A Fubini polynomial-based generalization of Szász-Baskakov operators

ABSTRACT

This work contains a generalization of the Szász-Baskakov operators with the help of Fubini polynomials. Firstly, we get the rate of convergence of our new operators and then we find some approximation results. Finally, Voronovskaya-type theorem and error table with comparisons are given.

KEYWORDS:

• Szász-Baskakov operators
• Fubini polynomials
• Steklov function

1. Introduction

Since the 1800s, approximation theory has been a significant part of mathematics. With approximation theory, the optimal method for estimating functions by utilizing smaller functions and computing the resulting errors may be quantitatively described. In addition, Appell, Bernoulli, Euler, Genocchi, Fubini, and Hermite polynomials are the most well-known polynomial families and are used in numerical analysis, asymptotic approximation, and special function theory. In recent years, special polynomials have attracted much attention and shed light on many fields, such as engineering and applied sciences. We can even construct new operators with the help of these polynomial families. The Szász operators are formed by extending the most well-known Bernstein operators to an infinite range. The definition of Szász operators in Szász (1950) is as follows,

(1.1) $S_n(\mathit{f};\mathit{x})=e^{-\mathit{nx}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{{(\mathit{nx})}^v}{\mathit{v}!}\mathit{f}\left(\frac{v}{n}\right).$(1.1)

Here, $\mathit{x}\ge 0$, $\mathit{n}\in \mathbb{N}\mathrm{and}\mathit{f}\in \mathit{C}[0,\mathrm{\infty })$. Baskakov (1957) introduced Baskakov operators in 1957 as follows:

(1.2) $B_n(\mathit{f};\mathit{x})=\frac{1}{{(1+\mathit{x})}^n}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{x^v}{{(1+\mathit{x})}^v}\mathit{f}\left(\frac{v}{n}\right),$(1.2)

$\mathit{n}\in {\mathbb{N}}^{+},\mathit{x}\in [0,\mathrm{\infty }).$

Szász-Mirakyan-Baskakov operators were examined in Prasad et al. (1983), which were later corrected and improved by Gupta (1993).

(1.3) $K_n(\mathit{f};\mathit{x})=(\mathit{n}-1)\sum _{\mathit{v}=0}^{\mathrm{\infty }}{e}^{-\mathit{nx}}\frac{{(\mathit{nx})}^v}{\mathit{v}!}{\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(\mathit{\tau }+1)}^{\mathit{n}+\mathit{v}}}\mathit{f}(\mathit{\tau })\mathit{d\tau }.$(1.3)

We refer to Govil et al. (2013), Acu and Gupta (2017), Gupta (2018), Mursaleen et al. (2019), Ansari et al. (2019, 2023), Nasiruzzaman et al. (2022, 2023), Usta et al. (2022), Sofyalıoğlu et al. (2022), Kanat and Erdal (2024) and Rahman and Ansari (2024) for some recent work on approximation by linear positive operators. The Appell polynomials $p_v(\mathit{x})$ generating functions are provided by

(1.4) $\mathit{A}(\mathit{k})e^{\mathit{kx}}=\sum _{\mathit{v}=0}^{\mathrm{\infty }}{p}_v(\mathit{x})k^v.$(1.4)

where $\mathit{A}(\mathit{k})={\sum }_{v=0}^{\mathrm{\infty }}{a}_v{k}^v$ $(a_0\ne 0)$ is an analytic function in the disc$|\mathit{k}|<\mathit{R}(\mathit{R}>1)$ with $\mathit{A}(1)\ne 0$. Additionally, some special polynomials $B_v$ Bernoulli polynomials, $E_v$ Euler polynomials and $G_v$ Genocchi polynomials were defined by Cheon (2003) and Horadam (1991) as follows,

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{B}_v(\mathit{x})\frac{k^v}{\mathit{v}!}=\frac{\mathit{k}{\mathit{e}}^{\mathit{xk}}}{e^k-1}(|\mathit{k}|<2\mathit{\pi }),$

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{E}_v(\mathit{x})\frac{k^v}{\mathit{v}!}=\frac{2{\mathit{e}}^{\mathit{xk}}}{e^k+1}(|\mathit{k}|<\mathit{\pi })$

and

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{G}_v(\mathit{x})\frac{k^v}{\mathit{v}!}=\frac{2\mathit{k}{\mathit{e}}^{\mathit{xk}}}{e^k+1}(|\mathit{k}|<\mathit{\pi }),$

respectively. Our upcoming destination is a recent study by Kilar and Simsek (2017), which gives the details of a family of special numbers and polynomials that are connected to Fubini numbers and polynomials. The following generating function defines the Fubini-type polynomials $a_v^{\alpha }(\mathit{x})$ of order $\mathit{\alpha }$,

(1.5) $F^{(\mathit{\alpha })}(\mathit{x},\mathit{k})=\frac{2^{\alpha }}{{(2-e^k)}^{2\mathit{\alpha }}}{e}^{\mathit{xk}}=\sum _{\mathit{v}=0}^{\mathrm{\infty }}{a}_v^{(\alpha )}(\mathit{x})\frac{k^v}{\mathit{v}!}(\mathit{\alpha }\in \mathbb{N};|\mathit{k}|<log2),$(1.5)

it produces the following generating function for $\mathit{x}=0$, for the Fubini-type numbers $a_v^{\alpha }$ of order $\mathit{\alpha }$ given by $a_v^{\alpha }=a_v^{\alpha }(0)$,

$F^{(\mathit{\alpha })}(0,\mathit{k})=\frac{2^{\alpha }}{{(2-e^k)}^{2\mathit{\alpha }}}=\sum _{\mathit{v}=0}^{\mathrm{\infty }}{a}_v^{(\alpha )}\frac{k^v}{\mathit{v}!}(\mathit{\alpha }\in \mathbb{N};|\mathit{k}|<log2).$

In the light of this information, we define the Fubini-type Szász-Baskakov operators of order $1$ as follows,

(1.6) ${\mathcal{Y}}_n(\mathit{f};\mathit{x})=\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}{\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{f}(\mathit{\tau })\mathit{d\tau },$(1.6)

where $\mathit{n}\in \mathbb{N}$, $\mathit{x}\ge 0$, and $\mathit{f}\in \mathit{C}[0,\mathrm{\infty }).$

2. Approximation properties of ${\mathcal{Y}}_n$ operators

Lemma 2.1.

The subsequent formulas are valid:

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}\mathit{v}\frac{a_v(\mathit{x})}{\mathit{v}!}=\frac{2x{e}^x}{{(2-\mathit{e})}^2}+\frac{4{\mathit{e}}^{\mathit{x}+1}}{{(2-\mathit{e})}^3},$

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{v}^2\frac{a_v(\mathit{x})}{\mathit{v}!}=\frac{2x(x+1)e^x}{{(2-\mathit{e})}^2}+\frac{8(\mathit{x}+1){\mathit{e}}^{\mathit{x}+1}}{{(2-\mathit{e})}^3}+\frac{12{\mathit{e}}^{\mathit{x}+2}}{{(2-\mathit{e})}^4},$

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{v}^3\frac{a_v(\mathit{x})}{\mathit{v}!}=\frac{2x(x^2+3\mathit{x}+1)e^x}{{(2-\mathit{e})}^2}+\frac{(12x^2+36\mathit{x}+20)e^{\mathit{x}+1}}{{(2-\mathit{e})}^3}+\frac{36(\mathit{x}+2){\mathit{e}}^{\mathit{x}+2}}{{(2-\mathit{e})}^4}+\frac{48{\mathit{e}}^{\mathit{x}+3}}{{(2-\mathit{e})}^5},$

$\sum _{\mathit{v}=0}^{\mathrm{\infty }}{v}^4\frac{a_v(\mathit{x})}{\mathit{v}!}=\frac{2x(x^3+6x^2+7\mathit{x}+1)e^x}{{(2-\mathit{e})}^2}+\frac{(16x^3+96x^2+144\mathit{x}+60)e^{\mathit{x}+1}}{{(2-\mathit{e})}^3}$

$+\frac{(72x^2+360\mathit{x}+384)e^{\mathit{x}+2}}{{(2-\mathit{e})}^4}+\frac{(192\mathit{x}+576){\mathit{e}}^{\mathit{x}+3}}{{(2-\mathit{e})}^5}+\frac{240{\mathit{e}}^{\mathit{x}+4}}{{(2-\mathit{e})}^6}.$

Proof.

We get the expected results if we substitute $\mathit{k}=1$ and $\mathit{\alpha }=1$ after considering the derivatives of Equation (1.5)(1.5) $F^{(\mathit{\alpha })}(\mathit{x},\mathit{k})=\frac{2^{\alpha }}{{(2-e^k)}^{2\mathit{\alpha }}}{e}^{\mathit{xk}}=\sum _{\mathit{v}=0}^{\mathrm{\infty }}{a}_v^{(\alpha )}(\mathit{x})\frac{k^v}{\mathit{v}!}(\mathit{\alpha }\in \mathbb{N};|\mathit{k}|<log2),$(1.5) with respect to $\mathit{k}$.□

Lemma 2.2.

