summaryrefslogtreecommitdiff
path: root/texmf-dist/doc/latex/eqexam/examples/hw01.tex
diff options
context:
space:
mode:
Diffstat (limited to 'texmf-dist/doc/latex/eqexam/examples/hw01.tex')
-rw-r--r--texmf-dist/doc/latex/eqexam/examples/hw01.tex166
1 files changed, 166 insertions, 0 deletions
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}