1 Introduction

Bernstein based functions have many applications not only in mathematics but also in other sciences. These functions and their generating functions are elated to various areas such as B-splines, splines, probability generating functions and distribution functions, the Bezier curves, computer geometric designs, etc. For this reason, they still attract the attention of researchers working in many different fields. Therefore, the motivation of this article is to reveal new formulas and relations by blending generating functions for Bernstein basis functions, moment generating functions containing these basis functions, and some other special number families and polynomial families. These formulas and relations are related to the Apostol–Euler numbers and polynomials, the Bernstein basis functions, the Stirling numbers, the central moments, the combinatorial numbers and polynomials.

The Bernstein operator is defined as follows:

$$\begin{aligned} B_{n}\left[ f\right] (x)=\sum _{j=0}^{n}B_{j}^{n}(x)f\left( \frac{j}{n} \right) , \end{aligned}$$

where \(f:\left[ 0,1\right] \rightarrow R\), \(n\in {\mathbb {N}}_{0}={\mathbb {N}} \cup \{0\}\), \({\mathbb {N}}=\left\{ {1,2,3,...}\right\} \), \(0\le j\le n\), and \(B_{j}^{n}(x)\) denotes the Bernstein basis functions:

$$\begin{aligned} B_{j}^{n}(x)=\left( {\begin{array}{c}n\\ j\end{array}}\right) x^{j}(1-x)^{n-j}, \end{aligned}$$
(1)

which are defined by means of the following generating function:

$$\begin{aligned} \frac{(tx)^{k}}{k!}e^{(1-x)t}=\sum \limits _{n=0}^{\infty }B_{k}^{n}(x)\frac{ t^{n}}{n!} \end{aligned}$$
(2)

(cf. [2, 3, 9]).

With the aid of q analysis, generating function in (2) was found by Simsek and Acikgoz [22]. Equation (2) was also given Acikgoz and Arici [1, 15] and Simsek [15].

There are many applications of these generating functions, see also for detail (cf. [3, 6, 8, 16, 17]).

The Bernstein operator has remained popular for the following reasons: this operator is given explicitly for rational values of the variable x because derivatives and integrals of the Bernstein basis functions are easily calculated. These basis functions are related to many special functions, special polynomials, and also integral transform (cf. [2, 3, 9]; see also the references cited in each of these earlier works).

Many different mathematical models can be studied by extending Bernstein basis functions from the range \(0\le x\le 1\) to any range \(a\le x\le b\). Simsek [21] give a relation between generating functions for the uniform B-splines and generating functions for the Bernstein basis functions.

Substituting \(x=\frac{x-a}{b-a}\) into (1) and (2), we have

$$\begin{aligned} B_{k}^{n}\left( x;a,b\right) =\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) }{\left( b-a\right) ^{n}} \left( x-a\right) ^{k}\left( b-x\right) ^{n-k}, \end{aligned}$$
(3)

which are defined by means of the following generating functions:

$$\begin{aligned} f_{\beta }(t,x;k,a,b)=\left( \frac{x-a}{b-a}t\right) ^{k}e^{\left( \frac{b-x }{b-a}\right) t}=\sum \limits _{n=0}^{\infty }k!B_{k}^{n}(x;a,b)\frac{t^{n}}{n! } \end{aligned}$$
(4)

(cf. [16]).

By using (3), pobability generating functions, moment generating functions and other applications were studied, see ( [6,7,8,9, 12,13,14, 16,17,20, 24]; see also the references cited in each of these earlier works).

Bernstein [2] gave the central moments involving the Bernstein basis function. This moment is defined by

$$\begin{aligned} M(x;r,n)=\sum _{k=0}^{n}\left( \frac{k}{n}-x\right) ^{r}B_{k}^{n}(x) \end{aligned}$$
(5)

where \(n\in {\mathbb {N}}\) and \(r\in {\mathbb {N}}_{0}\) (cf. [3, p. 106]).

There are many books, manuscripts, and papers for the the central moments given in (5) (cf. [2, 3, 9]; see also the references cited in each of these earlier works). Here we can use notation in [3]. There are also many other papers and books for the generating functions of the \(B_{k}^{n}(x)\). In [3, p. 106, Eq. (2.15)], the following polynomials are given as follows:

$$\begin{aligned} T(x;r,n)= & {} n^{r}M(x;r,n) \nonumber \\= & {} n^{r}\sum \limits _{k=0}^{n}\left( \frac{k}{n}-x\right) ^{r}B_{k}^{n}(x). \end{aligned}$$
(6)

In [3], generating function for M(x; r, n) is given by the following theorem.

Theorem 1

Let \(n\in {\mathbb {N}}\) and \(0\le x\le n\). Then we have

$$\begin{aligned} F_{n}(t,x)= & {} \left( e^{\frac{-tx}{n}}\left( 1-x+xe^{\frac{tx}{n}}\right) \right) ^{n} \nonumber \\= & {} \sum \limits _{r=0}^{\infty }M(x;r,n)\frac{t^{r}}{r!}. \end{aligned}$$
(7)

Proof of (7) was firstly given by Bernstein [2]. The other proof given by same line by Bustamante [3].

The goal of this article is to give moments with aid of the Bernstein basis functions on the interval \(a\le x\le b\). Generating functions for these moments with their properties are given. Some functional equations with their applications of these functions are given. A polynomial T(x; r, n; a, b) is also defined. This polynomial can be used in generalized Voronovskaya-type theorems. In order to give these results, we need the Apostol–Euler polynomials of the first kind \({\mathcal {E}} _{n}(x,\lambda )\), which are defined by means of the following generating function:

$$\begin{aligned} F_{{\mathcal {E}},P}(t,x;\lambda )=\frac{2e^{tx}}{\lambda e^{t}+1} =\sum _{n=0}^{\infty }{\mathcal {E}}_{n}(x,\lambda )\frac{t^{n}}{n!} \end{aligned}$$
(8)

(cf. [4, 10, 19, 23]).

Putting \(x=0\) in (8), generating function for Apostol–Euler numbers of the first kind is given as follows:

$$\begin{aligned} F_{{\mathcal {E}},N}(t;\lambda )=\frac{2}{\lambda e^{t}+1}=\sum _{n=0}^{\infty } {\mathcal {E}}_{n}(\lambda )\frac{t^{n}}{n!} \end{aligned}$$
(9)

(cf. [4, 10, 19, 23]).

