Exercise 6.  Show that the closed interval      is contained in the set      where   .  

Solution 6.

See text and/or instructor's solution manual.

        For the initial seed    we compute the iterates as follows:  

                    ,     ,     ,  ...  .  

Hence, the orbit of      generated by      is the set   .  

        For the initial seed    we compute the iterates as follows:  

                    ,     ,     ,  ...  .  

Hence, the orbit of      generated by      is the set   .  

        Since    is the set of points with bounded orbits for   ,   it follows that both     and  .

        We now show that  .   

If     and     then     and  .  

Hence      and it follows that      and   .
        
Now assume for some    that      and      then      and  .  

Hence      and it follows that     and  .

Thus, the iterates of    generated by    all lie in the interval    which is a bounded set,.

Hence   .

Therefore,  .

We are done.   

Aside.  We can use Mathematica to explore this situation.

                    The image of the interval      under      is contained in   .

                    Thus, each point      has a bounded orbit for    and therefore   .

We are really done.   

Remark.  It can be shown that   is the closed interval   .

The complete details of this fact are beyond the scope of this course.

We leave it to the reader to research this situation.  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell