I'm using \Theta for some concept in my work, then I need a variation of such concept.
I was thinking to use \Theta symbol with a square instead of a circle.
How can I obtain such symbol?
Thanks
[Tex/LaTex] \Theta with square
symbols
Related Solutions
The \textperthousand command is available both with the T1 encoding (used by classicthesis) and the TS1 encoding.
However, the Palatino font loaded via the mathpazo package by classicthesis hasn't the required glyph: in the T1 encoding \textperthousand is built by adding a small zero next to % and the small zero is missing (a black square is used to show this).
However the Palatino text companion font has the glyph \textperthousand, so all you need to do is to load textcomp: add
\usepackage{textcomp}
to your document preamble.
Note that \textperthousand is not legal in math mode and produces a warning. You can avoid it by using
\mbox{\textperthousand}
or, better, by loading also amsmath and using
\text{\textperthousand}
You may want to define a variant command that works both in text and math mode:
\usepackage{textcomp}
\usepackage{amsmath}
\DeclareRobustCommand{\perthousand}{%
\ifmmode
\text{\textperthousand}%
\else
\textperthousand
\fi}
Here, I just scaled the \square in the vertical direction by a factor of 1.5 times the width, and called it \tallqed. The \smash prevents it from affecting line spacing. EDITED to reduce size. Note that first argument of \scalebox is the horizontal scaling, while 2nd (optional) argument is the vertical scaling. These can be adjusted to suit.
\documentclass{article}
\usepackage{amssymb,graphicx}
\def\tallqed{\smash{\scalebox{.75}[1.125]{$\square$}}}
\begin{document}
\noindent In view of the choice we made for the orientation of $\partial D$, we conclude that
\[\int_{\partial \mathbf{D}} \iota^*\omega =
(-1)^n \int_{\mathbb{R}^{n-1}} a_n(\cdot,\cdot, \dots, \cdot, 0).\]
This completes the proof of the theorem. \tallqed
\end{document}
WChargin correctly points out that the symbol stretch makes the box border thickness non-uniform on the sides compared with the top/bottom. If that is an issue, the problem can be remedied with a slightly altered definition, by \ooaligning two of the stretched \squares with a slight kern.
\documentclass{article}
\usepackage{amssymb,graphicx}
\def\tallqedX{\smash{\scalebox{.75}[1.125]{$\square$}}}
\def\tallqed{\ooalign{\tallqedX\cr\kern.2pt\tallqedX}}
\begin{document}
\noindent In view of the choice we made for the orientation of $\partial D$, we conclude that
\[\int_{\partial \mathbf{D}} \iota^*\omega =
(-1)^n \int_{\mathbb{R}^{n-1}} a_n(\cdot,\cdot, \dots, \cdot, 0).\]
This completes the proof of the theorem. \tallqed
\end{document}
Best Answer
A solution with TikZ: