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From Fibonacci to Foxtrot: Investigating Recursion Relations with Geometric Sequences

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From Fibonacci to Foxtrot: Investigating Recursion Relations with Geometric Sequences 1 From Fibonacci to Foxtrot: 2 Investigating Recursion Relations with Geometric Sequences 3 4 5 The Algebra Standard from the Principles and Standards for School Mathematics 6 (NCTM, 2000) states that all students should “Understand patterns, relations, and 7 functions.” In particular, to meet the grades 9 – 12 expectations, students should 8 “generalize patterns using explicitly defined and recursively defined functions.” The 9 Foxtrot comic strip shown below (Amend) provides a wonderful starting point to create 10 lessons that address the algebra standard. 11 12 In the first three panels of the cartoon, we see that Marcus has scored a 13 touchdown by identifying Jason’s sequence 0, 1, 1, 2, 3, 5, 8, 13… as the “Fibonacci 14 series” (though it would be preferable to replace “series” with “sequence,” since in 15 mathematics terminology a series designates the sum of the terms in a sequence). The 16 Fibonacci sequence is one of the most widely known in all of mathematics, recursively =+ 17 defined by the recurrence relation FFFnnn++21. Thus, many readers of Foxtrot could 2 18 have emulated Marcus’ scoring success. However, there are at least two lingering 19 questions: 20 • Why did Jason begin his count with zero? It is more common to start the == == 21 Fibonacci sequence with FF121, 1 instead of FF010, 1. 22 • How can Jason score a touchdown? In the last panel of the comic strip, 23 Marcus has challenged Jason with the sequence 3, 0, 2, 3, 2, 5, … . What 24 is this sequence? 25 The key to our investigation of these questions is the geometric sequence 234 = n 26 aarar,, , arar , ,K, which is defined explicitly by xn ar , n = 0, 1, 2, … . = = 27 Alternatively, the sequence can be defined recursively by xnn+1 rx where x0 a . The x 28 constant r = n+1 is the common ratio of the sequence. xn 29 We begin by first investigating a connection between the Fibonacci sequence and 30 geometric sequences. The techniques we develop provide us with a method to investigate 31 the mysterious sequence 3, 0, 2, 3, 2, 5, … . 32 33 IS THE FIBONACCI SEQUENCE A GEOMETRIC SEQUENCE? 34 The short answer to the question just raised is no. After all, FF10/ is not defined, and == == 35 FF21/1/11 and FF32/2/12 are different ratios. Rather than being discouraged, 36 let’s examine several more ratios of successive Fibonacci numbers, as shown in the 37 following table. n 0123 45678 F F/nF 0112 3 5 8 1321 38 n+1 n - 1.0 2.0 1.5 1.667 1.600 1.625 1.615 1.619 3 39 As n becomes larger, the Fibonacci sequence increasingly resembles a geometric 40 sequence with a common ratio of about 1.6. Since the early integer values of the 41 Fibonacci sequence seem to cause difficulty, suppose we not worry yet about the initial = n 42 values of the sequence. Instead, we simply seek a geometric sequence xn ar that =+ 43 satisfies the Fibonacci recurrence xnnn++21xx. That is, we want to determine values of 44 r for which arnnn++21=+ ar ar , since it is clear that an arbitrary value of the constant a is 45 allowed. Thus, we want values of r ≠ 0 for which rrrnnn++21=+. The terms in this 46 equation can be divided by the common factor r n , giving us the quadratic equation 47 rr2 =+1. Using the quadratic formula, we discover there are two roots r of the quadratic 15+ 15− 48 equation: p = and q = . As a decimal, p is about 1.618, and we suspect we 2 2 49 have identified the value we encountered in the numerical table above. For arbitrary 50 choices of the constants a and b, both of the geometric sequences apn and bqn solve the =+ nnn++21=+ 51 Fibonacci recursion xnnn++21xx. Moreover, since ap ap ap and 52 bqnnn++21=+ bq bq , we can add these two equations to see that 53 ()()()apnn++22+=+++ bq ap nn ++ 11 bq ap nn bq . =+nn 54 That is, for any choices of the constants a and b, xn ap bq solves the Fibonacci =+ 55 recurrence relation xnnn++21xx. 56 To summarize our progress, we have proved the following result: =+ =+nn 57 Theorem. The Fibonacci recursion formula xnnn++21xx, is solved by xn ap bq 15+ 15− 58 where p = , q = , and a and b are arbitrary constants. 