{"id":326,"date":"2023-09-20T22:49:07","date_gmt":"2023-09-20T22:49:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/other-iterative-processes\/"},"modified":"2024-08-05T13:33:13","modified_gmt":"2024-08-05T13:33:13","slug":"newtons-method-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/newtons-method-learn-it-3\/","title":{"raw":"Newton\u2019s Method: Learn It 3","rendered":"Newton\u2019s Method: Learn It 3"},"content":{"raw":"<h2>Other Iterative Processes<\/h2>\r\n<p id=\"fs-id1165043396127\">As mentioned earlier, Newton\u2019s method is a type of iterative process. We now look at an example of a different type of iterative process.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Consider a function [latex]F[\/latex] and an initial number [latex]x_0[\/latex]. Define the subsequent numbers [latex]x_n[\/latex] by the formula [latex]x_n=F(x_{n-1})[\/latex].<\/p>\r\n<p>This process is an iterative process that creates a list of numbers [latex]x_0,x_1,x_2, \\cdots ,x_n, \\cdots[\/latex]. This list of numbers may approach a finite number [latex]x^{*}[\/latex] as [latex]n[\/latex] gets larger, or it may not.<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165043396129\">\u00a0In the next example, we see an example of a function [latex]F[\/latex] and an initial guess [latex]x_0[\/latex] such that the resulting list of numbers approaches a finite value.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Let [latex]F(x)=\\frac{1}{2}x+4[\/latex] and let [latex]x_0=0[\/latex]. For all [latex]n \\ge 1[\/latex], let [latex]x_n=F(x_{n-1})[\/latex]. Find the values [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. Make a conjecture about what happens to this list of numbers [latex]x_1,x_2,x_3, \\cdots,x_n, \\cdots[\/latex] as [latex]n\\to \\infty[\/latex]. If the list of numbers [latex]x_1,x_2,x_3, \\cdots[\/latex] approaches a finite number [latex]x^*[\/latex], then [latex]x^*[\/latex] satisfies [latex]x^*=F(x^*)[\/latex], and [latex]x^*[\/latex] is called a <em>fixed point<\/em> of [latex]F[\/latex].<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165042373851\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042373851\"]\r\n\r\n<p id=\"fs-id1165042373851\">If [latex]x_0=0[\/latex], then<\/p>\r\n<div id=\"fs-id1165042330836\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1=\\frac{1}{2}(0)+4=4 \\\\ x_2=\\frac{1}{2}(4)+4=6 \\\\ x_3=\\frac{1}{2}(6)+4=7 \\\\ x_4=\\frac{1}{2}(7)+4=7.5 \\\\ x_5=\\frac{1}{2}(7.5)+4=7.75 \\\\ x_6=\\frac{1}{2}(7.75)+4=7.875 \\\\ x_7=\\frac{1}{2}(7.875)+4=7.9375 \\\\ x_8=\\frac{1}{2}(7.9375)+4=7.96875 \\\\ x_9=\\frac{1}{2}(7.96875)+4=7.984375 \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165042980764\">From this list, we conjecture that the values [latex]x_n[\/latex] approach 8.<\/p>\r\n<p id=\"fs-id1165043378115\">Figure 6 provides a graphical argument that the values approach 8 as [latex]n\\to \\infty[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"418\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211343\/CNX_Calc_Figure_04_09_006.jpg\" alt=\"The function F(x) = (1\/2)x + 4 is graphed along with y = x. From x0, which appears to be at the origin, a line is drawn to the function F(x) at x1 = F(x0). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x2 = F(x1). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x3 = F(x2). Continuing this process would converge on the two lines\u2019 intersection point at x* = 8.\" width=\"418\" height=\"382\" \/> Figure 6. This iterative process approaches the value [latex]x^*=8[\/latex].[\/caption]\r\n\r\n<p>Starting at the point [latex](x_0,x_0)[\/latex], we draw a vertical line to the point [latex](x_0,F(x_0))[\/latex]. The next number in our list is [latex]x_1=F(x_0)[\/latex]. We use [latex]x_1[\/latex] to calculate [latex]x_2[\/latex]. Therefore, we draw a horizontal line connecting [latex](x_0,x_1)[\/latex] to the point [latex](x_1,x_1)[\/latex] on the line [latex]y=x[\/latex], and then draw a vertical line connecting [latex](x_1,x_1)[\/latex] to the point [latex](x_1,F(x_1))[\/latex]. The output [latex]F(x_1)[\/latex] becomes [latex]x_2[\/latex].