Newton-Raphson Method

Date: 02/28/2000 at 09:47:19
From: James Dyer
Subject: Direct-Search / Newton-Raphson Method

Using direct-search method and Newton-Raphson method, find the 1st, 
2nd, 3rd iterations and the parameters of the following:

     x^3-13.1x^2+48.48x-46.62

I'm not sure how to do this. Help, please?

Date: 02/28/2000 at 10:50:59
From: Doctor Mitteldorf
Subject: Re: Direct-Search / Newton-Raphson Method

Dear James,

You can help us help you by telling us what you know and exactly where 
you get stuck. Don't be afraid to flood us with information - let's 
see what you've tried already, even things that you figured out along 
the way weren't going to work. I'll do the best I can, but I can't be 
sure I'm aiming at the kind of answer that will be most helpful to 
you.

I don't know what your teacher means by the direct-search method. 
Perhaps you do. I do know Newton-Raphson, however, and here's an 
explanation of it that I recently wrote for another questioner:

   Inventing an Operation to Solve x^x = y
   http://mathforum.org/dr.math/problems/redsting.02.19.00.html   

In your case,

     f(x) = x^3 - 13.1x^2 + 48.48x - 46.62

and y isn't specified - presumably it's zero:

      x^3 - 13.1x^2 + 48.48x - 46.62 = 0

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/28/2000 at 10:53:40
From: Doctor Jerry
Subject: Re: Direct-Search / Newton-Raphson Method

Hi James,

Because it's a cubic, all you have to do is to determine one real 
root. You can locate that root by finding an interval in which the 
sign changes. At 0 and 1 this polynomial is negative and at x = 2 it 
is positive. So, there's a root between 1 and 2. We'll try x_1 = 1.5, 
where by x_1 I mean "x sub 1," the first guess for Newton's method.

     x_2 = 1.26679325896
     x_3 = 1.26102002205
     etc.

where

     x_{n+1} = x_n-f(x_n)/f'(x_n),   n = 1, 2, 3, ...

Now you can use synthetic division to remove this root, leaving a 
quadratic, which you can solve by hand if you like. Or you can locate 
roots near 4 and 7 (as above) and then use Newton's method.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/