1 Introduction

If we take the composition of some integral-type operator with a discrete operator, we get a discrete operator. More precisely, we take the composition \(U_{m,n}=V_{m}\circ W_{n}\) of two operators

$$\begin{aligned} \left( V_{m}f\right) \left( x\right) =\int _{I}k_{m}\left( x,t\right) f(t)dt, \end{aligned}$$

where I is a certain real interval and \(k_{m}\left( x,t\right) \) is a kernel function, and

$$\begin{aligned} \left( W_{n}f\right) \left( x\right) =\sum _{k}w_{n,k}\left( x\right) f\left( x_{n,k}\right) . \end{aligned}$$

For a suited class of functions f, we obtain a discrete operator

$$\begin{aligned} \left( U_{m,n}f\right) \left( x\right) =\sum _{k}u_{k,m,n}\left( x\right) f\left( x_{n,k}\right) . \end{aligned}$$

In several concrete cases it is possible to give a closed expression for the coefficients

$$\begin{aligned} u_{k,m,n}\left( x\right) =\int _{I}k_{m}\left( x,t\right) w_{n,k}\left( t\right) dt. \end{aligned}$$

Throughout the paper we tacitly assume that the functions f have the property that interchanging the order of the limit processes integration and summation is justified.

In this note we give three examples of the composition of an integral operator with a discrete operator to obtain a discrete operator. It turns out that the resulting operators are based on Laguerre polynomials. To each of the discrete operators so obtained we find the moment generating function providing the moments which are essential for studying the approximation properties of the new operators. Without a proof we present some basic convergence results for the new operators including a Voronovskaja-type formula.

2 Preliminaries

In this section we present some auxiliary results, in particular on Laguerre polynomials which play an important role in the following.

For nonnegative integers n and arbitrary real \(\alpha \), the polynomial solutions of the differential equation \(xy^{\prime \prime }+\left( \alpha +1-x\right) y^{\prime }+ny=0\) are called generalized Laguerre polynomials \( L_{n}^{\left( \alpha \right) }\), or associated Laguerre polynomials. A concise representation is the Rodrigues’ formula

$$\begin{aligned} L_{n}^{\left( \alpha \right) }\left( x\right) =\frac{1}{n!}x^{-\alpha }e^{x}\left( \frac{d}{dx}\right) ^{n}\left( x^{\alpha +n}e^{-x}\right) . \end{aligned}$$

The explicit form is given by

$$\begin{aligned} L_{n}^{\left( \alpha \right) }\left( x\right) =\sum _{k=0}^{n}\left( -1\right) ^{k}\left( {\begin{array}{c}n+\alpha \\ n-k\end{array}}\right) \frac{x^{k}}{k!}. \end{aligned}$$
(1)

The ordinary Laguerre polynomials \(L_{n}=L_{n}^{\left( 0\right) }\) are the special case \(\alpha =0\), where

$$\begin{aligned} L_{n}\left( x\right) =\sum _{k=0}^{n}\left( -1\right) ^{k}\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^{k}}{k!}. \end{aligned}$$
(2)

We gather some representations and properties of Laguerre functions which will be useful in the following. Note that the definition of Laguerre functions can be extended to real values of n and \(\alpha \) by writing them in terms of confluent hypergeometric functions

$$\begin{aligned} L_{n}^{\left( \alpha \right) }\left( x\right) =\left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) M\left( -n,\alpha +1,x\right) , \end{aligned}$$
(3)

[3, Eq. (13.6.9)], where \(\left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) =\Gamma \left( n+\alpha +1\right) /\left( \Gamma \left( n+1\right) \Gamma \left( n+\alpha +1\right) \right) \) and

$$\begin{aligned} M\left( a,b,x\right) =\sum _{k=0}^{\infty }\frac{\left( a\right) _{k}}{\left( b\right) _{k}}\frac{x^{k}}{k!} \end{aligned}$$
(4)

is the confluent hypergeometric function (also denoted by \(_{1}F_{1}\)) and \( \left( a\right) _{k}=\Gamma \left( a+k\right) /\Gamma \left( a\right) \), \( k=0,1,2,\ldots \), is the Pochhammer symbol. When n is an integer the function reduces to a polynomial of degree n. Recall the Kummer’s transformation [3, Eq. (13.1.27)]

