Download CSC 10400 Discrete Mathematical Structures Lecture 10

CSC 10400
Discrete Mathematical Structures
Lecture 10
Instructor: Pavel Rytir
Email: [email protected]
The City College of New York
Spring 2015
Section 10.1 The First-Order Linear Recurrence Relation
First-order linear recurrence relation
Homogeneous and nonhomogeneous recurrence relation
Geometric Progression
A geometric progression is an infinite sequence of numbers, such as
5, 15, 45, 135, . . ., where the division of each term, other than the
first, by its immediate predecessor is a constant. One can define
this geometric progression as:
a0 = 5
an = 3an−1 (n > 0)
Geometric Progression
A geometric progression is an infinite sequence of numbers, such as
5, 15, 45, 135, . . ., where the division of each term, other than the
first, by its immediate predecessor is a constant. One can define
this geometric progression as:
a0 = 5
an = 3an−1 (n > 0)
FHowever, if we change the value of a0 to a0 = 7, we obtain a
different geometric progression: 7, 21, 63, 189, . . ..
Recurrence Relation
Recurrence relation
A recurrence relation is a function a(n), preferably written as
an , where an depends on some of the prior terms
an−1 , an−2 , . . . , a1 , a0 .
Recurrence Relation
Recurrence relation
A recurrence relation is a function a(n), preferably written as
an , where an depends on some of the prior terms
an−1 , an−2 , . . . , a1 , a0 .
First-order recurrence relation
A recurrence relation is a first-order recurrence relation if an
only depends on its immediate predecessor, an−1 .
Recurrence Relation
Recurrence relation
A recurrence relation is a function a(n), preferably written as
an , where an depends on some of the prior terms
an−1 , an−2 , . . . , a1 , a0 .
First-order recurrence relation
A recurrence relation is a first-order recurrence relation if an
only depends on its immediate predecessor, an−1 .
Linear recurrence relation
A recurrence relation is linear if each subscripted term appears
to the first power.
Recurrence Relation
Recurrence relation
A recurrence relation is a function a(n), preferably written as
an , where an depends on some of the prior terms
an−1 , an−2 , . . . , a1 , a0 .
First-order recurrence relation
A recurrence relation is a first-order recurrence relation if an
only depends on its immediate predecessor, an−1 .
Linear recurrence relation
A recurrence relation is linear if each subscripted term appears
to the first power.
Boundary (initial) conditions
The value a0 , given in addition to the recurrence relation, is
called boundary condition, or initial condition.
Unique Solution and General Solution
Unique solution
The recurrence relation an = 3an−1 with a0 = 5 determines a
unique solution an = 5(3n ).
Unique Solution and General Solution
Unique solution
The recurrence relation an = 3an−1 with a0 = 5 determines a
unique solution an = 5(3n ).
General solution
The recurrence relation an = 3an−1 without specifying the
value of a0 determines a general solution an = A(3n ). This
general solution becomes unique when a0 = A is specified.
Solutions For Geometric Progression
For the recurrence relation defined as
a0 = A
an = dan−1 (n > 0)
The unique solution is
an = Ad n
Problem
Solve the recurrence relation
a2 = 98
an = 7an−1 (n > 0)
Answer
Problem
Solve the recurrence relation
a2 = 98
an = 7an−1 (n > 0)
Answer
we have a2 = a0 (72 ) = 98, implying a0 = 2. Therefore, the unique
solution is an = 2(7n ).
Problem
A bank pays 6% annual interest on savings, compounding the interest monthly. If Percy deposit 1000 dollars on the first day of May,
how much will this deposit be worth a year later?
Answer
Problem
A bank pays 6% annual interest on savings, compounding the interest monthly. If Percy deposit 1000 dollars on the first day of May,
how much will this deposit be worth a year later?
Answer
The monthly rate is 6%
12 = 0.005. For 0 ≤ n ≤ 12, let pn denote the
value of the deposit after the end of n months. We have p0 = 1000
and pn = pn−1 + 0.005pn−1 = 1.005pn−1 for n > 0. Therefore, the
value of this deposit a year later is p12 = 1000×1.00512 ≈ 1061.68.
Problem
A positive integer n can be represented as a composition
of positive integers. For example, 3 can be represented by
3 1+2 2+1 1+1+1 Let an count the number of compositions of n. How to compute an directly?
