# Tag Info

2

I think that, for this cases, it's better to give the command “active” control over the output. So I would use \inv{A}, read as inverse of A. That way you have full control over the output. \newcommand*\inv[1]{#1^{-1}} $\inv{A}$ The difference between A^\inv and \inv{A} is that in the first case you have to adjust \inv to be taken correctly by ^, whereas ...

5

Adding a pair of braces inside the definition of \inv should work. \newcommand{\inv}{{-1}} Now, when you do A^\inv, it will expand to A^{-1} instead of A^-1.

1

You should edit the entries and tell the maintainers of Mendeley to stop outputting such rubbish. Here's a workaround that assumes you don't need \textless anywhere else and that the tags are just <sup>...</sup> and <sub>...</sub>, without nesting. \begin{filecontents*}{\jobname.bib} @article{Testtest2016, author = {Testtest}, title ...

0

The stackengine package provides this feature directly, in the form of \stackanchor. If the stacking type is Short, the gap between the sub and superscript is set to a specified length. This mimics, in form, the natural behavior of the sub and superscripts in math mode. In the case of short stacks, the gap is centered about the math axis. On the other ...

2

With \rlap and some \raisebox because of ascenders and descenders in the us/superscript: \documentclass[12pt]{article} \begin{document} \textbf{Adv}\textsubscript{\raisebox{-1pt}{\rlap{$\Pi$, DP, X}}}\textsuperscript{\raisebox{1pt}{downgrade}} \end{document} Added: The code given above supposes the user knows which is shortest, of the superscript or ...

5

As discussed in numerous comments, the underlying problem was that ^ was invoked in text, not math mode. Here are two possibilities, depending on how you want it to behave. \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \newcommand*\circled[1]{\tikz[baseline=(char.base)]{ \node[shape=circle,fill=blue!20,draw,inner sep=0.5pt] (...

2

From the TeXbook, second doubly dangerous paragraph on page 150: Question: What happens if a subscript or superscript follows a large delimiter? Answer: That's a good question. After a \left delimiter, it is the first subscript or superscript of the enclosed subformula, so it is effectively preceded by {}. After a \right delimiter, it is a subscript or ...

1

not finding an earlier question addressing this situation, here is my suggestion: add a \vphantom to the numerator of the fraction on the left to make it appear to be the same size as the fraction on the right: N = \left( \frac{s\vphantom{d}}{2} \right)^2 - \left( \frac{d}{2} \right)^2

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