Power series. Abel's theorem, radius and interval of convergence
Elements of semantic structure
Semantic structure of a sentence.
(this question is on self-study!)
This type of analysis relates the semantic organization of a sentence to its formal organization. This direction put forward the concept of semantic structure of a sentence (primarily N.Yu. Shvedova).
A structural diagram has its own semantics, which is created by the formal meanings of the components, the rules of their lexical content and the relationship of the components to each other (in non-single-component schemes).
The linguistic meaning of a specific sentence constructed according to one or another pattern is formed by the mutual action of the semantics of this pattern and lexical semantics those words that took the positions of its components: The student writes; the child rejoices with the general semantics of MSS (“the relationship between the subject and its predicative attribute - an action or procedural state”) in the first case the meaning is “the relationship between the subject and his specific action”, in the second case - “the 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, henceforth, 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 specified series converges uniformly.
Proof. Since the series converges, its common term is therefore bounded, i.e. there is a constant such that
Let it be now. Then we will have
Because the geometric progression converges (), then by the first comparison theorem the series also converges. The first part of the theorem is proven.
Since, according to what has been proven, the series converges and it majorizes as (see) the series, then by Weierstrass’s theorem the last series converges uniformly at The theorem is completely proven.
It follows from Abel’s theorem that we can expand the interval until the moment comes when the series diverges at a point (or such a moment does not come at all, i.e.). Then the indicated 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. We'll give you more precise definition convergence interval.
Definition 2. The number is called radius of convergence series, if within the interval this series converges absolutely, and outside the segment it diverges. In this case the interval is called convergence interval row.
Note that at the indicated power series converges only at the point and at it converges at all real The following examples show that these cases are not excluded: An example of a series with a non-zero finite radius of convergence can be a geometric progression. Note also that on the boundary of the convergence interval a power series can both to converge and to diverge. For example, a series converges conditionally 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 derived.
Theorem 4.Let be the radius of convergence of the power series. Then the following statements hold:
1. The sum of a given power series is continuous in the interval of convergence;
2. If is the radius of convergence of a power series, then the series of derivatives will have the same radius of convergence. It follows that the power series can be differentiated as many times as desired (i.e., its sum is infinitely differentiable in the interval of convergence), and the equality holds
3. A power series can be integrated on any segment lying inside its convergence interval, i.e.
Proof, for example, the first property will be like this. Let an arbitrary point of the convergence interval . Let's surround this point with a symmetric segment. According to Abel's theorem, the series converges uniformly on the segment, so its sum is continuous on the indicated segment, and therefore continuous, in particular, at the point. Property 1 is proven. The rest of the 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 met:
a) there is a (finite or infinite) limit
b) there is a (finite or infinite) limit (it is assumed that there is a number such that).
Then the number is the radius of convergence of the series.
Proof Let's do it for case a). Let us apply the Cauchy test to the modular series: According to the specified test, the series converges absolutely if the number i.e. if If i.e. if then the indicated series diverges. Therefore, the radius of convergence of the series. The theorem has been proven.
Note 1. Theorem 1-4 can be transferred practically without changing the formulation to power series of the form (with the slight amendment that in this case the domain of convergence is the interval).
Example 1. Find the area of convergence of the series ( task 10, T.R., Kuznetsov L.A.)
Solution. Let us apply an analogue of a) Cauchy's theorem: the radius of convergence of a given series. This means that the series converges absolutely in the region
Let us study the convergence of the series at the ends of the interval. We have
diverges, because
diverges, because
Consequently, the area of convergence of the original series is the interval.
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 decomposability of a function in a Taylor series of elementary functions Table of expansions in a power series (Maclaurin series) of basic elementary functions.
