MAT335H1 Lecture : MAT335 Problem Set 5 Self Generated Solutions.pdf
Document Summary
Chapter 12: #1(a), 1(c), 1(e), 4, 5, 7, 8. 48 = 24 3 and 56 = 23 7, so 56 is before 48 in the sarkovskii ordering. By the theorem on page 138 it is possible for f to have a periodic point of period. 176 = 24 11, 96 = 25 3, so 176 is before 96 in the sarkovskii ordering. By sarkovskii"s theorem if f has a periodic point of order 176 it must also have a periodic point of period 96. The one on the left has cycles of all other periods; the one on the right has only cycles of periods 1, 2 and 4. For the graph on the left: the 4-cycle is 0 1 2 3 0. This graph has cycles of all other periods because it has a 3-cycle; one has been drawn in. F (3/4) = 7/4, f (7/4) = 11/4, f (11/4) = 3/4.
