下面写出的两个不等式（紧接着 \vdots 的不等式）超出了边界。下面是 MWE：
\documentclass[12pt,a4paper]{report}
\makeatletter
% these two slashes-commands for continuous chapter numbering in parts of the document
\makeatother
\usepackage{titlesec}
\titleformat{\chapter}{\normalfont\huge}{\thechapter.}{20pt}{\huge\bf} %for chapter headings
\RequirePackage{etoolbox}
\usepackage{tikz-cd} % and in the above line to use commutative diagrams
\usepackage{amsfonts}
\usepackage[utf8]{inputenc}
\usepackage{amsthm}
% below four lines to just one or two cyrillic letter
\DeclareFontFamily{U}{wncy}{}
\DeclareFontShape{U}{wncy}{m}{n}{<->wncyr10}{}
\DeclareSymbolFont{mcy}{U}{wncy}{m}{n}
\DeclareMathSymbol{\Sh}{\mathord}{mcy}{"58}
%Cyrillic letter command over
\usepackage{sagetex}
\usepackage[toc]{appendix} % to make appendi appear in the table of contents
\usepackage{amssymb} %to use direct sum symbol
\usepackage{graphicx} %toinsert images
\usepackage{amsmath} %to use matrices
\usepackage{hyperref}
\usepackage{bm} % to use bold font in math mode use \bm{ insert math mode text}
\usepackage{mathrsfs} %for scripted english characters
\usepackage{mathtools}
\usepackage{tikz-cd} % to draw commutative diagrams
\usepackage{enumitem}
\begin{align*}
\begin{split}
h(P_n) \leq \frac{1}{m^2}(2h(P_{n-1})+C'_1+C_2) & = \frac{2}{m^2}h(P_{n-1}) + \frac{C'_1+C_2}{m^2} \\
& \leq \frac{2}{m^2} \Big(\frac{1}{m^2}(2h(P_{n-2})\frac{C'_1+C_2}{m^2} \Big) + \frac{C'_1+C_2}{m^2} \\
& = (\frac{2}{m^2})^2 h(P_{n-2}) + \Big( \frac{1}{m^2}+\frac{2}{(m^2)^2} \Big)(C'_1+C_2) \\
\vdots \\
& \leq (\frac{2}{m^2})^n h(P) + \frac{1}{m^2} \Big( 1+ \frac{2}{m^2}+ \ldots + \frac{2^{n-1}}{(m^2)^{n-1}} \Big)(C'_1+C_2) \\
& \leq (\frac{2}{m^2})^n h(P) + \frac{1}{m^2} \Big( 1+ \frac{2}{m^2} + \frac{2^2}{(m^2)^2} + \ldots \Big)(C'_1+C_2) \\
& = (\frac{2}{m^2})^n h(P) + \frac{(C'_1+C_2)}{m^2-2} \\
& \leq \frac{1}{2^n}h(P)+ \frac{1}{2}(C'_1+C_2) \text{ $m \geq 2$}\\
\end{split}
\end{align*}
If $n$ is sufficiently large so that $2^n$ exceeds $h(P)$ and $\frac{h(P)}{2^n} < 1$ then
\[
h(P_n) \leq 1+ \frac{1}{2}(C'_1+C_2)
\]
\end{document}
（上图的垂直线表示文本块的边缘）
我该如何正确地书写它们以使它们不超出文本宽度？
答案1
为了实现您在帖子中陈述的格式化目标，您应该在 的第一个实例处引入（a）一个额外的对齐点\leq和（b）在 的第一个实例之前引入一个换行符&=。
此外，您可能希望对表达式进行修改，并提供更多有关各个步骤中发生的情况的信息。这样，您的读者会更容易理解您的论点。例如，如果您用\vdots实际的单词替换，例如，您的读者可能会很感激Applying $n-2$ additional backward subtitution steps, we further find。请参阅以下代码和屏幕截图以获取更多建议。
顺便说一句，我认为明确说明要求m^2>2无穷和收敛并因此达到所述极限是一种好的形式。（当然，该条件m^2>2由您的后续条件满足m\ge 2。）这意味着相关的弱不等式实际上是严格的不等式。
以下屏幕截图以红色突出显示了我应用的主要更改。为简单起见，下面显示的代码省略了所有\color和\textcolor命令。
\documentclass[12pt,a4paper]{report}
\usepackage{mathtools}
\allowdisplaybreaks
