Latex Ref [E]

本文详细介绍使用LaTeX和KaTeX进行数学公式和符号的排版技巧,涵盖希腊字母、箭头、运算符、积分、矩阵等各类数学元素的书写方式。

Latex

Katex1


Doublestruck Letters

$\mathbb{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z}$

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 \mathbb{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} ABCDEFGHIJKLMNOPQRSTUVWXYZ


Repeating fractions

$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} \equiv 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$

1 ( ϕ 5 − ϕ ) e 2 5 π ≡ 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} \equiv 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } (ϕ5 ϕ)e52π11+1+1+1+1+e8πe6πe4πe2π


Summation notation

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

( ∑ k = 1 n a k b k ) 2 ≤ ( ∑ k = 1 n a k 2 ) ( ∑ k = 1 n b k 2 ) \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) (k=1nakbk)2(k=1nak2)(k=1nbk2)


Sum of a Series

$\displaystyle\sum_{i=1}^{k+1}i$

∑ i = 1 k + 1 i \displaystyle\sum_{i=1}^{k+1}i i=1k+1i

$\displaystyle= \left(\sum_{i=1}^{k}i\right) +(k+1)$

= ( ∑ i = 1 k i ) + ( k + 1 ) \displaystyle= \left(\sum_{i=1}^{k}i\right) +(k+1) =(i=1ki)+(k+1)

$\displaystyle= \frac{k(k+1)}{2}+k+1$

= k ( k + 1 ) 2 + k + 1 \displaystyle= \frac{k(k+1)}{2}+k+1 =2k(k+1)+k+1

$\displaystyle= \frac{k(k+1)+2(k+1)}{2}$

= k ( k + 1 ) + 2 ( k + 1 ) 2 \displaystyle= \frac{k(k+1)+2(k+1)}{2} =2k(k+1)+2(k+1)

$\displaystyle= \frac{(k+1)(k+2)}{2}$

= ( k + 1 ) ( k + 2 ) 2 \displaystyle= \frac{(k+1)(k+2)}{2} =2(k+1)(k+2)

$\displaystyle= \frac{(k+1)((k+1)+1)}{2}$

= ( k + 1 ) ( ( k + 1 ) + 1 ) 2 \displaystyle= \frac{(k+1)((k+1)+1)}{2} =2(k+1)((k+1)+1)


Product notation

$\displaystyle\text{ for }\lvert q\rvert < 1.$

 for  ∣ q ∣ < 1. \displaystyle\text{ for }\lvert q\rvert < 1.  for q<1.

$= \displaystyle \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},$

= ∏ j = 0 ∞ 1 ( 1 − q 5 j + 2 ) ( 1 − q 5 j + 3 ) , = \displaystyle \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, =j=0(1q5j+2)(1q5j+3)1,

$\displaystyle1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots$

1 + q 2 ( 1 − q ) + q 6 ( 1 − q ) ( 1 − q 2 ) + ⋯ \displaystyle1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots 1+(1q)q2+(1q)(1q2)q6+


Inline math2

<span class="math">...</span>

And $k_{n+1} = n^2 + k_n^2 - k_{n-1}$, text.

And k n + 1 = n 2 + k n 2 − k n − 1 k_{n+1} = n^2 + k_n^2 - k_{n-1} kn+1=n2+kn2kn1​, text.


Greek Letters

$\Gamma\ \Delta\ \Theta\ \Lambda\ \Xi\ \Pi\ \Sigma\ \Upsilon\ \Phi\ \Psi\ \Omega$

Γ   Δ   Θ   Λ   Ξ   Π   Σ   Υ   Φ   Ψ   Ω \Gamma\ \Delta\ \Theta\ \Lambda\ \Xi\ \Pi\ \Sigma\ \Upsilon\ \Phi\ \Psi\ \Omega Γ Δ Θ Λ Ξ Π Σ Υ Φ Ψ Ω

$\alpha\ \beta\ \gamma\ \delta\ \epsilon\ \zeta\ \eta\ \theta\ \iota\ \kappa\ \lambda\ \mu\ \nu\ \xi\ \omicron\ \pi\ \rho\ \sigma\ \tau\ \upsilon\ \phi\ \chi\ \psi\ \omega$

α   β   γ   δ   ϵ   ζ   η   θ   ι   κ   λ   μ   ν   ξ   ο   π   ρ   σ   τ   υ   ϕ   χ   ψ   ω \alpha\ \beta\ \gamma\ \delta\ \epsilon\ \zeta\ \eta\ \theta\ \iota\ \kappa\ \lambda\ \mu\ \nu\ \xi\ \omicron\ \pi\ \rho\ \sigma\ \tau\ \upsilon\ \phi\ \chi\ \psi\ \omega α β γ δ ϵ ζ η θ ι κ λ μ ν ξ ο π ρ σ τ υ ϕ χ ψ ω

$\varepsilon\ \vartheta\ \varpi\ \varrho\ \varsigma\ \varphi$

ε   ϑ   ϖ   ϱ   ς   φ \varepsilon\ \vartheta\ \varpi\ \varrho\ \varsigma\ \varphi ε ϑ ϖ ϱ ς φ


Arrows

$\gets\ \to\ \leftarrow\ \rightarrow\ \uparrow\ \Uparrow\ \downarrow\ \Downarrow\ \updownarrow\ \Updownarrow$

←   →   ←   →   ↑   ⇑   ↓   ⇓   ↕   ⇕ \gets\ \to\ \leftarrow\ \rightarrow\ \uparrow\ \Uparrow\ \downarrow\ \Downarrow\ \updownarrow\ \Updownarrow          

$\Leftarrow\ \Rightarrow\ \leftrightarrow\ \Leftrightarrow\ \mapsto\ \hookleftarrow$

⇐   ⇒   ↔   ⇔   ↦   ↩ \Leftarrow\ \Rightarrow\ \leftrightarrow\ \Leftrightarrow\ \mapsto\ \hookleftarrow      

$\leftharpoonup\ \leftharpoondown\ \rightleftharpoons\ \longleftarrow\ \Longleftarrow\ \longrightarrow$

↼   ↽   ⇌   ⟵   ⟸   ⟶ \leftharpoonup\ \leftharpoondown\ \rightleftharpoons\ \longleftarrow\ \Longleftarrow\ \longrightarrow      

$\Longrightarrow\ \longleftrightarrow\ \Longleftrightarrow\ \longmapsto\ \hookrightarrow\ \rightharpoonup$

⟹   ⟷   ⟺   ⟼   ↪   ⇀ \Longrightarrow\ \longleftrightarrow\ \Longleftrightarrow\ \longmapsto\ \hookrightarrow\ \rightharpoonup      

LaTeX code:
$\rightharpoondown\ \leadsto\ \nearrow\ \searrow\ \swarrow\ \nwarrow$

⇁   ⇝   ↗   ↘   ↙   ↖ \rightharpoondown\ \leadsto\ \nearrow\ \searrow\ \swarrow\ \nwarrow      


Symbols

$\surd\ \barwedge\ \veebar\ \odot\ \oplus\ \otimes\ \oslash\ \circledcirc\ \boxdot\ \bigtriangleup$

√   ⊼   ⊻   ⊙   ⊕   ⊗   ⊘   ⊚   ⊡   △ \surd\ \barwedge\ \veebar\ \odot\ \oplus\ \otimes\ \oslash\ \circledcirc\ \boxdot\ \bigtriangleup          

$\bigtriangledown\ \dagger\ \diamond\ \star\ \triangleleft\ \triangleright\ \angle\ \infty\ \prime\ \triangle$

▽   †   ⋄   ⋆   ◃   ▹   ∠   ∞   ′   △ \bigtriangledown\ \dagger\ \diamond\ \star\ \triangleleft\ \triangleright\ \angle\ \infty\ \prime\ \triangle          


Calculus

$\int u \frac{dv}{dx}\,dx=uv-\int \frac{du}{dx}v\,dx$

∫ u d v d x   d x = u v − ∫ d u d x v   d x \int u \frac{dv}{dx}\,dx=uv-\int \frac{du}{dx}v\,dx udxdvdx=uvdxduvdx

$f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x}$

f ( x ) = ∫ − ∞ ∞ f ^ ( ξ )   e 2 π i ξ x f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} f(x)=f^(ξ)e2πiξx

$\oint \vec{F} \cdot d\vec{s}=0$

∮ F ⃗ ⋅ d s ⃗ = 0 \oint \vec{F} \cdot d\vec{s}=0 F ds =0


Lorenz Equations

$\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned}$

x ˙ = σ ( y − x ) y ˙ = ρ x − y − x z z ˙ = − β z + x y \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned} x˙y˙z˙=σ(yx)=ρxyxz=βz+xy

Cross Product

$\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix}$

V 1 × V 2 = ∣ i j k ∂ X ∂ u ∂ Y ∂ u 0 ∂ X ∂ v ∂ Y ∂ v 0 ∣ \mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix} V1×V2=iuXvXjuYvYk00


