测试数学公式

$$
\begin{equation}
\label{eq1}r = r_F+ \beta (r_M – r_F) + \epsilon
\end{equation}$$
$$
\begin{aligned}
\color{red}{\dot{x}} & =\color{blue}{ \sigma(y-x)}
\\
\dot{y} & = \rho x – y – xz
\\
\dot{z} & = -\beta z + xy
\end{aligned}
$$
\[
\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}
\]

$$x^2+y_1+z_{12}^{34}
\\
\sin^{-1} (x)
\\
\lim\limits_{h->0} \frac{f(x+h)-f(x)}{h}
\\
\lim\nolimits_{h\to 0} \frac{f(x+h)-f(x)}{h}
\\
f(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
\\
\int_0^1 f(x)dx
\\
\left[\begin{array} {lcr}
1 & 2 & 3
\\
4 & 5 & 6
\\
7 & 8 & 9
\end{array}\right]
\\
\left[\begin{array}{lcr}
a & b \\c & d\end{array}\right] \left(\begin{array}{lcr}n \\k\end{array}\right)$$
$$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\\\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0\end{vmatrix}\\P(E) = {n \choose k} p^k (1-p)^{ n-k}\\\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}} {1+\ldots} } } }\\1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|<1$}.\\\begin{aligned}\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$$

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测试数学公式》有 2 条评论

  1. rexdf说:
    Google Chrome 26.0.1410.64 Windows 8 x64 Edition
    Mozilla/5.0 (Windows NT 6.2; WOW64) AppleWebKit/537.31 (KHTML, like Gecko) Chrome/26.0.1410.64 Safari/537.31
    #include <stdio.h>
    #include <stdlib.h>
    int main()
    {
    printf("Hello\n");
    return 0;
    }
    

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

  2. rexdf说:
    Google Chrome 26.0.1410.64 Windows 8 x64 Edition
    Mozilla/5.0 (Windows NT 6.2; WOW64) AppleWebKit/537.31 (KHTML, like Gecko) Chrome/26.0.1410.64 Safari/537.31 AppEngine-Google; (+http://code.google.com/appengine; appid: s~bngfhjghfjgh)

    \(\sum\limits_{i=0}^{i=\infty}\frac{x^i}{i!}\)

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