The Stirling numbers of the second kind are given by

$$\begin{aligned} x^{r}=\sum \limits _{v=0}^{r}x_{(v)}S_{2}(r,v), \end{aligned}$$
(10)

where \(x_{(0)}=1\) and \(x_{(v)}=x(x-1)\cdots (x-v+1)\) (cf. [19, 23]).

Generating function for the following combinatorial numbers

$$\begin{aligned} y_{1}\left( n,k,\lambda \right) =\frac{1}{k!}\sum _{j=0}^{k}\left( {\begin{array}{c}k\\ j\end{array}}\right) j^{n}\lambda ^{j} \end{aligned}$$
(11)

are by

$$\begin{aligned} (\lambda e^{t}+1)^{k}=\sum _{n=0}^{\infty }k!y_{1}\left( n,k,\lambda \right) \frac{t^{n}}{n!} \end{aligned}$$
(12)

(cf. [18]).

The combinatorial numbers \(y_{4}(r,n;\tau ,u,w)\) are defined as follows:

$$\begin{aligned} F_{y_{4}}(t,n;u,w,\tau )= & {} \sum _{r=0}^{\infty }y_{4}(r,n;\tau ,u,w)\frac{ t^{r}}{n!} \nonumber \\= & {} e^{utn}\left( \frac{1+\tau e^{\left( w-u\right) t}}{u+w+1}\right) ^{r} \end{aligned}$$
(13)

(cf. [17]).

A brief summary of the sections of this article where the new results are exhibited is given as follows:

In Sect. 2, we give a new class of the polynomials T(x; r, n; a, b). These polynomials are also related to the Bernstein basis functions with real parameters and the central factorial moments. We construct generating functions for these polynomials with their properties. Using functional equations and derivative equations of these generating functions, we derive some formulas and relations.

In Sect. 3, by using partial derivative equations of the generating functions, we give higher-order derivative formulas for the polynomials T(x; t, n; a, b).

In Sect. 4, we introduce some integral formulas of the polynomials T(x; t, n; a, b). By applying the Laplace transformation to the generating functions, we derive infinite series representations for these polynomials and the Bernstein basis functions.

In Sect. 5, we give a conclusion section.

2 Generating functions for polynomials T(x; r, n; a, b)

In this section, we construct generating functions for the polynomials T(x; r, n; a, b), which involving the Bernstein basis functions with real parameters and the central factorial moments. Some properties of these functions can be given. We also give some identities for these polynomials with aid of functional equations of these generating functions.

Let a and b be any real parameters such that \(a\ne b\). Let \(a\le x\le b\).

Using (6) and (10), we give the following polynomial in terms of the Bernstein basis functions and the Stirling numbers of the second kind with real parameters a and b:

$$\begin{aligned} T(x;r,n;a,b)=\sum \limits _{v=0}^{r}\sum \limits _{k=0}^{n}n_{(v)}S_{2}(r,v)B_{k}^{n}(x;a,b)\left( \frac{k}{n}-\frac{ x-a}{b-a}\right) ^{r}. \end{aligned}$$
(14)

By using (14), we also have

$$\begin{aligned} T(x;r,n;a,b)=n^{r}M(x;r,n;a,b). \end{aligned}$$
(15)

Remark 1

Substituting \(a=0\) and \(b=1\), (15) reduces to (6).

Generating function for the polynomials T(x; r, n; a, b) is given by the following theorem.

Theorem 2

Let \(n\in {\mathbb {N}}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=e^{-tn\left( \frac{x-a}{b-a}\right) }\sum \limits _{k=0}^{n}e^{tk}B_{k}^{n}\left( x;a,b\right) , \end{aligned}$$
(16)

which yields

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=\sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r! }. \end{aligned}$$

Proof

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{r=0}^{ \infty }n^{r}M(x;r,n;a,b)\frac{t^{r}}{r!}. \end{aligned}$$

Joining the above equation with (14) yields

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{r=0}^{ \infty }n^{r}\sum \limits _{k=0}^{n}\left( \frac{k}{n}-\frac{x-a}{b-a}\right) ^{r}B_{k}^{n}(x;a,b)\frac{t^{r}}{r!}. \end{aligned}$$

After some elementary calculations, we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{k=0}^{n}B_{k}^{n}(x;a,b)\sum \limits _{r=0}^{\infty }\frac{ n^{r}t^{r}\left( \frac{k}{n}-\frac{x-a}{b-a}\right) ^{r}}{r!}. \end{aligned}$$

Therefore

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{k=0}^{n}B_{k}^{n}(x;a,b)e^{t\left( k-\frac{n(x-a)}{b-a}\right) }. \end{aligned}$$

Thus we arrive at the desired result. \(\square \)

Remark 2

By using (16), we get

$$\begin{aligned} e^{tn\left( \frac{x-a}{b-a}\right) }{\mathcal {A}}(x;t,n;a,b)=M_{X}(x;t,n;a,b). \end{aligned}$$

This equation was proved by Simsek ([12, 13], and see also [14]).

Combining (16) with the equation (6) in [12], we get the following result:

Corollary 3

Let \(n\in {\mathbb {N}}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=e^{-tn\left( \frac{x-a}{b-a}\right) }M_{x}(x;t,n;a,b). \end{aligned}$$
(17)

Joining (17), (11) with equation (13) in [12], we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{r=0}^{ \infty }\left( -n\left( \frac{x-a}{b-a}\right) \right) ^{r}\frac{t^{r}}{r!} \sum _{r=0}^{\infty }\left( n!\left( \frac{b-x}{b-a}\right) ^{n}y_{1}\left( r,n,\frac{x-a}{b-a}\right) \right) \frac{t^{r}}{r!}. \end{aligned}$$

By applying the Cauchy product rule to the right-hand side of the above equation, we obtain the following theorem:

Theorem 4

Let \(n,r\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} T(x;r,n;a,b)=\sum _{j=0}^{\infty }\left( {\begin{array}{c}r\\ j\end{array}}\right) n!\left( \frac{b-x}{b-a}\right) ^{n}y_{1}\left( j,n,\frac{x-a}{b-a}\right) (-n)^{r-j}\left( \frac{x-a}{b-a} \right) ^{r-j}. \end{aligned}$$

Alternating generating functions for (16) are given as follows:

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)= & {} \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \frac{x-a}{b-a }\right) ^{k}\left( \frac{b-x}{b-a}\right) ^{n-k}e^{tk}e^{-tn\left( \frac{x-a }{b-a}\right) }, \nonumber \\ {\mathcal {A}}(x;t,n;a,b)= & {} e^{-tn\left( \frac{x-a}{b-a}\right) }\left( \frac{x-a }{b-a}e^{t}+\frac{b-x}{b-a}\right) ^{n}, \end{aligned}$$
(18)
$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=\frac{e^{-tn\left( \frac{x-a}{b-a}\right) }}{(b-a)^{n} }\left( b-x+(x-a)e^{t}\right) ^{n} \end{aligned}$$

and

$$\begin{aligned} e^{tn\left( \frac{x-a}{b-a}\right) }{\mathcal {A}}(x;t,n;a,b)=\left( \frac{ b-x+(x-a)e^{t}}{b-a}\right) ^{n}. \end{aligned}$$
(19)

Combining (19) with (7), we get the following result:

$$\begin{aligned} {\mathcal {A}}(x;t,n;0,1)=F_{n}(x,t;n). \end{aligned}$$

Using the above functional equation, we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }M(x;r,n;a,b)\frac{t^{r}}{r!}=\sum \limits _{r=0}^{ \infty }\frac{T(x;r,n;a,b)t^{r}}{n^{r}r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!} \) on the both sides of the above equation with \(a=0 \) and \(b=1 \) yields (6).

By implementing (4) with (19 ), we get the following functional equation:

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=\left( \left( \frac{b-x}{b-a}\right) e^{-\left( \frac{ x-a}{b-a}\right) t}+\frac{k!}{\left( \frac{x-a}{b-a}\right) ^{k-1}t^{k}} f_{\beta }(t,x;k;a,b)\right) ^{n}. \end{aligned}$$
(20)

Using (20), we obtain the following functional equation:

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)= & {} \sum \limits _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \left( \frac{b-x}{b-a }\right) ^{n-j}e^{-(n-j)\left( \frac{x-a}{b-a}\right) t}\nonumber \\{} & {} \times \left( \frac{k!}{ \left( \frac{x-a}{b-a}\right) ^{k-1}t^{k}}\right) ^{j}\left( f_{\beta }(t,x;k,a,b)\right) ^{j}. \end{aligned}$$
(21)

By combining (21) with the following formula

$$\begin{aligned} \left( f_{\beta }(t,x;k,a,b)\right) ^{j}=\sum \limits _{r=0}^{\infty }\left( B_{k}^{r}(x;a,b)\right) ^{j}\frac{t^{r}}{r!} \end{aligned}$$

where

$$\begin{aligned} \left( B_{k}^{r}(x;a,b)\right) ^{j}= & {} \sum _{r_{1}+r_{2}+...+r_{j}=r}\left( {\begin{array}{c}r \\ r_{1},r_{2}-r_{1},r_{3}-r_{2},...,r-r_{j}\end{array}}\right) \\{} & {} B_{k}^{r_{1}}(x;a,b)B_{k}^{r_{2}-r_{1}}(x;a,b)...B_{k}^{r_{j}-r_{j-1}}(x;a,b)B_{k}^{r-r_{j}}(x;a,b), \\ \sum _{r_{1}+r_{2}+...+r_{j}=r}= & {} \sum _{r_{j}=0}^{r}\sum _{r_{j-1}=0}^{r_{j}}... \sum _{r_{2}=0}^{r_{3}}\sum _{r_{1}=0}^{r_{2}}, \end{aligned}$$

and

$$\begin{aligned} \left( {\begin{array}{c}r\\ r_{1},r_{2}-r_{1},r_{3}-r_{2},...,r-r_{j}\end{array}}\right) =\left( {\begin{array}{c}r\\ r_{j}\end{array}}\right) \left( {\begin{array}{c} r_{j}\\ r_{j-1}\end{array}}\right) ...\left( {\begin{array}{c}r_{2}\\ r_{1}\end{array}}\right) \end{aligned}$$

which were given by Simsek [17], we get

$$\begin{aligned}{} & {} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}\\ {}{} & {} \quad =\sum _{j=0}^{n}\left( \frac{b-x}{b-a}\right) ^{n-j}\sum _{r=0}^{\infty }(-1)^{r}\left( \frac{x-a}{ b-a}\right) ^{r}(n-j)^{r}\frac{t^{r-kj}}{r!}\left( \frac{k!}{\left( \frac{x-a }{b-a}\right) ^{k-1}}\right) ^{j}\\{} & {} \qquad \times \sum _{r=0}^{\infty }\left( B_{k}^{r}(x;a,b)\right) ^{j}\frac{t^{r}}{r!}. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}= & {} \sum _{j=0}^{n} \left( {\begin{array}{c}n\\ j\end{array}}\right) \left( \frac{b-x}{b-a}\right) ^{n-j}\left( \frac{k!}{\left( \frac{x-a}{b-a}\right) ^{k-1}}\right) ^{j} \\{} & {} \times \sum _{r=0}^{\infty }\sum _{v=0}^{r}\left( {\begin{array}{c}r\\ v\end{array}}\right) (-1)^{v}\left( \frac{ x-a}{b-a}\right) ^{v}(n-j)^{v}\left( B_{k}^{r-v}(x;a,b)\right) ^{j}\\{} & {} \times \frac{ t^{r-jk}}{r!}. \end{aligned}$$

After some calculations, we have

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n,a,b)\frac{t^{r}}{r!}= & {} \sum _{j=0}^{n} \left( {\begin{array}{c}n\\ j\end{array}}\right) \left( \frac{b-x}{b-a}\right) ^{n-j}\left( \frac{k!}{\left( \frac{x-a}{b-a}\right) ^{k-1}}\right) ^{j} \\{} & {} \times \sum _{r=0}^{\infty }\sum _{v=0}^{r+kj}\left( {\begin{array}{c}r+kj\\ v\end{array}}\right) (-1)^{v}\left( \frac{x-a}{b-a}\right) ^{v}(n-j)^{v} \\{} & {} \times \left( B_{k}^{r+kj-v}(x;a,b)\right) ^{j} \frac{t^{r}}{\left( {\begin{array}{c}r+kv\\ v\end{array}}\right) r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following theorem:

