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Posted 02 December 2010 - 10:37 PM
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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
Posted 03 December 2010 - 01:23 AM
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Original post by Hodgman
I'm assuming that "1/inf" is the smallest value that is still greater than zero.
Posted 03 December 2010 - 01:39 AM
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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.
Posted 03 December 2010 - 01:44 AM
Posted 03 December 2010 - 04:27 AM
Posted 03 December 2010 - 04:48 AM
Posted 03 December 2010 - 06:31 AM
Posted 03 December 2010 - 06:47 AM
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Original post by Platinum314
We created a shape with the same area as a circle, but not a circle.
Posted 03 December 2010 - 06:52 AM
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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.
Posted 03 December 2010 - 06:57 AM
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Original post by BitMaster 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.
Posted 03 December 2010 - 07:00 AM
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Original post by Talroth 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.
Posted 03 December 2010 - 07:57 AM
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Original post by JoeCooper
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.
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Thus we can quantify that squares are an inferior approximation of a circle compared to a regular polygon.
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Original post by Antheus
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.
Posted 03 December 2010 - 08:31 AM
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Original post by Way Walker
but the most basic definition of pi is a relation between the perimeter and diameter.
Quote:Except that in this case the curve does not define Pi, if my assumption about difference between perimeter of curve and area holds.
It's maybe more robust to derive it from the area,
Posted 03 December 2010 - 08:41 AM
Posted 03 December 2010 - 10:09 AM
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Original post by BitMaster Quote:
Original post by Talroth 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.
The problem is, as soon as limits are involved an argumentation like that just does not work anymore. ;) Just take a look at a simple sequence like S(n) = 1/n. Whichever n you put in, it's always bigger than 0. But the sequence is the example of a sequence converging to 0.
Posted 03 December 2010 - 10:49 AM
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Actually, that would produce a lower bound. The upper bound is found by placing the midpoints of the edges, not the vertices, of a regular polygon on the circle.
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One thing that bothers me about this analysis is that it singles out the worst points. If we just look at the best points, the number of points with zero deviation increases linearly for the regular polygons but exponentially for the square excavation.
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Perhaps the area enclosed between the two would be a better metric
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It's not so simple as showing that it is inferior. What we need is to show that it's inferior in a meaningful way.
Posted 03 December 2010 - 02:48 PM
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Original post by JoeCooper
First, singling out the worst points. I don't necessarily do that, actually. If I add in the valid vertices' deviation, all I do in either analysis is add 0 (and half it, I think. I did this earlier and now I can't remember exactly.)
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Secondly, the iteration # means nothing outside of a given procedure.
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Perhaps the area enclosed between the two would be a better metric
I disagree; the area doesn't predict the shape's perimeter.
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It's not so simple as showing that it is inferior. What we need is to show that it's inferior in a meaningful way.
This procedure iterates over a square, cutting parts away, so that the vertices are closer and closer to a circle.
My analysis is to examine their hypotenuse, and the deviation from a circle, and see if it actually converges on 0 or not.
It does not, so, it is not an approximation of a circle, and that makes it inappropriate the only meaningful way; if it isn't a circle, than any resemblance between its properties and those of a circle is purely coincidental.
Posted 03 December 2010 - 03:03 PM
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Original post by Antheus Quote:
Original post by Way Walker
but the most basic definition of pi is a relation between the perimeter and diameter.
The geometric definition is either that of circumference or that of area.Quote:Except that in this case the curve does not define Pi, if my assumption about difference between perimeter of curve and area holds.
It's maybe more robust to derive it from the area,
Posted 03 December 2010 - 08:06 PM
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Original post by Talroth Quote:
Original post by BitMaster Quote:
Original post by Talroth 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.
The problem is, as soon as limits are involved an argumentation like that just does not work anymore. ;) Just take a look at a simple sequence like S(n) = 1/n. Whichever n you put in, it's always bigger than 0. But the sequence is the example of a sequence converging to 0.
Like you said, 1/n, where n is any positive value, never actually equals zero. The area enclosed by the example shown in szecs's first post is that of one that neither equals the area of the circle, nor is it one who's shape is similar to that of a true circle.
I don't know what kind of kindergarten you went to, but back in my day circles didn't have a series of right angles on them.
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