Solving Recurrence Relations by Iteration

1. Solving Recurrence Relations by Iteration Lecture 41 Section 8.2 Fri, Apr 13, 2007

2. Solving Recurrence Relations • Our method will involve two steps. • Guess the answer. • Verify the guess, using mathematical induction.

3. Guessing the Answer • Write out the first several terms, as many as necessary. • Look for a pattern. • Two strategies • Do the arithmetic. • Spot the pattern in the resulting numbers. • Postpone the arithmetic. • Spot the pattern in the algebraic formulas.

4. Example: Do the Arithmetic • Define {an} by • a1 = 2, • an = 2an – 1 + 1, for all n 2. • Find a formula for an. • First few terms: 2, 5, 11, 23, 47, 95, 191.

5. Example: Do the Arithmetic • Define {an} by • a1 = 2, • an = 2an – 1 + 1, for all n 2. • Find a formula for an. • First few terms: 2, 5, 11, 23, 47, 95, 191. • Compare to: 1, 2, 4, 8, 16, 32, 64.

6. Example: Do the Arithmetic • Define {an} by • a1 = 2, • an = 2an – 1 + 1, for all n 2. • Find a formula for an. • First few terms: 2, 5, 11, 23, 47, 95, 191. • Compare to: 1, 2, 4, 8, 16, 32, 64. • Guess that an = 32n – 1 + 1.

7. Example: Postpone the Arithmetic • Define {an} by • a1 = 1, • an = 2an – 1 + 5, for all n 2. • Find a formula for an. • First few terms: 1, 7, 19, 43, 91. • What is an?

8. Example: Postpone the Arithmetic • Calculate a few terms • a1 = 1. • a2 = 2  1 + 5. • a3 = 22  1 + 2  5 + 5. • a4 = 23  1 + 22  5 + 2  5 + 5. • a5 = 24  1 + 23  5 + 22  5 + 2  5 + 5. • It appears that, in general, • an = 2n – 1 + (2n – 2 + 2n – 3 + … + 1)  5.

9. Lemma: Geometric Series • Lemma: Let r 1. Then

10. Example: Postpone the Arithmetic • an = 2n – 1 + (2n – 2 + 2n – 3 + … + 1)  5 = 2n – 1 + (2n – 1 – 1)/(2 – 1)  5 = 2n – 1 + (2n – 1 – 1)  5 = 2n – 1 + 5  2n – 1 – 5 = 6  2n – 1 – 5 = 3  2n – 5.

11. Example • Define {an} by • a0 = a, • an = ran – 1 + b, for all n 1. • Find a formula for an. • a1 = ra + b. • a2 = r(ra + b) + b = r2a + (rb + b). • a3 = r(r2a + (rb + b)) + b = r3a + (r2b + rb + b).

12. Example: Future Value of an Annuity • It appears that, in general,

13. Verifying the Answer • Use mathematical induction to verify the guess.

14. Solving First-Order Linear Recurrence Relations • A first-order linear recurrence relation with constant coefficients is a recurrence relation of the form an = san – 1 + t, n 1, with initial condition a0 = u, where s, t, and u are real numbers.

15. Solving First-Order Linear Recurrence Relations • Theorem: Depending on the value of s, the recurrence relation will have one of the following solutions: • If s = 0, the solution is a0 = u, an = t, for all n 1. • If s = 1, the solution is an= u + nt, for all n 0.

16. Solving First-Order Linear Recurrence Relations • If s  0 and s  1, then the solution is of the form an = Asn + B, for all n 0, for some real numbers A and B.

17. Solving First-Order Linear Recurrence Relations • To solve for A and B in the general case, substitute the values of a0 and a1 and solve the system for A and B. a0 = A + B = u a1 = As + B = su + t

18. Example • Solve the recurrence relation a0 = 1, an = 2an – 1 + 1, n 1. • Solve the recurrence relation a0 = 1, an = 2an – 1 + 5, n 1.