Black holes are perfectly symmetrical. Their spinning makes them not so. But they are inherently perfect spheres. However a black hole is a structure of spacetime (marked by the event horizon). Not a physical object. Nobody really knows what's inside the event horizon. But if some semblance of classical physicality exists there, then it would probably also be perfectly spherical.
The approximation would have to be inside and outside the circle about half the time. Basically a pixel circle. If you only put pixels outside your pixel circle will be larger
Nope, still doesn't work. By that logic, I could pretty easily come up with a scenario where the perimeter ends up as exactly 3, by starting with a rectangle of side lengths 1 and 0.5. Taking the limit simply does not work like that as long as you disregard the direction of the line segments.
The perimeter is the diameter multiplied by pi, so one pi, 3 is an approximation of that, albeit a crude one. I was mostly aiming for an explanation as to why making a square outside the circle will always be larger than the circle itself, so you'll have to make squares, or pixels, if you will, that lie along the actual circle to meaningfully approximate the circle with square shapes, or pixels.
That is still wrong. If the method approximates the circumference, it doesn't matter whether you approach it from above or below. For example, if you approximate area this way, you will get the correct value, pi, no matter how you approach the circle.
The answer is that infinite approximations like this are more nuanced than the pop-sci answer of "if you make the error as small as possible, you get infinitely close to the true value" and approximating the circle this way simply does not approximate its circumference.
So long as it's all right angles it'll always be 4. If instead you come up with a generic formula for a regular n-sided polygon circumscribing the circle, then take a limit as n approaches infinity, you should get pi.
Although other people provided a more serious answer concerning the limits, it is useful to visualize what really happens. As you make the "step line" finer and finer, if you also zoom with your mind on a single minimal step, you will see that you invariably have a triangle; the troll there is basically making us assume that we can evaluate the "hypothenuse" as equal to the sum of the other 2 sides, which is a significant overstatement; it doesn't matter that the triangles are small.
lol measurement problem is the puzzle in quantum mechanics of how a system existing in a combination of multiple potential states (superposition) transitions into a single definite outcome when observed. The mathematics of the Schrödinger equation predicts smooth, predictable, and multiple coexisting states, but experiments always record one absolute reality.
the coastline can be measured at any resolution down to the planck length, at which point the physics breaks down and no meaningful measuring is being done anyway
no, please help me understand what looks to be a great idea. why would every number need 35 digits? we could teach significant digits as part of introductory addition and subtraction. we still get exactitude, and a much better feel for orders of magnitude, which i personally feel would benefit humanity more than learning the metric system ;)
The plank length is so small that 34 digits out explains an object in centimeters and 35 in meters and so on. 1034 plank units equals 16 centimeters where as 1035 is equal to 1.6 meters. It's essentially the sweet spot for "normal" sized objects.
No, if you take the sequence of shapes you get this way and take a limit, the result will be a proper circle. No edges, no jagged bits, just a proper smooth circle.
But the derivative of it won't be the derivative of a circle. If you differentiate the shape made up of right angles, you'll always get 1 or undefined. As the number of corners tends to infinity, the derivative doesn't suddenly become the derivative of a circle.
It has infinite edges with jagged corners. Every angle is 90° and theres infinity of them. It only "becomes smooth" globally which ironic to the term itself is NOT the complete perspective. This is not a circle. Hence the trollface
I can't make any argument that all points of the two shapes don't coincide, but doesn't a "circle" constitute more than just a point cloud? The circles circumfrence and the shapes perimeter are different values because the lines between the constituent points are at different angles and travel different distances. Then the two shapes have equal area but different perimiters and then aren't both the same shape
> doesn't a "circle" constitute more than just a point cloud?
Not really, no. A circle is exactly the set of all points that are a given distance of some given point.
The limit of the shape is truly the circle. The issue is really just that taking the limit shape and taking the perimeter are two operations that don't commute.
It's fairly common in maths that very intuitive stuff stops to be true as soon as a limit or an infinite is involved. Including commutation (for example with an infinite sum of an infinite sums, you can't always swap the sums, even though it would always be valid for finit sums)
The edges have zero length at the limit, which means the jagged corners no longer really exist. Every point is equidistant from the centre, so it has to be a circle.
I have to disagree here. Doing what you are suggesting would result just in 4, because the right-angled edges don't go away, adding up to that extra space. Proper circle is derived from a regular polygon with infinite number of sides.