\begin{document}
Setting $\widetilde{C}=C'_1+C_2>0$, we find
\begin{align*}
h(P_n)
&\leq \frac{1}{m^2}\bigl(2h(P_{n-1})+\widetilde{C}\,\bigr)\\
&= \frac{2}{m^2}h(P_{n-1}) + \frac{1}{m^2}\widetilde{C}\\
&\leq \frac{2}{m^2} {\underbrace{\Bigl(\frac{2}{m^2}h(P_{n-2})
+\frac{1}{m^2}\widetilde{C} \Bigr)}_{\ge h(P_{n-1})}}
+\frac{1}{m^2}\widetilde{C} \\
&= \Bigl(\frac{2}{m^2}\Bigr)^{\!2} h(P_{n-2})
+\frac{1}{m^2}\Bigl[1 +\frac{2}{m^2} \Bigr]\widetilde{C}\,. \\
\intertext{Applying $n-2$ additional backward subtitution steps, we further find}
h(P_n)
&\leq \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+\frac{1}{m^2} \Bigl[\,\sum_{j=0}^{n-1}\Bigl(\frac{2}{m^2}\Bigr)^{\!j}
\,\Bigr]\widetilde{C} \\
&< \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+\frac{1}{m^2} {\underbrace{\Bigl[\,\sum_{j=0}^{\infty}
\Bigl(\frac{2}{m^2}\Bigr)^{\!j} \,\Bigr]}_{=m^2/(m^2-2)}}\widetilde{C}
\quad\text{for $m^2>2$}\\
&= \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+ \frac{1}{m^2-2}\,\widetilde{C} \\
&\leq \frac{1}{2^n}h(P)+ \frac{1}{2}\widetilde{C}
\quad\text{for $m\ge2$ and hence $2/m^2\le 1/2$.}
\intertext{If $n$ is sufficiently large so that $h(P)<2^n$ and hence $h(P)/2^n<1$,}
h(P_n) &< 1 + \frac{1}{2}\widetilde{C}= 1+\frac{1}{2}(C'_1+C_2)\,.
\end{align*}
\end{document}
\documentclass[12pt,a4paper]{report}
\usepackage{mathtools}
\allowdisplaybreaks
\begin{document}
\begin{align*}
h(P_n)
&\leq \frac{1}{m^2}\bigl(2h(P_{n-1})+\widetilde{C}\,\bigr)
\quad\text{where $\widetilde{C}\equiv C'_1+C_2>0$}\\
&= \frac{2}{m^2}h(P_{n-1}) + \frac{1}{m^2}\widetilde{C}\\
&\leq \frac{2}{m^2} {\underbrace{\Bigl(\frac{2}{m^2}h(P_{n-2})
+\frac{1}{m^2}\widetilde{C} \Bigr)}_{\ge h(P_{n-1})}}
+\frac{1}{m^2}\widetilde{C} \\
&= \Bigl(\frac{2}{m^2}\Bigr)^{\!2} h(P_{n-2})
+\frac{1}{m^2}\Bigl[1 +\frac{2}{m^2} \Bigr]\widetilde{C} \\
&\vdotswithin{=} \text{\footnotesize[$n-2$ further backward substitution steps]}\\
&\leq \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+\frac{1}{m^2} \Bigl[\,\sum_{j=0}^{n-1}\Bigl(\frac{2}{m^2}\Bigr)^{\!j}
\,\Bigr]\widetilde{C} \\
&< \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+\frac{1}{m^2} {\underbrace{\Bigl[\,\sum_{j=0}^{\infty}
\Bigl(\frac{2}{m^2}\Bigr)^{\!j} \,\Bigr]}_{=m^2/(m^2-2)}}\widetilde{C}
\quad\text{for $m^2>2$}\\
&= \Bigl(\frac{2}{m^2}\Bigr)^{\!n} h(P)
+ \frac{1}{m^2-2}\,\widetilde{C} \\
&\leq \frac{1}{2^n}h(P)+ \frac{1}{2}\widetilde{C}
\quad\text{for $m\ge2$ and hence $2/m^2\le 1/2$.}
\end{align*}
If $n$ is sufficiently large so that $2^n\ge h(P)$ and hence $h(P)/2^n \le 1$,
\[
h(P_n) < 1 + \frac{1}{2}\widetilde{C}\equiv 1+\frac{1}{2}(C'_1+C_2)\,.