Considering each $\mathit{x}\in [0,\mathrm{\infty })$ and $e_m={\tau }^m$ for $(\mathit{m}=0,1,2,3,4),$ we write

${\mathcal{Y}}_n(e_0,\mathit{x})=1,$

${\mathcal{Y}}_n(e_1,\mathit{x})=\frac{n}{\mathit{n}-2}\mathit{x}+\frac{2+\mathit{e}}{(\mathit{n}-2)(2-\mathit{e})},(\mathit{n}>2),$

${\mathcal{Y}}_n(e_2,\mathit{x})=\frac{n^2}{(\mathit{n}-2)(\mathit{n}-3)}{x}^2+4\mathit{n}\frac{2}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e})}\mathit{x}$

$+\frac{6e^2}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e}{)}^2}+\frac{10\mathit{e}}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e})}+\frac{2}{(\mathit{n}-2)(\mathit{n}-3)},(\mathit{n}>3),$

${\mathcal{Y}}_n(e_3,\mathit{x})=\frac{n^3}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)}{x}^3$

$+3n^2(\frac{2\mathit{e}}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}+\frac{3}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)})x^2$

$+\mathit{n}(\frac{18e^2}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}+\frac{42\mathit{e}}{(\mathit{n}-2)(2-\mathit{e})(\mathit{n}-3)(\mathit{n}-4)}$

$+\frac{18}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)})\mathit{x}+\frac{24e^3}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^3}$

$+\frac{72e^2}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}+\frac{56\mathit{e}}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$+\frac{6}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)},(\mathit{n}>4),$

${\mathcal{Y}}_n(e_4,\mathit{x})=\frac{n^4}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)}{x}^4$

$+n^3\left(\frac{8\mathit{e}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}+\frac{12}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)}\right)x^3$

$+n^2(\frac{36e^2}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}$

$+\frac{58\mathit{e}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}+\frac{47}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)})x^2$

$+\mathit{n}(\frac{96e^3}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^3}+\frac{270e^2}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}$

$+\frac{302\mathit{e}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}+\frac{91}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)})\mathit{x}$

$+\frac{120e^4}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^4}+\frac{528e^3}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^3}$

$+\frac{636e^2}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}+\frac{370\mathit{e}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$+\frac{24}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)},(\mathit{n}>5).$

Proof.

The Beta-Gamma function is given as

(2.1) $\mathcal{B}(\mathit{v},\mathit{n})={\int }_0^{\mathrm{\infty }}\frac{{\mathit{\tau }}^{\mathit{v}-1}}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{d\tau }=\frac{\mathrm{\Gamma }(\mathit{v})\mathrm{\Gamma }(\mathit{n})}{\mathrm{\Gamma }(\mathit{v}+\mathit{n})}=\frac{(\mathit{v}-1)!(\mathit{n}-1)!}{(\mathit{v}+\mathit{n}-1)!}.$(2.1)

By taking $\mathit{f}(\mathit{\tau })=1=e_0$ in the operator (1.6) and by using (2.1), we get

${\mathcal{Y}}_n(e_0;\mathit{x})=\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}{\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}+\mathit{n}-1\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(\mathit{\tau }+1)}^{\mathit{n}+\mathit{v}}}\mathit{d\tau }$

$=\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\left(\begin{array}{c}\mathit{v}+\mathit{n}-1\\ \mathit{v}\end{array}\right)\mathcal{B}(\mathit{v}+1,\mathit{n}-1)$

$=\frac{{(2-\mathit{e})}^2}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}$

$=1.$

$\mathit{f}(\mathit{\tau })=\mathit{\tau }=e_1$

${\mathcal{Y}}_n(e_1;\mathit{x})=\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\left(\begin{array}{c}\mathit{v}+\mathit{n}-1\\ \mathit{v}\end{array}\right)\mathcal{B}(\mathit{v}+2,\mathit{n}-2)$

$=\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}(\sum _{\mathit{v}=0}^{\mathrm{\infty }}\mathit{v}\frac{a_v(\mathit{nx})}{\mathit{v}!}+\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!})$

$=\frac{n}{\mathit{n}-2}\mathit{x}+\frac{\mathit{e}+2}{(\mathit{n}-2)(\mathit{e}-2)},(\mathit{n}>2).$

One can find ${\mathcal{Y}}_n(e_2;\mathit{x})$, ${\mathcal{Y}}_n(e_3;\mathit{x})$, ${\mathcal{Y}}_n(e_4;\mathit{x})$, respectively, in a similar way.□

Lemma 2.3

Considering each $\mathit{x}\in [0,\mathrm{\infty })$, we have

${\mathcal{Y}}_n(\mathit{\tau }-\mathit{x};\mathit{x})=(\frac{n}{\mathit{n}-2}-1)\mathit{x}+\frac{\mathit{e}+2}{(\mathit{n}-2)(\mathit{e}-2)},(\mathit{n}>2)$

${\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})=\left(\frac{n^2}{(\mathit{n}-3)(\mathit{n}-2)}-1)x^2+(\frac{4\mathit{n}(\mathit{e}+2)}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e})}-\frac{8-2\mathit{e}}{(\mathit{n}-3)(\mathit{n}-2)}\right)\mathit{x}$

$-\frac{2e^2-12\mathit{e}-8}{(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2},(\mathit{n}>3),$

${\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^4;\mathit{x})=(\frac{n^4}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)}-\frac{4n^3}{(\mathit{n}-2)(\mathit{n}-3)(\mathit{n}-4)}$

