Circles vs an infinitely n-sided polygon.

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  • #1
serp777
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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?
 

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  • #2
micromass
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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
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Is there some proof that says an infinitely sided polygon is different from a circle or does it just come down to definitions?

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).
 
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  • #4
serp777
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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.

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 isn't actually zero, it's just non zero. So take the limit as N--> infinity. Or similarly summing an infinite series.
 
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micromass
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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 isn't actually zero, it's just non zero. So take the limit as N--> infinity. Or similarly summing an infinite series.

Sure, I get how limits are defined for numbers. But how do you define limits for subsets of the plane?
 
  • #6
serp777
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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).

"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 wouldn't it hold that the smallest polygon that fits outside the circle is also just a circle which has the same perimeter and area? Thus wouldn't those two things actually be the same?
 
  • #7
micromass
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"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 wouldn't it hold that the smallest polygon that fits outside the circle is also just a circle which has the same perimeter and area? Thus wouldn't those two things actually be the same?

Right, it makes no sense since circles aren't polygons. So there is no "largest polygon" fitting in the circle.
 
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  • #8
pwsnafu
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Suppose you have a square, and you simply start increasing the number of vertices and edges proportionally, all the way to infinity.

You are going to need to be much stricter in your definitions other wise you are allowing koch curves.
 
  • #9
serp777
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You are going to need to be much stricter in your definitions other wise you are allowing koch curves.
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
 
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  • #10
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You can derive the expression for the area of a circle by approximating it as a regular polygon with "infinite" sides.
 
  • #11
micromass
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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?

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.

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

Sure, it sounds logical, but

troll_mathemathics_pi.jpg
 
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  • #12
willem2
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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.

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
micromass
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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.

I think this is called Kuratowski convergence: http://en.wikipedia.org/wiki/Kuratowski_convergence
 
  • #14
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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
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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?

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
mesa
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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.

Archimedes might disagree (as also pointed out by serp777).
 
  • #17
micromass
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Archimedes might disagree (as also pointed out by serp777).

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
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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.

Find a better definition and I will be happy to change my view.
 
  • #19
micromass
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Find a better definition and I will be happy to change my view.

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
mesa
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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.

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.
 
  • #21
micromass
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if the work of Archimedes isn't good enough for PF then none of us are.

That is obviously not what I said.

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

Have fun waiting!

Thread locked.
 

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