{"id":416,"date":"2021-02-04T02:05:33","date_gmt":"2021-02-04T02:05:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=416"},"modified":"2022-03-16T05:51:45","modified_gmt":"2022-03-16T05:51:45","slug":"approximating-with-newtons-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/approximating-with-newtons-method\/","title":{"raw":"Approximating with Newton\u2019s Method","rendered":"Approximating with Newton\u2019s Method"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the steps of Newton\u2019s method<\/li>\r\n \t<li>Explain what an iterative process means<\/li>\r\n \t<li>Recognize when Newton\u2019s method does not work<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165042613128\" class=\"bc-section section\">\r\n<h2>Describing Newton\u2019s Method<\/h2>\r\n<p id=\"fs-id1165043187457\">Consider the task of finding the solutions of [latex]f(x)=0[\/latex]. If [latex]f[\/latex] is the first-degree polynomial [latex]f(x)=ax+b[\/latex], then the solution of [latex]f(x)=0[\/latex] is given by the formula [latex]x=-\\frac{b}{a}[\/latex]. If [latex]f[\/latex] is the second-degree polynomial [latex]f(x)=ax^2+bx+c[\/latex], the solutions of [latex]f(x)=0[\/latex] can be found by using the quadratic formula. However, for polynomials of degree 3 or more, finding roots of [latex]f[\/latex] becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if [latex]f[\/latex] is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example, consider the function<\/p>\r\n\r\n<div id=\"fs-id1165042548842\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=x^5+8x^4+4x^3-2x-7[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043013826\">No formula exists that allows us to find the solutions of [latex]f(x)=0[\/latex]. Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of [latex] \\tan (x)-x=0[\/latex]. No simple formula exists for the solutions of this equation. In cases such as these, we can use <strong>Newton\u2019s method<\/strong> to approximate the roots.<\/p>\r\n<p id=\"fs-id1165043104195\">Newton\u2019s method makes use of the following idea to approximate the solutions of [latex]f(x)=0[\/latex]. By sketching a graph of [latex]f[\/latex], we can estimate a root of [latex]f(x)=0[\/latex]. Let\u2019s call this estimate [latex]x_0[\/latex]. We then draw the tangent line to [latex]f[\/latex] at [latex]x_0[\/latex]. If [latex]f^{\\prime}(x_0)\\ne 0[\/latex], this tangent line intersects the [latex]x[\/latex]-axis at some point [latex](x_1,0)[\/latex]. Now let [latex]x_1[\/latex] be the next approximation to the actual root. Typically, [latex]x_1[\/latex] is closer than [latex]x_0[\/latex] to an actual root. Next we draw the tangent line to [latex]f[\/latex] at [latex]x_1[\/latex]. If [latex]f^{\\prime}(x_1)\\ne 0[\/latex], this tangent line also intersects the [latex]x[\/latex]-axis, producing another approximation, [latex]x_2[\/latex]. We continue in this way, deriving a list of approximations: [latex]x_0, x_1, x_2, \\cdots[\/latex]. Typically, the numbers [latex]x_0,x_1,x_2, \\cdots[\/latex] quickly approach an actual root [latex]x*[\/latex], as shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"550\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211323\/CNX_Calc_Figure_04_09_001.jpg\" alt=\"This function f(x) is drawn with points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)) marked on the function. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x1. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x2. If a tangent line were drawn from (x2, f(x2)), it appears that it would come very close to x*, which is the actual root. Each tangent line drawn in this order appears to get closer and closer to x*.\" width=\"550\" height=\"439\" \/> Figure 1. The approximations [latex]x_0,x_1,x_2, \\cdots[\/latex] approach the actual root [latex]x*[\/latex]. The approximations are derived by looking at tangent lines to the graph of [latex]f[\/latex].[\/caption]\r\n<p id=\"fs-id1165042981157\">Now let\u2019s look at how to calculate the approximations [latex]x_0,x_1,x_2, \\cdots[\/latex]. If [latex]x_0[\/latex] is our first approximation, the approximation [latex]x_1[\/latex] is defined by letting [latex](x_1,0)[\/latex] be the [latex]x[\/latex]-intercept of the tangent line to [latex]f[\/latex] at [latex]x_0[\/latex]. The equation of this tangent line is given by<\/p>\r\n\r\n<div id=\"fs-id1165042526516\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(x_0)+f^{\\prime}(x_0)(x-x_0)[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">Therefore, [latex]x_1[\/latex] must satisfy<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1165043276857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x_0)+f^{\\prime}(x_0)(x_1-x_0)=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042328480\">Solving this equation for [latex]x_1[\/latex], we conclude that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_1=x_0-\\dfrac{f(x_0)}{f^{\\prime}(x_0)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042881142\">Similarly, the point [latex](x_2,0)[\/latex] is the [latex]x[\/latex]-intercept of the tangent line to [latex]f[\/latex] at [latex]x_1[\/latex]. Therefore, [latex]x_2[\/latex] satisfies the equation<\/p>\r\n\r\n<div id=\"fs-id1165043082009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_2=x_1-\\dfrac{f(x_1)}{f^{\\prime}(x_1)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043351477\">In general, for [latex]n&gt;0, \\, x_n[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1165043389758\" class=\"equation\" style=\"text-align: center;\">[latex]x_n=x_{n-1}-\\dfrac{f(x_{n-1})}{f^{\\prime}(x_{n-1})}[\/latex]<\/div>\r\nNext we see how to make use of this technique to approximate the root of the polynomial [latex]f(x)=x^3-3x+1[\/latex].\r\n<div id=\"fs-id1165043423917\" class=\"textbook exercises\">\r\n<h3>Example: Finding a Root of a Polynomial<\/h3>\r\nUse Newton\u2019s method to approximate a root of [latex]f(x)=x^3-3x+1[\/latex] in the interval [latex][1,2][\/latex]. Let [latex]x_0=2[\/latex] and find [latex]x_1,x_2,x_3,x_4[\/latex], and [latex]x_5[\/latex].\r\n\r\n[reveal-answer q=\"99095\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99095\"]\r\n\r\nFrom Figure 2, we see that [latex]f[\/latex] has one root over the interval [latex](1,2)[\/latex]. Therefore [latex]x_0=2[\/latex] seems like a reasonable first approximation. To find the next approximation, we use the equation we found for [latex]x_n[\/latex]. Since [latex]f(x)=x^3-3x+1[\/latex], the derivative is [latex]f^{\\prime}(x)=3x^2-3[\/latex]. Using the equation for\u00a0[latex]x_n[\/latex] with [latex]n=1[\/latex] (and a calculator that displays 10 digits), we obtain\r\n<p style=\"text-align: center;\">[latex]x_1=x_0-\\dfrac{f(x_0)}{f^{\\prime}(x_0)}=2-\\dfrac{f(2)}{f^{\\prime}(2)}=2-\\dfrac{3}{9} \\approx 1.666666667[\/latex]<\/p>\r\nTo find the next approximation, [latex]x_2[\/latex], we use (Figure) with [latex]n=2[\/latex] and the value of [latex]x_1[\/latex] stored on the calculator. We find that\r\n<p style=\"text-align: center;\">[latex]x_2=x_1=\\dfrac{f(x_1)}{f^{\\prime}(x_1)} \\approx 1.548611111[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Continuing in this way, we obtain the following results:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} x_1 \\approx 1.666666667 \\\\ x_2 \\approx 1.548611111 \\\\ x_3 \\approx 1.532390162 \\\\ x_4 \\approx 1.532088989 \\\\ x_5 \\approx 1.532088886 \\\\ x_6 \\approx 1.532088886 \\end{array}[\/latex]<\/p>\r\nWe note that we obtained the same value for [latex]x_5[\/latex] and [latex]x_6[\/latex]. Therefore, any subsequent application of Newton\u2019s method will most likely give the same value for [latex]x_n[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"425\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211326\/CNX_Calc_Figure_04_09_002.jpg\" alt=\"The function f(x) = x3 \u2013 3x + 1 is drawn. It has roots between \u22122 and \u22121, 0 and 1, and 1 and 2.\" width=\"425\" height=\"350\" \/> Figure 2. The function [latex]f(x)=x^3-3x+1[\/latex] has one root over the interval [latex][1,2][\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding a Root of a Polynomial.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6f9WGeq3oj0?controls=0&amp;start=115&amp;end=485&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.9NewtonsMethod115to485_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.9 Newton's Method\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165043433379\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLetting [latex]x_0=0[\/latex], use Newton\u2019s method to approximate the root of [latex]f(x)=x^3-3x+1[\/latex] over the interval [latex][0,1][\/latex] by calculating [latex]x_1[\/latex] and [latex]x_2[\/latex].\r\n\r\n[reveal-answer q=\"2661078\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2661078\"]\r\n\r\nUse the equation for [latex]x_n[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043345490\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043345490\"]\r\n\r\n[latex]x_1 \\approx 0.33333333, \\, x_2 \\approx 0.347222222[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043348404\">Newton\u2019s method can also be used to approximate square roots. Here we show how to approximate [latex]\\sqrt{2}[\/latex]. This method can be modified to approximate the square root of any positive number.<\/p>\r\n\r\n<div id=\"fs-id1165042936506\" class=\"textbook exercises\">\r\n<h3>example: Finding a Square Root<\/h3>\r\nUse Newton\u2019s method to approximate [latex]\\sqrt{2}[\/latex] (Figure 3). Let [latex]f(x)=x^2-2[\/latex], let [latex]x_0=2[\/latex], and calculate [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. (We note that since [latex]f(x)=x^2-2[\/latex] has a zero at [latex]\\sqrt{2}[\/latex], the initial value [latex]x_0=2[\/latex] is a reasonable choice to approximate [latex]\\sqrt{2}[\/latex].)\r\n<div id=\"fs-id1165043182674\" class=\"exercise\">[reveal-answer q=\"fs-id1165043433206\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043433206\"]\r\n<p id=\"fs-id1165043433206\">For [latex]f(x)=x^2-2, \\, f^{\\prime}(x)=2x[\/latex]. From the equation for [latex]x_n[\/latex], we know that<\/p>\r\n\r\n<div id=\"fs-id1165043010307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} x_n &amp; = x_{n-1}-\\frac{f(x_{n-1})}{f^{\\prime}(x_{n-1})} \\\\ &amp; = x_{n-1}-\\frac{(x_{n-1})^2-2}{2x_{n-1}} \\\\ &amp; = \\frac{1}{2}x_{n-1}+\\frac{1}{x_{n-1}} \\\\ &amp; = \\frac{1}{2}(x_{n-1}+\\frac{2}{x_{n-1}}) \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165042321204\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042705668\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1 = \\frac{1}{2}(x_0+\\frac{2}{x_0})=\\frac{1}{2}(2+\\frac{2}{2})=1.5 \\\\ x_2 = \\frac{1}{2}(x_1+\\frac{2}{x_1})=\\frac{1}{2}(1.5+\\frac{2}{1.5})\\approx 1.416666667 \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165043327685\">Continuing in this way, we find that<\/p>\r\n\r\n<div id=\"fs-id1165043384502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1=1.5 \\\\ x_2 \\approx 1.416666667 \\\\ x_3 \\approx 1.414215686 \\\\ x_4 \\approx 1.414213562 \\\\ x_5 \\approx 1.414213562 \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165043190853\">Since we obtained the same value for [latex]x_4[\/latex] and [latex]x_5[\/latex], it is unlikely that the value [latex]x_n[\/latex] will change on any subsequent application of Newton\u2019s method. We conclude that [latex]\\sqrt{2} \\approx 1.414213562[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"542\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211330\/CNX_Calc_Figure_04_09_003.jpg\" alt=\"The function y = x2 \u2013 2 is drawn. A dashed line comes up from x0 = 2, and a tangent line is drawn down from there. It touches x1 = 1.5, which is near x* = the square root of 2.\" width=\"542\" height=\"496\" \/> Figure 3. We can use Newton\u2019s method to find [latex]\\sqrt{2}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043098719\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse Newton\u2019s method to approximate [latex]\\sqrt{3}[\/latex] by letting [latex]f(x)=x^2-3[\/latex] and [latex]x_0=3[\/latex]. Find [latex]x_1[\/latex] and [latex]x_2[\/latex].\r\n\r\n[reveal-answer q=\"107083\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"107083\"]\r\n\r\nFor [latex]f(x)=x^2-3[\/latex], the equation for\u00a0[latex]x_n[\/latex] reduces to [latex]x_n=\\frac{x_{n-1}}{2}+\\frac{3}{2x_{n-1}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043395413\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043395413\"]\r\n\r\n[latex]x_1=2, \\, x_2=1.75[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]223631[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042317491\">When using Newton\u2019s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function [latex]F(x)=x-\\left[\\frac{f(x)}{f^{\\prime}(x)}\\right][\/latex], we can rewrite the equation for[latex]x_n[\/latex] as [latex]x_n=F(x_{n-1})[\/latex]. This type of process, where each [latex]x_n[\/latex] is defined in terms of [latex]x_{n-1}[\/latex] by repeating the same function, is an example of an <strong>iterative process<\/strong>. Shortly, we examine other iterative processes. First, let\u2019s look at the reasons why Newton\u2019s method could fail to find a root.<\/p>\r\n\r\n<h2>Failures of Newton\u2019s Method<\/h2>\r\n<p id=\"fs-id1165043086275\">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<ol id=\"fs-id1165042429145\">\r\n \t<li>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<li>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<li>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[caption id=\"\" align=\"aligncenter\" width=\"610\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211335\/CNX_Calc_Figure_04_09_004.jpg\" alt=\"A function is drawn with two roots, labeled root sought and root found. A point x0 is chosen such that when the tangent of x0 is taken, even though it is nearer to the root sought, the tangent points to the root found.\" width=\"610\" height=\"375\" \/> 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<div id=\"fs-id1165043091063\" class=\"textbook exercises\">\r\n<h3>example: When Newton\u2019s Method Fails<\/h3>\r\nConsider 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].\r\n<div id=\"fs-id1165042562908\" class=\"exercise\">[reveal-answer q=\"fs-id1165043327561\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043327561\"]\r\n<p id=\"fs-id1165043327561\">For [latex]f(x)=x^3-2x+2[\/latex], the derivative is [latex]f^{\\prime}(x)=3x^2-2[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042551991\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<p id=\"fs-id1165043092291\">In the next step,<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[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<p id=\"fs-id1165042926618\">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[caption id=\"\" align=\"aligncenter\" width=\"425\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211339\/CNX_Calc_Figure_04_09_005.jpg\" alt=\"The function f(x) = x3 \u2013 2x + 2 is drawn, which has a root between \u22122 and \u22121. The tangent from x = 0 goes to x = 1, and the tangent from x = 1 goes to x = 0.\" width=\"425\" height=\"313\" \/> Figure 5. The approximations continue to alternate between 0 and 1 and never approach the root of [latex]f[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: When Newton\u2019s Method Fails.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6f9WGeq3oj0?controls=0&amp;start=775&amp;end=911&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.9NewtonsMethod775to911_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.9 Newton's Method\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165043428456\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFor [latex]f(x)=x^3-2x+2[\/latex], let [latex]x_0=-1.5[\/latex] and find [latex]x_1[\/latex] and [latex]x_2[\/latex].\r\n\r\n[reveal-answer q=\"86000122\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"86000122\"]\r\n\r\nUse the equation for [latex]x_n[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042608726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042608726\"]\r\n\r\n[latex]x_1 \\approx -1.842105263, \\, x_2 \\approx -1.772826920[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042331529\">From the example above, we see that Newton\u2019s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton\u2019s method are included in texts on numerical analysis.