您是否知道如何减少方程环境中的边距,以确保很长的方程式适合一行(我已经减小了方程式的字体大小)?谢谢
这是我的代码:
\documentclass[11 pt,a4paper,oneside,openany, notitlepage]{article}
\input epsf
\usepackage{amsmath, amssymb, graphics}
\newcommand{\mathsym}[1]{{}}
\newcommand{\unicode}[1]{{}}
\newcounter{mathematicapage}
\usepackage{amsthm}
\usepackage{amsfonts}
\marginparwidth 0pt
\oddsidemargin 0pt
\evensidemargin 0pt
\marginparsep 0pt
\linespread{1.5}
\topmargin 0pt
\textwidth 6.5in
\textheight 8.5 in
\begin{document}
{\tiny
\begin{equation}
\begin{pmatrix}
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{l,t,R}^{(1)}(x_t,n;\theta)\hat{\mathbb{P}}_{x_t,n,T}-\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{1,x_t,n,T}\}-\sum_{x \in \mathcal{X}}^{}(\max\{0,H_{l,t}^{(1)}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)-\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(1)},X_t=x,N_t=n)\})\\
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{1,x_t,n,T}-\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{u,t,R}^{(1)}(x_t,n;\theta)\mathbb{P}_{x_t,n,T}\}-\sum_{x \in \mathcal{X}}^{}(\max\{0,\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(1)},X_t=x,N_t=n)-H_{u,t}^{(1)}(x,n;\theta)\mathbb{P}(X_t=x,N=n)\})\\
\vdots\\
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{l,t,R}^{(2^{n-1})}(x_t,n;\theta)\hat{\mathbb{P}}_{x_t,n,T}-\hat{\mathbb{P}}_{2^{n-1},x_t,n,T}\}-\sum_{x \in \mathcal{X}}^{}(\max\{0,H_{l,t}^{(2^{n-1})}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)-\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(2^{n-1})},X_t=x,N_t=n)\})\\
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{2^{n-1},x_t,n,T}-\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{u,t,R}^{(2^{n-1})}(x_t,n;\theta)\hat{\mathbb{P}}_{x_t,n,T}\}-\sum_{x \in \mathcal{X}}^{}(\max\{0,\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(2^{n-1})},X_t=x,N_t=n)-H_{u,t}^{(2^{n-1})}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)\})\\
\end{pmatrix}\\
\end{equation}
}
\end{document}
答案1
如果愿意的话,你可以把它当作\parenVectorstack
数学。\textstyle
\documentclass[11 pt,a4paper,oneside,openany, notitlepage]{article}
\input epsf
\usepackage{amsmath, amssymb, graphicx}
\newcommand{\mathsym}[1]{{}}
\newcommand{\unicode}[1]{{}}
\newcounter{mathematicapage}
\usepackage{amsthm}
\usepackage{amsfonts}
\marginparwidth 0pt
\oddsidemargin 0pt
\evensidemargin 0pt
\marginparsep 0pt
\linespread{1.5}
\topmargin 0pt
\textwidth 6.5in
\textheight 8.5 in
\usepackage[usestackEOL]{stackengine}
\stackMath
\begin{document}
\savestack{\rowA}{\Longstack[r]{%
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{l,t,R}^{(1)}(x_t,n;%
\theta)\hat{\mathbb{P}}_{x_t,n,T}-\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{1,x_t,n,T}\}\\
-\sum_{x \in \mathcal{X}}^{}(\max\{0,H_{l,t}^{(1)}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)\\
-\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(1)},X_t=x,N_t=n)\})%
}}
\savestack{\rowB}{\Longstack[r]{%
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{1,x_t,n,T}-\frac{1}%
{T}\sum_{t=1}^{T}\hat{H}_{u,t,R}^{(1)}(x_t,n;\theta)\mathbb{P}_{x_t,n,T}\}\\
-\sum_{x \in \mathcal{X}}^{}(\max\{0,\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(1)},X_t=x,N_t=n)\\
-H_{u,t}^{(1)}(x,n;\theta)\mathbb{P}(X_t=x,N=n)\})
}}
\savestack{\rowC}{\Longstack[r]{%
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{H}_{l,t,R}^{(2^{n-1})}(x_t,n;%
\theta)\hat{\mathbb{P}}_{x_t,n,T}-\hat{\mathbb{P}}_{2^{n-1},x_t,n,T}\}\\
-\sum_{x \in \mathcal{X}}^{}(\max\{0,H_{l,t}^{(2^{n-1})}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)\\
-\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(2^{n-1})},X_t=x,N_t=n)\})
}}
\savestack{\rowD}{\Longstack[r]{%
\max\{0,\frac{1}{T}\sum_{t=1}^{T}\hat{\mathbb{P}}_{2^{n-1},x_t,n,T}-\frac{1}%
{T}\sum_{t=1}^{T}\hat{H}_{u,t,R}^{(2^{n-1})}(x_t,n;\theta)\hat{\mathbb{P}}_{x_t,n,T}\}\\
-\sum_{x \in \mathcal{X}}^{}(\max\{0,\mathbb{P}(G_{\cdot j,t}=g_{\bullet}^{(2^{n-1})},X_t=x,N_t=n)\\
-H_{u,t}^{(2^{n-1})}(x,n;\theta)\mathbb{P}(X_t=x,N_t=n)\})
}}
\begin{equation}
\setstackgap{L}{4\baselineskip}
\parenVectorstack{\rowA \\ \rowB \\ \raisebox{\baselineskip}{\vdots} \\ \rowC \\ \rowD}
\end{equation}
\end{document}
将末尾附近的行更改为
\scalebox{.7}{$%
\parenVectorstack{\rowA \\ \rowB \\ \raisebox{\baselineskip}{\vdots} \\ \rowC \\ \rowD}$}
会减小物体的尺寸: