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 2009-08-31, 03:40 #12 CRGreathouse     Aug 2006 172516 Posts Here's the Ramsey theory version: $\pi<10^{10^{10^{963}}}$
 2009-08-31, 07:43 #13 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 2·2,861 Posts Perfect for solving quadratic equations: $\huge\frac{-b \pm \sqrt {b^2-4ac}}{2a}$
 2009-08-31, 07:48 #14 Kees     Dec 2005 22·72 Posts $V-E+F=2$
2009-08-31, 09:12   #15
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

4E916 Posts

Why not allow permanent edit feature to our own posts forever? I don't bother to moderate others' posts. But, what justice is it if we can't edit our own posts after 1 hour from the time of posting? Perhaps, that it is a feature of vBulletin, I would talk about that later on.

Quote:
 Originally Posted by Xyzzy 1. Post an interesting math snippet, using the forum's math rendering engine.
A magic square with all the entries being prime numbers!

$
\huge \begin{bmatrix} 17 & 113 & 47 \\ 89 & 59 & 29 \\ 71 & 5 & 101 \end{bmatrix}
$

Last fiddled with by Raman on 2009-08-31 at 09:33

 2009-08-31, 13:12 #16 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 Posts $\pi(x) \approx \frac{x}{ln(x) - 1}$
 2009-08-31, 16:14 #17 kar_bon     Mar 2006 Germany 2·1,423 Posts $ \frac{1}{\pi} = \frac{sqrt{8}}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{({n!})^4 396^{4n}}$
 2009-08-31, 16:41 #18 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 23·3·5·53 Posts $\exp(\pi \sqrt{58}) \approx f(396^{-4}) \! \mathrm{where} \! f(u) = u^{-1} - 104 - 4372u - 550944u^2$ (I believe this is related to the previous statement ...) Last fiddled with by fivemack on 2009-08-31 at 16:41
 2009-08-31, 20:08 #19 davar55     May 2004 New York City 108616 Posts $\int e^x = f(u,n)$
 2009-08-31, 20:30 #20 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·641 Posts Traditional Caltech sports cheer: $e^ududx$, pronounced: "e to the u dee u dee x!"
 2009-09-01, 15:50 #21 lycorn     Sep 2002 Oeiras, Portugal 101011110002 Posts Another contribution from Euler: $\pi^2/6=\sum_{n=1}^\infty^1/n^2$ Last fiddled with by lycorn on 2009-09-01 at 15:51
 2009-09-01, 19:50 #22 Jeff Gilchrist     Jun 2003 Ottawa, Canada 100100100012 Posts The last part of my secret proof. It is true, I swear. Trust me... $\therefore P=NP$

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