4.1 Recurrence Relations Dr Patrick Chan School of Computer Science and Engineering South China University of Technology Discrete Mathematic Chapter 4: Advanced Counting Techniques 4.2 Solving Linear Recurrence Relations 4.4 Generating Functions 2 Ch. 4.1, 4.2 & 4.5 Agenda Recurrence Relations Modeling with Recurrence Relations Linear Nonhomogeneous Recurrence Relations with Constant Coefficients Generating Functions Useful Facts About Power Series Extended Binomial Coefficient Extended Binomial Theorem Counting Problems and Generating Functions Using Generating Functions to Solve Recurrence Relations
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4.1
Recurrence Relations
Dr Patrick ChanSchool of Computer Science and Engineering
South China University of Technology
Discrete Mathematic
Chapter 4: Advanced Counting Techniques
4.2
Solving Linear Recurrence Relations4.4
Generating Functions
2Ch. 4.1, 4.2 & 4.5
Agenda
� Recurrence Relations
� Modeling with Recurrence Relations
� Linear Nonhomogeneous Recurrence Relations with Constant Coefficients
� Generating Functions
� Useful Facts About Power Series
� Extended Binomial Coefficient
� Extended Binomial Theorem
� Counting Problems and Generating Functions
� Using Generating Functions to Solve Recurrence Relations
3Ch. 4.1, 4.2 & 4.5
Recursion
How many i-products the
families of students in our
school have?
0 0 1 1 1 1 1 0 1 0 00 1 0
2 4 1
16
7How many i-products?
How many
i-pr oduc ts?How many i-pr oduc ts?
How m any
i -produc t s?
How m any
i- produc t s?
How m any
i -produc t s?
4Ch. 4.1, 4.2 & 4.5
Recurrence Relations
� A recurrence relation for a sequence {an} is an equation that expresses an in terms of one or more previous elements (a0, …, an−1)
� A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation
5Ch. 4.1, 4.2 & 4.5
Recurrence Relations
Example 1
� Let {an} be a sequence that satisfies the recurrence relation an = an-1 - an-2 for n = 2, 3, 4, . . . , and suppose that a0 = 3 and a1 = 5. What are a2 and a3?
� From the recurrence relation:
� a2
� a3
= a1 – a0 = 5 - 3 = 2
= a2 – a1 = 2 - 5 = -3
6Ch. 4.1, 4.2 & 4.5
Recurrence Relations
Example 2
� Consider the recurrence relation
an = 2an−1 − an−2, where n ≥ 2
� Which of the following are solutions?� an = 3n
� 2an−1 − an−2 = 2(3(n - 1)) - 3(n - 2) = 3n = an
� an = 2n
� 2an−1 − an−2 = 2(2n-1) – nn-2 = 2n ≠ an
� an = 5
� 2an−1 − an−2 = 2 x 10 – 5 = 5 = an
����
����
����
7Ch. 4.1, 4.2 & 4.5
Recurrence Relations
� The initial conditions for a sequence specify the terms that precede the first term where the recurrence relation takes effect
� For examplean = an-1 + an-2, what is the value of a3?
Answer depends on a0 and a1 (initial conditions)
� a0 = 3 and a1 = 5 :
� a0 = 1 and a1 = 2 :
� A sequence is determined uniquely by
� Recurrence relation
� Initial conditions
a2 = 8, a3 = 13
a2 = 3, a3 = 5
8Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Compound Interest
� Growth of saving in a bank account with r% interest per given period
� Sn = Sn-1 + r · Sn-1 = (r+1) · Sn-1
� Example:� Suppose that a person deposits $10,000 in a savings
account at a bank yielding 11 % per year with interest compounded annually. How much will be in the account after 30 years?
Fibonacci (Rabbits) Numbers � A young pair of rabbits
(one of each sex) is placed on an island
� A pair of rabbits does not breed until they are 2 months old
� After they are 2 months old, each pair of rabbits produces another pair each month
� Pn = Pn−1 + Pn−2
1
1
1.11
1 1.21.1
1 1.3
1.1.11.1
1.2
Younger than two months
At least two
months old
1
2
3
4
5
10Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of Hanoi
� Objective
� Get all disks from peg 1 to peg 3
� Rules
� Only move 1 disk at a time
� Never put a larger disk on a smaller one
1 2 3
11Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of HanoiMore than 1 steps
More than 1 steps 1 step More than 1 steps
Break Down
Recursive Call Recursive Call
12Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of HanoiMore than 1 steps
More than 1 steps 1 step More than 1 steps
Break Down
Recursive Call Recursive Call
13Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of HanoiMore than 1 steps
1 step 1 step 1 step
Break Down
Base Case Base Case
14Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of Hanoi
The solution
15Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Tower of Hanoi
� Let Hn be the number of moves for a stack of n disks.
� Strategy:
� Move top n−1 disks (Hn−1 moves)
� Move bottom disk (1 move)
� Move top n−1 to bottom disk (Hn−1 moves)
� Hn = 2Hn-1 + 1
16Ch. 4.1, 4.2 & 4.5
Modeling with Recurrence Relations
Example � Find a recurrence relation and give initial conditions for the
number of bit strings of length n that do not have two consecutive 0s.