Answer
Problem
A positive integer n can be represented as a composition
of positive integers. For example, 3 can be represented by
3 1+2 2+1 1+1+1 Let an count the number of compositions of n. How to compute an directly?
Answer
A composite of n can generate two different compositions of n + 1:
(1) by increasing the last summand by 1, e.g., from 1 + 2 to 1 + 3;
or (2) by adding a new summand 1, e.g., from 1 + 2 to 1 + 2 + 1.
Thus, we have an = 2an−1 for n > 1. Clearly, a1 = 1.
We can set a0 = a21 = 12 , and have the recurrence relation an =
2an−1 with a0 = 12 . Therefore, an = 12 2n = 2n−1 .
Homogeneous and Nonhomogeneous Recurrence Relations
The general first-order linear recurrence relations with constant
coefficients have the form an + can−1 = f (n) for n > 0, where c is
a constant and f (n) is a function with domain Z+ .
Homogeneous recurrence relation
A recurrence relation is homogeneous if f (n) = 0.
FExample: the geometric progression
Nonhomogenous recurrence relation
A recurrence relation is nonhomogeneous if f (n) 6= 0.
FExample: an = an−1 + 3n
Problem
The bubble sort algorithm shown on blackboard (you can also
google/wiki it) sorts a sequence of numbers. How many comparisons does the bubble sort algorithm execute to sort a sequence of
n numbers?
Answer
In the first iteration, the algorithm executes n−1 comparisons, places
the smallest number at the first position, and leaves n − 1 number
unsorted. Let an be the number of comparisons executed in sorting
n numbers. We have an = an−1 + (n − 1) for n ≥ 2, and a1 = 0.
Problem
The bubble sort algorithm shown on blackboard (you can also
google/wiki it) sorts a sequence of numbers. How many comparisons does the bubble sort algorithm execute to sort a sequence of
n numbers?
Answer
In the first iteration, the algorithm executes n−1 comparisons, places
the smallest number at the first position, and leaves n − 1 number
unsorted. Let an be the number of comparisons executed in sorting
n numbers. We have an = an−1 + (n − 1) for n ≥ 2, and a1 = 0.
Question
How can we solve the recurrence relation:
a1 = 0; an = an−1 + (n − 1), n ≥ 2
Problem
Solve the recurrence relation:
a1 = 0; an = an−1 + (n − 1), n ≥ 2
Answer
Problem
Solve the recurrence relation:
a1 = 0; an = an−1 + (n − 1), n ≥ 2
Answer
a1 = 0
a2 = a1 + (2 − 1) = 1
a3 = a2 + (3 − 1) = 1 + 2
a4 = a3 + (4 − 1) = 1 + 2 + 3
......
an = an−1 + (n − 1) = 1 + 2 + 3 + . . . + (n − 1)
Therefore, an = 1 + 2 + 3 + . . . + (n − 1) =
n(n−1)
.
2
Section 10.2 The Second-Order Linear Homogeneous
Recurrence Relation with Constant Coefficients
Second-order linear homogeneous recurrence relation
Start From Recursive Definition
Let us consider the following recursive definition.

n=0
 0
1
n=1
Fn =

Fn−1 + Fn−2 n ≥ 2
This is the definition of well-known Fibonacci numbers.
Start From Recursive Definition
Let us consider the following recursive definition.

n=0
 0
1
n=1
Fn =

Fn−1 + Fn−2 n ≥ 2
This is the definition of well-known Fibonacci numbers.
Question
How can we represent Fn directly?
Second-order Linear Recurrence Relation
General form of linear recurrence relations
Let k, n ∈ Z+ with k ≤ n. Consider the following recurrence
relation
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
where C0 , C1 , . . . , Ck are real numbers with C0 6= 0 and
Ck 6= 0. This equation is the general form of a linear
recurrence relation with constant coefficients of order k.
Second-order Linear Recurrence Relation
General form of linear recurrence relations
Let k, n ∈ Z+ with k ≤ n. Consider the following recurrence
relation
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
where C0 , C1 , . . . , Ck are real numbers with C0 6= 0 and
Ck 6= 0. This equation is the general form of a linear
recurrence relation with constant coefficients of order k.
Second-order linear homogeneous recurrence relation with
constant coefficients
In this section, we only consider the recurrence relations
defined as:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
Characteristic Equation
Because the second-order recurrence function is extended from the
first order recurrence relation (If C2 = 0, it becomes a first-order
recurrence relation), we guess a solution as an = cr n where
c, r 6= 0 (this guess may be not correct!).
