4.1. Functional series: basic concepts, area of ​​convergence

Definition 1. A series whose members are functions of one or
several independent variables defined on a certain set is called functional range.

Consider a functional series, the members of which are functions of one independent variable X. Sum of first n members of a series is a partial sum of a given functional series. General member there is a function from X, defined in a certain region. Consider the functional series at the point . If the corresponding number series converges, i.e. there is a limit on the partial sums of this series
(Where − sum of a number series), then the point is called convergence point functional range . If the number series diverges, then the point is called divergence point functional range.

Definition 2. Area of ​​convergence functional range is called the set of all such values X, at which the functional series converges. The convergence region, consisting of all convergence points, is denoted . Note that R.

The functional series converges in the region , if for any it converges like a number series, and its sum will be some function . This is the so-called limit function sequences : .

How to find the area of ​​convergence of a function series ? You can use a sign similar to d'Alembert's sign. For a row compose and consider the limit for a fixed X:
. Then is a solution to the inequality and solving the equation (we take only those solutions of the equation in
which corresponding number series converge).

Example 1. Find the area of ​​convergence of the series.

Solution. Let's denote , . Let's compose and calculate the limit
, then the region of convergence of the series is determined by the inequality and the equation . Let us further investigate the convergence of the original series at the points that are the roots of the equation:

and if , , then we get a divergent series ;

b) if , , then the series converges conditionally (by

Leibniz's criterion, example 1, lecture 3, section. 3.1).

Thus, the region of convergence series looks like: .

4.2. Power series: basic concepts, Abel's theorem

Let us consider a special case of a functional series, the so-called power series , Where
.

Definition 3. Power series is called a functional series of the form,

Where − constant numbers called coefficients of the series.

A power series is an “infinite polynomial” arranged in increasing powers . Any number series is
a special case of a power series for .

Let us consider the special case of a power series for :
. Let's find out what type it is
convergence region of this series .

Theorem 1 (Abel's theorem). 1) If the power series converges at a point , then it converges absolutely for any X, for which the inequality holds .

2) If the power series diverges at , then it diverges for any X, for which .

Proof. 1) By condition, the power series converges at the point ,

i.e. the number series converges

(1)

and according to the necessary criterion of convergence, its common term tends to 0, i.e. . Therefore, there is such a number that all members of the series are limited by this number:
.

Let us now consider any X, for which , and make a series of absolute values: .
Let's write this series in a different form: since , then (2).

From inequality
we get, i.e. row

consists of terms that are greater than the corresponding terms of series (2). Row represents a convergent series of a geometric progression with a denominator , and , because . Consequently, series (2) converges at . Thus, the power series absolutely matches.

2) Let the series diverges at , in other words,

number series diverges . Let us prove that for any X () the series diverges. The proof is by contradiction. Let for some

fixed ( ) the series converges, then it converges for all (see the first part of this theorem), in particular, when , which contradicts condition 2) of Theorem 1. The theorem is proven.

Consequence. Abel's theorem allows us to judge the location of the convergence point of a power series. If the point is the point of convergence of the power series, then the interval filled with convergence points; if the point of divergence is the point , That
infinite intervals filled with divergence points (Fig. 1).

Rice. 1. Intervals of convergence and divergence of the series

It can be shown that there is such a number that in front of everyone
power series converges absolutely, and when − diverges. We will assume that if the series converges only at one point 0, then , and if the series converges for all , That .

Definition 4. Convergence interval power series such an interval is called that in front of everyone this series converges and, moreover, absolutely, and for all X, lying outside this interval, the series diverges. Number R called radius of convergence power series.

Comment. At the ends of the interval the question of convergence or divergence of a power series is solved separately for each specific series.

Let us show one of the ways to determine the interval and radius of convergence of a power series.

Consider the power series and denote .

Let's make a series of absolute values ​​of its members:

and apply d'Alembert's test to it.

Let it exist

.

According to d'Alembert's test, a series converges if , and diverges if . Hence the series converges at , then the interval of convergence is: . When the series diverges, since .
Using the notation , we obtain a formula for determining the radius of convergence of a power series:

,

Where − power series coefficients.

If it turns out that the limit , then we assume .

To determine the interval and radius of convergence of a power series, you can also use the radical Cauchy test; the radius of convergence of the series is determined from the relation .

