Recurrence Relations Solving Linear Recurrence Relations€¦ · Recurrence Relations Dr Patrick...

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


� 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


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


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


� 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?

� S30 = 1.11S29 = 1.11(1.11S28) = …= (1.11)30 10,000

9Ch. 4.1, 4.2 & 4.5


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


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


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



More than 1 steps 1 step More than 1 steps

Break Down

Recursive Call Recursive Call

13Ch. 4.1, 4.2 & 4.5



1 step 1 step 1 step

Break Down

Base Case Base Case

14Ch. 4.1, 4.2 & 4.5


Tower of Hanoi

The solution

15Ch. 4.1, 4.2 & 4.5


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


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


� 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


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


� Given 2-LiHoReCoCo: an

= c1a

n−1+ c

2a

n−2 , and

a0 = c and a1 = d


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


� 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


� 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


� Given 2-LiHoReCoCo: an = c1an−1 + c2an−2 , and a0 = c and a1 = d


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


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


� 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


� 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


� 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


� 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


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


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



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



an = w1r0n + w2 n r0

n

34Ch. 4.1, 4.2 & 4.5


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


� the recurrence relation: an = - an + 6an-2

� Initial conditions a0 = 0 and a1 = 5

� Characteristic Equation:


� 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


� the recurrence relation: an = - 2an - an-2

� Initial conditions a0 = 5 and a1 = -6



� 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


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


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


� 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


� the recurrence relation: an = an-1 + an-2 - an-3

� Initial conditions a0 = 5, a1 = 5 and a2 = -6



� 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


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


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


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


� Find all solutions to an = 7an−1+(2n2+2)3n, which solution has a1 = 10?

58Ch. 4.1, 4.2 & 4.5


� Its associated 1-LiHoReCoCo� an = 7an−1 and root is 7

� Solution is = c7n

� The solutions of 1-LiNoReCoCo are in the form

� Need to do is find one

)(h

na

)()( h

n

p

nn aaa +=

)( p

na

an = 7an−1+(2n2+2)3n

a0 = 10

59Ch. 4.1, 4.2 & 4.5


� Now, F(n) = (2n2+2)3n

� = (an2 + bn + c)3n

an = 7an-1 + (2n2+2)3n

)( p

na

(an2+bn+c)3n = 7(a(n-1)2+b(n-1)+c)3n-1 + (2n2+2)3n

3an2+3bn+3c = 7an2 - 14an + 7a + 7bn – 7b + 7c + 6n2 + 6

0 = n2(4a+6)+n(4b-14a) +(4c +7a-7b+6 )

4c +7a-7b+6=04b-14a=04a+6=0

a=-3/2 b=-21/4 c=-129/16

np

n nna 3)6/1294/212/3(2)( −−−=

an = 7an−1+(2n2+2)3n

a0 = 10

0 = 4an2 - 14an + 7a + 4bn – 7b + 4c + 6n2 + 6

60Ch. 4.1, 4.2 & 4.5


� Therefore, we have

� By using a0 = 10

� a0 = 10 = -129/6 + c

� c = 189/6

� As a result,

)()( h

n

p

nn aaa +=nh

nca 7

)( =nn cnn 73)6/1294/212/3( 2 +−−−=

6/71893)6/1294/212/3(2 nn

nnna ⋅+−−−=

np

nnna 3)6/1294/212/3(

2)( −−−=

an = 7an−1+(2n2+2)3n

a0 = 10

61Ch. 4.1, 4.2 & 4.5

Generating Functions

� 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


� 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


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


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


� 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


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


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


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


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


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


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


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


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

81Ch. 4.1, 4.2 & 4.5


Example 6

a0 = 1

a1

= 8a0

+ 101-1 = 90

( )n

n

n

G x a x∞

=

=∑

0

1

( ) n

n

n

G x a a x∞

=

= +∑1

0 1

1

( ) (8 10 )n n

n

n

G x a a x∞

−−

=

= + +∑1

0 1

1 1

( ) 8 10n n n

n

n n

G x a a x x∞ ∞

−−

= =

= + +∑ ∑ 0

1

1

k k

k

a xax

∞

=

=−

∑

1 1 1

0 1

1 0

( ) 8 10n n n

n

n n

G x a x a x x x∞ ∞

− − −−

= =

= + +∑ ∑

0( ) 8 ( )

1 10

xG x a xG x

x= + +

−

an = 8an-1 + 10n-1 a1 = 9

82Ch. 4.1, 4.2 & 4.5


Example 6

1 9( )

(1 8 )(1 10 )

xG x

x x

−=

− −

1 1 1

2 1 8 1 10x x

= + − −

0 0

18 10

2

n n n n

n n

x x∞ ∞

= =

= +

∑ ∑

( )0

18 10

2

n n n

n

x∞

=

= +∑ ( )18 10

2

n n

na = +

0( ) 8 ( )1 10

xG x a xG x

x= + +

−a0 = 1

0

1

1

k k

k

a xax

∞

=

=−

∑

Recurrence Relations Solving Linear Recurrence Relations€¦ · Recurrence Relations Dr Patrick...

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