\]
\end{document}
答案2
只需移动对齐点。也可以使用\vdotswithin{=}。编辑：根据建议，我让所有括号都一致。有两个版本，一个是\Bigl(和\Bigr)，另一个是\left(和\right)。至于哪个更好看，就看个人喜好了。
\documentclass[12pt,a4paper]{report}
\usepackage{mathtools}
\begin{document}
\begin{align*}
h(P_n) &\leq \frac{1}{m^2}(2h(P_{n-1})+C'_1+C_2) = \frac{2}{m^2}h(P_{n-1}) + \frac{C'_1+C_2}{m^2} \\
& \leq \frac{2}{m^2} \Bigl(\frac{1}{m^2}(2h(P_{n-2})\frac{C'_1+C_2}{m^2} \Bigr) + \frac{C'_1+C_2}{m^2} \\
& = \Bigl(\frac{2}{m^2}\Bigr)^2 h(P_{n-2}) + \Big( \frac{1}{m^2}+\frac{2}{(m^2)^2} \Big)(C'_1+C_2) \\
&\vdotswithin{=} \\
& \leq \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{1}{m^2} \Bigl( 1+ \frac{2}{m^2}+
\ldots + \frac{2^{n-1}}{(m^2)^{n-1}} \Bigr)(C'_1+C_2) \\
& \leq \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{1}{m^2} \Bigl( 1+ \frac{2}{m^2}
+ \frac{2^2}{(m^2)^2} + \ldots \Bigr)(C'_1+C_2) \\
& = \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{(C'_1+C_2)}{m^2-2} \\
& \leq \frac{1}{2^n}h(P)+ \frac{1}{2}(C'_1+C_2) \quad\text{if } m \geq 2\\
\end{align*}
\begin{align*}
h(P_n) &\leq \frac{1}{m^2}(2h(P_{n-1})+C'_1+C_2) = \frac{2}{m^2}h(P_{n-1}) + \frac{C'_1+C_2}{m^2} \\
& \leq \frac{2}{m^2} \left(\frac{1}{m^2}(2h(P_{n-2})\frac{C'_1+C_2}{m^2} \right) + \frac{C'_1+C_2}{m^2} \\
& = \left(\frac{2}{m^2}\right)^2 h(P_{n-2}) + \Big( \frac{1}{m^2}+\frac{2}{(m^2)^2} \Big)(C'_1+C_2) \\
&\vdotswithin{=} \\
& \leq \left(\frac{2}{m^2}\right)^n h(P) + \frac{1}{m^2} \left( 1+ \frac{2}{m^2}+
\ldots + \frac{2^{n-1}}{(m^2)^{n-1}} \right)(C'_1+C_2) \\
& \leq \left(\frac{2}{m^2}\right)^n h(P) + \frac{1}{m^2} \left( 1+ \frac{2}{m^2}
+ \frac{2^2}{(m^2)^2} + \ldots \right)(C'_1+C_2) \\
& = \left(\frac{2}{m^2}\right)^n h(P) + \frac{(C'_1+C_2)}{m^2-2} \\
& \leq \frac{1}{2^n}h(P)+ \frac{1}{2}(C'_1+C_2) \quad\text{if } m \geq 2\\
\end{align*}
\end{document}
答案3
看看以下结果是否可以接受：
\documentclass[12pt,a4paper]{report}
\usepackage{amssymb} %to use direct sum symbol
\usepackage{mathtools}
%---------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\begin{document}
\begin{align*}
\MoveEqLeft
h(P_n) \leq \frac{1}{m^2}\bigl(2h(P_{n-1})+C'_1+C_2\bigr) \\
& = \frac{2}{m^2}h(P_{n-1}) + \frac{C'_1+C_2}{m^2} q\\
& \leq \frac{2}{m^2} \Bigl(\frac{1}{m^2}(2h(P_{n-2})\frac{C'_1+C_2}{m^2}\Bigr) + \frac{C'_1+C_2}{m^2} \\
& = \Bigl(\frac{2}{m^2}\bigr)^2 h(P_{n-2}) + \Bigl( \frac{1}{m^2}+\frac{2}{(m^2)^2}\Bigr)(C'_1+C_2) \\
& \hspace{0.25\linewidth}\vdots \\
& \leq \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{1}{m^2} \Bigl(1+ \frac{2}{m^2}+ \ldots + \frac{2^{n-1}}{(m^2)^{n-1}}\Bigr)(C'_1+C_2) \\
& \leq \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{1}{m^2} \Bigl(1+ \frac{2}{m^2} + \frac{2^2}{(m^2)^2} + \ldots \Bigr)(C'_1+C_2) \\
& = \Bigl(\frac{2}{m^2}\Bigr)^n h(P) + \frac{(C'_1+C_2)}{m^2-2} \\
& \leq \frac{1}{2^n}h(P)+ \frac{1}{2}(C'_1+C_2), \quad m\geq 2\\
\end{align*}
If $n$ is sufficiently large so that $2^n$ exceeds $h(P)$ and $\frac{h(P)}{2^n} < 1$ then
\[
h(P_n) \leq 1+ \frac{1}{2}(C'_1+C_2)
\]
\end{document}
编辑：考虑@Mico 关于括号的评论。