Exact upper bound of the proof set. Existence of an upper (lower) face

Rice. 1.12

Comment. It is proved in a similar way that the infimum is unique.

If E is not bounded above, then write sup E = +¥, similarly, if E is not bounded below, then write inf E= -¥ .

Chapter 2. Sequences

2.1. Basic Sequence Concepts

Number sequence And various concepts, associated with sequences. In particular, edges, limit, monotony.

OPR1.

OPR2. exact upper bound and is designated sup A.

OPR2'.

UTV. OPR2. ó OPR2’.

=> OPR2 is fulfilled, i.e. M = sup A – the smallest of all upper bounds => M – upper bound of set A => (i.e. 1) OPR2’ has been completed).

Dm 2) by contradiction, i.e. the upper bound of the set A, and M is not the smallest upper bound - a contradiction, since M is the upper bound => property 2) OPR2’ is satisfied.

<= выполнено ОПР2’, т.е.

Because M' on the top. Face of set A, sl-but, M – smallest upper bound of set A => OPR2 is fulfilled.

Ticket number 2 page 2

OPR3.

OPR4. exact bottom edge and is designated inf A.

OPR4'.

UTV. OPR4. ó OPR4’

The proof is similar with UTV. OPR2. ó OPR2’.

THEOREM!!!

DOC-VO!!!

Comment: if set A is not bounded above => it has no upper bounds =>

Ticket No. 1 “LIMITED AND UNLIMITED SETTS. EXAMPLES".

OPR1: number A name. bounded above, If . In this case, M is the top. edge of mn-va A.

Example: And it is limited from above. M = 3 – upper bound. Any number greater than 3 is the upper bound.

OPR2: number A name. bounded below, If . In this case, m is the lower one. edge of mn-va A.

Example:

N – bounded from below. m = 1 – lower bound. Any number less than 1 will be the lower bound.

OPR3: number A name. limited, if it is bounded above and below, i.e. .

OPR3': number A name. limited, If

WE PROVE THAT OPR3 ó OPR3’

=> N.D. OPR3 => OPR3’

We have: Let

Those. completed OPR3'

<= Н.Д. ОПР3’ =>OPR3

We have: ,i.e. completed OPR3.

OPR4. Mn – in A is called unlimited, If

Ticket No. 3 “NUMERIC SEQUENCES”.

OPR. If for each natural number we put in correspondence a number according to some law, then the number is the set of numbers , called the numerical sequence. let's denote the number of the last one. ; numbers - elements of the sequence

Example:

OPR. The number a is called the limit of the last. , if (for any positive number)

Indicated by:

Example:

Designation: neighborhood t.a.

Ticket No. 4 “B.M. THE LAST AND THEIR SAINTS (2 THEOREMS).”

OPR. The last is called infinitesimal (infinitesimal) if

Example: b.m.last

SV-VA:

THEOREM_1!!! let it be - b.m. afterbirth, then:

1) Afterbirth b.m.last

2) Afterbirth b.m.last

DOC-VO!!!

1) given: b.m, i.e.

Dm, what b.m. afterbirth, i.e.

Let's select and label it .

Because b.m. => for number ,

B.m. => for number

Because put number =>

2) Um, what b.m.last

Let's select and designate it.

B.m. => for number,

B.m. => for number

Ticket No. 4 page 2

Because put number => def. is performed. b.m. for , i.e. b.m.

THEOREM_2!!!

Let b.m.last, limited. positive afterbirth, then b.m.positive sequence

OPR. Afterbirth. limited If

DOC-VO!!!

We fix it.

Limit. =>

B.m.last => for

Consequence:

Let b.m.last. Then for last b.m.

Indeed, consider afterbirth.

Ogre afterbirth. b.m, t.k b.m.

Example:

THAT. according to THEOREM_2!!!

Comment:

From THEOREM_1!!! It follows that

1) the amount of any finite number b.m. afterbirth. there is a b.m.last.

2) the product of any finite number of b.m. afterbirth. there is b.m. afterbirth.

Ticket No. 5 “BB SEQUENCES AND THEIR RELATIONSHIP WITH BM SEQUENCES.”

OPR. let it be called b.b.last, if

Let's denote

THEOREM!!! Let b.b.last., Then b.m.last.

DOC-VO!!!

Fixed Afterbirth

THAT.
b.m. afterbirth.

CONNECTION OF BB WITH BM SEQUENCES.

B.b. afterbirth. b.m. afterbirth. Inverse relationship.

Ticket 18 properties of limits of functions (a) uniqueness of the limit. B) limited functions that have a limit.)

Uniqueness of the limit

THEOREM!!! If f-i has a limit at K®0, then it is unique

DOC-VO!!!(from the opposite side)

Let And

Rassm X n¹a " n

Because Þ for given (X n ) sequence

Þ for a given ( X n ) sequence

That. ( f(x)-ch.p-t)opposite because cannot have

b¹c 2 different limits Þ in = c

.With

Consequences

Question No. 22 2nd wonderful limit

Consequences

(an-no a x =lna)

Bil22str4
Ticket 23 properties bm functions

ticket 24 bb functions and their connection with bb

Ticket 26.equivalence bm f-ii.(table, t.)

ticket 26 page 2

Ticket 25. Comparison of bm f-y.

Ticket 28. Inaccurate f-ii at point.

beat.28

TICKET 30. classification of discontinuity points of a function (definition and examples)

Let f(x) def. in some U(a) (m.b. excluding t.a. itself). t.a. called break point functions f(x), if f is not constant in t.a. let t.a. be the discontinuity point of the function f(x).

Def. 1) t.a.-break point 1st kind, if (i.e. nouns are finite one-sided)

2) If, in addition, then t.a- removable break point.

3) t.a. - break point 2nd kind , if it is not a rupture of the 1st kind.

Examples. 1)y=sgn(x). x=0-t.r. of the 1st kind, because

2)y= , x=0 –t. device once, because

3) y= x=0 – t.r. of the 2nd kind, because

,

Discontinuity point of the 2nd kind.

3).

,

x=0 is a discontinuity point of the 2nd kind.

4).

There is no point x=0 - a discontinuity point of the 2nd kind.

, . Point x=0 is a discontinuity point of the 2nd kind.

Ticket No. 2 “UPPER AND LOWER BOUNDS OF THE NUMERIC SET. THEOREM ON THE EXISTENCE OF EXACT LOWER AND UPPER BOUNDS OF A SET.

OPR1. M – the upper bound of the set A ó if .

OPR2. the smallest of all the upper faces of the set A, called exact upper bound and is designated sup A.

OPR2'. The number M is called the exact upper edge of the number A if

UTV. OPR2. ó OPR2’.

=> OPR2 is fulfilled, i.e. M = sup A – the smallest of all upper bounds => M – upper bound of set A => (i.e. 1) OPR2’ has been completed).

Dm 2) by contradiction, i.e. the upper bound of the set A, and M is not the smallest upper bound - a contradiction, since M is the upper bound => property 2) OPR2’ is satisfied.

<= выполнено ОПР2’, т.е.

It is clear that M is the least upper bound.

Dm by contradiction, i.e. Let M be the non-smallest top face. Designation according to St. 2) for this contradiction.

Because M' on the top. Face of set A, sl-but, M – smallest upper bound of set A => OPR2 is fulfilled.

Ticket number 2 page 2

OPR3. m – lower bound of set A ó if .

OPR4. the largest of all the lower faces of the set A, called exact bottom edge and is designated inf A.

OPR4'. The number m is called the exact infimum of the set A if

UTV. OPR4. ó OPR4’

The proof is similar with UTV. OPR2. ó OPR2’.

THEOREM!!! Every nonempty set bounded above (below) has an exact upper (lower) bound.

