A NOTE ON STATISTICAL APPROXIMATION PROPERTIES OF COMPLEX Q-SZÂSZ- MIRAKJAN OPERATORS

C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat.

Volum e 68, N umb er 1, Pages 457–465 (2019) D O I: 10.1501/C om mua1_ 0000000921 ISSN 1303–5991 E-ISN N 2618-6470

http://com munications.science.ankara.edu.tr/index.php?series= A 1

A NOTE ON STATISTICAL APPROXIMATION PROPERTIES

OF COMPLEX Q-SZÁSZ- MIRAKJAN OPERATORS

DIDEM AYDIN ARI

Abstract. The complex q Szász-Mirakjan operator attached to analytic func- tions satisfying a suitable exponential type growth condition has been studied in [14]. In this paper, we consider the A-statistical convergence of complex q-Szász- Mirakjan operator.

1. Introduction

In 1996, Phillips de…ned a generalization of the Bernstein operators called q Bernstein operators by using the q binomial coe¢ cients and the q binomial the- orem [21]. In 2008, Aral introduced q Szász-Mirakjan operators and studied some approximation properties of them [12]. In 2008, Gal studied some approximation results of the complex Favard-Szász-Mirakjan operators on compact disks [17]. A di¤erent type complex q Szász-Mirakjan operator was introduced by Mahmudov in [20] for q > 1 as

Mn;q(f ; z) = P¹

k=0

f [k]

[n]

1 q^{k(k 1)=2}

[n]^kz^k

[k]! "q [n] q ^kz (1.1) for the functions which are continuous and bounded on [0; 1).

In this paper, we study some operators by taking statistical convergence instead of ordinary convergence. In 2002, Gadjiev and Orhan gave some approximation results by using statistical convergence [1]. And several authors have studied in approximation theory by using statistical convergence concept (see [3], [4] and [5]

[6], [7], [8], [13], [19]).

Now, we give some notations on q analysis given in [16] and [21].

The q integer [n] is de…ned by

Received by the editors: December 13, 2017; Accepted: January 31, 2018.

2010 Mathematics Subject Classi…cation. 40A35, 30E10.

Key words and phrases. A-Statistical approximation, complex q-Szász- Mirakjan operators, q-exponential functions, q-derivative.

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457

[n] := [n]_q =

1 qⁿ 1 q; q 6= 1

n; q = 1 for q > 0 and the q factorial [n]! by

[n]! := [1]_q[2]_q::: [n]_q; n = 1; 2; :::

1; n = 0:

We give the following two q analogues of the exponential function e^x which is appeared in the de…nition of the operator :

"_q(x) = X1 n=0

1

[n]_q!xⁿ= 1

((1 q)x; q)₁; jxj < 1

1 q; jqj < 1; (1.2)

Eq(x) = X1 n=0

q^{n(n2 1)}

[n]_q! xⁿ= ( (1 q)x ; q)₁ ; x 2 R; jqj < 1; (1.3) where (x; q)₁= Q¹

k=1

(1 xq^{k 1}) (see [15]).

It is clear from (1.2) and (1.3) that "q(x)Eq( x) = 1 and

qlim!1 "q(x) = lim

q!1 Eq(x) = e^x:

Let q 2 (0; 1) [ (1; 1): The q derivative of a function f (x) is de…ned as

Dqf (x) := f (x) f (qx)

(1 q)x for; x 6= 0:

D_qf (0) = lim

x!0D_qf (x); where D⁰_qf := f; D_qⁿf := D_q(D^{n 1}_q f ); n = 1; 2; :::

As a consequence of the de…nition of D_qf; we …nd Dqxⁿ = [n]_qx^{n 1}; D_q"_q(ax) = a"_q(ax);

D_qE_q(ax) = aE_q(qax):

Also the formula for the q-di¤erential of a product is

Dq(u(x)v(x)) = Dq(u(x))v(x) + u(qx)Dq(v(x)):

We know that

Dq(t; x)ⁿ_q (t) = [n]_q(t; x)^{n 1}_q ; where (t; x)ⁿ_q =

n 1Q

k=0

(t xq^k) (see [16]).

