Role of the dynamics of political stability in firm performance: Evidence from Bangladesh

Mohammad Abdullah; Mohammad Ashraful Ferdous Chowdhury; Uttam Karmaker; Md. Habibur Rahman Fuszder; Md. Asif Shahriar; Mohammad Abdullah; Mohammad Ashraful Ferdous Chowdhury; Uttam Karmaker; Md. Habibur Rahman Fuszder; Md. Asif Shahriar

doi:10.3934/QFE.2022022

Quantitative Finance and Economics

2022, Volume 6, Issue 4: 518-536. doi: 10.3934/QFE.2022022

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Research article

Role of the dynamics of political stability in firm performance: Evidence from Bangladesh

1.
Faculty of Business and Management, University Sultan Zainal Abidin, Malaysia
2.
IRC for Finance and Digital Economy, KFUPM Business School, King Fahd University of Petroleum and Minerals, Dhahran, Saudi Arabia
3.
Universal College, Bangladesh
4.
Independent University, Bangladesh

Received: 29 August 2022 Revised: 29 September 2022 Accepted: 09 October 2022 Published: 25 October 2022
JEL Codes: G3, G32, G38

This study examines the role of political stability in a firm's financial performance in Bangladesh. By considering 139 listed companies from the Dhaka Stock Exchange over the period of 2011 to 2020, we applied a dynamic generalized method of moments (GMM), dynamic quantile regression and dynamic threshold regression. The empirical evidence of this study shows a significant positive impact of political stability on Bangladeshi firms' financial performances. Using dynamic quantile regression, we found a positive impact of political stability in the firms' upper and lower quantiles. Additionally, we found the threshold effect of political stability on the firms' performance to have a score of 13.680. This study contributes theoretically and empirically by examining the importance of political stability on financial performance. For the investors, policymakers and other stakeholders, this study provides evidence of a threshold of political stability on a firm's financial performance.

Keywords:

Citation: Mohammad Abdullah, Mohammad Ashraful Ferdous Chowdhury, Uttam Karmaker, Md. Habibur Rahman Fuszder, Md. Asif Shahriar. Role of the dynamics of political stability in firm performance: Evidence from Bangladesh[J]. Quantitative Finance and Economics, 2022, 6(4): 518-536. doi: 10.3934/QFE.2022022

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Abstract

1. Introduction

Many researchers present a large number of studies to improve and generalize classical Hermite-Hadamard inequality. This double inequality suggests that the mean value of a continuous convex function $g: \left[ a, b\right] \subseteq \mathbb{R} \rightarrow \mathbb{R}$ lies between the value of $g$ in the midpoint of the interval $\left[ a, b \right]$ and the arithmetic mean of the values of $g$ at the endpoints of this interval such that

$\begin{equation} g\left( \frac{a+b}{2}\right) \leq \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx\leq \frac{g\left( a\right) +g\left( b\right) }{2}. \end{equation}$

(1.1)

In addition, each side of the mentioned inequality characterizes convexity in the sense that a real-valued continuous function $g$ defined on an interval $I$ is convex if its restriction to each compact subinterval $\left[ a, b\right] \subset I$ hold both inequalities. If $g$ is a concave function, then the inequality is interchanged 1.2 ^[3,23]. In the literature, Hermite-Hadamard inequality is frequently preferred because of its importance in nonlinear analysis. Recently, Jain et al. ^[10] established some new inequalities related to Hermite-Hadamard inequality for the functions whose absolute values of second derivatives are $log-$ convex. Mehrez and Agarwal ^[13] introduced new Hermit-Hadamard type integral inequalities for convex functions. Mo-hammed ^[16] presented some new Hermite-Hadamard inequalities for $MT-$ convex functions. Besides, some new integral inequalities for the logarithmically $p-$ preinvex functions via generalized beta function were established by Mohammed ^[17].

Many authors introduced a large number of studies to generalize various integral inequalities. Cerone et al. ^[4] presented the generalized trapezoid inequality. Ujević ^[27] derived some new perturbations of the trapezoid inequality. Drogomir et al. ^[6] improved quasi-trapezoid quadrature formula by using some well-known classical inequalities. Cerone ^[5] obtained explicit bounds for perturbed trapezoidal rules. Liu and Park ^[11] suggested some perturbed versions of generalized trapezoid inequality. On the other hand, some integral inequalities for the convex functions have been frequently investigated by many researchers. Sarikaya and Aktan ^[24] intoduced the generalization of some integral inequalities for convex functions. Tunç and Şanal ^[26] established some perturbed trapezoid inequalities for twice differentiable convex, $s-$ convex and $tgs-$ convex functions. Ardıc ^[1] presented some integral inequalities such as Hölder, Hermite-Hadamard and Jensen integral inequality for $n$ times differentiable convex functions.

The studies in recent years have focused on the fractional integral inequalities for convex functions. Agarwal et al. ^[2] established some fractional integral inequalities via new Pólya-Szegö type integral inequalities. Fernandez and Mohammed ^[9] used the fractional integrals to obtain the Hermite-Hadamard inequality and related results. Mohammed ^[15] introduced some new integral inequalities by using the $\left(k, h\right)$ , $\left(k, s\right)$ -Riemann Liouville fractional integrals. The author presented new Hermite-Hadamard's type inequalities for Riemann Liouville fractional integrals of convex function ^[18]. Besides, Hermite-Hadamard's type inequalities have been obtained via the fractional integrals for different type convex functions ^{[14,15,16,17,18,19,20,21,22,23,24,25,26,27,28]}.

The main aim of the present study is to obtain some new inequalities related to general perturbed trapezoid inequality. The considered classes of functions consist of the functions whose n th derivatives of absolute values are convex.

Definition 1. ^[14] A function $g:I\subset \mathbb{R} \rightarrow \mathbb{R}$ is said to be convex on $I$ if inequality

$\begin{equation} g\left( ta+\left( 1-t\right) b\right) \leq tg\left( a\right) +\left( 1-t\right) g\left( b\right) \end{equation}$

(1.2)

holds for all $a, b\in I$ and $t\in \left[ 0, 1\right]$ . We say that $g$ is concave if $(-g)$ is convex. For numerical integration, the trapezoid inequality is introduced as

$\begin{equation} \left\vert \int_{a}^{b}g\left( x\right) dx-\frac{1}{2}\left( b-a\right) \left( g\left( a\right) +g\left( b\right) \right) \right\vert \leq \frac{1}{ 12}M_{2}\left( b-a\right) ^{3} \end{equation}$

(1.3)

where $g:\left[ a, b\right] \rightarrow \mathbb{R}$ is supposed to be twice differentiable on the interval $\left(a, b\right)$ , with the second derivative bounded on $\left(a, b\right)$ by $M_{2} = \sup_{x\in \left(a, b\right) }\left\vert g^{\prime \prime }\left(x\right) \right\vert < +\infty$

(^{[5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24]}).

Theorem 1. Grüss inequality : Let $g$ and $z$ to be two functions defined and integrable on $\left[ a, b\right]$ . If $k\leq g\left(x\right) \leq l$ and $m\leq z\left(x\right) \leq n$ is to be $\forall x\in \left[ a, b \right]$ and for constants $k, l, m, n\in \mathbb{R}$ , then

$\begin{equation} \left\vert \frac{1}{b-a}\int_{a}^{b}g\left( x\right) z\left( x\right) dx- \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx\frac{1}{b-a}\int_{a}^{b}z\left( x\right) dx\right\vert \leq \frac{1}{4}\left( l-k\right) \left( n-m\right) . \end{equation}$

(1.4)

The above inequality is held where $\frac{1}{4}$ is the best constant ^[7]. For the perturbed trapezoid inequality, the inequality obtained by the application of the Grüss inequality is given as

$\begin{eqnarray} &&\left\vert \int_{a}^{b}g\left( x\right) dx-\frac{1}{2}\left( b-a\right) \left( g\left( a\right) +g\left( b\right) \right) +\frac{1}{12}\left( b-a\right) ^{2}\left( g^{\prime }\left( b\right) -g^{\prime }\left( a\right) \right) \right\vert \\ &\leq &\frac{1}{32}\left( \Gamma _{2}-\gamma _{2}\right) \left( b-a\right) ^{3} \end{eqnarray}$

(1.5)

by Dragomir et al. ^[6] where $g$ is supposed to be twice differentiable on the interval $\left(a, b\right)$ with the second derivative bounded on $\left(a, b\right)$ by $\Gamma _{2} = \sup_{x\in \left(a, b\right) }g^{\prime \prime }\left(x\right) < +\infty$ and $\gamma _{2} = \inf_{x\in \left(a, b\right) }g^{\prime \prime }\left(x\right) > -\infty$ .

In the light of this information, we establish some inequalities for n th order differentiable convex functions.

