44
$\begingroup$

Basically the same as Formatting Sandbox in Meta Stack Exchange, but since this and Statistical Analysis are the only two sites (I know) supporting $\TeX$ formatting, I believe we also need one here for testing it.

$\endgroup$
15
  • 2
    $\begingroup$ Theoretical computer science also supports $\mathrm{\TeX/\LaTeX}$ formatting. $\endgroup$
    – JeffE
    Jun 1, 2012 at 7:29
  • 1
    $\begingroup$ @JeffE: You can use $\TeX$ and $\LaTeX$ (\Tex and \LaTeX) for the text. $\endgroup$
    – Asaf Karagila Mod
    Jun 2, 2012 at 21:09
  • 3
    $\begingroup$ @JeffE: In 2010 only 'stats' and 'math' support TeX formatting. Of course now there is also 'cstheory', 'cs', 'chemistry', 'quant', etc. $\endgroup$
    – kennytm
    Jun 3, 2012 at 6:25
  • $\begingroup$ test $$\begin{align*}\text{middle line}\end{align*}$$ new line $\endgroup$
    – Ruslan
    Jan 30, 2014 at 13:06
  • $\begingroup$ test test $\not\in(1)\notin(2)$ Who's better??? $\endgroup$
    – user93957
    Jan 31, 2014 at 22:25
  • $\begingroup$ $m^n + m^x + m^n = 555555$ test test $\endgroup$
    – hichris123
    Feb 2, 2014 at 19:09
  • $\begingroup$ line $\begin{array}\phantom{i}\\\phantom{i} \end{array} $ line2 $\endgroup$ Apr 18, 2014 at 15:15
  • $\begingroup$ quotes test ${``}=\text{’’}$ $\endgroup$
    – Ruslan
    Oct 25, 2017 at 11:13
  • 1
    $\begingroup$ $a``=\!\!\text{’’}b$ $\endgroup$
    – Ruslan
    Oct 25, 2017 at 11:17
  • $\begingroup$ I would understand the newly covered elements at the time we choose $S_3$ should be $S_3\setminus(S_1\cup S_2)$ instead of $S_3\setminus(S_1\cap S_2\cap S_3)$. $\endgroup$
    – Apass.Jack
    Aug 25, 2018 at 20:51
  • $\begingroup$ $(\begin{smallmatrix}1\\-1\end{smallmatrix})$ $\endgroup$ Jun 26, 2019 at 20:45
  • $\begingroup$ yes$%-------------$ $\endgroup$
    – Ruslan
    Oct 12, 2020 at 9:58
  • $\begingroup$ yes$%-------------$ $\endgroup$
    – Ruslan
    Oct 12, 2020 at 10:00
  • 1
    $\begingroup$ @AnindyaPrithvi: this looks like an attempt at making hard to edit comments. You know how to do it, now it should be deleted. $\endgroup$
    – robjohn Mod
    Nov 4, 2020 at 13:15
  • 1
    $\begingroup$ @robjohn The comment can be edited. The option comes over the MathJAX (including the delete option). Although you may call it hard to flag .Also, funnily I did not receive the mention for the above comment... $\endgroup$ Nov 4, 2020 at 13:42

58 Answers 58

25
$\begingroup$

A suggestion: if you want to see you TeX previewed, pretend to type your question/answer. Then wait for 4 seconds. We have on the fly previewing for LaTeX here. This way we don't keep popping this question to the top of meta.

$\endgroup$
4
  • 5
    $\begingroup$ May be this (and the main sandbox) should be made special unbumpable question? $\endgroup$
    – Vi0
    Aug 24, 2012 at 15:06
  • 3
    $\begingroup$ Except that there's no preview for comments. $\endgroup$
    – shoover
    Sep 24, 2014 at 18:22
  • 1
    $\begingroup$ and no preview for bounty texts... $\endgroup$
    – draks ...
    Mar 15, 2017 at 7:35
  • $\begingroup$ I guess this works most of the times - but there are quite a few bug reports pointing out that in some cases MathJax is displayed differently in the preview and in the final post. For example, a few of them are mentioned here: MathJax escape sequences don't work. $\endgroup$ Mar 2 at 4:28
33
$\begingroup$

Testing alternate way of implementing spoiler

$$ \require{action} \require{enclose} \toggle{ x\cdot 0 = 0\quad\enclose{roundedbox}{\text{ Click this for derivation }} }{ \begin{array}{rll} x\cdot 0 &= \mathtip{x\cdot 0 + 0}{0 \text{ is additive identity}} \\ &= \mathtip{x\cdot 0 + (x\cdot 0 + -(x\cdot 0))}{ -(x\cdot 0) \text{ is additive inverse of } x\cdot 0}\\ &= \mathtip{(x\cdot 0 + x\cdot 0) + -(x\cdot 0)}{ \text{ addition is associative }\;}\\ &= \mathtip{x\cdot(0 + 0) + -(x\cdot 0) }{ \text{ mulitplication is distributive }\;}\\ &= \mathtip{x\cdot 0 + -(x\cdot 0) }{ 0 \text{ is additive identity}} \\ &= \mathtip{0}{ -(x\cdot 0) \text{ is additive inverse of } x\cdot 0} \end{array} \quad\quad \bbox[4pt,border: 1px solid red]{ \begin{array}{l} \text{If you cannot figure out why a line}\\ \text{is true, move your mouse over}\\ \text{RHS of that line for hint.} \end{array}} }\endtoggle $$


This test uses the MathJAX extension Action, Enclose and BBox. The BBox seems to be automatically loaded. To use Action and Enclose. put \require{action} and require{enclose} somewhere between the $$.

  • the \enclose{roundedbox}{...} draws a rounded text around ....
  • the \texttip{math}{tip} and \mathtip{math}{tip} add a tooltip tip to a piece of math. The difference between textip and mathtip is the tip will be rendered in text and math mode respectively.
  • the \bbox[4pt,border: 1px outset red]{...} draws a red border with 4pt as padding around a piece of ...

Observation

  • toggle works properly.
  • even tooltip works, sometimes it doesn't go away properly.
  • missing a good construct to put multi-line text in math mode. \parbox doesn't work???

Fulling list of above test given below.

$$
\require{action}
\require{enclose}
\toggle{ 
x\cdot 0 = 0\quad\enclose{roundedbox}{\text{ Click this for derivation }}
}{
\begin{array}{rll}
x\cdot 0 
&= \mathtip{x\cdot 0 + 0}{0 \text{ is additive identity}} \\
&= \mathtip{x\cdot 0 + (x\cdot 0 + -(x\cdot 0))}{ -(x\cdot 0) \text{ is additive inverse of } x\cdot 0}\\
&= \mathtip{(x\cdot 0 + x\cdot 0) + -(x\cdot 0)}{ \text{ addition is associative }\;}\\
&= \mathtip{x\cdot(0 + 0) + -(x\cdot 0) }{ \text{ mulitplication is distributive }\;}\\
&= \mathtip{x\cdot 0 + -(x\cdot 0) }{ 0 \text{ is additive identity}} \\
&= \mathtip{0}{ -(x\cdot 0) \text{ is additive inverse of } x\cdot 0}
\end{array}
\quad\quad
\bbox[4pt,border: 1px solid red]{
\begin{array}{l}
\text{If you cannot figure out why a line}\\
\text{is true, move your mouse over}\\
\text{RHS of that line for hint.}
\end{array}}
}\endtoggle
$$
$\endgroup$
0
10
$\begingroup$

This answer is free for anyone to edit.

$\endgroup$
3
  • 1
    $\begingroup$ a comment with overlaps $\rlap{\color{red}{\Rule{10em}{1em}{0.5em}}}$ $\endgroup$ Jun 14, 2012 at 21:56
  • $\begingroup$ The extension linked to this answer can be used to improve the situation. $\endgroup$ Jun 14, 2012 at 22:00
  • 3
    $\begingroup$ $\rlap{\color{grey}{\Rule{200em}{1em}{0.75em}}}$ $\endgroup$
    – user93957
    Jan 7, 2014 at 13:01
8
$\begingroup$

Move your mouse around each symbol to know which font was used:

$$\require{action} \overset{\rlap{\overset{\,{\rlap{\overset{\overset{\overset{\color{red}{\rlap{\color{\green}{\,\,\star}}{\Rule{1em}{0.5em}{0.25em}}}}{}}{}}{}}{\Huge|}}}{}}{\Rule{0.5em}{0.25em}{0.05em}}}{\overset{\Rule{2em}{0.05em}{0.05em}}{\overset{\Rule{5em}{0.05em}{0.05em}}{\overset{\Rule{9em}{0.05em}{0.05em}}{\overset{\Rule{14em}{0.05em}{0.05em}} {\overline{\left\rfloor\left\rfloor\overset{\underline{{\displaystyle \Huge {\scr F\sf o\rm n\cal t\frak s}} }}{\underline{\underline{\underline{\underline{\underline{\underline{\left[\overline{\begin{matrix} \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\rm K}}}{\text{\rm}}\,\overset{\displaystyle f(c)}{} &\qquad \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\cal K}}}{\text{\cal}} \,\overset{\displaystyle f(c)}{}&\qquad \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\sf K}}}{\text{\sf}}\,\overset{\displaystyle f(c)}{} \\ \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\tt K}}}{\text{\tt}}\,\overset{\displaystyle f(c)}{} &\qquad \mathtip{\overset{ \ \ \, \infty}{\underset{j=0}{\LARGE\it K}}}{\text{\it}}\,\overset{\displaystyle f(c)}{} &\qquad \mathtip{\overset{\quad\infty}{\underset{j=0}{\LARGE\scr K}}}{\text{\scr}}\,\overset{\displaystyle f(c)}{} \\ \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\bf K}}}{\text{\bf}}\,\overset{\displaystyle f(c)}{} &\qquad \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\frak K}}}{\text{\frak}}\,\overset{\displaystyle f(c)}{} &\qquad \mathtip{\overset{\infty}{\underset{j=0}{\LARGE\Bbb K}}}{\text{\Bbb}}\,\overset{\displaystyle f(c)}{} \\\end{matrix}}\right]}}}}}}}\right\lfloor\right\lfloor}}}}}} $$ where $\:\color{red}{\rlap{\color{\green}{\,\,\tiny\star}}{\Rule{1em}{0.5em}{0.25em}}}\:$ is the flag of my country. $\overset{\cdot\cdot}\smile$

$\endgroup$
1
  • 1
    $\begingroup$ Interestingly, the star is behind the red rectangle for me on Chromium@Linux, so the flag looks just plain red. $\endgroup$
    – Ruslan
    Nov 1, 2021 at 9:41
6
$\begingroup$

What a lovely diagram. $$\require{AMScd} \begin{CD} H \otimes M @>{\rho}>> M @>{\delta}>> H \otimes M \\ @V{\Delta^2 \otimes \delta}VV @. @AA{m^2 \otimes \rho}A \\ H^{\otimes 4} \otimes M @>>{\mathbb{1} \otimes T \otimes \mathbb{1}}> H^{\otimes 4} \otimes M @>>{\mathbb{1}\otimes\mathbb{1}\otimes S \otimes\mathbb{1}\otimes\mathbb{1}}> H^{\otimes 4} \otimes M \end{CD} $$

$\endgroup$
1
  • 1
    $\begingroup$ Could be used for the depiction of Hess cycle/Born-Haber cycle/other thermodynamic cycles at Chem.SE. $\endgroup$ Feb 12, 2018 at 13:31
6
$\begingroup$

Following this . . .

$$\begin{array}{|lc} 1., p\lor(q\land r) & \text{Assumption} \ \hline \rlap{\begin{array}{|lc} 2., p & \text{Assumption} \ \hline \rlap{\begin{array}{|lc} 3., q\land r & \text{Assumption} \ 4., p & \text{2} \ 5., (q\land r)\to p & \text{MP(3,4)} \ 6., ((q\land r)\to p)\to p & \to\text{-intro(5,2)} \end{array}}& \

s\to r & \ & \ q\to (s\to r) & \end{array}}&\ (s\to p)\to (q\to (s\to r)). & \end{array}$$

