Summation of trigonometric series using analytic functions. trigonometric fourier series trigonometric fourier series

Let us show that almost any periodic function can be represented as a series whose members are simple harmonics, using the so-called trigonometric series.

Definition. A trigonometric series is a functional series of the form

where are the real numbers a 0 , a n , b n are called the coefficients of the series.

The free term of the series is written in the form for uniformity of the formulas obtained later.

Two questions need to be addressed:

1) Under what conditions does the function f(x) with period 2π can be expanded in a series (5.2.1)?

2) How to calculate odds a 0 ,… a n , b n ?

Let's start with the second question. Let the function f(x) is continuous on the interval and has a period T=2π. We present the formulas that we will need in what follows.

For any integer , since the function is even.

For any whole .

(m and n whole numbers)

At ( m and n integers) each of the integrals (III, IV, V) is converted into the sum of the integrals (I) or (II). If , then in formula (IV) we obtain:

Equality (V) is proved similarly.

Let us now assume that the function turned out to be such that an expansion into a convergent Fourier series was found for it, that is,

(Note that the summation is over the index n).

If the series converges, then denote its sum S(x).

Termwise integration (legitimate due to the assumption of convergence of the series) in the range from to gives

since all terms except the first one are equal to zero (relations I, II). From here we find

Multiplying (5.2.2) by ( m=1,2,…) and integrating term by term within the range from to , we find the coefficient a n.

On the right side of the equality, all terms are equal to zero, except for one m=n(relations IV, V), Hence we obtain

Multiplying (5.2.2) by ( m\u003d 1,2, ...) and integrating term by term within the range from to , we similarly find the coefficient b n

Values ​​- determined by formulas (5.2.3), (5.2.4), (5.2.5) are called Fourier coefficients, and the trigonometric series (5.2.2) is the Fourier series for a given function f(x).

So, we got the decomposition of the function f(x) in a Fourier series

Let's return to the first question and find out what properties the function should have f(x), so that the constructed Fourier series is convergent, and the sum of the series would be exactly equal to f(x).

Definition. The function f(x) is called piecewise continuous, if it is continuous or has a finite number of discontinuity points of the first kind.

Definition. Function f(x), given on the interval is called piecewise monotonic, if the segment can be divided by points into a finite number of intervals, in each of which the function changes monotonously (increasing or decreasing).

We will consider functions f(x), having a period T=2π. Such functions are called 2π- periodic.

Let us formulate a theorem representing a sufficient condition for the expansion of a function into a Fourier series.

Dirichlet's theorem(accept without proof) . If 2π-periodic function f(x) on a segment is piecewise continuous and piecewise monotonic, then the Fourier series corresponding to the function converges on this segment and in this case:

1. At points of continuity of a function, the sum of the series coincides with the function itself S(x)=f(x);

2. At every point x 0 function break f(x) the sum of the series is ,

those. the arithmetic mean of the limits of the function to the left and right of the point x 0 ;

3. At points (at the ends of the segment) the sum of the Fourier series is ,

those. the arithmetic mean of the limit values ​​of the function at the ends of the segment, when the argument tends to these points from the inside of the interval.

Note: if the function f(x) with a period of 2π is continuous and differentiable in the entire interval and its values ​​​​at the ends of the interval are equal, i.e., due to the periodicity, this function is continuous on the entire real axis and for any X the sum of its Fourier series is the same as f(x).

Thus, if a function integrable on an interval f(x) satisfies the conditions of the Dirichlet theorem, then the equality takes place on the interval (expansion in a Fourier series):

The coefficients are calculated by formulas (5.2.3) - (5.2.5).

The Dirichlet conditions are satisfied by most of the functions that occur in mathematics and its applications.

Fourier series, like power series, are used for approximate calculation of function values. If the expansion of the function f(x) into a trigonometric series takes place, then you can always use the approximate equality , replacing this function with the sum of several harmonics, i.e. partial sum (2 n+1) term of the Fourier series.

Trigonometric series are widely used in electrical engineering, with their help they solve many problems of mathematical physics.

Expand in a Fourier series a function with a period of 2π, given on the interval (-π; π).

Solution. Find the coefficients of the Fourier series:

We got the expansion of the function in a Fourier series

At the points of continuity, the sum of the Fourier series is equal to the value of the function f(x)=S(x), at the point x=0 S(x)=1/2, at points x=π,2π,… S(x)=1/2.

Recall that in real analysis a trigonometric series is a series in cosines and sines of multiple arcs, i.e. row of the form

A bit of history. The initial period of the theory of such series is attributed to the middle of the 18th century in connection with the problem of string vibrations, when the desired function was sought as the sum of series (14.1). The question of the possibility of such a representation caused heated debate among mathematicians, which lasted for several decades. Disputes related to the content of the concept of function. At that time, functions were usually associated with their analytical assignment, but here it became necessary to represent a function next to (14.1), whose graph is a rather arbitrary curve. But the significance of these disputes is greater. In fact, they raised questions related to many fundamentally important ideas of mathematical analysis.

And in the future, as in this initial period, the theory of trigonometric series served as a source of new ideas. It was in connection with them, for example, that set theory and the theory of functions of a real variable arose.

In this concluding chapter, we will consider material that once again links real and complex analysis, but is little reflected in textbooks on the TFCT. In the course of analysis, they proceeded from a predetermined function and expanded it into a trigonometric Fourier series. Here we consider the inverse problem: for a given trigonometric series, establish its convergence and sum. For this, Euler and Lagrange successfully used analytic functions. Apparently, Euler for the first time (1744) obtained equalities

Below we follow Euler's footsteps, limiting ourselves only to special cases of series (14.1), namely, trigonometric series

Comment. The following fact will be essentially used: if the sequence of positive coefficients a p monotonically tends to zero, then these series converge uniformly on any closed interval containing no points of the form 2lx (to gZ). In particular, on the interval (0.2n -) there will be pointwise convergence. See about this in work, pp. 429-430.

Euler's idea of ​​summing the series (14.4), (14.5) is that, using the substitution z = e a go to power series

If inside the unit circle its sum can be found explicitly, then the problem is usually solved by separating the real and imaginary parts from it. We emphasize that, using the Euler method, one should check the convergence of the series (14.4), (14.5).

Let's look at some examples. In many cases, the geometric series will be useful

as well as the series obtained from it by term-by-term differentiation or integration. For instance,

Example 14.1. Find the sum of a series

Solution. We introduce a similar series with cosines

Both series converge everywhere, since majorized by the geometric series 1 + r + r 2+.... Assuming z = e"x, we get

Here the fraction is reduced to the form

where we get the answer to the question of the problem:

Along the way, we established equality (14.2): Example 14.2. Sum rows

Solution. According to the above remark, both series converge on the specified interval and serve as Fourier series for the functions they define f(x) 9 g(x). What are these functions? To answer the question, in accordance with the Euler method, we compose series (14.6) with coefficients a p= -. Agree-

but equality (14.7) we get

Omitting details (the reader should reproduce them), we point out that the expression under the logarithm sign can be represented as

The modulus of this expression is equal to -, and the argument (more precisely, its main value is

• 2sin-

value) is equal Therefore In ^ = -ln(2sin

Example 14.3. At -l sum rows

Solution. Both series converge everywhere, since they are dominated by the convergent

next to the common member -! . Row (14.6)

n(n +1)

directly

J_ _\_ __1_

/?(/? +1) P /1 + 1

ns will give a known amount. On the basis, we represent it in the form

equality

Here the expression in parentheses is ln(l + z) and the expression in square brackets is ^ ^ + ** ^--. Hence,

= (1 + -)ln(1 + z). Now

should be put here z = eLX and perform the same steps as in the previous example. Omitting details, we point out that

It remains to open the brackets and write the answer. We leave this to the reader.

Tasks for chapter 14

Calculate the sums of the following rows.

• 1.3.1. a) z = 0 and z-- 2;
• b) z = l and z=-1;
• v) z = i and z= -I am.
• 1.3.2. a) 1; 6)0; c) oo.
• 2.1.1. Arc of the parabola, r = at 2 running from the point (1;1) to the point (1;- 1) and back.
• 2.1.2. Segment with start a, end b.
• 2.1.3. Jordan rectified path in Fig. nineteen.

• 2.1.4. arc of a parabola y = x 2 with start (-1;0), end (1;1).
• 2.1.5. Circle dg 2 + (at - 1) 2 = 4.
• 2.2.1. Half plane Rez > .
• 2.2.2. Open circle C x ""^) 2 + Y 2
• 2.2.3. The interior of a parabola 2y = 1 - x 2 .
• 2.2.4. Vicious circle (d: - 2) 2 + at 2
• 2.2.5. The appearance of the parabola 2x \u003d - y 2.

3.1.a). If w=u + iv, then and= -r- -v = -^-^. Hence

l: 2 + (1-.g) 2 .t 2 + (1-d:) 2

The origin of coordinates should be excluded from this circle, since (m, v) 9* (0; 0) V* e R, tone and= lim v = 0.

x-yx>.v->oo

• b). Eliminate x,y from equalities x + y \u003d l, and \u003d x 2 - y, v = 2 xy. Answer: parabola 2v = l-and 2 .
• 3.2. The straight line l: = i (l^O) goes into a circle
• (w--) 2 + v 2 = (-) 2 with a punctured point (r/, v) = (0; 0). Apply it with
• 2a 2 a

a = 1, a = 2.

• 3.4. In cases a), b) use the "sign of non-existence of the limit". In case c), the limit exists and is equal to 2.
• 3.5. Is not. Consider function limits over two sequences with common terms respectively

z "=-! + -> z,=-l -

• 4.1. a) nowhere ns differentiable; b) differentiable everywhere.
• 4.2. a) has a derivative at all points of the line y = x, in each of

them w = 2x; is nowhere holomorphic;

• b) is holomorphic in C(0), and / = - j.
• 4.3. holomorphic in C, W=3z 2 .
• 4.4. From equalities / ; (z) = -- + i-/ / (z) = 0 it follows that w,v is not

St St

depend on the variable "t. The Cauchy-Riemann conditions imply that these functions are also independent of y.

4.5. Consider, for example, the case Re f(z) = i(x, y) = const. WITH

using the Cauchy-Riemann conditions, deduce from this that Im/(z) = v(x 9 y) = const.

• 5.1. a) because J=--=- =-* 0(z * -/) and according to the condition of the problem
• (l-/z) 2 (z+/) 2

the argument of the derivative is equal to zero, then its imaginary part is zero, and the real part is positive. From here derive the answer: straight at = -X-1 (X * 0).

b) circle z + i=j2.

• 5.3. Check that the function does not take zero value and its derivative exists everywhere and is equal to the given function.
• 6.1. From the definition of tangent as the ratio of sine to cosine, prove that tg(z + n^-tgz with valid argument values. Let T some other period tg(z + T) = tgz. From here and from the previous equality, deduce that sin(/r- T)= 0, whence it follows that T multiple To .
• 6.2. Use equalities (6.6).
• 6.3. The first formula is not correct, because not always arg(zH ,) = argz + argvv (take, for example, z = -1, w = -1). The second formula is also wrong. Consider, for example, the case z = 2.
• 6.4. From equality a a = e 01 "0 deduce that here the right side has the form |i|« , e ca(a^a+2 yak)? sli p r and some different integers to 19 to 2

the expression in parentheses took on the same meaning, then they would have

which is contrary to irrationality a .

• 6.5. z \u003d 2? / r- / "ln (8 ± V63).
• 7.1. a) angle - I am w
• b) circular sector | w2, | argvr|
• 7.2. In both cases, a circle of radius 1 centered at the origin.
• 7.3. We will move along the border of the semicircle so that its interior remains on the left. We use the notation z = x + yi, w = u + vi. Location on

at= 0, -1 x 1 we have and =--e [-1,1]" v = 0. Consider the second segment of the boundary - the semicircle z=e u,tg. In this section, the expression

is converted to the form w=u=-- ,/* -. In between. According to (8.6), the desired integral is equal to

b). The lower semicircle equation has the form z(t) = e“,t e[l, 2n). By formula (8.8), the integral is equal to

• 8.2. a). Divide the desired integral into the sum of integrals over the segment O A and along the segment AB. Their equations are respectively z= / + //,/ with and

z = t + i,te. Answer: - + - i.

• b). The integration curve equation can be written as z = e", t € . Then Vz has two different values, namely,

.1 .t+2/r

e 2 ,e 2. It follows from the conditions of the problem that we are talking about the main value of the root: Vz, i.e. about the first of these. Then the integral is

8.3. In solving the problem, the drawing is deliberately not given, but the reader should complete it. The equation of a straight line segment is used, connecting two given points i, /> e C (a - Start, b - end): z = (l - /)fl+ /?,/€ . Let's break the desired integral into four:

I = I AB + I BC + I CD +1 D.A. On the segment AB we have z- (1 -1) ? 1 +1 /, so the integral on this segment, according to (8.8), is equal to

Proceeding in a similar way, we find

• 9.1. a) 2n7; b) 0.
• 9.2. Make a substitution z = z0 + re 11.0 t2/g.
• 9.3 Function f(z)=J is holomorphic in some simply connected z-a

area D containing Г and ns containing a. By the integral theorem applied to /),/], the desired integral is equal to zero.

