【工具】Latex 公式用法积累

野性酷女 2023-08-17 17:13 164阅读 0赞

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​​−a12​λ−a22​…−an2​​…………​−a1n​−a2n​…λ−ann​​∣∣∣∣∣∣∣∣​

  1. \left|
  2. \begin{ array}{ cccc}
  3. { \lambda-a_{ 11}} & { -a_{ 12}} & { \dots} & { -a_{ 1 n}} \\
  4. { -a_{ 21}} & { \lambda-a_{ 22}} & { \dots} & { -a_{ 2 n}} \\
  5. { \dots} & { \dots} & { \dots} & { \dots} \\
  6. { -a_{ n 1}} & { -a_{ n 2}} & { \dots} & { \lambda-a_{ n n}}
  7. \end{ array}
  8. \right|

y = [ y 1 ⋮ y n ] y = \left[ \begin{array}{c} y_1 \\ \vdots \\ y_n \end{array} \right] y=⎣⎢⎡​y1​⋮yn​​⎦⎥⎤​

  1. y =
  2. \left[
  3. \begin{ array}{ c}
  4. y_1 \\ \vdots \\ y_n
  5. \end{ array}
  6. \right]

y = 1 = 2 \begin{aligned} y & = 1 \\ & = 2 \end{aligned} y​=1=2​

  1. \begin{ aligned}
  2. y
  3. & = 1 \\
  4. & = 2
  5. \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​

  1. \left\{
  2. \begin{ array}{ l}
  3. v_1 = 1 \\
  4. v_2 = -1 \\
  5. v_3 = -1
  6. \end{ array}
  7. \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​

  1. $$
  2. \begin{ cases}
  3. p, & x = 1 \\
  4. 1 - p, & x = 0
  5. \end{ cases}
  6. $$

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} 0jk1argmin $\underset{0\leq j \leq k-1}{\arg\min}$
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<jn0<inAij $\sum{0<i \leq n \atop 0<j \leq n} A{i j}$
∈ \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 to​g(λ,ν)λ≥0​

  1. $$
  2. \begin{array}{cl}
  3. \max & g(\lambda, \nu) \\
  4. \text { subject to } & \lambda \ge 0
  5. \end{array}
  6. $$

max ⁡ g ( λ , ν ) subject to λ ≥ 0 \begin{array}{ll} \max & g(\lambda, \nu) \\ \text {subject to} & \lambda \ge 0 \end{array} maxsubject to​g(λ,ν)λ≥0​

  1. $$
  2. \begin{array}{ll}
  3. \max & g(\lambda, \nu) \\
  4. \text { subject to } & \lambda \ge 0
  5. \end{array}
  6. $$

参考

http://blog.lisp4fun.com/2017/11/01/formula

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