{"id":3163,"date":"2024-06-13T17:07:54","date_gmt":"2024-06-13T17:07:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3163"},"modified":"2024-08-05T02:30:15","modified_gmt":"2024-08-05T02:30:15","slug":"newtons-method-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/newtons-method-learn-it-2\/","title":{"raw":"Newton\u2019s Method: Learn It 2","rendered":"Newton\u2019s Method: Learn It 2"},"content":{"raw":"

Approximating with Newton\u2019s Method Cont.<\/h2>\r\n

Failures of Newton\u2019s Method<\/h2>\r\n

Typically, Newton\u2019s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton\u2019s method might fail include the following:<\/p>\r\n

    \r\n\t
  1. At one of the approximations [latex]x_n[\/latex], the derivative [latex]f^{\\prime}[\/latex] is zero at [latex]x_n[\/latex], but [latex]f(x_n) \\ne 0[\/latex]. As a result, the tangent line of [latex]f[\/latex] at [latex]x_n[\/latex] does not intersect the [latex]x[\/latex]-axis. Therefore, we cannot continue the iterative process.<\/li>\r\n\t
  2. The approximations [latex]x_0,x_1,x_2, \\cdots[\/latex] may approach a different root. If the function [latex]f[\/latex] has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see Figure 4). This event most often occurs when we do not choose the approximation [latex]x_0[\/latex] close enough to the desired root.<\/li>\r\n\t
  3. The approximations may fail to approach a root entirely. In the example below, we provide an example of a function and an initial guess [latex]x_0[\/latex] such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values.<\/li>\r\n<\/ol>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"610\"]\"A Figure 4. If the initial guess [latex]x_0[\/latex] is too far from the root sought, it may lead to approximations that approach a different root.[\/caption]\r\n\r\n\r\n
    \r\n

    Consider the function [latex]f(x)=x^3-2x+2[\/latex]. Let [latex]x_0=0[\/latex]. Show that the sequence [latex]x_1,x_2, \\cdots[\/latex] fails to approach a root of [latex]f[\/latex].<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165043327561\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1165043327561\"]\r\n\r\n\r\n

    For [latex]f(x)=x^3-2x+2[\/latex], the derivative is [latex]f^{\\prime}(x)=3x^2-2[\/latex]. Therefore,<\/p>\r\n

    [latex]x_1=x_0-\\frac{f(x_0)}{f^{\\prime}(x_0)}=0-\\frac{f(0)}{f^{\\prime}(0)}=-\\frac{2}{-2}=1[\/latex].<\/div>\r\n

    In the next step,<\/p>\r\n

    [latex]x_2=x_1-\\frac{f(x_1)}{f^{\\prime}(x_1)}=1-\\frac{f(1)}{f^{\\prime}(1)}=1-\\frac{1}{1}=0[\/latex].<\/div>\r\n

    Consequently, the numbers [latex]x_0,x_1,x_2, \\cdots[\/latex] continue to bounce back and forth between 0 and 1 and never get closer to the root of [latex]f[\/latex] which is over the interval [latex][-2,-1][\/latex] (see Figure 5). Fortunately, if we choose an initial approximation [latex]x_0[\/latex] closer to the actual root, we can avoid this situation.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"425\"]\"The Figure 5. The approximations continue to alternate between 0 and 1 and never approach the root of [latex]f[\/latex].[\/caption]\r\n\r\n\r\n

    Watch the following video to see the worked solution to this example.<\/p>\r\n