EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 2, 2018, 400-409
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Local approximation results for Stancu variant of
modified Szász-Mirakjan operators

Ankita R Devdhara1, Vishnu Narayan Mishra2,∗

1 Applied Mathematics and Humanities Department, Sardar Vallabhbhai National Institute
of Technology, Surat 395 007, India
2 Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkan-
tak 484 887, Madhya Pradesh, India

Abstract. The aim of this paper is to obtain local approximation results for Stancu type gen-
eralization of modified Szász-Mirakjan operators. First, we calculate moments of the operators.
Some direct results of the operators are investigated. The rate of convergence of the operators is
evaluated. In last section of the paper, the Voronovskaya type result is obtained.

2010 Mathematics Subject Classifications: 41A10, 41A25, 41A36
Key Words and Phrases: Szász-Mirakjan operators, Modulus of continuity, Rate of Conver-
gence, Peetre K -functional

1. Introduction

In 1950, Otto and Mirakjan [5] introduced Szász-Mirakjan operators; generalization of
Bernstein operators defined by

Mn(f)(x) = e−nx
∞∑
k=0

(nx)k

k!
f

(
k

n

)
. (1)

In his paper D.D. Stancu [3] introduced a positive linear polynomial type operators defined
by

Bα,β
n (f)(x) =

n∑
k=0

(
n

k

)
xk (1− x)n−k f

(
k + α

n+ β

)
. (2)

where, 0 ≤ α ≤ β and 0 ≤ x ≤ 1.
Walczak [13] investigated generalization of Szász-Mirakjan operators defined by

Sn[f ; an, bn, q, x] =
∞∑
k=0

san,k(x)f

(
k

bn + q

)
., (3)

∗Corresponding author.

Email addresses: krishna.devdhara@gmail.com (A.R. Devdhara), vishnunarayanmishra@gmail.com
(V.N. Mishra)

http://www.ejpam.com 400 c© 2018 EJPAM All rights reserved.



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 401

where san,k(x) = e−anx (anx)k

k! , for k = 0, 1, 2, ...; q ≥ 0 is a fixed number, (an)∞1 and
(bn)∞1 are increasing and unbounded sequences such that 1 ≤ an ≤ bn, and (an/bn)∞1 is
non-decreasing and

an
bn

= 1 + o

(
1

bn

)
Walczask [13] derived pointwise and uniform convergence of the operators (3) in expo-
nential weight space. There are some other linear positive operators with Stancu type
modification, e.g. [1], [10], [11], [12].
Recently, Gandhi and Mishra [4] introduced modification of Szász-Mirakjan operators (1)
given by

Sn(f ;x) =
∞∑
k=0

e−bnx
(bnx)k

k!
f

(
k

bn

)
, (4)

where (bn)∞1 is an increasing sequence of positive real numbers, bn →∞ as n→∞, b1 ≥ 1.
For bn = n, we get the operators defined in (1).
Gandhi and Mishra discussed local and global approximation results of the operators (4)
in polynomial weighted space of polynomials. Indeed, the rapid development has led to
the discovery of new generalizations of approximation operators (one may refer to [2],[7],
[8], [9], [15]).
In second section of this paper, we introduce our Stancu variant of modified Szász-Mirakjan
operators and we evaluate moments of our operators. The uniform convergence of the
operators is derived. In third section, we discuss local approximation result and rate of
convergence of the operators. In the last section, we derive Voronovskaya type result for
the operators.

2. Construction of the operators

Motivated by Gandhi and Mishra [4], we introduce Stancu type generalization of mod-
ified Szász-Mirakjan operators for f ∈ C[0,∞) as follows:

Sα,βn (f ;x) =
∞∑
k=0

(bnx)k

k!
e−bnxf

(
k + α

bn + β

)
, (5)

where, 1/bn → 0 as n → ∞, bn ≥ 1. For α = β = 0 we get the modified Szász-Mirakjan
operators.
Now, we calculate moments of our operators (5).

Lemma 2.1. Let ej(t) = tj for j = 0, 1, 2, the followings are true:

Sα,βn (1;x) = 1, (6)

Sα,βn (t;x) =
bnx+ α

bn + β
, (7)

Sα,βn (t2;x) =
b2nx

2

(bn + β)2
+

(1 + 2α)bn
(bn + β)2

x+
α2

(bn + β)2
. (8)



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 402

Proof. For i = 0, the result is obvious.
For i = 1,

Sα,βn (t;x) =
∞∑
k=0

(bnx)k

k!
e−bnx

(
k + α

bn + β

)

=
1

bn + β

∞∑
k=0

(bnx)k

k!
e−bnxk +

α

bn + β
Sn(1;x)

=
bnx+ α

bn + β
.

For i = 2,

Sα,βn (t2;x) =
∞∑
k=0

(bnx)k

k!
e−bnx

(
k + α

bn + β

)2

=
1

(bn + β)2

∞∑
k=0

(bnx)k

k!
e−bnxk2 +

2α

(bn + β)2

∞∑
k=0

(bnx)k

k!
e−bnxk +

α2

(bn + β)2

=
b2nx

2

(bn + β)2
+

(1 + 2bnα)

(bn + β)2
bnx+

α2

(bn + β)2
.

Hence, lemma is proved.

Lemma 2.2. The central moments Φα,β
m (x) = Sα,βn ((t−x)m;x) for m = 1, 2 are as follows:

Φα,β
1 (x) =

bnx+ α

bn + β
− x, (9)

Φα,β
2 (x) =

(
bn

bn + β
− 1

)2

x2 +

(
(1 + 2α)bn
(bn + β)2

− 2α

bn + β

)
x+

α2

(bn + β)2
. (10)

Proof. Using lemma:(2.1) we get the result.

Now, we obtain the uniform convergence of the operators Sα,βn to f , f ∈ Cξ[0,∞),

Cξ[0,∞) = {f ∈ C[0,∞) : |f(x)| ≤M(1 + t)ξ}

for M > 0, ξ > 0.

Theorem 2.3. Sα,βn (f ;x) converges uniformly to f(x) for 0 ≤ x ≤ a, f ∈ Cξ[0,∞), ξ ≥
2, a > 0.

Proof. Using Korovkin theorem, it is sufficient to show that

lim
n→∞

‖Sα,βn (tj ;x)− xj‖Cξ[0,∞) = 0



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 403

for j = 0, 1, 2.
The result is trivial for the case j = 0 using (6). For j = 1, the result can be obtained
using (7), as follows:

lim
n→∞

‖Sα,βn (t;x)− x‖Cξ[0,∞) = lim
n→∞

∥∥∥∥bnx+ α

bn + β
− x
∥∥∥∥
Cξ[0,∞)

= 0.

Finally, for j = 2, using (8), we get

lim
n→∞

‖Sα,βn (t2;x)− x2‖Cξ[0,∞) = lim
n→∞

∥∥∥∥ b2nx
2

(bn + β)2
+

(1 + 2α)bn
(bn + β)2

x+
α2

(bn + β)2
− x2

∥∥∥∥
Cξ[0,∞)

= lim
n→∞

∥∥∥∥ b2nx
2

(bn + β)2
− x2

∥∥∥∥
Cξ[0,∞)

= 0.

3. Direct result

In this section, we give some local results for the operators. Let CB[0,∞) be the space
of all real valued continuous bounded functions defined on [0,∞). The norm on the space
CB[0,∞) is the supremum norm ‖f‖ = sup

x∈[0,∞)
|f(x)|. Further, Peetre’s K -functional is

defined by

K2(f, δ) = inf
g∈W 2

{‖f − g‖+ δ‖g′′‖},

here W 2 = {g ∈ CB[0,∞) : g′, g′′ ∈ CB[0,∞)}, By [14] there exists a positive constant
C > 0 such that K2(f, δ) ≤ Cω2(f, δ

1/2), δ > 0, where

ω2(f, δ
1/2) = sup

0<h<δ1/2
, x ∈ [0,∞)|f(x+ 2h)− 2f(x+ h) + f(x)|

is the second order modulus of continuity of function f ∈ CB[0,∞). Also. for f ∈ CB[0,∞)
the usual modulus of continuity is given by

ω(f, δ1/2) = sup
0<h<δ1/2,x∈[0,∞)

|f(x+ h)− f(x)|.

Theorem 3.1. Let f ∈ CB[0,∞), then for all n ∈ N, there exists an absolute constant
C > 0 such that

|Sα,βn (f ;x)− f(x)| ≤ Cω2(f, δn(x)) + ω(f, αn(x)), (11)

where



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 404

δn(x) = [Sα,βn ((t− x)2;x) + (Sα,βn ((t− x);x))2]1/2

and

αn(x) =

∣∣∣∣bnx+ α

bn + β
− x
∣∣∣∣.

