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Standard methods, but reached a dead end with another example.
What is the difficulty and where can there be a snag? Let's put aside the soapy rope, calmly analyze the reasons and get acquainted with the practical methods of solution.
First and most important: in the overwhelming majority of cases, to study the convergence of a series, it is necessary to apply some familiar method, but the common term of the series is filled with such tricky stuffing that it is not at all obvious what to do with it. And you go around in circles: the first sign does not work, the second does not work, the third, fourth, fifth method does not work, then the drafts are thrown aside and everything starts anew. This is usually due to a lack of experience or gaps in other sections of calculus. In particular, if running sequence limits and superficially disassembled function limits, then it will be difficult.
In other words, a person simply does not see the necessary solution due to a lack of knowledge or experience.
Sometimes “eclipse” is also to blame, when, for example, the necessary criterion for the convergence of the series is simply not fulfilled, but due to ignorance, inattention or negligence, this falls out of sight. And it turns out like in that bike where the professor of mathematics solved a children's problem with the help of wild recurrent sequences and number series =)
In the best traditions, immediately living examples: rows 
and their relatives - diverge, since in theory it is proved sequence limits. Most likely, in the first semester, you will be beaten out of your soul for a proof of 1-2-3 pages, but for now it is enough to show that the necessary condition for the convergence of the series is not met, referring to known facts. Famous? If the student does not know that the root of the nth degree is an extremely powerful thing, then, say, the series 
put him in a rut. Although the solution is like two and two: , i.e. for obvious reasons, both series diverge. A modest comment “these limits have been proven in theory” (or even its absence at all) is quite enough for offset, after all, the calculations are quite heavy and they definitely do not belong to the section of numerical series.
And after studying the next examples, you will only be surprised at the brevity and transparency of many solutions:
Example 1
Investigate the convergence of a series
Solution: first of all, check the execution necessary criterion for convergence. This is not a formality, but a great chance to deal with the example of "little bloodshed".
"Inspection of the scene" suggests a divergent series (the case of a generalized harmonic series), but again the question arises, how to take into account the logarithm in the numerator?
Approximate examples of tasks at the end of the lesson.
It is not uncommon when you have to carry out a two-way (or even three-way) reasoning:
Example 6
Investigate the convergence of a series 
Solution: first, carefully deal with the gibberish of the numerator. The sequence is limited: . Then: 
Let's compare our series with the series . By virtue of the double inequality just obtained, for all "en" it will be true: 
Now let's compare the series with the divergent harmonic series.
Fraction denominator less the denominator of the fraction, so the fraction itself – more fractions (write down the first few terms, if not clear). Thus, for any "en": 
So, by comparison, the series 
diverges along with the harmonic series.
If we change the denominator a little: 
, then the first part of the reasoning will be similar: 
. But to prove the divergence of the series, only the limit test of comparison is already applicable, since the inequality is false.
The situation with converging series is “mirror”, that is, for example, for a series, both comparison criteria can be used (the inequality is true), and for a series, only the limiting criterion (the inequality is false).
We continue our safari wild nature, where a herd of graceful and succulent antelopes loomed on the horizon:
Example 7
Investigate the convergence of a series
Solution: the necessary convergence criterion is satisfied, and we again ask the classic question: what to do? Before us is something resembling a convergent series, however, there is no clear rule here - such associations are often deceptive.
Often, but not this time. By using Limit comparison criterion Let's compare our series with the convergent series . When calculating the limit, we use wonderful limit 
, where as infinitesimal stands: 
converges together with next to .
Instead of using the standard artificial technique of multiplication and division by a "three", it was possible to initially compare with a convergent series.
But here a caveat is desirable that the constant-multiplier of the general term does not affect the convergence of the series. And just in this style the solution of the following example is designed:
Example 8
Investigate the convergence of a series
Sample at the end of the lesson.
Example 9
Investigate the convergence of a series
Solution: in the previous examples, we used the boundedness of the sine, but now this property is out of play. The denominator of a fraction of a higher order of growth than the numerator, so when the sine argument and the entire common term infinitely small. The necessary condition for convergence, as you understand, is satisfied, which does not allow us to shirk from work.
We will conduct reconnaissance: in accordance with remarkable equivalence 
, mentally discard the sine and get a series. Well, something like that….
Making a decision:
Let us compare the series under study with the divergent series . We use the limit comparison criterion: 
Let us replace the infinitesimal with the equivalent one: for 
.
A finite number other than zero is obtained, which means that the series under study diverges along with the harmonic series.
Example 10
Investigate the convergence of a series
This is a do-it-yourself example.
For planning further actions in such examples, the mental rejection of the sine, arcsine, tangent, arctangent helps a lot. But remember, this possibility exists only when infinitesimal argument, not so long ago I came across a provocative series:
Example 11
Investigate the convergence of a series 
.