$$\begin{aligned} M\left( a,b,z\right) =e^{z}M\left( b-a,b,-z\right) \end{aligned}$$
(5)

for the confluent hypergeometric function. Furthermore, we have the recursive formula [3, Eq. (22.7.31)]

$$\begin{aligned} L_{n-1}^{\left( \alpha +1\right) }\left( x\right) =\frac{1}{x}\left[ \left( n+\alpha \right) L_{n-1}^{\left( \alpha \right) }\left( x\right) -nL_{n}^{\left( \alpha \right) }\left( x\right) \right] \end{aligned}$$
(6)

and its special instance \(\alpha =0\), i.e.,

$$\begin{aligned} L_{n-1}^{\left( 1\right) }\left( x\right) =\frac{n}{x}\left[ L_{n-1}\left( x\right) -L_{n}\left( x\right) \right] . \end{aligned}$$
(7)

The following lemma will frequently be of use in the following.

Lemma 1

For \(\alpha ,n=0,1,2,\ldots \), it holds

$$\begin{aligned} \sum _{k=0}^{\infty }\left( {\begin{array}{c}k+n+\alpha \\ n\end{array}}\right) \frac{x^{k}}{k!} =e^{x}L_{n}^{\left( \alpha \right) }\left( -x\right) . \end{aligned}$$

Proof

Using the identity

$$\begin{aligned} \left( {\begin{array}{c}k+n+\alpha \\ n\end{array}}\right) {=}\left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) \frac{\left( n+\alpha +1\right) _{k}}{\left( \alpha +1\right) _{k}} \end{aligned}$$

and according to (4) we obtain

$$\begin{aligned} \sum _{k=0}^{\infty }\left( {\begin{array}{c}k+n+\alpha \\ n\end{array}}\right) \frac{x^{k}}{k!}=\left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) M\left( n+\alpha +1,\alpha +1,x\right) . \end{aligned}$$

Application of the Kummer transformation (5) leads to

$$\begin{aligned} \sum _{k=0}^{\infty }\left( {\begin{array}{c}k+n+\alpha \\ n\end{array}}\right) \frac{x^{k}}{k!}=\left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) e^{x}M\left( -n,\alpha +1,-x\right) . \end{aligned}$$

Now the desired formula follows by Eq. (3). \(\square \)

Throughout the paper we consider, for each complex A, the function \(\exp _{A}\) defined by \(\exp _{A}\left( x\right) =\exp \left( Ax\right) \). Furthermore, we denote by \(e_{r}\) \(\left( r=0,1,2,\ldots \right) \) the monomials \(e_{r}\left( x\right) =x^{r}\).

The moment generating function (m.g.f.) of a linear operator U on a space of functions, is defined by

$$\begin{aligned} \left( U\exp _{A}\right) \left( x\right) , \end{aligned}$$

provided that \(U\exp _{A}\) exists. It can be utilized to obtain the moments \(Ue_{r}\) of the operator U via

$$\begin{aligned} \left( Ue_{r}\right) \left( x\right) =\left[ \left( \frac{d}{dA}\right) ^{r}\left( U\exp _{A}\right) \left( x\right) \right] _{A=0}. \end{aligned}$$

The m.g.f. of several approximation operators are gathered in [4]. In this note we determine the m.g.f. of the new composition operators.

3 Composition operators based on Laguerre polynomials

The Szász–Mirakyan operator \(S_{n}\) associates to each function \(f: \left[ 0,+\infty \right) \rightarrow {\mathbb {R}}\) of at most exponential growth the function

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

where \(s_{k}\left( x\right) =e^{-x}x^{k}/k!\) \(\left( k=0,1,2,\ldots \right) \). Direct computation [4, Eq. (8)] confirms that its m.g.f. is given by

$$\begin{aligned} S_{n}\exp _{A}=\exp _{n\left( e^{A/n}-1\right) }. \end{aligned}$$
(8)

Now we list three compositions leading to positive linear approximation operators based on Laguerre polynomials.