� Let Pn denote the number of bit strings of length n that do not have two consecutive 0s
� Pn = Pn-1 + Pn-2 , n ≥ 3
� P1 = 2, P2 = 3
1
Any bit string of length n - 1 with no two consecutive 0s
10
Any bit string of length n - 2 with no two consecutive 0s
Pn-1
Pn-2
0101 0110
0111 1010
1011 1110
1111
For P4
P3
P21101
17Ch. 4.1, 4.2 & 4.5
Solving Linear Recurrence Relations
� Given Pn = Pn−1 + Pn−2, what is P100?
� It is not easy to calculate
� Need a better solutionwhich is not in
relation form
E.g. Pn = n*10 -1
18Ch. 4.1, 4.2 & 4.5
Solving Linear Recurrence Relations
� Linear Homogeneous Recurrence of Degree k with Constant Coefficients is a recurrence of the form
an = c1an−1 + … + ckan−k
where the ci are all real numbers, and ck ≠ 0
� Linear: the power of all a i term is one
� Homogeneo us: no constant term (no team without ai)
� Recurrence: a sequence {an} which a
nin terms of a
n-1, a
n-2,…
� Degree k: refer to k previous terms an-k
� Constant Coefficients: c1, c2, … independent from n
� The short name is “k-LiHoReCoCo”
∑=
−=k
i
iniac
1
19Ch. 4.1, 4.2 & 4.5
Solving Linear Recurrence Relations
Example
� Mn = Mn−1 + (1.11)Mn−1
� 1-LiHoReCoCo
� Pn = Pn−1 + Pn−2
� 2-LiHoReCoCo
� an = an-5
� 5-LiHoReCoCo
� an =an-1 + (an-2)2
� Not linear
� Hn = 2Hn−1 + 1
� Not homogeneous
� Bn = nBn-1
� Non-constant coefficient
(n is a variable)k-LiHoReCo Co
� Linear: the power of a ll a i term is one
� Homogeneous: no constant term (no team wi thout ai)
� Recurrence: a sequence {an} which an in terms of an-1 , an-2,…
� Degree k: refer to k previous terms an -k
� Constant Coefficients: c1, c2, … independent from n
20Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Given 2-LiHoReCoCo: an = c1an−1 + c2an−2
� Assume an = w1r1n + w2r2
n for n ≥ 0, r1 and r2 are different and some constants w1, w2
21Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Given 2-LiHoReCoCo: an
= c1a
n−1+ c
2a
n−2 , and
a0 = c and a1 = d
� Assume an = w1r1n + w2r2
n for
r1 and r2 are different and some constants w1, w2
an = w1r1n + w2r2
n
c1an-1 = c1(w1r1n-1 + w2r2
n-1)
c2an-2 = c2(w1r1n-2 + w2r2
n-2)
–
–
an - c1an−1 - c2an−2w1 (r1
n - c1r1n-1 - c2r1
n-2)
+ w2 (r2n - c1r2
n-1 - c2r2n-2)
=
w1 (r1n - c1r1
n-1 - c2r1n-2) + w2 (r2
n - c1r2n-1 - c2r2
n-2) = 0
Equal to 0
w1 r1n-2 (r1
2 - c1r1 - c2) + w2 r2n-2 (r2
2 - c1r2 - c2) = 0
22Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Assume w1, w2, r1 and r2 are not 0
� This formula is hold for any n, therefore,
w1 r1n-2 (r1
2 - c1r1 - c2) + w2 r2n-2 (r2
2 - c1r2 - c2) = 0
w1 r1n-2 (r1
2 - c1r1 - c2) + w2 r2n-2 (r2
2 - c1r2 - c2) = 0
w1 r1n-2
w2 r2n-2
(r12 - c1r1 - c2) + (r2
2 - c1r2 - c2) = 0
r12 - c1r1 - c2 = 0 r2
2 - c1r2 - c2 = 0and
an = w1r1n + w2r2
n
23Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� If w1 and w2 are both 0,
� an is a special sequence and all values are 0
� If either w1 or w2 is 0, or either r1 or r2 is 0 (as r1 and r2 are different)
an = w1r1n + w2r2
n
w1 r1n-2 (r1
2 - c1r1 - c2) + w2 r2n-2 (r2
2 - c1r2 - c2) = 0
w1 r1n-2 (r1
2 - c1r1 - c2) = 0
w2 r2n-2 (r2
2 - c1r2 - c2) = 0
r12 - c1r1 - c2 = 0
r22 - c1r2 - c2 = 0
w1 or r1 is 0 (w2 and r2 are non-zero)
w2 or r2 is 0 (w1 and r1 are non-zero)
(w1 and r2) or (w2 and r1) is 0
an is a special sequence and all values are 0
24Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Given 2-LiHoReCoCo: an = c1an−1 + c2an−2 , and a0 = c and a1 = d
� Assume an = w1r1n + w2r2
n for r
1and r
2are different and some constants w
1, w
2
� We know that r1
2 - c1r
1- c
2= 0 and r
22 - c
1r
2- c
2= 0
� Characteristic Equation: r2 - c1r - c2 = 0
� Characteristic Roots: r1 and r2
� w1 and w2 can be calculated by using c and d
a0 = c = w1r10 + w2r2
0
a1 = d = w1r11 + w2r2
1
25Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
Theorem
� Consider an arbitrary 2-LiHoReCoCo:
an = c1an−1 + c2an−2
� By substituting an = rn, we have the characteristic equation:
r2 − c1r − c2 = 0
� If there has two different roots r1 and r2, then an = w1r1
n + w2r2n
for n ≥ 0 and some constants w1, w2
26Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Proof
Given r1, r2 are the characteristic root
where and w1, w2 are constants
�Two steps for the proof
1. Show if an = w1r1n + w2r2
n , {an} is a solution of the recurrence relation
2. Show if {an} is the solution of the recurrence
relation, an = w1r1n + w2r2
n
an = c1an−1 + c2an−2 an = w1r1n + w2r2
n⇔⇔⇔⇔
27Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Step 1Show if an = w1r1
n + w2r2n , {an} is a solution of the
recurrence relation
2211 −− + nnacac ( ) ( )2
22
2
122
1
22
1
111
−−−− +++= nnnnrwrwcrwrwc
( ) ( )221
2
22211
2
11 crcrwcrcrwnn +++= −−
2
2
2
22
2
1
2
11 rrwrrwnn −− +=
nnrwrw 2211 +=
na=
021
2 =−− crcrr1 and r2 are the so lu tion of
28Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� Step 2Show if {an} is the solution of the recurrence relation, an = w1r1
n + w2r2n
� Suppose that {an} is a solution of the recurrence relation, and the initial conditions a0 = C0 and a1 = C1 hold
� We want to show that there are constants w1 and w2 such that the sequence {an} with an = w1r1
n + w2r2n satisfies these same initial
conditions
� a0 = C0 = w1 + w2 and a1 = C1 = w1r1 + w2r2
� By solving these two equations:
� When r1 ≠ r2, {an} with w1r1n + w2r2
n satisfy the 2 initial conditions
21
2011
rr
rCCw
−
−=
21
1202
rr
CrCw
−
−=
29Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
� We know that {an} and {α1r1n + α2r2
n} are both solutionsof the recurrence relation an = c1an−1 + c2an−2 and both
satisfy the initial conditions when n = 0 and n = 1
� Because there is a unique solution of 2-LiHoReCoCo
with two initial conditions, it follows that the two
solutions are the same, that is, an = α1r1n + α2r2
n for all nonnegative integers n
� We have completed the proof
30Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
Example 1� Solve the recurrence an = an−1 + 2an−2 given the initial
conditions a0 = 2, a1 = 7
� Characteristic Equation: r2 − r − 2 = 0
� Characteristic Root:
� r = (1 ± 3) / 2
� r = 2 or r = −1
� Therefore, an = w1 2n + w2 (−1)n
� By using a0 = 2, a1 = 7
� a0 = 2 = w120 + w2 (−1)0
� a1 = 7 = w121 + w2 (−1)1
� w1 = 3 and w2 = 1
� Therefore, an = 3·2n − (−1)n
a
acbbx
2
42 −±−
=
2-LiHoReCoCo: an = c1an−1 + c2an−2 ,
Characteristic Equation: r2 - c1r - c2 = 0
an = w1r1n + w2r2
n (r1 and r2 are different)
31Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
Example 2� Find an explicit formula for the Fibonacci numbers
� Recall fn = fn-1 + fn-2
� Characteristic equation: r2 - r - 1 = 0
� Characteristic roots:
� Initial conditions f0 = 0 and f 1 = 1
2/)51(1 +=r 2/)51(2 −=rnn
nwwf
−+
+=
2
51
2
5121
2100 wwf +==
−+
+==
2
51
2
511 211 wwf 5
12
−=w5
11=w
nn
nf
−−
+=
2
51
5
1
2
51
5
1
2-LiHoReCoCo: an = c1an−1 + c2an−2 ,
Characteristic Equation: r2 - c1r - c2 = 0
an = w1r1n + w2r2
n (r1 and r2 are different)
32Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos with two same roots
Theorem
� Let c1 and c2 be real numbers with c2 ≠ 0. Suppose that r2 – c
1r - c
2= 0 has only one
root r0
� A sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2 if and only if an = w1r0
n + w2nr0n, for n = 0, 1, 2, ...,
where w1 and w2 are constants
33Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos with two same roots
Example 1� What is the solution of the recurrence relation
an = 6an-1 - 9an-2 with initial conditions a0 = 1 and a1 = 6?
� Characteristic equation: r2 - 6r + 9 = 0
� Only one characteristic root: r = 3
� Hence, the solution to this recurrence relation is
an = w13n + w2n3n
for some constants α1 and α2
� By using the initial conditions,
a0 = 1 = w1, a1 = 6 = 3w1 + 3w2, so w1=1 and w2 = 1
� Consequently, an = 3n + n3n
2-LiHoReCoCo: an = c1an−1 + c2an−2 ,
Characteristic Equation: r2 - c1r - c2 = 0
an = w1r0n + w2 n r0
n
34Ch. 4.1, 4.2 & 4.5
Solving 2-LiHoReCoCos
Summary
� Given : an = c1an−1 + c2an−2 and a0 = c and a1 = d
1. Characteristic equation: r2 − c1r − c2 = 0
2a. If Characteristic Root (r1 and r2) are different
1. an = w1r1n + w2r2
n is the solution
2. Use a0 = w1 + w2 = c and a1 = w1r1 + w2r2 = d to
solve w1 and w2
2b. If Characteristic Root (r 1 and r2) are the same
1. an = w1rn + w2 n rn is the solution
2. Use a0 = w1 = c and a1 = w1r1 + w2r2 = d to solve w1 and w2
35Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� What is the solution of the recurrence relation a
n= - a
n + 6a
n-2with initial conditions a
0= 0
and a1 = 5?
� What is the solution of the recurrence relation an = - 2an - an-2 with initial conditions a0 = 5 and a1 = -6?
36Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� the recurrence relation: an = - an + 6an-2
� Initial conditions a0 = 0 and a1 = 5
� Characteristic Equation:
� Characteristic Root:
� Therefore,
� Using the initial condition� a0 = 0 = w1 + w2
� a1 = 5 = -3w1 + 2w2
� w1 = -1, w2 = 1
� Therefore,
(r + 3)(r – 2) = 0r2 + r - 6 = 0
r1 = -3, r2 = 2
an = w1 (-3)n + w2 (2)n
an = - (-3)n + (2)n
37Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� the recurrence relation: an = - 2an - an-2
� Initial conditions a0 = 5 and a1 = -6
� Characteristic Equation:
� Characteristic Root:
� Therefore,
� Using the initial condition� a0 = 5 = w1
� a1 = -6 = -w1 - w2
� w1 = 5 , w2 = 1
� Therefore,
(r + 1)(r + 1) = 0r2 + 2r + 1 = 0
r1 = -1
an = w1 (-1)n + w2 n (-1)n
an = 5 (-1)n + n (-1)n
38Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCos
� k-LiHoReCoCo:
� Characteristic Equation is:
� TheoremIf there are k distinct roots ri , then the solutions to the recurrence are of the form:
for all n ≥ 0, where the wi are constants
∑=
−=k
i
inin aca1
01
=−∑=
−k
i
ik
i
krcr
∑=
=k
i
n
iin rwa1
an = c1an−1 + c2an−2
r2 − c1r − c2 = 0
an = w1r1n + w2r2
n
2-kiHoReCoCo
39Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCos
Example� Find the solution to the recurrence relation
an = 6an-1 – 11an-2 + 6an-3
with the initial conditions a0 = 2, a1 = 5, and a2 = 15.
� The characteristic equation is:
r3 - 6r2 + 11r – 6 = (r - 1)(r - 2)(r - 3)� The characteristic roots are r = 1, r = 2, and r = 3
� an = w11n + w22
n + w33n
� By using the initial conditions� a0 = 2 = w1 + w2 + w3
� a1 = 5 = w1 + w2 x 2 + w3 x 3
� a2 = 15 = w1 + w2 x 4 + w3 x 9
� Therefore, w1 = 1, w2 = -1 and w3 = 2
� As a result, an = 1 - 2n + 2 x 3n
40Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCoswith same roots
� Let c1, c2, ..., ck be real numbers
� Suppose that the characteristic equation
rk – c1rk-1 – ··· – ck = 0
has t distinct roots r1, r2, ..., rt with multiplicities m1, m2, ..., mt
� i.e. ri appear mi times
� m1 + m2 + ·· · + m t = k
41Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCoswith same roots� A sequence {an} is a solution of the recurrence relation
� If and only if
for n = 0,1,2, ..., where wi,j
are constants for 1 ≤ i ≤ t and 0 ≤ j ≤ m
i-1
nm
mnrnwnwwa1
1
1,11,10,1)...( 1
1
−−+++=
nm
m rnwnww 2
1
1,21,20,2 )...( 2
2
−−++++
n
t
m
mttt rnwnww t
t)...(...
