FractionalGeometry-Chap2

So suppose i1 i2 i3 is a binary sequence that

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Unformatted text preview: sums to 1/10.) While the degree of overlap can cause some difficulty when computing dimensions, it will not affect our analysis of random IFS. Before beginning the proof, we also must discuss the aspect of randomness that is central to the random IFS algorithm. While randomness is a very subtle concept, we adopt an interpretation based on the idea of algorithmic complexity. We say a sequence {i1 , i2 , i3 , . . . } is random if it cannot be specified in any form more compact than listing all its elements. Here are examples of nonrandom binary infinite sequences, followed by their more compact specifications. {1, 1, 1, . . . } {1, 0, 1, 0, 1, 0, . . . } {1, 0, 1, 1, 0, 1, 1, 0, 1, . . . } all entries are 1 the pair 0,1 repeats forever the triple 1,0,1 repeats forever Must all nonrandom sequences be periodic? No: {1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1 . . . } Is the pattern apparent? So perhaps the notion of randomness depends in part on how clever we are. Turning to decimal sequences fo...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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