Hermite–Hadamard–Fejér-Type Inequalities and Weighted Three-Point Quadrature Formulae

Abstract

The goal of this paper is to derive Hermite–Hadamard–Fejér-type inequalities for higher-order convex functions and a general three-point integral formula involving harmonic sequences of polynomials and w-harmonic sequences of functions. In special cases, Hermite–Hadamard–Fejér-type estimates are derived for various classical quadrature formulae such as the Gauss–Legendre three-point quadrature formula and the Gauss–Chebyshev three-point quadrature formula of the first and of the second kind.

1. Introduction

The Hermite–Hadamard inequalities and their weighted versions, the so-called Hermite-Hadamard-Fejér inequalities, are the most well-known inequalities related to the integral mean of a convex function (see [1] (p. 138)).

Theorem 1

(The Hermite–Hadamard–Fejér inequalities). Let $h:[a,b]\to \mathbb{R}$ be a convex function. Then

$$h\left(\frac{a+b}{2}\right){\int }_a^{b}u\left(x\right)dx\le \underset{a}{\overset{b}{\int }}u\left(x\right)h\left(x\right)dx\le \left[\frac{1}{2}h\left(a\right)+\frac{1}{2}h\left(b\right)\right]{\int }_a^{b}u\left(x\right)dx,$$

(1)

where $u:[a,b]\to \mathbb{R}$ is nonnegative, integrable and symmetric about $\frac{a+b}{2}$. If h is a concave function, then the inequalities in (1) are reversed.

If $u\equiv 1$, then we are talking about the Hermite–Hadamard inequalities.

Hermite–Hadamard and Hermite–Hadamard–Fejér-type inequalities have many applications in mathematical analysis, numerical analysis, probability and related fields. Their generalizations, refinements and improvements have been an important topic of research (see [1,2,3,4,5,6,7,8,9,10,11,12,13], and the references listed therein). In the past few years, Hermite–Hadamard–Fejér-type inequalities for superquadratic functions [2], GA-convex functions [7], quasi-convex functions [11] and convex functions [13] have been largely investigated in the literature.

The importance and significance of our paper are reflected in the way in which we prove new Hermite–Hadamard–Fejér-type inequalities for higher-order convex functions and the general weighted three-point quadrature formula by using inequality (1), and a weighted version of the integral identity expressed by w-harmonic sequences of functions.

For this purpose, let us introduce the notations and terminology used in relation to w-harmonic sequences of functions (see [14]).

Let us consider a subdivision $\sigma =\{a=x_0<x_1<\cdots <x_m=b\}$ of the segment $[a,b]$, $m\in \mathbb{N}$. Let $w:[a,b]\to \mathbb{R}$ be an arbitrary integrable function. For each segment $[x_{j-1},x_j]$, $j=1,\dots ,m$, we define w-harmonic sequences of functions ${\left\{w_{jk}\right\}}_{k=1,\dots ,n}$ by:

$$\begin{array}{ccc}\hfill w_{j1}^{\prime }\left(t\right)& =& w\left(t\right),t\in [x_{j-1},x_j],\hfill \\ \hfill w_{jk}^{\prime }\left(t\right)& =& w_{j,k-1}\left(t\right),t\in [x_{j-1},x_j],k=2,3,\dots ,n.\hfill \end{array}$$

(2)

Further, the function $W_{n,w}$ is defined as follows:

$$W_{n,w}(t,\sigma )=\left\{\begin{array}{cc}{w}_{1n}\left(t\right),\hfill & t\in [a,x_1],\hfill \\ w_{2n}\left(t\right),\hfill & t\in (x_1,x_2],\hfill \\ .\hfill \\ .\hfill \\ .\hfill \\ w_{mn}\left(t\right),\hfill & t\in (x_{m-1},b].\hfill \end{array}\right.$$

(3)

The following theorem gives a general integral identity (see [14]).

Theorem 2.

Let $f:[a,b]\to \mathbb{R}$ be such that $f^{\left(n\right)}$ is piecewise continuous on $[a,b]$. Then, the following holds:

$$\begin{array}{ccc}\hfill \underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt& =& \sum _{k=1}^n{(-1)}^{k-1}\left[w_{mk}\left(b\right)f^{(k-1)}\left(b\right)\right.\hfill \\ & +& \left.\sum _{j=1}^{m-1}\left[w_{jk}\left(x_j\right)-w_{j+1,k}\left(x_j\right)\right]f^{(k-1)}\left(x_j\right)-w_{1k}\left(a\right)f^{(k-1)}\left(a\right)\right]\hfill \\ & +& {(-1)}^n\underset{a}{\overset{b}{\int }}{W}_{n,w}(t,\sigma )f^{\left(n\right)}\left(t\right)dt.\hfill \end{array}$$

(4)

In [15], the authors proved the following Fejér-type inequalities by using identity (4).

Theorem 3.

Let $f:[a,b]\to \mathbb{R}$ be $(n+2)$-convex on $[a,b]$ and $f^{\left(n\right)}$ piecewise continuous on $[a,b]$. Further, let us suppose that the function $W_{n,w}$, defined in (3), is nonnegative and symmetric about $\frac{a+b}{2}$ (i.e., $W_{n,w}(t,\sigma )=W_{n,w}(a+b-t,\sigma )$). Then

$$\begin{array}{ccc}& & U_n\left(\sigma \right)·f^{\left(n\right)}\left(\frac{a+b}{2}\right)\hfill \\ & & \le {(-1)}^n\left\{\underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt-\sum _{k=1}^n{(-1)}^{k-1}\left[w_{mk}\left(b\right)f^{(k-1)}\left(b\right)\right.\right.\hfill \\ & & \left.\left.+\sum _{j=1}^{m-1}\left[w_{jk}\left(x_j\right)-w_{j+1,k}\left(x_j\right)\right]f^{(k-1)}\left(x_j\right)-w_{1k}\left(a\right)f^{(k-1)}\left(a\right)\right]\right\}\hfill \\ & & \le U_n\left(\sigma \right)·\left[\frac{1}{2}{f}^{\left(n\right)}\left(a\right)+\frac{1}{2}{f}^{\left(n\right)}\left(b\right)\right],\hfill \end{array}$$

(5)

where

$$\begin{array}{cc}\hfill U_n\left(\sigma \right)& =\frac{{(-1)}^n}{n!}\underset{a}{\overset{b}{\int }}w\left(t\right)·t^{n}dt-{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^{k-1}}{(n-k+1)!}\hfill \\ \hfill & ·\left(w_{mk}\left(b\right)b^{n-k+1}+\sum _{j=1}^{m-1}\left(w_{jk}\left(x_j\right)-w_{j+1,k}\left(x_j\right)\right)x_j^{n-k+1}-w_{1k}\left(a\right)a^{n-k+1}\right).\hfill \end{array}$$

(6)

If $W_{n,w}(t,\sigma )\le 0$ or f is an $(n+2)$-concave function on $[a,b]$, then the inequalities in (5) hold with reversed inequality signs.