<\/p>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the steps of Newton\u2019s method<\/li>\n<li>Explain what an iterative process means<\/li>\n<li>Recognize when Newton\u2019s method does not work<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165042613128\" class=\"bc-section section\">\n<h2>Describing Newton\u2019s Method<\/h2>\n<p id=\"fs-id1165043187457\">Consider the task of finding the solutions of [latex]f(x)=0[\/latex]. If [latex]f[\/latex] is the first-degree polynomial [latex]f(x)=ax+b[\/latex], then the solution of [latex]f(x)=0[\/latex] is given by the formula [latex]x=-\\frac{b}{a}[\/latex]. If [latex]f[\/latex] is the second-degree polynomial [latex]f(x)=ax^2+bx+c[\/latex], the solutions of [latex]f(x)=0[\/latex] can be found by using the quadratic formula. However, for polynomials of degree 3 or more, finding roots of [latex]f[\/latex] becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if [latex]f[\/latex] is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example, consider the function<\/p>\n<div id=\"fs-id1165042548842\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=x^5+8x^4+4x^3-2x-7[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043013826\">No formula exists that allows us to find the solutions of [latex]f(x)=0[\/latex]. Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of [latex]\\tan (x)-x=0[\/latex]. No simple formula exists for the solutions of this equation. In cases such as these, we can use <strong>Newton\u2019s method<\/strong> to approximate the roots.<\/p>\n<p id=\"fs-id1165043104195\">Newton\u2019s method makes use of the following idea to approximate the solutions of [latex]f(x)=0[\/latex]. By sketching a graph of [latex]f[\/latex], we can estimate a root of [latex]f(x)=0[\/latex]. Let\u2019s call this estimate [latex]x_0[\/latex]. We then draw the tangent line to [latex]f[\/latex] at [latex]x_0[\/latex]. If [latex]f^{\\prime}(x_0)\\ne 0[\/latex], this tangent line intersects the [latex]x[\/latex]-axis at some point [latex](x_1,0)[\/latex]. Now let [latex]x_1[\/latex] be the next approximation to the actual root. Typically, [latex]x_1[\/latex] is closer than [latex]x_0[\/latex] to an actual root. Next we draw the tangent line to [latex]f[\/latex] at [latex]x_1[\/latex]. If [latex]f^{\\prime}(x_1)\\ne 0[\/latex], this tangent line also intersects the [latex]x[\/latex]-axis, producing another approximation, [latex]x_2[\/latex]. We continue in this way, deriving a list of approximations: [latex]x_0, x_1, x_2, \\cdots[\/latex]. Typically, the numbers [latex]x_0,x_1,x_2, \\cdots[\/latex] quickly approach an actual root [latex]x*[\/latex], as shown in the following figure.<\/p>\n<div style=\"width: 560px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211323\/CNX_Calc_Figure_04_09_001.jpg\" alt=\"This function f(x) is drawn with points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)) marked on the function. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x1. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x2. If a tangent line were drawn from (x2, f(x2)), it appears that it would come very close to x*, which is the actual root. Each tangent line drawn in this order appears to get closer and closer to x*.\" width=\"550\" height=\"439\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The approximations [latex]x_0,x_1,x_2, \\cdots[\/latex] approach the actual root [latex]x*[\/latex]. The approximations are derived by looking at tangent lines to the graph of [latex]f[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165042981157\">Now let\u2019s look at how to calculate the approximations [latex]x_0,x_1,x_2, \\cdots[\/latex]. If [latex]x_0[\/latex] is our first approximation, the approximation [latex]x_1[\/latex] is defined by letting [latex](x_1,0)[\/latex] be the [latex]x[\/latex]-intercept of the tangent line to [latex]f[\/latex] at [latex]x_0[\/latex]. The equation of this tangent line is given by<\/p>\n<div id=\"fs-id1165042526516\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(x_0)+f^{\\prime}(x_0)(x-x_0)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\">Therefore, [latex]x_1[\/latex] must satisfy<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165043276857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x_0)+f^{\\prime}(x_0)(x_1-x_0)=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042328480\">Solving this equation for [latex]x_1[\/latex], we conclude that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_1=x_0-\\dfrac{f(x_0)}{f^{\\prime}(x_0)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042881142\">Similarly, the point [latex](x_2,0)[\/latex] is the [latex]x[\/latex]-intercept of the tangent line to [latex]f[\/latex] at [latex]x_1[\/latex]. Therefore, [latex]x_2[\/latex] satisfies the equation<\/p>\n<div id=\"fs-id1165043082009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_2=x_1-\\dfrac{f(x_1)}{f^{\\prime}(x_1)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043351477\">In general, for [latex]n>0, \\, x_n[\/latex] satisfies<\/p>\n<div id=\"fs-id1165043389758\" class=\"equation\" style=\"text-align: center;\">[latex]x_n=x_{n-1}-\\dfrac{f(x_{n-1})}{f^{\\prime}(x_{n-1})}[\/latex]<\/div>\n<p>Next we see how to make use of this technique to approximate the root of the polynomial [latex]f(x)=x^3-3x+1[\/latex].