Theory and Modern Applications

# Unified Bernstein and Bleimann-Butzer-Hahn basis and its properties

## Abstract

In this paper we introduce the unification of Bernstein and Bleimann-Butzer-Hahn basis via the generating function. We give the representation of this unified family in terms of Apostol-type polynomials and Stirling numbers of the second kind. More generating functions of trigonometric type are also obtained to this unification.

MSC:11B65, 11B68, 41A10, 30C15.

## 1 Introduction

In this paper, we introduce a two-parameter generating function, which generates not only the Bernstein basis polynomials, but also the Bleimann-Butzer-Hahn basis functions. The generating function that we propose is given by

${\mathcal{G}}_{a,b}\left(t,x;k,m\right):={\left[\frac{{2}^{1-k}{x}^{k}{t}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{t\left[\frac{1+bx}{1+ax}\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}}{n!},$
(1)

where $k,m\in {\mathbb{Z}}^{+}:=\left\{1,2,\dots \right\}$, $a,b\in \mathbb{R}$, $t\in \mathbb{C}$. Here, $x\in I$ where I is a subinterval of such that the expansion in (1) is valid. The following two cases will be important for us.

1. 1.

The case $a=0$, $b=-1$. In this case, we let $x\in \left[0,1\right]$ and we see that

${\mathcal{G}}_{0,-1}\left(t,x;k,m\right)={\left[{2}^{1-k}{x}^{k}{t}^{k}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{t\left[1-x\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(0,-1\right)}\left(x;k,m\right)\frac{{t}^{n}}{n!}$

generates the unifying Bernstein basis polynomials ${\mathcal{P}}_{n}^{\left(0,-1\right)}\left(x;k,m\right):={\mathcal{B}}_{n}\left(mk,x\right)$ which were introduced and investigated in [1]. We should note further that ${\mathcal{G}}_{0,-1}\left(t,x;1,m\right)$ gives

${\mathcal{G}}_{0,-1}\left(t,x;1,m\right)={\left[xt\right]}^{m}\frac{1}{m!}{e}^{t\left[1-x\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_{n}\left(m,x\right)\frac{{t}^{n}}{n!}$

which generates the celebrated Bernstein basis polynomials (see [28])

${\mathcal{B}}_{n}\left(m,x\right):={B}_{m}^{n}\left(x\right)=\left(\genfrac{}{}{0}{}{n}{m}\right){x}^{k}{\left(1-x\right)}^{n-m}.$

Note that the Bernstein operators ${B}_{n}:C\left[0,1\right]\to C\left[0,1\right]$ are given by

${B}_{n}\left(f;x\right)=\sum _{m=0}^{n}f\left(\frac{m}{n}\right)\left(\genfrac{}{}{0}{}{n}{m}\right){x}^{k}{\left(1-x\right)}^{n-m},\phantom{\rule{1em}{0ex}}n\in \mathbb{N}:=\left\{1,2,\dots \right\}$

and by the Korovkin theorem, it is known that ${B}_{n}\left(f;x\right)⇉f\left(x\right)$ for all $f\in C\left[0,1\right]$, where $C\left[0,1\right]$ denotes the space of continuous functions defined on $\left[0,1\right]$, and the notation ‘’ denotes the uniform convergence with respect to the usual supremum norm on $C\left[0,1\right]$. Very recently, interesting properties of Bernstein polynomials were discussed in [7, 911] and [12].

1. 2.

The case $a=1$, $b=0$. In this case, we let $x\in \left[0,\mathrm{\infty }\right)$ and define

$\begin{array}{rcl}{\mathcal{G}}_{1,0}\left(t,x;k,m\right)& :=& {\left[\frac{{2}^{1-k}{x}^{k}{t}^{k}}{{\left(1+x\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{t\left[\frac{1}{1+x}\right]}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(1,0\right)}\left(x;k,m\right)\frac{{t}^{n}}{n!}.\end{array}$

We will see that this generating function produces the generalized Bleimann-Butzer-Hahn basis functions ${\mathcal{P}}_{n}^{\left(1,0\right)}\left(x;k,m\right):={\mathcal{H}}_{n}\left(mk,x\right)$. Furthermore, the special case

$\begin{array}{rcl}{\mathcal{G}}_{1,0}\left(t,x;1,m\right)& =& {\left[\frac{xt}{\left(1+x\right)}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{t\left[\frac{1}{1+x}\right]}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{H}}_{n}\left(m,x\right)\frac{{t}^{n}}{n!}\end{array}$

generates the well-known Bleimann-Butzer-Hahn basis functions:

