Spheres and cubes in high dimensions

For the discussion of math. Duh.

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antonfire
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Spheres and cubes in high dimensions

Here's a weird thing.

Consider a hypercube in n dimensions, and tuck 2n hyperspheres into the corners, so that they are tangent to each other and to the faces of the cube. Now put a hypersphere centered at the center of the cube, tangent to all those.

square-circle.png (28.51 KiB) Viewed 1315 times

What's the limit, as n goes to infinity, of the ratio of the volume of the central sphere to the volume of the cube?

Edited for picture and clarity.
Last edited by antonfire on Mon Apr 06, 2009 6:02 am UTC, edited 1 time in total.
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quintopia
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Re: Spheres and cubes in high dimensions

I don't understand your set up? What do you mean "tangent to all these?" If you put unit axis-aligned hypercubes centered at the vertices of an axis-aligned unit hypercube centered at the origin (which would satisfy "they all touch each other" since they all contain the origin) then there is no space left in the center of the cubes for a ball to fit so that it is tangent to them all. So. . .just be more specific please?

antonfire
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Re: Spheres and cubes in high dimensions

I added a picture which hopefully makes things more clear.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

GreedyAlgorithm
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Re: Spheres and cubes in high dimensions

I assume the answer is 0 but haven't done any work at all. Actually I'd say my priors are around 90% 0, 9% 1, and 1% something else.

ETA: Sweet, looks like the answer is going to be surprising.
Last edited by GreedyAlgorithm on Mon Apr 06, 2009 1:29 pm UTC, edited 1 time in total.
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HenryS
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Re: Spheres and cubes in high dimensions

Spoiler:
According to http://en.wikipedia.org/wiki/N-sphere, if [imath]V_n[/imath] is the volume of an [imath]n[/imath]-ball of radius [imath]R[/imath] then $\frac{V_n}{R^n} \rightarrow 0$ as [imath]n\rightarrow \infty[/imath]. The largest that any ball that fits inside a cube of side length 2 can be is the ball of radius 1, and so its volume goes to 0, whereas the volume of the cube is [imath]2^n[/imath]. So the ratio goes to 0 even faster.

Unless I'm missing something.

Edit: Ah ok, I had assumed that the central sphere had to be contained inside the cube.
Last edited by HenryS on Mon Apr 06, 2009 2:50 pm UTC, edited 1 time in total.

doogly
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Re: Spheres and cubes in high dimensions

It seems true that the radius of the inner sphere will always be
$R=\frac{1}{4}(\sqrt{n}-1)$
Because the big spheres will have radius 1/2, and the diagonal will have length [imath]\sqrt(n)[/imath]
So then you just need the volume of an n-sphere. This is
$V_n=\frac{\pi^{n/2}R^n}{\Gamma(\frac{n}{2}+1)}$
$V_n=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)} \left[\frac{1}{4}(\sqrt{n}-1)\right]^n$
This goes to infinity as n does. It has a minimum at 263.5! It is pretty neat. I am thinking this is probably wrong now. But, I think you do always want to look at the radius of the inner sphere, and figure it out using a diagonal. Diagonal always has the same length, sqrt(n). I am pretty sure about the radius of the outer spheres always being 1/2. It is of course 3 am.
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phlip
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Re: Spheres and cubes in high dimensions

I got the same result as doogly (except I get a radius of (sqrt(n)-1)/2, not (sqrt(n)-1)/4, but the result is the same).  Actually, I misread before the picture was added, that the corner spheres were centred on the points of the cube, not inscribed like that. But having them inscribed in a cube of side length 1 is the same as having them centred on the corners of a cube of side length 1/2, so (sqrt(n)-1)/4 is correct. [/edit]

Simplified to only the cases where the dimension is even, since the volume-of-a-hypersphere formula is simpler (and if there's a limit for all n, then there will be a limit for even n), and plugging numbers into Python, I get it seeming to increase without bound for the numbers I plugged in.

Of interest: in four  Again, misread: actually, nine [/edit] dimensions, the inner sphere is tangent to the 16  512 [/edit] corner spheres, and tangent to the cube. For higher dimensions, the sphere will protrude from the cube.

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doogly
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Re: Spheres and cubes in high dimensions

I got four because: half the diagonal is sqrt(n)/2. Diameter of the sphere is 1/2. So the half of the diagonal not in the sphere is 1/2(sqrt(n)-1). There are two parts to the not in the sphere part, which are of equal size. Half of that is in the radius of the inner sphere, half is in the corner, so the radius of the inner sphere is (sqrt(n)-1)/4.
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