将水平条移动到下方

Question

在我下面的图片中，分子项密集，但分母​​只有一项。因此，最好将水平线向下移动以节省空间并看起来美观。我不知道该怎么做，在其他地方也找不到有关它的任何信息。

我的代码：

\begin{equation*}
    \Rightarrow \alpha \mathlarger{\mathlarger{\sum}}_{c}A_c a_k\left( \frac{
    \scalemath{0.85}{
    \begin{aligned}
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(y_k^2+y_k y_{k+1} +y_{k+1}^2)\right) - \lambda_1\right]^2}_{\mathlarger{\mu_1}} +
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_2 \right]^2}_{\mathlarger{\mu_2}} + \\[1em]
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_3 \right]^2}_{\mathlarger{\mu_3}}+
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(x_k^2+x_k x_{k+1} +x_{k+1}^2)\right) - \lambda_4 \right]^2}_{\mathlarger{\mu_4}}
    \end{aligned}
    }
    }
    {\Lambda^2}
    \right)
\end{equation*}

Answer 1

我建议采用另一种呈现方式。

\documentclass{article}
\usepackage{amsmath,relsize,graphicx}
\newcommand\scalemath[2]{\scalebox{#1}{\mbox{\ensuremath{\displaystyle #2}}}}
\begin{document}
\begin{equation*}
    \Rightarrow \alpha \mathlarger{\mathlarger{\sum}}_{c}A_c a_k\left( \frac{
    B
    }
    {\Lambda^2}
    \right)
\end{equation*}
where
\begin{equation*}
    \begin{aligned}
        B = &\underbrace{\left[\left(\sum_{k=1}^{\nu-1}
        \frac{a_k}{12}(y_k^2+y_k y_{k+1} +y_{k+1}^2)\right) - 
          \lambda_1\right]^2}_{\mathlarger{\mu_1}} +\\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+
          2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_2 \right]^2}
          _{\mathlarger{\mu_2}} + \\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+
          2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_3 \right]^2}
          _{\mathlarger{\mu_3}}+\\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(x_k^2+x_k x_{k+1} +x_{k+1}^2)\right) - \lambda_4 \right]^2}_{\mathlarger{\mu_4}}
    \end{aligned}
\end{equation*}
\end{document}

这是另一种方法：

\documentclass{article}
\usepackage{amsmath,relsize,graphicx}
\newcommand\scalemath[2]{\scalebox{#1}{\mbox{\ensuremath{\displaystyle #2}}}}
\begin{document}
\begin{equation*}
    \begin{aligned}
        \Rightarrow \alpha \mathlarger{\mathlarger{\sum}}_{c}A_c a_k = 
          \frac{1}{\Lambda^2}\Biggl\{
        &\underbrace{\left[\left(\sum_{k=1}^{\nu-1}
        \frac{a_k}{12}(y_k^2+y_k y_{k+1} +y_{k+1}^2)\right) - 
          \lambda_1\right]^2}_{\mathlarger{\mu_1}} +\\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+
          2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_2 \right]^2}
          _{\mathlarger{\mu_2}} + \\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+
          2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_3 \right]^2}
          _{\mathlarger{\mu_3}}+\\[1ex]&
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(x_k^2+x_k x_{k+1} +x_{k+1}^2)\right) - \lambda_4 \right]^2}_{\mathlarger{\mu_4}}\Bigg\}
    \end{aligned}
\end{equation*}
\end{document}

OP 仍然希望将其放在单独的演示文稿中。 我强烈建议不要采用这种方法，但这里有一种方法：

\documentclass{article}
\usepackage{amsmath,relsize,graphicx,scalerel}
\newcommand\scalemath[2]{\scalebox{#1}{\mbox{\ensuremath{\displaystyle #2}}}}
\begin{document}
\begin{equation*}
    \Rightarrow \alpha \mathlarger{\mathlarger{\sum}}_{c}A_c a_k
    \vcenter{\hbox{$\scaleleftright[2ex]{(}{ \frac{
    \scalemath{0.85}{
    \begin{aligned}
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(y_k^2+y_k y_{k+1} +y_{k+1}^2)\right) - \lambda_1\right]^2}_{\mathlarger{\mu_1}} +
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_2 \right]^2}_{\mathlarger{\mu_2}} + \\[1em]
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k)\right) - \lambda_3 \right]^2}_{\mathlarger{\mu_3}}+
        \underbrace{\left[\left(\sum_{k=1}^{\nu-1}\frac{a_k}{12}(x_k^2+x_k x_{k+1} +x_{k+1}^2)\right) - \lambda_4 \right]^2}_{\mathlarger{\mu_4}}
    \end{aligned}
    }
    }
    {\Lambda^2}}
    {)}$}}
\end{equation*}
\end{document}

如果从等式中去掉\vcenter{\hbox{$和$}}，那么大分数就会移动，这样分数的分割线就会保留在数学轴上。

Answer 2

我建议根据命令采用以下布局，\splitfrac旨在处理此类情况，\mathllap包括来自的mathtools，以及medmath来自的命令nccmath（中型公式，约占的 80% \displaystyle）和flalign*环境：

\documentclass{article}
\usepackage[showframe]{geometry} \usepackage{mathtools, nccmath, relsize}
\usepackage{graphicx} \newcommand{\scalemath}[2]{\scalebox{#1}{\begin{math} {#2} \end{math}}}

\begin{document}

\begin{flalign*}
   ⇒ α\mathlarger{\mathlarger{\sum}}_{c}A_c a_k × {} \\[-4ex]
  & & & & &\mathllap{\frac{%
    \medmath{\splitfrac{
      \overbrace{\left[∑_{k=1}^{\nu-1}\frac{a_k}{12}(y_k²+y_k y_{k+1} +y_{k+1}²) - \lambda₁\right]²}^{\textstyle\mu₁} +
      \overbrace{\left[∑_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k) - \lambda₂ \right]²}^{\textstyle\mu₂}}%
      {\underbrace{\left[∑_{k=1}^{\nu-1}\frac{a_k}{24}(x_k y_{k+1}+2x_ky_k+2x_{k+1}y_{k+1} +x_{k+1}y_k) - \lambda₃ \right]²}_{\textstyle\mu₃}+
        \underbrace{\left[∑_{k=1}^{\nu-1}\frac{a_k}{12}(x_k²+x_k x_{k+1} +x_{k+1}²) - \lambda₄ \right]²}_{\textstyle\mu₄}}}}
    {\Lambda²}}
\end{flalign*}

\end{document}