${\mathcal{H}}_{n}\left(m,x\right):={H}_{m}^{n}\left(x\right)=\left(\genfrac{}{}{0}{}{n}{m}\right)\frac{{x}^{m}}{{\left(1+x\right)}^{n}}.$

The Bleimann-Butzer-Hahn operators were introduced in [5] and defined by

${L}_{n}\left(f;x\right)=\frac{1}{{\left(1+x\right)}^{n}}\sum _{m=0}^{n}f\left(\frac{m}{n}\right)\left(\genfrac{}{}{0}{}{n}{m}\right){x}^{m};\phantom{\rule{1em}{0ex}}x\in \left[0,\mathrm{\infty }\right),n\in \mathbb{N}.$

Denoting ${C}_{B}\left[0,\mathrm{\infty }\right)$ by the space of real-valued bounded continuous functions defined on $\left[0,\mathrm{\infty }\right)$, they proved that ${L}_{n}\left(f\right)\to f$ as $n\to \mathrm{\infty }$. On the other hand, the convergence is uniform on each compact subset of $\left[0,\mathrm{\infty }\right)$, where the norm is the usual supremum norm of ${C}_{B}\left[0,\mathrm{\infty }\right)$. For the review of the results concerning the Bleimann-Butzer-Hahn operators obtained in the period 1980-2009, we refer to [13].

The following theorem gives the explicit representation of the basis family defined in (1). Note that throughout the paper, we let ${\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right):=0$ for $n\le mk$.

Theorem 1 If $n\ge mk$, we have

${\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)={2}^{\left(1-k\right)m}{x}^{mk}\left(\genfrac{}{}{0}{}{n}{mk}\right)\frac{{\left(1+bx\right)}^{n-mk}}{{\left(1+ax\right)}^{n}}.$

Proof

Direct calculations give

$\begin{array}{rcl}{\mathcal{G}}_{a,b}\left(t,x;k,m\right)& =& {\left[\frac{{2}^{1-k}{x}^{k}{t}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{t\left[\frac{1+bx}{1+ax}\right]}\\ =& \frac{{2}^{\left(1-k\right)m}}{\left(mk\right)!}{\left(\frac{xt}{1+ax}\right)}^{mk}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1+bx}{1+ax}\right)}^{n}\frac{{t}^{n}}{n!}\\ =& {2}^{\left(1-k\right)m}{x}^{mk}\sum _{n=mk}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{n}{mk}\right)\frac{{\left(1+bx\right)}^{n-mk}}{{\left(1+ax\right)}^{n}}\frac{{t}^{n}}{n!}.\end{array}$
(2)

Comparing (1) and (2), we get the result. □

Corollary 2 By taking $a=0$, $b=-1$ in Theorem  1, we obtain the explicit representation of the unifying Bernstein basis polynomials [1]:

${\mathcal{P}}_{n}^{\left(0,-1\right)}\left(x;k,m\right):={\mathcal{B}}_{n}\left(mk,x\right)={2}^{\left(1-k\right)m}{x}^{mk}\left(\genfrac{}{}{0}{}{n}{mk}\right){\left(1-x\right)}^{n-mk}.$

Furthermore, ${\mathcal{B}}_{n}\left(m,x\right)={B}_{m}^{n}\left(x\right)$ is the well-known Bernstein basis.

Corollary 3 Taking $a=1$, $b=0$ in Theorem  1, we get the explicit representation of the generalized Bleimann-Butzer-Hahn basis:

${\mathcal{P}}_{n}^{\left(1,0\right)}\left(x;k,m\right):={\mathcal{H}}_{n}\left(mk,x\right)={2}^{\left(1-k\right)m}{x}^{mk}\left(\genfrac{}{}{0}{}{n}{mk}\right)\frac{1}{{\left(1+x\right)}^{n}}.$

Moreover, ${\mathcal{H}}_{n}\left(m,x\right)={H}_{m}^{n}\left(x\right)$ is the Bleimann-Butzer-Hahn basis function.

We organize the paper as follows. In Section 2, we obtain the representation of this unified family in terms of Apostol-type polynomials and Stirling numbers of the second kind. In Section 3, we give more trigonometric generating functions for this unification and obtain a certain summation formula. All the special cases are listed at the end of each theorem.

