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

Let $A_n$ be the regular $n$-gon inscribed in the unit circle.

It appears intuitively obvious that as $n$ grows, the resulting polygon approximates a circle ever closer.

Can it be shown that the limit as $n \rightarrow \infty $ of $A_n$ is a circle?

share|cite|improve this question
If your polygons have an area and circumference as usually understood, then the boundary and the interior of each $A_n$ would each have an uncountable number of points. – Henry Jan 10 '12 at 13:11
There are several regular $n$-gons centred at the origin. Ignoring orientation, you may want to constrain them somehow, such as the edge length, or the maximum or minimum distance of the boundary from the centre. – Henry Jan 10 '12 at 13:14
You could start with any of the equations given here. You should also find $\lim\limits_{x\to\infty}\frac{\lfloor k x\rfloor}{x}=k$ useful. – J. M. Jan 10 '12 at 13:32
up vote 2 down vote accepted

Suppose you take the $n$-gons $A_n$ to be inscribed in the unit circle $S^1$, so we're not talking about a sequence of $n$-gons that is growing without bound, or oscillating in size, or anything like that. There is a notion of distance between subsets of a metric space called the Hausdorff distance. In this case the Hausdorff distance is defined as $$d_H(X,Y)=\max \left\{\sup_{x\in X} \; \inf_{y\in Y} \; |x-y|, \sup_{y\in Y} \; \inf_{x\in X} \; |x-y| \right\}$$ where $X$ and $Y$ are two subsets of $\mathbb R^2$. (For a general metric space, replace $|x-y|$ by $d(x,y)$.) Now $A_n$ does indeed limit to $S^1$ in the sense that $d_H(A_n,S^1)\to 0$.

share|cite|improve this answer
Jim, I made some TeX changes in the post, and while at it, I also took the liberty to add a link to the Wikipedia article. Hope this is ok. – Srivatsan Jan 10 '12 at 19:49
@Srivatsan: much appreciated. – Grumpy Parsnip Jan 10 '12 at 21:00

Your question is vague, since you do not specify in which sense the sequence of n-gons is suspected to converge to the circle. You could for example view the n-gons as graphs of functions


and ask whether the limit of the sequence $(f_n)_n$ is a function $f:[0,2\pi]\rightarrow\mathbb{R}^2$ having the circle as its graph. Convergence can now be meant pointwise, that is:

$\forall t\in [0,2\pi]: \lim\limits_{n\rightarrow\infty}f_n(t)=f(t)$,

or uniform, that is

$\lim\limits_{n\rightarrow\infty}\sup (|f_n(t)-f(t)| : t\in [0,2\pi])=0$.

share|cite|improve this answer
How can the circle be the graph of any function? Graphs never have vertically aligned points. Certainly you are using "graph" in an unusual sense. (I can guess which, but you should clarify.) – Marc van Leeuwen Jan 10 '12 at 15:08
@Marc: Given the domain $[0,2\pi]$, it's pretty clear that he means a graph in polar coordinates. – mjqxxxx Jan 10 '12 at 15:21

Parametrize the unit circle $S^1$ as $t\in [0,2\pi] \mapsto (\cos t,\sin t) \in S^1$. Let $p$ be a point in $S^1$ and consider the corresponding $t$. Then for any $\varepsilon>0$, there is a rational $u/v$ such that $|t-u/v|<\varepsilon$. The point corresponding to $u/v$ belongs to the $v$-th regular polygon and is within $\varepsilon$ of $p$.

Also the distance between $A_n$ and the unit circle is given by the sagitta, which is the "co-apothem" $1-\cos(\pi/n)$, and this goes to $0$ as $n\to\infty$.

share|cite|improve this answer

Given a sequence of sets $(A_n)_{n\geq3}$ there is a natural $\lim\inf_{n\to\infty} A_n=:\underline{A}$ and a natural $\lim\sup_{n\to\infty}A_n=:\overline{A}$ of this sequence.

In the problem at hand the $A_n$ are closed regular $n$-gons inscribed in the unit circle, all sharing the point $P:=(1,0)$.

The set $\underline{A}$ consists of all points that are in all but finitely many of the $A_n$. It is easy to see that all points $z\in D:=\{(x,y)\ |\ x^2+y^2 < 1\}$ satisfy this condition and that in fact $\underline{A}=D\cup\{P\}$.

The set $\overline{A}$ consists of all points that are in infinitely many $A_n$. Obviously $\overline{A}\supset\underline{A}\ $, and $\overline{A}$ is contained in $\overline{D}=\{(x,y)\ |\ x^2+y^2 \leq 1\}$. In fact $\overline{A}\cap\partial D$ consists of all points on the unit circle whose argument is a rational multiple of $\pi$.

This is how much you can say on the pure set-theoretical level; an actual limit set $A_*$ does not exist.

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.