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I don't like how the wide mn index of the roots is being displayed and am looking to see how others have handled this problem. I am using the closed square root from the references listed below which allows me to tweak the locations slightly and hence am able to get the m and n aligned on the left hand side, and tried moving the indices on the right, but am not pleased with the result:

enter image description here

References:

Code:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}
\usepackage{letltxmacro}

\LetLtxMacro{\oldsqrt}{\sqrt}

\def\DHLhksqrt#1#2{%
  \setbox0=\hbox{$#1\oldsqrt{#2\,}$}\dimen0=\ht0\relax%
  \advance\dimen0-0.25\ht0\relax%
  \setbox2=\hbox{\kern-0.375pt\vrule height\ht0 depth -\dimen0}%
  {%
    \hbox{$#1\expandafter\oldsqrt\expandafter[\DHLindex]{#2\,}$}%
    \lower\ifx\math@version\bold@name0.60pt\else0.4pt\fi\box2%
  }%
}
\newcommand{\ClosedSqrt}[1][]{\def\DHLindex{#1}\mathpalette\DHLhksqrt}%

\RenewDocumentCommand{\sqrt}{O{\hphantom{3}} O{0} O{0}  m}{\ClosedSqrt[\leftroot{#2}\uproot{#3}#1]{#4}}%

\begin{document}
\begin{alignat*}{3}
    \oldsqrt[m]{\oldsqrt[n]{x}} &= \oldsqrt[mn]{x} &\quad\text{original sqrt}\\
    \sqrt[m][-3]{\sqrt[n][-1][2]{x}} &= \sqrt[mn]{x} = \sqrt[mn][-3][2]{x} &\quad\text{closed sqrt}
\end{alignat*}
\end{document}
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  • 1
    definitely not an answer, but I quite like to use rational exponents when the index gets too unruly :) – cmhughes Jul 5 '12 at 5:25
  • @cmhughes: Agreed, for general equations. But for the case where this radicals are being introduced to students, I think it is important to stick with radicals. – Peter Grill Jul 5 '12 at 5:57
  • The first example seems fine to me. Of course there's always $x^{1/(mn)}$. :) – egreg Jul 5 '12 at 8:43
  • Since you already have amsmath loaded the \uproot and \leftroot commands can be used for adjusting the position of the indices by small amounts – David Carlisle Jul 5 '12 at 13:59
  • @DavidCarlisle: I am already doing that in the above as the redefined \sqrt accepts two additional parameters which make use of the up and left movement. – Peter Grill Jul 5 '12 at 16:25
3

This is what I can offer.

\documentclass{article}
\usepackage{amsmath,array}
\usepackage{xparse}
\usepackage{letltxmacro}

\LetLtxMacro{\latexsqrt}{\sqrt}

\ExplSyntaxOn
\RenewDocumentCommand{\sqrt}{ O{\hphantom{3}} O{0} O{0}  m }
 {
  \grill_root:nnxx 
   { #1 } % index
   { #4 } % main argument
   { \int_eval:n { #2 - 1 } } % left shift (default -1)
   { \int_eval:n { #3 + 1 } } % up shift (default 1)
 }
\cs_new_protected:Npn \grill_root:nnnn #1 #2 #3 #4
 {
  \latexsqrt[ \leftroot{#3} \uproot{#4} #1 ] { #2 }
 }
\cs_generate_variant:Nn \grill_root:nnnn { nnxx }
\ExplSyntaxOff

\begin{document}
\renewcommand{\arraystretch}{1.5}

\begin{tabular}{>{$\displaystyle}r<{$}@{}>{$\displaystyle{}}l<{$}}

\multicolumn{2}{l}{Original} \\
\latexsqrt[m]{\latexsqrt[n]{x}} &= \latexsqrt[mn]{x}\\
\latexsqrt[n]{\latexsqrt[m]{x}} &= \latexsqrt[nm]{x}\\
\latexsqrt[p]{\latexsqrt[q]{x}} &= \latexsqrt[pq]{x}\\

\multicolumn{2}{l}{New} \\
\sqrt[m]{\sqrt[n]{x}} &= \sqrt[mn]{x}\\
\sqrt[n]{\sqrt[m]{x}} &= \sqrt[nm]{x}\\
\sqrt[p]{\sqrt[q]{x}} &= \sqrt[pq]{x}

\end{tabular}
\end{document}

The renewed \sqrt command applies by default a \leftroot{-1} and a \uproot{1} to which the optional arguments add. This seems a good compromise in normal cases: an index with a descender doesn't touch the root symbol.

I'm afraid that something that works well with all sizes of the radical is impossible.

If you say

\sqrt[3][1][-1]

you clear off the default adjustment, but the optional arguments can be any integers.

Of course I won't consider the "closed" root, for reasons you should know. ;-)

enter image description here

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