Latex公式

(by 亲持红叶)

CATLOG

基础推荐

https://zhuanlan.zhihu.com/p/261750408

一、基础

1.1、一元一次方程

$$
y = a bx
$$

$$y=abx\text{\hspace{1em}(1)}$$

1.2 、上下标

$$
y = a   b_{ij}^2
$$

y = a b i j 2 (2) y = a b_{ij}^2\tag 2 y=a bij2(2)

1.3、求和

$$
\sum_{i=0}^n(x_i^2 y_j^3)
$$

∑ i = 0 n ( x i 2 y j 3 ) (3) \sum_{i=0}^n(x_i^2 y_j^3)\tag 3 i=0∑n(xi2 yj3)(3)

1.4、分数

$$
\frac{1}{3}-\frac{x_i^2}{y_j^3}
$$

或者

$$
\cfrac{1}{3}-\cfrac{x_i^2}{y_j^3}
$$

1 3 − x i 2 y j 3 1 3 − x i 2 y j 3 (4) \frac{1}{3}-\frac{x_i^2}{y_j^3} \\ \cfrac{1}{3}-\cfrac{x_i^2}{y_j^3} \tag 4 31−yj3xi231−yj3xi2(4)

1.5、矩阵

$$
\left[ 
\begin{array} {cccc}
X_1&Y_1^2\\
X_2 & Y_2^2\\
\ldots \\
X_n & Y_n^2
\end{array} 
\right]
$$

[ X 1 Y 1 2 X 2 Y 2 2 … X n Y n 2 ] (5) \left[ \begin{array} {cccc} X_1&Y_1^2\\ X_2 & Y_2^2\\ \ldots \\ X_n & Y_n^2 \end{array} \right] \tag 5 ⎣ ⎡X1X2…XnY12Y22Yn2⎦ ⎤(5)

1.6、希腊字母

注意大小写的首字母是区分大小写的关键

1.7、省略号

-- \ldots 　与底线对齐的省略号，\cdots 与中线对齐省略号
-- 换行 "\\"

$$
f(x_1,x_2,\ldots,x_i) = x_1 x_2 \cdots x_i
$$

f ( x 1 , x 2 , … , x i ) = x 1 x 2 ⋯ x i (6) f(x_1,x_2,\ldots,x_i) = x_1 x_2 \cdots x_i \tag 6 f(x1,x2,…,xi)=x1 x2 ⋯ xi(6)

f ( x 1 , x 2 , … , x i ) = x 1 x 2 ⋯ x i (7) f(x_1,x_2,\ldots,x_i) \\ = x_1 x_2 \cdots x_i \tag 7 f(x1,x2,…,xi)=x1 x2 ⋯ xi(7)

1.8、大括号

" \{  \}""  (注意这里直接{}打是不行的,显示不出来的)

$$
\{a b\}
$$

{ a b } (8) \{a b\} \tag 8 {a b}(8)

1.8、大大括号

* "\left(  \right)"
* "\left[  \right]"

$$ 
\left(a b^2 (c*r)\right) 
$$

或者

$$ 
\left[a b^2 (x*r)\right] 
$$

( a b 2 ( c ∗ r ) ) [ a b 2 ( x ∗ r ) ] (9) \left(a b^2 (c*r)\right) \\ \left[a b^2 (x*r)\right] \tag 9 (a b2 (c∗r))[a b2 (x∗r)](9)

1.9、多行公式

$$
\begin{aligned}
\cos 2 \theta & = & \cos^2 \theta - \sin^2 \theta\\
&=& 2\cos^2 \theta - 1 
\end{aligned}
$$

或者

$$ 
\begin{aligned}
\cos 2 \theta & = & \cos^2 \theta - \sin^2 \theta\\
&=& 2\cos^2 \theta - 1 
\end{aligned}
$$

cos ⁡ 2 θ = cos ⁡ 2 θ − sin ⁡ 2 θ = 2 cos ⁡ 2 θ − 1 (10) \begin{aligned} \cos 2 \theta & = & \cos^2 \theta - \sin^2 \theta\\ &=& 2\cos^2 \theta - 1 \end{aligned} \tag {10} cos2θ==cos2θ−sin2θ2cos2θ−1(10)

cos ⁡ 2 θ = cos ⁡ 2 θ − sin ⁡ 2 θ = 2 cos ⁡ 2 θ − 1 (11) \begin{aligned} \cos 2 \theta & = & \cos^2 \theta - \sin^2 \theta\\ &=& 2\cos^2 \theta - 1 \end{aligned} \tag {11} cos2θ==cos2θ−sin2θ2cos2θ−1(11)

---

二、数学公式

• 下面的公式是从wps中套用过来的

2.1、傅里叶级数

$$
f(x) = \frac{a_0}{2}   \sum_{n=1}^\infty(a_n \cos {nx}   b_n \sin {nx}) 
$$

或者

$$
\{ f(x) = {{{a_0}} \over 2}   \sum\limits_{n = 1}^\infty  {({a_n}\cos {nx}   {b_n}\sin {nx})}  \}
$$

f ( x ) = a 0 2 ∑ n = 1 ∞ ( a n cos ⁡ n x b n sin ⁡ n x ) { f ( x ) = a 0 2 ∑ n = 1 ∞ ( a n cos ⁡ n x b n sin ⁡ n x ) } (12) f(x) = \frac{a_0}{2} \sum_{n=1}^\infty(a_n \cos {nx} b_n \sin {nx}) \\ \{ f(x) = {{{a_0}} \over 2} \sum\limits_{n = 1}^\infty {({a_n}\cos {nx} {b_n}\sin {nx})} \} \tag {12} f(x)=2a0 n=1∑∞(ancosnx bnsinnx){f(x)=2a0 n=1∑∞(ancosnx bnsinnx)}(12)

2.2、高斯公式

$$
\iiint _ { \Omega } \left( \frac { \partial {P} } { \partial {x} }   \frac { \partial {Q} } { \partial {y} }   \frac { \partial {R} }{ \partial {z} } \right) \mathrm { d } V = \oint _ { \partial \Omega } ( P \cos \alpha   Q \cos \beta   R \cos \gamma ) \mathrm{ d} S 
$$ 

∭ Ω ( ∂ P ∂ x ∂ Q ∂ y ∂ R ∂ z ) d V = ∮ ∂ Ω ( P cos ⁡ α Q cos ⁡ β R cos ⁡ γ ) d S (13) \iiint _ { \Omega } \left( \frac { \partial {P} } { \partial {x} } \frac { \partial {Q} } { \partial {y} } \frac { \partial {R} }{ \partial {z} } \right) \mathrm { d } V = \oint _ { \partial \Omega } ( P \cos \alpha Q \cos \beta R \cos \gamma ) \mathrm{ d} S \tag {13} ∭Ω(∂x∂P ∂y∂Q ∂z∂R)dV=∮∂Ω(Pcosα Qcosβ Rcosγ)dS(13)

2.3、定积分

$$
\lim\limits_ { n \rightarrow   \infty } \sum _ { i = 1 } ^ { n } f \left[ a   \frac { i } { n } ( b - a ) \right] \frac { b - a } { n } = \int _ { a } ^ { b } f ( x )\mathrm {d}x 
$$

lim ⁡ n → ∞ ∑ i = 1 n f [ a i n ( b − a ) ] b − a n = ∫ a b f ( x ) d x (14) \lim\limits_ { n \rightarrow \infty } \sum _ { i = 1 } ^ { n } f \left[ a \frac { i } { n } ( b - a ) \right] \frac { b - a } { n } = \int _ { a } ^ { b } f ( x )\mathrm {d}x \tag {14} n→ ∞limi=1∑nf[a ni(b−a)]nb−a=∫abf(x)dx(14)

2.4、和的展开式

$$  
( 1   x ) ^ { n } = 1   \frac { n x } { 1 ! }   \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! }   \cdots   
$$

( 1 x ) n = 1 n x 1 ! n ( n − 1 ) x 2 2 ! ⋯ (15) ( 1 x ) ^ { n } = 1 \frac { n x } { 1 ! } \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! } \cdots \tag {15} (1 x)n=1 1!nx 2!n(n−1)x2 ⋯(15)

2.5、三角恒等式

$$  
\sin \alpha \pm \sin \beta = 2 \sin \frac { 1 } { 2 } ( \alpha \pm \beta ) \cos \frac { 1 } { 2 } ( \alpha \mp \beta )  
$$



$$
\cos \alpha   \cos \beta = 2 \cos \frac { 1 } { 2 } ( \alpha   \beta ) \cos \frac { 1 } { 2 } ( \alpha - \beta ) 
$$

sin ⁡ α ± sin ⁡ β = 2 sin ⁡ 1 2 ( α ± β ) cos ⁡ 1 2 ( α ∓ β ) cos ⁡ α cos ⁡ β = 2 cos ⁡ 1 2 ( α β ) cos ⁡ 1 2 ( α − β ) (16) \sin \alpha \pm \sin \beta = 2 \sin \frac { 1 } { 2 } ( \alpha \pm \beta ) \cos \frac { 1 } { 2 } ( \alpha \mp \beta ) \\ \cos \alpha \cos \beta = 2 \cos \frac { 1 } { 2 } ( \alpha \beta ) \cos \frac { 1 } { 2 } ( \alpha - \beta ) \tag {16} sinα±sinβ=2sin21(α±β)cos21(α∓β)cosα cosβ=2cos21(α β)cos21(α−β)(16)

2.6、欧拉公式

$$ 
{e^{ix}} = \cos {x}   i\sin {x}  
$$

e i x = cos ⁡ x i sin ⁡ x (17) {e^{ix}} = \cos {x} i\sin {x} \tag {17} eix=cosx isinx(17)

2.7、格林公式

$$  
\int\!\!\!\int\limits_D {({{\partial Q} \over {\partial x}} - {{\partial P} \over {\partial y}})dxdy = \oint\limits_L {Pdx   Qdy} }  
$$

∫  ⁣ ⁣ ⁣ ∫ D ( ∂ Q ∂ x − ∂ P ∂ y ) d x d y = ∮ L P d x Q d y (18) \int\!\!\!\int\limits_D {({{\partial Q} \over {\partial x}} - {{\partial P} \over {\partial y}})dxdy = \oint\limits_L {Pdx Qdy} } \tag {18} ∫D∫(∂x∂Q−∂y∂P)dxdy=L∮Pdx Qdy(18)

2.8、伯努利方程

$$  
{{dy} \over {dx}}   P(x)y = Q(x){y^n}(n \ne 0,1) 
$$

d y d x P ( x ) y = Q ( x ) y n ( n ≠ 0 , 1 ) (19) {{dy} \over {dx}} P(x)y = Q(x){y^n}(n \ne 0,1) \tag {19} dxdy P(x)y=Q(x)yn(n=0,1)(19)

2.9 、全微分方程

$$   
du(x,y) = P(x,y)dx   Q(x,y)dy = 0  
$$

$$du(x,y)=P(x,y)dxQ(x,y)dy=0\text{\hspace{1em}(20)}$$

2.10 、非齐次微分方程通解

$$   
y = (\int {Q(x){e^{\int {P(x)dx} }}dx   C){e^{ - \int {P(x)dx} }}}   
$$

y = ( ∫ Q ( x ) e ∫ P ( x ) d x d x C ) e − ∫ P ( x ) d x (21) y = (\int {Q(x){e^{\int {P(x)dx} }}dx C){e^{ - \int {P(x)dx} }}} \tag {21} y=(∫Q(x)e∫P(x)dxdx C)e−∫P(x)dx(21)

2.11、柯西中值定理

 $$ 
 \frac{{f(b) - f(a)}}{{F(b) - F(a)}} = \frac{{f'(\xi )}}{{F'(\xi )}} 
 $$

f ( b ) − f ( a ) F ( b ) − F ( a ) = f ′ ( ξ ) F ′ ( ξ ) (22) \frac{{f(b) - f(a)}}{{F(b) - F(a)}} = \frac{{f'(\xi )}}{{F'(\xi )}} \tag {22} F(b)−F(a)f(b)−f(a)=F′(ξ)f′(ξ)(22)

2.12、拉格朗日中值定理

$$ 
f(b) - f(a) = f'(\xi )(b - a) 
$$

f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) (23) f(b) - f(a) = f'(\xi )(b - a) \tag{23} f(b)−f(a)=f′(ξ)(b−a)(23)

2.13、曲面面积

$$
A = \int\!\!\!\int\limits_D {\sqrt {1   {{\left( {{{\partial z} \over {\partial x}}} \right)}^2}   {{\left( {{{\partial z} \over {\partial y}}} \right)}^2}} dxdy}
$$  

A = ∫  ⁣ ⁣ ⁣ ∫ D 1 ( ∂ z ∂ x ) 2 ( ∂ z ∂ y ) 2 d x d y (24) A = \int\!\!\!\int\limits_D {\sqrt {1 {{\left( {{{\partial z} \over {\partial x}}} \right)}^2} {{\left( {{{\partial z} \over {\partial y}}} \right)}^2}} dxdy} \tag{24} A=∫D∫1 (∂x∂z)2 (∂y∂z)2 dxdy(24)

2.14、重积分

 $$  
 \int\!\!\!\int\limits_D {f(x,y)dxdy}  = \int\!\!\!\int\limits_{D'} {f(r\cos \theta ,r\sin \theta )rdrd\theta }   
 $$

∫  ⁣ ⁣ ⁣ ∫ D f ( x , y ) d x d y = ∫  ⁣ ⁣ ⁣ ∫ D ′ f ( r cos ⁡ θ , r sin ⁡ θ ) r d r d θ (25) \int\!\!\!\int\limits_D {f(x,y)dxdy} = \int\!\!\!\int\limits_{D'} {f(r\cos \theta ,r\sin \theta )rdrd\theta } \tag{25} ∫D∫f(x,y)dxdy=∫D′∫f(rcosθ,rsinθ)rdrdθ(25)

2.15、arcsinx 求导

$$ 
(\arcsin x)' = \frac{1}{{\sqrt {1 - {x^2}} }} 
$$

( arcsin ⁡ x ) ′ = 1 1 − x 2 (26) (\arcsin x)' = \frac{1}{{\sqrt {1 - {x^2}} }} \tag{26} (arcsinx)′=1−x2 1(26)

2.16、泰勒级数

 $$ p;ko[p';p[oooo]]
 e^x = 1   \frac{x}{1!}    \frac{x}{2!}  \frac{x}{3!}   \cdots   O(x^n),     -\infty < x < \infty 
 $$

e x = 1 x 1 ! x 2 ! x 3 ! ⋯ O ( x n ) , − ∞ < x < ∞ (27) e^x = 1 \frac{x}{1!} \frac{x}{2!} \frac{x}{3!} \cdots O(x^n), -\infty < x < \infty \tag{27} ex=1 1!x 2!x 3!x ⋯ O(xn),−∞<x<∞(27)

2.17、三角函数积分

$$
\int tgx \mathrm {d} x = -\ln {|\cos x|}   c
$$

$$\int tgx\mathrm{d}x=-\ln |\cos x|c\text{\hspace{1em}(28)}$$

2.18、二次曲面

 $$
 \frac{{{x^2}}}{{{a^2}}}   \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1 
 $$

x 2 a 2 y 2 b 2 − z 2 c 2 = 1 (29) \frac{{{x^2}}}{{{a^2}}} \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1 \tag{29} a2x2 b2y2−c2z2=1(29)

2.19、二阶微分

$$
\frac{d^2y}{dx^2} P(x)\frac{dy}{dx} Q(x)y=f(x)
$$

d 2 y d x 2 P ( x ) d y d x Q ( x ) y = f ( x ) (30) \frac{d^2y}{dx^2} P(x)\frac{dy}{dx} Q(x)y=f(x) \tag{30} dx2d2y P(x)dxdy Q(x)y=f(x)(30)

2.20、方向导数

$$
\frac{\partial f}{\partial l}=\frac{\partial f}{\partial x}\cos{\phi} \frac{\partial f}{\partial y}\sin{\phi}
$$

∂ f ∂ l = ∂ f ∂ x cos ⁡ ϕ ∂ f ∂ y sin ⁡ ϕ (31) \frac{\partial f}{\partial l}=\frac{\partial f}{\partial x}\cos{\phi} \frac{\partial f}{\partial y}\sin{\phi} \tag{31} ∂l∂f=∂x∂fcosϕ ∂y∂fsinϕ(31)

---

三、公式

• 下面的公式来自清风数学建模中有一张教latex语法的时候给出的4个例子

3.1、练习一

$$
\begin{aligned}
\min {f_1} &= \frac{\sqrt{\frac{1}{n-1}\sum(W_j-\overline{W}^2)}}{\overline{W}}
\\
\min{f_2} &= \max\limits_{1<j<n}{T_j}
\end{aligned}
\\
\operatorname{ s.t. }
\left\{
\begin{array}{}
x_{ij} &= 1, i = j \\
\sum\limits_{j=1}^n x_{ij}&=1, i \neq j \\
\frac{S_{ij}}{V} &\leq 3\\
x_{ij} & \in \{0,1\} \\
\overline{W} &= \frac{1}{n}\sum\limits_{j=1}^n W_j\\
&(i = 1,2,3,\cdots,m;j=1,2,\cdots,n)
\end{array}
\right.
$$

min ⁡ f 1 = 1 n − 1 ∑ ( W j − W ‾ 2 ) W ‾ min ⁡ f 2 = max ⁡ 1 < j < n T j s.t. ⁡ { x i j = 1 , i = j ∑ j = 1 n x i j = 1 , i ≠ j S i j V ≤ 3 x i j ∈ { 0 , 1 } W ‾ = 1 n ∑ j = 1 n W j ( i = 1 , 2 , 3 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) (32) \begin{aligned} \min {f_1} &= \frac{\sqrt{\frac{1}{n-1}\sum(W_j-\overline{W}^2)}}{\overline{W}} \\ \min{f_2} &= \max\limits_{1<j<n}{T_j} \end{aligned} \\ \operatorname{ s.t. } \left\{ \begin{array}{} x_{ij} &= 1, i = j \\ \sum\limits_{j=1}^n x_{ij}&=1, i \neq j \\ \frac{S_{ij}}{V} &\leq 3\\ x_{ij} & \in \{0,1\} \\ \overline{W} &= \frac{1}{n}\sum\limits_{j=1}^n W_j\\ &(i = 1,2,3,\cdots,m;j=1,2,\cdots,n) \end{array} \right. \tag{32} minf1minf2=Wn−11∑(Wj−W2) =1<j<nmaxTjs.t.⎩ ⎨ ⎧xijj=1∑nxijVSijxijW=1,i=j=1,i=j≤3∈{0,1}=n1j=1∑nWj(i=1,2,3,⋯,m;j=1,2,⋯,n)(32)

3.2、练习二

$$
\begin{aligned}
\int_{k-1}^k \frac{dx_0^{(1)}(t)}{dt}\mathrm{dt} & \approx  - \hat{a}\int_{k-1}^kx_0^{(1)}dt \int_{k-1}^k\hat b dt 
\\
& =  \int_{k-1}^k \left[-\hat a x_0 ^ {(1)} (t)   \hat b \right]dt
\end{aligned}
$$

∫ k − 1 k d x 0 ( 1 ) ( t ) d t d t ≈ − a ^ ∫ k − 1 k x 0 ( 1 ) d t ∫ k − 1 k b ^ d t = ∫ k − 1 k [ − a ^ x 0 ( 1 ) ( t ) b ^ ] d t (33) \begin{aligned} \int_{k-1}^k \frac{dx_0^{(1)}(t)}{dt}\mathrm{dt} & \approx - \hat{a}\int_{k-1}^kx_0^{(1)}dt \int_{k-1}^k\hat b dt \\ & = \int_{k-1}^k \left[-\hat a x_0 ^ {(1)} (t) \hat b \right]dt \end{aligned} \tag{33} ∫k−1kdtdx0(1)(t)dt≈−a^∫k−1kx0(1)dt ∫k−1kb^dt=∫k−1k[−a^x0(1)(t) b^]dt(33)

3.3、练习三

$$
\left\{
\begin{aligned}
z_1 & =  l_{11}x_1   l_{12}x_2   \cdots     l_{1p}x_p 
\\
z_2 & =  l_{21}x_1   l_{22}x_2   \cdots     l_{2p}x_p 
\\
& \vdots 
\\
z_m & =  l_{m1}x_1   l_{m2}x_2   \cdots     l_{mp}x_p 
\end{aligned}
\right.
$$

{ z 1 = l 11 x 1 l 12 x 2 ⋯ l 1 p x p z 2 = l 21 x 1 l 22 x 2 ⋯ l 2 p x p ⋮ z m = l m 1 x 1 l m 2 x 2 ⋯ l m p x p (34) \left\{ \begin{aligned} z_1 & = l_{11}x_1 l_{12}x_2 \cdots l_{1p}x_p \\ z_2 & = l_{21}x_1 l_{22}x_2 \cdots l_{2p}x_p \\ & \vdots \\ z_m & = l_{m1}x_1 l_{m2}x_2 \cdots l_{mp}x_p \end{aligned} \right. \tag{34} ⎩ ⎨ ⎧z1z2zm=l11x1 l12x2 ⋯ l1pxp=l21x1 l22x2 ⋯ l2pxp⋮=lm1x1 lm2x2 ⋯ lmpxp(34)

3.4、练习四

$$
x = 
x =  \left[ 
\begin{aligned} 
x_{11} & x_{12} & \cdots & x_{1p} \\ 
x_{21} & x_{22} & \cdots & x_{2p} \\ 
\vdots & \vdots & \ddots & \vdots \\ 
x_{n1} & x_{n2} & \cdots & x_{np} \\ 
\end{aligned} \right] 
=(x_1,x_2,\cdots,x_p)
$$

x = x = [ x 11 x 12 ⋯ x 1 p x 21 x 22 ⋯ x 2 p ⋮ ⋮ ⋱ ⋮ x n 1 x n 2 ⋯ x n p ] = ( x 1 , x 2 , ⋯   , x p ) (35) x = x = \left[ \begin{aligned} x_{11} & x_{12} & \cdots & x_{1p} \\ x_{21} & x_{22} & \cdots & x_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{np} \\ \end{aligned} \right] =(x_1,x_2,\cdots,x_p) \tag{35} x=x=⎣ ⎡x11x21⋮xn1x12x22⋮xn2⋯⋯⋱⋯x1px2p⋮xnp⎦ ⎤=(x1,x2,⋯,xp)(35)

四、其他的一些例子

• 博主平时学习、工作中用的，顺着写的就不一个一个地整理了

y = f ( x ) = W T ⋅ x b = [ x 1 x 2 ⋮ x n ] b y = f(x)=W^T\cdot x b = \left[\begin{array}{} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] b y=f(x)=WT⋅x b=⎣ ⎡x1x2⋮xn⎦ ⎤ b

χ A = { 1 ,  A occurs  , 0 ,  A doesn’t occur.  \chi_{A}=\left\{\begin{array}{lc} 1, & \text { A occurs }, \\ 0, & \text { A doesn't occur. } \end{array}\right. χA={1,0, A occurs , A doesn’t occur. 

A = ( a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a m 1 ⋯ a m n ) A = \left(\begin{array}{ccc} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots & \vdots \\ a_{m1} &\cdots & a_{mn} \end{array} \right) A=⎝ ⎛a11⋮am1⋯⋱⋯a1n⋮amn⎠ ⎞

ω ^ i = ∏ j = 1 n a i j n ∑ j = 1 n ∏ j = 1 n a i j n , i ∈ [ 1 , m ] (m为指标数量) \hat{\omega}_i = \frac{\sqrt[n]{\prod\limits_{j=1}^na_{ij}}}{\sum\limits_{j=1}^n\sqrt[n]{\prod\limits_{j=1}^na_{ij}}} , i \in [1,m]\tag{m为指标数量} ω^i=j=1∑nnj=1∏naij nj=1∏naij ,i∈[1,m](m为指标数量)

C R = C I R I = λ m a x − n R I ( n − 1 ) CR = \frac{CI}{RI} = \frac{\lambda_{max} - n}{RI(n-1)} CR=RICI=RI(n−1)λmax−n

B j = − ∑ i = 1 m p i j ln ⁡ p i j ln ⁡ m B_j = -\sum\limits_{i=1}^m p_{ij}\frac{\ln{p_{ij}}}{\ln{m}} Bj=−i=1∑mpijlnmlnpij

w j = 1 − B j ∑ j = 1 n ( 1 − B j ) w_{j} = \frac{1-B_j}{\sum\limits_{j=1}^n(1-B_j)} wj=j=1∑n(1−Bj)1−Bj

∑ j = 1 n w j = 1 , j ∈ [ 1 , n ] \sum_{j=1}^n w_j = 1,j \in [1,n] j=1∑nwj=1,j∈[1,n]

