Thread for posting tiny primes - Page 59

kar_bon · 2010-09-23, 11:30

corrected:

p = $e^{-lambda}$ for n=0 (because 0! = 1 by definition)

CRGreathouse · 2010-09-23, 13:19

So solve for lambda in terms of p (or the constant 0.01), then substitute the definition of lambda and solve further.

science_man_88 · 2010-09-23, 17:51

Quote:

Originally Posted by CRGreathouse

I was looking for a function that, given a probability p (and N, t, L), would give an n such that

estimatePrimes(N,t,n,L)

is 1 - p.

what happened to this one ?

CRGreathouse · 2010-09-23, 19:07

Quote:

Originally Posted by science_man_88

what happened to this one ?

That's what I was showing you in #634 etc.

science_man_88 · 2010-09-23, 20:24

lambda = ((t/log(N))*exp(Euler)*log(L))

p= $e^{-lambda}$

p= $exp^{-((t/log(N))*exp(Euler)*log(L))}$

CRGreathouse · 2010-09-23, 20:40

OK, so take the (natural) log of both sides. (TeX note: e^{\gamma}, not e^{(Euler)}. "Euler" is what Pari calls it, "\gamma" is what TeX calls it.)

Remember, you're solving for t.

science_man_88 · 2010-09-23, 20:46

actually no I'm not stop assuming.

CRGreathouse · 2010-09-23, 20:59

Quote:

Originally Posted by science_man_88

actually no I'm not stop assuming.

I don't know what that means. I didn't say you were assuming anything, nor that you were stopping any assumptions.

science_man_88 · 2010-09-23, 21:06

you say I'm solving for t without proof of that and my statement of the opposite yet you still claim it hence you assume it's true there's the assumption.

3.14159 · 2010-09-23, 21:18

Dammit.. I accidentally clicked away at 1 day's worth of work.. Now I lost about 10 hours of work..

science_man_88 · 2010-09-23, 21:21

Quote:

Originally Posted by 3.14159

Dammit.. I accidentally clicked away at 1 day's worth of work.. Now I lost about 10 hours of work..

that's better than 24 hours worth Pi !!!!!

2010-09-23, 11:30	#639
kar_bon Mar 2006 Germany 101101011100₂ Posts	corrected: p = $e^{-lambda}$ for n=0 (because 0! = 1 by definition) Last fiddled with by kar_bon on 2010-09-23 at 11:35

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2010-09-23, 13:19	#640
CRGreathouse Aug 2006 3·1,993 Posts	So solve for lambda in terms of p (or the constant 0.01), then substitute the definition of lambda and solve further.

2010-09-23, 20:24	#643
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	lambda = ((t/log(N))exp(Euler)log(L)) p= $e^{-lambda}$ p= $exp^{-((t/log(N))exp(Euler)log(L))}$

2010-09-23, 20:40	#644
CRGreathouse Aug 2006 3×1,993 Posts	OK, so take the (natural) log of both sides. (TeX note: e^{\gamma}, not e^{(Euler)}. "Euler" is what Pari calls it, "\gamma" is what TeX calls it.) Remember, you're solving for t.

2010-09-23, 20:46	#645
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	actually no I'm not stop assuming.

2010-09-23, 21:06	#647
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	you say I'm solving for t without proof of that and my statement of the opposite yet you still claim it hence you assume it's true there's the assumption.

2010-09-23, 21:18	#648
3.14159 May 2010 Prime hunting commission. 2⁴·3·5·7 Posts	Dammit.. I accidentally clicked away at 1 day's worth of work.. Now I lost about 10 hours of work..