Finding the coefficients of the Fourier series. Determination of the coefficients of a number by Fourier formulas

Fourier series of periodic functions with a period of 2π.

Fourier series allows you to study periodic functions, decomposing them into components. Variables and voltages, displacements, speed and acceleration of crank-connecting mechanisms and acoustic waves are typical practical examples of the use of periodic functions in engineering calculations.

Fourier decomposition is based on the assumption that all of the practical value of the function in the interval -π ≤x≤ π can be expressed in the form of converging trigonometric rows (the number is considered to be converging if the sequence of partial sums composed of its members converges:

Standard (\u003d Normal) Recording through the SINX and COSX amount

f (x) \u003d a o + a 1 cosx + a 2 cos2x + a 3 cos3x + ... + b 1 sinx + b 2 sin2x + b 3 sin3x + ...

where A O, A 1, A 2, ..., B 1, B 2, .. - Valid constants, i.e.

Where for the range from -π to π coefficients row Fourier Calculated by formulas:

The coefficients A O, A N and B N are called fourier coefficientsand if they can be found, then a number (1) is called near Fourier, The corresponding functions f (x). For a number (1), a member (A 1 COSX + B 1 SINX) is called first or main harmonic

Another way to record a number is to use the ACOSX + BSINX \u003d CSIN ratio (x + α)

f (x) \u003d a o + c 1 sin (x + α 1) + c 2 sin (2x + α 2) + ... + c n sin (nx + α n)

Where a o is a constant, C 1 \u003d (A 1 2 + B 1 2) 1/2, with n \u003d (a n 2 + b n 2) 1/2 - amplitude of the various components, and is a n \u003d arctg a n / b n.

For a number (1), a member (A 1 COSX + B 1 SINX) or C 1 SIN (X + α 1) is called the first or main harmonic (A 2 COS2X + B 2 SIN2X) or C 2 SIN (2x + α 2) called second harmonic etc.

For accurate presentation of the complex signal, an infinite number of members is usually required. However, in many practical tasks, it is enough to consider only a few first members.

Fourier series of non-periodic functions with a period of 2π.

Definition of non-periodic functions.

If the function f (x) is non-periodic, it means that it cannot be decomposed in a Fourier series for all values \u200b\u200bx. However, you can define a series of Fourier, representing a function in any range of 2π width.

If the non-periodic function is specified, you can create a new function, selecting the values \u200b\u200bof F (x) in a specific range and repeating them out of this range with an interval of 2π. Since the new function is periodic with a period of 2π, it can be decomposed into a Fourier series for all values \u200b\u200bx. For example, the function f (x) \u003d x is not periodic. However, if it is necessary to decompose it in a Fourier series at the interval from about 2π, then a periodic function with a period of 2π is built outside of this interval (as shown in Fig. Below).

For non-periodic functions, such as f (x) \u003d x, the sum of the Fourier series is equal to the value f (x) at all points of the specified range, but it is not equal to f (x) for points outside the range. To find a row of Fourier of the non-periodic function in the range of 2π, the Fourier coefficients formula is used.

Even and odd functions.

They say the function y \u003d f (x) evenif f (-x) \u003d f (x) for all values \u200b\u200bx. The graphs of even functions are always symmetrical with respect to the axis y (i.e. are mirror reflected). Two examples of even functions: y \u003d x 2 and y \u003d cosx.

It is said that the function y \u003d f (x) oddif f (-x) \u003d - f (x) for all values \u200b\u200bx. Charts of odd functions are always symmetrical relative to the start of coordinates.

Many functions are neither even nor odd.

Decomposition in a Fourier series by cosine.

The Fourier series of the even periodic function f (x) with a period of 2π contains only members with cosine (i.e., does not contain members with sinus) and may include a permanent member. Hence,

where the coefficients of the Fourier series

Fourier series of an odd periodic function f (x) with a period of 2π contains only members with sinus (i.e. does not contain members with cosine).

Hence,

where the coefficients of the Fourier series

Fourier row on half aode.

If the function is defined for a range, say from 0 to π, and not only from 0 to 2π, it can be decomposed into a row only on sines or tolo by cosine. The resulting Fourier series is called near Fourier on a half-period.

If you want to get a decomposition Fourier on a semiprode of cosinefunctions f (x) in the range from 0 to π, it is necessary to make an even periodic function. In fig. Below is the function f (x) \u003d x, built on the interval from x \u003d 0 to x \u003d π. Since the even function is symmetrical with respect to the axis F (x), carry out the AB line as shown in Fig. below. If you assume that outside the considered interval, the resulting triangular form is periodic with a period of 2π, then the final schedule has the form, show. In fig. below. Because it is required to obtain a Fourier decomposition by cosine, as before, calculate the Fourier coefficients A O and A N

If required to get fourier decomposition on a semi-period of sines Functions f (x) in the range from 0 to π, then it is necessary to make an odd periodic function. In fig. Below is the function f (x) \u003d x, built on the interval from from x \u003d 0 to x \u003d π. Since the odd function is symmetrical relative to the start of coordinates, we build a CD line, as shown in Fig. If we assume that outside the considered interval, the resulting sawn signal is periodic with a period of 2π, then the final schedule has the appearance shown in Fig. Since it is required to obtain a decomposition of fury on a semi-period of sines, as before, calculate the Fourier coefficient. B.

Fourier series for an arbitrary interval.

Decomposition of a periodic function with a period of L.

The periodic function f (x) is repeated by increasing x on L, i.e. f (x + l) \u003d f (x). The transition from previously discussed functions with a period of 2π to functions with a period L is quite simple, since it can be made by replacing the variable.

To find a Fourier series F (X) in the -L / 2≤X≤L / 2 range, we introduce a new variable U in such a way that the function f (x) occurs 2π relative to U. If u \u003d 2πx / l, then x \u003d -l / 2 at u \u003d -π and x \u003d l / 2 at u \u003d π. Also let f (x) \u003d f (lu / 2π) \u003d f (u). Fourier series F (U) has the view

(Integration limits can be replaced by any interval L length, for example, from 0 to L)

Fourier series on a half-period for functions specified in the interval L ≠ 2π.

For substitution U \u003d πh / l, the interval from x \u003d 0 to x \u003d L corresponds to the interval from U \u003d 0 to U \u003d π. Consequently, the function can be decomposed in a row only by cosine or only in sinus, i.e. in fourier series on half aode.

Decomposition of cosine in the range from 0 to L has the form

One of the types of functional series is a trigonometric series

The task is to choose the coefficients of the series so that it converges to the functions given in the interval [-π, π]; In other words, it is necessary to decompose this function in the trigonometric row. A sufficient condition of solvability of this problem is that the function is in the interval [-π, π] piecewise-continuous and piecewise differentiable, i.e., so that the interval [-π, π] can be divided into a finite number of partial intervals, in Each of which this function is continuous and has a derivative (at the ends of partial intervals, the function must have finite one-sided limits and one-sided derivatives, when calculating the function of the function at the end of the partial interval, its one-sided limit is taken). The condition of a piece of differentiability can be replaced by a condition of a piece monotony of the function, that is, the requirement to ensure that the function is monotonne in each of the partial intervals. A sufficient condition for decomposability of the function in the interval [-π, π] in the trigonometric series is also the requirement so that in this interval the function has a limited change. By definition of the function f (x), there is a limited change in the interval if with any partition of this interval to a finite number of intervals

value

limited from above the same number.

It is with such functions that have to deal with the solutions of practical tasks.

When performing any of the three specified sufficient conditions The function f (x) is represented in the interval [-π, π] trigonometric near, in which the coefficients are determined by formulas

With such coefficients, the trigonometric row is called near Fourier. This series converges to f (x) at each point of its continuity; At the break points, it converges to the average arithmetic left and right limit values, i.e. k, if x is a gap point (Fig. 1); At the borders of the segment a number converges to.

Picture 1.

Function expressed near Fourier, there is a periodic function, and therefore a number compiled for a function specified on the segment [-π, π] converges out of this segment to a periodic continuation of this function (Fig. 2).

Figure 2.

If the Fourier function f (x) is presented in an arbitrary interval [α, α + 2π] length 2π, then the coefficients of the row A 0, AK, BK (Fourier coefficients) can be determined by the specified formulas in which the integration limits are replaced by α and α + 2π. In general, since in formulas for a 0, a k, b k, there are functions with a period of 2π, integration can be carried out at any interval with a length of 2π.

Fourier series can be used for an approximate representation of the function, namely: the function f (x) is replaced by an approximately equal amount of the amount S n (x) of the first few members of the Fourier series:

Expression Sn (x), where a 0, ak, bk are Fourier coefficients F (X), compared with other expressions of the same species with the same value n, but with other coefficients, leads to a minimal average quadratic deviation Sn (x ) from f (x), which is defined as

Depending on the kind of symmetry of the function, some simplifications are possible. If the function is even, i.e. f (-x) \u003d f (x), then

and the function decomposes into a row of cosine. If the functions are odd, i.e. f (-m) \u003d - f (x), then

and the function decomposes into a row of sinus. If the function satisfies the condition f (x + π) \u003d - f (x), i.e. the curve relating to half the segment 2π is a mirror reflection of the other half of the curve,

The function can be specified not only on the segment of 2π long, but also on the segment of any length 2L. If it satisfies the above conditions on this segment, it is laid out in a Fourier series of the following type:

and the coefficients of the number are calculated by formulas

In tab. 1 Dana decomposition of some functions.

Table 1.

Trigonometric row can be recorded in this form:

Fourier series Function F (X) converges in the sooner, the smoother function. If the function f (x) and its derivatives f "(x), f" (x), ..., fk -1 (x) are continuous, and F (k) (x) allows only the gap points of the 1st genus Ultimately, the Fourier coefficients A N, BN functions f (x) will be

The symbol is indicated by such a magnitude that

The decomposition into the trigonometric row is called harmonic analysis, and trigonometric functionsincluded in this series - harmonics. The calculation of the components of harmonics is called harmonic synthesis.

When calculating structures, it is often necessary to decompose in the Fourier series various functionsdefined by charts, and above all depicting the load. In tab. 2 and 3 decompositions are given for some functions characteristic of loads, including the rows corresponding to the concentrated forces.

Table 2.

Schedule of functions	Fourier row	n.

Transcript.

