[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-657521-105":3,"detail-sidebar-cat-0-en-105":80,"doc-detail-657521-en":130},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":73,"head_meta":75,"extra_data":77,"updated_unix":79},105,"en","limitations-of-fixed-point-iteration-starter-iteration-process-graphs","Limitations of Fixed-Point Iteration - Starter - Iteration Process Graphs","","Lesson material on fixed-point iteration for solving equations, starting with reviewing a quartic polynomial. It demonstrates how to prove a root lies in an interval, derive the corresponding rearranged iteration formula, and use an initial guess to approximate the root to 3 d.p. It then explains iteration diagrams using cobweb and staircase constructions on the same axes, and states the convergence condition -1 \u003C g'(a) \u003C 1.",{"@graph":14,"@context":72},[15,34,55],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/limitations-of-fixed-point-iteration-starter-iteration-process-graphs/657521/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/limitations-of-fixed-point-iteration-starter-iteration-process-graphs/657521.png","ImageObject",300,407,{"name":42,"@type":43},"Genevieve","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-29","2026-10-26",true,{"@type":52,"interactionType":53,"userInteractionCount":22},"InteractionCounter",{"@type":54},"ViewAction",{"@type":56,"mainEntity":57},"FAQPage",[58,64,68],{"name":59,"@type":60,"acceptedAnswer":61},"How do you show that x^4 − 5x + 3 = 0 has a root between 1.4 and 1.5?","Question",{"text":62,"@type":63},"Define f(x)=x^4−5x+3 and evaluate f(1.4)\u003C0 and f(1.5)>0. Since the function is continuous, a sign change implies a root in the interval (1.4, 1.5).","Answer",{"name":65,"@type":60,"acceptedAnswer":66},"How is the fixed-point iteration formula derived for the polynomial example?",{"text":67,"@type":63},"Rearrange x^4 − 5x + 3 = 0 into the form x = 3√(5−3x)/3 (as given in the notes) so that xn+1 can be computed from xn using the derived iteration function g(x).",{"name":69,"@type":60,"acceptedAnswer":70},"What condition determines whether fixed-point iteration converges to the root a?",{"text":71,"@type":63},"For xn+1 = g(xn), the iterations converge to the root x=a when −1 \u003C g'(a) \u003C 1, i.e., |g'(a)| \u003C 1.","https://schema.org",{"og:url":32,"og:type":74,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":76,"canonical":32},"index,follow",{"doc_id":78,"site_id":7},657521,1792981953,{"code":4,"msg":81,"data":82},"success",[83,87,91,94,99,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":84,"show_sort_weight":85,"slug":86},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":88,"show_sort_weight":89,"slug":90},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":92,"slug":93},70,"exam",{"id":95,"doc_module":4,"doc_module_name":25,"category_name":96,"show_sort_weight":97,"slug":98},5,"Comic",60,"comic",{"id":100,"doc_module":4,"doc_module_name":25,"category_name":101,"show_sort_weight":102,"slug":103},6,"Technology",50,"technology",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},7,"Healthcare",40,"healthcare",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},8,"Research & Report",30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":95,"slug":129},19,"General","general",{"code":4,"msg":81,"data":131},{"doc_id":78,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":22,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":95,"language":139,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":12,"update_tm":79,"read_time":143},1374391974585,"https://ap-avatar.wpscdn.com/davatar_276721f389ce27ea32af1340a28f341c","Limitations of Fixed-Point Iteration  \nStarter  \n1 (Review of last lesson)  \n(a) Show that x 4 − 5x + 3 = 0 has a root in the interval 1.4 \u003C x \u003C 1.5.  \n(b) Show that x 4 − 5x + 3 = 0 can rearrange to the iteration formula  \nxn+1 = 3 5 − 3x~~ ~~  \nn  \n(c) Use the iterative formula xn+1 = 3 5 − 3x~~ ~~ with x0 = 1.45 to find an approximation  \nn  \nto the root to 3 d.p.  \nWorking: (a)  \n(b)  \n(c)  \nLet f (x) = x 4 − 5x + 3.  \nf (1.4) = 1.44 − 5 × 1.4 + 3 \u003C 0 f (1.5) = 1.54 − 5 × 1.5 + 3 > 0  \nSince there is a sign change and the curve is continuous, root between x = 1.4 and x = 1.5.  \nx 4 − 5x + 3 = 0  \nx 4 = 5x − 3  \nx 3 = 5 −  3  \nx  \nx = 3 5 − ~~ ~~3x then putting in the n and n + 1 , xn+1 = 3  \nthere is a  \n5 −  3  x  \nn  \nx0 = 1.45  \nx1 = 1.4311 x2 = 1.4267  \nx3 = 1.4256  \nx4 = 1.4253  \nx5 = 1.4253  \nSo the root is x = 1.425 to 3 d.p.  \nNotes  \nIn this lesson we will look at what the iteration process looks like graphically.  \nThe graphs of y = g (x) and y = x are drawn on the same axes, where g(x) is the iteration formula. The point where the graphs intersect is the root we are aiming for.  \nSuccess criteria for the iteration process  \n1. The iteration starts with the point x0 on the x−axis.  \n2. A vertical line goes up (or down) to the curve from x0.  \n3. At the curve the line becomes horizontal and goes to the line y = x. This converts they −value into an x−value —this is x1.  \n4. From x1 , a vertical line goes up (or down) to the curve  \n5. At the curve the line becomes horizontal and goes to the line y = x. This converts they −value into an x−value —this is x2.  \n6. The process repeats until the iterations converge, diverge or oscillate.  \n[www.mathspanda.com](www.mathspanda.com)  \nN.B. The key idea to remember is: up to the curve, across to the line y = x  \nOn the graph of the iteration, each vertical line represents an xn  \nE.g. 1 Show graphically the iteration process for the solution of x = cos x using the iteration formula xn+1 = cos xn. Use x0 = 0.1.  \nWorking: The iteration formula is xn+1 = cos xn so g(x) = cos x  \nHence, we draw y = cos x and y = x and on the same set of axes.  \nN.B. The graph of y = cos x must be in radians  \nx0 x2 x4 x3  \nx1  \nx5  \nAs can be seen from the graph, the iterations (blue lines) create a cobweb like structure. Hence, this type of diagram is called a cobweb diagram.  \nExtending the vertical blue lines down to the x−axis, gives us x0 , x1 , x2 etc.  \n[www.mathspanda.com](www.mathspanda.com)  \nE.g. 2 Show graphically the iteration process for the solution of x = x using the iteration formula xn+1 = x . Use x0 = 0.1.  \nn  \nWorking: The iteration formula is xn+1 = xn so g(x) = x  \nHence, we draw y = x and y = x and on the same set of axes.  \nx0 x1 x2 x3 x4  \nAs can be seen from the graph, the iterations (blue lines) create a staircase structure. Hence, this type of diagram is called a staircase diagram.  \nExtending the vertical blue lines down to the x−axis, gives us x0 , x1 , x2 etc.  \nDepending on the starting value the diagram of an iteration can form a cobweb or a staircase.  \nIteration diagrams  \nCobweb diagrams look like they’re spiralling in to or away from the root.  \nOn a convergent cobweb diagram the iterations alternate between being below and above the root, but are getting closer each time. A divergent cobweb diagram would have a similar shape, but each iteration would spiral away from the root.  \n[www.mathspanda.com](www.mathspanda.com)  \nStaircase diagrams form a set of steps going towards, or away from, the root.  \nE.g. 3 Complete questions (a)—(c) on the worksheet Sketching iteration diagrams.  \nIf you don’t have a printer, draw a rough sketch of the diagrams.  \nHints: Make sure you draw the line y = x on the diagram.  \nStart from the x−axis, go up/down to the curve, go across the line y = x, go up/ down to the curve, go across the line y = x, etc.  \nWorking:  \nWhen does the iteration converge to the root?  \nFixed-point iteration only finds the root x = a if the graph isn","cbCaigSEql0j6Tpb","https://ap.wps.com/l/cbCaigSEql0j6Tpb","pdf",21958662,"English","# Starter\n## Review of last lesson\n# Notes\n## Iteration process success criteria\n## Cobweb and staircase diagrams\n# When does the iteration converge?\n## Convergence condition for g'(a)\n# Exercise and summary","[{\"question\":\"How do you show that x^4 − 5x + 3 = 0 has a root between 1.4 and 1.5?\",\"answer\":\"Define f(x)=x^4−5x+3 and evaluate f(1.4)\\u003c0 and f(1.5)\\u003e0. Since the function is continuous, a sign change implies a root in the interval (1.4, 1.5).\"},{\"question\":\"How is the fixed-point iteration formula derived for the polynomial example?\",\"answer\":\"Rearrange x^4 − 5x + 3 = 0 into the form x = 3√(5−3x)/3 (as given in the notes) so that xn+1 can be computed from xn using the derived iteration function g(x).\"},{\"question\":\"What condition determines whether fixed-point iteration converges to the root a?\",\"answer\":\"For xn+1 = g(xn), the iterations converge to the root x=a when −1 \\u003c g'(a) \\u003c 1, i.e., |g'(a)| \\u003c 1.\"}]","Limitations of Fixed-Point Iteration - Starter - Iteration Process Graphs | PDF",13]