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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.

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    $\begingroup$ Theoretical computer science also supports $\mathrm{\TeX/\LaTeX}$ formatting. $\endgroup$
    – JeffE
    Jun 1, 2012 at 7:29
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    $\begingroup$ @JeffE: You can use $\TeX$ and $\LaTeX$ (\Tex and \LaTeX) for the text. $\endgroup$
    – Asaf Karagila Mod
    Jun 2, 2012 at 21:09
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    $\begingroup$ @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
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    $\begingroup$ $a``=\!\!\text{’’}b$ $\endgroup$
    – 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
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    $\begingroup$ @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

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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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  • $\begingroup$ \rm ${}{}{}{}{}$ $\endgroup$
    – user93957
    Jan 7, 2014 at 13:00
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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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Sudoku test $$\begin{array}{||c|c|c||c|c|c||c|c|c||} \hline\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 \\ \hline 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\hline 2 & 3 & 1 & 5 & 6 & 4 & 8 & 9 & 7 \\ \hline 5 & 6 & 4 & 8 & 9 & 7 & 2 & 3 & 1 \\ \hline 8 & 9 & 7 & 2 & 3 & 1 & 5 & 6 & 4 \\ \hline\hline 3 & 1 & 2 & 6 & 4 & 5 & 9 & 7 & 8 \\ \hline 6 & 4 & 5 & 9 & 7 & 8 & 3 & 1 & 2 \\ \hline 9 & 7 & 8 & 3 & 1 & 2 & 6 & 4 & 5 \\ \hline\hline \end{array}$$

Result: Doesn't work as expected. Actual $\rm\LaTeX$ output of the same code:

LaTeX output

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Replace an SEO unfriendly pic question

a sample pic question

with a markdown blockquote for site SEO (in order to avoid harming the content originality by enclosing cited contents with <blockquote> using markdown)

The following sum $$ \sqrt{25-\left(\frac5n\right)^2} \cdot \frac5n + \sqrt{25-\left(\frac{10}n\right)^2} \cdot \frac5n + \cdots + \sqrt{25-\left(\frac{5n}{n} \right)^2} \cdot \frac5n $$ is a right Riemann sum for the definite integral $$\int_a^b f(x)\,\mathrm{d}x$$ where $b = \:\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{5em}{1em}{0.1em}} {\tt 5}}$
and $f(x) = \:\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{1em}{1em}{0.1em}} {\tt sqrt(25 - x^2)}}$.
The limit of these Riemann sums as $n \to \infty$ is  $\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{8em}{1em}{0.1em}}}$ .

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Test for MathJax trickeries on hyperlinks where the tag appears on the LEFT. Also see the closely related post about footnotes using html.

The main objective is to get the tag (visual cue) on the left, at the beginning of a line, be it in mathmode or text (sectioning).

Does this work? Yes, and inline \ref{sec01} tag works \ref{sec02} just as well in a standard mode or \ref{keypoint}. Now the challenge is to $\hspace{600pt}\ref{sec03}$. Hell yeah! .... umm why is it one line off? See the thoughts on phantom lines later.

So let's see how these silly ones work. First there's the blank Eq.\eqref{05} and the 2nd Eq.\eqref{12}, both tags consisting of 3 white spaces. Then we have Eq.\eqref{13} and lastly Eq.\eqref{14}. oh how about not in eqref but jus ref like these Eq.\ref{05}, Eq.\ref{12}, Eq.\ref{13}, and Eq.\ref{14}?

\begin{align*} &\text{This is an} & &\text{align environment} & &\text{Seriously} \tag{ } \label{05} \\ \mathbb{E}[X] &\equiv \mathcal{X} & \frac{\mathscr{X}}{\mathbb{Y}} &\approx \frac45 \Theta & j &= 2 \tag*{ } \label{12}\\ A &\equiv B+C & \frac{x}y &\approx \frac12 \Theta & i &= 3 \tag*{no kidding} \label{13}\\ \tag*{tag and label in 1st column} \label{14} \alpha &\equiv \beta+\gamma & \frac{\eta}\xi &\approx \frac23 \Omega & k &= 4 \\ \end{align*}

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Vivamus bibendum a quam eu vulputate. Phasellus congue velit mi, non gravida neque lobortis ut.