DOC-VO!!! Non-empty set A – limited. from above, then the set A has at least one upper bound. Let Y be the set of all upper faces of the set A, i.e. , and the set Y is non-empty, because set A has at least one upper bound.

THAT. non-empty sequences A and Y and continuous according to the origin. valid numbers i.e. upper bound of mn-va A. M = sup A.

Comment: if set A is not bounded above => it has no upper bounds => there is no exact upper bound. In this case, it is sometimes believed that . Likewise, if the set A is not limited. from below, it is sometimes believed that

MATHEMATICAL ANALYSIS

Part I

LIMIT THEORY. Sequence limit and function limit. Existence theorem for an exact supremum.

Let the variable x n takes an infinite sequence of values

x 1 , x 2 , ..., x n , ..., (1)

and the law of change of variable is known x n, i.e. for everyone natural number n you can specify the appropriate value x n. Thus it is assumed that the variable x n is a function of n:

x n = f(n)

Let us define one of the most important concepts of mathematical analysis - the limit of a sequence, or, what is the same, the limit of a variable x n, running through the sequence x 1 , x 2 , ..., x n , ... . .

Definition. Constant number a called limit of the sequence x 1 , x 2 , ..., x n , ... . or the limit of a variable x n, if for an arbitrarily small positive number e there is such a natural number N(i.e. number N) that all values ​​of the variable x n, starting from x N, differ from a in absolute value less than by e. This definition briefly written like this:

| x n - a |< (2)

in front of everyone n  N, or, what is the same,

Determination of the Cauchy limit. A number A is called the limit of a function f (x) at a point a if this function is defined in some neighborhood of the point a, with the possible exception of the point a itself, and for every ε > 0 there exists δ > 0 such that for all x satisfying condition |x – a|< δ, x ≠ a, выполняется неравенство |f (x) – A| < ε.

Determination of the Heine limit. A number A is called the limit of a function f (x) at a point a if this function is defined in some neighborhood of the point a, with the possible exception of the point a itself, and for any sequence such that converging to a number a, the corresponding sequence of function values ​​converges to the number A.

If a function f (x) has a limit at point a, then this limit is unique.

The number A 1 is called the limit of the function f (x) on the left at point a if for every ε > 0 there exists δ >

The number A 2 is called the limit of the function f (x) on the right at point a if for each ε > 0 there is δ > 0 such that the inequality holds for all

The limit on the left is denoted by the limit on the right - These limits characterize the behavior of the function to the left and right of point a. These are often called one-way limits. In the designation of one-sided limits for x → 0, the first zero is usually omitted: and. So, for the function

If for every ε > 0 there exists a δ-neighborhood of a point such that for all x satisfying the condition |x – a|< δ, x ≠ a, выполняется неравенство |f (x)| >ε, then they say that the function f (x) has an infinite limit at point a:

Thus, the function has an infinite limit at the point x = 0. Limits equal to +∞ and –∞ are often distinguished. So,

If for every ε > 0 there is a δ > 0 such that for every x > δ the inequality |f (x) – A|< ε, то говорят, что предел функции f (x) при x, стремящемся к плюс бесконечности, равен A:

Existence theorem for an exact supremum

Definition:АR mR, m is the upper (lower) face of А, if аА аm (аm).

Definition: A set A is bounded from above (from below), if there exists an m such that aA, am (am) holds.

Definition: SupA=m, if 1) m is the supremum of A

2) m’: m’ m’ is not the supremum of A

InfA = n, if 1) n is the infimum of A

2) n’: n’>n => n’ is not the infimum of A

Definition: SupA=m is a number such that: 1)  aA am

2) >0 a  A, such that a  a-

InfA = n is a number such that: 1) 1)  aA an

2) >0 a  A, such that a E a+

Theorem: Any non-empty set AR bounded from above has an exact supremum, and a unique one.

