NT2009gausscircle - GAUSSS CIRCLE PROBLEM 1. Introduction...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: GAUSSS CIRCLE PROBLEM 1. Introduction We wish to study a very classical problem: how many lattice points lie on or inside the circle x 2 + y 2 = r 2 ? Equivalently, for how many pairs ( x,y ) Z 2 do we have x 2 + y 2 r 2 ? Let L ( r ) denote the number of such pairs. Upon gathering a bit of data, it becomes apparent that L ( r ) grows quadratically with r , which leads to consideration of L ( r ) r 2 . Now: L (10) / 10 2 = 3 . 17 . L (100) / 100 2 = 3 . 1417 . L (1000) / 1000 2 = 3 . 141549 . L (10 4 ) / 10 8 = 3 . 14159053 . The pattern is pretty clear! Theorem 1. As r , we have L ( r ) r 2 . Explicitly, lim r L ( r ) r 2 = 1 . Once stated, this result is quite plausible geometrically: suppose that you have to tile an enormous circular bathroom with square tiles of side length 1 cm. The total number of tiles required is going to be very close to the area of the floor in square centimeters. Indeed, starting somewhere in the middle you can do the vast major- ity of the job without even worrying about the shape of the floor. Only when you come within 1 cm of the boundary do you have to worry about pieces of tiles and so forth. But the number of tiles required to cover the boundary is something like a constant times the perimeter of the region in centimeters so something like Cr whereas the number of tiles in the interior is close to r 2 . Thus the contribution to the boundary is neglible: precisely, when divided by r 2 , b it approaches 0 as r . I myself find this heuristic convincing but not quite rigorous. More precisely, I believe it for a circular region and become more concerned as the boundary of the region becomes more irregularly shaped, but the heuristic doesnt single out exactly what nice properties of the circle are being used. Moreover the error bound is fuzzy: it would be useful to know an explicit value of C . To be more quantitative about it, we define the error E ( r ) = | L ( r )- r 2 | , so that Theorem 1 is equivalent to the statement lim r E ( r ) r 2 = 0 . 1 2 GAUSSS CIRCLE PROBLEM The above heuristic suggests that E ( r ) should be bounded above by a linear func- tion of r . The following elementary result was proved by Gauss in 1837. Theorem 2. For all r 7 , E ( r ) 10 r . Proof. Let P = ( x,y ) Z 2 be such that x 2 + y 2 r 2 . To P we associate the square S ( P ) = [ x,x + 1] [ y,y + 1], i.e., the unit square in the plane which has P as its lower left corner. Note that the diameter of S ( P ) i.e., the greatest distance between any two points of S ( P ) is 2. So, while P lies within the circle of radius r , S ( P ) may not, but it certainly lies within the circle of radius r + 2. It follows that the total area of all the squares S ( P ) which is nothing else than the number L ( r ) of lattice points is at most the area of the circle of radius r + 2, i.e., L ( r ) ( r + 2) 2 = r 2 + 2 2 r...
View Full Document

This note was uploaded on 10/26/2011 for the course MATH 4400 taught by Professor Staff during the Spring '11 term at University of Georgia Athens.

Page1 / 8

NT2009gausscircle - GAUSSS CIRCLE PROBLEM 1. Introduction...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online