Exercise 5 (b).  Lucas numbers.  Solve     with   .  
                          Hint.  Get  .  

Solution 5 (b).

See text and/or instructor's solution manual.

Answer.   .  

Remark.   It is our goal to learn Z-transforms and use residues.  

Solution.   

Method 1.  The characteristic equation  

                      

has roots and .  

The general solution is  

                    .  

Solve the linear system  

                      

and get    and  .

Therefore,  

                    .

Aside.  A few terms in the sequence are

                    

Method 2.  Take the z-transform of both sides and use the initial conditions      

                    ,  

and then get    

                    .  

New solve for    and obtain:    

                    .  

Using Tables.   Using Table 9.1 and the formula      we get:  

                      

Remark.  The details for the partial fraction expansion are at the bottom of the page.

Therefore,  

                    .  

We are done.   

Using Residues.   Calculate the residues    at the poles   .  

                      

and

                      

Thus,  

                       

Therefore,  

                    

We are done.   

Aside.  We can let Mathematica double check our work.

Aside.  The Maple commands are similar  

  

                                                            

  

                                                            
                                                            
                                                            
  

                                                            


  

                                                            

We are really done.   

Aside.  We can use Mathematica's Rsolve subroutine.

Aside.  The Maple command is similar  

  

                                                            

Remark. The Lucas numbers start with the subscript    and are

                    

We are really really done.   

Aside.  We can graph some of the terms in the sequence.

                    The sequence   .  

We are really really really done.   

The Details for the Partial Fractions.   

Aside.  How can we expand      into the proper partial fractions?

Method (i).   Use the following algebra steps  

                      

Method (ii).   Find the linear combination of   ,  

                    .  

Equate the numerators   ,  

and solve the linear system  

                    

and get   .   

Therefore, the desired partial fraction form is  

                    .  

Aside.   The Mathematica commands for Method (ii)  are

Therefore, the desired partial fraction form is  

                    .  

This solution is complements of the authors.

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