Solution: it is useless to use the limitedness of the arc tangent here, and the equivalence does not work either. The output is surprisingly simple: 
Study Series diverges, since the necessary criterion for the convergence of the series is not satisfied.
The second reason"Gag on the job" consists in a decent sophistication of the common member, which causes difficulties of a technical nature. Roughly speaking, if the series discussed above belong to the category of “figures you guess”, then these ones belong to the category of “you decide”. Actually, this is called complexity in the "usual" sense. Not everyone will correctly resolve several factorials, degrees, roots and other inhabitants of the savannah. Of course, factorials cause the most problems:
Example 12
Investigate the convergence of a series
How to raise a factorial to a power? Easily. According to the rule of operations with powers, it is necessary to raise each factor of the product to a power:
And, of course, attention and once again attention, the d'Alembert sign itself works traditionally: 
Thus, the series under study converges.
I remind you of a rational technique for eliminating uncertainty: when it is clear order of growth numerator and denominator - it is not at all necessary to suffer and open the brackets.
Example 13
Investigate the convergence of a series
The beast is very rare, but it is found, and it would be unfair to bypass it with a camera lens.
What is double exclamation point factorial? The factorial "winds" the product of positive even numbers:
Similarly, the factorial “winds up” the product of positive odd numbers: 
Analyze what is the difference between
Example 14
Investigate the convergence of a series
And in this task, try not to get confused with the degrees, wonderful equivalences and wonderful limits.
Sample solutions and answers at the end of the lesson.
But the student gets to feed not only tigers - cunning leopards also track down their prey:
Example 15
Investigate the convergence of a series 
Solution: the necessary criterion of convergence, the limiting criterion, the d'Alembert and Cauchy criteria disappear almost instantly. But worst of all, the feature with inequalities, which has repeatedly rescued us, is powerless. Indeed, comparison with a divergent series is impossible, since the inequality 
incorrect - the multiplier-logarithm only increases the denominator, reducing the fraction itself 
in relation to the fraction. And another global question: why are we initially sure that our series 
is bound to diverge and must be compared with some divergent series? Does he fit in at all?
Integral feature? Improper integral 
evokes a mournful mood. Now, if we had a row 
… then yes. Stop! This is how ideas are born. We make a decision in two steps:
1) First, we study the convergence of the series 
. We use integral feature:
Integrand continuous on the
Thus, a number 
diverges together with the corresponding improper integral.
2) Compare our series with the divergent series 
. We use the limit comparison criterion:
A finite number other than zero is obtained, which means that the series under study diverges along with side by side 
.
And there is nothing unusual or creative in such a decision - that's how it should be decided!
I propose to independently draw up the following two-move:
Example 16
Investigate the convergence of a series 
A student with some experience in most cases immediately sees whether the series converges or diverges, but it happens that a predator cleverly disguises itself in the bushes:
Example 17
Investigate the convergence of a series 
Solution: at first glance, it is not at all clear how this series behaves. And if we have fog in front of us, then it is logical to start with a rough check of the necessary condition for the convergence of the series. In order to eliminate uncertainty, we use an unsinkable multiplication and division method by adjoint expression:
The necessary criterion for convergence did not work, but led to clean water our Tambov comrade. As a result of the performed transformations, an equivalent series was obtained 
, which in turn strongly resembles a convergent series .
We write a clean solution:
Compare this series with the convergent series . We use the limit comparison criterion:
Multiply and divide by the adjoint expression:
A finite number other than zero is obtained, which means that the series under study converges together with next to .
Perhaps some have a question, where did the wolves come from on our African safari? Don't know. They probably brought it. You will get the following trophy skin:
Example 18
Investigate the convergence of a series 
Sample Sample solutions at the end of the lesson
And, finally, one more thought that visits many students in despair: instead of whether to use a rarer criterion for the convergence of the series? Sign of Raabe, sign of Abel, sign of Gauss, sign of Dirichlet and other unknown animals. The idea is working, but in real examples it is implemented very rarely. Personally, in all the years of practice, I have only 2-3 times resorted to sign of Raabe when nothing really helped from the standard arsenal. I reproduce the course of my extreme quest in full:
Example 19
Investigate the convergence of a series
Solution: Without any doubt a sign of d'Alembert. In the course of calculations, I actively use the properties of degrees, as well as second wonderful limit:
Here's one for you. D'Alembert's sign did not give an answer, although nothing foreshadowed such an outcome.
After going through the manual, I found a little-known limit proven in theory and applied a stronger radical Cauchy criterion:
Here's two for you. And, most importantly, it is not at all clear whether the series converges or diverges (an extremely rare situation for me). Necessary sign of comparison? Without much hope - even if in an unthinkable way I figure out the order of growth of the numerator and denominator, this still does not guarantee a reward.