<\/p>\r\n<p>Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines [latex]F(x)=\\frac{x}{2}+4[\/latex] and [latex]y=x[\/latex]. The line segments get closer to the intersection point of these two lines, which occurs when [latex]x=F(x)[\/latex].<\/p>\r\n<p>Solving the equation [latex]x=\\frac{x}{2}+4[\/latex], we conclude they intersect at [latex]x=8[\/latex]. Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers [latex]x_0,x_1,x_2, \\cdots[\/latex] approaches [latex]x^*=8[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6f9WGeq3oj0?controls=0&amp;start=926&amp;end=1123&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>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\/4.9NewtonsMethod926to1123_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>Consider the function [latex]F(x)=\\frac{1}{3}x+6[\/latex]. Let [latex]x_0=0[\/latex] and let [latex]x_n=F(x_{n-1})[\/latex] for [latex]n \\ge 2[\/latex]. Find [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. Make a conjecture about what happens to the list of numbers [latex]x_1,x_2,x_3, \\cdots, x_n, \\cdots[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\r\n<p>[reveal-answer q=\"5564830\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"5564830\"]<\/p>\r\n<p>Consider the point where the lines [latex]y=x[\/latex] and [latex]y=F(x)[\/latex] intersect.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1165042705588\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042705588\"]<\/p>\r\n<p>[latex]x_1=6, \\, x_2=8, \\, x_3=\\frac{26}{3}, \\, x_4=\\frac{80}{9}, \\, x_5=\\frac{242}{27}; \\, x^*=9[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]223596[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<h2>Other Iterative Processes<\/h2>\n<p id=\"fs-id1165043396127\">As mentioned earlier, Newton\u2019s method is a type of iterative process. We now look at an example of a different type of iterative process.<\/p>\n<section class=\"textbox example\">\n<p>Consider a function [latex]F[\/latex] and an initial number [latex]x_0[\/latex]. Define the subsequent numbers [latex]x_n[\/latex] by the formula [latex]x_n=F(x_{n-1})[\/latex].<\/p>\n<p>This process is an iterative process that creates a list of numbers [latex]x_0,x_1,x_2, \\cdots ,x_n, \\cdots[\/latex]. This list of numbers may approach a finite number [latex]x^{*}[\/latex] as [latex]n[\/latex] gets larger, or it may not.<\/p>\n<\/section>\n<p id=\"fs-id1165043396129\">\u00a0In the next example, we see an example of a function [latex]F[\/latex] and an initial guess [latex]x_0[\/latex] such that the resulting list of numbers approaches a finite value.<\/p>\n<section class=\"textbox example\">\n<p>Let [latex]F(x)=\\frac{1}{2}x+4[\/latex] and let [latex]x_0=0[\/latex]. For all [latex]n \\ge 1[\/latex], let [latex]x_n=F(x_{n-1})[\/latex]. Find the values [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. Make a conjecture about what happens to this list of numbers [latex]x_1,x_2,x_3, \\cdots,x_n, \\cdots[\/latex] as [latex]n\\to \\infty[\/latex]. If the list of numbers [latex]x_1,x_2,x_3, \\cdots[\/latex] approaches a finite number [latex]x^*[\/latex], then [latex]x^*[\/latex] satisfies [latex]x^*=F(x^*)[\/latex], and [latex]x^*[\/latex] is called a <em>fixed point<\/em> of [latex]F[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042373851\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042373851\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042373851\">If [latex]x_0=0[\/latex], then<\/p>\n<div id=\"fs-id1165042330836\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} x_1=\\frac{1}{2}(0)+4=4 \\\\ x_2=\\frac{1}{2}(4)+4=6 \\\\ x_3=\\frac{1}{2}(6)+4=7 \\\\ x_4=\\frac{1}{2}(7)+4=7.5 \\\\ x_5=\\frac{1}{2}(7.5)+4=7.75 \\\\ x_6=\\frac{1}{2}(7.75)+4=7.875 \\\\ x_7=\\frac{1}{2}(7.875)+4=7.9375 \\\\ x_8=\\frac{1}{2}(7.9375)+4=7.96875 \\\\ x_9=\\frac{1}{2}(7.96875)+4=7.984375 \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165042980764\">From this list, we conjecture that the values [latex]x_n[\/latex] approach 8.