Abel's theorem. Interval and radius of convergence of a power series A power series is a functional series of the form (o or type (2) where the coefficients are constants. Series (2) by the formal replacement x - x<> on x reduces to series (1). Power series (1) always converges at the point x = 0, and series (2) at the point x0, and their sum at these points is equal to ω. Example. The rows are laid rows. Let us find out the form of the convergence region of the power series. Theorem 1 (Abel). If a power series converges at, then it converges absolutely for all x such that if a power series diverges at x = xi, then it diverges at any x for which Let the power series CONVERGES 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 decomposability of a function in a Taylor series of elementary functions Table of expansions in a power series (Maclaurin series) of basic elementary functions. It follows that a means that there is a number such that M for all n. Consider the series where and estimate its common term. We have whered= . But the series is composed of terms of a geometric progression with a denominator q, where it means it converges. Based on the comparison criterion, row 2 |с„:гп| converges at any point x for which. Consequently, the power series is absolutely convergent FOR Let now the power series be the points O), which separate the intervals of divergence from the interval of convergence. The following theorem holds. Theorem 2. Let a power series converge at a point x Φ 0. Then either this series converges absolutely at every point on the number line, or there is a number R > O such that the series converges absolutely at and diverges at Diverge. Abs. converges diverges d Fig. 1 Definition. The interval of convergence of a power series is the interval (-R, R), where R > 0, such that at each point x € (-R, R) the series converges absolutely, and at points x such that |i| > R, the series diverges. The number R is called the radius of convergence of the power series. Comment. As for the ends of the convergence interval (-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) a power series converges at one end of the convergence interval and diverges at the other. Comment. The power series where hof 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 series (4) will converge , if and diverge, if. The power series converges absolutely for all x such that it diverges at. By determining the radius of convergence, we find that 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 a power series converges only at the point x = 0, then we say that its radius of convergence is R = 0 (this is possible, for example, when lim L^D = oo or If the power series converges at all points of the real axis, then we assume R = + oo (this occurs, for example, when lim n^p = 0 or The convergence region of the 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 radius of convergence R (for example, using one of the above formulas) and thereby find the interval of convergence 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 Ф 0. Then either this series converges absolutely at every point of the number line, or there is a number R > O such that the series converges absolutely at and diverges at | Diverges. Abs. converges diverges Definition. The interval of convergence of a power series is the interval (-R, R), where R > 0, such that at each point x € (-R, R) the series converges absolutely, and at points x such that |i| > R, the series diverges. The number R is called the radius of convergence of the power series. Comment. As for the ends of the convergence interval (-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) a power series converges at one end of the convergence interval and diverges at the other. Comment. The power series where hof 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 series (4) will converge , if \, and diverge if, that is, the power series converges absolutely for all x such that it diverges for \. By defining 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 decomposability of a function in a Taylor series of elementary functions Table of expansions in 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 test. If the power series converges only at the point x = 0, then we say that its radius of convergence is R = 0 (this is possible, for example, when lim b^D = oo or. If the power series converges at all points of the real axis, then we assume R = +oo (this occurs, 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 radius of convergence R (for example, using one of the above formulas) and thereby find the interval of convergence in which the series converges absolutely, then investigate the convergence of the series at the ends of the convergence interval - at the points x = xo - R, x = xq + R. Example 1. Find the region of convergence of the power series M 1) To find the radius of convergence R of this series, it is convenient to apply the formula (3).So somehow we will have The series converges absolutely on the interval 2) Let us study the convergence of the series (6) at the ends of the convergence interval. Putting x = -1, we obtain a number series whose divergence is obvious (the necessary criterion for convergence is not met: . For x - 1, we obtain a number series for which does not exist, which means this series diverges. So, the region of convergence of series (6) is an interval Example 2. Find the area of convergence of the series M 1) We find the radius of convergence using formula (3). We have Series (7) converges absolutely on the interval, from where When we obtain a numerical series that diverges (harmonic series). At x = 0 we will have a number series that is conditionally convergent. Thus, 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 This means that this series converges for all values of x, i.e. the region of convergence is the interval Example 4. Find the interval of convergence of the series, then we obtain The equality R = 0 means that