13 because the second iteration consists of 9 squares

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Unformatted text preview: traction factor, consider the scaling needed to take a side of the initial square to a side of the image square. √ The initial square has side length 2, the image square 1/ 2, giving √ 1/ 2 1 image length = =√ r= initial length 2 22 PrxsSol 2.1.3 Because the second iteration consists of 9 squares, the IFS consists of 3 transformations. Say T1 takes the unit square to S1 = [0, 1/2] × [0, 1/2], T2 takes the unit square to S2 = [1/2, 1] × [0, 1/2], and T3 takes the unit square to S3 = [0, 1/2] × [1/2, 1]. How care we to find the parameters of the three transformations? Calling the unit square S , T1 (S ) is the corner of the L in the first generation. 32 CHAPTER 2. ITERATED FUNCTION SYSTEMS In the second generation, the three corners of the L are T1 (T1 (L)), T2 (T1 (L)) and T3 (T1 (L)). Note that T1 (T1 (L)) is in the same orientation as L, so T1 is (i) rotation by π , (ii) reflection across the x-axis, (iii) reflection across the y -axis, or (iv) reflection across the y -axis and rotation by π/2. (Note from Eq 2.1 that (iv) is...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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