Accents

$\hat{x}\ \vec{x}\ \ddot{x}$

x ^   x ⃗   x ¨ \hat{x}\ \vec{x}\ \ddot{x} x^ x  x¨


Stretchy brackets

$\left(\frac{x^2}{y^3}\right)$

( x 2 y 3 ) \left(\frac{x^2}{y^3}\right) (y3x2)


Evaluation at limits

$\left.\frac{x^3}{3}\right|_0^1$

x 3 3 ∣ 0 1 \left.\frac{x^3}{3}\right|_0^1 3x301


Case definitions

$f(n) = \begin{cases} \frac{n}{2}, & \text{if } n\text{ is even} \\ 3n+1, & \text{if } n\text{ is odd} \end{cases}$

f ( n ) = { n 2 , if  n  is even 3 n + 1 , if  n  is odd f(n) = \begin{cases} \frac{n}{2}, & \text{if } n\text{ is even} \\ 3n+1, & \text{if } n\text{ is odd} \end{cases} f(n)={2n,3n+1,if n is evenif n is odd


Maxwell’s Equations

$\begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$

∇ × B ⃗ −   1 c   ∂ E ⃗ ∂ t = 4 π c j ⃗ ∇ ⋅ E ⃗ = 4 π ρ ∇ × E ⃗   +   1 c   ∂ B ⃗ ∂ t = 0 ⃗ ∇ ⋅ B ⃗ = 0 \begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned} ×B c1tE E ×E +c1tB B =c4πj =4πρ=0 =0

$\begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\[1em] \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\[0.5em] \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\[1em] \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$

∇ × B ⃗ −   1 c   ∂ E ⃗ ∂ t = 4 π c j ⃗ ∇ ⋅ E ⃗ = 4 π ρ ∇ × E ⃗   +   1 c   ∂ B ⃗ ∂ t = 0 ⃗ ∇ ⋅ B ⃗ = 0 \begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\[1em] \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\[0.5em] \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\[1em] \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned} ×B c1tE E ×E +c1tB B =c4πj =4πρ=0 =0


Statistics

$\frac{n!}{k!(n-k)!} = {^n}C_k$

n ! k ! ( n − k ) ! = n C k \frac{n!}{k!(n-k)!} = {^n}C_k k!(nk)!n!=nCk

${n \choose k}$

( n k ) {n \choose k} (kn)


Fractions on fractions

$\frac{\frac{1}{x}+\frac{1}{y}}{y-z}$

1 x + 1 y y − z \frac{\frac{1}{x}+\frac{1}{y}}{y-z} yzx1+y1


n-th root

$\sqrt[n]{1+x+x^2+x^3+\ldots}$

1 + x + x 2 + x 3 + … n \sqrt[n]{1+x+x^2+x^3+\ldots} n1+x+x2+x3+


Matrices

$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}$

( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ) \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix} a11a21a31a12a22a32a13a23a33

$begin{bmatrix} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0 \end{bmatrix}$

[ 0 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 0 ] \begin{bmatrix} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0 \end{bmatrix} 0000


Punctuation

$(x) = \sqrt{1+x} \quad (x \ge -1)$

f ( x ) = 1 + x ( x ≥ − 1 ) f(x) = \sqrt{1+x} \quad (x \ge -1) f(x)=1+x (x1)

$f(x) \sim x^2 \quad (x\to\infty)$

f ( x ) ∼ x 2 ( x → ∞ ) f(x) \sim x^2 \quad (x\to\infty) f(x)x2(x)

$f(x) = \sqrt{1+x}, \quad x \ge -1$

f ( x ) = 1 + x , x ≥ − 1 f(x) = \sqrt{1+x}, \quad x \ge -1 f(x)=1+x ,x1

$f(x) \sim x^2, \quad x\to\infty$

f ( x ) ∼ x 2 , x → ∞ f(x) \sim x^2, \quad x\to\infty f(x)x2,x


Γ \Gamma Γ function

$\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N$

Γ ( n ) = ( n − 1 ) ! ∀ n ∈ N \Gamma(n) = (n-1)!\quad\forall n\in\mathbb N Γ(n)=(n1)!nN


Euler integral

$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,.$

Γ ( z ) = ∫ 0 ∞ t z − 1 e − t d t   . \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,. Γ(z)=0tz1etdt.





  1. katext - ref ↩︎

  2. Elaboration needed ↩︎

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