$+\frac{6n^2}{(\mathit{n}-3)(\mathit{n}-2)}-\frac{-4\mathit{n}}{(\mathit{n}-2)}+1)x^4$

$+(\frac{24n^3-4\mathit{e}{n}^3}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}-\frac{72n^2-12\mathit{e}{n}^2}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$+\frac{48\mathit{n}}{(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}-\frac{8+4\mathit{e}}{(\mathit{n}-2)(2-\mathit{e})})x^3$

$+(\frac{94n^2+11\mathit{e}{n}^2}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$+\frac{36e^2{n}^2}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}-\frac{144\mathit{n}+96\mathit{en}}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$-\frac{72e^2\mathit{n}}{(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2}+\frac{28e^2+12\mathit{e}+8}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e}{)}^2})x^2$

$+(\frac{182\mathit{n}+211\mathit{en}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e})}$

$+\frac{540e^2\mathit{n}-174e^3\mathit{n}}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^3}$

$-\frac{576e^2-192e^3}{(\mathit{n}-2)(\mathit{n}-3)(\mathit{n}-4)(2-\mathit{e}{)}^3}-\frac{220\mathit{e}+48}{(\mathit{n}-2)(\mathit{n}-3)(\mathit{n}-4)(2-\mathit{e})})\mathit{x}$

$+\frac{1056e^3-408e^4}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^4}$

$+\frac{290e^2+644\mathit{e}+96}{(\mathit{n}-5)(\mathit{n}-4)(\mathit{n}-3)(\mathit{n}-2)(2-\mathit{e}{)}^2},(\mathit{n}>5).$

Proof.

By taking advantage of the operator’s linearity to determine the central moments, we can express it as,

${\mathcal{Y}}_n(\mathit{\tau }-\mathit{x};\mathit{x})={\mathcal{Y}}_n(e_1;\mathit{x})-\mathit{x}{\mathcal{Y}}_n(e_0;\mathit{x}),(\mathit{n}>2),$

${\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})={\mathcal{Y}}_n(e_2;\mathit{x})-2\mathit{x}{\mathcal{Y}}_n(e_1;\mathit{x})+x^2{\mathcal{Y}}_n(e_0;\mathit{x}),(\mathit{n}>3),$

${\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^4;\mathit{x})={\mathcal{Y}}_n(e_4;\mathit{x})-4\mathit{x}{\mathcal{Y}}_n(e_3;\mathit{x})+6x^2{\mathcal{Y}}_n(e_2;\mathit{x})-4x^3{\mathcal{Y}}_n(e_1;\mathit{x})+x^4{\mathcal{Y}}_n(e_0;\mathit{x}),(\mathit{n}>5),$

we obtain the desired outcome.

Remark 2.1 The outcome of the limit of central moments is as follows:

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}{\mathcal{Y}}_n(\mathit{\tau }-\mathit{x};\mathit{x})=2\mathit{x}-\frac{2+\mathit{e}}{2-\mathit{e}},$

(2.2) $\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})=\mathit{x}(\mathit{x}+2),$(2.2)

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}{n}^2{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^4;\mathit{x})=3x^2(\mathit{x}+2{)}^2.$

Theorem 2.1

Let $\mathit{f}$ be continuous on $[0,\mathrm{\infty })$ and a member of the class

$\mathit{E}=\{\mathit{f}:\frac{\mathit{f}(\mathit{x})}{1+x^2}\mathit{is}\mathit{convergent}\mathit{as}\mathit{x}\to \mathrm{\infty }\}.$

Then, ${\mathcal{Y}}_n(\mathit{f};\mathit{x})$ operators converge uniformly on each compact subset of $[0,\mathrm{\infty })$ as

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}{\mathcal{Y}}_n(\mathit{f};\mathit{x})=\mathit{f}(\mathit{x}).$

Proof.

From Lemma 2.2, we get

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathcal{Y}({\tau }^i;\mathit{x})=x^i,\mathit{i}=0,1,2.$

These convergences are consistently satisfied on the compact subset of the interval $[0,\mathrm{\infty })$. Therefore, the classical Korovkin theorem (Francesco and Campiti 2011) provides the proof.

Definition 2.1.

Let $\mathit{f}\in$ $C_D[0,\mathrm{\infty })$ and $\mathit{\delta }>0$. Modulus of continuity of the function $\mathit{f}$, denoted by $\mathit{\omega }(\mathit{f},\mathit{\delta })$ is

(2.3) $\mathit{\omega }(\mathit{f},\mathit{\delta })=\underset{\underset{|x-\tau |\le \delta }{\mathit{x},\mathit{\tau }\in [0,\mathrm{\infty })}}{sup}|\mathit{f}(\mathit{x})-\mathit{f}(\mathit{\tau })|,$(2.3)

where $C_D[0,\mathrm{\infty })$ represents the space of uniformly continuous functions on $[0,\mathrm{\infty })$ (Altomare and Campiti, 1994).

Definition 2.2.

