# Diagonal brace with nested parenthesis

I would to obtain a result similar to the image below. So far I only have \bigg( \Big( \big( a^q \big)^q \Big)^{\rotatebox{25}{$\cdots$}} \bigg)^q, without the brace.

I tried to follow this question, but there were no parenthesis and I was not able to circumvent this. • Hi, just a curiosity...for my opinion your question is good and I put +1. But do you have seen this image (I never I have seen this in a textbook)? :-) Feb 23 at 13:01
• @Sebastiano : thanks for your encouraging comment. I did the image myself, but admittedly I never saw this in a book – maybe in the future if we get nice answers to my question ;-) Feb 23 at 13:07
• I don't even really know how to do it properly with a horizontal brace. Feb 23 at 13:09
• Yes, I'm agree fully with you....In this moment for me it is very complicated done it in math-mode. I think with TiKZ or PSTricks it is very simple. I have understood your thought. :-) Feb 23 at 13:09
• @Sebastiano : a solution with TikZ (or PSTricks that I don't really know) would be fine. This is the approach used in the quesiton I linked. Feb 23 at 13:10

Very easy with horizontal brace; somewhat less easy with rotated brace. I have no time right now to implement a general solution, so I'll go with a brute-force version for now.

\documentclass{article}

\begin{document}

$\biggl(\Bigl(\bigl(a {% extra group for correct spacing \overbrace{ {}^q \bigr)^q \Bigr)^{\rotatebox{20}{\scriptstyle\dots}} \biggr)^{\!\!q\mathstrut} }^{\text{n times}} }$

$\biggl( \Bigl( \bigl(a^q\bigr)^{q} \Bigr)\quad \biggr)^{ \!\! q \setbox0=\hbox to 0pt{\hss \rotatebox{15}{ \overbrace{\kern1.8em\scriptstyle\mathstrut\dots\kern1.2em}^{\rotatebox{-15}{\scriptsizen times}} }% }% \lower1.5ex\box0 }$

\end{document} • Thanks for your answer :-). In the horizontal case, it is possible to make the overbrace closer to the math formula? I tried to use \overbracket from mathtools but I am not able to get what I want. I tried to implement solutions from here, but negative depth does not seem to work anyway... Feb 23 at 17:09
• Probably using your idea with "\lower3ex\box0" can make it work, but I am not sure how. Feb 23 at 17:13
• @Watson No, you need no \lower in the first case. I've put an extra \mathstrut to increase the distance; you can simply remove that. Feb 23 at 19:38
• Excellent works for every user! +1 for all. Feb 23 at 21:52

As given in my answer at Nested Mathematical Functions

The .4pt stacking gap added with each nesting will determine the growth rate of the parens.

\documentclass{article}
\usepackage{scalerel,stackengine,graphicx,amsmath}
\stackMath
\ignoremathstyle
\newcommand\rdots{\ThisStyle{.\raisebox{.1ex}{$\scriptstyle.$}%
\raisebox{.2ex}{$\scriptstyle.$}}}
\begin{document}
$\stackinset{r}{1pt}{t}{-14.5pt}{\rotatebox{6} {\overbrace{\rule{30pt}{0pt}}^{\textrm{\tinyn times}}}}{% \Comp{\Comp{\Comp{a^q}^q}^{\rdots}}^q}$
\end{document} Here's a version where the stackgap is increased to .6pt, to bring out the increasing size of parens: • Thanks but you are missing the diagonal brace above the parenthesis, so this does not really address my question. Feb 23 at 13:40
• @Watson Please see my revision. Feb 23 at 14:02

A solution with pstricks-add:

\documentclass{article}
$\biggl(\Bigl(\bigl(a^{\pnode[0,1ex]{E}{q}}\bigr)^{\! \rnode{F}{q}}\Bigr)\enspace \biggr)^{\!\!\rnode{G}{q}\pnode[0,1ex]{H}}% \psset{singleline, braceWidth=1.2pt, braceWidthInner=2pt, braceWidthOuter=2pt, nodesepB=-2pt} \psbrace[rot=14,ref=c, linewidth=0.3pt](H)(E){n\:\mathrm{times}}% \ncline[linestyle=dotted, dotsep=2.pt, nodesepA=4.5pt, nodesepB=3.2pt]{F}{G}$ 