Imageboard now has LaTeX.
Name:
Anonymous
2010-02-15 2:13
Text /sci/ is officially obsolete.
Good night, sweet prince. ;_;
/board
Name:
Anonymous
2010-02-15 5:13
Not necessarily, I mean still got the bible trolls there etc
\sqrt{x+2}
Name:
Anonymous
2010-02-15 5:13
Name:
test
2010-02-15 5:41
\mathBB{R}
Name:
Anonymous
2010-02-15 18:42
It's surprisingly active even after they took away the notes, even if half of it is tired trolling. Maybe America is finally shaping up again.
Name:
Anonymous
2010-02-16 0:45
Imageboard still crap, maybe two good threads at any given moment.
Name:
Anonymous
2010-02-16 3:08
The textboard isn't going anywhere. I mean at least threads here live on, and the threads on the imageboard don't.
Name:
Anonymous
2010-02-16 3:59
Name:
Anonymous
2010-02-16 8:22
>>8
01:42:34<@MrVacBob> you can use latex now in /sci/
01:42:39<@MrVacBob> in case you want to do that
Name:
Anonymous
2010-02-16 21:42
>>6
>implying there are any good threads here either>
Name:
Anonymous
2010-02-16 22:05
>>10
>implying this is how you greentext on the textboards
Name:
Anonymous
2010-02-17 4:51
'>green text on the text boards, whoda thunk it
Name:
Anonymous
2010-02-17 16:50
>>10-12
Back to the imageboards, please.
Name:
Anonymous
2010-02-24 8:57
Name:
Anonymous
2010-02-28 19:25
<latex>2x+1=0</latex>
Name:
Anonymous
2010-02-28 19:26
[latex]2x+1=0[/latex]
Name:
Anonymous
2011-03-09 4:53
[eqn] \left ( a+b \right )^{n}=\sum_{i=0}^{n} { }^{n}\textrm{C}_{r} a^{n-i} b^{i} [\eqn]
Name:
Anonymous
2011-03-19 20:51
do people actually go on this thing?
i'm only exploring....
Name:
Anonymous
2011-04-10 11:25
\sqrt{x+2}
Name:
Anonymous
2011-09-29 13:46
f(-4)=-2f(-3)+3=3f(-2)-3=-3f(-1)+6=\frac{3}{2}f(0)-3=\frac{3}{2}
Name:
Anonymous
2011-09-29 13:48
[tex] f(-4)=-2f(-3)+3=3f(-2)-3=-3f(-1)+6=\frac{3}{2}f(0)-3=\frac{3}{2} [/tex]
Name:
Anonymous
2011-09-29 13:50
f(-4)=-2f(-3)+3=3f(-2)-3=-3f(-1)+6=\frac{3}{2}f(0)-3=\frac{3}{2}
Name:
Anonymous
2011-09-29 20:29
\int
honestly i've never even been here.
my first thought was "where are the pictures?"
I really have no clue how I got here.
Name:
Anonymous
2011-10-12 8:24
<math>
\int_{0}^{\inf} e^{-st}.dt
</math>
Name:
Anonymous
2011-10-14 9:26
[tex]$\mathcal{FUCK U I WNT DO WUT U TELL ME}$[/tex]
Name:
Anonymous
2011-10-25 13:47
\frac{te}{st}
Name:
Anonymous
2012-01-23 14:21
<tex>\frac{e^z-e^{-z}}{e^z+e^{-z}}=\frac{\frac{e^{2z}-1}{e^z}}{\frac{e^{2z}+1}{e^z}}=\frac{e^{2z}-1}{e^{2z}+1}</tex>
Name:
Anonymous
2012-01-23 17:36
Yeah, you guys are really funny.
Name:
Anonymous
2012-09-11 17:52
\sqrt{x+2}
[latex]\sqrt{x+2}[/latex]
[tex]\sqrt{x+2}[/tex]
Name:
Anonymous
2012-09-11 17:56
\sqrt{x+2}
[green]test[/green]
Name:
Anonymous
2013-05-19 16:01
[latex] \forall P \in \mathbb{C} [X \] \; ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{n}(0)| [\latex]
Name:
Anonymous
2013-05-19 16:06
<span class="math"> f(-4)=-2f(-3)+3=3f(-2)-3=-3f(-1)+6=\frac{3}{2}f(0)-3=\frac{3}{2} </span>
Name:
Anonymous
2013-05-19 16:12
[math] \forall P \in \mathbb{C} [ X ] , ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{n}(0)| [\math]
[equation] \forall P \in \mathbb{C} [ X ] , ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{n}(0)| [\equation]
Name:
Anonymous
2013-05-19 16:14
[math] \forall P \in \mathbb{C} [ X ] , ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{n}(0)|
Name:
Anonymous
2013-05-19 16:14
\forall P \in \mathbb{C} [ X ] , ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{n}(0)|
Name:
Anonymous
2013-05-19 16:17
\forall P \in \mathbb{C} [ X ] , ||P|| = \sum_{n=0}^{+ \infty} a_n|P^{(n)}(0)|
This is a norm!
Name:
Anonymous
2013-06-03 17:23
It's now broken!
Name:
Anonymous
2013-06-19 6:51
$$n=n_2^4$$
Name:
Anonymous
2013-06-19 6:54
$$n=n_2^4$$
Name:
Anonymous
2013-06-19 6:55
hax my = a_n_u^s
Name:
Anonymous
2013-06-19 6:55
hax my = a_n(_u)^s
Name:
Anonymous
2013-06-19 6:56
hax my = a_n^us
Name:
Anonymous
2013-06-19 6:56
hax my = a_nus
Name:
Anonymous
2013-06-24 19:48
2^{2}
Name:
Anonymous
2013-06-24 19:59
\mathcal{Just~testing~LaTeX}
Name:
Anonymous
2013-08-14 6:32
$$a^2+b^2=c^2$$
Name:
Anonymous
2013-08-14 6:33
{latex} a^2 {/latex}
Name:
Anonymous
2013-09-06 18:34
\sqrt{2}-1