\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{graphicx}
\title{How I spent last summer}
\author{William Fayers\\
School of Mathematics and Physics,\\
University of Lincoln}
\date{October 17, 2023}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\begin{document}
\maketitle
\begin{abstract}
In this article, I shall discuss how I spent last summer.
\end{abstract}
\section*{Introduction}
Preparation is described in Section 1, beginning in Section 2, and summer proper in Section 3. Section 4 contains a photo showing a capybara that I saw when I visited the capybara cafe in Tokyo.
\textit{Note: Everything written about in this document is entirely fictional and not intentionally based on any true events.}
\section{Preparation}
Prepare for summer:
\begin{enumerate}
\item Get fit
\begin{enumerate}
\item Jogging
\item Press-ups
\item Learn to swim
\end{enumerate}
\item Buy summer clothes
\item Buy tickets
\end{enumerate}
\section{May}
Apart from recreational activities, in May I also did some maths, as described in Subsection 2.3.
\subsection{May weather}
It was raining \textit{a lot}. This phenomenon is explained in \cite[Section 4]{weather}.
\subsection{Reading books}
Because of what is described in Subsection 2.1, we stayed indoors and read the book \cite{book}.
\subsection{Maths}
I was also revising some maths for A-level exams, like the following.
\begin{theorem}
The solutions of a quadratic equation \( ax^2 + bx + c = 0 \) are given by the formulae
$$
x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a} \quad \text{and} \quad x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}.
$$
\end{theorem}
\begin{lemma}
A finite sum \( \sum_{i=0}^{k} r^i a_0 \) of a geometric progression with ratio \( r \neq 1 \) is equal to
$$
a_0\left(\frac{1-r^{k+1}}{1-r}\right).
$$
\end{lemma}
\begin{proof}
We have
$$
(1-r)(1+r+r^2+\cdots+r^k)=1-r^{k+1},
$$
which is verified by expanding brackets:
\begin{align*}
&(1+r+r^2+\cdots+r^k)-r(1+r+r^2+\cdots+r^k)\\
&\quad=1+r+r^2+\cdots+r^k-r-r^2-\cdots-r^k-r^{k+1}=1-r^{k+1}.
\end{align*}
Dividing both sides of formula (3) by \( r-1 \) and multiplying by \( a_0 \) we have
$$
a_0+r a_0+r^2 a_0+r^3 a_0+\cdots+r^k a_0=a_0\left(\frac{1-r^{k+1}}{1-r}\right).
$$
\end{proof}
\begin{theorem}
The infinite sum \( \sum_{i=0}^{\infty} r^i a_0 \) of a geometric progression with ratio \( r \) satisfying \( 0<r<1 \) is equal to \( \frac{a_0}{1-r} \).
\end{theorem}
\begin{proof}
The sum \( \sum_{i=0}^{\infty} r^i a_0 \) is equal to the limit of the partial sums \( \sum_{i=0}^{k} r^i a_0 \) as \( k \) tends to \( \infty \). By Lemma 1,
$$
\sum_{i=0}^{k} r^i a_0=a_0\left(\frac{1-r^{k+1}}{1-r}\right).
$$
When \( 0<r<1 \), the term \( r^{k+1} \) tends to 0 as \( k \) tends to \( \infty \). Hence the result.
\end{proof}
\section{Summer proper}
After exams, we watched football matches on TV. At the moment the table in group E looks as in Table 1.
\begin{table}[h]
\centering
\caption{Euro 2016. Group E}\label{tab:groupE}
\begin{tabular}{|c|l||c|c|c|c|c|c||c|}
\hline
Pos & Team & Pld & W & D & L & GD & Pts & Qualification \\
\hline\hline
1 & Italy & 3 & 2 & 0 & 1 & +2 & 6 & Qualified \\
\hline
2 & Belgium & 3 & 2 & 0 & 1 & +2 & 6 & Qualified \\
\hline
3 & Republic of Ireland & 3 & 1 & 1 & 1 & 0 & 4 & Round of 16 \\
\hline
4 & Sweden & 3 & 0 & 1 & 2 & -2 & 1 & \\
\hline
\end{tabular}
\end{table}
\section{Photo}
Figure 1 shows the capybara cafe from Tokyo
\begin{figure}[h]
\includegraphics[width=\textwidth]{CapybaraCafe}
\caption{Capybara Cafe}\label{fig:capybara}
\end{figure}
\begin{thebibliography}{9}
\bibitem{weather}
A. Nother, Recent advances in weather prediction, \textit{J. Adv. Weather} \textbf{76}, no. 3 (2013), 23-45.
\bibitem{book}
S. Someone, \textit{A great book}, Lincoln, 2014.
\bibitem{image}
C. Capybara, Cafe Capybara introduction page, \textit{Cafe Capybara}, 2023.
\end{thebibliography}
\end{document}