165 lines
5.4 KiB
TeX
165 lines
5.4 KiB
TeX
\documentclass{beamer}
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\usepackage[utf8]{inputenc}
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%\usepackage{polski}
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\newtheorem{twierdzenie}{Twierdzenie}
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\newtheorem{definicja}{Definicja}
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\usepackage{tikz}
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\usepackage{pgfplots}
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\usepackage{datapie}
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\usetikzlibrary{patterns}
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\usetikzlibrary{arrows}
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\usetikzlibrary{positioning}
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\usetikzlibrary{decorations.pathmorphing}
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\usetikzlibrary{decorations.pathreplacing}
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\usetikzlibrary{decorations.text}
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\usetikzlibrary{decorations.footprints}
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\usetikzlibrary{shapes.symbols}
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\usetikzlibrary{shapes.arrows}
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\pgfplotsset{compat=1.10}
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\DeclareMathOperator{\Hol}{Hol}
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\DeclareMathOperator{\LI}{LI}
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\DeclareMathOperator{\sgn}{sgn}
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\DeclareMathOperator{\ord}{ord}
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\DeclareMathOperator{\Cl}{Cl}
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\global\long\def\BC{\mathbb{C}}
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\global\long\def\BNZ{\mathbb{N}_{0}}
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\global\long\def\NPI{\mathbb{Z}\setminus\mathbb{N}}
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\global\long\def\BN{\mathbb{N}}
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\global\long\def\BZ{\mathbb{Z}}
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\global\long\def\BQ{\mathbb{Q}}
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\global\long\def\BR{\mathbb{R}}
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\global\long\def\BM{\mathbb{M}}
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\global\long\def\BP{\mathbb{P}}
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\global\long\def\BE{\mathbb{E}}
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\newcommand{\w}{\overline{w}}
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\usepackage{ragged2e}
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\newenvironment{tikzexample}[1]{\def\temp{#1}\centering}{\par\bigskip\temp\newpage}
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\newcommand{\nextexample}{\par\vspace{24pt}}
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\apptocmd{\frame}{}{\justifying}{}
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\makeatletter
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\newcommand{\cs}[1]{\texttt{\@backslashchar #1}}
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\makeatother
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% \usetheme{Frankfurt}
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\usetheme{Warsaw}
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% \usetheme{Madrid}
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% \usetheme{Bergen}
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% \usetheme{Antibes}
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% \usetheme{Montpellier}
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% \usetheme{Berkeley}
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% \usetheme{Ilmenau}
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% \usetheme{Singapore}
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% \usecolortheme{crane}
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\AtBeginSection[] % Do nothing for \section*
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{
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\begin{frame}<beamer>
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\frametitle{Spis treści}
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\tableofcontents[currentsection]
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\end{frame}
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}
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\begin{document}
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\title{Shiroin package - function \textit{prove}}
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\author{Grzegorz Adamski}
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\date{\today}
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\begin{frame}
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\maketitle
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\end{frame}
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\begin{frame}{Jensen and weighted AM-GM inequalities}\begin{block}{Jensen inequality}
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If $f:\BR^k\to \BR$ is a convex function, $v_1,...,v_n\in \BR^k$, $w_1,...,w_n>0$ and $w=\sum_{i=1}^n w_i$, then
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$$\frac{\sum_{i=1}^n w_i f(v_i)}{w}\ge f\left(\frac{\sum_{i=1}^n v_i}{w} \right).$$
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\end{block}
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If $f(w_1,w_2,...,w_n)=\prod_{i=1}^n x_i^{w_i}$ for some $x_1,...,x_n>0$, then we've got
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$$\frac{\sum_{i=1}^n w_i x_i}{w}\ge \sqrt[w]{\prod_{i=1}^n x_i^{w_i}}. $$
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This is called inequality of weighted arithmetic and geometric means (AM-GM inequality for short).
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We will use an equivalent inequality
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$$\sum_{i=1}^n w_i x_i\ge w\prod_{i=1}^n x_i^{w_i/w}. $$
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\end{frame}
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\begin{frame}{Example}
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\begin{block}{}
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Let $a,b>0$. Prove that
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$$30a^2b^2+60ab^4\le 48a^3+56b^6.$$
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\end{block}
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\pause
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We will use AM-GM inequality. Let $x_1=a^3$, $x_2=b^6$. We need to find such $w_1,w_2,w_3,w_4$, that
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$$30a^2b^2\le w_1a^3+w_2b^6$$
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$$60ab^4\le w_3 a^3+w_4 b^6$$
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To fulfill assumptions of AM-GM, we need:
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$w_1,w_2,w_3,w_4\ge 0$
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$$w_1+w_2=30$$
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$$w_3+w_4=60$$
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$$(a^3)^{w_1/30}(b^6)^{w_2/30}=a^2b^2$$
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$$(a^3)^{w_3/30}(b^6)^{w_4/30}=ab^4$$
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We also need $w_1+w_3\le 48$ and $w_2+w_4\le 56$
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\end{frame}
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\begin{frame}{Example}
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$$w_1,w_2,w_3,w_4\ge 0$$
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$$w_1+w_2=30$$
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$$w_3+w_4=60$$
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$$3w_1=30\cdot 2$$
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$$6w_2=30\cdot 2$$
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$$3w_3=30\cdot 1$$
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$$6w_4=30\cdot 4$$
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$$w_1+w_3\le 48$$
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$$w_2+w_4\le 56$$
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All these equations and inequalities are linear, so $w_1,w_2,w_3,w_4$ can be found using linear programming (in this case they can be found directly from equations, but this is not true in general case).
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\end{frame}
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\begin{frame}{Problem}
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\begin{block}{}
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Prove that for all $x>0$ $$4x^2 \le 4x+x^3.$$
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\end{block}
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This inequality follows directly from AM-GM. But we can't use the same algorithm as in the previous problem.
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Let's try to do this. We want to find $w_1,w_2\ge 0$ such that
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$$4x^2\le w_1x+w_2x^3$$
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$$w_1+w_2=4$$
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$$1w_1+3w_2=4\cdot 2$$
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$$w_1\le 4$$
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$$w_2\le 1$$
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From the equations we can find that $w_1=w_2=2$. But $w_2\le 1$, so there is no solution.
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\end{frame}
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\begin{frame}{Substitutions}
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This problem can be solved by substituting $x$ by $2y$. This way the problem can be reformulated.
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\begin{block}{}
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Prove that for all $y>0$ $$16y^2 \le 8x+8x^3.$$
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\end{block}
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This problem can be solved using the main algorithm.
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\end{frame}
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\begin{frame}{Dealing with real coefficients in the proof}
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\begin{block}{}
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Prove that for all $x>0$ $$4x^2 \le 1+2x+3x^3+4x^4.$$
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\end{block}
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This time the biggest problem is not finding a solution, but finding one with nice coefficients.
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For example, using weighted AM-GM we can show that
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$$4x^2\le 2x+2x^3$$
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or
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$$4x^2\le 1+x+x^3+x^4.$$
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but we can also find a less nice solution like
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$$4x^2\le 0.657238842+0.342761158x+0.342761158x^3+0.657238842x^4.$$
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\end{frame}
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\begin{frame}{Dealing with real coefficients in the proof}
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The simplest idea is to specify a goal function with random coefficients. Algorithm choose then (almost surely) an extreme point from the set of feasible solutions.
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\end{frame}
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\begin{frame}{Set of solutions}
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\begin{center}
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\begin{figure}
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\includegraphics[scale=2]{solset2.png}
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\caption{Set of solutions projected onto the plane $w_3w_4$. $w_1,w_2$ are defined as $w_1=2w_3+3w_4-4$, $w_2=8-3w_3-4w_4$}
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\end{figure}
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\end{center}
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\end{frame}
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\begin{frame}{Function \textit{prove} - algorithm}
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\begin{enumerate}
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\item Find a proof with real coefficients.\label{itm:first}
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\item Check which inequalities have integer coefficients.
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\item Subtract them from original inequality and go back to \ref{itm:first} with the new inequality.
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\end{enumerate}
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The loop breaks when all coefficients are integer or when it has run fixed amount of times.
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\end{frame}
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\end{document}
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