Exercise 12.  Suppose that    is a 2-cycle for   .   

12 (a).  Show that, if    is an attracting fixed point for  ,  then so is the point  .  

Hint.   Differentiate  ,  using the chain rule, and show that  .  

Solution 12 (a).

See text and/or instructor's solution manual.

        Since    is a 2-cycle for   ,  the set    has the property that     and  .

Hence, the function    has the property that

                    ,   

and it follows that

                    ,

so that    is a fixed point for  .  

Assuming that    is an attracting fixed point for    we have   ,  and

                    .

Then we get
                    
                      

Therefore,    is an attracting fixed point for  .

We are done.   

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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