MathCraft: PDF to LaTeX file: Sample-01
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\documentclass[10pt]{amsart} %\usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} %\usepackage[version=4]{mhchem} \usepackage{stmaryrd} \usepackage{bbold} \usepackage{hyperref} \usepackage{enumerate} \hypersetup{colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan,} \urlstyle{same} \title{ON STOLARSKY AND RELATED MEANS } \author{K. GACŠETIĆ AND SAIMA NAZ KHAN} \date{} %New command to display footnote whose markers will always be hidden \let\svthefootnote\thefootnote \newcommand\blfootnotetext[1]{% \let\thefootnote\relax\footnote{#1}% \addtocounter{footnote}{-1}% \let\thefootnote\svthefootnote% } %Overriding the \footnotetext command to hide the marker if its value is `0` \let\svfootnotetext\footnotetext \renewcommand\footnotetext[2][?]{% \if\relax#1\relax% \ifnum\value{footnote}=0\blfootnotetext{#2}\else\svfootnotetext{#2}\fi% \else% \if?#1\ifnum\value{footnote}=0\blfootnotetext{#2}\else\svfootnotetext{#2}\fi% \else\svfootnotetext[#1]{#2}\fi% \fi } \begin{document} \maketitle ABSTRACT. We give a simple proof of the Stolarsky means inequality as well as some related inequalities for similar means of Stolarsky type. \vspace{10mm} \begin{center} {\large\textbf{1. Introduction and Preliminaries}}\end{center} \vspace{2mm} Let us consider the following means $$ \begin{aligned} & E(x, y ; r, s)=\left\{\dfrac{r\left(y^{s}-x^{s}\right)}{s\left(y^{r}-x^{r}\right)}\right\}^{\dfrac{1}{s-r}} \\ & E(x, y ; r, 0)=E(0, r)=\left\{\dfrac{y^{r}-x^{r}}{r(\ln y-\ln x)}\right\}^{1 / r} \\ & E(x, y ; r, r)=e^{-\dfrac{1}{r}}\left(\dfrac{x^{x^{r}}}{y^{y^{r}}}\right)^{1 /\left(x^{r}-y^{r}\right)} \\ & E(x, y ; 0,0)=\sqrt{x y} \end{aligned} $$ where $x$ and $y$ are positive real numbers $x \neq y, r$ and $s$ are any real numbers but $0.$ These means, known in literature, are called Stolarsky means. Namely Stolarsky[1] in 1975 (see also [2, p.120]) introduced these means. Stolarsky proved that the function $E(r, s)$ is increasing in both $r$ and $s$ i.e. for $r \leq u$ and $s \leq v$, we have \begin{equation*} E(x, y ; r, s) \leq E(x, y ; u, v) . \tag{1} \end{equation*} \vspace{2mm} In this paper, first we shall give a simple proof of inequality (1). Further we shall introduce two new classes of means of Stolarsky type. \vspace{10mm} \begin{center} {\large\textbf{2. A Simple Proof of Stolarsky Means Inequality}}\end{center} \vspace{2mm} Note that $E(r, s)$ is continuous, this means it is enough to prove (1) in the case where $r, s, u, v \neq 0, r \neq s$ and $u \neq v$. \footnotetext{Key words and phrases. convex function, log-convex function, Stolarsky means. } \vspace{2mm} We consider the following function $f(x)=p^{2} \varphi_{r}(x)+2 p q \varphi_{t}(x)+q^{2} \varphi_{s}(x) \quad$ where $t=\dfrac{r+s}{2}$ and $p, q \in \mathbb{R}$, and $$ \varphi_{r}(x)= \begin{cases}x^{r} / r, & r \neq 0 \\ \ln x, & r=0\end{cases} $$ Now $$ \begin{aligned} f^{\prime}(x) & =p^{2} x^{r-1}+2 p q x^{t-1}+q^{2} x^{s-1} \\ & =\left(p x^{(r-1) / 2}+q x^{(s-1) / 2}\right)^{2} \geq 0 \end{aligned} $$ This implies $f$ is monotonically increasing. So for $x \neq y$ $$ \dfrac{f(x)-f(y)}{x-y} \geq 0 $$ i.e. $$ p^{2} \dfrac{\varphi_{r}(x)-\varphi_{r}(y)}{x-y}+2 p q \dfrac{\varphi_{t}(x)-\varphi_{t}(y)}{x-y}+q^{2} \dfrac{\varphi_{s}(x)-\varphi_{s}(y)}{x-y} \geq 0 . $$ Let $$ \phi(r)=\dfrac{\varphi_{r}(x)-\varphi_{r}(y)}{x-y} $$ then $$ p^{2} \phi(r)+2 p q \phi(t)+q^{2} \phi(s) \geq 0 $$ i.e. $$ \phi^{2}(t) \leq \phi(r) \cdot \phi(s) \quad \text { where } \quad t=\dfrac{r+s}{2} . $$ This implies $\phi$ is log-convex in Jensen sense. Also $\lim _{r \rightarrow 0} \phi(r)=\phi(0)$, which implies $\phi$ is continuous for all $r \in \mathbb{R}$. And therefore $\log$-convex. We need following lemma which proof can be found in [2]. \vspace{4mm} \noindent\textbf{Lemma 2.1.} Let $f$ be log-convex function and if, $x_{1} \leq y_{1}, x_{2} \leq$ $y_{2}, x_{1} \neq x_{2}, y_{1} \neq y_{2}$, then the following inequality is valid: \begin{equation*} \left(\dfrac{f\left(x_{2}\right)}{f\left(x_{1}\right)}\right)^{1 /\left(x_{2}-x_{1}\right)} \leq\left(\dfrac{f\left(y_{2}\right)}{f\left(y_{1}\right)}\right)^{1 /\left(y_{2}-y_{1}\right)} . \tag{2} \end{equation*} Applying Lemma 2.1 for $f=\phi$, (let $r, s, u, v \neq 0)$ we get an inequality $$ \left\{\dfrac{r\left(y^{s}-x^{s}\right)}{s\left(y^{r}-x^{r}\right)}\right\}^{1 /(s-r)} \leq\left\{\dfrac{u\left(y^{v}-x^{v}\right)}{v\left(y^{u}-x^{u}\right)}\right\}^{1 /(v-u)} . $$ Since $E(r, s)$ is continuous, we have (1). \section*{CONCLUSION} In the literature, many researchers have published so many results on different major generalizations of convex function. Many authors today focus on interval-valued functions, which is known as the $(h, m)$-convex interval-valued function. Additionally, we give the rigorous proof of the famous Hermite-Hadamard type inequality for $m$-convex in intervalvalued. \section*{REFERENCES} \begin{enumerate}[{[1]}] \item K. B. Stolarsky, Generilization of the logarithmic mean, Math. Mag. 48 (1975), 87-92. \item J. E. Pečarić, F. Proschan and Y. C. Tong, Convex functions, Partial Orderings and Statistical Applications, Academic Press, New York, 1992. \item J. E. Pečarić, I. Perić and H. M. Srivastava, A family of the Cauchy type meanvalue theorems, J. Math. Anal. Appl. 306 (2005) 730-739. \end{enumerate} \vspace{2mm} 1 Department of Mathematics, University of Mars, Venus Email address: \href{mailto:[email protected]}{[email protected]} \vspace{2mm} 2 School of Advanced Mathematical Sciences, Smith Town, WonDERLAND Email address: \href{mailto:[email protected]}{[email protected]} \end{document}