15
$\begingroup$

\begin{equation} \sigma(n) < e^\gamma n \log \log n \end{equation}

In 1984 Guy Robin proved that the inequality is true for all n ≥ 5,041 if and only if the Riemann hypothesis is true (Robin 1984).

I could not get a hold of his paper. I am trying to understand how did he derived the inequality. Can anyone can outline the steps, how Robin derived this criterion?

$\endgroup$
6
  • 2
    $\begingroup$ It turns out that the OP asked the same question on math.SE, where someone supplied the following link: mpim-bonn.mpg.de/preblob/2960 Double posting (especially without mention) is not cool, so I am voting to close now. $\endgroup$
    – Igor Rivin
    Commented Dec 25, 2011 at 18:14
  • $\begingroup$ When I click on Igor's link, I get a screen full of gobbledegook. To make it work, save the link (e.g. by right clicking on it). It's a PDF file, which you can then view in the normal way. $\endgroup$ Commented Dec 25, 2011 at 18:34
  • 1
    $\begingroup$ If you have access to a math library you should be able to find Robin's original paper. It's in French but, fairly readable. If I recall, the main idea of the paper was to sharpen some existing bounds on the sum-of-divisors function for highly composite numbers. I'm curious if there's a different derivation. $\endgroup$
    – Alex R.
    Commented Dec 25, 2011 at 18:46
  • $\begingroup$ @Tom: sorry about that, I just cut and pasted the url :( $\endgroup$
    – Igor Rivin
    Commented Dec 25, 2011 at 21:01
  • $\begingroup$ @Igor: Yes, I posted a similar question on math.SE. But I realized that this question is better suited here. $\endgroup$ Commented Dec 25, 2011 at 23:12

3 Answers 3

11
$\begingroup$

I have requested a pdf of Robin 1984 from campus scanning service. One highlight of the article that really should be mentioned is this:

For $n \geq 3,$ we have $$ \sigma(n) \; < \; \; e^\gamma \; n \log \log n \; + \; \frac{ \; 0.64821364942... \; \; n \; }{\log \log n},$$ with the constant in the numerator giving equality for $n=12.$

see:
Which $n$ maximize $G(n)=\frac{\sigma(n)}{n \log \log n}$?

That, at least, rests on effective bounds of Rosser and Schoenfeld (1962), which can be downloaded from ROSSER

Well, maybe not so directly. R+S do the unconditional bound for $n/\phi(n)$ in Theorem 15, pages 71-72, formulas (3.41) and (3.42). The treatment for $\sigma(n)$ is quite similar in spirit, maybe Robin was the first to write it down. The analogue of the primorials PRIMO and $n^{1-\delta}/\phi(n)$ is the colossally abundant CA numbers and $\sigma(n)/ n^{1 + \delta}.$

Well, I am not sure where it is written down, but it is easy enough to show that the maximum value, for some $0 < \delta \leq 1, $ of $$ \frac{ n^{1-\delta}}{\phi(n)} $$ occurs when the prime factor $p$ of $n$ has exponent $$ v_p(n) = \left \lfloor \frac{p^{1-\delta}}{p-1} \right \rfloor.$$ Since, for a fixed $\delta,$ this expression is either 0 or 1 and nonincreasing in $p,$ it turns out that the optima occur at the primorials, the products of the consecutive primes from 2 to something...

From Alaoglu and Erdos, the maximum value, for some $0 < \delta \leq 1, $ of $$ \frac{\sigma(n)}{ n^{1+\delta}} $$ occurs when the prime factor $p$ of $n$ has exponent $$ v_p(n) = \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. The results of this construction are the colossally abundant numbers. The construction is originally due to Ramanujan, but the part of his manuscript that dealt with ca numbers was not printed owing to paper shortages at the time.

Hardy and Wright use $d(n)$ for the number of divisors of $n.$ This is in the original paper by Ramanujan. For some $0 < \delta \leq 1, $ the maximum of $$ \frac{d(n)}{ n^{\delta}} $$ occurs when the prime factor $p$ of $n$ has exponent $$ v_p(n) = \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor. $$ The results are called the superior highly composite numbers SHC.

