Next Article in Journal
High-Accuracy Simulation of Rayleigh Waves Using Fractional Viscoelastic Wave Equation
Next Article in Special Issue
Further Hermite–Hadamard-Type Inequalities for Fractional Integrals with Exponential Kernels
Previous Article in Journal
On Constructing a Family of Sixth-Order Methods for Multiple Roots
Previous Article in Special Issue
Subordination Principle for Generalized Fractional Zener Models
 
 
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

Fractional Maclaurin-Type Inequalities for Multiplicatively Convex Functions

1
Faculty MISM, Department of Mathematics, University of 8 May 1945 Guelma, P.O. Box 401, Guelma 24000, Algeria
2
Laboratory of Analysis and Control of Differential Equations “ACED”, Faculty MISM, Department of Mathematics, University of 8 May 1945 Guelma, P.O. Box 401, Guelma 24000, Algeria
3
Department of Mathematics, College of Science, University of Ha’il, Ha’il 55473, Saudi Arabia
4
Mathematics Department, College of Sciences, King Khalid University, P.O. Box 9004, Abha 61413, Saudi Arabia
*
Author to whom correspondence should be addressed.
Fractal Fract. 2023, 7(12), 879; https://doi.org/10.3390/fractalfract7120879
Submission received: 26 October 2023 / Revised: 23 November 2023 / Accepted: 7 December 2023 / Published: 12 December 2023
(This article belongs to the Special Issue New Trends on Generalized Fractional Calculus)

Abstract

:
This paper’s major goal is to prove some symmetrical Maclaurin-type integral inequalities inside the framework of multiplicative calculus. In order to accomplish this and after giving some basic tools, we have established a new integral identity. Based on this identity, some symmetrical Maclaurin-type inequalities have been constructed for functions whose multiplicative derivatives are bounded as well as convex. At the end, some applications to special means are provided.

1. Introduction

The theory of inequality has seen a rise in research activity over the past 20 years in different fields of sciences, both theoretical and applied, including in the study of the qualitative properties of solutions to ordinary, partial, and integral differential equations as well as in numerical analysis, where this tool is essential for estimating quadrature errors, and in a variety of calculation types, including time scale calculus [1,2,3], fractional calculus [4,5,6,7], quantum calculus [8,9], and classical (Newtonian) calculus [10,11,12].
The term multiplicative calculus originates from the classical calculation of Newton and Leibniz, which was introduced by Grossman and Katz when they presented and examined the first non-Newtonian systems [13].
The multiplicative derivative and integral were presented by Bashirov et al. [14]. Its relationship to the classical derivative and integral, as well as some of its features, are mentioned below.
The multiplicative derivative of the function  G with the notation  G * is as follows:
Definition 1
([14]). For a positive function  G : R R + . The multiplicative derivative is
d * G d t = G * t = lim Δ 0 G t + Δ G t 1 Δ .
Remark 1.
If  G is positive and differentiable at t, then  G * exists and is related to the standard derivative  G as follows:
G * t = e ln G t = e G t G t .
The multiplicative integral or  * integral of the function  G noted  b a G t d t is as follows:
Proposition 1
([14]). Let  G   L 1 [ a , b ] . Then, the  * integral of the function  G is
b a G t d t = exp b a ln G t d t .
It is also practical to remember the integration-by-parts formula.
Theorem 1
([14]). Let  G , χ : [ a , b ] R , where  G is a multiplicative differentiable function and χ is a differentiable function. So, the function  G χ is a multiplicative integrable function that satisfies
b a G * t χ t d t = G b χ b G a χ a × 1 b a G t χ t d t .
Lemma 1
([15]). Let  G , k : [ a , b ] R , where  G is a differentiable multiplicative function and  k is a differentiable function. Suppose  χ : J R R is a differentiable function, then
b a G * k t k t χ t d t = G k b χ b G k a χ a × 1 b a G k t χ t d t .
The analogous multiplicative of the Hermite–Hadamard inequality was provided by Ali et al. in [16], as follows:
Theorem 2.
Let  G be a positive and multiplicatively convex function on the interval  [ α 1 , α 2 ] ; then, the following double inequality is true:
G α 1 + α 2 2 α 1 α 2 G x d x 1 α 2 α 1 G α 1 G α 2 .
Since the publication of the aforementioned paper, several works concerning multiplicative inequalities have been published (see, for instance, [15,17,18,19,20]).
In [21], Meftah investigated some Maclaurin-type inequalities for multiplicatively convex functions and established the following results.
Theorem 3.
Assume that  G :   α 1 , α 2   R + is a multiplicative differentiable map with multiplicative convex derivative  G * on  [ α 1 , α 2 ] . Then, we have
G 5 α 1 + α 2 6 3 G α 1 + α 2 2 2 G α 1 + 5 α 2 6 3 1 8 α 1 α 2 G x d x 1 α 1 α 2 G * α 1 64 G * 5 α 1 + α 2 6 379 G * α 1 + α 2 2 314 G * α 1 + 5 α 2 6 379 G * α 2 64 α 2 α 1 13 , 824 .
Theorem 4.
Assume that Theorem 3’s whole set of hypotheses is true. Then, we have
G 5 α 1 + α 2 6 3 G α 1 + α 2 2 2 G α 1 + 5 α 2 6 3 1 8 α 1 α 2 G x d x 1 α 1 α 2 G * α 1 221 G * 5 α 1 + α 2 6 379 G * α 1 + 5 α 2 6 379 G * α 2 221 α 2 α 1 13 , 824 .
Theorem 5.
Assume that Theorem 3’s whole set of hypotheses is true. Then, we have
G 5 α 1 + α 2 6 3 G α 1 + α 2 2 2 G α 1 + 5 α 2 6 3 1 8 α 1 α 2 G x d x 1 α 1 α 2 G * α 1 8 G * 5 α 1 + α 2 6 67 G * α 1 + 5 α 2 6 67 G * α 2 8 b a 1728 .
The multiplicative Riemann–Liouville fractional integrals were first introduced by Abdeljawad and Grossman in [4] and satisfies the following relations:
Definition 2.
The left and right multiplicative Riemann–Liouville fractional integral of order  α C , where  R e α > 0 , is given as follows:
a I * α φ ϰ = e J a + α ln φ ϰ
and
* I b α φ ϰ = e J b α ln φ ϰ ,
where  J r 1 + α and  J r 2 α are the left and right Riemann–Liouville fractional integrals, respectively, defined as follows:
J a + α φ ξ = 1 Γ α a ξ ξ μ α 1 φ μ d μ , a < ξ
and
J b α φ ξ = 1 Γ α ξ b μ ξ α 1 φ μ d μ , ξ < b .
Budak and Özçelik [22] proved some multiplicative fractional Hermite–Hadamard-type inequalities by combining the operators (2) and (3) with the definition of multiplicative convex functions. One can also consult [22,23,24,25,26,27,28] concerning fractional multiplicative inequalities.
Very recently, Peng and Du [29] established some non-symmetrical fractional Maclaurin-type inequalities as follows:
Theorem 6.
G :   α 1 , α 2   R + is an increasing multiplicative differentiable map. If  G * is multiplicative convex on  [ α 1 , α 2 ] , then for  α > 0 , the following inequality related to multiplicative RL-fractional integrals holds:
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 L G G * ε 1 ε 2 ε 1 6 Δ 1 G * 5 ε 1 + ε 2 6 ε 2 ε 1 3 1 2 Δ 2 + Δ 3 + Δ 4 × G * ε 1 + ε 2 2 ε 2 ε 1 3 Δ 5 + Δ 6 + Δ 7 + Δ 8 × G * ε 1 + 5 ε 2 6 ε 2 ε 1 3 Δ 9 + Δ 10 + 1 2 Δ 11 G * ε 2 ε 2 ε 1 6 Δ 12 ,
where
L G = * I 5 ε 1 + ε 2 6 α G ε 1 * I ε 2 α G ε 1 + 5 ε 2 6 6 α 1 Γ α + 1 ε 2 ε 1 α * I ε 1 + ε 2 2 α G 5 ε 1 + ε 2 6 * I ε 1 + 5 ε 2 6 α G ε 1 + ε 2 2 3 α 1 Γ α + 1 ε 2 ε 1 α
with
Δ 1 = 1 6 α + 1 α + 2 , 1 2 Δ 2 + Δ 3 + Δ 4 = 10 11 α 15 α 48 α + 1 α + 2 + 5 α 12 α + 1 5 8 1 α 5 α 24 α + 2 5 8 2 α , Δ 5 + Δ 6 + Δ 7 + Δ 8 = 16 8 α 8 α 2 48 α + 1 α + 2 + 5 α 24 α + 2 5 8 2 α + α 4 α + 1 3 8 1 α α 8 α + 2 3 8 2 α , Δ 9 + Δ 10 + 1 2 Δ 11 = 10 + 19 α + α 2 48 α + 1 α + 2 + α 8 α + 2 3 8 2 α , Δ 12 = α 12 α + 2 .
Theorem 7.
Under the assumptions of Theorem 6, if  G * M with  M > 0 , then we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 L G M ε 2 ε 1 6 5 α 12 α + 1 5 8 1 α + α 2 α + 1 3 8 1 α + 5 3 α 6 α + 1 .
The goal of the current study is to construct some new symmetrical fractional Maclaurin-type inequalities for multiplicatively convex functions, which are motivated by the previously stated papers. To address this, we provide a novel integral identity, from which the fractional Maclaurin inequality for bounded multiplicative derivatives is derived initially. The situation when the multiplicative derivatives are convex is then covered. Some applications to special means are provided at the end. The remainder of the current paper is organized as follows: Some symmetrical fractional Maclaurin inequalities are presented in Section 2. Section 3 provides some applications to special means. Section 4 draws the conclusion.

2. Main Results

We begin with the auxiliary result that follows.
Lemma 2.
Assume that  G :   ε 1 , ε 2    R + is a multiplicative differentiable mapping with multiplicative integrable derivative  G * on  ε 1 , ε 2 . Then, we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α = 1 0 G * 1 ε 1 + 5 ε 1 + ε 2 6 1 6 α d ε 2 ε 1 6 × 1 0 G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 1 24 3 8 1 α d ε 2 ε 1 3
× 1 0 G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 1 24 8 α 3 d ε 2 ε 1 3 × 1 0 G * 1 ε 1 + 5 ε 2 6 + ε 2 1 6 1 α d ε 2 ε 1 6 ,
where
Θ G = * I 5 ε 1 + ε 2 6 α G ε 1 ε 1 + 5 ε 2 6 I * α G ε 2 2 α 1 × 5 ε 1 + ε 2 6 I * α G ε 1 + ε 2 2 * I ε 1 + 5 ε 2 6 α G ε 1 + ε 2 2 .
Proof. 
Let
I 1 = 1 0 G * 1 ε 1 + 5 ε 1 + ε 2 6 1 6 α d ε 2 ε 1 6 ,
I 2 = 1 0 G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 1 24 3 8 1 α d ε 2 ε 1 3 ,
I 3 = 1 0 G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 1 24 8 α 3 d ε 2 ε 1 3
and
I 4 = 1 0 G * 1 ε 1 + 5 ε 2 6 + ε 2 1 6 1 α d ε 2 ε 1 6 .
By using the integration by parts for multiplicative integrals,  I 1 yields
I 1 = 1 0 G * 1 ε 1 + 5 ε 1 + ε 2 6 1 6 α d ε 2 ε 1 6 = 1 0 G * 1 ε 1 + 5 ε 1 + ε 2 6 ε 2 ε 1 6 1 6 α d = G 5 ε 1 + ε 2 6 1 6 1 . 1 1 0 G 1 ε 1 + 5 ε 1 + ε 2 6 α 6 α 1 d = G 5 ε 1 + ε 2 6 1 6 1 exp 1 0 α 6 α 1 ln G 1 ε 1 + 5 ε 1 + ε 2 6 d = G 5 ε 1 + ε 2 6 1 6 1 exp 6 α 1 α ε 2 ε 1 α 5 ε 1 + ε 2 6 ε 1 u ε 1 α 1 ln G u d u
= G 5 ε 1 + ε 2 6 1 6 1 exp 1 Γ α 5 ε 1 + ε 2 6 ε 1 u ε 1 α 1 ln G u d u 6 α 1 Γ α + 1 ε 2 ε 1 α = G 5 ε 1 + ε 2 6 1 6 * I 5 ε 1 + ε 2 6 α G ε 1 6 α 1 Γ α + 1 ε 2 ε 1 α .
Similarly, we have
I 2 = 1 0 G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 1 24 3 8 1 α d ε 2 ε 1 3 = 1 0 G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 ε 2 ε 1 3 1 24 3 8 1 α d = G ε 1 + ε 2 2 1 8 G 5 ε 1 + ε 2 6 5 24 . 1 1 0 G 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 α 3 1 α 1 d = G 5 ε 1 + ε 2 6 5 24 G ε 1 + ε 2 2 1 8 exp 1 0 α 3 1 α 1 ln G 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 d = G 5 ε 1 + ε 2 6 5 24 G ε 1 + ε 2 2 1 8 exp 3 α 1 Γ α + 1 ε 2 ε 1 α 1 Γ α ε 1 + ε 2 2 5 ε 1 + ε 2 6 ε 1 + ε 2 2 u α 1 ln G u d u = G 5 ε 1 + ε 2 6 5 24 G ε 1 + ε 2 2 1 8 exp 1 Γ α ε 1 + ε 2 2 5 ε 1 + ε 2 6 ε 1 + ε 2 2 u α 1 ln G u d u 3 α 1 Γ α + 1 ε 2 ε 1 α = G 5 ε 1 + ε 2 6 5 24 G ε 1 + ε 2 2 1 8 5 ε 1 + ε 2 6 I * α G ε 1 + ε 2 2 3 α 1 Γ α + 1 ε 2 ε 1 α ,
I 3 = 1 0 G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 1 24 8 α 3 d ε 2 ε 1 3 = 1 0 G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 ε 2 ε 1 3 1 24 8 α 3 d = G ε 1 + 5 ε 2 6 5 24 G ε 1 + ε 2 2 1 8 . 1 1 0 G 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 1 3 α α 1 d = G ε 1 + 5 ε 2 6 5 24 G ε 1 + ε 2 2 1 8 exp 1 0 1 3 α α 1 ln G 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 d
