f(x) + f(x − hφ(x))|,

φ(x) = \sqrt{(x + 1)(x + 2)} , x ⩾ 0. The corresponding K-functional is

K_{2,φ} (f, δ^{2}) = inf_{h∈W_{∞}^{2} (φ)} {∥f − h∥ + δ^{2}∥φ^{2}h′′∥},

where W_{∞}^{2} (φ) = {h ∈ C_{B}[0, ∞) : h′ ∈ AC_{loc}[0, ∞) : φ^{2}h′′ ∈ C_{B}[0, ∞)}. By [5, Thm. 2.1.1], it follows that

C^{−1} ω_{φ}^{2}(f, δ) ⩽ K_{2,φ} (f, δ^{2}) ⩽ Cω_{φ}^{2}(f, δ)

for some absolute constant C > 0. Also, the Ditzian–Totik modulus of the first order is given by

\overrightarrow{ω} _{φ} (f, δ) = sup_{0⩽h⩽δ} sup_{x±hφ(x)∈[0,∞)} |f(x + hφ(x)) − f(x)|,

where φ is an admissible step-weight function on [0, ∞).

Theorem 2.2. If f ∈ C_{B}[0, ∞) and n ∈ ℕ, then we have the inequality

∥D_{n}(f, x) − f(x)∥ ⩽ 4ω_{φ}^{2}(f, 1/\sqrt{n} ) + ω⃗_{φ} (f, 1/n).

Proof. We set φ(x) = \sqrt{(x + 1)(x + 2)} ,

W_{φ}^{2}[0, ∞) = {g ∈ AC_{loc}[0, ∞) : φ^{2}g′′∈ C_{B}[0, ∞)}
D̃_{n} (f, x) = D_{n}(f, x) − f(x + 1/n) + f(x).

Then \frac{|t−u|}{φ^{2} (u)} ⩽ \frac{|t−x|}{φ^{2} (x)} for u between x and t, D̃_{n} (t − x, x) = 0 by using (2.1), and for g ∈ W_{φ}^{2}[0, ∞), by Taylor’s formula, we have

g(t) = g(x) + g′(x)(t − x) + ∫_{x}^{t} g′′(u)(t − u) du.

Applying the operator D_{n} to above equality and then taking modulus, we get

270 GUPTA AND MALIK

|D̃_{n} (g, x) − g(x)|

⩽ D_{n}( | ∫_{x}^{t}(t − u) g′′(u) du|, x) + | ∫_{x}^{x+ \frac{1}{n}} ( x + \frac{1}{n} − u) g′′(u)du|

⩽ ∥φ^{2}g′′∥\frac{D_{n}((t − x)^{2}, x)}{(x + 1)(x + 2)} + ∥φ^{2}g′′∥| ∫_{x}^{x+ \frac{1}{n}} \frac{|x + \frac{1}{n} − u|}{(x + 1)(x + 2)} du|

⩽ ∥φ^{2}g′′∥\{ \frac{1}{n} + ( \frac{1}{n})^{2}\} ⩽ \frac{2}{n}∥φ^{2}g′′∥.

Now for f ∈ C_{B}[0, ∞), we have

|D_{n} (f, x) − f(x)|

= |D̃_{n} (f − g, x) − (f − g)(x)| + |D̃_{n} (g, x) − g(x)| + |f( x + \frac{1}{n}) − f(x)|

⩽ 4∥f − g∥ + \frac{2}{n}∥φ^{2}g′′∥ + |f( x + \sqrt{(x + 1)(x + 2)} \frac{1}{n\sqrt{(x + 1)(x + 2)}} ) − f(x)|

⩽ 4\{∥f − g∥ + \frac{1}{n}∥φ^{2}g′′∥\} + sup_{t⩾0} |f( t + φ(t)\frac{1}{n\sqrt{(x + 1)(x + 2)}} ) − f(t)|

⩽ 4\{∥f − g∥ + \frac{1}{n}∥φ^{2}g′′∥\} + ω⃗_{φ} ( f, \frac{1}{n\sqrt{(x + 1)(x + 2)}} )

⩽ 4\{∥f − g∥ + \frac{1}{n}∥φ^{2}g′′∥\} + ω⃗_{φ} ( f, \frac{1}{n}) .

Hence, by definition of K_{2,φ} (f, δ^{2}), we have the inequality

∥D_{n}(f, x) − f(x) ⩽ 4K_{2,φ} ( f, \frac{1}{n}) + \overrightarrow{ω} _{φ} ( f, \frac{1}{n}) ⩽ 4ω_{φ}^{2}( f, \frac{1}{\sqrt{n}}) + \overrightarrow{ω} _{φ} ( f, \frac{1}{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

μ_{n,m}^{a} (x) = ∑_{k=0}^{∞} b_{n,k}^{a} (x)( \frac{k}{n})^{m},

then, we have the following recurrence relation:

μ_{n,m+1}^{a} (x) = \frac{x(1 + x)}{n} [μ_{n,m}^{a} (x)]′+ ( x + \frac{ax}{n(1 + x)} ) μ_{n,m}^{a} (x).

Proof. Using the identity

x(1 + x)^{2}[b_{n,k}^{a} (x)]′ = [(k − nx)(1 + x) − ax]b_{n,k}^{a} (x),

we have

x(1 + x)^{2} [μ_{n,m}^{a} (x)]′

= ∑_{k=0}^{∞} [(k − nx)(1 + x) − ax]b_{n,k}^{a} (x)( \frac{k}{n})^{m}

= n(1 + x) ∑_{k=0}^{∞} b_{n,k}^{a} (x)( \frac{k}{n})^{m+1} − [nx(1 + x) + ax] ∑_{k=0}^{∞} b_{n,k}^{a} (x)( \frac{k}{n})^{m}

= n(1 + x)μ_{n,m+1}^{a} (x) − [nx(1 + x) + ax]μ_{n,m}^{a} (x).  □

Remark 3.1. Using Lemma 3.1, the first few moments are as follows:

μ_{n,0}^{a} (x) = 1, μ_{n,1}^{a} (x) = x + \frac{ax}{n(1 + x)} ,

μ_{n,2}^{a} (x) = x^{2} + \frac{1}{n}[x + x^{2} + \frac{2ax^{2}}{(1 + x)} ] + \frac{1}{n^{2}} [ \frac{a^{2}x^{2}}{(1 + x)^{2}} + \frac{ax}{(1 + x)} ],
μ_{n,3}^{a} (x) = x^{3} + \frac{1}{n}[3x^{3} + 3x^{2} + \frac{3ax^{3}}{(1 + x)} ]

+ \frac{1}{n^{2}} [2x^{3} + 3x^{2} + x + \frac{3ax^{3}}{(1 + x)} + \frac{6ax^{2}}{(1 + x)} + \frac{3a^{2}x^{3}}{(1 + x)^{2}} ]

+ \frac{1}{n^{3}} [ \frac{3a^{2}x^{2}}{(1 + x)^{2}} + \frac{ax}{(1 + x)} + \frac{a^{3}x^{3}}{(1 + x)^{3}} ]

μ_{n,4}^{a} (x) = x^{4} + \frac{1}{n}[6x^{4} + 6x^{3} + \frac{4ax^{4}}{(1 + x)} ]

+ \frac{1}{n^{2}} [11x^{4} + 18x^{3} + 7x^{2} + \frac{12ax^{4}}{(1 + x)} + \frac{18ax^{3}}{(1 + x)} + \frac{6a^{2}x^{4}}{(1 + x)^{2}} ]

+ \frac{1}{n^{3}} [6x^{4} + 12x^{3} + 7x^{2} + x + \frac{8ax^{4}}{(1 + x)} + \frac{18ax^{3}}{(1 + x)} + \frac{14ax^{2}}{(1 + x)}

+ \frac{3a^{2}x^{3}(x^{2} + 4x + 3)}{(1 + x)^{3}} + \frac{9a^{2}x^{3}}{(1 + x)^{2}} + \frac{4a^{3}x^{4}}{(1 + x)^{3}} + \frac{3a^{2}x^{4}}{(1 + x)^{2}} ]

+ \frac{1}{n^{4}} [ \frac{7a^{2}x^{2}}{(1 + x)^{2}} + \frac{ax}{(1 + x)} + \frac{6a^{3}x^{3}}{(1 + x)^{3}} + \frac{a^{4}x^{4}}{(1 + x)^{4}} ].

Lemma 3.2. [16, Lemma 2] For 0 ⩽ β < 1, we have

\frac{⟨L_{n,k}^{(β)} (t), t^{r}⟩}{⟨L_{n,k}^{(β)} (t), 1⟩}= P_{r}(k; β),

where ⟨f, g⟩ = ∫_{0}^{∞} f(t) g(t) dt and P_{r}(k; β) is a polynomial of order r in the variable k. In particular

P_{0}(k; β) = 1,
P_{1}(k; β) = \frac{1}{n}[(1 − β)k + \frac{1}{1 − β} ],
P_{2}(k; β) = \frac{1}{n^{2}} [(1 − β)^{2}k^{2} + 3k + \frac{2!}{1 − β} ],
P_{3}(k; β) = \frac{1}{n^{3}} [(1 − β)^{3}k^{3} + 6(1 − β)k^{2} + \frac{(11 − 8β)k}{1 − β} + \frac{3!}{1 − β} ],

P_{4}(k; β) = \frac{1}{n^{4}} [(1 − β)^{4}k^{4} + 10(1 − β)^{2}k^{3} + 5(7 − 4β)k^{2} + \frac{10(5 − 3β)k}{1 − β} + \frac{4!}{1 − β} ],
P_{5}(k; β) = \frac{1}{n^{5}} [(1 − β)^{5}k^{5} + 15(1 − β)^{3}k^{4} + 5(1 − β)(17 − 8β)k^{3}

+ \frac{15(15 − 20β + 6β^{2})k^{2}}{1 − β} + \frac{(274 − 144β)k}{1 − β} + \frac{5!}{1 − β} ].

