Decomposition in a comprehensive Fourier Theory. Fourier Rows in Examples and Tasks

07.09.2020

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 Fouriers.

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,

Fourier series is a representation of an arbitrarily taken function with a specific period in the form of a series. In general, this decision is called the decomposition of the element on the orthogonal basis. The decomposition of functions in a Fourier series is a rather powerful tools when solving a variety of tasks due to the properties of this transformation during integration, differentiation, as well as the shift of the expression on the argument and convolution.

A person who is not familiar with the highest mathematics, as well as with the works of French scientist Fourier, most likely will not understand what kind of "rows" and what they need. Meanwhile, this transformation has quite tightly entered our life. They enjoy not only mathematics, but also physicists, chemists, doctors, astronomers, seismologists, oceanographers and many others. Let us even get acquainted with the works of the great French scientist who has made the discovery that beat their time.

Fourier Man and Transformation

Fourier series are one of the methods (along with the analysis and other) this process occurs every time a person hears any sound. Our ear in automatic mode produces the conversion of elementary particles in an elastic medium is laid out into the rows (by the spectrum) of the sequential values \u200b\u200bof the volume level for the tones of different height. Next, the brain turns this data into the sounds familiar to us. All this happens, in addition to our desire or consciousness, in itself, but in order to understand these processes, it will take several years to study the highest mathematics.

Historical reference

The Father-Founder of this theory is French mathematician Jean Batist Joseph Fourier. With his name later, this transformation was named. Initially, the scientist applied his method for studying and explaining thermal conductivity mechanisms - heat distribution in solid bodies. Fourier suggested that the initial irregular distribution can be decomposed on the simplest sinusoids, each of which will have its own temperature minimum and maximum, as well as its phase. At the same time, each such component will be measured from a minimum to the maximum and back. The mathematical function that describes the upper and lower peaks of the curve, as well as the phase of each of the harmonics, was called the Fourier transformation from the expression of temperature distribution. The author of the theory reduced the overall distribution function, which is difficult to give in mathematical description, It is very convenient in circulation of a row of cosine and sinus, in the sum of the initial distribution.

Principle of conversion and views of contemporaries

Contemporaries of a scientist - leading mathematicians of the beginning of the nineteenth century - did not accept this theory. The main objection was the approval of Fourier that the discontinuous function describing the direct line or the burst curve can be represented as the sum of sinusoidal expressions that are continuous. As an example, you can consider the "step" of Heviside: its value is zero to the left of the break and the unit on the right. This function describes the dependence of the electric current from the time variable when the chain is closed. Contemporaries of theory at that time were never encountered with a similar situation, when the discontinuous expression would be described by a combination of continuous, conventional functions, such as exhibitor, sinusoid, linear or quadratic.

What embarrassed French mathematicians in Fourier's theory?

After all, if the mathematician was in their rights in his statements, then, by summing the endless trigonometric row of Fourier, it is possible to obtain an accurate representation of stepwise expression, even if it has many similar steps. At the beginning of the nineteenth century, such an approval seemed absurd. But despite all doubts, many mathematicians expanded the scope of study of this phenomenon, to bring it outside the research of thermal conductivity. However, most scientists continued to suffer a question: "Can the sum of the sinusoidal series converge accurate value discontinuous function? "

Fourier series convergence: Example

The question of convergence rises whenever the need to summarize the infinite series of numbers. To understand this phenomenon, consider a classic example. Can you ever reach the wall if each subsequent step is twice as smaller than the previous one? Suppose you are two meters from the target, the very first step brings to the mark on the half of the path, the next one to the mark of three quarters, and after the fifth you will overcome almost 97 percent of the path. However, no matter how much you have taken steps, you do not achieve the goal in a strict mathematical sense. Using numerical calculations, it can be proved that in the end it is possible to approach an arbitrarily small predetermined distance. This proof is equivalent to the demonstration of the fact that the total value of one second, one fourth, etc. will strive for unity.

The issue of convergence: the second coming, or the device of Lord Kelvin

Repeated this question rose at the end of the nineteenth century, when Fourier's ranks tried to apply to predict the intensity of tides and tides. At this time, Lord Kelvin invented the device that is an analog computing device that allowed the sailors of the military and the merchant fleet to track this natural phenomenon. This mechanism was determined by phase sets and amplitudes on the height table of the height of the tides and the corresponding time moments, carefully measured in this harbor during the year. Each parameter was a sinusoidal component of the expression of the height of the tide and was one of the regular components. The measurement results were introduced into the Lord Kelvin computing device, a synthesizing curve, which predicted the height of water as a temporary function for the next year. Very soon, such curves were drawn up for all the harbors of the world.