$\endgroup$
4
  • $\begingroup$ [Click on this][1] [1]: math.stackexchange.com/questions/1029759/… $\endgroup$
    – ahorn
    Mar 29, 2015 at 14:42
  • $\begingroup$ How do I create a hyperlink in a comment? $\endgroup$
    – ahorn
    Mar 29, 2015 at 14:43
  • 1
    $\begingroup$ @ahorn: Test. Put the link text in square brackets, and the link in normal brackets, like this: [text](link) $\endgroup$
    – Aryabhata
    Mar 29, 2015 at 15:44
  • $\begingroup$ @ahorn $[$...$]($...$)$ and then "..." becomes the hyperlink. $\endgroup$
    – Mr Pie
    Apr 8, 2019 at 6:45
5
$\begingroup$

$\def\col#1{\color{#1}{\text{#1}}}\col{white}$

I am testing whether there are any uses of the #rrggbb color notation to represent usefully distinguishable colors. Certainly $\col{#d10000}$ is distinguishable from $\col{#df0000}$, but the former is indistinguishable from $\col{#d00}$ and the latter from $\col{#e00}$.

Red

$$ \col{#000}\col{#080000}\col{#100}\\ \col{#100}\col{#190000}\col{#200}\\ \col{#200}\col{#2a0000}\col{#300}\\ \col{#300}\col{#3b0000}\col{#400}\\ \col{#400}\col{#4c0000}\col{#500}\\ \col{#500}\col{#5d0000}\col{#600}\\ \col{#600}\col{#6e0000}\col{#700}\\ \col{#700}\col{#7f0000}\col{#800}\\ \col{#800}\col{#900000}\col{#900}\\ \col{#900}\col{#a10000}\col{#a00}\\ \col{#a00}\col{#b20000}\col{#b00}\\ \col{#b00}\col{#c30000}\col{#c00}\\ \col{#c00}\col{#d40000}\col{#d00}\\ \col{#d00}\col{#e50000}\col{#e00}\\ \col{#e00}\col{#f60000}\col{#f00}\\ $$

Yellow

$$ \col{#000}\col{#080800}\col{#110}\\ \col{#110}\col{#191900}\col{#220}\\ \col{#220}\col{#2a2a00}\col{#330}\\ \col{#330}\col{#3b3b00}\col{#440}\\ \col{#440}\col{#4c4c00}\col{#550}\\ \col{#550}\col{#5d5d00}\col{#660}\\ \col{#660}\col{#6e6e00}\col{#770}\\ \col{#770}\col{#7f7f00}\col{#880}\\ \col{#880}\col{#909000}\col{#990}\\ \col{#990}\col{#a1a100}\col{#aa0}\\ \col{#aa0}\col{#b2b200}\col{#bb0}\\ \col{#bb0}\col{#c3c300}\col{#cc0}\\ \col{#cc0}\col{#d4d400}\col{#dd0}\\ \col{#dd0}\col{#e5e500}\col{#ee0}\\ \col{#ee0}\col{#f6f600}\col{#ff0}\\ $$

Green

$$ \col{#000}\col{#000800}\col{#010}\\ \col{#010}\col{#001900}\col{#020}\\ \col{#020}\col{#002a00}\col{#030}\\ \col{#030}\col{#003b00}\col{#040}\\ \col{#040}\col{#004c00}\col{#050}\\ \col{#050}\col{#005d00}\col{#060}\\ \col{#060}\col{#006e00}\col{#070}\\ \col{#070}\col{#007f00}\col{#080}\\ \col{#080}\col{#009000}\col{#090}\\ \col{#090}\col{#00a100}\col{#0a0}\\ \col{#0a0}\col{#00b200}\col{#0b0}\\ \col{#0b0}\col{#00c300}\col{#0c0}\\ \col{#0c0}\col{#00d400}\col{#0d0}\\ \col{#0d0}\col{#00e500}\col{#0e0}\\ \col{#0e0}\col{#00f600}\col{#0f0}\\ $$

Blue

$$ \col{#000}\col{#000008}\col{#001}\\ \col{#001}\col{#000019}\col{#002}\\ \col{#002}\col{#00002a}\col{#003}\\ \col{#003}\col{#00003b}\col{#004}\\ \col{#004}\col{#00004c}\col{#005}\\ \col{#005}\col{#00005d}\col{#006}\\ \col{#006}\col{#00006e}\col{#007}\\ \col{#007}\col{#00007f}\col{#008}\\ \col{#008}\col{#000090}\col{#009}\\ \col{#009}\col{#0000a1}\col{#00a}\\ \col{#00a}\col{#0000b2}\col{#00b}\\ \col{#00b}\col{#0000c3}\col{#00c}\\ \col{#00c}\col{#0000d4}\col{#00d}\\ \col{#00d}\col{#0000e5}\col{#00e}\\ \col{#00e}\col{#0000f6}\col{#00f}\\ $$

Gray

$$ \col{#000}\col{#080808}\col{#111}\\ \col{#111}\col{#191919}\col{#222}\\ \col{#222}\col{#2a2a2a}\col{#333}\\ \col{#333}\col{#3b3b3b}\col{#444}\\ \col{#444}\col{#4c4c4c}\col{#555}\\ \col{#555}\col{#5d5d5d}\col{#666}\\ \col{#666}\col{#6e6e6e}\col{#777}\\ \col{#777}\col{#7f7f7f}\col{#888}\\ \col{#888}\col{#909090}\col{#999}\\ \col{#999}\col{#a1a1a1}\col{#aaa}\\ \col{#aaa}\col{#b2b2b2}\col{#bbb}\\ \col{#bbb}\col{#c3c3c3}\col{#ccc}\\ \col{#ccc}\col{#d4d4d4}\col{#ddd}\\ \col{#ddd}\col{#e5e5e5}\col{#eee}\\ \col{#eee}\col{#f6f6f6}\col{#fff}\\ $$

Conclusion: on a typical LCD monitor, a half-step (#08) is perceptible in the lighter colors, but not in the darker ones. Even a full step (#11) is too small to be useful for distinguishing different text in a post on this web site.

$\endgroup$
8
  • 2
    $\begingroup$ Note that whether or not those colors are distinguishable depends upon the capabilities of the monitor and graphics system. Most consumer level displays have limited capabilities (8-bit,low gamut). For some discussion see e.g. here. $\endgroup$ Jan 16, 2014 at 1:47
  • 2
    $\begingroup$ #00e means #0000ee not #0000e0. $\endgroup$
    – kennytm
    Jan 16, 2014 at 8:12
  • 2
    $\begingroup$ @KennyTM Thanks! Of course it must be so, or else #FFF wouldn't be white. Thanks for pointing this out. $\endgroup$
    – MJD
    Jan 16, 2014 at 14:13
  • $\begingroup$ @Bill That is interesting, but not relevant to the issue of mathematical typesetting on this web site. $\endgroup$
    – MJD
    Jan 16, 2014 at 15:32
  • $\begingroup$ @MJD Sure it is. You are attempting to judge if color differences are perceptible on MSE. My point is that it is very difficult to accurately judge that unless one has professional-level graphics hardware and specialized knowledge in this area. For example, what you see as different may display the same to someone else using a monitor with less capability (e.g. one using dithering/interpolation from 8bit to 10bit color). $\endgroup$ Jan 16, 2014 at 15:48
  • $\begingroup$ I am trying to judge if color differences are usefully different. For example, you are fond of using colored text to highlight parts of equations, as here. The fact that $\color{#0000c3}{\text{#0000c3}}$ might be distinguishable from $\color{#0000cc}{\text{#0000cc}}$ on a professional-quality wide-gamut monitor is of absolutely no use to you in doing that. $\endgroup$
    – MJD
    Jan 16, 2014 at 15:52
  • $\begingroup$ My goal in writing this post was to decide if I should mention the #rrggbb notation in this post about typesetting colors, in addition to the #rgb notation. My conclusion is that there is no need to do that. $\endgroup$
    – MJD
    Jan 16, 2014 at 15:55
  • 11
    $\begingroup$ $\rlap{\color{#000}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#010}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#020}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#030}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#040}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#050}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#060}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#070}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#080}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#090}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#0a0}{\Rule{200em}{1em}{0.75em}}}\\ \rlap{\color{#0b0}{\Rule{200em}{1em}{0.75em}}}\\ $ $\endgroup$
    – user93957
    Jan 31, 2014 at 22:29
5
$\begingroup$

Country Flags (to be put in your profile)


Some codes are too long to fit in the location section, that's why I'm working on making less long codes. $\checkmark$ denotes ones that can fit in.


Example:

$\phantom{XXXXX}$enter image description here

$\checkmark$ Morocco Preview: (don't use the codes associated with the previews, instead use the ones written under)

$$\Huge\:\color{red}{\rlap{\color{\green}{\qquad\star}}{\Rule{2.1em}{1em}{0.5em}}}\:$$

\def\s{\space}\color{red}{\rlap{\color{\green}{\s\s\!\!\tiny\star}}{\Rule{1em}{0.5em}{0.25em}}}

$\checkmark$ France Preview:

$$\Huge\def\r{\Rule{.7em}{1em}{0.5em}}\color{#009}{\r}\color{#}{\r}\color{red}{\r}$$

\def\r{\Rule{.333em}{.5em}{.25em}}\color{#009}{\r}\color{#}{\r}\color{red}{\r}

$\checkmark$ Italy Preview:

$$\Huge{\color{green}{\Rule{0.7em}{1em}{0.5em}}}{\color{white}{\Rule{0.7em}{1em}{0.5em}}}{\color{red}{\Rule{0.7em}{1em}{0.5em}}}$$

\def\r{\Rule{.333em}{.5em}{.25em}}\color{#090}{\r}\color{#}{\r}\color{red}{\r}

Ireland Preview:

$$\Huge{\color{darkorange}{\Rule{0.7em}{1em}{0.5em}}}{\color{white}{\Rule{0.7em}{1em}{0.5em}}}{\color{green}{\Rule{0.7em}{1em}{0.5em}}}$$

\color{darkorange}{\Rule{0.333em}{0.5em}{0.25em}}\color{white}{\Rule{0.33em}{0.5em}{0.25em}}\color{green}{\Rule{0.33em}{0.5em}{0.25em}}

Mali Preview:

$$\Huge{\color{green}{\Rule{0.7em}{1em}{0.5em}}}{\color{yellow}{\Rule{0.7em}{1em}{0.5em}}}{\color{red}{\Rule{0.7em}{1em}{0.5em}}}$$

\color{green}{\Rule{0.333em}{0.5em}{0.25em}}\color{yellow}{\Rule{0.33em}{0.5em}{0.25em}}\color{red}{\Rule{0.33em}{0.5em}{0.25em}}

Senegal Preview:

$$\Huge\rlap{\qquad\color{green}{\star}}{\Huge{\color{green}{\Rule{0.7em}{1em}{0.5em}}}{\color{yellow}{\Rule{0.7em}{1em}{0.5em}}}{\color{red}{\Rule{0.7em}{1em}{0.5em}}}}$$

\rlap{\space\space\!\!\color{green}{\tiny\star}}{\color{green}{\Rule{0.333em}{0.5em}{0.25em}}\color{yellow}{\Rule{0.33em}{0.5em}{0.25em}}\color{red}{\Rule{0.33em}{0.5em}{0.25em}}}

Romania Preview:

$$\Huge{\color{blue}{\Rule{0.7em}{1em}{0.5em}}}{\color{yellow}{\Rule{0.7em}{1em}{0.5em}}}{\color{red}{\Rule{0.7em}{1em}{0.5em}}}$$