Lemma 3.3. If the r-th order moment with monomials e_{r}(t) = t^{r}, r = 0, 1, … of the operators (1.2) be defined by

T_{n,r}^{a,β} (x) := D_{n}^{a,β} (e_{r}, x) = ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(β)} (t), t^{r}⟩}{⟨L_{n,k}^{(β)} (t), 1⟩}b_{n,k}^{a} (x) = ∑_{k=0}^{∞} P_{r}(k; β)b_{n,k}^{a} (x).

The first few are

T_{n,0}^{a,β} (x) = 1, T_{n,1}^{a,β} (x) = x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} ,

T_{n,2}^{a,β} (x) = x^{2}(1 − β)^{2} + \frac{1}{n}[x^{2}(1 − β)^{2} + x(1 − β)^{2} + \frac{2ax^{2}(1 − β)^{2}}{1 + x} + 3x]

+ \frac{1}{n^{2}} [ \frac{a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{ax(1 − β)^{2}}{1 + x} + \frac{3ax}{1 + x} + \frac{2}{1 − β} ],

T_{n,3}^{a,β} (x)= x^{3}(1 − β)^{3}

+ \frac{1}{n}[3x^{3}(1 − β)^{3} + 3x^{2}(1 − β)^{3} + \frac{3ax^{3}(1 − β)^{3}}{(1 + x)} + 6x^{2}(1 − β)]

+ \frac{1}{n^{2}} [2x^{3}(1 − β)^{3} + 3x^{2}(1 − β)^{3} + x(1 − β)^{3} + \frac{3ax^{3}(1 − β)^{3}}{(1 + x)}

+ \frac{6ax^{2}(1 − β)^{3}}{(1 + x)} + \frac{3a^{2}x^{3}(1 − β)^{3}}{(1 + x)^{2}} + 6x^{2}(1 − β)

+ 6x(1 − β) + \frac{12ax^{2}(1 − β)}{(1 + x)} + \frac{(11 − 8β)x}{(1 − β)} ]

+ \frac{1}{n^{3}} [ \frac{3a^{2}x^{2}(1 − β)^{3}}{(1 + x)^{2}} + \frac{ax(1 − β)^{3}}{(1 + x)} + \frac{a^{3}x^{3}(1 − β)^{3}}{(1 + x)^{3}}

+ \frac{6a^{2}x^{2}(1 − β)}{(1 + x)^{2}} + \frac{6ax(1 − β)}{(1 + x)} + \frac{(11 − 8β)ax}{(1 − β)(1 + x)} + \frac{6}{(1 − β)} ]

T_{n,4}^{a,β} (x)= x^{4}(1 − β)^{4}

+ \frac{1}{n}[6x^{4}(1 − β)^{4} + 6x^{3}(1 − β)^{4} + \frac{4ax^{4}(1 − β)^{4}}{(1 + x)} + 10x^{3}(1 − β)^{2}]

+ \frac{1}{n^{2}} [11x^{4}(1 − β)^{4} + 18x^{3}(1 − β)^{4} + 7x^{2}(1 − β)^{4} + \frac{12ax^{4}(1 − β)^{4}}{(1 + x)}

+ \frac{18ax^{3}(1 − β)^{4}}{(1 + x)} + \frac{6a^{2}x^{4}(1 − β)^{4}}{(1 + x)^{2}} + 30x^{3}(1 − β)^{2}

+ 30x^{2}(1 − β)^{2} + \frac{30ax^{3}(1 − β)^{2}}{(1 + x)} + (35 − 20β)x^{2}]

+ \frac{1}{n^{3}} [6x^{4}(1 − β)^{4} + 12x^{3}(1 − β)^{4} + 7x^{2}(1 − β)^{4} + x(1 − β)^{4}

+ \frac{8ax^{4}(1 − β)^{4}}{(1 + x)} + \frac{18ax^{3}(1 − β)^{4}}{(1 + x)} + \frac{14ax^{2}(1 − β)^{4}}{(1 + x)}

+ \frac{3a^{2}x^{3}(x^{2} + 4x + 3)(1 − β)^{4}}{(1 + x)^{3}} + \frac{9a^{2}x^{3}(1 − β)^{4}}{(1 + x)^{2}} + \frac{4a^{3}x^{4}(1 − β)^{4}}{(1 + x)^{3}}

+ \frac{3a^{2}x^{4}(1 − β)^{4}}{(1 + x)^{2}} + 20x^{3}(1 − β)^{2} + 30x^{2}(1 − β)^{2} + 10x(1 − β)^{2}

+ \frac{30ax^{3}(1 − β)^{2}}{(1 + x)} + \frac{60ax^{2}(1 − β)^{2}}{(1 + x)} + \frac{30a^{2}x^{3}(1 − β)^{2}}{(1 + x)^{2}}

+ (35 − 20β)x^{2} + (35 − 20β)x + \frac{(70 − 40β)ax^{2}}{(1 + x)} + \frac{(50 − 30β)x}{(1 − β)} ]

+ \frac{1}{n^{4}} [ \frac{7a^{2}x^{2}(1 − β)^{4}}{(1 + x)^{2}} + \frac{ax(1 − β)^{4}}{(1 + x)} + \frac{6a^{3}x^{3}(1 − β)^{4}}{(1 + x)^{3}} + \frac{a^{4}x^{4}(1 − β)^{4}}{(1 + x)^{4}}

+ \frac{30a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{10ax(1 − β)^{2}}{(1 + x)} + \frac{10a^{3}x^{3}(1 − β)^{2}}{(1 + x)^{3}}

+ \frac{(35 − 20β)a^{2}x^{2}}{(1 + x)^{2}} + \frac{(35 − 20β)ax}{(1 + x)} + \frac{(50 − 30β)ax}{(1 − β)(1 + x)} + \frac{24}{(1 − β)} ].

Proof. Obviously by (1.2), we have T_{n,0}^{a,β} (x) = 1. Next, by definition of T_{n,r}^{a,β} (x), we have

T_{n,r}^{a,β} (x) = ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(β)} (t), t^{r}⟩}{⟨L_{n,k}^{(β)} (t), 1⟩}b_{n,k}^{a} (x) = ∑_{k=0}^{∞} P_{r}(k; β) b_{n,k}^{a} (x).

Thus, using Lemma 3.1 and Lemma 3.2, we have

T_{n,1}^{a,β} (x)= ∑_{k=0}^{∞} b_{n,k}^{a} (x)P_{1}(k; β) = ∑_{k=0}^{∞} b_{n,k}^{a} (x)\frac{1}{n}[(1 − β)k + \frac{1}{1 − β} ]

= (1 − β)μ_{n,1}^{a} (x) + \frac{1}{n(1 − β)} μ_{n,0}^{a} (x) = x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)}

T_{n,2}^{a,β} (x)= ∑_{k=0}^{∞} b_{n,k}^{a} (x)P_{2}(k; β) = ∑_{k=0}^{∞} b_{n,k}^{a} (x)\frac{1}{n^{2}} [(1 − β)^{2}k^{2} + 3k + \frac{2}{1 − β} ]

= (1 − β)^{2}μ_{n,2}^{a} (x) + \frac{3}{n}μ_{n,1}^{a} (x) + \frac{2}{n^{2}(1 − β)} μ_{n,0}^{a} (x)

= (1 − β)^{2}[x^{2} + \frac{x^{2}}{n} + \frac{x}{n} + \frac{a^{2}x^{2}}{n^{2}(1 + x)^{2}} + \frac{2ax^{2}}{n(1 + x)} + \frac{ax}{n^{2}(1 + x)} ]

+ \frac{3}{n}[x + \frac{ax}{n(1 + x)} ] + \frac{2}{n^{2}(1 − β)}

= x^{2}(1 − β)^{2} + \frac{1}{n}[x^{2}(1 − β)^{2} + x(1 − β)^{2} + \frac{2ax^{2}(1 − β)^{2}}{1 + x} + 3x]

+ \frac{1}{n^{2}} [ \frac{a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{ax(1 − β)^{2}}{1 + x} + \frac{3ax}{1 + x} + \frac{2}{1 − β} ].

A continuation of this process will provide T_{n,r}^{a,β} (x) for cases of r ⩾ 3. □

Remark 3.2. If we denote the central moment as U_{n,r}^{a,β} (x) = D_{n}^{a,β} ((t − x)^{r}, x), then

U_{n,1}^{a,β} (x) = −xβ + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} ,

U_{n,2}^{a,β} (x) = x^{2}β^{2} + \frac{1}{n}[3x + x(1 + x)(1 − β)^{2} − \frac{2ax^{2}β(1 − β)}{1 + x} − \frac{2x}{1 − β} ]

+ \frac{1}{n^{2}} [ \frac{x^{2}a^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{ax(1 − β)^{2}}{1 + x} + \frac{3ax}{1 + x} + \frac{2}{1 − β} ],