And if the process is broken by the discontinuous function?

At that time it was obvious that the device predicting a tidal wave, with a large number of account elements can calculate a large number of phases and amplitudes and so provide more accurate predictions. Nevertheless, it turned out that this pattern is not respected in cases where the tidal expression that should be synthesized, contained a sharp jump, that is, it was discontinuous. In the event that data from the time-mind table is entered into the device, it produces the calculation of several Fourier coefficients. The initial function is restored due to the sinusoidal components (in accordance with the coefficients found). The discrepancy between the original and reconstructed expression can be measured at any point. When performing repeated calculations and comparisons, it can be seen that the value of the greatest error does not decrease. However, they are localized in the area corresponding to the break point, and at any point they tend to zero. In 1899, this result was theoretically confirmed by Joshua Willard Gibbs from Yale University.

Convergence of Fourier series and the development of mathematics in general

Fourier Analysis is not applicable to expressions containing an infinite number of bursts at a certain interval. In general, the Fourier series, if the original function is represented by the result of the real physical dimension, always converge. Issues of the convergence of this process for specific classes of functions led to the emergence of new sections in mathematics, such as the theory of generalized functions. It is connected with such names as L. Schwartz, J. Mikusinsky and J. Temple. As part of this theory, a clear and accurate theoretical Foundation For such expressions, as a Dirac delta function (it describes the area of \u200b\u200ba single area, concentrated in an infinitely small neighborhood of the point) and the "Step" of Heviside. Thanks to this work, the Fourier series has become applicable to solving equations and tasks in which intuitive concepts appear: point charge, point weight, magnetic dipoles, as well as a concentrated load on the beam.

Fourier method

Fourier series, in accordance with the principles of interference, begin with decomposition complex shapes on simpler. For example, the change in the heat flux is explained by its passage through various obstacles from the heat insulating material of the irregular shape or a change in the surface of the earth - an earthquake, a change in the orbit of the celestial body - the effect of the planets. As a rule, such equations describing simple classical systems are elementary solved for each individual wave. Fourier showed that simple solutions can also be summarized to obtain solving more complex tasks. I am expressed by the language of mathematics, the Fourier series is a methodology for the expression of the sum of the harmonic - a cosineid and sinusoid. Therefore, this analysis is also known as "Harmonic Analysis".

Fourier series - the ideal method to the "computer era"

Before creating computer equipment, Fourier technique was the best weapon in the arsenal of scientists when working with the wave nature of our world. The Fourier series in a comprehensive form allows you to solve not only simple tasks that are amenable to directly apply the laws of Newton mechanics, but also fundamental equations. Most of the discoveries of Newtonian science of the nineteenth century have become only possible thanks to the Fourier technique.

Fourier series today

With the development of Fourier transform computers rose to high quality new level. This technique firmly entrenched in almost all spheres of science and technology. A digital audio and video signal can be brought as an example. Its implementation has become possible only thanks to the theory developed by the French mathematician at the beginning of the nineteenth century. So, a number of Fourier in a comprehensive form made it possible to take a breakthrough in the study outer space. In addition, it affected the study of semiconductor and plasma physics, microwave acoustics, oceanography, radar, seismology.

Trigonometric Fourier row

In mathematics, Fourier series is a way of representing arbitrary complex functions of the amount of simpler. In general cases, the number of such expressions may be infinite. At the same time, the more their number is taken into account when calculating, the more precisely the final result is obtained. Most often as the simplest use trigonometric functions of cosine or sinus. In this case, the Fourier series is called trigonometric, and the solution of such expressions is the decomposition of harmonics. This method plays an important role in mathematics. First of all, the trigonometric row gives means for the image, as well as studying functions, it is the main apparatus of the theory. In addition, it allows you to solve a number of tasks of mathematical physics. Finally, this theory contributed to the development caused a number of very important sections of mathematical science (the theory of integrals, the theory of periodic functions). In addition, served as a starting point for the development of the following functions of the actual variable, and also marked the beginning of harmonic analysis.

Function defined at all values x. called periodic, If there is such a number T (t ≠ 0)that in any meaning x.equality is performed f (x + t) \u003d f (x). Number T.in this case, is a function period.

Properties of periodic functions:

1) sum, difference, work and private periodic periodic functions T.there is a periodic function of the period T.

2) if the function f (x)has a period T., then function f (AX)has a period

In fact, for any argument h.:

(multiplication of the argument for a number means compression or stretching the graph of this function along the axis OH)

For example, a function has a period, a function period is

3) if f (x) Periodic function period T.are equal to any two integrals from this function, taken by the length of the length T. (It is assumed that these integrals exist).

Fourier series for a function with a period T \u003d .

Trigonometric number is called a series of types:

or, in short,

Where ,,,,,,,,,, ... - real numbers called the coefficients of a number.

Each term trigonometric series is a periodic function function (because - has any

the period, and the period () is equal, and therefore). Each category (), when n \u003d1,2,3 ... is an analytical expression of simple harmonic oscillation, where A.- amplitude,

The initial phase. Given the said, we obtain: if the trigonometric row converges on the length of the period length, then it converges on the entire numeric axis and its sum is the periodic function function.

Let the trigonometric series evenly converge on the segment (consequently, on any segment) and its sum is equal. To determine the coefficients of this series, we will use the following equalities:

As well as take advantage of the following properties.

1) As is known, the sum is evenly converging on some segment of a number composed of continuous functions, itself is a continuous function on this segment. Given this, we obtain that the sum is evenly converging on the segment of the trigonometric series - a continuous function on the entire numerical axis.

2) the uniform convergence of the row on the segment will not break if all members of the row multiply the function continuous on this segment.

In particular, uniform convergence on the segment of this trigonometric series will not break if all members of the series multiply or on.

By condition

As a result of the killed integration of a uniformly converging series (4.2) and considering the above equality (4.1) (orthogonality trigonometric functions), we get:

Consequently, the coefficient

Multiplying equality (4.2) on, integrating this equality ranging from before and, given the above expressions (4.1), we obtain:

Consequently, the coefficient

Similarly, multiplying equality (4.2) on and integrating it ranging from before, with regard to equalities (4.1) we have:

Consequently, the coefficient

Thus, the following expressions were obtained for the coefficients of the Fourier series:

Sufficient signs of decomposability of functions in a row of Fourier.Recall that the point x.o Gap function f (x) refer to the point of breaking the first kind if there are end limits Right and left functions f (x) in the neighborhood of the point.

The limit on the right

Limit on the left.

Theorem (Dirichlet).If the function f (x)it has a period and on the segment continuous or has a finite number of first-order rupture points and, in addition, the segment can be divided into a finite number of segments so that inside each of them f (x)monotonna, then Fourier series for function f (x)converges at all values x.. And at the continuity points of the function f (x) Its amount is equal f (x), and at break points f (x) Its amount is equal, i.e. The average arithmetic limit values \u200b\u200bon the left and right. In addition, Fourier series for function f (x)converges evenly on any segment, which, together with its ends, owns the continuity interval of the function f (x).

Example: decompose a function in Fourier

Satisfying condition.

Decision.Function f (x)satisfies the conditions of decomposability in a row of Fourier, so you can write:

In accordance with formulas (4.3), it is possible to obtain the following values \u200b\u200bof the Fourier series coefficients:

When calculating the coefficients of the Fourier series, the formula "integration in parts" was used.

And, therefore,

Fourier series for read and odd functions with a period T \u003d.

Use the following property of the integral by symmetrical relatively x \u003d 0.gap:

If a f (x) - odd function,

if a f (x) - Something function.

Note that the product of two-way or two odd functions is aware of the function, and the product of an integral function on an odd function is an odd function. Let now f (x)- Forged periodic function with a period , satisfying the conditions of decomposability in a row of Fourier. Then, using the above property of the integrals, we get:

Thus, a series of Fourier for an intended function contains only even functions - cosines and recorded as follows:

and coefficients Bn \u003d 0.

Arguing similarly, we get that if f (X) - The odd periodic function that satisfies the setup conditions in the Fourier series, then, therefore, the Fourier series for the function is odd contains only odd functions - sines and is written as follows:

wherein an \u003d 0for n \u003d 0, 1, ...

Example: expand the Fourier periodic function

Since the specified odd function f (x) satisfies the conditions of decomposability in a row of Fourier, then

or that the same

And Fourier series for this feature f (x)you can question this:

Fourier series for functions of any period t \u003d 2 l..