Given a function $\mathit{f}\in \mathit{C}[\mathit{a},\mathit{b}]$, its second modulus of continuity is given by (Devore and Lorentz, 1993).

(2.4) ${\omega }_2(\mathit{f},\mathit{\delta })=\underset{\begin{array}{c}0<\mathit{\tau }\le \mathit{\delta }\end{array}}{sup}\parallel \mathit{f}(.+2\mathit{\tau })-2\mathit{f}(.+\mathit{\tau })+\mathit{f}(.)\parallel ,$(2.4)

where $\parallel \mathit{f}\parallel ={max}_{\begin{array}{c}\mathit{\tau }\in [\mathit{a},\mathit{b}]\end{array}}|\mathit{f}(\mathit{\tau })|.$

Lemma 2.4.

(Gavrea and Rasa, 1993) Given a sequence of positive and linear operators with the condition $K_n(1;\mathit{x})=1,$ let $\mathit{f}\in C^2[0,\mathit{a}]$ and $(K_n{)}_{\mathit{n}\ge 0}$ be represented. Subsequently

(2.5) $|K_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \parallel \mathit{f}{ }^{\mathrm{\prime }}\parallel \sqrt{K_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})}+\frac{1}{2}\parallel \mathit{f}{ }^{\mathrm{\prime }}{ }^{\mathrm{\prime }}\parallel K_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x}).$(2.5)

Definition 2.3.

(Zhuk 1989) The function $f_{\varphi }$ is called Steklov function if the following holds:

(2.6) $f_{\varphi }(\mathit{\tau })=\frac{1}{\varphi }{\int }_{\tau -\frac{\varphi }{2}}^{\mathit{\tau }+\frac{\varphi }{2}}\mathit{f}(\mathit{u})\mathit{du}=\frac{1}{\varphi }{\int }_{-\frac{\varphi }{2}}^{\frac{\varphi }{2}}\mathit{f}(\mathit{\tau }+\mathit{v})\mathit{dv},$(2.6)

where $\mathit{f}$ is integrable function on an implicit and bounded interval $[\mathit{a},\mathit{b}]$. Derivative of this function at almost every point is given as:

(2.7) $\mathit{f}{{\mathit{ }}^{\mathrm{\prime }}}_{\mathit{\varphi }}(\mathit{\tau })=\frac{1}{\varphi }\mathit{f}\left(\mathit{\tau }+\frac{\varphi }{2}\right)-\mathit{f}\left(\mathit{\tau }-\frac{\varphi }{2}\right).$(2.7)

The following relationships exist if the derivative $\mathit{f}$ is uniformly continuous on the real axis.

$\underset{\mathit{\tau }\in (-\mathrm{\infty },\mathrm{\infty })}{sup}|\mathit{f}(\mathit{\tau })-f_{\varphi }(\mathit{\tau })|\le \mathit{\omega }\left(\frac{\varphi }{2},\mathit{f}\right),$

$\underset{\mathit{\tau }\in (-\mathrm{\infty },\mathrm{\infty })}{sup}|\mathit{f}{{\mathit{ }}^{\mathrm{\prime }}}_{\mathit{\varphi }}(\mathit{\tau })|\le \frac{1}{2}\mathit{\omega }(\mathit{\varphi },\mathit{f}).$

Lemma 2.5.

(Zhuk 1989) Let $\mathit{f}\in \mathit{C}[\mathit{a},\mathit{b}]$ and $\mathit{\varphi }\in \left(0,\frac{(\mathit{b}-\mathit{a})}{2}\right).$ Let $f_{\varphi }$ be the second-order Steklov function associated with the function $\mathit{f}$. Then, the inequalities listed below are met:

i.

$\parallel f_{\varphi }-\mathit{f}\parallel \le \frac{3}{4}{\omega }_2(\mathit{f},\mathit{\varphi }),$

ii.

$\parallel {f_{\varphi }}^{\prime }{ }^{\mathrm{\prime }}\parallel \le \frac{3}{2{\varphi }^2}{\omega }_2(\mathit{f},\mathit{\varphi }).$

Generally, we will use the first and second modulus of continuity to limit the margin of error in linear positive operators. We shall determine the rate of convergence in the two theorems that follow with the use of the earlier definitions and theorems.

Theorem 2.2.

Let $\mathit{f}\in C_D[0,\mathrm{\infty })\cap \mathit{E}.{\mathcal{Y}}_n$ operators confirm the inequality that follows:

(2.8) $|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le 2\mathit{\omega }(\mathit{f},\sqrt{{\delta }_n(\mathit{x})}).$(2.8)

Here,

$\mathit{\delta }:={\delta }_n(\mathit{x})=\left(\frac{n^2}{(\mathit{n}-2)(\mathit{n}-3)}+1)x^2+(\frac{4\mathit{n}(\mathit{e}+2)}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e})}-\frac{8-2\mathit{e}}{(\mathit{n}-3)(\mathit{n}-2)}\right)\mathit{x}$

$-\frac{2e^2-12\mathit{e}-8}{(\mathit{n}-2)(\mathit{n}-3)(2-\mathit{e}{)}^2},(\mathit{n}>3).$

Proof.

