Ngày tải lên :
13/02/2014, 04:20
... 0, 0, 1, 1, 1, 1, 0, 0, ) x2 = (1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) x3 = (1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) x4 = (1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ... [11 ], we get j nj 1 oj = ; lim = and lim = r j→∞ nj j→∞ nj j→∞ nj lim We define s ∈ supp µ by s = S(x), where x = (1, 1, 1, 1, 0, 0, 0, , 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ... = (1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) (7) are six sequenses in D∞ Put F6n+i = # xi 6n+i for i = 0, 1, 2, 3, 4, 5, n ∈ N Then we have (i) F6n = 9.9n 1 ; F6n +1 = 12 .9n 1 ;...