\color{blue}{\Rule{0.333em}{0.5em}{0.25em}}\color{yellow}{\Rule{0.33em}{0.5em}{0.25em}}\color{red}{\Rule{0.33em}{0.5em}{0.25em}}

Belgium Preview:

$$\Huge{\color{black}{\Rule{0.7em}{1em}{0.5em}}}{\color{yellow}{\Rule{0.7em}{1em}{0.5em}}}{\color{red}{\Rule{0.7em}{1em}{0.5em}}}$$

\color{black}{\Rule{0.333em}{0.5em}{0.25em}}\color{yellow}{\Rule{0.33em}{0.5em}{0.25em}}\color{red}{\Rule{0.33em}{0.5em}{0.25em}}

More are to be added, you can contribute by making any flag you want :-)

$\endgroup$
5
  • $\begingroup$ Consider replacing all of your \color{white} with \color{#}. The same can be done with other colors such as replacing \color{darkblue} with \color{#009}. $\endgroup$
    – Brad
    Jun 30, 2014 at 6:14
  • 8
    $\begingroup$ Looks like codegolf.SE material... $\endgroup$
    – kennytm
    Jun 30, 2014 at 10:20
  • $\begingroup$ @Brad Yes, thanks! $\endgroup$
    – Hakim
    Jun 30, 2014 at 23:02
  • $\begingroup$ Emojis work pretty well too, though they’re not as crisp $\endgroup$ Jun 4, 2020 at 6:10
  • 1
    $\begingroup$ The star that's supposed to be in the middle of the Morocco's flag is not showing for me on Chrome / Windows.... $\endgroup$ Mar 1, 2021 at 7:29
5
$\begingroup$

Testing spoiler:

Without newlines:

$$\lim_{x \rightarrow \infty} \dfrac{\ln(x^2+4)}{\ln(x+\sqrt{1+x^2})} = \lim_{x \rightarrow \infty} \dfrac{\ln(x^2) + \ln(1+4/x^2)}{\ln(x) + \ln(1+\sqrt{1+1/x^2})}$$ $$ = \lim_{x \rightarrow \infty} \dfrac{\ln(x^2)}{\ln(x)} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$ $$ = \lim_{x \rightarrow \infty} 2 \dfrac{\ln(x)}{\ln(x)} \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$ $$ = 2 \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)} =2 $$

With newlines:

! $$\lim_{x \rightarrow \infty} \dfrac{\ln(x^2+4)}{\ln(x+\sqrt{1+x^2})} = \lim_{x \rightarrow \infty} \dfrac{\ln(x^2) + \ln(1+4/x^2)}{\ln(x) + \ln(1+\sqrt{1+1/x^2})}\\ = \lim_{x \rightarrow \infty} \dfrac{\ln(x^2)}{\ln(x)} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}\\ = \lim_{x \rightarrow \infty} 2 \dfrac{\ln(x)}{\ln(x)} \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}\\ = 2 \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)} =2 $$

Now hiding a whole paragraph - again everything has to go in a single line for this to work:

Since $N$ is fixed, we have some amount of primes $$p_1 < p_2 < \ldots p_k \leq N $$ We also have for any $x\in A\setminus\{1\}$ $$x = p_1^{l_1}p_2^{l_2}\ldots p_k^{l_k} ,\quad l_i\geq 0$$ Now it gets rough: By fixing the pair $(m,n)$ we have: $$m=p_1^{s_1}p_2^{s_2}\ldots p_k^{s_k},\ s_i\geq 0\qquad n=p_1^{t_1}p_2^{t_2}\ldots p_k^{t_k}, t_i\geq 0 $$ So we start counting powers $$\left (\begin{array}{}s_1 & s_2 &\ldots & s_k\\t_1 & t_2 &\ldots &t_k \end{array}\right ) $$ For neither $m$ nor $n$ can we have all the primes represented with power $\geq 1$, since that immediately makes the other number equal to $1$, which we have omitted for now.

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    $\begingroup$ I was trying this here because of the problems the poster had with this answer. $\endgroup$ Jun 1, 2012 at 9:51
  • $\begingroup$ this is a test. $\endgroup$
    – ahorn
    Mar 30, 2015 at 9:20
  • $\begingroup$ AHA! Martin's name did not appear when I tagged him! $\endgroup$
    – ahorn
    Mar 30, 2015 at 9:21
  • $\begingroup$ @ahorn See meta.math.stackexchange.com/questions/6281/… and other related threads. (BTW I think that the correct word in this context is to ping and not to tag.) And if you are wonrdering, the notification from your comment reached me. $\endgroup$ Mar 30, 2015 at 9:36
  • $\begingroup$ Does [faq] and [tour] work in comments? Here: faq and tour. $\endgroup$ Sep 8, 2019 at 10:08
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No downvotes!


$\hskip -3.5em \color{red}{\Rule{3em}{4.5em}{0em}}$ $\overset{\cdot\cdot}\smile$

$$ \require{action} \require{enclose} \toggle{ \quad\enclose{roundedbox}{\text{ Why I cannot see the votes or vote? }} }{ \quad\quad \bbox[4pt,border: 1px solid red]{ \begin{array}{l} \text{Because the answer has said so that you should not be able to click the downvote button except if you remove the element using inspect element.}\\ \end{array}} }\endtoggle $$ $$\quad\enclose{roundedbox}{\text{ Back to top }}$$

