LaTeX BB Code tests

Profile · 05:21 PM

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

$\reverse \LARGE e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

$\reverse \reverse\opaque \LARGE e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

$\reverse \reverse e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

$\reverse \mathbf{C}_j = \frac{n!}{(n - j)!} \sum_{i = 0}^j \frac{(-1)^{i + j} \mathbf{P}_i}{i! (j - i)!} = \prod_{m = 0}^{j - 1} (n - m) \sum_{i = 0}^j \frac{(-1)^{i + j} \mathbf{P}_i}{i! (j - i)!}$

Chruser · Posted 2008-11-22, 05:11 PM in reply to Chruser's post "LaTeX BB Code tests"

Wikipedia rendering comparison:

Given ''n'' + 1 control points Pi, the rational Bézier curve can be described by:

$\reverse \mathbf{B}(t) = \frac{ \sum_{i=0}^n b_{i,n}(t) \mathbf{P}_{i}w_i } { \sum_{i=0}^n b_{i,n}(t) w_i } $

or simply

$\reverse \mathbf{B}(t) = \frac{ \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}\mathbf{P}_{i}w_i } { \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}w_i }. $

Chruser · 05:19 PM

Annoying aliasing.

$\reverse \mathbf{B}(t) = \frac{ \sum_{i=0}^n b_{i,n}(t) \mathbf{P}_{i}w_i } { \sum_{i=0}^n b_{i,n}(t) w_i } $

or simply

$\reverse \mathbf{B}(t) = \frac{ \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}\mathbf{P}_{i}w_i } { \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}w_i }. $

Chruser

I fixed the aliasing, I think. I reversed the colors by default to make it look better in the Synthesis themes, but the downside is that you won't be able to see the output well in lighter themes such as Ocean and Zephios. To make it work in all themes, something like \reverse\opaque should be used, to give the output an appearance similar to this:

$\reverse \reverse\opaque x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ( (2k-1)t \right )}\over(2k-1)}$

Also, I need to switch to a new TeX host soon.

Chruser · 05:42 PM

$\reverse \widetilde{It\: seems\: to\: be\: working!\quad Welcome\: to\: the\: 1940's!} $

Chruser · Posted 2008-11-23, 03:59 AM in reply to Chruser's post "LaTeX BB Code tests"

Combining tables and LaTeX...

Function	Abbreviation	Identities (using radians)
Sine	sin	$\reverse \sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}\,$
Cosine	cos	$\reverse \cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,$
Tangent	tan (or tg)	$\reverse \tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} \,$
Cosecant	csc (or cosec)	$\reverse \csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} \,$
Secant	sec	$\reverse \sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} \,$
Cotangent	cot (or ctg or ctn)	$\reverse \cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} \,$

Chruser

what the heck chruser

jamer123

Damn you all; start using it.

Chruser · 02:30 AM

Chruser said: [Goto]

Damn you all; start using it.

Hey $\reverse \:Chronic \:User$ , I actually found a use for this BB Code in another thread. I needed 10 and a half or $\reverse \:10 \frac{1}{2}$ .

~ - Tilde
$\reverse \tilde{\:Test \:Test \:Test \:Test}$
$\reverse \widetilde{\:Test \:Test \:Test \:Test}$

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