# Solving Recurrence Relations by Iteration Lecture 41 Section

Solving Recurrence Relations by Iteration Lecture 41 Section 8. 2 Fri, Apr 13, 2007

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

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

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

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

Example: Do the Arithmetic ¢ Define {an} by a 1 = 2, l an = 2 an – 1 + 1, for all n 2. l 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 2 n – 1 + 1. ¢

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

Example: Postpone the Arithmetic ¢ Calculate a few terms a 1 = 1. l a 2 = 2 1 + 5. l a 3 = 22 1 + 2 5 + 5. l a 4 = 23 1 + 22 5 + 5. l a 5 = 24 1 + 23 5 + 22 5 + 5. l ¢ It appears that, in general, l an = 2 n – 1 + (2 n – 2 + 2 n – 3 + … + 1) 5.

Lemma: Geometric Series ¢ Lemma: Let r 1. Then

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

Example ¢ Define {an} by a 0 = a, l an = ran – 1 + b, for all n 1. l ¢ Find a formula for an. a 1 = ra + b. l a 2 = r(ra + b) + b = r 2 a + (rb + b). l a 3 = r(r 2 a + (rb + b)) + b = r 3 a + (r 2 b + rb + b). l

Example: Future Value of an Annuity ¢ It appears that, in general,

Verifying the Answer ¢ Use mathematical induction to verify the guess.

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 a 0 = u, where s, t, and u are real numbers.

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 a 0 = u, an = t, for all n 1. l If s = 1, the solution is an = u + nt, for all n 0. l

Solving First-Order Linear Recurrence Relations l 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.

Solving First-Order Linear Recurrence Relations ¢ To solve for A and B in the general case, substitute the values of a 0 and a 1 and solve the system for A and B. a 0 = A + B = u a 1 = As + B = su + t

Example Solve the recurrence relation a 0 = 1, an = 2 an – 1 + 1, n 1. ¢ Solve the recurrence relation a 0 = 1, an = 2 an – 1 + 5, n 1. ¢

- Slides: 18