Originally posted by PalynkaArea is irrelevant. There are figures in mathematics with infinite perimeter and finite area, such as the area under the curve y=e^(-x).
The area between the two figures converges to zero, one can find a sub-sequence that will converges to any point on the circumference, one cannot find any sub-sequence that converges to a point outside the circumference. The limit is the circumference.
Not only that, it converges uniformly. That there are other sequences that converge to it is irrelevant.
Originally posted by PalynkaThe areas will agree, yes, but the arc lengths will not (seeing as pi does not equal 4). And we figure arc lengths as limits of Riemann sums of a specified kind. If two paths have different lengths, that's our clue that they are not the same path.
Why does that make it irrelevant? This is a bounded figure. 😕
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Originally posted by SoothfastThe set of points in the sequence converges to the set of points in the circle. That certainly is enough for pointwise convergence. (http://mathworld.wolfram.com/PointwiseConvergence.html)
The areas will agree, yes, but the arc lengths will not (seeing as pi does not equal 4). And we figure arc lengths as limits of Riemann sums of a specified kind. If two paths have different lengths, that's our clue that they are not the same path.
The mistake is in assuming that all properties of the approximating sequence must all converge. This is not true (two examples posted above). Of course, because they don't, there will be more efficient sequences for that approximation but that doesn't mean that this one doesn't converge.
Originally posted by PalynkaIt sounds like you are saying any two points do not connect by a straight line but rather makes a triangle, the apex of which will always be outside the line of the circle and therefore will be longer than the circumference.
The area between the two figures converges to zero, for any point in the circumference, one can find a sub-sequence that will converges it, one cannot find any sub-sequence that converges to a point outside the circumference. The limit is the circumference.
That there are other sequences that converge to it is irrelevant.
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Originally posted by sonhouseBut that is also true for the sequence: 0.9, 0.99, 0.999,... which is smaller than 1 at any finite step. That does not mean that this is true at the limit.
It sounds like you are saying any two points do not connect by a straight line but rather makes a triangle, the apex of which will always be outside the line of the circle and therefore will be longer than the circumference.
For any epsilon, I can find an Nth iteration such that the distance between any point in the circle and the figure is smaller than epsilon. Do you disagree?
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Originally posted by SoothfastIn brief, the polygon is not a circle.
I'd say no, not a polygon in the usual sense, since a polygon in the usual sense must have a finite number of sides. There are more esoteric definitions of the term "polygon" in mathematics that may call it a polygon, but it wouldn't be your grandfather's polygon.
Originally posted by SoothfastWhat Palynka is telling us is that the limit, is not necessarily a member of the series. In algebra, we can have a bounding point for a set that is not a member of the set eg (0,1] is bound on the left by zero, but zero is not a member of the set and may have properties that are not permissible in the set (eg division).
I'd say no, not a polygon in the usual sense, since a polygon in the usual sense must have a finite number of sides. There are more esoteric definitions of the term "polygon" in mathematics that may call it a polygon, but it wouldn't be your grandfather's polygon.
So although the limit of the series of polygons, is a circle, it is not a polygon itself. And none of the members of the series has an infinite number of sides, so they are in fact all polygons.
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Originally posted by PalynkaIt's been awhile since I've looked at sequences of functions, but I gave this some more thought whilst stuck on a plane today.
The set of points in the sequence converges to the set of points in the circle. That certainly is enough for pointwise convergence. (http://mathworld.wolfram.com/PointwiseConvergence.html)
The mistake is in assuming that all properties of the approximating sequence must all converge. This is not true (two examples posted above). Of course, because they do nt[/i] sequences for that approximation but that doesn't mean that this one doesn't converge.
Construct a sequence of functions {f_n} as follows: we take f_1 to be the square, f_2 to be the square with "inverted" corners (yielding a cross shape), f_3 to be the cross shape with "inverted" corners, and so on*. Well, what have we got? I agree: the sequence {f_n} must converge uniformly to a circle, and not any kind of fractal-type entity. But, while I haven't checked any of your links (so I might be repeating something already mentioned), it certainly is true that not all geometrical properties of a curve are faithfully preserved in a limit process whose limit is the curve. The idea of path length is defined using a very specific kind of limit process which the limit process we're considering here (using "collapsing squares" or "corner inversions" ) does not match.
One interesting idea that occurs to me in considering how "badly" the "collapsing squares" sequence of functions misses the mark when it comes to actually "capturing" the circle is to consider this: f_1 only has 4 points in common with the circle, f_2 has 8 points, f_3 has 16 points, and in general f_n has 2^(n+1) points in common. Thus, as n gets larger the number of points the approximating path has in common with the circle is finite. Even if we take n "to infinity," given that n runs through positive integer values, we can only expect to intersect the circle at a countably infinite number of points, while the circle itself consists of uncountably infinite points. So yes indeed, things are quite anomalous here. We cannot expect this process to faithfully hone in on the "true" circumference of the circle.
It's an interesting puzzle, but ultimately it abuses the defined means whereby arc length is computed. That is to say, the puzzle at memebase.com is essentially defining a different metric than the customary Euclidean metric we all know and love.
* To have true functions we can either deal strictly with the upper half of the picture, or else use parameterized curves f_n(t)=(x_n(t),y_n(t)).
Originally posted by SoothfastNice post.