A complete d'Alembert, but the worst thing is that the series needs to be solved. Need. After all, this will be the first time that I give up. And then I remembered that there seemed to be some more strong signs. Before me was no longer a wolf, not a leopard and not a tiger. This was huge elephant waving big trunk. I had to pick up a grenade launcher:
Sign of Raabe
Consider a positive number series.
If there is a limit 
, then:
a) At a row diverges. Moreover, the resulting value can be zero or negative.
b) At a row converges. In particular, the series converges for .
c) When Raabe's sign does not give an answer.
We compose the limit and carefully simplify the fraction: 
Yes, the picture is, to put it mildly, unpleasant, but I was no longer surprised. lopital rules, and the first thought, as it turned out later, turned out to be correct. But first, for about an hour, I twisted and turned the limit using “usual” methods, but the uncertainty did not want to be eliminated. And walking in circles, as experience suggests, is a typical sign that the wrong way of solving has been chosen.
I had to turn to the Russian folk wisdom: "If all else fails, read the instructions." And when I opened the 2nd volume of Fichtenholtz, to my great joy I found a study of an identical series. And then the solution went according to the model.
By cosines and sines of multiple arcs, i.e. a series of the form
or in complex form
where a k,b k or, respectively, c k called coefficients of T. r.
For the first time T. r. meet at L. Euler (L. Euler, 1744). He got expansions
All R. 18th century In connection with the study of the problem of the free vibration of a string, the question arose of the possibility of representing the function characterizing the initial position of the string as a sum of T. r. This question caused a heated debate that lasted for several decades, the best analysts of that time - D. Bernoulli, J. D "Alembert, J. Lagrange, L. Euler ( L. Euler). Disputes related to the content of the concept of function. At that time, functions were usually associated with their analytics. assignment, which led to the consideration of only analytic or piecewise analytic functions. And here it became necessary for a function whose graph is sufficiently arbitrary to construct a T. r. representing this function. But the significance of these disputes is greater. In fact, they discussed or arose in connection with questions related to many fundamentally important concepts and ideas of mathematics. analysis in general - the representation of functions by Taylor series and analytical. continuation of functions, use of divergent series, limits, infinite systems of equations, functions by polynomials, etc.
And in the future, as in this initial one, the theory of T. r. served as a source of new ideas in mathematics. Fourier integral, almost periodic functions, general orthogonal series, abstract . Researches on T. river. served as a starting point for the creation of set theory. T. r. are a powerful tool for representing and exploring features.
The question that led to controversy among mathematicians in the 18th century was resolved in 1807 by J. Fourier, who indicated formulas for calculating the coefficients of T. r. (1), which must. represent on the function f(x):
and applied them in solving heat conduction problems. Formulas (2) are called Fourier formulas, although they were encountered earlier by A. Clairaut (1754), and L. Euler (1777) came to them using term-by-term integration. T. r. (1), the coefficients of which are determined by formulas (2), called. near the Fourier function f, and the numbers a k , b k- Fourier coefficients.
The nature of the results obtained depends on how the representation of a function is understood as a series, how the integral in formulas (2) is understood. Modern theory T. r. acquired after the appearance of the Lebesgue integral.
The theory of T. r. can be conditionally divided into two large sections - the theory Fourier series,
in which it is assumed that the series (1) is the Fourier series of a certain function, and the theory of general T. R., where such an assumption is not made. Below are the main results obtained in the theory of general T. r. (in this case, sets and the measurability of functions are understood according to Lebesgue).
The first systematic research T. r., in which it was not assumed that these series are Fourier series, was the dissertation of V. Riemann (V. Riemann, 1853). Therefore, the theory of general T. r. called sometimes the Riemannian theory of thermodynamics.
To study the properties of arbitrary T. r. (1) with coefficients tending to zero B. Riemann considered the continuous function F(x) ,
which is the sum of a uniformly convergent series
obtained after two-fold term-by-term integration of series (1). If the series (1) converges at some point x to a number s, then at this point the second symmetric exists and is equal to s. F functions:
then this leads to the summation of the series (1) generated by the factors 
called by the Riemann summation method. Using the function F, the Riemann localization principle is formulated, according to which the behavior of the series (1) at the point x depends only on the behavior of the function F in an arbitrarily small neighborhood of this point.
If T. r. converges on a set of positive measure, then its coefficients tend to zero (Cantor-Lebesgue). Tendency to zero coefficients T. r. also follows from its convergence on a set of the second category (W. Young, W. Young, 1909).
One of the central problems of the theory of general thermodynamics is the problem of representing an arbitrary function T. r. Strengthening the results of N. N. Luzin (1915) on the representation of T. R. functions by Abel-Poisson and Riemann summable methods, D. E. Men’shov proved (1940) the following theorem, which refers to the most important case when the representation of the function f is understood as T. r. to f(x) almost everywhere. For every measurable and finite almost everywhere function f, there exists a T. R. that converges to it almost everywhere (Men'shov's theorem). It should be noted that even if f is integrable, then, generally speaking, one cannot take the Fourier series of the function f as such a series, since there are Fourier series that diverge everywhere.
