\noalign 改变方程式数字的位置

Question

我刚刚发现\noalign可以将数字改为底部。但通常人们希望它出现在中间。如果我注释\noalign{choose $\overline{\vec{x}}^{n-1}$ with $\mathcal{F}(\overline{\vec{x}}^{n-1} )\le\min(\vec{F}({\vec{x}'}^{n-1}),\vec{F}(\vec{x}'))$}\\，数字的位置就没问题。如何将数字保持在正确的位置并使用\noalign？

  \documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
  \left\{  
  \begin{align}
    \vec{x}&=\vec{x}^{n-1}+\alpha(\vec{u}-\vec{x}^{n-1})\\
   \overline{\vec{y}}&=\vec{y}+\alpha(\nabla\mathcal{F}(\vec{x})-\vec{y})\\
   \overline{\gamma}&=\gamma+\alpha(\mathcal{F}(\vec{x})-\avg{\nabla{F}(\vec{x}),\vec{x}}-\gamma)\\
    {\vec{x}'}^{n-1}&=\argmin\limits_{\vec{z}\in\bra{\vec{x},\vec{x}^{n-1}}}\mathcal{F}(\vec{z})\\
    {\gamma'}^{n-1}&=\overline{\gamma}-\mathcal{F}({\vec{x'}^{n-1}})\\
    \vec{u}'&=\mathcal{U}({{\gamma'}^{n-1}},\overline{\vec{y}})\\
    \vec{x}'&=\vec{x}^{n-1}+\alpha(\vec{u}'-\vec{x}^{n-1})\\
   \noalign{choose $\overline{\vec{x}}^{n-1}$ with $\mathcal{F}(\overline{\vec{x}}^{n-1} )\le\min(\vec{F}({\vec{x}'}^{n-1}),\vec{F}(\vec{x}'))$}\\
      \overline{\gamma}^n&=\overline{\gamma}-\mathcal{F}(\overline{\vec{x}}^{n-1})\\
      \overline{\vec{u}}&=\mathcal{U}(\overline{\gamma}^n,\overline{\vec{y}})\\
       \overline{\eta}&=\mathcal{E}(\overline{\gamma}^n,\overline{\vec{y}})\\
       \vec{x}^n&=\overline{\vec{x}}^{n-1}\\
   \end{align}
\right.
 \end{equation}

 \end{document}

Answer 1

主要错误是align不能在数学模式下使用。我还猜测了你未定义的命令的定义，并修复了双下标错误。

\noalign不是在文档中使用的用户级命令。通常情况下，align您可以使用\intertext，但在这里，aligned我只使用了\rlap。

 \documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator\avg{avg}
\DeclareMathOperator\argmin{argmin}
\newcommand\bra[1]{\langle#1\rangle}
\begin{document}
\begin{equation}
  \left\{  
  \begin{aligned}
    \vec{x}&=\vec{x}^{n-1}+\alpha(\vec{u}-\vec{x}^{n-1})\\
   \overline{\vec{y}}&=\vec{y}+\alpha(\nabla\mathcal{F}(\vec{x})-\vec{y})\\
   \overline{\gamma}&=\gamma+\alpha(\mathcal{F}(\vec{x})-\avg{\nabla{F}(\vec{x}),\vec{x}}-\gamma)\\
    {\vec{x}'{}}^{n-1}&=\argmin\limits_{\vec{z}\in\bra{\vec{x},\vec{x}^{n-1}}}\mathcal{F}(\vec{z})\\
    {\gamma'}^{n-1}&=\overline{\gamma}-\mathcal{F}({\vec{x'}^{n-1}})\\
    \vec{u}'&=\mathcal{U}({{\gamma'}^{n-1}},\overline{\vec{y}})\\
    \vec{x}'&=\vec{x}^{n-1}+\alpha(\vec{u}'-\vec{x}^{n-1})\\
   \rlap{\hspace{-2em}choose $\overline{\vec{x}}^{n-1}$ with $\mathcal{F}(\overline{\vec{x}}^{n-1} )\le
\min(\vec{F}({\vec{x}'{}}^{n-1}),\vec{F}(\vec{x}'))$}\\
      \overline{\gamma}^n&=\overline{\gamma}-\mathcal{F}(\overline{\vec{x}}^{n-1})\\
      \overline{\vec{u}}&=\mathcal{U}(\overline{\gamma}^n,\overline{\vec{y}})\\
       \overline{\eta}&=\mathcal{E}(\overline{\gamma}^n,\overline{\vec{y}})\\
       \vec{x}^n&=\overline{\vec{x}}^{n-1}\\
   \end{aligned}
\right.
 \end{equation}

 \end{document}