Comment by cubefox

Comment by cubefox 17 hours ago

6 replies

Also interesting:

The area of the Euclidean unit disk (area of a circle with radius 1) is equal to π. However, the volume of the unit ball (ball with radius 1, one dimension more) is larger, namely 4/3π ≈ 4.18879. So how does the hypervolume for unit n-balls change in higher dimensions, where the 1-ball is a circular disk and the 2-ball an ordinary ball?

Surprisingly, it first increases but then converges to 0. The maximum is achieved for a unit 5-ball with a hypervolume of about 5.2638. For higher dimensions the value decreases again.

However: If we allow fractional dimensions, the 5-ball isn't at the peak volume. The n-ball with the largest volume is achieved for n≈5.256946404860577, with a volume of approximately 5.277768021113401, which are slightly larger numbers.

These were computed by GPT-5-thinking, so take it with a grain of salt. But the fractional dimension for peak volume is also reported here on page 34: http://lib.ysu.am/disciplines_bk/8d6a1692e567ede24330d574ac3...

Curiously, the paper above says that the area of the hyper surface of the n-ball (rather than its volume) peaks at n≈7.2569464048, while ChatGPT calculated it as n≈6.256946404860577, so exactly one dimension less than the paper. I assume the paper is right?

Also curiously, as you can see from these numbers, that fractional dimension with the peak hyper surface area is exactly two (according to the paper) or one (according to ChatGPT) dimension larger than the fractional dimension of the peak volume.

evanb 17 hours ago

This is a nice analog but unfortunately I think it breaks down in a way the "π" calculation does not.

In the article 2π(d) = the ratio of the circumference to the radius. This is dimensionless, in the sense that the circumference and the radius are both lengths (measured in meters, or whatever), so 2π(d) is really just a number.

But the (hyper)volumes you're talking about depend on dimension, which is exactly why you say "hyper". In 2 dimensions the volume is the area, πr^2, which has dimensions L^2 [measured in m^2 or whatever]. But in 3 dimensions the volume is 4/3 πr^3, which has dimensions L^3. The 5 dimensional (hyper)volume has dimensions L^5, and so on.

So, "comparing" these to find out which is bigger and which smaller is not really meaningful---just like you shouldn't ask which is the bigger mass: a meter or a second? Neither is, they aren't masses.

  • cubefox 16 hours ago

    Yeah, though they are somewhat similar. We could perhaps say the volume of the unit ball is (in some sense) "larger" than the area of the unit disk, because both are measured relative to a radius of 1. So the area of the disk is fewer "units" than the volume of the ball.

    Anyway, it seems independently interesting that this value peaks for the 5-ball, or the ~5.2569-ball. The non-fractional difference between fractional dimension of peak hyper volume and peak hyper surface area seems also interesting. (I assume there is some trivial explanation for this though.)

    • srean 16 hours ago

      No it doesn't work that way because the units of hyper-volume and length are different.

      However, once you take appropriate roots of hypervolume to get same units you can safely compare. Or the otherway round take appropriate powers of length to get same units as hypervolume.

      • evanb 15 hours ago

        I agree; the fair comparison is the nth root of the hypervolume in n dimensions, (V(n))^(1/n). This monotonically decreases from n=0, which shows the counterintuitive point that people often want to make anyway: an n-sphere takes up less and less of an n-hypercube. The peak at n~=5 is illusory.

        Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.

      • cubefox 16 hours ago

        With that it would probably mean the units monotonically approach 0, rather than first increasing and then decreasing. At least for whole dimensions. I'm not sure about monotonicity with fractional dimensions.

NooneAtAll3 5 hours ago

I hate this result because it's not actually saying what it's advertized to be saying

sure, you already got the complaint about comparing values of different units - but observe HOW this question is actually sidestepped! We divide hyper-volume of n-sphere by hyper-volume of n-cube!

Now this raises the question: WHAT n-cube are we taking?

If you take hyper-cube with side-length of sphere's diameter, you'll have nice relation between cube and its inscribed sphere - and, predictably, as n goes up, number of cube's "corners" also goes up. So this ratio consistently goes down

But what about your numbers? Well that result happens when you take cube with side-length of sphere's RADIUS. That way you arbitrarily add a scaling factor 2^n - and there's nothing geometric about this behaviour