引用自:
[1]https://blog.csdn.net/qq_33532713/article/details/108602463
[2]https://www.cnblogs.com/Sinte-Beuve/p/6160905.html
[3]https://blog.csdn.net/happyday_d/article/details/83715440
LaTeX中数学模式有两种形式:inline和display。前者是指正文插入行间数学公式,后者独立排列,可以有或没有编号。
行间公式(inline):用$...$
将公式括起来
块间公式(display):用$$...$$
将公式括起来,默认显示在行中间
-
下标:
$x_i$
==>\(x_i\) -
上标:
$x^2$
==>\(x^2\)
注意:上下标如果多于一个字母或者符号,需要用一对{}括起来:
-
$x _ {i1}$
==>\(x _ {i1}\) -
$x^{\alpha t}$
==>\(x^{\alpha t}\) -
根号:\sqrt , eg:
$\sqrt[n]{5}$
==>\(\sqrt[n]{5}\) -
省略号:\dots \cdots 分别表示 \(\dots\) ,\(\cdots\)
- 基本运算符:+ - * /等可以直接输入,其他特殊的有:
-
求和:
$\sum_1^n$
==>\(\sum_1^n\) -
累乘:
$\prod_{n=1}^{99}x_n$
==>\(\prod_{n=1}^{99}x_n\) -
积分:
$\int_1^n$
==> \(\int_1^n\) -
极限:\lim\limits _ {x \to \infty}==> \(\lim\limits _ {x \to \infty}\)
分数的表示:\frac{}{},如$\frac{3}{8}$
==> \(\frac{3}{8}\)
使用&
分隔同行元素,\\
表示换行:
示例:
$$
\begin{matrix}
1&x&x^2\\
1&y&y^2\\
1&z&z^2\\
\end{matrix}
$$
结果:
\[\begin{matrix} 1&x&x^2\\ 1&y&y^2\\ 1&z&z^2\\ \end{matrix} \]行列式示例:
$$
X=\left|
\begin{matrix}
x_{11} & x_{12} & \cdots & x_{1d}\\
x_{21} & x_{22} & \cdots & x_{2d}\\
\vdots & \vdots & \ddots & \vdots\\
x_{m1} & x_{m2} & \cdots & x_{md} \\
\end{matrix}
\right|
$$
结果:
\[X=\left| \begin{matrix} x_{11} & x_{12} & \cdots & x_{1d}\\ x_{21} & x_{22} & \cdots & x_{2d}\\ \vdots & \vdots & \ddots & \vdots\\ x_{m1} & x_{m2} & \cdots & x_{md} \\ \end{matrix} \right| \]箭头$$
\begin{equation}
E=mc^2
\end{equation}
$$
\[\begin{equation}
E=mc^2
\end{equation}
\]分隔符
各种括号用 () [] { } \langle
、\rangle
等命令表示,注意花括号通常用来输入命令和环境的参数,所以在数学公式中它们前面要加 \
。可以在上述分隔符前面加 \big \Big \bigg \Bigg
等命令来调整大小。
$$
\max \limits_{a<x<b} \Bigg\{f(x)\Bigg\}
$$
\[\max \limits_{a<x<b} \Bigg\{f(x)\Bigg\}
\]分段函数
$$
f(n) =
\begin{cases}
n/2, & \text {if $n$ is even}\\
3n+1, & \text {if $n$ is odd}
\end{cases}
$$
\[f(n) =
\begin{cases}
n/2, & \text {if $n$ is even}\\
3n+1, & \text {if $n$ is odd}
\end{cases}
\]方程组
$$
\left\{
\begin{array}{c}
a_1x+b_1y+c_1z=d_1\\
a_2x+b_2y+c_2z=d_2\\
a_3x+b_3y+c_3z=d_3
\end{array}
\right. # 注意right后面有个小数点
$$
\[\left\{
\begin{array}{c}
a_1x+b_1y+c_1z=d_1\\
a_2x+b_2y+c_2z=d_2\\
a_3x+b_3y+c_3z=d_3
\end{array}
\right.
\]案例
线性模型
$$h(\theta)=\sum_{j=0}^n \theta_j x_j$$
\[h(\theta)=\sum_{j=0}^n \theta_j x_j
\]均方误差
$$J(\theta) = \frac{1}{2m} \sum_{i=0}^m (y^i-h_\theta (x^i))^2$$
\[J(\theta) = \frac{1}{2m} \sum_{i=0}^m (y^i-h_\theta (x^i))^2
\]批量梯度下降
$$
\frac{\partial J(\theta)}{\partial\theta_j} = -\frac{1}{m}\sum_{i=0}^m (y^i - h_\theta (x^i))x^i_j
$$
\[\frac{\partial J(\theta)}{\partial\theta_j} = -\frac{1}{m}\sum_{i=0}^m (y^i - h_\theta (x^i))x^i_j
\]推导过程
$$
\begin{align}
\frac{\partial J(\theta)}{\partial \theta_j}
& = - \frac{1}{m} \sum_{i=0}^m (y^i-h_\theta(x^i))\frac{\partial}{\partial\theta_j}(y^i-h_\theta(x^i))\\
& = -\frac{1}{m}\sum_{i=0}^m(y^i-h_\theta(x^i))\frac{\partial}{\partial\theta_j}(\sum_{j=0}^n\theta_j x^i_j - y^i)\\
& = - \frac{1}{m}\sum_{i=0}^m(y^i-h_\theta (x^i))x^i_j
\end{align}
$$
\[\begin{align}
\frac{\partial J(\theta)}{\partial \theta_j}
& = - \frac{1}{m} \sum_{i=0}^m (y^i-h_\theta(x^i))\frac{\partial}{\partial\theta_j}(y^i-h_\theta(x^i))\\
& = -\frac{1}{m}\sum_{i=0}^m(y^i-h_\theta(x^i))\frac{\partial}{\partial\theta_j}(\sum_{j=0}^n\theta_j x^i_j - y^i)\\
& = - \frac{1}{m}\sum_{i=0}^m(y^i-h_\theta (x^i))x^i_j
\end{align}
\]