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NT2009gausscircle

# NT2009gausscircle - GAUSS’S CIRCLE PROBLEM 1 Introduction...

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Unformatted text preview: GAUSS’S 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 ) ∼ πr 2 . Explicitly, lim r →∞ L ( r ) πr 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 Cπr – whereas the number of tiles in the interior is close to πr 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 doesn’t 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 )- πr 2 | , so that Theorem 1 is equivalent to the statement lim r →∞ E ( r ) r 2 = 0 . 1 2 GAUSS’S 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 ) ≤ π ( r + √ 2) 2 = πr 2 + 2 √ 2 πr...
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NT2009gausscircle - GAUSS’S CIRCLE PROBLEM 1 Introduction...

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