ω ‾ = { W 1 w 1 ∑ j = 1 n W j w j , W 2 w 2 ∑ j = 1 n W j w j , ⋯   , W n w n ∑ j = 1 n W j w j } = ( ω 1 , ω 2 , ⋯   , ω n ) \overline{\omega}=\left\{\frac{W_{1} w_{1}}{\sum_{j=1}^{n} W_{j} w_{j}}, \frac{W_{2} w_{2}}{\sum_{j=1}^{n} W_{j} w_{j}}, \cdots, \frac{W_{n} w_{n}}{\sum_{j=1}^{n} W_{j} w_{j}}\right\}=\left(\omega_{1}, \omega_{2}, \cdots, \omega_{n}\right) ω={∑j=1nWjwjW1w1,∑j=1nWjwjW2w2,⋯,∑j=1nWjwjWnwn}=(ω1,ω2,⋯,ωn)

 s.t.  ∑ j = 1 n ω j = 1 ; ω j > 0 \text { s.t. } \sum_{j=1}^{n} \omega_{j}=1 ; \omega_{j}>0  s.t. j=1∑nωj=1;ωj>0

w j ( j = 1 , 2 , ⋯   , n ) w_j(j=1,2,\cdots,n) wj(j=1,2,⋯,n)

X = [ x 11 x 12 ⋯ x 1 m x 21 x 22 ⋯ x 2 m ⋮ ⋮ ⋱ ⋮ x n 1 x n 2 ⋯ x n m ] X=\left[\begin{array}{cccc} x_{11} & x_{12} & \cdots & x_{1 m} \\ x_{21} & x_{22} & \cdots & x_{2 m} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n 1} & x_{n 2} & \cdots & x_{n m} \end{array}\right] X=⎣ ⎡x11x21⋮xn1x12x22⋮xn2⋯⋯⋱⋯x1mx2m⋮xnm⎦ ⎤
那么，对其标准化的矩阵记为Z， $Z$ 中的每一个元素:
z i j = x i j / ∑ i = 1 n x i j 2 z_{i j}=x_{i j} / \sqrt{\sum_{i=1}^{n} x_{i j}^{2}} zij=xij/i=1∑nxij2
Z − = ( Z 1 − , Z 2 − , ⋯   , Z m − ) = ( min ⁡ { z 11 , z 21 , ⋯   , z n 1 } , min ⁡ { z 12 , z 22 , ⋯   , z n 2 } , ⋯   , min ⁡ { z 1 m , z 2 m , ⋯   , z n m } ) \begin{aligned} Z^{-} &=\left(Z_{1}^{-}, Z_{2}^{-}, \cdots, Z_{m}^{-}\right) \\ &=\left(\min \left\{z_{11}, z_{21}, \cdots, z_{n 1}\right\}, \min \left\{z_{12}, z_{22}, \cdots, z_{n 2}\right\}, \cdots, \min \left\{z_{1 m}, z_{2 m}, \cdots, z_{n m}\right\}\right) \end{aligned} Z−=(Z1−,Z2−,⋯,Zm−)=(min{z11,z21,⋯,zn1},min{z12,z22,⋯,zn2},⋯,min{z1m,z2m,⋯,znm})

D i = ∑ j = 1 m ( Z j − z i j ) 2 D_i^ = \sqrt{\sum\limits_{j=1}^m(Z_j^ - z_{ij})^2} Di =j=1∑m(Zj −zij)2

S i = D i − D i D i − S i ∈ [ 0 , 1 ] S_i = \frac{D_i^-}{D_i^ D_i^-} S_i \in [0,1] Si=Di Di−Di−Si∈[0,1]

S i ~ = S i ∑ i = 1 n S i ~ \tilde{S_i} = \frac{S_i}{\sum\limits_{i=1}^n \tilde{S_i}} Si~=i=1∑nSi~Si
∑ i = 1 n S i ~ = 1 \sum\limits_{i=1}^n \tilde{S_i} = 1 i=1∑nSi~=1

min ⁡ f 1 = 1 n − 1 ∑ j = 1 n ( W j − W ‾ ) 2 W ‾ \min {f_1} = \frac{\sqrt{\frac{1}{n-1}\sum \limits _{j=1}^n (W_j - \overline {W})^2}}{\overline {W}} minf1=Wn−11j=1∑n(Wj−W)2

min ⁡ f 2 = max ⁡ 1 < j < n T j \min{f_2} = \max\limits _ {1<j<n}{T_j} minf2=1<j<nmaxTj

∫ k − 1 k d x 0 ( 1 ) ( t ) d t d t ≈ − a ^ ∫ k − 1 k x 0 ( 1 ) \begin{aligned} \int_{k-1}^k \frac{dx_0^{(1)}(t)}{dt} dt & \approx & - \hat{a}\int_{k-1}^kx_0^{(1)} \end{aligned} ∫k−1kdtdx0(1)(t)dt≈−a^∫k−1kx0(1)

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