On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >> So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) . However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it. The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly! The corresponding content V(4) = rt5/24 ~ 0.093169 . Fred Lunnon