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添加非对齐的换行符时如何防止对齐的方程式发生偏移？

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添加非对齐的换行符时如何防止对齐的方程式发生偏移？

我在对齐环境中有一个方程式，用于将等号对齐。但是，最后一个方程式太长，我想将其拆分。当我插入换行符时，最后一个方程式上方的所有方程式都会向右移动。代码：

\begin{equation}
\begin{aligned}
\log\left(p\left(s|\alpha\right)\right) & = \sum\limits_{i=1}^O\left(\log\left(\frac{1}{\prod\limits_{k=1}^K\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)}\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(-\log\left(\prod\limits_{k=1}^K\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(\sum\limits_{k=1}^K\left(-\log\left(\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right)\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(\sum\limits_{k=1}^K\left(-\log\left(\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right)\right) + \log\left(\Gamma\left(\alpha_0\right)\right) -\log\left(\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right)
\end{aligned}
\end{equation}

我想要的是

\begin{equation}
\begin{aligned}
\log\left(p\left(s|\alpha\right)\right) & = \sum\limits_{i=1}^O\left(\log\left(\frac{1}{\prod\limits_{k=1}^K\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)}\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(-\log\left(\prod\limits_{k=1}^K\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(\sum\limits_{k=1}^K\left(-\log\left(\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right)\right) + \log\left(\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)}\right) + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right) \\
&= \sum\limits_{i=1}^O\left(\sum\limits_{k=1}^K\left(-\log\left(\Gamma\left(\hat{\mathcal{S}}_k^i+1\right)\right)\right) + \log\left(\Gamma\left(\alpha_0\right)\right) -\log\left(\Gamma\left(\sum\limits_{k=1}^K\left(\hat{\mathcal{S}}_k^i+a_k\right)\right)\right) \\
\qquad\qquad + \log\left(\prod\limits_{k=1}^K \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma\left(\alpha_k\right)}\right)\right)
\end{aligned}
\end{equation}

我不想使用 &+，因为这样它会与等号对齐，而且我希望它缩进得更远一些。

答案1

&\qquad\qquad

但请记住，\left...\right不能被打破，因此必须重写最后一行以说明

  &=
    \sum\limits_{i=1}^O\Bigl(\sum\limits_{k=1}^K\left(-\log\left(\Gamma  \left(\hat{\mathcal{S}}_k^i+1\right)\right)\right)
      + \log\left(\Gamma\left(\alpha_0\right)\right)
      -\log\left(\Gamma\left(\sum\limits_{k=1}^K  \left(\hat{\mathcal{S}}_k^i+a_k\right)\right)\right)
      \\
      &\qquad\qquad + \log\left(\prod\limits_{k=1}^K
        \frac{\Gamma\left(\hat{\mathcal{S}}_k^i+\alpha_k\right)}{\Gamma   \left(\alpha_k\right)}\right)\Bigr)