3.1 Example 1

For each nonnegative integer \(\alpha \), the generalized Durrmeyer variant of the Szász–Mirakyan operator is defined by

$$\begin{aligned} (\overline{S}_{n}^{\left[ \alpha \right] }f)\left( x\right) =n\sum _{j=0}^{\infty }s_{j}\left( nx\right) \int _{0}^{\infty }s_{j+\alpha }\left( nt\right) f\left( t\right) dt, \end{aligned}$$

provided that all integrals exist. Note that \(\left| f\left( t\right) \right| =O\left( e^{\gamma t}\right) \) as \(t\rightarrow +\infty \) implies that \(\left( \overline{S}_{n}^{\left[ \alpha \right] }f\right) \left( x\right) \) is well-defined, for \(n>\gamma \). We take the composition of the generalized Szász–Durrmeyer operator \(\overline{S}_{m}^{\left[ \alpha \right] }\) with the Szász–Mirakyan operators \(S_{n}\), to obtain a new approximation operator \(A_{m,n}^{\left[ \alpha \right] }\) as follows:

$$\begin{aligned} \left( A_{m,n}^{\left[ \alpha \right] }f\right) \left( x\right) :=\left( \left( \overline{S}_{m}^{\left[ \alpha \right] }\circ S_{n}\right) f\right) \left( x\right) . \end{aligned}$$

It is possible to give a concise form of \(A_{m,n}^{\left[ \alpha \right] }\) in terms of generalized Laguerre polynomials (1).

Theorem 1

The operator \(A_{m,n}\) can be represented in the form

$$\begin{aligned} \left( A_{m,n}^{\left[ \alpha \right] }f\right) \left( x\right) =\sum _{k=0}^{\infty }a_{k,m,n}^{\left[ \alpha \right] }\left( x\right) f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} a_{k,m,n}^{\left[ \alpha \right] }\left( x\right) =\frac{m^{\alpha +1}n^{k}}{ \left( m+n\right) ^{k+\alpha +1}}\exp \left( -\frac{mn}{m+n}x\right) L_{k}^{\left( \alpha \right) }\left( -\frac{m^{2}x}{m+n}\right) . \end{aligned}$$

Remark 1

Since \(L_{k}^{\left( \alpha \right) }\left( x \right) \ge 0\), for \(x \le 0\), the operator \(A_{m,n}^{\left[ \alpha \right] }\) is positive.

Remark 2

In the special case \(m=n\) the operator takes the form

$$\begin{aligned} \left( A_{n,n}^{\left[ \alpha \right] }f\right) \left( x\right) =e^{-nx/2}\sum _{k=0}^{\infty }\frac{1}{2^{k+\alpha +1}}L_{k}^{\left( \alpha \right) }\left( -\frac{nx}{2}\right) f\left( \frac{k}{n}\right) . \end{aligned}$$

This operator was discussed in [6, Eq. (3.4)].

Proof of Theorem 1

We have

$$\begin{aligned} \left( A_{m,n}^{\left[ \alpha \right] }f\right) \left( x\right)= & {} m\sum _{j=0}^{\infty }s_{j}(mx)\int _{0}^{\infty }s_{j+\alpha }(mt)\sum _{k=0}^{\infty }s_{k}(nt)f\left( \frac{k}{n}\right) dt \\= & {} m\sum _{j=0}^{\infty }s_{j}\left( mx\right) \sum _{k=0}^{\infty }f\left( \frac{k}{n}\right) \int _{0}^{\infty }\frac{e^{-\left( m+n\right) t}\left( mt\right) ^{j+\alpha }\left( nt\right) ^{k}}{k!\left( j+\alpha \right) !}dt \\= & {} m\sum _{j=0}^{\infty }s_{j}\left( mx\right) m^{j+\alpha }\sum _{k=0}^{\infty }\frac{n^{k}\left( k+j+\alpha \right) !}{k!\left( j+\alpha \right) !\left( m+n\right) ^{k+j+\alpha +1}}f\left( \frac{k}{n} \right) , \end{aligned}$$

such that

$$\begin{aligned} a_{k,m,n}^{\left[ \alpha \right] }\left( x\right) =e^{-mx}\frac{m^{\alpha +1}n^{k}}{\left( m+n\right) ^{k+\alpha +1}}\sum _{j=0}^{\infty }{\left( {\begin{array}{c} k+j+\alpha \\ k\end{array}}\right) }\frac{1}{j!}\left( \frac{m^{2}x}{m+n}\right) ^{j}. \end{aligned}$$

By Lemma 1, we infer that

$$\begin{aligned} a_{k,m,n}^{\left[ \alpha \right] }\left( x\right) =e^{-mx}\frac{m^{\alpha +1}n^{k}}{\left( m+n\right) ^{k+\alpha +1}}\exp \left( \frac{m^{2}x}{m+n} \right) L_{k}^{\left( \alpha \right) }\left( -\frac{m^{2}x}{m+n}\right) . \end{aligned}$$

Noting that \(e^{-mx}\exp \left( \frac{m^{2}x}{m+n}\right) =\exp \left( - \frac{mn}{m+n}x\right) \) completes the proof. \(\square \)

Remark 3

It is easily verified that the generalized Szász–Mirakyan–Durrmeyer operator has the m.g.f.

$$\begin{aligned} (\overline{S}_{n}^{\left[ \alpha \right] }\exp _{A})\left( x\right) =\left( \frac{n}{n-A}\right) ^{\alpha +1}\exp \left( \frac{nAx}{n-A}\right) { \qquad }\left( n>\left| A\right| \right) . \end{aligned}$$

Thus, by Eq. (8),

$$\begin{aligned} (A_{m,n}^{\left[ \alpha \right] }\exp _{A})(x)= & {} \left( \overline{S}_{m}^{ \left[ \alpha \right] }\circ S_{n}\right) \exp _{A}=\overline{S}_{m}^{\left[ \alpha \right] }\exp _{n\left( e^{A/n}-1\right) } \\= & {} \left( \frac{m}{m-n\left( e^{A/n}-1\right) }\right) ^{\alpha +1}\exp \left( \frac{mn\left( e^{A/n}-1\right) }{m-n\left( e^{A/n}-1\right) } x\right) , \end{aligned}$$

provided that \(m>\left| n\left( e^{A/n}-1\right) \right| \). In the special case \(m=n\), the m.g.f. takes the form

$$\begin{aligned} \left( A_{n,n}^{\left[ \alpha \right] }\exp _{A}\right) \left( x\right) =\left( \frac{1}{2-e^{A/n}}\right) ^{\alpha +1}\exp \left( -\frac{e^{A/n}-1}{ e^{A/n}-2}nx\right) , \end{aligned}$$

for all n satisfying \(\left| e^{A/n}\right| <2\).

Without giving the standard proofs we present some convergence results.

Theorem 2

Let f be a bounded function f on \(\left[ 0,+\infty \right) \). Then, for \(x\ge 0\), it holds

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( A_{m,n}^{\left[ \alpha \right] }f\right) \left( x\right)= & {} \left( \overline{S}_{m}f\right) \left( x\right) , \\ \lim _{m\rightarrow \infty }\left( A_{m,n}^{\left[ \alpha \right] }f\right) \left( x\right)= & {} (S_{n}f)\left( x\right) , \\ \lim _{r\rightarrow \infty }\left( A_{n,n}f\left( \frac{\cdot }{r}\right) \right) (rx)= & {} f\left( x\right) . \end{aligned}$$

We state the following Voronovskaja-type result.

Theorem 3

Given \(x>0\), for each bounded function f on \(\left[ 0,+\infty \right) \) it holds

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( A_{n,n}^{\left[ \alpha \right] }f\right) \left( x\right) -f\left( x\right) \right) =\left( 1+\alpha \right) f^{\prime }\left( x\right) +\frac{3}{2}xf^{\prime \prime }\left( x\right) , \end{aligned}$$

provided that the second derivative \(f^{\prime \prime }\left( x\right) \) exists.

3.2 Example 2

The exponential-type operator \(U_{n}\) associated with \(2x^{3/2}\) [5, Eq. (3.16)] (see also [1, 2]) is given by

$$\begin{aligned} \left( U_{n}f\right) \left( x\right) =\frac{n}{\sqrt{x}}\sum _{j=1}^{\infty }s_{j}\left( n\sqrt{x}\right) \int _{0}^{\infty }s_{j-1}\left( \frac{nt}{ \sqrt{x}}\right) f\left( t\right) dt+e^{-n\sqrt{x}}f\left( 0\right) . \end{aligned}$$

The composition of the exponential-type operator \(U_{m}\) associated with \(2x^{3/2}\) and \(S_{n}\) yields

$$\begin{aligned} \left( B_{m,n}f\right) \left( x\right) =\left( \left( U_{m}\circ S_{n}\right) f\right) \left( x\right) . \end{aligned}$$

It is possible to give a concise form of \(B_{m,n}\) in terms of the classical Laguerre polynomials (2).

Theorem 4

The operator \(B_{m,n}\) can be represented in the form

$$\begin{aligned} \left( B_{m,n}f\right) \left( x\right) =\sum _{k=0}^{\infty }b_{k,m,n}\left( x\right) f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} b_{0,m,n}\left( x\right) =\exp \left( \frac{-mnx}{m+n\sqrt{x}}\right) \end{aligned}$$

and, for \(k\ge 1\),

$$\begin{aligned} b_{k,m,n}\left( x\right)= & {} \left( \frac{n\sqrt{x}}{m+n\sqrt{x}}\right) ^{k}\exp \left( \frac{-mnx}{m+n\sqrt{x}}\right) \\{} & {} \left[ L_{k}\left( -\frac{ m^{2}\sqrt{x}}{m+n\sqrt{x}}\right) -L_{k-1}\left( -\frac{m^{2}\sqrt{x}}{m+n \sqrt{x}}\right) \right] . \end{aligned}$$

Remark 4

In particular

$$\begin{aligned} \left( B_{n,n}f\right) (x)=\sum _{k=0}^{\infty }b_{k,n,n}\left( x\right) f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} b_{0,n,n}\left( x\right) =\exp \left( \frac{-nx}{1+\sqrt{x}}\right) \end{aligned}$$

and, for \(k\ge 1\),

$$\begin{aligned} b_{k,n,n}\left( x\right) =\frac{(\sqrt{x})^{k}}{\left( 1+\sqrt{x}\right) ^{k} }\exp \left( \frac{-nx}{1+\sqrt{x}}\right) \left[ L_{k}\left( -\frac{n\sqrt{x }}{1+\sqrt{x}}\right) -L_{k-1}\left( -\frac{n\sqrt{x}}{1+\sqrt{x}}\right) \right] . \end{aligned}$$

Proof of Theorem 4

By definition, we have

$$\begin{aligned} \left( B_{m,n}f\right) \left( x\right)= & {} \frac{m}{\sqrt{x}} \sum _{j=1}^{\infty }s_{j}\left( m\sqrt{x}\right) \int _{0}^{\infty }s_{j-1}\left( \frac{mt}{\sqrt{x}}\right) \left( S_{n}f\right) \left( t\right) dt+e^{-m\sqrt{x}}(S_{n}f)\left( 0\right) \\= & {} e^{-m\sqrt{x}}\sum _{k=0}^{\infty }\frac{n^{k}}{k!}f\left( \frac{k}{n} \right) \sum _{j=1}^{\infty }\frac{m^{2j}}{j!\left( j-1\right) !} \\{} & {} \int _{0}^{\infty }\exp \left( \frac{-mt}{\sqrt{x}}-nt\right) t^{j+k-1}dt+e^{-m\sqrt{x}}f\left( 0\right) \\= & {} e^{-m\sqrt{x}}\sum _{k=0}^{\infty }f\left( \frac{k}{n}\right) \frac{n^{k}}{ k!}\left( \frac{\sqrt{x}}{m+n\sqrt{x}}\right) ^{k}\\{} & {} \sum _{j=1}^{\infty }\frac{ m^{2j}(j+k-1)!}{j!\left( j-1\right) !}\left( \frac{\sqrt{x}}{m+n\sqrt{x}} \right) ^{j}+e^{-m\sqrt{x}}f\left( 0\right) \\= & {} \sum _{k=0}^{\infty }b_{k,m,n}\left( x\right) f\left( \frac{k}{n}\right) . \end{aligned}$$

When \(k=0\), then, we have

$$\begin{aligned} b_{0,m,n}\left( x\right)= & {} e^{-m\sqrt{x}}\sum _{j=1}^{\infty }\frac{\left( m^{2}\sqrt{x}\right) ^{j}}{j!\left( m+n\sqrt{x}\right) ^{j}}+e^{-m\sqrt{x} }\\= & {} \exp \left( -m\sqrt{x}+\frac{m^{2}\sqrt{x}}{m+n\sqrt{x}}\right) =\exp \left( \frac{-mnx}{m+n\sqrt{x}}\right) . \end{aligned}$$

For \(k\ge 1\),

$$\begin{aligned} b_{k,m,n}\left( x\right)= & {} e^{-m\sqrt{x}}\left( \frac{n\sqrt{x}}{m+n\sqrt{x} }\right) ^{k}\frac{1}{k!}\sum _{j=1}^{\infty }\frac{m^{2j}(j+k-1)!}{j!\left( j-1\right) !}\left( \frac{\sqrt{x}}{m+n\sqrt{x}}\right) ^{j} \\= & {} e^{-m\sqrt{x}}\left( \frac{n\sqrt{x}}{m+n\sqrt{x}}\right) ^{k}\sum _{j=1}^{\infty }\frac{1}{j!}\left( \frac{m^{2}\sqrt{x}}{m+n\sqrt{x}} \right) ^{j}\left( {\begin{array}{c}k+j-1\\ k\end{array}}\right) . \end{aligned}$$

By Lemma 1, we infer that

$$\begin{aligned}{} & {} \sum _{j=1}^{\infty }\frac{1}{j!}\left( \frac{m^{2}\sqrt{x}}{m+n\sqrt{x}} \right) ^{j}\left( {\begin{array}{c}k+j-1\\ k\end{array}}\right) \\{} & {} \quad =\frac{1}{k}\sum _{j=0}^{\infty }\frac{1}{j!}\left( \frac{m^{2}\sqrt{x}}{m+n \sqrt{x}}\right) ^{j+1}\left( {\begin{array}{c}j+k-1+1\\ k-1\end{array}}\right) \\{} & {} \quad =\frac{1}{k}\left( \frac{m^{2}\sqrt{x}}{m+n\sqrt{x}}\right) \exp \left( \frac{m^{2}\sqrt{x}}{m+n\sqrt{x}}\right) L_{k-1}^{\left( 1\right) }\left( - \frac{m^{2}\sqrt{x}}{m+n\sqrt{x}}\right) . \end{aligned}$$

Furthermore, note that \(e^{-m\sqrt{x}}\exp \left( \frac{m^{2}\sqrt{x}}{m+n \sqrt{x}}\right) =\exp \left( \frac{-mnx}{m+n\sqrt{x}}\right) \). Hence,

$$\begin{aligned} b_{k,m,n}\left( x\right) =\left( \frac{n\sqrt{x}}{m+n\sqrt{x}}\right) ^{k}\exp \left( \frac{-mnx}{m+n\sqrt{x}}\right) \frac{1}{k}\left( \frac{m^{2} \sqrt{x}}{m+n\sqrt{x}}\right) L_{k-1}^{\left( 1\right) }\left( -\frac{m^{2} \sqrt{x}}{m+n\sqrt{x}}\right) . \end{aligned}$$

Now an application of Eq. (7) completes the proof of the desired formula. \(\square \)

Remark 5

The exponential-type operator related to \(2x^{3/2}\) has the m.g.f

$$\begin{aligned} \left( U_{n}\exp _{A}\right) \left( x\right) =\exp \left( \frac{Anx}{n-A \sqrt{x}}\right) \ {\qquad }(n>\left| A\right| \sqrt{x}). \end{aligned}$$
(9)

Thus

$$\begin{aligned} \left( B_{m,n}\exp _{A}\right) \left( x\right)= & {} \left( \left( U_{m}\circ S_{n}\right) \exp _{A}\right) \left( x\right) =\left( U_{m}\exp _{n\left( e^{A/n}-1\right) }\right) \left( x\right) \\= & {} \exp \left( \frac{nm\left( e^{A/n}-1\right) x}{m-n\left( e^{A/n}-1\right) \sqrt{x}}\right) { \qquad }\left( m>n\left| e^{A/n}-1\right| \sqrt{x}\right) . \end{aligned}$$

Without giving the proofs we present some convergence results.

Theorem 5

Let f be a bounded function f on \(\left[ 0,+\infty \right) \). Then for \(x\ge 0\), we have

$$\begin{aligned} \lim _{m\rightarrow \infty }\left( B_{m,n}f\right) \left( x\right)= & {} (S_{n}f)\left( x\right) , \\ \lim _{n\rightarrow \infty }\left( B_{m,n}f\right) \left( x\right)= & {} (U_{n}f)\left( x\right) , \\ \lim _{r\rightarrow \infty }\left( B_{n,n}f\left( \frac{\cdot }{r}\right) \right) (rx)= & {} f\left( x\right) . \end{aligned}$$

We state the following Voronovskaja-type result.

