1 Introduction
Let \(\lbrace x_{j}\rbrace_{j=0}^{n}\in [a,b] \) and \(\lbrace f_{j}\rbrace_{j=0}^{n} \), which may be samples of a function, say \(f(x) \), be given. The main aim of an interpolation problem is to find an appropriate model to approximate \(f(x) \) at any arbitrary point of \([a,b] \) other than \(x_{j} \). In other words, if \(\Psi(x;a_{0},\ldots,a_{n}) \) is a family of functions of a single variable x with \(n+1 \) free parameters \(\lbrace a_{j}\rbrace_{j=0}^{n} \), then the interpolation problem for Ψ consists of determining \(\lbrace a_{j}\rbrace_{j=0}^{n} \) so that, for \(n+1 \) given real or complex pairs of distinct numbers \(\lbrace (x_{j},f_{j}) \rbrace_{j=0}^{n} \), we have
$$ \Psi(x_{j};a_{0},\ldots,a_{n})=f_{j}. $$
For a polynomial type interpolation problem, various classical methods, such as Lagrange, Newton, and Hermite interpolations, are used. Lagrange’s interpolation as a classical method for approximating a continuous function \(f:[a,b]\to\mathbb{R} \) at \(n+1\) distinct nodes \(a\le x_{0}<\cdots<x_{n}\le b\) is applied in several branches of numerical analysis and approximation theory. It is expressed in the form [1, pp. 39–40]
for
where \(w_{n}(x)=\prod_{j=0}^{n}(x-x_{j})\) is the node polynomial and \(\ell_{j}^{(n)}(x) \) are the Lagrange polynomials.
$$ {\mathrm{P}}_{n}(f;x)=\sum_{j=0}^{n} f(x_{j})\ell_{j}^{(n)}(x) $$
$$ \ell_{j}^{(n)}(x)=\frac{w_{n}(x)}{(x-x_{j})w'_{n}(x_{j})}, $$
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Then \({\mathrm{P}}_{n}(f;x) \) is a unique element in the space of all polynomials of degree at most n, say \(\mathcal{P}_{n}\), which solves the interpolation problem
$$ {\mathrm{P}}_{n}(f;x_{j})=f(x_{j}),\quad j=0,1,2,\ldots,n. $$
For non-polynomial type interpolation problems, an interpolating function of the form
is usually considered [2], where \(\lbrace u_{j}(x) \rbrace_{j=0}^{n} \) is a set of linearly independent real-valued continuous functions on \([a,b] \) and \(\lbrace a_{j}\rbrace_{j=0}^{n} \) are determined by the initial conditions
The function \(\psi_{n}(x) \) exists and is unique in the space of \(\text{span}\lbrace u_{j} \rbrace_{j=0}^{n} \) for all \(f\in C[a,b] \) if and only if the matrix \(\lbrace u_{j}(x_{k}) \rbrace_{j,k=0}^{n} \) is nonsingular.
$$ \psi_{n}(x)=\sum_{j=0}^{n} a_{j}u_{j}(x) $$
$$\psi_{n}(x_{j})=f(x_{j}), \quad j=0,1,\ldots,n. $$
The general case of an interpolation problem was proposed by Davis [3] containing all the above-mentioned cases. It is indeed concerned with reconstructing functions on a basis of certain functional information, which are linear in many cases. Let Π be a linear space of dimension \(n+1 \) and \(L_{0},L_{1},\ldots, L_{n} \) be \(n+1 \) given linear functionals defined on Π, which are independent in \(\Pi^{*} \) (the algebraic conjugate space of Π). For a given set of values \(w_{0},w_{1},\ldots , w_{n} \), we can find an element \(f\in \Pi \) such that
$$ L_{j}(f)=w_{j},\quad j=0,1,\ldots, n. $$
Hence, one can construct new interpolation formulae using linear operators [4]. For instance, considering \(\Pi=\mathcal{P}_{n} \) and linear independent functionals as
leads to Taylor’s interpolation problem.
$$L_{j}(f)=f^{(j)}(x_{0}),\quad\text{for } j=0,1,\ldots,n , $$
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Davis also mentioned that the expansion of a function based on a series of predetermined (basis) functions can be interpreted as an interpolation problem with an infinite number of conditions. See also [5] in this regard.
