测试Mathjax支持

发布于 2019-02-21  316 次阅读


When $ a \ne 0 $, there are two solutions to $ ax^2 + bx + c = 0 $ and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

\begin{equation}\label{eq1}r = r_F+ \beta (r_M - r_F) + \epsilon\end{equation}
\(E=mc^2\)
标量场 \( \varphi \) 的梯度:

$${\rm grad}\varphi=\frac{\partial\varphi}{\partial x}e_1+\frac{\partial\varphi}{\partial y}e_2+\frac{\partial\varphi}{\partial z}e_3=\nabla\varphi$$

Thus we can write a vector ${\bf a}\in\mathbb {R}^2$ as

$${\bf a}=\left(\begin{matrix}a_1\a_2\end{matrix}\right)$$

对应位置元素相加即可。

$$\left(\begin{matrix}
a_1\a_2
\end{matrix}\right)+\left(\begin{matrix}
b_1\b_2
\end{matrix}\right)=\left(\begin{matrix}
a_1+b_1\a_2+b_2
\end{matrix}\right)$$

范数以及标准化
欧几里得范数(Euclidean Vector Norm)
For a vector ${\bf v}\in\mathbb{ R}^n$,the euclidean norm of $\mathbf{v}$is defined as,
$$\parallel\mathbf{v}\parallel=\left(\sum_{i=1}^{n}{v_i^2}\right)^\frac{1}{2}=\sqrt{\left\langle \mathbf{v,v}\right\rangle} $$

$\boxed{\vec{s}}$ \, $\overrightarrow{suv}$

import matplotlib.pyplot as plt
import numpy as np

x=np.linspace(0,20,100)
plt.plot(x, np.sin(x))
plt.show()