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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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?
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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
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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
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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
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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
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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
+
.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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