Off Topic: The Flood
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  • Subject: prove
Subject: prove
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\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}

  • 02.02.2005 6:12 PM PDT

We’ve watched while the stars burned
Out, and creation played in reverse.
The Universe freezing in half-light.
Once I thought to escape.
To end a master, step out of the
Path of collapse. Escape would make us God.
Yet I cannot help but remember one enigma,
A hybrid, elusive destroyer.
This is the one mystery I have not solved.
The only element unaccounted for.

No. And stop trying to show off. This one looks like you just made it up, whether or not it is real.

[Edited on 2/2/2005 6:15:04 PM]

  • 02.02.2005 6:14 PM PDT
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just move on man, noone cares!

  • 02.02.2005 6:16 PM PDT
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my brain hurts now.. what the hell did he just say?

  • 02.02.2005 6:18 PM PDT
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yea quite lame
2+2 = 5 anyway

  • 02.02.2005 6:18 PM PDT

We’ve watched while the stars burned
Out, and creation played in reverse.
The Universe freezing in half-light.
Once I thought to escape.
To end a master, step out of the
Path of collapse. Escape would make us God.
Yet I cannot help but remember one enigma,
A hybrid, elusive destroyer.
This is the one mystery I have not solved.
The only element unaccounted for.

Posted by: Superbxen
\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}

What I get from this is
divided by the variable (greek for "z") equals divided by the sum (of?) ((the variable n which equals one) to infinity which still equals one) divided by the fraction (1 times the (variable n to the variable s power) which equals the variable n to the variable s power)

So the simplified load of crap is

/z(s)=/sum_1/frac(n^s)

Which is still a load of bs.

  • 02.02.2005 6:26 PM PDT
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*points and yells* SPAMMER!

  • 02.02.2005 6:31 PM PDT
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Posted by: un gato
Posted by: Superbxen
\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}

What I get from this is
divided by the variable (greek for "z") equals divided by the sum (of?) ((the variable n which equals one) to infinity which still equals one) divided by the fraction (1 times the (variable n to the variable s power) which equals the variable n to the variable s power)

So the simplified load of crap is

/z(s)=/sum_1/frac(n^s)

Which is still a load of bs.


Oww. Don't make my head hurt like that showoff.

  • 02.02.2005 6:43 PM PDT