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” .

Power series is called a functional series of the form

Here x – real variable. Numbers a n (n = 0, 1, 2, … ) are called coefficients of the series. In what follows we will restrict ourselves to the case when everything a n and magnitude x 0 – real numbers. Power series (9.5) is also called close by degrees of differencex  x 0 .

If x 0 = 0 , then we obtain a power series of the form

, (9.6)

which is called next in degreex .

The power series (9.5) is reduced to the form (9.6) using a simple transformation x  x 0 = t (transfer the origin on the number axis). Because of this, the theory of power

series (9.5) and (9.6) are common. Therefore, in the future we will limit ourselves to considering the main properties of series of the form (9.6).

When considering power series, the main issue is to determine them areas of convergence, i.e., the sets of those values x , at which the series converges.

This problem is solved on the basis Abel's theorems .

If the power series (9.6) converges at some valuex = x 1  0 , then it converges absolutely for all valuesx , satisfying the inequality  x  <  x 1  .

If the series diverges at a certain value x = x 2 , then it diverges and for allx ,satisfying the inequality  x  >  x 2 .

Abel's theorem allows us to judge the location of the points of convergence and divergence of the power series (9.6).

Indeed, if x 1 is the point of convergence, then the entire interval (   x 1  ,  x 1 ) is filled with points of absolute convergence.

If x 2 is the point of divergence, then the intervals (   ,   x 2 ) and ( x 2  , +  )consist of points of divergence.

From this we can conclude that there is such a number R , what at  x  < R the power series is absolutely convergent, and for  x  > R – diverges.

Interval (  R , R ) is called convergence interval power series (9.6). Number R called radius of convergence power series.

Note that the interval of convergence of some series represents the entire number line (in this case R =  ), for others it degenerates into one point (the case R = 0 ). At x =  R , i.e., at the ends of the convergence interval, the series can converge absolutely, conditionally, or diverge. To clarify the behavior of the series at the end points, it is necessary to substitute in the expression for the series x values  R and examine the resulting two number series for convergence. This issue is resolved for each specific row individually.

When applied to power series of the form (9.5), the results obtained are modified only in that the center of convergence is located at the point x = x 0 , not at the point x = 0 , i.e. the interval of convergence of the power series (9.5) is symmetric with respect to the point x = x 0 and represents the interval x 0  R < x < x 0 + R .

Note that to find the interval of convergence of the power series (9.6), we can examine the series

, (9.7)

composed of the modules of the members of a given series, since the convergence intervals of these series coincide.

To determine the convergence of series (9.7), the terms of which are positive, the D’Alembert or Cauchy convergence tests are usually used.

Let's assume there is a limit
.

Then, by d’Alembert’s criterion, series (9.7) converges for
, i.e. if
, and diverges at
, i.e. if
. Thus, this series converges within the interval
and diverges outside it, i.e., the radius of convergence is equal to
.

Notes.

1) If A = 0 , then the original series converges absolutely for all numerical values x , since in this case we have  x   A = 0 < 1 for anyone x . In this case, the radius of convergence R =  .

2) If A =  , That the original series converges at a single point x = 0 . It was previously accepted that in this case R = 0 .

3) Similarly, to determine the convergence interval, you can use the Cauchy test if there is
. In this case
.

4) The convergence interval can be found using D’Alembert’s or Cauchy’s tests directly.

Example 9.11. Determine the area of ​​convergence of the series
.

Solution. Here
. That's why,

.

So the interval
is the convergence interval of the given series.

Let us study the behavior of the series at the ends of the convergence interval. At
the series will take the form
. This is a harmonic series, it diverges. At
the series will take the form
. This alternating series converges conditionally, since it is easy to check that the conditions of the Leibniz criterion are satisfied, and the series of moduli diverges.

So, when
The series converges absolutely when
The series converges conditionally; at all other points the series diverges.

Example 9.12. Find the area of ​​convergence of the series
.

Solution. Let's use Cauchy's test. We have

Hence, the series converges absolutely only when x =  1 , and at all other points of the numerical axis the series diverges. Convergence radius R = 0 .

Example 1. Find the region of convergence of the power series:

A) ; b) ;

V) ; G)
;

d)
.

A) Let's find the radius of convergence R. Because
,
, That

.

x
, that is, the interval of convergence of the series
.

At
we get a number series . This series converges because it is a generalized harmonic series at
.

At
we get a number series
. This series is absolutely convergent, since a series composed of the absolute values ​​of its terms , convergent.

.

b) Let's find the radius of convergence R. Because
, That
.

So, the convergence interval of the series
.

We examine the convergence of this series at the ends of the convergence interval.

At
we have a number series

.

At
we have a number series
. This series is divergent because
does not exist.

So, the region of convergence of this series
.

V) Let's find the radius of convergence R. Because
,
That
.

So, the convergence interval
. The convergence region of this series coincides with the convergence interval, that is, the series converges for any value of the variable x.

G) Let's find the radius of convergence R. Because
,
That
.

Because
, then the series converges only at the point
. This means that the region of convergence of this series is one point
.

d) Let's find the radius of convergence R.

Because
,
, That

.

So, the series converges for absolutely everyone x, satisfying the inequality
, that is
.

From here
− convergence interval,
− radius of convergence.

Let us examine this series for convergence at the ends of the convergence interval.

At
we get a number series

,

which diverges (harmonic series).

At
we get a number series
, which converges conditionally (the series converges according to Leibniz’s criterion, and the series composed of the absolute values ​​of its members diverges, since it is harmonic).

So, the region of convergence of the series
.

2.3. Taylor and Maclaurin series.

Expansion of functions into power series.

Application of power series to approximate calculations

Examples of problem solving

Example 1. Expand the function into a power series:

A)
; b)
;

V)
; G)
.

A) Replacing in the formula
x on
, we obtain the desired expansion:

Where

b) Substituting in equality

Where
x on
, we obtain the desired expansion:

V) This function can be written like this:
. To find the desired series, it is enough to expand

Where
substitute
. Then we get:

G) This function can be rewritten like this: .

Function
can be expanded into a power series by putting in a binomial series
, we will receive .

Where
.

To obtain the desired expansion, it is enough to multiply the resulting series (due to the absolute convergence of these series).

Hence,

, Where
.

Example 2. Find approximate values ​​of these functions:

A)
with an accuracy of 0.0001;

b)
with an accuracy of 0.00001.

A) Because
, then into the expansion of the function, where
let's substitute
:

or

Because
, then the required accuracy will be ensured if we limit ourselves to only the first two terms of the resulting expansion.

.

We use the binomial series

Where
.

Believing
And
, we obtain the following expansion:

If in the last alternating series only the first two terms are taken into account and the rest are discarded, then the error in the calculation
will not exceed 0.000006 in absolute value. Then the error in the calculation
will not exceed the number. Hence,

Example 3. Calculate to the nearest 0.001:

A)
; b)
.

A)
.

Let us expand the integrand into a power series. To do this, let us substitute in the binomial series
and replace x on :

.

Since the segment of integration
belongs to the region of convergence of the resulting series
, then we will integrate term by term within the specified limits:

.

In the resulting alternating series, the fourth term is less than 0.001 in absolute value. Consequently, the required accuracy will be ensured if only the first three terms of the series are taken into account.

.

Since the first of the discarded terms has a minus sign, the resulting approximate value will be in excess. Therefore, the answer to within 0.001 is 0.487.

b) Let us first represent the integrand as a power series. Let us replace in the expansion of the function

Where

x on
, we get:

Then
.

The resulting alternating series satisfies the conditions of Leibniz's criterion. The fourth term of the series is less than 0.001 in absolute value. To ensure the required accuracy, it is enough to find the sum of the first three terms.

Hence,
.

Application

Power series on the site for practical exercises in order to consolidate the material covered. And honing students’ skills in order to learn how to uniquely determine the convergence of a power series. Practical exercises fully give the desired result if a sufficient number of lessons are allocated in the course of study. This will fully ensure high-quality training for students. But what to do when they are not there? In this case, our website or a similar resource will help you solve power series online. However, such calculators will not always be able to provide the correct answer to the problem. Just for this purpose, using the example of one condition, you need to compare the answers received between the solutions of similar sites. You can notice that the region of convergence of a series is sometimes calculated using different theorems and the answer, although correct, can be expressed in different forms of notation. Of course, this will not be considered a mistake; the whole point is how it will be more convenient for you to perceive it. In short, it’s up to you to find the convergence of a power series using one site or another, that is, how it will be convenient for you for further applications of the answer. Sometimes the solution of a power series itself is expressed through notations with inequality signs, and most often through the modulus sign. This is not accidental, since in practice the most common methods used are comparisons of common members of a series using modules. Through a series of transformations, a variable enclosed in a module is isolated, and a short notation remains that is normally perceived for understanding the solution. For a visual representation, the radius of convergence of a series can be represented on the numerical axis indicating the boundary points; this, by the way, is also welcome in a number of cases. There is no need to force yourself into any specific framework that will narrow your horizons. Generally speaking, power series is an important topic in mathematics because it is complex and you will have to take several courses to understand it. For example, the theory of passage to the limit and integral calculus, since to prove the convergence of a power series they often use precisely such techniques in which these actions are present. We offer you to take practical classes and test your knowledge of studying power series online directly on the site, since we guarantee that all solved problems are given with an accurate answer, in a matter of seconds and absolutely free in real time. In addition to the area of ​​convergence of the series, or as it is also called the radius of convergence of the series, we offer you many other related calculators, which you will certainly appreciate at the highest level. If you need to find the convergence of a power series, then let us do it for you, since the site is a guarantee of accuracy and a guarantee of an impeccable quality answer. Many students often ask such a question as how to quickly prepare for solving a power series, but not just a solution, but a high-quality and correct one. At all times, power series have had a more extensive meaning than is now taught to students. This is understandable, because this is explained by the fact that there is no time due to the need for in-depth study of more important topics. On the one hand - YES, but then does this mean that we can neglect the convergence of the power series? Most likely not, because without properly studying power series online, you simply will not be able to competently answer obvious questions in your coursework or dissertation defense. Let's say your subject area includes a discipline such as continuum mechanics or structural mechanics. It is obvious that the stability of systems is important when designing strategic objects, especially if it directly concerns the protection of human life. It would seem that what useful things can be learned if you learn or at least understand the essence of how to find the area of ​​convergence of a series? It is difficult to convey the importance of this definition in one sentence. But take my word for it, finding the convergence of a power series is as important and necessary a procedure as, for example, knowing the Pythagorean theorem. If the solution to a power series is carried out with an error, then in further calculations this will definitely play a cruel joke on the student. It sometimes happens that, due to an annoying inaccuracy in an error, the aircraft crashes during the first tests. Agree, this is a shame after the work done and the enormous investment of time. Therefore, learn and learn again to find the radius of convergence of a series, thereby instilling correctness and rigor in solving problems from the very beginning. Let's return to the topic of power series and tell you a little more about this section. In practice, many power series begin with the first term, although there are series in which the first term can begin with both the second and third terms. This is largely due to the fact that, for example, starting from the first term, the entire sum of the series immediately goes to infinity, which is of course trivial, in essence. The convergence of a power series, as a subject for studying the domain of its convergence, is not often used in practice, especially by students, if they do not study it at the department of mathematical analysis. The essence is clear and the tasks are all set. Our calculator calculates power series online, and also talks about the convergence of the series, by what criteria does a number series converge, in short, it can determine the convergence of power series. A variable can fall into the region of convergence of a series if it satisfies a specific single condition, that is, that the corresponding resulting numerical series converges to a finite real numerical value. Perhaps this is not just one condition; it is also necessary that all terms of the series for any ordinal natural value of the parameter n exist and are uniquely determined. Finding the convergence of a power series means determining the region of its convergence on the x-axis if we are talking about a Cartesian coordinate system. It seems possible to do this using D'Alembert's principle, however, you need to understand that it is only based on the criterion, since the principle itself establishes only the interval in which the variable will fall. Remember, d’Alembert’s test is not applicable for functional series, it is only for number series. Solving a power series is directly related to finding the radius of convergence of this series, but for brevity it is expressed this way. We will also use this term in order to keep up with the trend in the scientific world. Power series at boundary points are studied separately. Of course, this is part of the general problem of studying the convergence of a power series. At these boundary points, the series is examined as a numerical series - constant sign or alternating sign, depending on the type of the general term of the series. Series whose members are power functions are called power series, and the calculator can solve them online. When they say this, the following assumption immediately comes to mind, and if the members of the series are periodic functions, then such a series should probably be called a functional periodic series! It's a funny thing, but it's all very serious. When we have determined the area of ​​convergence of the series, it is then necessary to perform final calculations, namely, to examine numerical series for convergence, which are obtained by substituting the boundaries of a certain interval instead of the variable x of the power series. Then you can write a full answer with a solution. Let's consider an example of how you can find the convergence of a power series without using the main theorems, but only in a comparative way. In this case, we need to competently make comparisons of two functional series until we simplify the original series to the elementary one that has been studied for a long time. According to this principle, we will take as the answer the result that has long been known to everyone in advance. Based on the solution of a power series, it is not yet possible to unambiguously guess what exactly the radius of convergence of the series will be, since before this it is still necessary to study at least two number series on each of the boundaries of the interval. In appearance, all power series are the same in that their common term represents is a regular function of the argument. The essence of the study is precisely to determine the acceptable values ​​of this argument for the convergence of the series (conditional or unconditional), as well as at what intervals the corresponding numerical series will diverge. Studying power series for convergence takes a lot of time, and we recommend that you use a ready-made calculator on the website. It is necessary to examine the boundaries of the interval too, otherwise the task will not be completed completely, which means that two points are guaranteed to be deducted. On our website you can calculate the sum of power series online. Always fast, reliable, and most importantly free! Convenient interface and clear data request. By right, the area of ​​convergence of a series is a specific condition for the existence of the sum of a numerical series. If the value on the boundary of the interval gives a divergence of the resulting alternating series. then they say that the series converges conditionally, that is, it certainly converges in this region, but under certain conditions, which is important in any case. If we abstract from the concept of a power series, and for a moment simply imagine the sum of a power series as a certain function in the variable x, then we are no longer talking about finding the convergence of a power series, but about determining the conditions under which the value of the function will exist at different values ​​of its argument x. In short, we reduce the problem to the simplest finding of the domain of definition of a function. The truth is very simple and clear! Any solution to a power series always speaks about the radius of convergence of such a power series and is usually determined through the D’Alembert criterion, but not directly, but only with a condition. After this, the modulus of the resulting inequality is revealed and the numerical series are examined for absolute or conditional convergence. Then they draw a conclusion. It is very interesting when power series in their original form are integrated or differentiated, and then the sum of the series from the new power series is calculated. This leads to many options for how a series behaves under certain conditions. The found sum of the power series from the integrated terms of the original series is, in fact, the integrated sum of the original power series. Interesting and informative, isn't it? If you correctly formulate the text of the problem, it looks something like this: find the interval of convergence of the power series and study it on the boundaries of the found interval. From here the series can converge or diverge absolutely, which does not require additional research. Uniform convergence shows power series in online calculation, adding one by one all the terms of the original series, written in classical form, like at the university. Relying only on his instincts, a student risks falling into the trap of his self-confidence due to inexperience, when it is as easy as shelling pears to just take and use a website calculator at the very beginning of his studies. From the area of ​​convergence of the series, conclusions are drawn about the convergence of the functional, or more precisely, power series, namely, it is established that it converges either conditionally or absolutely. All this is necessary for the final recording of the final response. Without complicating the situation and without using the names of complex theorems, let’s say that finding the convergence of a power series will be easier to understand if you imagine a certain function as the sum of the series and study it. And this has long been clear to everyone and understands how to do it! The radius of convergence of a series and the solution of a power series are identical concepts, since they mean the same thing; more precisely, they uniquely define the region from which the values ​​of the variable are given by the convergence of the corresponding number series.