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使用宏在方程中增加更大的边界

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使用宏在方程中增加更大的边界

我怎样才能排列这个具有更大边界的方程，类似于下图中的例子？

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我尝试过的：

\[\begin{split}\max_{\{W^{*}_{h,t}\}}E_{t}\sum_{s=0}^{\infty}(\xi_w \beta)^s\Biggl[\phi_{t+s}(1-\tau^w_{t+s})\mu^s_t\Pi^s_t                             W^{\star}_{h,t}\Biggl(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_ {t+s}\\%-\frac{\zeta^L_{t+s}}{1+\sigma_l}\Biggl(\biggl(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t+s}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_{t+s}\Biggr)^{1+\sigma_l}\Biggr]\end{split}\]

但我得到了：

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答案1

\documentclass{article}

\usepackage{amsmath}

\begin{document}
\[
  \max_{\{W^{*}_{h,t}\}}E_{t}\sum_{s=0}^{\infty}(\xi_w \beta)^s\left[
    \begin{array}{c}
      \phi_{t+s}(1-\tau^w_{t+s})\mu^s_t\Pi^s_t W^{\star}_{h,t}\Biggl(\frac{\mu^s_t\Pi^s_t
      W^{\star}_{h,t}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_ {t+s}
    \\
      -\frac{\zeta^L_{t+s}}{1+\sigma_l}\Biggl(\biggl(\frac{\mu^s_t\Pi^s_t
      W^{\star}_{h,t+s}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_{t+s}\Biggr)^{1+\sigma_l}
    \end{array}
    \right]
\]

\end{document}

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答案2

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而是split使用multlinedinside \left[ ...\right]。

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\[
\max_{\{W^{*}_{h,t}\}}E_{t}\sum_{s=0}^{\infty}(\xi_w \beta)^s
    \left[\begin{multlined}[0.75\textwidth]
    \phi_{t+s}(1-\tau^w_{t+s})\mu^s_t\Pi^s_t W^{\star}_{h,t}
    \left(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t}}{W_{t+s}}
    \right)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_ {t+s}\\[1ex]
%
    -\frac{\zeta^L_{t+s}}{1+\sigma_l}
    \left(\left(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t+s}}{W_{t+s}}
    \right)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}
        H_{t+s}\right)^{1+\sigma_l}
    \end{multlined}\right]
\]
\end{document}

答案3

您可以使用来自的代码mathtools来定义可以跨行分隔的成对分隔符。我调用了命令\brbracket。带星号的版本添加了一个隐式\left … \right对。

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{lmodern}

\newcommand\MTkillspecial[1]{% helper macro
\bgroup
\catcode`\&=9
\let\\\relax%
\scantokens{#1}%
\egroup
}

\DeclarePairedDelimiter\brbracket \lbrack \rbrack
\reDeclarePairedDelimiterInnerWrapper\brbracket{star}{
\mathopen{#1\vphantom{\MTkillspecial{#2}}\kern-\nulldelimiterspace\right.}
#2
\mathclose{\left.\kern-\nulldelimiterspace\vphantom{\MTkillspecial{#2}}#3}}

\begin{document}

 \[ \begin{split}\max_{\{W^{*}_{h,t}\}}E_{t}\sum_{s=0}^{\infty}(\xi_w \beta)^s\brbracket*{\phi_{t+s}(1-\tau^w_{t+s})\mu^s_t\Pi^s_t W^{\star}_{h,t}\Biggl(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_ {t+s} & \\-\frac{\zeta^L_{t+s}}{1+\sigma_l}\Biggl(\biggl(\frac{\mu^s_t\Pi^s_t W^{\star}_{h,t+s}}{W_{t+s}}\biggr)^{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}H_{t+s}\Biggr)^{1+\sigma_l} & }\end{split}\]

\end{document}

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答案4

我认为您正在寻找的是 2 行multline*环境或 2 行bmatrix环境——您发布的屏幕截图和编写的代码对于什么是正确的留下了一些歧义。

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\documentclass{article}
\usepackage{amsmath}
% define macros for some recurring items
\newcommand\terma{\mu^s_t\Pi^s_t W^{\star}_{\!h,t}}
\newcommand\termb{\frac{\lambda_{w,t+s}}{1-\lambda_{w,t+s}}}
\newcommand\termc{\biggl(\frac{\terma}{W_{t+s}}
   \biggr)^{\!\!\termb}H_{t+s}}

\begin{document} 
\begin{multline*}
\max_{\{W^{*}_{\!h,t}\}} E_{t} \sum_{s=0}^{\infty}(\xi_w \beta)^s
\left[\phi_{t+s}(1-\tau^w_{t+s}) \terma\termc \right.\\
\left.{} -\frac{\zeta^L_{t+s}}{1+\sigma_l} \Biggl(\termc\Biggr)^{\!\!1+\sigma_l}\right]
\end{multline*}
\[
\max_{\{W^{*}_{\!h,t}\}} E_{t} \sum_{s=0}^{\infty}(\xi_w \beta)^s
\begin{bmatrix}
\displaystyle
\phi_{t+s}(1-\tau^w_{t+s}) \terma\termc \\[3ex]
\displaystyle
-\frac{\zeta^L_{t+s}}{1+\sigma_l} \Biggl(\termc\Biggr)^{\!\!1+\sigma_l}
\end{bmatrix}
\]
\end{document}

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