Lemma 1. ^[26] Let $g$ $:I\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable mapping on $I^{\circ }$ , $a, b\in I^{\circ }$ with $a < b$ . If $g^{\prime \prime }\in L[a, b]$ , then one obtains

$\begin{eqnarray} &&\int_{a}^{b}g\left( x\right) dx-\frac{1}{2}\left( b-a\right) \left( g\left( a\right) +g\left( b\right) \right) +\frac{5}{4}\left( b-a\right) ^{2}\left( g^{\prime }\left( b\right) -g^{\prime }\left( a\right) \right) \\ & = &\frac{\left( b-a\right) ^{3}}{4}\int_{0}^{1}\left( t+1\right) ^{2}\left[ g^{\prime \prime }\left( ta+\left( 1-t\right) b\right) +g^{\prime \prime }\left( tb+\left( 1-t\right) a\right) \right] dt. \end{eqnarray}$

(1.6)

Theorem 2. ^[12] Minkowski Inequality: Let $g^{p}$ , $z^{p}$ and $\left(g+z\right) ^{p}$ be integrable functions on $\left[ a, b \right]$ . If $p > 1$ , then

$\begin{equation} \left[ \int_{a}^{b}\left\vert g\left( x\right) +z\left( x\right) \right\vert ^{p}dx\right] ^{\frac{1}{p}}\leq \left[ \int_{a}^{b}\left\vert g\left( x\right) \right\vert ^{p}dx\right] ^{\frac{1}{p}}+\left[ \int_{a}^{b}\left \vert z\left( x\right) \right\vert ^{p}dx\right] ^{\frac{1}{p}}. \end{equation}$

(1.7)

Similarly, if $p > 1$ and $a_{k}$ , $b_{k} > 0$ , then Minkowski sum inequality is expressed as

$\begin{equation} \left[ \sum\limits_{k = 1}^{n}\left\vert a_{k}+b_{k}\right\vert ^{p}\right] ^{\frac{1 }{p}}\leq \left[ \sum\limits_{k = 1}^{n}\left\vert a_{k}\right\vert ^{p}\right] ^{ \frac{1}{p}}+\left[ \sum\limits_{k = 1}^{n}\left\vert b_{k}\right\vert ^{p}\right] ^{ \frac{1}{p}}. \end{equation}$

(1.8)

If the sequences $a_{1}$ , $a_{2}$ , ... and $b_{1}$ , $b_{2}$ , ... are proportional, the inequality is provided. We will use the following notations and conventions throughout this article. Let us consider as $I = \left[ 0, \infty \right) \subset \mathbb{R} = \left(-\infty, +\infty \right)$ and $a, b\in I$ with $0 < a < b$ and $g^{\left(n\right) }\in L\left[ a, b\right]$ and

$\begin{eqnarray*} A\left( a,b\right) = \frac{a+b}{2},G\left( a,b\right) = \sqrt{ab}, H\left( a,b\right) = \frac{2ab}{a+b}, \\ L\left( a,b\right) = \frac{b-a}{\ln b-\ln a},a\neq b, L_{p}\left( a,b\right) , \\ L_{p} = L_{p}\left( a,b\right) = \left\{ \begin{array}{ccc} a , a = b \\ \left[ \frac{b^{p+1}-a^{p+1}}{\left( p+1\right) \left( b-a\right) }\right] ^{ \frac{1}{p}} , a\neq b \end{array} \right. \;\; a,b\geq 0 \end{eqnarray*}$

are the arithmetic mean, geometric mean, harmonic mean, logarithmic mean, generalized $p$ -logarithmic mean for $a, b > 0,$ respectively ^[8].

In this study, we introduce some results related to the perturbed trapezoid inequality and prove some applications for special means of real numbers.

2. Main results

Lemma 2. Let $g:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be $n$ times differentiable mapping on $I^{\circ }$ , $a, b\in I^{\circ }$ with $a < b$ $\$ where $n$ is even number. If $g^{\left(n\right) }\in L[a, b]$ , then the following equality is obtained:

$\begin{eqnarray} &&\frac{1}{b-a}\int\limits_{b}^{a}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \\ &&+\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}} \\ &&\times \left( g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right) \\ &&+\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4.a_{2}\right] }{2.n!.a_{n}} \\ &&\times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&-\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{ 2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \\ &&+\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0}\right] }{ 2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \\ & = &\frac{\left( b-a\right) ^{n}}{2.n!.a_{n}} \\ &&\times \int\limits_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt \end{eqnarray}$

(2.1)

Proof. If the right-hand side of the equality is considered and the integration by parts is applied, then one obtains

$\begin{eqnarray*} I_{1} & = &\int\limits_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) dt \\ & = &\left. \left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \frac{g^{\left( n-1\right) }\left( ta+\left( 1-t\right) b\right) }{a-b}\right\vert _{0}^{1} \\ &&-\frac{1}{\left( a-b\right) }\int\limits_{0}^{1}\left( n.a_{n}t^{n-1}+\cdots +2a_{2}t+a_{1}\right) g^{\left( n-1\right) }\left( ta+\left( 1-t\right) b\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{a-b}g^{\left( n-1\right) }\left( a\right) -\frac{a_{0}}{a-b}g^{\left( n-1\right) }\left( b\right) \\ &&-\frac{1}{\left( a-b\right) }\int\limits_{0}^{1}\left( n.a_{n}t^{n-1}+\cdots +2a_{2}t+a_{1}\right) g^{\left( n-1\right) }\left( ta+\left( 1-t\right) b\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{a-b}g^{\left( n-1\right) }\left( a\right) -\frac{a_{0}}{a-b}g^{\left( n-1\right) }\left( b\right) \\ &&-\frac{n.a_{n}+\cdots +2a_{2}+a_{1}}{\left( a-b\right) ^{2}}g^{\left( n-2\right) }\left( a\right) +\frac{a_{1}}{\left( a-b\right) ^{2}}g^{\left( n-2\right) }\left( b\right) \\ &&+\frac{1}{\left( a-b\right) ^{2}}\times \int\limits_{0}^{1}\left( n.\left( n-1\right) .a_{n}t^{n-2}+\cdots +3.2.a_{3}t+2.a_{2}\right) \times g^{\left( n-2\right) }\left( ta+\left( 1-t\right) b\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{a-b}g^{\left( n-1\right) }\left( a\right) -\frac{a_{0}}{a-b}g^{\left( n-1\right) }\left( b\right) \\ &&-\frac{n.a_{n}+\cdots +2a_{2}+a_{1}}{\left( a-b\right) ^{2}}g^{\left( n-2\right) }\left( a\right) +\frac{a_{1}}{\left( a-b\right) ^{2}}g^{\left( n-2\right) }\left( b\right) \\ &&+\frac{n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+2.a_{2}}{ \left( a-b\right) ^{3}}g^{\left( n-3\right) }\left( a\right) -\frac{2.a_{2}}{ \left( a-b\right) ^{3}}g^{\left( n-3\right) }\left( b\right) \\ &&-\frac{n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +3.2.1.a_{3}}{ \left( a-b\right) ^{4}}g^{\left( n-4\right) }\left( a\right) +\frac{3.2.a_{3} }{\left( a-b\right) ^{4}}g^{\left( n-4\right) }\left( b\right) \\ &&+\cdots -\frac{n!.a_{n}+\left( n-1\right) !.a_{n-1}}{\left( a-b\right) ^{n} }g\left( a\right) +\frac{\left( n-1\right) !.a_{n-1}}{\left( a-b\right) ^{n}} g\left( b\right) \\ &&+\frac{n!a_{n}}{\left( a-b\right) ^{n}}\int\limits_{0}^{1}g\left( ta+\left( 1-t\right) b\right) dt \end{eqnarray*}$

$\begin{eqnarray*} I_{2} & = &\int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) dt \\ & = &\left. \left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \frac{g^{\left( n-1\right) }\left( tb+\left( 1-t\right) a\right) }{b-a}\right\vert _{0}^{1} \\ &&-\frac{1}{\left( b-a\right) }\int_{0}^{1}\left( n.a_{n}t^{n-1}+\cdots +2a_{2}t+a_{1}\right) g^{\left( n-1\right) }\left( tb+\left( 1-t\right) a\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{b-a}g^{\left( n-1\right) }\left( b\right) -\frac{a_{0}}{b-a}g^{\left( n-1\right) }\left( a\right) \\ &&-\frac{1}{\left( b-a\right) }\int_{0}^{1}\left( n.a_{n}t^{n-1}+\cdots +2a_{2}t+a_{1}\right) g^{\left( n-1\right) }\left( tb+\left( 1-t\right) a\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{b-a}g^{\left( n-1\right) }\left( b\right) -\frac{a_{0}}{b-a}g^{\left( n-1\right) }\left( a\right) \\ &&-\frac{n.a_{n}+\cdots +2a_{2}+a_{1}}{\left( b-a\right) ^{2}}g^{\left( n-2\right) }\left( b\right) +\frac{a_{1}}{\left( b-a\right) ^{2}}g^{\left( n-2\right) }\left( a\right) \\ &&+\frac{1}{\left( b-a\right) ^{2}}\int_{0}^{1}\left[ n.\left( n-1\right) a_{n}t^{n-2}+\cdots +3.2.a_{3}t+2.a_{2}\right] \times g^{\left( n-2\right) }\left( tb+\left( 1-t\right) a\right) dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{b-a}g^{\left( n-1\right) }\left( b\right) -\frac{a_{0}}{b-a}g^{\left( n-1\right) }\left( a\right) \\ &&-\frac{n.a_{n}+\cdots +2a_{2}+a_{1}}{\left( b-a\right) ^{2}}g^{\left( n-2\right) }\left( b\right) +\frac{a_{1}}{\left( b-a\right) ^{2}}g^{\left( n-2\right) }\left( a\right) \\ &&+\frac{n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+2.a_{2}}{ \left( b-a\right) ^{3}}g^{\left( n-3\right) }\left( b\right) -\frac{2.a_{2}}{ \left( b-a\right) ^{3}}g^{\left( n-3\right) }\left( a\right) \\ &&-\frac{n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +3.2.a_{3}}{ \left( b-a\right) ^{4}}g^{\left( n-4\right) }\left( b\right) +\frac{3.2.a_{3} }{\left( b-a\right) ^{4}}g^{\left( n-4\right) }\left( a\right) \\ &&+\cdots -\frac{n!.a_{n}+\left( n-1\right) !.a_{n-1}}{\left( b-a\right) ^{n} }g\left( b\right) +\frac{\left( n-1\right) !.a_{n-1}}{\left( b-a\right) ^{n}} g\left( a\right) \\ &&+\frac{n!a_{n}}{\left( b-a\right) ^{n}}\int_{0}^{1}g\left( tb+\left( 1-t\right) a\right) dt \end{eqnarray*}$

Summing $\ I_{1}$ and $I_{2}$ , then one obtains

$\begin{eqnarray*} I_{1}+I_{2} & = &\int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt \\ & = &\frac{\left( a_{n}+\cdots +a_{1}+a_{0}\right) }{b-a}\left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] + \frac{a_{0}}{b-a}\left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \\ &&-\frac{\left( n.a_{n}+\cdots +2a_{2}+a_{1}\right) }{\left( b-a\right) ^{2}} \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] +\frac{a_{1}}{\left( b-a\right) ^{2}}\left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \\ &&+\frac{\left( n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2a_{3}+2.a_{2}\right) }{\left( b-a\right) ^{3}}\left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&+\frac{2.a_{2}}{\left( b-a\right) ^{3}}\left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&-\frac{\left( n.\left( n-1\right) .\left( n-2\right) a_{n}+\cdots +3.2a_{3}\right) }{\left( b-a\right) ^{4}}\left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \\ &&+\frac{3.2.a_{3}}{\left( b-a\right) ^{4}}\left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] +\cdots \\ &&-\frac{n!.a_{n}+\left( n-1\right) !.a_{n-1}}{\left( b-a\right) ^{n}}\left[ g\left( a\right) +g\left( b\right) \right] +\frac{\left( n-1\right) !.a_{n-1} }{\left( b-a\right) ^{n}}\left[ g\left( a\right) +g\left( b\right) \right] \\ &&+\frac{\left( n\right) !.a_{n}}{\left( b-a\right) ^{n}}\left[ \int_{0}^{1} \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt\right] \\ & = &\frac{a_{n}+\cdots +a_{1}+2a_{0}}{b-a}\left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \\ &&-\frac{n.a_{n}+\cdots +2a_{2}}{\left( b-a\right) ^{2}}\left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \\ &&+\frac{n.\left( n-1\right) a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4a_{2}}{ \left( b-a\right) ^{3}}\left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&-\frac{n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}}{ \left( b-a\right) ^{4}}\left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \\ &&+\cdots -\frac{n!.a_{n}}{\left( b-a\right) ^{n}}\left[ g\left( a\right) +g\left( b\right) \right] +\frac{2.n!.a_{n}}{\left( b-a\right) ^{n+1}} \int_{a}^{b}g\left( x\right) dx \end{eqnarray*}$

$\begin{eqnarray*} &&\frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \\ &&+\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \\ &&+\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4.a_{2}\right] }{2.n!.a_{n}}\left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&-\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{ 2.n!.a_{n}}\left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \\ &&+\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0}\right] }{ 2.n!.a_{n}}\left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \\ & = &\frac{\left( b-a\right) ^{n}}{2.n!.a_{n}}\int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \times \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt \end{eqnarray*}$

Thus, the proof is completed.

