Beautiful math in all browsers - MathJax

终于配置好了满意的 MathJax,写一篇测试文档留作纪念。

字符

数学模式重音符

$$
\ \ \hat{a} \ \ \check{a} \ \ \tilde{a} \ \ \acute{a}
\ \ \grave{a} \ \ \dot{a} \ \ \ddot{a} \ \ \breve{a}
\ \ \bar{a} \ \ \vec{a} \ \ \widehat{A} \ \ \widetilde{A}
$$

小写希腊字母

$$
\ \ \alpha \ \ \theta \ \ \upsilon
\ \ \beta \ \ \vartheta \ \ \pi \ \ \phi
\ \ \gamma \ \ \iota \ \ \varpi \ \ \varphi
\ \ \delta \ \ \kappa \ \ \rho \ \ \chi
\ \ \epsilon \ \ \lambda \ \ \varrho \ \ \psi
\ \ \varepsilon \ \ \mu \ \ \sigma \ \ \omega
\ \ \zeta \ \ \nu \ \ \varsigma
\ \ \eta \ \ \xi \ \ \tau
$$

大写希腊字母

$$
\ \ \Gamma \ \ \Lambda \ \ \Sigma \ \ \Psi
\ \ \Delta \ \ \Xi \ \ \Upsilon \ \ \Omega
\ \ \Theta \ \ \Pi \ \ \Phi
$$

二元关系表示符

$$
\ \ <\ \ <\ \ >\ \ >\ \ =\ \
\ \ \leq o\ \ \le \ \ \geq o\ \ \ge \ \ \equiv
\ \ \ll \ \ \gg \ \ \doteq
$$
$$
\ \ \prec \ \ \succ \ \ \sim
\ \ \preceq \ \ \succeq \ \ \simeq
\ \ \subset \ \ \supset \ \ \approx
\ \ \subseteq \ \ \supseteq \ \ \cong
\ \ \sqsubset a \ \ \sqsupset \ \ \Join
\ \ \sqsubseteq \ \ \sqsupseteq
$$
$$
\ \ \bowtie
\ \ \in \ \ \ni \ \ \owns \ \ \propto
\ \ \vdash \ \ \dashv |\ \ \models
\ \ \mid \ \ \parallel \ \ \perp
\ \ \smile \ \ \frown \ \ \asymp
\ \ : \ \ \notin \ \ \neq \ \ \ne
$$

二元运算符
$$
+\ \ −\ \ -
\ \ \pm \ \ \mp \ \ \triangleleft
\ \ \cdot \ \ \div \ \ \triangleright
\ \ \times\ \ \setminus \ \ \star
\ \ \cup \ \ \cap \ \ \ast
\ \ \sqcup \ \ \sqcap \ \ \circ
\ \ \vee \ \ \lor \ \ \wedge
$$
$$
\land \ \ \bullet
\ \ \oplus \ \ \ominus \ \ \diamond
\ \ \odot \ \ \oslash \ \ \uplus
\ \ \otimes \ \ \bigcirc \ \ \amalg
\ \ \bigtriangleup \ \ \bigtriangledown \ \ \dagger
\ \ \lhd \ \ \rhd
\ \ \ddagger
\ \ \unlhd \ \ \unrhd
\ \ \wr
$$
大尺寸运算符
$$
\ \ \sum \ \ \bigcup \ \ \bigvee \ \ \bigoplus
\ \ \prod \ \ \bigcap \ \ \bigwedge \ \ \bigotimes
\ \ \coprod \ \ \bigsqcup \ \ \bigodot
\ \ \int \ \ \oint \ \ \biguplus
$$
箭头
$$
\leftarrow \ \ \gets \ \ \longleftarrow\ \ \uparrow\ \ \rightarrow \ \ \to \ \ \longrightarrow\ \ \downarrow\ \ \leftrightarrow \ \ \longleftrightarrow\ \ \updownarrow
$$
$$
\ \Leftarrow \ \ \Longleftarrow\ \ \Uparrow\ \ \Rightarrow\ \ \Longrightarrow\ \ \Downarrow \Leftrightarrow \ \ \Longleftrightarrow\ \ \Updownarrow
\ \ \mapsto \ \ \longmapsto\ \ \nearrow
$$
$$
\ \ \hookleftarrow \ \ \hookrightarrow\ \ \searrow\ \ \leftharpoonup\ \ \rightharpoonup\ \ \swarrow\ \ \leftharpoondown\ \ \rightharpoondown\ \ \nwarrow\ \ \rightleftharpoons \ \ \iff \ \ \leadsto
$$
定界符
$$
(\ \ ) \ \ \uparrow \ \ \Uparrow
\ \ \lbrack \ \ \rbrack \ \ \downarrow \ \ \Downarrow
\ \ { \ \ \lbrace \ \ } \ \ \rbrace \ \ \updownarrow \ \ \Updownarrow
\ \ \langle \ \ \rangle \ \ \vert \ \ \Vert
\ \ \lfloor \ \ \rfloor \ \ \lceil \ \ \rceil
\ \ \backslash
$$
其他符号
$$
\ \ \dots \ \ \cdots
\ \ \vdots
\ \ \ddots
\ \ \hbar \ \ \imath \ \ \jmath \ \ \ell
\ \ \Re \ \ \Im \ \ \aleph \ \ \wp
\ \ \forall \ \ \exists \ \ \mho \ \ \partial
\ \ \prime \ \ \emptyset
$$
$$
\infty
\ \ \nabla \ \ \triangle \ \ \Box \ \ \Diamond
\ \ \bot \ \ \top \ \ \angle \ \ \surd
\ \ \diamondsuit \ \ \heartsuit \ \ \clubsuit \ \ \spadesuit
\ \ \neg \ \ \lnot \ \ \ \ \natural \ \ \sharp
$$