The above Men'shov theorem admits the following refinement: if a function f is measurable and finite almost everywhere, then there exists such that 
almost everywhere and the term-by-term differentiated Fourier series of the function j converges to f(x) almost everywhere (N. K. Bari, 1952).
It is not known (1984) whether it is possible to omit the finiteness condition for the function f almost everywhere in Men'shov's theorem. In particular, it is not known (1984) whether T. r. converge almost everywhere
Therefore, the problem of representing functions that can take on infinite values on a set of positive measure was considered for the case when it is replaced by the weaker requirement - . Convergence in measure to functions that can take on infinite values is defined as follows: partial sums of T. p. s n(x) converges in measure to the function f(x) .
if where f n(x) converge to / (x) almost everywhere, and the sequence converges to zero in measure. In this setting, the problem of representation of functions has been solved to the end: for every measurable function, there exists a T. R. that converges to it in measure (D. E. Men'shov, 1948).
Much research has been devoted to the problem of the uniqueness of T. r.: Can two different T. diverge to the same function? in a different formulation: if T. r. converges to zero, does it follow that all the coefficients of the series are equal to zero. Here one can mean convergence at all points or at all points outside a certain set. The answer to these questions essentially depends on the properties of the set outside of which convergence is not assumed.
The following terminology has been established. Many names. uniqueness set or U- set if, from the convergence of T. r. to zero everywhere, except, perhaps, for points of the set E, it follows that all the coefficients of this series are equal to zero. Otherwise Enaz. M-set.
As G. Cantor (1872) showed, as well as any finite are U-sets. An arbitrary is also a U-set (W. Jung, 1909). On the other hand, every set of positive measure is an M-set.
The existence of M-sets of measure was established by D. E. Men'shov (1916), who constructed the first example of a perfect set with these properties. This result is of fundamental importance in the problem of uniqueness. It follows from the existence of M-sets of measure zero that, in the representation of functions of T. R. that converge almost everywhere, these series are defined invariably ambiguously.
Perfect sets can also be U-sets (N. K. Bari; A. Rajchman, A. Rajchman, 1921). Very subtle characteristics of sets of measure zero play an essential role in the problem of uniqueness. The general question about the classification of sets of measure zero on M- and U-sets remains (1984) open. It is not solved even for perfect sets.
The following problem is related to the uniqueness problem. If T. r. converges to the function 
then whether this series must be the Fourier series of the function /. P. Dubois-Reymond (P. Du Bois-Reymond, 1877) gave a positive answer to this question if f is integrable in the sense of Riemann and the series converges to f(x) at all points. From results III. J. Vallee Poussin (Ch. J. La Vallee Poussin, 1912) implies that the answer is positive even if the series converges everywhere except for a countable set of points and its sum is finite.
If a T. p converges absolutely at some point x 0, then the points of convergence of this series, as well as the points of its absolute convergence, are located symmetrically with respect to the point x 0
(P. Fatou, P. Fatou, 1906).
According to Denjoy - Luzin theorem from the absolute convergence of T. r. (1) on a set of positive measure, the series converges 
and, consequently, the absolute convergence of series (1) for all X. This property is also possessed by sets of the second category, as well as by certain sets of measure zero.
This survey covers only one-dimensional T. r. (one). There are separate results related to general T. p. from several variables. Here in many cases it is still necessary to find natural problem statements.
Lit.: Bari N. K., Trigonometric series, M., 1961; Sigmund A., Trigonometric series, trans. from English, vol. 1-2, M., 1965; Luzin N. N., Integral and trigonometric series, M.-L., 1951; Riemann B., Works, trans. from German, M.-L., 1948, p. 225-61.
S. A. Telyakovsky.
Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.
In science and technology, one often has to deal with periodic phenomena, i.e. those that are reproduced after a certain period of time T called the period. The simplest of the periodic functions (except for a constant) is a sinusoidal value: asin(x+ ), harmonic oscillation, where there is a “frequency” related to the period by the ratio: . From such simple periodic functions, more complex ones can be composed. Obviously, the constituent sinusoidal quantities must be of different frequencies, since the addition of sinusoidal quantities of the same frequency results in a sinusoidal quantity of the same frequency. If we add several values of the form
For example, we reproduce here the addition of three sinusoidal quantities: . Consider the graph of this function
This graph is significantly different from a sine wave. This is even more true for the sum of an infinite series composed of terms of this type. Let us pose the question: is it possible for a given periodic function of the period T represent as the sum of a finite or at least an infinite set of sinusoidal quantities? It turns out that with respect to a large class of functions, this question can be answered in the affirmative, but this is only if we include precisely the entire infinite sequence of such terms. Geometrically, this means that the graph of a periodic function is obtained by superimposing a series of sinusoids. If we consider each sinusoidal value as a certain harmonic oscillatory movement, then we can say that this is a complex oscillation characterized by a function or simply by its harmonics (first, second, etc.). The process of decomposition of a periodic function into harmonics is called harmonic analysis.
