Durrmeyer variant of certain approximation operators

Abstract

In the present article, we introduce a Durrmeyer variant of certain approximation operators. We estimate the moment-generating function and moments of these operators employing the Lambert W function and establish some direct results. We further provide a composition of these operators with Szász–Mirakjan operators and estimate direct results for the composition operator. Additionally, we provide a graphical comparison of the approximation properties of the operators.

1 Introduction

Theory of positive linear operators is a very active topic of research due to its significance in computer-aided graphics design, mathematical finance, differential equations, etc. In recent years, several new operators have been constructed by combining the existing approximation operators. In [2], Abel and Gupta gave some operators by combining certain integral-type operators with discrete operators. In [7], Govil et al. studied some new classes of Durrmeyer variants of certain operators. In [8], Gupta et al. discussed Baskakov type Pólya-Durrmeyer operators.

For any function $f: [0,\infty ) \rightarrow \textrm{R},$ the Szász–Mirakjan operators [12] are defined as

$$\begin{aligned} \left( S_\lambda f \right) (x)= \sum _{v=0}^{\infty } s_{\lambda ,v}(x) f\left( \frac{v}{\lambda }\right) , \end{aligned}$$

(1)

and the Szász–Mirakjan–Durrmeyer operators are defined as

$$\begin{aligned} \left( \overline{S}_\lambda f \right) (x)= \lambda \sum _{j=0}^{\infty } s_{\lambda ,j}(x) \int _{0}^{\infty } s_{\lambda ,j}(t) f(t) dt, \end{aligned}$$

(2)

where $s_{\lambda ,j}(x)= e^{-\lambda x} \frac{(\lambda x)^j}{j!},\; x \in [0, \infty )$ and $\lambda \in \textrm{N}.$

If we take $R_\lambda f:= \left( S_\lambda \circ \overline{S}_\lambda f\right)$, then we get integral operators of Durrmeyer-type, unlike the reverse order composition of $\overline{S}_\lambda \circ S_\lambda$, which is a discrete operator [1] and the new Durrmeyer-type operators are given by

$$\begin{aligned} (R_\lambda f)(x)= e^{-\lambda x} \sum _{j=0}^{\infty } \frac{1}{{(j!)}^2} \sum _{v=0}^{\infty } \frac{e^{-v} (\lambda x)^v v^j}{v!}\int _{0}^{\infty } e^{-u}u^j f \left( \frac{u}{\lambda } \right) du. \end{aligned}$$

(3)

By simple computation with $\exp _A (u)= e^{Au}$, we have

$$\begin{aligned}{} & {} \left( S_{\lambda }\exp _{A}\right) \left( x\right) =\exp \left( \lambda x\left( e^{A/\lambda }-1\right) \right) ,\\{} & {} (\overline{S}_{\lambda } \exp _{A})\left( x\right) = \frac{\lambda }{\lambda -A} \exp \left( \frac{\lambda Ax}{\lambda -A}\right) \qquad \left( \lambda >A \right) , \end{aligned}$$

and

$$\begin{aligned} (R_\lambda \exp _A)(x)= \frac{\lambda }{\lambda -A} \exp \bigg (\lambda x \left( e^{\frac{A}{\lambda -A}}-1\right) \bigg ) \qquad \left( \lambda >A \right) . \end{aligned}$$

Very recently, Gupta-Sharma [9] introduced a new discretely defined approximation operator (4), by combining the two exponential operators, namely the Ismail-May operator [11] and the Szász–Mirakjan operators, which are respectively connected to $x(1+x)^2$ and x.

$$\begin{aligned} \left( \mathcal {L}_{\lambda } f \right) (x) = \sum _{j=0}^{\infty } {\phi }_{\lambda ,j}(x) f\left( \frac{j}{\lambda }\right) , \end{aligned}$$

(4)

where

$$\begin{aligned} \phi _{\lambda , j}(x) = e^{ \frac{- \lambda x}{1+x}} \sum _{v=0}^{\infty } \frac{\lambda (\lambda +v)^{v-1}}{v!}\left( \frac{x}{1+x}\right) ^v e^{- v \left( \frac{1+2x}{1+x} \right) }\frac{v^j}{j!}. \end{aligned}$$

The MGF of this operator is given by

$$\begin{aligned} \left( \mathcal {L}_{\lambda } \exp _A \right) (x)= \exp \bigg (-\lambda \left\{ W\left( \frac{-x}{1+x} \exp \left( e^{A/\lambda }+\frac{1}{1+x}-2\right) \right) +\frac{x}{1+x}\right\} \bigg ), \end{aligned}$$

where W stands for the Lambert W function. These discrete operators $\mathcal {L}_{\lambda }$ are not suitable enough to approximate Lebesgue integrable functions. We overcome this issue by presenting the Durrmeyer variant of these operators, by taking Szász–Mirakjan weight function, in the following form:

$$\begin{aligned} \left( \mathcal {D}_\lambda f \right) (x) = \lambda \sum _{j=0}^{\infty } \phi _{\lambda , j}(x) \int _{0}^{\infty } s_{\lambda ,j}(t) f(t) dt, \end{aligned}$$

(5)

where $\phi _{\lambda , j}(x)$ and $s_{\lambda ,j}(t)$ are as defined above.

