对齐环境中的语法

Question

看看这个：

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
\begin{align*}
\Leftrightarrow \qquad P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} && (2.6)\\
\Leftrightarrow \qquad e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)&=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\end{align*}
\end{document}

是否有可能像方程式一样，将等价项放在彼此之下？

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
\begin{alignat*}{4}
P'_1(t)&=-\lambda P_1(t)+\lambda P_0(t)\\
\Leftrightarrow &&\qquad  P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} && (2.6)\\
\Leftrightarrow &&  e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)&=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\end{alignat*}
\end{document}

这几乎可行，但我提到了一个问题，即第一行被拉到最左边。

Answer 1

像这样吗？

\documentclass{article}

\usepackage{amsmath}
\begin{document}
\begin{align}
\Leftrightarrow&& P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} \tag{2.6} \\
\Leftrightarrow&& e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)&=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \notag
\end{align}
\end{document}

Answer 2

ArrowBetweenLines您可能还对以下内容感兴趣mathtools：

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{amssymb}

\numberwithin{equation}{section}

\begin{document}

\setcounter{section}{2}\setcounter{equation}{5}

\begin{equation}\label{equivalenteqs}
\begin{alignedat}{2}
 & & abc & =def \\
 \ArrowBetweenLines &\qquad& P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} \\
  \ArrowBetweenLines &\qquad& e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)
                           &=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\end{alignedat}
\end{equation}
As we see in equation~\eqref{equivalenteqs} we can blah blah.

\end{document} 

Answer 3

你不想要手动对方程式进行编号，并不是因为数字彼此不对齐，而是因为维持手动编号基本上是不可能的。

为此使用\label-机制。\ref

您的问题可以通过以下方式解决alignedat：

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}

\numberwithin{equation}{section}

\begin{document}

% let's emulate being in section 2 with five
% numbered equations before this one; a real document
% will have nothing like this
\setcounter{section}{2}\setcounter{equation}{5} 

\begin{equation}\label{equivalenteqs}
\begin{alignedat}{2}
  \Leftrightarrow &\qquad& P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} \\
  \Leftrightarrow &\qquad& e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)
                           &=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\end{alignedat}
\end{equation}
As we see in equation~\eqref{equivalenteqs} we can blah blah.

\end{document}

如果第一条线恰好没有箭头，只需添加适当数量的对齐点：

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}

\numberwithin{equation}{section}

\begin{document}

% let's emulate being in section 2 with five
% numbered equations before this one
\setcounter{section}{2}\setcounter{equation}{5} 

\begin{equation}
\begin{alignedat}{2}
                &\qquad&  P'_1(t)&=-\lambda P_1(t)+\lambda P_0(t)\\
\Leftrightarrow &      &  P'_1(t)+\lambda P_1(t)&=\lambda e^{-\lambda t} \\
\Leftrightarrow &      &  e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)&=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\Leftrightarrow &      & \frac{d}{dt}(e^{\lambda t}P_1(t))&=e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)=\lambda\\
\end{alignedat}
\end{equation}
As we see in equation~\eqref{equivalenteqs} we can blah blah.

\end{document}

另一方面，我认为没有理由在等号处对齐。

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}

\numberwithin{equation}{section}

\begin{document}

% let's emulate being in section 2 with five
% numbered equations before this one
\setcounter{section}{2}\setcounter{equation}{5} 

\begin{equation}\label{equivalenteqs}
\begin{aligned}
                && & P'_1(t)=-\lambda P_1(t)+\lambda P_0(t)\\
\Leftrightarrow && & P'_1(t)+\lambda P_1(t)=\lambda e^{-\lambda t} \\
\Leftrightarrow && & e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)=\lambda e^{\lambda t} e^{-\lambda t}=\lambda \\
\Leftrightarrow && & \frac{d}{dt}(e^{\lambda t}P_1(t))=e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t)=\lambda\\
\end{aligned}
\end{equation}
As we see in equation~\eqref{equivalenteqs} we can blah blah.

\end{document}

Answer 4

您还可以使用alignat：

代码：

\documentclass[10pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
\begin{alignat*}{4}
  \Leftrightarrow &&\qquad P'_1(t)+\lambda P_1(t) &=\lambda e^{-\lambda t} && (2.6) \\
  \Leftrightarrow &&\qquad e^{\lambda t}P'_1(t)+\lambda e^{\lambda t}P_1(t) &=\lambda e^{\lambda t} e^{-\lambda t}=\lambda 
\end{alignat*}
\end{document}