將水平條向下移動分數

Question

在下圖中，分子有很多項，但分母只有一項。因此，最好將水平條移低以節省空間並看起來不錯。我不知道該怎麼做，也無法在其他地方找到任何相關資訊。

我的程式碼：

\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

我提出一個替代的演示。

\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}

這是另一種方法：

\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 仍然希望將其放在一個簡報中。 我強烈建議反對這種方法，但這可能是一種方法：

\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}

如果從方程式中刪除\vcenter{\hbox{$和$}}，那麼大分數將被移動，因此分數的分割線將保留在數學軸上。

Answer 2

我建議基於命令的以下佈局，\splitfrac旨在處理此類情況，\mathllap，來自mathtools，以及medmath來自nccmath（中等大小的公式，約 80 % \displaystyle）和flalign*環境的命令：

\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}