我有以下表达:
\begin{equation}
\begin{aligned}
%C1
\sum_{i = 1}^{m}{y^{C_1}_i} &= m\beta_0^{C_1} + \beta_1^{C_1}\sum_{i = 1}^{m}{x_{i1}^{C_1}} + \cdots + \beta_q^{C_1}\sum_{i = 1}^{m}{x_{iq}^{C_1}}\\
\sum_{i = 1}^{m}{y^{C_1}_ix_{i1}^{C_1}} &= \beta_0^{C_1}\sum_{i = 1}^{m}{x_{i1}^{C_1}} + \beta_1^{C_1}\sum_{i = 1}^{m}{(x_{i1}^{C_1})^2} + \cdots + \beta_q^{C_1}\sum_{i = 1}^{m}{x_{iq}^{C_1}x_{i1}^{C_1}}\\
&\vdots\\
\sum_{i = 1}^{m}{y^{C_1}_ix_{iq}^{C_1}} &= \beta_0^{C_1}\sum_{i = 1}^{m}{x_{iq}^{C_1}} + \beta_1^{C_1}\sum_{i = 1}^{m}{x_{i1}^{C_1}x_{iq}^{C_1}} + \cdots + \beta_q^{C_1}\sum_{i = 1}^{m}{(x_{iq}^{C_1})^2}\\
%C2
\sum_{i = 1}^{m}{y^{C_2}_i} &= m\beta_0^{C_2} + \beta_1^{C_2}\sum_{i = 1}^{m}{x_{i1}^{C_2}} + \cdots + \beta_q^{C_2}\sum_{i = 1}^{m}{x_{iq}^{C_2}}\\
\sum_{i = 1}^{m}{y^{C_2}_ix_{i1}^{C_2}} &= \beta_0^{C_2}\sum_{i = 1}^{m}{x_{i1}^{C_2}} + \beta_1^{C_2}\sum_{i = 1}^{m}{(x_{i1}^{C_2})^2} + \cdots + \beta_q^{C_2}\sum_{i = 1}^{m}{x_{iq}^{C_2}x_{i1}^{C_2}}\\
&\vdots\\
\sum_{i = 1}^{m}{y^{C_2}_ix_{iq}^{C_2}} &= \beta_0^{C_2}\sum_{i = 1}^{m}{x_{iq}^{C_2}} + \beta_1^{C_2}\sum_{i = 1}^{m}{x_{i1}^{C_2}x_{iq}^{C_2}} + \cdots + \beta_q^{C_2}\sum_{i = 1}^{m}{(x_{iq}^{C_2})^2}\\
%R
\sum_{i = 1}^{m}{y^{R}_i} &= m\beta_0^{R} + \beta_1^{R}\sum_{i = 1}^{m}{x_{i1}^{R}} + \cdots + \beta_q^{R}\sum_{i = 1}^{m}{x_{iq}^{R}}\\
\sum_{i = 1}^{m}{y^{R}_ix_{i1}^{R}} &= \beta_0^{R}\sum_{i = 1}^{m}{x_{i1}^{R}} + \beta_1^{R}\sum_{i = 1}^{m}{(x_{i1}^{R})^2} + \cdots + \beta_q^{R}\sum_{i = 1}^{m}{x_{iq}^{R}x_{i1}^{R}}\\
&\vdots\\
\sum_{i = 1}^{m}{y^{R}_ix_{iq}^{R}} &= \beta_0^{R}\sum_{i = 1}^{m}{x_{iq}^{R}} + \beta_1^{R}\sum_{i = 1}^{m}{x_{i1}^{R}x_{iq}^{R}} + \cdots + \beta_q^{R}\sum_{i = 1}^{m}{(x_{iq}^{R})^2}
\end{aligned}
\label{eq:normals_equations}
\end{equation}
我想要放置这种形式的括号:
答案1
您可以使用的基本构建块是
\left\{
\begin{aligned}
....
\end{\aligned}
\right.