Now we de…ne the complex Szász-Mirakjan operator based on q integers.

Suppose that R_n;q := _{[n](1 q)}^bn ; where (b_n) is a sequence of positive numbers such that lim

n!1bn = 1 and that D^R = fz 2 C : jzj < Rg ; 1 < R < R^n;q: The complex Szász-Mirakjan operator based on q integers is obtained directly from the real version (see [12]) by taking z in place of x, namely

S_n^q(f ; z) = Sn(f ; q; z) (1.4)

= : Eq [n] z bn

P1 k=0

f [k]

[n]bn

([n] z)^k [k]! (bn)^k;

where n 2 N; 0 < q < 1 and f : [R; 1) [ DR! C has exponential growth and it has an analytical continuation into an open disk centered at the origin.

Throughout the paper we call the operator (1.4) as complex q Szász-Mirakjan operator.

It is clear that by using divided di¤erences S_n^q(f ; z) can be expressed as S_n^q(f ; z) = Sn(f; q; z) = P¹

j=0

q^j(j21)f 0;bn[1]

[n] ; :::;bn[j]

[n] z^j; (1.5) similar to the real version of the q Szász-Mirakjan operators (see [12]), where fh

0;^bn_[n]^[1]; :::;^bn_[n]^[j]i

denotes the divided di¤erence of f on the knots 0;^bn_[n]^[1]; :::;^bn_[n]^[j]:

2. Statistical Convergence of S_n^qn(f ; z)

First of all, we recall some de…nitions and notations which we use in this study.

Let A = (ajn) be a nonnegative regular matrix. The A density of K N given by

A(K) := lim

j

P1 n2K

ajn;

whenever the limit exists. A sequence x = (xn) is called A statistically convergent to a number L if for every " > 0;

A(fn 2 N : jxⁿ Lj "g) = 0: (2.1) It is not di¢ cult to see that (2.1) is equivalent to

jlim!1

P1 n:jxn Lj "

ajn= 0; for every " > 0:

This limit expression is denoted by stA lim xn = L ( see in [2], [9],[10], [11]).

Now, we give a lemma which we use in the proof of Theorem 1.

Lemma 1. Let D_R= fz 2 C : jzj < Rg ; 1 < R < Rn;q; where Rn;q = _[n]^bn

q(1 q) and

f : [R; 1) [ D^R! C

be continuous in [R; 1) [ D^R, analytic in DR, namely f (z) = P¹

k=0

ckz^k for all z 2 D^R and there exist M; C; B > 0 and A 2 R¹; 1 , with the property jc^kj ^{M A}k!^k

for all k = 0; 1; :::(which implies jf(z)j M e^Ajzj for all z 2 D^Rand jf(x)j Ce^Bx for all x 2 [R; 1)): Then Sn^q(f ; z) is well de…ned and analytic as function of z in DR (see [14]).

Theorem 1. Suppose that the conditions of Lemma 1 are satis…ed. Suppose also that A be a nonnegative regular summability matrix, q = q_n is a sequence such that 0 < q_n< 1 and st_A lim q_n = 1 and st_A lim_[n]^bn

qn = 0:

i. Let 1 r < _B¹ be arbitrary …xed. Then for all jzj r, we have st_A lim jSn^qn(f ; z) f (z)j = 0:

ii. For the simultaneous approximation by complex q Szász-Mirakjan operator, we have

stA lim D^(p)_qn (S_n^qn(f ; z)) D^(p)_qnf (z) = 0 where Cr1;A is given as in the case (i) :

Proof. i. From [14], by taking e_k(z) = z^k; it is clear that T_n;k(z) := S_n^qn(e_k; z) is a polynomial of degree k, k = 0; 1; 2; ::: and

