ajb Posted February 6, 2012 Share Posted February 6, 2012 I think your analysis is right DrRocket. For me, a great theorem/conjecture/hypothesis/lemma etc is one that links seemingly distinct topics in mathematics and requires new tools to be developed. These ideas and tools may well go on well beyond their initial applications. The Riemann hypothesis links complex analysis with number theory. The idea of solving the hypothesis in terms of an operator is, to my mind, very interesting. I know that Berry has worked on this. The idea is that the spectra of some quantum Hamiltonian will give the information needed. These Hamiltonians, if I recall do not look like those found in physics, but it has been suggested that one may be able to manufacture a physical system with the right Hamiltonian and then experimentally explore the hypothesis. I of course remain sceptical, but it is a great train of thought! Link to comment Share on other sites More sharing options...

PeterJ Posted February 6, 2012 Author Share Posted February 6, 2012 (edited) There is no such thing as a "correct heuristic proof". You either have a proof or you don't. If a "mathematician" told you differently then he is not much of a mathematician. Yes. this is what I thought to start with. It took me a while to figure out how it could be correct but unrigorous. This seemed to be a contradiction. In the end I saw that it is correct in principle but that I have not quite ruled out the possibility that it might go wrong eventually if the numbers behave in a very particular way. I can do this, I think, but I haven't got around to doing it yet. The mathematician in question is fairly senior and well respected. (Being at Bristol I imagine he knows Berry quite well). You can look him up if you want. I rather think he is right. In fact I know he is. The proof is too simple to be confusing. I'd put it up for comment but I want to finish it first. I still cannot grasp what the significance is of the non-trivial zeros. What does the output value of the Z-function actually tell us about the input value? Or is that another misguided question? To me the zeros are as meaningful as 42 is as an answer for 'life, the universe and everything'. Is this explicable to someone without a PhD? I've heard the zeros described as tuning forks, but why? Edited February 6, 2012 by PeterJ Link to comment Share on other sites More sharing options...

Shadow Posted February 6, 2012 Share Posted February 6, 2012 This might help answer a lot of your questions: http://plus.maths.org/content/music-primes Link to comment Share on other sites More sharing options...

PeterJ Posted February 6, 2012 Author Share Posted February 6, 2012 Many thanks. That's the most useful article I've ever read on the topic. My understanding of the primes is almost entirely musical, being more a musician than a mathematician. I wish this had been a chapter in his book. Now I'm starting to get it. Or maybe. Is it possible to say anything very simply about what the RH actually does? Probably not. I recognise all the chains of reciprocals from my attempts to count primes, but most of it is meaningless to me. What are its basic operations and stages? Can they be summarised? I appreciate your help by the way. This must be like going back to primary school. Link to comment Share on other sites More sharing options...

DrRocket Posted February 7, 2012 Share Posted February 7, 2012 Yes. this is what I thought to start with. It took me a while to figure out how it could be correct but unrigorous. This seemed to be a contradiction. In the end I saw that it is correct in principle but that I have not quite ruled out the possibility that it might go wrong eventually if the numbers behave in a very particular way. I can do this, I think, but I haven't got around to doing it yet. The mathematician in question is fairly senior and well respected. (Being at Bristol I imagine he knows Berry quite well). You can look him up if you want. I rather think he is right. In fact I know he is. The proof is too simple to be confusing. I'd put it up for comment but I want to finish it first. baloney Link to comment Share on other sites More sharing options...

PeterJ Posted February 7, 2012 Author Share Posted February 7, 2012 (edited) Oh dear. I'm not impressed that you have not asked even one question and just dismiss what I say out of hand. I'm beginning to realise this is a common practice on this forum. The arrogance is surprising and very difficult to cope with. Perhaps it's my inability to speak as a mathematician that leads you to this dismissive view. Where p1, p2 are consecutive primes there must always be a twin prime between p1^2 ans p2^2, and as p increases so, on average, does the quantity of twin primes in the range. This can be established by simply examining the combination wave of the products of the primes up to p1. It's not my fault. It's the way the primes work. You remind me that I once phoned a professional mathematician I knew slightly to discuss the primes and in the course of the discussion happened to mention that the primes >3 must always fall at 6n+/-1. To my utter astonishment he did not believe me. I had to spend five minutes trying to convince him this was not baloney. I did not succeed. Ridiculous. Exactly which bit was baloney? Do I believe you or the most prominent mathematician I have ever spoken to? Hell, another time I discussed this with a mathematician who was the wife of a Field Prize winner. The word 'baloney' did not come up. Anway, let's leave the TPC. The RH is the topic. I'd like to thank everybody for being so helpful on this. I've veen asking some of these questions for a long time elsewhere and this is the first time I've had some answers I can understand. Edited February 7, 2012 by PeterJ Link to comment Share on other sites More sharing options...

imatfaal Posted February 7, 2012 Share Posted February 7, 2012 Oh dear. I'm not impressed that you have not asked even one question and just dismiss what I say out of hand. I'm beginning to realise this is a common practice on this forum. The arrogance is surprising and very difficult to cope with. Perhaps it's my inability to speak as a mathematician that leads you to this dismissive view. Where p1, p2 are consecutive primes there must always be a twin prime between p1^2 ans p2^2, and as p increases so, on average, does the quantity of twin primes in the range. This can be established by simply examining the combination wave of the products of the primes up to p1. It's not my fault. It's the way the primes work. You remind me that I once phoned a professional mathematician I knew slightly to discuss the primes and in the course of the discussion happened to mention that the primes >3 must always fall at 6n+/-1. To my utter astonishment he did not believe me. I had to spend five minutes trying to convince him this was not baloney. I did not succeed. Ridiculous. Exactly which bit was baloney? Do I believe you or the most prominent mathematician I have ever spoken to? Hell, another time I discussed this with a mathematician who was the wife of a Field Prize winner. The word 'baloney' did not come up. Anway, let's leave the TPC. The RH is the topic. I'd like to thank everybody for being so helpful on this. I've veen asking some of these questions for a long time elsewhere and this is the first time I've had some answers I can understand. The difference is that they are being polite and not wishing you to stop your investigations and thoughts by being brusque. DrRocket on a forum is not held back by those limitations of society. Frankly I struggle to believe that a "professional mathematician" could not have sorted out the idea that primes over 3 must be of the form 6n+/-1 - I wonder if he was of the impression that you were asserting that all numbers (or a pattern of all numbers) of the form 6n+/-1 were primes this is very different. It is the work of 30 seconds and primary school maths: 6n - even (as six is even then 6n must be) 6n+1 6n+2 - even (as six is even then 6n must be - and so must that number plus 2) 6n+3 - must be divisible by 3 as 6 is divisible by 3 and +3 is also divisible by 3 6n+4 - even (as six is even then 6n must be - and so must that number plus 4) 6n+5 6(n+1) - back to 6n There are only two "spaces" left which could be prime. Although I am not a professional mathematician I agree with DocRock - maths proofs must be rigorous, a heuristic in maths is only of a bit of explanatory use. Edit; the 6n+/-1 bit I have put in is fairly convincing to me - but it is not even close to a proof. I am pretty sure all my assumptions are correct - but it is not rigorous (just wanted to be sure I wasn't claiming it was bearing in mind what I said immediately afterwards. Link to comment Share on other sites More sharing options...

Shadow Posted February 7, 2012 Share Posted February 7, 2012 Where p1, p2 are consecutive primes there must always be a twin prime between p1^2 ans p2^2 Could you elaborate on how you got this result? Link to comment Share on other sites More sharing options...

PeterJ Posted February 7, 2012 Author Share Posted February 7, 2012 (edited) The difference is that they are being polite and not wishing you to stop your investigations and thoughts by being brusque. DrRocket on a forum is not held back by those limitations of society. Frankly I struggle to believe that a "professional mathematician" could not have sorted out the idea that primes over 3 must be of the form 6n+/-1 - I wonder if he was of the impression that you were asserting that all numbers (or a pattern of all numbers) of the form 6n+/-1 were primes this is very different. It is the work of 30 seconds and primary school maths: How astonishingly patronising. But yes. That's why I was so surprised. His area is statistics, so perhaps he'd never thought about it. He probably figured it as soon as I put the phone down. There are only two "spaces" left which could be prime. Although I am not a professional mathematician I agree with DocRock - maths proofs must be rigorous, a heuristic in maths is only of a bit of explanatory use. I have not suggested otherwise. I do happen to believe otherwise, but I have not suggested it and will not here. Edit; the 6n+/-1 bit I have put in is fairly convincing to me - but it is not even close to a proof. I am pretty sure all my assumptions are correct - but it is not rigorous (just wanted to be sure I wasn't claiming it was bearing in mind what I said immediately afterwards. I don't know exactly what you would count as a proof and what not. Are you saying that you're not sure you can rigorously prove that the products of 2 and 3 are not prime? Could you elaborate on how you got this result? Thanks for asking. Despite the vast implausibility of it I'm not quite sure that my full argument won't end up being interesting in some way, so I'll keep one or two important bit back if that's okay. But here's the idea. It could hardly be more simple. I am not claiming it is a proof of anything. I am saying that I have been told by someone who should know that it is a correct heuristic proof but not yet rigorous. He did not suggest I had changed the world, and neither have I at any point. I know that an heuristic proof is not a mathematical proof. I use the 6n numbers as a metric against which to measure the behaviour of the products of the other primes. This means ignoring 4 numbers in every 6 and examing only the two possible twin prime locations. Two products of p (>3) occur at 6n+/-1 in every 6p numbers. These will occur at 6p+/-p. So, for instance, 2 locations in every 6 cannot be twin primes since they are products of 5. Etc etc. Some fiddly corrections terms are needed, but these become increasingly unnecessary as p increase. Roughly speaking... For the range p1 squared to p2 squared, call it R, there are R/6 possible locations of twin primes. We know how many products of any p will be created in R for each prime up to sqrt p1, so we can deduct one total from the other. Of course, very often these products will occur where there is already a product of a prime, i.e 35, but this doesn't matter. I do not count twin primes, just set a lower limit for twin primes in R. It is never less than 1, and the calculation always underestimates it. R increases with p, while the quantity of products of primes below sqrt p1 occurring in R decreases with p. Ergo, on average the quantity of twin primes in R increases with p. Quite where it goes to I'm not sure but it's always on the increase. Nothing very fancy. I treated the the number line as a vibrating string and examined the combination wave created by the music of the products of the primes. As a result I would happily bet my life that there an infinite quantity of twin primes. But I would not be able to put together what you guys would call a formal proof. An heuristic proof is easy enough though, since it just means describing the mechanics. Please note that it was not me who made such a big deal out of this. I mentioned it only because I was being accused of being a dunderhead. Edited February 7, 2012 by PeterJ -1 Link to comment Share on other sites More sharing options...

imatfaal Posted February 7, 2012 Share Posted February 7, 2012 How astonishingly patronising. But yes. That's why I was so surprised. His area is statistics, so perhaps he'd never thought about it. He probably figured it as soon as I put the phone down. Please do not post personal comments - I have been perfectly polite throughout this thread from explaining the complex plane thru to my last post and I do not expect to be called patronising because I have disagreed with you and have suggested an explanation for perceived reactions.I have not suggested otherwise. I do happen to believe otherwise, but I have not suggested it and will not here. Unfortunately the use of the phrase "I have managed a correct heuristic proof " is enough to show that you have suggested it here.I don't know exactly what you would count as a proof and what not. Are you saying that you're not sure you can rigorously prove that the products of 2 and 3 are not prime? I could and have proved it to my satisfaction - DocRock and the other teachers of maths at U level would probably give any rigorous proof I attempted an "F - See Me!!" The trouble would be that I would be unsure what I could take as axiomatic and that which I had to prove.Please note that it was not me who made such a big deal out of this. I mentioned it only because I was being accused of being a dunderhead. firstly - you opened this thread and later on claimed to have "a correct heuristic proof" and that " I think I can make it rigorous when I get around to it" of one of the oldest and most intractable problems of number theory - you cannot say things like that on a maths board and not pique people's curiosity. And secondly, no one called you a dunderhead (unless I missed it, in which case apologies) - inter alia they said your theory, and its reception was Baloney, I said I would bet dollars to donuts it was wrong - but no comments about you - merely about your theory. Don't make this personal when it isn't. I will have a look through the proof later. Link to comment Share on other sites More sharing options...

PeterJ Posted February 7, 2012 Author Share Posted February 7, 2012 (edited) Please do not post personal comments - I have been perfectly polite throughout this thread from explaining the complex plane thru to my last post and I do not expect to be called patronising because I have disagreed with you and have suggested an explanation for perceived reactions. Oh c'mon. You wrote... "The difference is that they are being polite and not wishing you to stop your investigations and thoughts by being brusque. DrRocket on a forum is not held back by those limitations of society." This is patronising and it is a personal comment. You have no reason to jump to this conclusion. You have no idea of what happened but just assume I am a fool. How did you expect me to react? Unfortunately the use of the phrase "I have managed a correct heuristic proof " is enough to show that you have suggested it here. What? When I said 'I have not suggested otherwise" it was in response to your statement "maths proofs must be rigorous, a heuristic in maths is only of a bit of explanatory use." Why are you persecuting me like this, with deception and trickery? Yes, I have managed a correct heuristic proof. And yes, I agree with you about maths proofs. I could and have proved it to my satisfaction - DocRock and the other teachers of maths at U level would probably give any rigorous proof I attempted an "F - See Me!!" The trouble would be that I would be unsure what I could take as axiomatic and that which I had to prove. Okay. This is not my problem. I did not attempt a rigorous maths proof. firstly - you opened this thread and later on claimed to have "a correct heuristic proof" and that " I think I can make it rigorous when I get around to it" of one of the oldest and most intractable problems of number theory - you cannot say things like that on a maths board and not pique people's curiosity. I did not pique your curiosity, as I intended, I only aroused your anger. I aroused Shadow's curiosity, but not yours. And secondly, no one called you a dunderhead (unless I missed it, in which case apologies) - inter alia they said your theory, and its reception was Baloney, I said I would bet dollars to donuts it was wrong - but no comments about you - merely about your theory. Don't make this personal when it isn't. For you it is clearly personal. I'd rather it wasn't. Your words are there for anyone to read. I will have a look through the proof later. I think maybe this would have been a better starting point than the assumption the people who I've talked to were being kind to me or are poor mathematicians. I'm not annoyed by the way. I know exactly why you reacted as you did. But it's not much fun to have to deal with such reactions, being continually forced to defend my sanity in this way, and I'd rather just deal with the issues. Edited February 7, 2012 by PeterJ -1 Link to comment Share on other sites More sharing options...

DrRocket Posted February 7, 2012 Share Posted February 7, 2012 The difference is that they are being polite and not wishing you to stop your investigations and thoughts by being brusque. DrRocket on a forum is not held back by those limitations of society. Frankly I struggle to believe that a "professional mathematician" could not have sorted out the idea that primes over 3 must be of the form 6n+/-1 - I wonder if he was of the impression that you were asserting that all numbers (or a pattern of all numbers) of the form 6n+/-1 were primes this is very different. It is the work of 30 seconds and primary school maths: 6n - even (as six is even then 6n must be) 6n+1 6n+2 - even (as six is even then 6n must be - and so must that number plus 2) 6n+3 - must be divisible by 3 as 6 is divisible by 3 and +3 is also divisible by 3 6n+4 - even (as six is even then 6n must be - and so must that number plus 4) 6n+5 6(n+1) - back to 6n There are only two "spaces" left which could be prime. Although I am not a professional mathematician I agree with DocRock - maths proofs must be rigorous, a heuristic in maths is only of a bit of explanatory use. Edit; the 6n+/-1 bit I have put in is fairly convincing to me - but it is not even close to a proof. I am pretty sure all my assumptions are correct - but it is not rigorous (just wanted to be sure I wasn't claiming it was bearing in mind what I said immediately afterwards. You are on the right track. A rigorous proof is surprisingly short: If P is prime and > 3 then P is odd so P+1 and P-1 are divisible by 2 For any integerr P one of P-1, P, P+1 is divisible by 3. Since P is prime it is not P. So one of P+1, P-1 is divisible by both 2 and 3 and hence divisible by 6. QED Note that this is pretty elementary and a curiosity rather than a useful theorem. I am not at all surprised that a mathematician might react skeptically if quoted this fact out of the blue and not given some tiime to think about it. It is not much more than a parlor trick. He probably had many more interesting things to think about. I reiterate -- there is no such thing as a correct heuristic proof. Period. 1 Link to comment Share on other sites More sharing options...