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  • $\begingroup$ $\hskip -2.7em \color{black}{\Rule{2em}{1.8em}{2em}}$ $\endgroup$ Mar 22, 2019 at 16:56
  • $\begingroup$ $\quad\enclose{roundedbox}{\text{ Back to top }}$ $\endgroup$ Mar 27, 2019 at 18:01
  • $\begingroup$ $\require{action}\require{enclose}\toggle{\quad\enclose{roundedbox}{\text{ A button }}}{\quad\quad\bbox[4pt,border: 1px solid red]{\begin{array}{l}\text{Some text}\\\end{array}}}\endtoggle$ $\endgroup$ Mar 27, 2019 at 18:02
  • $\begingroup$ $\rlap{\color{grey}{\Rule{20em}{1em}{0.75em}}}$ $\endgroup$ May 3, 2019 at 10:30
  • $\begingroup$ $\hskip -2.7em \color{black}{\Rule{2em}{1.8em}{2em}}$ $\endgroup$ Nov 4, 2020 at 11:16
  • $\begingroup$ Can I downvote it just to prove that it can be downvoted? :P $\endgroup$ Nov 4, 2020 at 13:10
  • $\begingroup$ @RajdeepSindhu Yes you can, but it is a waste of reputation. You can just inspect element it out. $\endgroup$ Nov 4, 2020 at 20:05
  • $\begingroup$ @smileycreations15 Even without doing that, a little portion of the downvote button is visible. Also, I was joking :D $\endgroup$ Nov 4, 2020 at 20:20
  • $\begingroup$ :D, Also I will fix that >:D $\endgroup$ Nov 5, 2020 at 20:12
  • $\begingroup$ When the website is loading, you can see the downvote button over the red rectangle for a second. Here is a photo. $\endgroup$ Jun 25, 2021 at 12:05
  • $\begingroup$ Happens to me too $\endgroup$ Jun 25, 2021 at 20:17
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$\require{AMScd}$

$$\begin{CD} c & & d\\ @A t AA \rlap{{_{\rlap{x}}\style{display: inline-block; transform: rotate(30deg)}{{\xleftarrow[\rule{2em}{0em}]{}}}}}{\style{display: inline-block; transform: rotate(150deg)}{{\xleftarrow[\rule{2em}{0em}]{}}}}{_{\llap{y}}} @AA z A\\ a & & b \end{CD}$$

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Let me try if there is a difference between single dollar signs $\sum_{i = 0}^n k^i$ and double dollar signs $$\sum_{i = 0}^n k^i$$

From comment suggestion of using \limits... first with sum $\sum\limits_{i = 0}^n k^i$ then with anything else $\mathop{\large\spadesuit}\limits_{i = 0}^n k^i$ - nice.

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    $\begingroup$ You can always use this: $\sum\limits_{i = 0}^n k^i$ $\endgroup$
    – Quixotic
    Sep 19, 2011 at 20:12
  • $\begingroup$ Adding to what @Quixotic said in his/her comment, just add \limits in front of \sum, of which can be applied to other similar commands like \prod (which generates $\prod$ lest you did not know). $\endgroup$
    – Mr Pie
    Apr 8, 2019 at 6:44
  • $\begingroup$ Add \limits after \sum- or whatever collective operation, you might need to \mathop novel symbols, $\endgroup$
    – Joffan
    Mar 24, 2021 at 17:30
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And now what does it look like to another user who doesn't suspect that the command has been redefined?

$$\sin x$$

Very interesting.

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you may also go to MathURL and write your formula there; just remember the dollar signs before putting it here.

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Testing matrix environments - are multiple backslashes needed?

The following (with each line ending only with two backslashes) renders ok for me: $$\begin{pmatrix} 1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 0 \\ 1/2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1/2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1/2 & 1 \end{pmatrix}$$

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$$\require{cancel}\cancelto{1}{\dfrac{\sqrt{7x^7-y^9}}{8x^3+1}}$$

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  • $\begingroup$ And how about backwards cancellation? $\bcancel{x}{}$ Yup, that works, too :P $^1\bcancel{x}$ and $\bcancel{x^1}$ and $\bcancel{x}^1$ $\endgroup$
    – Mr Pie
    Apr 8, 2019 at 6:48
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\begin{array}{| r | r | r |} \hline N & \frac{1}{\sqrt{N-2^{1/2}}} & \frac{1}{c_4(N)\sqrt{N-1}} \\ \hline 3 & 0,7941 & 0,7979 \\ 4 & 0,6219 & 0,6267 \\ 5 & 0,5281 & 0,5319 \\ \hline \end{array}

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I asked myself something like "can we know everything in Math?" and found an answer at Quora:

Mathematics is so great it can even answer this question :). Kurt Gödel answered this question almost a century ago with Gödel's incompleteness theorems. ... we will never be able to answer all questions.

Fine, so math is infinite and we will never know everything, but

what kind of infinity is it? Is math or the set of mathematical theorems a ordered infinite set?

If it is a countable infinite set, how to prove that?

I ask if it possible to apply the concept of (infinite) sets onto mathematical theorems. I look at mathematical theorems as elements of a set, and I ask is this set infinite...

EDIT A way to formalize this could look as follows:

Let $S=\{a_0,a_1,...;t_0,t_1,...\}$ be a set. It contains axioms $a_n$ and theorems $t_m$ that are unprovable with the given axioms. You'll examples as answers to this question.

From a Gödel point of view the set is incomplete, but it could be extended by a, lets call it Gödel operation $\mathfrak G$ that maps $$ \begin{array}{cl} \mathfrak G:& S\to S \\ & \{a_0,...,a_k;t_0,...t_m\} \mapsto \{a_0,...,a_k;t_0,...,t_m,t_{m+1}\} \end{array} $$ by extending $S$ with an axiom for fixes a hole in the landscape of proofs. Now if you apply $\mathfrak G$ several times it looks like you can enumerate the individuals elements of $S$, finally making it a countable infinite number of theorems that make up your set.

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This is the site's logo:

test <span class=$2$">

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This is actually standard notation, insofar as there is any for quantifiers, and I was immediately able to recognize it as stating the pumping lemma for regular languages.

It is using $$(\exists p>1)(X)$$ to mean “there exists a $p$, such that $p>1$ and $X$ is true. Sometime this might be written as $$\exists p>1. X$$ or as $$\exists p: (p>1) \land X$$ but the meaning is the same although the notation varies slightly.

$\Rightarrow$ is being used for logical implication. $\operatorname{Regular}(L)$ is an assertion that the language $L$ is regular.