Definition 5. Generalized power series is called a series of the form

. It is also called power series .
For such a series, the convergence interval has the form: , Where − radius of convergence.

Let us show how to find the radius of convergence for a generalized power series.

those. , Where .

If , That , and the convergence region R; If , That and convergence region .

Example 2. Find the area of ​​convergence of the series .

Solution. Let's denote . Let's make a limit

Solving the inequality: , , therefore, the interval

convergence has the form: , and R= 5. Additionally, we examine the ends of the convergence interval:
A) , , we get the series , which diverges;
b) , , we get the series , which converges
conditionally. Thus, the area of ​​convergence is: , .

Answer: convergence region .

Example 3. Row different for everyone , because at , radius of convergence .

Example 4. The series converges for all R, radius of convergence .

Functional series. Power series.
Range of convergence of the series

Laughter for no reason is a sign of d'Alembert

The hour of functional ranks has struck. To successfully master the topic, and, in particular, this lesson, you need to have a good understanding of ordinary number series. You should have a good understanding of what a series is and be able to apply comparison criteria to examine the series for convergence. Thus, if you have just started studying the topic or are a beginner in higher mathematics, necessary work through three lessons in sequence: Rows for dummies,D'Alembert's sign. Cauchy's signs And Alternating rows. Leibniz's test. Definitely all three! If you have basic knowledge and skills in solving problems with number series, then coping with functional series will be quite simple, since there is not a lot of new material.

In this lesson, we will look at the concept of a functional series (what it even is), get acquainted with power series, which are found in 90% of practical tasks, and learn how to solve a common typical problem of finding the radius of convergence, convergence interval and convergence region of a power series. Next, I recommend considering the material about expansion of functions into power series, and first aid will be provided to the beginner. After catching our breath a little, we move on to the next level:

Also in the section of functional series there are numerous of them applications to approximate computing, and in some ways stand out Fourier Series, which, as a rule, are given a separate chapter in educational literature. I only have one article, but it’s a long one and there are many, many additional examples!

So, the landmarks are set, let's go:

The concept of functional series and power series

If the limit turns out to be infinity, then the solution algorithm also finishes its work, and we give the final answer to the task: “The series converges at ” (or at either “). See case No. 3 of the previous paragraph.

If the limit turns out to be neither zero nor infinity, then we have the most common case in practice No. 1 - the series converges on a certain interval.

In this case, the limit is . How to find the interval of convergence of a series? We make up the inequality:

IN ANY task of this type on the left side of the inequality should be result of limit calculation, and on the right side of the inequality – strictly unit. I will not explain exactly why there is such an inequality and why there is one on the right. The lessons are practically oriented, and it is already very good that my stories did not hang the teaching staff and some theorems became clearer.

The technique of working with a module and solving double inequalities was discussed in detail in the first year in the article Function Domain, but for convenience, I will try to comment on all the actions in as much detail as possible. Expanding the inequality with the modulus according to the school rule . In this case:

Half the way is over.

At the second stage, it is necessary to investigate the convergence of the series at the ends of the found interval.

First, we take the left end of the interval and substitute it into our power series:

At

We have obtained a number series, and we need to examine it for convergence (a task already familiar from previous lessons).

1) The series is alternating.
2) – the terms of the series decrease in modulus. Moreover, each next member of the series is less than the previous one in absolute value: , which means the decrease is monotonous.
Conclusion: the series converges.

Using a series made up of modules, we will find out exactly how:
– converges (“standard” series from the family of generalized harmonic series).

Thus, the resulting number series converges absolutely.

at – converges.

! I remind you that any convergent positive series is also absolutely convergent.

Thus, the power series converges, and absolutely, at both ends of the found interval.

Answer: area of ​​convergence of the power series under study:

Another form of answer has the right to life: A series converges if

Sometimes the problem statement requires you to indicate the radius of convergence. It is obvious that in the considered example .

Example 2

Find the region of convergence of the power series

Solution: we find the interval of convergence of the series by using d'Alembert's sign (but not BY attribute! – such a attribute does not exist for functional series):

The series converges at

Left we need to leave only, so we multiply both sides of the inequality by 3:

– The series is alternating.
– – the terms of the series decrease in modulus. Each next member of the series is less than the previous one in absolute value: , which means the decrease is monotonous.