It is important to note that such expansions often turn out to be useful in the study of functions that are given only in a certain finite interval and are not at all generated by any oscillatory phenomena.
Definition. A trigonometric series is a series of the form:
Or 
(1).
The real numbers are called the coefficients of the trigonometric series. This series can also be written like this:
If a series of the type presented above converges, then its sum is a periodic function with period 2p.
Definition. The Fourier coefficients of a trigonometric series are called: 
(2)
(3)
(4)
Definition. Near Fourier for a function f(x) is called a trigonometric series whose coefficients are the Fourier coefficients.
If the Fourier series of the function f(x) converges to it at all its points of continuity, then we say that the function f(x) expands in a Fourier series.
Theorem.(Dirichlet's theorem) If a function has a period of 2p and is continuous on a segment or has a finite number of discontinuity points of the first kind, the segment can be divided into a finite number of segments so that the function is monotone inside each of them, then the Fourier series for the function converges for all values X, and at the points of continuity of the function, its sum S(x) is equal to , and at the discontinuity points its sum is equal to , i.e. the arithmetic mean of the limit values on the left and right.
In this case, the Fourier series of the function f(x) converges uniformly on any interval that belongs to the interval of continuity of the function .
A function that satisfies the conditions of this theorem is called piecewise smooth on the interval .
Let's consider examples on the expansion of a function in a Fourier series.
Example 1. Expand the function in a Fourier series f(x)=1-x, which has a period 2p and given on the segment .
Solution. Let's plot this function
This function is continuous on the segment , that is, on a segment with a length of a period, therefore it can be expanded into a Fourier series that converges to it at each point of this segment. Using formula (2), we find the coefficient of this series: .
We apply the integration-by-parts formula and find and using formulas (3) and (4), respectively:
Substituting the coefficients into formula (1), we obtain 
or .
This equality takes place at all points, except for the points and (gluing points of the graphs). At each of these points, the sum of the series is equal to the arithmetic mean of its limit values on the right and left, that is.
Let us present an algorithm for expanding the function in a Fourier series.
The general procedure for solving the problem posed is as follows.
In a number of cases, by examining the coefficients of series of the form (C) or it can be established that these series converge (perhaps excepting individual points) and are Fourier series for their sums (see, for example, the previous n°), but in all these cases, the question naturally arises
how to find the sums of these series or, more precisely, how to express them in the final form in terms of elementary functions, if they are expressed in such a form at all. Even Euler (and also Lagrange) successfully used analytic functions of a complex variable to sum up trigonometric series in a final form. The idea behind the Euler method is as follows.
Let us assume that, for a certain set of coefficients, the series (C) and converge to functions everywhere in the interval, excluding only individual points. Consider now a power series with the same coefficients, arranged in powers of a complex variable
On the circumference of the unit circle, i.e., at , this series converges by assumption, excluding individual points:
In this case, according to the well-known property of power series, series (5) certainly converges at, i.e., inside the unit circle, defining there a certain function of a complex variable. Using known to us [see. § 5 of Chapter XII] of the expansion of elementary functions of a complex variable, it is often possible to reduce the function to them. Then for we have:
and by the Abel theorem, as soon as the series (6) converges, its sum is obtained as a limit
Usually this limit is simply equal to which allows us to calculate the function in the final form
Let, for example, the series
The statements proved in the previous paragraph lead to the conclusion that both these series converge (the first one, excluding the points 0 and
serve as Fourier series for the functions they define. But what are these functions? To answer this question, we make a series
By similarity with the logarithmic series, its sum is easily established:
Consequently,
Now an easy calculation gives:
so the modulus of this expression is , and the argument is .
and thus finally
These results are familiar to us and were even once obtained with the help of "complex" considerations; but in the first case, we started from the functions and , and in the second - from the analytic function. Here, for the first time, the series themselves served as a starting point. The reader will find further examples of this kind in the next section.
We emphasize once again that one must be sure in advance of the convergence and series (C) and in order to have the right to determine their sums using the limiting equality (7). The mere existence of a limit on the right-hand side of this equality does not yet allow us to conclude that the mentioned series converge. To show this with an example, consider the series
Hölder's condition. We say that a function $f(x)$ satisfies the Hölder conditions at a point $x_0$ if there are one-sided finite limits $f(x_0 \pm 0)$ and such numbers $\delta > 0$, $\alpha \in ( 0,1]$ and $c_0 > 0$ such that $|f(x_0+t)-f(x_0+0)|\leq c_0t^( \alpha )$, $|f(x_0-t)-f(x_0-0)|\leq c_0t^(\alpha )$.
Dirichlet formula. The transformed Dirichlet formula is called a formula of the form:
$$S_n(x_0)= \frac(1)(\pi)\int\limits_(0)^(\pi)(f(x_0+t)+f(x_0-t))D_n(t)dt \quad (1),$$ where $D_n(t)=\frac(1)(2)+ \cos t + \ldots+ \cos nt = \frac(\sin(n+\frac(1)(2))t) (2\sin\frac(t)(2)) (2)$ — .
Using the formulas $(1)$ and $(2)$, we write the partial sum of the Fourier series in the following form:
$$S_n(x_0)= \frac(1)(\pi)\int\limits_(0)^(\pi)\frac(f(x_0+t)+f(x_0-t))(2\sin\ frac(t)(2))\sin \left (n+\frac(1)(2) \right) t dt$$
$$\Rightarrow \lim\limits_(n \to \infty )S_n(x_0) — \frac(1)(\pi)\int\limits_(0)^(\pi)\frac(f(x_0+t) +f(x_0-t))(2\sin\frac(t)(2)) \cdot \\ \cdot \sin \left (n+\frac(1)(2) \right)t dt = 0 \quad (3)$$
For $f \equiv \frac(1)(2)$ the formula $(3)$ becomes: $$ \lim\limits_(n \to \infty )\frac(1)(\delta)\frac(\ sin(n+\frac(1)(2))t)(2\sin\frac(t)(2))dt=\frac(1)(2), 0
Convergence of the Fourier series at a point
Theorem. Let $f(x)$ be a $2\pi$-periodic absolutely integrable function on $[-\pi,\pi]$ and satisfy the Hölder condition at the point $x_0$. Then the Fourier series of the function $f(x)$ at the point $x_0$ converges to the number $$\frac(f(x_0+0)+f(x_0-0))(2).$$
If at the point $x_0$ the function $f(x)$ is continuous, then at this point the sum of the series is equal to $f(x_0)$.
Proof
Since the function $f(x)$ satisfies the Hölder condition at the point $x_0$, then for $\alpha > 0$ and $0< t$ $ < \delta$ выполнены неравенства (1), (2).
For a given $\delta > 0$, we write the equalities $(3)$ and $(4)$. Multiplying $(4)$ by $f(x_0+0)+f(x_0-0)$ and subtracting the result from $(3)$, we get $$ \lim\limits_(n \to \infty) (S_n (x_0) - \frac(f(x_0+0)+f(x_0-0))(2) - \\ - \frac(1)(\pi)\int\limits_(0)^(\delta)\ frac(f(x_0+t)+f(x_0-t)-f(x_0+0)-f(x_0-0))(2\sin \frac(t)(2)) \cdot \\ \cdot \ sin \left (n + \frac(1)(2) \right)t \, dt) = 0. \quad (5)$$
It follows from Hölder's condition that the function $$\Phi(t)= \frac(f(x_0+t)+f(x_0-t)-f(x_0+0)-f(x_0-0))(2\sin \frac(t)(2)).$$ is absolutely integrable on $$. Indeed, applying Hölder's inequality, we obtain that the following inequality holds for the function $\Phi(t)$: $|\Phi(t)| \leq \frac(2c_0t^(\alpha ))(\frac(2)(\pi)t) = \pi c_0t^(\alpha - 1) (6)$, where $\alpha \in (0,1 ]$.
By virtue of the comparison criterion for improper integrals, the inequality $(6)$ implies that $\Phi(t)$ is absolutely integrable on $.$
By Riemann's lemma $$\lim\limits_(n \to \infty)\int\limits_(0)^(\delta)\Phi(t)\sin \left (n + \frac(1)(2) \ right)t\cdot dt = 0 .$$
From the formula $(5)$ it now follows that $$\lim\limits_(n \to \infty)S_n(x_0) = \frac(f(x_0+0)+f(x_0-0))(2) . $$
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Consequence 1. If a $2\pi$-periodic and absolutely integrable on $[-\pi,\pi]$ function $f(x)$ has a derivative at a point $x_0$, then its Fourier series converges at this point to $f(x_0) $.
Consequence 2. If a $2\pi$-periodic and absolutely integrable on $[-\pi,\pi]$ function $f(x)$ has both one-sided derivatives at the point $x_0$, then its Fourier series converges at this point to $\frac (f(x_0+0)+f(x_0-0))(2).$
Consequence 3. If a $2\pi$-periodic and absolutely integrable on $[-\pi,\pi]$ function $f(x)$ satisfies the Hölder condition at the points $-\pi$ and $\pi$, then, due to periodicity, the sum of the series The Fourier transform at the points $-\pi$ and $\pi$ is equal to $$\frac(f(\pi-0)+ f(-\pi+0))(2).$$
Dini sign
Definition. Let $f(x)$ be a $2\pi$-periodic function. The point $x_0$ will be a regular point of the function $f(x)$ if
- 1) there are finite left and right limits $\lim\limits_(x \to x_0+0 )f(x)= \lim\limits_(x \to x_0-0 )f(x)= f(x_0+0)= f(x_0-0),$
- 2) $f(x_0)=\frac(f(x_0+0)+f(x_0-0))(2).$
Theorem. Let $f(x)$ be a $2\pi$-periodic absolutely integrable function on $[-\pi,\pi]$ and the point $x_0 \in \mathbb(R)$ be a regular point of the function $f(x)$ . Let the function $f(x)$ satisfy the Dini conditions at the point $x_0$: there exist improper integrals $$\int\limits_(0)^(h)\frac(|f(x_0+t)-f(x_0+0) |)(t)dt, \\ \int\limits_(0)^(h)\frac(|f(x_0-t)-f(x_0-0)|)(t)dt,$$
then the Fourier series of the function $f(x)$ at the point $x_0$ has the sum $f(x_0)$, i.e. $$ \lim\limits_(n \to \infty )S_n(x_0)=f(x_0)=\frac(f(x_0+0)+f(x_0-0))(2).$$
Proof
The partial sum $S_n(x)$ of the Fourier series has an integral representation $(1)$. And due to the equality $\frac(2)(\pi )\int\limits_(0)^(\pi )D_n(t) \, dt=1,$
$$ f(x_0)= \frac(1)(\pi )\int\limits_(0)^(\pi )f(x_0+0)+f(x_0-0)D_n(t) \, dt$$
Then we have $$S_n(x_0)-f(x_0) = \frac(1)(\pi)\int\limits_(0)^(\pi)(f(x_0+t)-f(x_0+0)) D_n(t) \, dt + $$ $$+\frac(1)(\pi)\int\limits_(0)^(\pi)(f(x_0-t)-f(x_0-0))D_n (t) \, dt. \quad(7)$$
Obviously, the theorem will be proved if we prove that both integrals in the formula $(7)$ have limits as $n \to \infty $ equal to $0$. Consider the first integral: $$I_n(x_0)=\int\limits_(0)^(\pi)(f(x_0+t)-f(x_0+0))D_n(t)dt. $$
The Dini condition is satisfied at the point $x_0$: the improper integral $$\int\limits_(0)^(h)\frac(|f(x_0+t)-f(x_0+0)|)(t) \, dt .$$
Therefore, for any $\varepsilon > 0$, there exists $\delta \in (0, h)$ such that $$\int\limits_(0)^(\delta )\frac(\left | f(x_0+t) -f(x_0+0) \right |)(t)dt
Given the chosen $\varepsilon > 0$ and $\delta > 0$, the integral $I_n(x_0)$ can be represented as $I_n(x_0)=A_n(x_0)+B_n(x_0)$, where
$$A_n(x_0)=\int\limits_(0)^(\delta )(f(x_0+t)-f(x_0+0))D_n(t)dt ,$$ $$B_n(x_0)=\ int\limits_(\delta)^(\pi )(f(x_0+t)-f(x_0+0))D_n(t)dt .$$
Consider first $A_n(x_0)$. Using the $\left | D_n(t)\right |
for all $t \in (0, \delta)$.
Therefore $$A_n(x_0) \leq \frac(\pi)(2) \int\limits_(0)^(\delta ) \frac(|f(x_0+t)-f(x_0+0)|)( t)dt
Let us pass to estimating the integral $B_n(x_0)$ as $n \to \infty $. To do this, we introduce the function $$ \Phi (t)=\left\(\begin(matrix)
\frac(f(x_0+t)-f(x_0+0))(2\sin \frac(t)(2)), 0
$$B_n(x_0)=\int\limits_(-\pi)^(\pi)\Phi (t) \sin \left (n+\frac(1)(2) \right)t\,dt.$$ We get that $\lim\limits_(n \to \infty )B_n(x_0)=0$, which means that for an arbitrary $\varepsilon > 0$ chosen earlier, there exists $N$ such that for all $n> N$ the inequality $|I_n(x_0)|\leq |A_n(x_0)| + |B_n(x_0)|
It is proved in exactly the same way that the second integral of the formula $(7)$ has a zero limit as $n \to \infty $.
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Consequence If a $2\pi$ periodic function $f(x)$ is piecewise differentiable on $[-\pi,\pi]$, then its Fourier series at any point $x \in [-\pi,\pi]$ converges to the number $$\frac(f(x_0+0)+f(x_0-0))(2).$$
On the segment $[-\pi,\pi]$ find the trigonometric Fourier series of the function $f(x)=\left\(\begin(matrix)
1, x \in (0,\pi),\\ -1, x \in (-\pi,0),
\\ 0, x=0.
\end(matrix)\right.$
Investigate the convergence of the resulting series.
Periodically extending $f(x)$ to the entire real axis, we obtain the function $\widetilde(f)(x)$, whose graph is shown in the figure.