1 Ministry of Education and Science of the Russian Federation Novosibirsk State University Faculty of Faculty of Faculty of R. K. Belleeva Ryadov Fourier in Examples and Tasks Tutorial Novosibirsk 211

2 UDC BBK BL161 B44 B44 Belleeva R. K. Fourier Rows in Examples and Tasks: Tutorial / Novosib. State un-t. Novosibirsk, p. ISBN The study manual sets out basic information about the ranks of Fourier, there are examples for each studied topic. An example of the use of the Fourier method is disassembled in detail to solving the problem of string transverse oscillations. An illustrative material is given. There are tasks for an independent solution. Designed for students and teachers of the Physical Faculty of NSU. Printed by the decision of the Methodological Commission of the Physical Faculty of NSU. Reviewer Dr. Fiz.-Mat. science V. A. Aleksandrov allowance prepared as part of the implementation of the NiU-NSU program development program. ISBN C Novosibirski state University, 211 C Belleeva R. K., 211

3 1. Decomposition of a 2π-periodic function in a Fourier series definition. Near the Fourier function f (x) is called functional series A 2 + (An Cosnx + BN SIN NX), (1) where AN, BN coefficients are calculated using the formulas: an \u003d 1 π bn \u003d 1 π f (x) cosnxdx, n \u003d , 1, ..., (2) f (x) sin nxdx, n \u003d 1, 2, .... (3) Formulas (2) (3) are called Euler Fourier formulas. The fact that the functions f (x) correspond to the Fourier series (1) are recorded as a formula F (X) A 2 + (An Cosnx + BN SIN NX) (4) and it is said that the right-hand side of the formula (4) is a formal number Fourier function f (x). In other words, formula (4) means only that the coefficients A N, B N are found by formulas (2), (3). 3.

4 Definition. 2π-periodic function f (x) is called piecewise smooth, if in the interval [, π] there is a finite number of points \u003d x< x 1 <... < x n = π таких, что в каждом открытом промежутке (x j, x j+1) функция f(x) непрерывно дифференцируема, а в каждой точке x j существуют конечные пределы слева и справа: f(x j) = lim h + f(x j h), f(x j +) = lim h + f(x j + h), (5) f(x j h) f(x j) f(x j + h) f(x j +) lim, lim. h + h h + h (6) Отметим, что последние два предела превратятся в односторонние производные после замены предельных значений f(x j) и f(x j +) значениями f(x j). Теорема о представимости кусочно-гладкой функции в точке своим рядом Фурье (теорема о поточечной сходимости). Ряд Фурье кусочно-гладкой 2π-периодической функции f(x) сходится в каждой точке x R, а его сумма равна числу f(x), если x точка непрерывности функции f(x), f(x +) + f(x) и равна числу, если x точка разрыва 2 функции f(x). ПРИМЕР 1. Нарисуем график, найдем ряд Фурье функции, заданной на промежутке [, π] формулой, f(x) = x, предполагая, что она имеет период 2π, и вычислим суммы 1 1 числовых рядов (2n + 1) 2, n 2. n= Решение. Построим график функции f(x). Получим кусочно-линейную непрерывную кривую с изломами в точках x = πk, k целое число (рис. 1). 4

5 Fig. 1. Function Function F (X) Calculate Fourier coefficients a \u003d 1 π f (x) dx \u003d 1 π x 2 2 π \u003d π, an \u003d 1 π f (x) cosnxdx \u003d 2 π \u003d 2 () x SIN NX COS nx + π nn 2 \u003d 2 π (1) n 1 n 2 \u003d bn \u003d 1 π π \u003d 2 π f (x) cosnxdx \u003d cos nx cos n 2 \u003d 4 πn2, with N odd, with n even, f (x ) SIN NXDX \u003d, because the function f (x) is even. We write a formal Fourier series for the function f (x): f (x) π 2 4 π k \u003d 5 cos (2k + 1) x (2k + 1) 2.

6 Find out whether the function f (x) is a piecewise smooth. Since it is continuous, we calculate only the limits (6) at the finite points of the gap x \u003d ± π and at the point of the break x \u003d: and f (π h) f (π) π h π lim \u003d lim h + hh + h \u003d 1, f (+ h) f (+) + h () lim \u003d lim h + hh + hf (+ h) f (+) + h lim \u003d lim \u003d 1, h + hh + h \u003d 1, f (h) f () H () Lim \u003d lim \u003d 1. H + HH + H limits exist and are finite, therefore, a piece of piecewise smooth function. According to the theorem on current convergence, its Fourier series converges to the number F (x) at each point, that is, f (x) \u003d π 2 4 π k \u003d cos (2k + 1) + x (2k + 1) 2 \u003d \u003d π 2 4 (COSX + 19 π COS 3X) COS 5X (7) in Fig. 2, 3 shows the nature of the approximation of partial sums of the Fourier series S n (x), where S n (x) \u003d an 2 + (AK Coskx + BK Sin KX), k \u003d 1 to the function f (x) in the interval [, π] . 6.

7 Fig. 2. The graph of the function f (x) with the partial sums s (x) \u003d a 2 and s 1 (x) \u003d a 2 + a 1 cos x is superimposed on it. 3. Schedule F (X) function with a partial sum of S 99 (X) \u003d A 2 + A 1 COS X + + A 99 COS 99X 7

8 Substitting in (7) x \u003d We obtain: \u003d π 2 4 π k \u003d 1 (2k + 1) 2, from where we find the sum of the numerical series: \u003d π2 8. The knowing amount of this row, it is easy to find the following amount: S \u003d ( ) S \u003d () \u003d π, therefore s \u003d π2 6, that is, 1 n \u003d π sum of this famous row found Leonard Euler first. It is often found in mathematical analysis and its applications. Example 2. Draw a schedule, we find a Fourier series of the function of a given formula F (x) \u003d x for x< π, предполагая, что она имеет период 2π, и вычислим суммы числовых (1) n) рядов + n= ((2n + 1,) (k k + 1) Решение. График функции f(x) приведен на рис. 4. 8

9 Fig. 4. Schedule F (X) function F (x) function continuously differentiable in the interval (, π). At points x \u003d ± π, it has a finite limit (5): f () \u003d, f (π) \u003d π. In addition, there are end limits (6): f (+ h) f (+) lim \u003d 1 and h + hf (π h) f (π +) lim \u003d 1. H + H means f (x) piecewise smooth function. Since the function f (x) is odd, then a n \u003d. BN coefficients Find integration in parts: Bn \u003d 1 π f (x) sin πnxdx \u003d 1 [x cosnx π πn + 1 n \u003d 1 πn [(1) n π + (1) n π] \u003d 2 (1) n + one. n Make a formal Fourier series of functions 2 (1) N + 1 F (X) SIN NX. N 9 COSNXDX] \u003d

10 According to the theorem on the current convergence of a piecewise smooth 2π-periodic function, the Fourier series f (x) is converged to the amount: 2 (1) n + 1 sin nx \u003d n f (x) \u003d x, if π< x < π, = f(π) + f(π +) 2 =, если x = π, (8) f() + f(+) =, если x =. 2 На рис. 5 8 показан характер приближения частичных сумм S n (x) ряда Фурье к функции f(x). Рис. 5. График функции f(x) с наложенным на него графиком частичной суммы S 1 (x) = a 2 + a 1 cos x 1

11 Fig. 6. The graph of the function f (x) with the partial sum of S 2 (X) superimposed on it). 7. Schedule F (X) function with a partial sum of S 3 (x) 11 superimposed on it.

12 Fig. 8. The graph of the function f (x) with the partial sum of S 99 (x) superimposed on it) use the resulting Fourier series to find the sums of two numeric rows. Put in (8) x \u003d π / 2. Then 2 () + ... \u003d π 2, or \u003d n \u003d (1) n 2n + 1 \u003d π 4. We easily found the sum of the known row of the leibitus. Putting into (8) x \u003d π / 3, we will find () + ... \u003d π 2 3, or (1+ 1) () (k) 3π + ... \u003d 3K

13 Example 3. Draw a schedule, we find a Fourier series of the function f (x) \u003d SIN X, assuming that it has a period of 2π, and 1 calculate the sum of the numerical series 4N 2 1. The solution. The graph function F (X) is shown in Fig. 9. Obviously, f (x) \u003d sin x is a continuous even function with a period of π. But 2π is also a function of the function f (x). Fig. 9. Function Function F (X) Calculate Fourier coefficients. All b n \u003d because the function is even. Taking advantage of trigonometric formulas calculate AN at n 1: an \u003d 1 π \u003d 1 π sin x cosnxdx \u003d 2 π sin x cosnxdx \u003d (sin (1 + n) x sin (1 n) x) dx \u003d 1 () π cos ( 1 + n) x cos (1 n) x + \u003d 2 () 1 + (1) n \u003d π 1 + n 1 n π 1 n 2 (4 1, if n \u003d 2k, \u003d π n 2 1, if N \u003d 2K.

14 This calculation does not allow us to find the coefficient A 1, because with n \u003d 1, the denominator addresses to zero. Therefore, we calculate the coefficient A 1 directly: a 1 \u003d 1 π sin x cosxdx \u003d. Since F (x) is continuously differentiated by (,) and (, π) and at points Kπ, (K integer), there are end limits (5) and (6), then the Fourier range of the function converges to it at each point: \u003d 2 π 4 π sinx \u003d 2 π 4 π COS 2NX 4N 2 1 \u003d (1 1 1 COS 2X COS 4X + 1) COS 6X The nature of the approximation of the function f (x) by the partial sums of the Fourier range is shown. (9) Fig. 1. Schedule F (X) function with partial sum S (X) 14 superimposed on it

15 rice 11. The graph of the function f (x) with the partial sum s) with the partial sum of S 1 (x) fig. 12. The graph of the function f (x) with the partial sum of S 2 (X) superimposed on it). 13. Function Function F (X) with a partial sum of S 99 (x) 15 superimposed on it

16 1 Calculate the sum of the numerical series. For this, 4N 2 1 put in (9) x \u003d. Then Cosnx \u003d 1 for all n \u003d 1, 2, ... and therefore, 2 π 4 π 1 4n 2 1 \u003d. 1 4N 2 1 \u003d 1 \u003d 1 2. Example 4. We prove that if a piecewise smooth continuous function f (x) satisfies the condition F (x π) \u003d f (x) for all x (i.e. is π-periodic) , then a 2n 1 \u003d b 2n 1 \u003d for all N 1, and vice versa, if a 2n 1 \u003d b 2n 1 \u003d for all N 1, then f (x) π-periodic. Decision. Let the function f (x) be π-periodic. Calculate its Fourier coefficients A 2N 1 and B 2N 1: \u003d 1 π (A 2N 1 \u003d 1 π F (x) COS (2N 1) XDX + F (X) COS (2N 1) xDX \u003d) F (X) COS (2N 1) XDX. In the first integral, we will replace the variable x \u003d t π: f (x) cos (2n 1) xdx \u003d f (t π) cos (2n 1) (t + π) dt. sixteen