Theorem 5

Let \(n,k\in {\mathbb {N}}_{0}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} T(x;n;a,b)= & {} \sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) B_{0}^{n-j}(x;a,b)\left( \frac{x-a}{b-a} \right) ^{j(1-k)}(k!)^{j}\\{} & {} \times \sum _{v=0}^{r+kj}\frac{(-1)^{v}\left( {\begin{array}{c}r+kj\\ v\end{array}}\right) }{ \left( {\begin{array}{c}r+kv\\ v\end{array}}\right) }B_{v}^{v}(x;a,b)(n-j)^{v}\left( B_{k}^{r+kj-v}(x;a,b) \right) ^{j}. \end{aligned}$$

Theorem 6

Let \(k\in {\mathbb {N}}_{0}\) and \(n\in {\mathbb {N}}\). Then we have

$$\begin{aligned} \frac{\partial ^{k}}{\partial t^{k}}{\mathcal {A}} (x;t,n;a,b)|_{t=0}=T(x;r,n;a,b) \end{aligned}$$

or

$$\begin{aligned} \frac{\partial ^{k}}{\partial t^{k}}{\mathcal {A}}(x;t,n;a,b)|_{t=0}=n^{k}\sum \limits _{j=0}^{n}\left( \frac{j}{n}-\frac{x-a}{b-a}\right) ^{k}B_{j}^{n}(x;a,b). \end{aligned}$$

Proof

By applying the partial derivative operator \(\frac{\partial ^{k}}{\partial t^{k}}\) to both sides of the Eq. (16), we get

$$\begin{aligned} \frac{\partial ^{k}}{\partial t^{k}}{\mathcal {A}}(x;t,n;a,b)|_{t=0}=\frac{ \partial ^{k}}{\partial t^{k}}\left\{ \sum \limits _{r=0}^{\infty }T(x;r,n;a,b) \frac{t^{r}}{r!}\right\} |_{t=0}. \end{aligned}$$

After some elementary calculations on the right side of the above equation, we complete proof of theorem. \(\square \)

Theorem 7

Let \(r\in {\mathbb {N}}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} \frac{\partial }{\partial x}\left\{ T(x;r,n;a,b)\right\}= & {} n\sum \limits _{j=0}^{r}\sum \limits _{v=0}^{r-j-1}\left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}r-j\\ v\end{array}}\right) T(x;j,n;a,b) \\{} & {} \times \sum \limits _{s=1}^{v}\sum \limits _{k=0}^{s}(-1)^{k}\left( {\begin{array}{c}s\\ k\end{array}}\right) \left( \frac{x-a}{b-a}\right) ^{s}k^{v}-nrT(x;r-1,n;a,b). \end{aligned}$$

Proof

Taking derivative of (19) with respect to x, we get

$$\begin{aligned} \frac{\partial }{\partial x}{\mathcal {A}}(x;t,n;a,b)= & {} \frac{\partial }{ \partial x}\left( \frac{b-x}{b-a}+\frac{x-a}{b-a}e^{t}\right) ^{n}e^{-nt\left( \frac{x-a}{b-a}\right) } \nonumber \\= & {} \left( \frac{n(e^{t}-1)}{\frac{b-x}{b-a}+\frac{x-a}{b-a}e^{t}}-nt\right) {\mathcal {A}}(x;t,n;a,b). \end{aligned}$$
(22)

Joining (22) with (8) and (9), we get the following partial derivative equation:

$$\begin{aligned} \frac{\partial }{\partial x}\left\{ {\mathcal {A}}(x;t,n;a,b)\right\}= & {} n {\mathcal {A}}(x;t,n;a,b)\\{} & {} \times \left( \frac{b-a}{2(b-x)}\left( F_{{\mathcal {E}},P}\left( t,1;\frac{x-a}{b-x}\right) -F_{{\mathcal {E}},N}\left( t;\frac{x-a}{ b-x}\right) \right) -t\right) . \end{aligned}$$

By implementing the above equation with (16), (8) and (9) and using the Cauchy product rule in the related series, we obtain

$$\begin{aligned} \sum \limits _{r=0}^{\infty }\frac{\partial }{\partial x}\left\{ T(x;r,n;a,b)\right\} \frac{t^{r}}{r!}= & {} \frac{\left( b-a\right) n}{2(b-x)} \sum \limits _{r=0}^{\infty }\sum \limits _{j=0}^{r}\left( {\begin{array}{c}r\\ j\end{array}}\right) T(x;j,n;a,b) {\mathcal {E}}_{r-j}\left( 1;\frac{x-a}{b-x}\right) \frac{t^{r}}{r!} \\{} & {} -\frac{\left( b-a\right) n}{2(b-x)}\sum \limits _{r=0}^{\infty }\sum \limits _{j=0}^{r}\left( {\begin{array}{c}r\\ j\end{array}}\right) T(x;j,n;a,b){\mathcal {E}}_{r-j}\left( \frac{ x-a}{b-x}\right) \frac{t^{r}}{r!} \\{} & {} -n\sum \limits _{r=0}^{\infty }rT(x;r-1,n;a,b)\frac{t^{r}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get

$$\begin{aligned} \frac{\partial }{\partial x}\left\{ T(x;r,n;a,b)\right\}= & {} \frac{\left( b-a\right) n}{2(b-x)}\sum \limits _{j=0}^{r}\left( {\begin{array}{c}r\\ j\end{array}}\right) T(x;j,n;a,b)\\{} & {} \times \left( {\mathcal {E}}_{r-j}\left( 1;\frac{x-a}{b-x}\right) -{\mathcal {E}}_{r-j}\left( \frac{x-a}{b-x}\right) \right) \\{} & {} -nrT(x;r-1,n;a,b). \end{aligned}$$

After some elementary calculations, we obtain

$$\begin{aligned} \frac{\partial }{\partial x}\left\{ T(x;r,n;a,b)\right\}= & {} \frac{\left( b-a\right) n}{2(b-x)}\sum \limits _{j=0}^{r}\sum \limits _{v=0}^{r-j-1}\left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}r-j\\ v\end{array}}\right) T(x;j,n;a,b){\mathcal {E}}_{v}\left( \frac{x-a}{b-x}\right) \nonumber \\{} & {} -nrT(x;r-1,n;a,b). \end{aligned}$$
(23)

We now modify Theorem 11 in [4], we have

$$\begin{aligned} {\mathcal {E}}_{v}\left( \frac{x-a}{b-x}\right) =\frac{2\left( b-x\right) }{b-a} \sum \limits _{s=1}^{v}\sum \limits _{k=0}^{s}(-1)^{k}\left( {\begin{array}{c}s\\ k\end{array}}\right) \left( \frac{ x-a}{b-a}\right) ^{s}k^{v}. \end{aligned}$$

Combining the above equation with (23), we get the desired result. \(\square \)

Substituting \(a=0\) and \(b=1\) into (23), for \(T(x;r-1,n):=T(x;r-1,n;0,1),\) we also get the following corollary:

Corollary 8

Let \(r\in {\mathbb {N}}\) and \(0\le x\le 1\). Then we have

$$\begin{aligned} \frac{\partial }{\partial x}\left\{ T(x;r,n)\right\}= & {} \frac{n}{2(1-x)} \sum _{j=0}^{r}\sum _{v=0}^{r-j-1}\left( {\begin{array}{c}r-j\\ v\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) {\mathcal {E}}_{v}\\ {}{} & {} \times \left( \frac{x}{1-x} \right) T(x;j,n)-nrT(x;r-1,n). \end{aligned}$$

Taking derivative of (19) with respect to t, we get

$$\begin{aligned} \frac{\partial }{\partial t}{\mathcal {A}}(x;t,n;a,b)= & {} \frac{\partial }{ \partial t}\left\{ \frac{e^{-nt\left( \frac{x-a}{b-a}\right) }}{(b-a)^{n}} \left( b-x+(x-a)e^{t}\right) ^{n}\right\} \\= & {} \left[ \frac{n(x-a)e^{t}}{b-x+(x-a)e^{t}}-n\left( \frac{x-a}{b-a}\right) \right] {\mathcal {A}}(x;t,n;a,b). \end{aligned}$$

Therefore

$$\begin{aligned} \frac{\partial }{\partial t}\sum _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}= & {} \frac{n(x-a)}{2(b-a)}\sum _{r=0}^{\infty }{\mathcal {E}}_{r}\left( 1,\frac{x-a }{b-x}\right) \frac{t^{r}}{r!}\sum _{r=0}^{\infty }{T(x;r,n;a,b)}\frac{t^{r}}{ r!} \\{} & {} -n\frac{x-a}{b-a}\sum _{r=0}^{\infty }{T(x;r,n;a,b)}\frac{t^{r}}{r!}. \end{aligned}$$

Using Cauchy product rule in the above equation yields

$$\begin{aligned} \sum _{r=1}^{\infty }T(x;r,n;a,b)\frac{t^{r-1}}{(r-1)!}= & {} \frac{n(x-a)}{ 2(b-a)}\sum _{r=0}^{\infty }\sum _{l=0}^{r}\left( {\begin{array}{c}r\\ l\end{array}}\right) {\mathcal {E}}_{l}\left( 1, \frac{x-a}{b-x}\right) T(x;r-l,n;a,b)\frac{t^{r}}{r!} \\{} & {} -n\frac{x-a}{b-a}\sum _{r=0}^{\infty }{T(x;r,n;a,b)}\frac{t^{r}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following theorem:

Theorem 9

(Recurrance relation) Let \(r\in {\mathbb {N}}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} T(x;r+1,n;a,b)= & {} \frac{n(x-a)}{2(b-a)}\sum _{l=0}^{r}\left( {\begin{array}{c}r\\ l\end{array}}\right) {\mathcal {E}} _{l}\left( 1,\frac{x-a}{b-x}\right) T(x;r-l,n;a,b)\\{} & {} -n\frac{x-a}{b-a}T(x;r,n;a,b). \end{aligned}$$

Combining (19) and (13), we get the following functional equation:

$$\begin{aligned} {\mathcal {A}}(x;t,n;a,b)=2^{n}\left( \frac{b-x}{b-a}\right) ^{2n}F_{y_{4}}\left( t,n;-\frac{x-a}{b-a},\frac{b-x}{b-a},\frac{x-a}{b-x} \right) . \end{aligned}$$

By using the above functional equation, we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=2^{n}\left( \frac{b-x }{b-a}\right) ^{2n}\sum _{r=0}^{\infty }y_{4}\left( r,n;\frac{x-a}{b-x},- \frac{x-a}{b-a},\frac{b-x}{b-a}\right) \frac{t^{r}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following theorem:

Theorem 10

Let \(r\in {\mathbb {N}}\) and \(a\le x\le b\). Then we have

$$\begin{aligned} T(x;r,n;a,b)=2^{n}\left( \frac{b-x}{b-a}\right) ^{2n}y_{4}\left( r,n;\frac{ x-a}{b-x},-\frac{x-a}{b-a},\frac{b-x}{b-a}\right) . \end{aligned}$$

3 Higher-order derivative formulas for T(x; r, n; a, b)

In this section, we give some novel higher-order derivative formulas for the polynomials T(x; r, n; a, b) with the aid of the partial derivative equations of \(\frac{\partial ^{l}}{\partial x^{l}}\left\{ {\mathcal {A}} (x;t,n;a,b)\right\} \) and \(\frac{\partial ^{l}}{\partial x^{l}}\left\{ B_{k}^{n}\left( x;a,b\right) \right\} \).

