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Community: [Closed] A positive answer to the Riemann hypothesis: A new result predicting the location of zeros

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    • CommentAuthorzeraoulia
    • CommentTimeDec 15th 2012 edited
     
    I very confused about the behavior of mathematical community. I have worked 15 years in chaos theory and I have published more than 80 papers and 8 books in several international publishers and in each time I seek comments from other specialist, they answred my positively and they never refuse to read may papers. For this time I send a request for comments (for my paper: http://vixra.org/pdf/1210.0176v7.pdf) to more than 800 mathematician arround the world and I receive only 7 replies. The main objective is to see the opinions of experts before sending the paper to a journal. Is the mathematical community behaves like the logistic equation (chaotic). More than this, some of them attack me personaly with very bold words in despite they do not know me.
  2.  

    Do you have anything specifically about MathOverflow that you want to say? This site is mainly for discussing MathOverflow.

  3.  
    Sadly, this is starting to look like spam.
    • CommentAuthorYemon Choi
    • CommentTimeDec 15th 2012
     

    I agree with Andres

    • CommentAuthorquid
    • CommentTimeDec 15th 2012 edited
     

    Calling it spam seems a bit harsh. It should be noted this thread was created before the contribution in the other thread of which OP was likely unaware for context [Added: meanwhile not generally visible anymore, as deleted via flags.]

    But I agree this posting is not sufficiently MO related. Though it might be a misunderstanding of some advice I gave RH on main. (Not sure the advice was so good in the first place, but RH is experiencing some difficulty with various of their contributions on MO, so I thought to tell them to ask on meta about it, as opposed to me in the comments of a closed question would be a good idea.)

    • CommentAuthorfedja
    • CommentTimeDec 16th 2012
     
    >> For this time I send a request for comments (for my paper: http://vixra.org/pdf/1210.0176v6.pdf) to more than 800 mathematician arround the world and I receive only 7 replies.<< Here is what I can say. The chances that you have a valid proof there are 0.00000.....000001%. I just was proofreading an attempt to solve Navier-Stokes on 12 pages last week by a mathematician with much higher reputation than yours. The mistake was a trivial miscomputation in some Lemma in the middle. My advice: proofread your paper yourself doing one page per day, reading it aloud, and carrying out every "trivial" computation in honest. If after 8 days of such proofreading you still claim it to be correct, I'll read it. However, if I find a mistake or a part that makes no sense, you'll not try to patch it immediately or move it to some other place, or anything else like that. You'll just accept the verdict and wait for 3 months before trying to show anything like that to anyone again. Deal?
    • CommentAuthorzeraoulia
    • CommentTimeDec 16th 2012 edited
     
    @fedja: Ok and thank you very much. I will spent 8 dyas to verify all possible errors and I will send it again to you.
  8.  

    Dear zeraoulia: I do not understand why you sent me an email asking me to check your preprint about the Riemann Hypothesis. If you check my web page, you will see quite clearly that I have never done any work on zeta functions or the distribution of primes. In particular, it is unlikely that I could verify a correct proof of RH without investing months of time learning the necessary background. Do you think it is reasonable to expect me to set aside the research I am currently doing for such a long time?

    If people like me are counted among your 800, you should not be surprised that you are getting few replies. Is your time so precious that you can't be bothered to find out if you are emailing a person in the correct field?

    To bring us back on topic, I am fairly certain that the general consensus of the community is the following:

    1. Requests for verification of preprints are off-topic on MathOverflow. If you ask such a question, it will be closed.
    2. Multiple requests for verification are typically marked as spam. People who make multiple such requests are considered unwelcome.

    I see that you have already asked more than 20 questions that were not appropriate. Please refrain from doing so again. If you are not sure about what is acceptable, please re-read the FAQ.

    • CommentAuthorzeraoulia
    • CommentTimeDec 17th 2012 edited
     
    @Scott Carnahan: Thank you for your reply. I find in MO that you have answred a closely related question to RH. Sorry for that problem.
    • CommentAuthorzeraoulia
    • CommentTimeDec 24th 2012 edited
     
    @fedja: I put the revised version of the paper at:
    http://vixra.org/pdf/1210.0176v7.pdf
    Please let me know that you receive this message.
    • CommentAuthorfedja
    • CommentTimeDec 24th 2012
     
    OK, I'll stick to my words. However, one thing needs to be clarified before I say anything else. The main theorem in the paper says something about the description of roots with real part 1/2 but it tells ABSOLUTELY NOTHING about the case when the real part is not 1/2. How can anything like that help to establish the RH? Either the general logic here is hopelessly flawed, or I just misunderstand the formulation. What exactly should the statement of the theorem be (forget the proof for now) and how does RH follow from it?
    • CommentAuthorzeraoulia
    • CommentTimeDec 24th 2012 edited
     
