4 ms·

"very far" is of course relative: if we have tree vectors, two of length R and one of length 0.99*R, it's not outlandish to call the distance 2R between the two

"very far" is of course relative: if we have tree vectors, two of length R and one of length 0.99*R, it's not outlandish to call the distance 2R between the two vectors of equal magnitude "very large" compared to the distance 0.01R between two vectors of dissimilar magnitude. Your last comment is completely incorrect, for a point at (1,1,1,....) each extra dimension adds a constant 1 to the euclidean distance, so that in 500 dimensions a point at (1,1,1,....) is around 22.4 units away from the origin, while in two dimensions it is only 1.4 units away from the origin.

__s 3y ago
https://www.youtube.com/watch?v=zwAD6dRSVyI https://www.youtube.com/watch?v=zwAD6dRSVyI 3Blue1Brown on visualizing higher dimensions explains it well
thaumasiotes 3y ago
> Your last comment is completely incorrect How so? Your followup makes no sense. > for a point at (1,1,1,....) each extra dimension adds a constant 1 to the euclidean distance, so that in 500 dimensions a point at (1,1,1,....) is around 22.4 units away from the origin, while in two dimensions it is only 1.4 units away from the origin. You're comparing vectors of different magnitudes. You could equally object that (200, 0) is much farther away from the origin than (2, 0) is. That's true, but so what? You're still in a two-dimensional space. Are you under the impression that the "magnitude" of a vector and its "distance from the origin" are separate concepts? They aren't.
- AnthonyMouse 3y ago
  Consider simple two-dimensional space. A point at (1,0) is 1 unit away from the origin, as is a point at (0,1). But a point at (1,1) is approximately 1.4 away from the origin, i.e. sqrt(1^2 + 1^2). See Pythagorean theorem.
  - thaumasiotes 3y ago
    Yes, what's your point? Vectors with larger magnitudes have larger magnitudes than vectors with smaller magnitudes do? If you're going to defend the idea that something I said was incorrect, maybe you should have some idea of what it was?
    - AnthonyMouse 3y ago
      You keep referring to the magnitude of the vector itself rather than the magnitude of its components. > Vectors with larger magnitudes have larger magnitudes than vectors with smaller magnitudes do? Vectors with more dimensions have larger magnitudes than vectors with fewer components, for the same average magnitude of the components. The distance between the origin and (1,1) is less than the distance between the origin and (1,1,1) even though the components in both cases all have magnitude 1.
      - SiempreViernes 3y ago
        I think their point boils down to the fact that you can require that all vectors have the same magnitude, irrespective of the dimensionality of the space, which is of course true. The next step is them doing a black knight and pretending they didn't put in the requirement by hand.
        thaumasiotes 3y ago
        Here's what you said: > Your last comment is completely incorrect, [random gibberish] Here's what you were referring to: >> If you stick to the first quadrant / octant / whatever n-dimensional division of space where all coordinates are positive... I don't think the number of dimensions makes any difference there either? Any two vectors define a plane (or a line, or, if they're both zero, a point), so two vectors in a 500-dimensional space can't be farther apart from each other than is possible for two vectors in a 2-dimensional space. Those 500-dimensional vectors are already embedded in a 2-dimensional space. All of those statements are, obviously, true. What did you think was incorrect?
      - 3y ago

__s 3y ago

https://www.youtube.com/watch?v=zwAD6dRSVyI https://www.youtube.com/watch?v=zwAD6dRSVyI 3Blue1Brown on visualizing higher dimensions explains it well

thaumasiotes 3y ago

> Your last comment is completely incorrect How so? Your followup makes no sense. > for a point at (1,1,1,....) each extra dimension adds a constant 1 to the euclidean distance, so that in 500 dimensions a point at (1,1,1,....) is around 22.4 units away from the origin, while in two dimensions it is only 1.4 units away from the origin. You're comparing vectors of different magnitudes. You could equally object that (200, 0) is much farther away from the origin than (2, 0) is. That's true, but so what? You're still in a two-dimensional space. Are you under the impression that the "magnitude" of a vector and its "distance from the origin" are separate concepts? They aren't.

AnthonyMouse 3y ago
Consider simple two-dimensional space. A point at (1,0) is 1 unit away from the origin, as is a point at (0,1). But a point at (1,1) is approximately 1.4 away from the origin, i.e. sqrt(1^2 + 1^2). See Pythagorean theorem.
- thaumasiotes 3y ago
  Yes, what's your point? Vectors with larger magnitudes have larger magnitudes than vectors with smaller magnitudes do? If you're going to defend the idea that something I said was incorrect, maybe you should have some idea of what it was?
  - AnthonyMouse 3y ago
    You keep referring to the magnitude of the vector itself rather than the magnitude of its components. > Vectors with larger magnitudes have larger magnitudes than vectors with smaller magnitudes do? Vectors with more dimensions have larger magnitudes than vectors with fewer components, for the same average magnitude of the components. The distance between the origin and (1,1) is less than the distance between the origin and (1,1,1) even though the components in both cases all have magnitude 1.
    - SiempreViernes 3y ago
      I think their point boils down to the fact that you can require that all vectors have the same magnitude, irrespective of the dimensionality of the space, which is of course true. The next step is them doing a black knight and pretending they didn't put in the requirement by hand.
      - thaumasiotes 3y ago
        Here's what you said: > Your last comment is completely incorrect, [random gibberish] Here's what you were referring to: >> If you stick to the first quadrant / octant / whatever n-dimensional division of space where all coordinates are positive... I don't think the number of dimensions makes any difference there either? Any two vectors define a plane (or a line, or, if they're both zero, a point), so two vectors in a 500-dimensional space can't be farther apart from each other than is possible for two vectors in a 2-dimensional space. Those 500-dimensional vectors are already embedded in a 2-dimensional space. All of those statements are, obviously, true. What did you think was incorrect?
    - 3y ago