the series (8) converges only at a point. That is, the region of convergence of a 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 w satisfying the condition, and for any n =. will have. But since the number series converges, then, according to Weierstrass’s 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 a certain segment on which the given series converges uniformly. Since the terms of the series are continuous, then its sum S(x) will be continuous on the interval [-a, a], and therefore 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 convergence interval (-R, R ), R > O, 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 following formula holds: Any point x from the convergence interval (-D, R) can be enclosed in some segment [-a, a], where On this segment this series will converge uniformly, and since the terms of the series are continuous, it can be integrated term by term, for example, in the range from 0 to x. Then, according to Theorem 4 of Chapter XVIII, Let us find the radius of convergence R" of the resulting 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 decomposability of a function in a Taylor series of elementary functions Table of expansions in a power series (Maclaurin series) of basic elementary functions. under the additional condition of the existence of a finite limit R. Ime So, the radius of convergence of the power series does not change during integration. Comment. The statement of the theorem remains valid for R = +oo. §4. Differentiation of power series Theorem 4 (on term-by-term differentiation of power series). A power series can be differentiated term by term at any point x of its interval of convergence 4 Let R be the radius of convergence of the series and R" be the radius of convergence of the series Assume that there is a (finite or infinite) limit Let us find the radius B! of the series where We have Thus, the radii of convergence of the series ( 1) and (2) are equal. Let us denote the sum of series (2) by Series (1) and (2) uniformly converge on any segment [-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). Due to the arbitrariness of a, the last equality also holds on the interval Sledspie. Power series Definition. We will say that the function /(x) expands into a power series ]G) SpXn on an interval if on this interval the indicated series converges and its sum is equal to /(x): Let us first prove that the function /(x) cannot have two different expansions in a power series of the form Theorem 5. If the function f(x) on the interval (-R, R) is expanded into a power series (1), then this expansion is unique, i.e., the coefficients of series (1) are uniquely determined from its sum. Let the function in the interval be expanded into a convergent power series. Differentiating this series termwise n times, we find When x = 0 we obtain whence Thus, the coefficients of the power series (1) by formula (2) are uniquely determined. Comment. If the function /(x) is expanded into a power series in powers of the difference x-zq, then the coefficients c„ of this series are determined by the formulas. Let the function / have derivatives of all orders, i.e. is infinitely differentiable at the point w. Let us 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 (here. The coefficients of this series... are called the Taylor coefficients of the function. For xo = 0, the Taylor series is called the Maclaurin series. The following statement follows from Theorem 5. Theorem b. If on the interval the function /(x) expands into a power series, then this series is the Taylor series of the function /(x). Example 1. Consider a function and find its derivatives. For z O, this function has derivatives of all orders, which are found according to the usual rules and, in general, where Pjn (i) is a polynomial of degree 3n with respect to j. Now we 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 (when calculating the limit we applied the Rhopital rule) . In a similar way, we can prove that Thus, the given function has derivatives of all orders on the number axis. Let us construct a formal Taylor series of the original function with respect to the point z0 = We have. Obviously, the sum of this series is identically equal to zero, while the function itself f( x) identically equal to zero is not. ^ This example is worth remembering when discussing comprehensive analysis(analyticity): a function, outwardly completely decent, exhibits 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 Φ 0. In this regard, 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 decomposability of a function in a Taylor series For simplicity, we will consider a power series of the form, i.e., the 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 the term Rn(x) tended to zero for all m Necessity. Let on the interval (the function f(x) be expanded into a power series, i.e. series (2) converges and its sum is equal to f(x). Then, by Theorem 4 and its corollary, the function f(x) has on the interval (-R , R) derivatives /(n^(x) of all orders. By Theorem 5 (formula (2)), the coefficients of 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 oo for all x Sufficiency: Let the function f(r) on the interval (-R, R) have derivatives of all orders and in its Taylor formula the remainder term Rn(x) 0 at oo for any x € (-Δ, R). Since as n -» oo. Since the nth 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 (-Δ , R) and its sum is the function f(x). Sufficient conditions for the expansion of a function into a power series, convenient for practical application, are described by the following theorem. Theorem 8. In order for the function f(x) on the interval (-R, R) to be expanded into a power series, it is sufficient that the function f(x) has derivatives of all orders on this interval and that there exists a constant M > O such that What. Let the function f(x) have derivatives of all orders on the interval (-D, R). Then we can formally write a Taylor series for it. Let us prove that it converges to the function f(x). To do this, it is enough to show that the remainder term in Taylor's formula (1) tends to zero as n oo for all x € (-Δ, R). Indeed, considering that). A number series converges by virtue of D'Alembert's criterion: by virtue of the necessary criterion of convergence. From inequality (3) we obtain! Although the function of M, from § b. Taylor series of elementary functions Let us consider series expansions of basic elementary functions. 6 This function has derivatives of all orders on the interval (- any number, and Therefore, exponential function ex can be expanded into a Taylor series on any interval (-a, a) and, thus, on the entire Ox axis. Since, we obtain the series If in expansion (1) we replace w by -a*, then we will have This function has derivatives of any order, and. Thus, by Theorem 8 sin function x expands into a Taylor series converging to it on the interval (-oo, +oo). Since this series has the following form: Radius of convergence of the series We similarly obtain that - any real number This function satisfies the relation and condition We will look for a power series whose sum 5(x) satisfies relation (4) and the condition 5(0) = 1. Let us put From here we find Substituting relations (5) and (6) into formula (4), we will have Equating the coefficients for the same powers of x in the left and right sides of the equality, we obtain from where 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 decomposability of a function in a Taylor series of elementary functions Table of expansions in a power series (Maclaurin series) of basic elementary functions. Substituting these values of the coefficients into relation (5), we obtain the series. Find the radius of convergence of 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(g) of series (7) on the interval (-1,1) is equal to (1 + g)°. 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 therefore, or The resulting series is called binomial, and its coefficients - binomial coefficients. Note: In case a - natural number(o = z"), the function (1 + z)a will be a polynomial nth degree, and Dn(x) = 0 for all n > a. Let us note two more expansions. For a = -1 we will have. Replacing w by -z in the last equality, we obtain the expansion of this function in a Taylor series in powers of w. We will integrate equality (9) within o Equality (11) is valid in the interval. Replacing x with -z in it, we obtain a series. It can be proven that equality (11) is also true for x = 1: Table of power series expansions (Maclaurin series) of basic elementary functions. Using this table, you can obtain power series expansions of more than complex functions. Let us show with examples how this is done. Example 1. Expand the function of 4 into a power series 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 function We have. Replacing x in formula (10) with ^. we obtain I I This expansion is valid when any of the equivalent inequalities are satisfied. Example 2. Expand the function in powers of x using formula (10). 4 Expanding the denominator into factors, we present this rational function as the difference of two simple fractions. After simple transformations we obtain 1 To each term on the right side of equality (13) we apply formula (10), as a result we obtain power series Series (14) converges for \ and series (15) converges for 2. 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 obtain the desired power series whose radius of convergence is equal to R = 1. This series converges absolutely for Example 3. Expand the function arcsin x into a Taylor series in the vicinity of the point xo = 0. 4 It is known that Apply to the function (formula (8), replacing x in it with -x2. As a result, for we obtain Integrating both sides of the last equality from zero to x (term-by-term integration is legal, since the power series converges uniformly on any segment with endpoints at points 0 and x, lying in the interval (-1,1)), we find or Thus, we finally obtain that Remark: Expansion in power series can be used to calculate integrals that cannot be expressed in final form through elementary functions. Let us give several examples Example 4. Calculate the integral (integral sine), It is known that the antiderivative for the function ^ is not expressed in terms of elementary functions. Let us expand the integrand into a power series, using the fact that From equality (16) we find Note that dividing series (16) by t for t φ O is legal. Equality (17) is also preserved if we assume that for t = O the relation is - = 1. Thus, series (17) converges for all values. Integrating it term by term, we obtain The resulting series is alternating in sign, so that the error when replacing its sum with a partial sum is easily assessed. Example 5. Calculate the integral Here the antiderivative for the integrand e is also not elementary function. To calculate the integral, we replace in the formula We get We integrate both sides of this equality in the range from 0 to x: This series converges for any r (its radius of convergence R = +oo) and is alternating in sign for Exercises Find the region of convergence of power series: Expand the following functions into a series Macloreia and indicate the areas of convergence of the obtained series: Instruction. Use the table. Using the table, expand the given functions into a Taylor series in powers x - x0 and indicate the intervals of convergence of the resulting series.
Rows.
Basic definitions.
Definition. The sum of the terms is infinite number sequence called number series.
In this case, we will call the numbers members of the series, and u n– a common member of the series.
Definition. Amounts, n = 1, 2, … are called private (partial) amounts row.
Thus, it is possible to consider sequences of partial sums of the series S 1, S 2, …, S n, …
Definition. The series is called convergent, if the sequence of its partial sums converges. Sum of convergent series is the limit of the sequence of its partial sums.
Definition. If the sequence of partial sums of a series diverges, i.e. has no limit, or has an infinite limit, then the series is called divergent and no amount is assigned to it.
Properties of rows.
1) The convergence or divergence of the series will not be violated if you change, discard or add final number members of the series.
2) Consider two series and , where C is a constant number.
Theorem. If a series converges and its sum is equal to S, then the series also converges and its sum is equal to CS. (C#0)
3) Consider two rows and . Amount or difference these series will be called a series, where the elements are obtained as a result of the addition (subtraction) of the original elements with the same numbers.
Theorem. If the series and converge and their sums are equal to S and s, respectively, then the series also converges and its sum is equal to S + s.
The difference of two convergent series will also be a convergent series.
The sum of a convergent and a divergent series is a divergent series.
It is impossible to make a general statement about the sum of two divergent series.
When studying series, they mainly solve two problems: studying convergence and finding the sum of the series.
Cauchy criterion.
(necessary and sufficient conditions for the convergence of the series)
In order for a sequence to be convergent, it is necessary and sufficient that for any there exists a number N such that for n > N and any p > 0, where p is an integer, the inequality would hold:
Proof. (necessity)
Let , then for any number there is a number N such that the inequality
Executed when n>N. For n>N and any integer p>0 the inequality also holds. Taking into account both inequalities, we obtain:
The need has been proven. We will not consider the proof of sufficiency.
Let us formulate the Cauchy criterion for the series.
In order for a series to be convergent, it is necessary and sufficient that for any there exist a number N such that for n>N and any p>0 the inequality would hold
However, in practice, using the Cauchy criterion directly is not very convenient. Therefore, as a rule, simpler convergence tests are used:
1) If the row converges, then it is necessary that the common term u n tended to zero. However, this condition is not sufficient. We can only say that if the common term does not tend to zero, then the series definitely diverges. For example, the so-called harmonic series is divergent, although its common term tends to zero.
Example. Investigate the convergence of the series
Let's find - the necessary criterion for convergence is not satisfied, which means the series diverges.
2) If a series converges, then the sequence of its partial sums is bounded.
However, this sign is also not sufficient.
For example, the series 1-1+1-1+1-1+ … +(-1) n+1 +… diverges, because the sequence of its partial sums diverges due to the fact that
However, the sequence of partial sums is limited, because at any n.
Series with non-negative terms.
When studying series of constant sign, we will limit ourselves to considering series with non-negative terms, because simply multiplying by –1 from these series can yield series with negative terms.
Theorem. For a series with nonnegative terms to converge, it is necessary and sufficient that the partial sums of the series be bounded.
A sign for comparing series with non-negative terms.
Let two rows be given and at u n , v n ³ 0.
Theorem. If u n£ vn at any n, then from the convergence of the series it follows that the series converges , and from the divergence of the series the series diverges.
Proof. Let us denote by S n And s n partial sums of series And . Because According to the conditions of the theorem, the series converges, then its partial sums are limited, i.e. in front of everyone n s n< M, где М – некоторое число. Но т.к. u n£ vn, That S n£ s n then the partial sums of the series are also limited, and this is sufficient for convergence.
Example.
Because , and the harmonic series diverges, then the series diverges.
Example. Examine the series for convergence
Because , and the series converges (like a decreasing geometric progression), then the series also converges.
The following convergence sign is also used:
Theorem. If there is a limit , where h is a number different from zero, then the series behave identically in the sense of convergence.
D'Alembert's sign.
(Jean Leron d'Alembert (1717 - 1783) - French mathematician)
If for a series with positive terms there is such a number q<1, что для всех достаточно больших n выполняется неравенство
then the series converges if the condition is satisfied for all sufficiently large n
then the series diverges.
D'Alembert's limiting sign.
D'Alembert's limiting criterion is a consequence of the above D'Alembert criterion.
< 1 ряд сходится, а при r >1 – diverges. If r = 1, then the question of convergence cannot be answered.
Example. Determine the convergence of the series.
Conclusion: the series converges.
Example. Determine the convergence of the series
Conclusion: the series converges.
Cauchy's sign. (radical sign)
If for a series with non-negative terms there is such a number q<1, что для всех достаточно больших n выполняется неравенство
then the series converges if for all sufficiently large n the inequality
then the series diverges.
Consequence. If there is a limit, then for r<1 ряд сходится, а при r>Row 1 diverges.
Example. Determine the convergence of the series.
Conclusion: the series converges.
Example. Determine the convergence of the series.
Those. The Cauchy test does not answer the question of the convergence of the series. Let us check that the necessary convergence conditions are met. As mentioned above, if a series converges, then the common term of the series tends to zero.
Thus, the necessary condition for convergence is not satisfied, which means the series diverges.
Integral Cauchy test.
If j(x) is a continuous positive function decreasing on the interval and then the integrals behave identically in the sense of convergence.
Alternating series.
Alternating rows.
An alternating series can be written as:
Leibniz's sign.
If the absolute values of u i for an alternating series decrease and the common term tends to zero, then the series converges.
Absolute and conditional convergence of series.
Let's consider some alternating series (with terms of arbitrary signs).
and a series composed of the absolute values of the members of the series (1):
Theorem. From the convergence of series (2) follows the convergence of series (1).
Proof. Series (2) is a series with non-negative terms. If series (2) converges, then by the Cauchy criterion for any e>0 there is a number N such that for n>N and any integer p>0 the following inequality is true:
According to the property of absolute values:
That is, according to the Cauchy criterion, from the convergence of series (2) the convergence of series (1) follows.
Definition. The series is called absolutely convergent, if the series converges.
It is obvious that for series of constant sign the concepts of convergence and absolute convergence coincide.
Definition. The series is called conditionally convergent, if it converges and the series diverges.
D'Alembert's and Cauchy's tests for alternating series.
Let be an alternating series.
D'Alembert's sign. If there is a limit, then for r<1 ряд будет абсолютно сходящимся, а при r>
Cauchy's sign. If there is a limit, then for r<1 ряд будет абсолютно сходящимся, а при r>Row 1 will be divergent. When r=1, the test does not give an answer about the convergence of the series.
Properties of absolutely convergent series.
1) Theorem. For absolute convergence of a series, it is necessary and sufficient that it can be represented as the difference of two convergent series with non-negative terms.
Consequence. A conditionally convergent series is the difference of two divergent series with non-negative terms tending to zero.
2) In a convergent series, any grouping of the terms of the series that does not change their order preserves the convergence and magnitude of the series.
3) If a series converges absolutely, then the series obtained from it by any permutation of terms also converges absolutely and has the same sum.
By rearranging the terms of a conditionally convergent series, one can obtain a conditionally convergent series having any predetermined sum, and even a divergent series.
4) Theorem. For any grouping of members of an absolutely convergent series (in this case, the number of groups can be either finite or infinite, and the number of members in a group can be either finite or infinite), a convergent series is obtained, the sum of which is equal to the sum of the original series.
5) If the series and converge absolutely and their sums are equal, respectively S and s, then the series composed of all products of the type taken in any order also converges absolutely and its sum is equal to S×s- the product of the sums of the multiplied series.
If you multiply conditionally convergent series, you can get a divergent series as a result.
Functional sequences.
Definition. If the members of the series are not numbers, but functions of X, then the series is called functional.
The study of the convergence of functional series is more complicated than the study of numerical series. The same functional series can, with the same variable values X converge, and with others - diverge. Therefore, the question of convergence of functional series comes down to determining those values of the variable X, at which the series converges.
The set of such values is called area of convergence.
Since the limit of each function included in the convergence region of the series is a certain number, the limit of the functional sequence will be a certain function:
Definition. Subsequence ( fn(x)} converges to function f(x) on the segment if for any number e>0 and any point X from the segment under consideration there is a number N = N(e, x) such that the inequality
is fulfilled when n>N.
For the chosen value e>0, each point on the segment has its own number and, therefore, there will be an infinite number of numbers corresponding to all points on the segment. If you choose the largest of all these numbers, then this number will be suitable for all points of the segment, i.e. will be common to all points.
Definition. Subsequence ( fn(x)} converges uniformly to function f(x) on the segment , if for any number e>0 there is a number N = N(e), such that the inequality
is fulfilled for n>N for all points of the segment.
