# Circles vs an infinitely n-sided polygon.

1. Mar 15, 2014

### serp777

Suppose you have a square, and you simply start increasing the number of vertices and edges proportionally, all the way to infinity.

What, exactly, distinguishes this infinitely sided polygon from a circle?

Logically, an infinitesimal edge would be like a point on a circle, although I suppose this might be an incorrect definitional explanation.

Is there some proof that says an infinitely sided polygon is different from a circle or does it just come down to definitions?

2. Mar 15, 2014

### micromass

Staff Emeritus
You can have polygons with $n$ sides for $n$ arbitrary large. So $n$ can be $45$, or $1352$ or whatever integer you want. But I don't see how you can ever get a polygon with an infinite number of sides.

If you say "increase the number of sides" then that's clear. But "all the way to infinity" isn't so clear to me what that means.

3. Mar 15, 2014

### Staff: Mentor

Neither.

First, consider that there are two ways of relating a polygon to a circle:
- You can draw the largest polygon that fits inside the circle; all the vertices of the polygon lie on the circle
- Or you can draw the smallest polygon that fits outside the circle; the midpoint of every edge of the polygon lies on the circle

It should be clear that the first approach gives you a polygon whose perimeter and area are smaller than the perimeter and area of the circle, while the second approach gives you a polygon whose perimeter and area are larger than those of the circle.

Now, look at what happens when you let the number of edges of the two polygons approach infinity. The more edges you have, the smaller the difference between two polygons, so as the number of edges approaches infinity the perimeter (similar arguments work for the area) of the two polygons converges on a single value. But we know that that value is greater than or equal to the circumference of the circle (because it's the perimeter of the larger polygon) and also less than or equal to to the circumference of the circle (because it's the perimeter of the smaller polygon). There's only one way that it can be both less than or equal and greater than or equal, and that's if it's equal.

So, in the limit as the number of sides goes to infinity, the polygon IS the circle.

(A good exercise is to use this line of thinking to show that the area of a circle is $\pi{r}^2$ - I once watched a ten-year-old do this without realizing that he was doing anything special).

4. Mar 15, 2014

### serp777

all the way to infinity meaning taking the limit. All the way to infinity is done all the time like taking the limit of 1/x as x approaches infinity. That equals zero although you could argue like you have done by saying that X is always going to be some finite number, and so 1/x isnt actually zero, it's just non zero. So take the limit as N--> infinity. Or similarly summing an infinite series.

5. Mar 15, 2014

### micromass

Staff Emeritus
Sure, I get how limits are defined for numbers. But how do you define limits for subsets of the plane?

6. Mar 15, 2014

### serp777

"It should be clear that the first approach gives you a polygon whose perimeter and area are smaller than the perimeter and area of the circle, while the second approach gives you a polygon whose perimeter and area are larger than those of the circle."

Hmm, can you explain the "it should be clear that" portion? If the limit as the number of sides goes to infinity of a polygon is a circle like you say, then wouldn't it hold that the largest polygon that fits inside the circle is in fact just a circle, meaning it has the same perimeter and area, and wouldnt it hold that the smallest polygon that fits outside the circle is also just a circle which has the same perimeter and area? Thus wouldnt those two things actually be the same?

7. Mar 15, 2014

### micromass

Staff Emeritus
Right, it makes no sense since circles aren't polygons. So there is no "largest polygon" fitting in the circle.

8. Mar 15, 2014

### pwsnafu

You are going to need to be much stricter in your definitions other wise you are allowing koch curves.

9. Mar 16, 2014

### serp777

Thanks for all of your guys' responses so far.

I should specify that I am referring to all vertices being equidistant from the center and all edges being the same length.

Also micromass, could you explain why the largest polygon that fits inside the circle is a circle is contradictory to the fact that the smallest polygon that fits outside the circle is also a circle? These two things approach the same limit, and in infinite steps, both things will become a circle. I imagine you could use multivariable limits to describe the subset you were describing.