Proof:

Let's construct the number m on the number line and prove that this is the supremum of A.

[m]=max([a]:aA) [[m],[m]+1]A=>[m]+1 - upper bound of A

Segment [[m],[m]+1] - divided into 10 parts

m 1 =max:aA)]

m 2 =max,m 1:aA)]

m k =max,m 1 ...m K-1:aA)]

[[m],m 1 ...m K , [m],m 1 ...m K + 1 /10 K ]A=>[m],m 1 ...m K + 1/ 10 K - top edge A

Let us prove that m=[m],m 1 ...m K is the supremum and that it is unique:

k: then there is a point at which the function reaches its maximum, there is a point at which the function reaches its minimum.

Proof:

Let the function f(x) be continuous on , then by Theorem 1 it is bounded on this interval. Consequently, the set of function values ​​is limited. Then, by virtue of the supremum principle, this set has an exact upper and an exact lower bound.

Let us denote: and show that this will be the largest value of the function f(x) on the segment : .

Let's assume the opposite, that is, .

Since , then f(x)< .

let us introduce the function . The function is continuous on , since -f(x) 0. Then, by Weierstrass’s first theorem, the function is bounded on .

, where >0

Since this inequality holds, the number is not the exact upper bound of the set of function values. We arrive at a contradiction, which means our assumption is incorrect. Similarly, one can prove that a continuous function reaches its minimum value on a segment. The theorem is proven.

DIFFERENTIABLE FUNCTIONS Theorems of Rolle and Lagrange. Formula TEylor with a remainder term in Lagrange form.

Rolle's theorem. If the function f(x) is continuous on the closed interval [a, b], has a derivative inside the interval and if

f(a) = f(b)

then inside the interval [a, b] there is at least one such value x 0 (a< x 0 < b), что

f "(x 0 ) = 0.

Proof. Let's consider two cases.

1. Function f(x) is constant on the interval [ a, b]; Then f" (x) = 0 for anyone x(a< x < b) , i.e. the statement of Rolle's theorem is carried out automatically.

2. Function f(x) is not constant (Figure 1); then it reaches its largest or smallest or both of these values ​​at the inner point of the interval, because f(b) = f(a), and if f(a)- the smallest value, then the largest value value function f(x) will take inside the interval.

Since, by condition, f(x) has at the point x 0 derivative, then by the theorem on the necessary criterion for an extremum,

f "(x 0 ) = 0 ,

and Rolle's theorem is proven.

Rolle's theorem has a simple geometric interpretation: if an arc AB of a curve y = f(x) is given, at each point of which there is a tangent, and the ends A and B are at the same distance from the Ox axis, then on this arc there is at least one point at which the tangent t to the curve will be parallel to the chord contracting the arc, and therefore to the Ox axis(see figure 1).

If we rotate the coordinate axes by angle a, then the ends A And B arcs AB will no longer be at the same distance from the axis Ox", but tangent t will still be parallel to the chord AB(see figure 1). Therefore, it is natural to expect that the theorem holds: If an arc AB of a curve y = f(x) with a continuously varying tangent is given, then on this arc there is at least one point at which the tangent is parallel to the chord AB subtending it(Figure 2).

Lagrange's theorem. If the function f(x) is continuous on a closed interval[a, b] and inside it has a derivative f "(x), then there is at least one such value x 0 (a< x 0 < b), что

f(b) - f(a) = (b - a)f "(x).

Proof. Consider the helper function

F(x) = f(x) - k(x - a),

Where - angular coefficient of the chord AB(see figure 2).

This function satisfies all the conditions of Rolle's theorem.

In fact, when x = a we have F(a) = f(a) - k(a - a) = f(a), at x = b we have

Moreover, since the function f(x) And k(x - a) continuous on [ a, b] and differentiable in ( a, b), then the function F(x) = f(x) - k(x - a) continuous on [ a, b] and differentiable in ( a, b).

