Optional arguments in \def

Question

I want to re-define the \sqrt command (as detailed in Nice-looking p-th roots), and so far I have the following:

\documentclass{article}

\usepackage{fouriernc} % use the New Century Schoolbook font
\usepackage{amsmath}

\let\oldsqrt\sqrt
\def\sqrt[#1]{\oldsqrt[\leftroot{-3}\uproot{3}#1]}

\begin{document}
This $\sqrt[p]{a}$ looks better than $\oldsqrt[p]{a}$,
since the $p$ does not intersect the root symbol.
\end{document}

However, when I try writing \sqrt{a}, I get an error. How can I make the #1 argument optional, so that if it is not passed in, it leaves that area above the root symbol blank (as it would in the default \sqrt command)?

(This is a follow-up to the comments on David's and Harish's answers to above-mentioned question.)

Answer 1

The primitive \def does not understand optional arguments. That is provided in LaTeX only. What you have defined is a delimited argument:

  \def\sqrt[#1]{...}

which means that the input must always be provided as [...], you could have written anything instead of []. To have an optional argument you need to define it in LaTeX using the following method:

  \DeclareRobustCommand\sqrt{\@ifnextchar[{true statement}{false statement}}

The @\ifnextchar checks to see if there is an [ and branches accordingly. You will need to modify your script on this basis.

The commands are normally defined to call two macros, one delimited and another undelimited, as for example:

  \makeatletter
  \DeclareRobustCommand\sqrt{\@ifnextchar[{\@@sqrt}{\@sqrt}}
  \def\@@sqrt[#1]{I have a square bracket #1}  
  \def\@sqrt#1{I don't have a square bracket #1}
  \sqrt{1}
  \sqrt[3]

Now redefining amsmath macros, is like walking barefoot on a coal fire, so a few more items need to be taken care of. Firstly to let save the old command, we use Oberdiek's package letltxmacro

   \LetLtxMacro{\oldsqrt}{\sqrt}

The rest are shown in the minimal below:

\documentclass{article}
\usepackage{fouriernc} % use the New Century Schoolbook font
\usepackage{amsmath}
\usepackage{letltxmacro}
\makeatletter
\begin{document}
\LetLtxMacro{\oldsqrt}{\sqrt}
\DeclareRobustCommand\sqrt{\@ifnextchar[\@@sqrt\oldsqrt}
\newcommand\@@sqrt[2][]{\oldsqrt[\leftroot{-3}\uproot{3}#1]{#2}}  
\[
   \sqrt{\alpha}\qquad
   \sqrt[\beta]{k} \qquad
   \sqrt[c]{k} \qquad
   \sqrt[d]{k}
\]
\end{document}

Answer 2

If you want commands with at most one optional argument, LaTeX's built in \newcommand and \renewcommand are an easy way to define them. The syntax is

\(re)newcommand⟨\name⟩[⟨number of arguments⟩][⟨default value for the first argument⟩]{⟨code⟩]

If you specify the second optional argument, then the first argument to \name will be optional. So in your case one can simply use

\renewcommand\sqrt[1][]{\oldsqrt[\leftroot{-3}\uproot{3}#1]}

Note that (as in your question), we do not need to specify the second argument, because \oldsqrt will look for it anyway, so that this definition is functionally equivalent to

\renewcommand\sqrt[2][]{\oldsqrt[\leftroot{-3}\uproot{3}#1]{#2}}

If you need several optional arguments or an argument other than the first one should be optional, you can either chain \@ifnextchars (as in Yiannis' answer), or use xparse (as in Peter's answer).

Answer 3

Using a new command: \mysqrt:

This is what I was suggesting in the comments to Nice-looking p-th roots using the xparse package to make the first three parameters optional:

\NewDocumentCommand{\mysqrt}{O{} O{-2} O{2}  m}{\sqrt[\leftroot{#2}\uproot{#3}#1]{#4}}

Then $\mysqrt[p]{a}$ would produce the desired result:

Note that the first optional parameter in the index of the \sqrt. The following two options are the tweaks that get applied to the \leftroot (defaults to -2 in the above code) and the \uproot (defaults to 2).

The same examples from the linked question would be specified as where the left uses the default settings and the right uses the manual tweaks:

\[\mysqrt[\beta]{k}            \quad\mysqrt[\beta][-3][3]{k}          \]
\[\mysqrt[\beta]{\frac{k}{h}}  \quad\mysqrt[\beta][-2][6]{\frac{k}{h}}\]

yielding:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}
\NewDocumentCommand{\mysqrt}{O{} O{-2} O{2}  m}{%
    \sqrt[\leftroot{#2}\uproot{#3}#1]{#4}%
}%
\begin{document}\noindent
This $\mysqrt[p]{a}$ looks better than $\sqrt[p]{a}$,
since the $p$ does not intersect the root symbol.
\end{document}

---

Redefine the existing \sqrt macro:

If you wish to redfine the \sqrt globally then you have to use \LetLtxMacro from the letltxmacro package since \sqrt has optional parameters already.

References:

• A detailed description of \LetLtxMacro can be found at this question on closed square root symbol

The following produces results identical to the second image above:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}
\usepackage{letltxmacro}
\LetLtxMacro{\OldSqrt}{\sqrt}
\RenewDocumentCommand{\sqrt}{O{} O{-2} O{2}  m}{\OldSqrt[\leftroot{#2}\uproot{#3}#1]{#4}}%
\begin{document}\noindent
[\sqrt[\beta]{k}            \quad\sqrt[\beta][-3][3]{k}          ]
[\sqrt[\beta]{\frac{k}{h}}  \quad\sqrt[\beta][-2][6]{\frac{k}{h}}]
\end{document}

Answer 4

This answer is about solving the specific problem, that is, initializing \leftroot and \uproot to different values from the default, leaving intact the possibility of using them for particular cases.

The trick is to notice that amsmath uses the values given as arguments to \leftroot and \uproot to set two count registers, \leftroot@ and \uproot@ which are initialized to zero. This happens in the internal macro called \root. Thus one can do

\usepackage{etoolbox}
\makeatletter
\patchcmd{\root}{\uproot@\z@}{\uproot@3 }{}{}
\patchcmd{\root}{\leftroot@\z@}{\leftroot@-3 }{}{}
\makeatother

and now those initial values will be applied, but saying

\sqrt[\leftroot{1}x]{y}

would give the same result as \sqrt[\leftroot{1}\uproot{3}x]{y} without the patching, which might be useful in particular cases. Similarly, \sqrt[\leftroot{0}\uproot{0}x]{y} would give the same result as the original \sqrt[x]{y}.

As Stefan Lehmke points out in his comment, changing the meaning of \sqrt is not really recommended. However, setting a different default for \leftroot and \uproot could be justified with some document fonts. The important thing is to know how the new \sqrt behaves with respect to \leftroot and \uproot.

Making \leftroot and \uproot adding to the default requires a very extensive patching of \root:

\makeatletter
\def\default@leftroot{-3} % set the default value of leftroot
\def\default@uproot{3}    % set the default value of uproot
\renewcommand{\root}{\relaxnext@
  \DN@{\ifx\@let@token\uproot\let\next@\nextii@\else
   \ifx\@let@token\leftroot\let\next@\nextiii@\else
   \let\next@\plainroot@\fi\fi\next@}%
  \def\nextii@\uproot##1{\uproot@\numexpr(\default@uproot+##1)\relax\FN@\nextiv@}%
  \def\nextiv@{\ifx\@let@token\@sptoken\DN@. {\FN@\nextv@}\else
   \DN@.{\FN@\nextv@}\fi\next@.}%
  \def\nextv@{\ifx\@let@token\leftroot\let\next@\nextvi@\else
   \let\next@\plainroot@\fi\next@}%
  \def\nextvi@\leftroot##1{\leftroot@\numexpr(\default@leftroot+##1)\relax\plainroot@}%
   \def\nextiii@\leftroot##1{\leftroot@\numexpr(\default@leftroot+##1)\relax\FN@\nextvii@}%
  \def\nextvii@{\ifx\@let@token\@sptoken
   \DN@. {\FN@\nextviii@}\else
   \DN@.{\FN@\nextviii@}\fi\next@.}%
  \def\nextviii@{\ifx\@let@token\uproot\let\next@\nextix@\else
   \let\next@\plainroot@\fi\next@}%
  \def\nextix@\uproot##1{\uproot@\numexpr(\default@uproot+##1)\relax\plainroot@}%
  \bgroup\uproot@\default@uproot \leftroot@\default@leftroot\relax \FN@\next@}
\makeatother

(Thanks to Stefan Lehmke for spotting an error in the initial version.)