I'm wondering if there are generalizations of the sqrt function in Maxima? In particular, I'd like to be able to control whether x^(a/b) displays as $x^{\frac{a}{b}}$ or $\sqrt[b]{x^{a}}$.
I searched the index of the documentation for root but didn't find anything that I could identify as relevant, and there weren't any links from the sqrtdispflag entry.
This is an interesting question, showing, once again, that Maxima could benefit from a more flexible TeX output system, but I digress. Here's a possible solution.
This uses pattern matching (in defmatch) to identify something-to-a-quotient expressions for further processing, pulling out the numerator and denominator of the quotient if the match is successful. The pattern matcher fails to identify stuff like b/c^2 -- this should probably be considered a bug in the pattern matcher.
The default TeX output function is a Lisp function named TEX-MEXPT. I wrote a line of Lisp code (admittedly obscure) to call it.
/* tex_mexpt.mac -- look for quotient in exponent
* copyright 2021 by Robert Dodier
* I release this work under terms of the GNU General Public License
*/
matchdeclare (aa, all);
matchdeclare (bb, "#"(1));
defmatch (matches_quotient, aa/bb);
defmatch (matches_minus_quotient, -aa/bb);
:lisp (defun $tex_mexpt_default (e) (let (lop rop) (apply 'concatenate 'string (tex-mexpt e nil nil))))
my_tex_mexpt_quotient (base, expt_num, expt_denom) :=
if expt_num = 1
then printf(false, "\\sqrt[~a]{~a}", tex1 (expt_denom), tex1 (base))
else printf(false, "\\sqrt[~a]{~a}", tex1 (expt_denom), tex_mexpt_default (base^expt_num));
my_tex_mexpt_minus_quotient (base, expt_num, expt_denom) :=
if expt_num = 1
then printf(false, "\\sqrt[~a]{{1}\\over{~a}}", tex1 (expt_denom), tex1 (base))
else printf(false, "\\sqrt[~a]{~a}", tex1 (expt_denom), tex_mexpt_default (base^-expt_num));
my_tex_mexpt (e) :=
if tex_mexpt_look_for_quotient
then block ([base, expt],
[base, expt]: args(e),
if matches_quotient(expt) # false
then my_tex_mexpt_quotient (base, aa, bb)
elseif matches_minus_quotient(expt) # false
then my_tex_mexpt_minus_quotient (base, aa, bb)
else tex_mexpt_default (e))
else tex_mexpt_default (e);
/* examples */
stringdisp: true $
verbatim_and_equation (e) ::= printf (S, "\\begin{verbatim}~%~a~%\\end{verbatim}~%$$~a$$~%", string(e), tex1(e));
S: openw ("/tmp/foo.tex");
printf (S, "\\documentclass{article}~%\\begin{document}~%");
/* first without my_tex_mexpt at all */
verbatim_and_equation (a^(b/c));
verbatim_and_equation (a^-(b/c));
verbatim_and_equation (a^(-b/c));
verbatim_and_equation (a^(1/c));
verbatim_and_equation (a^-(1/c));
verbatim_and_equation (a^(-1/c));
verbatim_and_equation ((1 + (1 - x)^((y - z)/(y - w)))/((2*u - v)^(1/(n + 1))));
/* now enable my_tex_mexpt */
texput ("^", my_tex_mexpt);
tex_mexpt_look_for_quotient:true;
verbatim_and_equation (a^(b/c));
verbatim_and_equation (a^-(b/c));
verbatim_and_equation (a^(-b/c));
verbatim_and_equation (a^(1/c));
verbatim_and_equation (a^-(1/c));
verbatim_and_equation (a^(-1/c));
verbatim_and_equation ((1 + (1 - x)^((y - z)/(y - w)))/((2*u - v)^(1/(n + 1))));
/* verify disabled produces same output as originally */
tex_mexpt_look_for_quotient:false;
verbatim_and_equation (a^(b/c));
verbatim_and_equation (a^-(b/c));
verbatim_and_equation (a^(-b/c));
verbatim_and_equation (a^(1/c));
verbatim_and_equation (a^-(1/c));
verbatim_and_equation (a^(-1/c));
verbatim_and_equation ((1 + (1 - x)^((y - z)/(y - w)))/((2*u - v)^(1/(n + 1))));
printf (S, "\\end{document}~%");
close(S);
As you can see I put a number of examples in there to verify the output a little bit. You can execute it via maxima --batch=foo.mac or whatever the name of the saved file is. It generates output in /tmp/foo.tex. I processed that with latex and then viewed it with xdvi.
