Expansion of a periodic function into a trigonometric Fourier series. Fourier series expansion in cosines
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.
On the 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 the hamster and intrusive thoughts about the hardships of life aquarium fish. The Fourier series is not difficult to understand, however practical tasks they simply require increased concentration of attention - ideally, you 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. Truth.
Secondly, before flying into space, you need to study the dashboard spaceship. Let's start with the values of the functions that should be clicked on the machine:
For any natural value:
one) . And in fact, the sinusoid "flashes" the x-axis through each "pi":
. When negative values 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 conscientiously treat the task and draw up integrals 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 etc. Do not panic, this is not comparable to the excitement before going to outer space. 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 wording, please refer to the textbook on mathematical analysis (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)
In this way: 
. 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 specific 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 is the mean arithmetic sum"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 mathematical model thermal conductivity, and later the series named after him began to be used to study many periodic processes, which are apparently invisible in the surrounding 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 who wish can familiarize themselves with 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 amount). 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: .
In this way, 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 general view and it's 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:
In this way:
, 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:
How to paste mathematical formulas to the website?
If you ever need to add one or two mathematical formulas to a web page, then the easiest way to do this is as described in the article: mathematical formulas are easily inserted into the site in the form of pictures that Wolfram Alpha automatically generates. In addition to simplicity, this universal way will help improve the visibility of the site in search engines. It has been working for a long time (and I think it will work forever), but it is morally outdated.
If you are constantly using mathematical formulas on your site, then I recommend that you use MathJax, a special JavaScript library that displays mathematical notation in web browsers using MathML, LaTeX, or ASCIIMathML markup.
There are two ways to start using MathJax: (1) using a simple code, you can quickly connect a MathJax script to your site, which will be automatically loaded from a remote server at the right time (list of servers); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method is more complex and time consuming and will allow you to speed up the loading of the pages of your site, and if the parent MathJax server becomes temporarily unavailable for some reason, this will not affect your own site in any way. Despite these advantages, I chose the first method, as it is simpler, faster and does not require technical skills. Follow my example, and within 5 minutes you will be able to use all the features of MathJax on your website.
You can connect the MathJax library script from a remote server using two code options taken from the main MathJax website or from the documentation page:
One of these code options needs to be copied and pasted into the code of your web page, preferably between the tags
and or right after the tag . According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you paste the second code, then the pages will load more slowly, but you will not need to constantly monitor MathJax updates.The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the load code above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all , since the MathJax script is loaded asynchronously). That's all. Now learn the MathML, LaTeX, and ASCIIMathML markup syntax and you're ready to embed math formulas into your web pages.
Any fractal is built on certain rule, which is successively applied an unlimited number of times. Each such time is called an iteration.
The iterative algorithm for constructing a Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 cubes adjacent to it along the faces are removed from it. It turns out a set consisting of 20 remaining smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process indefinitely, we get the Menger sponge.
Lecture #60
6.21. Fourier series for even and odd functions.
Theorem: For any even function, its Fourier series consists only of cosines.
For any odd function: 
.
Proof: It follows from the definition of an even and odd function that if ψ(x) – even function, then
.
Really,
since by definition of an even function ψ(- x) = ψ(x).
Similarly, one can prove that if ψ(x) is an odd function, then
If an odd function ƒ(x) is expanded into a Fourier series, then the product ƒ(x) ·coskx is also an odd function, and ƒ(x) ·sinkx is even; Consequently,
(21)
i.e., the Fourier series of an odd function contains "only sines".
If an even function is expanded into a Fourier series, then the product ƒ(x) sinkx is an odd function, and ƒ(x) coskx is even, then:
(22)
i.e., the Fourier series of an even function contains "only cosines".
The obtained formulas make it possible to simplify calculations when searching for the Fourier coefficients in cases where the given function is even or odd, and also to obtain expansion in a Fourier series of a function given on a part of the interval .
In many tasks, the function 
is set in the interval 
. It is required to represent this function as an infinite sum of sines and cosines of angles that are multiples of natural numbers, i.e. it is necessary to expand the function in a Fourier series. Usually in such cases proceed as follows.
To expand a given function in terms of cosines, the function 
redefine in the interval 
in an even way, i.e. so that in the interval 
. Then, for the “extended” even function, all the arguments of the previous section are valid, and, consequently, the coefficients of the Fourier series are determined by the formulas
,
In these formulas, as we see, the values of the function appear 
, only given in the interval 
. To expand the function 
, specified in the interval 
, by sines, it is necessary to redefine this function in the interval 
in an odd way, i.e. so that in the interval 
.
Then the calculation of the coefficients of the Fourier series must be carried out according to the formulas
.
Theorem 1. A function given on an interval can be expanded in an infinite number of ways in a trigonometric Fourier series, in particular, in cos or in sin.
Comment. Function 
, specified in the interval 
can be extended in the interval 
in any way, and not just the way it was done above. But with an arbitrary extension of the function, the expansion in a Fourier series will be more complicated than that obtained by expanding in terms of sines or cosines.
