Number of vertices of an n-gon as n tends to infinity?

[cmos]

The following is a problem that I’m sure is considered “basic” for mathematicians. I would therefore be gracious if somebody could, at the least, point me in the right direction to some reference. Since I’m not a mathematician, the simpler the better.

In short, my question is: what happens to the number of vertices of an n-sided polygon as n tends to infinity?

Let us consider a regular n-sided polygon. Obviously, the n-gon also has n vertices. The interior angle subtended at the n vertices increases as we increase n. As n tends to infinity, the interior angles each tend to 180°. In this same limit, the n-gon tends to a circle. On one hand, we now have an infinity of vertices. On the other hand, since the interior angle subtended at each vertex is infinitesimally 180°, we actually have no vertices. It thus appears that the number of vertices increases monotonically with n except at infinity where the number of vertices suddenly drops from infinity to zero.

Another way of seeing the above statement is that the circle (which is the limit of the n-gon as n tends to infinity), being a smooth curve, has no vertices.

Am I seeing the problem the right way? Any help would be much appreciated!

 

[DaveC426913]

On the other hand, since the interior angle subtended at each vertex is infinitesimally 180°, we actually have no vertices. It thus appears that the number of vertices increases monotonically with n except at infinity where the number of vertices suddenly drops from infinity to zero.

No. It has vertices equal to n approaching infinity.

Another way of seeing the above statement is that the circle (which is the limit of the n-gon as n tends to infinity), being a smooth curve, has no vertices.

Am I seeing the problem the right way? Any help would be much appreciated!

Well, that's the point of limits. You can get arbitrarily close to infinity but never reach it.

Thus, your circle has an arbitrarily large number vertices, that are arbitrarily shy of 180 degrees, but you never get infinite vertices or 180 degree angles.

 

[homeomorphic]

You could define a vertex as a point where it's not smooth, ie. not differentiable. From that point of view, the circle has no vertices. But there's nothing strange about approximating a smooth thing with a non-smooth thing.

Number of vertices goes to infinity, it's infinity in the limit, but the function value doesn't equal the limit.

Mathematically, if you think of each figure as a point in a space, what you are seeing is that the "number of vertices" function has a discontinuity at the point that represents the circle. And there's nothing odd about having a discontinuous function. Happens all the time.