The problem of the representation of an arbitrary function by means of linear combinations of prescribed functions has received a lot of attention in approximation theory. It is well known that a special case of this problem directly leads to the Taylor series expansion where the prescribed functions are monomial bases [6].
The main aim of this paper is to introduce a new type of Taylor series expansion through a variant of the classical integration by parts. In the next section, we present the general form of this expansion and consider some interesting cases of it leading to new closed forms for integrals involving Jacobi and Laguerre polynomials. Also, an error analysis is given in Sect. 3 for the introduced expansion.
2 A new type of Taylor series expansion
Let F and G be two smooth enough functions such that repeated differentiation and repeated integration by parts are allowed for them. The rule of integration by parts [7] allows one to perform successive integrations on the integrals of the form \(\int F(t) G(t) \,dt\) without tedious algebraic computations.
By the general rule
one obtains
where \(G_{n} \) denotes the nth antiderivative of G.
$$\int u \,dv=uv- \int v \,du, $$
$$\begin{aligned} \int_{a}^{b} F(t) G(t) \,dt =& \bigl(F(t) G_{1}(t)-F'(t) G_{2}(t) +\cdots + (-1)^{n-1}F^{(n-1)}(t) G_{n}(t) \bigr) \big\vert _{a}^{b} \\ &{}+ (-1)^{n} \int_{a}^{b}F^{(n)}(t) G_{n}(t) \,dt \\ =& \sum_{k=0}^{n-1}(-1)^{k} \bigl(F^{(k)}(t) G_{k+1}(t) \bigr)\big\vert _{a}^{b} + (-1)^{n} \int_{a}^{b}F^{(n)}(t) G_{n}(t) \,dt, \end{aligned}$$
(1)
Formula (1) provides a straightforward proof for Taylor’s theorem with an integral remainder term, according to the following result.
Theorem 2.1
Let
\(f\in C^{n+1}[a,b] \)
and
\(x_{0}\in [a,b] \). Then, for all
\(a\leq x \leq b \), we have
$$ f(x)=\sum_{k=0}^{n} \frac{1}{k!}(x-x_{0})^{k} f^{(k)}(x_{0}) + \frac{1}{n!} \int_{x_{0}}^{x}f^{(n+1)}(t) (x-t)^{n} \,dt. $$
(2)
Proof
For a given function f, assume in (1) that \(G(t)=f^{(n+1)}(t) \) and \(F(t)=p_{n}(x-t) \), where \(p_{n}(x)=\sum_{k=0}^{n}c_{k}x^{k} \) is an arbitrary polynomial of degree n. So, we have
which is equivalent to
and can be written as
$$\begin{aligned} & \int_{a}^{x}p_{n}(x-t) f^{(n+1)}(t) \,dt \\ &\quad = \sum_{k=0}^{n-1}(-1)^{k} \biggl( \frac{{\mathrm{d}}^{k}}{{\mathrm{d}}t^{k}}p_{n}(x-t)_{\vert_{t=x}} f^{(n-k)}(x)- \frac{{\mathrm{d}}^{k}}{{\mathrm{d}}t^{k}}p_{n}(x-t)_{\vert_{t=a}} f^{(n-k)}(a) \biggr) \\ &\qquad{} +(-1)^{n} \int_{a}^{x}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}}t^{n}}p_{n}(x-t) f'(t) \,dt \\ &\quad =\sum_{k=0}^{n-1}(-1)^{k} \biggl((-1)^{k}k! c_{k}f^{(n-k)}(x)- \frac{{\mathrm{d}}^{k}}{{\mathrm{d}}t^{k}}p_{n}(x-t)_{\vert_{t=a}} f^{(n-k)}(a) \biggr) \\ &\qquad{} +(-1)^{n} \int_{a}^{x}(-1)^{n}n! c_{n}f'(t) \,dt \\ &\quad =\sum_{k=0}^{n} \biggl( k! c_{k}f^{(n-k)}(x)-(-1)^{k}\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}t^{k}}p_{n}(x-t)_{\vert_{t=a}} f^{(n-k)}(a) \biggr) \\ &\quad =\sum_{k=0}^{n} \bigl( k! c_{k}f^{(n-k)}(x)-p_{n}^{(k)}(x-a) f^{(n-k)}(a) \bigr) , \end{aligned}$$
