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from the article: The volume of the unit d-sphere goes to 0 as d grows! A high dimensional unit sphere encloses almost no volume!

Maybe I'm nitpicking, but this interpretation is not accurate IMO. Volume of N-dimensional sphere is measured is different units than that of (N-1)-dimensional sphere. E.g. one is m^5, and another is m^4. Comparing values measured in different units is not the best idea. It would be better worded as : the ratio of volume of N-dimensional sphere to the volume of same-dimensional cube approaches zero when N goes to infinity. But this is not a counter-intuitive statement.




A more precise, unit-respecting result is to plot the ratio of the unit sphere's volume to the unit cube's volume. This is a dimensionless quantity, so it makes sense to compare them, but happily the numeric values are unchanged because the unit cube always has volume 1 m^N in dimension N.


The ‘pithy’ version of this idea is that most of a high-dimensional orange is the peel.

It’s from this excellent article by Pedro Domingos: https://homes.cs.washington.edu/~pedrod/papers/cacm12.pdf


This is actually slightly different issue.

Regardless of the dimension, the volume of n-ball is V_n(r) = c_n * r^n, where r is radius, and c_n is a constant depending on n. The grandparent observation is that surprisingly enough, c_n goes (rather quickly) to 0 as n goes to infinity, so for unit n-ball, that is, a ball of radius 1, the volume is exactly c_n, which is very small. This is surprising to us, because in familiar case of n = 3, c_3 = 4/3 pi, which is moderately large.

As for the mass being concentrated around the peel, suppose we have an orange of radius R+e, and its peel has thickness of e. Then, the ratio of volume of the peel to the volume of the whole orange is exactly:

(V_n(R+e) - V_n(R))/V_n(R+e) = 1 - V_n(R)/V_n(R+e) = 1 - c_n R^n / c_n (R+e)^n = 1 - R^n/(R+e)^n = 1 - (R/(R+e))^n

Now, since R/(R+e) < 1, for large n, (R/(R+e))^n will be very small, and so the ratio of the volume of the peel to the volume of the whole orange will be 1 minus something very small, so close to 1.

Note that the peel argument works just as well with "square" oranges -- they also have most of their mass concentrated around the peel, but contrary to round oranges, their whole mass does not go to 0 as the dimension increases. To see that, note that the volume of n-dimensional square with side of R is exactly R^n, and if you do the above computation, it's exactly the same (note that the constant c_n cancelled out anyway).

In this sense, your comment and the grandparent ones are about two different phenomenons -- grandparent is talking specificly about the geometry of the sphere in L_2 norm, while you are talking generally about n-dimensional volumes.


That's what I thought reading that too. In general I don't think it's a good idea for maths people to just forget about units ("because it is for physics")


Speaking about math: I'm not even sure what the meaning of polynomial is. E.g. x^2+3x+1. Each component is measured in different units. Is it intuitive?


But such polynomials do turn up in physics, e.g. accelerated motion: x(t) = 1/2 a t^2 + v t + x0. The trick is that the coefficients aren't unitless either.


Such a polynomial only makes sense when x is a dimensionless quantity, or alternatively when the coefficients have appropriate dimensions to compensate.


The definition of a polynomial has nothing to do with units of measurement. The coefficients come from a ring and a ring has nothing g to do with units of measurement.


That’s fine as a pure mathematics perspective, but in physics dimensions come into play often, and one often has polynomials whose domain and coefficients come “tagged” with particular dimensions.


OP did say, “speaking about math...”. The operations on polynomials and dealing with polynomials doesn’t have anything to do with units of measurement. It may be the case that when used in some areas and in some contexts that the units matter but I think it clouds issues to bring them up.

The intuition for operating with polynomials is best obtained by not worrying about units.


I think we’re not really disagreeing about anything. In a pure math context, x would be, as you say, a quantity in which dimensions play no role.


As he said, x is part of a ring. In physics the multiplication of two elements from the same ring end up on another ring (meters^2 for example) in which, for example, you cannot do addition with elements from the original ring (meters)


The mathematical definition of a ring precludes this from happening. By definition, in a ring when you multiply 2 elements from the ring you get another element of the ring. Perhaps you are referring tensor products (geometric algebra for physicists).


I admit that I have no intuition of a product of 2 dimensionless quantities :)


That would just be like a double-rescaling. Like "a meter is now twice as long as double a normal meter."


In abstract algebra, polynomials are just tuples (a_n, ... a_0) with addition (simple elementwise) and multiplication (more involved) defined. The +s and xs in the "x²+3x+1" notation are just syntax without semantic relevance; in particular x is not a variable and the substitution a ↦ P(a), a ∈ 𝗦 for any a and 𝗦 is not defined. This is because most interesting things about polynomials can be reasoned about without assuming anything about what x is.


Yes. But I've learned from bitter experience that mathematicians don't believe in units. At least not in their gut. To them, it's all just numbers.

(Even more infuriating is that physicists do believe in units, except when writing code).


Would it be counter intuitive though, that the ratio increases until dimension 5 (I think) then starts decreasing?


It's a matter of one's individual intuition, but said ratio is a non-linear function, and non-linear functions most often are non-monotonous.


Except exponential, log, log-linear, sqrt, the time or space complexity of pretty much every problem in CS...


And all cumulative distribution functions, x^k where k is odd, all functions F(x) = integral(f(x') dx', -inf, x) where f(x) is continuous..


Exactly. The article is really bad and serves only to confuse those with no experience in high-dimensional mathematics.




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