我如何才能像下图这样排列这个等式?
这是我使用的代码和输出:
\[
\begin{split}
\Biggl\{ \biggl( {1 - \frac{1}{1 + \rho_{f, t + s}}} \biggr)
P^{\circ}^{X}_{t + s\textbar{t}} (i)^{-\frac{1}{1 + \rho_{f, t +
s}}} \Biggl( \frac{1}{\Pi^{\theta_{x}}_{x, t - 1 + s}
\bar{\Pi}^{1 - \theta_{x}} P^X_{t - 1 + s}} \Biggr)^{-\frac{1}{1
+ \rho_{f, t + s}}} - \\%
\frac{S^{-1}_{t + s}\mathcal{MC}_{x, t +
s}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1 -
\theta_{x}} P^X_{t - 1 + s}} \left( -\frac{1}{1 + \rho_{f, t +
s}} \right) \Biggl[ \frac{P^{\circ}^{X}_{t +
s\textbar{t}}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1
- \theta_{x}}P^X_{t - 1 + s}} \Biggr]^{-\frac{1}{1 + \rho_{f,
t + s}} - 1} \Biggr\} X_{t + s\textbar{t}}
\end{split}
\]
答案1
我将把这个等式重写成:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\[
\begin{multlined}[b][0.7\textwidth]
\left\{\left(1 - A\right) P^{\circ X}_{t + s | t} (i)^{-A}
\left(\frac{1}{B}\right)^{-A} \right.\\
\left. - \frac{S^{-1}_{t + s}\mathcal{MC}_{x, t + s}(i)}{B} \left(-A\right)
\left[ \frac{P^{\circ X}_{t + s | t}(i)}{B} \right]^{-A - 1}
\right\} X_{t + s | t} = ?% not specified
\end{multlined}
\]
where are:
\[
A = \frac{1}{1 + \rho_{f, t + s}}
\quad\text{and}\quad
B = \Pi^{\theta_x}_{x,t-1+s}\bar{\Pi}^{1-\theta_x}_{} P^{X}_{t-1+s}
\]
\end{document}
答案2
使用gathered
环境 from mathtools
,并移动大括号和随后的内容外部该环境产生下列结果。
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\[
\left\{
\begin{gathered}
\biggl( {1 - \frac{1}{1 + \rho_{f, t + s}}} \biggr)
P^{\circ X}_{t + s | t} (i)^{-\frac{1}{1 + \rho_{f, t +
s}}} \Biggl( \frac{1}{\Pi^{\theta_{x}}_{x, t - 1 + s}
\bar{\Pi}^{1 - \theta_{x}} P^X_{t - 1 + s}} \Biggr)^{-\frac{1}{1
+ \rho_{f, t + s}}} - \\%
\frac{S^{-1}_{t + s}\mathcal{MC}_{x, t +
s}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1 -
\theta_{x}} P^X_{t - 1 + s}} \left( -\frac{1}{1 + \rho_{f, t +
s}} \right) \Biggl[ \frac{P^{\circ X}_{t +
s | t}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1
- \theta_{x}}P^X_{t - 1 + s}} \Biggr]^{-\frac{1}{1 + \rho_{f,
t + s}} - 1}
\end{gathered}
\right\} X_{t + s | t}
\]
\end{document}
答案3
您可以使用lrcases
环境(cases
带有左括号和右括号,遵循@Gonzalo Medina 的代码)并nccmath
获得中等大小的分数:
\documentclass{article}
\usepackage{mathtools, nccmath}
\makeatletter
\newcases{dlrcases}{\quad}{%
$\m@th\displaystyle{##}$\hfil}{$\m@th\displaystyle{##}$\hfil}{\lbrace}{\rbrace}
\newcases{lrcases}{\quad}{%
$\m@th{##}$\hfil}{{##}\hfil}{\lbrace}{\rbrace}
\makeatother
\begin{document}
\[
\begin{lrcases} \Bigl( {1 - \mfrac{1}{1 + \rho_{f, t + s}}}\Bigr)
P^{ ∘ X}_{t + s\textbar{t}} (i)^{-\frac{1}{1 + \rho_{f, t +
s}}} \Bigl( \mfrac{1}{\Pi^{\theta_{x}}_{x, t - 1 + s}
\bar{Π}^{1 - \theta_{x}} P^X_{t - 1 + s}} \Bigr)^{-\frac{1}{1
+ \rho_{f, t + s}}} - \\%
\mfrac{S⁻¹_{t + s}\mathcal{MC}_{x, t +
s}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{Π}^{1 -
\theta_{x}} P^X_{t - 1 + s}} \Bigl( -\mfrac{1}{1 + \rho_{f, t +
s}} \Bigr) \biggl[ \mfrac{P^{ ∘ X}_{t +
s\textbar{t}}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{Π}^{1
- \theta_{x}}P^X_{t - 1 + s}} \biggr]^{-\frac{1}{1 + \rho_{f,
t + s}} - 1} \end{lrcases} X_{t + s\textbar{t}}
\]
\end{document}
答案4
下面是使用堆栈的一种方法(注意,我还必须将其更改^{\circ}^{X}
为^{\circ X}
):
\documentclass{article}
\usepackage{stackengine}
\begin{document}
\[
\left\{
\ensurestackMath{\stackanchor[10pt]{
\biggl( {1 - \frac{1}{1 + \rho_{f, t + s}}} \biggr)
P^{\circ X}_{t + s\textbar{t}} (i)^{-\frac{1}{1 + \rho_{f, t +
s}}} \Biggl( \frac{1}{\Pi^{\theta_{x}}_{x, t - 1 + s}
\bar{\Pi}^{1 - \theta_{x}} P^X_{t - 1 + s}} \Biggr)^{-\frac{1}{1
+ \rho_{f, t + s}}} -
}{
\frac{S^{-1}_{t + s}\mathcal{MC}_{x, t +
s}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1 -
\theta_{x}} P^X_{t - 1 + s}} \left( -\frac{1}{1 + \rho_{f, t +
s}} \right) \Biggl[ \frac{P^{\circ X}_{t +
s\textbar{t}}(i)}{\Pi^{\theta_{x}}_{x, t - 1 + s} \bar{\Pi}^{1
- \theta_{x}}P^X_{t - 1 + s}} \Biggr]^{-\frac{1}{1 + \rho_{f,
t + s}} - 1}
}}
\right\} X_{t + s\textbar{t}}
\]
\end{document}
\documentclass[final,1p,times,authoryear]{elsarticle}
按照 OP 在评论中引用的方式进行替换,仍然可以编译。注意。我正在使用stackengine 2016/10/04 v4.00
: