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> The angle and cosine start to lose their geometric intuition when we go beyond 3D ... they do? "Geometry" in general loses intuition beyond 3D, but apart fr

> The angle and cosine start to lose their geometric intuition when we go beyond 3D ... they do? "Geometry" in general loses intuition beyond 3D, but apart from that, angles between two vectors are probably the one thing that still remains intuitive in higher dimensions (since the two vectors can always be reduced to their common plane).

Certhas 5y ago
Even angles behave very counter-intuitively in high dimensions. E.g. in high dimensional spaces uniformaly randomly chosen vectors always have the same inner product. Why? Sum x_i y_i is a sum of iid random variables, so the variance goes to zero by the central limit theorem.
- bollu 5y ago
  I would say that this is intuitive. For any direction you pick, there are (n-1) orthogonal directions in nD space. It's only natural that the expected inner product drops to zero.
- atq2119 5y ago
  The variance goes to 0 only if you normalize, which is to say that two random high-dimensional vectors are very likely to be close to orthogonal (under mild assumptions on their distribution). I agree that that's one of those important but initially unintuitive facts about high dimensions. Just like almost all of the volume of a reasonably round convey body is near its surface. But it also doesn't really contradict the GP comment.
  - 10000truths 5y ago
    > Just like almost all of the volume of a reasonably round convey body is near its surface. I’d say that’s pretty intuitive for anyone who can see a pattern in surface area to volume ratios. 1D ball: 2 / (2 * r) = 1/r 2D ball: (2 * pi * r) / (pi * r^2) = 2/r 3D ball: (4 * pi * r^2) / (4/3 * pi * r^3) = 3/r nD ball: ... = n/r
    - atq2119 5y ago
      Most people don't find arguing from formulas intuitive unless the formulas themselves are intuitive. If you truly believe they are, I'd be curious to know why.
      - krastanov 5y ago
        There is an intuitive version of this. Volume in n dimensions is C*r^n (C is some constant) and surface is the first derivative, leading to a ratio of n/r (the C constant cancels out). Hmm... Maybe not that intuitive
        kazinator 5y ago
        But the former formula already tells you that most of the volume is near the high range of r: that which was to be shown. The surface area to volume concept adds nothing. Because the volume of a sphere is proportional to r cubed, you know there is much more volume between r in [0.9, 1.0] than in the same sized interval of r [0.0, 0.1]. You can find the break-even point almost in your head. At what r value is half the volume of a R = 1.0 sphere below that value? Why, that's just the cube root of 1/2 ~= 0.794. So almost half the volume is within 20% of the radius from the surface. That's far from the claim that almost all the volume is near the surface: half is not almost all, and 20% isn't all that near. However, you can see how it gets nearer and nearer for higher dimensions. For a ten dimensional sphere, the tenth root of 1/2 is ~ 0.933. So over half the volume of a ten dimensional sphere is within the 7% depth.
        10000truths 5y ago
        The surface area to volume ratio is just a limit of the shell volume to total volume ratio as the shell thickness goes to zero. So both should asymptotically scale with higher dimensions in the same way.

Certhas 5y ago

Even angles behave very counter-intuitively in high dimensions. E.g. in high dimensional spaces uniformaly randomly chosen vectors always have the same inner product. Why? Sum x_i y_i is a sum of iid random variables, so the variance goes to zero by the central limit theorem.

bollu 5y ago
I would say that this is intuitive. For any direction you pick, there are (n-1) orthogonal directions in nD space. It's only natural that the expected inner product drops to zero.
atq2119 5y ago
The variance goes to 0 only if you normalize, which is to say that two random high-dimensional vectors are very likely to be close to orthogonal (under mild assumptions on their distribution). I agree that that's one of those important but initially unintuitive facts about high dimensions. Just like almost all of the volume of a reasonably round convey body is near its surface. But it also doesn't really contradict the GP comment.
- 10000truths 5y ago
  > Just like almost all of the volume of a reasonably round convey body is near its surface. I’d say that’s pretty intuitive for anyone who can see a pattern in surface area to volume ratios. 1D ball: 2 / (2 * r) = 1/r 2D ball: (2 * pi * r) / (pi * r^2) = 2/r 3D ball: (4 * pi * r^2) / (4/3 * pi * r^3) = 3/r nD ball: ... = n/r
  - atq2119 5y ago
    Most people don't find arguing from formulas intuitive unless the formulas themselves are intuitive. If you truly believe they are, I'd be curious to know why.
    - krastanov 5y ago
      There is an intuitive version of this. Volume in n dimensions is C*r^n (C is some constant) and surface is the first derivative, leading to a ratio of n/r (the C constant cancels out). Hmm... Maybe not that intuitive
      - kazinator 5y ago
        But the former formula already tells you that most of the volume is near the high range of r: that which was to be shown. The surface area to volume concept adds nothing. Because the volume of a sphere is proportional to r cubed, you know there is much more volume between r in [0.9, 1.0] than in the same sized interval of r [0.0, 0.1]. You can find the break-even point almost in your head. At what r value is half the volume of a R = 1.0 sphere below that value? Why, that's just the cube root of 1/2 ~= 0.794. So almost half the volume is within 20% of the radius from the surface. That's far from the claim that almost all the volume is near the surface: half is not almost all, and 20% isn't all that near. However, you can see how it gets nearer and nearer for higher dimensions. For a ten dimensional sphere, the tenth root of 1/2 is ~ 0.933. So over half the volume of a ten dimensional sphere is within the 7% depth.
        10000truths 5y ago
        The surface area to volume ratio is just a limit of the shell volume to total volume ratio as the shell thickness goes to zero. So both should asymptotically scale with higher dimensions in the same way.