lecture 4,5 mathematical induction and fibonacci sequences

Lecture 4,5

Mathematical Induction and

Fibonacci Sequences

• Mathematical induction is a powerful, yet straight-forward method of proving statements whose domain is a subset of the set of integers.

• Usually, a statement that is proven by induction is based on the set of natural numbers.

• This statement can often be thought of as a function of a number n, where n = 1, 2, 3,. . .

• Proof by induction involves three main steps– Proving the base of induction

– Forming the induction hypothesis

– Proving that the induction hypothesis holds true for all numbers in the domain.

What is Mathematical Induction?

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Let P(n) be the predicate defined for any positive integers n, and let n0 be a fixed integer. Suppose the following two statements are true

1. P(n0) is true.

2. For any positive integers k, k n0,3. if P(k) is true then P(k+1)is true.

If the above statements are true then the statement:

n N, such that n n0, P(n) is also true

What is Mathematical Induction?

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Claim: P(n) is true for all n Z+, for n n0

1. Basis– Show formula is true when n = n0

2. Inductive hypothesis– Assume formula is true for an arbitrary n = k

where, k Z+ and k n0

3. To Prove Claim– Show that formula is then true for k+1

Note: In fact we have to prove

1) P(n0) and2) P(k) P(k+1)

Steps in Proving by Induction

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Example 1• Prove that n2 n + 100 n 11

Solution

Let P(n) n2 n + 100 n 11

1. P(11) 112 11 + 100 121 111, true

2. Suppose predicate is true for n = k, i.e.

P(k) k2 k + 100, true k 11

3. Now it can be proved that

P(k+1) (k+1)2 (k+1) + 100, k2 + 2k +1 k +1 + 100 k2 + k 100 (by 1 and 2)

Hence P(k) P(K+1)

Proof by Induction

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Example 1• Prove that n2 n + 100 n 11SolutionInitially, base case

Solution set = {11} By, P(k) P(K+1) P(11) P(12), taking k = 11

Solution set = {11, 12}Similarly, P(12) P(13), taking k = 12

Solution set = {11, 12, 13}And, P(13) P(14), taking k = 13

Solution set = {11, 12, 13, 14}And so on

Validity of Proof

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Reasoning of ProofExample 2

Use Mathematical Induction to prove that sum of the first n odd positive integers is n2.

Proof • Let P(n) denote the proposition that • Basis step : P(1) is true , since 1 = 12

• Inductive step : Let P(k) is true for a positive integer k, i.e., 1+3+5+…+(2k-1) = k2

• Note that: 1+3+5+…+(2k-1)+(2k+1) = k2+2k+1= (k+1)2

∴ P(k+1) true, by induction, P(n) is true for all n Z+

Another Proof 2

11

)1(2)12( nnnnniin

i

n

i

Another Easy Example

2

1

)12( nin

i

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Reasoning of ProofExample 3• Use mathematical Induction to prove that n < 2n for all n Z+

Proof• Let P(n) be the proposition that n < 2n

• Basis step : P(1) is true since 1 < 21 .• Inductive step : Assume that P(n) is true for a positive integer n = k,

i.e., k < 2k.• Now consider for P(k+1) : Since, k + 1 < 2k + 1 2k + 2k = 2.2k = 2k + 1 ∴ P(k+1) is true. It proves that P(n) is true for all n Z+.

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kH k

1...

3

1

2

11

21

2

nH n

The harmonic numbers Hk, k = 1, 2, 3, …, are

defined by

Use mathematical induction to show that

Proof

Let P(n) be the proposition that

Basis step :

P(0) is true, since,

Inductive step

Assume that P(k) is true for some k,

whenever n is a nonnegative integer.

2/12

nH n

12/011120 HH

2/12

kH k

Example 4: Harmonic Numbers

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∴P(k+1) is true. Hence the statement is true for all n Z+.

kkkkkkkH222

1

22

1

12

1

2

1

3

1

2

11

12 1

12 2

1

22

1

12

1

kkkkH

12

1

22

1

12

1)

21(

kkk

k

kkkkkk

k

22

1

22

1

22

1)

21(

kk

kk

22

2)

21(

2

11

2

1

21

kk

Now consider

Example 4: Harmonic Numbers

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Fibonacci Sequences

Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

In this lecture we will cover the following:

• Fibonacci Problem and its Sequence

• Construction of Mathematical Model

• Recursive Algorithms

• Generalizations of Rabbits Problem and Constructing its Mathematical Models

• Applications of Fibonacci Sequences

Today Covered

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• By studying Fibonacci numbers and constructing Fibonacci sequence we can imagine how mathematics is connected to apparently unrelated things in this universe.

• Even though these numbers were introduced in 1202 in Fibonacci’s book Liber abaci.

• Fibonacci, who was born Leonardo da Pisa gave a problem in his book whose solution was the Fibonacci sequence as we will discuss it today.

Fibonacci Sequence

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Statement:

• Start with a pair of rabbits, one male and one female, born on January 1.

• Assume that all months are of equal length and that rabbits begin to produce two months after their own birth.

• After reaching age of two months, each pair produces another mixed pair, one male and one female, and then another mixed pair each month, and no rabbit dies.

How many pairs of rabbits will there be after one year?

Answer: The Fibonacci Sequence!

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . .

Fibonacci’s Problem

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Construction of Mathematical Model

end of month 2

end of month 1

end of month 4

end of month 3

end of month 6

end of month 5

F2 = 1

F3 = 2

F5 = 5

F6 = 8

F7 = 13. . . . . . . . .

F1 = 1

F0 = 0

F4 = 3

end of month 12

end of month 7

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• Total pairs at level k = Total pairs at level k-1 + Total pairs born at level k (1)

• Since

Total pairs born at level k = Total pairs at level k-2 (2)• Hence by equation (1) and (2)

Total pairs at level k = Total pairs at level k-1 + Total pairs at level k-2

• Now let us denote

Fk = Total pairs at level k

• Now our recursive mathematical model will become

Fk = Fk-1 + Fk-2

Construction of Mathematical Model

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Since Fk = Fk-1 + Fk-2 F0 = 0, F1= 1

• F2 = F1 + F0= 1 + 0 = 1

• F3 = F2 + F1= 1 + 1 = 2

• F4 = F3 + F2= 2 + 1 = 3

• F5 = F4 + F3= 3 + 2 = 5

• F6 = F5 + F4= 5 + 3 = 8

• F7 = F6 + F5= 8 + 5 = 13

• F8 = F7 + F6= 13 + 8 = 21

• F9 = F8 + F7= 21 + 13 = 34

• F10 = F9 + F8= 34 + 21 = 55

• F11 = F10 + F9= 55 + 34 = 89

• F12 = F11 + F10= 89 + 55 = 144 . . .

Computing Values using Mathematical Model

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21 kkk FFF

1 condition initialwith

2

10

21

FF

kFFF kkk

Theorem: The fibonacci sequence F0,F1, F2,…. Satisfies the recurrence relation

Find the explicit formula for this sequence.Solution:

Let tk is solution to this, then characteristic equation

The given fibonacci sequence

Explicit Formula Computing Fibonacci Numbers

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2

51 ,

2

51

2

411

21

tt

t

Fibonacci Sequence

For some real C and D fibonacci sequence satisfies the relation

0 1 0

F

2

51

2

51

0

0 2

51

2

51

0

0

00

0

FDC

DC

DCF

n

nDCF

nn

n

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1 2 12

51

2

51

2

51

2

51F

1 Now

1

1

FDC

DC

n

nn

D

52

51

5

1

2

51

5

1F

Hence5

1,

5

1C

get usly wesimultaneo 2 and 1 Solving

n

Fibonacci Sequence

Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

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After simplifying we get

which is called the explicit formula for the Fibonacci sequence recurrence relation.

nn

2

51

5

1

2

51

5

1Fn

Fibonacci Sequence

5

1

5

1F

then 2

51 and

2

51Let

nnn

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Example: Compute F3

then 2

51 and

2

51 where

5

1

5

1F Since nn

n

33

3 2

51

5

1

2

51

5

1F

Verification of the Explicit Formula

8

555.1.35.1.31

5

1

8

555.1.35.1.31

5

1F Now,

22

3

555.1.35.1.315.8

1555.1.35.1.31

5.8

1F3

2555.1.35.1.31555.1.35.1.315.8

1F3

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Fibo-R(n)

if n = 0

then 0

if n = 1

then 1

else Fibo-R(n-1) + Fibo-R(n-2)

Recursive Algorithm Computing Fibonacci Numbers

Terminating conditions

Recursive calls

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• Least Cost: To find an asymptotic bound of computational cost of this algorithm.

Running Time of Recursive Fibonacci Algorithm

2 n )2()1(

2 if 1)(

nTnT

nnT

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Recursion Tree

Drawback in Recursive Algorithms

F(n)

F(n-1) F(n-2)

F(0) F(1)

F(n-2) F(n-3) F(n-3) F(n-4)

F(1) F(0)

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Statement:

• Start with a pair of rabbits, one male and one female, born on January 1.

• Assume that all months are of equal length and that rabbits begin to produce two months after their own birth.

• After reaching age of two months, each pair produces two other mixed pairs, two male and two female, and then two other mixed pair each month, and no rabbit dies.

How many pairs of rabbits will there be after one year?

Answer: Generalization of Fibonacci Sequence!

0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, . . .

Generalization of Rabbits Problem

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Construction of Mathematical Model

F0 = 0

F1 = 1

F2 = 1

F3 = 3

F4 = 5

F5 = 11

F6 = 21

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• Total pairs at level k = Total pairs at level k-1 + Total pairs born at level k (1)

• SinceTotal pairs born at level k =

2 x Total pairs at level k-2 (2)• By (1) and (2), Total pairs at level k =

Total pairs at level k-1 + 2 x Total pairs at level k-2• Now let us denote

Fk = Total pairs at level k • Our recursive mathematical model:

Fk = Fk-1 + 2.Fk-2

• General Model (m pairs production): Fk = Fk-1 + m.Fk-2

Construction of Mathematical Model

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• Recursive mathematical model

(one pair production)

Fk = Fk-1 + Fk-2

• Recursive mathematical model

(two pairs production)

Fk = Fk-1 + 2.Fk-2

• Recursive mathematical model

(m pairs production)

Fk = Fk-1 + m.Fk-2

Generalization

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Since Fk = Fk-1 + 2.Fk-2 F0 = 0, F1 = 1

• F2 = F1 + 2.F0= 1 + 0 = 1

• F3 = F2 + 2.F1= 1 + 2 = 3

• F4 = F3 + 2.F2= 3 + 2 = 5

• F5 = F4 + 2.F3= 5 + 6 = 11

• F6 = F5 + F4= 11 + 10 = 21

• F7 = F6 + F5= 21 + 22 = 43

• F8 = F7 + F6= 43 + 42 = 85

• F9 = F8 + F7= 85 + 86 = 171

• F10 = F9 + F8= 171 + 170 = 341

• F11 = F10 + F9= 341 + 342 = 683

• F12 = F11 + F10= 683 + 682 = 1365 . . .

Computing Values using Mathematical Model

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Statement:• Start with a different kind of pair of rabbits, one male and

one female, born on January 1. • Assume all months are of equal length and that rabbits

begin to produce three months after their own birth. • After reaching age of three months, each pair produces

another mixed pairs, one male and other female, and then another mixed pair each month, and no rabbit dies.

How many pairs of rabbits will there be after one year?

Answer: Generalization of Fibonacci Sequence!

0, 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, . . .

Another Generalization of Rabbits Problem

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Construction of Mathematical Model

F3 = 1

F4 = 2

F6 = 4

F8 = 9

F1 = 1

F0 = 0

F5 = 3

F2 = 1

F7 = 6

F9 = 13

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• Total pairs at level k =

Total pairs at level k-1 + Total pairs born at level k (1)

• Since

Total pairs born at level k = Total pairs at level k-3 (2)

• By (1) and (2)

Total pairs at level k =

Total pairs at level k-1 + Total pairs at level k-3

• Now let us denote

Fk = Total pairs at level k

• This time mathematical model: Fk = Fk-1 + Fk-3

Construction of Mathematical Model

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Since Fk = Fk-1 + Fk-3 F0 = 0, F1= F2= 1

• F3 = F2 + F0= 1 + 0 = 1

• F4 = F3 + F1= 1 + 1 = 2

• F5 = F4 + F2= 2 + 1 = 3

• F6 = F5 + F3= 3 + 1 = 4

• F7 = F6 + F4= 4 + 2 = 6

• F8 = F7 + F5= 6 + 3 = 9

• F9 = F8 + F6= 9 + 4 = 13

• F10 = F9 + F7= 13 + 6 = 19

• F11 = F10 + F8= 19 + 9 = 28

• F12 = F11 + F9= 28 + 13 = 41 . . .

Computing Values using Mathematical Model

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• Recursive mathematical model (one pair, production after three months)

Fk = Fk-1 + Fk-3

• Recursive mathematical model (two pairs, production after three months)

Fk = Fk-1 + 2.Fk-3

• Recursive mathematical model (m pairs, production after three months)

Fk = Fk-1 + m.Fk-3

• Recursive mathematical model (m pairs, production after n months)

Fk = Fk-1 + m.Fk-n

More Generalization

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Fibonacci sequences• Are used in trend analysis• By some pseudorandom number generators• Many plants show the Fibonacci numbers in the

arrangements of the leaves around the stems.• Seen in arrangement of seeds on flower heads • Consecutive Fibonacci numbers give worst case

behavior when used as inputs in Euclid’s algorithm.

Applications of Fibonacci Sequences

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