1. Solving Second-Order Recursive Relations Lecture 36 ½ Section 8.3 Wed, Apr 19, 2006

2. Second-Order Linear Homogeneous Recurrence Relations with Constant Coefficients What a title! Second-order – Each term is defined in terms of the previous two terms. Linear – The terms of the sequence appear to the first power in the relation. Homogeneous – There is no constant term. Constant coefficients – The coefficients of the terms are constants.

3. The Form of Such Recurrence Relations Such a recurrence relation is of the form a n = Aa n – 1 + Ba n – 2 for all integers n  2, with initial terms a 0 and a 1 given. The Fibonacci sequence is a very simple example.

4. The Characteristic Equation The characteristic equation of a n = Aa n – 1 + Ba n – 2 is t 2 – At – B = 0. For the Fibonacci sequence, the characteristic equation is t 2 – t – 1 = 0. The sequence a n = t n satisfies the recurrence relation.

5. Solving Such Recurrence Relations – Case I Theorem: If the characteristic equation has roots r and s which are distinct real numbers, then the recursive sequence is given by a n = Cr n + Ds n, where C and D are constants, determined by the values of a 0 and a 1.

6. Solving Such Recurrence Relations – Case I Proof: Since r n and s n satisfy the recurrence relation, it follows easily that Cr n + Ds n also satisfies it. By using the initial conditions, we can solve for C and D. The resulting formula satisfied the recurrence relation with the initial conditions.

7. Example Solve the recurrence relation a 0 = 2, a 1 = 3, a n = a n – 1 + 2a n – 2, for all n  2. The first few terms are 2, 3, 7, 13, 27, 53, 107, 213, …

8. Example The roots of the characteristic equation are r = 2 and s = -1. So the general form is a n = C(2 n ) + D(-1) n. Solve a 0 = C + D = 2 and a 1 = 2C – D = 3. We get C = 5/3 and D = 1/3. The sequence is

9. Example Find a non-recursive formula for the Fibonacci numbers. Find a non-recursive formula for the Lucas numbers: L 1 = 1 L 2 = 3 L n = L n – 1 + L n – 2, for all n  2.

10. Solving Such Recurrence Relations – Case II Theorem: If the characteristic equation has double root r, then the recursive sequence is given by a n = Cr n + Dnr n = (C + Dn)r n, where C and D are constants determined by the values of a 0 and a 1.

11. Example Solve the recurrence relation a 0 = 0, a 1 = 4, a n = a n – 1 – (¼)a n – 2, for all n  2. The first few terms are 0, 4, 4, 3, 2, …

12. Example The root of the characteristic equation is the double root r = ½. So the general form is a n = (C + Dn)(½) n. Solve a 0 = C = 0 and a 1 = (C + D)(½) = 4. We get C = 0 and D = 8. The sequence is

13. Example Find a nonrecursive formula for the recursive sequence a 0 = 1 a 1 = 10 a n = 2a n – 1 – a n – 2, for all n  2.

14. Case III – Complex Roots The sequences become more interesting when the roots of the characteristic equation are complex numbers. Write the first few terms of the sequence a 0 = 0 a 1 = 1 a n = 2a n – 1 – 5a n – 2, for all n  2.