Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a circle with center $(0, 0)$ and radius $r$. Let $n$ be the number of grid points inside or on the circle that at least one of its neighboring (up, down, left, right) grid points is outside the circle.

enter image description here

With my computer, I got some $r-n$ pairs: $$ \begin{array}{c|lcr} r&n\\ \hline 1&4\\ 2&8\\ 3&16\\ 4&20\\ 5&28\\ 10&56\\ 10^2&564\\ 10^3&5656\\ 10^4&56568\\ 10^5&565684\\ 10^6&5656852\\ 10^7&56568540\\ 10^8&565685424\\ 10^9&5656854248\\ 10^{10}&56568542492\\ \end{array} $$

And I found that $$\lim_{r\to\infty}\frac nr\approx5.6568542\approx4\sqrt2$$

My question is: how to prove the following equation? $$\lim_{r\to\infty}\frac nr=4\sqrt2$$

share|cite|improve this question
This is awesome. – Mike Miller Feb 21 '14 at 8:16
up vote 6 down vote accepted

Let's look at the 1st quadrant (and then multiply by 4). For each $x$ value from $1$ to $r\sqrt2/2$, there is exactly one $y$-value, and it's greater than $r\sqrt2/2$. For each $y$-value from $1$ to $\sqrt2/2$, there's exactly one $x$-value, and it's greater than $r\sqrt2/2$. So that gives you $r\sqrt2$ points in the first quadrant, and proves your observation.

share|cite|improve this answer

Here's a slightly different perspective. The $L^\infty$ arc length of the (ordinary) circle of radius $r$ is exactly $(4\sqrt2)r$. So when we define $n$ using a discrete grid, we get the approximation $n=(4\sqrt2)r+O(1)$, and $$\lim_{r\to\infty}\frac{(4\sqrt2)r+O(1)}{r}=4\sqrt2.$$

(This is a suggestive argument, rather than a full proof. I don't know the precise hypotheses we would need to work with more general curves.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.