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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.
by 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 agoHuh? 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 agoWhat 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.