我正在准备一个 beamer 演示文稿。编译使用 pdflatex 完成。其中一个框架包含几个方程式，其中一个方程式太长，无法放入页面宽度。我希望自动调整整个框架的字体大小以适合页面，同时保留原始行距。请注意，我不想更改其他框架中的字体大小。
编辑：这是代码：
\documentclass[aspectratio=43]{beamer}
\usepackage{tcolorbox}
\mode<presentation>
\hypersetup{pdfpagemode=FullScreen}
\usetheme{Goettingen}
\setbeamercovered{invisible}
\usefonttheme[onlymath]{serif}
\setbeamerfont{title in sidebar}{size=\fontsize{5.5pt}{6pt}\selectfont}
\setbeamerfont{author in sidebar}{size=\fontsize{5.5pt}{6pt}\selectfont}
\setbeamerfont{normal}{size=\fontsize{5pt}\selectfont}
\setbeamertemplate{frametitle}{\parbox[c][][c]{\linewidth}{\centering\small~\vspace{1.0em}\\%
~{\smash{\textbf{\Large\insertframetitle}}}~\\%
~{\smash{\rm\insertframesubtitle}}~}}
%
\makeatletter
\def\section@yeshadestyle{}%
\def\section@noshadestyle{\color{section in sidebar.fg}}%
\def\subsection@yeshadestyle{}%
\def\subsection@noshadestyle{}%\color{subsection in sidebar.fg}}%
\def\scondfont{\protect\@scondfont[\thesection]}
\def\@scondfont [#1]{\ifnum#1=\c@section%
\section@noshadestyle\else% yes section
\section@yeshadestyle\fi}% no section
\def\sscondfont{\protect\@sscondfont[\thesection.\thesubsection]}
\def\@sscondfont [#1.#2]{\ifnum#1=\c@section\ifnum#2=\c@subsection%
\subsection@noshadestyle\else% yes section & yes subsection
\subsection@yeshadestyle\fi\else% yes section & no subsection
\subsection@yeshadestyle\fi}% no section & no subsection
\makeatother
\def\safe\\{\texorpdfstring{\\}{,~}}%
\title[High-Order Embedded Finite Difference Schemes for Initial Boundary Value Problems]
{High-Order Embedded Finite Difference\safe\\Schemes for Initial Boundary Value Problems\safe\\Involving Mixed Derivative Terms in Complex Domains}
\def\SupervisorsGAP{\texorpdfstring{\scriptsize\vspace{1em}\\Supervised~by:~}{,~}}
\author[Mr. Tomer Livneh]
{Mr. Tomer Livneh\SupervisorsGAP%
Dr. Adi Ditkowski}
\date{September, 2015}
\gdef\ttx#1{\text{\tiny{#1}}}
\begin{document}
\begin{frame}[t]{The two-dimensional case}{The scheme}%
\onslide<1->
\begin{tcolorbox}
$\frac{d{\bf V}}{dt} = ({\cal M}^{(x)} + P^T {\cal M}^{(y)}P){\bf V} + {\bf G}^{(x)} + P^T{\bf G}^{(y)} + {\bf F} \;\; \nonumber$
\end{tcolorbox}
\onslide<2->
\resizebox{\textwidth}{!}{\parbox{\linewidth}{%
\begin{equation}
{\cal M}^{(x)} = \left [ \begin{array}{llll}
M^{(x)}_1\\
& M^{(x)}_2\\
& & \ddots\\
& & & M^{(x)}_{M_{R}}
\end{array}\right ]; {\cal M}^{(y)} = \left [ \begin{array}{llll} M^{(y)}_1 \\ & M^{(y)}_2\\ & & \ddots\\ & & & M^{(y)}_{M_{C}} \end{array}\right ] ; \nonumber
\end{equation}
\begin{flalign}
M^{(x)}_k &= D^{(x)}_k -\tau_{L_{k}}A_{L_{k}}-\tau_{R_{k}}A_{R_{k}} &\nonumber \\
M^{(y)}_{j} &= D^{(y)}_j - \tau_{B_{j}}A_{B_{j}}-\tau_{T_{j}}A_{T_{j}} &\nonumber \\
{\bf G}^{(x)} & = \left[ (\tau_{L_{1}}{\bf g}_{L_{1}}+\tau_{R_{1}}{\bf g}_{R_{1}}), \ldots , (\tau_{L_{k}}{\bf g}_{L_{k}}+\tau_{R_{k}}{\bf g}_{R_{k}}), \ldots ,(\tau_{L_{M_{R}}}{\bf g}_{L_{M_{R}}}+\tau_{R_{M_{R}}}{\bf g}_{R_{M_{R}}}) \right] &\nonumber\\
{\bf G}^{(y)}& = \left[ (\tau_{B_{1}}{\bf g}_{B_{1}}+\tau_{T_{1}}{\bf g}_{T_{1}}), \ldots ,(\tau_{B_{j}}{\bf g}_{B_{j}} +\tau_{T_{j}} {\bf g}_{T_{j}}),\ldots ,(\tau_{B_{M_{C}}}{\bf g}_{B_{M_{C}}}+\tau_{T_{M_{C}}}{\bf g}_{T_{M_{C}}}) \right] &\nonumber
\end{flalign}
}}
\end{frame}
\end{document}
答案1
我不确定保留行距是否是正确的做法。
和
\begin{frame}
\expandafter\small\expandafter\baselineskip\the\baselineskip
...
\end{frame}
你应该得到你想要的。\small你可以尝试\footnotesize