10 has attractor the lled in unit square from theorem

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Unformatted text preview: en (xk , yk ), generate (xk+1 , yk+1 ) by (xk+1 , yk+1 ) = Tik+1 (xk , yk ), where {i1 , i2 , ...} are chosen randomly from among {1, ..., n}. Then the sequence Σ = {(xi , yi ), i = 1, 2, 3 . . . } lies entirely in A, and in fact A is the closure of Σ. The final reason, then, is this. So far as we know Stewart [147] first posed this question. What happens if the Ti are selected in a non-random order? This perfectly natural question gives rise to driven IFS, one of the main topics of Ch 4. Of course, this question is natural, or even makes sense, only after the random IFS algorithm is familiar. So, let’s look into the details of this method. That A is the closure of Σ means A is closed and for every point (x′ , y ′ ) ∈ A, and for every ǫ > 0, there is some (xi , yi ) ∈ Σ with d((x′ , y ′ ), (xi , yi )) < ǫ. Let us see why this is so. We outline the steps of the argument now. Recall proofs are near the end of this section. First, the fixed points of the Ti are good starting points because they belong to A. Lemma 2.4.1 The fixed point (x0 , y0 ) of Ti belongs to A. Next, we note that if we start in A, applicatio...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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