= G ε 1 + 5 ε 2 6 5 24 G ε 1 + ε 2 2 1 8 exp 3 α 1 ε 2 ε 1 α α ε 1 + 5 ε 2 6 ε 1 + ε 2 2 u ε 1 + ε 2 2 α 1 ln G u d u = f ε 1 + 5 ε 2 6 5 24 f ε 1 + ε 2 2 1 8 exp 1 Γ α ε 1 + 5 ε 2 6 ε 1 + ε 2 2 u ε 1 + ε 2 2 α 1 ln G u d u 3 α 1 Γ α + 1 ε 2 ε 1 α = G ε 1 + 5 ε 2 6 5 24 G ε 1 + ε 2 2 1 8 . * I ε 1 + 5 ε 2 6 α G ε 1 + ε 2 2 3 α 1 Γ α + 1 ε 2 ε 1 α
and
I 4 = 1 0 G * 1 ε 1 + 5 ε 2 6 + ε 2 1 6 1 α d ε 2 ε 1 6 = 1 0 G * 1 ε 1 + 5 ε 2 6 + ε 2 1 6 ε 2 ε 1 6 1 α d = 1 G ε 1 + 5 ε 2 6 1 6 . 1 1 0 G 1 ε 1 + 5 ε 2 6 + ε 2 1 6 α 1 α 1 d = G ε 1 + 5 ε 2 6 1 6 exp 1 0 1 6 α 1 α 1 ln G 1 ε 1 + 5 ε 2 6 + ε 2 d = G ε 1 + 5 ε 2 6 1 6 exp 6 α 1 ε 2 ε 1 α α ε 2 ε 1 + 5 ε 2 6 ε 2 u α 1 ln G u d u = G ε 1 + 5 ε 2 6 1 6 exp 1 Γ α ε 2 ε 1 + 5 ε 2 6 ε 2 u α 1 ln G u d u 6 α 1 Γ α + 1 ε 2 ε 1 α = G ε 1 + 5 ε 2 6 1 6 ε 1 + 5 ε 2 6 I * α G ε 2 6 α 1 Γ α + 1 ε 2 ε 1 α .
Multiplying (5)–(8) yields the desired outcome. □
Theorem 8.
Assume that Lemma 2’s hypotheses are all true. If  ln G * ln M on  [ ε 1 , ε 2 ] , then we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α M b a 18 α + 1 7 3 α 2 + 3 α 3 8 1 α ,
where  Θ . is defined by (4).
Proof. 
According to Lemma 2, multiplicative integration, and the hypothesis that  ln G * ln M , we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α = 1 0 G * 1 ε 1 + 5 ε 1 + ε 2 6 1 6 α d ε 2 ε 1 6 × 1 0 G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 1 24 3 8 1 α d ε 2 ε 1 3 × 1 0 G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 1 24 8 α 3 d ε 2 ε 1 3 × 1 0 G * 1 ε 1 + 5 ε 2 6 + ε 2 1 6 1 α d ε 2 ε 1 6 exp 1 0 ε 2 ε 1 36 α ln G * 1 ε 1 + 5 ε 1 + ε 2 6 d × exp 1 0 ε 2 ε 1 72 3 8 1 α ln G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 d × exp 1 0 ε 2 ε 1 72 8 α 3 ln G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 d × exp 1 0 ε 2 ε 1 36 1 α ln G * 1 ε 1 + 5 ε 2 6 + ε 2 d exp ε 2 ε 1 36 ln M 1 0 α d exp ε 2 ε 1 72 ln M 1 0 3 8 1 α d × exp ε 2 ε 1 72 ln M 1 0 8 α 3 d exp ε 2 ε 1 36 ln M 1 0 1 α d = exp ε 2 ε 1 36 α + 1 ln M exp ε 2 ε 1 36 α + 1 5 3 α 2 + 3 α 3 8 1 α ln M × exp ε 2 ε 1 36 α + 1 5 3 α 2 + 3 α 3 8 1 α ln M exp ε 2 ε 1 36 α + 1 ln M = M ε 2 ε 1 18 α + 1 7 3 α 2 + 3 α 3 8 1 α .
where we have used
1 0 3 8 1 α d = 1 0 8 α 3 d = 5 3 α α + 1 + 6 α α + 1 3 8 1 α .
The proof is finished. □
Corollary 1.
By assuming that  α = 1 in Theorem 6, we obtain
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 ε 2 ε 1 G u d u 1 ε 1 ε 2 M 25 ε 2 ε 1 288 .
Theorem 9.
Assume that Lemma 2’s hypotheses are all true. If  G * is multiplicative convex on  [ ε 1 , ε 2 ] , then we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α G * ε 1 1 α + 1 α + 2 G * 5 ε 1 + ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α × G * ε 1 + ε 2 2 16 3 α + 1 α + 2 2 α + 1 α + 2 + 6 α α + 1 3 8 1 α 3 α α + 2 3 8 2 α × G * ε 1 + 5 ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α G * ε 2 1 α + 1 α + 2 ε 2 ε 1 36 ,
where  Θ . is defined by (4).
Proof. 
According to Lemma 2, multiplicative integration, and the multiplicative convexity of  G * , we have
