Lecture 1 C2

Mathematical Analysis I, a.a.
2024/25, 10 CFU
Politecnico di Torino
Department of Mathematical Sciences DISMA
General information
Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

General information
Course coordinator: Patrik Wahlberg

General information
Teachers: Patrik Wahlberg (lectures), Enrico Carlini (exercise classes)

General information
Literature: Mathematical Analysis I, C. Canuto, A. Tabacco, Pearson
2022

General information
2022
Analisi Matematica I, F. Nicola, CLUT 2020

General information
2022
Material: Portale della Didattica

General information
2022
Two courses: Course 2 FAV–OZJ, Course 3 OZK–ZZZ

General information
2022
This is Course 2

General information
2022
This is Course 2
Weekly timetable

General information
2022
This is Course 2
Weekly timetable
1. Monday 8:30–10:00 exercise group A

General information
2022
This is Course 2
Weekly timetable
2. Monday 10:00–11:30 exercise group B

General information
2022
This is Course 2
Weekly timetable
3. Tuesday 13:00–14:30 lecture

General information
2022
This is Course 2
Weekly timetable
4. Thursday 16:00–19:00 lecture

General information
2022
This is Course 2
Weekly timetable
4. Thursday 16:00–19:00 lecture
5. Friday 8:30–10:00 exercise for both group A and group B
Please note: Holidays on Friday 1/11 and on Monday 6/1.

Please note: Holidays on Friday 1/11 and on Monday 6/1.
Group A: surname FAV–KHU

Group B: surname KHV–OZJ

Exam

Exam
Four occasions:

Exam
Four occasions:
20–22 January

Exam
Four occasions:
20–22 January
6–7 February

Exam
Four occasions:
20–22 January
6–7 February
12–13 June (to be confirmed)

Exam
Four occasions:
20–22 January
6–7 February
September

Exam
Four occasions:
20–22 January
6–7 February
September
Registration necessary. Deadline ∼ one week before exam

Exam
Four occasions:
20–22 January
6–7 February
September
Structure of the exam:

Exam
Four occasions:
20–22 January
6–7 February
September

1 “Ongoing activities” during the course: max 3 credits

Exam
Four occasions:
20–22 January
6–7 February
September

2 Multiple choice test (45 min), 15 questions. Max 15 credits

Exam
Four occasions:
20–22 January
6–7 February
September

3 If pass (score ⩾ 8) on 2: written exam (90 minutes). Max 15 credits

Exam
Four occasions:
20–22 January
6–7 February
September

4 Pass: ⩾ 8 credits on the written exam, and sum ⩾ 18

Exam
Four occasions:
20–22 January
6–7 February
September

4 Pass: ⩾ 8 credits on the written exam, and sum ⩾ 18
5 Oral exam: optional

Ongoing activities (optional)

Ongoing activities (optional)
Four tests:
1 Test 1: Monday 4 November
2 Test 2: Monday 25 November
3 Test 3: Monday 9 December
4 Test 4: Monday 6 January

Contents of the course

Mathematics is an indispensible tool for a scientific study of
engineering, physics, economy, biology, medicine, . . .

Analysis is the part of mathematics that concerns quantities that
may vary continuously, e.g. the space coordinate of a particle as a
function of time

function of time
The real line R is a mathematical model of time and of space

function of time
1 Set theory and logic

function of time
2 Numerical sets

function of time
2 Numerical sets
3 Functions

function of time
2 Numerical sets
3 Functions
4 Limits and continuity

function of time
2 Numerical sets
3 Functions
5 Differential calculus

function of time
2 Numerical sets
3 Functions
6 Taylor expansions

function of time
2 Numerical sets
3 Functions
6 Taylor expansions
7 Integral calculus

function of time
2 Numerical sets
3 Functions
6 Taylor expansions
7 Integral calculus
8 Complex numbers

function of time
2 Numerical sets
3 Functions
6 Taylor expansions
7 Integral calculus
8 Complex numbers
9 Differential equations
Set theory

Set theory
If X is a set then
x ∈X
means that the element x belongs to X .

Set theory
If X is a set then
x ∈X
Example: 34 ∈ N where
N = {0, 1, 2, . . . }
is the set of natural numbers.

Set theory
If X is a set then
x ∈X
N = {0, 1, 2, . . . }
Example: π ∈
/N

Set theory
If X is a set then
x ∈X
N = {0, 1, 2, . . . }
Example: π ∈
/N
A subset A ⊆ X is a set of elements in X .

Set theory
If X is a set then
x ∈X
N = {0, 1, 2, . . . }
Example: π ∈
/N

Two ways:
A = {x1 , x2 , . . . , xn }

Set theory
If X is a set then
x ∈X
N = {0, 1, 2, . . . }
Example: π ∈
/N

Two ways:
A = {x1 , x2 , . . . , xn }
or
A = {x ∈ X : p(x)}
where p(x) is a property of x.
Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }
For a proper subset A ⊂ X it holds A ̸= X .

