Command for argmin or argmax?

Question

How can you get the argmin or argmax in Latex? Two solutions I found are:

\underset{x}{\operatorname{argmax}} 

\DeclareMathOperator*{\argmin}{arg\,min}

Any other ideas?

[1] http://www.breakthru.altervista.org/?p=27

[2] http://researchonsearch.blogspot.com/2007/05/enter-argmax-argmin-in-latex.html

How do these solutions (especially the second) feel not right? The \DeclareMathOperator is the way to go in this case. — Pieter, Nov 11 '10 at 17:57
Personally, I like the look of \text{arg}\,\max\limits_{\theta}\, — Jean-Paul, Sep 07 '16 at 08:11

score 343 · Accepted Answer · edited Feb 08 '18 at 11:52

343

As Pieter pointed out, the correct way to define argmin and argmax operators in LaTeX is:

\usepackage{amsmath}
\DeclareMathOperator*{\argmax}{arg\,max}
\DeclareMathOperator*{\argmin}{arg\,min}

Actually, using the amsopn package would be sufficient but it's loaded by amsmath internally, which is recommended for math typesetting anyway.

The * in \DeclareMathOperator* places the underscored argument underneath the word rather than to the bottom right of it.

edited Feb 08 '18 at 11:52

danijar

1,005

answered Nov 12 '10 at 04:05

Jan Hlavacek

19,242

57

If you use \DeclareMathOperator*{\argmin}{\arg\!\min} you get rid of the ugly space between arg and min. – rbp Apr 06 '11 at 21:39
37

@rbp Why should one use \arg\min when simply argmin does the same? – egreg Jul 08 '12 at 15:31
4

@egreg you're right. I arrived at this one from some experimentation with other options, and didn't look back :) – rbp Jul 09 '12 at 11:56
5

In ConTeXt that is: \definemathcommand [argmin] [limop] {\mfunction{arg\,min}} – mb21 Jul 30 '13 at 11:49
10

Explanation: I left the thin space (,) between the arg and min because that's what the original poster had in the question, so I assumed that they wanted the space there. Personally I would just do \DeclareMathOperator*{\argmin}{argmin}, without the space. – Jan Hlavacek Aug 28 '15 at 01:14
How can I make it work also in a context like $\argmin_x{v(x)}$ i.e. when the code is between $$ and not in equations? – Donbeo Mar 23 '16 at 15:07
what about how to make it work in Jupyter? – Charlie Parker Aug 12 '19 at 03:15
@CharlieParker This should work with MathJaX in Jupyter, and KaTeX already has \argmax. What does not work for you in Jupyter? – mforbes Dec 06 '22 at 18:15

score 173 · Answer 2 · edited Jun 10 '20 at 12:32

Summary

In the sequel, LaTeX is assumed.

Preliminary questions

1. Should a thin space separate “arg” from “min”?

This is a matter of personal preference. The default definition for \limsup has a thin space, on the other hand “arcsin” is usually written without a space between the two components. Consistency is, as always, the keyword: using a macro definition will ensure it and also easiness in modifying the typesetting in the whole document, if switching from one option to another is needed.

2. Should limits go below the operator in display style?

Also this is debatable and conventions used in the field the document is written for should be followed.

3. Should the operator name be typeset upright?

Definitely, like all other operator/function names such as sine and cosine.

Available tools

1. `amsmath`

With amsmath one can do, in the preamble,

\DeclareMathOperator*{\argmin}{arg\,min} % thin space, limits underneath in displays
\DeclareMathOperator*{\argmin}{argmin} % no space, limits underneath in displays
\DeclareMathOperator{\argmin}{arg\,min} % thin space, limits on side in displays
\DeclareMathOperator{\argmin}{argmin} % no space, limits on side in displays

Of course, only one of these should be used. In the document it's sufficient to type \argmin and the spacing around the operator will be the right one according to the common rules of math typesetting.

The seemingly different

\newcommand{\argmin}{\operatornamewithlimits{argmin}}

is not really so, because it's essentially the same as \DeclareMathOperator*{\argmin}{argmin}; moreover it uses a deprecated command, that should be \operatorname*.

2. No package

The following definitions are very similar to the ones above, in the same order

\newcommand{\argmin}{\mathop{\mathrm{arg\,min}}
\newcommand{\argmin}{\mathop{\mathrm{argmin}}
\newcommand{\argmin}{\mathop{\mathrm{arg\,min}\nolimits}
\newcommand{\argmin}{\mathop{\mathrm{argmin}\nolimits}

With `\underset`

\underset{x}{\mathrm{argmin}}

This also was mentioned in the question and in one answer, but it is wrong, as the visual comparison will show.

3. Differences

The “no package” version is much more rigid than the amsmath version, because this package can receive the nonamelimits option that will automatically change all operators such as \lim, \max that usually have limits underneath (in displays) to have them on the side. The same option would of course act also on newly defined operators, provided \DeclareMathOperator* has been used.

Test document

In the following test document, the proposed definitions or constructions will be compared.

\documentclass{article}
\usepackage{amsmath}
% limits underneath
\DeclareMathOperator{\argminA}{arg,min} % Jan Hlavacek
\DeclareMathOperator{\argminB}{argmin}   % Jan Hlavacek
\DeclareMathOperator*{\argminC}{\arg\min}   % rbp
\newcommand{\argminD}{\arg!\min} % AlfC
\newcommand{\argminE}{\mathop{\mathrm{argmin}}}          % ASdeL
\newcommand{\argminF}{\mathop{\mathrm{argmin}}\limits}   % ASdeL
% limits on side
\DeclareMathOperator{\argminG}{arg,min} % Jan Hlavacek
\DeclareMathOperator{\argminH}{argmin}   % Jan Hlavacek
\newcommand{\argminI}{\mathop{\mathrm{argmin}}\nolimits} % ASdeL
\newcommand{\cs}[1]{\texttt{\symbol{`\}#1}}
\begin{document}
\begin{align}
&\cs{argminA}  & \argminA_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminB}  & \argminB_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminC}  & \argminC_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminD}  & \argminD_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminE}  & \argminE_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminF}  & \argminF_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{underset} & \underset{x}{\mathrm{argmin}} f(x) &= {x \mid f(x) = \min_{x'} f(x')}\
&\cs{argminG}  & \argminG_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminH}  & \argminH_x f(x) &= {x \mid f(x) = \min_{x'} f(x')} \
&\cs{argminI}  & \argminI_x f(x) &= {x \mid f(x) = \min_{x'} f(x')}
\end{align}
\end{document}

Results

As predicted, many lines typeset the same. However, the macros defined with \mathop will not obey the nonamelimits option. Also the result of \argminD is clearly wrong, as the limit is typeset below “max” and not below the whole operator name.

The construction with \underset is wrong because it will not use the correct spacing after the operator: a thin space follows all others and it is the right way to typeset. Compare \sin x and \mathrm{sin}x to see this thin space in a different context.

Note that \argminC and \argminA give the same result, as well as \argminD and \argminB. Explanation: \arg and \max are already defined as operators, so TeX inserts a thin space if one directly follows another one. Using \argminC or \argminD just makes TeX spin its wheels a little more, with no advantage over \argminA or \argminB.

Isn't \newcommand{\argmin}{\operatornamewithlimits{argmin}} different in that it doesn't use amsmath package? — Homero Esmeraldo, May 15 '19 at 15:00
@HomeroEsmeraldo \operatornamewithlimits is an obsolete command defined by amsopn (a package loaded by amsmath, rarely used by itself). — egreg, May 15 '19 at 15:08
Thanks. For some reason only \underset gave satisfactory results for with (with x being under argmin, not adjacent to it). — Dan M., May 23 '19 at 02:15
@CharlieParker What's Jupyter? If it uses MathJax, then it's not LaTeX. — egreg, Aug 12 '19 at 08:44
In Python code documentation I used the underset method including a \; to get the right spacing. It is the best approach for Sphinx mathjax extension. — Attack68, Oct 25 '22 at 18:49
Indeed, but this answer and my comment are helpful to Mathjax users who may stumble across it as I did when searching. There are limited, if any, answers to the same question in a Mathjax context. — Attack68, Oct 26 '22 at 20:31
@Attack68 I just tried with MathJax on Math.SE and \DeclareMathOperator*{\argmax}{arg\,max} provides the expected result. Code (click) Output (click} — egreg, Oct 26 '22 at 20:49

score 40 · Answer 3 · answered May 25 '17 at 16:54

40

Use \usepackage{amsmath} and then:

\operatorname*{argmin}_\theta f(x)

Or similarly (for adding a space),

\operatorname*{arg\,max}_\theta f(x)

answered May 25 '17 at 16:54

iamaziz

601

concise and to the point! – Shawyan Azdam Aug 17 '20 at 07:54
1

It works when used in \begin{equation} but not using $$ – martin36 Nov 15 '23 at 08:26

score 33 · Answer 4 · answered May 22 '12 at 19:55

33

I use \newcommand{\argmin}{\operatornamewithlimits{argmin}}.

answered May 22 '12 at 19:55

Michael Litvin

431

4

+1 I like this - why overcomplicate things? This looks exactly the same as the \DeclareMathOperator* answer. – zelanix Aug 13 '14 at 13:52
5

This yields "argmin". To get "arg min" (arguably the most usual notation), set it to {arg\,min}. – Waldir Leoncio Jan 21 '16 at 16:38
3

Google gave me 669,000 answers for "arg min" and 520,000 answers for "argmin", so I'd call it a tie. ;-) – PatrickT Dec 03 '18 at 04:49
this does work...? is there a missing import statement? – Charlie Parker Aug 12 '19 at 03:16
1

Google Scholar got me 171,000 for "arg min" and 116,000 for "argmin". – Lori Feb 07 '21 at 06:02

alfC · Answer 5 · 2018-12-13T06:43:38.417

9

Just another alternative (in some sense the poorer solution around, see comment below), could be to define \argmin in terms of \min and \arg commands.

\newcommand{\argmin}{\arg\!\min}

In this way, 1) \argmin will behave ~~always~~ the same way as \min, 2) doesn't need amsmath or care about \operator... commands 3) yes, the variable in not centered (it is centered in the min part), but that may even be what you want (since it is centered in the same ways a \min, also the 'g' in \arg doesn't further lower the under-argument).

\documentclass[fleqn]{article}
\newcommand{\argmin}{\arg\!\min}
\begin{document}

\[ \argmin_x f(x) = \{x | f(x) = \min_{x'} f(x')\} \]
\[ \min_x f(x) = \{f(x) | f(x) < f(x_0) \forall x_0 \in R \]

\end{document}

argmin

edited Dec 13 '18 at 06:43

answered Nov 27 '13 at 23:15

alfC

14,350

5

This doesn't look like a good idea. \argmin_{x\in X\cap Y} f(x) will break it completely. – yo' Feb 02 '14 at 18:03
What is the \arg command for anyway? – Thomas Ahle Jun 03 '14 at 12:47
1

@ThomasAhle, arg is for the argument function (angle in the complex plane, $z = |z| e^{i \arg z}$. tohecz is right by the way. – alfC Jun 03 '14 at 15:33
@boycott.se-yo', good point. – alfC Dec 13 '18 at 06:44

score 6 · Answer 6 · answered Sep 19 '13 at 13:07

6

or you could use the underset command. For example:

    R = \underset{n} {\mathrm{argmax}} ~P(L_n|\mathbf{x})

answered Sep 19 '13 at 13:07

drben

77

13

Please don't use this. This makes argmin a mathord rather than a mathop and the spacing is incorrect. Moreover, this will not automatically adjust the limits in inline math mode. – Aditya Sep 19 '13 at 15:19

score 6 · Answer 7 · answered Feb 18 '21 at 05:03

Using \underset does not require any other package. An above answer says:

With \underset

This also was mentioned in the question and in one answer, but it is wrong, as the visual comparison will show.

If the only thing "wrong" is that there is no space between the argmax and the f(x), then just add \, (thin space) to fix the problem:

z = \underset{x}{\mathrm{argmax}}f(x)

z = \underset{x}{\mathrm{argmax}}\, f(x)

score 2 · Answer 8 · answered Sep 21 '15 at 13:35

I would use \mathop and \mathrm, and eventually \limits or \nolimits, depending on the desired behavior when combined with _ and ^. See below.

\documentclass{article}

\begin{document}

{
nothing:
\newcommand{\argmin}{\mathop{\mathrm{argmin}}}
$\argmin_a^b$
$$\argmin_a^b$$
}

{
limits:
\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}
$\argmin_a^b$
$$\argmin_a^b$$
}

{
nolimits:
\newcommand{\argmin}{\mathop{\mathrm{argmin}}\nolimits}
$\argmin_a^b$
$$\argmin_a^b$$
}

\end{document}

score -2 · Answer 9 · answered Dec 20 '15 at 20:12

-2

This works without any additional package. However it will create a new line.

$$\pi(s) = arg\max_\theta$$

answered Dec 20 '15 at 20:12

Miae Kim

107

1

$$...$$ is not recommended to use with LaTeX. Better use \[ ... \] for correct vertical spacing. – Stefan Kottwitz Dec 20 '15 at 20:17
5

This solution will not typeset argmax correctly. Here arg will be in italic (and with bad spacing), and max will be upright. – Håkon Marthinsen Dec 20 '15 at 20:17

Command for argmin or argmax?

9 Answers9

Summary

Preliminary questions

1. Should a thin space separate “arg” from “min”?

2. Should limits go below the operator in display style?

3. Should the operator name be typeset upright?

Available tools

1. `amsmath`

2. No package

With `\underset`

3. Differences

Test document

Results

Linked

Related

Command for argmin or argmax?

9 Answers9

Summary

Preliminary questions

1. Should a thin space separate “arg” from “min”?

2. Should limits go below the operator in display style?

3. Should the operator name be typeset upright?

Available tools

1. amsmath

2. No package

With \underset

3. Differences

Test document

Results

Linked

Related

1. `amsmath`

With `\underset`