行内公式

有 $n$ 只鼠,白兔用 $n - 1$ 根蓝色绳子把它们连成了一棵树,每根蓝色绳子连着两只鼠,白云用 $n - 1$ 根红色绳子把它们连成了一棵树,每根红色绳子连接着两只鼠。

白云要给予每只鼠一个数。这个数可以是 $[1, y]$ 中的任意一个整数。

白兔给了白云一个要求:对于两只鼠 $p, q$,若存在一条连接这两只鼠的路径同时属于这两棵树,则 $p$ 和 $q$ 必须被给予相同的整数。存在一条路径同时属于这两棵树指的是:存在一个序列 $(a_1 = p, a_2, \cdots , a_m = q)$,使得:对于所有 $i \in [1, m - 1]$,都有 $a_i$ 和 $a_{i+1}$ 既有一根红色绳子直接相连也有一根蓝色绳子直接相连。

块状公式

$$
\int \frac{dx}{cos^2x}=\int sec^2xdx=tan x+C\
$$

$$
\int \frac{dx}{sin^2x}=\int csc^2xdx=-cot x+C\
$$

$$
\sum\limits_{i = 0}^{E} w[i]{M \choose i} {N \choose iS} \frac{(iS)!}{(S!)^{i}} \sum\limits_{j = 0}^{E - i} (-1)^{j} {M - i \choose j} {N - iS \choose jS} \frac{(jS)!}{(S!)^{j}} (M - i - j)^{N - iS - jS}
$$
$$
= \sum\limits_{i = 0}^{E} w[i]{M \choose i} {N \choose iS} \frac{(iS)!}{(S!)^{i}} \sum\limits_{j = i}^{E} (-1)^{j - i} {M - i \choose j - i} {N - iS \choose jS - iS} \frac{(jS - iS)!}{(S!)^{j - i}} (M - j)^{N - jS}
$$

矩阵

$$
\begin{bmatrix}
(\omega_n^0)^0 & (\omega_n^0)^1 & \cdots & (\omega_n^0)^{n-1}\\ (\omega_n^1)^0 & (\omega_n^1)^1 & \cdots & (\omega_n^1)^{n-1}\\
\vdots & \vdots & \ddots & \vdots \\
(\omega_n^{n-1})^0 & (\omega_n^{n-1})^1 & \cdots & (\omega_n^{n-1})^{n-1}
\end{bmatrix}
\begin{bmatrix}
a_0 \\
a_1 \\
\vdots \\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
A(\omega_n^0) \\
A(\omega_n^1) \\
\vdots \\
A(\omega_n^{n-1})
\end{bmatrix}
$$

$$
\begin{bmatrix}
a_0 \\
a_1 \\
\vdots \\
a_{n-1}
\end{bmatrix}
=
\frac{1}{n}
\begin{bmatrix}
(\omega_n^{-0})^0 & (\omega_n^{-0})^1 & \cdots & (\omega_n^{-0})^{n-1} \\
(\omega_n^{-1})^0 & (\omega_n^{-1})^1 & \cdots & (\omega_n^{-1})^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
(\omega_n^{-(n-1)})^0 & (\omega_n^{-(n-1)})^1 & \cdots & (\omega_n^{-(n-1)})^{n-1}
\end{bmatrix}
\begin{bmatrix}
A(\omega_n^0) \\
A(\omega_n^1) \\
\vdots \\
A(\omega_n^{n-1})
\end{bmatrix}
$$