It's been awhile since I've looked at sequences of functions, but I gave this some more thought whilst stuck on a plane today.
Construct a sequence of functions {f_n} as follows: we take f_1 to be the square, f_2 to be the square with "inverted" corners (yielding a cross shape), f_3 to be the cross shape with "inverted" corners, and so on*. Well, what ...[text shortened]... of the picture, or else use parameterized curves f_n(t)=(x_n(t),y_n(t)).
There are no more points in a curve with a longer arclength than in a smaller one. Moreover, by Dirichlet's theorem, we can approximate any real by a sequence of rational approximations, which is what the bisection will do. So I don't see that as being a problem in itself.
The most standard metric for checking whether subsets of metric spaces are close to one another is the Hausdorff metric, and it's easy to prove that the sub inf d(x,y) (the maximal (smallest) distance between points of the two curves is converging uniformly. Also convergence in arclength requires convergence in derivatives, and it's a well known property of limits that these are not necessarily preserved. So, in my view, it is you who is rejecting the standard metrics.
Sure, there are metrics for which it doesn't converge under that metric, but to say it doesn't converge one needs to reject standard epsilon/delta or sup inf definitions of limits. If anyone is using a non-standard definitions, that is you, who will need to use a C1 or something.
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Originally posted by PalynkaAh, so, there is no making peace with you, then.
Nice post.
There are no more points in a curve with a longer arclength than in a smaller one. Moreover, by Dirichlet's theorem, we can approximate any real by a sequence of rational approximations, which is what the bisection will do. So I don't see that as being a problem in itself.
The most standard metric for checking whether subsets of metric space is using a non-standard definitions, that is you, who will need to use a C1 or something.
I just said I agree that {f_n} is converging uniformly to a circle. What more do you want? If you want to go around accusing someone who has a math degree and does math every day that he doesn't know even the fundamentals of his field, well, that's your lookout, but then I'm not going to spend much time around you.
The Dirichlet theorem is irrelevant here. Hausdorff metric? Where did that come from? Irrelevant also. All we have is a sequence of functions, and I agree that my initial impression that a fractal curve was being approached was not right. Convergence in arc length does NOT require convergence in derivatives, because it all depends on how you want to define arc length. There is a standard way found in calculus texts, and that's not the way that's being employed in the OP. That is why it arrives at the erroneous conclusion that pi is 4. Because the thing is, one COULD define the arclength of a curve to be the limit of the arclengths of approximating curves of the type given in the OP, but it's not the way mathematicians would normally do it because it's less natural and leads to less-than-desirable outcomes.
So what "standard metrics" am I supposedly rejecting? The standard way arc length is defined uses a limit process involving lengths of line segments that are arrived at using the Euclidean metric (i.e. the "distance formula" ). But in general to have a metric you don't even need any calculus whatsoever, much less derivatives. Metric spaces are discussed in all sorts of abstract settings and can be quite bizarre. In the OP, if you want to say that the length of the limit curve is just the limit of the lengths of the approximating curves, then there is a limit and it is 4. But as I said, it's not in agreement with the more usual definition.
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Originally posted by SoothfastWow, where did that come out from? I even complimented your post. And I don't really care if you have a math degree, nobody is right or wrong based on that.
Ah, so, there is no making peace with you, then.
I just said I agree that {f_n} is converging uniformly to a circle. What more do you want? If you want to go around accusing someone who has a math degree and does math every day that he doesn't know even the fundamentals of his field, well, that's your lookout, but then I'm not going to spend much time and it is 4. But as I said, it's not in agreement with the more usual definition.
You seem to imply that the "standard way" is the only way (or else why would it be relevant?)" which is obviously wrong. I also am baffled how you believe the Hausdorff metric is irrelevant when talking about distances between metric spaces. Also, the Dirichlet theorem is not irrelevant, because if you put the origin on the lower left corner of the square, the points coordinates will be always be rationals. So if it were not true that we can approximate reals with sequences of rationals, then there would be points for which the sequence wouldn't converge to.
I said that the limit of the arclength of the approximating figures was 4. And that the figure converges to a circle (afterwards I even specified what type of convergence). Some people, including you, argued that was not possible and that it didn't converge to a circle. Thank you for now agreeing with me, although I don't see why you take it so personal.
Originally posted by SoothfastMe to KN:
In fact, nowhere in my last post have I said anything about how something "doesn't converge," so I cannot fathom what you're on about. You're probably completely misreading what I was saying about the fact that {f_n} never attains anything that actually intersects with the circle at a significant number of points, which is neither here nor there. ...[text shortened]... circle that are never incorporated in the range of any function f_n in the sequence.
If your answers to these two questions are both "none", then clearly it converges to the circle.
Your reply:
Since the figure being attained by the limit process is a planar curve with length 4, clearly it is not a circle.
Originally posted by PalynkaIn fact, nowhere in my last post have I said anything about how something "doesn't converge," so I cannot fathom what you're on about. You're probably completely misreading what I was saying about the fact that {f_n} never attains anything that actually intersects with the circle at a significant number of points, which is neither here nor there. What I was trying to convey is that there will be points on the circle that are never incorporated in the range of any function f_n in the sequence.
Sure, there are metrics for which it doesn't converge under that metric, but to say it doesn't converge one needs to reject standard epsilon/delta or sup inf definitions of limits. If anyone is using a non-standard definitions, that is you, who will need to use a C1 or something.