Limit of the sequence of regular n-gons.

Question

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?

Answer 1

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 $d_H(A_n,S^1)\to 0$.

Answer 2

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

$f_n:[0,2\pi]\rightarrow\mathbb{R}^2$

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

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

Answer 3

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 $n\to\infty$.

Answer 4

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.