Let be f (x) - Periodic function of any period T \u003d 2L(l-semiphered), piecewise smooth or piecewise monotonous on a segment [ -L, L.]. Believed x \u003d at,we get a function f (AT) Argument t, The period of which is equal . Pick up butso that the period of the function f (AT)it was equal, i.e. T \u003d 2L

Decision.Function f (x) - odd, satisfying the conditions of decomposability in a row of Fourier, therefore, on the basis of formulas (4.12) and (4.13), we have:

(When calculating the integral, the formula "integration in parts") was used.

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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Fourier rows - A way to represent a complex function of the sum of simpler, well known.
Sine and cosine are periodic functions. They also form an orthogonal basis. This property can be explained by analogy with the axes. X X. X. and Y y. Y. on the coordinate plane. Just as we can describe the point coordinates relative to the axes, we can describe any function relative to siquies and cosine. Trigonometric functions are well studied and easy to apply them in mathematics.

Present sinuses and cosines can be in the form of such waves:

Blue is cosine, red - sines. More such waves are called harmonics. Cosinees - even, sinuses are odd. The term harmonic came still from antiquity and is associated with observations of the relationship of the heights of sounds in music.

What is a Fourier series

Such a number where the functions of the sinus and cosine are used as the simplest, called trigonometric. He is named after his inventor Jean Batista Joseph Fourier, at the end of the XVIII early XIX century. Proven that any function can be represented as a combination of such harmonics. And the more they take them, the more accurate the representation will be. For an example, the picture below: You can see that with a large number of harmonics, i.e., members of the Fourier series, the red schedule is becoming closer to the blue - the source function.

Practical application in the modern world

Do you need these rows now? Where can they apply practically and do someone use the theoretical mathematicians? It turns out that Fourier is because it is famous for the whole world that the practical benefits of his rows literally innumerable. It is convenient to apply where there are any oscillations or waves: acoustics, astronomy, radio engineering, etc. The easiest example of its use: the mechanism of operation of the camera or camcorder. If you are briefly explained, these devices are not just written pictures, but Fourier series coefficients. And it works everywhere - when viewing pictures on the Internet, movie or listening to music. It is thanks to the ranks of Fourier, you can now read this article from your mobile phone. Without the Fourier conversion, we would not have enough through no bandwidth of Internet connections to just watch the video on YouTube even in standard quality.

On this scheme, the two-dimensional Fourier transform, which is used to decompose the image on the harmonic, i.e., the basis components. On this scheme, the black is coded -1, white 1. Right and down the graph increases the frequency.

Fourier decomposition

Probably you have already been tired of reading, so we turn to the formulas.
For such a mathematical reception, as the decomposition of functions in a Fourier series, you will have to take integrals. Many integrals. In general, the Fourier series is recorded as an endless amount:

F (x) \u003d a + Σ n \u003d 1 ∞ (An cos \u2061 (nx) + bn sin \u2061 (nx)) f (x) \u003d a + \\ displaystyle \\ Sum_ (n \u003d 1) ^ (\\ infty) (A_N \\ COS (NX) + b_n \\ sin (NX))f (x) \u003dA +.n \u003d 1.∑ ∞ (a. n. cOS (N x) +b. n. sin (n x))
Where
A \u003d 1 2 π ∫ - π π f (x) d x a \u003d \\ frac (1) (2 \\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) f (x) dxA \u003d.2 π.1 − π ∫ π f (x) d x
An \u003d 1 π ∫ - π π f (x) cos \u2061 (nx) dx a_n \u003d \\ frac (1) (\\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) f (x) \\ f (x) cos (n x) d xa. n. = π 1 − π ∫ π bn \u003d 1 π ∫ - π π f (x) sin \u2061 (nx) dx b_n \u003d \\ frac (1) (\\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) f (x) \\ f (x) sin (n x) d x
If we somehow be able to calculate the infinite amountb. n. = π 1 − π ∫ π a N A_N.