<\/p>\n<div id=\"fs-id1165043423917\" class=\"textbook exercises\">\n<h3>Example: Finding a Root of a Polynomial<\/h3>\n<p>Use Newton\u2019s method to approximate a root of [latex]f(x)=x^3-3x+1[\/latex] in the interval [latex][1,2][\/latex]. Let [latex]x_0=2[\/latex] and find [latex]x_1,x_2,x_3,x_4[\/latex], and [latex]x_5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q99095\">Show Solution<\/span><\/p>\n<div id=\"q99095\" class=\"hidden-answer\" style=\"display: none\">\n<p>From Figure 2, we see that [latex]f[\/latex] has one root over the interval [latex](1,2)[\/latex]. Therefore [latex]x_0=2[\/latex] seems like a reasonable first approximation. To find the next approximation, we use the equation we found for [latex]x_n[\/latex]. Since [latex]f(x)=x^3-3x+1[\/latex], the derivative is [latex]f^{\\prime}(x)=3x^2-3[\/latex]. Using the equation for\u00a0[latex]x_n[\/latex] with [latex]n=1[\/latex] (and a calculator that displays 10 digits), we obtain<\/p>\n<p style=\"text-align: center;\">[latex]x_1=x_0-\\dfrac{f(x_0)}{f^{\\prime}(x_0)}=2-\\dfrac{f(2)}{f^{\\prime}(2)}=2-\\dfrac{3}{9} \\approx 1.666666667[\/latex]<\/p>\n<p>To find the next approximation, [latex]x_2[\/latex], we use (Figure) with [latex]n=2[\/latex] and the value of [latex]x_1[\/latex] stored on the calculator. We find that<\/p>\n<p style=\"text-align: center;\">[latex]x_2=x_1=\\dfrac{f(x_1)}{f^{\\prime}(x_1)} \\approx 1.548611111[\/latex]<\/p>\n<p style=\"text-align: left;\">Continuing in this way, we obtain the following results:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} x_1 \\approx 1.666666667 \\\\ x_2 \\approx 1.548611111 \\\\ x_3 \\approx 1.532390162 \\\\ x_4 \\approx 1.532088989 \\\\ x_5 \\approx 1.532088886 \\\\ x_6 \\approx 1.532088886 \\end{array}[\/latex]<\/p>\n<p>We note that we obtained the same value for [latex]x_5[\/latex] and [latex]x_6[\/latex]. Therefore, any subsequent application of Newton\u2019s method will most likely give the same value for [latex]x_n[\/latex].<\/p>\n<div style=\"width: 435px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211326\/CNX_Calc_Figure_04_09_002.jpg\" alt=\"The function f(x) = x3 \u2013 3x + 1 is drawn. It has roots between \u22122 and \u22121, 0 and 1, and 1 and 2.\" width=\"425\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The function [latex]f(x)=x^3-3x+1[\/latex] has one root over the interval [latex][1,2][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding a Root of a Polynomial.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6f9WGeq3oj0?controls=0&amp;start=115&amp;end=485&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.9NewtonsMethod115to485_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.9 Newton&#8217;s Method&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165043433379\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Letting [latex]x_0=0[\/latex], use Newton\u2019s method to approximate the root of [latex]f(x)=x^3-3x+1[\/latex] over the interval [latex][0,1][\/latex] by calculating [latex]x_1[\/latex] and [latex]x_2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2661078\">Hint<\/span><\/p>\n<div id=\"q2661078\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the equation for [latex]x_n[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043345490\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043345490\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x_1 \\approx 0.33333333, \\, x_2 \\approx 0.347222222[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043348404\">Newton\u2019s method can also be used to approximate square roots. Here we show how to approximate [latex]\\sqrt{2}[\/latex]. This method can be modified to approximate the square root of any positive number.<\/p>\n<div id=\"fs-id1165042936506\" class=\"textbook exercises\">\n<h3>example: Finding a Square Root<\/h3>\n<p>Use Newton\u2019s method to approximate [latex]\\sqrt{2}[\/latex] (Figure 3). Let [latex]f(x)=x^2-2[\/latex], let [latex]x_0=2[\/latex], and calculate [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. (We note that since [latex]f(x)=x^2-2[\/latex] has a zero at [latex]\\sqrt{2}[\/latex], the initial value [latex]x_0=2[\/latex] is a reasonable choice to approximate [latex]\\sqrt{2}[\/latex].)<\/p>\n<div id=\"fs-id1165043182674\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043433206\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043433206\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043433206\">For [latex]f(x)=x^2-2, \\, f^{\\prime}(x)=2x[\/latex]. From the equation for [latex]x_n[\/latex], we know that<\/p>\n<div id=\"fs-id1165043010307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} x_n & = x_{n-1}-\\frac{f(x_{n-1})}{f^{\\prime}(x_{n-1})} \\\\ & = x_{n-1}-\\frac{(x_{n-1})^2-2}{2x_{n-1}} \\\\ & = \\frac{1}{2}x_{n-1}+\\frac{1}{x_{n-1}} \\\\ & = \\frac{1}{2}(x_{n-1}+\\frac{2}{x_{n-1}}) \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165042321204\">Therefore,<\/p>\n<div id=\"fs-id1165042705668\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1 = \\frac{1}{2}(x_0+\\frac{2}{x_0})=\\frac{1}{2}(2+\\frac{2}{2})=1.5 \\\\ x_2 = \\frac{1}{2}(x_1+\\frac{2}{x_1})=\\frac{1}{2}(1.5+\\frac{2}{1.5})\\approx 1.416666667 \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165043327685\">Continuing in this way, we find that<\/p>\n<div id=\"fs-id1165043384502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1=1.5 \\\\ x_2 \\approx 1.416666667 \\\\ x_3 \\approx 1.414215686 \\\\ x_4 \\approx 1.414213562 \\\\ x_5 \\approx 1.414213562 \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165043190853\">Since we obtained the same value for [latex]x_4[\/latex] and [latex]x_5[\/latex], it is unlikely that the value [latex]x_n[\/latex] will change on any subsequent application of Newton\u2019s method. We conclude that [latex]\\sqrt{2} \\approx 1.414213562[\/latex].<\/p>\n<div style=\"width: 552px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211330\/CNX_Calc_Figure_04_09_003.jpg\" alt=\"The function y = x2 \u2013 2 is drawn. A dashed line comes up from x0 = 2, and a tangent line is drawn down from there. It touches x1 = 1.5, which is near x* = the square root of 2.\" width=\"542\" height=\"496\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. We can use Newton\u2019s method to find [latex]\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043098719\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use Newton\u2019s method to approximate [latex]\\sqrt{3}[\/latex] by letting [latex]f(x)=x^2-3[\/latex] and [latex]x_0=3[\/latex]. Find [latex]x_1[\/latex] and [latex]x_2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107083\">Hint<\/span><\/p>\n<div id=\"q107083\" class=\"hidden-answer\" style=\"display: none\">\n<p>For [latex]f(x)=x^2-3[\/latex], the equation for\u00a0[latex]x_n[\/latex] reduces to [latex]x_n=\\frac{x_{n-1}}{2}+\\frac{3}{2x_{n-1}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043395413\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043395413\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x_1=2, \\, x_2=1.75[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm223631\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223631&theme=oea&iframe_resize_id=ohm223631&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1165042317491\">When using Newton\u2019s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function [latex]F(x)=x-\\left[\\frac{f(x)}{f^{\\prime}(x)}\\right][\/latex], we can rewrite the equation for[latex]x_n[\/latex] as [latex]x_n=F(x_{n-1})[\/latex]. This type of process, where each [latex]x_n[\/latex] is defined in terms of [latex]x_{n-1}[\/latex] by repeating the same function, is an example of an <strong>iterative process<\/strong>. Shortly, we examine other iterative processes. First, let\u2019s look at the reasons why Newton\u2019s method could fail to find a root.<\/p>\n<h2>Failures of Newton\u2019s Method<\/h2>\n<p id=\"fs-id1165043086275\">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>\n<ol id=\"fs-id1165042429145\">\n<li>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>\n<li>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>\n<li>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>\n<\/ol>\n<div style=\"width: 620px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211335\/CNX_Calc_Figure_04_09_004.jpg\" alt=\"A function is drawn with two roots, labeled root sought and root found. A point x0 is chosen such that when the tangent of x0 is taken, even though it is nearer to the root sought, the tangent points to the root found.\" width=\"610\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\">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.<\/p>\n<\/div>\n<div id=\"fs-id1165043091063\" class=\"textbook exercises\">\n<h3>example: When Newton\u2019s Method Fails<\/h3>\n<p>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>\n<div id=\"fs-id1165042562908\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043327561\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043327561\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043327561\">For [latex]f(x)=x^3-2x+2[\/latex], the derivative is [latex]f^{\\prime}(x)=3x^2-2[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165042551991\" class=\"equation unnumbered\" style=\"text-align: center;\">[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>\n<p id=\"fs-id1165043092291\">In the next step,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[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>\n<p id=\"fs-id1165042926618\">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>\n<div style=\"width: 435px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211339\/CNX_Calc_Figure_04_09_005.jpg\" alt=\"The function f(x) = x3 \u2013 2x + 2 is drawn, which has a root between \u22122 and \u22121. The tangent from x = 0 goes to x = 1, and the tangent from x = 1 goes to x = 0.\" width=\"425\" height=\"313\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The approximations continue to alternate between 0 and 1 and never approach the root of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: When Newton\u2019s Method Fails.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6f9WGeq3oj0?controls=0&amp;start=775&amp;end=911&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.9NewtonsMethod775to911_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.9 Newton&#8217;s Method&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165043428456\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>For [latex]f(x)=x^3-2x+2[\/latex], let [latex]x_0=-1.5[\/latex] and find [latex]x_1[\/latex] and [latex]x_2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q86000122\">Hint<\/span><\/p>\n<div id=\"q86000122\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the equation for [latex]x_n[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042608726\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042608726\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x_1 \\approx -1.842105263, \\, x_2 \\approx -1.772826920[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042331529\">From the example above, we see that Newton\u2019s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton\u2019s method are included in texts on numerical analysis.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-416\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.9 Newton&#039;s Method. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":33,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.9 Newton\\'s Method\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-416","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/416","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":18,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/416\/revisions"}],"predecessor-version":[{"id":4853,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/416\/revisions\/4853"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/416\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=416"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=416"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=416"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=416"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}