New modification of the Post Widder operators preserving exponential functions

Abstract

The current article deals with a modification of the Post-Widder operators which reproduce the exponential functions both $e^{\mu x}$ and $e^{2\mu x}$ for $\mu >0.$ The central moments, uniform convergence of the operators and the rate of convergence of these operators with the help of modulus of continuity are estimated. Also, a Voronovskaja-type asymptotic formula is established. After that, numerical results are obtained to confirm the theoretical results. Finally, new constructed operators are compared with modified Post-Widder operators.

Keywords:

• Post-Widder operators
• exponential functions
• Voronovskaya-type theorem

MSC CLASSIFICATION:

• 41A25
• 41A36
• 47A58

1. Introduction

In 1976, May (May, 1976) introduces the Post-Widder operators for $\lambda \ge 1,f\in C[0,\infty )$ and $J=(0,\infty )$ as follows: $$P_{\lambda }(f;x)=\frac{{\lambda }^{\lambda }}{x^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{x}}f(t)\mathit{\text{dt}},x\in J,$$ where $\Gamma (\lambda )=(\lambda -1)!.$ In 2020, for $f\in C[0,\infty ),x\in J$ and $\lambda \ge 1,$ the modified form of the Post-Widder operators (Sofyalıoğlu & Kanat, 2020) is defined by (1) $$P_{\lambda ,\alpha }^{*}(f;x)=\frac{{(2a)}^{\lambda }}{{\left(1-e^{\frac{-2\mathit{\text{ax}}}{\lambda }}\right)}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{\frac{-2\mathit{\text{at}}}{1-e^{-2\mathit{\text{ax}}/\lambda }}}f(t)\mathit{\text{dt}}.$$(1)

Here, the operators given by (1)(1) $$P_{\lambda ,\alpha }^{*}(f;x)=\frac{{(2a)}^{\lambda }}{{\left(1-e^{\frac{-2\mathit{\text{ax}}}{\lambda }}\right)}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{\frac{-2\mathit{\text{at}}}{1-e^{-2\mathit{\text{ax}}/\lambda }}}f(t)\mathit{\text{dt}}.$$(1) reproduce constant and $e^{2\mathit{\text{ax}}}$ for fixed a > 0. For some recent studies on linear positive operators preserving exponential functions, we refer the readers to the literature (Acar, Aral, Cárdenas-Morales, & Garrancho, 2017; Acu, Aral, & Rasa, 2023; Aral, Cárdenas-Morales, & Garrancho, 2018; Aral, Limmam, and Ozsarac, 2019, Gupta & Agrawal, 2019; Gupta & Herzog, 2023; Gupta & López-Moreno, 2018; Gupta & Tachev, 2022a; 2022b, Kanat & Sofyalıoğlu, 2021; Sofyalıoğlu & Kanat, 2019 and Torun, 2022).

In the current paper, we construct a generalization of the Post-Widder operators, which preserve both $e^{\mu x}$ and $e^{2\mu x}$ for $\mu >0.$ The aim of this paper is to obtain better approximation results than the results in (1)(1) $$P_{\lambda ,\alpha }^{*}(f;x)=\frac{{(2a)}^{\lambda }}{{\left(1-e^{\frac{-2\mathit{\text{ax}}}{\lambda }}\right)}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{\frac{-2\mathit{\text{at}}}{1-e^{-2\mathit{\text{ax}}/\lambda }}}f(t)\mathit{\text{dt}}.$$(1) . Firstly, our starting point is the general sequence (2) $$P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\beta }_{\lambda }(x)}}f(t)\mathit{\text{dt}},x\in J,$$(2) for $f\in C[0,\infty )$ and $\lambda \in {\mathbb{N}}^{+}.$ Here, ${\alpha }_{\lambda }(x)$ and ${\beta }_{\lambda }(x)$ are positive functions to be calculated in such a way that the operators (2)(2) $$P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\beta }_{\lambda }(x)}}f(t)\mathit{\text{dt}},x\in J,$$(2) hold fixed the functions $e^{\mu x}$ and $e^{2\mu x}$ for $\mu >0.$ In order to define the sequences ${\alpha }_{\lambda }(x)$ and ${\beta }_{\lambda }(x)$ explicitly, we consider the following identities $$\begin{array}{c}{P}_{\lambda }^{\alpha ,\beta }(e^{\mu t};x)=e^{\mu x},\\ P_{\lambda }^{\alpha ,\beta }(e^{2\mu t};x)=e^{2\mu x},\end{array}$$ for every $x\in J$ and $\lambda \in {\mathbb{N}}^{+}.$ By choosing $f(t)=e^{\mu t}$ in (2)(2) $$P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\beta }_{\lambda }(x)}}f(t)\mathit{\text{dt}},x\in J,$$(2) , we achieve $$\begin{array}{l}{e}^{\mu x}=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\left(\frac{\lambda }{{\beta }_{\lambda }(x)}-\mu \right)t}{e}^{\mu t}\mathit{\text{dt}},\frac{\lambda }{{\beta }_{\lambda }(x)}>\mu \\ ={\left(\frac{\lambda }{{\alpha }_{\lambda }(x)\left[\frac{\lambda }{{\beta }_{\lambda }(x)}-\mu \right]}\right)}^{\lambda }.\end{array}$$

Similarly, by taking $f(t)=e^{2\mu t}$ in (2)(2) $$P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\beta }_{\lambda }(x)}}f(t)\mathit{\text{dt}},x\in J,$$(2) , we have $$\begin{array}{l}{e}^{2\mu x}=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\left(\frac{\lambda }{{\beta }_{\lambda }(x)}-2\mu \right)t}\mathit{\text{dt}},\frac{\lambda }{{\beta }_{\lambda }(x)}>2\mu \\ ={\left(\frac{\lambda }{{\alpha }_{\lambda }(x)\left[\frac{\lambda }{{\beta }_{\lambda }(x)}-2\mu \right]}\right)}^{\lambda }.\end{array}$$

After simple calculations, we get $$\begin{array}{c}{\alpha }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{2\mu x}{\lambda }}},\\ {\beta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu \left(2e^{\frac{\mu x}{\lambda }}-1\right)}.\end{array}$$

Attention that ${\alpha }_{\lambda }(x)$ and ${\beta }_{\lambda }(x)$ tend to x as $\lambda \to \infty .$ So, the operators $P_{\lambda }^{\alpha ,\beta }$ return to $P_{\lambda }.$ Now, we substitute the values ${\alpha }_{\lambda }(x)$ and ${\beta }_{\lambda }(x)$ in the operators (2)(2) $$P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }}{{({\alpha }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\beta }_{\lambda }(x)}}f(t)\mathit{\text{dt}},x\in J,$$(2) . After some rearranging, we obtain (3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) where $\mathit{\text{ex}}{p}_{\mu }(t)=e^{\mu t},x\in J,\lambda \in {\mathbb{N}}^{+}$ and (4) $${\theta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{\mu x}{\lambda }}}.$$(4)

Henceforth, we will use the notation $P_{\lambda }^{\theta }$ of our new construction for the Post-Widder operators throughout the article. The classical Post-Widder operators $P_{\lambda }$ (1)(1) $$P_{\lambda ,\alpha }^{*}(f;x)=\frac{{(2a)}^{\lambda }}{{\left(1-e^{\frac{-2\mathit{\text{ax}}}{\lambda }}\right)}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{\frac{-2\mathit{\text{at}}}{1-e^{-2\mathit{\text{ax}}/\lambda }}}f(t)\mathit{\text{dt}}.$$(1) and the new construction of the Post-Widder operators $P_{\lambda }^{\theta }$ (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) are related by $$P_{\lambda }^{\theta }(f;x)=\mathit{\text{ex}}{p}_{\mu }(x)P_{\lambda }\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }};{\theta }_{\lambda }(x)\right).$$

Currently, we mention some auxilary lemmas.

Lemma 1.

Let $K<\mu \left(1+\frac{e^{\mu x/\lambda }}{e^{\mu x/\lambda }-1}\right)$, then we have (5) $$P_{\lambda }^{\theta }(e^{\mathit{\text{Kt}}};x)=e^{\mu x}{\left(1-\frac{(K-\mu ){\theta }_{\lambda }(x)}{\lambda }\right)}^{-\lambda },$$(5) where ${\theta }_{\lambda }(x)$ is given by (4)(4) $${\theta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{\mu x}{\lambda }}}.$$(4) .

Proof.

From (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) , we have $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(e^{\mathit{\text{Kt}}};x)\hfill & =\hfill & \frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\frac{\mathit{\text{ex}}{p}_K}{\mathit{\text{ex}}{p}_{\mu }}\left(t\right)\mathit{\text{dt}}\hfill \\ \hfill & =\hfill & \frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\left(\frac{\lambda }{{\theta }_{\lambda }(x)}+\mu -K\right)t}\mathit{\text{dt}},\frac{\lambda }{{\theta }_{\lambda }(x)}+\mu >K\hfill \\ \hfill & =\hfill & \frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }}\frac{1}{{\left(-K+\mu +\frac{\lambda }{{\theta }_{\lambda }(x)}\right)}^{\lambda }}\hfill \\ \hfill & =\hfill & e^{\mu x}{\left(1-\frac{(K-\mu ){\theta }_{\lambda }(x)}{\lambda }\right)}^{-\lambda },\hfill \end{array}$$ where ${\theta }_{\lambda }(x)$ is as given by (4)(4) $${\theta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{\mu x}{\lambda }}}.$$(4) .

Lemma 2.

For $s\in \mathbb{N}$ the rising factorial is denoted by ${(\lambda )}_s=\lambda (\lambda +1)\cdots (\lambda +s-1),{(\lambda )}_0=1$. Then we have $$P_{\lambda }^{\theta }(t^s{e}^{\mathit{\text{Kt}}};x)=\frac{e^{\mu x} {(\lambda )}_s {\lambda }^{\lambda }}{{({\theta }_{\lambda }(x))}^{\lambda }}{\left(-K+\mu +\frac{\lambda }{{\theta }_{\lambda }(x)}\right)}^{-s-\lambda }.$$

Proof.

From (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) , we have $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(t^s{e}^{\mathit{\text{Kt}}};x)\hfill & =\hfill & \frac{e^{\mu x}{\lambda }^{\lambda }}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda +s-1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\frac{\mathit{\text{ex}}{p}_K}{\mathit{\text{ex}}{p}_{\mu }}\left(t\right)\mathit{\text{dt}}\hfill \\ \hfill & =\hfill & \frac{e^{\mu x}{\lambda }^{\lambda }}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda +s-1}{e}^{-\left(\frac{\lambda }{{\theta }_{\lambda }(x)}+\mu -K\right)t}\mathit{\text{dt}},\frac{\lambda }{{\theta }_{\lambda }(x)}+\mu >K\hfill \\ \hfill & =\hfill & \frac{e^{\mu x} {(\lambda )}_s {\lambda }^{\lambda }}{{({\theta }_{\lambda }(x))}^{\lambda }}{\left(-K+\mu +\frac{\lambda }{{\theta }_{\lambda }(x)}\right)}^{-s-\lambda },\hfill \end{array}$$ where ${\theta }_{\lambda }(x)$ is as given by (4)(4) $${\theta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{\mu x}{\lambda }}}.$$(4) .

Lemma 3.

For the operators (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) we have the following moments $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(e_0;x)\hfill & =\hfill & 1,\hfill \\ P_{\lambda }^{\theta }(e_1;x)\hfill & =\hfill & \frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }},\hfill \\ P_{\lambda }^{\theta }(e_2;x)\hfill & =\hfill & \frac{\lambda (\lambda +1){\left(e^{\mu x/\lambda }-1\right)}^2}{{\mu }^2{e}^{4\mu x/\lambda }},\hfill \\ P_{\lambda }^{\theta }(e_3;x)\hfill & =\hfill & \frac{\lambda (\lambda +1)(\lambda +2){\left(e^{\mu x/\lambda }-1\right)}^3}{{\mu }^3{e}^{6\mu x/\lambda }},\hfill \\ P_{\lambda }^{\theta }(e_4;x)\hfill & =\hfill & \frac{\lambda (\lambda +1)(\lambda +2)(\lambda +3){\left(e^{\mu x/\lambda }-1\right)}^4}{{\mu }^4{e}^{8\mu x/\lambda }},\hfill \end{array}$$ where $e_s(t)=t^s,s=0,1,2,3,4.$

Proof.

By choosing K = 0 and $s=0,1,2,3,4$ in Lemma 2, respectively, we obtain the desired results.

Lemma 4.

Let ${\psi }_x^s(t)={(t-x)}^s,s=0,1,2,4.$ Then we write the following central moments $$\begin{array}{ccc}{P}_{\lambda }^{\theta }({\psi }_x^1;x)\hfill & =\hfill & \frac{e^{2\mu x}(\lambda (e^{\mu x/\lambda }-1)-\mu (2e^{\mu x/\lambda }-1)x)}{\mu {(2e^{\mu x/\lambda }-1)}^{\lambda +1}},\hfill \\ P_{\lambda }^{\theta }({\psi }_x^2;x)\hfill & =\hfill & \frac{e^{2\mu x}(\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2-2\mu \lambda (e^{\mu x/\lambda }-1)(2e^{\mu x/\lambda }-1)x+{\mu }^2{(2e^{\mu x/\lambda }-1)}^2{x}^2)}{{\mu }^2{(2e^{\mu x/\lambda }-1)}^{\lambda +2}},\hfill \\ P_{\lambda }^{\theta }({\psi }_x^4;x)\hfill & =\hfill & \frac{e^{2\mu x}}{{\mu }^4{(2e^{\mu x/\lambda }-1)}^{\lambda +4}}\{\lambda (\lambda +1)(\lambda +2)(\lambda +3){(e^{\mu x/\lambda }-1)}^4\hfill \\ \hfill & \hfill & -4\left[\mu \lambda (\lambda +1)(\lambda +2){(e^{\mu x/\lambda }-1)}^3(2e^{\mu x/\lambda }-1)\right]x\hfill \\ \hfill & \hfill & +6\left[{\mu }^2\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2{(2e^{\mu x/\lambda }-1)}^2\right]x^2-4\left[{\mu }^3\lambda (e^{\mu x/\lambda }-1){(2e^{\mu x/\lambda }-1)}^3\right]x^3\hfill \\ \hfill & \hfill & +{\mu }^4{(2e^{\mu x/\lambda }-1)}^4{x}^4\}.\hfill \end{array}$$

Proof.

By using the linearity of the operators (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) , $$\begin{array}{lll}{P}_{\lambda }^{\theta }({\psi }_x^1;x)\hfill & =\hfill & P_{\lambda }^{\theta }(e_1;x)-x{P}_{\lambda }^{\theta }(e_0;x)\hfill \\ P_{\lambda }^{\theta }({\psi }_x^2;x)\hfill & =\hfill & P_{\lambda }^{\theta }(e_2;x)-2x{P}_{\lambda }^{\theta }(e_1;x)+x^2{P}_{\lambda }^{\theta }(e_0;x)\hfill \\ P_{\lambda }^{\theta }({\psi }_x^4;x)\hfill & =\hfill & P_{\lambda }^{\theta }(e_4;x)-4x{P}_{\lambda }^{\theta }(e_3;x)+6x^2{P}_{\lambda }^{\theta }(e_2;x)-4x^3{P}_{\lambda }^{\theta }(e_1;x)+x^4{P}_{\lambda }^{\theta }(e_0;x)\hfill \end{array}$$ the proof is completed.

