语法:https://www.jianshu.com/p/e74eb43960a1
H ( Y ∣ X ) = ∑ x ∈ X , y ∈ Y p ( x , y ) log ( p ( x ) p ( x , y ) ) H(Y | X)=\sum_{x \in \mathcal{X}, y \in \mathcal{Y}} p(x, y) \log \left(\frac{p(x)}{p(x, y)}\right) H(Y∣X)=x∈X,y∈Y∑p(x,y)log(p(x,y)p(x))
H ( x ) = ∑ x ∈ X , y ∈ Y p ( x , y ) ⋅ log ( p ( x ) p ( x , y ) ) H(x)=\sum_{x \in X, y \in Y} p(x, y) \cdot \log \left(\frac{p(x)}{p(x, y)}\right) H(x)=x∈X,y∈Y∑p(x,y)⋅log(p(x,y)p(x))
H ( x ) = ∑ x ∈ X , y ∈ Y p ( x , y ) ⋅ log ( p ( x ) p ( x , y ) ) H(x)=\sum_{x \in \mathcal{X}, y \in \mathcal{Y}} p(x, y) \cdot \log \left(\frac{p(x)}{p(x, y)}\right) H(x)=x∈X,y∈Y∑p(x,y)⋅log(p(x,y)p(x))
上标与下标: x 4 ⋅ x 1 x^4 \cdot x_1 x4⋅x1
多个字符组合: 16 8 O 2 + 2 {16}_{8}O{2+}_{2} 168O2+2 C k n + 1 = C k n + C k + 1 n C_{k}^{n+1}=C_{k}^{n}+C_{k+1}^{n} Ckn+1=Ckn+Ck+1n
结合汉字: V 初 始 V_{初始} V初始
显示风格与除法: x + y y + z \displaystyle \frac{x+y}{y+z} y+zx+y
下划线: x + y ‾ \underline{x+y} x+y
KaTeX parse error: \tag works only in display equations # KaTeX parse error: \tag works only in display equations
上大括号: a + b + c + d ⏞ 2.0 \overbrace{a+b+c+d}^{2.0} a+b+c+d 2.0
下大括号: a + b + c + d ⎵ 2.0 \underbrace{a+b+c+d}_{2.0} 2.0 a+b+c+d
上位符号: x ⃗ = d e f x 1 , x 2 , … , x n \vec{x}\stackrel{\mathrm{def}}{=}{x_1,x_2,\dots,x_n} x=defx1,x2,…,xn
上位符号,箭头: x ⃗ ⟶ d e f x 1 , x 2 , ⋯   , x n \vec{x} \stackrel {def}{\longrightarrow} x_{1}, x_{2}, \cdots, x_{n} x⟶defx1,x2,⋯,xn
占位符号,一个空格: x y x \quad y xy,两个空格: x y x \qquad y xy
大空格: x y x \ y x y; 中空格: x   y x \: y xy; 小空格: x   y x \, y xy;没空格: x y xy xy;紧贴: x ​ y x \! y xy
括号,依次为原始,big修饰,Big修饰,bigg修饰,Bigg修饰: ( ) ( ) ( ) ( ) ( ) () \big(\big) \Big(\Big) \bigg(\bigg) \Bigg(\Bigg) ()()()()()
中括号,符号:[],如: [ x + y ] [x+y] [x+y]
大括号,符号:{ },记得用转义符号\,如: { x + y } \{x+y\} {x+y}
自适应括号,符号:\left \right,如: ( x ) \left(x\right) (x), ( x y z ) \left(x{yz}\right) (xyz)
组合公式,符号:{上位公式 \choose 下位公式},如: ( n + 1 k ) = ( n k ) + ( n k − 1 ) {n+1 \choose k}={n \choose k}+{n \choose k-1} (kn+1)=(kn)+(k−1n)
组合公式,符号:{上位公式 \atop 下位公式},如: ∑ k 0 , k 1 , ⋯ > 0 k 0 + k 1 + ⋯ = n A k 0 A k 1 ⋯ \sum_{k_0,k_1,\dots>0 \atop k_0+k_1+\cdots=n}A_{k_0}A_{k_1}\cdots ∑k0+k1+⋯=nk0,k1,⋯>0Ak0Ak1⋯