This article deals with the convergence properties of the operators $\mathcal {D}_\lambda$. We estimate moment-generating function and moments of these operators via the Lambert W function and establish some direct results. In the next sections, we further consider composition of these operators with Szász–Mirakjan operators and estimate direct results. Finally, we provide a graphical comparison of their approximation properties.

2 Estimation of moments

Lemma 1

For $\lambda \in \textrm{N}$, the MGF of the operators $\mathcal {D}_\lambda$ is given by

$$\begin{aligned} \left( \mathcal {D}_\lambda \exp _{A}\right) (x)= \frac{\lambda }{\lambda -A} \exp \bigg ( \frac{-\lambda x}{1+x}-\lambda W\left( \frac{-x}{1+x} e^{\frac{\lambda }{\lambda -A}-\frac{(1+2x)}{1+x}}\right) \bigg ),\; \lambda >A, \end{aligned}$$

where W denotes the Lambert W function and $\exp _A(q)= e^{Aq}$.

Proof

From the definition of $\mathcal {D}_\lambda$, we have

$$\begin{aligned}&\left( \mathcal {D}_\lambda \exp _A \right) (x) \\&\quad = \lambda \sum _{j=0}^{\infty } e^{\frac{-\lambda x}{1+x}} \sum _{v=0}^{\infty } \frac{\lambda (\lambda + v)^{v-1}}{v!} \left( x(1+x)^{-1}\right) ^v e^{\frac{-v(1+2x)}{1+x}}\frac{v^j}{j!}\int _{0}^{\infty }\frac{e^{-(\lambda -A) t}(\lambda t)^j}{j!} dt\\&\quad =\frac{\lambda }{\lambda - A} \sum _{j=0}^{\infty } e^{\frac{-\lambda x}{1+x}}\sum _{v=0}^{\infty } \frac{\lambda (\lambda + v)^{v-1}}{v!}\left( x(1+x)^{-1}\right) ^v e^{\frac{-v(1+2x)}{1+x}}\frac{1}{j!}\left( \frac{\lambda v}{\lambda -A}\right) ^j\\&\quad = \frac{\lambda }{\lambda - A} e^{\frac{-\lambda x}{1+x}} \sum _{v=0}^{\infty } \frac{\lambda (\lambda + v)^{v-1}}{v!} \left( x(1+x)^{-1}\right) ^v \left( e^{\frac{-(1+2x)}{1+x}}\right) ^v \left( e^{ \frac{\lambda }{\lambda - A}}\right) ^v. \end{aligned}$$

Since for $x \ge 0$, we have $\frac{-x}{1+x} e^{\frac{-(1+2x)}{1+x}} e^{ \frac{\lambda }{\lambda - A}}>\frac{-1}{e}$, therefore there exists s with $|s|<1$, such that $\frac{-x}{1+x} e^{\frac{-(1+2x)}{1+x}} e^{ \frac{\lambda }{\lambda - A}} = -s e^{-s}$. By the definition of Lambert W function, $W\left( \frac{- x}{1+x} e^{\frac{-(1+2x)}{1+x}} e^{ \frac{\lambda }{\lambda - A}}\right) =-s$. Using the following inversion formula, given by Lagrange

$$\begin{aligned} e^{\alpha z} = \sum _{k=0}^{\infty }\frac{ \alpha (\alpha + k )^{k-1}}{k!} (z e^{-z})^k, \end{aligned}$$

with $0<\alpha <\infty$ and $|z|<1$, we get

$$\begin{aligned}&\left( \mathcal {D}_\lambda \exp _A \right) (x) \\&\quad = \frac{\lambda }{\lambda - A} e^{\frac{-\lambda x}{1+x}} \sum _{v=0}^{\infty } \frac{\lambda (\lambda + v)^{v-1}}{v!} (s e^{-s})^v\\&\quad =\frac{\lambda }{\lambda - A} e^{\frac{-\lambda x}{1+x}} e^{\lambda s}\\&\quad = \frac{\lambda }{\lambda - A} \exp \bigg ( \frac{-\lambda x}{1+x} - \lambda W \left( \frac{- x}{1+x} e^{\frac{-(1+2x)}{1+x}} e^{ \frac{\lambda }{\lambda - A}} \right) \bigg ) \qquad \left( \lambda >A \right) , \end{aligned}$$

hence the lemma follows. $\square$

Remark 1

Let us denote the q-th order moments for the operators $\mathcal {D}_\lambda$ by $\left( \mathcal {D}_\lambda e_q \right) (x)$, then these can be obtained by the following relation between them and moment-generating function:

$$\begin{aligned} \left( \mathcal {D}_\lambda e_q \right) (x)=\left[ \frac{\partial ^q}{\partial A^q} \left\{ \frac{\lambda }{\lambda -A}\exp \bigg ( \frac{-\lambda x}{1+x}-\lambda W\left( \frac{-x}{1+x} e^{\frac{\lambda }{\lambda -A}-\frac{(1+2x)}{1+x}}\right) \bigg )\right\} \right] _{A=0}, \end{aligned}$$

where $e_q(t)= t^q,\, q =0,1,2, \cdots$. Similarly, the central moments, denoted by $\mu _{\lambda ,q}(x)= \left( \mathcal {D}_\lambda (e_1-x e_0)^q\right) (x)$, may be obtained using the following relation:

$$\begin{aligned} \mu _{\lambda ,q}(x)=\left[ \frac{\partial ^q}{\partial A^q} \left\{ \frac{\lambda }{\lambda -A}\exp \bigg ( \frac{-\lambda x}{1+x}-\lambda W\left( \frac{-x}{1+x} e^{\frac{\lambda }{\lambda -A}-\frac{(1+2x)}{1+x}}\right) -Ax\bigg )\right\} \right] _{A=0}, \end{aligned}$$

where $q=0,1,2,\cdots .$

Lemma 2

The moments for $\mathcal {D}_\lambda$ follow this linear combination:

$$\begin{aligned}&\sum _{q\ge 0} c_q \left( \mathcal {D}_\lambda e_q \right) (x)\\&\quad = c_0+ \left( x+\frac{1}{\lambda }\right) c_1+\left( x^2+ \frac{x(5+2x+x^2)}{\lambda }+\frac{2}{\lambda ^2}\right) c_2\\&\quad + \left( x^3+\frac{3x^2(x^2+2x+4)}{\lambda } +\frac{x(3x^4+10x^3+21x^2+24x+28)}{\lambda ^2}+\frac{6}{\lambda ^3}\right) c_3\\&\quad + \left( x^4+ \frac{6x^5+12 x^4+ 22 x^3}{\lambda }+ \frac{15 x^6+52 x^5+114 x^4+ 132 x^3+ 127 x^2}{\lambda ^2} \right. \\&\quad \left. + \frac{15 x^7+ 70 x^6+ 179 x^5+ 284 x^4+ 325 x^3+ 254 x^2 + 185 x}{\lambda ^3}+ \frac{24}{\lambda ^2}\right) c_4+ ..... , \end{aligned}$$

where $c_q$’s are arbitrary constants and $q \in \textrm{N} \cup \{0\}$.

Proof

The proof follows by the application of Lemma 1 and Remark 1. $\square$

Lemma 3

The central moments for $\mathcal {D}_\lambda$ follow the linear combination as follows:

$$\begin{aligned} \sum _{q\ge 0} c_q \mu _{\lambda ,q}(x)&= c_0+ c_1 \frac{1}{\lambda } + c_2 \left( \frac{x(x^2+2x+3)}{\lambda }+ \frac{2}{\lambda ^2}\right) \\&+ c_3 \left( \frac{x(3x^4+10x^3+21x^2+24x+22)}{\lambda ^2}+\frac{6}{\lambda ^3}\right) \\&+ c_4 \left( \frac{15x^7+70 x^6+179x^5+284 x^4+325 x^3+254 x^2+161 x }{\lambda ^3 } \right. \\&\left. + \frac{ 3x^6+ 12 x^5+30 x^4+36 x^3+27x^2}{\lambda ^2 } +\frac{24}{\lambda ^4}\right) + ..... , \end{aligned}$$

where $c_q$’s are arbitrary constants and $q \in \textrm{N} \cup \{0\}$.

Proof

The proof follows by the application of Lemma 1 and Remark 1. $\square$

3 Approximation

Let us denote $C_B[0,\infty )= \{f \,| f:[0,\infty )\rightarrow \textrm{R},\;f \text { is continuous and bounded} \}$ and let $C^{* }\left[ 0,\infty \right) = \{f \,| f:[0,\infty )\rightarrow \textrm{R},\;f \text { is continuous and } \lim _{x \rightarrow \infty } f(x)< \infty \}$.

