LaTeX Spoken Here!

Published on December 12, 2006 in asymptotia, mathematics and science. 148 Comments

This is a test of LaTeX on the site*.

The first equation I shall try is the following (for more on unpacking this equation and its meaning, see this post and links therein):

$u{\cal R}^2-\frac12 {\cal R}{\cal R}^{\prime\prime}+\frac14({\cal R}^\prime)^2=\Gamma^2\ .$

Yay! It works. I have implemented it in the comments too. So now we can have a new, sharper tool for our discussions and arguments.

Use: Type simple LaTeX commands enclosed between [ tex ] and [ / tex] (remove the spaces between the things in the square brackets) and it should work once you submit.

Enjoy.

-cvj

(*I got around the problems of not being able to have LaTeX running on my host. Hurrah! The compromise I used means that the LaTeX is not as nicely formed as it could be, but it’s good enough! Learn more about latexrender and mimetex here.)

148 Responses to “LaTeX Spoken Here!”

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stevem
December 13, 2006 at 1:46 pm

Darn, my hodge star is out. Quite nice! Just needs a bit more practice for use in comments. The case $\mathbb{G}=SU(3)$ is really hard to prove though–took me hours one night
Kea
December 13, 2006 at 1:47 pm

WOW!!! Let me have a go. This is a real long shot:

$\xymatrix {A \ar[d]_{!} \ar@{{}*!/-8pt/@{>}->}[r]^{m} & B \ar[d]^{\chi_{m}} \\ 1 \ar@{{}*!/-8pt/@{>}->}[r]_{t} \Omega}$
Kea
December 13, 2006 at 1:48 pm

OK, that was too ambitious. Sigh. But THANKS, Clifford! Is there a free blogging setup where one could do this? Or maybe we should just ask Google nicely…
Euler
December 13, 2006 at 4:16 pm

$e^{i\pi}+1=0$
Garrett Lisi
December 13, 2006 at 4:21 pm

Hmm, how about some little arrows…
${\cal L}_{\vec{v}} \underrightarrow{f} = \left( \underrightarrow{\partial} \vec{v} \right) \underrightarrow{f} + \left( \vec{v} \underrightarrow{\partial} \right) \underrightarrow{f}$
Robert
December 13, 2006 at 4:31 pm

$\begin{equation*} |x|= \begin{cases} x & \text{if $xÃ¢â�°Â¥0$,} \\ -x &\text{if $x\le 0$.} \end{cases} \end{equation*}$
That’s just plain geekalicious!
Garrett Lisi
December 13, 2006 at 4:44 pm

OK, lets try again…

${\cal L}_{\overset{\rightharpoonup}{v}} \underset{\rightharpoondown}{f} = \left( \underset{\rightharpoondown}{\partial} \overset{\rightharpoonup}{v} \right) \underset{\rightharpoondown}{f} + \left( \overset{\rightharpoonup}{v} \underset{\rightharpoondown}{\partial} \right) \underset{\rightharpoondown}{f}$

(By the way, that’s the Lie derivative of a 1-form.)
Gebar
December 13, 2006 at 4:49 pm

Yes, this is really fun! And I have good news for people with blogs in Blogger or in servers that do not support latex. Have a look here. Peter Jipsen of Chapman University has written ASCIIMathML a great javascript program that converts ASCII notation (more or less the one used in math and physics newsgroups, and of course in blogs, although not this one any more:-)) to MathML. Firefox rendering is great, although you may need to download some fonts. IE needs a plugin and rendering is not so good. Now, you are supposed to upload the javascript file to your server, but I suspect that it will work also if you put the content of the file in the head of your Blogger template. (Be warned though that the file is 42K, and that I have not tried this, I just assume it will work.)

If you do not want to risk this, there are two more solutions mentioned in the page given above. The first one are the ASCIIMath Image Fallback Scripts which use a public mimetex server (meaning you do not have to have mimetex in your server), and the second is LaTeXMathML, which as you may have guessed translates latex notation to mathml using again a public server if you cannot upload the file in your own.

Enjoy!
Carl Brannen
December 13, 2006 at 7:49 pm

The posts here are like the doodles that psychologists fool you into making, when they want to figure out what kind of crazy you are.
Clifford
December 13, 2006 at 7:56 pm

Oh, I’ve got you all figured out, all of you!

-cvj
Biswajit
December 13, 2006 at 8:41 pm

$\boldsymbol{\Nabla}\cdot\boldsymbol{\sigma} + \rho~\mathbf{b} = \rho~\dot{\mathbf{v}}$
Biswajit
December 13, 2006 at 8:44 pm

[tex]
\mathbf{\nabla}\cdot\mathbf{\sigma} + \rho~\mathbf{b} = \rho~\dot{\mathbf{v}}
[\tex]
Biswajit
December 13, 2006 at 8:46 pm

Last try. Delete immediately.
$\mathbf{\nabla}\cdot\mathbf{\sigma} + \rho~\mathbf{b} = \rho~\dot{\mathbf{v}}$
Gebar
December 14, 2006 at 3:31 am