Conclusion: the series converges.

Let us examine it for the nature of convergence:

Let's compare this series with a divergent series.
We use the limiting comparison criterion:

A finite number is obtained that is different from zero, which means that the series diverges from the series.

Thus, the series converges conditionally.

2) When – diverges (according to what has been proven).

Answer: Area of ​​convergence of the power series under study: . When the series converges conditionally.

In the example considered, the region of convergence of the power series is a half-interval, and at all points of the interval the power series converges absolutely, and at the point , as it turned out – conditionally.

Example 3

Find the interval of convergence of the power series and investigate its convergence at the ends of the found interval

This is an example for you to solve on your own.

Let's look at a couple of examples that are rare, but do occur.

Example 4

Find the area of ​​convergence of the series:

Solution: Using d'Alembert's test we find the interval of convergence of this series:

(1) We compose the ratio of the next member of the series to the previous one.

(2) We get rid of the four-story fraction.

(3) According to the rule of operations with powers, we bring the cubes under a single power. In the numerator we cleverly expand the degree, i.e. We arrange it in such a way that in the next step we can reduce the fraction by . We describe factorials in detail.

(4) Under the cube, we divide the numerator by the denominator term by term, indicating that . In a fraction we reduce everything that can be reduced. We take the factor beyond the limit sign; it can be taken out, since there is nothing in it that depends on the “dynamic” variable “en”. Please note that the modulus sign is not drawn - for the reason that it takes non-negative values ​​for any “x”.

In the limit, zero is obtained, which means we can give the final answer:

Answer: The series converges at

But at first it seemed that this row with the “terrible filling” would be difficult to solve. Zero or infinity in the limit is almost a gift, because the solution is noticeably reduced!

Example 5

Find the area of ​​convergence of the series

This is an example for you to solve on your own. Be careful;-) The full solution is at the end of the lesson.

Let's look at a few more examples that contain an element of novelty in terms of the use of technical techniques.

Example 6

Find the convergence interval of the series and investigate its convergence at the ends of the found interval

Solution: The common term of the power series includes a factor that ensures sign alternation. The solution algorithm is completely preserved, but when drawing up the limit, we ignore (do not write) this factor, since the module destroys all the “minuses”.

We find the interval of convergence of the series using d'Alembert's test:

Let's create a standard inequality:
The series converges at
Left we need to leave module only, so we multiply both sides of the inequality by 5:

Now we open the module in a familiar way:

In the middle of the double inequality, you need to leave only “X”; for this purpose, we subtract 2 from each part of the inequality:

– interval of convergence of the power series under study.

We investigate the convergence of the series at the ends of the found interval:

1) Substitute the value into our power series :

Be extremely careful, the multiplier does not provide sign alternation for any natural “en”. We take the resulting minus outside the series and forget about it, since it (like any factor constant) does not in any way affect the convergence or divergence of the number series.

Please note again that in the course of substituting the value into the general term of the power series, our factor was reduced. If this did not happen, it would mean that we either calculated the limit incorrectly or expanded the module incorrectly.

So, we need to examine the number series for convergence. Here the easiest way is to use the limiting comparison criterion and compare this series with a divergent harmonic series. But, to be honest, I’m terribly tired of the limiting sign of comparison, so I’ll add some variety to the solution.

So, the series converges at

We multiply both sides of the inequality by 9:

We extract the root from both parts, while remembering the old school joke:


Expanding the module:

and add one to all parts:

– interval of convergence of the power series under study.

Let us investigate the convergence of the power series at the ends of the found interval:

1) If , then the following number series is obtained:

The multiplier disappeared without a trace, since for any natural value “en” .

Functional range is called a formally written expression

u1 (x) + u 2 (x) + u 3 (x) + ... + u n ( x) + ... , (1)

Where u1 (x), u 2 (x), u 3 (x), ..., u n ( x), ... - sequence of functions from the independent variable x.

Abbreviated notation of a functional series with sigma: .

Examples of functional series include :

(2)

(3)

Giving the independent variable x some value x0 and substituting it into the functional series (1), we obtain the numerical series

u1 (x 0 ) + u 2 (x 0 ) + u 3 (x 0 ) + ... + u n ( x 0 ) + ...

If the resulting numerical series converges, then the functional series (1) is said to converge for x = x0 ; if it diverges, what is said is that series (1) diverges at x = x0 .

Example 1. Investigate the convergence of a functional series(2) at values x= 1 and x = - 1 .
Solution. At x= 1 we get a number series

which converges according to Leibniz's criterion. At x= - 1 we get a number series

,

which diverges as the product of a divergent harmonic series by – 1. So, series (2) converges at x= 1 and diverges at x = - 1 .

If such a check for the convergence of the functional series (1) is carried out with respect to all values ​​of the independent variable from the domain of definition of its members, then the points of this domain will be divided into two sets: for the values x, taken in one of them, series (1) converges, and in the other it diverges.

The set of values ​​of the independent variable at which the functional series converges is called its area of ​​convergence .

Example 2. Find the area of ​​convergence of the functional series

Solution. The terms of the series are defined on the entire number line and form a geometric progression with a denominator q= sin x. Therefore the series converges if

and diverges if

(values ​​not possible). But for the values ​​and for other values x. Therefore, the series converges for all values x, except . The region of its convergence is the entire number line, with the exception of these points.

Example 3. Find the area of ​​convergence of the functional series

Solution. The terms of the series form a geometric progression with the denominator q=ln x. Therefore, the series converges if , or , whence . This is the region of convergence of this series.

Example 4. Investigate the convergence of a functional series

Solution. Let's take an arbitrary value. With this value we get a number series

(*)

Let's find the limit of its common term

Consequently, the series (*) diverges for an arbitrarily chosen, i.e. at any value x. Its convergence region is the empty set.

Uniform convergence of a functional series and its properties

Let's move on to the concept uniform convergence of the functional series . Let s(x) is the sum of this series, and sn ( x) - sum n the first members of this series. Functional range u1 (x) + u 2 (x) + u 3 (x) + ... + u n ( x) + ... is called uniformly convergent on the interval [ a, b] , if for any arbitrarily small number ε > 0 there is such a number N that in front of everyone n ≥ N inequality will be fulfilled

|s(x) − s n ( x)| < ε

for anyone x from the segment [ a, b] .

The above property can be geometrically illustrated as follows.

Consider the graph of the function y = s(x) . Let's construct a strip of width 2 around this curve ε n, that is, we will construct curves y = s(x) + ε n And y = s(x) − ε n(in the picture below they are green).

Then for any ε n graph of a function sn ( x) will lie entirely in the strip under consideration. The same strip will contain graphs of all subsequent partial sums.

Any convergent functional series that does not have the characteristic described above is unevenly convergent.

Let's consider another property of uniformly convergent functional series:

the sum of a series of continuous functions uniformly converging on a certain interval [ a, b] , there is a function continuous on this interval.

Example 5. Determine whether the sum of a functional series is continuous

Solution. Let's find the sum n the first members of this series:

If x> 0, then

,

If x < 0 , то

If x= 0, then

And therefore .

Our research has shown that the sum of this series is a discontinuous function. Its graph is shown in the figure below.

Weierstrass test for uniform convergence of functional series

We approach the Weierstrass criterion through the concept majorizability of functional series . Functional range

u1 (x) + u 2 (x) + u 3 (x) + ... + u n ( x) + ...

Lukhov Yu.P. Lecture notes on higher mathematics. Lecture No. 42 5

Lecture 42

SUBJECT: Functional series

Plan.

1. Functional series. Convergence region.
2. Uniform convergence. Weierstrass sign.
3. Properties of uniformly convergent series: continuity of the sum of the series, term-by-term integration and differentiation.
4. Power series. Abel's theorem. The region of convergence of the power series. Radius of convergence.
5. Basic properties of power series: uniform convergence, continuity and infinite differentiability of the sum. Term-by-term integration and differentiation of power series.

Functional series. Convergence region

Definition 40.1. Infinite amount of functions

u 1 (x) + u 2 (x) +…+ u n (x) +…, (40.1)

where u n (x) = f (x, n), is called functional range.

If you specify a specific numeric value X , series (40.1) will turn into a number series, and depending on the choice of value X such a series can converge or diverge. Only convergent series are of practical value, so it is important to determine those values X , at which the functional series becomes a convergent number series.

Definition 40.2. Multiple meanings X , when substituting them into the functional series (40.1) a convergent numerical series is obtained, is calledarea of ​​convergencefunctional range.