Characteristic Equation
Because the second-order recurrence function is extended from the
first order recurrence relation (If C2 = 0, it becomes a first-order
recurrence relation), we guess a solution as an = cr n where
c, r 6= 0 (this guess may be not correct!).
By guessing an = cr n , we have C0 cr n + C1 cr n−1 + C2 cr n−2 = 0, or
C0 r 2 + c1 r + C2 = 0
This quadratic equation is called the characteristic equation. The
roots r1 and r2 of this equation are called characteristic roots.
These roots r1 and r2 determine three cases:
Characteristic Equation
Because the second-order recurrence function is extended from the
first order recurrence relation (If C2 = 0, it becomes a first-order
recurrence relation), we guess a solution as an = cr n where
c, r 6= 0 (this guess may be not correct!).
By guessing an = cr n , we have C0 cr n + C1 cr n−1 + C2 cr n−2 = 0, or
C0 r 2 + c1 r + C2 = 0
This quadratic equation is called the characteristic equation. The
roots r1 and r2 of this equation are called characteristic roots.
These roots r1 and r2 determine three cases:
Case 1 (Distinct real roots): r1 and r2 are distinct real
numbers.
Characteristic Equation
Because the second-order recurrence function is extended from the
first order recurrence relation (If C2 = 0, it becomes a first-order
recurrence relation), we guess a solution as an = cr n where
c, r 6= 0 (this guess may be not correct!).
By guessing an = cr n , we have C0 cr n + C1 cr n−1 + C2 cr n−2 = 0, or
C0 r 2 + c1 r + C2 = 0
This quadratic equation is called the characteristic equation. The
roots r1 and r2 of this equation are called characteristic roots.
These roots r1 and r2 determine three cases:
Case 1 (Distinct real roots): r1 and r2 are distinct real
numbers.
Case 2 (Distinct complex roots): r1 and r2 are not real
numbers but complex numbers.
Characteristic Equation
Because the second-order recurrence function is extended from the
first order recurrence relation (If C2 = 0, it becomes a first-order
recurrence relation), we guess a solution as an = cr n where
c, r 6= 0 (this guess may be not correct!).
By guessing an = cr n , we have C0 cr n + C1 cr n−1 + C2 cr n−2 = 0, or
C0 r 2 + c1 r + C2 = 0
This quadratic equation is called the characteristic equation. The
roots r1 and r2 of this equation are called characteristic roots.
These roots r1 and r2 determine three cases:
Case 1 (Distinct real roots): r1 and r2 are distinct real
numbers.
Case 2 (Distinct complex roots): r1 and r2 are not real
numbers but complex numbers.
Case 3 (Repeated real root): r1 and r2 are real numbers but
r1 = r2 .
Case 1: Distinct Real Roots
Let r1 and r2 be two distinct real roots of the characteristic
equation C0 r 2 + C1 r + C2 = 0 corresponding to the recurrence
relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
Case 1: Distinct Real Roots
Let r1 and r2 be two distinct real roots of the characteristic
equation C0 r 2 + C1 r + C2 = 0 corresponding to the recurrence
relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
General Solution
The general solution of the recurrence relation is:
an = q1 r1n + q2 r2n
Problem
The Fibonacci number was defined recursively as:
F0 = 0; F1 = 1; Fn = Fn−1 + Fn−2 , n ≥ 2
How to compute Fn directly?
Answer
Problem
The Fibonacci number was defined recursively as:
F0 = 0; F1 = 1; Fn = Fn−1 + Fn−2 , n ≥ 2
How to compute Fn directly?
Answer
The corresponding recurrence relation is: an − an−1 − an−2 = 0 with
2
a0 = 0 and a1 = 1. The characteristic
√ relation is r √− r − 1 = 0.
The characteristic roots are r1 = 1+2 5 and r2 = 1−2 5 . Thus, the
√
√
general solution is:an = q1 ( 1+2 5 )n + q2 ( 1−2 5 )n . We use the initial
√
5
values and obtain a0 = 0 = q1 + q2 and a1 = 1 = q1 1+2
√
q2 1−2 5 , implying that q1
√
√
√1 ( 1+ 5 )n − √1 ( 1− 5 )n .
2
2
5
5
=
√1
5
andq2 =
− √15 .
+
Therefore, Fn =
Problem
Solve the recurrence relation:
a0 = −1; a1 = 8; an + an−1 − 6an−2 = 0, n ≥ 2
Answer
Problem
Solve the recurrence relation:
a0 = −1; a1 = 8; an + an−1 − 6an−2 = 0, n ≥ 2
Answer
The characteristic equation is r 2 + r − 6 = 0, with roots r1 = 2 and
r2 = −3. Thus, the general solution is an = q1 2n + q2 (−3)n . With
a0 = −1 and a1 = 8, we have q1 = 1 and q2 = −2. Therefore, the
solution is an = 2n − 2(−3)n .
Problem
In many programming languages, one may consider legal arithmetic
expressions. Suppose we only consider the arithmetic expression
made up of the digits 0, 1, . . . , 9 and binary operations +, ∗, / (no
parenthesis). For example, 3 + 4 and 2 + 3 ∗ 5 are legal but 8 + ∗9 is
not. We also consider the number 00735 a legal number as 00735 =
735. How many possible legal arithmetic expressions are made of n
symbols?
Answer
Let an be the number of legal arithmetic expressions made of n
symbols. We have a1 = 10 for 0, 1, . . . , 9 and a2 = 100 for
00, 01, . . . , 99. When n > 2, we construct a symbol by (1) adding a
digit at the end of a legal equation of n − 1 symbols; and (2) adding
a binary operator and a digit at the end of a legal equation of n − 2
symbols. For the first case, we have 10 choices. For the second case,
we have 29 choices: +0, +1, . . . , +9, ∗0, ∗1, . . . , ∗9, /1, /2, . . . , /9.
Thus, we have an = 10an−1 + 29an−2 .
Answer
Let an be the number of legal arithmetic expressions made of n
symbols. We have a1 = 10 for 0, 1, . . . , 9 and a2 = 100 for
00, 01, . . . , 99. When n > 2, we construct a symbol by (1) adding a
digit at the end of a legal equation of n − 1 symbols; and (2) adding
a binary operator and a digit at the end of a legal equation of n − 2
symbols. For the first case, we have 10 choices. For the second case,
we have 29 choices: +0, +1, . . . , +9, ∗0, ∗1, . . . , ∗9, /1, /2, . . . , /9.
Thus, we have an = 10an−1 + 29an−2 .
The characteristic
equation is √r 2 − 10r − 29 = 0 with root
√
r1 = 5 + 3 √
6 and r2 = 5 −√3 6. Thus, the general solution is
an = q1 (5 + 3 6)n + q2 (5 − 3 6)n . With a1 = 10 and a2 = 100, we
5
5
have a0 = (a2 − 10a1 )/29 = 0, implying q1 = 3√
and q2 = − 3√
.
6
6
√
√
5
5
Therefore, the solution is: an = 3√
(5 + 3 6)n − 3√
(5 − 3 6)n .
6
6
Problem
How many binary sequences of length n having no consecutive 0’s
can we have?
Answer
Problem
How many binary sequences of length n having no consecutive 0’s
can we have?
Answer
Let an , bn , and cn be the number of binary sequences of length n
with no consecutive 0’s which ends with either 0 or 1, ends with 0,
and ends with 1 respectively. Clearly, an = bn + cn .
First, a1 = 2 and a2 = 3. When n > 2, we can generate a sequence
of length n from a sequence x of length n − 1. If x ends with
0, we can only add 1 at the end of x. If x ends with 1, we can
either add 0 or 1 at the end of x. Thus, an = bn−1 + 2cn−1 .
In addition, any binary sequence of length n − 2 can generate a
sequence of length n−1 ending with 1, i.e., cn−1 = an−2 . Therefore,
an = bn−1 +2cn−1 = (bn−1 +cn−1 )+c
=a
+a . Solve
the
√ n−1 √ n−1 √ n−2
√
5+2
1+
5
5−2
1−
5
n
n
recurrence relation, we have an = √5 ( 2 ) + √5 ( 2 ) .
Case 2: Distinct Complex Roots
Let r1 and r2 be two distinct complex roots (not real) of the
characteristic equation C0 r 2 + C1 r + C2 = 0 corresponding to the
recurrence relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
Case 2: Distinct Complex Roots
Let r1 and r2 be two distinct complex roots (not real) of the
characteristic equation C0 r 2 + C1 r + C2 = 0 corresponding to the
recurrence relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
General Solution
The general solution of the recurrence relation is:
an = q1 r1n + q2 r2n
Problem
Solve the recurrence relation an = 2(an−1 − an−2 ) with a0 = 1 and
a1 = 2.
Answer
Problem
Solve the recurrence relation an = 2(an−1 − an−2 ) with a0 = 1 and
a1 = 2.
Answer
The characteristic equation is r 2 − 2r + 2 = 0 with roots r1 = 1 + i
and r2 = 1 − i. The general solution is an = q1 (1 + i)n + q2 (1 − i)n .
Apply this general solution to a0 = 1 and a1 = 2, we have q1 = 1−i
2
1−i
n + 1+i (1 − i)n .
and q2 = 1+i
.
Therefore,
a
=
(1
+
i)
n
2
2
2
DeMoivre’s Theorem
A complex number
x + yi can be rewritten as R(cos θ + i sin θ),
p
2
where R = x + y 2 and θ = arctan( yx ).
DeMoivre’s Theorem
A complex number
x + yi can be rewritten as R(cos θ + i sin θ),
p
2
where R = x + y 2 and θ = arctan( yx ).
DeMoivre’s Theorem
(cos θ + i sin θ)n = cos nθ + i sin nθ
Case 3: Repeated Real Roots
Let r1 = r2 be the unique real root of the characteristic equation
C0 r 2 + C1 r + C2 = 0 corresponding to the recurrence relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
Case 3: Repeated Real Roots
Let r1 = r2 be the unique real root of the characteristic equation
C0 r 2 + C1 r + C2 = 0 corresponding to the recurrence relation:
C0 an + C1 an−1 + C2 an−2 = 0, n ≥ 2
General Solution
The general solution of the recurrence relation is:
an = q1 r1n + q2 nr1n = (q1 + q2 n)r1n
Problem
Solve the recurrence relation
a0 = 2; a1 = 3; an = 4an−1 − 4an−2 , n ≥ 2
Answer
Problem
Solve the recurrence relation
a0 = 2; a1 = 3; an = 4an−1 − 4an−2 , n ≥ 2
Answer
The characteristic equation is r 2 − 4r + 4 = 0 with unique root
r1 = r2 = 2. Thus, the general solution is an = (q1 + q2 n)2n . Apply
this solution to a0 = 2 and a1 = 4, we have q1 = 2 and q2 = − 21 .
Therefore,
n
an = (2 − )(2n )
2
Section 10.3 The Nonhomogeneous Recurrence Relation
First-order and second-order nonhomogeneous recurrence
relation
Solve some nonhomogeneous recurrence relations
Nonhomogeneous Recurrence Relation
Consider the nonhomogeneous linear recurrence relation with
constant coefficients.
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
Nonhomogeneous Recurrence Relation
Consider the nonhomogeneous linear recurrence relation with
constant coefficients.
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
Bad New
There is no general method to solve all nonhomogeneous relations!
Nonhomogeneous Recurrence Relation
Consider the nonhomogeneous linear recurrence relation with
constant coefficients.
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
Bad New
There is no general method to solve all nonhomogeneous relations!
Good New
For certain functions f (n), we can find a successful technique!
The Recurrence Relation an − an−1 = f (n)
Consider that
a1 = a0 + f (1)
a2 = a1 + f (2)
a3 = a2 + f (3)
. . . . . . an
= an−1 + f (n)
Thus, the general solution is:
an = a0 + f (1) + f (2) + . . . + f (n) = a0 +
n
X
i=1
f (i)
Particular Solution
Given a linear nonhomogeneous recurrence relation as
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
A particular solution is a solution to make the above equation valid.
FExamples
an = 3(3n ) is a particular solution of the recurrence relation
an − an−1 = 2(3n ).
an = 3(3n ) + 2 and an = 3(3n ) − 4 are both particular
solutions of the recurrence relation an − an−1 = 2(3n ).
Linear Nonhomogeneous Recurrence Relation
Theorem
Given a linear nonhomogeneous recurrence relation as
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = f (n)
where C0 , C1 , . . . , Ck are constants with C0 , Ck 6= 0. The general
solution of this nonhomogeneous relation is an = hn +pn , where hn is
the general solution for the corresponding homogeneous recurrence
relation
C0 an + C1 an−1 + C2 an−2 + . . . + Ck an−k = 0
and pn is a solution of the nonhomogeneous recurrence relation,
called the particular solution.
