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7.2 Solving Recurrence Relations. Definition 1 (p. 460)- LHRR-K Def: A linear homogeneous recurrence relations of degree k with constant coefficients.

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Presentation on theme: "7.2 Solving Recurrence Relations. Definition 1 (p. 460)- LHRR-K Def: A linear homogeneous recurrence relations of degree k with constant coefficients."— Presentation transcript:

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 a n = c 1 a n-1 + c 2 a n-2 + … +c k a n-k where c 1, c 2, …c k are real numbers, and c k ≠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, a 0 =1, a 1 =2

4 Solving LHRR-K: Step 1: Find a characteristic equation: a n =r n is a solution of a n = c 1 a n-1 + c 2 a n-2 + … +c k a n-k iff r n = c 1 r n-1 + c 2 r n-2 + … +c k r n-k Divide by r n-k : Then r k = c 1 r k-1 + c 2 r k-2 + … +c k So: r k - c 1 r k-1 - c 2 r k-2 - … - c k =0 For degree 2: the characteristic equation is r 2 -c 1 r –c 2 =0 (roots are used to find explicit formula) Basic Solution: a n =α 1 r 1 n + α 2 r 2 n where r 1 and r 2 are roots of the characteristic equation

5 Thm. 1 (for 2 nd degree equations) Theorem 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has two distinct roots r 1 and r 2, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 1 n + α 2 r 2 n for n=0, 1, 2… where α 1 and α 2 are constants.---------------------------- First an example… Proof: later…

6 Ex. 1. Let a n =7a n-1 – 10 a n-2 for n≥2; a 0 =2, a 1 =1 Find the characteristic equation … get r=2, 5 Let a n = α 1 r 1 n + α 2 r 2 n …solve system and get α 1 = 3 and α 2 = -1 So, basic solution is:a n = 3*2 n -5 n

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 a n =α 1 r 1 n + α 2 r 2 n is a solution of a n =c 1 a n-1 +c 2 a n-2 c 1 a n-1 +c 2 a n-2 =c 1 (α 1 r 1 n-1 + α 2 r 2 n-1 )+c 2 (α 1 r 1 n-2 + α 2 r 2 n-2 ) why? = α 1 r 1 n-2 (c 1 r 1 + c 2 ) + α 2 r 2 n-2 (c 1 r 2 + c 2 )algebra =α 1 r 1 n-2 r 1 2 + α 2 r 2 n-2 r 2 2 reason: r 1 r 2 are roots of r 2 -c 1 r-c 2 =0 so r 1 2 =c 1 r 1 +c 2 and r 2 2 = c 1 r 2 +c 2 =α 1 r 1 n + α 2 r 2 n = a n

9 Pf- step 2 Step 2: Show that there exist constants α 1 α 2 such that a n =α 1 r 1 n + α 2 r 2 n satisfies the initial conditions a 0 =C 0 and a 1 =C 1. a 0 = C 0 = … a 1 = C 1 = … Next solve system of 2 equations and 2 variables and get… α 1 = (C 1 -C 0 r 2 )/(r 1 -r 2 ) α 2 = (C 0 r 1 – C 1 ) / (r 1 -r 2 )

10 Ex: 2. a n =6a n-1 -8a n-2 for n≥2; a 0 =4 and a 1 =10. Find characteristic equation Find solution

11 Ex 2 - Prove it is a solution

12 Ex: 3. Fibonacci numbers: f n =f n-1 +f n-2 for n≥2; f 0 =0 and f 1 =1. Find characteristic equation r 2 = r+1find r 1 and r 2 Thm. 1 says f n = α 1 r 1 n + α 2 r 2 n Solve for α 1, α 2 Solution:

13 Note: Thm. 1 is only for r 1 ≠r 2 Thm. 2 is for r 1 = r 2 Theorem 2: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has only one root r 0, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 0 n + α 2 n r 0 n for n=0, 1, 2… where α 1 and α 2 are constants.

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

15 Ex: 5. a n = - 6a n-1 -9a n-2 for n≥2; a 0 =3 and a 1 = - 3 Find characteristic equation Find solution Prove it is a solution

16 Ex: 6. a n =8a n-1 -16a n-2 for n≥2; a 0 =2 and a 1 =20. Find characteristic equation Find solution Prove it is a solution

17 Summarize k degree solution

18 Ex: (#12 in book) a n =2a n-1 +a n-2 -2a n-3 for n≥3; a 0 =3,a 1 =6, a 2 =0 Find characteristic equation r 3 – 2 r 2 - 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) a n =2a n-1 +5a n-2 -6a n-3 for n≥3; a 0 =7,a 1 = - 4, a 2 =8 Find characteristic equation Use synthetic division to get Find solution Prove it is a solution

20 Ex: 7. a n =5a n-2 -4a n-4 ; a 0 =3, a 1 =2, a 2 =6 and a 3 =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


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