{"id":325,"date":"2023-09-20T22:49:06","date_gmt":"2023-09-20T22:49:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/approximating-with-newtons-method\/"},"modified":"2025-06-04T16:17:09","modified_gmt":"2025-06-04T16:17:09","slug":"newtons-method-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/newtons-method-learn-it-1\/","title":{"raw":"Newton\u2019s Method: Learn It 1","rendered":"Newton\u2019s Method: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Explain how Newton\u2019s method uses repetition to find roots of equations<\/li>\r\n\t<li>Recognize when Newton\u2019s method does not work<\/li>\r\n\t<li>Apply methods that repeat steps to solve different types of mathematical problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Approximating with Newton\u2019s Method<\/h2>\r\n<p>In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form [latex]f(x)=0[\/latex]. For most functions, however, it is difficult\u2014if not impossible\u2014to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the <span class=\"no-emphasis\">zeroes of functions<\/span>. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.<\/p>\r\n<h3>Describing Newton\u2019s Method<\/h3>\r\n<p id=\"fs-id1165043187457\">Consider the task of finding the solutions of [latex]f(x)=0[\/latex].<\/p>\r\n<p>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].<\/p>\r\n<p>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.<\/p>\r\n<p>However, for polynomials of degree [latex]3[\/latex] 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 [latex]5[\/latex] or greater, it is known that no such formulas exist.<\/p>\r\n<p>Consider the function [latex]f(x)=x^5+8x^4+4x^3-2x-7[\/latex]. No formula exists that allows us to find the solutions of [latex]f(x)=0[\/latex].<\/p>\r\n<p>Similar difficulties exist for nonpolynomial functions. Consider the task of finding solutions of [latex] \\tan (x)-x=0[\/latex]. No simple formula exists for the solutions of this equation.<\/p>\r\n<p>In cases such as these, we can use <strong>Newton\u2019s method<\/strong> to approximate the roots.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Newton\u2019s method<\/h3>\r\n<p>Newton's Method is an efficient numerical technique used to find approximately accurate roots of a real-valued function. By starting from an initial guess, the method iteratively refines this guess using the function and its derivative, quickly converging to a root where the function value is zero.<\/p>\r\n<\/section>\r\n<p>Newton\u2019s method makes use of the following idea to approximate the solutions of [latex]f(x)=0[\/latex].<\/p>\r\n<p>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].<\/p>\r\n<p>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].<\/p>\r\n<p>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\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<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<div class=\"equation unnumbered\">Therefore, [latex]x_1[\/latex] must satisfy<\/div>\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<p id=\"fs-id1165042328480\">Solving this equation for [latex]x_1[\/latex], we conclude that<\/p>\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<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<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<p id=\"fs-id1165043351477\">In general, for [latex]n&gt;0, \\, x_n[\/latex] satisfies<\/p>\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\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>\r\n<section class=\"textbox example\">\r\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>\r\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.<\/p>\r\n<p>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>\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\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>\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\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>\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]\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>Watch the following video to see the worked solution to the example above.<\/p>\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\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\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.<\/p>\r\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 \"4.9 Newton's Method\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\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>\r\n<p>[reveal-answer q=\"2661078\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"2661078\"]<\/p>\r\n<p>Use the equation for [latex]x_n[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043345490\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043345490\"]<\/p>\r\n<p>[latex]x_1 \\approx 0.33333333, \\, x_2 \\approx 0.347222222[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>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<section class=\"textbox example\">\r\n<p>Use Newton\u2019s method to approximate [latex]\\sqrt{2}[\/latex]. 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].<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"221179\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"221179\"]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].[\/hidden-answer]<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165043433206\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043433206\"]\r\n\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<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<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<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]\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]223631[\/ohm_question]<\/p>\r\n<\/section>\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>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Explain how Newton\u2019s method uses repetition to find roots of equations<\/li>\n<li>Recognize when Newton\u2019s method does not work<\/li>\n<li>Apply methods that repeat steps to solve different types of mathematical problems<\/li>\n<\/ul>\n<\/section>\n<h2>Approximating with Newton\u2019s Method<\/h2>\n<p>In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form [latex]f(x)=0[\/latex]. For most functions, however, it is difficult\u2014if not impossible\u2014to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the <span class=\"no-emphasis\">zeroes of functions<\/span>. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.<\/p>\n<h3>Describing Newton\u2019s Method<\/h3>\n<p id=\"fs-id1165043187457\">Consider the task of finding the solutions of [latex]f(x)=0[\/latex].<\/p>\n<p>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].<\/p>\n<p>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.<\/p>\n<p>However, for polynomials of degree [latex]3[\/latex] 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 [latex]5[\/latex] or greater, it is known that no such formulas exist.<\/p>\n<p>Consider the function [latex]f(x)=x^5+8x^4+4x^3-2x-7[\/latex]. No formula exists that allows us to find the solutions of [latex]f(x)=0[\/latex].<\/p>\n<p>Similar difficulties exist for nonpolynomial functions. Consider the task of finding solutions of [latex]\\tan (x)-x=0[\/latex]. No simple formula exists for the solutions of this equation.<\/p>\n<p>In cases such as these, we can use <strong>Newton\u2019s method<\/strong> to approximate the roots.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Newton\u2019s method<\/h3>\n<p>Newton&#8217;s Method is an efficient numerical technique used to find approximately accurate roots of a real-valued function. By starting from an initial guess, the method iteratively refines this guess using the function and its derivative, quickly converging to a root where the function value is zero.<\/p>\n<\/section>\n<p>Newton\u2019s method makes use of the following idea to approximate the solutions of [latex]f(x)=0[\/latex].<\/p>\n<p>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].<\/p>\n<p>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].<\/p>\n<p>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<figure style=\"width: 550px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\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<div class=\"equation unnumbered\">Therefore, [latex]x_1[\/latex] must satisfy<\/div>\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 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 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 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<section class=\"textbox example\">\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<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.<\/p>\n<p>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<figure style=\"width: 425px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox watchIt\">\n<p>Watch the following video to see the worked solution to the example above.<\/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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/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<\/section>\n<section class=\"textbox example\">\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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q2661078\">Hint<\/button><\/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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043345490\">Show Solution<\/button><\/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<\/section>\n<p>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<section class=\"textbox example\">\n<p>Use Newton\u2019s method to approximate [latex]\\sqrt{2}[\/latex]. 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].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q221179\">Hint<\/button><\/p>\n<div id=\"q221179\" class=\"hidden-answer\" style=\"display: none\">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].<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043433206\">Show Solution<\/button><\/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<figure style=\"width: 542px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 3. We can use Newton\u2019s method to find [latex]\\sqrt{2}[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm223631\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223631&theme=lumen&iframe_resize_id=ohm223631&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\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","protected":false},"author":6,"menu_order":23,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"learn_it","content_attributions":null,"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/325"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/325\/revisions"}],"predecessor-version":[{"id":4719,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/325\/revisions\/4719"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/325\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=325"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=325"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=325"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}