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\begin{document}
\sffamily
\begin{center}
\textbf{
\begin{tabular}{ p{5cm} p{9cm} p{2cm} }
Vryan Gil Palma
& 01-00781 & Math 171 \\
Jane Doe
& 01-00000 & Project 1 \\ \hline
\end{tabular}
}
\end{center}
\rmfamily
\begin{center}
\Large \textbf{Just An Example of a Latex File}
\end{center}
\normalsize
\begin{enumerate}
\item The following text is just for demonstrating encoding in Latex. Recall that we assumed that the
objective function $\varphi$ has the form
\begin{equation*}
\dfrac{\partial u}{\partial
t^{*}}=\dfrac{1}{(r^{*})^{2}}\dfrac{\partial}{\partial
r^{*}}\left((r^{*})^{2}\dfrac{\partial u}{\partial r^{*}}\right).
\end{equation*}
Suppose $\varphi = (0,1) \times (0,1)$. Then
\begin{eqnarray*} \label{eq:normgrad }
\left|\left| g_t \right|\right|_{L^2(\Omega \times (0,T))} & = &
\left( \int_{ \Omega \times (0,T))} \left| g \right|^2 \right)^\frac{1}{2} \\
&=&
\left( \int_{ \Omega \times (0,T))} \left[ g^{(x)2} + g^{(y)2} \right] \right)^\frac{1}{2} \\
&=&
\left( \sum_{t=1}^T \frac{1}{mn} \sum_{i=2}^m\sum_{j=2}^n \left[ g^{(x)2} + g^{(y)2} \right]
\right)^\frac{1}{2} .
\end{eqnarray*}
\item We have here an example of a matrix in Latex. We observe
\begin{center}
$\left[ \begin{array}{c c c c c c}
1+2r & -r \\
-r & 1+2r & -r &
& \\
& -r & 1+2r & -r \\
& & -r & 1+2r & -r\\
&
& &-r
&1+2r
\end{array} \right]
\left[ \begin{array}{c }
u_{1,0} \\
u_{1,1} \\
u_{1,2} \\
u_{1,M-2}\\
u_{1,M-1}
\end{array}\right]
\=\
\left[ \begin{array}{c }
u_{0,1} \\
u_{0,2} \\
u_{0,3} \\
u_{0,M-2}\\
u_{0,M-1}
\end{array} \right].$
\end{center}
\item We can make tables in Latex. The codes gave us the following results found in Table \ref{table:
error}
\begin{table}[!h]
\begin{center}
\begin{tabular}{|c|c|c|c|c| }
\hline $\Delta x $
& error & $p$ & $C$ \\
\hline
\hline$0.1*(\frac{1}{2})^{0}$ & 0.00045188320 & & \\
\hline$0.1*(\frac{1}{2})^{1}$ & 3.9468e+076 & &\\
\hline$0.1*(\frac{1}{2})^{2}$ & 4.2362e+187 &-368.836 &NaN\\
\hline$0.1*(\frac{1}{2})^{3}$ & 6.1828e+287 &-332.738 &NaN \\
\hline
\end{tabular}
\end{center}
\caption{Estimates of the accuracy order $p$ in the grid size
$\Delta x $ with $\vartheta = 0$ showing that the convergence order
is $p = -333$.}
\label{table: error}
\end{table}
\item A program for computing
$\left[ \begin{array}{c}
x \\
y\\
z
\end{array}\right]$
in the nonlinear system
$\left\{\begin{array}{ccc}
x^2-x+y^2+z^2-5 & =&0 \\
x^2+y^2-y+z^2-4 & =&0 \\
x^2+y^2+z^2+z-6 & =&0
\end{array}\right.$.
Codes should be written as using the verbatim command. \\
\begin{verbatim}
clear
epsilon = 10^(-10);
z = [-1; 0.5; 1]
s = 1000;
function F = fp(z)
// using fixed point
F(1) = (z(1)^2 - z(1) + z(2)^2 + z(3)^2 - 5 + s*z(1))/s
F(3) = sqrt(0.25 - z(1)^2 - z(2)^2 + 6 ) - 0.5
F(2) = sqrt((z(1)^2 + z(2)^2 - z(2) + z(3)^2 - 4 + s*z(2)^2)/s)
endfunction
for i = 1:100000
c = fp(z);
errorr = norm(c-z);
if errorr < 10^(-10)
ans_using_fp = c
no_of_it = i
break
end
z = c;
end
\end{verbatim}
\item To test and verify the results, we aim to find information regarding the shape of $\varphi$ on a
certain neighborhood containing the stationary point outputted by our implementation. This is shown in
Figure \ref{figure: example}.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.60\textwidth]{examplefigure.jpg}
\end{center}
\caption{The shape of $\varphi$. The minimum value of $\varphi$ on this grid is 8.5945 obtained at
$\alpha^{(x)}=0.0485$ and $\alpha^{(y)}=0$. The minimum value of $||\varphi||$ is 37.184. }
\label{figure: example}
\end{figure}
\end{enumerate}
\end{document}