discrete mathematics and its...
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Discrete Mathematics and Its ApplicationsLecture 2: Basic Structures: Sequence, Cardinality, and Matrix
MING GAO
DaSE@ ECNU(for course related communications)
Apr. 3, 2020
Outline
1 Sequences and Summations
2 Summations
3 Cardinality of Set
4 Matrix
5 Matrix derivatives
6 Take-aways
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 2 / 45
Sequences and Summations
Sequence
Definition
A sequence is a function from a subset of the set of integers (usuallyeither the set 0, 1, 2, · · · or the set 1, 2, 3, · · · ) to a set S . Weuse the notation an to denote the image of integer n. We call an aterm of the sequence.
Some useful sequences
nth Term First 10 Terms
n2 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, · · ·n3 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, · · ·n4 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, · · ·2n 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, · · ·3n 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, · · ·n! 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, · · ·
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 3 / 45
Sequences and Summations
Sequence
Definition
A sequence is a function from a subset of the set of integers (usuallyeither the set 0, 1, 2, · · · or the set 1, 2, 3, · · · ) to a set S . Weuse the notation an to denote the image of integer n. We call an aterm of the sequence.
Some useful sequences
nth Term First 10 Terms
n2 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, · · ·n3 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, · · ·n4 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, · · ·2n 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, · · ·3n 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, · · ·n! 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, · · ·
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 3 / 45
Sequences and Summations
Progression
Geometric progression
A geometric progression is a sequence of form a, ar , ar2, · · · , arn, · · ·where initial term a and common ratio r are real numbers.
Arithmetic progression
An arithmetic progression is a sequence of form a, a + d , a +2d , · · · , a + nd , · · · where initial term a and common difference dare real numbers.
Examples
The sequences bn is a form of bn = (−1)n. The list of termsb0, b1, b2, b3, · · · begins with 1,−1, 1,−1, · · · .The sequences sn is a form of sn = −1 + 4n. The list ofterms s0, s1, s2, s3, · · · begins with −1, 3, 7, 11, · · · .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 4 / 45
Sequences and Summations
Progression
Geometric progression
A geometric progression is a sequence of form a, ar , ar2, · · · , arn, · · ·where initial term a and common ratio r are real numbers.
Arithmetic progression
An arithmetic progression is a sequence of form a, a + d , a +2d , · · · , a + nd , · · · where initial term a and common difference dare real numbers.
Examples
The sequences bn is a form of bn = (−1)n. The list of termsb0, b1, b2, b3, · · · begins with 1,−1, 1,−1, · · · .The sequences sn is a form of sn = −1 + 4n. The list ofterms s0, s1, s2, s3, · · · begins with −1, 3, 7, 11, · · · .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 4 / 45
Sequences and Summations
Progression
Geometric progression
A geometric progression is a sequence of form a, ar , ar2, · · · , arn, · · ·where initial term a and common ratio r are real numbers.
Arithmetic progression
An arithmetic progression is a sequence of form a, a + d , a +2d , · · · , a + nd , · · · where initial term a and common difference dare real numbers.
Examples
The sequences bn is a form of bn = (−1)n. The list of termsb0, b1, b2, b3, · · · begins with 1,−1, 1,−1, · · · .The sequences sn is a form of sn = −1 + 4n. The list ofterms s0, s1, s2, s3, · · · begins with −1, 3, 7, 11, · · · .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 4 / 45
Sequences and Summations
Recurrence relation
Definition
A recurrence relation for the sequence an is an equation that ex-presses an in terms of one or more of the previous terms of thesequence, namely, a0, a1, · · · , an−1, for all integers n with n ≥ n0,where n0 is a nonnegative integer. A sequence is called a solution ofa recurrence relation if its terms satisfy the recurrence relation.
Example
Let an be a sequence that satisfies the recurrence relationan = an−1 + 3 for n = 1, 2, 3, · · · , and suppose that a0 = 2.What are a1, a2, and a3?
Let an be a sequence that satisfies the recurrence relationan = an−1 − an−2 for n = 2, 3, 4, · · · , and suppose that a0 = 3and a1 = 5. What are a2 and a3?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 5 / 45
Sequences and Summations
Recurrence relation
Definition
A recurrence relation for the sequence an is an equation that ex-presses an in terms of one or more of the previous terms of thesequence, namely, a0, a1, · · · , an−1, for all integers n with n ≥ n0,where n0 is a nonnegative integer. A sequence is called a solution ofa recurrence relation if its terms satisfy the recurrence relation.
Example
Let an be a sequence that satisfies the recurrence relationan = an−1 + 3 for n = 1, 2, 3, · · · , and suppose that a0 = 2.What are a1, a2, and a3?
Let an be a sequence that satisfies the recurrence relationan = an−1 − an−2 for n = 2, 3, 4, · · · , and suppose that a0 = 3and a1 = 5. What are a2 and a3?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 5 / 45
Sequences and Summations
Recurrence relation Cont’d
Fibonacci sequence
Fibonacci sequence, f0, f1, f2, · · · , is defined by initial conditions f0 =0, f1 = 1, and recurrence relation fn = fn−1 + fn−2 for n = 2, 3, 4, · · ·
Example
Determine whether sequence an, where an = 3n for every nonneg-ative integer n, is a solution of recurrence relation an = 2an−1−an−2for n = 2, 3, 4, · · · Answer the same question where an = 2n andwhere an = 5.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 6 / 45
Sequences and Summations
Recurrence relation Cont’d
Fibonacci sequence
Fibonacci sequence, f0, f1, f2, · · · , is defined by initial conditions f0 =0, f1 = 1, and recurrence relation fn = fn−1 + fn−2 for n = 2, 3, 4, · · ·
Example
Determine whether sequence an, where an = 3n for every nonneg-ative integer n, is a solution of recurrence relation an = 2an−1−an−2for n = 2, 3, 4, · · · Answer the same question where an = 2n andwhere an = 5.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 6 / 45
Sequences and Summations
The Fibonacci sequence
Source:
https://en.wikipedia.org/wiki/
File:Fibonacci.jpg
In 1202, Leonardo Bonacci (known as Fibonacci)asked the following question.
“[A]ssuming that: a newly born pair of rabbits, onemale, one female, are put in a field; rabbits are ableto mate at the age of one month so that at the endof its second month a female can produce another pairof rabbits; rabbits never die and a mating pair alwaysproduces one new pair (one male, one female) everymonth from the second month on.”
“The puzzle that Fibonacci posed was: how many pairs
will there be in one year?”
From https://en.wikipedia.org/wiki/Fibonacci number
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 7 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.
Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 1
2 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 1
3 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥
♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 2
4 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥
♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 3
5 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥
♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 5
6 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥
♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 8
7 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥
♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have.
This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Let’s try to solve Fibonacci’s question.Let ♠ denote a newly born rabit pair, and ♥ denote a mature rabit pair.
Month Rabits
1 ♠ 12 ♥ 13 ♥ ♠ 24 ♥ ♥ ♠ 35 ♥ ♥ ♥ ♠ ♠ 56 ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ 87 ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♠ ♠ ♠ ♠ ♠ 13
How many rabit pairs do we have at the beginning of the 8th month?
Surely all 13 rabit pairs we have in the 7th month remain there and are allmature. So, the question is how many newly born rabbit pairs that we have.
The number of newly born rabbit pairs equals the number of mature rabbitpairs we have. This is also equal to the number of rabit pairs that we havein the 6th month: 8.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 8 / 45
Sequences and Summations
Thus, we will have 13+8 rabit pairs at the beginning of the 8th month.
If we write down the sequence, we get the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, . . .
Again, what’s the next number in this sequence? How can you compute it?21+13 = 34 is the answer. You take the last two numbers and add themup to get the next number. Why?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 9 / 45
Sequences and Summations
Thus, we will have 13+8 rabit pairs at the beginning of the 8th month.If we write down the sequence, we get the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, . . .
Again, what’s the next number in this sequence? How can you compute it?21+13 = 34 is the answer. You take the last two numbers and add themup to get the next number. Why?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 9 / 45
Sequences and Summations
Thus, we will have 13+8 rabit pairs at the beginning of the 8th month.If we write down the sequence, we get the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, . . .
Again, what’s the next number in this sequence? How can you compute it?
21+13 = 34 is the answer. You take the last two numbers and add themup to get the next number. Why?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 9 / 45
Sequences and Summations
Thus, we will have 13+8 rabit pairs at the beginning of the 8th month.If we write down the sequence, we get the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, . . .
Again, what’s the next number in this sequence? How can you compute it?21+13 = 34 is the answer.
You take the last two numbers and add themup to get the next number. Why?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 9 / 45
Sequences and Summations
Thus, we will have 13+8 rabit pairs at the beginning of the 8th month.If we write down the sequence, we get the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, . . .
Again, what’s the next number in this sequence? How can you compute it?21+13 = 34 is the answer. You take the last two numbers and add themup to get the next number. Why?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 9 / 45
Sequences and Summations
To be precise, let Fn be the n-th number in the Fibonacci sequence. (Thatis, F1 = 1,F2 = 1,F3 = 2,F4 = 3 and so on.) We can define the(n + 1)-th number as
Fn+1 = Fn + Fn−1,
for n = 2, 3, . . ..
Is this enough to completely specify the sequence?No, because we do not know how to start. To get the Fibonacci sequence,we need to specify two starting values: F1 = 1 and F2 = 1 as well.Now, you can see that the equation and these special values uniquelydetermine the sequence. It is also convenient to define F0 = 0 so that theequation works for n = 1.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 10 / 45
Sequences and Summations
To be precise, let Fn be the n-th number in the Fibonacci sequence. (Thatis, F1 = 1,F2 = 1,F3 = 2,F4 = 3 and so on.) We can define the(n + 1)-th number as
Fn+1 = Fn + Fn−1,
for n = 2, 3, . . .. Is this enough to completely specify the sequence?
No, because we do not know how to start. To get the Fibonacci sequence,we need to specify two starting values: F1 = 1 and F2 = 1 as well.Now, you can see that the equation and these special values uniquelydetermine the sequence. It is also convenient to define F0 = 0 so that theequation works for n = 1.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 10 / 45
Sequences and Summations
To be precise, let Fn be the n-th number in the Fibonacci sequence. (Thatis, F1 = 1,F2 = 1,F3 = 2,F4 = 3 and so on.) We can define the(n + 1)-th number as
Fn+1 = Fn + Fn−1,
for n = 2, 3, . . .. Is this enough to completely specify the sequence?No, because we do not know how to start. To get the Fibonacci sequence,we need to specify two starting values: F1 = 1 and F2 = 1 as well.Now, you can see that the equation and these special values uniquelydetermine the sequence. It is also convenient to define F0 = 0 so that theequation works for n = 1.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 10 / 45
Sequences and Summations
A recurrence
The equationFn+1 = Fn + Fn−1
and the initial values F0 = 0 and F1 = 1 specify all values of the Fibonaccisequence. With these two initial values, you can use the equation to findthe value of any number in the sequence.This definition is called a recurrence. Instead of defining the value ofeach number in the sequence explicitly, we do so by using the values ofother numbers in the sequence.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 11 / 45
Sequences and Summations
Tilings with 1x1 and 2x1 tiles
You have a walk way of length n units. The width of the walk way is 1unit. You have unlimited supplies of 1x1 tiles and 2x1 tiles. Every tile ofthe same size is indistinguishable. In how many ways can you tile the walkway?Let’s consider small cases.
When n = 1, there are 1 way.
When n = 2, there are 2 ways.
When n = 3, there are 3 ways.
When n = 4, there are 5 ways.
Let’s define Jn to be the number of ways you can tile a walk way of lengthn. From the example above, we know that J1 = 1 and J2 = 2.Can you find a formula for general Jn?
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 12 / 45
Sequences and Summations
Figuring out the recurrence for Jn
To figure out the general formula for Jn, we can think about the firstchoice we can make when tiling a walk way of length n. There are twochoices:
(1) We can start placing a 1x1 tile at the beginning, or
(2) We can start placing a 2x1 tile at the beginning.
In each of the cases, let’s think about how many ways we can tile the restof the walk way, provided that the first step is made.
Note that if we start by placing a 1x1 tile, we are left with a walk way oflength n − 1. From the definition of Jn, we know that there are Jn−1 waysto tile the rest of the walk way of length n− 1. Using similar reasoning, weknow that if we start with a 2x1 tile, there are Jn−2 ways to tile the rest ofthe walk way.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 13 / 45
Sequences and Summations
The recurrence for Jn
From the discussion, we have that
Jn = Jn−1 + Jn−2,
where J1 = 1 and J2 = 2.
Note that this is exactly the same recurrence as the Fibonacci sequence,but with different initial values. In fact, we have that
Jn = Fn+1.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 14 / 45
Sequences and Summations
Compound interest
Problem
Suppose that a person deposits 10, 000 in a savings account at abank yielding 11% per year with interest compounded annually. Howmuch will be in the account after 30 years?
Solution
Let Pn denote the amount in the account after n years. We seethat sequence Pn satisfies the recurrence relation
Pn = Pn−1 + 0.11Pn−1 = (1.11)Pn−1.
We can find a formula for Pn:
P1 = (1.11)P0 P2 = (1.11)P1 = (1.11)2P0
· · · Pn = (1.11)Pn−1 = · · · = (1.11)nP0.
Therefore, P30 = (1.11)3010, 000 = 228, 922.97.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 15 / 45
Sequences and Summations
Compound interest
Problem
Suppose that a person deposits 10, 000 in a savings account at abank yielding 11% per year with interest compounded annually. Howmuch will be in the account after 30 years?
Solution
Let Pn denote the amount in the account after n years. We seethat sequence Pn satisfies the recurrence relation
Pn = Pn−1 + 0.11Pn−1 = (1.11)Pn−1.
We can find a formula for Pn:
P1 = (1.11)P0 P2 = (1.11)P1 = (1.11)2P0
· · · Pn = (1.11)Pn−1 = · · · = (1.11)nP0.
Therefore, P30 = (1.11)3010, 000 = 228, 922.97.MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 15 / 45
Summations
Summations
Summation notations
n∑j=m
aj ,n∑
i=m
ai ,∑
m≤k≤nak ,
∑j∈I
aj ,
where I = k ∈ Z : m ≤ k ≤ n.
Some useful summation formulae
sum form sum form∑nk=0 ark arn+1−a
r−1 (r 6= 1)∑n
k=0 k n(n+1)2∑n
k=0 k2 n(n+1)(2n+1)6
∑nk=0 k3 n2(n+1)2
4∑∞k=0 xk , |x | < 1 1
1−x∑∞
k=0 kxk−1, |x | < 1 1(1−x)2
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 16 / 45
Summations
Summations
Summation notations
n∑j=m
aj ,n∑
i=m
ai ,∑
m≤k≤nak ,
∑j∈I
aj ,
where I = k ∈ Z : m ≤ k ≤ n.
Some useful summation formulae
sum form sum form∑nk=0 ark arn+1−a
r−1 (r 6= 1)∑n
k=0 k n(n+1)2∑n
k=0 k2 n(n+1)(2n+1)6
∑nk=0 k3 n2(n+1)2
4∑∞k=0 xk , |x | < 1 1
1−x∑∞
k=0 kxk−1, |x | < 1 1(1−x)2
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 16 / 45
Summations
Remark
Requires calculus
Let x be a real number with |x | < 1. Find
∞∑n=0
xn.
Solution
Let a = 1 and r = x we see that∑n
k=0 xk = xk+1−1x−1 . Because
|x | < 1, it follows that
∞∑k=0
xn = limk→∞
xk+1 − 1
x − 1=
0− 1
x − 1=
1
1− x.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 17 / 45
Cardinality of Set
Cardinality
Definition
Sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. When A and B have the samecardinality, we write |A| = |B|.
Definition
If there is a one-to-one function from A to B, the cardinality of A isless than or the same as the cardinality of B and we write |A| ≤ |B|.Moreover, when |A| ≤ |B| and A and B have different cardinality, wesay that the cardinality of A is less than the cardinality of B and wewrite |A| < |B|.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 18 / 45
Cardinality of Set
Cardinality
Definition
Sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. When A and B have the samecardinality, we write |A| = |B|.
Definition
If there is a one-to-one function from A to B, the cardinality of A isless than or the same as the cardinality of B and we write |A| ≤ |B|.Moreover, when |A| ≤ |B| and A and B have different cardinality, wesay that the cardinality of A is less than the cardinality of B and wewrite |A| < |B|.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 18 / 45
Cardinality of Set
Countable sets
Definition
A set that is either finite or has the same cardinality as Z+ is calledcountable. If an infinite set S is countable, then |S | = ℵ0.
Example
Show that the set of odd positive integers is a countable set.
Solution
In terms of the same cardinality, we will exhibit a one-to-onecorrespondence between this set and Z+. Consider function
f (n) = 2n − 1 : Z+ → 2k + 1 : k is an integer
One-to-one: f (n) = f (m)⇒ 2n − 1 = 2m − 1⇒ n = m.Onto: suppose that t is an odd positive integer. Then t is 1 lessthan 2k, where k is a natural number. Hence t = 2k − 1 = f (k).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 19 / 45
Cardinality of Set
Countable sets
Definition
A set that is either finite or has the same cardinality as Z+ is calledcountable. If an infinite set S is countable, then |S | = ℵ0.
Example
Show that the set of odd positive integers is a countable set.
Solution
In terms of the same cardinality, we will exhibit a one-to-onecorrespondence between this set and Z+. Consider function
f (n) = 2n − 1 : Z+ → 2k + 1 : k is an integer
One-to-one: f (n) = f (m)⇒ 2n − 1 = 2m − 1⇒ n = m.Onto: suppose that t is an odd positive integer. Then t is 1 lessthan 2k, where k is a natural number. Hence t = 2k − 1 = f (k).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 19 / 45
Cardinality of Set
Countable sets
Definition
A set that is either finite or has the same cardinality as Z+ is calledcountable. If an infinite set S is countable, then |S | = ℵ0.
Example
Show that the set of odd positive integers is a countable set.
Solution
In terms of the same cardinality, we will exhibit a one-to-onecorrespondence between this set and Z+. Consider function
f (n) = 2n − 1 : Z+ → 2k + 1 : k is an integer
One-to-one: f (n) = f (m)⇒ 2n − 1 = 2m − 1⇒ n = m.
Onto: suppose that t is an odd positive integer. Then t is 1 lessthan 2k, where k is a natural number. Hence t = 2k − 1 = f (k).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 19 / 45
Cardinality of Set
Countable sets
Definition
A set that is either finite or has the same cardinality as Z+ is calledcountable. If an infinite set S is countable, then |S | = ℵ0.
Example
Show that the set of odd positive integers is a countable set.
Solution
In terms of the same cardinality, we will exhibit a one-to-onecorrespondence between this set and Z+. Consider function
f (n) = 2n − 1 : Z+ → 2k + 1 : k is an integer
One-to-one: f (n) = f (m)⇒ 2n − 1 = 2m − 1⇒ n = m.Onto: suppose that t is an odd positive integer. Then t is 1 lessthan 2k, where k is a natural number. Hence t = 2k − 1 = f (k).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 19 / 45
Cardinality of Set
Hilberts Grand Hotel
Grand Hotel with ℵ0 rooms
How can we accommodatea new guest arriving at thefully occupied Grand Ho-tel without removing any ofthe current guests?