Further, let us recall the definition of the divided difference and the definition of an n-convex function (see [1] (p. 15)).

Definition 1.

Let f be a real-valued function defined on the segment $[a,b]$. The divided difference of order n of the function f at distinct points $x_0,\dots ,x_n\in [a,b]$ is defined recursively by

$$f\left[x_i\right]=f\left(x_i\right),(i=0,\dots ,n)$$

and

$$f[x_0,\dots ,x_n]=\frac{f[x_1,\dots ,x_n]-f[x_0,\dots ,x_{n-1}]}{x_n-x_0}.$$

The value $f[x_0,\dots ,x_n]$ is independent of the order of points $x_0,\dots ,x_n$.

Definition 2.

A function $f:[a,b]\to \mathbb{R}$ is said to be n-convex on $[a,b]$, $n\ge 0$, if, for all choices of $(n+1)$ distinct points $x_0,\dots ,x_n\in [a,b]$, the n-th order divided difference in f satisfies

$$f[x_0,\dots ,x_n]\ge 0.$$

From the previous definitions, the following property holds: if f is an $(n+2)$-convex function, then there exists the n-th order derivative $f^{\left(n\right)}$, which is a convex function (see, e.g., [1] (pp. 16, 293)).

The paper is organized as follows. After this introduction, in Section 2, we establish Hermite–Hadamard–Fejér-type inequalities for weighted three-point quadrature formulae by using the integral identity with w-harmonic sequences of functions, the properties of harmonic sequences of polynomials and the properties of $n$-convex functions. Since we deal with three-point quadrature formulae that contain values of the function in nodes x, $\frac{a+b}{2}$ and $a+b-x$ and values of higher-ordered derivatives in inner nodes, the level of exactness of these quadrature formulae is retained. In Section 3, we derive Hermite–Hadamard–Fejér-type estimates for a generalization of the Gauss–Legendre three-point quadrature formula, and a generalization of the Gauss–Chebyshev three-point quadrature formula of the first and of the second kind.

Throughout the paper, the symbol B denotes the beta function defined by

$$B(x,y)=\underset{0}{\overset{1}{\int }}{s}^{x-1}{(1-s)}^{y-1}ds,$$

$\mathsf{\Gamma }$ denotes the gamma function defined as:

$$\mathsf{\Gamma }\left(x\right)=2\underset{0}{\overset{\infty }{\int }}{s}^{2x-1}{e}^{-s^2}ds,$$

and

$$F\left(\alpha ,\beta ,\gamma ;z\right)=\frac{1}{B(\beta ,\gamma -\beta )}\underset{0}{\overset{1}{\int }}{t}^{\beta -1}{(1-t)}^{\gamma -\beta -1}{(1-zt)}^{-\alpha }dt$$

is a hypergeometric function with $\gamma >\beta >0$, $z<1$.

In the paper, we assume that all considered integrals exist and that they are finite.

2. Hermite–Hadamard–Fejér-Type Inequalities for Three-Point Quadrature Formulae

In this section, we establish Hermite–Hadamard–Fejér-type inequalities for the weighted three-point formula using a weighted version of the integral identity expressed by w-harmonic sequences of functions that are given in Theorem 2 and the method that originated in [15].

In [16] (p. 54), the authors proved the following theorem.

Theorem 4.

Let $w:[a,b]\to \mathbb{R}$ be an integrable function, $x\in [a,\frac{a+b}{2})$, and let ${\left\{L_{j,x}\right\}}_{j=0,1,\dots ,n}$, $n\in \mathbb{N}$, be a sequence of harmonic polynomials such that $deg{L}_{j,x}\le j-1$ and $L_{0,x}\equiv 0$. Further, let us suppose that ${\left\{w_{jk}\right\}}_{k=1,..,n}$ are w-harmonic sequences of functions on $[x_{j-1},x_j]$, for $j=1,2,3,4,$ defined by the following relations:

$$w_{1k}\left(t\right)=\frac{1}{(k-1)!}\underset{a}{\overset{t}{\int }}{\left(t-s\right)}^{k-1}w\left(s\right)ds,t\in [a,x],$$

$$w_{2k}\left(t\right)=\frac{1}{(k-1)!}\underset{x}{\overset{t}{\int }}{\left(t-s\right)}^{k-1}w\left(s\right)ds+L_{k,x}\left(t\right),t\in \left(x,\frac{a+b}{2}\right],$$

$$w_{3k}\left(t\right)=-\frac{1}{(k-1)!}\underset{t}{\overset{a+b-x}{\int }}{\left(t-s\right)}^{k-1}w\left(s\right)ds+{(-1)}^k{L}_{k,x}(a+b-t),t\in \left(\frac{a+b}{2},a+b-x\right],$$

$$w_{4k}\left(t\right)=-\frac{1}{(k-1)!}\underset{t}{\overset{b}{\int }}{\left(t-s\right)}^{k-1}w\left(s\right)ds,t\in (a+b-x,b].$$

If $f:[a,b]\to \mathbb{R}$ is such that $f^{\left(n\right)}$ is piecewise continuous on $\left[a,b\right]$, then we have

$$\begin{array}{ccc}\hfill \underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt& =& \sum _{k=1}^n{A}_k\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(a+b-x)\right)\hfill \\ & +& \sum _{k=1}^n{B}_k\left(x\right)f^{(k-1)}\left(\frac{a+b}{2}\right)+{(-1)}^n\underset{a}{\overset{b}{\int }}{W}_{n,w}(t,x)f^{\left(n\right)}\left(t\right)dt,\hfill \end{array}$$

(7)

where

$$A_k\left(x\right)={(-1)}^{k-1}\left[\frac{1}{(k-1)!}\underset{a}{\overset{x}{\int }}{\left(x-s\right)}^{k-1}w\left(s\right)ds-L_{k,x}\left(x\right)\right],k\ge 1,$$

(8)

$$B_k\left(x\right)=2\left[\frac{1}{(k-1)!}\underset{x}{\overset{\frac{a+b}{2}}{\int }}{\left(\frac{a+b}{2}-s\right)}^{k-1}w\left(s\right)ds+L_{k,x}\left(\frac{a+b}{2}\right)\right],for oddk\ge 1,$$

(9)

and

$$B_k\left(x\right)=0,for evenk\ge 1,$$

such that

$$W_{n,w}(t,x)=\left\{\begin{array}{cc}{w}_{1n}\left(t\right),\hfill & t\in [a,x],\hfill \\ w_{2n}\left(t\right),\hfill & t\in \left(x,\frac{a+b}{2}\right],\hfill \\ w_{3n}\left(t\right),\hfill & t\in \left(\frac{a+b}{2},a+b-x\right],\hfill \\ w_{4n}\left(t\right),\hfill & t\in (a+b-x,b].\hfill \end{array}\right.$$

(10)

Remark 1.

If we assume $w\left(t\right)=w(a+b-t)$, for each $t\in \left[a,b\right]$, then the following symmetry conditions hold for $k=1,\dots ,n$:

$$w_{1k}\left(t\right)={(-1)}^k{w}_{4k}(a+b-t),fort\in [a,x],$$

and

$$w_{2k}\left(t\right)={(-1)}^k{w}_{3k}(a+b-t),fort\in \left(x,\frac{a+b}{2}\right].$$

Using Theorems 1 and 4, the properties of both n-convex functions and w-harmonic sequences of functions, and the method that originated in [15], in the next theorem, we derive new Hermite–Hadamard–Fejér-type inequalities for the weighted three-point quadrature Formula (7).

Theorem 5.

Let $w:[a,b]\to \mathbb{R}$ be an integrable function such that $w\left(t\right)=w(a+b-t)$, for each $t\in \left[a,b\right]$ and $x\in [a,\frac{a+b}{2})$. Let the function $W_{2n,w}$, defined by (10), be nonnegative. If $f:[a,b]\to \mathbb{R}$ is $(2n+2)$-convex on $\left[a,b\right]$ and $f^{\left(2n\right)}$ is piecewise continuous on $\left[a,b\right]$, then

$$\begin{array}{ccc}& & U_{n,w}\left(x\right)·f^{\left(2n\right)}\left(\frac{a+b}{2}\right)\hfill \\ & & \le \underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt-\sum _{k=1}^{2n}{A}_k\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(a+b-x)\right)\hfill \\ & & -\sum _{k=1,kodd}^{2n}{B}_k\left(x\right)f^{(k-1)}\left(\frac{a+b}{2}\right)\le U_{n,w}\left(x\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}\left(a\right)+\frac{1}{2}{f}^{\left(2n\right)}\left(b\right)\right],\hfill \end{array}$$

(11)

where

$$\begin{array}{ccc}\hfill U_{n,w}\left(x\right)& =& \frac{1}{\left(2n\right)!}\underset{a}{\overset{b}{\int }}w\left(t\right)·t^{2n}dt\hfill \\ & -& \sum _{k=1}^{2n}{A}_k\left(x\right)\frac{x^{2n-k+1}+{(-1)}^{k-1}{(a+b-x)}^{2n-k+1}}{(2n-k+1)!}\hfill \\ & -& \sum _{k=1,kodd}^{2n}{B}_k\left(x\right)\frac{{(a+b)}^{2n-k+1}}{2^{2n-k+1}(2n-k+1)!},\hfill \end{array}$$

(12)

and $A_k$ and $B_k$ are defined as in Theorem 4. If ${\displaystyle W_{2n,w}(t,x)\le 0}$ or f is a $(2n+2)$-concave function, then inequalities (11) hold with reversed inequality signs.

Proof. 

Let us observe that the function f is $(2n+2)$-convex. Hence, $f^{\left(2n\right)}$ is a convex function. It follows from Remark 1 that the function $W_{2n,w}$ is symmetric about $\frac{a+b}{2}$, i.e., $W_{2n,w}(t,x)=W_{2n,w}(a+b-t,x)$. Thus, inequalities (11) follow directly from Theorem 1, replacing a nonnegative and symmetric function u by a nonnegative and symmetric function $W_{2n,w}$, and a convex function h by a convex function $f^{\left(2n\right)}$, and then using identity (7) in $\underset{a}{\overset{b}{\int }}{W}_{2n,w}(t,x)f^{\left(2n\right)}\left(t\right)dt$.

Identity (7) yields $U_{n,w}\left(x\right)$ by substituting n with $2n$ and putting $f\left(t\right)=\frac{t^{2n}}{\left(2n\right)!}$. Then, $f^{\left(2n\right)}\left(t\right)=1$ and $f^{(k-1)}\left(t\right)=\frac{1}{(2n-k+1)!}·t^{2n-k+1}$. On the other hand, if $W_{2n,w}\left(t,x\right)$ is nonpositive, then $-W_{2n,w}\left(t,x\right)$ is nonnegative, from where there follow reversed signs in (11).

Further, let us assume that f is a $(2n+2)$-concave function. Hence, the function $-f^{\left(2n\right)}$ is convex. Reversed signs in (11) are obtained by putting $-f^{\left(2n\right)}$ and the nonnegative function $W_{2n,w}\left(t,x\right)$ in (1). This completes the proof.  □

Remark 2.

The value of $U_{n,w}\left(x\right)$ can be obtained from Theorem 3 by taking an appropriate subdivision of the segment $\left[a,b\right]$ and applying the properties of functions $w_{1k},w_{2k},w_{3k}$ and $w_{4k}$.

To get a maximum degree of exactness of quadrature Formula (7) for fixed $x\in \left[a,\frac{a+b}{2}\right)$, we consider a sequence of harmonic polynomials ${\left\{L_{j,x}\right\}}_{j=0,1,\dots ,n}$ defined as follows:

$$\begin{array}{ccc}\hfill L_{0,x}\left(t\right)& =& 0,fort\in \left[x,\frac{a+b}{2}\right],\hfill \\ \hfill L_{1,x}\left(x\right)& =& \underset{a}{\overset{x}{\int }}w\left(s\right)ds-\frac{2}{{(a+b-2x)}^2}\underset{a}{\overset{b}{\int }}\left(s^2-{\left(\frac{a+b}{2}\right)}^2\right)w\left(s\right)ds,\hfill \\ \hfill L_{j,x}\left(x\right)& =& \frac{1}{(j-1)!}\underset{a}{\overset{x}{\int }}{\left(x-s\right)}^{j-1}w\left(s\right)ds,j=2,3,4,5,6,\hfill \\ \hfill L_{j,x}\left(t\right)& =& \sum _{k=1}^{6\wedge j}{L}_{k,x}\left(x\right)\frac{{(t-x)}^{j-k}}{(j-k)!},\mathrm{for}t\in \left[x,\frac{a+b}{2}\right],j=1,\dots ,n.\hfill \end{array}$$

(13)

Therefore, we have

$$A_1\left(x\right)=\frac{2}{{(a+b-2x)}^2}\underset{a}{\overset{b}{\int }}\left(s^2-{\left(\frac{a+b}{2}\right)}^2\right)w\left(s\right)ds,$$

(14)

$$B_1\left(x\right)=\underset{a}{\overset{b}{\int }}w\left(s\right)ds-2A_1\left(x\right),$$

$A_k\left(x\right)=0,$ for $k=2,3,4,5,6$ and $B_k\left(x\right)=0,$ for $k=2,3,4$.