Definition 40.3. Function s(x), defined in the region of convergence of the series, which for each value X from the convergence region is equal to the sum of the corresponding numerical series obtained from (40.1) for a given value x is called the sum of the functional series.

Example. Let us find the region of convergence and the sum of the functional series

1 + x + x² +…+ x n +…

When | x | ≥ 1 therefore the corresponding number series diverge. If

| x | < 1, рассматриваемый ряд представляет собой сумму бесконечно убывающей геометрической прогрессии, вычисляемую по формуле:

Consequently, the range of convergence of the series is the interval (-1, 1), and its sum has the indicated form.

Comment . Just as for number series, you can introduce the concept of a partial sum of a functional series:

s n = 1 + x + x² +…+ x n

and the remainder of the series: r n = s s n .

Uniform convergence of a functional series

Let us first define the concept of uniform convergence of a number sequence.

Definition 40.4. Functional sequence fn(x) is called uniformly converging to a function f on the set X if and

Note 1. We will denote the usual convergence of a functional sequence and the uniform convergence by .

Note 2 . Let us note once again the fundamental difference between uniform convergence and ordinary convergence: in the case of ordinary convergence, for a chosen value of ε, for each there is your number N, for which at n>N inequality holds:

In this case, it may turn out that for a given ε the general number N, ensuring the fulfillment of this inequality for any X , impossible. In the case of uniform convergence, such a number N, common to all x, exists.

Let us now define the concept of uniform convergence of a functional series. Since each series corresponds to a sequence of its partial sums, the uniform convergence of the series is determined through the uniform convergence of this sequence:

Definition 40.5. The functional series is calleduniformly convergent on the set X, if on X the sequence of its partial sums converges uniformly.

Weierstrass sign

Theorem 40.1. If a number series converges for both everyone and everyone n = 1, 2,... the inequality is satisfied then the series converges absolutely and uniformly on the set X.

Proof.

For any ε > 0 s there is such a number N, which is why

For remainders r n series the estimate is fair

Therefore, the series converges uniformly.

Comment. The procedure for selecting a number series that meets the conditions of Theorem 40.1 is usually called majorization , and this series itself majorante for a given functional range.

Example. For a functional series majorant for any value X is a convergent series with positive sign. Therefore, the original series converges uniformly to (-∞, +∞).

Properties of uniformly convergent series

Theorem 40.2. If functions u n (x) are continuous at and the series converges uniformly to X, then its sum s (x) is also continuous at a point x 0 .

Proof.

Let us choose ε > 0. Then, therefore, there is such a number n 0 that

- the sum of a finite number of continuous functions, socontinuous at a point x 0 . Therefore there is a δ > 0 such that Then we get:

That is, the function s (x) is continuous at x = x 0.

Theorem 40.3. Let the functions u n (x) continuous on the interval [ a, b ] and the series converges uniformly on this segment. Then the series also converges uniformly to [ a , b ] and (40.2)

(that is, under the conditions of the theorem, the series can be integrated term by term).

Proof.

By Theorem 40.2 the function s(x) = continuous on [a, b ] and, therefore, is integrable on it, that is, the integral on the left side of equality (40.2) exists. Let us show that the series uniformly converges to the function

Let's denote

Then for any ε there is such a number N , which for n > N

This means that the series converges uniformly, and its sum is equal to σ ( x) = .

The theorem has been proven.

Theorem 40.4. Let the functions u n (x) are continuously differentiable on the interval [ a, b ] and a series composed of their derivatives:

(40.3)

converges uniformly on [ a, b ]. Then, if a series converges at least at one point, then it converges uniformly throughout [ a , b ], its sum s (x )= is a continuously differentiable function and

(the series can be differentiated term by term).

Proof.

Let us define the function σ( X ) How. By Theorem 40.3, series (40.3) can be integrated term by term:

The series on the right side of this equality converges uniformly to [ a, b ] by Theorem 40.3. But according to the conditions of the theorem, the number series converges, therefore, the series also converges uniformly. Then Function σ( t ) is the sum of a uniformly convergent series of continuous functions on [ a, b ] and therefore is itself continuous. Then the function is continuously differentiable on [ a, b ], and that is what needed to be proved.