So, taking all three with $\delta = 1/2,$ we get lemmas $$ \phi(n) \geq \sqrt{\frac{n}{2}}, \; \; d(n) \leq \sqrt{3n}, \; \; \sigma(n) \leq 3 \left( \frac{n}{2} \right)^{3/2}. $$

In all three cases, if $\delta$ is such that more than one number $n$ achieves the maximum value of the ratio specified, we are choosing the largest of these $n$'s.

$\endgroup$
0
15
$\begingroup$

In view of OP's comment on Igor Rivin's answer it seems that the 'actual' question could be something else.

The inequality under RH $$ \sigma(n) < e^{\gamma} n \log \log n $$ for sufficiently large $n$ is not due to Robin, but due to Ramanujan. And still before that Grönwald (1913) showed (uncoditionally) $$ \limsup_{n\to \infty}\frac{\sigma(n)}{n \log \log n} = e^{\gamma} $$

As to why questions like this are linked to RH at all. For example, recall that if one defines $\sigma_y (n) = \sum_{d|n}d^y$ then for the asociated Dirichlet series one has $$ \sum_{n=1}^{\infty} \frac{\sigma_y(n)}{n^s} = \zeta(s)\zeta(s-y) $$ so
$$ \sum_{n=1}^{\infty} \frac{\sigma (n)}{n^s} = \zeta(s)\zeta(s-1). $$

Without having followed up on the precise historical deveopment it seems rather like so: one studies the growth of $\sigma$ as for plenty of other arithmetical functions. Somebody (Grönwald) shows a nice result, somebody else (Ramanujan) shows something more precise under RH. Then somebody (Robin) decdides to investigate whether this is in fact equivalent to RH (as some other results known under RH, most notably the asymptotic count of prime numbers).

This seems like a quite natural development to me.

$\endgroup$
4
  • $\begingroup$ This is certainly quite a reasonable view, but the statements (of Robin, etc) are still quite surprising... $\endgroup$
    – Igor Rivin
    Commented Dec 25, 2011 at 18:06
  • $\begingroup$ Yes, of course, the results are surprising and nice. All I meant to say is that this fits fairly naturally into some general development, and did not come out of nowhere. And, the inequality is not Robin's so the question 'how did he derive it' seems in some sense ill-posed; as he was not the (first) one to derive it. $\endgroup$
    – user9072
    Commented Dec 25, 2011 at 18:17
  • $\begingroup$ quid, I got the Robin pdf today. If you would like a copy email me, you could create an address such as [email protected] for this type of purpose. I do not see the explicit Dirichlet series you show above, but he has a long bibliography. In particular, he is a student of J. L. Nicolas, who has a later survey article on this area in a book called Ramanujan Revisited. $\endgroup$
    – Will Jagy
    Commented Jan 3, 2012 at 21:34
  • $\begingroup$ @Will Jagy, in case you still read this: thank you for the kind offer and sorry for not following up earlier; my activity was a bit spurious lately. While not at the time of writing, in general it would be reasonable easy for me to get the paper, so thank you but it is not needed. Also, thank you regarding the clarification with the series; I did not want to imply (but perhaps did not make this clear enough) that it is this what is used (I did not know). The intention was merely to say something that shows that it is not toally unexpected that properties of zete could play a role. $\endgroup$
    – user9072
    Commented Jan 9, 2012 at 13:10
10
$\begingroup$

I cannot find Robin's paper either (thank you, Elsevier), but a stronger theorem was proved by Jeff Lagarias in 2002:

J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Amer. Math. Monthly 109 (2002), 534–543.

Lagarias' statement is:

The RH is true if and only if $ \sigma(n) < H_n + \exp(H_n) \log(H_n), $ where $H_n$ are the usual harmonic numbers.

$\endgroup$
3
  • 1
    $\begingroup$ I have gone through Lagarias' paper already. It seems to me that he has just treated Robin's inequality to give a better bound. But I never quite got how Robin managed to relate RH and sigma and got this inequality. $\endgroup$ Commented Dec 25, 2011 at 16:54
  • 7
    $\begingroup$ Since it's an if and only if statement, isn't neither statement technically stronger? $\endgroup$
    – Will Sawin
    Commented Dec 25, 2011 at 17:57
  • $\begingroup$ @Will: very true... $\endgroup$
    – Igor Rivin
    Commented Dec 25, 2011 at 18:05

Not the answer you're looking for? Browse other questions tagged .