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diff --git a/texmf-dist/doc/latex/eqexam/examples/hw01.tex b/texmf-dist/doc/latex/eqexam/examples/hw01.tex new file mode 100644 index 00000000..c9489aa2 --- /dev/null +++ b/texmf-dist/doc/latex/eqexam/examples/hw01.tex @@ -0,0 +1,166 @@ +\documentclass[12pt]{article} +\usepackage[fleqn]{amsmath} + +% This set of parameters are used to distribute the assignment to the class (in paper form) +% and for posting on the class web site (for those who missed the class). +% With the pdf option the information contained in the keys below are placed +% in the document info of the PDF document. If you don't have the AcroTeX Bundle +% installed, remove the pdf option. +\usepackage[pdf,forpaper,cfg=hw,nopoints,nosolutions]{eqexam} + +% Note: When using a PDF option like pdf, you need to specify a driver +% that is passed to hyperref, web.sty etc. For example, +% \usepackage[pdf,pdftex,forpaper,cfg=hw,nopoints,nosolutions]{eqexam} + +% This set of parameters are used to publish the solutions on the class web site, if +% desired. +% \usepackage[pdf,forpaper,cfg=hw,pointsonleft,answerkey]{eqexam} + +% Try compiling the file with vspacewithsolns +% % \usepackage[pdf,forpaper,cfg=hw,pointsonleft,vspacewithsolns]{eqexam} + +% Note the use of the myconfigi parameter. This then inputs eqexami.cfg, there I +% have placed some definitions specific to a homework assignment. + +\subject[AC2]{Advanced Calculus II} +\title[HW1]{HW \#1} +\author{Dr.\ D. P. Story} +\date{Spring 2005} +\duedate{01/28/05} +\keywords{Homework due \theduedate} + +\solAtEndFormatting{\eqequesitemsep{3pt}} + + +\begin{document} + +\maketitle + +\begin{exam}{HW} + +\ifanswerkey +\begin{instructions}[Solutions] +Below, please find a set of solutions to this assignment. +\end{instructions} +\else +\begin{instructions}[] +Assignments should be neatly-written, well-organized and concise. +If you miss a class and need to get an assignment, see +\[ + \text{\url{http://www.math.uakron.edu/~dpstory/}} +\] +All class assignments and other announcements will be posted on +this web site. +\end{instructions} +\fi + +\begin{eqComments}[]\S4.3, page 155, in the text\end{eqComments} + +\begin{problem}[4] +Problem 15. Use the definition to prove $f(x) = x^2$ is convex on +$\mathbb{R}$. +\begin{solution} +Let $[c,d]$ be any interval and let $t\in[0,1]$, we need to prove +\begin{equation} + f\bigl( (1-t)c + td \bigr) \le (1-t) f(c) + tf(d)\label{eq0} +\end{equation} +or, +\begin{equation} + \bigl( ( 1-t )c + td \bigr)^2 \le (1-t) c^2 + t d^2\label{eq2} +\end{equation} +We show that the right-side minus the left-side in \eqref{eq2} is nonnegative. Indeed, +\begin{align*} + (1-t) c^2 + &t d^2 - \left( ( 1-t )c + td \right)^2 \\& + = (1-t) c^2 + t d^2 - \left( ( 1-t )^2 c^2 + 2t(1-t)cd + t^2d^2 \right)\\& + = (1-t)[1-(1-t)]c^2 - 2t(1-t)cd + t(1-t)d^2\\& + = t(1-t)c^2 - 2t(1-t)cd + t(1-t)d^2\\& + = t(1-t)( c - d )^2 \ge 0 +\end{align*} +From the first and last lines we have $(1-t) c^2 + t d^2 - \left( ( 1-t )c + td \right)^2\ge0$. This is equivalent +to the desired inequality~\eqref{eq2}. \eqfititin{$\square$} + +\medskip\noindent\textit{Alternate Solution}: +We apply the \textbf{Cauchy-Schwartz Inequality}, page.~16, to the expression on the left side of line~\eqref{eq2}. +For convenience, I paraphrase the \textbf{Cauchy-Schwartz Inequality}: +\[ + \left(\sum_{k=1}^n a_k b_k \right)^2 \le \left(\sum_{k=1}^n a_k^2 \right) \left(\sum_{k=1}^n b_k^2 \right) +\] +Applying this inequality, with $a_1 = \sqrt{1-t}$, $b_1 = +\sqrt{1-t}\,c$, $a_2 = \sqrt{t}$, $b_2 = \sqrt{t}\,d$ (here, +$n=2$, two terms), we obtain, +\begin{align*} +(1-t)^2 c^2 + t^2 d^2 & + \le \left( (\sqrt{1-t})^2 + (\sqrt{t})^2\right)\left((\sqrt{1-t}\,c)^2 + (\sqrt{t}\,d)^2\right)\\& + = (1-t)c^2 + td^2 +\end{align*} +Thus, +\[ + (1-t)^2 c^2 + t^2 d^2 \le (1-t)c^2 + td^2 +\] +which is line~\eqref{eq2}, what we wanted to prove. +\end{solution} +\end{problem} + +\begin{problem}[3] +Problem 18. Prove the sum of two convex functions is convex. +\begin{solution} +Seems simple enough. Suppose $f$ and $g$ be convex on $I$. Let $[\,c,d\,]\subseteq$ and let $t\in[\,0,1\,]$. Then +\begin{align*} + (f+g)\bigl( (1-t) c + td \bigr) & + = f\bigl( (1-t) c + td \bigr) + g\bigl( (1-t) c + td \bigr)\\& + \le (1-t) f(c) + tf(d) + (1-t) g(c) + tg(d)\\& + = (1-t) (f+g)(c) + t(f+g)(d) +\end{align*} +Thus, $(f+g)\bigl( (1-t) c + td \bigr) \le (1-t) (f+g)(c) + t(f+g)(d)$, which is what we wanted to prove. +\end{solution} +\end{problem} + +\begin{problem}[2] +Problem 20. Give an example of a function that is convex and unbounded on $(0,1)$. +\begin{solution} +Let $ f(x) = 1/x $, $ x \in (0,1) $. This function is clearly unbounded and since $ f''(x) = 1/x^3\ge 0$ on $(0,1)$, +it is convex on $(0,1)$. +\end{solution} +\end{problem} + +\begin{problem}[4] +Problem 21. Define +\[ + f(x) = \begin{cases} + 2, & x = -1;\\ + x^2, & -1 < x < 2;\\ + 5, & x = 2 + \end{cases} +\] +Show $f$ is convex on $[\,-1,2\,]$ but not continuous on $[\,-1,2\,]$. +\begin{solution} +Define $g(x) = x^2$, $x\in[\,-1,2\,]$. Then $g$ is twice differentiable on $[\,-1,2\,]$ and $ g''(x) = 2\ge 0$, hence, +$g$ is convex on $[\,-1,2\,]$. Note that $ g(x) \le f(x) $ for all $x\in[\,-1,2\,]$. + +Let $[\,c,d\,]\subseteq [\,-1,2\,]$, we need to show, $\forall t \in [\,0,1\,]$, +$$ + f\bigl( (1-t)c + td \bigr) \le (1-t) f(c) + tf(d) +$$ +This inequality is \emph{always true} for $t=0$ and $t=1$, so it suffices to assume +$t\in(0,1)$, this implies $(1-t)c \ne -1$ and $ td \ne 2$, hence, $(1-t)c + td\in(-1,2)$ . Thus, +$$ + f\bigl( (1-t)c + td \bigr) = g\bigl( (1-t)c + td \bigr)) \le (1-t) g(c) + tg(d) = (1-t) f(c) + tf(d) +$$ +As the assertion about the discontinuity of $f$ (at its endpoints) is obvious, this completes the proof. +\end{solution} +\end{problem} + +\begin{problem}[3] +Problem 23. Suppose $f$ is convex on $\mathbb R$, prove $f$ is continuous on $\mathbb R$. +\begin{solution} +This is an application of \textbf{Theorem~4.28}. Let $x\in\mathbb R$, enclose $x$ in a open +interval $(a,b)$, where $a$, $b\in\mathbb R$. Then $f$ is convex on $(a,b)$, since it is convex +on $\mathbb R$, so by \textbf{Theorem~4.28}, $f$ is continuous on $(a,b)$. Since $f$ is continuous +on $(a,b)$, it is, in particular, continuous at $x\in(a,b)$. + +We have shown that for any $x\in\mathbb R$, $f$ is continuous at +$x$, this means that $f$ is continuous on $\mathbb R$. +\end{solution} +\end{problem} +\end{exam} +\end{document} |