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MAT335 Problem Set 5 Self Generated Solutions.pdf

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Dietrich Burbulla

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Solutions to MAT335H1F Problem Set 5
Chapter 11: #1, 2, 4, 7, 8
Chapter 12: #1(a), 1(c), 1(e), 4, 5, 7, 8
Chapter 13: #1(c), 1(d), 6, 8
Ch 11: #1. Yes. 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 Fto have a periodic point of period
48 but not one of period 56.
Ch 11: #2. No. 176 = 24·11,96 = 25·3,so 176 is before 96 in the Sarkovskii ordering.
By Sarkovskii’s Theorem if Fhas a periodic point of order 176 it must also have a
periodic point of period 96.
Ch 11: #4. Both functions below have 4-cycles. 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.
0 1 2 3
0 1 2 3
For the graph on the left: the 4-cycle is 0 1230.This graph has
cycles of all other periods because it has a 3-cycle; one has been drawn in. The
equation of the graph is
F(x) = x+ 1,if 0 x2
3x+ 9,if 2 < x 3
and so
F(3/4) = 7/4, F (7/4) = 11/4, F (11/4) = 3/4.
Alternatively, you can use Observation 1 from page 134: F3([0,1]) = [0,3] [0,1],
so F3has a fixed point in [0,1],which is a point of period 3 for F.
For the graph on the right: one 4-cycle is 0 3120,so two 2-cycles
are 0 1 and 3 2.We claim it has no cycles of any other periods, because we
shall show it has no 8-cycle. Let Gbe the function; first check that G4(x) = xfor all
x[0,1] [2,3],which means every point in [0,1] [2,3] is on a 4-cycle or a 2-cycle.
A typical 4-cycle is shown on the graph. So if Ghas a point with period 8 it must
be in the interval (1,2).Suppose x(1,2) and G8(x) = x. If G(x)[0,1] [2,3]
then the orbit of xunder Gleaves the interval (1,2) for good, since
G([0,1]) = [2,3] and G([2,3]) = [0,1].
Thus the 8-cycle for Gmust be x, H(x), H2(x), . . . , H7(x),for H(x) = G(x),restricted to
(1,2).But His 1-1 so H8(x) = xxis the unique fixed point of H; contradiction.
Alternatively, H(x)=42xand H8(x) = x256x340 = xx= 4/3.which is the
fixed point of G, so not a point of period 8.
Note: the graphs for Questions 7 and 8 are not labeled, so you can label them in any
convenient way you wish. The results of these types of problems depend on the shape of
the graph and its relative position to the line y=x, not its actual equation.
Ch 11: #7. The graph is below; call the function F. A 6-cycle for Fis
22→ −1301→ −2
By Sarkovskii’s Theorem Fwill have cycles of
all possible even periods. We claim Fhas no
cycles of odd period. For any odd value of n
Fn([2,0]) = [1,3] and Fn([1,3]) = [2,0].
So if Fnhas a fixed point it must be in the
interval (0,1).Suppose the fixed point is x; as
in the second part of Question #4, the n-cycle
for Fmust be
x, H(x), H2(x), . . . , Hn1(x),
for H=F, restricted to the interval (0,1).
But His 1-1, so the only solution to Hn(x) = xis the fixed point of Hin (0,1).
Ch 11. #8. The graph is below; call the function G. G has no cycle of odd period since
G([2,0]) = [0,2] and G([0,2]) = [2,0];
so if Gn(x) = x, for odd n, then x= 0,which is the fixed point of G. But Gdoes
have cycles of all even periods. This will follow, from Sarkovskii’s Theorem, if we
can show that Ghas a cycle of period 6, or equivalently, G2has a cycle of period 3.
21 0 1 2
The graph of G2is to
the right, including one
21 0 1 2
Now it is true that (G2)3([0,1]) = [0,2] [0,1],so by Observation 1 on page 134,
G6has a fixed point in [0,1].But you still have to show its prime period is 6 . . . .

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