Practice 4 Solutions

2 3n 5 3 6 1 2 b a 2n 5 26 13 b a 17 17 b a 2n 3n

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ions) Linear Algebra, Dave Bayer, November , [1] Solve the recurrence relation f(0) = a, f(n + 1) f(n) f(1) = b, −5 −4 1 0 = n f(n) = b a = f(n) = − 5 f(n − 1) − 4 f(n − 2) (−4)n 3 4 4 −1 −1 b a + (−1)n 3 −1 −4 1 4 b a (−4)n (−1)n ( − b − a) + (b + 4a) 3 3 [2] Solve the recurrence relation f(0) = a, f(n + 1) f(n) = f(1) = b, n 6 −5 1 0 f(n) = f(n) = 6 f(n − 1) − 5 f(n − 2) b a = 1 4 −1 5 −1 5 b a + 5n 4 5 −5 1 −1 b a 5n 1 ( − b + 5a) + (b − a) 4 4 [3] Solve the recurrence relation f(0) = a, f(n + 1) f(n) = −6 −8 1 0 f(1) = b, n b a = f(n) = − 6 f(n − 1) − 8 f(n − 2) (−4)n 2 4 8 −1 −2 b a + (−2)n 2 −2 −8 1 4 (−2)n (−4)n ( − b − 2a) + (b + 4a) 2 2 f(n) = [4] Solve the recurrence relation f(0) = a, f ( n + 1) f(n) = f(1) = b, −4 5 10 n f(n) = b a f(n) = − 4 f(n − 1) + 5 f(n − 2) = (−5)n 6 5 −5 −1 1 b a (−5)n 1 ( − b + a) + (b + 5a) 6 6 + 1 6 15 15 b a b a , 2 × 2 Exercise Set J (recurrence relations), November [5] Solve the recurrence relation f(0) = a, f(n + 1) f(n) = f(1) = b, n −1 6 10 b a f(n) = − f(n − 1) + 6 f(n − 2) (−3)n 5 = 3 −6 −1 2 b a 2n 5 + 26 13 b a 17 17 b a 2n (−3)n ( − b + 2a) + (b + 3a) 5 5 f(n) = [6] Solve the recurrence relation f(0) = a, f(n + 1) f(n) = f(1) = b, n −6 7 10 b a f(n) = − 6 f(n − 1) + 7 f(n − 2) = (−7)n 8 7 −7 −1 1 b a + 1 8 (−7)n 1 ( − b + a) + (b + 7a) 8 8 f(n) = [7] Solve the recurrence re...
View Full Document

This document was uploaded on 03/24/2014 for the course MATH V2010 at Barnard College.

Ask a homework question - tutors are online