teacher

03.07.202327.05.2017

Let the real function satisfy the Dirichlet conditions on the interval - L, L. We write its expansion in the trigonometric Fourier series:

If in (10.1) we express and through the exponential function of the imaginary argument:

then we get the series

where due to (10.2)

The last three formulas can be combined:

The series (10.3) with coefficients (10.4) is called the trigonometric Fourier series in complex form.

Example 1 Expand the function, where is a complex number, into a Fourier series on the interval.

Solution . Find the Fourier coefficients:

Since then

The required decomposition will have the form

where it is taken into account that

Applying to the series (10.5) Parseval's equality

you can find the sum of another number series. Indeed, in our case

Then from (10.6) it follows

Exercise 1. Prove that

indication. Put in (10.5) X= 0 and X = .

Exercise 2. Prove that for

Fourier integral

Convergence of the Fourier integral

Let the function be defined on the entire real axis. Assuming that on an arbitrary finite interval - L, L the given function satisfies the Dirichlet conditions, we represent it as a trigonometric Fourier series in complex form:

Frequency k-th harmonic; .

Entering expressions (11.2) into (11.1), we obtain

When the value. The right side of formula (11.3) is similar to the integral sum for a function with respect to a variable in the interval. Therefore, we can expect that after passing in (11.3) to the limit at , instead of the series, we obtain the integral

Formula (11.4) is called the Fourier integral formula, and its right side is called the Fourier integral.

The reasoning by which formula (11.4) is obtained is not rigorous and has only a suggestive character. The conditions under which the Fourier integral formula is valid are established by the theorem, which we accept without proof.

Theorem. Let the function, first, be absolutely integrable on the interval, i.e. the integral converges, and, secondly, satisfies the Dirichlet conditions on each finite interval (- L, L). Then the Fourier integral converges (in the sense of principal value) everywhere to, i.e., equality (11.4) holds for all X from the interval. Here, as before, it is assumed that at the point of discontinuity the value of the function is equal to half the sum of its one-sided limits at this point.

Fourier transform

We transform the Fourier integral formula (11.4) as follows. Let's put

If a function is continuous and absolutely integrable on the entire axis, then the function is continuous on the interval. Indeed, since then

and since the integral on the right converges, the integral on the left converges. hence the integral in (12.1) converges absolutely. Equality (12.2) holds simultaneously for all, so integral (12.1) converges uniformly with respect to. Hence it follows that the function is continuous (just as the uniform convergence of a series composed of continuous functions implies the continuity of its sum).

From (11.4) we get

The complex function defined by formula (12.1) is called the Fourier transform or the Fourier transform of the function. In turn, the formula (12.3) defines as the inverse Fourier transform, or the inverse image of the function. Equality (12.3) for a given function can be considered as an integral equation with respect to the function, the solution of which is given by formula (12.1). And, conversely, the solution of the integral equation (12.1) with respect to the function for a given one is given by formula (12.3).

In formula (12.3), the expression specifies, conditionally speaking, a package of complex harmonics with frequencies continuously distributed over the gap and a total complex amplitude. The function is called the spectral density. Formula (12.2), written as

can be interpreted as the expansion of a function into the sum of packets of harmonics, the frequencies of which form a continuous spectrum distributed over the interval.

Parseval equalities. Let and be the Fourier images of real functions and, respectively. Then

those. inner products and norms of functions are invariants of the Fourier transform. Let's prove this statement. by definition of the scalar product, we have. Replacing the function with its expression (12.3) in terms of the Fourier transform, we obtain

Due to (12.1)

Therefore, i.e. formula (12.4) is proved. Formula (12.5) is obtained from (12.4) at.

Cosine and sine Fourier transforms. If a real function is even, then its Fourier transform, which we denote here, is also a real even function. Really,

The last integral, due to the oddness of the integrand, vanishes. Thus,

Here property (7.1) of even functions is used.

It follows from (12.6) that the function is real and evenly depends on, since it enters (12.6) only through the cosine.

Formula (12.3) of the inverse Fourier transform in this case gives

Since and are respectively even and odd functions of a variable, then

Formulas (12.6) and (12.7) define the Fourier cosine transform.

Similarly, if a real function is odd, then its Fourier transform, where is a real odd function of. Wherein

Equalities (12.8), (12.9) define the sine Fourier transform.

Note that formulas (12.6) and (12.8) include the values of the function only for. Therefore, the cosine and sine Fourier transforms can also be applied to a function defined on a semi-infinite interval. In this case, for , the integrals in formulas (12.7) and (12.9) converge to the given function, and for , to its even and odd extensions, respectively.

Which are already pretty fed up. And I feel that the moment has come when it is time to extract new canned food from the strategic reserves of theory. Is it possible to expand the function into a series in some other way? For example, to express a straight line segment in terms of sines and cosines? It seems incredible, but such seemingly distant functions lend themselves to
"reunion". In addition to the familiar degrees in theory and practice, there are other approaches to expanding a function into a series.

In this lesson, we will get acquainted with the trigonometric Fourier series, touch on the issue of its convergence and sum, and, of course, we will analyze numerous examples for expanding functions into a Fourier series. I sincerely wanted to call the article “Fourier Series for Dummies”, but this would be cunning, since solving problems will require knowledge of other sections of mathematical analysis and some practical experience. Therefore, the preamble will resemble the training of astronauts =)

First, the study of the page materials should be approached in excellent shape. Sleepy, rested and sober. Without strong emotions about the broken paw of a hamster and obsessive thoughts about the hardships of the life of aquarium fish. The Fourier series is not difficult from the point of view of understanding, however, practical tasks simply require an increased concentration of attention - ideally, one should completely abandon external stimuli. The situation is aggravated by the fact that there is no easy way to check the solution and the answer. Thus, if your health is below average, then it is better to do something simpler. Is it true.

Secondly, before flying into space, it is necessary to study the instrument panel of the spacecraft. Let's start with the values of the functions that should be clicked on the machine:

For any natural value:

1) . And in fact, the sinusoid "flashes" the x-axis through each "pi":
. In the case of negative values of the argument, the result, of course, will be the same: .

2). But not everyone knew this. The cosine "pi en" is the equivalent of a "flashing light":

A negative argument does not change the case: .

Perhaps enough.

And thirdly, dear cosmonaut corps, you need to be able to ... integrate.
In particular, sure bring a function under a differential sign, integrate by parts and be on good terms with Newton-Leibniz formula. Let's start the important pre-flight exercises. I strongly do not recommend skipping it, so that later you don’t flatten in zero gravity:

Example 1

Calculate definite integrals

where takes natural values.

Solution: integration is carried out over the variable "x" and at this stage the discrete variable "en" is considered a constant. In all integrals bring the function under the sign of the differential:

A short version of the solution, which would be good to shoot at, looks like this:

Getting used to:

The four remaining points are on their own. Try to treat the task conscientiously and arrange the integrals in a short way. Sample solutions at the end of the lesson.

After a QUALITY exercise, we put on spacesuits
and getting ready to start!

Expansion of a function in a Fourier series on the interval

Let's consider a function that determined at least on the interval (and, possibly, on a larger interval). If this function is integrable on the segment , then it can be expanded into a trigonometric Fourier series:
, where are the so-called Fourier coefficients.

In this case, the number is called decomposition period, and the number is half-life decomposition.

Obviously, in the general case, the Fourier series consists of sines and cosines:

Indeed, let's write it in detail:

The zero term of the series is usually written as .

Fourier coefficients are calculated using the following formulas:

I understand perfectly well that new terms are still obscure for beginners to study the topic: decomposition period, half cycle, Fourier coefficients and others. Don't panic, it's not comparable to the excitement before a spacewalk. Let's figure everything out in the nearest example, before executing which it is logical to ask pressing practical questions:

What do you need to do in the following tasks?

Expand the function into a Fourier series. Additionally, it is often required to draw a graph of a function, a graph of the sum of a series, a partial sum, and in the case of sophisticated professorial fantasies, do something else.

How to expand a function into a Fourier series?

Essentially, you need to find Fourier coefficients, that is, compose and compute three definite integrals.

Please copy the general form of the Fourier series and the three working formulas in your notebook. I am very glad that some of the site visitors have a childhood dream of becoming an astronaut coming true right in front of my eyes =)

Example 2

Expand the function into a Fourier series on the interval . Build a graph, a graph of the sum of a series and a partial sum.

Solution: the first part of the task is to expand the function into a Fourier series.

The beginning is standard, be sure to write down that:

In this problem, the expansion period , half-period .

We expand the function in a Fourier series on the interval:

Using the appropriate formulas, we find Fourier coefficients. Now we need to compose and calculate three definite integrals. For convenience, I will number the points:

1) The first integral is the simplest, however, it already requires an eye and an eye:

2) We use the second formula:

This integral is well known and he takes it piecemeal:

When found used method of bringing a function under a differential sign.

In the task under consideration, it is more convenient to immediately use formula for integration by parts in a definite integral :

A couple of technical notes. First, after applying the formula the entire expression must be enclosed in large brackets, since there is a constant in front of the original integral. Let's not lose it! Parentheses can be opened at any further step, I did it at the very last turn. In the first "piece" we show extreme accuracy in substitution, as you can see, the constant is out of business, and the limits of integration are substituted into the product. This action is marked with square brackets. Well, the integral of the second "piece" of the formula is well known to you from the training task ;-)

And most importantly - the ultimate concentration of attention!

3) We are looking for the third Fourier coefficient:

A relative of the previous integral is obtained, which is also integrated by parts:

This instance is a little more complicated, I will comment out the further steps step by step:

(1) The entire expression is enclosed in large brackets.. I did not want to seem like a bore, they lose the constant too often.

(2) In this case, I immediately expanded those big brackets. Special attention we devote to the first “piece”: the constant smokes on the sidelines and does not participate in substituting the limits of integration ( and ) into the product . In view of the clutter of the record, it is again advisable to highlight this action in square brackets. With the second "piece" everything is simpler: here the fraction appeared after opening large brackets, and the constant - as a result of integrating the familiar integral ;-)

(3) In square brackets, we carry out transformations, and in the right integral, we substitute the limits of integration.

(4) We take out the “flasher” from the square brackets: , after which we open the inner brackets: .

(5) We cancel 1 and -1 in parentheses, we make final simplifications.

Finally found all three Fourier coefficients:

Substitute them into the formula :

Don't forget to split in half. At the last step, the constant ("minus two"), which does not depend on "en", is taken out of the sum.

Thus, we have obtained the expansion of the function in a Fourier series on the interval :

Let us study the question of the convergence of the Fourier series. I will explain the theory in particular Dirichlet theorem, literally "on the fingers", so if you need strict formulations, please refer to a textbook on calculus (for example, the 2nd volume of Bohan; or the 3rd volume of Fichtenholtz, but it is more difficult in it).

In the second part of the task, it is required to draw a graph, a series sum graph and a partial sum graph.

The graph of the function is the usual straight line on the plane, which is drawn with a black dotted line:

We deal with the sum of the series. As you know, functional series converge to functions. In our case, the constructed Fourier series for any value of "x" converges to the function shown in red. This function is subject to breaks of the 1st kind in points , but also defined in them (red dots in the drawing)

Thus: . It is easy to see that it differs markedly from the original function , which is why in the notation a tilde is used instead of an equals sign.

Let us study an algorithm by which it is convenient to construct the sum of a series.

On the central interval, the Fourier series converges to the function itself (the central red segment coincides with the black dotted line of the linear function).

Now let's talk a little about the nature of the considered trigonometric expansion. Fourier series includes only periodic functions (constant, sines and cosines), so the sum of the series is also a periodic function.

What does this mean in our particular example? And this means that the sum of the series –necessarily periodic and the red segment of the interval must be infinitely repeated on the left and right.

I think that now the meaning of the phrase "period of decomposition" has finally become clear. Simply put, every time the situation repeats itself again and again.

In practice, it is usually sufficient to depict three decomposition periods, as is done in the drawing. Well, and more "stumps" of neighboring periods - to make it clear that the chart continues.

Of particular interest are discontinuity points of the 1st kind. At such points, the Fourier series converges to isolated values, which are located exactly in the middle of the discontinuity "jump" (red dots in the drawing). How to find the ordinate of these points? First, let's find the ordinate of the "upper floor": for this, we calculate the value of the function at the rightmost point of the central expansion period: . To calculate the ordinate of the “lower floor”, the easiest way is to take the leftmost value of the same period: . The ordinate of the mean value is the arithmetic mean of the sum of the "top and bottom": . Nice is the fact that when building a drawing, you will immediately see whether the middle is correctly or incorrectly calculated.

Let us construct a partial sum of the series and at the same time repeat the meaning of the term "convergence". The motive is known from the lesson about the sum of the number series. Let's describe our wealth in detail:

To make a partial sum, you need to write down zero + two more terms of the series. That is,

In the drawing, the graph of the function is shown in green, and, as you can see, it wraps around the total sum quite tightly. If we consider a partial sum of five terms of the series, then the graph of this function will approximate the red lines even more accurately, if there are a hundred terms, then the “green serpent” will actually completely merge with the red segments, etc. Thus, the Fourier series converges to its sum.

It is interesting to note that any partial sum is continuous function, but the total sum of the series is still discontinuous.

In practice, it is not uncommon to build a partial sum graph. How to do it? In our case, it is necessary to consider the function on the segment, calculate its values at the ends of the segment and at intermediate points (the more points you consider, the more accurate the graph will be). Then you should mark these points on the drawing and carefully draw a graph on the period , and then “replicate” it into adjacent intervals. How else? After all, approximation is also a periodic function ... ... its graph somehow reminds me of an even heart rhythm on the display of a medical device.

Of course, it is not very convenient to carry out the construction, since you have to be extremely careful, maintaining an accuracy of no less than half a millimeter. However, I will please readers who are at odds with drawing - in a "real" task, it is far from always necessary to perform a drawing, somewhere in 50% of cases it is required to expand the function into a Fourier series and that's it.

After completing the drawing, we complete the task:

Answer:

In many tasks, the function suffers rupture of the 1st kind right on the decomposition period:

Example 3

Expand in a Fourier series the function given on the interval . Draw a graph of the function and the total sum of the series.

The proposed function is given piecewise (and, mind you, only on the segment) and endure rupture of the 1st kind at point . Is it possible to calculate the Fourier coefficients? No problem. Both the left and right parts of the function are integrable on their intervals, so the integrals in each of the three formulas should be represented as the sum of two integrals. Let's see, for example, how this is done for a zero coefficient:

The second integral turned out to be equal to zero, which reduced the work, but this is not always the case.

Two other Fourier coefficients are written similarly.

How to display the sum of a series? On the left interval we draw a straight line segment , and on the interval - a straight line segment (highlight the axis section in bold-bold). That is, on the expansion interval, the sum of the series coincides with the function everywhere, except for three "bad" points. At the discontinuity point of the function, the Fourier series converges to an isolated value, which is located exactly in the middle of the “jump” of the discontinuity. It is not difficult to see it orally: left-hand limit:, right-hand limit: and, obviously, the ordinate of the midpoint is 0.5.

Due to the periodicity of the sum , the picture must be “multiplied” into neighboring periods, in particular, depict the same thing on the intervals and . In this case, at the points, the Fourier series converges to the median values.

In fact, there is nothing new here.

Try to solve this problem on your own. An approximate sample of fine design and drawing at the end of the lesson.

Expansion of a function in a Fourier series on an arbitrary period

For an arbitrary expansion period, where "el" is any positive number, the formulas for the Fourier series and Fourier coefficients differ in a slightly more complicated sine and cosine argument:

If , then we get the formulas for the interval with which we started.

The algorithm and principles for solving the problem are completely preserved, but the technical complexity of the calculations increases:

Example 4

Expand the function into a Fourier series and plot the sum.

Solution: in fact, an analogue of Example No. 3 with rupture of the 1st kind at point . In this problem, the expansion period , half-period . The function is defined only on the half-interval , but this does not change things - it is important that both parts of the function are integrable.

Let's expand the function into a Fourier series:

Since the function is discontinuous at the origin, each Fourier coefficient should obviously be written as the sum of two integrals:

1) I will write the first integral as detailed as possible:

2) Carefully peer into the surface of the moon:

Second integral take in parts:

What should you pay close attention to after we open the continuation of the solution with an asterisk?

First, we do not lose the first integral , where we immediately execute bringing under the sign of the differential. Secondly, do not forget the ill-fated constant before the big brackets and don't get confused by signs when using the formula. Large brackets, after all, it is more convenient to open immediately in the next step.

The rest is a matter of technique, only insufficient experience in solving integrals can cause difficulties.

Yes, it was not in vain that the eminent colleagues of the French mathematician Fourier were indignant - how did he dare to decompose functions into trigonometric series ?! =) By the way, probably everyone is interested in the practical meaning of the task in question. Fourier himself worked on a mathematical model of heat conduction, and subsequently the series named after him began to be used to study many periodic processes, which are apparently invisible in the outside world. Now, by the way, I caught myself thinking that it was no coincidence that I compared the graph of the second example with a periodic heart rhythm. Those interested can get acquainted with the practical application Fourier transforms from third party sources. ... Although it’s better not to - it will be remembered as First Love =)

3) Given the repeatedly mentioned weak links, we deal with the third coefficient:

Integrating by parts:

We substitute the found Fourier coefficients into the formula , not forgetting to divide the zero coefficient in half:

Let's plot the sum of the series. Let us briefly repeat the procedure: on the interval we build a line, and on the interval - a line. With a zero value of "x", we put a point in the middle of the "jump" of the gap and "replicate" the chart for neighboring periods:

At the "junctions" of the periods, the sum will also be equal to the midpoints of the "jump" of the gap.

Ready. I remind you that the function itself is conditionally defined only on the half-interval and, obviously, coincides with the sum of the series on the intervals

Answer:

Sometimes a piecewise given function is also continuous on the expansion period. The simplest example: . Solution (See Bohan Volume 2) is the same as in the two previous examples: despite function continuity at the point , each Fourier coefficient is expressed as the sum of two integrals.

In the breakup interval discontinuity points of the 1st kind and / or "junction" points of the graph may be more (two, three, and in general any final quantity). If a function is integrable on every part, then it is also expandable in a Fourier series. But from practical experience, I don’t remember such a tin. Nevertheless, there are more difficult tasks than just considered, and at the end of the article for everyone there are links to Fourier series of increased complexity.

In the meantime, let's relax, leaning back in our chairs and contemplating the endless expanses of stars:

Example 5

Expand the function into a Fourier series on the interval and plot the sum of the series.

In this task, the function continuous on the decomposition half-interval, which simplifies the solution. Everything is very similar to Example #2. You can't get away from the spaceship - you'll have to decide =) Sample design at the end of the lesson, the schedule is attached.

Fourier series expansion of even and odd functions

With even and odd functions, the process of solving the problem is noticeably simplified. And that's why. Let's return to the expansion of the function in a Fourier series on a period of "two pi" and arbitrary period "two ales" .

Let's assume that our function is even. The general term of the series, as you can see, contains even cosines and odd sines. And if we decompose an EVEN function, then why do we need odd sines?! Let's reset the unnecessary coefficient: .

Thus, an even function expands into a Fourier series only in cosines:

Because the integrals of even functions over a segment of integration symmetric with respect to zero can be doubled, then the rest of the Fourier coefficients are also simplified.

For span:

For an arbitrary interval:

Textbook examples that are found in almost any calculus textbook include expansions of even functions . In addition, they have repeatedly met in my personal practice:

Example 6

Given a function. Required:

1) expand the function into a Fourier series with period , where is an arbitrary positive number;

2) write down the expansion on the interval , build a function and graph the total sum of the series .

Solution: in the first paragraph, it is proposed to solve the problem in a general way, and this is very convenient! There will be a need - just substitute your value.

1) In this problem, the expansion period , half-period . In the course of further actions, in particular during integration, "el" is considered a constant

The function is even, which means that it expands into a Fourier series only in cosines: .

Fourier coefficients are sought by the formulas . Pay attention to their absolute advantages. First, the integration is carried out over the positive segment of the expansion, which means that we safely get rid of the module , considering only "x" from two pieces. And, secondly, integration is noticeably simplified.

Two:

Integrating by parts:

Thus:
, while the constant , which does not depend on "en", is taken out of the sum.

Answer:

2) We write the expansion on the interval, for this we substitute the desired value of the half-period into the general formula:

Trigonometric Fourier series is called a series

a0 /2 + a 1 cos x + b 1 sin x + a 2 cos2 x + b 2 sin2 x + ... + a ncos nx + b n sin nx + ...

where are the numbers a0 , a 1 , b 1 , a 2 , b 2 , ..., a n b n,... - Fourier coefficients.

A more concise notation of the Fourier series with the symbol "sigma":

As we have just established, in contrast to the power series, in the Fourier series, instead of the simplest functions trigonometric functions are taken

1/2 cos x, sin x, cos2 x, sin2 x, ..., cos nx, sin nx, ... .

The Fourier coefficients are calculated using the following formulas:

All of the above functions in the Fourier series are periodic functions with a period of 2 π . Each term of the trigonometric Fourier series is a periodic function with period 2 π .

Therefore, any partial sum of the Fourier series has a period of 2 π . It follows that if the Fourier series converges on the segment [- π , π ] , then it converges on the entire real line and its sum, being the limit of a sequence of periodic partial sums, is a periodic function with period 2 π .

Convergence of the Fourier series and the sum of the series

Let the function F(x) defined on the entire number line and periodic with period 2 π , is a periodic continuation of the function f(x) , if on the segment [- π , π ] occurs F(x) = f(x)

If on the interval [- π , π ] the Fourier series converges to the function f(x) , then it converges on the whole number line to its periodic continuation.

The answer to the question of under what conditions the Fourier series of the function f(x) converges to this function, is given by the following theorem.

Theorem. Let the function f(x) and its derivative f"(x) - continuous on the interval [- π , π ] or have a finite number of discontinuity points of the first kind on it. Then the Fourier series of the function f(x) converges on the entire number line, and at each point x belonging to the segment [- π , π ] , wherein f(x) is continuous, the sum of the series is f(x) , and at each point x0 discontinuity of the function, the sum of the series is equal to the arithmetic mean of the limits of the function f(x) right and left:

Where And .

At the ends of the segment [- π , π ] the sum of the series is equal to the arithmetic mean of the values of the function at the extreme left and extreme right points of the expansion period:

At any point x belonging to the segment [- π , π ] , the sum of the Fourier series is equal to F(x) , If x- point of continuity F(x) , and is equal to the arithmetic mean of the limits F(x) left and right:

If x- breaking point F(x) , Where F(x) - periodic continuation f(x) .

Example 1 Periodic function f(x) with period 2 π defined as follows:

It is easier to write this function as f(x) = |x| . Expand the function into a Fourier series, determine the convergence of the series and the sum of the series.

Solution. Let us define the Fourier coefficients of this function:

Now we have everything to get the Fourier series of this function:

This series converges at all points, and its sum is equal to the given function.

Solve the Fourier series problem yourself, and then see the solution

Fourier series for even and odd functions

Let the function f(x) is defined on the segment [- π , π ] and is even, i.e. f(- x) = f(x) . Then its coefficients bn are equal to zero. And for the coefficients an the following formulas are correct:

Now let the function f(x) defined on the segment [- π , π ] , odd, i.e. f(x) = -f(- x) . Then the Fourier coefficients an are equal to zero, and the coefficients bn is determined by the formula

As can be seen from the formulas above, if function f(x) is even, then the Fourier series contains only cosines, and if odd, then only sines.

Example 3

Solution. This is an odd function, so its Fourier coefficients , and to find , you need to calculate a definite integral :

This equality is true for any . At the points, the sum of the Fourier series, according to the theorem given in the second paragraph, does not coincide with the values of the function , but is equal to . Outside the segment, the sum of the series is a periodic continuation of the function , its graph was given above as an illustration of the sum of the series.

Example 4 Expand the function in a Fourier series.

Solution. This is an even function, so its Fourier coefficients, and to find, you need to calculate certain integrals:

We get the Fourier series of this function:

This equality is valid for any , since at the points the sum of the Fourier series in this case coincides with the values of the function , since .

Fourier series with respect to any orthogonal system of functions

The sequence of functions continuous on the interval [ a,b], is called orthogonal system of functions on the interval[a,b], if all functions of the sequence are pairwise orthogonal on this segment, i.e., if

The system is called orthogonal and normalized (orthonormal) on the interval ,

if the condition is met

Let now f(x) - any function continuous on the segment [ a,b]. Near Fourier such a function f(x) on the segment [ a,b] by orthogonal system the row is called:

whose coefficients are determined by the equality:

N=1,2,...

If an orthogonal system of functions on the segment [ a,b] is orthonormal, then in this case

Where n=1,2,...

Let now f(x) is any function that is continuous or has a finite number of discontinuity points of the first kind on the segment [ a,b]. Near Fourier of such a function f(x) on the same segment

according to the orthogonal system, the series is called:

If the Fourier series of the function f(x) according to system (1) converges to the function f(x) at each of its points of continuity belonging to the segment [ a,b]. In this case they say that f(x) on the segment [ a,b] expands into a series in terms of the orthogonal system (1).

Complex form of the Fourier series

The expression is called the complex form of the Fourier series of the function f(x) if defined by the equality

,Where

The transition from the Fourier series in complex form to the series in real form and vice versa is carried out using the formulas:

(n=1,2, . . .)

String Vibration Problem

Let a string of length be stretched in equilibrium l meet x= 0 and x=l. Assume that the string is out of equilibrium and oscillates freely. We will consider small vibrations of the string occurring in the vertical plane.

Under the assumptions made above, it can be shown that the function u(x,t) characterizing the position of the string at each moment of time t, satisfies the equation

(1) , where a is a positive number.

Our task is to find the function u(x,t) , whose graph gives the shape of the string at any given time t, i.e., find a solution to equation (1) at the boundary:

and initial conditions:

First, we will look for solutions to equation (1) that satisfy boundary conditions (2). It is easy to see that u(x,t) 0 is a solution to equation (1) that satisfies the boundary conditions (2). We will look for solutions that are not identically equal to 0, representable as a product u(x,t)=X(x)T(t), (4) , where , .

Substituting expression (4) into equation (1) gives:

From which our task is reduced to finding solutions to the equations:

Using this condition X(0)=0, X(l)=0, we will prove that it is a negative number by examining all cases.

a) Let Then X”=0 and its general solution is written as follows:

whence and , which is impossible, since we are considering solutions that do not vanish identically.

b) Let . Then solving the equation

we get , and, subordinating, we find that

c) If then

The equations have roots:

Where - arbitrary constants. From the initial condition we find:

from where, i.e.

(n=1,2,...)

(n=1,2,...).

Given this, we can write:

(N=1,2,...).

and therefore

, (n=1,2,...),

but since A and B are different for different values of n, we have

, (n=1,2,...),

where and are arbitrary constants, which we will try to determine in such a way that the series satisfies equation (1), boundary conditions (2), and initial conditions (3).

So let's assign a function u(x,t) to the initial conditions, i.e., we choose and so that the conditions

These equalities are, respectively, expansions of the functions and into segments in a Fourier series in terms of sines. (This means that the coefficients will be calculated as for odd functions). Thus, the solution for the vibration of a string with given boundary and initial conditions is given by the formula

(n=1,2,...)

Fourier integral

Sufficient conditions for the representability of a function in the Fourier integral.

In order to f(x) was represented by the Fourier integral at all points of continuity and regular points of discontinuity, it suffices:

1) absolute integrability on

(i.e. the integral converges)

2) on any finite segment [- L, L] function would be piecewise smooth

3) at the points of discontinuity of the function, its Fourier integral is determined by the half-sum of the left and right limits at these points, and at the points of continuity to the function itself f(x)

The Fourier integral of the function f(x) is an integral of the form:

Where ,

Fourier integral for an even and odd function

Let f(x)-even function satisfying the conditions of representability by a Fourier integral.

Taking into account that , as well as the property of integrals with respect to the symmetric with respect to the point x=0 interval of even functions, from equality (2) we obtain:

(3)