以下是生成如下图像的完整代码：
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}
\begin{document}
\title{Extra Credit}
\maketitle
\begin{definition}
If f is analytic at $z_0$, then the series
\begin{equation}
f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
\end{equation}
is called the Taylor series for f around $z_0$.
\end{definition}
\begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
\begin{equation}
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )
\end{equation}
\end{theorem}
\hrulefill
\begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
\end{theorem}
\begin{proof}
Suppose that the function \textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=\Big\{z:|z-z_0|=\frac{R + R'}{2}, 0<R< R' \Big\}.$$ Letting $\zeta$ be an arbitrary point on $C$ and applying $(2)$ to $(1)$, we get
\begin{equation}
\sum_{n=0}^{\infty} \frac{(z-z_0)^{n}}{2\pi i} \int_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta.
\end{equation}\\
Or equivalently, we have that
\begin{align*}
\sum_{n=0}^{\infty} \frac{(z-z_0)^{n}}{2\pi i} \int_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta
&= \frac{1}{2\pi i} \sum_{n=0}^{\infty} \int_{C} \frac{(z-z_0)^{n}f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta
\\ &= \frac{1}{2\pi i}\int_{C}\sum_{n=0}^{\infty} \frac{1}{\zeta - z_0}\frac{f(\zeta)}{\displaystyle\frac{(\zeta - z_0)^n}{(z-z_0)^n}}d\zeta
\\ &= \frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta - z_0}\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n} d\zeta
\
\end{align*}
Now since
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}.
\end{equation}
The last result comes from the observation that
\begin{flalign}
&\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\
\end{flalign}
\\
And so,
\begin{flalign}
&\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\
\end{flalign}
具体来说，
Now since
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}.
\end{equation}
The last result comes from the observation that
\begin{flalign}
&\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\
\end{flalign}
\\
And so,
\begin{flalign}
&\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\
\end{flalign}
\\
生成上面的图像。如何让And so图像中的后面的线条与下方对齐And so。如何只保留图中所示的公式 (8) 和公式 (11) 的编号？
答案1
这是你想要的吗？
基本上，您希望将左对齐的方程式与居中的（或多或少）方程式重叠。这使用\mathrlap（mathtools 包）来防止左对齐的方程式占用空间。使用 &&& 会跳过左对齐部分。
\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
%\usepackage[utf8]{inputenc}% all they do for me is slow down the build.
%\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{mathtools}
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}
\begin{document}
Now since
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}.
\end{equation}
The last result comes from the observation that
\begin{flalign}
\mathrlap{\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}
+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]} &&&&
\\&&&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n} \notag
\\&&&\quad-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}
-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&&&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\intertext{And so,}
\mathrlap{\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1}
+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]} &&&
\\&&&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&&&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\
\end{flalign}
\end{document}