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α exp 1 0 ε 2 ε 1 36 α ln G * 1 ε 1 + 5 ε 1 + ε 2 6 d × exp 1 0 ε 2 ε 1 72 3 8 1 α ln G * 1 5 ε 1 + ε 2 6 + ε 1 + ε 2 2 d × exp 1 0 ε 2 ε 1 72 8 α 3 ln G * 1 ε 1 + ε 2 2 + ε 1 + 5 ε 2 6 d × exp 1 0 ε 2 ε 1 36 1 α ln G * 1 ε 1 + 5 ε 2 6 + ε 2 d exp 1 0 ε 2 ε 1 36 α ln G * ε 1 1 G * 5 ε 1 + ε 2 6 d × exp 1 0 ε 2 ε 1 72 3 8 1 α ln G * 5 ε 1 + ε 2 6 1 G * ε 1 + ε 2 2 d × exp 1 0 ε 2 ε 1 72 8 α 3 ln G * ε 1 + ε 2 2 1 G * ε 1 + 5 ε 2 6 d × exp 1 0 ε 2 ε 1 36 1 α ln f * ε 1 + 5 ε 2 6 1 f * ε 2 d
= exp ε 2 ε 1 36 1 0 α 1 ln G * ε 1 + ln G * 5 ε 1 + ε 2 6 d × exp ε 2 ε 1 72 1 0 3 8 1 α 1 ln G * 5 ε 1 + ε 2 6 + ln G * ε 1 + ε 2 2 d × exp ε 2 ε 1 72 1 0 8 α 3 1 ln G * ε 1 + ε 2 2 + ln G * ε 1 + 5 ε 2 6 d × exp ε 2 ε 1 36 1 0 1 α 1 ln G * ε 1 + 5 ε 2 6 + ln G * ε 2 d = G * ε 1 ε 2 ε 1 36 α + 1 α + 2 G * 5 ε 1 + ε 2 6 ε 2 ε 1 36 α + 2 + b a 72 10 3 α 2 α + 2 + 3 α α + 2 3 8 2 α × G * ε 1 + ε 2 2 ε 2 ε 1 36 16 3 α + 1 α + 2 2 α + 1 α + 2 + 6 α α + 1 3 8 1 α 3 α α + 2 3 8 2 α × G * ε 1 + 5 ε 2 6 ε 2 ε 1 36 α + 2 + b a 72 10 3 α 2 α + 2 + 3 α α + 2 3 8 2 α G * ε 2 ε 2 ε 1 36 α + 1 α + 2 = G * ε 1 1 α + 1 α + 2 G * 5 ε 1 + ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α × G * ε 1 + ε 2 2 16 3 α + 1 α + 2 2 α + 1 α + 2 + 6 α α + 1 3 8 1 α 3 α α + 2 3 8 2 α × G * ε 1 + 5 ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α G * ε 2 1 α + 1 α + 2 ε 2 ε 1 36 .
The result follows from the calculation of the following integrals:
1 0 α 1 d = 1 0 1 α d = 1 α + 1 α + 2 ,
1 0 α + 1 d = 1 0 1 α + 1 d = 1 α + 2 ,
1 0 3 8 1 α 1 d = 1 0 8 α 3 d = 10 3 α 2 α + 2 + 3 α α + 2 3 8 2 α
and
1 0 3 8 1 α d = 1 0 8 α 3 1 d = = 16 3 α + 1 α + 2 2 α + 1 α + 2 + 6 α α + 1 3 8 1 α 3 α α + 2 3 8 2 α .
The proof is completed. □
Remark 2.
If we put  α = 1 , Theorem 7 may be simplified to Theorem 3.2 from [4].
Corollary 2.
Using the multiplicative convexity of  G * , i.e.,  G * ε 1 + ε 2 2 G * ε 1 G * ε 2 , Theorem 7 becomes
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α G * ε 1 14 3 α 2 9 α 4 α + 1 α + 2 + 6 α 2 α + 1 3 8 1 α 3 α 2 α + 2 3 8 2 α G * 5 ε 1 + ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α × G * ε 1 + 5 ε 2 6 14 3 α 4 α + 2 + 3 α 2 α + 2 3 8 2 α G * ε 2 14 3 α 2 9 α 4 α + 1 α + 2 + 6 α 2 α + 1 3 8 1 α 3 α 2 α + 2 3 8 2 α ε 2 ε 1 36 .
Remark 3.
Corollary 2 will be reduced to Corollary 3.3 from [4], if we take  α = 1 .
Corollary 3.
Using the multiplicative convexity of  G * , i.e.,  G * ε 1 + ε 2 2 G * 5 ε 1 + ε 2 6 G * ε 1 + 5 ε 2 6 , Theorem 7 becomes
G 5 ε 1 + ε 2 6 3 G ε 1 + ε 2 2 2 G ε 1 + 5 ε 2 6 3 1 8 Θ G 3 α 1 Γ α + 1 ε 2 ε 1 α G * ε 1 1 α + 1 α + 2 G * 5 ε 1 + ε 2 6 12 + α 3 α 2 2 α + 1 α + 2 + 6 α 2 α + 1 3 8 1 α × G * ε 1 + 5 ε 2 6 12 + α 3 α 2 2 α + 1 α + 2 + 6 α 2 α + 1 3 8 1 α G * ε 2 1 α + 1 α + 2 ε 2 ε 1 36 .
Remark 4.
Corollary 3 will be reduced to Corollary 3.4 from [4], if we take  α = 1 .