So this is a hyperlink to Eq.\,\eqref{01} which is supposed to be the normal tag, and then Eq.\,\eqref{02} is supposed to clash with Eq.\,\eqref{03}. Of course, the standard use is like Eq.\,\eqref{04}. oh damn it no thin space unless in mathmode so maybe just Eq.\eqref{04} unless one goes through the trouble of $\text{Eq.}\,\eqref{04}$? Yeah there IS a visible diffrence.

Quisque dictum erat eget neque imperdiet lobortis. Vivamus vel malesuada purus. Nunc vulputate sem sit amet pretium luctus. Duis interdum, magna id interdum eleifend, est nisl malesuada dolor, sed mollis metus metus ac magna.

Now this is getting interesting: \Eq.\eqref{21}, or should I call \Eq.\eqref{22}? Getting into the realm of absurdity with \Eq.\eqref{23} and oh btw \Eq.\eqref{24} also with negative that's gonna mess things up, and \Eq.\eqref{25} is no exception. With just \ \ref we have \Eq.\ref{21} ipsum oh I cannot remember \Eq.\ref{22} yes lorem ipsum it is \Eq.\ref{23} can I block call this thing \Eq.\ref{24} by \require \Eq.\ref{25}?

Ut nisl risus, tempor eu dolor vel, suscipit luctus elit. Pellentesque facilisis arcu vitae lacus rutrum viverra. Ut enim sem, imperdiet non volutpat at, ultricies id est. Aenean ex mauris, scelerisque id dolor nec, vehicula tincidunt ligula.

$$s(m-1,n)= 2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}{n\choose \frac{n}{2}-i} \tag{clash tag} \label{02} $$

$$s(m,n)= 2^{m-2}\sum\limits_{i=0}^{\frac{n}{2}}(-1)^{\frac{n}{2}-i}{m+n+2i-1 \choose 2i}{n\choose \frac{n}{2}-i} \tag{normal tag} \label{01} $$

$$ s(m,n-1)=2^{m-2}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-1 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{clash tag} \label{03} $$

$$ s(m-1,n-1)=2^{m-3}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{4} \label{04} $$ Proin non egestas dui, in tincidunt ante. Nunc placerat, nunc ac varius molestie, ipsum ipsum molestie arcu, id rhoncus tellus nunc commodo sapien.


$\Large\color{green}{\text{Oh my god I cannot believe this works!!}}$

Let's quote without the push \Eq.\eqref{31} and with the deliberate \Eq.$\hspace{100pt}\eqref{31}$ positive space paired with the equation labels \Eq.$\hspace{2000pt}\eqref{32}$, Oh my god I cannot believe this works. But is that the inevitable double parenthesis? How about \Eq.$\hspace{300pt}\eqref{33}$, is there still this thing? Now this \Eq.$\hspace{400pt}\eqref{34}$ and again \Eq.$\hspace{500pt}\eqref{35}$ and on and on to \Eq.$\hspace{600pt}\eqref{36}$. Hold on let's see $\hspace{100pt}\ref{31}$, $\hspace{200pt}\ref{32}$, $\hspace{300pt}\ref{33}$. Yeah, these show that $\hspace{400pt}\ref{34}$ tag position push $\hspace{500pt}\ref{35}$ is not affected by equation content, and we have $\hspace{600pt}\ref{36}$. Okay so there's no need to manually tag (label) the parenthesis just one-sided? Now do these $\hspace{700pt}\ref{37}$ go over the boarder $\hspace{800pt}\ref{38}$? Yes they totally do. So it is capped at 600? How about $\hspace{600pt}\ref{37}$ and $\hspace{600pt}\ref{38}$? Yes, exactly 600pt….. wait, there’s a HUGE problem to this system with any negative space: the page jumps to the middle (or below) the equation and cutting the letters in half, unlike the standard setup that jumps to show the entire equation regardless of the height of the expression. The immediate response to solve this is to always setup a phantom line just above $\tag*{wherever}\label{keypoint}$ you want to jump to (ha, not exactly like this), and maybe use some commands the squash the height of the phantom line. This is less than

$\Large\color{green}{\text{I guess for the sake of sectioning}\ldots} \tag*{Sec.1} \label{sec01}$

. . . it’s okay to setup a phantom line (the whole section heading is in mathmode anyway).