Furthermore, we obtain the limits of the central moments as follows: $$\begin{array}{lll}\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda P_{\lambda }^{\theta }({\psi }_x^1;x)\hfill & =\hfill & -\frac{3\mu }{2}{x}^2,\hfill \\ \underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda P_{\lambda }^{\theta }({\psi }_x^2;x)\hfill & =\hfill & x^2,\hfill \\ \underset{\lambda \to \infty }{\mathrm{\text{lim}}}{\lambda }^2{P}_{\lambda }^{\theta }({\psi }_x^4;x)\hfill & =\hfill & 3x^4.\hfill \end{array}$$

1.1. Preliminaries

In the current part, we give the main definitions which will be used in the paper. Let the subspace of real-valued continuous functions on $(0,\infty )$ which possess finite limit at infinity be denoted by $C^{*}(0,\infty ).$ The space $C^{*}(0,\infty )$ is equipped with the uniform norm. In the following remark, we will check that the operators $P_{\lambda }^{\theta }(f;x)$ belong to the space $C^{*}(0,\infty ).$

In 2010, Holhoş (Holhoş, 2010) come to grips with the uniform convergence of the linear positive operators and obtained the following theorem with the help of modulus of continuity (6) $${\omega }^{*}(f,\eta )=\underset{\underset{x,t>0}{|e^{-x}-e^{-t}|\le \eta }}{\mathrm{\text{sup}}}|f(t)-f(x)|.$$(6)

Theorem

(Holhoş, 2010). Let $B_{\lambda }:C^{*}[0,\infty )\to C^{*}[0,\infty )$ be a sequence of linear positive operators, then $${\Vert B_{\lambda }(f;x)-f(x)\Vert }_{[0,\infty )}\le {\Vert f\Vert }_{[0,\infty )}{\delta }_{\lambda }+(2+{\delta }_{\lambda }){\omega }^{*}(f,\sqrt{{\delta }_{\lambda }+2{\sigma }_{\lambda }+{\rho }_{\lambda }})$$ for every function $f\in C^{*}[0,\infty )$, where $$\begin{array}{l}{\Vert B_{\lambda }(e_0;x)-1\Vert }_{[0,\infty )}={\delta }_{\lambda },\hfill \\ {\Vert B_{\lambda }(e^{-t};x)-e^{-x}\Vert }_{[0,\infty )}={\sigma }_{\lambda },\hfill \\ {\Vert B_{\lambda }(e^{-2t};x)-e^{-2x}\Vert }_{[0,\infty )}={\rho }_{\lambda }\hfill \end{array}$$

As $\lambda \to \infty$ the variables ${\delta }_{\lambda },{\sigma }_{\lambda }$ and ${\rho }_{\lambda }$ tend to zero.

Let the class of all bounded and uniform continuous functions f on $(0,\infty )$ be denoted by $C_B(0,\infty )$, equipped with the norm ${\Vert f\Vert }_{C_B}=\underset{x>0}{\mathrm{\text{sup}}}|f(x)|$. The modulus of continuity of the function $f\in C_B(0,\infty )$ is defined by $$\omega (f,\delta ):=\underset{0<k<\delta }{\mathrm{\text{sup}}}\underset{\mathrm{ }x,x+k\in J}{\mathrm{\text{sup}}}|f(x+k)-f(x)|,$$

Moreover, for $f\in C_B(0,\infty )$, the second order modulus of continuity is described as $${\omega }_2(f,\delta ):=\underset{0<k<\sqrt{\delta }}{\mathrm{\text{sup}}}\underset{\mathrm{ }x,x+k\in J}{\mathrm{\text{sup}}}|f(x+2k)-2f(x+k)+f(x)|,$$ where $\delta >0$. Furthermore, Peetre’s $\mathcal{K}$-functionals are defined by $${\mathcal{K}}_2(f,\delta ):=\underset{h\in C_B^2(0,\infty )}{\mathrm{\text{inf}}}\{\left||f-h\right|{|}_{C_B(0,\infty )}+\delta \left||h\right|{|}_{C_B^2(0,\infty )}\}.$$ $C_B^2(0,\infty )$ indicates the space of the functions f, for which f, $f\prime$ and $f″$ be a member of $C_B(0,\infty )$. In 1993, DeVore (DeVore & Lorentz, 1993) represented the relation between the second order modulus of continuity and Peetre’s $\mathcal{K}$-functional as follows $${\mathcal{K}}_2(f,\delta )\le M{\omega }_2(f,\sqrt{\delta }),$$ where M > 0.

The structure of this paper is as follows: In the next section, we give uniform convergence theorem as a main result. In Section 3, we give approximation results by using the modulus of continuity. In Section 3, we mention Voronovskaya-type theorem for asymptotic estimation. In last section, we present the comparison of the newly constructed operators with different operators theoretically and graphically.

2. Main results

In this part, we will briefly study the approximation properties of the new constructed operators $P_{\lambda }^{\theta }.$

Remark 1.

Let $\lambda \in {\mathbb{N}}^{+},x,t\in (0,\infty )$ and $$S_{\lambda }^{\theta }(x,t)=\frac{{\lambda }^{\lambda }}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{t}^{\lambda -1}{e}^{\mu x-\left(\frac{\lambda }{{\theta }_{\lambda }(x)}-\mu \right)}t.$$

Then we write $$P_{\lambda }^{\theta }(f;x)={\int }_0^{\infty }{S}_{\lambda }^{\theta }(x,t)f(t)\mathit{\text{dt}}$$ and we already know from Lemma 3 that $$P_{\lambda }^{\theta }(e_0;x)=1.$$

We conclude from the exponential component that $S_{\lambda }^{\theta }(x,t)$ is bounded. So, we can write $S_{\lambda }^{\theta }(x,t)\le K$ for fixed K and $x,t>0.$ Let $f\ne 0,f\in C(0,\infty )$ and $\underset{x\to \infty }{\mathrm{\text{lim}}}f(x)=0.$ For a fixed positive ϵ, such a positive L exists such that $\left|f(x)\right|\le \frac{ϵ}{2}$ for all $x\ge L.$ Let $a:=\frac{ϵ}{2K\left|\right|f|{|}_{\infty }}$ and $x\ge L,$ then we write $$\begin{array}{ccc}{\int }_a^{\infty }{S}_{\lambda }^{\theta }(x,t)\mathit{\text{dt}}\hfill & \le \hfill & {\int }_0^{\infty }{S}_{\lambda }^{\theta }(x,t)\mathit{\text{dt}}\hfill \\ \hfill & =\hfill & P_{\lambda }^{\theta }(1;x)\hfill \\ \hfill & =\hfill & \frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\frac{1}{\mathit{\text{ex}}{p}_{\mu }}\left(t\right)\mathit{\text{dt}}\hfill \\ \hfill & =\hfill & 1\hfill \end{array}$$ and $t\ge a\ge L,$ hence $\left|f(x)\right|\le \frac{ϵ}{2},\forall t\ge a.$ $$\begin{array}{ccc}\left|P_{\lambda }^{\theta }(f;x)\right|\hfill & \le \hfill & {\int }_0^{\infty }{S}_{\lambda }^{\theta }(x,t)|f(t)|\mathit{\text{dt}}\hfill \\ \hfill & =\hfill & {\int }_a^{\infty }{S}_{\lambda }^{\theta }(x,t)|f(t)|\mathit{\text{dt}}+{\int }_0^a{S}_{\lambda }^{\theta }(x,t)|f(t)|\mathit{\text{dt}}\hfill \\ \hfill & \le \hfill & \frac{ϵ}{2}{\int }_a^{\infty }{S}_{\lambda }^{\theta }(x,t)\mathit{\text{dt}}+\left|\right|f|{|}_{\infty }K{\int }_0^{a}dt\hfill \\ \hfill & \le \hfill & \frac{ϵ}{2}+\mathit{\text{aK}}\left|\right|f|{|}_{\infty }\hfill \\ \hfill & =\hfill & ϵ.\hfill \end{array}$$

Thusly, for all $x\ge L$ we have $\left|P_{\lambda }^{\theta }(f;x)\right|\le ϵ,$ scilicet, $\underset{x\to \infty }{\mathrm{\text{lim}}}f(x)=0.$ This result denotes that the operators $P_{\lambda }^{\theta }(f;x)$ belong to $C^{*}(0,\infty ).$ Finally, we write $P_{\lambda }^{\theta }:C^{*}(0,\infty )\to C^{*}(0,\infty ).$

In 1970, Boyanov and Veselinov (Boyanov & Veselinov, 1970) built a theorem in order to show uniform convergence of the linear positive operators. For the new constructed operators (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) , we adapt this theorem in the next form.

Theorem 1.

Let the sequence of linear positive operators $P_{\lambda }^{\theta }:C^{*}(0,\infty )\to C^{*}(0,\infty )$ satisfy $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}{P}_{\lambda }^{\theta }(e^{-\xi t};x)=e^{-\xi x},\xi =0,1,2,$$ uniformly in $(0,\infty )$. Then for each $f\in C^{*}(0,\infty ),$ $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}{P}_{\lambda }^{\theta }(f;x)=f(x)$$ is satisfied uniformly in $(0,\infty ).$

Proof.

From Lemma 1 and Lemma 3, we achieve (7) $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(1;x)\hfill & =\hfill & 1,\hfill \\ P_{\lambda }^{\theta }(e^{-t};x)\hfill & =\hfill & e^{\mu x}{\left(1+\frac{(1+\mu ){\theta }_{\lambda }(x)}{\lambda }\right)}^{-\lambda }\hfill \\ \hfill & =\hfill & e^{-x}+\frac{(1+\mu )(1+2\mu )x^2{e}^{-x}}{2\lambda }-\frac{(1+\mu )(1+2\mu )(2+3\mu )x^3{e}^{-x}}{6{\lambda }^2}\hfill \\ \hfill & \hfill & +\frac{{(1+\mu )}^2{(1+2\mu )}^2{x}^4{e}^{-x}}{8{\lambda }^2}+O({\lambda }^{-3})\hfill \end{array}$$(7) and (8) $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(e^{-2t};x)\hfill & =\hfill & e^{\mu x}{\left(1+\frac{(2+\mu ){\theta }_{\lambda }(x)}{\lambda }\right)}^{-\lambda }\hfill \\ \hfill & =\hfill & e^{-2x}+\frac{(1+\mu )(1+2\mu )x^2{e}^{-2x}}{\lambda }-\frac{(1+\mu )(1+2\mu )(4-3\mu )x^3{e}^{-2x}}{3{\lambda }^2}\hfill \\ \hfill & \hfill & +\frac{{(1+\mu )}^2(1+2\mu )(2+\mu )x^4{e}^{-2x}}{2{\lambda }^2}+O({\lambda }^{-3}),\hfill \end{array}$$(8) where ${\theta }_{\lambda }(x)$ given in (4)(4) $${\theta }_{\lambda }(x)=\frac{\lambda \left(e^{\frac{\mu x}{\lambda }}-1\right)}{\mu e^{\frac{\mu x}{\lambda }}}.$$(4) . By taking limit as λ goes to infinity, we complete the proof. Hereby, we obtain that $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}{P}_{\lambda }^{\theta }(e^{-\xi t};x)=e^{-\xi x},\xi =0,1,2,$$ uniformly in $(0,\infty ).$ Then for each $f\in C^{*}(0,\infty ),$ $\underset{\lambda \to \infty }{\mathit{\text{lim}}}{P}_{\lambda }^{\theta }(f;x)=f\left(x\right)$ converges uniformly in $(0,\infty ).$

Now, we present a quantitive estimation of the new construction of the Post-Widder operators $P_{\lambda }^{\theta }$ according to Holhoş’s (Holhoş, 2010) theorem as follows:

Theorem 2.

Let $P_{\lambda }^{\theta }:C^{*}[0,\infty )\to C^{*}(0,\infty )$ be a sequence of linear positive operators, then we write the next inequality for every $f\in C^{*}(0,\infty )$ $${\Vert P_{\lambda }^{\theta }f-f\Vert }_{(0,\infty )}\le 2{\omega }^{*}(f,\sqrt{2{\sigma }_{\lambda }+{\rho }_{\lambda }}),$$ where $$\begin{array}{ccc}{\Vert P_{\lambda }^{\theta }(e^{-t};x)-e^{-x}\Vert }_{(0,\infty )}\hfill & =\hfill & {\sigma }_{\lambda },\hfill \\ {\Vert P_{\lambda }^{\theta }(e^{-2t};x)-e^{-2x}\Vert }_{(0,\infty )}\hfill & =\hfill & {\rho }_{\lambda }.\hfill \end{array}$$

Here, $P_{\lambda }^{\theta }f$ converges uniformly to f. Additionally, ${\sigma }_{\lambda }$ and ${\rho }_{\lambda }$ tend to zero as $\lambda \to \infty .$

Proof.

From Holhoş’s theorem (Holhoş, 2010) and Lemma 3, we obtain that $${\delta }_{\lambda }={\Vert P_{\lambda }^{\theta }(1;x)-1\Vert }_{(0,\infty )}=0.$$

In order to calculate ${\sigma }_{\lambda },$ we take into account the equality (7)(7) $$\begin{array}{ccc}{P}_{\lambda }^{\theta }(1;x)\hfill & =\hfill & 1,\hfill \\ P_{\lambda }^{\theta }(e^{-t};x)\hfill & =\hfill & e^{\mu x}{\left(1+\frac{(1+\mu ){\theta }_{\lambda }(x)}{\lambda }\right)}^{-\lambda }\hfill \\ \hfill & =\hfill & e^{-x}+\frac{(1+\mu )(1+2\mu )x^2{e}^{-x}}{2\lambda }-\frac{(1+\mu )(1+2\mu )(2+3\mu )x^3{e}^{-x}}{6{\lambda }^2}\hfill \\ \hfill & \hfill & +\frac{{(1+\mu )}^2{(1+2\mu )}^2{x}^4{e}^{-x}}{8{\lambda }^2}+O({\lambda }^{-3})\hfill \end{array}$$(7) , $J=(0,\infty ),$ $$\begin{array}{ccc}{\sigma }_{\lambda }\hfill & =\hfill & {\Vert P_{\lambda }^{\theta }(e^{-t};x)-e^{-x}\Vert }_{(0,\infty )}\hfill \\ \hfill & =\hfill & \underset{x\in J}{\mathit{\text{sup}}}\left|P_{\lambda }^{\theta }(e^{-t};x)-e^{-x}\right|\hfill \\ \hfill & =\hfill & \underset{x\in J}{\mathit{\text{sup}}}|\frac{(1+\mu )(1+2\mu )x^2{e}^{-x}}{2\lambda }-\frac{(1+\mu )(1+2\mu )(2+3\mu )x^3{e}^{-x}}{6{\lambda }^2}+\frac{{(1+\mu )}^2{(1+2\mu )}^2{x}^4{e}^{-x}}{8{\lambda }^2}+O({\lambda }^{-3})|.\hfill \end{array}$$

By using the supremum values $$\underset{x\in J}{\mathit{\text{sup}}} x^2{e}^{-x}=\frac{4}{e^2},\underset{x\in J}{\mathit{\text{sup}}} x^3{e}^{-x}=\frac{27}{e^3},\underset{x\in J}{\mathit{\text{sup}}} x^4{e}^{-x}=\frac{256}{e^4},$$ we achieve $${\sigma }_{\lambda }\le \frac{2(1+\mu )(1+2\mu )}{\lambda e^2}+\frac{9(1+\mu )(1+2\mu )(2+3\mu )}{2{\lambda }^2{e}^3}+\frac{32{(1+\mu )}^2{(1+2\mu )}^2}{{\lambda }^2{e}^4}+O({\lambda }^{-3}).$$

