MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 38, Friday, December 5

Example 4: Complex and Multiple Roots a n = -2a n-2 – a n-4, a 0 = 0, a 1 = 1, a 2 = 2, a 3 = 3. Note that the solution from the book reduces to: a 4k = -4k a 4k+1 = 1 – 8k a 4k+2 = 4k+2 a 4k+3 = 8k + 3

7.4. Solution of Inhomogeneous Recurrence Relations Homework (MATH 310#11F): Read 7.5 Do 7.4: all odd numbered problems Turn in 7.4: 2,4,10,12,18

Example 1: Summation Recurrence (See Example 3, 7.1, p. 275) Dividing the plane: a n = a n-1 + n, a 0 = 1. a n = 1 + ( n) n = C(1,1) + C(2,1) C(n,1). By identity (8) from 5.5: C(r,r) + C(r+1,r) +...+C(n,r) = C(n+1,r+1) n = C(n+1,2) = (n+1)n/2 a n = 1 + n(n+1)/2. Another approach: Undermined coefficients: a n = An 2 + Bn + C. Select three values for n, say n= 0, n= 1 and n= 2 a 0 = C a 1 = A + B + C a 2 = 4A + 2B + C. Solve for A,B,C.

Particular Solutions f(n)p(n) dB dnB 1 n + B 0 dn 2 B 2 n 2 + B 1 n + B 0 ed n Bd n

Example 2: Solving Tower of Hanoi Puzzle (See Example 4 of 7.1) a n = 2a n-1 + 1, a 1 = 1. General solution to the homogeneous equation a n = 2a n-1. is a n = A 2 n. Particular solution to inhomogeneous relation a n * = B. Determine B. B = 2B + 1. Hence B =-1. General solution to inhomogeneous relation: a n = A 2 n –1. Determine A. a 1 = A 2 1 – 1 = 2 A – 1 = 1. Hence A = 1. Answer: a n = 2 n –1.

Example 3: Compound Inhomogeneous Term.