PeterJ Posted February 7, 2012 Author Share Posted February 7, 2012 Okay. But there are clearly proofs that some mathematicans think are correct and heuristic. I'm sorry but I don't know anyone here or their background, so I must assume that Dr. Booker is right. I do not usually try to second guess such people. His letter to me was the best written piece I have ever read on the topic. Interesting though. Why do you say there is no such thing a s correct heuristic proof? Is it in the nature of heuristic proofs that they cannot be correct, or is it the nature of correct proofs that they cannot be heuristic? Link to comment Share on other sites More sharing options...

DrRocket Posted February 7, 2012 Share Posted February 7, 2012 Okay. But there are clearly proofs that some mathematicans think are correct and heuristic. I'm sorry but I don't know anyone here or their background, so I must assume that Dr. Booker is right. I do not usually try to second guess such people. His letter to me was the best written piece I have ever read on the topic. Interesting though. Why do you say there is no such thing a s correct heuristic proof? Is it in the nature of heuristic proofs that they cannot be correct, or is it the nature of correct proofs that they cannot be heuristic? The latter. Proof and "heuristic" are mutually exclusive. A heuristic argument is a plausibility argument, and nothing more. It is most certainly not a proof. How astonishingly patronising. But yes. That's why I was so surprised. His area is statistics, so perhaps he'd never thought about it. He probably figured it as soon as I put the phone down. I have not suggested otherwise. I do happen to believe otherwise, but I have not suggested it and will not here. I don't know exactly what you would count as a proof and what not. Are you saying that you're not sure you can rigorously prove that the products of 2 and 3 are not prime? Thanks for asking. Despite the vast implausibility of it I'm not quite sure that my full argument won't end up being interesting in some way, so I'll keep one or two important bit back if that's okay. But here's the idea. It could hardly be more simple. I am not claiming it is a proof of anything. I am saying that I have been told by someone who should know that it is a correct heuristic proof but not yet rigorous. He did not suggest I had changed the world, and neither have I at any point. I know that an heuristic proof is not a mathematical proof. I use the 6n numbers as a metric against which to measure the behaviour of the products of the other primes. This means ignoring 4 numbers in every 6 and examing only the two possible twin prime locations. Two products of p (>3) occur at 6n+/-1 in every 6p numbers. These will occur at 6p+/-p. So, for instance, 2 locations in every 6 cannot be twin primes since they are products of 5. Etc etc. Some fiddly corrections terms are needed, but these become increasingly unnecessary as p increase. Roughly speaking... For the range p1 squared to p2 squared, call it R, there are R/6 possible locations of twin primes. We know how many products of any p will be created in R for each prime up to sqrt p1, so we can deduct one total from the other. Of course, very often these products will occur where there is already a product of a prime, i.e 35, but this doesn't matter. I do not count twin primes, just set a lower limit for twin primes in R. It is never less than 1, and the calculation always underestimates it. R increases with p, while the quantity of products of primes below sqrt p1 occurring in R decreases with p. Ergo, on average the quantity of twin primes in R increases with p. Quite where it goes to I'm not sure but it's always on the increase. Nothing very fancy. I treated the the number line as a vibrating string and examined the combination wave created by the music of the products of the primes. As a result I would happily bet my life that there an infinite quantity of twin primes. But I would not be able to put together what you guys would call a formal proof. An heuristic proof is easy enough though, since it just means describing the mechanics. Please note that it was not me who made such a big deal out of this. I mentioned it only because I was being accused of being a dunderhead. This is just ridiculous. Was Booker drunk or was he just trying to make you go away ? 1 Link to comment Share on other sites More sharing options...

