Ratio of n-volumes of hypercube and hyperball

Higher-dimensional geometry (previously "Polyshapes").

Ratio of n-volumes of hypercube and hyperball

Postby quickfur » Tue Oct 31, 2017 12:54 am

It's well(?)-known that the volume of a unit n-ball (i.e., a "filled" (n-1)-hypersphere in n dimensional space) peaks between n=5 and n=6, with n=5 having the largest value for an integer number of dimensions. The problem with this, is that this result is dependent on the radius of the n-ball: for larger radii, the peak actually can happen in a higher dimension than 5.

So what does this peak actually signify, then? Perhaps a better formulation of the result is in terms of the ratio of the n-volumes of ab n-ball of radius r, and an n-cube whose edge length is equal to r. Note that this is not the same thing as an n-cube that inscribes the ball, because the diameter of the ball is 2r, whereas the edge length of the cube is r, so the ball always protrudes from the cube if both are placed at the origin, for all dimensions above 0. But using the unit cube whose edge length is equal to the ball's radius allows us to make a more meaningful interpretation of the unusual behaviour of the 5-ball's 5-volume in 5D, because we can think of the unit cube as a "unit of n-space" and ask, how many units of n-space can we fit into an n-ball whose radius equals the edge length of this unit?

In low dimensions, some rather interesting things happen: in 0D, since the 0-cube and the 0-ball are identical, their volume ratio is just 1. In all higher dimensions, the unit n-cube has n-volume 1, but the n-ball's volume changes in interesting ways.

In 1D, the 1-cube is just a unit line segment of length 1, but the 1-ball has a line segment of radius 1, so it has length 2. So the ratio of the 1-volume of the 1-ball to the 1-cube is 2.

In 2D, the 2-cube is the square of area 1, and the 2-ball is the circle of radius 1, and we know that the area of this circle is pi, which is approximately 3.14159....

What about 3D? The unit cube has volume 1, of course, and the unit 3-ball has volume (4/3)pi = approx 4.189....

Then in 4D, the 4-ball has volume 4.935...;

... and in 5D, the 5-ball has volume (8/15)*pi2 = approx 5.264.

So from 0D to 5D, we see a steady increase of the volume of the n-ball from 1 to about 5.264. One might expect that the volume of the n-ball would grow, but interestingly enough, it doesn't. After 5D, the volume ratio of the n-ball to the n-cube begins to shrink:

In 6D, the volume of the 6-ball is pi3/6 = approx 5.168, which is very close to the value for 5D, but somewhat smaller.

In 7D, the volume of the unit 7-ball begins to shrink further to (16/105)pi3 = approx 4.725. Now the ratio has dropped below the 4D value.

This strange behaviour is because as you ascend the dimensions, the n-cube's volume becomes more and more concentrated around its vertices, and the number of vertices is growing exponentially (2n). Initially, the larger diameter of the n-ball makes its volume beat the n-cube by quite a bit; but as the dimensions increase, this advantage begins to lose out to the rapidly growing n-volume accumulating around the vertices of the n-cube.

So far, though, the n-ball still has a larger volume than the n-cube. Even though it's slowing down, the n-cube hasn't quite caught up yet. As we go on to 8 D, the volume ratio drops to 4.059, then in 9D, it drops further to 3.299, and in 10D, it dips to 2.550, which is now below the ratio for the 2D case.

But something more interesting happens between 12D and 13D: in 12D, the 12-ball still has a larger volume (1.335) than the 12-cube, but once we get to 13D, we find the the 13-ball has a smaller volume than the 13-cube! Now the volume of the n-ball has dropped to approx 0.9106, below its starting point at 0D with 1 unit. IOW, the unit 13-sphere is too small to contain 1 unit of 13-space! This is quite interesting, because its radius is 1, meaning it has diameter 2, so it still protrudes from each of the unit 13-cube's 26 facets (if both are placed at the origin) by 1/2 units, but all of this extra volume is unable to compensate for the volume concentrated around the 213 = 8192 vertices of the 13-cube.

Then when we go to 14D, we find that the 14-ball's volume is approx 0.5993, only a little more than half the volume of the 14-cube!

And in 15D, the 15-ball's volume is approx 0.3814, now less than half of the volume of the 15-cube.

In 16D, the n-ball's volume has dropped to 0.2353, less than 1/4 of the volume of the 16-cube.

By the time we get to 24D, the 24-ball's volume is only about 0.0019 of the volume of the 24-cube! And to put this into perspective: the 24-ball protrudes by 1/2 unit from each of the 48 facets of the 24-cube. That's quite a lot of protrusion! But it is woefully inadequate to counteract the huge amount of volume concentrated around the 24-cube's 16.7 million vertices. :lol: :lol:

And of course, the further you go, the smaller the n-ball's volume becomes. Eventually, in the limiting case, the volume of the n-ball approaches 0 as n grows to infinity, while the unit n-cube's volume remains at 1.

Of course, all of this is well-known... except perhaps for the interesting fact that 13D is the first dimension in which the volume of the unit n-ball is smaller than the volume of the unit n-cube.
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Re: Ratio of n-volumes of hypercube and hyperball

Postby quickfur » Tue Oct 31, 2017 1:12 am

Also, another detail that's interesting:

In 1D, the vertices of the 1-cube (i.e., line segment of length 1) lies completely inside the 1-ball (i.e., line segment of length 2).

In both 2D and 3D, the n-cube's vertices still lie strictly inside the volume of the n-ball.

However, in 4D, something interesting happens: the circumradius of the tesseract of edge length 1 is exactly 1, meaning that the vertices now lie on the surface of the 4-ball! IOW, the 4-ball inscribes the 4-cube.

Then in 5D, the vertices of the 5-cube begin to protrude from the 5-ball. The circumradius of the 5-cube is about 1.118, now slightly larger than the radius of the 5-ball. So we see something interesting happen in going from 4D to 5D, in that the n-cube is now beginning to stick its corners out of the n-ball. :D

In 6D, the circumradius of the 6-cube is approx 1.225, and now the volume of the 6-cube begins to catch up with the volume of the 6-ball.

In 12D, just before the volume of the n-cube overtakes the volume of the n-ball, the circumradius of the n-cube is 1.732... (or, to be precise, exactly √3).

In 13D, the circumradius grows to approx 1.803, and now the volume of the n-cube surpasses the volume of the n-ball.

When we get to 24D, the circumradius is approx 2.449, more than twice the radius of the n-ball. And there are 16 million vertices protruding from the 24-ball here. No wonder the 24-ball's volume is only 0.0019 of the 24-cube!

And of course, as the number of dimensions grows without bound, the circumradius of the n-cube approaches infinity. The n-ball has no hope to catch up. :XD: This is all very interesting, because the facets of the n-cube remain at exactly 1/2 units from the origin (measured to the centroid of the facet). The n-ball still protrudes out of each of the 2n facets by 1/2 units each, but it's just that the n-cube has so many more vertices (2n of them!) protruding out of the n-ball. :lol:
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