Fibonacci - 262 Chapter 6 Eigenvalues and Eigenvectors...

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Unformatted text preview: 262 Chapter 6 Eigenvalues and Eigenvectors Fiiaoriaccé Numbers We present a famous example, which leads to powers of matrices. The Fibonacci numbers start with F0 2 0 and F1 2 1. Then every new F is the sum ofthe two previous F’s: The sequence 0, 1, l, 2, 3, 5, 8, 13, comes from F“; = Fk+1+ Fk. These numbers tum up in a fantastic variety of applications. Plants and trees grow in a spiral pattern, and a pear tree has 8 growths for every 3 turns. For a willow those numbers can be 13 and 5. The champion is a sunflower of Daniel O’Connell, which had 233 seeds in 144 loops. Those are the Fibonacci numbers F13 and F12. Our problem is more basic. Problem: Find the Fibonacci number F100 The slow way is to apply the rule Fk+2 = Fk+1 + Fk one step at a time. By adding F6 2 8 to F7 : 13 we reach F8 2 21. Eventually we come to F100. Linear algebra gives a better way. The key is to begin with a matrix equation uk+1 = Auk. That is a one—step rule for vectors, while Fibonacci gave a two—step rule for scalars. We match them by putting Fibonacci numbers into the vectors: F F = F + F l 1 Let uk 2 [ 2+1] . The rule k+2 k+l k becomes uk+1 = [I 0] uk- (5) k Fk+l = Fk+1 After 100 steps we reach u100 = Amour): mu mu mu mu Wu The Fibonacci numbers come from the powers of Av—and we can compute A100 with— out multiplying 100 matrices. This problem is just right for eigenvalues. Subtract A from the diagonal of A: 1—}t l 1 —)\. Every step multiplies by A = [ A—k[=|: :| leadsto det(A—}t1):}t2~}t—I. The eigenvalues solve the equation A2 — A a 1 = 0. They come from the quadratic formula (~b :l: vbz — 4ac)/2a: z —.618. _1+¢§ 1—fi ‘ 2 2 Al Q1 1.618 and X2: These eigenvalues XI and k2 lead to eigenvectors x1 and x2. This completes step 1: l—AI 1 _ 0 w M l 1 «113611101 will 1—i2 1 ' _ 0 _ A2 1 1 1le - lol x2 - 1.1 l- 6.2 Diagonalizing a Matrix 263 Step 2 finds the combination of those eigenvectors that gives u0 = (1,0): 1 1 M 112 x1 — x2 = ~ or u = . 6 [Oi M—quli [1 0 Al—Az H The final step multiplies uo by A100 to find uloo. The eigenvectors stay separate! They are multiplied by (Ammo and (A2)'00: (A 100x _ A2 100x moo = (7) X1 - K2 We want F100 2 second component of uloo. The second components of x1 and x2 are 1. Substitute the numbers M and A2 into equation (7), to find A1 —)tz = «5 and F100: 100 g 100 F100: 1 [(Hfi) (1 ‘5) ]~3.54.1020. (8) fi 2 2 Is this a whole number? Yes. The fractions and square roots must disappear, because Fibonacci’s rule Fk+2 = Fk+1 + Fk stays with integers. The second term in (8) is less 1 than 3, so it must move the first term to the nearest whole number: kth Fibonacci number 1 (Ha/3)"- (9) nearest integer to —— 2 fl The ratio of F6 to F5 is 8/5 = 1.6. The ratio Flol/Fmo must be very close to (l + / 2. The Greeks called this number the “golden mean." For some reason a rectangle with sides 1.618 and 1 looks especially graceful. 6.2 Diagonalizing a Matrix 267 Questions 9—14 are about Fibonacci and Gibonacci numbers. 9 10 11 12 13 14 For the Fibonacci matrix A = compute A2 and A3 and A4. Then use the text and a calculator to find F20. Suppose each number Gk+2 is the average of the two previous numbers Gk+1 and Gk. Then Gk+2 =%(Gk+1+ Gk): 132:?l=1 A 113?]- (a) Find the eigenvalues and eigenvectors of A. Gk+2 = %Gk+l + %Gk Gk+l = Gk+l (b) Find the limit as n —> 00 of the matrices A’1 = SA’IS”. (c) If G0 = O and G1 = 1 show that the Gibonacci numbers approach Diagonalize the Fibonacci matrix by completing S“: 1 1 __ M X2 X1 0 1 0 — 1 1 0 k2 ' Do the multiplication SAkS_l[(1)] to find its second component. This is the kth Fibonacci number Fk = (A1; — AIS/(Al — A2). The numbers A]; and A]; satisfy the Fibonacci rule Fk+2 = Fk+1 + Fk: k+2__ k+1 k k+2__ k+l k A, —)t1 +Al and i2 _i2 +A2. Prove this by using the original equation for the A’s. Then any combination of A]; and Ag satisfies the rule. The combination Fk : (Alf — AID/(Al — A2) gives the right start F0 2 0 and F1 2 1. Suppose Fibonacci had started with F0 2 2 and F1 2 1. The rule Fk+2 = Fk+1 + Fk is the same so the matlix A is the same. Its eigenvectors add to x1 +x2= [%(l:fi)]+[%(l :fi)] = = After 20 steps the second component of A20(x1+x2) is ( )20+( )20. Compute that number F20. Prove that every third Fibonacci number in 0, 1, 1,2, 3, . .. is even. ...
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This note was uploaded on 04/26/2010 for the course CEG 616 taught by Professor Taylor during the Winter '10 term at Wright State.

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Fibonacci - 262 Chapter 6 Eigenvalues and Eigenvectors...

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