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Let $f(r)$ be the number of integral points inside circle of radius $r$ and center at origin,then $\lim_{r\to \infty}\frac{f(r)}{\pi r^2}$

I know the formula for number of lattice points inside the boundary of a circle of radius $r$ with center at the origin is given by $f(r)=1+4\lfloor r\rfloor+4\sum_{i=1}^{\lfloor r\rfloor}\lfloor \sqrt{r^2-i^2}\rfloor$

But i am not able to find $\lim_{r\to \infty}\frac{f(r)}{\pi r^2}$.

Brahmagupta
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3 Answers3

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Let $r$ be large. If $P$ is any lattice point inside the circle, colour blue the $1\times 1$ square which has $P$ as its centre and whose sides are parallel to the coordinate axes. Then the area $A$ coloured blue is the number of lattice points.

Note that every point in the circle with centre the origin and radius $r-10$ is coloured blue, and every point that is coloured blue is in a circle of radius $r+10$ with centre the origin. (We can do better than $10$.) It follows that $$\pi(r-10)^2\lt A\lt \pi(r+10)^2.$$ Thus $$\frac{\pi(r-10)^2}{r^2}\lt \frac{A}{r^2}\lt \frac{\pi(r+10)^2}{r^2}.$$ Now let $r\to\infty$.

André Nicolas
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  • Very nice. $,$ – Brian M. Scott Jan 24 '16 at 07:49
  • Very elegant proof. +1 – RFZ Jan 24 '16 at 12:27
  • What does it mean by $1\times 1$ square with center at $P$.I think if $P$ is a lattice point and if we construct a square around $P$ with center at $P$,it will be $2\text{units}\times 2\text{units}$ square.And also i do not understand why you have taken radius as $r-10$ and $r+10$?@AndreNicolas – Brahmagupta Jan 24 '16 at 13:58
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    The post says the square is $1\times 1$. So for example if $P=(2,6)$ then the bottom left corner of the square is $(1.5,5.5)$, the bottom right corner is $(2.5,5.5)$, and so on. I did it that way because of symmetry. But for the proof, we could instead use the $1\times 1$ square with bottom left corner $P$. (More) – André Nicolas Jan 24 '16 at 14:36
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    (More) I chose circles of radius $r-10$ and $r+10$ because the smaller one is for sure all blue, and all the blue stuff is contained in the larger one. This gives us upper and lower bounds for $A$. I think we can use $r-\sqrt{2}$ and $r+\sqrt{2}$ instead. It makes no difference to the argument, since we are only finding limiting behaviour. – André Nicolas Jan 24 '16 at 14:42
  • I got it,thanks! – Brahmagupta Jan 24 '16 at 14:53
  • You are welcome. – André Nicolas Jan 24 '16 at 14:54
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Let $g(r)$ be the maximum number of squares of side 1 having integral vertices contained inside the circle of radius $r$.

Let $G(r)$ be the minimum number of squares of side 1 having integral vertices that contain the circle of radius $r$.

Consider a bijective mapping of every such square onto the lower left vertex. Then simply counting all of them will give you the inequality,

$$g(r)<f(r)<G(r)$$

When we use second order integrals to find out the area of a figure we essentially divide it into infinitesimally small squares.This is exactly what we have done here. Hence, $\lim_{r\rightarrow \infty} \dfrac{g(r)}{\pi r^2}=\lim_{r\rightarrow \infty} \dfrac{G(r)}{\pi r^2}=1$

By sandwich theorem $\lim_{r\rightarrow \infty} \dfrac{f(r)}{\pi r^2}=1$

Shaswata
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Using Euler-Mclaurin summation formula we can get that: $$f(r)=\pi r^2+O(r).$$ Hence $$\lim \limits_{r\to \infty}\dfrac{f(r)}{\pi r^2}=1.$$

RFZ
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