
Na imagem abaixo, o numerador é denso em termos, mas o denominador possui apenas um termo. Assim, seria bom deslocar a barra horizontal para baixo para economizar espaço e ter uma boa aparência. Não tenho ideia de como fazer isso e não consegui encontrar nenhuma informação sobre isso em nenhum outro lugar.
meu código:
\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*}
Responder1
Proponho uma apresentação alternativa.
\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}
Aqui está outra maneira:
\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}
O OP ainda gostaria de colocá-lo em uma única apresentação. Eu recomendo fortemente contra essa abordagem, mas aqui pode haver uma maneira:
\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}
Se removermos \vcenter{\hbox{$
e $}}
da equação, a fração grande será deslocada, de modo que a linha de divisão da fração permanecerá no eixo matemático.
Responder2
Proponho o seguinte layout, baseado no \splitfrac
comando, projetado para lidar com tais situações, \mathllap
, tanto de mathtools
, quanto do medmath
comando de nccmath
(fórmulas de tamanho médio, ~80% de \displaystyle
) e do flalign*
ambiente:
\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}