0

Earlier this evening (morning), I posted a question about showing that a finite number of dyadic squares can fill up an arbitrary proportion of the unit disk. I'm sure there are better ways to prove this, but the way I came up with is to look at each square in the first quadrant — or, more specifically, each square's upper-right-hand corner, which has coordinates of the form $\left( \frac{p}{2^k}, \frac{q}{2^k}\right)$. Since this is the unit disk, I know that

$$\sqrt{\left(\frac{p}{2^k} \right)^2 + \left( \frac{q}{2^k} \right)^2 }\leq 1$$

or, equivalently, that $p^2 + q^2 \leq 4^k$.

What I want to do is count the number of solutions there are in $\mathbb{Z}^+\times \mathbb{Z}^+$, but I don't have any good idea on how to go about doing that. I have a lower bound, $\frac{\left(2^k - 1 \right)\left(2^k\right)}{2}$, since $(1,1), (1,2), (2,1), \ldots (1, 2^k - 1), (2, 2^k - 2), \ldots (2^k-1, 1)$ are all solutions. But this isn't nearly strong enough for what I need. Is there a closed form for the size of the solution set for any given $k$? I'm thinking that it would be useful if I had something I could compare to $4^k$ via a limit, though if there are other options, I'd certainly be interested in them.

Community
  • 1
dmk
  • 2,416
  • If you only consider the first quadrant of $\mathbb Z^2$ you the solution is basically $\left|{(p,q) \in \mathbb Z^{+2} : p^2+q^2 \leq 4^k}\right| = \sum\limits_{0\leq p \leq 4^k} \sum\limits_{0\leq q \leq \sqrt{4^k-p^2}} 1$, isn't it? – flawr Aug 24 '14 at 08:50
  • Here the $4^k$ plays no role. You want to count the number of integer point inside a very large disk, $p^2+q^2 \leq n^2$. When $n$ tends to infinity, it is well-known that this number is asymptotically equivalent to the area $\pi{n^2}$. Do you know about Riemann sums and integration ? This is a two-dimensional version of it. – Ewan Delanoy Aug 24 '14 at 09:04
  • @EwanDelanoy: Thanks for your response. I know what Riemann sums are; I've never done them, like, rigorously, though. (The problem is from Chap. 1 of Pugh's Real Mathematical Analysis.) I'm not sure it'll fly with my professor if I tie up my proof by referring to some "well-known" result :). Especially since, to be frank, I'm a little skeptical: I wouldn't know how to avoid writing $\lim_{n\rightarrow\infty} S(n) = n\cdot\infty^2$. Suggestions? – dmk Aug 24 '14 at 13:30
  • @flawr: Thank you. I'll edit my post: What you write is certainly true, and it "systematic," to use my original word; what I meant, though -- at four in the morning :) -- was: Is there a closed form for the answer? – dmk Aug 24 '14 at 13:35
  • This is known as the Gauss circle problem – Ross Millikan Aug 24 '14 at 14:32
  • @dmk: Sure, I just wanted to make sure that I've understeood your question correctly! I think you can 'simplify' that expression to $\sum\limits_{0\leq p \leq 4^k} \left\lceil \sqrt{4^k-p^2} \right\rceil$ but I do not think that this leads to anywhere. – flawr Aug 24 '14 at 14:32

0 Answers0