Transcript Chapter 8 Recursion 8.2 Solving Recurrence Relations by Iteration Recursion • A Sequence can be defined as: – informally be providing a few terms to demonstrate.

Chapter 8
Recursion
8.2
Solving Recurrence Relations by
Iteration
Recursion
• A Sequence can be defined as:
– informally be providing a few terms to
demonstrate the pattern, i.e. 3, 5, 7, ….
– give an explicit
formula for it nth term, i.e.
n
an 

1
n 1
– or, by recursion which requires a recurrence
relation.
Recursion by Iteration
• Method of Iteration
• most basic method for finding explicit formula
• given a sequence a0, a1, … defined by a recurrence relation
and initial conditions, calculate successive terms until a
pattern emerges
• define a formula based on the pattern
Example
• Finding explicit formula
– Let a0, a1, a2,… be the sequence defined as (k≥1):
1. ak = ak-1+ 2 (recurrence relation)
2. a0 = 1 (initial condition)
– Use iteration to guess formula.
– Solution
• ak = ak-1+ 2 k≥1,
• a1 = a0 + 2, a2 = a1 + 2, a3 = a2 + 2, a4 = a3 + 2 …
• an = an-1 + 2
Example
– Solution
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•
•
•
•
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ak = ak-1+ 2 k≥1,
a0 = 1
a1 = a0 + 2, (1) + 2, 1+ 1x2
a2 = a1 + 2, (1 + 2) + 2, 1 + 2x2
a3 = a2 + 2, (1 + 2 + 2) + 2, 1 + 3x2
a4 = a3 + 2, (1 + 2 + 2 + 2) + 2, 1 + 3x2
an = an-1 + 2 = 1 + nx2
– Guess
• an = 2n + 1
– Check by solving for some known number of n’s.
Definition
• Arithmetic Sequence
– A sequence is called an arithmetic sequence if,
and only if, there is a constant d such that
• ak = ak-1 + d, k≥1, or
• an = a0 + dn , n≥0
– Arithmetic sequence is a sequence in which the
current term equals the previous term plus a fixed
constant. (Note: illustrated in previous example)
Example
• Arithmetic sequence
– In a vacuum an object under gravity will fall 9.8
meters farther from one second to the next.
– A skydiver falls 4.9 meters b/n 0 & 1 second, 4.9 +
9.8=14.7 meters b/n 1 & 2 sec, how far will the
skydiver had falling b/n 60 & 61 secs?
– Solution
•
•
•
•
dk = dk-1 + 9.8 meters, k≥1
since dk is an arithmetic sequence it can be rewritten as:
dn = d0 + nx(9.8 meters), n≥0 , d0 = 4.9 meters
d60 = 4.9 + 60(9.8) = 592.9 meters
Geometric Sequence
• Geometric sequence is a sequence where each
term equals the previous term times a fixed
constant.
• Geometric sequences are found in population
growth models, compounding interest (financial),
number of operations for an algorithm, etc.
• Definition
– A sequence is called a geometric sequence if, and only
if, there is a constant r such that
• ak = r ak-1 , k≥1 Or
• an = a0 rn , n≥0
Example
• Geometric sequence
– A bank pays 4% per year of compounded interest, If the
initial amount deposited is $100,000, how much will the
account be worth in 21 years? In how many years will the
account be worth $1,000,000?
– Solution
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an = a0 rn , n≥0, a0 = $100,000 and r = 1.04 (1 + int rate)
a21 = $100,000 x (1.04)21 = $227,876.81
at = $100,000 x (1.04)t = $1,000,000, solve for t
(1.04)t = 1,000,000/100,000 = 10
t ln(1.04) = ln(10)
t = ln(10) / ln (1.04) = 58.7,
hence it would take 58.8 years to compound $100,000 into
$1,000,000 at 4% interest rate.
Formula Simplification
• Geometric sequence
–1+r+
r2
+…+
n 1
n
r
1
r =
r 1
• Sum of first n integers

,n  Z | n  0
– 1 + 2 + 3 + … + n = nn 1 ,n  Z | n 1
2

Example
• Tower of Hanoi using Iteration
– mk = 2mk-1 + 1 (recurrence)
– m1 = 1
– By Iteration
•
•
•
•
•
•
•
m1 = 1
m2 = 2m1 + 1 = 2(1) + 1 = 21 + 20
m3 = 2m2 + 1 = 2(2+1) + 1 = 22 + 21 + 20
m4 = 2m3 + 1 = 2(2*2 + 2*1 + 1)+1 = 23 + 22 + 21 + 20
m5 = 2m4 + 1 = 2(2*2*2 + 2*2*1 + 2*1 + 1) + 1 =24+23+22+21+20
…
mn = 2n-1 + 2n-2 + 2n-3 + … + 21 + 20
– From prior slide (sum of a geometric sequence, r=2)
• 2n-1 + 2n-2 + 2n-3 + … + 21 + 20 = 2 n 1
2 1
n
• mn = 2 - 1

Checking with Mathematical Induction
• Verifying Tower Hanoi Solution
– Problem: verify that sequence (m1, … , mn) is defined by mk
= 2mk-1 + 1, k≥2 and m1=1, then mn = 2n – 1.
• Given: mk = 2mk-1 + 1, k≥2 (recurrence) m1=1 (initial)
• Show: mn = 2n – 1
• Proof
1. Basis (n=1) m1=1 (initial) and m1 = 21 – 1 = 2 – 1 = 1
2. Inductive (k≥1) holds for n=k, then it holds for n=k+1
–
–
–
–
–
mk = 2k – 1 (k≥1), show that mk+1 = 2k+1 – 1
mk+1 = 2mk+1-1 + 1 (mk = 2mk-1 + 1, apply k+1+
= 2mk + 1 (subs in mk = 2k – 1 )
= 2(2k – 1 ) + 1 = 2k+1 – 2 + 1 = 2k+1 – 1
hence, left side (mk=2mk-1 + 1) is equal to right side (mk+1 =2k+1 – 1)