我正在海报中编制一组方程式,我想用\vdots
表示延续,代码如下:
\documentclass[portrait,a0b,final]{a0poster}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{mathtools}
\begin{itemize}
\item \emph{Calculation of Steady State with Successive Transition Matrices}
\begin{itemize}
\item Starting with the matrix power averaging technique, the time inhomogeneous steady state can be determined by the following steps:
\end{itemize}
\begin{align*}
P\,(\tau) &= T_{1} \, P(0) \\
\\
P\,(2\tau) &= T_{2} \, T_{1} \, P(0) \\
\\
P\,(3\tau) &= T_{3}\, T_{2}\, T_{1} \, P(0) \\
\shortvdotswithin{=}
\end{align*}
\end{itemize}
\end{document}
但编译后的版本看起来又小又挤:
如何解决这个问题,让它们的间距更长,并且点的尺寸也更大?
答案1
的原始定义\vdots
:
\vbox {\baselineskip 4\p@ \lineskiplimit \z@ \kern 6\p@ \hbox {.}\hbox {.}\hbox {.}}
使用正常的句号(参见\hbox
命令的参数)。
如果您希望所有内容\vdots
都更大并且具有更多的垂直空间,您可以重新定义它,例如使用:
\documentclass[portrait,a0b,final]{a0poster}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{mathtools}
\makeatletter
\DeclareRobustCommand{%
\vdots}{\vbox {\huge\baselineskip 12\p@ \lineskiplimit \z@ \kern 6\p@
\hbox {.}\hbox {.}\hbox {.}}%
}
\makeatother
\begin{document}
\begin{itemize}
\item \emph{Calculation of Steady State with Successive Transition Matrices}
\begin{itemize}
\item Starting with the matrix power averaging technique, the time inhomogeneous steady state can be determined by the following steps:
\end{itemize}
\begin{align*}
P\,(\tau) &= T_{1} \, P(0) \\
\\
P\,(2\tau) &= T_{2} \, T_{1} \, P(0) \\
\\
P\,(3\tau) &= T_{3}\, T_{2}\, T_{1} \, P(0) \\
&\shortvdotswithin{=}
\end{align*}
\end{itemize}
\end{document}
除了使用,\huge
您还可以.
用您想要的任何东西来替换,例如$\bullet$
(但在我看来$\bullet$
太多了)。
除了重新定义,\vdots
您还可以定义\vdotswithin
和的新版本\shortvdotswithin
:
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{mathtools}
\makeatletter
\DeclareRobustCommand\hugevdots{%
\vbox {\huge\baselineskip 12\p@ \lineskiplimit \z@ \kern 6\p@
\hbox {.}\hbox {.}\hbox {.}}%
}
\newcommand\hugevdotswithin[1]{%
{\mathmakebox[\widthof{\ensuremath{{}#1{}}}][c]{{\hugevdots}}}}
\ExplSyntaxOn
\def\shorthugevdotswithin{\relax
\@ifstar{\MT_svwi_aux_huge:nn{00}}{\MT_svwi_aux_huge:nn{01}}}
\def\MT_svwi_aux_huge:nn #1#2{
\MTFlushSpaceAbove
\if#1 \hugevdotswithin{#2}& \else &\hugevdotswithin{#2} \fi
\MTFlushSpaceBelow
}
\ExplSyntaxOff
\makeatother
\begin{document}
\begin{itemize}
\item \emph{Calculation of Steady State with Successive Transition Matrices}
\begin{itemize}
\item Starting with the matrix power averaging technique, the time inhomogeneous steady state can be determined by the following steps:
\end{itemize}
\begin{align*}
P\,(\tau) &= T_{1} \, P(0) \\
\\
P\,(2\tau) &= T_{2} \, T_{1} \, P(0) \\
\\
P\,(3\tau) &= T_{3}\, T_{2}\, T_{1} \, P(0) \\
&\shorthugevdotswithin{=}
\end{align*}
\end{itemize}
\end{document}
两个示例的结果都是:
如果需要,您还可以增加值\kern
以增加第一个点上方的空间,或者\kern
在最后一个点后添加一个额外的点\hbox{…}
以增加下方的空间。
答案2
我相信您只需进行修复\vdots
,它就会考虑当前的字体大小。
\documentclass[portrait,a0b,final]{a0poster}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{mathtools}
\makeatletter
\renewcommand{\vdots}{%
\vbox{%
\baselineskip \fpeval{0.3*\f@baselineskip}\p@
\lineskiplimit\z@
\kern\fpeval{0.5*\f@baselineskip}\p@
\hbox{.}\hbox{.}\hbox{.}%
}%
}
\makeatother
\begin{document}
\begin{align*}
P(\tau) &= T_{1} \, P(0) \\
\\
P(2\tau) &= T_{2} T_{1} \, P(0) \\
\\
P(3\tau) &= T_{3} T_{2} T_{1} \, P(0) \\
\shortvdotswithin{=}
P(n\tau) &= T_{n} \dotsm T_{1} \, P(0)
\end{align*}
\end{document}
问题在于该标准\vdots
使用固定尺寸。
我移除了很多放错位置的\,
令牌。