Why couldn't the largest polygon fitting in the circle actually just be a circle itself?

Furthermore, it would make sense that the ratio of the perimeter to the equidistant diameters of the n sided polygon would approach the value of PI as n approaches infinity

Last edited: Mar 16, 2014
10. Mar 16, 2014

### HomogenousCow

You can derive the expression for the area of a circle by approximating it as a regular polygon with "infinite" sides.

11. Mar 16, 2014

### micromass

Staff Emeritus
A circle is not a polygon. The definitions don't match. So there is no largest polygon fitting inside the circle.

Second, how do you take the limit of subsets of the plane? I feel that this needs to be specified too.

Sure, it sounds logical, but

Last edited: Mar 22, 2014
12. Mar 16, 2014

### willem2

I couldn't google this, so I tried it myself

Suppose you have an infinite sequence of subsets of the plane {S1,S2,S3,......}

a subset of the plane L is the limit of this set if:

For every point p in L and for every ε>0 there exists an n such that for all sets Sk with k>=n, a point q in Sk exists with distance(p,q) < ε.

For every point p not in L, there exists an n and an ε>0 such that for all sets Sk with k>=n no point q exists in Sk with distance(p,q) < ε

Basically, every point in the limit set must have all sets past some Sn come arbitrarily close to it. Every point not in the limit set must have a finite neighbourhood, that does not contain any points in any set in the sequence past some Sn.

With this definition of limit, it isn't true that the limit of the lengths of a sequence of curves is equal to the length of the limit curve. Take the approximation of a diagonal line with staircases of ever smaller steps.

13. Mar 16, 2014

### micromass

Staff Emeritus
I think this is called Kuratowski convergence: http://en.wikipedia.org/wiki/Kuratowski_convergence

14. Mar 20, 2014

### Whovian

Would the Vietoris Topology over R2, along with the net definition of convergence, give the "correct" limit of these sequences of polygons? And would the limit of the lengths of a sequence of curves be the length of the limiting curve?

15. Mar 21, 2014

### PeroK

Suppose you have a sequence of shapes. You could say that the sequence tends to a limit shape if the area between the nth shape and the limit shape tends to 0 as n tends to ∞.

In this case, your sequence of polygons would tend towards a circle. And, so, by this definition, the circle would be the limit shape.

Also, the total length of the sides of the polygon would converge to the circumference of the circle. So, the polygons would approximate the circumference closer and closer. THis is, in fact, one way to estimate Pi.

Now, consider approximating an isosceles right-angle triangle by a sequence of stepped shapes. Each shape would be like a staircase with n vertical steps and n horizontal steps. If the triangle's sides have length 1, then each step would be 1/n up and 1/n across.

Now, by the above definition, this set of shapes clearly converges to the triangle: the area of the difference gets less and less as the number of steps increases.

But, the length of the steps remains 2 for all n, which does not converge to the length of the hypotenuse.

So, you have to be careful about what properties are preserved by limits of sequences of shapes.

16. Mar 24, 2014

### mesa

Archimedes might disagree (as also pointed out by serp777).

17. Mar 24, 2014

### micromass

Staff Emeritus
Archimedes was a genius and probably one of the smartest people ever to live. But he lived so long ago that I don't think you should take him as a reference on mathematical definitions and practice. Math nowadays is far more careful and rigorous.

18. Mar 24, 2014

### mesa

Find a better definition and I will be happy to change my view.

19. Mar 24, 2014

### micromass

Staff Emeritus
This is a forum about math. So we take definitions from current mathematical practice. And current mathematics says that a circle is not a polygon. Disagree all you want, but this forum is about math and not personally invented definitions.

20. Mar 24, 2014

### mesa

Absolutely agree, and if the work of Archimedes isn't good enough for PF then none of us are.

In the meantime, still waiting on a better 'modern' mathematical definition.