3

From the guide MathTımeProfessional fonts mt2pro[lite] we known that in addition to the \sqrt command, which uses an extensible symbol, called \SQRT that produces individually designed root signs up to 4 inches high:

enter image description here

Actually I'm using into the preamble

\usepackage{baskervald}
\usepackage[baskervaldx]{newtxmath}

and I'm not sure if these packages used can produce a good root like the one framed in green. The MWE that I add provides a bad square root like the one framed in red that I don't like.

\documentclass[12pt,a4paper,oneside]{book}
\usepackage[lmargin=7cm,rmargin=.7cm,bmargin=2cm,marginparwidth=5.5cm,marginparsep=2em]{geometry}
\usepackage{sidenotes,tabularx}
\usepackage{afterpage}
\usepackage{baskervald}
\usepackage[baskervaldx]{newtxmath}
\usepackage[scr, scaled=1.1]{rsfso}
\usepackage{bm}
\usepackage{amsmath,amssymb}  
\usepackage[italian]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc} 
\usepackage{mathtools}
\usepackage{mathrsfs,lipsum}

\parindent 0cm

\begin{document}
\begin{equation}
d =r \sqrt{1+\left(\frac{r'}r\right)^2-2\left(\frac{r'}r\right)\cos\theta}
\end{equation}
\end{document}

enter image description here

I ask if exists a strategy to use the same square root (cube root, etc.) as the green rectangle. Thank you for your help(s).

  • 1
    I'd give \sqrt{1+\Bigl(\frac{r'}r\Bigr)^{\!2}-2\Bigl(\frac{r'}r\Bigr)\cos\theta} a try. By not making the parentheses overly large, the slope of the surd diminishes automatically. – Mico Aug 6 '18 at 20:44
  • @Mico had no doubts about your skill. It works! But I have to follow this option not making the parentheses too big or can there be a macro, a code that always applies to the roots with index n when I realize that I don't like the root? – Sebastiano Aug 6 '18 at 20:49
  • I've gone ahead and posted the earlier comment as a answer, with the additional suggestion to consider switching to inline-style fraction notation. – Mico Aug 6 '18 at 21:00
  • @Mico With a lot of sincerity I also had a triple integral with the square root at the numerator and I used your setting to consider ratios as (r'/r) and not as fractions. – Sebastiano Aug 6 '18 at 21:06
  • What is the reason, again for my questions, of a downvote? My question is not very clear? Surely someone will love you very much :-). – Sebastiano Aug 6 '18 at 21:56
2

Not a direct answer to your question (mainly because I don't know how to provide a direct answer!), but a suggestion to take a different approach: Don't autosize the parentheses via \left and \right. By using \Bigl( and \Bigr) throughout, the material under (inside?) the square root is much less tall, and this results in a less-steep surd as well. Better still, switch to inline-fraction notation and the appearance of the square root symbol pretty much becomes a non-issue.

enter image description here

\documentclass[12pt,a4paper,oneside]{book}
\usepackage[lmargin=7cm,rmargin=.7cm,bmargin=2cm,
     marginparwidth=5.5cm,marginparsep=2em]{geometry}
\usepackage{sidenotes,tabularx}
\usepackage{afterpage}
\usepackage{baskervald}
\usepackage[baskervaldx]{newtxmath}
\usepackage[scr, scaled=1.1]{rsfso}
\usepackage{bm}
\usepackage{amsmath,amssymb}  
\usepackage[italian]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc} 
\usepackage{mathtools}
\usepackage{mathrsfs,lipsum}

\parindent 0cm

\begin{document}
\begin{align*}
\quad d &=r \sqrt{1+\left(\frac{r'}r\right)^2 -2\left(\frac{r'}r\right)\cos\theta}\\
        &=r \sqrt{1+\Bigl(\frac{r'}r\Bigr)^{\!2} -2\Bigl(\frac{r'}r\Bigr)\cos\theta} \\
        &=r \sqrt{1+(r'/r)^2 -2(r'/r)\cos\theta}
\end{align*}
\end{document}
  • Mico you see a reason of a downvote for my question? – Sebastiano Aug 6 '18 at 21:50
  • @Sebastiano - I certainly do not. – Mico Aug 6 '18 at 22:04
  • But is possible that when I ask a question there is always a downvote? Before the users have right, now no. – Sebastiano Aug 6 '18 at 22:07
  • @Sebastiano - Have you flagged this issue, for the moderators' attention? – Mico Aug 6 '18 at 22:09
  • I have not use any flag! – Sebastiano Aug 6 '18 at 22:10
4

You can steal the \SQRT from mtpro2.

Attention! I didn't copy all of the definition so you won't be able to use an optional argument (\SQRT[3])

\documentclass[12pt,a4paper,oneside]{book}
\usepackage[lmargin=7cm,rmargin=.7cm,bmargin=2cm,marginparwidth=5.5cm,marginparsep=2em]{geometry}
\usepackage{sidenotes,tabularx}
\usepackage{afterpage}
\makeatletter
\normalsize
\dimen@\f@size pt
\font\MTEXA@=mt2exa at \the\dimen@
\multiply\dimen@\tw@
\font\MTEXE@=mt2exe at \the\dimen@
\multiply\dimen@\tw@
\font\MTEXF@=mt2exf at \the\dimen@
\multiply\dimen@\tw@
\font\MTEXG@=mt2exg at \the\dimen@

\newbox\preSbox@
\newbox\Sbox@
\newif\ifSQEX@
\def\SQEX@#1{\setbox\Sbox@\vbox{$$\radical"270370{\copy\preSbox@}$$}%
\setbox\Sbox@\vbox{\unvbox\Sbox@\unskip\unpenalty
\setbox\Sbox@\lastbox\setbox\Sbox@\hbox{\unhbox\Sbox@\setbox\Sbox@\lastbox
\setbox\Sbox@\hbox{\unhbox\Sbox@\setbox\Sbox@\lastbox\setbox\Sbox@\lastbox
\setbox0\hbox{#1}%
\ifdim\dp\Sbox@>\dp0\global\SQEX@true\else
\global\SQEX@false\fi}}}}
\newcount\SQcount@
\def\SQtest@#1{\setbox\preSbox@\hbox{$\displaystyle{#1}$}%
\SQEX@{\MTEXA@ s}%
\ifSQEX@
{\textfont3=\MTEXE@\SQEX@{\MTEXE@ u}}%
\ifSQEX@
{\textfont3=\MTEXF@\SQEX@{\MTEXF@ u}}%
\ifSQEX@
\def\SQtest@@{\textfont3=\MTEXG@}\global\SQcount@3
\else
\def\SQtest@@{\textfont3=\MTEXF@}\global\SQcount@2
\fi
\else
\def\SQtest@@{\textfont3=\MTEXE@}\global\SQcount@1
\fi
\else
\def\SQtest@@{\textfont3=\MTEXA@}\global\SQcount@0
\fi}
\newbox\SQRTbox@
\def\SQR@@T#1{\setbox\SQRTbox@\hbox{$\displaystyle{#1}$}%
\SQtest@{#1}%
\hbox{\SQtest@@$\displaystyle\radical"270370{\box\SQRTbox@}$}}
\DeclareRobustCommand\SQRT{\SQR@@T}
\makeatother
\usepackage[baskervaldx]{newtxmath}
\usepackage{amsmath,amssymb}
\usepackage[italian]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\usepackage{mathrsfs,lipsum}

\parindent 0cm

\begin{document}
\begin{equation}
d =r \SQRT{\sum_i^n 1+\left(\frac{r'}r\right)^2-2\left(\frac{r'}r\right)\cos\theta}
\end{equation}

\begin{equation}
d =r \sqrt{\sum_i^n1+\left(\frac{r'}r\right)^2-2\left(\frac{r'}r\right)\cos\theta}
\end{equation}
\end{document}

enter image description here

  • I have seen now you wonderful and complicate answer. Don't be worry \sqrt[3]{x}. At the moment it is not important for my document. +1 surely. Thank you very much. – Sebastiano Aug 7 '18 at 19:32
  • When you have time, could you generalize the roots, with index 3,4, etc? Thank you very much. – Sebastiano Aug 11 '18 at 20:51

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