Theorem 6

Given \(x>0\), for each bounded function f on \(\left[ 0,+\infty \right) \) it holds

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( B_{n,n}f\right) \left( x\right) -f\left( x\right) \right) = \left( x^{3/2}+\frac{1}{2}x \right) f^{\prime \prime }\left( x\right) , \end{aligned}$$

provided that the second derivative \(f^{\prime \prime }\left( x\right) \) exists.

3.3 Example 3

The Phillips operator is given by

$$\begin{aligned} \overline{P}_{n}\left( x\right) =\int _{0}^{\infty }e^{-n\left( x+t\right) }\sum _{j=1}^{\infty }\frac{\left( n^{2}x\right) ^{j}}{j!\left( j-1\right) !} t^{j-1}f\left( t\right) dt+e^{-nx}f\left( 0\right) . \end{aligned}$$

The composition of Phillips operator \(\overline{P}_{m}\) and Szász–Mirakyan operator \(S_{n}\) yields

$$\begin{aligned} \left( C_{m,n}f\right) \left( x\right) =\left( \left( \overline{P}_{m}\circ S_{n}\right) f\right) \left( x\right) . \end{aligned}$$

It is possible to give a concise form of \(C_{m,n}\) in terms of Laguerre polynomials (2).

Theorem 7

A concise form of \(C_{m,n}\) is given by

$$\begin{aligned} (C_{m,n}f)(x)=\sum _{k=0}^{\infty }c_{k,m,n}\left( x\right) f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} c_{0,m,n}\left( x\right) =\exp \left( \frac{-mnx}{m+n}\right) \end{aligned}$$

and for \(k\ge 1\)

$$\begin{aligned} c_{k,m,n}\left( x\right) =\left( \frac{n}{m+n}\right) ^{k}\exp \left( \frac{ -mn}{m+n}x\right) \left[ L_{k}\left( -\frac{m^{2}x}{m+n}\right) -L_{k-1}\left( -\frac{m^{2}x}{m+n}\right) \right] . \end{aligned}$$

Remark 6

In particular

$$\begin{aligned} (C_{n,n}f)(x)=\sum _{k=0}^{\infty }c_{k,n,n}\left( x\right) f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} c_{0,n,n}\left( x\right) =\exp \left( \frac{-nx}{2}\right) \end{aligned}$$

and for \(k\ge 1\)

$$\begin{aligned} c_{k,n,n}\left( x\right) =\left( \frac{1}{2}\right) ^{k}\exp \left( \frac{-n }{2}x\right) \left[ L_{k}\left( -\frac{nx}{2}\right) -L_{k-1}\left( -\frac{nx }{2}\right) \right] . \end{aligned}$$

Proof of Theorem 7

By definition, we have

$$\begin{aligned} \left( C_{m,n}f\right) \left( x\right)= & {} \int _{0}^{\infty }e^{-m\left( x+t\right) }\sum _{j=1}^{\infty }\frac{\left( m^{2}x\right) ^{j}}{j!\left( j-1\right) !}t^{j-1}\left( S_{n}f\right) \left( t\right) dt+e^{-mx}\left( S_{n}f\right) \left( 0\right) \\= & {} e^{-mx}\sum _{k=0}^{\infty }\frac{n^{k}}{k!}f\left( \frac{k}{n}\right) \sum _{j=1}^{\infty }\frac{\left( m^{2}x\right) ^{j}}{j!\left( j-1\right) !}\\{} & {} \int _{0}^{\infty }e^{-mt}t^{j-1}e^{-nt}t^{k}dt+e^{-mx}f\left( 0\right) \\= & {} \sum _{k=0}^{\infty }c_{k,m,n}\left( x\right) f\left( \frac{k}{n}\right) +e^{-mx}f\left( 0\right) , \end{aligned}$$