In tincidunt finibus porttitor. Integer lobortis dignissim iaculis. Mauris metus lectus, posuere eu cursus a, tempus sed nunc. Donec ut purus efficitur, pellentesque purus id, auctor erat.

\begin{gather*} s(m-1,n-1)=2^{m-3}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{21} \label{21} \\ =2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-1-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n}{2}-1-i} \tag{$\color{red}{red tag}$} \label{22} \\ =-2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}\frac{(n-1)!}{(\frac{n}{2}-1-i)!(\frac{n}{2}+i)!} \tag{white $\color{white}{tag}$} \label{23} \\ =-2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}\frac{n!}{(\frac{n}{2}-i)!(\frac{n}{2}+i)!}\frac{\frac{n}{2}-i}{n} \tag{$\hspace{-120pt}\color{blue}{\Large blue Large}$ tag negative space} \label{24}\\ n\equiv 0 \pmod 2 \tag{$\hspace{-120pt}\color{magenta}{\large magenta large}$ same -120pt} \label{25} \end{gather*}

Curabitur accumsan sed nisi vitae finibus. Quisque vitae nisl eu eros egestas sollicitudin in vel turpis. Donec dapibus, nibh nec fringilla suscipit, turpis leo tempor nibh, eleifend accumsan nisi mauris quis eros.

\begin{gather*} D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-100pt}$(-100pt)} \label{31} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-200pt}$(-200pt)} \label{32} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-300pt}$(-300pt)} \label{33} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-400pt}$Eq.(-400pt)} \label{34} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-500pt}$Eq.(-500pt)} \label{35} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-600pt)} \label{36} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-700pt)} \label{37} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-800pt)} \label{38} \end{gather*} Nunc malesuada suscipit quam non viverra. Praesent ante justo, semper non efficitur non, ullamcorper eget nulla. Pellentesque quis mi sit amet nulla consequat iaculis. Morbi consectetur augue nec leo aliquet, at venenatis est malesuada. Sed dapibus vehicula velit a rhoncus. Morbi a mi sollicitudin, accumsan odio ut, viverra turpis.

$\tag*{$\large\color{green}{\textit{Section.2}}$} \label{sec02}$

Nullam vulputate fermentum vestibulum. Integer dictum libero et sem consectetur, id consequat odio vestibulum. Nullam vel efficitur mi. Phasellus turpis massa, volutpat id blandit vitae, lacinia vel quam. Proin vitae ex quis mauris blandit molestie fringilla vestibulum augue. Phasellus euismod dictum eros sit amet vulputate. Nam condimentum convallis nisi, id tristique libero porttitor sit amet.

$\tag*{$\hspace{-600pt}\large\color{blue}{\textbf{Section.3}}$} \label{sec03}$

Praesent eleifend nisi sed eros bibendum, ut tincidunt nunc mattis. Pellentesque mi metus, ultrices congue tellus sit amet, semper bibendum ligula. In risus eros, tempor et tempus sit amet, imperdiet at dolor. Nam quis tristique elit. Praesent venenatis varius euismod. Fusce dignissim lorem sed odio dignissim, eu congue ex malesuada.

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Functions are generally $\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]\rightarrow\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]$ with input matrices $\in\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]$.