Finally, from identity (7), for $x\in \left[a,\frac{a+b}{2}\right)$, we obtain the following three-point weighted integral formula:

$$\begin{array}{ccc}\hfill \underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt& =& A_1\left(x\right)\left[f\left(x\right)+f(a+b-x)\right]+\left(\underset{a}{\overset{b}{\int }}w\left(s\right)ds-2A_1\left(x\right)\right)f\left(\frac{a+b}{2}\right)\hfill \\ & +& T_{n,w}\left(x\right)+{(-1)}^n\underset{a}{\overset{b}{\int }}{W}_{n,w}(t,x)f^{\left(n\right)}\left(t\right)dt,\hfill \end{array}$$

(15)

where

$$\begin{array}{ccc}\hfill T_{n,w}\left(x\right)& =& \sum _{k=7}^n{A}_k\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(a+b-x)\right)\hfill \\ & +& \sum _{k=5,oddk}^n{B}_k\left(x\right)f^{(k-1)}\left(\frac{a+b}{2}\right).\hfill \end{array}$$

(16)

Now, applying results from Theorem 5 to identity (15), we get the following results.

Corollary 1.

Let $w:[a,b]\to \mathbb{R}$ be an integrable function such that $w\left(t\right)=w(a+b-t)$, for each $t\in \left[a,b\right]$ and let $x\in [a,\frac{a+b}{2})$. Let the function $W_{2n,w}$, defined by (10), be nonnegative and let $L_{j,x}$ be defined by (13). If $f:[a,b]\to \mathbb{R}$ is $(2n+2)$-convex on $\left[a,b\right]$ and $f^{\left(2n\right)}$ is piecewise continuous on $\left[a,b\right]$, then

$$\begin{array}{ccc}& & U_{n,w}\left(x\right)·f^{\left(2n\right)}\left(\frac{a+b}{2}\right)\hfill \\ & & \le \underset{a}{\overset{b}{\int }}w\left(t\right)f\left(t\right)dt-A_1\left(x\right)\left[f\left(x\right)+f(a+b-x)\right]\hfill \\ & & -\left(\underset{a}{\overset{b}{\int }}w\left(s\right)ds-2A_1\left(x\right)\right)f\left(\frac{a+b}{2}\right)-T_{2n,w}\left(x\right)\hfill \\ & & \le U_{n,w}\left(x\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}\left(a\right)+\frac{1}{2}{f}^{\left(2n\right)}\left(b\right)\right],\hfill \end{array}$$

(17)

where

$$\begin{array}{ccc}\hfill U_{n,w}\left(x\right)& =& \frac{1}{\left(2n\right)!}\underset{a}{\overset{b}{\int }}w\left(t\right)·t^{2n}dt-A_1\left(x\right)\frac{x^{2n}+{(a+b-x)}^{2n}}{\left(2n\right)!}\hfill \\ & -& \left(\underset{a}{\overset{b}{\int }}w\left(s\right)ds-2A_1\left(x\right)\right)\frac{{(a+b)}^{2n}}{2^{2n}\left(2n\right)!}\hfill \\ & -& \sum _{k=7}^{2n}{A}_k\left(x\right)\frac{x^{2n-k+1}+{(-1)}^{k-1}{(a+b-x)}^{2n-k+1}}{(2n-k+1)!}\hfill \\ & -& \sum _{k=5,kodd}^{2n}{B}_k\left(x\right)\frac{{(a+b)}^{2n-k+1}}{2^{2n-k+1}(2n-k+1)!}.\hfill \end{array}$$

(18)

If ${\displaystyle W_{2n,w}(t,x)\le 0}$ or f is a $(2n+2)$-concave function, then inequalities (17) hold with reversed inequality signs.

Proof. 

The proof follows from Theorem 5 for the special choice of the polynomials $L_{j,x}$.  □

Remark 3.

If we assume $B_5\left(x\right)=0$, then we get

$$x=\frac{a+b}{2}-\sqrt{\frac{\underset{a}{\overset{b}{\int }}{\left(s-\frac{a+b}{2}\right)}^{4}w\left(s\right)ds}{\underset{a}{\overset{b}{\int }}\left(s^2-{\left(\frac{a+b}{2}\right)}^2\right)w\left(s\right)ds}}.$$

Therefore, for such a choice of x, we obtain the quadrature formula with three nodes, which is accurate for the polynomials of degree at most 5, and the approximation formula includes derivatives of order 6 and more.

3. Special Cases

Considering some special cases of the weight function w, in our results given in the previous section, we obtain estimates for the Gauss–Legendre three-point quadrature formula and for the Gauss–Chebyshev three-point quadrature formula of the first and of the second kind.

3.1. Gauss–Legendre Three-Point Quadrature Formula

Let us assume that $w\left(t\right)=1$, $t\in [a,b]$ and $x\in \left[a,\frac{a+b}{2}\right)$.

Now, from Theorem 4, we calculate

$$W_n^{GL}(t,x)=\left\{\begin{array}{cc}{w}_{1n}\left(t\right)=\frac{{(t-a)}^n}{n!},\hfill & t\in [a,x],\hfill \\ w_{2n}\left(t\right)=\frac{{(t-x)}^n}{n!}+L_{n,x}\left(t\right),\hfill & t\in \left(x,\frac{a+b}{2}\right],\hfill \\ w_{3n}\left(t\right)=\frac{{(t-a-b+x)}^n}{n!}+{(-1)}^n{L}_{n,x}(a+b-t),\hfill & t\in \left(\frac{a+b}{2},a+b-x\right],\hfill \\ w_{4n}\left(t\right)=\frac{{(t-b)}^n}{n!},\hfill & t\in (a+b-x,b],\hfill \end{array}\right.$$

(19)

and

$$A_k^{GL}\left(x\right)={(-1)}^{k-1}\left[\frac{{\left(x-a\right)}^k}{k!}-L_{k,x}\left(x\right)\right],\mathrm{for}k\ge 1,$$

$$B_k^{GL}\left(x\right)=2\left[\frac{{\left(\frac{a+b}{2}-x\right)}^k}{k!}+L_{k,x}\left(\frac{a+b}{2}\right)\right],\mathrm{for} \mathrm{odd}k\ge 1,$$

and

$$B_k^{GL}\left(x\right)=0,\mathrm{for} \mathrm{even}k>1.$$

Corollary 2.

