 My Math Forum A proof of Robin's inequality (and so of the Riemann Hypothesis)

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 November 21st, 2014, 09:05 AM #1 Newbie   Joined: Nov 2014 From: Italy Posts: 9 Thanks: 0 A proof of Robin's inequality (and so of the Riemann Hypothesis) I have been endorsed as well to submit to arXiv, and having fixed the few small errors pointed out to me, I am quite confident. I would be very grateful if someone else read through it. Before giving the article to my endorser, I submitted it to viXra, so you can find it here: viXra.org e-Print archive, viXra:1411.0206, Proof of the Riemann Hypothesis and an Upper Bound for the Divisor Sum Function . I combine: prime factorization, a result of Alaoglu and Erdos, the Akbary-Friggstad theorem and Mertens' third theorem. Any remarks are welcome. Last edited by Vincenzo Oliva; November 21st, 2014 at 09:58 AM. November 21st, 2014, 05:55 PM #2 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Lemma 2 is incorrect. The unnumbered formula following "implies" in the proof is unjustified. I'll leave it to you to find a counterexample (it's not hard). It seems that the first major mistake is the first inequality on page 5, where you absorb the factor of $1/2$ without justification. In fact I can't find any number for which the assertion holds -- I searched to 10 million. But the coup de grâce is the line following "we conclude", which sneakily keeps the i=1 factor both in the limit expression (5) and outside, essentially squaring its contribution. Thanks from MarkFL November 22nd, 2014, 09:29 AM   #3
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Quote:
 Originally Posted by CRGreathouse Lemma 2 is incorrect. The unnumbered formula following "implies" in the proof is unjustified. I'll leave it to you to find a counterexample (it's not hard). It seems that the first major mistake is the first inequality on page 5, where you absorb the factor of $1/2$ without justification. In fact I can't find any number for which the assertion holds -- I searched to 10 million. But the coup de grâce is the line following "we conclude", which sneakily keeps the i=1 factor both in the limit expression (5) and outside, essentially squaring its contribution.

As for Lemma 2, please let me know your thoughts on this alternative proof (still with the assumption that the limit of the first general product as $x\to \infty$ exists; I am not too familiar yet with asymptotics, I might have to come up with something else without it):

$g(m) \sim h(m)$ implies $g(m)=h(m) + o(h(m))$ and thus $$\displaystyle \lim_{m\to \infty } \frac{\prod_{n\le g(m)} f(n)}{\prod_{n< h(m)} f(n)}= \lim_{m \to \infty} \frac{\prod_{n\le h(m) + o(h(m))} f(n)}{\prod_{n< h(m)} f(n)} = \lim_{m \to \infty} \prod_{n\le o(h(m))} f(n) = 1.$$

With regard to your other observations, I think perhaps it is necessary to clarify (and write) better what I do on page 5. I will first do it here:
$$\displaystyle e^\gamma \log \log SA_m \prod_{i=1}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1}=e^\gamma \log \log SA_m \frac{2^{\alpha_1}}{2^{\alpha_1+1}-1} \prod_{i=2}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1} > \\ e^\gamma \log \log SA_m \frac{2^{\alpha_1+1}}{2^{\alpha_1+1}-1} \prod_{i=1}^l \frac{p_i-1}{p_i}.$$ You said you searched to 10 million and found no numer satisfying this? You probably misinterpreted what I wrote and I should have avoided those $1/2$'s. Or you were referring to something else, in which case could you explain yourself? Anyway, the inequality above is true: simplifying $e^\gamma \log \log SA_m$ we get $$\displaystyle \frac{2^{\alpha_1}}{2^{\alpha_1+1}-1} \prod_{i=2}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1} > \frac{2^{\alpha_1+1}}{2^{\alpha_1+1}-1} \prod_{i=1}^l \frac{p_i-1}{p_i}= \frac{2^{\alpha_1}}{2^{\alpha_1+1}-1} \prod_{i=2}^l \frac{p_i-1}{p_i} \\ \prod_{i=2}^l \frac{p_i ^{\alpha_i+1}}{p_i ^{\alpha_i +1}-1}>1.$$
In the line following "we conclude" I simply used $y_i>x_i \implies \prod_{i=1}^\infty y_i \ge \prod_{i=1}^\infty x_i$ (even though I recognize I wrote ">" instead of "$\ge$"). Here $y_i=\frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1}$ and $x_i= \frac{p_i-1}{p_i}$. Hence, $$\displaystyle \lim_{m\to \infty} e^\gamma \log \log SA_m \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1} \ge \lim_{m \to \infty} e^\gamma \log \log SA_m \frac{2^{\alpha_1+1}}{2^{\alpha_1+1}-1} \prod_{i=1}^l \frac{p_i-1}{p_i} \\ \lim_{m\to \infty} e^\gamma \log \log SA_m \prod_{i=1}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1} \ge \frac{2^{\alpha_1+1}}{2^{\alpha_1+1}-1} \lim_{m\to \infty} e^\gamma \log \log SA_m \prod_{i=1}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1}.$$ Thus, recalling Mertens' third theorem, we find $$\displaystyle \lim_{m\to \infty} e^\gamma \log \log SA_m \prod_{i=1}^l \frac{p_i ^{\alpha_i}(p_i-1)}{p_i ^{\alpha_i +1}-1} \ge \frac{2^{\alpha_1+1}}{2^{\alpha_1+1}-1} > 1.$$ November 22nd, 2014, 11:14 AM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Lemma 2 is incorrect -- it's not just that it has a faulty proof (which it does). You need more conditions if you want to keep the same conclusion. Your new proof, as it happens, retains the original flaw. I searched to 10 million and found no examples of what you claimed. Whether this was what you intended, I can't say. The third error remains in what you write above. I can't spare any more effort to correct your work, though; if I thought it was likely to be correctable I might be willing to donate more of my time to the cause. From your large number of revisions on that paper I expect you'll just paper over your mistakes once again. November 22nd, 2014, 11:23 AM #5 Newbie   Joined: Nov 2014 From: Italy Posts: 9 Thanks: 0 Actually most revisions were due to silly mistakes of writing or typing. Sometimes it was to correct some minor mistakes pointed out to me. And actually I don't care of the number of versions, as I've submitted to viXra only as a proof of existence. I see, as to Lemma 2. Though, out of curiosity, could you please tell me what the error is on this second proof? Consider I'm very new to asymptotics. But I can't even see for the other two errors. Explicitly, what are they? I really appreciate your help. Last edited by Vincenzo Oliva; November 22nd, 2014 at 11:33 AM. November 23rd, 2014, 03:40 AM #6 Newbie   Joined: Nov 2014 From: Italy Posts: 9 Thanks: 0 I've just noticed I made a typo in my explanation, in the last line before "thus". The second product should have $\frac{p_i−1}{p_i}$ as general factor. November 23rd, 2014, 09:23 AM   #7
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Quote:
 Originally Posted by Vincenzo Oliva Consider I'm very new to asymptotics.
You say you're very new to asymptotics, but you're proposing an asymptotic proof of (arguably) the most famous open problem in mathematics. This is like a trainee pilot on a Cessna saying that he could make an engine-out landing of a B2 on an aircraft carrier without using his tailhook, if only he could figure out how to work the aileron.

