Math 221: LINEAR ALGEBRA 3-4. Application to Linear Recurrences Le - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA §3-4. Application to Linear Recurrences Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

, we’d like to fjnd a formula for Instead of using the recurrence to compute that holds for all . Linear Recurrences Example The Fibonacci Numbers are the numbers in the sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , . . . and can be defined by the linear recurrence relation f n +2 = f n +1 + f n for all n ≥ 0 , with the initial conditions f 0 = 1 and f 1 = 1 .

, we’d like to fjnd a formula for Instead of using the recurrence to compute that holds for all . Linear Recurrences Example The Fibonacci Numbers are the numbers in the sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , . . . and can be defined by the linear recurrence relation f n +2 = f n +1 + f n for all n ≥ 0 , with the initial conditions f 0 = 1 and f 1 = 1 . Problem Find f 100 .

Linear Recurrences Example The Fibonacci Numbers are the numbers in the sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , . . . and can be defined by the linear recurrence relation f n +2 = f n +1 + f n for all n ≥ 0 , with the initial conditions f 0 = 1 and f 1 = 1 . Problem Find f 100 . Instead of using the recurrence to compute f 100 , we’d like to fjnd a formula for f n that holds for all n ≥ 0 .

Definitions A sequence of numbers x 0 , x 1 , x 2 , x 3 , . . . is defined recursively if each number in the sequence is determined by the numbers that occur before it in the sequence.

Definitions A sequence of numbers x 0 , x 1 , x 2 , x 3 , . . . is defined recursively if each number in the sequence is determined by the numbers that occur before it in the sequence. A linear recurrence of length k has the form x n + k = a 1 x n + k − 1 + a 2 x n + k − 2 + · · · + a k x n , n ≥ 0 , for some real numbers a 1 , a 2 , . . . , a k .

Example The simplest linear recurrence has length one, so has the form x n +1 = ax n for n ≥ 0 , with a ∈ R and some initial value x 0 .

Example The simplest linear recurrence has length one, so has the form x n +1 = ax n for n ≥ 0 , with a ∈ R and some initial value x 0 . In this case, x 1 = ax 0 ax 1 = a 2 x 0 x 2 = ax 2 = a 3 x 0 x 3 = . . . . . . . . . ax n − 1 = a n x 0 x n = Therefore, x n = a n x 0 .

Example Find a formula for x n if x n +2 = 2 x n +1 + 3 x n for n ≥ 0 , with x 0 = 0 and x 1 = 1 .

Example Find a formula for x n if x n +2 = 2 x n +1 + 3 x n for n ≥ 0 , with x 0 = 0 and x 1 = 1 . � � x n Solution. Define V n = for each n ≥ 0 . Then x n +1 � x 0 � 0 � � V 0 = = , x 1 1 and for n ≥ 0 , � x n +1 � � � x n +1 V n +1 = = 2 x n +1 + 3 x n x n +2

det Example (continued) � � x n +1 Now express V n +1 = as a matrix product: 2 x n +1 + 3 x n

det Example (continued) � � x n +1 Now express V n +1 = as a matrix product: 2 x n +1 + 3 x n � 0 � � � � � x n +1 1 x n V n +1 = = = AV n 2 x n +1 + 3 x n 3 2 x n +1

det Example (continued) � � x n +1 Now express V n +1 = as a matrix product: 2 x n +1 + 3 x n � 0 � � � � � x n +1 1 x n V n +1 = = = AV n 2 x n +1 + 3 x n 3 2 x n +1 This is a linear dynamical system, so we can apply the techniques from §3.3, provided that A is diagonalizable.

Example (continued) � � x n +1 Now express V n +1 = as a matrix product: 2 x n +1 + 3 x n � 0 � � � � � x n +1 1 x n V n +1 = = = AV n 2 x n +1 + 3 x n 3 2 x n +1 This is a linear dynamical system, so we can apply the techniques from §3.3, provided that A is diagonalizable. � � x − 1 � = x 2 − 2 x − 3 = ( x − 3)( x + 1) � � c A ( x ) = det ( xI − A ) = � � − 3 x − 2 � Therefore A has eigenvalues λ 1 = 3 and λ 2 = − 1 , and is diagonalizable.

