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 –

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 
a3 = 22  
a4 = 23   
a5 = 24    
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 –  2n – 1 – 5
= 6  2n – 1 – 5
= 3  2n – 5.

11. Example Define {an} by Find a formula for an. 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.