FractionalGeometry-Chap2

Similarly t3 2 2 3 3 and t1 3 3 1 1

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Unformatted text preview: athematical notation, we say for all k ≥ 1 and for all sequences {i1 , . . . , ik } and {j1 , . . . , jk }, int(Aik ...i1 ) ∩ int(Ajk ...j1 ) = ∅. The situation illustrated in Example 2.4.2 holds 57 2.4. CONVERGENCE OF RANDOM IFS in this generality: Proposition 2.4.1 If i1 → · · · → in is forbidden, then for all in+1 → · · · → in+m , we have int(Ain+m ...in+1 in ...i1 ) = ∅ Another way for an infinite sequence to fail to be random is by being periodic. What happens to the random IFS algorithm if the transformations are applied in a sequence of the form (recall address order is composition order) (in · · · i2 i1 )∞ ? We begin with a simple example. Example 2.4.3 Cyclic addresses. Suppose the gasket transformations are applied in the order (321)∞ . To provide a background for comparison both sides of Fig. 2.28 show 2000 points generated by the random IFS rules for the gasket. The left side shows 50 (boldface) points, starting from (0.8, 0.8), and then applying the transformations in the order T1 , T2 , T3 . Can you determine th...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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