4 ms·
Because the location of the peak falls at a different dimension depending on the underlying units (i.e., the radius), I don’t think the numerical location of th
by mturmon 4y ago
Because the location of the peak falls at a different dimension depending on the underlying units (i.e., the radius), I don’t think the numerical location of the peak is fundamental. So maybe it’s best that you don’t have intuition for it?
- dmurray 4y agoThe ratio of the volume of the sphere to the volume of the enclosing cube (the proportion of the box that it takes up, or the probability that a randomly chosen point inside the box is in the ball) is dimensionless, though. It's not dependent on the size of the cube.
- mturmon 4y agoYou can extract dimensionless quantities, but depending on the normalization, the peak will land in different "n" values. E.g., normalize by the enclosing cube (sides of length 2R) versus the enclosed cube. So I don't think there's anything special about the turnover point.