I'm working on putting together multiple different articles and I want them to be put together you know like normal articles would, the only problem is I'm trying to do it on one Latex document and I'm struggling with titles and what not.
Maybe there's an easy fix I'm missing but I'm really struggling. Here's what I have so far.
\documentclass{book}
\usepackage[svgnames]{xcolor}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{titling}
\usepackage{blindtext}
\usepackage{amssymb}
\usepackage{amsthm}
%----------------------------------------------------------------------------------------
% TITLE PAGE
%----------------------------------------------------------------------------------------
\newcommand*{\titleAT}{\begingroup % Create the command for including the title page in the document
\newlength{\drop} % Command for generating a specific amount of whitespace
\drop=0.1\textheight % Define the command as 10% of the total text height
\rule{\textwidth}{1pt}\par % Thick horizontal line
\vspace{2pt}\vspace{-\baselineskip} % Whitespace between lines
\rule{\textwidth}{0.4pt}\par % Thin horizontal line
\vspace{\drop} % Whitespace between the top lines and title
\centering % Center all text
\textcolor{Black}{ % Red font color
{\Huge High School Journal}\\[0.5\baselineskip] % Title line 1
{\Large OF}\\[0.75\baselineskip] % Title line 2
{\Huge \ Mathematics}} % Title line 3
\vspace{0.25\drop} % Whitespace between the title and short horizontal line
\rule{0.3\textwidth}{0.4pt}\par % Short horizontal line under the title
\vspace{\drop} % Whitespace between the thin horizontal line and the author name
{\Large \textsc{Vol. 1, June 1, 2017}}\par % Author name
\vspace*{\drop} % Whitespace under the publisher text
\rule{\textwidth}{0.4pt}\par % Thin horizontal line
\vspace{2pt}\vspace{-\baselineskip} % Whitespace between lines
\rule{\textwidth}{1pt}\par % Thick horizontal line
\endgroup}
%----------------------------------------------------------------------------------------
% BLANK DOCUMENT
%----------------------------------------------------------------------------------------
\begin{document}
\pagestyle{empty}
\titleAT % This command includes the title page
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem*{remark}{Remark}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\vspace{100mm}
\tableofcontents
\newpage
HS JOURNAL MATH FELLOWSHIP:
We at the High School Journal of Mathematics are very excited to announce the establishment of the first official High School Journal of Mathematics Fellowship. This Fellowship gives five high school students from around the world the opportunity to undertake high quality mathematical research under the tutelage of professional mathematicians or grad students in math.
This fellowship allows high school students to spend seven months studying, conversing with their mentors, and creating new mathematics which will eventually culminate in a final paper which will be published in the High School Journal of Mathematics.
The fellowship begins on October 1st with the research projects to be submitted on May 1st of the following year.
The advantages of this program are many. Firstly, it allows students to conduct mathematical research at an advanced age and prepares them for mathematical research which they will conduct later in life. Secondly, students can submit their projects to Siemens, Davidson Fellowship, Google Science fair, and the Regeneron Science Talent Search. Thirdly, research projects can greatly help with college admissions.
An application can be filled out and submitted at any time between now and September 1st, with all decisions and parrings to be made during the month of September. The application consists of the following:
APPLICATION:
Your name:
Your city of residence (does not affect decision):
Your age (must be between 14 and 18 years old to apply):
Grade (must be between 9 and 12 next school year):
School:
Email:
Gender:
List all of your relevant math experiences (Articles published, blogs, contests won, math camps, etc.).
List the three specific areas of mathematics in which you are most interested in pursuing research.
Most advanced math courses taken (with grades, indicating where you took them):
Your favorite proof/theorem in mathematics and why:
Why you think you're qualified to undertake this project (approximately 300 words):
All of these will factor into acceptance. Note that each application will be checked thoroughly and any cases of lying or deceit will result in immediate disqualification and the notification of your school and/or parents.
It should be noted that neither gender nor race will bear any part in whether or not you're accepted into the program, it will be strictly a merit-based program. Also note that if an applicant is not accepted into the program this year he/she should feel free to apply in future years. We may favor older students who are nearing the end of their high school career.
\newpage
\title{A Recursive Way to Calculate P(n)}
\author{Shaun Schoeman}
\date{April 2017}
\maketitle
\section{A Recursive Way to Calculate P(n)}
The partition function, $p(n)$ , is a function which takes an input $n$ and returns the number of partitions of $n$ i.e. the number of different ways one could write out $n$ as the sum of other non-negative integers without regarding the order of the summands or added summands of $0$.
Conventionally, any partition of $n$ is written out as $n=t_1+t_2+...+t_k$ with $(t_k\leq...\leq t_2\leq t_1\leq n)$. For this paper we will be focusing on a specific type of partition, namely di-partitions.
\begin{definition}
A di-partition of $n$ consists of two positive integers summing to $n$.
\end{definition}
Let $\Delta(n,i)$ denote the $i^{th}$ di-partition of $n$. Then
$n=\Delta(n,i)=s_1+s_2$ and let $(s_2\leq s_1<n)$ for convention.
At the moment it means nothing to talk about the $i^{th}$ di-partition of $n$ as we have not yet ordered di-partitions. So, let us order them according to decreasing values of $s_1$, which also means increasing values of $s_2$. Ordering di-partitions this way is an arbitrary choice, and choosing to order them differently would only result in different expressions of $s_1$ and $s_2$ in terms of $n$ and $i$. With our current way of ordering, we obtain {\Delta}(n,i)=(n-i)+i.
\begin{theorem}
$i\leq n-i$ must be true according to the bounds on $i$.
\end{theorem}
\begin{proof}
By definition, it is obvious that the minimum of $i$ be at least $1$ and the maximum be equal to the total number of unique di-partitions of $n$. We can then see that for any even $n$, this number is $\frac{n}{2}$
and for any odd $n$, $\frac{n-1}{2}$. Thus, $i$'s bounds are $1\leq i \leq \lfloor{\frac{n}{2}}\rfloor$. Multiplying through by $2$ gives
$$2\leq 2i \leq 2\lfloor\frac{n}{2}\rfloor$$
Observe $n-1\leq 2\lfloor \frac{n}{2}\rfloor \leq n$. Since $2i$ increases from either $n-1$ to $n+1$ or from $n$ to $n+2$ as $2i$ extends $2\lfloor \frac{n}{2}\rfloor$, we see that $n$ achieves the same restriction on $2i$ as $2\lfloor \frac{n}{2}\rfloor$. Thus, we can accurately say that $2\leq 2i \leq n$ which gives
$$2-i\leq i\leq n-i$$
completing the proof.
\end{proof}
With the fact that $1\leq i$, we clearly see that $2-i\leq 1$. As earlier said, the lower bound of $i$ must be at least $1$, and so here again it must be $1$. Thus,
\begin{equation}
\label{i's_bound}
1\leq i \leq n-i
\end{equation}
\subsection{Breaking down di-partitions}
Observe $n=\Delta(n,i) = (n-i) +i$. This is considered as the breaking down of $n$ into its di-partitions. To break it down further all terms formed must be broken down the same way.
\begin{equation}
\label{next_break}
n=\Delta(n-i,a)+\Delta(i,b)=(n-i-a)+a+(i-b)+b
\end{equation}
with some non-negative integers $a,b$.
There are now more, and obviously smaller, terms which have been formed. This process of breaking down $n$ can be repeated until we are inevitably left with $n$ many terms of $1$, as $1$ cannot break down any further. This is called completely breaking down $n$, and can be summarized by
\begin{equation}
\label{completely_break}
\Delta(n,i)=\sum_{a=1}^{n-i-a}[\Delta(n-i,a)] + \sum_{b=1}^{i-b}[\Delta(i,b)]
\end{equation}
This equation undoubtedly finds many duplicates as well as unique partitions of $n$.
\begin{definition}
A duplicate is a partition which has been previously formed during the process of completely breaking down $n$, whereas a unique partition has not.
\end{definition}
\begin{theorem}
All partitions of $n$, not necessarily all unique, are revealed when completely breaking down $n$.
\end{theorem}
The significance of this theorem is to give us reasonable motivation to explore further. We need to know that equation \ref{completely_break} at least gives us all possible partitions of $n$ before we deem it worthwhile to investigate eliminating the duplicates formed.
\begin{proof}
Each time we break down a term according to equation \ref{completely_break}, another partition of $n$ is revealed; whether new or a duplicate of one previously formed.
The argument follows: take all partitions of $n$ with $x$ many terms. Within each said partition, break each term down into all of its di-partitions; one term at a time, one partition at a time. This will definitely reveal, in total, all possible (including duplicate) partitions of $n$ with $x+1$ many terms. Setting $0<x<n$ and applying the argument for all set values of $x$ clearly completes the proof.
\end{proof}
\begin{lemma}
\label{T2}
If $\Delta(n,i_1)=\Delta(n,i_2)$, then whichever di-partition has a greater value between $i_1,i_2$ is a duplicate.
\end{lemma}
\begin{proof}
The way in which we have chosen to order di-partitions can be described as according to increasing values of the second input in the $\Delta$ function. The process of completely breaking down $n$ follows this same order, which can be clearly seen. Thus, the di-partition with a lesser value between $i_1,i_2$ will have been formed before the one with a greater value, which concludes the proving argument.
\end{proof}
\begin{theorem}
\label{T3}
Any partition formed during the process of completely breaking down $n$ must appear in descending order for it to not be a duplicate.
\end{theorem}
\begin{proof}
In other words, this theorem states that within equation \ref{next_break}, $b\leq i-b\leq a\leq n-i-a$ must be true so that $\Delta(n,i)$, given by equation \ref{completely_break}, does not reveal a duplicate partition of $n$. We already know from equation \ref{i's_bound} that $b\leq i-b$ and $a\leq n-i-a$. What is then left for us to prove is that $i-b\leq a$.
Observe equation \ref{next_break}, $\Delta(n,i)=(n-i-a)+a+(i-b)+b$. Summing the first and third terms then the second and fourth gives
$$\Delta(n,i)=(n-a-b)+(a+b)=\Delta(n,a+b)$$
From lemma \ref{T2} we see that $i\leq a+b$ which completes the proof.
\end{proof}
\begin{theorem}
\label{term_right}
The term on the right of each di-partition must not break down or else will reveal a duplicate.
\end{theorem}
\begin{proof}
To prove this theorem we simply need to prove that within equation \ref{completely_break}, $b=0$. This can be done similary to proving the previous theorem. Observe equation \ref{next_break}, $\Delta(n,i)=(n-i-a)+a+(i-b)+b$. Summing the first three terms gives
$$\Delta(n,i)=(n-b)+b=\Delta(n,b)$$
From lemma \ref{T2}, we require $i\leq b$, which clearly contradicts equation \ref{i's_bound}. Therefore, at least one of $a,b=0$ in order to avoid this contradiction. If we set $b\neq 0$, then $a=0$ from what we previously proved, meaning $b>a$ since $b$ is non-negative. This would contradict theorem \ref{T3} and so
\begin{equation}
\label{b's_value}
b=0
\end{equation}
\end{proof}
We can end off this section by assuring that our bounds on $a,b$ are optimal to achieve no duplicates formed within equation \ref{completely_break}. Note that $b$ is proved to be a constant ($b=0$), which is of course an optimal bound. Observe our current bounds for $a$ (from theorem \ref{T3}), $i\leq a\leq n-a-i$. We obtained this bound from summing the first and third terms within equation \ref{next_break}, $\Delta(n,i)=(n-i-a)+a+i$. The only other options we had were summing the first two or the last two terms. Summing the first two gives us the redundant result of $\Delta(n,i)=\Delta(n,i)$. Summing the last two would give us $\Delta(n,i)=\Delta(n,a+i)$ which, from lemma \ref{T2}, gives $i\leq a+i$ meaning $0\leq a$ which is also a non-contributing fact.
\section{Computing the partition function}
We can alter equation \ref{completely_break} to count the number of times a term is broken down by adding a term of $1$ to the equation. Whenever a term is broken down, a unique partition should be revealed using our optimal bounds on $a,b$ (theorems \ref{T3} and \ref{term_right}). Denoting this altered equation as $\nu(n,i)$,
$$\nu(n,i)=1+\sum_{a=i}^{n-i-a}[\nu(n-i,a)]$$
Keep in mind that $1\leq a$. Since $n\equiv\Delta(n,0)$, we have $$p(n)=\nu(n,0)$$
\\
\end{document}
Best Answer
As the
combineis better documented, here are an example of a volume journal with a few articles using thepaperclass with cover and a table of contents of articles.Indeed the cover is simple, the first article start in even page, there are not fancy headers, etc. but is only a minimal working example to show how to join articles in a volume, right? Anything else is left as exercise to the reader.
;)Output of pages 1-2:
Note: I have done that MWE in a single file but in practice you may want maintain the contents of each article in child documents. To construct a MWE with child document cut from
% ARTICLE 1until% ARTICLE 2, save it asarticle1.tex, and replace this chunk of text by\input{article1}or\include{article1}(there are no difference in this case). If you want render single articles, use\includeonlyor see thedocmutepackage to maintain child documents as complete LaTeX files.Reminder: Do not forget check look also the
combineclass.