JSH: Twin primes
in [Math]
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From: master1729 on 5 Feb 2010 21:15 > On Fri, 05 Feb 2010 02:44:43 -0800, Rotwang wrote: > > > On 5 Feb, 05:07, MichaelW <ms...(a)tpg.com.au> wrote: > >> > >> [a whole load of stuff, then this] > >> > >> As you can see the ratio of the prediction to the > actual value stays > >> pretty close to about 1.12. > > > > This number is rather familiar to those of use who > were following > > James's threads a few years back. See the following > article[1], and if > > you're after more info then it's worth looking at > Tim Peters' posts from > > around August 2006, especially those that mention > "Mertens". > > > > [1] > http://groups.google.co.uk/group/alt.math/msg/95688f21 > 176359b1 > i mentioned " Mertens " first in the JSH thread. i dont have more time now , will probably post more about primes twins later ... > Thanks Rotwang, that was an interesting read. I am > still wondering if > there is any relationship between the number 1.123... > and the twin primes > constant? > > I am not surprised that James is claiming victory > with an idea that a > simple program can and has disposed of long before. > This is a known > behaviour. > > I found an excellent quote from that era by Tim > Peters which I will > reproduce here since it really expanded my > understanding of what is going > on: > > <TimP> > The fatal flaws in your "naive" probability models > are explained in Hans > Riesel's book "Prime Numbers and Computer Methods for > Factorization", in > chapter "Subtleties in the Distribution of Primes". > You're not the first > person to get sucked into building on a fatally naive > model, and won't be > the last. The primes don't actually act the way you > assume they act (to > get the probability of N events occurring > simultaneously you simply > multiply the probabilities of the N individual > events, but that's valid > only if the events are /independent/, and while I > grant that it's not / > intuitively/ obvious, the events here are not > independent; Mertens's > Theorem is in fact a rigorous proof that they're not > independent in the > sense you need them to be). > </TimP> > > It is sad that posts such as yours and Tim's > (informative, providing > references, clear, the product of hard work and a > passion for rigorous > thought) are so rare these days; certainly beyond my > lowly skills. > > Thanks, Michael W.
From: JSH on 6 Feb 2010 10:59 On Feb 5, 5:20 pm, spudnik <Space...(a)hotmail.com> wrote: > well, that seems rather to dyspoze > of the whole issue, viz-a-vu. and, like I said, > in January, that Magadin said, > about primes of the form 4n +/- 1. > > thus quoth: > Mertens's Theorem is in fact a rigorous proof that they're not > independent in the sense you need them to be). They ARE independent. There is just clipping behavior for the bigger primes for their higher residues which is so easy to see. Between 5^2 and 7^2, there are 6 primes. The probability then is given by: prob = ((5-2)/(5-1))*((3-2)/(3-1) = (3/4)*(1/2) = 0.375 And 6*0.375 = 2.25 so you expect 2 twin primes in that interval. The primes are 29, 31, 37, 41, 43, 47 and you'll notice, two twin primes as predicted: 29,31 and 41, 43. However, there is an issue which shifts the probability slightly. If you go into the actual residues it jumps out at you: 29, 31, 37, 41, 43, 47 mod 3: 2, 1, 1, 2, 1, 2 mod 5: 4, 1, 2, 1, 3, 2 Here all the residues for 5 were in evidence so the count came out right, but for random it should have been possible for ALL the residues mod 5 to be 4, but it's not because with 6 primes there isn't enough space in the interval--4*5 = 20, but 48-25=24, where only 12 are odd and only 6 are primes. So the probability is actually off! A scenario where all residues are 4 is precluded by the size of the interval for the larger prime. That will tend to over-count because the higher residues are less likely to occur because they cannot fit. They cannot fit. So some probabilities are dropped to zero because they're impossible in the interval. That means that 2 as a lower residue is more likely to occur which means the twin prime count will be lower. Trivial to a real researcher. So the over count is easily explained and probabilistic behavior is still the only answer that makes sense. Of course if you're some loser who can't do real research to save your life it's convenient to claim otherwise for government grants, but then you're just on white collar welfare. White collar welfare. But if you ARE some pretender trying to be a mathematician what else can you do? If you tell the truth your only option is to quit pretending and leave the field as you can't do real research, so primes are a draw for the fakes. James Harris
From: Mark Murray on 6 Feb 2010 16:04 On 06/02/2010 15:59, JSH wrote: > On Feb 5, 5:20 pm, spudnik<Space...(a)hotmail.com> wrote: >> well, that seems rather to dyspoze >> of the whole issue, viz-a-vu. and, like I said, >> in January, that Magadin said, >> about primes of the form 4n +/- 1. >> >> thus quoth: >> Mertens's Theorem is in fact a rigorous proof that they're not >> independent in the sense you need them to be). > > They ARE independent. There is just clipping behavior for the bigger > primes for their higher residues which is so easy to see. I thought you knew what a proof was, and what it meant? M --
From: William Hughes on 6 Feb 2010 17:43 On Feb 6, 11:59 am, JSH <jst...(a)gmail.com> wrote: > They ARE independent. <snip> >it should have been possible for ALL the > residues mod 5 to be 4, but it's not So the residues are NOT independent. Does it not bother you that using the exact same reasoning you use to show there are an infinite number of double primes we can show there are an infinite number of triple primes? - William Hughes
From: JSH on 6 Feb 2010 18:47
On Feb 6, 2:43 pm, William Hughes <wpihug...(a)hotmail.com> wrote: > On Feb 6, 11:59 am, JSH <jst...(a)gmail.com> wrote: > > > They ARE independent. > > <snip> > > >it should have been possible for ALL the > > residues mod 5 to be 4, but it's not > > So the residues are NOT independent. If there are only 6 primes in an interval from 25 to 49, then it's just not possible for all the primes to be 4 mod 5, so when you calculate the probability you overstate the likelihood of higher residues. 29, 31, 37, 41, 43, 47 mod 3: 2, 1, 1, 2, 1, 2 mod 5: 4, 1, 2, 1, 3, 2 Here the actual pattern averages out ok, so you get the right number though notice that 3 doesn't show up. There actually is just enough room for 4 and 3, only if all the other residues are 1 mod 5 for the primes. (Curious readers can notice that where you do NOT have twin primes, it's ALWAYS the case that the prime minus 2 is 0 mod 5 or 0 mod 3.) So the overstatement of the prediction is easily explained. Some prime possibilities are removed by size constraints for bigger primes relative to the interval for their bigger residues. > Does it not bother you that using > the exact same reasoning you use > to show there are an infinite number > of double primes we can show there > are an infinite number of triple primes? > > - William Hughes The prime residue axiom states that the primes do not care about their residue modulo other primes, which means you can calculate twin primes probability easily enough, but some noticed that you get a prediction that is higher than the actual count which is explained by the size of the interval clipping off possible permutations for higher residues. It's mathematics William Hughes. You may not like it, but it will not change because of your disdain. I've noticed that posters seem to get giddy with agreement, as if that matters. But if you say 2+2 = 5, it doesn't matter how many people agree with you. You're still all wrong. James Harris |