ProofOfCalculusTheoremUsedInTheLagrangeMethod 2003-03-04 12:29:12 2004-03-23 10:35:18 Proof Lagrange multiplier method 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $f(\mathbf{x})$ and $g_i(\mathbf{x}), i=0,{\ldots},m$ be differentiable scalar functions; $\mathbf{x} \in R^n$. We will find local extremes of the function $f(\mathbf{x})$ where $\nabla f=0$. This can be proved by contradiction: \[ \nabla f \neq 0 \] \[ \Leftrightarrow \exists \epsilon_0 > 0, \forall \epsilon; 0<\epsilon<\epsilon_0: f(\mathbf{x}-\epsilon \nabla f) < f(\mathbf{x}) < f(\mathbf{x+\epsilon \nabla f}) \] but then $f(\mathbf{x})$ is not a local extreme. Now we put up some conditions, such that we should find the $\mathbf{x} \in S \subset R^n$ that gives a local extreme of $f$. Let $S=\bigcap_{i=1}^m S_i$, and let $S_i$ be defined so that $g_i(\mathbf{x})=0 \forall \mathbf{x} \in S_i$. Any vector $\mathbf{x} \in R^n$ can have one component perpendicular to the subset $S_i$ (for visualization, think $n=3$ and let $S_i$ be a flat surface). $\nabla g_i$ will be perpendicular to $S_i$, because: \[ \exists \epsilon_0>0, \forall \epsilon; 0<\epsilon<\epsilon_0: g_i(\mathbf{x}-\epsilon \nabla g_i)<g_i(\mathbf{x})< g_i(\mathbf{x}+\epsilon \nabla g_i) \] But $g_i(\mathbf{x})=0$, so any vector $\mathbf{x}+\epsilon \nabla g_i$ must be outside $S_i$, and also outside $S$. (todo: I have proved that there might exist a component perpendicular to each subset $S_i$, but not that there exists only one; this should be done) %We define the local maximum of $f(\mathbf{x})$ within $S$ to be all values such %that %\[ \exists \epsilon_0, \forall \mathbf{x_0} \in S; |\mathbf{x}-\mathbf{x_0}|<\epsilon_0: f(\mathbf{x_0})<f(\mathbf{x}) \] %and the local minimum similarly. By the argument above, $\nabla f$ must be zero - but now we can ignore all components of $\nabla f$ perpendicular to $S$. (todo: this should be expressed more formally and proved) So we will have a local extreme within $S_i$ if there exists a $\lambda_i$ such that \[ \nabla f = \lambda_i \nabla g_i \] We will have local extreme(s) within $S$ where there exists a set $\lambda_i, i=1,{\ldots},m$ such that \[ \nabla f = \sum \lambda_i \nabla g_i \]