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Math Help - Gauss' Circle Problem

  1. #1
    MHF Contributor Swlabr's Avatar
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    Gauss' Circle Problem

    I encountered this in a book yesterday (de la Harpe, if you're interested) when I was a tad bored and I decided to "understand" the proof. It stated,

    If we take R(t) = |\{ (a,b) \in \mathbb{Z}^2 : a^2+b^2 \leq t\}| then R(t) - \pi t = E(\sqrt{t}).

    The first part of the proof is simple: take the unit square with (a,b) in its south-west corner. Then in even a worst case we see that the square is contained in the circle with center (0,0) and radius \sqrt(2)+\sqrt(t) (just applying Pythagoras to the two squares that we find ourselves with and summing). Applying the area of a circle formula, we get R(t) \leq \pi (\sqrt(2) + \sqrt(t))^2 = \pi (2+2 \sqrt{2t} + t) \Rightarrow R(t) - \pi t \leq 2 \pi (1 + \sqrt{t}) for all t \geq 0.

    This is where I think we should stop. I mean, that is the statement in the Theorem, isn't it? However, the author then says "if the square corresponding to (a,b) touches the disc of radius \sqrt{t} - \sqrt{2}, then a^2+b^2 \leq t; hence

    R(t) \geq \pi (\sqrt{t} - \sqrt{2})^2

    for all t \geq 0. Thus

    |R(t) - \pi t| \leq 2 \pi (1+\sqrt{2t})

    for all t \geq 0. Ends."

    and I don't really see what we are doing continuing. I don't see what we are trying to prove, or how we get the absolute value sings.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    I encountered this in a book yesterday (de la Harpe, if you're interested) when I was a tad bored and I decided to "understand" the proof. It stated,

    If we take R(t) = |\{ (a,b) \in \mathbb{Z}^2 : a^2+b^2 \leq t\}| then R(t) - \pi t = E(\sqrt{t}).

    The first part of the proof is simple: take the unit square with (a,b) in its south-west corner. Then in even a worst case we see that the square is contained in the circle with center (0,0) and radius \sqrt(2)+\sqrt(t) (just applying Pythagoras to the two squares that we find ourselves with and summing). Applying the area of a circle formula, we get R(t) \leq \pi (\sqrt(2) + \sqrt(t))^2 = \pi (2+2 \sqrt{2t} + t) \Rightarrow R(t) - \pi t \leq 2 \pi (1 + \sqrt{t}) for all t \geq 0.

    This is where I think we should stop. I mean, that is the statement in the Theorem, isn't it? However, the author then says "if the square corresponding to (a,b) touches the disc of radius \sqrt{t} - \sqrt{2}, then a^2+b^2 \leq t; hence

    R(t) \geq \pi (\sqrt{t} - \sqrt{2})^2

    for all t \geq 0. Thus

    |R(t) - \pi t| \leq 2 \pi (1+\sqrt{2t})

    for all t \geq 0. Ends."

    and I don't really see what we are doing continuing. I don't see what we are trying to prove, or how we get the absolute value sings.

    Thanks in advance!

    Hi there:

    1) What book by de Harpe?

    2) What is E(Sqrt(t))?

    Tonio
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    Hi there:

    1) What book by de Harpe?

    2) What is E(Sqrt(t))?

    Tonio
    The book is "Topics in Geometric Group Theory", and the E(\sqrt{t})-term is error (I haven't quite worked out how to write the O for error properly in LaTeX).
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