FractionalGeometry-Chap2

# Simply take k large enough that 2 2k then construct v

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Unformatted text preview: tic IFS algorithms, is that of address. The general concept is illustrated by the next example. Example 2.4.1 Addresses of regions in the unit square. Recall the IFS Ti (x, y ) = (x/2, y/2) + (ai , bi ), with (ai , bi ) = (0, 0), (1/2, 0), (0, 1/2), (1/2, 1/2), i = 1, 2, 3, 4, (2.10) has attractor the ﬁlled-in unit square. From Theorem 2.3.1 we have A = T (A) = T 2 (A) = T 3 (A) = . . . . Now write A = T (A) = T1 (A) ∪ T2 (A) ∪ T3 (A) ∪ T4 (A) as A = A1 ∪ A2 ∪ A3 ∪ A4 . Think of the Ai as regions speciﬁed by length 1 addresses. The relation A = T 2 (A) can be written as A = T (A1 ∪ · · · ∪ A4 ) = T1 (A1 ∪ · · · ∪ A4 ) ∪ · · · ∪ T4 (A1 ∪ · · · ∪ A4 ) = A11 ∪ · · · ∪ A14 ∪ · · · ∪ A41 ∪ · · · ∪ A44 where we think of the Aij as regions speciﬁed by length 2 addresses. See Fig. 2.24. 33 43 44 32 41 42 13 14 23 24 11 1 34 31 3 12 21 22 4 2 Figure 2.24: Length 1 and 2 addresses determined by the IFS of Example 2.4.1. Denote the diameter of a compact set S ⊂ R2 by diam(S ) = |S | = max{d((x1 , y1 ), (x2 ,...
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