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Topic: Volume of n-Simplex (Read 4800 times) |
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ThudnBlunder
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Volume of n-Simplex
« on: Nov 30th, 2006, 5:03pm » |
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An n-simplex is an n-dimensional equivalent of the regular tetrahedron. That is, it has n+1 vertices, all of which are equidistant from each other. Thus a 1-simplex is a line segment, a 2-simplex is an equilateral triangle (area = sqrt(3)/4), and a 3-simplex is a regular tetrahedron (volume = sqrt(2)/12), etc.
For which values of n is the hypervolume of an n-simplex with unit side length rational?
(Jock would like this one but he seems to have been absent for a while.)
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| « Last Edit: Nov 30th, 2006, 6:53pm by ThudnBlunder » |
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Barukh
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Re: Volume of n-Simplex
« Reply #1 on: Dec 1st, 2006, 8:51am » |
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n = 1?
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ThudnBlunder
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Re: Volume of n-Simplex
« Reply #2 on: Dec 2nd, 2006, 6:01am » |
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In fact, there are more.
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Eigenray
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Re: Volume of n-Simplex
« Reply #3 on: Dec 3rd, 2006, 12:52am » |
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Let the vertices of an (n-1)-simplex Sn-1 in Rn-1 be x1,...,xn, and suppose x1+...+xn=0. By symmetry, say |xi|=r for all i, and <xi,xj> = s whenever i != j. Then from
0 = |x1+...+xn|2 = [sum]i,j <xi,xj> = n*r2 + n(n-1)*s
we get s = -r2/(n-1), and then
1 = |xi - xj|2 = 2r2 - 2s = 2r2(1+1/(n-1))
gives r2 = (n-1)/(2n). Let d = sqrt(1-r2) = sqrt((n+1)/(2n)). Then the origin, together with the points {(xi, d)} in Rn, are the vertices of an n-simplex Sn with unit side lengths.
For t in [0, d], the intersection of Sn with the hyperplane { (x,t) : x in Rn-1 } is a copy of Sn-1, scaled by a factor of (t/d), so has (n-1)-volume (t/d)n-1Vn-1, where Vn-1 is the volume of the original (n-1)-simplex. Hence the n-volume of Sn is given by
Vn = [int]0d (t/d)n-1 Vn-1 dt
= Vn-1*d/n = Vn-1*sqrt[(n+1)/(2n)]/n,
and since V1 = 1, we find inductively
Vn = sqrt[(n+1)/2n]/n!.
This is rational iff (n+1)2n is a perfect square. If n is even, then we need n+1 to be a square, and if n is odd, then we need n+1 to be twice a square. Hence n is of the form (2k+1)2-1 or 2k2-1. That is,
n=1, 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, ....
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Barukh
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Re: Volume of n-Simplex
« Reply #4 on: Dec 3rd, 2006, 4:09am » |
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Yes. Here is a slightly different approach.
The simplest way to put an n-simplex into coordinate system, is to put it into an (n+1)-space, like in this thread. Then, the i-th vertex will have all coordinates 0 except the i-th, which will equal 1. The center of n-simplex is then at (1/(n+1), …, 1/(n+1)), the side length is 21/2, and the circum-radius is Rn = [n/(n+1)]1/2. Therefore, the circum-radius of the unit-side simplex is Rn = [n/2(n+1)]1/2
Next, the in-radius rn satisfies the identity rn2 = Rn2 - Rn-12, and therefore equals [1/2n(n+1)]1/2.
Finally, we have the hyper-volume Vn = rnVn-1(n+1)/n = Vn-1[(n+1)/2]1/2/n, arriving at Eigenray’s formula.
on Dec 3rd, 2006, 12:52am, Eigenray wrote:| n=1, 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, .... |
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BTW, this sequence is not in Sloane’s database. Do you think it’s worth to file an addition form?
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