Math rendering engine

Xyzzy · 2005-08-29, 19:44

$\2^{57,885,161}-1$

This is using our new $\small \TeX$ tags... Look at the last entry here:

http://www.mersenneforum.org/misc.php?do=bbcode

Manual:

http://www.mersenneforum.org/mimetex.htm

paulunderwood · 2005-08-29, 20:06

Great!! now we can write things like $e^{i\pi}+1=0$ and $\sum_{i=1}^{n}x_i=\int_0^{1}f$ .

We need a few good links to online tex guides...

xilman · 2005-08-29, 20:08

Quote:

Originally Posted by Xyzzy

$\2^p-1$

This is using our new $\small \TeX$ tags... Look at the last entry here:

http://www.mersenneforum.org/misc.php?do=bbcode

Excellent!

I've long wanted to be able to use $\TeX$ formatting.

Paul

Xyzzy · 2005-08-29, 20:18

There are several reasons why this took so long to get set up. The first reason is our old server didn't run Linux so a lot of software wasn't available. Secondly, we had very limited access to the server which prevented us from doing really cool stuff. We were also was hesitant to dive into a solution that didn't have a workable license. As it is, it pains us that the forum software itself isn't GPLed.

We have no idea how much of a load this will put on the server, but at any time we can redirect the image generation to another box running this. For now, especially if you are going to test some really crazy stuff, use the interactive editor thing.

Now that we have this set up, we realize we could have had this a long time ago, by redirecting the image generation to another box. We apologize that it took us this long to figure it out.

Who is going to post the most interesting example formula?

Have fun!

Numbers · 2005-08-29, 23:30

Just testing.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Could be bigger, don't you think?

Xyzzy · 2005-08-29, 23:39

Quote:

Originally Posted by Numbers

Could be bigger, don't you think?

$\LARGE \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Numbers · 2005-08-30, 00:00

$\displaystyle2\left(\frac12+\frac23\right)=\frac53$

Xyzzy, I think you're an absolute genius.
But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.

Xyzzy · 2005-08-30, 00:03

Quote:

Originally Posted by Numbers

But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.

Just hit "Quote" to see the source and you will see we added a \LARGE command. The documentation covers this. We have a sneaky feeling we could spend a lot of time learning all this. It seems to be very powerful. The interactive editor we linked to earlier makes it easier to try new stuff quickly.

Wacky · 2005-08-30, 00:14

Quote:

Originally Posted by Numbers

$\displaystyle2\left(\frac12+\frac23\right)=\frac53$

Big, or small, I don't like the math.

$\LARGE\displaystyle2\left(\frac12+\frac23\right)=\frac73$

jinydu · 2005-08-30, 02:04

So there is more than one type of Latex?

$\displaystyle{f(x)=f(a)+f'(a)x+\frac{1}{2}f''(a)x^2+\frac{1}{6}f'''(a)x^3+...}$

(I guess that answers my question. What I just typed would surely have worked on a certain other forum).

Dougy · 2005-08-30, 04:32

Often I use congruences, so I'll just pick a theorem at random...

Theorem: Let $p$ be a prime such that $p^2$ divides $M_q=2^q-1$ for some prime $q$ . Then $2^{\frac{p-1}{2}} \equiv 1 \pmod {p^2}$ and so $p$ is a Wieferich prime.

Proof: We know that if $p$ divides $M_q=2^q-1$ , then $p=2kq+1$ for some $k \in \mathbb{N}$ . Then $q=\frac{p-1}{2k}$ , and if $p^2$ divides $M_q=2^q-1$ then $2^q \equiv 2^{\frac{p-1}{2k}} \equiv 1 \pmod {p^2}$ .

2005-08-29, 19:44	#1
Xyzzy Aug 2002 2³·1,069 Posts	Math rendering engine $\2^{57,885,161}-1$ This is using our new $\small \TeX$ tags... Look at the last entry here: http://www.mersenneforum.org/misc.php?do=bbcode Manual: http://www.mersenneforum.org/mimetex.htm

2005-08-29, 23:30	#5
Numbers Jun 2005 Near Beetlegeuse 2²×97 Posts	Just testing. $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Could be bigger, don't you think? Last fiddled with by Numbers on 2005-08-29 at 23:31

2005-08-30, 02:04	#10
jinydu Dec 2003 Hopefully Near M48 3336₈ Posts	So there is more than one type of Latex? $\displaystyle{f(x)=f(a)+f'(a)x+\frac{1}{2}f''(a)x^2+\frac{1}{6}f'''(a)x^3+...}$ (I guess that answers my question. What I just typed would surely have worked on a certain other forum). Last fiddled with by jinydu on 2005-08-30 at 02:05

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2005-08-29, 20:06	#2
paulunderwood Sep 2002 Database er0rr 2·3³·83 Posts	Great!! now we can write things like $e^{i\pi}+1=0$ and $\sum_{i=1}^{n}x_i=\int_0^{1}f$ . We need a few good links to online tex guides...

2005-08-29, 20:18	#4
Xyzzy Aug 2002 2³×1,069 Posts	There are several reasons why this took so long to get set up. The first reason is our old server didn't run Linux so a lot of software wasn't available. Secondly, we had very limited access to the server which prevented us from doing really cool stuff. We were also was hesitant to dive into a solution that didn't have a workable license. As it is, it pains us that the forum software itself isn't GPLed. We have no idea how much of a load this will put on the server, but at any time we can redirect the image generation to another box running this. For now, especially if you are going to test some really crazy stuff, use the interactive editor thing. Now that we have this set up, we realize we could have had this a long time ago, by redirecting the image generation to another box. We apologize that it took us this long to figure it out. Who is going to post the most interesting example formula? Have fun!

2005-08-30, 00:00	#7
Numbers Jun 2005 Near Beetlegeuse 110000100₂ Posts	$\displaystyle2\left(\frac12+\frac23\right)=\frac53$ Xyzzy, I think you're an absolute genius. But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.

2005-08-30, 04:32	#11
Dougy Aug 2004 Melbourne, Australia 98₁₆ Posts	Often I use congruences, so I'll just pick a theorem at random... Theorem: Let $p$ be a prime such that $p^2$ divides $M_q=2^q-1$ for some prime $q$ . Then $2^{\frac{p-1}{2}} \equiv 1 \pmod {p^2}$ and so $p$ is a Wieferich prime. Proof: We know that if $p$ divides $M_q=2^q-1$ , then $p=2kq+1$ for some $k \in \mathbb{N}$ . Then $q=\frac{p-1}{2k}$ , and if $p^2$ divides $M_q=2^q-1$ then $2^q \equiv 2^{\frac{p-1}{2k}} \equiv 1 \pmod {p^2}$ .