**Edit (8/18/11)**: renamed the announcement to centralize new developments instead of having multiple, space-wasting announcements.LaTex is now available, thanks to our generous host at ProPhpBB.com.

To insert a LaTex equation, simply bracket your equation with the BBCode as with below:

**Code:**

[latex]e=mc^2[/latex]

Your result should be the graphical equivalent of the equation!

We'll get a FAQ or How To put together in the next day or so. But, in the mean time, here are some examples that I've borrowed from Dr. Rocket, along with the corresponding BBCode + the equation to get you started:

**Code:**

[latex]\frac{d}{dx} (2x-1)^{\frac{1}{2}} = (2x-1)^{\frac {-1}{2}}[/latex]

[latex]\frac{d^2}{dx^2} (2x-1)^{\frac{1}{2}} = -(2x-1)^{\frac {-3}{2}}[/latex]

[latex]\frac{d^3}{dx^3} (2x-1)^{\frac{1}{2}} = 3(2x-1)^{\frac {-5}{2}}[/latex]

[latex]\frac{d^n}{dx^n} (2x-1)^{\frac{1}{2}} = ( (-1)^{n}\displaystyle\prod_{k=0}^{n-1} (2k-1) )(2x-1)^{\frac {-2n+1}{2}}[/latex]

Produces:

$ \displaystyle \frac{d}{dx} (2x-1)^{\frac{1}{2}} = (2x-1)^{\frac {-1}{2}}$

$ \displaystyle \frac{d^2}{dx^2} (2x-1)^{\frac{1}{2}} = -(2x-1)^{\frac {-3}{2}}$

$ \displaystyle \frac{d^3}{dx^3} (2x-1)^{\frac{1}{2}} = 3(2x-1)^{\frac {-5}{2}}$

$ \displaystyle \frac{d^n}{dx^n} (2x-1)^{\frac{1}{2}} = ( (-1)^{n}\displaystyle\prod_{k=0}^{n-1} (2k-1) )(2x-1)^{\frac {-2n+1}{2}}$

**Code:**

[latex]\displaystyle \sum_{n=0}^N x^n = 1 + x \displaystyle \sum_{n=o}^N x^n - x^{N+1}[/latex]

[latex](1-x)\displaystyle \sum_{n=0}^N x^n = 1-x^{N+1}[/latex]

[latex]\displaystyle \sum_{n=0}^N x^n = \dfrac {1-x^{N+1}}{1-x}[/latex]

Produces:

$ \displaystyle \displaystyle \sum_{n=0}^N x^n = 1 + x \displaystyle \sum_{n=o}^N x^n - x^{N+1}$

$ \displaystyle (1-x)\displaystyle \sum_{n=0}^N x^n = 1-x^{N+1}$

$ \displaystyle \displaystyle \sum_{n=0}^N x^n = \dfrac {1-x^{N+1}}{1-x}$

**Code:**

[latex]\displaystyle \sum_{n=1}^N x^n = \dfrac {1-x^{N+1}}{1-x} -1[/latex] [latex]= \dfrac {x-x^{N+1}}{1-x}[/latex]

So, if [latex]|x|<1[/latex]

[latex]\displaystyle \sum_{n=0}^\infty x^n[/latex] [latex]=\displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=0}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {1-x^{N+1}}{1-x}[/latex] [latex]= \dfrac {1}{1-x}[/latex]

And

[latex]\displaystyle \sum_{n=1}^\infty x^n[/latex] [latex]= \displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=1}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {x-x^{N+1}}{1-x}[/latex] [latex]= \dfrac {x}{1-x}[/latex]

[latex]0.99999........ = \displaystyle \sum_{n=1}^\infty 9 (\dfrac{1}{10})^n[/latex] [latex]= 9 \displaystyle \sum_{n=1}^\infty (\dfrac{1}{10})^n[/latex] [latex]= 9 \dfrac {\frac {1}{10}}{1- \frac{1}{10}}[/latex] [latex]= 9 \dfrac {1}{9} =1[/latex]

Produces:

$ \displaystyle \displaystyle \sum_{n=1}^N x^n = \dfrac {1-x^{N+1}}{1-x} -1$ $ \displaystyle = \dfrac {x-x^{N+1}}{1-x}$

So, if $ \displaystyle |x|<1$

$ \displaystyle \displaystyle \sum_{n=0}^\infty x^n$ $ \displaystyle =\displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=0}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {1-x^{N+1}}{1-x}$ $ \displaystyle = \dfrac {1}{1-x}$

And

$ \displaystyle \displaystyle \sum_{n=1}^\infty x^n$ $ \displaystyle = \displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=1}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {x-x^{N+1}}{1-x}$ $ \displaystyle = \dfrac {x}{1-x}$

$ \displaystyle 0.99999........ = \displaystyle \sum_{n=1}^\infty 9 (\dfrac{1}{10})^n$ $ \displaystyle = 9 \displaystyle \sum_{n=1}^\infty (\dfrac{1}{10})^n$ $ \displaystyle = 9 \dfrac {\frac {1}{10}}{1- \frac{1}{10}}$ $ \displaystyle = 9 \dfrac {1}{9} =1$