Therefore, by Rolle’s theorem, in the interval ( a, b) there is such a point x 0 , What

F"(x 0 ) = 0 ,

f "(x 0 ) - k = 0

From here we have

f(b) - f(a) = (b - a)f " (x 0 ) ,

Q.E.D.

Because a + (b - a) = b, then the value a+(b - a), where Q is a proper positive fraction (0 <  < 1) , is equal to some number in the interval ( a, b), therefore Lagrange’s formula can be written in the form

f(b) - f(a) = (b - a)f "

If you put a = x, b = x +x, where b - a =x, then the Lagrange formula will be written in the form

y = f(x +x) - f(x) =xf"(x+x).

It was previously proven that if a function is equal to a constant C at any value x in the interval (a, b), then its derivative is equal to zero.

Let us now prove the converse theorem, which is a consequence of Lagrange’s theorem:

If the derivative f "(x) vanishes for any values ​​of x in the interval (a, b), then in this interval f(x) = C.

In fact, if x 1 And x 2 - any two values ​​in the interval (a, b), then by Lagrange’s theorem, we have

f(x 2 ) - f(x 1 ) = (x 2 - x 1 )f"(x 0 ),

Where, x 1 < x 0 < x 2 . But since f"(x 0 ) = 0 , That

f(x 2 ) - f(x 1 ) = 0,

which proves our theorem.

An important theorem follows directly from this:

If two functions f 1 (x) and f 2 (x) have the same derivative in the interval (a, b), then they differ from each other by a constant value on this interval.

Indeed, consider the function

(x) = f 2 (x)-f 1 (x).

Then for any value x from the interval (a, b)

"(x) = f 2 "(x)-f 1 "(x) = 0.

But this means that  (x) = C and therefore

f 2 (x)-f 1 (x) = C.

Taylor's formula. Let on the intervalthe function f(x) is differentiable n times and the following equalities hold:

f(a) = f(b) = f "(a) = f ""(a)= ... = f (n-1) (a)=0

Then inside the intervalthere is at least one value with,at which

f (n) (c) = 0

Proof. By Rolle's theorem we have

f "(x 0 ) = 0 ,

Where a< x 0 < b . Then f "(x) on the interval satisfies Rolle’s theorem, since, by condition, f "(a) = 0 And f "(x 0 ) = 0 , and therefore

f ""(x 1 ) = 0 ,

Where a< x 1 < x 0 .

Applying Rolle's theorem successively to the functions f ""(x), f """(x), ..., f (n-1) (x), we finally find:

f (n) (c) = 0,

Where a< c < x n-1 < b . The theorem is proven.

Let us now derive Taylor formula with remainder term in Lagrange form.

Let the function f(x) differentiable n times on the interval.

Consider the helper function

(x) = f(x) - P(x),

Let's differentiate n times the function  (x). Then we will have

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

 (n-1) (x) = f (n-1) (x)-A n-1 - A n (x - a),

 (n) (x) = f (n) (x)-A n

We require that the function  (x) satisfied the conditions of the generalized Rolle theorem. Then we will have

(1) .

Since the function  (x) satisfies the conditions of the generalized Rolle theorem, then there is such a value with (a< c < b) , What

 (n) (c) = f (n) (c) - A n = 0 (2)

Let us prove another theorem, which is based on the property of continuity of real numbers.

Theme about the existence of an upper (lower) face. First, let's introduce a few definitions.

Definition. Numerical set X is called bounded above if there is a number M such that x ≤ M for any element x from many X .

Definition. Numerical set X is called bounded below if there is a number m such that x ≥ m for any element x from many X .

Definition. Numerical set X is called bounded if it is bounded above and below.

In symbolic notation, these definitions would look like this:

many X bounded above if ∃M ∀x ∈ X: x ≤ M ,

bounded below if ∃m ∀x ∈ X: x ≥ m And

limited if ∃m, M ∀x ∈ X: m ≤ x ≤ M .

Definition. For any number a R non-negative number

it's called absolute value or module. For absolute values numbers the inequality is true |a+b| < |a|, which follows from the definition of the modulus of a number and from the axioms of addition and order.

Theorem 4.3.1. Numerical set X is bounded if and only if there is a number C such that for all elements x from this set the inequality ≤ C.

Proof. Let the set X limited. Let's put C =max(m, M)- the largest of the numbers m and M. Then, using the properties of the modulus of real numbers, we obtain the inequalities x ≤M≤M ≤C and x≥m≥ −m≥ −C, which implies that ≤ C .

Conversely, if the inequality ≤ C holds, then −C ≤ x ≤ C . This is what is required if we put M = C and m = −C .◄

Number M, limiting the set X on top, called upper bound of the set. If M- upper bound of the set X, then any number M′, which is greater M, will also be the upper bound of this set. Thus, we can talk about the set of upper bounds for the set X. Let us denote the set of upper bounds by . Then, ∀x ∈ X and ∀M ∈ the inequality will be satisfied x ≤M, therefore, by the axiom of continuity there is a number such that x ≤ ≤ M. This number is called the exact upper bound of a number set X or the upper bound of this set or the supremum of the set X and is designated =sup X. Thus, we have proven that every non-empty number set that is bounded above always has an upper bound.

It is obvious that equality = sup X is equivalent to two conditions:

1) ∀x ∈ X the inequality x ≤ holds, i.e. - upper bound of the set X ;

2) ∀ε > 0 ∃xε ∈ X so that the inequality xε > −ε holds, i.e. this limit cannot be improved (reduced).

Similarly, one can prove that if a set is bounded below, then it has an infimum that is also called the infimum of the set X and is denoted by inf X. The equality =inf X is equivalent to the conditions:

1) ∀x ∈ X inequality holds x ≥ ;

2) ∀ε > 0 ∃xε ∈ X so that the inequality holds xε< + ε .

If a set X has the largest element, then we will call it

the maximum element of the set X and denote = max X . Then

supX =. Similarly, if there is a smallest element in a set, then we will call it minimal, denote minX and it will be the infimum of the set X .

Let us formulate several properties of the upper and lower faces:

Property 1. Let X- some numerical set. Let us denote by −X many (− x| x ∈ X ). Then sup (− X) = − inf X And inf (− X) = − sup X .

Property 2. Let X- some number set λ – a real number. Let us denote by λX many (λx | x ∈ X). Then if λ ≥ 0, then sup(λX) = λ supX , inf(λ X)= λ infX and if λ < 0, то sup(λ X)=λ infX , inf(λ X)=λ supX .

Property 3. Let X1 and X2- numerical sets. Let us denote by X1+X2 many ( x1+ x2 | x1 ∈ X1, x2 ∈ X2 ) and through X1 − X2 many (x1 − x2 | x1 ∈ X1, x2 ∈ X2). Then sup(X1 + X2)=supX1+supX2, inf(X1+X2)=infX1 +inf X2 , sup(X1 − X2) = sup X1 − inf X2 and inf (X1 − X2) = inf X1 − sup X2 .

Property 4. Let X1 and X2 be number sets all of whose elements are non-negative. Then sup (X1*X2) = sup X1 *sup X2 , inf (X1*X2) = inf X1* inf X2 .

Let us prove, for example, the first equality of Property 3. Let x1 ∈ X1, x2 ∈ X2 and x=x1+x2. Then x1 ≤ sup X1, x2 ≤ sup X2 And x ≤ sup X1 + sup X2, where sup(X1 + X2) ≤ sup X1 + sup X2 .

To prove the opposite inequality, take the number y . Then we can find the elements ∈ X1 and ∈ X2 such that ^{y . This means that there is an element = + ∈ X1+X2, which is greater than the number y and sup X1 + sup X2 = sup (X1 + X2). The remaining relations are proved similarly.}