3 ms·
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 counterin
by evanb 1y 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 1y agoImagine you want to compare sizes of video game levels, where some are 2D, made from pixels, and some are 3D, made from voxels. You could stipulate that n pixels are equivalent to sqrt(n) voxels. But you could also stipulate that n pixels are equivalent to n voxels. I don't think either is more correct than the other.
- evanb 1y agoWhich is bigger: a meter, a hectare, or a liter?
- cubefox 1y agoA hectare
- evanb 1y agoWhat if the property with 1 hectare of area were 0.1m x 10^5 m? The meter is bigger than the field is wide. Or, more to the point, suppose someone came with a different system of units and said: look, one of our standard lengths is 10^-6 of one of yours, but one of our standard areas is defined just like yours: 1 ha = (100m)^2 and (1 of our areas) = (100 of our lengths)^2. In other words, their standard length = 10^-6m; their area = 10^-12 ha = 10^-8 m^2. Is their standard area bigger or smaller than their standard length? Notice that in proportion the lengths and areas are the same.