Linear Nonhomogeneous Recurrence Relation
Proof.
Because pn is a particular solution, we have
C0 pn + C1 pn−1 + C2 pn−2 + . . . + Ck pn−k = f (n)
Because bn be another particular solution, we have
C0 bn + C1 bn−1 + C2 bn−2 + . . . + Ck bn−k = f (n)
Subtract the first of equation from the second equation, we have
C0 (bn −pn )+C1 (bn−1 −pn−1 )+C2 (bn−2 −pn−2 )+. . .+Ck (bn−k −pn−k ) = 0
If follows that bn − pn is a solution of the corresponding
homogeneous recurrence relation, say, hn . Consequently,
bn = hn + pn for all n.
Determine the Particular Solution pn
Question
How to determine pn ?
Determine the Particular Solution pn
Question
How to determine pn ?
Answer
Make a reasonable guess suggested by f (n).
Specified Solution
Let an be a general solution. We say that bn is a specified solution
of an , or a solution of an , if bn can be written in form of an by
specifying appropriate constants in an .
Specified Solution
Let an be a general solution. We say that bn is a specified solution
of an , or a solution of an , if bn can be written in form of an by
specifying appropriate constants in an . FExample
3n is a specified solution of α(3n ) by setting α = 1.
√
3 2
2
n
2 n √is a specified solution of αn + β3 bye
α = − 23 and β = 0.
3n3 is NOT a specified solution of αn2 + β3n .
−
setting
Determine Particular Solution For an + C1 an−1 = kr n .
If r n is not a specified solution on the general solution hn
pn = βr n
where β is a constant.
If r n is a specified solution on the general solution hn
pn = βnr n
where β is a constant.
Problem
Solve the recurrence relation
a0 = 2; an − 3an−1 = 5(7n ), n ≥ 1
Answer
Problem
Solve the recurrence relation
a0 = 2; an − 3an−1 = 5(7n ), n ≥ 1
Answer
The corresponding homogeneous relation is an − 3an−1 = 0 with
general solution hn = α(3n ). Because 7n is not a specified solution
of α(3n ), we can set pn = β(7n ). Apply the particular solution pn on
the relation an − 3an−1 = 5(7n ), we have β(7n ) − 3β(7n−1 ) = 5(7n ),
implying 7β − 3β = 5(71 ), i.e., β = 35
4 . Therefore, the general
n
solution is an = hn + pn = α(3n ) + β(7n ) = α(3n ) + 35
4 7 . Apply
27
this general solution with a0 = 2, we have α = − 4 . Therefore,
35 n
n
an = − 27
4 (3 ) + 4 (7 ).
Problem
Solve the recurrence relation
a0 = 2; an − 3an−1 = 5(3n ), n ≥ 1
Answer
Problem
Solve the recurrence relation
a0 = 2; an − 3an−1 = 5(3n ), n ≥ 1
Answer
The corresponding homogeneous relation is an − 3an−1 = 0 with
general solution hn = α(3n ). Because 3n is a specified solution of
α(3n ), we can set pn = β(n3n ). Apply the particular solution pn on
the relation an −3an−1 = 5(3n ), we have β(n3n )−3β((n−1)3n−1 ) =
5(3n ), implying βn−β(n−1) = 5, i.e., β = 5. Therefore, the general
solution is an = hn + pn = α(3n ) + β(n3n ) = α(3n ) + 5n(3n ). Apply
this general solution with a0 = 2, we have α = 2. Therefore,
an = (2 + 5n)(3n ).
Problem
Consider the game “The Tower of Hanoi” shown on blackboard (you
can google/wiki it). If we want to move n disks from leg 1 to leg 3,
what is the minimum number of moves needed to do?
Answer
Problem
Consider the game “The Tower of Hanoi” shown on blackboard (you
can google/wiki it). If we want to move n disks from leg 1 to leg 3,
what is the minimum number of moves needed to do?
Answer
Let an be the minimum number of moves to move n disks from
one leg to another leg. It is easy to show that an = 2an−1 + 1,
by moving n − 1 disks to an auxiliary leg. We have the recurrence
relation an −2an−1 = 1. hn = α2n . Since 1 = 1(1n ) is not a specified
solution of hn , pn = β(1n ) = β. Apply pn into an − 2an−1 = 1, we
have β = −1. Thus, the general solution is an = α2n − 1. Apply
this general solution with a1 = 1, we have α = 1. Therefore, the
solution is an = 2n − 1.