## 2 Representation in terms of Apostol-type polynomials and Stirling numbers

Recently [14], the first author introduced the unification of the Apostol-Bernoulli, Euler and Genocchi polynomials by

(3)

For the convergence of the series in (3), we refer to [[14], p.2453].

Some of the well-known polynomials included by ${Q}_{n,\beta }^{\left(\alpha \right)}\left(x;k,a,b\right)$ are listed below.

Remark 4 Having $k=a=b=1$ and $\beta =\lambda$ in (3), we get

${Q}_{n,\lambda }^{\left(\alpha \right)}\left(x;1,1,1\right)={\mathcal{B}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right).$

Note that ${\mathcal{B}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right)$ are the generalized Apostol-Bernoulli polynomials defined through the following generating relation:

where α and λ are arbitrary real or complex parameters and $x\in \mathbb{R}$. Note that when $\lambda \ne 1$, the order α should be restricted to nonnegative integer values. These polynomials were introduced by Luo and Srivastava [15] and investigated in [16, 17] and [18]. The Apostol-Bernoulli polynomials and numbers are obtained by the generalized Apostol-Bernoulli polynomials, respectively, as follows:

${B}_{n}\left(x;\lambda \right)={\mathcal{B}}_{n}^{\left(1\right)}\left(x;\lambda \right),\phantom{\rule{2em}{0ex}}{B}_{n}\left(\lambda \right)={B}_{n}\left(0;\lambda \right)\phantom{\rule{1em}{0ex}}\left(n\in {\mathbb{N}}_{0}\right).$

Taking $\lambda =1$ in the above relations, we obtain the classical Bernoulli polynomials ${B}_{n}\left(x\right)$ and Bernoulli numbers ${B}_{n}$.

Remark 5 Letting $k=-2a=b=1$ and $2\beta =\lambda$ in (3), we get

${Q}_{n,\frac{\lambda }{2}}^{\left(\alpha \right)}\left(x;1,\frac{-1}{2},1\right)={\mathcal{G}}_{n}^{\alpha }\left(x;\lambda \right),$

the Apostol-Genocchi polynomial of order α (arbitrary real or complex) which was defined by [19, 20]. Here the parameter λ is arbitrary real or complex. These polynomials are given as follows:

Note that when $\lambda \ne -1$, the order α should be restricted to nonnegative integer values. The Apostol-Genocchi polynomials and numbers are respectively given by

${G}_{n}\left(x;\lambda \right)={\mathcal{G}}_{n}^{1}\left(x;\lambda \right),\phantom{\rule{2em}{0ex}}{G}_{n}\left(\lambda \right)={G}_{n}\left(0;\lambda \right).$

When $\lambda =1$, the above relations give the classical Genocchi polynomials ${G}_{n}\left(x\right)$ and Genocchi numbers ${G}_{n}$.

Although our results do not contain the Apostol-Euler polynomials, for the sake of completeness, we give their definitions as a special case of the polynomial family ${Q}_{n,\beta }^{\left(\alpha \right)}\left(x;k,a,b\right)$.

Remark 6 Setting $k+1=-a=b=1$ and $\beta =\lambda$ in (3), we get

${Q}_{n,\lambda }^{\left(\alpha \right)}\left(x;0,-1,1\right)={\mathcal{E}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right).$

Recall that the Apostol-Euler polynomials ${\mathcal{E}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right)$ are generalized by Luo [21] and given by the generating relation

for arbitrary real or complex parameters α and λ and $x\in \mathbb{R}$. The Apostol-Euler polynomials and numbers are given respectively by

${E}_{n}\left(x;\lambda \right)={\mathcal{E}}_{n}^{1}\left(x;\lambda \right),\phantom{\rule{2em}{0ex}}{E}_{n}\left(\lambda \right)={E}_{n}\left(1;\lambda \right).$

When $\lambda =1$, the above relations give the classical Euler polynomials ${E}_{n}\left(x\right)$ and Euler numbers ${E}_{n}$.

Now, recall that the Stirling numbers of the second kind are denoted by $S\left(j,i\right)$ and defined by (see [[22], p.58 (15)])

${\left({e}^{t}-1\right)}^{i}=i!\sum _{j=i}^{\mathrm{\infty }}S\left(j,i\right)\frac{{t}^{j}}{j!}.$

The following theorem states an interesting explicit representation of the unified basis in terms of Apostol-type polynomials and relation between Stirling numbers of the second kind.

Theorem 7 The following representation:

$\begin{array}{rcl}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)& =& \frac{1}{\left(mk\right)!}{\left(\frac{x}{1+ax}\right)}^{mk}\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{d}-{c}^{d}\right)}^{m-i}{\beta }^{id}i!\\ ×\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){Q}_{n-j,\beta }^{\left(m\right)}\left(\frac{1+bx}{1+ax};k,c,d\right)\end{array}$

holds true between the unified Bernstein and Bleimann-Butzer-Hahn basis and Apostol-type polynomials.

Proof We get, using (1), that

(4)

On the other hand, since

$\begin{array}{rcl}{\left({\beta }^{d}-{c}^{d}+{\beta }^{d}\left[{e}^{t}-1\right]\right)}^{m}& =& \sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{d}-{c}^{d}\right)}^{m-i}{\beta }^{id}{\left[{e}^{t}-1\right]}^{i}\\ =& \sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{d}-{c}^{d}\right)}^{m-i}{\beta }^{id}i!\sum _{j=i}^{\mathrm{\infty }}S\left(j,i\right)\frac{{t}^{j}}{j!},\end{array}$

we can write from (4) that

$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}}{n!}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(mk\right)!}{\left(\frac{x}{1+ax}\right)}^{mk}{\left[\frac{{2}^{1-k}{t}^{k}}{{\beta }^{b}{e}^{t}-{a}^{b}}\right]}^{m}{e}^{t\left[\frac{1+bx}{1+ax}\right]}\hfill \\ \phantom{\rule{2em}{0ex}}×\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{b}-{a}^{b}\right)}^{m-i}{\beta }^{ib}i!\sum _{j=i}^{\mathrm{\infty }}S\left(j,i\right)\frac{{t}^{j}}{j!}.\hfill \end{array}$

Now, using (3) in the above relation, we get

$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}}{n!}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(mk\right)!}{\left(\frac{x}{1+ax}\right)}^{mk}\sum _{n=0}^{\mathrm{\infty }}{Q}_{n,\beta }^{\left(m\right)}\left(\frac{1+bx}{1+ax};k,c,d\right)\frac{{t}^{n}}{n!}\hfill \\ \phantom{\rule{2em}{0ex}}×\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{d}-{c}^{d}\right)}^{m-i}{\beta }^{id}i!\sum _{j=i}^{\mathrm{\infty }}S\left(j,i\right)\frac{{t}^{j}}{j!}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(mk\right)!}{\left(\frac{x}{1+ax}\right)}^{mk}\sum _{n=0}^{\mathrm{\infty }}\left\{\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left({\beta }^{d}-{c}^{d}\right)}^{m-i}{\beta }^{id}i!\hfill \\ \phantom{\rule{2em}{0ex}}×\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){Q}_{n-j,\beta }^{\left(m\right)}\left(\frac{1+bx}{1+ax};k,c,d\right)\right\}\frac{{t}^{n}}{n!}.\hfill \end{array}$

Whence the result. □

Now, we list some important corollaries of the above theorem.

Corollary 8 Since ${\mathcal{P}}_{n}^{\left(0,-1\right)}\left(x;1,m\right)={B}_{m}^{n}\left(x\right)$ and ${Q}_{n,\lambda }^{\left(\alpha \right)}\left(x;1,1,1\right)={\mathcal{B}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right)$, we obtain the following [1]:

${B}_{m}^{n}\left(x\right)=\frac{{x}^{m}}{m!}\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left(\lambda -1\right)}^{m-i}{\lambda }^{i}i!\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){\mathcal{B}}_{n-j}^{\left(m\right)}\left(1-x;\lambda \right).$

Furthermore, for $\lambda =1$, we have the following known relation:

${B}_{m}^{n}\left(x\right)={x}^{m}\sum _{j=m}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,m\right){B}_{n-j}^{\left(m\right)}\left(1-x\right).$

Corollary 9 Since ${\mathcal{P}}_{n}^{\left(0,-1\right)}\left(x;1,m\right)={B}_{m}^{n}\left(x\right)$ and ${Q}_{n,\frac{\lambda }{2}}^{\left(\alpha \right)}\left(x;1,\frac{-1}{2},1\right)={\mathcal{G}}_{n}^{\alpha }\left(x;\lambda \right)$, we get

${B}_{m}^{n}\left(x\right)=\frac{{x}^{m}}{{2}^{m}m!}\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left(\lambda +1\right)}^{m-i}{\lambda }^{i}i!\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){\mathcal{G}}_{n-j}^{m}\left(1-x;\lambda \right).$