还有更合适的尺寸。

此外，那些庞大的（）真的需要吗？对我来说，它们在准备公式时没有多大帮助，相反，它们真的很烦人。

此外，我甚至会使用mathtools下面的代码。没有必要使用所有这些自动缩放功能，因为最终看起来很糟糕。也不需要那些\limits。

\begin{equation}
  \begin{aligned}
    \MoveEqLeft
    \log\bigl(p(s|\alpha)\bigr) 
    \\
    & =
    \sum\limits_{i=1}^O\left(
      \log\Bigl(
        \frac{1}{\prod_{k=1}^K
          \Gamma(\hat{\mathcal{S}}_k^i+1)}
      \Bigr)
      +
      \log\Bigl(
        \frac{  
          \Gamma(\alpha_0)
        }{
          \Gamma\bigl(
          \sum_{k=1}^K(\hat{\mathcal{S}}_k^i+a_k)
          \bigr)
        }
      \Bigr)
      + \log\Bigl(
        \prod_{k=1}^K
        \frac{
          \Gamma(\hat{\mathcal{S}}_k^i+\alpha_k)
        }{
          \Gamma(\alpha_k)
        }
      \Bigr)
    \right)
    \\
    &=
    \sum\limits_{i=1}^O
    \left(
      -\log\Bigl(
        \prod_{k=1}^K
        \Gamma(\hat{\mathcal{S}}_k^i+1)
      \Bigr)
      +
      \log\Bigl(
        \frac{
          \Gamma(\alpha_0)
        }{
          \Gamma\bigl(
            \sum_{k=1}^K
            (\hat{\mathcal{S}}_k^i+a_k)
          \bigr)
        }
      \Bigr)
      + \log\Bigl(
        \prod_{k=1}^K
        \frac{
          \Gamma(\hat{\mathcal{S}}_k^i+\alpha_k)
        }{
          \Gamma(\alpha_k)
        }
      \Bigr)
    \right)
    \\
    &=
    \sum_{i=1}^O
    \left(
      \sum_{k=1}^K
      \Bigl(
        -\log
        \bigl(\Gamma(\hat{\mathcal{S}}_k^i+1) \bigr)
      \Bigr)
      +
      \log\Bigl(
        \frac{
          \Gamma(\alpha_0)
        }{
          \Gamma\bigl(
          \sum_{k=1}^K
            (\hat{\mathcal{S}}_k^i+a_k)
            \bigr)
          }
        \Bigr)
      + \log\Bigl(
        \prod_{k=1}^K
        \frac{
          \Gamma(\hat{\mathcal{S}}_k^i+\alpha_k)
        }{
          \Gamma(\alpha_k)
        }
      \Bigr)
    \right)
    \\
    &=
    \sum_{i=1}^O\Biggl(
    \sum_{k=1}^K
    \Bigl(
      -\log\bigl(
      \Gamma(\hat{\mathcal{S}}_k^i+1)
      \bigr)
    \Bigr)
    + \log\left(
      \Gamma(\alpha_0 )
    \right)
    -\log\biggl(
      \Gamma\Bigl(
      \sum_{k=1}^K
      (\hat{\mathcal{S}}_k^i+a_k)
      \Bigr
      )
    \biggr)
    \\
    &\qquad\qquad + \log\Bigl(
    \prod_{k=1}^K
    \frac{
      \Gamma(\hat{\mathcal{S}}_k^i+\alpha_k)
    }{
      \Gamma(\alpha_k)
    }
    \Bigr)
    \Biggr)
  \end{aligned}
\end{equation}

或者更好的办法是，不要把所有内容都写出来。这样对读者没有帮助。定义助手：

So ease notation, we define:
\begin{align*}
  A &= \prod_{k=1}^K  \Gamma(\hat{\mathcal{S}}_k^i+1) \\
  B &= \sum_{k=1}^K(\hat{\mathcal{S}}_k^i+a_k)\\
  C &= \prod_{k=1}^K  \frac{ \Gamma(\hat{\mathcal{S}}_k^i+\alpha_k)}{ \Gamma(\alpha_k) }
\end{align*}



\begin{equation}
  \begin{aligned}
    \MoveEqLeft
    \log\bigl(p(s|\alpha)\bigr) 
    \\
    & =
    \sum\limits_{i=1}^O\left(
      \log\Bigl( \frac{1}{A}  \Bigr)
      +
      \log\Bigl(
        \frac{  
          \Gamma(\alpha_0)
        }{
          \Gamma\bigl(
          B
          \bigr)
        }
      \Bigr)
      + \log(C)
    \right)
    \\
    &=
    \sum\limits_{i=1}^O
    \left(
      -\log( A )
      +
      \log\Bigl(
        \frac{ \Gamma(\alpha_0) }{ \Gamma( B ) }
      \Bigr)
      + \log( C )
    \right)
    \\
    &=
    \sum_{i=1}^O
    \left(
      \sum_{k=1}^K
      \Bigl(
        -\log
        \bigl(\Gamma(\hat{\mathcal{S}}_k^i+1) \bigr)
      \Bigr)
      +
      \log\Bigl(
        \frac{ \Gamma(\alpha_0) }{ \Gamma( B ) }
        \Bigr)
      + \log( C  )
    \right)
    \\
    &=
    \sum_{i=1}^O\biggl(
    \sum_{k=1}^K
    \Bigl(
      -\log\bigl(
      \Gamma(\hat{\mathcal{S}}_k^i+1)
      \bigr)
    \Bigr)
    + \log\left(
      \Gamma(\alpha_0 )
    \right)
    -\log\bigl(
      \Gamma( B )
    \bigr)
    + \log( C )
    \biggr)
  \end{aligned}
\end{equation}

在此处输入图片描述

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