By applying Leibnitz’s derivative formula to (16) with respect to x, we get the following higher-order partial derivative equation:

$$\begin{aligned} \frac{\partial ^{l}}{\partial x^{l}}\left\{ {\mathcal {A}}(x;t,n;a,b)\right\} =\sum _{k=0}^{n}\sum _{c=0}^{l}\left( {\begin{array}{c}l\\ c\end{array}}\right) \frac{\partial ^{c}}{\partial x^{c}} \left\{ B_{k}^{n}\left( x;a,b\right) \right\} \frac{\partial ^{l-c}}{ \partial x^{l-c}}\left\{ e^{tk-tn\left( \frac{x-a}{b-a}\right) }\right\} . \end{aligned}$$
(24)

By combining modified Theorem 3.8 in Simsek’s paper [15],

$$\begin{aligned} \frac{\partial ^{l}}{\partial x^{l}}\left\{ B_{k}^{n}\left( x;a,b\right) \right\} =\frac{n!}{(n-l)!}\sum _{j=0}^{l}(-1)^{l-j}\left( {\begin{array}{c}l\\ j\end{array}}\right) B_{k-j}^{n-l}(x;a,b) \end{aligned}$$

with (24), after some elementary calculations, we get

$$\begin{aligned} \sum _{r=0}^{\infty }\frac{\partial ^{l}}{\partial x^{l}}\left\{ {T(x;r,n;a,b) }\right\} \frac{t^{r}}{r!}= & {} \sum _{k=0}^{n}\sum _{c=0}^{l}\left( {\begin{array}{c}l\\ c\end{array}}\right) \frac{ n!}{(n-c)!} \\{} & {} \times \sum _{j=0}^{c}(-1)^{l-j}\left( {\begin{array}{c}c\\ j\end{array}}\right) \frac{n^{l-c}t^{l-c}}{\left( b-a\right) ^{l-c}}B_{k-j}^{n-c}(x)e^{t\left( k-n\left( \frac{x-a}{b-a} \right) \right) }. \end{aligned}$$

Therefore

$$\begin{aligned} \sum _{r=0}^{\infty }\frac{\partial ^{l}}{\partial x^{l}}\left\{ {T(x;r,n;a,b) }\right\} \frac{t^{r}}{r!}= & {} \sum _{k=0}^{n}\sum _{c=0}^{l}\left( {\begin{array}{c}l\\ c\end{array}}\right) \frac{ n!}{(n-c)!}\sum _{j=0}^{c}(-1)^{l-j}\left( {\begin{array}{c}c\\ j\end{array}}\right) B_{k-j}^{n-c}(x)\frac{ (n)^{l-c}}{\left( b-a\right) ^{l-c}} \\{} & {} \times \sum _{r=0}^{\infty }\frac{(k-n\left( \frac{x-a}{b-a}\right) )^{r}t^{r+l-c}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following theorem:

Theorem 11

Let \(l\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} \frac{\partial ^{l}}{\partial x^{l}}{T(x;r,n;a,b)}= & {} \sum _{k=0}^{n} \sum _{c=0}^{l}\sum _{j=0}^{c}(-1)^{l-j}\left( {\begin{array}{c}l\\ c\end{array}}\right) \left( {\begin{array}{c}n\\ c\end{array}}\right) \left( {\begin{array}{c}c\\ j\end{array}}\right) \frac{n^{l-c}c!r_{(l-c)}B_{k-j}^{n-c}(x)}{\left( b-a\right) ^{l-c}} \\{} & {} \times \left( k-n\left( \frac{x-a}{b-a}\right) \right) ^{r+c-l}. \end{aligned}$$

By taking \(l^{th}\) order derivative of (16) with respect to t, we get the following PDE:

$$\begin{aligned} \frac{\partial ^{l}}{\partial t^{l}}{\mathcal {A}}({x;t,n;a,b})= & {} \frac{ \partial ^{l}}{\partial t^{l}}\left\{ \sum \limits _{k=0}^{n}e^{t\left( k- \frac{n(x-a)}{b-a}\right) }B_{k}^{n}(x;a,b)\right\} \\= & {} \sum \limits _{k=0}^{n}\left( k-\frac{n(x-a)}{b-a}\right) ^{l}e^{t\left( k- \frac{n(x-a)}{b-a}\right) }B_{k}^{n}(x;a,b) \\= & {} \sum \limits _{j=0}^{l}\left( {\begin{array}{c}l\\ j\end{array}}\right) \left( \frac{n(x-a)}{b-a}\right) ^{l-j}\sum \limits _{k=0}^{n}k^{j}e^{t\left( k-\frac{n(x-a)}{b-a}\right) }B_{k}^{n}(x;a,b). \end{aligned}$$

By using the above equation, we get

$$\begin{aligned} \sum _{r=l}^{\infty }T(x;r,n;a,b)\frac{t^{r-l}}{\left( r-l\right) !}= & {} \sum \limits _{j=0}^{l}\left( {\begin{array}{c}l\\ j\end{array}}\right) \left( \frac{n(x-a)}{b-a}\right) ^{l-j}\sum \limits _{k=0}^{n}k^{j}B_{k}^{n}(x;a,b) \\{} & {} \times \sum _{r=0}^{\infty }\frac{\left( k-\frac{n(x-a)}{b-a}\right) ^{r}t^{r}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following theorem:

Theorem 12

Let \(l,r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} T(x;r+l,n;a,b)=\sum \limits _{j=0}^{l}\left( {\begin{array}{c}l\\ j\end{array}}\right) \left( \frac{n(x-a)}{b-a} \right) ^{l-j}\sum \limits _{k=0}^{n}k^{j}B_{k}^{n}(x;a,b)\left( k-\frac{n(x-a) }{b-a}\right) ^{r}. \end{aligned}$$

If we take \(l=1\), we get the following equation

$$\begin{aligned} \frac{\partial }{\partial t}\left\{ {\mathcal {A}}({x;t,n;a,b})\right\}= & {} \frac{n(a-x)}{b-a}\sum \limits _{k=0}^{n}e^{t\left( k-\frac{n(x-a)}{b-a} \right) }B_{k}^{n}(x;a,b) \\{} & {} +\sum \limits _{k=0}^{n}ke^{t\left( k-\frac{n(x-a)}{b-a}\right) }B_{k}^{n}(x;a,b). \end{aligned}$$