Example. Expand in a Fourier series in cosines the function 
, specified in the interval 
(Fig. 2a).
Solution. We extend the function 
in the interval 
in an even way (the graph is symmetrical about the axis 
)
,
Because 
, then
at 
,
at 
6.22. Fourier series for a function defined on an arbitrary interval
So far, we have considered a function defined in the interval 
, considering it to be periodic outside this interval, with a period 
.
Consider now the function 
, whose period is equal to 2 l, i.e. 
on the interval 
, and show that in this case the function 
can be expanded in a Fourier series.
Let's put 
, or 
. Then when changing 
from - l before l new variable 
changes from 
before 
and hence the function 
can be considered as a function given in the interval from 
before 
and periodic outside this interval, with a period 
.
So, 
.
Having decomposed 
into a Fourier series, we get
,
.
Moving on to the old variables, i.e. assuming 
, we get 
,
and 
.
That is, the Fourier series for the function 
, given in the interval 
, will look like:
,
,
.
If the function 
is even, then the formulas for determining the coefficients of the Fourier series are simplified:
,
,
.
In case the function 
odd:
,
,
.
If the function 
set in the interval 
, then it can be continued in the interval 
either even or odd. In the case of an even continuation of the function in the interval 
,
.
In the case of an odd redefinition of a function in the interval 
the coefficients of the Fourier series are found by the formulas
,
.
Example. Expand the function in a Fourier series
along the sines of multiple arcs.
Solution. Schedule given function shown in Fig.3. Let's continue the function in an odd way (Fig. 4), i.e. we will expand in terms of sines.
All odds 
,
We introduce the replacement 
. Then at 
we get 
, at 
we have 
.
In this way
.
6.23. .The concept of expansion in a Fourier series of non-periodic functions
The function given in the main area (-ℓ, ℓ) can be periodically extended beyond the main area using the functional relation ƒ(x+2 ℓ) = ƒ(x).
For a non-periodic function ƒ(x) (-∞ φ(x)= Formula (2.18) will be true on the entire axis -∞< x< ∞ . Можно написать подобное разложение
для функции ƒ(x)= Formula (2.19) will be true only on a finite interval (-ℓ, ℓ), since ƒ(x) and φ(x) coincide on this interval. Thus, a non-periodic function can be expanded into a Fourier series on a finite interval. Function f(x) defined on a segment and being piecewise monotonic and bounded on this segment can be expanded into a Fourier series in two ways. To do this, it suffices to represent the extension of the function to the interval [– l, 0]. If continued f(x) on the [- l, 0] is even (symmetric about the y-axis), then the Fourier series can be written using formulas (1.12–1.13), that is, using cosines. If we continue the function f(x) on the [- l, 0] in an odd way, then the expansion of the function in a Fourier series will be represented by formulas (1.14–1.15), that is, in terms of sines. In this case, both series will have in the interval (0, l) the same amount. Example. Expand the function in a Fourier series y
= x, given on the interval (see Fig.1.4). Solution. a). Cosine expansion. We construct an even continuation of the function to the neighboring interval [–1, 0]. The graph of the function, together with its even continuation by [–1, 0 ] and subsequent continuation (over the period T= 2) for the entire axis 0 x shown in Figure 1.5. Because l= 1, then the Fourier series for this function with an even expansion will have the form As a result, we get On the whole axis 0 x the series converges to the function shown in Figure 1.4. 2). Sine expansion. We construct an odd continuation of the function to the neighboring interval [–1, 0]. Graph of the function along with its odd continuation on [–1, 0] and subsequent periodic continuation on the entire numeric axis 0 x shown in Figure 1.6. For an odd expansion Therefore, the Fourier series in sines for this function at At the point It follows from a comparison of expressions (1.19) and (1.21) that the rate of convergence of series (1.19) is higher than that of series (1.21): it is determined in the first case by the factor In general, it can be shown that if the function f(x) does not vanish at least at one end of the interval , then it is preferable to expand it into a series in cosines. This is due to the fact that for an even continuation into the neighboring interval Functions In doing so, it is assumed that Consider the expansion of the function f(x), which is defined on the interval [ a,
b], in a series with respect to the system of orthogonal functions where coefficients To determine the expansion coefficients Due to the orthogonality of the functions Series (1.23) in a system of orthogonal functions whose coefficients are determined by formula (1.24) is called generalized Fourier series for function f(x). To simplify the formulas for the coefficients, the so-called function rationing. Function system φ
0 (x),
φ
1 (x),…,
φ
n (x),… is called normalized on the interval [ a,
b], if The theorem is true: any orthogonal system of functions can be normalized. This means that you can choose constant numbers μ
0 ,
μ
1 ,…,
μ
n,… so that the system of functions μ
0 φ
0 (x),
μ
1 φ
1 (x),…,
μ
n φ
n (x),… was not only orthogonal, but also normalized. Indeed, from the condition we get that called the norm
functions
If the system of functions is normalized, then, obviously, For an orthonormal system of functions, the coefficients of the generalized Fourier series are Example. Expand function y
= 2 – 3x on the segment after checking them for quadratic integrability and orthogonality. Comment. They say that the function Solution. First, we solve the eigenvalue problem. The general solution to the equation of this problem will be and its derivative will be written in the form Therefore, from the boundary conditions it follows: For a nontrivial solution to exist, it is necessary to take whence it follows and the eigenfunctions corresponding to them, up to a factor, will be Let's check the obtained eigenfunctions for orthogonality on the interval : since for integer Therefore, the found eigenfunctions are orthogonal on the interval . Let us expand the given function into a generalized Fourier series in terms of the system of orthogonal eigenfunctions (1.27): whose coefficients are calculated by (1.24): Substituting (129) into (1.28), we finally obtain Many processes occurring in nature and technology have the property of repeating themselves at regular intervals. Such processes are called periodic and are mathematically described by periodic functions. These features include sin(x)
,
cos(x)
,
sin(wx),
cos(wx)
. The sum of two periodic functions, for example, a function of the form ,
in general, is no longer periodic. But it can be shown that if the relation w 1
/
w 2
is a rational number, then this sum is a periodic function. The simplest periodic processes - harmonic oscillations - are described by periodic functions sin(wx)
and cos(wx).
