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 UAE_Math_Grade_12_Vol_1_SE_718383_ch3
                                  GO01962-Smith-v1.cls
                                     EXAMPLE 1.6    July 4, 2016  13:38
                                                     Using Newton’s Method to Approximate a Cube Root
                    y
                                                                      √
                                                                      3
                                     Use Newton’s method to approximate  7.
                  30
                                     Solution Since Newton’s method is used to solve equations of the form f(x) = 0, we
                                                                                √
                                                                                 3
                                     first rewrite the problem, as follows. Suppose x =  7. Then, x = 7, which can be
                                                                                          3
                                 x   rewritten as
                         2
                                                                       3
                                                                 f(x) = x − 7 = 0.
                                            ′
                 -30                 Here, f (x) = 3x and we obtain an initial guess from a graph of y = f(x). (See Fig-
                                                   2
                                     ure 4.10.) Notice that there is a zero near x = 2 and so we take x = 2. Newton’s
                                                                                              0
                 FIGURE 4.10         method then yields
                      3
                  y = x − 7                                      3
                                                         x = 2 −  2 − 7  =  23  ≈ 1.916666667.
                                                         1
                                                                 3(2 )  12
                                                                   2
                                     Continuing this process, we have
                                                                 x ≈ 1.912938458
                                                                  2
                                     and                       x ≈ 1.912931183 ≈ x .
                                                                3
                                                                                 4
          NOTES
                                     Further,                    f(x ) ≈ 1 × 10 −13
                                                                    4
          Examples 1.3 and 1.6 highlight
          two approaches to the same  and so, x is an approximate zero of f. This also says that
                                             4
          problem. Take a few moments to
                                                                √
          compare these approaches.                              3  7 ≈ 1.912931183,
                                                                              √
                                                                               3
                                     which compares very favorably with the value of  7 produced by your calculator.
                                     REMARK 1.1
                                       Although it is very efficient in examples 1.5 and 1.6, Newton’s method does not
                                       always work. Make sure that the values of x are getting progressively closer and
                                                                            n
                                       closer together (zeroing in, we hope, on the desired solution). Continue until
                                       you’ve reached the limits of accuracy of your computing device. Also, be sure to
                                       compute the value of the function at the suspected approximate zero; if this is not
                                       close to zero, do not accept the value as an approximate zero.


                                         As we illustrate in example 1.7, Newton’s method needs a good initial guess to find
                                     an accurate approximation.


                                     EXAMPLE 1.7     The Effect of a Bad Guess on Newton’s Method
               y
                                                                                              2
                                                                                         3
                                     Use Newton’s method to find an approximate zero of f(x) = x − 3x + x − 1.
              8
                                     Solution From the graph in Figure 4.11, there is a zero on the interval (2, 3). Using
                                     the (not particularly good) initial guess x = 1, we get x = 0,x = 1,x = 0 and so
                                                                                   1
                                                                                               3
                                                                       0
                                                                                         2
                                 x   on. Try this for yourself. Newton’s method is sensitive to the initial guess and
                       2   3         x = 1 is just a bad initial guess. If we instead start with the improved initial guess
                                      0
                                     x = 2, Newton’s method quickly converges to the approximate zero 2.769292354.
                                      0
                                     (Again, try this for yourself.)
             -8
                                         As we see in example 1.7, making a good initial guess is essential with Newton’s  Copyright © McGraw-Hill Education
                 FIGURE 4.11         method. However, this alone will not guarantee rapid convergence (meaning that it
              y = x − 3x + x − 1     takes only a few iterations to obtain an accurate approximation).
                       2
                  3
        234 | Lesson 4-1 | Linear Approximations and Newton’s Method
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