
En la imagen que tengo a continuación, el numerador está lleno de términos pero el denominador solo tiene un término. Por lo tanto, sería bueno mover la barra horizontal hacia abajo para ahorrar espacio y lucir bien. No tengo idea de cómo hacerlo y no pude encontrar información al respecto en ningún otro lugar.
mi 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*}
Respuesta1
Propongo una presentación 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}
Aquí hay otra manera:
\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}
Al OP todavía le gustaría colocarlo en una sola presentación. Recomiendo encarecidamente contra este enfoque., pero aquí podría haber una manera:
\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}
Si se elimina \vcenter{\hbox{$
y $}}
de la ecuación, entonces la fracción grande se desplazará, de modo que la línea de división de la fracción permanecerá en el eje matemático.
Respuesta2
Propongo el siguiente diseño, basado en el \splitfrac
comando, diseñado para manejar tales situaciones, \mathllap
tanto de mathtools
, como el medmath
comando de nccmath
(fórmulas de tamaño mediano, ~80 % de \displaystyle
) y el flalign*
entorno:
\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}