我正在整理多篇不同的文章，我希望它们像普通文章一样放在一起，唯一的问题是我试图在一个 Latex 文档上进行这项工作，但我在标题和其他方面却遇到了困难。
也许我错过了一个简单的解决方案，但我真的很挣扎。以下是我目前所拥有的。
\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
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\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}
答案1
由于combine有更好的记录，这里有一个卷杂志的例子，其中包含几篇文章，使用paper带有封面和文章目录的类别。
封面确实很简单，第一篇文章从偶数页开始，没有花哨的标题等，但只是一个最小示例展示了如何将文章合并到卷中，对吗？其他内容留给读者练习。;)
第 1-2 页的输出：
\documentclass[journal,noindent,headline]{paper}
\usepackage[rmargin=1.5cm]{geometry}
\usepackage[english,latin]{babel}
\usepackage{microtype}
\usepackage{xcolor}
\usepackage{lipsum} % for dummy text only
\usepackage[colorlinks,linkcolor=blue!50!black]{hyperref}
\usepackage{graphicx}
\begin{document}
\oddrunhead{J. Hiber. Res., Vol. \textsc{\romannumeral27}}
\begin{titlepage} \begin{center}
{\huge\sf Journal of Hibernation Research} \vfill
{\Large 2017, Vol. 27, No. 4} \vfill
{\large\bf Whesleepinwinter University Press}
\end{center} \end{titlepage}
\journalcontents
\twocolumn
\journalpart[Research articles]{}
% ARTICLE 1
\title{Do nothing improve a lot the maintenance of the winter's fat layer}
\subtitle{Sleep more and move less is good for prevent fat depletion before spring.}
\shorttitle{Do nothing improve the fat layer}
\author{Jon Marmot,\,%
\thanks{Corresponding author: \href{mailto:[email protected]}{[email protected]}.}
Teddy Bear\,%
\thanks{Do not disturb.},
and Ted Squirrel\,%
\thanks{Do not disturb me too.}}
\shortauthor{J. Marmot, T. Bear and T. Squirrel}
\institution{Fat Research Center}
\date{\today}
\maketitle
\begin{abstract}
\lipsum[5]
\end{abstract}
\begin{keywords}
fat, hibernation, activiity
\end{keywords}
\section{Introduction}
\begin{figure}[b!]
\includegraphics[width=\linewidth]{example-image-a}
\caption{The A of hibernAtion}
\end{figure}
\lipsum[10-13]
\section{Material and methods}\lipsum[14-17]
\section{Results and discussion}\lipsum[18-22]
% ARTICLE 2
\title{Torpor decrease one half of body weight}
\subtitle{Eat more, eat faster that anybody is yo want survive.}
\shorttitle{Torpor decrease body weight}
\institution{Center of High Metabolism Research}
\author{Walter Shrew\,
\thanks{Corresponding author: \href{mailto:[email protected]}{[email protected]}.}}
\shortauthor{W. Shrew}
\date{\today}
\maketitle
\section{Introduction}\lipsum[1-3]
\section{Material and methods}\lipsum[4-5]
\section{Results and discussion}\lipsum[6-9]
\journalpart[Review articles]{}
% ARTICLE 3
\title{Evolution of hibernation strategies}
\subtitle{Sleep more or less? That's the question.}
\shorttitle{Evolution of hibernation strategies}
\institution{University of Sleepingbeauty}
\author{David Searcher\,
\thanks{Corresponding author: \href{mailto:[email protected]}{[email protected]}.}}
\shortauthor{D. Searcher}
\date{\today}
\maketitle
\section{Introduction}\lipsum[23-24]
\section{Hibernation in tropical climates}\lipsum[25-32]
\section{Torpor as adaptation to global warming}\lipsum[33-39]
% Book reviews
\newpage
\journalpart{New published books}
\review[Tutorial]{Larry Lazy}{How-two sleep well four months}
Eleven edition, 2017, Ed. J. Marmot, Winter Press, Manitoba.
\bigskip
\lipsum[40-44]
\revauthor{Edouard Masterly\\Stanford University}
\end{document}
笔记：我已经在单个文件中完成了 MWE，但实际上您可能希望在子文档中维护每篇文章的内容。要构建一个 MWE，其中子文档从 到 剪切% ARTICLE 1，% ARTICLE 2将其另存为article1.tex，并将此文本块替换为\input{article1}或 \include{article1}（在这种情况下没有区别）。如果您想要呈现单个文章，请使用\includeonly或查看将docmute子文档维护为完整 LaTeX 文件的包。
提醒：不要忘记检查一下班级combine。