Stacking Symbols for a Math Operator [duplicate]
This question already has an answer here:
Superimpose letter on integral symbol
3 answers
I want to create a math operator that is an integral with a square (square) stack onto it such that the integral is still formated the way that the operator int is. How do I do this?
math-operators
marked as duplicate by egreg, Kurt, Stefan Pinnow, TeXnician, Circumscribe Jan 24 at 18:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Superimpose letter on integral symbol
3 answers
I want to create a math operator that is an integral with a square (square) stack onto it such that the integral is still formated the way that the operator int is. How do I do this?
math-operators
marked as duplicate by egreg, Kurt, Stefan Pinnow, TeXnician, Circumscribe Jan 24 at 18:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Related:subseteq+circas a single symbol ("open subset")
– Werner
Jan 24 at 17:30
I took tex.stackexchange.com/a/171445/4427 and changedRintosquare, getting what probably is your desired symbol.
– egreg
Jan 24 at 17:39
add a comment |
This question already has an answer here:
Superimpose letter on integral symbol
3 answers
I want to create a math operator that is an integral with a square (square) stack onto it such that the integral is still formated the way that the operator int is. How do I do this?
math-operators
This question already has an answer here:
Superimpose letter on integral symbol
3 answers
I want to create a math operator that is an integral with a square (square) stack onto it such that the integral is still formated the way that the operator int is. How do I do this?
This question already has an answer here:
Superimpose letter on integral symbol
3 answers
math-operators
math-operators
edited Jan 24 at 17:36
Steven B. Segletes
154k9194402
154k9194402
asked Jan 24 at 17:20
LinearGuyLinearGuy
61
61
marked as duplicate by egreg, Kurt, Stefan Pinnow, TeXnician, Circumscribe Jan 24 at 18:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by egreg, Kurt, Stefan Pinnow, TeXnician, Circumscribe Jan 24 at 18:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Related:subseteq+circas a single symbol ("open subset")
– Werner
Jan 24 at 17:30
I took tex.stackexchange.com/a/171445/4427 and changedRintosquare, getting what probably is your desired symbol.
– egreg
Jan 24 at 17:39
add a comment |
Related:subseteq+circas a single symbol ("open subset")
– Werner
Jan 24 at 17:30
I took tex.stackexchange.com/a/171445/4427 and changedRintosquare, getting what probably is your desired symbol.
– egreg
Jan 24 at 17:39
Related:
subseteq + circ as a single symbol ("open subset")– Werner
Jan 24 at 17:30
Related:
subseteq + circ as a single symbol ("open subset")– Werner
Jan 24 at 17:30
I took tex.stackexchange.com/a/171445/4427 and changed
R into square, getting what probably is your desired symbol.– egreg
Jan 24 at 17:39
I took tex.stackexchange.com/a/171445/4427 and changed
R into square, getting what probably is your desired symbol.– egreg
Jan 24 at 17:39
add a comment |
3 Answers
3
active
oldest
votes
documentclass{article}
usepackage{esint}
begin{document}
[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}

add a comment |
Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward
documentclass[a5paper]{article}
usepackage{graphicx}
usepackage{amsmath}
usepackage{amssymb}
makeatletter
letDOTSIrelax % amsmath support for dots
newcommand*{sqint}{%
DOTSI
mathop{%
mathpalette@LetterOnInt{square}%
}%
mkern-thinmuskip % thin space is inserted between two mathop
int
}
newcommand*{@LetterOnInt}[2]{%
sbox0{$#1intm@th$}%
sbox2{$%
ifx#1displaystyle
textstyle
else
scriptscriptstyle
fi
#2%
m@th$}%
dimen@=.4dimexprht0+dp0relax
ifdimdimexprht2+dp2relax>dimen@
sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
fi
dimen@=wd0 %
ifdimwd2>dimen@
dimen@=wd2 %
fi
rlap{hbox to dimen@{hfil
$#1vcenter{copy2}m@th$%
hfil}}%
ifdimdimen@>wd0 %
kern.5dimexprdimen@-wd0relax
fi
}
makeatother
begin{document}
Take the regulated function $f$, we want to compare the Riemann integral,
$int_a^b f$
to the regulated integral $sqint_a^b f$ by taking
sequences of step
functions.
We will prove that
[
sqint_a^b f = int_a^b f
]
[
displaystyle int_a^b f dots sqint_a^b f qquad
textstyle int_a^b f dots sqint_a^b f qquad
scriptstyle int_a^b f dots sqint_a^b f qquad
scriptscriptstyle int_a^b f dots sqint_a^b f
]
end{document}

