(A) As defined, your simplex has only n vertices, not n+1 ?! (B) If the missing simplex vertex was intended to be supplied by the origin, then the contents when n = 2 are respectively 4 and 1/2 ; so f(2) = 8 ?! (C) If also the hypercube was intended to have unit edges, then the contents when n = 3 are respectively 1 and 1/6 ; so f(3) >= 6 ?! Whatever the shapes and edge lengths, their content ratio must inevitably approach zero as n -> oo ; so f(n) -> 0 ?! Beyond my ken, I'm afraid ... WFL On 2/28/19, Dan Asimov <dasimov@earthlink.net> wrote:
Let C(n) denote the 2^n vertices of the n-cube [-1, 1]^n in R^n:
C(n) = {-1, 1}^n
Let S(n) denote the n+1 vertices of the n-simplex in R^(n+1):
S(n) = {e_j | 1 <= j <= n}
where { 0, i ≠ j <e_j, e_k> = delta(i, j) = { { 1, i = j
(so all interpoint distances = sqrt(2)).
Question: --------- For n >= 2, what is the smallest number f(n) of isometric copies of S(n) needed to cover C(n) ???
E.g., f(2) = 2; f(3) = 2, f(4) = ???
Question: --------- What is a nice function asymptotic to f(n) as n —> oo ???
—Dan
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