MATH NOTES

1 - NATURAL NUMBERS

Axioms of Peano; 1. 0 is a natural number; 2. to every natural number n there is associated a natural number σ(n) different from 0, called successor of n; different naturals have different successors; 3. if a set A of natural numbers contains 0 and the successor of each element, then A = ℕ

Starting from the notion of successor we can define addition, and on the basis of addition the multiplication between natural numbers

Commutative property, a+b = b+a, a⋅b = b⋅a; associative property, (a+b)+c = a+(b+c), (a⋅b)c = a(b⋅c); existence of the neutral element, a+0 = a, a⋅1 = a; distributive property, a(b+c) = a⋅b+a⋅c

Induction principle; if the family of propositions P(n), where n is a natural number, checks the conditions, 1. P(0) is true, 2. from the hypothesis that P(n) is true it follows that P(n+1) is true for every n, then P(n) is true for every natural n

The induction principle stems from Peano's axioms and states that if the base of induction P(0) is true, and the inductive step P(n+1) is also true, then the proposition P(n) is true for every natural number n

The sum of natural numbers not exceeding n is n(n+1)/2; 0+1+2+...+n = (n(n+1))/2, if n = 0 then (0(0+1))/2 = 0, the base of induction is true; 0+1+...+n+(n+1) = (n(n+1))/2+(n+1) = (n(n+1)+2(n+1))/2 = ((n+1)(n+2))/2, this is the inductive step; if n = n+1 then (n(n+1))/2 becomes (n+1(n+1+1))/2 = ((n+1)(n+2))/2, therefore the inductive step is true; the base of the induction is true, the inductive step is true, therefore the proposition is true for every natural number n

Calculate the number of parts of the plane divided by n straight lines; r(0) = 1; r(n+1) = r(n)+n+1; r(n) = 1+(n(n+1))/2

Recursive definitions; a⁰ := 1, aⁿ⁺¹ := a⋅aⁿ; 0! := 1, (n+1)! := (n+1)n!

2 - COMBINATORICS

Ordered lists = Dispositions

Dispositions with repetitions of n elements in groups of k, ^rD_n,k = n^k

The number of different bytes is 256 because they are dispositions with repetitions of 2 elements, bit 0 and bit 1, in groups of 8, ^rD_2,8 = 2⁸ = 256

A simple disposition is a disposition without repetitions; simple dispositions of n elements in groups of k, D_n,k = n(n-1)...(n-(k-1)) = n(n-1)...(n-k+1) = n!/(n-k)!, k ≤ n because repetitions are not allowed

D_3,2 = n!/(n-k)! = 3!/(3-2)! = 3!/1! = 3⋅2/1 = 6/1 = 6 simple dispositions of 3 elements in groups of 2

Simple dispositions of n elements in groups of n are called permutations of n elements, P_n = D_n,n = n(n-1)...n(n-(n-1)) = n(n-1)...2 = n!

4 people can sit in 4 chairs in 4! = 4⋅3⋅2 = 24 different ways

Unordered lists = Combinations

Simple combinations of n objects in groups of k, C_n,k = D_n,k/k! = (n(n-1)...(n-k+1))/k! = (n(n-1)...(n-k+1)(n-k)!)/(k!(n-k)!) = n!/(k!(n-k)!)

C_n,k = C_n,n-k, C_5,2 = C_5,3; C_n,n = 1

Binomial coefficient with upper index n and lower index k, C(n,k) := C_n,k = n!/(k!(n-k)!), C(n,k) ∈ ℕ

Formula of the binomial, (a+b)ⁿ = ⁿΣ_k=0(C(n,k)(a^n-kb^k))

Symmetry property of binomial coefficients, C(n,k) = C(n,n-k)

The binomial coefficients form the arithmetic triangle

Each row of the arithmetic triangle can be calculated using the previous row; C(n,0) = C(n,n) = 1; C(n,k) = C(n-1,k)+C(n-1,k-1), 1 ≤ k ≤ n-1

Each row of the arithmetic triangle can be calculated from the first element; C(n,0) = 1; C(n,k) = n!/(k!(n-k)!) = ((n-k+1)/k)(n!/((k-1)!(n-k+1)!) = ((n-k+1)/k)C(n,k-1)

3 - FROM NATURAL NUMBERS TO INTEGER NUMBERS

The set of integer numbers ℤ has several properties; a+b = b+a, a⋅b = b⋅a, commutative property; (a+b)+c = a+(b+c), (a⋅b)c = a(b⋅c), associative property; a+0 = a, a⋅1 = a, neutral element; a(b+c) = ab+ac, distributive property

The equation a+x = b is generally not solvable in the context of ℕ, when it is solvable it is said that a is less than or equal to b; the equation a+x = 0 can be solved in the context of ℕ if and only if a = 0, therefore only 0 admits an element opposite to itself in the set of natural numbers ℕ

The set ℤ of the integers is obtained from ℕ by adding to the numbers 0, 1, 2, 3, ... the numbers -1, -2, -3, ...

On the oriented line of integers ℤ, addition is a shift to the right, subtraction is a shift to the left

Once the addition in ℤ is defined, the equation a+x = b admits the solution x = b-a = b+(-a), the sum of b with the opposite of a

If the equation a+x = b admits a natural solution x, we say that a is less than or equal to b, and if it admits a natural x different from 0 as a solution, we say that a is less than b

ℤ is totally ordered by the minor or equal relation, the law of trichotomy applies; if a and b are 2 integers, one and only one of the 3 alternatives is true, a = b, a < b, a > b

If with the symbol x indicate any integer, with the symbol -x we indicate the opposite of x; if x is negative, then -x is positive

The symbol x indicates any integer, the symbol -x indicates the opposite of x; if x is negative, then -x is positive

If x is an integer, the absolute value of x, called the modulus of x, is x itself if x is natural, otherwise it is the opposite of x; |x| = {x, se x ∈ ℕ; -x, se x ∉ ℕ}

A natural number is a positive integer

The absolute value of a natural number is equal to the natural number, |3| = 3; the absolute value of a negative number is the opposite of the negative number, |-3| = 3

The product of 2 integers a and b is equal to the product of their absolute values if a and b have the same sign, but it is the opposite if they have opposite sign; the product of 2 numbers is 0 if one of the 2 numbers is 0

The equation a⋅x = 1 can be solved in the context of ℕ if and only if a = 1, in the context of ℤ if and only if a = 1 or a = -1; the set of natural numbers ℕ and the set of integer numbers ℤ are not satisfactory for the measurement of geometric quantities

The fraction n/m, where n is an integer and m an integer other than 0, has n as its numerator and m as its denominator; if n is a multiple of m, n = q⋅m, then the fraction is apparent, n/m = q

The fraction n/m is equal, better said equivalent, to the fraction p/q if and only if n⋅q = m⋅p

n/m+p/q := (nq+pm)/(mq), (n/m)(p/q) := (np)/(mq), these are the operations defined in the set of rational numbers ℚ

4 - FROM INTEGER NUMBERS TO RATIONAL NUMBERS

n/m+p/q := (nq+pm)/(mq), this is the sum of 2 rational numbers, mq is the common denominator, but if possible, it is better to use the least common denominator which is the least common multiple of the denominators; (n/m)(p/q) := (np)/(mq), this is the product of 2 rational numbers; in the set of rational numbers ℚ the sum and product operations are defined

ℚ is an ordered field; the fraction n/m is less than the fraction p/q, with the denominators m and q positive, if nq is less than pm, n/m < p/q ⇔ nq < pm

The set of rational numbers ℚ is dense, between two rational numbers there are infinitely many other rational numbers

A majorant of a set A is a number M greater than or equal to each element of A; a not empty set A is said to be bounded above if it has majorants

M is a majorant of the set A if ∀ a ∈ A, a ≤ M

A proper fraction is a fraction with a numerator smaller than the denominator, p < q ⇒ p/q < 1; the number 1 is a majorant of the set of proper fractions

A minorant of a set A is a number m less than or equal to any element of A; a not empty set A is said to be bounded below if it has minorants

m is a minorant of the set A if ∀ a ∈ A, a ≥ m

The maximum is the smallest of the majorants; the minimum is the largest of the minorants; the maximum of the set A is indicated by max(A); the minimum of the set A is indicated by min(A)

In ℤ, a set bounded above always has a maximum, and a set bounded below always has a minimum; in ℚ, a set can be bounded above without having a maximum, and it can be bounded below without having a minimum

The smallest of the majorants of the set A is called supremum of A, sup(A); the largest of the minorants of the set A is called the infimum of A, inf(A)

The set of proper fractions is a set bounded above and below, 1 is the supremum but there is no maximum, 0 is the infimum but there is no minimum; the supremum is the smallest of the majorants, the infimum is the largest of the minorants

There is no rational number whose square is equal to 2; the double of a perfect square cannot be a perfect square; in ℚ we cannot solve the equation x² = 2

Studying the equilateral hyperbola (2x+2)/(x + 2), we understand that there are sets that have no maximum and sets that have no minimum

The set of rational numbers ℚ is dense but not complete; in ℚ a set can be bounded above without having a supremum; in ℚ a set can be bounded below without having an infimum; there are infinite points on the oriented line that have no rational abscissa

5 - DECIMAL REPRESENTATION

The set of rational numbers ℚ is dense; between two rational numbers there are infinite rational numbers

A majorant of a set A is a number M greater than or equal to each element of A, and a not empty set A is bounded above if it has majorants

A minorant of a set A is a number m less than or equal to each element of A, and a not empty set A is bounded below if it has minorants

The supremum is the smallest of the majorants, and if it belongs to the set it is called the maximum; the infimum is the largest of the minorants, and if it belongs to the set it is called the minimum

The smallest of the majorants of the set A is the supremum of A, denoted by sup(A); if the supremum of A belongs to A, then the supremum of A is the maximum of A, denoted by max(A); sup(A) ∈ A ⇒ sup(A) = max(A)

The largest of the minorants of the set A is the infimum of A, indicated by inf(A); if the infimum of A belongs to A, then the infimum of A is the minimum of A, denoted by min(A); inf(A) ∈ A ⇒ inf(A) = min(A)

A proper fraction is a fraction with a numerator smaller than the denominator; the set of proper fractions has 1 as supremum, but it has no maximum, and it has 0 as infimum, but it has no minimum

In ℚ a set can be bounded above without having a supremum and a maximum, and can be bounded below without having an infimum and a minimum

There is no rational number whose square is equal to 2, or 3, or 5, or any other natural number that is not a perfect square

To every positive rational number x we associate the rational number x' := (2x+2)/(x+2); x² < 2 ⇒ x < x', x² > 2 ⇒ x' < x

The set of rational numbers ℚ is dense but not complete; in ℚ there are sets bounded above without supremum, and sets bounded below without infimum

There are infinite points on the oriented line that have no rational abscissa

Rational numbers have a periodic decimal representation, including period 0, and excluding period 9; when the period is 0 the number can be expressed as a decimal fraction that is a fraction whose denominator is a power of 10

Transform the periodic number x into a fraction, x = 1.235 = 1.2353535...; multiply by a power of 10 with exponent equal to the number of digits of the period, in this case 2, so we multiply the number x by 10² = 100, 100x = 123.5353535..., 100x-x = 123.5353535...-1.2353535... ⇒ 99x = 122.3 ⇒ x = 122.3/99 = 1223/990 = (990+233)/990 = 990/990+233/990 = 1+233/990

Irrational numbers are represented by unlimited and non-periodic decimal alignments

6 - THE FIELD OF REAL NUMBERS

The set ℝ of real numbers is the union of the set of rational numbers with the set of irrational numbers

ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

The absolute value of a non-negative real number is the number itself; the absolute value of a negative real number is its opposite

x ∈ ℝ, |x| := {x, x ≥ 0; -x, x < 0}

The set ℝ of real numbers is complete; in ℝ, every set bounded above has a supremum, and every set bounded below has an infimum

2 sets A and B of real numbers are separated if each element a of A is less than each element b of B, and this implies that sup(A) ≤ inf(B); separate sets A and B are contiguous if sup(A) = inf(B)

The set A and the set B are separated if ∀ a ∈ A ∧ ∀ b ∈ B, a < b; a ≤ sup(A) ≤ inf(B) ≤ b; if sup(A) = inf(B), sets A and B are separated and contiguous

The notion of contiguity between sets can be used to define operations in ℝ

The notion of contiguity between sets can be used to define π, ratio between the length of a circumference and the length of its diameter, ratio between the length of the semicircle and the length of its radius, length of the semicircle with radius 1

Archimedes in 240 BC calculated the value of π as 3+10/71 < π < 3+1/7; Johann Heinrich Lambert in 1767 proved that π is an irrational number; Ferdinand von Lindemann in 1882 proved that π is a transcendental number

√(2) is an irrational number, but it is not transcendental because it is the solution of the algebraic equation with integer coefficients x²-2 = 0; π is a transcendental number, it is irrational and it is not the solution of any algebraic equation with integer coefficients

7 - INEQUALITIES

The set ℝ of real numbers is the union of the set of rational numbers with the set of irrational numbers

|x+y| ≤ |x|+|y|, |xy| = |x||y|, these are the properties of the absolute value

|x+y| ≤ |x|+|y| is the triangular inequality and is used for the study of complex numbers

The geometric mean of 2 non-negative numbers does not exceed the arithmetic mean, √(xy) ≤ (x+y)/2

x ≥ 0, y ≥ 0, √(xy) ≤ (x+y)/2; x ≥ 0, y ≥ 0, x = y, √(xy) = (x+y)/2

If x is a real number of any sign, the square root of its square is the absolute value of x, √(x²) = |x|

In a right triangle the height relative to the hypotenuse is h, and the projections of the catheti on the hypotenuse are x and y; for the second theorem of Euclid, in a right triangle, the height relative to the hypotenuse is the proportional mean between the projections of the catheti on the hypotenuse, x/h = h/y ⇒ xy = h² ⇒ √(xy) = h, the geometric mean of the projections x and y of the catheti is the height relative to the hypotenuse of the right triangle, and considering the right triangle inscribed in a circle of radius r = (x+y)/2, h ≤ r, √(xy) ≤ (x+y)/2, √(xy) = (x+y)/2 only if x = y, when the right triangle is isosceles

The geometric mean of n ≥ 2 non-negative numbers does not exceed the arithmetic mean, ⁿ√(x₁x₂...x_n) ≤ (x₁+x₂+...+x_n)/n

ⁿ√(x₁x₂...x_n) ≤ (x₁+x₂+...+x_n)/n, if x₁ = x₂ = x_n then ⁿ√(x₁x₂...x_n) = (x₁+x₂+...+x_n)/n

If n ≥ 2 non-negative real numbers have n as their sum, their product does not exceed 1, x₁+x₂+...+x_n = n ⇒ x₁x₂...x_n ≤ 1

a > 1, 1 < ⁿ√(a) ≤ (a+n-1)/n = 1+(a-1)/n

For every x > -1 and for every natural n, (1+x)ⁿ ≥ 1+nx

∀ x ∈ ℝ, x > -1, ∀ n ∈ ℕ, (1+x)ⁿ ≥ 1+nx, this is the inequality of Jakob Bernoulli; (1+x)ⁿ is a geometric progression with ratio 1+x and the graph is an exponential curve, 1+nx is an arithmetic progression with common difference x and the graph is a straight line

Using the inductive method we can prove the Bernoulli inequality (1+x)ⁿ ≥ 1+nx, ∀ x ∈ ℝ, x > -1, ∀ n ∈ ℕ; base of induction, n = 0, (1+x)⁰ ≥ 1+0⋅x ⇒ 1 ≥ 1+0 ⇒ 1 ≥ 1 that is true; inductive step, n = n+1, (1+x)ⁿ⁺¹ = (1+x)(1+x)ⁿ, (1+x)ⁿ⁺¹ ≥ (1+nx)(1+x) ⇒ (1+x)ⁿ⁺¹ ≥ 1+x+nx+nx² ⇒ (1+x)ⁿ⁺¹ ≥ 1+x(1+n)+nx² ⇒ (1+x)ⁿ⁺¹ ≥ 1+(n+1)x+nx² ⇒ (1+x)ⁿ⁺¹ ≥ 1+(n+1)x+nx² ≥ 1+(n+1)x ⇒ (1+x)ⁿ⁺¹ ≥ 1+(n+1)x that coincides with the hypothesis (1+x)ⁿ ≥ 1+nx when n = n+1, we verified the base of induction and the inductive step, therefore we proved the Bernoulli inequality using the inductive method

8 - REAL FUNCTIONS AND SEQUENCES

A function f: A → ℝ associates to every number x of A a well determined real number

A function transforms an input that is a value of the domain or definition set, into an output that is a value of the image of the function contained in the codomain; the domain of the function, or set of definition, is the set of values that the function can accept, the image of the function is the set of values assumed by the function, dom(f) → im(f) = f(A), dom(f(A)) → im(f(A)) = f(A)

f: A → ℝ, the set A is the domain or definition set of the function f, denoted by dom(f), ℝ is the codomain, set of arrival, of the function f

The set of values assumed by the function f is called image of f and is indicated by im(f) or f(A)

The domain of a function can be an interval or the union of different intervals

[a,b], a ≤ x ≤ b, bounded and closed interval; (a,b], a < x ≤ b, left-open bounded interval; [a,b), a ≤ x < b, right-open bounded interval; (a,b), a < x < b, open bounded interval

With the letter a we indicate the extreme left of the set A, with the letter b we indicate the extreme right of the set A, therefore inf(A) = a, sup(A) = b

[a,b], a ≤ x ≤ b, inf(A) = a, min(A) = a, sup(A) = b, max(A) = b; (a,b], a < x ≤ b, inf(A) = a, sup(A) = b, max(A) = b; [a,b), a ≤ x < b, inf(A) = a, min(A) = a, sup(A) = b; (a,b), a < x < b, inf(A) = a, sup(A) = b

[a,+∞), x ≥ a, a ≤ x, unbounded above and closed interval; (a,+∞), x > a, a < x, unbounded above and open interval

[a,+∞), x ≥ a, a ≤ x, inf(A) = a, min(A) = a, sup(A) = +∞; (a,+∞), x > a, a < x, inf(A) = a, sup(A) = +∞

An unbounded above set has no majorants; an unbounded below set has no minorants

(-∞,b], x ≤ b, b ≥ x, unbounded below and closed interval; (-∞,b), x < b, b > x, unbounded below and open interval

(-∞,b], x ≤ b, b ≥ x, inf(A) = -∞, sup(A) = b, max(A) = b; (-∞,b), x < b, b > x, inf(A) = -∞, sup(A) = b

x₁, x₂ ∈ A, with the letter A we indicate an interval, x₁ < x₂, if x₁ < x < x₂, then x ∈ A, this is the connection property that all intervals have

A function defined in the set ℕ of natural numbers, or in the set ℕ* of natural numbers > 0, is a sequence

n → f → f(n) ∈ ℝ, this is a sequence; n → 1+1/n = (n+1)/n, this is an example of a sequence with n ∈ ℕ*

An arithmetic progression is a sequence in which, given an initial term, each term is obtained from the previous one by adding a constant; a geometric progression is a sequence in which, given an initial term, each term is obtained from the previous one by multiplying by a constant

The Bernoulli's inequality compares the terms of a geometric progression with the terms of an arithmetic progression

a₀, common difference d ≠ 0, d = a_n+1-a_n, a_n+1 := a_n+d, a₀, a₁ = a₀+d, a₂ = a₁+d = a₀+d+d = a₀+2d, a₃ = a₂+d = a₀+2d+d = a₀+3d, ..., a_n = a_n-1+d = a₀+nd, this is an arithmetic progression

a₀, ratio r ≠ 1, r = a_n+1/a_n, a_n+1 := a_n⋅r, a₁ = a₀⋅r, a₂ = a₁⋅r = a₀⋅r², a₃ = a₂⋅r = a₀⋅r²⋅r = a₀⋅r³, ..., a_n = a_n-1⋅r = a₀⋅rⁿ, this is a geometric progression

Sum of n terms in geometric progression, the initial term is 1 and the reason is r, the geometric progression is 1, r, r², r³, ..., the sum is 1+r+...+r^n-1, if r = 1 then the sum is n, if r ≠ 1 then 1+r+...+r^n-1 = (1-rⁿ)/(1-r), if 0 < r < 1 then (1-rⁿ)/(1-r) < 1/(1-r), 0 < 1+r+...+r^n-1 < 1/(1-r), if r = 1/2 then 1+1/2+1/4+...+1/2^n-1 < 1/(1-1/2) = 1/(1/2) = 2, 1+1/2+1/4+1/8+1/16+... < 2

A sequence is a function defined in the set of natural numbers

We want to understand if for sufficiently large indices a sequence assumes a value arbitrarily close to a trend value

A geometric progression is constant if r = 1, an arithmetic progression is constant if d = 0

(-1)ⁿ = 1,-1,1,-1,1,-1,... this sequence has no trend value

1/n, n ∈ ℕ*; n = 1, 1/n = 1; n = 2, 1/n = 1/2 = 0.5; n = 3, 1/n = 1/3 = 0.3; n = 4, 1/n = 1/4 = 0.25; n = 5, 1/n = 1/5 = 0.2; n = 10, 1/n = 1/10 = 0.1; n = 11, 1/n = 1/11 = 0.09; n = 100, 1/n = 1/100 = 0.01; n = 101, 1/n = 1/101 = 0.0099; in the sequence 1/n, as n increases, the integer part and the decimal digits, one after the other, become 0, therefore the trend value of the sequence 1/n is 0; ε > 0, |1/n-0| = 1/n < ε, n > 1/ε, n > n_ε ≥ 1/ε

9 - LIMIT OF SEQUENCES - PART 1

n ↦ c/n, ∀ ε > 0, 0 < c/n < ε ⇒ n/c > 1/ε ⇒ n > c/ε ⇒ n > n_ε ≥ c/ε, lim_n→∞(c/n) = 0

A function defined in the set ℕ of natural numbers, or in the set ℕ* of natural numbers > 0, is a sequence

The value towards which a sequence tends is the limit of the sequence

1 < ⁿ√(a) < 1+(a-1)/n, 1 < ⁿ√(a) < 1+c/n, |ⁿ√(a)-1| = ⁿ√(a)-1 < c/n

a_n → L, lim_n→+∞(a_n) = L, ∀ ε > 0 ∃ n_ε : ∀ n > n_ε ⇒ |a_n-L| < ε ⇒ -ε < a_n-L < ε ⇒ L-ε < a_n < L+ε, this is the definition of limit of a sequence

Definition of limit of a sequence, ∀ ε > 0, ∃ n_ε : n > n_ε ⇒ |a_n-L| < ε

I(L,ε) := (L-ε,L+ε), I(L,ε) is the neighborhood of center L and radius ε and is by definition the open interval (L-ε,L+ε)

Increasing sequence, ∀ n : a_n ≤ a_n+1; decreasing sequence, ∀ n : a_n ≥ a_n+1

∀n : a_n ≤ a_n+1, increasing sequence; ∀n : a_n < a_n+1, strictly increasing sequence; ∀n : a_n ≥ a_n+1, decreasing sequence; ∀n : a_n > a_n+1, strictly decreasing sequence

n ↦ c/n is a strictly decreasing sequence that tends to 0

n ↦ ⁿ√(a), a > 1, is a strictly decreasing sequence that tends to 1; n ↦ ⁿ√(a), a = 1, is a constant sequence, always equal to 1; n ↦ ⁿ√(a), 0 < a < 1, is a strictly increasing sequence that tends to 1

Increasing and decreasing sequences are also called monotonous sequences; a monotone sequence is an increasing or decreasing sequence

Each increasing sequence, bounded above, admits its own supremum as its limit; every decreasing sequence, bounded below, admits its own infimum as its limit

If a sequence is monotone increasing its limit is finite or infinite and coincides with the supremum; if a sequence is monotonous and decreasing its limit is finite or infinite and coincides with the infimum

a_n ≤ a_n+1, ∀ n a_n ≤ sup(a_n) := sup{a_n, n ∈ ℕ} := L, L-ε < a_nε ≤ a_n, L-ε < a_n ≤ L < L+ε, this is the proof for an increasing sequence bounded above

∀ M > 0 ∃ n_M : ∀ n > n_M ⇒ a_n > M, lim_n→+∞(a_n) = +∞, this is the definition of sequence which has as limit +∞, it diverges positively

∀ M > 0 ∃ n_M : ∀ n > n_M ⇒ a_n < -M, lim_n→+∞(a_n) = -∞, this is the definition of sequence which has as limit -∞, it diverges negatively

Let us consider the geometric progression of first element 1 and reason a, 1,a,a²,a³,...,aⁿ,...; if a = 1 the sequence is constant and is always 1, if a > 1 the sequence tends to +∞ or diverges positively; a = 1+d, d > 0, for the Bernoulli's inequality aⁿ = (1+d)ⁿ ≥ 1+nd, 1+nd is an arithmetic progression that diverges positively, therefore also the geometric progression (1+d)ⁿ diverges positively; ∀ M > 0 ∃ n_M : ∀ n > n_M ⇒ 1+nd > M ⇒ nd > M-1 ⇒ n > (M-1)/d, n_m ≥ (M-1)/d, if n > n_m ≥ (M-1)/d then the sequence 1+nd tends to +∞ and therefore also the sequence aⁿ tends to +∞, lim_n→+∞(aⁿ) = +∞ if a > 1

lim_n→+∞(aⁿ) = 0 if 0 < a < 1; (1/2)ⁿ = 1/2ⁿ, 2ⁿ tends to +∞, therefore the sequence 1/2ⁿ tends to 0

n ↦ ⁿ√(n) = n^1/n, n ∈ ℕ; n = 1, ⁿ√(n) = 1; n = 2, ⁿ√(n) = ²√(2); n = 3, ⁿ√(n) = ³√(3); n = 4, ⁿ√(n) = ⁴√(4); 2^1/2 < 3^1/3, 3^1/3 > 4^1/4; the sequence ⁿ√(n) decreases from the term n = 3

ⁿ√(n) is a monotone sequence decreasing from n = 3, and tends to 1; n = √(n)⋅√(n)⋅1...1, where the number 1 is repeated n-2 times, 1 ≤ ⁿ√(n) = ⁿ√(√(n)⋅√(n)⋅1...1) ≤ (2√(n)+n-2)/n < 1+(2√(n))/n = 1+2/√(n), 1 ≤ ⁿ√(n) < 1+2/√(n), lim_n→+∞(2/√(n)) = 0, 1 ≤ ⁿ√(n) < 1, lim_n→+∞(ⁿ√(n)) = 1, 0 ≤ ⁿ√(n)-1 < 2/√(n) < ε, 2/√(n) < ε ⇒ 4/n < ε² ⇒ n/4 > 1/ε² ⇒ n > 4/ε² ⇒ n_ε > 4/ε²

ⁿ√(n) is a monotone decreasing sequence from the term n = 3, n ≥ 3, n^1/n > (n+1)^1/(n+1) ⇒ n^{(1/n)(n(n+1))} > (n+1)^{(1/(n+1))(n(n+1))} ⇒ nⁿ⁺¹ > (n+1)ⁿ ⇒ n⋅nⁿ = nⁿ⁺¹ > (n+1)ⁿ ⇒ n > (n+1)ⁿ/nⁿ ⇒ n > ((n+1)/n)ⁿ ⇒ n > ((n/n)+(1/n))ⁿ ⇒ n > (1+1/n)ⁿ; n ≥ 3, (1+1/n)ⁿ < 3 ≤ n, ⁿΣ_k=0(C(n,k)(a^n-k·b^k)), ⁿΣ_k=0(C(n,k)(1/n^k)) = ⁿΣ_k=0(((n(n-1)(...)(n-k+1))/(n·n...n))(1/k!)) < ⁿΣ_k=0(1/k!) = 1+1+1/2!+1/3!+...+1/n! < 3

10 - LIMIT OF SEQUENCES - PART 2

An increasing sequence always has a limit that is the supremum, finite or infinite; a decreasing sequence always has a limit that is the infimum, finite or infinite

∀ n : a_n ≤ a_n+1, increasing sequence; ∀ n : a_n ≥ a_n+1, decreasing sequence

n ↦ ⁿ√(n) = n^1/n, this sequence is monotone decreasing from n = 3 and tends to 1; n^1/n > (n+1)^1/(n+1), (1+1/n)ⁿ < n, (1+1/n)ⁿ < 3 ≤ n, ⁿΣ_k=0(C(n,k)(1^n-k·1/n^k) = ⁿΣ_k=0(((n(n-1)...(n-k+1))/(n·n...n))(1/k!)) < ⁿΣ_k=0(1/k!) = 1+1+1/2!+1/3!+...+1/n! < 1+1+1/2+1/2²+...+1/2^n-1 < 3

The limit of an increasing sequence bounded above is the supremum of the sequence; the limit of a decreasing sequence bounded below is the infimum of the sequence

Monotony is a sufficient condition for the existence of the limit, but it is not a necessary condition, in fact there are sequences that are not monotonous but are convergent

F₀ := 0, F₁ := 1, F_n+2 := F_n+F_n+1, this is the Fibonacci sequence; r_n := F_n+1/F_n, n ≥ 1; n = 0, F_n = 0, r_n = ...; n = 1, F_n = 1, r_n = 1; n = 2, F_n = 1, r_n = 2; n = 3, F_n = 2, r_n = 1.5; n = 4, F_n = 3, r_n = 1.6; n = 5, F_n = 5, r_n = 1.6; n = 6, F_n = 8, r_n = 1.625; n = 7, F_n = 13, r_n = 1.615...; n = 8, F_n = 21, r_n = 1.619...; n = 9, F_n = 34, r_n = 1.617...; n = 10, F_n = 55, r_n = 1.618; n = 11, F_n = 89, r_n = 1.617...; n = 12, F_n = 144, r_n = 1.618...; n = 13, F_n = 233, r_n = ...; the Fibonacci sequence is monotonous increasing, diverges positively; the sequence of odd terms, r₁, r₃, r₅, ..., is an increasing sequence, the sequence of even terms, r₂, r₄, r₆, ..., is a decreasing sequence, but these 2 sequences tend towards a common value which is 1.61 ...; r_n+1 = F_n+2/F_n+1 = (F_n+1+F_n)/F_n+1 = 1+1/r_n; r₁ := 1, r₂ := 2, r_n+1 := 1+1/r_n, 1 ≤ r₁ < r₂ ⇒ 1/r₁ > 1/r₂ ⇒ 1+1/r₁ > 1+1/r₂ ⇔ r₂ > r₃; r_n+1-r_n = 1+1/r_n-(1+1/r_n-1) = 1/r_n-1/r_n-1 = (r_n-1-r_n)/(r_n⋅r_n-1) = -((r_n-r_n-1)/(r_n⋅r_n-1)), these differences change sign for each increment of n; |r_n+1-r_n| = |r_n-r_n-1|/(r_n⋅r_n-1) ⇒ |r_n+1-r_n| ≤ |r_n-r_n-1|/2, with each increment of n, the intervals are reduced in amplitude by at least a factor of 2; lim_n→+∞(r_n) = L, r_n+1 = 1+1/r_n ⇒ L = 1+1/L ⇒ L = (L+1)/L ⇒ L² = L+1 ⇒ L²-L-1 = 0, a = 1, b = -1, c = -1, Δ = b²-4ac = (-1)²-4⋅1(-1) = 1+5 = 5, x = (-b±√(Δ))/(2a) = (1±√(5))/(2⋅1) = (1±√(5))/2, x₁ = (1-√(5))/2 that is not acceptable, x₂ = (1+√(5))/2 that is acceptable; (1+√(5))/2 = 1.618... this number is called golden ratio; the Greeks considered the rectangles whose sides had this relationship more aesthetically beautiful; this example demonstrates that a non-monotone sequence can be convergent

a_n := (1+1/n)ⁿ, this sequence is monotonous increasing; a_n = (1+1/n)ⁿ < 3; b_n := (1+1/n)ⁿ⁺¹, this sequence is monotonous decreasing; b_n = (1+1/n)ⁿ⁺¹ = a_n(1+1/n) = a_n((n+1)/n), b_n > a_n; the sequence a_n, increasing monotone, and the sequence b_n, decreasing monotone, converge at the same limit whose value is the Euler number indicated with the letter e; a_n < a_n+1, 1(1+1/n)...(1+1/n), the term (1+1/n) is repeated n times, geometric mean ≤ arithmetic mean, ⁿ⁺¹√(1+1/n)ⁿ < (n+1+n/n)/(n+1) = (n+1+1)/(n+1) = (n+1)/(n+1)+1/(n+1) = 1+1/(n+1), ⁿ⁺¹√(1+1/n)ⁿ < 1+1/(n+1) ⇒ (ⁿ⁺¹√(1+1/n)ⁿ)ⁿ⁺¹ < (1+1/(n+1))ⁿ⁺¹ ⇒ (1+1/n)ⁿ < (1+1/(n+1))ⁿ⁺¹ ⇒ a_n < a_n+1; b_n = a_n(1+1/n) = a_n((n+1)/n) ⇒ a_n = b_n(n/(n+1)), 0 < b_n-a_n = b_n-b_n(n/(n+1)) = b_n(1-n/(n+1)) = b_n((n+1-n)/(n+1)) = b_n(1/(n+1)) ≤ 4/(n+1) < 4/n, therefore the difference b_n-a_n tends to 0

Number e = 2.718281..., limn→∞(1+1/n)ⁿ = limn→∞(1+1/n)ⁿ⁺¹ = e, the number e is not rational

(1+1/n)ⁿ is an increasing sequence, limn→+∞(1+1/n)ⁿ = e; (1+1/n)ⁿ⁺¹ is a decreasing sequence, limn→+∞(1+1/n)ⁿ⁺¹ = e; e = 2.718281... is an irrational number; this is the most famous example of sequences of rational numbers converging to an irrational number

a > 1, x ↦ a^x; the exponential function a^x, with a > 1, is increasing and intersects the y-axis at the point (0,1); x = 0, a^x = a⁰ = 1; x = 1, a^x = a¹ = a; the exponential function a^x, with a = 1, is the straight line y = 1, parallel to the x-axis; the rate of growth of the exponential function a^x is directly proportional to the value of a; the tangent to the exponential curve at the point (0,1) has a slope directly proportional to the value of a; the angular coefficient m of the straight line y = mx+q tangent to the exponential curve a^x at the point (0,1) is m > 1 if a = 3, and is m < 1 if a = 2; Leonhard Euler found the value of a to obtain a tangent line at the point (0,1) with angular coefficient m = 1, and this value is Euler's number, e = 2.718281...

11 - LIMIT OF FUNCTIONS

f(x), f: A → ℝ, lim_x→x0(f(x)) = L; the function does not need to be defined at point x₀, but it must be defined close to x₀

A neighborhood of a real number is an open interval centered on the number itself

I(x₀,r) := (x₀-r,x₀+r), a neighborhood of a real number x₀ is an open interval with center x₀ and radius r

A real number is called accumulation point of a set A if every neighborhood of this number contains infinite elements of A

x₀ is an accumulation point of A if ∀ r > 0, the intersection I(x₀,r)∩A contains infinite elements

A finite set has no accumulation points; +∞ is the only accumulation point of the natural numbers

A = (0,1], all the points of the set A are accumulation points, the point 0 is also an accumulation point; when an accumulation point belongs to the set it is called non-isolated point of the set

f(x) = x², the graph of this function is a parabola that passes through the points (-1,1), (0,0), (1,1); a secant line passes through the points (x₀,f(x₀)) = (1,1) and (x,f(x)) = (x,x²), the slope of the secant line is r(x) := (x²-1)/(x-1); the function f(x) = x² is defined in ℝ, the function r(x) = (x²-1)/(x-1) is defined in ℝ\{1}; the point x₀ does not belong to the defining set of r(x) but is an accumulation point, therefore we can calculate lim_x→x0(r(x)) that we need to calculate the angular coefficient of the tangent line at the point (1,1), and this is the derivative of f(x) computed for x = 1; the limit of the angular coefficient of the secant is the angular coefficient of the tangent, and is the instantaneous rate of change of the function; r(x) := (x²-1)/(x-1) = ((x-1)(x+1))/(x-1) = x+1 =: r*(x); x ≠ 1, |r(x)-2| = |x+1-2| = |x-1|, |r(x)-2| < ε ⇒ |x-1| < ε, |f(x)-L| < ε, 0 < |x-x₀| < δ_ε

Definition of limit of a function, ∀ ε > 0, ∃ δ_ε : x ∈ (A\{x₀}∩I(x₀,δ_ε) ⇒ |f(x)-L| < ε

|f(x)-L| < ε ⇔ L-ε < f(x) < L+ε ⇔ f(x) ∈ I(L,ε)

An accumulation point that belongs to the set is called a non-isolated point

If the limit of a function at a point is equal to the value of the function at that point, then the function is continuous

A function is continuous if the limit of the function at a point coincides with the value of the function at that point, lim_x→x0(f(x)) = L = f(x₀); a function is continuous if |f(x)-f(x₀)| < ε quando |x-x₀| < δ_ε

All polynomial functions are continuous functions

A constant function, f(x) = c, is a continuous function

f(x) = x, |f(x)-f(x₀)| = |x-x₀|, δ_ε = ε, f(x) = x is a continuous function

f(x) = 2x, |2x-2x₀| = 2|x-x₀| < ε, |x-x₀| < ε/2 = δ_ε; f(x) = 2x is a continuous function

f(x) = mx, |mx-mx₀| = m|x-x₀| < ε, |x-x₀| < ε/m = δ_ε; f(x) = mx is a continuous function

x → f → f(x), the function f only accepts values of x ∈ dom (f), x₀ is a non-isolated point of the set A because x₀ ∈ A and it is an accumulation point, therefore we can calculate f(x₀) e lim_x→x0(f(x)), if lim_x→x0(f(x)) = L = f(x₀) then the function is continuous; if a function is continuous, a small change in x₀ corresponds to a small change in f(x₀), x is the input, f(x) is the output, ε is the tolerance of the output, δ_ε is the tolerance of the input, therefore the variation of the output must not exceed ε; first we have to choose ε and then δ_ε

The exponential function a^x, with a > 0, is a continuous function

The tangent at the point (0,1) to the exponential curve e^x has angular coefficient 1, therefore it is parallel to the straight line y = x that is the bisector of the first and third quadrant; log_e(x) = ln(x) is the inverse function of e^x, therefore its graph is symmetrical to the graph of e^x with respect to the straight line y = x that is the bisector of the first and third quadrant; the function e^x is continuous in dom(e^x) = ℝ, the function ln(x) is continuous in dom(ln(x)) = (0,+∞)

Trigonometric functions, where they are defined, are continuous

All elementary functions are continuous

It makes no sense to speak of discontinuity of a function where it is not defined, because continuity occurs when lim_x→x0(f(x)) = L = f(x₀), if x₀ is an accumulation point but does not belong to the function definition set then it is not a non-isolated point, therefore f(x₀) does not exist, and the equation lim_x→x0(f(x)) = L = f(x₀) it does not make sense

f(x) = [x] ⇔ [x] = max{z ∈ ℤ, z ≤ x}, this function is called integer part or rounding down; the integer part of a decimal number is the number consisting of the digits before the decimal point; f(x) = [x] defined in [0,1], x = 1 is a non-isolated point of f(x), f(1) = 1, lim_x→1(f(x)) = 0, lim_x→1(f(x)) = 0 ≠ 1 = f(x), therefore the function f(x) = [x] is not continuous at x = 1, x = 1 is a point of discontinuity

12 - EXTENSION OF THE CONCEPT OF LIMIT

f: A → ℝ, L ∈ ℝ, x → x₀, lim_x→x0(f(x)) = L, ∀ ε > 0, ∃ δ_ε > 0 : x ∈ (A\{x₀)∩I(x₀,δ_ε) ⇒ |f(x)-L| < ε ⇔ L-ε < f(x) < L+ε ⇔ f(x) ∈ I(L,ε)

Convergent function at the limit L for x tending to an accumulation point of domain A: ∀ ε > 0, ∃ δ_ε > 0 : x ∈ (A\{x₀})∩I(x₀,δ_ε) ⇒ |f(x)-L| < ε

A*(x₀,δ) := (A\{x₀})∩I(x₀,δ); A(x₀,δ) := A∩I(x₀,δ); I(L,ε), f(A*(x₀,δ)) ⊆ I(L,ε); I(f(x₀),ε), f(A(x₀,δ)) ⊆ I(f(x₀),ε), this is the definition of continuous function

The notion of convergent function for x tending to an accumulation point in the domain A, and the notion of continuous function at the same point are related

r(x) := (x²-1)/(x-1), the limit of this function is 2 as x tends to 1 because the function r*(x): = x+1 coincides with r(x) for every x different from 1 and its limit is 2 as x tends to 1; a function f is convergent as x approaches x₀ if f*(x) := {f(x) when x ≠ x₀, L when x = x₀}, f*(x) is like f(x) but also has value in x₀, therefore the function f*(x) is continuous at the point x₀

f(x) = mx+q is a linear function, first degree polynomial function; the graph of f(x) = mx+q is a straight line, q is the ordinate of the intersection point of the line with the y-axis, m is the angular coefficient or slope of the line; f(x) = mx+q, m ≠ 0, f(x₀) = mx₀+q, f(x)-f(x₀) = mx+q-(mx₀+q) = mx+q-mx₀-q = mx-mx₀ = m(x-x₀), f(x)-f(x₀) = m(x-x₀) ⇒ |f(x)-f(x₀)| = |m(x-x₀)| ⇒ |f(x)-f(x₀)| = |m||x-x₀| < ε ⇒ |x-x₀| < ε/|m| =: δ_ε

f(x) = √(x), A = [0,+∞), verify that this function is continuous in all points of its domain; y = √(x) ⇒ y² = x; 0 ≤ x₁ < x₂, √(x₂)-√(x₁), 0 ≤ x₁ < x₂ ⇒ 0 < √(x₂)-√(x₁) ≤ √(x₂-x₁), 0 < x₁ < x₂, √(x₂)-√(x₁) ≤ √(x₂-x₁) ⇒ (√(x₂)-√(x₁))² ≤ (√(x₂-x₁))² ⇒ x₂-2√(x₁)√(x₂)+x₁ ≤ x₂-x₁ ⇒ x₂-2√(x₁x₂)+x₁ ≤ x₂-x₁ ⇒ x₁+x₁ ≤ 2√(x₁x₂) ⇒ 2x₁ ≤ 2√(x₁x₂) ⇒ x₁ ≤ √(x₁x₂) ⇒ (x₁)² ≤ (√(x₁x₂))² ⇒ (x₁)² ≤ x₁x₂ ⇒ x₁ ≤ x₂; x₀ < x, 0 < f(x)-f(x₀) = √(x)-√(x₀) ≤ √(x-x₀) < ε, √(x-x₀) < ε ⇒ (√(x-x₀))² < ε² ⇒ x-x₀ < ε² =: δ_{ε, we have proved the continuity of the function f(x) = √(x), A = [0,+∞)}

f(x) = 1/x², x ≠ 0; is an even function, f(x) = f(-x), the graph is symmetrical with respect to the ordinate axis; the function assumes large values as x approaches 0; ∀ M > 0, f(x) > M ⇒ 1/x² > M ⇒ x² < 1/M ⇒ |x| < 1/√(M) =: δ_M, f(x) > M if 0 < |x| < δ_M; ∀ x ∈ A*(x₀,δ_M), f(x) > M

Positively diverging function for x tending to an accumulation point of domain A: ∀ M > 0 ∃ δ_M > 0 : x ∈ {A\{x₀})∩I(x₀,δ_{M) ⇒ f(x) > M}

Negatively diverging function for x tending to an accumulation point of domain A: ∀ M > 0 ∃ δ_M > 0 : x ∈ {A\{x₀})∩I(x₀,δ_{M) ⇒ f(x) < -M}

Let us consider a function f defined in the set A which is unbounded above, it has no majorants, sup(A) = +∞; lim_x→+∞(f(x)) = L, the function is convergent; ∀ ε > 0 ∃ δ_ε > 0 : x ∈ A∩(δ_ε,+∞) ⇒ |f(x)-L| < ε

Convergent function at the limit L as x tends to +∞: ∀ ε > 0 ∃ δ_ε > 0 : x ∈ A∩(δ_ε,+∞) ⇒ |f(x)-L| < ε

∀ ε > 0 ∃ δ_ε > 0 : x > δ_ε ⇒ |f(x)-L| < ε, this is the definition of a convergent function at the limit L as x tends to +∞

f(x) = 1/x, A = ℝ* = ℝ\{0}; the graph is an equilateral hyperbola, with one branch in the first quadrant, and another branch in the third quadrant; lim_x→+∞(1/x) = 0, ∀ ε > 0 ∃ δ_ε > 0 : x > δ_ε ⇒ 0 < 1/x < ε ⇒ x > 1/ε =: δ_ε; lim_x→0^-(1/x) = -∞, lim_x→0⁺(1/x) = +∞, lim_x→0(1/x) = ∄

If a function has a limit, the limit is unique

The notions of limit on the right and limit on the left are related to the presence of an order relation in ℝ

f: A → ℝ, A⁺(x₀) := {x ∈ A, x > x₀}, A^-(x₀) := {x ∈ A, x < x₀}, x₀ is the accumulation point of the set A, but also of the set A⁺(x₀), and of the set A^-(x₀); the limit of the restricted function in A⁺(x₀) is the limit on the right, lim_x→x0⁺ = L⁺; the limit of the restricted function in A^-(x₀) is the limit on the left, lim_x→x0^- = L^-

If the limit on the right and the limit on the left exist and coincide, their common value is the limit, and vice versa

If a function admits a limit, this limit is also the right limit and the left limit

f(x) := [x] = max{z ∈ ℤ, z ≤ x}, integer part function, rounding down function; x₀ = 1, lim_x→1⁺([x]) = 1 = [1], the function is continuous on the right because the value of the limit on the right coincides with the value of the function; x₀ = 1, lim_x→1^-([x]) = 0 ≠ [1], the function is not continuous on the left because the value of the limit on the left does not coincide with the value of the function; in this example the left limit is different from the right limit, the limit on the right coincides with the value of the function at the point, the function is continuous on the right and discontinuous on the left

13 - LIMIT THEOREMS - PART 1

f(x) = 1/x², x ≠ 0, this function tends to +∞ as x tends to 0, ∀ M > 0, 1/x² > M ⇒ x² < 1/M ⇒ |x| < 1/√(M)

The limit of the sum is equal to the sum of the limits

f,g: A → ℝ, x₀ is the accumulation point of the common domain of functions f and g, lim_x→x0(f(x)) = α, lim_x→x0(g(x)) = β, lim_x→x0(f(x)+g(x)) = α+β; in this case x₀ is a non-isolated point, f and g are continuous functions, f(x₀) = α, g(x₀) = β; the sum of 2 continuous functions is a continuous function; the function f converges to the limit α, therefore ∀ ε > 0 ∃ δ' : ∀ x ∈ A*(x₀,δ') ⇒ |f(x)-α| < ε; the function g converges to the limit β, therefore ∀ ε > 0 ∃ δ'' : ∀ x ∈ A*(x₀,δ'') ⇒ |g(x)-β| < ε; δ = min(δ',δ''), A*(x₀,δ) = A*(x₀,δ')∩A*(x₀,δ''); |f(x)+g(x)-(α+β)| = |f(x)-α+g(x)-β|, for the triangular inequality the absolute value of the sum is less than or equal to the sum of the absolute values, |f(x)+g(x)-(α+β)| = |f(x)-α+g(x)-β| ≤ |f(x)-α|+|g(x)-β| < 2ε

If a function f(x) is convergent, as x tends to an accumulation point of its domain, it is bounded in a suitable neighborhood of the same point

A function is bounded when the image of the function is a bounded set that is a set that has majorants and minorants

A function that has a finite limit is a convergent function; a function that has an infinite limit is a divergent function; a regular function is a function that has a limit, finite or infinite; an irregular function is a function that has no limit; when a function is bounded in a set, the function image is bounded, therefore the values assumed by the function are included between a minorant and a majorant

f(x) → α, g(x) → β; lim_{x→x0(f(x)) = α, limx→x0(g(x)) = β, limx→x0(f(x)g(x)) = αβ; the function g is convergent, therefore it is bounded because in proximity to x0 β-ε < g(x) < β+ε, |g(x)| = |g(x)-β+β| ≤ ε+|β| ≤ |β|+1 = c; if a function is convergent, then it is bounded in the set A*(x0,δ), therefore the image of the function is a set with minorants and majorants; |f(x)g(x)-αβ| = |f(x)g(x)-αβ-αg(x)+αg(x)| = |g(x)(f(x)-α)+α(g(x)-β)|, for triangular inequality the absolute value of the sum is less than or equal to the sum of the absolute values, and the absolute value of the product is equal to the product of the absolute values, |f(x)g(x)-αβ| = |f(x)g(x)-αβ-αg(x)+αg(x)| = |g(x)(f(x)-α)+α(g(x)-β)| ≤ |g(x)||f(x)-α|+|α||g(x)-β| ≤ cε+|α|ε = (c+|α|)ε; we have shown that the limit of the product is equal to the product of the limits}

The limit of a product is equal to the product of the limits

The sum of continuous functions is a continuous function; the product of continuous functions is a continuous function

A polynomial function, also called integer rational function, is a continuous function; a polynomial function, also called simply a polynomial, is a sum of monomials; if every monomial function is continuous, then the polynomial function is continuous; a monomial function is x ↦ cxⁿ, x ↦ c is the constant function that is continuous, x ↦ x is the identity function that is continuous, therefore the monomial function x ↦ cxⁿ is continuous; every monomial function is continuous, therefore every polynomial function is continuous

A fractional rational function is a ratio between polynomials, and is defined in all values of x such that the denominator is different from 0; a fractional rational function is continuous in all points such that the denominator is different from 0

Property of permanence of the sign: if a function f(x) converges to a limit other than 0, as x tends to an accumulation point of the domain, it maintains the same sign of the limit in an appropriate neighborhood of the point itself

β > 0, β-ε < g(x) < β+ε, 0 < ε ≤ β/2, 0 < β/2 ≤ β-ε < g(x) < β+ε, this is the demonstration of the property of permanence of the sign

The limit of the quotient is equal to the quotient of the limits, but the limit of the denominator function must be other than 0

The reciprocal function of g(x) is 1/g(x); 1/g(x) → 1/β, β > 0; |1/g(x)-1/β| = |(β-g(x))/(βg(x))| < ε/(β⋅β/2) = 2ε/β² = (2/β²)ε; lim_x→x0(g(x)) = β, lim_x→x0(1/g(x)) = 1/β

f(x) → α, g(x) → β, β ≠ 0; lim_{x→x0(f(x)) = α, limx→x0(g(x)) = β, limx→x0(f(x)/g(x)) = α/β; f(x) → α, g(x) → β, 1/g(x) → 1/β, f(x)/g(x) = f(x)(1/g(x)) → α(1/β) = α/β}

Polynomial functions, also called integer rational functions, are continuous; fractional rational functions are continuous in all points where they are defined, in all points where the denominator is different from 0

lim_x→x0(f(x)) = 0, lim_x→x0(1/f(x)) = +∞ if f(x) > 0 ∀ x ∈ A*(x₀,δ), lim_x→x0(1/f(x)) = -∞ if f(x) < 0 ∀ x ∈ A*(x₀,δ)

lim_x→0(x²) = 0, x² > 0 for x ≠ 0, therefore lim_x→0(1/x²) = +∞

lim_x→0^-(1/x) = -∞, left-hand limit; lim_x→0⁺(1/x) = +∞, right-hand limit

f(x) > 0, lim_x→x0(1/f(x)) = +∞; ∀ M > 0, 1/f(x) > M ⇔ 0 < f(x) < 1/M =: ε

f(x) is divergent, study 1/f(x); lim_x→x0(|f(x)|) = +∞ ⇒ lim_x→x0(1/f(x)) = 0; |1/f(x)| = 1/|f(x)| < ε ⇒ |f(x)| > 1/ε =: M

a > 1, lim_n→+∞(aⁿ) = +∞; aⁿ = (1+d)ⁿ ≥ 1+nd for Bernoulli's inequality, 1+nd is an arithmetic progression and diverges positively, therefore also aⁿ, which is equal to the geometric progression (1+d)ⁿ, diverges positively; a > 1, n < 0 ⇒ n = -|n|, lim_n→-∞(aⁿ) = lim_|n|→+∞(a^-|n|), |n| = k, lim_|n|→+∞(a^-|n|) = lim_k→+∞(a^-k) = lim_k→+∞(1/a^k) = 0; the sequence aⁿ, with a > 1, tends to +∞ when n tends to +∞, and tends to 0 when n tends to -∞

tg(x) = tan(x) = sin(x)/cos(x), the tangent function is defined where cos(x) ≠ 0, therefore it is not defined in π/2+kπ with k ∈ ℤ; we consider the tangent function in the open interval -π/2 < x < π/2 because the tangent is a periodic function of period π; the tangent function is an odd function, tan(-x) = -tan(x), therefore its graph is symmetrical with respect to the origin of the Cartesian axes; lim_x→(π/2)^-(tan(x)) = +∞, tan(x) = sin(x)/cos(x) = 1/cos(x)/sin(x), x→(π/2)^-, the left neighborhood of π/2 is between π/2-δ and π/2, lim_x→π/2(sin(x)) = 1, lim_x→(π/2)^-(cos(x)) = 0⁺, therefore considering the set to the left of π/2 A^-(π/2,δ), tan(x) = sin(x)/cos(x) = 1/cos(x)/sin(x) = 1/0⁺ = +∞; lim_{x→(-π/2)⁺}(tan(x)) = -∞

14 - LIMIT THEOREMS - PART 2

f₁: A₁ → ℝ, f₂: A₂ → ℝ, f₁(A₁) ⊆ A₂, x ∈ A₁ → f₁ → f₁(x) =: y → f₂ → f₂(y) = f₂(f₁(x)), x ↦ f₂(f₁(x)) = (f₂∘f₁)(x), f₁ is the function that acts first, f₂ acts after f₁; f₂(f₁(x)) = (f₂∘f₁)(x) is possible if f₁(A₁) ⊆ A₂; f₁(f₂(x)) = (f₁∘f₂)(x) is possible if f₂(A₂) ⊆ A₁; f₂(f₁(x)) = (f₂∘f₁)(x) ≠ f₁(f₂(x)) = (f₁∘f₂)(x), the composition of functions does not have the commutative property

f₁: x ↦ x+1, f₂: x ↦ 2x; x → f₁ → x+1 → f₂ → 2x+2; x → f₂ → 2x → f₁ → 2x+1; therefore f₂(f₁(x)) = (f₂∘f₁)(x) ≠ f₁(f₂(x)) = (f₁∘f₂)(x)

By composing continuous functions we obtain continuous functions

f₁: A₁ → ℝ, x₀ is a non-isolated point of the set A₁, f₁ is continuous in x₀, f₁(x₀) = y₀ ∈ A₂, y₀ is a non-isolated point of the set A₂, f₂: A₂ → ℝ, f₁(A₁) ⊆ A₂, f₂ is continuous in y₀; x₀ → f₁ → f₁(x₀) =: y₀ → f₂ → f₂(f₁(x₀)) = (f₂∘f₁)(x), the compound function (f₂∘f₁)(x) is continuous; intuitively, a function is continuous if with a small variation of the input we obtain a small variation of the output, therefore a compound function is continuous if with a small variation of the input of the first function we obtain a small variation of the output of the second function

x ↦ √(x²+x+1); x ↦ x²+x+1, this function is continuous throughout its domain, the polynomial x²+x+1 assumes only positive values because it has no real roots, therefore we can apply the square root function; f₁(x) = x²+x+1 = y, f₂(y) = √(y), f₂(f₁(x)) = (f₂∘f₁)(x) = √(x²+x+1); polynomial functions are continuous functions, the root function is a continuous function, therefore the root of a polynomial is a continuous function

The function f is a minorant of the function g, and therefore the function g is a majorant of the function f, in the set A, if f(x) ≤ g(x) ∀ x ∈ A

f,g: A → ℝ, f(x) ≤ g(x), ∀ x ∈ A, function f is a minorant of function g, and therefore function g is a majorant of function f

If the function f(x) is greater than or equal to c in a neighborhood of an accumulation point of its domain, the limit of f, if any, is greater than or equal to c

f(x) ≥ c ∀ x ∈ A*(x₀,δ); if f(x) ≥ c it can never tend to a limit L < c

If f(x)> c the limit can be c; f(x) = x² > 0 for x ≠ 0, but lim_x→0(x²) = 0; the function 1/x is defined for x > 0, f(x) = 1/x > 0, but lim_x→+∞(1/x) = 0

If the function f(x) converges to the limit L > c, as x tends to an accumulation point of its domain, it is greater than c in a suitable neighborhood of the same point

f*(x) = f(x)-c, f* → L-c > 0; if the limit L exists and is greater than c, then the function is greater than c

Theorem of the 2 carabinieri: a function between 2 functions converging to the same limit also converges to the same limit

f,g,h: A → ℝ, x₀ is the accumulation point, f(x) ≤ g(x) ≤ h(x) ∀ x ∈ A*(x₀,δ), lim_x→x0(f(x)) = L ∧ lim_x→x0(h(x)) = L ⇒ lim_x→x0(g(x)) = L; ∀ x ∈ A*(x₀,δ), L-ε < f(x) ≤ g(x) ≤ h(x) < L+ε ⇒ L-ε < f(x) < L+ε, therefore the theorem of the 2 carabinieri is proved

If a function is a majorant of a function that diverges positively, then it also diverges positively

If a function is a minorant of a function that diverges negatively, then it also diverges negatively

f(x) ≤ g(x), lim_x→x0(f(x)) = +∞ ⇒ lim_x→x0(g(x)) = +∞

f(x) ≤ g(x), lim_x→x0(g(x)) = -∞ ⇒ lim_x→x0(f(x)) = -∞

a > 1, n ∈ ℕ*, the sequence ⁿ√(a) tends to 1, 1 < ⁿ√(a) < 1+(a-1)/n, 1+(a-1)/n tends to 1, for the theorem of the 2 carabinieri, and also ⁿ√(a) tends to 1

ⁿ√(n) tends to 1, and this can be proved with the 2 carabinieri theorem

a > 1, lim_n→+∞(aⁿ) = +∞, for Bernoulli's inequality, aⁿ = (1+d)ⁿ ≥ 1+nd, 1+nd is an arithmetic progression that diverges positively, therefore the geometric progression (1+d)ⁿ diverges positively because it is a minorant of a positively divergent sequence

