Talk:Divisor function

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Field: Number theory

Divisor function

Hmmm - the divisor function that springs to mind is the function d(n) = number of divisors of n (which is σ0). Needs a check here on terminology, and an inclusive page.

Charles Matthews 12:38, 24 Feb 2004 (UTC)

That's just σ0(n), isn't it? Dysprosia 12:40, 24 Feb 2004 (UTC)

Yes - you beat me to it posting ... occurs in the Dirichlet divisor problem, so may sometimes be called Dirichlet divisor function.

Charles Matthews 12:44, 24 Feb 2004 (UTC)

In mi opinion there are still obscure points in definition, for example should we always in the article suppose that.

$\sigma _{0}(n)=d(n)$

$\sigma _{1}(n)= \sigma (n)$

for example when relating Riemann Hypothesis to divisor function , they write $\sigma (n)$ what are they referring to ? ..---- 85.85.100.144 (talk) 21:09, 16 November 2007 (UTC)

Minor error

After the sentence, "The consequence of this is that, if we write..." the formula should show the uppercase pi not the summation symbol.

you're right, i'll correct it immediatly Xmlizer 08:05, 6 Jul 2004 (UTC)

equation

On my screen, the first equation looks like what I expect from < math >. The second equation also uses < math >, but it looks like the non-math text. Why is that (I don't see any reason). Bubba73 4 July 2005 00:38 (UTC)

Someone fixed it. 167.193.40.62 5 July 2005 17:15 (UTC)
On my screen, the second summation is different. It is a different font, larger and bold. The summation symbol doesn't have the serifs (sans serif font.) Bubba73 5 July 2005 19:05 (UTC) PS That is, it is like the regular text instead of the LaTeX. Bubba73 5 July 2005 22:27 (UTC)
Look at your wikipedia preferences setting; it changes how math renders for your browser. linas 5 July 2005 22:39 (UTC)
Thanks, that did it! (I didn't know about that.) It was set to "HTML if very simple...". I changed it to always PNG and that fixed it. Thanks again. Bubba73 July 5, 2005 23:12 (UTC)

A new article?

In my opinion, the sigma function should be a new article whose name could be sum-of-divisor function with redirects linking it from sigma function and sigma functions. What do you think ? Thanks. MathNT July 20, 2005.

Why? Why "a new article"? Something wrong with the current article? My mathbooks call it "the divisor function" not the "sum-of-divisors function" and not the "sigma function" either; so this article seems correctly titled to me. linas 14:02, 20 July 2005 (UTC)
The CRC Concise Encyclopedia of Mathmatics uses "Divisor function". I just added an external link to it (Mathworld). So I'm iclined to leave it as it is. It would be good to have redirects from those other names to this article, though. Bubba73 15:42, July 20, 2005 (UTC)
Done. I further improved the introduction lines and the last formula. In the old version the first line was inconsistent with the definition section and the last formula for the asymptotics, as appeared, was not correct. Feel free to send me some comments, if any.MathNT July 26, 2005.
Oh my god, this article is simply a shit, mixing a lot of functions that deserve separate articles. In mathematic-related articles, you should use ideas what are mathematically definiable. "A function that somehow relates with the divisors of the elements of domain" is not mathematics, this is amateurist babbling only, whether poor-quality encyclopaedieas use it or not. :-( Gubbubu 08:52, 22 July 2011 (UTC)

Mathworld

A recent remark cited mathworld ... please be careful; we don't want to end up with copyright violations by copying or even modeling after mathworld ... linas 20:30, 26 July 2005 (UTC)

Mathematical facts and equations are not copyrightable, so that wouldn't be a violation of copyright. Bubba73 20:52, July 26, 2005 (UTC)
PS - the reason I made a reference to MathWorld and the Encyclopedia of Mathematics is because I thought having the e^gamma in the constant term on the right (instead of on the left) was more standard in the literature. I looked at several statements of the limit on the web and they all did the same thing. It is in Hardy & Wright that way too. Bubba73 04:40, July 27, 2005 (UTC)

Inequalities

Hi,

I was just wondering, but I'm not sure enough to edit myself :

I thought that voor every e>0 sigma(n ) /n^(1+e) has limit 0, so in other words, every power of n greater than n^1 beats sigma

also, I read that the last equality cannot be improved, so that lim sup (sigma( n) *phi( n) /n/n ) = 1 and lim inf (sigma( n) *phi( n) /n/n ) = 6/Pi^2

Can this be included?

