Document 1 - Basics (1-4)

\documentclass[14pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{ulem}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{multirow}

\title{Document 1 - Basics}
\author{Thomas Boufikos}
\date{September 2021}



\begin{document}

\maketitle
\section{Text Manipulation}
Normal text size. \\
{\LARGE Large text size.} \\
{\LARGE This text contains only one {\LARGE LARGER} word (LARGE in LARGE).} \\
Back to normal size \\
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\textbf{Bold only} \\
\textit{Italic only} \\
\underline{Underline only} \\
\textbf{\textit{Bold and Italic}} \\
\underline{\textit{Italic and Underline}} \\
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First one with identation.

\noindent Second one without.

Third one with.

\noindent Fourth one without



\section{Math manipulation} \\

There are 2 ways for this: a) display mode and b) inline mode. \\ \\ \\
\underline{{\LARGE a) Display math mode}} \\
Notice that by substitution we have the following equation:
\[
f(x) = a_2x^2 + a_1x + a_0 \]
\[
     = 4x^2 + 4x + 1
\]
By completing the squares, we have:
\[
f(x) = (2x+1)^2
\]
\\ \\
\underline{{\LARGE Second way with align}} \\ \\
Notice that by substitution we have the following equation: \\
\begin{align}
  f(x) & = a_2x^2 + a_1x + a_0 \\
       & = 4x^2 + 4x + 1
\end{align}
\\ \\
\underline{\LARGE Align* method with 3 equations (ams math)}}
\\ \\
\begin{align*}
    2x + 1 & = 9  &  3y + 5 & = 11  &  4z - 6 & = 24 \\
        2x & = 8  &      3y & = 6   &      4z & = 28 \\
         x & = 4  &       y & = 2   &       z & = 7 \\
\end{align*}
\\ \\ \\ \\
\underline{{\LARGE b) Inline math mode}} \\

Notice that by substitution we have the following equation: \( f(x) =  a_2x^2 + a_1x + a_0 = 4x^2 + 4x + 1 \). That means we can also complete the square by this way: $ f(x) = (2x+1)^2 $. These are the 2 solutions. \\

The area of a circle is:
\[ E = \pi \cdot r^2\] \\
where \( E \) is the area of the circle, \( \pi \) = 3.14159.... and $ r $ is the radius.
\\ \\
\underline{{\LARGE Math methods}} \\ \\
We know that Euler proved that (display math mode):
\[
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\]
We know that (inline math mode): \( \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \)
\\ \\ \\
\underline{{\LARGE Basic Arithmetic operations}} \\ \\
2 + 3 = 5 \\
4 - 1 = 3 \\
5 \cdot 4 = 20 \\
12 \times 4 = 48 \\
25 \div 5 = 5 \\
8 \cdot \frac{3}{4} = 6 \\

Here we are! We will make the following prove:
\[
f(x) = (a_1^{\frac{1}{b_1}} \cdot a_2^{\frac{1}{b_2}}) ^ {b_1b_2x} = (a_1^{b_2}a_2^{b_1})^x
\]
Nice done! Let's move on the next task.
If $ a $ belongs to $\mathbb{R}$ and $   a > 1 $ , we have the following:
\[
\left( \sum_{n=0}^\infty \left( \frac{1}{a} \right) \right)^2 = \left( \sum_{n=0}^\infty \left( a^{-1} \right) \right)^2 = \left( \frac{1}{1-a^{-1}} \right)^2 = \frac{1}{1-2a+a^2}
\]
\\ \\ \\
\section{Tables}
\begin{tabular}{l|c|r}
    \hline
    1 & 2 & 3 \\
    \hline
     Element 1 & Room 2 & $a = 3$ \\
     \hline
     4 & 5 & 6 \\
     \hline
     5 \cdot 4 &  $ \mathbb{C} $ & $ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}$\\
     \hline

\end{tabular}
\\ \\ \\
\begin{tabular}{|c|c||c|c|}
    \hline
    Distance(km) & Velocity(km/hr) & Time \\
    \hline
    10 & 5 & 2 hours = 120 minutes \\
    \hline
    20 & 80 & 0.25 hrs = 15 minutes \\
    \hline
    1 & 60 & $\frac{1}{60}$ hrs = 60 seconds \\
    \hline

\end{tabular}
\\ \\ \\
\begin{tabular}{|c|c|c|}
    \hline
    X & & \\
    \hline
    O & X & \\
    \hline
    O & O & X \\
    \hline
\end{tabular}
\\ \\ \\
\begin{tabular}{|l|c|c|r|}
     \hline
     \multicolumn{3}{|r|}{Right 3 columns} & Right a_{14} \\
     \hline
     \multirow{2}{*}{Top 2} & Center a_{22} & Center a_{23} & Also Right a_{24} \\
      & Center a_{32} & Center a_{33} & Also Right a_{34} \\
     \hline
     \multicolumn{2}{|l|}{Left a_{41}, a_{42}} & Center a_{43} & Right a_{44} \\
     \hline
\end{tabular}
\\ \\ \\
\section{Arrays} \\
\underline{\LARGE Arrays must be in math mode}
\\

$
\begin{array}{ccc}
     a_{11} & a_{12} & a_{13} \\
     a_{21} & a_{22} & a_{23} \\
\end{array}
$
\\ \\ \\
$
\begin{array}{c_cc}
     a_{11} & a_{12} & a_{13} \\
     \hline
     a_{21} & a_{22} & a_{23} \\
\end{array}
$
\\ \\

There are 2 ways for matrix creation: a) array with left parenthesis and b) pmatrix or bmatrix or vmatrix. \\ \\ \\
\underline{{\LARGE a) With parenthesis}} \\ \\
$
\left(
\begin{array}{cccc}
    a_{11} & a_{12} & a_{13} & a_{14} \\
    a_{21} & a_{22} & a_{23} & a_{24} \\
    a_{31} & a_{32} & a_{33} & a_{34} \\
\end{array}
\right)
$
\\ \\ \\
\underline{{\LARGE b) With pmatrix}} \\ \\
$
\begin{pmatrix}
    a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\
    a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\
    a_{31} & a_{32} & a_{33} & a_{34} & a_{35} \\
\end{pmatrix}
$
\\ \\ \\ \\
\underline{{\LARGE Matrices multiplication}} \\ \\
$
\begin{pmatrix}
    a_{11} & a_{12} & a_{13} & a_{14} \\
    a_{21} & a_{22} & a_{23} & a_{24} \\
    a_{31} & a_{32} & a_{33} & a_{34} \\
\end{pmatrix}
\cdot
\begin{pmatrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
    a_{11} & a_{12} & a_{13} & a_{14} \\
    a_{21} & a_{22} & a_{23} & a_{24} \\
    a_{31} & a_{32} & a_{33} & a_{34} \\
\end{pmatrix}
$
\\ \\ \\
\underline{{\LARGE Dots}} \\ \\
$
\begin{pmatrix}
    a_{11} & a_{12} & \cdots \\
    a_{21} & a_{22} & \cdots \\
    \vdots & \vdots & \ddots \\
\end{pmatrix}
$

\end{document}