Remark 1. Using the change of the variable $x = ta+\left(1-t\right) b$ where $t\in \left[ 0, 1\right]$ , Eq.(2.1) can be written as

$\begin{eqnarray*} &&\frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \\ &&+\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \\ &&+\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \\ &&-\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{ 2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \\ &&+\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0}\right] }{ 2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \\ & = &\frac{-\left( b-a\right) ^{n-1}}{2.n!.a_{n}}\times \int_{0}^{1}\left( a_{n}\left( \frac{x-b}{a-b}\right) ^{n}+\cdots +a_{1}\left( \frac{x-b}{a-b} \right) +a_{0}\right) \left[ g^{\left( n\right) }\left( x\right) +g^{\left( n\right) }\left( a+b-x\right) \right] dx \end{eqnarray*}$

Theorem 3. Let $g:I\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be $n$ times differentiable mapping on $I^{\circ }$ , $a, b\in I^{\circ }$ with $a < b$ where $n$ is even number. If $\left\vert g^{\left(n\right) }\right\vert$ is convex on $[a, b]$ , then the inequality in the following holds:

(2.2)

Proof. From Lemma 2, it is concluded that

$\begin{eqnarray*} &&\left\vert \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+2.a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+...+a_{1}+2a_{0}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \right\vert \\ & = &\left\vert \frac{\left( b-a\right) ^{n}}{2.n!.a_{n}}\times \left\{ \int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt\right\} \right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \times \left\{ \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left[ \left\vert g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) \right\vert +\left\vert g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right\vert \right] dt\right\} \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \\ &&\times \left\{ \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left[ t\left\vert g^{\left( n\right) }\left( a\right) \right\vert +\left( 1-t\right) \left\vert g^{\left( n\right) }\left( b\right) \right\vert +t\left\vert g^{\left( n\right) }\left( b\right) \right\vert +\left( 1-t\right) \left\vert g^{\left( n\right) }\left( a\right) \right\vert \right] dt\right\} \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert }\left[ \left\vert g^{\left( n\right) }\left( a\right) \right\vert +\left\vert g^{\left( n\right) }\left( b\right) \right\vert \right] \times \int_{0}^{1} \left[ \left\vert a_{n}t^{n}\right\vert +\cdots +\left\vert a_{1}t\right\vert +\left\vert a_{0}\right\vert \right] dt \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \times \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+1}\times \left[ \left\vert g^{\left( n\right) }\left( a\right) \right\vert +\left\vert g^{\left( n\right) }\left( b\right) \right\vert \right] \end{eqnarray*}$

The theorem is proved.

Theorem 4. Let $g:I\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be $n$ times differentiable mapping on $I^{\circ }$ , $a, b\in I^{\circ }$ with $a < b$ , and $p > 1$ with $1/p+1/q = 1$ where $n$ is even number. If the mapping $\left\vert g^{\left(n\right) }\right\vert ^{q}$ is convex on $\left[ a, b\right]$ , then we obtain:

(2.3)

Proof. Using Lemma 2, Hölder's integral inequality and Minkowsky's integral inequality, we establish

$\begin{eqnarray} &&\left\vert \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+2.a_{2}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0} \right] }{2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \right\vert \\ & = &\left\vert \frac{\left( b-a\right) ^{n}}{2.n!.a_{n}}\right. \\ &&\left. \times \left\{ \int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt\right\} \right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \\ &&\times \left[ \begin{array}{c} \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left\vert g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) \right\vert dt \\ +\int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left\vert g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right\vert dt \end{array} \right] \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \\ &&\times \left[ \left( \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\times \left( \int_{0}^{1}\left\vert g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}}\right. \\ &&\left. +\left( \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\times \left( \int_{0}^{1}\left\vert g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}}\right] \\ &\leq &\frac{\left( b-a\right) ^{n}}{n!.\left\vert a_{n}\right\vert }\times \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( kp+1\right) ^{1/p}}\right] \times \left[ \frac{\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{q}+\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{q}}{2}\right] ^{1/q} \end{eqnarray}$

(2.4)

such that $\frac{1}{p}+\frac{1}{q} = 1$ . Considering the convexity of $\left\vert g^{\left(n\right) }\right\vert ^{q}$ , then we find

$\begin{eqnarray} &&\int_{0}^{1}\left\vert g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) \right\vert ^{q}dt \\ &\leq &\int_{0}^{1}\left[ t\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{q}+\left( 1-t\right) \left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{q}\right] dt = \frac{\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{q}+\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{q}}{2} \\ &&\int_{0}^{1}\left\vert g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right\vert ^{q}dt \\ &\leq &\int_{0}^{1}\left[ t\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{q}+\left( 1-t\right) \left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{q}\right] dt = \frac{\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{q}+\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{q}}{2} \end{eqnarray}$

(2.5)

