考虑以下
\documentclass{文章}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}
\begin{document}
\begin{lem}
\begin{align}%
\begin{split}%
&\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) = \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\
& = \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}- \sum\limits_{k=u+1}^{u+s-z} \sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{split}\nonumber
\end{align}
\end{lem}
\end{document}
答案1
如果想要达到类似的效果,\raggedright
就必须将对齐字符放在环境&
内每一行的第一位split
。
\documentclass{article}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}
\begin{document}
\begin{lem}
\begin{align}%
\begin{split}%
&\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) = \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\
& = \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}- \sum\limits_{k=u+1}^{u+s-z} \sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{split}\nonumber
\end{align}
\end{lem}
\end{document}
答案2
这使用了进一步的线,因为最终的公式非常宽。
如果文本宽度允许,您可以在第一行向左移动对齐点,这样也许可以避免拆分最后一个公式。
\documentclass{article}
%\usepackage{geometry}
\usepackage{amsthm,amsmath,amssymb}
\newtheorem{lem}{Lemma}
\begin{document}
\begin{lem}
\begin{equation}
\hspace{0pt}
\begin{aligned}[b]
\mathbb{P}_u(\tau^0=s&, -R_{\tau^0}=z)
=\sum_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k}
\\
&= \sum_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}
\\
&\qquad-\sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j}
p^{*(j-u)}_j p_{u+s+z-k}
\end{aligned}
\hspace{10000pt minus 1fil}
\end{equation}
\end{lem}
\end{document}
我使用了教科书练习 19.8 中解释的一个技巧。为了使对齐对齐到左侧,必须在开头有一个零粘连点。这个\hspace{100000pt minus 1fil}
技巧让 TeX 认为对齐太宽了(但实际上可以收缩很多以允许排版)。
答案3
geometry
在和的帮助下mathtools
,我提出了以下两种变体:
\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{amsthm, amssymb}
\newtheorem{lem}{Lemma}
\usepackage{mathtools}
\begin{document}
\begin{lem}
\begin{multline} \mathbb{P}_u(\tau^0=s , -R_{\tau^0}=z) = \smash{\sum_{k=0}^{u+s-z}} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\ = \sum_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}-\!\! \sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{multline}
\end{lem}
\vskip2em
\begin{lem}$\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) = \displaystyle\smash[b]{\smashoperator{\sum_{k=0}^{u+s-z}}} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k}$
\begin{equation} =\smashoperator{\sum_{k=0}^{u+s-z}} p^{*(s-1)}_k p_{u+s+z-k} -\!\! \sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} = \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{equation}
\end{lem}
\end{document}
答案4
在你的情况下,我会考虑multline*
数学环境(由定义amsmath
):
\documentclass{article}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}
\begin{document}
\begin{lem}
\begin{multline*}
\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) = \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\
= \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}-
\sum\limits_{k=u+1}^{u+s-z}
\sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{multline*}
\end{lem}
\end{document}