Lecture Notes, 19 Nov 2010
Last Update: 22 November 2010
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§§7.1–7.2: Recurrence Relations
- Recurrence Relations:
- A sequence can be defined in two different ways:
- non-recursively ("explicitly"),
in terms of its current I/P:
by giving initial conditions (first few terms)
& a recurrence relation that defines the sequence in terms of
its previous O/P:
- Question: Given initial conditions & recurrence relation
an = h(a0 = C0, …,
(how) can we compute the explicit formula f(n)?
- This is called "solving" the recurrence relation.
an = 3an–1 – 2an–2,
∀ n ≥ 2
- Lots of different sequences share this pattern,
differing only in their initial conditions
- Given the initial conditions,
we can compute the nth term (n ≥ 2)
without knowing what the function does to its I/P!
- We compute it on the basis of what it did:
we compute it on the basis of what its previous O/P was!
- 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)
|a0 ||0 ||0 ||1 ||1 ||1 ||2 ||2|
|a1 ||0 ||1 ||0 ||1 ||2 ||1 ||2|
|a2 ||0 ||3 ||–2 ||1 ||4 ||–1 ||2|
|a3 ||0 ||7 ||–6 ||1 ||8 ||–5 ||2|
|a4 ||0 ||15 ||–14 ||1 ||16 ||–13 ||2|
|a5 ||0 ||… ||–30 ||1 ||… ||… ||…|
|an ||0 ||2n–1 ||2–2n ||1 ||2n ||3–2n ||2|
- What good are recurrence relations?
- They describe similar patterns of growth,
based on differing
initial conditions or "seeds"
- E.g.) compound interest:
- So the question is: How to "solve" a recurrence relation.
A linear, homogeneous, recurrence relation of degree 2
a recurrence relation of the form
where c1, c2 ∈ R & c2 ≠
- See text for complete def of linear homogeneous recurrence
relation of degree k with constant coefficients.
- "linear": no exponents
- "homogeneous": all terms are multiples of the
- Note: pronounced "homoJEENee-us",
not "hoMOJenus", with 5 syllables
- "constant coefficients":
they are not functions of n;
they are constants
- This determines a family of sequences
that differ only in
their initial conditions.
- They are solvable!
- How do you solve them?
- The trick:
- Given a0=C0,
look for solutions of the form an = rn, for
- Because, in the simplest case, when
the ratio of sucessive terms is constant;
∴ it's a geometric sequence
Given the sequence
a0 = C0
a1 = c1C0
a2 = c1²C0
an = c1nC0
∴ an = a0c1n
an = rn is a solution
(i.e., an "explicit", non-recursive formula)
an = c1an–1 +
(an =) rn = c1rn–1
(by substituting rn for an)
rn c1rn–1 + c2rn–2
___ = _____________ (for r ≠ 0)
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)
- Thm 1 (p. 462):
- In a theorem, you have to say where everything comes
i.e., you must give their data types.
Let C0, C1 ∈ N be constants.
Let a0 = C0 and a1 = C1 be
the initial conditions of a recurrence relation.
Let c1, c2 ∈ R be such that
an = c1an–1
+ c2an–2 is the recurrence relation.
Let r1 ≠ r2 be 2 distinct roots of the "characteristic
of the recurrence relation.
(∃α1, α2 ∈ R)(∀n ∈ N)[an =
- 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
- (Outline of) procedure (i.e., algorithm)
for solving a (linear homogeneous)
(of degree 2 with constant coefficients):
- Details will be given in the next lecture.
- I/P: recurrence relation
an = c1an–1
- Set up the characteristic equation:
- Solve the characteristic eqn for r1, r2
- if r1 = r2,
then begin O/P "no solution"; halt end
else goto (2b)
Find α1, α2 such that
an = α1r1n
- Use initial conditions to produce 2 simultaneous eqns in 2
- Solve these for α1 & α2
- O/P: explicit formula for an, namely:
Text copyright © 2010 by William J. Rapaport
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