{"id":863,"date":"2025-06-20T17:18:43","date_gmt":"2025-06-20T17:18:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=863"},"modified":"2025-09-18T14:40:49","modified_gmt":"2025-09-18T14:40:49","slug":"basics-of-differential-equations-apply-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-apply-it-2\/","title":{"raw":"Sequences and Their Properties: Apply It","rendered":"Sequences and Their Properties: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the pattern in a sequence and write a formula for its terms<\/li>\r\n \t<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\r\n \t<li>Figure out if a sequence converges or diverges<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 style=\"text-align: left;\" data-type=\"title\">The Fibonacci Sequence: From Recursion to the Golden Ratio<\/h2>\r\nThe Fibonacci sequence is one of mathematics' most fascinating patterns, appearing everywhere from flower petals to spiral galaxies. Named after Leonardo Fibonacci, who introduced it to Western mathematics in 1202, this sequence demonstrates how simple recursive rules can produce profound mathematical relationships.\r\n<p id=\"fs-id1169739202389\">The <span class=\"no-emphasis\" data-type=\"term\">Fibonacci numbers<\/span> are defined recursively by the sequence [latex]\\left\\{{F}_{n}\\right\\}[\/latex] where [latex]{F}_{0}=0[\/latex], [latex]{F}_{1}=1[\/latex] and for [latex]n\\ge 2[\/latex], [latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex].<\/p>\r\n<p id=\"fs-id1169739344200\">In this exploration, you'll discover how to move from the recursive definition of Fibonacci numbers to a closed formula, and ultimately connect this ancient sequence to the golden ratio\u2014a number that has captivated mathematicians, artists, and architects for centuries.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Write out the first twenty Fibonacci numbers.[reveal-answer q=\"457784\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"457784\"]\r\n\r\nStarting with [latex]F_0 = 0[\/latex] and [latex]F_1 = 1[\/latex], and using [latex]F_n = F_{n-1} + F_{n-2}[\/latex]:\r\n\r\n[latex]\r\n\\begin{array}{rcl}\r\nF_0 &amp; = &amp; 0 \\\\\r\nF_1 &amp; = &amp; 1 \\\\\r\nF_2 &amp; = &amp; F_1 + F_0 = 1 + 0 = 1 \\\\\r\nF_3 &amp; = &amp; F_2 + F_1 = 1 + 1 = 2 \\\\\r\nF_4 &amp; = &amp; F_3 + F_2 = 2 + 1 = 3 \\\\\r\nF_5 &amp; = &amp; F_4 + F_3 = 3 + 2 = 5 \\\\\r\nF_6 &amp; = &amp; F_5 + F_4 = 5 + 3 = 8 \\\\\r\nF_7 &amp; = &amp; F_6 + F_5 = 8 + 5 = 13 \\\\\r\nF_8 &amp; = &amp; F_7 + F_6 = 13 + 8 = 21 \\\\\r\nF_9 &amp; = &amp; F_8 + F_7 = 21 + 13 = 34 \\\\\r\nF_{10} &amp; = &amp; F_9 + F_8 = 34 + 21 = 55 \\\\\r\nF_{11} &amp; = &amp; F_{10} + F_9 = 55 + 34 = 89 \\\\\r\nF_{12} &amp; = &amp; F_{11} + F_{10} = 89 + 55 = 144 \\\\\r\nF_{13} &amp; = &amp; F_{12} + F_{11} = 144 + 89 = 233 \\\\\r\nF_{14} &amp; = &amp; F_{13} + F_{12} = 233 + 144 = 377 \\\\\r\nF_{15} &amp; = &amp; F_{14} + F_{13} = 377 + 233 = 610 \\\\\r\nF_{16} &amp; = &amp; F_{15} + F_{14} = 610 + 377 = 987 \\\\\r\nF_{17} &amp; = &amp; F_{16} + F_{15} = 987 + 610 = 1597 \\\\\r\nF_{18} &amp; = &amp; F_{17} + F_{16} = 1597 + 987 = 2584 \\\\\r\nF_{19} &amp; = &amp; F_{18} + F_{17} = 2584 + 1597 = 4181 \\\\\r\n\\end{array}\r\n[\/latex]\r\n\r\nThe first twenty Fibonacci numbers are: [latex]0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find a closed formula for the Fibonacci sequence by using the following steps.\r\n<ol id=\"fs-id1169739344213\" type=\"a\">\r\n \t<li>Consider the recursively defined sequence [latex]\\left\\{{x}_{n}\\right\\}[\/latex] where [latex]{x}_{o}=c[\/latex] and [latex]{x}_{n+1}=a{x}_{n}[\/latex]. Show that this sequence can be described by the closed formula [latex]{x}_{n}=c{a}^{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/li>\r\n \t<li>Using the result from part a. as motivation, look for a solution of the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739344309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nof the form [latex]{F}_{n}=c{\\lambda }^{n}[\/latex]. Determine what two values for [latex]\\lambda [\/latex] will allow [latex]{F}_{n}[\/latex] to satisfy this equation.<\/li>\r\n \t<li>Consider the two solutions from part b.: [latex]{\\lambda }_{1}[\/latex] and [latex]{\\lambda }_{2}[\/latex]. Let [latex]{F}_{n}={c}_{1}{\\lambda }_{1}{}^{n}+{c}_{2}{\\lambda }_{2}{}^{n}[\/latex]. Use the initial conditions [latex]{F}_{0}[\/latex] and [latex]{F}_{1}[\/latex] to determine the values for the constants [latex]{c}_{1}[\/latex] and [latex]{c}_{2}[\/latex] and write the closed formula [latex]{F}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"926812\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"926812\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">For [latex]n = 0[\/latex]: [latex]x_0 = c[\/latex] (given) [latex]ca^0 = c \\cdot 1 = c[\/latex] So the formula holds for [latex]n = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Assume [latex]x_k = ca^k[\/latex] for some [latex]k \\geq 0[\/latex]. We need to show [latex]x_{k+1} = ca^{k+1}[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">From the recurrence relation: [latex]x_{k+1} = ax_k[\/latex] Using: [latex]x_{k+1} = a(ca^k) = ca^{k+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, by mathematical induction, [latex]x_n = ca^n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">If [latex]F_n = c\\lambda^n[\/latex], then:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]F_{n-1} = c\\lambda^{n-1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]F_{n-2} = c\\lambda^{n-2}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">Substituting into the Fibonacci recurrence relation: [latex]c\\lambda^n = c\\lambda^{n-1} + c\\lambda^{n-2}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Dividing by [latex]c\\lambda^{n-2}[\/latex] (assuming [latex]c \\neq 0[\/latex] and [latex]\\lambda \\neq 0[\/latex]): [latex]\\lambda^2 = \\lambda + 1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Rearranging: [latex]\\lambda^2 - \\lambda - 1 = 0[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Using the quadratic formula: [latex]\\lambda = \\frac{1 \\pm \\sqrt{1 + 4}}{2} = \\frac{1 \\pm \\sqrt{5}}{2}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore, the two values are: [latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex] and [latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">We have [latex]F_n = c_1\\lambda_1^n + c_2\\lambda_2^n[\/latex] where:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Using the initial conditions:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]n = 0[\/latex]: [latex]F_0 = 0[\/latex] [latex]0 = c_1\\lambda_1^0 + c_2\\lambda_2^0 = c_1 + c_2[\/latex] So [latex]c_1 + c_2 = 0[\/latex], which means [latex]c_2 = -c_1[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]n = 1[\/latex]: [latex]F_1 = 1[\/latex] [latex]1 = c_1\\lambda_1 + c_2\\lambda_2[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Substituting [latex]c_2 = -c_1[\/latex]: [latex]1 = c_1\\lambda_1 - c_1\\lambda_2 = c_1(\\lambda_1 - \\lambda_2)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Now, [latex]\\lambda_1 - \\lambda_2 = \\frac{1 + \\sqrt{5}}{2} - \\frac{1 - \\sqrt{5}}{2} = \\frac{2\\sqrt{5}}{2} = \\sqrt{5}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore: [latex]1 = c_1\\sqrt{5}[\/latex], so [latex]c_1 = \\frac{1}{\\sqrt{5}}[\/latex] and [latex]c_2 = -\\frac{1}{\\sqrt{5}}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The closed formula for the Fibonacci sequence is: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^n - \\frac{1}{\\sqrt{5}}\\left(\\frac{1 - \\sqrt{5}}{2}\\right)^n[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">This can also be written as: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left[\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^n - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^n\\right][\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the answer in the third part of the previous question to show that:\r\n<p style=\"text-align: center;\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{F}_{n+1}}{{F}_{n}}=\\frac{1+\\sqrt{5}}{2}[\/latex].