Gauss' Circle Problem

**Swlabr** · October 1st 2009, 12:15 AM

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!

**tonio** · October 6th 2009, 03:36 AM

Originally Posted by Swlabr

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

**Swlabr** · October 7th 2009, 10:40 AM

Originally Posted by tonio

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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