1
1,1,0,
−−+++++
∑ ∑=
−
=
=
t
i
n
i
m
j
j
ji rnwi
1
1
0
,
Special case for k=2, One distinct rootan = w1r0
n + w2nr0n
an = c1an-1 + c2an-2 + ... + ckan-k
No. of distinct roots
Multiplicities for ri
an = (w1 + w2n)r0n
42Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCos with same roots
Example� Find the solution to the recurrence relation
with the initial conditions
� Characteristic equation:
� Roots: −1, −1, −1, 2
� Therefore:
� By initial conditions:
4321253 −−−− +++−=
nnnnnHHHHH
210,1 3210 ==== ,HH,HH
0253234 =−−−+ xxxx
nn
ncncnccH 2)1)((
4
2
321+−++=
0 1 4
1 1 2 3 4
2 1 2 3 4
3 1 2 3 4
1
2 0
2 4 4 1
3 9 8 2
H c c
H c c c c
H c c c c
H c c c c
= + = = − − − + =
= + + + = = − − − + =
9
2,0,
3
1,
9
74321 ==−== cccc
nnn
n nH 29
2)1(
3
1)1(
9
7+−−−=
0)1)(2( 3 =+− xx
nm
mnrnwnwwa1
1
1,11,10,1)...( 1
1
−−+++=
nm
m rnwnww 2
1
1,21,20,2 )...( 2
2
−−++++
n
t
m
mtttrnwnww t
t
)...(...1
1,1,0,
−
−+++++
43Ch. 4.1, 4.2 & 4.5
Solving k-LiHoReCoCos
Summary
� Given : and ai = ci , where i = 1, 2, …, k
1.Characteristic equation:
2. Characteristic Root (r1, r2, …, rk)
1. is the solution of k-LiHoReCoCos
where m i is the multiplicity of ri
2. solve wi by ap = cp
where p = 1, 2, ..., k
01
=−∑=
−k
i
ik
i
krcr
∑=
−=k
i
inin aca1
∑ ∑=
−
=
=
t
i
n
i
m
j
j
jin rnwai
1
1
0
,
∑ ∑=
−
=
=
t
i
p
i
m
j
j
jirpw
i
1
1
0
,
44Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� What is the solution of the recurrence relation a
n= a
n-1+ a
n-2- a
n-3with initial conditions a
0=
5, a1 = 5 and a2 = -6?
nm
mnrnwnwwa1
1
1,11,10,1)...( 1
1
−−+++=
nm
mrnwnww
2
1
1,21,20,2)...( 2
2
−−++++
n
t
m
mtttrnwnww t
t)...(... 1
1,1,0,
−−+++++
45Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� the recurrence relation: an = an-1 + an-2 - an-3
� Initial conditions a0 = 5, a1 = 5 and a2 = -6
� Characteristic Equation:
� Characteristic Root:
� Therefore,
� Using the initial condition� a0 = 0 = c1 + c3
� a1 = 8 = c1 + c2 – c3
� a1 = 4 = c1 + 2c2 + c3
� c1 = 3, c2 = -3, c3 = 2
� Therefore,
(r – 1)(r – 1)(r + 1) = 0r3 – r2 – r + 1 = 0
r1 = 1, r2= 1, r3 = -1
an = (c1 + c2 n) (1)n + c3 (-1)n
an = 3 (1 – n) (1)n + 2 (-1)n
46Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
� Linear nonhomogeneous recurrence of degree k with constant coefficients(k-LiNoReCoCos) contain some terms F(n)
that depend only on n but not ai
� General form:
an = c1an−1 + … + cka
n−k + F(n)
Associated Homogeneo us Recurrence Relation
47Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
� If { } is a particular solution of the nonhomogeneous linear recurrence relation with constant coefficients
an = c1an-1 + c2an-2 + ... + ckan-k + F(n)
� Then every solution is of the form { }, where { } is a solution of the associated homogeneous recurrence relation
an = c1an-1 + c2an-2 + ... + ckan-k
)( p
na
)()( h
n
p
n aa +)(h
na
48Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
� Proof
� As { } is a particular solution for LiNoReCoCos
� Suppose that { } is an another solution
� { } is a solution of the associated homogeneous linear recurrence, named { }
� Consequently, for all n.
)(...)()(
22
)(
11
)(nFacacaca
p
knk
p
n
p
n
p
n ++++= −−−
)(...2211 nFbcbcbcb knknnn ++++= −−−
( ) ( ) ( ))()(
222
)(
111
)( ... p
knknk
p
nn
p
nn
p
nnabcabcabcab −−−−−− −++−+−=−
)( p
nn ab −)( h
na)()( h
n
p
nn aab +=
)( p
na
nb
)()(
22
)(
11
)(...
h
knk
h
n
h
n
h
n acacaca −−− +++=
49Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
Example 1
� Find all solutions to an = 3an−1+2n, which solution has a1 = 3?
� Notice this is a 1-LiNoReCoCo.
� Its associated 1-LiHoReCoCo
� an = 3an−1 and root is 3
� Solution is = c3n
� The solutions of LiNoReCoCo are in the form
)(h
na
)()( h
n
p
nn aaa +=
an = 3an−1+2n
a1= 3
Next step
50Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
Example
� If F(n) is a degree-u polynomial in n, a degree-upolynomial should be tried as the particular solution
� Now, F(n) = 2n
� Try = cn + d, c and d are constants
� Solution is: = -n – 3/2
an = 3an-1 + 2n
cn+d = 3(c(n−1)+d) + 2n
(2c+2)n + (3c−2d) = 0
c = −1 and d = −3/2
)( p
na
)( p
na
)( p
na
an = 3an−1+2n
a1= 3
51Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos
Example
� Therefore, we have
� By using a1 = 3
� 3 = -1 - 3/2 + 3c
� c = 11/6
� As a result,
)()( h
n
p
nnaaa +=
nh
nca 3
)( =
( ) 3
2
p
na n= − −
33
2
nn c= − − +
6
311
2
3n
nna
⋅++−=
an = 3an−1+2n
a1= 3
52Ch. 4.1, 4.2 & 4.5
� Suppose {an} satisfies the LiNoReCoCo
where ci (i = 1,2,…k) are real numbers and
where b0, b
1, …, b
tand s are real numbers
� When s is not a root of the characteristic equation of the associated li near homogeneous RR, there is a particular solution( ) of the form
� When s is a root of the characteristic equation of the associated linear homogeneo us RR with multiplicity m, there is a particular solution ( ) of the form
)(1
nFacak
i
inin+
= ∑
=−
nt
t
t
t sbnbnbnbnF )...()( 01
1
1 ++++= −−
nt
t
t
tspnpnpnp )...(
01
1
1++++ −
−
nt
t
t
t
m spnpnpnpn )...(01
1
1++++ −
−
Solving LiNoReCoCos
Particular Solution
)( p
na
)( p
na
53Ch. 4.1, 4.2 & 4.5
� F(n) = 3n
� F(n) = n3n
� F(n) = n22n
� F(n) = (n2 + 1)3n
� Consider the associated ho mogeneous RR:
an
= 6an-1
- 9an-2
� Characteristic Equation
� Characteristic Root is 3, of multiplicity m=2
Solving LiNoReCoCos : Particular Solution
Example 1� What form does a particular solution of the linear
nonhomogeneous recurrence relation an = 6an-1 - 9a
n-2+F(n) have when ...
0)3(9622 =−=+− rrr
npn 3)(
0
2
npnpn 3)(
01
2 +
npnpnp 2)(
01
2
2++
npnpnpn 3)(
01
2
2
2 ++
nt
t
t
tspnpnpnp )...(
01
1
1++++ −
−
nt
t
t
t
m spnpnpnpn )...(01
1
1++++ −
−
s is not a root
s is a root
nt
t
t
tsbnbnbnbnF )...()(
01
1
1++++= −
−
54Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos : Particular Solution
Example 2� Let an be the sum of the first n positive integers, so that
� an satisfies the linear nonhomogeneous RR
� Associated linear homogeneous RR is an = an-1
� Root is 1. The solution is , c is a constant
� Since F(n) = n = n x (1)n, and s = 1 is a root of degree one of the characteristic equation of the associated linear homogeneous RR
� So the particular solution has the form n(p1n+p0)
nh
n ca )1()( =
∑=
=n
k
n ka1
nt
t
t
t spnpnpnp )...( 01
1
1 ++++ −−
nt
t
t
t
mspnpnpnpn )...( 01
1
1 ++++ −−
s is not a root
s is a root
ntt
tt sbnbnbnbnF )...()( 01
11 ++++= −−
an
= an-1
+ n
55Ch. 4.1, 4.2 & 4.5
Solving LiNoReCoCos : Particular Solution
Example 2
� By solving p1n2 + p0n = p1 (n - 1)2 + p0(n - 1) + n
� We have p0 = p1 =1/2
� Recall,
� By using a1 = 1, so c = 0
� Therefore,
2
)1( +=
nna
n
)()( h
n
p
nnaaa +=
cnnan
++= 2/)1(ca h
n=)(
2/)1()( += nna
p
n
an = an-1 + n( )
1 0( )p
na n p n p= +
56Ch. 4.1, 4.2 & 4.5
Solving Linear Recurrence Relations
Summary� k-LiHoReCoCos with m same roots (without F(x))
� Find the root of characteristic equation
�
� Use initial terms to find alphas
� k-LiNoReCoCos with m same roots (with F(x))
� Find the solution of charac teristic equationof Associated linear homogeneo us RR
� Find the particular solution of LiNoReCoCo using
� Fina lly
� Use initial terms to find alphas
nm
mnrnna1
1
1,11,10,1)...( 1
1
−−+++= ααα
nm
mrnn2
1
1,21,20,2)...( 2
2
−−++++ ααα
n
t
m
mttt rnn t
t)...(...
1
1,1,0,
−−+++++ ααα
nt
t
t
t
mspnpnpnpn )...( 01
1
1 ++++ −−
)()( h
n
p
n aa +
=)( p
na
)(h
na
57Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� Find all solutions to an = 7an−1+(2n2+2)3n, which solution has a1 = 10?
58Ch. 4.1, 4.2 & 4.5
☺☺☺☺ Small Exercise ☺☺☺☺
� Its associated 1-LiHoReCoCo� an = 7an−1 and root is 7
� Generating functions ( G(x) ) are used to represent sequences efficiently by coding the terms of a sequence as coefficients of powers of a variable x in a formal power series
� Generating function for the sequence
a0, a1, a2, …, ak, … of real numbers is the infinite series
0 1
0
0 1( )k
k k
k kG x x x xa a a xa∞
=
= + + + + =∑⋯ ⋯
62Ch. 4.1, 4.2 & 4.5
Example
� What is the generating function for the
following sequence?
� {0, 2, …, 2k,…}
� {1, 1, 1, 1, 1}
∑∞
=0
2k
kkx
∑=
4
0k
kx1 + x + x2 + x3 + x4 =
0 + 2x + … + 2k · xk + … =
63Ch. 4.1, 4.2 & 4.5
Useful Facts About Power Series
�
is generating functionof the sequence 1,1,1,1, ...
�
is generating functionof the sequence 1, a, a2, a3, ...
� Recall, in Chapter 2.4,
( )x
xf−
=1
1 ( )ax
xf−
=1
1
∑∞
=0k
kx ∑
∞
=0k
kkxa
x−=
1
1
ax−=
1
1
64Ch. 4.1, 4.2 & 4.5
Useful Facts About Power Series
� Given: ( ) ∑∞
=
=0k
k
k xaxf ( ) ∑∞
=
=0k
k
k xbxg
( ) ( )=+ xgxf
( ) ( )=xgxf
( )∑∞
=
+0k
k
kk xba
( )∑∑∞
= =−=
0 0k
k
j
k
jkj xba
∑∑∞
=
∞
= 00 k
k
k
k
k
k xbxa
...)...)((1
1
0
0
1
1
0
0 ++++= xbxbxaxa
+++ )()( 0110
1
00
0babaxbax=
...)( 021120
2 +++ bababax00ba 0110 baba +
021120 bababa ++
65Ch. 4.1, 4.2 & 4.5
Useful Facts About Power Series
Example
� Let ,
� Find the coefficients a0, a1, a2, ... in the expansion
∑∞
==
0)(
k
k
kxaxh
2)1(
1)(
xxh
−=
( ) ∑∞
=
=−
=01
1
k
kx
xxf
( ) ∑∞
=
=0k
k
kxaxf ( ) ∑
∞
=
=0k
k
kxbxg
( ) ( ) ( )∑∞
=
+=+0k
k
kkxbaxgxf
( ) ( ) ( )∑∑∞
= =−=
0 0k
k
j
k
jkjxbaxgxf
)(xh)1(
1
)1(
1
xx −−=
2)1(
1
x−=
0 0
kk
k j
x∞
= =
=∑∑
∑∞
=
+=0
)1(k
kxkak=k+1
0 0
k k
k k
x x∞ ∞
= =
= ∑ ∑
66Ch. 4.1, 4.2 & 4.5
Extended Binomial Coefficient
� Recall, the binomial coefficient,
where n is a nonnegative integer and r is an integer with 0 ≤ r ≤ n
� The extended binomial coefficients is defined by replacing nonngeative integer n by real number u
!
)1)...(1(
r
rnnn +−−=
)!(!
!
rnr
n
−=
r
n
67Ch. 4.1, 4.2 & 4.5
Extended Binomial Coefficient
Example
� Find the values of the extended binomial coefficients
−
3
2
3
2/1
!
)1)...(1(
r
rnnn
r
n +−−=
3
5 ( ) ( )123
25155
⋅⋅
−⋅−⋅= 10=
( ) ( )123
22122
⋅⋅
−−⋅−−⋅−= 4−=
( ) ( ) ( )123
22/112/12/1
⋅⋅
−−⋅−−⋅−=
16
1=
(Standard binomial coefficients)
68Ch. 4.1, 4.2 & 4.5
Extended Binomial Coefficient
� When the top parameter is a negative integer, the extended binomial coefficient can be expressed in terms of an ordinary binomial coefficient
−
r
n ( )( ) ( )!
11
r
rnnn +−−⋅⋅⋅−−−=
( ) ( )( ) ( )!
111
r
rnnnr −+⋅⋅⋅+−
=
( ) ( )( ) ( )!
211
r
nrnrnr ⋅⋅⋅−+−+−
=
( )
−+−=
r
rnr 11
69Ch. 4.1, 4.2 & 4.5
Extended Binomial Theorem
� Recall, the ordinary binomial theorem
� Extended binomial theorem
where x is a real number with |x|<1, u is a real number
k
k
u xk
ux ∑
∞
=
=+
0
)1(
( )nyx+ jjn
n
j
yxj
n −
=∑
=
0
70Ch. 4.1, 4.2 & 4.5
Extended Binomial Theorem
Example
� Find the generating functions for (1+x)-n and (1-x)-n, where n is a positive integer, using the extended binomial theorem.
nx
−− )1(
nx
−+ )1(
( )kk
xk
n−
−=∑
∞
=0
k
k
xk
n∑∞
=
−=
0
∑∞
=
−+=0
),1(k
kxkknC
∑∞
=
−+−=0
),1()1(k
kkxkknC
−
r
n ( )
−+−=
r
rnr 11∑
∞
=
=0
)(k
k
k xaxG
ak
k
k
u xk
ux ∑
∞
=
=+
0
)1(
71Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions� How can we solve the counting problems,
including the recurrence relation, by using the Generating Functions?
