Basically same as Formatting Sandbox in meta.SO, but since this and Statistical Analysis are the only 2 sites (I know) supporting TeX formatting, I believe we also need one here for testing it.

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1  
Theoretical computer science also supports $\mathrm{\TeX/\LaTeX}$ formatting. –  JeffE Jun 1 '12 at 7:29
1  
@JeffE: In 2010 only 'stats' and 'math' support TeX formatting. Of course now there is also 'cstheory', 'cs', 'chemistry', 'quant', etc. –  kennytm Jun 3 '12 at 6:25

31 Answers 31

up vote 14 down vote accepted

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.

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4  
May be this (and the main sandbox) should be made special unbumpable question? –  Vi0 Aug 24 '12 at 15:06
1  
Except that there's no preview for comments. –  shoover Sep 24 '14 at 18:22

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.

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

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This is a 1e1ea2ce-0342-4835-a7cc-ee70fbdfe27d
bug

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$\hskip -3em \color{red}{\Rule{2em}{1em}{1em}}$Testing of negative skips to overlap the buttons on the left.

$\rlap{\smash{\lower 3em{\color{red}{\Rule{8em}{2em}{0em}}}}}$Testing overlapping on the bottom. OK, both seem to be problems.

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$\rlap{\color{grey}{\Rule{200em}{1em}{0.75em}}}$ –  user93957 Jan 7 '14 at 13:01

How does newcommand and renewcommand work?

$$\newcommand{\sin}{FOO} \sin x$$

$$\sin y$$

$$\renewcommand{\nonexistent}{QUX} \nonexistent$$


hello$\renewcommand{\sin}{hello}$world.

hello$$\renewcommand{\sin}{world}$$world.

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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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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:

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

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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 $$\sum_{i = 0}^n k^i$$

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$\hskip 36em {\require{cancel}\require{cancelto} _\text{psst! over here!}\cancelto{\hspace{1pt}}{\hspace{20pt}}}$

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

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.

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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. –  Bill Dubuque Jan 16 '14 at 1:47
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#00e means #0000ee not #0000e0. –  kennytm Jan 16 '14 at 8:12
1  
@KennyTM Thanks! Of course it must be so, or else #FFF wouldn't be white. Thanks for pointing this out. –  MJD Jan 16 '14 at 14:13
3  
$\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}}}\\ $ –  user93957 Jan 31 '14 at 22:29

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 :-)

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1  
Looks like codegolf.SE material... –  kennytm Jun 30 '14 at 10:20

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

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$\rm{\bf Hint}\:\ (p\!-\!1)^2\! \mid p^q\!-1 \!\iff\! p\!-\!1\ \bigg|\ \dfrac{p^q\!-1}{p\!-\!1} = p^{q-1}\! +\cdots\!+p\! +\! 1$ $\rm\equiv q\ (mod\ p\!-\!1)$

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

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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  
See here –  t.b. Sep 18 '11 at 22:10

$ (\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!

$$¡^IGNORE\ \ M_e!$$

$$¡^IGNORE\ \ M_e!$$

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

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

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1  
Okay, only one of three variants work... –  Guess who it is. Dec 3 '11 at 6:53

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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MathJax properly handles nested dollars, but they are not protected from MarkDown when used in comments. They are when used in questions and answers. –  Davide Cervone Jul 14 '12 at 18:30

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!

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Testing striking out:

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

tag: text text

url: text math.SE text

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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): $\AnotherRidiculouslyLongName$

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$\varnothing$ -- porton says this does not work.

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The following gives [Math Processing Error] on iOS Safari, but OK on iOS Chrome and on desktop browsers.

$ \newcommand{\ket}[1]{\left| #1 \right\rangle} \newcommand{\Psih}{\hat{\Psi}} \newcommand{\Psihd}{\hat{\Psi}^\dagger} \newcommand{\bx}{\mathbf{x}} \newcommand{\Hh}{\hat{H}} \newcommand{\Hsp}{\Hh_{\mathrm{sp}}} \newcommand{\cn}{\chi(\bx_1,\dots\bx_n)} $

$\frac12 \sum_{i,j} V(\bx_i-\bx_j)\cn$

$$\ket{\chi_n} = \int d\bx_1 \dots d\bx_n \cn\prod_k \Psihd(\bx_k)\ket{0}$$

$$\begin{align} \hat{V}\ket{\chi_n} &= \frac12 \int d\bx d \bx'd\bx_1 \dots d\bx_n V(\bx-\bx')\cn\Psihd(\bx) \Psihd(\bx') \Psih(\bx')\Psih(\bx)\prod_k \Psihd(\bx_k)\ket{0} \\ &= \frac12 \int d\bx d \bx'd\bx_1 \dots d\bx_n V(\bx-\bx') \sum_{i,j:i\ne j}\delta(\bx-\bx_i) \delta(\bx'-\bx_j) \cn\prod_k \Psihd(\bx_k)\ket{0}\\ &= \frac12 \int d\bx_1 \dots d\bx_n \sum_{i,j:i\ne j} V(\bx_i-\bx_j) \cn\prod_k \Psihd(\bx_k)\ket{0} \end{align} $$ which is the interaction we expect.

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This was done in an attempt (so far unsuccessful) to replicate the problem mentioned in line breaks in incorrect places

However, when I was trying this, I stumbled upon some strange behavior of MathJax, which I can't explain. (And I am unable to spot a mistake in the TeX code below.)


Two separate formulas - they work fine:

$\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|=$ $\sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+\frac{\sup_{|x|}\|A^2x\|}{2!}+\frac{\sup_{|x|}\|A^3x\|}{3!}+\ldots$

Now without any changes they are put together and everything is broken: $\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+\frac{\sup_{|x|}\|A^2x\|}{2!}+\frac{\sup_{|x|}\|A^3x\|}{3!}+\ldots$

This first part of the fromula still works ok: $\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq$

When I add the next bit, it becomes weird: $\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+$


$\|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| \|A\| \|B\| \|C\| $

$\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|+\|A\|+\|B\|+\|C\|$


I'm trying to prove this inequality:

$\|e^A\|\le e^{\|A\|}$, where $A$ is a matrix and $\|A\|:=\sup_{|x|=1} |Ax|$.

My attempt of solution:

Since $e^A:=I+A+A^2/2!+A^3/3!+\ldots$

we have

$\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|=$ $\sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+\frac{\sup_{|x|}\|A^2x\|}{2!}+\frac{\sup_{|x|}\|A^3x\|}{3!}+\ldots$

Am I right so far? I couldn't go further

I need help!

Thanks a lot.

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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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$$[Donkeys][1]$$

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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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This line $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ should be $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[\frac{1}{f} f\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ so $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[1\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ and $\left[1\right]_a^b=0$

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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. $\overset{\cdot\cdot}\smile$

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This space intentionally left blank. (Free to use again)

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To put multiple "hats" over a letter $f$:

$$\hat{\hat{\hat{f}}}\colon X\rightarrow Y \qquad \hat{\hat{\hat{\mbox{$f$}}}}\colon X\rightarrow Y$$ The attempt on the right isn't perfect (because the hats aren't aligned horizontally with the top of the $f$) but it's much better than the attempt on the left (where the hats are not aligned with each other).

To indicate an indefinite number of hats atop the letter $f$ by putting three dots between the topmost hat and the bottommost hat: $$\hat{\dot{\dot{\dot{\hat{\mbox{$f$}}}}}}\colon X\rightarrow Y$$

This is the best I've done so far; it would be interesting to know if it can be done better. Less vertical space would be nice, for one thing, both for the three-hat case and the many-hats case. And of course fixing the horizontal alignment.

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