Pi

This was never funny

Limits are to actual values as undefined is to defined.

youtu.be/CMP9a2J4Bqw

Not related.

this is like saying that it is the same to go from california to Florida and then to new York, than going straight from California to NY, knowamsayin

>repeat to infinity
define it and explain the process

mathematics with vague notions will always result in vague, if not incorrect, results

> pi = 24

But that's exactly what a circle is

1)
Induction only works for a finite number of steps. You cannot induct to show that the infinite case hold true. For example, you cannot induct on the rationality of 3, 3.1, 3.14, etc. to show that pi is rational.

2) induction aside, the resultant figure is not a circle. A circle contain uncountably many points along it's perimeter, whereas that figure contains countable many points on the perimeter (each point touching the meme circle was bent at some finite step in the process, and each step adds in a finite number of points) at which it was

the black lines are an incredibly poorly drawn picture of a circle made up of a bunch of line segments
the red lines are what the method in OP gives
You will note that the red lines have a larger perimeter than the black lines

have you ever looked up the definition of a circle?

forgot pic

but it approximates the circumference of a circle

TOC tic board missing a line. Xxx

mimicked shapes do not necessarily share more attributes (with an original) but the mimicked shape. also, two same sized catheti will always be longer than the corresponding hypotenuse - no matter how "infinite small" the triangle gets.

4 is the wrong answer

A circle is just one dot with a bunch of lines / vectors coming from it with the same R

>entire thread is brainlets whining about limits and infinity
this is NOT the problem. the problem is the assumption that just because you have a sequence of curves f_n that converges to some curve f, THIS DOES NOT IMPLY that the arclengths of the f_ns converge to the arclength of f.

inb4 angry 10 year old who dont know the definition of arclength disagree with me

theres a better version of that picture, btw: take a square and indent one corner inside, repeat for smaller corners etc and you get 2 = sqrt(2)
it's better because it's simpler

The area, not the perimeter.

>after infinite iteration this shape contains infinitesimal edges that all add up to made a perimeter of 4.
that was so dumb I think it gave me cancer. the given sequence of curves converges (uniformly even) to the circle.

tell me what you dont understand about this and we'll have a talk

Congratulations, you've determined the circumference of that weird shape in the taxicab metric

Assume it converges to the circle. In every step you get 4 more points that intersect with the square, so after infinte iterations you would still have countable intersections, but this is a contradiction to the fact that R is uncountable.

This would be so easy to explain with Landau symbols. Why does nobody use Landau symbols?

use them and explain, then

I don't understand what this math thing is. Like... why math?

Here's an explanation, brainlets.
Imagine a different example, where instead you placed points around a circle and made tiny tiny triangles with roughly negligible area. You then take these triangles, and stretch them out far away from the circumference of the circle, but at the same time adjust the closeness of the base points such that the area of the triangles remains constant as they stretch out (similar to the dirac delta function). You would end up with a shape that has a massive perimeter, but almost identical area to before. This is why this doesn't work. Pi is by definition, the ratio of circumference to diameter for a circle. This is not a circle.

You have a sequence of paths [math]\gamma_n(t)[/math] representing the perimeter with each time corners removed. E.g [math]\gamma_1(t)[/math] is a square and [math]\gamma_2(t)[/math] and [math]\gamma_3(t)[/math] represents the perimeters as in the pictures 3 and 4. Now for each natural number [math]n[/math], the length of [math]\gamma_n(t)[/math] is [math]\int_0^1|\gamma_n'(t)| dt=4[/math]. But if we let [math]n[/math] approach to infinity, we get a circle [math]\gamma_(t)[/math] which lenght is [math]\int_0^1|\gamma'(t)| dt=\pi[/math].

so in this case [math] \lim_{n \rightarrow \infty} \int_0^1 |\gamma_n'(t)|dt= 4 \neq \int_0^1 |\frac{d}{dt} \lim_{n \rightarrow \infty} \gamma_n(t)|dt=\pi[/math]

Correct answer.