수평 막대를 분수 아래로 이동

Question

아래 그림에서 분자에는 용어가 밀집되어 있지만 분모에는 용어가 하나만 있습니다. 따라서 공간을 절약하고 보기 좋게 하려면 가로 막대를 아래로 이동하는 것이 좋을 것입니다. 나는 그것을 수행하는 방법을 전혀 모르고 다른 곳에서는 이에 대한 정보를 찾을 수 없습니다.

내 코드:

\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

나는 대안적인 프레젠테이션을 제안합니다.

\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}

또 다른 방법은 다음과 같습니다.

\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}

OP는 여전히 단일 프리젠테이션에 배치하고 싶어합니다. 나는 이 접근 방식을 강력히 권장합니다., 하지만 다음과 같은 방법이 있을 수 있습니다.

\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}

\vcenter{\hbox{$방정식에서 및를 제거하면 $}}큰 분수가 이동되어 분수의 구분선이 수학 축에 유지됩니다.

Answer 2

나는 이러한 \splitfrac상황을 처리하도록 설계된 명령을 기반 으로 다음 과 같은 레이아웃을 제안 \mathllap합니다 .mathtoolsmedmathnccmath\displaystyleflalign*

\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}