Thus we get

$$\begin{aligned} \frac{\partial }{\partial t}\left\{ {\mathcal {A}}({x;t,n;a,b})\right\}= & {} \frac{n(a-x)}{b-a}{\mathcal {A}}({x;t,n;a,b}) \\{} & {} +\sum \limits _{k=0}^{n}ke^{t\left( k-\frac{n(x-a)}{b-a}\right) }B_{k}^{n}(x;a,b). \end{aligned}$$

By using the above equation, we get

$$\begin{aligned} \sum _{r=0}^{\infty }T(x;r+1,n;a,b)\frac{t^{r}}{r!}= & {} -\frac{n(x-a)}{b-a} \sum _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!} \\{} & {} +\sum _{r=0}^{\infty }\sum \limits _{k=0}^{n}k\left( k-\frac{n(x-a)}{b-a} \right) ^{r}B_{k}^{n}(x;a,b)\frac{t^{r}}{r!}. \end{aligned}$$

Comparing coefficient \(\frac{t^{r}}{r!}\) on the both sides of the above equation, we get the following corollary:

Corollary 13

Let \(l,r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} T(x;r+1,n;a,b)= & {} -\frac{n(x-a)}{b-a}T(x;r,n;a,b)\nonumber \\ {}{} & {} +\sum \limits _{k=0}^{n}k\left( k-\frac{n(x-a)}{b-a}\right) ^{r}B_{k}^{n}(x;a,b). \end{aligned}$$
(25)

When \(a=0\) and \(b=1\), (25) reduces to following result:

Corollary 14

Let \(l,r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} T(x,r+1;n)=-nxT(x,r;n)+\sum \limits _{k=0}^{n}k(k-nx)^{r}B_{k}^{n}(x). \end{aligned}$$

4 Integral formulas and application Laplace transformation to the generating functions for T(x; r, n; a, b)

In this section, we give some novel integral formulas for the polynomials T(x; r, n; a, b) with the aid of the Euler gamma function and beta function. By using these formulas, we derive some finite sums. By applying the Laplace transform to the generating function for the polynomials T(x; r, n; a, b), we also give infinite series representation for these polynomials and the Bernstein basis functions.

$$\begin{aligned} \int _{a}^{b}T(x;r,n;a,b)dx=\int _{a}^{b}n^{r}\sum _{k=0}^{n}\left( \frac{k}{n}- \frac{x-a}{b-a}\right) ^{r}B_{k}^{n}(x;a,b)dx. \end{aligned}$$

Therefore

$$\begin{aligned} \int _{a}^{b}T(x;r,n;a,b)dx=\sum _{k=0}^{n}\sum _{j=0}^{r}(-1)^{r-j}\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \frac{k^{j}}{n^{j-r}}\frac{b-a}{(n+r-j+1)\left( {\begin{array}{c}n+r-j\\ k+r-j\end{array}}\right) }. \end{aligned}$$

After some calculations, we get

$$\begin{aligned} \int _{a}^{b}T(x;r,n;a,b)dx=\sum _{k=0}^{n}\sum _{j=0}^{r}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) (-1)^{r-j}(b-a)k^{j}}{n^{j-r}(n+r-j+1)\left( {\begin{array}{c}n+r-j\\ k+r-j\end{array}}\right) }. \end{aligned}$$
(26)

Other value of the integral (26) is also given by

$$\begin{aligned}{} & {} \int _{a}^{b}T(x;r,n;a,b)dx \\{} & {} \quad =\sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{2r-2j+n-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}k+r-j\\ w\end{array}}\right) a^{k+r-j-w}b^{n-k}}{n^{j-r}(b-a)^{r-j+n}} \\{} & {} \qquad \times \int _{a}^{b}x^{r+n-j-v}dx. \end{aligned}$$

Thus

$$\begin{aligned}{} & {} \int _{a}^{b}T(x;r,n;a,b)dx \nonumber \\{} & {} \quad =\sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{n-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}k+r-j\\ w\end{array}}\right) a^{k+r-j-w}b^{n-k} }{ (b-a)^{r-j+n}(r+n-j-v+1)n^{j-r}} \nonumber \\{} & {} \qquad \times \left( b^{r+n-j-v+1}-a^{r+n-j-v+1}\right) . \end{aligned}$$
(27)

Comparing (26) and (27) yields the following theorem:

Theorem 15

Let \(r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned}{} & {} \sum _{k=0}^{n}\sum _{j=0}^{r}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) (-1)^{r-j}(b-a)k^{j}}{n^{j-r}(n+r-j+1)\left( {\begin{array}{c}n+r-j\\ k+r-j\end{array}}\right) }\\{} & {} \quad = \sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{n-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}k+r-j\\ w\end{array}}\right) a^{k+r-j-w}b^{n-k} }{ (b-a)^{r-j+n}(r+n-j-v+1)n^{j-r}} \nonumber \\{} & {} \qquad \times \left( b^{r+n-j-v+1}-a^{r+n-j-v+1}\right) . \end{aligned}$$

By using the integrals of

$$\begin{aligned}{} & {} \int _{a}^{b}M(x;r,n;a,b)dx,\\{} & {} \int _{a}^{b}\left( \frac{x-a}{b-a}\right) ^{k}T(x;r,n;a,b)dx \end{aligned}$$

and

$$\begin{aligned} \int _{a}^{b}\left( \frac{x-a }{b-a}\right) ^{k}M(x;r,n;a,b)dx \end{aligned}$$

following corollary is obtained, respectively.

