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Disjoint dyadic squares in unit disk

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Given ε > 0, show that there is a collection of disjoint dyadic squares in the unit disk that has a total area which exceeds π - ε.

    2. Relevant equations

    Define a dyadic interval as an interval of the form [a, b] such that a = p/2k and b = (p + 1)/2k, p and k are integers. A dyadic square is the product of two of these intervals.

    This problem is ch1 22b of Pugh's Real Mathematical Analysis.

    3. The attempt at a solution

    I figured this had something to do with aliasing on an circular arc, so I wrote [tex]A(n) = { 1 \over 4^n } \sum_{i=1}^{2^n} \lfloor \sqrt{4^n - i^2} \rfloor[/tex] as an expression where A is the area of a quadrant of the unit circle aliased down to a [itex]{ 1 \over 2^n }[/itex] granular dyadic grid. Then I noted that A → π/4 as n → ∞. What I am stuck on is the disjoint condition, as that makes requiring larger chunks of squares mandatory for the collection's total area to approach π I think, but it feels like to me that the supremum I want is less than π.
    Last edited: May 18, 2013
  2. jcsd
  3. May 19, 2013 #2


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    I don't understand your difficulty. The squares in A(n) are disjoint. What do you mean by "makes requiring larger chunks of squares mandatory"?
  4. May 19, 2013 #3
    I believe the squares include their boundaries, and by definition of the disjoint condition two adjacent squares would have a line/vertex (nonempty) intersection and so would not be disjoint. If I interpreted the question wrong and things are fine, then ... haha yeah, there isn't much difficulty after all. On the other hand, if the disjoint condition is that strict, then intuitively, I'm not even sure the question can be resolved.
  5. May 19, 2013 #4


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    I don't think anyone's going to worry about line and point overlaps, which are clearly sets of measure zero. But if it bothers you, use intervals like [a, b) or (a, b).
  6. May 20, 2013 #5
    Alright, thank you haruspex. I'm starting to think the stricter interpretation has an interesting upper bound in its own right, but I don't think this would not be the forum for that discussion. :)
  7. May 21, 2013 #6
    I asked the author for some clarification and he has chosen to grace me with a response! Here's the quote:

    In short, I'm not quite done yet.
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