Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

Very counterintuitive. I understand how the proportional space of an n-ball bounded in an n-cube can go to zero, but I don't understand how the volume of an n+1-ball can be of a smaller volume than an n-ball.


> Very counterintuitive. I understand how the proportional space of an n-ball bounded in an n-cube can go to zero, but I don't understand how the volume of an n+1-ball can be of a smaller volume than an n-ball.

It can't be, in the same sense that it's not true that π square meters (the area of a unit circle) is less than (4/3)π cubic meters (the volume of a unit sphere); they are simply not comparable quantities. What is true, as wging very nicely put it (https://news.ycombinator.com/item?id=31349002), is that the (n + 1)-ball (of radius 1) eventually takes up a smaller fraction of the (n + 1)-cube (of side 2) than the n-ball does of the n-cube.

I agree that this is still probably sufficiently unintuitive that the best one can do is to argue why one shouldn't disbelieve it, not to try to make it intuitive; but nonetheless I will share one of the closest things I've seen to an explanation, which is that the number of corners in a cube grows exponentially with the dimension of the cube, so that, sooner or later, the cube is "mostly corners"—and the ball does not poke into the corners.


> volume of an n+1-ball can be of a smaller volume than an n-ball

Volume is the ratio of the n-ball to the n-cube.

Saying that n+1-ball has “smaller volume” than the n-ball is equivalent to saying that the ratio of the n+1-ball to the n+1-cube is smaller than the ratio of the n-ball to the n-cube.

> I understand how the proportional space of an n-ball bounded in an n-cube can go to zero

The volume comment is saying that the ratio of the n-ball to its bounding n-cube goes to zero faster than (base-2) exponential:

The bounding n-cube is 2^n times the unit n-cube.


How about this example:

For r=1

A two-dimensional n-ball is a circle. If you extend it into 3 dimensions, by default it is a cylinder, of h=1 and r=1. That’s not a 3-dimensional n-ball however.

If you increase the dimensions of that 2-dimensional n-ball into 3 dimensions, it becomes a 3-dimensional n-ball—-a sphere of r=1.

And now you have a sphere that is smaller than the circle that was extended into 3 dimensions as a cylinder. The ball-ness of the n-ball formula chopped off the corners of the cylinder.

And that chopping off continues in higher dimensions.


The same was as how no circles can, not matter how many of them there are, fill a cylinder.


I mean, we can directly calculate the volume of a cylinder by modeling it as a big stack of circles and adding up their areas. That's the whole idea of integral calculus.

If you have enough circles, they will form a cylinder.




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: