1. MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 37, Wednesday, December 3

2. Theorem
Let an = can/k + f(n) be recurrence relation with positive constant c and the positive function f(n).
(a) If for large n, f(n) grows proportional to n logk c, then an grows proportional to nlogk c log2 c.
(b) If for large n f(n) · pnq, where p is a positive constant and q < logk c, then an grows at most at rate proportional to nlogk c.

3. 7.3. Solution of Linear Recurrence Relations
Homework (MATH 310#11W):
Read 7.4
Do 7.3: all odd numbered problems
Turn in 7.3: 2,4,6,8,10

4. Homogeneous Linear Recurrence
an = c1an-1 + c2an cran-r.
ar - c1ar-1- c2ar cr= 0 is called the charateristic equation.
General solution:
an = A1a1n + A2a2n Ararn =
The constants are determined by solving the linear system
A1a1k + A2a2k Arark = ak, 0 · k · r – 1.
If the root a has multiplicity m, then the individual solutions an, nan, ..., nm-1an should be used.

5. Example 1: Doubling Rabbit Population
an = 2an-1. a0 = 6.
Solution: an = 6 £ 2n.

6. Example 2: Second-Order Recurrence Relation
Solve the recurrence relation an = 2an-1 + 3an-2 with a0 = a1 = 1.
Solution:
an = A13n + A2(-1)n where A1 = A2 = ½.

7. Example 3: Fibonacci Relation
an = an-1 + an-2, a0 = a1 = 1.
Solution:
an = (½ + ½ sqrt(5))n+1/sqrt(5) + (½ - ½ sqrt(5))n+1/sqrt(5)

8. Example 4: Complex and Multiple Roots
an = -2an-2 – an-4,
a0 = 0, a1 = 1, a2 = 2, a3 = 3.
Note that the solution from the book reduces to:
a4k = -4k
a4k+1 = 1 – 8k
a4k+2 = 4k+2
a4k+3 = 8k + 3