3. Applications to Special Means

Consider the following means of arbitrary real number  η 1 , η 2 , , η n :
The arithmetic mean:  A η 1 , η 2 , , η n = η 1 + η 2 + + η n n .
The harmonic mean:  H η 1 , η 2 , , η n = n 1 η 1 + 1 η 2 + + 1 η n .
The logarithmic means:  L η 1 , η 2 = η 2 η 1 ln η 2 ln η 1 η 1 , η 2 > 0 , and  η 1 η 2 .
The k-logarithmic mean:  L k η 1 , η 2 = η 2 k + 1 η 1 k + 1 k + 1 η 2 η 1 1 k η 1 , η 2 > 0 , η 1 η 2 , and  k R \ 1,0 .
Proposition 2.
For two positive real numbers  0 < η 1 < η 2 , we have
e 3 8 A p η 1 , η 1 , η 1 , η 1 , η 1 , η 2 + 1 4 A p η 1 , η 2 + 3 8 A p η 1 , η 2 , η 2 , η 2 , η 2 , η 2 L p p η 1 , η 2 e 25 η 2 η 1 p η 2 p 1 288 .
Proof. 
It suffices to apply Corollary 2, taking  G = e p as a function with  p 2 where  G * = e p p 1 , M = e p η 2 p 1 , and  η 2 η 1 G u d u 1 η 1 η 2 = exp L p p η 1 , η 2 . □
Proposition 3.
For two positive real numbers  0 < η 1 < η 2 and  n > 0 , we have
H 3 n 8 η 2 , η 2 , η 2 , η 2 , η 2 , η 1 H n 4 η 2 , η 1 H 3 n 8 η 2 , η 1 , η 1 , η 1 , η 1 , η 1 η 2 n η 1 η 1 η 2 η 1 n η 2 η 1 η 2 e 1 n η 2 η 1 864 4 η 1 + η 2 + 67 η 1 η 2 3 η 1 η 2 .
Proof. 
It suffices to apply Corollary 3 with  α = 1 on the interval  1 η 2 , 1 η 1 to the function  G t = 1 n , whose  G * = e n and  1 η 1 1 η 2 f u d u η 1 η 2 η 1 η 2 = η 1 n η 2 η 1 η 2 η 2 n η 1 η 1 η 2 e n . □