Let $w_{2,2n}\left(t\right)\ge 0$, for all $t\in \left(x,\frac{a+b}{2}\right]$ and for $n\in \mathbb{N}$. If $f:[a,b]\to \mathbb{R}$ is a $(2n+2)$-convex function and $f^{\left(2n\right)}$ is piecewise continuous on $\left[a,b\right]$, then

$$\begin{array}{ccc}\hfill & & U_n^{GL}\left(x\right)·f^{\left(2n\right)}\left(\frac{a+b}{2}\right)\hfill \\ & & \le \underset{a}{\overset{b}{\int }}f\left(t\right)dt-\sum _{k=1}^{2n}{A}_k^{GL}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(a+b-x)\right)\hfill \\ & & -\sum _{k=1,kodd}^{2n}{B}_k^{GL}\left(x\right)f^{(k-1)}\left(\frac{a+b}{2}\right)\le U_n^{GL}\left(x\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}\left(a\right)+\frac{1}{2}{f}^{\left(2n\right)}\left(b\right)\right],\hfill \end{array}$$

(20)

where

$$\begin{array}{ccc}\hfill U_n^{GL}\left(x\right)& =& \frac{b^{2n+1}-a^{2n+1}}{(2n+1)!}\hfill \\ & -& \sum _{k=1}^{2n}{A}_k^{GL}\left(x\right)\frac{x^{2n-k+1}+{(-1)}^{k-1}{(a+b-x)}^{2n-k+1}}{(2n-k+1)!}\hfill \\ & -& \sum _{k=1,kodd}^{2n}{B}_k^{GL}\left(x\right)\frac{{(a+b)}^{2n-k+1}}{2^{2n-k+1}(2n-k+1)!}.\hfill \end{array}$$

(21)

If f is a $(2n+2)$-concave function, then inequalities (20) hold with reversed inequality signs.

Proof. 

A special case of Theorem 5 for $w\left(t\right)=1$, $t\in \left[a,b\right]$, and a nonnegative function $W_{2n}^{GL}$ defined by (19).  □

If we assume that the polynomials $L_{j,x}\left(t\right)$ are such that

$$\begin{array}{ccc}\hfill L_{0,x}\left(t\right)& =& 0,\mathrm{for}t\in \left[x,\frac{a+b}{2}\right],\hfill \\ \hfill L_{1,x}\left(x\right)& =& x-a-\frac{{(b-a)}^3}{6{(a+b-2x)}^2},\hfill \\ \hfill L_{j,x}\left(x\right)& =& \frac{{(x-a)}^j}{j!},j=2,3,4,5,6,\hfill \\ \hfill L_{j,x}\left(t\right)& =& \sum _{k=1}^{6\wedge j}{L}_{k,x}\left(x\right)\frac{{(t-x)}^{j-k}}{(j-k)!},\mathrm{for}t\in \left[x,\frac{a+b}{2}\right],j=1,\dots ,n,\hfill \end{array}$$

(22)

we get $A_1^{GL}\left(x\right)=\frac{{(b-a)}^3}{6{(a+b-2x)}^2}$, $A_k^{GL}\left(x\right)=0,$ for $k=2,3,4,5,6$, $B_1^{GL}\left(x\right)=b-a-2A_1^{GL}\left(x\right)$ and $B_3^{GL}\left(x\right)=0$. Thus, we obtain the following non-weighted three-point quadrature formulae:

$$\begin{array}{ccc}\hfill \underset{a}{\overset{b}{\int }}f\left(t\right)dt& =& \frac{{(b-a)}^3}{6{(a+b-2x)}^2}\left[f\left(x\right)+f(a+b-x)\right]\hfill \\ & +& \left(b-a-\frac{{(b-a)}^3}{3{(a+b-2x)}^2}\right)f\left(\frac{a+b}{2}\right)\hfill \\ & +& T_n^{GL}\left(x\right)+{(-1)}^n\underset{a}{\overset{b}{\int }}{W}_n^{GL}(t,x)f^{\left(n\right)}\left(t\right)dt,\hfill \end{array}$$

(23)

where

$$\begin{array}{ccc}\hfill T_n^{GL}\left(x\right)& =& \sum _{k=7}^n{A}_k^{GL}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(a+b-x)\right)\hfill \\ & +& \sum _{k=5,oddk}^n{B}_k^{GL}\left(x\right)f^{(k-1)}\left(\frac{a+b}{2}\right).\hfill \end{array}$$

(24)

In particular, according to Remark 3, for $[a,b]=[-1,1]$ and $x=\frac{-\sqrt{15}}{5}$, we get $B_5^{GL}\left(x\right)=0$, and there follows a generalization of the Gauss–Legendre three-point formula. Now, we derive Hermite–Hadamard–Fejér-type estimates for this generalization of the Gauss–Legendre three-point formula.

If the assumptions of Corollary 1 hold for $w\left(t\right)=1$, $t\in [-1,1]$, and if $f:[-1,1]\to \mathbb{R}$ is a $(2n+2)$-convex function, we derive:

$$\begin{array}{ccc}& & U_n^{GL}\left(\frac{-\sqrt{15}}{5}\right)·f^{\left(2n\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}f\left(t\right)dt-\frac{1}{9}\left[5f\left(\frac{-\sqrt{15}}{5}\right)+8f\left(0\right)+5f\left(\frac{\sqrt{15}}{5}\right)\right]-T_{2n}^{GL}\left(\frac{-\sqrt{15}}{5}\right)\hfill \\ & & \le U_n^{GL}\left(\frac{-\sqrt{15}}{5}\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}(-1)+\frac{1}{2}{f}^{\left(2n\right)}\left(1\right)\right],\hfill \end{array}$$

