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Department of Mathematics, Josip Juraj Strossmayer University of Osijek, Trg Ljudevita Gaja 6, 31000 Osijek, Croatia
Mathematics 2021, 9(15), 1720; https://doi.org/10.3390/math9151720
Received: 28 June 2021 / Revised: 17 July 2021 / Accepted: 20 July 2021 / Published: 22 July 2021

## 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  (p. 138)).
Theorem 1
(The Hermite–Hadamard–Fejér inequalities). Let $h : [ a , b ] → R$ be a convex function. Then
$h a + b 2 ∫ a b u ( x ) d x ≤ ∫ a b u ( x ) h ( x ) d x ≤ 1 2 h ( a ) + 1 2 h ( b ) ∫ a b u ( x ) d x ,$
where $u : [ a , b ] → R$ is nonnegative, integrable and symmetric about $a + b 2$. If h is a concave function, then the inequalities in (1) are reversed.
If $u ≡ 1$, 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 , GA-convex functions , quasi-convex functions  and convex functions  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 ).
Let us consider a subdivision $σ = { a = x 0 < x 1 < ⋯ < x m = b }$ of the segment $[ a , b ]$, $m ∈ N$. Let $w : [ a , b ] → R$ be an arbitrary integrable function. For each segment $[ x j − 1 , x j ]$, $j = 1 , … , m$, we define w-harmonic sequences of functions ${ w j k } k = 1 , … , n$ by:
$w j 1 ′ ( t ) = w ( t ) , t ∈ [ x j − 1 , x j ] , w j k ′ ( t ) = w j , k − 1 ( t ) , t ∈ [ x j − 1 , x j ] , k = 2 , 3 , … , n .$
Further, the function $W n , w$ is defined as follows:
$W n , w ( t , σ ) = w 1 n ( t ) , t ∈ [ a , x 1 ] , w 2 n ( t ) , t ∈ ( x 1 , x 2 ] , . . . w m n ( t ) , t ∈ ( x m − 1 , b ] .$
The following theorem gives a general integral identity (see ).
Theorem 2.
Let $f : [ a , b ] → R$ be such that $f ( n )$ is piecewise continuous on $[ a , b ]$. Then, the following holds:
$∫ a b w ( t ) f ( t ) d t = ∑ k = 1 n ( − 1 ) k − 1 w m k ( b ) f ( k − 1 ) ( b ) + ∑ j = 1 m − 1 w j k ( x j ) − w j + 1 , k ( x j ) f ( k − 1 ) ( x j ) − w 1 k ( a ) f ( k − 1 ) ( a ) + ( − 1 ) n ∫ a b W n , w ( t , σ ) f ( n ) ( t ) d t .$
In , the authors proved the following Fejér-type inequalities by using identity (4).
Theorem 3.
Let $f : [ a , b ] → R$ be $( n + 2 )$-convex on $[ a , b ]$ and $f ( n )$ piecewise continuous on $[ a , b ]$. Further, let us suppose that the function $W n , w$, defined in (3), is nonnegative and symmetric about $a + b 2$ (i.e., $W n , w ( t , σ ) = W n , w ( a + b − t , σ )$). Then
$U n ( σ ) · f ( n ) a + b 2 ≤ ( − 1 ) n ∫ a b w ( t ) f ( t ) d t − ∑ k = 1 n ( − 1 ) k − 1 w m k ( b ) f ( k − 1 ) ( b ) + ∑ j = 1 m − 1 w j k ( x j ) − w j + 1 , k ( x j ) f ( k − 1 ) ( x j ) − w 1 k ( a ) f ( k − 1 ) ( a ) ≤ U n ( σ ) · 1 2 f ( n ) ( a ) + 1 2 f ( n ) ( b ) ,$
where
$U n ( σ ) = ( − 1 ) n n ! ∫ a b w ( t ) · t n d t − ( − 1 ) n ∑ k = 1 n ( − 1 ) k − 1 ( n − k + 1 ) ! · w m k ( b ) b n − k + 1 + ∑ j = 1 m − 1 w j k ( x j ) − w j + 1 , k ( x j ) x j n − k + 1 − w 1 k ( a ) a n − k + 1 .$
If $W n , w ( t , σ ) ≤ 0$ 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  (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 $x 0 , … , x n ∈ [ a , b ]$ is defined recursively by
$f [ x i ] = f ( x i ) , ( i = 0 , … , n )$
and
$f [ x 0 , … , x n ] = f [ x 1 , … , x n ] − f [ x 0 , … , x n − 1 ] x n − x 0 .$
The value $f [ x 0 , … , x n ]$ is independent of the order of points $x 0 , … , x n$.
Definition 2.
A function $f : [ a , b ] → R$ is said to be n-convex on $[ a , b ]$, $n ≥ 0$, if, for all choices of $( n + 1 )$ distinct points $x 0 , … , x n ∈ [ a , b ]$, the n-th order divided difference in f satisfies
$f [ x 0 , … , x n ] ≥ 0 .$
From the previous definitions, the following property holds: if f is an $( n + 2 )$-convex function, then there exists the n-th order derivative $f ( n )$, which is a convex function (see, e.g.,  (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, $a + b 2$ 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
$B ( x , y ) = ∫ 0 1 s x − 1 ( 1 − s ) y − 1 d s ,$
$Γ$ denotes the gamma function defined as:
$Γ ( x ) = 2 ∫ 0 ∞ s 2 x − 1 e − s 2 d s ,$
and
$F α , β , γ ; z = 1 B ( β , γ − β ) ∫ 0 1 t β − 1 ( 1 − t ) γ − β − 1 ( 1 − z t ) − α d t$
is a hypergeometric function with $γ > β > 0$, $z < 1$.
In the paper, we assume that all considered integrals exist and that they are finite.

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 .
In  (p. 54), the authors proved the following theorem.
Theorem 4.
Let $w : [ a , b ] → R$ be an integrable function, $x ∈ [ a , a + b 2 )$, and let $L j , x j = 0 , 1 , … , n$, $n ∈ N$, be a sequence of harmonic polynomials such that $d e g L j , x ≤ j − 1$ and $L 0 , x ≡ 0$. Further, let us suppose that ${ w j k } k = 1 , . . , n$ are w-harmonic sequences of functions on $[ x j − 1 , x j ]$, for $j = 1 , 2 , 3 , 4 ,$ defined by the following relations:
$w 1 k ( t ) = 1 ( k − 1 ) ! ∫ a t t − s k − 1 w ( s ) d s , t ∈ [ a , x ] ,$
$w 2 k ( t ) = 1 ( k − 1 ) ! ∫ x t t − s k − 1 w ( s ) d s + L k , x ( t ) , t ∈ x , a + b 2 ,$
$w 3 k ( t ) = − 1 ( k − 1 ) ! ∫ t a + b − x t − s k − 1 w ( s ) d s + ( − 1 ) k L k , x ( a + b − t ) , t ∈ a + b 2 , a + b − x ,$
$w 4 k ( t ) = − 1 ( k − 1 ) ! ∫ t b t − s k − 1 w ( s ) d s , t ∈ ( a + b − x , b ] .$
If $f : [ a , b ] → R$ is such that $f ( n )$ is piecewise continuous on $a , b$, then we have
$∫ a b w ( t ) f ( t ) d t = ∑ k = 1 n A k ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( a + b − x ) + ∑ k = 1 n B k ( x ) f ( k − 1 ) a + b 2 + ( − 1 ) n ∫ a b W n , w ( t , x ) f ( n ) ( t ) d t ,$
where
$A k ( x ) = ( − 1 ) k − 1 1 ( k − 1 ) ! ∫ a x x − s k − 1 w ( s ) d s − L k , x ( x ) , k ≥ 1 ,$
and
such that
$W n , w ( t , x ) = w 1 n ( t ) , t ∈ [ a , x ] , w 2 n ( t ) , t ∈ x , a + b 2 , w 3 n ( t ) , t ∈ a + b 2 , a + b − x , w 4 n ( t ) , t ∈ ( a + b − x , b ] .$
Remark 1.
If we assume $w ( t ) = w ( a + b − t )$, for each $t ∈ a , b$, then the following symmetry conditions hold for $k = 1 , … , n$:
$w 1 k ( t ) = ( − 1 ) k w 4 k ( a + b − t ) , f o r t ∈ [ a , x ] ,$
and
$w 2 k ( t ) = ( − 1 ) k w 3 k ( a + b − t ) , f o r t ∈ x , a + b 2 .$
Using Theorems 1 and 4, the properties of both n-convex functions and w-harmonic sequences of functions, and the method that originated in , in the next theorem, we derive new Hermite–Hadamard–Fejér-type inequalities for the weighted three-point quadrature Formula (7).
Theorem 5.
Let $w : [ a , b ] → R$ be an integrable function such that $w ( t ) = w ( a + b − t )$, for each $t ∈ a , b$ and $x ∈ [ a , a + b 2 )$. Let the function $W 2 n , w$, defined by (10), be nonnegative. If $f : [ a , b ] → R$ is $( 2 n + 2 )$-convex on $a , b$ and $f ( 2 n )$ is piecewise continuous on $a , b$, then
$U n , w ( x ) · f ( 2 n ) a + b 2 ≤ ∫ a b w ( t ) f ( t ) d t − ∑ k = 1 2 n A k ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( a + b − x ) − ∑ k = 1 , k o d d 2 n B k ( x ) f ( k − 1 ) a + b 2 ≤ U n , w ( x ) · 1 2 f ( 2 n ) ( a ) + 1 2 f ( 2 n ) ( b ) ,$
where
$U n , w ( x ) = 1 ( 2 n ) ! ∫ a b w ( t ) · t 2 n d t − ∑ k = 1 2 n A k ( x ) x 2 n − k + 1 + ( − 1 ) k − 1 ( a + b − x ) 2 n − k + 1 ( 2 n − k + 1 ) ! − ∑ k = 1 , k o d d 2 n B k ( x ) ( a + b ) 2 n − k + 1 2 2 n − k + 1 ( 2 n − k + 1 ) ! ,$
and $A k$ and $B k$ are defined as in Theorem 4. If $W 2 n , w ( t , x ) ≤ 0$ or f is a $( 2 n + 2 )$-concave function, then inequalities (11) hold with reversed inequality signs.
Proof.
Let us observe that the function f is $( 2 n + 2 )$-convex. Hence, $f ( 2 n )$ is a convex function. It follows from Remark 1 that the function $W 2 n , w$ is symmetric about $a + b 2$, i.e., $W 2 n , w ( t , x ) = W 2 n , w ( a + b − t , x )$. Thus, inequalities (11) follow directly from Theorem 1, replacing a nonnegative and symmetric function u by a nonnegative and symmetric function $W 2 n , w$, and a convex function h by a convex function $f ( 2 n )$, and then using identity (7) in $∫ a b W 2 n , w ( t , x ) f ( 2 n ) ( t ) d t$.
Identity (7) yields $U n , w ( x )$ by substituting n with $2 n$ and putting $f ( t ) = t 2 n ( 2 n ) !$. Then, $f ( 2 n ) ( t ) = 1$ and $f ( k − 1 ) ( t ) = 1 ( 2 n − k + 1 ) ! · t 2 n − k + 1$. On the other hand, if $W 2 n , w t , x$ is nonpositive, then $− W 2 n , w t , x$ is nonnegative, from where there follow reversed signs in (11).
Further, let us assume that f is a $( 2 n + 2 )$-concave function. Hence, the function $− f ( 2 n )$ is convex. Reversed signs in (11) are obtained by putting $− f ( 2 n )$ and the nonnegative function $W 2 n , w t , x$ in (1). This completes the proof.  □
Remark 2.
The value of $U n , w ( x )$ can be obtained from Theorem 3 by taking an appropriate subdivision of the segment $a , b$ and applying the properties of functions $w 1 k , w 2 k , w 3 k$ and $w 4 k$.
