答案1
有人可能会说这不符合 OP 的描述,但由于该符号是非标准的,因此人们可以在定义它时具有一定的灵活性。
\documentclass{article}
\usepackage{stackengine,mathtools}
\stackMath
\newcommand\ssqrt[2][\relax]{%
\ifx\relax#1\relax%
\stackengine{0pt}{\sqrt{\phantom{#2}}}{\mkern3mu\sqrt{#2}}{O}{c}{F}{F}{L}%
\else%
\stackengine{0pt}{\sqrt[#1]{\phantom{#2}}}{\mkern3mu\sqrt[\phantom{#1}]{#2}}{O}{c}{F}{F}{L}%
\fi%
}
\begin{document}
\[
\ssqrt{z}\quad\ssqrt[n]{z}
\]
\[
\ssqrt{\frac{x}{y}}\quad\ssqrt[3]{\frac{x}{y}}
\]
\end{document}
对于更复杂的东西,当然不是完美的,并且宏的尺寸取决于字体和字体大小,
\documentclass{article}
\usepackage{stackengine,mathtools,trimclip,scalerel}
\stackMath
\newcommand\ssqrt[2][\relax]{\ThisStyle{%
\ifx\relax#1\relax%
\setbox0=\hbox{$\SavedStyle\sqrt{\phantom{\cramped{#2}}}$}%
\def\tmpwd{\dimexpr2.5pt + .18\ht0\relax}% FONT DEPENDENT
\stackengine{0pt}{%
\SavedStyle\mkern1.5mu\sqrt{#2}%
}{%
\kern\tmpwd\clipbox{\tmpwd{} 2pt 0pt 0pt}{\copy0}%
}{O}{l}{F}{F}{L}%
\else%
\setbox0=\hbox{$\SavedStyle\sqrt[\phantom{#1}]{\phantom{\cramped{#2}}}$}%
\setbox2=\hbox{$\scriptscriptstyle#1$}%
\def\tmpwd{\dimexpr+3.2pt +.5\wd2 + .08\ht0\relax}% FONT DEPENDENT
\stackengine{0pt}{%
\SavedStyle\mkern1.5mu\sqrt[#1]{#2}%
}{%
\kern\tmpwd\clipbox{\tmpwd{} 0pt 0pt 0pt}{\copy0}%
}{O}{l}{F}{F}{L}%
\fi%
}}
\begin{document}
\[
\ssqrt{z}\quad\ssqrt[3]{z}\quad\ssqrt[n]{z^2 + 37}
\]
\[
\ssqrt{\frac{x}{y}}\quad\ssqrt[3]{\frac{x}{y}}\quad\ssqrt[n]{\frac{x^2 + 37}{y}}
\]
\[\textstyle
\ssqrt{z}\quad\ssqrt[3]{z}\quad\ssqrt[n]{z^2 + 37}
\]
\[\textstyle
\ssqrt{\frac{x}{y}}\quad\ssqrt[3]{\frac{x}{y}}\quad\ssqrt[n]{\frac{x^2 + 37}{y}}
\]
\[\scriptstyle
\ssqrt{z}\quad\ssqrt[3]{z}\quad\ssqrt[n]{z^2 + 37}
\]
\[\scriptstyle
\ssqrt{\frac{x}{y}}\quad\ssqrt[3]{\frac{x}{y}}\quad\ssqrt[n]{\frac{x^2 + 37}{y}}
\]
\end{document}
答案2
为了娱乐:首次尝试并得到@marmot's
很好的答案:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{tikzmark,calc}
\newcounter{stuff}
%https://tex.stackexchange.com/questions/461193/floating-point-square-root-symbol
\tikzset{toughtoroot/.style={path picture={\draw
let \p1=($(path picture bounding box.north west)-(path picture bounding box.south west)$) in ($(path picture bounding box.south west)+(0.4em, -0.1em)$)
-- ($(path picture bounding box.north)-({0.25em+\x1/10},0)-(0.3em, 0.1em)$);}}}
\begin{document}
$\tikzmarknode[toughtoroot]{1}{\sqrt{a+b}}$
\end{document}
这将给你
免责声明:这远非成熟的解决方案。工作正在进行中
;-)
。请注意,这仅适用于这种情况,而不是一般情况(然而!)。