(25)

where

$$\begin{array}{ccc}\hfill U_n^{GL}\left(\frac{-\sqrt{15}}{5}\right)& =& \frac{2·5^{n-1}-2(2n+1)·3^{n-2}}{5^{n-1}(2n+1)!}\hfill \\ & -& \sum _{k=7}^{2n}{A}_k^{GL}\left(\frac{-\sqrt{15}}{5}\right)\frac{{(-\sqrt{15})}^{2n-k+1}+{(-1)}^{k-1}{\left(\sqrt{15}\right)}^{2n-k+1}}{5^{2n-k+1}(2n-k+1)!}.\hfill \end{array}$$

In a special case, for $n=3$, we get

$$\begin{array}{ccc}& & \frac{1}{15,750}·f^{\left(6\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}f\left(t\right)dt-\frac{1}{9}\left[5f\left(\frac{-\sqrt{15}}{5}\right)+8f\left(0\right)+5f\left(\frac{\sqrt{15}}{5}\right)\right]\hfill \\ & & \le \frac{1}{15,750}·\left[\frac{1}{2}{f}^{\left(6\right)}(-1)+\frac{1}{2}{f}^{\left(6\right)}\left(1\right)\right].\hfill \end{array}$$

(26)

3.2. Gauss–Chebyshev Three-Point Quadrature Formula of the First Kind

Let us assume that $w\left(t\right)=\frac{1}{\sqrt{1-t^2}}$, $t\in (-1,1)$ and $x\in \left[-1,0\right)$.

From Theorem 4, there follow:

$$W_{n,w}^{GC1}(t,x)=\left\{\begin{array}{cc}{w}_{1n}\left(t\right)=\frac{1}{(n-1)!}\underset{-1}{\overset{t}{\int }}\frac{{(t-s)}^{n-1}}{\sqrt{1-s^2}}ds,\hfill & t\in [-1,x],\hfill \\ w_{2n}\left(t\right)=\frac{1}{(n-1)!}\underset{x}{\overset{t}{\int }}\frac{{(t-s)}^{n-1}}{\sqrt{1-s^2}}ds+L_{n,x}\left(t\right),\hfill & t\in (x,0],\hfill \\ w_{3n}\left(t\right)=-\frac{1}{(n-1)!}\underset{t}{\overset{-x}{\int }}\frac{{(t-s)}^{n-1}}{\sqrt{1-s^2}}ds+{(-1)}^n{L}_{n,x}(-t),\hfill & t\in (0,-x],\hfill \\ w_{4n}\left(t\right)=-\frac{1}{(n-1)!}\underset{t}{\overset{1}{\int }}\frac{{(t-s)}^{n-1}}{\sqrt{1-s^2}}ds,\hfill & t\in (-x,1],\hfill \end{array}\right.$$

(27)

$$A_k^{GC1}\left(x\right)={(-1)}^{k-1}\left[\frac{{\left(x+1\right)}^{k-1/2}\sqrt{\pi }}{\sqrt{2}\mathsf{\Gamma }\left(\frac{1}{2}+k\right)}F\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}+k,\frac{x+1}{2}\right)-L_{k,x}\left(x\right)\right],k\ge 1,$$

and

$$B_k^{GC1}\left(x\right)=2\left[\frac{{(-1)}^{k-1}}{(k-1)!}\underset{x}{\overset{0}{\int }}\frac{s^{k-1}}{\sqrt{1-s^2}}ds+L_{k,x}\left(0\right)\right],\mathrm{for} \mathrm{odd}k\ge 1,$$

and

$$B_k^{GC1}\left(x\right)=0,\mathrm{for} \mathrm{even}k>1.$$

Corollary 3.

Let $w_{2,2n}\left(t\right)\ge 0$, for all $t\in (x,0]$ and for $n\in \mathbb{N}$. If $f:[-1,1]\to \mathbb{R}$ is a $(2n+2)$-convex function and $f^{\left(2n\right)}$ is piecewise continuous on $\left[-1,1\right]$, then

$$\begin{array}{ccc}\hfill & & U_n^{GC1}\left(x\right)·f^{\left(2n\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}\frac{f\left(t\right)}{\sqrt{1-t^2}}dt-\sum _{k=1}^{2n}{A}_k^{GC1}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(-x)\right)\hfill \\ & & -\sum _{k=1,kodd}^{2n}{B}_k^{GC1}\left(x\right)f^{(k-1)}\left(0\right)\le U_n^{GC1}\left(x\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}(-1)+\frac{1}{2}{f}^{\left(2n\right)}\left(1\right)\right],\hfill \end{array}$$

(28)

where

$$\begin{array}{ccc}\hfill U_n^{GC1}\left(x\right)& =& \frac{1}{\left(2n\right)!}B\left(\frac{1}{2},\frac{1}{2}+n\right)\hfill \\ & -& \sum _{k=1}^{2n}{A}_k^{GC1}\left(x\right)\frac{x^{2n-k+1}+{(-1)}^{k-1}{(-x)}^{2n-k+1}}{(2n-k+1)!}.\hfill \end{array}$$

(29)

If f is a $(2n+2)$-concave function, then inequalities (28) hold with reversed inequality signs.

Proof. 

A special case of Theorem 5 for $w\left(t\right)=\frac{1}{\sqrt{1-t^2}}$, $t\in \left(-1,1\right)$, and a nonnegative function $W_{2n,w}^{GC1}$ defined by (27).  □

If we assume that the polynomials $L_{j,x}\left(t\right)$ are such that

$$\begin{array}{ccc}\hfill L_{0,x}\left(t\right)& =& 0,\mathrm{for}t\in \left[x,0\right],\hfill \\ \hfill L_{1,x}\left(x\right)& =& arcsinx+\frac{\pi }{2}-\frac{\pi }{4x^2},\hfill \\ \hfill L_{j,x}\left(x\right)& =& \frac{{\left(x+1\right)}^{j-1/2}\sqrt{\pi }}{\sqrt{2}\mathsf{\Gamma }\left(\frac{1}{2}+j\right)}F\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}+j,\frac{x+1}{2}\right),j=2,3,4,5,6,\hfill \\ \hfill L_{j,x}\left(t\right)& =& \sum _{k=1}^{6\wedge j}{L}_{k,x}\left(x\right)\frac{{(t-x)}^{j-k}}{(j-k)!},\mathrm{for}t\in \left[x,0\right],j=1,\dots ,n,\hfill \end{array}$$

we calculate $A_1^{GC1}\left(x\right)=\frac{\pi }{4x^2}$, $A_k^{GC1}\left(x\right)=0,$ for $k=2,3,4,5,6$, $B_1^{GC1}\left(x\right)=\pi -\frac{\pi }{2x^2}$ and $B_3^{GC1}\left(x\right)=0$.

