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

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

(1)

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

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

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

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

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

which generates the celebrated Bernstein basis polynomials (see [2–8])

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

Note that the Bernstein operators $B_n:C[0,1]\to C[0,1]$ are given by

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

and by the Korovkin theorem, it is known that $B_n(f;x)\rightrightarrows f(x)$ for all $f\in C[0,1]$, where $C[0,1]$ denotes the space of continuous functions defined on $[0,1]$, and the notation ‘⇉’ denotes the uniform convergence with respect to the usual supremum norm on $C[0,1]$. Very recently, interesting properties of Bernstein polynomials were discussed in [7, 9–11] and [12].

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

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

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

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

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

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

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

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

Denoting $C_B[0,\mathrm{\infty })$ by the space of real-valued bounded continuous functions defined on $[0,\mathrm{\infty })$, they proved that $L_n(f)\to f$ as $n\to \mathrm{\infty }$. On the other hand, the convergence is uniform on each compact subset of $[0,\mathrm{\infty })$, where the norm is the usual supremum norm of $C_B[0,\mathrm{\infty })$. 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 ${\mathcal{P}}_n^{(a,b)}(x;k,m):=0$ for $n\le mk$.

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

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

Proof

Direct calculations give

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

(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]:

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

Furthermore, ${\mathcal{B}}_n(m,x)=B_m^n(x)$ 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:

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

Moreover, ${\mathcal{H}}_n(m,x)=H_m^n(x)$ 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 $Q_{n,\beta }^{(\alpha )}(x;k,a,b)$ are listed below.

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

$$Q_{n,\lambda }^{(\alpha )}(x;1,1,1)={\mathcal{B}}_n^{(\alpha )}(x;\lambda ).$$

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

$$\begin{array}{c}{\left(\frac{t}{\lambda e^t-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_n^{(\alpha )}(x;\lambda )\frac{t^n}{n!}\hfill \\ (|t|<2\pi \text{ when }\lambda =1;|t|<|log\lambda |\text{ when }\lambda \ne 1),\hfill \end{array}$$

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:

$$B_n(x;\lambda )={\mathcal{B}}_n^{(1)}(x;\lambda ),B_n(\lambda )=B_n(0;\lambda )(n\in {\mathbb{N}}_0).$$

Taking $\lambda =1$ in the above relations, we obtain the classical Bernoulli polynomials $B_n(x)$ and Bernoulli numbers $B_n$.

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

$$Q_{n,\frac{\lambda }{2}}^{(\alpha )}(x;1,\frac{-1}{2},1)={\mathcal{G}}_n^{\alpha }(x;\lambda ),$$

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:

$$\begin{array}{c}{\left(\frac{2t}{\lambda e^t+1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{G}}_n^{\alpha }(x;\lambda )\frac{t^n}{n!}\hfill \\ (|t|<\pi \text{ when }\lambda =1;|t|<|log(-\lambda )|\text{ when }\lambda \ne 1).\hfill \end{array}$$

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

$$G_n(x;\lambda )={\mathcal{G}}_n^1(x;\lambda ),G_n(\lambda )=G_n(0;\lambda ).$$

When $\lambda =1$, the above relations give the classical Genocchi polynomials $G_n(x)$ 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 $Q_{n,\beta }^{(\alpha )}(x;k,a,b)$.

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

$$Q_{n,\lambda }^{(\alpha )}(x;0,-1,1)={\mathcal{E}}_n^{(\alpha )}(x;\lambda ).$$

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

$$\begin{array}{c}{\left(\frac{2}{\lambda e^t+1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_n^{\alpha }(x;\lambda )\frac{t^n}{n!}\hfill \\ (|t|<\pi \text{ when }\lambda =1;|t|<|log(-\lambda )|\text{ when }\lambda \ne 1;1^{\alpha }:=1)\hfill \end{array}$$

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

$$E_n(x;\lambda )={\mathcal{E}}_n^1(x;\lambda ),E_n(\lambda )=E_n(1;\lambda ).$$

When $\lambda =1$, the above relations give the classical Euler polynomials $E_n(x)$ and Euler numbers $E_n$.

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

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

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:

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

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

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

we can write from (4) that

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

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

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

Whence the result. □

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

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

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

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

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

Corollary 9 Since ${\mathcal{P}}_n^{(0,-1)}(x;1,m)=B_m^n(x)$ and $Q_{n,\frac{\lambda }{2}}^{(\alpha )}(x;1,\frac{-1}{2},1)={\mathcal{G}}_n^{\alpha }(x;\lambda )$, we get

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

Corollary 10 Since ${\mathcal{P}}_n^{(1,0)}(x;1,m)=H_m^n(x)$ and $Q_{n,\lambda }^{(\alpha )}(x;1,1,1)={\mathcal{B}}_n^{(\alpha )}(x;\lambda )$, we obtain

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

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

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

Corollary 11 Since ${\mathcal{P}}_n^{(1,0)}(x;1,m)=H_m^n(x)$ and $Q_{n,\frac{\lambda }{2}}^{(\alpha )}(x;1,\frac{-1}{2},1)={\mathcal{G}}_n^{\alpha }(x;\lambda )$, we get

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

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:

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

(5)

and

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

(6)

Finally,

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

(7)

Proof Writing $k=2l$ ($l\in {\mathbb{N}}_0$) in (1), we get

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

Letting $t\to it$, we get

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

and hence

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

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

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

Putting $t\to it$,

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

Therefore, we get

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

which is precisely (6).

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

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

Taking $t\to it$,

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

Thus,

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

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:

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

and

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

(8)

Finally,

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

(9)

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

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

and

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

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:

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

and

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

(10)

Finally,

(11)

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

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

Finally,

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

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:

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

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

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

(12)

we get

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

(13)

and hence

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

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

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

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

(14)

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

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

(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:

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

(16)

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

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

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:

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

(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:

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