Rosen, 5.4 Binomial Coefficients

Discrete Structures Lecture 27
CMSC 2123
8.2 Solving Linear Recurrence Relations
DEFINITION 1
EXAMPLE 1
EXAMPLE 2
A linear homogeneous recurrence relation of degree k with constant
coefficients is a recurrence relation of the form
aa nn = cc 1 aa nn−1 + cc 2 aa nn−2 + ⋯ + cc kk aa nn−kk
where cc 1 , cc 2 , ⋯ , cc kk are real numbers and cc kk ≠ 0.
The recurrence relation PP nn = (1.11)PP nn−1 is a linear homogenous recurrence
relation of degree one. The recurrence relations ff nn = ff nn−1 + ff nn−2 is a linear
homogeneous recurrence relation of degree two. The recurrence relation
aa nn = aa nn−5 is a linear homogeneous recurrence relation of degree five.
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The recurrence relation aa nn = aa nn−1 + aa nn−2 is not linear. The recurrence
relation HH nn = 2HH nn−1 + 1 is not homogeneous. The recurrence relation BB nn =
nnBB nn−1 does not have constant coefficients.
Solving Linear Homogeneous Recurrence Relations with Constant Coefficients
The basic approach for solving linear homogeneous recurrence relations is to look for solutions of
the form aa nn = rr nn , where rr is a constant. Note that aa nn = rr nn is a solution of the recurrence
relation aa nn = cc 1 aa nn−1 + cc 2 aa nn−2 + ⋯ + cc kk aa nn−kk if and only if
rr nn = cc 1 rr nn−1 + cc 2 rr nn−2 + ⋯ + cc kk rr nn−kk
When both sides are divided by rr nn−kk and the right-hand side is subtracted from the left, we obtain
the equations
rr kk − cc 1 rr kk−1 − cc 2 rr kk−2 − ⋯ − cc kk−1 rr − cc kk = 0
Consequently, the sequence {aa nn } with aa nn = rr nn is a solution if and only if rr is a solution of this last
equation. We call this the characteristic equation of the recurrence relation. The solutions of
this equation are called the characteristic roots of the recurrence relation.
THEOREM 1 Let cc 1 and cc 2 be real numbers. Suppose rr 2 − cc 1 rr − cc 2 = 0 has two distinct
roots rr 1 and rr 2 . Then the sequence {aa nn }is a solution of the recurrence relation
aa nn = cc 1 aa nn−1 + cc 2 aa nn−2 if and only if aa nn = αα 1 rr nn nn
1 + αα 2 rr 2 for nn = 1, 2, …,
where αα 1 and αα 2 are constants.
EXAMPLE 3
What is the solution of the recurrence relation
aa nn = aa nn−1 + 2aa nn−2
with aa 0 = 2 and aa 1 = 7?
Solution: Theorem 1 can be used to solve this problem. The characteristic
equation of the recurrence relation is rr 2 − rr − 2 = 0. Its roots are rr = 2
and rr = −1. Hence the sequence {aa nn } is a solution to the recurrence relation
if and only if
aa nn = αα 1 2 nn + αα 2 (−1) nn
for some constants αα 1 and αα 2 . From the initial conditions, it follows that
aa 0 = 2 = αα 1 + αα 2
aa 1 = 7 = αα 1 ∙ 2 + αα 2 ∙ (−1)
Solving these two equations show that αα 1 = 3 and αα 2 = −1. Hence, the
solution to the recurrence relation and initial conditions is the sequence {aa nn }
with
aa nn = 3 ∙ 2 nn − (−1) nn
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Discrete Structures Lecture 27
CMSC 2123
8.2 Solving Linear Recurrence Relations
EXAMPLE 3.1
What is the solution of the recurrence relation
aa kk = 1 2 aa kk−1
with aa 0 = NN?
Solution: Theorem 1 can be used to solve this problem. The characteristic
equation of the recurrence relation is rr − 1 = 0. Its root is = 1 . Hence the
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sequence {aa nn } is a solution to the recurrence relation if and only if
aa kk = αα 1 1 kk
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for a constants αα 1 . From the initial conditions, it follows that
aa 0 = NN = αα 1
Solving these two equations show that αα 1 = NN. Hence, the solution to the
recurrence relation and initial conditions is the sequence {aa kk } with
aa kk = NN kk
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EXAMPLE 4
Find an explicit formula for the Fibonacci numbers.
Solution: Type equation here.
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