Proof. For 0 ≤ x <∞, we consider the auxiliary operators Ŝα,βn (f ;x) defined by

Ŝα,βn (f ;x) = Sα,βn (f ;x) + f(x)− f(
bnx+ α

bn + β
).

Using above operators and (9), we get

Ŝα,βn (t− x;x) = Sα,βn (t− x;x)− (
bnx+ α

bn + β
− x)

= 0.

Now, 0 ≤ x <∞ and g ∈W 2. Applying Taylor’s formula, we get

g(t) = g(x) + (t− x)g′(x) +

∫ t

x
(t− u)g′′(u)du.

Applying Ŝα,βn on the both sides of the above equation, we obtain

Ŝα,βn ((g;x)− g(x)) = Ŝα,βn ((t− x)g′(x);x) + Ŝα,βn

(∫ t

x
(t− u)g′′(u)du;x

)
= g′(x)Ŝα,βn ((t− x);x) + Sα,βn

(∫ t

x
(t− u)g′′(u)du;x

)
−
∫ bnx+α

bn+β

x

(
bnx+ α

bn + β
− u
)
g′′(u)du

= Sα,βn

(∫ t

x
(t− u)g′′(u)du;x

)
−
∫ bnx+α

bn+β

x

(
bnx+ α

bn + β
− u
)
g′′(u)du.

Also, ∣∣∣∣ ∫ t

x
(t− u)g′′(u)du

∣∣∣∣ ≤ ∫ t

x
|t− u||g′′(u)|du ≤ ‖g′′‖

∫ t

x
|t− u|du ≤ (t− x)2‖g′′‖.

and ∣∣∣∣ ∫ bnx+α
bn+β

x

(
bnx+ α

bn + β
− u
)
g′′(u)du

∣∣∣∣ ≤ (bnx+ α

bn + β
− x
)2

‖g′′‖.

Therefore, we can conclude that

|Ŝα,βn ((g;x)− g(x))| =
∣∣∣∣Sα,βn

(∫ t

x
(t− u)g′′(u)du;x

)∣∣∣∣



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 405

+

∣∣∣∣ ∫ bnx+α
bn+β

x

(
bnx+ α

bn + β
− u
)
g′′(u)du

∣∣∣∣
≤ ‖g′′‖Sα,βn ((t− x)2;x) +

(
bnx+ α

bn + β
− x
)2

‖g′′‖

= δ2n‖g′′‖.

Also, we get

|Ŝα,βn ((g;x)| ≤ |Sα,βn (f ;x)|+ 2‖f‖ ≤ 3‖f‖.

Therefore,

|Sα,βn (f ;x)− f(x)| ≤ |Ŝα,βn ((f − g);x)− (f − g)(x)|+ |Ŝα,βn ((g;x)− g(x))|

+

∣∣∣∣f(x)− f(
bnx+ α

bn + β
)

∣∣∣∣
≤ 4‖f − g‖+ ‖g′′‖δ2n(x) + ω

(
f :

∣∣∣∣bnx+ α

bn + β
− x
∣∣∣∣).

Hence, taking the infimum on the right hand side over all g ∈W 2, we obtain

|Sα,βn (f ;x)− f(x)| ≤ 4K2(f, δ
2
n(x)) + ω(f, αn(x)).

By using property of K -functional, we have

|Sα,βn (f ;x)− f(x)| ≤ Cω2(f, δn(x)) + ω(f, αn(x)).

Hence the result is obtained.

Now, we consider the following class of functions:
Hx2 [0,∞) = {f : [0,∞)→ R : |f(x)| ≤Mf (1 + x2) here Mf is constant depending on the
function f},
Cx2 [0,∞) = {f ∈ Hx2 [0,∞) : f is continuous},
C∗x2 [0,∞) = {f ∈ Cx2 [0,∞) : lim|x|→∞

f(x)
(1+x2)

is finite}.
The norm on the space C∗x2 [0,∞)∗ is defined by ‖f‖x2 = supx∈[0,∞) |

f(x)
1+x2
|.

We denote the modulus of continuity of f on closed interval [0, a], a > 0 by:

ωa(f ; δ) = sup|t−x|≤δ,x,t∈[0,a] |f(t)− f(x)|.

