1. RECURRENCE Sequence Recursively defined sequence
Finding an explicit formula for recurrence relation

2. Learning Outcomes You should be able to solve
first-order and second-order linear homogeneous recurrence relation with constant coefficients

3. Preamble
What is recurrence and how does it relate to a sequence?

4. Sequences
A sequence is 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 ≥ 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, …
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

7. Explicit formula and recurrence relation
Show that the sequence 1,-1!,2!,-3!,4!,…,(-1)nn!,… for n≥0, satisfies the recurrence relation
sk = (-k)sk-1 for all integers k≥1.
The general term of the sequence: sn=(-1)nn!
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

8. 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

9. Tower of Hanoi

13. Application of recurrence
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 algo.
Recurrence relation is used to express the no. of operation in the algorithm.

14. Solving Recurrence relation 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 is iteration
- start from the initial condition
- calculate successive terms until a pattern can be seen
- guess an explicit formula

15. Some examples
Let a0,a1,a2,… be the sequence defined recursively as follows: For all integers k≥1,
(1) ak = ak-1+2
(2) a0 = 1
Use iteration to guess an explicit formula for the sequence.
a0=1
a1=a0+2
a2=a1+2=(1+2)+2 = 1+2.2
a3=a2+2=(1+2.2)+2 = 1+3.2
a4=a3+2=(1+3.2)+2 = 1+4.3 ….
Guess: an=1+n.2=1+2n
The above sequence is an arithmetic sequence.

16. Geometric Sequence
Let r be a fixed nonzero constant, and suppose a sequence a0,a1,a2,… is defined as follows:
ak = rak-1 for all integers k ≥ 1
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.

18. Explicit formula for tower of Hanoi
mn = 2n – 1. (exponential order)
To move 1 disk takes 1 second
m64 = 264 –1 = * 1019 seconds
= * 1011 years
= billion years.

19. Second-Order Linear Homogeneous with constant coefficients
A second-order linear homogeneous recur. relation with c.c. is a recur. 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.

20. Terminology ak = Aak-1 + Bak-2
Second order: ak contains the two previous terms
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

21. Examples Second-Order Linear Homogeneous with constant coefficients
ak = 3ak-1 + 2ak yes.
bk = bk-1 + bk-2 + bk-3 - no
dk = (dk-1)2 + dk-1dk no; not linear
ek = 2ek yes; A = 0, B = 2.
fk = 2fk no; not homogeneous

22. Characteristic equation
ak = Aak-1 + Bak for k>=2 ….. (1)
Suppose that sequence 1, t, t2, t3,… 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.

23. 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, …,(-1)n, …

24. 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 c.c., 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.
C and D can be calculated using initial conditions.

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

26. 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, …

28. Steps for finding explicit formula
1. 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.

32. Single-Root Case Two sequences that satisfy the relation
ak = A.ak-1+B.ak-2
S1: 1,r,r2,r3,…,rn,…
S2: 0,r,2r2,3r3,…,nrn,…
where r is the root of t2 - A.t - B = 0.
Explicit formula for the new sequence
ak = C.rk + D.nrn

33. Example A sequence b0, b1, b2,… satisfies the rec. 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
Charac eq: t2 – 4t +4 =0
(t-2)2=0. t=2.
Seq: 1,2,22, …, 2n,..
0,2,2.22,3.23,…,n.2n,…
Explicit formula: bn = C.rn + D.nrn = C.2n + D.n2n

34. Example bn = C.2n + D.n2n b0 = 1 = C + D.0 = C b1 = 3 = C.2 + D.2
Hence D = ½.
Therefore bn = 2n + (1/2).n.2n
= 2n (1+ n/2) for integer n>=0.

35. Question
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36. Summary How to express the sequence
Find explicit formula for first-order relation
Find explicit formula for second-order relation (distinct and single root)

37. THANK YOU