The first thing you say may be true for sufficiently large n. Do you have a reference? The best I know is that it can be as large as the constant times n to the 3/2 power. I can put it in and make the other change too, but I need to be sure about the correct statement. Bubba73 17:37, September 1, 2005 (UTC)

Technicality

Eequor - if you are now happy with divisor function, how would you feel about removing the "too technical" tag ? Or do you think there are further problems with the article ? Gandalf61 14:48, September 10, 2005 (UTC)

I'm not sure. I think the article could still use some improvement, especially near the end, and the amount of text seems a bit low in relation to the mathematical formulas. The material in Series relations and Inequalities seems closer to a list of recipes than truly encyclopedic information, and would be more helpful if expanded. What do you think of the Prerequisites box? ᓛᖁ 02:05, 11 September 2005 (UTC)
I think that one paragraph (or so) at the beginning can be not too technical, but after that, I don't think there is much that can be done to make it less technical. Bubba73 (talk) 02:39, September 11, 2005 (UTC)

The lead section mentions some things that are discussed very briefly in the article. ᓛᖁ 02:59, 11 September 2005 (UTC)

The Series relations section does give the basic identity linking the sigma functions to a product of the Riemann zeta and a translate; and a link to the Eisenstein series page where the sigma function is explicit enough. Charles Matthews 06:40, 11 September 2005 (UTC)

Removal by WAREL

The following text was removed by WAREL due to lack of reference:

I really am interested in this result. It's just that I'm not sure if I can trust it. Anyone who knows about this, please let us know.WAREL 23:59, 10 March 2006 (UTC)

Inequalities

For the number of divisors function,

$d(n) < n^{\frac {2} {3}}$, for n > 12.

Another bound on the number of divisors is

$\log d(n) < 1.066 \frac {\log n} {\log \log n}$, for n ≥ 3.

For the sum of divisors function,

$\sigma(n) < \frac {6n^\frac {3} {2}} {\pi^2}$, for n > 12.

A pair of inequalities combining the divisor function and the φ function are:

$\frac {6 n^2} {\pi^2} < \varphi(n) \sigma(n) < n^2$, for n > 1.

Dmharvey 13:04, 10 March 2006 (UTC)

The inequality $6/\pi^2<\varphi(n)\sigma(n)/n^2<1$ is a standard easy fact. It can be found in any number theory textbook, such as Hardy and Wright. -- EJ 03:03, 11 March 2006 (UTC)

I do have Hardy and Wright "The Theory of Numbers".Which page is that on? WAREL 02:47, 13 March 2006 (UTC)

Theorem 329 in section 18.3 on page 265. For reasons unknown, they leave the constant $\prod_p(1-p^{-2})=1/\zeta(2)=6/\pi^2$ unevaluated, and call it A. I could only find the 2nd edition of the book in our library, if you have a newer edition the page number may differ. You can also find the inequality as Exersize 9(a) of chapter 3 on page 71 in Apostol's "Introduction to Analytic Number Theory". -- EJ 18:37, 13 March 2006 (UTC)

I found it.It was written in Hardy's book. However, for some reason, it doesn't say " for n > 1 ".Why is that? Were they just lazy? Or does it have any deeper reasons?WAREL 20:15, 14 March 2006 (UTC)

Well, for n=1 we have $\varphi(n)\sigma(n)=1=n^2$, thus this case has to be excluded if the result is formulated with strict inequality. I can't imagine any deeper reason behind it. -- EJ 22:12, 14 March 2006 (UTC)

Bounds

I cannot edit the following into article as I personally derived it but it approximates sigma to the half-harmonic for highly composite numbers. I have the supporting calc/data

$ln(sigma)~=1.0604*Q + B$,


where ln=natl log and Q=H(N/2) ie the harmonic # of N/2 N=HCN (highly composite)thru 963761198400. The form structure relates to a companion proof requiring B=ln(2)