$$ \sum_{k=0}^{n}k! c_{k} f^{(n-k)}(x)=\sum_{k=0}^{n}p_{n}^{(k)}(x-a)f^{(n-k)}(a)+ \int_{a}^{x}p_{n}(x-t)f^{(n+1)}(t) \,dt, $$
(3)
$$ f(x)=\frac{1}{n! c_{n}} \Biggl( \sum_{k=0}^{n}p_{n}^{(k)}(x-a)f^{(n-k)}(a)- \sum_{k=0}^{n-1}k! c_{k} f^{(n-k)}(x)+ \int_{a}^{x}p_{n}(x-t)f^{(n+1)}(t) \,dt \Biggr). $$
Remark 1
If \(p_{n}(x-t)=\frac{1}{n!}(x-t)^{n} \), then \(c_{j}=0 \) for every \(j=0,1,\ldots,n-1 \) and \(c_{n}=\frac{1}{n!} \). In this case, (3) is reduced to
which is the same as formula (2).
$$ f(x)=\sum_{k=0}^{n}\frac{(x-a)^{n-k}}{(n-k)!}f^{(n-k)}(a)+ \frac{1}{n!} \int_{a}^{x}(x-t)^{n}f^{(n+1)}(t) \,dt, $$
Now, let us consider some particular examples of the main formula (3). We would like to notice here that the closed forms for the integrals involving Jacobi and Laguerre polynomials in the following examples are new in the literature (see, e.g., [8, 9]) to the best of our knowledge, and they can be computed only for specific values of the parameters by using symbolic computation.
Example 2.2
Let \(p_{n}(x)=P_{n}^{(\alpha,\beta)}(x) \) be the Jacobi polynomials [10]. It is known that, for \(\alpha,\beta>-1\),
satisfies the orthogonality relation
where
Moreover, they satisfy the important relation
Now, according to (3), we obtain
For instance, if \(f(x)=\frac{1}{1-x} \), then
and relation (4) for \(a=0 \) and \(x<1 \) reads as
$$ P_{n}^{(\alpha,\beta)}(x)=\sum_{k=0}^{n}C_{k}^{(\alpha,\beta,n)} x^{k}, \quad C_{k}^{(\alpha,\beta,n)}=\sum _{j=k}^{n}(-1)^{j-k} 2^{-j} \begin{pmatrix} n+\alpha+\beta +j\\ j \end{pmatrix} \begin{pmatrix} n+\alpha\\ n-j \end{pmatrix} \begin{pmatrix} j\\ k \end{pmatrix} $$
$$\int_{-1}^{1}(1-x)^{\alpha}(1+x)^{\beta}P_{m}^{(\alpha,\beta)}(x)P_{n}^{(\alpha,\beta)}(x) \,dx = \biggl( \int_{-1}^{1}(1-x)^{\alpha}(1+x)^{\beta} \bigl(P_{n}^{(\alpha,\beta)}(x) \bigr)^{2} \,dx \biggr) \delta_{m,n}, $$
$$\delta_{n,m}= \textstyle\begin{cases} 1, & m=n,\\ 0, & m\neq n. \end{cases} $$
$$ \frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}}P_{n}^{(\alpha ,\beta)}(x)= \frac{\Gamma(\alpha+\beta+n+1+k)}{2^{k} \Gamma(\alpha+\beta+n+1)}P_{n-k}^{(\alpha+k,\beta+k)}(x),\quad k\leq n. $$
$$\begin{aligned} &\sum_{k=0}^{n}k! C_{k}^{(\alpha , \beta , n)} f^{(n-k)}(x) \\ &\quad =\frac{1}{\Gamma(\alpha+\beta+n+1)}\sum_{k=0}^{n}(x-a)^{k} \sum_{j=0}^{n-k}\frac{1}{2^{j}}f^{(n-j)}(a) \Gamma(\alpha+\beta+n+1+j)C_{k}^{(\alpha+j,\beta+j,n-j)} \\ &\qquad {}+\sum_{k=0}^{n}C_{k}^{(\alpha ,\beta, n)} \int_{a}^{x}(x-t)^{k}f^{(n+1)}(t) \,dt. \end{aligned}$$
(4)
$$f^{(n-k)}(x)=(n-k)! (1-x)^{-(n-k+1)}, $$
$$\begin{aligned} & \int_{0}^{x}P_{n}^{(\alpha,\beta)}(t) (t-x+1)^{-(n+2)} \,dt\\ &\quad =\frac{(1-x)^{-(n+1)}}{n+1}\sum_{k=0}^{n}C_{k}^{(\alpha,\beta,n)} \begin{pmatrix} n\\ k \end{pmatrix}^{-1}(1-x)^{k} \\ &\qquad {}-\frac{1}{(n+1)! \Gamma(n+\alpha+\beta+1)}\sum_{k=0}^{n}x^{k} \sum_{j=0}^{n-k}2^{-j}(n-j)! \Gamma(n+\alpha+\beta+1+j) C_{k}^{(\alpha+j,\beta+j,n-j)}. \end{aligned}$$
Also if for example \(f(x)=e^{x} \), then for \(a=0 \) we obtain
$$\begin{aligned} e^{x}={}& \frac{1}{\Gamma(\alpha+\beta+n+1)\sum_{k=0}^{n}k! C_{k}^{(\alpha , \beta , n)}}\sum_{k=0}^{n} \sum_{j=0}^{n-k}\frac{1}{2^{j}}\Gamma( \alpha+\beta+n+1+j)C_{k}^{(\alpha+j,\beta+j,n-j)}x^{k} \\ &{} +\frac{1}{\sum_{k=0}^{n}k! C_{k}^{(\alpha , \beta , n)}}\sum_{k=0}^{n}C_{k}^{(\alpha ,\beta, n)} \int_{0}^{x}e^{t}(x-t)^{k} \,dt. \end{aligned}$$
Remark 2
In Example 2.2, choosing \(\alpha=\beta=-\frac{1}{2} \) gives the first kind of Chebyshev polynomials [10] as
with
This means that
Hence, replacing \(p_{n}(x)=T_{n}(x) \) in (3) gives
For instance, if \(f(x)=e^{x} \), then
and for \(x=a \) we obtain
$$ T_{n}(x)=\cos (n\arccos x)=\sum_{k=0}^{n}C_{k}^{(n)}x^{n}, $$
$$ C_{k}^{(n)}=2^{2n} \begin{pmatrix} 2n\\ n \end{pmatrix}^{-1} \sum_{j=k}^{n}(-1)^{j-k} 2^{-j} \begin{pmatrix} n +j-1\\ j \end{pmatrix} \begin{pmatrix} n-\frac{1}{2}\\ n-j \end{pmatrix} \begin{pmatrix} j\\ k \end{pmatrix}. $$
$$ T_{n}(x)=\cos (n\arccos x)=2^{2n}{\binom{2n}{n}}^{-1}P_{n}^{(-\frac{1}{2},-\frac{1}{2})}(x). $$
$$\begin{aligned} &\sum_{k=0}^{n}k! C_{k}^{(n)} f^{(n-k)}(x) \\ &\quad =\sum_{k=0}^{n}\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}}\cos \bigl(n\arccos (x-a) \bigr) f^{(n-k)}(a) + \int_{a}^{x}\cos \bigl(n\arccos(x-t) \bigr)f^{(n+1)}(t) \,dt \\ &\quad =\frac{2^{2n}}{(n-1)!}\sum_{k=0}^{n}(x-a)^{k} \sum_{j=0}^{n-k}2^{-j}(n+j-1)! C_{k}^{(j-\frac{1}{2},j-\frac{1}{2},n-j)}f^{(n-j)}(a) \\ & \qquad{}+ \int_{a}^{x}\cos \bigl(n\arccos(x-t) \bigr)f^{(n+1)}(t) \,dt,\quad n\geq1 . \end{aligned}$$
$$ \sum_{k=0}^{n}k! C_{k}^{(n)}=e^{a-x} \sum_{k=0}^{n}\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}}\cos \bigl(n \arccos (x-a) \bigr)+ \int_{0}^{x-a}e^{-t} \cos (n\arccos t ) \,dt, $$
$$\begin{aligned} \sum_{k=0}^{n}k! C_{k}^{(n)}&= \sum_{k=0}^{n}\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}}\cos \bigl(n \arccos (x) \bigr)_{\vert_{x=0}} \\ &=\frac{2^{2n}}{(n-1)!}\sum_{j=0}^{n}2^{-j}(n+j-1)! C_{0}^{(j-\frac{1}{2},j-\frac{1}{2},n-j)}. \end{aligned}$$
Another special case of Jacobi polynomials is the Legendre polynomials, which are directly derived from the definition of \(P_{n}^{(\alpha,\beta)}(x) \) for \(\alpha=\beta=0 \) and has the explicit representation [10]
Hence, according to (4), we obtain
in which
For instance, replacing \(f(x)=\cos x \) for \(a=0 \) in (5) gives
generating many new identities for different values of x.
$$ P_{n}(x)=\frac{1}{2^{n}}\sum_{l=0}^{[\frac{n}{2}]}(-1)^{l} \frac{(2n-2l)!}{l! (n-l)! (n-2l)!} x^{n-2l}. $$
$$\begin{aligned} \sum_{l=0}^{[\frac{n}{2}]}(n-2l)! s_{l}^{(n)}f^{(2l)}(x)={}& \frac{1}{n!}\sum_{k=0}^{n}(x-a)^{k} \sum_{j=0}^{n-k}2^{-j} f^{(n-j)}(a) (n+j)! C_{k}^{(j,j,n-j)} \\ &{} + \int_{a}^{x} P_{n}(x-t) f^{(n+1)}(t) \,dt, \end{aligned}$$
(5)
$$ s_{l}^{(n)}=(-1)^{l} \frac{(2n-2l)!}{2^{n} l! (n-l)! (n-2l)!}. $$
$$\begin{aligned} &\Biggl( \sum_{k=0}^{[\frac{n}{4}]}(n-4k)! s_{2k}^{(n)} - \sum_{k=0}^{[\frac{[\frac{n}{2}]-1}{2}]}(n-2-4k)! s_{2k+1}^{(n)} \Biggr) \cos x \\ &\quad =\frac{1}{n!}\sum_{k=0}^{n}x^{k} \sum_{j=0}^{n-k}2^{-j} (n+j)! \cos \biggl((n-j)\frac{\pi}{2} \biggr) C_{k}^{(j,j,n-j)}\\ &\qquad {}+ \int_{0}^{x} P_{n}(x-t) \cos \biggl(t+(n+1)\frac{\pi}{2} \biggr) \,dt, \end{aligned}$$
Example 2.3
For \(\alpha>-1\), let \(p_{n}(x)=L_{n}^{(\alpha)}(x) \) be the Laguerre polynomials [10] given by
It is known that
Hence, according to (3), we have
As a special case, assume that \(f(x)=xe^{x} \). Since \(\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}} (xe^{x})=(x+k)e^{x}\), we get
For instance, if \(x=1 \) and \(a=0 \) in (6), then
For \(\alpha=0\), the right-hand side of (7) can be expressed in terms of hypergeometric series and evaluations of Laguerre polynomials as
For \(\alpha \neq 0\), the right-hand side of (7) can be written as
where the latter sum can be expressed in terms of hypergeometric series as
$$ L_{n}^{(\alpha)}(x)=\sum_{k=0}^{n}C_{k}^{(\alpha,n)} x^{k}, \quad C_{k}^{(\alpha ,n)}=(-1)^{k} \frac{1}{k!} \begin{pmatrix} n+\alpha\\ n-k \end{pmatrix} . $$
$$ \frac{{\mathrm{d}}^{k}}{{\mathrm{d}}x^{k}}L_{n}^{(\alpha)}(x)=(-1)^{k} L_{n-k}^{(\alpha+k)}(x)\quad\text{for any }k\leq n. $$
$$\begin{aligned} \sum_{k=0}^{n}(-1)^{k} \begin{pmatrix} n+\alpha\\ n-k \end{pmatrix} f^{(n-k)}(x)= {}&\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}(x-a)^{k}\sum_{j=0}^{n-k}(-1)^{j} \begin{pmatrix} n+\alpha\\ n-k-j \end{pmatrix} f^{(n-j)}(a) \\ &{}+ \int_{a}^{x}L_{n}^{(\alpha)}(x-t) f^{(n+1)}(t) \,dt. \end{aligned}$$
$$\begin{aligned} &e^{x}\sum_{k=0}^{n}(-1)^{k} \binom{n+\alpha}{n-k}(x+n-k) \\ &\quad =e^{a} \sum_{k=0}^{n} \frac{(-1)^{k}}{k!}(x-a)^{k}\sum_{j=0}^{n-k}(-1)^{j} \begin{pmatrix} n+\alpha\\ n-k-j \end{pmatrix} (a+n-j) \\ &\qquad {}+\int_{a}^{x}(t+n+1) e^{t} L_{n}^{(\alpha)}(x-t) \,dt. \end{aligned}$$
(6)
$$\begin{aligned} &\int_{0}^{1}(n+2-t) e^{-t} L_{n}^{(\alpha)}(t) \,dt \\ &\quad =\sum_{k=0}^{n}(-1)^{k} \begin{pmatrix} n+\alpha\\ n-k \end{pmatrix} (n-k+1) -e^{-1}\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}\sum_{j=0}^{n-k}(-1)^{j} \begin{pmatrix} n+\alpha\\ n-k-j \end{pmatrix} (n-j) . \end{aligned}$$
(7)
$$n _{1}F_{0}(1-n;\ ;1)+ _{1}F_{0}(-n;\ ;1)+ \frac{n^{2} _{1}F_{1}(1-n;2;1)-L_{n-1}^{(0)}(1)}{e (n-1)}. $$
$$ \frac{ (\alpha n+\alpha +n^{2}+n-1 ) \Gamma (\alpha +n-1)}{\Gamma (\alpha ) \Gamma (n+1)} - e^{-1} \sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \frac{(n (\alpha +n)-k) \Gamma (\alpha +n-1)}{\Gamma (\alpha +k) \Gamma (-k+n+1)}, $$
$$\begin{aligned} &\sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \frac{(n (\alpha +n)-k) \Gamma (\alpha +n-1)}{\Gamma (\alpha +k) \Gamma (-k+n+1)} \\ &\quad = \frac{ _{1}F_{1}(1-n;\alpha +1;1) \Gamma (\alpha +n-1)}{\Gamma (\alpha +1) \Gamma (n)} +\frac{(\alpha +n) \Gamma (\alpha +n-1)}{\Gamma (\alpha ) \Gamma (n)} {}_{1}F_{1}(-n; \alpha ;1) . \end{aligned}$$
For more certain new, interesting, and useful integrals and expansion formulas involving the hypergeometric function and the Laguerre polynomials, see [11].
3 Error analysis
It is clear that relation (3) can be considered as an approximation. This means that the expression \(\sum_{k=0}^{n}k! c_{k} f^{(n-k)}(x)\) can be approximated by \(\sum_{k=0}^{n}p_{n}^{(k)}(x-a)f^{(n-k)}(a) \), which is indeed a polynomial of degree n. Hence, the exact remainder [3] of this approximation reads as
Now, if \(f\in C^{n+1}[a,b] \), a direct result for the corresponding error term is that
where \(M_{n}=\max_{a\leq t \leq x}\vert f^{(n+1)}(t)\vert \).
$$ E_{n}(x;f)= \int_{a}^{x}p_{n}(x-t)f^{(n+1)}(t) \,dt. $$
$$\begin{aligned} \bigl\vert E_{n}(x;f) \bigr\vert &= \biggl\vert \int_{a}^{x}p_{n}(x-t)f^{(n+1)}(t) \,dt \biggr\vert \leq \int_{a}^{x} \bigl\vert p_{n}(x-t)f^{(n+1)}(t) \bigr\vert \,dt \\ &\leq M_{n} \int_{a}^{x} \bigl\vert p_{n}(x-t) \bigr\vert \,dt, \end{aligned}$$
Moreover, if the polynomial \(p_{n}(\cdot) \) is nonnegative on \([0,x-a] \), e.g., when the coefficients \(\lbrace c_{k}\rbrace_{k=0}^{n} \) are all nonnegative, then we have
and therefore
For instance, let us consider the function \(f\in C^{(n+1)}[0,1] \) and choose the polynomial as
Then we obtain
$$\int_{a}^{x} \bigl\vert p_{n}(x-t) \bigr\vert \,dt= \int_{0}^{x-a} p_{n}(t) \,dt =\sum _{k=0}^{n}\frac{c_{k}}{k+1}(x-a)^{k+1} , $$
$$\bigl\vert E_{n}(x;f) \bigr\vert \leq M_{n} \sum _{k=0}^{n}\frac{c_{k}}{k+1}(x-a)^{k+1} . $$
$$p_{n}(x)=\sum_{k=0}^{n} \frac{1}{2^{k}} x^{k}. $$
$$\bigl\vert E_{n}(x;f) \bigr\vert \leq \max_{0\leq t \leq x} \bigl\vert f^{(n+1)}(t) \bigr\vert \sum_{k=0}^{n} \frac{1}{2^{k}(k+1)} x^{k+1}. $$
As another example, consider the polynomial \(p_{n}(x)=\sum_{k=0}^{n} \frac{(m)_{k}}{k!} x^{k} \) for \(0< m<1 \), where \((m)_{k}=\prod_{j=0}^{k-1}(m+j) \) is the Pochhammer symbol. If \(f\in C^{(n+1)}[0,1] \), then for any \(x\in [0,1] \) we obtain
Now if \(n\rightarrow\infty \), then we get
where \(_{2}F_{1}(m,1;2;x) \) denotes the Gauss hypergeometric function [12]. For instance, replacing \(f(x)=e^{x} \) in (8) yields
and the error bound for \(x=1 \) can be computed as
where we have used the Gauss formula [13, 14]
$$\bigl\vert E_{n}(x;f) \bigr\vert \leq \max_{0\leq t \leq x} \bigl\vert f^{(n+1)}(t) \bigr\vert \sum_{k=0}^{n} \frac{(m)_{k}}{k! (k+1)} x^{k+1} = \max_{0\leq t \leq x} \bigl\vert f^{(n+1)}(t) \bigr\vert x \sum_{k=0}^{n} \frac{(m)_{k} (1)_{k}}{(2)_{k}} \frac{x^{k}}{k!}. $$
$$ \bigl\vert E_{n}(x;f) \bigr\vert \leq \max _{0\leq t \leq x} \bigl\vert f^{(n+1)}(t) \bigr\vert x \sum _{k=0}^{\infty}\frac{(m)_{k} (1)_{k}}{(2)_{k}} \frac{x^{k}}{k!} = \max_{0\leq t \leq x} \bigl\vert f^{(n+1)}(t) \bigr\vert x _{2}F_{1}(m,1;2;x), $$
(8)
$$\bigl\vert E_{n}\bigl(x;e^{x}\bigr) \bigr\vert \leq x e^{x} _{2}F_{1}(m,1;2;x), $$
$$\bigl\vert E_{n}\bigl(1;e^{x}\bigr) \bigr\vert \leq e _{2}F_{1}(m,1;2;1)=\frac{\Gamma(1-m)}{\Gamma(2-m)} e, $$
$$_{2}F_{1}(a,b;c;1)=\frac{\Gamma(c) \Gamma(c-b-a)}{\Gamma(c-b) \Gamma(c-a)}. $$
Acknowledgements
The work of the first author has been supported by the Alexander von Humboldt Foundation under the grant number: Ref 3.4–IRN–1128637–GF-E. The third and fourth authors thank for the financial support from the Agencia Estatal de Innovación (AEI) of Spain under grant MTM2016–75140–P, co-financed by the European Community fund FEDER and Xunta de Galicia, grants GRC 2015–004 and R 2016/022.
Competing interests
The authors declare that they have no competing interests.
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