knitr::opts_chunk$set(echo = FALSE, message=FALSE, warning=FALSE, fig.width=5, fig.height=3, fig.align="center")
Setting problem as a section
\secprob{1} This is a problem
\subprob{(a)} This is a subsection of this problem
\sol
We can also make problem as a theorem
\begin{problem} this is a problem \end{problem}
\begin{exercise} this is a exercise \end{exercise}
\begin{enumerate} \item $T(n) = 2 T(\lfloor n/4 \rfloor) + n^{1/2}$. \item $T(n) = 3 T(\lfloor n/2 \rfloor) + n \lg n$. \item $T(n) = 5 T(\lfloor n/5 \rfloor) + \frac{n}{\lg n}$. \item $T(n) = 4 T(\lfloor n/2 \rfloor) + n^2 \sqrt{n}$. \item $T(n) = 2 T(\lfloor n/2 \rfloor) + n \lg n$. \end{enumerate}
$a = 3, b = 2$ implies a reference function $g(n) = n^{\log_2 3}$. Converting as follows, \begin{eqnarray} y & = & \log_2 3 \ 2^y & = & 3 \ y \ln 2 & = & \ln 3 \ y & = & \frac{\ln 3}{\ln 2} = 1.585, \end{eqnarray} we have $g(n) = n^{1.585}$. The ``glue'' function is $f(n) = n \lg n$. Let $g_\epsilon (n) = n^{1.585 - \epsilon}$, for $0 < \epsilon < 0.5$. Since \begin{eqnarray} \frac{f(n)}{g_\epsilon (n)} & = & \frac{n \lg n}{n^{1.585 - \epsilon}} = \frac{\lg n}{n^{0.585 - \epsilon}} \ & \leq & \frac{\lg n}{n^{0.085}} \rightarrow 0 \end{eqnarray} as $n \rightarrow \infty$, we have $f(n) = o(g_\epsilon (n))$, which implies $f(n) = O(g_\epsilon (n))$ and allows case (1) of the master template. Therefore $T(n) = \Theta(g(n)) = \Theta(n^{1.585})$.
\vspace*{0.5in} \noindent Answers to incidental LaTeX question may be found at: \begin{verbatim} http://www.tug.org/begin.html \end{verbatim}
\begin{table}[H] \centering \caption{"Caption"} \resizebox{0.8\textwidth}{0.4\textheight}{\input{"table.tex"}} \end{table}
\begin{figure} \caption{"Caption"} \begin{center}\includegraphics[width=0.8\textwidth]{Figure1.pdf} \end{center}
{\footnotesize \textit{Notes}: Figure notes ........................} \end{figure}
summary(cars) # command plot(pressure)
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