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Chapter 2




                                                                                 Root Approximations




                T     his chapter is an introduction to Newton’s and bisection methods for approximating roots

                      of linear and non−linear equations. Several examples are presented to illustrate practical
                      solutions using MATLAB and Excel spreadsheets.


                2.1 Newton’s Method for Root Approximation

                Newton’s (or Newton−Raphson) method can be used to approximate the roots of any linear or
                non−linear equation of any degree. This is an iterative (repetitive procedure) method and it is
                derived with the aid of Figure 2.1.

                                      y                      Tangent line (slope) to the curve
                                                             y =  f()    at point  x fx(,{ x  1  1 )  }
                                          y =  f x()

                                                  • x fx( ,{  )  }
                                                      1   1
                                                •              x
                                                (  x 0, )  2
                                 Figure 2.1. Newton’s method for approximating real roots of a function


                We assume that the slope is neither zero nor infinite. Then, the slope (first derivative) at x =  x 1
                is
                                                              y  f x (–  )
                                                                    1
                                                     f ' x (  1 )  =  ---------------------
                                                                 –
                                                               xx
                                                                    1
                                                y  f x (–  1 )  f ' x ) ( =  1  (  xx )  –  1           (2.1)
                                                                                                   ,
                The slope crosses the x –  axis  at x =  x 2  and y =  0 . Since this point  x fx(,[  2  2 )  ]  ( =  x 0 )   lies on
                                                                                                  2
                the slope line, it satisfies (2.1). By substitution,
                                                  0  f x (–  1 )  f ' x ) ( =  1  (  x –  x )  1
                                                                     2
                                                              fx ) (
                                                                 1
                                                    x =   x –  --------------- )                        (2.2)
                                                     2
                                                           1
                                                              f ' x (
                                                                  1
                and in general,
                                                                fx ) (
                                                                   n
                                                   x n +  1  =  x –  --------------- )                  (2.3)
                                                            n
                                                               f ' x (
                                                                   n
               Numerical Analysis Using MATLAB® and Excel®, Third Edition                               2−1
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