To get a maximum degree of exactness of quadrature Formula (7) for fixed $x ∈ a , a + b 2$, we consider a sequence of harmonic polynomials ${ L j , x } j = 0 , 1 , … , n$ defined as follows:
$L 0 , x ( t ) = 0 , f o r t ∈ x , a + b 2 , L 1 , x ( x ) = ∫ a x w ( s ) d s − 2 ( a + b − 2 x ) 2 ∫ a b s 2 − a + b 2 2 w ( s ) d s , L j , x ( x ) = 1 ( j − 1 ) ! ∫ a x x − s j − 1 w ( s ) d s , j = 2 , 3 , 4 , 5 , 6 , L j , x ( t ) = ∑ k = 1 6 ∧ j L k , x ( x ) ( t − x ) j − k ( j − k ) ! , for t ∈ x , a + b 2 , j = 1 , … , n .$
Therefore, we have
$A 1 ( x ) = 2 ( a + b − 2 x ) 2 ∫ a b s 2 − a + b 2 2 w ( s ) d s ,$
$B 1 ( x ) = ∫ a b w ( s ) d s − 2 A 1 ( x ) ,$
$A k ( x ) = 0 ,$ for $k = 2 , 3 , 4 , 5 , 6$ and $B k ( x ) = 0 ,$ for $k = 2 , 3 , 4$.
Finally, from identity (7), for $x ∈ a , a + b 2$, we obtain the following three-point weighted integral formula:
$∫ a b w ( t ) f ( t ) d t = A 1 ( x ) f ( x ) + f ( a + b − x ) + ∫ a b w ( s ) d s − 2 A 1 ( x ) f a + b 2 + T n , w ( x ) + ( − 1 ) n ∫ a b W n , w ( t , x ) f ( n ) ( t ) d t ,$
where
$T n , w ( x ) = ∑ k = 7 n A k ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( a + b − x ) + ∑ k = 5 , o d d k n B k ( x ) f ( k − 1 ) a + b 2 .$
Now, applying results from Theorem 5 to identity (15), we get the following results.
Corollary 1.
Let $w : [ a , b ] → R$ be an integrable function such that $w ( t ) = w ( a + b − t )$, for each $t ∈ a , b$ and let $x ∈ [ a , a + b 2 )$. Let the function $W 2 n , w$, defined by (10), be nonnegative and let $L j , x$ be defined by (13). If $f : [ a , b ] → R$ is $( 2 n + 2 )$-convex on $a , b$ and $f ( 2 n )$ is piecewise continuous on $a , b$, then
$U n , w ( x ) · f ( 2 n ) a + b 2 ≤ ∫ a b w ( t ) f ( t ) d t − A 1 ( x ) f ( x ) + f ( a + b − x ) − ∫ a b w ( s ) d s − 2 A 1 ( x ) f a + b 2 − T 2 n , w ( x ) ≤ U n , w ( x ) · 1 2 f ( 2 n ) ( a ) + 1 2 f ( 2 n ) ( b ) ,$
where
$U n , w ( x ) = 1 ( 2 n ) ! ∫ a b w ( t ) · t 2 n d t − A 1 ( x ) x 2 n + ( a + b − x ) 2 n ( 2 n ) ! − ∫ a b w ( s ) d s − 2 A 1 ( x ) ( a + b ) 2 n 2 2 n ( 2 n ) ! − ∑ k = 7 2 n A k ( x ) x 2 n − k + 1 + ( − 1 ) k − 1 ( a + b − x ) 2 n − k + 1 ( 2 n − k + 1 ) ! − ∑ k = 5 , k o d d 2 n B k ( x ) ( a + b ) 2 n − k + 1 2 2 n − k + 1 ( 2 n − k + 1 ) ! .$
If $W 2 n , w ( t , x ) ≤ 0$ or f is a $( 2 n + 2 )$-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 $B 5 ( x ) = 0$, then we get
$x = a + b 2 − ∫ a b s − a + b 2 4 w ( s ) d s ∫ a b s 2 − a + b 2 2 w ( s ) d s .$
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 $w ( t ) = 1$, $t ∈ [ a , b ]$ and $x ∈ a , a + b 2$.
Now, from Theorem 4, we calculate
$W n G L ( t , x ) = w 1 n ( t ) = ( t − a ) n n ! , t ∈ [ a , x ] , w 2 n ( t ) = ( t − x ) n n ! + L n , x ( t ) , t ∈ x , a + b 2 , w 3 n ( t ) = ( t − a − b + x ) n n ! + ( − 1 ) n L n , x ( a + b − t ) , t ∈ a + b 2 , a + b − x , w 4 n ( t ) = ( t − b ) n n ! , t ∈ ( a + b − x , b ] ,$
and
$A k G L ( x ) = ( − 1 ) k − 1 x − a k k ! − L k , x ( x ) , for k ≥ 1 ,$
and
Corollary 2.
