Finding the sum of a series through integration and differentiation. Integration and differentiation of power series

Therefore, based on the comparison criterion, we conclude that the series converges, which means that the series

POWER SERIES Abel's theorem. Interval and radius of convergence of a power series Uniform convergence of a power series and continuity of its sum Integration of power series Differentiation of power series Taylor series Conditions for the expansion of a function into a Taylor series of elementary functions Table of expansions into a power series (Maclaurin series) of basic elementary functions.

Abel's theorem. The interval and radius of convergence of a power series A power series is a functional series of the form (o or of the form (2) where the coefficients are constants. Series (2) by a formal replacement x - x<> on x reduces to the series (1). The power series (1) always converges at the point x = 0, and the series (2) converges at the point x0, and their sum at these points is equal to co. Example. The rows are stacked rows. Let us find out the form of the region of convergence of the power series. Theorem 1 (Abel). If a power series converges at, then it converges absolutely for all x such that if the power series diverges at x = xi, then it diverges at any x for which Let the power series CONVERGATE at. number series converges POWER SERIES Abel's theorem. Interval and radius of convergence of a power series Uniform convergence of a power series and continuity of its sum Integration of power series Differentiation of power series Taylor series Conditions for the expansion of a function into a Taylor series of elementary functions Table of expansions into a power series (Maclaurin series) of basic elementary functions. It follows from this that, and hence, there exists a number such that M for all n. Consider the series where and estimate its common term. We have whered = . But the series is made up of members of a geometric progression with the denominator q, where it means converges. Based on the sign of comparison series 2 |с„:гп| converges at any point x for which. Therefore, the power series converges absolutely FOR Let now the power series of the point O), which separate the intervals of divergence from the interval of convergence. The following theorem holds. Theorem 2. Let the power series converge at the point x Φ 0. Then either this series absolutely converges at each point of the real line, or there exists a number R > 0 such that the series converges absolutely at and diverges at Divergent. Abs. converges divergent d Fig. 1 Definition. An interval of convergence of a power series is an interval (-R, R), where R > 0, such that at each point x € (-A, R) the series converges absolutely, and at points x such that |n| > R, the series diverges. The number R is called the radius of convergence of the power series. Comment. As for the ends of the interval of convergence (-R, R), the following three cases are possible: I) the power series converges both at the point x = -R and at the point x = R, 2) the power series diverges at both points, 3) the power series converges at one end of the convergence interval and diverges at the other. Comment. The power series where x φ 0 has the same radius of convergence as the series To prove formula (3), consider a series composed of the absolute values ​​of the terms of this series Applying the d'Alembert test to this series, we find It follows that the series (4) will converge , if and diverge if. the power series converges absolutely for all x such that and diverges at. By definition of the radius of convergence, we find that the radius of convergence of a power series can also be found by the formula if there is a finite limit Formula (5) can be easily obtained using the Cauchy criterion. If the power series converges only at the point x = 0, then they say that its radius of convergence is R = 0 (this is possible, for example, when lim b^A = oo or If the power series converges at all points of the real axis, then we put R = + oo (this takes place, for example, when lim n^p = 0 or The domain of convergence of a power series can be either the interval (, or the segment [, or one of the half-intervals (x0 - R, x0 + D) or [. If R = + oo, then the region of convergence of the series will be the entire numerical axis, i.e., the interval (-oo, + oo).To find the region of convergence of a power series, you must first calculate its convergence radius R (for example, using one of the above formulas) and find the interval of convergence of the point O) that separate the intervals of divergence from the interval of convergence. The following theorem holds: Theorem 2. Let the power series converge at the point x Φ 0. Then either this series converges absolutely at every point on the real line, or there exists a number R > O such that the series converges absolutely at and diverges at | Consumption it. Abs. converges divergent Definition. An interval of convergence of a power series is an interval (-R, R), where R > 0, such that at each point x € (-A, R) the series converges absolutely, and at points x such that |n| > R, the series diverges. The number R is called the radius of convergence of the power series. Comment. As for the ends