# 测试数学公式

 $$$\label{eq1}r = r_F+ \beta (r_M - r_F) + \epsilon$$$ \begin{aligned} \color{red}{\dot{x}} & =\color{blue}{ \sigma(y-x)} \ \dot{y} & = \rho x - y - xz \ \dot{z} & = -\beta z + xy \end{aligned} $x^2+y1+z{12}^{34} \ \sin^{-1} (x) \ \lim\limits{h->0} \frac{f(x+h)-f(x)}{h} \ \lim\nolimits{h\to 0} \frac{f(x+h)-f(x)}{h} \ f(x)=\sum\limits{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \ \int0^1 f(x)dx \ \left[ \begin{array} {lcr} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{array} \right] \ \left[ \begin{array}{lcr} a & b \ c & d \end{array} \right] \left( \begin{array}{lcr} n \ k \end{array} \right)$ \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) \ \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} \ P(E) = {n \choose k} p^k (1-p)^{ n-k} \ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } \ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for  q <1}. \ \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}$$$\label{eq1}r = r_F+ \beta (r_M - r_F) + \epsilon$$$% $\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}$$% 0} \frac{f(x+h)-f(x)}{h} \\\ \lim\nolimits_{h\to 0} \frac{f(x+h)-f(x)}{h} \\\ f(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \\\ \int_0^1 f(x)dx \\\ \left[\begin{array} {lcr} 1 & 2 & 3 \\\ 4 & 5 & 6 \\\ 7 & 8 & 9 \end{array}\right] \\\ \left[\begin{array}{lcr} a & b \\\c & d\end{array}\right] \left(\begin{array}{lcr}n \\\k\end{array}\right) %]]>$\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)\\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}\P(E) = {n \choose k} p^k (1-p)^{ n-k}\\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}} {1+\ldots} } } }\1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =\prod{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for  q <1}.\\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}

rexdf:

#include <stdio.h>
#include <stdlib.h>
int main()
{
printf("Hello\n");
return 0;
}


rexdf: $\sum\limits_{i=0}^{i=\infty}\frac{x^i}{i!}$