3 ms·

> It turns out you can take an arbitrary shape and split it into tiny rectangles Citation needed?

by jvanderbot 3y ago

Filligree 3y ago
> Citation needed? Pixels.
- mannykannot 3y ago
  Hm... What about a fractal shape with no right angles? Is that feasible? Or do you just approximate it with a fractal arrangement of rectangles?
  - deleted 3y ago
    [deleted]
  - calamari4065 3y ago
    I can't say I've ever done calculus on a fractal, but my guess is that most fractals have a fixed surface area ratio. Since fractals have infinite textural depth, there's a hard limit to their surface area compared to the overall area. But what you're asking about is the inaccuracies in the Reimann Integral https://en.m.wikipedia.org/wiki/Riemann_integral https://en.m.wikipedia.org/wiki/Riemann_integral The short version is that you pick a reasonably small size for your rectangles, and it mostly averages out. The discrepancy is so small that it can largely be ignored. But for argument's sake, you can simply imagine infinitely many infinitely small rectangles. That will fit the curve perfectly, but is impractical to compute. So we accept the slightly less accurate value.
  - fspeech 3y ago
    A fractal has measure zero in the space it's embedded in. It obviously can not contain any object of nonzero measure of that space.
    - Filligree 3y ago
      Does that include space-filling curves? …do those count as fractals, if the fractal dimension would be an integer?
Horffupolde 3y ago
Riemannian calculus.
- jvanderbot 3y ago
  Of course. My mistake.
calamari4065 3y ago
This is pretty much how integration works, and calculus shows us that integrating any arbitrary function can be done this way. But really isn't not that hard to intuit. Take any arbitrary shape and draw any number of arbitrarily small rectangles inside it. It should be easy to see that by adding the area of each rectangle, you approach the area of the shape. As the rectangles get smaller, the combined area of the rectangles gets closer to the true area of the shape.

> It turns out you can take an arbitrary shape and split it into tiny rectangles Citation needed?

Filligree 3y ago

> Citation needed? Pixels.

mannykannot 3y ago
Hm... What about a fractal shape with no right angles? Is that feasible? Or do you just approximate it with a fractal arrangement of rectangles?
- deleted 3y ago
  [deleted]
- calamari4065 3y ago
  I can't say I've ever done calculus on a fractal, but my guess is that most fractals have a fixed surface area ratio. Since fractals have infinite textural depth, there's a hard limit to their surface area compared to the overall area. But what you're asking about is the inaccuracies in the Reimann Integral https://en.m.wikipedia.org/wiki/Riemann_integral https://en.m.wikipedia.org/wiki/Riemann_integral The short version is that you pick a reasonably small size for your rectangles, and it mostly averages out. The discrepancy is so small that it can largely be ignored. But for argument's sake, you can simply imagine infinitely many infinitely small rectangles. That will fit the curve perfectly, but is impractical to compute. So we accept the slightly less accurate value.
- fspeech 3y ago
  A fractal has measure zero in the space it's embedded in. It obviously can not contain any object of nonzero measure of that space.
  - Filligree 3y ago
    Does that include space-filling curves? …do those count as fractals, if the fractal dimension would be an integer?

Horffupolde 3y ago

Riemannian calculus.

jvanderbot 3y ago
Of course. My mistake.

calamari4065 3y ago

This is pretty much how integration works, and calculus shows us that integrating any arbitrary function can be done this way. But really isn't not that hard to intuit. Take any arbitrary shape and draw any number of arbitrarily small rectangles inside it. It should be easy to see that by adding the area of each rectangle, you approach the area of the shape. As the rectangles get smaller, the combined area of the rectangles gets closer to the true area of the shape.