## ASSIGNMENT 6 SOLUTIONS

Assignment: Page 151: Nos. 1,2,3,4,8 Page 161: Nos. 1-8

Page 151

• 1. Can a continuous functon on R have a periodic point of period 48 and not of of period 56? Why?
Yes. Observe that 56=(2^3)*7 precedes 48=(2^4)*3 in the Sarkovskii ordering.
Thus, if a continuous function F has a cycle of period 56, then it also has a cycle of period 48. But, a continuous function F with a periodic point of period 48 may or may not have a periodic point of period 56.

• 2.Can a continuous functon on R have a periodic point of period 176 and not of of period 96? Why?
No. In this case, 176=(2^4)*11 precedes 96=(2^5)*3 in the Sarkovskii ordering, and so, by Sarkovskii's theorem, a continuous function with a periodic point of period 176 must also have a periodic point of period 96.

• 3.Give and example of a funciton F:[0,1] --> [0,1] that has a periodic point of period 3 and no other periods. Can this happen?
Sarkovskii's theorem applies only to continuous functions and so it follows that such a function must be discontinuous. Consider the function,

This function has lots of three cycles:
0->1/3->2/3->0
1/6->1/2->6/6->1/6
1/12->5/12->9/12->1/12
Indeed, the graph clearly shows that
f[0,1/3]=[1/3,2/3]
f[1/3,2/3]=[2/3,1)
f[2/3,1]=[0,1/3]
and so every point is of period 3 (except the endpoint 1 which is eventually period 3)

• 4.The graphs shown each have a cyle of period 4 given by {0,1,2,3}. One of these functions has a cycle of all other periods, and one has only period 1, 2, and 4. Identify which function has each of these properties.
To see the graphs under consideration, click
here.
Note: Each of these functions is contuniuous and so we need only find a 3-cycle to identify which has cycles of all periods.
From the graph of Fig. A {naming this function F}, it's clear that
F[0,1]=[1,2]
F[1,2]=[2,3]
F[2,3]=[0,3]
and so we have [0,1] ( [0,3] = F^3[0,1]. Therefore, F has a period 3 point in [0,1].
Furthermore the equation of the piecewise linar graph in Fig. B {naming this function G} is G(x) = { 3-x if 0<=x<=1, 4-2x if 1<=x<=2, x-2 if 2<=x<=3
Now let 0<=x<=1, Then -1<=-x<=0 and so 2<=3-x<=3. In other words, G[0,1]=[2,3], and similarly, G[2,3]=[0,1]. (These facts are easily verified by inspection fo the graph of G.) This means that no point in [0,1] U [2,3] can have odd period. Now let's see what happens when we take and arbitary x in [0,1] and iterate G.
G(x)=3-x , since x is in [0,1]
G^2(x)=G(3-x)=(3-x)-2=1-x , since 3-x is in [2,3]
G^3(x)=G(1-x)=3-(1-x)=2+x , since 1-x is in [0,1]
G^4(x)=G(2+x)=(2+x)-2=x , since 2+x is in [2,3]
So every point in [0,1] U [2,3] is of period 4. Now G[1,2]=[0,2] and G[0,2]=[0,3]. But as soon as a point leaves the closed interval [1,2], it's locked into a 4-cycle. So the question is: are there points in [1,2] that remain in [1,2] for all time? The answer is no! The orbits of points close to the fiexed point x=4/3 oscillate away from the fixed point and eventually enter [0,1], an interval of 4-cycles.

• 8) Consider the function whose graph is displayed in fig. 11.17b on page 152. Prove that this function has cycles of all even periods but no odd periods (except 1).

First observe that 2 is fixed by F. (In fact any multiple of a power of 2 in [0,4] is eventually fixed.) Inspection of the graph shows that
[3,4]->[0,1]->[2,4] and [2,3]->[1,2]->[2,4]
But [2,4]->[0,2] and vice versa. In general,
F^(2n+1)[0,1]=F^(2n+1)[1,2]=[2,4] and
F^(2n+1)[2,3]=F^(2n+1)[3,4]=[0,2]
Hence there can be no odd periodic points.

On the other hand, all even periodic orbits exist because
F^(2n)[0,1]=F^(2n)[1,2]=[0,2] and
F^(2n)[2,3]=F^(2n)[3,4]=[2,4].
Therefore, since both [0,1] and [1,2] are contained in [0,2] and both [2,3] and [3,4] are contained in [2,4] even periodic orbits are guaranteed to exist.

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