我想强调并放大此对齐环境的第一行和最后一行。这是因为中间的两行当然很重要,但不如结果重要。我正在做一个投影仪演示,因此
\documentclass{beamer}
\begin{document}
\begin{frame}
{\scriptsize
\begin{align*}
\mathcal{I}(n) &= \int_\mathbb{R} \exp\{-n g(y)\} dy\\
&= \int_{\mathbb{R}} \exp \left\{-n\left[\tilde{g} + \frac{1}{2}(y-\tilde{y})^{2} \tilde{g}^{\prime \prime} + \frac{1}{6}(y - \tilde{y})^3 \tilde{g}^{\prime \prime \prime} + \frac{1}{24}(y - \tilde{y})^4 \tilde{g}^{I V} + O((y-\tilde{y})^{5})\right]\right\} d y \\
&= \exp\{-n \tilde{g}\} \int_{\mathbb{R}} \exp \left\{-\frac{n}{2}(y-\tilde{y})^{2} \tilde{g}^{\prime \prime} - \frac{n}{6}(y - \tilde{y})^3 \tilde{g}^{\prime \prime \prime} - \frac{n}{24}(y - \tilde{y})^4 \tilde{g}^{I V}+ nO(y-\tilde{y})^{5}\right\} d y \\
&=\frac{\exp\{-n \tilde{g}\} \sqrt{2 \pi}}{\sqrt{n \tilde{g}^{\prime \prime}}}\left\{1-\frac{1}{8} \frac{\tilde{g}^{IV}}{n(\tilde{g}^{\prime \prime})^2} + \frac{5}{24n} \frac{(\tilde{g}^{\prime \prime \prime})^2}{(\tilde{g}^{\prime \prime})^3} + O (n^{-2})\right\},
\end{align*}}
\end{frame}
\end{document}
有人可以帮忙吗?
答案1
我不会这么做:如果计算在演示中不重要,就省略它们。你的观众没有时间解析这些长公式,更不用说遵循这些步骤了。
无论如何,你可以这样做。
\documentclass{beamer}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\REDUCE}[1]{\mbox{\tiny$\displaystyle#1$}}
\begin{document}
\begin{frame}
\begin{align*}
\mathcal{I}(n) &= \int_{\RR} \exp\{-n g(y)\} \, dy\\[2ex]
&= \REDUCE{
\int_{\RR} \exp \left\{
-n\left[
\tilde{g} + \frac{1}{2}(y-\tilde{y})^{2} \tilde{g}''
+ \frac{1}{6}(y - \tilde{y})^3 \tilde{g}'''
+ \frac{1}{24}(y - \tilde{y})^4 \tilde{g}^{IV} + O((y-\tilde{y})^{5})
\right]
\right\} \, dy
} \\[1ex]
&= \REDUCE{
\exp\{-n \tilde{g}\} \int_{\RR} \exp \left\{
-\frac{n}{2}(y-\tilde{y})^{2} \tilde{g}''
- \frac{n}{6}(y - \tilde{y})^3 \tilde{g}'''
- \frac{n}{24}(y - \tilde{y})^4 \tilde{g}^{I V}
+ nO(y-\tilde{y})^{5}
\right\} \, dy
} \\[2ex]
&=\frac{\exp\{-n \tilde{g}\} \sqrt{2 \pi}}{\sqrt{n \tilde{g}''}}
\left\{
1-\frac{1}{8} \frac{\tilde{g}^{IV}}{n(\tilde{g}'')^2}
+ \frac{5}{24n} \frac{(\tilde{g}''')^2}{(\tilde{g}'')^3}
+ O (n^{-2})
\right\},
\end{align*}
\end{frame}
\end{document}