We represent the convergence rate of ${\sigma }_{\lambda }$ in Table 1 for $\mu \in \{0.25,0.5,0.75,1,1.25\},x\in [0.1,20]$ and $\lambda \in \{50,75,100,125\}.$ In the same manner, $$\begin{array}{ccc}{\rho }_{\lambda }\hfill & =\hfill & {\Vert P_{\lambda }^{\theta }(e^{-2t};x)-e^{-2x}\Vert }_{(0,\infty )}\hfill \\ \hfill & =\hfill & \underset{x\in J}{\mathit{\text{sup}}}\left|P_{\lambda }^{\theta }(e^{-2t};x)-e^{-2x}\right|\hfill \\ \hfill & =\hfill & \underset{x\in J}{\mathit{\text{sup}}}|\frac{(1+\mu )(1+2\mu )x^2{e}^{-2x}}{\lambda }-\frac{(1+\mu )(1+2\mu )(4-3\mu )x^3{e}^{-2x}}{3{\lambda }^2}\hfill \\ \hfill & \hfill & +\frac{{(1+\mu )}^2(1+2\mu )(2+\mu )x^4{e}^{-2x}}{2{\lambda }^2}+O({\lambda }^{-3})|.\hfill \end{array}$$

By using $$\underset{x\in J}{\mathit{\text{sup}}} x^2{e}^{-2x}=\frac{1}{e^2},\underset{x\in J}{\mathit{\text{sup}}} x^3{e}^{-2x}=\frac{27}{8e^3},\underset{x\in J}{\mathit{\text{sup}}} x^4{e}^{-2x}=\frac{16}{e^4},$$ we obtain the inequality $${\rho }_{\lambda }\le \frac{(1+\mu )(1+2\mu )}{\lambda e^2}+\frac{9(1+\mu )(1+2\mu )(4-3\mu )}{8{\lambda }^2{e}^3}+\frac{8{(1+\mu )}^2(1+2\mu )(2+\mu )}{{\lambda }^2{e}^4}+O({\lambda }^{-3}).$$

We investigate the convergence rate of ${\rho }_{\lambda }$ in Table 2 numerically.

Finally, we analyze the results of Figures 1(a), (b) and Tables 1 and 2. Then we achieve that ${\sigma }_{\lambda }$ and ${\rho }_{\lambda }$ tend to zero as $\lambda \to \infty .$

Figure 1. About ${\sigma }_{\lambda }$ and ${\rho }_{\lambda }.$

3. Modulus of continuity

In this part, we mention the rate of convergence with the help of usual modulus of continuity.

Lemma 5

. Let $f\in C_B(0,\infty )$. Then we have $$|P_{\lambda }^{\theta }(f;x)|\le \left|\right|f\left|\right|.$$

Proof.

From (3)(3) $$P_{\lambda }^{\theta }(f;x):=P_{\lambda }^{\alpha ,\beta }(f;x)=\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\left(\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\right)\left(t\right)\mathit{\text{dt}},$$(3) , we get $$\begin{array}{ccc}|P_{\lambda }^{\theta }(f;x)|\hfill & =\hfill & \left|\frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\frac{\lambda t}{{\theta }_{\lambda }(x)}}\frac{f}{\mathit{\text{ex}}{p}_{\mu }}\left(t\right)\mathit{\text{dt}}\right|\hfill \\ \hfill & \le \hfill & \frac{{\lambda }^{\lambda }{e}^{\mu x}}{{({\theta }_{\lambda }(x))}^{\lambda }\Gamma (\lambda )}{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-\left(\frac{\lambda }{{\theta }_{\lambda }(x)}+\mu \right)t}\left|f(t)\right|\mathit{\text{dt}}\hfill \\ \hfill & \le \hfill & \left|\right|f\left|\right|P_{\lambda }^{\theta }(1;x)=\left|\right|f\left|\right|.\hfill \end{array}$$

Theorem 3.

Let $f\in C_B(0,\infty )$. Then for every $x\in J$, there is a constant M > 0, such that $$|P_{\lambda }^{\theta }(f;x)-f(x)|\le M{\omega }_2(f,\sqrt{{\eta }_{\lambda ,\mu }})+\omega \left(f,{\delta }_{\lambda ,\mu }\right),$$ where $$\begin{array}{ccc}{\eta }_{\lambda ,\mu }\hfill & =\hfill & \frac{e^{2\mu x}(\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2-2\mu \lambda (e^{\mu x/\lambda }-1)(2e^{\mu x/\lambda }-1)x+{\mu }^2{(2e^{\mu x/\lambda }-1)}^2{x}^2)}{{\mu }^2{(2e^{\mu x/\lambda }-1)}^{\lambda +2}}\hfill \\ \hfill & \hfill & +{\left(x-\frac{\lambda (e^{\mu x/\lambda }-1)}{\mu e^{2\mu x/\lambda }}\right)}^2,\hfill \\ {\delta }_{\lambda ,\mu }\hfill & =\hfill & \left|\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right|.\hfill \end{array}$$

Proof.

For the proof, we describe the auxiliary operators ${\tilde{P}}_{\lambda }^{\theta }:C_B(0,\infty )\to C_B[0,\infty )$ (9) $${\tilde{P}}_{\lambda }^{\theta }(h;x)=P_{\lambda }^{\theta }(h;x)+h(x)-h\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right).$$(9)

By using the Taylor expansion for a function $h\in C_B^2(0,\infty ),$ we have (10) $$h(t)=h(x)+(t-x)h\prime (x)+{\int }_x^t(t-u)h″(u)\mathit{\text{du}},x,t\in J.$$(10)

We apply ${\tilde{P}}_{\lambda }^{\theta }$ operators to the Eqn. (10)(10) $$h(t)=h(x)+(t-x)h\prime (x)+{\int }_x^t(t-u)h″(u)\mathit{\text{du}},x,t\in J.$$(10) . By using Lemma 4, we get (11) $$\begin{array}{l}\left|{\tilde{P}}_{\lambda }^{\theta }(h;x)-h(x)\right|=\left|{\tilde{P}}_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\\ \le \left|P_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\\ +\left|{\int }_x^{\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}}\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-u\right)h″(u)\mathit{\text{du}}\right|.\end{array}$$(11)

Further, (12) $$\left|P_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\le P_{\lambda }^{\theta }\left({\int }_x^t|t-u\left|\right|h″(u)|\mathit{\text{du}};x\right)\le \left|\right|h″\left|\right|P_{\lambda }^{\theta }({\psi }_x^2;x)$$(12) and (13) $$\left|{\int }_x^{\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}}\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-u\right)h″(u)\mathit{\text{du}}\right|\le \left|\right|h″\left|\right|{\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right)}^2.$$(13)

Using the inequalities (11)(11) $$\begin{array}{l}\left|{\tilde{P}}_{\lambda }^{\theta }(h;x)-h(x)\right|=\left|{\tilde{P}}_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\\ \le \left|P_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\\ +\left|{\int }_x^{\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}}\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-u\right)h″(u)\mathit{\text{du}}\right|.\end{array}$$(11) , (12)(12) $$\left|P_{\lambda }^{\theta }\left({\int }_x^t(t-u)h″(u)\mathit{\text{du}};x\right)\right|\le P_{\lambda }^{\theta }\left({\int }_x^t|t-u\left|\right|h″(u)|\mathit{\text{du}};x\right)\le \left|\right|h″\left|\right|P_{\lambda }^{\theta }({\psi }_x^2;x)$$(12) and (13)(13) $$\left|{\int }_x^{\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}}\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-u\right)h″(u)\mathit{\text{du}}\right|\le \left|\right|h″\left|\right|{\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right)}^2.$$(13) , we obtain (14) $$\begin{array}{l}\left|{\tilde{P}}_{\lambda }^{\theta }(h;x)-h(x)\right|\le \left|\right|h″||\left(P_{\lambda }^{\theta }({\psi }_x^2;x)+{\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right)}^2\right)\\ =\left|\right|h″||(\frac{e^{2\mu x}(\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2-2\mu \lambda (e^{\mu x/\lambda }-1)(2e^{\mu x/\lambda }-1)x+{\mu }^2{(2e^{\mu x/\lambda }-1)}^2{x}^2)}{{\mu }^2{(2e^{\mu x/\lambda }-1)}^{\lambda +2}}\\ +{\left(x-\frac{\lambda (e^{\mu x/\lambda }-1)}{\mu e^{2\mu x/\lambda }}\right)}^2)\\ :=\left|\right|h″\left|\right|{\eta }_{\lambda ,\mu },\end{array}$$(14) where $$\begin{array}{c}{\eta }_{\lambda ,\mu }=\frac{e^{2\mu x}(\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2-2\mu \lambda (e^{\mu x/\lambda }-1)(2e^{\mu x/\lambda }-1)x+{\mu }^2{(2e^{\mu x/\lambda }-1)}^2{x}^2)}{{\mu }^2{(2e^{\mu x/\lambda }-1)}^{\lambda +2}}\\ +{\left(x-\frac{\lambda (e^{\mu x/\lambda }-1)}{\mu e^{2\mu x/\lambda }}\right)}^2.\end{array}$$

From auxiliary operators (9)(9) $${\tilde{P}}_{\lambda }^{\theta }(h;x)=P_{\lambda }^{\theta }(h;x)+h(x)-h\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right).$$(9) and Lemma 5, we write (15) $$\left|\right|{\tilde{P}}_{\lambda }^{\theta }(f;x)\left|\right|\le \left|\right|P_{\lambda }^{\theta }(f;x)\left|\right|+2\left|\right|f\left|\right|\le 3\left|\right|f\left|\right|.$$(15)

By using (9)(9) $${\tilde{P}}_{\lambda }^{\theta }(h;x)=P_{\lambda }^{\theta }(h;x)+h(x)-h\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right).$$(9) , (14)(14) $$\begin{array}{l}\left|{\tilde{P}}_{\lambda }^{\theta }(h;x)-h(x)\right|\le \left|\right|h″||\left(P_{\lambda }^{\theta }({\psi }_x^2;x)+{\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right)}^2\right)\\ =\left|\right|h″||(\frac{e^{2\mu x}(\lambda (\lambda +1){(e^{\mu x/\lambda }-1)}^2-2\mu \lambda (e^{\mu x/\lambda }-1)(2e^{\mu x/\lambda }-1)x+{\mu }^2{(2e^{\mu x/\lambda }-1)}^2{x}^2)}{{\mu }^2{(2e^{\mu x/\lambda }-1)}^{\lambda +2}}\\ +{\left(x-\frac{\lambda (e^{\mu x/\lambda }-1)}{\mu e^{2\mu x/\lambda }}\right)}^2)\\ :=\left|\right|h″\left|\right|{\eta }_{\lambda ,\mu },\end{array}$$(14) , (15)(15) $$\left|\right|{\tilde{P}}_{\lambda }^{\theta }(f;x)\left|\right|\le \left|\right|P_{\lambda }^{\theta }(f;x)\left|\right|+2\left|\right|f\left|\right|\le 3\left|\right|f\left|\right|.$$(15) for every $h\in C_B^2(0,\infty )$ and by choosing ${\delta }_{\lambda ,\mu }=\left|\frac{\lambda (e^{\mu x/\lambda }-1)}{\mu e^{2\mu x/\lambda }}-x\right|,$ we have $$\begin{array}{ccc}|P_{\lambda }^{\theta }(f;x)-f(x)|\hfill & =\hfill & \left|{\tilde{P}}_{\lambda }^{\theta }(f;x)-f(x)+f\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right)-f(x)+{\tilde{P}}_{\lambda }^{\theta }(h;x)-{\tilde{P}}_{\lambda }^{\theta }(h;x)+h(x)-h(x)\right|\hfill \\ \hfill & \le \hfill & \left|{\tilde{P}}_{\lambda }^{\theta }(f-h;x)-(f-h)(x)\right|+\left|f\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right)-f(x)\right|+\left|{\tilde{P}}_{\lambda }^{\theta }(h;x)-h(x)\right|\hfill \\ \hfill & \le \hfill & 4\left|\right|f-h\left|\right|+\left|\right|h″\left|\right|{\eta }_{\lambda ,\mu }+\left|f\left(\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}\right)-f(x)\right|\hfill \\ \hfill & \le \hfill & {\mathcal{K}}_2(f,{\eta }_{\lambda ,\mu })+\omega \left(f,\left|\frac{\lambda \left(e^{\mu x/\lambda }-1\right)}{\mu e^{2\mu x/\lambda }}-x\right|\right)\hfill \\ \hfill & \le \hfill & M{\omega }_2(f,\sqrt{{\eta }_{\lambda ,\mu }})+\omega \left(f,{\delta }_{\lambda ,\mu }\right).\hfill \end{array}$$

Remark 2.

One can check that ${\eta }_{\lambda ,\mu }\to 0$ and ${\delta }_{\lambda ,\mu }\to 0$ as $\lambda \to \infty .$ This result guarantees the convergence of the Theorem 3.

4. Voronovskaya-type theorem

In this part, we mention some asymptotic estimation results of the pointwise convergence in the case of the functions with exponential growth.

Theorem 4.

For $f,f\prime ,f″\in C^{*}(0,\infty )$ and $x\in J,$ we write $$\begin{array}{ccc}\left|\lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)+\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)-\frac{x^2}{2}f″(x)\right|\hfill & \le \hfill & |q_{\lambda }(x)||f\prime (x)|+|s_{\lambda }(x)||f″(x)|\hfill \\ \hfill & \hfill & +2(2s_{\lambda }(x)+x^2+t_{\lambda }(x)){\omega }^{*}(f″,{\lambda }^{-1/2}),\hfill \end{array}$$ where $$\begin{array}{ccc}{q}_{\lambda }(x)\hfill & =\hfill & \lambda P_{\lambda }^{\theta }({\psi }_x^1;x)+\frac{3\mu }{2}{x}^2,\hfill \\ s_{\lambda }(x)\hfill & =\hfill & \frac{\lambda }{2}{P}_{\lambda }^{\theta }({\psi }_x^2;x)-\frac{x^2}{2},\hfill \\ t_{\lambda }(x)\hfill & =\hfill & {\lambda }^2\sqrt{P_{\lambda }^{\theta }({(e^{-x}-e^{-t})}^4;x)}\sqrt{P_{\lambda }^{\theta }({\psi }_x^4;x)}.\hfill \end{array}$$

Proof.

From Taylor’s formula, we can write (16) $$f(t)=f(x)+(t-x)f\prime (x)+\frac{{(t-x)}^2}{2}f″(x)+r(t,x){(t-x)}^2,$$(16) where (17) $$r(t,x):=\frac{f″(\xi )-f″(x)}{2}$$(17) is the remainder term. ξ is a number between x and t. If we apply the $P_{\lambda }^{\theta }$ operators to (16)(16) $$f(t)=f(x)+(t-x)f\prime (x)+\frac{{(t-x)}^2}{2}f″(x)+r(t,x){(t-x)}^2,$$(16) , we achieve $$P_{\lambda }^{\theta }(f;x)-f(x)=f\prime (x)P_{\lambda }^{\theta }({\psi }_x^1;x)+\frac{f″(x)}{2}{P}_{\lambda }^{\theta }({\psi }_x^2;x)+P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x).$$

After that, $$\begin{array}{ccc}\left|\lambda \left[P_{\lambda }^{\theta }(f;x)-f(x)\right]+\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)-\frac{x^2}{2}f″(x)\right|\hfill & \le \hfill & \left|\lambda P_{\lambda }^{\theta }({\psi }_x^1;x)+\frac{3\mu }{2}{x}^2\right|\left|f\prime (x)\right|+\frac{1}{2}\left|\lambda P_{\lambda }^{\theta }({\psi }_x^2;x)-x^2\right|\left|f″(x)\right|\hfill \\ \hfill & \hfill & +\left|\lambda P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x)\right|\hfill \\ \hfill & =\hfill & |q_{\lambda }(x)||f\prime (x)|+|s_{\lambda }(x)||f″(x)|+\left|\lambda P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x)\right|.\hfill \end{array}$$

We briefly denote that $q_{\lambda }(x):=\lambda P_{\lambda }^{\theta }({\psi }_x^1;x)+\frac{3\mu }{2}{x}^2$ and $s_{\lambda }(x):=\frac{1}{2}(\lambda P_{\lambda }^{\theta }({\psi }_x^2;x)-x^2).$ So, $$\left|\lambda \left[P_{\lambda }^{\theta }(f;x)-f(x)\right]+\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)-\frac{x^2}{2}f″(x)\right|\le |q_{\lambda }(x)||f\prime (x)|+|s_{\lambda }(x)||f″(x)|+\left|\lambda P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x)\right|.$$

Note that $q_{\lambda }(x)$ and $s_{\lambda }(x)$ tend to zero as $\lambda \to \infty$ from Eqn. (6)(6) $${\omega }^{*}(f,\eta )=\underset{\underset{x,t>0}{|e^{-x}-e^{-t}|\le \eta }}{\mathrm{\text{sup}}}|f(t)-f(x)|.$$(6) and Eqn. (6)(6) $${\omega }^{*}(f,\eta )=\underset{\underset{x,t>0}{|e^{-x}-e^{-t}|\le \eta }}{\mathrm{\text{sup}}}|f(t)-f(x)|.$$(6) . Right now, we concern about the part $\left|\lambda P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x)\right|$ with the reminder term. (18) $$\left|f(t)-f(x)\right|\le \left(1+\frac{{(e^{-x}-e^{-t})}^2}{{\eta }^2}\right){\omega }^{*}(f,\eta ),$$(18) where ${\omega }^{*}(f,\eta )$ is given by (6)(6) $${\omega }^{*}(f,\eta )=\underset{\underset{x,t>0}{|e^{-x}-e^{-t}|\le \eta }}{\mathrm{\text{sup}}}|f(t)-f(x)|.$$(6) . Using (17)(17) $$r(t,x):=\frac{f″(\xi )-f″(x)}{2}$$(17) and the inequality (18)(18) $$\left|f(t)-f(x)\right|\le \left(1+\frac{{(e^{-x}-e^{-t})}^2}{{\eta }^2}\right){\omega }^{*}(f,\eta ),$$(18) , we write $\left|r(t,x)\right|\le (1+\frac{{(e^{-x}-e^{-t})}^2}{{\eta }^2}){\omega }^{*}(f″,\eta ).$ For $\eta >0,$ if $|e^{-x}-e^{-t}|>\eta ,$ then $1<\frac{e^{-x}-e^{-t}}{\eta }<\frac{{(e^{-x}-e^{-t})}^2}{{\eta }^2},$ so $\left|r(t,x)\right|\le \frac{2{(e^{-x}-e^{-t})}^2}{{\eta }^2}{\omega }^{*}(f″,\eta )$ and if $|e^{-x}-e^{-t}|\le \eta ,$ then we get $\left|r(t,x)\right|\le 2{\omega }^{*}(f″,\eta ).$ Thusly, we have $$\left|r(t,x)\right|\le 2\left(1+\frac{{(e^{-x}-e^{-t})}^2}{{\eta }^2}\right){\omega }^{*}(f″,\eta ).$$

So, $$\begin{array}{ccc}\left|\lambda P_{\lambda }^{\theta }(r(t,x){\psi }_x^2;x)\right|\hfill & \le \hfill & \lambda P_{\lambda }^{\theta }(|r(t,x)|{\psi }_x^2;x)\hfill \\ \hfill & \le \hfill & 2\lambda {\omega }^{*}(f″,\eta )P_{\lambda }^{\theta }({\psi }_x^2;x)+\frac{2\lambda }{{\eta }^2}{\omega }^{*}(f″,\eta )P_{\lambda }^{\theta }\left({(e^{-x}-e^{-t})}^2{\psi }_x^2;x\right)\hfill \\ \hfill & \le \hfill & 2\lambda {\omega }^{*}(f″,\eta )P_{\lambda }^{\theta }({\psi }_x^2;x)+\frac{2\lambda }{{\eta }^2}{\omega }^{*}(f″,\eta )\sqrt{P_{\lambda }^{\theta }\left({(e^{-x}-e^{-t})}^4;x\right)}\sqrt{P_{\lambda }^{\theta }\left({\psi }_x^4;x\right)}.\hfill \end{array}$$

If we choose $\eta =1/\sqrt{\lambda }$ and $t_{\lambda }:=\sqrt{{\lambda }^2{P}_{\lambda }^{\theta }({(e^{-x}-e^{-t})}^4;x)}\sqrt{{\lambda }^2{P}_{\lambda }^{\theta }({\psi }_x^4;x)},$ we get $$\begin{array}{c}\left|\lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)+\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)-\frac{x^2}{2}f″(x)\right|\le |q_{\lambda }(x)||f\prime (x)|+|s_{\lambda }(x)||f″(x)|\\ +(4s_{\lambda }(x)+2x^2+2t_{\lambda }(x)){\omega }^{*}(f″,{\lambda }^{-1/2}).\end{array}$$

Remark 3.

We obtain the following result by direct calculations $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}{\lambda }^2{P}_{\lambda }^{\theta }({\psi }_x^4;x)=3x^4.$$

Additionally, we get the following result $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}{\lambda }^2{P}_{\lambda }^{\theta }({(e^{-t}-e^{-x})}^4;x)=3x^4{e}^{-4x}.$$

We present the next corollary as an important result of Theorem 4 and Remark 3.

Corollary 5.

Let $f,f\prime ,f″\in C^{*}(0,\infty )$ and $x\in J$. Then we have (19) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)=-\frac{3\mu }{2}{x}^{2}f\prime (x)+\frac{x^2}{2}f″(x).$$(19)

5. Comparison with Post-Widder operators preserving $e^{2\mathit{\text{ax}}},a>0$

Now, we give a theorem that the $P_{\lambda }^{\theta }(f;x)$ operators preserving both $e^{\mu x}$ and $e^{2\mu x},\mu >0$ approximate better than the Post-Widder operators $P_{\lambda ,\alpha }^{*}(f;x)$ preserving $e^{2\mathit{\text{ax}}},a>0$ (Sofyalıoğlu & Kanat, 2020).

Corollary 6.

Let $f\in C^2(0,\infty )$ be a decreasing and convex function. Then there is a natural number m₀ such that for $m\ge m_0$, we write $f(x)<P_{\lambda }^{\theta }(f;x)$ for all x > 0.

Theorem 7.

Let $f\in C^2(0,\infty )$. Assume that there exists $m_0\in \mathbb{N}$ such that for all $m\ge m_0,a>\frac{3\mu }{2},x\in (0,\infty )$ (20) $$f(x)\le P_{\lambda }^{\theta }(f;x)\le P_{\lambda ,\alpha }^{*}(f;x).$$(20)

Then (21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21)

In particular, $f″(x)\ge 0.$

Conversely, if (21)(21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21) holds with strict at a point $x\in (0,\infty )$, then there is a natural number m₀ such that for $m\ge m_0$ $$f(x)<P_{\lambda }^{\theta }(f;x)<P_{\lambda ,\alpha }^{*}(f;x).$$

Proof.

From (20)(20) $$f(x)\le P_{\lambda }^{\theta }(f;x)\le P_{\lambda ,\alpha }^{*}(f;x).$$(20) we have for all $m\ge m_0$ and $x\in (0,\infty )$ that $$0\le \lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)\le \lambda \left(P_{\lambda ,\alpha }^{*}(f;x)-f(x)\right).$$

By taking into consideration the Voronovskaya-type theorem given in (Sofyalıoğlu & Kanat, 2020), we write (22) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda ,\alpha }^{*}(f;x)-f(x)\right)=-a{x}^2{f}^{\mathrm{\prime }} (x)+\frac{x^2}{2}f″(x).$$(22)

From Eqn. (19)(19) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)=-\frac{3\mu }{2}{x}^{2}f\prime (x)+\frac{x^2}{2}f″(x).$$(19) and Eqn. (22)(22) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda ,\alpha }^{*}(f;x)-f(x)\right)=-a{x}^2{f}^{\mathrm{\prime }} (x)+\frac{x^2}{2}f″(x).$$(22) , we have the following inequality $$0\le -\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)+\frac{x^2}{2}f″(x)\le -a{x}^2{f}^{\mathrm{\prime }}(x)+\frac{x^2}{2}f″(x).$$

So, we directly obtain inequality (21)(21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21) .

