我想通过将左对齐的等号向左移动来适应长方程的推导。
\begin{align*}
\MoveEqLeft
p(y = 1\mid x; \phi, \mu_0, \mu_1, \Sigma) \\
&= \frac{p(x\mid y=1)p(y=1)}{p(x)} \\
&= \frac{p(x\mid y=1)p(y=1)}{p(x\mid y=1)p(y=1)+p(x\mid y=0)p(y=0)}&\\
&= \frac{\frac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi}{\frac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi + \frac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_{0})^T \Sigma^{-1} (x-\mu_{0})\right) (1-\phi)} \\
&= \frac{1}{1+\exp \left( -\frac{1}{2}(x-\mu_{0})^T \Sigma^{-1} (x-\mu_{0}) + \frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1)\right) \frac{1-\phi}{\phi}} & (\text{divide by numerator}) \\
&= \frac{1}{1+\exp \left( \log \frac{1-\phi}{\phi} +\frac{1}{2}
\left[ (x-\mu_1)^T \Sigma^{-1} (x-\mu_1) - (x-\mu_{0})^T \Sigma^{-1} (x-\mu_{0})
\right]\right)}
\end{align*}
这是屏幕截图。您可以看到，第三个等号太长了，无法容纳。我该如何将所有对齐的等号进一步向左移动以容纳这个长等式。
答案1
一种方法是将最后一个等式中的分母写成两行：
\documentclass{article}
\usepackage{mathtools, nccmath}
\newcommand\di{\mathrm{d}}
%---------------- show page layoutdon't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\begin{document}
\begin{align}
\MoveEqLeft[1]
p(y = 1\mid x; \phi, \mu_0, \mu_1, \Sigma) \\
& = \frac{p(x\mid y=1)p(y=1)}{p(x)} \\
& = \frac{p(x\mid y=1)p(y=1)}{p(x\mid y=1)p(y=1)+p(x\mid y=0)p(y=0)} \\
& = \frac{\mfrac{1}{(2\pi)^{\di/2} |\Sigma|^{1/2}}\exp\left(-\frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi}
{\left(\begin{multlined}[0.85\hsize]
\mfrac{1}{(2\pi)^{\di/2} |\Sigma|^{1/2}} \exp\left(-\mfrac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi \\
+ \mfrac{1}{(2\pi)^{\di/2} |\Sigma|^{1/2}} \exp\left(-\mfrac{1}{2}(x-\mu_0)^T \Sigma^{-1} (x-\mu_0) \right) (1-\phi)
\end{multlined}\right)}
\end{align}
\end{document}
注意，您没有提供 是什么的信息\di。请重新定义它或将其定义的包添加到序言中。
答案2
您可以将其设置为常规格式align*，左侧为零宽度框 ( \mathrlap)。然后您可以根据自己的喜好插入空格，这将根据您的需要移动右侧。下面我使用了5em，因此请更改它以满足您的需要。
\documentclass{article}
\usepackage[
margin=1in,
landscape
]{geometry}% Just for this example
\usepackage{mathtools}
\newcommand{\di}{\delta}
\begin{document}
\begin{align*}
\mathrlap{p(y = 1 \mid x; \phi, \mu_0, \mu_1, \Sigma)}
\hspace{5em} & \\ % <---------- Change to suit your needs
&= \frac{p(x \mid y = 1) p(y = 1)}{p(x)}\\
&= \frac{p(x \mid y = 1) p(y = 1)}{p(x \mid y = 1) p(y = 1) + p(x \mid y = 0) p(y = 0)} \\
&= \frac{\frac{1}{(2 \pi)^{\di / 2} |\Sigma|^{1 / 2}}
\exp\bigl(-\frac{1}{2} (x - \mu_1)^T \Sigma^{-1} (x - \mu_1) \bigr) \phi}{\frac{1}{(2 \pi)^{\di / 2} |\Sigma|^{1 / 2}}
\exp\bigl(-\frac{1}{2} (x - \mu_1)^T \Sigma^{-1} (x - \mu_1) \bigr) \phi + \frac{1}{(2 \pi)^{\di / 2} |\Sigma|^{1 / 2}}
\exp\bigl(-\frac{1}{2} (x - \mu_0)^T \Sigma^{-1} (x - \mu_0) \bigr) (1 - \phi)}
\end{align*}
\end{document}
一般来说，这种大型方程可以通过使用变量来表示常见元素（如 前面的分数）来简化exp。
答案3
您可以将\MoveEqLeft第一行的命令（我将前两行分组）与geometry包结合起来，以获得更合理的默认 matgins（除非您使用\marginpar 和来自的中等大小的分数nccmath：
\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{mathtools, amssymb, nccmath}
\begin{document}
\begin{align*}
\MoveEqLeft p(y = 1\mid x; \phi, \mu_0, \mu_1, \Sigma) = \frac{p(x\mid y=1)p(y=1)}{p(x)} \\
&= \frac{p(x\mid y=1)p(y=1)}{p(x\mid y=1)p(y=1)+p(x\mid y=0)p(y=0)} \\
&= \mfrac{\cfrac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi}{\cfrac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_1)^T \Sigma^{-1} (x-\mu_1) \right) \phi + \cfrac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp\left(-\frac{1}{2}(x-\mu_{0})^T \Sigma^{-1} (x-\mu_{0})\right) (1-\phi)}
\end{align*}
\end{document}