<\/p>\n<p id=\"fs-id1165043378115\">Figure 6 provides a graphical argument that the values approach 8 as [latex]n\\to \\infty[\/latex].<\/p>\n<figure style=\"width: 418px\" 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\/11211343\/CNX_Calc_Figure_04_09_006.jpg\" alt=\"The function F(x) = (1\/2)x + 4 is graphed along with y = x. From x0, which appears to be at the origin, a line is drawn to the function F(x) at x1 = F(x0). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x2 = F(x1). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x3 = F(x2). Continuing this process would converge on the two lines\u2019 intersection point at x* = 8.\" width=\"418\" height=\"382\" \/><figcaption class=\"wp-caption-text\">Figure 6. This iterative process approaches the value [latex]x^*=8[\/latex].<\/figcaption><\/figure>\n<p>Starting at the point [latex](x_0,x_0)[\/latex], we draw a vertical line to the point [latex](x_0,F(x_0))[\/latex]. The next number in our list is [latex]x_1=F(x_0)[\/latex]. We use [latex]x_1[\/latex] to calculate [latex]x_2[\/latex]. Therefore, we draw a horizontal line connecting [latex](x_0,x_1)[\/latex] to the point [latex](x_1,x_1)[\/latex] on the line [latex]y=x[\/latex], and then draw a vertical line connecting [latex](x_1,x_1)[\/latex] to the point [latex](x_1,F(x_1))[\/latex]. The output [latex]F(x_1)[\/latex] becomes [latex]x_2[\/latex].<\/p>\n<p>Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines [latex]F(x)=\\frac{x}{2}+4[\/latex] and [latex]y=x[\/latex]. The line segments get closer to the intersection point of these two lines, which occurs when [latex]x=F(x)[\/latex].<\/p>\n<p>Solving the equation [latex]x=\\frac{x}{2}+4[\/latex], we conclude they intersect at [latex]x=8[\/latex]. Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers [latex]x_0,x_1,x_2, \\cdots[\/latex] approaches [latex]x^*=8[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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=926&amp;end=1123&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/4.9NewtonsMethod926to1123_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>Consider the function [latex]F(x)=\\frac{1}{3}x+6[\/latex]. Let [latex]x_0=0[\/latex] and let [latex]x_n=F(x_{n-1})[\/latex] for [latex]n \\ge 2[\/latex]. Find [latex]x_1,x_2,x_3,x_4,x_5[\/latex]. Make a conjecture about what happens to the list of numbers [latex]x_1,x_2,x_3, \\cdots, x_n, \\cdots[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q5564830\">Hint<\/button><\/p>\n<div id=\"q5564830\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider the point where the lines [latex]y=x[\/latex] and [latex]y=F(x)[\/latex] intersect.<\/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-id1165042705588\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042705588\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x_1=6, \\, x_2=8, \\, x_3=\\frac{26}{3}, \\, x_4=\\frac{80}{9}, \\, x_5=\\frac{242}{27}; \\, x^*=9[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm223596\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223596&theme=lumen&iframe_resize_id=ohm223596&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":25,"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\/326"}],"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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/326\/revisions"}],"predecessor-version":[{"id":4560,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/326\/revisions\/4560"}],"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\/326\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=326"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=326"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=326"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=326"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}