OK, I did not mention the most important thing (for hardcore latex users). All the above programs (ASCIIMathML etc.) also recognize latex notation.
Plato
December 14, 2006 at 8:21 am

I=I_0\:{\rm e}^{-\mu x}

Photons are attenuated in matter via the processes of the photoelectric effect, Compton scattering and pair production. The intensity of a photon beam varies in matter according to

$I=I_0\:{\rm e}^{-\mu x}$

.
Clifford
December 14, 2006 at 8:24 am

fixed it for you plato. -cvj
Basudeb Dasgupta
December 15, 2006 at 8:26 pm

The new tree knows…

[tex] \nu_{e},\nu_{\mu},\nu_{\tau}[\tex]
Basudeb Dasgupta
December 15, 2006 at 8:28 pm

The new tree knowsâ€¦

$\nu_{e}, \nu_{\mu}, \nu_{\tau}$
Count Iblis
December 16, 2006 at 7:10 am

$\exp\left(\pi\sqrt{163}\right)\approx 262537412640768743.9999999999992\ldots$
Youssef Faltas
December 17, 2006 at 11:41 am

$T^{\mu \nu} = – \nabla^{\mu} \phi (x) \frac{\nabla L}{\nabla (\nabla_{\nu} \phi)}+ g_{\mu \nu} L\\ \nabla_{\nu} V^{\mu} = \partial_{\nu} V^{\mu} + \Gamma^{\mu}{}_{\sigma \nu} V^{\sigma}$
Plato
December 18, 2006 at 9:05 am

The converse of the Principle,

$x=y \rightarrow \foral F(Fx \leftrightarrow Fy)$ ,

is called the Indiscernibility of Identicals. Sometimes the conjunction of both principles, rather than the Principle by itself, is known as Leibniz’s Law.
Plato
December 19, 2006 at 11:48 pm

http://en.wikipedia.org/wiki/Perfect_fluid

In physics, a perfect fluid is a fluid that can be completely characterized by its rest frame energy density Ï and isotropic pressure p.

In tensor notation, the energy-momentum tensor of a perfect fluid can be written in the form

$T^{\mu\nu} = (\rho + p) \, U^\mu U^\nu + p \, \eta^{\mu\nu}\,$

where U is the velocity vector field of the fluid and where Î·Î¼Î½ is the metric tensor of Minkowski spacetime

How did you get the latex to show on wordpress without having it downloaded to a server?
Warren
December 22, 2006 at 4:48 am

Here’s a test:

$\def\bigface#1{ \setlength{\unitlength}{.5mm} \begin{figure}[t] \begin{picture}(0,0)(-320,-20) \put(0,0){\circle{24}} #1 \end{picture} \vskip-.3in \end{figure}} \bigface{\put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]}}$
Warren
December 22, 2006 at 4:52 am

another test:

$\setlength{\unitlength}{.5mm} \begin{figure}[t] \begin{picture}(0,0)(-320,-20) \put(0,0){\circle{24}} \put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]} \end{picture} \vskip-.3in \end{figure}$
Warren
December 22, 2006 at 4:54 am

Failed to rasterize \begin{picture}(0,0)(-320,-20) \put(0,0){\circle{24}} \put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]} \end{picture}
yet another test:

$\begin{picture}(0,0)(-320,-20) \put(0,0){\circle{24}} \put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]} \end{picture}$
Warren
December 22, 2006 at 4:58 am

Failed to rasterize \begin{picture}(0,0) \put(0,0){\circle{24}} \put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]} \end{picture}
hope I don’t get banned for too much failed garbage:

$\begin{picture}(0,0) \put(0,0){\circle{24}} \put(-4,3){\circle{4}} \put(4,3){\circle{4}} \put(-3,3){\circle*{2}} \put(3,3){\circle*{2}} \put(0,-1){\oval(12,6)[b]} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)[b]} \end{picture}$
Warren
December 22, 2006 at 5:03 am

Nobody’s reading this thread anymore, right?

$\begin{picture}(30,30) \put(15,15){\circle{24}} \put(11,18){\circle{4}} \put(19,18){\circle{4}} \put(12,18){\circle*{2}} \put(18,18){\circle*{2}} \put(15,14){\oval(12,6)[b]} \put(15,8){\line(0,1){3}} \put(15,11){\oval(4,10)[b]} \end{picture}$
Warren
December 22, 2006 at 5:06 am

I guess mimetex isn’t really latex

$\begin{picture}(300,300) \put(150,150){\circle{240}} \put(110,180){\circle{40}} \put(190,180){\circle{40}} \put(120,180){\circle*{20}} \put(180,180){\circle*{20}} \put(150,140){\oval(120,60)[b]} \put(150,80){\line(0,10){30}} \put(150,110){\oval(40,100)[b]} \end{picture}$
Warren
December 22, 2006 at 5:09 am

zzzzzzzzz………..

$\begin{picture}(300,300) \put(150,150){\circle{240}} \put(110,180){\circle{40}} \put(190,180){\circle{40}} \put(120,180){\circle*{20}} \put(180,180){\circle*{20}} \put(150,140){\oval(120,60)} \put(150,80){\line(0,10){30}} \put(150,110){\oval(40,100)} \end{picture}$
Warren
December 22, 2006 at 5:10 am

$\begin{picture}(300,300)\put(150,150){\circle{240}}\put(110,180){\circle{40}}\put(190,180){\circle{40}}\put(120,180){\circle*{20}}\put(180,180){\circle*{20}}\put(150,140){\oval(120,60)}\put(150,80){\line(0,10){30}}\put(150,110){\oval(40,100)}\end{picture}$
Warren
December 22, 2006 at 5:13 am

ignore me

$\begin{picture}(300,300) \put(150,150){\circle{240}} \end{picture}$
Warren
December 22, 2006 at 5:24 am

Maybe circles are the problem?

$\setlength{\unitlength}{.5mm} \begin{picture}(0,0)(-320,-20) \put(0,0){\oval(24,24)} \put(-4,3){\oval(4,4)} \put(4,3){\oval(4,4)} \put(-3,3){\oval*(2,2)} \put(3,3){\oval*(2,2)} \put(0,-1){\oval(12,6)} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)} \end{picture}$
Warren
December 22, 2006 at 5:25 am

Failed to rasterize \begin{picture}(0,0)(-320,-20) \put(0,0){\oval(24,24)} \put(-4,3){\oval(4,4)} \put(4,3){\oval(4,4)} \put(-3,3){\oval*(2,2)} \put(3,3){\oval*(2,2)} \put(0,-1){\oval(12,6)} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)} \end{picture}
maybe not

$\begin{picture}(0,0)(-320,-20) \put(0,0){\oval(24,24)} \put(-4,3){\oval(4,4)} \put(4,3){\oval(4,4)} \put(-3,3){\oval*(2,2)} \put(3,3){\oval*(2,2)} \put(0,-1){\oval(12,6)} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)} \end{picture}$
Warren
December 22, 2006 at 5:27 am

latex for pictures is obsolete anyway

$\begin{picture}(20,20) \put(0,0){\oval(24,24)} \put(-4,3){\oval(4,4)} \put(4,3){\oval(4,4)} \put(-3,3){\oval*(2,2)} \put(3,3){\oval*(2,2)} \put(0,-1){\oval(12,6)} \put(0,-7){\line(0,1){3}} \put(0,-4){\oval(4,10)} \end{picture}$
Warren
December 22, 2006 at 5:30 am

time to read the manual

$\begin{picture}(300,300) \put(150,150){\oval(240,240)} \put(110,180){\oval(40,40)} \put(190,180){\oval(40,40)} \put(120,180){\oval*(20,20)} \put(180,180){\oval*(20,20)} \put(150,140){\oval(120,60)} \put(150,80){\line(0,10){30}} \put(150,110){\oval(40,100)} \end{picture}$
Peter Fred
January 2, 2007 at 6:43 am

$x^2$
Peter Fred
January 2, 2007 at 9:44 am

$P_c=g_{s}\rho R$

$F_{nite}=g_s\rho R (\pi R^2)$

$F_{day}=(g_s-\Delta g)\rho R(\pi R^2)$

$F_{net}=\Delta g \rho\pi R^3$

$\Delta g \rho\pi R^3=({GM_\odot m_\oplus}/{R_{AU}^2})*(3/4)$

$({GM_\odot m_\oplus}/{R_{AU}^2})*(3/4)$
Peter Fred
January 2, 2007 at 9:51 am

$(\frac{GM_\odot m_\oplus}{r_{AU}^2})*(3/4)$
Peter Fred
January 5, 2007 at 2:19 am

$\Delta g \rho\pi r^3={\frac{GM_\odot m_\oplus}{r_{AU}^2}*(3/4)$
Peter Fred
January 5, 2007 at 2:35 am

${\frac {V^2} {r_{AU}}}g \rho\pi r^3 \approx {\frac{GM_\odot \oplus}{r_{AU}^2}*(3/4)$
Peter Fred
January 5, 2007 at 2:43 am

${\[\frac {V^2} {r_{AU}}\]}\g \rho\pi r^3 \approx {\frac{GM_\odot \oplus}{r_{AU}^2})$
Tester John
February 1, 2007 at 7:50 am

test
tex

\bfseries{Hello}
$\delta_{5} \times \alpha^{\sum i}$
/tex
Tester John
February 1, 2007 at 7:51 am

$\bfseries{Hello} $\delta_{5} \times \alpha^{\sum i}$$
Tester John
February 1, 2007 at 7:52 am

$\bfseries{Hello} $ \delta_{5} \times \alpha^{\sum i}$$
Guido
November 4, 2007 at 6:33 pm

$e^{i \pi} + 1 = 0$
Guido
November 4, 2007 at 6:34 pm

$\sum_{k=0}^{\infty} \cfrac {k!} {(2k + 1)!!} = cfrac \pi 2$

Cool
Guido
November 4, 2007 at 6:37 pm

Messed up

$\sum_{k=0}^{\infty} {\cfrac {k!} {(2k + 1)!!}} = {\cfrac {\pi} {2}}$
test
August 27, 2008 at 10:42 pm

$\sum_{k=1}^{n} \frac{1}{n^3}$

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