This is silly. It's like looking at a parabola and stating that since it has an asymptote of x = 0, it must at some point into infinity have x = 0 actually lying on the parabola's curve.

Can we assume that x^2+y^2=r^2 is a valid test for whether or not a vertex lies on a circle? Is that a given?

If so, in a regular polygon (the method this spoofs that - I think - we're trying to differentiate it from), every specified vertex lies on a circle.

But you draw a line from that vertex to the next and the midpoint of that line must deviate from a circle; x^2+y^2 will be less than r^2.

Clearly, it's not a circle, many points on it will do not satisfy the equation of a circle. But if I do something like this...



This shows the sum total of the mid-point's hypotenuse divergence from r in a regular polygon. As we add vertices, the divergence from a circle approaches zero; the difference between our circle approximation and a hypothetical true circle is reduced.

The excavation method does not have this characteristic.

Both the square excavation method and regular polygon contain vertices which satisfy the points, alternating with ones that do not. But as we iterate the square excavation method, the sum total of the vertice's deviation increases.

I'm not really sure how to make a function for this one - I could try if necessary - but for now I'll but I'll run some numbers here manually.

In the first iteration, showing the square, we have four specified vertices for which their hypotenuse is sqrt(2) (1.41etc) each, so we can say they deviate from r^2 (1) by 0.41etc, or 1.65ish total. (There are also four points that do fit on the circle, at the mid-points.)

On our second iteration, the deviant vertices' average hypotenuse is sqrt(cos(45)^2+1) or 1.22ish, so the deviation is that, minus one; 0.22ish. There are now 8 of these, so the total sum of the midpoint vertices' deviation from r is now increased to 1.76.

This contrasts with the regular polygon which reduces deviation from the definition of a circle by adding vertices.

Thus we can quantify that squares are an inferior approximation of a circle compared to a regular polygon.

Or something.

Quote: You've given no good reason why a valid "iterative" approximation of a quantity should necessarily converge (you said "reduce"?) indefinitely to some limiting value

In the above pitch, a correct procedure should converge onto zero when measured this way because a non-zero value shows failure to satisfy the equation of a circle. It should be more like a circle because the goal is to approximate a circle's properties by creating a shape to approximate a circle, then measuring its properties.

Is any of this relevant or am I barking up the wrong tree?

[Edited by - JoeCooper on December 3, 2010 8:37:55 AM]

Quote: Original post by Hodgman
I'm assuming that "1/inf" is the smallest value that is still greater than zero.

Such a value does not exist. Whenever you believe you have found such a value x, then x/2 is smaller, positive and not zero either (this works regardless of whether we are working with real or rational numbers).
In actual textbook math (not school-math, university-level math) there are preciously few points were an "infinity" can be. There are some more specialized branches of mathematics which allow actually using infinity as a more regular symbol, but even there you have to be extremely careful what you are allowed to do with it. There is a Wikipedia article about that.
The rest of the post kinda dies with that.


Maybe I should not have skimmed the thread like that but so far it seems arguments are on the "drawing sketches, waving hands"-level largely.

The only proper definition for the convergence of a sequence S(n) to the limit C I know is

and I don't think I have seen anyone trying to apply this to the problem. That would of course first require us to put the happy sketch into an actual formula you can work with but I feel a bit lazy today.
My personal guess is the result will be that the happy sketch idea simply does not converge because for every finite value the circumference is 4 while the limit would have to be pi. So even for it's not possible to find an as required.

Quote: My personal guess is the result will be that the happy sketch idea simply does not converge because for every finite value the circumference is 4 while the limit would have to be pi.

Isn't that just saying that it doesn't work because the result isn't Pi? Or, it should converge on Pi and it doesn't?

I just got my rear handed to me on a platter for that.

Well, as I said that is my personal guess. Actually proofing that would require a bit of work I currently do not have the time (and honestly, also not the motivation) for.

I'd be happy though if the rest of the post steers the topic toward the slightly more mathematical. A lot of the things I read while I glanced over the thread made me wince really, really badly even though my last 'real' math is seven, eight years behind me now. It's a bit frustrating, like watching people who are unwilling to believe that but do not have the mathematical background to actually argue their point or even understand the proves people who do know what they are talking about are supplying. :(

In my first post I asserted that the limit curve is not a circle. I was wrong. It does converge to a circle, and this is clear just from the definitions.

The root issue here is that limits and derivatives do not generally commute, and they fail to do so here for a very clear geometric reason.

Let c_n(t) : [0,1] --> R^2 designate the curve after the nth iteration with c_0 being the square, and let c(t) designate the unit circle.

The length of the limit curve is the integral over [0,1] of |(limn-->infty c_n)'(t)|. The limit of the lengths of the curves is limn-->infty of the integral over [0,1] of |c_n'(t)|. The limits don't commute with the integrals. The reason in this case is that c'(t) is a vector that could point in any direction depending on t whereas c_n'(t) is always either horizontal or vertical. Observe that this does *not* happen when one uses the traditional circumscribed n-gons that we're all familiar with.

[Edited by - nilkn on December 3, 2010 12:27:13 PM]

I think the trick also comes from fractal nature of the problem - unlike typical shapes, the perimeter and area of a fractal are not necessarily correlated.

Koch snowflake has infinite perimeter, but finite area.

I'd say that the problem here comes from implication that value of Pi follows directly from perimeter, whereas even when using the curve it must be derived from area.

The limit curve is 'infinitely' squiggly/pointy (Non-differentiable everywhere).

The area is being monotonically reduced each time with the limit converging to PI*r^2 as any area outside of the circle eventually gets cut out. The perimeter is not getting reduced each time, and has a limit of 4. We created a shape with the same area as a circle, but not a circle.

The limit need not equal the value of the function. For example the piece wise function { 1 if x = 0; 0 otherwise } the limit as x->0 exists and equals 0. The value at 0 is 1.

Quote: Original post by Platinum314
We created a shape with the same area as a circle, but not a circle.

But, you didn't do that. You always have a saw tooth edge that comes near the circle, but must always include an area that exists outside the boundary of the circle. It therefore must have an area Larger than the circle itself, not equal to.

Quote: Original post by Platinum314
The limit curve is 'infinitely' squiggly/pointy (Non-differentiable everywhere).

The area is being monotonically reduced each time with the limit converging to PI*r^2 as any area outside of the circle eventually gets cut out. The perimeter is not getting reduced each time, and has a limit of 4. We created a shape with the same area as a circle, but not a circle.

The limit need not equal the value of the function. For example the piece wise function { 1 if x = 0; 0 otherwise } the limit as x->0 exists and equals 0. The value at 0 is 1.

In this case however the curves not only converge to the circle but they do so uniformly. Let c_n(t) be the curve after the nth iteration. For any epsilon > 0 draw an annulus a_epsilon of width epsilon whose central circle is the unit circle. For sufficiently large n the curve c_n(t) is contained entirely within this annulus a_epsilon. This is the condition for uniform convergence to the limit curve (the circle in this case).

The issue in this case is that two functions can be close yet their derivatives can differ by a lot. More precisely, the derivative of a limit of a sequence of functions need not coincide with the limit of the derivatives. I actually think you can say something stronger in this case case: despite the fact that the limit curve is everywhere differentiable the sequence of derivatives c_n'(t_0) for a fixed t_0 in [0,1] does not have a limit.