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Solving Recurrence Relations by Iteration: Lecture Overview

Learn to solve recurrence relations step by step through guessing the answer, verifying it using induction, and applying arithmetic and algebraic patterns.

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Solving Recurrence Relations by Iteration: Lecture Overview

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  1. Solving Recurrence Relations by Iteration Lecture 36 Section 8.2 Mon, Apr 17, 2006

  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, 3, 5, 9, 17, 33, 65. • Compare to: 1, 2, 4, 8, 16, 32, 64. • Guess that an = 2n – 1 + 1.

  5. 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?

  6. 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.

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

  8. 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.

  9. Example: Future Value of an Annuity • Define {an} by • a0 = d, • an = (1 + r)an – 1 + d, for all n 1. • Find a formula for an. • a1 = (1 + r)d + d. • a2 = (1 + r)2d + (1 + r)d + d. • a3 = (1 + r)3d + (1 + r)2d + (1 + r)d + d.

  10. Example: Future Value of an Annuity • It appears that, in general, an = (1 + r)nd + … + (1 + r)d + d = d((1 + r)n + 1 – 1)/((1 + r) – 1) = d((1 + r)n + 1 – 1)/r.

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

  12. Verifying the Answer • Define {an} by • a1 = 1, • an = 2an – 1 + 5, for all n 2. • Verify, by induction, the formula an = 3  2n – 5, for all n 1.

  13. Future Value of an Annuity • Verify the formula an = d((1 + r)n + 1 – 1)/r for all n 0, for the future value of an annuity.

  14. Solving First-Order Linear Recurrence Relations • A first-order linear recurrence relation 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. • 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.

  16. Solving First-Order Linear Recurrence Relations • To solve for A and B in the general case, substitute the values of a1 and a2 and solve the system for A and B. a0 = A + B = u a1 = As + B = su + t

  17. Example • Solve the recurrence relation a1 = 1, an = 2an – 1 + 5, n 2. • Solve the recurrence relation a0 = d, an = (1 + r)an – 1 + d, n  1.

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