B N b_n. (they are called Fourier decomposition coefficients a. n. and A A. b. n. A. - It is simply constant of this decomposition), then the resulting series as a result will be 100% coincided with the source function f (x) f (x) F (X) On the cut from - π - \\ pi before π \\ pi. − π . Such a segment is due to the properties of the integration of sine and cosine. The bigger N N. π For which we will calculate the decomposition coefficients of the function in a row, the more accurate will be this decomposition. Example n.Take a simple feature

y \u003d 5 x y \u003d 5x

y \u003d. 5 X. a \u003d 1 2 π ∫ - π π f (x) dx \u003d 1 2 π ∫ - π π 5 xdx \u003d 0 a \u003d \\ frac (1) (2 \\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) f (x) dx \u003d \\ FRAC (1) (2 \\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) 5xdx \u003d 0f (x) d x \u003d
5 x D x \u003dA \u003d.2 π.1 − π ∫ π a 1 \u003d 1 π ∫ - π π f (x) cos \u2061 (x) dx \u003d 1 π ∫ - π π 5 x cos \u2061 (x) dx \u003d 0 a_1 \u003d \\ frac (1) (\\ pi) \\ displaystyle \\ \\ COS (X) DX \u003d 02 π.1 − π ∫ π f (x) cos (x) d x \u003d0
5 x COS (x) D x \u003da. 1 = π 1 − π ∫ π b 1 \u003d 1 π ∫ - π π f (x) sin \u2061 (x) dx \u003d 1 π ∫ - π π 5 x sin \u2061 (x) dx \u003d 10 b_1 \u003d \\ frac (1) (\\ pi) \\ displaystyle \\ \\ sin (x) dx \u003d 10π 1 − π ∫ π f (x) sin (x) d x \u003d0
5 x sin (x) d x \u003db. 1 = π 1 − π ∫ π a 2 \u003d 1 π ∫ - π π f (x) cos \u2061 (2 x) dx \u003d 1 π ∫ - π π 5 x cos \u2061 (2 x) dx \u003d 0 a_2 \u003d \\ FRAC (1) (\\ pi) \\ ) 5x \\ COS (2X) DX \u003d 0π 1 − π ∫ π f (x) cos (2 x) d x \u003d1 0
5 x COS (2 x) D x \u003da. 2 = π 1 − π ∫ π {!LANG-1ceac906cb4e21979a29bafaac122428!}π 1 − π ∫ π {!LANG-32a03ec80536f343933b7534a9524914!}0
B 2 \u003d 1 π ∫ - π π f (x) sin \u2061 (2 x) dx \u003d 1 π ∫ - π π 5 x sin \u2061 (2 x) dx \u003d - 5 b_2 \u003d \\ FRAC (1) (\\ PI) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ pi) f (x) \\ sin (2x) dx \u003d \\ frac (1) (\\ pi) \\ displaystyle \\ int \\ limits _ (- \\ pi) ^ (\\ F.b. 2 = π 1 − π ∫ π sin.(x.) d.(2 x.) etc. In the case of such a function, we can immediately say that allx.= π 1 − π ∫ π 5 x.d.(2 x.) etc. In the case of such a function, we can immediately say that allx.= − 5

a n \u003d 0 a_n \u003d 0 $a._{, coefficients = 0 You have to calculate. If we take the first four member of the decomposition in a Fourier series for a function b._{, coefficients y. we will get: = 5 x. 5 x \approx 10 \cdot sin \u2061 (x) - 5 \cdot sin \u2061 (2 \cdot x) + 10 3 \cdot sin \u2061 (3 \cdot x) - 5 2 \cdot sin \u2061 (4 \cdot x) 5x \\ Approx 10 \\ Cdot \\ Sin (x) - 5 \\ Cdot \\ sin (2 \\ Cdot X) + \\ FRAC (10) (3) \\ CDOT \\ SIN (3 \\ CDOT X) - \\ FRAC (5) (2) \\ CDOT \\ SIN (4 \\ The graph of the resulting function will look like this:}}$

$5 x. \approx 10 \cdot d. (x.) - 5 \cdot d. (2 \cdot x.) + 3 1 0 \cdot d. ( 3 \cdot x. ) - 2 5 \cdot d. ( 4 \cdot x. )$

The more decomposition members in a row, the higher the accuracy.

If we change the scale of the graph slightly, we can notice another transformation feature: Fourier series is a periodic function with a period

2 π 2 \\ pi
Thus, you can represent any function that is continuous on the segment $\\ pi; \\ pi]$

. All this is necessary in order to facilitate the analysis of some phenomena, which are described by complex functions. It is not always possible analytically (i.e. according to the formula) to calculate the derivative, and in the case of a set of sinuses and cosine, such a problem will not arise. The actual decomposition in the Fourier series shows that often the tasks can be solved analytically on simplified models, one of the examples of which is a number of Fouriers. $[- π; π] Save on Facebook.$

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