Solution 13.

Answer.      is valid for   .   

Solution.  Start with the series expansion for     which is  

                      

Use the familiar long division scheme for fractions:


          
          


Therefore,      is valid for   .   

We are done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

Alternate solution.  In Section 7.4 we will formally state that    has a simple zero at the origin.

Then we will define a simple pole, which means that the negative powers in the Laurent series only involve the one term .

As a consequence, we will be permitted to express    as

          .

The Cauchy product formula can easily be extended for this situation by multiplying    in two places:

                    ,  

with      and   ,   where   .  

The coefficients    

are all known and so are  .

Then it is easy to recursively solve for the unknown coefficients  .  

        Start with the series expansion for     which is  

                      

Use the coefficients  ,  which are

                       

Write out the equations corresponding to  ,  which are

                    

                    

                    

                    

                      

                    

                    

                    

                      

                    

                    

These equations can be summarized as follows:

                      

This system of equations can be solved recursively as follows:  

                    

Then

                    
                    
                    
Therefore,      is valid for   .   

We are really really 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