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

The empty set ∅ ⊆ X contains no elements.

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

The power set P(X ) is the set of all subsets of X .

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

Example: X = {1, 3}

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

Example: X = {1, 3}
P(X ) = {∅, {1}, {3}, {1, 3} = X }

Examples:
A = {n ∈ N : n ⩽ 7} = {0, 1, . . . , 7}
B = {n ∈ N : n > 2} = {3, 4, . . . }

Example: X = {1, 3}
P(X ) = {∅, {1}, {3}, {1, 3} = X }
If X has n elements (cardinality) then P(X ) has 2n elements.

Complement of A ⊆ X :
CA = {x ∈ X : x ∈
/ A}

CA = {x ∈ X : x ∈
/ A}
CX = ∅ C∅ = X C(CA) = A

CA = {x ∈ X : x ∈
/ A}
CX = ∅ C∅ = X C(CA) = A
Intersection of A ⊆ X and B ⊆ X :
A ∩ B = {x ∈ X : x ∈ A and x ∈ B}

CA = {x ∈ X : x ∈
/ A}
CX = ∅ C∅ = X C(CA) = A
Intersection of A ⊆ X and B ⊆ X :
A ∩ B = {x ∈ X : x ∈ A and x ∈ B}
Union of A ⊆ X and B ⊆ X :
A ∪ B = {x ∈ X : x ∈ A or x ∈ B}

Properties:
A∩B =B ∩A A∪B =B ∪A
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
Take x ∈ (A ∩ B) ∪ C .

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
Take x ∈ (A ∩ B) ∪ C . Then x ∈ A ∩ B or x ∈ C .

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
Take x ∈ (A ∩ B) ∪ C . Then x ∈ A ∩ B or x ∈ C . Thus x ∈ A ∪ C and
x ∈ B ∪ C.

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
x ∈ B ∪ C . This shows x ∈ (A ∪ C ) ∩ (B ∪ C ).

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
x ∈ B ∪ C . This shows x ∈ (A ∪ C ) ∩ (B ∪ C ). We have shown
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
Take x ∈ (A ∪ C ) ∩ (B ∪ C ).

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
Take x ∈ (A ∪ C ) ∩ (B ∪ C ). Then x ∈ A ∪ C and x ∈ B ∪ C .

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
Take x ∈ (A ∪ C ) ∩ (B ∪ C ). Then x ∈ A ∪ C and x ∈ B ∪ C . Either
x ∈ C.

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
x ∈ C . If x ∈
/ C then x ∈ A ∩ B.

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
x ∈ C . If x ∈
/ C then x ∈ A ∩ B. Thus x ∈ (A ∩ B) ∪ C .

Properties:
(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )
(A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C )
Proof: 1. (A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
(A ∩ B) ∪ C ⊆ (A ∪ C ) ∩ (B ∪ C )
2. (A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
x ∈ C . If x ∈
/ C then x ∈ A ∩ B. Thus x ∈ (A ∩ B) ∪ C . We have shown
(A ∪ C ) ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ C
(A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C )

(A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C )
De Morgan laws:
C(A ∩ B) = CA ∪ CB

(A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C )
De Morgan laws:
C(A ∩ B) = CA ∪ CB
C(A ∪ B) = CA ∩ CB

(A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C )
De Morgan laws:
C(A ∩ B) = CA ∪ CB
C(A ∪ B) = CA ∩ CB
Relative complement:
A \ B = A ∩ CB

Logic
A formula p is true or false.

Logic
Example: p = “2 > 1” is true, p = “1 > 2” is false.

Logic
The negation ¬p of p has opposite truth value to p.

Logic
Conjunction: p and q
p∧q

Logic
p∧q
Disjunction: p or q
p∨q

Logic
p∧q
Disjunction: p or q
p∨q
Implication:
P⇒Q

Logic
p∧q
Disjunction: p or q
p∨q
Implication:
P⇒Q
If P holds then Q holds.

Logic
p∧q
Disjunction: p or q
p∨q
Implication:
P⇒Q
If P holds then Q holds.

Example: P = “n ⩾ 5”, Q = “n ⩾ 0”
Equivalence: P ⇒ Q and Q ⇒ P
P⇔Q

P⇔Q
Contrapositive implication:
P⇒Q ⇔ ¬Q ⇒ ¬P

P⇔Q
P⇒Q ⇔ ¬Q ⇒ ¬P
Proof of P ⇒ Q by contradiction:
P⇒Q ⇔ (P ∧ ¬Q) ⇒ ¬P

P⇔Q
P⇒Q ⇔ ¬Q ⇒ ¬P
Proof of P ⇒ Q by contradiction:
P⇒Q ⇔ (P ∧ ¬Q) ⇒ ¬P
or
P⇒Q ⇔ (P ∧ ¬Q) ⇒ r ∧ ¬r
for some formula r

Quantifiers:
∀x ∈ X : p(x)

Quantifiers:
∀x ∈ X : p(x)
reads “for all x ∈ X , p(x) holds”

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)
reads “there exists x ∈ X such that p(x) holds”

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)
Examples:
∀n ∈ N : 2n is an even number

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)
Examples:
∃n ∈ N : n > 273

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)
Examples:
∃n ∈ N : n > 273
¬(∀x ∈ X : p(x)) ⇐⇒ ∃x ∈ X : ¬p(x)

Quantifiers:
∀x ∈ X : p(x)
∃x ∈ X : p(x)
Examples:
∃n ∈ N : n > 273
¬(∀x ∈ X : p(x)) ⇐⇒ ∃x ∈ X : ¬p(x)
∀x ∈ N : ∃y ∈ N : x < y true
∃y ∈ N : ∀x ∈ N : x < y false

Sets of numbers

Sets of numbers
The set of natural numbers
N = {0, 1, 2, . . . }

Sets of numbers
N = {0, 1, 2, . . . }
The set of integers

Z = {0, ±1, ±2, . . . }

Sets of numbers
N = {0, 1, 2, . . . }
The set of integers

Z = {0, ±1, ±2, . . . }
The set of rational numbers

p
Q= : p ∈ Z, q ∈ N+
q

Sets of numbers
N = {0, 1, 2, . . . }
The set of integers

Z = {0, ±1, ±2, . . . }

p
Q= : p ∈ Z, q ∈ N+
q
p
Decimal expansions for q ∈ Q:

Sets of numbers
N = {0, 1, 2, . . . }
The set of integers

Z = {0, ±1, ±2, . . . }

p
Q= : p ∈ Z, q ∈ N+
q
p
Decimal expansions for q ∈ Q: Either it is finite
p
= 1.134123
q

Sets of numbers
N = {0, 1, 2, . . . }
The set of integers

Z = {0, ±1, ±2, . . . }

p
Q= : p ∈ Z, q ∈ N+
q
p
Decimal expansions for ∈ Q: Either it is finite
q
p
= 1.134123
q
or it is infinite and eventually periodic
p z}|{ z}|{ z}|{
= 2.17845 321 321 321 · · ·
q
Decimal expansions are not unique.

1 = 0.9999 · · ·

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2∈
/Q

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2∈
/Q
Proof.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2 ∈ /Q
Proof. By contradiction.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2 ∈ /Q √ p
Proof. By contradiction. Suppose 2 = q where p ∈ N and q ∈ N+ have
no common factors.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2 ∈ /Q √ p
Proof. By contradiction. Suppose 2 = q where p ∈ N and q ∈ N+ have
no common factors. Then p 2 = 2q 2 .

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
Theorem: 2 ∈ /Q √
Proof. By contradiction. Suppose 2 = qp where p ∈ N and q ∈ N+ have
no common factors. Then p 2 = 2q 2 . Thus p 2 is even,

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
no common factors. Then p 2 = 2q 2 . Thus p 2 is even, and p cannot be
odd:

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
odd: (2k + 1)2 = 4k 2 + 4k + 1 is odd.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
odd: (2k + 1)2 = 4k 2 + 4k + 1 is odd. Hence p = 2k is even.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
odd: (2k + 1)2 = 4k 2 + 4k + 1 is odd. Hence p = 2k is even. This gives
4k 2 = 2q 2 that is q 2 = 2k 2 .

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
4k 2 = 2q 2 that is q 2 = 2k 2 . Thus q 2 is even

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
4k 2 = 2q 2 that is q 2 = 2k 2 . Thus q 2 is even ⇒ q = 2n is even.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
4k 2 = 2q 2 that is q 2 = 2k 2 . Thus q 2 is even ⇒ q = 2n is even. This
contradicts the assumption of no common factors in p and q.

1 = 0.9999 · · ·
Proof:
1
= 0.333 · · ·
3
1
⇒1=3· = 0.999 · · ·
3
√
4k 2 = 2q 2 that is q 2 = 2k 2 . Thus q 2 is even ⇒ q = 2n is even. This
contradicts
√ the assumption of no common factors in p and q. Thus
p
2 = q cannot hold. 2

Lecture 1 C2

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Mathematical Analysis I, a.a.

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 2 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 3 / 16

Group A: surname FAV–KHU

Mathematical Analysis I Lecture 1 Politecnico di Torino 3 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Structure of the exam:

Mathematical Analysis I Lecture 1 Politecnico di Torino 4 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 5 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 5 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 6 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 7 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 7 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 7 / 16

Mathematical Analysis I Lecture 1 Politecnico di Torino 7 / 16

A subset A ⊆ X is a set of elements in X .

Mathematical Analysis I Lecture 1 Politecnico di Torino 7 / 16

A subset A ⊆ X is a set of elements in X .