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UNCONSTRAINED OPTIMIZATION  323

              2

                                          a 3  c 3  d 3  b 3
              1                   a 2    c 2  d 2    b 2
                                              c 1     d 1       b 1   h 1  = b 1  − a 1  = rh
                                  a 1
                                                             2
                a                  c          d   (1 − r)h = rh 1  = r h  b  h = b − a
              0
              −1
                0      0.5      1       1.5      2       2.5      3
                           Figure 7.1 Process for the golden search method.

                                                                       2
              ž The golden ratio r is fixed so that a point c 1 = b 1 − rh 1 = b − r h in the
                new interval [c, b] conforms with d = a + rh = b − (1 − r)h,thatis,
                                                      √             √
                                                 −1 +   1 + 4  −1 +   5
                  2           2
                 r = 1 − r,  r + r − 1 = 0,  r =             =           (7.1.3)
                                                      2            2
            7.1.2  Quadratic Approximation Method
            The idea of this method is to (a) approximate the objective function f(x) by a
            quadratic function p 2 (x) matching the previous three (estimated solution) points
            and (b) keep updating the three points by replacing one of them with the minimum
            point of p 2 (x). More specifically, for the three points

                        {(x 0 ,f 0 ), (x 1 ,f 1 ), (x 2 ,f 2 )}  with x 0 <x 1 <x 2

            we find the interpolation polynomial p 2 (x) of degree 2 to fit them and replace

            one of them with the zero of the derivative—that is, the root of p (x) = 0[see
                                                                    2
            Eq. (P3.1.2) in Problem 3.1]:
                                      2
                                                              2
                                                          2
                                              2
                                                  2
                                 2
                             f 0 (x − x ) + f 1 (x − x ) + f 2 (x − x )
                                             2
                                                              1
                                                  0
                                                          0
                                 1
                                      2
                    x = x 3 =                                            (7.1.4)
                            2{f 0 (x 1 − x 2 ) + f 1 (x 2 − x 0 ) + f 2 (x 0 − x 1 )}
            In particular, if the previous estimated solution points are equidistant with an
            equal distance h (i.e., x 2 − x 1 = x 1 − x 0 = h), then this formula becomes
                               2   2       2    2      2    2
                                                            1
                                                       0
                                   2
                               1
                                               0
                                           2
                           f 0 (x − x ) + f 1 (x − x ) + f 2 (x − x )
                     x 3 =
                          2{f 0 (x 1 − x 2 ) + f 1 (x 2 − x 0 ) + f 2 (x 0 − x 1 )} x 1 =x +h

                                                               x 2 =x 1 +h
                                 3f 0 − 4f 1 + f 2
                       = x 0 + h                                         (7.1.5)
                               2(−f 0 + 2f 1 − f 2 )
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