PeterJ Posted February 7, 2012 Author Share Posted February 7, 2012 (edited) The latter. Proof and "heuristic" are mutually exclusive. A heuristic argument is a plausibility argument, and nothing more. It is most certainly not a proof. I think you've gone right to the heart of our mutual irritableness. In philosophy, where I am more comfortable, an argument is sometimes called a proof, albeit that it might be a failed proof. That is, a proof does not have to succeed in order to be called a proof. A syllogistic argument may be called a proof and also be invalid, unrigorous or successful. A proof or argument may may be correct and yet still be unrigorous and fail. I assumed the same would be true in mathematics. Apparently not. My proof is correct in the sense that it works, that it would be a way of proving, or perhaps demonstrating would be a better word, the TPC, but it is not yet successful becase it is unrigorous, there are one or two loopholes. Dr. Booker was of the opinion that mathematicians already knew of this method of proof, but had never managed to close the loopholes. I think I can see how to do it. We'll see. This is just ridiculous. Was Booker drunk or was he just trying to make you go away ? No. He was polite, and took the trouble to reply with a very thoughful and helpful letter, perhaps because he heads up an external mathematics education project and is used to talking to amateurs, congratulating me on my correct heuristic proof. He clearly did not expect me to make this proof sucessful, or to derive from it the sort of abstract proof that you would call a proper proof. I don't think he was either drunk or patronising, but you'd have to ask him. I'm more suspicious of you, mate, to be honest. Edited February 7, 2012 by PeterJ Link to comment Share on other sites More sharing options...

DrRocket Posted February 8, 2012 Share Posted February 8, 2012 I think you've gone right to the heart of our mutual irritableness. In philosophy, where I am more comfortable, an argument is sometimes called a proof, albeit that it might be a failed proof. That is, a proof does not have to succeed in order to be called a proof. A syllogistic argument may be called a proof and also be invalid, unrigorous or successful. A proof or argument may may be correct and yet still be unrigorous and fail. I assumed the same would be true in mathematics. Apparently not. If that is your concept of a "proof" one wonders what is needed to qualify as gibberish. But since you talk in the context of philosophy, I concede that there may be no difference. In themathematical world you don't have a proof of anything, just babbling. You have not even addressed the twin prime conjecture at all. For clarity, the twin prime conjecture is that there are infinitely many numbers n such that n and n+2 are prime. In terms of your "characterization" of primes as being form 6n +/- 1 the conjecture is that there are infinitely many numbers n such that 6n-1 and 6n+1 are both prime. (like 5 and 7 are 6x1-1 and 6x1+1). In your so-called heuristic proof you failed to show the existence of even one set of twin primes, let alone infinitely many of them. You don't have a proof. You don't have a heuristic proof. You don't even have a sensible sentence. 2 Link to comment Share on other sites More sharing options...

PeterJ Posted February 8, 2012 Author Share Posted February 8, 2012 My apologies. At one point I wrote 6p+/-p when it should have been 6np+/-p. This rendered one sentence a bit meaningless. If that is your concept of a "proof" one wonders what is needed to qualify as gibberish. But since you talk in the context of philosophy, I concede that there may be no difference. In themathematical world you don't have a proof of anything, just babbling. You have not even addressed the twin prime conjecture at all. For clarity, the twin prime conjecture is that there are infinitely many numbers n such that n and n+2 are prime. In terms of your "characterization" of primes as being form 6n +/- 1 the conjecture is that there are infinitely many numbers n such that 6n-1 and 6n+1 are both prime. (like 5 and 7 are 6x1-1 and 6x1+1). In your so-called heuristic proof you failed to show the existence of even one set of twin primes, let alone infinitely many of them. You don't have a proof. You don't have a heuristic proof. You don't even have a sensible sentence. Ho ho. I won't rise to this. I've talked to maybe fifteen or so mathematicians about all this and you're the first to interpret my words in such a way. I'll go with the statistics and assume you have some axe to grind. But sorry about the silly mistake. That confused the issues. -1 Link to comment Share on other sites More sharing options...

