This document discusses recurrence relations and their use in defining sequences. It introduces key concepts like recurrence relations, initial conditions, explicit formulas, and solving recurrence relations using techniques like backtracking or finding the characteristic equation. As examples, it examines the Fibonacci sequence and linear homogeneous recurrence relations of varying degrees.

1. Counting

2. Recurrence Relations
 Sequences are ordered lists of elements.
• 1, 2, 3, 5, 8
• 1, 3, 9, 27, 81, …
 Sequences arise throughout mathematics, computer
science, and in many other disciplines, ranging from
botany to music.
 We will introduce the terminology to represent
sequences.
© S. Turaev, CSC 1700 Discrete Mathematics 2

3. Recurrence Relations
Definition: A sequence is a function from a subset of the
integers (usually either the set 0, 1, 2, 3, 4, … or
1, 2, 3, 4, … ) to a set 𝑆𝑆.
 The notation 𝑎𝑎𝑛𝑛 is used to denote the image of the
integer 𝑛𝑛. We can think of 𝑎𝑎𝑛𝑛 as the equivalent of
𝑓𝑓 𝑛𝑛 where 𝑓𝑓 is a function from 0,1,2, … to 𝑆𝑆. We
call 𝑎𝑎𝑛𝑛 a term of the sequence.
Example: Consider the sequence 𝑎𝑎𝑛𝑛 where 𝑎𝑎𝑛𝑛 = 1/𝑛𝑛.
Then
1,
1
2
,
1
3
,
1
4
, …
© S. Turaev, CSC 1700 Discrete Mathematics 3

4. Recurrence Relations
Definition: A recurrence relation for the sequence 𝑎𝑎𝑛𝑛 is
an equation that expresses 𝑎𝑎𝑛𝑛 in terms of one or more of
the previous terms of the sequence, namely,
𝑎𝑎0, 𝑎𝑎1, … , 𝑎𝑎𝑛𝑛−1, for all integers 𝑛𝑛 with 𝑛𝑛 ≥ 𝑛𝑛0, where 𝑛𝑛0
is a nonnegative integer.
 A sequence is called a solution of a recurrence relation
if its terms satisfy the recurrence relation.
 The initial conditions for a sequence specify the terms
that precede the first term where the recurrence
relation takes effect.
© S. Turaev, CSC 1700 Discrete Mathematics 4

5. Recurrence Relations
Example: Let 𝑎𝑎𝑛𝑛 be a sequence that satisfies the
recurrence relation 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 3 for 𝑛𝑛 = 1, 2, 3, 4, …
and suppose that 𝑎𝑎0 = 2. What are 𝑎𝑎1, 𝑎𝑎2 and 𝑎𝑎3?
[Here the initial condition is 𝑎𝑎0 = 2.]
Example: Let 𝑎𝑎𝑛𝑛 be a sequence that satisfies the
recurrence relation 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 − 𝑎𝑎𝑛𝑛−2 for 𝑛𝑛 = 2, 3, 4, …
and suppose that 𝑎𝑎0 = 3 and 𝑎𝑎1 = 5. What are 𝑎𝑎2 and
𝑎𝑎3?
[Here the initial conditions are 𝑎𝑎0 = 3 and 𝑎𝑎1 = 5.]
© S. Turaev, CSC 1700 Discrete Mathematics 5

6. Fibonacci Numbers
Definition: the Fibonacci sequence, 𝑓𝑓0, 𝑓𝑓1, 𝑓𝑓2, … , is
defined by:
 Initial Conditions: 𝑓𝑓0 = 0, 𝑓𝑓1 = 1
 Recurrence Relation: 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2
Example: Find 𝑓𝑓4, 𝑓𝑓5, 𝑓𝑓6, 𝑓𝑓7 and 𝑓𝑓8.
© S. Turaev, CSC 1700 Discrete Mathematics 6

7. Recurrence Relations
 Finding a formula for the 𝑛𝑛th term of the sequence
generated by a recurrence relation is called solving the
recurrence relation.
 Such a formula is called a explicit (closed) formula.
 One technique for finding an explicit formula for the
sequence defined by a recurrence relation is
backtracking.
© S. Turaev, CSC 1700 Discrete Mathematics 7

8. Recurrence Relations
Example: Let 𝑎𝑎𝑛𝑛 be a sequence that satisfies the
recurrence relation 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 3 for 𝑛𝑛 = 2, 3, 4, … and
suppose that 𝑎𝑎1 = 2. Find an explicit formula for the
sequence.
𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 3
= 𝑎𝑎𝑛𝑛−2 + 3 + 3 = 𝑎𝑎𝑛𝑛−2 + 2 ⋅ 3
= 𝑎𝑎𝑛𝑛−3 + 3 + 2 ⋅ 3 = 𝑎𝑎𝑛𝑛−3 + 3 ⋅ 3
⋯
= 𝑎𝑎2 + 𝑛𝑛 − 2 ⋅ 3 = 𝑎𝑎1 + 3 + 𝑛𝑛 − 2 ⋅ 3
= 𝑎𝑎1 + 𝑛𝑛 − 1 ⋅ 3 = 2 + 𝑛𝑛 − 1 ⋅ 3
= 3 ⋅ 𝑛𝑛 − 1
© S. Turaev, CSC 1700 Discrete Mathematics 8

9. Recurrence Relations : Backtracking
Example 1: 𝑏𝑏𝑛𝑛 = 5𝑏𝑏𝑛𝑛−1 + 3, 𝑏𝑏1 = 3.
Example 2: 𝑐𝑐𝑛𝑛 = 𝑐𝑐𝑛𝑛−1 + 𝑛𝑛, 𝑏𝑏1 = 4.
Example 3: 𝑑𝑑𝑛𝑛 = 𝑛𝑛 ⋅ 𝑑𝑑𝑛𝑛−1, 𝑑𝑑1 = 6.
© S. Turaev, CSC 1700 Discrete Mathematics 9

10. Fibonacci Numbers
Definition: the Fibonacci sequence, 𝑓𝑓0, 𝑓𝑓1, 𝑓𝑓2, … , is
defined by:
 Initial Conditions: 𝑓𝑓0 = 0, 𝑓𝑓1 = 1
 Recurrence Relation: 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2
Example: Find 𝑓𝑓4, 𝑓𝑓5, 𝑓𝑓6, 𝑓𝑓7 and 𝑓𝑓8.
© S. Turaev, CSC 1700 Discrete Mathematics 10

11. Linear Homogeneous Recurrence Relations
Definition: A linear homogeneous recurrence relation of
degree 𝑘𝑘 with constant coefficients is a recurrence
relation of the form
𝑎𝑎𝑛𝑛 = 𝑟𝑟1 𝑎𝑎𝑛𝑛−1 + 𝑟𝑟2 𝑎𝑎𝑛𝑛−2 + ⋯ + 𝑟𝑟𝑘𝑘 𝑎𝑎𝑛𝑛−𝑘𝑘
where 𝑟𝑟1, 𝑟𝑟2, … , 𝑟𝑟𝑘𝑘 are real numbers, and 𝑟𝑟𝑘𝑘 ≠ 0.
© S. Turaev, CSC 1700 Discrete Mathematics
• it is linear because the right-hand side is a sum of the previous
terms of the sequence each multiplied by a function of 𝑛𝑛.
• it is homogeneous because no terms occur that are not
multiples of the 𝑎𝑎𝑗𝑗s. Each coefficient is a constant.
• the degree is 𝑘𝑘 because 𝑎𝑎𝑛𝑛 is expressed in terms of the
previous 𝑘𝑘 terms of the sequence.
11

12. Linear Homogeneous Recurrence Relations
 𝑝𝑝𝑛𝑛 = 1.11 𝑝𝑝𝑛𝑛−1
 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2
 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 𝑎𝑎𝑛𝑛−2
2
 ℎ𝑛𝑛 = 2ℎ𝑛𝑛−1 + 1
 𝑏𝑏𝑛𝑛 = 𝑛𝑛𝑏𝑏𝑛𝑛−1
© S. Turaev, CSC 1700 Discrete Mathematics 12

13. Linear Homogeneous Recurrence Relations
 𝑝𝑝𝑛𝑛 = 1.11 𝑝𝑝𝑛𝑛−1
 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2
 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 𝑎𝑎𝑛𝑛−2
2
 ℎ𝑛𝑛 = 2ℎ𝑛𝑛−1 + 1
 𝑏𝑏𝑛𝑛 = 𝑛𝑛𝑏𝑏𝑛𝑛−1
© S. Turaev, CSC 1700 Discrete Mathematics
linear homogeneous recurrence
relation of degree 1
linear homogeneous recurrence
relation of degree 2
not linear
not homogeneous
coefficients are not constants
13

14. Linear Homogeneous Recurrence Relations
 The basic approach is to look for solutions of the form
𝑎𝑎𝑛𝑛 = 𝑥𝑥 𝑛𝑛
, where 𝑥𝑥 is a constant.
 Note that 𝑎𝑎𝑛𝑛 = 𝑥𝑥 𝑛𝑛
is a solution to the recurrence
relation
𝑎𝑎𝑛𝑛 = 𝑟𝑟1 𝑎𝑎𝑛𝑛−1 + 𝑟𝑟2 𝑎𝑎𝑛𝑛−2 + ⋯ + 𝑟𝑟𝑘𝑘 𝑎𝑎𝑛𝑛−𝑘𝑘
if and only if
𝑥𝑥 𝑛𝑛
= 𝑟𝑟1 𝑥𝑥 𝑛𝑛−1
+ 𝑟𝑟2 𝑥𝑥 𝑛𝑛−2
+ ⋯ + 𝑟𝑟𝑘𝑘 𝑥𝑥 𝑛𝑛−𝑘𝑘
.
© S. Turaev, CSC 1700 Discrete Mathematics 14

15. Linear Homogeneous Recurrence Relations
 Algebraic manipulation yields the characteristic
equation:
𝑥𝑥 𝑘𝑘
− 𝑟𝑟1 𝑥𝑥 𝑘𝑘−1
− 𝑟𝑟2 𝑥𝑥 𝑘𝑘−2
− ⋯ − 𝑟𝑟𝑘𝑘 = 0
 The sequence 𝑎𝑎𝑛𝑛 with 𝑎𝑎𝑛𝑛 = 𝑥𝑥 𝑛𝑛
is a solution if and
only if 𝑥𝑥 is a solution to the characteristic equation.
© S. Turaev, CSC 1700 Discrete Mathematics 15

16. Linear Homogeneous Recurrence Relations
Theorem: If the characteristic equation
𝑥𝑥2
− 𝑟𝑟1 𝑥𝑥 − 𝑟𝑟2 = 0
of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑟𝑟1 𝑎𝑎𝑛𝑛−1 + 𝑟𝑟2 𝑎𝑎𝑛𝑛−2 has two
distinct roots, 𝑥𝑥1 and 𝑥𝑥2, then
𝑎𝑎𝑛𝑛 = 𝑝𝑝𝑥𝑥1
𝑛𝑛
+ 𝑞𝑞𝑥𝑥2
𝑛𝑛
where 𝑝𝑝 and 𝑞𝑞 depend on the initial conditions, is the
explicit formula for the sequence.
© S. Turaev, CSC 1700 Discrete Mathematics 16

17. Linear Homogeneous Recurrence Relations
Theorem: If the characteristic equation
𝑥𝑥2
− 𝑟𝑟1 𝑥𝑥 − 𝑟𝑟2 = 0
of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑟𝑟1 𝑎𝑎𝑛𝑛−1 + 𝑟𝑟2 𝑎𝑎𝑛𝑛−2 has a
single root, 𝑥𝑥, then
𝑎𝑎𝑛𝑛 = 𝑝𝑝𝑥𝑥 𝑛𝑛
+ 𝑞𝑞𝑞𝑞𝑥𝑥 𝑛𝑛
where 𝑝𝑝 and 𝑞𝑞 depend on the initial conditions, is the
explicit formula for the sequence.
© S. Turaev, CSC 1700 Discrete Mathematics 17

18. Linear Homogeneous Recurrence Relations
Example 1: Find an explicit formula for the sequence
defined by 𝑎𝑎𝑛𝑛 = 4𝑎𝑎𝑛𝑛−1 + 5𝑎𝑎𝑛𝑛−2 with the initial
conditions 𝑎𝑎1 = 2 and 𝑎𝑎2 = 6.
Example 2: Find an explicit formula for the sequence
defined by 𝑏𝑏𝑛𝑛 = −6𝑏𝑏𝑛𝑛−1 − 9𝑏𝑏𝑛𝑛−2 with the initial
conditions 𝑏𝑏1 = 2.5 and 𝑏𝑏2 = 4.7.
Example 3 (Fibonacci sequence): Find an explicit formula
for the sequence defined by 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2 with the
initial conditions 𝑓𝑓0 = 0 and 𝑓𝑓1 = 1.
© S. Turaev, CSC 1700 Discrete Mathematics 18