Example. Consider the sequence
This sequence converges on the entire number line to the function f(x)=0, because
Let's build graphs of this sequence:
As can be seen, with increasing number n the sequence graph approaches the axis X.
Functional series.
Definition. Private (partial) amounts functional series are called functions
Definition. The functional series is called convergent at point ( x=x 0), if the sequence of its partial sums converges at this point. The limit of the sequence is called amount row at a point x 0.
Definition. Set of all values X, for which the series converges is called area of convergence row.
Definition. The series is called uniformly convergent on the interval if the sequence of partial sums of this series converges uniformly on this interval.
Theorem. (Cauchy criterion for uniform convergence of series)
For the series to converge uniformly, it is necessary and sufficient that for any number e>0 there exists a number N(e) such that for n>N and any integer p>0 the inequality
would hold for all x on the interval .
Theorem. (Weierstrass test for uniform convergence)
(Karl Theodor Wilhelm Weierstrass (1815 – 1897) – German mathematician)
A series converges uniformly and, moreover, absolutely on an interval if the moduli of its members on the same interval do not exceed the corresponding terms of the convergent number series with positive terms:
those. there is an inequality:
They also say that in this case the functional series is majorized number series.
2) Theorem on term-by-term integration of a series.
A series with continuous terms that uniformly converges on an interval can be integrated term by term on this interval, i.e. a series composed of integrals of its terms over the segment converges to the integral of the sum of the series over this segment.
3) Theorem on term-by-term differentiation of a series.
If the terms of a series converging on a segment are continuous functions that have continuous derivatives, and the series composed of these derivatives converges uniformly on this segment, then this series converges uniformly and can be differentiated term by term.
Based on the fact that the sum of the series is some function of the variable X, you can perform the operation of representing a function in the form of a series (expansion of a function into a series), which is widely used in integration, differentiation and other operations with functions.
(Nils Henrik Abel (1802 – 1829) – Norwegian mathematician)
Theorem. If a power series converges at x = x 1, then it converges and, moreover, absolutely for everyone.
Proof. According to the conditions of the theorem, since the terms of the series are limited, then
Where k- some constant number. The following inequality is true:
From this inequality it is clear that when x
Therefore, based on the comparison criterion, we conclude that the series converges, which means the series
Consider the functional series$\sum \limits _(n=1)^(\infty )u_(n) (x)=u_(1) (x)+u_(2) (x)+u_(3) (x) +...$, whose members are functions of one independent variable x. The sum of the first n terms of the series $S_(n) (x)=u_(1) (x)+u_(2) (x)+...+u_(n) (x)$ is the partial sum of this functional series. The general term $u_(n) (x)$ is a function of x defined in some domain. Let's consider the functional series at the point $x=x_(0) $. If the corresponding number series $\sum \limits _(n=1)^(\infty )u_(n) (x_(0))$converges, i.e. there is a limit of partial sums of this series$\mathop(\lim )\limits_(n\to \infty ) S_(n) (x_(0))=S(x_(0))$(where $S(x_(0) )
Definition 2
Area of convergence of a functional series $\sum \limits _(n=1)^(\infty )u_(n) (x)$ is the set of all values of x for which the functional series converges. The convergence region, consisting of all convergence points, is denoted by $D(x)$. Note that $D(x)\subset $R.
A function series converges in the domain $D(x)$ if for any $x\in D(x)$ it converges as a number series, and its sum is some function $S(x)$. This is the so-called limit function of the sequence $\left\(S()_(n) (x)\right\)$: $\mathop(\lim )\limits_(n\to \infty ) S_(n) (x) =S(x)$.
How to find the region of convergence of the functional series $D(x)$? You can use a sign similar to d'Alembert's sign. For the series $\sum \limits _(n=1)^(\infty )u_(n) (x)$ we compose $u_(n+1) (x)$ and consider the limit for a fixed x: $\mathop(\ lim )\limits_(n\to \infty ) \left|\frac(u_(n+1) (x))(u_(n) (x)) \right|=\left|l(x)\right| $. Then $D(x)$ is a solution to the inequality $\left|l(x)\right|
Example 1
Find the area of convergence of the series $\sum \limits _(n=1)^(\infty )\, \frac(x^(n) )(n) \, $.