Using the Theorem 2, we have

$\begin{eqnarray} &&\left( \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert ^{p}dt\right) ^{\frac{1}{p}} \\ &\leq &\left[ \int_{0}^{1}\left( a_{n}t^{n}\right) ^{p}dt\right] ^{\frac{1}{p }}+\cdots +\left[ \int_{0}^{1}\left( a_{1}t\right) ^{p}dt\right] ^{\frac{1}{p }}+\left[ \int_{0}^{1}\left( a_{0}\right) ^{p}dt\right] ^{\frac{1}{p}} \\ & = &\frac{\left\vert a_{n}\right\vert }{\left( np+1\right) ^{1/p}}+\frac{ \left\vert a_{n-1}\right\vert }{\left\vert \left( n-1\right) p+1\right\vert ^{1/p}}+\cdots +\frac{\left\vert a_{1}\right\vert }{\left( p+1\right) ^{1/p}} +\left\vert a_{0}\right\vert \\ & = &\left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( kp+1\right) ^{1/p}}\right] \end{eqnarray}$

(2.6)

By using (2.5) and (2.6), the inequality (2.3) is obtained.

Theorem 5. Let $g:I\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be $n$ times differentiable mapping on $I^{\circ }$ , $a, b\in I^{\circ }$ with $a < b$ , and $p > 1$ such that $1/p+1/q = 1$ where $n$ is even number. If the mapping $\left\vert g^{\left(n\right) }\right\vert ^{p}$ is convex on $\left[ a, b\right]$ , then the inequality in the following holds:

$\begin{eqnarray} &&\left\vert \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4.a_{2}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0} \right] }{2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert }\left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+1}\right] ^{1-\frac{1}{p} } \\ &&\times \left\{ \left( \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+2}\right] \left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{p}+\left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( k+1\right) \left( k+2\right) }\right] \left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p}\right) ^{\frac{1}{p} }\right. \\ &&\left. +\left( \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+2 }\right] \left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p}+ \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( k+1\right) \left( k+2\right) }\right] \left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{p}\right) ^{\frac{1}{p}}\right\} \end{eqnarray}$

(2.7)

Proof. Using Lemma 2 and power mean integral inequality, one obtains

$\begin{eqnarray*} &&\left\vert \frac{1}{b-a}\int_{a}^{b}g\left( x\right) dx-\frac{g\left( a\right) +g\left( b\right) }{2}+\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) \left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-4\right) }\left( a\right) +g^{\left( n-4\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +4.3.a_{4}+3.2.a_{3}+4.a_{2}\right] }{2.n!.a_{n}}\right. \\ &&\left. \times \left[ g^{\left( n-3\right) }\left( b\right) -g^{\left( n-3\right) }\left( a\right) \right] \right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] }{2.n!.a_{n}}\times \left[ g^{\left( n-2\right) }\left( a\right) +g^{\left( n-2\right) }\left( b\right) \right] \right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +a_{1}+2a_{0} \right] }{2.n!.a_{n}}\times \left[ g^{\left( n-1\right) }\left( b\right) -g^{\left( n-1\right) }\left( a\right) \right] \right\vert \\ & = &\left\vert \frac{\left( b-a\right) ^{n}}{2.n!.a_{n}}\right. \\ &&\left. \times \left\{ \int_{0}^{1}\left( a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right) \left[ g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) +g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right] dt\right\} \right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \\ &&\times \left[ \begin{array}{c} \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left\vert g^{\left( n\right) }\left( ta+\left( 1-t\right) b\right) \right\vert dt \\ +\int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \left\vert g^{\left( n\right) }\left( tb+\left( 1-t\right) a\right) \right\vert dt \end{array} \right] \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert } \left( \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert dt\right) ^{^{1-\frac{1}{p}}} \\ &&\times \left\{ \left( \begin{array}{c} \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \\ \left( t\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{p}+\left( 1-t\right) \left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p}\right) dt \end{array} \right) ^{\frac{1}{p}}\right. \\ &&\left. +\left( \begin{array}{c} \int_{0}^{1}\left\vert a_{n}t^{n}+\cdots +a_{1}t+a_{0}\right\vert \\ \left( t\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p}+\left( 1-t\right) \left\vert g^{n}\left( a\right) \right\vert ^{p}\right) dt \end{array} \right) ^{\frac{1}{p}}\right\} \\ &\leq &\frac{\left( b-a\right) ^{n}}{2.n!.\left\vert a_{n}\right\vert }. \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+1}\right] ^{1- \frac{1}{p}} \\ &&\times \left\{ \left( \begin{array}{c} \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+2}\right] .\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{p} \\ +\left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( k+1\right) \left( k+2\right) }\right] .\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p} \end{array} \right) ^{\frac{1}{p}}\right. \\ &&\left. +\left( \begin{array}{c} \left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+2}\right] .\left\vert g^{\left( n\right) }\left( b\right) \right\vert ^{p} \\ +\left[ \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( k+1\right) \left( k+2\right) }\right] .\left\vert g^{\left( n\right) }\left( a\right) \right\vert ^{p} \end{array} \right) ^{\frac{1}{p}}\right\} \end{eqnarray*}$

The proof is completed.

3. Applications to special means

In this section, we consider the results of Section 2 to verify the new proposed inequalities.

Proposition 1. Let $a, b\in \mathbb{R}$ , $0 < a < b$ , $n > 2$ where $n$ is even number. Then, the inequality in the following holds:

$\begin{eqnarray} &&\left\vert L_{n}^{n}\left( a,b\right) -A\left( a^{n},b^{n}\right) +\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] \left( a^{4}+b^{4}\right) }{ 2.4!.a_{n}}\right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +3.2.a_{3}+4.a_{2}\right] \left( b^{3}-a^{3}\right) }{ 2.3!.a_{n}}\right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] \left( a^{2}+b^{2}\right) }{2.2!.a_{n}}\right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +2.a_{0}\right] \left( b-a\right) }{2.a_{n}}\right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{\left\vert a_{n}\right\vert }\times \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+1} \end{eqnarray}$

(3.1)

Proof. The proof is clearly obtained from Theorem 3 for $g(x) = x^{n},$ $x\in \mathbb{R}$ .

Proposition 2. Let $a, b\in \mathbb{R}$ , $0 < a < b$ , $n > 2$ where $n$ is even number. For all $p > 1$ , one obtains

$\begin{eqnarray} &&\left\vert L_{n}^{n}\left( a,b\right) -A\left( a^{n},b^{n}\right) +\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] \left( a^{4}+b^{4}\right) }{ 2.4!.a_{n}}\right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots +3.2.a_{3}+4.a_{2}\right] \left( b^{3}-a^{3}\right) }{ 2.3!.a_{n}}\right. \\ &&\left. -\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] \left( a^{2}+b^{3}\right) }{2.2!.a_{n}}\right. \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+\cdots +2.a_{0}\right] \left( b-a\right) }{2.a_{n}}\right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{\left\vert a_{n}\right\vert }\times \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( kp+1\right) ^{1/p}} \end{eqnarray}$

(3.2)

Proof. The proof is completed from Theorem 4 applied for $g(x) = x^{n},$ $x\in \mathbb{R}$ .

Proposition 3. Let $a, b\in \mathbb{R}$ , $0 < a < b$ , $n > 2$ where $n$ is even number. Then, we obtain for all $p > 1$ ,

$\begin{eqnarray} &&\left\vert L_{n}^{n}\left( a,b\right) -A\left( a^{n},b^{n}\right) +\cdots \right. \\ &&\left. +\frac{-\left( b-a\right) ^{n-4}\left[ n.\left( n-1\right) .\left( n-2\right) .a_{n}+\cdots +4.3.2.a_{4}\right] \left( a^{4}+b^{4}\right) }{ 2.4!.a_{n}}\right. \\ &&\left. +\frac{\left( b-a\right) ^{n-3}\left[ n.\left( n-1\right) .a_{n}+\cdots ++3.2.a_{3}+4.a_{2}\right] \left( b^{3}-a^{3}\right) }{ 2.3!.a_{n}}\right. \\ &&-\frac{\left( b-a\right) ^{n-2}\left[ n.a_{n}+\cdots +2.a_{2}\right] \left( a^{2}+b^{3}\right) }{2.2!.a_{n}} \\ &&\left. +\frac{\left( b-a\right) ^{n-1}\left[ a_{n}+...+2.a_{0}\right] \left( b-a\right) }{2.a_{n}}\right\vert \\ &\leq &\frac{\left( b-a\right) ^{n}}{\left\vert a_{n}\right\vert }\times \left( \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+1}\right) ^{1-1/p}\times \left( \sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{k+2} +\sum\limits_{k = 0}^{n}\frac{\left\vert a_{k}\right\vert }{\left( k+1\right) \left( k+2\right) }\right) ^{\frac{1}{p}} \end{eqnarray}$

(3.3)

Proof. The proof is obtained from Theorem 5 such that $g(x) = x^{n},$ $x\in \left[ a, b\right]$ .

Conflict of interest

The authors declare that there is no conflict of interest.

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Role of the dynamics of political stability in firm performance: Evidence from Bangladesh

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