Exercise 9 (b).  Show that if    then    is not in the Mandelbrot set.  

Solution 9 (b).

        Applying inequality (1-24) in Section 1.3 we have

                      
                    
                    
                    
                    
                    
                    

Now start with    and compute

                    ,   and    

                    ,

and since   we have

                    .

Also,  

                    ,   and

                     ,  

                     ,  

                     .  

Next, compute    

                    ,

and

                       

Thus,

                     .  

Suppose for some    we have

                    ,    and  

                         for    .

Then  

                        

Therefore,  for all integers    we have

                    ,    and  

                         for    .

And since  

                    ,  

                    ,  

                    ,  

it will also follow that  

                    .  

Thus,    

                    .  

Hence, the orbit of  0  under    is not bounded and therefor    is not in the Mandelbrot set.     

We are done.   

Aside.  We can use Mathematica to explore this situation.

This solution is complements of the authors.

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