# Diagonal underbrace

For didactic purposes I would like to build a underbrace that is diagonal as in the drawing below. How could I do this?

• If you load mathdots package x^{x^{\iddots^{\raisebox{-.4ex}{$\scriptscriptstyle x$}}}} gives you the expression without the brace. It seems to me that TikZ type of way is the easiest to go. But n-1 times text is too long and big to be fit in the exponent. I would suggest exp() notation. Oct 25, 2013 at 13:44
• @percusse Very good your solution. Simple and elegant. I accept this answer as a solution to my problem. Oct 25, 2013 at 14:23

Another TikZ variant, with a variable number of things stacked into the exponent:

## Code

\documentclass{scrartcl}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing,calc}
\usepackage{graphicx}

%\documentclass[tikz,border=2mm]{standalone}

\begin{document}

\newcommand{\veryhigh}[3]% base, exponent, text
{   \begin{tikzpicture}
\node (tempnode-0) at (0,0) {$#1$};
\foreach \mytext [count=\c] in {#2}
{ \pgfmathtruncatemacro{\b}{\c-1}
\node[above right,font=\tiny,inner sep=3pt] (tempnode-\c) at (tempnode-\b) {$\mytext$};
\xdef\maxexp{\c}
}
\draw [decoration={brace,amplitude=4pt,mirror,raise=2pt},decorate] ($(tempnode-1.south east)+(-0.13,0.13)$) -- node[below right=1mm,font=\tiny] {#3} ($(tempnode-\maxexp.south east)+(-0.13,0.13)$);
\end{tikzpicture}
}

\veryhigh{x}{x,{\rotatebox{45}{$\cdots$}},x}{$(n-1)$ times}

\veryhigh{\pi}{a,b,c,d,e,f,g,h}{too much!}

\end{document}


## Output

• An alternative for your problem with better spacing would be \veryhigh{x}{x,.,.,.,x}{$(n-1)$ times} Oct 26, 2013 at 12:45

This is a rough and ready answer, done mostly by eye-balling rather than by thinking through general measurements. But, for a one-off case, or series of similar ones, it should serve your purposes. (Input from more experienced people, who are doubtless rubbing their eyes at my code, is appreciated.)

\documentclass[12pt]{report}
\usepackage{mathptmx, mathdots, rotating}

\begin{document}

\newlength{\Width}
\settowidth{\Width}{% the width of the exponent x...x
${}^{\,\,x% \hbox{\raisebox{.8ex}{%$\iddots$\raisebox{1.3ex}{${\,}^x$}}}}$}

$x^{\,\,x% %$x^x$looks a bit too tight to me, needs some extra space % \makebox[0pt][l]{% %pretend the brace has no width % \raisebox{0ex}[0pt][0pt]{% %pretend the brace has no height or depth % \hspace{.2em}% %give it a bit of space % \rotatebox{40}{% %oddly, 45 degree rotation doesn't look right %$\underbrace{\hspace{.9\Width}}$}}}% %make the brace a bit small than the exponent % \hbox{\raisebox{.8ex}% %raise the dots and the following x % {$\iddots$% %here are the dots % \raisebox{1.3ex}% %raise the x even more % {${\,}^x$}}}}$%
%here's the x, with a bit of extra space
%
\kern-.2em%
%reduce space between x^{x...x} and n-1
%
\raisebox{.35ex}%
%raise n-1 a bit
%
{\scriptsize $n-1$ times}%

\end{document}


This is my try with Tikz. Since the command has some code in it, I made a \newcommand. Here you can see the result.

I fixed the code in order to make the characters stay "horizontal", borrowing the code from morbusg's answer.

\documentclass{article}
\usepackage{tikz}
\usepackage{rotating}

\usetikzlibrary{decorations.pathreplacing}

\tikzset{liltext/.style={font=\tiny}}

\newcommand{\diagbrace}[2]{
\textsuperscript{
\hspace*{-6mm}\raisebox{-1mm}{
\begin{tikzpicture}
\node[rotate=45] at (0,0) (text1) {\rotatebox{-40}{#1}};
\draw [decorate,decoration={brace,amplitude=4pt,mirror,raise=2pt}]
(text1.west) -- (text1.east) node [right,midway,xshift=1mm, yshift=-2mm,liltext] {#2};
\end{tikzpicture}}}}

\def\rddots#1{\cdot^{\cdot^{\cdot^{#1}}}}

\begin{document}

$X$\diagbrace{${x^{\rddots x}}$}{$n-1$ times}

\end{document}

• @Alenanno In this case had the drawback that the 'x' characters were also rotated. Oct 25, 2013 at 14:21
• @Elias I fixed the code. :) Oct 25, 2013 at 15:05

It's not exactly easy to use, but it is adjustable.

\documentclass{article}
\usepackage{mathtools}
\usepackage{mathdots}
\usepackage{tikz}

\newcommand{\dbrace}[1]{
\begin{tikzpicture}
\draw (0,0) node[rotate=45]{$\underbrace{\rule{#1}{0pt}}$};
\end{tikzpicture}}

\begin{document}

$x^{x^{\iddots^x}} \raisebox{-5pt}[0pt][0pt]{\hspace*{-20pt}\dbrace{10pt}} \raisebox{2pt}[0pt][0pt]{\hspace*{-10pt}\scriptstyle{n-1}}$

\end{document}


• Harbour's solution came out after I started but before I finished. Oct 25, 2013 at 15:12