The exponential function is continuous on ℝ

The exponential function is continuous in every point of ℝ; x ↦ a^x is a continuous function, therefore lim_x→x0(a^x) = a^x₀, x₀ ∈ ℝ, a > 1, x > x₀; if a > 1 the function a^x is increasing, the curve of the exponential function intersects the y-axis at the point (0,1), the point of abscissa 1 has the ordinate equal to the base a; a^x₁+x₂ = a^x₁⋅a^x₂ this is the fundamental property of the exponential function; 0 < a^x-a^x₀ = a^x-x₀+x₀-a^x₀ = a^x-x₀⋅a^x₀-a^x₀ = a^x₀(a^x-x₀-1), ⁿ√(a) = a^1/n tends to 1 when n tends to +∞, 0 < x-x₀ ≤ 1/n ⇔ n ≤ 1/(x-x₀), n = n(x) = [1/(x-x₀)], square brackets indicate the integer part, lim_x→x0⁺(1/(x-x₀)) = +∞, therefore n tends to +∞, 0 < a^x₀(a^x-x₀-1) ≤ a^x₀(a^1/n-1), 1 < ⁿ√(a) = a^1/n < 1+(a-1)/n ⇒ 1-1 < ⁿ√(a)-1 = a^1/n-1 < 1+(a-1)/n-1 ⇒ 0 < ⁿ√(a)-1 = a^1/n-1 < (a-1)/n, 0 < a^x₀(a^x-x₀-1) ≤ a^x₀(a^1/n-1) ≤ a^x₀((a-1)/n), when x tends to x₀⁺ n tends to +∞, and (a-1)/n tends to 0, therefore a^x tends to a^x₀, this is the proof of continuity to the right of the function a^x

a_n = (1+1/n)ⁿ is an increasing sequence, b_n = (1+1/n)ⁿ⁺¹ is a decreasing sequence, the sequences a_n and b_n tend to the number e; f(x) = (1+1/x)^x = ((x+1)/x)^x, this function is defined when (x+1)/x > 0, for x > 0 ∧ x < -1; lim_x→+∞((1+1/x)^x) = e, lim_x→-∞((1+1/x)^x) = e, therefore lim_x→±∞((1+1/x)^x) = e ≈ 2.71828

15 - LIMIT THEOREMS - PART 3

The exponential function is continuous on ℝ

x ↦ a^x, 0 < a ≠ 1, the exponential function is a continuous function

a_n = (1+1/n)ⁿ, n ∈ ℕ, it is an increasing monotonous sequence that tends to the number e; lim_x→+∞(1+1/x)^x = e, lim_x→-∞(1+1/x)^x = e; the function (1+1/x)^x is defined in (-∞,-1)∪(0,+∞), the base 1+1/x = (x+1)/x is positive for x > 0 ∧ x < -1

lim_x→+∞(1+1/x)^x; we consider the integer part, also called rounded down part, n = [x] = ⌊x⌋, n ∈ ℕ is the smallest integer that does not exceed x, and when x tends to +∞ also n tends to +∞; [x] ≤ x < [x]+1, (1+1/(n+1))ⁿ = (1+1/([x]+1))^[x] < (1+1/x)^x ≤ (1+1/[x])^[x]+1 = (1+1/n)ⁿ⁺¹ ⇒ a_n+1/(1+1/(n+1)) = (1+1/(n+1))ⁿ⁺¹/(1+1/(n+1)) < (1+1/x)^x < (1+1/n)ⁿ⁺¹ = b_n ⇒ e/1 < (1+1/x)^x < e ⇒ e < (1+1/x)^x < e, therefore the function (1+1/x)^x converges to the number e; we have proved lim_x→+∞(1+1/x)^x = e, and this is an important limit because it is useful for solving other limits

The circular functions, sine and cosine, are continuous on ℝ

The measure in radians of an angle is the ratio between the length of the arc associated with the angle and the length of the radius of the circumference; the ratio between the length of the arc and the radius of the circumference does not vary as the radius of the circumference varies; an angle measures 1 radian when the length of the arc is equal to the length of the radius; the length of the semicircle of radius 1 is π ≈ 3.14, therefore the angle associated with the semicircle of radius 1 is π/1 = π radians; 180/π ≈ 57, an angle of π radians corresponds to an angle of approximately 57°

|sin(x)| ≤ |x|; 0 < x < 1; a trigonometric circumference is a circumference with radius 1 and the center is the origin; x is the measure of the angle in radians, which in a circumference of radius 1 corresponds to the measure of the arc associated with the angle; the measure of an arch is the supremum of the lengths of the inscribed polygons; observing the trigonometric circumference, we understand that sin(x), which is the length of the cathetus opposite the angle, cannot be greater than the length of the arc associated with the angle, therefore sin(x) ≤ x; using a prosthaphaeresis formula, sin(x)-sin(x₀) = 2sin((x-x₀)/2)cos((x+x₀)/2) ⇒ |sin(x)-sin(x₀)| = 2|sin((x-x₀)/2)|⋅|cos((x+x₀)/2)| ⇒ |sin(x)-sin(x₀)| ≤ 2|(x-x₀)/2|⋅1 ⇒ |sin(x)-sin(x₀)| ≤ |x-x₀|, when x tends to x₀, sin(x) tends to sin(x₀), therefore δ_ε = ε; we have proved the continuity of the function sin(x)

cos(x) = sin(x+π/2), x ↦ x+π/2 =: t ↦ sin(t), composing 2 continuous functions we obtain a continuous compound function, therefore the function cos(x) is a continuous function

tan(x) = sin(x)/cos(x) is a continuous function because it is the ratio of 2 continuous functions

cot(x) = cos(x)/sin(x) is a continuous function because it is the ratio of 2 continuous functions

The ratio sin(x)/x tends to 1 as x tends to 0

lim_x→0(sin(x)/x) = 1 is an important limit because it is useful for resolving other limits; the function sin(x)/x is defined for x ≠ 0; we want to calculate the value of lim_x→0(sin(x)/x); sin(x) is an odd function, x is an odd function, the ratio of 2 odd functions is an even function, therefore sin(x)/x is an even function; lim_x→0(sin(x)/x) = 1 = lim_x→0^-(sin(x)/x) = lim_x→0⁺(sin(x)/x) because the function is even; we cannot apply the quotient limit theorem because in this case the denominator tends to 0; lim_x→0(sin(x)/x), the numerator tends to 0 and the denominator tends to 0, therefore it is a limit of the type 0/0 that is an indeterminate form; lim_x→0((1-cos(x))/x) = 0, this is also a limit of the indeterminate form 0/0; the ratio of 2 functions both converging to 0 must be studied case by case, because it depends on how the numerator and denominator tend to 0; studying the geometry of the trigonometric circumference we understand that 0 < sin(x) < x < tan(x) ⇒ sin(x)/sin(x) < x/sin(x) < tan(x)/sin(x) ⇒ 1 < x/sin(x) < (sin(x)/cos(x))/sin(x) ⇒ 1 < x/sin(x) < sin(x)/(cos(x)sin(x)) ⇒ 1 < x/sin(x) < 1/cos(x), lim_x→0(cos(x)) = cos(0) = 1 because the function cos(x) is continuous, therefore for the theorem of the 2 carabinieri lim_x→0(x/sin(x)) = 1 = lim_x→0(sin(x)/x); this function becomes continuous on all ℝ if f(x) = {sin(x)/x for x ≠ 0, 1 for x = 0}

A monotone sequence, or a monotone function, increasing or decreasing, is always regular, always has a limit; A monotone sequence, or a monotone function, increasing or decreasing, cannot be irregular, or oscillating

An increasing monotone sequence or function tends to its supremum, finite or +∞; a decreasing monotone sequence or function tends to its infimum, finite or -∞

If f is an increasing monotone function on the unbounded above set A, the limit of f as x tends to +∞ exists and coincides with sup(f(A))

If f is a decreasing monotone function on the unbounded below set A, the limit of f as x tends to +∞ exists and coincides with inf(f(A))

a^x, a > 1, is an increasing monotone function and is unbounded above; the sequence aⁿ diverges positively, the values assumed by the sequence aⁿ constitute an unbounded above set, which is a set without majorants; the image of the function a^x has no majorants, therefore lim_x→+∞(a^x) = +∞; lim_n→-∞(aⁿ) = 0⁺, lim_x→-∞(a^x) = 0⁺, the infimum of the image of the function a^x is 0, the function a^x is monotone and therefore as x tends to -∞, it tends to 0; a > 1, lim_x→+∞(a^x) = +∞, lim_x→-∞(a^x) = 0

x₁ < x₂, x₁,x₂ ∈ I, x₁ < x < x₂, x ∈ I, this is the connection property of the intervals

If a function is continuous on a bounded and closed interval, also called compact interval, its image is bounded above and below; if f: [a,b] → ℝ, and f is continuous, the image of the function is a set that is bounded below and above, it is a set that has minorants and majorants

The function f is continuous on a bounded and closed interval, f: [a,b] → ℝ, the image of the function is a set bounded below and above, therefore the image has minorants and majorants; f: [a,b] → ℝ, f is continuous ⇒ sup(f([a,b])) ∈ ℝ; we make a demonstration by absurdity, considering the negation of the thesis as a hypothesis and showing the contradiction that follows; we must prove that the image of the function is bounded above, the supremum is a real number; absurdly we suppose that the image of the function is unbounded above, and we show that this leads to the contradiction with the hypothesis of the continuity of the function; we suppose that the image of the function f in [a,b] is unbounded above, we divide the interval [a,b] into 2 equal parts by finding the central point c, and we try to understand if the function can be bounded in the intervals [a,c] and [c,b], and this cannot be true; let us consider the false hypothesis sup(f([a,b])) = +∞, therefore in [a,c], or in [c,b], the function is unbounded above; we rename the interval [a,c] to [a₁,b₁] and we take the central point c₁, therefore for the false hypothesis in [a₁,c₁], or in [c₁,b₁], the function is unbounded above; we consider [a_n,b_n], each interval is obtained by halving the previous interval, and by hypothesis the function is unbounded on each of these intervals, the length of these intervals is halved with each step; the sequence a_n increases and tends to its supremum, the sequence b_n decreases and tends to its infimum; the difference b_n-a_{n/sub> is halved at each step, therefore these 2 sequences converge at the same limit, therefore the set of elements of an and the set of elements of bn are two contiguous sets, therefore there exists an x0 which is the common limit of the sequences an and bn, x0 is the intersection of all intervals, the only element common to all intervals, we know that at point x0 the function is continuous, therefore the values that the function assumes in a neighborhood I(x0-δ,x0+δ) are included between f(x0)-ε and f(x0)+ε, f(x0)-ε < f(x) < f(x0)+ε; in the neighborhood I(x0-δ,x0+δ) the function is bounded above, the assumed values are minorants of f(x0)+ε; an increases and tends to x0, and bn decreases and tends to x0, therefore for large values of n, the interval [an,bn] is certainly contained within I(x0-δ,x0+δ), and therefore we have an absurdity, in the neighborhood I(x0-δ,x0+δ) the function is bounded above because the values it assumes, for the continuity at the point x0 are minorants of f(x0)+ε, while in all intervals [an,bn], for large values of n such that the interval [an,bn] is contained in the neighborhood I(x0,δ), the function is unbounded above, and all this is absurd; on a larger set the function is bounded above, and on a smaller set, contained in the previous one, the function is unbounded above, this is a contradiction because if a function is bounded on a set, it is bounded on a subset, therefore the continuous function f on the interval [a,b] is bounded above; with an analogous reasoning we prove that the continuous function f on the interval [a,b] is bounded below; sup(f([a,b])) = C ∈ ℝ, inf(f([a,b])) = c ∈ ℝ, we want to understand if the function assumes the values C and c, if C is a value assumed by the function then C is the maximum of the function, if c is a value assumed by the function then c is the minimum of the function}

16 - PROPERTIES OF CONTINUOUS FUNCTIONS ON AN INTERVAL

A continuous function in the bounded and closed interval [a,b] is bounded

f(x) = 1/x, (0,1], this function is continuous, unbounded above, bounded below, the graph is a branch of equilateral hyperbola

Weierstrass theorem: a continuous function f on a bounded and closed interval [a,b] has a maximum and a minimum

To prove Weierstrass theorem we must show that the supremum belongs to the function, and therefore is the maximum of the function, E := sup(f([a,b])), E ∈ f([a,b]); the supremum is the smallest of the majorants, therefore E is a majorant, f(x) ≤ E ∀ x; if E is the maximum of the function, there exists at least one value of x such that f(x) = E; we can prove Weierstrass theorem by absurdity, assuming f(x) < E ∀ x and showing that this leads to a contradiction, f(x) < E ⇒ E-f(x) > 0, g(x) := 1/(E-f(x)), the denominator is never equal to 0 because E-f(x) > 0, g(x) is a continuous function, by the previous theorem the function g must be bounded above, but in this case we observe that the function g is unbounded above; E is the smallest of the majorants, ∀ ε > 0 ∃ x_ε : E-ε < f(x_ε) < E ⇒ E-f(x_ε) < ε ⇒ 1/(E-f(x_ε)) > 1/ε ⇒ g(x_ε) > 1/ε, but this is a contradiction because ε is any positive number, 1/ε = M is any positive number, therefore ∀ M ∃ x_ε in which g(x_ε) > M, and therefore the function g would be unbounded above, but by hypothesis the function g, being a continuous function on the bounded and closed interval [a,b], by the previous theorem, should be bounded above, therefore the hypothesis f(x) < E ∀ x is not true because it is contradictory, and therefore we have proved Weierstrass theorem; if a function is continuous over a bounded and closed interval, the function has maximum and minimum

Study the function 1/x in the interval [1,+∞); [1,+∞) → 1/x → (0,1]; the function 1/x in the interval [1,+∞) has maximum but no minimum, 1 is the maximum, 0 is the infimum but it is not the minimum because the function 1/x does not assume the value 0

A continuous function on an interval cannot pass from negative to positive values, or vice versa, without assuming the value 0 at least in one point

If a function is continuous over an interval [a,b], a < b, f(a) < 0, f(b) > 0, there must exist a point x where f(x) = 0; we consider the point c intermediate between point a and point b, if f(c) = 0 we have found the point where f(x) = 0; if f(c) < 0 or f(c) > 0, we have to consider the interval in which the function changes sign and we rename it [a₁,b₁] and consider the intermediate point c₁; if f(c₁) = 0, we found the point where f(x) = 0, if f(c₁) < 0 or f(c₁) > 0 we must continue with the procedure considering the interval [a₂,b₂], and continuing we construct a sequence of intervals [a_n,b_n], each interval is nested in the previous one and has a width that is half of the previous interval, therefore we can affirm the existence of a point x₀ which is the common limit of the increasing sequence a_n and the decreasing sequence b_n, and at the extremes of the interval [a_n,b_n] the function assumes values of opposite sign, therefore f(a_n)⋅f(b_n) < 0, when n tends to +∞, a_n and b_n tend to x₀, the function is continuous; f(a_n) tends to f(x₀), f(b_n) tends to f(x₀), therefore f(a_n)⋅f(b_n) tends to (f(x₀))², if a sequence assumes only negative values and has a limit, the limit is negative or null, f(a_n)⋅f(b_n) < 0 ⇒ (f(x₀))² ≤ 0; the square of a real number is always a number greater than or equal to 0, 0 ≤ (f(x₀))² ≤ 0, therefore f(x₀) = 0 is the point we were looking for

A continuous function on an interval cannot assume 2 values without assuming all the intermediate values

Considering the continuous function f: I → ℝ, f(x₁) = y₁ < y₂ = f(x₂), y₁ < y < y₂, g(x) := f(x)-y, g(x₁) = f(x₁)-y = y₁-y < 0, g(x₂) = f(x₂)-y = y₂-y > 0, therefore by the previous theorem there is at least one point x in which g(x) = 0 = f(x)-y ⇒ f(x) = y

The sin(x) function is continuous, sin(π/2) = 1, sin(-π/2) = -1, -1 < sin(x) < 1, im(sin(x)) = [-1,1], the function sin(x) assumes all the values between -1 and 1; dom(sin(x)) = ℝ, im(sin(x)) = [-1,1], ℝ → sin(x) → [-1,1]

Continuous functions transform intervals into intervals, in particular they transform bounded and closed intervals into intervals of the same type

We consider the function tan(x) defined in the open interval (-π/2,π/2); lim_x→(π/2)^- = +∞, lim_{x→(-π/2)⁺} = -∞; tan(x) is an odd function; the image of the tan(x) function is unbounded above and below, but it must be an interval because it is the image of an interval, and the only interval that is unbounded above and below is all ℝ, dom(tan(x)) = (-π/2,π/2) → tan(x) → im(tan(x)) = (-∞,+∞) = ℝ

x ↦ a^x, if a > 1 the exponential function a^x is monotone increasing, if a < 1 the exponential function a^x is monotone decreasing, dom(a^x) = ℝ, im(a^x) = (0,+∞), +∞ is the supremum and 0 is the infimum, 0 does not belong to the image because the function a^x does not assume the value 0, ℝ → (0,+∞) = ℝ⁺ = ℝ*₊

√(x), the square root function is the inverse function of the squaring function, dom(√(x)) = [0,+∞), im(√(x)) = [0,+∞) = ℝ₊, 0 is the minimum of the function, it belongs to the image because the function √(x) assumes the value 0, and this is also true for the function ⁿ√(x)

f(x) is a function, 1/f(x) is the reciprocal function, f^-1(x) is the inverse function; x ∈ I → f → f(x); a function f is 1-1 or injective if it associates different output values to different input values; monotonous strictly increasing functions and monotonous strictly decreasing functions are injective; if the function f is strictly increasing, x₁ → f → f(x₁), x₂ → f → f(x₂), x₁ < x₂ ⇒ f(x₁) < f(x₂); if the function f is strictly decreasing, x₁ → f → f(x₁), x₂ → f → f(x₂), x₁ > x₂ ⇒ f(x₁) > f(x₂); a function is injective when any parallel line intersects the graph of the function in no more than 1 point; the function sin(x), without specifying the domain, is not injective, in fact a parallel line intersects the graph of sin(x) in an infinite number of points, and this is due to the periodicity of the function sin(x); if a function is injective then it is invertible; x → f → f(x) = y → f^-1 → f^-1(y) = x

y = f(x) = mx+q, m ≠ 0, it is a strictly monotonous function, strictly increasing if m > 0, strictly decreasing if m < 0, therefore it is an injective function; inverse function means expressing x as a function of y, f(x) = y = mx+q ⇒ mx = y-q ⇒ x = (1/m)(y-q) = f^-1(y)

The square root function is the inverse function of the square function, but it is restricted to the set of non-negative values; the graph of the function f(x) = y = x² is a parabola, it is not an injective function in fact every line parallel to the x-axis, for y > 0, intersects the graph of the parabola in 2 points, but if we restrict the domain to [0,+∞) the function f(x) = y = x² becomes strictly increasing, therefore injective, and consequently invertible, y = x² ⇔ x = √(y); exchanging the name of the variables x and y is irrelevant; f: x ↦ x² with x ≥ 0, f^-1: x ↦ √(x) with x ≥ 0

The graph of the inverse function is symmetrical to the graph of the direct function with respect to the bisector of the first and third quadrant

If f is a continuous and strictly monotone function that transforms the interval I into the interval J, the inverse function is continuous and strictly monotone on J

If a function is strictly increasing, its inverse function is strictly increasing; if a function is strictly decreasing, its inverse function is strictly decreasing

A function is surjective when the codomain is the image of the function

If a function is continuous and strictly monotone over an interval I, then it is injective; if the codomain of the function coincides with its image J, then it is surjective; if a function is injective and surjective, then it is invertible; f: I → J, f^-1: J → I

The exponential function a^x has ℝ as domain and ℝ⁺ = (0,+∞) as image; if a > 1, the exponential function is strictly increasing; the exponential function is a continuous and strictly monotone function on ℝ, therefore it is injective and invertible; x ↦ a^x with x ∈ ℝ, x ↦ log_a(x) with x ∈ ℝ⁺; the logarithm function is the inverse function of the exponential function; x → f → a^x → f^-1 → x = a^log_a(x); the logarithm of base a, of a number, is the exponent to be given to the base a to obtain that number; the logarithm function is a continuous function, is defined in ℝ⁺ = (0,+∞) and ℝ is the image; the graph of the logarithm function is symmetrical to the graph of the exponential function with respect to the bisector of the first and third quadrant; the exponential function has the x-axis as its horizontal asymptote, but has no vertical asymptote; the logarithm function has the y-axis as its vertical asymptote, but it has no horizontal asymptote

Circular functions are not injective because they are periodic, therefore they cannot be inverted globally

It is not very correct to say that the square root function is the inverse function of the square function, because the squaring function is not injective; the square root function is the inverse function of the squaring function after it has been restricted to the set of non-negative numbers

sin(x) is a continuous and strictly increasing function in the bounded and closed interval [-π/2,π/2], therefore the function sin(x) is invertible if we consider the restricted domain [-π/2,π/2], dom(sin(x)) = [-π/2,π/2] → sin(x) → im(sin(x)) = [-1,1]; the inverse function of sin(x) restricted to [-π/2,π/2] is arcsin(x) that is a continuous and strictly increasing function, dom(arcsin(x)) = [-1,1] → arcsin(x) → im(arcsin(x)) = [-π/2,π/2]; the graph of the function arcsin(x) is symmetric to the graph of the function sin(x) restricted to [-π/2,π/2] with respect to the bisector of the first and third quadrant; the arcsine function is the inverse of the sine function restricted to the interval [-π/2,π/2]; the arcsin(x) function accepts as input values between -1 and 1, and the outputs values are between -π/2 and π/2

cos(x) is a continuous and strictly decreasing function in the bounded and closed interval [0,π], therefore the function cos(x) is invertible if we consider the restricted domain [0,π], dom(cos(x)) = [[0,π]] → cos(x) → im(cos(x)) = [-1,1]; the inverse function of cos(x), restricted to [0,π], is arccos(x), which is a continuous and strictly decreasing function, dom(arccos(x)) = [-1,1] → arccos(x) → im(arccos(x)) → [0,π]; the graph of the function arccos(x) is symmetrical to the graph of the function cos(x), restricted to [0,π], with respect to the bisector of the first and third quadrant; the arccosine function is the inverse of the cosine function restricted to the interval [0,π]; the arccos(x) function accepts as input values between -1 and 1, and the outputs are values between 0 and π

tan(x) = sin(x)/cos(x), the tangent function is not defined in the values of x such that cos(x) = 0, therefore x = k(π/2), with k ∈ ℤ, are vertical asymptotes of the function tan(x); the tan(x) function is continuous and strictly increasing in the bounded and open interval (-π/2,π/2), therefore the function tan(x) is invertible if we consider the restricted domain (-π/2,π/2), dom(tan(x)) = (-π/2,π/2) → tan(x) → im(tan(x)) = (-∞,∞) = ℝ; the inverse function of tan(x) is arctan(x) that is a continuous and strictly increasing function, dom(arctan(x)) = (-∞,∞) = ℝ → arctan(x) → im(arctan(x)) = (-π/2,π/2); the graph of the function arctan(x) is symmetrical to the graph of the function tan(x), restricted to (-π/2,π/2), with respect to the bisector of the first and third quadrant; the arctangent function is the inverse of the tangent function restricted to the interval (-π/2,π/2); the arctan(x) function accepts as input any value of ℝ, and returns as output values between -π/2 and π/2; the supremum of arctan(x) is π/2, the infimum of arctan(x) is -π/2; the distance between the supremum of arctan(x), which is π/2, and the infimum of arctan(x), which is -π/2, is called the oscillation of arctan(x) and its value is π

17 - INTRODUCTION TO THE CONCEPT OF VECTOR SPACE

In physics, forces, velocities, and acceleration are vectors; F = m⋅a

The gravity force G and the gravity acceleration g are vectors facing the center of the Earth; G = m⋅g

A vector is an arrow, or oriented segment OP, O → P, OP = v; the characteristics of a vector are the direction that is the straight line passing through O and P, the verse that goes from O to P, the modulus or length that is the positive or null number that measures the length of the OP segment

The sum of 2 vectors is obtained with the parallelogram rule

2 vectors are opposite when they have the same direction, same modulus, but opposite directions

For the sum of vectors the commutative property holds: v+w = w+v

For the sum of vectors the associative property holds: (v+w)+z = v+(w+z)

The product of a real number by a vector, a⋅v, has the same direction as v, has length |a||v|, has the same verse as that of v if a is positive, opposite to that of v if a is negative, a⋅v is the null vector if a = 0 or if v is the null vector

First distributive property: a(v+w) = av+aw

Second distributive property: (a+b)v = av+bv

1 is the neutral number for multiplication: 1⋅v = v

Associative property for multiplication: (ab)v = a(bv) = b(av)

The set of vectors of the plane, or of the space, is endowed with the sum operation and the product operation for a number, and it is endowed with the commutative property, the associative property, the distributive property, the existence of the opposite, the neutral element, the null element

ℝ² is the set of all pairs (a,b) of real numbers

Sum of pairs: (a,b)+(c,d) (a+c,b+d)

Product of a number for a pair: m(a,b) = (ma,mb)

Properties of the sum in R²; S1, commutative property, (a,b)+(c,d) = (c,d)+(a,b); S2, associative property, ((a,b)+(c,d))+(e,f) = (a,b)+((c,d)+(e,f)); S3, existence of 0, there exists the pair (0,0) such that (a,b)+(0,0) = (a,b); S4, existence of the opposite, for each pair (a,b) there exists the opposite pair (-a,-b) such that (a,b)+(-a,-b) = (0,0)

Properties of the product in ℝ²; P1, first distributive property, m((a,b)+(c,d)) = m(a,b)+m(c,d); P2, second distributive property, (m+n)(a,b) = m(a,b)+n(a,b); P3, 1(a,b) = (a,b); P4, (m⋅n)(a,b) = m(n(a,b))

The properties of vectors are the same as the properties of tuples in ℝⁿ

The vectors of the plane ℝ², but in general the vectors of ℝⁿ, have sum, product for a number, and 8 properties

18 - VECTOR SPACES - LINEAR DEPENDENCE - LINEAR INDEPENDENCE

A vector space is a set where it is possible to make the sum of vectors and the product of a number by a vector

Vector spaces have 8 properties, also called axioms; S1, commutative property, v+w = w+v; S2, associative property, (v+w)+z = v+(w+z); S3, existence of 0, v+0_v = v; S4, existence of the opposite, v+(-v) = 0_v; P1, first distributive property, a(v+w) = av+aw; P2, second distributive property, (a+b)v = av+bv; P3, 1⋅v = v; P4, (a⋅b)v = a(b⋅v) = b(a⋅v)

ℝ, ℝ², ℝⁿ are vector spaces; the vectors of the 2-dimensional space form a vector space

{(x,x) ∈ ℝ²} = (1,1),(2,2),(-1,-1),(3,3); adding 2 pairs of this set we obtain a pair belonging to the set, (1,1)+(2,2) = (3,3); multiplying a number by a pair of this set we obtain a pair belonging to the set, α(x,x) = (αx,αx); all 8 properties are satisfied, commutative property, associative property, distributive property, (0+0)+(x,x) = (x,x), -(x,x) = (-x,-x)

0 = 0_v = null vector; -v is the vector opposite to the vector v, v+(-v) = v-v = 0_v

(x+y)+(-(x,y)) = (x,y)-(x-y)

a ∈ ℝ, v ∈ V, a⋅v = 0_v, 0⋅v = 0_v, a⋅0_v = 0_v, a⋅v = 0_v ⇒ a = 0 ∨ v = 0_v, this is the product cancellation law

-(a⋅v) = (-a)v = a(-v); -(4(2,3)) = -4(2,3) = (-8,-12), -(4(2,3)) = 4(-2,-3) = (-8,-12)

W subspace of V, W ⊂ V, W is a vector space with the sum and product properties of V

{(x,x) ∈ ℝ²} is a vector subspace of ℝ²

W = {(x,y,x+y) ∈ ℝ³}; a triple of this subspace is for example (3,2,5); the triple (3,2,0) does not belong to this subspace; adding 2 vectors of W we obtain another vector of W, multiplying a number by a vector of W we obtain another vector of W, and all 8 properties are maintained, therefore W is a vector subspace of the vector space ℝ³

W = {(x,x+1)}; (3,3+1) = (3,4) is a pair of the set W; (3,3) is not a pair of the set W; (3,4)+(5,6) = (8,10) which is not a pair of the set W, in the set W the operation of the sum is not defined, therefore the set W is not a vector subspace; 2(3,4) = (6,8) which is not a pair of the set W, in the set W the operation of the product is not defined, therefore the set W is not a vector subspace; W is an example of a subset of ℝ² which is not a vector subspace; W is a set that is not a vector space

a₁v₁+a₂v₂+...+a_nv_n is a linear combination; 3(2,-2)+(-4)(0,1)+7(1,7) is a linear combination in ℝ²

2(1,1,1)+(2,2,2)-(4,4,4) = (2⋅1+1⋅2+(-1)4,2⋅1+1⋅2+(-1)4,2⋅1+1⋅2+(-1)4) = (2+2-4,2+2-4,2+2-4) = (0,0,0); (1,1,1),(2,2,2),(4,4,4) are linearly dependent vectors because their linear combination is zero when at least one coefficient is not zero, and in this case the coefficients are a = 2, b = 1, c = -1

a(1,0)+b(1,1) = (0,0), a⋅1+b⋅1 = 0 ⇒ a+b = 0, a⋅0+b⋅1 = 0 ⇒ b = 0, a+0 = 0 ⇒ a = 0; (1,0),(1,1) are linearly independent vectors because their linear combination is zero when all coefficients are zero, and in this case the coefficients are a = 0, b = 0

Linearly independent vectors: a₁v₁+a₂v₂+...+a_nv_n = 0 ⇒ a₁ = a₂ = ... = a_n = 0

If we can obtain a null linear combination with at least one non-zero coefficient then the vectors are linearly dependent

The vectors v₁, ..., v_m are linearly independent if and only if: v₁ ≠ 0_v, v₂ is not a multiple of v₁, v₃ is not linear combination of v₁ and v₂, v_m is not linear combination of v₁, ..., v_m-1

The fundamental versors of ℝ³ are e₁ = (1,0,0), e₂ = (0,1,0), e₃ = (0,0,1), and are linearly independent in ℝ³; e₁ = (1,0,0) ≠ (0,0,0), ∄ a ∈ ℝ : (0,1,0) = a(1,0,0), (0,0,1) = a(1,0,0)+b(0,1,0), 0 = a⋅1+b⋅0 ⇒ 0 = a, 0 = a⋅0+b⋅1 ⇒ 0 = b, 1 = a⋅0+b⋅0 ⇒ 1 = 0 that is false, therefore the fundamental versors of ℝ³ are linearly independent

The fundamental versors of ℝⁿ are e₁ = (1,0,0,...,0), e₂ = (0,1,0,...,0), ..., e_n = (0,0,0,...,1), and are linearly independent in ℝⁿ

Vector v is the linear combination of n linearly independent vectors, v = a₁v₁+...+a_nv_n, v = b₁v₁+...+b_nv_n, a₁ = b₁, ..., a_n = b_n; a linear combination of linearly independent vectors has unique coefficients

(1,0,0),(0,1,0),(0,0,1) are linearly independent vectors, therefore a linear combination of them has unique coefficients; (1,2,1) = a(1,0,0)+b(0,1,0)+c(0,0,1), a = 1, b = 2, c = 1

The coefficients of a linear combination are also called vector components

19 - GENERATORS - BASES - DIMENSION OF A VECTOR SPACE

Generators of a vector space: {v₁,v₂,...,v_n} generate V such that for every v in V we have v = a₁v₁+a₂v₂+...+a_nv_n

From the generating vectors it is possible to obtain all the vectors of the vector space

The fundamental versors e₁ = (1,0,0,...,0), e₂ = (0,1,0,...,0), ..., e_n = (0,0,0,...,n), generate ℝⁿ

The fundamental versors of ℝⁿ are generators of ℝⁿ, each element of ℝⁿ can be written as a linear combination of these vectors

n = 3, (1,0,0),(0,1,0),(0,0,1), these 3 vectors are generators of the vector space ℝ³, (a,b,c) = a(1,0,0)+b(0,1,0)+c(0,0,1), a(1,0,0)+b(0,1,0)+c(0,0,1) = (a⋅1+b⋅0+c⋅0,a⋅0+b⋅1+c⋅0,a⋅0+b⋅0+c⋅1) = (a+0+0,0+b+0,0+0+c) = (a,b,c)

The fundamental versors are generators, but generators may also not be fundamental versors

Vectors (1,1),(1,0) in ℝ² are not versors of ℝ², but they are generators of ℝ², in fact with their linear combinations a(1,1)+b(1,0) we can write any vector of ℝ²; a(1,1)+b(1,0) = (α,β) ⇒ (a⋅1+b⋅1,a⋅1+b⋅0) = (α,β) ⇒ (a+b,a) = (α,β) ⇒ a+b = α ∧ a = β ⇒ β+b = α ⇒ b = α-β, a = β ∧ b = α-β; to write the pair (3,5) as a linear combination, a = β ⇒ a = 5, b = α-β ⇒ b = 3-5 = -2; this is a system of generators for the vector space ℝ²

v₁ = (1,1), v₂ = (1,0), v₃ = (0,1), these 3 vectors are a set of generators for ℝ², with the linear combinations a(1,1)+b(1,0)+c(0,1) = (α,β) we obtain all vectors, a+b = α, a+c = β, there are infinite solutions, the coefficients a, b, c can have infinite values, a = 0 ⇒ b = α ∧ c = β, a = 1 ⇒ b = α-1 ∧ c = β-1

Bases of vector spaces: {v₁,v₂,...,v_n} base of V ⇒ linearly independent generators

A set of vectors is a base of a vector space when the vectors are linearly independent generators

When vectors are linearly independent generators they can be written as a linear combination v = a₁v₁+...+a_nv_n, and the coefficients a₁, a_{n are unique; a base is a set of generators such that each vector v of the vector space is written as a linear combination of these vectors in a unique way}

Bases of n-tuple: e₁ = (1,0,...,0), e₂ = (0,1,...,0), ..., e_n = (0,0,...,1); the fundamental versors of ℝⁿ are a base of ℝⁿ

The fundamental versors of ℝⁿ are a system of generators of ℝⁿ and also a base of ℝⁿ because they are linearly independent, that is, each element of ℝⁿ can be written uniquely as a linear combination of the fundamental versors of ℝⁿ

Fundamental versors of ℝ² (1,0),(0,1) are a base of ℝ²; in ℝ² there are also other bases, for example the vectors (1,1),(1,0) are a base of ℝ² because they generate ℝ² and are linearly independent because the vector (1,1) is not null and the vector (1,0) is not a multiple of the vector (1,1), and the coefficients of the linear combination are unique

The vectors v₁ = (1,1), v₂ = (1,0), v₃ = (0,1) are generators of the vector space ℝ² but they are not a base because they are non linearly independent generators, if they were linearly independent each vector of the vector space v could be written in a unique way as a linear combination of these vectors; these vectors allow to represent any other vector v by linear combination, but there is no uniqueness because they are linearly dependent vectors, therefore this is a set of generators that is not a base

Components: v = a₁v₁+a₂v₂+...+a_nv_n, the numbers a₁,a₂,...,a_n, considering the order, are the components of v with respect to the base (v₁,v₂,...,v_n)

A base consists of ordered vectors, the vectors that form a base have a very precise order, changing the order of the vectors we get another base because the components change

Steinitz theorem: x₁,x₂,...,x_n are generators, y₁,y₂,...,y_m are linearly independent, then m ≤ n

Steinitz's theorem serves us to count the number of elements of a base of a vector space; if we have a system of n generators of a vector space and a system of m linearly independent vectors in the vector space, then the number m of linearly independent vectors can never exceed the number n of generators; a set of independent vectors is always less than or equal to the number of vectors of any other generator system; generators are more than linearly independent vectors

Steinitz's theorem states that in a vector space generators are more than linearly independent vectors

Corollary of Steinitz's theorem: all bases have the same number of elements

The corollary of Steinitz's theorem states that all the bases of a vector space have the same number of elements; if a base is v₁,...,v_m, and another base is w₁,...,w_r, r and m must be the same, in fact the Steinitz theorem states that the number m is the number of vectors of the set v₁,...,v_m, and it is also the number of vectors of a set of linearly independent vectors because a base is formed by linearly independent vectors, r is the number of vectors of another base or is the number of vectors of a set of generators because the bases are also generators, therefore m ≤ r, for the same reason r ≤ m, therefore m = r, therefore all the bases of a vector space have the same number of elements

By the corollary of Steinitz's theorem we know that all the bases of a vector space have the same number of elements, and this is a property of vector spaces that allows us to arrive at the definition of the dimension of a vector space

Dimension: (v₁,v₂,...,v_n) any base of V, n = dim (V); all bases of V have n elements

The dimension of a vector space is the number of elements of its base, and dimension is a characteristic of any vector space

The dimension of ℝ² is 2 because the base (1,0),(0,1) is formed by 2 elements; all the bases of ℝ² are formed by 2 elements, therefore ℝ² has dimension 2; dim(ℝ²) = 2

ℝ³ is a vector space that has a base formed by 3 vectors, so its dimension is 3, dim(ℝ³) = 3

ℝ⁴ is a vector space that has a base formed by 4 vectors, so its dimension is 4, dim(ℝ⁴) = 4

ℝⁿ is a vector space that has a base formed by n fundamental versors, therefore its dimension is n, dim(ℝⁿ) = n

Consequences of Steinitz's theorem: dim (V) = n, and this implies that n linearly independent vectors form a base, and n generators form a base

If a vector space V admits a vector subspace W, then W also has its dimension and its bases

Dimension of subspaces: W subspace of V, dim (W) ≤ dim (V); if dim (V) = n, then 0 ≤ dim (W) ≤ n; dim (W) = 0 ⇒ W = {0 v}; dim (W) = n ⇒ W = V

The dimension of the vector subspace cannot exceed the dimension of the vector space; if a vector space, or a vector subspace, has dimension 0 then it is formed by the null vector only; if the dimension of the vector subspace is equal to the dimension of the vector space then the vector subspace coincides with the vector space

In a vector space ℝⁿ there cannot exist a subspace with dimension n+1; if a vector subspace contains some element other than 0, then its dimension cannot be 0

In a vector space V we have the subspace W = L(v₁,...,v_m), the vectors v₁,...,v_m are the generators of the subspace W; we want to find a base for this vector subspace

Method of rejects: from a system of generators v₁,v₂,...,v_m we have to discard the null vectors, the second vector must not be a multiple of the first, the third vector must not be a linear combination of the preceding vectors, we must discard any vector that is linear combination of the previous vectors, when we have eliminated all the vectors that are linear combinations of the previous ones we obtain a base of the vector space

Generators are usually not a base because there are too many, therefore they are not linearly independent; vectors are not linearly independent when there are too many

From a system of generators we can obtain a base of the vector space by eliminating all the vectors which are linear combinations of the previous ones, this is the method of rejects

v₁,v₂,v₃ is a set of generators of ℝ², ℝ² = L(v₁,v₂,v₃), v₁ = (1,1), v₂ = (1,0), v₃ = (0,1), we want to find a base with the method of rejects; we must eliminate the null vectors, but there is no null vector here; the first vector, v₁ = (1,1) is not zero therefore v₁ = (1,1) is part of the base; the second vector, v₂ = (1,0) is not a linear combination of the first vector v₁ = (1,1), and the second vector v₂ = (1,0) is not a multiple of the first vector v₁ = (1,1), therefore v₂ = (1,0) is part of the base; v₃ = (0,1) is linear combination of v₁ = (1,1) and v₂ = (1,0 ) because 3 vectors are too many to form a base of ℝ², therefore we have to discard v₃ = (0,1); starting from ℝ² = L(v₁,v₂,v₃), using the method of rejects, we obtained the base (v₁,v₂) that consists of the 2 independent vectors v₁ and v₂; if we consider the 3 generating vectors of the vector space ℝ² in the order v₃,v₁,v₂ the procedure remains the same but the result changes, in fact v₃ = (0,1) is not zero and therefore is part of the base, the vector v₁ = (1,1) is not a multiple of the vector v₃ = (0,1), therefore we are obliged to discard the vector v₂ = (1,0) because it is certainly a linear combination of the preceding vectors by Steinitz's theorem; starting from ℝ² = L(v₃,v₁,v₂), using the method of rejects, we obtained the base (v₃,v₁) that consists of the 2 independent vectors v₃ and v₁; this case is also simple because we knew the dimension of the vector space dim(ℝ²) = 2

We have a vector space V with generating vectors v₁,...,v_m, but we don't want to find a base from these generators; w₁,w₂,...,w_r are other linearly independent vectors of the vector space, we want to find a base of the vector space V formed by the vectors w₁,w₂,...,w_r and possibly by other vectors, we must not discard but add, therefore we have to effect the completion of the vectors

Completion of the linearly independent vectors w₁,w₂,...,w_r to a base; v₁,v₂,...,v_m are generators of V, w₁,w₂,...,w_r,v₁,v₂,...,v_m are still generators of V, and from these we can extract a base with the method of rejects, we must discard any vector that is a linear combination of the previous ones

In ℝ³ we are interested in a base which contains the vector (1,1,1); we know that a base of the vector space ℝ³, or a system of generators, is formed by the vectors (1,1,1),(1,0,0),(0,1,0),(0,0,1), and we apply the method of rejects; (1,1,1) is not 0 therefore we do not discard it, (1,0,0) is not multiple of (1,1,1) and we do not discard it, we have to figure out whether to discard (0,1,0), (0,1,0) = a(1,1,1)+b(1,0,0) ⇒ 0 = a+b ∧ 1 = a ∧ 0 = a that are 3 incompatible equations, therefore the vector (0,1,0) is not a linear combination of the vectors (1,1,1) and (1,0,0); the fourth vector (0,0,1) is to be discarded because ℝ³ has dimension 3, therefore a base of the space ℝ³ is formed by 3 vectors; the base of the vector space ℝ³ containing the vector (1,1,1) is formed by the vectors (1,1,1),(1,0,0),(0,1,0); with the completion method, starting from a vector, we obtained a base of ℝ³

Base = linearly independent generators

Dimension = number of elements of each base

A vector space base consists of linearly independent generators

The dimension of a vector space is the number of elements of each base

A system of generators is a set of vectors from whose linear combination we can obtain every vector v of the vector space V, {v₁,v₂,...,v_n} is a system of generators of the vector space V if ∀ v ∈ V occurs v = a₁v₁+a₂v₂+...+a_nv_n, therefore from a system of generators we can obtain all the vectors of the vector space

A base is a system of linearly independent generators, therefore a base is a set of generators such that each vector v of the vector space V can be written as a linear combination of these vectors in a unique way, the coefficients of the linear combination have a unique value; as a consequence of Steinitz's theorem, if dim(V) = n, then n linearly independent vectors form a base, and n generators form a base; a base is a set of linearly independent generators; the dimension of a vector space is the number of vectors constituting each base

Determine if the vector (0,1,0) is a linear combination of the vectors (1,1,1) and (1,0,0); (0,1,0) = a(1,1,1)+b(1,0,0) ⇒ {0 = a+b, 1 = a, 0 = 0}, these equations are incompatible, therefore the vector (0,1,0) is not a linear combination of the vectors (1,1,1) and (1,0,0)

Determine if the vectors (1,1,3), (0,1,1), (1,0,2) are linearly dependent or independent; a(1,1,3)+b(0,1,1)+c(1,0,2) = (0,0,0) ⇒ {a+c = 0, a+b = 0, 3a+b+2c = 0} ⇒ {c = -a, b = -a, 3a+b+2c = 0} ⇒ {c = -a, b = -a, 3a+(-a)+2(-a) = 0} ⇒ {c = -a, b = -a, 3a-a-2a = 0} ⇒ {c = -a, b = -a, 3a-3a = 0} ⇒ {c = -a, b = -a, 0 = 0}, therefore the system admits infinite solutions, any triplet (a,-a,-a) is a solution, for example with a = 1, b = -1, c = -1 we get 1(1,1,3)-1(0,1,1)-1(1,0,2) = (1-0-1,1-1-0,3-1-2) = (0,0,0), we have obtained the null vector with the coefficients of the linear combination that are not all null, therefore the vectors are linearly dependent; the vectors (1,1,3),(0,1,1),(1,0,2) are not linearly independent because the vector (1,1,3) is a linear combination of (0,1,1) and (1,0,2), (1,1,3) = (0,1,1)+(1,0,2); 3 linearly dependent vectors are coplanar, they are on the same plane; 3 linearly independent vectors are not coplanar, they are not on the same plane

The vectors v₁,v₂,...,v_n are a system of generators of V if any vector of V can be obtained as a linear combination of v₁,v₂,...,v_n that is if V = L(v₁,v₂,...,v_n), if the vector subspace generated by vectors v₁,v₂,...,v_n coincides with the entire vector space V

The vectors v₁,v₂,...,v_n are a base of the vector space V if they are linearly independent and if they are a system of generators; a set of linearly independent generators are a base

If the vector space V has a base consisting of n vectors, then every other base of V is made up of n vectors, therefore V has dimension n, dim (V) = n

The decomposition of a vector v as a linear combination of the vectors of a base is unique, if v₁,v₂,...,v_n are a base of the vector space V then ∀ v ∈ V the coefficients a₁,a₂,...,a_n of the linear combination are unique; v = a₁v₁+a₂v₂+...+a_nv_n, the coefficients a₁,a₂,...,a_n are the components of v with respect to the base v₁,v₂,...,v_n

In ℝ² a possible base is formed by vectors v₁ = (1,0) and v₂ = (0,1), and these 2 vectors are the fundamental versors of ℝ², they are linearly independent so they are not parallel, and they are generators because ∀ v = (a,b) ∈ ℝ² : v = a(1,0)+b(0,1)

In ℝ³ a possible base is v₁ = (1,0,0), v₂ = (0,1,0), v₃ = (0,0,1), these 3 vectors are the fundamental versors of ℝ³, are linearly independent so they are not coplanar, and they are generators because ∀ v = (a,b,c) ∈ ℝ³ : v = a(1,0,0)+b(0,1,0)+c(0,0,1)

A base made up of the fundamental versors is called a canonical base

If dim(V) = n and v₁,v₂,...,v_n are a set of linearly independent vectors, then these vectors are a base; 2 linearly independent vectors in ℝ² are a base, we do not have to verify that they are also generators; 3 linearly independent vectors in ℝ³ are a base, we do not have to verify that they are also generators

If dim(V) = n and v₁,v₂,...,v_n are a generator system, then these vectors are a base and therefore they are linearly independent

Prove that the vectors v₁ = (1,2) and v₂ = (5,1) form a base of ℝ², and find the components of v₃ = (11,4) with respect to this base; to prove that v₁ and v₂ are a base of ℝ² we have to prove that they are 2 linearly independent vectors and that they are a system of generators; a(1,2)+b(5,1) = (0,0) ⇒ {a+5b = 0, 2a+b = 0} ⇒ {a = -5b, 2a+b = 0} ⇒ {a = -5b, 2(-5b)+b = 0} ⇒ {a = -5b, -10b+b = 0} ⇒ {a = -5b, -9b = 0} ⇒ {a = -5b, b = 0} ⇒ {a = 0, b = 0}, the null vector (0,0) is obtained when in the linear combination a(1,2)+b(5,1) the coefficients a and b are equal to 0, therefore the vectors v₁ = (1,2) and v₂ = (5,1) are linearly independent; to prove that v₁ = (1,2) and v₂ = (5,1) are a system of generators we must prove that any vector of ℝ² can be written as a linear combination of v₁ = (1,2) and v₂ = (5,1), so we must show that a(1,2)+b(5,1) = (α,β) ⇒ {a+5b = α, 2a+b = β} ⇒ {a = α-5b, 2a+b = β} ⇒ {a = α-5b, 2(α-5b)+b = β} ⇒ {a = α-5b, 2α-10b+b = β} ⇒ {a = α-5b, 2α-9b = β} ⇒ {a = α-5b, -9b = β-2α} ⇒ {a = α-5b, b = (β-2α)/-9} ⇒ {a = α-5b, b = (2α-β)/9} ⇒ {a = α-5((2α-β)/9), b = (2α-β)/9} ⇒ {a = α+((-10α+5β)/9), b = (2α-β)/9} ⇒ {a = (9α-10α+5β)/9, b = (2α-β)/9} ⇒ {a = (-α+5β)/9, b = (2α-β)/9}, we found the coefficients of the linear combination of vectors v₁ = (1,2) and v₂ = (5,1) that allow to obtain the components (α,β) of a generic vector of ℝ², a(1,2)+b(5,1) = ((-α+5β)/9)(1,2)+((2α-β)/9)(5,1) = (α,β) is the linear combination to obtain any vector of ℝ² starting from the generator system {v₁ = (1,2), v₂ = (5,1)}; the vectors v₁ = (1,2) and v₂ are linearly independent and are a system of generators, therefore they are a base of ℝ²; for the corollary of the Steinitz Theorem, we know that n linearly independent vectors of a vector space of dimension n form a base, and we know that a base is a set of linearly independent generators, moreover if we have n generators in a vector space of dimension n, then these generators are a base and therefore they are linearly independent vectors; the vector space ℝ² has dimension 2, the vectors v₁ = (1,2) and v₂ are linearly independent, therefore they form a base, therefore they are generators; the simplest method to prove that a set of vectors is a system of generators is to show that their number is equal to the dimension of the vector space and that they are linearly independent; we have to find the components of v₃ = (11,4) with respect to the base constituted by the vectors v₁ = (1,2) and v₂ = (5,1); the components of a vector with respect to the base are the coefficients of the linear combination of the base resulting in the vector we are interested in; to find the components a and b of the vector (11,4) we must impose α = 11 and β = 4, a = (-α+5β)/9 = (-11+5⋅4)/9 = (-11+20)/9 = 9/9 = 1, b = (2α-β)/9 = (2⋅11-4)/9 = (22-4)/9 = 18/9 = 2, a(1,2)+b(5,1) = (α,β), 1(1,2)+2(5,1) = (11,4) ⇒ (1+10,2+2) = (11,4) ⇒ (11,4) = (11,4), 1⋅v₁+2⋅v₂ = v₃, a = 1 and b = 2 are the components of the vector v₃ with respect to the base given by v₁ and v₂

After verifying that v₁ = (1,2,3) and v₂ = (2,0,3) are linearly independent, complete the set {v₁,v₂} so you get a base of ℝ³; to verify the linear independence of vectors v₁ = (1,2,3) and v₂ = (2,0,3) we must show that the null vector can be obtained from the linear combination of these vectors only when the coefficients are all 0, a(1,2,3)+b(2,0,3) = (0,0,0) ⇒ {a+2b = 0, 2a = 0, 3a+3b = 0} ⇒ {a+2b = 0, a = 0, 3a+3b = 0} ⇒ {2b = 0, a = 0, 3b = 0} ⇒ {a = 0, b = 0}, the coefficients of the linear combination of vectors v₁ = (1,2,3) and v₂ = (2,0,3) are a = 0 and b = 0, therefore we have proved that the vectors v₁ = (1,2,3) and v₂ = (2,0,3) are linearly independent; 2 vectors are linearly independent when their components are not proportional, when a vector is not a multiple of the other; 2 linearly independent vectors do not lie on the same line; the vector space ℝ³ has dimension 3 and therefore its bases are made up of 3 vectors, therefore to complete the set {v₁,v₂} to obtain a base we must add a third vector, and these 3 vectors must be linearly independent; we have to find a vector v₃ that is not linear combination of vectors v₁ and v₂, v₃ ≠ a⋅v₁+b⋅v₂, a vector not coplanar to v₁ and v₂; as the third vector we choose v₃ = (0,0,1) and we verify that v₁,v₂,v₃ are 3 linearly independent vectors, a(1,2,3)+b(2,0,3)+c(0,0,1) = (0,0,0) ⇒ {a+2b = 0, 2a = 0, 3a+3b+c = 0} ⇒ {a+2b = 0, a = 0, 3a+3b+c = 0} ⇒ {2b = 0, a = 0, 3b+c = 0} ⇒ {b = 0, a = 0, c = -3b} ⇒ {b = 0, a = 0, c = 0} ⇒ {a = 0, b = 0, c = 0}, the coefficients of the linear combination of vectors v₁ = (1,2,3), v₂ = (2,0,3), v₃ = (0,0,1) are a = 0, b = 0, c = 0, therefore we have proved that the vectors v₁ = (1,2,3), v₂ = (2,0,3), v₃ = (0,0,1) are linearly independent and therefore they form a base of the vector space of ℝ³; the vector v₃ = (0,0,1) is not coplanar of v₁ = (1,2,3) and v₂ = (2,0,3), therefore these 3 vectors are linearly independent

Determine the dimension of the subspace of ℝ⁵ generated by v₁ = (1,2,-1,1,1), v₂ = (1,2,-1,0,0), v₃ = (0,0,-1,1,0), v₄ = (3,6,-3,3,3); if these 4 vectors were linearly independent they would form a base of the subspace of dimension 4; v₄ = (3,6,-3,3,3) = 3⋅v₁ = 3(1,2,-1,1,1), therefore v₁,v₂,v₃,v₄ are linearly dependent; we verify the linear independence of v₁,v₂,v₃, a(1,2,-1,1,1)+b(1,2,-1,0,0)+c(0,0,-1,1,0) = (0,0,0,0,0) ⇒ {a+b = 0, 2a+2b = 0, -a-b-c = 0, a+c = 0, a = 0} ⇒ {b = -a, 2b = -2a, c = -a-b, c = -a, a = 0} ⇒ {b = 0, b = 0, c = 0, c = 0, a = 0} ⇒ {a = 0, b = 0, c = 0}, the coefficients of the linear combination of vectors v₁,v₂,v₃ are a = 0, b = 0, c = 0, therefore the 3 vectors v₁,v₂,v₃ are linearly independent; the subspace generated by v₁,v₂,v₃,v₄ is equal to the subspace generated by v₁,v₂,v₃, L(v₁,v₂,v₃,v₄) = L(v₁,v₂,v₃); the subspace generated by vectors v₁,v₂,v₃ has dimension 3 and v₁,v₂,v₃ are a base of this subspace

A set of vectors is a system of generators of a vector space if it contains a suitable number of linearly independent vectors to form a base of the vector space; a set of vectors is a system of generators for ℝ² if it contains 2 linearly independent vectors; a set of vectors is a system of generators for ℝ³ if it contains 3 linearly independent vectors; a set of vectors is a system of generators for ℝⁿ if it contains n linearly independent vectors

20 - MATRICES - PART 1 - RANK AND REDUCTION

Matrix with m rows and n columns: A =

|(a_1,1,...,a_1,n)|

|(...,...,...)|

|(a_m,1,...,a_m,n)|

a_i,j = row i, column j

Row space: L(R₁,...,R_m) = vector space contained in ℝⁿ and generated by R₁ = (a_1,1,...,a_1,n), ..., R_m = (a_m,1,...,a_m,n)

A =

|1,1,1|

|2,1,0|, this is a matrix formed by 2 rows that are 2 vectors with 3 elements, therefore 2 vectors of ℝ³; the row space is the vector subspace generated by vectors (1,1,1) and (2,1,0) inside ℝ³

Column space: L(C₁,...,C_m) = vector space contained in ℝ^m and generated by C₁ = (a_1,1,...,a_1,n), ..., C_m = (a_1,n,...,a_m,n)

Row space and column space are completely different spaces, generated by different vectors that can be in different environments

A =

|1,1,1|

|2,1,0|, the column space is formed by the vectors (1,2),(1,1),(1,0) that are 3 vectors of ℝ²

There are matrices that have the row space and the column space contained in the same environment;

|1,1,1|

|2,2,2|

|0,0,3|, this is a matrix with 3 rows and 3 columns, the row space is generated by vectors of ℝ³, the column space is generated by vectors of ℝ³; row space and column space are 2 subspaces of ℝ³, and they can be completely different

A matrix with the number of rows equal to the number of columns, m = n, is a square matrix

Rank of a matrix: dim(L(R₁,...,R_m)) = dim(L(C₁,...,C_n)) = rank of the matrix A = ρ(A)

The row space is a subspace of ℝⁿ, the column space is a subspace of ℝ^m; the dimension of the row space and of the column space is the same, even if the spaces are completely different, and this dimension is called the rank of the matrix, ρ(A); the dimension of the row space is equal to the dimension of the column space

A =

|1,2,1|

|0,1,3|, A is a matrix with 2 rows and 3 columns; each row is an element of ℝ³, the row space is composed by 2 vectors of ℝ³, L((1,2,1),(0,1,3)) ⊆ ℝ³, (1,2,1) is not null and (0,1,3) is not multiple of (1,2,1), therefore they are two linearly independent vectors, and the dimension of the row space is 2 and coincides with the rank of the matrix A, 2 = ρ(A); each column is an element of ℝ², the column space is generated by 3 vectors of ℝ², L((1,0),(2,1),(1,3)) ⊆ ℝ²; for the matrix rank theorem, the dimension of the vector space L((1,0),(2,1),(1,3)) is equal to the rank of the matrix, the vector (1,0) is not zero, the vector (2,1) is not a multiple of (1,0), vector (1,3) is linear combination of (1,0) and (2,1), (1,3) = -5(1,0)+3(2,1), in ℝ² a base consists of 2 vectors, vectors (1,0) and (2,1) form a base of ℝ², therefore the rank of the matrix is equal to the dimension of the column space, 2 = ρ(A)

We want to calculate the dimension of the row space, the dimension of the column space, the rank of matrix B =

|1,2,3,-1|

|4,3,2,1|

|5,5,-1,7|, matrix B is made up of 3 rows and 4 columns; considering the row space, the first row (1,2,3,-1) is not null, the second row (4,3,2,1) is not a multiple of the first row, but to determine if the third line is a linear combination of the first line and the second line, we must take a long time; we need to find a technique to calculate in a simple way the dimension of the row space and the dimension of the column space, therefore the rank of a matrix, even if the number of rows and columns is large

B =

|1,2,1|

|0,1,3|

|0,5,0|, in this matrix it is easy to calculate the dimension of the row space and the dimension of the column space; R₁ = (1,2,1) is not null; R₂ = (0,1,3) it is not a multiple of the first row for the zero in the first position; we have to understand if R₃ = (0,5,0) is linear combination of R₁ and R₂, R₃ = a⋅R₁+b⋅R₂, in this case it is impossible to find the value of a and b, the presence of 0s in those positions allows us to easily calculate the rank of the matrix, in fact the matrix has 3 linearly independent lines; (1,2,1),(0,1,3),(0,5,0) are 3 linearly independent vectors that generate a vector space, therefore they are a base of this vector space, ρ(B) = 3; it is useless to calculate the dimension of the column space, and for the column space the calculations are more complicated because the 0s are not in good position; in the first row the number 1 has 2 0s below, in the second line the number 3 has 1 zero below, in the third line the number 5 has nothing below; for a matrix of this type the dimension of the row space is calculated by counting the rows, these 3 lines form a base of the row space; a matrix like this is called row reduced matrix

Row reduced matrix A =

|a,b,a,a,a|

|b,0,a,a,a|

|0,0,b,a,a|

|0,0,0,a,b|

|0,0,0,a,0|, a = any number, b = number other than 0 which has only 0 below

If a matrix is row reduced, the rank of the matrix is equal to the number of non-zero rows; if there are null rows in a matrix, the matrix remains reduced; null rows do not affect the rank of the matrix; the rank of a matrix is always equal to the number of non-zero rows

C =

|1,0,0|

|2,4,0|

|4,3,-1|, is a column reduced matrix; this matrix has 3 non-zero columns, the dimension of the column space is 3, the dimension of the row space is 3, the rank of the matrix is 3

D =

|1,0,0,0|

|2,4,0,0|

|4,3,0,-1|, the column of 0s does not affect the rank of the matrix

E =

|1,0,0,0,0|

|2,0,4,0,0|

|4,0,3,0,-1|, columns of 0s do not affect the rank of the matrix

To calculate the rank of a matrix we can use a row reduced matrix, or a column reduced matrix, it depends on what suits us best

A =

|1,1,1|

|2,1,1|

|3,1,-1|, this is a 3x3 matrix, 3 rows and 3 columns that is not reduced neither by rows nor by columns, but we can reduce this matrix through a series of operations; A → A', A is the starting matrix, A' is the reduced matrix either by rows or by columns, and after the reduction ρ(A) = ρ(A'); to reduce a matrix we must carry out operations such as to preserve the rank of the matrix

Elementary transformation on the rows: multiply a row by a scalar, R_i → a⋅R_i, a ≠ 0; swap a row with another row, R_i ↔ R_j, i ≠ j; replace a row with the sum of the row with the multiple of another row, R_i → R_i+a⋅R_j, a ≠ 0, i ≠ j; these operations do not change the row space and therefore do not change the rank ρ(A)

To transform any matrix into a reduced matrix, keeping the rank, we can perform operations called elementary transformations on the rows, which are multiplying a row by a number other than 0, exchange two rows, replace a row with the sum of the row with the multiple of another row; these operations do not affect the row space, therefore they do not change the rank ρ(A)

To calculate the rank of a matrix we operate with these elementary transformations, we get a new matrix which is row reduced, and the rank of the new matrix is equal to the rank of the starting matrix

A =

|1,1,1|

|2,1,1|

|3,1,-1|, R₂ → R₂-R₁,

|1,1,1|

|1,0,0|

|3,1,-1|, R₃ → R₃+R₁,

|1,1,1|

|1,0,0|

|4,2,0|, R₃ → R₃-4R₂,

|1,1,1|

|1,0,0|

|0,2,0|, the matrix is row reduced and has 3 rows, the rank of the row reduced matrix is 3, therefore the starting matrix has rank 3; another method to reduce this matrix is

|1,1,1|

|2,1,1|

|3,1,-1|, R₂ → R₂-R₁,

|1,1,1|

|1,0,0|

|3,1,-1|, R₃ → R₃+R₁,

|1,1,1|

|1,0,0|

|4,2,0|, R₃ ↔ R₂,

|1,1,1|

|4,2,0|

|1,0,0|, the matrix is row reduced and has 3 rows, therefore the rank of this matrix is 3; another method to reduce this matrix is

|1,1,1|

|2,1,1|

|3,1,-1|, R₂ → R₂-R₁,

|1,1,1|

|1,0,0|

|3,1,-1|, R₃ → R₃+R₁,

|1,1,1|

|1,0,0|

|4,2,0|, R₂ → 4R₂,

|1,1,1|

|4,0,0|

|4,2,0|, R₃ → R₃-R₂,

|1,1,1|

|4,0,0|

|0,2,0|, the matrix is reduced by rows and has 3 rows, therefore the rank of this matrix is 3

In the process of reduction, null lines are ignored; a null row is a row containing only 0s

To find the rank of a matrix we need to reduce the matrix by rows, or we have to reduce the matrix by columns; a matrix with 50 rows and 3 columns should be column reduced; the rank of a matrix is useful for calculating the dimension of a vector space

v₁ = (a_1,1,...,a_1,n), ..., v_m = (a_m,1,...,a_m,n) ⇒ A =

|a_1,1,...,a_1,n|

|...,...,...|

|a_m,1,...,a_m,n|

We write the vectors as rows of a matrix, we calculate the rank of the matrix, the rank of the matrix is the dimension of the row space, the row space is the vector space that has these vectors as generators, therefore we have calculated the dimension of the vector space which has as generators the vectors that are the rows of the matrix

In the vector space ℝ⁴ we want to calculate the dimension of the subspace W generated by the vectors (1,1,2,1),(2,1,0,3),(4,4,1,0); using these 3 vectors we create a matrix with 3 rows and 4 columns, and reducing this matrix by rows we obtain the dimension of the row space which is the dimension of the vector subspace,

|1,1,2,1|

|2,1,0,3|

|4,4,1,0|,R₃ → 2R₃-R₁,

|1,1,2,1|

|2,1,0,3|

|7,7,0,-1|,R₃ → R₃-7R₂,

|1,1,2,1|

|2,1,0,3|

|-7,0,0,-22|, these 3 lines are linearly independent vectors and are a base of the row space, the rank of the matrix is 3, therefore the dimension of the vector space W is 3, ρ(A) = 3 = dim (W); after these operations, the row space W has not changed, the generators have changed, but the 3 vectors (1,1,2,1),(2,1,0,3),(-7,0,0,-22) are a base of the space W that is a subspace of ℝ⁴

Rank = dim(row space) = dim(column space) = number of non-zero rows of a reduced matrix; the dimension of a subspace is the rank of its matrix

21 - MATRICES - PART 2 - OPERATIONS

Sum of 2 matrices:

|a_1,1,...,a_1,n|

|....,...,....|

|a_m,1,...,a_m,n|

+

|b_1,1,...,b_1,n|

|....,...,....|

|b_m,1,...,b_m,n|

=

|a_1,1+b_1,1,...,a_1,n+b_1,n|

|.......,.......,.......|

|a_m,1+b_m,1,...,a_m,n+b_m,n|

Product of a number by a matrix:

a⋅A =

a⋅

|a_1,1,...,a_1,n|

|....,...,....|

|a_m,1,...,a_m,n|

=

|a⋅a_1,1,...,a⋅a_1,n|

|.....,....,.....|

|a⋅a_m,1,...,a⋅a_m,n|

A =

|2,1,0|

|3,0,2|,

B =

|1,4, 1|

|2,1,-1|,

A+B =

|2+1,1+4,0+1|

|3+2,0+1,2-1|

=

|3,5,1|

|5,1,1|

2 matrices can be added when they have the same number of rows and columns; we must add the elements that have the same position in the matrix

A =

|2,1,0|

|3,0,2|,

a = -2,

a⋅A = -2A =

|-2⋅2,-2⋅1,-2⋅0|

|-2⋅3,-2⋅0,-2⋅2|

=

|-4,-2, 0|

|-6, 0,-4|, we can multiply any number by any matrix

A =

|2,1,0|

|3,0,2|,

b = 0,

b⋅A = 0⋅A =

|0⋅2,0⋅1,0⋅0|

|0⋅3,0⋅0,0⋅2|

=

|0,0,0|

|0,0,0|, multiplying 0 by any matrix we always get the null matrix

Properties of the sum: A+B = B+A, commutative property; A+(B+C) = (A+B)+C, associative property; A+0 = 0+A = A, existence of the null matrix; A+(-A) = A-A = 0, existence of the opposite matrix

The null matrix has 0 in all positions, 0 =

|0,.,0|

|.,.,.|

|0,.,0|

If a matrix is added to its opposite matrix, the null matrix is obtained, A+(-A) =

|0,.,0|

|.,.,.|

|0,.,0|

A =

| 1, 2, 0|

|-1, 4, 2|,

-A =

|-1,-2, 0|

| 1,-4,-2|,

A+(-A) =

| 1, 2, 0|

|-1, 4, 2|

+

|-1,-2, 0|

| 1,-4,-2|

=

|0,0,0|

|0,0,0|

Properties of the product of a number by a matrix: a(A+B) = a⋅A+a⋅B, first distributive property; (a+b)A = a⋅A+b⋅A, second distributive property; 1⋅A = A, 1 is the neutral of the multiplication; (a⋅b)A = a(b⋅A) = b(a⋅A), associative property

With ℝ^m,n we denote the matrices of real numbers with m rows and n columns; in ℝ^m,n the sum A+B is defined, and the product a⋅A is defined, therefore ℝ^m,n is a vector space; we can add two matrices only when they have the same number of rows and the same number of columns; 2 matrices with different number of rows and columns cannot be added, therefore they are not a vector space; ℝ^2,3 is the vector space of matrices with 2 rows and 3 columns; ℝ^4,2 is the vector space of matrices with 4 rows and 2 columns; ℝ^n,n is the vector space of square matrices with n rows and n columns

Product of matrices:

C = A⋅B =

|a,b,c|

|.,.,.|

|.,.,.|

⋅

|.,.,d|

|.,.,e|

|.,.,f|, c_1,3 = a⋅d+b⋅e+c⋅f

The product between 2 matrices can be made only when the number of columns of the first matrix is equal to the number of rows of the second matrix, and the resulting matrix has the number of rows of the first matrix and the number of columns of the second matrix, A^m,p⋅B^p,n = C^m,n, the matrix A^m,p has m rows and p columns, the matrix B^p,n has p rows and n columns, the matrix C^m,n has m rows and n columns

A =

|2,1,0|

|3,3,0|,

B =

|4,4,2|

|2,2,4|, multiplication A⋅B cannot be performed because matrix A^2,3 has 3 columns and matrix B^2,3 has 2 rows; the multiplication between 2 matrices can be performed only when the number of columns of the first matrix is equal to the number of rows of the second matrix

A =

|2,1|

|3,3|,

B =

|4,4,2|

|2,2,4|, the multiplication A⋅B can be performed because the matrix A^2,2 has 2 columns and the matrix B^2,3 has 2 rows, A⋅B = C; c_1,1 = 2⋅4+1⋅2 = 8+2 = 10, c_1,2 = 2⋅4+1⋅2 = 8+2 = 10, c_1,3 = 2⋅2+1⋅4 = 4+4 = 8, c_2,1 = 3⋅4+3⋅2 = 12+6 = 18, c_2,2 = 3⋅4+3⋅2 = 12+6 = 18, c_2,3 = 3⋅2+3⋅4 = 6+12 = 18; A_2,2⋅B_2,3 = C_2,3 =

|10,10, 8|

|18,18,18|, the resulting matrix has 2 rows like matrix A, and 3 columns like matrix B

A =

|1, 1,1|

|3, 3,0|

|0,-1,4|,

B =

|-1|

| 2|

| 1|,

A⋅B =

|1(-1)+1⋅2+1⋅1|

|3(-1)+3⋅2+0⋅1|

|0⋅1+(-1)2+4⋅1|

=

|-1+2+1|

|-3+6+0|

|0-2+4|

=

|2|

|3|

|2|, A^3,3⋅B^3,1 = C^3,1, matrix C has 3 rows as matrix A and 1 column as matrix B

Matrix A can be multiplied by matrix B when the number of columns in A equals the number of rows in B, and the result is a matrix C which has the number of rows of A and the number of columns of B, A^m,p⋅B^p,n = C^m,n

Identity matrix with 3 rows and 3 columns: I =

|1,0,0|

|0,1,0|

|0,0,1|, the elements a_1,1,a_2,2,a_3,3 form the main diagonal and are equal to 1, and all other elements are equal to 0

Identity matrix with 4 rows and 4 columns: I =

|1,0,0,0|

|0,1,0,0|

|0,0,1,0|

|0,0,0,1|, the elements a_1,1,a_2,2,a_3,3,a_4,4 form the main diagonal and are equal to 1, and all other elements are equal to 0

A⋅I = A, multiplying a matrix A by the corresponding identity matrix I we get the matrix A back; the identity matrix I must have the same number of rows as the number of columns of matrix A, without this condition multiplication A⋅I could not be performed

A =

|1,2,3|

|4,4,2|

⋅

|1,0,0|

|0,1,0|

|0,0,1|

=

|1⋅1+2⋅0+3⋅0,1⋅0+2⋅1+3⋅0,1⋅0+2⋅0+3⋅1|

|4⋅1+4⋅0+2⋅0,4⋅0+4⋅1+2⋅0,4⋅0+4⋅0+2⋅1|

=

|1+0+0,0+2+0,0+0+3|

|4+0+0,0+4+0,0+0+2|

=

|1,2,3|

|4,4,2| = A

Foods: fats, carbohydrates, proteins; A: 30% fats, 10% carbohydrates, 10% proteins; B: 20% fats, 20% carbohydrates, 5% proteins; C: 15% fats, 15% carbohydrates, 10% proteins; 100 grams of food A = (30,10,10); 100 grams of food B = (20,20,5); 100 grams of food C = (15,15,10); arranging each food in a column we obtain the matrix A =

|30,20,15|

|10,20,15|

|10, 5,10|;

if we want to calculate the amount of fat, carbohydrates and proteins ingested by eating 120 grams of A, 50 grams of B and 150 grams of C, we have to make the product between 2 matrices,

|30,20,15|

|10,20,15|

|10, 5,10|

⋅

|1.20|

|0.50|

|1.50|

=

|68.5|

|44.5|

|29.5|

Properties of the product of matrices: A(B⋅C) = (A⋅B)C, associative property; A(B+C) = A⋅B+A⋅C, distributive property of the product with respect to the sum; A⋅I = I⋅A = A, property of the identical matrix

Cancellation of the product of matrices: sometimes it can happen that A ≠ 0 and B ≠ 0, but A⋅B = 0

The product of 2 real numbers is 0 if one factor is 0, or both factors are 0; the product between 2 matrices can give 0 even when they are both different from 0;

A =

|0,-1|

|0,-1|,

B =

|0,-1|

|0, 0|, 2 square matrices can be multiplied with each other if they have the same order, in this case A^2,2⋅B^2,2 = C^2,2,

|0,-1|

|0,-1|

⋅

|0,-1|

|0, 0|

=

|0,0|

|0,0|, C_1,1 = R₁ of A multiplied by C₁ of B, C_1,2 = R₁ of A multiplied by C₂ of B, C_2,1 = R₂ of A multiplied by C₁ of B, C_2,2 = R₂ of A multiplied by C₂ of B; from 2 non-zero square matrices we have obtained a null matrix

For matrices, the product cancellation law does not apply, because the product of 2 non-zero matrices can be zero; obviously if one of the two matrices is null, or both are null, their product is null,

|0,0|

|0,0|

⋅

|a,b|

|c,d|

=

|0,0|

|0,0|

Reduction of matrix A that contains a parameter, A =

|1,1,h|

|2,1,3|, we reduce this matrix by rows avoiding to use h as pivot because it could be 0, R₂ → R₂-R₁,

|1,1, h|

|1,0,3-h| = B, after this reduction, the matrix B, which has the same rank as matrix A, has row R₁ not null, and R₂ is not null, therefore the rank of matrix A and the rank of matrix B is 2, ρ(A) = ρ(B) = 2

Reduction of matrix C that contains a parameter, C =

|1,1,h|

|1,1,3|, R₂ → R₂-R₁,

|1,1, h|

|0,0,3-h| = D, h = 3 ⇒ R₂ is a null row and the rank of the matrix is 1, h ≠ 3 ⇒ R₂ is a non-zero row and the rank of the matrix is 2

Matrices^m,n: sum operation, 2 matrices can be added if they have the same number of rows and columns, A^m,n+B^m,n = C^m,n; product operation, 2 matrices can be multiplied row by column if the number of columns of A equals the number of rows of B, A^m,n⋅B^n,p = C^m,p

22 - MATRICES - PART 3 - INVERSE AND TRANSPOSE

Inverse matrix: A⋅A^-1 = A^-1⋅A = I

A matrix A is invertible if exists a matrix A^-1 such that A⋅A^-1 = A^-1⋅A = I, where A^-1 is the inverse matrix, and I is the identity matrix; if matrix A has n rows and n columns, then the inverse matrix A^-1 has n rows and n columns, the inverse matrix A^-1 is a square matrix of order n; A⋅A^-1 = A^-1⋅A = I, ma A⋅B ≠ B⋅A, because for matrices there is no commutative property of the product

We need to understand if the matrix A is invertible, A =

|2, 1|

|3,-1|, therefore we must look for a matrix X =

|a,b|

|c,d|, such that A⋅X = I,

|2, 1|

|3,-1|

⋅

|a,b|

|c,d|

=

|1,0|

|0,1|, from the product A^2,2⋅X^2,2 we get a matrix with 2 rows and 2 columns,

|2,1|

|3,-1|

⋅

|a,b|

|c,d|

=

|2a+c,2b+d|

|3a-c,3b-d|,

⇒

|2a+c,2b+d|

|3a-c,3b-d|

=

|1,0|

|0,1| ⇒ 2a+c = 1 ∧ 2b+d = 0 ∧ 3a-c = 0 ∧ 3b-d = 1, are 4 equations with 4 unknowns a, b, c, d, which form 2 separate systems of equations, {2a+c = 1, 3a-c = 0} ∧ {2b+d = 0, 3b-d = 1}; we solve the system {2a+c = 1, 3a-c = 0} by adding together the 2 equations, 2a+c+3a-c = 1+0 ⇒ 5a = 1 ⇒ a = 1/5, 3a-c = 0 ⇒ c = 3a ⇒ c = 3(1/5) = 3/5; we solve the system {2b+d = 0, 3b-d = 1} by adding together the 2 equations, 2b+d+3b-d = 0+1 ⇒ 5b = 1 ⇒ b = 1/5, 2b+d = 0 ⇒ d = -2b ⇒ d = -2(1/5) = -2/5; matrix A is invertible, therefore there exists the inverse matrix which is X = A^-1 =

|1/5, 1/5|

|3/5,-2/5|; we verify that A⋅A^-1 = I,

|2, 1|

|3,-1|

⋅

|1/5, 1/5|

|3/5,-2/5|

=

|2(1/5)+1(3/5),2(1/5)+1(-2/5)|

|3(1/5)-1(3/5),3(1/5)-1(-2/5)|

=

|2/5+3/5,2/5-2/5|

|3/5-3/5,3/5+2/5|

=

|5/5, 0|

| 0,5/5|

=

|1,0|

|0,1| = I; to find the inverse matrix of a matrix with 2 rows and 2 columns we had to find the value of 4 unknowns by solving 4 equations

To find the inverse matrix of a matrix A^3,3 we have to find the value of 9 unknowns by solving 9 equations,

A =

|2,1,-1|

|3,0, 4|

|0,0, 2|,

X =

|a,b,c|

|d,e,f|

|g,h,i|,

A⋅A^-1 = A^-1⋅A = A⋅X = I,

I =

|1,0,0|

|0,1,0|

|0,0,1|;

A⋅X =

|2,1,-1|

|3,0, 4|

|0,0, 2|

⋅

|a,b,c|

|d,e,f|

|g,h,i|

=

|2a+d-g,2b+e-h,2c+f-i|

| 3a+4g, 3b+4h, 3c+4i|

| 2g, 2h, 2i|;

A⋅X = I,

|2a+d-g,2b+e-h,2c+f-i|

| 3a+4g, 3b+4h, 3c+4i|

| 2g, 2h, 2i|

=

|0,1,0|

|1,0,0|

|0,0,1|, we have to solve a system of 9 first degree equations in 9 unknowns, {2a+d-g = 1, 2b+e-h = 0, 2c+f-i = 0, 3a+4g = 0, 3b+4h = 1, 3c+4i = 0, 2g = 0, 2h = 0, 2i = 1} ⇒ 2g = 0 ⇒ g = 0, 2h = 0 ⇒ h = 0, 2i = 1 ⇒ i = 1/2, 3c+4i = 0 ⇒ 3c = -4i ⇒ c = (-4/3)i = (-4/3)(1/2) = -2/3, 3b+4h = 1 ⇒ 3b = 1-4h ⇒ b = 1/3-(4/3)h = 1/3-(4/3)0 = 1/3, 3a+4g = 0 ⇒ 3a = -4g ⇒ a = (-4/3)g = (-4/3)0 = 0, 2c+f-i = 0 ⇒ f = -2c+i = -2(-2/3)+1/2 = 4/3+1/2 = (8+3)/6 = 11/6, 2b+e-h = 0 ⇒ e = -2b+h = -2(1/3)+0 = -2/3, 2a+d-g = 1 ⇒ d = -2a+g+1 = -2⋅0+0-1 = 0+0-1 = -1, a = 0 ∧ b = 1/3 ∧ c = -2/3 ∧ d = 1 ∧ e = -2/3 ∧ f = 11/6 ∧ g = 0 ∧ h = 0 ∧ i = 1/2, we have shown that the matrix A is invertible because the inverse matrix exists and is A^-1 =

|0, 1/3,-2/3|

|1,-2/3,11/6|

|0, 0, 1/2|

A =

|2,3|

|0,0|,

X =

|a,b|

|c,d|,

A⋅X =

|2,3|

|0,0|

⋅

|a,b|

|c,d|

=

|2a+3c,2b+3d|

|0a+0c,0b+0d|

=

|2a+3c,2b+3d|

| 0+0, 0+0|

=

|2a+3c,2b+3d|

| 0, 0|,

A⋅X = I =

|1,0|

|0,1|, matrix A does not admit inverse because

|1,0|

|0,1|

≠

|2a+3c,2b+3d|

| 0, 0|; matrix A is an example of matrix 2⋅2 that is not invertible

Matrix A with m rows and n columns:

A^m,n =

|a_1,1,...,a_1,n|

|....,...,....|

|a_m,1,...,a_m,n|

Transpose matrix: transpose of matrix A = A^T =

|a_1,1,...,a_m,1|

|....,...,....|

|a_1,n,...,a_m,n|, A^T has n rows and m columns; the transpose matrix A^T is obtained from matrix A by swapping the rows with the columns

The rows of matrix A become the columns of the transpose matrix A^T, the columns of the matrix A become the rows of the transpose matrix A^T; if matrix A has m rows and n columns, the transpose matrix A^T has m columns and n rows

Example of transpose matrix:

A =

|4,5,6|

|7,8,9|

|0,4,3|,

A^T =

|4,7,0|

|5,8,4|

|6,9,3|

Example of transpose matrix:

A =

|1,0,7,4|

|3,3,2,2|,

A^T =

|1,3|

|0,3|

|7,2|

|4,2|

Properties of the transpose matrix: (A+B)^T = A^T+B^T; (A⋅B)^T = B^T⋅A^T; (a⋅A)^T = a⋅A^T

(A⋅B)^T = B^T⋅A^T, the product of matrices is not commutative; the product A⋅B can be made when the number of columns of A is equal to the number of rows of B; the transposition operation swaps the rows with the columns, therefore (A⋅B)^T = A^T⋅B^T is true only when A and B are square matrices, (A⋅B)^T = B^T⋅A^T is always true; the transpose of the product is the product of the transposes but by changing the order of the matrices, because the commutative property does not hold for matrices

Symmetric matrix:

|a,x,y|

|x,b,z|

|y,z,c|

If a matrix A coincides with its transpose matrix A^T, then the matrix A is called symmetric matrix; a symmetric matrix has the elements symmetrical with respect to the diagonal; a symmetric matrix is always a square matrix; a matrix must be square to be symmetric

A =

|1,0, 3|

|0,2, 4|

|3,4,-1|, A is a symmetric matrix;

B =

|1,0, 2|

|0,2, 4|

|3,4,-1|, B is not a symmetric matrix

If a matrix coincides with the opposite of the transpose matrix it is called antisymmetric matrix; A is an antisymmetric matrix if A = -A^T

Antisymmetric matrix:

| 0, x,y|

|-x, 0,z|

|-y,-z,0|

A =

| 0, 1,5|

|-1, 0,2|

|-5,-2,0|, A is an antisymmetric matrix

Orthogonal matrices: orthogonal square matrix A, A^T = A^-1; a matrix is orthogonal when the inverse matrix and the transpose matrix coincide

An orthogonal matrix is an invertible square matrix such that its transpose matrix is equal to its inverse matrix; an orthogonal matrix is a particular invertible square matrix such that its inverse matrix coincides with its transpose matrix

Orthogonal matrix:

A =

|cos(α),-sin(α)|

|sin(α), cos(α)|, α is any number, A is an orthogonal matrix

A =

|cos(α),-sin(α)|

|sin(α), cos(α)|, A is an orthogonal matrix because its inverse matrix is equal to its transpose matrix,

A^T =

| cos(α),sin(α)|

|-sin(α),cos(α)|;

A⋅A^T = A^T⋅A because in this case the commutative property holds,

| cos(α),sin(α)|

|-sin(α),cos(α)|

⋅

|cos(α),-sin(α)|

|sin(α), cos(α)|

=

|cos²(α)+sin²(α), cos(α)(-sin(α))+sin(α)cos(α)|

| -sin(α)cos(α)+cos(α)sin(α),sin²(α)+cos²(α)|

=

|1,0|

|0,1| = I, the transpose matrix of matrix A coincides with the inverse matrix of matrix A, A^T = A^-1, therefore the matrix A is an orthogonal matrix

Study for which real values of the parameter h the matrix A is invertible,

A =

|1,h|

|2,0|;

A⋅X = I,

|1,h|

|2,0|

⋅

|a,b|

|c,d|

=

|1,0|

|0,1|,

A⋅X =

|1,h|

|2,0|

⋅

|a,b|

|c,d|

=

|a+hc,b+hd|

| 2a, 2b|

=

|1,0|

|0,1| ⇒ {a+hc = 1, b+hd = 0, 2a = 0, 2b = 1}, 2a = 0 ⇒ a = 0, 2b = 1 ⇒ b = 1/2, a+hc = 1 ⇒ hc = 1-a ⇒ c = (1-a)/h = (1-0)/h = 1/h, b+hd = 0 ⇒ hd = -b ⇒ d = -b/h = -1/(2h), a = 0 ∧ b = 1/2 ∧ c = 1/h ∧ d = -1/(2h); if h = 0 the matrix A is not invertible, therefore we must suppose that h ≠ 0; the matrix A is invertible ∀ h ≠ 0 and the inverse matrix is A^-1 =

| 0, 1/2|

|1/h,-1/(2h)|; if h = 0 the matrix becomes A =

|1,0|

|2,0|, and this matrix has rank 1, less than order 2; if h ≠ 0 the matrix A has rank 2; the rank of a matrix indicates if the matrix is invertible

Inverse matrix: A⋅A^-1 = A^-1⋅A = I

Symmetric matrix: A = A^T

Antisymmetric matrix: A = -A^T

Orthogonal matrix: A^-1 = A^T

Multiplying the matrix A by its inverse matrix we obtain the identity matrix; the inverse matrix A^-1 exists only when matrix A is a square matrix; a matrix is invertible only if it is square; a matrix is orthogonal when its inverse matrix is equal to the transpose matrix

23 - THE CONCEPT OF LINEAR APPLICATION

Application is synonymous with function, therefore a linear application is a linear function

Let us consider the linear application between the vector space ℝ² and the vector space ℝ, f: ℝ² → ℝ : f(x,y) = x+y; we consider the pair (x,y) and the pair (x',y'), (x,y) → f → x+y, (x',y') → f → x'+y', (x,y)+(x',y') = (x+x',y+y'), (x+x',y+y') → f → (x+x')+(y+y'); f(x,y) = x+y, f(x',y') = x'+y', f(x+x',y+y') = (x+x')+(y+y') = (x+y)+(x'+y') = f(x,y)+f(x',y'), this function f preserves the sum of pairs; a(x,y) = (ax,ay), this is the definition of multiplying a number by a pair, (ax,ay) → f → ax+ay, f(ax,ay) = ax+ay, a(x,y) → f → a(x+y), f(a(x,y)) = a(x+y) = ax+ay; this function f of ℝ² in ℝ has 2 fundamental properties, keep the sum and the product for a number, and it is an example of linear application between 2 vector spaces

A linear application keeps the sum and keeps the product for a number

Properties of linear applications: f: V → W linear application ⇒ f(v+v') = f(v)+f(v'), f(a⋅v) = a⋅f(v), f(0_v) = 0_w, f(-v) = -f(v)

f: V → W, f is a linear application between the vector space V and the vector space W if f(v+v') = f(v)+f(v') e f(a⋅v) = a⋅f(v); for a linear application between 2 vector spaces, the sum and the product for a number are conserved; f(0_v) = 0_w, the 0 of the vector space V is transformed into the 0 of the vector space W; f(-v) = -f(v), the opposite of v in the vector space V coincides with the opposite of f(v) in the vector space W

Let us consider the function from ℝ³ to ℝ², f: ℝ³ → ℝ², f(x,y,z) = (x,y+z); f(x,y,z) = (x,y+z), f(x',y',z') = (x',y'+z'), (x,y,z)+(x',y',z') = (x+x',y+y',z+z'), f(x+x',y+y',z+z') = (x+x',(y+y')+(z+z')), (x,y+z)+(x',y'+z') = (x+x',(y+z)+(y'+z')) = (x+x',(y+y')+(z+z')), the function applied to a sum of elements of ℝ³ gives as a result the sum of the functions applied to the single elements, the function keeps the sum; a(x,y,z) = (ax,ay,az), f(ax,ay,az) = (ax,ay+az) = a(x,y+z) = a⋅f(x,y,z), the product of a number by a pair is by definition a(x,y+z) = (ax,ay+az), therefore f(ax,ay,az) = a⋅f(x,y,z), the function preserves the product; f(0,0,0) = (0,0+0) = (0,0), the function keeps 0, transforms 0 into 0, f of the triad (0,0,0) transforms into the pair (0,0); f(-x,-y,-z) = (-x,-y-z) = -(x,y+z) = -f(x,y,z), f(-v) = -f(v), the function that has the opposite as argument is equal to the opposite of the function; this function is a linear application because it preserves the sum, the product, the zero, and the function of the opposite is equal to the opposite of the function

Let us study an example of a function between vector spaces that is not linear, f(x,y) = x², f: ℝ² → ℝ; (x,y)+(x',y') = (x+x')+(y+y'), (1,2)+(3,1) = (4,3), f(1,2) = 1² = 1, f(3,1) = 3² = 9, f(1,2)+f(3,1) = 1+9, f(4,3) = 4² = 16, 1+9 = 10 ≠ 16, f(1,2)+f(3,1) ≠ f(4,3), therefore this function does not preserve the sum, and it is enough to show this case to affirm that the function is not linear; f(2,1) = 4, 5f(2,1) = 5⋅4 = 20, 5(2,1) = (10,5), f(10,5) = 100, 20 ≠ 100, 5f(2,1) ≠ f(10,5), therefore this function does not preserve the product, and this further demonstrates that the function is not linear

f(x,y) = x+1, f: ℝ² → ℝ; this function is not a linear application because it does not preserve the sum and does not preserve the product, but the simplest way to prove that this function is not linear is that 0 is not maintained, f(0,0) = 0+1 = 1 ≠ 0

Properties of linear applications: f: V → W; f(v) = 0_w ⇔ v ∈ Ker(f); w = f(v) ⇔ w ∈ Im(f)

f: V → W, ℝⁿ → ℝ^m; f(v) = 0_w ⇔ v ∈ Ker(f), f(0) = 0, but there can exist non-zero vectors in the vector space V whose image is 0, and these vectors form a set called Ker(f); w = f(v) ⇔ w ∈ Im(f), vectors that are images of some vector form a subset of W called Im(f)

f(x,y) = x+y, f: ℝ² → ℝ; f(0,0) = 0, but there are also other couples (x,y) such that f(x,y) = 0, f(x,y) = x+y = 0 ⇒ x = -y, therefore all the pairs (x,-x) are transformed into 0 by f, f(x,-x) = 0, Ker(f) is the subset of ℝ² formed by the pairs (x,-x) which is a vector subspace formed by the multiples of the vector (1,-1), (x,-x) = x(1,-1)

If we have an application f: V → W, the set of v which through f is transformed into 0 of W, v → f → 0_w, is Ker(f) ⊆ V; Ker(f) is a vector subspace that contains 0, but typically also contains other vectors besides 0, and it is not forbidden that it contains only 0; the vector space Ker(f) can contain only the null vector

Another important concept is Im(f); f: V → W, f(v) = w and the set of these vectors is Im(f) ⊆ W, Im(f) is a subspace of W

f(x,y) = (x,x) is a linear application of ℝ² in ℝ², ℝ² → ℝ²; the pairs (0,0),(1,1),(2,2) are images of the function, the pair (1,0) is not an image of the function, some couples are in the image, but others are not, precisely the image is the set of all the pairs obtained by multiplying the pair (1,1) by an arbitrary number x, x(1,1), the image is the vector subspace of ℝ² formed by multiples of (1,1), it has as base (1,1), and this image is different from ℝ² because it is only a part of ℝ², Im(f) ≠ ℝ²

Ker(f), nucleus of f, is a subspace of V formed by all the elements that transform into 0; Im(f), image of f, is a subspace of W formed by the vectors which are f(v)

Ker is the abbreviation of the English word Kernel which means nucleus; Ker(f) is formed by the vectors that transform into 0; Ker(f) can contain only 0_v or also contain other vectors; if Ker(f) is formed by only 0 v then the function f is an injective linear application; a linear application is injective when v ≠ v' ⇒ f(v) ≠ f(v'); to understand if a linear application is injective, just check that there is only 0 in the nucleus

Determine if the linear application f(x,y) = (x,x) is injective; the linear application f(x,y) = (x,x) is injective if its nucleus contains only 0; to find the nucleus of the linear application f(x,y) = (x,x) we must find the pairs (x,y) such that (x,x) = (0,0), all pairs with x = 0 and y is any value, (0,y) = y(0,1) that is the vector subspace of ℝ² generated by the vector (0,1) and all its multiples, therefore Ker(f) does not contain only 0, so this application is not injective

Determine the nucleus of the linear application f(x,y) = (2x+y,x-y); to find the nucleus we must find x and y such that 2x+y = 0 and at the same time x-y = 0, x-y = 0 ⇒ x = y, 2x+y = 0 ⇒ 2y+y = 0 ⇒ 3y = 0 ⇒ y = 0, x = 0 ∧ y = 0, there are no pairs other than the pair (0,0) that have the pair (0,0) as an image, this function is a linear injective application because its nucleus contains only the pair (0,0)

To understand if a linear application is injective we have to study its Ker(f); the Ker(f) of a linear application is useful to understand the injectivity

With the image of f we can study the surriectivity of a linear application; the function f: V → W is a surjective linear application if ∀ w ∈ W ∃ v ∈ V : f(v) = w, this is the definition of surectivity for any function; ∀ w ∈ W ∃ v ∈ V : f(v) = w means that Im(f) = W; to establish if a linear application is surjective we have to see the image

The linear application f(x,y) = (x,x) is not surjective because the image of f is different from ℝ², Im(f) ≠ ℝ², because there are elements of ℝ² which are not of the form (x,x)

We consider the linear application f(x,y) = 3x, ℝ² → ℝ; we must understand if for any real number m there exists a pair (x,y) such that f(x,y) = m, ∀ m ∈ ℝ ∃ (x,y) : f(x,y) = m ⇒ 3x = m ⇒ x = m/3 ⇒ f(m/3,y) = 3x = 3(m/3) = m, therefore every number m admits a counterimage in ℝ²; this is a surjective linear application of ℝ² in ℝ; studying the vector subspace which is the image of f we can understand if the application is surjective

A linear application f: V → W can be simultaneously injective and surjective; it is a function because for every v there is an image in W, it is an injective function because different v and v' have different images, it is a surjective function because every w comes from a v, therefore between V and W there is a one-to-one correspondence, all this is expressed by stating that the function f is a bijective linear application, also called isomorphism

Linear applications that are isomorphisms are bijective linear applications, linear applications that are simultaneously injective and surjective, therefore the nucleus is reduced to only 0 and the image is all W

The linear application f(x,y) = (2x+y,x-y) is injective and surjective, therefore it is bijective, also called isomorphism; given 2 numbers (α,β) in the image space that is ℝ², there are x and y such that (α,β) = (2x+y,x-y) ⇒ {2x+y = α, x-y = β}, x-y = β ⇒ y = x-β, 2x+y = α ⇒ 2x+x-β = α ⇒ 3x = α+β ⇒ x = (α+β)/3, y = x-β = (α+β)/3-β = (α+β-3β)/3 = (α-2β)/3, therefore for each value of α and β we can find the corresponding values of x and y, so this linear application is not only injective but also surjective, in fact this is an example of isomorphism that is a bijective linear application

To find the nucleus of the linear application f: ℝ³ → ℝ³, f(x,y,z) = (x+y,x-y,x+2z), we have to find triples (x,y,z) such that x+y = 0 ∧ x-y = 0 ∧ x+2z = 0, therefore we have to solve these 3 first degree equations; to find the nucleus of a linear application we have to solve a system of first degree equations without a known term; a linear system is a system of first degree equations; to find the image we have to see if the triple (α,β,γ) can be obtained by solving the linear system {α = x+y, β = x-y, γ = x+2z}

If f: V → W and g: V → W are linear applications, then the sum operation and the product operations for a number are defined; sum of linear applications f+g: (f+g)(v) = f(v)+g(v); product of a number for a linear application α⋅f: v → α⋅f (v); if f: V → W and g: W → Z, the product operation g∘f of linear applications is defined, v → f → f (v) → g → g(f(v)), applying successively first the linear application f and after the linear application g we get a linear application V → Z that is called product or compisition of linear applications

f: V → W linear application f(v+w) = f(v)+f(w), f(a⋅v) = a⋅f(v); f(v) = 0_w ⇔ v ∈ Ker(f), f is injective ⇔ Ker(f) = {0_v}; w = f(v) ⇔ w ∈ Im(f), f is surjective ⇔ Im(f) = W

24 - LINEAR APPLICATIONS AND MATRICES

We can associate a matrix to each linear application, and to each matrix we can associate a linear application, therefore there is a correspondence between matrices and linear applications; the matrices can be used to study the surjectivity, the injectivity, and the invertibility of linear applications

Matrix associated with a linear application: f(x,y,z) = (ax+by+cz,dx+ey+fz);

M_f =

|a,b,c|

|d,e,f|

f: ℝ³ → ℝ², f(x,y,z) = (ax+by+cz,dx+ey+fz),

the matrix associated with this linear application is M_f =

|a,b,c|

|d,e,f|, the matrix M_f has on the first row the coefficients a,b,c of the first component, and on the second line the coefficients d,e,f of the second component

f(x,y,z) = (ax+by+cz,dx+ey+fz),

M_f =

|a,b,c|

|d,e,f|, f: ℝ³ → ℝ², the 3 fundamental versors of ℝ³ are e₁ = (1,0,0), e₂ = (0,1,0), e₃ = (0,0,1), we apply f to these 3 vectors that form the basis of ℝ³, e₁ = (1,0,0) →^f (a,d), e₂ = (0,1,0) →^f (b,e), e₃ = (0,0,1) →^f (c,f), f(e₁) = (a,d) = C₁, the first column of the matrix M_f, f(e₂) = (b,e) = C₂, the second column of the matrix M_f, f(e₃) = (c,f) = C₃, the third column of the matrix M_f; the elements of the matrix M_f have a double meaning, R₁ = (a,b,c) is the first row of the matrix M_f and it is composed by the coefficients of x,y,z of the first component, R₂ = (d,e,f) is the second row of the matrix M_f and it is composed by the coefficients of x,y,z of the second component, C₁ = (a,d) is the first column of the matrix M_f and it is f(e₁), C₂ = (b,e) is the second column of the matrix M_f and it is f(e₂), C₃ = (c,f) is the third column of the matrix M_f and it is f(e₃)

Considering a linear application between vector spaces ℝⁿ → ℝ^m, we have to build the matrix associated with the linear application, in the way that the rows are composed by the coefficients of the components, and the columns are composed by the images of the fundamental versors

Considering the linear application f: ℝⁿ → ℝ^m, n and m are any 2 numbers; e₁ = (1,0,...,0) is the first fundamental versor, e₂ = (0,1,...,0) is the second fundamental versor, e_n = (0,0,...,n) is the last fundamental versor; with the linear application f, each of these vectors is transformed into an element of ℝ^m, e₁ = (1,0,...,0) →^f (a_1,1,a_2,1,...,a_m,1), e₂ = (0,1,...,0) →^f (a_1,2,a_2,2,...,a_m,2), e_n = (0,0,...,n) →^f (a_1,n,a_2,n,...,a_m,n), we have constructed n vectors which are the images of the fundamental versors of ℝⁿ, and we obtained the matrix associated with the linear application, M_f =

|a_1,1,a_1,2,...,a_1,n|

|a_2,1,a_2,2,...,a_2,n|

|....,....,...,....|

|a_m,1,a_m,2,...,a_m,n|; C₁ = f(e₁), the first column is the function applied to the first fundamental versor; C₂ = f(e₂), the second column is the function applied to the second fundamental versor; C_n = f(e_n), the nth column is the function applied to the nth fundamental versor; the columns are the images of the fundamental versors; f: ℝⁿ → ℝ^m, the result of the function f(x₁,x₂,...,x_n) is an element of ℝ^m, so it must have m components and each component is a row of the matrix M_f, f(x₁,x₂,...,x_n) = (a_1,1⋅x₁+a_1,2⋅x₂+...+a_1,n⋅x_n,a_2,1⋅x₁+a_2,2⋅x₂+...+a_2,n⋅x_n,...,a_m,1⋅x₁+a_m,2⋅x₂+...+a_m,n⋅x_n), the rows of the matrix M_f are formed by the coefficients of x₁,x₂,x_n, the first row corresponds to the coefficients of the first component, the second row corresponds to the coefficients of the second component, the nth row corresponds to the coefficients of the nth component; the matrix M_f associated with the linear application f: ℝⁿ → ℝ^m has n columns and m rows, the number of columns of the matrix is equal to the dimension of the space ℝⁿ, the number of rows of the matrix is equal to the dimension of the space ℝ^m; the components of f(x₁,x₂,...,x_n) are a_1,1⋅x₁+a_1,2⋅x₂+...+a_1,n⋅x_n, a_2,1⋅x₁+a_2,2⋅x₂+...+a_2,n⋅x_n, a_m,1⋅x₁+a_m,2⋅x₂+...+a_m,n⋅x_n, each of these m components is a linear combination of (x₁,x₂,...,x_n), that is a homogeneous polynomial of first degree in x₁,x₂,...,x_n with no known term

A linear application is characterized by the fact that the function f applied to a tuple has m components and each is a linear combination, that is a homogeneous polynomial of first degree in x₁,x₂,...,x_n with coefficients deducible from the matrix

f: ℝ² → ℝ³, f(x,y) = (x+2y,x-2y,x+y), the associated matrix M_f has 2 columns and 3 rows,

M_f =

|1, 2|

|1,-2|

|1, 1|; f(1,0) = (1,1,1) = C₁, if we apply the function to the fundamental versor (1,0) we obtain the first column of the matrix M_f; f(0,1) = (2,-2,1) = C₂, if we apply the function to the fundamental unit vector (0,1) we obtain the second column of the matrix M_f; this application is linear because x+2y,x-2y,x+y are 3 homogeneous polynomials of first degree in the variables x and y of ℝ²

f: ℝ³ → ℝ², from 3 to 2 variables, g(x,y,z) = (x+z,y+1), x+z is a homogeneous polynomial of first degree in x,y,z, but y+1 is not a homogeneous polynomial of first degree in x,y,z due to the presence of the known term, so this is not a linear application

An application is linear if its components are homogeneous polynomials of the first degree

To write the matrix associated with a linear application we must put the coefficients of each component on each row, or we write each column as an image of the function applied to each fundamental versor

From a matrix we can obtain the associated linear application;

A =

|1,-1,3|

|2, 2,0|, this matrix A has 3 columns and 2 rows, therefore the corresponding linear application f must operate from ℝ³ to ℝ², f: ℝ³ → ℝ², f(x,y,z) = (x-y+3z,2x+2y), and this is a linear application because x-y+3z and 2x+2y are homogeneous polynomials of first degree in x,y,z; from the linear application f(x,y,z) = (x-y+3z,2x+2y) we can get the associated matrix that is M_f =

|1,-1,3|

|2, 2,0| = A; from matrix A we have constructed the associated linear application, and from this linear application we have re-obtained A as an associated matrix, therefore the correspondence between linear application and matrix is one-to-one

To a linear application f: ℝⁿ → ℝ^m corresponds a matrix with n columns and m rows; to a matrix with n columns and m rows corresponds a linear application f: ℝⁿ → ℝ^m

A linear application f: ℝⁿ → ℝ^m is injective if v ≠ v' ⇒ f(v) ≠ f(v'), is surjective if Im(f) = ℝ^m which is like saying that every vector of ℝ^m has a counterimage, it is invertible if it is both injective and surjective and therefore there exists the inverse linear application f^-1; a linear application is invertible if n = m, therefore an invertible linear application is f: ℝⁿ → ℝⁿ, and the inverse linear application is f^-1: ℝⁿ → ℝⁿ, v →^f v' = f(v), v'→^f-1 v = f^-1(v')

The linear application f: ℝⁿ → ℝ^m is surjective if Im(f) = ℝ^m; column 1 of the matrix M_f is C₁ = f(e₁), column 2 of the matrix M_f is C₂ = f(e₂), column n of the matrix M_f is C_n = f(e_n); a vector v is a linear combination of e₁,e₂,...,e_n, therefore f(v) is a linear combination of f(e₁),f(e₂),...,f(e_n), so f(e₁),f(e₂),...,f(e_n) are generators of the image of f which is a vector subspace of ℝ^m; Im(f) is a vector subspace of ℝ^m and is generated by vectors f(e₁),f(e₂),...,f(e_n); if the vectors f(e₁),f(e₂),...,f(e_n) generate all the vector space ℝ^m then the linear application is surjective; the vector space ℝ^m has dimension m, and we need to determine if the space generated by the columns f(e₁),f(e₂),f(e_n) has dimension m, therefore we must calculate the rank of the matrix; if ρ(M_f) = m, then the space of the columns has dimension m, so the image coincides with all ℝ^m

An easy way to know if a linear application is surjective is to establish if ρ(M_f) = m, where m is the number of rows of the matrix M_f

A linear application is injective if and only if Ker(f) = {0_ℝⁿ}; we know that dim(Ker(f)) = n-dim(Im(f)) = n-ρ(M_f) because dim(Im(f)) = ρ(M_f), Ker(f) = 0 ⇒ dim(Ker(f)) = 0, dim(Ker(f)) = n-ρ(M_f), 0 = n-ρ(M_f) ⇒ ρ(M_f) = n, where n is the number of columns; the linear application is injective if the rank of the associated matrix is equal to the number of columns

The columns of a matrix give information on the injectivity of the associated linear application, the rows of a matrix give information on the surriectivity of the associated linear application

f: ℝⁿ → ℝⁿ, if ρ(M_f) = n, the rank of the matrix M_f is equal to the number of columns that is equal to the number of rows, therefore the linear application is simultaneously injective and surjective, so it is a linear invertible application

Injective and surjective linear applications: a linear application f: ℝⁿ → ℝ^m is injective if ρ(M_f) = n, is surjective if ρ(M_f) = m, is invertible if ρ(M_f) = m = n

A matrix is invertible if the rank of the matrix is equal to the dimension of the row space and the dimension of the column space, therefore to be invertible the matrix must be square

f: ℝ⁴ → ℝ², f(x,y,z,t) = (x-y+z,x+y-t); the associated matrix M_f has 4 columns and 2 rows,

M_f =

|1,-1,1, 0|

|1, 1,0,-1|, M_f is reduced by lines, ρ(M_f) = 2 = dim (ℝ²), therefore the matrix M_f represents a surjective linear application; to be injective the number of columns must be equal to the rank of the matrix, but this is false because the columns are 4 and ρ(M_f) < 4, precisely the rank of the matrix is 2, therefore this matrix represents a non-injective linear application; going from ℝⁿ to ℝ^m, the rank of the associated matrix cannot exceed the smaller of the numbers n and m; if n > m it is impossible for the rank to be n, therefore it is impossible for the linear application to be injective; if n < m it is impossible for the rank to be m, therefore it is impossible for the linear application to be surjective

f: ℝ² → ℝ³, f(x,y) = (x-y,x+y,x); the matrix M_f associated with this linear application has 2 columns and 3 rows,

M_f =

|1,-1|

|1, 1|

|1, 0|, ρ(M_f) = 2 = dim(ℝ²), therefore this linear application is injective, but it is not surjective because the rank of the matrix is not 3

f: ℝ² → ℝ², f(x,y) = (x-y,x+y); the initial vector space and the final vector space have the same dimension, the matrix M_f associated with this linear application has 2 columns and 2 rows,

M_f =

|1,-1|

|1, 1|, ρ(M_f) = 2 = dim(ℝ²), the rank of the matrix is 2 because the 2 rows are linearly independent, and the rank of the matrix is equal to the number of columns and the number of rows, therefore this linear application is injective and surjective, so it is invertible; the inverse linear application is an application from ℝ² to ℝ², f^-1: ℝ² → ℝ²; M_f is the matrix associated with the linear application f, M_f^-1 is the matrix associated with the inverse linear application f^-1, M_f^-1 = (M_f)^-1; to find the inverse linear application we have to compute the inverse of the associated matrix, M_f⋅M_f^-1 = I,

M_f =

|1,-1|

|1, 1|

⋅

|a,b|

|c,d|

=

|1,0|

|0,1|

⇒

|a-c,b-d|

|a+c,b+d|

=

|1,0|

|0,1| ⇒ {a-c = 1, b-d = 0, a+c = 0, b+d = 1} ⇒ {a-c = 1, a+c = 0} ∧ {b-d = 0, b+d = 1}; a-c = 1 ⇒ a = c+1, a+c = 0 ⇒ a = -c, a = c+1 ⇒ -c = c+1 ⇒ 2c = -1 ⇒ c = -1/2, a = -c ⇒ a = -(-1/2) = 1/2; b-d = 0 ⇒ b = d, b+d = 1 ⇒ d+d = 1 ⇒ 2d = 1 ⇒ d = 1/2, d = 1/2 ⇒ b = 1/2; the coefficients of the matrix (M_f)^-1 = M_f^-1 are a = 1/2 ∧ b = 1/2 ∧ c = -1/2 ∧ d = 1/2,

(M_f)^-1 = M_f^-1 =

| 1/2,1/2|

|-1/2,1/2|, which is the matrix associated with the inverse linear application f^-1(x,y) = ((1/2)x+(1/2)y,(-1/2)x+(1/2)y)

If a linear application is invertible, the inverse linear application is found by calculating the inverse matrix, M_f^-1 = (M_f)^-1

Let us consider the linear application f and the linear application g, M_f is the matrix associated with the linear application f, M_g is the matrix associated with the linear application g, M_g·M_f = M_g∘f; if it is possible to calculate the compound linear application g∘f, the matrix of the linear compound application g∘f is given by the product of the matrix M_g and the matrix M_f

f^-1·f = identity, M_f^-1·M_f = I, because M_f^-1 is the inverse of the matrix M_f

f: ℝⁿ → ℝ^m, it is injective if ρ(M_{f^{) = n ⇒ Ker(f) = {0v}, it is surjective if ρ(Mf) = m ⇒ Im(f) = ℝm, it is invertible if ρ(Mf) = n = m}}

In order to be invertible, a matrix must be square

25 - LINEAR SYSTEMS - PART 1

Find the nucleus of the linear application f: ℝ³ → ℝ², f(x,y,z) = (x+y,x-z); to find the nucleus of the linear application we have to solve the linear system of 2 equations {x+y = 0, x-z = 0}; x+y = 0 ⇒ x = -y ⇒ y = -x, x-z = 0 ⇒ x = z ⇒ z = x, x = z = -y, x is the free unknown, z and y are functions of x, the most general solution is the triad of numbers (x,-x,x)

An example of a linear system is {x+3y-2z+t = 5, 2x-8y+4z-t = 0, -x+5y-3t = 12}, system of 3 equations with 4 unknowns and known terms

Solve the system {x+y+2z = 1, 2y-z = 0, 4y = 5}; this system is simple to solve because there are many null coefficients, many zeros; 4y = 5 ⇒ y = 5/4, 2y-z = 0 ⇒ 2(5/4)-z = 0 ⇒ 5/2-z = 0 ⇒ z = 5/2, x+y+2z = 1 ⇒ x = -y-2z+1 ⇒ x = -5/4-2(5/2)+1 ⇒ x = -5/4-5+1 = -5/4-4 = (-5-16)/4 = -21/4; the system solutions are x = -21/4 ∧ y = 5/4 ∧ z = 5/2; x+y+2z = 1 ⇒ -21/4+5/4+2(5/2) = 1 ⇒ -16/4+5 = 1 ⇒ -4+5 = 1 ⇒ 1 = 1; 2y-z = 0 ⇒ 2(5/4)-5/2 = 0 ⇒ 5/2-5/2 = 0 ⇒ 0 = 0; 4y = 5 ⇒ 4(5/4) = 5 ⇒ 5 = 5

The most general system is a system with m equations and n unknowns, {a_1,1⋅x₁+a_1,2⋅x₂+...+a_1,n⋅x_n = b₁, a_2,1⋅x₁+a_2,2⋅x₂+...+a_2,n⋅x_n = b₂, a_m,1⋅x₁+a_m,2⋅x₂+...+a_m,n⋅x_n = b_m}; this general system is connected to a matrix, indeed, to be precise, the connected matrices are 2; the coefficients of the x are real numbers indicated by the letter a, and they form a matrix with m rows and n columns, A =

|a_1,1,a_1,2,...,a_1,n|

|a_2,1,a_2,2,...,a_2,n|

|....,....,...,....|

|a_m,1,a_m,2,...,a_m,n|, we must also consider the known terms b₁,b₂,b_m that form the column B =

|b₁|

|b₂|

|...|

|b_m|; the matrix A of the coefficients of the x with the matrix B of the known terms form the complete matrix (A|B)

Linear systems and matrices: A⋅X = B; complete matrix (A|B) =

|a_1,1,...,a_1,n|b₁|

|....,...,....|..|

|a_m,1,...,a_m,n|b_m|

A linear system can be represented by its complete matrix (A|B) that is the union of the matrix A of the coefficients of the x with the matrix B of the known terms; A is a matrix with m rows and n columns, B is a matrix consisting of only 1 column; the system can be written in the synthetic form A⋅X = B, X is the matrix of unknowns, X =

|x₁|

|x₂|

|..|

|x_n|; X is a matrix with 1 column, therefore from the product A⋅X we obtain a matrix with 1 column that is the matrix B, the column of known terms; in this way we can write a linear system in the form of a matrix, linear system written in matrix form; matrices are very useful tools because thanks to their use we can solve systems of linear equations

Transform the linear system {x+y-z = 1, 2x+y-2z = 0, x+y = 4} into the matrix (A|B);

(A|B) =

|1,1,-1|1|

|2,1,-2|0|

|1,1, 0|4|,

A =

|1,1,-1|

|2,1,-2|

|1,1, 0|,

X =

|x|

|y|

|z|,

B =

|1|

|0|

|4|, the linear system is A⋅X = B

A⋅X = B is any linear system of m equations with n unknowns; a reduced system is easier to solve; a linear system is reduced when the matrix A is reduced by rows; to reduce a linear system we have to reduce by rows the matrix A of the coefficients

{2x-y+z = 1, 4y-3z = 0, 5y = 10} is a system of 3 equations with 3 unknowns x,y,z; A⋅X = B,

A =

|2,-1,1|

|0,4,-3|

|0,5, 0|,

X =

|x|

|y|

|z|,

B =

| 1|

| 0|

|10|,

A|B =

|2,-1, 1| 1|

|0, 4,-3| 0|

|0, 5, 0|10|; the matrix A of the coefficients is reduced by rows, therefore this is a reduced system; to solve a reduced system we have to start from the last equation because it contains more zeros, therefore fewer unknowns; 5y = 10 ⇒ y = 10/5 = 2, 4y-3z = 0 ⇒ 3z = 4y ⇒ z = (4/3)y = (4/3)2 = 8/3, 2x-y+z = 1 ⇒ 2x = y-z+1 ⇒ x = (y-z+1)/2 = (2-8/3+1)/2 = (3-8/3)/2 = (1/3)/2 = (1/3)(1/2) = 1/6, the system solutions are x = 1/6 ∧ y = 2 ∧ z = 8/3; this system is easy to solve because it is reduced by rows, the matrix A of the coefficients is reduced by rows

Reduced systems: reduced system A⋅X = B;

(A|B) =

|a,b,a,a,a|a|

|b,0,a,a,a|a|

|0,0,b,a,a|a|

|0,0,0,a,b|a|

|0,0,0,b,0|a|, a are any elements, b are special elements or non-null elements that have only 0 below; a matrix is reduced by rows when there is a special element in each row; a system A⋅X = B is reduced when the matrix A is reduced

The general rule for solving a system A⋅X = B reduced by rows is to solve the last equation that contains the greatest number of zeros and therefore the least number of unknowns, and we must solve this equation with respect to one unknown as a function of the other unknowns; repeat for the penultimate equation; repeat until the first equation

In more complicated cases the number of equations and the number of unknowns are not the same; we consider the matrix (A|B) =

|2,1, 4,1|1|

|0,2,-1,1|1|

|0,0, 2,1|0|; matrix (A|B) is reduced because matrix A is reduced; the matrix (A|B) is the complete matrix of the system A⋅X = B which is formed by 3 equations with 4 unknowns; the system associated with the matrix is {2x+y+4z+t = 1, 2y-z+t = 1, 2z+t = 0}; the last equation of the system is 2z+t = 0, therefore we get an unknown as a function of the other, 2z+t = 0 ⇒ t = -2z, 2y-z+t = 1 ⇒ 2y = z-t+1 ⇒ y = (z-t+1)/2 = (z-(-2z)+1)/2 = (z+2z+1)/2 = (3z+1)/2 = (3/2)z+1/2, 2x+y+4z+t = 1 ⇒ 2x = -y-4z-t+1 ⇒ x = (-y-4z-t+1)/2 = (-((3/2)z+1/2)-4z-(-2z)+1)/2 = ((-3/2)z-1/2-4z+2z+1)/2 = ((-3/2-4+2)z+1-1/2)/2 = ((-3/2-2)z+1/2)/2 = ((-7/2)z+1/2)/2 = (-7/2)z+1/2)(1/2) = (-7/4)z+1/4, we have found the unknowns x,y,t as a function of z that is a free unknown, x = (-7/4)z+1/4 ∧ y = (3/2)z+1/2 ∧ t = -2z; for the reason that z is a free unknown, this system has infinite solutions; the values of x,y,t change for each different value of z; z = 0, x = (-7/4)z+1/4 ⇒ (-7/4)0+1/4 = 1/4, y = (3/2)z+1/2 ⇒ (3/2)0+1/2 = 1/2, t = -2z ⇒ t = -2⋅0 = 0; if a system is reduced and an unknown is free then the system has infinite solutions

To solve a linear system we have to write the complete associated matrix, reduce the matrix, and start solving the last equation

A reduced system may have no solutions; we consider the matrix (A|B) =

|2,1, 4,1|1|

|0,2,-1,1|1|

|0,0, 2,1|0|

|0,0, 0,0|2|, the last equation is 0x+0y+0z+0t = 2 that has no solution, therefore this system has no solutions, it is an irresoluble system, also called incompatible system

If a system has no solutions, it is called an irresoluble system or an incompatible system; if the matrix (A|B) is reduced it is easy to understand if the system has no solutions, the last row of matrix A is composed by only 0, but the last known term of column B is not 0

Considering a linear system A⋅X = B, we must establish if the linear system is solvable and find the solutions that are all the tuples of numbers that lead to equality between A⋅X and B; if the matrix A of the coefficients is reduced by rows, the resolution of the linear system is simple; if the matrix A of the coefficients is not reduced by rows, the resolution of the linear system is more complicated

If we have an unreduced system A⋅X = B, we transform it into the row reduced system A'⋅X = B' that has the same solutions as the system A⋅X = B; solving the system reduced by rows A'⋅X = B' we find the same solutions as for the system A⋅X = B

A⋅X = B → A'⋅X = B':

(A|B) =

|a,a,a,a,a|a|

|a,a,a,a,a|a|

|a,a,a,a,a|a|

|a,a,a,a,a|a|

|a,a,a,a,a|a|

→

(A'|B') =

|a,b,a,a,a|a|

|b,0,a,a,a|a|

|0,0,b,a,a|a|

|0,0,0,a,b|a|

|0,0,0,b,0|a|, a = any element, b = special element other than 0, 0 = zero; with the elementary transformations on the rows we reduce by rows matrix A, therefore from the matrix (A|B) we obtain the reduced matrix (A'|B'); after having obtained the reduced matrix (A'|B') we have to solve the system

