fix in powerprove explanation

.ipynb_checkpoints/examples-checkpoint.ipynb

@@ -2558,7 +2558,9 @@
     "newproof()\n",
     "#this is equivalent to\n",
     "#prove(makesubs('(x-1)^4','[1,oo]'))\n",
-    "#prove(makesubs('(x-1)^4','[-oo,1]'))\n",
+    "#prove(makesubs('(x-1)^4','[1,0]')) \n",
+    "#you can write ends of interval in any order, so [1,0] is the same as [0,1]\n",
+    "#but the substitution is slightly simpler when 0 is on the right side\n",
     "powerprove('(x-1)^4')"
    ]
   },
@@ -3408,16 +3410,16 @@
    ],
    "source": [
     "newproof()\n",
-    "formula=Sm('-(3a + 2b + c)(2a^3 + 3b^2 + 6c + 1) + (4a + 4b + 4c)(a^4 + b^3 + c^2 + 3)')\n",
+    "formula=Sm('-(3a+2b+c)(2a^3+3b^2+6c+1)+(4a+4b+4c)(a^4+b^3+c^2+3)')\n",
     "#this is equivalent to\n",
     "#prove(makesubs(formula,'[1,oo],[1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[-1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[-1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[-1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[-1,oo],[-1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,oo],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,0],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,0],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,oo],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,oo],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,0],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,0],[1,0]'))\n",
     "powerprove(formula)"
    ]
   },

examples.ipynb

@@ -2558,7 +2558,9 @@
     "newproof()\n",
     "#this is equivalent to\n",
     "#prove(makesubs('(x-1)^4','[1,oo]'))\n",
-    "#prove(makesubs('(x-1)^4','[-oo,1]'))\n",
+    "#prove(makesubs('(x-1)^4','[1,0]')) \n",
+    "#you can write ends of interval in any order, so [1,0] is the same as [0,1]\n",
+    "#but the substitution is slightly simpler when 0 is on the right side\n",
     "powerprove('(x-1)^4')"
    ]
   },
@@ -3408,16 +3410,16 @@
    ],
    "source": [
     "newproof()\n",
-    "formula=Sm('-(3a + 2b + c)(2a^3 + 3b^2 + 6c + 1) + (4a + 4b + 4c)(a^4 + b^3 + c^2 + 3)')\n",
+    "formula=Sm('-(3a+2b+c)(2a^3+3b^2+6c+1)+(4a+4b+4c)(a^4+b^3+c^2+3)')\n",
     "#this is equivalent to\n",
     "#prove(makesubs(formula,'[1,oo],[1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[-1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[-1,oo],[1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[1,oo],[-1,oo],[-1,oo]'))\n",
-    "#prove(makesubs(formula,'[-1,oo],[-1,oo],[-1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,oo],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,0],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,0],[1,oo]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,oo],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,oo],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,oo],[1,0],[1,0]'))\n",
+    "#prove(makesubs(formula,'[1,0],[1,0],[1,0]'))\n",
     "powerprove(formula)"
    ]
   },

prove.tex

@@ -0,0 +1,164 @@
+\documentclass{beamer}
+\usepackage[utf8]{inputenc}
+%\usepackage{polski}
+\newtheorem{twierdzenie}{Twierdzenie}
+\newtheorem{definicja}{Definicja}
+\usepackage{tikz}
+\usepackage{pgfplots}
+\usepackage{datapie}
+\usetikzlibrary{patterns}
+\usetikzlibrary{arrows}
+\usetikzlibrary{positioning}
+\usetikzlibrary{decorations.pathmorphing}
+\usetikzlibrary{decorations.pathreplacing}
+\usetikzlibrary{decorations.text}
+\usetikzlibrary{decorations.footprints}
+\usetikzlibrary{shapes.symbols}
+\usetikzlibrary{shapes.arrows}
+\pgfplotsset{compat=1.10}
+\DeclareMathOperator{\Hol}{Hol}
+\DeclareMathOperator{\LI}{LI}
+\DeclareMathOperator{\sgn}{sgn}
+\DeclareMathOperator{\ord}{ord}
+\DeclareMathOperator{\Cl}{Cl}
+\global\long\def\BC{\mathbb{C}}
+\global\long\def\BNZ{\mathbb{N}_{0}}
+\global\long\def\NPI{\mathbb{Z}\setminus\mathbb{N}}
+\global\long\def\BN{\mathbb{N}}
+\global\long\def\BZ{\mathbb{Z}}
+\global\long\def\BQ{\mathbb{Q}}
+\global\long\def\BR{\mathbb{R}}
+\global\long\def\BM{\mathbb{M}}
+\global\long\def\BP{\mathbb{P}}
+\global\long\def\BE{\mathbb{E}}
+\newcommand{\w}{\overline{w}}
+\usepackage{ragged2e}
+\newenvironment{tikzexample}[1]{\def\temp{#1}\centering}{\par\bigskip\temp\newpage}
+\newcommand{\nextexample}{\par\vspace{24pt}}
+\apptocmd{\frame}{}{\justifying}{}
+\makeatletter
+\newcommand{\cs}[1]{\texttt{\@backslashchar #1}}
+\makeatother
+
+% \usetheme{Frankfurt}
+ \usetheme{Warsaw}
+% \usetheme{Madrid}
+% \usetheme{Bergen}
+% \usetheme{Antibes}
+% \usetheme{Montpellier}
+% \usetheme{Berkeley}
+% \usetheme{Ilmenau}
+% \usetheme{Singapore}
+
+% \usecolortheme{crane}
+
+\AtBeginSection[] % Do nothing for \section*
+{
+\begin{frame}<beamer>
+  \frametitle{Spis treści}
+  \tableofcontents[currentsection]
+\end{frame}
+}
+\begin{document}
+
+\title{Shiroin package - function \textit{prove}}
+\author{Grzegorz Adamski}
+\date{\today}
+
+\begin{frame}
+  \maketitle  
+\end{frame}
+\begin{frame}{Jensen and weighted AM-GM inequalities}\begin{block}{Jensen inequality}
+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 
+$$\frac{\sum_{i=1}^n w_i f(v_i)}{w}\ge f\left(\frac{\sum_{i=1}^n v_i}{w} \right).$$
+\end{block}
+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
+$$\frac{\sum_{i=1}^n w_i x_i}{w}\ge \sqrt[w]{\prod_{i=1}^n x_i^{w_i}}. $$
+This is called inequality of weighted arithmetic and geometric means (AM-GM inequality for short).
+We will use an equivalent inequality
+$$\sum_{i=1}^n w_i x_i\ge w\prod_{i=1}^n x_i^{w_i/w}. $$
+\end{frame}
+\begin{frame}{Example}
+\begin{block}{}
+Let $a,b>0$. Prove that
+$$30a^2b^2+60ab^4\le 48a^3+56b^6.$$
+\end{block}
+\pause
+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
+$$30a^2b^2\le w_1a^3+w_2b^6$$
+$$60ab^4\le w_3 a^3+w_4 b^6$$
+To fulfill assumptions of AM-GM, we need:
+$w_1,w_2,w_3,w_4\ge 0$
+$$w_1+w_2=30$$
+$$w_3+w_4=60$$
+$$(a^3)^{w_1/30}(b^6)^{w_2/30}=a^2b^2$$
+$$(a^3)^{w_3/30}(b^6)^{w_4/30}=ab^4$$
+We also need $w_1+w_3\le 48$ and $w_2+w_4\le 56$
+\end{frame}
+\begin{frame}{Example}
+$$w_1,w_2,w_3,w_4\ge 0$$
+$$w_1+w_2=30$$
+$$w_3+w_4=60$$
+$$3w_1=30\cdot 2$$
+$$6w_2=30\cdot 2$$
+$$3w_3=30\cdot 1$$
+$$6w_4=30\cdot 4$$
+$$w_1+w_3\le 48$$
+$$w_2+w_4\le 56$$
+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).
+\end{frame}
+\begin{frame}{Problem}
+\begin{block}{}
+Prove that for all $x>0$ $$4x^2 \le 4x+x^3.$$
+\end{block}
+This inequality follows directly from AM-GM. But we can't use the same algorithm as in the previous problem.
+
+Let's try to do this. We want to find $w_1,w_2\ge 0$ such that
+$$4x^2\le w_1x+w_2x^3$$
+$$w_1+w_2=4$$
+$$1w_1+3w_2=4\cdot 2$$
+$$w_1\le 4$$
+$$w_2\le 1$$
+From the equations we can find that $w_1=w_2=2$. But $w_2\le 1$, so there is no solution.
+\end{frame}
+\begin{frame}{Substitutions}
+This problem can be solved by substituting $x$ by $2y$. This way the problem can be reformulated.
+\begin{block}{}
+Prove that for all $y>0$ $$16y^2 \le 8x+8x^3.$$
+\end{block}
+This problem can be solved using the main algorithm.
+\end{frame}
+\begin{frame}{Dealing with real coefficients in the proof}
+\begin{block}{}
+Prove that for all $x>0$ $$4x^2 \le 1+2x+3x^3+4x^4.$$
+\end{block}
+This time the biggest problem is not finding a solution, but finding one with nice coefficients.
+For example, using weighted AM-GM  we can show that
+$$4x^2\le 2x+2x^3$$
+or 
+$$4x^2\le 1+x+x^3+x^4.$$
+but we can also find a less nice solution like
+$$4x^2\le 0.657238842+0.342761158x+0.342761158x^3+0.657238842x^4.$$
+\end{frame}
+\begin{frame}{Dealing with real coefficients in the proof}
+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.
+\end{frame}
+\begin{frame}{Set of solutions}
+\begin{center}
+\begin{figure}
+\includegraphics[scale=2]{solset2.png}
+
+\caption{Set of solutions projected onto the plane $w_3w_4$. $w_1,w_2$ are defined as $w_1=2w_3+3w_4-4$, $w_2=8-3w_3-4w_4$}
+\end{figure}
+\end{center}
+\end{frame}
+\begin{frame}{Function \textit{prove} - algorithm}
+\begin{enumerate}
+    \item Find a proof with real coefficients.\label{itm:first}
+    \item Check which inequalities have integer coefficients.
+    \item Subtract them from original inequality and go back to \ref{itm:first} with the new inequality.
+\end{enumerate}
+The loop breaks when all coefficients are integer or when it has run fixed amount of times.
+\end{frame}
+\end{document}
+