Definition 41.1. Power series is called a functional series of the form

(41.1)

Comment. Using replacement x x 0 = t series (41.1) can be reduced to the form, therefore it is enough to prove all the properties of power series for series of the form

(41.2)

Theorem 41.1 (Abel's 1st theorem).If the power series (41.2) converges at x = x 0, then for any x: | x |< | x 0 | series (41.2) converges absolutely. If series (41.2) diverges at x = x 0, then it diverges for any x: | x | > | x 0 |.

Proof.

If the series converges, then there is a constant c > 0:

Consequently, and the series for | x |<| x 0 | converges because it is the sum of an infinitely decreasing geometric progression. This means that the series at | x |<| x 0 | absolutely matches.

If it is known that series (41.2) diverges at x = x 0 , then it cannot converge at | x | > | x 0 | , since from what was previously proven it would follow that it converges at the point x 0 .

Thus, if you find the largest number x 0 > 0 such that (41.2) converges for x = x 0, then the region of convergence of this series, as follows from Abel’s theorem, will be the interval (- x 0, x 0 ), possibly including one or both boundaries.

Definition 41.2. The number R ≥ 0 is called radius of convergencepower series (41.2), if this series converges and diverges. Interval (- R, R) is called convergence interval series (41.2).

Examples.

1. To study the absolute convergence of a series, we apply the d’Alembert test: . Therefore, the series converges only when X = 0, and its radius of convergence is 0: R = 0.
2. Using the same d'Alembert test, we can show that the series converges for any x, that is
3. For a series using d'Alembert's criterion we obtain:

Therefore, for 1< x < 1 ряд сходится, при

x< -1 и x > 1 diverges. At X = 1 we obtain a harmonic series, which, as is known, diverges, and when X = -1 series converges conditionally according to the Leibniz criterion. Thus, the radius of convergence of the series under consideration R = 1, and the convergence interval is [-1, 1).

Formulas for determining the radius of convergence of a power series.

1. d'Alembert's formula.

Let's consider a power series and apply d'Alembert's criterion to it: for the series to converge, it is necessary that. If exists, then the region of convergence is determined by the inequality, that is

- (41.3)

• d'Alembert's formulato calculate the radius of convergence.

1. Cauchy-Hadamard formula.

Using the radical Cauchy test and reasoning in a similar way, we find that we can define the region of convergence of a power series as a set of solutions to the inequality, subject to the existence of this limit, and, accordingly, find another formula for the radius of convergence:

(41.4)

• Cauchy-Hadamard formula.

Properties of power series.

Theorem 41.2 (Abel's 2nd theorem). If R radius of convergence of the series (41.2) and this series converges at x = R , then it converges uniformly on the interval (- R, R).

Proof.

A positive series converges by Theorem 41.1. Consequently, series (41.2) converges uniformly in the interval [-ρ, ρ] by Theorem 40.1. From the choice of ρ it follows that the interval of uniform convergence (- R, R ), which was what needed to be proven.

Corollary 1 . On any segment that lies entirely within the interval of convergence, the sum of the series (41.2) is a continuous function.

Proof.

The terms of the series (41.2) are continuous functions, and the series converges uniformly on the interval under consideration. Then the continuity of its sum follows from Theorem 40.2.

Corollary 2. If the limits of integration α, β lie within the interval of convergence of the power series, then the integral of the sum of the series is equal to the sum of the integrals of the terms of the series:

(41.5)

The proof of this statement follows from Theorem 40.3.

Theorem 41.3. If series (41.2) has a convergence interval (- R, R), then the series

φ (x) = a 1 + 2 a 2 x + 3 a 3 x ² +…+ na n x n- 1 +…, (41.6)

obtained by term-by-term differentiation of the series (41.2) has the same interval of convergence (- R, R). Wherein

φ΄(x) = s΄ (x) for | x |< R , (41.7)

that is, within the interval of convergence, the derivative of the sum of a power series is equal to the sum of the series obtained by its term-by-term differentiation.

Proof.

Let's choose ρ: 0< ρ < R и ζ: ρ < ζ < R . Then the series converges, therefore, that is, If| x | ≤ ρ, then

Where Thus, the terms of the series (41.6) are smaller in absolute value than the terms of the positive-sign series, which converges according to D’Alembert’s criterion:

that is, it is a majorant for the series (41.6) for Therefore, the series (41.6) converges uniformly on [-ρ, ρ]. Therefore, by Theorem 40.4, equality (41.7) is true. From the choice of ρ it follows that series (41.6) converges at any interior point of the interval (- R, R).

