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  #21  
Old 06-24-2010, 11:04 PM
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dont make me start talking about infinite series lol, I might short circuit peoples brains cells
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  #22  
Old 06-24-2010, 11:22 PM
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Seriously, does anyone here have any idea what a 'good troll' is?
All the trolls I've seen here are so lame, for example the Peter Griffin guy, and now this guy.

But I guess it's good that all the trolls I've seen on here are laughable at best, and none of them have been irritating.
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  #23  
Old 06-24-2010, 11:53 PM
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Quote:
Originally Posted by Drake498 View Post
and yet no one is...
exactly! he failed
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  #24  
Old 06-24-2010, 11:58 PM
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I could go on forever about infinite series! I got stuck in a loop in my time machine one time, and I made like a million copies of myself trying to get back to the present. Somewhere in the late 13th century, there are 1,273 VVV's, all trying to figure out how to get back to now. I would go back and tell them all how I eventually do it, but it would likely create a universe-unraveling paradox. And the universe being destroyed would be bad. Why? Because Alizée is there!
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  #25  
Old 06-25-2010, 12:07 AM
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I can't believe this stupid thread got more than one page
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  #26  
Old 06-25-2010, 06:24 AM
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I can't believe it's not butter!
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  #27  
Old 06-25-2010, 09:32 AM
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Quote:
Originally Posted by Jalen View Post
I can't believe this stupid thread got more than one page
Then make your post per page longer, I hate constantly clicking for different pages
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  #28  
Old 06-25-2010, 10:35 AM
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  #29  
Old 06-25-2010, 10:48 AM
Merci Alizée
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Just to make you go insane
Quote:
Originally posted by Future Raptor Ace
Lets do a simple Geometric Series
1+ 1/2 + 1/4 +1/16 + 1/256 + ....... ∞-0Σ1/(2^n)
the limit as n goes to ∞ = 2 Now to prove this said function convergent (meaning there is a limit to as how high or low the function can go) we can do an integral test but first we have to set up our integral. Let F(x) our starting function ratio 1/(2^n) = un..... F(x)=un fun! In order to prove that this is convergent F(x) has to be a decreasing function as 1/(2^n) is a decreasing function. If F(x)=un than 1/(2^n) = 1/(2^x). Now take the integral of 1/(2^x) from 0 to ∞. When done doing so you will find it converges so long as x is greater than 1, for all else it is divergent. In the series 1 + 1/2 + 1/4 +1/16 + 1/256 n starts at 0 and then goes to 1 and then to 2 and then to 3 making the n... now x to be greater than the said 1 we just proved. Whoever reads this, I feel bad for you, almost as much as I feel bad for myself for writing this.... you wanted spam, why not make it spam that makes people go insane
Ha, you were feeling bad because your this message got deleted accidently by you. Don't worry, I have brought it back.

Sorry, I couldn't recall the positions of "yay" and some other words like that and the formatting that you did with your message, but I guess this is what you wanted

Last edited by Merci Alizée; 06-25-2010 at 10:51 AM..
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  #30  
Old 06-25-2010, 11:43 AM
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but yeah. stop feeding the guy. let's just move to a different thread :3
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