<\/p>\r\n\r\n[reveal-answer q=\"310782\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"310782\"]\r\n<p class=\"whitespace-normal break-words\">From part 2(c), we have: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left[\\lambda_1^n - \\lambda_2^n\\right][\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex] and [latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]F_{n+1} = \\frac{1}{\\sqrt{5}}\\left[\\lambda_1^{n+1} - \\lambda_2^{n+1}\\right][\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Computing the ratio: [latex]\\frac{F_{n+1}}{F_n} = \\frac{\\frac{1}{\\sqrt{5}}[\\lambda_1^{n+1} - \\lambda_2^{n+1}]}{\\frac{1}{\\sqrt{5}}[\\lambda_1^n - \\lambda_2^n]} = \\frac{\\lambda_1^{n+1} - \\lambda_2^{n+1}}{\\lambda_1^n - \\lambda_2^n}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Factoring out [latex]\\lambda_1^n[\/latex] from both numerator and denominator: [latex]\\frac{F_{n+1}}{F_n} = \\frac{\\lambda_1^n(\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n)}{\\lambda_1^n(1 - (\\frac{\\lambda_2}{\\lambda_1})^n)} = \\frac{\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n}{1 - (\\frac{\\lambda_2}{\\lambda_1})^n}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Now we need to examine [latex]\\frac{\\lambda_2}{\\lambda_1}[\/latex]: [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{\\frac{1 - \\sqrt{5}}{2}}{\\frac{1 + \\sqrt{5}}{2}} = \\frac{1 - \\sqrt{5}}{1 + \\sqrt{5}}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Rationalizing the denominator: [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{1 - \\sqrt{5}}{1 + \\sqrt{5}} \\cdot \\frac{1 - \\sqrt{5}}{1 - \\sqrt{5}} = \\frac{(1 - \\sqrt{5})^2}{1 - 5} = \\frac{1 - 2\\sqrt{5} + 5}{-4} = \\frac{6 - 2\\sqrt{5}}{-4} = \\frac{\\sqrt{5} - 3}{2}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]\\sqrt{5} \\approx 2.236[\/latex], we have [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{\\sqrt{5} - 3}{2} \\approx \\frac{-0.764}{2} \\approx -0.382[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore [latex]\\left|\\frac{\\lambda_2}{\\lambda_1}\\right| &lt; 1[\/latex], which means [latex]\\left(\\frac{\\lambda_2}{\\lambda_1}\\right)^n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Taking the limit: [latex]\\lim_{n \\to \\infty} \\frac{F_{n+1}}{F_n} = \\lim_{n \\to \\infty} \\frac{\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n}{1 - (\\frac{\\lambda_2}{\\lambda_1})^n} = \\frac{\\lambda_1 - \\lambda_2 \\cdot 0}{1 - 0} = \\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\lim_{n \\to \\infty} \\frac{F_{n+1}}{F_n} = \\frac{1 + \\sqrt{5}}{2}[\/latex], which is indeed the golden ratio [latex]\\varphi[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox connectIt\" aria-label=\"Connect It\">The number [latex]\\varphi =\\frac{\\left(1+\\sqrt{5}\\right)}{2}[\/latex] is known as the <span class=\"no-emphasis\" data-type=\"term\">golden ratio<\/span> (Figures 7 and 8).\r\n<figure id=\"CNX_Calc_Figure_09_01_004\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234259\/CNX_Calc_Figure_09_01_004.jpg\" alt=\"This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number.\" width=\"325\" height=\"244\" data-media-type=\"image\/jpeg\" \/> Figure 7. The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number\u2014always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)[\/caption]<\/figure>\r\n<span data-type=\"newline\">\u00a0<\/span>\r\n<figure id=\"CNX_Calc_Figure_09_01_005\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234302\/CNX_Calc_Figure_09_01_005.jpg\" alt=\"This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple\u2019s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof.\" width=\"488\" height=\"300\" data-media-type=\"image\/jpeg\" \/> Figure 8. The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)[\/caption]<\/figure>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the pattern in a sequence and write a formula for its terms<\/li>\n<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\n<li>Figure out if a sequence converges or diverges<\/li>\n<\/ul>\n<\/section>\n<h2 style=\"text-align: left;\" data-type=\"title\">The Fibonacci Sequence: From Recursion to the Golden Ratio<\/h2>\n<p>The Fibonacci sequence is one of mathematics&#8217; most fascinating patterns, appearing everywhere from flower petals to spiral galaxies. Named after Leonardo Fibonacci, who introduced it to Western mathematics in 1202, this sequence demonstrates how simple recursive rules can produce profound mathematical relationships.<\/p>\n<p id=\"fs-id1169739202389\">The <span class=\"no-emphasis\" data-type=\"term\">Fibonacci numbers<\/span> are defined recursively by the sequence [latex]\\left\\{{F}_{n}\\right\\}[\/latex] where [latex]{F}_{0}=0[\/latex], [latex]{F}_{1}=1[\/latex] and for [latex]n\\ge 2[\/latex], [latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex].<\/p>\n<p id=\"fs-id1169739344200\">In this exploration, you&#8217;ll discover how to move from the recursive definition of Fibonacci numbers to a closed formula, and ultimately connect this ancient sequence to the golden ratio\u2014a number that has captivated mathematicians, artists, and architects for centuries.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Write out the first twenty Fibonacci numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q457784\">Show Answer<\/button><\/p>\n<div id=\"q457784\" class=\"hidden-answer\" style=\"display: none\">\n<p>Starting with [latex]F_0 = 0[\/latex] and [latex]F_1 = 1[\/latex], and using [latex]F_n = F_{n-1} + F_{n-2}[\/latex]:<\/p>\n<p>[latex]\\begin{array}{rcl}  F_0 & = & 0 \\\\  F_1 & = & 1 \\\\  F_2 & = & F_1 + F_0 = 1 + 0 = 1 \\\\  F_3 & = & F_2 + F_1 = 1 + 1 = 2 \\\\  F_4 & = & F_3 + F_2 = 2 + 1 = 3 \\\\  F_5 & = & F_4 + F_3 = 3 + 2 = 5 \\\\  F_6 & = & F_5 + F_4 = 5 + 3 = 8 \\\\  F_7 & = & F_6 + F_5 = 8 + 5 = 13 \\\\  F_8 & = & F_7 + F_6 = 13 + 8 = 21 \\\\  F_9 & = & F_8 + F_7 = 21 + 13 = 34 \\\\  F_{10} & = & F_9 + F_8 = 34 + 21 = 55 \\\\  F_{11} & = & F_{10} + F_9 = 55 + 34 = 89 \\\\  F_{12} & = & F_{11} + F_{10} = 89 + 55 = 144 \\\\  F_{13} & = & F_{12} + F_{11} = 144 + 89 = 233 \\\\  F_{14} & = & F_{13} + F_{12} = 233 + 144 = 377 \\\\  F_{15} & = & F_{14} + F_{13} = 377 + 233 = 610 \\\\  F_{16} & = & F_{15} + F_{14} = 610 + 377 = 987 \\\\  F_{17} & = & F_{16} + F_{15} = 987 + 610 = 1597 \\\\  F_{18} & = & F_{17} + F_{16} = 1597 + 987 = 2584 \\\\  F_{19} & = & F_{18} + F_{17} = 2584 + 1597 = 4181 \\\\  \\end{array}[\/latex]<\/p>\n<p>The first twenty Fibonacci numbers are: [latex]0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find a closed formula for the Fibonacci sequence by using the following steps.<\/p>\n<ol id=\"fs-id1169739344213\" type=\"a\">\n<li>Consider the recursively defined sequence [latex]\\left\\{{x}_{n}\\right\\}[\/latex] where [latex]{x}_{o}=c[\/latex] and [latex]{x}_{n+1}=a{x}_{n}[\/latex]. Show that this sequence can be described by the closed formula [latex]{x}_{n}=c{a}^{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/li>\n<li>Using the result from part a. as motivation, look for a solution of the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739344309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nof the form [latex]{F}_{n}=c{\\lambda }^{n}[\/latex]. Determine what two values for [latex]\\lambda[\/latex] will allow [latex]{F}_{n}[\/latex] to satisfy this equation.<\/li>\n<li>Consider the two solutions from part b.: [latex]{\\lambda }_{1}[\/latex] and [latex]{\\lambda }_{2}[\/latex]. Let [latex]{F}_{n}={c}_{1}{\\lambda }_{1}{}^{n}+{c}_{2}{\\lambda }_{2}{}^{n}[\/latex]. Use the initial conditions [latex]{F}_{0}[\/latex] and [latex]{F}_{1}[\/latex] to determine the values for the constants [latex]{c}_{1}[\/latex] and [latex]{c}_{2}[\/latex] and write the closed formula [latex]{F}_{n}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q926812\">Show Answer<\/button><\/p>\n<div id=\"q926812\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">For [latex]n = 0[\/latex]: [latex]x_0 = c[\/latex] (given) [latex]ca^0 = c \\cdot 1 = c[\/latex] So the formula holds for [latex]n = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Assume [latex]x_k = ca^k[\/latex] for some [latex]k \\geq 0[\/latex]. We need to show [latex]x_{k+1} = ca^{k+1}[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">From the recurrence relation: [latex]x_{k+1} = ax_k[\/latex] Using: [latex]x_{k+1} = a(ca^k) = ca^{k+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, by mathematical induction, [latex]x_n = ca^n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">If [latex]F_n = c\\lambda^n[\/latex], then:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]F_{n-1} = c\\lambda^{n-1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]F_{n-2} = c\\lambda^{n-2}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">Substituting into the Fibonacci recurrence relation: [latex]c\\lambda^n = c\\lambda^{n-1} + c\\lambda^{n-2}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Dividing by [latex]c\\lambda^{n-2}[\/latex] (assuming [latex]c \\neq 0[\/latex] and [latex]\\lambda \\neq 0[\/latex]): [latex]\\lambda^2 = \\lambda + 1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Rearranging: [latex]\\lambda^2 - \\lambda - 1 = 0[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Using the quadratic formula: [latex]\\lambda = \\frac{1 \\pm \\sqrt{1 + 4}}{2} = \\frac{1 \\pm \\sqrt{5}}{2}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore, the two values are: [latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex] and [latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">We have [latex]F_n = c_1\\lambda_1^n + c_2\\lambda_2^n[\/latex] where:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Using the initial conditions:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]n = 0[\/latex]: [latex]F_0 = 0[\/latex] [latex]0 = c_1\\lambda_1^0 + c_2\\lambda_2^0 = c_1 + c_2[\/latex] So [latex]c_1 + c_2 = 0[\/latex], which means [latex]c_2 = -c_1[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]n = 1[\/latex]: [latex]F_1 = 1[\/latex] [latex]1 = c_1\\lambda_1 + c_2\\lambda_2[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Substituting [latex]c_2 = -c_1[\/latex]: [latex]1 = c_1\\lambda_1 - c_1\\lambda_2 = c_1(\\lambda_1 - \\lambda_2)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Now, [latex]\\lambda_1 - \\lambda_2 = \\frac{1 + \\sqrt{5}}{2} - \\frac{1 - \\sqrt{5}}{2} = \\frac{2\\sqrt{5}}{2} = \\sqrt{5}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore: [latex]1 = c_1\\sqrt{5}[\/latex], so [latex]c_1 = \\frac{1}{\\sqrt{5}}[\/latex] and [latex]c_2 = -\\frac{1}{\\sqrt{5}}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">The closed formula for the Fibonacci sequence is: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^n - \\frac{1}{\\sqrt{5}}\\left(\\frac{1 - \\sqrt{5}}{2}\\right)^n[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">This can also be written as: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left[\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^n - \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^n\\right][\/latex]<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the answer in the third part of the previous question to show that:<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{F}_{n+1}}{{F}_{n}}=\\frac{1+\\sqrt{5}}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q310782\">Show Answer<\/button><\/p>\n<div id=\"q310782\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-normal break-words\">From part 2(c), we have: [latex]F_n = \\frac{1}{\\sqrt{5}}\\left[\\lambda_1^n - \\lambda_2^n\\right][\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]\\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex] and [latex]\\lambda_2 = \\frac{1 - \\sqrt{5}}{2}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]F_{n+1} = \\frac{1}{\\sqrt{5}}\\left[\\lambda_1^{n+1} - \\lambda_2^{n+1}\\right][\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Computing the ratio: [latex]\\frac{F_{n+1}}{F_n} = \\frac{\\frac{1}{\\sqrt{5}}[\\lambda_1^{n+1} - \\lambda_2^{n+1}]}{\\frac{1}{\\sqrt{5}}[\\lambda_1^n - \\lambda_2^n]} = \\frac{\\lambda_1^{n+1} - \\lambda_2^{n+1}}{\\lambda_1^n - \\lambda_2^n}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Factoring out [latex]\\lambda_1^n[\/latex] from both numerator and denominator: [latex]\\frac{F_{n+1}}{F_n} = \\frac{\\lambda_1^n(\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n)}{\\lambda_1^n(1 - (\\frac{\\lambda_2}{\\lambda_1})^n)} = \\frac{\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n}{1 - (\\frac{\\lambda_2}{\\lambda_1})^n}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Now we need to examine [latex]\\frac{\\lambda_2}{\\lambda_1}[\/latex]: [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{\\frac{1 - \\sqrt{5}}{2}}{\\frac{1 + \\sqrt{5}}{2}} = \\frac{1 - \\sqrt{5}}{1 + \\sqrt{5}}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Rationalizing the denominator: [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{1 - \\sqrt{5}}{1 + \\sqrt{5}} \\cdot \\frac{1 - \\sqrt{5}}{1 - \\sqrt{5}} = \\frac{(1 - \\sqrt{5})^2}{1 - 5} = \\frac{1 - 2\\sqrt{5} + 5}{-4} = \\frac{6 - 2\\sqrt{5}}{-4} = \\frac{\\sqrt{5} - 3}{2}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]\\sqrt{5} \\approx 2.236[\/latex], we have [latex]\\frac{\\lambda_2}{\\lambda_1} = \\frac{\\sqrt{5} - 3}{2} \\approx \\frac{-0.764}{2} \\approx -0.382[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore [latex]\\left|\\frac{\\lambda_2}{\\lambda_1}\\right| < 1[\/latex], which means [latex]\\left(\\frac{\\lambda_2}{\\lambda_1}\\right)^n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Taking the limit: [latex]\\lim_{n \\to \\infty} \\frac{F_{n+1}}{F_n} = \\lim_{n \\to \\infty} \\frac{\\lambda_1 - \\lambda_2(\\frac{\\lambda_2}{\\lambda_1})^n}{1 - (\\frac{\\lambda_2}{\\lambda_1})^n} = \\frac{\\lambda_1 - \\lambda_2 \\cdot 0}{1 - 0} = \\lambda_1 = \\frac{1 + \\sqrt{5}}{2}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\lim_{n \\to \\infty} \\frac{F_{n+1}}{F_n} = \\frac{1 + \\sqrt{5}}{2}[\/latex], which is indeed the golden ratio [latex]\\varphi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">The number [latex]\\varphi =\\frac{\\left(1+\\sqrt{5}\\right)}{2}[\/latex] is known as the <span class=\"no-emphasis\" data-type=\"term\">golden ratio<\/span> (Figures 7 and 8).<\/p>\n<figure id=\"CNX_Calc_Figure_09_01_004\">\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234259\/CNX_Calc_Figure_09_01_004.jpg\" alt=\"This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number.\" width=\"325\" height=\"244\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 7. The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number\u2014always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)<\/figcaption><\/figure>\n<\/figure>\n<p><span data-type=\"newline\">\u00a0<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_09_01_005\">\n<figure style=\"width: 488px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234302\/CNX_Calc_Figure_09_01_005.jpg\" alt=\"This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple\u2019s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof.\" width=\"488\" height=\"300\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 8. The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. 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