Solution. Let us denote $u_(n) (x)=\frac(x^(n) )(n) $, $u_(n+1) (x)=\frac(x^(n+1) )(n+1 ) $. Let's compose and calculate the limit $\mathop(\lim )\limits_(n\to \infty ) \left|\frac(u_(n+1) (x))(u_(n) (x)) \right|=\ mathop(\lim )\limits_(n\to \infty ) \left|\frac(x^(n+1) \cdot n)(x^(n) \cdot (n+1)) \right|=\ left|x\right|$, then the region of convergence of the series is determined by the inequality $\left|x\right|
if $x=1$, $u_(n) (1)=\frac(1)(n) $, then we get a divergent series $\sum \limits _(n=1)^(\infty )\, \frac (1)(n) \, $;
if $x=-1$, $u_(n) (-1)=\frac((-1)^(n) )(n) $, then the series $\sum \limits _(n=1)^( \infty )\frac((-1)^(n) )(n) \, \, $ converges conditionally (using the Leibniz criterion).
Thus, the region of convergence $D(x)$ of the series $\sum \limits _(n=1)^(\infty )\, \frac(x^(n) )(n) \, $has the form:$- 1\le x
Properties of power series
Consider the power series $\sum \limits _(n=0)^(\infty )a_(n) x^(n) $, whose convergence interval is $(-R;\, R)$, then the sum of the power series $ S(x)$ is defined for all $x\in (-R;R)$ and we can write the equality $S(x)=\sum \limits _(n=0)^(\infty )a_(n) x^ (n)$.
Property 1. The power series $\sum \limits _(n=0)^(\infty )a_(n) x^(n) $ converges absolutely in any interval $\, \, \subset \, (-R;R)$, lying in the convergence interval, and the sum of the power series $S(x)$ is a continuous function for all $x\in $.
Property 2. If the segment is $\, \, \subset \, (-R;R)$, then the power series can be integrated termwise from a to b, i.e. If
$S(x)=\sum \limits _(n=0)^(\infty )a_(n) x^(n) =a_(0) +a_(1) x+a_(2) x^(2 ) +...$, then
$\int \limits _(a)^(b)S(x)\, (\rm d)x =\sum \limits _(n=0)^(\infty )\int \limits _(a)^ (b)a_(n) x^(n) \, (\rm d)x=\int \limits _(a)^(b)a_(0) (\rm d)x +\int \limits _( a)^(b)a_(1) x\, (\rm d)x +...+\int \limits _(a)^(b)a_(n) x^(n) \, (\rm d)x +...$.
In this case, the radius of convergence does not change:
where $a"_(n) =\frac(a_(n) )(n+1) $ are the coefficients of the integrated series.
Property 3. The sum of a power series is a function that has derivatives of any order within the convergence interval. The derivatives of the sum of a power series will be the sums of series obtained from a given power series by term-by-term differentiation the appropriate number of times, and the radii of convergence of such series will be the same as those of the original series.
If $S(x)=a_(0) +a_(1) x+a_(2) x^(2) +...+a_(n) x^(n) +...=\sum \limits _(n=0)^(\infty )\, a_(n) \cdot x^(n) $,then $S"(x)=a_(1) +2a_(2) x+...+na_( n) x^(n-1) +...=\sum \limits _(n=1)^(\infty )\, n\cdot a_(n) \cdot x^(n-1) $,$ S""(x)=2a_(2) +6a_(3) x+...+n(n-1)a_(n) x^(n-2) +...=\sum \limits _(n =2)^(\infty )\, n\cdot (n-1)\cdot a_(n) \cdot x^(n-2) $, ... , etc.
Examples
Series $\sum \limits _(n=1)^(\infty )n!\; x^(n) $ converges only at the point $x=0$; the series diverges at all other points. $V:\left\(0\right\).$
Series $\sum \limits _(n=1)^(\infty )\frac(x^(n) )(n $ сходится во всех точках оси, $V=R$.!}
The series $\sum \limits _(n=1)^(\infty )\frac((-1)^(n) x^(n) )(n) $ converges in the region $V=(-1,\, 1]$.
The series $\sum \limits _(n=1)^(\infty )\frac(1)(n+\cos x) $ diverges at all points of the $V=$$\emptyset$ axis.
