What happened to my underset?

Question

\hat{y} = \hat{f}(x) = \overset{C}{argmax}\underset{c=1}\> p(y = c|x, D)

produces this:

How can I use \underset to get c=1 under argmax?

score 3 · Accepted Answer · answered May 05 '22 at 16:43

3

I would suggest not using \underset and \overset but instead declaring argmax as an operator. In that way, argmax is also typeset correctly.

\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator*{\argmax}{argmax}
\begin{document}
[
\hat{y} = \hat{f}(x) = \argmax_{c=1}^{C}&gt; p(y = c|x, D)
]
\end{document}

answered May 05 '22 at 16:43

1

@kb9alpp This is the correct answer. In KaTeX and LaTeX you can do \operatorname*{argmax}_{c=1}^C that will do the right thing as well. Unfortunately KaTeX doesn't know \DeclareMathOperator. Anyway, KaTeX specific questions are off-topic on the site. – egreg May 05 '22 at 16:54
Perfect!! Thanks! The one line \hat{y} = \hat{f}(x) = \argmax_{c=1}^{C}\> p(y = c|x, D) works. – Ben May 05 '22 at 16:57
I didn't realize the katex tag, sorry. I can't provide a suitable answer for KaTeX. If anyone else can, feel free :) – May 05 '22 at 16:57

score 0 · Answer 2 · answered May 05 '22 at 16:43

Just put the \underset part inside the argument of the \overset:

\documentclass{article}
\usepackage{mathtools}
\begin{document}
[
\hat{y} = \hat{f}(x) = \overset{C}{\underset{c=1}{argmax}}&gt; p(y = c|x, D)
]
\end{document}

Edit: also, argmax isn't typeset correctly, it's an operator and shouldn't be rendered with slanted letters.

What happened to my underset?

2 Answers2