Volumes of Hypercubes

1. Dec 10, 2011

gmax137

I am reading Julian Havil’s book Nonplussed, and in one chapter he’s discussing hypercubes, he says that the volume of an n-dimensional cube of side length L is L^n; then he goes on to note that as n-> infinity, the volume goes to zero if L<1; volume goes to 1 if L=1, and volume goes to infinity if L > 1. Ok that makes sense to me until I ask the units of L. I mean if I tell you that the side length is one meter, then 1*1*1*…1 =1 alright. Then I say, “oops, I meant one yard, so L= 0.914 meter” so now as n goes to infinity the volume is zero (0.914 * 0.914 * ....-> 0). I can see everything is OK as long as n is some finite number, because then we can say the volume is XXX (meters^n) which is equal to YYY (yards^n) and the difference is just a units conversion (=(m/y)^n). But what happens to the conversion factor “when n goes to infinity”?

2. Dec 10, 2011

homeomorphic

Exercise:

If you have a square of side length L and you scale up lengths by a factor of k, then how does the area get scaled?

That's the issue.

Yes, the limit will depend on what your units are. If this seems strange, you might think of a measurement as telling you how big the ratio of something is with respect to the thing that you decide has a length of 1. So, it is based on an arbitrary choice. Your choice of units of length will determine a choice of units for area, volume, etc, which, in turn, determines how big volumes are, which, in turn, determines what will happen when you take the limit.

3. Dec 10, 2011

homeomorphic

Here's another way of thinking of it. Take one of the edges of a cube and chop in up into k pieces. Then, chop up the big cubes into little cubes with the corresponding side length. The number of cubes will go to infinity as you go to higher dimensions. That is the case when the length is greater than your chosen unit. If it is less than the units you chose--let's say half as big, you can do the same kind of thing. As the dimension goes to infinity, you will need more and more little cubes to build a cube of unit hypervolume, so the ratio is going to zero.