Heady with success in 4-space, I embarked enthusiastically upon a general program for n-space --- only to recall, yet again, that it is better to travel hopefully than to arrive. The 5-space case is truly horrible: there are evidently multiple local maxima all with side approximately 1.5, corresponding to simplex vertices lying on various combinations of boundary elements of the hypercube. While one simplex vertex always nestles securely in a hypercube vertex, just deciding whether other simplex vertices actually meet a hypercube line, plane etc. is in practice difficult because of ill-conditioning. My most persuasive candidate so far in 20 attempts, by reason of symmetry, has also the second largest known side = 1.510898923478 ; its vertices are [1, 1, 1, a, b, c], [1, c, b, 1, a, 1], [1, a, c, b, 1, 1], [1, 1, a, c, 1, b], [1, b, 1, 1, c, a], [1, 0, 0, 0, 0, 0], where a := 0.003086685229, b := 0.089689781542, c := 0.524177233794; though these constants don't appear to be anything simply algebraic. But alas for elegance --- a much gnarlier specimen has just turned up with side = 1.511084, making it fairly certain that the last word remains to be spoken ... Fred Lunnon On 9/21/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
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