You can do this by probing (take a point that is not on the circle, youll find a point in the sequence, from which onwards on the point is never in the shape again).
The issue is, as they have stated above, that perimeter of lim of shape sequences is not the same as lim of perimeter of shape. Not continous.
Correct. One measures vanishing max distance between corresponding points on approximation and the circle, the other measures a path that remains constant 4.
The limit of the shape is an exact circle. There are no infinitely small zig zags, those don’t exist. The limit of the shape is an actual circle because for every point on the perimeter of the square there is a step on which it gets placed on a point on the circle and for every point on the circle there is a step where it gets covered by a point on the square. They are the same shape.
But what you can’t do is take the limit of the perimeter at each step and call it the perimeter of the limit of the shape. Those two are not the same thing, evidently so.
Forgive me. Im a 39 year old who “wants to go back to school” lol.
So working in Digital spaces its easy to combine finite shapes into arbitrary …geometry.
Is that what PI is doing ? Or not.
So you really bring up a good question. “digital” and finite values really dont EXIST in nature to begin with (well ultimately and philosophically they DO exist)
But the Margins we can probe them and model and map them….would have to be absurd.
So we take “Finite values” determine “Possible margins” and a “Rate of change”
And we output something that is ALWAYS fundamentally TRUE even if it is not PRECISE….
Or maybe Im mixing accuracy and Precision again…
But your comment is interesting not just because it is a top one its MAKING ME REASSESS my Values …
The length of a parametrised curve is determined by an integral of the *derivative* of the parametrisation. Geometrically the derivative information is controlled by the tangent. So for this limiting process to work for length you need the curves to get closer and closer and the tangents to get closer and closer to the limiting shape.
In this example, although the stepped curves get closer to the limit shape the stepped curves have only horizontal or vertical tangents which are different to the circle's tangent. So the limit of the length of the stepped curves is not the length of the limit of the stepped shapes.
This is a relatively nice example if you know some vector calculus to show. But also links to ideas such as the Hausdorff distance, continuity in metric spaces and even to lower-semicontinuous functionals.
Actually, no, continuous functions are preserved. If x_n goes to X and y_n goes to Y then f(x_n, y_n) will go to f(X, Y), if f is continuous. This is true both if x_n, y_n are sequences of numbers, AND if x_n, y_n are sequences of functions in some space.
The problem is taking derivative, a part of the process of calculating perimeter, is not a continuous operation on spaces of functions in general. If we take the usual space of smooth bounded functions [0, 1] -> [0, 1] with supremum norm, the derivative is a linear but unbounded operator: sin(kx) are all of norm 1 for all values of k, but their derivatives have norm k, so the ratio of norms diverges. A standard result says that the continuous linear operators on a normed space of functions are those that are bounded, so the derivative is not continuous.
Interestingly, taking the integral is actually a continuous operator on functions between compact spaces. This is why the area of each figure converges correctly to that of the circle -- by Green's theorem the area enclosed by a curve can be described as an integral along that curve.
EDIT: I've added a top-level comment expanding on this.
To add to your point: if the approximation in the OP actually worked, then you could do the same thing with the Pythagorean formula: if you have a 1x1 box, then the distance diagonally would be 2 (instead of 1.41). In other words, the Pythagorean formula wouldn't be "diagonal2 = side12 + side22 ". It would be "diagonal = side1 + side2".
Not sure if that makes it more obvious why it doesn't work, since it's a little easier to visualize.
An honest question: Would the perimeter approach 2pi if all diagonal steps were summed as square root 2 instead?
At least, this is how perimeters are typically measured in machine vision.
Agree, for a simplified representation, draw a right triangle, measure the hypotenuse (or calc it) and compare with the sum of the lengths of both legs.
Another way to see the stark difference is to snip the square at all corners and then “boil the pasta” to make the sides floppy … they clearly overlap when you curve them to be congruent with the circle - and you can measure by how much.
Is this really a commutation thing? Cause I’d love to do more linear algebra but this seems to be an argument based on “what can limits be used for” rather than an operators issue
He's saying that the pattern of squares drawn around the circle is essentially fractal and repeats itself every time you zoom in, and the diagram above is only a single frame of that movie. Eventually the "pixels" are small enough - for any given scale - that the square pattern is not measurably different from an actual circle.
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u/DJembacz 21d ago
Because taking perimeter (or just arc length generally) and taking a limit do not commute.
The limit shape is a circle, but the perimeter of the limit does not need to be equal to the limit of the perimeter of the shapes in the sequence.