Let us prove that outside this interval series (41.6) diverges. Indeed, if it converged at x 1 > R , then, integrating it term by term on the interval (0, x 2 ), R< x 2 < x 1 , we would get that series (41.2) converges at the point x 2 , which contradicts the conditions of the theorem. So, the theorem is completely proven.

Comment . The series (41.6), in turn, can be differentiated term by term and this operation can be performed as many times as desired.

Conclusion: if the power series converges on the interval (- R, R ), then its sum is a function that has derivatives of any order inside the convergence interval, each of which is the sum of a series obtained from the original one using term-by-term differentiation the corresponding number of times; Moreover, the convergence interval for a series of derivatives of any order is (- R, R).

Department of Informatics and Higher Mathematics KSPU

Let the function be defined in the domain

Definition. Expression

Called functional near.

Example.

For some values ​​the series may converge, for other values ​​it may diverge.

Example.

Find the region of convergence of the series. This series is defined for the values

If then , the series diverges, since the necessary criterion for the convergence of the series is not satisfied; if the series diverges; if is an infinitely decreasing geometric progression.

Comparison of this series with the convergent series at gives the region of convergence of the series under study.

With values ​​from the functional series, a numerical series is obtained

If the number series converges, then the point is called convergence point functional range.

The set of all points of convergence of a series forms its region of convergence. The convergence region is usually some interval of the axis.

If the number series converge at each point, then the functional series is called convergent in area .

The sum of a functional series is some function of a variable defined in the region of convergence of the series

What properties do the functions have if the properties of the members of the series are known, that is.

Continuity of functions is not sufficient to draw a conclusion about continuity.

The convergence of a series of continuous functions to a continuous function is ensured by an additional condition that expresses one important feature of the convergence of a functional series.

Definition. A functional series is called convergent in the region if there is a limit of partial sums of this series, that is.

Definition. A functional series is said to be uniformly convergent in a domain if, for any positive number, there is a number such that the inequality holds for all.

Geometric meaning of uniform convergence

If you surround the graph of a function with a strip”, determined by the relation then the graphs everyone functions, starting from a sufficiently large value, entirely lie in this “- strip” surrounding the graph of the limit function.

Properties of a uniformly convergent series .

1. The sum of a uniformly convergent series in a certain domain composed of continuous functions is a continuous function in this domain.

2. Such a series can be differentiated term by term

3. The series can be integrated term by term

In order to determine whether a functional series is uniformly convergent, one must use the sufficient Weierstrass convergence test.

Definition. The functional series is called majorized in some region of change, if there is a convergent number series with positive terms such that the inequalities are satisfied for everyone from this region.

Weierstrass sign(uniform convergence of the functional series).

Functional range converges uniformly in the region of convergence if it is majorizable in this region.

In other words, if functions in a certain region do not exceed the corresponding positive numbers in absolute value and if the number series converges, then the functional series in this region converges uniformly.

Example. Prove the uniform convergence of the functional series.

Solution. . Let us replace the common member of this series with a common member of the numerical series, but exceeding each member of the series in absolute value. To do this, it is necessary to determine , at which the total term of the series will be maximum.

The resulting number series converges, which means that the functional series converges uniformly according to the Weierstrass criterion.

Example. Find the sum of the series.

To find the sum of a series, we use the well-known formula for the sum of a geometric progression

Differentiating the left and right sides of formula (1), we obtain sequentially

Let us select in the sum to be calculated the terms proportional to the first and second derivatives:

Let's calculate the derivatives:

Power series.

Among functional series there is a class of power and trigonometric series.

Definition. Functional series of the form

is called power by powers. Expressions are constant numbers.

If the series is a power series in powers of .

The region of convergence of the power series. Abel's theorem.

Theorem. If a power series converges at a point, then it converges and, moreover, absolutely for any value that is smaller in absolute value, that is, or in the interval.

Proof.

Due to the convergence of rad, its common term must tend to zero, therefore all terms of this series are uniformly limited: there is such a constant positive number that for each the inequality ., which for all with center at the point