(完整word版)LATEX 数学公式总结
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勾股定理~\(a^2+b^2=c^2\)~也称商高定理.
勾股定理~$a^2+b^2=c^2$~也称商高定理.
\section{行间公式}
\subsection{单行公式}
\begin{displaymath}
a^2+b^2=c^2.
\end{displaymath}
\[
a^2+b^2 = c^2.
\]
\newcommand{\liuhao}{\fontsize{7.875pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\qihao}{\fontsize{5.25pt}{\baselineskip}\selectfont} %字号设置
%%%%%%%%% END %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\renewcommand{\baselinestretch}{1.3}
\begin{document}
\begin{CJK*}{GBK}{song}
\CJKtilde\CJKindent
{\hei\sanhao数学公式举例:}
\bigskip
\section{概述}
数学模式中的普通文本必须放入一个~LR盒子里.如:
\setcounter{equation}{1}
\begin{eqnarray}
d(uv) & = & (uv)' dx \\
& = & (u'v+uv') dx\\
& = & v(u'dx)+u(v'dx) \nonumber\\
\setcounter{equation}{5}
& = & v du+u dv \label{leibniz}
\newcommand{\xiaosihao}{\fontsize{12pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\wuhao}{\fontsize{10.5pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaowuhao}{\fontsize{9pt}{\baselineskip}\selectfont} %字号设置
数学公式小结
请运行以下程序:
\documentclass[11pt]{article}
\usepackage{CJK}
\usepackage{indentfirst}
\usepackage{latexsym}
\usepackage{bm}
\usepackage{amsmath,amssymb,amsfonts}
\textbackslash{}mathstrut\footnote{宽度为~0,高度与圆括号相同},例:
$\sqrt{a}, \sqrt{b},\quad \sqrt{a\mathstrut}, \sqrt{b\mathstrut}$.
\section{求和与积分}
\newcommand{\dx}{\mathrm{d}\,x}
$ x^2+\sin(x)=0 is a nonlinear equation$.
$ x^2+\sin(x)=0 \mbox{ is a nonlinear equation} $.
$ x^2+\sin(x)=0 \mbox{是一个非线性方程}$.
\section{行内公式}
勾股定理~\begin{math}a^2+b^2=c^2\end{math}~也称商高定理.
\newcommand{\sanhao}{\fontsize{15.75pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaosanhao}{\fontsize{15pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\sihao}{\fontsize{14pt}{\baselineskip}\selectfont} %字号设置
x^{m^n}, \quad
{x^x}^{x^x}
$$
中文角标:\qquad
$ x^{\mbox{\scriptsize平方}},\quad x^{y^{\mbox{\tiny平方}}} $
导数符号:\qquad
$ f^{\prime} \quad\mbox{或者}\quad f' $
\section{分式}
\end{eqnarray}
这样就得到了公式~(\ref{leibniz}).
\section{角标:上标与下标}
注意:这里的角标命令必须在数学模式下使用!!
$$
x_1, \quad
x_{11}, \quad
x_{11}^{22}, \quad
x_{m}^{(k)},\quad
{}^* x ^*, \quad
\ml{breve}\{a\}~$\to \breve{a}$ & \ml{check}\{a\}~$\to \check{a}$\\
\ml{acute}\{a\}~$\to \acute{a}$ & \ml{grave}\{a\}~$\to \grave{a}$\\
\ml{mathring}\{a\}~$\to \mathring{a}$ & \\
\bigskip
$$ \underbrace{a+\overbrace{b+\dots+b}^{m\mbox{\scriptsize个}}+ c}_
{20\mbox{\scriptsize个}}
\newcommand{\hei}{\CJKfamily{hei}} %黑体(Windows自带simhei.ttf)
\newcommand{\li}{\CJKfamily{li}} %隶书(Windows自带simli.ttf)
\newcommand{\you}{\CJKfamily{you}} %幼圆(Windows自带simyou.ttf)
\subsection{多行公式}
\begin{eqnarray*}
x^2 + y^2 = R^2 \\
2x + 3y = b
\end{eqnarray*}
\begin{eqnarray}
x^2 + y^2 & = & R^2 \\
2x + 3y & = & b
\end{eqnarray}
\setlength{\arraycolsep}{2.5pt}
出现在行内的分式: $ (x+y)/2 $和~$ \frac{x+y}{2} $,第二个分式用的是一级角标字体.
分式中的分式: $\frac{\frac{x}{x+y}}{x+y+z}$,字体会更小,但最小为二级角标字体.
行间公式
$$
\frac{x+y}{2},\qquad \frac{\frac{x}{x+y}}{x+y+z}
公式~\ref{eq:square}~表示的是一个圆的标准方程.
\setcounter{equation}{5}
\begin{equation}\label{lap}
-\triangle u(x,y) = f(x,y),\quad (x,y)\in\Omega .
\end{equation}
方程~\eqref{lap}~则是一个椭圆型的偏微分方程.
\begin{equation}
a^2+b^2=c^2.
\end{equation}
$$
a^2+b^2=c^2. \eqno (*)
$$
$$
a^2+b^2=c^2. \eqno (4a)
$$
\begin{equation}\label{eq:square}
x^2+y^2=R^2.
\end{equation}
\newcommand{\chuhao}{\fontsize{42pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaochuhao}{\fontsize{36pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\yichu}{\fontsize{32pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\yihao}{\fontsize{28pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\erhao}{\fontsize{21pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaoerhao}{\fontsize{18pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\ml}[1]{\texttt{\textcolor{blue}{\char`\\ #1}}}
\renewcommand{\arraystretch}{1.2}
\setlength{\tabcolsep}{6pt}
\begin{tabular}{|p{0.4\textwidth}|p{0.4\textwidth}|}\hline
$$
\section{根式}
$ \sqrt{x},\quad \sqrt{1+\sqrt{2}} $
$ \surd{x},\quad \surd{1+\sqrt{2}} $
当被开方式字符高度不同时,根号线会在不同水平线上,如:
$\sqrt{a}, \sqrt{b}$.
解决办法:加入{\hei数学支柱}~
\hline
\end{tabular}
\bigskip
加宽的帽子和波浪号: $\widehat{hello},\quad \widetilde{good}$
\section{上划线、下划线及类似符号}
$$ \overline{\overline{a}^2 + \underline{ab} + \bar{b}^2} $$
$$Baidu Nhomakorabea
\varint_a^b f(x)\dx, \quad
\iint_a^b f(x)\dx, \quad
\iiint_a^b f(x)\dx,\quad
\varoint_a^b f(x)\dx,\quad
\oiint_a^b f(x)\dx,\quad
$$
$$
\varint\nolimits_a^b f(x)\dx, \quad
\iint\nolimits_a^b f(x)\dx, \quad
\iiint\nolimits_a^b f(x)\dx,\quad
\varoint\nolimits_a^b f(x)\dx,\quad
\oiint\nolimits_a^b f(x)\dx,\quad
$$
\section{数学重音符号}
\newcommand{\song}{\CJKfamily{song}} %宋体(Windows自带simsun.ttf)
\newcommand{\fs}{\CJKfamily{fs}} %仿宋体(Windows自带simfs.ttf)
\newcommand{\kai}{\CJKfamily{kai}} %楷体(Windows自带simkai.ttf)
\ml{hat}\{a\}~$\to \hat{a}$ & \ml{bar}\{a\}~$\to \bar{a}$\\
\ml{dot}\{a\}~$\to \dot{a}$ & \ml{ddot}\{a\}~$\to \ddot{a}$\\
\ml{tilde}\{a\}~$\to \tilde{a}$ & \ml{vec}\{a\}~$\to \vec{a}$\\
\usepackage{wasysym}
\usepackage{xcolor}
\usepackage{cases}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%重定义字体、字号命令%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$
\int_a^b f(x)\mathrm{d}x,\quad
\oint_a^b f(x)\mathrm{d}x,\quad
$$
$$
\int\limits_a^b f(x)\mathrm{d}x,\quad
\oint\limits_a^b f(x)\mathrm{d}x,\quad
$$
直立的积分号:
勾股定理~$a^2+b^2=c^2$~也称商高定理.
\section{行间公式}
\subsection{单行公式}
\begin{displaymath}
a^2+b^2=c^2.
\end{displaymath}
\[
a^2+b^2 = c^2.
\]
\newcommand{\liuhao}{\fontsize{7.875pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\qihao}{\fontsize{5.25pt}{\baselineskip}\selectfont} %字号设置
%%%%%%%%% END %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\renewcommand{\baselinestretch}{1.3}
\begin{document}
\begin{CJK*}{GBK}{song}
\CJKtilde\CJKindent
{\hei\sanhao数学公式举例:}
\bigskip
\section{概述}
数学模式中的普通文本必须放入一个~LR盒子里.如:
\setcounter{equation}{1}
\begin{eqnarray}
d(uv) & = & (uv)' dx \\
& = & (u'v+uv') dx\\
& = & v(u'dx)+u(v'dx) \nonumber\\
\setcounter{equation}{5}
& = & v du+u dv \label{leibniz}
\newcommand{\xiaosihao}{\fontsize{12pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\wuhao}{\fontsize{10.5pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaowuhao}{\fontsize{9pt}{\baselineskip}\selectfont} %字号设置
数学公式小结
请运行以下程序:
\documentclass[11pt]{article}
\usepackage{CJK}
\usepackage{indentfirst}
\usepackage{latexsym}
\usepackage{bm}
\usepackage{amsmath,amssymb,amsfonts}
\textbackslash{}mathstrut\footnote{宽度为~0,高度与圆括号相同},例:
$\sqrt{a}, \sqrt{b},\quad \sqrt{a\mathstrut}, \sqrt{b\mathstrut}$.
\section{求和与积分}
\newcommand{\dx}{\mathrm{d}\,x}
$ x^2+\sin(x)=0 is a nonlinear equation$.
$ x^2+\sin(x)=0 \mbox{ is a nonlinear equation} $.
$ x^2+\sin(x)=0 \mbox{是一个非线性方程}$.
\section{行内公式}
勾股定理~\begin{math}a^2+b^2=c^2\end{math}~也称商高定理.
\newcommand{\sanhao}{\fontsize{15.75pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaosanhao}{\fontsize{15pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\sihao}{\fontsize{14pt}{\baselineskip}\selectfont} %字号设置
x^{m^n}, \quad
{x^x}^{x^x}
$$
中文角标:\qquad
$ x^{\mbox{\scriptsize平方}},\quad x^{y^{\mbox{\tiny平方}}} $
导数符号:\qquad
$ f^{\prime} \quad\mbox{或者}\quad f' $
\section{分式}
\end{eqnarray}
这样就得到了公式~(\ref{leibniz}).
\section{角标:上标与下标}
注意:这里的角标命令必须在数学模式下使用!!
$$
x_1, \quad
x_{11}, \quad
x_{11}^{22}, \quad
x_{m}^{(k)},\quad
{}^* x ^*, \quad
\ml{breve}\{a\}~$\to \breve{a}$ & \ml{check}\{a\}~$\to \check{a}$\\
\ml{acute}\{a\}~$\to \acute{a}$ & \ml{grave}\{a\}~$\to \grave{a}$\\
\ml{mathring}\{a\}~$\to \mathring{a}$ & \\
\bigskip
$$ \underbrace{a+\overbrace{b+\dots+b}^{m\mbox{\scriptsize个}}+ c}_
{20\mbox{\scriptsize个}}
\newcommand{\hei}{\CJKfamily{hei}} %黑体(Windows自带simhei.ttf)
\newcommand{\li}{\CJKfamily{li}} %隶书(Windows自带simli.ttf)
\newcommand{\you}{\CJKfamily{you}} %幼圆(Windows自带simyou.ttf)
\subsection{多行公式}
\begin{eqnarray*}
x^2 + y^2 = R^2 \\
2x + 3y = b
\end{eqnarray*}
\begin{eqnarray}
x^2 + y^2 & = & R^2 \\
2x + 3y & = & b
\end{eqnarray}
\setlength{\arraycolsep}{2.5pt}
出现在行内的分式: $ (x+y)/2 $和~$ \frac{x+y}{2} $,第二个分式用的是一级角标字体.
分式中的分式: $\frac{\frac{x}{x+y}}{x+y+z}$,字体会更小,但最小为二级角标字体.
行间公式
$$
\frac{x+y}{2},\qquad \frac{\frac{x}{x+y}}{x+y+z}
公式~\ref{eq:square}~表示的是一个圆的标准方程.
\setcounter{equation}{5}
\begin{equation}\label{lap}
-\triangle u(x,y) = f(x,y),\quad (x,y)\in\Omega .
\end{equation}
方程~\eqref{lap}~则是一个椭圆型的偏微分方程.
\begin{equation}
a^2+b^2=c^2.
\end{equation}
$$
a^2+b^2=c^2. \eqno (*)
$$
$$
a^2+b^2=c^2. \eqno (4a)
$$
\begin{equation}\label{eq:square}
x^2+y^2=R^2.
\end{equation}
\newcommand{\chuhao}{\fontsize{42pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaochuhao}{\fontsize{36pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\yichu}{\fontsize{32pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\yihao}{\fontsize{28pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\erhao}{\fontsize{21pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\xiaoerhao}{\fontsize{18pt}{\baselineskip}\selectfont} %字号设置
\newcommand{\ml}[1]{\texttt{\textcolor{blue}{\char`\\ #1}}}
\renewcommand{\arraystretch}{1.2}
\setlength{\tabcolsep}{6pt}
\begin{tabular}{|p{0.4\textwidth}|p{0.4\textwidth}|}\hline
$$
\section{根式}
$ \sqrt{x},\quad \sqrt{1+\sqrt{2}} $
$ \surd{x},\quad \surd{1+\sqrt{2}} $
当被开方式字符高度不同时,根号线会在不同水平线上,如:
$\sqrt{a}, \sqrt{b}$.
解决办法:加入{\hei数学支柱}~
\hline
\end{tabular}
\bigskip
加宽的帽子和波浪号: $\widehat{hello},\quad \widetilde{good}$
\section{上划线、下划线及类似符号}
$$ \overline{\overline{a}^2 + \underline{ab} + \bar{b}^2} $$
$$Baidu Nhomakorabea
\varint_a^b f(x)\dx, \quad
\iint_a^b f(x)\dx, \quad
\iiint_a^b f(x)\dx,\quad
\varoint_a^b f(x)\dx,\quad
\oiint_a^b f(x)\dx,\quad
$$
$$
\varint\nolimits_a^b f(x)\dx, \quad
\iint\nolimits_a^b f(x)\dx, \quad
\iiint\nolimits_a^b f(x)\dx,\quad
\varoint\nolimits_a^b f(x)\dx,\quad
\oiint\nolimits_a^b f(x)\dx,\quad
$$
\section{数学重音符号}
\newcommand{\song}{\CJKfamily{song}} %宋体(Windows自带simsun.ttf)
\newcommand{\fs}{\CJKfamily{fs}} %仿宋体(Windows自带simfs.ttf)
\newcommand{\kai}{\CJKfamily{kai}} %楷体(Windows自带simkai.ttf)
\ml{hat}\{a\}~$\to \hat{a}$ & \ml{bar}\{a\}~$\to \bar{a}$\\
\ml{dot}\{a\}~$\to \dot{a}$ & \ml{ddot}\{a\}~$\to \ddot{a}$\\
\ml{tilde}\{a\}~$\to \tilde{a}$ & \ml{vec}\{a\}~$\to \vec{a}$\\
\usepackage{wasysym}
\usepackage{xcolor}
\usepackage{cases}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%重定义字体、字号命令%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$
\int_a^b f(x)\mathrm{d}x,\quad
\oint_a^b f(x)\mathrm{d}x,\quad
$$
$$
\int\limits_a^b f(x)\mathrm{d}x,\quad
\oint\limits_a^b f(x)\mathrm{d}x,\quad
$$
直立的积分号: