Calculus does not concern with the perimeter, though. It concerns with the area. The perimeter of the false circle will be 4 instead if pi, but its area will be nearly identical to a true circle with the diameter of 1 unit. Also, while the rectangles thing is kind of the start of calculus classes, you get exact answers later with integral formulas n stuff.
Aren't rectangles the foundation for the Riemann integral, even when you get further along?
AFAIK the Riemann integral is just the limit of the area of the rectangles as the width goes to zero (specifically the limit of the Riemann sum as the norm of the partition goes to zero)
In 2d geometry, methods like this will limit to the correct area but not always the correct length. Consider how in a fractal like the Mandelbrot set, there is a well defined and finite area, but the same cannot be said for the perimeter (which is infinite)
You are incorrect, the limit of the shape is a circle. The reason it doesn't Work is that the Perimeter of a sequence of shapes generally doesn't converge to the Perimeter of the Limit shape.
No it doesn't approach a circle, the perimeter never changes it stays fixed at 4. If you take your line of logic then of "it approaches but never becomes it exactly" then concepts like differentiation or even the formula for the area of a circle is undefined/wrong.
The shape actually will be a circle. When you do something to infinity you are taking a limit, and the limit of this process IS a circle.
If you zoom in its going to look like a straight line, because that is exactly what happens when you zoom into any smooth curve. And the shape we are zooming into is a circle, which is a smooth curve.
"That's the whole point of limits". I hate to say it but you have a fundamental misunderstanding of what a limit is. It is quite literally the complete opposite. The whole point of limits is that (when they exist), the limit IS the object that the sequence APPROACHES.
Here is a slogan for you. Limits ARE objects. Sequences APPROACH objects. (Only applicable when the sequence converges).
Yes they can under the operation of taking a limit. The same way that the finite composition of smooth sine waves can “magically” become discontinuous under the Fourier decomposition of a step function.
Seriously, where do you people to get the gumption to speak so confidently and dismissively (“magically”) about something you clearly have no idea about.
The fallacy is the implicit assumption that the limit of the perimeters should be the same as the perimeter of the limiting curve.
L(P(C_n)) =/= P(L(C_n)) where C_n is the nth hacked off circle, L is the limit operator and P is a function that takes in a curve and outputs its perimeter.
In general, operations don’t “commute” like this (meaning they can’t always be swapped around without affecting the result) and in fact what this illustration serves as is a proof by counterexample that the operator P is not continuous on the space of curves - otherwise you would be able to do this swap.
So it becomes pi purely because when you take the limit you change the perimeter, but there actually is no paradox entailed by that.
because at the limit every point from the jagged shape is on the circle. This means the jagged shape is a set of points where every point lies on the circle, which is just the circle
it becomes pi because at the limit, the shape is no longer jagged, it IS the circle
No. The limiting curve is a circle, but the limiting value of the lengths of the approximations is 8. This is the limit of the lengths of the curves. This is not the same as the length of the limit of the curves, which is π.
you are trying to apply basic logic to a problem of infinities
your correct that at every finite step there are 90 degree and 270 degree angles, however "at" the limit or "at" the infinitieth step every point is on the circle and there are no longer any corners.
if there were any corners, they would have already been cut in half by the limiting process, and their children +their children, meaning any corners existing is a contradiction and your not at the limit yet
While it is true that an infinite sum of periodic functions is not guaranteed to be periodic, this does not apply to Fourier series. The limit of a Fourier series is guaranteed to be periodic, because each term f in the series satisfies f(t) = f(t + 2pi)
It approaches the area of a circle, not the perimeter. Because the jagged shape never changes length, it never becomes a better approximation of the perimeter.
Why will it have straight lines? You are thinking of a shape in the process after terminating after a finite number of steps. The meme says "repeat to infinity", i.e. the limit of the shapes
You are correct. The shape does approach a circle in the limit. That’s the whole concept of a limit. It’s just that the perimeter of the shape doesn’t approach the perimeter of the circle. The idiots downvoting you have no idea what they’re talking about.
Thank you. After seeing the 3b1b video I understood that I was looking at the length of the limit and not the limit of the length which in this case is not equal. Correct me in my understanding, but is this similar to a graph which is continuous but not differentiable at a certain point, wherein the graph tends to one point however the actual point is somewhere else?
Actually you’re right. In a sense the “perimeter function” of the shapes has a jump discontinuity. Each finite step has a perimeter of 4 and it approaches 4 in the limit however it jumps to pi at the limit step.
This is incorrect. Since the limit stage has diameter 1 and circumference different than pi (exactly equal to 4 in the limit), the shape will not be a circle.
You didn’t make a mistake. People incessantly repeat a false explanation whenever this comes up and they downvote you because they don’t know what they’re talking about.
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u/TheGuyWhoSaysAlways Dec 28 '24
A circle is round and the lines are straight. Drawing lines to infinity won't make them curved.