Let $w 2 , 2 n ( t ) ≥ 0$, for all $t ∈ x , a + b 2$ and for $n ∈ N$. If $f : [ a , b ] → R$ is a $( 2 n + 2 )$-convex function and $f ( 2 n )$ is piecewise continuous on $a , b$, then
$U n G L ( x ) · f ( 2 n ) a + b 2 ≤ ∫ a b f ( t ) d t − ∑ k = 1 2 n A k G L ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( a + b − x ) − ∑ k = 1 , k o d d 2 n B k G L ( x ) f ( k − 1 ) a + b 2 ≤ U n G L ( x ) · 1 2 f ( 2 n ) ( a ) + 1 2 f ( 2 n ) ( b ) ,$
where
$U n G L ( x ) = b 2 n + 1 − a 2 n + 1 ( 2 n + 1 ) ! − ∑ k = 1 2 n A k G L ( x ) x 2 n − k + 1 + ( − 1 ) k − 1 ( a + b − x ) 2 n − k + 1 ( 2 n − k + 1 ) ! − ∑ k = 1 , k o d d 2 n B k G L ( x ) ( a + b ) 2 n − k + 1 2 2 n − k + 1 ( 2 n − k + 1 ) ! .$
If f is a $( 2 n + 2 )$-concave function, then inequalities (20) hold with reversed inequality signs.
Proof.
A special case of Theorem 5 for $w ( t ) = 1$, $t ∈ a , b$, and a nonnegative function $W 2 n G L$ defined by (19).  □
If we assume that the polynomials $L j , x ( t )$ are such that
$L 0 , x ( t ) = 0 , for t ∈ x , a + b 2 , L 1 , x ( x ) = x − a − ( b − a ) 3 6 ( a + b − 2 x ) 2 , L j , x ( x ) = ( x − a ) j j ! , j = 2 , 3 , 4 , 5 , 6 , L j , x ( t ) = ∑ k = 1 6 ∧ j L k , x ( x ) ( t − x ) j − k ( j − k ) ! , for t ∈ x , a + b 2 , j = 1 , … , n ,$
we get $A 1 G L ( x ) = ( b − a ) 3 6 ( a + b − 2 x ) 2$, $A k G L ( x ) = 0 ,$ for $k = 2 , 3 , 4 , 5 , 6$, $B 1 G L ( x ) = b − a − 2 A 1 G L ( x )$ and $B 3 G L ( x ) = 0$. Thus, we obtain the following non-weighted three-point quadrature formulae:
$∫ a b f ( t ) d t = ( b − a ) 3 6 ( a + b − 2 x ) 2 f ( x ) + f ( a + b − x ) + b − a − ( b − a ) 3 3 ( a + b − 2 x ) 2 f a + b 2 + T n G L ( x ) + ( − 1 ) n ∫ a b W n G L ( t , x ) f ( n ) ( t ) d t ,$
where
$T n G L ( x ) = ∑ k = 7 n A k G L ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( a + b − x ) + ∑ k = 5 , o d d k n B k G L ( x ) f ( k − 1 ) a + b 2 .$
In particular, according to Remark 3, for $[ a , b ] = [ − 1 , 1 ]$ and $x = − 15 5$, we get $B 5 G L ( x ) = 0$, 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 $w ( t ) = 1$, $t ∈ [ − 1 , 1 ]$, and if $f : [ − 1 , 1 ] → R$ is a $( 2 n + 2 )$-convex function, we derive:
$U n G L − 15 5 · f ( 2 n ) 0 ≤ ∫ − 1 1 f ( t ) d t − 1 9 5 f − 15 5 + 8 f ( 0 ) + 5 f 15 5 − T 2 n G L − 15 5 ≤ U n G L − 15 5 · 1 2 f ( 2 n ) ( − 1 ) + 1 2 f ( 2 n ) ( 1 ) ,$
where
$U n G L − 15 5 = 2 · 5 n − 1 − 2 ( 2 n + 1 ) · 3 n − 2 5 n − 1 ( 2 n + 1 ) ! − ∑ k = 7 2 n A k G L − 15 5 ( − 15 ) 2 n − k + 1 + ( − 1 ) k − 1 ( 15 ) 2 n − k + 1 5 2 n − k + 1 ( 2 n − k + 1 ) ! .$
In a special case, for $n = 3$, we get
$1 15 , 750 · f ( 6 ) 0 ≤ ∫ − 1 1 f ( t ) d t − 1 9 5 f − 15 5 + 8 f ( 0 ) + 5 f 15 5 ≤ 1 15 , 750 · 1 2 f ( 6 ) ( − 1 ) + 1 2 f ( 6 ) ( 1 ) .$

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

Let us assume that $w ( t ) = 1 1 − t 2$, $t ∈ ( − 1 , 1 )$ and $x ∈ − 1 , 0$.
From Theorem 4, there follow:
$W n , w G C 1 ( t , x ) = w 1 n ( t ) = 1 ( n − 1 ) ! ∫ − 1 t ( t − s ) n − 1 1 − s 2 d s , t ∈ [ − 1 , x ] , w 2 n ( t ) = 1 ( n − 1 ) ! ∫ x t ( t − s ) n − 1 1 − s 2 d s + L n , x ( t ) , t ∈ ( x , 0 ] , w 3 n ( t ) = − 1 ( n − 1 ) ! ∫ t − x ( t − s ) n − 1 1 − s 2 d s + ( − 1 ) n L n , x ( − t ) , t ∈ ( 0 , − x ] , w 4 n ( t ) = − 1 ( n − 1 ) ! ∫ t 1 ( t − s ) n − 1 1 − s 2 d s , t ∈ ( − x , 1 ] ,$
$A k G C 1 ( x ) = ( − 1 ) k − 1 x + 1 k − 1 / 2 π 2 Γ 1 2 + k F 1 2 , 1 2 , 1 2 + k , x + 1 2 − L k , x ( x ) , k ≥ 1 ,$
and
and
Corollary 3.
Let $w 2 , 2 n ( t ) ≥ 0$, for all $t ∈ ( x , 0 ]$ and for $n ∈ N$. If $f : [ − 1 , 1 ] → R$ is a $( 2 n + 2 )$-convex function and $f ( 2 n )$ is piecewise continuous on $− 1 , 1$, then
$U n G C 1 ( x ) · f ( 2 n ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − ∑ k = 1 2 n A k G C 1 ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( − x ) − ∑ k = 1 , k o d d 2 n B k G C 1 ( x ) f ( k − 1 ) 0 ≤ U n G C 1 ( x ) · 1 2 f ( 2 n ) ( − 1 ) + 1 2 f ( 2 n ) ( 1 ) ,$
where
$U n G C 1 ( x ) = 1 ( 2 n ) ! B 1 2 , 1 2 + n − ∑ k = 1 2 n A k G C 1 ( x ) x 2 n − k + 1 + ( − 1 ) k − 1 ( − x ) 2 n − k + 1 ( 2 n − k + 1 ) ! .$
If f is a $( 2 n + 2 )$-concave function, then inequalities (28) hold with reversed inequality signs.
Proof.
A special case of Theorem 5 for $w ( t ) = 1 1 − t 2$, $t ∈ − 1 , 1$, and a nonnegative function $W 2 n , w G C 1$ defined by (27).  □
If we assume that the polynomials $L j , x ( t )$ are such that
$L 0 , x ( t ) = 0 , for t ∈ x , 0 , L 1 , x ( x ) = arcsin x + π 2 − π 4 x 2 , L j , x ( x ) = x + 1 j − 1 / 2 π 2 Γ 1 2 + j F 1 2 , 1 2 , 1 2 + j , x + 1 2 , j = 2 , 3 , 4 , 5 , 6 , L j , x ( t ) = ∑ k = 1 6 ∧ j L k , x ( x ) ( t − x ) j − k ( j − k ) ! , for t ∈ x , 0 , j = 1 , … , n ,$
we calculate $A 1 G C 1 ( x ) = π 4 x 2$, $A k G C 1 ( x ) = 0 ,$ for $k = 2 , 3 , 4 , 5 , 6$, $B 1 G C 1 ( x ) = π − π 2 x 2$ and $B 3 G C 1 ( x ) = 0$.
Now, we derive
$∫ − 1 1 f ( t ) 1 − t 2 d t = π 4 x 2 f ( x ) + π − π 2 x 2 f 0 + π 4 x 2 f ( − x ) + T n , w G C 1 ( x ) + ( − 1 ) n ∫ − 1 1 W n , w G C 1 ( t , x ) f ( n ) ( t ) d t ,$
where
$T n , w G C 1 ( x ) = ∑ k = 7 n A k G C 1 ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( − x ) + ∑ k = 5 , o d d k n B k G C 1 ( x ) f ( k − 1 ) 0 .$
In particular, there follows a generalization of the Gauss–Chebyshev three-point quadrature formula of the first kind for $x = − 3 2$. 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 $w ( t ) = 1 1 − t 2$, $t ∈ ( − 1 , 1 )$, and if $f : [ − 1 , 1 ] → R$ is a $( 2 n + 2 )$-convex function, we get
$U n G C 1 − 3 2 · f ( 2 n ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − π 3 f − 3 2 + f ( 0 ) + f 3 2 − T 2 n , w G C 1 − 3 2 ≤ U n G C 1 − 3 2 · 1 2 f ( 2 n ) ( − 1 ) + 1 2 f ( 2 n ) ( 1 ) ,$
where
$U n G C 1 − 3 2 = 1 ( 2 n ) ! B 1 2 , 1 2 + n − π · 3 n − 1 2 2 n − 1 ( 2 n ) ! − ∑ k = 7 2 n A k G C 1 − 3 2 ( − 3 ) 2 n − k + 1 + ( − 1 ) k − 1 ( 3 ) 2 n − k + 1 2 2 n − k + 1 ( 2 n − k + 1 ) ! .$
In a special case, for $n = 3$, we obtain
$π 23 , 040 · f ( 6 ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − π 3 f − 3 2 + f ( 0 ) + f 3 2 ≤ π 23 , 040 · 1 2 f ( 6 ) ( − 1 ) + 1 2 f ( 6 ) ( 1 ) .$

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

Assuming $w ( t ) = 1 − t 2$, $t ∈ [ − 1 , 1 ]$ and $x ∈ − 1 , 0$ and using Theorem 4, we obtain
$W n , w G C 2 ( t , x ) = w 1 n ( t ) = 1 ( n − 1 ) ! ∫ − 1 t ( t − s ) n − 1 1 − s 2 d s , t ∈ [ − 1 , x ] , w 2 n ( t ) = 1 ( n − 1 ) ! ∫ x t ( t − s ) n − 1 1 − s 2 d s + L n , x ( t ) , t ∈ ( x , 0 ] , w 3 n ( t ) = − 1 ( n − 1 ) ! ∫ t − x ( t − s ) n − 1 1 − s 2 d s + ( − 1 ) n L n , x ( − t ) , t ∈ ( 0 , − x ] , w 4 n ( t ) = − 1 ( n − 1 ) ! ∫ t 1 ( t − s ) n − 1 1 − s 2 d s , t ∈ ( − x , 1 ] ,$
$A k G C 2 ( x ) = ( − 1 ) k − 1 x + 1 k + 1 / 2 2 π Γ 3 2 + k F − 1 2 , 3 2 , 3 2 + k , x + 1 2 − L k , x ( x ) , k ≥ 1 ,$
and
Corollary 4.
Let $w 2 , 2 n ( t ) ≥ 0$, for all $t ∈ ( x , 0 ]$ and for $n ∈ N$. If $f : [ − 1 , 1 ] → R$ is a $( 2 n + 2 )$-convex function and $f ( 2 n )$ is piecewise continuous on $− 1 , 1$, then
$U n G C 2 ( x ) · f ( 2 n ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − ∑ k = 1 2 n A k G C 2 ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( − x ) − ∑ k = 1 , k o d d 2 n B k G C 2 ( x ) f ( k − 1 ) 0 ≤ U n G C 2 ( x ) · 1 2 f ( 2 n ) ( − 1 ) + 1 2 f ( 2 n ) ( 1 ) ,$
where
$U n G C 2 ( x ) = 1 ( 2 n ) ! B 3 2 , 1 2 + n − ∑ k = 1 2 n A k G C 2 ( x ) x 2 n − k + 1 + ( − 1 ) k − 1 ( − x ) 2 n − k + 1 ( 2 n − k + 1 ) ! .$
If f is a $( 2 n + 2 )$-concave function, then inequalities (35) hold with reversed inequality signs.