add a comment |
Tweaked for CM font.
documentclass{article}
usepackage{amssymb,graphicx}
newcommandsqint{mathchoice
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
{scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
{scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
int}
begin{document}
$displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$sqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
end{document}

add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
documentclass{article}
usepackage{esint}
begin{document}
[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}

add a comment |
documentclass{article}
usepackage{esint}
begin{document}
[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}

add a comment |
documentclass{article}
usepackage{esint}
begin{document}
[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}

documentclass{article}
usepackage{esint}
begin{document}
[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}

answered Jan 24 at 17:49
HerbertHerbert
273k24412725
273k24412725
add a comment |
add a comment |
Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward
documentclass[a5paper]{article}
usepackage{graphicx}
usepackage{amsmath}
usepackage{amssymb}
makeatletter
letDOTSIrelax % amsmath support for dots
newcommand*{sqint}{%
DOTSI
mathop{%
mathpalette@LetterOnInt{square}%
}%
mkern-thinmuskip % thin space is inserted between two mathop
int
}
newcommand*{@LetterOnInt}[2]{%
sbox0{$#1intm@th$}%
sbox2{$%
ifx#1displaystyle
textstyle
else
scriptscriptstyle
fi
#2%
m@th$}%
dimen@=.4dimexprht0+dp0relax
ifdimdimexprht2+dp2relax>dimen@
sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
fi
dimen@=wd0 %
ifdimwd2>dimen@
dimen@=wd2 %
fi
rlap{hbox to dimen@{hfil
$#1vcenter{copy2}m@th$%
hfil}}%
ifdimdimen@>wd0 %
kern.5dimexprdimen@-wd0relax
fi
}
makeatother
begin{document}
Take the regulated function $f$, we want to compare the Riemann integral,
$int_a^b f$
to the regulated integral $sqint_a^b f$ by taking
sequences of step
functions.
We will prove that
[
sqint_a^b f = int_a^b f
]
[
displaystyle int_a^b f dots sqint_a^b f qquad
textstyle int_a^b f dots sqint_a^b f qquad
scriptstyle int_a^b f dots sqint_a^b f qquad
scriptscriptstyle int_a^b f dots sqint_a^b f
]
end{document}

add a comment |
Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward
documentclass[a5paper]{article}
usepackage{graphicx}
usepackage{amsmath}
usepackage{amssymb}
makeatletter
letDOTSIrelax % amsmath support for dots
newcommand*{sqint}{%
DOTSI
mathop{%
mathpalette@LetterOnInt{square}%
}%
mkern-thinmuskip % thin space is inserted between two mathop
int
}
newcommand*{@LetterOnInt}[2]{%
sbox0{$#1intm@th$}%
sbox2{$%
ifx#1displaystyle
textstyle
else
scriptscriptstyle
fi
#2%
m@th$}%
dimen@=.4dimexprht0+dp0relax
ifdimdimexprht2+dp2relax>dimen@
sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
fi
dimen@=wd0 %
ifdimwd2>dimen@
dimen@=wd2 %
fi
rlap{hbox to dimen@{hfil
$#1vcenter{copy2}m@th$%
hfil}}%
ifdimdimen@>wd0 %
kern.5dimexprdimen@-wd0relax
fi
}
makeatother
begin{document}
Take the regulated function $f$, we want to compare the Riemann integral,
$int_a^b f$
to the regulated integral $sqint_a^b f$ by taking
sequences of step
functions.
We will prove that
[
sqint_a^b f = int_a^b f
]
[
displaystyle int_a^b f dots sqint_a^b f qquad
textstyle int_a^b f dots sqint_a^b f qquad
scriptstyle int_a^b f dots sqint_a^b f qquad
scriptscriptstyle int_a^b f dots sqint_a^b f
]
end{document}

add a comment |
Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward
documentclass[a5paper]{article}
usepackage{graphicx}
usepackage{amsmath}
usepackage{amssymb}
makeatletter
letDOTSIrelax % amsmath support for dots
newcommand*{sqint}{%
DOTSI
mathop{%
mathpalette@LetterOnInt{square}%
}%
mkern-thinmuskip % thin space is inserted between two mathop
int
}
newcommand*{@LetterOnInt}[2]{%
sbox0{$#1intm@th$}%
sbox2{$%
ifx#1displaystyle
textstyle
else
scriptscriptstyle
fi
#2%
m@th$}%
dimen@=.4dimexprht0+dp0relax
ifdimdimexprht2+dp2relax>dimen@
sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
fi
dimen@=wd0 %
ifdimwd2>dimen@
dimen@=wd2 %
fi
rlap{hbox to dimen@{hfil
$#1vcenter{copy2}m@th$%
hfil}}%
ifdimdimen@>wd0 %
kern.5dimexprdimen@-wd0relax
fi
}
makeatother
begin{document}
Take the regulated function $f$, we want to compare the Riemann integral,
$int_a^b f$
to the regulated integral $sqint_a^b f$ by taking
sequences of step
functions.
We will prove that
[
sqint_a^b f = int_a^b f
]
[
displaystyle int_a^b f dots sqint_a^b f qquad
textstyle int_a^b f dots sqint_a^b f qquad
scriptstyle int_a^b f dots sqint_a^b f qquad
scriptscriptstyle int_a^b f dots sqint_a^b f
]
end{document}

Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward
documentclass[a5paper]{article}
usepackage{graphicx}
usepackage{amsmath}
usepackage{amssymb}
makeatletter
letDOTSIrelax % amsmath support for dots
newcommand*{sqint}{%
DOTSI
mathop{%
mathpalette@LetterOnInt{square}%
}%
mkern-thinmuskip % thin space is inserted between two mathop
int
}
newcommand*{@LetterOnInt}[2]{%
sbox0{$#1intm@th$}%
sbox2{$%
ifx#1displaystyle
textstyle
else
scriptscriptstyle
fi
#2%
m@th$}%
dimen@=.4dimexprht0+dp0relax
ifdimdimexprht2+dp2relax>dimen@
sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
fi
dimen@=wd0 %
ifdimwd2>dimen@
dimen@=wd2 %
fi
rlap{hbox to dimen@{hfil
$#1vcenter{copy2}m@th$%
hfil}}%
ifdimdimen@>wd0 %
kern.5dimexprdimen@-wd0relax
fi
}
makeatother
begin{document}
Take the regulated function $f$, we want to compare the Riemann integral,
$int_a^b f$
to the regulated integral $sqint_a^b f$ by taking
sequences of step
functions.
We will prove that
[
sqint_a^b f = int_a^b f
]
[
displaystyle int_a^b f dots sqint_a^b f qquad
textstyle int_a^b f dots sqint_a^b f qquad
scriptstyle int_a^b f dots sqint_a^b f qquad
scriptscriptstyle int_a^b f dots sqint_a^b f
]
end{document}

answered Jan 24 at 17:45
community wiki
egreg
add a comment |
add a comment |
Tweaked for CM font.
documentclass{article}
usepackage{amssymb,graphicx}
newcommandsqint{mathchoice
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
{scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
{scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
int}
begin{document}
$displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$sqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
end{document}

add a comment |
Tweaked for CM font.
documentclass{article}
usepackage{amssymb,graphicx}
newcommandsqint{mathchoice
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
{scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
{scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
int}
begin{document}
$displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$sqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
end{document}

add a comment |
Tweaked for CM font.
documentclass{article}
usepackage{amssymb,graphicx}
newcommandsqint{mathchoice
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
{scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
{scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
int}
begin{document}
$displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$sqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
end{document}

Tweaked for CM font.
documentclass{article}
usepackage{amssymb,graphicx}
newcommandsqint{mathchoice
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
{raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
{scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
{scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
int}
begin{document}
$displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$sqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip
$scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
end{document}

answered Jan 24 at 17:36
Steven B. SegletesSteven B. Segletes
154k9194402
154k9194402
add a comment |
add a comment |
Related:
subseteq+circas a single symbol ("open subset")– Werner
Jan 24 at 17:30
I took tex.stackexchange.com/a/171445/4427 and changed
Rintosquare, getting what probably is your desired symbol.– egreg
Jan 24 at 17:39