Tn;0(z) = 1; Tn;1(z) = z for all z 2 C Also, using q derivative of Tn;k(z) for z 6= 0, we get

D_qT_n;k(z)

= [n]_qn

zb_n Tn;k+1(z) [n]_qn

bn

E_q [n]_q

n q_n z bn

P1 j=0

[j]qn

[n]_qnb_n

!k

[n]_qnqnz ^j

[j]qn! (b_n)^j (2.2) for all z 2 C; k = 0; 1; 2; :::: Therefore, we obtain

Tn;k(z) = zTn;k 1(qnz) + zbn

[n]_q

n

Dq(Tn;k 1(z)) : The last equality implies that

T_n;k(z) z^k

= zbn

[n]_qnDq Tn;k 1(z) z^{k 1} + z Tn;k 1(qnz) (qnz)^{k 1} +[k 1]

qn

[n]_q

n

b_nz^{k 1} z^k(1 q_n) [k 1]_qn: (2.3)

From the Bernstein inequality in D_r= fz 2 C: jzj rg ; we have jDq(P_k(z)j kPk⁰k k

rkPkkr; (2.4)

where k:kr= max

z2D^r jf(z)j (see [18, p. 55]): From (2.3) and (2.4), we obtain that Tn;k(z) z^k

rb_n

[n]_qn T_{n;k 1}(z) z^{k 1} _rk 1 r

+r Tn;k 1(qnz) (qnz)^{k 1} + r^{k 1}[k 1]_qn

[n]_qn bn+ r^k[k 1]_qnj1 qnj : By passing to norm we reach to

Tn;k(z) z^k

r

(k 1)bn

[n]_q

n

+ r

!

T_{n;k 1}(z) z^{k 1} _r+ r^kk 1 q_n+ bn

[n]_q

n

! : By using mathematical induction with respect to k; the above recurrence formula gives that

T_n;k(z) z^k

r

(k + 1)!r^k

2 1 q_n+ b_n [n]_qn

!

for all k 2 and …xed an arbitrary n n0: There exists an n0 such that for all n > n0; then _[n]^bn

qn < 1: Assume that it is true for k: Since [k]_qn (k + 1) is satis…ed for all 0 < qn< 1; the recurrence formula reduces to

Tn;k+1(z) z^k+1

r

1 q_n+ b_n [n]_qn

!r^k+1 2

(

(k + 1)!k b_n

[n]_qn + (k + 1)! + 2(k + 1) )

for all k 2 and for all n > n₀: By this inequality, it follows Tn;k+1(z) z^k+1

r

(k + 2)!

2 r^k+1 1 qn+ bn

[n]_qn

! : for k 2 and for all n > n0:

Now, we show that

S_n^qn(f ; z) = P¹

k=0

c_kS_n^qn(e_k; z) = P¹

k=0

c_kT_n;k(z) (2.5) for all z 2 DR: For any m 2 N, let us de…ne

f_m(z) = Pm j=0

c_jz^j if jzj r < R and f_m(x) = f (x) if x 2 (r; 1):

From the hypothesis on f; it is clear that for any m 2 N, jfm(x)j C_me^Bmx for all x 2 [0; 1): Ratio test implies that for each …xed m; n 2 N and z;

jSn^qn(f_m; z)j C_m E_q [n]_qn z b_n

P1 k=0

[n]_qn

k

jzj^k [k]_q

n! (b_n)^k e^Bm

[k]qn [n]qnbn

< 1:

Therefore, S_n^qn(fm; z) is well de…ned. Now, we set

f_m;k(z) = c_ke_k(z) if jzj r and f_m;k(x) = f (x)

m + 1 if x 2 (r; 1):

It is clear that each f_m;k is of exponential growth on [0; 1) and that f_m(z) =

Pm k=0

f_m;k(z):

Since S_n^qn is linear, we have S^q_nⁿ(fm; z) =

Pm k=0

ckS^q_nⁿ(ek; z) for all jzj r;

which proves that

mlim!1S_n^qn(f_m; z) = S_n^qn(f ; z) for any …xed n 2 N and jzj r: But this is immediate from

mlim!1kfm f kr= 0 and from the inequality

jSn^qn(fm) S_n^qn(f )j Eq [n]_qn z bn

"q [n]_qnjzj

bn kf^m f kr

Mr;nkf^m f kr,

for all jzj r: Consequently the statement (2.5) is satis…ed.

In this way, from the hypothesis on ck, this implies for all jzj r jSn^qn(f ; z) f (z)j

1 qn+ bn

[n]_qn

!

Cr;B; (2.6)

where

Cr;B=M A 2

P1 k=2

(k + 1) (rA)^{k 1} is …nite for all 1 r < _B¹: Note that the series P¹

k=2

u^k+1and its derivative P1

k=2

(k + 1)u^k are uniformly and absolutely convergent in any compact disk included in the open unit disk.

As st_A lim q_n= 1 there exists n₁(") and K₁ N of density 1 such that 1 qn< "

for all n 2 K1 and n > n₁("): On the other hand, since st_A lim_[n]^bn

qn = 0 there exists n2(") and K2 N of density 1 such that _[n]^bn

qn < " for all n 2 K² and n > n2("): Now de…ne K = K1\ K²: Note that A(K1\ K²) = 1 and for all " > 0 and for all n > n0(") = max fn¹; n2g

1 qn+ bn

[n]_qn < " (2.7)

Hence (2.7) and (2.6) imply that

stA lim jSn^qn(f ; z) f (z)j = 0:

ii. Let be the circle of radius r1 > r with centered 0, since for any jzj r and v 2 ; we have jv zj r1 r; by Cauchy’s formulas it follows that for all jzj r and n 2 N

D^(p)_qn (S_n^qn(f ; z)) D^(p)_qnf (z) Sn^q(p)n (f ; z) f^(p)(z)

= p!

2

Z S_n^qn(f ; v) f (v) (v z)^p+1 dv

1 qn+ bn

[n]_qn

! Cr1;B

p!

2

2 r1

(r₁ r)^p+1

= 1 qn+ b_n [n]_qn

! Cr1;B

p!r₁ (r₁ r)^p+1: Similarly we get from hypothesis that for all " > 0 there exists a subset K N of density 1 and n0= n0(") such that D^(p)_qn (Sn^qn(f ; z)) D^(p)_qnf (z) < " for all n > n0

and n 2 K:

Remark 1. Consider the matrix method C = (c_jn) which is called Cesåro matrix and de…ned as

c_jn=

1 j; 0;

n j

otherwise :

In this case A-statistical convergence reduces to statistical convergence. Now de…ne a sequence q = (q_n) as

q_n =

1 n;

n n+1;

n = m² (m 2 N) otherwise :

It is obvious that q is not convergent but it is statistically convergent to 1:

Remark 2. Let d_n be a sequence of positive numbers such that d_n ! 1 and

dn

[n]_qn ! 0 as n ! 1. Note that the sequence de…ned as bn = n;

dn;

n = m² (m 2 N) otherwise ;

bn

[n]_qn is statistically convergent to zero.

Note that these examples do not satisfy the hypothesis of Theorem 2.3 in [14], but they satisfy the hypothesis of our theorem.

Remark 3. If we take A = I identity matrix, we get ordinary convergence. There- fore when we take A = I;we have Theorem 2.3 in [14].

References

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Current address : K¬r¬kkale Universty Faculty of Arts and Sciences, Department of Mathemat- ics, Yah¸sihan,K¬r¬kkale Turkey

E-mail address : didemaydn@hotmail.com

ORCID Address: http://orcid.org/0000-0002-5527-8232

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