Now, we derive

$$\begin{array}{ccc}\hfill \underset{-1}{\overset{1}{\int }}\frac{f\left(t\right)}{\sqrt{1-t^2}}dt& =& \frac{\pi }{4x^2}f\left(x\right)+\left(\pi -\frac{\pi }{2x^2}\right)f\left(0\right)+\frac{\pi }{4x^2}f(-x)\hfill \\ & +& T_{n,w}^{GC1}\left(x\right)+{(-1)}^n\underset{-1}{\overset{1}{\int }}{W}_{n,w}^{GC1}(t,x)f^{\left(n\right)}\left(t\right)dt,\hfill \end{array}$$

(30)

where

$$\begin{array}{ccc}\hfill T_{n,w}^{GC1}\left(x\right)& =& \sum _{k=7}^n{A}_k^{GC1}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(-x)\right)\hfill \\ & +& \sum _{k=5,oddk}^n{B}_k^{GC1}\left(x\right)f^{(k-1)}\left(0\right).\hfill \end{array}$$

(31)

In particular, there follows a generalization of the Gauss–Chebyshev three-point quadrature formula of the first kind for $x=-\frac{\sqrt{3}}{2}$. Now, we derive Hermite–Hadamard-type estimates for the Gauss–Chebyshev three-point quadrature formula of the first kind.

If the assumptions of Corollary 1 hold for $w\left(t\right)=\frac{1}{\sqrt{1-t^2}}$, $t\in (-1,1)$, and if $f:[-1,1]\to \mathbb{R}$ is a $(2n+2)$-convex function, we get

$$\begin{array}{ccc}& & U_n^{GC1}\left(\frac{-\sqrt{3}}{2}\right)·f^{\left(2n\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}\frac{f\left(t\right)}{\sqrt{1-t^2}}dt-\frac{\pi }{3}\left[f\left(\frac{-\sqrt{3}}{2}\right)+f\left(0\right)+f\left(\frac{\sqrt{3}}{2}\right)\right]-T_{2n,w}^{GC1}\left(\frac{-\sqrt{3}}{2}\right)\hfill \\ & & \le U_n^{GC1}\left(\frac{-\sqrt{3}}{2}\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}(-1)+\frac{1}{2}{f}^{\left(2n\right)}\left(1\right)\right],\hfill \end{array}$$

(32)

where

$$\begin{array}{ccc}\hfill U_n^{GC1}\left(\frac{-\sqrt{3}}{2}\right)& =& \frac{1}{\left(2n\right)!}B\left(\frac{1}{2},\frac{1}{2}+n\right)-\frac{\pi ·3^{n-1}}{2^{2n-1}\left(2n\right)!}\hfill \\ & -& \sum _{k=7}^{2n}{A}_k^{GC1}\left(\frac{-\sqrt{3}}{2}\right)\frac{{(-\sqrt{3})}^{2n-k+1}+{(-1)}^{k-1}{\left(\sqrt{3}\right)}^{2n-k+1}}{2^{2n-k+1}(2n-k+1)!}.\hfill \end{array}$$

In a special case, for $n=3$, we obtain

$$\begin{array}{ccc}& & \frac{\pi }{23,040}·f^{\left(6\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}\frac{f\left(t\right)}{\sqrt{1-t^2}}dt-\frac{\pi }{3}\left[f\left(\frac{-\sqrt{3}}{2}\right)+f\left(0\right)+f\left(\frac{\sqrt{3}}{2}\right)\right]\hfill \\ & & \le \frac{\pi }{23,040}·\left[\frac{1}{2}{f}^{\left(6\right)}(-1)+\frac{1}{2}{f}^{\left(6\right)}\left(1\right)\right].\hfill \end{array}$$

(33)

3.3. Gauss–Chebyshev Three-Point Quadrature Formula of the Second Kind

Assuming $w\left(t\right)=\sqrt{1-t^2}$, $t\in [-1,1]$ and $x\in \left[-1,0\right)$ and using Theorem 4, we obtain

$$W_{n,w}^{GC2}(t,x)=\left\{\begin{array}{cc}{w}_{1n}\left(t\right)=\frac{1}{(n-1)!}\underset{-1}{\overset{t}{\int }}{(t-s)}^{n-1}\sqrt{1-s^2}ds,\hfill & t\in [-1,x],\hfill \\ w_{2n}\left(t\right)=\frac{1}{(n-1)!}\underset{x}{\overset{t}{\int }}{(t-s)}^{n-1}\sqrt{1-s^2}ds+L_{n,x}\left(t\right),\hfill & t\in (x,0],\hfill \\ w_{3n}\left(t\right)=-\frac{1}{(n-1)!}\underset{t}{\overset{-x}{\int }}{(t-s)}^{n-1}\sqrt{1-s^2}ds+{(-1)}^n{L}_{n,x}(-t),\hfill & t\in (0,-x],\hfill \\ w_{4n}\left(t\right)=-\frac{1}{(n-1)!}\underset{t}{\overset{1}{\int }}{(t-s)}^{n-1}\sqrt{1-s^2}ds,\hfill & t\in (-x,1],\hfill \end{array}\right.$$

(34)

$$A_k^{GC2}\left(x\right)={(-1)}^{k-1}\left[\frac{{\left(x+1\right)}^{k+1/2}\sqrt{2\pi }}{\mathsf{\Gamma }\left(\frac{3}{2}+k\right)}F\left(-\frac{1}{2},\frac{3}{2},\frac{3}{2}+k,\frac{x+1}{2}\right)-L_{k,x}\left(x\right)\right],k\ge 1,$$

$$B_k^{GC2}\left(x\right)=2\left[\frac{{(-1)}^{k-1}}{(k-1)!}\underset{x}{\overset{0}{\int }}{s}^{k-1}\sqrt{1-s^2}ds+L_{k,x}\left(0\right),\right],\mathrm{for} \mathrm{odd}k\ge 1,$$

and

$$B_k^{GC2}\left(x\right)=0,\mathrm{for} \mathrm{even}k>1.$$

Corollary 4.

