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

    The first part of the proof is simple: take the unit square with $\displaystyle (a,b)$ 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 $\displaystyle \sqrt(2)+\sqrt(t)$ (just applying Pythagoras to the two squares that we find ourselves with and summing). Applying the area of a circle formula, we get $\displaystyle R(t) \leq \pi (\sqrt(2) + \sqrt(t))^2 = \pi (2+2 \sqrt{2t} + t) \Rightarrow R(t) - \pi t \leq 2 \pi (1 + \sqrt{t})$ for all $\displaystyle t \geq 0$.

    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 $\displaystyle (a,b)$ touches the disc of radius $\displaystyle \sqrt{t} - \sqrt{2}$, then $\displaystyle a^2+b^2 \leq t$; hence

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

    for all $\displaystyle t \geq 0$. Thus

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

    for all $\displaystyle t \geq 0$. 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 $\displaystyle R(t) = |\{ (a,b) \in \mathbb{Z}^2 : a^2+b^2 \leq t\}|$ then $\displaystyle R(t) - \pi t = E(\sqrt{t})$.

    The first part of the proof is simple: take the unit square with $\displaystyle (a,b)$ 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 $\displaystyle \sqrt(2)+\sqrt(t)$ (just applying Pythagoras to the two squares that we find ourselves with and summing). Applying the area of a circle formula, we get $\displaystyle R(t) \leq \pi (\sqrt(2) + \sqrt(t))^2 = \pi (2+2 \sqrt{2t} + t) \Rightarrow R(t) - \pi t \leq 2 \pi (1 + \sqrt{t})$ for all $\displaystyle t \geq 0$.

    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 $\displaystyle (a,b)$ touches the disc of radius $\displaystyle \sqrt{t} - \sqrt{2}$, then $\displaystyle a^2+b^2 \leq t$; hence

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

    for all $\displaystyle t \geq 0$. Thus

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

    for all $\displaystyle t \geq 0$. 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 $\displaystyle E(\sqrt{t})$-term is error (I haven't quite worked out how to write the $\displaystyle O$ for error properly in LaTeX).
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