    @fedja: Theorem 1 proves that the Riemann hypothesis is true for the function η(s) if and only if θ(s)≠0 (mod2π), x(s)=u(s) and y(s)=-v(s) for every root s∈D of the equation η(s)=0.
    • CommentAuthorfedja
    • CommentTimeDec 24th 2012
     
    OK, I just interpreted the statement the other way (it is a bit ambiguous). All right, I'll look at the rest now (I have a long flight tomorrow,
    so don't expect a quick response but be sure that I'll come back eventually). :).
    • CommentAuthorfedja
    • CommentTimeDec 24th 2012
     
    OK, the next question (I'm a slow reader, you know...) The conditions of the theorem are $x=u$, $y=-v$. However, just a few lines below you talk about the fixed point of $B$, which would require $x=u$, $y=v$ to the best of my understanding. This little minus is either misplaced or forgotten. Am I missing something again?
    • CommentAuthorzeraoulia
    • CommentTimeDec 24th 2012 edited
     
    The fixed points of B do not require $x=u$, $y=-v$ . The condition $x=u$, $y=-v$ implies that alpha=0.5.
    • CommentAuthorfedja
    • CommentTimeDec 25th 2012
     
    Now I don't understand you at all. Where does the statement that (u,v) is a fixed point come from then? The only thing you do is to say that (x,y)-vector is the (normalized) image of the (u,v)-vector. If no relation between (x,y) and (u,v) is used in the proof of $s$ being a root, then you effectively claim that the condition $\theta(s)\ne 0\mod 2\pi$ alone implies that $s$ is a root of $\eta$, which is absurd. So, here is the question again: we have the assumptions $x=u$, $y=-v$, $\theta(s)\ne 0\mod 2\pi$ and nothing else. How does it follow that $s$ is a root? This is claimed on the top of page 5 already. Assuming that you didn't intend to derive it immediately from the fixed point argument but rather from (8), I see no reason why (8) implies anything of this sort. All you say there is that if $\alpha=\frac 12$, then SOMETHING, after which the claim is made that $\alpha=\frac 12$ is the only possible option. Equation (8) is certainly correct by itself, but the short paragraph between it and the proof of the reverse implication seems to make no sense either as a demonstration that $\eta(s)=0$, or as a demonstration that $\alpha=\frac 12$. Note also that $\theta$ doesn't enter (8) in any way but is mentioned in this paragraph, thus making me think again that the fact that $s$ is a root was claimed BEFORE we arrived at (8) and is merely recalled now. Whether it all makes sense or not, the presentation is extremely confusing. So, please clarify what is the last line of the proof that $s$ is a root (forget about the part proving $\alpha=\frac 12$ for a while).
    • CommentAuthorzeraoulia
    • CommentTimeDec 25th 2012 edited
     
    @fedja: "Where does the statement that (u,v) is a fixed point come from then? "

    Rotations have single fixed point with respect to (u,v). This is a geometrical result. I have used this fact to determine that if $\theta(s)\ne 0\mod 2\pi$, then s is a root.

    "This is claimed on the top of page 5 already. Assuming that you didn't intend to derive it immediately from the fixed point argument but rather from (8), I see no reason why (8) implies anything of this sort".

    Since, we speak about a single and isolated root s, and you have (8), then by intuition you obtain that if alpha=0.5 then ro=1 and (8) holds for one beta since again the root is isolated.

    "thus making me think again that the fact that $s$ is a root was claimed BEFORE we arrived at (8) and is merely recalled now".

    No, if you read the proof, firstly, I assume that s is in D and then I collect the two conditions to claim that s is a root with alpha=0.5.

    "the presentation is extremely confusing."

    You have two implications see item (1) for the second one and item (2) for the first one.
    • CommentAuthorfedja
    • CommentTimeDec 25th 2012
     
    OK, so how do you show that $(u,v)$ is a fixed point of B? There seems to be no equation anywhere that would imply it. Of course, if it is, many things start making sense and I do not disagree with that. However, to the best of my understanding, the only thing that could be used for showing that $(u,v)$ is a fixed point of B is (7) and that would require a different sign in the relation y=-v, as I claimed above.

    By the way, I can easily write one line whose meaning nobody will ever understand. "Short+structured" doesn't always make "clear" :).
    • CommentAuthorzeraoulia
    • CommentTimeDec 25th 2012 edited
     
    @fedja:

    "OK, so how do you show that $(u,v)$ is a fixed point of B? "

    The fixed point of the rotation in (7) must satisfies (I₂-B(s))(u(s),v(s))=0 where I₂ is the 2×2 unit matrix (this is a linear system of algebraic equation). The determinant of the matrix (I₂-B(s)) is -2(cosθ(s)-1) and it is not zero since θ(s)≠0 (mod2π), this means that s is a solution of η(s)=0 by using (3).

