Extra Solution 1.

See text and/or instructor's solution manual.

Answer.   .  

Alternative Answer.   ,  

                                      where for all , and for all .   

Another Answer.   .  

Solution.   Find the Fourier Series   ,

by computing the coefficients with Euler's formulae:  

(12.2)        ,  

        and  

(12.3)        .  

First, calculate   .  

                      

Then

                      

Second, calculate   .  

                      

Alternately,    is an odd function so that     is an odd function, and   

                        for all    .

Then Theorem 12.4 shows that

                    ,  

where the coefficients can be computed with the special formula     

                    .  

Now calculate

                      

Get The Answer.

Therefore,

                    .  

We are done.   

It is easy to rewrite    and    using the well known facts        and  

                     .  

Therefore,

                    .  

We are really done.   

We have shown that     for all  .  

The even coefficients    can be written

                        for all   .   
                    
The even coefficients    can be written

                        for all   .   
                    
The odd coefficients    can be written

                        for all   .   
                    
Therefore,  

                    ,  
                    
                    where      for all  ,   and      for all  .

We are really really done.   

Aside.  We can calculate a few terms in the sequences to confirm that they are the same  ( up to   ).  

Aside.  We can calculate a few terms in these series to verify that they are the same  ( up to    ).  

We are really done.   

Aside.  We can let Mathematica double check our work.

Aside.  The Maple commands are similar  



                                                              




                                                              




                                                              

The partial lists of coefficients    and    are:

For illustration, we can sum the first few terms in these series  ( up to   ).  

Aside.  The function    can be expressed in the form   .  

We can compute a few terms of the Fourier series of    using Mathematica's built in procedure  FourierTrigSeries.

                    .  

Remark.  Since    is a point of discontinuity of  ,  we know that    is not defined at  ,  and

                        ,  

where    and    denote the left-hand and right-hand limits, respectively.   

Convergence is not uniform on the closed interval  ,  and the overshooting of   is referred to as "Gibbs phenomenon."  

We are really really done.   

Aside.  There are at least four solutions to this problem.  

We can use Mathematica to sum the infinite series.  

However, these sum might not be as familiar as those studied in calculus.

Aside.  The Maple commands are similar  

  

                                                              


  

                                                              


  

                                                              

We are really really really done.   

Aside.  We can graph these functions to verify they are correct.  

We are really really really really done.   

Aside.  Let us announce that in Section 12.2 we interpret the Fourier Series   

                      

as boundary values on the circle , and construct the harmonic function inside the unit disk with  .

          The harmonic function  ,  with  .  

          A contour graph of the harmonic function  .  

This solution is complements of the authors.

(c) 2010 John H. Mathews, Russell W. Howell