\sqrt 对于一行来说太长了

Question

我有一个很长的 SQRT() 方程，一行都写不下。我无法简化它，因为它涉及加法项。有没有其他方法可以将根表示为“SQRT{某个函数}”？

\documentclass{article}
\begin{document}
\begin{equation}
    \Delta\sigma_{V,ij} = \sqrt{\Delta\sigma_{x,ij}^2 + \Delta\sigma_{y,ij}^2 + \Delta\sigma_{z,ij}^2 - (\Delta\sigma_{x,ij}\cdot\Delta\sigma_{y,ij} + \Delta\sigma_{y,ij}\cdot\Delta\sigma_{z,ij} + \Delta\sigma_{z,ij}\cdot\Delta\sigma_{x,ij}) + 3 \cdot (\Delta\tau_{xy,ij}^2 + \Delta\tau_{yz,ij}^2 + \Delta\tau_{zx,ij}^2)}
\end{equation}
\end{document}

Answer 1

有关详细信息，请参阅如何继续下一行的等式。

\documentclass{article}
\usepackage{mathtools}
\begin{document}

\begin{equation*}
    \Delta\sigma_{V,ij} = \sqrt{\Delta\sigma_{x,ij}^2 + \Delta\sigma_{y,ij}^2 + \Delta\sigma_{z,ij}^2 - (\Delta\sigma_{x,ij}\cdot\Delta\sigma_{y,ij} + \Delta\sigma_{y,ij}\cdot\Delta\sigma_{z,ij} + \Delta\sigma_{z,ij}\cdot\Delta\sigma_{x,ij}) + 3 \cdot (\Delta\tau_{xy,ij}^2 + \Delta\tau_{yz,ij}^2 + \Delta\tau_{zx,ij}^2)}
\end{equation*}

% https://tex.stackexchange.com/questions/364116/
\begin{equation*}
    \Delta\sigma_{V,ij} =
    \sqrt{\begin{multlined}[b] 
    \Delta\sigma_{x,ij}^2 + \Delta\sigma_{y,ij}^2 + \Delta\sigma_{z,ij}^2 - \\
    (\Delta\sigma_{x,ij}\cdot\Delta\sigma_{y,ij} + \Delta\sigma_{y,ij}\cdot\Delta\sigma_{z,ij} + \Delta\sigma_{z,ij}\cdot\Delta\sigma_{x,ij}) + \\ 
    3 \cdot (\Delta\tau_{xy,ij}^2 + \Delta\tau_{yz,ij}^2 + \Delta\tau_{zx,ij}^2)
    \end{multlined}}
\end{equation*}

\end{document}

Answer 2

我会把平方根符号全部去掉：

\documentclass{article}
\usepackage{amsmath}

\newcommand{\Diff}{\mathop{}\!\Delta}

\begin{document}

\begin{equation}
\begin{split}
\Diff\sigma_{V,ij} =\Bigl(
&  \Diff\sigma_{x,ij}^2 + \Diff\sigma_{y,ij}^2 + \Diff\sigma_{z,ij}^2 \\
&- \bigl(\Diff\sigma_{x,ij}\Diff\sigma_{y,ij} +
    \Diff\sigma_{y,ij}\Diff\sigma_{z,ij} +
    \Diff\sigma_{z,ij}\Diff\sigma_{x,ij}\bigr) \\
&+ 3\bigl(\Diff\tau_{xy,ij}^2 + \Diff\tau_{yz,ij}^2 + \Diff\tau_{zx,ij}^2\bigr)
\Bigr)^{\!1/2}
\end{split}
\end{equation}

\begin{equation}
\begin{split}
\bigl(\Diff\sigma_{V,ij})^2 
={}& \Diff\sigma_{x,ij}^2 + \Diff\sigma_{y,ij}^2 + \Diff\sigma_{z,ij}^2 \\
&- \bigl(\Diff\sigma_{x,ij}\Diff\sigma_{y,ij} +
    \Diff\sigma_{y,ij}\Diff\sigma_{z,ij} +
    \Diff\sigma_{z,ij}\Diff\sigma_{x,ij}\bigr) \\
&+ 3\bigl(\Diff\tau_{xy,ij}^2 + \Diff\tau_{yz,ij}^2 + \Diff\tau_{zx,ij}^2\bigr)
\end{split}
\end{equation}

\end{document}

您\Diff不需要\cdot，因为间距就足够了。

Answer 3

我建议用一种 hack 的方式将平方根符号分成两行，有两种变体：

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{mathtools, nccmath}
\usepackage{dashrule} 

\begin{document}

\begin{fleqn}[1em]
\begin{equation}
\begin{split}
\\[-1ex]
   & Δ\sigma_{V,ij} = \\
    &\begin{multlined}[0.9\linewidth]
    √{Δ\sigma_{x,ij}² + Δ\sigma_{y,ij}² + Δ\sigma_{z,ij}² - (Δ\sigma_{x,ij} · \Delta\sigma_{y,ij} + Δ\sigma_{y,ij} · \Delta\sigma_{z,ij} + Δ\sigma_{z,ij} · \Delta\sigma_{x,ij})} \\[0.5ex]
     \overline{\rule{0pt}{2.4ex}+ 3 · (Δ\tau_{xy,ij}² + Δ\tau_{yz,ij}² + Δ\tau_{zx,ij}²)}
    \end{multlined}
    \end{split}
\end{equation}

\begin{equation}
\begin{split}
\\[-1ex]
   & Δ\sigma_{V,ij} = \\
    &\begin{multlined}[0.9\linewidth]
    √{Δ\sigma_{x,ij}² + Δ\sigma_{y,ij}² + Δ\sigma_{z,ij}² - (Δ\sigma_{x,ij} · \Delta\sigma_{y,ij} + Δ\sigma_{y,ij} · \Delta\sigma_{z,ij} + Δ\sigma_{z,ij} · \Delta\sigma_{x,ij})}\hdashrule[2.66ex]{1.8em}{0.4pt}{2pt} \\[0.5ex]
     {\hdashrule[2.67ex]{1.8em}{0.4pt}{2pt}}\mkern -4mu\overline{\rule{0pt}{2.4ex}+ 3 · (Δ\tau_{xy,ij}² + Δ\tau_{yz,ij}² + Δ\tau_{zx,ij}²)}
    \end{multlined}
    \end{split}
\end{equation}
\end{fleqn}

\end{document}