Solution 9 (a).

Answer.   Series I.       for    ,   

Answer.   Series II.      for    .

Solution.   First series.

Start by writing the function in the form

                    .  

Series I.   Differentiate the geometric series      which is valid for     to establish the identity  

                        which is valid for   .     (See Exercise 8 for the proof.)   

Make the substitution    and get:

                       
                     

Therefore,        is valid for  .     

We are done.   

Aside.  We can let Mathematica double check our work.

The first series.

We are really done.   

Aside.  As an optional experiment, we can plot some of the partial sums of the series   .  

The mapping    is not one-to-one in the punctured disk  ,

and so we will choose the small portion  .  

                    The images of    under    for  .

                    The image of      under the mapping   .  

Recall Exercise 8.   The formula      is valid for   .     (From Exercise 8.)   

Use the geometric series      which is valid for      and differentiate termwise to get:

                       
                    

Hence,        is valid for   .  

This solution is complements of the authors.

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