17 Taking this way that COS (2N 1) (T + π) \u003d COS (2N 1) T and F (T π) \u003d F (T), we obtain: a 2n 1 \u003d 1 π (F (X) COS (2N 1) X DX +) F (X) COS (2N 1) x DX \u003d. Similarly, it is proved that b 2n 1 \u003d. On the contrary, let a 2n 1 \u003d b 2n 1 \u003d. Since the function f (x) is continuous, then by the theorem on the representability of the function at the point in its next Fourier, we have then f (x π) \u003d \u003d f (x) \u003d (A 2N COS 2NX + B 2N SIN 2NX). (A2N COS 2N (x π) + b 2n sin 2n (x π)) \u003d (A2N COS 2NX + B 2N SIN 2NX) \u003d F (x), which means that F (x) is a π-periodic function. Example 5. We prove that if a piecewise smooth function f (x) satisfies the condition f (x) \u003d f (x) for all x, then a \u003d and a 2n \u003d b 2n \u003d for all N 1, and vice versa, if a \u003d a 2n \u003d b 2n \u003d, then f (x π) \u003d f (x) for all x. Decision. Let the function f (x) satisfies the condition F (x π) \u003d f (x). Calculate its Fourier coefficients: 17

18 \u003d 1 π (a n \u003d 1 π f (x) COS NXDX + F (X) Cosnxdx \u003d) f (x) Cosnxdx. In the first integral, we will replace the variable x \u003d t π. Then f (x) cosnxdx \u003d f (t π) Cosn (T π) DT. Taking this way that cos n (t π) \u003d (1) n cosnt and f (t π) \u003d f (t), we obtain: an \u003d 1 π ((1) n) f (t) Cosnt DT \u003d, if N even, \u003d 2 π f (t) COS NT DT, if N is odd. π Similarly, it is proved that b 2n \u003d. On the contrary, let a \u003d a 2n \u003d b 2n \u003d, for all n 1. Since the function f (x) is continuous, then by the theorem on representability, the equality F (x) \u003d (A 2N 1 COS (x) \u003d (A 2N 1 COS ( 2N 1) X + B 2N 1 SIN (2N 1) x). eighteen

19 Then \u003d f (x π) \u003d \u003d \u003d f (x). Example 6. We study how to continue the function f (x) integrated on the interval [, π / 2] to the interval of [, π] so that its Fourier series believes: A 2N 1 COS (2N 1) x. (1) Decision. Let the function graph be viewed in fig. 14. Since in a number (1) a \u003d a 2n \u003d b 2n \u003d for all n, then from Example 5 it follows that the function f (x) must satisfy the equality f (x π) \u003d f (x) for all x. This observation gives a way to continue the function f (x) to the interval [, / 2]: f (x) \u003d f (x + π), fig. 15. From the fact that the row (1) contains only cosines, we conclude that the continued function f (x) should be even (i.e. its schedule should be symmetrical with respect to the Oy axis), rice

20 Fig. 14. Function graph F (X) Fig. 15. Schedule to continue the function f (x) for the interval [, / 2] 2

21 So, the desired function is viewed in Fig. 16. Fig. 16. Schedule to continue the function f (x) for the interval [, π] summing up, we conclude that the function should be continued as follows: f (x) \u003d f (x), f (π x) \u003d f (x), that is, The interval [π / 2, π], the graph of the function f (x) is centrally symmetrical with respect to the point (π / 2), and on the interval [, π], its graph is symmetrical with respect to the OY axis. 21.

22 Generalization of examples 3 6 Let L\u003e. Consider two conditions: a) f (l x) \u003d f (x); b) f (l + x) \u003d f (x), x [, l / 2]. From a geometrical point of view, condition (a) means that the graph function f (x) is symmetrical with respect to the vertical direct x \u003d L / 2, and the condition (b) that the graph F (x) is centrally symmetrical with respect to the point (L / 2;) on the axis abscissa. Then the following statements are true: 1) if the function f (x) is even condition (a), then b 1 \u003d b 2 \u003d B 3 \u003d ... \u003d, a 1 \u003d a 3 \u003d a 5 \u003d ... \u003d; 2) if the function f (x) is even condition (b), then b 1 \u003d b 2 \u003d b 3 \u003d ... \u003d, a \u003d a 2 \u003d a 4 \u003d ... \u003d; 3) if the function f (x) is odd and condition (a) is satisfied, then a \u003d a 1 \u003d a 2 \u003d ... \u003d, b 2 \u003d b 4 \u003d b 6 \u003d ... \u003d; 4) If the function f (x) is odd and condition (b) is satisfied, then a \u003d a 1 \u003d a 2 \u003d ... \u003d, b 1 \u003d b 3 \u003d b 5 \u003d ... \u003d. Tasks in tasks 1 7 Draw graphs and find the Fourier series for functions (assuming that they have a period of 2π :, if< x <, 1. f(x) = 1, если < x < π. 1, если < x < /2, 2. f(x) =, если /2 < x < π/2, 1, если π/2 < x < π. 3. f(x) = x 2 (< x < π). 4. f(x) = x 3 (< x < π). { π/2 + x, если < x <, 5. f(x) = π/2 x, если < x < π. 22

23 (1, if / 2< x < π/2, 6. f(x) = 1, если π/2 < x < 3π/2. {, если < x <, 7. f(x) = sin x, если < x < π. 8. Как следует продолжить интегрируемую на промежутке [, π/2] функцию f(x) на промежуток [, π], чтобы ее ряд Фурье имел вид: b 2n 1 sin (2n 1)x? Ответы sin(2n 1)x sin(2n + 1)x. π 2n 1 π 2n + 1 n= 3. 1 (1) n () 12 3 π2 + 4 cosnx. 4. (1) n n 2 n 2π2 sin nx. 3 n 5. 4 cos(2n + 1)x π (2n + 1) (1) n cos(2n + 1)x. π 2n + 1 n= n= 7. 1 π sin x 2 cos 2nx. 8. Функцию следует продолжить следующим образом: f(x) = f(x), f(π x) = f(x), π 4n 2 1 то есть на промежутке [, π], график функции f(x) будет симметричен относительно вертикальной прямой x = π/2, на промежутке [, π] ее график центрально симметричен относительно точки (,). 23

24 2. Decomposition of the function specified in the interval [, π], only by sinus or only by cosine. Let the function f set in the interval [, π]. Wanting to decompose it in this gap in a row of Fourier, we will first continue the F into the interval [, π] randomly, and then use Euler Fourier formulas. The arbitrariness in the continuation of the function leads to the fact that for the same function f: [, π] r we can get different Fourier rows. But you can use this arbitrariness to get decomposition only in sines or only by cosine: in the first case, it is enough to continue f an odd way, and in the second even. The solution algorithm 1. Continue the function to an odd (even) way on (,), and then periodically with a period of 2π continue the function on the entire axis. 2. Calculate Fourier coefficients. 3. Make a Fourier series F (X). 4. Check the conditions for the convergence of the row. 5. Specify the function to which this series will converge. Example 7. Spread the function f (x) \u003d COSX,< x < π, в ряд Фурье только по синусам. Решение. Продолжим функцию нечетным образом на (,) (т. е. так, чтобы равенство f(x) = f(x) выполнялось для всех x (, π)), а затем периодически с периодом 2π на всю ось. Получим функцию f (x), график которой приведен на рис

25 Fig. 17. The schedule of the continued function is obvious that the function f (x) is a piecewise smooth. Calculate Fourier coefficients: a n \u003d for all n because the function f (x) is odd. If N 1, then bn \u003d 2 π f (x) sin πnxdx \u003d 2 π cosx sin nxdx \u003d 2 π dx \u003d 2 π cos (n + 1) x COS (n 1) x + \u003d π n + 1 n 1 \u003d 1 (1) n (1) N 1 1 \u003d π n + 1 n 1 \u003d 1, if n \u003d 2 k + 1, (1) n + 1 (n 1) + (n + 1) \u003d π ( N + 1) (N 1) 2 2N, if n \u003d 2k. π N 2 1 at n \u003d 1 In previous calculations, the denominator refers to zero, so the coefficient B 1 is calculated directly- 25

26: B 1 \u003d 2 π COSX SIN XDX \u003d. We will form a row of Fourier function f (x): f (x) 8 π k \u003d 1 k 4k 2 1 sin 2kx. Since the function f (x) is a piecewise smooth, then according to the detective convergence theorem, the Fourier range of the F (X) function converges to the amount: COSX, if π< x <, S(x) =, если x =, x = ±π, cosx, если < x < π. В результате функция f(x) = cosx, заданная на промежутке (, π), выражена через синусы: cosx = 8 π k=1 k 4k 2 1 sin 2kx, x (, π). Рис демонстрируют постепенное приближение частичных сумм S 1 (x), S 2 (x), S 3 (x) к разрывной функции f (x). 26

27 Fig. 18. The graph of the function f (x) with the partial sum of S 1 (x) is superimposed on it. 19. Schedule F (X) function with a partial sum of S 2 (X) 27 superimposed

28 Fig. 2. The graph of the function f (x) with the partial sum of S 3 (x) superimposed on it) in Fig. 21 shows the graphs of the function f (x) and its partial sum S 99 (x). Fig. 21. Schedule F (X) function with a partial sum of S 99 (x) 28 superimposed on it

29 Example 8. Defaulting the function f (x) \u003d e ax, a\u003e, x [, π], in a row of Fourier only by cosine. Decision. We continue the function in an even way on (,) (i.e., so that the equality f (x) \u003d f (x) was performed for all x (, π)), and then periodically with a period of 2π per numerical axis. We obtain the function f (x), the graph of which is presented in Fig. 22. Function F (x) at points Fig. 22. The schedule of the continued function f (x) x \u003d kπ, k is an integer, has a fox. Calculate Fourier coefficients: b n \u003d, since F (x) is even. Integrating in parts we get 29

30 An \u003d 2 π a \u003d 2 π \u003d 2 Cosnxd (E AX) \u003d 2 πA E AX DX \u003d 2 π A (EAπ 1), F (X) COS πnxdx \u003d 2 π πa EAX COSNX \u003d 2 πA (EAπ COSNπ 1 ) + 2n πa 2 π E AX COS NXDX \u003d + 2N E AX SIN NXDX \u003d πA SIN NXDE AX \u003d 2 πA (EAπ COS N π 1) + 2N π SIN NX π A 2EAX 2N2 E AX COS NXDX \u003d 2 π a 2 π A (EAπ COS N π 1) N2 AA N. 2 Therefore, a n \u003d 2a E Aπ COS N π 1. π A 2 + N 2 Since f (x) is continuous, then according to the current of the current convergence, its Fourier row converges to f (x). So, for all x [, π], we have f (x) \u003d 1 π a (Eaπ 1) + 2a π k \u003d 1 E Aπ (1) k 1 A 2 + k 2 coskx (x π). Rice demonstrate the gradual approximation of the partial sums of the Fourier series to a given bursting function. 3.

