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17 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

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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?

mashhoood.webs.com

2


Let P(n) be the predicate defined for any positive
integers n, and let n
0

be a fixed integer. Suppose the
following two statements are true

1.
P(n
0
) is true.

2.
For any positive integers k, k


n
0
,

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


n
0
, P(n) is also true

What is Mathematical Induction?

mashhoood.webs.com

3

Claim
:
P(n)

is true for all n


Z
+
, for
n



n
0

1.
Basis


Show formula is true when
n

=
n
0


2.
Inductive hypothesis


Assume formula is true for an arbitrary
n = k





where, k


Z
+

and
k



n
0

3.
To Prove Claim


Show that formula is then true for
k+1

Note: In fact we have to prove


1)
P(n
0
) and

2)
P(k)


P(k+1)


Steps in Proving by Induction

mashhoood.webs.com

4

Example 1


Prove that n
2



n + 100




n


11

Solution

Let P(n)



n
2



n + 100




n


11

1.
P(11)



11
2



11 + 100


121


111,
true

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


P(k)



k
2



k + 100,
true



k



11

3.
Now it can be proved that



P(k+1)



(k+1)
2



(k+1) + 100,




k
2

+ 2k +1


k +1 + 100


k
2

+ k


100
(by 1 and 2)



Hence
P(k)


P(K+1)

Proof by Induction

mashhoood.webs.com

5

Example 1


Prove that n
2



n + 100




n


11

Solution

Initially, 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

mashhoood.webs.com

6

Reasoning of Proof

Example 2


Use Mathematical Induction to prove that sum of the
first
n

odd positive integers is
n
2
.


Proof



Let
P
(
n
) denote the proposition that


Basis step

:
P
(1) is true , since 1 = 1
2



Inductive step

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


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




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


Z
+


Another Proof

2
1
1
)
1
(
2
)
1
2
(
n
n
n
n
n
i
i
n
i
n
i











Another Easy Example


2
1

)
1
2
(
n
i
n
i




mashhoood.webs.com

7

Reasoning of Proof

Example 3


Use mathematical Induction to prove that


n
< 2
n

for all
n


Z
+

Proof


Let
P
(
n
) be the proposition that
n

< 2
n


Basis step

:

P
(1) is true since 1 < 2
1
.


Inductive step :


Assume that
P
(
n
) is true for a positive integer n =
k
,


i.e.,
k

< 2
k
.


Now consider for
P
(
k
+1) :


Since,
k

+ 1 < 2
k

+ 1


2
k

+ 2
k
= 2.2
k

= 2
k

+ 1





P
(
k
+1) is true.


It proves that
P
(
n
) is true for all
n


Z
+
.

mashhoood.webs.com

8

k
H
k
1
...
3
1
2
1
1





2
1
2
n
H
n


The harmonic numbers
H
k
,
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
/
1
2
n
H
n


1
2
/
0
1
1
1
2
0





H
H
2
/
1
2
k
H
k


Example 4: Harmonic Numbers

mashhoood.webs.com

9


P
(
k
+1)

is true.

Hence the statement is true for all
n


Z
+
.

k
k
k
k
k
k
k
H
2
2
2
1
2
2
1
1
2
1
2
1
3
1
2
1
1
1
2
1

















1
2
2
1
2
2
1
1
2
1








k
k
k
k
H

1
2
1
2
2
1
1
2
1
)
2
1
(









k
k
k
k

k
k
k
k
k
k
k
2
2
1
2
2
1
2
2
1
)
2
1
(










k
k
k
k
2
2
2
)
2
1
(




2
1
1
2
1
2
1






k
k
Now consider

Example 4: Harmonic Numbers

mashhoood.webs.com

10






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

mashhoood.webs.com

12


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

mashhoood.webs.com

13

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

mashhoood.webs.com

14

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
F
2
=
1
F
3
=
2
F
5
=
5
F
6
=
8
F
7
=
13
. . .
. . .
. . .
F
1
=
1
F
0
=
0
F
4
=
3
end of month
12
end of month
7
mashhoood.webs.com

15


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



F
k

= Total pairs at level k


Now our recursive mathematical model will become



F
k

= F
k
-
1

+ F
k
-
2

Construction of Mathematical Model

mashhoood.webs.com

16

Since

F
k

= F
k
-
1

+ F
k
-
2



F
0

= 0, F
1
= 1


F
2

= F
1

+ F
0
= 1 + 0 = 1


F
3

= F
2

+ F
1
= 1 + 1 = 2


F
4

= F
3

+ F
2
= 2 + 1 = 3


F
5

= F
4

+ F
3
= 3 + 2 = 5


F
6

= F
5

+ F
4
= 5 + 3 = 8


F
7

= F
6

+ F
5
= 8 + 5 = 13


F
8

= F
7

+ F
6
= 13 + 8 = 21


F
9

= F
8

+ F
7
= 21 + 13 = 34


F
10

= F
9

+ F
8
= 34 + 21 = 55


F
11

= F
10

+ F
9
= 55 + 34 = 89


F
12

= F
11

+ F
10
= 89 + 55 = 144 . . .

Computing Values using Mathematical Model

mashhoood.webs.com

17



2
1




k
k
k
F
F
F
1

condition

initial
with
2


1
0

2
1








F
F
k
F
F
F
k
k
k
Theorem
:

The fibonacci sequence F
0
,F
1
, F
2
,…. Satisfies
the recurrence relation

Find the explicit formula for this sequence.

Solution
:

Let t
k

is solution to this, then characteristic equation

The given fibonacci sequence

Explicit Formula Computing Fibonacci Numbers

0
1
2



t
t
mashhoood.webs.com

18

2
5
1

,
2
5
1
2
4
1
1
2
1







t
t
t
Fibonacci Sequence

For some real C and D fibonacci sequence satisfies the relation



0

1

0
F

2
5
1
2
5
1

0

0


2
5
1
2
5
1

0
0
0
0
0


















































F
D
C
D
C
D
C
F
n
n
D
C
F
n
n
n

mashhoood.webs.com

19



1

2

1
2
5
1
2
5
1

2
5
1
2
5
1
F

1

Now
1
1





























F
D
C
D
C
n





n
n
D



































5
2
5
1
5
1
2
5
1
5
1
F
Hence
5
1
,
5
1
C
get

usly we
simultaneo

2

and

1

Solving
n
Fibonacci Sequence

Dr Nazir A. Zafar






Advanced Algorithms Analysis and Design

mashhoood.webs.com

20


After simplifying we get





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




n
n




















2
5
1
5
1
2
5
1
5
1
F
n
Fibonacci Sequence


5
1


5
1
F
then
2
5
1

and

2
5
1
Let
n
n
n




























mashhoood.webs.com

21

Example
:
Compute F
3








then
2
5
1

and

2
5
1

where
5
1


5
1
F

Since
n
n
n




























3
3
3
2
5
1
5
1
2
5
1
5
1
F




















Verification of the Explicit Formula

























8
5
5
5
.
1
.
3
5
.
1
.
3
1
5
1
8
5
5
5
.
1
.
3
5
.
1
.
3
1
5
1
F

Now,
2
2
3




5
5
5
.
1
.
3
5
.
1
.
3
1
5
.
8
1
5
5
5
.
1
.
3
5
.
1
.
3
1
5
.
8
1
F
3










2
5
5
5
.
1
.
3
5
.
1
.
3
1
5
5
5
.
1
.
3
5
.
1
.
3
1
5
.
8
1
F
3









mashhoood.webs.com

22

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

mashhoood.webs.com

23


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
)
(









n
T
n
T
n
n
T
mashhoood.webs.com

24

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)

mashhoood.webs.com

25

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

mashhoood.webs.com

26

Construction of Mathematical Model






F
0
=
0
F
1
=
1
F
2
=
1
F
3
=
3
F
4
=
5
F
5
=
11
F
6
=
21
mashhoood.webs.com

27


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 =






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



F
k

= Total pairs at level k


Our recursive mathematical model:




F
k

= F
k
-
1

+ 2.F
k
-
2


General Model (m pairs production):
F
k

= F
k
-
1

+ m.F
k
-
2

Construction of Mathematical Model

mashhoood.webs.com

28


Recursive mathematical model


(one pair production)




F
k

= F
k
-
1

+ F
k
-
2


Recursive mathematical model


(two pairs production)




F
k

= F
k
-
1

+ 2.F
k
-
2


Recursive mathematical model


(m pairs production)




F
k

= F
k
-
1

+ m.F
k
-
2


Generalization

mashhoood.webs.com

29

Since

F
k

= F
k
-
1

+ 2.F
k
-
2


F
0

= 0, F
1
= 1


F
2

= F
1

+ 2.F
0
= 1 + 0 = 1


F
3

= F
2

+ 2.F
1
= 1 + 2 = 3


F
4

= F
3

+ 2.F
2
= 3 + 2 = 5


F
5

= F
4

+ 2.F
3
= 5 + 6 = 11


F
6

= F
5

+ F
4
= 11 + 10 = 21


F
7

= F
6

+ F
5
= 21 + 22 = 43


F
8

= F
7

+ F
6
= 43 + 42 = 85


F
9

= F
8

+ F
7
= 85 + 86 = 171


F
10

= F
9

+ F
8
= 171 + 170 = 341


F
11

= F
10

+ F
9
= 341 + 342 = 683


F
12

= F
11

+ F
10
= 683 + 682 = 1365 . . .

Computing Values using Mathematical Model

mashhoood.webs.com

30

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

mashhoood.webs.com

31

Construction of Mathematical Model

F
3
=
1
F
4
=
2
F
6
=
4
F
8
=
9
F
1
=
1
F
0
=
0
F
5
=
3
F
2
=
1
F
7
=
6
F
9
=
13
F
10
=
19
mashhoood.webs.com

32


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



F
k

= Total pairs at level k


This time mathematical model
:

F
k

= F
k
-
1

+ F
k
-
3

Construction of Mathematical Model

mashhoood.webs.com

33

Since

F
k

= F
k
-
1

+ F
k
-
3



F
0

= 0, F
1
= F
2
= 1


F
3

= F
2

+ F
0
= 1 + 0 = 1




F
4

= F
3

+ F
1
= 1 + 1 = 2


F
5

= F
4

+ F
2
= 2 + 1 = 3


F
6

= F
5

+ F
3
= 3 + 1 = 4


F
7

= F
6

+ F
4
= 4 + 2 = 6


F
8

= F
7

+ F
5
= 6 + 3 = 9


F
9

= F
8

+ F
6
= 9 + 4 = 13


F
10

= F
9

+ F
7
= 13 + 6 = 19


F
11

= F
10

+ F
8
= 19 + 9 = 28


F
12

= F
11

+ F
9
= 28 + 13 = 41 . . .

Computing Values using Mathematical Model

mashhoood.webs.com

34


Recursive mathematical model


(one pair, production after three months)




F
k

= F
k
-
1

+ F
k
-
3


Recursive mathematical model


(two pairs, production after three months)




F
k

= F
k
-
1

+ 2.F
k
-
3


Recursive mathematical model


(m pairs, production after three months)





F
k

= F
k
-
1

+ m.F
k
-
3


Recursive mathematical model


(m pairs, production after n months)





F
k

= F
k
-
1

+ m.F
k
-
n

More Generalization

mashhoood.webs.com

35

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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36