Discrete Structures

# Lecture Notes, 19 Nov 2010

 Last Update: 22 November 2010 Note: or material is highlighted

### §§7.1–7.2: Recurrence Relations

1. Recurrence Relations:

1. A sequence can be defined in two different ways:

1. non-recursively ("explicitly"),
in terms of its current I/P:

an = f(n)

2. recursively,
by giving initial conditions (first few terms)

a0 = C0, …, an–1 = Cn–1

& a recurrence relation that defines the sequence in terms of its previous O/P:

an = h(a0 = C0, …, an–1)

2. Question: Given initial conditions & recurrence relation an = h(a0 = C0, …, an–1),
(how) can we compute the explicit formula f(n)?

• This is called "solving" the recurrence relation.

3. E.g.)

1. initial conditions:

a0 = C0
a1 = C1

recurrence relation:

an = 3an–1 – 2an–2, ∀ n ≥ 2

differing only in their initial conditions

1. Given the initial conditions,
we can compute the nth term (n ≥ 2)
without knowing what the function does to its I/P!

2. We compute it on the basis of what it did:
we compute it on the basis of what its previous O/P was!

3. Here are some examples:
(initial conditions are in the first 2 rows;
last row shows "explicit" "solution",
i.e.) def in terms of I/P)

4. What good are recurrence relations?

1. They describe similar patterns of growth,
based on differing initial conditions or "seeds"

2. E.g.) compound interest:

• 2 people deposit different amounts of \$ in same bank;
∴ same recurrence relation computes their interest;

But the actual interest depends on their initial deposit!

2. So the question is: How to "solve" a recurrence relation.

1. Def:

A linear, homogeneous, recurrence relation of degree 2
with constant coefficients
isdef
a recurrence relation of the form

an = c1an–1 + c2an–2

where c1, c2R & c2 ≠ 0

1. See text for complete def of linear homogeneous recurrence relation of degree k with constant coefficients.
2. "linear": no exponents
3. "homogeneous": all terms are multiples of the ai
• Note: pronounced "homoJEENee-us",
not "hoMOJenus", with 5 syllables

4. "constant coefficients":
they are not functions of n;
they are constants

2. This determines a family of sequences
that differ only in their initial conditions.

• E.g.) The Fibonacci recurrence relation:

fn = fn–1 + fn–2

is a linear homegeneous recurrence relation of degree 2
and can have differing intial conditions,
yielding different Fibonacci sequences:

1. f0 = 0 & f1 = 1 yields:

0,1,1,2,3,5,…

2. f0 = 1 & f1 = 1 yields:

1,1,2,3,5,…

3. f0 = 1 & f1 = 2 yields:

1,2,3,5,…

3. They are solvable!

3. How do you solve them?

1. The trick:

• Given a0=C0, a1=C1, & an=c1an–1+c2an–2
look for solutions of the form an = rn, for constant r

2. Why?

1. Because, in the simplest case, when an=c1an–1,
the ratio of sucessive terms is constant;
∴ it's a geometric sequence

2. Given the sequence
a0 = C0
an = c1an–1

we have:

a0 = C0
a1 = c1C0
a2 = c1²C0

an = c1nC0
∴ an = a0c1n

3. an = rn is a solution
(i.e., an "explicit", non-recursive formula)
for an = c1an–1 + c2an–2
(an =) rn = c1rn–1 + c2rn–2 (by substituting rn for an)

rn         c1rn–1 + c2rn–2
___ = _____________ (for r ≠ 0)
rn–2           rn–2

r² = c1r + c2
r² – c1r – c2 = 0 [the characteristic equation of the recurrence relation]
r is a solution of this equation [the characteristic root]
(i.e., makes the equation come out T)

4. Thm 1 (p. 462):

• In a theorem, you have to say where everything comes from;
i.e., you must give their data types.

Let C0, C1N be constants.

Let a0 = C0 and a1 = C1 be the initial conditions of a recurrence relation.

Let c1, c2R be such that an = c1an–1 + c2an–2 is the recurrence relation.

Let r1 ≠ r2 be 2 distinct roots of the "characteristic equation"

r² – c1r – c2

of the recurrence relation. Then:

(∃α1, α2R)(∀n ∈ N)[an = α1r1n + α2r2n]

• i.e.) the recurrence relation for the nth term
can be computed non-recursively
using the formula in terms of αi and ri

• This is a non-constructive existence claim!

• We need an algorithm to show how to find the αi

5. (Outline of) procedure (i.e., algorithm)
for solving a (linear homogeneous) recurrence relation
(of degree 2 with constant coefficients):

• Details will be given in the next lecture.

• I/P: recurrence relation an = c1an–1 + c2an–2

1. Set up the characteristic equation:

r² – c1r – c2

2. Solve the characteristic eqn for r1, r2

1. if r1 = r2,
then begin O/P "no solution"; halt end
else goto (2b)

2. Find α1, α2 such that an = α1r1n + α2r2n:

1. Use initial conditions to produce 2 simultaneous eqns in 2 unknowns:

2. Solve these for α1 & α2

• O/P: explicit formula for an, namely:

an = α1r1n + α2r2n