Theorem 3.2. For f ∈ Cx2 [0,∞);ωa(f ; δ) be its modulus of continuity on the interval
[0, a+ 1] ⊂ [0,∞), a > 0, we have

‖Sα,βn (f ;x)− f(x)‖ ≤ 6Mf (1 + a2)λn + 2ωa+1(f ;
√
λn),

here

λn =

(
1− b2n

(bn + β)2

)
a2 +

(
bn − 2αβ

bn + β

)
a

bn + β
+

α2

(bn + β)2
.



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 406

Proof. For 0 ≤ x ≤ a and t ≥ 0 , we have [6]

|f(t)− f(x)| ≤ 6Mf (1 + a2)(t− x)2 + ωa+1(f ; δn)

(
|t− x|
δn

+ 1

)
.

Applying above inequality and Cauchy-Schwarz inequality, we have

‖Sα,βn (f(t);x)− f(x)‖C[0,a] ≤ Sα,βn (|f(t)− f(x)|;x)

≤ 6Mf (1 + a2)Sα,βn ((t− x)2;x)

+ ωa+1(f ; δn)

(
1 +

1

δ2n
Sα,βn ((t− x)2;x

)1/2

.

For 0 ≤ x ≤ a, using lemma (2.2),

Sα,βn ((t− x)2;x) =

(
bn

bn + β
− 1

)2

x2 +

(
(1 + 2α)bn
(bn + β)2

− 2α

bn + β

)
x+

α2

(bn + β)2

≤
(

bn
bn + β

− 1

)2

a2 +

(
(1 + 2α)bn
(bn + β)2

− 2α

bn + β

)
a+

α2

(bn + β)2

≤
(

1− b2n
(bn + β)2

)
a2 +

(
bn − 2αβ

bn + β

)
a

bn + β
+

α2

(bn + β)2

= λn.

Taking δn =
√
λn, we will get the theorem.

4. Voronovskaya type result

Theorem 4.1. For f ∈ Cξ[0,∞) such that f ′, f ′′ ∈ Cξ[0,∞), we have

lim
n→∞

bn[Sα,βn (f ;x)− f(x)] = (α− βx)f ′(x) +
x

2
f ′′(x). (12)

where 0 ≤ x ≤ a, a > 0.

Proof. From Taylor’s formula, we have

f(t) = f(x) + (t− x)f ′(x) + 1
2(t− x)2f ′′(x) + (t− x)2r(t, x),

here, r(t, x) is reminder term and lim
t→x

r(t, x) = 0.

Therefore,

bn[Sα,βn (f ;x)− f(x)] = bnf
′(x)Sα,βn ((t− x);x)

+ bn
f ′′(x)

2
Sα,βn ((t− x)2;x)

+ bnS
α,β
n (r(t, x)(t− x)2;x).

By the Cauchy-Schwarz inequality, we get



A.R. Devdhara, V.N. Mishra / Eur. J. Pure Appl. Math, 11 (2) (2018), 400-409 407

Sα,βn (r(t, x)(t− x)2;x) ≤
√
Sα,βn (r2(t, x);x)

√
Sα,βn (t− x)4;x).

As r(t, x) ∈ Cξ[0,∞), therefore by Theorem (2.3) and from the fact that limt→x r(t, x) = 0,
we obtain

lim
n→∞

Sα,βn (r2(t, x);x) = r2(x, x) = 0.

Therefore,

lim
n→∞

bn[Sα,βn (f ;x)− f(x)] = lim
n→∞

bnf
′(x)Sα,βn ((t− x);x)

+ lim
n→∞

bn
f ′′(x)

2
Sα,βn ((t− x)2;x).

Now,

lim
n→∞

bnS
α,β
n ((t− x);x) = limn→∞bn

(
bnx+ α

bn + β
− x
)

= lim
n→∞

bn

(
bn

bn + β
− 1

)
x+ lim

n→∞
bn

(
α

bn + β

)
= α− βx,

and

lim
n→∞

bnS
α,β
n ((t− x)2;x) = lim

n→∞
bn

((
bn

bn + β
− 1

)2

x2

+

(
(1 + 2α)bn
(bn + β)2

− 2α

bn + β

)
x+

α2

(bn + β)2

)
= x.

Hence, from above equations, we have

lim
n→∞

bn[Sα,βn (f ;x)− f(x)] = (α− βx)f ′(x) +
x

2
f ′′(x).

Hence the theorem is proved.

Acknowledgements

The authors would like to express their deep gratitude to the anonymous learned
referee(s) and the editor for their valuable suggestions and constructive comments, which
resulted in the subsequent improvement of this research article.



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