More complex periodic processes are described by functions that are composed either of a finite or an infinite number of terms of the form sin(wx)
and cos(wx).
Consider a functional series of the form: This row is called trigonometric; numbers a 0
,
b 0
,
a 1
,
b 1
,a 2
,
b 2
…,
a n ,
b n ,…
called coefficients trigonometric series. Series (1) is often written as follows: Since the members of the trigonometric series (2) have a common period Let's assume that the function f(x)
is the sum of this series: In this case, the function is said to be f(x)
expands into a trigonometric series. Assuming that this series converges uniformly over the interval The coefficients of the series determined by these formulas are called Fourier coefficients. The trigonometric series (2), whose coefficients are determined by the Fourier formulas (4), are called near Fourier corresponding to the function f(x).
Thus, if the periodic function f(x)
is the sum of a convergent trigonometric series, then this series is its Fourier series. Formulas (4) show that the Fourier coefficients can be calculated for any interval integrable Recall that the function f(x),
defined on the segment [
a;
b]
, is called piecewise smooth if it and its derivative have at most a finite number of discontinuity points of the first kind. The following theorem gives sufficient conditions for the expansion of a function into a Fourier series. Dirichlet's theorem.
Let Example1. Expand the function in a Fourier series f(x)=
x, given on the interval Solution. This function satisfies the Dirichlet conditions and hence can be expanded into a Fourier series. Applying formulas (4) and the method of integration by parts Thus, the Fourier series for the function f(x)
has a look.
(2.18)
(2.19)
(1.18)
,
,
(1.20)
.
will look like
the sum of the series will be equal to zero, although the original function is equal to 1. This is due to the fact that with such a periodic continuation, the point x= 1 becomes the break point.
, and in the second case by the factor 1/ n. Therefore, expansion in a series in cosines is preferable in this case.
the function will be continuous (see Fig. 1.5), and the rate of convergence of the resulting series will be higher than that of the sine series. If the function given on vanishes at both ends of the interval, then it is preferable to expand it into a series in terms of sines, since in this case not only the function itself will be continuous f(x), but also its first derivative.1.6. Generalized Fourier Series
and 
(n,
m= 1, 2, 3,…) are called orthogonal on the segment [ a,
b], if n
≠ m
.
(1.22)
and 
.
(i= 0,1,2...) are constant numbers.
we multiply equality (1.23) by 
and integrate term by term on the segment [ a,
b]. We get equality
all integrals on the right side of the equality will be equal to zero, except for one (for 
). Hence it follows that
(1.24)
.
(1.25)
.
and is denoted by 
.
. Function Sequence φ
0 (x),
φ
1 (x),…,
φ
n (x),… defined on the interval [ a,
b], is orthonormal on this interval, if all functions are normalized and mutually orthogonal on [ a,
b].
.
(1.26)
into the generalized Fourier series with respect to the system of orthogonal functions on this interval, for which we take the eigenfunctions of the eigenvalue problem
, given on the segment 
, is a square-integrable function if it and its square are integrable on 
, that is, if there are integrals 
and 
.
,
Therefore, the eigenvalues of the parameter 
equal
,
.
(1.27)
.Wherein
,
(1.28)
.
(1.29)3.2. trigonometric series. Fourier coefficients
.
(2)
, then the sum of the series, if it converges, is also a periodic function with period 
.
.
(3)
, you can determine its coefficients by the formulas:
, 
,
.
(4)3.3. Fourier Series Convergence
-periodic function, i.e. for such a function one can always compose a Fourier series. But will this series converge to the function f(x)
and under what conditions?
-periodic function f(x)
is piecewise smooth on 
. Then its Fourier series converges to f(x)
at each of its points of continuity and to the value 0,5(f(x+0)+
f(x-0))
at the breaking point.
.
, we find the Fourier coefficients:
Loading...