Since the function $f(x)$ is odd, then $$a_k=\frac(1)(\pi)\int\limits_(-\pi)^(\pi)f(x)\cos kx dx =0; $$
$$b_k=\frac(1)(\pi)\int\limits_(-\pi)^(\pi)f(x)\sin kx \, dx = $$ $$=\frac(2)(\ pi)\int\limits_(0)^(\pi)f(x)\sin kx \, dx =$$ $$=-\frac(2)(\pi k)(1- \cos k\pi) $$
$$b_(2n)=0, b_(2n+1) = \frac(4)(\pi(2n+1)).$$
Therefore, $\tilde(f)(x)\sim \frac(4)(\pi)\sum_(n=0)^(\infty)\frac(\sin(2n+1)x)(2n+1 ).$
Since $(f)"(x)$ exists for $x\neq k \pi$, then $\tilde(f)(x)=\frac(4)(\pi)\sum_(n=0)^ (\infty)\frac(\sin(2n+1)x)(2n+1)$, $x\neq k \pi$, $k \in \mathbb(Z).$
At the points $x=k \pi$, $k \in \mathbb(Z)$, the function $\widetilde(f)(x)$ is not defined, and the sum of the Fourier series is equal to zero.
Setting $x=\frac(\pi)(2)$, we obtain the equality $1 — \frac(1)(3) + \frac(1)(5)- \ldots + \frac((-1)^n) (2n+1)+ \ldots = \frac(\pi)(4)$.
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Find the Fourier series of the following $2\pi$-periodic and absolutely integrable on $[-\pi,\pi]$ function:
$f(x)=-\ln |
\sin \frac(x)(2)|$, $x \neq 2k\pi$, $k \in \mathbb(Z)$, and examine the resulting series for convergence.
Since $(f)"(x)$ exists for $ x \neq 2k \pi$, the Fourier series of the function $f(x)$ will converge at all points of $ x \neq 2k \pi$ to the value of the function. Obviously that $f(x)$ is an even function and therefore its Fourier series expansion must contain cosines. \sin \frac(x)(2)dx = $$ $$= -2 \int\limits_(0)^(\frac(\pi)(2))\ln \sin \frac(x)(2) dx \,- \, 2\int\limits_(\frac(\pi)(2))^(\pi)\ln \sin \frac(x)(2)dx =$$ $$= -2 \int \limits_(0)^(\frac(\pi)(2))\ln \sin \frac(x)(2)dx \, — \, 2\int\limits_(0)^(\frac(\pi )(2))\ln\cos \frac(x)(2)dx=$$ $$= -2 \int\limits_(0)^(\frac(\pi)(2))\ln (\frac (1)(2)\sin x)dx =$$ $$= \pi \ln 2 \, — \, 2 \int\limits_(0)^(\frac(\pi)(2))\ln \ sin x dx =$$ $$= \pi \ln 2 \, — \, \int\limits_(0)^(\pi)\ln \sin \frac(t)(2)dt = \pi\ln 2 + \frac(\pi a_0)(2),$$ whence $a_0= \pi \ln 2$.
Let us now find $a_n$ for $n \neq 0$. We have $$\pi a_n = -2 \int\limits_(0)^(\pi)\cos nx \ln \sin \frac(x)(2)dx = $$ $$ = \int\limits_(0) ^(\pi) \frac(\sin(n+\frac(1)(2))x+\sin (n-\frac(1)(2))x)(2n \sin\frac(x)(2) )dx=$$ $$= \frac(1)(2n) \int\limits_(-\pi)^(\pi) \begin(bmatrix)
D_n(x)+D_(n-1)(x)\\ \end(bmatrix)dx.$$
Here $D_n(x)$ is the Dirichlet kernel defined by formula (2) and we get that $\pi a_n = \frac(\pi)(n)$ and, consequently, $a_n = \frac(1)(n) $. So $$-\ln |
\sin \frac(x)(2)| = \ln 2 + \sum_(n=1)^(\infty ) \frac(\cos nx)(n), x \neq 2k\pi, k \in \mathbb(Z).$$
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Literature
- Lysenko Z.M., lecture notes on mathematical analysis, 2015-2016
- Ter-Krikorov A.M. and Shabunin M.I. Course of mathematical analysis, pp. 581-587
- Demidovich B.P., Collection of tasks and exercises in mathematical analysis, edition 13, revised, CheRo Publishing House, 1997, pp. 259-267
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Task 1 of 5
1 .
Number of points: 1If a $2\pi$ -periodic and absolutely integrable on $[−\pi,\pi]$ function $f(x)$ has a derivative at the point $x_0$, then what will its Fourier series converge to at this point?
Task 2 of 5
2 .
Number of points: 1If all the conditions of the Dini test are satisfied, then to what number does the Fourier series of the function $f$ converge at the point $x_0$?