Corollary 10 Since ${\mathcal{P}}_{n}^{\left(1,0\right)}\left(x;1,m\right)={H}_{m}^{n}\left(x\right)$ and ${Q}_{n,\lambda }^{\left(\alpha \right)}\left(x;1,1,1\right)={\mathcal{B}}_{n}^{\left(\alpha \right)}\left(x;\lambda \right)$, we obtain

$\begin{array}{rcl}{H}_{m}^{n}\left(x\right)& =& \frac{1}{m!}{\left(\frac{x}{1+x}\right)}^{m}\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left(\lambda -1\right)}^{m-i}{\lambda }^{i}i!\\ ×\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){\mathcal{B}}_{n-j}^{\left(m\right)}\left(\frac{1}{1+x};\lambda \right).\end{array}$

Furthermore, when $\lambda =1$, we have the following:

${H}_{m}^{n}\left(x\right)={\left(\frac{x}{1+x}\right)}^{m}\sum _{j=m}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,m\right){B}_{n-j}^{\left(m\right)}\left(\frac{1}{1+x}\right).$

Corollary 11 Since ${\mathcal{P}}_{n}^{\left(1,0\right)}\left(x;1,m\right)={H}_{m}^{n}\left(x\right)$ and ${Q}_{n,\frac{\lambda }{2}}^{\left(\alpha \right)}\left(x;1,\frac{-1}{2},1\right)={\mathcal{G}}_{n}^{\alpha }\left(x;\lambda \right)$, we get

$\begin{array}{rcl}{H}_{m}^{n}\left(x\right)& =& \frac{1}{{2}^{m}m!}{\left(\frac{x}{1+x}\right)}^{m}\sum _{i=0}^{m}\left(\genfrac{}{}{0}{}{m}{i}\right){\left(\lambda -1\right)}^{m-i}{\lambda }^{i}i!\\ ×\sum _{j=i}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)S\left(j,i\right){\mathcal{G}}_{n-j}^{m}\left(\frac{1}{1+x};\lambda \right).\end{array}$

## 3 Generating functions of trigonometric type

In this section, we obtain a trigonometric generating relation for the unified Bernstein and Bleimann-Butzer-Hahn basis. Furthermore, we give a certain summation formula for this unification. We start with the following theorem.

Theorem 12 For the unified family, we have the following implicit summation formulae:

$\begin{array}{r}{\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+ax\right)}^{2l}}\right]}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}cost\left(\frac{1+bx}{1+ax}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{t}^{2n}}{\left(2n\right)!},\\ {\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+ax\right)}^{2l}}\right]}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}sint\left(\frac{1+bx}{1+ax}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}\end{array}$
(5)

and

$\begin{array}{r}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}cost\left(\frac{1+bx}{1+ax}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{t}^{2n}}{\left(2n\right)!},\\ {\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}sint\left(\frac{1+bx}{1+ax}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\end{array}$
(6)

Finally,

$\begin{array}{r}\begin{array}{r}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{\left(2lj+l+j\right)}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}tsint\left(\frac{1+bx}{1+ax}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{t}^{2n}}{\left(2n\right)!},\end{array}\\ \begin{array}{r}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{\left(2lj+l+j\right)}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}tcost\left(\frac{1+bx}{1+ax}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\end{array}\end{array}$
(7)

Proof Writing $k=2l$ ($l\in {\mathbb{N}}_{0}$) in (1), we get

${\left[\frac{{2}^{1-2l}{x}^{2l}{t}^{2l}}{{\left(1+ax\right)}^{2l}}\right]}^{m}\frac{1}{\left(2lm\right)!}{e}^{t\left[\frac{1+bx}{1+ax}\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{t}^{n}}{n!}.$

Letting $t\to it$, we get

${\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+ax\right)}^{2l}}\right]}^{m}\frac{{\left(it\right)}^{2lm}}{\left(2lm\right)!}{e}^{it\left[\frac{1+bx}{1+ax}\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{\left(it\right)}^{n}}{n!}$

and hence

$\begin{array}{c}{\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+ax\right)}^{2l}}\right]}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}\left\{cost\left(\frac{1+bx}{1+ax}\right)+isint\left(\frac{1+bx}{1+ax}\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{\left(it\right)}^{2n}}{\left(2n\right)!}+\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{\left(it\right)}^{2n+1}}{\left(2n+1\right)!}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{t}^{2n}}{\left(2n\right)!}\hfill \\ \phantom{\rule{2em}{0ex}}+i\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l,m\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\hfill \end{array}$

Equating real and imaginary parts, we get (5).

