Talk:Divisor summatory function

Merge

I suggest merging this article, and Dirichlet divisor problem, as these are on essentially the same problem. I did not discover that article until, of course, I'd started this. linas 05:20, 13 July 2006 (UTC)

Confusion over formulas

There is some confusion over formulas. The multitude of sources say $x\log x+x(2\gamma -1)$ in the asymptotic behaviour. However, actual numerics makes it pretty darned incredibly clear that the true asymptotic behaviour has a $x\log x-2x(1-\gamma )$ instead. At first, I thought this was a typo in the textbook, but now I am not sure, I haven't checked. Perhaps there's an error in my source code. All this needs double checking; for the moment I left it all hanging. linas 05:20, 13 July 2006 (UTC)

Anyone with a numerics package, please check this for me. My source code passes a variety of double-checks, I am just not seeing the error, and yet the numerical behaviour seems incontrovertible, at least for N less than a million (!) ... yes, a million is a small number, (I had to compute the exponential sum on the von Mangoldt function to two billion!! terms before the divergent behaviour became clear. But really, I am quite surprised. linas 05:39, 13 July 2006 (UTC)

I have had a quick look at the standard proof of the result ($x\log x+x(2\gamma -1)+error$) and cannot see anything wrong with it. Also checked the statement in a variety of sources, so I believe it is right despite the numerics. Since we are dealing with an asymptotic formula, I would not be too surprised to see numerical diffences at small numbers - how high can your calculations realistically go? As for the merger, I would have no problem with the proposed merger - if it is done carefully, I am sure it would improve things. Madmath789 06:34, 13 July 2006 (UTC)

Just done some calculations (10^2, 10^3, ... 10^13) and found that the $x\log x+x(2\gamma -1)+error$ version seems accurate, and (with these small values of x) the error term seems to be pretty close to a small mutlipe of $x^{1/4}$ - even though it has been proved that the true error must be greater than $O(x^{1/4}).$ Madmath789 11:53, 13 July 2006 (UTC)