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Fix error: use radius=1/2 instead of unity n-ball

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edited Dec 10 2009 at 8:42

A sphere packing argument (and some kissing number construction), because having smaller n-balls is equivalent to be able to pack more of them in the unity n-cube.

edit: This needed to use r=1/2, not unity n-balls. Fixing the associated values. This hurts the usefulness of the argument, but still give some geometric insight on how the n-ball fills the n-cube and the shape of the space between them.

First, we take n=16 n=4 and show how to put two unity n-balls $B_n$ having a radius of 1/2 into the unity n-cube $C_n$: we place the first at the center of the n-cube, and use this as origin. Then we place the second at (1/2,1/2,...,1/2) 1/2,1/2,1/2,1/2) and wrap it into the n-cube. These two $B_n$ are now into $C_n$ (even if one can be considered as split in parts) and are disjoints because the distance between them is $\sqrt{16(1/2)^2} \sqrt{4(1/2)^2} = 2$1$.

This proves that the volume $V(B_{16}) V(B_4) < 1/2$.

Now, when n=16kn=4k, notice that we can place k other $B_n$ plus the one at center by using the same type of translation as before. For the first, we only set the 16 4 first coordinates as 1/2 and the rest as 0. For the second, only the 16 4 next ones, etc... All these additional $B_n$ are at distance 2 1 from the centered one, and at distance 4 $\sqrt{2}>1$ from each other, making them all disjoint. Thus, we have

$V(B_{16k}) V(B_{4k}) < 1/(k+1)$,

which goes to 0 when n=16k n=4k increases. Of course, when n=16k+pn=4k+p, the same trick still work, but only with k+1 n-balls, which is not a problem.

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answered Dec 9 2009 at 16:57

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A sphere packing argument (and some kissing number construction), because having smaller n-balls is equivalent to be able to pack more of them in the unity n-cube.

First, we take n=16 and show how to put two unity n-balls $B_n$ into the unity n-cube $C_n$: we place the first at the center of the n-cube, and use this as origin. Then we place the second at (1/2,1/2,...,1/2) and wrap it into the n-cube. These two $B_n$ are now into $C_n$ (even if one can be considered as split in parts) and are disjoints because the distance between them is $\sqrt{16(1/2)^2} = 2$.

This proves that the volume $V(B_{16}) < 1/2$.

Now, when n=16k, notice that we can place k other $B_n$ plus the one at center by using the same type of translation as before. For the first, we only set the 16 first coordinates as 1/2 and the rest as 0. For the second, only the 16 next ones, etc... All these additional $B_n$ are at distance 2 from the centered one, and at distance 4 from each other, making them all disjoint. Thus, we have

$V(B_{16k}) < 1/(k+1)$,

which goes to 0 when n=16k increases. Of course, when n=16k+p, the same trick still work, but only with k+1 n-balls, which is not a problem.