Now, taking $k=2l+1$and $m=2j$ ($l,j\in {\mathbb{N}}_{0}$) in (1), we obtain

${\left[\frac{{2}^{1-\left(2l+1\right)}{x}^{2l+1}{t}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j}\frac{1}{\left(2j\left(2l+1\right)\right)!}{e}^{t\left[\frac{1+bx}{1+ax}\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{t}^{n}}{n!}.$

Putting $t\to it$,

${\left[\frac{{2}^{-2l}{x}^{2l+1}{\left(it\right)}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j}\frac{1}{\left(2j\left(2l+1\right)\right)!}{e}^{it\left[\frac{1+bx}{1+ax}\right]}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{\left(it\right)}^{n}}{n!}.$

Therefore, we get

$\begin{array}{c}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}\left\{cost\left(\frac{1+bx}{1+ax}\right)+isint\left(\frac{1+bx}{1+ax}\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{t}^{2n}}{\left(2n\right)!}+i\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l+1,2j\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!},\hfill \end{array}$

which is precisely (6).

Finally, for $k=2l+1$, $m=2j+1$,

${\left[\frac{{2}^{-2l}{x}^{2l+1}{t}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j+1}\frac{{e}^{t\left[\frac{1+bx}{1+ax}\right]}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{t}^{n}}{n!}.$

Taking $t\to it$,

${\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j+1}\frac{{\left(it\right)}^{\left(2l+1\right)\left(2j+1\right)}{e}^{it\left[\frac{1+bx}{1+ax}\right]}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{\left(it\right)}^{n}}{n!}.$

Thus,

$\begin{array}{c}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+ax\right)}^{2l+1}}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{\left(2lj+l+j\right)}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}\left[-tsint\left(\frac{1+bx}{1+ax}\right)+itcost\left(\frac{1+bx}{1+ax}\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{t}^{2n}}{\left(2n\right)!}\hfill \\ \phantom{\rule{2em}{0ex}}+i\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{P}}_{2n+1}^{\left(a,b\right)}\left(x;2l+1,2j+1\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\hfill \end{array}$

Equating real and imaginary parts we get (7). □

Since we obtain the unified Bernstein family in the case $a=0$, $b=-1$, we have the following corollary at once.

Corollary 13 For the unified Bernstein family, we have the following implicit summation formulae:

$\begin{array}{c}{\left({2}^{1-2l}{x}^{2l}\right)}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}cost\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n}\left(2lm,x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {\left({2}^{1-2l}{x}^{2l}\right)}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}sint\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n+1}\left(2lm,x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}\hfill \end{array}$

and

$\begin{array}{r}{\left({2}^{-2l}{x}^{2l+1}\right)}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}cost\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n}\left(\left(2l+1\right)\left(2j\right),x\right)\frac{{t}^{2n}}{\left(2n\right)!},\\ {\left({2}^{-2l}{x}^{2l+1}\right)}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}sint\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n+1}\left(\left(2l+1\right)\left(2j\right),x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\end{array}$
(8)

Finally,

$\begin{array}{r}\begin{array}{r}{\left[{2}^{-2l}{x}^{2l+1}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{\left(2lj+l+j\right)}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}tsint\left(1-x\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n}\left(\left(2l+1\right)\left(2j+1\right),x\right)\frac{{t}^{2n}}{\left(2n\right)!},\end{array}\\ \begin{array}{r}{\left[{2}^{-2l}{x}^{2l+1}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{\left(2lj+l+j\right)}}{\left[\left(2j+1\right)\left(2l+1\right)\right]!}tcost\left(1-x\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{B}}_{2n+1}\left(\left(2l+1\right)\left(2j+1\right),x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\end{array}\end{array}$
(9)

On the other hand, taking $l=0$ in (8) and (9), we get the following relations for the Bernstein basis:

$\begin{array}{c}{x}^{2j}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j\right)!}cost\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{B}_{2j}^{2n}\left(x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {x}^{2j}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j\right)!}sint\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{B}_{2j}^{2n+1}\left(x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}\hfill \end{array}$

and

$\begin{array}{c}{x}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j+1\right)!}tsint\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{B}_{2j+1}^{2n}\left(x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {x}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j+1\right)!}tcost\left(1-x\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{B}_{2j+1}^{2n+1}\left(x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\hfill \end{array}$

Since the case $a=1$, $b=0$ gives the unified Bleimann-Butzer-Hahn family, we immediately obtain the following corollary.

Corollary 14 For the unified Bleimann-Butzer-Hahn family, we have the following implicit summation formulae:

$\begin{array}{c}{\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+x\right)}^{2l}}\right]}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}cos\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{H}}_{2n}\left(2lm,x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {\left[\frac{{2}^{1-2l}{x}^{2l}}{{\left(1+x\right)}^{2l}}\right]}^{m}\frac{{\left(-{t}^{2}\right)}^{lm}}{\left(2lm\right)!}sin\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{H}}_{2n+1}\left(2lm,x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}\hfill \end{array}$

and

$\begin{array}{r}{\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+x\right)}^{2l+1}}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}cos\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{H}}_{2n}\left(\left(2l+1\right)\left(2j\right),x\right)\frac{{t}^{2n}}{\left(2n\right)!},\\ {\left[\frac{{2}^{-2l}{x}^{2l+1}}{{\left(1+x\right)}^{2l+1}}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{\left(2l+1\right)j}}{\left(2j\left(2l+1\right)\right)!}sin\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\mathcal{H}}_{2n+1}\left(\left(2l+1\right)\left(2j\right),x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\end{array}$
(10)

Finally,

(11)

Taking $l=0$ in (10) and (11), we get the following relations for the Bleimann-Butzer-Hahn basis:

$\begin{array}{c}{\left[\frac{x}{1+x}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j\right)!}cos\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{H}_{2j}^{2n}\left(x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {\left[\frac{x}{1+x}\right]}^{2j}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j\right)!}sin\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{H}_{2j}^{2n+1}\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\hfill \end{array}$

Finally,

$\begin{array}{c}{\left[\frac{x}{1+x}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j+1\right)!}tsin\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{H}_{2j+1}^{2n}\left(x\right)\frac{{t}^{2n}}{\left(2n\right)!},\hfill \\ {\left[\frac{x}{1+x}\right]}^{2j+1}\frac{{\left(-{t}^{2}\right)}^{j}}{\left(2j+1\right)!}tcos\left(\frac{t}{1+x}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{H}_{2j+1}^{2n+1}\left(x\right)\frac{{t}^{2n+1}}{\left(2n+1\right)!}.\hfill \end{array}$

Finally, we obtain a summation formula for the unified Bernstein and Bleimann-Butzer-Hahn basis as follows.

Theorem 15 For all $n,l\in {\mathbb{N}}_{0}$; $a,b\in \mathbb{R}$, the following implicit summation formula holds true:

${\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(y;k,m\right)=\sum _{p,r=0}^{l,n}\left(\genfrac{}{}{0}{}{n}{r}\right)\left(\genfrac{}{}{0}{}{l}{p}\right){\mathcal{P}}_{n+l-r-p}^{\left(a,b\right)}\left(x;k,m\right){\left[\frac{1+by}{1+ay}-\frac{1+bx}{1+ax}\right]}^{r+p}.$

Proof Letting $t\to t+u$ in (1) and then using the fact that

$\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^{\mathrm{\infty }}A\left(l,n\right)=\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^{n}A\left(l,n-l\right),$
(12)

we get

$\begin{array}{rcl}{\left[\frac{{2}^{1-k}{x}^{k}{\left(t+u\right)}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{\left(t+u\right)\left[\frac{1+bx}{1+ax}\right]}& =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)\frac{{\left(t+u\right)}^{n}}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_{n}^{\left(a,b\right)}\left(x;k,m\right)\sum _{l=0}^{n}\frac{{t}^{n-l}{u}^{l}}{l!\left(n-l\right)!}\\ =& \sum _{n,l=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}{u}^{l}}{n!l!}\end{array}$
(13)

and hence

${\left[\frac{{2}^{1-k}{x}^{k}{\left(t+u\right)}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}={e}^{-\left(t+u\right)\left[\frac{1+bx}{1+ax}\right]}\sum _{n,l=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}{u}^{l}}{n!l!}.$

Multiplying both sides by ${e}^{\left(t+u\right)\left[\frac{1+by}{1+ay}\right]}$ and then expanding the function ${e}^{\left(t+u\right)\left[\frac{1+by}{1+ay}-\frac{1+bx}{1+ax}\right]}$, we get, after using (12) twice, that