Let $\mathit{f}\in \mathit{C}[0,\mathrm{\infty })$. Next, using the modulus of continuity property,

(2.9) $|\mathit{f}(\mathit{x})-\mathit{f}(\mathit{\tau })|\le (\frac{|\tau -x|}{\delta }+1)\mathit{\omega }(\mathit{f},\mathit{\delta })$(2.9)

is provided. Using the inequality stated above, we get

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le |\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}{\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{f}(\mathit{\tau })\mathit{d\tau }$

$-\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}{\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{f}(\mathit{x})\mathit{d\tau }|$

$\le \frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}$

$\times {\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}|\mathit{f}(\mathit{\tau })-\mathit{f}(\mathit{x})|\mathit{d\tau }$

$\le \frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}$

$\times {\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{v}-1+\mathit{n}\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}(\frac{|\tau -x|}{\delta }+1)\mathit{\omega }(\mathit{f},\mathit{\delta })\mathit{d\tau }$

$\le \left\{1+\frac{1}{\delta }\left(\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\times {\int }_0^{\mathrm{\infty }}\left(\frac{\mathit{v}-1+\mathit{n}}{v}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}|\mathit{\tau }-\mathit{x}|\mathit{d\tau }\right)\right\}\mathit{\omega }(\mathit{f},\mathit{\delta }).$

The integral can be calculated using the Cauchy-Schwarz inequality to get the following result:

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \left\{1+\left(\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\right)\frac{1}{\delta }\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\right.$

$\times {\left({\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{n}+\mathit{v}-1\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{d\tau }\right)}^{\frac{1}{2}}$

$\left.\times {\left({\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{n}+\mathit{v}-1\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}|\mathit{\tau }-\mathit{x}{|}^2\mathit{d\tau }\right)}^{\frac{1}{2}}\right\}\mathit{\omega }(\mathit{f},\mathit{\delta }).$

By examining the Cauchy-Schwarz inequality in summation, one can easily reach the following conclusion

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \left\{1+\frac{1}{\delta }\left(\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\right.\right.$

${\left.\times {\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{n}+\mathit{v}-1\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}\mathit{d\tau }\right)}^{\frac{1}{2}}$

$\times \left(\frac{{(2-\mathit{e})}^2(\mathit{n}-1)}{2e^{\mathit{nx}}}\sum _{\mathit{v}=0}^{\mathrm{\infty }}\frac{a_v(\mathit{nx})}{\mathit{v}!}\right.$

$\left.{\left.\times {\int }_0^{\mathrm{\infty }}\left(\begin{array}{c}\mathit{n}+\mathit{v}-1\\ \mathit{v}\end{array}\right)\frac{{\tau }^v}{{(1+\mathit{\tau })}^{\mathit{n}+\mathit{v}}}|\mathit{\tau }-\mathit{x}{|}^2\mathit{d\tau }\right)}^{\frac{1}{2}}\right\}\mathit{\omega }(\mathit{f},\mathit{\delta })$

$=\left\{1+\frac{1}{\delta }{({\mathcal{Y}}_n(e_0;\mathit{x}))}^{\frac{1}{2}}{({\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x}))}^{\frac{1}{2}}\right\}\mathit{\omega }(\mathit{f},\mathit{\delta })$

$=\left\{1+\frac{1}{\delta }{(Y_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x}))}^{\frac{1}{2}}\right\}\mathit{\omega }(\mathit{f},\mathit{\delta })$

the desired result is achieved.

(2.10) $|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \left\{1+\frac{1}{\delta }{\left({\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})\right)}^{\frac{1}{2}}\right\}\mathit{\omega }(\mathit{f},\mathit{\delta }).$(2.10)

By choosing $\mathit{\delta }:={\delta }_n(\mathit{x})=\sqrt{{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})}$, we have the desired result.

Theorem 2.3.

The following estimate applies to $\mathit{f}\in \mathit{C}[0,\mathit{a}]$,

(2.11) $|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \frac{2}{a}\parallel \mathit{f}\parallel {\varphi }^2+\frac{3}{4}(\mathit{a}+2+{\varphi }^2){\omega }_2(\mathit{f},\mathit{\varphi }),$(2.11)

where

(2.12) $\mathit{\varphi }:={\varphi }_n(\mathit{x})=\sqrt[4]{{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})}.$(2.12)

Proof.

