Lecture Week 9

# Discrete Mathematics and Its Applications with MathZone

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UCI ICS/Math 6A, Summer 2007 9-AdvancedCounting -1 [email protected] , after [email protected] Recurrence Relations (RRs) A “Recurrence Relation ” for a sequence {a n } is an equation that expresses a n in terms of one or more of the previous terms in the sequence (i.e., a 0 ,a 1 ,a 2 ,…,a n-1 ) for all n≥n 0 . Examples: a 0 =1, a 1 =3, a 2 =4; for n ≥3, a n = a n-1 +a n-2 -a n-3 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,… a 0 =2, a 1 =4, a 2 =3; for n ≥3, a n = a n-1 +a n-2 -a n-3 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,… a 0 =4, a 1 =3, a 2 =3; for n ≥3, a n = a n-1 +a n-2 -a n-3 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,…

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UCI ICS/Math 6A, Summer 2007 9-AdvancedCounting -2 [email protected] , after [email protected] Solutions to Recurrence Relations A sequence {a n } is called a “solution ” of the recurrence relation if its terms satisfy the recurrence relation. Examples: For n ≥3, a n = a n-1 +a n-2 -a n-3 ; Initial conditions: a 0 =1, a 1 =3, a 2 =4. Solution: 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,… For n ≥3, a n = a n-1 +a n-2 -a n-3 ; Initial conditions: a 0 =2, a 1 =4, a 2 =3. Solution: 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,… For n ≥3, a n = a n-1 +a n-2 -a n-3 ; Initial conditions: a 0 =4, a 1 =3, a 2 =3. Solution: 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,… Every recurrence relationship has many solutions each determined uniquely by its own initial conditions. We say a function f: N→R is a “solution ” to a recurrence relation if the sequence {f(n)} is a solution to it. Example: f(n)=5 ∙2 n is a solution to the recurrence relation a n =2∙a n-1 .
UCI ICS/Math 6A, Summer 2007 9-AdvancedCounting -3 [email protected] , after [email protected] Solving Recurrence Relations If a n =4a n-1 and a 0 =3, find a function f such that f(n)=a n . a n =4a n-1 =4 2 a n-2 =4 3 a n-3 =...=4 n a 0 =3 4 n . If a n =a n-1 +n and a 0 =4, find a function f such that f(n)=a n . a n =a n-1 +n=a n-2 +n+(n-1) =a n-2 +n+(n-1)+(n-2)=... =a 0 +n+(n-1)+(n-2)+...+1=4+n(n+1)/2=(n 2 +n+8)/2. If a n =2na n-1 and a 0 =5, find a function f such that f(n)=a n . a n =2na n-1 =2 2 n(n-1)a n-2 =2 3 n(n-1)a n-3 =...=2 n n!a 0 =5 2 n n!. If a n =2a n-1 +1 and a 0 =0, find a function f such that f(n)=a n . a n =2a n-1 +1=2 2 a n-2 +2+1=2 3 a n-3 +4+2+1=2 4 a n-4 +8+4+2+1=... =2 n-1 +2 n-2 +2 n-3 +...+8+4+2+1=2 n -1.

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UCI ICS/Math 6A, Summer 2007 9-AdvancedCounting -4 [email protected] , after [email protected] Modeling with Recurrence Relations An initial deposit of P 0 dollars deposited at 7% annual interest. P n =(1+0.07)P n-1 is the value after n years. P n =(1+0.07) n P 0 Rabbits on an island. Each pair producing a new pair every month. Fibonacci Numbers: f 0 =0, f 1 =1; for n ≥2, f n = f n-1 +f n-2 .
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