... 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 {y : y = XX_1 yiwi, 0 < y < 1) then S is contained in the translated parallelepiped vo 4 Po. Now there is a matrix U e GL(n, Z) taking the basis matrix to the lower-triangular form (Hermite normal form): Q 11 al al (3.1) ...

... 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 ...

... 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, (ii) the six midpoints of the segments joining the centre of the parallelepiped to the centre of each face, (iii) the six centres of the faces, (iv) the eight vertices, we obtain formula (34) which has fifth degree ...

... 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 ...

... 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. If *1. * * * * *m are given vectors we define the parallelepiped P(x, - - - - *m' to In consist of all vectors of the form 2 cox. with co real and l i i-1 0 < c < 1 for all i . The high-school formula for m-dimensional ...

... 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 ...