![投影仪中对齐方程式的方程式中断](https://linux22.com/image/416666/%E6%8A%95%E5%BD%B1%E4%BB%AA%E4%B8%AD%E5%AF%B9%E9%BD%90%E6%96%B9%E7%A8%8B%E5%BC%8F%E7%9A%84%E6%96%B9%E7%A8%8B%E5%BC%8F%E4%B8%AD%E6%96%AD.png)
问题:
在这里,我正在研究一个例子,其中我有一些正在使用对齐的方程式。
问题是,我可以在不编写新的对齐环境的情况下将下一个方程式放在下一张幻灯片上吗?第二个问题:我的最后一个方程式很长,我可以将其拆分吗?
梅威瑟:
\documentclass[12pt,fleqn]{beamer}
\usetheme{AnnArbor}
\usepackage{setspace}
\usepackage{amsmath}
\usepackage[utf8]{inputenc}
%\setstretch{1.0}
\usecolortheme{beaver}
\usefonttheme{professionalfonts} % using non standard fonts for beamer
\usefonttheme{serif} % default family is serif
\addtobeamertemplate{frametitle}{}{\vspace{-0.4em}} % decrease
\makeatletter
\newcommand{\Pause}[1][]{\unless\ifmeasuring@\relax
\pause[#1]%
\fi}
\makeatother
\title[pqr lmn] %optional
{Chapter}
\subtitle{Lecture - 0}
\author[X. Y. Z] % (optional, for multiple authors)
{x.~y.~z \\ abc \\ def}
%\date[\today] % (optional)
\begin{document}
\frame{\titlepage}
\begin{frame}
%\setstretch{1.0}
\textbf{Find nth order derivative of} $\boldsymbol{\cos x \cos 2x \cos 3x}$
\begin{align*}
y&=\cos x \cos 2x \cos 3x\\[8pt]
&=\dfrac{1}{2}\cos x \left(2\,\cos 3x \cos 2x\right)\\[12pt]
&=\dfrac{1}{2}\,\cos x \,\left(\cos 5x + \cos x\right)\\[12pt]
&=\dfrac{1}{4}\,\left[2 \cos 5x \cos x + 2\cos^2 x\right]\\[12pt]
&=\dfrac{1}{4}\,\left[\cos 6x + \cos 4x + 1 + \cos 2x\right]
\end{align*}
\end{frame}
\begin{frame}
%\setstretch{1.0}
$$\therefore\;\;y_n=\dfrac{1}{4}\,\left[\left\{\cos 6x\right\}_n + \left\{\cos 4x\right\}_n + \left\{1\right\}_n + \left\{\cos 2x\right\}_n\right]$$
$$=\dfrac{1}{4}\,\left[6^n\,\left\{\cos 6x+\dfrac{n\pi}{2}\right\} + 4^n\,\left\{\cos 4x+\dfrac{n\pi}{2}\right\} + 0 + 2^n\,\left\{\cos 2x+\dfrac{n\pi}{2}\right\}\right]$$
\end{frame}
\end{document}
答案1
通过将所有实例更改为\dfrac
,\tfrac
删除所有\left
和\right
指令,可以在一个框架上编写整个表达式。哦,我会去掉\,
细空格。它们根本不需要很好地排版数学材料。
\documentclass[12pt,fleqn]{beamer}
\usetheme{AnnArbor}
\usepackage{setspace}
%\setstretch{1.0}
% \usepackage{amsmath} % is loaded automatically by 'beamer' class
\usepackage[utf8]{inputenc}
\usecolortheme{beaver}
\usefonttheme{professionalfonts} % use non standard fonts
\usefonttheme{serif} % default family is CM
\begin{document}
\begin{frame}
\textbf{\boldmath Find the $n$th order derivative of $\cos x \cos 2x \cos 3x$}
\begin{align*}
y&=\cos x \cos 2x \cos 3x\\[4pt]
&=\tfrac{1}{2}\cos x (2\cos 3x \cos 2x)\\[4pt]
&=\tfrac{1}{2}\cos x (\cos 5x + \cos x)\\[4pt]
&=\tfrac{1}{4}[2 \cos 5x \cos x + 2\cos^2 x]\\[4pt]
&=\tfrac{1}{4}[\cos 6x + \cos 4x + 1 + \cos 2x]\\[8pt]
\therefore\quad
y_n &=\tfrac{1}{4}\bigl[\{\cos 6x\}_n + \{\cos 4x\}_n + \{1\}_n + \{\cos 2x\}_n\bigr] \\[4pt]
&=\tfrac{1}{4}\bigl[6^n\bigl\{\cos 6x+\tfrac{n\pi}{2}\bigr\} + 4^n\bigl\{\cos 4x+\tfrac{n\pi}{2}\bigr\} +0 \\
&\qquad + 2^n\bigl\{\cos 2x+\tfrac{n\pi}{2}\bigr\}\bigr]
\end{align*}
\end{frame}
\end{document}
附录回应 OP 的后续请求:frame
如果有必要,我会这样将材料分成两部分。
\documentclass[12pt,fleqn]{beamer}
\usetheme{AnnArbor}
\usepackage{setspace}
%\usepackage{amsmath} % is loaded automatically by 'beamer'
\usepackage[utf8]{inputenc}
\usecolortheme{beaver}
\usefonttheme{professionalfonts} % use non standard fonts
\usefonttheme{serif} % default family is CM
\begin{document}
\begin{frame}[t]
\frametitle{Find the $n$th order derivative of $\cos x \cos 2x \cos 3x$}
\begin{align*}
y &=\cos x \cos 2x \cos 3x\\[4pt]
&=\tfrac{1}{2}\cos x (2\cos 3x \cos 2x)\\[4pt]
&=\tfrac{1}{2}\cos x (\cos 5x + \cos x)\\[4pt]
&=\tfrac{1}{4}[2 \cos 5x \cos x + 2\cos^2 x]\\[4pt]
&=\tfrac{1}{4}[\cos 6x + \cos 4x + 1 + \cos 2x]
\end{align*}
\end{frame}
\begin{frame}[t]
\frametitle{(finding the $n$th order derivative, cont'd)}
\begin{align*}
\therefore\quad
y_n &=\tfrac{1}{4}\bigl[\{\cos 6x\}_n + \{\cos 4x\}_n + \{1\}_n + \{\cos 2x\}_n\bigr] \\[4pt]
&=\tfrac{1}{4}\bigl[6^n\bigl\{\cos 6x+\tfrac{n\pi}{2}\bigr\} + 4^n\bigl\{\cos 4x+\tfrac{n\pi}{2}\bigr\} +0 \\
&\qquad + 2^n\bigl\{\cos 2x+\tfrac{n\pi}{2}\bigr\}\bigr]
\end{align*}
\end{frame}
\end{document}