参数:

markup: pandoc

基本公式

$$ 0 \le b, g, r, \alpha \le 1 $$

$$ \forall (b,g,r) \in (\hat{I}_1, \hat{I}_2, O): 0 \le b, g, r \le 1 $$

\[ 0 \le b, g, r, \alpha \le 1 \]

\[ \forall (b,g,r) \in (\hat{I}_1, \hat{I}_2, O): 0 \le b, g, r \le 1 \]

行内公式

This $ \alpha=4 $ will not work in pandoc.

This works: $\alpha = \frac{1}{3}$, then $\beta=1-\alpha=\frac{2}{3}$

This $ $ will not work in pandoc.

This works: \(\alpha = \frac{1}{3}\), then \(\beta=1-\alpha=\frac{2}{3}\)

换行-begin{array}

$$\begin{array}{c}
I_1 &= O * A + BG_1 * (1 - A) \\
I_2 &= O * A + BG_2 * (1 - A) \\
\end{array}$$

\[\begin{array}{c} I_1 &= O * A + BG_1 * (1 - A) \\ I_2 &= O * A + BG_2 * (1 - A) \\ \end{array}\]

复杂示例1

来自 github.com/waylonflinn/markdown-it-katex#block

\[ \begin{array}{c} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{array} \]

$$
\begin{array}{c}
\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} &
= \frac{4\pi}{c}\vec{\mathbf{j}}    \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\
\nabla \cdot \vec{\mathbf{B}} & = 0
\end{array}
$$

复杂示例2

来自 混合图片生成原理分析

\[\begin{aligned} \\ \alpha' &= 1 - \cfrac{b'_o \cdot b_b + g'_o \cdot g_b + r'_o \cdot r_b}{b_b^2 + g_b^2 + r_b^2} \\ &= 1 - \cfrac{ \sum\limits_{c=b, g, r} (A + B_1 c_{out1} - B_2 c_{out2}) c_b }{b_b^2 + g_b^2 + r_b^2} \\ &= 1 - \cfrac{ A(b_b+g_b+r_b)+ \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b }{b_b^2 + g_b^2 + r_b^2} \\ &= 1 - \cfrac{ A(b_b+g_b+r_b)+ \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b }{b_b^2 + g_b^2 + r_b^2} \\ &= 1 - A \cfrac{b_b+g_b+r_b}{b_b^2 + g_b^2 + r_b^2} + \cfrac{1}{b_b^2 + g_b^2 + r_b^2} \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b \\ \end{aligned}\]

$$\begin{aligned} \\
\alpha' &= 1 - \cfrac{b'_o \cdot b_b + g'_o \cdot g_b + r'_o \cdot r_b}{b_b^2 + g_b^2 + r_b^2} \\
 &= 1 - \cfrac{
 \sum\limits_{c=b, g, r} (A + B_1 c_{out1} - B_2 c_{out2}) c_b
 }{b_b^2 + g_b^2 + r_b^2} \\
 &= 1 - \cfrac{
  A(b_b+g_b+r_b)+
 \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b
 }{b_b^2 + g_b^2 + r_b^2} \\
 &= 1 - \cfrac{
  A(b_b+g_b+r_b)+
 \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b
 }{b_b^2 + g_b^2 + r_b^2} \\
 &= 1 - A \cfrac{b_b+g_b+r_b}{b_b^2 + g_b^2 + r_b^2}  + \cfrac{1}{b_b^2 + g_b^2 + r_b^2} 
  \sum\limits_{c=b, g, r} (B_1 c_{out1} - B_2 c_{out2}) c_b \\
\end{aligned}$$