MathJax实例

本篇文章基于MathJax编写了\(15\)个实例，尽量涵盖了各种常用符号，以熟悉各种符号的用法。

扩展欧几里得算法（Python）

$\begin{aligned}
& \texttt{def ex_gcd(a, b):}\\
& \qquad \texttt{if b == 0: }\\
& \qquad \qquad \texttt{return 1, 0, a}\\
& \qquad \texttt{else: }\\
& \qquad \qquad \texttt{x, y, g = ex_gcd(b, a % b)}\\
& \qquad \qquad \texttt{return y, x - (a // b) * y, g}\\
\\
\\
& \texttt{N = 5}\\
& \texttt{for n in range(1, N + 1):}\\
& \qquad \texttt{for m in range(1, N + 1):}\\
& \qquad \qquad \texttt{print(ex_gcd(n, m))}
\end{aligned}$

\(\begin{aligned} & \texttt{def ex_gcd(a, b):}\\ & \qquad \texttt{if b == 0: }\\ & \qquad \qquad \texttt{return 1, 0, a}\\ & \qquad \texttt{else: }\\ & \qquad \qquad \texttt{x, y, g = ex_gcd(b, a % b)}\\ & \qquad \qquad \texttt{return y, x - (a // b) * y, g}\\ \\ \\ & \texttt{N = 5}\\ & \texttt{for n in range(1, N + 1):}\\ & \qquad \texttt{for m in range(1, N + 1):}\\ & \qquad \qquad \texttt{print(ex_gcd(n, m))} \end{aligned}\)

范德蒙德行列式

$$\begin{vmatrix}
1   & 1   & \dots & 1\\
x_1 & x_2 & \dots & x_n\\
\vdots & \vdots & \ddots & \vdots\\
x_1^{n-1} & x_2^{n-2} & \dots & x_n^{n-1}
\end{vmatrix} = \prod_{1\le j\le i\le n} (x_i-x_j)$$

\[\det \begin{bmatrix} 1 & 1 & \dots & 1\\ x_1 & x_2 & \dots & x_n\\ \vdots & \vdots & \ddots & \vdots\\ x_1^{n-1} & x_2^{n-2} & \dots & x_n^{n-1} \end{bmatrix} = \prod_{1\le j\le i\le n} (x_i-x_j)\]

碳酸氢钙溶液加热

$$\mathrm{Ca(HCO_3)_2\xrightarrow{\triangle}CaCO_3\downarrow+CO_2\uparrow+H_2O}$$

\[\mathrm{Ca(HCO_3)_2\xrightarrow{\triangle}CaCO_3\downarrow+CO_2\uparrow+H_2O}\]

薛定谔方程

$$i\hbar\dfrac{\partial}{\partial t}|\psi (t)\rangle = \hat{H}|\psi(t)\rangle$$

$$i\hbar\dfrac{\partial}{\partial t}\Psi (x,t) = \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x,t)\right]\Psi(x,t)$$

\[i\hbar\dfrac{\partial}{\partial t}|\psi (t)\rangle = \hat{H}|\psi(t)\rangle\]

\[i\hbar\dfrac{\partial}{\partial t}\Psi (x,t) = \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x,t)\right]\Psi(x,t)\]

麦克斯韦方程组

$$\begin{cases}
\nabla \cdot \mathbf{E} = \dfrac{\rho}{\varepsilon_0} & \text{Gauss's law}\\
\nabla \cdot \mathbf{B}=0 & \text{Gauss's law for magnetism}\\
\nabla \times \mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t} & \text{Maxwell–Faraday equation}\\
\nabla \times \mathbf{B} =  \mu_0 \left(\mathbf{J} + \varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}\right) & \text{Ampère's circuital law}
\end{cases}$$

\[\begin{cases} \nabla \cdot \mathbf{E} = \dfrac{\rho}{\varepsilon_0} & \text{Gauss's law}\\ \nabla \cdot \mathbf{B}=0 & \text{Gauss's law for magnetism}\\ \nabla \times \mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t} & \text{Maxwell–Faraday equation}\\ \nabla \times \mathbf{B} = \mu_0 \left(\mathbf{J} + \varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}\right) & \text{Ampère's circuital law} \end{cases}\]

异或运算\(S=x \oplus y\)的真值表

$$\begin{array}{|l|l|l|}
\hline
x & y & S\\
\hdashline
0 & 0 & 0\\
\hline
0 & 1 & 1\\
\hline
1 & 0 & 1\\
\hline
1 & 0 & 0\\
\hline
\end{array}$$

\[\begin{array}{|l|l|l|} \hline x & y & S\\ \hdashline 0 & 0 & 0\\ \hline 0 & 1 & 1\\ \hline 1 & 0 & 1\\ \hline 1 & 0 & 0\\ \hline \end{array}\]

