1. Discrete Structures Chapter 6 Recurrence Relations
Nurul Amelina Nasharuddin
Multimedia Department

2. Objectives On completion of this topic, student should be able to:
Solve first-order linear homogeneous recurrence relation
Solve second-order linear with constant coefficients

3. Outline First-order linear homogeneous recurrence relation
Second-order linear with constant coefficients
Divide and conquer algorithm example

4. Sequence
An ordered list of objects (or events). Like a set, it contains members (also called terms)
Sequences can be finite or infinite
2,4,6,8,… for i ≥ 1, ai = 2i (explicit formula)
infinite sequence with infinite distinct values
-1,1,-1,1,… for i ≥ 1, bi = (-1)i
infinite sequence with finite distinct values
For 1  i  ci = i+5
finite sequence (with finite distinct values)
6,7,8,9,10,11

5. Ways to Define Sequence
Write the first few terms:
3,5,7,…
Use explicit formula for its nth term
an = 2n for n ≥ 1
Use recursion. How to define a sequence using a recursion?

6. Recursively Defined Sequences
Recursion can be used to defined a sequence
This requires:
A recurrence relation: a formula that relates each term ak to some previous terms ak-1, ak-2, …
ak = ak-1 + 2ak-2
The initial conditions: the values of the first few terms a0, a1, …

7. Example For all integers k ≥ 2, find the terms b2, b3 and b4:
bk = bk-1 + bk-2 (recurrence relation)
b0 = 1 and b1 = 3 (initial conditions)
Solution:
b2 = b1 + b0 = = 4
b3 = b2 + b1 = = 7
b4 = b3 + b2 = = 11

8. Explicit Formula and Recurrence Relation
Eg: Show that the sequence 1, -1!, 2!, -3!, 4!,…, (-1)n n!, … for n ≥ 0, satisfies the recurrence relation
sk = (-k)sk-1 for all integers k ≥ 1
The general term of the sequence: sn=(-1)n n!
Substitute k and k-1 for n to get sk=(-1)kk!, sk-1=(-1)k-1(k-1)!
Substitute sk-1 into recurrence relation:
(-k)sk-1 = (-k)(-1)k-1(k-1)!
= (-1)k(-1)k-1(k-1)!
= (-1)(-1)k-1 k(k-1)!
= (-1)k k! = sk

9. Examples of Recursively Sequence
Famous recurrences
Arithmetic sequences: ak = ak-1 + d
e.g. 1,4,7,10,13,…
Geometric sequences: ak = ark-1
e.g. 1,3,9,27,…
Factorial: f(n) = nf(n-1)
Fibonacci numbers: fk = fk-1+fk-2
1,1,2,3,5,8,…
Tower of Hanoi problem

14. Application of Recurrence
Factorial Notation
Analysis of algorithm containing recursive function such as factorial function
Algorithm f(n)
/input: A nonnegative integer
/output: The value of n!
If n = 0 return 1
else return f(n-1)*n
No. of operations (multiplication) determines the efficiency of algorithm
Recurrence relation is used to express the no. of operation in the algorithm

15. Solving Linear Recurrence Relations by Iteration
It is helpful to know an explicit formula for a sequence
An explicit formula is called a solution to the recurrence relation
Most basic method to find explicit formula is iteration
start from the initial condition
calculate successive terms until a pattern can be seen
guess an explicit formula

16. Example of Arithmetic Sequence (pg 476)
Let a0, a1, a2,… be the sequence defined recursively as follows:
(1) ak = ak-1+2 for all integers k ≥ 1
(2) a0 = 1
Use iteration to guess an explicit formula for the sequence
a0 =1
a1 =a0+2=1+2
a2 =a1+2=(1+2) = = 1+2.2
a3 =a2+2=(1+2+2) = = 1+3.2
a4 =a3+2=( )+2 = = 1+4.2 ….
Guess: an = 1+ n.2 = 1 + 2n
The above sequence is an arithmetic sequence

17. Arithmetic Sequence
A sequence a0, a1, a2, … is called arithmetic sequence iff there is a constant d such that
ak = ak-1 + d for all integers k  1
or equivalently
an = a0 + dn for all integers n  0

18. Example of Geometric Sequence
Let r be a fixed nonzero constant, and suppose a sequence a0,a1,a2,… is defined as follows:
(1) ak = rak-1 for all integers k ≥ 1
(2) a0 = a
Use iteration to guess an explicit formula for the sequence
a0=a
a1=ra0 =ra
a2=ra1 =r(ra) =r2a
a3=ra2 =r(r2a) =r3a …
Guess: an= rna = arn for all integers n≥0
The above sequence is geometric sequence and r is a common ratio

