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PROBLEMS   201
             4.3 Recursive (Self-Calling) Routine for Bisection Method
                As stated in Section 1.3, MATLAB allows us to make nested (recursive) rou-
                tines which call itself. Modify the MATLAB routine “bisct()” (in Section
                4.2) into a nested routine “bisct_r()” and run it to solve Eq. (P4.2.1).

             4.4 Newton Method and Secant Method
                As can be seen in Fig. 4.5, the secant method introduced in Section 4.5
                was devised to remove the necessity of the derivative/gradient and improve
                the convergence. But, it sometimes turns out to be worse than the Newton
                method. Apply the routines “newton()”and “secant()” to solve
                                             3
                                                 2
                                  f p44 (x) = x − x − x + 1 = 0          (P4.4)
                starting with the initial point x 0 =−0.2 one time and x 0 =−0.3 for another
                shot.

             4.5 Acceleration of Aitken–Steffensen Method
                                              o
                A sequence converging to a limit x can be described as
                         o
                                                   o
                        x − x k+1 = e k+1 ≈ Ae k = A(x − x k )
                                                   o
                                                  x − x k+1
                                         with lim          = A(|A| < 1) (P4.5.1)
                                                    o
                                              k→∞ x − x k
                In order to think about how to improve the convergence speed of this
                sequence, we define a new sequence p k as
                     o                o
                    x − x k+1       x − x k    o        o            o     2
                             ≈ A ≈          ; (x − x k+1 )(x − x k−1 ) ≈ (x − x k )
                      o              o
                     x − x k       x − x k−1
                                   o
                                           o
                                                                 o
                                                          o 2
                         o 2
                       (x ) − x k+1 x − x k−1 x + x k+1 x k−1 ≈ (x ) − 2x x k + x 2
                                                                       k
                                      x k+1 x k−1 − x 2
                                 o               k
                                x ≈                  = p k              (P4.5.2)
                                    x k+1 − 2x k + x k−1
                (a) Check that the error of this sequence p k is as follows.
                                     x k+1 x k−1 − x 2
                       o        o               k
                      x − p k = x −
                                    x k+1 − 2x k + x k−1
                                    x k−1 (x k+1 − 2x k + x k−1 ) − x 2  + 2x k−1 x k − x 2
                                o                           k−1            k
                             = x −
                                                x k+1 − 2x k + x k−1
                                            (x k − x k−1 ) 2
                                o
                             = x − x k−1 +
                                          x k+1 − 2x k + x k−1
                                                   o
                                                              o
                                               (−(x − x k ) + (x − x k−1 )) 2
                                o
                             = x − x k−1 +
                                             o
                                                          o
                                                                     o
                                          −(x − x k+1 ) + 2(x − x k ) − (x − x k−1 )
                                                       o
                                                    2
                                            (−A + 1) (x − x k−1 ) 2
                                o
                             = x − x k−1 +                        = 0   (P4.5.3)
                                                          o
                                              2
                                          (−A + 2A − 1)(x − x k−1 )
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