The second-order Steklov function associated with function $\mathit{f}$ is denoted by $f_{\varphi }$. With regard to the individual ${\mathcal{Y}}_n(1;\mathit{x})=1$, we achieve

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le |{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})+{\mathcal{Y}}_n(f_{\varphi };\mathit{x})-{\mathcal{Y}}_n(f_{\varphi };\mathit{x})+f_{\varphi }(\mathit{x})-f_{\varphi }(\mathit{x})|$

$\le |{\mathcal{Y}}_n(\mathit{f};\mathit{x})-{\mathcal{Y}}_n(f_{\varphi };\mathit{x})|+|{\mathcal{Y}}_n(f_{\varphi };\mathit{x})-f_{\varphi }(\mathit{x})|+|f_{\varphi }(\mathit{x})-\mathit{f}(\mathit{x})|$

$\le \parallel \mathit{f}-f_{\varphi }\parallel {\mathcal{Y}}_n(1;\mathit{x})+|{\mathcal{Y}}_n(f_{\varphi };\mathit{x})-f_{\varphi }(\mathit{x})|+\parallel f_{\varphi }-\mathit{f}(\mathit{x})\parallel$

$\le 2\parallel f_{\varphi }-\mathit{f}\parallel +|{\mathcal{Y}}_n(f_{\varphi };\mathit{x})-f_{\varphi }(\mathit{x})|.$

With the help of Lemma 2.3 and Lemma 2.4,

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \sqrt{{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})}\parallel \mathit{f}{{\mathit{ }}^{\mathrm{\prime }}}_{\mathit{\varphi }}\parallel +\frac{1}{2}\parallel {f_{\varphi }}^{\prime }{ }^{\mathrm{\prime }}\parallel {\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x}).$

The Landau inequality is defined from Fink (1982) as follows,

$\parallel \mathit{f}{ }^{\mathrm{\prime }}\parallel \le 2\parallel \mathit{f}{\parallel }^{\frac{1}{2}}\parallel \mathit{f}{ }^{\mathrm{\prime }}{ }^{\mathrm{\prime }}{\parallel }^{\frac{1}{2}}.$

When Lemma 2.5 and Landau inequality are combined,

$\parallel {f_{\varphi }}^{\prime }\parallel \le \frac{2}{a}\parallel f_{\varphi }\parallel +\frac{a}{2}\parallel {f_{\varphi }}^{\prime }{ }^{\mathrm{\prime }}\parallel$

(2.13) $\le \frac{2}{a}\parallel \mathit{f}\parallel +\frac{3\mathit{a}}{4}\frac{1}{{\varphi }^2}{\omega }_2(\mathit{f},\mathit{\varphi }).$(2.13)

If we specifically assume $\mathit{\varphi }=\sqrt[4]{{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})}$, then we write

(2.14) $|{\mathcal{Y}}_n(f_{\varphi };\mathit{x})-f_{\varphi }(\mathit{x})|\le \frac{2}{a}\parallel \mathit{f}\parallel {\varphi }^2+\frac{3\mathit{a}}{4}{\omega }_2(\mathit{f};\mathit{\varphi })+\frac{3}{4}{\varphi }^2{\omega }_2(\mathit{f},\mathit{\varphi }).$(2.14)

From Lemma 2.4,

$|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})|\le \frac{2}{a}\parallel \mathit{f}\parallel {\varphi }^2+\frac{3}{4}(\mathit{a}+2+{\varphi }^2){\omega }_2(\mathit{f},\mathit{\varphi })$

is obtained and the proof is completed.

3. Voronovskaya-type theorem

Theorem 3.1.

For $\mathit{f},\mathit{f}{ }^{\mathrm{\prime }},\mathit{f}{ }^{\mathrm{\prime }}{ }^{\mathrm{\prime }}\in \mathit{C}[0,\mathrm{\infty })\cap \mathit{E}$ and $\mathit{x}\in [0,\mathrm{\infty })$, we receive

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}({\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x}))=2\mathit{x}-\frac{2+\mathit{e}}{2-\mathit{e}}\mathit{f}{ }^{\mathrm{\prime }}(\mathit{x})+\frac{1}{2}\mathit{x}(\mathit{x}+2)\mathit{f}{ }^{\mathrm{\prime }}{ }^{\mathrm{\prime }}(\mathit{x})$

uniformly in each $[0,\mathrm{\infty })$ compact subset of.

Proof.

When we apply the function $\mathit{f}$‘s classical Taylor expansion, we get

(3.1) $\mathit{f}(\mathit{\tau })=\mathit{f}(\mathit{x})+\mathit{f}{ }^{\mathrm{\prime }}(\mathit{x})(\mathit{\tau }-\mathit{x})+\mathit{f}{ }^{\mathrm{\prime }}{ }^{\mathrm{\prime }}(\mathit{x})\frac{{(\mathit{\tau }-\mathit{x})}^2}{2}+(\mathit{\tau }-\mathit{x}{)}^2\mathit{r}(\mathit{\tau },\mathit{x}).$(3.1)

Here, the phrase ‘reminder’ $\mathit{r}(\mathit{\tau },\mathit{x})\in \mathit{C}[0,\mathrm{\infty })\cap \mathit{E}$ and ${lim}_{\mathit{\tau }\to \mathit{x}}\mathit{k}(\mathit{\tau },\mathit{x})=0.$ By putting the ${\mathcal{Y}}_n$ operators on Equation (3.1) two sides, we get

${\mathcal{Y}}_n(\mathit{f};\mathit{x})=\mathit{f}(\mathit{x})+{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x});\mathit{x})\mathit{f}{ }^{\mathrm{\prime }}(\mathit{x})+\frac{\mathit{f}{\mathit{ }}^{\mathrm{\prime }}{ }^{\mathrm{\prime }}(\mathit{x})}{2}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})+{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2\mathit{r}(\mathit{\tau },\mathit{x});\mathit{x}).$

Then

$\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}({\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x}))=\mathit{f}{ }^{\mathrm{\prime }}(\mathit{x})\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x});\mathit{x})+\frac{\mathit{f}{\mathit{ }}^{\mathrm{\prime }}{ }^{\mathrm{\prime }}(\mathit{x})}{2}\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2;\mathit{x})$

$+\underset{\mathit{n}\to \mathrm{\infty }}{lim}\mathit{n}{\mathcal{Y}}_n(\mathit{r}(\mathit{\tau },\mathit{x})(\mathit{\tau }-\mathit{x}{)}^2;\mathit{x}).$

Utilizing the Cauchy-Schwarz inequality ${lim}_{\mathit{n}\to \mathrm{\infty }}\mathit{n}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2\mathit{r}(\mathit{\tau },\mathit{x});\mathit{x})$ provides

$\mathit{n}{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^2{r}^2(\mathit{\tau },\mathit{x});\mathit{x})\le \sqrt{{\mathcal{Y}}_n(r^2(\mathit{\tau },\mathit{x});\mathit{x})}\sqrt{n^2{\mathcal{Y}}_n((\mathit{\tau }-\mathit{x}{)}^4;\mathit{x})}.$

Considering that $\mathit{r}(\mathit{\tau },\mathit{x})\to 0$ as $\mathit{\tau }\to \mathit{x}$,

(3.2) $\underset{\mathit{n}\to \mathrm{\infty }}{lim}{\mathcal{Y}}_n(r^2(\mathit{\tau },\mathit{x});\mathit{x})=r^2(\mathit{x},\mathit{x})=0.$(3.2)

is equably confirmed in every case $[0,\mathrm{\infty })$ compact subset of. That way, we get the desired outcome from (2.2) and (3.2).

Example 3.1.

By choosing $\mathit{f}(\mathit{x})=\frac{log({\mathit{e}}^{-\mathit{x}}+1)}{1000}$, $0\le \mathit{x}\le 1$ we show the error estimation of the Fubini-type Szász-Baskakov operators ${\mathcal{Y}}_n$ using modulus of continuity in Table 1.

Table 1. Error approximation for ${\mathcal{Y}}_n$ by using the modulus of continuity.

$\mathit{n}$	$\left|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})\right|$
$10$	$0.000729999$
${10}^2$	$0.0000712712$
${10}^3$	$7.0828\times {10}^{-6}$
${10}^4$	$7.08232\times {10}^{-7}$
${10}^5$	$7.12622\times {10}^{-8}$
${10}^6$	$1.26728\times {10}^{-8}$
${10}^7$	$1.16623\times {10}^{-9}$

Example 3.2.

For $\mathit{f}(\mathit{x})=\frac{log(e^{-x^2}+1)}{1000}$, $0\le \mathit{x}\le 2$, we show the comparison of the error estimates of the Fubini-type Szász-Baskakov operators $({\mathcal{Y}}_n)$ with the Szász-Baskakov Appell $A^2$ operators $({\mathcal{M}}_n)$ by Kanat et al. (2024) by using the modulus of continuity in Table 2. Here we specially choose $\mathit{A}(\mathit{\omega })=e^{8\mathit{\omega }}$ and $\mathit{B}(\mathit{\omega })=0$ in Kanat et al. (2024).

Table 2. Error approximation for ${\mathcal{Y}}_n$ and ${\mathcal{M}}_n$ by using the modulus of continuity.

$\mathit{n}$	$\left|{\mathcal{M}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})\right|$	$\left|{\mathcal{Y}}_n(\mathit{f};\mathit{x})-\mathit{f}(\mathit{x})\right|$
$10$	$0.00116314$	$0.000929913$
${10}^2$	$0.000112691$	$0.0000806178$
${10}^3$	$0.0000110306$	$7.89009\times {10}^{-6}$

As one can see in Table 2, the Fubini-type Szász-Baskakov operator has less error than the Szász-Baskakov Appell $A^2$ under the current choices.

4. Conclusion

The Fubini-type Szász-Baskakov operators are created in this study. The central moments of the newly created ${\mathcal{Y}}_n$ operator were examined. The modulus of continuity is investigated. Then, the Voronovskaya-type theorem for Fubini-type Szász-Baskakov operators is established. Lastly, numerical examples are presented to show and compare the error estimations.

Disclosure statement

No potential conflict of interest was reported by the author(s).