On the contrary, if inequality (21)(21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21) holds with strict at a given point $x\in (0,\infty ),$ then $$0<-\frac{3\mu }{2}{x}^2{f}^{\mathrm{\prime }}(x)+\frac{x^2}{2}f″(x)<-a{x}^2{f}^{\mathrm{\prime }}(x)+\frac{x^2}{2}f″(x).$$

After that, by using Eqn. (19)(19) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda }^{\theta }(f;x)-f(x)\right)=-\frac{3\mu }{2}{x}^{2}f\prime (x)+\frac{x^2}{2}f″(x).$$(19) and Eqn. (22)(22) $$\underset{\lambda \to \infty }{\mathrm{\text{lim}}}\lambda \left(P_{\lambda ,\alpha }^{*}(f;x)-f(x)\right)=-a{x}^2{f}^{\mathrm{\prime }} (x)+\frac{x^2}{2}f″(x).$$(22) ve get the desired result.

Now, we illustrate various examples to verify our theoretical results. When specifying the functions for the following two examples, we take into account the inequality (21)(21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21) . Then we compare the operators $P_{\lambda }^{\theta }$ with $P_{\lambda ,\alpha }^{*}$ (Sofyalıoğlu & Kanat, 2020), and we present the attractive events.

Example 1.

Let we choose $f(x)=e^{-5x},[0.1,5].$ In Figure 2, we illustrate the convergence of the new constructed $P_{\lambda }^{\theta }(f;x)$ operators to the function $f(x)=e^{-5x}$ for different $\mu \in \{0.25,0.5,1\}$ and $\lambda =10,40,100$ values. We see that we obtain pleasant convergence results from these choices. In Figure 3, we compare $P_{\lambda }^{\theta }(f;x),\mu =0.25$ (red), $P_{\lambda ,\alpha }^{*}(f;x),a=2$ (green) and the original function $f(x)=e^{-5x}$ (blue).

Figure 2. Behaviour of $P_{\lambda }^{\theta }$ operators for various values of λ and μ.

Figure 3. Comparison of operators $P_{\lambda }^{\theta }(f;x)$ and $P_{\lambda ,\alpha }^{*}(f;x)$ for different λ values.

From the Figure 3 and Table 3, one can easily arrive that, the operators $P_{\lambda }^{\theta }(f;x)$ converge more rapidly than $P_{\lambda ,\alpha }^{*}(f;x)$ to the function $f(x)=e^{-5x}.$

Table 1. The values of ${\sigma }_{\lambda }$ for various λ and μ in the interval $x\in [0.1,20].$

λ	$\mu =0.25$	$\mu =0.5$	$\mu =0.75$	μ =  1	$\mu =1.25$
50	0.0101506913953	0.0164649246670	0.0244687401054	0.034373606654	0.114654279
75	0.0067620928581	0.0109199853118	0.0161298883207	0.022484666940	0.030102290
100	0.0050695332037	0.0081681916317	0.0120276616287	0.016699839271	0.022250028
125	0.0040546102332	0.0065239445991	0.0095883173189	0.013280805769	0.017642789

Table 2. The values of ${\rho }_{\lambda }$ for various λ and μ in the interval $x\in [0.1,20].$

λ	$\mu =0.25$	$\mu =0.5$	$\mu =0.75$	μ =  1	$\mu =1.25$
50	0.0075337890598	0.0101015913706	0.0130553239149	0.0164151540271	0.0202047397
75	0.0050244938197	0.0067247763708	0.0086713306489	0.0108731469581	0.0133407054
100	0.0037690406841	0.0050397960288	0.0064910512850	0.0081278703462	0.0099561368
125	0.0030155365685	0.0040299724812	0.0051867714986	0.0064891769777	0.0079409496

Table 3. Comparison Of the errors for $P_{\lambda ,\alpha }^{*},a=2$ and $P_{\lambda }^{\theta }$ with respect to various values of $\mu \in \{0.25,0.5,1\},$ for function $f(x)=e^{-5x}$ in $x\in [0.1,4].$

λ	$P_{\lambda ,\alpha }^{*},a=2$	$P_{\lambda }^{\theta },\mu =0.25$	$P_{\lambda }^{\theta },\mu =0.5$	$P_{\lambda }^{\theta },\mu =1$
10	0.0211165837883	0.0104160822281	0.0128387920967	0.0193125431808
25	0.0060255257096	0.0034631541717	0.0040218632162	0.0055365258294
50	0.0027102082215	0.0017414012825	0.0019895931130	0.0025308291489
75	0.0018110450521	0.0011631367734	0.0013290402874	0.0016908893096
100	0.0013598799550	0.0008731813121	0.0009977755384	0.0012695484868
125	0.0010886717521	0.0006989435513	0.0007986990840	0.0010163032756

Example 2.

Let $f(x)=(2x^4+3x^2+6)e^{-2x},[0.1,1.5].$ In Figure 4, we compare $P_{\lambda }^{\theta }(f;x),\mu =0.25$ (red), $P_{\lambda ,\alpha }^{*}(f;x),a=2$ (green) and the original function $f(x)=(2x^4+3x^2+6)e^{-2x}$ (blue) for $\lambda \in \{25,50,75\}.$

Figure 4. Comparison of operators $P_{\lambda }^{\theta }(f;x)$ and $P_{\lambda ,\alpha }^{*}(f;x)$ for different λ values.

From the Figure 4 and Table 4, one can see that, the operators $P_{\lambda }^{\theta }(f;x)$ converge more rapidly than $P_{\lambda ,\alpha }^{*}(f;x)$ to the selected function $f(x)=(2x^4+3x^2+6)e^{-2x}.$

Table 4. Comparison of the errors for $P_{\lambda ,\alpha }^{*},a=2$ and $P_{\lambda }^{\theta }$ with respect to various values of $\mu \in \{0.25,0.5,0.6\},$ for function $f(x)=(2x^4+3x^2+6)e^{-2x}$ in $x\in [0.1,5].$

λ	$P_{m,\alpha }^{*},a=2$	$P_{\lambda }^{\theta },\mu =0.25$	$P_{\lambda }^{\theta },\mu =0.5$	$P_{\lambda }^{\theta },\mu =0.6$
10	0.3580078363242	0.1636569640768	0.2428086366319	0.2812895777007
25	0.1356249060185	0.0645460265532	0.0944989489807	0.1090011457796
50	0.0660639319941	0.0320722662244	0.0466945717810	0.0537644561336
75	0.0436121407380	0.0213329668881	0.0309963808286	0.0356665216378
100	0.0325414463369	0.0159809493762	0.0231958027528	0.0266818157290
125	0.0259510498680	0.0127755988459	0.0185315429465	0.0213122775669

6. Conclusion

In this paper, we construct a new modification of Post-Widder operators which preserve both $e^{\mu x}$ and $e^{2\mu x}$ for positive μ. After that, we submit the uniform convergence theorem, rate of convergence theorems and Voronovskaya-type theorem. Moreover, we give an comparison theorem by using the results of Voronovskaya-type theorem of the new operators $P_{\lambda }^{\theta }$ and the reference operators $P_{\lambda ,\alpha }^{*}.$ According to the conditions (21)(21) $$\frac{1}{2}f″(x)\ge \left(\frac{3\mu }{2}-a\right)f\prime (x)\ge 0.$$(21) of the comparison theorem, we determine the functions that are used in examples. Finally, the effectiveness of the newly constructed operators is shown with several graphs and error comparison tables. For the future work, Stancu-type generalization of the Post-Widder operators preserving two exponential functions can be investigated.

Disclosure statement

No potential conflict of interest was reported by the authors.