Proof.
A special case of Theorem 5 for $w ( t ) = 1 − t 2$, $t ∈ − 1 , 1$, and a nonnegative function $W 2 n , w G C 2$ defined by (34).  □
If the polynomials $L j , x ( t )$ are such that
$L 0 , x ( t ) = 0 , for t ∈ x , 0 , L 1 , x ( x ) = 1 2 arcsin x + π 2 − π 8 x 2 + x 1 − x 2 2 , L j , x ( x ) = x + 1 j + 1 / 2 2 π Γ 3 2 + j F − 1 2 , 3 2 , 3 2 + j , x + 1 2 , j = 2 , 3 , 4 , 5 , 6 , L j , x ( t ) = ∑ k = 1 6 ∧ j L k , x ( x ) ( t − x ) j − k ( j − k ) ! , for t ∈ x , 0 , j = 1 , … , n ,$
we have $A 1 G C 2 ( x ) = x 1 − x 2 4 − π 16 x 2$, $A k G C 2 ( x ) = 0 ,$ for $k = 2 , 3 , 4 , 5 , 6$, $B 1 G C 2 ( x ) = π 2 − x 1 − x 2 2 + π 8 x 2$ and $B 3 G C 2 ( x ) = 0$, so we obtain
$∫ − 1 1 f ( t ) 1 − t 2 d t = A 1 G C 2 ( x ) f ( x ) + f ( − x ) + B 1 G C 2 ( x ) f 0 + T n , w G C 2 ( x ) + ( − 1 ) n ∫ − 1 1 W n , w G C 2 ( t , x ) f ( n ) ( t ) d t ,$
where
$T n , w G C 2 ( x ) = ∑ k = 7 n A k G C 2 ( x ) f ( k − 1 ) ( x ) + ( − 1 ) k − 1 f ( k − 1 ) ( − x ) + ∑ k = 5 , o d d k n B k G C 2 ( x ) f ( k − 1 ) 0 .$
In particular, a generalization of the Gauss–Chebyshev three-point quadrature formula of the second kind follows for $x = − 2 2$. Now, we derive Hermite–Hadamard-type estimates for the Gauss–Chebyshev three-point quadrature formula of the second kind.
Applying Corollary 1 to $w ( t ) = 1 − t 2$, $t ∈ [ − 1 , 1 ]$, $x = − 2 2$, and a $( 2 n + 2 )$-convex function f, we obtain
$U n G C 2 − 2 2 · f ( 2 n ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − π 8 f − 2 2 + 2 f ( 0 ) + f 2 2 − T 2 n , w G C 2 − 2 2 ≤ U n G C 2 − 2 2 · 1 2 f ( 2 n ) ( − 1 ) + 1 2 f ( 2 n ) ( 1 ) ,$
where
$U n G C 2 − 2 2 = 1 ( 2 n ) ! B 3 2 , 1 2 + n − π 2 n + 2 ( 2 n ) ! − ∑ k = 7 2 n A k G C 2 − 2 2 ( − 2 ) 2 n − k + 1 + ( − 1 ) k − 1 ( 2 ) 2 n − k + 1 2 2 n − k + 1 ( 2 n − k + 1 ) ! .$
As a special case, for $n = 3$, we obtain
$π 92 , 160 · f ( 6 ) 0 ≤ ∫ − 1 1 f ( t ) 1 − t 2 d t − π 8 f − 2 2 + 2 f ( 0 ) + f 2 2 ≤ π 92 , 160 · 1 2 f ( 6 ) ( − 1 ) + 1 2 f ( 6 ) ( 1 ) .$

## Funding

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## Conflicts of Interest

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