2 2 4 15+ 59 The constant p ==1.6180 is the famous Golden Ratio. It was known 2 K 60 (though not by that name*) in ancient Greek mathematics, since it solved this question: 61 determine the point C on line segment AB so that AB/AC = AC/BC. 15− 62 The second solution q ==−0.6180 of the quadratic equation 2 K 63 rr2 −−=10 can be investigated by factoring the quadratic polynomial. That is, since 64 rr22−−=1( rprqr − )() − = −() pqrpq + + , 65 then equating coefficients shows us that p +=q 1 and pq = –1. It’s interesting to see how 66 these relationships can be found without requiring the explicit formulas for p and q that 67 involve √5. 68 69 THE LUCAS AND FIBONACCI SEQUENCES =+nn 70 A simple choice of constants for xn ap bq in the theorem above is a = 1 and nn ⎛⎞⎛⎞15+− 15 71 b = 1. This gives us the solution Lpq=+=nn + . For example, n ⎜⎟⎜⎟ ⎝⎠⎝⎠22 =+=+=00 =+= 72 Lpq0 11 2 and Lpq1 1, and we are delighted that no 5 term appears. =+ == 73 Moreover, since LLLnnn++21and LL012 and 1, the terms of the sequence are all 74 positive integers. We have rediscovered the Fibonacci-like sequence 2, 1, 3, 4, 7, 11, 18, 75 29, … named for Edouard Lucas (1842-1891). Though this sequence is less well known * The entry for “Golden Ratio” in the MathWorld encylopedia (Weisstein) informs us that “The term "golden section" (in German, goldener Schnitt or der goldene Schnitt) seems to first have been used by Martin Ohm in the 1835 2nd edition of his textbook Die Reine Elementar-Mathematik (Livio 2002, p. 6). The first known use of this term in English is in James Sulley's 1875 article on aesthetics in the 9th edition of the Encyclopedia Britannica.” 5 76 than the Fibonacci sequence, the Lucas sequence also has many properties of interest (see 77 the concluding section of this article.) 78 To obtain the Fibonacci numbers, we need to see if it possible to choose the =+nn 00+=+ 79 constants a and b so that Fapbqn . Since F0 = 0 and ap bq a b , we will ⎛⎞1515+− 80 want b = – a. Also, since F =1 and ap11+= bq a() p −= q a − = a 5 , 1 ⎜⎟ ⎝⎠22 1 81 we must choose a = . Thus, the nth Fibonacci number is given by the explicit formula 5 nn nn−+−⎡⎛⎞⎛⎞⎤ ==pq 115 − 15 82 Fn ⎢⎜⎟⎜⎟⎥ . 55⎢⎜⎟⎜⎟22⎥ ⎣⎝⎠⎝⎠⎦ 83 This formula is known as Binet’s formula for the Fibonacci numbers, named after 84 Jacques Binet (1786-1865) who discovered it in 1843. However, the formula was 85 discovered first in 1718 by Abraham DeMoivre (1667-1754). 86 Since Binet’s formula gives F0 = 0, we now know why Jason started his list of 87 Fibonacci numbers with 0. The formula also explains why the Fibonacci sequence, 88 though not geometric, is increasingly close to one. Since q =−()1 5 / 2 ≈− 0.618 we see 89 that |qn| is increasingly small as n becomes large. Therefore, we have the approximate 1 90 equality Fp≈ n . This explains why the Fibonacci sequence is nearly a geometric n 5 91 series, as we noticed in the table of values computed earlier. If we let {x} denote the n = ⎧⎫p ≥ 92 “round to the integer nearest to x” function, it is easy to check that Fn ⎨⎬ for all n ⎩⎭5 = n ≥ 93 0. Similarly, the Lucas numbers can be written as Lpn { } when n 1. 6 94 INVESTIGATING MARCUS’ SEQUENCE 3, 0, 2, 3, 2, 5, … 95 It seems reasonable to guess that the new sequence is related to the Fibonacci sequence. 96 Therefore, suppose that we add consecutive pairs of the sequence 3, 0, 2, 3, 2, 5. The first 97 three sums are 3 + 0 = 3, 0 + 2 = 2, and 2 + 3 = 5 We seem to be on the right track, since 98 we get the next three terms 3, 2, and 5 of the sequence. This suggests that the term xn+ 3 is 99 the sum of the two consecutive terms xn and xn + 1. That is, Marcus’ sequence is defined 100 by the recursion formula =+ 101 xnnn++31xx. = n 102 As before, we can search for a geometric sequence of the form xn ar that 103 solves the recurrence relation. As before, the constant a is arbitrary, but now we want r to 104 satisfy the equation rrrnnn++31=+. Dividing each term by the common factor r n , we get 105 the cubic equation rr3 =+1. We could turn to a computer algebra system or a graphing 106 calculator to solve this cubic and discover there is one real root u =1.32472 … and a pair 107 of complex conjugate roots v = –0.662359+0.56228 i and w = – 0.662359 – 0.56228 i. 108 However, let us just suppose we factor the cubic polynomial to get 109 rr3 −−=1( rurvrw − )()() − − . 110 By expanding the product of the three binomials on the right side, we get the equation 111 rr33−−=1 r −()() uvwruvvwuwruvw ++ 2 + + + − . 112 Since these two polynomials in the variable r are equal if and only if their coefficients are 113 equal, we obtain the equations 114 u + v + w = 0, uv + vw + uw = –1, and uvw = 1.
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