5
$\begingroup$

Could Riemann' Hypothesis be proven true using Robin's Inequality and that a counter-example to Riemann's Hypothesis can not have a divisor that is a prime number to the exponent 5 ,according to some of Robin's Theories? Also I think it can be proven the product of two numbers A and B that are counter-examples to R.H. is also a counter-example. Note a number $n$ is a counter-example to R.H. if $\sigma(n)\gt e^{\gamma}n\ln\ln n$ (forgive notation).

$\endgroup$
  • 3
    $\begingroup$ Language remark: such an $n$ would not be a "counterexample to RH," it would be a counterexample to an equivalent to the RH. A counterexample to RH would be a nontrivial zero with real part off the critical line. $\endgroup$ – anon Sep 8 '14 at 6:44
  • $\begingroup$ According to Lagharius ( forgive spelling) if R.H. is false for n then 𝜎(n) ≥ H𝗇 + eᴴ𝗇 l n (H𝗇) , because H𝗇 > l n ( n+1) ∴ 𝜎(n) ≥ l n ( n+1) + ( n+1) l n( l n (n+1)) (∀ n > 1) If l n (l n ( n+1)) ≥ 3 then 𝜎( n) ≥ 3(n+1) + l n ( n+1) $\endgroup$ – 201044 Mar 30 '16 at 2:49
3
$\begingroup$

Robin's inequality was proved to be true if the Riemann Hypothesis holds, so disproving Robin's inequality would be one way of disproving the Riemann Hypothesis. However, the inequality has been around for a long time (Ramanujan got the result in 1915), this is probably just as hard as any other way of settling the Riemann Hypothesis.

I am not quite sure what you mean by numbers that are counter-examples to the Riemann Hypothesis. The Hypothesis itself is about the complex zeros to a complex function (the claim is that the zeros with positive real part all have real part 1/2).

[Added later] Yes, Robin proved in 1984 that the inequality holding for all $n>5040$ is equivalent to the RH. So in principle you can disprove RH by finding a single integer which fails Robin's inequality. But thirty years later, no one has succeeded by that route.

[Added later] @user128932 Sorry, I failed to answer your point about $a,b$. So suppose $a,b$ (both $>5040$) failed to satisfy Robin's inequality, so that $\sigma(a)>e^\gamma a \ln\ln a$ and $\sigma(b)>e^\gamma b \ln\ln b$, does that mean $ab$ also fails to satisfy it?

$\sigma(n)$ is a "multiplicative function" which means (in this context) that $\sigma(mn)=\sigma(m)\sigma(n)$ provided that $m,n$ are relatively prime. So if $a,b$ were relatively prime exceptions, we would have $\sigma(ab)=\sigma(a)\sigma(b)>e^{2\gamma}(ab)(\ln\ln a)(\ln\ln b)$.

For $ab$ to be another exception we want $\sigma(ab)>e^\gamma ab\ln\ln(ab)$. Now $e^\gamma>1.78$, so $e^{2\gamma}>1.78e^\gamma$, which is helpful. We have $\ln(a)\ln(b)>\ln(a)+\ln(b)=\ln(ab)$ for large $a,b$, which is also helpful, eg $\ln(5051)\ln(5059)=72.73>17.06=\ln(5051\cdot5059)$. Of course this effect is reduced for $\ln\ln$, eg $(\ln\ln 5051)(\ln\ln 5059)=4.59>2.84=\ln\ln(5051\cdot5059)$. But $(\ln\ln a)(\ln\ln b)>\ln\ln(ab)$ obviously holds in general for sufficiently large $a,b$ (and maybe for all $a,b$ I have not really thought about it).

So yes in the $a,b$ relatively prime case it looks as though $ab$ would be another exception. The case where they are not relatively prime looks more doubtful, eg $\sigma(16)=5<8=\sigma(2)\sigma(8)$.

$\endgroup$
  • $\begingroup$ Robin's inequality says if sigma(n) < $e^{gamma}$ n lnlnn for any n > 5040 then R.H. is true. ( I looked this up in this website) So if there exists m such that sigma(m) >= $e^{gamma}$ m lnlnm where m > 5040 then m could be 'called' a counter-example to R.H.; using this with the fact the product of two 'counter-example integers' is also a 'counter-example integer' maybe one can show such 'counter-examples' can't exist. $\endgroup$ – user128932 Sep 8 '14 at 6:33
  • $\begingroup$ Actually, you can. RH is equivalent to universal statements about the integers. There is a thread on MO where this is discussed $\endgroup$ – Andrés E. Caicedo Sep 8 '14 at 6:51
  • $\begingroup$ @user128932 Thank you. Yes a single integer $m$ could be used to disprove RH. I have corrected my answer. $\endgroup$ – almagest Sep 8 '14 at 6:52
  • 1
    $\begingroup$ Note if a is a counterexample then sigma(a)/a >= e^{gamma} lnlna ; if a was deficient then 2 > sigma(a)/a , so 2 > e^{gamma} lnlna . So if one lets a > e^{e^2} and b > e^{e^2} then a and b can not be deficient counterexamples. Therefore if a and b > e^{e^2} and they are coprime counterexamples then so is (a b). Assume there is a least squarefree C, that is not a counterexample; C = (a b) for some a and b (C > a, b > 1); a and b are coprime and both squarefree. Assume a is a counterexample then b can't be, yet if b is not a counterexample it is < C. This shows no counterexample is squarefree. $\endgroup$ – user128932 Sep 15 '14 at 6:32
  • 1
    $\begingroup$ @user128932 It does take a bit of getting used to. But the basic idea is easy. Put symbols between a pair of dollar signs \$ ... \$. Use the backslash for "special words", such as sigma. So to get sigma(a) as $\sigma(a)$ you need "\$\ sigma(a)\$" (without the outer quotes). Sorry the spacing is slightly awry there, defeating the system is harder than using it! :) $\endgroup$ – almagest Sep 15 '14 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.