\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}

\title{Brief Article}
\author{The Author}
\begin{document}
\maketitle






\begin{center}
\bf{MATH 120\\HOMEWORK\\Assignment 3\\9-30-02}
\end{center}
\vskip.5in


Let $x_1=1,\  x_2=2,\  x_3 = 2+\varepsilon,\  \smallint(x) = x^2$

Find pol. of degree $1\  p(x) = mx + c$

such that $max \left\{  |f (x_i) -  p(x_i)| : i = 1,2,3, \right\}$ is minimal possible.

\underline{Solution}

$p(x_1) = m + c + B = 1 = f(1)$

$p(x_2) = 2m + c - B = 4 = f(2)$

$p(x_3) - 2m + \varepsilon m+ c + B = (2+\varepsilon)^2 = 4 + 4\varepsilon + \varepsilon^2$

\hskip.5 in  $f(x) = x^2$

$x_1 = 1$\hskip.5in$x_2 = 1 + \varepsilon$

$c + B = 1$

$c - B = (1+\varepsilon)^2$

$2C=1+(1+\varepsilon)^2$

$C=\frac{1 + (1+\varepsilon)^2}{2}$

$2B=1 - (1+\varepsilon)^2$

$\Rightarrow B = \frac{1 - (1+\varepsilon)^2}{2}$

\underline{Problem 1.}

Given $\varepsilon > 0, x_1 = 1, x_2 = 1 + \varepsilon, f(x) = x^2,$ find a constant $C$ such that $max \left\{  |f (x_i) - C|, |f(x_2) - C| \right\}$ is
minimal possible.  What is minimal error?  What happens when $\varepsilon \rightarrow 0?$

\underline{Problem 2.}

Let $a$ be a real parameter.  Consider the following optimization problem:

$$x_1 + x_2 \rightarrow min$$
$$x_1^2 + 3x_2^2 \leqslant a$$

Find the global minimizer for this problem as a function of the parameter $a$.

\underline{Problem 3.}  

Describe normal cone $N_{\mathbb{R}^n_+} (x)$ for any point $x\epsilon\mathbb{R}^1_+$ (positive orthant).  Using this description write down necessary and
sufficient optimality conditions for the problem

\hskip1in $f(x)\rightarrow min$ 

\hskip1in $x\epsilon\mathbb{R}^1_+$,

where $f$ is a convex G$\hat{a}$teaux differentiable function.

\underline{Problem 4.} 

Let ${A,B}$ be convex subsets in $E$ such that $A \cap B=\oslash$.   Prove that there exists $a\epsilon E, a\neq, 0$ such that $$inf <a,x> Let {A,B} be convex
subsets in $E$ such that $A \cap B=\oslash}$ Let ${A,B}$ be convex subsets in $E$ such that $A \cap B=\oslash$$$

 








\end{document} 