$$\begin{align} \mu(a) &= b \Leftrightarrow b_{ij} = a_{8i + j}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \gamma(a) &= b\Leftrightarrow b_{ij} = S[a_{ij}]\text{, }0\leqslant i\text{, }j\leqslant 7\\ \pi(a) &= b\Leftrightarrow b_{ij} = a_{(i-j)\bmod 8,j}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \sigma[k](a) &= b \Leftrightarrow b_{ij} = a_{ij}\oplus k_{ij}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \\ c^{r}_{0j} &\equiv 0\text{, }1\leqslant i\leqslant 7\text{, }0\leqslant j\leqslant 7\\ c^{r}_{0j} &\equiv S[8(r - 1) + j]\text{, }0\leqslant j\leqslant 7\\ \\ \rho[k] &\equiv \sigma[k]\circ\theta\circ\mu\circ\gamma\\ \\ K^0 &= K\\ K^r &= \rho[c^r](K^{r-1})\text{, } r \ge 0\\ \\ W[K] &= \biggl(\underset{1}{\overset{r=R}{\bigcirc}} \rho[K^r]\biggr)\circ \sigma[K^0]\\ \\ \eta_i &= \mu(m_i)\\ H_0 &= \mu(IV)\\ H_i &= W[H_{i-1}](\eta_i)\oplus H_{i-1}\oplus\eta_i\text{, }1\leqslant i\leqslant t\\ \\ \operatorname{Whirlpool}(M) &\equiv \mu^{-1}(H_t) \end{align}$$

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I am unable to find some examples on the main site right now, but I think that I have seen some instances where MathJax formulas with slightly higher vertical height are hidden behind the text. Let us try some experiments here:

It seems that $$ a=\frac{A}{K}, b=\frac{B}{K}, c=\frac{C}{K}, d=\frac{dK}{K}, n=\frac{N}{K}. $$ works fine, but using \text inside math mode might lead to problems: $$ a=\frac{A}{K}, b=\frac{B}{K}, c=\frac{C}{K}, d=\frac{dK}{K} \text{ and } n=\frac{N}{K}. $$

However, this contains \text{ and } and it is displayed fine: $$A=\frac{a}{a+b+c}, B=\frac{b}{a+b+c}, \text{ and } C=\frac{c}{a+b+c}$$ In this way we can transform the problem to a problem about a homogeneous polynomial.

Since both sides are multiplicative functions, it suffices to verify the equality for $n=p^a$. In this case the LHS equals $$\frac{n}{n\left(1-\frac1p\right)}=\frac1{1-\frac1p}=\frac{p}{p-1}.$$ The RHS is equal to $$1+\frac1{p-1}=\frac{p}{p-1}.$$

It is also known that $A, B, C, K$ and $N$ have a common factor $K$, and so dividing by $K$ gives the equivalent scalar equation $$ ax + by + cxy + d = n, $$ where $a, b, c, d$ and $n$ are respectively equal to$$ a=\frac{A}{K},\; b=\frac{B}{K},\; c=\frac{C}{K},\; d=\frac{dK}{K}\;\text{ and }\; n=\frac{N}{K}. $$ Therefore given these two equations supplied with constants $A, B, C, a, b, c, D$ and constants $K, N$ and $n$, is it possible to solve for $x, y$ and, $z = xy$. (This was copied from here.)

We put $$\overline D=\limsup_{n\to\infty}\frac{w_n}{\sum\limits_{k=1}^n w_k},$$ where $w_k\ge0$ are the given weights.

We put $$\overline D=\limsup_{n\to\infty}\frac{w_n}{\sum\limits_{k=1}^n w_k}\text{ and }\underline D=\liminf_{n\to\infty}\frac{w_n}{\sum\limits_{k=1}^n w_k}$$ where $w_k\ge0$ are the given weights.

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For your concern about Kodaira embedding theorem:

This follows from the inverse function theorem and is completely local (being an analytic subvariety is an local condition).

Let $k\le n$, $U\subset \mathbb C^k$ be open and $f : U \to \mathbb C^n$ be a holomorphic map so that it is immersed at $p\in U$. By composing with a complex linear map $\mathbb C^n \to \mathbb C^n$, we assume $$ \operatorname{Im} df(p) = \mathbb C^k \times \{0\} \subset \mathbb C^n.$$

Then the function $$ F: U\times \mathbb C^{n-k} \to \mathbb C^n, \ \ \ F(z, u) = f(z) + (0,u)$$

has invertible $dF (p, 0)$. Thus there is $V\subset U \times \mathbb C^{n-k}$, W\subset \mathbb C^n$ so that $F : V \to W$ is a biholomorphism, and in particular,

\begin{align} \operatorname{Im} f (U\cap V) &= \{ F(z, u) : u_1 = \cdots u_{n-k} = 0\}\\ &= \{ w \in W : F^{-1}\circ u_1 = \cdots =F^{-1}\circ u_{n-k} = 0\}. \end{align}

and $F^{-1} \circ u_i$ are holomorphic functions on $W$. Thus the image is an analytic subvariety.

If your concern is why

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\begin{align} xXyYzZ &= iIjJkK & aAbBcC &= fFgGhH & \alpha^2\beta^3\gamma^5 &= \boxed{\Lambda \eta\Gamma_1 } \\ aAbBcC &= fFgGhH & \boxed{xXyYzZ} &= iIjJkK & \alpha^7\beta^{11}\gamma^{13} &= \Lambda \eta\Gamma_2 \\ iIjJkK &= \boxed{aAbBcC} & xXyYzZ &= fFgGhH & \beta^{-2}\gamma^{3\alpha}\eta^5 &= \Omega \xi\Gamma_3 \\ fFgGhH &= xXyYzZ & iIjJkK &= aAbBcC & \delta^{\log 2}\sqrt\rho\mu^{-1} &= \epsilon \chi\Theta_4 \\ xXyYzZ &\begin{aligned}[t] &= fFgGhH_1 \\ &= fFgGhH_2 \end{aligned} & aAbBcC &\!\begin{aligned}[t] &= iIjJkK_1 \\ &= iIjJkK_2\end{aligned} & (1/2)\Delta\nu^{-1} &\!\begin{aligned}[t] &= \Psi \phi\Xi_{5_1} \\ &= \Psi \phi\Xi_{5_2} \end{aligned} \\ xXyYzZ &\!\begin{aligned}[t] &= fFgGhH_1 \\ &= fFgGhH_2 \end{aligned} & aAbBcC &\boxed{\!\begin{aligned}[t] &= iIjJkK_1 \\ &= iIjJkK_2\end{aligned} } & (1/2)\Delta\nu^{-1} &\boxed{\!\!\begin{aligned}[t] &= \Psi \phi\Xi_{6_1} \\ &= \Psi \phi\Xi_{6_2} \end{aligned} } \\ xXyYzZ &\!\begin{aligned}[t] &= fFgGhH_1 \\ &= fFgGhH_2 \end{aligned} & aAbBcC &\boxed{ \hspace{-2.4pt} \! \begin{aligned}[t] &= iIjJkK_1 \\ &= iIjJkK_2\end{aligned} } & (1/2)\Delta\nu^{-1} &\!\begin{aligned}[t] &= \Psi \phi\Xi_{7_1} \\ &= \Psi \phi\Xi_{7_2} \end{aligned} \\ xXyYzZ &\!\begin{aligned}[t] &= fFgGhH_1 \\ &= fFgGhH_2 \end{aligned} & aAbBcC &\!\begin{aligned}[t] &= iIjJkK_1 \\ &= iIjJkK_2\end{aligned} & (1/2)\Delta\nu^{-1} &\!\begin{aligned}[t] &= \Psi \phi\Xi_{8_1} \\ &= \Psi \phi\Xi_{8_2} \end{aligned} \end{align}

Urna porttitor rhoncus dolor purus non enim praesent elementum facilisis. Velit laoreet id donec ultrices tincidunt. Donec ultrices tincidunt arcu non sodales. Cursus risus at ultrices mi tempus.

$\require{begingroup}\begingroup\require{color} \definecolor{blue}{RGB}{3,169,244} \definecolor{green}{RGB}{76,175,80} \definecolor{red}{RGB}{244,67,54} \newcommand{\k}[1]{\color{black}{\textsf{#1}}} \newcommand{\b}[1]{\color{blue}{\textsf{#1}}} \newcommand{\g}[1]{\color{green}{\textsf{#1}}} \newcommand{\r}[1]{\color{red}{\textsf{#1}}}$Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. $\g{Nibh tellus molestie nunc non bla}\b{ndit massa enim. Vitae justo eget}$ magna fermentum iaculis eu non diam phasellus. Sit amet volutpat consequat mauris nunc congue nisi vitae suscipit. Eleifend mi in nulla posuere. Quis enim lobortis scelerisque fermentum dui faucibus in ornare quam. Magna fringilla urna porttitor rhoncus dolor purus non enim praesent.

$$\color{red}\cos\left(\frac{2\pi}{7}\right) + \color{green}\cos \left(\frac{4\pi}{7}\right) + \color{teal}\cos \left(\frac{6\pi}{7}\right) = -\frac{1}{2} \\ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right) = -\frac{1}{2} \\ \color{orange}\cos\left(\frac{2\pi}{7}\right) + \color{#800020}\cos \left(\frac{4\pi}{7}\right) + \color{brown}\cos \left(\frac{6\pi}{7}\right) = -\frac{1}{2}$$

Consequat ac felis donec et odio pellentesque. Cras ornare arcu dui vivamus. Et egestas quis ipsum suspendisse ultrices gravida dictum fusce ut. Turpis egestas maecenas pharetra convallis posuere morbi leo urna.

$$\cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right) = -\frac{1}{2} \\ \color{purple}\cos\left(\frac{2\pi}{7}\right) + \color{#C0A000}\cos \left(\frac{4\pi}{7}\right) + \color{#674EA7}\cos \left(\frac{6\pi}{7}\right) = -\frac{1}{2}$$ Fermentum odio eu feugiat pretium nibh ipsum consequat. Faucibus a pellentesque sit amet porttitor eget dolor. $\endgroup$

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Markdown input: dollar-a-space-backslash-dollar-space-b-dollar

Preview: a-dollar-b in math mode

Output: a-dollar-b in math mode (matches the preview)

$a \$ b$


Markdown input: backtick-dollar-a-space-backslash-dollar-space-b-dollar-backtick

Preview: dollar-a-space-backslash-backslash-dollar-space-b-dollar in code format

Output: dollar-a-space-backslash-backslash-backslash-dollar-space-b-dollar in code format

$a \\\$ b$

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A simple mental model of a PDA or NPDA is a finite state machine plus a stack.

Both parts can be used to remember which kind has been the input taken by the automaton so far. The finite-state-machine part can remember bounded-many information. The stack can remember unbounded-many information (but for any particular moment, it is only readable/changeable at the very top).

When there is some information that cannot be remembered by the stack easily, the finite-state-machine part can help remember.


The transitions of the wanted NPDA can be as simple as the following.

# when next input is $a$, the state becomes $q_a$. So, after these transitions, the subscript "${\,}_a$" in state "$q_a$" means the last input symbol was an $a$.
$q_1, a,\$\to q_a, a\$$
$q_1, a, a\to q_a, aa$
$q_a, a, a\to q_a, aa$
$q_1, a, b\to q_a, \lambda$
$q_a, a, b\to q_a, \lambda$

# when next input is $b$, the state becomes $q_1$. So $q_1$ implies the last input symbol was not an $a$. $q_1, b,\$\to q_1, b\$$
$q_1, b, b\to q_1, bb$
$q_a, b, b\to q_1, bb$
$q_1, b, a\to q_1, \lambda$
$q_a, b, a\to q_1, \lambda$

# Only at state $q_1$, the NPDA may transition to accept.
$q_a, \lambda, \$\to q_f, \$$

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  • $\begingroup$ What if Victor continues to next round when he has been convinced, then leave the protocol after step 1? He would have gained extra knowledge! $\endgroup$
    – Apass.Jack
    Apr 13, 2022 at 5:16
  • $\begingroup$ [mcve] does it work? $\endgroup$
    – Apass.Jack
    May 23, 2022 at 18:35
  • $\begingroup$ Short.` a ` ` b` ` c ` $\endgroup$
    – Apass.Jack
    Jun 10, 2022 at 13:00
  • $\begingroup$ Can I do strike through? <strike>bad words</strike> <del>wrong way></del> $\endgroup$
    – Apass.Jack
    Dec 22, 2022 at 16:44
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\begin{equation} \Delta(t) =\frac{\partial O}{\partial S} \end{equation}

\begin{equation} \Delta(t) =\frac{\partial O}{\partial S} \tag{1} \end{equation}

$O(t)$

$\displaystyle\frac{\partial O}{\partial S}\in(0,1]$

$\partial O / \partial S$

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Testing a CR bug.

","abc","
","abc"," $abc$
$$abc$$


When writing a MathJax heavy post I noticed a couple of bugs. For reference, here is an example of how the MathJax should look:

$$ a_{i+1} = \begin{cases} \frac{a_i}{2} & \text{if $a_i \% 2 = 0$ (even)} \\ 3a_i + 1 & \text{if $a_i \% 2 = 1$ (odd)} \end{cases} $$

An unclosed math-mode delimiter swallows some backslashes

Related: https://codereview.meta.stackexchange.com/q/7036

Code

$$
a_{i+1} = \begin{cases}
    \frac{a_i}{2}  & \text{if $a_i \% 2 = 0$ (even)} \\
    3a_i + 1 & \text{if $a_i \% 2 = 1$ (odd)}
\end{cases}

Live

$$ a_{i+1} = \begin{cases} \frac{a_i}{2} & \text{if $a_i % 2 = 0$ (even)} \ 3a_i + 1 & \text{if $a_i % 2 = 1$ (odd)} \end{cases}

Preview

How the live is displayed in the preview.

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We have $G$ solvable, so that we can find a tower of subgroups $$ G = G_0 \supset G_1 \supset\cdots\supset G_m = 1 $$ so that the conditions on solvability are met. We then take the quotient so that $$ G/H = (G/H)_0\supset (G/H)_1 \supset\cdots\supset (G/H)_m = 1 $$ where $(G/H)_j = G_j/H$. It is then immediately verified that all factor groups in this tower are abelian, as $(G_i/H)/(G_{i+1}/H) = G_i/G_{i+1}$, which is abelian by assumption.

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This answer is free for editing.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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This answer is free for anyone to edit.

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$$\begin{align} \{0,1\}^\ast\rightarrow\{0,1\}^n\\ \{0,1\}^n\rightarrow\{0,1\}^\ast \end{align}$$

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Links with tooltips.

$2^k-1$ an interesting result?">Link to a question - this looks ok in the preview; but not in the final post.

After removing MathJax from the tooltip: Link to a question

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$$ \begin{array}{c|c|c} \begin{array}{c} R_{−+}, x_0 < −e_0, x_1 > e_1 \\ C − e_0U_0 + e_1U_1 \\ (x_0 + e_0)^2 + (x_1 − e_1)^2 + x^2_2 \end{array} & \begin{array}{c} R_{0+}, |x_0| ≤ e_0, x_1 > e_1 \\ C + x_0U_0 + e_1U_1 \\ (x_1 − e_1)^2 + x^2_2 \end{array} & \begin{array}{c} R_{++}, x_0 > e_0, x_1 > e_1 \\ C + e_0U_0 + e_1U_1 \\ (x_0 - e_0)^2 + (x_1 − e_1)^2 + x^2_2 \end{array} \\\hline \begin{array}{c} R_{−0}, x_0 < −e_0, |x_1| ≤ e_1 \\ C − e_0U_0 + x_1U_1 \\ (x_0 + e_0)^2 + x^2_2 \end{array} & \begin{array}{c} R_{00}, |x_0| ≤ e_0, |x_1| ≤ e_1 \\ C + x_0U_0 + x_1U_1 \\ x^2_2 \end{array} & \begin{array}{c} R_{+0}, x_0 > e_0, |x_1| ≤ e_1 \\ C + e_0U_0 + x_1U_1 \\ (x_0 - e_0)^2 + x^2_2 \end{array} \\\hline \begin{array}{c} R_{−-}, x_0 < −e_0, x_1 < -e_1 \\ C − e_0U_0 - e_1U_1 \\ (x_0 + e_0)^2 + (x_1 + e_1)^2 + x^2_2 \end{array} & \begin{array}{c} R_{0-}, |x_0| ≤ e_0, x_1 < -e_1 \\ C + x_0U_0 - e_1U_1 \\ (x_1 + e_1)^2 + x^2_2 \end{array} & \begin{array}{c} R_{+-}, x_0 > e_0, x_1 < -e_1 \\ C + e_0U_0 - e_1U_1 \\ (x_0 - e_0)^2 + (x_1 + e_1)^2 + x^2_2 \end{array} \end{array} $$

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