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.
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.
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)
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
On the other hand, all even periodic orbits exist because
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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.
[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.
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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