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68      2 Nonlinear algebraic systems






                             initia esses n  w cn vere t     2
                         2
                                                      −  −  2



                        2
                      stin            1    1  initia ess  2
                                                 2
                         2
                    atins  1  1

                        1
                         1
                      e
                    ner  iter  t a

                                      1    1     2    2
                                             initia ess
                   Figure 2.5 Convergence of a simple factoring method to find remaining root at x = 2 for f (x) =
                   (x − 3) (x − 2) (x − 1).



                   erratically. Let us assume now that the current estimate of the solution is indeed near a
                   solution. Then, can we say anything about the rate at which successive Newton iterations
                   converge upon the true solution value? First, we write a Taylor series of f (x) about the
                                 [k]
                   current estimate x ,
                                                    1             1
                         [k]         [k]    (1)     [k]    2  (2)    [k]     3  (3)    [k]
                     f x  + ε = f x    + εf  x   +    ε f  x   +    ε f  x   + ···    (2.27)
                                                    2!            3!
                   We define the error at iteration k as


                                                 ε k ≡ x s − x [k]                    (2.28)

                   If the current estimate is very close to the true solution, ε k is very small and


                                            |ε k |  ε    ε    ···                     (2.29)
                                                    2
                                                    k     k
                                                          3
                   Therefore, if we can neglect terms of third order and higher in ε k , we can approximate
                             [k]                    [k]     (1)    [k]    1 2  (2)     [k]
                         f x  + ε k = f (x s ) = 0 ≈ f x  + ε k f  x  + ε f  x        (2.30)
                                                                    2 k
                   Now, from Newton’s method (2.19), we have

                                                                    [k]
                                                                f x
                                         [k+1]        [k]
                                       x    − x s = x   − x s −                       (2.31)
                                                               f  (1)  x [k]
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