    " However, to the best of my understanding, the only thing that could be used for showing that $(u,v)$ is a fixed point of B is (7) and that would require a different sign in the relation y=-v, as I claimed above."

    No, relations x=u and y=-v are used to show that alpha=0.5 in the first part of theorem 1.

    "By the way, I can easily write one line whose meaning nobody will ever understand. "Short+structured" doesn't always make "clear" :). "

    Can you specify the matter.
    • CommentAuthorfedja
    • CommentTimeDec 25th 2012
     
    Yes, the fixed point of the rotation in (7) must be (0,0). However, what makes you think that (u(s),v(s)) is a fixed point of that rotation? I see no proof of this anywhere.
    • CommentAuthorzeraoulia
    • CommentTimeDec 25th 2012
     
    The objective is not the fixed point. The objective is to find conditions to get a zero s for eta. I made this approach to find possible conditions to get zeros of the eta function. The only condition is when $\theta(s)\ne 0\mod 2\pi$.
    • CommentAuthorfedja
    • CommentTimeDec 25th 2012
     
    So, do you claim that $\theta(s)\ne 0\mod 2\pi$ alone implies that $\eta(s)=0$? As I said, this is just absurd.
    • CommentAuthorzeraoulia
    • CommentTimeDec 25th 2012 edited
     
    "Now I don't understand you at all. Where does the statement that (u,v) is a fixed point come from then? The only thing you do is to say that (x,y)-vector is the (normalized) image of the (u,v)-vector. If no relation between (x,y) and (u,v) is used in the proof of $s$ being a root, then you effectively claim that the condition $\theta(s)\ne 0\mod 2\pi$ alone implies that $s$ is a root of $\eta$, which is absurd. So, here is the question again: we have the assumptions $x=u$, $y=-v$, $\theta(s)\ne 0\mod 2\pi$ and nothing else. "

    You know that all non-trivial rotations have a single and unique fixed point. If we apply this result to the rotation in (7) we obtain only the condition $\theta(s)\ne 0\mod 2\pi$. The condition $x=u$, $y=-v$, $ is used only to deduce that aplha=0.5 for a single beta.

    I write simply: It is well known that a non trivial rotation must have a unique fixed point, its rotocenter. The rotation in (5) is non trivial if ϕ(s)≠1 (here we assumed that θ(s)≠0 (mod2π) in the second part of Theorem 1). The reason is that the trivial rotation corresponding to the identity matrix, in which no rotation takes place. The fixed point of the rotation in (7) must satisfies (I₂-B(s))(u(s),v(s))=0 where I₂ is the 2×2 unit matrix. The determinant of the matrix (I₂-B(s)) is -2(cosθ(s)-1) and it is not zero since θ(s)≠0 (mod2π), this means that s is a solution of η(1-s)=0 and by using (3) we obtain that η(s)=0 .
  24.  

    Um, excuse me for interrupting, but why is this discussion being held on meta? I thought meta was for discussions about the operation of MathOverflow, not about mathematics.

    • CommentAuthorgilkalai
    • CommentTimeDec 26th 2012
     
    This is a very nice service to the community by Fedja. Let's not interupt and welcome it on this thread.
  26.  

    In my opinion, it sets a bad precedent: that purported solutions to RH and discussion thereof are now welcomed on meta. But I am happy to have this be my last comment on the matter, and let the moderators decide.

    • CommentAuthorHenry Cohn
    • CommentTimeDec 26th 2012
     

    I agree with Gil that it is a nice thing Fedja is doing, but I'm also a little worried about creating the appearance that if you want mathematicians to evaluate an unconventional proof of a famous conjecture, you can do this by asking on meta.MO (or that this is a particularly interesting or promising approach to RH).

    I don't want to get caught up in a long discussion of the proof myself, but it has fundamental issues even aside from what Fedja and Zeraoulia are currently discussing. For example, page 9 of the current version makes no sense to me, even taking for granted the earlier assertions about characterizing roots of eta on the critical line. The first half of page 9 describes a trivial equivalence relation on points with theta not equal to an integral multiple of 2\pi (it simply defines them all to be equivalent). Then the paper exhibits one root of eta with theta not a multiple of 2\pi and concludes using the equivalence relation that all the roots of eta have this property. This is an elaboration on page 7 of the previous version, which simply made the same assertion without the equivalence relation explanation, but it's not a proof. My conclusion is that this paper does not give a proof of the Riemann hypothesis, and I can't imagine any way of correcting it.

    • CommentAuthorzeraoulia
    • CommentTimeDec 26th 2012 edited
     
    @Henry Cohn:

    No, the eta function has infinitely many roots. The equivalence relation is only a caracterisation of them. If We follow your opinion, then the set of multiples of an interger of the form n-m has a single element and the half of algebra must be omitted.

    Copied from Wikipedia:

    In mathematics, an equivalence relation is a relation that, loosely speaking, partitions a set so that every element of the set is a member of one and only one cell of the partition. Two elements of the set are considered equivalent (with respect to the equivalence relation) if and only if they are elements of the same cell. The intersection of any two different cells is empty; the union of all the cells equals the original set.
    • CommentAuthorYemon Choi
    • CommentTimeDec 26th 2012 edited
     

    I agree with Todd Trimble's sentiments.

    (Separately: I feel that the issues arising from this discussion show that all the previous MO questions about fancy gadgets in number theory were a complete waste of time, and that MO questions should not be judged quickly based on their vocabulary. "O Americano, Outra Vez!" to quote from a celebrated raconteur.)

  30.  
    Yes, this thread's amusement value wore out quickly. Moderators, please close.
  31.  
    I think, by now Fedja fulfilled his civic duty reading the paper (going far beyond of what most of us would do in this situation), while Henry nailed down the main logical flaw in the paper. Thus, I second Andres' request for this thread to be closed. Dear Elhadj (Zeraoulia): If you still think that your proof is valid, why don't you just send it to the journal of your choice and wait for referee's report, since MO (including meta) is not the right place for requesting opinions on your papers.
    • CommentAuthorfedja
    • CommentTimeDec 27th 2012
     
    Yeah, I completely agree that the paper is beyond repair. So, I'll stop here too. What I disagree with is the attitude of Scott: yes, sometimes the cranks should be talked away politely, but zeraoulia is not a crank and the story of the Navier Stokes paper I told is real, though, obviously, I'm not telling any names here. Sometimes we all indulge in wishful thinking and want things to be correct too much to be able to see the flaws in our own arguments. There is nothing shameful about it, but there is nothing good about it either. So, I'm just asking zeraoulia to stick to his part of the deal now. :)
    • CommentAuthorzeraoulia
    • CommentTimeDec 28th 2012 edited
     
    @fedja: But I am still not convainced that using "equivalent relation" would affect that proof. The Henry claim must have strong evidences and it makes no sense to me. Here he just said "My conclusion is that this paper does not give a proof of the Riemann hypothesis, and I can't imagine any way of correcting it" without giving any argument on his claim. This is not acceptable unless he give strong arguments!
    • CommentAuthorabatkai
    • CommentTimeDec 28th 2012
     
    @zeraoulia: You may put pupils of a class in an equivalence relation by being in the same schoolclass and prove that one of them has black hair in a class. Does this prove that all of them in the same class have black hair?
    • CommentAuthorolga
    • CommentTimeDec 28th 2012
     
    @abatkai: Yes it does, with some pretty high probability. Explanation: given the worldwide distribution of various hair colors, often concentrated over countries, etc, having someone in a class with black hair proves at about 70-80% or so that all the other pupils in the class have black hair too!
    • CommentAuthorfedja
    • CommentTimeDec 28th 2012
     
    @zeraoulia Well, I can claim merely that the part I discussed with you is beyond repair. Note that I asked you for a bug-free paper, so the first crude mistake counts whether it is crucial for the argument as a whole or not. Also, you should understand that it is your duty to convince reasonably qualified people that your proof is correct, not their duty to convince you that it is incorrect. Of course, convincing an idiot of anything is beyond the wisest man's abilities, but I suggest you take it for granted that neither I, nor Henry are idiots, so, if you want to ever come public with this again, please take care of the flaws we found first...
    • CommentAuthorzeraoulia
    • CommentTimeDec 28th 2012
     
    @Fedja and Henry: Ok, ad thank you very much for valuable discussions. I will repair the paper.
    • CommentAuthorAngelo
    • CommentTimeDec 30th 2012
     
    Dear Zeraoulia,

    it seems to me that you are threading on very slippery ground.

    A long time ago I thought I had proved a major conjecture (nothing like the Riemann hypothesis, but still a big deal). I wrote a paper, but before making it public I showed it to a friend, an extremely competent mathematician, who thought it was ok. Fortunately I decided to think about it some more before distributing it, because I found a fatal error after a couple of days. I still remember the heady feeling when I thought I had a proof, and the huge letdown when I found the mistake.

    The heady feeling can be addictive; it's very hard to give it up. I have seen very good mathematicians fall prey to it. The fact that you take it for granted that you can repair the paper makes me think that you might be falling into this pit.
    • CommentAuthorzeraoulia
    • CommentTimeDec 30th 2012 edited
     
    @Angelo:

    Thank you very much for your letter.
    • CommentAuthorYemon Choi
    • CommentTimeDec 30th 2012 edited
     

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