KaTex 编写示例公式数学公式

例1. 展示风格

编码：
$$
c(t) = \displaystyle\sum_{i=0}^nb**\scriptscriptstyle** i \displaystyle B \scriptscriptstyle i \displaystyle(t),0 \le t \le1, \tag{1}
$$
效果：
c(t)=∑i=0nbiBi(t),0≤t≤1,(1)c(t) = \displaystyle\sum_{i=0}^nb\scriptscriptstyle i \displaystyle B\scriptscriptstyle i \displaystyle(t),0 \le t \le1, \tag{1} c(t)=i=0∑nbiBi(t),0≤t≤1,(1)

例2. 穿插风格

编码：
式中，n为次数，$b \scriptscriptstyle i$为第$i$个控制点，$B \scriptscriptstyle i,n \displaystyle (t)$为Bernstein基多项式，如式(2)所示:
效果：
式中，n为次数，bib\scriptscriptstyle ibi为第iii个控制点，Bi,n(t)B\scriptscriptstyle i,n \displaystyle (t)Bi,n(t)为Bernstein基多项式，如式(2)所示:

例3. 混合

编码：
$$
B \scriptscriptstyle i,n \displaystyle (t)=\dbinom{n}{i} t^i(1-t){n-i},i = 0,…,n, \tag{2}
$$
式中，$\binom{n}{i}$是二项式系数
效果：
Bi,n(t)=(ni)ti(1−t)n−i,i=0,...,n,(2)B\scriptscriptstyle i,n \displaystyle (t)=\dbinom{n}{i} t^i(1-t)^{n-i},i = 0,...,n, \tag{2} Bi,n(t)=(in)ti(1−t)n−i,i=0,...,n,(2)
式中，(ni)\binom{n}{i}(in)是二项式系数

例4. 矩阵

编码：
$$
\left[
\begin{matrix}
B_{0,3}(t_0) & \cdots & B_{3,3}(t_0) \\
B_{0,3}(t_1) & \cdots & B_{3,3}(t_1) \\
\vdots & \ddots & \vdots \\
B_{0,3}(t_m) & \cdots & B_{3,3}(t_m)
\end{matrix}
\right]
\left[
\begin{matrix}
b_{x0} & b_{y0} \\
b_{x1} & b_{y1} \\
b_{x2} & b_{y2} \\
b_{x3} & b_{y3}
\end{matrix}
\right] =
\left[
\begin{matrix}
p_{x0} & p_{y0} \\
p_{x1} & p_{y1} \\
\vdots & \vdots \\
p_{xm} & p_{ym}
\end{matrix}
\right]
\tag{4}
$$
效果：
[B0,3(t0)⋯B3,3(t0)B0,3(t1)⋯B3,3(t1)⋮⋱⋮B0,3(tm)⋯B3,3(tm)][bx0by0bx1by1bx2by2bx3by3]=[px0py0px1py1⋮⋮pxmpym](4)\left[ \begin{matrix} B_{0,3}(t_0) & \cdots & B_{3,3}(t_0) \\ B_{0,3}(t_1) & \cdots & B_{3,3}(t_1) \\ \vdots & \ddots & \vdots \\ B_{0,3}(t_m) & \cdots & B_{3,3}(t_m) \end{matrix} \right] \left[ \begin{matrix} b_{x0} & b_{y0} \\ b_{x1} & b_{y1} \\ b_{x2} & b_{y2} \\ b_{x3} & b_{y3} \end{matrix} \right] = \left[ \begin{matrix} p_{x0} & p_{y0} \\ p_{x1} & p_{y1} \\ \vdots & \vdots \\ p_{xm} & p_{ym} \end{matrix} \right] \tag{4} ⎣⎢⎢⎢⎡B0,3(t0)B0,3(t1)⋮B0,3(tm)⋯⋯⋱⋯B3,3(t0)B3,3(t1)⋮B3,3(tm)⎦⎥⎥⎥⎤⎣⎢⎢⎡bx0bx1bx2bx3by0by1by2by3⎦⎥⎥⎤=⎣⎢⎢⎢⎡px0px1⋮pxmpy0py1⋮pym⎦⎥⎥⎥⎤(4)

例5. 其他

编码：
$$
op = bp \cdot \dfrac {g_{iw}} {h_{out}}+tp \cdot
(1-\dfrac {g_{ih}}{h_{out}})
$$
效果：
op=bp⋅giwhout+tp⋅(1−gihhout)op = bp \cdot \dfrac {g_{iw}} {h_{out}}+tp \cdot (1-\dfrac {g_{ih}}{h_{out}}) op=bp⋅houtgiw+tp⋅(1−houtgih)

参考文档

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