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Newton’s method for a single equation                                 65



                  a    2

                       1                  stin at





                      1
                                    1    1      2    2



                       2       at ints
                      dd  1



                      1
                                    1    1      2    2


                  Figure 2.1 Plot of (a) the function and (b) the first derivative with a single real root, f (x) = (x −
                  3) (x − i)(x + i).




                     vered rest dts  a initia esses               ve et  ve rit
                                cn vere t




                     cn
                                     1    1     2     2
                                             initia ess
                      1
                    atins  1                        ere  are neativ  e
                                                    carries e wtn s etd
                       1                             ar awa  r
                                                    e re cn  verin
                     e
                    ner  iter  t a  2

                                     1    1     2     2
                                             initia ess

                  Figure 2.2 Update function and effect of initial guess on convergence, f (x) = (x − 3) (x − i)(x +
                  i). In the upper plot, the line is u(x) vs. x and the dots indicate the converged result of Newton’s
                           [0]
                  method vs. x .
                    We next consider the performance of Newton’s method for a system with multiple real
                  roots to ask the question: is there any clear relation between the choice of initial guess and
                  the identity of the root that is found? Consider the cubic polynomial

                                                            3
                                                                  2
                                 f (x) = (x − 3)(x − 2)(x − 1) = x − 6x + 11x − 6    (2.24)
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