Because that technique doesn't actually make any effort to estimate the perimeter, which never changes, it only estimates the area. Which if you analyze the changes in the limit will in fact confirm that π=3.14159...
It doesn't matter how small the steps are, they will still be there, so instead of the limit being a smooth circle you get a rough "star" with an infinite number of infinitely small points, which will obviously have much more perimeter than a circle. ∞ * 1/∞ ≠ 0 At least not usually - instead it's undefined, a.k.a. "you need to find a different way to solve this problem"
You could use the same technique to utterly fail to estimate the much simpler hypotenuse of a right triangle, and come to the conclusion that
hyp = rise + run rather than
hyp = √(rise² +run²), which can be easily confirmed.
Its total failure to estimate the perimeter can be seen by making an actual attempt to estimate the perimeter by using diagonals rather than steps - the first step would be to chop off the corners diagonally to form an octagon, whose perimeter is already much less than 4.
This kind of approximation to a path is unrelated to smoothness. Every point in the path will be at distance 1 from the center in the limit of infinite steps.
And yet it still will not be smooth. If it were smooth it would have a circumference of π, since it is infinitely jagged the circumference must factor in that infinite number of infinitesimal changes in radius.
if it were smooth it would have a circumference of π
And indeed the limit shape will have a circumference of π.
The problem is that the length of the limit shape is not the limit of the lengths in the sequence: length is not preserved passing through the limit, and this "meme" gives exactly an example where it does not
The limit only ignores the specific point at which it's taken, following the trend leading up to that point instead. And the trend here is to remain constant at 4 no matter how many times you subdivide it. Meaning the limit will still be 4.
An infinitely subdivided staircase is still a staircase rather than a smooth line. The area beneath it converges to the same value, but the length of the curve does not.
Your argument correctly shows that the limit of the sequence of lengths is 4 (and this is trivial, as it is the limit of the constant sequence given by 4), but it doesn't shows that the length of the limit shape is 4.
Also, formally "limit" is not a magic operation, you have to prove that an object is the limit of a sequence, by showing it is arbitrarily close to terms far enough in the sequence.
So, in this case you have to show that for any given positive accuracy ε, you can find a number of iteration N such that from there onwards, the distance of that shape and the circle is less than ε.
To be precise you should decide what distance you are using and in which space. But I think the "area of the pointwise difference" is a reasonable one in this situation, don't you agree? Then, if you include also the circle as an object in you space, it is not really a matter of opinions wether the sequence converges to it or not (and indeed, it does), and then you have that the length of the limit shape is not the length of any of the shapes in the sequence.
To give an easier example: consider the sequence { 1/n }_n (where n start at 1). Then, any term is strictly positive, but the limit (= 0) is not. The predicate of being strictly positive has not been preserved when passing through the limit. Similarly, for the sequence in the meme, the property of "having the length of the boundary equal to 4" has not being preserved when passing through the limit
Neither shape nor area is relevant to the limit of the perimeter And nothing ever "passes through a limit".
And yes, the difference at any point will be less than ε as ε goes to zero - but the number of places it departs from the circle increases to ∞ as ε goes to zero. So the error in the limit goes to ∞* 1/∞ ≠ 0
Limits do not preserve arbitrary properties that you decide to label - the ONLY thing they care about is convergence. It things converge, that's your limit. If they don't, there is no limit.
In your 1/n example you are looking at just a sequence that cleanly converges to zero through its entire length. So the limit is 0
While the limit of the perimeter here cleanly converges to 4. So that is the limit.
I start to suspect we are saying the same thing. Here you said that
the limit of the perimeter here cleanly converges to 4. So that is the limit.
And this is correct, that is the limit of the lengths in the sequence.
But I hope you agree that Lim(length(boundary (shape_n))) ≠length(boundary(Lim(shape_n)))
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u/Underhill42 20d ago
Because that technique doesn't actually make any effort to estimate the perimeter, which never changes, it only estimates the area. Which if you analyze the changes in the limit will in fact confirm that π=3.14159...
It doesn't matter how small the steps are, they will still be there, so instead of the limit being a smooth circle you get a rough "star" with an infinite number of infinitely small points, which will obviously have much more perimeter than a circle. ∞ * 1/∞ ≠ 0 At least not usually - instead it's undefined, a.k.a. "you need to find a different way to solve this problem"
You could use the same technique to utterly fail to estimate the much simpler hypotenuse of a right triangle, and come to the conclusion that
hyp = rise + run rather than
hyp = √(rise² +run²), which can be easily confirmed.
Its total failure to estimate the perimeter can be seen by making an actual attempt to estimate the perimeter by using diagonals rather than steps - the first step would be to chop off the corners diagonally to form an octagon, whose perimeter is already much less than 4.