This question is related to this and this ones. The Gauss Circle problem asks for the number $N(r)$ of integer points within a sphere of radius $r$ centered at the origin. It is well known that $N(r) = \pi r^2+ E(r)$, where $|E(r)| \leq 2 \sqrt{2} \pi r$. On the other hand, more sophisticated techniques show that (Van der Corput) $E(r) = O(r^{2/3})$ and (Huxley) $E(r) = O(r^\theta)$, where $\theta$ is something like 0.63.

Edit: Try to fix the question so that it makes sense.

My question is how "asymptotic" are the more recent results on the problem. Is it possible to estimate the hidden constants behind these results? For example, we know that

$$\lim_{r \to \infty} \frac{\pi r^2}{\pi r^2+E(r)} = 1.$$

If I were to find $r_0$ such that the for $r \geq r_0$ the "relative error" of the limit is $$\left| \frac{\pi r^2 }{\pi r^2 + E(r)} - 1 \right| \leq 10^{-2} $$ what would be the best estimate to use? (for the problem above, $|E(r)| \leq 2 \sqrt{2} \pi r$ gives us that $r$ should be greater than $280$ while numerical computations suggest that $r > 13$ suffices).

[*Huxley paper, given by an answer to this post] Huxley, M. N.(4-WALC) Area, lattice points, and exponential sums. London Mathematical Society Monographs. New Series, 13. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1996. xii+494 pp. ISBN: 0-19-853466-3 11L07 (11J54 11P21)

  • $\begingroup$ Note that $E(r)$ is often negative. Do you want the least $R$ so that $\frac{\pi r^2}{\pi r^2+|E(r)|} \gt 0.9$ for all (or all integer) $r \ge R?$ You certainly aren't looking for the least $r$ since $N(2)=13$ while $4\pi \gt 12.$ $\endgroup$ Aug 19, 2014 at 20:43
  • $\begingroup$ You are right. I need to re-think the statement in a way that it makes sense. I was just thinking about the best "approximation" for reasonably small R. $\endgroup$
    – Campello
    Aug 19, 2014 at 21:11
  • $\begingroup$ For what it is worth, the last time your inequality fails is for $r=\sqrt{10}$ when $\pi r^2 \lt 32$ is less than $9/10$ of $N(r)=37$. $\endgroup$ Aug 19, 2014 at 21:17
  • $\begingroup$ Done! I changed the question so it works with "relative error" instead. $\endgroup$
    – Campello
    Aug 19, 2014 at 21:25
  • $\begingroup$ My vague understanding of these methods is that they are completely effective. For example, if you understand the $2/3$ exponent (which, I believe, is due to Sierpinski), then you can make the error term completely explicit. $\endgroup$
    – GH from MO
    Aug 19, 2014 at 23:48

2 Answers 2


I wouldn't expect any of these bounds to give really superb estimates of the last time your ratio achieves a certain record because $E(r)$ and your ratio undergo rapid fluctuations. Also, the exciting places are unlikely to happen at integers. The local minima of $E(r)$ occur at the places where $r=\sqrt{a^2+b^2}.$ Then $E(r)$ increases (at a rate of $2\pi r$) until the next local minimum.

I had a program calculate out to $r=1000.$ There $216342$ values $r=\sqrt{a^2+b^2}$ in that range of which $87483$ have $N(r) \lt \pi r^2$ and the other $128859$ have $N(r) \gt \pi r^2.$ (I'm not sure why the imbalance. New question!)

This turns out to yield ten guaranteed records with $\frac{\pi r^2 }{N(r)} \gt 1.003.$ They are:

[49, 1.0331], [144, 1.0258], [288, 1.0177], [576, 1.0092], [722, 1.0068], [1152, 1.0061], [1444, 1.0052], [1844, 1.0042], [2592, 1.0037], [2593, 1.0031]

The third entry [288, 1.0177] says that at $r=\sqrt{288}$ the ratio is $1.0177$ and it is never that large again. Most have $r$ an integer or an integer times $\sqrt{2}.$

There are 32 guaranteed records with $\frac{\pi r^2 }{N(r)} \lt 0.997.$ Namely:

[0, 0.], [1, .62832], [2, .69813], [5, .74800], [10, .84908], [13, .90757], [20, .91061], [26, .91777], [29, .93924], [41, .94018], [53, .94070], [65, .95870], [85, .96403], [90, .96499], [130, .97009], [149, .97318], [170, .97995], [185, .98009], [205, .98025], [234, .98149], [340, .98446], [377, .98616], [425, .98683], [586, .98924], [986, .99124], [1325, .99370], [1700, .99473], [1781, .99541], [1885, .99544], [2260, .99593], [3146, .99662], [3400, .99668]

Notice that the last record mentioned is at $r=\sqrt{3400}$ while the calculation went out to $r=\sqrt{1000000}.$ I'm sure that there are many more records in that data.

What I mean by guaranteed is that With the $|\pi r^2 -N(r)|=|E(r)| \lt \sqrt{8}\pi r$ result we can be sure that for all $r \gt 1000$ we have $$0.997179550 \lt \frac{\pi r^2 }{N(r)} \lt 1.0028364498.$$

  • $\begingroup$ I think it makes sense to allow all $r$'s such that $r^2$ is a sum of two squares (incidentally, in my computations I was assuming that $r$ is integer). Anyway, how did you prove that these are the record (and it never goes down again, for some "exciting" large $r$)? $\endgroup$
    – Campello
    Aug 20, 2014 at 10:55
  • $\begingroup$ I updated to give better details. The claims I made before about when the ratio is outside $(0.99,1.01)$ involved calculations out far enough to be morally sure I had found all the records (and it turns out I was right.) But the results claimed now are guaranteed. $\endgroup$ Aug 21, 2014 at 8:37

It is well known that giving good constants for estimates obtained via exponential sums is difficult and a lot of work. I remember Kraetzel giving a talk on effective versions of the exponent pair $(1/6, 2/3)$, which gives the exponent $2/3$ for the circle problem. The constants involved were quite sensible, so they should imply the bound you look for for all $n$ larger than a few millions.


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