Let $w_{2,2n}\left(t\right)\ge 0$, for all $t\in (x,0]$ and for $n\in \mathbb{N}$. If $f:[-1,1]\to \mathbb{R}$ is a $(2n+2)$-convex function and $f^{\left(2n\right)}$ is piecewise continuous on $\left[-1,1\right]$, then

$$\begin{array}{ccc}\hfill & & U_n^{GC2}\left(x\right)·f^{\left(2n\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}f\left(t\right)\sqrt{1-t^2}dt-\sum _{k=1}^{2n}{A}_k^{GC2}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(-x)\right)\hfill \\ & & -\sum _{k=1,kodd}^{2n}{B}_k^{GC2}\left(x\right)f^{(k-1)}\left(0\right)\le U_n^{GC2}\left(x\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}(-1)+\frac{1}{2}{f}^{\left(2n\right)}\left(1\right)\right],\hfill \end{array}$$

(35)

where

$$\begin{array}{ccc}\hfill U_n^{GC2}\left(x\right)& =& \frac{1}{\left(2n\right)!}B\left(\frac{3}{2},\frac{1}{2}+n\right)\hfill \\ & -& \sum _{k=1}^{2n}{A}_k^{GC2}\left(x\right)\frac{x^{2n-k+1}+{(-1)}^{k-1}{(-x)}^{2n-k+1}}{(2n-k+1)!}.\hfill \end{array}$$

(36)

If f is a $(2n+2)$-concave function, then inequalities (35) hold with reversed inequality signs.

Proof. 

A special case of Theorem 5 for $w\left(t\right)=\sqrt{1-t^2}$, $t\in \left[-1,1\right]$, and a nonnegative function $W_{2n,w}^{GC2}$ defined by (34).  □

If the polynomials $L_{j,x}\left(t\right)$ are such that

$$\begin{array}{ccc}\hfill L_{0,x}\left(t\right)& =& 0,\mathrm{for}t\in \left[x,0\right],\hfill \\ \hfill L_{1,x}\left(x\right)& =& \frac{1}{2}\left(arcsinx+\frac{\pi }{2}-\frac{\pi }{8x^2}+\frac{x\sqrt{1-x^2}}{2}\right),\hfill \\ \hfill L_{j,x}\left(x\right)& =& \frac{{\left(x+1\right)}^{j+1/2}\sqrt{2\pi }}{\mathsf{\Gamma }\left(\frac{3}{2}+j\right)}F\left(-\frac{1}{2},\frac{3}{2},\frac{3}{2}+j,\frac{x+1}{2}\right),j=2,3,4,5,6,\hfill \\ \hfill L_{j,x}\left(t\right)& =& \sum _{k=1}^{6\wedge j}{L}_{k,x}\left(x\right)\frac{{(t-x)}^{j-k}}{(j-k)!},\mathrm{for}t\in \left[x,0\right],j=1,\dots ,n,\hfill \end{array}$$

we have $A_1^{GC2}\left(x\right)=\frac{x\sqrt{1-x^2}}{4}-\frac{\pi }{16x^2}$, $A_k^{GC2}\left(x\right)=0,$ for $k=2,3,4,5,6$, $B_1^{GC2}\left(x\right)=\frac{\pi }{2}-\frac{x\sqrt{1-x^2}}{2}+\frac{\pi }{8x^2}$ and $B_3^{GC2}\left(x\right)=0$, so we obtain

$$\begin{array}{ccc}\hfill \underset{-1}{\overset{1}{\int }}f\left(t\right)\sqrt{1-t^2}dt& =& A_1^{GC2}\left(x\right)\left[f\left(x\right)+f(-x)\right]+B_1^{GC2}\left(x\right)f\left(0\right)\hfill \\ & +& T_{n,w}^{GC2}\left(x\right)+{(-1)}^n\underset{-1}{\overset{1}{\int }}{W}_{n,w}^{GC2}(t,x)f^{\left(n\right)}\left(t\right)dt,\hfill \end{array}$$

(37)

where

$$\begin{array}{ccc}\hfill T_{n,w}^{GC2}\left(x\right)& =& \sum _{k=7}^n{A}_k^{GC2}\left(x\right)\left(f^{(k-1)}\left(x\right)+{(-1)}^{k-1}{f}^{(k-1)}(-x)\right)\hfill \\ & +& \sum _{k=5,oddk}^n{B}_k^{GC2}\left(x\right)f^{(k-1)}\left(0\right).\hfill \end{array}$$

(38)

In particular, a generalization of the Gauss–Chebyshev three-point quadrature formula of the second kind follows for $x=-\frac{\sqrt{2}}{2}$. Now, we derive Hermite–Hadamard-type estimates for the Gauss–Chebyshev three-point quadrature formula of the second kind.

Applying Corollary 1 to $w\left(t\right)=\sqrt{1-t^2}$, $t\in [-1,1]$, $x=-\frac{\sqrt{2}}{2}$, and a $(2n+2)$-convex function f, we obtain

$$\begin{array}{ccc}& & U_n^{GC2}\left(\frac{-\sqrt{2}}{2}\right)·f^{\left(2n\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}f\left(t\right)\sqrt{1-t^2}dt-\frac{\pi }{8}\left[f\left(-\frac{\sqrt{2}}{2}\right)+2f\left(0\right)+f\left(\frac{\sqrt{2}}{2}\right)\right]-T_{2n,w}^{GC2}\left(\frac{-\sqrt{2}}{2}\right)\hfill \\ & & \le U_n^{GC2}\left(\frac{-\sqrt{2}}{2}\right)·\left[\frac{1}{2}{f}^{\left(2n\right)}(-1)+\frac{1}{2}{f}^{\left(2n\right)}\left(1\right)\right],\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill U_n^{GC2}\left(\frac{-\sqrt{2}}{2}\right)& =& \frac{1}{\left(2n\right)!}B\left(\frac{3}{2},\frac{1}{2}+n\right)-\frac{\pi }{2^{n+2}\left(2n\right)!}\hfill \\ & -& \sum _{k=7}^{2n}{A}_k^{GC2}\left(\frac{-\sqrt{2}}{2}\right)\frac{{(-\sqrt{2})}^{2n-k+1}+{(-1)}^{k-1}{\left(\sqrt{2}\right)}^{2n-k+1}}{2^{2n-k+1}(2n-k+1)!}.\hfill \end{array}$$

As a special case, for $n=3$, we obtain

$$\begin{array}{ccc}& & \frac{\pi }{92,160}·f^{\left(6\right)}\left(0\right)\hfill \\ & & \le \underset{-1}{\overset{1}{\int }}f\left(t\right)\sqrt{1-t^2}dt-\frac{\pi }{8}\left[f\left(-\frac{\sqrt{2}}{2}\right)+2f\left(0\right)+f\left(\frac{\sqrt{2}}{2}\right)\right]\hfill \\ & & \le \frac{\pi }{92,160}·\left[\frac{1}{2}{f}^{\left(6\right)}(-1)+\frac{1}{2}{f}^{\left(6\right)}\left(1\right)\right].\hfill \end{array}$$