【工具】Latex 公式用法积累
公式更建议用 Mathpix snipping tool 这个软件识别哈，没办法才看我这个哈
∣ λ − a 11 − a 12 … − a 1 n − a 21 λ − a 22 … − a 2 n … … … … − a n 1 − a n 2 … λ − a n n ∣ \left| \begin{array}{cccc} {\lambda-a_{11}} & {-a_{12}} & {\dots} & {-a_{1 n}} \\ {-a_{21}} & {\lambda-a_{22}} & {\dots} & {-a_{2 n}} \\ {\dots} & {\dots} & {\dots} & {\dots} \\ {-a_{n 1}} & {-a_{n 2}} & {\dots} & {\lambda-a_{n n}} \end{array} \right| ∣∣∣∣∣∣∣∣λ−a11−a21…−an1−a12λ−a22…−an2…………−a1n−a2n…λ−ann∣∣∣∣∣∣∣∣
\left|\begin{ array}{ cccc}{ \lambda-a_{ 11}} & { -a_{ 12}} & { \dots} & { -a_{ 1 n}} \\{ -a_{ 21}} & { \lambda-a_{ 22}} & { \dots} & { -a_{ 2 n}} \\{ \dots} & { \dots} & { \dots} & { \dots} \\{ -a_{ n 1}} & { -a_{ n 2}} & { \dots} & { \lambda-a_{ n n}}\end{ array}\right|
y = [ y 1 ⋮ y n ] y = \left[ \begin{array}{c} y_1 \\ \vdots \\ y_n \end{array} \right] y=⎣⎢⎡y1⋮yn⎦⎥⎤
y =\left[\begin{ array}{ c}y_1 \\ \vdots \\ y_n\end{ array}\right]
y = 1 = 2 \begin{aligned} y & = 1 \\ & = 2 \end{aligned} y=1=2
\begin{ aligned}y& = 1 \\& = 2\end{ aligned}
{ v 1 = 1 v 2 = − 1 v 3 = − 1 \left\{ \begin{array}{l} {v_1 = 1} \\ {v_2 = -1} \\ {v_3 = -1} \end{array} \right. ⎩⎨⎧v1=1v2=−1v3=−1
\left\{\begin{ array}{ l}v_1 = 1 \\v_2 = -1 \\v_3 = -1\end{ array}\right.
{ p , x = 1 1 − p , x = 0 \begin{cases} p, & x = 1 \\ 1 - p, & x = 0 \end{cases} { p,1−p,x=1x=0
$$\begin{ cases}p, & x = 1 \\1 - p, & x = 0\end{ cases}$$
箭头
| 效果 | 写法 |
|---|---|
| → \rightarrow → | \rightarrow |
| ⇒ \Rightarrow ⇒ | \Rightarrow |
| ↔ \leftrightarrow ↔ | \leftrightarrow |
| ⟶ \longrightarrow ⟶ | \longrightarrow |
空格
| 效果 | 写法 |
|---|---|
| a b a \qquad b ab | a \qquad b |
| a b a \quad b ab | a \quad b |
| a b a \ b a b | a \ b |
符号
| 效果 | 写法 |
|---|---|
| ≥ \geq ≥ | $\geq$ |
| ≤ \leq ≤ | $\leq$ |
| ≠ \neq = | $\neq$ |
| ∝ \propto ∝ | \propto |
| ∀ \forall ∀ | \forall |
| ∃ \exists ∃ | \exists |
| a ^ b c \hat abc a^bc | $\hat a$ |
| a b c ^ \widehat{abc} abc | $\widehat{a}$ |
| ≜ \triangleq ≜ | $\triangleq$ |
| ∑ i = 1 \sum{i=1} ∑i=1 | $\sum{i=1}$ |
| ∑ i = 1 \sum\limits{i=1} i=1∑ | $\sum\limits{i=1}$ |
| ∏ i = 1 \prod{i=1} ∏i=1 | $\prod{i=1}$ |
| ∏ i = 1 \prod\limits{i=1} i=1∏ | $\prod\limits{i=1}$ |
| arg min 0 ≤ j ≤ k − 1 \underset{0\leq j \leq k-1}{\arg\min} 0≤j≤k−1argmin | $\underset{0\leq j \leq k-1}{\arg\min}$ |
| B A \underset{A}{B} AB | $\underset{A}{B}$ |
| a = ? b a\overset{?}=b a=?b | $a \overset{?}= b$ |
| ∑ 0 < i ≤ n 0 < j ≤ n A i j \sum{0<i \leq n \atop 0<j \leq n} A{i j} ∑0<j≤n0<i≤nAij | $\sum{0<i \leq n \atop 0<j \leq n} A{i j}$ |
| ∈ \in ∈ | $\in$ |
| ∉ \notin ∈/ | $\notin$ |
| A B C D E F G H I J K L M N O P Q R S T U V W X Y Z \mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ} ABCDEFGHIJKLMNOPQRSTUVWXYZ | $\mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ |
多行公式对齐
可以控制左右的对齐方式
max g ( λ , ν ) subject to λ ≥ 0 \begin{array}{cl} \max & g(\lambda, \nu) \\ \text {subject to} & \lambda \ge 0 \end{array} maxsubject tog(λ,ν)λ≥0
$$\begin{array}{cl}\max & g(\lambda, \nu) \\\text { subject to } & \lambda \ge 0\end{array}$$
max g ( λ , ν ) subject to λ ≥ 0 \begin{array}{ll} \max & g(\lambda, \nu) \\ \text {subject to} & \lambda \ge 0 \end{array} maxsubject tog(λ,ν)λ≥0
$$\begin{array}{ll}\max & g(\lambda, \nu) \\\text { subject to } & \lambda \ge 0\end{array}$$
参考
http://blog.lisp4fun.com/2017/11/01/formula
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布满荆棘的人生
Bertha 。
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