We could write this out in English, but it might not be “more readable” or “easier to understand”. I will give it a try.

$$ \begin{array}{r|l} (\forall L\subseteq \Sigma^*) & \text{For any language $L$ over some alphabet $\Sigma$} \\ (\mbox{regular}(L) & \text{we will say that $L$ is “regular”} \\ \Rightarrow & \text{if: }\\ \quad ((\exists p\geq 1) & \text{there is some positive number $p$ (the ‘pumping constant for $L$’)} \\ ( (\forall w\in L) ((|w|\geq p) \Rightarrow & \text{such that for every word $w$ in $L$, of length is at least $p$,} \\ \quad ((\exists x,y,z \in \Sigma^*)(w=xyz & \text{We can always break $w$ into three strings, $x,y,$ and $z$} \\ \qquad\land (|y|\geq 1 \land |xy|\leq p & \text{where $y$ is nonempty and $xy$ is no longer than the pumping constant,} \\ \land (\forall n\geq 0)(xy^nz\in L) & \text{so that $xy^nz$ is also in $L$ for each non-negative number $n$.} \\ ))))))) & \text{(P.S. the author is excessively concerned with notation.)} \end{array} $$

I do not want to suggest that the meaning should be obvious or even clear from the English translation. The pumping lemma can be hard to understand until you have seen a few examples of how it works, and the nested quantifiers are one of the difficulties.

It often happens that statements with many nested quantifiers are easier to understand as games. You are trying to prove that $L$ is regular, and some adversary is trying to foil your proof. You make a move for each $\exists$ quantifier, and the adversary makes a move for each $\forall$ quantifier. (Or vice versa, if you think it's not regular.) The game goes like this:

  1. You say “$L$ is regular. Here is my proposal for the pumping constant $p$.”
  2. The adversary says “Here is my string $w$.” (It must have length at least $p$ to be a legal move in the game.)
  3. You say “Here is how I want to break $w$ into three strings $x,y,z$. (Your $x$ and $y$ must satisfy the constraints $|y|>0$ and $|xy|≤p$.)
  4. The adversary says “here is my choice of $n$
  5. At this point, if $xy^nz$ is in $L$, you win, and if not, the adversary wins.

If you can present a strategy that is guaranteed to win the game for any possible moves by your adversary, that is your proof that $L$ is regular. If the adversary can always win, regardless of what you do, that is a proof that $L$ is not regular.

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  • $\begingroup$ $\hskip -2.8em \color{red}{\Rule{2em}{2.5em}{1em}}$ isnt this to be in the answers sandbox? $\hskip +28em \color{red}{\Rule{2em}{2.5em}{1em}}$ $\endgroup$ Nov 4, 2020 at 11:23
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    $\begingroup$ @AnindyaPrithvi Yes, in confusion, I originally went to the wrong sandbox. Also, I was tardy for leaving it up for so long. I have just released the answer slot. $\endgroup$ Nov 6, 2020 at 20:13
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Demonstrating the issue from this question

$\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}$

$\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} + \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} - \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} = \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\times\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}$

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Some stairs you can walk on. Don't worry! It's stable!

$$ n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n^{n}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} $$

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  • $\begingroup$ $$\text{reverse_bwt("aba", 0)="aaa", which is not an anagram.}$$ $$\text{reverse_bwt("aba", 0)="aba", which is an anagram. bwt("aba")=("baa", 1)}$$ $$\text{reverse_bwt("aba", 0)="bab", which is not an anagram.}$$ $\endgroup$
    – Apass.Jack
    Mar 26, 2022 at 1:33
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Adding MathJax in tooltips seems to mess up the rendering in the submitted post, but not in the preview. Here is a test:


See [this][1] paper.

  [1]: https://hrcak.srce.hr/clanak/399487 "Kraljević, Hrvoje. The structure of the algebra $(\mathcal{U}(\mathfrak{g})\otimes C(\mathfrak{p}))^K$ for the groups $\mathrm{SU}(n,1)$ and $\mathrm{SO}_e (n,1)$. Math. Commun. 27, No. 1, 11-18 (2022). https://zbmath.org/?q=an:7544424"

See $(\mathcal{U}(\mathfrak{g})\otimes C(\mathfrak{p}))^K$ for the groups $\mathrm{SU}(n,1)$ and $\mathrm{SO}_e (n,1)$. Math. Commun. 27, No. 1, 11-18 (2022). https://zbmath.org/?q=an:7544424">this paper.


This is the rendered preview of the above portion that I see:

Screenshot of rendered preview.

Including the tooltip, this is what I see in the preview:

Screenshot of rendered preview with tooltip.


This is what I see in the submitted post:

Screenshot of submitted post.

This is what I see in the submitted post along with the tooltip:

Screenshot of submitted post with tooltip.


Just to confirm, escaping the dollars fixes this problem:

See [this][6] paper.

  [6]: https://hrcak.srce.hr/clanak/399487 "Kraljević, Hrvoje. The structure of the algebra \\\$(\mathcal{U}(\mathfrak{g})\otimes C(\mathfrak{p}))^K\\\$ for the groups \\\$\mathrm{SU}(n,1)\\\$ and \\\$\mathrm{SO}_e (n,1)\\\$. Math. Commun. 27, No. 1, 11-18 (2022). https://zbmath.org/?q=an:7544424"

See this paper.


Of course, the MathJax is never rendered in the tooltip itself, so all this just helps to understand how to add a dollar sign to a tooltip and not break the post (since the preview does not show that the formatting actually breaks).

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  • $\begingroup$ For those who notice, there is a bug related to escaping the dollar signs which is visible in the last piece of source code. This is also tested in a different answer on this thread. $\endgroup$ Jun 21, 2022 at 12:52
  • 1
    $\begingroup$ Additionally, I haven't mentioned it in the post itself, but there is another difference between the preview and final output when escaping the dollar signs: the tooltip in the preview displays "backslash+dollars" everywhere, whereas the tooltip in the final output displays only "dollars" everywhere. The bottomline seems to be: diligently escape any and all dollar signs in the tooltip, and the final output will render just fine. $\endgroup$ Jun 21, 2022 at 12:56
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$\hskip 36em {\require{cancel}\require{cancelto} _\text{psst! over here!}\cancelto{\hspace{1pt}}{\hspace{20pt}}}$

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Can we do pictures?

\begin{picture}(2,2) \put(0,0){\line(1,0){1}} \put(0,0){\line(0,1){1}} \end{picture}

\begin{math} 2 \end{math}

aw dang..

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  • 1
    $\begingroup$ See here $\endgroup$
    – t.b.
    Sep 18, 2011 at 22:10
  • $\begingroup$ @TheoBuehler: Aw, dang. Thanks for the link. $\endgroup$
    – user541686
    Sep 18, 2011 at 22:28
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$ (\not \in \notin) 1 \times 2 \in S \implies S \notin S$

$$ \lim_ {k\to\infty}^{\diamond \circ \square \sum \int} \sum_{j=1}^k {j^{2^j_k}_3}_{x_i} \int_2^3x\ dx $$

$C3^\#_\flat\natural\colon$ musical stuff!

$$&iexcl;^IGNORE\ \ M_e!$$

$$¡^IGNORE\ \ M_e!$$

$$!`^IGNORE\ \ M_e!$$

$$\unicode{xA1}^IGNORE\ \ M_e!$$

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  • $\begingroup$ Hm... odd... &iexcl;^IGNORE\ \ M_e! works in the preview. $\endgroup$
    – Ry-
    Dec 3, 2011 at 6:31
  • 1
    $\begingroup$ Okay, only one of three variants work... $\endgroup$ Dec 3, 2011 at 6:53
  • $\begingroup$ Make that two.${}$ $\endgroup$ Dec 6, 2011 at 14:23
  • $\begingroup$ Test: $\mathbb{R}^{n+1}$ $\endgroup$
    – Dan Moore
    Jul 12, 2012 at 15:27
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Can we enter nested math inside \text now, and have it saved? $$ \{\,p\mid\text{$p$ and $p+2$ are prime}\,\} $$ Edit: it seems we can.

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  • $\begingroup$ But will Markdown respect it? $\text{This should not become a hyperlink: $[test](google.com/)$}$ $\endgroup$
    – celtschk
    Jul 14, 2012 at 10:28
  • $\begingroup$ The changes don't affect comments, only questions and answers. Comments seem to be processed quite differently. $\endgroup$ Jul 14, 2012 at 13:57
  • $\begingroup$ That's interesting. So $\text{does this $x^2$ also not work properly?}$ Well, it seems to work. $\endgroup$
    – celtschk
    Jul 14, 2012 at 16:06
  • 2
    $\begingroup$ MathJax properly handles nested dollars, but they are not protected from MarkDown when used in comments. They are when used in questions and answers. $\endgroup$ Jul 14, 2012 at 18:30
  • $\begingroup$ You can also use \mbox $\endgroup$
    – Mr Pie
    Apr 8, 2019 at 6:47
1
$\begingroup$

The preview recognizes $\rm\LaTeX$ environments and protects the contents from Markdown. Does that work once saved?

\begin{equation} x _1 = y_ 1 \end{equation}

Edit: It seems that it does!

$\endgroup$
0
1
$\begingroup$

Testing striking out:

math: text $a^2-b^2=(a-b)(a+b)$ text

tag: text text

url: text math.SE text

$\endgroup$
7
  • $\begingroup$ What about comments? math: <s> text $a^2-b^2=(a-b)(a+b)$ text </s> tag: <s> text tex text </s> url: <s> text math.SE text </s> $\endgroup$ Jun 20, 2012 at 12:53
  • $\begingroup$ Any concave function $f\colon[0,\infty)\to\mathbb R$ such that $f(0)=0$ is subadditive. $\endgroup$ Jun 28, 2012 at 14:10
  • $\begingroup$ Any concave function $f\colon[0,\infty]\to\mathbb R$ such that $f(0)=0$ is subadditive. $\endgroup$ Jun 28, 2012 at 14:12
  • $\begingroup$ What about \left[\right)? Any concave function $f\colon\left[0,\infty\right)\to\mathbb R$ such that $f(0)=0$ is subadditive. $\endgroup$ Jun 28, 2012 at 14:33
  • $\begingroup$ $\left[0,\infty\right)$ and [link](http://math.stackexchange.com) produces $\left[0,\infty\right)$ and link. $\endgroup$ Jun 28, 2012 at 14:43
  • $\begingroup$ Do you mean that in one dimension the definition of a manifold with boundary is: "A topological space such that each point has a neighbourhood homeomorphic to an open subset of $(-\infty, 0]$ or $[0, \infty)"? Or do you mean you can define a one dimensional manifold with boundary as a topological space such that each point has a neighbourhood homeomorphic to an open subset of $(-\infty, 0]$, or equivalently, $[0, \infty)$? Surely you mean the latter as $\phi : [0, \infty) \to (-\infty, 0]$, $\phi(x) = -x$ is a homeomorphism. $\endgroup$ Oct 12, 2012 at 13:36
  • $\begingroup$ What happens if I put here link to a cooment? $\endgroup$ Apr 5, 2013 at 17:19
1
$\begingroup$

Preview seems to "leak" macro definitions, even to before the macro is defined. Let's see what happens with saving.

This macro should be undefined here (and thus show the name in red): $\NobodyWouldCreateSuchALongMacroName$

So should this: $\AnotherRidiculouslyLongName$

Now start a local group. $\require{begingroup}\begingroup$

Define the first macro to $1$ $\def\NobodyWouldCreateSuchALongMacroName{1}$ and use it: $\NobodyWouldCreateSuchALongMacroName$ — This should display as $1$.

Now define the second one, this time using gdef, to the value $2$. $\gdef\AnotherRidiculouslyLongName{2}$ Again, use it: $\AnotherRidiculouslyLongName$ — This should show up as $2$.

Now end the group. $\endgroup$

Now the first macro should be undefined again: $\NobodyWouldCreateSuchALongMacroName$

The second macro, however, should still be defined (or I've misunderstood something): $\AnotherRidiculouslyLongName$

$\endgroup$
2

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