... parallelepipeds IIoo(u), and by 9 the point set x1 > 1, x2 > l, ... , xn > 1. The difference set S = Q — IIo of all points of Q which are not in IIo, is evidently closed, since IIo, as a sum of open sets, is open, and since Q is closed ...

... parallelepiped. Observe that to each product set there corresponds a unique parallelepiped which is equivalent to it. THEOREM 1. Let J . . . . , Øto be product sets in Z" and let 91,..., 9, be the parallelepipeds which are equivalent ...

... parallelepiped of all points X = (xi, x2, ... , xn) which satisfy the inequalities | x1 - a,u, s#o(2a,u),| x2 ... parallelepipeds IIo(u), and by P = {Po)(u)} the set of all points Po(u). Since, from (4), p(2#.) s #., because both ...

... parallelepiped. Such an equation can be effectively solved by a collocation method. In this case the parallelepiped G is partitioned into small parallelepipeds (cells) and the approximate solution is searched in the form of a function ...

... parallelepipeds, but differ from those of vitamin C in that they are not rectangular. Soluble in water, they taste salty Merck & Co.. Rahw ALUMINUM New Plants and Processes Swing into Action to Meet. VITAMIN C (ascorbic acid) is a white ...

... parallelepiped P, having one vertex at the origin; say. ft,= u. (X. +. P,),. irs/sk. xe Ai The dimensions of P ... parallelepipeds P, and thus on C) each ft,p is a bounded Lipschitz domain; in fact (x + Pj) n (y + P/) * 0 for every ...

... parallelepiped. For the solution of Eq. (1) we shall use the piecewise polynomial collocation method (cf. [1]): the set G will be partitioned into small parallelepipeds (cells) and the approximate solution will be searched in the form ...

... parallelepiped in \(n;pi /;„). Furthermore, any subgroup H C G is of the form H = Q\ X ... X Qn where each Q( is a ... parallelepipeds and disjoint covering systems, Discrete Math, (in press). 2. Burshtein. N.. On natural covering ...

... parallelepipeds" of the form (1.1) P = {an, +an2 + ... + anr : r 6 N, n\ <n2<...<nr}, (an)£L, being a sequence of integers [4, Def. 2.3]. The elements of P can be ordered in a natural way to form a sequence as follows. If / — Y^k=\ e*2 ...

... parallelepipeds such that the ratio between the longest and shortest edges is bounded, and also lim d (t) = 0 (cf. [23], p. 538). In the case r - 1, / is a family of countable partitions of [0, 1] into intervals such that the length of ...