31 Fig. 23. Graphs of functions f (x) and s (x) Fig. 24. Graphs of functions f (x) and s 1 (x) Fig. 25. Graphs of functions f (x) and s 2 (x) Fig. 26. Graphs of functions f (x) and s 3 (x) 31

32 fig. 27. Function graphs F (x) and s 4 (x) Fig. 28. Charts of functions f (x) and s 99 (x) Tasks 9. Explore the function f (x) \u003d cos x, x π, in the Fourier row only by cosine. 1. Explore the function f (x) \u003d e ax, a\u003e, x π, in a Fourier series only in sinus. 11. Explore the function f (x) \u003d x 2, x π, in the Fourier series only in sinuses. 12. Spread the function f (x) \u003d sin ax, x π, in the Fourier series by only cosine. 13. Spread the function f (x) \u003d x sin x, x π, in a Fourier series only in sinus. Answers 9. COSX \u003d COSX. 1. E AX \u003d 2 [1 (1) K E Aπ] K SIN KX. π A 2 + k2 k \u003d 1 11. x 2 2 [π 2 (1) N 1 π n + 2] N 3 ((1) N 1) SIN NX. 32.

33 12. If a is not an integer, then Sin AX \u003d 1 Cosaπ (1 + 2A COS 2NX) + π A 2 (2N) 2 + 2A 1 + Cosaπ COS (2N 1) x π A 2 (2N 1) 2; If a \u003d 2m is an even number, then sin 2mx \u003d 8m cos (2n 1) x π (2m) 2 (2N 1) 2; If a \u003d 2m 1 is a positive odd number, then sin (2m 1) x \u003d 2 (COS 2NX) 1 + 2 (2M 1). π (2m 1) 2 (2N) π 16 N SIN X SIN 2NX. 2 π (4N 2 1) 2 3. Fourier series of functions with an arbitrary period Suppose that the function f (x) is set in the interval [L, L], L\u003e. By making the substitution X \u003d LY, Y π, we obtain the function G (y) \u003d f (Ly / π), determined in the gap π [, π]. This function G (y) corresponds to (formal) Fourier series () LY F \u003d G (Y) A π 2 + (An Cosny + BN SIN NY), the coefficients of which are located according to Euler Fourier formulas: An \u003d 1 π G (y) Cosny DY \u003d 1 π F (LY π) COS NY DY, N \u003d, 1, 2, ..., 33

34 BN \u003d 1 π G (y) sinny dy \u003d 1 π f () LY SIN NY DY, N \u003d 1, 2, .... π returning to the old variable, i.e. believing in the written formulas y \u003d πx / L, we get for the function f (x) trigonometric series several modified type: where f (x) a 2 + an \u003d 1 lbn \u003d 1 LLLLL (AN COS πNX LF (X) COS πNX LF (X) SIN πNX L + BN sin πnx), (11) l dx, n \u003d, 1, 2, ..., (12) dx, n \u003d 1, 2, .... (13) It is said that formulas (11) (13) are specified Decomposition in a Fourier series of functions with an arbitrary period. Example 9. We will find a Fourier series of the function specified in the interval (L, L) of the expression (A, if L< x, f(x) = B, если < x < l, считая, что она периодична с периодом 2l. Решение. Продолжим функцию периодически, с периодом 2l, на всю ось. Получим функцию f (x), кусочно-постоянную в промежутках (l + 2kl, l + 2kl), и претерпевающую разрывы первого рода в точках x = lk, k целое число. Ее коэффициенты Фурье вычисляются по формулам (12) и (13): 34

35 A \u003d 1 LLF (X) DX \u003d 1 L A DX + 1 LL B DX \u003d A + B, LLAN \u003d 1 LLLF (X) COS πNX L DX \u003d 1 L \u003d 1 LL A COS πNX L \u003d A + B π NLBN \u003d 1 L DX + 1 LL B COS πNX L SIN πN \u003d, if n, ll a sin πnx lf (x) sin πnx l dx + 1 ll dx \u003d b sin πnx l \u003d ba (1 cosπn). πN will make a row of Fourier function f (x): f (x) a + b π (b a since cosπn \u003d (1) n, then n dx \u003d dx \u003d (1 cosπn) sin πnx). l for n \u003d 2k we obtain b n \u003d b 2k \u003d, with n \u003d 2k 1 b n \u003d b 2k 1 \u003d 35 2 (b a) π (2k 1).

36 From here f (x) A + B (Ba) π (sin πx + 1 3πx sin + 1 5πx sin + ... L 3 L 5 L According to the theorem on the degenerate convergence, the Fourier series F (X) is converged to the amount A, If L.< x, S(x) = A + B, если x =, x = ±l, 2 B, если < x < l. Придавая параметрам l, A, B конкретные значения получим разложения в ряд Фурье различных функций. Пусть l = π, A =, B = 3π. На рис. 29 приведены графики первых пяти членов ряда, функции f (x) и частичной суммы S 7 (x) = a 2 + b 1 sin x b 7 sin 7x. Величина a является средним значением функции на промежутке. Обратим внимание на то, что с возрастанием ча- 2 стоты гармоники ее амплитуда уменьшается. Для наглядности графики трех высших гармоник сдвинуты по вертикали. На рис. 3 приведен график функции f(x) и частичной суммы S 99 (x) = a 2 + b 1 sin x b 99 sin 99x. Для наглядности на рис. 31 приведен тот же график в другом масштабе. Последние два графика иллюстрируют явление Гиббса. 36).

37 Fig. 29. The graph of the function f (x) with the charts superimposed by the harmonics s (x) \u003d a 2 and s 1 (x) \u003d b 1 sinx. For clarity graphics of three higher harmonics S 3 (x) \u003d B 3 SIN 3πX, S L 5 (x) \u003d B 5 SIN 5πX L and S 7 (X) \u003d B 7 SIN 7πX shifted up vertical Up L 37

38 Fig. 3. The graph of the function f (x) with the partial sum of S 99 (x) is superimposed on it. 31. Fragment Fig. 3 on another scale 38

39 Tasks in tasks to decompose the specified functions in the specified intervals. 14. f (x) \u003d x 1, (1, 1). 15. F (x) \u003d CH2x, (2, 2] f (x) \u003d x (1 x), (1, 1]. 17. f (x) \u003d cos π x, [1, 1] f (x ) \u003d sin π x, (1, 1). (2 1, if 1< x < 1, 19. f(x) = 2l = 4., если 1 < x < 3; x, если x 1, 2. f(x) = 1, если 1 < x < 2, 2l = 3. { 3 x, если 2 x < 3;, если ωx, 21. f(x) = 2l = 2π/ω. sin ωx, если ωx π; Разложить в ряды Фурье: а) только по косинусам; б) только по синусам указанные функции в заданных промежутках (, l) { 22. f(x) = { 23. f(x) = ax, если < x < l/2, a(l x), если l/2 < x < l. 1, если < x 1, 2 x, если 1 x 2. Ответы 14. f(x) = 4 cos(2n 1)πx. π 2 (2n 1) f(x) = sh sh4 (1) n nπx cos 16 + π 2 n f(x) = cos 2nπx. π 2 n f(x) = 2 π + 8 π (1) n n 1 4n 2 cosnπx. 39

40 18. F (x) \u003d 8 (1) N n sin nπx. π 1 4N (1) N 2N + 1 COS πX. π 2n πn 2πnx π 2 sin2 cos n π sin ωx 2 cos 2nωx π 4N 2 1. (L 22. a) f (x) \u003d Al 4 2) 1 (4N 2) πx cos, π 2 (2n 1) 2 L b) f (x) \u003d 4al (1) n 1 (2n 1) πx sin. π 2 (2n 1) 2 l 23. a) f (x) \u003d (cos π π 2 2 x 2 2 cos 2π 2 2 x cos 3π 2 2 x cos 5π), 2 2 x ... b) f ( x) \u003d 4 (sin π π 2 2 x 1 3 sin 3π) + 2 2 x (sin π π 2 x cos 2π) 2 x complex form of a series of Fourier decomposition f (x) \u003d CNE Inx, where cn \u003d 1 2π f (x) E Inx dx, n \u003d ± 1, ± 2, ... is called a complex form of a series of Fourier. The function decomposes into a complex Fourier series when performing the same conditions under which it decomposes into the Fourier real series. four

41 Example 1. Find a series of Fourier in the complex form of a function specified by the formula F (x) \u003d E AX, in the interval [, π), where A is a real number. Decision. Calculate the coefficients: \u003d C n \u003d 1 2π F (x) E Inx DX \u003d 1 2π E (A IN) X DX \u003d 1 ((1) N E Aπ (1) N E Aπ) \u003d (1) N sh Aπ. 2π (a in) π (A IN) The complex Fourier series of the function F has the form f (x) SH Aπ π n \u003d (1) N a in EINX. It is convinced that the function f (x) is a piecewise smooth: in the interval (, π) it is continuously differentiable, and at points x \u003d ± π exist a finite limits (5), (6) Lim H + Ea (+ H) \u003d E Aπ, LIM H + EA (π H) \u003d E Aπ, EA (+ H) EA (+) Lim H + H \u003d AE Aπ EA (π H) Ea (π), Lim H + H \u003d AE Aπ. Consequently, the function f (x) is representing the Fourier SH Aπ π n \u003d (1) n A in EINX, which converges to the amount: (E S (x) \u003d AX, if π< x < π, ch a, если x = ±π. 41

42 Example 11. We will find a Fourier series in the complex and real form of a function given by the formula F (X) \u003d 1 A 2 1 2A COSX + A2, where a< 1, a R. Решение. Функция f(x) является четной, поэтому для всех n b n =, а a n = 2 π f(x) cosnxdx = 2 (1 a2) π cos nxdx 1 2a cosx + a 2. Не будем вычислять такой сложный интеграл, а применим следующий прием: 1. используя формулы Эйлера sin x = eix e ix 2i = z z 1, cosx = eix + e ix 2i 2 = z + z 1, 2 где z = e ix, преобразуем f(x) к рациональной функции комплексной переменной z; 2. полученную рациональную функцию разложим на простейшие дроби; 3. разложим простейшую дробь по формуле геометрической прогрессии; 4. упростим полученную формулу. Итак, по формулам Эйлера получаем = f(x) = 1 a 2 1 a(z + z 1) + a 2 = (a 2 1)z (z a)(z a 1) = a z a az. (14) 42

43 Recall that the sum of the infinite geometric progression with the denominator Q (q< 1) вычисляется по формуле: + n= q n = 1 1 q. Эта формула верна как для вещественных, так и для комплексных чисел. Поскольку az = a < 1 и a/z = a < 1, то az = + a n z n = a n e inx, a z a = a z 1 1 a/z = a z n= + n= a n z = + n n= n= a n+1 z = + a n+1 e i(n+1)x. n+1 После замены переменной (n + 1) = k, < k < 1, получим: 1 a z a = a k e ikx. Следовательно, f(x) + n= k= c n e inx, где c n = n= { a n, если n, a n, если n <, то есть c n = a n. Поскольку функция f(x) непрерывна, то в силу теоремы о поточечной сходимости имеет место равенство: f(x) = + n= a n e inx. Тем самым мы разложили функцию f(x) в ряд Фурье в комплексной форме. 43

44 Now we find a Fourier series in real form. To do this, we grouped the terms with numbers N and N for n: a n e Inx + a n e Inx \u003d 2a neinx + E Inx because C \u003d 1, then 2 \u003d 2a N cos nx. f (x) \u003d 1 A 2 1 2a Cosx + a \u003d a n cosnx. 2 These are a Fourier series in the real form F (X) function. Thus, without having a single integral, we found a number of Fourier functions. At the same time, we calculated a difficult integral, depending on the parameter COS NXDX 1 2A COSX + A \u003d 2 π AN 2 1 A2, A< 1. (15) ПРИМЕР 12. Найдем ряд Фурье в комплексной и вещественной форме функции, заданной формулой a sin x f(x) = 1 2a cosx + a2, a < 1, a R. Решение. Функция f(x) является нечетной, поэтому для всех n a n = и b n = 2 π f(x) sin nxdx = 2a π sin x sin nxdx 1 2a cosx + a 2. Чтобы записать ряд Фурье нужно вычислить сложные интегралы или воспользоваться приемом, описанным выше. Поступим вторым способом: 44

45 A (zz 1) f (x) \u003d 2i (1 A (zz 1) + a 2) \u003d i 2 + i (a + a 1) z 2 2 (za) (za 1) \u003d \u003d i 2 + i () A 2 ZA + A 1. Za 1 Each of the simple fractions decompose according to the formula of geometric progression: + aza \u003d a 1 z 1 a \u003d aanzzn, n \u003d za 1 za \u003d az \u003d anz n. n \u003d this is possible because Az \u003d A / Z \u003d A< 1. Значит + ia n /2, если n <, f(x) c n e inx, где c n =, если n =, n= ia n /2, если n >, or, more short, C n \u003d 1 2i a n SGNN. Thus, Fourier series in a comprehensive form found. Grouped the terms with numbers n and n we obtain a series of Fourier functions in a real form: \u003d f (x) \u003d + a sin x 1 2a Cosx + A + 2 (1 2i An E Inx 1 2i An e Inx n \u003d +) \u003d CNE Inx \u003d An SIN NX. We again managed to calculate the following complex integral: SIN X SIN NXDX 1 2A COSX + A 2 \u003d π AN 1. (16) 45

46 Tasks 24. Using (15), calculate the COS NXDX 1 2A COSX + A 2 integral for real A, A,\u003e Using (16), calculate the SIN X SIN NXDX integral for real A, A\u003e A COSX + A2 in tasks, find the rows Fourier in a comprehensive form for functions. 26. F (x) \u003d SGN x, π< x < π. 27. f(x) = ln(1 2a cosx + a 2), a < 1. 1 a cosx 28. f(x) = 1 2a cosx + a2, a < Докажите, что функция f, определенная в промежутке [, π], вещественнозначна, если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является четной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n = ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является нечетной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2,.... Ответы 1 2π 24. a n a π a n i + e 2inx, где подразумевается, что слагаемое, соответствующее n =, пропущено. π n n= a n n cosnx. 28. a n cosnx. n= 46

47 5. The equality of Lyapunov Theorem (Lyapunov Equality). Let the function f: [, π] r it is such that f 2 (x) dx< +, и пусть a n, b n ее коэффициенты Фурье. Тогда справедливо равенство, a (a 2 n + b2 n) = 1 π называемое равенством Ляпунова. f 2 (x) dx, ПРИМЕР 13. Напишем равенство Ляпунова для функции { 1, если x < a, f(x) =, если a < x < π и найдем с его помощью суммы числовых рядов + sin 2 na n 2 и + Решение. Очевидно, 1 (2n 1) 2. 1 π f 2 (x) dx = 1 π a a dx = 2a π. Так как f(x) четная функция, то для всех n имеем b n =, a = 2 π f(x) dx = 2 π a dx = 2a π, 47

48 a n \u003d 2 π f (x) cosnxdx \u003d 2 π A COS NXDX \u003d 2 SIN NA πN. Therefore, the equality of Lyapunov for the function f (x) takes the form: 2 A 2 π + 4 SIN 2 Na \u003d 2a 2 π 2 N 2 π. From the last equality for a π we find SIN 2 Na N 2 \u003d A (π a) 2, assuming A \u003d π 2, we obtain SIN2 Na \u003d 1 at n \u003d 2k 1 and Sin 2 Na \u003d at n \u003d 2k. Consequently, k \u003d 1 1 (2k 1) 2 \u003d \u003d π2 8. Example 14. We will write the lapunov equality for the function f (x) \u003d x cosx, x [, π], and we will find the sum of the numerical row (4N 2 + 1) 2 (4N 2 1) 4. 1 π solution. Direct calculations give \u003d π π f 2 (x) dx \u003d 1 π x 2 cos 2 xdx \u003d 1 π x sin 2xdx \u003d π π x cos x \u003d π x 21 + cos 2x dx \u003d 2 π 1 4π cos 2xdx \u003d

49 Since f (x) is an even function, then for all n we have bn \u003d, an \u003d 2 π \u003d 1 π 1 \u003d π (n + 1) \u003d f (x) cosnxdx \u003d 2 π 1 cos (n + 1) x π (n + 1) 2 x cosxcosnxdx \u003d x (cos (n + 1) x + cos (n 1) x) dx \u003d 1 π sin (n + 1) xDX sin (n 1) xDX \u003d π (N 1) π π 1 + COS (N 1) x \u003d π (n 1) 2 1 (\u003d (1) (n + 1) 1) 1 (+ (1) (n + 1) 1) \u003d π (n + 1) 2 π (N 1) 2 () \u003d (1) (n + 1) 1 1 π (n + 1) + 1 \u003d 2 (n 1) 2 \u003d 2 (1) (n + 1) 1 nk π (n 2 1) \u003d π (4K 2 1) 2, if n \u003d 2k, 2, if n \u003d 2k + 1. The coefficient A 1 must be calculated separately, since in the general formula for n \u003d 1, the denomoter is drawn to zero. \u003d 1 π a 1 \u003d 2 π f (x) cosxdx \u003d 2 π x (1 + cos 2x) dx \u003d π 2 1 2π 49 x cos 2 xdx \u003d sin 2xdx \u003d π 2.

50 Thus, Lyapunov equality for the function f (x) has the form: 8 π + π (4N 2 + 1) 2 π 2 (4N 2 1) \u003d π, from where we find the sum of the numerical row (4N 2 + 1) 2 (4N 2 1) \u003d π π tasks 32. Write a Lyapunov equality for a function (xf (x) \u003d 2 πx, if x< π, x 2 πx, если π < x. 33. Напишите равенства Ляпунова для функций f(x) = cos ax и g(x) = sin ax, x [, π]. 34. Используя результат предыдущей задачи и предполагая, что a не является целым числом, выведите следующие классические разложения функций πctgaπ и (π/ sin aπ) 2 по рациональным функциям: πctgaπ = 1 a + + 2a a 2 n 2, (π) = sin aπ (a n) 2. n= 35. Выведите комплексную форму обобщенного равенства Ляпунова. 36. Покажите, что комплексная форма равенства Ляпунова справедлива не только для вещественнозначных функций, но и для комплекснозначных функций. 5

51 π (2n + 1) \u003d π sin 2απ 2απ \u003d 2sin2 απ α 2 π 2 Answers + 4 sin2 απ π 2 α 2 (α 2 N 2) 2; SIN 2απ 1 2απ \u003d απ n 2 4sin2 π 2 (α 2 n 2) 2. 1 π 35. f (x) g (x) dx \u003d CNDN, where Cn Fourier coefficient 2π functions f (x), and DN Fourier coefficient functions G (x). 6. Differentiation of Fourier series Let F: R R continuously differentiable 2π-periodic function. Her Fourier series has the form: F (x) \u003d a 2 + (a n cos nx + b n sin nx). The derivative F (x) of this function will be a continuous and 2π-periodic function for which the Fourier formal range can be written: F (x) a 2 + (AN COS NX + BN SIN NX), where A, AN, BN, N \u003d 1 , 2, ... Fourier coefficients F (X). 51.

