Exercise 3.  Prove that Newton's method always works for polynomials of degree  1  (functions of the form  ,  where  .  

How many iterations are necessary before Newton's method produces the solution    to   ?  

Solution 3.

See text and/or instructor's solution manual.

        For   ,   we have   ,   and

if our initial guess is  ,   then  

                    .  

But this is the solution to the equation  ,  so our iteration either stops here or with    if by chance we had set  .  

We are done.   

Caveat.   It seems that Newton's method will also work for some polynomials of degree greater than  1.  

Aside.  We can use Mathematica to explore this situation.  

Extra Example. (For a Polynomial)  Iteration using a polynomial of degree greater than  1.

Consider using Newton's method to find the complex roots of the polynomial   

                    .

Let us explore what happens when we use the starting value  .

Newton's iteration is converging to the complex root   .

                    Some iterates for Newton's method for  ,

                    starting with   we have  
          
                    
          
                    This sequence is converging to the complex root   .

We are really done.   

Consider using Newton's method to find the complex roots of the polynomial   .

Let us explore what happens when we use the starting value  .

Newton's iteration is converging to the complex root   .

                    Some iterates for Newton's method for  ,

                    starting with   we have
          
                    

                    This sequence is converging to the complex root   .

We are really really done.   

        Let us explore what happens when we use the starting value  .

In this case we have chosen an unlucky starting value.  The iteration produces an oscillating sequence that is divergent.

                    Some iterates for Newton's method for  ,

                     starting with      we have

                                

                    This oscillating sequence is divergent.

We are really really really done.   

Aside.  We can use Mathematica to find the roots.

We are really really really  really done.   

        It is tempting to mimic the proof of Newton's method for real numbers and try to make a proof

for complex numbers which would show that the Newton iteration will construct a convergent sequence.  

But the proof for real numbers uses the Mean Value Theorem for derivatives.  In Exercise 13 in Section 3.1

we asked you to show that the Mean Value Theorem for derivatives does not extend to complex functions.  

So a proof that    converges is not straight forward and is beyond the scope of this course.

Caveat.  

Newton's method can sometimes be successful in locating the real solution to  ,  

even though the starting value    is a complex number,

but proving why it will work or when it will work is not an easy task.    

Section 3.1, Exercise 13.

This solution is complements of the authors.

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