U_{n,3}^{a,β} (x) = x^{3}(1 − β)^{3} − x^{3} − 3x^{3}(1 − β)^{2} + 3x^{3}(1 − β)

+ \frac{1}{n}[3x^{3}(1 − β)^{3} + 3x^{2}(1 − β)^{3} + \frac{3ax^{3}(1 − β)^{3}}{(1 + x)}

+ 6x^{2}(1 − β) − 3x^{3}(1 − β)^{2} − \frac{6ax^{3}(1 − β)^{2}}{(1 + x)}

− 3x^{2}(1 − β)^{2} − 9x^{2} + \frac{3ax^{3}(1 − β)}{(1 + x)} + \frac{3x^{2}}{(1 − β)} ]

+ \frac{1}{n^{2}} [2x^{3}(1 − β)^{3} + 3x^{2}(1 − β)^{3} + x(1 − β)^{3} + \frac{3ax^{3}(1 − β)^{3}}{(1 + x)}

+ \frac{6ax^{2}(1 − β)^{3}}{(1 + x)} + \frac{3a^{2}x^{3}(1 − β)^{3}}{(1 + x)^{2}} + 6x^{2}(1 − β)

+ 6x(1 − β) + \frac{12ax^{2}(1 − β)}{(1 + x)} + \frac{(11 − 8β)x}{(1 − β)}

− \frac{3a^{2}x^{3}(1 − β)^{2}}{(1 + x)^{2}} − \frac{3ax^{2}(1 − β)^{2}}{(1 + x)} − \frac{9ax^{2}}{(1 + x)} − \frac{6x}{(1 − β)} ]

+ \frac{1}{n^{3}} [ \frac{3a^{2}x^{2}(1 − β)^{3}}{(1 + x)^{2}} + \frac{ax(1 − β)^{3}}{(1 + x)} + \frac{a^{3}x^{3}(1 − β)^{3}}{(1 + x)^{3}}

+ \frac{6a^{2}x^{2}(1 − β)}{(1 + x)^{2}} + \frac{6ax(1 − β)}{(1 + x)} + \frac{(11 − 8β)ax}{(1 − β)(1 + x)} + \frac{6}{(1 − β)} ]

U_{n,4}^{a,β} (x) = x^{4}(1 − β)^{4} + x^{4} + 6x^{4}(1 − β)^{2} − 4x^{4}(1 − β) − 4x^{4}(1 − β)^{3}

+ \frac{1}{n}[6x^{4}(1 − β)^{4} + 6x^{3}(1 − β)^{4} + \frac{4ax^{4}(1 − β)^{4}}{1 + x}

+ 10x^{3}(1 − β)^{2} + 6x^{4}(1 − β)^{2} + \frac{12ax^{4}(1 − β)^{2}}{1 + x}

+ 6x^{3}(1 − β)^{2} + 18x^{3} − \frac{4ax^{4}(1 − β)}{1 + x} − \frac{4x^{3}}{1 − β}

− 12x^{4}(1 − β)^{3} − 12x^{3}(1 − β)^{3} − \frac{12ax^{4}(1 − β)^{3}}{1 + x} − 24x^{3}(1 − β)]

+ \frac{1}{n^{2}} [11x^{4}(1 − β)^{4} + 18x^{3}(1 − β)^{4} + 7x^{2}(1 − β)^{4} + \frac{12ax^{4}(1 − β)^{4}}{1 + x}

+ \frac{18ax^{3}(1−β)^{4}}{1 + x} + \frac{6a^{2}x^{4}(1−β)^{4}}{(1 + x)^{2}} + 30x^{3}(1−β)^{2} + 30x^{2}(1−β)^{2}

+ \frac{30ax^{3}(1−β)^{2}}{1 + x} + (35 − 20β)x^{2} + \frac{6a^{2}x^{4}(1−β)^{2}}{(1 + x)^{2}} + \frac{6ax^{3}(1−β)^{2}}{1 + x}

+ \frac{18ax^{3}}{1 + x} + \frac{12x^{2}}{1 − β} − 8x^{4}(1 − β)^{3} − 12x^{3}(1 − β)^{3} − 4x^{2}(1 − β)^{3}

− \frac{12ax^{4}(1 − β)^{3}}{1 + x} − \frac{24ax^{3}(1 − β)^{3}}{1 + x} − \frac{12a^{2}x^{4}(1 − β)^{3}}{(1 + x)^{2}}

− 24x^{3}(1−β) − 24x^{2}(1−β) − \frac{48ax^{3}(1−β)}{1 + x} − \frac{4(11 − 8β)x^{2}}{1 − β} ]

+ \frac{1}{n^{3}} [6x^{4}(1−β)^{4} + 12x^{3}(1−β)^{4} + 7x^{2}(1−β)^{4} + x(1−β)^{4} + \frac{8ax^{4}(1−β)^{4}}{1 + x}

+ \frac{18ax^{3}(1 − β)^{4}}{1 + x} + \frac{14ax^{2}(1 − β)^{4}}{1 + x} + \frac{3a^{2}x^{3}(x^{2} + 4x + 3)(1 − β)^{4}}{(1 + x)^{3}}

+ \frac{9a^{2}x^{3}(1 − β)^{4}}{(1 + x)^{2}} + \frac{4a^{3}x^{4}(1 − β)^{4}}{(1 + x)^{3}} + \frac{3a^{2}x^{4}(1 − β)^{4}}{(1 + x)^{2}} + 20x^{3}(1 − β)^{2}

+ 30x^{2}(1 − β)^{2} + 10x(1 − β)^{2} + \frac{30ax^{3}(1 − β)^{2}}{1 + x} + \frac{60ax^{2}(1 − β)^{2}}{1 + x}

+ \frac{30a^{2}x^{3}(1 − β)^{2}}{(1 + x)^{2}} + (35 − 20β)x^{2} + (35 − 20β)x + \frac{(70 − 40β)ax^{2}}{1 + x}

+ \frac{(50 − 30β)x}{1 − β} − \frac{12a^{2}x^{3}(1 − β)^{3}}{(1 + x)^{2}} − \frac{4ax^{2}(1 − β)^{3}}{1 + x} − \frac{4a^{3}x^{4}(1 − β)^{3}}{(1 + x)^{3}}

− \frac{24a^{2}x^{3}(1 − β)}{(1 + x)^{2}} − \frac{24ax^{2}(1 − β)}{1 + x} − \frac{4(11 − 8β)ax^{2}}{(1 − β)(1 + x)} − \frac{24x}{1 − β} ]

+ \frac{1}{n^{4}} [ \frac{7a^{2}x^{2}(1 − β)^{4}}{(1 + x)^{2}} + \frac{ax(1 − β)^{4}}{1 + x} + \frac{6a^{3}x^{3}(1 − β)^{4}}{(1 + x)^{3}} + \frac{a^{4}x^{4}(1 − β)^{4}}{(1 + x)^{4}}

+ \frac{30a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{10ax(1 − β)^{2}}{1 + x} + \frac{10a^{3}x^{3}(1 − β)^{2}}{(1 + x)^{3}}

+ \frac{(35 − 20β)a^{2}x^{2}}{(1 + x)^{2}} + \frac{(35 − 20β)ax}{1 + x} + \frac{(50 − 30β)ax}{(1 − β)(1 + x)} + \frac{24}{1 − β} ].

4. Direct estimates for general case

In this section, we establish the following direct results:

Proposition 4.1. Let f be a continuous function on [0, ∞) and β = β_{n} → 0 as n → ∞, the sequence {D_{n}^{a,β} (f, x)} converges uniformly to f(x) in [a, b] ⊂ [0, ∞).

Proof. Using Lemma 3.3, we have

D_{n}^{a,β} (e_{0}, x) = 1, D_{n}^{a,β} (e_{1}, x) = x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)}

and

D_{n}^{a,β} (e_{2}, x) = x^{2}(1 − β)^{2} + \frac{1}{n}[x^{2}(1 − β)^{2} + x(1 − β)^{2} + \frac{2ax^{2}(1 − β)^{2}}{1 + x} + 3x]

+ \frac{1}{n^{2}} [ \frac{a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{ax(1 − β)^{2}}{1 + x} + \frac{3ax}{1 + x} + \frac{2}{1 − β} ].

Obviously, D_{n}^{a,β} (e_{0}, x), D_{n}^{a,β} (e_{1}, x) and D_{n}^{a,β} (e_{2}, x) converges uniformly to 1, x and x^{2} 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 = β_{n} → 0 as n → ∞, it follows that

lim_{n→∞} n[D_{n}^{a,β_{n}} (f, x) − f(x)] = [1 + \frac{ax}{1 + x} ]f′(x) + [ \frac{x^{2} + 2x}{2} ]f′′(x).

Proof. By Taylor’s expansion of f, we have

(4.1)  f(t) = f(x) + f′(x)(t − x) + \frac{1}{2}f′′(x)(t − x)^{2} + r(t, x)(t − x)^{2},

where r(t, x) is the remainder term and lim_{t→x} r(t, x) = 0. Operating D_{n}^{a,β} to the equation (4.1), we obtain

D_{n}^{a,β} (f, x) − f(x) = D_{n}^{a,β} (t − x, x)f′(x) + D_{n}^{a,β} ((t − x)^{2}, x)\frac{f′′(x)}{2}
+ D_{n}^{a,β} (r(t, x)(t − x)^{2}, x)

Using the Cauchy–Schwarz inequality, we have

(4.2)  D_{n}^{a,β} (r(t, x)(t − x)^{2}, x) ⩽ \sqrt{D_{n}^{a,β} (r^{2}(t, x), x)} \sqrt{D_{n}^{a,β} ((t − x)^{4}, x)} .

As r^{2}(x, x) = 0 and r^{2}(t, x) ∈ C_{2}^{∗}[0, ∞), we have

(4.3)  lim_{n→∞} D_{n}^{a,β_{n}} (r^{2}(t, x), x) = r^{2}(x, x) = 0

uniformly with respect to x ∈ [0, A], for some A > 0. Now from (4.2), (4.3) and from Remark 3.2, we get lim_{n→∞} nD_{n}^{a,β_{n}} (r(t, x)(t − x)^{2}, x) = 0. Thus

lim_{n→∞} n(D_{n}^{a,β_{n}} (f, x) − f(x))

= lim_{n→∞} n[D_{n}^{a,β_{n}} (t−x, x)f′(x)+ \frac{1}{2}f′′(x)D_{n}^{a,β_{n}} ((t−x)^{2}, x)+D_{n}^{a,β_{n}} (r(t, x)(t−x)^{2}, x)]

= [1 + \frac{ax}{1 + x} ]f′(x) + [ \frac{x^{2} + 2x}{2} ]f′′(x). □

We denote the norm ∥f∥ = sup_{x∈[0,∞)} |f(x)|. For f ∈ C_{B}[0, ∞) and δ > 0, the m-th order modulus of continuity is defined as

ω_{m}(f, δ) = sup_{0⩽h⩽δ} sup_{x∈[0,∞)} |Δ_{h}^{m}f(x)|,

where Δ 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

K_{2}(f, δ) = inf_{g∈C_{B}^{2} [0,∞)} {∥f − g∥ + δ∥g′′∥ : g ∈ C_{B}^{2}[0, ∞)},

where

C_{B}^{2}[0, ∞) = {g ∈ C_{B}[0, ∞) : g′, g′′∈ C_{B}[0, ∞)}.

Theorem 4.2. Let f ∈ C_{B}[0, ∞) and 0 < β < 1, then

|D_{n}^{a,β} (f, x) − f(x)| ⩽ Cω_{2}(f, \sqrt{δ_{n}}) + ω( f, \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} − xβ)

where C is a positive constant and δ_{n} is given by

δ_{n} = 2x^{2}β^{2} + \frac{1}{n}[3x + x(x + 1)(1 − β)^{2} − \frac{4ax^{2}β(1 − β)}{1 + x} − \frac{2x}{1 − β} − \frac{2xβ}{1 − β} ]

+ \frac{1}{n^{2}} [ \frac{2a^{2}x^{2}(1 − β)^{2}}{(1 + x)^{2}} + \frac{5ax}{1 + x} + \frac{ax(1 − β)^{2}}{1 + x} + \frac{2}{1 − β} + \frac{1}{(1 − β)^{2}} ].

with β = β_{n} → 0 as n → ∞.

Proof. We introduce the auxiliary operators D̄_{n}^{a,β} : C_{B}[0, ∞) → C_{B}[0, ∞) by

(4.4) D̄_{n}^{a,β} (f, x) = D_{n}^{a,β} (f, x) − f( x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} ) + f(x).

These operators are linear and preserve the linear functions in view of Lemma 3.3. Let g ∈ C_{B}^{2}[0, ∞) and x, t ∈ [0, ∞). By Taylor’s expansion, we have

g(t) = g(x) + (t − x) g′(x) + ∫_{x}^{t}(t − u) g′′(u) du,

Hence,

|D̄_{n}^{a,β} (g, x) − g(x)|

⩽ D̄_{n}^{a,β} ( | ∫_{x}^{t}(t − u)g′′(u)du|, x) ⩽ D_{n}^{a,β} ( | ∫_{x}^{t}(t − u)g′′(u)du|, x)

+ | ∫_{x}^{x(1−β)+ \frac{ax(1−β)}{n(1+x)} + \frac{1}{n(1−β)}} ( x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} − u) g′′(u)du|

⩽ D_{n}^{a,β} ((t − x)^{2}, x)∥g′′∥

+ | ∫_{x}^{x(1−β)+ \frac{ax(1−β)}{n(1+x)} + \frac{1}{n(1−β)}} ( \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} − xβ) du|∥g′′∥

Next, using Remark 3.2, we have

(4.5) \begin{aligned}|D̄_{n}^{a,β} (g, x) − g(x)| & ⩽ [D_{n}^{a,β} ((t − x)^{2}, x) \\ & + ( \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} − xβ)^{2}]∥g′′∥ = δ_{n}∥g′′∥. \end{aligned}

Since

|D_{n}^{a,β} (f, x)| ⩽ ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(β)} (t), |f(t)|⟩}{⟨L_{n,k}^{(β)} (t), 1⟩}b_{n,k}^{a} (x) ⩽ ∥f∥ ∑_{k=0}^{∞} \frac{⟨L_{n,k}^{(β)} (t), 1⟩}{⟨L_{n,k}^{(β)} (t), 1⟩}b_{n,k}^{a} (x) ⩽ ∥f∥.

Now by (4.4), we have

(4.6)  |∥D̄_{n}^{a,β} (f, x)|| ⩽ ∥|D_{n}^{a,β} (f, x)∥ + 2∥f∥ ⩽ 3∥f∥, f ∈ C_{B}[0, ∞).

Using (4.4), (4.5) and (4.6), we have

|D_{n}^{a,β} (f, x) − f(x)| ⩽ |D̄_{n}^{a,β} (f − g, x) − (f − g)(x)| + |D̄_{n}^{a,β} (g, x) − g(x)|

+ |f( x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} ) − f(x)|

⩽ 4∥f − g∥ + δ_{n}∥g′′∥

+ |f( x(1 − β) + \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} ) − f(x)|

⩽ C\{∥f − g∥ + δ_{n}∥g′′∥\} + ω( f, \frac{ax(1 − β)}{n(1 + x)} + \frac{1}{n(1 − β)} − xβ) .

Taking infimum over all g ∈ C_{B}^{2}[0, ∞), and using the inequality

K_{2}(f, δ) ⩽ Cω_{2}(f, \sqrt{δ}), δ > 0

due to [4], we get the desired assertion. □

References

1. P. N. Agrawal, V. Gupta, A. S. Kumar, A. Kajla, Generalized Baskakov–Szász type operators, Appl. Math. Comput. 236 (2014), 311–324.

2. A. Aral, T. Acar, Weighted approximation by new Bernstein–Chlodowsky–Gadjiev operators, Filomat 27(2) (2013), 371–380.

3. S. S. Dragomir, Approximation of continuous linear functionals in real normed spaces, Rend. Mat. Appl., VII. Ser. 12(2) (1992), 357–364.

4. R. A. DeVore, G. G. Lorentz, Constructive Approximation, Springer, Berlin, 1993.

5. Z. Ditzian, V. Totik, Moduli of Smoothness, Springer-Verlag, New York, 1987.

6. J. L. Durrmeyer, Une formule d’ inversion de la Transformee de Laplace: Applications a la Theorie des Moments, These de 3e Cycle, Faculte des Sciences de l’ Universite de Paris, 1967.

7. A. Erencin, Durrmeyer type modification of generalized Baskakov operators, Appl. Math. Comput. 218(3) (2011), 4384–4390.

8. N. K. Govil, V. Gupta, Simultaneous Approximation for Stancu-Type Generalization of Certain Summation Integral-Type Operators, in: G. V. Milovanović and Th. M. Rassias (eds.) Analytic Number Theory, Approximation Theory, and Special Functions, Springer, 2014, 531– 548.

9. V. Gupta, R. P. Agarwal, Convergence Estimates in Approximation Theory, Springer, 2014.

10. V. Gupta, R. P. Agarwal, D. K. Verma, Approximation for a new sequence of summationintegral type operators, Adv. Math. Sci. Appl. 23(1) (2013), 35–42.

11. V. Gupta, J. Sinha, Simultaneous Approximation for Generalized Baskakov–Durrmeyer-Type Operators, Mediterr. J. Math. 4(4)(2007), 483–495.

12. V. Gupta, G. S. Srivastava, Simultaneous approximation by Baskakov–Szász type operators, Bull. Math. Soc. Sci. Math. Roum., Nouv. Sér. 37(85)(3–4) (1993), 73–85.

13. G. C. Jain, Approximation of functions by a new class of linear operators, J. Austral. Math. Soc. 13(3) (1972), 271–276.

14. M.-J. Luo, G. V. Milovanović, P. Agarwal, Some results on the extended beta and extended hypergeometric functions, Appl. Math. Comput. 248 (2014), 631–651.

15. V. Mihesan, Uniform approximation with positive linear operators generated by generalized Baskakov method, Automat. Comput. Appl. Math. 7(1) (1998), 34–37.

16. V. Gupta, G. C. Greubel, Moment Estimations of New Szász–Mirakyan–Durrmeyer Operators, Appl. Math. Comput. 271 (2015), 540–547.

17. N. Ispir, On modified Baskakov operators on weighted spaces, Turk. J. Math. 26(3) (2001), 355–365.

18. H. M. Srivastava, V. Gupta, A certain family of summation-integral type operators, Math. Comput. Modelling 37(12)(2003), 1307–1315.

Department of Mathematics, (Received 12 01 2015) Netaji Subhas Institute of Technology (Revised 04 07 2015) New Delhi India

    vijaygupta2001@hotmail.com
    neha.malik_nm@yahoo.com