of the interval of convergence (-R, R), the following three cases are possible: I) the power series converges both at the point x = -R and at the point x = R, 2) the power series diverges at both points, 3) the power series converges at one end of the convergence interval and diverges at the other. Comment. The power series where x φ 0 has the same radius of convergence as the series To prove formula (3), consider a series composed of the absolute values ​​of the terms of this series Applying the d'Alembert test to this series, we find It follows that the series (4) will converge , if \, and diverge if, i.e., the power series converges absolutely for all x such that and diverges for \. By definition of the radius of convergence, we obtain that R = £, i.e., POWER SERIES. Abel's theorem. Interval and radius of convergence of a power series Uniform convergence of a power series and continuity of its sum Integration of power series Differentiation of power series Taylor series Conditions for the expansion of a function into a Taylor series of elementary functions Table of expansions into a power series (Maclaurin series) of basic elementary functions. The radius of convergence of a power series can also be found using the formula if there is a finite limit Formula (5) can be easily obtained using the Cauchy criterion. If the power series converges only at the point x = 0, then they say that its radius of convergence is R = 0 (this is possible, for example, when lim b^A = oo or. If the power series converges at all points of the real axis, then we assume R = + oo (this takes place, for example, when The region of convergence of a power series can be either the interval (, or the segment ], or one of the half-intervals (x0 - R, x0 + D) or [. If R = + oo, then the region of convergence of the series will be the entire numerical axis, i.e., the interval (-oo, + oo).To find the region of convergence of a power series, you must first calculate its convergence radius R (for example, using one of the above formulas) and thereby find the convergence interval in which the series converges absolutely, then - to investigate. (3) Since we will have The series converges absolutely on the interval 2) Let us investigate We estimate the convergence of series (6) at the ends of the interval of convergence. Putting x = -1, we get a number series whose divergence is obvious (the necessary convergence criterion is not met: . For x - 1 we get a number series for which does not exist, which means that this series diverges. So, the convergence area of ​​series (6) is an interval Example 2. Find the region of convergence of the series M 1) The radius of convergence is found by formula (3). We have Row (7) converges absolutely on the interval, whence When we get a numerical series that diverges (harmonic series). For x = 0, we will have a number series that converges conditionally. Thus, the series (7) converges in the region Example 3. Find the interval of convergence of the series Since = , then to find the radius of convergence, we apply the formula the area of ​​convergence is the interval Example 4. Find the interval of convergence of the series, then we get Equality R = 0 means that the series (8) converges only at a point. i.e. the region of convergence of the given power series consists of one point §2. Uniform convergence of a power series and continuity of its sum Theorem 1. A power series converges absolutely and uniformly on any segment contained in the interval of convergence of the series Let. Then for all x satisfying the condition and for any n =. will have. But since the number series converges, then, according to the Weierstrass criterion, this power series converges absolutely and uniformly on the segment. Theorem 2. The sum of a power series is continuous at each point x of its convergence interval (4) Any point x from the convergence interval (-D, R) can be enclosed in some segment on which this series converges uniformly. S(x) will be continuous on the segment [-a, a], and hence also at the point x. Integration of power series Theorem 3 (on term-by-term integration of a power series) A ​​power series can be integrated term-by-term in its interval of convergence (-R, R ), R > 0, and the radius of convergence of the series obtained by term-by-term integration is also equal to R. In particular, for any x from the interval (-R, R) the formula is valid Any point x from the interval of convergence (-D, R) can be concluded in some segment [-a, a], where. On this segment, the given series will converge uniformly, and since the terms of the series are continuous, it can be integrated term by term, for example, within the range from 0 to x. Then, according to Theorem 4 of Chapter XVIII, Let us find the radius of convergence R" of the obtained series POWER P POISONS Abel's theorem. Interval and radius of convergence of a power series Uniform convergence of a power series and continuity of its sum Integration of power series Differentiation of power series Taylor series Conditions for the expansion of a function into a Taylor series of elementary functions Table of expansions into a power series (Maclaurin series) of basic elementary functions. under the additional condition of the existence of a finite limit R. So, the radius of convergence of the power series does not change during integration. Comment. The assertion of the theorem remains valid for H = +oo. §4. Derivation of power series Theorem 4 (on term-by-term differentiation of a power series). A power series can be differentiated term by term at any point x of its convergence interval 1) and (2) are equal Let us denote the sum of series (2) by Series (1) and (2) converge uniformly on any interval [-a, a|, where. Moreover, all terms of series (2) are continuous and are derivatives of the corresponding terms of the series (1). Therefore, according to Theorem 5 of Chapter XVIII, the equality holds on the interval [-a, a) By virtue of the arbitrariness of a, the last equality also holds on the interval C. Power series Definition. We will say that the function f(x) expands into a power series ]Γ) CnXn on an interval if the indicated series converges on this interval and its sum is equal to f(x): Let us first prove that the function f(x) cannot have two different power series expansions of the form Theorem 5. If the function /(x) on the interval (-R, R) is expanded into a power series (1), then this expansion is unique, i.e., the coefficients of the series (1) are uniquely determined by its sum. Let the function in the interval be expanded into a convergent power series. Differentiating this series term-by-term n times, we find For x = 0 we get from whence Thus, the coefficients of the power series (1) are uniquely determined by formula (2). Comment. If the function /(x) is expanded into a power series in powers of the difference x-zq, then the coefficients cn of this series are determined by formulas. Let the function / have derivatives of all orders. is infinitely differentiable at the point jo. Let's compose a formal power series for this function by calculating its coefficients using formula (3). §5. Definition. The Taylor series of the function /(x) with respect to the point x0 is called a power series of the form the function /(x) expands into a power series, then this series is the Taylor series of the function /(x). where Pjn(i) is a polynomial of degree 3n with respect to j. Let us now show that at the point 2 = 0 this function also has derivatives of any order, and all of them are equal to zero. Based on the definition of the derivative, we have In a similar way, we can prove that Thus, the given function has derivatives of all orders on the real axis.Construct a formal Taylor series of the original function with respect to the point z0 = We have. the sum of this series is identically equal to zero, while the function f(x) itself is not identically equal to zero. ^ This example is worth remembering when discussing complex analysis (analyticity): a function that is outwardly completely decent, shows a capricious character on the real axis, which is a consequence of troubles on the imaginary axis. The series formally constructed in the example for a given infinitely differentiable function converges, but its sum does not coincide with the values ​​of this function for x Ф 0. In connection with this, a natural question arises: what conditions should the function f(x) satisfy on the interval (xo - R, xo + R) so that it can be expanded into a Taylor series converging to it? Conditions for the expansion of a function into a Taylor series For simplicity, we will consider a power series of the form m. e. Maclaurin series. Theorem 7. In order for the function f(x) to be expanded into a power series on the interval (-R, R), it is necessary and sufficient that on this interval the function f(x) has derivatives of all orders and that in its Taylor formula the residual term Rn(x) tends to zero as for all m Necessity. Let on the interval (the function f(x) is expandable into a power series, i.e., the series (2) converges and its sum is equal to f(x). Then, by Theorem 4 and the corollary from it, the function f(x) has on the interval (-R , R) derivatives f(n^(x) of all orders. By Theorem 5 (formula (2)) the coefficients of the series (2) have the form i.e. we can write the equality Due to the convergence of this series on the interval (-R, R ) its remainder 0 tends to zero as n oo for all x Sufficiency Let the function f(xr) on the interval (-R, R) have derivatives of all orders and in its Taylor formula the remainder term Rn(x) 0 as n oo for any x € (-D, R). Since for n -» oo. Since the n-th partial sum of the Taylor series is written in square brackets, formula (4) means that the Taylor series of the function f (x) converges on the interval (-D , R) and its sum is the function f(x). Sufficient conditions for the expansion of a function into a power series, convenient for practical use, are described by the following theorem: Theorem 8. In order for the function f(x) it was possible It is sufficient to add to a power series so that the function f(x) has derivatives of all orders on this interval and that there exists a constant M > 0 such that. Let the function f(x) have derivatives of all orders on the interval (-D, R). Then we can formally write the Taylor series for it. Let us prove that it converges to the function f(x). To do this, it suffices to show that the remainder term in Taylor's formula (1) tends to zero as n oo for all x € (-A, R). Indeed, given that). The number series converges by virtue of the d'Alembert criterion: by virtue of the necessary convergence criterion. From inequality (3) we obtain! Taylor Series of Elementary Functions Consider expansions into a series of basic elementary functions. 6 This function has derivatives of all orders on the interval (- any number, and Therefore, the exponential function ex expands into a Taylor series on any interval (-a, a) and, thus, on the entire Ox axis. Since, then we get the series If in expansion (1) replace x by -a*, then we have This function has derivatives of any order, and moreover, by Theorem 8, the function sin x expands into a Taylor series converging to it on the interval (-oo, +oo). Since then this series has the following form Radius of convergence of the series Similarly, we obtain that - any real number This function satisfies the relation and condition We will look for a power series whose sum 5(g) satisfies relation (4) and the condition 5(0) = 1. We set From here we find Substituting relations (5) and (6) into formula (4), we will have Equating the coefficients at the same powers of x in the left and right parts of the equality, we obtain from which we find POWER SERIES Abel's theorem. Interval and radius of convergence of a power series Uniform convergence of a power series and continuity of its sum Integration of power series Differentiation of power series Taylor series Conditions for the expansion of a function into a Taylor series of elementary functions Table of expansions into a power series (Maclaurin series) of basic elementary functions. Substituting these values ​​of the coefficients into relation (5), we obtain the series Let us find the radius of convergence of the series (7) in the case when a is not a natural number. We have So, series (7) converges at. e. on the interval Let us prove that the sum 5(x) of series (7) on the interval (-1,1) is equal to (1 + x)°. To do this, consider the relation Since 5(x) satisfies the relation (then for the derivative of the function φ(x) we obtain: for. It follows that. In particular, for x = 0 we have and hence, or The resulting series is called binomial, and its coefficients - binomial coefficients. Remark. If a is a natural number (o = z"), the function (1 + z) a will be a polynomial of degree n, and Dn (x) = 0 for all n > a. We also note If a = -1, we will have Replacing w by -x in the last equality, we obtain an expansion of this function in a Taylor series in powers of x, we integrate equality (9) within o Equality (11) is valid in the interval. We can prove that equality (11) is also valid for x = 1: Table of power series expansions (Maclaurin series) of basic elementary functions. on examples of how this is done.Example 1. Expand the function of 4 in a power p poison in the vicinity of the point xq = 2, i.e., in powers of the difference z -2. Let's transform this function so that we can use series (10) for the We have function. Replacing x in formula (10) by ^. we get I I This expansion is valid when any of the equivalent inequalities is satisfied. Example 2. Expand the function in powers of x using formula (10). 4 Decomposing the denominator into factors, we represent this rational function as the difference of two simple fractions. After simple transformations, we obtain Both series (14) and (15) will converge simultaneously for \. Since series (14) and (15) converge in the interval (-1,1), they can be subtracted term by term. As a result, we get the desired power series whose convergence radius is R = 1. This series converges absolutely for Example 3. Expand the function arcsin x in the Taylor series in the vicinity of the point x0 = 0. 4 It is known that Let us apply to the function (formula (8). replacing x by -x2 in it. As a result, for we obtain Integrating both parts of the last equality from zero to x (term-by-term integration is legal, since the power series converges uniformly on any segment with ends at the points 0 and x lying in the interval (-1,1)), we find or Thus, we finally obtain that Example 4. Calculate the integral (integral sine) , It is known that the antiderivative for the function ^ is not expressed in terms of elementary functions.We expand the integrand in a power series, using the fact that From equality (16) we find Note that dividing the series (16) by t at t f 0 is legal Equality (17) is also preserved at if we assume that at t = 0 the ratio - = 1. Thus, the series (17) converges for all values ​​Integrating it term by term, we obtain so that the error in replacing its sum by a partial sum is easily estimated. Example 5. Calculate the integral Here, the antiderivative for the integrand e is also not an elementary function. To calculate the integral, we replace in the formula We get Let's integrate both parts of this equality in the range from 0 to x: This series converges for any r (its convergence radius R \u003d + oo) and is alternating at Exercises Find the area of ​​​​convergence of power series: Expand the following functions in a series Makloreya and indicate the areas of convergence of the obtained series: Indication. Use the table. Using the table, expand the given functions in a Taylor series in powers of x - x0 and indicate the intervals of convergence of the resulting series.

Elements of the semantic structure

The semantic structure of the sentence.

(This question is for independent study!)

This type of analysis relates the semantic organization of a sentence to its formal organization. This direction put forward the concept of the semantic structure of a sentence (primarily N.Yu. Shvedova).

A block diagram has its own semantics, which is created by the formal values ​​of the components, the rules for their lexical content, and the relationship of the components to each other (in non-single-component diagrams).

The linguistic meaning of a particular sentence constructed according to one or another pattern is formed by the mutual action of the semantics of this pattern and the lexical semantics of those words that have taken the positions of its components: The student writes; the child rejoices with the general semantics of the MSS (“relationship between the subject and its predicative feature - action or procedural state”) in the first case, the meaning “relationship between the subject and his specific action” is presented, in the second case - “relationship between the subject and his emotional state” .

Functional series of the form where (coefficients of the series) and (center of the series) are constants, a variable, are called power series. It is clear that if we learn to calculate the region of convergence of a power series (with a center), then we can easily find the region of convergence of the original series. Therefore, from now on, unless otherwise stated, we will consider power series of the form.

Abel's theorem.If a power series converges at a point, then it converges absolutely and in the interval On any segment, the indicated series converges uniformly.

Proof. Since the series converges, its common term is therefore bounded, i.e. there is a constant such that

Let now. Then we will have

Since the geometric progression converges (), then the first comparison theorem converges and the series The first part of the theorem is proved.

Since the series converges by what has been proved and it majorizes as (see) the series, then by the Weierstrass theorem the last series converges uniformly as . The theorem is completely proved.

It follows from Abel's theorem that we can expand the interval until the moment comes when the series diverges at the point (or such a moment does not come at all, i.e.). Then the specified interval will be the region of convergence of the series. Thus, any power series has as its region of convergence not an arbitrary set, but precisely an interval. Let us give a more precise definition of the interval of convergence.

Definition 2. The number is called radius of convergence series, if inside the interval this series converges absolutely, and outside the segment it diverges. In this case, the interval is called convergence interval row.

Note that for , the indicated power series converges only at the point and for , it converges for all real values. The following examples show that these cases are not excluded: An example of a series with a nonzero finite radius of convergence can be a geometric progression. both converge and diverge. For example, the series conditionally converges at a point and diverges at a point

From the properties of uniformly convergent functional series (Theorems 1-3), the following properties of power series are easily deduced.

Theorem 4.Let be the convergence radius of the power series. Then the following statements take place:

1. The sum of a given power series is continuous in the interval of convergence;

2. If is the radius of convergence of the power series, then the series of derivatives will have the same radius of convergence. This implies that the power series can be differentiated any number of times (i.e., its sum is infinitely differentiable in the interval of convergence), and the equality

3. A power series can be integrated on any interval lying inside its interval of convergence, i.e.

Proof, for example, the first property will be like this. Let an arbitrary point of the interval of convergence . Surround this point with a symmetrical segment. According to Abel's theorem, the series converges uniformly on the segment, so its sum is continuous on the specified segment, and therefore continuous, in particular, and at the point Property 1 is proved. The remaining properties of our theorem are proved similarly.

Now let's calculate the radius of convergence of a power series from its coefficients.

Theorem 4 . Let at least one of the following conditions be satisfied:

a) there is a (finite or infinite) limit

b) there is a (finite or infinite) limit (it is assumed that there exists a number such that).

Then the number is the convergence radius of the series.

Proof we will carry out for case a). Let's apply the Cauchy test to the modular series: According to the indicated test, the series converges absolutely if the number i.e. if If, i.e. if then the indicated series diverges. Hence, the convergence radius of the series. The theorem has been proven.

Remark 1. Theorem 1-4 can be carried over to power series of the form almost without changing the wording (with a slight correction that in this case the region of convergence is the interval).

Example 1 Find the area of ​​convergence of the series ( task 10, T.R., Kuznetsov L.A.)

Decision. We apply an analog of a) of Cauchy's theorem: the radius of convergence of a given series. So the series converges absolutely in the region

We investigate the convergence of the series at the ends of the interval. We have

diverges, because

diverges, because

Therefore, the area of ​​convergence of the original series is the interval.

Definition. Functional series of the form

where ... are real numbers, is called a power series.

The region of absolute convergence of the series is the interval , where the number R is the radius of convergence.

Let the power series have a radius of convergence R > 0. Then the following statements are true:

1. The sum of the series is a continuous function of x throughout the convergence interval.

2. The series converges uniformly on any segment where .

3. The series can be integrated term by term over any interval lying inside the interval .

4. A series can be differentiated term by term at any point any time.

Notes:

1. When integrating or differentiating a power series term by term, new power series are obtained, while their radius of convergence remains the same.

2. The radius of convergence of a power series can be found using one of the formulas:

, (10)

(11)

provided that the indicated limits exist, is the coefficient of the series.

Task 17.31

Find the sum of a series .

Decision:

I way. Find the interval of convergence of the series:

, , .

Simplify the rational fraction , .

Then the series can be represented by the difference of two series:

The convergence of each of them remains the same (see for yourself). So there is equality. Denote the sums of the series by and , respectively, and the desired sum by , .

Let's find the sum of the first row:

Differentiating term by term the series inside the interval of convergence , we obtain: ; is a geometric progression with denominator .

When the progression converges, , , and the sum is: ; . Now, integrating on the interval lying inside the convergence interval , we obtain:

.

Find the sum of the second row:

Let's do the transformation:

Let us denote the sum of the series in parentheses by and differentiate in the interval :

This is also a geometric progression.

, , ;

.

So the sum of the original series is:

or
for .

II method. Without repeating the details of the first method related to the interval of convergence of this series, we offer the second option for solving the problem. Let's denote the sum of the series by: .

Multiply by this row: . Differentiate the twice obtained series:

,

Represents a geometric progression with a denominator , then . Let's integrate on the interval:

Integrating by parts, we get:

for .

Task 18.31

Find the sum of a series .

Decision:

This series converges in the interval (see for yourself). Let's rewrite it, presenting it as the sum of three rows:

This is possible, since each of the series has the same area of ​​convergence - the interval. Denote the sums of the three series by , , , respectively, and the desired sum by .

as the sum of the terms of a geometric progression with a denominator

Let's do the transformation:

Denote by the sum of the series .

Integrating term by term this series on a segment inside the interval of convergence , we obtain:

To find , we need to differentiate the fraction:

.

Hence, .

Now let's find:

Let's take it out of brackets:

Denote by the sum of the series in parentheses. Then

In these brackets there is a series, the sum of which is found: . We get: .

But , . Then the sum of the original series

So, for .

Taylor series

Definition. Row

is called the Taylor series in powers of the function .

A function can be expanded into a Taylor series if it has derivatives of all orders at the point under consideration and if the remainder term at the point at tends to zero. The Taylor series is sometimes called the Maclaurin series.

Theorem

If a function expands into a power series, then this series is unique for it and is a Taylor series.

Note. By finding successively the derivatives of functions and their values ​​at the point , one can write down the Taylor series. But at the same time, the study of the residual term presents great difficulties. Therefore, they often go the other way: they use ready-made expansions of basic elementary functions into power series in combination with the rules of addition, subtraction, multiplication of series and theorems on their integration and differentiation, as, for example, was shown in problems 17.31 and 18.31.

Task 19.31

Expand function in a Taylor series in powers of .

Decision:

X 0 = 0. Let's use the note. As

then the function is simplified if we apply the method of indefinite coefficients:

.

The sum of the terms of a geometric progression with a denominator is: . In our case . is the radius of convergence of this series. Term ,

Adding the rows, we get: or , where is the general region of convergence. lies entirely in the region of convergence of the series .

To calculate this integral with an accuracy of 0.001, we need to take two of its terms in the resulting series (0.0005<0,001) (см. задачу 9.31).

Thus,

Questions for self-examination

Number series

1. Give definitions of convergent and divergent series.

2. Formulate the necessary criterion for the convergence of the series.

3. Formulate sufficient signs of convergence of series with positive terms: comparison of series with positive terms; sign of d'Alembert; radical Cauchy sign, integral Cauchy sign.

4. Define an absolutely convergent series. State the properties of absolutely convergent series.

5. Formulate the Leibniz sign.

functional rows

6. Define the area of ​​convergence of the functional series.

7. What series is called uniformly convergent?

8. Formulate the Weierstrass test.

9. Conditions for the expansion of a function into a Taylor series.

10. Formulate theorems on integration and differentiation of power series.

11. State the method of approximate calculation of definite integrals using series.

1. Kudryavtsev L.D. A short course in mathematical analysis. – M.: Nauka, 1989. – 736 p.

2. Bugrov Ya.S. Differential and integral calculus / Ya.S. Bugrov, S.M. Nikolsky. – M.: Nauka, 1984. – 432 p.

3. Shmelev P.A. Theory of series in tasks and exercises. - M.: Higher School, 1983. - 176 p.

4. Piskunov N.S. Differential and integral calculus for technical colleges. T. 2. - M.: Nauka, 1985. - 576 p.

5. Fikhtengolts G.M. Course of differential and integral calculus. T. 2. - M.: Fizmatgiz, 1962. - 808 p.

6. Zaporozhets G.I. Guide to solving problems in mathematical analysis. - M.: Higher school, 1966. - 460 p.

7. Kuznetsov L.A. Collection of tasks in higher mathematics (TR). - M.: Higher school, 1983. - 174 p.

8. Danko P.E. Higher mathematics in exercises and tasks. Part 2 / P.E. Danko, A.G. Popov, T.Ya. Kozhevnikov. - M.: Higher School, 1986. - 415 p.

9. Bronstein I.N. Handbook of mathematics for engineers and students of higher educational institutions / I.N. Bronstein, K.A. Semendyaev. – M.: Nauka, 1986. – 544 p.

Educational edition

Borodin Nikolai Pavlovich

Millstone Varvara Viktorovna

Shumetova Lyudmila Viktorovna

Shorkin Vladimir Sergeevich

ROWS

Teaching aid

Editor T.D. Vasiliev

Technical editor T.P. Prokudin

Orel State Technical University

License ID No. 00670 dated 01/05/2000

Signed for publication on August 26, 2004. Format 60 x 84 1/16.

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