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The answer is that the perimeter “at infinity” matches the triangle. There’s nothing that says the perimeter of the final “infinite” shape has to equal the lim

The answer is that the perimeter “at infinity” matches the triangle. There’s nothing that says the perimeter of the final “infinite” shape has to equal the limit of the perimeters as they go to infinity. In order for this to work, you have to prove that they’re equal. Typically you’d come up with two limits, one which you prove is always greater than the value in question, and one which is always less. Then you show that they’re equal in the limit. That squeezes the number you’re interested in, proving that it’s also equal to that number. For example, the typical proof of the circumference of a circle involves making two polygons, one inside the circle and one outside. The perimeter of the one inside is always less than the circumference, and the one outside is always greater. Then you take the limit of the polygon perimeters as the number of sides goes to infinity. They converge on the same number, 2 pi r.

pif 2y ago
Sorry, you are completely wrong. The perimeter is equal to the square one at every iteration, thus its limit towards infinity is still the square one.
- wat10000 2y ago
  Huh? I never said otherwise. I said that the perimeter of the limit doesn’t have to equal the limit of the perimeter. In order for this to work, you have to bound the value from above and below and show that they meet at the limit. This only bounds it from above. Obviously you’ll never find a sequence that bounds the perimeter from below that has a limit equal to the square one, so you’ll never be able to prove this equality.
  - pif 2y ago
    > This only bounds it from above AND from below. At every iteration, the two perimeters are the same.
    - wat10000 2y ago
      What two perimeters? There’s only one. And it hasn’t been shown to bound the triangle’s perimeter from below. It is plainly greater than the triangle’s perimeter at every step, so it just establishes an upper bound. To derive the triangle’s perimeter, you also need another thing that’s always less than the triangle’s perimeter, then show that the lower and upper bounds converge on the same value.
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pif 2y ago

Sorry, you are completely wrong. The perimeter is equal to the square one at every iteration, thus its limit towards infinity is still the square one.

wat10000 2y ago
Huh? I never said otherwise. I said that the perimeter of the limit doesn’t have to equal the limit of the perimeter. In order for this to work, you have to bound the value from above and below and show that they meet at the limit. This only bounds it from above. Obviously you’ll never find a sequence that bounds the perimeter from below that has a limit equal to the square one, so you’ll never be able to prove this equality.
- pif 2y ago
  > This only bounds it from above AND from below. At every iteration, the two perimeters are the same.
  - wat10000 2y ago
    What two perimeters? There’s only one. And it hasn’t been shown to bound the triangle’s perimeter from below. It is plainly greater than the triangle’s perimeter at every step, so it just establishes an upper bound. To derive the triangle’s perimeter, you also need another thing that’s always less than the triangle’s perimeter, then show that the lower and upper bounds converge on the same value.

deleted 2y ago

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