\documentclass[a4paper,12pt]{article}
\usepackage[english,vietnamese]{babel}
\usepackage{amsmath}
\usepackage{lmodern}
\usepackage{hyperref}
\usepackage{tikz}

\newcommand{\exercise}[1]{\noindent\textbf{#1.}}
\renewcommand{\thesection}{\Roman{section}}
\renewcommand*{\thefootnote}{\fnsymbol{footnote}}

\title{Numerical Method: Labwork 2 Report}
\author{Nguyễn Gia Phong--BI9-184}
\date{Fall 2019}

\begin{document}
\maketitle
\setcounter{section}{2}
\section{Polynomial}
\exercise{1.c} At the time of writing, function \verb|fzero|
in Octave have not support the \verb|Display| option
just yet\footnote{Bug report: \url{https://savannah.gnu.org/bugs/?56954}}.
However, the implementation of this option is rather trivial,
thus I made a quick patch (which is also attached at the bug report).
Using this, one can easily display all the iterations as followed:

\begin{verbatim}
octave:1> fzero (@(x) x.^2 - 9, 0, optimset ('display', 'iter'))

Search for an interval around 0 containing a sign change:
Func-eval 1,  how = initial,  a = 0,  f(a) = -9,  b = 0,  f(b) = -9
Func-eval 2,  how = search,  a = 0,  f(a) = -9,  b = 0.099,  f(b) = -8.9902
Func-eval 3,  how = search,  a = 0,  f(a) = -9,  b = 0.1025,  f(b) = -8.98949
Func-eval 4,  how = search,  a = 0,  f(a) = -9,  b = 0.095,  f(b) = -8.99098
Func-eval 5,  how = search,  a = 0,  f(a) = -9,  b = 0.11,  f(b) = -8.9879
Func-eval 6,  how = search,  a = 0,  f(a) = -9,  b = 0.075,  f(b) = -8.99437
Func-eval 7,  how = search,  a = 0,  f(a) = -9,  b = 0.15,  f(b) = -8.9775
Func-eval 8,  how = search,  a = 0,  f(a) = -9,  b = 0,  f(b) = -9
Func-eval 9,  how = search,  a = 0,  f(a) = -9,  b = 0.35,  f(b) = -8.8775
Func-eval 10,  how = search,  a = 0,  f(a) = -9,  b = -0.4,  f(b) = -8.84
Func-eval 11,  how = search,  a = 0,  f(a) = -9,  b = 1.1,  f(b) = -7.79
Func-eval 12,  how = search,  a = 0,  f(a) = -9,  b = -4.9,  f(b) = 15.01

Search for a a zero in the interval [-4.9, 0]:
Func-eval 13,  how = initial,  x = 0,  f(x) = -9
Func-eval 14,  how = interpolation,  x = -1.83673,  f(x) = -5.62641  (NaN%)
Func-eval 15,  how = interpolation,  x = -3.36837,  f(x) = 2.3459  (141.7%)
Func-eval 16,  how = interpolation,  x = -3.19097,  f(x) = 1.1823  (-49.6%)
Func-eval 17,  how = interpolation,  x = -2.99725,  f(x) = -0.0164972  (-101.4%)
Func-eval 18,  how = interpolation,  x = -3.00258,  f(x) = 0.0154927  (193.9%)
Func-eval 19,  how = interpolation,  x = -3,  f(x) = 3.07975e-07  (-100.0%)
Func-eval 20,  how = interpolation,  x = -3,  f(x) = -7.10543e-15  (-100.0%)
Func-eval 21,  how = interpolation,  x = -3,  f(x) = 5.32907e-15  (169.7%)

Algorithm converged

ans = -3.0000
\end{verbatim}

To answer the question in part b, (since I believe these parts are linked
to each other), the current implementation of \verb|fzero| search for
the second bracket over quantitative chages below if \verb|X0| if it is a
single scalar, thus $[-4.9, 0]$ is gotten and the found solution is negative:

\begin{verbatim}
[-.01 +.025 -.05 +.10 -.25 +.50 -1 +2.5 -5 +10 -50 +100 -500 +1000]
\end{verbatim}

\section{Non-linear Systems}
\exercise{1.a} These statements were used to plot the given functions:
\begin{verbatim}
ezplot(@(x1, x2) x1 .^ 2 + x1 .* x2 - 10)
hold on
ezplot(@(x1, x2) x2 + 3 .* x1 .* x2 .^ 2 - 57)
\end{verbatim}

As shown in the graphs (where $x_1^2 + x_1 x_2 = 10$ are the blue lines
and $x_2 + 3 x_1 x_2 = 57$ are the yellow ones), the solutions of $(x_1, x_2)$
are quite close to $(2, 3)$ and $(4.5, -2)$.

\begin{figure}[!h]
  \centering
  \scalebox{0.37}{\input{2a.tikz}}
\end{figure}

I would also like to note that I am personally impressed how gnuplot
(which is utilised by Octave) is able to export to TikZ graphics with ease.
\end{document}