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--- a/macros/latex/contrib/easybook/doc/chapter2.tex
+++ b/macros/latex/contrib/easybook/doc/chapter2.tex
@@ -1,80 +1,111 @@
% Chapter 2
-\chapter{定理盒子}
+\chapter{常用环境}
-\begin{outline}
-\index{zhaiyao@摘要盒子}
- \item 这是一个摘要盒子。
- \item 它的标题是可选参数,默认标题是摘要。
-\end{outline}
+\section{公式}
+\index{g@公式}
+\zhlipsum*[42][name = aspirin]
+\begin{equation}\label{eq:matrix}
+ \bm{A} = \left[
+ \begin{matrix}
+ a_{11} & a_{12} & \cdots & a_{1n} \\
+ a_{21} & a_{22} & \cdots & a_{2n} \\
+ \vdots & \vdots & \ddots & \vdots \\
+ a_{m1} & a_{m2} & \cdots & a_{mn} \\
+ \end{matrix}
+ \right] = \left[
+ \begin{matrix}
+ \bm{a}_1 & \bm{a}_2 & \cdots & \bm{a}_n
+ \end{matrix}
+ \right]
+\end{equation}
+可以通过添加标签在正文中引用公式,如带有中文括号的引用\ref{eq:matrix}。
-\section{定理}
-\index{d@定理}
-\begin{theorem}[(高斯公式\footnote{选自高等数学公式。})]
-\label{theorem:gauss formula}
-\index{d@定理!g@高斯公式}
-设空间闭区域$\Omega$是由分片光滑的闭曲面$\Sigma$围成,若函数$P(x,y,z)$,$Q(x,y,z)$,$R(x,y,z)$在$\Omega$上具有一节连续偏导数,则有
-\begin{align}
- \iiint_\Omega\left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial x}\right)\mathrm{d}V & =\oiint_\Sigma P\mathrm{d}y\mathrm{d}z + Q\mathrm{d}z\mathrm{d}x + R\mathrm{d}x\mathrm{d}y
- \\
- & =\oiint_\Sigma \left(P\cos\alpha + Q\cos\beta + R\cos\gamma\right)\mathrm{d}S
-\end{align}
-这里$\Sigma$是整个边界曲面$\Omega$的外侧,$\cos\alpha,\cos\beta,\cos\gamma$是$\Sigma$在点$(x,y,z)$处的法向量的方向余弦。引用这个公式如\ref{theorem:gauss formula}。
-\end{theorem}
+\zhlipsum*[56][name = aspirin]
+\begin{equation}
+ \begin{split}
+ \iint_{A}E\rho(\bm{v}\cdot\bm{n})\mathrm{d}A + \frac{\partial}{\partial t}\iiint_{V}E\rho \mathrm{d}V &= \frac{\delta Q}{\mathrm{d}t} - \frac{\delta W}{\mathrm{d}t}
+ \\
+ \iint_{A}\left(U + \frac{v^{2}}{2} + gz + \frac{P}{\rho}\right)\rho(\bm{v}\cdot\bm{n})\mathrm{d}A + \frac{\partial}{\partial t}\iiint_{V}E\rho\mathrm{d}V &= \frac{\delta Q}{\mathrm{d}t} - \frac{\delta W_{s}}{\mathrm{d}t}
+ \end{split}
+\end{equation}
-\begin{definition}[(Stokes formula)]
-\index{d@定理!s@Stokes formula}
-Let $\Gamma$ be a piecewise smooth directed closed curve, $\Sigma$ is a piecewise smooth directed surface bounded by $\Gamma$, the side of $\Gamma$ and the positive direction of $\Sigma$ According to the right-hand rule, if the function $P(x,y,z)$, $Q(x,y,z)$, $R(x,y,z)$ has a first-order continuous deviation on the curve $\Sigma$ Derivative, then
-\begin{multline}
- \iint_\Sigma\left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathrm{d}y\mathrm{d}z + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)\mathrm{d}z\mathrm{d}x
+\zhlipsum*[42][name = aspirin]
+\begin{subnumcases}{y =}
+ C_1e^{r_1x} + C_2e^{r_2x} & \quad $p^2 - 4q > 0$
\\
- + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y = \oint_\Gamma P\mathrm{d}x + Q\mathrm{d}y + R\mathrm{d}z
-\end{multline}
-The Stokes formula is an extension of the basic calculus formula in the case of surface integration. It is also an extension of the Green formula. This formula gives the second type of surface integral on the surface block and the second type of curve on the boundary curve Link between points.