4. Conclusions

The conclusions produced in this work are based on a novel identity. We have constructed certain fractional Maclaurin-type integral inequalities for functions whose multiplicative derivatives are both bounded and multiplicatively convex. We have also discussed some particular cases. A few applications of our findings to special means are given. Our results improve those established in [29], and they also recover those established in [21].

Author Contributions

Conceptualization, M.M., B.M., A.M. and M.B.; Methodology, M.M., B.M., A.M. and M.B.; Writing—original draft, M.M., B.M., A.M. and M.B.; Writing—review & editing, M.M., B.M., A.M. and M.B. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by King Khalid University through large research project under grant number R.G.P.2/252/44.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Acknowledgments

The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University for funding this work.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Du, T.; Peng, Y. Hermite–Hadamard type inequalities for multiplicative Riemann–Liouville fractional integrals. J. Comput. Appl. Math. 2024, 440, 115582. [Google Scholar] [CrossRef]
  2. Rashid, S.; Noor, M.A.; Noor, K.I.; Safdar, F.; Chu, Y.-M. Hermite-Hadamard type inequalities for the class of convex functions on time scale. Mathematics 2019, 7, 956. [Google Scholar] [CrossRef]
  3. Shen, J.-M.; Rashid, S.; Noor, M.A.; Ashraf, R.; Rehana; Chu, Y.M. Certain novel estimates within fractional calculus theory on time scales. AIMS Math. 2020, 5, 6073–6086. [Google Scholar] [CrossRef]
  4. Grossman, M.; Katz, R. Non-Newtonian Calculus; Lee Press: Pigeon Cove, MS, USA, 1972. [Google Scholar]
  5. Rashid, S.; Akdemir, A.O.; Jarad, F.; Noor, M.A.; Noor, K.I. Simpson’s type integral inequalities for κ-fractional integrals and their applications. AIMS Math. 2019, 4, 1087–1100. [Google Scholar] [CrossRef]
  6. Tariq, M.; Ntouyas, S.K.; Shaikh, A.A. New Variant of Hermite-Hadamard, Fejér and Pachpatte-Type Inequality and Its Refinements Pertaining to Fractional Integral Operator. Fractal Fract. 2023, 7, 405. [Google Scholar] [CrossRef]
  7. Zhou, S.-S.; Rashid, S.; Dragomir, S.S.; Latif, M.A.; Akdemir, A.O.; Liu, J.-B. Some new inequalities involving κ-fractional integral for certain classes of functions and their applications. J. Funct. Spaces 2020, 2020, 5285147. [Google Scholar]
  8. Chiheb, T.; Boumaza, N.; Meftah, B. Some new Simpson-like type inequalities via preqausiinvexity. Transylv. J. Math. Mech 2020, 12, 1–10. [Google Scholar]
  9. Sitthiwirattham, T.; Ali, M.A.; Budak, H. On some new Maclaurin’s type inequalities for convex functions in q-calculus. Fractal Fract. 2023, 7, 572. [Google Scholar] [CrossRef]
  10. Ali, M.A.; Budak, H.; Sarikaya, M.Z.; Zhang, Z. Ostrowski and Simpson type inequalities for multiplicative integrals. Proyecciones 2021, 40, 743–763. [Google Scholar] [CrossRef]
  11. Ciurdariu, L.; Grecu, E. Several Quantum Hermite-Hadamard-Type Integral Inequalities for Convex Functions. Fractal Fract. 2023, 7, 463. [Google Scholar] [CrossRef]
  12. Kashuri, A.; Sahoo, S.K.; Aljuaid, M.; Tariq, M.; Sen, M.D.L. Some New Hermite–Hadamard Type Inequalities Pertaining to Generalized Multiplicative Fractional Integrals. Symmetry 2023, 15, 868. [Google Scholar] [CrossRef]
  13. Fu, H.; Peng, Y.; Du, T. Some inequalities for multiplicative tempered fractional integrals involving the λ-incomplete gamma functions. AIMS Math. 2021, 6, 7456–7478. [Google Scholar] [CrossRef]
  14. Alomari, M.W.; Dragomir, S.S. Various error estimations for several Newton-Cotes quadrature formulae in terms of at most first derivative and applications in numerical integration. Jordan J. Math. Stat. 2014, 7, 89–108. [Google Scholar]