where

$$\begin{aligned} c_{0,m,n}\left( x\right)= & {} e^{-mx}\sum _{j=1}^{\infty }\frac{\left( m^{2}x\right) ^{j}}{j!\left( m+n\right) ^{k+j}}+e^{-mx}=e^{-mx}\left( \exp \left( \frac{m^{2}x}{m+n}\right) -1\right) +e^{-mx}\\= & {} \exp \left( \frac{-mnx}{m+n}\right) \end{aligned}$$

and, for \(k\ge 1\),

$$\begin{aligned} c_{k,m,n}\left( x\right)= & {} e^{-mx}\frac{n^{k}}{k!}\sum _{j=1}^{\infty }\frac{ \left( m^{2}x\right) ^{j}\left( k+j-1\right) !}{j!\left( j-1\right) !\left( m+n\right) ^{k+j}}\\= & {} \frac{n^{k}e^{-mx}}{\left( m+n\right) ^{k}} \sum _{j=1}^{\infty }\frac{1}{j!}\left( \frac{m^{2}x}{m+n}\right) ^{j}\left( {\begin{array}{c}k+j-1\\ k\end{array}}\right) . \end{aligned}$$

As in the proof of Theorem 4 it follows, by Lemma 1, that

$$\begin{aligned} \sum _{j=1}^{\infty }\frac{1}{j!}\left( \frac{m^{2}x}{m+n}\right) ^{j}\left( {\begin{array}{c} k+j-1\\ k\end{array}}\right) =\frac{m^{2}x}{m+n}\exp \left( \frac{m^{2}x}{m+n}\right) \frac{1}{k} L_{k-1}^{\left( 1\right) }\left( -\frac{m^{2}x}{m+n}\right) . \end{aligned}$$

We get

$$\begin{aligned} c_{k,m,n}(x)=\frac{n^{k}m^{2}x}{\left( m+n\right) ^{k+1}}\exp \left( \frac{ -mn}{m+n}x\right) \frac{1}{k}L_{k}^{\left( 1\right) }\left( -\frac{m^{2}x}{m+n}\right) . \end{aligned}$$

Now an application of Eq. (7) completes the proof of the desired formula. \(\square \)

Remark 7

For each complex number A, the moment generating function of the operator \(C_{m,n}\) is given by

$$\begin{aligned} \left( C_{m,n}\exp _{A}\right) \left( x\right) =\exp \left( \frac{mn\left( e^{A/n}-1\right) }{m-n\left( e^{A/n}-1\right) }x\right) { \qquad } \left( m>n\left| e^{A/n}-1\right| \right) . \end{aligned}$$

This is a consequence of the combination of equation (8) and the m.g.f. of the Phillips operator

$$\begin{aligned} \left( \overline{P}_{m}\exp _{A}\right) \left( x\right) =\exp \left( \frac{ mAx}{m-A}\right) { \qquad }\left( m>\left| A\right| \right) , \end{aligned}$$

which implies that

$$\begin{aligned} C_{m,n}\exp _{A}=\left( \overline{P}_{m}\circ S_{n}\right) \exp _{A}= \overline{P}_{m}\exp _{n\left( e^{A/n}-1\right) }=\exp _{mn\left( e^{A/n}-1\right) /\left( m-n\left( e^{A/n}-1\right) \right) }. \end{aligned}$$

Without giving the proofs we present some convergence results.

Theorem 8

Let f be a bounded function f on \(\left[ 0,+\infty \right) \). Then for \(x\ge 0\), we have

$$\begin{aligned} \lim _{m\rightarrow \infty }\left( C_{m,n}f\right) \left( x\right)= & {} (S_{n}f)\left( x\right) , \\ \lim _{n\rightarrow \infty }\left( C_{m,n}f\right) \left( x\right)= & {} ( \overline{P}_{n}f)\left( x\right) , \\ \lim _{r\rightarrow \infty }\left( C_{n,n}f\left( \frac{\cdot }{r}\right) \right) (rx)= & {} f\left( x\right) . \end{aligned}$$

We state the following Voronovskaja-type result.

Theorem 9

Given \(x>0\), for each bounded function f on \(\left[ 0,+\infty \right) \) it holds

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( C_{n,n}f\right) \left( x\right) -f\left( x\right) \right) =\frac{3}{2}xf^{\prime \prime }\left( x\right) , \end{aligned}$$

provided that the second derivative \(f^{\prime \prime }\left( x\right) \) exists.

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