Theorem 1

If $f \in C_B[0,\infty )$, then

• (i) The operator $\mathcal {D}_{\lambda }$ satisfies the following property with operator $R_\lambda$ defined in Eq. (3)

  $$\begin{aligned} \lim _{\lambda \rightarrow \infty } \left( \mathcal {D}_{n\lambda } f(\lambda t)\right) \left( \frac{x}{\lambda }\right) = \left( R_n f(t)\right) (x). \end{aligned}$$

• (ii) For operator $\mathcal {L}_{\lambda }$ defined in Eq. (4), we have

  $$\begin{aligned} \lim _{\lambda \rightarrow \infty } \left( \mathcal {L}_{n\lambda } f(\lambda t)\right) \left( \frac{x}{\lambda }\right) = \left( S_n \circ S_n f(t)\right) (x), \end{aligned}$$

  where $n \ge 1$ and $x \ge 0$.

Proof

For $\lambda \in \textrm{N}$, we have

• (i) By simple calculations,

  $$\begin{aligned}&\lim _{\lambda \rightarrow \infty } \left( \mathcal {D}_{n\lambda } \exp _{is\lambda }\right) \left( \frac{x}{\lambda }\right) \\&\quad = \frac{n}{n-is} \lim _{\lambda \rightarrow \infty } \exp \bigg ( \frac{-n\lambda x}{\lambda +x}-n\lambda W\left( \frac{-x}{\lambda +x} e^{\frac{n}{n -is}-1-\frac{x}{\lambda +x}}\right) \bigg )\\&\quad =\frac{n}{n-is} \exp \bigg (nx \left( e^{\frac{is}{n-is}}-1\right) \bigg ) = \left( R_n \exp _{is}\right) (x), \end{aligned}$$

  where $\exp _{is\lambda }(u) = \cos (s\lambda u) + i \sin (s \lambda u)$ and $s \in \textrm{R}$.

• (ii) In a similar manner, we have

  $$\begin{aligned}&\lim _{\lambda \rightarrow \infty } \left( \mathcal {L}_{n\lambda } \exp _{is\lambda }\right) \left( \frac{x}{\lambda }\right) \\&\quad =\lim _{\lambda \rightarrow \infty }\exp \bigg (-n\lambda W\left( \frac{-x}{\lambda +x} \exp \left( e^{is/n}-1-\frac{x}{\lambda +x}\right) \right) -\frac{n\lambda x}{\lambda +x}\bigg )\\&\quad = \exp \bigg (nx \left( e^{e^\frac{is}{n}-1}-1\right) \bigg ) = \left( S_n \circ S_n\exp _{is}\right) (x). \end{aligned}$$

Now, the proof concludes from [4, Theorem 1] and [5, Theorem 2.1]. $\square$

Now, we establish Korovkin–type theorem, similar to the one given in [6, 10], as follows:

Theorem 2

[10] Let $A_{\lambda }:C^{* }\left[ 0,\infty \right) \rightarrow C^{* }\left[ 0,\infty \right)$ be endowed with uniform norm $\left\| A_{\lambda }\exp _{-q} -\exp _{-q}\right\| _{\left[ 0,\infty \right) }=C_{\lambda }^q,$ $q \in \{0,1,2\}$ and $C_{\lambda }^q \rightarrow 0$ as $\lambda \rightarrow \infty$, then

$$\begin{aligned} \left\| A_{\lambda }f-f\right\| _{\left[ 0,\infty \right) }\le C_{\lambda }^0\left\| f\right\| _{\left[ 0,\infty \right) }+\left( C_{\lambda }^0+ 2\right) {\omega }^{*}\left( f;\sqrt{ C_{\lambda }^0 +2C_{\lambda }^1 +C_{\lambda }^2 } \right) , \end{aligned}$$

where $\omega ^{*} (f; \sigma ) =\underset{\underset{\left| e^{-x_1}-e^{-x_2}\right| \le \sigma }{ x_1,x_2 \ge 0}}{\sup }\left| f\left( x_1\right) -f\left( x_2\right) \right|$ is the modulus of continuity.

Theorem 3

For $f\in C^{* }[ 0,\infty )$ and $\lambda \in \textrm{N},$ let $\left\| \mathcal {D}_{\lambda } \exp _{-q}-\exp _{-q}\right\| _{[ 0,\infty ) }= B_{\lambda }^q$, where $q \in \{0,1,2\}$ and $\lim _{\lambda \rightarrow \infty } B_{\lambda }^q =0$, then

$$\begin{aligned} \left\| \mathcal {D}_{\lambda } f-f\right\| _{[ 0,\infty ) }\le 2 {\omega }^{*}\left( f;\sqrt{2 B_{\lambda }^1+B_{\lambda }^2}\right) . \end{aligned}$$

