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

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()