Formatting Sandbox

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

$\hskip -3em \color{red}{\Rule{5em}{1em}{1em}}$. Testing of negative skips to overlap the buttons on the left.

$\rlap{\smash{\lower 3em{\color{red}{\Rule{5em}{2em}{0em}}}}}$Testing overlapping on the bottom. OK, both seem to be problems.

Answer 4

How does newcommand and renewcommand work?

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

$$\sin y$$

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

Answer 5

And now what does it look like to another user who doesn't suspect that the command has been redefined?

$$\sin x$$

Very interesting.

Answer 6

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:

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

Answer 7

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

Answer 8

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

Answer 9

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..

Answer 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!

$$¡^IGNORE\ \ M_e!$$

$$¡^IGNORE\ \ M_e!$$

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

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

Answer 11

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.

Answer 12

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!

Answer 13

Testing striking out:

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

tag: text tex text

url: text math.SE text

Answer 14

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): $\AnotherRidiculouslyLongName$

Answer 15

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

Answer 16

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

Answer 17

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.