Proof

Since $\mathcal {D}_{\lambda }$ preserves constants, therefore $B_{\lambda }^0=0$. With the help of software Mathematica, we get

$$\begin{aligned}&\left( \mathcal {D}_{\lambda } \exp _{-1}\right) \left( x\right) - \exp _{-1}\left( x\right) \\&\quad = \frac{1}{2 \lambda }e^{-x}\left( x^3+2x^2+3x-2\right) +\frac{1}{24\lambda ^{2}}e^{-x}\left( 3x^{6}-10x^4-48x^{3} \right. \\&\quad \left. -69x^{2}-88x+24\right) +O\left( \lambda ^{-3}\right) . \end{aligned}$$

Next, we have

$$\begin{aligned} \displaystyle&\sup _{x \ge 0} e^{-x} = 1 \text { and } \sup _{x\ge 0} x^{m} e^{-x} = m^m e^{-m},\; m=1,2,3,\cdots , \end{aligned}$$

whence, we get

$$\begin{aligned} B_\lambda ^1(x)&=\sup _{x\ge 0}\left| \left( \mathcal {D}_{\lambda }\exp _{-1}\right) \left( x\right) -\exp _{-1}\left( x\right) \right| \\&\le \frac{1}{\lambda }\left( \frac{27e^{-3}}{2}+4e^{-2}+\frac{3}{2}e^{-1}+1\right) \\&+\frac{1}{\lambda ^{2}}\left( 5832e^{-6}+\frac{320e^{-4}}{3}+54 e^{-3}+\frac{69}{6} e^{-2}+\frac{11}{3}e^{-1}+1\right) \\&+O\left( \lambda ^{-3}\right) \rightarrow 0\;\text {as }\lambda \rightarrow \infty . \end{aligned}$$

In similar manner, we have

$$\begin{aligned}&\left( \mathcal {D}_{\lambda } \exp _{-2}\right) \left( x\right) - \exp _{-2}\left( x\right) \\&\quad = \frac{2}{\lambda }e^{-2x}\left( x^3+2x^2+3x-1\right) +\frac{2}{3\lambda ^{2}}e^{-2x}\left( 3x^{6}+6x^5+10x^4-6x^{3} \right. \\&\quad \left. -21x^{2}-44x+6\right) +O\left( \lambda ^{-3}\right) . \end{aligned}$$

Also,

$$\begin{aligned}&\sup _{x \ge 0} e^{-2x} = 1\;\text {and}\;\sup _{x\ge 0}x^{m}e^{-2x} =\left( \frac{m}{2}\right) ^m e^{-m},\; m=1,2,3,\cdots , \end{aligned}$$

whence, we get

$$\begin{aligned} B_\lambda ^2(x)&=\sup _{x\ge 0}\left| \left( \mathcal {D}_{\lambda }\exp _{-2}\right) \left( x\right) -\exp _{-2}\left( x\right) \right| \\&\le \frac{1}{\lambda }\left( \frac{27e^{-3}}{4}+4{e}^{-2}+3e^{-1}+2\right) \\&\quad +\frac{1}{\lambda ^{2}}\left( 1458e^{-6}+\frac{3125 e^{-5}}{8}+\frac{320 e^{-4}}{3}+\frac{27e^{-3}}{2}+14 e^{-2}+\frac{44 e^{-1}}{3}+4\right) \\&\quad +O\left( \lambda ^{-3}\right) \rightarrow 0\;\text {as }\lambda \rightarrow \infty . \end{aligned}$$

The proof readily follows from Theorem 2. $\square$

Theorem 4

Let $f, f^{\prime }, f^{\prime \prime } \in C^{* }[ 0,\infty )$, then

$$\begin{aligned}&\left| \lambda \left[ \left( \mathcal {D}_{\lambda }f\right) \left( x\right) -f\left( x\right) \right] -f^{\prime }(x)-\frac{1}{2}x\left( x^2+ 2x+3\right) f^{\prime \prime }\left( x\right) \right| \\&\quad \le 2\biggl [\frac{1}{2\lambda } \left| f^{\prime \prime } (x)\right| + \frac{2}{\lambda }+ x(x^2+2x+3)\\&\quad +\lambda ^2\left( \left( \mathcal {D}_{\lambda } \left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^4\right) (x)\cdot \mu _{\lambda ,4} (x)\right) ^{\frac{1}{2}} \biggr ] {\omega }^{*}\left( f^{\prime \prime };\frac{1}{\sqrt{\lambda }}\right) . \end{aligned}$$