Solution
Because the rooms of the Grand Hotel are countable, we can list them asRoom 1, Room 2, Room 3, and so on. When a new guest arrives, wemove the guest in Room 1 to Room 2, the guest in Room 2 to Room 3,and in general, the guest in Room n to Room n + 1, for all positiveintegers n. This frees up Room 1, which we assign to the new guest, andall the current guests still have rooms.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 20 / 45
Cardinality of Set
Hilberts Grand Hotel
Grand Hotel with ℵ0 rooms
How can we accommodatea new guest arriving at thefully occupied Grand Ho-tel without removing any ofthe current guests?
Solution
Because the rooms of the Grand Hotel are countable, we can list them asRoom 1, Room 2, Room 3, and so on. When a new guest arrives, wemove the guest in Room 1 to Room 2, the guest in Room 2 to Room 3,and in general, the guest in Room n to Room n + 1, for all positiveintegers n. This frees up Room 1, which we assign to the new guest, andall the current guests still have rooms.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 20 / 45
Cardinality of Set
Countable sets
Integer set
Show that the set of all integers is countable.
Solution
f (n) =
n/2, n is an even number;−(n − 1)/2, otherwise.
Positive rational numbers
Show that all positive rationalnumbers is countable.
Solution
A rational numbers can belisted in p/q first withp + q = 2, then p + q = 3, · · ·
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 21 / 45
Cardinality of Set
Countable sets
Integer set
Show that the set of all integers is countable.
Solution
f (n) =
n/2, n is an even number;−(n − 1)/2, otherwise.
Positive rational numbers
Show that all positive rationalnumbers is countable.
Solution
A rational numbers can belisted in p/q first withp + q = 2, then p + q = 3, · · ·
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 21 / 45
Cardinality of Set
An uncountable set
Real numbers
Show that the set of real numbers is an uncountable set.
Solution
We suppose that the set of real numbers is countable and arrive ata contradiction.Then, ∀x ∈ (0, 1) would also be countable (because any subset of acountable set is also countable).Under this assumption, ∀x ∈ (0, 1) can be listed in some order, say,r1, r2, r3, · · · Let the decimal representation of these real numbers be
r1 = 0.d11d12d13d14 · · ·r2 = 0.d21d22d23d24 · · ·r3 = 0.d31d32d33d34 · · ·· · ·
where dij ∈ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 22 / 45
Cardinality of Set
An uncountable set
Real numbers
Show that the set of real numbers is an uncountable set.
Solution
We suppose that the set of real numbers is countable and arrive ata contradiction.
Then, ∀x ∈ (0, 1) would also be countable (because any subset of acountable set is also countable).Under this assumption, ∀x ∈ (0, 1) can be listed in some order, say,r1, r2, r3, · · · Let the decimal representation of these real numbers be
r1 = 0.d11d12d13d14 · · ·r2 = 0.d21d22d23d24 · · ·r3 = 0.d31d32d33d34 · · ·· · ·
where dij ∈ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 22 / 45
Cardinality of Set
An uncountable set
Real numbers
Show that the set of real numbers is an uncountable set.
Solution
We suppose that the set of real numbers is countable and arrive ata contradiction.Then, ∀x ∈ (0, 1) would also be countable (because any subset of acountable set is also countable).Under this assumption, ∀x ∈ (0, 1) can be listed in some order, say,r1, r2, r3, · · · Let the decimal representation of these real numbers be
r1 = 0.d11d12d13d14 · · ·r2 = 0.d21d22d23d24 · · ·r3 = 0.d31d32d33d34 · · ·· · ·
where dij ∈ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 22 / 45
Cardinality of Set
An uncountable set Cont’d
Solution
For example, if r1 = 0.23794 · · · , we have d11 = 2, d12 = 3, etc.)
Then, form a new real number r = 0.d1d2d3d4 · · · , where the decimaldigits are determined by the following rule:
di =
4, if dii 6= 4;5, otherwise.
For example, let r1 = 0.23794 · · · , r2 = 0.44590 · · · , r3 =0.09118 · · · , r4 = 0.80553 · · · , etc. Then we have r =0.d1d2d3d4 · · · = 0.4544 · · · , where d1 = 4 since d11 6= 4, d2 = 5since d22 = 4, d3 = 4 since d33 6= 4, d4 = 4 since d44 6= 4, etc.Every real number has a unique decimal expansion. Therefore, r isnot equal to any of r1, r2, · · · because the decimal expansion of rdiffers from the decimal expansion of ri in the i-th place to the rightof the decimal point, for each i .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 23 / 45
Cardinality of Set
An uncountable set Cont’d
Solution
For example, if r1 = 0.23794 · · · , we have d11 = 2, d12 = 3, etc.)Then, form a new real number r = 0.d1d2d3d4 · · · , where the decimaldigits are determined by the following rule:
di =
4, if dii 6= 4;5, otherwise.
For example, let r1 = 0.23794 · · · , r2 = 0.44590 · · · , r3 =0.09118 · · · , r4 = 0.80553 · · · , etc. Then we have r =0.d1d2d3d4 · · · = 0.4544 · · · , where d1 = 4 since d11 6= 4, d2 = 5since d22 = 4, d3 = 4 since d33 6= 4, d4 = 4 since d44 6= 4, etc.
Every real number has a unique decimal expansion. Therefore, r isnot equal to any of r1, r2, · · · because the decimal expansion of rdiffers from the decimal expansion of ri in the i-th place to the rightof the decimal point, for each i .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 23 / 45
Cardinality of Set
An uncountable set Cont’d
Solution
For example, if r1 = 0.23794 · · · , we have d11 = 2, d12 = 3, etc.)Then, form a new real number r = 0.d1d2d3d4 · · · , where the decimaldigits are determined by the following rule:
di =
4, if dii 6= 4;5, otherwise.
For example, let r1 = 0.23794 · · · , r2 = 0.44590 · · · , r3 =0.09118 · · · , r4 = 0.80553 · · · , etc. Then we have r =0.d1d2d3d4 · · · = 0.4544 · · · , where d1 = 4 since d11 6= 4, d2 = 5since d22 = 4, d3 = 4 since d33 6= 4, d4 = 4 since d44 6= 4, etc.Every real number has a unique decimal expansion. Therefore, r isnot equal to any of r1, r2, · · · because the decimal expansion of rdiffers from the decimal expansion of ri in the i-th place to the rightof the decimal point, for each i .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 23 / 45
Cardinality of Set
Results about cardinality
Theorem
If A and B are countable sets, then A ∪ B is also countable.
Proof.
Without loss of generality, we can assume that A and B aredisjoint. (If they are not, we can replace B by B − A, becauseA ∩ (B − A) = ∅ and A ∪ (B − A) = A ∪ B.)
C1: A and B are both finite;
C2: A is infinite and B is finite;
C3: A and B are both countably infinite.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 24 / 45
Cardinality of Set
Results about cardinality
Theorem
If A and B are countable sets, then A ∪ B is also countable.
Proof.
Without loss of generality, we can assume that A and B aredisjoint. (If they are not, we can replace B by B − A, becauseA ∩ (B − A) = ∅ and A ∪ (B − A) = A ∪ B.)
C1: A and B are both finite;
C2: A is infinite and B is finite;
C3: A and B are both countably infinite.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 24 / 45
Cardinality of Set
Results about cardinality
Theorem
If A and B are countable sets, then A ∪ B is also countable.
Proof.
Without loss of generality, we can assume that A and B aredisjoint. (If they are not, we can replace B by B − A, becauseA ∩ (B − A) = ∅ and A ∪ (B − A) = A ∪ B.)
C1: A and B are both finite;
C2: A is infinite and B is finite;
C3: A and B are both countably infinite.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 24 / 45
Cardinality of Set
SCHRODER-BERNSTEIN theorem and the continuumhypothesis
Theorem
If A and B are sets with |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|,i.e., if there are one-to-one functions f from A to B and g from Bto A, then there is a one-to-one correspondence between A and B.
Hypothesis
Show that the power set of Z+ and R have the same cardinality, i.e.,|P(Z+)| = |R| = c .Furthermore, |P(Z+)| = |R| can be expressed as 2ℵ0 = c .
Cantor theorem
The cardinality of a set is always less than the cardinality of its powerset, i.e., |Z+| < |P(Z+)|.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 25 / 45
Cardinality of Set
SCHRODER-BERNSTEIN theorem and the continuumhypothesis
Theorem
If A and B are sets with |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|,i.e., if there are one-to-one functions f from A to B and g from Bto A, then there is a one-to-one correspondence between A and B.
Hypothesis
Show that the power set of Z+ and R have the same cardinality, i.e.,|P(Z+)| = |R| = c .Furthermore, |P(Z+)| = |R| can be expressed as 2ℵ0 = c .
Cantor theorem
The cardinality of a set is always less than the cardinality of its powerset, i.e., |Z+| < |P(Z+)|.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 25 / 45
Cardinality of Set
SCHRODER-BERNSTEIN theorem and the continuumhypothesis
Theorem
If A and B are sets with |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|,i.e., if there are one-to-one functions f from A to B and g from Bto A, then there is a one-to-one correspondence between A and B.
Hypothesis
Show that the power set of Z+ and R have the same cardinality, i.e.,|P(Z+)| = |R| = c .Furthermore, |P(Z+)| = |R| can be expressed as 2ℵ0 = c .
Cantor theorem
The cardinality of a set is always less than the cardinality of its powerset, i.e., |Z+| < |P(Z+)|.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 25 / 45
Matrix
Linear equations and matrix
Linear equations
The subject of algebra arose from studying equations.
If x1, x2, · · · , xn are variables and a1, a2, · · · , an and c are constant,then the equation a1x1 + a2x2 + · · ·+ anxn = c is said to be a linearequation, where ai are the coefficient.
More generally, a linear system consisting of m equations in nunknowns will look like:
a11x1 + a12x2 + · · ·+ a1nxn = c1
a21x1 + a22x2 + · · ·+ a2nxn = c2
· · ·am1x1 + am2x2 + · · ·+ amnxn = cm
The main problem is to find the solution set of a linear system(Gaussian reduction). While that is not the focus of this course
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 26 / 45
Matrix
Vector
An n-tuple (pair, triple, quadruple, ...) of scalars can be written as ahorizontal row or vertical column. A column is called a vector. For
example x =
x1x2· · ·xn
and xT = [x1, x2, · · · , xn]
Operations
Addition: x + y = [x1 + y1, x2 + y2, · · · , xn + yn]
x + y = y + x
cx = [cx1, cx2, · · · , cxn] and 0x =−→0
x + (y + z) = (x + y) + z
Manipulation: xT y =∑n
i=1 xiyixT y = yT x(x + y)T z = xT z + yT z
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 27 / 45
Matrix
Vector
An n-tuple (pair, triple, quadruple, ...) of scalars can be written as ahorizontal row or vertical column. A column is called a vector. For
example x =
x1x2· · ·xn
and xT = [x1, x2, · · · , xn]
Operations
Addition: x + y = [x1 + y1, x2 + y2, · · · , xn + yn]
x + y = y + x
cx = [cx1, cx2, · · · , cxn] and 0x =−→0
x + (y + z) = (x + y) + z
Manipulation: xT y =∑n
i=1 xiyixT y = yT x(x + y)T z = xT z + yT z
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 27 / 45
Matrix
Examples of vector
Examples
Entity: an entity can be modeled as a vector x = [x1, x2, · · · , xn],where n denotes the number of features and xi denotes the value ofthe i−th feature. For example, patients, email, student, and user, etc.
Set: given a universal set, a subset of the universal set can be modelas a binary vector x = [0, 1, 1, · · · , 0]T , where the dimension of X isthe size of universal set and xi = 1 means that the i−th item exists inthe set. For example, document VS. words, vertex VS. neighbors,string VS. n-gram, user VS. products, etc.
Distribution: given a discrete sample space Ω, PMF can be modeledas a vector p = [p1, p2, · · · , pn], where n is the cardinality of Ω,0 ≤ pi ≤ 1 and
∑ni=1 pi = 1. For example, document VS. topics,
Markov chain VS. states, tweet VS. polarity, entity VS. classes, etc.
Latent vector: matrix factorization, topic modeling, and wordembedding, etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 28 / 45
Matrix
Examples of vector
Examples
Entity: an entity can be modeled as a vector x = [x1, x2, · · · , xn],where n denotes the number of features and xi denotes the value ofthe i−th feature. For example, patients, email, student, and user, etc.
Set: given a universal set, a subset of the universal set can be modelas a binary vector x = [0, 1, 1, · · · , 0]T , where the dimension of X isthe size of universal set and xi = 1 means that the i−th item exists inthe set. For example, document VS. words, vertex VS. neighbors,string VS. n-gram, user VS. products, etc.
Distribution: given a discrete sample space Ω, PMF can be modeledas a vector p = [p1, p2, · · · , pn], where n is the cardinality of Ω,0 ≤ pi ≤ 1 and
∑ni=1 pi = 1. For example, document VS. topics,
Markov chain VS. states, tweet VS. polarity, entity VS. classes, etc.
Latent vector: matrix factorization, topic modeling, and wordembedding, etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 28 / 45
Matrix
Examples of vector
Examples
Entity: an entity can be modeled as a vector x = [x1, x2, · · · , xn],where n denotes the number of features and xi denotes the value ofthe i−th feature. For example, patients, email, student, and user, etc.
Set: given a universal set, a subset of the universal set can be modelas a binary vector x = [0, 1, 1, · · · , 0]T , where the dimension of X isthe size of universal set and xi = 1 means that the i−th item exists inthe set. For example, document VS. words, vertex VS. neighbors,string VS. n-gram, user VS. products, etc.
Distribution: given a discrete sample space Ω, PMF can be modeledas a vector p = [p1, p2, · · · , pn], where n is the cardinality of Ω,0 ≤ pi ≤ 1 and
∑ni=1 pi = 1. For example, document VS. topics,
Markov chain VS. states, tweet VS. polarity, entity VS. classes, etc.
Latent vector: matrix factorization, topic modeling, and wordembedding, etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 28 / 45
Matrix
Examples of vector
Examples
Entity: an entity can be modeled as a vector x = [x1, x2, · · · , xn],where n denotes the number of features and xi denotes the value ofthe i−th feature. For example, patients, email, student, and user, etc.
Set: given a universal set, a subset of the universal set can be modelas a binary vector x = [0, 1, 1, · · · , 0]T , where the dimension of X isthe size of universal set and xi = 1 means that the i−th item exists inthe set. For example, document VS. words, vertex VS. neighbors,string VS. n-gram, user VS. products, etc.
Distribution: given a discrete sample space Ω, PMF can be modeledas a vector p = [p1, p2, · · · , pn], where n is the cardinality of Ω,0 ≤ pi ≤ 1 and
∑ni=1 pi = 1. For example, document VS. topics,
Markov chain VS. states, tweet VS. polarity, entity VS. classes, etc.
Latent vector: matrix factorization, topic modeling, and wordembedding, etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 28 / 45
Matrix
How far from two vectors
Distance
Distance is a numerical description of how far apart vectors are. Given twovectors x and y , there are many ways to measure distance of two vectors.
Distance in Euclidean space: the Minkowski distance of order p(p-norm distance) is defined as
1-norm distance (Manhattan distance):∑n
i=1 |xi − yi |2-norm distance (Euclidean distance):
(∑ni=1(xi − yi )
2)1/2
p-norm distance:(∑n
i=1(xi − yi )p)1/p
Infinity norm distance:
limp→∞(∑n
i=1(xi − yi )p)1/p
= maxxi − yi |i = 1, 2, · · · , nMahalanobis distance: it is defined as a dissimilarity between tworandom vectors X and Y of the same distribution with covariancematrix S : D(X ,Y ) =
((X − Y )TS−1(X − Y )
)1/2. If S = I , the
Mahalanobis distance reduces to the Euclidean distance
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 29 / 45
Matrix
Matrix
Definition
An m× n matrix A = (aij) (1 ≤ i ≤ m, 1 ≤ j ≤ n) is a rectangular array ofmn scalars in m rows and n columns, such as
A =
a11 a12 · · · a1na21 a22 · · · a2n· · · · · · · · · · · ·am1 am2 · · · amn
The i-th row of A is the 1× n matrix [ai1, ai2, · · · , ain]. The j−th column
of A is the m × 1 matrix
a1ja2j· · ·amj
.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 30 / 45
Matrix
Matrix
Concepts
Boldface uppercase letters will be used to represent matrices;
The set of all m × n matrices with real entries will be denoted byRm×n;
A is called the identity matrix if
aij =
1, if i = j ;0, otherwise.
If m = n, A is called a square matrix;
AT = (aji )(∈ Rn×m) is a n ×m matrix;
If A is a n × n square matrix, Trace(A) =∑n
i aii = Trace(AT );
AIn = ImA = A;
A square matrix A is called symmetric if A = AT , i.e., aij = aji for ∀iand ∀j .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 31 / 45
Matrix
Examples of matrix
Homogeneous graph
Vertex can be Web page, user, protein,road, route, etc. Edge can be directed,undirected, weighted or labeled.
Heterogeneous graph
Vertex can be user-Web page,user-product, user-service, user-paper,etc.
Signed graph
user friend and enemy networks, likeand dislike networks, etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 32 / 45
Matrix
Examples of matrix Cont.
Image processing
color, texture and shape, etc
Feature representation
Many applications can be modeled inthis manner, such as search engine,email classification, disease diagnosis,churn predication, anomaly detection,etc.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 33 / 45
Matrix
Operations of matrix
Operations
Addition: Let A = [aij ] and B = [bij ] be m × n matrices,A + B = (aij + bij)
A + cB = (aij + cbij), especially A− B = (aij − bij), where c is a scalarA + B = B + AA + (B + C ) = (A + B) + C
Manipulation: If A is an n×m matrix and B is an m× p matrix, thenAB = (
∑mk=1 aikbkj)
Not commutative: AB 6= BADistributive over matrix addition: (A + B)C = AC + BCScalar multiplication is compatible with matrix multiplication:λAB = (λA)B = A(λB)(AB)T = BTAT
Trace(AB) = Trace(BA) andTrace(ABCD) = Trace(BCDA) = Trace(CDAB) = Trace(DABC ). Ingeneral, Trace(ABC ) 6= Trace(ACB)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 34 / 45
Matrix
Power of matrices
Definition
Let A be an n × n matrix, we have
A0 = In,Ak = AA · · ·A︸ ︷︷ ︸
k times
Definition of diagonalizable
A square matrix A is said to be diagonalizable if it is similar to a diagonalmatrix, i.e., there exists an invertible matrix P and a diagonal matrix Dsuch that A = PDP−1.
Why useful? If A is diagonalizable, then Ak = PDkP−1 for k > 0.
How to find diagonal matrix? If v1, · · · , vn are linearly independenteigenvectors of A and λi are their corresponding eignevalues, thenA = PDP−1, where P = [v1 · · · vn] and D = Diag(λ1, · · · , λn).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 35 / 45
Matrix
Power of matrices
Definition
Let A be an n × n matrix, we have
A0 = In,Ak = AA · · ·A︸ ︷︷ ︸
k times
Definition of diagonalizable
A square matrix A is said to be diagonalizable if it is similar to a diagonalmatrix, i.e., there exists an invertible matrix P and a diagonal matrix Dsuch that A = PDP−1.
Why useful? If A is diagonalizable, then Ak = PDkP−1 for k > 0.
How to find diagonal matrix? If v1, · · · , vn are linearly independenteigenvectors of A and λi are their corresponding eignevalues, thenA = PDP−1, where P = [v1 · · · vn] and D = Diag(λ1, · · · , λn).
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 35 / 45
Matrix
Matrix inverseDefinition
Suppose two n × n matrices A and B have the property that AB =BA = In. Then we say A is an inverse of B (and B is an inverse ofA). Matrix A is invertible if A has inverse.
Properties
Suppose A ∈ Rn×n and A has an inverse B. Then B is unique.
A ∈ Rn×n is nonsingular (i.e., has rank n), if and only if there exists amatrix B ∈ Rn×n such that BA = In.
If A is nonsingular, then the system Ax = b has the unique solutionx = A−1b.
(A|In)→ (In|B) (row reduce), then B is the inverse of A.
(A|b)→ (In|c) (row reduce), where the components of c are certainlinear combinations of the components of b. The coefficients in theselinear combinations give us the entries of A−1.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 36 / 45
Matrix derivatives
Matrix derivatives
Type scalar vector matrix
scalar ∂y∂x
∂y∂x
∂Y∂x
vector ∂y∂x
∂y∂x
matrix ∂y∂X
Derivatives by scalar
Assume that x , y , a ∈ R, x ∈ Rn×1, X ,Y ,A ∈ Rn×m, and y, a ∈Rm×1. Let a, a and A be constant scalar, vector and matrix.
∂y∂x and ∂a
∂x = 0
∂y∂x =
∂y1∂x· · ·∂ym∂x
and ∂a∂x = 0 (vector)
∂Y∂x =
∂y11∂x · · · ∂y1m
∂x· · · · · · · · ·∂yn1∂x · · · ∂ynm
∂x
and ∂A∂x = 0 (matrix)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 37 / 45
Matrix derivatives
Matrix derivatives
Type scalar vector matrix
scalar ∂y∂x
∂y∂x
∂Y∂x
vector ∂y∂x
∂y∂x
matrix ∂y∂X
Derivatives by scalar
Assume that x , y , a ∈ R, x ∈ Rn×1, X ,Y ,A ∈ Rn×m, and y, a ∈Rm×1. Let a, a and A be constant scalar, vector and matrix.
∂y∂x and ∂a
∂x = 0
∂y∂x =
∂y1∂x· · ·∂ym∂x
and ∂a∂x = 0 (vector)
∂Y∂x =
∂y11∂x · · · ∂y1m
∂x· · · · · · · · ·∂yn1∂x · · · ∂ynm
∂x
and ∂A∂x = 0 (matrix)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 37 / 45
Matrix derivatives
Matrix derivatives Cont.
Derivatives by vector
∂y∂x =
∂y∂x1· · ·∂y∂xn
T
and ∂a∂x = 0T (vector)
∂y∂x =
∂y1∂x1
· · · ∂y1∂xn
· · · · · · · · ·∂ym∂x1
· · · ∂ym∂xn
, ∂a∂x = 0 (matr.) and ∂x
∂x = I (matr.)
Derivatives by matrix
∂y∂X =
∂y∂x11
· · · ∂y∂xn1
· · · · · · · · ·∂y∂x1m
· · · ∂y∂xnm
and ∂a∂X = 0T (matrix)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 38 / 45
Matrix derivatives
Matrix derivatives Cont.
Derivatives by vector
∂y∂x =
∂y∂x1· · ·∂y∂xn
T
and ∂a∂x = 0T (vector)
∂y∂x =
∂y1∂x1
· · · ∂y1∂xn
· · · · · · · · ·∂ym∂x1
· · · ∂ym∂xn
, ∂a∂x = 0 (matr.) and ∂x
∂x = I (matr.)
Derivatives by matrix
∂y∂X =
∂y∂x11
· · · ∂y∂xn1
· · · · · · · · ·∂y∂x1m
· · · ∂y∂xnm
and ∂a∂X = 0T (matrix)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 38 / 45
Matrix derivatives
Common properties of matrix derivatives
Properties
c1 ∂aT x∂x = ∂xT a
∂x = aT
c2 ∂xT x∂x = 2xT
c3 ∂(xT a)2
∂x = 2xTaaT
c4 ∂Ax∂x = A and ∂xTA
∂x = AT
c5 ∂xTAx∂x = xT (A + AT )
Proof c2. Let s = xTx =∑n
i=1 x2i . Then, ∂s
∂xi= 2xi . So, ∂xT x
∂x =
2xT .c3. Let s = xTa. Then ∂s2
∂xi= 2s ∂s
∂xi= 2sai . Thus, ∂(xT a)2
∂x =
2xTaaT .
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 39 / 45
Matrix derivatives
Properties of matrix derivatives cont.Properties for scalar by scaler
ss1 ∂(u+v)∂x = ∂u
∂x + ∂v∂x
ss2 ∂uv∂x = u ∂v
∂x + v ∂u∂x (product rule)
ss3 ∂g(u)∂x = ∂g(u)
∂u∂u∂x (chain rule)
ss4 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
Properties for vector by scaler
vs1 ∂(au+v)∂x = a ∂u
∂x + ∂v∂x , where a is not a function of x .
vs2 ∂Au∂x = A∂u
∂x where A is not a function of x .
vs3 ∂uT
∂x =(∂u∂x )T
vs4 ∂g(u)∂x = ∂g(u)
∂u∂u∂x (chain rule)
vs5 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 40 / 45
Matrix derivatives
Properties of matrix derivatives cont.Properties for scalar by scaler
ss1 ∂(u+v)∂x = ∂u
∂x + ∂v∂x
ss2 ∂uv∂x = u ∂v
∂x + v ∂u∂x (product rule)
ss3 ∂g(u)∂x = ∂g(u)
∂u∂u∂x (chain rule)
ss4 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
Properties for vector by scaler
vs1 ∂(au+v)∂x = a ∂u
∂x + ∂v∂x , where a is not a function of x .
vs2 ∂Au∂x = A∂u
∂x where A is not a function of x .
vs3 ∂uT
∂x =(∂u∂x )T
vs4 ∂g(u)∂x = ∂g(u)
∂u∂u∂x (chain rule)
vs5 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 40 / 45
Matrix derivatives
Properties of matrix derivatives cont.Properties for matrix by scaler
ms1 ∂aU∂x = a∂U
∂x , where a is not a function of x .