19. Geometric Sequence
A sequence a0, a1, a2, … is called geometric sequence iff there is a constant r such that
ak = rak-1 for all integers k  1
or equivalently
an = a0rn for all integers n  0

21. Second-Order Linear Homogeneous with Constant Coefficients
A second-order linear homogeneous recurrence relation with constant coefficients is a recurrence relation of the form
ak = Aak-1 + Bak-2
for all integers k ≥ some fixed integer,
where A and B are fixed real numbers with B ≠ 0

22. Terminology ak = Aak-1 + Bak-2
Second order: ak contains the two previous terms (ak-1 and ak-2)
Linear: ak-1 and ak-2 appear in separate terms and to the first power
Homogeneous: total degree of each term is the same (no constant term)
Constant coefficients: A and B are fixed real numbers

23. Example Second-Order Linear Homogeneous with constant coefficients
ak = 3ak-1 + 2ak-2 (Yes)
bk = bk-1 + bk-2 + bk-3 (No, not second order)
dk = (dk-1)2 + dk-1dk-2 (No, not linear)
ek = 2ek-2 (Yes, A = 0, B = 2)
fk = 2fk (No, not homogeneous)

24. Characteristic equation
ak = Aak-1 + Bak for k  2 ….. (1)
Suppose that sequence 1, t, t2, t3,…, tn,.. satisfies relation (1) where t is a nonzero real number
General term for the sequence an= tn
Hence, ak-1= tk-1 and ak-2= tk-2
Substitute ak-1 and ak-2 into relation (1)
tk = Atk-1 + Btk-2
Divide the equation by tk-2:
t2 = At + B or t2 – At – B = 0
This equation is called the characteristic equation of the
relation. Recurrence relation (1) is satisfied by the sequence 1,t,t2,t3,… iff t satisfies the characteristic equation

25. Using the Characteristic Equation to Find Sequences
Example: Consider the following recurrence relation
ak = ak-1+2ak-2 for all k  2
Find sequences that satisfy the relation
Solution: For the given relation, A=1 and B=2
Relation is satisfied by the sequence 1,t,t2,t3,… iff t satisfies the characteristic equation
t2 – At – B = 0 or
t2 – t – 2 = 0
(t – 2)(t + 1) = 0.
t = 2 or t = -1.
Sequences: 1,2,22,23,… and 1,-1,(-1)2,(-1)3, … or ,-1,1,-1, …,(-1)n, …

26. Linear Combination of Sequences
Lemma
If r0, r1, r2,…, rn,.. and s0, s1, s2,…, sn,… are sequences that satisfy the same second-order linear homogeneous recurrence relation with constant coefficients, and
if C and D are any numbers, then
the sequence a0, a1, a2,…defined by the formula
an = C.rn + D.sn for all integer n  0 also satisfies the same recurrence relation and
D can be calculated using initial conditions

27. Two Possible Solutions
For the characteristic equation t2 – At – B = 0
there are two possible solutions:
Distinct-roots case
Single-root case

28. Explicit Formula for Second-Order Relation – Distinct-roots case
Find a sequence (explicit formula) that satisfies the recurrence relation
ak=ak-1 + 2ak-2 for k  2
and initial conditions
a0=1 and a1=8
Solution:
A=1 and B=2
Characteristic equation : t2 – At – B =0
Substitute A and B, t2 – t – 2 = 0
Sequences: 1,2,22,23,… and 1,-1,1,-1, …,(-1)n, …

30. Steps for Finding Explicit Formula
Form the characteristic equation
2. Solve the equation – let r and s be the roots. ( r ≠ s)
3. Set up an explicit formula:
ak = C.rk + D.sk
4. Find C and D using initial conditions.

34. Single-Root Case Two sequences that satisfy the relation
ak = A.ak-1+B.ak-2
where r is the root of t2 - A.t - B = 0
Explicit formula for the new sequence
an = C.rn + D.nrn

35. Example (1)
A sequence b0, b1, b2,… satisfies the reccurence relation bk = 4bk-1 – 4bk-2 for k>=2 with initial conditions b0=1 and b1=3
Find explicit formula for the sequence
Solution: A = 4 and B = -4
Characteristic equation: t2 – 4t +4 =0
(t-2)2=0; t=2
Sequence: 1,2,22, …, 2n,..
0,2,2.22,3.23,…,n.2n,…

36. Example (1) Explicit formula: bn = C.rn + D.nrn = C.2n + D.n2n
bn = C.2n + D.n2n
n=0, b0 = 1 = C.20 + D.0.20 = C
n=1, b1 = 3 = C.2 + D.2
3 = D
D = ½
Therefore bn = 2n + (½).n.2n
= 2n (1+ n/2) for integer n  0
Sequence: 1,3,8,20,…