Corollary 16

Let \(r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned}{} & {} \sum _{k=0}^{n}\sum _{j=0}^{r}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) (-1)^{r-j}(b-a)k^{j}}{n^{j}(n+r-j+1)\left( {\begin{array}{c}n+r-j\\ k+r-j\end{array}}\right) } \\{} & {} \quad = \sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{n-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}k+r-j\\ w\end{array}}\right) a^{k+r-j-w}b^{n-k} }{ (b-a)^{r-j+n}(r+n-j-v+1)n^{j}} \nonumber \\{} & {} \qquad \times \left( b^{r+n-j-v+1}-a^{r+n-j-v+1}\right) , \\{} & {} \sum _{k=0}^{n}\sum _{j=0}^{r}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) (-1)^{r-j}(b-a)k^{j}}{n^{j-r}(k+n+r-j+1)\left( {\begin{array}{c}k+n+r-j\\ 2k+r-j\end{array}}\right) } \\{} & {} \quad = \sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{n+k-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}2k+r-j\\ w\end{array}}\right) a^{2k+r-j-w}b^{n-k} }{ (b-a)^{k+r-j+n}(k+r+n-j-v+1)n^{j-r}}\\{} & {} \qquad \times \left( b^{k+r+n-j-v+1}-a^{k+r+n-j-v+1}\right) . \end{aligned}$$

and

$$\begin{aligned}{} & {} \sum _{k=0}^{n}\sum _{j=0}^{r}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) (-1)^{r-j}(b-a)k^{j}}{n^{j}(k+n+r-j+1)\left( {\begin{array}{c}k+n+r-j\\ 2k+r-j\end{array}}\right) } \\{} & {} \quad = \sum _{k=0}^{n}\sum _{j=0}^{r}\sum _{v=0}^{n-k}\sum _{w=0}^{k+r-j}\frac{ (-1)^{2r-2j+n+k-v-w}k^{j}\left( {\begin{array}{c}n-v\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}r\\ j\end{array}}\right) \left( {\begin{array}{c}2k+r-j \\ w\end{array}}\right) a^{2k+r-j-w}b^{n-k} }{ (b-a)^{k+r-j+n}(k+r+n-j-v+1)n^{j}}\\{} & {} \qquad \times \left( b^{k+r+n-j-v+1}-a^{k+r+n-j-v+1}\right) . \end{aligned}$$

By applying the Laplace transformation to the generating function of \({\mathcal {A}}(x;t,n;a,b)\), we obtain the following novel formula involving T(x; r, n; a, b) and \(B_{v}^{n}(x;a,b)\):

Theorem 17

Let \(n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)=\left( b-a\right) \sum _{v=0}^{n} \frac{B_{v}^{n}(x;a,b)}{n\left( x-a\right) +\left( 1-v\right) \left( b-a\right) }. \end{aligned}$$
(28)

Proof

From the Eq. (18), we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}=e^{-tn\left( \frac{ x-a}{b-a}\right) }\sum _{v=0}^{n}\left( {\begin{array}{c}n\\ v\end{array}}\right) \left( \frac{b-x}{b-a}\right) ^{n-v}\left( \frac{x-a}{b-a}\right) ^{v}e^{tv}. \end{aligned}$$

Multiplying both sides of the above equation by \(e^{-t}\), we have

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!}e^{-t}=\sum _{v=0}^{n} \left( {\begin{array}{c}n\\ v\end{array}}\right) \left( \frac{b-x}{b-a}\right) ^{n-v}\left( \frac{x-a}{b-a} \right) ^{v}e^{-t+tv-tn\left( \frac{x-a}{b-a}\right) }. \end{aligned}$$

Integrate both sides of the above equation with respect to t from 0 to \( \infty \), we obtain

$$\begin{aligned} \int _{0}^{\infty }\sum \limits _{r=0}^{\infty }T(x;r,n;a,b)\frac{t^{r}}{r!} e^{-t}= & {} \sum _{v=0}^{n}\left( {\begin{array}{c}n\\ v\end{array}}\right) \left( \frac{b-x}{b-a}\right) ^{n-v}\left( \frac{x-a}{b-a}\right) ^{v}\nonumber \\ {}{} & {} \times \int _{0}^{\infty }e^{-t\left( n\left( \frac{x-a}{ b-a}\right) +1-v\right) }dt. \end{aligned}$$
(29)

In order to guarantee the convergence of the above integral, we assume that \( t>0\) and \(t\left( n\left( \frac{x-a}{b-a}\right) +1-v\right) >0\), \( t,x\in {\mathbb {R}}\) and \(n,v\in {\mathbb {N}}\). By using these appropriate the conditions, the following Laplace transform of the function \(f(t)=t^{r}\) give us

$$\begin{aligned} {\mathcal {L}}\left[ f\right] (t)=\int _{0}^{\infty }t^{r}e^{-t}=r! \end{aligned}$$

on the left side of equation (29), we get

$$\begin{aligned} \sum \limits _{r=0}^{\infty }T(x;r,n;a,b)=\sum _{v=0}^{n}\left( {\begin{array}{c}n\\ v\end{array}}\right) \frac{ \left( \frac{b-x}{b-a}\right) ^{n-v}\left( \frac{x-a}{b-a}\right) ^{v}}{ n\left( \frac{x-a}{b-a}\right) +1-v}\lim _{R\longrightarrow \infty }\int _{0}^{R}e^{-u}du. \end{aligned}$$

After some elementary calculations, we arrive the assertion of the theorem. \(\square \)

Substituting \(a=0\) and \(b=1\) into (28), we get the following result:

Corollary 18

Let \(r,n\in {\mathbb {N}}_{0}\). Then we have

$$\begin{aligned} \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \frac{x}{1-x}\right) ^{k}\sum \limits _{r=0}^{\infty }(k-nx)^{r}=\sum _{v=0}^{n}\left( {\begin{array}{c}n\\ v\end{array}}\right) \frac{ x^{v}}{\left( nx+1-v\right) \left( 1-x\right) ^{v}}. \end{aligned}$$

Applications of the Laplace transform and other integral applications to the generating functions for the certain family of special polynomials were also given in ([4, 5, 11,12,13,14, 16,17,18, 20, 21]; see also the references cited in each of these earlier works).

5 Conclusion

The content of this paper was constructing generating functions for central moments which is involving Bernstein basis functions. This work is mainly geared to give derivative formulas and a recurrence relation of central moments with the help of their generating functions. Many formulas and identities were derived from these generating functions. Furthermore, we have showed relations between combinatorial polynomials and central moments. Not only some derivative and integral formulas for the polynomials T(x; r, n; a, b), but also some finite sums were given. By applying the Laplace transform to the generating function for the polynomials T(x; r, n; a, b), some infinite series representation for the polynomials T(x; r, n; a, b) and the Bernstein basis functions were given.

As we mentioned in the introduction, in addition to all the results discussed in this article, our results have a high potential to be used in applied sciences. This may provide us with new projects on the research and application of the Bernstein basis functions with their generating functions.