  15. Ali, M.A.; Abbas, M.; Zafer, A.A. On some Hermite-Hadamard integral inequalities in multiplicative calculus. J. Ineq. Special Func. 2019, 10, 111–122. [Google Scholar]
  16. Abdeljawad, T.; Grossman, M. On geometric fractional calculus. J. Semigroup Theory Appl. 2016, 2016, 2. [Google Scholar]
  17. Özcan, S. Some integral inequalities of Hermite-Hadamard type for multiplicatively preinvex functions. AIMS Math. 2020, 5, 1505–1518. [Google Scholar] [CrossRef]
  18. Özcan, S. Hermite-Hadamard type inequalities for multiplicatively h-convex functions. Konuralp J. Math. 2020, 8, 158–164. [Google Scholar] [CrossRef]
  19. Özcan, S. Hermite-Hadamard type inequalities for multiplicatively h-preinvex functions. Turk. J. Anal. Number Theory 2021, 9, 65–70. [Google Scholar] [CrossRef]
  20. Ali, M.A.; Abbas, M.; Zhang, Z.; Sial, I.B.; Arif, R. On integral inequalities for product and quotient of two multiplicatively convex functions. Asian Res. J. Math. 2019, 12, 1–11. [Google Scholar] [CrossRef]
  21. Meftah, B. Fractional Ostrowski type inequalities for functions whose first derivatives are s-preinvex in the second sense. Int. J. Anal. Appl. 2017, 15, 146–154. [Google Scholar]
  22. Budak, H.; Özçelik, K. On Hermite-Hadamard type inequalities for multiplicative fractional integrals. Miskolc Math. Notes 2020, 21, 91–99. [Google Scholar] [CrossRef]
  23. Bashirov, A.E.; Kurpinar, E.M.; Özyapici, A. Multiplicative calculus and its applications. J. Math. Anal. Appl. 2008, 337, 36–48. [Google Scholar] [CrossRef]
  24. Dragomir, S.S.; Torebek, B.T. Some Hermite-Hadamard type inequalities in the class of hyperbolic p-convex functions. Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 2019, 113, 3413–3423. [Google Scholar] [CrossRef]
  25. Fagbemigun, B.O.; Mogbademu, A.A.; Olaleru, J.O. Hermite-Hadamard inequality for a certain class of convex functions on time scales. Honam Math. J. 2022, 44, 17–25. [Google Scholar]
  26. Hyder, A.A.; Budak, H.; Barakat, M.A. New Versions of Midpoint Inequalities Based on Extended Riemann–Liouville Fractional Integrals. Fractal Fract. 2023, 7, 442. [Google Scholar] [CrossRef]
  27. Peng, Y.; Fu, H.; Du, T. Estimations of bounds on the multiplicative fractional integral inequalities having exponential kernels. Commun. Math. Stat. 2022, 1–25. [Google Scholar] [CrossRef]
  28. Peng, Y.; Du, T. Hermite-Hadamard-type inequalities for * differentiable multiplicative m-preinvexity and (s,m)-reinvexity via the multiplicative tempered fractional integrals. J. Math. Inequal. 2023, 17, 1179–1201. [Google Scholar] [CrossRef]
  29. Peng, Y.; Du, T. Fractional Maclaurin-type inequalities for multiplicatively convex functions and multiplicatively P-functions. Filomat 2023, 37, 9497–9509. [Google Scholar]
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Share and Cite

MDPI and ACS Style

Merad, M.; Meftah, B.; Moumen, A.; Bouye, M. Fractional Maclaurin-Type Inequalities for Multiplicatively Convex Functions. Fractal Fract. 2023, 7, 879. https://doi.org/10.3390/fractalfract7120879

AMA Style

Merad M, Meftah B, Moumen A, Bouye M. Fractional Maclaurin-Type Inequalities for Multiplicatively Convex Functions. Fractal and Fractional. 2023; 7(12):879. https://doi.org/10.3390/fractalfract7120879

Chicago/Turabian Style

Merad, Meriem, Badreddine Meftah, Abdelkader Moumen, and Mohamed Bouye. 2023. "Fractional Maclaurin-Type Inequalities for Multiplicatively Convex Functions" Fractal and Fractional 7, no. 12: 879. https://doi.org/10.3390/fractalfract7120879

Article Metrics

Back to TopTop