Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $C(r)$ be the origin-centered circle of radius $r$, and let $\beta(r)$ be the exterior buffer around $C(r)$: the distance from $C(r)$ to the closest lattice point exterior to $C(r)$:
     BetaBuffer
For example, $\beta(2) = \sqrt{5}-2 \approx 0.24$ because there are no lattice points strictly between $C(2)$ and $C(\sqrt{5})$, and this is the largest buffer around $C(2)$.

I am interested in the behavior of $\beta(r)$ for large $r \in \mathbb{R}$, as I believe understanding that behavior will answer my question concerning ratchet spirals, Lattice radial-step (ratchet) spirals.

I'll pose a specific question before formulating the general question.

Q1. Is there an $R$ such that, for all $r > R$, $\beta(r) < \frac{1}{2}$ ?

If so, then, for example, the spiral $S(3,\frac{1}{2})$ depicted in that question is unbounded.

Q2. Is there an $R(\epsilon)$ such that, for all $r > R(\epsilon)$, $\beta(r) < \epsilon$, where $0 < \epsilon < 1$ ?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

Bambah and Chowla ("On numbers which can be expressed as a sum of two squares") proved that there is a constant $C$ such that for any $x > 0$, there is an integer between $x$ and $x+ Cx^{1/4}$ which is the sum of two squares. From this it easily follows that the answer to the more general Q2 is in the affirmative. (in fact $x + Cx^{1/2 - \epsilon}$ would do for any $\epsilon$).

share|improve this answer
    
I don't understand. Doesn't this give only a bound $\beta(r) \leq Cr^{\frac{1}{4}}$? –  Vít Tuček Oct 30 '13 at 15:53
    
You apply it with $x = r^2$. It gives a lattice point $(a,b)$ with $n = a^2 + b^2$, where $n$ is between $r$ and $\sqrt{r^2 + C\sqrt{r}}$. Therefore the distance to the circle is $\sqrt{n} - r$, but $r = \sqrt{r^2} \leq \sqrt{n} \leq \sqrt{r^2 + Cr^{1/2}} = r(1 + r^{-3/2}/2 ) = r + r^{-1/2}/2$. –  Abhinav Kumar Oct 30 '13 at 16:13
    
I see. Much obliged. –  Vít Tuček Oct 30 '13 at 16:39
    
Great---Thanks, Abhinav! This resolves my other question as well. –  Joseph O'Rourke Oct 30 '13 at 17:20

This is really just a comment, not a complete answer.

Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $$\sqrt{1+r^2}-r$$ which has asymptotics $\frac{1}{2r}$ and hence for integral diameters the answers are affirmative.

If $r$ is nonintegral, then the point $[0,\lceil r \rceil]$ is closer to $C(r)$ than $[0,r]$ and its distance to $C(r)$ $$ \lceil r \rceil - r $$ gets arbitrarily close to 1 (as Abhinav Kumar pointed out in the comments). So for nonintegral diameters one really has to pick a good lattice point inside the first quadrant.

share|improve this answer
    
But for nonintegral $r$, the point $(0,r)$ is not integral ... –  Abhinav Kumar Oct 30 '13 at 12:38
    
Sorry, I meant to take the ceiling of $r$. Should be fixed now. Thanks! –  Vít Tuček Oct 30 '13 at 13:48
    
That doesn't quite work - the distance from that point to $C(r)$ is $\lceil r \rceil - r$ which can be arbitrarily close to $1$ (and in particular is not $O(1/r)$). –  Abhinav Kumar Oct 30 '13 at 14:07
    
Right. There's the stupid mistake I've been talking about. :) I'll edit my "answer" accordingly. Thanks again. –  Vít Tuček Oct 30 '13 at 15:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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