Determine Particular Solution For
an + C1 an−1 + C2 an−2 = kr n .
If r n is not a specified solution of the general solution hn
pn = βr n
If r n is a specified solution of the general solution hn
If hn = c1 r1n + c2 r2n with r1 6= r2
pn = βnr n
If hn = (c1 + c2 n)r1n
pn = βn2 r n
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 5an−1 + 6an−2 = 10, n ≥ 2
Answer
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 5an−1 + 6an−2 = 10, n ≥ 2
Answer
The corresponding homogeneous recurrence relation an − 5an−1 +
6an−2 = 0 has a general solution hn = α1 2n + α2 3n . Because 10 =
10(1n ) is not a specified solution of hn , we can set pn = β(1n ) = β.
Apply the particular solution pn = β on an − 5an−1 + 6an−2 = 10,
we have β − 5β + 6β = 10, i.e., β = 5. Thus, the general solution
is an = α1 2n + α2 3n + 5. Apply this general solution with a0 = 1
and a1 = 2, we have α1 = −9 and α2 = 5. Therefore, an =
−9(2n ) + 5(3n ) + 5.
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 5an−1 + 6an−2 = 3(2n ), n ≥ 2
Answer
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 5an−1 + 6an−2 = 3(2n ), n ≥ 2
Answer
The corresponding homogeneous recurrence relation an − 5an−1 +
6an−2 = 0 has a general solution hn = α1 2n + α2 3n . Because 3(2n )
is a specified solution of hn , we can set pn = βn(2n ). Apply the
particular solution pn = βn2n on an − 5an−1 + 6an−2 = 3(2n ), we
have βn2n − 5β(n − 1)2n−1 + 6β(n − 2)2n−2 = 3(2n ), implying
4βn − 10β(n − 1) + 6β(n − 2) = 12, i.e., β = −6. Thus, the
general solution is an = α1 2n + α2 3n − 6n2n . Apply this general
solution with a0 = 1 and a1 = 2, we have α1 = −11 and α2 = 12.
Therefore, an = −11(2n ) + 12(3n ) − 6n2n .
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 4an−1 + 4an−2 = 5(3n ), n ≥ 2
Answer
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 4an−1 + 4an−2 = 5(3n ), n ≥ 2
Answer
The corresponding homogeneous recurrence relation an − 4an−1 +
4an−2 = 0 has a general solution hn = α1 2n +α2 n2n . Because 5(3n )
is not a specified solution of hn , we can set pn = β(3n ). Apply the
particular solution pn = β3n on an −4an−1 +4an−2 = 5(3n ), we have
β3n − 4β3n−1 + 4β3n−2 = 5(3n ), implying 9β − 12β + 4β = 45, i.e.,
β = 45. Thus, the general solution is an = α1 2n + α2 n2n + 45(3n ).
Apply this general solution with a0 = 1 and a1 = 2, we have α1 =
45
n
n
n
−44 and α2 = − 45
2 . Therefore, an = −44(2 ) − 2 n(2 ) + 45(3 ).
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 4an−1 + 4an−2 = 5(2n ), n ≥ 2
Answer
Problem
Solve the recurrence relation
a0 = 1; a1 = 2; an − 4an−1 + 4an−2 = 5(2n ), n ≥ 2
Answer
The corresponding homogeneous recurrence relation an − 4an−1 +
4an−2 = 0 has a general solution hn = α1 2n +α2 n2n . Because 5(2n )
is a specified solution of hn , we can set pn = βn2 (2n ). Apply the
particular solution pn = βn2 2n on an − 4an−1 + 4an−2 = 5(2n ), we
have βn2 2n − 4β(n − 1)2 2n−1 + 4β(n − 2)2 2n−2 = 5(2n ), implying
4βn2 − 8β(n − 1)2 + 4β(n − 2)2 = 20, i.e., β = 25 . Thus, the general
solution is an = α1 2n + α2 n2n + 52 n2 2n . Apply this general solution
with a0 = 1 and a1 = 2, we have α1 = 1 and α2 = −frac52.
Therefore, an = 2n − 52 n2n + 25 n2 2n .