如果你想要q+1
在前面,那么就把它写在前面\left\{
。你想要三个这样的组在不同的行上,显然只有一个方程编号,所以你可以把这些组放在
\begin{equation}
\begin{gathered}
...\\
---\\
\end{gathered}
\end{equation}
给予
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{gathered}
\left\{
\begin{aligned}
% C1
\sum_{i=1}^{m}{y^{C_1}_i}
&= m\beta_0^{C_1}
+ \beta_1^{C_1}\sum_{i=1}^{m}{x_{i1}^{C_1}} + \dots
+ \beta_q^{C_1}\sum_{i=1}^{m}{x_{iq}^{C_1}}\\
\sum_{i=1}^{m}{y^{C_1}_ix_{i1}^{C_1}}
&= \beta_0^{C_1}\sum_{i=1}^{m}{x_{i1}^{C_1}}
+ \beta_1^{C_1}\sum_{i=1}^{m}{(x_{i1}^{C_1})^2} + \dots
+ \beta_q^{C_1}\sum_{i=1}^{m}{x_{iq}^{C_1}x_{i1}^{C_1}}\\
&\vdots\\
\sum_{i=1}^{m}{y^{C_1}_ix_{iq}^{C_1}}
&= \beta_0^{C_1}\sum_{i=1}^{m}{x_{iq}^{C_1}}
+ \beta_1^{C_1}\sum_{i=1}^{m}{x_{i1}^{C_1}x_{iq}^{C_1}}
+ \dots + \beta_q^{C_1}\sum_{i=1}^{m}{(x_{iq}^{C_1})^2}\\
\end{aligned}
\right.\\
% C2
\left\{
\begin{aligned}
\sum_{i=1}^{m}{y^{C_2}_i}
&= m\beta_0^{C_2}
+ \beta_1^{C_2}\sum_{i=1}^{m}{x_{i1}^{C_2}} + \dots
+ \beta_q^{C_2}\sum_{i=1}^{m}{x_{iq}^{C_2}}\\
\sum_{i=1}^{m}{y^{C_2}_ix_{i1}^{C_2}}
&= \beta_0^{C_2}\sum_{i=1}^{m}{x_{i1}^{C_2}}
+ \beta_1^{C_2}\sum_{i=1}^{m}{(x_{i1}^{C_2})^2} + \dots
+ \beta_q^{C_2}\sum_{i=1}^{m}{x_{iq}^{C_2}x_{i1}^{C_2}}\\
&\vdots\\
\sum_{i=1}^{m}{y^{C_2}_ix_{iq}^{C_2}}
&= \beta_0^{C_2}\sum_{i=1}^{m}{x_{iq}^{C_2}}
+ \beta_1^{C_2}\sum_{i=1}^{m}{x_{i1}^{C_2}x_{iq}^{C_2}}
+ \dots + \beta_q^{C_2}\sum_{i=1}^{m}{(x_{iq}^{C_2})^2}\\
\end{aligned}
\right.\\
% R
\left\{
\begin{aligned}
\sum_{i=1}^{m}{y^{R}_i}
&= m\beta_0^{R} + \beta_1^{R}\sum_{i=1}^{m}{x_{i1}^{R}}
+ \dots + \beta_q^{R}\sum_{i=1}^{m}{x_{iq}^{R}}\\
\sum_{i=1}^{m}{y^{R}_ix_{i1}^{R}}
&= \beta_0^{R}\sum_{i=1}^{m}{x_{i1}^{R}}
+ \beta_1^{R}\sum_{i=1}^{m}{(x_{i1}^{R})^2} + \dots
+ \beta_q^{R}\sum_{i=1}^{m}{x_{iq}^{R}x_{i1}^{R}}\\
&\vdots\\
\sum_{i=1}^{m}{y^{R}_ix_{iq}^{R}}
&= \beta_0^{R}\sum_{i=1}^{m}{x_{iq}^{R}}
+ \beta_1^{R}\sum_{i=1}^{m}{x_{i1}^{R}x_{iq}^{R}}
+ \dots + \beta_q^{R}\sum_{i=1}^{m}{(x_{iq}^{R})^2}
\end{aligned}
\right.
\label{eq:normals_equations}
\end{gathered}
\end{equation}
\end{document}
我已经将您的改为,\cdots
因为\dots
知道在加号amsmath
之间使用哪些点。+
您可以考虑使用\vdotswithin{=}
该mathtools
包来使垂直点位于等号的中心。
如果要跨组垂直对齐各个元素,请参阅跨“对齐”方程块进行对齐