52 Theorem (about the killed differentiation of Fourier series). With the assumptions made above, equality a \u003d, a n \u003d nb n, b n \u003d na n, n 1. Example 15. Let a piecewise smooth function f (x) continuously in the interval [, π]. We prove that when performing the condition F (x) DX \u003d, there is an inequality 2 DX 2 Dx, called glass inequality, and make sure that the equality in it is carried out only for functions of the form F (x) \u003d a COSX. In other words, the inequality of Steklov gives conditions, when performing the smallness of the derivative (in the rms), follows the smallness of the function (in the mean-of-mean-square). Decision. We will continue the function f (x) for the interval [,] evenly. Denote the continued function of the same symbol F (X). Then the continued function will be continuous and piecewise smooth on the segment [, π]. Since the function f (x) is continuous, then f 2 (x) is continuous on the segment and 2 dx< +, следовательно, можно применить теорему Ляпунова, согласно которой имеет место равенство 1 π 2 dx = a () a 2 n + b 2 n. 52

53 Since the continued function is even, then b n \u003d, a \u003d under the condition. Consequently, Lyapunov's equality takes the form 1 π 2 dx \u003d a 2 π n. (17) It is convinced that for F (X) the conclusion of the theorem on the unimaginal differentiation of the Fourier series is concluded, that is, that a \u003d, an \u003d nb n, bn \u003d na n, n 1. Let the derivative F (X) undergo the breaks at the points x 1, x 2, ..., x n in the interval [, π]. Denote x \u003d, x n + 1 \u003d π. We divide the interval of integration [, π] on n +1 gap (x, x 1), ..., (x n, x n + 1), on each of which f (x) is continuously differentiable. Then, using the property of the additivity of the integral, and then integrating in parts, we obtain: bn \u003d 1 π \u003d 1 π \u003d 1 π f (x) sin nxdx \u003d 1 π n f (x) sin nx j \u003d n f (x) sin nx j \u003d x j + 1 xjx j + 1 xjnn π n j \u003d x j + 1 xjx j + 1 xjf (x) sin nxdx \u003d f (x) cosnxdx \u003d f (x) cosnxdx \u003d \u003d 1 π [(f (x 1) SIN NX 1 F (X) SIN NX) + + (F (X 2) SINNX 2 F (X 1) SIN NX 1)

54 + (f (x n + 1) sin nx n + 1 f (x n) sin nx n)] na n \u003d 1 π na n \u003d \u003d 1 π na n \u003d na n. x j + 1 a \u003d 1 f (x) dx \u003d 1 n f (x) dx \u003d π π j \u003d xj \u003d 1 n x j + 1 f (x) π \u003d 1 (f (π) f ()) \u003d . X j π J \u003d The last equality takes place due to the fact that the function f (x) was continued evenly, and therefore f (π) \u003d f (). Similarly, we obtain a n \u003d nb n. We have shown that the theorem on the renovation differentiation of Fourier series for a continuous piecewise smooth 2π-periodic function, the derivative of which in the interval [, π] undergoes the first kind breaks, is correct. So f (x) a 2 + (An cosnx + bn sin nx) \u003d (na n) sin nx, as a \u003d, an \u003d nb n \u003d, bn \u003d na n, n \u003d 1, 2, .... Since 2 dx.< +, то по равенству Ляпунова 1 π 2 dx = 54 n 2 a 2 n. (18)

55 Since each member of the row in (18) is greater than or equal to the corresponding member of the row in (17), then 2 DX 2 DX. Recalling that f (x) is an even continuation of the original function, we have 2 DX 2 DX. What proves the equality of Steklov. Now we investigate for what functions in the inequality of Steklov, there is equality. If at least for one n 2, the coefficient a n is different from zero, then a 2 n< na 2 n. Следовательно, равенство a 2 n = n 2 a 2 n возможно только если a n = для n 2. При этом a 1 = A может быть произвольным. Значит в неравенстве Стеклова равенство достигается только на функциях вида f(x) = A cosx. Отметим, что условие πa = f(x)dx = (19) существенно для выполнения неравенства Стеклова, ведь если условие (19) нарушено, то неравенство примет вид: a a 2 n n 2 a 2 n, а это не может быть верно при произвольном a. 55

56 Tasks 37. Let a piecewise smooth function f (x) continuously in the interval [, π]. Prove that when performing the condition F () \u003d F (π) \u003d, there is a 2 DX 2 DX inequality, also called glassware inequality, and make sure that the equality in it takes place only for functions of the form F (x) \u003d b SIN x. 38. Let the function f are continuous in the interval [, π] and has in it (with the exception of the final number of points) the derivative F (x) integrated with the square. Prove that if the conditions f () \u003d f (π) and f (x) dx \u003d are satisfied, then there is an inequality 2 DX 2 DX, called the inequality of Virginger, and the equality in it takes place only for the functions of the form F (x ) \u003d A COSX + B SIN x. 56.

57 7. The use of Fourier series to solve differential equations in private derivatives in the study of a real object (the phenomenon of nature, production process, management systems, etc.) two factors are essential: the level of accumulated knowledge about the object under study and the degree of development of the mathematical apparatus. At the present stage of scientific research, the next chain was developed: the physical model of the mathematical model. The physical formulation (model) of the task is as follows: the conditions for the development of the process and the main factors affecting it are detected. The mathematical formulation (model) is to describe the factors selected in the physical formulation in the form of a system of equations (algebraic, differential, integral, etc.). The task is called correctly set if in a specific functional space the solution of the problem exists, the only and continuous depends on the initial and boundary conditions. The mathematical model is not identical to the object under consideration, but is its approximate description of the output of the equation of loose small transverse oscillations of the string will follow the textbook. Let the ends of the strings are fixed, and the string itself is tight. If you bring the string from the equilibrium position (for example, to delay or hit it), the string will start 57

58 fluctuate. We assume that all the points of the string move perpendicular to its equilibrium position (transverse oscillations), and at each time the string lies in the same plane. Take in this plane the XOU rectangular coordinate system. Then, if at the initial moment of time T \u003d the string was located along the OX axis, then u would mean the deflection of the string from the equilibrium position, that is, the position of the point of the string with the abscissa x into an arbitrary moment T corresponds to the function u (x, t). With each fixed value of T, the graph function U (x, t) represents the shape of the oscillating string at time t (Fig. 32). With a constant value of X, the function u (x, t) gives the law of motion of the point with an abscissa x along the straight, parallel OU axis, the derivative of U T the speed of this movement, and the second derivative 2 U t 2 acceleration. Fig. 32. The forces applied to an infinitely small stretch of strings will be the equation to which the function u (x, t) must satisfy. To do this, we will make some more simplifying assumptions. We will consider the string absolutely power - 58

59 koy, that is, we will assume that the string does not resist the bending; This means that the stresses arising in the string are always aimed at a tangent of its instant profile. The string is assumed to be an elastic and obese law of the throat; This means that the change in the magnitude of the tension force is in proportion to the change in the length of the string. We will assume that the string is homogeneous; This means that its linear density ρ is constant. We neglect external forces. This means that we consider free oscillations. We will study only small string fluctuations. If you designate Φ (x, t) an angle between the abscissa axis and tangent to the string at the abscissa point x at the time T, the condition of the oscillations is due to the value of φ 2 (x, t) can be neglected compared to φ (x, t), i.e. φ 2. Since the angle φ is small, then cosφ 1, φ sin φ tg φ u is therefore, the value (UXX,) 2 can also be neglected. From here it immediately follows that in the process of oscillations we can neglect the change in the length of any string. Indeed, the length of the string M 1 m 2, which is designed to the abscissa axis gap, where x 2 \u003d x 1 + x is equal to L \u003d x 2 x () 2 u dx x. x We show that with our assumptions the value of the tension force T will be constant along the entire string. Take for this any section of the string M 1 m 2 (Fig. 32) at time t and replace the action of the discarded part - 59

60 pieces of tension forces T 1 and T 2. Since under the condition all the string points are moving parallel to the OU axes and the external forces are missing, the sum of the projections of the tension forces on the OX axis should be zero: T 1 cosφ (x 1, t) + T 2 cosφ (x 2, t) \u003d. From here, due to the smallness of the angles φ 1 \u003d φ (x 1, t) and φ 2 \u003d φ (x 2, t), we conclude that T 1 \u003d T 2. Denote the total value of T 1 \u003d T 2 through T. Now we calculate the amount of projections F u of the same forces on the OU axis: f U \u003d t sin φ (x 2, t) t sin φ (x 1, t). (2) Since for small angles sin φ (x, t) Tg φ (x, t), and Tg φ (x, t) u (x, t) / x, then equation (2) can be rewritten so f u T (Tg φ (x 2, T) TG φ (x 1, T)) (U t x (x 2, t) u) x (x 1, t) xx t 2 ux 2 (x 1, t) x . Since the point x 1 is selected arbitrarily, then F U T 2 U x2 (x, t) x. After all the forces acting on the M 1 m 2 section are found, we will apply to him the second Newton law, according to which the product of the mass to accelerate is equal to the sum of all current forces. The mass of the string m 1 m 2 is m \u003d ρ l ρ x, and the acceleration is 2 U (x, t). The Newton equation T 2 takes the form: 2 U t (x, t) x \u003d u 2 α2 2 x2 (x, t) x, where α 2 \u003d t ρ is a constant positive number. 6.

61 Cutting to X, we obtain 2 U t (x, t) \u003d u 2 α2 2 x2 (x, t). (21) As a result, we obtained a linear homogeneous differential equation with second-order private derivatives with constant coefficients. It is called the string oscillation equation or one-dimensional wave equation. Equation (21) is essentially reformulating Newton's law and describes the string movement. But in the physical formulation of the task, the requirements were present that the end of the strings are fixed and the position of the string at some time is known. These conditions will be recorded by these conditions: a) we will assume that the ends of the strings are fixed at points x \u003d and x \u003d l, i.e. we assume that for all T, the ratios u (, t) \u003d, u (l, t ) \u003d; (22) B) We will assume that at the time T \u003d the position of the string coincides with the graph function F (x), i.e., we assume that the equality u (x,) \u003d f (x,) \u003d f ( x); (23) B) We will assume that at the time T \u003d point of the string with the abscissa x is given the speed G (x), i.e. we will assume that u (x,) \u003d g (x). (24) T ratios (22) are called boundary conditions, and relations (23) and (24) are called initial conditions. Mathematical model of loose small transverse 61

62 The fluctuations of the string is to solve equation (21) with boundary conditions (22) and initial conditions (23) and (24), the solution of the equation of free small transverse oscillations of the string by the Fourier method of solving equation equation (21) in the field x L,< t <, удовлетворяющие граничным условиям (22) и начальным условиям (23) и (24), будем искать методом Фурье (называемым также методом разделения переменных). Метод Фурье состоит в том, что частные решения ищутся в виде произведения двух функций, одна из которых зависит только от x, а другая только от t. То есть мы ищем решения уравнения (21), которые имеют специальный вид: u(x, t) = X(x)T(t), (25) где X дважды непрерывно дифференцируемая функция от x на [, l], а T дважды непрерывно дифференцируемая функция от t, t >. Substituting (25) in (21), we obtain: x t \u003d α 2 x t, (26) or t (t) α 2 t (t) \u003d x (x) x (x). (27) They say that the variables separation occurred. Since x and t do not depend on each other, the left side in (27) does not depend on x, and the right from T and the total value of these relations 62