• 9.4. a) 2/n(cosl2 + /sinl2); b) 34l-/.
• 9.5. In case a) the singular points ±2/ lie inside the given circle, so the integral is equal to

• b). The singular points ±3/ also lie inside the circle. The solution is similar. Answer: 0.
• 10.1. Represent the function as /(z) = -----use
• 3 1 + -

geometric series 1 + q + q2 (||

• 1 -h
• 10.2. Differentiate term by term a geometric series.
• 10.3. a) | z+/1t = z2. Answer: z .
• 11.1. Use power expansions of exponent and sine. In case a) the order is 3, in case b) it is 2.
• 11.2. Up to an obvious change of variable, the equation can be

represent in the form /(z) = /(-^z). Without loss of generality, we can assume that

the radius of convergence of the Taylor series of the function centered at the point 0 is greater than one. We have:

The values ​​of the function are the same on a discrete set with a limit point belonging to the circle of convergence. By the uniqueness theorem /(z) = const.

11.3. Let us assume that the desired analytic function /(z) exists. Let's compare its values ​​with the function (z) = z2 on the set E,

consisting of dots z n = - (n = 2,3,...). Their meanings are the same, and since E

has a limit point belonging to the given circle, then by the uniqueness theorem /(z) = z 2 for all arguments of the given circle. But this contradicts the condition /(1) = 0. Answer: ns does not exist.

• 11.4. Yes, /(*) = -L
• 2 + 1
• 11.5. There is no contradiction, since the limit point of unit values ​​does not lie in the domain of the function.
• - 1 1
• 12.1. a) 0 ; b) 2

  12.2. a). Represent the function in the form and expand the parentheses.

  • b). Swap the terms, use the standard cosine and sine expansions.
  • 12.3.
  
  • 12.4. a) points 0, ± 1 are simple poles;
  • b) z = 0 - removable point;
  • c) z = 0 is an essentially singular point.
  • 13.1. a). The points a = 1, a = 2 are the poles of the integrand. The residue with respect to the first (simple) pole is found according to (13.2), it is equal to 1. The residue with respect to the second pole is found by the formula (13.3) with the order of multiplicity u = 2 and is equal to -1. The sum of the residues is zero, so the integral is zero by the fundamental residue theorem.
  • b). Inside the rectangle with the indicated vertices are three

  simple poles 1,-1,/. The sum of the residues in them is equal to --, and the integral is equal to

  v). Among the poles 2 Trki(kGZ) of the integrand, only two lie inside the given circle. It's 0 and 2 I am both of them are simple, the residues in them are equal in 1. Answer: 4z7.

  multiply it by 2/r/. Omitting details, we indicate the answer: / = -i .

  13.2. a). Let us put e"=z, then e"idt =dz , dt= - . Ho

  e" - e~" z-z~ x

  sin / =-=-, the intefal will be reduced to the form

  

  Here the denominator is factorized (z-z,)(z-z 2), where z, = 3 - 2 V2 / lies inside the circle at , a z,=3 + 2V2 / lies above. It remains to find the residue with respect to the simple pole z, using the formula (13.2) and

  b) . Assuming, as above, e" = z , we reduce the intefal to the form

  

  The subintephal function has three simple poles (which ones?). Leaving the reader to calculate the residues in them, we indicate the answer: I= .

  • v) . The subintegral function is equal to 2(1--=-), the desired integral
  • 1 + cos t

  equals 2(^-1- h-dt). Denote the integral in brackets by /.

  Applying the equality cos "/ = - (1 + cos2f) we get that / = [- cit .

  By analogy with cases a), b) make a substitution e 2,t = z, reduce the integral to the form

  

  where the integration curve is the same unit circle. Further arguments are the same as in case a). Answer: the original, sought-for integral is equal to /r(2-n/2).

  13.3. a). Consider the auxiliary complex integral

  /(/?)= f f(z)dz, where f(z) = - p-, G (I) - a contour composed of

  semicircles y(R): | z |= R> 1, Imz > 0 and all diameters (make a drawing). Let's split this integral into two parts - according to the interval [-/?,/?] and according to y(R).

  to. Ya.

  Only simple poles lie inside the circuit z 0 \u003d e 4, z, = e 4 (Fig. 186). We find with respect to their residues:

  It remains to verify that the integral over y(R) tends to zero as R. From the inequality |g + A|>||i|-|/>|| and from the estimate of the integral for z e y(R) it follows that

In a number of cases, by investigating 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 ie 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:

hence,

Now an easy calculation gives:

so the modulus of this expression is , and the argument is .

and thus ultimately

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 rows (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 above series converge. To show this with an example, consider the series

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 defined only in a certain finite interval and are not generated at all 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 monotonic 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.