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TeX

Name: Anonymous 2010-10-11 17:02

I'm new here :D
HOW I TEX?
$\alpha \sim B(\alpha,\beta)$

Name: Anonymous 2010-10-12 15:33

>>1
Get out.

Name: Anonymous 2010-10-12 16:20

>>2
RUDE!

Name: Anonymous 2010-10-12 18:55

[math] \Sigma \vec{F} = m \vec{a} [\math}

Name: Anonymous 2010-10-12 18:55

[math] \Sigma \vec{F} = m \vec{a} [\math]

Name: Anonymous 2010-10-12 18:56

\Sigma \vec{F} = m \vec{a}

Name: Anonymous 2010-10-15 2:51

\Sigma \vec{F} = m \vec{a}

Name: Anonymous 2010-10-15 2:58

[\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{n}} \] $\forall n : n>2 \in \mathbf{I}:

Name: Anonymous 2010-10-15 2:59

[\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{n}} \] $\forall n : n>2 \in \mathbf{I}:

Name: Anonymous 2010-10-15 3:00

sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{n}}  $\forall n : n>2 \in \mathbf{I}:

Name: Anonymous 2010-10-15 3:01

[\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{n}} \] : n>2

Name: Anonymous 2010-10-15 3:01

\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{n}} : n>2

Name: Anonymous 2010-10-15 3:31

Prove:

\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{i}} : n \geq 2

First lets prove this:
\sum_{i=1}^n \frac{1}{\sqrt[2]{n}}<\sum_{i=1}^n \frac{1}{\sqrt[2]{i}} : n \geq 2

This should be self evident.  We sum the same number of terms on each side, but on the right hand side we are summing terms of decreasing value, whereas on the left hand side we are summing terms that have the same value as the minimal right hand side term.

However the previous is the same as:
\frac{n}{\sqrt[2]{n}}<\sum_{i=1}^n \frac{1}{\sqrt[2]{i}} : n \geq 2

Which is obviously:
\sqrt[2]{n}<\sum_{i=1}^n \frac{1}{\sqrt[2]{i}} : n \geq 2

Name: Anonymous 2010-10-15 3:37

Prove:

\begin{equation} \sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

First lets prove this:

\begin{equation} \sum{i=1}^n \frac{1}{\sqrt{n}} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

This should be self evident.  We sum the same number of terms on each side, but on the right hand side we are summing terms of decreasing value, whereas on the left hand side we are summing terms that have the same value as the minimal right hand side term.

However the previous is the same as:

\begin{equation} \frac{n}{\sqrt{n}} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Which is obviously:

\begin{equation} \sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Name: Anonymous 2010-10-15 3:38

Prove:

\begin{equation} \sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

First lets prove this:

\begin{equation} \sum{i=1}^n \frac{1}{\sqrt{n}} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

This should be self evident.  We sum the same number of terms on each side, but on the right hand side we are summing terms of decreasing value, whereas on the left hand side we are summing terms that have the same value as the minimal right hand side term.

However the previous is the same as:

\begin{equation} \frac{n}{\sqrt{n}} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Which is obviously:

\begin{equation} \sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Name: Anonymous 2010-10-15 3:39

Prove:

\begin{equation} \sqrt{n} < \sum_{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

First lets prove this:

\begin{equation} \sum_{i=1}^n \frac{1}{\sqrt{n}} < \sum_{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

This should be self evident.  We sum the same number of terms on each side, but on the right hand side we are summing terms of decreasing value, whereas on the left hand side we are summing terms that have the same value as the minimal right hand side term.

However the previous is the same as:

\begin{equation} \frac{n}{\sqrt{n}} < \sum_{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Which is obviously:

\begin{equation} \sqrt{n} < \sum_{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}

Name: Anonymous 2010-10-25 15:25

\sum_0^\infty\frac{1}{i}=\text{4chan}

Name: Anonymous 2010-10-25 15:26

\int e^x\ dx

Name: Anonymous 2010-11-05 14:59

[tex]\begin{equation} \sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2 \end{equation}[/tex]

Name: Anonymous 2010-11-07 6:33

\sqrt{n} < \sum{i=1}^n \frac{1}{\sqrt{i}} : n \geq 2

Name: Anonymous 2010-11-12 14:22

\prod_{n}^{1}\cdot cx^{n-d}

Name: Anonymous 2010-12-01 12:01

(i\echelon \mathbb{Z})

Name: Anonymous 2010-12-04 13:22

[tex]1=2[/tex]

Name: Anonymous 2010-12-04 13:25

x = 1

Name: Anonymous 2010-12-04 13:25

\Sigma

Name: Anonymous 2010-12-04 13:25

\Sigma

Name: Anonymous 2010-12-04 13:34

$a~B(a,\beta)$

Name: Anonymous 2010-12-04 13:35

$\alpha ~ B(\alpha,\beta)$

Name: Anonymous 2010-12-12 8:02

sqrt(65)

Name: Anonymous 2011-01-07 15:35

\int_a^b \! f(x) \, \mathrm{d}x.

Name: Anonymous 2011-01-07 15:36

[math]\int_a^b \! f(x) \, \mathrm{d}x.[\math]

Name: Anonymous 2011-01-07 15:37

\int_a^b \! f(x) \, \mathrm{d}x.

Name: Anonymous 2011-01-07 15:39

\int_0^x \! f(x^3) \, \mathrm{d}x.

Name: Anonymous 2011-01-07 18:53

\int_a^b \! f(\lambda) \, \mathrm{d}x.

Name: Anonymous 2011-01-14 9:05

[math]\lambda \in \sigma (A)[\math]

Name: Anonymous 2011-01-14 17:11

Can't someone change it to not use [math], how annoying. At least make it $$x$$ and [x\]

Name: ab 2011-01-22 13:26

ab

Name: Anonymous 2011-01-23 6:01

\int_0^\infty \! x^{2n} e^{\frac{-(x^2)}{a^2}} \, \mathrm{d}x = \sqrt{\pi} \frac{(2n)!}{n!} \frac{a}{2}^{2n+1}

test

Name: Anonymous 2011-01-23 6:13

\int_0^\infty \! x^{2n} e^{-(x^2) \over{a^2}} \, \mathrm{d}x = \sqrt{\pi} \frac{(2n)!}{n!} { ( }a/2{ ) }^{2n+1}

V={1,2,3,4},E={12,23,32,34,41}

A =[math]\pmatrix {0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 1\cr 1 & 0 & 0 & 0\cr}

test2

Name: Anonymous 2011-01-23 6:14

\int_0^\infty \! x^{2n} e^{-(x^2) \over{a^2}} \, \mathrm{d}x = \sqrt{\pi} \frac{(2n)!}{n!} { ( }a/2{ ) }^{2n+1}

V={1,2,3,4},E={12,23,32,34,41}

A =\pmatrix {0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 1\cr 1 & 0 & 0 & 0\cr}

Name: Anonymous 2011-01-23 6:15

\int_0^\infty \! x^{2n} e^{-(x^2) \over{a^2}} \, \mathrm{d}x = \sqrt{\pi} \frac{(2n)!}{n!} { ( }a/2{ ) }^{2n+1}

V={1,2,3,4},E={12,23,32,34,41}

A =\pmatrix {0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 1\cr 1 & 0 & 0 & 0\cr}

Name: Anonymous 2012-07-04 4:46

R_{\alpha}^t \cdot (\vec{a} t + \vec{b}) + R_\beta^t \cdot \vec{c} - d = 0
\Leftrightarrow    \left(\begin{matrix}
\cos(\alpha t)
\sin(\alpha t)\end{matrix}\right) \cdot \left(\begin{matrix}
a_x t + b_x
a_y t + b_y\end{matrix}\right) + \left(\begin{matrix}
\cos(\beta t)
\sin(\beta t)\end{matrix}\right) \cdot \left(\begin{matrix}
c_x
c_y\end{matrix}\right) - d = 0

Name: Anonymous 2012-07-04 4:48

Red button:
100% 10^6 = 10^6

Green button:
50% 10^{8}, 50% 10^{-Infinity} = 10^\frac{8 - Infinity}{2} = 10^{-Infinity} = 0


Of course red button.

Name: Anonymous 2012-07-04 4:49

Red button:
100\% 10^6 = 10^6

Green button:
50\% 10^{8}, 50\% 10^{-Infinity} = 10^\frac{8 - Infinity}{2} = 10^{-Infinity} = 0


Of course red button.

Name: Anonymous 2012-07-04 4:52

Red button: 100%\ 10^6 = 10^6
Green button: 50%\ 10^{8}, 50%\ 10^{-Infinity} = 10^\frac{8 - Infinity}{2} = 10^{-Infinity} = 0

Of course red button.

Name: Anonymous 2012-07-04 4:52

Red button: 100\%\ 10^6 = 10^6
Green button: 50\%\ 10^{8}, 50\%\ 10^{-Infinity} = 10^\frac{8 - Infinity}{2} = 10^{-Infinity} = 0

What am I wrong?

Name: Anonymous 2012-07-04 5:22

\bbox[background-image:url('http://www.seinfeld-fan.net/pictures/george/george_costanza006.jpg');pos
ition:fixed;top:0;left:0;width :100%;height:100%]~

Name: Anonymous 2012-07-09 0:25

different person here

\oint_0^\infty n(c)\ dc \rightarrow \oint_0^\infty x(c) + m(c)\ dc
Don't change these.
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