[[1]] comparative plot


in addition, the number of divisors, tau is related as follows:

     $ln(tau)~=0.9762*[Q **(0.6639)]$
w/ $R^2=0.9991$


reconciling the two above yields the approx sigma-tau relation (thru 146th HCN)as:

     $ln(sigma)~=1.311*[ln(tau)**(1.4294)]$
w/ $R^2=0.996$


--Billymac00 19:33, 15 May 2006 (UTC)

Picture mistake

The picture about "sigma function" doesn't show the real sigma function. Sigma(n) is greater than n+1 for all n>1 values, so it can't be 0 anywhere. Gubb     2006. September 12 19:45 (CEST) 19:45, 12 September 2006 (UTC)

Do you mean ? That picture is supposed to be for $\sigma_0(n)$. However, it doesn't look correct either since it hits 1 many times and that function is >1 for n>1. JoshuaZ 20:28, 12 September 2006 (UTC)
Ooops sorry. Thats odd, as I remember both creating and fixing an off-by-one bug the day I created that image, and, by golly, well, must have uploaded the image at the wrong time. Uploaded a new file to wikimedia commons but am waiting for it to percolate to here. linas 03:15, 13 September 2006 (UTC)
I believe he actually meant . I suspect the problem with both pictures is the same, namely that the sum was computed only over proper divisors. -- EJ 21:00, 12 September 2006 (UTC)
Ok, making rough sketches for the proper sum graphs up to 20 or so they both seem to agree. Maybe we should ask whatever kind user who made these to remake them? JoshuaZ 21:07, 12 September 2006 (UTC)
Will fix shortly. linas 03:17, 13 September 2006 (UTC)
Thanks. JoshuaZ 03:23, 13 September 2006 (UTC)
Image move request at [2] This is soo embarassing, I *never* make mistakes. :-( linas 03:52, 13 September 2006 (UTC)
Thanks. -- EJ 15:58, 13 September 2006 (UTC)

What is ''the'' divisor function?

The header defines "the divisor function" as σ0(), and then mentions the "related" divisor summatory function, but does not mention σ1(). The articles on perfect number, deficient number, abundant number, etc use "the divisor function" as σ1(). Sumbuddy who's a better mathemetician than me should make this consistent. — Randall Bart 08:26, 11 February 2007 (UTC)

The article does mention $\sigma_1$: it is in the first few sentences under "definition". I changed some mentions of "the divisor function" to "the sum-of-divisors function" in a few articles for clarity, though $\sigma$ is a divisor function, so it is proper to say "the divisor function $\sigma$". Hopefully this change will make things more clear. Doctormatt 21:14, 26 April 2007 (UTC)

Nielsen

For positive numbers n and d, let N be an odd positive number such that :$\sigma_{1}(N)/ N = n/d$ and ω(N)= k. Then, $N <(d+1)^{4^{k}}$ (Nielsen, 2003). —Preceding unsigned comment added by 218.133.184.93 (talk) 22:10, 15 September 2007 (UTC)

Guy robin proof

Could someone please cite the article stating that Guy Robin proved that this is equivalent to RH :$\sigma(n) for n > 5,040
20:25, 10 November 2008 (UTC) —Preceding unsigned comment added by Stdazi (talkcontribs)

A 1984 paper by Robin is already in Divisor function#References which says "Original publication of Robin's theorem". Is that not what you want? PrimeHunter (talk) 23:04, 10 November 2008 (UTC)

This is valuable information; unfortunately, I had to remove it due to the only source being the German-language Wikipedia, where it isn't sourced at all. Because of the propensity of mathematics to lend itself to original research, I felt it best not to leave a {{cite}} note. If anyone knows if this has been published...

Series expansion

The divisor function can be written as a finite trigonometric series

$\sigma_x(n)=\sum_{\mu=1}^{n} \mu^{x-1}\sum_{\nu=1}^{\mu}\cos\frac{2\pi\nu n}{\mu}$

without an explicit reference to the divisors of $n$.

