我想显示数学结果的分步证明。我希望在amsmath
的align
环境中有步骤遵循其他步骤,但我想要\pause
证明中每一行后面的幻灯片即一次展示一个步骤,以便我的学生不会因为看幻灯片而分心或感到困惑。
\documentclass{beamer}
\usepackage[protrusion,expansion,kerning,tracking,spacing]{microtype}
\usepackage[utf8]{inputenc}
\usepackage{multicol}
\usepackage{framed}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\microtypecontext{spacing=nonfrench}
\nonfrenchspacing
\usetheme{Berkeley}
\usefonttheme{serif}
\begin{document}
\section{Example 1}
\begin{frame}[allowframebreaks]{\(u\)-substitution example}
\begin{framed}
Find the antiderivative of the function defined by \(f(x) = \frac{x}{\sqrt{x^2 + 4}}\).
\end{framed}
\begin{align}
\intertext{Introduce the auxiliary variable $u$.}
u &= x^2 + 4\\
\intertext{Differentiate with respect to $x$.}
\frac{du}{dx} &= 2x\\
\intertext{Multiply through by the differential.}
du &= 2x\,dx\\
\intertext{Substitute $u$ into $f$.}
f(x) &= \frac{x}{\sqrt{u}}\\
\intertext{Substitute into antiderivative expression.}
\int f(x)\,dx &= \int \frac{x}{\sqrt{u}}\,dx\\
\intertext{Extract a constant.}
&= \frac{1}{2}\int \frac{2x}{\sqrt{u}}\,dx
\intertext{Rewrite differential in terms of $u$.}
&= \frac{1}{2}\int \frac{1}{\sqrt{u}}\left(2x\,dx\right)\\
&= \frac{1}{2}\int \frac{1}{\sqrt{u}}\,du
\intertext{Apply the power rule for indefinite integration.}
&= \frac{1}{2}\int u ^ {-\frac{1}{2}}\,du\\
&= \frac{1}{2}\frac{\sqrt{u}}{1/2} + C\\
&= \sqrt{u} + C
\intertext{Substitute $u$.}
\int f(x)\,dx &= \sqrt{x^2 + 4} + C
\end{align}
\end{frame}
\end{document}
答案1
您\uncover<x-y>
很可能可以\uncover<+->
自由使用(即“从这里开始揭开”)。它没有以任何方式影响格式。只需确保不要引入虚假空格(行尾是 (La)TeX 的空格,除非行以以 开头的注释结尾%
,否则“受保护”行尾后的空格将被忽略)。例如,我会写类似的东西:
\uncover<+->{%
This is a step in my proof%
}%
\uncover<+->{%
, continued here.
}
这样,逗号前就不会出现多余的空格。
答案2
我不确定我是否理解了这个问题。看起来证明不能放在一张幻灯片上,但可以将第二行作为的参数,\uncover<2->{.....}
将第三行作为的参数,\uncover<3->{...}
依此类推。这是我最近发表的一次演讲中的一个实际框架:
\begin{frame}{Some properties of $G$ and $K$}
\begin{itemize}[<+->]
\item\hiblue{$G(D)$ is a subring of the perfect closure of the field of
fractions of $D$.}
\item\hiblue{The inner adjunction $R\to K(R)$ is an injection.}
\item\hiblue{The inner adjunction $R\to K(R)$ is epic in semiprime
rings.}
\item\hiblue{The induced $\spec(K(R))\to\spec(R)$ is a bijection.}
\end{itemize}
\uncover<5->{\hired{$K$ is {\bf domain reflective} if it takes
domains to domains.}}
\uncover<6->{$K_\fld$ is not domain reflective.}
\end{frame}
前四项也一一展示出来。