Example (continued) � 1 � � x 1 = is a basic eigenvector corresponding to λ 1 = 3 , and 3 � − 1 � � x 2 = is a basic eigenvector corresponding to λ 2 = − 1 . 1

Example (continued) � 1 � � x 1 = is a basic eigenvector corresponding to λ 1 = 3 , and 3 � − 1 � � x 2 = is a basic eigenvector corresponding to λ 2 = − 1 . 1 � 1 � − 1 � � Furthermore P = � x 1 � x 2 = is invertible and is the 3 1 � 3 � 0 diagonalizing matrix for A, and P − 1 AP = D = 0 − 1

Example (continued) � 1 � � x 1 = is a basic eigenvector corresponding to λ 1 = 3 , and 3 � − 1 � � x 2 = is a basic eigenvector corresponding to λ 2 = − 1 . 1 � 1 � − 1 � � Furthermore P = � x 1 � x 2 = is invertible and is the 3 1 � 3 � 0 diagonalizing matrix for A, and P − 1 AP = D = 0 − 1 � b 1 � Writing P − 1 V 0 = , we get b 2 � � 0 � b 1 1 � = 1 � 1 1 � � � 4 = − 3 1 1 1 b 2 4 4

Example (continued) Therefore, � � x n b 1 λ n x 1 + b 2 λ n V n = = 1 � 2 � x 2 x n +1 � 1 � − 1 1 � + 1 � 43 n 4( − 1) n = , 3 1 and so x n = 1 43 n − 1 4( − 1) n .

Example Solve the recurrence relation x k +2 = 5 x k +1 − 6 x k , k ≥ 0 with x 0 = 0 and x 1 = 1 .

Example Solve the recurrence relation x k +2 = 5 x k +1 − 6 x k , k ≥ 0 with x 0 = 0 and x 1 = 1 . Solution. Write � x k +1 � � � � � � � x k +1 0 1 x k V k +1 = = = x k +2 5 x k +1 − 6 x k − 6 5 x k +1

Example Solve the recurrence relation x k +2 = 5 x k +1 − 6 x k , k ≥ 0 with x 0 = 0 and x 1 = 1 . Solution. Write � x k +1 � � � � � � � x k +1 0 1 x k V k +1 = = = x k +2 5 x k +1 − 6 x k − 6 5 x k +1 Find the eigenvalues and corresponding eigenvectors for � 0 1 � A = − 6 5

Example (continued) � 1 � A has eigenvalues λ 1 = 2 with corresponding eigenvector � x 1 = , and 2 � 1 � λ 2 = 3 with corresponding eigenvector � x 2 = . 3

Example (continued) � 1 � A has eigenvalues λ 1 = 2 with corresponding eigenvector � x 1 = , and 2 � 1 � λ 2 = 3 with corresponding eigenvector � x 2 = . 3 � 1 � � � 1 3 − 1 , P − 1 = P = , 2 3 − 2 1 and � � 0 � b 1 � − 1 � � � � 3 − 1 = P − 1 V 0 = = − 2 1 1 1 b 2

Example (continued) � 1 � A has eigenvalues λ 1 = 2 with corresponding eigenvector � x 1 = , and 2 � 1 � λ 2 = 3 with corresponding eigenvector � x 2 = . 3 � 1 � � � 1 3 − 1 , P − 1 = P = , 2 3 − 2 1 and � � 0 � b 1 � − 1 � � � � 3 − 1 = P − 1 V 0 = = − 2 1 1 1 b 2 Finally, � 1 � 1 � � � � x k = b 1 λ k x 1 + b 2 λ k x 2 = ( − 1)2 k + 3 k V k = 1 � 2 � x k +1 2 3

Example � 1 � 1 � � � � x k = ( − 1)2 k + 3 k 2 3 x k +1 and therefore x k = 3 k − 2 k .