ms2 ∂AUB∂x = A∂U
∂x B where A and B are not a function of x .
ms3 ∂(U+V )∂x = ∂U
∂x + ∂V∂x
ms4 ∂UV∂x = U ∂V
∂x + ∂U∂x V (product rule)
Properties for scalar by vector
sv1 ∂(au+v)∂x = a∂u
∂x + ∂v∂x , where a is not a function of x .
sv2 ∂uv∂x = u ∂v
∂x + v ∂u∂x (product rule)
sv3 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
sv4 ∂uT v∂x = uT ∂v
∂x + vT ∂u∂x (product rule)
sv5 ∂uTAv∂x = uTA∂v
∂x + vTAT ∂u∂x , where A is not a function of x
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 41 / 45
Matrix derivatives
Properties of matrix derivatives cont.Properties for matrix by scaler
ms1 ∂aU∂x = a∂U
∂x , where a is not a function of x .
ms2 ∂AUB∂x = A∂U
∂x B where A and B are not a function of x .
ms3 ∂(U+V )∂x = ∂U
∂x + ∂V∂x
ms4 ∂UV∂x = U ∂V
∂x + ∂U∂x V (product rule)
Properties for scalar by vector
sv1 ∂(au+v)∂x = a∂u
∂x + ∂v∂x , where a is not a function of x .
sv2 ∂uv∂x = u ∂v
∂x + v ∂u∂x (product rule)
sv3 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
sv4 ∂uT v∂x = uT ∂v
∂x + vT ∂u∂x (product rule)
sv5 ∂uTAv∂x = uTA∂v
∂x + vTAT ∂u∂x , where A is not a function of x
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 41 / 45
Matrix derivatives
Properties of matrix derivatives cont.
Properties for scalar by matrix
sm1 ∂au∂X = a ∂u
∂X , where a is not a function of x .
sm2 ∂(u+v)∂X = ∂u
∂X + ∂v∂X
sm3 ∂uv∂X = u ∂v
∂X + v ∂u∂X (product rule)
sm4 ∂f (g(u))∂X = ∂f (g)
∂g∂g(u)∂u
∂u∂X (chain rule)
Properties for vector by vector
vv1 ∂(au+v)∂x = a∂u
∂x + ∂v∂x , where a is not a function of x .
vv2 ∂Au∂x = A∂u
∂x , where A is not a function of x.
vv3 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 42 / 45
Matrix derivatives
Properties of matrix derivatives cont.
Properties for scalar by matrix
sm1 ∂au∂X = a ∂u
∂X , where a is not a function of x .
sm2 ∂(u+v)∂X = ∂u
∂X + ∂v∂X
sm3 ∂uv∂X = u ∂v
∂X + v ∂u∂X (product rule)
sm4 ∂f (g(u))∂X = ∂f (g)
∂g∂g(u)∂u
∂u∂X (chain rule)
Properties for vector by vector
vv1 ∂(au+v)∂x = a∂u
∂x + ∂v∂x , where a is not a function of x .
vv2 ∂Au∂x = A∂u
∂x , where A is not a function of x.
vv3 ∂f (g(u))∂x = ∂f (g)
∂g∂g(u)∂u
∂u∂x (chain rule)
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 42 / 45
Matrix derivatives
Zero-one matrices
Definition
A matrix all of whose entries are either 0 or 1 is called a zero-onematrix.
Operations
∧: Let A = [aij ] and B = [bij ] be m × n zero-one matrices,A ∧ B = [aij ∧ bij ], i.e.,
(A ∧ B)ij =
1, if aij = bij = 1;0, otherwise.
∨: Let A = [aij ] and B = [bij ] be m × n zero-one matrices,A ∨ B = [aij ∨ bij ], i.e.,
(A ∨ B)ij =
1, if aij = 1 or bij = 1;0, otherwise.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 43 / 45
Matrix derivatives
Zero-one matrices
Definition
A matrix all of whose entries are either 0 or 1 is called a zero-onematrix.
Operations
∧: Let A = [aij ] and B = [bij ] be m × n zero-one matrices,A ∧ B = [aij ∧ bij ], i.e.,
(A ∧ B)ij =
1, if aij = bij = 1;0, otherwise.
∨: Let A = [aij ] and B = [bij ] be m × n zero-one matrices,A ∨ B = [aij ∨ bij ], i.e.,
(A ∨ B)ij =
1, if aij = 1 or bij = 1;0, otherwise.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 43 / 45
Matrix derivatives
Boolean product
Definition
If A is an n×m zero-one matrix and B is an m× p zero-one matrix,then A
⊙B = [cij ] =
∨mk=1(aik ∧ bkj).
Example
A =
1 00 11 0
,B =
(1 1 00 1 1
)Solution
A⊙
B = (1 ∧ 1) ∨ (0 ∧ 0) (1 ∧ 1) ∨ (0 ∧ 1) (1 ∧ 0) ∨ (0 ∧ 1)(0 ∧ 1) ∨ (1 ∧ 0) (0 ∧ 1) ∨ (1 ∧ 1) (0 ∧ 0) ∨ (1 ∧ 1)(1 ∧ 1) ∨ (0 ∧ 0) (1 ∧ 1) ∨ (0 ∧ 1) (1 ∧ 0) ∨ (0 ∧ 1)
=
1 1 00 1 11 1 0
.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 44 / 45
Matrix derivatives
Boolean product
Definition
If A is an n×m zero-one matrix and B is an m× p zero-one matrix,then A
⊙B = [cij ] =
∨mk=1(aik ∧ bkj).
Example
A =
1 00 11 0
,B =
(1 1 00 1 1
)Solution
A⊙
B = (1 ∧ 1) ∨ (0 ∧ 0) (1 ∧ 1) ∨ (0 ∧ 1) (1 ∧ 0) ∨ (0 ∧ 1)(0 ∧ 1) ∨ (1 ∧ 0) (0 ∧ 1) ∨ (1 ∧ 1) (0 ∧ 0) ∨ (1 ∧ 1)(1 ∧ 1) ∨ (0 ∧ 0) (1 ∧ 1) ∨ (0 ∧ 1) (1 ∧ 0) ∨ (0 ∧ 1)
=
1 1 00 1 11 1 0
.
MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Apr. 3, 2020 44 / 45
Take-aways
Take-aways
Conclusions
Sequence and summations
Cardinality of set
Matrix
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