狄利克雷函数

$$ \mathbf{1}_{\mathbb{Q}}(x)=
{\begin{cases}
1&x\in \mathbb {Q} \\
0&x\notin \mathbb {Q}
\end{cases}}$$

\[ \mathbf{1}_{\mathbb{Q}}(x)= {\begin{cases} 1&x\in \mathbb {Q} \\ 0&x\notin \mathbb {Q} \end{cases}}\]

雅可比符号

$$\left(\dfrac{a}{p}\right) =
\begin{cases}
0 & \text{if $a\equiv 0\pmod{p}$}\\
1 & \text{if }a\not\equiv 0\pmod p \wedge \exists  x: x^2\equiv a\pmod p \\
-1 & \text{if }a\not\equiv 0\pmod p \wedge \nexists  x: x^2\equiv a\pmod p
\end{cases}$$

\[\left(\dfrac{a}{p}\right) = \begin{cases} 0 & \text{if $a\equiv 0\pmod{p}$}\\ 1 & \text{if }a\not\equiv 0\pmod p \wedge \exists x: x^2\equiv a\pmod p \\ -1 & \text{if }a\not\equiv 0\pmod p \wedge \nexists x: x^2\equiv a\pmod p \end{cases}\]

斯托克斯公式

$$\oint_{\Gamma}Pdx+Qdy+Rdz=\iint_{\Sigma}\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right) dydz+\left(\dfrac{\partial P}{\partial z}-\dfrac{\partial R}{\partial x}\right) dzdx+\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) dxdy$$

\[\oint_{\Gamma}Pdx+Qdy+Rdz=\iint_{\Sigma}\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right) dydz+\left(\dfrac{\partial P}{\partial z}-\dfrac{\partial R}{\partial x}\right) dzdx+\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) dxdy\]

\(\Gamma\)函数与黎曼\(\zeta\)函数

$$\begin{aligned}
\Gamma(z)=&\int_{0}^{\infty} t^{z-1} \mathrm{e}^{-t}dt &\\
\zeta(s)=&\sum_{n=1}^{\infty} \dfrac{1}{n^s}
\end{aligned}$$

\[\begin{aligned} \Gamma(z)=&\int_{0}^{\infty} t^{z-1} \mathrm{e}^{-t}dt &\\ \zeta(s)=&\sum_{n=1}^{\infty} \dfrac{1}{n^s} \end{aligned}\]

德摩根律

$$\begin{align}
& \neg (P\lor Q)\Longleftrightarrow \neg P\land\neg Q \tag{propositional logic}\\
& \neg (P\land Q)\Longleftrightarrow \neg P\lor\neg Q \\
& (A \cup B)^{\complement}=A^{\complement}\cap B^{\complement} \tag{set theory}\\
& (A \cap B)^{\complement}=A^{\complement}\cup B^{\complement} \\
& \overline{A\cup B}=\overline{A}\cap\overline{B} \tag{probability theory}\\
& \overline{A\cap B}=\overline{A}\cup\overline{B}
\end{align}$$

\[\begin{align} & \neg (P\lor Q)\Longleftrightarrow \neg P\land\neg Q \tag{propositional logic}\\ & \neg (P\land Q)\Longleftrightarrow \neg P\lor\neg Q \\ & (A \cup B)^{\complement}=A^{\complement}\cap B^{\complement} \tag{set theory}\\ & (A \cap B)^{\complement}=A^{\complement}\cup B^{\complement} \\ & \overline{A\cup B}=\overline{A}\cap\overline{B} \tag{probability theory}\\ & \overline{A\cap B}=\overline{A}\cup\overline{B} \end{align}\]

贝叶斯公式

$$P(A_i|B)=\dfrac{P(B|A_i)\cdot P(A_i)}{\sum_{j=1}^n P(B|A_j)\cdot P(A_j)}$$

\[P(A_i|B)=\dfrac{P(B|A_i)\cdot P(A_i)}{\sum_{j=1}^n P(B|A_j)\cdot P(A_j)}\]

麦克劳林级数

$$\begin{align}
&e^x=\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\\
&\sin x = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}\\
&\cos x = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n}}{(2n)!}\\
&\ln(x+1) = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{n+1}}{(n+1)!}\\
\end{align}$$

\[\begin{align} &e^x=\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\\ &\sin x = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}\\ &\cos x = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n}}{(2n)!}\\ &\ln(x+1) = \sum_{n=0}^{\infty} (-1)^n\dfrac{x^{n+1}}{(n+1)!}\\ \end{align}\]

自然对数\(e\)的定义

$$e=\lim_{x\rightarrow\infty}\left(1+\dfrac{1}{x}\right)^x$$

\[e=\lim_{x\rightarrow\infty}\left(1+\dfrac{1}{x}\right)^x\]

二项式定理

$$(x+y)^n=\sum_{k=0}^n \dbinom{n}{k} x^ky^{n-k}$$

\[(x+y)^n=\sum_{k=0}^n \dbinom{n}{k} x^ky^{n-k}\]