ankush04 Posted March 2, 2012 Share Posted March 2, 2012 Hi, I know a little bit on the Riemann hypothesis which is a long standing conjecture of Riemann. Basically the conjecture is about the non-trivial (complex) zeroes of the Riemann Zeta function, Re(s)>1 , which Riemann claims to lie in the critical line that is on the line for which the real part equals 1/2. The conjecture is not yet proved nor it is disproved. But the assertion of the hypothesis is checked for a large number of complex numbers which ensures its validity. However the conjecture if it is solved attempts to answer a large number of mysteries including the distribution of primes. Purely a analytical number theoretic problem, several attempts have been made to solve this conjecture from the point of view of non-commutative geometry. The problem remains a challenge to the entire mathematical community and in particular to the number theorist. P.S. I haven't discussed any kind of technicalities pertaining to the zeta function, which has in itself much richer theory, one example is being its applicability in proving the infinitude of primes using Euler product of Dirichlet series, of which the above series is a type. Another important aspect of the zeta function is that the inverse of zeta function evaluated on natural number say n gives the probability that set of n integers selected at random are pairwise coprime. Link to comment Share on other sites More sharing options...

PeterJ Posted March 2, 2012 Author Share Posted March 2, 2012 (edited) Hello Ankush04 I ended up in some trouble here so left the thread to fade away, but I do want to understand this better so will keep going. Let's come back to the twin primes issue. Is what follows correct? The nontrivial zeros of Z correlate with correction-terms or maybe 'harmonics' that can modify Euler's dice-based prediction for the density of primes so as to make it more accurate. This would allow us to predict the density to infinity, if only we could prove that all the nontrivial zeros of Z are on the R's line, where the real part of the input is 1/2. Is that about it? Edited March 2, 2012 by PeterJ Link to comment Share on other sites More sharing options...

PeterJ Posted March 8, 2012 Author Share Posted March 8, 2012 Yes? No? Is that sentence way off or about right? If it's nonsense just say so. Link to comment Share on other sites More sharing options...

DrRocket Posted March 9, 2012 Share Posted March 9, 2012 Yes? No? Is that sentence way off or about right? If it's nonsense just say so. Total nonsense. Link to comment Share on other sites More sharing options...

PeterJ Posted March 9, 2012 Author Share Posted March 9, 2012 (edited) Total nonsense. Amazing. I'm not generally a stupid person. I wonder if you giving it a generous reading, or getting very picky about the words. If it really is nonsense then I would be genuinely surprised. Are you sure it isn't just that it needs some tweaks? Are you actually prepared to loosen up a bit, or must I fight you every inch of the way? Edited March 9, 2012 by PeterJ Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted March 10, 2012 Share Posted March 10, 2012 ! Moderator Note Apparently no one caught what was said in this thread when it was being said a month ago, but since recent comments look to be going down the same track I will now tell you once and once only to keep this discussion civil and void of personal comments. This goes for everybody.Do not respond to this modnote. Link to comment Share on other sites More sharing options...

DrRocket Posted March 10, 2012 Share Posted March 10, 2012 If it really is nonsense then I would be genuinely surprised. It is time for you to be surprised. Link to comment Share on other sites More sharing options...

PeterJ Posted March 10, 2012 Author Share Posted March 10, 2012 DrRocket - Could you have a break and let me talk to someone who might be helpful? There really is no point in deliberately trying to wind me up, and I don't want to get into trouble with the moderators. If you could re-write that sentence of mine so that it is correct that would help. Don't worry, I'll give you a chance to shoot down my ideas about the twin primes when I have time to go back and remind myself what they were and can start a thread. Link to comment Share on other sites More sharing options...

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