Twin Bird (talk) 03:20, 10 July 2009 (UTC)

This follows at once from the fact that
$\sum_{\nu=1}^\mu\cos\frac{2\pi\nu n}{\mu}=\left\{\begin{array}{ll} \mu & \mu|n \\ 0 &\mu\nmid n\end{array}\right.$
When μ|n, the summand is always 1, so the sum is clearly μ. Otherwise, the sum is over complete sets of roots of unity (using that cos x = (exp(ix)+exp(-ix))/2), so the sum vanishes because of the standard fact that
$\sum_{m=1}^{r}\zeta^m=0\,$
where ζ is an rth root of unity. So, a reference for the latter fact would do I would think. A book on analytic number theory should have this, or even the original statement. What I've found quickly is Ireland–Rosen "A classical introduction to modern number theory" Proposition 8.2.1 on Gauss sums. Hope this helps. RobHar (talk) 05:16, 10 July 2009 (UTC)
In other words, the formula is just a cheap trick replacing the characteristic function of the divisibility relation by the sum over cosines. The main problem is thus not finding a reference, but establishing notability. I for one seriously doubt that it is useful or interesting. — Emil J. 09:36, 10 July 2009 (UTC)

Carella's result

I don't see why people think Carella's result is a col orally of Gronwall's theorem.Orera (talk) 03:32, 15 January 2011 (UTC)

Why don't you? See Limit superior and limit inferior#Interpretation. See also ja:ノート:約数関数. --白駒 (talk) 06:36, 15 January 2011 (UTC)
I just don't see it.　Gronwall's theorem　means $e^\gamma$ is the smallest real number b such that, for any positive real number $\varepsilon$, there exists a natural number $N$ such that $\sigma(n) / (n \log \log n ) for all $n>N$. How does this imply Carella's result?Orera (talk) 03:19, 16 January 2011 (UTC)
You claim Carella's result is:
• $\ \sigma(N) \le (1+\epsilon)e^\gamma N \log \log N$ for all sufficiently large integers N.
The lim sup result, forward, (substituting $e^\gamma$ for b), is:
• $\sigma(n) / (n \log \log n ) < e^\gamma+\varepsilon$ for all sufficiently large integers n.
Looks the same to me, with $\epsilon \approx e^{-\gamma} \varepsilon$. — Arthur Rubin (talk) 06:53, 16 January 2011 (UTC)

Unconditional upper bound

The article gives

$\ \sigma(n) < e^\gamma n \log \log n + \frac{0.6482\ n}{\log \log n}$

but I've also seen it cited as

$\ \sigma(n) < e^\gamma n \log \log n + \frac{0.6483\ n}{\log \log n}$

(different only in the constants). Which is correct?

I thought it better to bring this up here than to put a {{fact}} tag on the main page.

CRGreathouse (t | c) 19:41, 15 February 2011 (UTC)

Only the latter is correct.

For the constant in the numerator of the last term, the precise value is

$\ (7/3 - e^\gamma \log \log 12) \log \log 12 = 0.64821\dotsc.$

Robin writes the inequality as

$\ \frac{\sigma(n)}{n} \le e^\gamma \log \log n + \frac{0.6482\dots}{\log \log n}$

and says that equality holds for n = 12.

Thus, in the article, 0.6483 is correct and 0.6482 is wrong (e.g., for n = 12). Jsondow (talk) 15:39, 3 December 2011 (UTC)

Negative divisors

According to the article Divisor, divisors can be negative. When the 'proper divisors' are mentioned in this article, it is not added that only the positive divisors are counted, but when following the mentioned link , it's also not mentioned there that only the positive divisors are counted. So what is the consensus about this? Samaritaan (talk) 19:12, 21 October 2011 (UTC)

Definition

For a non-mathematician, the definition is a problem. What does d represent? Can that be made more explicit as has been done for the other elements of the definition? Thanks Ben (talk) 13:37, 6 December 2011 (UTC)

The text and the formula are independent of each other. You can read the first sentence of the definition on its own, and then you don't need to think about the letter d at all. In the formula, d is a dummy variable; the notation ${d|n}$ underneath the sigma sign tells you that d ranges over all the divisors of n. Jowa fan (talk) 00:25, 17 September 2012 (UTC)