Exercise 13.  Prove that      in Theorem 4.10.  

Solution 13.

See text and/or instructor's solution manual.

        The starting place for this proof is to use the inequality that was derived in the proof of Theorem 4.10, i.e.

There exists an    and a disk    so that if   ,  then

                    .

The difference quotient    can be made continuous at    if we define  ,

and since    is an attracting fixed point we have  .

Without loss of generality, we can assume that    has been chosen small enough so that

we can find a real number for    with    and

                    ,    for all    .

Hence, we have

                    ,    for all    .

Given    we consider the sequence    where      for   ,   then

                    .    

Assume that for some    we have  ,  then

                    .    

Thus, by mathematical induction we have

                        for all    .

Since  ,  this implies   .  

Therefore,     .  

This solution is complements of the authors.

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