By reducing the matrix A of the coefficients, the complete matrix (A|B) is automatically reduced

Considering the system {x+y+z-t = 1, x-y+2z+t = 0} that has 2 equations and 4 unknowns; the matrix associated with the system is (A|B) =

|1, 1,1,-1|1|

|1,-1,2, 1|0|, matrix A is not reduced, therefore this system is not reduced; we reduce by rows matrix A,

(A|B) =

|1, 1,1,-1|1|

|1,-1,2, 1|0| → R₂ = R₂+R₁ →

|1,1,1,-1|1|

|2,0,3, 0|1| = (A'|B'), the reduced system associated with the reduced matrix (A'|B') is {x+y+z-t = 1, 2x+3z = 1}, 2x+3z = 1 ⇒ 2x = -3z+1 ⇒ x = (-3z+1)/2 = (-3/2)z+1/2, x+y+z-t = 1 ⇒ y = -x-z+t+1 ⇒ -((-3/2)z+1/2)-z+t+1 = (3/2)z-1/2-z+t+1 = (3/2-1)z+t+1-1/2 = (1/2)z+t+1/2, z and t are the free unknowns and the solutions are x = (-3/2)z+1/2 ∧ y = (1/2)z+t+1/2; this is the simplest possible technique to solve a system with row reduction; this procedure does not work with column reduction; the reduction by columns is not used for the resolution of linear systems

To solve a linear system we have to use the reduction by rows; the reduction by columns does not allow the resolution of a linear system

A linear system can be easily solved if the matrix A of the coefficients is reduced, if the matrix A of the coefficients is not reduced then we have to reduce it by rows

26 - LINEAR SYSTEMS - PART 2

Solve the linear system AX = B,

A =

| 1, 1,-1,1,1|

|-1,-1, 0,1,1|

|-3,-1, 0,1,2|,

B =

|0|

|1|

|1|; matrix A has 3 rows and 5 columns, therefore the linear system AX = B is a system of 3 equations with 5 unknowns, and to solve this system we have to write the complete matrix (A | B) and reduce it by rows,

(A|B) =

| 1, 1,-1,1,1|0|

|-1,-1, 0,1,1|1|

|-3,-1, 0,1,2|1| → R₃ = R₃-R₂ →

| 1, 1,-1,1,1|0|

|-1,-1, 0,1,1|1|

|-2, 0, 0,0,1|0|,

| x, y, z,t,u|, now the complete matrix (A|B) is reduced by rows and from the last row we get the equation -2x+u = 0, -2x+u = 0 ⇒ u = 2x, -x-y+t+u = 1 ⇒ -x-y+t+2x = 1 ⇒ x-y+t = 1 ⇒ t = -x+y+1, x+y-z+t+u = 0 ⇒ x+y-z-x+y+1+2x = 0 ⇒ 2x+2y-z+1 = 0 ⇒ z = 2x+2y+1; we must always specify the variable that does not appear in the lines below; z = 2x+2y+1, z is a function of x and y; u = 2x, u is a function of x; t = -x+y+1, t is a function of x and y; the unknowns x and y can be assigned at will, x and y are 2 free unknowns; the unknowns z,u,t cannot be assigned at will but depend on x and y; x,y,z,t,u, are the 5 unknowns, x,y are 2 the free unknowns, z,t,u are the 3 unknowns that depend on the 2 free unknowns x,y;

A' =

| 1, 1,-1,1,1|

|-1,-1, 0,1,1|

|-2, 0, 0,0,1|, the matrix A' is reduced and has 3 non-zero rows, therefore ρ(A') = 3 ⇒ ρ(A'|B') = 3, the matrix of the coefficients A and the complete matrix (A|B) have the same rank and this is the first property of the Rouché-Capelli theorem; this system has 5 unknowns of which 2 are the free unknowns, 5-2 = 3 = ρ(A') = ρ(A'|B'); the number of free unknowns is linked to the total number of unknowns by the rank of the coefficient matrix A which is equal to the rank of the complete matrix (A|B); 5-2 = 3 ⇒ 2 = 5-3, the number of free unknowns is equal to the total number of unknowns minus the rank of the matrix, number of free unknowns = number of unknowns - rank of the matrix, number of free unknowns = n-ρ(A), the letter n indicates the total number of unknowns which is equal to the number of columns in the matrix, and all this is true for the hypothesis that ρ(A) = ρ(A|B)

The first property of the Rouché-Capelli theorem states that a system that is solvable has the rank of the coefficient matrix equal to the rank of the complete matrix, this is the condition of solvability of the Rouché-Capelli theorem; the Rouché-Capelli theorem states that the number of free unknowns is equal to the total number of unknowns minus the rank of the matrix

AX = B, (A|B) =

|1,1,1|1|

|2,1,1|0|

|3,2,2|4|; we perform a row reduction of the system,

|1,1,1|1|

|2,1,1|0|

|3,2,2|4| → R₂ = R₂-R₁ →

|1,1,1| 1|

|1,0,0|-1|

|3,2,2| 4| → R₃ = R₃-2R₁ →

|1,1,1| 1|

|1,0,0|-1|

|1,0,0| 2| → R₃ = R₃-R₂ →

|1,1,1| 1|

|1,0,0|-1|

|0,0,0| 3|, therefore the last row of the matrix becomes the equation 0 = 3, this means that the system has no solutions; ρ(A') = ρ(A) because the rank of the reduced matrix is equal to the rank of the original matrix, and the rank of a reduced matrix is equal to the number of non-zero rows, tehrefore ρ(A') = ρ(A) = 2, but ρ(A'|B') = ρ(A|B) = 3, ρ(A) = 2 < ρ(A|B) = 3, the rank of the matrix of the coefficients A is different from the rank of the complete matrix (A|B) and this indicates that the system AX = B is not solvable

When a linear system has no solutions, the rank of the coefficient matrix A is different from the rank of the complete matrix (A|B); if the rank of the coefficient matrix A is different from the rank of the complete matrix (A|B), the associated linear system has no solutions

Any system can be reduced by leaving the solutions unchanged

Rouché-Capelli theorem: linear system AX = B of m equations in n unknowns with ρ(A) = p and ρ(A|B) = q; solvable ⇔ p = q, then there are ∞^n-p solutions or n-p unknowns are free

The Rouché-Capelli theorem is fundamental for understanding linear systems; considering the linear system AX = B, A is the coefficient matrix, X is the column of unknowns, B is the column of known terms; the lines are m, so the equations are m; the columns are n, therefore the unknowns are n; if ρ(A) = ρ(A|B) ⇒ solvable system, if the rank of the coefficient matrix A is equal to the rank of the complete matrix (A|B), then the system AX = B is solvable; a system is solvable, also called compatible system, when it admits at least one solution; the number of free unknowns is equal to the total number of unknowns minus the rank of the matrix, number of free unknowns = n-ρ(A) = n-ρ(A|B); we must not confuse the rank of the matrix with the number of equations of the matrix; the rank of a matrix is ​​the number of non-zero rows after the matrix has been reduced; if ρ(A) ≠ ρ(A|B) it is useless to calculate the number of free unknowns because the system has no solutions

The Rouché-Capelli theorem allows us to understand if a linear system admits solutions before solving it; if a system has no solutions and we proceed with the resolution we find absurd results like 0 = 1, therefore the Rouché-Capelli theorem allows us to save time; the Rouché-Capelli theorem allows to study a linear system without solving it

The Rouché-Capelli theorem states that, considering a linear system AX = B, if ρ(A) = ρ(A|B) the system is solvable and we have to find the solutions and the free unknowns are n-ρ(A), if ρ(A) ≠ ρ(A|B) the system is not solvable and it is useless to find the solutions because the system has no solutions; to find ρ(A) and ρ(A|B) we must perform the row reduction on the matrix (A|B); if ρ(A) = ρ(A|B) we proceed with the solution of the system considering that the number of free unknowns is n-ρ(A); if we find a number of free unknowns other than n-ρ(A) there was an error in the procedure, therefore the Rouché-Capelli theorem is useful to check the resolution procedure of a linear system

We study the linear system {x+y = 1, x-3y = 2, x+7y = 0} that has 3 equations and 2 unknowns; the number of equations is not important, the rank is important; the complete matrix of the linear system is (A|B) =

|1, 1|1|

|1,-3|2|

|1, 7|0|, the coefficient matrix A has 2 columns, therefore ρ(A) ≤ 2;

(A|B) =

|1, 1|1|

|1,-3|2|

|1, 7|0| → R₂ = R₂-R₁ →

|1, 1|1|

|0,-4|1|

|1, 7|0| → R₃ = R₃-R₁ →

|1, 1| 1|

|0,-4| 1|

|0, 6|-1| → R₃ = (4/6)R₃+R₂ →

|1, 1| 1|

|0,-4| 1|

|0, 0|1/3|, therefore the matrix of the coefficients A has rank 2, the complete matrix (A|B) has rank 3, ρ(A) = 2, ρ(A|B) = 3, ρ(A) ≠ ρ(A|B), therefore the system is not solvable because 2 < 3; the system has no solutions for the Rouché-Capelli theorem, and it is useless to continue with the calculations

Homogeneous systems: AX = 0 always has n-ρ(A) free unknowns

A homogeneous linear system is composed by equations with a known term equal to 0; a homogeneous linear system AX = 0 always has solutions, and the number of free unknowns is always n-ρ(A); for the Rouché-Capelli theorem, a homogeneous linear system always has solutions, because if column B is a column of 0, the complete matrix (A|B) cannot have a rank higher than the rank of the coefficient matrix A, therefore ρ(A) = ρ(A|B); a homogeneous linear system always has solutions, also because x₁ = x₂ = ... = x_n = 0 is a solution; in a homogeneous linear system there is always the null solution, but we must find if there are solutions other than the null solution by counting the number of solutions; if we have a system with n unknowns, m equations, ρ(A) = p = ρ(A|B) = q, the number of free unknowns is n-p; a linear system has only one solution when there are no free unknowns, or when each unknown has only one value, and this happens when n = p, n = ρ(A) = ρ(A|B); when tehre are no free unknowns, the solution is unique; if n = p, we can symbolically write ∞^n-p = 1 = ∞⁰; if a homogeneous system has only one solution, then the only solution is the null solution; a homogeneous linear system, like all linear systems that have solutions, has ∞^n-p solutions; a homogeneous linear system has only one solution, the null solution, when n = p, or when the number of unknowns coincides with the rank of the matrix of the coefficients A that is equal to the rank of the complete matrix (A|B); a homogeneous linear system has other solutions, besides the null solution, when n-p > 0, when the number of unknowns is greater than the rank of the matrix

A linear system may have no solutions; a homogeneous linear system has at least one solution; a linear system has only one solution when all the unknowns has a single value; if there is no single solution then n-p > 0, therefore the linear system has a free unknown and the solutions are infinite

A linear system may have no solutions, may have a solution, may have infinite solutions

If there is no single solution, then n-p > 0, therefore there is at least one free unknown, or infinite solutions

It is not possible for a linear system to have a finite number of solutions other than 1 or 0; a linear system can have 0 solutions, 1 solution, ∞ solutions; the number of free unknowns indicates the degree of infinity

Homogeneous linear systems have always or only 1 solution taht is zero, or they have infinite solutions; for a homogeneous linear system, searching for non-zero solutions is equivalent to searching for infinite solutions; there are infinite solutions when n-p > 0; if the number of unknowns is greater than the rank of the matrix, the linear system has infinite solutions

We study the homogeneous linear system {x-y+z = 0, 2x+3y-5z = 0} of 2 equations with 3 unknowns; ρ(A) ≤ 2, the unknowns are 3, so there is at least one free unknown, therefore there are infinite solutions; to find the solutions we have to reduce the matrix by rows

We study the homogeneous linear system {x-y+z = 0, 2x+3y-5z = 0, x+5y-z = 0, y+2z = 0} of 4 equations with 3 unknowns; when a system has many equations and few unknowns, there are usually no solutions, but a homogeneous system always has solutions; we must understand if this homogeneous linear system has non-zero solutions, therefore we write the complete matrix of the coefficients A =

|1,-1,1|

|2,3,-5|

|1,5,-1|

|0,1, 2|, we have to calculate ρ(A), so we have to proceed with the row reduction;

A =

|1,-1, 1|

|2, 3,-5|

|1, 5,-1|

|0, 1, 2| → R₂ = R₂-2R₁ →

|1,-1, 1|

|0, 5,-7|

|1, 5,-1|

|0, 1, 2| → R₃ = R_{3/sub>-R1 →}

|1,-1, 1|

|0, 5,-7|

|0, 6,-2|

|0, 1, 2| → R₃ = 5R_{3/sub>-6R2 →}

|1,-1, 1|

|0, 5,-7|

|0, 0,32|

|0, 1, 2| → R₄ = 5R_{4/sub>-R2 →}

|1,-1, 1|

|0, 5,-7|

|0, 0,32|

|0, 0,17| → R₄ = 32R_{4/sub>-17R3 →}

|1,-1, 1|

|0, 5,-7|

|0, 0,32|

|0, 0, 0|, ρ(A) = 3 = n, this homogeneous linear system has only 1 null solution because the number of unknowns is equal to the rank of the matrix

Let us consider the linear application f: ℝⁿ → ℝ^m; to find the elements that form the nucleus of the linear application we have to solve a homogeneous linear system; if A is the matrix associated with the linear application, A = M_f, the system we have to solve to find the nucleus of the linear application is AX = 0,

X =

|x₁|

|..|

|x_n|,

f(x₁,...,x_n) = (a_1,1⋅x₁+...)

Systems and nucleus: M_{f = A, Ker(f) = solutions of AX = 0, dim(Ker(f)) = n-ρ(Mf)}

The dimension of the nucleus is related to the rank of the matrix M_f, dim(Ker(f)) = n-ρ(M_f), the dimension of the nucleus is equal to the dimension of the starting space minus the rank of the matrix which is the dimension of the image

The nucleus is the set of solutions of a homogeneous system AX = 0; M_f = A, ρ(A) = ρ(M_f) = p, the nucleus is the set of solutions of a homogeneous system that has ρ(M_f) = p, n-p is the number of free unknowns, n-p = dim(Ker(f))

To solve the system AX = B we have to reduce the matrix A by rows; AX = B solvable ⇒ ρ(A) = ρ(A|B) = p; if the system AX = B is solvable there are n-p free unknowns

27 - LINEAR SYSTEMS - PART 3

Counter-images: f: ℝⁿ → ℝ^m, M_f = A, AX = [a₁,...,a_m] gives the counter-image f^-1(a₁,...,a_m)

f(x,y,z) = (x+y,x-z), ℝ³ → ℝ²; we choose an element of ℝ², we choose a vector v = (1,-1), and we look for the vectors of ℝ³ that are the triples (x,y,z) such that f(x,y,z) = (1,-1);

M_f = A =

|1,1, 0|

|1,0,-1|, {x+y = 1, x-z = -1}, x-z = -1 ⇒ x = z-1, x+y = 1 ⇒ y = -x+1 = -(z-1)+1 = -z+1+1 = -z+2, x = z-1 ∧ y = -z+2, z is the free unknown, therefore the most general solution is the triple (z-1,-z+2,z); the counter-images of the vector (1,-1) are all the infinite triples (z-1,-z+2,z), if z = 0 the counter-image is the triple (-1,2,0), if z = 10 the counter-image is the triple (9,-8,10), the triples change for each value of z, so there are infinite counter-images

It may happen that there is no counter-image; let us consider the linear application ℝ³ → ℝ³, f(x,y,z) = (2x-y,2x-y,0), and the associated matrix is A =

|2,-1,0|

|2,-1,0|

|0, 0,0|; to find the counter-image of the vector represented by the triple (4,4,1) we have to solve the system {2x-y = 4, 2x-y = 4, 0 = 1}, but the absurd equation 0 = 1 indicates that this system has no solutions; the counter image of the vector (4,4,1) does not exist, therefore the linear application is not surjective because not all the elements of ℝ³ have a counter-image; to understand that this linear application is not surjective it is enough to observe that there is a null row and also a null column; the linear application has an associated matrix with rank that is not 3, even the 3 columns are all linearly dependent therefore the rank of matrix A is 1, ρ(A) = 1 < 3, therefore the linear application is neither injective nor surjective, so it is possible that there are vectors with no counter-image; to find the counter image of 0, of the triple (0,0,0), f^-1(0,0,0), we have to solve the system of 3 equations {2x-y = 0, 2x-y = 0, 0 = 0} which is actually a single equation, 2x-y = 0 ⇒ y = 2x; therefore x and z are 2 free unknowns and can be assigned at will, x is the non-free unknown; the most general solution is the triple (x,2x,z), therefore there are infinite elements in the core of this linear application; the nucleus is the counter-image of the null vector; in this case the nucleus is the triple (2,2x,z) and contains infinite elements depending on the 2 free unknowns x,z; the nucleus is represented by the triple (x,2x,z), therefore (1,2,0), (0,0,1), (10,20,85) are elements of the nucleus, there are ∞² elements in the nucleus because the free unknowns are 2; the nucleus is formed by the triplets (x,2x,z) and we want a base of the nucleus, 2 free unknowns means that the core has dimension 2, if x = 1 and z = 0 we find the vector (1,2,0), if x = 0 and z = 1 we find the vector (0,0,1), therefore the 2 vectors (1,2,0),(0,0,1) are a basis of the nucleus; when in f^-1(0,0,0) we have free unknowns we have to put them equal to 1 and 0 in all possible ways and we find a basis of the nucleus

The counter image of 0 is the nucleus of the linear application

If A is a square matrix of order n, with n rows and n columns, the inverse matrix A^-1 follows the formula A^-1·A = A·A^-1 = I, where the letter I indicates the identity matrix of order n; to find the inverse of the matrix A =

|1,1|

|2,3|, we have to solve

|1,1|

|2,3|

·

|a,b|

|c,d|

=

|1,0|

|0,1|, therefore we have to find the solutions of the linear system of equations {a+c = 1, b+d = 0, 2a+3c = 0, 2b+3d = 1}; to find the inverse matrix we have to solve a system of linear equations in which the unknowns are the elements of the inverse matrix; the inverse matrix of a 2x2 square matrix leads to 2⋅2 = 4 unknowns, the inverse matrix of a 3x3 square matrix leads to 3⋅3 = 9 unknowns, the inverse matrix of a matrix of order 25 leads to 25⋅25 = 625 unknowns, if n is the order of the matrix the number of unknowns and equations is n²; this is not the best way to calculate the inverse matrix

Inverse of matrix A:

|a₁,b₁,c₁|

|a₂,b₂,c₂|

|a₃,b₃,c₃|

⋅

X =

|1,0,0|

|0,1,0|

|0,0,1|,

X =

|X₁|

|X₂|

|X₃|; X₁,X₂,X₃ = righe di X = A^-1

To find the inverse matrix A^-1 of the matrix A of order 3, we must not consider the matrix X as a set of 9 coefficients, but as 3 rows; the unknown becomes the line, in this way we don't have to solve 9 equations with 9 unknowns, but only 3 equations with 3 unknowns, and each unknown is a triple because it is a row of the matrix

We can consider the rows x₁,x₂,x₃ like 3 unknowns, a₁⋅X₁+b₁⋅X₂+c₁⋅X₃ = (1,0,0), a₂⋅X₁+b₂⋅X₂+c₂⋅X₃ = (0,1,0), a₃⋅X₁+b₃⋅X₂+c₃⋅X₃ = (0,1,0); we have obtained a linear system in which the unknowns are triples, the coefficients are the elements of the matrix A, each known term is a triple, the column of known terms has become the identical matrix; the complete matrix is (A|B) =

|a₁,b₁,c₁|1,0,0|

|a₂,b₂,c₂|0,1,0|

|a₃,b₃,c₃|0,0,1| = (A|I), this system can be solved by reducing the complete matrix by rows, in the end we get a reduced system, the unknowns are the rows of the inverse matrix, and this is the easiest way to calculate the inverse matrix

Find the inverse matrix X = A^-1 of the matrix A =

|2,1|

|0,3|; the inverse matrix X = A^-1 must have 2 rows and 2 columns like matrix A; the matrix A has 2 linearly independent rows, ρ(A) = 2; matrix A is definitely an invertible matrix, A is a matrix associated with an invertible linear application; X₁ and X₂ are the 2 rows of the matrix X = A^-1, and are the unknowns of the linear system; {2X₁+X₂ = (1,0), 3X₂ = (0,1)}, we have to solve this system of 2 equations in the 2 unknowns X₁ and X₂ that are 2 lines with 2 elements, 3X₂ = (0,1) ⇒ X₂ = (0,1/3), 2X₁+X₂ = (1,0) ⇒ 2X₁ = -X₂+(1,0) ⇒ X₁ = (-(0,1/3)+(1,0))/2 = (1,-1/3)/2 = (1/2,-1/6), X₁ = (1/2,-1/6) ∧ X₂ = (0,1/3),

X = A^-1 =

|1/2,-1/6|

| 0, 1/3|; we verify that A⋅A^-1 = I,

|2,1|

|0,3|

⋅

|1/2,-1/6|

| 0, 1/3|

=

|2(1/2)+1⋅0,2(-1/6)+1(1/3)|

|0(1/2)+3⋅0,0(-1/6)+3(1/3)|

=

|1+0,-1/3+1/3|

|0+0, 0+1|

=

|1,0|

|0,1|

Find the inverse matrix X = A^-1 of the matrix A =

|2,1|

|4,2|; in this case the inverse matrix X = A^-1 does not exist because ρ(A) = 1; matrix A corresponds to a non-invertible linear application, therefore the inverse matrix A^-1 does not exist; we proceed anyway by obtaining the system {2X₁+X₂ = (1,0), 4X₁+2X₂ = (0,1)}, we write the complete matrix and we reduce it by rows,

|2,1|1,0|

|4,2|0,1| → R₂ = R₂-2R₁ →

|2,1| 1,0|

|0,0|-2,1|, the second line leads to the absurd 0 = (-2,1), therefore the inverse matrix A^-1 does not exist because the matrix A is not invertible

To linear systems that are useful for finding the inverse of a matrix we can apply the Rouché-Capelli theorem, which states that a linear system admits solutions if the rank of the matrix of the coefficients A is equal to the rank of the complete matrix (A|B), ρ(A) = ρ(A|B), and when we look for an inverse matrix the formula becomes ρ(A) = ρ(A|I), I is the identity matrix and is formed by linearly independent rows because they are fundamental vectorors, therefore the rank of the complete matrix (A|I) is the maximum possible, so the linear system that allows to find the inverse is solvable if and only if the rank of the matrix A is also the maximum possible, and A is a square matrix of order n, therefore ρ(A) = n

With a matrix of order 3 we get 9 coefficients, but if we work with the rows we get only 3 equations; let us consider the matrix A 3x3, to find the inverse matrix X = A^-1 we use the formula AX = I,

X =

|a,b,c|

|d,e,f|

|g,h,i|, we get 9 equations with 9 unknowns, if instead we consider the rows X₁,X₂,X₃ we only get 3 unknowns and each is a triple

Find the inverse matrix of the 3x3 matrix A =

|1,2, 1|

|2,0, 1|

|0,0,-1|; to find the inverse matrix we have to write the linear system that has the 3 rows of the inverse matrix as unknowns, each row is a triple, A is the matrix of the coefficients and the known terms are the 3 rows of the 3x3 identity matrix I;

(A|I) =

|1,2, 1|1,0,0|

|2,0, 1|0,1,0|

|0,0,-1|0,0,1|, matrix A is already reduced by rows, ρ(A) = 3, therefore the inverse matrix A^-1 exists; starting from the last row of the matrix, |0,0,-1|0,0,1| ⇒ -X₃ = (0,0,1) ⇒ X₃ = (0,0,-1), |2,0, 1|0,1,0| ⇒ 2X₁+X₃ = (0,1,0) ⇒ 2X₁ = (0,1,0)-X₃ ⇒ X₁ = ((0,1,0)-X₃)/2 = ((0,1,0)-(0,0,-1))/2 = (0,1,1)/2 = (0,1/2,1/2), |1,2, 1|1,0,0| ⇒ X₁+2X₂+X₃ = (1,0,0) ⇒ 2X₂ = (1,0,0)-X₁-X₃ ⇒ X₂ = ((1,0,0)-X₁-X₃)/2 ⇒ X₂ = ((1,0,0)-(0,1/2,1/2)-(0,0,-1))/2 = ((1,-1/2,1/2))/2 = (1/2,-1/4,/1/4), we found the 3 rows of the inverse matrix, X₁ = (0,1/2,1/2) ∧ X₂ = (1/2,-1/4,/1/4) ∧ X₃ = (0,0,-1), therefore the inverse matrix is A^-1 =

| 0, 1/2,1/2|

|1/2,-1/4,1/4|

| 0, 0, -1|; to verify the inverse matrix we use the formula A⋅A^-1 = I,

A =

|1,2, 1|

|2,0, 1|

|0,0,-1|

⋅

| 0, 1/2,1/2|

|1/2,-1/4,1/4|

| 0, 0, -1|

=

| 1⋅0+2(1/2)+1⋅0, 1(1/2)+2(-1/4)+1⋅0, 1(1/2)+2(1/4)+1(-1)|

| 2⋅0+0(1/2)+1⋅0, 2(1/2)+0(-1/4)+1⋅0, 2(1/2)+0(1/4)+1(-1)|

|0⋅0+0(1/2)+(-1)0,0(1/2)+0(-1/4)+(-1)0,0(1/2)+0(1/4)+(-1)(-1)|

=

|0+1+0,1/2-1/2+0,1/2+1/2-1|

|0+0+0, 1+0+0, 1+0-1|

|0+0+0, 0+0+0, 0+0+1|

=

|1,0,0|

|0,1,0|

|0,0,1|

Find for which real values of the parameter h the matrix A is invertible, A =

|1, h|

|3,-1|; the matrix A depends on the parameter h, and we want to study for which values of h the matrix A is invertible, therefore we must understand for which values of h the system (A|I) admits solutions, (A|I) =

|1, h|1,0|

|3,-1|0,1|, this system admits solutions only if ρ(A) = 2; to calculate ρ(A) we have to reduce the matrix A =

|1, h|

|3,-1| → R₂ = R₂-3R₁ →

|1, h|

|0,-1-3h|, the matrix is invertible for the values of h such that ρ(A) = 2, for those values of h such that the 2 lines are linearly independent, therefore -1-3h ≠ 0 ⇒ 3h ≠ -1 ⇒ h ≠ -1/3, if h = -1/3 the matrix is not invertible, if h ≠ -1/3 the matrix is invertible; we reduce by rows the complete matrix (A|I) =

|1, h|1,0|

|3,-1|0,1| → R₂ = R₂-3R₁ →

|1, h|1 ,0|

|0,-1-3h|-3,1|, therefore we look for the inverse of the matrix A for h = -1,

(A|I) =

|1, -1|1 ,0|

|0,-1-3(-1)|-3,1|

=

|1, -1| 1,0|

|0,-1+3|-3,1|

=

|1,-1| 1,0|

|0, 2|-3,1|, and from the last row of the matrix we get |0, 2|-3,1| ⇒ 2X₂ = (-3,1) ⇒ X₂ = (-3/2,1/2), |1,-1| 1,0| ⇒ X₁-X₂ = (1,0) ⇒ X₁ = (1,0)+X₂ = (1,0)+(-3/2,1/2) = (-1/2,1/2), X₁ = (-1/2,1/2) ∧ X₂ = (-3/2,1/2), therefore the inverse matrix for h = -1 is A^-1 =

|-1/2,1/2|

|-3/2,1/2|; we can find the inverse matrix A^-1 for any value of h except for h = -1/3, in fact for h = -1/3 the inverse matrix A^-1 does not exist because the matrix A is not invertible

Linear systems are useful for finding the counter-image, in fact the counter-image of an element in a linear application is found by solving a linear system; linear systems are useful for finding Ker(f), if f: ℝⁿ → ℝ^m is a linear application, and if M_f = A is the matrix, AX = 0 is the linear system to be solved; to find the counter-image and Ker(f) we have to solve ordinary linear systems because the unknowns are real numbers; if we want to find the inverse of a matrix, the unknowns are not numbers but tuples, rows of a matrix, and the known terms are not numbers but tuples that correspond to the rows of the identity matrix

Solving linear systems we find: Ker(f), counter-images, inverse matrices

28 - THE DETERMINANT OF A SQUARE MATRIX

The generic 2x2 square matrix is A =

|a_1,1,a_1,2|

|a_2,1,a_2,2|; the determinant of this matrix is denoted by |A| or with det(A); the determinant of a square matrix is always a real number which is calculated from the coefficients of the matrix; det(A) = a_1,1⋅a_2,2-a_1,2⋅a_2,1

Let us consider the square matrix A =

|1, 3|

|5,-2|, det(A) = 1(-2)-3⋅5 = -2-15 = -17

The generic 3x3 square matrix is A =

|a_1,1,a_1,2,a_1,3|

|a_2,1,a_2,2,a_2,3|

|a_3,1,a_3,2,a_3,3|; the complementary matrix of an element is found excluding the row and column of the element; the complementary matrix of the element a_1,1 is A_1,1 =

|a_2,2,a_2,3|

|a_3,2,a_3,3|; the complementary matrix of the element a_1,2 is A_1,2 =

|a_2,1,a_2,3|

|a_3,1,a_3,3|; the complementary matrix of the element a_1,3 is A_1,3 =

|a_2,1,a_2,2|

|a_3,1,a_3,2|; det(A)= a_1,1⋅A_1,1-a_1,2⋅A_1,2+a_1,3⋅A_1,3, the sign of each addendum is given by a rule, the number of the row is added with the number of the column that are excluded to obtain the complementary matrix, the sign is positive if the sum is an even number, the sign is negative if the sum is an odd number

Let us consider a square matrix of order n, A =

|a_{1,1,a1,2,...,a1,n|}

|a_{2,1,a2,2,...,a2,n|}

|....,....,...,....|

|a_{n,1,an,2,...,an,n|; for each element of the matrix we must define its algebraic complement; the algebraic complement of an element is the determinant of the matrix which is obtained by deleting the row and the column that cross in that element and with a sign; the algebraic complement of the element a2,1 is obtained by deleting row 2 and column 1, therefore we obtain the submatrix Aa2,1 =}

|a_1,2,...,a_1,n|

|a_3,2,...,a_3,n|

|....,...,....|

|a_n,2,...,a_n,n|, A_2,1 is a submatrix having one row less and one column less than the matrix A; the determinant of submatrix A_2,1, det(A_2,1), preceded by a sign, is the algebraic complement of the element a_2,1; to obtain A_2,1 we deleted row R₂ and column C₁, 2+1 = 3 which is an odd number, therefore det(A_2,1) is preceded by the negative sign; the algebraic complement of the element a_2,2 has the positive sign because 2+2 = 4 which is an even number

Let us consider a 4x4 square matrix, A =

|1,3, 5,-1|

|2,2, 0, 1|

|1,0, 1, 4|

|1,1,-1, 0|; to find the algebraic complement of the element a_2,3 = 0 we have to delete R₂ and C₃, and we get det(A_2,3) =

|1,3,-1|

|1,0, 4|

|1,1, 0|, we deleted R₂ and C₃, 2+3 = 5 which is an odd number, therefore the determinant must be preceded by the negative sign, the algebraic complement of a_2,3 = 0 is -det(A_2,3)

We define the determinant of a 4x4 matrix; we take the elements of a row, for example we take the elements of R₁, a_1,1,a_1,2,a_1,3,a_1,4, and we multiply each element of R₁ by its algebraic complement, a_1,1⋅A_1,1+a_1,2⋅A_1,2+a_1,3⋅A_1,3+a_1,4⋅Aa_1,4 = det(A) when matrix A is 4x4; to define the determinant we can also use the elements of a column, such as C₁, a_1,1,a_2,1,a_3,1,a_4,1, and we multiply each element of C₁ by its algebraic complement, a_1,1⋅A_1,1+a_2,1⋅A_2,1+a_3,1⋅A_3,1+a_4,1⋅A_4,1 = det(A)

We take the elements of a row or a column and multiply them by their algebraic complements, and we must add them together, in this way we can calculate the determinant of a square matrix of any order

Considering a matrix 1x1, A = |a|, det(A) = a

Laplace's rule allows to calculate the determinant of a matrix

We calculate the determinant of a 3x3 square matrix using Laplace's rule, A =

|1,0, 1|

|2,1,-1|

|0,0, 2|; we delete R₂, det(A_2,1) =

|0,1|

|0,2| = 0⋅2-1⋅0 = 0-0 = 0,

det(A_2,2) =

|1,1|

|0,2| = 1⋅2-1⋅0 = 2-0 = 2,

det(A_2,3) =

|1,0|

|0,0| = 1⋅0-0⋅0 = 0-0 = 0, det(A) = -2⋅det(A_2,1)+1⋅det(A_2,2)-(-1)det(A_2,3) = -2⋅0+1⋅2+1⋅0 = 0+2+0 = 2, using Laplace's rule we calculated the determinant of the matrix A, det(A) = 2

By choosing a row that contains many 0s, Laplace's rule can be simplified

We consider a square matrix A of order n, and its determinant is det(A); in some cases the determinant of a matrix is null; if a matrix A has a null row or a null column, det(A) = 0; if a matrix A has equal rows, R_i = R_j, det(A) = 0; if a matrix A has equal columns, C_i = C_j, det(A) = 0; A' = a⋅A ⇒ det(A') = aⁿ⋅det(A); if we multiply only a row of the matrix A by a number a, obtaining the matrix A ', then det(A') = a⋅det(A); if a row of matrix A is the sum of 2 rows, each element is the sum of 2 elements,

A =

|a_1,1+b_1,1,...,a_1,n+b_1,n|

|.........,...,.........|

|a_m,1+b_m,1,...,a_m,n+b_m,n|,

A_a =

|a_1,1,...,a_1,n|

|....,...,....|

|a_m,1,...,a_m,n|,

A_b =

|b_1,1,...,b_1,n|

|....,...,....|

|b_m,1,...,b_m,n|,

det(A) = det(A_a)+det(A_b); all these are fundamental properties of a determinant of a matrix

If matrix A is a square matrix of order n, and if we apply elementary transformations to the rows or columns of matrix A, the determinant of the matrix does not change, or only its sign changes; if we exchange 2 rows R_i ↔ R_j obtaining the matrix A', det(A') = -det (A); if R_i → a⋅R_i obtaining the matrix A', det(A') = a⋅det(A); if R_i → R_i+a⋅R_j obtaining the matrix A', det(A') = det(A); therefore we can simplify the calculations of Laplace's rule by using elementary transformations

Find the determinant of the square matrix 4x4 A =

| 1, 1,1,1|

| 1,-1,0,2|

| 0, 1,0,0|

|-1, 4,2,1|; we can apply Laplace's rule using R₃ which contains 3 zeros, therefore the only element that interests us is a_3,2 = 1,

A_3,2 =

| 1,1,1|

| 1,0,2|

|-1,2,1|, det(A) = -1⋅det(A_3,2),

| 1,1,1|

| 1,0,2|

|-1,2,1| → C₁ = C₁-(1/2)C₃ →

| 1-(1/2)1,1,1|

| 1-(1/2)2,0,2|

|-1-(1/2)1,2,1| =

| 1-1/2,1,1|

| 1-1,0,2|

|-1-1/2,2,1| =

| 1/2,1,1|

| 0,0,2|

|-3/2,2,1| = A'_3,2, after this reduction the determinant has not changed;

det(A) =

-1⋅2(-1)⋅

| 1/2,1|

|-3/2,2| = 2((1/2)2-1(-3/2)) = 2(1+3/2) = 2(5/2) = 5, det(A) = 5; the simplest way to calculate the determinant is to bring up zeros, paying attention to any changes in the sign of the determinant

An interesting exercise is to find the determinant of the matrix A =

| 1, 1,1,2|

| 1, 3,2,1|

| 4, 3,2,1|

|-1,-1,2,5|

If A is a square matrix of order n, det(A) = 0 ⇒ non-invertible matrix, det(A) ≠ 0 ⇒ invertible matrix

A square matrix A of order n is invertible if det(A) ≠ 0, it is not invertible if det(A) = 0

Determine if matrix A is invertible, A =

|1, 2|

|3,-1|; det(A) = 1(-1)-2⋅3 = -1-6 = -7, therefore the matrix A is invertible because det(A) = -7 ≠ 0

Determine if matrix B is invertible,

B =

|1,4|

|2,8|; det(B) = 1⋅8-4⋅2 = 8-8 = 0, therefore the matrix B is not invertible because det(B) = 0

If det(A) ≠ 0, the matrix A is invertible and ρ(A) = n, an invertible matrix has the highest possible rank; if det(A) = 0, the matrix A is not invertible and ρ(A) < n

If det(A) ≠ 0, the matrix is invertible; the inverse matrix A^-1 can be calculated from the elements of matrix A and their algebraic complements;

A^-1 =

1/det(A)

⋅

|A_{1,1,A2,1,...,An,1|}

|A_1,2,A_2,2,...,A_n,2|

|A_1,3,A_2,3,...,A_n,3|

|....,....,...,....|

|A_1,n,A_2,n,...,A_n,n|, the matrix is composed of the algebraic complements after having swapped the rows with the columns

Find the inverse matrix of matrix A using the determinant, A =

| 1,3|

|-1,5|; det(A) = 1⋅5-3(-1) = 5+3 = 8, therefore the matrix A is invertible because det(A) = 8 ≠ 0; A_1,1 = 5, A_1,2 = -(-1) = 1, A_2,1 = -3, A_2,2 = 1, A^-1 =

1/8

⋅

|5,-3|

|1, 1|

=

|5/8,-3/8|

|1/8, 1/8|;

A⋅A^-1 = I,

| 1,3|

|-1,5|

⋅

|5/8,-3/8|

|1/8, 1/8|

=

| 1(5/8)+3(1/8), 1(-3/8)+3(1/8)|

|-1(5/8)+5(1/8),-1(-3/8)+5(1/8)|

=

| 5/8+3/8,-3/8+3/8|

|-5/8+5/8, 3/8+5/8|

=

|8/8,0|

|0,8/8|

=

|1,0|

|0,1| = I

A square matrix A of order n is orthogonal if A^T = A^-1, therefore an orthogonal matrix is an invertible matrix; the determinant of an orthogonal matrix A is 1 or -1; if the matrix A is orthogonal, then det(A) = ±1

Orthogonal matrices: A =

|a_1,1,...,a_1,n|

|....,...,....|

|a_n,1,...,a_n,n|; orthogonal square matrix P, P^T⋅P = I ⇔ P^-1 = P^T; det(P) = ±1; if det(P) = 1, then the orthogonal matrix P is called special

P =

|cos(a),-sin(a)|

|sin(a), cos(a)|, P is a special orthogonal matrix

P =

|cos(a),-sin(a)|

|sin(a), cos(a)|, det(P) = cos(a)⋅cos(a)-(-sin(a)⋅sin(a)) = cos²(a)+sin²(a) = 1, therefore P is a special orthogonal matrix

Laplace's rule: det(A) = a_i,1⋅A_i,1+a_i,2⋅A_i,2+...+a_i,n⋅A_i,n = a_1,j⋅A_1,j+a_2,j⋅A_2,j+...+a_n,j⋅A_n,j; A is invertible if and only if det(A) ≠ 0; det(A^-1) = 1/det(A); se det(A) ≠ 0,

A^-1 =

1/det(A)

⋅

|A_1,1,...,A_n,1|

|...,...,...|

|A_1,n,...,A_n,n|

det(A⋅B) = det(A)⋅det(B)

---

PROPERTIES OF EXPONENTS

a^m⋅aⁿ = a^m+n

a^m/aⁿ = a^m-n

(a^m)ⁿ = a^m⋅n

aⁿ⋅bⁿ = (a⋅b)ⁿ

aⁿ/bⁿ = (a/b)ⁿ

PYTHAGOREAN THEOREM

In a right triangle, the area of the square constructed on the hypotenuse is equal to the sum of the areas of the squares constructed on the catheti; in a right-angled triangle ABC, with a right angle A, AB and AC are the catheti, BC is the hypotenuse, BC² = AB²+AC²

FIRST THEOREM OF EUCLID

In a right triangle, the area of the square constructed on a cathetus is equal to the area of the rectangle which has as sides the hypotenuse and the projection of the cathetus on the hypotenuse; in a right-angled triangle ABC, with a right angle A, AB and AC are the catheti, BC is the hypotenuse, BH is the projection of the cathetus AB on the hypotenuse BC, CH is the projection of the cathetus AC on the hypotenuse BC, AB² = BC⋅BH, AC² = BC⋅CH

In a right triangle, each cathetus is the proportional mean between the hypotenuse and the projection of the cathetus on the hypotenuse; in a right-angled triangle ABC, with a right angle A, AB and AC are the catheti, BC is the hypotenuse, BH is the projection of the side AB on the hypotenuse BC, CH is the projection of the side AC on the hypotenuse BC, BC/AB = AB/BH, BC/AC = AC/CH

SECOND THEOREM OF EUCLID

In a right triangle, the area of the square constructed on the height relative to the hypotenuse is equal to the area of the rectangle which has as sides the projections of the catheti on the hypotenuse; in a right-angled triangle ABC, with a right angle A, AB and AC are the catheti, BC is the hypotenuse, BH is the projection of the side AB on the hypotenuse BC, CH is the projection of the side AC on the hypotenuse BC, AH is the height relative to the hypotenuse, AH² = BH⋅CH

In a right triangle, the height relative to the hypotenuse is the proportional mean between the projections of the catheti on the hypotenuse; in a right-angled triangle ABC, with a right angle A, AB and AC are the catheti, BC is the hypotenuse, BH is the projection of the side AB on the hypotenuse BC, CH is the projection of the side AC on the hypotenuse BC, AH is the height relative to the hypotenuse, BH/AH = AH/CH ⇒ AH² = BH⋅CH, CH/AH = AH/BH ⇒ AH² = BH⋅CH

IMPORTANT ALGEBRAIC FORMULAS

(a+b)² = (a+b)(a+b) = a²+ab+ab+b² = a²+2ab+b²

(a-b)² = (a-b)(a-b) = a²-ab-ab+b² = a²-2ab+b²

(a+b)(a-b) = a²-ab+ab-b² = a²-b²

(a+b)³ = (a+b)²(a+b) = (a+b)(a+b)(a+b) = (a²+ab+ab+b²)(a+b) = (a²+2ab+b²)(a+b) = a³+a²b+2a²b+2ab²+ab²+b³ = a³+3a²b+3ab²+b³

(a-b)³ = (a-b)²(a-b) = (a-b)(a-b)(a-b) = (a²-ab-ab+b²)(a-b) = (a²-2ab+b²)(a-b) = a³-a²b-2a²b+2ab²+ab²-b³ = a³-3a²b+3ab²-b³

(a+b)²(a-b) = (a+b)(a+b)(a-b) = (a²+ab+ab+b²)(a-b) = (a²+2ab+b²)(a-b) = a³-a²b+2a²b-2ab²+ab²-b³ = a³+a²b-ab²-b³ = a²(a+b)-b²(a+b) = (a+b)(a²-b²)

(a-b)²(a+b) = (a-b)(a-b)(a+b) = (a²-ab-ab+b²)(a+b) = (a²-2ab+b²)(a+b) = a³+a²b-2a²b-2ab²+ab²+b³ = a³-a²b-ab²+b³ = a²(a-b)-b²(a-b) = (a-b)(a²-b²)

(a+b)(a²-ab+b²) = a³-a²b+ab²+a²b-ab²+b³ = a³+b³

(a-b)(a²+ab+b²) = a³+a²b+ab²-a²b-ab²-b³ = a³-b³

BINOMIAL COEFFICIENT

C(n,k) = C(n,n-k) = n!/(k!(n-k)!), combinations of n elements in groups of k elements

Binomial formula: (a+b)ⁿ = C(n,0)aⁿ+C(n,1)a^n-1b+C(n,2)a^n-2b²+...+C(n,n-2)a²b^n-2+C(n,n-1)ab^n-1+C(n,n)bⁿ = ⁿΣ_k=0(C(n,k)a^n-kb^k)

(a+b)³ = (a+b)(a+b)(a+b) = (a²+ab+ab+b²)(a+b) = (a²+2ab+b²)(a+b) = a³+a²b+2a²b+2ab²+ab²+b³ = a³+3a²b+3ab²+b³; C(3,2), C(n,k) = C(n,n-k), C(3,2) = C(3,3-2) = C(3,1); [1],[1,1],[1,2,1],[1,3,3,1]; C(n,k) = n!/(k!(n-k)!) = 3!/(2!(3-2)!) = 3⋅2/2 = 3; abc, ab, ac, bc, 3 elements in groups of 2 give 3 combinations

[1], (a+b)⁰ = 1

[1,1], (a+b)¹ = a+b

[1,2,1], (a+b)² = a²+2ab+b²

[1,3,3,1], (a+b)³ = a³+3a²b+3ab²+b³

[1,4,6,4,1], (a+b)⁴ = a⁴+4a³b+6a²b²+4ab³+b⁴

[1,5,10,10,5,1], (a+b)⁵ = a⁵+5a⁴b+10a³b²+10a²b³+5ab⁴+b⁵

[1,6,15,20,15,6,1], (a+b)⁶ = a⁶+6a⁵b+15a⁴b²+20a³b³+15a²b⁴+6ab⁵+b⁶

[1,7,21,35,35,21,7,1], (a+b)⁷ = a⁷+7a⁶b+21a⁵b²+35a⁴b³+35a³b⁴+21a²b⁵+7ab⁶+b⁷

[1,8,28,56,70,56,28,8,1], (a+b)⁸ = a⁸+8a⁷b+28a⁶b²+56a⁵b³+70a⁴b⁴+56a³b⁵+28a²b⁶+8ab⁷+b⁸

[1,9,36,84,126,126,84,36,9,1], (a+b)⁹ = a⁹+9a⁸b+36a⁷b²+84a⁶b³+126a⁵b⁴+126a⁴b⁵+84a³b⁶+36a²b⁷+9ab⁸+b⁹

[1,10,45,120,210,252,210,120,45,10,1], (a+b)¹⁰ = a¹⁰+10a⁹b+45a⁸b²+120a⁷b³+210a⁶b⁴+252a⁵b⁵+210a⁶b⁴+120a⁷b³+45a⁸b²+10a⁹b+b¹⁰

SEQUENCE

A sequence of real numbers is a function a: ℕ → ℝ, and the independent variable is usually denoted by the letter n

a_n = 2n-3; n = 0, a_n = 2⋅0-3 = 0-3 = -3; n = 1, a_n = 2⋅1-3 = 2-3 = -1; n = 2, a_n = 2⋅2-3 = 4-3 = 1; n = 3, a_n = 2⋅3-3 = 6-3 = 3; n = 4, a_n = 2⋅4-3 = 8-3 = 5

A sequence is bounded below if ∃ m ∈ ℝ : a_n ≥ m ∀ n ∈ ℕ

A sequence is bounded above if ∃ M ∈ ℝ : a_n ≤ M ∀ n ∈ ℕ

A sequence is bounded if ∃ m ∈ ℝ ∧ ∃ M ∈ ℝ : m ≤ a_n ≤ M ∀ n ∈ ℕ

A sequence is monotone increasing if a_n ≤ a_n+1 ∀ n ∈ ℕ

A sequence is monotone strictly increasing if a_n < a_n+1 ∀ n ∈ ℕ

A sequence is monotone decreasing if a_n ≥ a_n+1 ∀ n ∈ ℕ

A sequence is monotone strictly decreasing if a_n > a_n+1 ∀ n ∈ ℕ

SERIES

The series ^∞Σ_n=0(a_n) is the sum of the infinite terms of the sequence a_n

s_n is the sequence of the partial sums of the series; s₀ = a₀, s₁ = a₀+a₁, s₂ = a₀+a₁+a₂, ..., s_n = a₀+a₁+a₂+...+a_n = ⁿΣ_k=0(a_k)

^∞Σ_n=0(a_n) = lim_n→+∞(s_n) = lim_n→+∞(ⁿΣ_k=0(a_k))

The series converges to the real number l if lim_n→+∞(s_n) = l

The series diverges to +∞ if lim_n→+∞(s_n) = +∞

The series diverges to -∞ if lim_n→+∞(s_n) = -∞

The series is indeterminate or irregular if lim_n→+∞(s_n) does not exist

GEOMETRIC SERIES

A geometric series is the sum of infinite terms, and each term has a common ratio with the adjacent term; a geometric series is ^∞Σ_k=0(a⋅r^k) = a+a⋅r+a⋅r²+a⋅r³+..., where a is the coefficient of each term and r is the common ratio between adjacent terms

A geometric series is ^∞Σ_n=0(qⁿ) with q ∈ ℝ

s_n = 1+q+q²+...+qⁿ; s_n = n+1 if q = 1; s_n = (1-qⁿ⁺¹)/(1-q) if q ≠ 1

lim_n→+∞(s_n) = 1/(1-q) if |q| < 1; lim_n→+∞(s_n) = +∞ if q ≥ 1; lim_n→+∞(s_n) does not exist if q ≤ -1

The geometric series ^∞Σ_n=0(qⁿ) is convergent with sum 1/(1-q) if |q| < 1, is divergent if q ≥ 1, is indeterminate if q ≤ -1

FUNCTION

A function is a relationship between two sets, called domain and codomain of the function, which associates to each element of the domain one and only one element of the codomain

∀ x ∈ X ∃! y ∈ Y : (x,y) ∈ f ⇔ y = f(x), for each x belonging to X there is only one y belonging to Y such that (x,y) belongs to f, which is equivalent to y = f(x)

y: X → Y, x → f(x), the set X is the domain of the function f, the set Y is the codomain of the function f; y = f(x) is an element of the codomain and is the image of x; x is an element of the domain and is the counterimage of y

f^-1(y) = {x ∈ X : f(x) = y}, f^-1(y) = x, f^-1 is the inverse function of f, x is the counterimage of y; if f^-1(y) ≠ Ø ∀ y ∈ Y, then f is surjective; se f^-1(y) contains at most one element ∀ y, then f is injective; if f is surjective and injective at the same time, then it is bijective

{y ∈ Y : ∃ x ∈ X : y = f(x)} = Im(f) = f(X), the set of elements y of the codomain for which there exists at least one x in the domain that has y as an image is called the image of f and is denoted by Im(f) or by f(X)

A function is injective when it associates distinct elements of the domain to distinct elements of the codomain; a function f: X → Y is injective if 2 distinct elements of the domain have distinct images, that is a₁ ≠ a₂ ⇒ f(a₁) ≠ f(a₂), therefore if 2 elements of the domain have the same image then they coincide, that is f(a₁) = f(a₂) ⇒ a₁ = a₂; a function is injective if ∀ x,y ∈ X, f(x) = f(y) ⇒ x = y, that is if ∀ x,y ∈ X, x ≠ y ⇒ f(x) ≠ f(y); any straight line parallel to the x-axis intersects the graph of an injective function in at most one point; if an injective function is defined and continuous over an interval, then it is strictly monotone, that is strictly increasing or strictly decreasing

A function is surjective when each element of the codomain is the image of at least one element of the domain, that is the image of the function coincides with its codomain; a function f: X → Y is surjective if ∀ y ∈ Y, ∃ x ∈ X : f(x) = y; any straight line parallel to the x-axis intersects the graph of a surjective function in at least one point

If a function is injective and surjective then it is bijective; any straight line parallel to the x-axis intersects the graph of a bijective function at one point only

In a function to each x corresponds only one y; if to each y corresponds no more than one x, the function is injective; if to each y corresponds at least one x, the function is surjective; an injective and surjective function is bijective; a bijective function is invertible

f: A → B is an injective function if a₁ ≠ a₂ ⇒ f(a₁) ≠ f(a₂)

f: A → B is a surjective function if ∀ b ∈ B, ∃ a ∈ A : f(a) = b

f: A → B is a bijective function if ∀ b ∈ B, ∃! a ∈ A : f(a) = b

The identity function id_X: X → X, x → x, is injective and surjective; the identity function is a function that returns the same value used as argument

A constant function f_y: X → Y, x → y, is neither injective, nor surjective

For a, b ∈ ℝ and a ≠ 0, the function f: ℝ → ℝ, x → ax+b, is injective and surjective; any function that describes a straight line is injective and surjective

The exponential function f: ℝ → ℝ, x → a^x, is injective, but is not surjective

The logarithm function f: (0,+∞) → ℝ, x → log(x), is injective and surjective

The natural logarithm function f: (0,+∞) → ℝ, x → ln(x), is injective and surjective

A differentiable real function, which has a strictly positive or strictly negative derivative, is injective

A differentiable real function, which has derivative that changes sign, is not injective

The square function f: ℝ → ℝ, x → x², is neither injective, nor surjective

The square function restricted in the domain f: [0,+∞) → ℝ, x → x², is injective, but is not surjective

The square function restricted in the codomain f: ℝ → [0,+∞), x → x², is not injective, but is surjective

The square function restricted in the domain and in the codomain f: [0,+∞) → [0,+∞), x → x², is injective and surjective, therefore is invertible

The square root function f: [0,+∞) → ℝ, x → √(x), is injective, but not surjective

The cube function f: ℝ → ℝ, x → x³, is injective and surjective

A periodic function is not injective

A function is invertible if it is bijective, that is, at the same time injective and surjective

f^-1(f(x)) = x, f^-1(f(x)) is the identity function

Bijective function f: X → Y, f(x) = y, ∃! g: Y → X, g(y) = x : {g(f(x)) = x, f(g(y)) = y} : f(x) = y ⇔ x = f^-1(y)

f: [0,+∞) → [0,+∞), x → x²; f^-1: [0,+∞) → [0,+∞), y → √(y); f(x) = x² = y, f^-1(y) = √(y) = x; y = x² ⇔ x = √(y)

The direct function f(x) = y, and its inverse function f^-1(y) = x, are symmetrical with respect to the bisector of the first and third quadrant, that is the straight line y = x; the point (x,y) in the curve of the direct function becomes the point (y,x) in the curve of the inverse function

If f is an even function then f(-x) = f(x); the graph of an even function is symmetrical with respect to the y-axis; f(x) = x² is a quadratic function and its graph is a parabola that is symmetrical with respect to the ordinate axis

If f is an odd function then f(-x) = -f(x); the graph of an odd function is symmetrical with respect to the origin of the Cartesian axes; f(x) = x³ is a cubic function and its graph is symmetrical with respect to the point (0,0)

The functions f(x) and -f(x) are symmetrical with respect to the x-axis

The functions f(x) and f(-x) are symmetrical with respect to the y-axis, characteristic of the even functions

The functions f(x) and -f(-x) are symmetrical with respect to the origin of the x and y axes, characteristic of the odd functions

The direct function f(x) and its inverse function f^-1(y) are symmetrical with respect to the bisector of the first and third quadrant

A monotone function is an increasing or decreasing function

A monotone function is increasing if f: X ⊆ ℝ → ℝ, ∀ x₁,x₂ ∈ X : x₁ ≤ x₂ ⇒ f(x₁) ≤ f(x₂)

A monotone function is strictly increasing if f: X ⊆ ℝ → ℝ, ∀ x₁,x₂ ∈ X : x₁ < x₂ ⇒ f(x₁) < f(x₂)

A monotone function is decreasing if f: X ⊆ ℝ → ℝ, ∀ x₁,x₂ ∈ X : x₁ ≤ x₂ ⇒ f(x₁) ≥ f(x₂)

A monotone function is strictly decreasing if f: X ⊆ ℝ → ℝ, ∀ x₁,x₂ ∈ X : x₁ < x₂ ⇒ f(x₁) > f(x₂)

TRIGONOMETRIC FUNCTIONS

The trigonometric functions, or circular functions, are real functions which relate an angle of a right-angled triangle to ratios of two side lengths

All right-angled triangles with the same acute angle θ are similar to each other

The amplitude of an angle is expressed in radians, indicated by rad

The width of the subtended angle is the length of the arc in the circumference of unit radius

The angle subtended to the arc of length 1 in the circumference of unit radius is 1 rad ≈ 57.3°, an angle of 180° = π ≈ 3.14 rad, an angle of 360° = 2π ≈ 6.28 rad

degrees = radians(180/π)

radians = degrees(π/180)

sin(θ) = (opposite cathetus)/(hypotenuse), the sine is the ratio between the cathetus opposite the acute angle θ and the hypotenuse

csc(θ) = (hypotenuse)/(opposite cathetus) = 1/sin(θ), the cosecant is the ratio between the hypotenuse and the cathetus opposite the acute angle θ, therefore it is the reciprocal of the sine

cos(θ) = (adjacent cathetus)/(hypotenuse), the cosine is the ratio between the cathetus adjacent to the acute angle θ and the hypotenuse

sec(θ) = (hypotenuse)/(adjacent cathetus), 1/cos(θ), the secant is the ratio between the hypotenuse and the cathetus adjacent to the acute angle θ, therefore it is the reciprocal of the cosine

tan(θ) = (opposite cathetus)/(adjacent cathetus) = sin(θ)/cos(θ), the tangent is the ratio between the cathetus opposite the acute angle θ and the cathetus adjacent to the acute angle θ, therefore it is the ratio between sine and cosine

cot(θ) = (adjacent cathetus)/(opposite cathetus) = cos(θ)/sin(θ), the cotangent is the ratio between the cathetus adjacent to the acute angle θ and the cathetus opposite the acute angle θ, therefore it is the ratio between cosine and sine

In a right-angled triangle, the sum of the two acute angles is a right angle, that is, 90° or π/2 radians

sin(θ) = cos(π/2-θ)

cos(θ) = sin(π/2-θ)

tan(θ) = sin(θ)/cos(θ) = cos(π/2-θ)/sin(π/2-θ) = cot(π/2-θ)

csc(θ) = 1/sin(θ) = 1/cos(π/2-θ) = sec(π/2-θ)

sec(θ) = 1/cos(θ) = 1/sin(π/2-θ) = csc(π/2-θ)

cot(θ) = cos(θ)/sin(θ) = sin(π/2-θ)/cos(π/2-θ) = tan(π/2-θ)

sin(0) = 0

cos(0) = 1

tan(0) = 0

sin(π/12) = (√(6)-√(2))/4

cos(π/12) = (√(6)+√(2))/4

tan(π/12) = 2-√(3)

sin(π/6) = 1/2

cos(π/6) = √(3)/2

tan(π/6) = √(3)/3

sin(π/4) = √(2)/2

cos(π/4) = √(2)/2

tan(π/4) = 1

sin(π/3) = √(3)/2

cos(π/3) = 1/2

tan(π/3) = √(3)

sin(5π/12) = (√(6)+√(2))/4

cos(5π/12) = (√(6)-√(2))/4

tan(5π/12) = 2+√(3)

sin(π/2) = 1

cos(π/2) = 0

tan(π/2) = undefined

The cosine and the secant are even functions, and the other trigonometric functions are odd functions

sin(-x) = -sin(x), sine is an odd function

cos(-x) = cos(x), cosine is an even function

tan(-x) = -tan(x), tangent is an odd function

csc(-x) = -csc(x), cosecant is an odd function

sec(-x) = sec(x), secant is an even function

cot(-x) = -cot(x), cotangent is an odd function

All trigonometric functions are periodic functions of period 2π, except for the tangent and the cotangent, which have period π

sin(x+2kπ) = sin(x), ∀ k ∈ ℤ

cos(x+2kπ) = cos(x), ∀ k ∈ ℤ

tan(x+π) = tan(x), ∀ k ∈ ℤ

csc(x+2kπ) = csc(x), ∀ k ∈ ℤ

sec(x+2kπ) = sec(x), ∀ k ∈ ℤ

cot(x+kπ) = cot(x), ∀ k ∈ ℤ

sin²(x)+cos²(x) = 1, the Pythagorean identity is the expression of the Pythagorean theorem in terms of trigonometric functions

sin²(x) = 1-cos²(x)

cos²(x) = 1-sin²(x)

sin²(x) = 1-cos², sin²(x)/cos²(x) = (1-cos²)/cos²(x), tan²(x) = (1-cos²)/cos²(x) = 1/cos²(x)-cos²(x)/cos²(x) = (1-cos²)/cos²(x) = 1/cos²(x)-1 = sec²(x)-1, tan²(x) = sec²(x)-1, sec²(x) = tan²(x)+1

sin²(x)+cos²(x) = 1, (sin²(x)+cos²(x))/cos²(x) = 1/cos²(x), sin²(x)/cos²(x)+cos²(x)/cos²(x) = 1/cos²(x), tan²(x)+1 = 1/cos², tan²(x)+1 = sec²(x), sec²(x) = tan²(x)+1, tan²(x) = sec²(x)-1

tan²(x) = sec²(x)-1, tan²(x) = 1/cos²(x)-1

sec²(x) = tan²(x)+1, 1/cos²(x) = tan²(x)+1, cos²(x) = 1/(tan²(x)+1)

cos²(x) = 1-sin²(x), cos²(x)/sin²(x) = (1-sin²(x))/sin²(x), cot²(x) = (1-sin²(x))/sin²(x) = 1/sin²(x)-sin²(x)/sin²(x) = 1/sin²(x)-1 = csc²(x)-1, cot²(x) = csc²(x)-1, csc²(x) = cot²(x)+1

sin²(x)+cos²(x) = 1, (sin²(x)+cos²(x))/sin²(x) = 1/sin²(x), sin²(x)/sin²(x)+cos²(x)/sin²(x) = 1/sin²(x), 1+cot²(x) = 1/sin²(x), 1+cot²(x) = csc²(x), csc²(x) = cot²(x)+1, cot²(x) = csc²(x)-1

cot²(x) = csc²(x)-1, 1/tan²(x) = 1/sin²(x)-1

csc²(x) = cot²(x)+1, 1/sin²(x) = 1/tan²(x)+1

sec²(x)+csc²(x) = sec²(x)csc²(x)

sin(x+y) = sin(x)cos(y)+cos(x)sin(y)

sin(x-y) = sin(x)cos(y)-cos(x)sin(y)

sin(2x) = 2sin(x)cos(x) = 2tan(x)/(tan²(x)+1)

sin(2x) = 2sin(x)cos(x), sin(2x) = 2sin(x)cos(x)(cos(x)/cos(x)) = 2(sin(x)/cos(x))cos²(x) = 2tan(x)cos²(x) = 2tan(x)/(tan²(x)+1)

cos(x+y) = cos(x)cos(y)-sin(x)sin(y)

cos(x-y) = cos(x)cos(y)+sin(x)sin(y)

cos(2x) = cos²(x)-sin²(x) = 1-2sin²(x) = 2cos²-1 = (1-tan²(x))/(1+tan²(x))

cos(2x) = 1-2sin²(x), cos(2x)/cos²(x) = (1-2sin²(x))/cos²(x) = 1/cos²(x)-2sin²/cos²(x) = sec²(x)-2tan²(x) = tan²(x)+1-2tan²(x) = 1-tan²(x), cos(2x)/cos²(x) = 1-tan²(x), cos(2x)sec²(x) = 1-tan²(x), cos(2x) = (1-tan²(x))/sec²(x) = (1-tan²(x))/(1+tan²(x)), cos(2x) = (1-tan²(x))/(1+tan²(x))

tan(x+y) = (tan(x)+tan(y))/(1-tan(x)tan(y))

tan(x-y) = (tan(x)-tan(y))/(1+tan(x)tan(y))

tan(2x) = 2tan(x)/(1-tan²(x))

sin(α)+sin(β) = 2sin((α+β)/2)cos((α-β)/2); sin(α)+sin(β) = sin((α+β)/2+(α-β)/2)+sin((α+β)/2+(β-α)/2) = sin((α+β)/2)cos((α-β)/2)+cos((α+β)/2)sin((α-β)/2)+sin((α+β)/2)cos((β-α)/2)+cos((α+β)/2)sin((β-α)/2) = sin((α+β)/2)cos((α-β)/2)+cos((α+β)/2)sin((α-β)/2)+sin((α+β)/2)cos((α-β)/2)-cos((α+β)/2)sin((α-β)/2) = sin((α+β)/2)cos((α-β)/2)+sin((α+β)/2)cos((α-β)/2) = 2sin((α+β)/2)cos((α-β)/2)

sin(α)+sin(β) = sin(α)cos(β)+cos(α)sin(β) = 2sin((α+β)/2)cos((α-β)/2)

sin(α)-sin(β) = 2sin((α-β)/2)cos((α+β)/2); sin(α)-sin(β) = sin((α+β)/2+(α-β)/2)-sin((α+β)/2+(β-α)/2) = sin((α+β)/2)cos((α-β)/2)+cos((α+β)/2)sin((α-β)/2)-sin((α+β)/2)cos((β-α)/2)-cos((α+β)/2)sin((β-α)/2) = sin((α+β)/2)cos((α-β)/2)+cos((α+β)/2)sin((α-β)/2)-sin((α+β)/2)cos((α-β)/2)+cos((α+β)/2)sin((α-β)/2) = 2cos((α+β)/2)sin((α-β)/2) = 2sin((α-β)/2)cos((α+β)/2)

sin(α)-sin(β) = sin(α)cos(β)-cos(α)sin(β) = 2sin((α-β)/2)cos((α+β)/2)

cos(α)+cos(β) = 2cos((α+β)/2)cos((α-β)/2); cos(α)+cos(β) = cos((α+β)/2+(α-β)/2)+cos((α+β)/2+(β-α)/2) = cos((α+β)/2)cos((α-β)/2)-sin((α+β)/2)sin((α-β)/2)+cos((α+β)/2)cos((β-α)/2)-sin((α+β)/2)sin((β-α)/2) = cos((α+β)/2)cos((α-β)/2)-sin((α+β)/2)sin((α-β)/2)+cos((α+β)/2)cos((α-β)/2)+sin((α+β)/2)sin((α-β)/2) = 2cos((α+β)/2)cos((α-β)/2)

cos(α)+cos(β) = cos(α)cos(β)-sin(α)sin(β) = 2cos((α+β)/2)cos((α-β)/2)

cos(α)-cos(β) = -2sin((α+β)/2)sin((α-β)/2); cos(α)-cos(β) = cos((α+β)/2+(α-β)/2)-cos((α+β)/2+(β-α)/2) = cos((α+β)/2)cos((α-β)/2)-sin((α+β)/2)sin((α-β)/2)-cos((α+β)/2)cos((β-α)/2)+sin((α+β)/2)sin((β-α)/2) = cos((α+β)/2)cos((α-β)/2)-sin((α+β)/2)sin((α-β)/2)-cos((α+β)/2)cos((α-β)/2)-sin((α+β)/2)sin((α-β)/2) = -2sin((α+β)/2)sin((α-β)/2)

cos(α)-cos(β) = cos(α)cos(β)+sin(α)sin(β) = -2sin((α+β)/2)sin((α-β)/2)

tan(α)+tan(β) = sin(α+β)/(cos(α)cos(β)), α and β ≠ (2k+1)π/2 and k ∈ ℤ; tan(α)+tan(β) = sin(α)/cos(α)+sin(β)/cos(β) = (sin(α)cos(β)+cos(α)sin(β))/(cos(α)cos(β)) = sin(α+β)/(cos(α)cos(β))

tan(α)-tan(β) = sin(α-β)/(cos(α)cos(β)), α and β ≠ (2k+1)π/2 and k ∈ ℤ; tan(α)-tan(β) = sin(α)/cos(α)-sin(β)/cos(β) = (sin(α)cos(β)-cos(α)sin(β))/(cos(α)cos(β)) = sin(α-β)/(cos(α)cos(β))

cot(α)+cot(β) = sin(α+β)/(sin(α)sin(β)), α and β ≠ kπ and k ∈ ℤ; cot(α)+cot(β) = cos(α)/sin(α)+cos(β)/sin(β) = (cos(α)sin(β)+sin(α)cos(β))/(sin(α)sin(β)) = sin(α+β)/(sin(α)sin(β))

cot(α)-cot(β) = sin(β-α)/(sin(α)sin(β)), α and β ≠ kπ and k ∈ ℤ; cot(α)-cot(β) = cos(α)/sin(α)-cos(β)/sin(β) = (cos(α)sin(β)-sin(α)cos(β))/(sin(α)sin(β)) = -(sin(α)cos(β)-cos(α)sin(β))/(sin(α)sin(β)) = -sin(α-β)/(sin(α)sin(β)) = sin(β-α)/(sin(α)sin(β))

sin(A)/a = sin(B)/b = sin(C)/c = 2Δ/abc, law of sines, where A, B, C, are the opposite angles to the sides a, b, c, and Δ is the area of the triangle

a/sin(A) = b/sin(B) = c/sin(C) = 2R, another version of the law of sines, where A, B, C, are the opposite angles to the sides a, b, c, and R is the radius of the circumcircle

c² = a²+b²-2ab⋅cos(C), cos(C) = (a²+b²-c²)/2ab, law of cosines, which is an extension of the Pythagorean theorem, where a, b, c, are the sides of a triangle and C is the angle opposite to the side c; the law of cosines can be used to determine a side of a triangle if two sides and the angle between them are known, and it can also be used to find the cosines of an angle, and consequently the angles themselves, if the lengths of all the sides are known

tan((A-B)/2)/tan((A+B)/2) = (a-b)/(a+b), law of tangents, where A and B are the opposite angles to sides a and b

The trigonometric functions are periodic, so not injective, therefore they do not have an inverse function; however, on each interval on which a trigonometric function is monotonic, it is possible to define an inverse function; to define a true inverse function, the domain must be restricted to an interval where the function is monotonic, and is thus bijective from this interval to its image by the function; the inverse trigonometric functions are denoted with the prefix arc

y = sin(x), -π/2 ≤ x ≤ π/2, x = arcsin(y), -1 ≤ y ≤ 1

y = arcsin(x), -1 ≤ x ≤ 1, x = sin(y), -π/2 ≤ y ≤ π/2

y = cos(x), 0 ≤ x ≤ π, x = arccos(y), -1 ≤ y ≤ 1

y = arccos(x), -1 ≤ x ≤ 1, x = cos(y), 0 ≤ y ≤ π

y = tan(x), -π/2 < x < π/2, x = arctan(y), -∞ < y < ∞

y = arctan(x), -∞ < x < ∞, x = tan(y), -π/2 < y < π/2

y = csc(x), -π/2 ≤ x ≤ π/2, x ≠ 0, x = arccsc(y), y ≤ -1 ∪ y ≥ 1

y = arccsc(x), x ≤ -1 ∪ x ≥ 1, x = csc(y), -π/2 ≤ y ≤ π/2, y ≠ 0

y = sec(x), 0 ≤ x ≤ π, x ≠ π/2, x = arcsec(y), y ≤ -1 ∪ y ≥ 1

y = arcsec(x), x ≤ -1 ∪ x ≥ 1, x = sec(y), 0 ≤ y ≤ π, y ≠ π/2

y = cot(x), 0 < x < π, x = arccot(y), -∞ < y < ∞

y = arccot(x), -∞ < x < ∞, x = cot(y), 0 < y < π

The derivatives of trigonometric functions result from those of sine and cosine by applying quotient rule

(sin(x))' = cos(x)

(cos(x))' = -sin(x)

(tan(x))' = 1/cos²(x) = sec²(x) = 1+tan²(x)

(csc(x))' = -cos(x)/sin²(x) = -cos(x)csc²(x) = -csc(x)cot(x)

(sec(x))' = sin(x)/cos²(x) = sin(x)sec²(x) = sec(x)tan(x)

(cot(x))' = -csc²(x) = -1-cot²(x)

LOGARITHMIC FUNCTIONS

log_a(b) = c, a^c = b

log_a(b) = c, a^c = b; log_a(c) = b, a^b = c; log_b(a) = c, b^c = a; log_b(c) = a, b^a = c; log_c(a) = b, c^b = a; log_c(b) = a, c^a = b

log_a(a) = 1

log_a(1) = 0

a^log_a(x) = log_a(a^x) = x

log_a(x·y) = log_a(x)+log_a(y)

log_a(x/y) = log_a(x)-log_a(y)

log_a(x^k) = k·log_a(x)

log_a(^k√(x)) = log_a(x^1/k) = (1/k)log_a(x)

log_a(1/x) = log_a(x^-1) = -log_a(x)

log_b(x) = log_k(x)/log_k(b)

log_b(x) = 1/log_x(b)

COMPLETING THE SQUARE

ax²+bx+c = a(x-h)²+k, h = -b/(2a), k = c-ah² = c-a(b²/(4a²)) = c-b²/(4a) = (4ac-b²)/(4a) = -Δ/(4a)

DISCRIMINANT

The discriminant of a polynomial is a quantity that depends on the coefficients and determines various properties of the roots; the discriminant is often indicated with the symbol Δ

The discriminant of a quadratic polynomial ax²+bx+c with a ≠ 0 is Δ = b²-4ac, quantity that appears under the square root in the quadratic formula

In a quadratic polynomial with real coefficients, if Δ > 0 there are 2 distinct real roots, if Δ = 0 there is a double real root that are 2 coincident real roots, if Δ < 0 there are 2 distinct complex conjugate roots

In a cubic polynomial with real coefficients, if Δ > 0 there are 3 distinct real roots, if Δ = 0 there is a triple real root that are 3 coincident real roots, if Δ < 0 there is a real root and 2 distinct complex conjugate roots

QUADRATIC FORMULA

ax²+bx+c = 0, a ≠ 0, second degree equation, also called quadratic equation, in standard form

Δ = b²-4ac

x = (-b±√(b²-4ac))/(2a) = (-b±√(Δ))/(2a), quadratic formula

ax²+bx+c = 0, ax²+bx = -c, x²+(b/a)x = -(c/a), using the algebraic technique of completing the square, x²+(b/a)x+(b/(2a))² = -(c/a)+(b/(2a))², (x+(b/(2a)))² = (b/(2a))²-(c/a), (x+(b/(2a)))² = (b²/(4a²))-(c/a), (x+(b/(2a)))² = (b²-4ac)/(4a²) = x+(b/(2a)) = ±√((b²-4ac)/(4a²)), x+(b/(2a)) = (±√(b²-4ac))/2a, x = ((±√(b²-4ac))/2a)-(b/(2a)) = (-b±√(b²-4ac))/(2a)

PARABOLA

The parabola is the geometric locus of points equidistant from a straight line called the directrix and from a point called focus; the focus does not lie on the directrix; the line perpendicular to the directrix passing through the focus is the axis of symmetry; the point where the parabola intersects its axis of symmetry is called vertex and is the point where the parabola is most curved; the distance between the vertex and the focus, measured along the axis of symmetry, is the focal length

The parabola is a set of points P where the distance between a point P of the parabola and its focus F is equal to the distance between a point P of the parabola and its directrix d, {P: |PF| = |Pd|}; the straight line perpendicular to the directrix passing through the focus is the symmetry axis of the parabola; the point of intersection between the axis of symmetry and the parabola is the vertex

ax²+bxy+cy²+dx+ey+f = 0, b²-4ac = 0, implicit equation of a parabola on the Cartesian plane

ax+by+c = 0 is the directrix, F(x₀,y₀) is the focus, the parabola is Ax²+Bxy+Cy²+Dx+Ey+F = 0; A = b²/(a²+b²); B = -((2ab)/(a²+b²)); C = a²/(a²+b²); D = -2(x₀+((ac)/(a²+b²))); E = -2(y₀+((bc)/(a²+b²))); F = x₀²+y₀²-(c²/(a²+b²))

y = ax²+bx+c, equation of a parabola with axis of symmetry parallel to the y-axis

f(x) = ax²+bx+c, a, b, c, ∈ ℝ, a ≠ 0; f(x) = ax²+bx+c = a(x²+(b/a)x)+c = a(x²+(b/a)x)+c+((ab²)/(4a²))-((ab²)/(4a²)) = a(x²+(b/a)x+((b²)/(4a²)))+c-((ab²)/(4a²)) = a(x²+(b/a)x+(b/(2a))²)+c-(b²/(4a)) = a(x+(b/(2a))²+((4ac-b²)/(4a)); the axis of symmetry is x = -b/(2a); the focal length is 1/(4a); the vertex is V = (-b/(2a),(4ac-b²)/(4a)) = (-b/(2a),-Δ/(4a)); the focus is F = (-b/(2a),(4ac-b²+1)/(4a)) = (-b/(2a),(-Δ+1)/(4a); the directrix is y = (4ac-b²-1)/(4a) = (-Δ-1)/(4a); the intersection point of the parabola with the y-axis is (0,c); the tangent line at the point of intersection of the parabola with the y-axis is y = bx+c

y = ax², equation of a parabola which has the y-axis as its axis of symmetry and the point (0,0) as its vertex; the vertex is (0,0); the focus is (0,1/(4a)); the directrix is y = -1/(4a)

CIRCLE

d = 2r, d is the diameter which is twice the length of the radius r

C = 2πr, C is the length of the circumference, r is the length of the radius, π is the constant which expresses the ratio between any circumference and its diameter

π = C/2r = C/d, π is a transcendental number, therefore irrational, and is equal to 3,14159...

A = πr², A is the area of the circle

x²+y²-2axz-2byz+cz² = 0, implicit equation of a circle on the Cartesian plane

(x-x_c)²+(y-y_c)² = r², equation of the circle with center of coordinates (x_c,y_c) and radius r

x²+y² = r², equation with center in the origin of the Cartesian axes

{x = x_c+r·cos(t), y = y_c+r·sin(t)}, 0 ≤ t ≤ 2π, parametric form of the equation of the circle of center (x_c,y_c), and t is a parametric variable between 0 e 2π

ELLIPSE

An ellipse is the geometrical locus of the points of the plane for which the sum of the distances from two fixed points called foci remains constant

The equation of the ellipse is found by equating the sum of the distances between a point P(x,y) and the two foci F₁(x₁,y₁) and F₂(x₂,y₂) with twice the semi-major axis, PF₁+PF₂ = 2a, {P ∈ ℝ² : |PF₁|+|PF₂| = 2a}, √((x-x₁)²+(y-y₁)²)+√((x-x₂)²+(y-y₂)²) = 2a

ax²+bxy+cy²+dx+ey+f = 0, b²-4ac < 0, implicit equation of an ellipse in the Cartesian plane

(x²/a²)+(y²/b²) = 1, equation of a standard ellipse centered at the origin with width 2a and height 2b

Assuming a ≥ b, the foci are (±c,0) for √(a²-b²)

{x = a⋅cos(t), y = b⋅sin(t)}, 0 ≤ t ≤ 2π, standard parametric equation of an ellipsis, and t is a parametric variable between 0 e 2π

HYPERBOLE

y = 1/x, equation of the simplest hyperbola, the asymptotes are the Cartesian axes

ax²+bxy+cy²+dx+ey+f = 0, b²-4ac > 0, implicit equation of a hyperbola on the Cartesian plane

(x²/a²)-(y²/b²) = 1, equation of the hyperbola if its axes coincide with the Cartesian axes, if its center is in the origin of the axes, if the two branches of the hyperbola intersect the x-axis

(x²/a²)-(y²/b²) = -1, equation of the hyperbola if its axes coincide with the Cartesian axes, if its center is in the origin of the axes, if the two branches of the hyperbola intersect the y-axis

y = ±(b/a)x, equations of the 2 asymptotes of the hyperbola if its axes coincide with the Cartesian axes, and if its center is in the origin of the axes

If a = b, the asymptotes of the hyperbola are perpendicular, and the hyperbola is called equilateral

((x-x_c)²/a²)-((y-y_c)²/b²) = 1, equation of hyperbola which has axes of symmetry parallel to the Cartesian axes and center (x_c,y_c)

((y-y_c)²/b²)-((x-x_c)²/a²) = 1, equation of hyperbola which has axes of symmetry parallel to the Cartesian axes and center (x_c,y_c), and is rotated 90° with respect to the previous hyperbola

{x = a⋅cosh(t), y = b⋅sinh(t)}, parametric equation of the right branch of a hyperbola

{x = -a⋅cosh(t), y = b⋅sinh(t)}, parametric equation of the left branch of a hyperbola

cosh(x) = (e^x+e^-x)/2, hyperbolic cosine

sinh(x) = (e^x-e^-x)/2, hyperbolic sine

LIMIT

A limit is the value that a function, or sequence, approaches as the input, or index, approaches some value

lim_x→c(f(x)) = L, is read as the limit of f of x as x approaches c equals L

At the beginning of the 19th century, Augustin-Louis Cauchy and Karl Weierstrass formalized the definition of the limit of a function with the concept of numbers ε and δ, that is, the limit of a function is expressed as lim_x→c(f(x)) = L when |f(x)-L| < ε, where ±ε is a vertical neighborhood of L, and ±δ is a horizontal neighborhood of c that is the point towards which the limit tends

f: A → ℝ, lim_{x→x0(f(x)) = L, the function does not need to be defined at point x0, but it is important that it is defined close to x0}

A neighborhood of a real number x₀ is an open interval centered in the number x₀ with radius r, U(x₀,r) := (x₀-r,x₀+r)

A real number is an accumulation point of a set A if every neighborhood of this number contains infinite elements of A; if x₀ is an accumulation point of A, then ∀ r > 0, U(x₀,r) ⋂ A, the intersection between the neighborhood U(x₀,r) and the set A contains infinite elements

A finite set has no accumulation points

The set ℕ of natural numbers is infinite but has no accumulation points; +∞ can be thought of as the only accumulation point of ℕ

A non-isolated point of the set is an accumulation point that belongs to the set

lim_x→x0(f(x)) = L ⇒ |f(x)-L| < ε ⇒ 0 < |x-x₀| < δ_ε

∀ ε > 0, ∃ δ_ε : x ∈ (A\{x₀}) ⋂ U(x₀,δ_ε) ⇒ |f(x)-L| < ε; in this definition of limit, U(x₀,δ_ε) is the neighborhood of x₀ with radius δ_ε, U(x₀,δ_ε) = (x₀-δ_ε,x₀+δ_ε); |f(x)-L| < ε ⇒ f(x) ∈ V(L,ε) ⇒ L-ε < f(x) < L+ε, V(L,ε) is the neighborhood of the limit L with radius ε, V(L,ε) = (L-ε,L+ε); the function converges to the limit L when x approaches x₀ which is the accumulation point of the set A where the function is defined

|f(x)-L| < ε, f(x) ∈ V(L,ε), L-ε < f(x) < L+ε

lim_x→x0(f(x)) = L, |f(x)-f(x₀)| < ε, |x-x₀| < δ_ε

f: A ⊆ ℝ → ℝ, x₀ ∈ ℝ, lim_x→x0(f(x)) = L ∈ ℝ, if ∀ V(L,ε) ∃ U(x₀,δ_ε) : x ∈ U(x₀,δ_ε) ⋂ A\{x₀} ⇒ f(x) ∈ V(L,ε); in this definition of limit, V(L,ε) is the neighborhood of the limit L with radius ε, V(L,ε) = (L-ε,L+ε), and U(x₀,δ_ε) is the neighborhood of x₀ with radius δ_ε, U(x₀,δ_ε) = (x₀-δ_ε,x₀+δ_ε)

If the accumulation point x₀ = +∞ and the limit L = +∞, then f: A ⊆ ℝ → ℝ, lim_x→+∞(f(x)) = +∞, if ∀ M > 0 ∃ K > 0 : x ∈ (K,+∞) ∩ A ⇒ f(x) > M

The function is continuous if, in a non-isolated point, that is an accumulation point belonging to the set, the limit of the function coincides with the value of the function; the function is continuous if lim_x→x0(f(x)) = f(x₀)

If f(x) is a continuous function, then lim_x→x0(f(x)) = f(x₀)

All polynomial functions are continuous functions

f(x) = c, |f(x)-f(x₀)| = |x-x₀| = 0, f(x) is a continuous function

f(x) = x, |f(x)-f(x₀)| = |x-x₀|, δ_ε = ε, f(x) is a continuous function

f(x) = 2x, |f(x)-f(x₀)| = |2x-2x₀| = 2|x-x₀| < ε, |x-x₀| < ε/2 = δ_ε, f(x) is a continuous function

f(x) = mx, |f(x)-f(x₀)| = |mx-mx₀| = m|x-x₀| < ε, |x-x₀| < ε/m = δ_ε, f(x) is a continuous function

f(x) = x+1, lim_x→1(f(x)) = 2, the limit of the function coincides with the value of the function at that point x = 1, so the function is continuous at the point x = 1

f(x) = {x+1 if x ≠ 1, 3 if x = 1}, lim_x→1(f(x)) = 2, the limit of the function does not coincide with the value of the function at the point x = 1, so the function is not continuous at the point x = 1

f: A ⊆ ℝ → ℝ, x₀ ∈ A is an accumulation point of A, f is a continuous function in x₀ if lim_x→x0(f(x)) = f(x₀); if a function is continuous at any point of an interval then it is continuous on the interval; if a function is continuous throughout its domain, then it is simply called a continuous function

Power functions are continuous; y = x² is a continuous function, lim_x→1(x²) = 1² = 1; y = x³ is a continuous function, lim_x→1(x³) = 1³ = 1; y = x^1/2 is a continuous function, lim_x→1(x^1/2) = 1^1/2 = √(1) = 1

Exponential functions are continuous; y = 2^x is a continuous function, lim_x→1(2^x) = 2¹ = 2; y = 3^x is a continuous function, lim_x→1(3^x) = 3¹ = 3; y = e^x is a continuous function, lim_x→1(e^x) = e¹ = e

Logarithmic functions are continuous; y = log(x) is a continuous function, lim_x→10(log(x)) = log(10) = 1; y = ln(x) is a continuous function, lim_x→e(ln(x)) = ln(e) = 1; y = log₂(x) is a continuous function, lim_x→2(log₂(x)) = log₂(2) = 1

Trigonometric functions are continuous; y = sin(x) is a continuous function, lim_x→π/2(sin(x)) = sin(π/2) = 1; y = cos(x) is a continuous function, lim_x→0(cos(x)) = cos(0) = 1; y = tan(x) is a continuous function, lim_x→0(tan(x)) = tan(0) = 0

Power functions, exponential functions, logarithmic functions, and goniometric functions, are called elementary functions, and are continuous in all the points of their set of definition; all the functions that can be obtained as sum, product, quotient, and composition, of elementary functions are continuous in their set of definition; y = e^x2 is a continuous function; y = 3x²-x-1 is a continuous function; y = sin(x)cos(x) is a continuous function

f(x) = x; lim_x→+∞(f(x)) = +∞; lim_x→-∞(f(x)) = -∞

f(x) = xⁿ, n is an even natural number ≥ 2; lim_x→+∞(f(x)) = +∞; lim_x→-∞(f(x)) = +∞

f(x) = xⁿ, n is an odd natural number ≥ 3; lim_x→+∞(f(x)) = +∞; lim_x→-∞(f(x)) = -∞

f(x) = 1/x; lim_x→+∞(f(x)) = 0⁺; lim_x→-∞(f(x)) = 0^-; lim_x→0⁺(f(x)) = +∞; lim_x→0^-(f(x)) = -∞; lim_x→0(f(x)) = ∄

When at a point the right limit is different from the left limit, then the limit of the function at that point does not exist

f(x) = a^x, a > 1; lim_x→+∞(f(x)) = +∞; lim_x→-∞(f(x)) = 0⁺

f(x) = a^x, 0 < a < 1; lim_x→+∞(f(x)) = 0⁺; lim_x→-∞(f(x)) = +∞

f(x) = log_a(x), a > 1; lim_x→+∞(f(x)) = +∞; lim_x→0⁺(f(x)) = -∞

f(x) = log_a(x), 0 < a < 1; lim_x→+∞(f(x)) = -∞; lim_x→0⁺(f(x)) = +∞

f(x) = sin(x), periodic function of period 2π that oscillates infinite times between -1 and 1; lim_x→+∞(f(x)) = ∄; lim_x→-∞(f(x)) = ∄

f(x) = cos(x), periodic function of period 2π that oscillates infinite times between -1 and 1; lim_x→+∞(f(x)) = ∄; lim_x→-∞(f(x)) = ∄

f(x) = tan(x), periodic function of period π that oscillates infinite times between -∞ e +∞; lim_x→+∞(f(x)) = ∄; lim_x→-∞(f(x)) = ∄; lim_x→(π/2)^-(f(x)) = +∞; lim_{x→(-π/2)⁺}(f(x)) = -∞

lim_x→x0(f(x)+g(x)) = lim_x→x0(f(x))+lim_x→x0(g(x)), the limit of the sum of functions is equal to the sum of the limits of the functions

lim_x→x0(f(x)-g(x)) = lim_x→x0(f(x))-lim_x→x0(g(x)), the limit of the difference of functions is equal to the difference of the limits of the functions

lim_x→x0(c·f(x)) = c·lim_x→x0(f(x)), the limit of the product of a constant for a function is equal to the product of the constant for the limit of the function

lim_x→x0(f(x)·g(x)) = lim_x→x0(f(x))·lim_x→x0(g(x)), the limit of the product of functions is equal to the product of the limits of functions

lim_x→x0(f(x)/g(x)) = lim_x→x0(f(x))/lim_x→x0(g(x)), g(x₀) ≠ 0, the limit of the ratio of functions is equal to the ratio of the limits of functions

lim_x→x0(g(f(x))) = g(lim_x→x0(f(x))), the limit of a compound function is equal to the outer function with argument the limit of the inner function

A rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials

lim_x→2((x+1)/(x-1)); lim_x→2((x+1)/(x-1)) = (2+1)/(2-1) = 3/1 = 3, in this limit of rational function neither the numerator nor the denominator is equal to 0, and the function is continuous in x₀

lim_x→1((x²-x)/(x+2)); lim_x→1((x²-x)/(x+2)) = (1-1)/(1+2) = 0/3 = 0, in this limit of rational function the numerator is equal to 0, and the function is continuous in x₀

lim_x→2((x+1)/(2-x)); lim_x→2((x+1)/(2-x)) = (2+1)/(2-2) = 3/0, in this limit of rational function the denominator is equal to 0, but not the numerator, then we have to calculate the right limit and the left limit; lim_x→2⁺((x+1)/(2-x)) = (2⁺+1)/(2-2⁺) = 3/0^- = -∞, the right limit at the point x = 2 is -∞; lim_x→2^-((x+1)/(2-x)) = (2^-+1)/(2-2^-) = 3/0⁺ = +∞, the left limit at the point x = 2 is +∞; the right limit and the left limit at the point x = 2 do not coincide therefore the limit at the point x = 2 does not exist, lim_x→2((x+1)/(2-x)) = ∄; x = 2 is a vertical asymptote

lim_x→2((x+1)/(2-x)²); lim_x→2((x+1)/(2-x)²) = (2+1)/(2-2)² = 3/0² = 3/0, in this limit of rational function the denominator is equal to 0, but not the numerator, then we have to calculate the right limit and the left limit; lim_x→2⁺((x+1)/(2-x)²) = (2⁺+1)/(2-2⁺)² = 3/(0^-)² = 3/0⁺ = +∞, the right limit at the point x = 2 is +∞; lim_x→2^-((x+1)/(2-x)²) = (2^-+1)/(2-2^-)² = 3/(0⁺)² = 3/0⁺ = +∞, the left limit at the point x = 2 is +∞; the right limit and the left limit at x = 2 is +∞, so the limit at the point x = 2 is +∞; x = 2 is a vertical asymptote

lim_x→1((x²-3x+2)/(x²-1)); lim_x→1((x²-3x+2)/(x²-1)) = (1²-3⋅1+2)/(1²-1) = (1-3+2)/(1-1) = 0/0, in this limit of rational function the numerator and denominator are equal to 0, so we have to decompose and simplify; lim_x→1((x²-3x+2)/(x²-1)) = lim_x→1(((x-1)(x-2))/((x+1)(x-1))) = lim_x→1((x-2)/(x+1)) = -1/2; this function is not defined in x = 1, because for x = 1 the denominator is equal to 0, so x = 1 is a point excluded from the domain of the function; the calculation of this limit allows us to understand that, even if in x = 1 the function is not defined, in the vicinity of x = 1 the function has a value of -1/2

lim_x→+∞(x³-5x²+1); lim_x→+∞(x³-5x²+1) = +∞-∞, to solve the limit of a polynomial we have to collect the term of higher degree and calculate the limit of each term; lim_x→+∞(x³-5x²+1) = lim_x→+∞(x³(x³/x³-5x²/x³+1/x³)) = lim_x→+∞(x³(1-5/x+1/x³)) = +∞(1-5/+∞+1/+∞) = +∞(1-0+0) = +∞⋅1 = +∞

lim_x→+∞((x³-4x-3)/(2x²+5)); lim_x→+∞((x³-4x-3)/(2x²+5)) = +∞/+∞, indeterminate form; to solve the limit of a rational function, that is a function composed by the ratio of two polynomials, we must collect the term of major degree of the polynomial in the numerator, collect the term of major degree of the polynomial in the denominator, simplify, and calculate the limit of each term; lim_x→+∞((x³-4x-3)/(2x²+5)) = lim_x→+∞((x³(x³/x³-4x/x³-3/x³))/(2x²(2x²/2x²+5/2x²))) = lim_x→+∞((x³(1-4/x²-3/x³))/(2x²(1+5/2x²))) = lim_x→+∞((x(1-4/x²-3/x³))/(2(1+5/2x²))) = (+∞(1-4/+∞-3/+∞))/(2(1+5/+∞)) = (+∞(1-0-0))/(2(1+0)) = (+∞⋅1)/(2⋅1) = +∞/2 = +∞

lim_x→-∞((x³-4x)/(5x⁶-1)); lim_x→-∞((x³-4x)/(5x⁶-1)) = -∞/+∞, indeterminate form; lim_x→-∞((x³-4x)/(5x⁶-1)) = lim_x→-∞(x³(x³/x³-4x/x³)/(5x⁶(5x⁶/5x⁶-1/5x⁶))) = lim_x→-∞((x³(1-4/x²))/(5x⁶(1-1/5x⁶))) = lim_x→-∞((1-4/x²)/(5x³(1-1/5x⁶))) = (1-4/(-∞)²)/(5(-∞)³(1-1/5(-∞)⁶) = (1-4/+∞)/(5(-∞)(1-1/5(+∞))) = (1-0)/(-∞(1-0)) = 1/(-∞⋅1) = 1/-∞ = 0^-

lim_x→-∞((2x²-x)/(x²+1)); lim_x→-∞((2x²-x)/(x²+1)) = +∞/+∞, indeterminate form; lim_x→-∞((2x²-x)/(x²+1)) = lim_x→-∞((2x²(2x²/2x²-x/2x²))/(x²(x²/x²+1/x²))) = lim_x→-∞((2x²(1-1/2x))/(x²(1+1/x²))) = lim_x→-∞((2(1-1/2x))/(1+1/x²)) = (2(1-1/2(-∞)))/(1+1/(-∞)²) = (2(1-1/-∞))/(1+1/+∞) = (2(1-0))/(1+0) = 2⋅1/1 = 2

The limit of a rational function is +∞ or -∞ when the polynomial in the numerator has a higher degree than the polynomial in the denominator; the limit of a rational function is 0 when the polynomial in the numerator has a lower degree than the polynomial in the denominator; the limit of a rational function is a finite number other than 0 when the polynomial in the numerator has the same degree as the polynomial in the denominator

log_a(x) << x^b << c^x << x^x, ∀ a > 1, ∀ b > 1, ∀ c > 1, asymptotic control scale for x → +∞; the logarithm function grows less rapidly than the power function; the power function grows less rapidly than the exponential function; the exponential function grows less rapidly than the function x^x

y = log₁₀(x); x = 10, y = 1; x = 100, y = 2; x = 1000, y = 3

y = x²; x = 10, y = 10²; x = 100, y = 10⁴; x = 1000, y = 10⁶

y = 10^x; x = 10, y = 10¹⁰; x = 100, y = 10¹⁰⁰; x = 1000, y = 10¹⁰⁰⁰

y = x^x; x = 10, y = 10¹⁰; x = 100, y = 10²⁰⁰; x = 1000, y = 10³⁰⁰⁰

lim_x→+∞(x⁶-6^x); lim_x→+∞(x⁶-6^x) = +∞-∞, indeterminate form; lim_x→+∞(x⁶-6^x) = lim_x→+∞(6^x(x⁶/6^x-6^x/6^x)) = lim_x→+∞(6^x(x⁶/6^x-1)) = +∞(0-1) = +∞(-1) = -∞

lim_x→+∞((e^x-x²)/(3x+ln(x))); lim_x→+∞((e^x-x²)/(3x+ln(x))) = (+∞-∞)/(+∞+∞) = (+∞-∞)/+∞, indeterminate form; lim_x→+∞((e^x-x²)/(3x+ln(x))) = lim_x→+∞(e^x(e^x/e^x-x²/e^x)/(3x(3x/3x+ln(x)/3x))) = lim_x→+∞(e^x(1-x²/e^x)/(3x(1+ln(x)/3x))) = (+∞(1-0))/(1+0) = +∞⋅1/1 = +∞

lim_x→+∞((√(x²+3))/(log₂(x)-2x)); lim_x→+∞((√(x²+3))/(log₂(x)-2x)) = +∞/(∞-∞), indeterminate form; lim_x→+∞((√(x²+3))/(log₂(x)-2x)) = lim_x→+∞((√(x²(x²/x²+3/x²)))/(2x(log₂(x)/2x-2x/2x))) = lim_x→+∞((√(x²(1+3/x²)))/(2x(log₂(x)/2x-1))) = lim_x→+∞((|x|√(1+3/x²))/(2x(log₂(x)/2x-1))) = lim_x→+∞((x√(1+3/x²))/(2x(log₂(x)/2x-1))) = √(1+0)/(2(0-1)) = √(1)/(2(-1)) = 1/-2 = -1/2

lim_x→+∞(cos(3/e^x)); lim_x→+∞(cos(3/e^x)) = cos(3/+∞) = cos(0) = 1

lim_x→+∞(e^{(x2+1)/(2x2-x)}); lim_x→+∞(e^{(x2+1)/(2x2-x)}), y = (x²+1)/(2x²-x), lim_x→+∞(y) = lim_x→+∞((x²+1)/(2x²-x)) = lim_x→+∞((x²(x²/x²+1/x²))/(2x²(2x²/2x²-x/2x²))) = lim_x→+∞((1+1/x²)/(2(1-1/2x))) = (1+0)/(2(1-0) = 1/2⋅1 = 1/2, lim_x→+∞(e^{(x2+1)/(2x2-x)}) = lim_y→1/2(e^y) = e^1/2 = √(e)

lim_x→+∞((sin(7^x+ln(x)))/(5x²+1)); lim_x→+∞(sin(7^x+ln(x))) = ∄, this limit does not exist because it is a quantity that oscillates infinitely between -1 and 1; we can solve this limit using the squeeze theorem, also called the theorem of carabinieri; -1 ≤ sin(7^x+ln(x)) ≤ 1, -1/(5x²+1) ≤ (sin(7^x+ln(x)))/(5x²+1) ≤ 1/(5x²+1); lim_x→+∞(-1/(5x²+1)) = lim_x→+∞(-1/(5x²(5x²/5x²+1/5x²))) = lim_x→+∞(-1/(5x²(1+1/5x²))) = -1/(+∞(1+0)) = -1/(+∞⋅1) = -1/+∞ = 0; lim_x→+∞(1/(5x²+1)) = lim_x→+∞(1/(5x²(5x²/5x²+1/5x²))) = lim_x→+∞(1/(5x²(1+1/5x²))) = 1/(+∞(1+0)) = 1/(+∞⋅1) = 1/+∞ = 0; so for the squeeze theorem, lim_x→+∞((sin(7^x+ln(x)))/(5x²+1)) = 0

lim_x→4((√(x)-2)/(3x-12)); lim_x→4((√(x)-2)/(3x-12)) = (√(4)-2)/((3·4)-12) = (2-2)/(12-12) = 0/0, indeterminate form; lim_x→4((√(x)-2)/(3x-12)) = lim_x→4((√(x)-2)/(3(3x/3-12/3))) = lim_x→4((√(x)-2)/(3(x-4))) = lim_x→4(((√(x)-2)(√(x)+2))/((3(x-4))(√(x)+2))) = lim_x→4((x-4)/(3(x-4)(√(x)+2))) = lim_x→4(1/(3(√(x)+2))) = 1/(3(√(4)+2)) = 1/(3(2+2)) = 1/(3·4) = 1/12

The squeeze theorem, or carabinieri theorem, states that, if a function is enclosed between two functions that converge to the same limit, then it converges to that limit; h(x), f(x), g(x), are 3 functions defined in the neighborhood I(x₀,δ), h(x) ≤ f(x) ≤ g(x), lim_x→x0(h(x)) = lim_x→x0(g(x)) = l ⇒ lim_x→x0(f(x)) = l; lim_x→x0(h(x)) = l, |h(x)-l| < ε, ∀ x ∈ I_h; lim_x→x0(g(x)) = l, |g(x)-l| < ε, ∀ x ∈ I_g; ∀ I = I_h ∩ I_g, ∀ x ∈ I {l-ε < h(x) < l+ε, l-ε < g(x) < l+ε}; h(x) ≤ f(x) ≤ g(x) ⇒ l-ε < h(x) ≤ f(x) ≤ g(x) < l+ε, ∀ x ∈ I ⇒ l-ε < f(x) < l+ε, ∀ x ∈ I ⇒ |f(x)-l| < ε, ∀ x ∈ I ⇒ lim_x→x0(f(x)) = l

lim_x→0(x⋅sin(1/x)); -|x| ≤ x⋅sin(1/x) ≤ |x|; lim_x→0(-|x|) = 0; lim_x→0(|x|) = 0; for the squeeze theorem, g(x) ≤ f(x) ≤ h(x), lim_x→x0(g(x)) = lim_x→x0(h(x)) = l ⇒ lim_x→x0(f(x)) = l, then -|x| ≤ x⋅sin(1/x) ≤ |x|, lim_x→0(-|x|) = lim_x→0(|x|) = 0 ⇒ lim_x→0(x⋅sin(1/x)) = 0

lim_x→0(sin(x)/x); lim_x→0(sin(x)/x) = 0/0, indeterminate form; to calculate this limit we use the squeeze theorem, g(x) ≤ f(x) ≤ h(x), lim_x→x0(g(x)) = lim_x→x0(h(x)) = l ⇒ lim_x→x0(f(x)) = l; sin(x) < x < tan(x), ∀ x ∈ (0,π/2), sin(x) < x < sin(x)/cos(x), sin(x)/sin(x) < x/sin(x) < sin(x)/(cos(x)sin(x)), 1 < x/sin(x) < 1/cos(x), 1 > sin(x)/x > cos(x), cos(x) < sin(x)/x < 1, lim_x→0⁺(cos(x)) = 1, lim_x→0⁺(sin(x)/x) = 1; sin(x) > x > tan(x), ∀ x ∈ (-π/2,0), sin(x) > x > tan(x), sin(x) > x > sin(x)/cos(x), sin(x)/sin(x) > x/sin(x) > sin(x)/(cos(x)sin(x)), 1 > x/sin(x) > 1/cos(x), 1 < sin(x)/x < cos(x), lim_0^-(cos(x)) = 1, lim_x→0^-(sin(x)/x) = 1; lim_x→0^-(sin(x)/x) = lim_x→0⁺(sin(x)/x) = 1 ⇒ lim_x→0(sin(x)/x) = 1

lim_x→∞((1+1/x)^x); lim_x→∞((1+1/x)^x) = 1^∞, indeterminate form; lim_x→∞((1+1/x)^x), y = (1+1/x)^x, ln(y) = ln((1+1/x)^x), ln(y) = x·ln(1+1/x), lim_x→∞(y) = lim_x→∞((1+1/x)^x), lim_x→∞(ln(y)) = lim_x→∞(ln((1+1/x)^x)), lim_x→∞(ln(y)) = lim_x→∞(x·ln(1+1/x)) = lim_x→∞(ln(1+1/x)/(1/x)), applying L'Hôpital's rule, lim_x→∞(ln(1+1/x)/(1/x)) = lim_x→∞((ln(1+1/x))'/(1/x)') = lim_x→∞(((1/(1+1/x)(-1/x²))/(-1/x²)) = lim_x→∞(1/(1+1/x)) = 1/(1+0) = 1/1 = 1, lim_x→∞(ln(y)) = 1, ln(lim_x→∞(y)) = 1, lim_x→∞(y) = e, lim_x→∞((1+1/x)^x) = e

lim_x→0((1+x)^1/x); lim_x→0((1+x)^1/x), x = 1/y, y = 1/x, if x → 0 then y → ∞, lim_x→0((1+x)^1/x) = lim_y→∞(1+1/y)^y = e, lim_x→0((1+x)^1/x) = e

lim_x→0(tan(x)/x); lim_x→0(tan(x)/x) = 0/0, indeterminate form; lim_x→0(tan(x)/x) = lim_x→0((sin(x)/cos(x))(1/x)) = lim_x→0((sin(x)/x)(1/cos(x))) = 1⋅1 = 1, lim_x→0(tan(x)/x) = 1

lim_x→0((1-cos(x))/x²); lim_x→0((1-cos(x))/x²)= 0/0, indeterminate form; lim_x→0((1-cos(x))/x²) = lim_x→0(((1-cos(x))/(x²))((1+cos(x))/(1+cos(x)))) = lim_x→0((1-cos²(x))/((x²)(1+cos(x)))) = lim_x→0(sin²(x)/((x²)(1+cos(x)))) = (1²)(1/2) = (1)(1/2) = 1/2, lim_x→0((1-cos(x))/x²) = 1/2

lim_x→0((2sin(x)+4tan(x))/(x⋅cos(x)+2sin(x))); lim_x→0((2sin(x)+4tan(x))/(x⋅cos(x)+2sin(x))) = 0/0, indeterminate form; lim_x→0((2sin(x)+4tan(x))/(x⋅cos(x)+2sin(x))) = lim_x→0((x(2sin(x)/x+4tan(x)/x))/(x(x⋅cos(x)/x+2sin(x)/x))) = lim_x→0((2sin(x)/x+4tan(x)/x)/(cos(x)+2sin(x)/x)) = (2⋅1+4⋅1)/(1+2⋅1) = (2+4)/(1+2) = 6/3 = 2, lim_x→0((2sin(x)+4tan(x))/(x⋅cos(x)+2sin(x))) = 2

lim_x→0(√((cos(x)-cos²(x))/(2x²))); lim_x→0(√((cos(x)-cos²(x))/(2x²))) = 0/0, indeterminate form; lim_x→0(√((cos(x)-cos²(x))/(2x²))) = lim_x→0(√((cos(x)(1-cos(x)))/(2x²))) = lim_x→0(√((cos(x)/2)((1-cos(x)/x²)))) = √((1/2)(1/2)) = √(1/4) = 1/2

lim_x→0((ln(1+x))/x); lim_x→0((ln(1+x))/x) = 0/0, indeterminate form; lim_x→0((ln(1+x))/x) = lim_x→0(ln(1+x)^1/x), y = 1/x, x = 1/y, if x → 0 then y → ∞, lim_x→0(ln(1+x)^1/x) = lim_y→∞(ln(1+1/y)^y) = ln(e) = 1, lim_x→0(ln(1+x)/x) = 1

lim_x→0((e^x-1)/x); lim_x→0((e^x-1)/x) = 0/0, indeterminate form; lim_x→0((e^x-1)/x), y = e^x-1, e^x = y+1, ln(e^x) = ln(y+1), x = ln(y+1), if x → 0 then y → 0, lim_x→0((e^x-1)/x) = lim_y→0(y/ln(y+1) = 1/1 = 1, because lim_x→0((ln(1+x))/x) = 1, lim_x→0((e^x-1)/x) = 1

lim_x→0(log_a(1+x)/x); lim_x→0(log_a(1+x)/x) = 0/0, indeterminate form; lim_x→0(log_a(1+x)/x) = lim_x→0((1/x)log_a(1+x)) = lim_x→0(log_a(1+x)^1/x), y = 1/x, x = 1/y, if x → 0 then y → ∞, lim_y→∞(log_a(1+1/y)^y), log_a(e) = ln(e)/ln(a) = 1/ln(a), lim_x→0(log_a(1+x)/x) = 1/ln(a)

lim_x→0((a^x-1)/x); lim_x→0((a^x-1)/x) = 0/0, indeterminate form; lim_x→0((a^x-1)/x), y = a^x-1, a^x = y+1, ln(a^x) = ln(y+1), x⋅ln(a) = ln(y+1), x = ln(y+1)/ln(a), if x → 0 then y → 0, lim_y→0(y/(ln(y+1)/ln(a))) = lim_y→0(y⋅ln(a)/ln(y+1)) = ln(a)lim_y→0(y/ln(y+1)) = ln(a), because lim_y→0(y/ln(y+1)) = 1, lim_x→0((a^x-1)/x) = ln(a)

lim_x→0(((1+x)^a-1)/(ax)); lim_x→0(((1+x)^a-1)/(ax)) = 0/0, indeterminate form; lim_x→0(((1+x)^a-1)/(ax)) =^H lim_x→0((a(1+x)^a-1)/a) = lim_x→0((1+x)^a-1) = 1^a-1 = 1, lim_x→0(((1+x)^a-1)/(ax)) = 1

lim_x→0(sin(x)/x) = 1; lim_x→0(tan(x)/x) = 1; lim_x→0((1-cos(x))/x²) = 1/2; lim_x→∞((1+1/x)^x) = e; lim_x→0(ln(1+x)/x) = 1; lim_x→0((e^x-1)/x) = 1; lim_x→0(log_a(1+x)/x) = 1/ln(a); lim_x→0((a^x-1)/x) = ln(a); lim_x→0(((1+x)^a-1)/(ax)) = 1

lim_x→0(tan(3x)/x); lim_x→0(tan(3x)/x) = 0/0, indeterminate form; lim_x→0(tan(3x)/x) = lim_x→0(3tan(3x)/(3x)) = 3lim_x→0(tan(3x)/(3x)), y = 3x, if x → 0 then y → 0, 3lim_x→0(tan(3x)/(3x)) = 3lim_y→0(tan(y)/y) = 3·1 = 3, because lim_x→0(tan(x)/x) = 1, lim_x→0(tan(3x)/x) = 3

lim_x→0(ln(1+√(2)x)/x); lim_x→0(ln(1+√(2)x)/x) = 0/0, indeterminate form; lim_x→0(ln(1+√(2)x)/x) = lim_x→0(√(2)ln(1+√(2)x)/(√(2)x)) = √(2)lim_x→0(ln(1+√(2)x)/(√(2)x)), y = (2)x, if x → 0 then y → 0, √(2)lim_x→0(ln(1+√(2)x)/(√(2)x)) = √(2)lim_y→0(ln(1+y)/y) = √(2)·1 = √(2), because lim_x→0(ln(1+x)/x) = 1, lim_x→0(ln(1+√(2)x)/x) = √(2)

lim_x→0(ln(1+sin(x))/sin(x)); lim_x→0(ln(1+sin(x))/sin(x)) = 0/0, indeterminate form; lim_x→0(ln(1+sin(x))/sin(x)), y = sin(x), if x → 0 then y → 0, lim_x→0(ln(1+sin(x))/sin(x)) = lim_y→0(ln(1+y)/y) = 1, because lim_x→0(ln(1+x)/x) = 1, lim_x→0(ln(1+sin(x))/sin(x)) = 1

lim_x→0(ln(1+sin(x))/x); lim_x→0(ln(1+sin(x))/x) = 0/0, indeterminate form; lim_x→0(ln(1+sin(x))/x) = lim_x→0((ln(1+sin(x))/x)(sin(x)/sin(x))) = lim_x→0((ln(1+sin(x))/sin(x))(sin(x)/x)) = 1⋅1 = 1, because lim_x→0(sin(x)/x) = 1, and because lim_x→0(ln(1+x)/x) = 0, lim_x→0(ln(1+sin(x))/x) = 1

lim_x→0((e^x-cos(√(x)))/x); lim_x→0((e^x-cos(√(x)))/x) = 0/0, indeterminate form; lim_x→0((e^x-cos(√(x)))/x) = lim_x→0((e^x-1+1-cos(√(x)))/x) = lim_x→0(((e^x-1)/x)+((1-cos(√(x)))/x)) = 1+1/2 = 3/2, because lim_x→0((e^x-1)/x) = 1, and because lim_x→0((1-cos(x))/x²) = 1/2, lim_x→0((e^x-cos(√(x)))/x) = 3/2

lim_x→∞((x-1)/(x+1)+sin(x)/x); lim_x→∞((x-1)/(x+1)+sin(x)/x) = 1+0 = 1, because lim_x→∞((x-1)/(x+1)) = lim_x→∞((x(1-1/x))/(x(1+1/x))) = lim_x→∞((1-1/x)/(1+1/x)) = (1-0)/(1+0) = 1/1 = 1, and lim_x→∞(sin(x)/x) = 0 because the numerator is -1 ≤ sin(x) ≤ 1 and the denominator is x → ∞, lim_x→∞((x-1)/(x+1)+sin(x)/x) = 1

lim_x→0(ln(1+cos(x)/cos(x)); lim_x→0(ln(1+cos(x)/cos(x)) = ln(2)/1 = ln(2)

lim_x→0((sin(x)-x+2x⁵)/(3x³)); lim_x→0((sin(x)-x+2x⁵)/(3x³)) = 0/0; lim_x→0((sin(x)-x+2x⁵)/(3x³)) =^H lim_x→0((cos(x)-1+10x⁴)/(9x²)) =^H lim_x→0((-sin(x)+40x³)/(18x)) =^H lim_x→0((-cos(x)+120x²)/18) = (-1+0)/18 = -1/18, lim_x→0((sin(x)-x+2x⁵)/(3x³)) = -1/18

Two functions f(x) and g(x) are asymptotically equivalent for x → x₀ if lim_x→x0(f(x)/g(x)) = 1, and it is written f(x) ~ g(x) for x → x₀

lim_x→0(sin(x)/x) = 1, sin(x) ~ x for x → 0

lim_x→0((1-cos(x))/x²) = 1/2, 1-cos(x) ~ (1/2)x² for x → 0

lim_x→0(tan(x)/x) = 1, tan(x) ~ x for x → 0

lim_x→0((e^x-1)/x) = 1, e^x-1 ~ x for x → 0

lim_x→0(ln(1+x)/x) = 1, ln(1+x) ~ x for x → 0

lim_x→0(((1+x)^a-1)/(ax)) = 1, (1+x)^a-1 ~ ax for x → 0

lim_x→0(sin(5x)/5x); lim_x→0(sin(5x)/5x) = 0/0, indeterminate form; lim_x→0(sin(5x)/5x), y = 5x, if x → 0 then y → 0, lim_x→0(sin(5x)/5x) = lim_y→0(sin(y)/y) = 1, because lim_x→0(sin(x)/x) = 1, lim_x→0(sin(5x)/5x) = 1; we can also solve this limit using the l'Hôpital's rule, lim_x→0(sin(5x)/5x) =^H = lim_x→0((5cos(5x))/5) = lim_x→0(cos(5x)) = cos(0) = 1, lim_x→0(sin(5x)/5x) = 1; lim_x→0(sin(5x)/5x) = 1, sin(5x) ~ 5x for x → 0

lim_x→∞((e^1/x-1)/(1/x)); lim_x→∞((e^1/x-1)/(1/x)) = 0/0, indeterminate form; lim_x→∞((e^1/x-1)/(1/x)), y = 1/x, if x → ∞ then y → 0, lim_x→∞((e^1/x-1)/(1/x)) = lim_y→0((e^y-1)/(y)) = 1, because lim_x→0((e^x-1)/(x)) = 1, lim_x→∞((e^1/x-1)/(1/x)) = 1; we can also solve this limit using the l'Hôpital's rule, lim_x→∞((e^1/x-1)/(1/x)) = lim_x→∞((-e^1/x/x²)/(-1/x²)) = lim_x→∞((-e^1/x/x²)(-x²)) = lim_x→∞(e^1/x) = e⁰ = 1, lim_x→∞((e^1/x-1)/(1/x)) = 1; lim_x→∞((e^1/x-1)/(1/x)) = 1, e^1/x-1 ~ 1/x for x → ∞

If f₁(x) ~ g₁(x) for x → x₀, and f₂(x) ~ g₂(x) for x → x₀, then f₁(x)⋅f₂(x) ~ g₁(x)⋅g₂(x) for x → x₀, therefore lim_x→x0(f₁(x)⋅f₂(x)) = lim_x→x0(g₁(x)⋅g₂(x))

If f₁(x) ~ g₁(x) for x → x₀, and f₂(x) ~ g₂(x) for x → x₀, then f₁(x)/f₂(x) ~ g₁(x)/g₂(x) for x → x₀, therefore lim_x→x0(f₁(x)/f₂(x)) = lim_x→x0(g₁(x)/g₂(x))

If f(x) ~ g(x) for x → x₀, then (f(x))^a ~ (g(x))^a for x → x₀, therefore lim_x→x0((f(x))^a) = lim_x→x0((g(x))^a)

lim_x→0(((e^3x-1)sin(4x))/tan(2x²)); lim_x→0(((e^3x-1)sin(4x))/tan(2x²)) = 0/0, indeterminate form; e^3x-1 ~ 3x for x → 0; sin(4x) ~ 4x for x → 0; tan(2x²) ~ 2x² for x → 0; ((e^3x-1)sin(4x))/tan(2x²) ~ (3x⋅4x)/(2x²) for x → 0; lim_x→0(((e^3x-1)sin(4x))/tan(2x²)) = lim_x→0((3x⋅4x)/(2x²)) = 6, lim_x→0(((e^3x-1)sin(4x))/tan(2x²)) = 6

lim_x→0^-(√(1-cos(x²))/ln(1+2x)); lim_x→0^-(√(1-cos(x²))/ln(1+2x)) = 0/0, indeterminate form; 1-cos(x) ~ (1/2)x² for x → 0, 1-cos(x²) ~ (1/2)x⁴ for x → 0, √(1-cos(x²) ~ √((1/2)x⁴) = x²/√(2) for x → 0; ln(1+x) ~ x for x → 0, ln(1+2x) ~ 2x for x → 0; √(1-cos(x²))/ln(1+2x) ~ (x²/√(2))/(2x) = x/(2√(2)) for x → 0; lim_x→0^-(√(1-cos(x²))/ln(1+2x)) = lim_x→0^-(x/(2√(2))) = 0, lim_x→0^-(√(1-cos(x²))/ln(1+2x)) = 0

DERIVATIVE

The derivative of a function is the slope of the tangent line to the graph of the function at that point