63 should be permanent, which is denoted by λ: T (T) α 2 T (T) \u003d x (x) x (x) \u003d λ. From here we get two ordinary differential equations: X (x) λx (x) \u003d, (28) t (t) α 2 λt (t) \u003d. (29) In this case, the boundary conditions (22) will take the form x () t (t) \u003d and x (l) t (t) \u003d. Since they must be performed for all T, T\u003e, then x () \u003d x (L) \u003d. (3) We find solutions of equation (28) that satisfies the boundary conditions (3). Consider three cases. Case 1: λ\u003e. Denote λ \u003d β 2. Equation (28) takes the form x (x) β 2 x (x) \u003d. Its characteristic equation k 2 β 2 \u003d has the roots k \u003d ± β. Hence, common decision Equations (28) has the form x (x) \u003d c e βx + de βx. We must choose constant C and D so that the boundary conditions are observed (3), i.e. x () \u003d c + d \u003d, x (L) \u003d c e βl + de βl \u003d. Since β, this system of equations has a single solution C \u003d D \u003d. Consequently, x (x) and 63

64 U (x, t). Thus, in the case of 1, we received a trivial decision, which we will not further consider. Case 2: λ \u003d. Then equation (28) takes the form x (x) \u003d and its solution is obviously defined by the formula: x (x) \u003d c x + d. Substituting this solution in the boundary conditions (3), we obtain x () \u003d d \u003d and x (L) \u003d CL \u003d, it means C \u003d D \u003d. Consequently, x (x) and u (x, t), and we again received a trivial solution. Case 3: λ<. Обозначим λ = β 2. Уравнение (28) принимает вид: X (x)+β 2 X(x) =. Его характеристическое уравнение имеет вид k 2 + β 2 =, а k = ±βi являются его корнями. Следовательно, общее решение уравнения (28) в этом случае имеет вид X(x) = C sin βx + D cosβx. В силу граничных условий (3) имеем X() = D =, X(l) = C sin βl =. Поскольку мы ищем нетривиальные решения (т. е. такие, когда C и D не равны нулю одновременно), то из последнего равенства находим sin βl =, т. е. βl = nπ, n = ±1, ±2,..., n не равно нулю, так как сейчас мы рассматриваем случай 3, в котором β. Итак, если β = nπ (nπ) 2, l, т. е. λ = то существуют l решения X n (x) = C n sin πnx, (31) l C n произвольные постоянные, уравнения (28), не равные тождественно нулю. 64

65 In the future, we will give N only positive values \u200b\u200bof n \u003d 1, 2, ..., since with negative N, the solutions of that (species. Nπ) of λ n \u003d are called their own numbers, and the functions x n (x) \u003d c N SIN πNX with its own functions of a differential equation (28) with boundary conditions (3). Now let's solve equation (29). For it, the characteristic equation has the form K 2 α 2 λ \u003d. (32) L 2 Since we found out that the non-trivial solutions x (x) of equation (28) are available only for negative λ, equal to λ \u003d n2 π 2, then we will consider such λ. The roots of equation (32) are k \u003d ± iα λ, and solutions of equation (29) have the form: t n (t) \u003d a n sin πnαt + b n cos πnαt, (33) L l where a n and b n arbitrary constants. Substituting formulas (31) and (33) in (25), we will find private solutions of equation (21) that satisfy the edge conditions (22): (un (x, t) \u003d b n cos πnαt + a n sin πnαt) c n sin πnx. LLL introducing a multiplier C n into the bracket and introducing the designation C n a n \u003d bn and b n c n \u003d an, write un (x, t) in the form (un (x, t) \u003d an cos πnαt + bn sin πnαt) sin πnx. (34) L L L 65

66 string fluctuations corresponding to solutions u n (x, t) are called their own string fluctuations. Since equation (21) and boundary conditions (22) are linear and homogeneous, then a linear combination of solutions (34) (U (x, t) \u003d An cos πnαt + bn sin πnαt) SIN πNX (35) LLL will be solved by equation (21 ) satisfying boundary conditions (22) with a special selection of the coefficients AN and BN, which ensures the uniform convergence of the row. Now we will now select the coefficients AN and BN solutions (35) so that it satisfies not only boundary, but also initial conditions (23) and (24), where f (x), G (x) specified functions (and f () \u003d f (L) \u003d G () \u003d G (L) \u003d). We believe that the functions f (x) and g (x) satisfy the decomposition conditions in the Fourier series. Substituting in (35) the value of T \u003d, we obtain u (x,) \u003d a n sin πnx l \u003d f (x). Differentiating the series (35) by t and substituting T \u003d, we obtain u t (x,) \u003d πnα b n sin πnx l L \u003d G (x), and this is the decomposition of functions F (x) and G (x) in the Fourier series. Therefore, a n \u003d 2 L l f (x) sin πnx l dx, b n \u003d 2 l g (x) sin πnx dx. πnα L (36) 66

67 Substituting examinationsThe cells A N and B N in a row (35), we obtain the solution of equation (21) that satisfies the boundary conditions (22) and initial conditions (23) and (24). Thus, we solved the problem of free small transverse vibrations of the string. Find out the physical meaning of the eigenfunctions u n (x, t) the tasks of the free fluctuations of the strings defined by formula (34). We rewrite it in the form where u n (x, t) \u003d α n cos πnα l α n \u003d a 2 n + b2 n, (t + δ n) sin πnx, (37) l πnα δ n \u003d arctg b n. L A N of formula (37) shows that all the points of the string make harmonic oscillations with the same frequency ω n \u003d πnα and the πnα Δ n phase. The amplitude of oscillation depends on the L L of the abscissa X of the point of the string and is equal to α n sin πnx. With such a oscillation, all the points of the string simultaneously reach their l maximum deviation in one direction or another and at the same time pass the equilibrium position. Such oscillations are called standing waves. Standing wave will have n + 1 still point defined by the roots of the sin πnx equation \u003d in the interval [, L]. Fixed dots are called the standing wave nodes. In the middle between nodes, there are points in which the deviations reach the maximum; Such points are called Puffs. Each string can have its own oscillations of strictly defined frequencies ω n \u003d πnα, n \u003d 1, 2, .... These frequencies are called their own string frequencies. The lowest L tone that can produce strings is determined by 67

68 Low own frequency ω 1 \u003d π T and is called the main tone of the string. The remaining tones corresponding to the L ρ frequencies ω n, n \u003d 2, 3, ... are called obramstones or harmonics. For clarity, we will show typical string profiles, emitting the basic tone (Fig. 33), the first oberton (Fig. 34) and the second oberton (Fig. 35). Fig. 33. String profile, publishing the basic tone. 34. String profile publishing the first Operton rice. 35. String profile published by the second obton if the string performs free oscillations determined by the initial conditions, the function u (x, t) seems to be seen from formula (35), as a sum of individual harmonics. Thus, an arbitrary fluctuation 68

69 Strings is a superposition of standing waves. At the same time, the nature of the string (tone, the power of sound, the timbre) will depend on the ratio between the amplitudes of individual harmonics force, the height and the voice of the sound of the oscillating string excites the fluctuations of the air, perceived by the human ear as the sound, published by the string. The power of sound is characterized by energy or amplitude of oscillations: the greater the energy, the greater the power of the sound. The height of the sound is determined by its frequency or period of oscillations: the greater the frequency, the higher the sound. Sound timbre is determined by the presence of overtones, the distribution of energy by harmonics, i.e., the method of excitation of oscillations. The amplitudes of overtones, generally speaking, less amplitude of the main tone, and the OPERTON Phase can be arbitrary. Our ear is not sensitive to the oscillation phase. Compare, for example, two curves in Fig. 36 borrowed from. This is a recording of sound with the same main tone extracted from clarinet (a) and piano (b). Both sounds do not constitute simple sinusoidal oscillations. The main frequency of the sound in both cases is the same and creates the same tone. But the drawings of curves are different because different obhrothon is applied to the main tone. In a sense, these drawings show what the timbre is. 69.

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should:

1) draw schedule f (x) on the gap at least a length of two periods to show that this function is periodic;

2) Draw a schedule S (x) Similarly, to be seen in what points f (x) ¹s (x);

3) Calculate Fourier coefficients and write a row of Fourier.

Tasks

№1. Furiate

Decision. notice, that f (x) set on the length of the length T \u003d 4.. Because f (x) It is assumed to be periodic, then this number is its period, then - l \u003d 2.

1) Graph f (x):

2) graph S (x):

The arrows at the ends of the lines show that the function does not accept the values \u200b\u200bdetermined from the expression specified on the interval. When comparing graphs f (x) and S (x) It is clearly visible that at the break points f (x) ¹s (x).

3) Calculate Fourier coefficients. This can be done by formulas (3 *) :; ; . Exactly: ; so,

Decomposition f (x) Fourier has the form:

Comments. 1) when integrating on [-1;3] This segment was broken on and because On these segments f (x) Set different values.

2) when calculating the coefficients used integrals: and, where a \u003d const.

№2 . Furiate

Decision. Here T \u003d 2., l \u003d 1..

Fourier series has the form: where; ; because l \u003d 1..

1) Graph f (x):

2) graph S (x):

№3. Decompose Fourier on sines

Decision. Note that only odd functions are launched in the Fourier sinuses. Because f (x) Defined only for x\u003e 0, xî (0; 2) è (2; 3)then this means that on a symmetrical interval (-3; -2) è (-2; 0) F (x)need to continue so that equality f (-x) \u003d -f (x). Therefore, the length of the gap on which f (x) It is set as an odd function, equal to 6. So T \u003d 6, L \u003d 3. Fourier series for f (x) It has the form:, where, n \u003d 1, 2, 3, (according to formulas (5 ")).

1) Graph f (x).

To draw graph f (x) as an odd function, draw first the schedule on (0; 2) è (2; 3), and then take advantage of the fact that the schedule of the odd function is symmetrical on the start of the coordinates. From these considerations we get a schedule f (x) on the (-3; -2) è (-2; 0). Then continue f (x) T \u003d 6..

2) graph S (x).

Schedule S (x) Differs from schedule f (x) at break points f (x). For example, in t. x \u003d 2 f (x)not defined, and S (x) has at x \u003d 2. Meaning equal to half-sided function limits f (x), exactly: where,.

So, then decomposition f (x) Fourier has the form :.

№4 . Dispatch Fourier on cosine.