For the record, here is the foo.tex output I get.
\documentclass{article}
\begin{document}
\begin{verbatim}
a^(b/c)
\end{verbatim}
$$a^{{{b}\over{c}}}$$
\begin{verbatim}
a^-(b/c)
\end{verbatim}
$${{1}\over{a^{{{b}\over{c}}}}}$$
\begin{verbatim}
a^((-b)/c)
\end{verbatim}
$${{1}\over{a^{{{b}\over{c}}}}}$$
\begin{verbatim}
a^(1/c)
\end{verbatim}
$$a^{{{1}\over{c}}}$$
\begin{verbatim}
a^-(1/c)
\end{verbatim}
$${{1}\over{a^{{{1}\over{c}}}}}$$
\begin{verbatim}
a^((-1)/c)
\end{verbatim}
$${{1}\over{a^{{{1}\over{c}}}}}$$
\begin{verbatim}
(1+(1-x)^((y-z)/(y-w)))/(2*u-v)^(1/(n+1))
\end{verbatim}
$${{\left(1-x\right)^{{{y-z}\over{y-w}}}+1}\over{\left(2\,u-v\right)^{{{1}\over{n+1}}}}}$$
\begin{verbatim}
a^(b/c)
\end{verbatim}
$$\sqrt[c]{a^{b}}$$
\begin{verbatim}
a^-(b/c)
\end{verbatim}
$${{1}\over{\sqrt[c]{a^{b}}}}$$
\begin{verbatim}
a^((-b)/c)
\end{verbatim}
$${{1}\over{\sqrt[c]{a^{b}}}}$$
\begin{verbatim}
a^(1/c)
\end{verbatim}
$$\sqrt[c]{a}$$
\begin{verbatim}
a^-(1/c)
\end{verbatim}
$${{1}\over{\sqrt[c]{a}}}$$
\begin{verbatim}
a^((-1)/c)
\end{verbatim}
$${{1}\over{\sqrt[c]{a}}}$$
\begin{verbatim}
(1+(1-x)^((y-z)/(y-w)))/(2*u-v)^(1/(n+1))
\end{verbatim}
$${{\sqrt[y-w]{\left(1-x\right)^{y-z}}+1}\over{\sqrt[n+1]{2\,u-v}}}$$
\begin{verbatim}
a^(b/c)
\end{verbatim}
$$a^{{{b}\over{c}}}$$
\begin{verbatim}
a^-(b/c)
\end{verbatim}
$${{1}\over{a^{{{b}\over{c}}}}}$$
\begin{verbatim}
a^((-b)/c)
\end{verbatim}
$${{1}\over{a^{{{b}\over{c}}}}}$$
\begin{verbatim}
a^(1/c)
\end{verbatim}
$$a^{{{1}\over{c}}}$$
\begin{verbatim}
a^-(1/c)
\end{verbatim}
$${{1}\over{a^{{{1}\over{c}}}}}$$
\begin{verbatim}
a^((-1)/c)
\end{verbatim}
$${{1}\over{a^{{{1}\over{c}}}}}$$
\begin{verbatim}
(1+(1-x)^((y-z)/(y-w)))/(2*u-v)^(1/(n+1))
\end{verbatim}
$${{\left(1-x\right)^{{{y-z}\over{y-w}}}+1}\over{\left(2\,u-v\right)^{{{1}\over{n+1}}}}}$$
\end{document}