A function f(x) is differentiable at a point a of its domain if the limit of the incremental ratio exists, L = lim_h→0((f(a+h)-f(a))/h), that is, for every positive real number ε there exists a positive real number δ such that |h| < δ, and |L-((f(a+h)-f(a))/h)| < ε, and this limit is called the derivative of f at point a and is denoted by f'(a), or (df/dx)(a) and is read as the derivative of f with respect to x at point a

f'(a) = lim_h→0((f(a+h)-f(a))/h)

f'(x) = lim_h→0((f(x+h)-f(x))/h)

f'(x) = lim_x→x0((f(x)-f(x₀))/(x-x₀))

f(x) = x², f'(x) = lim_h→0((f(x+h)-f(x))/h) = lim_h→0(((x+h)²-x²)/h) = lim_h→0((x²+2hx+h²-x²)/h) = lim_h→0((2hx+h²)/h) = lim_h→0(h(2x+h)/h) = lim_h→0(2x+h) = 2x

f(x) = x², f'(x) = lim_x→x0((f(x)-f(x₀))/(x-x₀)) = lim_x→x0((x²-x₀²)/(x-x₀)) = lim_x→x0((x+x₀)(x-x₀)/(x-x₀)) = lim_x→x0((x+x₀)) = 2x₀

(xⁿ)' = nx^n-1; f'(x) = lim_h→0((f(x+h)-f(x))/h); (xⁿ)' = lim_h→0(((x+h)ⁿ-xⁿ)/h) = lim_h→0((xⁿ+nx^n-1h+C(n,2)x^n-2h²+...+hⁿ-xⁿ)/h) = lim_h→0((nx^n-1h+C(n,2)x^n-2h²+...+hⁿ)/h) = lim_h→0(nx^n-1+C(n,2)x^n-2h+...+h^n-1) = nx^n-1

(f(x)+g(x))' = f'(x)+g'(x); f'(x) = lim_x→x0((f(x)-f(x₀)/(x-x₀)); (f(x)+g(x))' = lim_x→x0((f(x)+g(x)-f(x₀)-g(x₀))/(x-x₀)) = lim_x→x0((f(x)-f(x₀))/(x-x₀))+lim_x→x0((g(x)-g(x₀))/(x-x₀)) = f'(x)+g'(x)

(c⋅f(x))' = c⋅f'(x); (c⋅f(x))' = lim_x→x0((c⋅f(x)-c⋅f(x₀))/(x-x₀)) = c⋅lim_x→x0(f(x)-f(x₀)/(x-x₀) = c⋅f'(x)

(f(x)g(x))' = f'(x)g(x)+f(x)g'(x); (f(x)g(x))' = lim_x→x0((f(x)g(x)-f(x₀)g(x₀))/(x-x₀)) = lim_x→x0((f(x)g(x)-f(x₀)g(x₀)+f(x₀)g(x)-f(x₀)g(x))/(x-x₀)) = lim_x→x0((g(x)(f(x)-f(x₀))+f(x₀)(g(x)-g(x₀)))/(x-x₀)) = lim_x→x0(g(x)(f(x)-f(x₀))/(x-x₀)+f(x₀)(g(x)-g(x₀))/(x-x₀)) = g(x₀)f'(x₀)+f(x₀)+g'(x₀) = f'(x₀)g(x₀)+f(x₀)g'(x₀) = f'(x)g(x)+f(x)g'(x)

(1/f(x))' = -f'(x)/(f(x))²; (1/f(x))' = lim_x→x0((1/f(x)-1/f(x₀))/(x-x₀)) = lim_x→x0((1/f(x)-1/f(x₀))(1/(x-x₀))) = lim_x→x0(((f(x₀)-f(x))/(f(x)f(x₀)))(1/(x-x₀))) = lim_x→x0(((f(x₀)-f(x))/(x-x₀))(1/(f(x)f(x₀)))) = lim_x→x0((-1/(f(x)f(x₀)))((f(x)-f(x₀))/(x-x₀))) = (-1/(f(x₀))²)f'(x₀) = -f'(x₀)/(f(x₀))² = -f'(x)/(f(x))²

(f(x)/g(x))' = (f'(x)g(x)-f(x)g'(x))/(g(x))²; (f(x)/g(x))' = (f(x)(1/g(x)))' = f'(x)(1/g(x)+f(x)(1/g(x))' = f'(x)/g(x)+f(x)(-g'(x)/(g(x))²) = f'(x)/g(x)-f(x)(g'(x)/(g(x))²) = (f'(x)g(x)-f(x)g'(x))/(g(x))²

(f(g(x)))' = g'(x)f'(g(x)), the derivative of a compound function is the product of the derivative of the internal function times the derivative of the external function, and this is the chain rule; y = f(g(x)), dy/dx = (f(g(x)))' = g'(x)f'(g(x)) = lim_h→0((f(g(x+h))-f(g(x)))/h) = lim_h→0(((f(g(x+h))-f(g(x)))/h)((g(x+h)-g(x))/(g(x+h)-g(x)))) = lim_h→0(((f(g(x+h))-f(g(x)))/(g(x+h)-g(x)))((g(x+h)-g(x))/h)) = lim_h→0((f(g(x+h))-f(g(x)))/(g(x+h)-g(x)))lim_h→0((g(x+h)-g(x))/h) = lim_h→0((f(g(x+h))-f(g(x)))/(g(x+h)-g(x)))g'(x), k = g(x+h)-g(x), if h→0 then k→0, g(x+h) = g(x)+k, lim_h→0((f(g(x+h))-f(g(x)))/(g(x+h)-g(x)))g'(x) = lim_k→0((f(g(x)+k))-f(g(x)))/(k))g'(x), f'(a) = lim_h→0((f(a+h)-f(a))/h), a = g(x), lim_k→0((f(g(x)+k))-f(g(x)))/(k))g'(x) = f'(g(x))g'(x), d/dx(f(g(x))) = (f(g(x)))' = g'(x)f'(g(x))

f'(x) = 1/(f^-1(y))', the derivative of the direct function y = f(x) is equal to the reciprocal of the derivative of the inverse function x = f^-1(y)

(f^-1(y))' = 1/f'(x), the derivative of the inverse function x = f^-1(y) is equal to the reciprocal of the derivative of the direct function y = f(x)

y = f(x) is a definite and invertible function in the interval I and its inverse function is x = f^-1(y); if f(x) is differentiable at the point x ∈ I and f'(x) ≠ 0, then f^-1(y) is differentiable at the point y = f(x) and (f^-1(y))' = 1/f'(x); considering the incremental ratio of the inverse function f^-1(y), (f^-1(y)-f^-1(y₀))/(y-y₀), y ≠ y₀, (f^-1(y)-f^-1(y₀))/(y-y₀) = (x-x₀)/(f(x)-f(x₀)), x ≠ x₀, if f(x) is a continuous function then its inverse function f^-1(y) is also continuous, therefore (f^-1(y₀))' = lim_y→y0((f^-1(y)-f^-1(y₀))/(y-y₀)) = lim_x→x0((x-x₀)/(f(x)-f(x₀))) = 1/f'(x₀), (f^-1(y))' = 1/f'(x)

(sin(x))' = cos(x); f'(x) = lim_h→0((f(x+h)-f(x))/h), (sin(x))' = lim_h→0((sin(x+h)-sin(x))/h) = lim_h→0((sin(x)cos(h)+cos(x)sin(h)-sin(x))/h) = lim_h→0((sin(x)(cos(h)-1)+cos(x)sin(h))/h) = lim_h→0(((sin(x)(cos(h)-1))/h+(cos(x)sin(h))/h) = lim_h→0((sin(x)(cos(h)-1))/h)+lim_h→0((cos(x)sin(h))/h) = sin(x)lim_h→0((cos(h)-1)/h)+cos(x)lim_h→0(sin(h)/h) = sin(x)⋅0+cos(x)⋅1 = cos(x); lim_x→0((cos(x)-1)/x) =^H lim_x→0(-sin(x)/1) = 0/1 = 0; lim_x→0(sin(x)/x) =^H lim_x→0(cos(x)/1) = 1/1 = 1

(sin(x))' = cos(x); f'(x) = lim_x→x0((f(x)-f(x₀))/(x-x₀)), (sin(x))' = lim_x→x0((sin(x)-sin(x₀))/(x-x₀)) = lim_x→x0(2sin((x-x₀)/2)cos((x+x₀)/2)/(x-x₀)), t = (x-x₀)/2, x → x₀ ⇒ t → 0, lim_x→x0(2sin((x-x₀)/2)cos((x+x₀)/2)/(x-x₀)) = lim_t→0(sin(t)/t)lim_x→x0(cos((x+x₀)/2)) = 1cos((x₀+x₀)/2) = cos((2x₀)/2) = cos(x₀) = cos(x); lim_x→0(sin(x)/x) =^H lim_x→0(cos(x)/1) = 1/1 = 1

(cos(x))' = -sin(x); f'(x) = lim_h→0((f(x+h)-f(x))/h), (cos(x))' = lim_h→0((cos(h+x)-cos(x))/h) = lim_h→0((cos(h)cos(x)-sin(h)sin(x)-cos(x))/h) = lim_h→0((cos(x)(cos(h)-1)-sin(h)sin(x))/h) = lim_h→0((cos(x)(cos(h)-1))/h)-lim_h→0((sin(h)sin(x))/h) = cos(x)lim_h→0((cos(h)-1)/h)-sin(x)lim_h→0(sin(h)/h) = cos(x)⋅0-sin(x)⋅1 = -sin(x); lim_x→0((cos(x)-1)/x) =^H lim_x→0(-sin(x)/1) = 0/1 = 0; lim_x→0(sin(x)/x) =^H lim_x→0(cos(x)/1) = 1/1 = 1

(tan(x))' = 1/cos²(x) = sec²(x) = 1+tan²(x); (f(x)/g(x))' = (f'(x)g(x)-f(x)g'(x))/(g(x))²; (tan(x))' = (sin(x)/cos(x))' = (cos(x)cos(x)-sin(x)(-sin(x)))/cos²(x) = (cos²(x)+sin²(x))/cos²(x) = 1/cos²(x) = sec²(x)

(csc(x))' = -cos(x)/sin²(x) = -cos(x)csc²(x) = -csc(x)cot(x); (1/f(x))' = -f'(x)/(f(x))²; (csc(x))' = (1/sin(x))' = -cos(x)/sin²(x) = -cos(x)csc²(x) = -csc(x)cot(x)

(sec(x))' = sin(x)/cos²(x) = sin(x)sec²(x) = sec(x)tan(x); (1/f(x))' = -f'(x)/(f(x))²; (sec(x))' = (1/cos(x))' = -(-sin(x))/cos²(x) = sin(x)/cos²(x) = sin(x)sec²(x) = sec(x)tan(x)

(cot(x))' = -csc²(x) = -1-cot²(x); (f(x)/g(x))' = (f'(x)g(x)-f(x)g'(x))/(g(x))²; (cot(x))' = (cos(x)/sin(x))' = (-sin(x)sin(x)-cos(x)cos(x))/sin²(x) = (-sin²(x)-cos²(x))/sin²(x) = -(sin²(x)+cos²(x))/sin²(x) = -1/sin²(x) = -csc²(x)

(arcsin(x))' = 1/√(1-x²); f(x) = arcsin(x) = y, x ∈ [-1,1]; f^-1(y) = sin(y) = x, y ∈ [-π/2,π/2]; (f^-1(y))' = (sin(y))' = cos(y), y ∈ (-π/2,π/2); (f^-1(y))' = 1/f'(x), f'(x) = 1/(f^-1(y))' = 1/cos(y); sin²(x)+cos²(x) = 1, cos²(x) = 1-sin²(x), cos(x) = ±√(1-sin²(x)); f'(x) = 1/(f^-1(y))' = 1/cos(y) = 1/√(1-sin²(y)) = 1/√(1-x²), f'(x) = (arcsin(x))' = 1/√(1-x²)

f(x) = e^x, f'(x) = e^x; f'(x) = lim_h→0((f(x+h)-f(x))/h); (e^x)' = lim_h→0((e^x+h-e^x)/h) = lim_h→0(e^h⋅e^x-e^x)/h) = lim_h→0((e^x(e^h-1))/h) = e^x⋅lim_h→0((e^h-1)/h) = e^x⋅1 = e^x, because lim_x→0((e^x-1)/x) = 1, (e^x)' = e^x

(e^x)' = D(e^x) = d/dx(e^x) = e^x; y = e^x, ln(y) = ln(e^x), ln(y) = x, (ln(y))' = (x)', (1/y)(dy/dx) = 1, dy/dx = y, dy/dx = e^x, d/dx(e^x) = e^x

(ln(x))' = D(ln(x)) = d/dx(ln(x)) = 1/x; y = f(x), f^-1(y) = x; y = e^x, ln(y) = x; f(x) = y = e^x, f^-1(y) = x = ln(y); D(f^-1(y)) = 1/D(f(x)); D(ln(y)) = 1/D(e^x), D(ln(y)) = 1/e^x, D(ln(y)) = 1/y, D(ln(x)) = 1/x

(e^-x)' = -e^-x; (e^-x)' = (1/e^x)', (1/f(x))' = -f'(x)/(f(x))², (1/e^x)' = -e^x/e^2x = -e^xe^-2x = -e^-x

EULER'S NUMBER

lim_x→∞(1+1/x)^x = e

lim_x→0(1+x)^1/x = e

^∞Σ_n=0(1/n!) = e

f(x) = e^x, natural exponential function

f(x) = e^x, f^-1(x) = ln(x); e^x is the inverse function of ln(x), their graphs are symmetrical with respect to the bisector of the first and third quadrant; the tangent line to the function e^x at the point (0,1) has an angular coefficient equal to 1; the tangent line to the function ln(x) at the point (1,0) has an angular coefficient equal to 1

(e^x)' = e^x; d/dx(e^x) = e^x; e^x is the only non zero function whose slope of the tangent line in a point is equal to the value of the ordinate of the point

_-∞∫^x(e^xdx) = e^x

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EXPONENTIATION

Exponentiation is a mathematical operation, written as bⁿ, involving two numbers, the base b and the exponent or power n, and pronounced as "b raised to the power of n"; when n is a positive integer, exponentiation corresponds to repeated multiplication of the base, bⁿ is the product of multiplying n bases

bⁿ is called "b raised to the nth power", "b raised to the power of n", "the nth power of b", "b to the nth power", or most briefly as "b to the nth"

b¹ = b

b⁰ = 1

b^-1 = 1/b, the reciprocal of b

b^-n = 1/bⁿ

bⁿ⋅b^m = b^n+m

bⁿ/b^m = b^n-m

bⁿ⋅aⁿ = (b⋅a)ⁿ

bⁿ/aⁿ = (b/a)ⁿ

(ⁿ√(b))ⁿ = b

b^1/n = ⁿ√(b)

(b^1/n)ⁿ = b^(1/n)⋅n = b¹ = b

b^n/m = ^m√(bⁿ)

e^x-e = e(e^x/e-e/e) = e(e^x-1-1)

(a+b)^c/d = a^c/d(1+b/a)^c/d

ALGEBRA EXERCISES

A monomial, also called power product, is a product of powers of variables with non-negative integer exponents

A polynomial is an expression consisting of variables, also called indeterminates, and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables

Examples of monomial: 8x; 5x²; 3x²y

Examples of binomial: 5x+6; 7x-3; x+2

Examples of trinomial: x²+6x+5, 2x²-3x+1, -5x²+7x-3

5x+4x = 9x

3x+4y+5x+8y = 8x+12y

3√(2)+5√(7)+8√(2)+3√(7) = 11√(2)+8√(7)

7x+4x²+5x+9x² = 13x²+12x

(9x²+6x+5)+(3x²-5x-9) = 12x²+x-4

(3x²+7x-4)-(8x²-5x+7) = 3x²+7x-4-8x²+5x-7 = -5x²+12x-11

7x(x²+2x-3) = 7x³+14x²-21x

(5x²)(3x⁴-6x³+5x-8) = 15x⁶-30x⁵+25x³-40x²

(3x-4)(2x+7) = 6x²+21x-8x-28 = 6x²+13x-28

(2x-5)(4x+7) = 8x²-20x+14x-35 = 8x²-6x-35

(2x-3)² = (2x-3)(2x-3) = 4x²-6x-6x+9 = 4x²-12x+9

(5x-9)(2x²-3x+4) = 10x³-18x²-15x²+27x+20x-36 = 10x³-33x²+47x-36

x³x⁴ = x⁷

x⁹/x⁴ = x⁵

(x⁷)⁶ = x⁴²

x²x³ = (x⋅x)(x⋅x⋅x) = x⁵

(x²)³ = (x²)(x²)(x²) = (x⋅x)(x⋅x)(x⋅x) = x⁶

x⁵/x² = (x⋅x⋅x⋅x⋅x)/(x⋅x) = x⋅x⋅x = x³

x⁴/x⁷ = x^-3 = (x⋅x⋅x⋅x)/(x⋅x⋅x⋅x⋅x⋅x⋅x) = 1/(x⋅x⋅x) = 1/x³

(3x⁴y⁵)(5x⁶y⁷) = 15x¹⁰y¹²

(8x³y^-2)(7x^-8y⁵) = 56x^-5y³ = 56y³/x⁵

(24x⁷y^-2)/(6x⁴y⁵) = 2³⋅3x³y^-7/2⋅3 = 4x³/y⁷

(32x⁵y^-3z⁴)/(40x^-8y^-7z^-8) = (2⁵x¹³y⁴z¹²)/(2³5) = 2²x¹³y⁴z¹²/5 = 4x¹³y⁴z¹²/5

(3x³)² = 3²x⁶ = 9x⁶

(5x)² = 5²x² = 25x²

(5+x)² = (5+x)(5+x) = 25+5x+5x+x² = x²+10x+25

(4x²y³)³ = 4³x⁶y⁹ = 64x⁶y⁹

(8x²y⁵z⁶)⁰ = 1

-2(5xy³)⁰ = -2⋅1 = -2

(5x^-2/y^-3)(8x⁴/y^-5) = (5y³/x²)(8x⁴y⁵) = 40x²y⁸

((35x^-3)/(40xy⁵))((24x²y²)/(42y^-4)) = (5⋅7/2³⋅5x⁴y⁵)(2³⋅3x²y⁶)/(2⋅3⋅7) = y/(2x²)

((24xy)/(27x^-2))/((36x²y^-3)/(45xy⁴)) = ((2³⋅3x³y)/(3³))((3²5xy⁷)/(2²3²x²)) = (2⋅5x²y⁸)/(3²) = 10x²y⁸/9

(3x/5)/(7xy/9) = (3x/5)(9/(7xy)) = 27/(35y)

(7+2/x)/(5-3/y) = ((7+2/x)xy)/((5-3/y)xy) = (7xy+2y)/(5xy-3x) = (y(7x+2))/(x(5y-3))

x+4 = 9, x = 5; verify: 5+4 = 9, 9 = 9

3x+5 = 11, 3x = 6, x = 6/3, x = 2; verify: 3⋅2+5 = 11, 6+5 = 11, 11 = 11

2(x-1)+6 = 10, 2(x-1) = 4, x-1 = 2, x = 3; verify: 2(3-1)+6 = 10, 2⋅2+6 = 10, 4+6 = 10, 10 = 10

5-3(x+4) = 7+2(x-1), -3(x+4) = 2+2(x-1), -3x-12 = 2+2x-2, -3x-12 = 2x, -5x = 12, x = -12/5; verify: 5-3(-12/5+4) = 7+2(-12/5-1), -3(-12/5+4) = 2+2(-12/5-1), -3((-12+20)/5) = 2+2((-12-5)/5), -3(8/5) = 2+2(-17/5), -24/5 = 2-34/5, -24/5 = (10-34)/5, -24/5 = -24/5

(2/3)x+5 = 8, (2/3)x = 3, 2x = 9, x = 9/2; verify: (2/3)(9/2)+5 = 8, 3+5 = 8, 8 = 8

(3/4)x-1/3 = 1, 12((3/4)x-1/3) = 12, 9x-4 = 12, 9x = 16, x = 16/9; verify: (3/4)(16/9)-1/3 = 1, 4/3-1/3 = 1, 3/3 = 1, 1 = 1

(x+2)/5 = 7/8, 8(x+2) = 7⋅5, 8x+16 = 35, 8x = 19, x = 19/8; verify: ((19/8)+2))/5 = 7/8, (35/8)(1/5) = 7/8, 7/8 = 7/8

0.04x+0.15 = 0.09x-0.25, 100(0.04x+0.15) = (0.09x-0.25)100, 4x+15 = 9x-25, -5x = -40, x = 40/5, x = 8; verify: 0.04⋅8+0.15 = 0.09⋅8-0.25, 0.32+0.15 = 0.72-0.25, 0.47 = 0.47

x²-25 = 0, x² = 25, √(x²) = √(25), x = ±5, x = 5 or x = -5; x²-25 = 0, (x+5)(x-5) = 0, x+5 = 0, x = -5, or x-5 = 0, x = 5

2x²-18 = 0, 2x² = 18, x² = 18/2, x² = 9, √(x²) = √(9), x = ±3, x = 3 or x = -3; 2x²-18 = 0, 2(x²-9) = 0, 2(x+3)(x-3) = 0, x+3 = 0, x = -3, or x-3 = 0, x = 3

3x²-48 = 0, 3x² = 48, x² = 48/3, x² = 16, √(x²) = √(16), x = ± 4, x = 4 or x = -4; 3x²-48 = 0, 3(x²-16) = 0, 3(x+4)(x-4) = 0, x+4 = 0, x = -4, or x-4 = 0, x = 4

x⁴-81 = 0, (x²+9)(x²-9) = 0, (x²+9)(x+3)(x-3), x+3 = 0, x = -3 or x-3 = 0, x = 3, the real solutions are x = 3 or x = -3; √(-9) = ±3i

x²-5x+6 = 0, (x-3)(x-2) = 0, x-3 = 0, x = 3, or x-2 = 0, x = 2

x²-2x-15 = 0, (x+3)(x-5) = 0, x+3 = 0, x = -3, or x-5 = 0, x = 5

x²+3x-28 = 0, (x+7)(x-4) = 0, x+7 = 0, x = -7, or x-4 = 0, x = 4

2x²+3x-2 = 0, 2x²+4x-x-2 = 0, 2x(x+2)-1(x+2) = 0, (x+2)(2x-1) = 0, x+2 = 0, x = -2, or 2x-1 = 0, 2x = 1, x = 1/2; 2x²+3x-2 = 0, ax²+bx+c = 0, x = (-b±√(b²-4ac))/(2a), x = (-3±√(3²-4⋅2⋅-2))/(2⋅2), x = (-3±√(9+16))/4, x = (-3±√(25))/4, x = (-3±5)/4, x = (-3+5)/4, x = 2/4 = 1/2, or x = (-3-5)/4, x = -8/4 = -2

6x²+7x-3 = 0, 6x²+9x-2x-3 = 0, 3x(2x+3)-1(2x+3) = 0, (2x+3)(3x-1) = 0, 2x+3 = 0, 2x = -3, x = -3/2, or 3x-1 = 0, 3x = 1, x = 1/3; 6x²+7x-3 = 0, ax²+bx+c = 0, x = (-b±√(b²-4ac))/(2a), x = (-7±√(7²-4⋅6⋅-3))/(2⋅6), x = (-7±√(49+72))/12, x = (-7±√(121))/12, x = (-7±11)/12, x = (-7+11)/12 = 4/12 = 1/3, or x = (-7-11)/12 = -18/12 = -3/2

x³-4x²-x+4 = 0, x²(x-4)-1(x-4) = 0, (x-4)(x²-1) = 0, (x-4)(x+1)(x-1), x-4 = 0, x = 4, or x+1 = 0, x = -1, or x-1 = 0, x = 1

Slope-intercept form of a linear equation: y = mx+b, m is the slope that is rise/run, b is the y-intercept that is the point 0,b

y = 2x-1; y = mx+b, m = 2, b = -1

y = (3/4)x-2; y = mx+b, m = 3/4, b = -2

Standard form of a linear equation: ax+by = c; y = 0, ax = c, x = c/a, x-intercept = (c/a,0); x = 0, by = c, y = c/b, y-intercept = (0,c/b)

2x-3y = 6; y = 0, 2x = 6, x = 6/2 = 3, x-intercept = (3,0); x = 0, -3y = 6, y = -6/3 = -2 = y-intercept = (0,-2)

Point-slope form of a linear equation: y-y₁ = m(x-x₁)

Find the equation of the line knowing the angular coefficient and a point: m = 2, P(1,3); y-y₁ = m(x-x₁), y-3 = 2(x-1) is the solution in point-slope form; y-3 = 2(x-1), y-3 = 2x-2, y = 2x-2+3 = 2x+1, y = 2x+1 is the solution in slope-intercept form; y = 2x+1, -2x+y = 1 is the solution in standard form

Find the equation of the line passing through two points: P₁(2,4), P₂(-1,5); m = (y₂-y₁)/(x₂-x₁) = (5-4)/(-1-2) = 1/-3 = -1/3, m = -1/3; y-y₁ = m(x-x₁), y-4 = -1/3(x-2) is the solution in point-slope form; y-4 = -1/3(x-2), y-4 = (-1/3)x+2/3, y = (-1/3)x+2/3+4 = (-1/3)x+2/3+12/3 = (-1/3)x+14/3, y = (-1/3)x+14/3 is the solution in slope-intercept form; y = (-1/3)x+14/3, 3y = 3((-1/3)x+14/3), 3y = -x+14, x+3y = 14 is the solution in standard form

Find the equation of the line passing through the point P(1,3) and parallel to 2x-3y-5 = 0; 2x-3y-5 = 0, -3y = -2x+5, y = (-2/-3)x+5/-3 = (2/3)x-5/3, m = 2/3; y-y₁ = m(x-x₁), y-3 = 2/3(x-1) is the solution in point-slope form; y-3 = 2/3(x-1), y-3 = (2/3)x-2/3, y = (2/3)x-2/3+3 = (2/3)x+7/3, y = (2/3)x+7/3 is the solution in slope-intercept form

Find the equation of the line passing through the point P(-2,1) and perpendicular to 3x+2y-7 = 0; 3x+2y-7 = 0, 2y = -3x+7, y = (-3/2)x+7/2, the slope of the perpendicular line is the negative reciprocal of the slope of the starting line, so m = 2/3; y-y₁ = m(x-x₁), y-1 = (2/3)(x+2) is the solution in point-slope form; y-1 = (2/3)(x+2), y-1 = (2/3)x+4/3, y = (2/3)x+4/3+1 = (2/3)x+7/3, y = (2/3)x+7/3 is the solution in point-intercept form; point of intersection of the two perpendicular lines: y = (-3/2)x+7/2, y = (2/3)x+7/3, (-3/2)x+7/2 = (2/3)x+7/3, (-3/2)x-(2/3)x = 7/3-7/2, 6((-3/2)x-(2/3)x) = 6(7/3-7/2 ), -9x-4x = 14-21, -13x = -7, x = -7/-13 = 7/13, x = 7/13 = 0,538..., y = (2/3)x+7/3, y = (2/3)(7/13)+7/3 = 14/39+7/3 = (14+91)/39 = 105/39 = 35/13 = 2,692..., the point of intersection is P(7/13,35/13)

x > 2; (2,+∞)

x ≥ -1; [-1,+∞)

x < 4; (-∞,4)

x ≤ -2; (-∞,-2]

1 < x ≤ 4; (1,4]

x < -2 or x ≥ 3; (-∞,-2)∪[3,+∞)

x+4 > 5; x > 5-4, x > 1, (1,∞)

3x-5 < -8; 3x < -8+5, 3x < -3, x < -3/3, x < -1, (-∞,-1)

7-2x ≤ 12; -2x ≤ 12-7, -2x ≤ 5, x ≥ 5/(-2), x ≥ -5/2, [-5/2,+∞)

4-2x ≥ 3x+19; -2x-3x ≥ 19-4, -5x ≥ 15, x ≤ 15/(-5), x ≤ -3, (-∞,-3]

5-2(x-3) > 3+4(2x-1); 5-2x+6 > 3+8x-4, -2x+11 > 8x-1, -8x-2x > -1-11, -10x > -12, x < -12/-10, x < 12/10, x < 6/5, (-∞,6/5)

(3/2)x < -9; 2(3/2)x < -9⋅2, 3x < -18, x < -18/3, x < -6, (-∞,-6)

(1/3)x+4 ≥ 8; (1/3)x ≥ 8-4, (1/3)x ≥ 4, 3(1/3)x ≥ 4⋅3, x ≥ 12, [12,+∞)

(5/4)x-2 ≤ (3/2)x+1/3; 12((5/4)x-2) ≤ ((3/2)x+1/3)12, 15x-24 ≤ 18x+4, 15x-18x ≤ 4+24, -3x ≤ 28, x ≥ 28/-3, x ≥ -28/3, [-28/3,+∞)

(2/3)x+5 > (1/5)x+3/4; 60((2/3)x+5) > 60((1/5)x+3/4), 40x+300 > 12x+45, 40x-12x > 45-300, 28x > -255, x > -255/28, (-255/28,+∞)

|x| < 4; x < 4, x> -4, (-4,4)

|2x-3| ≥ 8; 2x-3 ≥ 8, 2x ≥ 8+3, 2x ≥ 11, x ≥ 11/2; 2x-3 ≤ -8, 2x ≤ -8+3, 2x ≤ -5, x ≤ -5/2; x ≥ 11/2 or x ≤ -5/2, (-∞,-5/2]∪[11/2,+∞)

5-3|4x+1| ≥ -9; -3|4x+1| ≥ -9-5, -3|4x+1| ≥ -14, |4x+1| ≤ -14/-3, |4x+1| ≤ 14/3; 4x+1 ≤ 14/3, 4x ≤ 14/3-1, 4x ≤ (14-3)/3, 4x ≤ 11/3, x ≤ 11/3⋅4, x ≤ 11/12; 4x+1 ≥ -14/3, 4x ≥ -14/3-1, 4x ≥ -17/3, x ≥ -17/12; x ≤ 11/12 or x ≥ -17/12, [-17/12,11/12], -17/12 ≤ x ≤ 11/12

-3 ≤ 2x+5 < 19; -3-5 ≤ 2x+5-5 < 19-5, -8 ≤ 2x < 14, -8/2 ≤ 2x/2 < 14/2, -4 ≤ x < 7, [-4,7)

x²-x ≥ 12; x²-x-12 ≥ 0, (x-4)(x+3) ≥ 0, x-4 = 0, x = 4, x+3 = 0, x = -3; if x = 5 then (5-4)(5+3) = 1⋅8 = 8 > 0; if x = 0 then (0-4)(0+3) = -4⋅3 = -12 < 0; if x=-4 then (-4-4)(-4+3)=-8⋅-12=96> 0; x ≤ -3 or x ≥ 4, (-∞,-3]∪[4,+∞)

x²+9 > 6x; x²-6x+9 > 0, (x-3)(x-3) > 0, (x-3)² > 0, x-3 = 0, x = 3; if x = 4 then (4-3)² = 1² = 1 > 0; if x = 2 then (2-3)² = -1² = 1 > 0; x < 3 or x> 3, (-∞,3)∪(3,+∞)

2x² > x+6; 2x²-x-6 > 0, 2x²-4x+3x-6 > 0, 2x(x-2)+3(x-2) > 0, (x-2)(2x+3) > 0, x-2 = 0, x = 2, 2x+3 = 0, 2x = -3, x = -3/2; if x = -2 then (-2-2)(2⋅-2+3) = -4(-4+3) = -4⋅-1 = 4 > 0; if x = 0 then (0-2)(2⋅0+3) = -2(0+3) = -2⋅3 = -6 < 0; if x=3 then (3-2)(2⋅3+3)=1(6+3)=1⋅9=9>0; x < -3/2 or x> 2, (-∞,-3/2)∪(2,+∞); if 2x² < x+6 then -3/2 < x < 2, (-3/2,2)

FUNCTIONS EXERCISES

Find the range of the linear function y = x+3; the range of any linear function is R(-∞,+∞)

y = x²; R[0,+∞); when the leading term has an even power, the range is limited

y = x²-3; R[-3,+∞); the coefficient of the leading term is positive, therefore the parabola opens upwards

y = 4-x²; R(-∞;4]; the coefficient of the leading term is negative, therefore the parabola opens downwards

y = x²-4x+5; calculating the coordinates of the vertex of the parabola: x_v = -b/2a = -(-4)/2⋅1 = 4/2 = 2, y_v = 2²-4⋅2+5, y_v = 4-8+5 = 1, vertex(2,1); the coefficient of the leading term is positive, therefore the parabola opens upwards; R[1,+∞)

y = x³; R(-∞,+∞)

y = x³+5x²-8; for any cubic polynomial the range is always R(-∞,+∞)

y = x⁵-x³+6x²+8; when the function is a polynomial and the leading term has an odd exponent, the range is always R(-∞,+∞); polynomial functions are continuous functions

y = |x|; the v shape opens upwards because the absolute value function is positive; R[0,+∞)

y = |x|+3; the graph is shifted up by 3 and the v shape opens upwards because the absolute value function is positive; R[3,+∞)

y = |x-2|-3; the graph is shifted right by 2 and down by 3 and the v shape opens upwards because the absolute value function is positive; R[-3,+∞)

y = 2-|x-3|; the graph is shifted up by 2 and right by 3 and the v shape opens downwards because the absolute value function is negative; R(-∞,2]

y = √(x); the graph is in the first quadrant, x ≥ 0 and y ≥ 0; R[0,+∞)

y = -√(x); the graph is in the fourth quadrant, x ≥ 0 and y ≤ 0; R(-∞,0]

y = √(-x); the graph is in the second quadrant, x ≤ 0 and y ≥ 0; R[0,+∞)

y = -√(-x); the graph is in the third quadrant, x ≤ 0 and y ≤ 0; R(-∞,0]

y = -√(x-3)+4; the graph is shifted up by 4 and right by 3; P₁(3,4), P₂(4,3), P₃(7,2); R(-∞,4]

SET

In mathematics, a set is a well-defined collection of distinct elements

Sets are conventionally denoted with capital letters such as A, B, C

The elements of a set are defined with lowercase letters such as a, b, c

= : equal sign; equality

≠ : not equal sign; inequality

∈ : element of; belongs to; set membership

∉ : not element of; no set membership

a ∈ A : "a is an element of A" or "a belongs to A"

∃ : exists

∄ : does not exist

! : unique

∃! : exists and it is unique

⇒ : implies

P ⇒ Q : P implies Q; if P is true implies that Q is true

⇔ : equivalent; if and only if (iff)

P ⇔ Q : if P is true then Q is true and vice versa

⊥ : contradiction

↯ : contradiction

P:= {n ∈ N | n is a prime number} : representation of a set by property

⊆ : subset

A ⊆ B : subset; A is a subset of B; set A is included in set B; {9,14,28} ⊆ {9,14,28}

⊂ : proper subset

A ⊂ B : proper subset / strict subset; A is a subset of B, but A is not equal to B; {9,14} ⊂ {9,14,28}

⊄ : not subset

A ⊄ B : not subset; set A is not a subset of set B; {9,66} ⊄ {9,14,28}

⊇ : superset

A ⊇ B : superset; A is a superset of B; set A includes set B; {9,14,28} ⊇ {9,14,28}

A ⊃ B : proper superset / strict superset; A is a superset of B, but B is not equal to A; {9,14,28} ⊃ {9,14}

⊅ : not superset

A ⊅ B : not superset; set A is not a superset of set B; {9,14,28} ⊅ {9,66}

A = {1,3,7,a} : representation of a set by enumeration

|A| : cardinality; the number of elements of set A

#A : cardinality; the number of elements of set A

Sets can be represented with diagrams of Euler and Venn

David Hilbert (1862 - 1943) was a German mathematician and one of the most influential and universal mathematicians of the 19th and early 20th centuries

Bertrand Russell (1872 - 1970) was a British polymath, philosopher, logician, mathematician, historian, writer, social critic, political activist, and Nobel laureate

Russell's paradox: A = {X | X ∉ X}; A ∈ A ⇒ A ∉ A : ↯; A ∉ A ⇒ A ∈ A : ↯

Kurt Gödel (1906 - 1978) was a German-Austrian logician, mathematician, and analytic philosopher. Considered along with Aristotle and Gottlob Frege to be one of the most significant logicians in history

Gödel's incompleteness theorems are two theorems of mathematical logic that demonstrate the inherent limitations of every formal axiomatic system capable of modelling basic arithmetic. These results, published by Kurt Gödel in 1931, are important both in mathematical logic and in the philosophy of mathematics. The theorems are widely, but not universally, interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all mathematics is impossible

{ } : set; a collection of elements; A = {3,7,9,14}, B = {9,14,28}

∪ : union

A ∪ B : union; objects that belong to set A or set B; A ∪ B = {3,7,9,14,28}

A ∪ B = {c | c ∈ A or c ∈ B}

∩ : intersection

A ∩ B : intersection; objects that belong to set A and set B; A ∩ B = {9,14}

A ∩ B = {c | c ∈ A and c ∈ B}

A \ B : relative complement; objects that belong to A and not to B; A = {3,9,14}, B = {1,2,3}, A \ B = {9,14}

A - B : relative complement; objects that belong to A and not to B; A = {3,9,14}, B = {1,2,3}, A - B = {9,14}

A \ B = {a ∈ A | a ∉ B}

A × B : cartesian product; set of all ordered pairs from A and B; A × B = {(a,b) | a ∈ A , b ∈ B}; (a,b) ≠ (b,a)

A ∪ B = B ∪ A : on the union of sets the commutative property is valid

A ∩ B = B ∩ A : on the intersection of sets the commutative property is valid

A \ B ≠ B \ A : on the difference of sets the commutative property is not valid

A x B ≠ B x A : on the cartesian product of sets the commutative property is not valid

A = {1,2,3}, B = {3,5}; A ∪ B = {1,2,3,5}; A ∩ B = {3}; A \ B = {1,2}; B \ A = {5}; A x B = {(1,3),(1,5),(2,3),(2,5),(3,3),(3,5)}

Card(A) = n, Card(B) = m, Card(A x B) = nm

A x A = A²

A x B x C = {(a,b,c) | a ∈ A, b ∈ B, c ∈ C}

A x A x ... x A = Aⁿ

NUMERICAL SETS

Natural numbers set: ℕ = {0,1,2,3...}

Integer numbers set: ℤ = {0,1,-1,2,-2,3,-3...}

Rational numbers set: ℚ = {p/q | p,q ∈ ℤ, q ≠ 0}

Real numbers set: ℝ = {p,α₀,α₁,α₂... | p ∈ ℤ, α_i ∈ {0,1,2...9}i, i ∈ ℕ}

ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

Irrational numbers set: ℝ \ ℚ

Examples of irrational numbers are: π, e, √2, √3

In mathematics, a field is a set on which addition, subtraction, multiplication, and division are defined and behave as the corresponding operations on rational and real numbers do

Informally, a field is a set, along with two operations defined on that set: an addition operation written as a + b, and a multiplication operation written as a ⋅ b, both of which behave similarly as they behave for rational numbers and real numbers, including the existence of an additive inverse −a for all elements a, and of a multiplicative inverse b⁻¹ for every nonzero element b

Subtraction: a − b = a + (−b)

Division: a / b = a · b⁻¹

Formally, a field is a set F together with two binary operations on F called addition and multiplication

A binary operation on F is a mapping F × F → F, that is, a correspondence that associates with each ordered pair of elements of F a uniquely determined element of F

The result of the addition of a and b is called the sum of a and b, and is denoted a + b

The result of the multiplication of a and b is called the product of a and b, and is denoted ab or a ⋅ b

Addition and multiplication must satisfy some properties called field axioms: commutativity, associativity, identity, inverse, distributivity

Commutativity of addition: a + b = b + a, ∀ a,b ∈ F

Commutativity of multiplication: a · b = b · a, ∀ a,b ∈ F

Associativity of addition: a + (b + c) = (a + b) + c, ∀ a,b,c ∈ F

Associativity of multiplication: a · (b · c) = (a · b) · c, ∀ a,b,c ∈ F

Additive identiy: a + 0 = a, ∀ a ∈ F

Multiplicative identity: a · 1 = 1, ∀ a ∈ F

Additive inverse: a + (-a) = 0, ∀ a ∈ F

Multiplicative inverse: a · a^-1 = 1, ∀ a ≠ 0 ∈ F

Distributivity of multiplication over addition: a · (b + c) = a·b + a·c, ∀ a,b,c ∈ F

(ℚ, +, ·) is a field

(ℚ, +, ·, ≤) is an ordered field

∀ a,b ∈ ℚ, a ≤ b

(ℝ, +, ·) is a field

(ℝ, +, ·, ≤) is an ordered field

∀ a,b ∈ ℝ, a ≤ b

The real numbers can be defined synthetically as an ordered field satisfying the completeness axiom

Intuitively, completeness implies that there are not any “gaps” (in Dedekind's terminology) or “missing points” in the real number line; this contrasts with the rational numbers, whose corresponding number line has a “gap” at each irrational value

INTERVAL

In mathematics, a (real) interval is a set of real numbers that contains all real numbers lying between any two numbers of the set

Open interval of extremes a and b: (a,b) = {x ∈ ℝ | a < x < b}

Closed interval of extremes a and b: [a,b] = {x ∈ ℝ | a ≤ x ≤ b}

Left-closed and right-open interval: [a,b) = {x ∈ ℝ | a ≤ x < b}

Left-open and right-closed interval: (a,b] = {x ∈ ℝ | a < x ≤ b}

(a,+∞) = {x ∈ ℝ | a < x}

[a,+∞) = {x ∈ ℝ | a ≤ x}

(-∞,b) = {x ∈ ℝ | x < b}

(-∞,b] = {x ∈ ℝ | x ≤ b}

(a,a) = {x ∈ ℝ | a < x < a} = Ø

[a,a] = {x ∈ ℝ | a ≤ x ≤ a} = {a}

MAJORANT - MINORANT - SUPREMUM - INFIMUM - MAXIMUM - MINIMUM

In mathematics, particularly in order theory, an upper bound or majorant of a subset S of some preordered set (K, ≤) is an element of K which is greater than or equal to every element of S

Dually, a lower bound or minorant of S is defined to be an element of K which is less than or equal to every element of S

A set with an upper (respectively, lower) bound is said to be bounded from above or majorized (respectively bounded from below or minorized) by that bound

The terms bounded above (bounded below) are also used in the mathematical literature for sets that have upper (respectively lower) bounds

A ⊆ ℝ, A ≠ Ø, s ∈ ℝ, s = upper bound or majorant, a ≤ s ∀ a ∈ A

A ⊆ ℝ, A ≠ Ø, s ∈ ℝ, s = lower bound or minorant, a ≥ s ∀ a ∈ A

A = [1,3) = {x ∈ ℝ | 1 ≤ x < 3}, 3 = upper bound or majorant, 3 ≥ a ∀ a ∈ A

A = [-1,4] = {x ∈ ℝ | -1 ≤ x ≤ 4}, -2 = lower bound or minorant, -2 ≤ a ∀ a ∈ A

A = [-1,4] = {x ∈ ℝ | -1 ≤ x ≤ 4}, -1 = lower bound or minorant, -1 ≤ a ∀ a ∈ A

A = ℕ, ∀ s ∈ ℝ, ∃ n ∈ ℕ | s < n ⇒ ℕ does not have a majorant or upper bound

The minorant or lower bound of ℕ is 0

Every subset of the natural numbers has a lower bound since the natural numbers have a least element (0 or 1, depending on convention)

An infinite subset of the natural numbers cannot be bounded from above

ℤ is not bounded above or below

ℤ \ ℕ is limited above but not below; 0 is a majorant or upper bound

(-∞,1] is limited above but not below

Definition of majorant or upper bound: A ⊆ ℝ, s ∈ ℝ, s ≥ a ∀ a ∈ A

Definition of minorant or lower bound: A ⊆ ℝ, s ∈ ℝ, s ≤ a ∀ a ∈ A

A = [0,1), s ≥ 1 = majorant or upper bound of A, s < 1 ≠ majorant or upper bound of A

A = (-1,5], s ≤ -1 = minorant or lower bound of A, s > -1 ≠ minorant or lower bound of A

A ⊆ ℝ, A ≠ Ø, S ∈ ℝ is the supremum of A if it is the least upper bound of A

S = sup A ⇔ {S ≥ a ∀ a ∈ A; ∀ ε > 0, ∃ a ∈ A | S - ε < a}

S = sup A ⇔ {S ≥ a ∀ a ∈ A (S = majorant or upper bound of A); ∀ ε > 0, ∃ a ∈ A | S - ε < a (S - ε ≠ majorant or upper bound of A)}

A ⊆ ℝ, A ≠ Ø, s ∈ ℝ is the infimum of A if it is the greatest lower bound of A

s = inf A ⇔ {s ≤ a ∀ a ∈ A; ∀ ε > 0, ∃ a ∈ A | s + ε > a}

s = inf A ⇔ {s ≤ a ∀ a ∈ A (s = minorant or lower bound of A); ∀ ε > 0, ∃ a ∈ A | s + ε > a (s + ε ≠ minorant or upper bound of A)}

Supremum is the least upper bound; infimum is the greatest lower bound

Definition of supremum: A ⊆ ℝ, A ≠ Ø, s = sup A ⇔ {s ≥ a ∀ a ∈ A; ∀ ε > 0 ∃ a ∈ A | s - ε < a}

Definition of infimum: A ⊆ ℝ, A ≠ Ø, i = inf A ⇔ {i ≤ a ∀ a ∈ A; ∀ ε > 0 ∃ a ∈ A | i + ε > a}

A = [0,1) = {x ∈ ℝ | 0 ≤ x < 1}; sup A = 1 (1 = majorant or upper bound); ∀ ε > 0 ∃ x ∈ [0,1) | 1 - ε < x (1 = least upper bound = supremum)

A = [0,1) = {x ∈ ℝ | 0 ≤ x < 1}; inf A = 0 (0 = minorant or lower bound); ∀ ε > 0 ∃ x ∈ [0,1) | 0 + ε > x (0 = greatest lower bound = infimum)

If the supremum belongs to the set, then it is the maximum of the set

A ⊆ ℝ, A ≠ Ø, if sup A ∈ A, then sup A = max A

If the infimum belongs to the set, then it is the minimum of the set

A ⊆ ℝ, A ≠ Ø, if inf A ∈ A, then inf A = min A

Intuitively, completeness implies that there are not any "gaps" (in Dedekind's terminology) or "missing points" in the real number line; this contrasts with the rational numbers, whose corresponding number line has a “gap” at each irrational value

In mathematics, the least-upper-bound property (sometimes called completeness or supremum property) is a fundamental property of the real numbers

An ordered set X has the least-upper-bound property if every non-empty subset of X with an upper bound has a least upper bound (supremum) in X

Not every ordered set has the least upper bound property; for example, the set ℚ of all rational numbers with its natural order does not have the least upper bound property

Completeness of ℝ Theorem: ℝ is complete because every upper-bounded subset of ℝ admits a supremum, and every lower-bounded subset of ℝ admits an infimum

Completeness of ℝ Theorem: ℝ is complete because every upper-bounded subset of ℝ admits a supremum (that is the least upper bound), and every lower-bounded subset of ℝ admits an infimum (that is the greatest lower bound)

The subset A is upper-bounded if it admits an upper bound or majorant

The subset A is lower-bounded if it admits a lower bound or minorant

The property of completeness is true in ℝ, but is false in ℚ

A = {x ∈ ℝ | x² < 2} = (-√2,√2); sup A = √2, inf A = -√2; √2 ∈ ℝ \ ℚ

A = {x ∈ ℚ | x² < 2} = (-√2,√2) ∩ ℚ; sup A ≠ √2 (√2 ∉ ℚ); inf A ≠ -√2 (-√2 ∉ ℚ)

Density of ℚ in ℝ: ∀ x,y ∈ ℝ, x < y, ∃ q ∈ ℚ | x < q < y

Considering the property of the rational numbers, the density of ℚ in ℝ, it is impossible to find a supremum or an infimum, because there is always a rational number between two real numbers; there is no completeness in ℚ

The property of completeness is false in ℚ; in the set of rational numbers the sumpremum (the least upper bound or least majorant) does not exist, and the infimum (the greatest lower bound or greatest minorant) does not exist

(ℚ, +, ⋅, ≥); ℚ is not complete because every subset of ℚ does not admit a supremum or an infimum

(ℝ, +, ⋅, ≥); ℝ is complete because every subset of ℝ admits a supremum or an infimum

If A ≠ Ø and it is upper-bounded, S = Sup A always exists (completeness); if S ∈ A, then S = sup A = max A

A = (1,2), sup A = 2 and 2 ∉ A then ∄ max A, inf A = 1 and 1 ∉ A then ∄ min A

If A is not upper-bounded, sup A = +∞

If A is not lower-bounded, inf A = -∞

Sup ℕ = +∞

Inf ℤ = -∞

Sup ℤ = +∞

A = [-π,3] ∪ {0} ∪ [1,+∞], inf A = -π = min A, sup A = +∞

A = {1/n | n ∈ ℕ \ {0}}; A = {1\n | n ∈ ℕ}; sup A = max A = 1; inf A = 0; 0 ≤ 1/n ∀ n ∈ ℕ \ {0}; ∀ ε > 0 ∃ n ∈ ℕ | 0 + ε > 1/n; ε > 1/n ⇔ n > 1/ε; if 0 = min A ⇔ 0 ∈ A ⇔ ∃ n ∈ ℕ | 0 = 1/n and it does not admit solutions; 0 ≠ 1/n

A = {1 - 1/n | n ∈ ℕ \ {0}}; {0, 1/2, 2/3, 3/4, 4/5...}; 0 ≤ 1 - 1/n ≤ 1; inf A = 0, 0 ∈ A, min A = 0; sup A = 1, 1 - 1/n ≤ 1 ∀ n ∈ ℕ (1 is an upper bound, a majorant), ∀ ε > 0 ∃ n ∈ ℕ \ {0} | 1 - ε < 1 - 1/n (1 is the least upper bound, the supremum, ∀ ε > 0 ∃ a ∈ A | S - ε < a), 1 - ε < 1 - 1/n ⇔ - ε < - 1/n ⇔ ε > 1/n ⇔ n > 1/ε, sup A = 1; if max A = 1 ⇔ ∃ n ∈ ℕ \ {0} | 1 - 1/n = 1 ⇔ - 1/n = 0 (never)

A = {n + 2/n | n ∈ ℕ}; 0 ≤ n + 2/n (0 is a minorant or lower bound); n + 2/n ≥ n ∀ n ∈ ℕ, sup A = +∞ (no majorant or upper bound); n = 1: 1 + 2/1 = 3; n = 2: 2 + 2/2 = 3; n = 3: 3 + 2/3 > 3; n = 4: 3 + 2/4 > 3; ∀ n ≥ 3, n + 2/n ≥ n ≥ 3; min A = 3; A = {n + 2/n | n = 1, n = 2} ∪ {n + 2/n | n ≥ 3} = {3} ∪ {n + 2/n | ∀ n ≥ 3}

FACTORIAL

The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n: n! = n⋅(n-1)⋅(n-2)⋅(n-3)⋅...⋅3⋅2⋅1

The value of 0! is 1, according to the convention for an empty product

The factorial counts the possible distinct sequences (the permutations) of n distinct objects

Factorial: n ∈ ℕ, n! = {1 (n=0); 1⋅2⋅3⋅n (n≠0)}

3! = 1⋅2⋅3 = 6

5! = 1⋅2⋅3⋅4⋅5 = 120

n! is the number of possible orders of n objects

3! = 6; abc, acb, bac, bca, cab, cba = 6 possible distinct sequences called permutations

In general, n objects can be sorted into n! different ways

BINOMIAL COEFFICIENT

Binomial coefficient: n, k ∈ ℕ | n ≥ k ≥ 0; C(n, k) = n!/k!(n-k)!

The binomial coefficient expresses the number of unordered subset of k elements from a fixed set of n elements

C(3, 2) = 3!/2!(3-2)! = 6/2 = 3

C(5, 2) = 5!/2!(5-2)! = 120/12 = 10

The Binomial Coefficient C(n, k) represents the number of subsets of k objects of the n given objects

C(3, 2) = 3!/2!(3-2)! = 6/2 = 3; {a, b, c}: {a, b}, {a, c}, {b, c}; For 3 objects there are 3 subsets of 2 objects

C(90, 6) = 90!/6!(90-6)! = 90!/6!⋅84! = 622.614.630; 1/C(90, 6) = 10^-9

Binomial Coefficient Properties: C(n, k) ∈ ℕ, ∀ n, k ∈ ℕ; C(n, 0) = C(n, n) = 1; C(n, k) = C(n, n-k); C(n, k) = C(n-1, k) + (n-1, k-1)

1

1, 1

1, 2, 1

1, 3, 3, 1

1, 4, 6, 4, 1

1, 5, 10, 10, 5, 1

1, 6, 15, 20, 15, 6, 1

1, 7, 21, 35, 35, 21, 7, 1

1, 8, 28, 56, 70, 56, 28, 8, 1

SUMMATION

Summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total

For simple patterns, summation of long sequences may be represented with most summands replaced by ellipses; summation of the first 100 natural numbers may be written as 1 + 2 + 3 + 4 + ⋅⋅⋅ + 99 + 100

Summation can be described using the symbol Σ, the capital Greek letter sigma

The sum of the first n natural integers can be denoted as ⁿΣ_i=1 i

ⁿΣ_i=m a_i = a_m + a_m+1 + a_m+2 + ... + a_n-1 + a_n; where i is the index of summation; a_i is an indexed variable representing each term of the sum; m is the lower bound of summation, and n is the upper bound of summation; the index starts from m and is incremented by one for each successive term, stopping when i = n

1²+2²+3²+...+100² = ¹⁰⁰Σ_k=1 k²

{a_i | a_i ∈ ℝ, i ∈ {1,...,N}}, ^NΣ_i=1 a_i = a₁ + a₂ + a₃ + a_N

^NΣ_i=1 a_i = ^NΣ_k=1 a_k

⁶Σ_k=3 k³ = 3³ + 4³ + 5³ + 6³

⁶Σ_i=3 i³ = 3³ + 4³ + 5³ + 6³

¹⁰Σ_k=1 1/k = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10

^NΣ_k=1 c ⋅ a_k = c ⋅ ^NΣ_k=1 a_k (c ∈ ℝ)

c ⋅ a₁ + c ⋅ a₂ + ... + c ⋅ a_n = c ⋅ (a₁ + a₂ + ... + a_n)

^NΣ_k=1 a_k = ^MΣ_k=1 a_k + ^NΣ_M+1 a_k, M,N ∈ ℕ | M < N

^NΣ_k=1 (a_k + b_k) = ^NΣ_k=1 a_k + ^NΣ_k=1 b_k

BINOMIAL THEOREM

Newton's binomial formula: a,b ∈ ℝ, n ∈ ℕ, (a+b)ⁿ = ⁿΣ_k=0 C(n,k)a^n-kb^k

Newton's binomial formula for n = 2: (a+b)² = ²Σ_k=0 C(2,k)a^2-kb^k = C(2,0) a²b⁰ + (C2,1) ab + C(2,2) a⁰b² = a² + 2ab + b²

1

1 1

1 2 1

MATHEMATICAL INDUCTION

Mathematical induction is a mathematical proof technique; it is essentially used to prove that a statement P(n) holds for every natural number n = 0, 1, 2, 3, ...

A proof by induction consists of two cases: the first, the base case (or basis), proves the statement for n=0 without assuming any knowledge of other cases; the second case, the induction step, proves that if the statement holds for any given case n=k, then it must also hold for the next case n=k+1; these two steps establish that the statement holds for every natural number n

The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n

The proof consists of two steps: The initial or base case proves that the statement holds for 0, or 1; The induction step, inductive step, or step case proves that for every n, if the statement holds for n, then it holds for n+1; in other words, assume that the statement holds for some arbitrary natural number n, and prove that the statement holds for n+1

The hypothesis in the inductive step, that the statement holds for a particular n, is called the induction hypothesis or inductive hypothesis; to prove the inductive step, one assumes the induction hypothesis for n and then uses this assumption to prove that the statement holds for n+1

P(n) is a property that depends on an index n ∈ ℕ: ∃ n₀ ∈ ℕ | P(n₀) true (basis of induction); P(n) true ⇒ P(n+1) true, ∀ n ≥ n₀ (induction step); then P(n) true ∀ n ≥ n₀

Considering the Newton's binomial formula, P(n): (a + b)ⁿ = ⁿΣ_k=0 C(n,k) a^n-kb^k

P(n): ⁿΣ_k=1 k = n(n+1)/2 ∀ n ≥ 1; n = 10, ¹⁰Σ_k=1 k = 10·11/2; n = 100, ¹⁰⁰Σ_k=1 k = 100·101/2; Basis of induction: n = 1, ¹Σ_k=1 k = 1·2/2, 1 = 1 true; inductive step: hypothesis ⁿΣ_k=1> k = n(n+1)/2, thesis ⁿ⁺¹Σ_k=1 k = (n+1)((n+1)+1)/2 = (n+1)(n+2)/2; ⁿ⁺¹Σ_k=1 k = ⁿΣ_k=1 k+ⁿ⁺¹Σ_k=n+1 k = n(n+1)/2 + (n+1) = (n(n+1) + 2(n+1))/2 = (n+1)(n+2)/2

P(n): 2ⁿ ≥ n, ∀ n ∈ ℕ, n ≥ 1; Basis of induction: n = 1, 2¹ ≥1 true; Inductive step: hypothesis 2ⁿ ≥ n, thesis 2ⁿ⁺¹ ≥ n+1; 2ⁿ⁺¹ = 2⋅2ⁿ ≥ 2n ≥ n+1

P(n): n²+n even ∀ n ∈ ℕ, n ≥ 1; Basis of induction: n = 1, 1²+1 = 2 even; Inductive step: hypothesis n²+n even, thesis (n+1)²+(n+1) even; (n+1)²+(n+1) = n²+2n+1+n+1 = n²+n+2n+2 = n²+n+2(n+1) even, n²+n even for hypothesis and 2(n+1) even, and the sum of two even numbers is an even number ⇒ (n+1)²+(n+1) even

Bernoulli's inequality, (1+x)ⁿ ≥ 1+nx, ∀ x ∈ ℝ, x > -1, ∀ n ∈ ℕ; Basis of induction: n=0, (1+x)⁰ ≥ 1+0x, 1 ≥ 1 true; Inductive step: hypothesis (1+x)ⁿ ≥ 1+nx, thesis (1+x)ⁿ⁺¹ ≥ 1+(n+1)x; (1+x)ⁿ⁺¹ = (1+x)ⁿ(1+x) ≥ [1] (1+nx)(1+x) = (1+nx)+x(1+nx) = 1+nx+x+nx² = 1+(n+1)x+nx² ≥ [2] 1+(n+1)x; [1] (1+x) > 0, (1+x)ⁿ(1+x) ≥ (1+nx)(1+x); [2] 1+(n+1)x+nx² ≥ 1+(n+1)x, if b ≥ 0 then a+b ≥ a

FUNCTION

A function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set

Functions were originally the idealization of how a varying quantity depends on another quantity

A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y, the codomain of the function

A function is usually denoted by letters such as f, g and h

If the function is called f, this relation is denoted by y = f(x), where the element x is the argument or input of the function, and y is the value of the function, the output, or the image of x by f

The symbol that is used for representing the input is the variable of the function (f is a function of the variable x)

A function is uniquely represented by the set of all pairs (x,f(x)), called the graph of the function

When the domain and the codomain are sets of real numbers, each such pair may be thought of as the Cartesian coordinates of a point in the plane; the set of these points is called the graph of the function

A function is a process that associates each element of a set X, to a single element of a set Y

Formally, a function f from a set X to a set Y is defined by a set G of ordered pairs (x,y) such that x ∈ X, y ∈ Y, and every element of X is the first component of exactly one ordered pair in G

For every x in X, there is exactly one element y such that the ordered pair (x,y) belongs to the set of pairs defining the function f; the set G is called the graph of the function

In the definition of function, X and Y are respectively called the domain and the codomain of the function f

If (x,y) belongs to the set defining f, then y is the image of x under f, or the value of f applied to the argument x

In the context of numbers in particular, one also says that y is the value of f for the value x of its variable, or, more concisely, that y is the value of f of x, denoted as y = f(x)

Two functions f and g are equal, if their domain and codomain sets are the same and their output values agree on the whole domain; more formally, f = g if f(x) = g(x) ∀ x ∈ X, where f: X → Y and g: X → Y

A = domain of the function, B = codomain of the function; f: A → B, a ∈ A → b = f(a) ∈ B

Image of the function: Im(f) = {b ∈ B | ∃ a ∈ A | b = f(a)} = f(A)

Graph of the function: G(f) = {(a,b) ∈ AxB | b = f(a)}

EXERCISES - SUP - INF - MAX - MIN

A = {(-1)ⁿ/(2+n²) | n ∈ ℕ}; (-1)ⁿ/(2+n²) = {1/(2+n²) | n = 2k, k ∈ ℕ; -1/(2+n²) | n = 2k+1, k ∈ ℕ; A = A_even ∪ A_odd; A_even = {1/(2+n²) | n = 2k, k ∈ ℕ}; A_odd = {-1/(2+n²) | n = 2k+1, k ∈ ℕ}; sup A = max {sup A_even, sup A_odd}; inf A = min {inf A_even, inf A_odd}; 1/(2+n²) ≥ 0 | n = 2k, k ∈ ℕ; -1/(2+n²) ≤ 0 | n = 2k+1, k ∈ ℕ; sup A = sup A_even; inf A = inf A_odd; 0 ≤ 1/(2+n²) ≤ 1 | n = 2k, k ∈ ℕ; A_even = {1/2,1/6,1/18...}, 1/(2+n²) decreases when n increases ⇒ sup A_even = max A_even = 1/2 ⇒ sup A = sup A_even = 1/2 = max A; -1 ≤ -1/(2+n²) ≤ 0 | n = 2k+1, k ∈ ℕ; A_odd = {-1/3,-1/11,-1/27...}, -1/(2+n²) increases when n increases; inf A_odd = inf A = -1/3 = min A

A = {|3-n|/(3+n) | n ∈ ℕ}; |x| = {x if x ≥ 0 ∀ x ∈ ℝ; -x if x < 0 ∀ x ∈ ℝ}; |3-n|/(3+n) = {(3-n)/(3+n) if 3-n ≥ 0 ⇔ -n ≥ -3 ⇔ n ≤ 3; (n-3)/(n+3) if 3-n < 0 ⇔ -n < -3 ⇔ n > 3}; 0 ≤ |3-n|/(n+3); |x| ≥ 0 ∀ x ∈ ℝ, |x| > 0 if x ≠ 0; n = 3 ⇒ (3-n)/(n+3) = 0, 0 is a minorant, 0 ∈ A, min A = 0; n = 0 → (3-n)(3+n) = 1 (max), n = 1 → (3-n)(3+n) = 1/2, n = 2 → (3-n)(3+n) = 1/5, n = 3 → (3-n)(3+n) = 0; 0 ≤ (n-3)/(n+3) ≤ 1; n-3 ≤ n+3 ⇒ (n-3)/(n+3) < 1; sup A = max A = 1

A {(-1)ⁿ+1/2ⁿ | n ∈ ℕ}; (-1)ⁿ+1/2ⁿ = 1+1/2ⁿ | n = 2k, k ∈ ℕ; -1+1/2ⁿ | n = 2k+1, k ∈ ℕ; A = A_even ∪ A_odd; 1+1/2ⁿ ≥ 0; -1+1/2ⁿ ≤ 0; sup A = sup A_even; inf A = inf A_odd

; 1+1/2ⁿ ≤ 2; A_even = {2,5/4,17/16...}, sup A_even = 2 ⇒ sup A = max A = 2; -1 ≤ -1+1/2ⁿ ≤ 0; A_odd = {-1/2,-7/8,-31/32...}; inf A_odd = -1 ⇒ inf A = -1; -1 is a minorant; ∀ ε > 0 ∃ n = 2k+1 | -1+ε > -1+1/2ⁿ (-1+ε is not a minorant), -1+ε > -1+1/2ⁿ ⇔ ε > 1/2ⁿ ⇔ 2ⁿ > 1/ε true for large values of n; -1 = min A ? ⇔ ∃ n = 2k+1 | -1 = -1+1/2ⁿ ⇔ 0 = 1/2ⁿ false, ∄ min A

A ⊆ (1,4]; a) inf A = 1; b) A has no minimum; c) A has a majorant; d) A is composed by a finite number of elements; a) false, A = [2,3], A ⊆ (1,4], inf A = 2; b) false, A = [2,3], A ⊆ (1,4], min A = 2; d) false, A = [2,3] is not composed by a finite number of elements; c) true, A ⊆ (1,4] = {x ∈ ℝ | 1 < x ≤ 4} ⇒ 4 is a majorant of A

EXERCISES - MATHEMATICAL INDUCTION

ⁿΣ_k=0 q^k = (1-qⁿ⁺¹)/(1-q), ∀ n ∈ ℕ, q ∈ ℝ, q ≠ 1; q = 2, 1+2+4+8+16+32; Basis of induction: n = 0, ⁰Σ_k=0 q^k = (1-q)/(1-q), q⁰ = 1, 1 = 1 true; Inductive step - Hypothesis: ⁿΣ_k=0 q^k = (1-qⁿ⁺¹)/(1-q); Inductive step - Thesis: ⁿ⁺¹Σ_k=0 q^k = (1-q^(n+1)+1)/(1-q) = (1-q⁽ⁿ⁺²⁾)/(1-q); ⁿ⁺¹Σ_k=0 q^k = ⁿΣ_k=0 q^k + ⁿ⁺¹Σ_k=n+1 q^k = (1-qⁿ⁺¹)/(1-q) + qⁿ⁺¹ = (1-qⁿ⁺¹+(1-q)qⁿ⁺¹)/(1-q) = (1-qⁿ⁺¹+qⁿ⁺¹-qⁿ⁺²)/(1-q) = (1-qⁿ⁺²)/(1-q)

The sum of the internal angles of a convex polygon of n+2 sides is equal to n flat angles (n·π); Basis of induction: n = 1, a convex polygon of n+2 sides with n=1 is a triangle, the sum of the interior angles of a triangle is π, 1·π = π, true; Inductive step - Hypothesis: the sum of the internal angles of a convex polygon of n+2 sides is n·π; Inductive step - Thesis: the sum of the internal angles of a convex polygon of (n+1)+2 = n+3 sides is (n+1)π; Considering a convex polygon of 5 sides, n = 2, n+3 sides convex polygon, it can be divided in a triangle and a quadrangle; the triangle is n+1 sides and the sum of the angles is π; the quadrangle is n+2 sides and the sum of the angles is n·π; angles of the triangle (π) + angles of the quadrangle (n·π) = π + n·π = (n+1)π

COMPLEX NUMBER

(ℝ²,+,⋅) is the field of Complex Numbers, identified by the symbol ℂ.

ℝ² = {(a,b) | a,b ∈ ℝ}

(a,b)+(c,d) = (a+c,b+d)

(a,b)⋅(c,d) = (ac-bd,ad+bc)

(0,0) is the neutral element of addition; (a,b)+(0,0) = (a+0,b+0) = (a,b)

∀ (a,b) ∈ ℝ², ∃ the additive inverse (the opposite number) (-a,-b) | (a+b)+(-a,-b) = (a-a,b-b) = (0,0)

(1,0) is the neutral element of multiplication; (a,b)⋅(1,0) = (a⋅1-b⋅0,a⋅0+b⋅1) = (a,b)

∀ (a,b) ∈ ℝ², (a,b) ≠ (0,0), ∃ the multiplicative inverse (the inverse number) (a/(a²+b²),-b/(a²+b²)) | (a,b)⋅(a/(a²+b²),-b/(a²+b²)) = (1,0)

ℂ₀ = {(a,0) | a ∈ ℝ} ⊆ ℂ

(a,0)+(b,0) = (a+b,0)

(a,0)⋅(b,0) = (a⋅b-0⋅0,a⋅0+0⋅b) = (ab,0)

(a,0) ∈ ℂ₀ → a ∈ ℝ

ℂ₀ = ℝ ⊆ ℂ

(a,b) = (a,0)+(0,1)⋅(b,0) = a + i⋅b

(a,0) = a; (0,1) = i; (b,0) = b

Imaginary unit: i = (0,1)

Algebraic form of complex numbers: (a,b) = a+i⋅b

i⋅i = i² = (0,1)⋅(0,1) = (0⋅0-1⋅1,0⋅1+1⋅0) = (-1,0) = -1 ⇔ i² = -1

(a+i⋅b)+(c+i⋅d) = a+c+i(b+d) = (a+c,b+d)

(a+i⋅b)⋅(c+i⋅d) = a⋅c+i⋅a⋅d+i⋅b⋅c+i²⋅b⋅d = a⋅c+i⋅a⋅d+i⋅b⋅c-b⋅d = a⋅c-b⋅d+i(a⋅d+b⋅c) = (a⋅c-b⋅d,a⋅d+b⋅c)

(3+i)+(6+2i) = (3+6)+i(1+2) = 9+3i

(4+2i)+3i = 4+i(2+3) = 4+5i

(1+i)⋅(2+3i) = 2+3i+2i+3i² = 2+3i+2i-3 = 2-3+i(3+2) = -1+5i

2i⋅(3+6i) = 6i+12i² = -12+6i

Algebraic form of complex numbers: z = a+ib; real part of z: a = Re(z); imaginary part of z: b = Im(z)

z = 4-7i; Re(z) = 4; Im(z) = -7

z = 2i; Re(z) = 0; Im(z) = 2

A real number a can be regarded as a complex number a+0i, whose imaginary part is 0

A purely imaginary number bi is a complex number 0+bi, whose real part is zero

Purely imaginary: z ∈ ℂ | Re(z) = 0

z = (a,b) = a+ib

1 = (1,0); i = (0,1); 1+i = (1,1)

(a+ib)+(c+id) = a+c+i(b+d)

The complex conjugate of the complex number z = x+yi is given by x−yi; it is denoted by either z or z*

Geometrically, z is the "reflection" of z about the real axis.

Considering z ∈ ℂ, z = a+ib, the complex conjugate of z is z = a-ib; Re(z) = Re(z), Im(z) = -Im(z)

z = 2+6i ⇒ z = 2-6i

z = -2i ⇒ z = 2i

z = 4 ⇒ z = 4

The conjugate of a sum is the sum of the conjugates: z+w = z+w, ∀ z,w ∈ ℂ

The conjugate of a product is the product of the conjugates: z⋅w = z⋅w, ∀ z,w ∈ ℂ

The conjugate of a ratio is the ratio of the conjugates: (z/w) = z/w, ∀ z,w ∈ ℂ, w ≠ 0

z⋅z = a²+b² if z = a+ib

(1+3i)(1-3i) = 1-3i+3i-9i² = 1+9 = 10

Conjugating twice gives the original complex number

(1+i)/(1+2i) = (1+i)/(1+2i) ⋅ (1-2i)/(1-2i) = (1+i)(1-2i)/(1+2i)(1-2i) = (1-2i+i-2i²)/(1-2i+2i-4i²) = (1-2i+i+2)/(1-2i+2i+4) = (3-i)/5 = 3/5 - (1/5)i

Modulus of a complex number: z ∈ ℂ, z = a+ib, the modulus is the real number |z| = √(a²+b²)

z = 1+i; |z| = √(1²+1²) = √2

z = a, |z| = √(a²) = |a|

|z| ≥ 0, the modulus of a complex number is always ≥ 0

|z| = 0 ⇔ z = 0, the modulus of z is 0 when a and b are 0, therefore z is 0

|w⋅z| = |w|⋅|z|, the modulus of the product is the product of the moduli

|z+w| ≤ |z|+|w|, triangle inequality.

z⋅z = |z|²

ρ (rho): length of the vector connecting the point to the origin; ρ ∈ [0,+∞]

θ (theta): angle that the vector forms with the x axis; θ ∈ [0,2π)

ρ and θ are polar coordinates.

From Polar coordinates to Cartesian coordinates: a = ρ⋅cos(θ); b = ρ⋅sin(θ)

Fron Cartesian coordinates to Polar coordinates: ρ = √(a²+b²); tan(θ) = b/a

z ∈ ℂ, z = a+ib = ρ⋅cos(θ)+i⋅ρ⋅sin(θ) = ρ(cos(θ)+i⋅sin(θ))

Algebraic form of complex numbers: z = a+ib

Trigonometric form of complex numbers: z = ρ(cos(θ)+i⋅sin(θ))

ρ = √(a²+b²) = |z|

θ = arg(z)

z = 1+i; ρ = |z| = √(a²+b²) = √(1²+1²) = √2; tan(θ) = 1 ⇔ θ = arg(z) = π/4; z = ρ(cos(θ)+i⋅sin(θ)) = √2(cos(π/4)+i⋅sin(π/4)) = √2(√2/2+i√2/2) = 1+i

z = -5; ρ = |z| = √(a²+b²) = √(-5²+0²) = √25 = 5; z = ρ(cos(θ)+i⋅sin(θ)) = 5(cos(π)+i⋅sin(π)) = 5(-1+i⋅0) = -5

z = i; ρ = |z| = √(a²+b²) = √(0²+1²) = 1; z = ρ(cos(θ)+i⋅sin(θ)) = 1(cos(π/2)+i⋅sin(π/2)) = 1(0+i⋅1) = 1(i) = i

z = a+ib, algebraic form of a complex number; a,b ∈ ℝ; i² = -1

z = ρ(cos(θ)+i⋅sin(θ)), trigonometric form of a complex number; a = ρ⋅cos(θ), b = ρ⋅sin(θ), ρ ∈ [0,+∞), θ ∈ [0,2π)

z₁,z₂ ∈ ℂ, z_i = ρ_i(cos(θ_i)+i⋅sin(θ_i)), i = 1,2

z₁⋅z₂ = ρ₁⋅ρ₂(cos(θ₁+θ₂)+i⋅sin(θ₁+θ₂))

ρ = √(a²+b²) = |z|; θ = arg(z)

|z₁⋅z₂| = |z₁|⋅|z₂|; arg(z₁⋅z₂) = arg(z₁)+arg(z₂)

z₁⋅z₂ = ρ₁(cos(θ₁)+i⋅sin(θ₁))⋅ρ₂(cos(θ₂)+i⋅sin(θ₂)) = ρ₁⋅ρ₂[cos(θ₁)cos(θ₂)-sin(θ₁)sin(θ₂)+i(sin(θ₁)cos(θ₂)+cos(θ₁)sin(θ₂))] = ρ₁⋅ρ₂(cos(θ₁+θ₂)+i⋅sin(θ₁+θ₂))

z₁/z₂ = ρ₁/ρ₂(cos(θ₁-θ₂)+i⋅sin(θ₁-θ₂))

|z₁/z₂| = |z₁|/|z₂|; arg(z₁/z₂) = arg(z₁)-arg(z₂)

De Moivre's formula: zⁿ = ρⁿ(cos(nθ)+i⋅sin(nθ)), z = ρ(cos(θ)+i⋅sin(θ)), n ∈ ℕ

(1+i)⁵; a+ib, a = 1, b = 1; ρ = √(a²+b²) = √2; θ = π/4; z = ρ(cos(θ)+i⋅sin(θ)) = √2(cos(π/4)+i⋅sin(π/4)); z⁵ = 2^5/2(cos(5π/4)+i⋅sin(5π/4)) = 2^5/2(√2/2+i√2/2) = 2^5/2(1/√2+i/√2) = 2^5/2(-1/2^1/2-i/2^1/2) = -2^4/2-2^4/2i = -2²-2²i = -4-4i ⇔ (1+i)⁵ = -4-4i

Euler's formula: e^iθ = cos(θ)+i⋅sin(θ), ∀ θ ∈ ℝ; cos(θ) = (e^iθ+e^-iθ)/2, sin(θ) = (e^iθ-e^-iθ)/2i

e^iθ = cos(θ)+i⋅sin(θ); e^-iθ = cos(-θ)+i⋅sin(-θ) = cos(θ)-i⋅sin(θ); e^iθ+e^-iθ = cos(θ)+i⋅sin(θ)+cos(θ)-i⋅sin(θ) = 2⋅cos(θ)

e^iπ = cos(π)+i⋅sin(π) = -1; e^iπ = -1 ⇔ e^iπ + 1 = 0

w ∈ ℂ, n ∈ ℕ\{0}, z ∈ ℂ is the n-th root of w if zⁿ = w

w ∈ ℂ, w = ρ(cos(θ)+i⋅sin(θ)), n ∈ ℕ\{0}, there are n different complex roots z_k, k = 1, ..., n, of w

z_k = r(cos(θ_k)+i⋅sin(θ_k)); r = ρ^1/n = ⁿ√ρ; θ_k = (θ+2kπ)/n, k = 1, ..., n

w = ρ(cos(θ)+i⋅sin(θ)), z = r(cos(φ)+i⋅sin(φ)) | zⁿ = w; using the De Moivre's formula zⁿ = rⁿ(cos(nφ)+i⋅sin(nφ)); zⁿ = w, rⁿ(cos(nφ)+i⋅sin(nφ)) = ρ(cos(θ)+i⋅sin(θ)) ⇔ rⁿ = ρ ⇔ r = ρ^1/n, nφ = θ+2kπ ∀ k ∈ ℤ ⇔ φ = (θ+2kπ)/n ∀ k ∈ ℤ; k = 1, φ₁ = (θ+2π)/n; k = n+1, φ_n+1 = (θ+2(n+1)π)/n = (θ+2π)/n + 2nπ/n = (θ+2π)/n + 2π

√-1; ρ(cos(θ)+i⋅sin(θ)), ρ = 1, θ = π; -1 = 1(cos(π)+i⋅sin(π)); z_k = ρ^1/2(cos((θ+2kπ)/2)+i⋅sin((θ+2kπ)/2) k = 0, 1; z₀ = cos(π/2)+i⋅sin(π/2) = 0+i⋅1 = i; z₁ = cos((π+2π)/2)+i⋅sin((π+2π)/2) = cos(3π/2)+i⋅sin(3π/2) = 0+i(-1) = -i

The complex roots of degree 2 of -1 are i and -i; i² = -i² = -1

The n-th roots of a complex number are the vertices of a regular polygon of n sides inscribed in the circumference of radius ρ^1/n where ρ is the modulus of the number

(ℂ,+,⋅) is a field containing the field of real numbers (ℝ,+,⋅) and is algebraically closed; algebraically closed means that a polynomial equation can always be solved in ℂ; the field of real numbers is not algebraically closed because many polynomial equations are not solvable in ℝ

The field of complex numbers ℂ is not orderable

If ℂ were orderable: z ∈ ℂ ⇒ z² ≥ 0 and z ∈ ℂ, z ≥ 0 ⇔ -z ≤ 0; i² = -1 ≥ 0 and 1 ≥ 0 ⇒ -1 ≤ 0 and it is an absurd result, a contradiction

FUNDAMENTAL THEOREM OF ALGEBRA

For an algebraic equation a_n⋅zⁿ+a_n-1⋅z^n-1+...+a₁⋅z+a₀ = 0, a_i ∈ ℂ, a_n ≠ 0, n ∈ ℕ, there are n solutions in ℂ, considering the multiplicity

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root; this includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero

Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed

The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots

z²(z+1) = 0, 0 is a root of multiplicity 2

In mathematics, a zero (also sometimes called a root) of a real-, complex-, or generally vector-valued function f, is a member x of the domain of f such that f(x) vanishes at x; that is, the function f attains the value of 0 at x, or equivalently, x is the solution to the equation f(x) = 0; a "zero" of a function is thus an input value that produces an output of 0

A root of a polynomial is a zero of the corresponding polynomial function

The fundamental theorem of algebra shows that any non-zero polynomial has a number of roots at most equal to its degree, and that the number of roots and the degree are equal when one considers the complex roots (or more generally, the roots in an algebraically closed extension) counted with their multiplicities

If the function maps real numbers to real numbers, then its zeros are the x-coordinates of the points where its graph meets the x-axis; an alternative name for such a point (x,0) in this context is an x-intercept

Every equation in the unknown x may be rewritten as f(x) = 0, by regrouping all the terms in the left-hand side; it follows that the solutions of such an equation are exactly the zeros of the function f; in other words, a "zero of a function" is precisely a "solution of the equation obtained by equating the function to 0", and the study of zeros of functions is exactly the same as the study of solutions of equations

Every real polynomial of odd degree has an odd number of real roots (counting multiplicities); likewise, a real polynomial of even degree must have an even number of real roots; consequently, real odd polynomials must have at least one real root (because the smallest odd whole number is 1), whereas even polynomials may have none; this principle can be proven by reference to the intermediate value theorem: since polynomial functions are continuous, the function value must cross zero, in the process of changing from negative to positive or vice versa (which always happens for odd functions)

The fundamental theorem of algebra states that every polynomial of degree n has n complex roots, counted with their multiplicities; the non-real roots of polynomials with real coefficients come in conjugate pairs

EXERCISES - COMPLEX NUMBERS

Write in algebraic form (1+i)/(1-i); z⋅z = |z|² = a²+b²; (1+i)/(1-i)⋅(1+i)/(1+i) = (1+i)²/(1²+1²) = (1-1+2i)/(1+1) = 2i/2 = i

z²+z = 0; z = a+ib; (a+ib)²+(a-ib) = 0 ⇔ a²-b²+i2ab+a-ib = 0 ⇔ (a²-b²+a)+i(2ab-b) = 0 ⇔ a²-b²+a = 0, 2ab-b = 0; 2ab-b = 0 ⇔ b(2a-1) = 0 ⇔ b = 0, a = 1/2; b = 0, a² + a = 0 ⇔ a(a+1) = 0 ⇔ a = 0, a = -1; z₁ = 0, z₂ = -1; a = 1/2, 1/4 - b² + 1/2 = 0 ⇔ -b² + 3/4 = 0 ⇔ -b² = - 3/4 ⇔ b² = 3/4 ⇔ b = ±√(3/4) = ±√(3)/2; z₃ = 1/2+i√(3)/2, z₄ = 1/2-i√(3)/2

(z)⁴ = |z|; z = ρ(cos(θ)+i⋅sin(θ)); |z| = ρ; z = ρ(cos(θ)-i⋅sin(θ)); Considering De Moivre's formula, ρ⁴(cos(4θ)-i⋅sin(4θ)) = ρ(cos(0)-i⋅sin(0)); ρ⁴ = ρ; 4θ = 0+2kπ k ∈ ℤ; ρ⁴ - ρ = 0 ⇔ ρ(ρ³-1) = 0 ⇔ ρ = 0, ρ = 1; θ = kπ/2 k ∈ ℤ; ρ = 0, z₁ = 0; ρ = 1 and k = 0, z₂ = 1(cos(0)+i⋅sin(0)) = 1; ρ = 1 and k = 1, z₃ = 1(cos(π/2)+i⋅sin(π/2)) = i; ρ = 1 and k = 2, z₄ = 1(cos(π)+i⋅sin(π)) = -1; ρ = 1 and k = 3, z₅ = 1(cos(3π/2)+i⋅sin(3π/2)) = -i

³√(i-1); z = -1+i; |z| = √(a²+b²) = √(-1²+1²) = √2; arg(z) = 3π/4; z_k = ρ^1/n(cos((θ+2kπ)/n)+i⋅sin((θ+2kπ)/n), k = 0, 1, ..., n-1; z_k = (2^1/2)^1/3(cos(((3/4)π+2kπ)/3)+i⋅sin(((3/4)π+2kπ)/3)); k = 0, z₀ = 2^1/6(cos(π/4)+i⋅sin(π/4)) = 2^1/6(√2/2+i⋅√2/2); k = 1, z₁ = 2^1/6(cos(((3/4)π+2π)/3)+i⋅sin(((3/4)π+2π)/3)) = 2^1/6(cos((11/12)π)+i⋅sin((11/12)π)); k = 2, z₂ = 2^1/6(cos(((3/4)π+4π)/3)+i⋅sin(((3/4)π+4π)/3)) = 2^1/6(cos((19/12)π)+i⋅sin((19/12)π))

z³-z|z|²+z = 0; z(z²-|z|²+1) = 0; z = 0; z²-|z|²+1 = 0; z = a+ib a,b ∈ ℝ; |z| = √(a²+b²); (a+ib)²-(a²+b²)+1 = 0 ⇔ a²+2abi-b²-a²-b²+1 = 0 ⇔ -2b²+1+2abi = 0 ⇔ {-2b²+1 = 0 ⇔ b = ± 1/√2, 2ab = 0 ⇔ {a = 0, b = 0}}; a = 0, b = ± 1/√2, z₁ = (1/√2)i, z₂ = -(1/√2)i; b = 0, b = ± 1/√2 impossible; z₀ = 0, z₁ = (1/√2)i, z₂ = -(1/√2)i

z⁵ = 1/(1+i)³; z = ρ(cos(θ)+i⋅sin(θ)); z = ρ(cos(θ)-i⋅sin(θ)); Considering De Moivre's formula, z⁵ = ρ⁵(cos(5θ)-i⋅sin(5θ)); 1/(1+i) = (1/(1+i))((1-i)/(1-i)) = (1-i)/(1²+1²) = (1-i)/2 = (1/2)-(1/2)i; ρ = |1/(1+i)| = √((1/2)²+(1/2)²) = √((1/4)+(1/4)) = √(2/4) = √(1/2) = 1/√2; arg(1/(1+i)) = (7/4)π; 1/(1+i) = 1/√2(cos((7/4)π)+i⋅sin((7/4)π)); (1/(1+i))³ = (1/2^1/2)³(cos(3⋅(7/4)π)+i⋅sin(3⋅(7/4)π)) = 1/2^3/2(cos((21/4)π)+i⋅sin((21/4)π)) = 1/2^3/2(cos((5/4)π)+i⋅sin((5/4)π)), (21/4)π = (16/4)π + (5/4)π = 2(2π)+(5/4)π, φ₁ = φ+2kπ, k ∈ ℤ ⇒ cos(φ₁) = cos(φ), sin(φ₁) = sin(φ); z⁵ = 1/(1+i)³ ⇔ ρ⁵(cos(5θ)-i⋅sin(5θ)) = 1/2^3/2(cos((5/4)π)+i⋅sin((5/4)π)); ρ⁵(cos(-5θ)+i⋅sin(-5θ)) = 1/2^3/2(cos((5/4)π)+i⋅sin((5/4)π)); ρ⁵ = 1/2^3/2; -5θ = (5/4)π+2kπ, k ∈ ℤ ⇔ θ = -(5/20)π-(2/5)kπ = -(5/20)π+(2/5)kπ, k ∈ ℤ; k = 0, θ₀ = -(5/20)π; k = 1, θ₁ = -(5/20)π+(2/5)π = (3/20)π; k = 2, θ₂ = -(5/20)π+(2/5)2π = -(5/20)π+(4/5)π = -(5/20)π+(16/20)π = (11/20)π; k = 3, θ₃ = -(5/20)π+(2/5)3π = -(5/20)π+(6/5)π = -(5/20)π+(24/20)π = (19/20)π; k = 4, θ₄ = -(5/20)π+(2/5)4π = -(5/20)π+(8/5)π = -(5/20)π+(32/20)π = (27/20)π; ∃ k | -(5/20)π+2kπ = (27/20)π ?, ∃ k | -(5/20)π+(40/20)kπ = (27/20)π ?, False; k = 5, θ₅ = -(5/20)π+(2/5)5π = -(5/20)π+(10/5)π = -(5/20)π+(40/20)π = (35/20)π; ∃ k | -(5/20)π+2kπ = (35/20)π ?, ∃ k | -(5/20)π+(40/20)kπ = (35/20)π ?, true for k = 1; There are 5 different solutions: ρ⁵ = 1/2^3/2, k = 0 → θ₀ = -(5/20)π, k = 1 → θ₁ = (3/20)π, k = 2 → θ₂ = (11/20)π, k = 3 → θ₃ = (19/20)π, k = 4 → θ₄ = (27/20)π

SEQUENCE

A sequence is a function from ℕ to ℝ

A sequence is a function that has a domain of natural numbers, and a codomain of real numbers

f: ℕ → ℝ

n → f(n) ∈ ℝ

f: ℕ → ℝ, n → f(n) = n²

f: ℕ → ℝ, n → f(n) = (-1)ⁿ

f: ℕ → ℝ, n → f(n) = 1/n, n ≠ 0

f: ℕ → ℝ, n → f(n) = a_n ⇔ {a_n}_n∈ℕ

f: ℕ → ℝ, n → f(n) = n² ⇔ {n²}_n∈ℕ = {1,4,9...}

A sequence {a_n}_n∈ℕ satisfies a certain property definitively if ∃ n₀ ∈ ℕ | the property is true ∀ n > n₀

{n-100}_n∈ℕ is definitively positive, in fact n-100 > 0 ∀ n > 100, where 100 is n₀

{(-1)ⁿ}_n∈ℕ is not definitively positive; ∀ n₀ ∃ n > n₀, n = 2k+1 | (-1)ⁿ = -1 < 0

Geometric sequence or geometric progression: {qⁿ}_n∈ℕ, q ∈ ℝ

{2ⁿ}_n∈ℕ is a geometric sequence with q = 2

{(1/3)ⁿ}_n∈ℕ is a geometric sequence with q = 1/3

{(-1)ⁿ}_n∈ℕ is a geometric sequence with q = -1

Harmonic sequence or harmonic progression: {n^α}_n∈ℕ, α ∈ ℝ

{n^1/2}_n∈ℕ is an harmonic sequence with α = 1/2

{n⁴}_n∈ℕ is an harmonic sequence with α = 4

{1/n²}_n∈ℕ is an harmonic sequence with α = -2, 1/n² = n^-2

A sequence {a_n}_n∈ℕ is monotonic increasing if a_n ≤ a_n+1 ∀ n ∈ ℕ

A sequence {a_n}_n∈ℕ is monotonic decreasing if a_n ≥ a_n+1 ∀ n ∈ ℕ

A sequence {a_n}_n∈ℕ is strictly monotonic increasing if a_n < a_n+1 ∀ n ∈ ℕ

A sequence {a_n}_n∈ℕ is strictly monotonic decreasing if a_n > a_n+1 ∀ n ∈ ℕ

{n}_n∈ℕ is a monotonic increasing sequence

{1/n}_n∈ℕ is a monotonic decreasing and bounded sequence

{-1/n}_n∈ℕ is a monotonic increasing and bounded sequence

{√(n)}_n∈ℕ is a monotonic increasing sequence

{-n²}_n∈ℕ is a monotonic decreasing sequence

{(-1)ⁿ}_n∈ℕ is not a monotonic sequence

LIMIT OF A SEQUENCE

{a_n}_n∈ℕ, l ∈ ℝ, the sequence converges to l as n tends to +∞, written as lim_n→+∞ a_n = l, if ∀ ε > 0, ∃ n₀ = n₀(ε) ∈ ℕ | ∀ n ≥ n₀ then |a_n-l| < ε

|x| < r ⇔ -r < x < r; |a_n-l| < ε ⇔ -ε < a_n-l < ε ⇔ l-ε < a_n < l+ε; ∀ ε > 0, |a_n-l| < ε definitely

lim_n→+∞ 1/n = 0 ⇔ ∀ ε > 0, ∃ n₀ | ∀ n > n₀, |(1/n)-0| < ε; |1/n| < ε ⇔ -ε < 1/n < ε ⇔ 1/n < ε ⇔ n > 1/ε; n₀ = [1/ε] + 1 that is the first natural number greater than 1/ε; [x] = n, if n ≤ x < n+1, [x] = integer part of x

If lim_n→∞ a_n = 0 then {a_n}_n∈ℕ is an infinitesimal sequence

Theorem of uniqueness of the limit of a sequence: if the limit of a sequence exists, then it is unique

Proof by contradiction of the theorem of uniqueness of the limit of a sequence: {a_n}_n∈ℕ, l₁, l₂ ∈ ℝ, l₁ ≠ l₂ | lim_n→∞ a_n = l₁ and lim_n→∞ a_n = l₂; l₁ ≠ l₂ ⇒ |l₁-l₂| ≠ 0; ε := |l₁-l₂| > 0; ∃ n₁ | ∀ n > n₁, |a_n-l₁| < ε/4; ∃ n₂ | ∀ n > n₂, |a_n-l₂| < ε/4; n₀ = max{n₁,n₂}; ∀ n > n₀ = max{n₁,n₂}, ε = |l₁-l₂| = |l₁-a_n+a_n-l₂| [1] ≤ |l₁-a_n|+|a_n-l₂| = |a_n-l₁|+|a_n-l₂| < (ε/4)+(ε/4) = ε/2 ⇒ ε < ε/2 ⇒ 1 < 1/2 that is a contradiction; [1] Triangle inequality, |x+y| ≤ |x|+|y|

lim_n→∞ 1/n = 0; {1/n}_n∈ℕ = {1, 1/2, 1/3, 1/4, ...}

lim_n→∞ n²; {n²}_n∈ℕ = {0, 1, 4, 9, ...}

{a_n}_n∈ℕ diverges to +∞ for n→+∞, lim_n→+∞ a_n = +∞, if ∀ M ∈ ℝ, ∃ n₀ = n₀(M) ∈ ℕ | ∀ n ≥ n₀ then a_n > M

{a_n}_n∈ℕ diverges to -∞ for n→+∞, lim_n→+∞ a_n = -∞, if ∀ M ∈ ℝ, ∃ n₀ = n₀(M) ∈ ℕ | ∀ n ≥ n₀ then a_n < M

{a_n}_n∈ℕ is divergent if it diverges to ±∞

lim_n→+∞ a_n = +∞ if ∀ M ∈ ℝ, a_n > M definitely

lim_n→+∞ a_n = -∞ if ∀ M ∈ ℝ, a_n < M definitely

lim_n→∞ n² = +∞; ∀ M ∈ ℝ, ∃ n₀ = n₀(M) ∈ ℕ | ∀ n ≥ n₀, n² > M; n² > M ⇔ {∀ n ∈ ℕ, M < 0; n > √M, M ≥ 0}; n > √M ⇔ n ≥ [√M]+1 = n₀(M)

lim_n→∞ -n³ = -∞

{(-1)ⁿ}_n∈ℕ

If {a_n}_n∈ℕ neither converges to l ∈ ℝ, nor diverges, then it is irregular or oscillating

{a_n} is regular if it is convergent or divergent, or it is irregular if it is oscillating

(-1)ⁿ⋅n = {n, n = 2k; -n, n = 2k+1}; this sequence is irregular or oscillating {0, -1, 2, -3, 4, -5, 6, -7, 8, -9, ...}

(-n)ⁿ = (-1)ⁿ⋅nⁿ; this sequence is irregular or oscillating {0, -1, 4, -27, 256, -3125, ...}

a_n = {1, n = 2k; 1/n, n = 2k+1}; this sequence is irregular or oscillating {1, 1, 1, 1/3, 1, 1/5, ...}

{sin(n)}_n∈ℕ is an irregular sequence, because for n that tends to infinity it does not converge or diverge; ∄ lim_n→∞ sin(n)

{cos(n)}_n∈ℕ is an irregular sequence, because for n that tends to infinity it does not converge or diverge; ∄ lim_n→∞ cos(n)

lim_n→∞ (nⁿ+n!)/(eⁿ+2ⁿ) = +∞

Geometric sequence {qⁿ}_n∈ℕ, q ∈ ℝ; lim_n→∞ qⁿ = {+∞, q > 1; 1, q = 1; 0, -1 < q < 1; ∄, q ≤ -1}; lim_n→∞ 1ⁿ = lim_n→∞ 1 = 1; q > 1 ⇒ q = 1+(q-1), q-1 > 0, qⁿ = (1+(q-1))ⁿ [1] ≥ 1+n(q-1) ∀ n ∈ ℕ, [1] Bernoulli's inequality (1+x)ⁿ ≥ 1+nx ∀ x > -1, lim_n→∞ 1+n(q-1) = +∞ ⇒ lim_n→∞ qⁿ = +∞; q < -1 ⇒ -q > 1, qⁿ = (-1(-q))ⁿ = (-1)ⁿ(-q)ⁿ, that is +∞ for n = 2k and -∞ n = 2k+1

Harmonic sequence {n^α}_n∈ℕ, α ∈ ℝ; lim_n→∞ n^α = {+∞, α > 0; 1, α = 0; 0, α < 0}; lim_n→∞ n^3/2 = +∞; lim_n→∞ 1/n² = 0, 1/n² = n^-2; lim_n→∞ n⁰ = lim_n→∞ 1 = 1

If {a_n}_n∈ℕ is convergent, then it is bounded, that is ∃ m, M ∈ ℝ | m ≤ a_n ≤ M ∀ n ∈ ℕ

{(-1)ⁿ} is bounded, -1 ≤ (-1)ⁿ ≤ 1 ∀ n ∈ ℕ, but ∄ lim_n→∞ (-1)ⁿ

How can we calculate the limit of a sequence? To apply the definition of limit lim_n→∞ a_n = l, we need to know l ∈ ℝ

The limit of a sum is the sum of the limits: lim_n→∞ (a_n+b_n) = l+m; lim_n→∞ a_n = l, lim_n→∞ b_n = m, l,m ∈ ℝ

The limit of a difference is the difference of the limits: lim_n→∞ (a_n-b_n) = l-m; lim_n→∞ a_n = l, lim_n→∞ b_n = m, l,m ∈ ℝ

The limit of a product is the product of the limits: lim_n→∞ (a_n⋅b_n) = l⋅m; lim_n→∞ a_n = l, lim_n→∞ b_n = m, l,m ∈ ℝ

The limit of a ratio is the ratio of the limits: lim_n→∞ (a_n/b_n) = l/m, m ≠ 0; lim_n→∞ a_n = l, lim_n→∞ b_n = m, l,m ∈ ℝ

The limit of a power is the power of the limits: lim_n→∞ (a_n)^b_n = l^m, l > 0; lim_n→∞ a_n = l, lim_n→∞ b_n = m, l,m ∈ ℝ

The limit of a module is the module of the limit: lim_n→∞ |a_n| = |l|; lim_n→∞ a_n = l, l ∈ ℝ

lim_n→∞ (n²-n)/(n²-1) = lim_n→∞ n²(1-(1/n))/n²(1-(1/n²)) = lim_n→∞ (1-(1/n))/(1-(1/n²)) = 1/1 = 1; lim_n→∞ 1/n = lim_n→∞ n^-1 = 0; lim_n→∞ 1/n² = lim_n→∞ n^-2 = 0

lim_n→∞ a_n = l ∈ ℝ, lim_n→∞ b_n = ±∞, lim_n→∞ a_n+b_n = ±∞; l±∞ = ±∞

Extended set of real numbers: ℝ = ℝ ∪ {±∞}

l±∞ = ±∞ ∀ l ∈ ℝ

+∞+∞ = +∞; lim_n→∞ a_n = +∞, lim_n→∞ b_n = +∞ ⇒ lim_n→∞ a_n+b_n = +∞

-∞-∞ = -∞; lim_n→∞ a_n = -∞, lim_n→∞ b_n = -∞ ⇒ lim_n→∞ a_n+b_n = -∞

+∞⋅l = {+∞, l > 0; -∞, l < 0}

l/±∞ = 0 ∀ l ∈ ℝ

q^+∞ = {+∞, q > 1; 0, q < 1}

+∞⋅+∞ = +∞

-∞⋅-∞ = +∞

+∞⋅-∞ = -∞

-∞⋅+∞ = -∞

lim_n→∞ (n²)+(1/n)+(3ⁿ) = +∞+0+∞ = +∞

lim_n→∞ (n²+2n-1)/(n⁵+3n+2) = lim_n→∞ (n²(1+(2/n)-(1/n²)))/(n⁵(1+(3/n⁴)+(2/n⁵))) = (1+(2/n)-(1/n²))/(n³(1+(3/n⁴)+(2/n⁵))) = 1/+∞ = 0

Indeterminate Forms: +∞-∞; ±∞/±∞; 0/0; 0⋅(±∞); 1^±∞; 0⁰; ∞⁰; l/0, l ∈ ℝ

lim_n→∞ a_n = +∞, lim_n→∞ b_n = -∞, lim_n→∞ a_n+b_n = ?

a_n = n+1, b_n = -n; lim_n→∞ a_n = +∞, lim_n→∞ b_n = -∞ ⇒ lim_n→∞ a_n+b_n = lim_n→∞ 1 = 1

a_n = n, b_n = -n; lim_n→∞ a_n = +∞, lim_n→∞ b_n = -∞ ⇒ lim_n→∞ a_n+b_n = 0

An indeterminate form of a limit must be transformed into a determinate form: a limit is always either convergent, or divergent, or irregular

a_n = n³, b_n = n²; lim_n→∞ a_n/b_n = lim_n→∞ n³/n² = ∞/∞ = lim_n→∞ n = +∞; this sequence was initially considered indeterminate, but it turned out to be divergent

a_n = 2n³, b_n = n³; lim_n→∞ a_n/b_n = lim_n→∞ 2n³/n³ = ∞/∞ = lim_n→∞ 2 = 2; this sequence was initially considered indeterminate, but it turned out to be convergent

a_n = n, b_n = n⁴; lim_n→∞ a_n/b_n = lim_n→∞ n/n⁴ = ∞/∞ = lim_n→∞ 1/n³ = 1/∞ = 0; this sequence was initially considered indeterminate, but it turned out to be infinitesimal

l/0, l ≠ 0, is an indeterminate form

a_n = 1, b_n = (-1)ⁿ/n; lim_n→∞ a_n = lim_n→∞ 1 = 1; lim_n→∞ b_n = lim_n→∞ (-1)ⁿ/n = 0; lim_n→∞ a_n/b_n = lim_n→∞ 1/(-1)ⁿ/n = 1/0 = lim_n→∞ n/(-1)ⁿ = lim_n→∞ (-1)ⁿn = ∄; this sequence was initially considered indeterminate, but it turned out to be irregular

∞⁰ is an indeterminate form; a^b = e^b·ln(a); ∞⁰ = e^0·ln(∞) = e^0·∞

If a_n is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ a_n ≤ M ∀ n; lim_n→∞ b_n = 0 ⇒ lim_n→∞ a_nb_n = 0; a bounded sequence, multiplied by an infinitesimal sequence, is equal to zero

If a_n is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ a_n ≤ M ∀ n; lim_n→∞ b_n = ±∞ ⇒ lim_n→∞ a_n/b_n = 0; a bounded sequence, divided by a divergent sequence, is equal to zero

lim_n→∞ sin(n)/n = 0; sin(n) is irregular and bounded because -1 ≤ sin(x) ≤ 1 ∀ x ∈ ℝ

lim_n→∞ cos(2ⁿ)/2ⁿ = 0; cos(2ⁿ) is irregular and bounded because -1 ≤ cos(x) ≤ 1 ∀ x ∈ ℝ ⇒ -1 ≤ cos(2ⁿ) ≤ 1 ∀ n ∈ ℕ; lim_n→∞ 1/2ⁿ = lim_n→∞ (1/2)ⁿ = 0, geometric sequence qⁿ with -1 < q < 1

Comparison theorem: {a_n}_n∈ℕ, {b_n}_n∈ℕ | {a_n} ≤ {b_n} ∀ n ∈ ℕ, lim_n→∞ a_n = l, lim_n→∞ b_n = m, l ≤ m

Squeeze theorem or sandwich theorem or theorem of carabinieri: {a_n}_n∈ℕ, {b_n}_n∈ℕ, {c_n}_n∈ℕ | a_n ≤ c_n ≤ b_n ∀ n ∈ ℕ, lim_n→∞ a_n = lim_n→∞ b_n = l, lim_n→∞ c_n = l

Permanence of the sign: {a_n} | lim_n→∞ a_n = l; if a_n ≥ 0 ∀ n, then l ≥ 0; if a_n ∈ [α,β] ∀ n, then l ∈ [α,β]

If a_n > 0, then l > 0? No. For example {1/n}_n∈ℕ, 1/n > 0, lim_n→∞ 1/n = 0

lim_n→∞ 1/(5ⁿ+cos(n)); 5ⁿ-1 ≤ 5ⁿ+cos(n) ≤ 5ⁿ+1; 1/(5ⁿ+1) ≤ 1/(5ⁿ+cos(n)) ≤ 1/(5ⁿ-1); 0 ≤ 1/(5ⁿ+cos(n)) ≤ 0; for the squeeze theorem lim_n→∞ 1/(5ⁿ+cos(n)) = 0

lim_n→∞ ⁿ√(a) = 1 ∀ a > 0; for a > 1, lim_n→∞ ⁿ√(a) = 1 ⇔ lim_n→∞ ⁿ√(a)-1 = 0; d_n := ⁿ√(a)-1; we have to prove that lim_n→∞ d_n = 0; d_n = ⁿ√(a)-1 ⇔ ⁿ√(a) = 1+d_n ⇔ a = (1+d_n)ⁿ [1] ≥ 1+nd_n ⇔ a ≥ 1+nd_n ⇔ a-1 ≥ nd_n ⇔ (a-1)/n ≥ d_n ≥ 0 [2] ⇒ lim_n→∞ d_n = 0; [1] Bernoulli's inequality, (1+x)ⁿ ≥ 1+nx ∀ x > -1, ∀ n ∈ ℕ; [2] for the squeeze theorem; for 0 < a < 1, lim_n→∞ ⁿ√(a) = lim_n→∞ 1/(1/ⁿ√(a)) = lim_n→∞ 1/ⁿ√(1/a)) [1] = 1/1 = 1; [1] 0 < a < 1 ⇒ 1/a > 1, applying the previous case; the nth roots of the real numbers converge to 1

{a_n}_n∈ℕ | lim_n→∞ a_n = 0; if a_n ≥ 0 ∀ n ∈ ℕ, then lim_n→∞ a_n = 0⁺; if a_n ≤ 0 ∀ n ∈ ℕ, then lim_n→∞ a_n = 0^-

l/0⁺ = {+∞, l > 0; -∞, l < 0}

l/0^- = {+∞, l < 0; -∞, l > 0}

lim_n→∞ a_n = l, lim_n→∞ b_n = 0⁺ ⇒ lim_n→∞ a_n/b_n = {+∞, l > 0; -∞, l < 0}

lim_n→∞ 1/((1/2)ⁿ+1/n) = 1/0⁺ = +∞; (1/2)ⁿ+1/n ≥ 0

Monotone convergence theorem or theorem of regularity of monotone sequence: if {a_n}_n∈ℕ is monotone, then it is regular, that is convergent or divergent; if {a_n}_n∈ℕ is increasing, then lim_n→∞ a_n = sup{a_n | n ∈ ℕ}; if {a_n}_n∈ℕ is decreasing, then lim_n→∞ a_n = inf{a_n | n ∈ ℕ}

A monotonous succession is always regular, it cannot be oscillating; if it is increasing converges to a real number or diverges to + ∞; if it is decreasing converges to a real number or diverges to -∞

Demonstration of the monotone convergence theorem: suppose that l = sup{a_n | n ∈ ℕ} ∈ ℝ; thesis l = lim_n→∞ a_n ⇔ ∀ ε > 0, ∃ n₀ | ∀ n > n₀, |a_n-l| < ε ⇔ -ε < a_n-l < ε; by definition of supremum, l is a majorant or upper bound of {a_n | n ∈ ℕ}, therefore a_n ≤ l ∀ n ∈ ℕ ⇔ a_n-l ≤ 0 < ε ∀ n ∈ ℕ; by definition of supremum, l-ε is not a majorant or upper bound of {a_n | n ∈ ℕ} ⇒ ∃ n₀ ∈ ℕ | l-ε < a_n0, because {a_n}_n∈ℕ is increasing, a_n ≥ a_n0 ∀ n > n₀, therefore -ε ≤ a_n-l ∀ n ≥ n₀

{(1+1/n)ⁿ}_n∈ℕ is the Euler's sequence

The Euler's sequence {(1+1/n)ⁿ}_n∈ℕ is bounded, 2 ≤ (1+1/n)ⁿ ≤ 3, and it is increasing

lim_n→∞ (1+1/n)ⁿ := e, the Euler's number

For the permanence of the sign 2 ≤ e ≤ 3

(1+1/n)ⁿ, increasing the value of n, there is a better approximation of the value of e

e = 2.71828..

e ∈ ℝ \ ℚ, the Euler's number is irrational

{a_n}_n∈ℕ, {b_n}_n∈ℕ are asymptotic sequences if lim_n→∞ a_n/b_n = 1; a_n ~ b_n n→∞

a_n ~ b_n ⇒ lim_n→∞ a_n = lim_n→∞ b_n

a_n = n²-3n+2, b_n = n²+1; lim_n→∞ a_n/b_n = lim_n→∞ (n²-3n+2)/(n²+1) = lim_n→∞ n²(1-3/n+2/n²)/n²(1+1/n²) = lim_n→∞ (1-3/n+2/n²)/(1+1/n²) = 1/1 = 1 ⇒ n²-3n+2 ~ n²+1

a_n = n^5/2+n²+n^1/2+2, b_n = n^5/2; lim_n→∞ a_n/b_n = lim_n→∞ (n^5/2+n²+n^1/2+2)/(n^5/2) = lim_n→∞ 1+1/n^1/2+1/n²+2/n^5/2 = 1

lim_n→∞ a_n = lim_n→∞ b_n !⇒ a_n ~ b_n

a_n = n², b_n = n; lim_n→∞ a_n = lim_n→∞ b_n = +∞; lim_n→∞ a_n/b_n = lim_n→∞ n²/n = +∞

The asymptotic relation is an equivalence relation: it is reflexive, symmetrical, and transitive; reflexive: a_n ~ a_n ⇔ lim_n→∞ a_n/a_n = 1; symmetrical: a_n ~ b_n ⇔ b_n ~ a_n, lim_n→∞ b_n/a_n = lim_n→∞ 1/a_n/b_n = lim_n→∞ 1/1 = 1; transitive: a_n ~ b_n, b_n ~ c_n ⇒ a_n ~ c_n, lim_n→∞ a_n/c_n = lim_n→∞ (a_n/b_n)(b_n/c_n) = lim_n→∞ 1⋅1 = 1

Substitution principle: if a_n ~ a_n^', b_n ~ b_n^', c_n ~ c_n^', then lim_n→∞ a_n⋅b_n/c_n = lim_n→∞ a_n^'⋅b_n^'/c_n^'

lim_n→∞ (n³+2n+1)/(n⁴+4n²+3); n³+2n+1 ~ n³, n⁴+4n²+3 ~ n⁴; lim_n→∞ (n³+2n+1)/(n⁴+4n²+3) = lim_n→∞ n³/n⁴ = lim_n→∞ 1/n = 0

The substitution principle applies only to products and ratios: a_n ~ a_n^', b_n ~ b_n^' ⇒ lim_n→∞ a_n/b_n = lim_n→∞ a_n^'/b_n^', lim_n→∞ a_nb_n = lim_n→∞ a_n^'⋅b_n^'

The substitution principle does not generally apply to sums, differences and powers: a_n ~ a_n^', b_n ~ b_n^' !⇒ lim_n→∞ a_n±b_n = lim_n→∞ a_n^'±b_n^', lim_n→∞ a_n^b_n = lim_n→∞ a_n^'^b_n^'

a_n = n+1 ~ a_n^' = n, b_n = -n ~ b_n^' = -n; lim_n→∞ a_n+b_n = lim_n→∞ n+1-n = 1; lim_n→∞ a_n^'+b_n^' = lim_n→∞ n-n = 0

lim_n→∞ a_n = l ⇔ lim_n→∞ a_n/l = 1; a_n ~ l

a_n = 1+1/n ~ a_n^' = 1, b_n = n ~ b_n^' = n; lim_n→∞ a_n^b_n = lim_n→∞ (1+1/n)ⁿ = e; lim_n→∞ a_n^'^b_n^' = lim_n→∞ 1ⁿ = lim_n→∞ 1 = 1

lim_n→∞ ln(n)/n^α = 0, α > 0; it would be ∞/∞, but the harmonic sequence at the denominator grows faster than the logarithm sequence at the numerator

lim_n→∞ n^α/qⁿ = 0, α > 0, q > 1; it would be ∞/∞, but the geometric sequence at the denominator grows faster than the harmonic sequence at the numerator

lim_n→∞ qⁿ/n! = 0, q > 1; it would be ∞/∞, but the factorial sequence at the denominator grows faster than the geometric sequence at the numerator

lim_n→∞ n!/nⁿ = 0; it would be ∞/∞, but the nⁿ sequence at the denominator grows faster than the factorial sequence at the numerator

Infinity hierarchy from slowest to fastest: ln(n), n^α, qⁿ, n!, nⁿ

lim_n→∞ qⁿ/nⁿ = lim_n→∞ (qⁿ/n!)(n!/nⁿ) = 0

lim_n→∞ nⁿ/qⁿ = lim_n→∞ 1/qⁿ/nⁿ = 1/0⁺ = +∞

lim_n→∞ ⁿ√(n) = 1; a^b = e^b⋅ln(a); lim_n→∞ ⁿ√(n) = lim_n→∞ n^1/n = lim_n→∞ e^(ln(n))/n = lim_n→∞ e⁰ = 1

lim_n→∞ (1+x/a_n)^a_n = e^x, ∀ x ∈ ℝ, ∀ {a_n}_n∈ℕ | lim_n→∞ a_n = +∞; lim_n→∞ (1+1/n)ⁿ = e, x = 1, a_n = n; lim_n→∞ (1+2/(n²+1))ⁿ²⁺¹ = e², x = 2, a_n = n²+1

lim_n→∞ sin(a_n)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ (e^a_n-1)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ ln(1+a_n)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ (1-cos(a_n))/a_n² = 1/2, lim_n→∞ a_n = 0

lim_n→∞ n⋅sin(1/n) = ∞⋅0; lim_n→∞ sin(1/n) = sin(0) = 0; lim_n→∞ n⋅sin(1/n) = lim_n→∞ (sin(1/n))/1/n = 1

lim_n→∞ (sin(n))/n = 0; a bounded sequence, divided by a divergent sequence goes to zero

lim_n→∞ (1+1/n)ⁿ = 1^∞ = e; lim_n→∞ 1ⁿ = 1^∞ = lim_n→∞ 1 = 1; indeterminate forms can be resolved

Cesàro criterion: {a_n}_n∈ℕ | a_n ≥ 0 ∀ n ∈ ℕ, lim_n→∞ a_n+1/a_n = l ∈ ℝ, lim_n→∞ ⁿ√(a_n) = l

If a sequence has a limit it implies that it is bounded, but if a sequence is bounded it does not imply that it has a limit, like sin(n)

A monotone sequence is always regular, if it is bounded it admits limit, if it is not bounded and increasing it diverges to +∞, if it is not bounded and decreasing it diverges to -∞

In a sequence defined by recurrence, each element depends on the preceding ones

In the Fibonacci sequence each number is the sum of the two preceding ones, starting from 0 and 1; a₀ = 0, a₁ = 1, a_n = a_n-1+a_n-2, n > 1; the beginning of the sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144...; under some older definitions, the value a₀ = 0 is omitted, so that the sequence starts with a₁ = a₂ = 1, and the recurrence a_n = a_n−1+a_n−2 is valid for n > 2; the Fibonacci sequence can also be defined as a_n+2 = a_n+1+a_n, a₀ = a₁ = 1

EXERCISES - LIMIT OF SEQUENCES

lim_n→∞ (n+4)/n = 1; ∀ ε > 0, ∃ n₀ = n₀(ε) | ∀ n ≥ n₀ ⇒ |a_n-l| < ε; ∀ ε > 0, ∃ n₀ = n₀(ε) | ∀ n ≥ n₀ ⇒ |((n+4)/n)-1| < ε; |(n+4-n)/n| < ε; |4/n| < ε; 4/n < ε; n > 4/ε; n₀ > [4/ε]+1

lim_n→∞ (n^3/2+1)/(2n-1); n^3/2+1 ~ n^3/2, 2n-1 ~ 2n; lim_n→∞ (n^3/2+1)/(2n-1) = lim_n→∞ n^3/2/2n = lim_n→∞ n^1/2/2 = lim_n→∞ √(n)/2 = +∞; √(n)/2 is a harmonic sequence with α = 1/2; {n^α}_n∈ℕ, α ∈ ℝ; lim_n→∞ n^α = {+∞, α > 0; 1, α = 0; 0, α < 0}

lim_n→∞ (n⁴+3n-1/n+2)/(n²+1/n²) = lim_n→∞ n⁴/n² = lim_n→∞ n² = +∞

lim_n→∞ (1+√(n))/(n+1)²n^α, α ∈ ℝ; 1+√(n) ~ n^1/2; (n+1)² ~ n²; n^α ~ n^α; lim_n→∞ (1+√(n))/(n+1)²n^α = lim_n→∞ n^1/2/n²n^α = lim_n→∞ n^1/2-2-α = lim_n→∞ n^-3/2-α = {+∞, -3/2-α > 0 ⇔ α < -3/2; 1, -3/2-α = 0 ⇔ α = -3/2; 0, -3/2-α < 0 ⇔ α > -3/2}

Geometric sequence: lim_n→∞ qⁿ = {+∞, q > 1; 1, q = 1; 0, -1 < q < 1; ∄, q ≤ -1}

Harmonic sequence: lim_n→∞ n^α = {+∞, α > 0; 1, α = 0; 0, α < 0}

lim_n→∞ 3ⁿ+4ⁿ-5ⁿ = +∞-∞ = lim_n→∞ 5ⁿ(3ⁿ/5ⁿ+4ⁿ/5ⁿ-5ⁿ/5ⁿ) = lim_n→∞ 5ⁿ((3/5)ⁿ+(4/5)ⁿ-1) = -∞; 5ⁿ→∞, (3/5)ⁿ→0, (4/5)ⁿ→0

lim_n→∞ (2ⁿ⁺¹+1)/(3ⁿ+n) = lim_n→∞ 2ⁿ⁺¹/3ⁿ = lim_n→∞ 2ⁿ2/3ⁿ = lim_n→∞ 2(2/3)ⁿ = 0

lim_n→∞ (2ⁿ-4ⁿ)/(3ⁿ-n!) = lim_n→∞ 4ⁿ(2ⁿ/4ⁿ-4ⁿ/4ⁿ)/n!(3ⁿ/n!-n!/n!) = lim_n→∞ 4ⁿ((1/2)ⁿ-1)/n!(3ⁿ/n!-1) = 0; (1/2)ⁿ-1→-1, 3ⁿ/n!-1→-1

lim_n→∞ ((n²+n)/(n²-n+2))ⁿ = 1^∞ [1] = lim_n→∞ (1+(n²+n)/(n²-n+2)-1)ⁿ = lim_n→∞ (1+(n²+n-n²+n-2)/(n²-n+2))ⁿ = lim_n→∞ (1+1/(n²-n+2)/(2n-2))ⁿ [2] = lim_n→∞ ((1+1/(n²-n+2)/(2n-2))^{(n2-n+2)/(2n-2)})^{((2n-2)/(n2-n+2))n} [3] = e²; [1] lim_n→∞ (1+x/a_n)^a_n = e^x, lim_n→∞ a_n = +∞; [2] lim_n→∞ (n²-n+2)/(2n-2) = +∞; [3] lim_n→∞ (1+1/(n²-n+2)/(2n-2))^{(n2-n+2)/(2n-2)} = e, lim_n→∞ = n(2n-2)/(n²-n+2) = lim_n→∞ 2n²/n² = 2

lim_n→∞ n⋅tan(1/n) = ∞⋅0 = lim_n→∞ n⋅sin(1/n)/cos(1/n) = lim_n→∞ (sin(1/n)/1/n)(1/cos(1/n)) = 1; lim_n→∞ sin(a_n)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ (1-cos(3/n))/sin(3/n²) = 0/0 = lim_n→∞ ((1-cos(3/n))/(3/n)²)((3/n)²3/n²/sin(3/n²)3/n²) = lim_n→∞ ((1-cos(3/n))/(3/n)²)(3/n²/sin(3/n²))(9/n²)(n²/3) = 3/2; lim_n→∞ (1-cos(a_n))/a_n² = 1/2, lim_n→∞ a_n = 0; lim_n→∞ sin(a_n)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ (2^1/n-1)/1/n = lim_n→∞ ((e^(1/n)ln(2)-1)/(1/n)ln(2))ln(2) = ln(2); a^b = e^b⋅ln(a); lim_n→∞ (e^a_n-1)/a_n = 1, lim_n→∞ a_n = 0

lim_n→∞ (2ⁿ-1)/n = lim_n→∞ 2ⁿ(1-1/2ⁿ)/n = +∞; A geometric sequence diverges to +∞ more rapidly than a harmonic sequence

lim_n→∞ (1+1/nⁿ)^n! = 1^∞ = lim_n→∞ ((1+1/nⁿ)ⁿⁿ)^{n!/nn = e0 = 1; lim_n→∞ (1+x/a_n)a_n = ex, lim_n→∞ a_n = +∞; (ab)c = abc}

lim_n→∞ √(n+2)-√(n+1) = +∞-∞ = lim_n→∞ (√(n+2)-√(n+1))(√(n+2)+√(n+1))/(√(n+2)+√(n+1)) = lim_n→∞ ((n+2)-(n+1))/√(n)(√(1+2/n)+√(1+1/n)) = 1/2√(n) = 1/∞ = 0; (a-b)(a+b) = a²-b²

lim_n→∞ ⁿ√(n)

Applying the Cesàro criterion: lim_n→∞ ⁿ√(a_n), a_n = n

lim_n→∞ a_n+1/a_n = lim_n→∞ (n+1)/n = 1 ⇒ lim_n→∞ ⁿ√(n) = 1

lim_n→∞ ⁿ√(n!/(2ⁿ+1)); applying the Cesàro criterion: a_n = n!/(2ⁿ+1); lim_n→∞ a_n+1/a_n = lim_n→∞ (n+1)!/(2ⁿ⁺¹+1)(2ⁿ+1)/n! [1] = lim_n→∞ (n!(n+1)/(2ⁿ⁺¹+1))((2ⁿ+1)/n!) = lim_n→∞ ((n+1)/(2ⁿ⁺¹+1))(2ⁿ+1) = lim_n→∞ (n+1)(2ⁿ(1+1/2ⁿ)/2ⁿ⁺¹(1+1/2ⁿ⁺¹)) = lim_n→∞ (n+1)(2ⁿ(1+1/2ⁿ)/2⋅2ⁿ(1+1/2ⁿ⁺¹)) = lim_n→∞ (n+1)/2 = +∞; lim_n→∞ ⁿ√(n!/(2ⁿ+1)) = +∞; [1] (n+1)! = n!(n+1)

lim_n→∞ ⁿ√(C(2n, n))); binomial coefficient: n, k ∈ ℕ | n ≥ k ≥ 0; C(n, k) = n!/k!(n-k)!; applying the Cesàro criterion: a_n = C(2n, n); lim_n→∞ a_n+1/a_n = lim_n→∞ C(2(n+1), n+1)/C(2n, n) = lim_n→∞ C(2n+2, n+1)/C(2n, n) = lim_n→∞ ((2n+2)!/(n+1)!(2n+2-n-1)!)/((2n)!/n!(2n-n)!) = lim_n→∞ ((2n+2)!/(n+1)!(n+1)!)(n!n!/(2n)!) [1][2] = lim_n→∞ ((2n)!(2n+1)(2n+2)/n!(n+1)n!(n+1))(n!n!/(2n)!) = lim_n→∞ (2n+1)(2n+2)/(n+1)(n+1) = lim_n→∞ (2n+1)(2n+2)/(n+1)² = lim_n→∞ (2n)²(1+1/2n)(1+2/2n)/n²(1+1/n)² = lim_n→∞ 4n²/n² = 4; lim_n→∞ ⁿ√(C(2n, n)) = 4; [1] (n+1)! = n!(n+1); [2] (2n+2)! = (2n)!(2n+1)(2n+2)

lim_n→∞ ⁿ²√(n!); 1 = ⁿ²√(1) ≤ ⁿ²√(n!) ≤ ⁿ²√(nⁿ) = (nⁿ)^1/n2 = n^1/n; lim_n→∞ 1 ≤ lim_n→∞ ⁿ²√(n!) ≤ lim_n→∞ n^1/n; 1 ≤ lim_n→∞ ⁿ²√(n!) ≤ 1; lim_n→∞ ⁿ²√(n!) = 1

{a_n}_n∈ℕ | {|a_n|}_n∈ℕ is decreasing, then: a) {a_n}_n∈ℕ is monotone, b) lim_n→∞ a_n ∃, c) lim_n→∞ a_n ∄, d) {a_n}_n∈ℕ is bounded; a) {a_n}_n∈ℕ is monotone, false, considering {a_n}_n∈ℕ = {(-1)ⁿ}, {|(-1)ⁿ|} = {1}_n∈ℕ, a constant sequence is both increasing and decreasing, {|(-1)ⁿ|} is decreasing, but (-1)ⁿ is not a monotone sequence; b) lim_n→∞ a_n ∃, false, considering {a_n}_n∈ℕ = {(-1)ⁿ}, {|(-1)ⁿ|} is decreasing, but lim_n→∞ (-1)ⁿ ∄; c) lim_n→∞ a_n ∄, false, considering {a_n}_n∈ℕ = {1/n} | {|1/n|}_n∈ℕ = {1/n} is decreasing and lim_n→∞ 1/n = 0; d) {a_n}_n∈ℕ is bounded, true, {|a_n|}_n∈ℕ is decreasing ⇒ |a₀| ≥ |a₁| ≥ ... ≥ |a_n| ∀ n ∈ ℕ ⇔ |a_n| ≤ |a₀| ∀ n ∈ ℕ ⇔ -|a₀| ≤ a_n ≤ |a₀| ∀ n ∈ ℕ, {a_n}_n∈ℕ is bounded if ∃ m, M ∈ ℝ | m ≤ a_n ≤ M ∀ n ∈ ℕ

{a_n} = (2n+3)^(-1)n: a) {a_n} converges, b) {a_n} diverges, c) {a_n} is bounded but does not converge, d) {a_n} is not bounded, but does not diverge; (2n+3)^(-1)n = {(2n+3), n = 2k; 1/(2n+3), n = 2k+1}; the sequence diverges to +∞ with n even, converges to zero with n odd; the sequence is not bounded and it is irregular, so is true d) {a_n} is not bounded, but does not diverge

{a_n}_n∈ℕ is a bounded sequence: a) |a_n| ≤ 1000 ∀ n ∈ ℕ, b) {a_n}_n∈ℕ is not increasing, c) ∃ α > 0 | 2^-a_n < α ∀ n, d) lim_n→∞ a_n ∃ as a finite number, the sequence converges; a) |a_n| ≤ 1000 ∀ n ∈ ℕ, false, a_n = 1001 ∀ n is bounded, but it does not verify a); b) {a_n}_n∈ℕ is not increasing, false, a_n = -1/n is bounded, -1 ≤ -1/n ≤ 0, and it is increasing; d) lim_n→∞ a_n ∃ as a finite number, the sequence converges, false, {sin(n)}_n∈ℕ is bounded, but it is irregular, in fact if ∃ lim_n→∞ a_n ⇒ {a_n} is bounded, but if {a_n} is bounded !⇒ ∃ lim_n→∞ a_n; c) ∃ α > 0 | 2^-a_n < α ∀ n, true

a₀ = 1, a_n+1 = √(1+a_n): a) verify by induction that {a_n} is increasing, b) verify by induction that 1 ≤ a_n ≤ 2 ∀ n ∈ ℕ, c) calculate lim_n→∞ a_n; a) verify by induction that {a_n} is increasing; thesis: a_n ≤ a_n+1 ∀ n ∈ ℕ, basis of induction (n = 0): a₀ ≤ a₁, a₀ = 1, a₁ = √(1+a₀) = √(1+1) = √(2), 1 ≤ √(2); inductive step: hypothesis a_n ≤ a_n+1, thesis a_n+1 ≤ a_n+2, a_n+2 = √(1+a_n+1) ≥ √(1+a_n) = a_n+1, a_n+2 ≥ a_n+1; b) verify by induction that 1 ≤ a_n ≤ 2 ∀ n ∈ ℕ; considering a_n ≥ 1 ∀ n ∈ ℕ; basis of induction (n = 0): a₀ ≥ 1 ⇔ 1 ≥ 1; inductive step: hypothesis a_n ≥ 1, thesis a_n+1 ≥ 1, a_n+1 = √(1+a_n) ≥ √(1+1) = √(2) ≥ 1; c) calculate lim_n→∞ a_n; a) + b) ⇒ ∃ lim_n→∞ a_n = l ⇒ {lim_n→∞ a_n+1 = l; lim_n→∞ a_n+1 = lim_n→∞ √(1+a_n) = lim_n→∞ √(1+l)} ⇔ l = √(1+l) ⇔ l² = 1+l ⇔ l²-l-1 = 0, using the quadratic formula ax²+bx+c = 0, x = (-b±√(b²-4ac))/2a, l²-l-1 = 0, (1±√(1+4))/2, (1±√(5))/2, 1 ≤ a_n ≤ 2 ⇒ 1 ≤ l ≤ 2 for the sign permanence theorem, but (1-√(5))/2 < 0 ⇒ l = (1+√(5))/2 that is the golden ratio

SERIES

A series is the sum of the elements of a sequence

{a_n}_n∈ℕ, {S_n}_n∈ℕ is a sequence of nth partial sums , S_n = ⁿΣ_k=0 a_k, {S₀ = a₀, S₁ = a₀+a₁, S₂ = a₀+a₁+a₂, ..., S_n = a₀+a₁+...+a_n = S_n-1+a_n}

{a_n}_n∈ℕ, ⁿΣ_k=0 a_k: 1) it converges if ∃ lim_n→∞ S_n = S ∈ ℝ, ^∞Σ_k=0 a_k = S that is the sum of the series; 2) it diverges if lim_n→∞ S_n = ±∞; 3) it is irregular if ∄ lim_n→∞ S_n, that is {S_n}_n∈ℕ is irregular

^+∞Σ_k=0 a_k = lim_n→∞ ⁿΣ_k=0 a_k = S

A geometric series is the sum of an infinite number of terms that have a constant ratio between successive terms

Geometric series: ^+∞Σ_k=0 q^k, q ∈ ℝ

^+∞Σ_k=0 q^k = ^+∞Σ_n=0 qⁿ

S_n = ⁿΣ_k=0 q^k = (1-qⁿ⁺¹)/(1-q), q ≠ 1

S_n = ⁿΣ_k=0 1^k = n+1, q = 1

lim_n→∞ qⁿ = {+∞, q > 1; 1, q = 1; 0, |q| < 1; ∄, q ≤ -1}; q ≠ 1, lim_n→∞ S_n = lim_n→∞ (1-qⁿ⁺¹)/(1-q) = {+∞, q > 1; 1/(1-q), |q| < 1; ∄, q ≤ -1}; q = 1, lim_n→∞ S_n = lim_n→∞ n+1 = +∞

q = 1/2, ^+∞Σ_k=0 (1/2)^k = 1+1/2+1/4+1/8+1/16+...; ^+∞Σ_k=0 (1/2)^k = 1/(1-1/2) = 2

q = 3, ^+∞Σ_k=0 3^k = 1+3+9+27+81 = +∞

^+∞Σ_k=0 (-1)^k = 1-1+1-1+...; S₀ = ⁰Σ_k=0 (-1)^k = 1; S₁ = ¹Σ_k=0 (-1)^k = 1-1 = 0; S₂ = ²Σ_k=0 (-1)^k = 1-1+1 = 1

Generalized harmonic series: ^+∞Σ_k=1 1/k^α = ^+∞Σ_k=1 k^-α, α ∈ ℝ

Harmonic series: ^+∞Σ_k=1 1/k = +∞

S_n = ⁿΣ_k=1 1/k = 1+1/2+1/3+1/4+...+1/n

^+∞Σ_k=1 1/k = 1+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16)+... ≥ 1+1/2+1/2+1/2+1/2+...; 1/3+1/4 ≥ 1/4+1/4 = 2/4 = 1/2, 1/5+1/6+1/7+1/8 ≥ 1/8+1/8+1/8+1/8 = 4/8 = 1/2, ...; it can be proved by induction that S_2ⁿ ≥ 1+n/2 → +∞

A telescoping series is a series whose general term t_n can be written as t_n = a_n-a_n+1

The Mengoli series is a telescopic series: ^+∞Σ_k=1 1/k(k+1) = 1/1⋅2+1/2⋅3+1/3⋅4+...; S_n = ⁿΣ_k=1 1/k(k+1) = ⁿΣ_k=1 (1/k-1/(k+1)) = (1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/(n-1)-1/n)+(1/n-1/(n+1)) = 1-1/(n+1) ⇒ S_n = 1-1/(n+1); lim_n→∞ S_n = 1 ⇒ ^+∞Σ_k=1 1/k(k+1) = 1

^+∞Σ_k=0 a_k, ^+∞Σ_k=n a_k, these two series have the same behavior even if the sum could be different

^+∞Σ_k=1 q^k, |q| < 1; ^+∞Σ_k=0 q^k = 1/(1-q); ^+∞Σ_k=1 q^k = ^+∞Σ_k=0 q^k-q⁰ = 1/(1-q)-1 = 1/(1-q)-(1-q)/(1-q) = q/(1-q)

It is important to study methods to understand the behavior of the series; behavior means convergence, divergence or irregularity

Necessary but not sufficient condition for convergence: if the series ^+∞Σ_k=0 a_k is convergent, then lim_n→∞ a_n = 0

By hypothesis: lim_n→∞ S_n = S ∈ ℝ ⇒ lim_n→∞ S_n-1 = S; lim_n→∞ (S_n-S_n-1) = lim_n→∞ S_n - lim_n→∞ S_n-1 = 0; S_n-S_n-1 = (a₀+a₁+...+a_n-1+a_n)-(a₀+a₁+...+a_n-1) = a_n; lim_n→∞ a_n = lim_n→∞ (S_n-S_n-1) = 0

^+∞Σ_k=0 a_k = S ∈ ℝ ⇒ lim_n→∞ a_n = 0; lim_n→∞ a_n = 0 !⇒ ^+∞Σ_k=0 a_k = S ∈ ℝ; lim_n→∞ 1/n = 0, ^+∞Σ_k=1 1/k = +∞

A necessary condition is useful only when it is not verified; if it is verified I cannot say anything about the corresponding property

^+∞Σ_k=1 (1+1/k)²; a_n = (1+1/n)²; lim_n→∞ a_n = 1 ⇒ ^+∞Σ_k=1 (1+1/k)² does not converge

^+∞Σ_k=0 sin(k); ∄ lim_n→∞ sin(n) ⇒ ^+∞Σ_k=0 sin(k) does not converge; if the necessary condition is not verified, the series does not converge

^+∞Σ_k=0 sin(1/k); lim_n→∞ sin(1/n) = 0; if the necessary condition is verified, it is not possible to predict the behavior of the series, that is, the series could converge, diverge or be oscillating

Positive term series: ^+∞Σ_k=0 a_k, a_n ≥ 0 ∀ n ∈ ℕ

If ^+∞Σ_k=0 a_k is a positive term series, then {S_n}_n∈ℕ is increasing and therefore regular

S_n+1 = a₀+...+a_n+a_n+1 = S_n+a_n+1 ≥ S_n ⇒ S_n+1 ≥ S_n ∀ n ∈ ℕ

A positive term series is not irregular, so S_n is an increasing sequence therefore converges or diverges to +∞; an increasing monotone sequence converges to its supremum which can be a real number or + ∞

^+∞Σ_k=1 1/k² is a positive term series

^+∞Σ_k=1 (-1)^k/k is not a positive term series

A positive term series is always regular, it is convergent or divergent to +∞

There are 4 criteria to understand the behavior of a positive term series: comparison, asymptotic comparison, ratio, root

Comparison criterion: ^+∞Σ_k=0 a_k, ^+∞Σ_k=0 b_k are two positive term series | a_k ≤ b_k ∀ k ∈ ℕ. Then, 1) ^+∞Σ_k=0 a_k = +∞ ⇒ ^+∞Σ_k=0 b_k = +∞, 2) ^+∞Σ_k=0 b_k < +∞ ⇒ ^+∞Σ_k=0 a_k < +∞

= +∞ means that it diverges; < +∞ means that it converges

a_k ≤ b_k ∀ k ∈ ℕ ⇒ ⁿΣ_k=0 a_k ≤ ⁿΣ_k=0 b_k; if lim_n→∞ ⁿΣ_k=0 a_k = +∞ ⇒ lim_n→∞ ⁿΣ_k=0 b_k = +∞

If ^+∞Σ_k=0 a_k < +∞, it is impossible to predict the behavior of ^+∞Σ_k=0 b_k; if the minorant series converges, it is impossible to predict the behavior of the majorant series

If ^+∞Σ_k=0 b_k = +∞, it is impossible to predict the behavior of ^+∞Σ_k=0 a_k; if the majorant series diverges, it is impossible to predict the behavior of the minorant series

If the minorant series diverges, or if the majorant series converges, it is possible to understand the behavior of the series

If the minorant series converges, or if the majorant series diverges, it is not possible to understand the behavior of the series

^+∞Σ_k=0 1/k! = 1/0!+1/1!+1/2!+1/3!+1/4!+...; if the necessary condition is not verified, the series does not converge and therefore diverges to +∞; the necessary condition is verified because lim_n→∞ 1/n! = 1/∞ = 0, so we need more information to understand the behavior of the series; k! = 1⋅2⋅3⋅...⋅k ≥ 1⋅2⋅2⋅...⋅2 ⇒ k! ≥ 2^k-1 ⇔ 1/k! ≤ 1/2^k-1; ^+∞Σ_k=0 1/2^k-1 = ^+∞Σ_k=0 2/2^k = 2 ^+∞Σ_k=0 (1/2)^k, it is a geometric series with q = 1/2; ^+∞Σ_k=0 q^k = {+∞, q ≥ 1; 1/(1-q), |q| < 1; ∄, q ≤ -1}; the majorant series is convergent and therefore, for the comparison criterion, the series converges, ^+∞Σ_k=0 1/k! < +∞; ^+∞Σ_k=0 1/k! = e

Asymptotic comparison criterion: ^+∞Σ_k=0 a_k, ^+∞Σ_k=0 b_k are two positive term series; 1) if lim_k→∞ a_k/b_k = l ≠ 0, then ^+∞Σ_k=0 a_k < +∞ ⇔ ^+∞Σ_k=0 b_k < +∞, the two series have the same behavior; 2) if lim_k→∞ a_k/b_k = 0, then ^+∞Σ_k=0 b_k < +∞ ⇒ ^+∞Σ_k=0 a_k < +∞, ^+∞Σ_k=0 a_k = +∞ ⇒ ^+∞Σ_k=0 b_k = +∞

^+∞Σ_k=1 1/k^α, α ∈ ℝ, generalized harmonic series; α = 1, ^+∞Σ_k=1 1/k = +∞; α = 2, ^+∞Σ_k=1 1/k²; 1/k² ~ 1/k(k+1) ⇔ lim_k→∞ (1/k²)/(1/k(k+1)) = 1; lim_k→∞ (1/k²)/(1/k(k+1)) = lim_k→∞ k(k+1)/k² = lim_k→∞ k²/k² = 1; ^+∞Σ_k=1 1/k(k+1) = 1, telescopic series; considering the asymptotic comparison criterion with a_k = 1/k² and b_k = 1/k(k+1), then ^+∞Σ_k=1 1/k² < +∞

α ≥ 2, k^α ≥ k² ⇔ 1/k^α ≤ 1/k² ⇒ ^+∞Σ_k=1 1/k^α ≤ ^+∞Σ_k=1 1/k² < +∞; considering the criterion of comparison ^+∞Σ_k=1 1/k^α < +∞ ∀ α ≥ 2

α ≤ 1, k^α ≤ k ⇒ 1/k ≤ 1/k^α ⇒ 1/k^α ≥ 1/k ⇒ ^+∞Σ_k=1 1/k^α ≥ ^+∞Σ_k=1 1/k = +∞; considering the criterion of comparison ^+∞Σ_k=1 1/k^α = +∞

^+∞Σ_k=1 1/k^α {< +∞, α > 1; = +∞, α ≤ 1}

^+∞Σ_k=1 1/k^3/2 < +∞, α = 3/2 > 1

^+∞Σ_k=1 1/k^2/5 = +∞, α = 2/5 < 1

^+∞Σ_k=1 sin(1/k), lim_k→∞ sin(1/k) = 0, the necessary condition is verified; sin(1/k) ≥ 0 ∀ k, it is a positive term series so it converges or diverges; sin(a_n) ~ a_n if lim_n→∞ a_n = 0, sin(a_n) ~ a_n ⇔ lim_n→∞ sin(a_n)/a_n = 1; sin(1/k) ~ 1/k for k → ∞, sin(1/k) ⇔ lim_k→∞ sin(1/k)/(1/k) = 1; ^+∞Σ_k=1 1/k = +∞ ⇒ for the asymptotic comparison, ^+∞Σ_k=1 sin(1/k) = +∞

Ratio criterion: ^+∞Σ_k=0 a_k is a positive term series that is a_k ≥ 0; if ∃ lim_k→∞ a_k+1/a_k = l ∈ ℝ ∪ {+∞}, then: 1) if l < 1 ⇒ ^+∞Σ_k=0 a_k < +∞, 2) if l > 1 ⇒ ^+∞Σ_k=0 a_k = +∞, 3) if l = 1 continue studying

Root criterion: ^+∞Σ_k=0 a_k is a positive term series that is a_k ≥ 0; if ∃ lim_k→∞ ^k√(a_k) = l ∈ ℝ ∪ {+∞}, then: 1) if l < 1 ⇒ ^+∞Σ_k=0 a_k < +∞, 2) if l > 1 ⇒ ^+∞Σ_k=0 a_k = +∞, 3) if l = 1 continue studying

Considering the Cesàro criterion, if ∃ lim_k→∞ a_k+1/a_k = l ⇒ lim_k→∞ ^k√(a_k) = l

^+∞Σ_k=0 x^k/k!, x ∈ ℝ, x > 0; it is a positive term series, lim_k→∞ x^k/k! = 0, the necessary condition is verified; a_k = x^k/k!, lim_k→∞ a_k+1/a_k = lim_k→∞ (x^k+1/(k+1)!)/(x^k/k!) = lim_k→∞ (x^k+1/(k+1)!)(k!/x^k) = lim_k→∞ (x^k⋅x/k!(k+1))(k!/x^k) = lim_k→∞ x/(k+1) = 0 = l ⇒ for the ratio criterion, ^+∞Σ_k=0 x^k/k! < +∞, ∀ x ∈ ℝ, x > 0

^+∞Σ_k=1 x^k/k^k, x ∈ ℝ, x > 0; it is a positive term series, lim_k→∞ x^k/k^k = 0, the necessary condition is verified; a_k = x^k/k^k, lim_k→∞ ^k√(a_k) = lim_k→∞ ^k√(x^k/k^k) = lim_k→∞ x/k = 0 ⇒ for the root criterion,^+∞Σ_k=1 x^k/k^k < +∞

For the ratio criterion and the root criterion, if l = 1 continue studying; in this example the ratio criterion if l = 1; ^+∞Σ_k=1 1/k = +∞, a_k = 1/k, lim_k→∞ a_k+1/a_k = lim_k→∞ (1/(k+1))/(1/k) = lim_k→∞ k/(k+1) = 1; ^+∞Σ_k=1 1/k² < +∞, a_k = 1/k², lim_k→∞ a_k+1/a_k = lim_k→∞ (1/(k+1)²)/(1/k²) = lim_k→∞ k²/(k+1)² = 1

Alternating series: ^+∞Σ_k=0 (-1)^k·a_k, a_k ≥ 0 ∀ k ∈ ℕ

^+∞Σ_k=0 (-1)^k·a_k = a₀-a₁+a₂-a₃+..

^+∞Σ_k=1 1/k = 1+1/2+1/3+1/4+...

^+∞Σ_k=1 (-1)^k·1/k = -1+1/2-1/3+1/4+...

Leibniz criterion: ^+∞Σ_k=0 (-1)^k⋅a_k if, 1) lim_k→∞ a_k = 0, 2) a_k+1 < a_k ∀ k ∈ ℕ, then ^+∞Σ_k=0 (-1)^ka_k < +∞

^+∞Σ_k=1 (-1)^k1/k^α α ∈ ℝ, generalized harmonic series with alternating signs; a_k = 1/k^α; 1) lim_k→∞ 1/k^α = 0 if α > 0; 2) a_k+1 < a_k ⇔ 1/(k+1)^α < 1/k^α ⇔ k^α < (k+1)^α, true ∀ k

^+∞Σ_k=1 (-1)^k1/k^α converges if α > 0

^+∞Σ_k=1 1/k^α converges if α > 1

Positive term series: comparison criterion, asymptotic comparison criterion, ratio criterion, root criterion

Alternating series: Leibniz criterion

^+∞Σ_k=0 a_k absolutely converges if ^+∞Σ_k=0 |a_k| converges

If ^+∞Σ_k=0 a_k absolutely converges, then it converges

|a+b| ≤ |a|+|b|; |ⁿΣ_k=0 a_k| ≤ ⁿΣ_k=0 |a_k|; |S_n| ≤ ⁿΣ_k=0 |a_k|

Absolute convergence is a sufficient condition but not a necessary condition for the convergence of the series

If the series converges absolutely, then it converges, but it can also converge without converging absolutely

Absolute convergence implies convergence, viceversa a series can converge without converging absolutely

Absolute convergence ⇒ convergence

Convergence !⇒ absolute convergence

Convergence does not imply absolute convergence; a series can converge without converging absolutely

The alternating harmonic series ^+∞Σ_k=1 (-1)^k1/k converges for the Leibniz criterion, but it does not converge absolutely; ^+∞Σ_k=1 |(-1)^k1/k| = ^+∞Σ_k=1 |(-1)^k||1/k| = ^+∞Σ_k=1 1/k = +∞

^+∞Σ_k=0 |a_k| is a positive term series; if lim_k→∞ ^k√(a_k) = l < 1, then ^+∞Σ_k=0 |a_k| converges ⇒ ^+∞Σ_k=0 a_k converges absolutely, then converges

EXERCISES - SERIES

^+∞Σ_k=0 k/(k²+1); necessary condition lim_k→∞ a_k = lim_k→∞ k/(k²+1) = 0; it is a positive term series, a_k = k/(k²+1) ≥ 0 ∀ k ∈ ℕ; lim_k→∞ a_k/b_k = l ≠ 0, then ^+∞Σ_k=0 a_k < +∞ ⇔ ^+∞Σ_k=0 b_k < +∞, the two series have the same behavior; k/(k²+1) ~ k/k² = 1/k ⇔ lim_k→∞ (k/(k²+1))/(1/k) = 1; ^+∞Σ_k=1 1/k = +∞ ⇒ ^+∞Σ_k=1 k/(k²+1) = +∞

^+∞Σ_k=1 1/(3^k-k); necessary condition lim_k→∞ 1/(3^k-k) = lim_k→∞ 1/(3^k(1-k/3^k)) = 0; lim_k→∞ k/3^k = 0 ⇒ 3^k grows faster than k ⇒ 3^k-k > 0, it is a positive term series; (3^k-k) ~ 3^k ⇔ 1/(3^k-k) ~ 1/3^k; ^+∞Σ_k=1 1/3^k = ^+∞Σ_k=1 (1/3)^k < +∞ ⇒ ^+∞Σ_k=1 1/(3^k-k) < +∞

^+∞Σ_k=1 ln(k)/k³; lim_k→∞ ln(k)/k³ = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k³ ~ ?/k³; according to the second part of the asymptotic comparison criterion, if lim_k→∞ a_k/b_k = 0, then ^+∞Σ_k=0 b_k < +∞ ⇒ ^+∞Σ_k=0 a_k < +∞; lim_k→∞ (ln(k)/k³)/(1/k²) = lim_k→∞ (ln(k)/k³)k² = 0; ^+∞Σ_k=1 1/k² < +∞ ⇒ ^+∞Σ_k=1 ln(k)/k³ < +∞; ^+∞Σ_k=1 1/k^α = {< +∞, α > 1; = +∞, α ≤ 1}

^+∞Σ_k=1 ln(k)/k² < +∞; lim_k→∞ ln(k)/k² = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k² ~ ?/k²; according to the second part of the asymptotic comparison criterion, if lim_k→∞ a_k/b_k = 0, then ^+∞Σ_k=0 b_k < +∞ ⇒ ^+∞Σ_k=0 a_k < +∞; lim_k→∞ (ln(k)/k²)/(1/k^3/2) = lim_k→∞ ln(k)/k^1/2 = 0; ^+∞Σ_k=1 1/k^3/2 < +∞ ⇒ ^+∞Σ_k=1 ln(k)/k² < +∞; lim_k→∞ ln(k)/k^α = 0 ∀ α > 0; ^+∞Σ_k=1 1/k^α = {< +∞, α > 1; = +∞, α ≤ 1}

^+∞Σ_k=1 k!/k^k; lim_k→∞ k!/k^k = 0, the necessary condition is verified; it is a positive term series; k!/k^k ~ ?/k^k; it is possible to use the Stirling's approximation or Stirling's formula, but the most logical way is to use the ratio criterion; a_k = k!/k^k, lim_k→∞ a_k+1/a_k = lim_k→∞ ((k+1)!/(k+1)^k+1)/(k!/k^k) = lim_k→∞ (k!(k+1)/(k+1)^k(k+1))(k^k/k!) = lim_k→∞ k^k/(k+1)^k = lim_k→∞ (k/(k+1))^k = lim_k→∞ 1/((k+1)/k)^k = lim_k→∞ 1/(1+1/k)^k = 1/e < 1, according to the ratio criterion, the series converges

^+∞Σ_k=1 x^k/k, x ∈ ℝ; considering the absolute convergence, if the series absolutely converges, then it converges; ^+∞Σ_k=1 |x^k/k| = ^+∞Σ_k=1 |x|^k/k, it is a positive term series, so it is possible to apply the ratio criterion; |x^k| = |x⋅x⋅...⋅x| = |x|⋅|x|⋅...⋅|x| = |x|^k; lim_k→∞ ^k√(|x|^k/k) = lim_k→∞ |x|/^k√(k) = |x| < 1 ⇒ ^+∞Σ_k=1 |x^k/k| converges, then ^+∞Σ_k=1 x^k/k absolutely converges and therefore converges; |x| = 1, case of doubt; |x| > 1, ^+∞Σ_k=1 |x^k/k| diverges, that is, it does not absolutely converge, therefore the starting series cannot be defined; |x| > 1 ⇔ x < -1 and x > 1; x > 1, lim_k→∞ x^k/k = +∞, the necessary condition is not verified, the series does not converge; x < -1, lim_k→∞ x^k/k = lim_k→∞ ((-1)(-x))^k/k, lim_k→∞ ((-1)^k(-x)^k)/k ∄, the necessary condition is not verified, then the series does not converge; |x| = 1 ⇔ { x = 1, ^+∞Σ_k=1 1/k = +∞; x = -1, ^+∞Σ_k=1 (-1)^k/k < +∞}; ^+∞Σ_k=1 x^k/k = {absolutely converges if |x| < 1; converges if x = -1; does not converge if |x| > 1 and x = 1}

^+∞Σ_k=1 (3k-3^k)/(5^k-5k) = - ^+∞Σ_k=1 (3^k-3k)/(5^k-5k); (3^k-3k)/(5^k-5k) ~ 3^k/5^k = (3/5)^k; ^+∞Σ_k=1 (3/5)^k < +∞ ⇒ ^+∞Σ_k=1 (3k-3^k)/(5^k-5k) converges

^+∞Σ_k=1 (√(k+1)-√(k))/k; √(k+1)-√(k) > 0, it is a positive term series; lim_k→∞ (√(k+1)-√(k))/k = 0, the necessary condition is verified; ((√(k+1)-√(k))/k)((√(k+1)+√(k))/(√(k+1)+√(k))) = (k+1-k)/(k(√(k+1)+√(k))) = 1/(k(√(k+1)+√(k))) ~ 1/2k√(k) = 1/2k^3/2, considering that √(k+1)+√(k) = √(k)(√((k+1)/k)+1); ^+∞Σ_k=1 1/k^3/2 < +∞ ⇒ ^+∞Σ_k=1 (√(k+1)-√(k))/k < +∞

^+∞Σ_k=1 (k!+2)/(k+2)!; (k!+2)/(k+2)! ~ k!/(k+2)! = k!/k!(k+1)(k+2) = 1/(k+1)(k+2) ~ 1/k²; ^+∞Σ_k=1 1/k² < +∞ ⇒ ^+∞Σ_k=1 (k!+2)/(k+2)! < +∞

^+∞Σ_k=1 (-1)^k/log(k+1); according to the Leibniz criterion, ^+∞Σ_k=1 (-1)^k a_k, a_k = 1/log(k+1), if 1) lim_k→∞ a_k = 0 and 2) a_k > a_k+1 ∀ k ⇒ ^+∞Σ_k=1 (-1)^k a_k converges; 1) lim_k→∞ 1/log(k+1) = 0 and 2) a_k > a_k+1 ⇔ 1/log(k+1) > 1/log(k+2) ⇔ log(k+2) > log(k+1), then ^+∞Σ_k=1 (-1)^k/log(k+1) converges

{a_k}_k∈ℕ | lim_k→∞ a_k = 1/2, b_k = (a_k)^k, then, a) ^+∞Σ_k=0 b_k is irregular, b) ^+∞Σ_k=0 b_k diverges, c) ^+∞Σ_k=0 b_k converges, d) none of the previous answers; a_k = 1/2 ∀ k ∈ ℕ, b_k = (1/2)^k ⇒ ^+∞Σ_k=0 (1/2)^k < +∞; lim_k→∞ a_k = 1/2, then a_k ≤ 2/3 ∀ k ∈ ℕ, 0 ≤ b_k = (a_k)^k ≤ (2/3)^k, the series b_k is upper bounded by a converging geometric series; ^+∞Σ_k=0 (2/3)^k converges ⇒ according to the comparison theorem ^+∞Σ_k=0 b_k < +∞

REAL FUNCTION

A function is a law that associates an element of one set with only one element of a second set

f: A → B, a → b = f(a)

A real function of a real variable is a function f: X ⊆ ℝ → ℝ, x → y = f(x)

f: ℝ → ℝ, x → f(x) = x²

X domain of f; f(X) = {y ∈ ℝ, y = f(x)} image of f; G(f) = {(x,y) ∈ ℝ², x ∈ X, y = f(x)} ⊆ ℝ² graph of f

f: X → ℝ, g: X → ℝ; f+g: X → ℝ, x → (f+g)(x) = f(x)+g(x); f-g: X → ℝ, x → (f-g)(x) = f(x)-g(x); f⋅g: X → ℝ, x → (f⋅g)(x) = f(x)⋅g(x); f/g: X\{x ∈ X: g(x) = 0} → ℝ, x → (f/g)(x) = f(x)/g(x)

f: X → Y, g: Y → ℝ; compound function g∘f: X → ℝ, x → (g∘f)(x) = g(f(x)); the codomain of f must coincide with the domain of g

f: ℝ → ℝ, x → f(x) = cos(x); g: ℝ → ℝ, x → g(x) = x³; h: ℝ → ℝ, x → h(x) = e^x; (g∘f)(x) = g(f(x)) = g(cos(x)) = (cos(x))³; (h∘g∘f)(x) = h(g(f(x))) = h(cos³(x)) = e^cos3(x)

f: ℝ → ℝ, x → f(x) = 1-x²; g: [0,+∞) → ℝ, x → g(x) = √(x); f(x) ∈ [0,+∞), f(X) must be in the domain of g ⇔ 1-x² ≥ 0 ⇔ x ∈ [-1,1]; f: [-1,1] → ℝ, x → f(x) = 1-x²; g: [0,+∞] → ℝ, x → g(x) = √(x); g∘f: [-1,1] → ℝ, x → g(f(x)) = √(1-x²)

The codomain of the internal function must be contained in the domain of the external function

f⋅g = g⋅f ⇔ f(x)g(x) = g(x)f(x), the product of two functions is commutative

f∘g ≠ g∘f, the composition product is not commutative

f(x) = |x|, x ∈ ℝ, f: ℝ → ℝ, x → f(x) = |x|; g(x) = sin(x), x ∈ ℝ, g: ℝ → ℝ, x → g(x) = sin(x); g∘f(x) = sin|x| → g∘f(3π/2) = sin|3π/2| = -1; f∘g(x) = |sin(x)| → f∘g(3π/2) = |sin(3π/2)| = |-1| = 1

f: A → B; 1) injective function if a₁ ≠ a₂ ⇒ f(a₁) ≠ f(a₂); 2) surjective function if ∀ b ∈ B, ∃ a ∈ A | f(a) = b; 3) bijective function if injective and surjective, that is ∀ b ∈ B ∃! a ∈ A | f(a) = b

An injective function, also known as injection, or one-to-one function, is a function that maps distinct elements of its domain to distinct elements of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain

A surjective function, also known as onto, or a surjection, is a function f from a set X to a set Y, that for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that f(x) = y; it is not required that x be unique; the function f may map one or more elements of X to the same element of Y

A bijective function, also known as bijection, or one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set; there are no unpaired elements; a bijective function f: X → Y is a injective and surjective mapping of a set X to a set Y

f: ℝ → ℝ, x → f(x) = x²; this function is non-injective and non-surjective; -1 = x² has no solutions in ℝ, so -1 is not image of any x through f, it is non-surjective; 1 = x² ⇔ x = ± 1, it is non-injective; f: ℝ → [0,+∞), x → f(x) = x², it is surjective but non-injective; f: [0,+∞) → [0,+∞), x → f(x) = x², it is bijective

Injective function ⇔ a line parallel to the x axis meets G(f) at most once

Surjective function ⇔ a line parallel to the x axis meets G(f) at least once

Bijective function ⇔ a line parallel to the x axis meets G(f) once and only once

An inverse function, or anti-function, is a function that reverses another function; if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x; g(y) = x if and only if f(x) = y; the inverse function of f is also denoted as f^-1

f: X → Y, bijective, ∃! g: Y → X | {g(f(x)) = x; f(g(y)) = y}; the function g is the inverse function of f and it is written as f^-1; the inverse function is the function that binds to y ∈ Y the only x ∈ X | f(x) = y ⇔ x = f^-1(y)

f: [0,+∞) → [0,+∞), x → f(x) = x²; f^-1: [0,+∞) → [0,+∞), x → f(x) = √(x); x² = y ⇔ x = √(y)

A function must be bijective to have an inverse function

(x,y) ∈ G(f) ⇔ y = f(x) ⇔ x = f^-1(y) ⇔ (y,x) ∈ G(f^-1); (x,y) and (y,x) are symmetric to the bisector of the 1° and 3° quadrant

G(f) and G(f^-1) are symmetric to y = x

f^-1(x) ≠ f(x)^-1, inverse does not mean reciprocal; f(x)^-1 = 1/f(x)

f: X → ℝ, 1) bounded from above if ∃ M ∈ ℝ | f(x) ≤ M ∀ x ∈ X, 2) bounded from belove if ∃ m ∈ ℝ | f(x) ≥ m ∀ x ∈ X, 3) bounded if ∃ m,M ∈ ℝ | m ≤ f(x) ≤ M ∀ x ∈ X

Symmetry: X | if x ∈ X ⇒ -x ∈ X and f: X → ℝ, 1) f is even if f(x) = f(-x) ∀ x ∈ X, 2) f is odd if f(x) = -f(-x) ∀ x ∈ X

f(x) = xⁿ n ∈ ℕ; 1) if n is even, the function is even, x² = (-x)²; 2) if n is odd, the function is odd, x³ = -(-x)³

If the function is even, the graph is symmetrical to the y axis; for example f(x) = x² that is the quadratic function or parabola

If the function is odd, the graph is symmetrical to the origin; for example f(x) = x³ that is the cubic function

Monotonic function: f: X ⊆ ℝ → ℝ, 1) increasing if ∀ x₁,x₂ ∈ X | x₁ ≤ x₂ ⇒ f(x₁) ≤ f(x₂), 2) decreasing if ∀ x₁,x₂ ∈ X | x₁ ≤ x₂ ⇒ f(x₁) ≥ f(x₂)

f(x) = 1/x, it is a hyperbola; decreasing in (-∞,0) and decreasing in (0,+∞), but it is not decreasing in (-∞,0) ∪ (0,+∞); if it were decreasing ⇒ -1 = f(-1) ≥ f(1) = 1, but this is absurd

A periodic function is a function that repeats its values at regular intervals, for example, the trigonometric functions, which repeat at intervals of 2π radians; any function that is not periodic is called aperiodic

f: X → ℝ is a periodic function if ∃ T ≠ 0 | f(x+T) = f(x) ∀ x ∈ X | x+T ∈ X; the smallest T > 0 is called period of the function

cos(x) = cos(x+2π) ∀ x ∈ ℝ, T = 2π; cos(x) = cos(x+4π) ∀ x ∈ ℝ

The fractional part (also known as mantissa) is a function that associates to each real number x its value minus its integer part, {x} = x-[x]; the mantissa function assumes all the values of the interval [0,1), it is periodic with a period equal to 1, it is neither even nor odd, and it is not an injective function, so it is not invertible

A polynomial is an expression consisting of variables, also called indeterminates, and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables

A polynomial is an elementary function

Polynomial: p(x) = a_nxⁿ+...+a₁x+a₀ = ⁿΣ_k=0 a_kx^k, a_i ∈ ℝ, a_n ≠ 0, n is the degree of the polynomial

A rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials

Rational function: f(x) = p(x)/q(x), p and q are polynomials; X = ℝ \ {x ∈ ℝ: q(x) = 0}; degree(f) = degree(p)-degree(q); for example f(x) = (x²+1)/(x³+3x+2), the degree is -1

Roots: xⁿ = {bijective from [0,+∞) to [0,+∞), n even; bijective from ℝ to ℝ, n odd}; f^-1(x) = ⁿ√(x), inverse function of f(x) = xⁿ; ⁿ√(x) = {defined from [0,+∞) to [0,+∞), n even; defined from ℝ to ℝ, n odd}

Real power and exponential: xⁿ = x⋅x⋅...⋅x n-times; 3⁴ = 3⋅3⋅3⋅3; 3^1/4 = ⁴√(3); 2^π = ?; π^√(2) = ?; a^r, a,r ∈ ℝ; 1) r ∈ ℕ a^r = a⋅...⋅a r-times ∀ a ∈ ℝ; 2) r = p/q ∈ ℚ q ≠ 0 a^r = a^p/q = ^q√(a^p) = (^q√(a))^p ∀ r ∈ ℚ if a > 0; r ∈ ℝ, r = p,α₀α₁α₂... p ∈ ℤ, α_i ∈ {0, 1, ..., 9}; 1.32 = 132/100; it is possible to construct a sequence of rational numbers that approximates r, considering a > 1 and r > 0, for a ∈ (0, 1) and r < 0 it is similar; r_n = p.α₀α₁α₂...α_n, π = 3.141592, r₀ = 3.1, r₁ = 3.14, r₂ = 3.141; 1) r_n ∈ ℚ ∀ n, r_n = (p.α₀α₁α₂...α_n)/10ⁿ; 2) {r_n} is an increasing sequence; 3) {r_n} is bounded, p ≤ r_n < p+1; an increasing and bounded sequence is convergent; 4) lim_n→∞ r_n = r, 0 ≤ r-r_n = 0.00...0α_n+1α_n+2 ≤ 0.00...10000 = 1/10ⁿ → 0 ⇒ lim_n→∞ r-r_n = 0 ⇔ lim_n→∞ r_n = r; defining the sequence {a^r_n}_n∈ℕ; 1) the sequence is well defined because r_n ∈ ℚ, considering that a > 0; 2) {a^r_n}_n∈ℕ is increasing because {r_n}_n is increasing and a > 1, if 0 < a < 1 it would be decreasing; 3) a^p ≤ a^r_n ≤ a^p+1 because p ≤ r < p+1 and a > 1; considering the regularity theorem of monotonous sequences, if a sequence is monotone and increasing, then it admits a limit, so ∃ lim_n→∞ a^r_n, and for definition a^r := lim_n→∞ a^r_n; if lim_n→∞ r_n = lim_n→∞ s_n = r, with {r_n} and {s_n} ⊆ ℚ, then a^r = lim_n→∞ a^r_n = lim_n→∞ a^s_n; a^ra^s = a^r+s, (a^r)^s = a^rs, valid for r,s ∈ ℚ but also for r,s ∈ ℝ

Real power: f(x) = x^r ∀ x ∈ (0,+∞), r ∈ ℝ; if r > 0, f(x) = x^r is defined for x = 0 with f(0) = 0; r ∈ ℕ, f(x) = x^r is defined ∀ x ∈ ℝ

Exponential: f(x) = a^x ∀ x ∈ ℝ, a > 0; f(x) = e^x, ∀ x ∈ ℝ, exponential; f(x) = 3^x, ∀ x ∈ ℝ, exponential of base 3; f(x) = π^x, ∀ x ∈ ℝ, exponential of base π; a^x, a > 1, is an increasing function, bounded below; a^x, a ∈ (0,1), is a decreasing function, bounded below

Hyperbolic functions: cosh(x) = (e^x+e^-x)/2, ∀ x ∈ ℝ; sinh(x) = (e^x-e^-x)/2, ∀ x ∈ ℝ; tanh(x) = sinh(x)/cosh(x) = (e^x-e^-x)/(e^x+e^-x); they are similar to Euler's formulas: e^ix = cos(x)+i⋅sin(x), x ∈ ℂ, cos(x) = (e^ix+e^-ix)/2, sin(x) = (e^ix-e^-ix)/2i; cosh(x) is an even function and it is bounded below by the point x=0 y=1; sinh(x) is an odd function; tanh(x) is an odd function and it is bounded above by the point x=0 y=1 and bounded below by the point x=0 y=-1; the fundamental identity of hyperbolic functions is cosh²(x)-sinh²(x) = 1, that is similar to cos²(x)+sin²(x) = 1; the graph of the hyperbolic cosine function is called catenary

Trigonometric functions: a trigonometric function is a real function which relate an angle of a right-angled triangle to ratios of two side lengths; one radian is defined as the angle subtended from the center of a circle which intercepts an arc equal in length to the radius of the circle; the magnitude in radians of a subtended angle is equal to the ratio of the arc length to the radius of the circle, that is θ = s/r, where θ is the subtended angle in radians, s is arc length, and r is radius; 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2, 120° = 2π/3, 135° = 3π/4, 150° = 5π/6, 180° = π, 210° = 7π/6, 225° = 5π/4, 240° = 4π/3, 270° = 3π/2, 300° = 5π/3, 315° = 7π/4, 330° = 11π/6, 360° = 2π; sin(x) is a periodic function with period of 2π, odd, bounded, -1 ≤ sin(x) ≤ 1 ∀ x ∈ ℝ; cos(x) is a periodic function with period of 2π, even, bounded, -1 ≤ cos(x) ≤ 1 ∀ x ∈ ℝ; tan(x) = sin(x)/cos(x) ∀ x ∈ ℝ \ {x ∈ ℝ | cos(x) = 0} ⇔ ∀ x ∈ ℝ \ {π/2 + kπ, k ∈ ℤ}, tan(x) is a periodic function with period of π, odd, not bounded; fundamental identity: cos²(x)+sin²(x) = 1

LIMIT OF A REAL FUNCTION

f(x) = sin(x)/x ∀ x ∈ ℝ \ {0}, the limit in 0 exists, it is a regular case; g(x) = sin(1/x) ∀ x ∈ ℝ \ {0}, the limit in 0 does not exist, it is an irregular case

Limit point or cluster point or accumulation point: c ∈ ℝ = ℝ ∪ {±∞}, X ⊆ ℝ, c is an accumulation point for X if ∃ {x_n}_n∈ℕ | 1) x_n ∈ X ∀ n ∈ ℕ, 2) x_n ≠ c ∀ n ∈ ℕ, 3) lim_n→∞ x_n = c; c ∈ ℝ can be approximated by points contained in X; X = ℝ \ {0}, c = 0 is an accumulation point for X, x_n = 1/n; X = (a,b) or [a,b], the set of accumulation points of X, identified as X, is [a,b]; the set of accumulation points of an open or closed interval is always the whole closed interval; x_n = c+1/n², c=1, x_n = 1+1/n², c=2, x_n = 2-1/n; X = ℕ, the only accumulation point is c = +∞, x_n = n; the only accumulation point of the natural number set is +∞; X = {-1} ∪ {1} ∪ [3,+∞), X = [3,+∞) ∪ {+∞}, -1,1 are isolated point of X; X = {1/n, n ∈ ℕ}, c = 0 is the only accumulation point, x_n = 1/n

What is the set of limit points for ℚ?

f: X ⊆ ℝ → ℝ, c ∈ ℝ is an accumulation point for X, then f tends to l as x tends to c, lim_x→c f(x) = l, where l ∈ ℝ if ∀ {x_n}_n∈ℕ | x_n ∈ X ∀ n ∈ ℕ, x_n ≠ c ∀ n ∈ ℕ, lim_n→∞ x_n = c ⇒ lim_n→∞ f(x_n) = l; because c is an accumulation point for X, and exists at least a sequence {x_n}_n ⊆ X | x_n → c for n → ∞; l must not depend on the sequence {x_n}_n∈ℕ that converges to the accumulation point c

f(x) = sin(x)/x ∀ x ∈ ℝ \ {0}; lim_x→0 sin(x)/x = 1 ⇔ ∀ {x_n}_n∈ℕ | x_n ≠ 0 and lim_n→∞ x_n = 0 ⇒ lim_n→∞ f(x_n) = lim_n→∞ sin(x_n)/x_n = 1, because lim_n→∞ sin(a_n)/a_n = 1 if lim_n→∞ a_n = 0

∄ lim_x→0 sin(1/x); to prove that a limit does not exist, it is necessary to find two sequences that converge at the accumulation point, in this case 0, and the corresponding limits of the images must be different from each other; x_n = 1/2πn ≠ 0, lim_n→∞ x_n = lim_n→∞ 1/2πn = 0; lim_n→∞ g(x_n) = lim_n→∞ sin(1/x_n) = lim_n→∞ sin(2πn) = 0; y_n = 1/(π/2+2πn) ≠ 0, lim_n→∞ y_n = lim_n→∞ 1/(π/2+2πn) = 0, lim_n→∞ g(y_n) = lim_n→∞ sin(1/y_n) = lim_n→∞ sin(π/2+2πn) = 1; the limit of the images is different and it depends on the chosen sequence

This definition of limit is well defined for c ∈ ℝ ∪ {±∞} and l ∈ ℝ ∪ {±∞}

f: X ⊆ ℝ → ℝ, c ∈ ℝ, l ∈ ℝ, then lim_x→c f(x) = l if ∀ ε > 0 ∃ δ = δ(ε) > 0 | if x ∈ X and 0 < |x-c| < δ ⇒ |f(x)-l| < ε ⇔ l-ε < f(x) < l+ε

The definition of a limit of a real function that uses the sequences is unique, instead the definition of a limit of a real function using ε and δ must be formulated in 6 different cases, c = ℝ, c = -∞, c = + ∞, l = ℝ, l = -∞, l = + ∞

l = +∞, c ∈ ℝ, lim_x→c f(x) = +∞ ⇔ ∀ M ∈ ℝ ∃ δ = δ(M) > 0 | if x ∈ X, 0 < |x-c| < δ ⇒ f(x) > M

The limit value does not depend on the value at the accumulation point; the function could be defined or undefined at the accumulation point

Calculating a limit we do not see what happens in the accumulation point, but we see what happens in the neighboring points that tend to the accumulation point, in fact x_n → 0, x_n ≠ 0

f(x) = {1, x = 0; 0, x ≠ 0}; lim_x→0 f(x) = 0 ⇒ f(x_n) = 0 ⇒ lim_n→∞ f(x_n) = 0

f(x) = x/|x| = {1, x > 0; -1, x < 0}; f(x) = sgn(x) ∀ x ∈ ℝ \ {0}; lim_x→0 f(x) ∄; x_n = 1/n → 0 for n → ∞ and lim_n→∞ f(x_n) = lim_n→∞ (1/n)/|1/n| = 1, y_n = -1/n → 0 for n → ∞ and lim_n→∞ f(y_n) = lim_n→∞ (-1/n)/|-1/n| = -1, ⇒ ∄ lim_x→0 f(x)

One-sided limit: {x_n}_n∈ℕ, c ∈ ℝ, 1) lim_n→∞ x_n = c⁺ if {lim_n→∞ x_n = c, x_n ≥ c ∀ n ∈ ℕ}, 2) lim_n→∞ x_n = c^- if {lim_n→∞ x_n = c, x_n ≤ c ∀ n ∈ ℕ}; limit from right x_nn→∞ → c⁺, limit from left x_nn→∞ → c^-; limit from right: c ∈ ℝ, l ∈ ℝ, lim_x→c⁺ f(x) = l if ∀ {x_n}_n∈ℕ | x_n ∈ X, x_n ≠ c, lim_n→∞ x_n = c⁺, then lim_n→∞ f(x_n) = l; limit from left: c ∈ ℝ, l ∈ ℝ, lim_x→c^- f(x) = l if ∀ {x_n}_n∈ℕ | x_n ∈ X, x_n ≠ c, lim_n→∞ x_n = c^-, then lim_n→∞ f(x_n) = l

lim_x→0⁺ x/|x| = 1; lim_x→0^- x/|x| = -1; this function admits limits from right and left, but the limits from right and left are different and therefore does not admit limit

Necessary and sufficient condition for lim_x→c f(x) = l is lim_x→c^- f(x) = lim_x→c⁺ f(x) = l; if the limit exists, the limits from the right and from the left exist and coincide with the value l, vice versa if the limits from the right and from the left exist and coincide, then the limit of the function exists and has value l

lim_x→0 1/x; x_n = 1/n → 0, lim_n→∞ 1/x_n = lim_n→∞ n = +∞; y_n = -1/n → 0, lim_n→∞ 1/y_n = lim_n→∞ (-n) = -∞

lim_x→0⁺ 1/x = +∞, {x_n} | lim_n→∞ x_n = 0⁺ ⇒ lim_n→∞ 1/x_n = 1/0⁺ = +∞; lim_x→0^- 1/x = -∞

If lim_x→±∞ f(x) = l ∈ ℝ, then the line y = l is a horizontal asymptote in ±∞

If lim_x→c f(x) = ±∞, c ∈ ℝ, then the line x = c is a vertical asymptote for f

Limit and algebraic operations: lim_x→c f(x) = l, lim_x→c g(x) = m, m,l ∈ ℝ; lim_x→c (f(x)+g(x)) = l+m; lim_x→c (f(x)-g(x)) = l-m; lim_x→c f(x)⋅g(x) = l⋅m; lim_x→c f(x)/g(x) = l/m, m ≠ 0; lim_x→c f(x)^g(x) = l^m, l > 0; lim_x→c |f(x)| = |l|; if l,m ∈ ℝ these rules are useful for the determinate forms, in the case of an indeterminate form this is useless

Limit of compound function: f: X ⊆ ℝ → Y, g: Y → ℝ, c is an accumulation point for X, 1) lim_x→c f(x) = y₀, 2) lim_y→y0 g(y) = l, 3) ∃ δ > 0 | 0 < |x-c| < δ, x ∈ X ⇒ f(x) ≠ y₀, then lim_x→c (g∘f)(x) = lim_x→c g(f(x)) = l

The limit of a compound function is the limit of the external function with accumulation point given by the limit of the internal function

lim_x→+∞ sin(1/x²); h(x) = sin(1/x²) = g(f(x)) = {f(x) = 1/x²; g(y) = sin(y)}; lim_x→+∞ 1/x² = 0, lim_x→+∞ sin(1/x²) = lim_y→0 sin(y) = 0

f(x) = 0 ∀ x ∈ ℝ; g(y) = {1, x ≠ 0; 0, x = 0}; lim_x→0 g(f(x)) = lim_x→0 g(0) = 0; lim_y→0 g(y) = 1; the condition f(x) ≠ y₀ is not met

Order of limits for real functions: f,g: X ⊆ ℝ → ℝ, c is the accumulation point for X, f(x) ≤ g(x) ∀ x ∈ X, lim_x→c f(x) = l, lim_x→c g(x) = m, then l ≤ m

Squeeze theorem or theorem of carabinieri for real functions: f,g,h: X → ℝ, f(x) ≤ h(x) ≤ g(x) ∀ x ∈ X, lim_x→c f(x) = lim_x→c g(x) = l, then lim_x→c h(x) = l

Theorem of permanence of the sign for real functions: f: X ⊆ ℝ → ℝ, f(x) ≥ 0 ∀ x ∈ X, lim_x→c f(x) = l, then l ≥ 0

Generalizing the permanence of the sign, if f(x) ∈ [α,β] then l ∈ [α,β]

The theorem of permanence of the sign is true for not strict inequalities which are ≥ and ≤, but it is false for strict inequalities which are > and <; f(x) = 1/x > 0 ∀ x ∈ (0,+∞), lim_x→+∞ 1/x = 0

lim_x→0 sin(x)/x = 1; x > 0 ⇒ sin(x) > 0, 0 ≤ sin(x) ≤ x ≤ tan(x) ≤ sin(x)/cos(x); sin(x)/sin(x) ≤ x/sin(x) ≤ (sin(x)/cos(x))(1/sin(x)); 1 ≤ x/sin(x) ≤ 1/cos(x); cos(x) ≤ sin(x)/x ≤ 1, for x→0⁺ cos(x) goes to 1, so for the squeeze theorem or theorem of carabinieri lim_x→0⁺ sin(x)/x = 1