$\begin{array}{c}{\left[\frac{{2}^{1-k}{x}^{k}{\left(t+u\right)}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{\left(t+u\right)\left[\frac{1+by}{1+ay}\right]}\hfill \\ \phantom{\rule{1em}{0ex}}={e}^{\left(t+u\right)\left[\frac{1+by}{1+ay}-\frac{1+bx}{1+ax}\right]}\sum _{n,l=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(x;k,m\right)\frac{{t}^{n}{u}^{l}}{n!l!}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n,l=0}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(x;k,m\right)\frac{{\left[\frac{1+by}{1+ay}-\frac{1+bx}{1+ax}\right]}^{r}}{r!}{\left(t+u\right)}^{r}\frac{{t}^{n}{u}^{l}}{n!l!}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n,l,p,r=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(x;k,m\right){\left[\frac{1+by}{1+ay}-\frac{1+bx}{1+ax}\right]}^{r+p}\frac{{t}^{n+r}{u}^{p+l}}{n!l!r!p!}.\hfill \end{array}$

Now, using (12) with the index pairs $\left(n,r\right)$ and $\left(l,p\right)$, we get

(14)

Since the left-hand side is equal by (13) to

${\left[\frac{{2}^{1-k}{x}^{k}{\left(t+u\right)}^{k}}{{\left(1+ax\right)}^{k}}\right]}^{m}\frac{1}{\left(mk\right)!}{e}^{\left(t+u\right)\left[\frac{1+by}{1+ay}\right]}=\sum _{n,l=0}^{\mathrm{\infty }}{\mathcal{P}}_{n+l}^{\left(a,b\right)}\left(y;k,m\right)\frac{{t}^{n}{u}^{l}}{n!l!},$
(15)

the proof is completed by comparing the coefficients of $\frac{{t}^{n}{u}^{l}}{n!l!}$ in (14) and (15). □

In the case $a=0$, $b=-1$, we obtain the following result for the unified Bernstein family at once.

Corollary 16 For all $n,l\in {\mathbb{N}}_{0}$, the following implicit summation formula:

${\mathcal{B}}_{n+l}\left(mk,y\right)=\sum _{p,r=0}^{l,n}\left(\genfrac{}{}{0}{}{n}{r}\right)\left(\genfrac{}{}{0}{}{l}{p}\right){\mathcal{B}}_{n+l-r-p}\left(mk,x\right){\left[x-y\right]}^{r+p}$
(16)

holds true for the unified Bernstein family. Taking $k=1$ in (16), we get the following relation for the Bernstein basis:

${B}_{m}^{n+l}\left(y\right)=\sum _{p,r=0}^{l,n}\left(\genfrac{}{}{0}{}{n}{r}\right)\left(\genfrac{}{}{0}{}{l}{p}\right){B}_{m}^{n+l-r-p}\left(x\right){\left[x-y\right]}^{r+p}.$

Since the case $a=1$, $b=0$ gives the unified Bleimann-Butzer-Hahn family, we have the following result.

Corollary 17 For all $n,l\in {\mathbb{N}}_{0}$, the following implicit summation formula:

${\mathcal{H}}_{n+l}\left(mk,y\right)=\sum _{p,r=0}^{l,n}\left(\genfrac{}{}{0}{}{n}{r}\right)\left(\genfrac{}{}{0}{}{l}{p}\right){\mathcal{H}}_{n+l-r-p}\left(mk,x\right){\left[x-y\right]}^{r+p}$
(17)

holds true for the unified Bleimann-Butzer-Hahn family. Upon taking $k=1$ in (17), we get the following relation for the Bleimann-Butzer-Hahn basis:

${H}_{m}^{n+l}\left(y\right)=\sum _{p,r=0}^{l,n}\left(\genfrac{}{}{0}{}{n}{r}\right)\left(\genfrac{}{}{0}{}{l}{p}\right){H}_{m}^{n+l-r-p}\left(x\right){\left[x-y\right]}^{r+p}.$

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## Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Correspondence to Mehmet Bozer.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors completed the paper together. All authors read and approved the final manuscript.

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Özarslan, M.A., Bozer, M. Unified Bernstein and Bleimann-Butzer-Hahn basis and its properties. Adv Differ Equ 2013, 55 (2013). https://doi.org/10.1186/1687-1847-2013-55

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• DOI: https://doi.org/10.1186/1687-1847-2013-55

### Keywords

• generating function
• Bernstein polynomials
• Bernoulli polynomials
• Euler polynomials
• Genocchi polynomials
• Stirling numbers of the second kind