Chapter 8

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# Chapter 8 - PowerPoint PPT Presentation

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.

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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.
• 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):
• ak = ak-1+ 2 (recurrence relation)
• 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
• 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 = a0rn , 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
• an = a0rn , 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 + … + rn=
• Sum of first n integers
• 1 + 2 + 3 + … + n =
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 =
• mn = 2n - 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
• Basis (n=1) m1=1 (initial) and m1 = 21 – 1 = 2 – 1 = 1
• 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)