Decision. Note that only even functions are launched in a row of Fourier on cosine. Because f (x) Set only for x\u003e 0, xî (0; 2) è (2; 3], then this means that on a symmetrical interval [-3; -2) è (-2; 0) F (x)it is necessary to continue so that equality is performed: f (-x) \u003d f (x). Therefore, the length of the gap on which f (x) set as an even function is equal to 6, then T \u003d 6, L \u003d 3. Fourier series in this case looks:

where; ; n \u003d 1,2, ... (according to formulas (4 ")).

1) Graph f (x).

To draw graph f (x) as even functions, draw a schedule first f (x) on the (0; 2) è (2; 3]And then take advantage of the fact that a graph of an even function is symmetrical about the axis of the ordinate. From these considerations we get a schedule f (x) on the [-3; -2) è (-2; 0). Then continue f (x) on the whole numerical direct as a periodic function with a period T \u003d 6..

Here is a schedule f (x) Drawn on two full periods of function.

2) graph S (x).

Schedule S (x) Differs from schedule f (x) at break points f (x). For example, in t. x \u003d 0 f (x)not defined, and S (x) has the meaning: , so schedule S (x) Not interrupted in t. x \u003d 0., unlike schedule f (x).

Decomposition f (x) In a row of Fourier on cosineses, it has the form :.

№5. Furiate f (x) \u003d | x |, xî (-2; 2)..

Decision. By condition, f (x) is an even function on (-2;2) ; those. her Fourier series contains only cosines, while T \u003d 4, L \u003d 2, ,

where; ; n \u003d 1, 2,

1) Graph f (x):

2) graph S (x):

3), because | X | \u003d X.for x\u003e 0.; .

Then decay f (x) Fourier has the form :. Note that when integrating expressions or applies the integration formula in parts: where u \u003d x; DV \u003d COS (AX) DXor DV \u003d SIN (AX) DX.

№6. Decompose the function in Fourier series: a) in the interval (-?,?); b) in the interval (0, 2?); c) in the interval (0 ,?) in a row of sinuses.

Decision. a) Function graph with 2? - periodic continuation is

The function satisfies the conditions of the Dirichlet theorem and therefore it can be decomposed into a series of Fourier.

Calculate Fourier coefficients. Since the function is even, then bn \u003d 0 (n \u003d 0, 1, 2, ...) and (n \u003d 0, 1, 2, ...).

To calculate this integral, the integration formula in parts in a specific integral is used. Receive

Fourier series of this function has the form. Due to the sign of Dirichle, this series represents the function x2 in the interval (-?,?).

b) the interval (0, 2?) is not symmetrical on the start of the coordinates, and its length 2 l. \u003d 2?. Calculate Fourier coefficients by formulas:

Therefore, Fourier has a form. By virtue of the Dirichlet Theorem, a series converges to generating functions at points x? (0.2?), And at points 0 and 2? to meaning. The sum of the amount of the row has the form

c) The function decomposed into a row of sinus should be odd. Therefore, we will provide the specified function x2 in (-π, π) an odd way, i.e. We consider the function. For this function f (x) we have a \u003d 0 (n \u003d 0, 1, 2, ...) and

The desired decomposition has the form.

The sum of the amount of the row has the form

Note that at points x \u003d (-π, π), the Fourier row converges to zero.

№7 Lay a function specified graphically in Fourier:

Decision . We obtain an explicit expression for F (X). The graph of the function is a straight line, we use the equation direct in the form. As can be seen from the drawing,, i.e. f (x) \u003d x - 1 (-1< x < 1) и период Т = 2.

This feature satisfies the terms of the Dirichlet, so it decomposes into the Fourier series. Calculate Fourier coefficients ( l. = 1):

; (n \u003d 1, 2, ...);

Fourier series for function f (x) has the view

It represents the function f (x) at -1< x < 1, а в точках х0 = -1 и х0 = 1 ряд сходится к -1.

№8. Dispatch the function in the trigonometric row of Fourier on the segment and specify the function to which the resulting series converges.

Decision.Draw a schedule of a function by continuing it periodically with a period or on the entire axis. The continued function has a period.

Check the conditions of sufficient signs of the convergence of the Fourier series (Dini-Lipschitsa, Jordan, Dirichlet).

Function piecewise monotonna on the segment: it increases on and on. At points, the function has the first kind of gaps.

Find out the parity or oddness of the function: the function is neither even or odd.

a) if the function is specified on

b) if the function is set on

Create a row Fourier function :.

Specify the function to which this series will converge, using the current signs of convergence: according to the sign of Dirichle, a Fourier series of functions converges to the amount:

№9. Eliminate the function in a series of Fourier on sines on and with the help of this decomposition to find the sum of the numerical series.

Decision.Continue the function is even (odd) way on (- p.0) or (- l.0), and then periodically with a period of 2 p. or 2 l. Continue the function on the entire axis.

We continue the function to an odd way on, and then periodically, with a period, will continue it to the entire axis.

Draw a graph of a periodic continuation. We will get the function of the form:

Check the conditions of sufficient signs of the convergence of the Fourier series (Dini-Lipica, Jordan, Dirichlet).

Function piecewise constant in the interval: it is equal to -1 on and 1 on. At points, the function has the first kind of gaps.

Calculate Fourier coefficients:

Its Fourier coefficients are calculated by formulas:

Make a row Fourier function. .

Specify the function to which this series will converge, using the detective signs of convergence.

According to the sign of Dirichle, a number of Fourier function converges to the amount:

Consequently, for

Substituting the values, specify the sum of the specified numeric series.

Believing in the resulting decomposition, we will find

where, because.

№10. Write the equality of parceivers for a function, and, based on this equality, find the amount of the numerical series.

Decision.Install whether this feature is a function with an integrable square.

The function is continuous, and, therefore, integrated on. For the same reason, its square integrate on.

Calculate Fourier coefficients by formulas:

Since an odd function, then its Fourier coefficients are calculated by formulas:

Calculate the integral.

Write a formula Parseval:

Thus, the formula Parseval has the form

Arranged, if required, arithmetic actions in the right and left parts, to obtain the amount of this numeric series.

Sharing both parts of the equality obtained by 144, we find :.

№11. Find Fourier Integrated Functions

and build it schedule.

Decision.Build a graph of a function.

Check the fulfillment of the conditions of sufficient signs of the convergence of the Fourier integral (Dini, Dirichlet-Jordan or the consequences of them).

The function is absolutely integrated in the gap, continuous when and, and at the point there is a gap of the first kind. Further, with and the function has a finite derivative, and the final right and left derivatives exist in zero. Find out the parity or oddness of the function. The function is neither even nor odd. ; .

So, or,

Near Fourier The functions f (x) on the interval (-π; π) are called a trigonometric series of the form:
where

Near the Fourier function f (x) on the interval (-l; L) is called a trigonometric series of the form:
where

Purpose. Online calculator Designed to decompose the function f (x) in a row of Fourier.

For module functions (for example, | x |), use decomposition by cosine.

Rules for entering functions:

For module functions, use the cosine decomposition. For example, for | x | It is necessary to enter a function without a module, i.e. x.

Fourier series piecewise continuous, piecewise monotonous and limited on the interval (- l.;l.) Functions converges on the entire numeric axis.

The sum of the Fourier series S (X):

• it is a periodic function with a period of 2 l.. The function u (x) is called periodic with a period t (or t-periodic), if for all x region R, U (X + T) \u003d U (x).
• on the interval (- l.;l.) coincides with the function f.(x.), except for breaking points
• at break points (first kind, because function is limited) functions f.(x.) And at the ends of the interval takes averages:

.
It is said that the function is laid out in a Fourier series on the interval (- l.;l.): .

If a f.(x.) - even function, then only even functions are involved in its decomposition, that is, b N.=0.
If a f.(x.) - an odd function, then only odd functions are involved in its decomposition, that is, a N.=0

Near Fourier Functions f.(x.) on the interval (0; l.) on cosine dougses called row:
where
.
Near Fourier Functions f.(x.) on the interval (0; l.) on sines of multiple arc called row:
where .
The sum of the Fourier series by cosine of multiple arcs is an even periodic function with a period of 2 l.Coinciding S. f.(x.) on the interval (0; l.) At continuity points.
The sum of the Fourier series in sines of multiple arcs is an odd periodic function with a period of 2 l.Coinciding S. f.(x.) on the interval (0; l.) At continuity points.
Fourier series for this function at this interval has the property of uniqueness, that is, if the decomposition is obtained in any other way than the use of formulas, for example, using the selection of coefficients, then these coefficients coincide with the formulas calculated by the formulas.

Example number 1. Lay the function f (x) \u003d 1:
a) in the full range of Fourier on the interval(-π ;π);
b) in a row on sines of multiple arcs on the interval(0;π); build a graph of the resulting Fourier series
Decision:
a) decomposition in a Fourier series on the interval (-π; π) has the form:
,
and all coefficients b N.\u003d 0, because This feature is even; in this way,

Obviously, equality will be performed if you take
but 0 =2, but 1 =but 2 =but 3 =…=0
By virtue of the properties of uniqueness, this is the desired coefficients. Thus, the desired decomposition: or just 1 \u003d 1.
In this case, when a series identically coincides with its function, the schedule of a series of Fourier coincides with the schedule of the function on the entire numerical line.
b) decomposition on the interval (0; π) on sines of multiple arcs has the form:
It is impossible to choose the coefficients so that equality is identically impossible. We use the formula for calculating the coefficients:


Thus, for even n. (n.=2k.) Have b N.\u003d 0, for odd ( n.=2k.-1) -
Finally, .
We construct a graph of the resulting Fourier series by using its properties (see above).
First of all, we build a graph of this function at a specified interval. Next, using the accuracy of the sum of the number, we continue the schedule symmetrically the beginning of the coordinates:

We continue in a periodic manner on the entire numeric axis:


And finally, at the gap points fill the average (between the right and left limit) of the value:

Example number 2. Dismiss the function On the interval (0; 6) on sines of multiple arcs.
Decision: The desired decomposition has the form:

Since the left, and the right part of equality contain only functions sin. From various arguments, check whether they match any values \u200b\u200bof N (natural!) The arguments of sinuses in the left and right parts of equality:
Or, from where n \u003d 18. It means that such a term is contained on the right side and the coefficient with it must coincide with the coefficient in the left side: b. 18 =1;
Or, from where n \u003d 4. It means b. 4 =-5.
Thus, with the help of the selection of coefficients, it was possible to obtain the desired decomposition.