0 1
0
0 1( )k
k k
k kG x x x xa a a xa∞
=
= + + + + =∑⋯ ⋯
72Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 1� Find the number of solutions of e1 + e2 + e3 = n when n = 17,
where e1 ,e2, e3 are nonnegative integers with 2 ≤≤≤≤ e1 ≤≤≤≤ 5, 3 ≤≤≤≤ e2 ≤≤≤≤ 6, 4 ≤≤≤≤ e3 ≤≤≤≤ 7
� By considering this generating function for the sequence {an}, where an is the number of solution for n
� As n = 17, a17, which is the coefficient of x17, is the solution
� Answer is 3
=∑∞
=0k
k
kxa
⋅+++ )( 5432 xxxx
⋅+++ )(6543
xxxx
)(7654
xxxx +++
73Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 2
� In how many different ways can eight identical cookies be distributed among three distinct childrenif each child receives at least two cookies and no more than four cookies?
� By considering this generating function for the sequence {an}, where an is the number of solution for n
� The coefficient of x8 is 6
))()((432432432
xxxxxxxxx ++++++=∑∞
=0k
k
kxa
74Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 3
� Use generating functions to determine the number of ways to insert tokens worth $1, $2, and $5 into a vending machine to pay for an item that costs r dollars in both the cases when the order in which
� the order of inserting tokens does not matter
� the order does matter
� For example,
� Item that costs $3
� Order does not matter (2)
� Three $1 tokens
� $1 token and $2 token
� Order does matter (3)
� Three $1 tokens
� $1 token and a $2 token
� $2 token and a $1 token
75Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 3
� Order does not matter
� By considering:
� $1:
� $2:
� $5:
� The answer is the coefficient of xr in
...)(321 +++ xxx
...)(642 +++ xxx
...)( 15105 +++ xxx
1 2 3
2 4 6
5 10 15
( . ..)( . ..)( ...)
x x xx x xx x x
+ + + ⋅+ + + ⋅+ + +
=∑∞
=0k
k
kxa
76Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 3
� Order does matter
� The number of ways to insert n tokens to produce a total of r dollars is the coefficient of xr in
� Answer is the coefficient of xr in
nxxx )( 521 ++
xx
k
k
−=∑
∞
= 1
1
0
1 2 5
0
( )n
n
x x x∞
=
= + +∑
)(1
152
xxx ++−=
521
1
xxx −−−=
1 2 5 1 2 5 21 ( ) ( ) ...x x x x x x+ + + + + + +
E.g. r = 5
1 2 5 2( )x x x+ +
1 2 5 3( )x x x+ +1 2 5 4
( )x x x+ +1 2 5 5
( )x x x+ +
11 2 5
( )x x x+ +
0
3
4
1
Coef. of x5
=∑∞
=0k
k
kxa
77Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 4� Use generating functions to find the number of ways
to select r objects of n different kinds if we must select at least one object of each kind.
� Each kind of objects contributes
to generating function G(x) for the sequence {xr}, where xr is the number of ways to select r objects of n different kinds
� Hence, there are n kinds:
At least one object of each kind
nxxx ...)( 321 +++=
...)(321 +++ xxx
)(xGnn xxx ...)1( 21 +++=
n
n
x
x
)1( −=
78Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 4
)(xGnn xx )1/( −=
(1 )n nx x −= −
r
r
nx
r
nx )(
0
−
−= ∑
∞
=
rr
r
rnxrrnCx )()1(),1()1(
0
−−+−= ∑∞
=
rn
r
xrrnC+
∞
=∑ −+= )(),1(
0
t
nt
xnttC )(),1(∑∞
=
−−=
r
nr
xnrrC )(),1(∑∞
=
−−=
There are C(r - 1, r - n) ways
to select r objects of n different kinds if we must
select at least one object of
each kind.
Extended binomial theore m
k
k
ux
k
ux ∑
∞
=
=+
0
)1(
−
r
n( )
−+−=
r
rnr 11
79Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 5
� Solve the recurrence relation ak = 3ak-1 for k=1, 2, 3,
… and initial condition a0=2
� Let G(x) be the generating function for the sequence {ak}, that is
∑∞
=
=0
)(k
k
kxaxG
0
1
( ) k
k
k
G x a a x∞
=
= +∑
0 1
1
( ) 3 k
k
k
G x a a x∞
−=
= +∑1
0 1
1
( ) 3 k
k
k
G x a x a x∞
−−
=
= + ∑
0( ) 3 ( )G x a xG x= +
2( )
1 3G x
x=
−
0
( ) 2 3k k
k
G x x∞
=
= ⋅∑
∑∞
=0k
kk xaax−
=1
1
2 3k
ka = ⋅
80Ch. 4.1, 4.2 & 4.5
Counting Problems and Generating Functions
Example 6
� The sequence {an} satisfies the recurrence relation
an
= 8an-1 + 10n -1
and the initial condition a1 = 9
� Use generating functions to find an explicit formula for an