Markdown公式(二)

来源:互联网 发布:csi网络犯罪调查第二季 编辑:程序博客网 时间:2024/05/21 06:38

参考资料https://gavin_nicholas.coding.me/archives/

1. 如何输入括号和分隔符

()[]| 表示自己, {} 表示 {} 。当要显示大号的括号或分隔符时,要用 \left\right 命令。

例子:$$f(x,y,z) = 3y^2z \left( 3+\frac{7x+5}{1+y^2} \right)$$ ,显示:
\[f(x,y,z) = 3y^2z \left( 3+\frac{7x+5}{1+y^2} \right)\]

有时候要用\left.\right.进行匹配而不显示本身。

例子:$$\left. \frac{ {\rm d}u}{ {\rm d}x} \right| _{x=0}$$,显示:
\[\left. \frac{ {\rm d}u}{ {\rm d}x} \right| _{x=0}\]

2. 运算符:

关系运算符markdown语言集合运算符markdown语言对数运算符markdown语言戴帽符号markdown语言\(\pm\)$\pm$\(\emptyset\)$\emptyset$\(\log\)$\log$\(\hat{y}\)$\hat{y}$\(\times\)$\times$\(\in\)$\in$\(\lg\)$\lg$\(\check{y}\)$\check{y}$\(\div\)$\div$\(\notin\)$\notin$\(\ln\)$\ln$\(\breve{y}\)$\breve{y}$\(\mid\)$\mid$\(\subset\)$\subset$\(\nmid\)$\nmid$\(\supset\)$\supset$\(\cdot\)$\cdot$\(\subseteq\)$\subseteq$\(\circ\)$\circ$\(\supseteq\)$\supseteq$\(\ast\)$\ast$\(\bigcap\)$\bigcap$\(\bigodot\)$\bigodot$\(\bigcup\)$\bigcup$\(\bigotimes\)$\bigotimes$\(\bigvee\)$\bigvee$\(\bigoplus\)$\bigoplus$\(\bigvee\)$\bigvee$\(\leq\)$\leq$\(\bigwedge\)$\bigwedge$\(\geq\)$\geq$\(\biguplus\)$\biguplus$\(\neq\)$\neq$\(\bigsqcup\)$\bigsqcup$\(\approx\)$\approx$\(\equiv\)$\equiv$\(\sum\)$\sum$\(\prod\)$\prod$\(\coprod\)$\coprod$

三角运算符markdown语言微积分运算符markdown语言逻辑运算符markdown语言\(\bot\)$\bot$\(\prime\)$\prime$\(\because\)$\because$\(\angle\)$\angle$\(\int\)$\int$\(\therefore\)$\therefore$\(30^\circ\)$30^\circ$\(\iint\)$\iint$\(\forall\)$\forall$\(\sin\)$\sin$\(\iiint\)$\iiint$\(\exists\)$\exists$\(\cos\)$\cos$\(\iiiint\)$\iiiint$\(\not=\)$\not=$\(\tan\)$\tan$\(\oint\)$\oint$\(\not>\)$\not>$\(\cot\)$\cot$\(\lim\)$\lim$\(\not\subset\)$\not\subset$\(\sec\)$\sec$\(\infty\)$\infty$\(\csc\)$\csc$\(\nabla\)$\nabla$箭头符号markdown语言\(\uparrow\)$\uparrow$\(\downarrow\)$\downarrow$\(\Uparrow\)$\Uparrow$\(\Downarrow\)$\Downarrow$\(\rightarrow\)$\rightarrow$\(\leftarrow\)$\leftarrow$\(\Rightarrow\)$\Rightarrow$\(\Leftarrow\)$\Leftarrow$\(\longrightarrow\)$\longrightarrow$\(\longleftarrow\)$\longleftarrow$\(\Longrightarrow\)$\Longrightarrow$\(\Longleftarrow\)$\Longleftarrow$