7.2 Solving Recurrence Relations

1. 7.2 Solving Recurrence Relations

2. Definition 1 (p. 460)- LHRR-K Def: A linear homogeneous recurrence relations of degree k with constant coefficients (referred to as “LHRR-K”) is a recurrence relation of the form an = c1an-1 + c2 an-2 + … +ckan-k where c1, c2, …ck are real numbers, and ck≠0. ------------------------------ Note: These can be explicitly solved in a systematic way.

3. First, are the following examples or non-examples? • a n = 3 a n-1 • a n = a n-1 + a n-2 2 • f n = f n-1 + f n-2 • H n = 2H n-1 +1 • B n = n B n-1 • a n = a n-1 +3 a n-2 , a0 =1, a1 =2

4. Solving LHRR-K: Step 1: Find a characteristic equation: an=rn is a solution of an= c1an-1+ c2 an-2 + … +ckan-k iff rn = c1 rn-1 + c2 rn-2 + … +ckrn-k Divide by rn-k: Then rk= c1 rk-1 + c2 rk-2 + … +ck So: rk - c1 rk-1 - c2 rk-2 - … - ck =0 For degree 2: the characteristic equation is r2-c1r –c2=0 (roots are used to find explicit formula) Basic Solution: an=α1r1n+ α2r2n where r1 and r2 are roots of the characteristic equation

5. Thm. 1 (for 2nd degree equations) Theorem 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has two distinct roots r1 and r2, Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2 iff an=α1r1n+ α2r2n for n=0, 1, 2… whereα1 and α2 are constants.---------------------------- First an example… Proof: later…

6. Ex. 1. Let an=7an-1 – 10 an-2 for n≥2; a0=2, a1=1 Find the characteristic equation … get r=2, 5 Let an= α1r1n+ α2r2n … solve system and get α1 = 3 and α2 = -1 So, basic solution is: an= 3*2n-5n

7. Prove this is a solution (as we did in sec 7.1):

8. Sketch of Proof of Thm. 1: (p. 462) Step 1: Show that an=α1 r1n+ α2 r2n is a solution of an=c1an-1+c2an-2 c1an-1+c2an-2 =c1(α1 r1n-1+ α2 r2n-1 )+c2 (α1 r1n-2+ α2 r2n-2) why? = α1r1n-2(c1 r1 + c2 ) + α2r2n-2(c1 r2 + c2 ) algebra =α1r1n-2r12 + α2r2n-2r22 reason: r1 r2are roots of r2-c1r-c2=0 so r12=c1r1 +c2 and r22= c1r2 +c2 =α1r1n + α2r2n =an

9. Pf- step 2 Step 2: Show that there exist constants α1 α2 such that an=α1 r1n+ α2 r2n satisfies the initial conditions a0=C0 and a1=C1. a 0 = C 0 = … a 1 = C 1 = … Next solve system of 2 equations and 2 variables and get… α1 = (C1-C0r2)/(r1-r2) α2 = (C0r1 – C1) / (r1-r2)

10. Ex: 2. an =6an-1 -8an-2 for n≥2; a0=4 and a1=10. Find characteristic equation Find solution

11. Ex 2 - Prove it is a solution

12. Ex: 3. Fibonacci numbers: fn =fn-1 +fn-2 for n≥2; f0=0 and f1=1. Find characteristic equation r2 = r+1 find r1 and r2 Thm. 1 says fn= α1r1n+ α2r2n Solve for α1, α2 Solution:

13. Note: Thm. 1 is only for r1≠r2Thm. 2 is for r1 =r2 Theorem 2: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has only one root r0 , Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2iff an=α1r0n+ α2 n r0n for n=0, 1, 2… whereα1 and α2 are constants.

14. Ex: 4. (Recall this ex from section 5.1) an =2an-1 -an-2 for n≥2; a0=0 and a1=3 Find characteristic equation Find solution Prove it is a solution

15. Ex: 5. an = - 6an-1 -9an-2 for n≥2; a0=3 and a1= - 3 Find characteristic equation Find solution Prove it is a solution

16. Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20. Find characteristic equation Find solution Prove it is a solution

17. Summarize k degree solution

18. Ex: (#12 in book) an =2an-1 +an-2 -2an-3 for n≥3; a0=3 ,a1=6, a2=0 Find characteristic equation r3 – 2 r2- r +2=0 Use synthetic division to get (r-1)(r+1)(r-2)=0 Find solution Prove it is a solution

19. Ex: (#15 bk) an =2an-1 +5an-2 -6an-3 for n≥3; a0=7 ,a1= - 4, a2=8 Find characteristic equation Use synthetic division to get Find solution Prove it is a solution

20. Ex: 7. an =5an-2 -4an-4 ; a0=3, a1=2, a2=6 and a3=8 Find characteristic equation Find solution Prove it is a solution

21. Ex: (#8 in book—modeling number of lobsters caught) Find characteristic equation Find solution Prove it is a solution