-\end{definition}
-
-\clearpage
-\begin{lemma}
-定理环境的边框和习题标题盒子可使用 \lstinline{thmbox = false} 关闭。
-\end{lemma}
+ \left(C_1 + C_2\right)e^{r_1x} & \quad $p^2 - 4q = 0$
+ \\
+ e^{\alpha x}\left(C_1\cos\beta x + C_2\sin\beta x\right) & \quad $p^2 - 4q < 0$
+\end{subnumcases}
-\begin{corollary}
-这是一个推论。
-\end{corollary}
+\zhlipsum*[57][name = aspirin]
-\begin{proposition}
-这是一个性质。
-\end{proposition}
+\clearpage
+\section{插图}
+\index{c@插图}
+当我们需要排列一组子图共享标题的图片时,可以使用依赖于 \lstinline{caption} 宏包的 \lstinline{subcaption} 宏包的功能,效果见\ref{fig:example}。
+\begin{figure}[htbp]
+ \centering
+ \subcaptionbox{并排子图1}
+ {\includegraphics[width = 0.4\textwidth]{example-image-4x3}}
+ \hspace{1.2em}
+ \subcaptionbox{并排子图2}
+ {\includegraphics[width = 0.4\textwidth]{example-image-4x3}}
+ \bicaption[使用 subcaption 排版子图]{使用 subcaption 宏包的 subcaptionbox 命令排版子图}[Use subcaption to typeset subgraphs]{Use the subcaptionbox command of the subcaption package to typeset sub-pictures}\label{fig:example}
+\end{figure}
-\begin{example}
-这是一个例。
-\end{example}
+\section{表格}
+\index{b@表格}
+合并表格列使用 \lstinline{\multicolumn} 命令,合并行可以使用 \lstinline{multirow} 宏包的 \lstinline{\multirow} 命令。当插入的表格内容过长以至于一行放不下的情况可以使用 \lstinline{tabularx} 环境,文档类设置了\textsf{L、C和R}三个列对齐选项,一个例子如\ref{tab:example} 所示。
+\begin{table}[htbp]
+ \centering
+ \bicaption{使用 tabularx 创建内容过长表格}{Use tabularx to create a table with too long content}\label{tab:example}\small
+ \begin{tabularx}{0.87\textwidth}{@{}llL@{}}
+ \toprule
+ Aliquam & Integer & Pellentesque tincidunt purus
+ vel magna. \\
+ \midrule
+ viverra & \multirow[c]{3}*{metus} & Nulla malesuada porttitor diam. Vestibulum lectus. Proin mauris. Proin eu nunc eu urna hendrerit faucibus. \\
+ semper & & Nullam elementum, urna vel imperdiet sodales, elit ipsum pharetra ligula, ac pretium ante justo a nulla. Curabitur tristique arcu eu metus. \\
+ \multicolumn{2}{c}{\multirow{3}*[0ex]{convallis}} & Nunc elementum fermentum wisi. Aenean placerat. Ut imperdiet, enim sed gravida sollicitudin, felis odio placerat quam, ac pulvinar elit purus eget enim. \\
+ \bottomrule
+ \end{tabularx}
+\end{table}
-\begin{remark}
-这是一个注。
-\end{remark}
+\clearpage
+\section{列表}
+\index{l@列表}
-\begin{proof}
-这是一个证明,末尾自动添加证明结束符。
-\end{proof}
+\subsection{排序列表}
+\index{l@列表!p@排序列表}
+\begin{enumerate}
+ \item \zhlipsum[41-42][name = aspirin]
+ \begin{enumerate}
+ \item \zhlipsum*[42][name = aspirin]
+ \begin{enumerate}
+ \item \zhlipsum*[42][name = aspirin]
+ \end{enumerate}
+ \end{enumerate}
+\end{enumerate}
-\begin{easybox}*[MintCream](Aspirin)
-\index{z@自定义盒子}
- \zhlipsum*[3][name = aspirin]
- \tcblower
- \zhlipsum*[8][name = aspirin]
-\end{easybox}
+\subsection{常规列表}
+\index{l@列表!c@常规列表}
+\begin{itemize}
+ \item \zhlipsum[41-42][name = aspirin]
+ \begin{itemize}
+ \item \zhlipsum*[42][name = aspirin]
+ \begin{itemize}
+ \item \zhlipsum*[42][name = aspirin]
+ \end{itemize}
+ \end{itemize}
+\end{itemize}
-\zhlipsum*[3][name = aspirin]
-\begin{exercise}[LightYellow][1.](习题)
-\index{x@习题环境}
- \item 设$w = f(x + y + z,xyz)$,$f$具有二阶连续偏导数,求$\dfrac{\partial w}{\partial x}$和$\dfrac{\partial ^2 w}{\partial x\partial z}$。
- \item 已知$y = y(x)$在任意点$x$处的增量$\Delta y = \dfrac{y\Delta x}{1 + x^2} + \alpha$,其中$\alpha$是$\Delta x$的高阶无穷小($\Delta x\to 0$时),$y(0) = \pi$,则$y(1) = \uline{\mbox{\hspace{2em}}}$。
- \item 设函数$f(x)$在$(-\infty,+\infty)$上有定义,则下述命题中正确的是 \mbox{(\hspace{1.5em})}
- \begin{tasks}
- \task 若$f(x)$在$(-\infty,+\infty)$上可导且单调增加,则对一切$x\in (-\infty,+\infty)$,都有$f'(x) > 0$。
- \task 若$f(x)$在点$x_0$处取得极值,则$f'(x_0) = 0$。
- \task 若$f''(x_0) = 0$,则$(x_0,f(x_0))$是曲线$y = f(x)$的拐点坐标。
- \task 若$f'(x_0) = 0$, $f''(x_0) = 0$,$f'''(x_0)\ne 0$,则$x_0$一定不是$f(x)$的极值点。
- \end{tasks}
-\end{exercise} \ No newline at end of file
+\subsection{主题列表}
+\index{l@列表!z@主题列表}
+\begin{description}
+ \item[主题一] \zhlipsum*[42][name = aspirin]
+ \begin{description}
+ \item[主题二] \zhlipsum*[42][name = aspirin]
+ \end{description}
+\end{description} \ No newline at end of file