Proof

Applying Taylor’s formula on f, we have for $x,l \in [0,\infty )$,

$$\begin{aligned} f(l)= f(x)+ (l-x)f^{\prime } (x)+ \frac{1}{2} (l-x)^2 f^{\prime \prime } (x)+ (l-x)^2 \zeta (l;x), \end{aligned}$$

where $\lim _{l \rightarrow x} \zeta (l;x)=0$. Operating $\mathcal {D}_{\lambda }$ and using Lemma 3, we have

$$\begin{aligned}&\left| \lambda \left[ \left( \mathcal {D}_{\lambda }f\right) \left( x\right) -f\left( x\right) \right] -f^{\prime }(x)-\frac{1}{2}x\left( x^2+ 2x+3\right) f^{\prime \prime }\left( x\right) \right| \nonumber \\&\quad \le \frac{1}{\lambda } \left| f^{\prime \prime } (x)\right| + \lambda \left| \left( \mathcal {D}_\lambda \zeta (l;x) \left( l-x\right) ^2\right) (x)\right| . \end{aligned}$$

(6)

For $\delta >0$, the modulus of continuity satisfies the following property [3]

$$\begin{aligned} \zeta \left( l;x\right) \le 2\left( 1+\frac{\left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^{2}}{\delta ^{2}}\right) \omega ^{* }\left( f^{\prime \prime };\delta \right) . \end{aligned}$$

Applying Cauchy–Schwarz inequality on the last term in R.H.S. of (6) gives

$$\begin{aligned}&\lambda \left| \left( \mathcal {D}_\lambda \zeta (l;x) \left( l-x\right) ^2\right) (x)\right| \\&\quad \le 2\lambda {\omega }^{* }\left( f^{\prime \prime };\delta \right) \mu _{\lambda ,2} (x) \\&\quad +\frac{2\lambda }{\delta ^{2}}{\omega }^{* }\left( f^{\prime \prime };\delta \right) \left( \mathcal {D}_\lambda \left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^4\right) ^{1/2} \displaystyle \sqrt{\mu _{\lambda ,4} (x)}. \end{aligned}$$

The proof follows by selecting $\delta = \frac{1}{\sqrt{\lambda }}$. $\square$

4 Further composition with Szász–Mirakyan operator

Combining the operators $\mathcal {D}_{\lambda }$ and Szász–Mirakjan operators yields a new operator, denoted by $E_\lambda$ and represented as

$$\begin{aligned} \left( E_\lambda f \right) (x):= \left( \mathcal {D}_{\lambda } \circ S_\lambda f\right) (x). \end{aligned}$$

Lemma 4

The MGF of the operators $E_\lambda$ is

$$\begin{aligned} (E_\lambda \exp _A) (x)= \frac{1}{2-e^{A/\lambda }} \exp \bigg ( \frac{-\lambda x}{1+x}-\lambda W\left( (-x)(1+x)^{-1} e^{\frac{1}{2-e^{A/\lambda }}-\frac{(1+2x)}{1+x}}\right) \bigg ). \end{aligned}$$

Furthermore, let us denote the moments of q-th order by $(E_\lambda e_q)(x),$ where $e_q(x)=x^q$ and $q=0,1,2,\cdots$, then

$$\begin{aligned}&\sum _{q \ge 0} d_q (E_\lambda e_q)(x) \\&\quad = d_0 + d_1 \left( x+\frac{1}{\lambda }\right) + d_2 \left( x^2+\frac{x^3+2x^2+6x}{\lambda }+\frac{3}{\lambda ^2}\right) \\&\quad +d_3 \left( x^3+\frac{3x^4+ 6x^3+ 15x^2}{\lambda }+\frac{3x^5+10x^4+ 24x^3+ 30x^2+44x}{\lambda ^2}+\frac{13}{\lambda ^3} \right) \\&\quad + d_4 \left( x^4 + \frac{6x^5+ 12x^4+ 28x^3}{\lambda }+\frac{15x^6+ 52 x^5+ 132 x^4+ 168x^3+ 206 x^2 }{\lambda ^2} \right. \\&\quad \left. +\frac{15x^7+ 70x^6+ 197x^5+344x^4+458x^3+412x^2+389x}{\lambda ^3}+\frac{75}{\lambda ^4}\right) + \cdots , \end{aligned}$$

where $d_q$’s, $q=0,1,2, \cdots$ are certain constants.

Lemma 5

For the central moments of q-th order, which are denoted by $\tilde{\mu }_{\lambda ,q}(x)= \left( E_\lambda (e_1-x e_0)^q\right) (x)$, we have

$$\begin{aligned}&\sum _{q \ge 0} d_q \tilde{\mu }_{\lambda ,q}(x) \\&\quad = d_0 +\frac{d_1}{\lambda }+d_2\left( \frac{x^3+2x^2+4x}{\lambda }+ \frac{3}{\lambda ^2}\right) \\&\quad + d_3\left( \frac{3x^5+10x^4+24x^3+30x^2+35x}{\lambda ^2}+ \frac{13}{\lambda ^3}\right) \\&\quad + d_4\left( \frac{3x^6+12x^5+36x^4+48x^3+48x^2}{\lambda ^2}\right. \\&\quad \left. +\frac{15x^7+70x^6+197x^5+344x^4+458x^3+412x^2+337x}{\lambda ^3} + \frac{75}{\lambda ^4}\right) + \cdots , \end{aligned}$$

where $d_q$’s, $q=0,1,2, \cdots$ are certain constants.

Now, we present some theorems analogous to those for the operator $\mathcal {D}_\lambda .$

Theorem 5

If $f \in C_B[0,\infty )$ and $n \ge 1$, then

$$\begin{aligned} \lim _{\lambda \rightarrow \infty } \left( E_{n\lambda } f(\lambda t)\right) \left( \frac{x}{\lambda }\right) =\left( V_n f(t) \right) (x), \end{aligned}$$

where $V_\lambda f:= (R_\lambda \circ S_\lambda f)$ and $x \ge 0$.

Proof

For $\lambda \in \textrm{N}$ and $s \in \textrm{R}$, we have

$$\begin{aligned}&\lim _{\lambda \rightarrow \infty } \left( E_{n\lambda } \exp _{is\lambda }\right) \left( \frac{x}{\lambda }\right) \\&\quad = \lim _{\lambda \rightarrow \infty } \frac{1}{2-e^{is/n}} \exp \bigg ( \frac{-n\lambda x}{\lambda +x}-n\lambda W\left( \frac{-x}{\lambda +x} e^{\frac{1}{2-e^{is/n}}-1-\frac{x}{\lambda +x}}\right) \bigg )\\&\quad =\frac{1}{2-e^{is/n}} \exp \bigg (nx \left( e^{\frac{1}{2-e^{is/n}}-1}-1\right) \bigg ) = \left( V_n \exp _{is} \right) (x). \end{aligned}$$

Now, the conclusion follows from [4, Theorem 1] and [5, Theorem 2.1]. $\square$

Theorem 6

For $f\in C^{* }[ 0,\infty )$ and $\lambda \in \textrm{N}$, let $\left\| E_{\lambda } \exp _{-q}-\exp _{-q}\right\| _{[ 0,\infty ) }= M_{\lambda }^q$, where $q \in \{0,1,2\}$ and $\lim _{\lambda \rightarrow \infty } M_{\lambda }^q =0$, then

$$\begin{aligned} \left\| E_{\lambda } f-f\right\| _{[ 0,\infty ) }\le 2 {\omega }^{*}\left( f;\sqrt{2 M_{\lambda }^1+M_{\lambda }^2}\right) . \end{aligned}$$

Proof

Since $\left( E_{\lambda } 1\right) (x)=1$, therefore $M_{\lambda }^0=0$. Next, we have

$$\begin{aligned}&\left( E_{\lambda } \exp _{-1}\right) \left( x\right) - \exp _{-1}\left( x\right) \\&\quad = \frac{1}{2 \lambda }e^{-x}\left( x^3+2x^2+4x-2\right) +\frac{1}{24\lambda ^{2}}e^{-x}\left( 3x^{6}-4x^4-48x^{3} \right. \\&\quad \left. -72x^{2}-140x+36\right) +O\left( \lambda ^{-3}\right) , \end{aligned}$$

and

$$\begin{aligned}&\left( E_{\lambda } \exp _{-2}\right) \left( x\right) - \exp _{-2}\left( x\right) \\&\quad = \frac{2}{ \lambda }e^{-2x}\left( x^3+2x^2+4x-1\right) +\frac{2}{3\lambda ^{2}}e^{-2x}\left( 3x^{6}+6x^5+16x^4 \right. \\&\quad \left. -12x^{2}-70x+9\right) +O\left( \lambda ^{-3}\right) . \end{aligned}$$

Using

$$\begin{aligned} \displaystyle&\sup _{x \ge 0} e^{-mx} = 1,\; \sup _{x\ge 0} x^{m} e^{-x} = m^m e^{-m},\;\text {and}\;\\&\sup _{x\ge 0}x^{m}e^{-2x} =\left( \frac{m}{2}\right) ^m e^{-m},\; m=1,2,3,\cdots , \end{aligned}$$

we get

$$\begin{aligned} M_\lambda ^1(x)&=\sup _{x\ge 0}\left| \left( E_{\lambda }\exp _{-1}\right) \left( x\right) -\exp _{-1}\left( x\right) \right| \\&\le \frac{1}{\lambda }\left( \frac{27e^{-3}}{2}+4e^{-2}+2e^{-1}+1\right) \\&+\frac{1}{ \lambda ^{2}}\left( 5832e^{-6}+\frac{128e^{-4}}{3}+54 e^{-3}+12 e^{-2}+\frac{35}{6}e^{-1}+1.5\right) \\&+O\left( \lambda ^{-3}\right) \rightarrow 0\;\text {as }\lambda \rightarrow \infty . \end{aligned}$$

and

$$\begin{aligned} M_\lambda ^2(x)&=\sup _{x\ge 0}\left| \left( E_{\lambda }\exp _{-2}\right) \left( x\right) -\exp _{-2}\left( x\right) \right| \\&\le \frac{1}{\lambda }\left( \frac{27e^{-3}}{4}+4{e}^{-2}+4e^{-1}+2\right) \\&+\frac{1}{\lambda ^{2}}\left( 1458e^{-6}+\frac{3125 e^{-5}}{8}+\frac{512 e^{-4}}{3}+8 e^{-2}+\frac{70 e^{-1}}{3}+6\right) \\&+O\left( \lambda ^{-3}\right) \rightarrow 0\;\text {as }\lambda \rightarrow \infty . \end{aligned}$$

The proof readily follows from Theorem 2. $\square$

Theorem 7

Let $f, f^{\prime }, f^{\prime \prime } \in C^{* }[ 0,\infty )$, then

$$\begin{aligned}&\left| \lambda \left[ \left( E_{\lambda }f\right) \left( x\right) -f\left( x\right) \right] -f^{\prime }(x)-\frac{1}{2}x\left( x^2+ 2x+4\right) f^{\prime \prime }\left( x\right) \right| \\&\quad \le 2\biggl [\frac{3}{4\lambda } \left| f^{\prime \prime } (x)\right| + \frac{3}{\lambda }+ x(x^2+2x+4)\\&\quad +\lambda ^2\left( \left( E_{\lambda } \left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^4\right) (x)\cdot \tilde{\mu }_{\lambda ,4} (x)\right) ^{\frac{1}{2}} \biggr ] {\omega }^{*}\left( f^{\prime \prime };\frac{1}{\sqrt{\lambda }}\right) . \end{aligned}$$

Proof

Applying Taylor’s formula to the operator $E_\lambda$ and using Lemma 5, we have

$$\begin{aligned}&\left| \lambda \left[ \left( E_{\lambda }f\right) \left( x\right) -f\left( x\right) \right] -f^{\prime }(x)-\frac{1}{2}x\left( x^2+ 2x+4\right) f^{\prime \prime }\left( x\right) \right| \\&\quad \le \frac{3}{2\lambda } \left| f^{\prime \prime } (x)\right| + \lambda \left| \left( E_\lambda \zeta (l;x) \left( l-x\right) ^2\right) (x)\right| . \end{aligned}$$

Now, for $\delta >0$, applying Cauchy–Schwarz inequality on the last term from above and using the property $\zeta \left( l;x\right) \le 2\left( 1+\frac{\left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^{2}}{\delta ^{2}}\right) \omega ^{* }\left( f^{\prime \prime };\delta \right)$, we have

$$\begin{aligned}&\lambda \left| \left( E_\lambda \zeta (l;x) \left( l-x\right) ^2\right) (x)\right| \\&\quad \le 2\lambda {\omega }^{* }\left( f^{\prime \prime };\delta \right) \tilde{\mu }_{\lambda ,2} (x) \\&\quad +\frac{2\lambda }{\delta ^{2}}{\omega }^{* }\left( f^{\prime \prime };\delta \right) \left( E_\lambda \left( \exp _{-1}\left( x\right) -\exp _{-1}\left( l\right) \right) ^4\right) ^{1/2} \displaystyle \sqrt{\tilde{\mu }_{\lambda ,4} (x)}. \end{aligned}$$

The proof follows by selecting $\delta = \frac{1}{\sqrt{\lambda }}$. $\square$

5 Graphical representation

We present following graphs to give a comparison among the rate of approximations of the operators $\mathcal {D}_{\lambda },E_\lambda$ and $R_\lambda$.

In Fig. 1, the approximations of exponential function $f(x)=e^{-4x}$, by these operators are compared (see Fig.a and Fig.b).

Fig. 1

Comparison among graphs of $\mathcal {D}_{\lambda },E_\lambda$ and $R_\lambda$ for $f(x)=e^{-4x}$

Likewise, in Fig. 2, the graphs (Fig.c, Fig.d) compare the approximations of cubic polynomial $f(x)=x^3+2x^2+6x+2$.

Fig. 2

Comparison among graphs of $\mathcal {D}_{\lambda },E_\lambda$ and $R_\lambda$ for $f(x)=x^3+2x^2+6x+2$

We observe that $R_{\lambda }$ yields the best approximation, followed by $\mathcal {D}_\lambda$, with $E_\lambda$ being the least precise; which indicates that higher order compositions produce less precise approximations. Moreover, as $\lambda$ increases, the approximations become more precise.