在 Latex 中对齐方程

Question

我有以下等式：

    \begin{equation}\label{eq:5}
  \frac{\partial s}{\partial b} = \frac{\partial \sum_{i=1}^{n} (y_i -(W\cdot x_i+b))^2}{\partial b}\\

   =\sum_{i=1}^{n} 2(y_i -W\cdot x_i-b)\cdot(-1))

   =2\cdot \sum_{i=1}^{n} (-y_i + a \cdot x_i + b )  \stackrel{!}{=} 0} \\

   \Longleftrightarrow a\cdot (\sum_{i=1}^{n} x_i) + \sum_{i=1}^{n}b = \sum_{i=1}^{n} y_i

   \Longleftrightarrow a\cdot (\sum_{i=1}^{n} x_i) + n \cdot b = \sum_{i=1}^{n}y_i
\end{equation}

结果如下：

我坚持认为使用 align 可以解决这个问题，这是我尝试过的：

    \begin{equation}\label{eq:5}
  \begin{align}

  \frac{\partial s}{\partial b} = \frac{\partial \sum_{i=1}^{n} (y_i -(W\cdot x_i+b))^2}{\partial b}\\

   &=\sum_{i=1}^{n} 2(y_i -W\cdot x_i-b)\cdot(-1))

   =2\cdot \sum_{i=1}^{n} (-y_i + a \cdot x_i + b )  \stackrel{!}{=} 0} \\

   \Longleftrightarrow a\cdot (\sum_{i=1}^{n} x_i) + \sum_{i=1}^{n}b = \sum_{i=1}^{n} y_i

   \Longleftrightarrow a\cdot (\sum_{i=1}^{n} x_i) + n \cdot b = \sum_{i=1}^{n}y_i


 \end{align}
\end{equation}

现在结果发生了变化！我的问题是，尽管开始了新的一行，如何对齐方程式？

Answer 1

不太清楚您喜欢如何格式化您的方程式...像这样？

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{equation}\label{eq:5}
    \begin{split}
\frac{\partial s}
     {\partial b} 
    & = \frac{\partial \sum_{i=1}^{n} (y_i -(W\cdot x_i+b))^2}
                         {\partial b}\\
    & = \sum_{i=1}^{n} 2(y_i -W\cdot x_i-b)\cdot(-1))
    = 2\cdot \sum_{i=1}^{n} (-y_i + a \cdot x_i + b )  \stackrel{!}{=} 0    \\
    &   \Longleftrightarrow a\cdot \sum_{i=1}^{n} x_i + \sum_{i=1}^{n}b = \sum_{i=1}^{n} y_i  \\
    &   \Longleftrightarrow a\cdot \sum_{i=1}^{n} x_i + n \cdot b = \sum_{i=1}^{n}y_i
    \end{split}
\end{equation}
\end{document}

Answer 2

正如侧面的相关链接所示（比如这个https://tex.stackexchange.com/a/74822/132800)，每行都必须放置对齐字符。对于您的情况，

\begin{align}
\frac{\partial s}{\partial b} =& \frac{\partial \sum_{i=1}^{n} (y_i -(W\cdot x_i+b))^2}{\partial b}\\
=&\sum_{i=1}^{n} 2(y_i -W\cdot x_i-b)\cdot(-1))\\
=&2\cdot \sum_{i=1}^{n} (-y_i + a \cdot x_i + b )  \stackrel{!}{=} 0} \\
\Longleftrightarrow& a\cdot (\sum_{i=1}^{n} x_i) + \sum_{i=1}^{n}b = \sum_{i=1}^{n} y_i\\
\Longleftrightarrow& a\cdot (\sum_{i=1}^{n} x_i) + n \cdot b = \sum_{i=1}^{n}y_i
\end{align}