第一段代码
documentclass{article} usepackage{ctex} egin{document} section{文字} 特可爱模板 section{数学} [ a^2=b^2+c^2 ] end{document}
第二段代码
documentclass[UTF8]{ctexart}
itle{杂谈勾股定理}
author{张三}
date{ oday}
ibliographystyle{plain}
egin{document}
maketitle
ableofcontents
section{勾股定理在古代}
section{勾股定理的近代形式}
ibliography{math}
end{document}
第三段代码
documentclass[UTF8]{ctexart}
usepackage{amsmath}
usepackage{amsfonts}
usepackage{amsthm}
usepackage{graphicx}
usepackage{amssymb}
usepackage{bm}
usepackage{titlesec}
usepackage{graphicx}
usepackage{booktabs}
usepackage{multirow}
usepackage{booktabs}
usepackage{subfig}
usepackage{hyperref}
usepackage{fancyhdr}
usepackage{pdfpages}
usepackage{hyperref}
usepackage{geometry}
usepackage{CJK}
itle{杂谈勾股定理}
author{张三}
date{ oday}
ibliographystyle{plain}
egin{document}
maketitle
ableofcontents
section{勾股定理在古代}
section{勾股定理的近代形式}
ibliography{math}
end{document}
《latex入门》第一章
documentclass[UTF8]{ctexart} %导言区开始 usepackage{graphicx} usepackage{float} usepackage{amsmath} usepackage{geometry} %usepackage[format=hang,font=small,textfont=it]{caption} %geometry{a6paper,centering,scale=0.8} ewtheorem{thm}{定理}%声明 itle{heiti 杂谈勾股定理} author{kaishu 张三} date{ oday} ibliographystyle{plain} %导言区结束 egin{document} maketitle egin{abstract} 这是一篇关于勾股定理的小短文 end{abstract} ableofcontents section{勾股定理在古代} section{勾股定理的近代形式} ibliography{math} ......见于欧几里得footnote{欧几里得,约公元前330--275年。}《几何原本》的..... ......的整数成为emph{勾股数} ......答周公问: egin{quote} 勾广三,股修四,径隅五。 end{quote} 又载陈子(约公元前7--6世纪)答荣方问: egin{quote} zihao{-5}kaishu 若求邪至日者,以日下为沟,日高为股,勾股各自乘,并而开方除之,得邪至日。 end{quote} 都较古希腊更早。...... egin{thm}[勾股定理] 直角三角形斜边的平方等于两腰的平方和。 可以用符号语言表述为...... end{thm} egin{equation} a(b+c)=ab+ac end{equation} $angle ACB =pi /2$ egin{equation} AB^2=BC^2+AC^2 end{equation} $2^{10}=1024$ $90^circ$ $A_{5}$ egin{tabular}{|rrr|} hline 直角边$a$&直角边$b$ &斜边$c$ \ hline 3 & 4 & 5\ 5 & 12 & 13\ hline end{tabular} egin{table}[H]%usepackage{float},不浮动 egin{tabular}{|rrr|} hline 直角边$a$&直角边$b$ &斜边$c$ \ hline 3 & 4 & 5\ 5 & 12 & 13\ hline end{tabular} qquad ($a^2+b^2=c^2$) end{table} egin{equation}label{eq:gougu} AB^2=BC^2+AC^2 end{equation} 满足式eqref{eq:gougu}的整数称为emph{勾股数} café quad Gödel quad Antonín Dvořák χα ``\,`A' or `B?'\,'' he asked. She $dots$ she got it. egin{enumerate} item 中文 item English item Francais end{enumerate} egin{quote} 学而时习之,不亦说乎? 有朋自远方来,不亦乐乎? end{quote} egin{quote} 学而时习之,不亦说乎? 有朋自远方来,不亦乐乎? end{quote} egin{itemize} item 中文 item English item Francais end{itemize} egin{description} item[中文] 中国的语言文字 item[English] The language of England item[Francais] La lanue de France end{description} end{document}
第4段
documentclass[UTF8]{ctexart} usepackage{amsmath} usepackage{mathdots} usepackage{mathtools} usepackage{breqn} usepackage{amsfonts} egin{document} [ A= egin{pmatrix} a_{11} & a_{22} & a_{13} \ 0 & a_{22} & a_{23} \ 0 & 0 & a_{33} end{pmatrix} ] [ A=egin{bmatrix} a_{11} & dots & a_{1n} \ & ddots & vdots \ 0 & 0 & a_{nn} end{bmatrix}_{n imes n} ] 复数$z=(x,y)$也可用矩阵( ( egin{smallmatrix} x & -y\ y &x end{smallmatrix} ) )来表示 [ sum_{substack{0<i<n \0<j<i}} A_{ij} ] [ ordermatrix{ & 1 & 2 & 3 cr 1 & A & B & C cr 2 & D & E & F cr} ] $e^{pi i}+1=0$ [ mathcal{F}(x) =sum_{k=0}^infty oint_0^1 f_k(x,t) \,mathrm{d}t ] [ int f(x) \,mathrm{d} x ] ewcommanddiff{\,mathrm{d}} [ iintlimits_{0<x,y,z<1} f(x,y,z) diff x diff y diff z] $cos 2x = cos(x+x) = cos^2x -sin^2x$ ewcommanddefeq{stackrel{ ext{d}}{=}} %stackrel产生堆叠的效果 $f(x) defeq ax^2+bx+c$ [ A xleftarrow{0<x<1} B xrightarrow[x leq 0]{x geq 1} C ] \ $x=y implies x+a=y+a$ \ $x=y impliedby x+a=y+a$ \ $x=y iff x le y And x ge y$ [ a mathbin{heartsuit} b = b mathbin{heartsuit} a ] ewcommandvarnotin{% %这里为什么要加%呢?需要查阅一下,应该是间隔的问题 mathrel{overline{in}}} $forall x$,$forall S$,$xvarnotin S$ ewcommand*abs[1]{lvert#1 vert} $abs{x+y} le abs{x} +abs{y}$ [ partial_x partial_y left[ frac12 left( x^2+y^2 ight)^2 +xy ight] ] [ left. int_0^x f(t,lambda) \,mathrm{d}t ight |_{x=1},qquad lambda in left[frac12,infty ight). ] [ Pr left( X>frac12 middle| Y=0 ight) = left. int_0^1 p(t)\,mathrm{d}t middle/ (N^2+1) ight. ] [ iggl( sum_{i=1}^n A_i iggr) cdot iggl( sum_{i=1}^n B_i iggr) >0 ] $1+ Bigl( 2-igl(3 imes (4 div 5) igr) Bigr)$ [ P= iggl< frac12 iggr>,qquad M=left< egin{matrix} a&b \ c & d\ end{matrix} ight> ] $a:b=ac:bc$ [ Pr(xcolon g(x)>5) = 0.25, qquad gcolon x mapsto x^2 ] [ (1,dots,n) qquad 1+dots+n qquad a=dots=z ] egin{align} &(a+b)(a^2-ab+b^2) otag \ ={} & a^3-a^2b+ab^2+a^2b-ab^2 +b^2 otag \ ={} & a^3 +b^3 end{align} egin{flalign} x &= t &x &=2 \ y &= 2t & y &=4 end{flalign} egin{alignat}{2} x&=sin t&quad & ext{水平方向} \ y&=cos t& & ext{垂直方向} end{alignat} egin{align*}{2} x^2+2x&=-1 intertext{移项得} x^2+2x+1&=0 end{align*} 设$G$是一个带有运算$*$的集合,则$G$是emph{群},当且仅当: egin{subequations}label{eq:group} egin{alignat}{2} forall a,b,c &in G, &qquad (a*b)*c &=a*(b*c);label{subeq:assoc} \ exists e, forall a &in G, & 3*a &= a;\ forall a ,exists b &in G, & b*a&=e. end{alignat} end{subequations} 式~eqref{eq:group} 的三个条件中,eqref{subeq:assoc}~又称为结合律 egin{multline} a+b+c \ +d+e+f\ +g+h+i\ +j+k+l end{multline} egin{equation} egin{split} cos 2x &= cos^2x-sin^2x\ &=2cos^2x-1 end{split} end{equation} egin{dmath} frac12(sin(x+y) +sin(x-y)) =frac12(sin xcos y+cos x sin y)+frac12(sin xcos y-cos xsin y)=sin x cos y) end{dmath} egin{equation} D(x)=egin{cases} 1,& ext{if } x in mathbb{Q};\ 0,& ext{if } x in mathbb{R} setminus mathbb{Q}. end{cases} end{equation} [ leftlvert x-frac12 ight vert =egin{dcases} x-frac12,& x geq frac12;\ frac12-x,& x<frac12. end{dcases} ] end{document}