Math rendering engine - Page 6

MattcAnderson · 2013-11-10, 22:49

Hi Mersenne forum,

I want to try -

hello world

you did a good thing

http://mathforum.org/kb/message.jspa?messageID=6744802

2³ + 1 = 3*3

ok, that was fun

*grinz*

Regards,
Matt

R.D. Silverman · 2013-11-11, 11:59

Quote:

Originally Posted by MattcAnderson

Hi Mersenne forum,

I want to try -

hello world

you did a good thing

http://mathforum.org/kb/message.jspa?messageID=6744802

2³ + 1 = 3*3

ok, that was fun

*grinz*

Regards,
Matt

Do you have a urge for producing content free posts?
You do so repeatedly.

xilman · 2013-11-11, 12:24

Quote:

Originally Posted by R.D. Silverman

Do you have a urge for producing content free posts?
You do so repeatedly.

Ooh, Bob, that's a bit harsh. "Homeopathic" might be more appropriate.

R.D. Silverman · 2013-11-11, 13:31

Quote:

Originally Posted by xilman

Ooh, Bob, that's a bit harsh. "Homeopathic" might be more appropriate.

Is that homeopathic or homeopathetic?

only_human · 2013-11-11, 20:51

(Just checking out the renderer...)
http://www.mathjax.org/demos/tex-samples/

Quote:

The Lorenz Equations
$\dot{x} = \sigma(y-x) \\ \dot{y} = \rho x - y - xz \\ \dot{z} = -\beta z + xy $

The Cauchy-Schwarz Inequality
$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

A Cross Product Formula
$\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix}$

The probability of getting k heads when flipping n coins is
$P(E) = {n \choose k} p^k (1-p)^{ n-k}$

An Identity of Ramanujan
$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } }$

A Rogers-Ramanujan Identity
$1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for $|q|<1$}.$

Maxwell’s Equations
$\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0$

Tex command inline here: $\sqrt{3x-1}+(1+x)^2$ in this sentence.
The code for the quoted region above is:

Quote:

[SIZE="4"]The Lorenz Equations
[TEX]\dot{x} = \sigma(y-x) \\
\dot{y} = \rho x - y - xz \\
\dot{z} = -\beta z + xy
[/TEX]

The Cauchy-Schwarz Inequality
[TEX]\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)[/TEX]

A Cross Product Formula
[TEX]\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}[/TEX]

The probability of getting k heads when flipping n coins is
[TEX]P(E) = {n \choose k} p^k (1-p)^{ n-k}[/TEX]

An Identity of Ramanujan
[TEX]\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } }[/TEX]

A Rogers-Ramanujan Identity
[TEX]1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|<1$}.[/TEX]

Maxwell’s Equations
[TEX]\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\
\nabla \cdot \vec{\mathbf{B}} & = 0[/TEX]
[/SIZE]

Gamma function from https://stackedit.io/ welcome document:
The Gamma function satisfying $\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N$ is via through the Euler integral
$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,$
Trying something here. This is from How to draw a commutative diagram? I changed \xrightarrow to \rightarrow because \xrightarrow seemed to cause a syntax error. The small letters are supposed to label arrows (not follow them) so this is not right as yet:
$\begin{array}{ccccccccc} 0 & \rightarrow{i} & A & \rightarrow{f} & B & \rightarrow{q} & C & \rightarrow{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \downarrow\\ 0 & \rightarrow{j} & D & \rightarrow{g} & E & \rightarrow{r} & F & \rightarrow{e} & 0\end{array}$

CRGreathouse · 2013-11-12, 04:34

Quote:

Originally Posted by only_human

The small letters are supposed to label arrows (not follow them) so this is not right as yet:
$\begin{array}{ccccccccc} 0 & \rightarrow{i} & A & \rightarrow{f} & B & \rightarrow{q} & C & \rightarrow{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \downarrow\\ 0 & \rightarrow{j} & D & \rightarrow{g} & E & \rightarrow{r} & F & \rightarrow{e} & 0\end{array}$

Subscripting seems to work, is there something you wanted to avoid here?

$\begin{array}{ccccccccc} 0 & \rightarrow_{i} & A & \rightarrow_{f} & B & \rightarrow_{q} & C & \rightarrow_{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \downarrow\\ 0 & \rightarrow_{j} & D & \rightarrow_{g} & E & \rightarrow_{r} & F & \rightarrow_{e} & 0\end{array}$

LaurV · 2013-11-12, 06:11

Why, the superscripting doesn't work for you??

$\begin{array}{ccccccccc} 0 & \rightarrow^{i} & A & \rightarrow^{f} & B & \rightarrow^{etc}&C \end{array}$

(extended arrows use 2 parameters, if you replaced to normal arrow, the second one is not a parameter anymore, so you must use indexing, _ or ^ symbols)

only_human · 2013-11-12, 06:35

Either would be fine. I copied it from the linked reference late in my editing because I was looking for something more graphical and I ran out of time. I'm not that TeX savvy and was just massaging the text that I copied to get rid of syntax errors. I have LyX and MiKTeX on my system but haven't been using them. I was also distracted by not wanting to use Category Theory words to talk about that entry because I didn't feel like being a dog trying to bark Greek.

LaurV · 2013-11-12, 06:49

Quote:

Originally Posted by only_human

[offtopic]Now I am getting angry with Xyzzy! Even Raman got his own smiley.... And I am stuck with the big dog, which I can't use...

It even doen't look like me, it is tall and slender and I am small and fat

[/offtopic]

Xyzzy · 2014-09-30, 05:46

Here is an interesting page: http://detexify.kirelabs.org/classify.html

only_human · 2014-09-30, 08:52

Quote:

Originally Posted by Xyzzy

Here is an interesting page: http://detexify.kirelabs.org/classify.html

Cool. It works from my tablet, which is never a certainty.

2013-11-12, 06:11	#62
LaurV Romulan Interpreter Jun 2011 Thailand 7·1,373 Posts	Why, the superscripting doesn't work for you?? $\begin{array}{ccccccccc} 0 & \rightarrow^{i} & A & \rightarrow^{f} & B & \rightarrow^{etc}&C \end{array}$ (extended arrows use 2 parameters, if you replaced to normal arrow, the second one is not a parameter anymore, so you must use indexing, _ or ^ symbols) Last fiddled with by LaurV on 2013-11-12 at 06:16 Reason: quote to prec post deleted

2013-11-12, 06:35	#63
only_human "Gang aft agley" Sep 2002 2·1,877 Posts	Either would be fine. I copied it from the linked reference late in my editing because I was looking for something more graphical and I ran out of time. I'm not that TeX savvy and was just massaging the text that I copied to get rid of syntax errors. I have LyX and MiKTeX on my system but haven't been using them. I was also distracted by not wanting to use Category Theory words to talk about that entry because I didn't feel like being a dog trying to bark Greek. Last fiddled with by only_human on 2013-11-12 at 06:44 Reason: added smilies

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2013-11-10, 22:49	#56
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2⁵×5² Posts	Hi Mersenne forum, I want to try - hello world you did a good thing http://mathforum.org/kb/message.jspa?messageID=6744802 2³ + 1 = 33 ok, that was fun grinz* Regards, Matt

2014-09-30, 05:46	#65
Xyzzy "Mike" Aug 2002 3·2,741 Posts	Here is an interesting page: http://detexify.kirelabs.org/classify.html