But to show you there are no hard feelings, I'll walk you through your proof. I really don't have the time for it but that's a separate issue.

Quote:
 Originally Posted by Vincenzo Oliva $g(m) \sim h(m)$ implies $g(m)=h(m) + o(h(m))$
This is true -- it's essentially the definition of ~. But you need to be careful about what this means, because the = sign is essentially a lie. It's more like $\in$ -- there is some function, which isn't named here, but which is in o(h(m)).

Quote:
 Originally Posted by Vincenzo Oliva $$\displaystyle \lim_{m\to \infty } \frac{\prod_{n\le g(m)} f(n)}{\prod_{n< h(m)} f(n)}= \lim_{m \to \infty} \frac{\prod_{n\le h(m) + o(h(m))} f(n)}{\prod_{n< h(m)} f(n)}$$
With the understanding above about o(h(m)) meaning an anonymous function, this is correct. You haven't said anything yet, though -- you're still just unpacking definitions.

Quote:
 Originally Posted by Vincenzo Oliva $$\lim_{m \to \infty} \frac{\prod_{n\le h(m) + o(h(m))} f(n)}{\prod_{n< h(m)} f(n)} = \lim_{m \to \infty} \prod_{n\le o(h(m))} f(n)$$
This is the first line where you do something, but it's wrong. You've shifted the interval. It's like writing
$$\sum_{n=1}^{103}n-\sum_{n=1}^{100}n=\sum_{n=1}^{3}n$$

Quote:
 Originally Posted by Vincenzo Oliva $$\lim_{m \to \infty} \prod_{n\le o(h(m))} f(n) = 1.$$
This is also wrong (and will remain wrong when you shift the interval to where it belongs). Can you see why? If not, break this down into individual steps and try to justify each. At some point you'll have to use a new assumption not in your original, or you'll have to make a false leap. November 23rd, 2014, 12:56 PM   #8
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That asymptotic "proof" is not mine, I did not want to state it, but I think I better say it now. It has been proposed to me. And anyway, it is just a lemma, not the theorem. Speaking of which, what requires the lemma later on in the paper, that limit equality, is most lilely correct, both intuitively and for my calculation with very large SA numbers. And note that my original "proof" is that in the PDF you read

I am still having dinner here, so I have not looked thoroughly at your last message. I thank you anyway for your time. I have attached a PDF with a "corrected" (by another person from the first) version of the asymptotic "proof". I would be very grateful if you could point out any mistakes in it.
Attached Images Screenshot_2014-11-23-21-47-20.jpg (10.4 KB, 2 views) November 23rd, 2014, 01:54 PM #9 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The jpeg you attached is unreadable, but that's OK as I do not have time to review the proof yet again. If you think your proof is valid, I suggest hiring a grad student to review it for you. November 23rd, 2014, 02:25 PM #10 Newbie   Joined: Nov 2014 From: Italy Posts: 9 Thanks: 0 Alright. I wrote PDF without thinking, of course it (unluckily) was a jpeg. Tags hypothesis, inequality, proof, riemann, robin ,

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